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https://mathoverflow.net/questions/433120 | 6 | In the German Wikipedia entry for $L^p$-Raum it is stated ([Link](https://de.wikipedia.org/wiki/Lp-Raum))
>
> Das $L$ in der Bezeichnung geht auf den französischen Mathematiker Henri Léon Lebesgue zurück, da diese Räume über das Lebesgue-Integral
> definiert werden.
>
>
>
>
>
> "The $L$ in the name goes back to the French mathematician Henri Léon Lebesgue, since these spaces are defined by the Lebesgue integral."
>
>
>
However, in the English Wikipedia version,
>
> They are sometimes called Lebesgue spaces, named after Henri Lebesgue (Dunford & Schwartz 1958, III.3), although according to the Bourbaki group (Bourbaki 1987) they were first introduced by Frigyes Riesz (Riesz 1910)
>
>
>
If we check the Riesz paper (1910) paper, this is how he introduces...
>
> In der vorliegenden Arbeit wird die Voraussetzung der quadratischen Integrierbarkeit durch jene der Integrierbarkeit von $|f(x)|^p$ ersetzt; $p$ bedeutet eine beliebige, rationale oder irrationale Zahl $>1$.\*) Jede Zahl $p$ bestimmt eine Funktionenklasse $\left[L^p\right]$. Die Rolle der Klasse $\left[L^2\right]$ übernehmen hier je zwei Klassen $\left[L^p\right]$ und $\left[\frac{p}{L^{p-1}}\right] ;$ sie haben die Eigenschaft, dass jede Funktion, die mit allen Funtetionen der einen Klasse integrierbare Produkte ergibt, sicher der andern Klasse angehört. Die Untersuchung dieser Funktionenklassen wird auf die wirklichen und scheinbaren Vorteile des Exponenten $p=2$ ein ganz besonderes Licht werfen; und man kann auch behaupten, dab sie für eine axiomatische Untersuchung der Funktionenräume brauchbares Material liefert.
>
>
>
>
> In the present work, the condition of quadratic integrability is replaced by that of integrability of $|f(x)|^p$; $p$ means any
> rational or irrational number $>1$.\*) Each number $p$ determines a
> class of functions $\left[L^p\right]$. The role of the class
> $\left[L^2\right]$ is played here by two classes each
> $\left[L^p\right]$ and $\left[\frac{p}{L^{p-1}}\right] ;$ they have the property that any function yielding integrable products with all functions of one class certainly belongs to the other class. The study of these classes of functions will throw a very special light on the real and apparent advantages of the exponent $p=2$; and one can also claim that it provides useful material for an axiomatic study of function spaces.
>
>
>
*Translated with [www.DeepL.com/Translator](http://www.DeepL.com/Translator) (free version)*
The author does not associate the function class $L$ with the name Lebesgue. Is it possible that the explanation of the choice of $L$ is just like the $\operatorname{sinc}$ function which is popularly known as *sinus cardinalis* but without any solid source ([Origin of $\operatorname{sinc}$ function](https://mathoverflow.net/questions/341436/origin-of-the-term-sinc-function))?
I mean the original author never explained his choice of $L$ but later people associated $L$ with Lebesgue.
| https://mathoverflow.net/users/142414 | Origin of $L$ in $L^1$ and $L^2$ norms | On page 453 of his 1910 paper [Untersuchungen über Systeme integrierbarer Funktionen](https://eudml.org/doc/158473) Riesz writes (in a footnote)
>
> \*) H. Lebesgue [Sur les intégrales singulières](https://eudml.org/doc/72816) (1909) appeared after the completion of this work. That paper touches on many of the points raised in sections 2 and 3 of the
> present paper, but only for the case $p=2$.
>
>
>
So it is not a stretch to assume that Riesz used $L^2$ in deference to Lebesgue, then generalized to $L^p$.
| 5 | https://mathoverflow.net/users/11260 | 433128 | 175,216 |
https://mathoverflow.net/questions/433125 | 9 | There is some work which generalises the usual [Wilson loop](https://en.wikipedia.org/wiki/Wilson_loop) in QFT to higher dimensions and constructs non-abelian [Wilson surface functionals](https://www.sciencedirect.com/science/article/pii/S0001870810003804) in the context of [non-abelian gerbes](https://www.sciencedirect.com/science/article/pii/S0001870805002513).
It seems to me that the context of this work is mostly geometry/topological in nature, where the aim is ultimately to try and rigorously define four-dimensional invariants of knotted surfaces.
However, I am not really clear on what the exact physical application is of these Wilson surface observables besides knot invariants. Do they play any more concrete or direct physical role in calculations of amplitudes similar to regular Wilson loops in gauge theory, or are they used to study topological QFTs or string theory?
Edit: Besides the works mentioned in the answer, I also found [this article](https://arxiv.org/abs/0804.1561) of Gukov and Witten which mentions quite a large number of applications of these surfaces.
| https://mathoverflow.net/users/119114 | Physics application of Wilson surface observables | Wilson surfaces have been used to describe non-Abelian quasiparticles in topological states of matter, see
* [Framed Wilson Operators,
Fermionic Strings, and Gravitational Anomaly in 4d](https://arxiv.org/abs/1404.4385)
* [Bosonic Topological
Insulators and Paramagnets: a view from cobordisms](https://arxiv.org/abs/1404.6659)
| 7 | https://mathoverflow.net/users/11260 | 433129 | 175,217 |
https://mathoverflow.net/questions/433130 | 3 | A fact about triangulated categories is that (exact) localisation functors and so-called colocalisation functors come in pairs, making an exact localisation triangle. I've tried to come up with less traditional examples.
Let $A$ be a commutative ring, and $I \subset A$ an ideal. Then tensoring by $A / I$ is idempotent and so we can look at the localisation functor on $D(A)$, the derived category. This is exact in the triangulated sense, as tensoring is right-exact. Am I correct in thinking the colocalisation should just be the kernel of this tensor map, i.e. multiplying by $I$? If so, I expect for every complex of modules $M^\bullet$ that we should get a quasi-isomorphism $$I^2 M^\bullet \to I M^\bullet$$ from the natural inclusion. Is there an elementary way of seeing this? I'm finding it difficult to follow the machinery that guarantees a natural isomorphism $\Gamma^2 \to \Gamma$ over $D(A)$ for whichever colocalisation functor $\Gamma$ is actually associated to the derived tensor $A / I \otimes \\_ $ defined over $D(A)$.
| https://mathoverflow.net/users/488857 | Is there an elementary reason that this colocalisation map of complexes is a quasi-isomorphism? | The derived tensor product with $A/I$ is typically not idempotent (because ${\rm Tor}\_{\*}^A(A/I, A/I)$ is nontrivial), so this won't give you a localization sequence.
One case where it is idempotent is when $I = eA$ for $e \in A$ an idempotent. In this case, tensoring with $A/I$ is the same as localizing at $1-e$, and the localization sequence does take the form $eM \to M \to M/eM$.
| 4 | https://mathoverflow.net/users/131945 | 433134 | 175,219 |
https://mathoverflow.net/questions/433138 | 3 | Let $\mathcal{O}$ be a compact good orbifold, where we understand a *good orbifold* to be an orbifold obtained as a global quotient $M/G$, where $M$ is a manifold and $G$ is a discrete group. Are there a compact manifold $\widetilde{M}$ and a discrete group $\widetilde{G}$ acting on $\widetilde{M}$ such that $\mathcal{O}=\widetilde{M}/\widetilde{G}$?
| https://mathoverflow.net/users/151395 | Can a compact good orbifold be realized as a global quotient of a compact manifold? | This is true in dimension two (work of Fox from the 1950’s) and in dimension three (the so-called orbifold theorem). I don’t know the status in dimension four (or higher).
| 3 | https://mathoverflow.net/users/1650 | 433141 | 175,223 |
https://mathoverflow.net/questions/433135 | 2 | We are given a $d$-dimensional convex shape $S$ inscribed in the hypercube $[-1,1]^d$. We want find an approximation of its volume based only on a set of curves given by the intersection of the $S$ boundary and a finite number of $2$-planes.
We denote by $\gamma\_{i,j}$ the curve given by the intersection of the $S$ boundary with the $2$-plane $\mathrm{span}\left(\mathbf{e}\_i,\mathbf{e}\_j\right)$ for all $1\le i<j\le d$, where $\mathbf{e}\_1,\mathbf{e}\_2,\ldots,\mathbf{e}\_d$ are the standard basis vectors.
---
**Question:** Knowing only the curves $\gamma\_{i,j}$ for all $1\le i<j\le d$, is it possible to provide an estimation $V'$ of the volume $V$ of $S$ such that there exist two positive constants $\alpha$ and $\beta$ for which $\alpha\le\frac{V'}{V}\le\beta$?
| https://mathoverflow.net/users/115803 | Estimating the volume of a convex shape in higher dimensions based only on normal sections | Those constants don't exist for any $d\geq4$, here is an idea of why.
For each $\varepsilon>0$ let $A\_\varepsilon=\{(x\_1,\dots,x\_d)\in[-1,1]^d;\lvert (d-1)x\_d-\sum\_{i=1}^{d-1} x\_i\rvert\leq\varepsilon\text{ and }\lvert\frac{(d-1)(d-2)}{2}x\_d-\sum\_{i=1}^{d-1}ix\_i\rvert\leq\varepsilon\}$. Note that there is a big constant $K$ independent of $\varepsilon$ such that for all $i,j$, the intersection of $A\_\varepsilon$ with span$(e\_i,e\_j)$ is contained in the ball $B(0,K\varepsilon)$.
Also note that $v=(1,1,\dots,1)\in A\_\varepsilon$, and let $B\_\varepsilon=\{x\in A\_\varepsilon;d(x,span(v))\leq K\varepsilon\}$.
Then both $A\_\varepsilon$ and $B\_\varepsilon$ are convex and inscribed in the cube, and their intersection with span$(e\_i,e\_j)$ is the same $\forall i,j$, but when $\varepsilon\to0$, the volume of $A\_\varepsilon$ is, up to some constant, proportional to $\varepsilon^2$ and the volume of $B\_\varepsilon$ is proportional up to some constant to $\varepsilon^{d-1}$.
| 2 | https://mathoverflow.net/users/172802 | 433158 | 175,228 |
https://mathoverflow.net/questions/432148 | 11 | $\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}$Let $ p $ be a prime for which $ \PSL(2,p) $ is simple (so $ p \neq 2,3 $).
Is the minimal irrep of $ \PSL(2,p) $ defined over a quadratic extension? In particular I wish to ask:
for $ p $ congruent to $ 1 $ mod $ 4 $ is $ \PSL(2,p) $ always a subgroup of
$$
\SO\_{d\_{\min}}(\mathbb{Q}(\sqrt{p}))
$$
and for $ p $ congruent to $ 3 $ mod $ 4 $ if $ \PSL(2,p) $ is always a subgroup of
$$
\SU\_{d\_{\min}}(\mathbb{Q}(\sqrt{-p}))
$$
For example, this is true for $ p=5 $ with the "icosahedral" $ A\_5\cong \PSL(2,5) $ subgroup of $ \SO\_3 $. Appropriate generators are given in the section "Coxeter group generators" of <https://en.wikipedia.org/wiki/Icosahedral_symmetry>
Background:
The minimal degree $ d\_{\min} $ of a nontrivial irrep of $ \PSL(2,p) $ is
$$
d\_{\min}=\frac{p+1}{2}
$$
if $ p $ is congruent to $ 1 $ mod $ 4 $ and is
$$
d\_{\min}=\frac{p-1}{2}
$$
if $ p $ is congruent to $ 3 $ mod $ 4 $.
The value of the characters for degree $ d\_{\min} $ irreps are mostly $ 0,1,-1 $s also of course $ d\_{\min} $ and finally either
$$
\frac{1}{2} \pm \frac{\sqrt{p}}{2}
$$
if $ p $ is congruent to $ 1 $ mod $ 4 $ or
$$
-\frac{1}{2} \pm \frac{\sqrt{-p}}{2}
$$
if $ p $ is congruent to $ 3 $ mod $ 4 $. There are always exactly two irreps of degree $ d\_{\min} $ and their characters are related exactly by conjugation in the corresponding quadratic extension. All this information is in the first two nontrivial characters which can be found, for example, here
[http://www2.math.umd.edu/~jda/characters/psl2/](http://www2.math.umd.edu/%7Ejda/characters/psl2/)
Update: At this point I feel confident that, for $ p $ congruent to $ 1 $ mod $ 4 $, there is a $ PSL(2,p) $ subgroup of
$$
\SO\_{d\_{\min}}(\mathbb{R})
$$
and a $ PSL(2,p) $ subgroup of
$$
\SL\_{d\_{\min}}(\mathbb{Q}(\sqrt{p}))
$$
I'm still trying to understand if there is a $ PSL(2,p) $ subgroup of
$$
\SO\_{d\_{\min}}(\mathbb{Q}(\sqrt{p})) = \SL\_{d\_{\min}}(\mathbb{Q}(\sqrt{p})) \cap \SO\_{d\_{\min}}(\mathbb{R})
$$
Similarly, for $ p $ congruent to $ 1 $ mod $ 4 $, there is a $ PSL(2,p) $ subgroup of
$$
\SU\_{d\_{\min}}(\mathbb{R})
$$
and a $ PSL(2,p) $ subgroup of
$$
\SL\_{d\_{\min}}(\mathbb{Q}(\sqrt{-p}))
$$
I'm still trying to understand if there is a $ PSL(2,p) $ subgroup of
$$
\SU\_{d\_{\min}}(\mathbb{Q}(\sqrt{-p})) = \SL\_{d\_{\min}}(\mathbb{Q}(\sqrt{-p})) \cap \SU\_{d\_{\min}}
$$
Every representation is a determinant 1 since $ PSL(2,p) $ is simple ( for $ p \neq 2,3 $).
And we can always find a representation defined over the appropriate quadratic extension $ \mathbb{Q}(\sqrt{\pm p}) $ since every character of $ PSL(2,p) $ has Schur index 1 (thanks to @BenjaminSteinberg for this fact that an irrep with Schur index 1 can always be realized over the field generated by its character values for the fact that $ PSL(2,p) $ always has Schur index 1 see part (I) of Theorem 6.1 of <https://link.springer.com/content/pdf/10.1007/BF02762888.pdf>).
For $ p $ congruent to $ 1 $ mod $ 4 $ we can always find an orthogonal representation since the Frobenius-Schur indicator is $ 1 $ (indeed every irrep of $ PSL(2,p) $ for $ p $ congruent to $ 1 $ mod $ 4 $ has FS indicator $ 1 $, this is related to the fact that for such $ p $ every element of $ PSL(2,p) $ is strongly real, meaning conjugate to its inverse by an involution, and thus the group as a whole is totally orthogonal, see <https://arxiv.org/abs/1811.05343> for details).
| https://mathoverflow.net/users/387190 | Minimal irrep of $\mathrm{PSL}(2,p) $ | $\def\QQ{\mathbb{Q}}\def\FF{\mathbb{F}}\def\Sp{\text{Sp}}\def\SL{\text{SL}}\def\GL{\text{GL}}$Okay, time to do both cases using the Weil representation. This is going to rely on some Key Facts which I can check from my [explicit matrices for the Weil representation](https://mathoverflow.net/questions/432407) and which I do not know how to prove abstractly.
**Notation:** Let $\zeta$ be a primitive $p$-th root of unity. Let $L = \QQ(\zeta)$. Let $p^{\ast} = (-1)^{(p-1)/2} p$ and let $K = \QQ(\sqrt{p^{\ast}})$, this is the unique quadratic subfield of $L$. Let $\chi(a)$ be the quadratic residue symbol $\left( \tfrac{a}{p} \right)$.
For $a \in \FF\_p^{\times}$, let $\sigma\_a$ be the element of $\text{Gal}(L/\QQ)$ with $\sigma\_a(\zeta) = \zeta^a$.
For $a \in \FF\_p^{\times}$, let $t(a)$ be the matrix $\left[ \begin{smallmatrix} a & 0 \\ 0 & a^{-1} \end{smallmatrix} \right]$ in $\SL\_{2}(\FF\_p)$. I'll write $\gamma : \SL\_{2}(\FF\_p) \to \SL\_{p}(\mathbb{C})$ for the Weil representation. (This argument will also extend to the Weil representation of $\Sp\_{2k}(\FF\_p)$ for $k>1$.)
Here are the key facts:
**Key Facts** We can choose the map $\gamma$ such that all of the following are simultaneously true:
(1) All the matrices $\gamma(g)$ are in $\SL\_{p}(L)$.
(2) For $a \in \FF\_p^{\times}$, we have $\sigma\_{a^2}(\gamma(g)) = \gamma(t(a)^{-1} g t(a))$.
(3) The matrices $t(\gamma(a))$ are $\pm 1$ times the permutation matrices with permute $\FF\_p$ by multiplication by $a$. Specifically, the sign is $\chi(a)$.
**Writing down the minimal representation:**
The involution $t(-1)$ is central in $\SL\_2(\FF\_p)$. Using Key Fact (3) above, it acts on the Weil representation with eigenvalues $1$ and $-1$, having multiplicities $\tfrac{p + \chi(-1)}{2}$ and $\tfrac{p-\chi(-1)}{2}$ respectively; the $1$-eigenspace is the representation which factors through $\text{PSL}\_2(\FF\_p)$. I'll write $\gamma\_+(g)$ for the action of $g \in \SL\_2(\FF\_p)$ on this $1$-eigenspace. We will need to understand this more explicitly:
When $p \equiv 1 \bmod 4$, the matrix $\gamma(t(-1))$ is the permutation matrix $b \mapsto -b$ on $\FF\_p$. So we can think of the $1$-eigenspace as even functions on $\FF\_p$, or in other words as functions on $\FF\_p/\{ \pm 1 \}$. We can think of $\gamma\_+(t(a))$ as the permutation $a$ acting on $\FF\_p/\{\pm 1\}$, times $\chi(a)$.
When $p \equiv 3 \bmod 4$, the matrix $\gamma(t(-1))$ is $-1$ times the permutation matrix $b \mapsto -b$ on $\FF\_p$. So we can think of the $1$-eigenspace as odd functions on $\FF\_p$. Write $S$ for the group of quadratic residues in $\FF\_p^{\times}$. For any $a \in \FF\_p^{\times}$, exactly one of $a$ and $-a$ is in $S$, so an odd function is determined by its value on $S$, so we can think of the $1$-eigenspace as functions on $S$. We can think of $\gamma\_+(t(a))$ as the permutation $\chi(a) a$ acting on $S$.
**Conjugating into $\SL(K)$**
Let $\delta$ be a matrix in $\GL(L)$. We'd like to have $\delta \gamma\_+(g) \delta^{-1} \in \SL(K)$ for all $g$ in $\SL\_2(\FF\_p)$. Now, $K$ is the fixed field of the subgroup $\{ \sigma\_{a^2}: a \in \FF\_p \}$, so we want to have
$$\sigma\_{a^2}(\delta \gamma\_+(g) \delta^{-1}) = \delta \gamma\_+(g) \delta^{-1}.$$
We can rewrite the LHS as
$$\sigma\_{a^2}(\delta) \sigma\_{a^2}(\gamma\_+(g)) \sigma\_{a^2}(\delta)^{-1} =
\sigma\_{a^2}(\delta) \gamma\_+(t(a))^{-1} \gamma\_+(g) \gamma\_+(t(a)) \sigma\_{a^2}(\delta)^{-1}.$$
So we want
$$\sigma\_{a^2}(\delta) \gamma\_+(t(a))^{-1} \gamma\_+(g) \gamma\_+(t(a)) \sigma\_{a^2}(\delta)^{-1} = \delta \gamma\_+(g) \delta^{-1}$$
or, in other words,
we want
$$\delta^{-1} \sigma\_{a^2}(\delta) \gamma\_+(t(a))^{-1}$$
to commute with all $\gamma\_+(g)$.
Since the representation $\gamma\_+$ is irreducible, by Schur's lemma, this will happen if and only if
$$\sigma\_{a^2}(\delta) = c\_a \delta \gamma\_+(t(a)) \qquad (\ast)$$
for all $a \in \FF\_p^{\times}$, for some scalar $c\_a \in L^{\times}$.
Our goal, therefore, is to write down an invertible matrix $\delta$ with entries in $L$, satisfying $(\ast)$.
**Finding $\delta$**
In the case that $p = 3 \bmod 4$, we think of the rows and columns of $\delta$ as indexed by $S$. We define the matrix $\delta$ by $\delta\_{yx} = \zeta^{y^2 x^2}$. Since $x \mapsto x^2$ is a permutation of $S$, this is a Vandermonde matrix and hence invertible. (Really, $y^2$ could be replaced by any bijection $S \to S$, but the $x^2$ term is important, and I thought it was most elegant to use the same formula for both.)
For $a \in S$, we have $\sigma\_{a^2}(\zeta^{y^2 x^2}) = \zeta^{a^2 (y^2x^2)} = \zeta^{y^2 (ax)^2}$. The latter is the $(y,x)$ entry in $\delta \gamma\_+(t(a))$. This checks $(\ast)$ for $a \in S$; if $-a \in S$ instead, then note that $\sigma\_{a^2} = \sigma\_{(-a)^2}$ and $\gamma\_+(t(a)) = \gamma\_+(t(-a))$. So we have checked $(\ast)$ with $c\_a = 1$.
Now, let's do the $p \equiv 1 \bmod 4$ case. In this case, we think of the rows and columns of $\delta$ as indexed by $\FF\_p/ \{ \pm 1 \}$. Once again, we define $\delta$ by $\delta\_{yx} = \zeta^{y^2 x^2}$. Once again, this is a Vandermonde matrix, and thus invertible. If $\gamma\_+(t(a))$ acted by the permutation of multiplication by $a$ on $\FF\_p/ \{ \pm 1 \}$, we would once again verify $(\ast)$ with $c\_a = 1$. Since, in fact, $\gamma\_+(t\_(a))$ is $\chi(a)$ times this permutation matrix, we still win, but with $c\_a = \chi(a)$.
| 3 | https://mathoverflow.net/users/297 | 433161 | 175,229 |
https://mathoverflow.net/questions/433117 | 3 | Let $G$ be a regular $n$-vertex graph which is edge transitive. How large can the degrees of $G$ be if it is not a complete $r$-partite graph?
The best I can do is about $n/2$ by considering two disjoint cliques of the same size. A relatively easy argument shows that any $n$-vertex edge transitive graph with degree $n-o(\sqrt{n})$ must be complete $r$-partite, so one can not hope to do better than this. I'd be interested to see improvements to either of these two bounds, and in particular in knowing whether there can exist such graphs with degrees $n-o(n)$.
| https://mathoverflow.net/users/106377 | Super dense edge-transitive graphs | Take $G=\overline{K\_m\square K\_m}$, the complement of the Cartesian product of two copies of the complete graph on $m$ vertices. It has $m^2$ vertices, valency $(m-1)^2$ and is edge-transitive. (In fact, we have $\mathrm{Aut}(G)=S\_m\wr S\_2= (S\_m\times S\_m)\rtimes S\_2$.)
| 5 | https://mathoverflow.net/users/22377 | 433163 | 175,230 |
https://mathoverflow.net/questions/432834 | 0 | Given $n$ balls, all of which are initially in the first of $n$ numbered boxes, $a(n)$ is the number of steps required to get one ball in each box when a step consists of moving to the next box every second ball from the highest-numbered box that has more than one ball.
Not so long ago I proposed this sequence to the OEIS, where during the discussion [**Jon E. Schoenfield**](https://oeis.org/wiki/User:Jon_E._Schoenfield) proposed a formula for $n=2^k$ (already proven [here](https://mathoverflow.net/a/432747/231922)), as well as the following recurrent formula:
$$a(n)=\frac{n(n-1)}{2}-b(n)$$
$$b(n)=[n>3](b\left(\left\lfloor\frac{n}{2}\right\rfloor\right)+c\left(\left\lceil\frac{n}{2}\right\rceil-2\right)+\left\lfloor\frac{n}{2}\right\rfloor-1)$$
Here $c(n)$ is [A181132](https://oeis.org/A181132), i.e., $c(0)=0$; thereafter $c(n) = $ total number of $0$'s in binary expansions of $1, \cdots, n$.
It looks very likely that this formula is correct. To verify it one may use this PARI prog:
```
a(n)=my(A, B, v); v=vector(n, i, 0); v[1]=n; A=0; while(v[n]==0, B=n; while(v[B]<2, B--); v[B+1]+=v[B]\2; v[B]-=v[B]\2; A++); A
c(n)=n + sum(k=1, n, logint(k, 2) - hammingweight(k))
b(n)=if(n>3, b(n\2) + c(ceil(n/2) - 2) + n\2 - 1)
a1(n)=n*(n-1)/2 - b(n)
```
Is there a way to prove it?
| https://mathoverflow.net/users/231922 | Recurrence for the number of steps required to get one ball in each box | Generalise $a$: $a(n, k)$ is the number of steps to perform this process with $n+k$ boxes and balls starting with $n \ge 1$ balls in the first box and one ball each in the next $k$ boxes. Then the original $a(n)$ is $a(n, 0)$.
Clearly $a(1, k) = 0$. If $n > 1$ then we propagate $\lfloor \tfrac n2 \rfloor$ balls right one box, process that tail, and then return to the head. So we have $$a(n, k) = \begin{cases}
0 & \textrm{if } n = 1 \\
1 + a(\lfloor \tfrac n2 \rfloor, 0) + a(\lceil \tfrac n2 \rceil, \lfloor \tfrac n2 \rfloor) & \textrm{if } n > 1 \wedge k = 0 \\
1 + a(1+\lfloor \tfrac n2 \rfloor, k-1) + a(\lceil \tfrac n2 \rceil, k + \lfloor \tfrac n2 \rfloor) & \textrm{if } n > 1 \wedge k > 0
\end{cases}$$
---
If we were to propagate the balls one at a time, it's easy to see that we would require $\frac{n(n-1)}2$ steps. Therefore if we define $b(n, k) = \frac{n(n-1)}2 - a(n, k)$ it counts the total "excess" over one ball per group of the groups propagated. Using the same structural argument as for the cases of $a(n,k)$ we therefore find
$$b(n, k) = \begin{cases}
0 & \textrm{if } n = 1 \\
\lfloor \tfrac n2 \rfloor - 1 + b(\lfloor \tfrac n2 \rfloor, 0) + b(\lceil \tfrac n2 \rceil, \lfloor \tfrac n2 \rfloor) & \textrm{if } n > 1 \wedge k = 0 \\
\lfloor \tfrac n2 \rfloor - 1 + b(1+\lfloor \tfrac n2 \rfloor, k-1) + b(\lceil \tfrac n2 \rceil, k + \lfloor \tfrac n2 \rfloor) & \textrm{if } n > 1 \wedge k > 0
\end{cases}$$
Then the conjecture is equivalent to $$b(\lceil \tfrac n2 \rceil, \lfloor \tfrac n2 \rfloor) \stackrel?= c(\lceil \tfrac n2 \rceil - 2)$$
---
$c(m)$ is noted in <https://oeis.org/A181132> to have recurrence (which I've relabelled)
>
> $$c(m) = \begin{cases} 0 & \textrm{if } m = 0 \\ c(j) + c(j-1) + j & \textrm{if } m = 2j \\ c(j) + c(j) + j & \textrm{if } m = 2j+1 \end{cases}$$
>
>
>
which we could rewrite as $$c(n) = \begin{cases} 0 & \textrm{if } n = 0 \\
c(\lfloor \tfrac n2 \rfloor) + c(\lceil \tfrac n2 \rceil - 1) + \lfloor \tfrac n2 \rfloor & \textrm{otherwise} \end{cases}$$
---
Finally, we prove by strong induction over $n$ that $$\forall n > 1, k \ge \lfloor \log\_2 n \rfloor - 1 : b(n, k) = c(n - 2)$$
Base cases: $n \in \{2,3\}$. We only ever propagate one ball at a time, so $b(2, k) = b(3, k) = 0$.
Inductive step: let $n \ge 4$ and suppose the given property holds for all $n' < n$. Let $k \ge \lfloor \log\_2 n \rfloor - 1$. Then $$\begin{eqnarray\*}b(n, k) &=& \lfloor \tfrac n2 \rfloor - 1 + b(1+\lfloor \tfrac n2 \rfloor, k-1) + b(\lceil \tfrac n2 \rceil, k + \lfloor \tfrac n2 \rfloor) \\
&=& \lfloor \tfrac n2 \rfloor - 1 + c(\lfloor \tfrac n2 \rfloor - 1) + c(\lceil \tfrac n2 \rceil - 2) \\
&=& c(n - 2)
\end{eqnarray\*}$$ as desired.
---
Going beyond what was asked in the question, this then lets us derive $$b(n) = b(\lfloor \tfrac n2 \rfloor) + c(n-2) - c(\lfloor \tfrac n2 \rfloor - 1)$$ from which, by induction, we have $$b(n) = c(n-1) - \sum\_{a \ge 0} \textrm{A080791}\left(\left\lfloor \frac n{2^a} \right\rfloor - 1\right)$$ where [A080791](https://oeis.org/A080791) is the sequence summed by $c$.
| 1 | https://mathoverflow.net/users/46140 | 433195 | 175,238 |
https://mathoverflow.net/questions/433153 | 6 | I found myself trying to prove the following, but I had to compute everything explicitly.
It is well known that if $u:\mathbb{R}^n\to\mathbb{R}$ is an harmonic function on $\mathbb{R}^n$, then the so-called Kelvin transform of $u$
$$(Ku)(x):=\frac{1}{|x|^{n-2}}u\biggl(\frac{x}{|x|^2}\biggr)$$
is harmonic in $\mathbb{R}^n\setminus\{0\}$.
Question: Is there a way to prove that $Ku$ is harmonic in $\mathbb{R}^n\setminus\{0\}$ WITHOUT making the explicit computations?
I know that there are tricks to soften the calculations (for example noticing that $1/|x|^{n-2}$ is harmonic in $\mathbb{R}^n\setminus\{0\}$), however I wasn't able to find a "clever" way to do it. I noticed that $x\mapsto x/|x|^2$ is a special transformation but I wasn't able to go anywhere, not even using the mean value property.
| https://mathoverflow.net/users/170893 | Why is $\frac{1}{|x|^{n-2}}u(\frac{x}{|x|^2})$ harmonic if $u$ is harmonic? | An explanation can be the following. Take a harmonic function of the form $u(x)=r^\alpha P(\omega)$, with $r=|x|$, $|\omega|=1$ and $P$ a polynomial. Then $0=\Delta u=r^{\alpha-2}\left (\alpha(N-2+\alpha)P(\omega)+\Delta\_S P(\omega)\right)$ with $\Delta\_S$ the Laplace-Beltrami on the unit sphere $S$. Then $\Delta\_S P=-\alpha(N-2+\alpha)P$. However we know that the eigenvalues of $\Delta\_S$ are given by $-k(N-2+k)$ with $k$ a nonegative integer but the equation $\alpha(N-2+\alpha)=k(N-2+k)$ has another root $\alpha=2-N-k$ which corresponds to the Kelvin transform $Ku(x)=|x|^{2-N}u(x/|x|^2)$. This shows that $Ku$ is harmonic whenever $u$ is an harmonic polynomial and then (approximating locally) for any harmonic function $u$.
| 7 | https://mathoverflow.net/users/150653 | 433212 | 175,246 |
https://mathoverflow.net/questions/125878 | 14 | I'm sure this is a fairly basic question, but I can't seem to find a solid answer:
My primary question is: Is there a reasonably nice subsystem of second-order arithmetic corresponding essentially to "primitive recursive comprehension?" I'm interested only in $\omega$-models - that is, I don't care about how much induction the system allows. In fact, I'd prefer it to have full induction, so that I know the weakness of the system lies squarely in its comprehension axioms. I do mean "comprehension" here: I'd like this system to be a two-sorted system, like $RCA$ itself, so e.g. $PRA$ is not what I'm looking for - although maybe it can be tweaked into a satisfactory system in an easy way?
My secondary question is: assuming a positive answer to the first question, what are some natural statements which are equivalent (over this base system) to $RCA$ (or $RCA\_0$)?
(I'm almost certain this is written up nicely somewhere easily accessible, and my google-fu is simply failing me; if this is the case, and this question is therefore inappropriate for mathoverflow, please feel free to close it.)
| https://mathoverflow.net/users/8133 | Reverse mathematics below RCA | It looks like there is now a paper explicitly treating such a system! [Bazhenov/Fiori-Carones/Liu/Melnikov, *Primitive Recursive Reverse Mathematics*](https://arxiv.org/pdf/2210.13080.pdf) just appeared on the arxiv. Here are a couple key points:
* Models of their base theory, $\mathsf{PRA}^2$, are precisely pairs $(M,X)$ where $M\models\mathsf{PRA}$ and $X$ is a family of functions on $M$ closed under composition and primitive recursion and containing the constants.
* Over $\mathsf{PRA^2}$, there are two natural versions of recursive comprehension: $\mathsf{RCA\_0}$ itself and the somewhat-weaker $2^\mathbb{N}$-$\mathsf{RCA\_0}$. The difference is whether the objects considered belong to Baire or Cantor space.
| 2 | https://mathoverflow.net/users/8133 | 433216 | 175,248 |
https://mathoverflow.net/questions/433081 | 3 | For $s\in(0,1],$ consider the following non-local fractional laplacian:
$$(-\Delta)^sv= f ~~\text{on } \mathbb{R}^n.$$
Then how to use "the standard elliptic estimate" to obtain:
* for $p\in[1, \frac{n}{n-2s}),$
$$\|v\|\_{L^p(B\_2\setminus B\_1)} \lesssim \|f\|\_{L^1(B\_3\setminus B\_{\frac{1}{2}})}+\|v\|\_{L^1(B\_4\setminus B\_{\frac{1}{2}})}.$$
When I see the index $\frac{n}{n-2s}$, I immediately think it is due to the Sobolev inequality and
the Calderon-Zygmund inequality, but the Calderon-Zygmund inequality fails at the endpoint index $1$ even in the case $s=1$. Any insights or references are appreciated!
* Update:
When $s=1$, consider a cut-off function $\eta$ equals $1$ in $B\_2\setminus B\_1$, and $0$ out of $B\_4\setminus B\_{\frac{1}{2}}.$ Then we have
$$-\Delta \eta v = f\eta -v\Delta \eta -2\nabla v. \nabla \eta$$
For any $\epsilon>0$, $h\in L^{\frac{n}{2}+\epsilon}(B\_4\setminus B\_{\frac{1}{2}})$, let $\phi\in H\_0^1(B\_4\setminus B\_{\frac{1}{2}})$ be the solution of
$$-\Delta \phi=h.$$
By Calderon-Zygmund inequality and Sobolev embedding we have
$$\|\phi\|\_{C^1(B\_4\setminus B\_{\frac{1}{2}})}\leq \|h\|\_{L^{\frac{n}{2}+\epsilon}(B\_4\setminus B\_{\frac{1}{2}})}$$
Test $-\Delta \eta v = f\eta -v\Delta \eta -2\nabla v. \nabla \eta$ by $\phi,$ we obtain:
$$\int\eta v h \lesssim (\|f\|\_{L^1(B\_3\setminus B\_{\frac{1}{2}})}+\|v\|\_{L^1(B\_4\setminus B\_{\frac{1}{2}})})\|\phi\|\_{C^1(B\_4\setminus B\_{\frac{1}{2}})}.$$
Hence
$$\int\eta v h \lesssim (\|f\|\_{L^1(B\_3\setminus B\_{\frac{1}{2}})}+\|v\|\_{L^1(B\_4\setminus B\_{\frac{1}{2}})}) \|h\|\_{L^{\frac{n}{2}+\epsilon}(B\_4\setminus B\_{\frac{1}{2}})}.$$
By dual we obtain that for every $p\in[1, \frac{n}{n-2})$:
$$\|\eta v\|\_{L^p} \lesssim \|f\|\_{L^1(B\_3\setminus B\_{\frac{1}{2}})}+\|v\|\_{L^1(B\_4\setminus B\_{\frac{1}{2}})}.$$
But I have no idea to deal with the non-local case $s<1.$
| https://mathoverflow.net/users/166368 | Endpoint Calderon-Zygmund inequality of nonlocal fractional laplacian | I think the estimate is false, at least for $n=1$ and $0 < s < 1/2$, due to the non-locality you mention (I imagine similar arguments would work in other non-local cases). If it held, then one would have
$$ \|v\|\_{L^p([1,2])} \leq C \| v \|\_{L^1([-4,4])} \quad (1)$$
for some fixed constant $C$ whenever $(-\Delta)^s v$ vanished on $[-3,3]$. In particular the estimate (1) should hold when $v$ is a translated Riesz potential $(x-x\_0)^{2s-1}$ for any $x\_0$ outside of say $[-5,5]$ (one can mollify first and take limits if one wants $f$ to be a nice function rather than a Dirac mass). By taking difference quotients and passing to the limit, the same estimate (1) must then hold for $\frac{\partial}{\partial x\_0} (x-x\_0)^{2s-1}$, and similarly for higher derivatives. Thus (1) holds for all functions of the form $P \left(\frac{1}{x-10} \right) (x-10)^{2s-1}$ (say) for any polynomial $P$, hence by the Weierstrass approximation theorem it holds for all $v \in C([-4,4])$ (say). But one can easily contradict (1) in this class by considering a $v$ that is concentrated in a small subinterval of $[1,2]$.
| 9 | https://mathoverflow.net/users/766 | 433221 | 175,250 |
https://mathoverflow.net/questions/433217 | 3 | Suppose $X$ is a Banach space with the following property: For any $x\in X$ there exists a two dimensional subspace $E$ *isometric* with $l\_2^2$ such that $x\in E$. Does this property characterize a (separable) Hilbert space?
What about the stronger property: For any $x$ and any $n\in\mathbb{N}$ there exists a $n$-dimensional subspace $E$ isometric with $l\_2^n$ such that $x\in E$?
| https://mathoverflow.net/users/69275 | Do these properties characterize Hilbert spaces? | For any Banach space $X$ you can consider $X\oplus l^2$, with norm $||(x,y)||:=(||x||^2+||y||^2)^\frac{1}{2}$. Then for each $x\in X$, span$(x)\oplus l^2$ is isometric to $l^2$, so $X\oplus l^2$ is covered by isometric copies of $l^2$
| 6 | https://mathoverflow.net/users/172802 | 433225 | 175,251 |
https://mathoverflow.net/questions/433181 | 8 | Let $X$ and $Y$ be metric space, $X$ be compact, $C(X,Y)$ denote the set of continuous functions from $X$ to $Y$ with uniform convergence on compacts topology, and $\operatorname{Lip}(X,Y)$ denote the subspace of Lipschitz functions from $X$ to $Y$. Under what conditions is $\operatorname{Lip}(X,Y)$ a dense subset of $C(X,Y)$?
---
*I expect that there should be some “topological compatibility condition” between $X$ and $Y$, but this is purely (and likely unfounded) intuition.*
| https://mathoverflow.net/users/491352 | Uniform density of Lipschitz maps is space of continuous function — for general metric spaces | Let $(X,\rho)$ be a compact metric space, and let $(Y,d)$ be a separable metric space. Then I claim that one can endow $X$ with a compatible metric $d$ such that every continuous $f:X\rightarrow Y$ can be uniformly approximated by a Lipschitz function.
Let $(f\_n)\_{n\geq 0}$ be a sequence of continuous functions such that every continuous $f:X\rightarrow Y$ can be uniformly approximated by some $f\_n$. Let $(\alpha\_n)\_{n\geq 0}$ be a sequence of positive real numbers such that $\alpha\_n\cdot\operatorname{Diam}(f\_n[X])\rightarrow 0$. Then define a metric $d$ on $X$ by setting $$d(x,y)=\rho(x,y)+\sup\_n\alpha\_n\cdot d(f\_n(x),f\_n(y)).$$ Then the metric $d$ is compatible with the original topology on $X$. Furthermore, for each $n$, we have
$d(f\_n(x),f\_n(y))\leq\alpha\_n^{-1}\cdot d(x,y)$, so each $f\_n$ is Lipschitz.
| 14 | https://mathoverflow.net/users/22277 | 433227 | 175,252 |
https://mathoverflow.net/questions/433226 | 7 | Are the categories of sets, abelian groups, and commutative rings unique? Independence results like the independence of the generalized continuum hypothesis, the Whitehead problem, and the global dimension of $\prod\_{n = 1}^\infty \mathbb{F}\_2$ from ZFC seem to indicate no. And yet we don't say "Let $\mathbf{Set}$ be a category of sets" instead of "Let $\mathbf{Set}$ be the category of sets," etc. If these categories are not unique, then *could* they be if we wanted them to? And *shouldn't* they be unique, if the terms "set," "function," "abelian group," "commutative ring," etc., are to be well-defined?
EDIT: Questions have arisen as to what I mean by "unique". Unique as in one and only one? Unique as in unique up to equivalence of categories? Unique as in unique up to isomorphism of categories? Unique as in unique up to unique isomorphism of categories? Honestly, any of those four senses of "unique" is fine with me. Just pick one and answer that particular question. Does it make sense to consider $\mathbf{Set}$ as a category that is unique in some sense described above?
EDIT (SUMMARY): Answers to this question seem to have been of one of the following flavors.
1. Monism allows $\mathbf{Set}$ to be unique, but pluralism does not. Some mathematicians are monists, some are pluralists, while others think that both monism and pluralism are respectable philosophies of mathematics. Call these views "monism," "pluralism," and "monist-pluralist dualism."
2. Just as $\mathbb{Z}$ is unique up to iso in the category of rings, $\mathbf{Set}$ and $\mathbf{Ring}$ are unique up to category equivalence in the meta-category of categories, but there is no sense of uniqueness in an absolute sense. $\mathbb{Z}$ and $\mathbf{Set}$ and $\mathbf{Ring}$ are not unique in an absolute sense. If this is the case, then I'd argue that mathematical truth, too, is not absolute, but rather relative to a theory. Call this view "relativism." This view is new to me. I am not a relativist, at least as yet.
3. $\mathbb{Z}$ is unique up to iso, $\mathbf{Set}$ and $\mathbf{Ring}$ are unique up to category equivalence, etc. Mathematical truth is absolute and does not mean truth within a theory, and uniqueness is also absolute. Call this view "absolutism."
4. Conventionally, mathematicians today are operating under the assumptions of ZFC, and ZFC doesn't answer the question in the positive or in the negative, and so the question doesn't have a mathematical answer. Call this view "conventionism".
The answer one gives to the question apparently depends on what philosophical views one holds. But the same is true about questions like, "Is the axiom of choice true," "Does every nontrivial commutative ring have a maximal ideal," "Does every vector space have a basis?" A relativist would say, yes, if one is working under the assumptions of ZFC and the ordinary rules of mathematical proof. An absolutist would just say yes (if they believed that the axioms and theorems of ZFC were true). Others might just say yes, under the tacit assumption that most mathematicians today are operating under the assumptions of ZFC, while remaining agnostic about the status of mathematical truth.
Absolutists might see the question I posted as a valid mathematical question. A problem with asbolutism is that many important questions, like GCH and the Whitehead problem, have not been settled, at least as yet. A problem with relativism is, why are "if-then" statements true? What makes Boolean first-order logic absolute but not the rest of mathematics? Why are any mathematical proofs valid at all? Why not assume some non-Boolean constructive or intuitionistic logic? If my question is not a purely mathematical one and is partly "philosophical" and "open to interpretation", then isn't every mathematical question thus?
These aren't additional questions I'm asking for discussion here. I'm just trying to explain why I thought my question was a valid mathematical question and not a purely philosophical one. Thank you to those who tried to answer it, as your answers have much clarified my thinking about the problem. I've accepted Hamkins' answer because it was least biased, but I welcome other answers to the question and can always change what answer I accept. I felt pressured to accept an answer because of a vote to close the question as inappropriate for MO.
| https://mathoverflow.net/users/17218 | Are the categories of sets, abelian groups, and commutative rings unique? | *Introduction to pluralism*
A version of this question lies at the heart of the ongoing dispute on pluralism in the philosophy of mathematics. Is there at bottom just one mathematical reality? Does every mathematical question, whether about arithmetic, about the real continuum, or about set theory, have a definite mathematical answer?
Most mathematicians (but not all) take the view that arithmetic assertions, for example, assertions such as the Riemann hypothesis or the question whether there are infinitely many prime pairs, have a definite answer. Either there are infinitely many prime pairs or there are not, full stop. According to this view, arithmetic questions have a determinate answer, whether we shall ever come to know it.
We know by Gödel's theorem, of course, that no effective axiomatic system (whether formalized in set theory, category theory, HoTT, or what have you) will be able to establish the truth of all the true arithmetic assertions. But this observation can be taken to be merely about the weakness of our formal theories, rather than necessarily about any kind of pluralism in the arithmetic facts of the matter. One can admit that any given theory is weak, even if one holds that true arithmetic is meaningful. Peano arithmetic PA is incomplete, but ZFC proves more arithmetic truths, and ZFC + large cardinals settles still more.
*Characterizing structures in second-order logic*
Support for this determinate view of arithmetic truth is often taken from the categoricity results. Dedekind proved, namely, that the natural number structure $\langle\mathbb{N},S,0\rangle$ is uniquely determined up to isomorphism by the three Dedekind axioms, that $0$ is not a successor, that the successor function is one-to-one, and that every number is generated from $0$ by successor, in the sense that every set of numbers containing $0$ and closed under successor contains all the numbers. Once one knows that the axioms determine the structure, then one knows that those axioms determine all arithmetic truth. I have argued that Dedekind's categoricity result is the beginning of structuralism in mathematics.
Other mathematicians observe that we have such categorical accounts of essentially all our familiar mathematical structures. The integer ring is uniquely determined up to isomorphism from the natural numbers, and the rational field. The real field is the unique complete ordered field. The complex numbers are the unique algebraic closure of the real field, or alternatively, they are the unique algebraically closed field of size continuum.
Thus, all our familiar mathematical structures are characterized uniquely in second-order logic. This can be taken as support for the view that mathematical truth generally is determinate in nature. The structures are determined by the (second-order) axioms, and thus the truths are determined.
*Characterizing structures in first-order set theory*
All these familiar categoricity arguments can be viewed as taking place in first-order set theory, provable as theorems of ZFC. In a sense, the first-order theory of sets provides a natural interpretation of the second-order logic of any particular structure, by providing the sets that will constitute the second-order part. So ZFC proves that there is a unique structure of arithmetic up to isomorphism, a unique real-closed field with a unique algebraic closure. Similarly, ZFC proves that the category of groups and the category of sets and so on is unique up to suitable isomorphism.
The philosophical difficulty here is that, meanwhile, ZFC is a first-order theory and thus subject to the incompleteness phenomenon. We know, for example, that if ZFC is consistent, then there can be models $M\_0$ and $M\_1$ of ZFC whose natural number structures $\mathbb{N}^{M\_0}$ and $\mathbb{N}^{M\_1}$ are not isomorphic to each other, even though each of them is thought to be the unique natural number structure in those respective set-theoretic universes. The philosophical difficulty is that, although ZFC proves that arithmetic truth is definite, nevertheless different models of ZFC can think that it is different arithmetic truths that come to be part of the definitely true arithmetic theory.
*Characterizing structures in second-order set theory*
In light of this, some logicians and philosophers of mathematics seek to apply the second-order categoricity results to set theory itself. Indeed, Zermelo proved that the models of second-order ZFC${}\_2$ enjoy a quasi-categoricity result—they all agree on their common initial segments. Specifically, Zermelo proved that the models of second-order ZFC${}\_2$ are exactly the models $V\_\kappa$ for an inaccessible cardinal $\kappa$, or in other words, exactly the Zermelo-Grothendieck universes. These set-theoretic worlds are linearly ordered and all agree on the assertions expressible in their common parts.
Kreisel famously pointed out that indeed essentially all questions of classical mathematics are expressible as sentences that are absolute to low-level ranks of the cumulative hierarchy, which have the same truth value in all these Zermelo-Grothendieck universes. Thus, according to Kreisel, the continuum hypothesis has a determinate truth value in second-order set theory. It is either definitely true or definitely false, as a matter of (second-order) logic.
*The universe view of sets*
This is the beginning of the *universe view*, which holds that there is a unique set-theoretic reality underlying mathematics, and all statements have a definite truth value in this unique set-theoretic realm. This is the set-theoretic universe arising from the cumulative set-building process, where one iteratively computes the $V\_\alpha$ hierarchy by adding all subsets at each stage and iterating through the ordinals. On this view, the answer to your question is Yes, there is a unique category of all groups, and it is the category of groups as defined in this final true set-theoretic universe $V$, and similarly with the category of rings and what have you.
Critics point out that second-order logic is simply a species of set theory. How can we establish the definiteness of our concept of finiteness by appealing to the comparatively murky concept of *arbitrary set* required in the second-order induction axiom? It seems hopeless to ground our concept of the finite this way. See my essay, [A question for the mathematics oracle](http://jdh.hamkins.org/question-for-the-math-oracle/).
*The multiverse view of sets*
Set-theoretic pluralism offers an alternative perspective. According to the *multiverse view*, there are many concepts of set, each giving rise to a different set-theoretic realm. The continuum hypothesis might hold in some and not in others, and this situation is itself a kind of answer to the CH question—it holds and fails throughout the set-theoretic multiverse in a way that is quite deeply understood.
In regard to your question, these different set-theoretic universes each have their own (unique in that universe) categories of groups and categories of rings and so on, and these categories are not always isomorphic to each other across the universes. On the pluralist view, the answer to your question is negative.
Indeed, the question of whether they are isomorphic or not presumes a certain degree of set theory in the metatheory where those universes exist, and in this way one is led to the idea that there is a hierarchy of metatheoretic contexts. Indeed, every model of set theory provides a meta-theoretic context for the theories and models and categories which exist within that model. In this way, the traditional object-theory/meta-theory distinction is seen to break down as naive or crude, for we actually have a rich hierarchy of theories, each serving also as a metatheory.
*More extreme and more moderate alternatives*
Strong forms of the pluralist view extend to pluralism even in arithmetic as well as higher set theory. Many mathematicians prefer a kind of compromise position, taking arithmetic truth as definite, but allowing indeterminacy in higher set-theoretic truths. The universists hold that the set-theoretic universe is determinate all the way up, and the large cardinal hierarchy is pointing the way toward the one road upward.
So there are philosophical positions taken on all sides of this issue. I have written at length on these topics in various venues, but perhaps you might look at:
* My paper: *Hamkins, Joel David*, [**The set-theoretic multiverse**](http://dx.doi.org/10.1017/S1755020311000359), Rev. Symb. Log. 5, No. 3, 416-449 (2012). [ZBL1260.03103](https://zbmath.org/?q=an:1260.03103).
* Chapter 8 of my book: *Hamkins, Joel David*, [Lectures on the philosophy of mathematics](https://mitpress.mit.edu/9780262542234/), MIT Press, 2021.
| 23 | https://mathoverflow.net/users/1946 | 433235 | 175,256 |
https://mathoverflow.net/questions/433239 | 5 | Leap years are determined by a scheme in which every $4$th year is a leap year, but every $4\cdot 25$th year is exempted, but every $4\cdot 25 \cdot 4$th year is reinstated $\ldots $ and there we stop, because that's good enough in practice to approximate the actual length of a year in terms of days. But what if we wanted to approximate arbitrarily well? In other words, can any real number $r$, $0<r<1$, be written in the form
$$
r=\sum\_{n=0}^{\infty} \frac{(-1)^n }{\prod\_{i=0}^{n} a\_i }
$$
with $a\_i $ a sequence of positive integers? For the earth at its current rate of rotation, where a fraction $r=0.242375$ of a day has to be represented, the existing sequence $a\_0 =4$, $a\_1 =25$, $a\_2 =4$ merely has to be supplemented by $a\_3 =20$ to reproduce this $r$ to the full known precision.
| https://mathoverflow.net/users/134299 | Leap year formula to arbitrary precision | The greedy algorithm always produces a suitable expansion. The proof follows.
**Lemma**: If $0<x<1$, then $x$ may be written as $\frac 1n(1-y)$ for some $n$ with $0\le y<\frac 1{n+1}$.
Proof: Let $n$ be such that $\frac 1{n+1}<x\le \frac 1n$. Then $0\le \frac 1n-x< \frac 1{n(n+1)}$. In particular, writing $y=n(\frac 1n-x)$, we see $0\le y<\frac 1{n+1}$ as required.
Note that in the above, $n=\lfloor \frac 1x\rfloor$ so that $y=\lfloor\frac 1x\rfloor (1/\lfloor \frac 1x\rfloor-x)=1-x\lfloor \frac 1x\rfloor$. Define the function $f(x)=\lfloor \frac 1x\rfloor$ and $g(x)=1-xf(x)$.
We now apply this inductively: let $x\_0=r$, and $x\_{n+1}=g(x\_n)$ for each $n$ and $a\_n=f(x\_n)$. We claim that for each $n$,
$$
r=\sum\_{j=0}^n \frac{(-1)^j}{\prod\_{i=0}^j a\_i}+\frac{(-1)^{n+1}}{\prod\_{i=0}^n a\_i}x\_{n+1}.
$$
Since $x\_{n+1}=\frac 1{a\_{n+1}}(1-x\_{n+2})$, the formula holds immediately by induction, proving the claim.
| 8 | https://mathoverflow.net/users/11054 | 433243 | 175,258 |
https://mathoverflow.net/questions/433142 | 5 | Cross-posted from [MSE](https://math.stackexchange.com/questions/4539025/can-solvable-lie-groups-have-maximal-subgroups).
Many interesting manifolds can be expressed as $ G/H $ for $ G $ a connected Lie group and $ H $ a maximal closed subgroup. Examples include the projective spaces $ \mathbb{C}P^n \cong \operatorname{SU}\_n/U\_{n-1} $ where $ U\_{n-1} $ is maximal for $ n \geq 3 $, and $ \mathbb{R}P^n \cong \operatorname{SO}\_n/O\_{n-1} $, again $ O\_{n-1} $ is maximal for $ n \geq 3 $. Another example is the Poincare homology sphere $ \operatorname{SO}\_3(\mathbb{R})/A\_5 $.
Solvmanifolds provide many interesting examples of manifolds, especially of torus bundles over tori (a solvmanifold is a manifold of the form $ G/H $ for $ G $ a solvable Lie group).
The examples I list above of manifolds $ G/H $, $ H $ maximal, all have the property that $ G $ is connected semisimple (indeed simple).
This leads me to wonder about the opposite case: maximal closed subgroups $ H $ of connected solvable Lie groups $ G $. Do they even exist?
Let $ G $ be a connected Lie group.
If $ G $ is abelian then certainly $ G $ does not have any maximal closed subgroups. Does the same hold for $ G $ solvable?
$\DeclareMathOperator\Ab{Ab}$Comment: Let $ G' $ be the commutator subgroup of the connected group $ G $. Let
$$
\Ab: G \to G/G'
$$
be the abelianization map. If $ H $ is a maximal closed subgroup of $ G $ then we must have
$$
\Ab(H)=G/G'
$$
because if $ \Ab(H) $ was properly contained then $ \Ab(H) $ would be a maximal closed subgroup of the connected abelian group $ G/G' $ which is impossible. In particular that implies that $ H $ does not contain $ G' $ (because if $ \Ab(H)=G/G' $
and $ H $ contained $ G' $ that would imply that $ H $ is all of $ G $, contradicting maximality).
Update: Recall that $ G $ is always a connected Lie group.
If $ G $ is nilpotent then there does not exist any maximal proper closed subgroup (proved in the original answer of [LSpice](https://mathoverflow.net/a/433144)).
If $ G $ is non-nilpotent then there does exist some maximal proper closed subgroup. We prove this with two cases.
$\DeclareMathOperator\Lie{Lie}$If the non-nilpotent group $ G $ is moreover non-solvable then we appeal to basically a Levi decomposition. $ \Lie(G) $ can be written as
$$
\Lie(G)= \mathfrak{g}\_\text{solv} \rtimes \mathfrak{g}\_\text{ss}.
$$
Let $ G\_\text{solv} $ be a maximal solvable closed connected subgroup of $ G $ corresponding to the Lie subalgebra $ \mathfrak{g}\_\text{solv} $. Let $ G\_\text{ss} $ be a maximal semisimple closed connected subgroup of $ G $, corresponding to the Lie subalgebra $ \mathfrak{g}\_\text{ss} $. Pick $ H\_\text{max} $ to be a maximal proper closed subgroup of $ G\_\text{ss} $ (there are lots of fairly well known maximal closed subgroups of semisimple groups). Then the group generated by $ G\_\text{solv} $ and $ H\_\text{max} $ should be roughly $ G\_\text{solv} \rtimes H\_\text{max} $ and should be a maximal proper closed subgroup of $ G $.
For the case that the non-nilpotent group $ G $ is solvable, then apply the [answer](https://mathoverflow.net/a/433250) from YCor (accepted below) which shows that a solvable non-nilpotent Lie group $ G $ must have a quotient which is one of the four solvable non-nilpotent subgroups of
$$
\operatorname{AGL}\_1(\mathbb{C}) \cong \mathbb{C}^\* \ltimes \mathbb{C}
$$
that YCor lists below. In that case there is a maximal proper closed subgroup of the quotient so we can pullback through the quotient map to get a maximal proper closed subgroup of the solvable non-nilpotent Lie group $ G $.
This proves the claim, from YCor's [comment](https://mathoverflow.net/questions/433142/can-solvable-connected-lie-groups-have-maximal-subgroups#comment1115150_433142), that a connected Lie group $ G $ has a maximal proper closed subgroup if and only if $ G $ is non-nilpotent.
| https://mathoverflow.net/users/387190 | Can solvable connected Lie groups have maximal subgroups? | $\newcommand{\g}{\mathfrak{g}}$We can obtain the characterization by classifying just non-nilpotent connected Lie groups.
>
> Let $G$ be a solvable, non-nilpotent connected Lie group. Suppose that every quotient Lie group of $G$ of dimension $<\dim(G)$ is nilpotent. Then $G$ has the form $H\ltimes V$ with both $H,V$ abelian of dimension $\le 2$ and $H$ acting nontrivially irreducibly on $V$ (see below for a list of possibilities).
>
>
>
First suppose that $G$ has a trivial center.
Let $\g$ be its Lie algebra, $(\g^i)\_{i\ge 1}$ the lower central series and $\g^\infty=\bigcap\_{i\ge 1}\g^i$, and $(G^i)$ the corresponding subgroups (a priori not closed). Note that $G^\infty$ is nilpotent (being contained in the derived subgroup $G^2$). Consider the adjoint representation of $G$. Since we can triangulate it (after complexification), we see that $\g^\infty$ maps to upper triangular matrices with zero diagonal. Hence $G^\infty$ maps to upper unipotent matrices. Thus $G^\infty$ is closed in $G$, and is simply connected. The action of $G$ on $G^\infty/[G^\infty,G^\infty]$ is not unipotent, since otherwise the action on $G^\infty$ would also be, and we would deduce that $G$ is nilpotent. So $G/[G^\infty,G^\infty]$ is a non-nilpotent Lie quotient of $G$, and hence, by the assumption, we deduce that $G^\infty$ is abelian. We claim that $\g^\infty$ is an irreducible module over $G/G^\infty$. Indeed, if it has a nonzero submodule, this corresponds to a connected nontrivial normal subgroup $N$ of $G$, contained in $G^\infty$. By the initial assumption, $G/N$ is nilpotent. But $G/G^\infty$ is the largest nilpotent quotient of $G$. So $N=G$. This proves irreducibility. Thus $G^\infty$ has dimension 1 or 2.
Let $\mathfrak{h}$ be a Cartan subalgebra in $\g$ (in the sense of Bourbaki: self-normalizing nilpotent subalgebra), and $H$ the corresponding subgroup (a priori possibly not closed). Then $\g=\mathfrak{h}+\g^\infty$. The intersection $\mathfrak{h}\cap\g^\infty$ is normalized by $\mathfrak{h}$, hence by $\mathfrak{g}$ (since $\g^\infty$ is abelian). By irreducibility, it is equal to either $\{0\}$ or $\g^\infty$. The latter is impossible, since $\mathfrak{h}$ is nilpotent. So $\g=\mathfrak{h}\ltimes\g^\infty$. By irreducibility, $[\mathfrak{h},\mathfrak{h}]$ centralizes $\g^\infty$. Let $\mathfrak{z}$ be the center of $\mathfrak{h}$. Then $[\mathfrak{h},\mathfrak{h}]\cap\mathfrak{z}$ is central in $\mathfrak{g}$, hence is zero. Since every nonzero ideal of a nilpotent Lie algebra meets the center, we deduce that $[\mathfrak{h},\mathfrak{h}]=0$: so $\mathfrak{h}$ is abelian. Also, the centralizer of $\g^\infty$ in $\mathfrak{h}$ is trivial, for the same reason.
Finally, we have one of the following
1. $\dim(\g^\infty)=1$: then $\g=\mathbf{R}\ltimes\mathbf{R}$, so $G$ is the non-abelian $\mathbf{R}\_{>0}\ltimes\mathbf{R}$.
2. $\dim(\g^\infty)=2$, $\dim(\mathfrak{h})=1$. Then $G=\mathrm{U}(1)\ltimes\mathbf{C}$, or $G=\mathbf{R}\ltimes\mathbf{C}$ (acting by an unbounded 1-parameter subgroup of $\mathbf{C}^\*$)
3. $\dim(\g^\infty)=2$, $\dim(\mathfrak{h})=2$: then $G=\mathbf{C}^\*\ltimes\mathbf{C}$.
If the center of $G$ is nontrivial, then it is discrete and the quotient has a trivial center. We obtain the possible connected covers of the previous examples (i.e., of $\mathrm{U}(1)\ltimes\mathbf{C}$ and of $\mathbf{C}^\*\ltimes\mathbf{C}$).
(Note that $H$ is eventually closed: we see this by first viewing the cover $H\ltimes G^\infty$ of $G$, and then observing that this cover is trivial.)
---
Turning back to the issue of maximal subgroups: in all cases we have obtained a group of the form $H\ltimes V$ where $H$ acts irreducibly on $V$. So $H$ is a maximal proper subgroup, which is closed.
If $G$ is an arbitrary non-nilpotent connected Lie group, then it has a quotient of this form, and hence admits a maximal proper subgroup that is closed.
| 3 | https://mathoverflow.net/users/14094 | 433250 | 175,261 |
https://mathoverflow.net/questions/433066 | 7 | A beautiful and surprising (to me at least) result around the axiom of choice is that, while full $\mathsf{AC}$ is preserved by forcing, a model of $\mathsf{ZF}$ + "There are no amorphous sets" may have a (set-)generic extension in which there do exist amorphous sets. This was proved by Monro, *[On generic extensions without the axiom of choice](https://www.jstor.org/stable/2273318#metadata_info_tab_contents)*; see also Asaf Karagila's summary [here](https://mathoverflow.net/a/77002/8133).
I'm interested in not-too-strong sufficient conditions on a c.t.m. $\mathcal{M}\models\mathsf{ZF}$ to have no generic extensions in which amorphous sets exist. Specifically, I'm curious if the following model-theoretic condition does the job:
>
> Say that a c.t.m. $\mathcal{M}\models\mathsf{ZF}$ is **expansive** iff for every first-order theory $T\in \mathcal{M}$ in a finite language with infinite models, every infinite set in $\mathcal{M}$ is the underlying set of a model of $T$ in $\mathcal{M}$.
>
>
>
*(Note that such an $\mathcal{M}$ *does* correctly compute whether such a $T$ has infinite models.)* For instance, expansiveness prevents the existence of infinite Dedekind-finite sets, since we can take $T$ to be the theory of a discrete linear order. My question, then, is how expansiveness, amorphousness in particular, and forcing interact:
* Is every set-generic extension of an expansive c.t.m. also expansive?
* If not, is there a set-generic extension of an expansive c.t.m. which has amorphous sets?
(Incidentally, it's not immediately clear to me that expansiveness is any stronger than "Infinite = Dedekind-infinite"!)
| https://mathoverflow.net/users/8133 | How hard is it to get "absolutely" no amorphous sets? | Turning my comment into an answer, an $X$ which is the universe of any finitely axiomatized theory with an infinite model must be orderable, and there must be a bijection between between $X$ and $X^2.$ Both of these follow from $X$ satisfying a large finite fragment of PA. In fact this is a characterization of such $X$ by the E.M. model construction, treating $X$ equipped with an arbitrary order as the generating set of indiscernibles. The resulting model is equinumerous with $X$ since $|X|=|X|^2$ implies $|X|=|X|^{<\omega}.$
Just from $|X|=|X|^2,$ we can deduce that a model is expansive precisely if it models choice, by Tarski's equivalence of AC with every infinite set being in bijection with its square. Trivially, such models satisfy generic non-existence of amorphous sets.
There are still a lot of open questions regarding generic non-existence of amorphous sets. Asaf's posts summarize the state of knowledge regarding this principle: it's strictly stronger than nonexistence of amorphous sets, since it fails in the Cohen model, and it's implied by "all sets are almost well-orderable." I don't even know how to show the latter is strictly stronger.
| 6 | https://mathoverflow.net/users/109573 | 433255 | 175,264 |
https://mathoverflow.net/questions/433254 | 7 | Let $K$ be a global field (ie either a number field or the function field of a curve over a finite field). Let $A,B$ be abelian varieties over $K$ and let $\phi:A\to B$ be an isogeny. Associated with $\phi$ is the well-known Selmer group ${\rm Sel}(\phi)\subseteq H^1(K,{\rm ker}(\phi))$, which contains $B(K)/\phi(A(K))$. It is classical that this group is finite but I couldn't find a reference in the situation where $K$ has positive characteristic. I have it in my memory that J. Milne proved this in the early 1970s but I wasn't able to find the corresponding article. I would be grateful for any help.
| https://mathoverflow.net/users/17308 | Reference request. Finiteness of the Selmer group | The paper is Milne, J. S. Elements of order p in the Tate-Šafarevič group. Bull. London Math. Soc. 2 (1970), 293–296. He deduces his statement about the Tate-Shafarevich group from a statement about the Selmer group. He also notes that if you omit any places in the definition of these groups, then you may get infinitely many elements of order p.
| 8 | https://mathoverflow.net/users/492821 | 433263 | 175,267 |
https://mathoverflow.net/questions/433264 | 3 | I am interested in the following congruence
$$\binom{ap^n}{bp^n}\equiv \binom{a}{b}\pmod{p^n}$$
I am aware that by some reference in a book the above it should actually hold modulo $p^{3n}$; the reference in question is Zieve (1999) but could not find any trace of the paper.
I am also aware of the various generalizations of Lucas's theorem, but I do not see how it follows directly from one of them.
By standard tricks you can reduce the above to $$\binom{ap^n-1}{p^n-1}\equiv 1 \pmod{p^n}.$$
| https://mathoverflow.net/users/41010 | Binomial coefficient congruence modulo $p^n$ | This is false in general.
To show this, I will use the congruence
$$ \binom{p^k a}{p^k b} \equiv \binom{p^{k-1} a}{p^{k-1}b} \bmod p^{3k}$$
for $p\ge 5$ and $k \ge 1$.
This is originally due to Ljunggren and Jacobsthal in "On the divisibility of the difference between two binomial coefficients" in Skand. Mat.-Kongr., Trondheim 1949, 42-54 (1952). The case $k=1$ was established in the 19th century by Wolstenholme, with earlier works (with smaller exponents) by Cabbage and Lucas. See also Meštrović's survey mentioned in Ira Gessel's answer.
Let us consider the case $a=2$, $b=1$, $n=4$ of your congruence. You claim
$$\binom{2p^4}{p^4} \equiv \binom{2}{1} \bmod p^4.$$
Here is why it fails. By Ljunggren--Jacobsthal,
$$\binom{2p^4}{p^4} \equiv \binom{2p^3}{p^3} \bmod p^{12},$$
$$\binom{2p^3}{p^3} \equiv \binom{2p^2}{p^2} \bmod p^{9},$$
$$\binom{2p^2}{p^2} \equiv \binom{2p}{p} \bmod p^{6}.$$
In particular, we do know that
$$\binom{2p^4}{p^4} \equiv \binom{2p}{p} \bmod p^4.$$
Your congruence then implies that
$$\binom{2p}{p} \equiv 2 \bmod p^4.$$
Primes that satisfy this have a name: they are called Wolstenholme primes. The only such primes up to $10^9$ are 16843 and 2124679... One can show that $p$ is such a prime iff $p$ divides the numerator of the Bernoulli number $B\_{p-3}$.
The proofs of Ljunggren--Jacobsthal suggest that for any given $a,b$, your congruence fails if $n > 3+\nu\_p(\binom{a}{b}ab(a-b))$ where $\nu\_p$ is the $p$-adic valuation.
---
For $n\le 3$ your congruence is true and directly follows from Ljunggren-Jacobsthal. Indeed, applying L-J with $k=1,2,3$ we have, in particular, that
$$\binom{ap^3}{bp^3} \equiv \binom{ap^2}{bp^2} \equiv \binom{ap}{bp} \equiv \binom{a}{b} \bmod p^{3}.$$
---
A more refined version of Ljunggren--Jacobsthal was proved by G. S. Kazandzidis:
$$ \binom{ap}{bp} \equiv \binom{a}{b} \bmod p^{3+\delta}$$
where $\delta=\nu\_p(ab(a-b)\binom{a}{b})$.
This allows you to extend the range of validity $n \le 3$ if one of $a$, $b$, $a-b$ or $\binom{a}{b}$ is divisible by $p$.
This result (and variations) is proved in:
* "On a congruence and on a practical method for finding the highest power of a prime p which divides the binomial coefficient (AB)", Bull. Soc. Math. Grèce, N. Sér. 6, No. 2, 358-360 (1965).
* "Congruences on the binomial coefficients", Bull. Soc. Math. Grèce, N. Ser. 9, No. 1, 1-12 (1968).
* "On congruences in number-theory", Bull. Soc. Math. Grèce, N. Sér. 10, No. 1, 35-40 (1969).
All the mentioned papers are elementary. The modern treatment is a bit less so, but is more illuminating. Indeed, $p$-adic analysis can lead to a proof. See Chapter 7 of Alain M. Robert's book "A course in p-adic analysis" (GTM 198, Springer, 2000). The proof given there is based on the material in the paper "The Kazandzidis supercongruences. A simple proof and an application", Rend. Semin. Mat. Univ. Padova 94, 235-243 (1995), by Alain M. Robert and Maxime Zuber.
Some remarks:
* The congruence $\binom{ap^k}{bp^k} \equiv \binom{ap^{k-1}}{bp^{k-1}} \bmod p^{2k}$ has a combinatorial proof (as opposed to algebraic proofs mod $p^{3n}$). A reference for a combinatorial proof when the modulus is $p^n$ is Gian-Carlo Rota and Bruce Sagan's paper "Congruences derived from group action", Eur. J. Comb. 1, 67-76 (1980). I am not sure if this is the earliest reference of this kind of argument, though.
* There are various $q$-analogues of these $p$-adic congruences.
* Some of the above papers include also the cases $p=2$ and $p=3$ for which the exponent of $p$ in the modulus is slightly smaller.
| 7 | https://mathoverflow.net/users/31469 | 433268 | 175,269 |
https://mathoverflow.net/questions/433237 | 0 | Let $e\_{1}$ and $e\_{2}$ be involutions in the algebraic group $G=\operatorname{PGL}\_{n}(\mathbb{C})$. Do we have $$C\_{G}(\langle e\_{1},e\_{2}\rangle)^{\circ} = C\_{G}(e\_{1})^{\circ}\cap C\_{G}(e\_{2})^{\circ}.$$ It's obvious $C\_{G}(\langle e\_{1},e\_{2}\rangle)^{\circ} \leqslant C\_{G}(e\_{1})^{\circ}\cap C\_{G}(e\_{2})^{\circ}.$ If yes, can it be generalised to two elementary abelian $2$-subgroups $E\_{1}, E\_{2}$.
Here $H^{\circ}$ denotes the identity component of the group $H$.
I can only see this through some examples. For instance, in $G=\operatorname{PGL}\_{4}(\mathbb{C})$, let $$ x=\left[ \begin{array}{cc}
-1 & 0 & 0 & 0\\
0 & 1& 0 & 0\\
0 & 0 & -1 & 0\\
0 & 0 & 0 & 1
\end{array} \right],f= \left[ \begin{array}{cc}
0 & 1 & 0 & 0\\
1 & 0 & 0 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 1 & 0
\end{array} \right].$$ Then $$C\_{G}(x) = \left[ \begin{array}{cc}
a\_{1} & 0 & a\_{3} & 0\\
0 & b\_{2} & 0 & b\_{4}\\
c\_{1} & 0 & c\_{3} & 0\\
0 & d\_{2} & 0 & d\_{4}
\end{array} \right] \sqcup f\left[ \begin{array}{cc}
a\_{1} & 0 & a\_{3} & 0\\
0 & b\_{2} & 0 & b\_{4}\\
c\_{1} & 0 & c\_{3} & 0\\
0 & d\_{2} & 0 & d\_{4}
\end{array} \right].$$
Let $$ y=\left[ \begin{array}{cc}
1 & 0 & 0 & 0\\
0 & -1& 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{array} \right]. \mbox{So } C\_{G}(y) = \left[ \begin{array}{cc}
u\_{1} & 0 & u\_{3} & u\_{4}\\
0 & v\_{2} & 0 & 0\\
s\_{1} & 0 & s\_{3} & s\_{4}\\
t\_{1} & 0 & t\_{3} & t\_{4}
\end{array} \right].$$
So $C\_{G}(\langle x,y\rangle)^{\circ} = \left[ \begin{array}{cc}
\* & 0 & \* & 0\\
0 & \* & 0 & 0\\
\* & 0 & \* & 0\\
0 & 0 & 0 & \*
\end{array} \right]\cong A\_{1}T\_{2}$ which is, indeed, the intersection of the identity components of the two centralisers.
| https://mathoverflow.net/users/488802 | Intersection of identity components | $\DeclareMathOperator\Cent{C}\newcommand\oG{\overline G}\newcommand\oe{\overline e}$Put $\oG = \operatorname{PGL}\_n(\mathbb C)$. I hope you will permit me to denote the involutions by $\oe\_i$ instead of $e\_i$.
At first, it matters only that we are dealing with semisimple elements (which is implied by being an involution in characteristic $\ne 2$). It doesn't even matter at first that we are dealing with $\oG = \operatorname{PGL}\_n(\mathbb C)$; I will put $G = \operatorname{GL}\_n(\mathbb C)$, but there is a general theory of $z$-extensions, of which every connected, reductive group $\oG$ admits one, which are to $\oG$ as $\operatorname{GL}\_n(\mathbb C)$ is to $\operatorname{PGL}\_n(\mathbb C)$.
If $t$ is any semisimple element of $G$, then $\Cent\_G(t)$ is connected, $\Cent\_G(t) = \Cent\_G(t)^\circ \to \Cent\_{\oG}(\overline t)^\circ$ is a surjection, and $\Cent\_G(t)$ is the full pre-image in $G$ of $\Cent\_{\oG}(\overline t)^\circ$, where $\overline t$ is the image of $t$ in $\oG$. In particular, if $e\_i$ is a lift of $\oe\_i$, then $\Cent\_{\oG}(\oe\_1)^\circ \cap \Cent\_{\oG}(\oe\_2)^\circ$ is the image in $\oG$ of $\Cent\_G(e\_1) \cap \Cent\_G(e\_2) = \Cent\_G(e\_1, e\_2)$. Thus, the question is whether $\Cent\_G(e\_1, e\_2)$ is connected. If so, then $\Cent\_{\oG}(\oe\_1)^\circ \cap \Cent\_{\oG}(\oe\_2)^\circ$ is the image $\Cent\_{\oG}(\oe\_1, \oe\_2)^\circ$ in $\oG$ of $\Cent\_G(e\_1, e\_2) = \Cent\_G(e\_1, e\_2)^\circ$.
In the full generality that I have considered so far (where $G$ has connected centre and simply connected derived group, and the $e\_i$ can be any semisimple elements), this need not be true. Even in your setting, where $\oG = \operatorname{PGL}\_n(\mathbb C)$ and the $\oe\_i$ are involutions, I do not know whether $C\_G(e\_1, e\_2)$ is always connected. However, you have [indicated](https://mathoverflow.net/questions/433237/intersection-of-identity-components#comment1115408_433237) that you are interested in the case where $\oe\_1$ and $\oe\_2$ belong to a common maximal torus $\overline T$ (which is the same as $\oe\_2$ belonging to $\Cent\_{\oG}(\oe\_1)^\circ$). I now specialise to that case, without requiring that the $\oe\_i$ are involutions. Let $T$ be the pre-image in $G$ of $\overline T$. Then $T$ is a maximal torus in $G$.
In general (not just for $G = \operatorname{GL}\_n$), the map from $\operatorname{stab}\_W(e\_1, e\_2)/\langle s\_\alpha : \alpha(e\_1) = \alpha(e\_2) = 1\rangle$ to $\Cent\_G(e\_1, e\_2)/\Cent\_G(e\_1, e\_2)^\circ$ is an isomorphism, where $W$ is the Weyl group of $T$ in $G$. Now, finally specialising to $G = \operatorname{GL}\_n$ and so identifying $W$ with $\operatorname S\_n$, and conjugating $T$ if necessary so that it is the diagonal torus of $G$, an explicit computation shows that $\operatorname{stab}\_W(e\_1, e\_2)$ is the product of permutation groups that respect the decomposition of $\{1, \dotsc, n\}$ into maximal subsets $I$ such that all diagonal entries of $e\_1$ with entries in $I$ are equal, and all diagonal entries of $e\_2$ with entries in $I$ are equal; and that this is just $\langle s\_\alpha : \alpha(e\_1) = \alpha(e\_2) = 1\rangle$. Thus, $\Cent\_G(e\_1, e\_2)$ is connected, so your desired conclusion follows.
| 1 | https://mathoverflow.net/users/2383 | 433275 | 175,274 |
https://mathoverflow.net/questions/433274 | 4 | Suppose $p$ is a prime, that $F$ is a finite extension of the field $\mathbb{Q}\_p$, $D$ is the division quaternion algebra over $F$ and $\mathcal{O}\_D$ is the valuation ring of $D$. What is the abelianisation of the group of units $\mathcal{O}\_D^\times$? I'd also appreciate a reference.
Apologies that this may look like a homework problem. It isn't, for me at least. It is clear to me that there is a surjective group homomorphism from $\mathcal{O}\_D^\times$ to $\mathcal{O}\_F^\times$ given by reduced norm and a surjective homomorphism from $\mathcal{O}\_D^\times$ to the group of units in the quadratic field extension of the residue field of $F$ with kernel $1+P\_D$ where $P\_D$ is the unique maximal ideal in $\mathcal{O}\_D$. What isn't clear to me is whether the derived subgroup is precisely the intersection of the kernels of these two homomorphisms or it is smaller than that.
| https://mathoverflow.net/users/345 | Abelianization of unit quaternions over a p-adic field | The answer is yes: it is a result of Riehm, Corollary to Theorem 21 in The norm 1 group of $\mathfrak{p}$-adic division algebras *Amer. J. Math.* **92** 2 (1970), 499--523, see also Theorem 1.9 p.33 and the following Remark in Platonov and Rapinchuk, Algebraic groups and number theory *Pure and Applied Math.* **139** (1994).
The precise statement is as follows: define $C\_i = \ker(\mathrm{nrd}\colon U\_i \to \mathcal{O}\_F^\times)$ ($H\_r$ in Riehm's notation), where $U\_0 = \mathcal{O}\_D^\times$ and $U\_i = 1+P\_D^i$ is the usual filtration. Then $C\_1 = [C\_0,C\_0]$.
| 5 | https://mathoverflow.net/users/40821 | 433291 | 175,280 |
https://mathoverflow.net/questions/322023 | 3 | An algebra $(X,\*)$ is said to be self-distributive if it satisfies the identity $x\*(y\*z)=(x\*y)\*(x\*z)$ for all $x,y,z\in X$. If $(X,\*)$ is an algebra, then a subset $L\subseteq X$ is said to be a left-ideal if $x\*y\in L$ whenever $y\in L$. An element $x$ is said to be a left-identity if $x\*y=y$ for $y\in X$. Let $\mathrm{Li}(X)$ denote the set of all left-identities in the algebra $(X,\*)$.
A finite self-distributive algebra $X$ is said to be Laver-like if $\mathrm{Li}(X)$ is a left ideal and if there exists some finite linearly ordered set $L$ and a surjective function $\mathrm{crit}:X\rightarrow L$ such that
1. $\mathrm{crit}(x)=\max(L)$ if and only if $x\in\mathrm{Li}(X)$,
2. $\mathrm{crit}(x\*y)>\mathrm{crit}(y)$ whenever $\mathrm{crit}(y)\geq\mathrm{crit}(x)$, and
3. $\mathrm{crit}(x\*y)=\mathrm{crit}(y)$ whenever $\mathrm{crit}(y)<\mathrm{crit}(x)$.
The function $\mathrm{crit}$ is unique up to isomorphism. We shall say that the algebra $X$ has $|L|$ critical points.
Suppose that $X$ is a finite Laver-like algebra with $n$ critical points generated $(x\_{a})\_{a\in A}$. Then does there necessarily exist
a finite Laver-like algebra $Y$ with $n+1$ critical points generated by $(y\_{a})\_{a\in A}$ and surjective homomorphism
$\phi:Y\rightarrow X$ where $\phi(y\_{a})=x\_{a}$ for $a\in A$?
| https://mathoverflow.net/users/22277 | Can Laver tables go extinct? | **Multigenic Laver tables can go extinct.** Examples of multigenic Laver tables that go extinct with 2 generators are rare; one is not likely to find a multigenic Laver table that goes extinct unless one is looking for a multigenic Laver table that goes extinct, and one needs to use computer calculations to produce such an example and prove that it works.
Let us hold to the convention that the implied parentheses are always on the left. For example, $a\*b\*c\*d=((a\*b)\*c)\*d$. If $(X,\*)$ is a Laver-like algebra generated by the set $(x\_a)\_{a\in A}$, then the multigenic Laver table associated with $(X,\*)$ and generating set $(x\_a)\_{a\in A}$ is the algebra $M=M(X,(x\_a)\_{a\in A})\subseteq A^+$ where
$a\_1\dots a\_r\in M$ if and only if whenever $1\leq s<r$, we have $x\_{a\_1}\*\dots \*x\_{a\_r}\not\in\text{Li}(X)$. The multigenic Laver table $M$ can be endowed with a unique self-distributive operation $\*$ such that $\mathbf{x}\*a=\mathbf{x}a$ whenever $\mathbf{x}a\in M,a\in A,\mathbf{x}\in M,|\mathbf{x}|>0$ and where $\mathbf{x}\*\mathbf{y}=\mathbf{y}$ whenever $\mathbf{x},\mathbf{y}\in M,a\in A,\mathbf{x}a\not\in M$.
Let $X$ be a Laver-like algebra generated by $(x\_a)\_{a\in A}$, and let $M=M(X,(x\_a)\_{a\in A})$ denote its multigenic Laver table. Then let $\text{Cov}(M)$ denote the collection of all multigenic Laver tables of the form $N=M(Y,(y\_a)\_{a\in A})$ where $Y$ is generated by $(y\_a)\_{a\in A}$ and where there is a homomorphism $\phi:Y\rightarrow X$ with $\phi(y\_a)=x\_a$ for $a\in A$ and where $Y$ has one more critical point than $X$. It suffices to produce a multigenic Laver table $M$ where $\text{Cov}(M)=\emptyset$. Let $\text{Cov}^\*(M)$ be the smallest collection of multigenic Laver tables with $M\in\text{Cov}^\*(M)$ and where if $N\in\text{Cov}^\*(M)$, then $\text{Cov}(N)\subseteq\text{Cov}^\*(N)$. Let $\text{Cov}^+(M)=\text{Cov}^\*(M)\setminus\{M\}$.
Let $X$ be a Laver-like algebra generated by $(x\_a)\_{a\in A}$ and where $\text{crit}:X\rightarrow\{0,\dots,n\}$. Then let $c\_0,\dots,c\_n\in X$ be elements with $c\_i\*c\_i\in\text{Li}(X)$ and $\text{crit}(c\_i)=i$ whenever $0\leq i\leq n$. For $0\leq i<n$, let $p\_i((z\_a)\_{a\in A})$ (haha, that spells pizza) be the non-commutative polynomial defined by
$$p\_i((z\_a)\_{a\in A})=\sum\{z\_{a\_0}\dots z\_{a\_s}:\text{crit}(x\_{a\_0}\*\dots\*x\_{a\_s})=i,\forall r<s,\text{crit}(x\_{a\_0}\*\dots\*x\_{a\_r})<i\}.$$
If $X$ is a multigenic Laver table over the alphabet $A$ and $x\_a=a$ for $a\in A$, then let $\text{Poly}(X)=(p\_0((z\_a)\_{a\in A}),\dots,p\_{n-1}((z\_a)\_{a\in A})).$
If $M$ is a multigenic Laver table, then $M$ can be easily recovered from $\text{Poly}(X)$ since $$\sum\{z\_{a\_1}\dots z\_{a\_r}\mid a\_1\dots a\_r\in M\}=\big(1+p\_{n-1}((z\_a)\_{a\in A})\big)\dots\big(1+p\_0((z\_a)\_{a\in A})\big)\sum\_{a\in A}z\_a.$$
If $X$ is a Laver-like algebra, then let $\preceq$ be smallest partial ordering where if $x\in X\setminus\text{Li}(X),y\in X$, then $x\preceq x\*y$.
Example: Let $M$ be the multigenic Laver table generated by two elements with the following non-commutative polynomial sequence:
$$\text{Poly}(M)=( x, x^2, y, x^4, x\cdot y+x^2\cdot y+x^8, x^3\cdot y, x^5\cdot y+x^6\cdot y+x^9\cdot y+x^{10}\cdot y+x^{16}, y^2, y\cdot x^4, y\cdot x\cdot y,
(y\cdot x^4)^2, y\cdot x^3\cdot y+y\cdot x^3\cdot (x\cdot y)^2, y\cdot x^2\cdot y+(y\cdot x^4)^4 ).$$
Then $M$ is a multigenic Laver table with $\text{Cov}(M)=\emptyset$ where $|M|=2^{12}\cdot 3^3$.
We found the multigenic Laver table $M$ in two steps. We first search for a multigenic Laver table $N$ with alphabet $\{0,1\}$ with $|\text{Cov}(N)|<3$. After we do this, we search for the multigenic Laver table $M\in\text{Cov}^+(N)$ with $|\text{Cov}(M)|=0$.
**A glimpse of the algorithm for computing $\text{Cov}(M)$.**
We observe that $|X|\leq|M(X,(x\_a)\_{a\in A})|$ whenever $X$ is a Laver-like algebra generated by $(x\_a)\_{a\in A}$. In particular, the algebra $M(X,(x\_a)\_{a\in A})$ is generally too large to hold in memory when doing computations. On the other hand, given a multigenic Laver table $M$, the smallest algebra $(X,(x\_a)\_{a\in A})$ with $M=M(X,(x\_a)\_{a\in A})$ is insufficient for computational purposes; if $N\in\text{Cov}(M)$, then we cannot in general find a $(Y,(y\_a)\_{a\in A})$ with $N=M(Y,(y\_a)\_{a\in A})$ where $X$ embeds in $Y$ (we can write $X$ as a quotient of $Y$, but I have found it easier to undo the process of taking a subalgebra than to undo the process of taking a quotient algebra).
Give Laver-like algebra $X$, let $\mathcal{I}\_X$ be the smallest congruence on $X$ where if $c,d\in X,\text{crit}(c)=\text{crit}(d),c\*c,d\*d\in\mathrm{Li}(X)$, then $(c,d)\in\mathcal{I}\_X$. If $M$ is a multigenic Laver table, then the quotient algebra $M/\mathcal{I}\_M$ is typically small enough to hold in memory and $M=M(M/\mathcal{I}\_M,([a])\_{a\in A})$ while retaining the algebraic properties that allow for efficient computer calculations.
Suppose that $M$ is a multigenic Laver table over the alphabet $A$ and $N\in\text{Cov}(M)$. Now let $\mathbf{c}\in \text{Li}(M)\setminus\text{Li}(N)$. Then there are homomorphisms $\iota\_\mathbf{c}:M\rightarrow N,j:N\rightarrow M$ where $\iota(\mathbf{x})=\mathbf{cx}$ for $\mathbf{x}\in M$ and $j(a)=a$ for $a\in A$. The mappings $\iota\_\mathbf{c},j$ induce mappings $\iota^\sharp:M/\mathcal{I}\_M\rightarrow N/\mathcal{I}\_N,j^\sharp:N/\mathcal{I}\_N\rightarrow M/\mathcal{I}\_M$ where
$\iota^\sharp([\mathbf{x}]\_M)=[\mathbf{cx}]\_N$ and $j^\sharp([\mathbf{x}]\_N)=[j(\mathbf{x})]\_M$. The mapping $\iota^\sharp$ is an embedding. The algorithm for computing $N/\mathcal{I}\_N$ from $M/\mathcal{I}\_M$ consists of extending the algebra $M/\mathcal{I}\_M$ to $N/\mathcal{I}\_N$; we use backtracking to find all possible extensions of $M/\mathcal{I}\_M$ that could possibly be extended to $N/\mathcal{I}\_N$ which are isomorphic to the subalgebras of the form $\{[\mathbf{w}]\_N:\mathbf{w}\in N,\mathbf{z}\preceq\mathbf{w}\}\subseteq N/\mathcal{I}\_N$ for some $\mathbf{z}\in M$.
[Here is the code](https://github.com/jvanname/Laver-tables/blob/main/GAP/multigeniccreate) that I have written in the language GAP for which one can compute $\text{Cov}(M)$. For 2 generators, this backtracking algorithm for computing $\text{Cov}(M)$ runs quickly (as long as $\text{Cov}(M)$ and each element in $\text{Cov}(M)$ are not very large) even though I have not been able to obtain any theoretical results on the speed of such a backtracking algorithm. I suspect that this algorithm is not the best algorithm for computing $\text{Cov}(M)$ since $M/\mathcal{I}\_M$ can still be quite large, but this algorithm is sufficient for our purposes.
**More information about $M$**
The algebra $M/\mathcal{I}\_M$ has $260$ elements. If $X$ is a Laver-like algebra, then let $\simeq\_{\text{cmx}}$ be the congruence on $X$ where we have $x\simeq\_{\text{cmx}}y$ if and only if
$$c\*x\*a\_1\*\dots\*a\_r\in\text{Li}(X)\Leftrightarrow c\*y\*a\_1\*\dots\*a\_r\in\text{Li}(X)$$ whenever $c,a\_1,\dots,a\_r\in X$. Then
$|M/\simeq\_{\text{cmx}}|=86$. The quotient $M/\simeq\_{\text{cmx}}$ is the smallest Laver-like algebra where $M(M/\simeq\_{\text{cmx}})=M$. I have not been able to find a Laver-like algebra $X$ with 2 generators and fewer than 86 elements with $\text{Cov}(X)=\emptyset$. This procedure also produces many other examples of multigenic Laver tables $P$ with $\text{Cov}(P)=\emptyset.$
| 2 | https://mathoverflow.net/users/22277 | 433293 | 175,281 |
https://mathoverflow.net/questions/433290 | 1 | Given two random variables $X,Y$ which are both $\mathbb{N}$-valued and have the same expected value (which is some fixed positive constant), and denote their probability mass functions by ${\bf p} = (p\_0,p\_1,\ldots)$ and ${\bf q} = (q\_0,q\_1,\ldots)$, respectively. We also assume that $q\_n > 0$ for all $n \in \mathbb N$. I am pretty sure one can upper bound the Wasserstein distance (of order 1) $W\_1({\bf p},{\bf q})$ by the so-called $\chi^2$ distance defined via $$\chi^2({\bf p},{\bf q}) = \sum\limits\_{n\geq 0} \frac{|p\_n-q\_n|^2}{q\_n} = \sum\limits\_{n\geq 0} \frac{p^2\_n}{q\_n} - 1.$$ However, I fail to found any specific reference containing a statement like $$W\_1({\bf p},{\bf q}) \leq f\left(\chi^2({\bf p},{\bf q})\right)$$ for some appropriate non-negative function $f$ which vanishes at the origin. I will appreciate any help in locating a suitable reference for the advertised bound...
---
Remark: I forgot to mention that ${\bf p}$ and ${\bf q}$ have the same mean value in my original post.
| https://mathoverflow.net/users/163454 | Upper bound Wasserstein distance by $\chi^2$ distance | Such a real-valued function $f$ does not exist.
Indeed, for any natural $N$, let
$$(p\_N,p\_{2N},p\_{3N})=\tfrac18(1,4,2),\ (q\_N,q\_{2N},q\_{3N})=\tfrac18(2,2,3),$$
so that $p\_N+p\_{2N}+p\_{3N}=q\_N+q\_{2N}+q\_{3N}=\frac78$ and $Np\_N+2Np\_{2N}+3Np\_{3N}=Nq\_N+2Nq\_{2N}+3Nq\_{3N}$. Next,
for $n\in J:=\{0,1,\dots\}\setminus\{N,2N,3N\}$, let $q\_n$ be any positive real numbers such that $\sum\_{n\in J}q\_n=1-\frac78$, and let $p\_n=q\_n$ for $n\in J$.
Then $EX=EY$ and the $\chi^2$ distance between the distributions of $X$ and $Y$ is a certain positive real number, not depending on $N$.
On the other hand, the Wasserstein distance between the distributions of $X$ and $Y$ is $\sim cN\to\infty$ (as $N\to\infty$) for a certain positive real number $c$. (this easily follows from, say, the [known expression for such Wasserstein distance][1]
So, your desired inequality cannot hold for any real-valued function $f$.
| 1 | https://mathoverflow.net/users/36721 | 433297 | 175,284 |
https://mathoverflow.net/questions/433292 | 26 | The Riemann hypothesis for finite fields can be stated as follows: take a smooth projective variety X of finite type over the finite field $\mathbb{F}\_q$ for some $q=p^n$. Then the eigenvalues $\alpha\_j$ of the action of the Frobenius automorphism on the $i$th $\ell$-adic étale cohomology (are algebraic numbers and) have norm $q^{i/2}$. This is part of the general philosophy that led to the proof of the Weil conjectures. What I don't understand, and pardon me for saying this, is why we care about these eigenvalues.
As a homotopy theorist, I've found the other three conjectures (rationality, Betti numbers, and the functional equation) to be helpful in understanding zeta functions from a geometric perspective. Together, they describe the relationship between the combinatorics and the cohomology of a variety. The Riemann hypothesis, however, doesn't seem to admit a direct interpretation of this kind; it's unclear what role the $\alpha$s play in the analogy. Do they have an interpretation as the arithmetic version of a classical geometric/topological object, like how the degrees are Betti numbers? If not, how should I understand them?
| https://mathoverflow.net/users/158123 | Why do we care about the eigenvalues of the Frobenius map? | The Riemann hypothesis is very important for the relationship between the cohomology and combinatorics of the variety.
First, the Riemann hypothesis lets us read off the Betti numbers from the point counts over finite fields, i.e. the $i$'th Betti number is the number of zeroes/poles of $$e^{ \sum\_j \# X(\mathbb F\_q^j) u^j / j }$$ of absolute value $q^{-i/2}$.
Without the Riemann hypothesis, and with just the other Weil conjectures, it's not possible to calculate the Betti numbers in this way, because you can't distinguish which zeroes or poles are coming from which $P\_i$s or, worse, rule out the case that zeroes and poles will cancel. Without the Riemann hypothesis, one can only calculate the Euler characteristic.
Second, the Riemann hypothesis lets us get information about point counts over finite fields from the Betti numbers. The simplest of these is the upper bound $$|X(\mathbb F\_q)| \leq \sum\_{i=0}^{2n} \dim H^i(X) q^{i/2}.$$ Without the Riemann hypothesis, only much weaker results of this form could be proven (maybe one could replace $q^{i/2} $ with $q^{ \max(i,n)}$ or something like that). Without even a crude bound, even knowing exactly the Betti numbers won't typically rule out any particular value for the number of points over a given field.
I would even say this is much more direct than the relationship between geometry and combinatorics obtained from the remaining Weil conjectures.
---
In terms of an analogue in classical geometry / topology, the obvious thing would be the eigenvalues of the action of a map on the cohomology! Of course, one usually doesn't have an a priori exact formula for the absolute value of the eigenvalues, but if one did, it would certainly be useful for understanding the fixed points of the map.
So the Riemann hypothesis is a new phenomenon that doesn't have an analogue in topology (except for Serre's analogue of the Weil conjectures for Kahler manifolds), but the eigenvalues of operators acting on cohomology were a pre-existing notion. Lefschetz certainly wasn't thinking about the Frobenius when he proved his original fixed point formula!
Maybe one should mention also that the eigenvalues of the mapping class of a surface acting on its cohomology give you information on where that mapping class sits in the Nielsen-Thurston classification.
---
There is one aspect to classical analogues that I think deserves mentioning because it's of great importance:
The Riemann hypothesis in the Weil conjectures tells us that *calculating the high-degree (compactly-supported) or low-degree (if the variety is smooth) cohomology groups of a variety in topology is analogous to obtaining an approximate estimate for the number of points in arithmetic*.
This is the starting point for deep connections between stable homology and other topological methods for calculating the low-degree cohomology groups without necessarily calculating every cohomology group, and analytic number theory or other fields where quantities are calculated approximately!
So RH is not an analogue of anything classical in topology but it tells us what the analogues of some classical statements in topology are.
| 23 | https://mathoverflow.net/users/18060 | 433299 | 175,285 |
https://mathoverflow.net/questions/433308 | 4 | Let $f:x\to z$ and $g:y\to z$ be morphisms in an $\infty$-category $\mathcal C$. It seems that the square
$$\require{AMScd}
\begin{CD}
\operatorname{Map}\_{\mathcal C\_{/z}}(f,g) @>>> \operatorname{Map}\_{\mathcal C^{\Delta^1}}(f,g)\\
@VVV @VVV \\
\Delta^0 @>{\operatorname{id}\_z}>> \operatorname{Map}\_{\mathcal C}(z,z)
\end{CD}$$
should be (homotopy-)cartesian in the $\infty$-category of spaces. If this is indeed true, does anyone have a reference or a proof?
| https://mathoverflow.net/users/37110 | Maps in the slice category vs. maps in the arrow category | Let us use the fat slice $\mathcal{C}^{z/}$ (See HTT, $\S$4.2.1) and the model $\operatorname{Hom}\_{\mathcal{C}}(x,y)=\operatorname{Fun}(\Delta^1,\mathcal{C})\times \_{\mathcal{C}\times \mathcal{C}}\{(x,y)\}$ of the mapping space. By computation, we can check that the square
$$\require{AMScd}
\begin{CD}
\operatorname{Hom}\_{\mathcal{C}^{/z}}(f,g) @>>> \operatorname{Hom}\_{\operatorname{Fun}(\Delta^1,\mathcal{C})}(f,g)\\
@VVV @VVV \\
\Delta^0 @>{\operatorname{id}\_z}>> \operatorname{Hom}\_{\mathcal{C}}
(z,z)\end{CD}$$
is cartesian. The right vertical arrow is a Kan fibration, because $\operatorname{ev}\_1:\operatorname{Fun}(\Delta^1,\mathcal{C})\to\mathcal{C}$ is an inner fibration. (In general, inner fibrations induce Kan fibrations between various mapping spaces. See, e.g., Lemma 2.4.4.1. The lemma talks about the usual slice, but the proof applies to fat slice as well.) So the above square is homotopy cartesian.
| 6 | https://mathoverflow.net/users/144250 | 433312 | 175,289 |
https://mathoverflow.net/questions/359754 | 4 | Let $f= f(t,x) : \mathbb{R}\_+ \times \mathbb{R}^d \to \mathbb{R}$ be a Lipschitz function such that
$$
\partial\_t f - |\nabla f|^2 = 0 \qquad \text{almost everywhere in } \mathbb{R}\_+ \times \mathbb{R}^d.
$$
As is well-known, this condition does not determine the function $f$ uniquely in terms of the initial condition $f(0,\cdot)$; but uniqueness is restored if we impose in addition that for every $t \ge 0$, the mapping $x \mapsto f(t,x)$ is convex (or locally semiconvex). I want to assume instead that, for every $t \ge 0$, the mapping $x \mapsto f(t,x)$ is *concave*. Under this condition, is the function $f$ determined uniquely in terms of $f(0,\cdot)$? The answer is "no", because for instance (say in $d = 1$) the function $(t,x) \mapsto t-|x|$ satisfies the required properties, but one can check that this is not the solution given by the Hopf-Lax formula. However, I struggle to find a counter-example if there is no kink in the initial condition to start with. So here is my question.
>
> Assume that $f$ is a Lipschitz function that satisfies the equation above at every point of differentiability of $f$; that for every $t \ge 0$, the mapping $x \mapsto f(t,x)$ is concave; and that the mapping $x \mapsto f(0,x)$ is smooth. Does the initial condition $f(0,\cdot)$ determine the function $f$ uniquely?
>
>
>
| https://mathoverflow.net/users/56892 | Uniqueness condition for Hamilton-Jacobi equation? | Proposition A.2 of <https://arxiv.org/abs/2104.05360> shows that the answer is "yes". In fact, this is also valid for a general nonlinearity in the equation in place of the squared norm here.
| 0 | https://mathoverflow.net/users/56892 | 433317 | 175,290 |
https://mathoverflow.net/questions/433160 | 4 | I preface this by saying that I am fairly new to the enveloping von Neumann algebra scene, so there may be some gaps in my understanding.
Given a $C^\*$-algebra $A$ and a state $\phi$ on $A$, one may consider $\phi$ as a normal state on the universal enveloping von Neumann algebra $A''$ of $A$. In this case, there should be a support projection $\text{supp}(\phi) \in A''$ for $\phi$, that is, a minimal projection such that $\phi(\text{supp}(\phi))=1$.
For a (concerning) example, if $A = C([0,1])$ and $\phi$ is the Lebesgue integral, this seems to give us a projection in $C([0,1])''$ which is smaller than the operator of multiplication by the indicator function of any set of full Lebesgue measure. This is a confronting possibility, so have I made a mistake somewhere or is this just evidence of how complicated $C([0,1])''$ is?
On the other hand, we could restrict to closed projections (in the sense of Akemann, *The General Stone-Weierstrauss problem*, 1969). Then for commutative $C^\*$-algebras there is a smallest closed projection of full "measure", since (normal) states correspond to regular Borel probability measures and closed projections correspond to closed subsets of the spectrum. Furthermore, these "closed support projections" are much less pathological than the support projection in the enveloping von Neumann algebra.
I have a proof sketch via the universal representation that "closed support projections" do exist in the non-commutative case too, but nowhere do I use closedness, so I will not feel confident in its validity until I know what is going on with the support projection of the Lebesgue integral on $C([0,1])$.
I would also be interested to know if people have already thought about closed support projections, for example if it is known whether they are the closure of the support projection.
Thanks in advance.
| https://mathoverflow.net/users/480800 | Support projection vs closed support projection of a normal state in enveloping von Neumann algebra | I am not quite sure what the question is, so let me try to understand what your unease is. In the example we take $A=C([0,1])$ and for the state $\phi$ take normalised Lebesgue measure. What is the support projection in $A''$? When you write the following:
>
> ... which is smaller than the operator of multiplication by the indicator function of any set of full Lebesgue measure.
>
>
>
I think your intuition is proceeding as follows: we think of $A''$ as somehow being a space of functions on $[0,1]$, and then guess that the support projection of $\phi$ is the indicator function of $[0,1]$. But then we could adjust this function to be zero at some points, and it would still be a projection, and still give all of $\phi$. In this way, we seem to conclude that there can be no *minimal* projection.
The fault in this reasoning is to believe that $A''$ can in any way be thought of as functions on $[0,1]$. As $A''$ is a commutative von Neumann algebra, it is isomorphic to $C(X)$ for some compact hyperstonian $X$. In the literature, $X$ is called the *hyperstonian cover* of $[0,1]$. A recent book on this subject is "Banach spaces of continuous functions as dual spaces" by Dales, Dashiell, Jun, Lau and Strauss [Zbl 1368.46003](https://zbmath.org/?q=an%3A1368.46003)
How to think about $A''$? Let $z$ be the support projection of $\phi$, a central projection. One can show that $A'' \cong zA'' \oplus\_\infty (1-z)A''$ the $\ell^\infty$-direct sum of two von Neumann algebras. In this case, $zA'' \cong L^\infty([0,1])$ while $(1-z)A''$ is very complicated, given by all probability measures on $[0,1]$ (= states on $A$) which are singular with respect to Lebesgue measure. As $zA''\cong L^\infty([0,1])$ we see that the indicator function of $[0,1]$ *is minimal* in $L^\infty([0,1])$, with the property that it has full Lebesgue measure.
Thus one picture of $A''$ is to take some maximal family $(\mu\_i)$ of mutually singular probability measures on $[0,1]$ and then $A''$ is isomorphic to the $\ell^\infty$-direct sum of $L^\infty([0,1], \mu\_i)$.
I do not know about closed projections, and so will let someone else answer there.
| 3 | https://mathoverflow.net/users/406 | 433321 | 175,292 |
https://mathoverflow.net/questions/433320 | 3 | This might seem like a silly question considering my relatively elementary knowledge of representation theory.
The question is regarding Eugen Hellman 's paper titled ["On the derived category of the Iwahori-Hecke algebra"](https://arxiv.org/abs/2006.03013). Specifically on the paragraph above Lemma 2.18. First, let $G$ be a reductive group over a field $C$ of characteristic zero, $B$ a Borel subgroup and $U$ the unipotent radical of $B$. The paper states:
>
> ... recall that an algebraic representation of $B$ defines a $G$-equivariant vector bundles on $G/B$. We write $\mathcal{U}^\vee$ for the $G$-equivaraiant vector bundles on $G/B$ defined by the canonical $B$-representation on $\mathfrak{u}^\vee$. Here $\mathfrak{u}$ denotes the Lie algebra of $U$ (considered as a $C$-vector space), and $\mathfrak{u}^\vee$ denotes it's dual. In particular $\mathcal{U}^\vee$ admits a filtration who graded pieces are line bundles $\mathcal{L}\_\alpha$ on $G/B$ associated to negative roots (with respect to $B$) roots $\alpha$ of $G$.
>
>
>
My questions are the following:
1. Firstly, I believe by an algebraic representation, it just means a homomorphism of group schemes $B\rightarrow \operatorname{GL}\_n$. Is that correct?
2. How does such a representation define a vector bundle on $G/B$ ? I am guessing it has something to do with constructing a $\operatorname{GL}\_n$-torsor using the representation, but I am not sure how.
3. Why does $\mathcal{U}^\vee$ admit a filtration into line bundles? (this question might seem silly considering I don't understand how $\mathcal{U}^\vee$ is constructed in the first place)?
Thanks in advance.
| https://mathoverflow.net/users/157428 | Algebraic representations and vector bundles | The definition of algebraic representation should be exactly what you suggested in 1).
For what concerns 2), you can construct a $\operatorname{Gl}\_n$-torsor over $G/B$ given a representation $B \to \operatorname{Gl}\_n$ as you suggested. The reasoning should be the following: from the homomorphism $B \to \operatorname{Gl}\_n$ you get an action of $B$ over $V$ a complex vector space of dimension $n$. You then consider the product variety $G \times V$ with the $B$ (right) action $$b \cdot(g,x)=(gb,b^{-1}x) .$$
You can check that this action is free and there is a quotient variety $G \times V/B$ usually denoted $G \times\_B V$. The projection onto the first factor $p:G \times V \to G$ pasisng to the quotient induces a well defined map $p: G\times\_B V \to G/B$ which can be checked to be a vector bundle. You can see that the fiber of $p$ is actually isomorphic to $V$ and so the vector bundle has dimension $n$.
Notice that given a $B$-invariant subspace $W$ you get an embedding of vector bundles $G \times\_B W \subseteq G \times\_B V$ such that the following isomorphism of vector bundles holds $$G \times\_B V/W \cong (G \times\_B V)/ (G \times\_B W) .$$
You then need to find a filtration of $B$ representations of $\mathfrak{u}^{\vee}$ such that the associated graded piece is isomorphic to the $1$-dimensional space $\mathfrak{u}\_{\alpha}$ for $\alpha$ negative roots. This should come from reprentation theory of reductive Lie algebras and reductive groups.
| 4 | https://mathoverflow.net/users/146464 | 433322 | 175,293 |
https://mathoverflow.net/questions/433325 | 2 | I started trying to learn about Gromov-Witten invariants by reading the book "$J$-holomorphic curves and Symplectic Topology" and I have a doubt in an example the authors provide. It's example $7.1.3$ where we consider the homology class $A=0$.
So when we are focusing in the case where $k=3$ and we want to compute the invariant $GW^{M}\_{0,3}(a\_1,a\_2,a\_3)$ the authors claim that this will be equal to $\int\_Ma\_1\cup a\_2\cup a\_3$.
However I'm having some difficulty seeing why this result is true. So by definition we have that $GW^{M}\_{0,3}(a\_1,a\_2,a\_3)$ is equal to $f\cdot ev\_{J}$ where $f$ is a pseudo-cycle Poincaré dual to $\pi\_1^\*a\_1\cup \pi\_2^\*a\_2\cup \pi\_3^\*a\_3$ and $ev\_{J}$ is the evaluation pseudo-cycle. By definition of Poincaré-dual we have that for any $2n-$submanifold $X\subset M^3$, $\int\_{X}a=f\cdot X$. So i'm guessing the idea would be to take $X=M\times \{pt\}\times \{pt\}$? However how can I know that the pseudocycle $ev\_{J}$ can represent this submanifold ? Is it because $k=3$ it does not matter which points we choose?
Any insight is appreciated. I'm fairly new to this so if anyone knows of any reference for this or other examples I would appreciate it. Thanks in advance.
| https://mathoverflow.net/users/155363 | Question on Gromov-Witten invariants when $A=0$ | In this case, the J-holomorphic curves are all constant, so the
evaluation pseudocycle is the tridiagonal $\{(x,x,x) : x\in
M\}$. You take cycles $A\_1,A\_2,A\_3$ Poincare dual to
$a\_1,a\_2,a\_3$ respectively and intersect the tridiagonal with
$A\_1\times A\_2 \times A\_3$. This gives you precisely
$\{(x,x,x) : x\in A\_1\cap A\_2\cap A\_3\}$, which is the same
as the triple intersection of $A\_1,A\_2,A\_3$ (i.e. the integral $\int\_M a\_1\cup a\_2 \cup a\_3$).
| 4 | https://mathoverflow.net/users/10839 | 433329 | 175,297 |
https://mathoverflow.net/questions/433335 | 8 | Let $u \in C^{\infty}(\mathbb R^3)$ be harmonic. Suppose that $u$ has no critical points outside the unit ball but that it has at least one critical point inside the unit ball.
Does it follow that $u$ is a polynomial?
| https://mathoverflow.net/users/50438 | On critical points of harmonic functions | This is not true in dimension 2.
Function $f(z)=ze^z$ is entire, and $f'(z)=0$ at one point,
$z=-1$. It follows that the function
$$u(x,y)=\mathrm{Re}f(x+iy)=e^x(x\cos y-y\sin y)$$
is harmonic, has one critical point $(x,y)=(-1,0)$, and evidently not a polynomial.
Now one can construct a similar example in even dimension, for example:
$$v(x\_1,x\_2,x\_3,x\_4)=u(x\_1,x\_2)+u(x\_3,x\_4).$$
But for $n=3$ I could not figure out.
| 10 | https://mathoverflow.net/users/25510 | 433339 | 175,301 |
https://mathoverflow.net/questions/433218 | 2 | My question may be simple to an expert, but I'm not:
Let's consider $u \in C^{s}(\mathbb{R}^d)$ be a Hölder function sor some $s\in [0,1/2)$ which we may take very close to $0$.
Of course, $u^2 \in C^{s}(\mathbb{R}^d)$ so that $(u^2)' \in B^{s-1}\_{\infty,\infty}$, is a well-defined distribution, with some explicit negative regularity.
So it means that in this case one can define $u'u = (u^2)'$ as a distribution. But if one wants to define the product $u'u$ by using some general rule, like defining $vu$ for $u\in C^s$, $v\in C^{s-1}$, when $s>1/2$. But for $s<1/2$, this is not possible this way.
I know that there are critera like if $\{(x,\xi) \vert (x,\xi) \in WF(u), (x,-\xi)\in WF(v)\}=\varnothing$ then $uv$ is well-defined as a distribution. So my question is the following:
>
> Is there a way to apply these microlocal tools to define $u'u$ when $u\in C^s$, $s \ll 1$?
>
>
>
| https://mathoverflow.net/users/94414 | Microlocal approach to definition of product of distributions | Too long for a comment. For $u$ in $C^s$, $s\in (0,1)$, you can indeed define $u^2$ and then the distribution-derivative of $u^2$, which belongs to $B^{s-1}\_{\infty,\infty}$. Now that does not define the product $uu'$, unless you decide to define that product as $\frac12(u^2)'$. Assuming for instance that $u$ is also compactly supported, you can mollify everything and consider $u\_\epsilon=u\ast \rho\_\epsilon$
and you will have trouble at proving that
$$
u u'\_\epsilon
$$
has a (weak) limit in the distribution sense when $s<1/2$.
About your question, I believe that the answer is negative.
One reason is that you may consider only real-valued functions defined on the real line: in that case the wave-front-set is trivially deduced from the singular support and is
$$
\text{singsupp} u\times \mathbb R^\*.
$$
So microlocal analysis is of no help in that situation, whereas your problem of defining $uu'$ in that simple situation remains the same.
| 2 | https://mathoverflow.net/users/21907 | 433345 | 175,303 |
https://mathoverflow.net/questions/433337 | 2 | Assume that $C$ is a projective curve and $X$ is an elliptic fibration over $C$.
>
> What is the picard group of $X$? can we say something about it?
>
>
>
I think it should be generated by multisections and (components of ) fibres. How can I see it formally?
| https://mathoverflow.net/users/nan | Picard group of an elliptic fibration is generated by multisection and fibres | Do you know about the [Shioda-Tate formula](https://planetmath.org/shiodatateformula). (I'm surprised there isn't a Wikipedia article about it.) Anyway, it says that if the elliptic fibration does not split as a product and if you have at least one section, then the Neron-Severi group is generated by the following divisors:
1. One irreducible fiber.
2. For each reducible fiber, take all but one of
the components.
3. The zero-section.
4. Generators for the group of sections.
| 6 | https://mathoverflow.net/users/11926 | 433346 | 175,304 |
https://mathoverflow.net/questions/430142 | 4 | I'm not sure if this is completely relevant to MO, let me know if this would be better on MSE.
I have been told today by a professor of mine that the following is a classic result of Cartan. Suppose $M$ is a closed parallelizable smooth $n$-manifold and $X\_1, \ldots, X\_n$ are everywhere linearly independent vector fields with the property that for each $i < j$,
$$ [X\_i, X\_j] = \sum\_{k=1}^n \alpha\_{ij}^k X\_k $$ for real constants $\alpha\_{ij}^k$. Then this is equivalent to saying that $M \cong G/H$ for some (I don't know which) Lie group $G$ and some discrete subgroup $H$.
My professor claims this can be found in Cartan's 1936 book "La topologie des groupes de Lie". I have yet to obtain this reference and check myself but I believe him.
**EDIT**
I've obtained the reference and skimmed through it but I did not find any indication of this theorem. Do you know another reference that actually proves this?
**End EDIT**
Anyway, assuming this, I had this extremely overly ambitious idea to use this result as an alternative approach to the Poincaré conjecture. As a necessary disclaimer I do not make any claims to success in this regard, of course, and this is merely a curiosity of mine. The approach would be as follows. Let $M$ be a simply connected closed $3$-manifold. Since it is orientable it is parallelizable, and the goal would be to obtain three vector fields satisfying the above condition.
One might start by replacing the constants $\alpha$ with functions $\alpha(p) \in C^\infty (M)$ and try to solve for the $9$ equations $d\alpha\_{ij}^k = 0$. If successful, this would imply that $M$ is a homogeneous space, and the only closed simply connected $3$ dimensional homogeneous space is $S^3$.
Now, my professor also claimed that this approach is no easier than the original formulation of the conjecture, if not harder. I believe him, but would like to know why. In particular, how would one begin to approach such a problem, or something similar involving a system of a large number of first order PDEs on a closed manifold?
Putting it more specifically, my question is:
>
> Forgetting about the relevance to the Poincaré conjecture, how would one naively begin approaching this (very hard) problem, and to what field of study is this most relevant? For similar yet easier problems, could you recommend a good reference on the subject?
>
>
>
At this point I'm just hoping to learn something interesting and useful while playing around with this.
| https://mathoverflow.net/users/143629 | On a result of Cartan for homogeneous manifolds arising from a quotient of discrete subgroups | The result that you are looking for is not in Élie Cartan's 1936 book *La topologie des groupes de Lie* because it was not known to be true at the time the book was written. Indeed, as Cartan remarks in the book (which is the lecture notes from an October 1935 conference where he spoke), it was not even known at that time that every Lie algebra over the reals was the Lie algebra of a Lie group. At least, it was not known by him.
However, I. D. Ado, then a student at Kazan State University, had, in 1935, published (in Russian) a proof of his now famous result that every Lie algebra over the reals has a faithful, finite dimensional representation, which implies the existence of a (unique up to isomorphism) connected and simply-connected Lie group with any given Lie algebra. (Apparently, Cartan was not aware of Ado's work when his book went to press, but he was certainly aware of it sometime before 1938; it's hard to say exactly when.)
Assuming the existence of a connected, simply-connected Lie group $G$ whose left-invariant vector fields have a basis $Y\_i$ satisfying $[Y\_i,Y\_j]= c^k\_{ij} Y\_k$ (I omit the summation sign), one can sketch the proof of the result you want (assuming that $M$ itself is connected) as follows:
On the product manifold $P = M\times G$ consider the vector fields $Z\_i = X\_i + Y\_i$ (where I am identifying the tangent space $P\_{(m,g)}$ with $T\_mM\oplus T\_gG$ in the obvious way), these $n$ linearly independent vector fields satisfy $[Z\_i,Z\_j]= c^k\_{ij} Z\_k$ and hence span an $n$-plane field that is integrable. Thus, by the Frobenius theorem, $P$ is foliated by the $n$-dimensional 'leaves' that are everywhere tangent to the $Z\_i$. Fix an $m\in M$ and let $L\subset M\times G$ be the leaf of this foliation that passes through $(m,e)$.
The projection of $L$ to $G$ is a local diffeomorphism; we want to show that it is a covering map, i.e., that $\pi:L\to G$ has the homotopy lifting property.
Here is where the compactness of $M$ comes in. Since $M$ is compact, the vector fields $X\_i$ are complete (i.e., their flows exist for all time). Using this, it is easy to show any differentiable curve $\gamma:[0,1]\to G$ such that $\gamma(0) = \pi(p)$ for $p\in P$ can be lifted uniquely to a curve $\tilde\gamma:[0,1]\to P$ such that $\tilde\gamma(0)=p$ and $\pi\_2\circ\tilde\gamma = \gamma$ *and* $\tilde\gamma$ is everywhere tangent to the $n$-plane field spanned by the $Z\_i$. (In fact, you just write $\gamma'(t) = a^i(t) Y\_i(\gamma(t))$ for some functions $a^i$ and then let $\tilde\gamma$ satisfy $\tilde\gamma(0)=p$ and $\tilde\gamma'(t) = a^i(t) Z\_i(\tilde\gamma(t))$.) The fact that $L$ is a covering map follows immediately.
Since $G$ is connected and simply connected, $\pi$ must be a diffeomorphism, and, hence, the graph of a smooth mapping $f:G\to M$ such that $X\_i$ is $f$-related to $Y\_i$. Now, we can use the same trick as above to show that for any smooth curve $\gamma:[0,1]\to M$ with $\gamma(0) = f(g)$, there is a smooth curve $\tilde\gamma:[0,1]\to G$ such that $\tilde\gamma(0)=g$ and $f\circ\tilde\gamma = \gamma$. Since $M$ is connected, every point can be connected to $f(e)$ by a smooth curve, so $f$ is surjective and is a covering map.
Finally, let $H = f^{-1}(m)\subset G$. Then $H$ is discrete because $f$ is a covering map. Moreover, if $h\in H$ and $\lambda\_h:G\to G$ is left-multiplication by $h$, then $f\circ \lambda\_h(e) = f(h) = m$ and, since the vector fields $Y\_i$ are left-invariant, it easily follows that the graph of $f\circ\lambda\_h$ in $P$ is a leaf of the $n$-plane field spanned by the $Z\_i$ that contains $(m,e)$, so it must be equal to $L$. Consequently, $\lambda\_h(H) = H$ (implying that $H$ is a subgroup of $G$) and $\lambda\_h$ is a deck transformation for the covering map $f:G\to M$. Thus, $H$ is a discrete subgroup of $G$, and $M$ is diffeomorphic to $H\backslash G$.
You might want to think about whether you can extend this result to the case that $M$ is not connected (in which case, of course, $G$ would have to be disconnected). Consider the case where $M$ is the disjoint union of two circles and the flow of $X\_1$ has period 1 on one of the circles and period $\pi$ on the other.
Now, on to your second question, which is more subtle. We know that not every compact, connected, orientable $3$-manifold $M$ is homogeneous, so you can't hope to find a frame field $X\_i$ for which the $\alpha^i\_{jk}$ are constants in that generality. You might hope to impose some weaker condition $C$ on the $\alpha^i\_{jk}$ so that you could guarantee that a frame field satisfying $C$ would always exist and yet that $C$ would be 'geometric' enough that you could use it to recognize pieces into which $M$ could be cut as 'geometric'. Or you might hope to find some expression in the $\alpha^i\_{jk}$ that you could integrate over $M$ (using the volume form for which the $X\_i$ are unimodular) to give you some measure of the 'complexity' of the frame field $X\_i$ that you could try to minimize by some kind of gradient flow. If the minima of this functional were well-enough behaved, maybe that would give you a clue as to how to use such a minimizer to cut $M$ into geometric pieces. (Or you could follow the strategy of Perelman's argument and try to understand the singularities that develop when you attempt to flow to a minimizer.)
You could even give yourself a 'headstart' by imposing some vanishing at the start. For example, it's not hard to show that you can always choose a frame field $X\_i$ so that $\sum\_i\alpha^i\_{ij} =0$ for all $j$, which gets rid of $3$ of the $9$ components right off the bat.
The main problem, though, will be finding conditions on the $\alpha$ that are somehow attainable or approachable and yet give you geometric information. The nice thing about metrics, as in Hamilton's and Perelman's approach(es), is that we know a lot about Riemannian geometry. We know relatively little about how to interpret the invariants of frame fields geometrically except in very special circumstances (such as the $\alpha$s being constant), and they are (so far) just too special to be of much use in tackling something like the Poincaré Conjecture or the Geometrization Theorem.
I'm *not* saying that 'frame field geometry' won't get you anywhere in studying the topology of $3$-manifolds, I just think that the necessary work of developing the foundations of such a geometric theory hasn't been sufficiently pursued that we can have a good sense of what specific things to try. (And you can see, from what I wrote above, that there are many possible things to try.)
| 5 | https://mathoverflow.net/users/13972 | 433360 | 175,306 |
https://mathoverflow.net/questions/433375 | 7 | I asked this on Math Stack Exchange, but apparently no one paid attention to it. So, I am asking it again, filling in the background necessary to understand it.
Consider a countably infinite set $P$ of propositional atoms, indexed by the positive integers like so: $p\_1,p\_2,p\_3,p\_4,...$. We also have the connectives $\land$, $\vee$, and $\neg$ as well as the parentheses $($ and $)$. We define the set $W$ of well-formed formulas from the alphabet $P \cup \{\land,\vee,\neg,(,)\}$ in the standard recursive way given in almost any mathematical logic textbook. Over the set $W$, I define an equivalence relation $E$ by saying that two wffs $A$ and $B$ are related by $E$ iff they are logically equivalent, where logical equivalence is defined in the standard way in any logic textbook. Now, let $B$ be the set of equivalence classes of $W$ under $E$, and consider the Boolean algebra structure $(B;\land,\vee,\neg,0,1)$, where $0$ is the set of contradictions, $1$ is the set of tautologies, and $\land$, $\vee$, and $\neg$ are operations defined by passing to representatives (and it is easy to check that they are well-defined).
I previously asked if the set of propositional atoms, or even any non-empty subset of it, is definable without parameters in the structure $B$. The answer was negative. Now, if we use parameters, certainly any finite subset of the set of propositional atoms is definable. But I believe that is the best possible. So, my current question is, is the set of propositional atoms, or even any infinite subset of it, definable even with parameters? I believe it is not, but how to prove it?
Edit: As a bonus question, can anyone give a complete classification of all the parameter-definable subsets of the structure $B$?
Edit 2: As a matter of fact, I now conjecture that the parameter-definable subsets are finite and cofinite subsets of $B$, that is, $B$ is a minimal structure. Can anyone confirm or deny this conjecture?
| https://mathoverflow.net/users/43439 | Are no infinite subsets of the set of all propositional atoms definable in this structure, even with parameters? | It's a nice question. This Boolean algebra, known as the [*Lindenbaum algebra*](https://en.wikipedia.org/wiki/Lindenbaum%E2%80%93Tarski_algebra), is a countable atomless Boolean algebra — it is atomless because we can always take the conjunction of any formula with a new atom or its negation — and all such Boolean algebras are isomorphic by a standard back-and-forth argument. So your particular presentation of the algebra is inessential, since there are many presentations of this algebra.
The back-and-forth argument also shows that the structure is homogeneous — any finite partial isomorphism will extend to an automorphism of the whole algebra. Any finitely many parameters are contained in a finite Boolean subalgebra, and so any two elements that stand in the same relation to the atoms of that subalebra will be automorphic images. This severely limits the definable sets and will support a classification. Every definable set from parameters will be a finite union of the sets of points that all relate to the atoms of the finite subalgebra generated by those parameters in the same way. (Perhaps someone can provide a fully detailed account.)
In particular, the set of propositional atoms will not be definable with parameters, since any atom $p$ not appearing in the expressions of the parameters will be automorphic with the conjunction of two other such atoms $q\wedge r$ over those parameters. Similarly, no infinite set of atoms will be definable with parameters, since we can always move one of them automorphically to something that isn't an atom.
| 11 | https://mathoverflow.net/users/1946 | 433377 | 175,313 |
https://mathoverflow.net/questions/433378 | 10 | Consider a polynomial in one variable $p(x)$ with $p(0)>0$, and that is not a polynomial in $x^m$ for any $m>1$ (that is, the $gcd$ of the exponents appearing in $p(x)$ is 1). I would like to find necessary and sufficient conditions so that a power of $p$ has no negative coefficients. I know of a sufficient condition: if the degree is $d$, then some power of $p$ will have no negative coefficients if the coefficients of $x, x^d$ and $x^{d-1}$ are all positive, AND the inequality $|p(z)|<p(|z|)$ holds for all complex numbers $z$ other than positive real numbers. This condition is obviously not necessary because $p(x)=1+x^2+x^3$ does not satisfy it, yet its first power has no negative coefficients. On the other hand the condition $|p(z)|<p(|z|)$ for non-real and positive $z$ by itself is surely necessary (because it is satisfied by polynomials without negative coefficients, so if $p^n$ satisfies it, $p$ itself must satisfy it, by taking the $n$-th root), but not sufficient, as simple examples will show. Other necessary conditions can be obtained by identifying some crucial coefficients towards the beginning and the end of $p(x)$ that must be positive (see my article [On the inequality $|p(z)|\leq p(|z|)$ for polynomials](http://webusers.xula.edu/vdeangel/publications/ineq.pdf)). But there are examples that show that even if all these crucial coefficients are positive and the $|p(z)|<p(|z|)$ condition holds, the polynomial may still have some negative coefficients in all of its powers. So except for the rather special case when the coefficients of $x,x^d, x^{d-1}$ are positive, I do not know how to predict when a power of a polynomial will have no negative coefficients. Obviously my list of "crucial coefficients" that must be positive is not enough but I do not know how to enlarge it.
As a related note, David Handelman has proved that in the (much more difficult) context of polynomials in several variables, if one power of $p$ has no negative coefficients, then $p^n$ will have no negative coefficients for all sufficiently large $n$.
| https://mathoverflow.net/users/66323 | Conditions for a power of a polynomial to have no negative coefficients | Marcus Michelen, Julian Sahasrabudhe, A characterization of polynomials whose high powers have non-negative coefficients,
<https://arxiv.org/abs/1910.06890>
| 11 | https://mathoverflow.net/users/25510 | 433383 | 175,314 |
https://mathoverflow.net/questions/433364 | 3 | This is a generalisation of an older [question inspired by a football tournament](https://mathoverflow.net/questions/432988/on-a-combinatorial-design-inspired-by-a-football-soccer-tournament) (which does not have an answer yet).
Let $\frak P$ be a partition of $\omega$ into blocks, that is, pairwise disjoint sets, such that every block has at least $2$ elements. (Blocks may be infinite.) Assume that $\frak P$ consists of infinite many blocks.
Is there for every $n\in\mathbb{N}$ a bijection $\varphi\_n:\omega\to\omega$, such that the following condition is met?
>
> Whenever $a\neq b\in\omega$ there is a unique $n\in\mathbb{N}$ such that $\{\varphi\_n(a), \varphi\_n(b)\}\subseteq B$ for some $B\in{\frak P}$.
>
>
>
| https://mathoverflow.net/users/8628 | Cycling through a general combinatorial design on $\omega$ | If there are infinitely many blocks, then indeed you can find the permuations $\pi\_n$ as desired. Indeed, there are computable such $\pi\_n$. Furthermore, you don't need that every block has size at least two, but rather only that there is at least one block of size at least two.
We build the sequence of permuations $\pi\_n$ in stages, so that at each stage, only finitely much of finitely many permutations is specified. We want gradually to fulfill the following requirements:
* for each $n$ and $k$, the value of $\pi\_n(k)$ is specified
* for each $n$ and $k'$, there is some $k$ such that $\pi\_n(k)=k'$
* for each pair $a\neq b$, there is a unique $n$ such that $\pi\_n(a)$ and $\pi\_n(b)$ appear in the same block.
Now, the point is that we can gradually fulfill these requirements in a computable process that specifies more and more of the system of permuations. We can always fulfill the first requirement, for example, without any tension with the other requirements by specifying $\pi\_n(k)$ to have a value appearing in a totally new block not yet used. We can fulfill the second requirement by using a value $k$ not yet mentioned in the construction. Infinitely often, we can give attention to the third requirement by selecting for a given pair $a\neq b$ a totally new $n$ not yet considered at all and placing $\pi\_n(a)$ and $\pi\_n(b)$ to have values in the same block. At each later stage, we can preserve this uniqueness property for this pair.
In this way, the permutations are gradually defined, while fulfilling the properties, and so there are indeed such permuations.
| 4 | https://mathoverflow.net/users/1946 | 433387 | 175,315 |
https://mathoverflow.net/questions/433261 | 4 | Suppose that the closed, piecewise $C^1$-curve $f(\mathbb T)$ has exactly $n$ points that are run through twice, all other points are run through once. Is it true that the compact set $f(\mathbb T)$ has exactly $n+1$ holes? For $n=1$ this is true and is based on the well-known fact that a crosscut in a Jordan domain determines two Jordan subdomains (Proposition 2.12 in Pommerenke's book).
More generally, if for a finite decomposition $ \mathbb T=\cup\_{j=1}^n I\_j$ of $\mathbb T$ into closed arcs with pairwise disjoint interiors, $f$ is injective on the interior of each $I\_j$, what is the maximum number of holes of the closed set $f(\mathbb T)$?
| https://mathoverflow.net/users/61993 | Curves in the plane and their number of holes | Here's a slightly more hands-on point of view, using just the Jordan curve theorem (although of course in the end it does come down to the same thing as Euler's formula somehow, as described by Alex.)
Take a closed curve $\gamma\colon [0,1]\to S^2$, and assume that there are only finitely many pairs of points in $[0,1]$ that have the same image under $\gamma$. (Here $S^2$ is the sphere, i.e. the one-point compactification of the plane.)
Now define a finite sequence $(x\_n)\_{n=0}^{N}$ in $[0,1]$ by letting $x\_0 = 0$ and letting $x\_{n+1}>x\_n$ be minimal with $\gamma(x\_{n+1})\in \gamma([0,x\_{n+1}))$. By assumption, this yields a finite sequence with $x\_N = 1$.
Now, I claim that $\gamma([0,x\_n])$ separates the sphere $S^2$ (compactification of the plane) into $n+1$ connected components, which are all simply-connected (hence topologically equivalent to the plane). This is trivial for $n=0$. If true for $n<N$, then it follows for $n+1$. Indeed, $\gamma( (x\_n,x\_{n+1}) )$ lies in some complementary component $U$ of $\gamma([0,x\_n])$. Either this curve is injective, in which case it tends to $\partial U$ in both directions. Under the homeomorphism of $U$ to the plane, it is thus a Jordan curve in the plane, and separates $U$ in exactly two components.
If the curve is not injective, then there is a point $x'\_n \in (x\_n,x\_{n+1})$ with $\gamma(x'\_n) = \gamma(x\_{n+1})$. Then $\gamma( [x\_n', x\_{n+1}])$ is a Jordan curve in $U$, and $\gamma([x\_n,x\_n'])$ is an arc connecting a point of $\partial U$ to this Jordan curve. Again, we see that $\gamma(x\_n,x\_{n+1})$ separates $U$ into two components, and the total number of components increases by $1$.
So we have proved that $\gamma$ separates the plane into $N+1$ components. In your setting, we easily see that $N=n+1$ (there is one $x\_n$ for each of your multiple points, plus $x\_0 = 0$ and $x\_{n+1}=1$). So, there are n+2 components, or n+1 "holes" in your terminology.
Nb. You will notice that this is essentially the same argument as in the proof of Euler's formula: adding a new edge with one existing and one new vertex does not create new faces, while adding a an edge between two existing vertices DOES create an extra face.
| 1 | https://mathoverflow.net/users/3651 | 433393 | 175,318 |
https://mathoverflow.net/questions/433366 | 0 | Let $G:(0,1)\to(0,1)$ be the Gauss map, i.e., $G(x)=\left\{\frac1{x}\right\}$, which is known to act as the shift on the space of continued fraction expansions.
**Question.** Is there an explicit expression for the natural extension of $G$ in $\mathbb R^2$? What about its density?
| https://mathoverflow.net/users/8131 | Natural extension of the Gauss map | As it was already mentioned by Christophe Leuridan for classical continued fraction
we have extended Gauss measure
$$d\overline{\nu}(x,y)=\frac{1}{\log2}\cdot\frac{dx\,dy}{(1+xy)^2}=\frac{1}{\log2}\cdot\begin{vmatrix} 1 & \mp x\\
\pm y & 1
\end{vmatrix}^{-2} dx\,dy,\qquad (x,y)\in[0,1]^2,$$
which is invariant under the map
$$\overline{T}:(x,y)\to\begin{cases}(\{1/x\},([1/x]+y)^{-1},&\text{ if }x\ne 0;\\
(x,y),&\text{ if }x= 0.
\end{cases}$$
Details can be found in the book
*M. Iosifescu and C. Kraaikamp, Metrical theory of continued fractions, 2002.*
From geometrical point of view this extended Gauss measure describes the distribution of reduced Voronoi bases $(b\_1,b\_2)$ with $b\_1=(1,\mp x)$, $b\_2=(\pm y,1)$ on the boundary of the unit square.
In dimension $3$ something similar is knowт for Voronoi-Minkowski continued fractions(see [Three-dimensional continued fractions and Kloosterman sums](https://iopscience.iop.org/article/10.1070/RM2015v070n03ABEH004953/meta)). The measure has the form
$$\mathrm{const} \int\_{\Pi}\frac{d\alpha'd\beta'd\gamma'}{(\det X)^3},$$
where $X=\begin{pmatrix}
1 & \alpha\_2' & \pm\alpha\_3'\\
-\beta\_1' & 1 & \beta\_3'\\
\gamma\_1'&-\gamma\_2'&1\\
\end{pmatrix}$ are matricies of reduced Minkowski bases and
$\alpha'=(\alpha\_2',\alpha\_3')$,
$\beta'=(\beta\_1',\beta\_3')$,
$\gamma'=(\gamma\_1',\gamma\_2')$.
| 3 | https://mathoverflow.net/users/5712 | 433400 | 175,320 |
https://mathoverflow.net/questions/433200 | 4 | Let $M$ be a compact manifold and $\varphi:M\rightarrow M$ a diffeomorphism. The invariant differential forms
$$
\Omega^{k}\_{inv}(M)=\{\alpha\in\Omega^{k}(M):\varphi^{\*}\alpha=\alpha\}
$$
form a subcomplex $(\Omega\_{inv}(M),d)$ of the de Rham complex. Denote by $H\_{inv}(M)$ its cohomology.
**Question:** Is there an example where the map $H^{1}\_{inv}(M)\rightarrow H^{1}(M)$ is not injective?
It seems like the map is injective when forms can be averaged with respect to $\varphi$ in some way. For instance if $\varphi$ has finite order, or more generally if $\varphi$ is an isometry for some Riemannian metric $g$, in which case one can average over the closure $\overline{\{\varphi^{n}\}}$ in the compact Lie group $Isom(M,g)$. But I think there should be an example where the map is not injective.
| https://mathoverflow.net/users/150945 | Cohomology of invariant differential forms | There is no such thing: any $\phi$-invariant exact
1-form is a differential of $\phi$-invariant function.
Indeed, let $\alpha$ be an exact $\phi$-invariant
form, $\alpha=df$, where $f$ is not $\phi$-invariant.
Then $d(\phi^\* f -f)=0$, hence
$\phi^\* f = f + C$, where $C$ is a constant.
This gives $\sup f = \sup \phi^\* f = \sup f + C$, which is
a contradiction.
For higher cohomology I think there are some examples.
| 4 | https://mathoverflow.net/users/3377 | 433405 | 175,322 |
https://mathoverflow.net/questions/433294 | 2 | For the sake of simplicity, assume $f$ is a non-cm eigenform of weight $k$ on the group $\mathrm{SL}(2, \mathbb{Z})$. Are there any known results or conjectures regarding any special values of the associated $L$-function $L(f, n)$ for any integers $n$ for weight $k > 2$? If so, what can be said about the rational factor appearing said special values? In the case of elliptic curve $L$-functions, BSD predicts the factor of $\#E(\mathbb{Q}\_{\mathrm{tors}})^{2}$ appearing in the denominator for $L(E, 1)$. I'd like to know if something similar in the form of known results or conjectures exist for $L$-functions satisfying the criteria above. I do believe they are quite mysterious, and I have been rather curious about Ramanujan's $L$-function $L(\Delta, s)$ for $s = 11$ in particular. This is something I'm far from an expert in!
| https://mathoverflow.net/users/138669 | Special values of non-cm $L$-functions | Since the OP is interested in particular by $L(\Delta,11)$, the relevant
theorems come into the framework of Deligne's theory of special points and
special values, while Bloch-Kato would be for $s>11$. Here the theorems are
due to Shimura and especially Manin, which for $\Delta$ (and for general eigenforms similarly) states the following: set $r\_j=(-2i\pi)^{-j-1}j!L(\Delta,j+1)$.
There exist real numbers $\omega^+$ and $\omega^-$ such that
$$(r\_1,r\_3,r\_5,r\_7,r\_9)=(48,-25,20,-25,48)\omega^-$$
$$(r\_0,r\_2,r\_4,r\_6,r\_8,r\_{10})=(22680/691,-14,9,-9,14,-22680/691)\omega^+i$$
$$\omega^-\omega^+=(4096/691)<\Delta,\Delta>$$
| 3 | https://mathoverflow.net/users/81776 | 433407 | 175,323 |
https://mathoverflow.net/questions/433406 | 2 | I am working on a problem that involves an iterative application of a function I think might be a trapdoor function.
Formally, I have a function $f:X \to X$ that can be described as
$$
[x\_{1,N+1}, ..., x\_{s,N+1}]=f([x\_{1,N},...,x\_{s,N}])\\
\forall\_{i<s-1}x\_{i,N}=x\_{i+1,N+1}
$$
and $x\_{s, N+1}$ is probabilistically sampled and corresponds to an integer vector in high dimension.
If we know the starting state of the function ($[x\_{1,0},...,x\_{s,0}]$), calculating the forward propagation of the function is straightforward and we can get to $[x\_{1,N},...,x\_{s,N}]$ by performing $f \circ f$ $N$ times, meaning that for N<s, we can easily verify that the sequence [x\_{1,N}, ..., x\_{s,N}] was indeed sampled by $f$.
However, without ($[x\_{1,0},...,x\_{s,0}]$), there is no trivial way to revert the function but to sample all the possible values for $x\_{1,N}$.
I have a hunch that if $f$ was deterministic, it would correspond to a trapdoor function, with $[x\_{1,0},...,x\_{s,0}]$ acting as a secret $t$, but I have no clear idea as to how to prove such an equivalence.
Coming from a CS background, if I wanted to prove a problem was $NP$-hard, I would try to prove an equivalence to a known $NP$-hard problem. Could a similar approach work for trapdoor functions?
If yes, is there a list of known/suspected trapdoor functions?
| https://mathoverflow.net/users/44865 | Proving that a function is a trapdoor function | First, there is a site crypto.stackexchange.com that is typically better for questions like this.
Second,
>
> Coming from a CS background, if I wanted to prove a problem was NP-hard, I would try to prove an equivalence to a known NP-hard problem. Could a similar approach work for trapdoor functions?
>
>
>
The answer is *sort of*.
There are no known ways of building trapdoor functions that are secure assuming the *worst case* hardness of an NP-hard problem.
There are two lines of work that are formally "close" (not in the sense with "given enough time they will achieve this goal though").
1. *Lattice-based Cryptography*: [Regev established](https://cims.nyu.edu/%7Eregev/papers/lwesurvey.pdf) that a certain useful cryptographic assumption (the "Learning with Errors problem") is *average-case hard* as long as certain lattice problems are hard in the *worst case*.
These lattice problems are not NP hard (in fact, they are in NP and coNP), but they are still somehow "close" --- they are approximation problems for approximation factor $\gamma$ where for:
a. $\gamma = O(1)$ (actually slightly higher, but still) the problem is NP hard
b. for $\gamma = O(\sqrt{n})$ the problem is in NP and coNP, but we can (start to) build useful cryptography for it
c. we can solve the problems in polynomial time for sub-exponential approximation factors.
So we are somehow much closer to the "NP hard" side of things than the "known poly-time algorithms" side of things, but it seems unlikely $\gamma$ can be reduced further.
2. Recently, Pass and Liu have had some work ([this](https://drops.dagstuhl.de/opus/volltexte/2022/16597/) and [this](https://drops.dagstuhl.de/opus/volltexte/2022/16598/)) that shows that the existence of one-way functions (not equivalent to the existence of trapdoor functions, but sufficient to build much of cryptography, known as "minicrypt" primitives) is equivalent to establishing average-case hardness of an explicit NP hard problem, i.e. it suffices to prove worst-case to average-case reductions for this explicit problem.
That all being said, most cryptography doesn't concern itself with connections to worst-case hardness.
Instead, average-case hardness is the main thing needed.
Most books in cryptography will cover how to construct various cryptosystems assuming the average-case hardness of underlying computational problems.
A popular (undergraduate-level) cryptography textbook is Katz and Lindell's *Introduction to Modern Cryptography*.
I'd suggest reading through that.
If you want a list of "standard" cryptographic problems, it of course changes semi-frequently, but there was a time period ~10 years ago when there were *many more* used commonly than there are now.
This led to the following [paper](https://eprint.iacr.org/2015/907) criticizing the practice, which is perhaps a good place to look for what some cryptographers view as "standard" (and "non-standard") assumptions.
| 5 | https://mathoverflow.net/users/101207 | 433411 | 175,324 |
https://mathoverflow.net/questions/278822 | 9 | Natural theories extending EFA (exponential/elementary function arithmetic) are well-ordered by $Π^0\_1$ provability, and we would like a formal definition of the well-ordering that is robust yet as fine as possible. Here is the definition I came up with.
Fix an elementary ordinal notation system such that the predecessor partial function (given $α+1$ return $α$) is elementarily definable. Formally, the definition (and the ordinals it gives) depends on the ordinal notation system. However, at least if the ordinal is not too large, natural ordinal notation systems (such as Veblen normal form) are elementarily equivalent, and provably so in EFA, which should suffice.
Set $f\_0(n)=2^2\_n$, $f\_{α+1}(n) = f\_α(2^2\_n)$ ($2^2\_n$ is iterated exponentiation), and for limit α, $f\_α(n)=\max(f\_β(n): β<α ∧ \mathrm{code}(β)<n)$ ($\mathrm{code}(β)$ is the integer coding $β$).
(Alternatively, if we are given reasonable fundamental sequences, set $f\_α(n) = f\_{α[n]}(n)$.)
Con$\_α$ = ∀n Con$\_f$(EFA + $f\_α(n)$ exists) where Con$\_f$ is cut-free consistency.
Th($α$) = EFA + {Con$\_β$: $β<α$}
For a theory $T$ extending EFA,
$|T|\_{Π^0\_1} = \sup(α+1: T⊢\mathrm{Con}\_α)$
$T$ is $Π^0\_1$-regular iff $T$ and $\mathrm{Th}(|T|\_{Π^0\_1})$ prove the same $Π^0\_1$ statements.
The intended meaning of Con$\_α$ is to iterate the cut-free consistency of EFA $1+α$ times. If I understand correctly, in EFA cut-free consistency is robust but more fine grained than ordinary consistency, and EFA ⊢ Con(EFA) ⇔ Con$\_f$(EFA+Con$\_f$(EFA)) (and also Con$\_f$(EFA)⇔Con(Q)).
Is this definition equivalent to the standard definition of $Π^0\_1$ ordinals?
If I understand correctly, the standard definition is
$|T|\_{Π^0\_1} = \sup(α+1:T ⊢ \mathrm{Con}\_f(A(α)))$
where $A$ is an elementary formula with two free variables such that $\mathrm{EFA} ⊢ ∀α \, A(α) = \mathrm{EFA}∪\{\mathrm{Con}\_f(A(β)):β<α\}$
$A(α)$ means $\{n:A(α,n)\}$, and statements are coded by their Gödel numbers.
Informally, the standard definition is based on direct iteration of consistency (using fixed-point theorem to get the desired predicate) while the definition I gave is based on consistency of existence of large numbers.
Also, do the following intuitively true identities hold:
|EFA|$\_{Π^0\_1}$ = 0
|EFA+Con$\_f$(EFA)|$\_{Π^0\_1}$ = 1
|EFA+Con(EFA)|$\_{Π^0\_1}$ = 2
|PA|$\_{Π^0\_1}$ = $ε\_0$
|EFA+Con(PA)|$\_{Π^0\_1}$ = $ε\_0+1$
|EFA+Con(ACA$\_0$)|$\_{Π^0\_1}$ = $ε\_0+2$
|PRA+Con(PA)|$\_{Π^0\_1}$ = $ε\_0+ω^ω$
|PA+Con(PA)|$\_{Π^0\_1}$ = $ε\_0·2$
with each of these theories $Π^0\_1$-regular.
| https://mathoverflow.net/users/113213 | $Π^0_1$ Proof Ordinals | Modulo the fact that Beklemishev [1] considers consistency with cuts as the basic consistency notion, his $\mathsf{Con}(\mathsf{EA}\_\alpha)$ are equivalent to your's $\mathsf{Con}\_{\alpha}$. It is quite easy to account for the effects of the switch between cut-free and full consistency using the result of Visser [2] that for finite extensions $T$ of $\mathsf{EA}$, we have $\mathsf{EA}\vdash \mathsf{Con}(T)\mathrel{\leftrightarrow} \mathsf{Con}\_f(\mathsf{EA}+\mathsf{Con}\_f(T))$.
Beklemishev fixes an ordinal notation system and using Diagonal Lemma defines $\mathsf{EA}\_\alpha$ as a uniformly $\Sigma\_1$-axiomatizable family of theories for $\alpha$ from the ordinal notation system such that provably in $\mathsf{EA}$:
$$\mathsf{EA}\_\alpha=\mathsf{EA}+\{\mathsf{Con}(\mathsf{EA}\_\beta)\mid \beta<\alpha\}.$$
Using Löb's theorem it is easy to prove that in fact there is $\mathsf{EA}$-provably unique family of theories $\mathsf{EA}\_\alpha$ satisfying the condition above (unique in extensional sense as a family of sets of theorems). To make this definition matching with your's let
$$\mathsf{EA}\_\alpha'=\mathsf{EA}+\{\mathsf{Con}\_f(\mathsf{EA}\_\beta')\mid \beta<\alpha\}.$$
Using Löb's theorem (for cut-free $\mathsf{EA}$-provability) it is rather routine to prove that $\mathsf{EA}$-provably for all notation systems $\alpha$ we have $\mathsf{Con}\_f(\mathsf{EA}\_\alpha')\mathrel{\leftrightarrow}\mathsf{Con}\_\alpha$. Reasoning in $\mathsf{EA}$ and using additional assumption that there is a cut-free proof of $\forall \alpha(\mathsf{Con}\_f(\mathsf{EA}\_\alpha')\mathrel{\leftrightarrow}\mathsf{Con}\_\alpha)$ we need to prove the eqivalence between $\mathsf{Con}\_f(\mathsf{EA}\_\alpha')$ and $\mathsf{Con}\_\alpha$. We use the aforementioned cut-free proof to show equivalence between $\mathsf{Con}(\mathsf{EA}\_\alpha')$ and $\mathsf{Con}\_f(\mathsf{EA}+\{\mathsf{Con}\_\beta\mid \beta<\alpha\})$. Now we just need to establish the equivalence
$$\mathsf{Con}\_f(\mathsf{EA}+\{\mathsf{Con}\_\beta\mid \beta<\alpha\})\mathrel{\leftrightarrow}\mathsf{Con}\_\alpha.$$
From left to right it is rather straightforward by proving that for $\alpha\ne 0$ the formula to the left implies that for any $n$ we have $\mathsf{Con}\_f(\mathsf{EA}+\mathsf{Con}\_f(\mathsf{EA}+\text{$f\_\alpha(n)$ exists}))$. The idea of a proof of the implication from right to left, is to use purely universal axiomatization of $\mathsf{EA}$ in the language with all Kalmar elementary functions and next to employ Herbrand's theorem to reformulate the consistency assertion to the left.
Now analogously to what Beklemishev have done in his paper, we have
$$|\mathsf{EA}'\_\alpha|\_{\Pi^0\_1}=\alpha.$$
and again analogously to what he have done (see Section 7 from [1]), we immediately get most of the ordinals that you mentioned. For the other ordinals we would need a bit of additional work:
1. $|\mathsf{EA}+\mathsf{Con}(\mathsf{EA})|\_{\Pi^0\_1}=2$: follows immediatelly from the mentioned theorem of [2];
2. $|\mathsf{EA}+\mathsf{Con}(\mathsf{ACA}\_0)|\_{\Pi^0\_1}=\varepsilon\_0+2$, since $$\mathsf{EA}+\mathsf{Con}(\mathsf{ACA}\_0)=\mathsf{EA}+\mathsf{Con}\_f(\mathsf{EA}+\mathsf{Con}\_f(\mathsf{ACA}\_0))=\mathsf{EA}+\mathsf{Con}\_f(\mathsf{EA}+\mathsf{Con}(\mathsf{PA})),$$ where in the last equality we are using $\mathsf{EA}$-formalized conservativity proof of $\mathsf{ACA}\_0$ over $\mathsf{PA}$ via the transformation of cut-free $\mathsf{ACA}\_0$-proofs to $\mathsf{PA}$-proofs with cuts;
3. $|\mathsf{PRA}+\mathsf{Con}(PA)|\_{\Pi^0\_1}=\varepsilon\_0+\omega^\omega$: it wasn't mentioned by Beklemishev, but easily could be established by the same technique.
P.S. I very much like your equivalent definition of $\Pi^0\_1$-ordinal, I didn't knew this precise formulation. Curiously, almost at the same time you asked the question, in the fall of 2017, I was exploring in a somewhat similar direction. Namely, my idea was to measure how much consistency a model of arithmetic could prove in terms of how much longer than the original model could its end-extensions obtained by the formalized completeness theorem be. However, I haven't managed to make a nice formulation out of this.
[1] Beklemishev, Lev D. "Proof-theoretic analysis by iterated reflection." Archive for Mathematical Logic 42.6 (2003).
[2] A. Visser. Interpretability logic. In Mathematical Logic, Proceedings of
the 1988 Heyting Conference. Plenum Press, 1990
| 3 | https://mathoverflow.net/users/36385 | 433414 | 175,325 |
https://mathoverflow.net/questions/433422 | 1 | Does there exist a simple smooth closed curve $\gamma:S^1\to \mathbb C$ such that
$$ \int\_{0}^{2\pi} e^{\gamma(e^{it})} \, |\gamma'(e^{it} )|\,dt =0?$$
| https://mathoverflow.net/users/50438 | On a property of complex exponentials | After the change of the variable $w=\gamma(t)$ your integral
becomes
$$\int\_\gamma e^wds,$$
where $ds$ is the length element, $ds=\sqrt{dx^2+dy^2}$, and $\gamma$ is a Jordan curve of zero index about the origin. Now notice that on every vertical segment of length $2\pi k$, where $k$ is an integer,
this integral is zero. Now construct a simple closed curve, of zero index about the origin, consisting of 8 straight segments, 4 vertical and 4 horizontal. All vertical segments have length $2\pi k\_j$
and all hosizontal segments lie on the lines $y=\pi/2+\pi k\_j$, and the curve is symmetric with respect to complex conjugation. Such a curve is easy to draw, and the integral along it is 0. Indeed it is zero on each vertical segment,
while contributions from horizontal segments cancel.
Let me add how to make the curve smooth. We will slightly modify the original curve, keeping the vertical segments unchanged, so their contribution to
the integral is zero. We will also keep the symmetry
with respect to the complex conjugation, so that imaginary part of the integral will be 0.
First we choose the vertical segments $\mathrm{Re} z=x\_k$ so that for some segments $\cos x\_j<0$ and for
some other $\cos x\_j>0$, so that the contribution to
the integral of some segments is positive and for other segments it is negative. Then, to make the curve smooth, we replace some small pieces of horizontal segments near the corners by pieces which make the curve smooth. Since the contribution of some of these pieces is negative and others positive, it is clear that the pieces can be chosen so that their contribution cancels.
| 4 | https://mathoverflow.net/users/25510 | 433434 | 175,330 |
https://mathoverflow.net/questions/433435 | 0 | In Chapter 3 of the textbook: [An Introduction to Random Matrices](https://www.google.com.hk/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&cad=rja&uact=8&ved=2ahUKEwiC8aL504P7AhW_k2oFHTv9BK8QFnoECBUQAQ&url=http%3A%2F%2Fwww.wisdom.weizmann.ac.il%2F%7Ezeitouni%2Fcupbook.pdf&usg=AOvVaw0wbMPwbj788zuq0QFfT4Cp&cshid=1666985383466164), we have that for normalized GUE/GOE/GSE and ordering its eigenvalues $\lambda\_1\le \lambda\_2\le \cdots \le \lambda\_n$, we have that
$$
\lim\_{n\to \infty}P(n^{2/3}(\lambda\_n-2)\le x)\to \mbox{Tracy-Widom law}
$$
Thereby, we have $\lambda\_2-\lambda\_1\approx O\_p(n^{-2/3})$.
Question: Can we weaken the conditions of "GOE" (Gaussian orthogonal ensemble)? For example for any symmetric Wigner matrix ensembles with some conditions on the moment?
Here symmetric Wigner matrix ensembles mean the upper triangular coefficients $\xi\_{ij}, j\ge i$ are jointly independent and real with $\xi\_{ij}=\xi\_{ji}$, and the strictly upper triangular coefficients will be iid, as will the diagonal coefficients, but the diagonal classes' of coefficients may have a different distribution.
| https://mathoverflow.net/users/168083 | Do we have the universal property of the edge of the spectrum for the Wigner matrix? | There is an extensive literature on the universality of the Tracy-Widom distribution. Here are some pointers:
* [Quantitative Tracy-Widom
laws for the largest eigenvalue of generalized Wigner matrices](https://arxiv.org/abs/2207.00546)
studies the general case of a Wigner matrix.
* [A necessary and
sufficient condition for edge universality at the largest singular
values of covariance matrices](https://www.jstor.org/stable/26542349%20) considers Wishart matrices, of the
form $XX^\ast$.
* [Local law and Tracy–Widom
limit for sparse random matrices](https://arxiv.org/abs/1605.08767) considers sparse random
matrices.
* [Tracy-Widom distribution
for the edge eigenvalues of Gram type random matrices](https://arxiv.org/abs/2008.04166) considers
Gram matrices.
| 1 | https://mathoverflow.net/users/11260 | 433437 | 175,331 |
https://mathoverflow.net/questions/433425 | 3 | I am reading a paper where they refer to a certain algebra as a **PBW algebra**. What does this mean exactly? I would infer from the $U(\frak{g})$ setting that this means the existence of an ordered generating set $\{x\_1, \dots, x\_n\}$ such that
$$
x\_1^{d\_1} \cdots x\_n^{d\_n}, ~~~~~~ d\_k \in \mathbb{N}\_0.
$$
forms a basis for the algebra. Can somebody point me to a reference?
| https://mathoverflow.net/users/491434 | What is a PBW algebra? (I.e., an algebra generalising properties of $U(\frak{g})$) | This concept, and the name Poincare Birkhoff Witt algebra, is due to Stewart Priddy in his paper [Koszul resolutions, Trans. Amer. Math. Soc. 152 (1970), 39–60] that also first defined Koszul algebras. See section 5 of his paper for the PBW criterion, which is a bit more flexible than what you wrote. (The basis might not need all of those words.)
Added later: The reference to the published version of the nice survey paper that Vladimir Dotsenko mentions is [Shepler, Anne V.; Witherspoon, Sarah Poincaré-Birkhoff-Witt theorems. Commutative algebra and noncommutative algebraic geometry. Vol. I, 259–290, Math. Sci. Res. Inst. Publ., 67, Cambridge Univ. Press, New York, 2015].
| 6 | https://mathoverflow.net/users/102519 | 433442 | 175,332 |
https://mathoverflow.net/questions/433278 | 73 | It seems like the article ["The Twin Primes Conjecture is True in the Standard Model of Peano Arithmetic: Applications of Rasiowa–Sikorski Lemma in Arithmetic (I)"](https://link.springer.com/article/10.1007/s11225-022-10017-2) by Janusz Czelakowski published in *Studia Logica* yesterday, claims to have proven that the twin prime conjecture holds in the standard model of Peano arithmetic using the technique of [forcing](https://en.wikipedia.org/wiki/Forcing_(mathematics)).
This seems like a very significant achievement (if the claim is not erroneous) but I am by no means an expert in logic or number theory, and therefore I'm not qualified enough to understand and evaluate the contents of this paper. So I would appreciate others' inputs on whether this claim has merit.
| https://mathoverflow.net/users/156061 | Czelakowski's claimed proof of the Twin Prime Conjecture | The error in the paper is in the proof of Theorem 7.2. The proof of Theorem 7.2 is immediately suspicious because of how vague it is in places and because of how lofty the expository text before and after it is. In the proof, the author claims that because we can identify the set of variables $(v\_i)\_{i \in \mathbb{N}}$ with the natural numbers, the induction scheme
* $(\beta)$ If $\mathbf{P} \Vdash \varphi(0)$ and for every variable $v\_i$, $\mathbf{P} \Vdash \varphi(v\_i)\Rightarrow \mathbf{P} \Vdash \varphi(S(v\_i))$, then for every variable $v\_i$, $\mathbf{P} \Vdash \varphi(v\_i)$.
is just an instance of ordinary induction in $\text{ZFC}$, but this is ridiculous. $0$ and $S(v\_i)$ are not variables; they're terms. Furthermore, even if the assumptions of Theorem 7.2 were enough to ensure that for every $i$, $\textbf{P}\Vdash S(v\_i) \approx v\_j$ for some $j$ (and they're not), that would in no way ensure that $\mathbf{P}\Vdash S(v\_i) \approx v\_{i+1}$.
---
That said, we need to step through a fair amount of the paper if we want to show conclusively that Theorem 7.2 is wrong. After all, maybe the assumptions of the theorem are inconsistent or otherwise overly strong and the error is really elsewhere in the paper. What makes this really tedious though is the forcing machinery, which only manages to make the proof more confusing and technical. It's pretty clear given the forcing posets being used (discrete posets and *singleton* posets) that the forcing can't really be doing anything that couldn't be described more simply in some other way.
We have the Lindenbaum-Tarski algebra of $\text{PA}$, written $\mathbf{B}\_{\text{PA}}(L)$, which is the Boolean algebra of formulas in some fixed countable collection of variables modulo logical equivalence over $\text{PA}$. We write $[\varphi]\_{\text{PA}}$ for the set of formulas that are logically equivalent to $\varphi$ over $\text{PA}$. The forcing posets $\mathbf{P}=(P,\subseteq)$ considered are certain families $P$ of non-empty subsets of $\mathbf{B}\_{\text{PA}}(L)$ (page 14). *But* the poset considered in Theorem 7.2 is a singleton, which means that all of the forcing machinery isn't really doing anything in Theorem 7.2. Nevertheless, let's go through some of the paper and track what the assumption that $P = \{p\}$ means (where $p$ is some non-empty subset of $\mathbf{B}\_{\text{PA}}(L)$).
On page 15 we get to the definition of a condition $p$ forcing an atomic formula $\sigma$. Again, since $P = \{p\}$, what this definition collapses to is just $p \Vdash \sigma$ if and only if $[\sigma]\_{\text{PA}} \in p$. We then extend this to arbitrary formulas in the standard way, but again everything collapses:
* $p \Vdash \neg \varphi$ if and only if $p \Vdash \varphi$ fails.
* $p \Vdash \varphi \wedge \psi$ if and only if $p\Vdash \varphi$ and $p\Vdash \psi$.
* $p \Vdash \exists x \varphi$ if and only if there is a variable $y$ such that $p \Vdash \varphi(x//y)$ (where $\varphi(x//y)$ is $\varphi$ with instances of $x$ substituted by $y$ and existing instances of $y$ in $\varphi$ changed to some fresh variable to avoid binding).
* $p \Vdash \varphi \vee \psi$ if and only if $p\Vdash \varphi$ or $p \Vdash \psi$.
* $p \Vdash \varphi \to \psi$ if and only if $p\Vdash\neg \varphi$ or $p\Vdash \psi$.
* $p\Vdash (\forall x)\varphi$ if and only if $p\Vdash\varphi(x//y)$ for all variables $y$.
(The author mentions this simplification at the end of page 15 and the beginning of page 16.) Finally, we write $\mathbf{P}\Vdash \varphi$ to mean that $p \Vdash \varphi$ for all $p \in P$, which in our case is just equivalent to $p \Vdash \varphi$.
We say that $\mathbf{P}$ is *compatible with equality axioms* if
* $[x \approx x]\_{\text{PA}} \in p$ for some variable $x$,
* whenever $[x \approx y]\_{\text{PA}}$ and $[R(...,x,...)]\_{\text{PA}}$ are in $p$, then $[R(...,y,...)]\_{\text{PA}}$ is in $p$ for any relation symbol $R$, and
* if $[x \approx y]\_{\text{PA}} \in p$, then $[F(...,x,...)\approx F(...,y,...)]\_{\text{PA}} \in p$ for any function symbol $F$.
This is essentially just what you need to ensure that $p$ forces the standard axioms of equality. (Transitivity and symmetry follow from special cases of the second bullet point.)
We say that $\mathbf{P}$ is *standard* if $\mathbf{P}$ is compatible with equality axioms and has that for any atomic formula $\sigma$, if $\text{PA}\vdash \neg \sigma$, then $[\sigma]\_{\text{PA}} \notin p$. (Remember, we're assuming $P = \{p\}$.)
This is all we need to understand the statement of Theorem 7.2, which claims that if $\mathbf{P} = (\{p\},\subseteq)$ is standard, then $\mathbf{P}\Vdash \mathrm{Ind}(x;\varphi)$ for every formula $\varphi$ (where $\mathrm{Ind}(x;\varphi)$ is $\forall \bar{z}[\varphi(0,\bar{z}) \wedge \forall x(\varphi(x,\bar{z})\to\varphi(S(x),\bar{z})) \to \forall x\varphi(x,\bar{z})]$, which is induction for the formula $\varphi(x,\bar{z})$). There is a typo in the statement of Theorem 7.2, but it's clear from the proof that the statement is meant to be $\mathbf{P} \Vdash \mathrm{Ind}(x;\varphi)$, not $\mathbf{P} \Vdash \mathrm{Ind}(x;\sigma)$.
The argument (suppressing the other free variables) proceed by showing that $\mathbf{P} \Vdash \mathrm{Ind}(x;\varphi)$ if and only if the following holds:
* $(\beta)$ If $\mathbf{P} \Vdash \varphi(0)$ and for every variable $y$, $\mathbf{P} \Vdash \varphi(y)\Rightarrow \mathbf{P} \Vdash \varphi(S(y))$, then for every variable $y$, $\mathbf{P} \Vdash \varphi(y)$.
As discussed above, $(\beta)$ does not work.
Let's see a concrete example of Theorem 7.2 failing. Fix an enumeration $(v\_i)\_{i \in \mathbb{N}}$ of our variable symbols. From now on we'll write $\varphi(y)$ for $\varphi(x//y)$, where $x$ is established by context to be the relevant free variable of $\varphi$. Let $p$ be
$$\{[\sigma]\_{\text{PA}}: \text{PA} \cup \{v\_0 \approx 0\}\vdash \sigma,~\sigma~\text{atomic}\}.$$
It is easy to check that $\mathbf{P} = (\{p\},\subseteq)$ is standard. Consider the formula
$$\varphi(v\_1) = \exists v\_2(v\_2 + v\_2 \approx v\_1 \vee S(v\_2 + v\_2) \approx v\_1),$$
i.e., "$v\_1$ is either even or odd."
First, let's see that $\mathbf{P} \Vdash \varphi(0)$ (i.e., $p\Vdash \varphi(0)$). We have that $[v\_0+v\_0\approx 0]\_{\text{PA}} \in p$, so $p \Vdash v\_0+v\_0 \approx 0$. Therefore $p \Vdash v\_0+v\_0\approx 0 \vee S(v\_0+v\_0) \approx 0$ and $p \Vdash \exists v\_2( v\_2+v\_2\approx 0 \vee S(v\_2+v\_2) \approx 0)$.
Now fix a variable $v\_i$. There are two cases. Either $i = 0$ or $i \neq 0$.
If $i = 0$, then we have that $[S(v\_0+v\_0) \approx S(v\_0)]\_{\text{PA}} \in p$, so $p \Vdash S(v\_0+v\_0)\approx S(v\_0)$ and $p \Vdash v\_0 + v\_0 \approx S(v\_0) \vee S(v\_0+v\_0) \approx S(v\_0)$. Therefore $p \Vdash \exists v\_2(v\_2+v\_2 \approx S(v\_0) \vee S(v\_2+v\_2) \approx S(v\_0)$, i.e., $p \Vdash \varphi(S(v\_0))$.
If $i \neq 0$, then I claim that $p \not \Vdash \varphi(v\_i)$ (i.e., $p \not \Vdash \exists v\_2(v\_2 + v\_2 \approx v\_i \vee S(v\_2+v\_2) \approx v\_i)$). Fix a variable $v\_j$. Since $i \neq 0$, we have that $v\_j+v\_j \not \approx v\_i\wedge S(v\_j+v\_j)\not\approx v\_i$ is consistent with $\text{PA} \cup \{v\_0 \approx 0\}$ (even if $j = 0$), so $[v\_j+v\_j \approx v\_i]\_{\text{PA}} \notin p$ and $[S(v\_j+v\_j)\approx v\_i]\_{\text{PA}} \notin p$. Since we can do this for any $j$, we have that $p \not \Vdash \varphi(v\_i)$.
So in any case, we have that if $p\Vdash \varphi(v\_i)$, then $p\Vdash \varphi(S(v\_i))$, but as we just established, $p \not \Vdash \varphi(v\_1)$, contradicting Theorem 7.2. Incidentally, this also shows that the assumptions of Theorem 7.2 are not enough to ensure that for every $i$, $\mathbf{P} \Vdash S(v\_i)\approx v\_j$ for some $j$.
| 118 | https://mathoverflow.net/users/83901 | 433444 | 175,333 |
https://mathoverflow.net/questions/432969 | 1 | Let $R$ be a commutative Noetherian ring, and let $\text{mod } R$ denote the abelian category of finitely generated $R$-module. Consider the bounded derived category $D^b(\text{mod } R) $ which is a triangulated category. Every object $X \in \text{mod } R$ can be naturally identified in $D^b(\text{mod } R) $. For an object $X \in D^b(\text{mod } R)$, let $\text{Thick}\_{D^b(\text{mod } R)} X$ denote the intersection of all thick subcategories (<https://ncatlab.org/nlab/show/thick+subcategory>) of $D^b(\text{mod } R)$ containing $X$. For example, note that $\text{Thick}\_{D^b(\text{mod } R)} R$ is the collection of all perfect complexes, hence $M \in \text{mod } R$ belongs to $\text{Thick}\_{D^b(\text{mod } R)} R$ if and only if $M$ has finite projective dimension.
Following 4.1 of <https://doi.org/10.4171/cmh/56>, we say $X\in D^b(\text{mod } R)$ is Virtually small if there exists a non-exact complex $Y \in \text{Thick}\_{D^b(\text{mod } R)} X \cap \text{Thick}\_{D^b(\text{mod } R)} R$.
My question is: If $M \in \text{mod } R$ is Virtually small in $D^b(\text{mod } R)$, then does $M$ embed into a finitely generated module of finite projective dimension?
| https://mathoverflow.net/users/493381 | Finitely generated module, which is a virtually small complex, embeds into a module of finite projective dimension? | For every $M$, $M\oplus R$ is virtually small, so your question is equivalent to the question: Does every finitely generated $R$-module embed in a finitely generated module of finite projective dimension?
The answer is no. For example, let $R$ be a local commutative finite dimensional algebra over a field $k$, that is not self-injective, such as $R=k[x,y]/(x^2,xy,y^2)$. Then the only finitely generated modules of finite projective dimension are the finitely generated projective modules, and the injective generator $\operatorname{Hom}\_k(R,k)$ does not embed in a projective module.
| 2 | https://mathoverflow.net/users/22989 | 433460 | 175,335 |
https://mathoverflow.net/questions/433315 | 1 | Let $\mathbf{C}\_S \subset \mathbf{R}^{2n}$ be a Simons cone, where the dimension is large enough that it is area-minimizing: $n \geq 4$. Let $T$ be a leaf of the Hardt–Simon foliation with $\operatorname{dist}(T,0) = 1$.
For all $R > 1$, $T$ intersects $\partial B\_R$ transversely. Thus $T \cap \partial B\_R$ is a smoothly embedded submanifold of $\partial B\_R$; rescale this homothetically to $\Sigma\_R \subset \partial B\_1$.
* ~~When $R$ is close to $1$, $\Sigma\_R$ is diffeomorphic to the boundary of an $2n-1$-dimensional disk, so $\Sigma\_R \simeq \mathbf{S}^{2n-2}$.~~
* As $R \to \infty$, $\frac{1}{R} T \to \mathbf{C}\_S$, so when $R$ is large enough then $\Sigma\_R \simeq \mathbf{S}^{n-1} \times \mathbf{S}^{n-1}$.
**Question.** The paradox is that the family $(\Sigma\_R)$ gives an isotopy from $\mathbf{S}^{2n-2}$ to $\mathbf{S}^{n-1} \times \mathbf{S}^{n-1}$, where we let $R$ vary across $[1+\frac{1}{N},N]$ for example. What am I missing here?
*Edit.* As pointed out in the answers below, the first point is incorrect. I mixed up two arguments I thought up: the first working with balls centered around a point of the surface—this guarantees the validity of *both* bullet points—, and the second working with balls centered at the origin—guaranteeing the transversality. Presumably in the former approach the transversality would be *false* for some radii, which would allow the topology to change.
| https://mathoverflow.net/users/103792 | A paradox based on Simons cones | The issue here is that (I suspect) your intuition is failing you as your first bullet point is incorrect.
A good first visualization is the Clifford torus in $\mathbf{S}^3$ all the parallel surfaces are tori, until a critical distance is reached and one gets two circles on each side.
Another, way to think about the Simons' cone is to reduce by the symmetry group. After take the appropriate quotient one is left with the (closed) upper quadrant. Distance to the origin is the same in either picture. Any point away from the axes corresponds to a torus in the original picture, but the points on the axis are collapsed and are either collapsed version so the tori (i.e. like the circles above). Obviously at the origin everything is collapsed.
In the simple picture, the Simons' cones are just the line $y=x$. However the curves in the foliation lie on one side of this curve and look (qualitatively) like $y=x^2/(x+1)$ so meet the axis perpendicularly. The point is that the intersection with the sphere of radius 1 is just a point (in the reduced space) i.e. a "circle" in the original picture.
| 2 | https://mathoverflow.net/users/127803 | 433464 | 175,337 |
https://mathoverflow.net/questions/433475 | 2 | The categories $\mathbf{Top}$ of topological spaces, $\mathbf{sSet}$ of simplicial sets and $\mathbf{Cat}$ of small categories are all equipped with a functor $\pi\_0$ into the category $\mathbf{Set}$ of sets, which is a left adjoint and measures the number of connected components. There are also plenty of functors between the upper categories, which gives rise to the question, if they preserve the number of connected components.
I have already proven this for $|-|\colon\mathbf{sSet}\leftrightarrows\mathbf{Top}\colon \operatorname{Sing}$, for $\tau\colon\mathbf{sSet}\leftrightarrows\mathbf{Cat}\colon N$ and for $\operatorname{Sd}\colon\mathbf{sSet}\leftrightarrows\mathbf{sSet}\colon\operatorname{Ex}$. The functor $\operatorname{Ex}^\infty=\varinjlim\_{n\in\mathbb{N}}\operatorname{Ex}^n\colon\mathbf{sSet}\rightarrow\mathbf{sSet}$ still works fine since the colimit commutes with the left adjoint $\pi\_0$. The problem is, that $\operatorname{Sd}^\infty=\varprojlim\_{n\in\mathbb{N}}\operatorname{Sd}^n\colon\mathbf{sSet}\rightarrow\mathbf{sSet}$ does not work fine because of the limit and since it is not the left adjoint of $\operatorname{Ex}^\infty$ and also writing either $X$ or $\operatorname{Sd}^n(X)$ as a colimit doesn't seem to work out. Does $\operatorname{Sd}^\infty$ also preserve the number of connected components? If yes, how is it proven and if no, is there a counterexample?
The fact, that there are many other categories with a forgetful functor to $\mathbf{Set}$ and left adjoints ([free object functor](https://ncatlab.org/nlab/show/free+functor)) or right adjoints (cofree object functor) like we have for the three categories above, gives also rise to some more questions: Are there more categories with a functor like $\pi\_0$ (like for example the category $\mathbf{Graph}$ of graphs) giving the number of connected components and functors to other categories with one, which I have not yet considered? If yes, is the number of connected components preserved? Are there other categories with functors similar to $\pi\_0$ in the sense that they measure something similar, so we can look at more possible preservations?
| https://mathoverflow.net/users/479945 | Which functors preserve the number of connected components? | Ignoring issues with what TOP really should be, let me focus on the question whether $\pi\_0(Sd^\infty(X))=\pi\_0(X)$ for a simplicial set $X$. If we look at $X=\Delta^1$, we should get a counterexample. Here $\pi\_0$ should denote the equivalence class of vertices, where two vertices are equivalent, if there is a finite path (a list of one simplices ignoring the direction) joining them.
In the inverse limit, we have uncountably many zero simplices, but only countably many (non-degenerate) one simplices. Thus $\pi\_0(Sd^\infty(\Delta^1))$ must be uncountable.
A long time ago I thought a bit about these situations. Morally, I think there category of sets is somehow the wrong category for such questions. We can view any set as a discrete topological space. But then that functor $SET\to TOP$ is not compatible with limits.
I believe I could show a statement of the form that if $(X\_n)\_{n\in \mathbb{B}}$ is a inverse system of m-dimensional, simplicial (totally disconnected compact Hausdorff-spaces), then the canonical map $|\lim\_n X\_n|\to \lim\_n |X\_n|$ is a homeomorphism, where $m$-dimensional means that $Sk^m(X\_n)\to X\_n$
is a homeomorphism for all $n$ and $|-|$ is the non-fat geometric realization (the one that people would call the wrong one).
My motivation was just to write the Hawaiian earrings or compatifications of a tree as the geometric realization of something. That seemes to work, as long as we also allow a topology on the $n$-simplices. Funnily then one can also write $\Delta^1$ as the geometric realization with the Cantor-Set as its Zero-Skeleton.
I haven't published that but I might look it up if there is interest.
| 8 | https://mathoverflow.net/users/3969 | 433485 | 175,343 |
https://mathoverflow.net/questions/433487 | 5 | A propeller is a chiral structure. Propellers can have low or large pitch. Pitch differences can be seen as distinguishing propellers with low chirality from those with large chirality.
Is there a general way, given a closed 3d shape, to quantify its chirality? Is there some sort of integral over the shape that yields a number quantifying the degree of chirality of that shape?
| https://mathoverflow.net/users/493794 | Is there a way to quantify the chirality of a 3d shape? | This is a topic of some research, summarized in [On quantifying chirality — Obstacles and problems towards unification](https://link.springer.com/article/10.1007/BF01277559). One metric is the Hausdorff chirality, which quantifies the chirality of a geometric representation of an object by measuring the degree of coincidence of the object with its mirror image, see [The Hausdorff chirality measure and a proposed Hausdorff structure measure.](https://www.semanticscholar.org/paper/The-Hausdorff-chirality-measure-and-a-proposed-Yewande-Neal/0a7f7b2433ff23383fbf247e8a50381b0c8efae5)
| 7 | https://mathoverflow.net/users/11260 | 433488 | 175,344 |
https://mathoverflow.net/questions/433489 | 12 | **Context**: I am currently working on a rather important paper for my career, in the sense that it is a culmination of the past 5 years of (post Ph.D.) research. I started this particular article with 3 other members but only 1 other member (senior) has contributed. The other two likely cannot even explain our contribution non-superficially.
**Trouble**: The other two are very senior researchers and influential community members.
However, I should note that they are in no way my boss (i.e. I am a junior prof.).
**Question**: How can I signal that the other contributor and myself are the main (if not only) contributors to the researchers, without changing the author ordering or removing the other two?
*I cannot do the latter for political issues and the former is bad practice in analysis….
Any advice? What about putting an asterisk and footnote on our name, flagging principal investigation? I've seen this done in computer science….*
| https://mathoverflow.net/users/36886 | How to indicate when another author has done nothing significant | It seems clear to me that you need to talk to the senior author who actually did contribute and see if they can talk to the other two. You wrote that "one of the authors said that they were experts who are so great (bla bla)" so it sounds like it wasn't really your choice to add them in the first place.
Let's call the good author G. Let's assume G actually cares about your career (e.g., wants to keep working with you in the futute, wants you to get a permanent job, etc.). Then, G should be receptive if you TACTFULLY express your concern to him/her that the paper won't help your career as much with 4 authors as it would with 2. While you feel like you haven't seen any "substantial" contribution from the other 2, it is always possible they have been communicating with G and that some of G's contributions are in fact due to them.
If G agrees with you that the contributions of the other two don't warrant authorship, then G can fight this battle on your behalf, rather than having you try to signal it some way in the writing, which is most likely going to come off looking petulant and unprofessional. If G fights the battle on your behalf, you won't face the professional consequences you fear.
It's also possible that G won't be able to get them off the paper. In that case, G can still soften the blow by giving talks that state that you did most of the work, and writing the same in letters of recommendation for you. At all costs, do not alienate G.
For context, in my first paper, I invited my advisor to be a co-author because he'd made substantial contributions. He declined, telling me that I needed the paper more than he did. I still put a huge acknowledgment to thank him for his help. I have since paid this forward, sometimes helping people a lot but not ending up as an author (especially post-tenure). That's how a good field should function. If G is good, they will understand this and find a way to help you be successful.
| 31 | https://mathoverflow.net/users/11540 | 433492 | 175,345 |
https://mathoverflow.net/questions/433448 | 2 | Suppose that $\mathcal{D}$ is a Johnson-Lindenstrauss (JL) distribution on $\mathbb{R}^{r\times n}$ ($1 \le r \le n$), meaning that there exist constants $\epsilon, \delta \in(0,1)$ such that
$$
\mathbb{P}\_{A \sim \mathcal{D}}((1 - \epsilon)\|x\| \le \|Ax\|\le (1 + \epsilon)\|x\|) \ge \delta
\quad \text{for all}\quad x \in \mathbb{R}^n.
$$
Question: Do there exist constants $\epsilon', \delta' \in(0,1)$ such that
$$
\mathbb{P}\_{A \sim \mathcal{D}}((1 - \epsilon')\|x\| \le \|S(A)\, x\|\le (1 + \epsilon')\|x\|) \ge \delta'
\quad \text{for all}\quad x \in \mathbb{R}^n,
$$
where $S(A) = A/\|A\|$ (scaled JL transform)? In other words, if $A$ follows a JL distribution, is the distribution of $A/\|A\|$ also a JL distribution?
Here, all the norms are the Euclidean norm or spectral norm.
Thanks.
| https://mathoverflow.net/users/170208 | Distribution of scaled Johnson-Lindenstrauss transforms | $\newcommand\ep\epsilon\newcommand{\de}{\delta}\newcommand{\R}{\mathbb R}$We have
\begin{equation\*}
P((1-\ep)\|x\|\le\|Ax\|\le(1+\ep)\|x\|)\ge\de \tag{1}\label{1}
\end{equation\*}
for some $\ep,\de$ in $(0,1)$ and all $x\in\R^n$.
The OP asks if then
\begin{equation\*}
P((1-\ep')\|x\|\le\|S(A)x\|\le(1+\ep')\|x\|)\ge\de' \tag{2}\label{2}
\end{equation\*}
for some $\ep',\de'$ in $(0,1)$ and all $x\in\R^n$, where $S(A):=A/\|A\|$.
To avoid the division by $\|A\|$ when $\|A\|$ takes the value $0$, rewrite \eqref{2} as
\begin{equation\*}
P((1-\ep')\|A\|\|x\|\le\|Ax\|\le(1+\ep')\|A\|\|x\|)\ge\de'. \tag{2a}\label{2a}
\end{equation\*}
Let us now show that \eqref{2a} indeed holds for some $\ep',\de'$ in $(0,1)$ and all $x\in\R^n$.
There is some real $c>1$ such that
\begin{equation\*}
P(\|A\|>c)\le\de/2.
\end{equation\*}
Let $\ep':=1-\dfrac{1-\ep}c$ and $\de':=\de/2$, so that $1-\ep'=\dfrac{1-\ep}c\in(0,1)$, $\ep'\in(0,1)$, and $\de'\in(0,1)$.
Then for any $x\in\R^n$
\begin{equation\*}
\begin{aligned}
&P((1-\ep')\|A\|\|x\|\le\|Ax\|\le(1+\ep')\|A\|\|x\|) \\
=&P((1-\ep')\|A\|\|x\|\le\|Ax\|) \\
\ge&P(\|A\|\le c,(1-\ep')c\|x\|\le\|Ax\|) \\
=&P(\|A\|\le c,(1-\ep)\|x\|\le\|Ax\|) \\
\ge&P((1-\ep)\|x\|\le\|Ax\|)-P(\|A\|>c) \\
\ge&P((1-\ep)\|x\|\le\|Ax\|\le(1+\ep)\|x\|)-P(\|A\|>c) \\
\ge&\de-\de/2=\de',
\end{aligned}
\end{equation\*}
as claimed.
| 2 | https://mathoverflow.net/users/36721 | 433499 | 175,347 |
https://mathoverflow.net/questions/433470 | 1 | I know that the wave equation doesn't satisfy a maximum principle but I have also heard that hyperbolic equations do not satisfy any maximum principle. But I don't know any reference or proof regarding that. It would be really helpful if I get a help or resource regarding the same.
| https://mathoverflow.net/users/493046 | Maximum principle for hyperbolic PDEs | Fundamentally, the classical maximum principles for second order elliptic PDEs are based on the simple facts that
1. The Hessian of a function at a local maximum is positive semidefinite.
2. The full contraction (Hilbert-Schmidt product) between two (symmetric) positive semidefinite matrices is non-negative.
In point 2, one of the PSD matrices is the Hessian, and the other is the leading order coefficients of your second order PDE.
For hyperbolic PDEs you cannot invoke point 2, as the symbol of the equation is required to be indefinite. This is what people usually mean when they say that there is no maximum principle for hyperbolic PDEs: that the basic heuristic scheme can't even be started. This certainly does not rule out individual equations/solutions exhibiting something that looks a bit like a maximum principle.
(For an example of a fake maximum principle: the kernel for the linear wave equation in dimension 1, 2, and 3 are signed. This means that if you solve $\Box u = f$ and $\Box v = g$, both with vanishing data and with $f \geq g$ pointwise, you will find the solutions satisfy also $u \geq v$ pointwise. This looks sort of like a maximum principle, but is true for completely different reasons.)
| 4 | https://mathoverflow.net/users/3948 | 433500 | 175,348 |
https://mathoverflow.net/questions/433508 | 5 | I get the follow equation in a paper. Let $A \in \mathbb{R}^{2 \times 2}$, then $M = A^TA$ is a positive semi-definite matrix, the nuclear norm of $A$ is:
$$ \Vert A \Vert\_\* = \sqrt{\operatorname{tr}(M) + 2\sqrt{\det(M)}}.$$
Are there any analytical form of nuclear norm for $n \times n$ matrix like the above form? Or just for a $3 \times 3$ matrix?
| https://mathoverflow.net/users/493804 | Analytical form for the nuclear norm of an $n \times n$ matrix | **As Gro-Tsen pointed out, I had not computed the Galois group of the governing polynomial, and, in fact, my original answer was wrong. I believe that the following answer is correct, though.**
If by 'like the above form' you mean an expression in terms of radicals of the traces of the powers of $M$, the answer is 'no' for $n>5$. Even for $n=3$ and $n=4$, when there is such an expression, it is very ugly and will almost certainly be useless for any practical purpose:
Let $p\_i = \mathrm{tr}(M^i)$ for $1\le i\le n$, and let $q = \|A\|\_\*$. If $\lambda\_1,\ldots,\lambda\_n\ge 0$ are the eigenvalues of $M$, then $p\_i = {\lambda\_1}^i+\cdots+{\lambda\_n}^i$ and $q = \sqrt{\lambda\_1}+\cdots+\sqrt{\lambda\_n}$. The problem is to compute the Galois group of the minimal polynomial of $q$ in the field generated by the $p\_i$ and determine whether it is solvable, since this will determine whether $q$ can be expressed algebraically in terms of the $p\_i$ using only radicals.
To clarify things, write $\mu\_i = \sqrt{\lambda\_i}$, so that $q = \mu\_1 + \cdots + \mu\_n$ while $p\_i = {\mu\_1}^{2i}+\cdots+{\mu\_n}^{2i}$.
Let $\mathbb{S}\_n\subset\mathrm{O}(n)$ be the signed permutation group acting linearly on the span of the $\mu\_i$. Thus, $\sigma\in \mathbb{S}\_n$ satisfies $\sigma(\mu\_i) = \epsilon\_i\,\mu\_{\pi(i)}$ where $\epsilon\_i=\pm1$ and $\pi$ is a permutation of $\{1,\ldots,n\}$. Then $\sigma$ extends to an automorphism of the field $\mathbb{Q}(\mu\_1,\ldots,\mu\_n)$ whose fixed field is easily seen to be $\mathbb{Q}(p\_1,\ldots,p\_n)$.
Let $(\mathbb{Z}\_2)^n\subset\mathbb{S}\_n$ be the abelian normal subgroup consisting of the $\sigma$ of the form $\sigma(\mu\_i) = \epsilon\_i\,\mu\_{i}$ where $\epsilon\_i=\pm1$.
Consider the polynomial
$$
f(t) = \prod\_{\sigma\in(\mathbb{Z}\_2)^n}\bigl(t-\sigma(q)\bigr).
$$
Then $f(q) = 0$. The $t$-degree of $f$ is $2^n$, and it is easy to see that the $t$-coefficients of $f$ lie in $\mathbb{Q}(p\_1,\ldots,p\_n)$ (since they are symmetric polynomials in ${\mu\_1}^2,\ldots,{\mu\_n}^2$). Moreover, the set $R=\{\sigma(q)\ |\ \sigma\in (\mathbb{Z}\_2)^n\}$ consists of $2^n$ distinct elements, which are the roots of $f$ in $\mathbb{Q}(\mu\_1,\ldots,\mu\_n)$. The group $\mathbb{S}\_n$ acts transitively on $R$, and the $\mathbb{Q}$-linear span of $R$ contains $\mu\_1,\ldots,\mu\_n$, so $\mathbb{Q}(\mu\_1,\ldots,\mu\_n)$ is the splitting field of $f$. (Hence $f$ must be irreducible in $\mathbb{Q}(p\_1,\ldots,p\_n)[t]$.)
If $\alpha$ is an automorphism of $\mathbb{Q}(\mu\_1,\ldots,\mu\_n)$ that preserves the set $R$, then it must preserve its $\mathrm{Q}$-linear span and it must preserve the sum of the squares of the elements of $R$, which is clearly $2^n({\mu\_1}^2 + \cdots + {\mu\_n}^2)$. From this, it is easy to see that $\alpha$ must preserve the set $\{\pm\mu\_1,\ldots,\pm\mu\_n\}$ and hence must belong to $\mathbb{S}\_n$. Thus, the Galois group of $f$ is $\mathbb{S}\_n$.
Since $\mathbb{S}\_n/(\mathbb{Z}\_2)^n$ is isomorphic to the permutation group $S\_n$ and since $(\mathbb{Z}\_2)^n$ is abelian, it follows that $\mathbb{S}\_n$ is solvable if and only if $S\_n$ is solvable. Of course, $S\_n$ is solvable if and only if $n<5$.
| 6 | https://mathoverflow.net/users/13972 | 433517 | 175,355 |
https://mathoverflow.net/questions/432732 | 4 | Let $X$ be a subset of a group $G$. We say that $X$ is *left amenable* with respect to $G$ if there is a function $\mu:\mathcal P(G)\to [0,\infty]$ with the following three properties.
1. $\mu(A\cup B)=\mu(A)+\mu(B)$ for every pair of disjoint subsets $A, B$ of $G$.
2. $\mu(gA)=\mu(A)$ for all $g\in G$ and $A\in \mathcal{P}(G)$.
3. $\mu(X)=1$.
The subset $X$ is *right amenable* with respect to $G$ if there is a function $\mu:\mathcal P(G)\to [0,\infty]$ that has properties (1) and (3) above as well as property (2') below.
2') $\mu(Ag)=\mu(A)$ for all $g\in G$ and $A\in \mathcal{P}(G)$.
It is easy to see that $X$ is right amenable with respect to $G$ if and only if $X^{-1}$ is left amenable with respect to $G$. Moreover, $G$ being either left or right amenable with respect to $G$ is equivalent to $G$ being an amenable group in the usual sense.
**Question.** What is an example of a subset $X$ of an amenable group $G$ such that $X$ is left amenable with respect to $G$ but not right amenable with respect to $G$?
Tarski proved that $X$ is left (right) amenable with respect to $G$ if and only if $X$ fails to admit a left (right) $G$-paradoxical decomposition. One current reference for this result is the book "The Banach-Tarski Paradox" by G. Tomkowicz and S. Wagon (Corollary 11.2, page 197 in the second edition).
Hence the question posed is asking for an example of a subset $X$ of an amenable group $G$ that admits a right $G$-paradoxical decomposition but no left $G$-paradoxical decomposition.
| https://mathoverflow.net/users/132356 | Amenable subsets of groups | Here is a slightly modified version of an example that was communicated to the author of the question by Nicolas Monod.
Let $k>1$ be an integer and $G$ the Baumslag-Solitar group $\mathrm{BS}(1,k)$; that is,
$$G:=\langle a, b\ :\ bab^{-1}=a^k\rangle.$$
Let $A$ and $B$ be the infinite cyclic subgroups generated by $a$ and $b$, respectively. Notice that $G=A^G\rtimes B$, where $A^G\cong \mathbb Z[1/k]$ is the normal closure of $A$ in $G$.
Since $G$ is metabelian, it is an amenable group.
Put $X:=AB$. We claim that, relative to $G$, the subset $X$ is right amenable but not left amenable.
To prove the latter assertion, write $X\_0:=\langle a^k\rangle B$. Observe that $X\_0$ and $aX\_0$ are disjoint subsets of $X$, and that $bX=X\_0$. Hence the existence of a measure $\mu:\mathcal{P}(G)\to [0,\infty]$ satisfying properties (1), (2), and (3) above would imply
$$2=\mu(X)+\mu(X)=\mu(X\_0)+\mu(X\_0)=\mu(X\_0)+\mu(aX\_0)\leq \mu(X)=1.$$
Therefore $X$ is not left amenable with respect to $G$.
To establish the right amenability of $X$, we invoke the following proposition, due to J.M. Rosenblatt and appearing as Corollary 1.5 in his paper "A generalization of Følner's condition (*Math. Scand*. **33** (1973), 153-170)." Rosenblatt states the proposition in terms of left amenability; we have repurposed it for right amenability. Following Rosenblatt, we employ the following notation: for any $n$-tuple $(u\_1,\dots,u\_n)$ and any set $X$,
$$\|X\cap(u\_1,\dots,u\_n)\|:=|\{ i=1,\dots, n:u\_i\in X\}|,$$
where the '$|\ \ |$' notation on the right is for cardinality.
**Proposition** (Rosenblatt)**.** Let $G$ be an amenable group. A subset $X$ of $G$ is right amenable with respect to $G$ if and only if, for every $m$-tuple $(u\_1,\dots, u\_m)$ and $n$-tuple $(v\_1,\dots,v\_n)$ from $G$ such that $m<n$, there exists $g\in G$ such that
$$\|gX\cap (u\_1,\dots,u\_m)\|<\|gX\cap (v\_1,\dots, v\_n)\|.$$
Let $\{t\_{\alpha}:\alpha\in I\}$ be a complete set of coset representatives of $A$ in $A^G$. Then
$\{t\_{\alpha}X:\alpha \in I\}$ is a partition of $G$. In order to apply Rosenblatt's proposition, we let $(u\_1,\dots,u\_m)$ and $(v\_1,\dots, v\_n)$ be finite sequences from $G$ with $m<n$. Then
$$m=\sum\_{\alpha\in I} \|t\_{\alpha}X\cap (u\_1,\dots u\_m)\|\ \ \mbox{and}\ \ n=\sum\_{\alpha\in I} \|t\_{\alpha}X\cap (v\_1,\dots, v\_n)\|.$$
Thus it must be the case that, for some $\alpha\in I$,
$$\|t\_{\alpha}X\cap (u\_1,\dots,u\_m)\|<\|t\_{\alpha}X\cap (v\_1,\dots, v\_n)\|.$$
Hence $X$ is right amenable with respect to $G$.
The set $X^{-1}$ is, then, left amenable but not right amenable, answering the question as it was formulated.
| 4 | https://mathoverflow.net/users/132356 | 433527 | 175,357 |
https://mathoverflow.net/questions/433521 | 3 | Assume that $\Omega$ is a bounded connected domain and $\partial \Omega \in C^{\infty}$. Denote $\Gamma\_1,\Gamma\_2,\cdots,\Gamma\_n$ are $n$ connected components of $\partial\Omega$. This notation leads to $\partial \Omega=\cup^n\_{i=1}\Gamma\_i$. Consider the following problem.
\begin{cases}
\Delta \phi&=0\\
\phi|\_{\partial \Omega}&=g(x),\quad g\in C^{\infty}
\end{cases}
Denote $\mathcal{S}$ is subset of $\{1,2,\cdots,n\}$ and $\mathcal{S}^c$ is $\{1,2,\cdots,n\}\setminus \mathcal{S} $. If we know that
$$
\min\_{i\in \mathcal{S}}\inf\_{x\in \Gamma\_i}\phi(x)\ge \max\_{i\in \mathcal{S}^c} \sup\_{x\in \Gamma\_i} \phi(x)+\delta,
$$ where $\delta$ is a positive constant,
how to prove that
\begin{align\*}
\sum\_{i\in \mathcal{S}}\int\_{\Gamma\_i}\nabla \phi\cdot \vec{n}d\sigma>0?
\end{align\*}
where $\vec{n}$ denotes the outer normal of $\partial \Omega$.
**My effort:** I meet this question when I read a paper. This paper say that it is a standard comparison principle exercise, but I still don't know how to solve this question. When $g(x)$ is a step function, we may consider Hopf Lemma or strong maximum principle. However, $g$ is a function. Any comments will be welcome.
[the equation (2.51) of page 2992 in this paper](https://projecteuclid.org/journals/duke-mathematical-journal/volume-170/issue-13/Symmetry-in-stationary-and-uniformly-rotating-solutions-of-active-scalar/10.1215/00127094-2021-0002.full)
| https://mathoverflow.net/users/176379 | How to use comparison principle to prove the following inequality about Laplace equation? | Let $\psi$ be harmonic in $\Omega$, with $\psi=\phi$ on $\cup \_{i \in S} \Gamma\_i$, $\psi=m$ on $\cup\_{i \not \in S}\Gamma\_i$, where $m=\max\_{i \not \in S} \max\_{\Gamma\_i} \phi$. By comparison, $\psi \geq \phi$ and then $\nabla \psi \cdot n \leq \nabla \phi \cdot n$ on $\cup \_{i \in S} \Gamma\_i$. Then
$$
0=\int\_{\partial \Omega}\nabla \psi\cdot n=\sum\_{i \in S}\int\_{\partial \Gamma\_i} \nabla \psi\cdot n+\sum\_{i \not \in S}\int\_{\partial \Gamma\_i} \nabla \psi\cdot n.
$$
The function $\psi$ attains its minimum at any point of $\cup\_{i \not \in S} \Gamma\_i$, hence Hopf's lemma gives $\nabla \psi\cdot n<0$ and each integral on $\partial \Gamma\_i, i \not \in S$ is negative. Then
$$
\sum\_{i \in S}\int\_{\partial \Gamma\_i} \nabla \phi\cdot n \geq \sum\_{i \in S}\int\_{\partial \Gamma\_i} \nabla \psi\cdot n >0.
$$
| 5 | https://mathoverflow.net/users/150653 | 433536 | 175,361 |
https://mathoverflow.net/questions/433535 | 0 | I have been thinking about this for the last few days but I was not able to produce a definitive answer.
Take an integrable function $g$ that maps in $\mathbb{R}$ and with domain contained in $[0,M]$ (not necessarily equal to this interval). Consider now a family of probability measures $\mu\_{\theta}$ such that $\mu\_{\theta} \ll \lambda$ (Lebesgue measure) for every $\theta$. These measures have strictly positive density $f\_{\theta}(z) > 0 \: \: \forall \: 0 \leq z < \theta \leq M$. The densities are not defined for $z \geq \theta$. In symbols:
$$ \mathcal{M}\_{\lambda} = \bigg\{ \mu\_{\theta} \: : \: \theta \in [0,M] \: \: \text{and} \: \: \frac{d\mu\_{\theta}}{d\lambda} = f\_{\theta} \: : \: f\_{\theta}(z) > 0 \: \: \forall \: 0 \leq z < \theta \leq M, \:\forall \: \theta \in [0,M] \bigg\}$$
**Question:** is it true that:
$$ 0 = \int\_{0}^{\theta} g(z) d \mu\_{\theta} (z) = \int\_{0}^{\theta} g(z) \underbrace{f\_{\theta}(z)}\_{> 0} dz \: \: \: \: \: \: \: \forall \: \theta \in [0,M] \: \: \: \: \: \: \stackrel{?}{\implies} \: \: \: \: \: \: g(z) = 0 \: \: \: \: \: \: \: \text{a.e. for} \: z \in [0,M]$$
As an example, you can take as $f\_{\theta}(z) = \frac{\mathbb{1}\_{[0,\theta)}(z)}{2 \sqrt{\theta^2 - \theta z}}$ for fixed $\theta$.
**Proposal:** I have at the moment a proof of this statement that assumes that $g$ is continuous and have finitely many zeros. Basically I use the finitely many zeros assumption to say that $g$ cannot have infinitely many fluctuations from above to below zero and then I consider each one of these finitely many neighborhoods where $g$ is strictly positive or strictly negative (by continuity) and I show via contradiction that on that neighborhood actually $g$ must be $0$ because otherwise, we can find a $\theta$ in the middle of a such neighborhood so that the integral is nonzero. Therefore I would like to know if the claim holds more generally for not continuous functions. Or for all continuous functions, not necessarily with finitely many zeros.
Any help is extremely appreciated!
| https://mathoverflow.net/users/493038 | Deduce that a function is zero on interval $[0,M]$ | As you say, it is not difficult to prove that $g=0$ when there are finitely many fluctuations in sign.
More precisely, suppose we have points $0=a\_0,a\_1,\dots,a\_n=M$ such that $g\geq0$ or $g\leq0$ in each interval $[a\_i,a\_{i+1}]$. Then, if we have measures $\mu\_\theta$ as you say and $\int\_0^\theta g(z) \, d \mu\_\theta (z)=0$ for all $\theta$, then we can prove by induction on $i$ that $g=0$ almost everywhere in $[a\_i,a\_{i+1}]$.
More in general, if the function $g$ satisfies that for every $x\in[0,M]$ there is some $\varepsilon>0$ such that $g\geq0$ or $g\leq0$ in $[x,x+\varepsilon]$, we can prove in the same way that if $\int\_{0}^{\theta} g(z) d \mu\_{\theta} (z)=0$ for all $\theta$, then $g=0$ (consider the maximum $x$ such that $g\neq0$ almost everywhere in $[0,x]$, and if $x<M$ obtain a contradiction).
If $g$ is not as in the previous paragraph (this of course implies $g\neq 0$), then we can find some measures $\mu\_\theta$ as in the question so that $\int\_0^\theta g(z) \, d \mu\_\theta (z)=0$ for all $\theta\in[0,M]$.
To do it, let $k=\inf\{x\in[0,M];\not\exists\varepsilon>0\text{ such that } g\geq0\text{ or }g\leq0\text{ in }[x,x+\varepsilon]\}$. Then we can prove as before that $g=0$ a.e. in $[0,k]$. Moreover, by definition of $k$, for each $n\in\mathbb{N}$ there are sets of positive measure $A\_n,B\_n$, such that $A\_n\subseteq[k,k+\frac{1}{n}],B\_n\subseteq[k,k+\frac{1}{n}]$, $g>0$ in $A\_n$ and $g<0$ in $B\_n$.
Now let's create the measures $\mu\_\theta$: for $\theta\leq k$ we can choose any measures we want. For $\theta>k+\frac{1}{n}$, consider for each $s,t>0$ the density functions $f\_{\theta,s,t}=s\chi\_{A\_n}+t\chi\_{B\_n}+\chi\_{[0,\theta]\setminus(A\_n\cup B\_n)}$.
Then $\int\_0^\theta g(z) f\_{\theta,s,t}(z) \, dz = \int\_{[0,\theta] \setminus(A\_n\cup B\_n)} g(z) \, dz + s\int\_{A\_n}g(z) \, dz+t\int\_{B\_n}g(z) \, dz$. As $\int\_{A\_n}g(z) \, dz>0$ and $\int\_{B\_n}g(z) \, dz<0$, you can adjust $s,t$ so that $\int\_0^\theta g(z) f\_{\theta,s,t}(z) \, dz=0$.
| 1 | https://mathoverflow.net/users/172802 | 433545 | 175,364 |
https://mathoverflow.net/questions/433547 | 14 | For a little more context: I'm currently an undergrad (sophomore) at a small liberal arts college with a (from my experience so far) solid math program. So far, I've taken Calc I, II, III, linear algebra, and an introduction to logic class. This semester I'm taking graph theory and differential equations, an intro to statistics class, and an intro to proofs class. And outside of class, I'm (very slowly) working through an abstract algebra textbook.
What are some ways I can find questions that are accessible to me as a pretty mathematically inexperienced undergraduate? I don't even mean questions that are open or unsolved, just any questions that would be accessible that I could play around with and get a taste of working through.
| https://mathoverflow.net/users/493850 | How do I, as an undergraduate, find interesting, accessible questions to work on to see if I'd be interested in research mathematics? | Welcome to MathOverflow! I am a professor at a liberal arts college. The best place to start is by talking to your favorite math professor, or your advisor in the department. The graph theory course will have plenty of accessible problems that will give you a taste of research mathematics. Abstract algebra does, too, and I'll bet your independent study will be smoother if a faculty member is involved helping you see the "big picture", suggesting good exercises to work on, etc.
Most faculty at liberal arts colleges welcome the opportunity to work with undergraduates, because it's a lot of fun, and liberal arts colleges usually don't have graduate students. It also looks good to the university when a professor works with an undergraduate student, so don't be shy to ask your professors.
It may happen that when you talk to your favorite faculty member, they suggest doing a "directed study" with them in the spring, e.g., reading an interesting book covering material not normally taught in the curriculum (e.g., hypergraphs, or Galois theory). They might also have research problems in mind already, accessible to students. For example, [I list a few on my webpage](http://personal.denison.edu/%7Ewhiteda/research.html), in case students want to work with me. As a rising junior, you can also apply to REU programs (that stands for "research experience for undergraduates"), though you're more likely to get in as a rising senior. Such an experience would provide you housing and a stipend on the campus of some university, where you'd work alongside other undergraduates on a research problem suggested and supervised by a professor. It's a great way to get an idea of whether or not you like research and to improve your resume during the summer. Deadlines are often in February. The same faculty member you ask about directed studies can suggest REUs and write you a letter of recommendation. Based on your background, I think you have a good shot, even for this summer. Good luck!
| 18 | https://mathoverflow.net/users/11540 | 433550 | 175,366 |
https://mathoverflow.net/questions/433533 | 4 | Consider the left-regular representation $\lambda : G \to B(L^2(G))$, $\lambda\_g f(h) = f(g^{-1}h)$, for a locally compact group.
It is well-known that this is a unitary faithful and strongly-continuous representation, but is it also a homeomorphism onto its image $\lambda(G)$ (equipped with the strong-operator topology?
It would suffice to show that for any net $g\_\alpha$, convergence in the strong operator topology of $\lambda\_{g\_\alpha}$ to the identity $I\_{L^2(G)}$ implies convergence of $g\_\alpha$ to the neutral element $e$ in $G$.
| https://mathoverflow.net/users/485160 | Is the left-regular representation of a locally compact group a homeomorphism onto its image? | Yes. It's more generally true for every faithful $C^0$ unitary representation $\pi$ of $G$. (Recall that a unitary representation $\pi$ is $C^0$ if for all $v,w$ in the Hilbert space, one has $\langle \pi(g)v,w\rangle\to 0$ when $g$ leaves compact subsets of $G$.)
Indeed if this is not a homeomorphism onto its image, there exists an ultrafilter $\eta$ on $G$, not converging to $1$, such that $\lim\_{g\to\eta}\lambda\_g=\mathrm{id}$ (for the strong topology, i.e., $\lim\_{g\to\eta}\pi(g)v=v$ for every $v$ in the Hilbert space.
If $\eta$ is unbounded (i.e. no compact subset of $G$ is in $\eta$), then we get a contradiction, since the $C^0$ property implies $\lim\_{g\to\eta}\langle\pi(g)v,v\rangle=0$ for every $v$ in the Hilbert space.
If $\eta$ is bounded, then $\eta$ has a limit $g\_0$ (which by assumption is not $1\_G$), and we deduce $\lim\_{g\to\eta}\pi(g)=\pi(g\_0)$. So $\pi(g\_0)=\mathrm{id}$, contradicting the faithfulness assumption.
| 7 | https://mathoverflow.net/users/14094 | 433555 | 175,369 |
https://mathoverflow.net/questions/433556 | 2 | Let $\mu$ be the Haar measure defined on the space of unimodular lattices, identified with $\text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$.
The classical Siegel's formula in geometry of numbers states that for $f\in L^1(\mathbb R^d)$, let $\Lambda$ be a unimodular lattice, and let
$$\hat{f}(\Lambda):= \sum\_{v\in \Lambda\setminus 0} f(v).$$
Then $$\int\_X \hat{f} d\mu = \int\_{\mathbb R^d} f dv.$$
I wonder if there is such a generalization to this result:
For $g\in L^1((\mathbb R^d)^k)$, $1 \le k \le d$ and unimodular lattice $\Lambda$, let $\hat{g}^k(\Lambda):=\sum\_{(v\_1,\dots,v\_k)\in \Lambda^k \setminus 0} g(v\_1,\dots, v\_k)$. I speculate there might be such a generalization of Siegel's mean value formula:
$$\int\_X \hat{g}^k d\mu = \int\_{\mathbb R^d} \dots \int\_{\mathbb R^d} g(v\_1,\cdots,v\_k) dv\_1 \cdots dv\_k$$
Is this correct and proved somewhere in the literature? Or can anyone suggest a proof for it below if not too complicated from the classical Siegel's formula?
| https://mathoverflow.net/users/489992 | Proof of generalized Siegel's mean value formula in geometry of numbers | Such a generalization (roughly) exists, known as *Rodger's Integration Formula*.
See [Section 1.2 of Seungki Kim's Dissertation](https://stacks.stanford.edu/file/druid:ms598tw2644/thesis%20%281%29-augmented.pdf) for a reference.
Theorems 1.2 and 1.3 are of interest.
>
> **Theorem 1.2**: (Rogers). Let $k < n$, and $\rho : (\mathbb{R}^n)^kk \to \mathbb{R}$ be a compactly supported,
> bounded, and Borel measurable function. Then
> $$
> \int\_{X}\sum\_{\substack{v\_1,\dots,v\_k\in L\setminus\{0\}\\\mathsf{rank}(\langle v\_1,\dots,v\_k\rangle) = k}}\rho(v\_1,\dots,v\_k)d\mu = \int\_{\mathbb{R}^n}\dots \int\_{\mathbb{R}^n}\rho(x\_1,\dots,x\_k)dx\_1\dots dx\_k.
> $$
>
>
>
Up to the requirement that one sums over rank $k$ collections of vectors $(v\_1,\dots,v\_k)$, this is what you want.
One can remove this restriction (to get the generalization that you want), but the formula becomes more complex --- there is an additional (additive) term on the right-hand side.
See Theorem 1.3 of the dissertation for details.
| 4 | https://mathoverflow.net/users/101207 | 433557 | 175,370 |
https://mathoverflow.net/questions/433539 | 4 | Given any linear space $L$ over an ordered field $F$, consider the *equiproportion relation* $${\sim}=\{((x,y,z),(a,b,c))\in L^3\times L^3: \exists t\in[0,1]\subseteq F\;(y{-}x=t(z{-}x)\wedge b{-}a=t(c{-}a))\},$$called the *standard equiproportion relation* on $L$.
This relation seems to describe the affine geometry of $L$.
Observe that for any points $x,y,z\in L$ with $(x,y,z)\sim (x,y,z)$ the point $y$ lies *between* the points $x$ and $z$. Therefore the betweenness relation is encoded in the equiproportion relation.
>
> **Problem.** Is there any characterization of the affine spaces $\mathbb{R}^n$ in terms of the equiproportion relation? More precisely, given a set $X$ endowed with a relation ${\sim}\subseteq X^3\times X^3$ I would like to have a reasonably short list of axioms guaranteeing that the relation algebra $(X,\sim)$ is isomorphic to the space $\mathbb R^n$ endowed with the standard equiproportion relation.
>
>
>
| https://mathoverflow.net/users/61536 | Is the affine geometry a geometry of proportions? | The midpoint of $a$ and $b$ can be defined from equiproportion as the unique $m$ for which $(a,m,b)$ and $(b,m,a)$ are equiproportional.
Similarly, the collinearity of $a,b,c$ can be defined as one of $(a,b,c)$, $(b,c,a)$ or $(c,a,b)$ being equiproportional to itself.
Using these, $\mathbb{R}^n$ should be characterized by axioms for midpoints and collinearity plus the second-order axiom of continuity suggested in the comments.
For axioms of affine geometry from midpoint and collinearity, see Patrick [Suppes](https://www.jstor.org/stable/20117084), “Quantifier-Free Axioms for Constructive Affine Plane Geometry”, and the book referred to there: Wanda Szmielew, *From Affine to Euclidean Geometry: An Axiomatic Approach*.
| 1 | https://mathoverflow.net/users/nan | 433562 | 175,373 |
https://mathoverflow.net/questions/433551 | 3 | I don't know much (anything) about sieves, but as I read the section on the Selberg upper bound sieve from Greaves's *Sieves in Number Theory*, there is a theorem 4 which says that
If $Y\ge X \ge 2$, then
\begin{equation}
\pi(Y)-\pi(Y-X) \leq \frac{2X}{\log X} + O\left(\frac{X}{\log^2 X}\right).
\end{equation}
1. I would be interested in knowing what the best available bound is.
2. Also, I notice that this bound holds irrespective of the value of $Y$. Are there ways to improve it for say $Y = 2X$ or $X= \sqrt{Y}$?
Thank you in advance!
Here, $\pi(X) = \sum\_{p\leq x} 1$.
| https://mathoverflow.net/users/392272 | Best available bounds for $\pi(Y)-\pi(Y-X)$? | If $Y\geq X\geq 2$, then as Daniel Johnston wrote, [Montgomery and Vaughan](https://deepblue.lib.umich.edu/bitstream/handle/2027.42/152543/mtks0025579300004708.pdf) proved that
$$\pi(Y)-\pi(Y-X)\leq \frac{2X}{\log X}.$$
Whether this constitutes a "best bound" requires a definition of what you consider to be "best". Expounding on Noam Elkies' comment, it follows from work of [Heath-Brown](https://gdz.sub.uni-goettingen.de/id/PPN243919689_0389?tify=%7B%22view%22:%22info%22,%22pages%22:%5B26%5D%7D) (building on Hoheisel, Tchudakov, Heath-Brown and Iwaniec, Huxley, and many others) that if $Y^{7/12}\leq X\leq Y$, then
$$\pi(Y)-\pi(Y-X)\sim\frac{X}{\log Y}.$$
Therefore, for all $\epsilon>0$ there exists $Y\_0(\epsilon)>0$ such that if $Y\geq Y\_0(\epsilon)$ and $Y^{7/12}\leq X\leq Y$, then
$$\Big|\pi(Y)-\pi(Y-X)-\frac{X}{\log Y}\Big|\leq \epsilon \frac{X}{\log Y}.$$
(This can be made more precise using known error terms in the prime number theorem for short intervals.) The notion of "best" depends on context.
| 6 | https://mathoverflow.net/users/111215 | 433570 | 175,377 |
https://mathoverflow.net/questions/433581 | 12 | I would like to read Pincus' article [**Adding dependent choice**](https://doi.org/10.1016/0003-4843(77)90011-0), where he proves, among other things, the consistency of the theory $\mathsf{ZF+DC+O+\neg AC}$, where $\mathsf{DC}$ stands for Dependent Choice and $\mathsf{O}$ is the linear ordering principle, i.e. the statement "Every set can be linearly ordered". But in this paper Pincus uses Cohen's original presentation of forcing, which makes it (at least to me) hard to read.
Is there any newer account on his proof (in a thesis, paper etc.), which uses a more modern approach to forcing?
Thanks!
| https://mathoverflow.net/users/141146 | Is there a more modern account of the main results of "Adding Dependent Choice" by D. Pincus? | No.
Most of the work of Pincus that involved these sort of iterated constructions, where one "pushes the counterexamples out" has never been reworked into modern terms.
In my Ph.D. one of the reasons to develop the notion of an iteration of symmetric extensions was to help and simplify these proofs (also the works of Gershon Sageev). If you look closely, you will see that this is not just presented in the "old method of forcing", but it is also from an era where the distinction between generic and symmetric extension wasn't fully explained in texts (indeed, you can find even papers from the early '80s that refer to what is clearly a symmetric extension as a "generic extension").
Unfortunately, so far, nobody has wanted to take on this mantle, and I have a lot on my hands as it is. I am happy to collaborate or give advice, if anyone wants to go for it.
---
Let me give a broad reason as to *why* I expect these results to fit so well into my work. Just in case it's not very clear.
Many of the complicated choice-related constructions of the time (Morris' model, Pincus' work, Sageev's work, and others) were written before a very clear understanding of forcing, iterations, and symmetric extensions. Morris' thesis, with this regards, is particularly amazing, since he's quite literally iterating symmetric extension is a very non-trivial fashion.
But when you read these, it becomes very apparent that the constructions are *usually* of the form "take a symmetric extension to fix some problems, then repeat", with some kind of understanding of a symmetry-based structure at limit steps.
In Sageev's work, each step is itself an iteration of symmetric extensions.
So, yes, without looking *very closely*, I do expect that Pincus' work will fit into the iteration framework. It may very well be necessary to extend the current understanding of the framework, which is something that I am slowly doing with one of my postdocs, Jonathan Schilhan, but at the end, if we have a general technique for iterating symmetric extensions, then a result constructed as an iteration of symmetric extensions should fit into the framework.
| 15 | https://mathoverflow.net/users/7206 | 433582 | 175,381 |
https://mathoverflow.net/questions/433572 | 4 | Let $h=[\Bbb CP^1]\in H\_2(\Bbb CP^2;\Bbb Z)$. By a theorem of Kronheimer and Mrowka (Theorem 1 of this paper: [https://people.math.harvard.edu/~kronheim/thomconj.pdf](https://people.math.harvard.edu/%7Ekronheim/thomconj.pdf)), a class $nh \in H\_2(\Bbb CP^2;\Bbb Z)$ can be represented by a smoothly embedded 2-sphere if and only if $|n|\leq 2$.
Is there a similar result for connected sum of $\Bbb CP^2$s? Considering generators $h\_1,\dots,h\_n$ of $H\_2(n\sharp \Bbb CP^2;\Bbb Z)\cong H\_2(\Bbb CP^2;\Bbb Z)\oplus \cdots \oplus H\_2(\Bbb CP^2;\Bbb Z)$, can we answer that whether, for example, the class $3h\_1+4h\_3+6h\_4$ can be represented by a smoothly embedded 2-sphere?
| https://mathoverflow.net/users/164671 | Homology classes in connected sum of $\Bbb CP^2$'s that can be represented by smoothly embedded spheres | As Mike says in his comment, the answer is not known in full generality, not even for $n=2$, so for classes in $\mathbb{CP}^2\#\mathbb{CP}^2$. I think the paper that he links (<https://arxiv.org/abs/2210.12486>, by Marengon, Miller, Ray, and Stipsicz) describes the state of the art for $n=2$.
Some results are known for certain specific families of homology classes (for arbitrary $n$).
* Characteristic classes: these are homology classes in which all coefficients are odd (with respect to the standard basis). Since the complement of any surface in this homology class is spin, you can use the Rokhlin invariant/Rokhlin's signature theorem to exclude the existence of spheres in these homology classes. This is sometimes known as the Kervaire-Milnor obstruction (*On 2-spheres in 4-manifolds*, Proc. Nat.
Acad. Sci. U.S.A. **47** (1961)), and it says that $A^2 \equiv n \pmod{16}$ if the class $A \in H\_2(n\mathbb{CP}^2)$ is characteristic and represented by a smoothly embedded sphere.
* Divisible classes: these are classes that are (non-trivial) multiples of another class. Here you can use cyclic branched covers (and the G-signature theorem) to give lower bounds. (And these bounds also work in the topological category, by the way.) This is probably explained in Gilmer's paper on configurations of surfaces, together with much more general results (*Configurations of surfaces in 4-manifolds*, Trans. Amer. Math. Soc. **264** (1981)).
| 6 | https://mathoverflow.net/users/13119 | 433588 | 175,382 |
https://mathoverflow.net/questions/433554 | 31 | I've often seen Lurie's *Higher Topos Theory* praised as the next "great" mathematical book. As someone who isn't particularly up-to-date on the state of modern homotopy theory, the book seems like a lot of abstract nonsense and the initial developments unmotivated. I'm interested in what the tools developed concretely allow us to do.
**What does HTT let us do that we previously were unable to?**
Please note that I'm looking for concrete examples or theorems that can be expressed in terms of math that one doesn't need higher topos theory to understand. I'd also be interested in ways that the book has changed pre-existing perspectives on homotopy theory.
| https://mathoverflow.net/users/99902 | Higher Topos Theory- what's the moral? | It seems there are really two questions here:
1. Why higher category theory? What questions can you pose without the language of higher category theory which are best answered using higher category theory?
2. Why does Lurie's work specifically set the standard for the foundations of higher category theory?
These are really distinct questions. I'll leave it to others to address (1), and focus on (2). For this, I will refer back to an [old answer of mine](https://mathoverflow.net/questions/311158/what-parts-of-the-theory-of-quasicategories-have-been-simplified-since-the-publi/311273#311273) for a summary of some of the contents of HTT and HA. There, I said:
>
> In *Higher Topos Theory*, Lurie accomplishes many things. Let me highlight a few:
>
>
> * A study of the Joyal model structure and comparison to the Bergner model structure.
> * A study of cartesian fibrations and straightening / unstraightening, the $\infty$-categorical analog of the Grothendieck construction. This is often viewed as the technical heart of Lurie's theory, since cartesian fibrations are used systematically to avoid writing down all the higher coherence data involved in $qCat$-valued functors.
> * A development of the fundamental notions of category theory -- (co)limits, Kan extensions, cofinality, etc, allowing one to "do category theory" in the $\infty$-categorical setting.
> * A development of the theory of presentable $\infty$-categories. The point here is to get access to (the most important instances of) Freyd's adjoint functor theorem in the $\infty$-categorical setting, and in particular the theory of localizations.
> * The theory of (Grothendieck) $\infty$-toposes.
>
>
> In the context of foundations, maybe it's worth also mentioning some of the contents of *Higher Algebra*:
>
>
> * The Barr-Beck monadicity theorem. I tend to think of this, along with the adjoint functor theorem as "the only real theorems" of basic ordinary category theory.
> * A theory of operads, allowing one to "do algebra" $\infty$-categorically.
> * The theory of stable $\infty$-categories, playing roughly the roles of abelian categories and triangulated categories in the $\infty$-categorical setting.
>
>
>
So the reason that HTT and HA set the standard for the foundations of category theory is pretty self-evident. Nowhere else can one find such a comprehensive treatment! The 2500 pages in these books are there for a reason. Some pieces of HTT/HA were previously available in various sources, but some were not, and moreover HTT/HA synthesize them in coherent account. So you don't have to spend as much time as you otherwise would have to patching together results proven in slightly different frameworks using model-comparison results.
This is particularly striking from a historical perspective: in the days when HTT first appeared (almost 10 years ago now -- to call it "the next great mathematical book" is already a little behind the times I think: it's a *current* great mathematical book!), all of this was a dream. Lurie made it a reality.
I'll add on a personal note that my own most common mode of doing higher category theory is to pretend that everything is an ordinary category and freely use all the tools available there, until I've worked out a complete argument. After that I go through the process of looking up $\infty$-categorical analogs of each of the 1-categorical tools I've used in my argument. This works better than one might expect, because it's reasonable today to trust that most of these tools will indeed be available in the literature. That's thanks in large part to Lurie's work. Before Lurie, you could do something like this, but only if you were content to end up with incomplete arguments contingent on the dream of higher category theory working out. Today, I'd argue that higher category theory, among various mathematical disciplines, actually has a relatively *high* standard of rigor. That's thanks in part to Lurie's work setting the standard.
---
Let me close by sharing a sort of testimonial from Clark Barwick ([originally from the homotopy theory chat room](https://chat.stackexchange.com/transcript/message/41977621#41977621) here on MO, in the context of [another MO question](https://mathoverflow.net/questions/289259/the-derived-drift-is-pretty-unsatisfying-and-dangerous-to-category-theory-or)). Thanks to user1092847 for digging this up in the [comments](https://mathoverflow.net/questions/433554/higher-topos-theory-whats-the-moral/433590#comment1116677_433590) below!
**Clark Barwick on Lurie's impact with HTT:**
>
> ... I feel a need to defend Jacob Lurie's writing. Let me take a rather selfish perspective, because I grew up alongside higher categories in some sense. I read preprints and papers of Rezk, Hirschowitz-Simpson, Simpson, Tamsamani, Toen, Joyal, Jacob's HTT-prototype on the arXiv, and others as a grad student (2001-05).
>
>
> All of these works had the same feature: they were all organised around a specific goal, leaving the more serious work of a complete theory for a later time. There were all sorts of homotopy coherence issues that were left hanging.
>
>
> So I developed my own point of view about these things and started writing a manuscript, the first little bit of which was my thesis.
> By the time I was halfway through my first postdoc, I'd written a pile of 'prenotes' that did enough foundational work to ensure, e.g., that there was no confusion over 'how unique' an adjoint between ∞-categories is, how to prove the existence of all colimits in an ∞-category from, say, geometric realisations and coproducts, a theory of what we now call ∞-operads, etc., etc.
>
>
> It all involved layer upon layer of giant combinatorial gadgets, and they were often fragile enough that I wasn't sure I had them layered correctly.
>
>
> At around that time I met Jacob at a conference, and he mentioned that he'd revised the text he put on the arXiv to add a little more detail. I said that I'd love to see it. He sent me a PDF of 600 pages or so of HTT.
>
>
> To my surprise and horror, he'd done everything I'd done, but more of it and far far better. He'd understood issues like cofinality in a way I didn't have access to with the models I was using. In his text, the proofs worked because of some very compact, very robust models he chose early on, following Joyal.
>
>
> Those models required him to do a lot of pretty tedious technical labour in the first few sections, but it ensured that if something existed up to homotopy, it 'really' existed. (This always came down to selecting a section of a trivial fibration.) This meant that it was genuinely easy to understand the arguments.
>
>
> When you look at a proof in HTT or HA or SAG, it's all there. He doesn't tell you that you 'can' find the argument – he gives you the argument! That's the real advantage of Jacob's arguments (and Joyal's before him) – they're completely convincing. You can actually check (and in rare cases, yes, correct) his proofs, because every individual object is so concrete. (Cf. claims about $A\_{\infty}$-categories like the Fukaya category.)
>
>
> After a few sleepless nights, I just gave up on what I was trying to develop. I was not going to try to compete. On the other hand, I didn't feel comfortable enough in the Joyal/Lurie perspective to really use their model, so I tried to do things in a model-independent way, as Rune suggests.
>
>
> But even simple things, like constructing a symmetric monoidal functor between two symmetric monoidal ∞-categories when there isn't one for formal reasons, is very difficult from that perspective: the only path I saw was to check an infinite hierarchy of coherences.
> It took me a long time to realise that the fibrational perspective was exactly designed to make it easy (or at least convincing) to write these things down. Jacob is actually providing you with the tools to perform explicit, nontrivial, non-formal constructions with higher categories in a precise, legible, and convincing way. That's what results like HTT.3.2.2.13 are all about.
>
>
> Jacob's done this continually: at every turn, he's done an incredible service to the community by carving out not just a narrow path to a desired application, but an expansive tunnel through which a lot of us can travel. He offers incredibly refined, interlinking technologies that are ideal for people like me, at least. I'm in a particularly good position to appreciate that kind of labour, because I attempted it and failed where he succeeded.
>
>
> Does he solve every problem or define every conceivable object? No, of course not. (And if it's tough to read now, what would it be like if he did? (However, I will point out that he does deal with general pro-objects in SAG.E.2.)) Is it possible to sharpen his results or use little techniques to get improvements on his results? Sure. But overall, I think that the precision, clarity, and thoroughness of Jacob's writing is something to which homotopy theory should aspire.
>
>
>
| 33 | https://mathoverflow.net/users/2362 | 433590 | 175,383 |
https://mathoverflow.net/questions/433579 | 0 | Let $X=(X\_t:t\ge 0)$ be a stochastic process (martingale in general) starting at $X\_0=0$. For $T>0$ and $a<b$, let $U^T\_{a,b}(X)$ be the number of upcrossings of $X$ across the interval $[a,b]$ over $[0,T]$, i.e. $U^T\_{a,b}(X)$ is the supremum of the nonnegative integers $n\in\mathbb N$ such that there exist times $s\_k,t\_k\in [0,T]$ satisfying
$$s\_1<t\_1<s\_2<t\_2<\cdots<s\_n<t\_n$$
and for which $X\_{s\_k}\le a<b\le X\_{t\_k}$. My question is the following: For any continuous and non-decreasing function $h:\mathbb R\_+\to [0,\infty]$ with $h(0)=0$ (may not be strictly increasing), set $Y=(Y\_t:=X\_{h(t)}: t\ge 0)$. Does $U^T\_{a,b}(X)=U^{h(T)}\_{a,b}(Y)$ hold?
| https://mathoverflow.net/users/493556 | Is the number of uncrossing invariant under time-change? | We have $U\_{a,b}^T(Y) = U\_{a,b}^{h(T)}(X)$.
By construction, the function $h$ maps $[0,T]$ onto $[0,h(T)]$. More precisely, each $t' \in [0,h(T)]$ can be written $h(h^\leftarrow(t'))$,
where $h^\leftarrow(t') := \inf\{t \in [0,T] : h(t) \ge t'\} \in [0,T]$.
The function $h^\leftarrow$ thus defined is strictly increasing (but not necessarily continuous).
Hence, if $s'\_1<t'\_1<\ldots<s'\_n<t'\_n$ provide $n$ upcroassings of $X$ in $[0,h(T)]$ then $h^\leftarrow(s'\_1)<h^\leftarrow(t'\_1)<\ldots<h^\leftarrow(s'\_n)<h^\leftarrow(t'\_n)$ provide $n$ upcroassings of $Y$ in $[0,T]$.
Conversely, if $s\_1<t\_1<\ldots<s\_n<t\_n$ provide $n$ upcroassings of $Y$ in $[0,T]$,
then $h(s\_1)<h(t\_1)<\ldots<h(s\_n)<h(t\_n)$ provide $n$ upcroassings of $X$ in $[0,h(T)]$ (the inequalities cannot be equalities because of the conditions $X(h(s\_i)) \le a < b \le X(h(t\_i))$.
| 0 | https://mathoverflow.net/users/169474 | 433592 | 175,384 |
https://mathoverflow.net/questions/433597 | 1 | From the paper of [Ambrosio-Crippa](https://www.cambridge.org/core/journals/proceedings-of-the-royal-society-of-edinburgh-section-a-mathematics/article/abs/continuity-equations-and-ode-flows-with-nonsmooth-velocity/DD2A02C300757F63F7F0A96FAD8A70C5), it is known that if $\beta:\mathbb R^d\times[0, T[\longrightarrow\mathbb R^d$ is suitably regular, then the system
$$
\begin{cases}
\dfrac{\partial\mu}{\partial t}(x, t)+\operatorname{div}(\beta(x, t)\mu(x, t))=0,&(x, t)\in\mathbb R^d\times]0, T]\\
\mu(\cdot, 0)=\mu\_0
\end{cases}
$$
is well-posed. In the case when $\mu\_0$ is absolutely continuous, that is $\mu\_0=m\_0\mathcal L^d$, where $m\_0:\mathbb R^d\longrightarrow\mathbb R$ is the density, all the measures $\mu(\cdot, t)$ are absolutely continuous too, and their density $m(\cdot, t)$ can be explicitly computed as
$$
m(\cdot, t)=\frac{m\_0(\cdot)}{\operatorname{det}J\Phi\_t(\cdot)}\ \circ\ \Phi\_t^{-1}(\cdot),\label{1}\tag{$\triangle$}
$$
where $\Phi\_t$ is the flow associated to
$$
\begin{cases}
y'(s)=\beta(y(s), s),&s\in]0, T[\\
y(0)=x
\end{cases}.
$$
**Question.** Is it true that $m$ satisfies (in the sense of distributions) the continuity equation
$$
\begin{cases}
\dfrac{\partial m}{\partial t}(x, t)+\operatorname{div}(\beta(x, t)m(x, t))=0,&(x, t)\in\mathbb R^d\times]0, T]\\
m(\cdot, 0)=m\_0
\end{cases}\quad?\label{2}\tag{$\star$}
$$
I proved that, with suitable regularity hypotheses on the field $\beta$, if $m\_0\in H^1(\mathbb R^d)\cap W^{1, \infty}(\mathbb R^d)$ then $m(\cdot, t)$ belongs to the same space for every $t$. I think that \eqref{2} is true but, to be honest, i cannot prove that using directly \eqref{1}. Can you help me or give me some suggestion?
| https://mathoverflow.net/users/160186 | Continuity equation for a density of a measure | I am not so sure to understand the problem, maybe I am missing something. You should not use $(\triangle)$ but instead go back to the equation satisfied by the measure. Indeed, since $\mu$ is solution of your conservative transport equation, you have for any test function $\varphi\in\mathscr{D}(\mathbf{R}\_+\times\mathbf{R}^d)$, noting $\mu\_t:=\mu(t,\cdot)$
$$
\int\_0^{+\infty} \int\_{\mathbf{R}^d} (\partial\_t \varphi+\beta\cdot\nabla\_x \varphi) \mathrm{d}\mu\_t \,\mathrm{d}t = -\int\_{\mathbf{R}^d} \varphi(0)\mathrm{d}\mu\_0.
$$
If you did prove what you claim then you get directly your formulation.
| 2 | https://mathoverflow.net/users/27767 | 433602 | 175,385 |
https://mathoverflow.net/questions/433402 | 5 | I have a monoidal (not symmetric) triangulated category $(A,\otimes, 1)$ with unit 1.
Define $C$ the localizing subcategory of $A$ generated by the unit 1.
* is $(C, \otimes, 1) $ a symmetric monoidal triangulated category?
* Do we have a natural isomorphism $c\otimes a\cong a\otimes c$ for any $a\in A$ and any $c\in C$ ?
| https://mathoverflow.net/users/141114 | Monoidal triangulated categories | The answer to question 2, even with the assumption that $a\in C$, is no in general. It follows that the answer to question 1 is also no.
For a counterexample, one can consider for example a (derived) category of bimodules. Let $k$ be a base commutative ring, and $R$ a (say) flat $k$-algebra. Consider the derived category $A= D(R\otimes\_k R^{op})$ of $(R,R)$-bimodules over $k$. This is monoidal under $\otimes^\mathbb L\_R$, with unit $R$ viewed as an $R$-bimodule in the canonical way.
An explicit example is given by the polynomial ring $R= k[t]$. In this case, $R\otimes\_k R^{op} = k[t\_0,t\_1]$.
There is $k[t\_0,t\_1]$-linear extension $0\to k[t]\to k[t\_0,t\_1]/(t\_0-t\_1)^2\to k[t]\to 0$, where the rightmost map is the algebra map $t\_0,t\_1\mapsto t$, while the leftmost map is $P\mapsto (t\_0-t\_1)P(t\_0,t\_0)$. One easily checks that this is an extension: first, the rightmost map is clearly a bimodule map. For the leftmost map, one checks that $(t\_0-t\_1)t\_1P(t\_0,t\_0) = (t\_0-t\_1)t\_0P(t\_0,t\_0)$. Indeed it suffices to check it when $P = 1$ and then the relation is $t\_0t\_1-t\_1^2 = t\_0^2-t\_1t\_0$ which is a rephrasing of the relation $(t\_0-t\_1)^2 = 0$.
Second, suppose $P(t,t) = 0$. Then $P$ is divisible by $(t\_0-t\_1)$, to $P = (t\_0-t\_1)Q$ for some $Q$, and now $(t\_0-t\_1)Q = (t\_0-t\_1)Q(t\_0,t\_0)$ modulo $(t\_0-t\_1)^2$, because $Q-Q(t\_0,t\_0)$ is divisible by $(t\_0-t\_1)$. It follows that $P$ is in the image of the leftmost map.
This extension corresponds to a morphism $k[t]\to \Sigma k[t]$ in $A$, with fiber $E= k[t\_0,t\_1]/(t\_0-t\_1)^2$. In particular, $E$ belongs to $C$.
Another object in $C$ is $k$ where $t\_0,t\_1$ act trivially. Indeed, it is the mapping cone of $k[t]\xrightarrow{t}k[t]$ (this is a bimodule map, as $t$ is central).
I claim that $E\otimes^\mathbb L\_{k[t]}k \not\simeq k\otimes\_{k[t]}^\mathbb L E$.
To see this, we first observe that these derived tensor products are underived. The key lemma for this is:
*Lemma*: Suppose $(t\_0-t\_1)$ divides $t\_0 P$. Then $(t\_0-t\_1)$ divides $P$.
*Proof*: The assumption implies that $tP(t,t) = 0$ in $k[t]$. But $t$ is not a zero divisor, so $P(t,t) = 0$ and the conclusion follows.
Now we use this lemma twice : suppose $t\_0 P = 0$ in $E$. This means $(t\_0-t\_1)^2\mid t\_0P$ in $k[t\_0,t\_1]$. It follows that $(t\_0-t\_1)$ divides $t\_0 P$, and thus by the lemma, it divides $P$. Write $P= (t\_0-t\_1)Q$, and it follows easily that $(t\_0-t\_1) \mid t\_0 Q$, so that by the lemma, $(t\_0-t\_1)$ divides $Q$, and so $(t\_0-t\_1)^2$ divides $P$. So $P = 0$ in $E$.
By symmetry, the same holds for $t\_1$, and so the action of $t\_0$ and $t\_1$ on $E$ is injective, which proves the claim that the tensor products are underived.
Now, $E\otimes\_{k[t]}k$ is therefore $E/t\_1 E\cong k[t\_0,t\_1]/((t\_0-t\_1)^2, t\_1) = k[t\_0,t\_1]/(t\_0^2,t\_1)\cong k[t\_0]/(t\_0)^2$, on which $t\_0$ acts nontrivially while $t\_1$ acts trivially. Conversely, $k\otimes\_{k[t]}E\cong k[t\_1]/(t\_1^2)$, on which $t\_0$ acts trivially but $t\_1$ nontrivially. It follows that they are not isomorphic as bimodules, hence not isomorphic in the derived category either.
| 5 | https://mathoverflow.net/users/102343 | 433606 | 175,386 |
https://mathoverflow.net/questions/433589 | 1 |
>
> Define $\mathbb{W} = H\_{0}^{1}(-1,1) \times H\_{0}^{1}(-1,1)$, where $u \in H\_{0}^{1}(-1,1)$ if $u,u^{\prime} \in L^{2}(-1,1)$ and $u(-1) = u(1) = 0$. Consider the sesquilinear form $a: \mathbb{W} \times \mathbb{W} \to \mathbb{C}$ given by
> $$
> a((u,v),(w,z)) = \int\_{-1}^{1}u^{\prime}\overline{w}^{\prime}dx + \int\_{-1}^{1}v^{\prime}\overline{z}^{\prime}dx + \int\_{-1}^{1}u\overline{z}dx + \int\_{-1}^{1}v\overline{w}dx
> $$
> for $(u,v),(w,z) \in \mathbb{W}$. I want to show that $a$ is coercive, that is, there is $C > 0$ such that
>
>
>
$$
\text{Re}~a((u,v),(u,v)) \geq C \|(u,v)\|\_{W}^{2}, \ \ \forall (u,v) \in \mathbb{W}.
$$
Where $\|(u,v)\|\_{W} = \|u\|\_{H\_{0}^{1}} + \|v\|\_{H\_{0}^{1}}$, such that $\|u\|\_{H\_{0}^{1}} = \|u^{\prime}\|\_{L^{2}}^{2} + \|u\|\_{L^{2}}^{2} $
Let $(u,v) \in \mathbb{W}$ and
$$
a((u,v),(u,v)) = \int\_{-1}^{1}|u^{\prime}|^{2}dx + \int\_{-1}^{1}|v^{\prime}|^{2}dx + \int\_{-1}^{1}u\overline{v}dx + \int\_{-1}^{1}v\overline{u}dx
$$
But
$$
\int\_{-1}^{1}u\overline{v}dx + \int\_{-1}^{1}v\overline{u}dx = 2\text{Re}\int\_{-1}^{1}u\overline{v}dx
$$
Then, by Poincaré’s inequality
$$
a((u,v),(u,v)) = \|u^{\prime}\|\_{L^{2}}^2 + \|v^{\prime}\|\_{L^{2}}^2 + 2\text{Re}\int\_{-1}^{1}u\overline{v}dx \geq 2^{-3/2}\|u\|\_{L^{2}(-1,1)}^2 + 2^{-3/2}\|v\|\_{L^{2}(-1,1)}^2 + 2\text{Re}\int\_{-1}^{1}u\overline{v}dx
$$
I know that
$$
0 \leq (\|u\|\_{L^{2}} - \|v\|\_{L^{2}})^{2} =
(\|u\|\_{L^{2}}^2 + \|v\|\_{L^{2}}^2) - 2\|u\|\_{L^{2}} \|v\|\_{L^{2}}
$$
Then
$$
2\text{Re}\int\_{-1}^{1}u\overline{v}dx \leq 2\bigg|\int\_{-1}^{1}u\overline{v}dx \bigg| \leq 2\|u\|\_{L^{2}} \|v\|\_{L^{2}} \leq \|u\|\_{L^{2}}^2 + \|v\|\_{L^{2}}^2
$$
But I need a contrary inequality and not the one above. Help me, please!!
| https://mathoverflow.net/users/481556 | Proof that sesquilinear form in is coercive | The first eigenvalue of the second derivative with Dirichlet b.c. on $(-1,1)$ is $\pi^2/4$ (with eigenfunction $\cos \frac{\pi x}{2}$) and then Poincare' inequality with optimal constant is $\|u\|\_2^2 \leq \frac{4}{\pi^2}\|u'\|\_2^2$. It follows that
$$2| Re \int\_{-1}^1 u\bar v | \leq 2\|u\|\_2\|v\|\_2 \leq \|u\|\_2^2+\|v\|\_2^2\leq \frac{4}{\pi^2}(\|u'\|\_2^2+\|v'\|\_2^2)$$ which gives $C \geq 1-4/\pi^2$, using on $W$ the norm $\|(u,v)\|^2\_W=\|u'\|\_2^2+\|v'\|\_2^2$.
| 1 | https://mathoverflow.net/users/150653 | 433610 | 175,388 |
https://mathoverflow.net/questions/433604 | 2 | Setting and definitions
-----------------------
Let $X = \{X(t), t \in T \}$, $T \subset \mathbb{Z}$, be an infinite-variance associated stochastic process, i.e.
$$
\text{Cov}(f(X(I)), g(X(J))) \geq 0
$$
for all finite disjoint subsets $I, J \subset T$ and bounded, coordinate-wise increasing Borel functions $f: \mathbb{R}^{\vert I \vert} \rightarrow \mathbb{R}$, $g: \mathbb{R}^{\vert J \vert} \rightarrow \mathbb{R}$.
A stochastic process $Y$ is called (BL, $\theta$)-dependent if if there exists
a non-increasing sequence $\theta = (\theta\_r)\_{r \in \mathbb{Z}}$ with $\theta\_r \rightarrow 0$ as $r \rightarrow \infty$ and
$$
\Big\vert
\text{Cov}
\Big(
f\big(Y(I)\big), g\big(Y(J)\big)
\Big)
\Big\vert
\leq
\text{Lip}(f)\text{Lip}(g)(\vert I \vert \wedge \vert J \vert)
\theta\_r
$$
for any bounded Lipschitz-continuous functions $f: \mathbb{R}^{\vert I \vert} \rightarrow \mathbb{R}$, $g: \mathbb{R}^{\vert J \vert} \rightarrow \mathbb{R}$ and finite disjoint subsets $I, J \subset T$ such that $\text{dist}(I, J) := \min\{ \vert i - j \vert : i \in I, j \in J \} = r$.
Question
--------
Is $X$ (BL, $\theta$)-dependent? The finite-variance case can be proven as seen below. But the proof relies on covariances of the process and I don't know how to generalize the main inequality that was used in the proof.
Proof for finite-variance case
------------------------------
If $X$ had a finite-variance, then Theorem 5.3. in [Bulinski & Shashkin (2007)](https://www.worldscientific.com/worldscibooks/10.1142/6555#t=aboutBook) states that
$$
\Big\vert
\text{Cov}
\Big(
f\big(X(I)\big), g\big(X(J)\big)
\Big)
\Big\vert
\leq
\sum\_{i \in I} \sum\_{j \in J} \text{Lip}\_i(f)\text{Lip}\_j(g)
\text{Cov}(X(i), X(j))
$$
for all any bounded Lipschitz-continuous functions $f: \mathbb{R}^{\vert I \vert} \rightarrow \mathbb{R}$, $g: \mathbb{R}^{\vert J \vert} \rightarrow \mathbb{R}$ and finite disjoint subsets $I, J \subset T$.
Hence, $X$ is (BL, $\theta$)-dependent with
$$
\theta\_r := \sup\_{i \in I} \sum\_{j \in \mathbb{Z} : \vert i - j \vert \geq r} \vert \text{Cov}(X(i), X(j)) \vert
$$
under the assumption that these quantities exist and tend to zero.
| https://mathoverflow.net/users/302666 | Infinite-variance associated processes are (BL, $\theta$)-dependent | The answer is no. E.g., Let $X(t)=Z$ for all $t$, where $Z$ is any random variable with infinite variance. Then the process $(X(t)\colon t\in\mathbb Z)$ is positively associated. On the other hand, for $f\_n(x):=\min(n,\max(-n,x))$ and each natural $j$ we have
$$Cov(f\_n(X\_0),f\_n(X\_r))\to Var\,Z=\infty$$
as $n\to\infty$, whereas
$$\text{Lip}(f\_n)\text{Lip}(f\_n)(|\{0\}|\wedge|\{r\}|)=1.$$
| 2 | https://mathoverflow.net/users/36721 | 433615 | 175,389 |
https://mathoverflow.net/questions/433598 | 0 | I have one more question about the Example (I.5.1) on page 7 from
Rick Miranda's the basic theory of elliptic surfaces:
Let $C\_1$ be a smooth cubic curve in $\mathbb{P^2}$ and let $C\_2$
be any other cubic. Let $F\_1, F\_2 \in \mathbb{C}[X,Y,Z]$ the homogeneous cubic polynomials
generating the vanishing ideals of $C\_1$, respectively $C\_2$. With intersection theory and Bezout's lemma these curves intersect in
$9$ points. We form a pencil $X \subset \mathbb{P^2} $ generated
by $C\_1$ and $C\_2$; in detail the pencil is defined as the union
$$ X = \bigcup\_{[\lambda: \mu] \in \mathbb{P}^1} V(\lambda F\_1 + \mu F\_2) $$
of subschemes $V(\lambda F\_1 + \mu F\_2) $ in $\mathbb{P^2}$ running over $\mathbb{P}^1$.
This gives only a rational map to $\mathbb{P^1}$, since this map is not
defined in the nine intersection points of $C\_1$ and $C\_2$.
Next is claimed that after blowing up $X$ in these points,
we obtain a honest morphism $\pi: \widetilde{X} \to \mathbb{P^1}$ where
$\widetilde{X}= \text{Bl}(X)\_{x\_1,..., x\_9}$ is the blowup of the pencil $P$
in these $ 9 $ points.
Question: how can I calculate "by hand" the blow-up scheme $\widetilde{X}$ of $X$ along these nine points?
I know how to construct basically a blowup of $X$ along $Z \subset X$ corresponding to quasi-coherent sheaf of ideals
$ \mathcal{J} \subset \mathcal{O}\_ X $ *assuming !!!* I know the structure sheaf $\mathcal{O}\_ X $ of $X$ and sheaf of ideals
$ \mathcal{J}$.
In this case the most general way to construct the blowup is by
$$ \text{Bl}\_Z(X):=
\text{Proj}\_X (\bigoplus\_{n \ge 0} \mathcal{J}^n) $$
endowed with canonical projection $p: \text{Bl}\_Z(X) \to X$ and with exceptional divisor $p^{-1}(Z)$. Since this construction behaves well with respect
taking affine covers $(U\_i =\text{Spec} A\_i)\_{i \in I}$ it is also possible
construct it for these affine pieces $ U\_i= \text{Spec} A\_i $, ideal $J $ of $A\_i$ and
closed $Z = \text{Spec}(A\_i/J) = U\_i \cap Z
\subset U\_i$ separately by setting
$$ \text{Bl}\_Z(U\_i):= \text{Proj}(\bigoplus\_{n \ge 0} I^n) $$
and gluing after that these pieces together.
Moreover, if we consider $X
\subset \mathbb{P}^n$ as sitting as closed subscheme inside
a projective space and want to blowup $X$ along
a subscheme $Y \subset X$ *assuming* that I know the associated
sheaf ideals $\mathcal{I} \subset
\mathcal{J} $ of $X$ respectively $Y$, then it can be done even more concrete:
As before since blowups behave well with respect to affine covers we can work locally
by passing to any affine chart $\mathbb{A}^n$ of $\mathbb{P}^n$ and glue at the end
the pieces together. Therefore we land in affine the situation where
$ \operatorname{Spec}(A/J)=Y \cap \mathbb{A}^n
\subset \operatorname{Spec}(A)=X \cap \mathbb{A}^n \subset \mathbb{A}^n$
where $A= \mathbb{C}[x\_1,.., x\_n]/I$ and and there exist an finitely generated ideal
$\widetilde{J} = (g\_1,..., g\_m) \subset \mathbb{C}[x\_1,.., x\_n]$ such that
$I \subset \widetilde{J} $ with
$J = \widetilde{J}/I$.
We ignore $X$ for the moment define for each $g\_i$ the ring
$\mathbb{C}[x\_1,.., x\_n, g\_1/g\_i,..., g\_m/g\_i]$ which becomes later the $i$-th
affine chart of the blowup. Since this ring contains $\mathbb{C}[x\_1,.., x\_n]$
the ideal $\widetilde{I}:= I \cdot \mathbb{C}[x\_1,.., x\_n, g\_1/g\_i,..., g\_m/g\_i]$
makes sense and we can form the quotient
$B\_i:=\mathbb{C}[x\_1,.., x\_n, g\_1/g\_i,..., g\_m/g\_i]/\widetilde{I}$.
We glue the $B\_i$'s together and we obtain the blowup of the affine piece
$X \cap \mathbb{A}^n =\operatorname{Spec}(A) $ along $Y$. Then
be glue second time the obtained blowups obtain over the affine charts
of $ \mathbb{P}^n$ toghether.
To carry out that all 'by hand' is of course very laborious, but at the end of
the day we have constucted explicitly a blowup of $X$ along $Y$.
*Back* to the blowup of the pencil
$X= \bigcup\_{[\lambda: \mu] \in \mathbb{P}^1} V(\lambda F\_1 + \mu F\_2)
\subset \mathbb{P}^2$ along nine points $Y:= \{p\_1,..., p\_9 \}$
in Miranda's notes. I not know how to calculate here the blowup
of $X$ along $Y$ explicitly, since the pencil $X \subset \mathbb{P}^2$ is not described explicitly by an
associated ideal sheaf determining the structure sheaf $\mathcal{O}\_X$ which
is neccessary to know for the calculation of the blowup by all method above.
So seemingly my aproach to calculate the blowup on affine charts and
then patch the pieces together cannot applied here due to lack of knowledge of the ideal sheaf $\mathcal{I}$ associated to $X \subset \mathbb{P}^2$ even if I know the ideal sheaf $\mathcal{J}$ associated to the nine points as subscheme in $\mathbb{P}^2$ along which $X$ is blowed up. Indeed for the blowup constructions above I need to know the structure sheaf of $X$ and $\mathcal{J}$ as ideal sheaf with respect $X$, not $\mathbb{P}^2$.
Therefore I'm stuck at this point and my question is how to calculate the blowup "by hand" of this pencil along
the nince points and if this strategy could be generalized to arbitrary
pencils?
note I posted [identical question](https://math.stackexchange.com/questions/4563963/by-hand-calculation-of-a-blowup-of-a-pencil-of-curves) a week ago on MSE without getting any resonance.
| https://mathoverflow.net/users/108274 | Calculate blowup of a pencil of cubics "by hand" | If the two cubics $C\_1=V(F\_1)$ and $C\_2=V(F\_2)$ do not share a common component, then the ideal $I = (F\_1,F\_2)$ defines a $0$-dimensional subscheme $V(I)\subset \mathbb{P}^2$ length 9, which is a complete intersection of codimension 2.
In the fancy language above, the graded ring $\bigoplus\_{n\geq0}I^n$ is generated as an algebra over $\mathcal{O}\_{\mathbb{P}^2}$ by two elements $\xi\_1,\xi\_2$ in degree $1$, corresponding to the two equations $F\_1,F\_2$ generating $I$. They satisfy one relation $\xi\_1F\_2-\xi\_2F\_1=0$ corresponding to the syzyzy that holds between them.
Thus the blowup of $Z:=V(I)\subset \mathbb P^2$ is explicitly given by the projective variety
$$ \overline X = V(\xi\_1F\_2 - \xi\_2F\_1) \subset \mathbb P^1\_{\xi\_1,\xi\_2}\times \mathbb P^2\_{x,y,z}$$
and the morphism $\overline \pi\colon \overline X\to \mathbb{P}^1$ that you want to consider is simply the projection onto the first factor. (The blowup itself is given by projection onto the second factor.)
Unfortunately, if $Z$ is not reduced then this blowup $\overline X$ is going to be rather singular, so at this point you probably want to replace $\overline X$ with its minimal resolution $\mu\colon X\to \overline X$ and consider the induced morphism $\pi\colon X\to \mathbb P^1$.
I want to note that this construction is probably not exactly the approach Miranda had in mind. Judging from what he wrote he considers an iterative construction:
1. first blow up the (reduced) intersection points of $C\_1\cap C\_2$
2. look at the strict transform of $C\_1$ and $C\_2$ under this blowup and then blow up their intersection points,
3. repeat until $C\_1$ and $C\_2$ are disjoint.
If you think hard enough about it, then I think it will follow from (a) the universal property of blowing up, and (b) the existence of minimal resolutions of surfaces, that this construction gives the same surface as $X$ above. The advantage of the first approach is it is obvious how to get the morphism to $\mathbb P^1$. The advantage of the second approach is that it gives a smooth surface directly, without having to go through some desingularisation process at the end.
Lastly, by some abstract nonsense there should exist *some* sheaf of ideals $\mathcal J$ on $\mathbb P^2$ such that the morphism $\pi\colon X\to \mathbb{P}^2$ is given by blowing up $\mathcal J$. Perhaps you had in mind to construct $\pi\colon X\to \mathbb{P}^2$ all in one go by working out how to write down this $\mathcal J$. I would think this is highly unlikely to be a pleasant calculation.
| 2 | https://mathoverflow.net/users/104695 | 433616 | 175,390 |
https://mathoverflow.net/questions/433603 | 4 | Stallings' celebrated Fibration Theorem states that if a closed irreducible $3$-manifold $M$ admits a short exact sequence
\begin{equation}
1 \to N \to \pi\_1(M) \to \mathbb{Z} \to 1,
\end{equation}
where $N$ is finitely generated, then $M$ fibers over $S^1$.
My question is that whether it is possible to give the exact form of $M$ if we already know that $N$ is a surface group. More precisely, assume there is a short exact sequence
\begin{equation}
1 \to \pi\_1(S) \to \pi\_1(M) \to \mathbb{Z} \to 1,
\end{equation}
where $S$ is a closed surface. Let $t \in \pi\_1(M)$ be a preimage of $1 \in \mathbb{Z}$. Conjugation with $t$ induces an automorphism $\varphi$ of $\pi\_1(S)$ whose projection $[\varphi] \in Out(\pi\_1(S))$ in the outer automorphism group does not depend on the choice of $t$. By the Dehn-Nielsen-Baer Theorem the extended mapping class group $MCG^{\pm}(S)$ is isomorphic to $Out(\pi\_1(S))$. Let $f:S \to S$ be a homeomorphism so that $[f] \in MCG^{\pm}(S)$ corresponds to $[\varphi] \in Out(\pi\_1(S))$ under the Dehn-Nielsen-Baer isomorphism. My question is whether $M$ is diffeomorphic to the mapping torus $T\_f$ of $f:S \to S$. If this is true, does it just follow from Stallings' proof? Or, in case it is known but requires more work, is there a good reference for it?
More concretely, I am interested in the following situation (which might simplify things). If $M$ is a closed hyperbolic $3$-manifold so that $\pi\_1(M)$ is (abstractly) isomorphic to the fundamental group $\pi\_1(T\_f)$ of a mapping torus, then $M$ is diffeomorphic to $T\_f$. This is, of course, an easy consequence of the Geometrization Conjecture. However, I would like to know whether there is a proof of this fact that does *not* depend on the solution of the Geometrization Conjecture *nor* on Thurston's mapping torus theorem.
| https://mathoverflow.net/users/493880 | Stallings' fibration theorem - Explicit description | Your question ("is $M$ homeomorphic to $T\_f$?") is answered in the affirmative by Theorem 2 of Stallings' paper \*On fibering certain 3-manifolds". You will also need his Theorem 1. Here are the statements (slightly simplified).
**Theorem 1**: Suppose that $M$ is a compact connected three-manifold. Suppose that $\Gamma$ is a finitely generated normal subgroup of $\pi\_1(M)$ whose quotient group is $\mathbb{Z}$. Then there is a surface $F$ properly embedded in $M$ so that $\Gamma = \pi\_1(F)$.
**Theorem 2**: With hypotheses as in Theorem 1. Suppose that $M$ is irreducible. Suppose that $\Gamma$ is not $\mathbb{Z}/2\mathbb{Z}$. Then $M$ is a surface bundle over the circle, with $F$ isotopic to a fibre.
That is, your hypothesis on the short exact sequence (plus Theorem 1) gives the surface $F$. Your hypothesis that the manifold $M$ is hyperbolic then gives the additional hypotheses of Theorem 2.
Note that Stallings does not cite Waldhausen. I suppose that this is because his situation is a very very simple case of a Haken hierarchy. Once you have $F$ in your hands (and all the group theory hypotheses), it is "easy" to show that $M - F$ is homeomorphic to a product.
| 3 | https://mathoverflow.net/users/1650 | 433619 | 175,391 |
https://mathoverflow.net/questions/432563 | 1 | I am coming from [this question](https://mathoverflow.net/questions/336149/about-the-cartans-moving-frame-method?noredirect=1&lq=1), which has not being completely answered but I think is very interesting.
In several works ([Chern], [Griffiths] and [Clelland]) the Maurer-Cartan form for $E(n)$ is worked out in the following manner. They consider maps from $G=E(n)$ to $\mathbb{R}^n$, $x, e\_1,\ldots, e\_n$, and express their differentials in terms of the frame in which we are. But for me that doesn't seem natural because is something very particular of this example: the frame itself can be described in terms of the objects it describe. I consider more natural the general approach: the group $E(n)$ can be seen like a matrix group of a special type, that one with elements of the form
$$
\begin{pmatrix}
A & v\\
0 & 1\\
\end{pmatrix}
$$
with $A\in O(n)$ and $v\in \mathbb{R}^n$. And now you only have to apply the formula for MC form for a matrix group, $\theta=g^{-1}dg$, obtaining the same 1-forms.
>
> **Question 1 (solved)**
>
> Is this true for every Lie group of this type? That is, whenever we have a group $G\approx \mathbb{R}^n \rtimes H$ it can be seen as a subgroup of $GL(n+1)$ as above (see [this QA in MSE](https://math.stackexchange.com/questions/4554039/semidirect-product-and-matrix-groups/4554089#4554089)) and we can interpret the columns as vectors in the homogeneous space $G/H\approx \mathbb{R}^n$. Then, does the Maurer-Cartan form tell us the variation of these vectors expressed in the current frame?
>
>
>
Back to the case of $E(2)$, for simplicity. The MC form is
$$
\theta=g^{-1}dg=\begin{pmatrix}
0&-d\theta&cos(\theta)da+sin(\theta)db\\
d\theta&0&-sin(\theta)da+cos(\theta)db\\
0&0&0&\\
\end{pmatrix}
$$
If we consider the basis of $\mathfrak{e}(2)$ given by
$$
B=\left\{
\begin{pmatrix}
0&0&1\\
0&0&0\\
0&0&0
\end{pmatrix},
\begin{pmatrix}
0&0&0&\\
0&0&1\\
0&0&0&\\
\end{pmatrix},
\begin{pmatrix}
0&-1&0&\\
1&0&0\\
0&0&0&\\
\end{pmatrix}
\right\}\equiv
$$
$$
\equiv\{\partial\_a|\_e,\partial\_b|\_e,\partial\_{\theta}|\_e\}
$$
we have
$$
\theta=\mu\_1 \otimes\partial\_a|\_e+\mu\_2 \otimes\partial\_b|\_e +\mu\_3\otimes \partial{\theta}|\_e
$$
with
$$
\mu\_1=cos(\theta)da+sin(\theta)db
$$
$$
\mu\_2=-sin(\theta)da+cos(\theta)db
$$
$$
\mu\_3=d\theta
$$
In this case the Maurer-Cartan form has "two parts": $\mu\_1, \mu\_2$ on the one hand, and $\mu\_3$ on the other hand. I think that $(\mu\_1, \mu\_2)$ corresponds to the canonical solder form and $\mu\_3$ is the connection form of the Levi-Civita connection.
>
> **Question 2**
>
> Why is this the Levi-Civita connection? What relationship does it have (if any) with the group reduction of $GL(2)$ to $O(2)$ by means of the standard metric?
>
>
>
I have an intuition about some relation but I cannot grasp what it is... I know that the frame bundle for $\mathbb{R}^2$ is $\mathbb{R} \rtimes GL(2)$ and that the standard metric let us reduce the structure group of this principal bundle to $O(2)$...
**References**
[Chern]: Chapter 6 of S.S. Chern's book "Lectures on differential geometry"
[Griffiths]: On Cartan's method of Lie groups and moving frames as applied to uniqueness and existence questions in differential geometry
[Clelland]: From Frenet to Cartan: The Method of Moving Frames
| https://mathoverflow.net/users/129995 | Maurer-Cartan form and Levi-Civita connection | I have been working in question 2 and I think I have a good explanation. At the end is a triviality, but that's what (almost) always happens in math when you understand something.
I am going to discuss here the Euclidean plane from two different perspectives.
Euclidean plane
===============
From the point of view of classical differential geometry
---------------------------------------------------------
The Euclidean plane is the manifold $M=\mathbb{R}^2$, with coordinates $(x\_1,x\_2)$, together with the natural [Riemannian metric](https://ajpancollantes.github.io/%E2%99%BE%EF%B8%8F%20CONCEPTS/GEOMETRY/Riemannian%20metric.md.html)
$$
g=dx\_1\otimes dx\_1+dx\_2\otimes dx\_2.
$$
It is, therefore, a Riemannian manifold.
Seen as a plain manifold we can consider that $M$ is endowed with a natural linear connection $\nabla$, such that $\nabla\_{\partial\_{x\_i}}\partial\_{x\_j}=0\partial\_{x\_1}+0\partial\_{x\_2}$. The [induced](https://ajpancollantes.github.io/%E2%99%BE%EF%B8%8F%20CONCEPTS/GEOMETRY/associated%20connection.md.html#Example) principal connection on the frame bundle is given by the 1-form
$$
\omega=
\begin{pmatrix}
c\_{11}&c\_{12}\\
c\_{21}&c\_{22}\\
\end{pmatrix}^{-1}\cdot\begin{pmatrix}
dc\_{11}&dc\_{12}\\
dc\_{21}&dc\_{22}\\
\end{pmatrix}\in \Omega(\mathbb R^2,\mathfrak{gl}(2))
$$
at
$$
f=
\begin{pmatrix}
c\_{11} & c\_{12}& x\_1\\
c\_{21} & c\_{22}& x\_2\\
0 & 0& 1\\
\end{pmatrix}\in FM
$$
[Here](https://ajpancollantes.github.io/%E2%99%BE%EF%B8%8F%20CONCEPTS/GEOMETRY/associated%20connection.md.html#Example) it is shown how to construct $\omega$ from an arbitrary $\nabla$.
On the other hand, if we think of $M$ as a Riemannian manifold we can consider the [Levi-Civita connection](https://ajpancollantes.github.io/%E2%99%BE%EF%B8%8F%20CONCEPTS/GEOMETRY/Levi-Civita%20connection.md.html) $\nabla\_{LC}$. Since the metric is constant, the covariant derivative $\nabla\_{LC}$ coincides with the natural covariant derivative $\nabla$, so it induces the same connection $\omega$ on $FM$.
But the metric $g$ also specifies a orthonormal frame bundle $OM$ ([see here why](https://ajpancollantes.github.io/%E2%99%BE%EF%B8%8F%20CONCEPTS/orthonormal%20frame%20bundle.md.html)). The elements of this principal bundle are
$$
f=
\begin{pmatrix}
c & -\sqrt{1-c^2}& x\_1\\
\sqrt{1-c^2} & c& x\_2\\
0 & 0& 1\\
\end{pmatrix}\in OM
$$
with $c\in [-1,1]$.
Since
$$
\omega|\_{OM}=\begin{pmatrix}
c&\sqrt{1-c^2}\\
-\sqrt{1-c^2}&c\\
\end{pmatrix}\cdot\begin{pmatrix}
dc&\frac{c}{\sqrt{1-c^2}}dc\\
\frac{-c}{\sqrt{1-c^2}}dc&dc\\
\end{pmatrix}=
$$
$$
=\begin{pmatrix}
0&\frac{dc}{\sqrt{1-c^2}}\\
-\frac{dc}{\sqrt{1-c^2}}&0\\
\end{pmatrix} \in \Omega(\mathbb R^2,\mathfrak{o}(2)),
$$
according to Proposition 4.7. in [vicenteBundles](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&cad=rja&uact=8&ved=2ahUKEwj-w6a2xYj7AhVJ3hoKHVDpAMQQFnoECAsQAQ&url=http%3A%2F%2Fwww.mat.ucm.es%2F%7Evmunozve%2FNotas-fibrados.pdf&usg=AOvVaw2EmcybsLZEb16epRLqS8z8), this connection can be reduced to a connection on the orthonormal frame bundle determined by the metric $g$.
If we parameterize this principal bundle $OM$ with
$$
(x\_1,x\_2,\theta)\mapsto \begin{pmatrix}
\cos(\theta) & -\sin(\theta)& x\_1\\
\sin(\theta) & \cos(\theta)& x\_2\\
0 & 0& 1\\
\end{pmatrix}
$$
we obtain the more famous expression for $\omega$:
$$
\begin{pmatrix}
0&d\theta\\
-d\theta&0\\
\end{pmatrix}
$$
Remember: this 1-form **tells us how much the bases at $f$ and $f'$ "fail to be constant"** when we pass from the frame $f$ to a nearby frame $f'$, but expressing this mistake with respect to the frame $f$ itself.
From the point of view of Cartan geometry
-----------------------------------------
The Euclidean plane is a Cartan geometry modeled over $(E(2),O(2))$ , indeed **is** the Klein geometry $(E(2),O(2))$. Moreover, it is a [reductive Klein geometry](https://ajpancollantes.github.io/%E2%99%BE%EF%B8%8F%20CONCEPTS/reductive%20Cartan%20geometry.md.html#Interpretation) since
$$
\mathfrak e(2)=\left\{\begin{pmatrix}
C & v\\
0 & 0\\
\end{pmatrix}
:C\in \mathfrak{o}(2), v\in \mathbb R^2\right\}=\mathfrak o(2)\oplus \mathfrak p
$$
We have a natural choice for $\mathfrak p$
$$
\mathfrak{p}=\left\{\begin{pmatrix}
0 & p\\
0 & 0\\
\end{pmatrix}
:p\in \mathbb R^2\right\}.
$$
With this in mind, remember that the Maurer-Cartan form describes all possible "infinitesimal displacements" of the frame $f$, but from the point of view of the frame $f$ itself. That is, if we pass from the frame $f$ to another frame $f'$, the Maurer-Cartan form at $f$ applied to the "vector" $\vec{ff'}=(dx\_1,dx\_2,d\theta)$ is a packet of information
$$
A=\begin{pmatrix}
0&-d\theta&cos(\theta)dx\_1+sin(\theta)dx\_2\\
d\theta&0&-sin(\theta)dx\_1+cos(\theta)dx\_2\\
0&0&0&\\
\end{pmatrix}\in \mathfrak e(2)
$$
Here is encoded, on the one hand, how much have we moved the base point of $f$ to the base point of $f'$ and, on the other, how much have we changed the basis itself. The natural choice of $\mathfrak p$ let us think that the information about the change of base point is in the $v$ part (the projection of the Maurer-Cartan form on $\mathfrak p$), and so the projection of the Maurer-Cartan form on $\mathfrak o(2)$, $\begin{pmatrix}0&d\theta\\-d\theta&0\\\end{pmatrix}$, tell us how much has the basis changed. That is, the same as the connection 1-form of the connection on the orthonormal bundle induced by the metric $g$ (which is the Levi-Civita connection).
**To summarize:**
In the orthonormal frame bundle induced by the metric $g$ we consider a displacement from a frame $f=\begin{pmatrix}C & p\\0 & 1\\\end{pmatrix}$ to a frame $f'=\begin{pmatrix}C' & p'\\0 & 1\\\end{pmatrix}$.
The principal connection $\omega$ induced by the Levi-Civita connection measures the change from $C$ to $C'$ as an element of $\mathfrak o(2)$.
The Cartan connection (Maurer-Cartan form) measures the change from $f$ to $f'$ as an element of $\mathfrak e(2)$. This change can be decomposed like the union of a change from $C$ to $C'$ and a change from $p$ to $p'$. This is reflected in the fact that $\mathfrak e(2)=\mathfrak o(2)\oplus \mathfrak p$. If we focus on the change from $C$ to $C'$ we have the principal connection $\omega$.
| 0 | https://mathoverflow.net/users/129995 | 433620 | 175,392 |
https://mathoverflow.net/questions/433611 | 3 | Some years ago, I found a paper with all the formulas for the balls into bins problem when the "areas" (i.e., probabilities to capture a ball) of the bins are all different. However, the formulas looked quite involved and cumbersome in the general case. Now, I am instead trying to solve an elementary version of the balls into bins problem with a non-uniform probability of capturing balls, which I firmly believe has a simple and clean answer.
---
We are given $n$ bins $b\_1, b\_2, \ldots, b\_n$. In a sequential fashion, at each time step, one ball is placed into bin $b\_i$ with probability $p\_i$, where $\sum\_{i=1}^{n} p\_i=1$, and $p\_i=\alpha i p\_1$ for a given constant $\alpha\ge 1$ for all integer $i\in\{2,3\ldots,n\}$.
---
**Question**: What is expected number $m$ of balls that we need to throw to have that all $n$ bins contains at least one ball?
---
---
***Edit:*** *For any given fixed value of $n\in\mathbb{N}$, as $\alpha$ grows, the required expected number of balls $m p\_1$ cannot increase. For the minimum value of $\alpha$ in the problem, which is $1$ ($\alpha\ge 1$), it seems that $m=\frac{\beta}{p\_1}$ for some constant $\beta$. Since $\frac{1}{p\_1}$ balls are always necessary to make bin $b\_1$ non-empty, I guess that there is a constant $\gamma(\alpha)\in [1,\beta]$ depending on $\alpha$ such that $\frac{\gamma(\alpha)}{p\_1}$ is the expected number of balls required, but I do not know how $\gamma$ varies with $\alpha$. Anyway for $n\to\infty$ we always have $m\in\Theta\left(\frac{1}{p\_1}\right)$ (which is equal to $\Theta(n^2)$ for $\alpha=1$).*
| https://mathoverflow.net/users/115803 | A linearly distributed version of the balls into bins problem | From the referenced paper, I am writing in terms of their variables, $k$ is the number of bins or type of coupons:
Let $n\_1$ be the time where the last of the missing events is observed. Let $n\_2$ be the time where the second last of the missing events is observed, etc. until $n\_k$:
Define
$$
S\_kf(p\_1,\ldots,p\_m)=\sum\_{1\leq i\_1<i\_2<\cdots<i\_k\leq m}f(p\_{i\_1},\ldots,p\_{i\_m}),
$$
which means that the function is $f$ is to be formed for all $\binom{k}{m}$ combinations of the $k$ indices indicated in the subscript of $S$ and that all these terms are to be added.
For example
$$
S\_3 \frac{1}{p\_1+p\_2}= \frac{1}{p\_1+p\_2}+\frac{1}{p\_2+p\_3}+\frac{1}{p\_1+p\_3}
$$
The general equation given below is still not really clean, but you can simplify using your constraint of linear probabilities. The paper has the distribution and the cumulative distribution as well. However we have the expectation which follows in inclusion-exclusion manner thus
$$
\mathbb{E}(n\_m)=S\_k
\left\{
\binom{m-1}{m-1} \frac{1}{p\_1+p\_2+\cdots+p\_m}+\right.
$$
$$
-\binom{m}{m-1}\frac{1}{p\_1+p\_2+\cdots+p\_{m+1}}
+\cdots
$$
$$
\left.
+(-1)^{k-m} \binom{k-1}{m-1}\frac{1}{p\_1+p\_2+\cdots+p\_k}.
\right\}
$$
| 4 | https://mathoverflow.net/users/17773 | 433621 | 175,393 |
https://mathoverflow.net/questions/433418 | 7 | Let $X$ be a compact complex manifold, $L$ be a holomorphic line bundle on $X$, then the exponential exact sequence $0\to \mathbb Z\hookrightarrow \mathcal O\to \mathcal O^\*\to 0$ induces the map $c:H^1(X,\mathcal O^\*)\to H^2(X,\mathbb Z)$, it is well-known that the line bundle $L$ can be seen as an element in $H^1(X,\mathcal O^\*)$, the image $c(L)\in H^2(X,\mathbb Z)$ is called the Chern class of $L$.
According to Atiyah's [paper](https://www.ams.org/journals/tran/1957-085-01/S0002-9947-1957-0086359-5/S0002-9947-1957-0086359-5.pdf) in 1957, p.196, the exact sequence $0\to \mathbb C\hookrightarrow \mathcal O\stackrel{d}\to \Omega^1\to 0$ together with the exponential exact sequence above induce the map $\mathcal O^\*\to \Omega^1:f\mapsto \frac{1}{2\pi i}d\text{log}f$, which induces the map $\sigma:H^1(X,\mathcal O^\*)\to H^1(X,\Omega)$, we call the image $\sigma(L)\in H^1(X,\Omega^1)$ the Atiyah class of the line bundle $L$.
In Atiyah's paper, the author assumed $X$ to be a compact Kähler manifold, then he concludes that the Atiyah class equals to Chern class, here my question is: can this condition be weakened a bit? For example, $X$ is a $\partial\bar\partial$-manifold or the Frölicher spectral sequence degenerates at $E\_1$? Or equivalently, what's the sufficient and necessary condition of the Atiyah class coincides with the Chern class?
| https://mathoverflow.net/users/99826 | When Atiyah class and Chern class coincide? | $\def\ZZ{\mathbb{Z}}\def\CC{\mathbb{C}}\def\cO{\mathcal{O}}$To spell out my comment a little more, let $Z^1$ be the sheaf of $\partial$-closed holomorphic $(1,0)$-forms. Since "holomorphic" means $\overline{\partial}$-closed, this can also be described as the space of closed $(1,0)$forms. Then we have a commutative diagram, with exact rows and columns:
$$\begin{matrix}
&& && 0 && 1 && \\
&& && \downarrow && \downarrow && \\
0 &\longrightarrow& \mathbb{Z} &\overset{2 \pi i}{\longrightarrow}& \mathbb{C} &\overset{\exp}{\longrightarrow}& \mathbb{C}^{\times} &\longrightarrow& 1 \\
& &=& &\downarrow& &\downarrow& \\
0 &\longrightarrow& \mathbb{Z} &\overset{2 \pi i}{\longrightarrow}& \cO &\overset{\exp}{\longrightarrow}& \cO^{\times} &\longrightarrow& 1 \\
& && &\phantom{\partial} \downarrow \partial& &\phantom{\partial \log}\downarrow \partial \log& \\
&& && Z^1 &=& Z^1 && \\
&& && \downarrow && \downarrow && \\
&& && 0 && 0 && \\
\end{matrix}$$
Using each of the $4$ short exact sequences, we have a diagram
$$
\begin{matrix}
H^1(\cO^{\times}) &\longrightarrow& H^2(\ZZ) \\
\downarrow && \downarrow \\
H^1(Z^1) &\longrightarrow& H^2(\CC) \\
\end{matrix}$$
and, with care, one can check that it commutes.
We also have a short exact sequence $0 \to Z^1 \to \Omega^1 \overset{\partial}{\longrightarrow} Z^2 \to 0$, where $Z^2$ is the closed $(2,0)$-forms. So we can extend this diagram to
$$
\begin{matrix}
H^1(\cO^{\times}) &\longrightarrow& H^2(\ZZ) \\
\downarrow && \downarrow \\
H^1(Z^1) &\longrightarrow& H^2(\CC) \\
\downarrow && \\
H^1(\Omega^1) && \\
\end{matrix}$$
The Atiyah class is the image of a class from $H^1(\cO^{\times})$ in $H^1(\Omega^1)$; the Chern class is the image in $H^2(\CC)$.
This much is true without assuming anything about your complex manifold.
If you want to consider the Atiyah class and the Chern class to be "the same", then it seem like you want to ask is "when does it make sense to think of $H^1(\Omega^1)$ as a subspace of $H^2(\CC)$?" If your spectral sequence degenerates at $E^1$, then the map $H^1(Z^1) \to H^2(\CC)$ is an injection and $H^1(Z^1) \to H^1(\Omega^1)$ is a surjection, so $H^1(\Omega^1)$ is a subquotient of $H^2(\CC)$. But I don't know of anything which lets you identify $H^1(\Omega^1)$ with a subspace of $H^2(\CC)$ without Hodge theory. (With Hodge theory, on a compact Kahler manifold, $H^1(Z^1)$ is the piece $H^{20} \oplus H^{11}$ in the Hodge filtration, and the maps $H^1(Z^1) \to H^2(\CC)=H^{20} \oplus H^{11} \oplus H^{02}$ and $H^1(Z^1) \to H^1(\Omega^1)=H^{11}$ are the obvious ones.)
| 4 | https://mathoverflow.net/users/297 | 433622 | 175,394 |
https://mathoverflow.net/questions/433486 | 10 | For the purposes of this question, a *$T$-interpretation with arity $n$* will be a tuple $\Phi=(\delta,\eta,F)$ where
* $\delta$ and $\eta$ are individual formulas of arity $n$ and $2n$ respectively,
* $T$ proves that $\eta$ defines an equivalence relation on the set picked out by $\delta$, and
* $F$ is a (possibly infinite) set of formulas each of which has arity a multiple of $n$ such that for each $\varphi(x\_1,...,x\_{kn})\in F$ we have that $T$ proves that $\varphi$ is $\eta$-invariant on $\delta$.
Say that $\Phi=(\delta,\eta, F)$ is **sharp** in $T$ iff - letting $\Sigma$ be a language with a $k$-ary relation symbol corresponding to each $kn$-ary formula in $F$ - the following "substitution" principle holds:
>
> If $\mathcal{A}$ is a $\Sigma$-structure satisfying (the translation to $\Sigma$ of) every sentence $T$ proves about $\Phi$, then there is some $\mathcal{B}\models T$ with $\Phi^\mathcal{B}\cong\mathcal{A}$.
>
>
>
It's easy to show that not all interpretations are sharp. However, I don't have a good sense of how rare sharpness is, especially in "rich" theories like $T=\mathsf{ZFC}$ or $T=\mathsf{PA}$. The following seems like a natural question to get a better understanding of this:
>
> Suppose $\Phi=(\delta,\eta,F)$ is a $\mathsf{ZFC}$-interpretation. Must there be a set of sentences $G\supseteq F$ and a consistent extension $T\supseteq\mathsf{ZFC}$ such that $(\delta,\eta,G)$ is sharp in $T$?
>
>
>
I suspect the answer is **negative**, even if we restrict attention throughout to *countable* models, but I don't see how to prove this. (I also vaguely recall a paper on this topic from the 80s, maybe by Friedman or Mostowski, but I can't track it down.)
| https://mathoverflow.net/users/8133 | Can we always "sharpen" interpretations? | If you allow to add to an interpretation infinitely many predicate symbols (infinite $G\setminus F$) then any interpretation in $\mathsf{ZFC}$ could be extended to a sharp interpretation in $\mathsf{ZFC}+L=V$. So you original question has a positive answer.
However, if you restrict your attention to extensions by finitely many predicate symbols then it wouldn't be possible to extend some interpetations in $\mathsf{ZFC}$ to sharp interpretations. But actually all interpretations in $\mathsf{PA}$ admit finite extension to a sharp interpretation in some extension of $\mathsf{PA}$.
Now let me sketch the proofs of the claims that I mentioned.
Since in $\mathsf{ZFC}+V=L$ any interpretation is definably isomorphic to a one-dimensional interpretation with absolute equality, without loss of generality I'll assume that I am given a one-dimensional iterpretation $\Phi$ with absolute equality in $\mathsf{ZFC}+V=L$. Let the interpretation $\Phi'$ extend $\Phi$ by all possible relations on the domain of interpretation that are definable in set-theoretic language. Any model $\mathcal{M}$ satisfying all sentences whose $\Phi'$-translations are $\mathsf{ZFC}+V=L$-provable, contains a complete first-order theory in the language with parameters from $\mathcal{M}$ of a set-theoretic universe that should lead to $\mathcal{M}$ as the result of application of $\Phi'$. Since $\mathsf{ZFC}+V=L$ admits definable Skolem functions, we could recover from $\mathcal{M}$ a model $\mathcal{A}$ of $\mathsf{ZFC}+L=V$ that leads to $\Phi'^\mathcal{A}$ being an isomorphic copy of $\mathcal{M}$, by constructing it as a model consisting exactly of the definable sets with parameters from $\mathcal{M}$.
If we look just on the extensions of interpretations by finitely many relations, then the point is that set-size interpretations in non-$\omega$ models of $\mathsf{ZFC}$ always lead to recursively-saturated models. And for example, it is easy to see that for any finite extension $\Phi'$ of the standard interpretation of arithmetic in $\mathsf{ZFC}$ and any consistent $T\supseteq \mathsf{ZFC}$, we always will have non-recursively saturated, non-$\omega$ model of all sentences whose $\Phi'$-translations are $T$-provable.
For interpretations in $\mathsf{PA}$ we could exploit the dichotomy that in the standard model the domain of the interpretation is either finite or definably isomorphic to the set of naturals. And hence for a given interpretation $\Phi$ in $\mathsf{PA}$ provably in an appropriate extension $T$ of $\mathsf{PA}$, we have a parameter-free definable bijection between the domain of $\Phi$ and either a standard natural $n$ or the class of all elements. In the case of the finite domain there is nothing to prove. In the case of the domain isomorphic to all naturals, we simply expand the interpretation by the addition and multiplication.
| 5 | https://mathoverflow.net/users/36385 | 433625 | 175,395 |
https://mathoverflow.net/questions/433516 | 6 | Let $D$ be a small category. Does the category of diagrams $\mathsf{Top}^{D^{\text{op}}}$ have a classifier of (strong?) subobjects? I tried following the "sieve construction" for the category of presheaves, but I don't see what topology to put on the set of sieves on an object in $D$ (or perhaps this won't work anyway).
I'm also wondering whether the existence of (strong?) subobject classifier has anything at all to do with the existence of a (strong) subobject classifier in $\mathsf{Top}$, or whether this is more or less unrelated.
| https://mathoverflow.net/users/78650 | Subobject classifier in $\mathsf{Top}^{D^{\text{op}}}$? | If any of these categories had a subobject classifier, every monomorphism would be regular, so that's not happening.
The indiscrete two-point space is a strong subobject classifier in $\mathsf{Top}.$ Similarly, the subobject classifier for $\mathsf{Set}^{D^{\mathrm{op}}},$ equipped with the indiscrete topology on its values, is a strong subobject classifier for $\mathsf{Top}^{D^{\mathrm{op}}}.$ This is because the forgetful functor to presheaves induces an isomorphism between strong subobjects of a topological presheaf and arbitrary subobjects of a set-valued presheaf.
| 5 | https://mathoverflow.net/users/43000 | 433634 | 175,399 |
https://mathoverflow.net/questions/433652 | 0 | Consider a product of projective varieties $X\times Y$ and two cohernet sheaves $\mathcal{F},\mathcal{G}$ such that
$$\mathcal{F}|\_{x\times Y}\cong\mathcal{G}|\_{x\times Y}$$
for any $x\in X$. Thanks to user 40297, we know that in general we do not have $\mathcal{F}\cong q^\*\mathcal{M}\otimes\mathcal{G}$ for some line bundle $\mathcal{M}$. Now I wonder, will we have $$q^\*\mathcal{M}\otimes\mathcal{F}\cong q^\*\mathcal{N}\otimes\mathcal{G}$$
for some locally free sheaves $\mathcal{M},\mathcal{N}$ on $X$ of the same rank?
| https://mathoverflow.net/users/nan | Does $\mathcal{F}|_{x\times Y}\cong\mathcal{G}|_{x\times Y}\Rightarrow q^*\mathcal{M}\otimes\mathcal{F}\cong q^*\mathcal{N}\otimes\mathcal{G}$ | No. For instance, take $Y$ to be a point, $X = \mathbb{P^1}$, $F = \mathcal{O} \oplus \mathcal{O}(1)$, and $G = \mathcal{O} \oplus \mathcal{O}$.
| 2 | https://mathoverflow.net/users/4428 | 433657 | 175,406 |
https://mathoverflow.net/questions/429284 | 4 | Let $V$ be an $n$-dimensional vector space over a finite field $F$ (of order $q$). Denote by $\mathrm{AGL}(V)$ the group of invertible affine transformations of $V$; so $\mathrm{AGL}(V)$ consists of all mappings $x\mapsto Tx+v$ where $T\in GL(V)$ and $v\in V$.
Clearly, $\mathrm{AGL}(V)$ acts on the poset of all affine subspaces of $V$ where an affine subspace is, of course, a coset of a linear subspace. Let $P$ be the subposet of proper affine subspaces and let $\Delta(P)$ be the order complex of $P$. Then $\mathrm{AGL}(V)$ acts simplicially on $\Delta(P)$ and hence acts on the reduced homology of $\Delta(P)$.
Now $P$ is, it seems to me, the proper part of a geometric lattice. If we declare a subset of $V$ to be independent if its elements are affinely independent, then I believe this gives a matroid whose corresponding lattice of flats is the set of affine subspaces (and I guess the empty set). So $P$ should be the proper part of this and hence its reduced homology should be concentrated in the top dimension, which I guess is $n-1$ if I count correctly. Does anybody know any literature describing the decomposition of $\widetilde{H}\_{n-1}(\Delta(P),\mathbb C)$ into irreducible $\mathbb C\mathrm{AGL}(V)$-modules? Is it irreducible?
Of course, the easy case is when dimension $n$ of $V$ is $1$. Then $AGL(V)$ consists of all invertible $ax+b$-maps and the proper affine subspaces are points. So the reduced homology in dimension $0$ is the augmentation submodule of the permutation module of $F\rtimes F^\times$ acting on $F$. This action is doubly transitive, and so the augmentation submodule is irreducible of degree $q-1$ (and is the unique irreducible representation of degree greater than $1$).
| https://mathoverflow.net/users/15934 | Representation of $\mathrm{AGL}(V)$ on the homology of the poset of affine subspaces of $V$ | I suspect that
Solomon, Louis The affine group. I. Bruhat decomposition.
proves what you are looking for.
Let $A\_n(q)$ denote the poset of proper affine subspaces of $\mathbf{F}\_q^n$. The only non-vanishing reduced homology group
of $A\_n(q)$ is in degree $n-1$ and it is free abelian of rank $|E\_n(q)|$ where $E\_n(q) = -\widetilde{\chi}(A\_n(q))$ (minus the reduced Euler characteristic). This is because $A\_n(q)$ is the proper part of a geometric lattice with $\widehat 0 = \emptyset$ and $\widehat 1 = \mathbf{F}\_q^n$.
See papers by Folkman and Björner on homology of geometric lattices.
The poset $A\_0(q)$ is empty and $A\_1(q)$ is discrete on $q$ points so $E\_0(q)=1$ and $E\_1(q)=1-q$. The recursion
\begin{equation\*}
1 = \sum\_{0 \leq k \leq n} E\_k(q) \binom{n}{k}\_q q^{n-k}
\end{equation\*}
can be obtained from Corollary 3.8 in [Homotopy
equivalences between $p$-subgroup categories](https://web.math.ku.dk/%7Emoller/reprints/mgjmm2015.pdf) applied to
$A\_n(q)$. This formula features a Gaussian binomial coefficient. The
solution to the recursion is
\begin{equation\*}
E\_n(q) = \prod\_{1 \leq j \leq n} (1-q^j), \qquad n >0
\end{equation\*}
This shows that your representation and the one discussed by Solomon
are representations of the affine group of the same degree.
I suspect the two representations are identical. See also
Section 8.2 in Brown: The coset poset and Probabilistic Zeta function
of a Finite Group, Journal of Algebra 225 (2000) for the case where $q$ is a prime.
| 2 | https://mathoverflow.net/users/50509 | 433663 | 175,408 |
https://mathoverflow.net/questions/433671 | 0 | So, if we have system of differential equations obtained from Lagrange function, by means of Noether theoerem (if we know some one-parameter symmetry group), we can derive conserved quantity.
But how does it work with arbitrary system (not derived from Euler-Lagrange equations).
Does it mean that knowledge of symmetry give us somehow conserved quantity?
| https://mathoverflow.net/users/493428 | Conserved quantities | If the dynamics does not have a variational formulation a conservation law is not necessarily related to a symmetry. A more general approach than starting from the Euler-Lagrange equations is described in [Construction of conservation laws: how the direct method generalizes Noether’s theorem](https://personal.math.ubc.ca/~bluman/cyprus%20proceedings%20paper.pdf).
| 1 | https://mathoverflow.net/users/11260 | 433673 | 175,411 |
https://mathoverflow.net/questions/433594 | 4 | Let $K$ be a field, $K\_s$ its separable closure, $K$ $\subseteq$ $F$ $\subseteq$ $K\_s$ an extension with $[F:K]$ $=$ $n$, $R$ $\subseteq$ $K$ a Dedekind domain with quotient field $K$, $S$ the integral closure of $R$ in $F$, and $\mathfrak{p}$ a maximal ideal of $R$.
> Can we find a $d>n$ with $(d,n)=1$ and an extension $K$ $\subseteq$ $\widetilde{K}$ $\subseteq$ $K\_s$ with $[\widetilde{K}:K]$ $=$ $d$ such that $\widetilde{K}$ and $F$ are linearly disjoint over $K$ (in $K\_s$) and, if $\widetilde{R}$ denotes the integral closure of $R$ in $\widetilde{K}$, the integral closure of $R$ in $\widetilde{K}F$ is equal to $\widetilde{R}S$, and $\mathfrak{p}$ is inert in $\widetilde{K}$ (i.e. $\mathfrak{p}\widetilde{R}$ is a prime ideal of $\widetilde{R}$) ?
>
It is okay to assume that $K$ and $F$ are algebraic number fields, or even that $R$ $=$ $\mathbb{Z}$.
In the case where $K$ and $F$ are algebraic number fields, if $d$ $>$ $n$ is a prime number, any extension $K$ $\subseteq$ $\widetilde{K}$ of degree $d$ is linearly disjoint to $F$. And when the discriminants of $S$ and $\widetilde{R}$ over $R$ are relatively prime in $R$, the integral closure of $R$ in $\widetilde{K}F$ equals $\widetilde{R}S$. (Cf. Fröhlich and Taylor, *Algebraic Number Theory*, Ch. III, 2.13). If $\mathfrak{p}\cap\mathbb{Z}$ $=$ $p\mathbb{Z}$, by [Zsigmondy's theorem](https://en.wikipedia.org/wiki/Zsigmondy%27s_theorem) we can find a prime number $q$ such that $p$ has order $d$ in $\mathbb{F}\_q^{\times}$. Then $p$ splits as a product of $(q-1)/d$ primes of degree $d$ in $\mathbb{Q}(\zeta\_q)$. So if $d$ and $(q-1)/d$ are relatively prime, $p$ is inert in the unique subfield $D$ of $\mathbb{Q}(\zeta\_q)$ of degree $d$ over $\mathbb{Q}$. Starting out with a prime $d$ that is larger than the residue class degree of $\mathfrak{p}$ over $p$ as well, $\mathfrak{p}$ will be inert in the compositum $\widetilde{K}$ $=$ $KD$. And taking $d$ also larger than the absolute value of the discriminant of $F$ over $\mathbb{Q}$, $q$ is relatively prime to the discriminant of $S$ over $R$, since $q$ $>$ $d$. So this field $\widetilde{K}$ will do the trick.
However, there is no guarantee that $(d,(q-1)/d)$ $=$ $1$, and I would be much obliged to learn how such a construction could be made to work (without appealing to extended Riemann hypotheses).
The existence of such a $\widetilde{K}$ in the general (Dedekind domain) setup is used in [this Journal of Algebra paper](https://www.sciencedirect.com/science/article/pii/S0021869398980887) by Ilaria Del Corso and Roberto Dvornicich. In the paragraph following their Lemma 4, the authors optimistically "let $\widetilde{K}$ be an unramified extension of $K$ of degree $d$ such that $\mathfrak{p}$ is inert in $\widetilde{K}$" (with any $d$ that is large enough and relatively prime to $n$). They then show (Lemma 5) that $\widetilde{K}$ and $F$ are linearly disjoint and that the integral closure of $R$ in $\widetilde{K}F$ is $\widetilde{R}S$, using the fact that $\widetilde{K}$ is unramified over $K$. But, aside from $\mathbb{Q}$, various other fields $K$ exist that do not admit any non-trivial unramified extensions whatsoever (see [this answer](https://mathoverflow.net/questions/26491/is-there-a-ring-of-integers-except-for-z-such-that-every-extension-of-it-is-ram/26526#26526)).
| https://mathoverflow.net/users/31923 | Finding field extensions in which a given prime is inert | Thanks to Arno Fehm's observation above, the answer in the general case is **NO**.
In the number field case, $R$ $=$ $\mathfrak{O}\_K$ and $S$ $=$ $\mathfrak{O}\_F$ (or overrings thereof in $K$ resp. $F$), and we can argue as follows. Let $p\mathbb{Z}$ = $\mathfrak{p}\cap\mathbb{Z}$. Let $q\_1,\cdots,q\_s$ be the rational primes that ramify in $F$. If not on the list, include $p$ as well. Take a prime number $d$ $>$ $[F:\mathbb{Q}]$. Then $d$ $>$ $n$ and, with $q$ $:=$ $q\_i$ and $\mathfrak{q}$ an $\mathfrak{O}\_K$-prime lying over $q$, $d$ is larger than the residue class degree $r$ of $\mathfrak{q}$ over $q$. The smallest field containing $\mathbb{F}\_{q^d}$ and $\mathbb{F}\_{q^r}$ is $\mathbb{F}\_{q^{dr}}$, which is of degree $d$ over $\mathbb{F}\_{q^r}$. So when $\mathbb{Q}\subseteq D$ is any extension of degree $d$ in which all $q\_i$ are inert, the $\mathfrak{q}$ will be inert, and in particular unramified in the compositum $DK$. This goes in particular for $\mathfrak{q}$ $=$ $\mathfrak{p}$, and so $\widetilde{K}$ $:=$ $DK$ meets the requirements.
Now for $1\leq i\leq s$, pick a monic polynomial $f\_i$ $\in$ $\mathbb{Z}[X]$ of degree $d$ and irreducible mod $q\_i$. By the Chinese Remainder theorem, there is a monic $f$ in $\mathbb{Z}[X]$ of degree $d$ that is congruent to $f\_i$ mod $q\_i$ for every $i$. So each of the $q\_i$ is inert in $D$ $:=$ $\mathbb{Q}[X]/(f)$.
**Note.** This answer replaces two earlier attempts, which were flawed. Apologies to anyone who may have wasted time on them.
| 0 | https://mathoverflow.net/users/31923 | 433674 | 175,412 |
https://mathoverflow.net/questions/433584 | 13 | Let $K$ be an algebraically closed field, $A$ and $B$ two finite type $K$-algebras which are assumed to be UFD. Is $A \otimes\_K B$ again a UFD?
This question has been already asked [here](https://math.stackexchange.com/questions/3578801/is-the-tensor-product-of-ufds-again-ufd) and [here](https://math.stackexchange.com/questions/4293027/tensor-product-of-ufds), but no (convincing) answer was given. I would like to add a few examples to show that the answer might be slightly less standard than it appears.
$\bullet$ ~~If we drop the *finite type* assumption, the result is wrong. Indeed, take $\require{enclose}
\enclose{horizontalstrike}{A = K[[T]]}$ and $\require{enclose}
\enclose{horizontalstrike}{B = K[X,Y,Z]\Big/ \left(X^5 + Y^7-Z^2 \right)}$. Then Samuel proves in his lecture notes [Lecture on unique factorization domains](http://www.math.tifr.res.in/%7Epubl/ln/tifr30.pdf) that $A$ and $B$ are UFD (see Theorem 8.1 in chapter 1 and Theorem 2.1 in chapter 2). On the other hand, he also proves that $\require{enclose}
\enclose{horizontalstrike}{A \otimes\_K B = B[[T]]}$ is not a UFD (see Theorem 9.1 in chapter 1).~~
**Edit** : as mentionned by Friedrich Knop, $A \otimes\_K B$ is only a subalgebra of $B[[T]]$ (my stupid mistake) and so it's not clear wether it is a UFD or not.
$\bullet$ Even with the finite type assumption, the result doesn't seem to be known (at least at the time when Samuel wrote his lecture notes). Indeed, he states his Theorem 8.1 as:
*Let $A$ be a UFD and $A[X\_1, \ldots, X\_n]$ be graded by assigning weights $\omega\_i$ to $X\_i$ ($\omega\_i >0)$. Let $F(X\_1, \ldots, X\_n)$ be an irreducible homogeneous (for the grading) polynomial, $c$ be a positive integer, prime to the degree of $F$, and*:
$$B = A[X\_1,\ldots,X\_n,Z] \Big/ \left(Z^c-F \right).$$
*Then $B$ is a UFD if one of the following conditions is satisfied:*
1. $c \equiv 1 \ [\omega]$
2. *Every finitely generated projective $A$-module is free.*
If Samuel knew that the tensor product of two finite type $K$-algebras (with $K$ algebraically closed) which are UFD is again a UFD, he would certainly have mentionned that:
$$B = A[X\_1,\ldots,X\_n,Z] \Big/ \left(Z^c-F \right) = A \otimes\_K K[X\_1, \ldots, X\_n,Z]\Big/(Z^c-F)$$
is always a UFD when $A$ is a UFD which is a finite type algebra over $K$ algebraically closed (and he doesn't). At the end of the lecture notes, he gives various examples of such rings which are not UFD, but I can't find one where $K$ is an algebraically closed field and the second condition is the one that is not satisfied.
All in all, I would be very grateful to have a definitive answer on the initial question (even with a highly technological argument or counter-example).
| https://mathoverflow.net/users/37214 | Tensor product of finite type UFD algebras over an algebraically closed field is again UFD? | My [previous answer](https://mathoverflow.net/a/433662/82179) gives a partial result under some additional hypotheses, but these are not needed.
**Theorem** (Boissière–Gabber–Serman). *If $X$ and $Y$ are locally factorial varieties over an algebraically closed field $k$, then so is $X \times Y$. If $\operatorname{Cl}(X) = \operatorname{Cl}(Y) = 0$, then $\operatorname{Cl}(X \times Y)=0$.*
See [BGS, Thm. 5.1 and Cor. 5.4]. $\square$
Taking $X = \operatorname{Spec} A$ and $Y = \operatorname{Spec} B$ for unique factorisation domains $A$ and $B$ shows that $A \otimes\_k B$ is a unique factorisation domain.
---
**References.**
[BGS] S. Boissière, O. Gabber, O. Serman, *Sur le produit de variétés factorielles ou $\mathbf Q$-factorielles*. Preprint (2011). arXiv:[1104.1861](https://arxiv.org/abs/1104.1861).
| 7 | https://mathoverflow.net/users/82179 | 433678 | 175,414 |
https://mathoverflow.net/questions/433680 | 5 | In an abelian category $\mathcal{A}$, for a system $\{F\_i,\phi\_{ij}\}$ we have an exact sequence
$0\to \lim F\_i\to \prod F\_i \to \prod F\_i$
where the second map is given by $id-\prod\phi\_{ij}$. Is there a version of this for stable $\infty$-categories? Meaning, if $\mathcal{C}$ is a stable $\infty$-category and $F\_i$ is a system in $\mathcal{C}$, then is there a fiber sequence in $\mathcal{C}$ given by
$\lim F\_i\to \prod F\_i \to \prod F\_i$?
In case $\mathcal{C}=\mathcal{D}^+(A)$ where $A$ is an abelian category with exact products, enough invectives and the system is countable, then this is true, see for example [Stacks Project Lemma 0BK7](https://stacks.math.columbia.edu/tag/0BK7). I was wondering if one can conclude the same for a more general $\mathcal{C}$
| https://mathoverflow.net/users/197402 | limits and products stable $\infty$-category | In the case of an $\mathbb N^{op}$-indexed system specifically, the answer is yes (note that this is implicit in the Stacks project link you gave); in fact if you replace "fiber sequence" by "equalizer", this holds in an arbitrary $\infty$-category with the appropriate limits (namely products and equalizers). The description in terms of fibers does not hold in general though (in an arbitrary $\infty$-category with limits, the "fiber" without specifying a basepoint does not even make sense)
There are several ways a proof could go. One approach is to use the fact that $\mathbb N$, as an $\infty$-category, is the infinite pushout $[1]\coprod\_{[0]}[1]\coprod\_{[0]}[1] \dots$. Another approach is to use the Yoneda lemma to reduce to the case of the $\infty$-category of spaces, and there, use explicit models for homotopy limits (this is probably simpler to actually write down, if not as conceptual).
For a general filtered poset $I$, however, homotopy limits over $I^{op}$ can be more complicated.
There is always the general "Bousfield-Kan formula", which expresses $\lim\_{I^{op}}$ in the form of the totalization of a cosimplicial object (the $\infty$-analogue of an equalizer): namely, $\lim\_{I^{op}}F$ can be described as the limit over $\Delta$ of a functor that looks like $[n]\mapsto \prod\_{i\_0\to ... \to i\_n} F(i\_0)$ .
| 7 | https://mathoverflow.net/users/102343 | 433681 | 175,416 |
https://mathoverflow.net/questions/433675 | 13 | I've seen various fast algorithms for computing the first few, or directly the $n$-th, digits of $\pi$.
However, it seems to me that all these algorithms assume (see [last sentence here](https://en.wikipedia.org/wiki/Bailey%E2%80%93Borwein%E2%80%93Plouffe_formula#BBP_digit-extraction_algorithm_for_%CF%80)) that there are no very long carries in the computation, which I'm sure would follow from certain conjectures in number theory, but are not proven yet. So my question is:
>
> What is the best known $F(n)$ for which the first $n$ digits of $\pi$ can be computed in $F(n)$ steps?
>
>
>
As was pointed by Arno in a comment, some computable $F(n)$ needs to exist, but I would like a specific bound.
Of course, if anyone has a polynomial bound on $F(n)$, that would be best.
| https://mathoverflow.net/users/955 | Can we compute the first $n$ digits of $\pi$ in $F(n)$ time? | Mahler proved(1) that for any $p,q$, $\left | \pi -\frac{p}{q} \right| > \frac{1}{q^{42}}$. It follows that if one can compute $2^{ 42n} \pi$ to within an error of at most $1$, one can compute the first $n$ digits of $\pi$, since the only way the knowledge of $2^{42n}\pi$ to within an error of $1$ would not be determinative is if $ \left|\pi - \frac{p}{2^n } \right| \leq \frac{1}{ 2^{42n}}$.
So running an algorithm "assuming there are no long carries" to compute the first $42n$ digits will suffice unconditionally to compute the first $n$ digits.
(1): K. Mahler. [On the approximation of π.](https://www.elibm.org/ft/10012144000) Nederl. Akad. Wetensch.
Proc. Ser. A 56=Indag. Math., 15:30–42, 1953.
| 18 | https://mathoverflow.net/users/18060 | 433683 | 175,418 |
https://mathoverflow.net/questions/433623 | 3 | The *total variation distance* between (say discrete) probability distributions, represented as vectors over their support, is defined to be
$$\Delta(\vec p,\vec q) = \frac{1}{2}\lVert \vec p-\vec q\rVert\_1.$$
The (squared) Hellinger distance is then defined to be
$$H^2(\vec p, \vec q) = \frac{1}{2}\lVert \sqrt{\vec p}-\sqrt{\vec q}\rVert\_2^2,$$
where we use the convention that $\sqrt{\vec p} := \left(\sqrt{p\_1},\dots,\sqrt{p\_k}\right)$ is the (coordinate-wise) square root.
These quantities satisfy the following bounds
$$H^2(\vec p, \vec q) \leq \Delta(\vec p, \vec q) \leq \sqrt{2} H(\vec p, \vec q).$$
Here, $H(\vec p, \vec q) := \sqrt{H^2(\vec p, \vec q)}$.
Constant factors might be slightly off based on my normalizations, but I don't particularly care about this.
Instead, it would be convenient for me if $\Delta(\vec p, \vec q) = \Theta(H^\alpha(\vec p,\vec q))$ for some $\alpha \in [1,2]$, i.e. if the lower and upper bounds were tight (up to constant factors).
I doubt this holds in general.
I would be interested if it even held for some large subset of all distributions — for example if it held for $\vec p, \vec q$ such that $\forall i: 0<c < \vec p\_i/\vec q\_i < C$, or some other "simple" condition on $\vec p, \vec q$ (I'm being intentionally somewhat vague by what this means).
| https://mathoverflow.net/users/101207 | When are the total variation distance and Hellinger distance comparable? | $\newcommand{\R}{\mathbb R}\newcommand{\al}{\alpha}\newcommand{\be}{\beta}
\newcommand{\De}{\Delta}$What you want is impossible for any reasonable class of probability distributions, including the class defined by your condition that $0<c<p\_i/q\_i<C$ for some real $c,C$ and all $i$.
Indeed, for simplicity of writing, let $p=(p\_1,\dots,p\_k):=\vec p$ and $q=(q\_1,\dots,q\_k):=\vec q$. Let $P\_k$ denote the set of all probability vectors $p=(p\_1,\dots,p\_k)\in\R^k$. Let us say that a neighborhood $N\_q$ of $q\in P\_k$ is *$c$-good* for some $c\in(0,1)$ if the vector $p$ defined by the conditions
\begin{equation}
p\_1:=q\_1+h,\quad p\_2:=q\_2-h,\quad p\_i:=q\_i\text{ if }i\ge3 \tag{0}\label{0}
\end{equation}
for $h:=c\min(q\_1,q\_2)$ is in $N\_q$.
The first sentence of this answer is formalized as
>
> **Claim:** Suppose that for some real $\al$ and $\be$
> there is a constant $c\in(0,1)$ such that for each natural $k$ there is some $q\in P\_k$ such that
> \begin{equation}
> c\le kq\_i\le1/c \tag{1}\label{1}
> \end{equation}
> for all $i\in[k]:=\{1,\dots,k\}$ and
> \begin{equation}
> cH^\al(p,q)\le\De(p,q)\le H^\be(p,q)/c \tag{2}\label{2}
> \end{equation}
> for all $p\in P\_k$ in some $c$-good neighborhood $N\_q$ of $q$. Then $\al\ge2$ and $\be\le1$. So,
> it is impossible for a relation of the form $cH^\al(p,q)\le\De(p,q)\le H^\al(p,q)/c$ to hold for some real $\al$ and all such $q,p$.
>
>
>
*Proof:* For all $i\in[k]$, let
\begin{equation}
p\_i:=(q\_i+(k-m)h)\,1(i\le m)+(q\_i-mh)\,1(i>m),
\end{equation}
where $k\to\infty$, $m\sim k/2$, and $h>0$ is small enough so that $mh<q\_i$ for $i>m$ and $p\in N\_q$. Then $\De(p,q)=(k-m)mh\asymp k^2h$ and $H(p,q)\asymp k^2h$, so that $\De(p,q)\asymp H(p,q)$. Letting now $h\downarrow0$ fast enough so that $k^2h\to0$, we see that the inequality $\De(p,q)\le H^\be(p,q)/c$ in \eqref{2} can hold only if $\be\le1$.
Now let $p$ be as in \eqref{0}. Then $\De(p,q)=h\asymp1/k$ and $H^2(p,q)\asymp1/k$, so that $\De(p,q)\asymp H(p,q)^2$. Letting now $k\to\infty$, we see that the inequality $cH^\al(p,q)\le\De(p,q)$ in \eqref{2} can hold only if $\al\ge2$.
$\Box$
| 2 | https://mathoverflow.net/users/36721 | 433684 | 175,419 |
https://mathoverflow.net/questions/433686 | 40 | Do editors for top math journals ever read a submitted paper, agree that there are no mistakes and the result is new, yet still reject it on the basis that this is a top math journal and someone could've done that before but chose not to? Maybe some arrogant mathematician goes "I could've proven that in a day or week but didn't because there's better stuff to do."
I'm wondering because this seems to potentially fall into the category of results that are correct but not important enough. It appears the importance of a theorem depends not just on how many people care about it, how much it connects to other results, and how it can be applied, but also as a byproduct how many people have tried to prove it and failed. This last point is where the previous paragraph is relevant.
Note that I'm only counting attempts by mathematicians (let's say at least a degree in math or peer reviewed research for starters) since some of the most famous conjectures receive tons of crackpot attempts after becoming famous, in which case cause and effect are reversed. In fact, most problems in the scope of this question would be slightly famous at best.
If only a few people (or perhaps just 1) have tried and failed, does that discount whoever eventually succeeds? There are way more questions than there are people and hours around to answer them, so perhaps lots of people would like an answer (in the sense that we would like an answer to many questions but cannot attempt every question we're interested in) but only a few people are putting in the time. In the case where few people try because they believe it's too difficult, the paper probably will be accepted. However, if people think it's within their reach and don't try for other reasons, we may end up with a situation similar (but more respectful) than the one in the 1st paragraph.
| https://mathoverflow.net/users/127521 | Math papers where the only issue is that someone else could've done it but didn't | As Sam Hopkins comments, the short answer to the stated question is "yes, all the time." You'd be hard-pressed to find a professional mathematician who *hasn't* received a referee report that basically boils down your first paragraph. Often, the referee or editor can't find anything mathematically wrong with the result, but they reject the paper on the basis that it's not at the right level for the journal, meaning the result or techniques used are not interesting or novel enough, in their view. Essentially, this means they think the work could have been done by many people but wasn't really worth the effort.
The rest of the OP seems to be asking about whether or not it matters if someone else has tried and failed. In fact, it does matter, and it makes the paper *more likely* to be published if someone else has tried and failed, rather than less likely as the OP suggests.
Let me give you a concrete example. In 2017, my coauthor Donald Yau and I wrote the paper [Arrow Categories of Monoidal Model Categories](https://arxiv.org/abs/1703.05359). This paper was published in 2019 in *Math Scandinavica*. In it, we proved a fact that I would not normally have thought would be worthy of a paper in its own right. However, because the statement had been left as an Open Question by a well-known mathematician in the field (Mark Hovey), we were able to frame the paper as "answering a question of Mark Hovey" and I think that probably helped it get published.
For an example in the other direction, my co-author Michael Batanin and I wrote a paper, [Left Bousfield localization without left properness](https://arxiv.org/abs/2001.03764), that I think is definitely worthy of a publication. It shows how to side-step a problem that has bedeviled mathematicians in the field for a long time, and has zillions of examples illustrating the power of the approach. However, because it was left as a remark (4.13) in a [paper by Clark Barwick](https://projecteuclid.org/journals/homology-homotopy-and-applications/volume-12/issue-2/On-left-and-right-model-categories-and-left-and-right/hha/1296223884.full), it has been much harder to get this paper published. I got a rejection that essentially boiled down to "Clark Barwick knew how to prove this and didn't think it was worth writing down."
It is worth noting that the paper in question was one of Clark's earliest, and he later wrote a great essay about [The Future of Homotopy Theory](https://ncatlab.org/nlab/files/BarwickFutureOfHomotopyTheory.pdf) where he lamented this kind of thing. He wrote:
>
> We do not have a good culture of problems and conjectures. The people at the top of our field do not, as a rule, issue problems or programs of conjectures that shape our subject for years to come. In fact, in many cases, they simply announce results with only an outline of proof – and never generate a complete proof. Then, when others work to develop proofs, they are not said to have solved a problem of So-and-So; rather, they have completed the write-up of So-and-So’s proof or given a new proof of So-and-So’s theorem. The ossification of a caste system – in which one group has the general ideas and vision while another toils to realize that vision(6) – is no way for the subject to flourish. Other subjects have high-status visionaries who are no sketchier in details than those in homotopy theory, but whose unproved insights are nevertheless known a
> conjectures, problems, and programs.
>
>
>
He even includes a side-note saying
>
> (6) only to have their paper rejected with lines like the following, from a colleague: "After So-and-So’s [sketchy] work, it was essentially obvious that such a result would be possible, given the right framework."
>
>
>
So, based on that, I have to conclude that if he had a time machine, Clark probably would have written his Remark 4.13 as a Conjecture and then I could have published my paper saying I "proved a conjecture of Clark Barwick." I confess that I'm guilty of very much the same kind of behavior. I put a paper on arxiv in 2014 announcing a result that wasn't on arxiv till 2017 and one researcher told me my remark discouraged him from working on the project. I regret that. Nowadays I try to put many more Questions, Conjectures, and Problems in my papers, e.g., [this](https://arxiv.org/abs/1703.05377) one that just got accepted for publication.
So, to conclude, I call upon anyone who has read this far to include named/numbered Conjectures, Questions, and Problems, and at all costs avoid Remarks where you claim things are true but don't write out the proof. Let's make the field friendlier to young people and help them get their work published, while at the same time incentivizing them to build on our work by answering questions we explicitly leave. I [wrote something before](https://mathoverflow.net/a/375330/11540) to this effect here.
| 91 | https://mathoverflow.net/users/11540 | 433691 | 175,422 |
https://mathoverflow.net/questions/432871 | 2 | Let $M\_t$ be a continuous time martingale, and assume its quadratic variation is identically zero with some positive probability less than $1$.
To be more precise, assume there exists some event $E$ with $0 < \mathbb P(E) < 1$ such that for almost every $\omega \in E$, $\langle M, M\rangle\_t (\omega) = 0$ for all $t \geq 0$.
**Question:** Does it follow that $M$ is almost surely constant in time on $E$? That is, $M\_t = M\_0$ for all $t > 0$ a.s. on $E$.
*Remark: For continuous martingales, this follows directly from the Dambis-Dubins-Schwartz theorem.*
| https://mathoverflow.net/users/173490 | Is a martingale constant on the event that its quadratic variation is zero? | To answer your question, the following Lengart's inequality is useful: (Please refer to
S. W. He *et al., Semimartingale Theory and Stochastic Calculus*, Sci. Press and CRC(1992), p.239, Theorem 9.23.)
**Theorem** Let $ X $ be an adapted cadlag process, dominated by an predictable process $ A $. Then for arbitrary constants $C>0, d>0$, stopping time $ T $ and measurable
set $ H $ we have
\begin{equation\*}
\mathsf{P}(H\cap [X^\ast\_T\ge C])\le \frac{1}{C}\mathsf{E}[A\_T\wedge d]+
\mathsf{P}(H\cap [A\_T\ge d]). \tag{1}
\end{equation\*}
Hence for the continuous time martingale $M$(if $ M $ is also a locally square integrable martingale), $M^2$ is dominated by its pridictable quadratic variation $ \langle M \rangle $. Using (1) for $ H=E $, it follows
\begin{align\*}
\mathsf{P}(E\cap [M^{\ast2}\_T\ge C])&\le \frac{1}{C}\mathsf{E}[\langle M \rangle\_T\wedge d]+
\mathsf{P}(H\cap [\langle M \rangle\_T\ge d])\\
& = \frac{1}{C}\mathsf{E}[\langle M \rangle\_T\wedge d] \le \frac{d}{C}, \quad
\forall C>0, d>0.
\end{align\*}
Now let $d\downarrow0$ to get
\begin{equation\*}
\mathsf{P}(E\cap [M^{\ast2}\_T\ge C])=0, \qquad \forall C>0.
\end{equation\*}
Further more letting $ C\downarrow 0 $ to get that $ M=0 $ almost surely on $ E $.
| 2 | https://mathoverflow.net/users/103256 | 433723 | 175,430 |
https://mathoverflow.net/questions/433709 | 8 | The following question was [asked years ago on MSE](https://math.stackexchange.com/q/54761/53976), but let me recap it:
>
> **Question**: Is there anything currently known about the exact consistency strength of "$\mathsf{ZF}$ + both $\omega\_1$ and $\omega\_2$ are singular?" Or could we find a better bound?
>
>
>
It is well-known that if both $\omega\_1$ and $\omega\_2$ are singular, (of course, without choice) then $0^\sharp$ exists. It is quite famous that we can make all uncountable cardinals singular from a proper class of strongly compact cardinals.
However, if we only consider $\omega\_1$ and $\omega\_2$ then Gitik's result is overkill: Apter proved in
>
> Arthur W. Apter. **$\mathsf{AD}$ and patterns of singular cardinals below $\Theta$**. *J. Symbolic Logic* 61 (1996), no. 1, 225–235.
>
>
>
that if we start from $L(\mathbb{R})\models\mathsf{ZF+AD+DC}$ then we can find an extension $N$ of $V$ whose cardinals are the same with that of $V$, and both of $\omega\_1$ and $\omega\_2$ have cofinality $\omega$. Thus we have "$\mathsf{ZFC}$ + there are $\omega$ many Woodin cardinals" as an upper bound.
(Apter's result for $\omega\_1$ and $\omega\_2$ is further generalized in a paper by [Apter, Jackson, and Löwe](https://www.jstor.org/stable/23513464). Also, note that what Apter proved is stronger than I stated: he actually proved that if we work over $V=L(\mathbb{R})\models\mathsf{ZF+AD+DC}$ and if $A$, $B$ are sets partitioning the set of regular cardinals below $\Theta^{L(\mathbb{R})}$ then there is an extension $N$ of $V$ such that cardinals in $V$ and those in $N$ are the same and cardinals in $A$ has cofinality $\omega$, but cardinals in $B$ are still regular in $N$.)
On the other hand, Schindler proved in
>
> Ralf-Dieter Schindler, **Successive weakly compact or singular cardinals.** *J. Symbolic Logic* 64 (1999), no. 1, 139–146.
>
>
>
that if $\kappa$ is measurable in $\mathsf{HOD}$ and inaccessible in $V\models \mathsf{ZF}$ and there is $\delta<\kappa$ such that $\delta^+<\kappa$ and both of $\delta$ and $\delta^+$ are singular, then there is an inner model of a Woodin cardinal.
---
It brings possibly related questions that are also curious for me: First, it is easy to see that if we start from $\mathsf{ZF}$ with both $\omega\_2$ and $\omega\_3$ are singular, then there is a generic extension of $V$ that thinks $\omega\_1$ and $\omega\_2$ are singular.
>
> **Question.** How about the opposite direction? So, can we derive the consistency of "$\mathsf{ZF}$ + both $\omega\_2$ and $\omega\_3$ are singular" from that of "$\mathsf{ZF}$ + both $\omega\_1$ and $\omega\_2$ are singular"?
>
>
>
Regarding the upper bound, the role of $\mathsf{AD+DC}$ in Apter's proof is a normal measure over $\omega\_1$ and $\omega\_2$. He then uses a product of Prikry forcings and constructs an inner model of a generic extension. It seems like providing normal measures is the only role of $\mathsf{AD}$.
>
> **Question.** Can we find a model of $\mathsf{ZF}$ + "both of $\omega\_1$ and $\omega\_2$ have a normal measure" from large cardinal hypotheses weaker than $\omega$ many Woodin cardinals?
>
>
>
(The paper by Apter, Jackson, and Löwe pointed out in Theorem 32 that if we start from
$\mathsf{ZFC}$ two measurable cardinals then we have an extension $N\models\mathsf{ZF}$ of $V$ that thinks $\omega\_1$ and $\omega\_3$ are measurable and $\omega\_2$ is regular. I have no idea whether it is related to an answer to the above question, and probably not.)
| https://mathoverflow.net/users/48041 | Some relevant questions about the consistency strength of singularity of $\omega_1$ and $\omega_2$ | We know little to nothing about the exact strength. Here are the facts, as you already know them.
1. To get $\omega\_1$ and $\omega\_2$ both singular we need to have *at least* a Woodin cardinal.
2. If there are infinitely many Woodin cardinals, then it is consistent that $\omega\_1$ and $\omega\_2$ are singular (go to a model of $\sf AD$ and collapse $\omega\_1$ and $\omega\_2$ or shoot Prikry sequences through their measures to singularise them). Indeed, in this case we have infinitely many consecutive singular cardinals.
The prevailing conjecture, if so, is that successive singular cardinals *should* have the consistency strength of "roughly a Woodin".1
We also similar things about consecutive measurable cardinals. Essentially it boils down to failure of suitable covering theorems of core models (i.e., the successor of weakly compact cardinals is computed incorrectly, and therefore a Woodin cardinal must hide somewhere in an inner model).
The problem here, which I will use to advertise some of my thoughts on these matters, is that there are two levels here.
1. Consistency strength, which is what we *often* measure. This requires some sort of canonical inner models, or "slightly less canonical inner models" (à la $L(\Bbb R)$ or so) for us to establish these results.
2. Forcing strength, which is what we *often* can do "by hand". In other words, we want to construct a generic (or symmetric in the choiceless cases) extension in which the statement holds.
These can be very different. For example, "A singular $\kappa$ is regular in an inner model", and $\kappa$ can always be taken to be $\aleph\_\omega$ too and the inner model can be assumed to be $L$, has the consistency strength of exactly "$0^\#$ exists", but if you wanted to do it by forcing the consistency strength jumps to a measurable cardinal.
In order to create a symmetric extension with successive measurable cardinals or successive singular cardinals we need to venture into the zone of "higher compactness". Normally a supercompact will be used, but the proofs can be analysed to some bounded level of sufficient compactness if one insists on doing that. These are of course, only the known bounds, rather than the exact strength of these statement.
It seems to me that we have a phenomenon here of the same flavour as in the case of a singular cardinal that is regular in $L$. It seems to me that in order to construct these things "by hand", i.e. to force them, we probably need to rely on much stronger hypotheses.
Now we run into a problem. Since the consistency strength is often analysed by inner models + forcing combined, if you wish to "start from optimal hypothesis and produce a model", this means that the model produced is not produced with forcing, but we hardly have any other method at hand right now for these results.
So, finding the optimal results is hard and remains a mystery. Finding the "forcing strength" is also a mystery and will probably remain open a longer time. One can conjecture that once suitable inner model theory is developed for compact cardinals, we can try the following method "$V$ is a symmetric extension in which something happens, force choice in a 'minimally destructive way' and show that some cardinals are generically such and such". However, there is much more work before these are even close to being feasible and working results.
Now. You may argue, and you'd have a significant point there as well, that since $L(\Bbb R)$ is a symmetric extension of a model of $\sf ZFC$, and perhaps even of ${\rm HOD}^{L(\Bbb R)}$, it is in fact absolutely the case that consequences of $\sf AD$ should be forceable with "roughly the same consistency strength".
Alas, when we look at the results about ${\rm HOD}^{L(\Bbb R)}$, we see that $\omega\_1^V=\omega\_1^{L(\Bbb R)}$ is in fact the least measurable there, and that the Woodin cardinals actually appear relatively "high". This, in comparison to the forcing based construction where we collapse infinitely many Woodin cardinals to be countable and consider $L(\Bbb R)$ of the model containing only bounded reals (i.e. reals added by collapsing of a bounded part below the limit of the Woodins), in which case there are Woodin cardinals hiding well below $\omega\_1$.
This means that any construction attempting "optimal assumptions", or even "a symmetric extension from reasonably optimal assumptions" is bound to end up as being incredibly complicated. Which, again, puts us fairly far away from being able to understand these optimal results. If you even want to under $L(\Bbb R)$ as such a symmetric extension, matters become much worse, since we need to understand the collapsing of intervals of cardinals or singularising *and* collapsing simultaneously.
And since our main tool for symmetric extensions is homogeneity arguments, it seems that the forcing involved here may or may not have these homogeneity properties. Indeed, the obvious culprit would be some sort of a stationary tower forcing which is notoriously non-homogeneous.
So perhaps the question about "forcing strength" should be about forcing with reasonably behaved partial orders (e.g. sufficiently homogeneous ones). But much, much more needs to be understood before we can even address these questions of optimal strengths.
---
Footnotes.
1. When I was a student I thought that I saw a paper by a prominent set theorist that proved that, having spent years telling people about that, Yair Hayut eventually suggested we try to read it, having taken some basic inner model theory and stationary tower forcing courses with Menachem (Magidor). It turned out that this was a dream that I had. The result, of course, was that a sharp for a single Woodin cardinal is enough. I asked the alleged author, and they said it would have been a great result, and if I know how to prove it they'd be happy to read or write that paper. Alas, I have no idea how to prove something like that.
| 8 | https://mathoverflow.net/users/7206 | 433724 | 175,431 |
https://mathoverflow.net/questions/433740 | 3 | What is an example of a Frobenius algebra that is not Koszul? Are there reasonable requirements for a Frobenius to be Koszul?
| https://mathoverflow.net/users/491434 | What is an example of a Frobenius algebra that is not Koszul? | If an algebra A is Koszul with Koszul daul B, one has the formula gldim A=Loewy Length B-1.
Thus when a Frobenius algebra is Koszul (and not semisimple), the quadratic dual must be infinite dimensional.
For example for preprojective algebras of Dynkin type, the quadratic dual is finite dimensional and thus they can not be Koszul, see <https://link.springer.com/article/10.1023/A:1020146502185> .
| 5 | https://mathoverflow.net/users/61949 | 433741 | 175,438 |
https://mathoverflow.net/questions/433752 | 11 | I wanted to compute $\mathit{KO}^{-1}(\mathbb{R}P^3)$ and regrettably I could only think of using the Atiyah Hirzebruch spectral sequence, which seemed like a big overkill but looking at similar computations of $K^0(\mathbb{R}P^n)$ done in Atiyah's book do not look particularly simpler as he uses equivariant $\mathit{K}$-theory to get some short exact sequences involving the representation ring $R(\mathbb{Z}/2\mathbb{Z})$, which means having to deal in our case with real representations...
So is there a very simple way of computing $\mathit{KO}^{-1}(\mathbb{R}P^3)$? If not perhaps there is some trick which greatly simplifies the AHSS computation?
Thanks
| https://mathoverflow.net/users/494010 | Computing KO^-1 of RP^3 without AHSS | The $KO$-theory of all truncated real projective spaces (including the calculation you want) was carried out very systematically by Frank Adams in his famous paper on the vector fields on spheres: see section 7 of [J.F.Adams, Ann.Math. (1962)]. He also invented/constructed `Adams operations' (not his term!) in this paper, and carefully calculates what they are doing too. His calculation proceeds by going through complex K theory and complex projective spaces enroute, so one has a good sense of every element that is constructed.
| 17 | https://mathoverflow.net/users/102519 | 433753 | 175,439 |
https://mathoverflow.net/questions/433725 | 9 | Let $A$ be a selfadjoint operator on some Hilbert space $H$, let $U(t)=e^{itA}$ be the corresponding continuous group, and let $f\in H$ be orthogonal to all eigenvectors of $A$. Are there examples such that $U(t)f$ does not converge weakly to 0 as $t\to+\infty$?
(From the RAGE Theorem, if $K$ is a compact operator on $H$ and $f$ as above, then
$$
\lim\_{T\to+\infty}\frac1T\int\_0^T\|KU(t)f\|\_H^2dt=0.
$$
This implies $\liminf\_{t\to+\infty}\|KU(t)f\|\_H=0$ and suggests weak convergence of $U(t)f$ to 0, at least for suitable sequences of times.)
| https://mathoverflow.net/users/7294 | Counterexamples to weak dispersion for the Schrödinger group | The answer is yes.
A measure-preserving invertible shift $T: X \to X$ on a probability space $(X,\mu)$ is said to be [weakly mixing](https://en.wikipedia.org/wiki/Mixing_(mathematics)) if $\lim\_{N \to \infty} \frac{1}{N} \sum\_{n=1}^N |\langle f \circ T^{-n}, g \rangle|^2 = 0$ for all $f,g$ in the Hilbert space $L^2(X)\_0$ of mean zero square-integrable functions, and [strongly mixing](https://en.wikipedia.org/wiki/Mixing_(mathematics)) if $\lim\_{n \to \infty} |\langle f \circ T^{-n}, g \rangle| = 0$ for all such functions. (The minus sign is not important here, and one can replace $T^{-n}$ by $T^n$ if one wishes.) There are examples of systems that are weakly mixing but not strongly mixing; see for instance [this previous MathOverflow post](https://mathoverflow.net/questions/47080/examples-of-transformations-which-are-weak-mixing-but-not-strong-mixing) for some examples (indeed in certain technical senses a "generic" shift is of this form). Note that weakly mixing shifts have no eigenfunctions in $L^2(X)\_0$ (indeed this is an if and only if, by the discrete version of the RAGE theorem).
Such systems $(X,\mu,T)$ are discrete flows, but they can be converted into continuous flows by the standard device of taking a [suspension](https://en.wikipedia.org/wiki/Suspension_(topology)). Namely, let $\tilde X$ be $X \times {\bf R}/\sim$ where we quotient by the equivalence relation $(x,t) \sim (T^{-n} x, t+n)$ and endow this space with the product measure $\tilde \mu$ of $\mu$ and Lebesgue measure on the unit interval, and the continuous shift $\tilde T^t (x,s) := (x,s+t)$. If one then lets $H \equiv L^2([0,1]; L^2(X)\_0)$ be the Hilbert space of functions $f \in L^2(\tilde X)$ that are of mean zero on every time slice $X \times \{t\}$, and lets $U(t): H \to H$ be the [Koopman operator](https://en.wikipedia.org/wiki/Composition_operator) $U(t) f(x,s) := f \circ \tilde T^{-t}(x,s) = f(x,s-t)$, one can easily verify that $U(t)$ is a strongly continuous unitary flow (and thus of the form $e^{itA}$ by [Stone's theorem](https://en.wikipedia.org/wiki/Stone%27s_theorem_on_one-parameter_unitary_groups)) that has no eigenfunctions, but such that $U(t)$ fails to weakly converge to zero (even if we restrict $t$ to the integers, in which case the continuous flow basically collapses back to the discrete flow $T^n$).
UPDATE: Here is another example. Let $\mu$ be an atomless compactly supported probability measure on ${\bf R}$ whose Fourier transform does not decay to zero at infinity (for instance one can take the standard Cantor set measure). If we let $U(t): L^2(\mu) \to L^2(\mu)$ be the modulation flow $U(t) f(x) = e^{itx} f(x)$, then the flow has no eigenfunctions (because of the atomless condition) but $U(t) 1$ does not converge weakly to zero (because $\langle U(t) 1, 1 \rangle$ is basically the Fourier transform of $\mu$).
| 10 | https://mathoverflow.net/users/766 | 433756 | 175,442 |
https://mathoverflow.net/questions/433396 | 4 | Let $f: \mathbb R \to \mathbb R$ be a $C^1$ convex function, satisfying the growth conditions
$$\lim\_{x \to -\infty} \nabla f(x) = -\infty, \lim\_{x \to \infty} \nabla f(x) = \infty.$$
and let $\gamma\_t: [0, \infty) \to \mathbb R$ be a deterministic, Borel measurable process satisfying the following conditions:
1. $\gamma\_t > 0$ for all $t \in [0, \infty)$.
2. $\gamma\_t \to 0$ as $t \to \infty$.
3. $\int\_0^\infty \gamma\_t^2 \, dt < \infty$.
Consider the solution $X$ to the one dimensional SDE
$$dX\_t = -\nabla f(X\_t) \, dt + \gamma\_t \, dW\_t, \, \, X\_0 = x\_0$$
with $W$ a standard Brownian motion, and $x\_0 \in \mathbb R$ an arbitrary initial condition.
**Question:** Is it true that for all initial conditions $x\_0$, we have $\nabla f(X\_t) \to 0$ as $t \to \infty$ almost surely?
| https://mathoverflow.net/users/173490 | Convergence of a continuous time stochastic gradient descent algorithm | here "[An SDE perspective on stochastic convex optimization](https://arxiv.org/abs/2207.02750#:%7E:text=We%20analyze%20the%20global%20and,problems%20with%20noisy%20gradient%20input.)" they study these type of SDEs (Stochastic gradient descent) and basically ask bounded and an L2 condition for gamma\_t in theorem 3.1.
Here in (H0) they also have extra constraints to $f$ that might be sharp.
As shown [here](https://math.stackexchange.com/questions/2842884/c1-convex-function-but-not-c1-1) a function can be convex and C1 but its gradient not be Lipschitz
>
> The map $f:\Bbb R\to \Bbb R$, $f(x)=\frac23\lvert x\rvert^{3/2}=\int\_0^x \lvert t\rvert^{1/2}\operatorname{sgn}t\,dt$ is convex and $C^1$, but $f'(x)=\lvert x\rvert^{1/2}\operatorname{sgn}x$ is not Lipschitz continuous in any neighbourhood of $0$. More generally, integrate your favourite monotone increasing continuous function which is not locally Lipschitz and you'll obtain a counterexample in $\Bbb R$.
>
>
>
And as shown [here](https://math.stackexchange.com/questions/1507138/examples-where-constant-step-size-gradient-descent-fails-everywhere?rq=1) without this Lipschitz condition, we don't even have convergence for the deterministic gradient descent.
| 1 | https://mathoverflow.net/users/99863 | 433758 | 175,443 |
https://mathoverflow.net/questions/433685 | 18 | The question was cross-posted from Math.SE: <https://math.stackexchange.com/questions/4566017/strengthening-ax-grothendieck>
---
The question is simple. The Ax-Grothendieck theorem says a polynomial map $p\colon\mathbb C^n\to\mathbb C^n$ that is injective is also surjective.
>
> Is assuming $p$ has finite fibers enough?
>
>
>
I can't come up with any easy counter-examples, but the proof I know using finite fields does not work.
| https://mathoverflow.net/users/123673 | Strengthening Ax-Grothendieck | A counterexample for $n = 2$ is the map $(x\_1, x\_2) \mapsto (x\_1x\_2 - 1, x\_2(x\_1x\_2 - 1) + x\_1)$. This example, which I learned from a paper of Zbigniew Jelonek, was mentioned in an [earlier answer of mine](https://mathoverflow.net/a/343994/1508).
| 17 | https://mathoverflow.net/users/1508 | 433760 | 175,445 |
https://mathoverflow.net/questions/433763 | 3 | Circulant matrices are very useful in digital image processing.
I found the general formula for determinant of circulant matrix.
But I think it is not suitable for block-circulant matrices.
For example, consider the formula for $\det(K)$,
where $$K = \left(\begin{array}{cccc} A & B & C & D \\
D & A & B & C \\
C & D & A & B \\
B & C & D & A
\end{array}\right) $$
and $A$ , $B$ , $C$ and $D$ are size $n \times n$.
| https://mathoverflow.net/users/369335 | One question on block-circulant matrices | The formula for the specific case is
$$\det K=\det(A+B+C+D)\det(A-B+C-D)\det(A+iB-C-iD)\det(A-iB-C+iD).$$
More generally, for a block-circulant matrix with $n$ square blocks $A\_0,\ldots,A\_{n-1}$, the formula is
$$\det K=\prod\_{\omega^n=1}\det(A\_0+\omega A\_1+\cdots+\omega^{n-1}A\_{n-1}).$$
To see this, observe that $K$ is block-diagonalisable,
$$K=U^\*{\rm diag}(A\_0+\alpha A\_1+\cdots+\alpha^{n-1}A\_{n-1},A\_0+\alpha^2 A\_1+\cdots+\alpha^{2(n-1)}A\_{n-1},\ldots)U$$
where $\alpha=\exp\frac{2i\pi}n$ and
$$U=\frac1{\sqrt n}((\alpha^{(i-1)(j-1)}I\_d))\_{1\le i,j\le n}.$$
Hereabove, the blocks $A\_j$ are $d\times d$. This shows the formula, up to the factor $|\det U|^2$, which is easily seen to be equal to $1$.
| 6 | https://mathoverflow.net/users/8799 | 433765 | 175,446 |
https://mathoverflow.net/questions/433757 | 1 | Let $d,n\ge 1$ be fixed integers. Given some compact subset $E\subset \mathbb R^d$, consider the function $f: E^n\ni (x\_1,\ldots, x\_n) \longrightarrow f(x\_1,\ldots, x\_n)\in \mathbb R$ defined by
$$f(x\_1,\ldots, x\_n):= \max\_{(c\_1,\ldots,c\_n)\in\mathbb R^n}\left\{\int\_E \left(\min\_{1\le i\le n}|y-x\_i|^2-c\_i\right)p(y)dy + \sum\_{i=1}^n \alpha\_i c\_i\right\},$$
where $p:E\to \mathbb R\_+$ is a probability density on $E$ and $\alpha\_1,\ldots, \alpha\_n>0$ s.t. $\sum\_{i=1}^n \alpha\_i =1$. Under which conditions (on $E, \rho$) $f$ is differentiable (almost everywhere) on $E^n$?
| https://mathoverflow.net/users/493556 | Differentiability of some function defined as the maximum | Suppose that there is an open subset $U$ of $E$ such that the Lebesgue measure of $E\setminus U$ is $0$. Since $E$ is compact, the function $E^n\ni(x\_1,\dots,x\_n)\mapsto|y-x\_i|^2$ is $L$-Lipschitz for some real $L>0$ and each $i\in\{1,\dots,n\}$ and each $y\in E$. Therefore and because the $\max$, $\min$, and integration (with respect to a probability measure) operations preserve the $L$-Lipschitz condition, $f$ is $L$-Lipschitz.
So, by [Rademacher's theorem](https://en.wikipedia.org/wiki/Rademacher%27s_theorem), $f$ is differentiable almost everywhere (a.e.) on $U^n$ and hence a.e. on $E^n$.
| 1 | https://mathoverflow.net/users/36721 | 433767 | 175,448 |
https://mathoverflow.net/questions/433764 | 8 | Let $ X\_N = \text{span} \{\cos(2\pi lx): l=0, \cdots, N-1 \} $ with $ x \in [0, 1] $ and $ Y\_N = \{v =(v\_0, \cdots, v\_{N-1}): v\_j \in \mathbb{C}\} = \mathbb{C}^N $. Then $ X\_N $ is the space of trigonometric polynomials. We equip $ X\_N $ with the usual $ L^p (1 \leq p \leq \infty ) $ norm and equip $ Y\_N $ with the discrete $ l^p (1 \leq p \leq \infty)$:
$$ \| v \|\_{l^p} = \left( \frac{1}{N}\sum\_{j=0}^{N-1} |v\_j|^p \right)^\frac{1}{p}, \quad 1 \leq p < \infty, $$
$$ \| v \|\_{l^\infty} = \max\_{0 \leq j \leq N-1} |v\_j|, \quad v \in Y\_N. $$
Let $ R\_N:X\_N \rightarrow Y\_N $ be the restriction of the trigonometric polynomials to some discrete points: for $ u \in X\_N $,
$$ v = R\_N u, \quad v\_j = u(x\_{j+1/2}), \quad j=0, \cdots, N-1, $$
where $ x\_{j+1/2} = (j+1/2)/N $.
**Question**:
can we find a constant $ C $ independent of $ N $ such that
$$ \| R\_N u \|\_{l^4} \leq C\| u \|\_{L^4}, \quad u \in X\_N. $$
Moreover, does it hold for higher dimensions though currently it is only stated in 1D?
Note that by Parseval's identity, one has,
$$ \| R\_N u \|\_{l^2} = \| u \|\_{L^2}. $$
| https://mathoverflow.net/users/484187 | Bounding the discrete $l^p$ norm by the continuous $L^p$ norm for trigonometric polynomials | Yes, this goes back to the work of
*Plancherel, M.; Pólya, George*, [**Fonctieres entières et intégrales de Fourier multiples**](http://dx.doi.org/10.1007/BF01258191), Comment. Math. Helv. 9, 224-248 (1937). [ZBL0016.36004](https://zbmath.org/?q=an:0016.36004).
(see for instance Theoreme III). Nowadays one would usually prove such a result using Littlewood-Paley projections or similar devices to express $R\_N u$ as a suitable integral expression of $u$ to which Schur's test (for instance) can be applied.
| 12 | https://mathoverflow.net/users/766 | 433774 | 175,449 |
https://mathoverflow.net/questions/433583 | 2 | Let $D \subset \mathbb{R}^d$ be a bounded $C^1$ domain. We consider a reflected Brownian motion $X=(\{X\_t\}\_{t \ge 0},\{P\_x\}\_{x \in \overline{D}})$ on $\overline{D}$. Let $\{p\_t\}\_{t>0}$ denote the semigroup of $X$ (in other words, $\{p\_t\}\_{t>0}$ is the Neumann semigroup). It is known that $\{p\_t\}\_{t>0}$ is a feller semigroup. That is, we have $p\_t f \in C(\overline{D})$ for any $t>0$ and $f \in C(\overline{D})$. Here, $C(\overline{D})$ is the space of continuous functions on $\overline{D}.$ We always equip $C(\overline{D})$ with the sup-norm so that it becomes a real Banach space.
Let $(L,D(L))$ be the Neumann Laplacian on $C(\overline{D})$. That is,
\begin{align\*}
D(L)&=\left\{ f \in C(\overline{D})\mid \lim\_{t \to 0}\frac{p\_t f-f}{t} \text{ exists in }C(\overline{D})\right\},\\
Lf&=\lim\_{t \to 0}\frac{p\_t f-f}{t}.
\end{align\*}
Let $\nu$ denote the inward unit normal vector on $\partial D$, and define \begin{align\*}
\mathcal{C}=\{f \in C^2(\mathbb{R}^d)|\_{\overline{D}} \mid \langle \nabla f(x),\nu(x) \rangle=0\text{ for every $x \in \partial D$}\}.
\end{align\*}
Can we prove that $\mathcal{C}$ is a core for $(L,D(L))$?
| https://mathoverflow.net/users/68463 | On a core for Neumann Laplacian on $C(\overline{D})$ | I think $\mathcal C$ is not even dense in the space of continuous functions!
To be specific: consider a 2-D domain $D$ lying above the graph of a $C^1$ function $\phi : \mathbb R \to \mathbb R$, and assume that $\phi'$ is continuous, but nowhere differentiable — say, a generic sample path of the Wiener process.
Let $f$ be a $C^2$ function on $\mathbb R^2$ such that $\nabla f$ is orthogonal to the normal vector at each boundary point. This means that $$\partial\_y f(x, \phi(x)) = \phi'(x) \partial\_x f(x, \phi(x)).$$ On every interval where $\partial\_x f(x, \phi(x)) \ne 0$, we find that $$\phi'(x) = \frac{\partial\_y f(x, \phi(x))}{\partial\_x f(x, \phi(x))}$$ is a $C^1$ function, a contradiction. Thus, $\partial\_x f(x, \phi(x)) = 0$ for all $x$, and consequently also $\partial\_y f(x, \phi(x)) = 0$ for all $x$. In particular, $f(x, \phi(x))$ is constant.
But that means that $\mathcal C$ only contains functions $f$ which are constant on the boundary, so $\mathcal C$ is not even dense in the space $C(\overline D)$ of continuous functions.
The same argument works for $C^{1,\alpha}$ domains, at least when $\alpha < 1$.
| 3 | https://mathoverflow.net/users/108637 | 433776 | 175,450 |
https://mathoverflow.net/questions/433637 | 3 | Are there any connective $E\_1$ rings $R$ over $\mathbb{F}\_p$ satisfying the following?
1. $\pi\_\*(R)$ is a finite dimensional $\mathbb{F}\_p$ vector space
2. $R$ is compact as a module over $R \otimes R^{op}$
3. $R$ is not concentrated in degree 0
Motivation: I am looking for a smooth proper category with negative Hochschild cohomology and someone suggested $\mathit{Perf}(R)$ with $R$ satisfying the above if it existed. In fact they suggested to look for a finite CW complex $X$ with
$H\_\*(\Omega X,\mathbb{F}\_p)$ finite over $\mathbb{F}\_p$ (then we can take $R=H\_\*(\Omega X,\mathbb{F}\_p)$) but I have not found any of those either.
Edit: As Shaul Barkan explained to me, such an example cannot come from a space X as described above, because the conditions will imply that $H\_\*(\Omega X,\mathbb{F}\_p)$ is in degree 0.
| https://mathoverflow.net/users/136287 | Are there strictly connective smooth proper algebras over $\mathbb{F}_p$? | Here's an example: consider the $\infty$-category $Fun(\Delta^1,Perf(\mathbb F\_p))$. It has two canonical generators $A= \mathbb F\_p\to \mathbb F\_p$ and $B=\mathbb F\_p\to 0$; and I claim that $R= End(A\oplus \Sigma B)$ is an example of what you're looking for (derived endomorphisms - more generally, everything here is "derived").
Note that $\hom(A,B) = \mathbb F\_p$, $\hom(B,A)=0$, $\hom(A,A) = \mathbb F\_p, \hom(B,B)= \mathbb F\_p$.
In particular, $End(A\oplus\Sigma B) = \hom(A,\Sigma B)\oplus \hom(A,A)\oplus \hom(\Sigma B,A)\oplus \hom(\Sigma B,\Sigma B)$.
Note that $\hom(\Sigma B,\Sigma B)\simeq \hom(B,B)$, $\hom(\Sigma B,A) = 0$, so really $End(A\oplus \Sigma B)= \hom(A,A)\oplus \hom(B,B)\oplus \Sigma\hom(A,B)\simeq \mathbb F\_p\oplus\mathbb F\_p\oplus\Sigma\mathbb F\_p$.
In particular, $End(A\oplus\Sigma B)$ is connective and perfect over $\mathbb F\_p$, so it remains to argue that it is smooth. But smoothness is *Morita invariant*, i.e. it only depends on the category of modules.
Now, $A,B$ are generators of $Fun(\Delta^1,Perf(\mathbb F\_p))$, so $Perf(End(A\oplus\Sigma B))\simeq Fun(\Delta^1,Perf(\mathbb F\_p))$ by the Schwede-Shipley theorem (I'm using implicitly here that $Ind(C^{\Delta^1}) = Ind(C)^{\Delta^1}$ for any stable $C$), so it suffices to argue that $Fun(\Delta^1,Perf(\mathbb F\_p))$ is smooth.
Now, for an indexing diagram $C$, $Fun(C,Mod\_R)^\omega$ is smooth over $R$ if and only if $R[map\_C(-,-)]$ is compact in $Fun(C^{op}\times C,Mod\_R)$, and one can show that $map(R[map\_C(-,-)], F)\simeq \int\_{c\in C}F$, the *end* of $F$. It follows that $Fun(C,Mod\_R)^\omega$ is smooth if and only if $\int\_{c\in C}: Fun(C^{op}\times C,Mod\_R)\to Mod\_R$ preserves filtered colimits.
Recall that ends are a special case of limits over $Tw(C)$, so it suffices (but is not equivalent, as far as I know) for $\lim\_{Tw(C)}$ to preserve filtered colimits. In this case, however, $Tw(\Delta^1)$ is a *finite* $\infty$-category, so this stronger condition is satisfied, and $Fun(\Delta^1,Perf(\mathbb F\_p))$ is smooth, hence so is $R$.
| 3 | https://mathoverflow.net/users/102343 | 433806 | 175,455 |
https://mathoverflow.net/questions/433782 | 4 | Let $\mathcal{F}\_N$ be the set of all strictly increasing sequences of positive integers. For every two $F\_1, F\_2\in\mathcal{F}\_N$, if we use $\delta(F\_1,F\_2)$ to denote the first $n$-th coordinate where $F\_1(n)\neq F\_2(n)$, then $d(F\_1, F\_2) = \exp[-\delta(F\_1, F\_2)]$ defines a metric on the space $\mathcal{F}\_N$. One can observe that $d$ is an ultra-metric because, for each $F\_1, F\_2, F\_3\in\mathcal{F}\_N$,
$$
d(F\_1, F\_3)\leq \max\Big[ d(F\_1, F\_2), d(F\_2, F\_3) \Big].
$$
Now fix $r\in\big(\frac{1}{2}, 1\big)$. My questions are:
1. Will there exists a finite set of positive integers $M$ such that $\sum\_{i\in M}r^i = 1$? Does the existence of such a finite set depend on $r$?
2. Notice that we can view $\mathcal{F}\_N$ as the family of all **infinite** subset of $\mathbb{N}$ (viewing each element in $\mathcal{F}\_N$ as a sequence is more compatible to the given metric). For the fix $r\in\big(\frac{1}{2}, 1\big)$, one can observe that the set $\Big\{ \sum\_{i\in F}r^i\,\vert\, F\in\mathcal{F}\_N \Big\}$ is bounded below by $0$ (which is also the infimum) and bounded above by $\frac{r}{1-r}$. Is the following function surjective?
$$
S\_r:\mathcal{F}\_N\rightarrow \Big(0, \frac{r}{1-r}\Big], \quad F\mapsto \sum\_{i\in F}r^i
$$
In particular, I wonder, when **part 1)** is not true for all $r\in\big(\frac{1}{2}, 1\big)$, if it is true that, for each $r\in\big(\frac{1}{2}, 1\big)$, I can always find $F\_r\in\mathcal{F}\_N$ such that $S\_r(F\_r)=1$. One can easily check that $S\_r$ is continuous for each $r\in\big(\frac{1}{2}, 1\big)$. If I instead let $\mathcal{F}\_N$ be the set of all **non-negative** integers, then I suppose $S\_{\frac{1}{n}}$ will be surjective for each $n\in\mathbb{N}$; however, under this assumption, I do not know if $S\_r$ will be surjective even when $r\neq\frac{1}{n}$ for each $n\in\mathbb{N}$, but my current biggest concern is on the those two questions above.
Any hints or thoughts will be appreciated! The same question is posted in **MS** and I would like thank [Ryszard Sszwarc](https://math.stackexchange.com/users/715896/ryszard-szwarc) for his help with this question. He proved that numbers that meet the condition in **part 1)** must be algebraic and irrational. However, for a fixed algebraic and irrational number, whether or not a necessary condition that guarantees the existence of such a finite set exists remain unclear.
| https://mathoverflow.net/users/151332 | Could the range of $\sum_{k\geq 1}r^{n(k)}$ for $r\in \big(\frac{1}{2}, 1\big)$ be continuous? | For (1), I'm not sure you're going to find a better condition than "$r$ is an algebraic number which is the root of some polynomial with coefficients $0$ and $1$"; I certainly don't think there's an intrinsic necessary and sufficient condition there.
For (2), the answer is yes, the function is always surjective. You could prove this with a greedy algorithm. For a fixed $x \in (0, \frac{r}{1-r}]$, define $x\_0 = x$, and for all $n > 0$, define $x\_{n}$ to be $0$ if $x\_{n-1} = 0$, and if $x\_{n-1} > 0$, define $x\_n = x\_{n-1} - r^m$, where $m = m\_n$ is the minimal positive integer for which $r^m \leq x\_{n-1}$.
Since $m$ was minimal, $x\_{n-1} < r^{m-1}$, and so $x\_n < x\_{n-1}(1-r)$. Therefore, $x\_n \rightarrow 0$. This means that $x = \sum\_n r^{m\_n}$. Clearly
$m\_0 > 0$, since $x < 1$. So it remains only to check that the $m\_n$ are strictly increasing. But this is easy; if $m\_{n+1} \leq m\_n$, then $x\_{n-1} \geq r^{m\_n}$ and $x\_n = x\_{n-1} - r^{m\_n} \geq r^{m\_{n+1}}$ , so $x\_{n-1} \geq r^{m\_n} + r^{m\_{n+1}}
\geq 2r^{m\_n} \geq r^{m\_n-1}$, contradicting minimality of $m\_n$. Therefore $m\_n$ is strictly increasing, and you have a representation $x = \sum\_n r^{m\_n}$.
By the way, this proof works even if there is a finite expansion, it will just end at that point (when $x\_n = 0$).
Another point is that even when your (1) is true, i.e. $1$ has a finite expansion, you can still find an infinite expansion of $1$! This is because, if $1 = r^{n\_1} + \ldots + r^{n\_k}$, then $1 - r^{n\_k} = r^{n\_1} + \ldots + r^{n\_{k-1}}$. But
$1 + r^{n\_k} + r^{2n\_k} + r^{3n\_k} + \ldots = \frac{1}{1 - r^{n-k}}$. Multiplying these yields
$1 = (r^{n\_1} + \ldots + r^{n\_{k-1}})(1 + r^{n\_k} + r^{2n\_k} + \ldots) =
r^{n\_1} + \ldots + r^{n\_{k-1}} + r^{n\_1 + n\_k} + r^{n\_2 + n\_k} + \ldots + r^{n\_{k-1} + n\_k} + r^{n\_1 + 2n\_k} + r^{n\_2 + 2n\_k} + \ldots + r^{n\_{k-1} + 2n\_k} + \ldots$.
So, you don't have to have the clause "when (1) is not true"; for every $r \in (1/2, 1)$, $1$ can be written as $S\_r(F)$ for some $F \in \mathcal{F}\_N$.
Also, something you said is confusing; you said if you allow $0$ exponents, then $S\_{1/n}$ would always be surjective, but this is clearly false. For instance, the range of $S\_{1/3}$ would be the set of numbers with ternary expansion (including a possible $(1/3)^0 = 1$ term) containing only $0$s and $1$s, which is clearly a Cantor set and not an interval.
| 5 | https://mathoverflow.net/users/116357 | 433820 | 175,461 |
https://mathoverflow.net/questions/433779 | 2 | I'm trying to check that certain examples of Young functions in the harmonic analysis literature are actually Young functions, and in doing so need to prove the following convexity-like inequality for $p > 1, \delta > 0$ and $0 \leq a \leq 1 < b$:
\begin{equation}\displaystyle \left( \frac{a+b}{2} \right)^p\left[ 1 + \ln\left( \frac{a+b}{2} \right) \right]^{p-1+\delta} \leq \frac{1}{2}a^p + \frac{1}{2} b^p[1+ \ln b]^{p-1+\delta} \label{ToProve} \end{equation}.
Clearly for fixed $b > 1$ this is true if $a = 0$ and $a = 1$ (the latter by the convexity of $\displaystyle x^p \left[1 + \ln x \right]^{p-1+\delta}$ for $x \geq 1$). Also, this is easy for all $0 \leq a \leq 1 < b$ assuming $p - 2 + \delta \geq 0$:
For fixed $0 \leq a \leq 1$ let $L(b)$ and $R(b)$ denote the left and right hand sides. Then then by convexity and the fact that $1 + \ln\left( \frac{a+1}{2} \right) \in (1-\ln 2, 1]$ we have \begin{align\*} L(1) \leq \left( \frac12 a^p + \frac12 \right)\left[ 1 + \ln\left( \frac{a+1}{2} \right) \right]^{p-1+\delta} \leq \frac12 a^p + \frac12 = R(1). \end{align\*}using basic calculus and recalling that $a < b$, we see that
\begin{align\*}
L'(b)
&= \frac{1}{2} \left(\frac{a+b}{2}\right)^{p-1}\left[p \left( 1+\ln\left(\frac{a+b}{2}\right)\right)^{p-1+\delta} + (p-1+\delta)\left( 1+\ln\left(\frac{a+b}{2}\right)\right)^{p-2+\delta} \right] \\
&\leq \frac12b^{p-1} \left[p \left( 1+\ln b\right)^{p-1+\delta} + (p-1+\delta)\left( 1+\ln b\right)^{p-2+\delta} \right] \\
&= R'(b),
\end{align\*}
as desired. Here, we clearly need $p-2+\delta\geq0$ to use the fact that $\ln$ is increasing.
Any thoughts on how to remove this assumption? Does the inequality I seek follow from anything out in the literature?
P.S. For anyone curious, the Young functions in question are $\Phi(x) = x^p \left[1 + \ln^+ x \right]^{p-1+\delta}$ for $p > 1$ and $\delta > 0$, which induce Orlicz spaces that appear very naturally in modern harmonic analysis (see C. Perez "On sufficient conditions for...", Proc. London Math. Soc., (1995), and other papers that cite this).
| https://mathoverflow.net/users/15280 | Elementary convexity example | Here is how to remove the assumption that $p-2+\delta\ge0$.
Let
\begin{equation\*}
s:=p-1+\delta.
\end{equation\*}
The conditions $p>1$ and $\delta>0$ imply $s\ge0$. No other conditions on $p$ and $s$ will be used or needed in what follows.
The inequality in question will follow from the inequality
\begin{equation\*}
L\le R(s)
\end{equation\*}
for
\begin{equation\*}
s\in[0,\infty),\ p\in[1,\infty),\ 0<a\le1\le b,
\end{equation\*}
where
\begin{equation\*}
L:=2^{1-p} (a+b)^p,\\
R(s):=R(s,p):=a^p \Big(1+\ln\frac{a+b}2\Big)^{-s}+b^p \Big(\frac{1+\ln b}{1+\ln\frac{a+b}2}\Big)^s.
\end{equation\*}
Clearly, $R(s)$ is convex in $s$.
>
> **Lemma 1:** $R'(0)\ge0$, and hence $R(s)$ is increasing in $s$.
>
>
>
By Lemma 1 and the power-means inequality,
\begin{equation\*}
R(s)\ge R(0)=a^p + b^p\ge 2^{1-p} (a+b)^p=L,
\end{equation\*}
as desired.
It remains to provide
*Proof of Lemma 1:* Let
\begin{equation\*}
R\_1(p,a,b):=R'(0)/a^p \\
=
\Big(\frac{b}{a}\Big)^p
\ln\frac{1+\ln b}{1+\ln\frac{a+b}{2}}
-\ln\Big(1+\ln\frac{a+b}{2}\Big),
\tag{0}\label{0}
\end{equation\*}
which is clearly nondecreasing in $p$. So,
\begin{equation\*}
aR\_1(p,a,b)\ge aR\_1(1,a,b) \\
=b \ln (1+\ln b)
-(a+b) \ln\Big(1+\ln\frac{a+b}{2}\Big),
\end{equation\*}
which latter is clearly decreasing in $a$. So,
\begin{equation\*}
aR\_1(p,a,b)\ge aR\_1(1,a,b)\ge R\_1(1,1,b)=bg(b),
\tag{1}\label{1}
\end{equation\*}
where
\begin{equation\*}
g(b):=\ln (1+\ln b) -\frac{1+b}b\,\ln\Big(1+\ln\frac{1+b}2\Big).
\end{equation\*}
So, in view of \eqref{0}, it remains to prove that
\begin{equation\*}
g(b)\overset{\text{(?)}}\ge0 \tag{2}\label{2}
\end{equation\*}
(for real $b>1$).
To begin the proof of \eqref{2}, let
\begin{equation\*}
g\_1(b):=g'(b)b^2 \\
= b \Big(\frac{1}{1+\ln b}-\frac{1}{1+\ln\frac{1+b}2}\Big)
+\ln\Big(1+\ln\frac{1+b}{2}\Big). \tag{3}\label{3}
\end{equation\*}
Now use the substitution $b=2e^t-1$, so that $t>0$ and $\ln\frac{1+b}{2}=t$. Let
\begin{equation\*}
\begin{aligned}
G\_2(t)&:=\frac{dg\_1(2e^t-1)}{dt}\,
\frac{(1+t)^2 (1+\ln (2 e^t-1))^2}{(2 e^t-1)t} \\
& =\frac{2 (t+e^t (1+t^2)) \ln (2e^t-1)}{(2 e^t-1) t}-1-\ln ^2(2 e^t-1).
\end{aligned}
\tag{4}\label{4}
\end{equation\*}
Next, note that
\begin{equation\*}
h(t):=t^3+t^2+t+2 e^t \left(t^2+1\right)-1>0
\end{equation\*}
(for $t>0$) and let
\begin{equation\*}
\begin{aligned}
G\_3(t)&:=G\_2'(t)\frac{t^2(2 e^t-1)^2}{2e^t h(t)} \\
& =\frac{2 t (t + e^t (1 + t^2))}{h(t)}- \ln(2e^t-1),
\end{aligned}
\tag{5}\label{5}
\end{equation\*}
\begin{equation\*}
G\_4(t):=G'\_3(t)\frac{(2e^t-1)h(t)^2}{2t} \\
=2 - t + t^3 + 2 e^{2 t} (1 - 7 t - 4 t^2 - 2 t^3 - t^4 + t^5) -
e^t (5 - 8 t + 6 t^3 + 3 t^4 + 2 t^5),
\tag{5.5}\label{5.5}
\end{equation\*}
\begin{equation\*}
G\_5:=G\_4',\ G\_6:=G\_5',\ G\_7:=G\_6', \tag{6}\label{6}
\end{equation\*}
\begin{equation\*}
G\_8(t):=G\_7'(t)\frac{e^{-t}}{P(t)}
=16 e^t \frac{Q(t)}{P(t)}-1,
\tag{7}\label{7}
\end{equation\*}
where
\begin{equation\*}
P(t):=189 + 736 t + 768 t^2 + 294 t^3 + 43 t^4 + 2 t^5,\ \\
Q(t):=-65 - 91 t - 8 t^2 + 40 t^3 + 18 t^4 + 2 t^5.
\end{equation\*}
Next,
\begin{equation\*}
G'\_8(t)=16 e^t \frac{S(t)}{P(t)^2}>0,
\end{equation\*}
where
\begin{equation\*}
S(t):=18356 + 31777 t + 25602 t^2 + 49850 t^3 + 84244 t^4 + 72903 t^5 +
34758 t^6 + 9548 t^7 + 1492 t^8 + 122 t^9 + 4 t^{10}.
\end{equation\*}
So, $G\_8(t)$ is increasing (in $t>0$). Also, $G\_8(0)<0$ and $G\_8(\infty-):=\lim\_{t\to\infty}G\_8(t)=\infty>0$. So, $G\_8$ is $-+$ (on $(0,\infty)$) -- that is, for some real $c>0$ we have $G\_8<0$ on $(0,c)$ and $G\_8>0$ on $(c,\infty)$.
So, by \eqref{7}, $G\_7$ is down-up (on $(0,\infty)$) -- that is, for some real $c>0$ we have that $G\_7$ is decreasing on $(0,c)$ and increasing on $(c,\infty)$.
Also, $G\_7(0)<0$ and $G\_7(\infty-)=\infty>0$. So, $G\_7$ is $-+$.
So, by \eqref{6}, $G\_6$ is down-up.
Also, $G\_6(0)<0$ and $G\_6(\infty-)=\infty>0$. So, $G\_6$ is $-+$.
So, by \eqref{6}, $G\_5$ is down-up.
Also, $G\_5(0)<0$ and $G\_5(\infty-)=\infty>0$. So, $G\_5$ is $-+$.
So, by \eqref{6}, $G\_4$ is down-up.
Also, $G\_4(0)<0$ and $G\_4(\infty-)=\infty>0$. So, $G\_4$ is $-+$.
So, by \eqref{5.5}, $G\_3$ is down-up.
Also, $G\_3(0)=0$ and $G\_3(\infty-)=-\ln2<0$. So, $G\_3<0$.
So, by \eqref{5}, $G\_2$ is decreasing.
Also, $G\_2(0+):=\lim\_{t\downarrow0}G\_2(t)=3>0$ and $G\_2(\infty-)=-\infty<0$. So, $G\_2$ is $+-$ (on $(0,\infty)$) -- that is, for some real $c>0$ we have $G\_2>0$ on $(0,c)$ and $G\_2<0$ on $(c,\infty)$.
So, by \eqref{4}, $g\_1$ is up-down -- that is,
for some real $c>1$ we have that $g\_1$ is increasing on $(1,c)$ and decreasing on $(c,\infty)$.
Also, $g\_1(1)=0$ and $g\_1(\infty-)=-\infty<0$. So, $g\_1$ is $+-$ (on $(1,\infty)$).
So, by \eqref{3}, $g$ is up-down.
Also, $g(1+)=0$ and $g(\infty-)=0$. Thus, \eqref{2} follows. $\quad\Box$
| 2 | https://mathoverflow.net/users/36721 | 433821 | 175,462 |
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