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https://mathoverflow.net/questions/418549 | 3 | Consider the $D\_{\mathbb{A}^1}$-module $M:=D\_{\mathbb{A}^1}/(x)$, and the map $f:z\mapsto z^k$. I want to know $f^\*(M)$. I believe it only has a single non-zero cohomology, namely in degree $0$, which is equal to
$$\mathbb{C}[z]\otimes\_{\mathbb{C}[z^k]}(D/(x)),$$
with connection
$$\partial(z^q\otimes \partial^w)=qz^{q-1}\otimes \partial^w+kz^{q+k-1}\otimes \partial^{w+1}.$$
Now by general theory this should be holonomic, and in particular finitely generated. However, I have some trouble finding a generating set. For example, it is not clear to me how the elements $1\otimes \partial^{w}, w=1,2,\dots,$ can be generated from a finite generating set.
Any help or hints would be appreciated.
| https://mathoverflow.net/users/64302 | Explicit computation of D-modules pullback | If you write the degree of $z^q \otimes \delta^w$ as $ kw -q$ then there is a one-dimensional vector space of elements of each degree $\geq (1-k)$ (since if $q \geq k$ we can reduce by bring $z^q$ over to the other side) and $\partial$ takes elements of a given degree to elements of the same degree plus one.
So to check that every element can be generated from a finite generating set, it suffices to check that there are only finitely many degrees where $\partial$ applied to the generator in that degree produces the zero multiple of the generator in the next degree (instead of a nonzero multiple).
In fact, I claim there are no such degrees, and the whole thing is generated by $z^{k-1} \otimes 1$. I calculated this by passing to constructible sheaves, computing the pullback there, and going back but let's check this explicitly in D-modules.
To check this, note that for $q\geq 1$,
$$\partial(z^q\otimes \partial^w)=qz^{q-1}\otimes \partial^w+kz^{q+k-1}\otimes \partial^{w+1} $$ $$= qz^{q-1}\otimes \partial^w+kz^{q-1}\otimes x\partial^{w+1}$$ $$ = qz^{q-1}\otimes \partial^w+kz^{q-1}\otimes (\partial^{w+1} x - (w+1) \partial^w) $$ $$ = (q- k (w+1)) z^{q-1}\otimes \partial^w $$
and we have $q \leq k-1$ for the generator $q- k (w+1)>0$.
For the $q=0$ case, the argument is simpler - the first term vanishes and the second term is nontrivial.
| 3 | https://mathoverflow.net/users/18060 | 418559 | 170,420 |
https://mathoverflow.net/questions/418556 | 2 | My question is about the proof of Proposition A.6.1.6 in Lurie's Spectral Algebraic Geometry, which says the following:
Let $\mathcal{X}$ be any $\infty$-topos and denote by $\mathcal{X}^{coh}$ the full subcategory of the coherent objects. Then $\mathcal{X}^{coh}$ is a local $\infty$-pretopos.
In the proof, to show that $\mathcal{X}^{coh}$ is closed under geometric realizations of groupoid object, Lurie in a crucial way uses Proposition A.2.1.5 which states that if the pullback of a morphism along some effective epimorphism is relatively $n$-coherent, then the original morphism is already relatively $n$-coherent. This proposition, however, is only applicable if one assumes furthermore local $n$-coherence of $\mathcal{X}$, which again enters the proof in a very crucial way (no find an $n$-coherent cover of the object $U$ in the notation there).
Thus, it seems like the proof only works if $\mathcal{X}$ is already locally coherent, unless I just misunderstood the argument. Can it still somehow be salvaged for a general $\infty$-topos, as was claimed? I unfortunately also didn't find a proof of a classical analogon of this statement in the literature.
| https://mathoverflow.net/users/156537 | Subcategory of coherent objects in an $\infty$-topos forming a local $\infty$-pretopos | If $X\_0\to X$ is an effective epimorphism and $X\_0$ is locally $n$-coherent, then $X$ is also locally $n$-coherent: every $Y$ over $X$ is covered by $Y\times\_XX\_0$, which is in turn covered by a coproduct of $n$-coherent objects. So in the proof of A.6.1.6 we know beforehand that $X$ is locally $n$-coherent for all $n$, hence all the results that assume local $n$-coherence apply.
| 2 | https://mathoverflow.net/users/20233 | 418574 | 170,422 |
https://mathoverflow.net/questions/418534 | 9 | In MacLane's *Categories for the working mathematician*, the author shows that the evaluation at 1 gives an equivalence of categories $\mathrm{hom}\_{\mathrm{BMC}}(B,M)\simeq M\_0$ where $B$ is the braid category, $M$ is a braided monoidal category and $M\_0$ is the underlying (ordinary) category to $M$. As a result of that, he states a second theorem (coherence theorem) claiming that each composite of canonical maps in $M$ induces a braiding (element of the braid group), and that two such composites are equal for all $M$ if and only they induce the same element braiding.
I understand how to construct the braiding from a given composition of canonical morphisms, but I fail to write it properly using the previous theorem and I don't know how to prove that two such maps are equal if and only if they induce the same braiding.
I also read the original preprint of Joyal and Street which appears in the references given by MacLane ([http://maths.mq.edu.au/~street/JS1.pdf](http://maths.mq.edu.au/%7Estreet/JS1.pdf)) but I still don't see how to write a proper proof.
| https://mathoverflow.net/users/479089 | Coherence theorem in braided monoidal categories | Theorem 1 says precisely that for every braided monoidal category $M$ and every object $V \in M$, any isotopy class of braid on $n$ strands induces an isomorphism
$$
V^{\otimes n}\longrightarrow V^{\otimes n}.
$$
In particular this theorem also says this isomorphism is well defined, i.e. it does not depend one the particular braid one choose in the equivalence class. In other words, it does indeed says that if two braids are representatives of the same element in the braid group $B\_n$, then they define the same morphism for all pair $(M,V)$. Conversely, applying this to the case $M=B$ and $V=\bullet$, i.e. to the identity functor $B\rightarrow B$, one sees that this is an "only if": if two braids are **not** isotopic, then there exists at least one braided monoidal category (e.g. $B$ itself) for which these two braids induce non-equal morphisms.
So to me theorem 2 is really just a slightly weaker formulation of theorem 1 but they're still pretty much saying the same thing..
Also as Donald says, theorem 1 stats precisely that $B$ is the free braided monoidal category on one object. As a useful analogy, if $A$ is an abelian group and $A\_0$ its underlying set, the statement that there is a natural bijection
$$
Hom\_{Ab}(\mathbb{Z},A)\cong A\_0$$
states precisely that $\mathbb{Z}$ is the free abelian group on one generator.
| 2 | https://mathoverflow.net/users/13552 | 418583 | 170,424 |
https://mathoverflow.net/questions/418596 | -1 |
>
> Does $g\in C([0,1],[0,1])=A$ exist such that $\{g^n ,n\in\mathbb N\}$ is dense in $A$ provided with the uniform norm?
>
>
>
with $g^2=g \circ g $
If we can find $g$ then $F$ a closed of $A$, $id \in F$ with $g\circ F\subset F$, we have $F=A$?
| https://mathoverflow.net/users/110301 | The grail of functional analysis? | The answer is no. If there was, then for some $N$ we would have $\|g^N\|\leq 1/2$, in order for $g^N$ to be within distance $1/2$ of the constant $0$ function. This means that the range of $g^N$ is contained in $[0,1/2]$. But then the same holds for $g^n$ for all $n\geq N$, since $g^n=g^N\circ g^{n-N}$. In particular $\{g^n\}$ cannot be dense in $A$.
Essentially the same argument shows that if some constant function is a limit point of the sequence $g^n$, then it is in fact the limit of this sequence.
| 13 | https://mathoverflow.net/users/30186 | 418599 | 170,429 |
https://mathoverflow.net/questions/418509 | 2 | I am trying to understand this paper by [Chapelle and Li "An Empirical Evaluation of Thompson Sampling" (2011)](https://papers.nips.cc/paper/2011/file/e53a0a2978c28872a4505bdb51db06dc-Paper.pdf). In particular, I am failing to derive the equations in algorithm 3 (page 6). The first equation looks like an NLL of $p(x|w) \, p(w)$ where the latter is the prior shown in the second line of the algorithm; modulo the constant term that does not depend on $w$. The question is: what is the likelihood? Obviously, it is not the canonical cross-entropy with $y\_j \in \{0, 1\}$ but almost the cross-entropy with $y\_j \in \{-1, +1\}$?
Furthermore, I don't understand the update step for $q\_j$ in the last line: I can derive something that comes close using the Laplace approximation,
$$q\_j \leftarrow -\frac{\partial^2}{\partial w\_j^2} \ln p(x, w),$$
and discarding correlations... but it is not the same and there are still some $y\_j$ and other terms floating around.
Can someone tell me, how to derive these equations?
Thanks a lot!
| https://mathoverflow.net/users/177845 | Derive equation for regularized logistic regression with batch updates | I found the solution (with the help of a friend: cudos!). The posterior is
$$\begin{align\*}
-\log p(\boldsymbol{w}|\boldsymbol{x}) &= -\log p(\boldsymbol{x}|\boldsymbol{w}) - \log p(\boldsymbol{w}) + \text{const.} \\
&= \sum\limits\_j \log \left( 1 + \exp(-y\_j \boldsymbol{w}^\top \boldsymbol{x}\_j) \right) + \sum\limits\_i \frac{q\_i (w\_i - m\_i)^2}{2} + \text{const.}'
\end{align\*}$$
where the constant terms do not depend on $\boldsymbol{w}$, and with the NLL
$$\begin{align\*}
-\log p(\boldsymbol{x}|\boldsymbol{w}) &= \sum\_j \log \left[ \frac{1 + y\_j}{2} \left( 1 + \mathrm{e}^{-\boldsymbol{w}^\top \boldsymbol{x}\_j} \right) + \frac{1 - y\_j}{2} \left( 1 + \mathrm{e}^{+\boldsymbol{w}^\top \boldsymbol{x}\_j} \right) \right] \\
&= \sum\limits\_j \log \left( 1 + \mathrm{e}^{-y\_j\boldsymbol{w}^\top \boldsymbol{x}\_j} \right)
\end{align\*}$$
for $y\_j = \{-1, +1\}$ **(!)**.
Ignoring correlations, the Laplace approximation then yields
$$\begin{align\*}
q\_i &\leftarrow - \left. \frac{1}{p(\boldsymbol{\hat w} | \boldsymbol{x})} \frac{\partial^2 p(\boldsymbol{w} | \boldsymbol{x})}{\partial w\_i^2} \right|\_{\boldsymbol{w} = \boldsymbol{\hat w}} \\
&= \left. -\frac{\partial^2 \log p(\boldsymbol{w} | \boldsymbol{x})}{\partial w\_i^2} \right|\_{\boldsymbol{w} = \boldsymbol{\hat w}} \\
&= q\_i + \sum\limits\_j x\_{ij}^2 \frac{1}{1 + \mathrm{e}^{-y\_j \boldsymbol{\hat w}^\top \boldsymbol{x}\_j}} \frac{1}{1 + \mathrm{e}^{+y\_j \boldsymbol{\hat w}^\top \boldsymbol{x}\_j}} \\
&= q\_i + \sum\limits\_j x\_{ij}^2 \frac{1}{1 + \mathrm{e}^{-\boldsymbol{\hat w}^\top \boldsymbol{x}\_j}} \frac{1}{1 + \mathrm{e}^{+\boldsymbol{\hat w}^\top \boldsymbol{x}\_j}} \\
&= q\_i + \sum\limits\_j x\_{ij}^2 \, p\_j (1-p\_j)
\end{align\*}$$
with $x\_{ij} = (\boldsymbol{x}\_j)\_i$ and $$\boldsymbol{\hat w} = \underset{\boldsymbol{w}}{\operatorname{argmax}} p(\boldsymbol{x} | \boldsymbol{w}) \,.$$
| 3 | https://mathoverflow.net/users/177845 | 418600 | 170,430 |
https://mathoverflow.net/questions/418361 | 5 | Let $X$ be a connected compact complex manifold, $U$ an open subset of $X$ such that the complement of $U$ in $X$ is an analytic subset of codimension at least 2 in $X$. Let $O\_X$ (resp. $O\_U$) be the sheaf of holomorphic functions on $X$ (resp. on $U$). If $n$ is a nonnegative integer then there is a natural homomorphism of complex vector spaces
$$r\_n: H^n(X,O\_X) \to H^n(U,O\_U).$$
The second Riemann extension theorem actually asserts that $r\_n$ is an isomorphism for $n=0$. I am looking for a reference where it is proven that $r\_n$ is an isomorphism, say, for $n=1$ or $2$ (may be, under some additional assumptions). Thanks!
| https://mathoverflow.net/users/9658 | A cohomological variant of the second Riemann's extension theorem | For the first cohomology statement, you need the codimension to be at least three and for the first and second cohomology, codimension four. This theorem was proved by G Scheja in [1]. You can also find a proof in the book by Banica and Stanasila ([2] Chapter II, §II.3 pages 66-67).
**References**
[1] Constantin Banica, Octavian Stanasila, *Algebraic methods in the global theory of complex spaces*. Rev. English ed. (English) Bucuresti: Editura Academiei; London-New York-Sydney: John Wiley&Sons, pp. 296 (1976), [MR0463470](https://mathscinet.ams.org/mathscinet-getitem?mr=MR0463470), [Zbl 0334.32001](https://zbmath.org/?q=an%3A0334.32001).
[2] Günter Scheja, "Riemannsche Hebbarkeitssätze für Cohomologieklassen" (German) Mathematische Annalen 144, 345-360 (1961), [MR0148941](https://mathscinet.ams.org/mathscinet-getitem?mr=MR0148941), [Zbl 0112.38001](https://zbmath.org/?q=an%3A0112.38001).
| 3 | https://mathoverflow.net/users/4696 | 418609 | 170,435 |
https://mathoverflow.net/questions/418607 | 4 | I am looking for a smooth curve $C$ of genus $g=2k+1 \geq 5$ over the complex numbers, endowed with a *free* $\mathbb{Z}/2$-action such that the following condition is satisfied: denoting by $$H^0(C, \, \omega\_C) = V^+ \oplus V^{-},$$ the decomposition of $H^0(C, \, \omega\_C)$ into invariant and anti-invariant subspaces, there are bases
$\{ \omega\_1, \ldots, \omega\_{k+1} \}$ of $V^+$
and $\{\omega\_{k+2}, \ldots, \omega\_g \}$ of $V^{-}$ such that the divisors $\mathrm{div}({\omega\_i})$, $\mathrm{div}({\omega\_j})$ have disjoint supports for all $1 \leq i < j \leq g$.
>
> **Question.** Does such a curve exist?
>
>
>
**What I have tried.** Let me explain one example that does *not* work. I considered a hyperelliptic curve $C$ of affine equation $$y^2=(x^2-a\_1^2)\ldots (x^2-a\_{2g+2}^2),$$ where the scalars $a\_i$ are general. Since $g$ is assumed to be odd, such a curve admits a free $\mathbb{Z}/2$-action given by $(x, \, y) \mapsto (-x, \, -y)$. A basis for $H^0(C, \, \omega\_C)$ is given by $$\left \{ \omega\_i \; \; | \; \; 0 \leq i \leq g-1 \right \},$$ where $\omega\_i=x^i \frac{dx}{y}$. Therefore we have $$V^+=\{\omega\_i \, | \, i\textrm{ even} \}, \quad V^-=\{\omega\_i \, | \, i\textrm{ odd} \}.$$ This shows that all the $1$-forms in $V^-$ vanish at the points of the curve with $x=0$, hence the condition above cannot be satisfied when $g \geq 5$ (that means $\dim V^- \geq 2$).
| https://mathoverflow.net/users/7460 | Looking for a curve with a special, free $\mathbb{Z}/2$-action | Your examples are the *only* curves with fixed-point-free involution that don't have such a basis.
Let $C\_0$ be a curve of genus $k+1$ and let $L$ be a nontrivial line bundle on $C\_0$ with $L^2\cong \mathcal O\_{C\_0}$. Then $\mathcal O\_{C\_0}+L$ has a natural algebra structure defined using that isomorphism. The relative Spec is a double cover $C$ of genus $2k+1$. Since it's a double cover, it has a natural involution. All curves with involution arise this way.
This involution acts on $H^0(C, \omega\_C)$. The invariant subspace consists of 1-forms that are equal on each of the two fibers of $C \to C\_0$, hence are pullbacks from $C\_0$, and thus is isomorphic to $ H^0(C\_0 , \omega\_{C\_0})$, while the anti-invariant subspace consits of 1-forms with opposite values on the two fibers, hence locally pullbacks of 1-forms from $C\_0$ multiplied by section of $L$, and thus is isomorphic to $H^0(C\_0, \omega\_{C\_0}\otimes L)$.
For your criterion, it suffices that $\omega\_{C\_0}$ and $\omega\_{C\_0}\otimes L$ are both base-point free. Indeed, for a base-point free line bundle, a generic section avoids any given finite set of points, so it is easy to choose a basis where each section in the basis successively avoids the vanishing locus of the previous ones.
By Serre duality, $\omega\_{C\_0}$ is always base-point-free and $\omega\_{C\_0} \otimes L$ is base-point free if and only if the divisor class of $L$ can be expressed as the difference of two points of $C\_0$.
If $P-Q$ is a two-torsion divisor, then $(2P, 2Q)$ are two divisors of degree $2$ equivalent to each other, thus defining a $g^1\_2$, so $C\_0$ is hyperelliptic. Even in the hyperelliptic case, this can only happen if $P$ and $Q$ are Weierestrass, so there are $\binom{ 2k+4}{2}$ two-torsion line bundles where such a basis does not exist on the cover, out of $2^{2k+2}-1$ nontrivial two-torsion line bundles total, so there are plenty of line bundles that do work for any $k \geq 2$. For the ones that don't, the cover can be obtained from the double cover of $\mathbb P^1$ branched at the $x$ coordinates of $P$ and $Q$, meaning that, after a change of variables, it's given by your construction.
| 7 | https://mathoverflow.net/users/18060 | 418612 | 170,436 |
https://mathoverflow.net/questions/418603 | 2 | $\newcommand{\pt}{\mathit{pt}}$For $d>1$ is it possible to understand $\text{Aut}(\text{Sym}^n(\mathbb{P}^d\_{\mathbb{C}}))$? Automorphisms mean biregular morphisms from the variety to itself. Not let $\pt$ be a point on the projective space. Consider the point on $\text{Sym}^n(\mathbb{P}^d\_{\mathbb{C}})$ corresponding to the point $\pt$ but with degree $n$. Let's denote it by $\pt^n$. What is the orbit of $\pt^n$ under $\text{Aut}(\text{Sym}^n(\mathbb{P}^d\_{\mathbb{C}}))$? Do they necessarily get mapped to another point of the form $\pt\_2^n$?
| https://mathoverflow.net/users/127776 | Automorphisms of symmetric powers of projective space | $\DeclareMathOperator\PGL{PGL}$Yes, it's possible to understand this automorphism group, and yes, such a point is necessarily mapped to another point of the same form.
I claim the automorphism group is given by $\PGL\_{d+1}$, acting in the way induced by its action on $\mathbb P^d$. The statement about orbits follows immediately.
To prove this, let's consider the smooth locus of $\operatorname{Sym}^n(\mathbb P^d)$. I claim it consists of only distinct unordered tuples of points. Since the singular locus is closed, it suffices to check that the point $(x\_1,x\_1, x\_2,\dots, x\_{n-1})$ is not smooth, for which it suffices to check that its tangent space has dimension $d (n-2) + d+ d (d+1)/2> nd$.
Now the map $(\mathbb P^d)^n \to \operatorname{Sym}^n(\mathbb P^d)$, restricted to the smooth locus, is a finite etale covering from $(\mathbb P^d)^n$ minus the big diagonal to the smooth locus. Since $\mathbb P^d$ is simply-connected, and the big diagonal has codimension $d >1$, the complement of the big diagonal is simply-connected and thus equal to the universal cover of the smooth locus.
Now every automorphism of $\operatorname{Sym}^n(\mathbb P^d)$ restricts to an automorphism of the smooth locus, hence lifts to an automorphism of its universal cover, thus extends to an automorphism of the normalization of the variety in that cover, which is $(\mathbb P^d)^n$. The automorphism group of $(\mathbb P^d)^n$ is the semidirect product $(\PGL\_{d+1})^n \rtimes S\_n$, so it must arise from one of those, but only the subgroup $\PGL\_{d+1} \times S\_n$ preserve the map to $\operatorname{Sym}^n(\mathbb P^d)$ and give automorphisms of $\operatorname{Sym}^n(\mathbb P^d)$. Since $S\_n$ gives trivial automorphisms of $\operatorname{Sym}^n(\mathbb P^d)$, every automorphism must come from $\PGL\_{d+1}$.
| 7 | https://mathoverflow.net/users/18060 | 418613 | 170,437 |
https://mathoverflow.net/questions/418592 | 9 | I am currently studying the applications of games in quantum information theory and related fields and I am aware of its uses in places like model theory and set theory. So I was curious, what are some other somewhat surprising places(somewhat because I am aware of the subjectivity of the term) where games were/are useful tools?
| https://mathoverflow.net/users/467143 | Surprising applications of the theory of games? | I think evolutionary biology is a major application, if you're accepting answers outside of math. The central notion of an [evolutionarily stable strategy (ESS)](https://www.siue.edu/%7Eevailat/ev-gt-simple.htm) is a Nash equilibrium.
| 9 | https://mathoverflow.net/users/23141 | 418616 | 170,438 |
https://mathoverflow.net/questions/417015 | 2 | Let $f: [0, 1] \to \mathbb R$ be a bounded, continuous function, and $W$ a standard Brownian motion.
Denote $Y := \int\_0^1 f(t) \, dW\_t$.
For each $\varepsilon > 0$, consider the conditioned random variable $Y\_\varepsilon := \varepsilon Y | \{W\_1 \geq \frac{1}{\epsilon}\}.$
**Question:** Is it true that $Y\_\varepsilon$ converges in law to the deterministic random variable $\int\_0^1 f(t) \, dt$ as $\varepsilon \to 0$?
| https://mathoverflow.net/users/173490 | A large noise limit | Let $\varphi$ be the standard normal density. Since
$P[W\_1 \ge x] =(1+o(1))\varphi(x)/x$ as $ x \to \infty$ by [1], we obtain for fixed $\delta>0$ that as $\epsilon \to 0$,
$$P[W\_1 \ge \epsilon^{-1}+\delta \,| \,W\_1 \ge \epsilon^{-1}] \le 2 \varphi(\epsilon^{-1}+\delta)/ \varphi(\epsilon^{-1}) \to 0 \,, $$
so in particular,
$(\*)$ given $W\_1 \ge \epsilon^{-1}$, we have $\epsilon W\_1 \to 1$ in probability as $\epsilon \to 0$.
Write $W\_t=tW\_1+B\_t$, where
$$\{B\_t: 0 \le t \le 1\}=\{W\_t-tW\_1 : 0 \le t \le 1\}$$ is a standard Brownian bridge in $[0,1]$, independent of $W\_1$. Then
$$Y = \int\_0^1 f(t) \, dW\_t = W\_1 \int\_0^1 f(t) \, dt + \int\_0^1 f(t) \, dB\_t \,,
$$
so by $(\*)$, given $W\_1 \ge \epsilon^{-1} \,,$ we have
$$ \epsilon Y = \epsilon W\_1 \int\_0^1 f(t) \, dt + \epsilon \int\_0^1 f(t) \, dB\_t \to \int\_0^1 f(t) \, dt
$$
as $\epsilon \to 0$ in probability (and hence also in law.)
[1] <https://en.wikipedia.org/wiki/Mills_ratio#cite_note-S-4>
| 4 | https://mathoverflow.net/users/7691 | 418622 | 170,441 |
https://mathoverflow.net/questions/418403 | 9 | One characteristic of the surreal numbers is that they are a monster model of the first-order theory of real numbers, according to Joel David Hamkins in [this post](https://mathoverflow.net/questions/126158/a-mother-of-all-groups-what-kind-of-structures-have-mother-of-alls). Thus they are real-closed, and every other real-closed field embeds into them with very nice properties. So we may ask if one can similarly create a monster model of the first-order theory of natural numbers as a kind of "surnaturals."
One natural question to ask is if the non-negative elements in Conway's ring of "omnific integers" would fit the bill, given that they are often promoted as the "surreal version of the integers" or something like that. However, it is rather easy to see that this is not true: the omnific integers have the interesting property that their field of fractions is the entire surreal number field, thus there are two omnific integers whose quotient is the square root of 2, whereas no such integers exist in any model of the naturals.
Thus in some sense, the omnific integers are not quite the most direct correspondant of the natural numbers within the surreal numbers. So one question is if the monster model can be built constructively, similarly to the surreals. I would suspect that such a construction exists, given that Joel David Hamkins was able to explicitly construct a monster model of all groups (!) in the above post, which would seem to suggest a monster model of the integers also exists, and thus the naturals as those non-negative integers.
How can one build such a model?
EDIT: I initially asked if the Grothendieck ring of the ordinals with commutative addition/multiplication could be a monster model for the integers, but it is apparently too small, since for $\omega$ we should have that $\omega$ is either even or odd. This means there should be some element $x$ such that $x + x = \omega$, or $x + x = \omega + 1$, so either $\omega/2$ or $(\omega + 1)/2$ should be in our set. Also, there will need to be some element in the monster model which is divisible by every standard finite number, and if that were $\omega$, we'd thus need to have $\omega \cdot q$ for every rational q. Thanks to Noah Schweber and Emil Jeřábek for pointing this out, and also to Emil for clearing up some confusion I had about whether the non-unique factorization of omnific integers necessarily implies they are not a model of the naturals (apparently it does not, but there are other reasons why, such as there are two omnifics whose ratio is sqrt(2), which is provably not true in any model of PA).
| https://mathoverflow.net/users/24611 | A "surnatural numbers" as a largest model of the natural numbers | I asked (and also answered) [a more general version of this question](https://mathoverflow.net/questions/304290/for-which-theories-does-zfc-without-global-choice-prove-the-existence-of-a-prope) a while ago. To summarize the answer, some [results of Kanovei and Shelah](https://arxiv.org/abs/math/0311165) have the following corollary:
**Fact.** In $\mathsf{ZFC}$ there is a uniform procedure for building 'set-saturated,' class-sized elementary extensions of arbitrary structures. That is to say there are formulas $S(M,L,x)$ and $F(M,L,f,x)$ in the language of set theory such that in any model $V \models \mathsf{ZFC}$ if $L \in V$ is a language and $M \in V$ is an $L$-structure, then the following hold (where $M^\ast = \{x \in V : V \models S(M,L,x)\}$):
* $M \subseteq M^\ast$,
* if $\varphi \in V$ is an $L$-formula with free variables $x\_0,\dots,x\_n$ and $\bar{a} \in M^\ast$ is an $n$-tuple, then $V \models F(M,L,\exists x\_n\varphi,\bar{a})$ if and only if $V \models (\exists x \in M^\ast) F(M,L,\varphi,\bar{a}x)$ (where we are using some fixed coding of tuples in $\mathsf{ZFC}$),
* furthermore, if $\bar{c} \in M$ is an $(n+1)$-tuple, then $V \models F(M,L,\varphi,\bar{c})$ if and only if $V \models “M \models \varphi(\bar{c})”$ (in particular, if $\varphi$ is a sentence, then $V \models F(M,L,\varphi,\varnothing)$ if and only if $V \models “M \models \varphi”$),
* $F$ is compatible with Boolean combinations (i.e., $V\models F(M,L,\varphi\wedge \psi,\bar{a})$ if and only if $V\models F(M,L,\varphi,\bar{a})\wedge F(M,L,\psi,\bar{a})$, etc.), and
* if $A \subseteq M^\ast$ is a set and $p(x)$ is a finitely satisfiable set of $L\_A$-formulas with free variable $x$, then there is $b \in M^\ast$ such that for any $\varphi(x,\bar{a}) \in p(x)$, $V \models F(M,L,\varphi,b\bar{a})$.
So to state it informally, $S(M,L,x)$ defines the universe of a class-sized elementary extension of $M$ and $F(M,L,f,x)$ is its truth predicate.
Applying this to the naturals tells us that there is a formula that defines a proper class monster model of $\mathrm{Th}(\mathbb{N})$ in any model of $\mathsf{ZFC}$.
One thing to note, though, is that without global choice (which makes my original question trivial), it's unclear whether there's always a definable isomorphism between different set-saturated class-sized models of a given theory. I believe this is related to [an unanswered MathOverflow question of Hamkins](https://mathoverflow.net/questions/227849/is-the-universality-of-the-surreal-number-line-a-weak-global-choice-principle). That said, if $M$ and $N$ are $L$-structures and $M \equiv N$, then there will be an isomorphism between $M^\ast$ and $N^\ast$ that is definable with certain parameters.
Another thing to note is that some constructions that model theorists commonly use with the monster model are unclear in the context of these class monster models. There isn't necessarily a good way to talk about arbitrary global types, for instance. You do, however, get a good homogeneity property: There is a subgroup $G$ of $\mathrm{Aut}(M^\ast)$ that can be represented as a class in a definable way which has the property that if $\bar{a}$ and $\bar{b}$ are set-sized tuples that realize the same type, then there is a $\sigma \in G$ such that $\sigma \bar{a} = \bar{b}$.
| 5 | https://mathoverflow.net/users/83901 | 418630 | 170,443 |
https://mathoverflow.net/questions/418629 | 1 | We consider the sum
$$ \sum\_{m \in \mathbb Z^2} \frac{1}{(3 m\_1^2+3m\_2^2+3(m\_1+m\_1m\_2+m\_2)+1)^2}. $$
Numerically, it is not particularly hard to see that the value of this series is well below $4$, indeed one gets numerically an upper bound of roughly $3.43$
I wonder if there is analytically a quick argument that the value of this double sum is less than $4$?
EDIT: One could observe that
$$ \sum\_{m\_1=-1}^1\sum\_{m\_2=-1}^1 \frac{1}{(3 m\_1^2+3m\_2^2+3(m\_1+m\_1m\_2+m\_2)+1)^2} = \frac{40545}{12544}. $$
So it could suffice to show that the remaining terms are sufficiently small.
| https://mathoverflow.net/users/457901 | Upper bound on double series | As noted in the comment by Beni Bogosel, the sum in question is
\begin{equation}
s:=\sum\_{x=-\infty}^\infty\sum\_{y=-\infty}^\infty\frac1{f(x,y)^2},
\end{equation}
where
\begin{equation}
f(x,y):=\frac32\, ((x + 1)^2 + (y + 1)^2 + (x + y)^2) - 2.
\end{equation}
Note that
\begin{equation}
f(x,y)\ge x^2+y^2+2\ge2\sqrt{x^2+1}\sqrt{y^2+1}\text{ if } \max(|x|,|y|)\ge3.
\end{equation}
So,
\begin{equation}
s\le s\_9+r\_{10},
\end{equation}
where
\begin{equation}
s\_9:=\sum\_{x=-9}^9\sum\_{y=-9}^9\frac1{f(x,y)^2}<3.42256
\end{equation}
and
\begin{equation}
\begin{aligned}
r\_k&:=4\sum\_{x\ge k}\sum\_{y=-\infty}^\infty\frac1{4(x^2+1)(y^2+1)} \\
&=\sum\_{x\ge k}\frac1{x^2+1}\sum\_{y=-\infty}^\infty\frac1{y^2+1} \\
&\le\int\_{k-1/2}^\infty\frac{dx}{x^2}\,\Big(1+2\int\_2^\infty\frac{dy}{y^2}\Big)
=\frac5{k-1/2}<0.52632
\end{aligned}
\end{equation}
for $k=10$.
Thus,
\begin{equation}
s<3.42256 + 0.52632<4,
\end{equation}
as desired.
| 3 | https://mathoverflow.net/users/36721 | 418636 | 170,445 |
https://mathoverflow.net/questions/418634 | 1 | A real random variable $X$ is said to be *subgaussian* if there exists an $a > 0$ such that $\mathbb{E}[e^{\lambda X}] < e^{a^2 \lambda^2}$ for all $\lambda \in \mathbb{R}$. The space of such random variables admits a Banach space structure, with an Orlicz norm given by $$\| X \|\_{\psi\_2} = \inf\left\{ t > 0 : \mathbb{E}\left[ \psi\_2\left( \frac{|X|}{t} \right) \right] \leq 1 \right\},$$ with $\psi\_2(x) = e^{x^2} - 1$.
The main question here is a simple one to ask, but I've been unable to find an answer. Suppose I have a sequence of i.i.d. subgaussian random variables. Is it known whether or not the normalized sums $$S\_n =
\frac{1}{\sqrt{n}} \sum\_{i=1}^n X\_i,$$
converge in this norm to a normally distributed random variable? Moreover, if they do not do so in general, is a sufficient condition known?
| https://mathoverflow.net/users/479167 | The central limit theorem in the subgaussian Orlicz norm | $\newcommand\ep\varepsilon$No. If this were so, then (by Lemma 1 below) $S\_n$ would converge to a normally distributed random variable $Y$ in probability, which is [false](https://math.stackexchange.com/questions/2145140/clt-cannot-be-enhanced-to-convergence-in-probability) for any iid $X\_i$'s.
---
>
> **Lemma 1:** If $\|S\_n-Y\|\_{\psi\_2}\to0$ (as $n\to\infty$), then $S\_n\to Y$ in probability.
>
>
>
*Proof:* Suppose that $\|S\_n-Y\|\_{\psi\_2}\to0$. Then
$$E\psi\_2(|S\_n-Y|/t\_n)\le1$$
for all natural $n$, where $t\_n:=\|S\_n-Y\|\_{\psi\_2}+1/n\to0$.
So, by Markov's inequality, for each real $\ep>0$,
$$P(|S\_n-Y|\ge\ep)\le\frac{E\psi\_2(|S\_n-Y|/t\_n)}{\psi\_2(\ep/t\_n)}
\le\frac1{\psi\_2(\ep/t\_n)}\to0.$$
So, $S\_n\to Y$ in probability. $\quad\Box$
| 0 | https://mathoverflow.net/users/36721 | 418637 | 170,446 |
https://mathoverflow.net/questions/418579 | 5 | Let $X$ be a finite type scheme over $\mathbb{Z}\_p$ for some prime $p$. Assume that $X\_{\mathbb{Q}\_p}$ is smooth of dimension $n$, but not necessarily irreducible. Then is
$$X(\mathbb{Z}/p^k\mathbb{Z}) = O(p^{kn})$$
as $k \to \infty?$
| https://mathoverflow.net/users/5101 | Number of points on schemes modulo $p^k$ | Yes, this is true, for elementary reasons (i.e. not to do with the Igusa zeta function or something).
By passing to an open cover, we may assume $X$ is affine, say $X = \operatorname{Spec} \mathbb Z\_p[x\_1,\dots, x\_N]/ (f\_1,\dots, f\_m)$. The Jacobian of this system of equations is an $N \times m$ matrix.
Consider for each $x \in X(\mathbb Z/p^k)$ the minimum $p$-adic valuation of the determinant of an $N-n \times N-n$ submatrix of this Jacobian, considering the $p$-adic valuation of $0$ in $\mathbb Z/p^k$ to be $k$. I claim this is bounded by some constant $c$. If not, we can choose a sequence where this increases to $\infty$ (necessarily with increasing $k$) and then a subsequence which converges $p$-adically, which must converge to a singular point of $X$.
So we may assume that one of these determinants has $p$-adic valuation at most $c$. Summing over the finitely many possible choices, it suffices to bound the number of points assuming a particular determinant divides $p^c$. We may as well throw away all the equations not involved in our fixed submatrix. We will show that, setting all variables not involved in this submatrix to specific values, the number of solutions is $O(1)$. This suffices as the number of ways to assign the other variables is $p^{kn}$.
To do this, consider two solutions $x\_1,\dots, x\_{N-n}$ and $y\_1,\dots, y\_{N-n}$. Let $\ell$ be the minimum $p$-adic valuation of $x\_j-y\_j$. We have $f\_i (x\_1,\dots, x\_{N-n}) = 0= f\_i (y\_1,\dots, y\_{N-n}) = f\_{i} (x\_1,\dots, x\_{N-n}) + \sum\_j \frac{\partial f\_i}{\partial x\_j} (y\_j-x\_j) + O \max\_j ( y\_j -x\_j )^2$. Since the determinant divides $p^c$, for some $i$ the middle term divides $p^{\ell+c}$. So we must either have $\ell +c \geq k$ or we have the middle term cancelled by the following term, meaning $\ell +c \geq 2\ell$, i.e. $\ell \leq c$. So $\ell$ is either in $[0,c]$ or in $[k-c k]$.
This means that, if $x\_1,\dots, x\_{N-n}$ and $y\_1, \dots, y\_{N-n}$ are congruent modulo $p^{c+1}$, they are congruent modulo $p^{k-c}$. So the number of solutions in each congruence class modulo $p^{c+1}$ is at most $p^{(N-n)c}$, and thus the total number of solutions is at most $p^{2 (N-n)c}$. This is $O(1)$, and we're done.
| 5 | https://mathoverflow.net/users/18060 | 418638 | 170,447 |
https://mathoverflow.net/questions/417993 | 2 | Recall that $M\subseteq\omega$ is **maximal** if it is c.e., and can be only trivially extended by other c.e. sets, i.e. if $M\subseteq N$ and $N$ is c.e., then either $\overline{N}$ or $N\setminus M$ is finite. Similarly say a set $M$ is **$A$-maximal** if it is $A$-c.e. and only trivially extended by other $A$-c.e. sets.
I am interested in sets $A$ such that all $A$-maximal sets are $\Delta^0\_2$. If $A$ has this property, then relativizing Yates' construction of a complete maximal set gives an $A$-maximal $M$ with $M\oplus A\equiv\_T A'$, so that $A$ is $\mathrm{GL}\_1$.
Thus among $A\in\Delta^0\_2$, the sets with this property are exactly the low ones - one direction is above, and for the reverse, if $A$ is low then all $A$-c.e. sets are $\Delta^0\_2$.
My question is - is this property enjoyed by any non-$\Delta^0\_2$ set?
~~By Martin's high domination theorem, this condition is equivalent to asking for a set $A$ such that all $A$-high, $A$-c.e. sets are $\Delta^0\_2$ and high, i.e. if $B\in\Sigma^0\_1(A)$ and $(B\oplus A)'\equiv\_T A''$, then $B\in\Delta^0\_2$ and $B'\equiv\_T \emptyset''$.~~ **Edit**: My understanding of the relativization of Martin's result was incorrect (thanks to Emma Harper on Twitter for spotting this!). The correct statement is that if a set $B$ is $A$-c.e. and $A$-high, there is an $A$-maximal $M$ with $M\oplus A\equiv\_T B\oplus A$, so what I struck through is not obviously equivalent to the property I am concerned with. Indeed, it does not hold outside of the $\Delta^0\_2$ degrees, as Noah correctly points out below.
My suspicion is that this requirement is too strong, that if $A\not\in\Delta^0\_2$ *some* $A$-maximal real will always fail to be in $\Delta^0\_2$. But I do not have a proof, nor can I rule out that some 'weak' set (like a hyperimmune-free or a bi-immune-free) might somehow have this property.
| https://mathoverflow.net/users/172527 | Sets $A$ such that $A$-maximal sets are $\Delta^0_2$ | Turns out (as I suspected!) all such sets are $\Delta^0\_2$, so that this property exactly characterizes lowness.
Fix an $A$-maximal $M$. For any $B\leq\_T A$, one can build $C = \overline{M}\oplus\_B\emptyset = \{n \mid p\_B(n)\in\overline{M}\}$. Then any $W\_e^A$ you might wish to intersect with $C$ can be transformed into a $W\_{f(e)}^A$ (to be intersected with $\overline{M}$) by putting $n$ into $W\_{f(e)}^A$ whenever $p\_B(n)$ enters $W\_e^A$.
Now the $A$-cohesiveness of $\overline{M}$ carries over nicely to $C$, since $|W\_e^A\cap C| = |W\_{f(e)}^A\cap\overline{M}|$ (and similarly for $\overline{W\_e^A}$). So since $\overline{C}$ is $A$-c.e. and $A$-cohesive, it is $A$-maximal and thus $\Delta^0\_2$.
Finally any $B\leq\_T A$ has a $\Delta^0\_2$ subset, which happens only when $A\in\Delta^0\_2$.
| 0 | https://mathoverflow.net/users/172527 | 418641 | 170,448 |
https://mathoverflow.net/questions/416977 | 7 | Let $B$ be a paracompact space with the property that any (topological) vector bundle $E \to B$ is trivial. What are some non-trivial examples of such spaces, and are there any interesting properties that characterize them?
For simple known examples we of course have contractible spaces, as well as the 3-sphere $S^3$. This one follows from the fact that its rank $n$ vector bundles are classified by $\pi\_3 (BO(n)) = \pi\_2 (O(n)) = 0$. I'm primarily interested in the case where $B$ is a closed manifold. Do we know any other such examples?
There is this [nice answer](https://math.stackexchange.com/a/1108129/354855) to a MSE question which talks about using the Whitehead tower of the appropriate classifying space to determine whether a bundle is trivial or not. This seems like a nice tool (of which I am not familiar with) to approaching this problem. As a secondary question, could I ask for some insight/references to this approach?
**EDIT** Now that we know from the answers all the examples for closed $3$-manifolds, I guess I can now update the question to the case of higher odd dimensions. Does there exist a higher dimensional example?
| https://mathoverflow.net/users/143629 | Examples and properties of spaces with only trivial vector bundles | Let $B$ be a closed manifold with such that every vector bundle is trivial. Then $H^1(B; \mathbb{Z}\_2) = 0$, otherwise there would be a non-trivial line bundle. Therefore every bundle over $B$ is orientable and $B$ itself is orientable. Orientable rank two bundles over $B$ are classified by $H^2(B; \mathbb{Z})$, so we must have $H^2(B; \mathbb{Z}) = 0$. It follows these conditions that there are no examples with $\dim B = 1, 2$. If $\dim B > 2$ we then have to consider the possibility of non-trivial bundles of rank at least three.
Suppose now that $\dim B = 3$. As $B$ is closed and orientable we have $H\_1(B; \mathbb{Z}) \cong H^2(B; \mathbb{Z}) = 0$ by Poincaré duality and $H^3(B; \mathbb{Z}) \cong \mathbb{Z}$. It follows that $B$ is an integral homology sphere - note, the condition $H^1(B; \mathbb{Z}\_2) = 0$ is superfluous in this case as $H^1(B; \mathbb{Z}) \cong \operatorname{Hom}(H\_1(B; \mathbb{Z}), \mathbb{Z}\_2)$ by the Universal Coefficient Theorem. We still have to consider the possibility of non-trivial bundles of rank at least three.
Suppose $E \to B$ has rank greater than three, then $E\cong E\_0\oplus\varepsilon^k$ for some rank three bundle $E\_0$, see [this answer](https://mathoverflow.net/a/417820/21564); in particular, we only need to consider the possibility of a non-trivial vector bundle of rank three. Suppose then that $\operatorname{rank} E = 3$. As $E$ is orientable, it has an Euler class $e(E)$ which is the first obstruction to a nowhere-zero section (it is also the only obstruction because $\operatorname{rank}E = \dim B$). As $E$ has odd rank, the Euler class of $E$ is two-torsion, but $e(E) \in H^3(B; \mathbb{Z}) \cong \mathbb{Z}$ which is torsion-free, so $e(E) = 0$. Therefore $E \cong E\_0\oplus\varepsilon^1$ where $\operatorname{rank}E\_0 = 2$. As we already know rank two bundles over $B$ are trivial, we see that $E$ is also trivial.
In conclusion, we have the following:
>
> Let $B$ be a closed three-manifold. Every vector over $B$ is trivial if and only if $B$ is an integral homology sphere.
>
>
>
Note, we only considered real vector bundles above, but the same is true for complex vector bundles since every such bundle is the direct sum of a trivial bundle and a complex line bundle, but the latter are classified by $H^2(B; \mathbb{Z}) = 0$.
The fact that every vector bundle over a three-dimensional integral homology sphere is trivial can also be seen using [Quillen's plus construction](https://encyclopediaofmath.org/index.php?title=Plus-construction) (also see section $\mathrm{IV}.1$ of Weibel's *An Introduction to Algebraic K-Theory*). As $B$ is an integral homology sphere, its fundamental group $\pi\_1(B)$ is perfect. By the plus construction, there is a simply connected CW complex $B^+ = B^+\_{\pi\_1(B)}$ and a map $q : B \to B^+$ inducing isomorphisms on homology satisfying the following: if $f : B \to X$ is a map with $\ker f\_\* : \pi\_1(B) \to \pi\_1(X)$ equal to $\pi\_1(B)$, then there is a map $g : B^+ \to X$, unique up to homotopy, such that $f = g\circ q$. As $H^1(B; \mathbb{Z}\_2) = 0$, every bundle over $B$ is orientable and hence classified by a map $f : B \to BSO(n)$. Since $\pi\_1(BSO(n)) \cong \pi\_0(SO(n)) = 0$, the kernel of $f\_\* : \pi\_1(B) \to \pi\_1(BSO(n))$ is $\pi\_1(B)$ so $f = g\circ q$ for some map $g : B^+ \to BSO(n)$. Note that $B^+$ is a simply connected CW complex with $H\_\*(B^+) \cong H^+(S^3)$, so $B^+$ is homotopy equivalent to $S^3$ by the homological Whitehead Theorem. As $\pi\_3(BSO(n)) \cong \pi\_2(SO(n)) = 0$, the maps $g$ and $f$ are nullhomotopic, so $f$ classifies the trivial bundle. Replacing $BSO(n)$ with $BU(n)$ yields the same result for complex vector bundles.
As for higher-dimensional examples, note that they must have odd dimension. To see this, suppose $\dim B = 2m$. Choose a degree one map $\varphi : B \to S^{2m}$ and consider the bundle $\varphi^\*TS^{2m} \to B$. As $e(TS^{2m}) \neq 0$ and $\varphi^\* : H^{2m}(S^{2m}; \mathbb{Z}) \to H^{2m}(B; \mathbb{Z})$ is an isomorphism, we see that $e(\varphi^\*TS^{2m}) = \varphi^\*e(TS^{2m}) \neq 0$ and hence $\varphi^\*TS^{2m}$ is non-trivial. If one considers complex bundles instead, the same argument works by replacing $TS^{2m}$ with a complex vector bundle $E$ with $e(E) = c\_n(E) \neq 0$ - such a bundle always exists as $\operatorname{ch} : K(S^{2m})\otimes\mathbb{Q}\to H^{\text{even}}(S^{2m}; \mathbb{Q})$ is an isomorphism.
One last comment about dimension five. As was established by Jason DeVito [here](https://math.stackexchange.com/a/4409768/39599), $B$ must be a rational homology sphere. If a five-dimensional example exists, I claim it is also a $\mathbb{Z}\_2$ homology sphere. To see this, first note that we have $H^1(B; \mathbb{Z}\_2) = 0$ from above. Now, if $v \in H^2(B; \mathbb{Z}\_2)$, then there is a bundle $E \to B$ with $w\_2(E) = v$ if and only if $\beta(v^2) \in H^5(B; \mathbb{Z})$ is zero where $\beta$ denotes the mod $2$ Bockstein, see [this answer](https://mathoverflow.net/a/344172/21564). As $\beta(v^2)$ is two-torsion and $H^5(B; \mathbb{Z}) \cong \mathbb{Z}$ is torsion-free, we see that $\beta(v^2) = 0$ for every $v \in H^2(B; \mathbb{Z}\_2)$, so every such class arises as $w\_2(E)$ for some $E$. Therefore, we must have $H^2(B; \mathbb{Z}\_2) = 0$. The claim now follows by Poincaré duality.
| 7 | https://mathoverflow.net/users/21564 | 418642 | 170,449 |
https://mathoverflow.net/questions/418656 | 0 | Let $s$ and $d$ be non-negative integers with $0\leq s<d$ and let $v,u\in \mathbb{R}^d$ be vectors satisfying the sparsity estimate
$$
\max\{\|u\|\_0,\|v\|\_0\}\leq d-s,
$$
where, as usual, for any vector $x \in \mathbb{R}^d$ we define $\|x\|\_0:=\sum\_{i=1}^d\,I\_{x\_i\neq 0}$.
Let $A$ be an $d\times d$-matrix solving $Au=v$. How sparse can $A$ be? I.e.: what is
$$
\inf\_{A\in L(\mathbb{R}^d),\,Au=v}\, \|A\|\_0,
$$
where as above $\|A\|\_0:=\sum\_{i,j=1}^d\,I\_{A\_{i,j}\neq 0}$.
| https://mathoverflow.net/users/36886 | How sparse can a matrix mapping between sparse vectors be? | $||v||\_0$, so $\leq d-s$: The matrix $A$ needs to have at least one entry for every entry of $v$ (otherwise it can't obtain that entry). It is also sufficient to have so many entries, as if we consider one $j$ with $u\_j \not= 0$ (which exists otherwise the system has a solution (the zero matrix) only if $v\_i = 0$ for all $i$) then we can set for every entry $v\_i \not= 0$ the matrix entry $A\_{i,j} = \frac{v\_i}{u\_j}$.
| 1 | https://mathoverflow.net/users/101157 | 418657 | 170,454 |
https://mathoverflow.net/questions/418660 | 3 | I was reading [this answer](https://mathoverflow.net/a/32914/146831), which says that:
>
> In his Master's Thesis, Merlin Carl has computed a polynomial that is solvable in the integers iff ZFC is inconsistent. A joint paper with his advisor Boris Moroz on this subject can be found at <http://www.math.uni-bonn.de/people/carl/preprint.pdf>.
>
>
>
Note that the link is dead. Emil Jeřábek provided an alternate link here: [*A polynomial encoding provability in pure mathematics (outline of an explicit construction)*](https://doi.org/10.36045/bbms/1366306724).
That phrase "solvable in the integers iff $\mathsf{ZFC}$ is inconsistent". If $\mathsf{ZFC}$ is inconsistent, then of course the polynomial is solvable in the integers - every statement in the model is true! So it seems to be just a fancy way of saying that the polynomial is not solvable in the integers.
I believe I'm missing some subtleties here, so I would like to have someone address this confusion of mine.
| https://mathoverflow.net/users/146831 | What does it really mean for a polynomial to be solvable in $\mathbb{Z}$ iff $\mathsf{ZFC}$ is inconsistent? | Edit: After I wrote this, the paper was posted in the commentary. What I wrote below is what is meant, however they use $\mathsf{GBC}$ instead of $\mathsf{ZFC}$. Since these two theories are equiconsistent and have the same first order consequences, this technical difference is immaterial.
However, when people say things like this there are a number of ways to interpret it, none of which are trivial. If I had to guess, probably what is meant is something like ``there is a polynomial $p(x)$ (often concretely computed) and $\mathsf{ZFC}$ proves ($\exists x \in \mathbb Z$ $p(x) = 0$ iff $\mathsf{ZFC}$ is inconsistent) ". By the second incompleteness theorem even if $\mathsf{ZFC}$ is consistent it cannot prove this fact so the equivalence is not trivial. Moreover, there are models on $\mathsf{ZFC}$ which think $\mathsf{ZFC}$ is consistent, and in such models there are no integer solutions to $p(x)$ and other models of $\mathsf{ZFC}$ which think that $\mathsf{ZFC}$ is inconsistent and in such models some (necessarily nonstandard) integer will satisfy $p(x)$.
The point is usually that the kinds of coding that lead to the incompleteness theorems can in fact be coded into surprisingly concrete polynomials. In the case of $\mathsf{PA}$ there are many examples of such coming from MRDP/Matiyasevich's theorem. In the wikipedia article on diophantine equations this is mentioned: <https://en.wikipedia.org/wiki/Diophantine_set>, see ``further applications".
| 4 | https://mathoverflow.net/users/114946 | 418663 | 170,455 |
https://mathoverflow.net/questions/418628 | 1 | Given a symmetric Hessenberg matrix $A = \left[\begin{matrix}\ddots& \vdots & \vdots\\\dotsb & a & b\\\dotsb& b & c\end{matrix}\right]$, the Wilkinson shift $\mu$ employed in some eigenvalue solvers is given by
$$\mu = c - \frac{b^{2} \operatorname{sign}{\left(\delta \right)}}{\sqrt{b^{2} + \delta^{2}} + \left|{\delta}\right|}$$
where $\delta = (a-c)/2$, and $\operatorname{sign}{\left(0 \right)}$ is $1$ or $-1$ (it doesn't matter which).
This is discontinuous in $A$, as is easy to verify. And it always produces one of the two eigenvalues of the bottom-right $2 \times 2$ submatrix of $A$. Does the shift need to be discontinuous? Could it not simply take the smaller of the two eigenvalues of the bottom-right submatrix, which would indeed be continuous in $A$?
| https://mathoverflow.net/users/75761 | Does Wilkinson's shift need to be discontinuous? | **It must be discontinuous to ensure that all fixed points are attractive**.
We make the following assumptions about iterative eigenvalue algorithms over PSD matrices:
1. They consist of iterating some function $f$,
2. $f(M)$ is orthogonally similar to $M$
3. The sequence $(f^n(M))\_{n\in\mathbb N}$ converges for every $M$. Obviously, this converges to a fixed point of $f$.
4. A fixed point can only be a diagonal matrix.
We now make the following additional assumptions:
5. All fixed points of $f$ are attractive. (Note that this is a highly desirable property because it bounds the computation time.)
6. $f$ is continuous. (For critical analysis.)
We restrict the domain and codomain of $f$ to the set of matrices orthogonally similar to some PSD matrix $M$, where $M$ is not a multiple of the identity matrix. The fact that we can do this follows from condition 2. **We show that $f$ has at least two fixed points under assumptions 5 and 6**: Assume it only has one fixed point $x$. It must be attractive by condition 5. By condition 3, all points attract to $x$. This results in the space of matrices orthogonally similar to $M$ being a contractible space, which it surely isn't. We get a contradiction. Therefore $f$ has at least two fixed points orthogonally similar to $M$.
**Now we show that under assumptions 5 and 6, one of these two fixed points is not attractive, so that 5 and 6 cannot hold simultaneously**. The space of matrices orthogonally similar to some arbitrary matrix $M$ is connected. The set of points which attract to some attractive fixed point is always open. The basins of attraction of the attractive fixed points are disjoint, so by topological connectivity there are points not in their union, which are the points which don't attract to an attractive fixed point. Therefore *the Wilkinson shift must be discontinuous* to make all fixed points attractive.
| 2 | https://mathoverflow.net/users/75761 | 418664 | 170,456 |
https://mathoverflow.net/questions/418632 | 4 | I think the question as expressed in the title should be clear. I do not know whether there is a known "characterization" of the weakly compact convex sets in $c\_0(\mathbb N\_0)$ but testing examples has lead me to conjecture that sequences in $\ell^1(\mathbb N\_0)$ converging to zero uniformly on these sets also converge to zero in the norm topology. Is this conjecture correct/know/well-know/is there an explicit reference?
| https://mathoverflow.net/users/12643 | Are sequences in $\ell^1(\mathbb N_0)$ converging uniformly on convex weakly compact subsets of $c_0(\mathbb N_0)$ norm convergent? | The answer is yes.
*Proof.* Assume to the contrary that a sequence $(x\_n)$ in $\ell^1$ converges, say to $0$, uniformly on convex weakly compact subsets of $c\_0$, but is not norm convergent and hence not norm convergent to $0$.
Note that the sequence even converges uniformly on all weakly compact subsets of $c\_0$, be they convex or not (since [the closed convex hull of a weakly compact set is again weakly compact](https://mathoverflow.net/a/218164/102946)). So if we construct a sequence $(y\_n)$ in $c\_0$ that converges weakly to $0$ but such that $\langle y\_n, x\_n \rangle \not\to 0$, then we have a contradiction.
After replacing $(x\_n)$ with a subsequence we may assume that $\|x\_n\| \ge \varepsilon$ for some $\varepsilon > 0$ and all $n$.
Moreover, since $(x\_n)$ converges, in particular, weakly to $0$, it also converges pointswise to $0$. So after replacing $(x\_n)$ with yet another subsequence we may assume that, for all $n$, we have $\sum\_{k=1}^{n-1} |x\_n(k)| < \varepsilon/3$. Hence, there exist numbers $N\_n \ge n$ such that $\sum\_{k=n}^{N\_n} |x\_n(k)| \ge \varepsilon / 3$ for each $n$.
Now simply choose $y\_n(k)$ to be zero for $k$ outside the set $\{n,\dots, N\_n\}$ and to be the complex conjugate of the (complex) sign of $x\_n(k)$ for $k$ inside this set. Then we have $\langle y\_n, x\_n \rangle \ge \varepsilon / 3$ for each $n$. Moreover, the sequence $(y\_n)$ in $c\_0$ is bounded and converges pointwise to $0$; hence, it also converges weakly to $0$.
| 3 | https://mathoverflow.net/users/102946 | 418669 | 170,459 |
https://mathoverflow.net/questions/418554 | 51 | $\DeclareMathOperator\SO{SO}\newcommand{\R}{\mathbb{R}}\newcommand{\S}{\mathbb{S}}$The periodic table of elements has row lengths $2, 8, 8, 18, 18, 32, \ldots $, i.e., perfect squares doubled. The group theoretic explanation for this that I know (forgive me if it is an oversimplification) is that the state space of the hydrogen atom is made of functions on $\R^3$, which we can decompose as functions on $\S^2$ times functions on $\R^+$. Then $\SO(3)$ acts on the functions on $\S^2$ and commutes with the action of the Hamiltonian, so we can find pure states inside irreducible representations of $\SO(3)$. The orbital lengths depend on how these representations line up by energy, which is a function on $\R^+$. It happens that the spaces with the same energy have the form $(V\_0 \oplus V\_2 \oplus V\_4 \oplus \cdots ) \otimes W$, where $V\_i$ is the irreducible representation of $\SO(3)$ of dimension $i+1$, and $W$ is a $2$-dimensional space that represents the spin. So the orbital length is the dimension of $V\_0 \oplus V\_2 \oplus V\_4 \oplus \cdots$ is the sum of the first $k$ odd numbers, which is a perfect square, and you double it because of the spin.
So, in short, the perfect squares arise as the sums of the first $k$ odd numbers, and the invariant subspaces arrange themselves into energy levels that way because... well, here I get stuck. Factoring out the spin, which explains the doubling, can anyone suggest a more conceptual (symmetry-based?) explanation for why perfect squares arise here?
| https://mathoverflow.net/users/478715 | Is there a good mathematical explanation for why orbital lengths in the periodic table are perfect squares doubled? | In this answer, I'm going to crib from [this](https://math.ucr.edu/home//baez/hydrogen/4d/hydrogen_4d.pdf) presentation by @JohnBaez and the paper [On the Regularization of the Kepler Problem](https://arxiv.org/abs/1007.3695). Milnor's [paper](https://www.maa.org/programs/maa-awards/writing-awards/on-the-geometry-of-the-kepler-problem) includes a lot of the same information.
First, I'm going to state a few facts without proof. One can compose the stereographic projection of $\mathbb{R}^3$ to $S^3$ with the symplectomorphism swapping $p$s and $q$s on $T^\*(\mathbb{R}^3)$ to get a symplectomorphism from the punctured $T^\*(S^3)$ to $T^\*(\mathbb{R}^3)$.
Furthermore, one can show that the Hamiltonian flow of $p^2$ on a a constant energy surface in $T^\*(S^3)$ maps to the Hamiltonian flow of the Kepler potential on $T^\*(\mathbb{R}^3)$. This maps the constant energy classical mechanics of a negative energy state in the Kepler potential to a free particle on $S^3$ with fixed energy. This also exhibits the $SO(4)$ symmetry as rotations on $S^3$. Thus (and I'm still undecided if there's some handwaving here), the energy eigenstates in the quantum theory should be irreps of $SO(4)$. You can also exhibit the $SO(4)$ symmetry directly in the quantum theory, so any handwaving isn't a problem.
To see what the representations are, the $SO(4)$ action on $S^3$ can be exhibited by the two $SU(2)$ factors in $\operatorname{Spin}(4)$ acting on both sides of $S^3 \cong SU(2)$. The element $(-I,-I) \in SU(2) \times SU(2)$ acts trivially, so you get an $SO(4)$ action.
With this, we can decompose a la the Peter–Weyl theorem:
$$
L^2(S^3) \cong \bigoplus\_i \rho\_i \otimes \rho\_i^\star
$$
Each $\rho\_i$ is an irrep of $SU(2)$, and those irreps can be labelled by an integer $n$. Thus, the problem decomposes into $n^2$-dimensional irreps of $SO(4)$, which explains the question asked.
[N.B. -- I'd be interested in understanding if this can be done "all at once" as opposed to working with constant energy surfaces and arguing by scaling as I see in the references. If I have time, I'd also want to show that the different irreps of SO(4) have different energies, or maybe I'm missing something obvious. This can all be done by looking at the symmetry explicitly in the quantum theory, I'm sure, but it would be nice to see it geometrically.]
| 10 | https://mathoverflow.net/users/947 | 418677 | 170,462 |
https://mathoverflow.net/questions/418672 | 7 | In many areas of mathematics it is informative to conduct numerical experiments.
But, it not uncommon that the searches do not lead to the examples or data one was hoping for. Since the numerical searches can be quite time consuming, it seems useful to share these negative results, so that others avoid spending time attempting the same searches.
>
> What would be best practices in this regard? And would setting up and maintaining an open-access database be a realistic prospect?
>
>
>
As an example, consider the question of the (non-)existence of $n$-body choreographies when allowing the bodies to have different masses and different time-lags (see ["$n$-body choreographies" by Montgomery](http://dx.doi.org/10.4249/scholarpedia.10666) for background as of 2013, and [Minton's fine applet](http://gminton.org/choreo.html) for the type of numerical searches that one could try to adapt). I'd be interested to know what has been already attempted by others. I could ask experts, but it may well overlook work by other experts and by Ph.D. students.
| https://mathoverflow.net/users/469 | Reporting inconclusive experimental searches | An easy and reliable way to share code is via [Zenodo](https://en.wikipedia.org/wiki/Zenodo) --- works much like arXiv, you get a DOI, can update your files, and it's free. We use it regularly to document computer simulations in physics, I imagine computational mathematics is not that different.
Note that Zenodo explicitly welcomes both positive and negative results.
| 6 | https://mathoverflow.net/users/11260 | 418678 | 170,463 |
https://mathoverflow.net/questions/418578 | 2 | The current post comes from my previous post at [stackexchange](https://math.stackexchange.com/questions/4408614/sqrtf-has-bounded-tangential-derivatives-if-f-in-c1-1). However, I have not get any comment yet.
In a celebrated paper written by [Guan, Trudinger, and Wang](https://projecteuclid.org/journals/acta-mathematica/volume-182/issue-1/On-the-dirichlet-problem-for-degenerate-Monge-Amp%C3%A8re-equations/10.1007/BF02392824.full), authors proved the existence and uniqueness of convex $C^{1,1}$-solution to the Dirichlet problem for degenerate Monge-Ampere equations, and they also provided a global $C^{1,1}$-control to the solution. They first reduced the global $C^2$-estimates to the boundary estimates for the second-order derivatives, and then did the estimates on the boundary. To this end, a "simple" inequality was used for several times, which can be stated precisely as follows.
>
> Let $n\geq2$ be an integer and $\Omega\subset\mathbf R^n$ be a bounded
> domain with smooth boundary. Assume $\Omega$ is uniformly convex if necessary. Given $f\in C^{1,1}(\bar\Omega)$
> such that $$f>0\quad\hbox{in $\Omega$.}$$ Does there exist a constant
> $C>0,$ which is at most dependent on $n,~\Omega,$ and
> $\|f\|\_{C^{1,1}(\bar\Omega)},$ such that \begin{equation}|\nabla
> f(x)\cdot\tau(x)|^2\leq Cf(x)\quad\hbox{for all $x\in\partial\Omega$},\label{1}\tag{1}\end{equation}
> where $\tau(x)$ is a unit tangent vector of $\partial\Omega$ at $x.$
>
>
>
Or equivalently, does the tangential derivative $\partial\_{\tau}\sqrt f$ have a uniform upper bound?
Actually, authors used in above reference an incorrect inequality as
$$
|\nabla f|^2\leq Cf\quad\hbox{in $\Omega$.}
$$
Near the boundary, it is clear that the distance function $d(x):=\mathrm{dist}(x,\partial\Omega)$ is a counterexample since $|\nabla d|=1$. Fortunately, it seems that all consequences still hold if \eqref{1} is true. If $f(x\_0)=0$ for some boundary point $x\_0,$ then obviously we can take above $C=1$ at $x\_0.$ The remaining difficulty is to control the tangential derivative of $\sqrt f$ if $f(x)>0$ but it is very small. Probably above inequality is reasonable in the following sense. For $x\_0\in\partial\Omega.$ Roughly, if $f$ behaves like some positive power of $d\_{x\_0}:=d(x,x\_0)$ near $x\_0,$ for example we say it as $d\_{x\_0}^\alpha.$ Then, $|\nabla\_\tau f|$ might behave as $d\_{x\_0}^{\alpha-1}$ near $x\_0.$ If \eqref{1} is not true, then $\alpha-1<\alpha/2,$ and thus $\alpha<2.$ However, this is impossible as $f$ is of $C^{1,1}.$ I hope someone could give me some comment on this topic.
| https://mathoverflow.net/users/105893 | Estimates on the second-order derivatives for degenerate Monge-Ampere equations | This inequality comes from scaling. Assume that $f$ satisfies $|D^2f| \leq 1$ on $\mathbb{R}^n$ and $f \geq 0$. It suffices to prove that
$$|\nabla f(0)|^2 \leq 2f(0).$$
Equality holds if $f(0) = 0$, so assume that $f(0) > 0$. We may assume that $f(0) = 1$ after taking the rescaling $\tilde{f}(x) = \lambda^{-2}f(\lambda x)$, with $\lambda^2 = f(0)$, since this rescaling preserves the Hessian of $f$, the sign of $f$, and the ratio of interest:
$$|\nabla \tilde{f}(0)|^2 = |\nabla f(0)|^2/f(0).$$
In the situation $f(0) = 1$ it is clear that $|\nabla f(0)|^2 \leq 2$, otherwise the Hessian bound would imply that $f < 0$ somewhere (follow a ray in the direction $-\nabla f(0)$).
The situation when $f$ is a nonnegative $C^{1,\,1}$ function on a compact manifold ($\partial \Omega$) is similar. One can e.g. make a nonnegative extension of $f$ to $\mathbb{R}^n$ with comparable $C^{1,1}$ norm and apply the previous reasoning.
| 3 | https://mathoverflow.net/users/16659 | 418684 | 170,464 |
https://mathoverflow.net/questions/418676 | 7 | Suppose $\mathcal{A}$ is a Grothendieck abelian category with enough projectives, then $\mathcal{A}$ is tensored and cotensored over $\mathrm{Ab}$ with $\mathbb{Z}^{\oplus S}\otimes X\cong \bigoplus\_S X$ for any set $S$, and for any abelian group $A$, with presentation $\mathbb{Z}^{\oplus R}\xrightarrow{f} \mathbb{Z}^{\oplus G}\to A \to 0$ we define
$$A\otimes X := \mathrm{coker}(\mathbb{Z}^{\oplus G}\otimes X \xrightarrow{f\otimes A} \mathbb{Z}^{\oplus G}\otimes X).$$
Now if we have some $\mathrm{Ab}$-enriched category $Q$ and functors $P:Q^{op}\to \mathrm{Ab}$ and $X:Q\to \mathcal{A}$ we can define the functor tensor product of $P$ and $X$ to be
$$
P\otimes\_Q X:= \int^{p \in Q}P(p)\otimes X(p) \in \mathcal{A}.
$$
This is clearly functorial in both variables. I am wondering does anyone know a balancing result, similarly to balancing Tor. That is is $\mathbb{L}\_i(P\otimes\_Q -)(X)\cong \mathbb{L}\_i(-\otimes\_Q X)(P)$.
In general I would be very interested in any result on these derived functors or their relation to $\mathrm{Ext}$ in $\mathrm{Fun}(Q,\mathcal{A})$.
I am interested in these, as they relate to certain "homology" functors related to $\mathrm{Fun}(Q,\mathcal{A})$.
| https://mathoverflow.net/users/479208 | Derived functor of functor tensor product | The answer is yes if you assume enough things. In particular, the notion of a left flat object of $\mathcal A$ comes up :
**Definition:** An object $L\in\mathcal A$ is left flat if $-\otimes L$ is exact.
My assumption will be that $\mathcal A$ has enough left flat objects. This allows you to even define $\mathbb L(P\otimes\_Q -)$ in terms of left flat resolutions as I will explain below.
(*Note* : Here I'm using a definition of $\mathbb L$ which is slightly more general than "take projective resolutions", namely I'm using the one using "left defomations", see Riehl's book on homotopical algebra. If you want to use projective resolutions, the assumption is that $\mathcal A$ has enough left flat projectives)
*Warning*: Note that if $\mathcal A$ has projectives that are not left flat, then the answer is no even for $Q=\*$, as is easy to convince yourself of. In particular, if $\mathcal A$ has enough projectives (to define $\mathbb L$ in the classical homological algebra sense), the assumption is that enough (equivalently all) projectives are left flat.
In this case, the answer is yes, and basically the proof is the same as in the ordinary case. Let me work in the bounded below case because there are the same subtleties as in the ordinary case for the unbounded case.
In particular, up to shifting, I will work in the connective case.
Let me sketch a proof below (convention : my functor categories are categories of $Ab$-enriched functors, I'm assuming this is what you meant - otherwise, replace $\hom\_Q$ with $\mathbb Z[\hom\_Q]$):
**Definition:** $P$ is right flat if $P\otimes\_Q -$ is exact, and same for $X$ being left flat.
**Lemma:** Projectives in $Fun(Q^{op},Ab)$ are right flat, and there are enough left flats in $Fun(Q,\mathcal A)$.
*Proof*: By the enriched Yoneda lemma and smallness of $Q$, every projective is a summand of $\bigoplus\_i \hom\_Q(-,q\_i)$ for some family of $q\_i$'s. Therefore it suffices to prove it for those ones, because flatness is stable under retracts. Flatness is also stable under direct sums because they are exact ($\mathcal A$ is Grothendieck, in particular AB5), so it suffices to prove it for $\hom\_Q(-,q)$. But now by the enriched Yoneda lemma again (the "canonical colimit of representable presheaves" version), we have $\hom\_Q(-,q)\otimes\_Q X \cong X(q)$, which is manifestly an exact functor.
The dual case is dual, you simply need to allow $\hom\_Q(q,-)\otimes L$ for enough flat objects $L$ of $\mathcal A$.
In particular this lemma tells you that the two things you want to compare are well-defined.
**Corollary:** If $C$ is a chain complex of right flat functors $Q^{op}\to Ab$, $C\otimes\_Q -$ preserves quasi-isomorphisms (where you define $\otimes\_Q$ on chain complexes in the obvious way); and dually.
*Proof* : Let $n\in \mathbb N$, and let $C\_{\leq n}$ denote the so-called stupid truncation of $C$, i.e. $0\to C\_n\to ... \to C\_0$. Because $-\otimes\_Q -$ manifestly preserves filtered colimits in each variable, because filtered colimits are exact ($\mathcal A$ is Grothendieck, in particular AB5), and finally because $C= \mathrm{colim}\_n C\_{\leq n}$, it suffices to prove that each $C\_{\leq n}\otimes\_Q -$ preserves quasi-isomorphisms.
Now we have a short exact sequence $0\to C\_{\leq n-1}\to C\_{\leq n}\to C\_n\to 0$ which is split *as a sequence of graded objects*. Note that the underlying graded object of $A\otimes\_Q B$ only depends on the underlying graded objects of $A,B$ respectively. It follows that this exact sequence remains exact after tensoring (over $Q$) with anything.
In particular, by the long exact sequence in homology, and by induction, we reduce to proving that $C\_n\otimes\_Q-$ preserves quasi-isomorphisms for each $n$. But this is by the assumption that each $C\_n$ is right flat.
The dual case is completely dual.
*Note* : I am implicitly using the fact that $-\otimes -: Ab\times\mathcal A\to\mathcal A$ preserves colimits in each variable. This is an exercise I'll leave to you :)
*Note* : This lemma is essentially what allows you to define $\mathbb L$ using left/right flat resolutions.
*Note*: This lemma is often packaged as a spectral sequence argument, but it's really about the filtration.
**Corollary:** $\mathbb L(P\otimes\_Q -)(X)\simeq \mathbb L(-\otimes\_Q X)(P)$.
*Proof* : Pick a right flat resolution $\tilde P$ of $P$, and a left flat resolution $\tilde X$ of $X$ (those exist by the first lemma).
You have a zigzag $$\mathbb L(P\otimes\_Q -)(X) =P\otimes\_Q \tilde X \to \tilde P\otimes\_Q \tilde X \leftarrow \tilde P\otimes\_Q X = \mathbb L(-\otimes\_Q X)(\tilde P)$$
The extreme equalities are by definition, and the two middle arrows are quasi-isomorphisms by the previous corollary.
*Note* : this quasi-isomorphism can be made as natural as left flat resolutions in $\mathcal A$.
| 5 | https://mathoverflow.net/users/102343 | 418688 | 170,465 |
https://mathoverflow.net/questions/418438 | 2 | Let $K$ be a symmetric convex body in $\mathbb{R}^n$ (that is the unit ball of a norm). Let $h\_K$ be its support function, that is $h\_K(u) = \sup\_{x \in K}\langle x,u \rangle$. The quantity $w(K) = \int\_{S^{n-1}} h\_K(\theta)\,d\theta$ is called the mean width of $K$ and is well studied.
Assume that $\operatorname{vol}(K) = \operatorname{vol}(B\_n)$ where $B\_n$ is the unit euclidean ball. Since $K$ does not contain $B\_n$, its polar $K^\circ = \{ y \in \mathbb{R^n}, h\_K(y) \leq 1 \}$ is not contained in $B\_n$ (polarity exchange inclusions and $B\_n$ is its polar), so that there is $y$ in $\mathbb{R^n}$ such that $\vert y \vert\_2 \geq 1 $ and $h\_K(y) \leq 1$. Setting $\theta = \frac{y}{\vert y \vert\_2}$, we have $h\_K(\theta) \leq 1$.
What can be said about the level-sets of the restriction $h\_K : S^{n-1} \to \mathbb{R}^+$, that is the sets:
$$W\_c^K = \{ \theta \in S^{n-1}, h\_K(\theta) \leq c\} \quad ?$$
The argument above shows that for all $c\geq 1$, $W\_c^K$ is non-empty.
Can we say something for instance about the (Haar) measure of $W\_{100}^K$, where 100 is an example of a dimension independent constant?
In particular, it is known that in some cases, such as the cross-polytope, the average of $h\_K$ is large, that is if we normalize the cross-polytope $C\_n$ so that it has the same volume as the ball then $w(C\_n)$ is of order $\log(n)$. However can we still hope that for instance $\mu(W\_{100}^K) \geq \frac{1}{2}$ for every $K$ of volume $1$, where $\mu$ is the Haar measure?
| https://mathoverflow.net/users/125325 | Distribution of the support function of convex bodies: beyond mean width | Actually the statement is pretty strong, in particular it implies the Bourgain-Milman (reverse Blaschke Santalo inequality) theorem, which says that the Mahler conjecture is true up to constant, that is, there exists $c\_0>0$ such that for all convex body $K$ :
$$\operatorname{vol}(K)\operatorname{vol}(K^\circ) \geq c\_0^n w\_n^2 $$
where $w\_n = \operatorname{vol}(B\_n)$ is the volume of the euclidean unit ball. Note that the theorem is stated in an equivalent unusual way.
Now, let $K$ be a convex body with $\operatorname{vol}(K) = w\_n$ assume that there exists $c>0$ such that $\mu(W\_K^c) \geq \frac{1}{2}$, then by integrating in polar :
\begin{align\*}
\operatorname{vol}(K^\circ) &= nw\_n\int\_{S^{n-1}}\int\_{0}^{\vert\vert \theta \vert\vert\_{K^\circ}^{-1}}t^{n-1}\, dt\,d\theta \\
& = w\_n \int\_{S^{n-1}}\vert\vert \theta \vert\vert\_{K^\circ}^{-n}\,d\theta \\
&= w\_n \int\_{S^{n-1}} h\_K^{-n}(\theta) \\
& \geq w\_n \,\mu(W\_K^c)\frac{1}{c^n} \geq \left(\frac{1}{2c}\right)^nw\_n
\end{align\*}
So that we indeed get $$ \operatorname{vol}(K)\operatorname{vol}(K^\circ) \geq\left(\frac{1}{2c}\right)^n w\_n^2 $$
(Note that the volume product is affine invariant, hence the assumption about the volume of K does not lose generality).
However, one cannot (a priori) invert the implication, so that it would be a stronger statement than Bourgain-Milman.
| 1 | https://mathoverflow.net/users/125325 | 418689 | 170,466 |
https://mathoverflow.net/questions/418671 | 4 | In "M. B. Nathanson - Elementary Methods in Number Theory" is shown (Theorem 7.14) that if $A$ is a set of positive integers such that $\sum\_{a \in A} 1 / a$ converges then the set of multiples of $A$ has a natural density (a set of positive integers $S$ has a natural density if exists $\lim\_{x \to \infty} |S \cap [1,x]| / x$).
Equivalently, the set of positive integers $n$ such that $a \nmid n$ for all $a \in A$ has a natural density.
I am looking for any reference for the analogous result that if $b$ is some fixed positive integer and $A$ is the same set as above with additional condition that $1\notin A$ then the relative density of primes $p=1+b\cdot m$ where $m$ is not divisible by any $a\in A$ is positive if there exist at least one such prime. I have a proof of this result but I am curious if this result is already in literature.
Thanks in advance for any assistance.
| https://mathoverflow.net/users/160943 | Relative density of primes in certain congruence classes | This is too long for a comment and possibly it still answers your question.
Fix a positive integer $b \in \mathbb{N}$ and set $A\subseteq \mathbb{N}$ such that $\sum\_{n \in A}1/n<\infty$ and $1\notin A$. Let $\{a\_1,a\_2,\ldots\}$ be the increasing enumeration of $A$. Let also $\mathbb{P}$ be the set of primes and define
$$
X:=\left\{p \in \mathbb{P}: p\equiv 1\bmod{b}\,\, \text{ and }\,\,p\not\equiv 1\bmod{ab} \text{ for all }a\in A\right\}.
$$
In other words, $X=\bigcap\_{N\ge 1}X\_N$, where
$$
X\_N:=\left\{p \in \mathbb{P}: p\equiv 1\bmod{b}\,\, \text{ and }\,\,p\not\equiv 1\bmod{a\_nb} \text{ for all }n=1,\ldots,N\right\}
$$
It follows by the main result [here](https://drive.google.com/file/d/1OWj4RFYwZNUAfDOSZVCjpXMxw1Dz0EgN/view) (at least for $b=1$) that $X\_N$ admits asymptotic density relative to $\mathbb{P}$ and, in addition, such value is at least
$$
\frac{1}{\varphi(b)}\prod\_{n=1}^N\left(1-\frac{1}{\varphi(a\_nb)}\right).
$$
With the same proof of Theorem 7.14 given by Nathanson, also the set $X$ admits asymptotic density relative to $\mathbb{P}$.
The fact that the latter value is positive should follow by the fact that
$$
\lim\_{N\to \infty}\lim\_{n\to \infty}\frac{|X\_N \cap [1,n]|}{|\mathbb{P}\cap [1,n]|}
\ge \frac{1}{\varphi(b)}\lim\_{N\to \infty}\prod\_{n=1}^N\left(1-\frac{1}{\varphi(a\_nb)}\right)
$$
and since $\varphi(a\_nb)\ge \sqrt{a\_nb}\to \infty$ as $n\to \infty$, we obtain a lower bound
$$
\gg \lim\_{N\to \infty}\mathrm{exp}\left(\sum\_{n=1}^N \log\left(1-\frac{1}{\varphi(a\_nb)}\right)\right)
\gg \mathrm{exp}\left(-\lim\_{N\to \infty}\sum\_{n=1}^N\frac{1}{\varphi(a\_nb)} \right);
$$
hence to conclude the proof, it is sufficient to show that $\sum\_{n=1}^\infty\frac{1}{\varphi(a\_nb)}$ is finite. Now
$$
\sum\_{n=1}^\infty\frac{1}{\varphi(a\_nb)}\le \sum\_{n=1}^\infty\frac{1}{\varphi(a\_n)\varphi(b)}\ll \sum\_{n=1}^\infty\frac{1}{\varphi(a\_n)}.
$$
However, why the latter sum should be finite? I mean, potentially every estimate is optimal if $b=1$ and $A$ has pairwise coprime elements. Still, it is known that there exists infinitely many $n$ such that
$$
\varphi(n)<\frac{n}{\log \log n}.
$$
Hence, how do you prove that $\sum\_n \frac{1}{\varphi(a\_n)}<\infty$, assuming that $\sum\_n \frac{1}{a\_n}<\infty$?
| 3 | https://mathoverflow.net/users/32898 | 418691 | 170,468 |
https://mathoverflow.net/questions/418680 | 6 | Write $\psi(x) = \sum\_{n\le x} \Lambda(n)$. The classical omega theorem says that
$\psi(x) - x = \Omega\_{\pm}(x^{1/2})$.
Question: How often does this hold? For example, what do we know about the size of the set
$\{ n\le x: \psi(x) - x > c x^{1/2} \}$
for some $c$? Ditto for $\psi(x) - x < c x^{1/2}$. What about replacing e.g. $x^{1/2}$ by $x^{\alpha}$ for some fixed $0 < \alpha < 1/2$? What's a good reference for such results? Thanks!
| https://mathoverflow.net/users/66397 | How often does the omega theorem hold? | For any $\varepsilon>0$, there exist $c(\varepsilon)>0$ and $X\_0(\varepsilon)>0$ such that for any $X>X\_0(\varepsilon)$ we have
$$\sup\_{X\leq x\leq X^{1+\varepsilon}}\frac{\psi(x)-x}{\sqrt{x}\log\log x}>c(\varepsilon)\qquad\text{and}\qquad
\inf\_{X\leq x\leq X^{1+\varepsilon}}\frac{\psi(x)-x}{\sqrt{x}\log\log x}<-c(\varepsilon).$$
More precisely, [Ingham (1935)](https://bibliotekanauki.pl/articles/1395100) proved a stronger result for the case when the real parts of the zeta zeros have a maximum, while [Pintz (1980)](https://bibliotekanauki.pl/articles/1393142) proved a stronger result for the case when the real parts of the zeta zeros do not have a maximum. There might be even stronger results in the literature, please check.
| 7 | https://mathoverflow.net/users/11919 | 418713 | 170,475 |
https://mathoverflow.net/questions/418714 | 4 | Let $P$ be a "nice" distribution on $\mathbb R^m$ (e.g., multivariate Gaussian, etc.), with density $p$. Let $H := \{x \in \mathbb R^m \mid x^\top w = b\}$ be a hyperplane in $\mathbb R^m$ with unit-normal $w \in \mathbb R^m$. Let $R$ be the *Radon transform* of $p$ w.r.t $H$ by
$$
R := \int\_H p(x)\,ds(x),
$$
where $ds(x)$ is the surface-area element on $H$. Finally, let $X\_1,\ldots,X\_n$ be an iid sample from $P$.
>
> **Question.** *Is there a simple statistical estimator $\widehat R\_n := s(X\_1,\ldots,X\_n)$ which converges to $R$ in the limit $n \to \infty$ ?*
>
>
>
| https://mathoverflow.net/users/78539 | Consistent empirical estimation of Radon transform of a multivariate density function | $\newcommand\th\theta\newcommand\R{\mathbb R}$Suppose that we have a parametric setting, that is, the unknown distribution $P$ belongs to a known parametric family $(P\_\th)$ of distributions parameterized by a sufficiently low-dimensional parameter $\th$; this may be the case if $P$ is Gaussian.
Then you can get (say) a maximum likelihood estimate (MLE) $\hat\th\_n$ of $\th$ and estimate $R$ by
$$\hat R\_n:=\int\_H p\_{\hat\th\_n}\,ds,$$
where $p\_{\hat\th\_n}$ is the density of $P\_{\hat\th\_n}$.
The MLE is usually consistent, so that you will have $\hat\th\_n\to\th$ in probability (as $n\to\infty$). If now $p\_\th$ is continuous on $\th$ uniformly on compact subsets of $\R^m$, you will get $p\_{\hat\th\_n}\to p\_\th$ uniformly on compact subsets of $\R^m$. If, moreover, you can control the tails of the densities $p\_\th$ so as to have their local uniform integrability (as you would have in the Gaussian case), then you will end up with the desired conclusion $\hat R\_n\to R$ in probability.
---
If the setting is nonparametric but the dimension $m$ is small, then you can use nonparametric (say kernel) estimators of the unknown density $p$, instead of the parametric estimators. However, then, naturally, the convergence $\hat R\_n\to R$ will be substantially more problematic.
If the setting is completely nonparametric and the dimension $m$ is not small (say $m\ge10$), then the situation will of course be even worse. Here you will need the sample size $n$ to be at least as big as something like $100^m\ge10^{20}$.
| 4 | https://mathoverflow.net/users/36721 | 418715 | 170,476 |
https://mathoverflow.net/questions/418402 | 16 | In Conway's "On Numbers And Games," page 44, he writes:
>
> **NON-STANDARD ANALYSIS**
>
>
> We can of course use the Field of all numbers, or rather various small
> subfields of it, as a vehicle for the techniques of non-standard
> analysis developed by Abraham Robinson. Thus for instance for any
> reasonable function $f$, we can define the derivative of $f$ at the
> real number $x$ to be the closest real number to the quotient
>
>
> $$\frac{f[x + (1/\omega)] - f(x)}{1/\omega}$$
>
>
> The reason is that *any* totally ordered real-closed field is a model
> for the elementary states about the real numbers. But for precisely
> this reason, there is little point in using subfields of $\mathbf{No}$
> when so many more visible fields will do. So we can say in fact the
> field $\mathbf{No}$ is really irrelevant to non-standard analysis.
>
>
>
Conway here makes clear that you could, if you wanted, use the surreal numbers for non-standard analysis, because they are a real-closed field and thus a model for the theory of all elementary (first-order) statements about the reals. Conway does also make the point, in the last two sentences, that he doesn't view nonstandard analysis as the ultimate application of the surreals. But, for the sake of curiosity, I'm quite interested in understanding how you could do what Conway is hinting at above.
However, the statement that you can use the surreals for nonstandard analysis is really quite strong, and much stronger than just the field being real-closed. The real meat of the claim being made is the expression $f(x + 1/\omega)$ even exists at all. This would demand some kind of "transfer principle" for $f(x)$ to the surreals. Just being real-closed wouldn't be enough for this: the real algebraic numbers are real closed, but that doesn't mean we can use the real algebraic numbers for nonstandard analysis. But Conway says this is "of course" possible with the surreals.
**So the main question is: how would such a transfer principle work?**
Or an even stronger question: **do we need anything like an ultrafilter lemma for Conway's claim to be true?** Unlike the hyperreals, the surreals don't require any ultrafilter at all -- which is a pretty significant achievement, really, since we are at least certain something like $1/\omega$ exists even in ZF, and that the resulting field is real-closed. It would seem to be possible to just go through the motions with taking the nonstandard derivative of, for instance, $\frac{\exp(x + 1/\omega) - \exp(x)}{1/\omega} = \exp{x}\left(\frac{\exp^{1/\omega} - 1}{1/\omega}\right) \approx \exp(x)$, treating $1/\omega$ as a formal expression satisfying the first-order properties of the reals and deriving $\exp(x)$ as the closest real number to the result, none of which seems to have required any kind of ultrafilter. (Or has it, implicitly?)
I've tried to keep this short but there is an enormous amount of subtlety to this question, so I will go into some of that below.
---
Some later results have clarified the relationship between the surreals and hyperreals, so some additional detail regarding what is being asked is probably necessary.
There has been a little bit of prior discussion about this, for instance in [this post](https://mathoverflow.net/questions/91646/surreal-numbers-vs-non-standard-analysis), where it is talked about the much more modern result that the surreals are isomorphic to the proper-class sized ultrapower of the reals. These isomorphisms can be thought of as various ways to transfer real functions to the surreal numbers. So in one sense, the answer is yes, a transfer principle exists in theory. But the pitfall with this approach is that everything requires ultrafilters, and is non-constructive, and there is no canonical choice of isomorphism. This is very different from the way that the surreals are built, which do not require ultrafilters.
On the other hand, Conway's book was written before any of the above results were published (with possibly an exception regarding one paper of Keisler). So partly the question is informal - what did Conway have in mind? But the other part of it is to formally ask if there is some other way to do this that doesn't involve this very particular method of using these isomorphisms, or even to use ultrafilters at all. For instance, what if we don't have the ultrafilter lemma? Then the hyperreals don't necessarily exist at all, but we can still build the surreals, which don't even require choice. Even if we don't have the ultrafilter lemma, can we still just go ahead anyway and say that $\frac{f(x + 1/\omega) - f(x)}{1/\omega}$ is a well-defined expression, and look for the closest real number to it, using some other way to derive a transfer principle?
The other part of the question is admittedly a soft question, but still well worth answering. The ultrafilter construction makes it very easy to see how such a transfer principle would work. Every hyperreal is a set of reals (or an equivalence class thereof), and to transfer any first-order predicate to some hyperreal, you simply ask the predicate of every real in the set and see if it's true of "most" of them (where "most" means "in the ultrafilter"). Thus you have a real-closed field, a "transfer principle," and all of that. Conway, on the other hand, has a very interesting way of building up the surreals in his book which is somewhat agnostic to the choice of set theory, using "birthdays," "left and right sets," etc. I am curious if there is some way to interpret Conway's assertion regarding the existence of $f(x+1/\omega)$ using his own machinery for the surreals, perhaps doing something clever and inductive with the left and right sets, rather than using these later developments involving isomorphisms with the ultrapower.
The last subtlety involves a philosophical point that has sometimes been raised with the topic of surreals vs hyperreals, but it is also worth addressing. There is, for instance, some debate regarding how functions like $\sin$ and $\cos$ should be transferred to the surreals. In theory, you could say that since we have these isomorphisms to the hyperreals, which have a transfer principle, these guarantee the existence of some kind of function on the surreals with the required first-order properties. But the surreals are very tangible in a very constructive sort of way, whereas these isomorphisms are typically totally non-constructive, so there is no way to use them to see what $\sin(\omega)$ should be, if it's positive or negative, etc. On the other hand, you could raise the same philosophical issue with the hyperreals, because there also is no real answer regarding what $\sin((1,2,3,4,...))$ should be, where $(1,2,3,4...)$ is a particular hyperreal number. The answer depends on the ultrafilter, which determines what $(1,2,3,4...)$ even means to begin with, or what properties it has, or if you like, which hyperreal it's referring to.
But what you *can* do with the hyperreals, which is part of the appeal, is you can kind of get "part of the way there" in a totally constructive manner. You know, for instance, that whatever $\sin((1,2,3,4,...))$ is, in some sense it's $(\sin(1), \sin(2), \sin(3), \sin(4), ...)$. Since we don't know what the ultrafilter is we don't know exactly what properties that has, but we do know *something*: we know that for *any* ultrafilter this value will not be an integer, for instance, or any rational number. So, we have some idea of what the transferred sin function would have to look like on the surreals as a result, at least given that $\omega$ is some hypernatural. So even though the ultrafilter is non-constructive, you can at least get "part of the way there" in an entirely constructive manner, which is part of what makes the entire thing interesting. And of course you don't really need to know much more than these few constructive things to actually do nonstandard analysis, just kind of happily plodding along formally doing nonstandard derivatives, with the understanding that the ultrafilter handles all of the various pathological, undefinable sets of indices in some logically consistent way or another.
So the last question is if there is some way for us to do something similar with the surreals, to get "part of the way there," in this sense. That is, to at least have enough constructive "transfer" for us to play around with all of this stuff, but using the framework of the surreals rather than the hyperreals, so that we can see that $f(1 + 1/\omega)$ even makes sense to begin with and play around with it. Something like Terry Tao's "[cheap nonstandard analysis](https://terrytao.wordpress.com/2012/04/02/a-cheap-version-of-nonstandard-analysis/)", perhaps.
| https://mathoverflow.net/users/24611 | Interpreting Conway's remark about using the surreals for non-standard analysis | Conway was of course correct in saying that NSA is irrelevant to the surreals, but like Mike I found Conway’s further remarks about NSA
puzzling and I am not sure what he had in mind. What I think Conway might have said is: "$\mathbf{No}$ is really irrelevant to nonstandard analysis", and, vice versa. After all, whereas the transfer property of hyperreal number systems, a property not possessed by $\mathbf{No}$, is central to the development of nonstandard analysis, the s-hierarchical (i.e the algebraico-tree-theoretic) structure of $\mathbf{No}$, which is absent from hyperreal number systems, is central to the theory of surreal numbers and is responsible for its canonical nature. On the other hand, as I have noted on occasion, I do not rule out the possibility that down the road there might be cross-fertilization between the two theories. However, even if not, it seems to me that to not appreciate that both theories—which are used for quite different purposes--are remarkably powerful and beautiful is a demonstration of ignorance.
While surrealists have thus far shown little interest in applying surreal numbers to nonstandard analysis or in providing an infinitesimalist approach to classical analysis based on surreal numbers more generally, a number of surrealists beginning with Norton, Kruskal, and Conway have fostered the idea of extending analysis to the entire surreal domain. While this has been a comparatively small focal point of the theory thus far and progress has been slow, I am happy to say that, contrary to the assertions of Sam and Emanuele, there is now a reasonably strong theory of surreal integration. After laying dormant for quite some time and in need of much revision, work on surreal integration has recently made substantial progress and I expect that a paper on the subject by Ovidiu Costin and myself will be posted before too long. The theory extends integration from the reals to the surreals for a large portion of Écalle’s class of resurgent functions, in particular, a large subclass of the resurgent functions that are found in applied analysis. On the other hand, there are considerations in the foundation of mathematics that preclude the theory from being extended very much further. So, for example, the theory cannot be applied in general to the class of smooth functions.
Thus far, a sizable portion of the literature on the surreals has dealt with the theory of ordered algebraic systems and (as Emanuele suggests) model-theoretic issues thereof—issues involving ordered abelian groups, ordered domains, ordered fields, ordered exponential fields and ordered differential fields. In the latter two cases, there has been considerable work on Hardy fields and Écalle’s ordered differential field of transseries and generalizations thereof. Some of the work on ordered exponential fields and ordered differential fields, however, go well beyond ordered algebraic and model-theoretic considerations and are concerned with developing asymptotic differential algebra--the subject that aims at understanding the asymptotics of solutions to differential equations from an algebraic point of view—for the surreals. In their 2017 ICM talk (*On Numbers, Germs, and Transseries*), Aschenbrenner, van den Dries and van der Hoeven outline the program they (along with Mantova, Berarducci, Bagayoko and Kaplan) are engaged in for developing an ambitious theory of asymptotic differential algebra for all of the surreals, though one that would require a derivation on $\mathbf{No}$ having compositional properties not enjoyed by the derivation introduced by Mantova and Berarducci in the paper cited by Emanuele. Such a program, if successful, would provide the most dramatic advance towards interpreting growth rates as numbers since the pioneering work of Paul du Bois-Reymond, G. H. Hardy and Felix Hausdorff on "orders of infinity" in the decades bracketing the turn of the 20th century. Unlike the surreals, the framework on NSA does not appear to be particularly well suited to this end, despite the fact that Robinson and Lightstone did make modest contributions to the theory of asymptotics in their nice monograph that applies NSA to the subject.
In his response, Sam noted that (as far as he knew) the surreals do not have a visible copy of surnatural numbers. Sam is, of course, correct; but given the recursive nature of the construction of the surreals no one should expect to find one, and for the purposes for which it has been used it has not proven to be a limitation. On the flip side, of course, unlike the surreals, the number systems employed in NSA do not have a canonical copy of the ordinals or significant initial segments thereof. However, for the purpose of developing an infinitesimalist approach to classical analysis, I don’t see this to be a problem either. Given all the areas and questions to which mathematicians apply finite, infinite and infinitesimal numbers, it is extremely unlikely there will ever be one theory ideally suited for all applications. An assertion like "For all its weaknesses" made by Sam about the surreals, fails to appreciate this. What would be a weakness in one context need not necessarily be a weakness in another.
All of the people I know who are working in the theory of surreal numbers are familiar with, and have great respect for, NSA. Sadly, however, there appears to be a small segment of the contemporary NSA community who, while repeatedly demonstrating their lack of knowledge of the subject and its applications, attack it time and again, one person (as is evident from the comments) even describing those who work on it as members of a cult. During the 19th century, Cantor repeatedly attacked the works of du Bois-Reymond, Stolz, and Veronese on their non-Cantorian theories of the infinite (and infinitesimal), theories designed to deal with issues not addressed by Cantor’s theory—non-Archimedean geometry, the rates of growth of real functions, and non-Archimedean ordered algebraic systems. Abraham Robinson, who was as gracious a person as he was a great and knowledgable mathematician, attempted to soften his well-deserved implicit critique of Cantor’s misguided and narrow-minded attacks by noting: "It may be recalled that, at that time, Cantor was fighting hard in order to obtain recognition for his own theory" (*The Metaphysics of the calculus*, p. 39). I wonder to what extent some of the aforementioned attacks are motivated by similar considerations, despite the fact that NSA is already widely, albeit not universally, regarded as a major contribution. However, whatever the motivation may be, I believe losing sight of the lesson of Cantor or the humanity of Robinson would be an unfortunate mistake indeed.
P.S. One of the longstanding bugaboos in the aforementioned attacks on the surreals has been that there is no natural sine and cosine functions for the surreals. For a proof that this contention is mistaken, see Section 11 of Kaplan and the author’s recent *[Surreal Ordered Exponential Fields](https://doi.org/10.48550/arXiv.2002.07739)*, The Journal of Symbolic Logic, 86 (2021) pp.1066-1115.
| 8 | https://mathoverflow.net/users/18939 | 418718 | 170,478 |
https://mathoverflow.net/questions/418727 | 8 | It is well known that the only solution is $f$ a constant function. However, by putting some restrictions on the functional equation, we might get other solutions, with potential implications to solving Diophantine equations or factoring an integer.
The restrictions are as follows: $f(xy)=f(x+y)$ if $x,y\geq 3$ are positive integers and $\gcd(x,y)=1$. Now my question is whether or not there is a non-constant function satisfying these requirements. I consider that $f(u)$ is defined for integers $u\geq 7$.
Let $f$ be a solution. I use the following notation: $u\sim v\bmod{f}$ if and only if $f(u)=f(v)$. This defines equivalence classes, just like modulo operators define residue classes. Various functional equations similar to the one discussed here, produce various sets of equivalence classes, in the same way that various moduli produce various sets of residue classes.
**Link to Diophantine equations and factoring**
This is just a trivial example to illustrate the concept. Using various non-constant $f$ solutions of various related functional equations, we can replace the equation $x^2 + y^2 = z^2$ by $(xy)^2 \sim z^2 \bmod{f}$ if we could come up with a large class of $f$'s that make the two equations equivalent. In short, we reduce an additive equation with $k$ variables (here $k=3$) to one having $k-1$ variables.
Likewise, factoring $z$ consists of finding $x,y$ such that $xy=z$. It is equivalent to solving the additive equation $x+y \sim z$ for various $f$'s, hopefully easier to handle once turned from a multiplicative into an additive problem.
**But do non-constant solutions $f$ really exist?**
That is my question. A corollary, assuming such functions exist, is how difficult they are to handle, and even whether finding such a function could be an unsolved problem for years to come.
If such non-constant functions exist and are rather simple, then Fermat's theorem would have been proved long ago. This makes me think that either they don't exist, or if they exist, they are a tough nut to crack.
**Possible approach to finding a non-constant $f$**
I define $f(u)$ as the smallest integer $v\geq 7$ such that $u\sim v \bmod{f}$. Again, see the analogy with residue classes.
Now let's start with $u=7$. We have $7=3+4$, thus $7\sim 12$. We have $12=5+7$, thus $7\sim 12\sim 35$. Apply the principle recursively, and you will get an infinite equivalence class for $7$. Does it cover all integers $\geq 7$? If yes the only solution for $f$ is $f(u)=7$ (a constant function).
Now let's start with $u=11$. We also end up with an infinite class, seemingly larger than the previous one, and apparently containing all prime numbers $>7$ and many non-primes. In particular
$$11 \sim 13 \sim 17\sim 19 \sim 23 \sim 24\sim 25 \sim 28 \sim 29 \sim 30\sim 31 \sim 36 \sim 37 \sim 40\sim 41$$
How many equivalence classes do we end up with? One? (then $f$ is constant.) Two? Three? More than three? Note that finding the equivalence classes consists of finding all the connected components of an undirected (infinite) graph.
If you prove that $7\sim 11$, that won't answer my question, but it will be a big blow, big enough that I may stop doing research on this topic. And I would accept the answer. Likewise, if you prove that $7$ and $11$ are not equivalent ($\bmod{f}$) it will show that a non-constant $f$ exists, and this would be a big encouragement.
**Update**
See solutions provided by readers. The possibility of the existence of a non-constant function $f$ was "too good to be true". That said, the functional equation discussed here is a particular case of $a f(xy)= b f(x+y) +c$ with $a,b,c$ integers. Or something more general like (say) $f(xy)=f(g(x+y))$ for some function $g$. Maybe a non-constant solution can be found in this setting. In the end, the goal is to transform an additive problem into a multiplicative one, or the other way around. And to do it in such a way that it leads to simplifications.
| https://mathoverflow.net/users/140356 | Solving functional equation $f(xy)=f(x+y)$ and Diophantine equations | Too long for a comment. My guess is that for $x\geq 7$ we that that $f(x)=c$ by strong induction. We need to check the cases $7\leq x\leq 20$ by hand .
Let us now suppose we have $y\geq 20$ and integer. If it can be factored as $a\cdot b$ with $a,b\geq 3$ coprime then obviously $6\leq a+b<ab=y$ and we are done. Thus the only cases left are primes $p^k$ and $2p^k$ both being bigger than $20$.
For $2p^k$ we have two cases $p^k\equiv 1\pmod{3}$ and $p^k\equiv 2\pmod{3}$. For $p^k\equiv 1 \pmod{3}$ note that $$f(2p^k)=f(2p^k-5+5)=f(5(2p^k-5))=f(5\cdot 3^{a} (2p^k-5)/3^a)=f(5\cdot 3^a+(2p^k-5)/3^a)$$
where $a=v\_3(2p^k-5)$. We are done unless $2p^k-5=3^a$. Of course we can now switch and do $f(2p^k-11+11)$ and similarly we will get that $2p^k-11$ is a power of $3$. It is easy to see that the only only solution to $3^a+6$ being also a power of $3$ is $a=1$ but this $3^2$.
The above argument violates the coprime condition when $p=5$ and $5^k\equiv 1\pmod{3}$ and $p=11$ and $11^k\equiv 1\pmod 3$. But for the first option this leads to $2\cdot 5^k-11$ being a power of $3$, and the second leads to $2\cdot 11^k-5$ being a power of $3$. But we can further write $2\cdot 5^k=2\cdot 5^k-17+17$ and same for $2\cdot 11^k$ and we get a contradiction.
For $p^k\equiv 2\pmod{3}$ we do the same trick $f(2p^k)=f(2p^k-7+7)$ and again unless $2p^k-7$ is a power of $3$ we win. Swiching with $13$ we obtain than also $2p^k-13$ is a power of $3$ and like above $p^k=8$ but this is not in the range.
For $p^k$ we have $f(p^k)=f(p^k-3+3)=f(3(p^k-3))$. Note that $3$ does not divide $p^k-3$. If $p\equiv 1\pmod{4}$ then $f(3(p^k-3))=f(6(p-3)/2)=f(6+(p^k-3)/2)$ and we have $6+(p^k-3)/2<p^k$. If $p^k\equiv 3\pmod{4}$ then $f(p^k-5+5)=f(5(p^k-5))=f(10 (p^k-5)/2)=f(10+(p^k-5)/2)$ and we are done.
We miss again the coprimality when $p=3$. We have $f(3^k)=f(3^k−5+5)=f(5(3^k−5))=f(5⋅2^a⋅(3^k−5)/2^a)$ so we are done unless $3^k−5$ is a power of 2. We can iterate the argument with 7 and we get $3^k−7$ is a power of 2 and this forces $k=2$.
| 9 | https://mathoverflow.net/users/41010 | 418730 | 170,480 |
https://mathoverflow.net/questions/418721 | 2 | I am a bit confused by the following question and I hope someone could help me out.
Let $u$ be the solution of the following initial value problem
$$
u''(t) = g(t) \; \text{ in } (0,\infty), \quad\quad u(0)=a, \quad u'(0)=0. \label{1}\tag{1}
$$
Let $U$ and $G$ be the extensions of $u$ and $g$ as zero to $(-\infty, \infty)$, that is,
$$
U(t) = \left\{
\begin{array}{ll}
u(t) & t \geq 0 \\
0 & t < 0,
\end{array}
\right.
\quad\quad\quad
G(t) = \left\{
\begin{array}{ll}
g(t) & t \geq 0 \\
0 & t < 0,
\end{array}
\right.
$$
For any compactly supported smooth function $\varphi$, $$
(U'',\varphi) = (U,\varphi'') = \int\limits^\infty\_0 u(t)\varphi''(t) \,dt = - a \varphi'(0) + (g,\varphi).
$$ This means
$$
U'' = a \delta' + G \quad \text{ as distributions in } \mathbb{R}. \label{2}\tag{2}
$$
It is clear that \eqref{2} restricted to $(0,\infty)$ yields the equation in \eqref{1}, but what conditions are needed to restore the initial conditions (other than the trivial condition that $U(0)=a$ and $U'(0)=0$)? I feel that the initial condition should have been included in \eqref{2} as the constant $a$ appears, but I don't know how to restore them.
Thank you.
| https://mathoverflow.net/users/479159 | Restore initial condition for distributions | $\newcommand{\R}{\mathbb R}$In accordance with comments by Willie Wong and the OP, let us extend $u$ by $a$ to the left of $0$:
\begin{equation\*}
U(t) :=
\begin{cases}
u(t) & \text{ if }t\ge0, \\
a & \text{ if }t<0.
\end{cases}
\tag{3}\label{3}
\end{equation\*}
Then
\begin{equation\*}
U''=G, \tag{4}\label{4}
\end{equation\*}
where $U$ and $G$ are identified with the corresponding distributions on $\R$.
Let $V$ be another distribution on $\R$ such that \eqref{4} holds with $V$ in place of $U$. Let $X:=V-U$.
Then $X''=0$. Hence, the distribution $X$ can be identified with an affine function on $\R$ (please let me know if you need details on this claim). We want to impose an additional condition on $X$ to ensure that $X=0$. Since $X$ and its derivatives are distributions, they do not have definite values at any point in $\R$. So, we cannot impose any condition on such values. What we can do instead is require that $X$ be $0$ on some nonempty finite open interval $I\subset\R$ -- i.e., that $(X,\psi)=0$ for all smooth functions $\psi$ with support $S\_\psi\subset I$.
Thus, $U$ will be uniquely determined by condition \eqref{4} together with the condition
\begin{equation\*}
(U,\psi)=a\int\_\R\psi
\end{equation\*}
for all smooth functions $\psi$ with $S\_\psi\subset I$, where $I$ is a nonempty finite open subinterval of the interval $(-\infty,0)$.
| 2 | https://mathoverflow.net/users/36721 | 418755 | 170,490 |
https://mathoverflow.net/questions/418706 | 1 | It is [well known](https://i.stack.imgur.com/EJ9gG.png) that for the diffusions
\begin{align\*}
dX&=f(X)dt+&cdB\\
dY&=&cdB
\end{align\*}
the density of the law of $X$ with respect to the law of $Y$ is
\begin{align\*}
\frac{d\mu\_X}{d\mu\_Y}(c B)&=\exp\left(\int\_0^T\frac{f(cB(t))}{c}dB(t)-\frac12\int\_0^T\frac{f^2(cB(t))}{c^2}dt\right)\\
&=\exp\left(\frac{1}{c^2}\left(\int\_0^Tf(cB(t))d(cB(t))-\frac12\int\_0^Tf^2(cB(t))dt\right)\right)\\
&:=\exp\left(\frac{1}{c^2}L(cB(t))\right),\end{align\*}
where $L(cB)=\int\_0^Tf(cB(t))d(cB(t))-\frac12\int\_0^Tf^2(cB(t))dt$. My interest now is in making $\frac{1}{c^2}L(cB(t))$ defined pathwise. Applying integration by parts yields that
$$\frac{1}{c^2}L(cB)=\frac{1}{c^2}\left(F(cB(T))-\frac{c^2}{2}\int\_0^T f'(cB(t))dt-\frac12\int\_0^T f^2(cB(t))dt\right):=\frac{1}{c^2}L^c(cB).$$
I guess my question is - why does the "pathwise" functional depend on $c$ where the nonpathwise functional doesn't? What is the term $\int\_0^T f'(B(t))dt$ in relation to the original $L$? Is there a nice expression?
If we define $L\_1(B)=\int\_0^Tf(B(t))d(B(t))-\frac12\int\_0^Tf^2(B(t))dt$ and $L\_2(B)= F(B(T))-\frac{1}{2}\int\_0^T f'(B(t))dt-\frac12\int\_0^T f^2(B(t))dt\ $. Then $L\_1(B)=L\_2(B)$ however seemingly $L\_1(cB)\neq L\_2(cB)$. Edit:actually this comes from the fact that Ito's lemma only holds a.s. and cB is singular with B for $c\neq \pm 1$.
| https://mathoverflow.net/users/479223 | Where does the extra term in the density of a diffusion with respect to $c B(t)$ come from? | In a nutshell you are asking the following:
When $F(x)=\int\_0^xf(y)\,dy$ we get from Ito
$$
F(cB(t))=\int\_0^Tf(cB(t))\,d(cB(t))+\frac{1}{2}\int\_0^Tf'(cB(t))\,c^2\,dt\,
$$
so that the expressions
\begin{align}
&F(cB(t))-\frac{c^2}{2}\int\_0^Tf'(cB(t))\,dt\,,\tag{1}\\
&\int\_0^Tf(cB(t))\,d(cB(t))\tag{2}
\end{align}
are equal. Then why (2) depends only on $cB(t)$ and not separately on $c^2$ and $cB(t)$ like (1) seems to do?
Good question. I think the answer is that (1) can also be written as
\begin{align}
&F(cB(t))-\frac{1}{2}\int\_0^Tf'(cB(t))\,d\big\langle cB\big\rangle\_t\,,\tag{1'}
\end{align}
and gone is the isolated $c^2$.
| 2 | https://mathoverflow.net/users/340165 | 418756 | 170,491 |
https://mathoverflow.net/questions/418746 | 5 | Let $A$ be a unital complex algebra with the unit $\bf1$. Let $\mathcal{N}$ be the family of all norms on $A$ making it a unital normed algebra with the same unit $\bf1$. Let us put
$B\_{\|\cdot\|}=\{x\in A : \|x-{\bf1}\|<1\}$ where $\|\cdot\|\in \mathcal{N}$.
Clearly, the intersection $\bigcap\_{\mathcal{N}}B\_{\|\cdot\|}$ is contained in $A^{-1}\_{\|\cdot\|}$ where $A\_{\|\cdot\|}$ is the completion of $A$ with respect to norm $\|\cdot\|$ in $\mathcal{N}$.
>
> Q. Does there exist any non-scalar element in the intersection $\bigcap\_{\mathcal{N}}B\_{\|\cdot\|}$?
>
>
>
| https://mathoverflow.net/users/84390 | Permanent invertible elements | Here is a non-trivial condition:
>
> If $x$ belongs to this intersection, then $x$ commutes with every nilpotent element.
>
>
>
*Proof*. For every invertible $a\in A$, the function
$$N(z):=\lVert a^{-1}za\rVert$$
is a norm of unital algebra. Let $n$ be a nilpotent element, of order $k$, and choose $a\_t=\mathbf1-tn$ in the construction above, with $t\in\mathbb R$ a parameter. Then
$$a\_t^{-1}xa\_t=(\mathbf 1+tn+\dotsb+t^{k-1}n^{k-1})x(\mathbf1-n)=x+t(nx-xn)+\cdots-t^kn^{k-1}xn$$
is a polynomial function of $t$.
By assumption, $t\mapsto\lVert a\_t^{-1}xa\_t\rVert$ is a bounded (by $1$) function, hence the polynomial above needs to be constant. In particular $nx-xn=0$.
As a corollary, the answer for the case of $A=M\_r({\mathbb C})$ with $r\ge2$ is as you expected. Write $X$ instead of $x$ (it is a matrix). Let $u\in{\mathbb C}^r$ be a non-zero vector. Choose $v$ such that $v^Tu=0$ (it exists). Then $uv^T$ is nilpotent, hence $Xuv^T=uv^TX$, which implies that $Xu$ is parallel to $u$. Thus every vector is an eigenvector, which implies that $X$ is scalar: $X=\alpha I\_r$.
**Addition**. Suppose now that $(A,\|\cdot\|)$ is a unital Banach algebra. The spectral radius
$$r(u)=\lim\inf\|u^k\|^{1/k}$$
is well-defined. As above, $\cal N$ contains all norms $N\_a=\|a^{-1}\cdot a\|$ for $a\in A^\times$. If $x$ belongs to the OP's intersection, then the set
$$\{a^{-1}xa\;|a\in A^\times\}$$
is bounded. Consider an element $u\in A$ for which $r(u)=0$ ($u$ can be be nilpotent, but this is not necessary if $A$ is infinite dimensional). Then ${\bf1}-zu$ is invertible for every $z\in\mathbb C$. Thus
$$z\mapsto({\bf1}-zu)^{-1}x({\bf1}-zu)$$
is a bounded entire function, hence a constant function, $\equiv x$. In other words $x({\bf1}-zu)\equiv ({\bf1}-zu)x$, that is $xu=ux$. Thus the property mentionned above extends to:
>
> If $x$ belongs to this intersection, then $x$ commutes with every element of spectral radius $0$.
>
>
>
| 8 | https://mathoverflow.net/users/8799 | 418762 | 170,494 |
https://mathoverflow.net/questions/418726 | 3 | An operator $T$ on a separable Hilbert space $H$ is called *unicellular* if any two closed invariant subspaces $M$ and $N$ are comparable; that is either $M\subseteq N$ or $N\subseteq M$. There are many examples of (compact) unicellular operators, for example, the Volterra operator.
I am looking at a weaker property. Namely, the existence of a closed nontrivial invariant subspace $M$ that is *comparable* to all the other closed, invariant subspaces. Obviously, any closed invariant subspace of a unicellular operator satisfies this property. On the other hand, any diagonal operator will not be unicellular and won't satisfy this weaker property either.
In particular, I am interested in compact operators that do not have eigenvalues. My questions are:
1. Are there examples of compact operators without eigenvalues having this weaker property, but are not unicellular? (I suspect the answer is *Yes*)
2. Does any compact operator without eigenvalues have this weaker property? (I suspect the answer is *No*)
Edit: The question is edited for clarity.
| https://mathoverflow.net/users/69275 | Unicellular compact operators | $V\oplus V\in B(L^2(0,1)\oplus L^2(0,1))$, with $V$ being the Volterra operator, is an operator without eigenvalues that does not have your property. The invariant subspaces are $L^2(0,a)\oplus L^2(0,b)$, and for any such space $M=A\oplus B$, we can find another one $N=C\oplus D$, with $C\subsetneq A$, $D\supsetneq B$, unless $M=0, H$.
| 1 | https://mathoverflow.net/users/48839 | 418769 | 170,498 |
https://mathoverflow.net/questions/418770 | 5 | Currently I study the mathematical formulation of the (classical) standard model of particle physics using the language of gauge theory and spin geometry. One of the central objects in the standard model are "charged spinors", which are fermionic particles, which transform under a non-trivial representation. Now, the point is that I stumbled over two appearently different definitions and I would like to know if they are somehow related.
To fix notation, let us take a (suffiently nice) pseudo-Riemannian manifold $(\mathcal{M},g)$ with signature $(s,t)$ and a principal $G$-bundle $P\stackrel{\pi}{\to}\mathcal{M}$. Furthermore, let $\mathrm{Spin}(\mathcal{M})$ be a spin-structure on $\mathcal{M}$ and $(\rho,V)$ be a finite-dimensional representation of $G$.
1. The first definition seems to be a little bit more standard in the mathematical literature. It is for example used in the textbook "*Mathematical Gauge Theory*" by M. J. D. Hamilton from Springer. First of all, consider the spinor bundle
$$S:=\mathrm{Spin}(\mathcal{M})\times\_{\kappa}\Delta\_{n},$$
where $(\kappa,\Delta\_{n})$ denotes the spinor representation of $\mathrm{Spin}(s,t)$ (the vector space $\Delta\_{n}$ is $\Delta\_{n}\cong\mathbb{C}^{N}$ where $N=2^{n/2}$ if $n$ is even and $N=2^{(n-1)/2}$ otherwise). Then, a "*charged spinor*" is defined to be a section of the twisted bundle
$$E\otimes S,$$
where $E$ denotes the associated vector bundle $E:=P\times\_{\rho}V$.
2. The second definition is used in [nlab](https://ncatlab.org/nlab/show/The+Dirac+Electron) and also in the book "*Gauge Theory and Variational Principles*" from D. Bleecker. Here, one defines first of all the concept of "bundle splicing": Take two principal bundles $P\_{i}\stackrel{\pi\_{i}}{\to}\mathcal{M}$ with structure group $G\_{i}$ for $i\in\{1,2\}$. Then, the fibre product $P\_{1}\times\_{\mathcal{M}}P\_{2}$ is a principal $(G\_{1}\times G\_{2})$-bundle denoted by $P\_{1}\circ P\_{2}$. Now, take a representation $R:\mathrm{Spin}(s,t)\to\mathrm{Aut}(V)$, which commutes with $\rho$. Then, one defines "*charged spinors*" to be sections of
$$(\mathrm{Spin}(\mathcal{M})\circ P)\times\_{(R\times\rho)}V$$.
So, is there any relations between the two? I expect that there should be one, since in the end both of them are used to model the same physical object. However, I cannot see how they are related. If they are different, is one of them more general in some sense?
| https://mathoverflow.net/users/199422 | Different definitions of "charged spinors": "bundle splicing" vs. "twisted spinor bundles" | I'll assume that the vector space "$V$" occuring in constructions (1) and (2) doesn't have to be the same. In that case I'll rename vector space in construction (2) to "$W$."
Then I claim that construction (1) is a special case of construction (2) with $W=V\otimes\Delta\_n$. That follows from the general fact that if $V\_1$ and $V\_2$ are representations of $G\_1$ and $G\_2$, and $P\_1$ and $P\_2$ are $G\_1$- and $G\_2$-principal bundles over $M$, then the tensor product of the vector bundles $P\_1\times\_{G\_1}V\_1$ and $P\_2\times\_{G\_2}V\_2$ is isomorphic to $(P\_1\circ P\_2)\times\_{(G\_1\times G\_2)}(V\_1\otimes V\_2)$.
I think the easiest way to see that is to use classifying spaces. Then the bundle $P\_i$ are maps $f\_i:M\rightarrow BG\_i$, the representations determine maps $\rho\_i:BG\_i\rightarrow B\text{Aut}(V\_i)$, the bundle $P\_1\circ P\_2$ is the map $M\xrightarrow{\text{diag}} M\times M\xrightarrow{f\_1\times f\_2}BG\_1\times BG\_2$, the vector bundle $P\_i\times\_{G\_i}V\_i$ is the map $M\xrightarrow{f\_i} BG\_i\xrightarrow{\rho\_i}B\text{Aut}(V\_i)$, and we find that the vector bundles we're trying to compare are both given by the map
$$M\xrightarrow{\text{diag}} M\times M\xrightarrow{f\_1\times f\_2}BG\_1\times BG\_2\xrightarrow{\rho\_1\times\rho\_2}B\text{Aut}(V\_1)\times B\text{Aut}(V\_2)\rightarrow B\text{Aut}(V\_1\otimes V\_2)$$
| 6 | https://mathoverflow.net/users/163893 | 418781 | 170,503 |
https://mathoverflow.net/questions/418778 | 5 | Let $G$ be a reductive group over $\mathbf{C}$. It acts on the dual of its Lie algebra $\mathfrak{g}^\*$ by conjugation.
1. One can describe the orbits of $\mathfrak{g}^\*$ explicitly (e.g. using Jordan blocks for $\operatorname{GL}\_n$),
2. There is an especially interesting class, of *nilpotent* orbits. There are finitely many, labelled by combinatorial data attached to $G$ (e.g. partitions of $n$), and are related to a lot of interesting maths.
My **question** is: what when you replace $\mathfrak{g}^\*$ by a general finite dimensional representation $V$ (with some conditions, if you like)? What can be said about their orbits: are they interesting, and can they be classified?
Possibly the answer might just be no, and that $\mathfrak{g}^\*$ is special because it's Poisson/has other special properties, but hopefully at least some other $V$'s are interesting.
| https://mathoverflow.net/users/119012 | Interesting properties of "coadjoint" orbits inside $V\in \operatorname{Rep}G$ | Generally they can be classified when the action of $G$ on $V$ is *visible*, which by definition means there are finitely many orbits in the nullcone. Irreducible visible pairs $(G, V)$ were classified in [Kac - Some remarks on nilpotent orbits](https://doi.org/10.1016/0021-8693(80)90141-6) (with some corrections in [Dadok–Kac - Polar representations](https://doi.org/10.1016/0021-8693(85)90136-X)). Most of them come from Vinberg's theta-groups, which are graded Lie algebras/groups (graded by the integers or integers mod $n$). The visible action in those cases is then to consider the adjoint action of the Lie group on its Lie algebra, restricted to the zero-graded piece $G\_0$ of the group acting on the 1-graded piece $\mathfrak{g}\_1$ of the Lie algebra. In those cases being in the nullcone and being nilpotent in the usual sense coincide, and the Jordan decomposition respects the grading.
| 5 | https://mathoverflow.net/users/38434 | 418788 | 170,504 |
https://mathoverflow.net/questions/418780 | 6 | How do we deduce the following statement from the Chebotarev density theorem? The statement is from Ngo's Fundamental Lemma paper.
>
> Let $U$ be a scheme of finite type over $\mathbb{F}\_q$. Let $\mathcal{L}\_1$, $\mathcal{L}\_2$ be two local systems of pure weight $i$. Suppose for any $u \in U$ and $k \in \mathbb{N}$, we have
> $Tr(\sigma\_u^k,\mathcal{L}\_1)=Tr(\sigma\_u^k,\mathcal{L}\_2)$, where $\sigma\_u$ is the Frobenius conjugacy class of $u$ in $\pi\_1(U)$.
> Then $\mathcal{L}\_1$ and $\mathcal{L}\_2$ are isomorphic up to semisimplification.
>
>
>
| https://mathoverflow.net/users/130879 | Chebotarev density theorem and pure weight local systems | As Piotr says, we must assume $U$ normal. The purity assumption is not needed.
There are two steps to this proof
(1) Suppose for any $u \in U$ and $k \in \mathbb{N}$, we have
$Tr(\sigma\_u^k,\mathcal{L}\_1)=Tr(\sigma\_u^k,\mathcal{L}\_2)$, where $\sigma\_u$ is the Frobenius conjugacy class of $u$ in $\pi\_1(U)$. Then for any $\sigma \in \pi\_1(U)$, we have $Tr(\sigma^k,\mathcal{L}\_1)=Tr(\sigma^k,\mathcal{L}\_2)$
(2) Suppose that for any $\sigma \in \pi\_1(U)$, we have $Tr(\sigma^k,\mathcal{L}\_1)=Tr(\sigma^k,\mathcal{L}\_2)$ Then $\mathcal{L}\_1$ and $\mathcal{L}\_2$ are isomorphic up to semisimplification.
The first one follows from the Chebotarev density theorem. Because the traces are $\ell$-adic numbers, it suffices to prove the identity mod $\ell^n$ for each $n$. Whether this is true or not for a given $\sigma$ depends only on the image of $\sigma$ in some finite group (a subgroup of $GL\_m (\mathbb Z/\ell^n\mathbb Z)^2$ where $m$ is the rank of these representations).
But in a finite quotient of a Galois group (here is where we use normalcy, to get that the fundamental group is a quotient of a Galois group), every element arises from a Frobenius conjugacy class. Since we have the identity on every Frobenius element, we have it on every element.
The second one is a general fact about representations of groups over fields of characteristic zero. In particular, we do not use the $\ell$-adic topology, so we may assume the base field is $\mathbb C$. Let $G$ be the Zariski closure of the image of $\pi\_1$ in $GL\_m \times GL\_m$ acting on the two (semisimplified) representations. Because $G$ has a faithful semisimple representation, it is reductive. Take a maximal compact subgroup. If two representations are isomorphic on restriction to this compact subgroup they are isomorphic (Weyl). Now use character theory on this compact subgroup.
| 11 | https://mathoverflow.net/users/18060 | 418793 | 170,506 |
https://mathoverflow.net/questions/418801 | 4 | Assume $0^\#$ exists and there is an inaccessible cardinal.
Are there two transitive sets $M,N$ s.t. $M\in N,M\vDash ZF+V=L[0^\#],N\vDash ZF+V=L$?
| https://mathoverflow.net/users/170286 | Can local $0^\#$ exists in L? | Take a countable elementary submodel of $L\_\kappa[0^\#]$, code that into a real, note that "There is a real coding a well-founded model of $V=L[0^\#]$" is a $\Sigma^1\_2$ statement, remember what Shoenfield said about such statements: they are also true in $L$.
Now take $M$ to be a transitive collapse of some real coding that in $L$ and $N=L\_\kappa$, or some transitive collapse of a countable elementary submodel thereof.
| 8 | https://mathoverflow.net/users/7206 | 418803 | 170,508 |
https://mathoverflow.net/questions/417589 | 1 | Consider a normal complex analytic space $X$ and a projective birational resolution $f:Y\rightarrow X$. Let $T$ be a closed positive current of bi-dimension $(p,p)$ on $X$. Is there always a closed positive current $S$ on $Y$ such that $f\_\*S=T$?
| https://mathoverflow.net/users/58308 | Do all closed positive currents lift to a resolution? | I believe the answer is no.
If I am reading the article below by Méo correctly, he shows that if
$\pi : Y \to X$ is the blowup of the polydisc $X=D^m$ along $Z := \{ z\_1 = \dots = z\_k = 0 \}$, and if $k \leq p \leq n-2$, then there exists a closed positive $(p,p)$-current $T$ on $X$ such that $T' := \pi^\* (T|\_{X\setminus Z})$ does not have locally finite mass near $\pi^{-1}(Z)$. If such $S$ exists, then since $\pi$ is a biholomorphism on $Y' := \pi^{-1}(X\setminus Z)$, one would have $S|\_{Y'} = T'$, which would thus have locally finite mass.
**Reference**
Michel Méo, "[Image inverse d'un courant positif fermé par une application analytique surjective](https://gallica.bnf.fr/ark:/12148/bpt6k9800442g/f35.item)", C. R. Acad. Sci. Paris Sér. I Math. 322 (1996), no. 12, 1141–1144, [MR1396655](https://mathscinet.ams.org/mathscinet-getitem?mr=MR1396655), [Zbl 0858.32012](https://zbmath.org/?q=an%3A0858.32012).
| 2 | https://mathoverflow.net/users/49151 | 418811 | 170,510 |
https://mathoverflow.net/questions/418789 | 3 | Let $p $ be an odd prime. Assume that we have the following perfect pattern:
all the primes below $p$ are successively quadratic residues and quadratic non-residues. What can we say about $p$? Is it possible that only finitely many such $p$ exist?
Edit: Mathematically I mean: denote by $p\_n$ the $n$-th smallest prime. Then is it true that there are infinitely many $N$ with the two properties
$$1\leq n < N \& \ n \text{ odd } \Rightarrow (\frac{p\_n}{p\_N})=(\frac{p\_1}{p\_N})$$
and
$$1\leq n < N \& \ n \text{ even } \Rightarrow (\frac{p\_n}{p\_N})=(\frac{p\_2}{p\_N}).$$
| https://mathoverflow.net/users/9232 | Perfect equidistribution for the Legendre symbol | It is very likely that there are only finitely many primes with this property. Heuristically, the probability of this happening for the $N$th prime is $2^{2-N}$, and $\sum\_{N \geq 2} 2^{2-N} =2$, so we expect only a couple primes where this happens.
I can't see how one would ever hope to prove this. However, we can prove there are not many such primes in some large intervals.
If $p$ and $q$ are two such primes with $p<q$, then $\left( \frac{ \ell}{pq}\right)$ is either the constant function $1$ for $\ell<p$ or the constant function $-1$. Thus $\left( \frac{m}{pq} \right)$ is either the constant function $1$ for $m < pq$ or the Möbius function. The second case is a Siegel zero-type phenomenon, but the first case is easier to rule out.
Since most numbers $< p^{\sqrt{e}}$ have only prime factors $<p$, the average of the Legendre symbol $\left( \frac{m}{pq} \right)$ over numbers $< p^{\sqrt{e}}$ is positive and large in the first case - there are more $1$s then $-1$s. By Burgess, this is impossible unless $pq > (p^{\sqrt{e}})^{4-\epsilon}$, or $q> p^{ 4 \sqrt{e}-1 -o(1)}$.
So every perfectl equidistributed prime between $p$ and $ p^{ 4 \sqrt{e}-1 -o(1)}$ must have the opposite sign (of the Legendre symbol applied to 2) from $p$. Since these perfectly equidistributed primes must also have opposite signs from each other, there can be at most one more perfectly equidistributed prime in that range.
This means the number of perfectly equidistributed primes $<n$ is at most $$ \left(\frac{2}{ \log (4 \sqrt{e}-1)} + o(1)\right) \log \log n.$$
So there can be very few such primes in a given range.
| 6 | https://mathoverflow.net/users/18060 | 418819 | 170,514 |
https://mathoverflow.net/questions/418749 | 3 | Suppose that $r>1$ and $s>1$ are irrational numbers, and let $a\_n=\lfloor nr \rfloor$ and $b\_n=\lfloor ns \rfloor$. Assume that $r$ and $s$ are numbers for which $\{a\_n\}\cap\{b\_n\}$ is infinite, and let $(c\_n)$ be the increasing sequence of numbers in $\{a\_n\}\cap\{b\_n\}$. It appears that the number
$$t = \lim\_{n \to \infty}\frac{c\_n}{n}$$
exists. Can someone determine its value?
Here are a few estimates:
$$\begin{array}{c|c|c|}
r & s & t \\ \hline
\sqrt{3} & \sqrt{3}+1 & 4.098 \\ \hline
\sqrt{3} & \sqrt{5} & 3.864 \\ \hline
\sqrt{5} & \sqrt{5}+1 & 6.792 \\ \hline
\sqrt{5} & \sqrt{5}-1 & 6.696 \\ \hline
\pi & e & 8.539 \\ \hline
\end{array}$$
| https://mathoverflow.net/users/172419 | Limit associated with two Beatty sequences that are not a Beatty pair | I will give a complete answer when $1,1/r$ and $1/s$ are linearly independent over $\mathbb Q$ and a recipe to compute your $t$ otherwise.
First of all, notice that $n$ lies in a Beatty sequence $\lfloor m\alpha\rfloor$, where $\alpha>1$ if and only if $\{n/\alpha\}>1-\frac{1}{\alpha}$. Indeed, if $n=\lfloor m\alpha \rfloor$, then
$$
\frac{n}{\alpha}=\frac{m\alpha-\{m\alpha\}}{\alpha}=m-\frac{\{m\alpha\}}{\alpha}, \text{ so }\{n/\alpha\}>1-\frac{1}{\alpha}.
$$
On the other hand, if $\{n/\alpha\}>1-\frac{1}{\alpha}$, then one can set $m=\lfloor n/\alpha \rfloor+1=n/\alpha+1-\{n/\alpha\}$ and get
$$
m\alpha=n+\alpha\left(1-\{n/\alpha\}\right),
$$
so $\lfloor m\alpha \rfloor=n$, as needed.
Now, Case 1: $1,1/r$ and $1/s$ are linearly independent over $\mathbb Q$. In this case, it is known (say, by Weyl's criterion), that the points
$$
t\_n=(n/r \mod 1, n/s \mod 1)\in \mathbb R^2 \diagup \mathbb Z^2
$$
are uniformly distributed on the torus. This means that for $N\to +\infty$ the proportion of $n\leq N$ with
$$
\{n/r\}>1-\frac{1}{r}\text{ and }\{n/s\}>1-\frac{1}{s}
$$
is equal to the measure of the set $(x,y) \in \mathbb R^2\diagup \mathbb Z^2$ with $x>1-\frac{1}{r}, y>1-\frac{1}{s}$, hence the inverse $t$ of this limiting proportion is equal to $rs$. For example, for $\pi$ and $e$ we get $\pi e\approx 8.5397$ and for $\sqrt{3}$ and $\sqrt{5}$ we get $\sqrt{15}\approx 3.873$.
Case 1.5 is when both $r$ and $s$ are rational. In this case, your problem reduces to enumeration of certain congruence classes $\mod \mathrm{num}(r)\mathrm{num}(s)$.
Finally, Case 2 is when $1/r$ is irrational, but for some $a,p,q\in \mathbb Z$ we have $1/s=p/q\cdot1/r+a/q$. Without loss of generality one can assume that $p,q>0$, $a\geq 0$. Next, let us divide $\mathbb N$ into congruence classes $\mod q$ and evaluate the (relative) density $\delta\_b$ of all $n\equiv b\pmod q$ such that
$$
\{n/r\}>1-\frac{1}{r}\text{ and }\{n/s\}>1-\frac{1}{s}.
$$
In this case, the answer would be
$$
t=\frac{q}{\delta\_0+\delta\_1+\ldots+\delta\_{q-1}}.
$$
Fix $b$. The congruence $n\equiv b\pmod q$ means that $n=qm+b$. Now, we need
$$
\{qm/r+b/r\}>1-\frac{1}{r}\text{ and }\{pm/r+b/s\}>1-1/s.
$$
Since $1/r$ is irrational, $\{m/r\}$ is equidistributed $\mod 1$. This means that our relative density $\delta\_b$ is equal to the measure of all $x\in [0,1]$ such that
$$
\{qx+b/r\}>1-\frac{1}{r}\text{ and }\{px+b/s\}>1-1/s.
$$
(By relative density I mean the density of corresponding $n$ in the residue class, i.e. the actual density of your $n$ with $n\equiv b\pmod q$ is $\delta\_b/q$)
| 4 | https://mathoverflow.net/users/101078 | 418824 | 170,518 |
https://mathoverflow.net/questions/418753 | 3 | Let $\mu$ be a centred Radon Gaussian measure on a locally convex space $X$ and $q : X \to \mathbb{R}$ a seminorm that is $\mathcal{B}(X)\_\mu$-measurable, where $\mathcal{B}(X)\_\mu$ is the Lebesgue completion of the Borel $\sigma$-algebra on $X$.
Consider a linear functional $\phi : X \to \mathbb{R}$ with $\vert \phi \vert \le q$. Is it also necessarily $\mathcal{B}(X)\_\mu$-measurable?
It follows from theorem 2.10.11 in Bogachev's monograph "Gaussian measures" that there is a unique (up to equivalence $\mu$-almost everywhere) linear functional $\psi : X \to \mathbb{R}$ with $\phi = \psi$ on the Cameron-Martin space of $\mu$.
But I do not see how to prove whether the equality would hold $\mu$-almost everywhere.
| https://mathoverflow.net/users/18936 | Is a functional bounded by a measurable seminorm also measurable? | I have found an answer exploiting the equivalence of Lusin and Borel measurability as stated in [Radon Measures on Arbitrary Topological Spaces and Cylindrical Measures](https://books.google.de/books/about/Radon_measures_on_arbitrary_topological.html?id=19zuAAAAMAAJ) on page 6, theorem 5:
>
> Let $H : X \to Y$. If $H$ is Lusin $\mu$-measurable, then
> $H$ is Borel $\mu$-measurable, and conversely, if $Y$ is metrizable and
> separable, then every Borel $\mu$-measurable function is also Lusin
> $\mu$-measurable.
>
>
>
By assumption $q : X \to \mathbb{R}$ is Borel $\mu$-measurable such that $q$ is also Lusin $\mu$-measurable. Consequently, there is a compact set $K$ of positive measure on which $q$ is continuous and in particular, bounded. Hence, $\phi$ is also bounded on $K$ and the proof is now actually given in theorem 3.11.3 of Bogachev's book.
The idea is to prove that $\phi$ has measurable minorants and majorants that are linear.
For completeness, the definitions of measurability from Schwartz's book on page 25:
>
> Definition 9. The mapping $H$ is said to be Lusin $\mu$-measurable
> if for every compact set $K \subseteq X$ and every $\delta > 0$,
> there exists a compact set $K\_\delta \subseteq K$ with
> $p(K \setminus K\_\delta) < \delta$ and such that $H$ restricted to
> $K\_\delta$ is continuous.
>
>
> We recall here the weaker notion of Borel $\mu$-measurable functions:
> $H$ is Borel $\mu$-measurable if, the inverse image under $H$ of
> every Borel set in $Y$, is a $\mu$-measurable set.
>
>
>
| 1 | https://mathoverflow.net/users/18936 | 418826 | 170,520 |
https://mathoverflow.net/questions/418479 | 5 | Let $G$ be a reductive group scheme over a normal ring $A$. Then, we know that Zariski locally it admits a maximal torus.
Let us assume that it admits a maximal torus after a finite surjective (resp. finite flat) cover, is it possible to replace it by a finite étale cover?
| https://mathoverflow.net/users/27398 | Group scheme with an isotrivial maximal torus | **Edit.** I realized after the original post that there are even easier examples. These examples also show that the Quillen–Suslin Theorem fails already for smooth affine quadric hypersurfaces (in some sense, the "simplest" smooth affine varieties after affine spaces, where Quillen–Suslin does hold).
Let $\overline{B}$ be a Grassmannian parameterizing $2$-dimensional quotient vector spaces of a fixed vector space $V$ of dimension $n\geq 4$, $$\overline{B}=\operatorname{Grass}\_k(V,2),$$ let the rank $2$, locally free $\mathcal{O}\_{\overline{B}}$-module be the tautological quotient, $$V\otimes\_k \mathcal{O}\_{\overline{B}} \twoheadrightarrow \mathcal{E},$$ and let $D\subset \overline{B}$ be a general hyperplane section for the Plücker embedding. For the open complement $B=\overline{B}\setminus D$, all of the arguments in the original post still apply. Moreover, by van Kampen's theorem, the (tame) fundamental group of $B$ is trivial. So that is definitely a better example than in the original post. For completeness, I include below the original post.
**Original post.**
Sorry for the delay. I had an overly elaborate example starting with an Enriques surface in characteristic $2$. In fact, sufficiently general K3 surfaces in characteristic $0$ give even simpler examples.
Let $\overline{B}$ be a K3 surface over $\mathbb{C}$. Recall by Mumford's Theorem that the group $\operatorname{CH}\_0(\overline{B})$ is uncountably generated, in fact, it cannot even be parameterized by any (finite dimensional) algebraic variety, in an appropriate and precise sense.
MR0249428
Mumford, D.
Rational equivalence of 0-cycles on surfaces.
J. Math. Kyoto Univ. 9 (1968), 195–204.
<https://projecteuclid.org/journals/journal-of-mathematics-of-kyoto-university/volume-9/issue-2/Rational-equivalence-of-0-cycles-on-surfaces/10.1215/kjm/1250523940.full>
Moreover, in every ample divisor class on $\overline{B}$, there exists a divisor $D$ whose irreducible components have normalization isomorphic to $\mathbb{P}^1$, cf. Beauville's exposition of the Yau–Zaslow theorem.
MR1682284
Beauville, Arnaud
Counting rational curves on K3 surfaces.
Duke Math. J. 97 (1999), no. 1, 99–108.
<https://arxiv.org/abs/alg-geom/9701019>
In fact, by a theorem of Beauville–Voisin, the subgroup of $\operatorname{CH}\_0(\overline{B})$ generated by all pushforwards of all $0$-cycles on (possibly singular) rational curves in $\overline{B}$ is a cyclic subgroup commensurate with the cyclic subgroup generated by the cycle class $c\_2(T\_{\overline{B}})$ and containing all cup products of first Chern classes of divisors.
MR2047674
Beauville, Arnaud; Voisin, Claire
On the Chow ring of a K3 surface.
J. Algebraic Geom. 13 (2004), no. 3, 417–426.
<https://arxiv.org/abs/math/0111146>
Now assume that $\overline{B}$ is a very general K3 surface, so that the Picard group is generated by an ample $\mathcal{O}\_{\overline{\mathcal{B}}}$-module $\mathcal{L}$. For definiteness, assume that the degree of $c\_1(\mathcal{L})\cap c\_1(\mathcal{L})$ equals $2$.
Let $D$ be an irreducible, nodal curve in the linear system of $\mathcal{L}$ such that the normalization of $D$ is a rational curve, i.e., an irreducible, nodal curve of arithmetic genus $2$ and geometric genus $0$. For the affine surface $B=\overline{B}\setminus D$, there exists a closed point $p\in B$ such that the cycle class of $\{p\}$ in $\operatorname{CH}\_0(\overline{B})$ is not contained in the cyclic subgroup coming from pushforwards of $0$-cycles from $D$. In particular, the cycle class of $\{p\}$ in $\operatorname{CH}\_0(B)$ does not equal a cup product of two divisor classes.
Apply Serre's construction to the ideal sheaf $\mathcal{I}\_p\subset \mathcal{O}\_{\overline{B}}$ with determinant invertible sheaf $\mathcal{L}=\mathcal{O}\_{\overline{B}}(D)$, i.e., consider the local-to-global spectral sequence computing $\operatorname{Ext}^\*\_{\mathcal{O}\_{\overline{B}}}(\mathcal{I}\_p,\mathcal{L})$. The long exact sequence of low degree terms is $$ 0 \to H^1(\overline{B},\mathcal{L}) \to \operatorname{Ext}^1\_{\mathcal{O}\_{\overline{B}}}(\mathcal{I}\_p,\mathcal{L}) \to L\rvert\_p \to H^2(\overline{B},\mathcal{L}) .$$
Here $L\rvert\_p$ is the $1$-dimensional vector space that is, essentially, the fiber at $p$ of $\mathcal{L}$. Since the dualizing sheaf on $\overline{B}$ is isomorphic to the structure sheaf, Kodaira vanishing gives that $H^q(\overline{B},\mathcal{L})$ is the zero vector space for $q=1,2$. Thus, there is a unique isomorphism class of a nontrivial extension $\mathcal{E}$ of $\mathcal{I}\_p$ by $\mathcal{L}$.
The sheaf $\mathcal{E}$ is a locally free $\mathcal{O}\_{\overline{B}}$-module of rank $2$ whose first Chern class equals $c\_1(\mathcal{L})$ and whose second Chern class equals the cycle class of $\{p\}$. In particular, the restriction $\mathcal{E}\rvert\_B$ is a locally free $\mathcal{O}\_B$-module of rank $2$ whose first Chern class is zero and whose second Chern class is the nonzero class of $\{p\}$.
Consider the group scheme $G$ over $\overline{B}$ parameterizing $\mathcal{O}\_B$-automorphisms of $\mathcal{E}$ such that the induced automorphism of $\det(\mathcal{E}) \cong \mathcal{L}$ is the identity automorphism. This is an inner form of $\mathbf{SL}\_2$ over $\overline{B}$. Since $\mathcal{E}$ is locally free, this inner form is Zariski locally isomorphic to $\mathbf{SL}\_2$, and thus has split maximal tori Zariski locally. However, the restriction over $B$ does not have a maximal torus. If it did, the corresponding representation $\mathcal{E}\rvert\_B$ would split as a direct sum of locally free sheaves of rank $1$. Then, by the Whitney sum formula, the second Chern class of $\mathcal{E}\rvert\_B$ would be in the image of the cyclic subgroup of Beauville–Voisin, i.e., the second Chern class would be a torsion class in $\operatorname{CH}\_0(B)$, contrary to the choice of $p$.
Now consider the projective space bundle over $\overline{B}$ that parameterizes invertible quotients of the pullback of $\mathcal{E}$. This is a projective scheme. By Bertini theorems, an intersection of this projective scheme with a sufficiently general hyperplane is a smooth closed subscheme $\overline{A}$ whose projection to $\overline{B}$ is finite surjective, and thus finite flat. In particular, the restriction $A$ over $B$ is a finite, flat cover of $B$. The pullback of $\mathcal{E}\rvert\_B$ to $A$ has an invertible quotient. Since $A$ is affine, thus surjection automatically splits, so that the pullback of $\mathcal{E}\rvert\_B$ to $A$ is isomorphic to a direct sum of two invertible sheaves. Thus, the pullback of $G$ to $A$ does have a maximal torus.
| 6 | https://mathoverflow.net/users/13265 | 418827 | 170,521 |
https://mathoverflow.net/questions/418809 | 8 | For $n\in \mathbb{N}$ with prime decomposition $n=p\_1^{r\_1}\cdots p\_k^{r\_k},p\_i\neq p\_j$, let $A=\{p\_1,\cdots,p\_k\}$; then the following holds:
\begin{equation}
|\{q\in \mathbb{N},q<Q: \text{all prime factors of } q\,\in A\}|<Ce^{c\frac{\log Q}{\log \log Q}}d(n,Q)
\end{equation}
where $d(n,Q):=|\{m\in \mathbb{N},m<Q: m\mid n\}|$.
This lemma appears below (3.46) in Bourgain's article "Fourier transform restriction phenomena for certain lattice subsets and applications to nonlinear evolution equations".
>
> I'm not familiar with number theory; could you please explain how to
> deduce this conclusion and give some references on number theory?
>
>
>
| https://mathoverflow.net/users/172051 | Estimates about prime numbers: a lemma in Bourgain's article | Let $q<Q$ be such that all its prime divisors are in $A$. We can write $q=q\_0\times r\_1^{n\_1}\cdots r\_s^{n\_s}$, where all prime factors of $q\_0$ are $<\log Q$ and $\log Q\leq r\_1<\cdots<r\_s\in A$. Clearly, $q$ is uniquely determined once we have fixed a) $q\_0$, b) the set $\{r\_1,\ldots,r\_s\}$, and c) the integer vector $(n\_1,\ldots,n\_s)$.
The number of possible $q\_0$ is at most the number of $q<Q$ which have all prime factors $<\log Q$. This is a well-studied quantity in analytic number theory (counting 'smooth' or 'friable' numbers), and is $\leq \exp(O(\log Q/\log\log Q))$, which can be shown via elementary methods, see any textbook on analytic number theory.
There are clearly at most $d(n,Q)$ choices for the set $\{r\_1,\ldots,r\_k\}$.
Finally, to estimate the number of choices for (c), note that
$$\sum n\_i\log r\_i < \log Q,$$
and $\log r\_i\gg \log \log Q$ by construction, hence it suffices to bound the number of positive integers $n\_1,\ldots,n\_s\geq 1$ (where $s$ may vary) such that $\sum n\_i \ll \log Q/\log\log Q$. This is at most $\exp(O(\log Q/\log\log Q))$, using the elementary fact (simple combinatorics, 'stars and bars') that the the number of sequences of positive integers $n\_i$ with $\sum n\_i \leq M$ is exactly $2^M$.
Thus combining these three estimates, the number of possible $q$ is at most
$$ (\exp(O(\log Q/\log\log Q)))^2\times d(n,Q) \ll \exp(O(\log Q/\log\log Q))d(n,Q).$$
| 10 | https://mathoverflow.net/users/385 | 418839 | 170,524 |
https://mathoverflow.net/questions/418837 | 1 | Let $S$ be a K3 surface and $h:=c\_1(i^\*\mathcal{O}\_{\mathbb{P^3}}(1))$, then we can compute that $c\_1(S)=0,c\_2(S)=6h^2$. Hence
\begin{align}
\sqrt{\text{td}(S)}=1+\frac{c\_2(S)}{24}=1+\frac{1}{4}h^2
\end{align}
According to materials e.g. *Lectures on K3 Surfaces*, the Mukai vector of a vector bundle $E$ on $S$ is defined by
$$\nu(E):=\text{ch}(E).\sqrt{\text{td}(S)}=(\text{rk}(E),c\_1(E),\text{ch}\_2(E)+\text{rk}(E))$$
I am a bit confused about the operation $.$ between $\text{ch}(E)$ and $\sqrt{\text{td}(S)}$. By the appearance, I guess that it was given by
$$v\_k=\sum\_{i+j=k}\text{ch}\_i(E)\bullet(\sqrt{\text{td}(S)})\_j$$
by some operation $\bullet$. But then it turns out that $\text{rk}(E)\bullet\frac{1}{4}h^2=\text{rk}(E)$. Here I get $h=4$ and $c\_2(S)=24$.
>
> But $h\in H^2(S,\mathbb{Z})$ is a cohomology class, how should I understand $h=4$?
>
>
>
| https://mathoverflow.net/users/nan | Understand the Mukai vector | $c\_2(S)$ should not depend on the choice of polarization. See Corollary 3.3 of Huybrecht's Lectures on K3 surfaces.
$c\_2(S) = 24 c\_s$ where $c\_s$ is the generator of $H^2(S, \mathbb{Z})$. Then note that $\sqrt{td(S)} = (1+2c\_S)^{\frac{1}{2}} = 1 + c\_S$.
| 2 | https://mathoverflow.net/users/164620 | 418840 | 170,525 |
https://mathoverflow.net/questions/418705 | 4 | Let $X$ be a complex variety containing some point $x$. Then $X$ is naturally a complex-analytic space, and we have an inclusion of rings $\mathbb{C}[X]\_x\hookrightarrow\mathbb{C}\{X\}\_x\hookrightarrow\widehat{\mathbb{C}[X]\_x}$; that is, the ring of holomorphic germs is sandwiched between the ring of polynomial germs and the ring of formal power series at $x$. A number of important algebraic properties of formal power series can also be shown to hold for holomorphic germs; one useful method of proof is to do a construction in $\widehat{\mathbb{C}[X]\_x}$, then show that it doesn't leave $\mathbb{C}\{X\}\_x$. The most fundamental of these results is probably the Weierstrass Preparation Theorem, which can be proven for holomorphic germs in this way (see Zariski-Samuel vol II, Ch 7.1). Another important such result is Hensel's lemma, which can again be proven algebraically as well as by using the implicit function theorem. (So the ring of holomorphic germs is actually sandwiched between the Henselization of the regular germs and the formal power series.)
I've been trying to find an appropriate way to describe this relationship, and I was wondering if it can be done with model theory. Of the four rings (polynomials, algebraic power series, holomorphic germs, power series) describing the local behavior of the variety, what is the relationship between their first-order theories (in the language of $\mathbb{C}$-algebras)? Are any of the inclusions elementary—and, if so, how should this be interpreted geometrically? Which of these substructures are definable? Is model theory even the right way to approach this question?
| https://mathoverflow.net/users/158123 | Comparing the first-order theories of different kinds of local rings of a complex variety | It turns out that all of these except the ring of polynomial germs are elementary equivalent. The following answer is from Elliot Kaplan (posted with permission):
>
> I think it's hard to say in general what an elementary extension of commutative rings looks like, but a necessary condition is certainly that the smaller ring R is existentially closed in S (that is, any system of polynomial equations and inequations over R which has a solution in S also has a solution in R itself). Hans Schoutens has done a lot of work on the intersection of model theory and commutative algebra, and you may be able to find better answers in some of his papers: http://websupport1.citytech.cuny.edu/Faculty/hschoutens/index.html
>
>
> Here's a specific instance where we know an extension to be elementary: suppose that R and S are both henselian valuation rings with S a local ring extension of R. Suppose also that R and S have the same residue field, that this residue field has characteristic zero, and that the fraction fields of R and S have the same value group (that is Frac(R)/U(R) = Frac(S)/U(S), where U(R) is the multiplicative group of units of R). Then the Frac(R) is an elementary extension of Frac(S) as valued fields, and so R is an elementary extension of S as commutative rings (this is an instance of the Ax-Kochen-Ershov theorem for henselian valued fields of residue characteristic zero)
>
>
> This gives a positive answer when your ring of holomorphic germs is isomorphic to the ring of convergent power series over C in one variable, since this ring is henselian. In more than one variable, I don't know whether the extension is even existentially closed (I have a gut feeling that the extension is not elementary, but I don't have a good reason why not)
>
>
>
The ring of algebraic power series, convergent power series, and formal power series are all henselian valued fields and, as a corollary of Weierstrass Preparation, have the same value group. Consequently, these extensions are in fact elementary.
| 1 | https://mathoverflow.net/users/158123 | 418841 | 170,526 |
https://mathoverflow.net/questions/418806 | 5 | I recently learned about the [MIP^\*=RE result](https://arxiv.org/abs/2001.04383). I have to admit that I don't understand big parts of this paper and I am barely familiar with quantum physics. I hope my questions below make sense.
I copy the basic definitions:
IP= the class of languages with an randomized interactive proof system (that runs in polynomial time)
MIP= the class of languages with an multiprover interactive proof system
MIP^\* = the class of languages with a classical polynomial-time verifier interacting with multiple quantum provers sharing entanglement.
R.E= the class of recursively enumerable languages
It was known previously that MIP=NEXP (non-deterministic exponential time) and that MIP$\subsetneq$ MIP$^\*$, but it was expected that MIP$^\*$ wouldn't be "very far" from MIP, definitely within the computable.
My questions:
1. Since the 1930's there have been some very clear bounds on what is computable (e.g. by a Turing machine) and what is not. Later developments, e.g. multi-tape Turing machines, non-deterministic Turing machines, Turing machines with randomization etc. can be seen as efforts to "speed-up the computations" but without changing what is computable. A function computable by a non-deterministic Turing machine is also computable by a deterministic Turing machine, although the former may run in polynomial time while the latter in exponential time. I always thought that quantum computations will bring in some sort of "exponential speed-up" but again without changing the underlying notion of computable, i.e. quantum computable = computable by a classical Turing machine.
>
> In view of this new result is there a reason to believe otherwise? In particular, is it possible that quantum computable languages = r.e. languages?
>
>
>
2. Although proof assistants are in infancy right now, it is expected that in the future will play a vital role in the development of Mathematics, much like the role computers play in chess today.
>
> Are there any consequences of the above result for the development of proof assistants?
> At least in theory, is there any hope that problems which are intractable by classical computers will be tractable by quantum computers?
>
>
>
This is more like a dream, but can quantum computations give proof assistants the boost they need to come into play in everyday Mathematics?
| https://mathoverflow.net/users/13694 | MIP^*=RE and quantum computation | **Q1:** *"Do you have a reference for `quantum computable = computable by a classical Turing machine' ?*
**A1:** This follows immediately from the fact that any quantum computation can be simulated by a classical computer with exponential overhead.
**Q2:** Is there any result (or conjecture) that quantum machines can be (exponentially) faster than classical machines?
**A2:** The conjecture is that the complexity class BQP is a strict superset of BPP.
BQP is the class of decision problems solvable by a quantum computer in polynomial time, while BPP is the class of decision problems solvable by a probabilistic Turing machine in polynomial time. Both BPP and BQP are defined in terms of randomized algorithms, which
are allowed to make random choices and need to reach the desired result with probability greater than 2/3.
If BQP is strictly larger than BPP, a quantum computer can provide an exponential speed-up (from non-polynomial to polynomial). This would imply that the "strong" (or "extended") Church-Turing thesis is false: a probabilistic Turing machine (which models BPP) cannot *efficiently* (= in polynomial time) simulate a quantum computer (which models BQP). The simulation is certainly possible if we allow for an exponentially long time, so the basic Church-Turing thesis still holds.
| 5 | https://mathoverflow.net/users/11260 | 418854 | 170,531 |
https://mathoverflow.net/questions/418857 | 3 | The problem is to embed Cayley graph of free group with $n\geq2$ generators (the same as Bethe lattice with coordination number $2n$) into any model of $\mathbb{H}^2$ (we have no model preference, the only condition is to preserve the metric structure of the graph). Any numerical algorithms like MDS are not suitable. Unfortunately, I can't find any explicit formulas. But my guess is that insofar as embedding is unique, formula must exist. I would be glad if someone help with useful papers or provide any useful reasoning.
**UPD**
Two additions: 1) embedding must be isomorphic, 2) I meant the embedding into $\mathbb{H}^n$ – I guess that for $n>2$ generators Cayley graph cannot be embedded into hyperbolic plane.
| https://mathoverflow.net/users/479332 | Explicit formula for embedding Cayley graph of free group into hyperbolic space | The subgroup $\Gamma < \mathrm{SL}(2, \mathbb{R})$ generated by the matrices
$
a =
\begin{pmatrix}
1 & 2 \\
0 & 1
\end{pmatrix}
$
and
$
b =
\begin{pmatrix}
1 & 0 \\
2 & 1
\end{pmatrix}
$ is free of rank two. It acts on the upper half plane model of $\mathbb{H}^2$ via Mobius transformations. The orbit of $i$ gives the vertices of the Cayley graph; translates that differ by $a$, $b$, $a^{-1}$, or $b^{-1}$ are connected by a geodesic edge. It is an exercise to show that this gives the desired embedding. [Hint: build the Voronoi domain about $i$.]
---
A few remarks.
1. As YCor points out, no embedding can be isometric.
2. There are "quasi-isometric" embeddings of the Cayley graph into $\mathbb{H}^2$, but this one is not, as the generators are parabolic.
3. By taking finite index subgroups you can obtain equally nice actions of higher rank free groups on the hyperbolic plane.
| 7 | https://mathoverflow.net/users/1650 | 418860 | 170,532 |
https://mathoverflow.net/questions/418863 | 7 | It is known that the kernel of the (non-negative) Laplacian operator on a closed manifold consists of constant functions. I would like to ask if some similar phenomena happens for the modified operator:
$$ Lu=\Delta u+ fu,$$
where $f$ is a smooth function.
**More specifically:** If $f$ equals to minus an eigenvalue of $\Delta$, then $Lu=0$ has non-trivial solutions. Are these the only $f$ with non-trivial solutions? Can we conclude that $Lu=0$ has only zero (or constant solutions) if we assume $f$ non-constant? Otherwise, can you parametrize its kernel (as you parametrize constant functions by their integrals, or by their value in one point)?
Thank you very much.
| https://mathoverflow.net/users/81414 | Kernel of the Laplacian + a function | **Q:** Can we conclude that $Lu=\Delta u+ fu=0$ has only zero (or constant solutions) if we assume $f$ non-constant?
**A:** No, a counter example in one dimension is the [Mathieu equation](https://en.wikipedia.org/wiki/Mathieu_function), which has non-constant $\pi$-periodic or $2\pi$-periodic solutions $u(x)$ when $f(x)=a-2q\cos 2x$ for any given real $q$ at an infinite sequence of values of $a\_n(q)$, $n=1,2,3,\ldots$.
More generally, $-L$ is the Hamiltonian of a particle in the potential $-f$, and we can readily adjust the potential so that it has a bound state at zero energy – simply by adding a constant to the potential to shift the bound state up or down.
Do note that the answer to the question would be affirmative for any generic $f$. To have a nonzero $u$ with $Lu=0$ requires fine tuning of the function $f$, for a generic $f$ such a solution will not exist.
| 11 | https://mathoverflow.net/users/11260 | 418867 | 170,533 |
https://mathoverflow.net/questions/418829 | 8 | There is so much literature on the relation between the multiplicative structure of a finite field and elements having zero trace, that I am hoping that the following is known.
Let $q$ be a prime power, let $n$ be an odd prime number, let $\mathbb{F}\_{q^{2n}}$ be the field of cardinality $q^{2n}$ and let $\mathrm{Tr}:\mathbb{F}\_{q^{2n}}\to \mathbb{F}\_{q^2}$ be the trace map. Let $\mathcal{A}$ be the subgroup of $\mathbb{F}\_{q^{2n}}^\times$ having order $(q^n+1)/(q+1)$.
Is there an element $a\in \mathcal{A}$ and an element $y$ in the multiplicative group $\mathbb{F}\_{q^n}^\ast$ of the field $\mathbb{F}\_{q^n}$ with
$\mathrm{Tr}(y)\ne 0$ and $\mathrm{Tr}(ay)=0$?
It seems to me that $(n,q)=(3,2)$ is the only exception.
| https://mathoverflow.net/users/45242 | Zero trace elements in finite fields | I can show that exceptions occur at most for $n=3$. (Primality of $n$ is never used.)
Since $n$ is odd, $\mathbb F\_{q^{2n}} = \mathbb F\_{q^2} \otimes\_{\mathbb F\_q} \mathbb F\_{q^n}$. The trace map $\operatorname{Tr}:\mathbb F\_{q^{2n}} \to \mathbb F\_{q^2}$ is obtained by tensoring the identity map $\mathbb F\_{q^2} \to \mathbb F\_{q^2}$ with the trace map $\operatorname{tr} : \mathbb F\_{q^n} \to \mathbb F\_q$.
Thus, choosing an arbitrary basis of $\mathbb F\_{q^2}$, we can write any $a$ as a pair of elements $a\_1,a\_2 \in \mathbb F\_{q^n}$, and your condition that $y \in \mathbb F\_{q^n}$ satisfies $\operatorname{Tr}(y)\neq 0$ but $\operatorname{Tr}(ay)=0$ is equivalent to the condition that $\operatorname{tr} (y) \neq 0$ but $\operatorname{tr} (a\_1 y ) =\operatorname{tr}(a\_2 y)=0$. (We can ignore the condition that $y\neq 0$ as it is implied by the condition that $y$ has trace zero.)
Since the trace map of a product is a perfect $\mathbb F\_q$-linear pairing on $\mathbb F\_q^n$, such a $y$ exists unless $1$ is an $\mathbb F\_q$-linear combination of $a\_1$ and $a\_2$.
I will show there must exist a member of $\mathcal A$ that has this unusual property by bounding the number of members of $\mathcal A$ that do have this unusual property.
Note that every member of $\mathcal A$ is in the subgroup of order $q^n+1$, thus has norm to $\mathbb F\_{q^n}$ equal to $1$. This is a nonsingular quadratic equation in $a\_1,a\_2$. For each $\lambda\_1,\lambda\_2$ in $\mathbb F\_q$, not both zero, $\lambda\_1 a\_1 + \lambda\_2 a\_2 =1$ is a linear equation. There can be at most two solutions to a linear equation together with an nonsingular quadratic equation in two variables, since it gives a nontrivial quadratic equation in one variable.
Summing over possible choices of $\lambda\_1,\lambda\_2$, the number of members of $\mathcal A $ with this unusual property is at most $2 (q^2-1)$. So we can only have all members of $\mathcal A$ with this property if
$$ \frac{q^n+1}{q+1} \geq 2 (q^2-1)$$
i.e.
$$q^n+1 \geq 2 (q^2-1) (q+1).$$
For $n\geq 5$, the left side dominates the right side for any $q$.
| 5 | https://mathoverflow.net/users/18060 | 418868 | 170,534 |
https://mathoverflow.net/questions/418875 | 4 | I am reading a [paper](https://link.springer.com/content/pdf/10.1007/s11222-015-9598-x.pdf) about constructing a non-reversible Metropolis Hastings Markov chain from a reversible one as described at a high level in paragraph $3$ of page $1$.
>
> But I don't understand how, given a reversible Markov chain $P$ with stationary distribution $\pi$, a non-reversible Markov chain is constructed with the same stationary distribution. This is explained in Section $2.3$.
>
>
>
It says the idea is, given a non-reversible Markov chain $P$ with stationary distribution $\pi$, to consider the vorticity matrix $\Gamma$ of $P$ defined (in equation $(1)$) as:
$$\Gamma(x,y):=\pi(x)P(x,y)-\pi(y)P(y,x)$$
In Section $2.3$, it says consider a Markov chain $Q$ and a vorticity matrix $\Gamma$, then it defines the non-reversible Hastings ratio as
$$R\_{\Gamma}(x,y):=\begin{cases}
\frac{\Gamma(x,y)+\pi(y)Q(y,x)}{\pi(x)Q(x,y)}\quad\text{if}\quad \pi(x)Q(x,y)\neq 0 \\
1 \quad\text{otherwise}
\end{cases}
$$
and the acceptance probabilities as $A\_\Gamma(x,y) = \min\left(1,R\_\Gamma(x,y) \right)$
and finally the transition probabilities $P\_{\Gamma}$ of the non-reversible Metropolis Hastings in $(9)$. Let me breakdown my misunderstanding in two questions.
>
> $(1)$ Firstly, is $\Gamma$ the vorticity matrix of $Q$ as it is defined in Section $2.3$?
>
>
>
I think it is because the inequality under $(8)$ says $\Gamma(x,y)\leq \pi(x)Q(x,y)$ which would be true only if $\Gamma$ is the vorticity matrix of $Q$ because if $\Gamma(x,y):=\pi(x)P(x,y)-\pi(y)P(y,x)$, then $\Gamma(x,y)\leq \pi(x)Q(x,y)$.
>
> $(2)$ If the answer to $(1)$ is yes, then how are we constructing a non-reversible Markov chain from a reversible one? Because $\Gamma\neq 0$ iff $Q$ is non-reversible and thus we are constructing a non-reversible chain $P\_{\Gamma}$ from a non-reversible chain $Q$. Isn't that correct?
>
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| https://mathoverflow.net/users/479350 | About non-reversible Metropolis Hastings Markov chain | As defined after lemma 2.1, any skew-symmetric matrix $\Gamma$ (i.e. $\Gamma=-\Gamma^T$) that satisfies $\Gamma {\mathbb 1}=0$ is called a vorticity matrix. It need not be of the form $\pi(x)P(xy)-\pi(y)P(y,x)$.
| 4 | https://mathoverflow.net/users/7691 | 418879 | 170,540 |
https://mathoverflow.net/questions/418880 | 4 | Helmut Wielandt [discussed](https://books.google.com/books/about/Finite_Permutation_Groups.html?id=npviBQAAQBAJ) an old question (Chap. 2, Section 15, which can be dated back to Camille Jordan):
Let $g\neq 1$ be a permutation in some finite primitive permutation group $G$ of degree $n$. The minimal degree $m$ is defined to be the least number of points that $g$ permutes. It is known that if $m>3$, then $n$ is bounded by $$\frac{m^2}{4}\log\frac{m}{2}+m\left(\log\frac{m}{2}+\frac{3}{2}\right).$$
**Question:** Do we know a better upper bound today (with CFSG and O'Nan-Scott etc.)?
| https://mathoverflow.net/users/18286 | Minimal degree of primitive permutation group | You seem to be aware of the answer to your own question, since you give the reference to the paper of Guralnick and Magaard, which classifies groups of minimal degree $\leq n/2$. Therefore $n \leq 2m$ with explicit exceptions list in the G--M paper. See also the previous paper of Liebeck and Saxl, which is easier and does $m < n/3$.
Even *without* CFSG and O'Nan--Scott, we know a sharp bound today. It's a result of Babai that $m \geq c \sqrt{n}$. Babai's proof is purely combinatorial and extends to primitive coherent configurations (e.g., strongly regular graphs), and it's close to sharp for the groups $S\_m \wr S\_2 \leq S\_{m^2}$ and $S\_m \leq \mathrm{Sym} (\binom{m}{2})$. An exactly sharp bound follows from recent work of Sun and Wilmes (<https://arxiv.org/abs/1510.02195>), which implies a classification of primitive coherent configurations where the minimal degree is $\leq n^{2/3} (\log n)^{-C}$.
| 5 | https://mathoverflow.net/users/20598 | 418891 | 170,543 |
https://mathoverflow.net/questions/418892 | 1 | I want to generate all strongly connected tournament of size $n \in \{4, 11\}$.
As a strongly connected tournament has an hamiltonian path I may assume that $v\_i v\_{i+1}$ is always an arc, and $v\_n v\_1$ is an arc. Thus, I have to decide the outcome of $m:= n\frac{(n-1)}{2} -n$ games, ie I have $2^m$ tournaments to study. \n
However, it is clear some tournaments are generated multiple times.
From what I read here on similar questions it seems that finding an optimal way to do it is probably a very complicated question; but:
1. Do you have guesses on what additional assumptions I may take?
2. A completely different idea on how to do abord the question?
Bonus question: if I do not change my algorithm, can I know how many times each tournament is generated?
| https://mathoverflow.net/users/342793 | Generate all strongly connected tournament | Brendan McKays program “gentourng” can generate all tournaments up to isomorphism - I recently used it for all 11-vertex tournaments.
It is part of the nauty package:
[https://users.cecs.anu.edu.au/~bdm/nauty/](https://users.cecs.anu.edu.au/%7Ebdm/nauty/)
| 1 | https://mathoverflow.net/users/1492 | 418896 | 170,545 |
https://mathoverflow.net/questions/418898 | 3 | In [1], Huybrechts and Mauri argue that a holomorphic Lagrangian fibration $f: X \to B$ with smooth base $B$ is flat. This is an application of so called miracle flatness [2, Thm 23.1], because Lagrangian fibrations have equidimensional fibers. Then they write
>
> *Remark 1.18.* Note that the conclusion that $f$ is flat really needs the base to be smooth. In fact, by miracle flatness, $f$ is flat if and only if $B$ is smooth.
>
>
>
I see that smoothness of $B$ is needed to apply miracle flatness, but I don't see the converse. Why does flatness necessarily fail if $B$ is not smooth?
[1] Huybrechts, Mauri, [*Lagrangian fibrations*](https://link.springer.com/article/10.1007/s00032-022-00349-y), [arXiv](https://arxiv.org/abs/2108.10193), 2022
[2] Matsumura, *Commutative Ring Theory*, 1986
| https://mathoverflow.net/users/111897 | Does miracle flatness always fail for a non-regular base? | The answer lies in *Theorem 23.7* from Matsumura's *Commutative Ring Theory*:
>
> *Theorem 23.7.* Let $(A, \mathfrak m, k)$ and $(B, \mathfrak n, k')$ be local Noetherian local rings, and $A \to B$ a local homomorphism [...]. We assume that $B$ is flat over $A$. (i) If $B$ is regular then so is $A$.
>
>
>
As $X$ is always assumed to be smooth, the flatness of $f$ will imply the smoothness of the base $B$.
| 2 | https://mathoverflow.net/users/111897 | 418904 | 170,547 |
https://mathoverflow.net/questions/418866 | 3 | Quantum groups are useful for making knot/link invariants: for example, $U\_q(\mathfrak{sl}\_2$) you get the Jones polynomial. This boils down to the fact that $\mathcal C = \operatorname{rep }U\_q(\mathfrak{sl}\_2)$ is a braided monoidal category, which is **not** symmetric, hence gives us interesting knot invariants.
A deformation theoretic path to the quantum group $U\_q(\mathfrak g)$ from a Lie algebra $\mathfrak g$ is as follows. Fixing a Casimir element $\Omega\in\mathfrak g\otimes \mathfrak g$ endows the category $\mathcal C=\operatorname{rep }\mathfrak g$ with the structure of an infinitesimally braided symmetric monoidal category. We can infinitesimally deform $\mathcal C$ to an honest braided monoidal category $\mathcal C[\hbar]/(\hbar^2)$ with braiding given by $\sigma\circ(1+\hbar\Omega/2)$. The next step is usually to use a Drinfeld associator to integrate this infinitesimal deformation to a formal one $\mathcal C[[\hbar]]$ with braiding given by $\sigma\circ\exp(\hbar\Omega/2)$. After some twisting/scrootching/variable replacement we get $\mathcal C[[\hbar]]=\operatorname{ rep}U\_q(\mathfrak g)$.
>
> **Question** From the vantage point of knot invariants, is there any benefit to integrating the infinitesimal deformation $\mathcal C[\hbar]/(\hbar^2)$ to the formal one? The infinitesimal deformation is already a non-symmetric braided monoidal category.
>
>
>
| https://mathoverflow.net/users/5323 | Motivating quantum groups from knot invariants | Let $\mathcal{C}$ be the category of finite-dimensional representations of a semisimple Lie algebra $\mathfrak{g}$ and $\mathcal{C}[\![\hbar]\!]$ the ribbon category you mention (which depends on the choice of a Drinfeld associator).
Given an irreducible $\mathfrak{g}$-representation $V$, the corresponding knot invariant obtained from $\mathcal{C}[\![\hbar]\!]$ is the Kontsevich integral evaluated using the weight system coming from $\mathfrak{g}$, $\Omega$ and $V$; this is Theorem 10 in Lê--Murakami's ``The universal Vassiliev-Kontsevich invariant for framed oriented links'' (<https://arxiv.org/abs/hep-th/9401016>).
The $n$-th Taylor coefficient (with respect to the $\hbar$ expansion) of the Kontsevich integral is a finite type (Vassiliev) invariant of degree less than $n + 1$. But the first nontrivial finite type invariant of oriented knots has degree 2 and so requires working at least with $\mathcal{C}[\hbar]/\hbar^3$.
So, you could consider oriented knot invariants working modulo $\hbar^2$, but they are all independent of the knot.
| 7 | https://mathoverflow.net/users/18512 | 418906 | 170,548 |
https://mathoverflow.net/questions/418894 | 6 | I have been interested in the following paper ("On systems of linear indeterminate equations and congruences" by
Henry J. Stephen Smith,
Philosophical Transactions of the Royal Society of London, Vol. 151 (1861), pp. 293-326) [JSTOR open link](https://www.jstor.org/stable/108738?seq=1).
On page 300, the author states "if we apply Euler's method for the resolution of indeterminate equations to the system (16.)" The latter is a system of $n$ linear equations in $n+m$ variables, with no constants. I am not following what he means by Euler's method in this context. Does someone know what it means?
| https://mathoverflow.net/users/84272 | What is Euler's method in linear algebra? | Euler's method is described, for example, by James Fogo in [Linear indeterminate problems](https://www.jstor.org/stable/27957030?seq=1). This applies to systems of equations where there are more unknowns than there are equations, and a solution can be found by restricting the solutions to integers. The method was published by Euler in his book [Elements of Algebra](https://en.wikipedia.org/wiki/Elements_of_Algebra) (1770). For the historical context, see [The historical background of a famous indeterminate problem and some teaching perspectives.](https://www.researchgate.net/publication/358022392_The_Historical_Background_of_a_Famous_Indeterminate_Problem_and_Some_Teaching_Perspectives)
Here is an example described by Euler, for the case of one equation with two unknowns: express the two unknowns in terms of a
single auxiliary variable and then use the integer condition to restrict that single variable.
[page 312 of Euler's Algebra](https://books.google.nl/books?id=X8yv0sj4_1YC), describing the *Regula Caeci* ("blind man's rule") also known as the [The Rule of False Position.](https://en.wikipedia.org/wiki/Regula_falsi)
I'm not sure why this name is appropriate here; also note that the English translation from 1822 reproduced above is corrupted, for "Position, or The Rule of False" read "or The Rule of False Position"
| 9 | https://mathoverflow.net/users/11260 | 418908 | 170,549 |
https://mathoverflow.net/questions/418903 | 2 | Given an absolutely integrable function $f:\mathbb R^n \to \mathbb R$, let $R[f]$ be its Radon transform defined for every $(w,b) \in (\mathbb R^n \setminus \{0\}) \times \mathbb R$ by
$$
R[f](w,b) := \int\_H f(x)ds(x) = \int\_{\mathbb R^n}f(x)\delta(b-x^\top w)\,dx,
$$
where $\delta$ is the Dirac distribution and $ds$ is area element on the hyperplane
$$
H:=\{x \in \mathbb R^n \mid x^\top w = b\}.
$$
In particular, if $P$ is a probability distribution on $\mathbb R^n$ with density $f$ and $z \sim P$, then we can inteprete $R[f](w,b)$ as the density of the random variable $z^\top w$ evaluated at the point $b$.
Now, let $g:\mathbb R^n \to \mathbb R$ be the density of the multivariate Gaussian distribution $N(\mu,\Sigma)$ with mean $\mu \in \mathbb R^n$ and covariance matrix $\Sigma \in \mathbb R^{n \times n}$. Since $z^\top w \sim N(\mu^\top w,w^\top \Sigma w)$ for $z \sim N(\mu,\Sigma)$, it is clear by virtue of the previous remark that,
$$
R[g](w,b) = \varphi\left(\frac{b-\mu^\top w}{\|w\|\_\Sigma}\right),
\tag{1}
$$
where $\varphi$ is the density of the standard Gaussian distribution $N(0,1)$, and $\|w\|\_\Sigma := (w^\top \Sigma w)^{1/2}$.
Consider the function $h:\mathbb R^n \to \mathbb R$ defined by $h(x)=x\_1 g(x)$ for every $x=(x\_1,\ldots,x\_n) \in \mathbb R^n$.
>
> **Question.** *In the spirit of (1), what is an analytic formula for $R[h](w,b)$ ?*
>
>
>
In the special case where $d=1$ so that $\Sigma=\sigma>0$ is just a scalar, [a simple computation](https://www.wolframalpha.com/input?i=integrate%20x%20*%20exp%28-%28x-%5Cmu%29%5E2%2F%282*sigma%5E2%29%29*delta%28b-w*x%29%2Fsqrt%282*pi*sigma%5E2%29%20dx%20from%20-inf%20to%20inf) gives
$$
R[h](w,b) = \int\_{-\infty}^{+\infty} x\varphi(x)\delta(b-wx)\,dx = \frac{b}{w}\varphi(\frac{b-w\mu}{2\sigma w}) = \frac{b}{w}R[g](w,b).
$$
---
Update
------
The accepted answer shows that
$$
R[h](w,b) = (\alpha(b-\mu\_2)+\mu\_1)R[g](w,b),
$$
where $\mu\_1 := e\_1^\top \mu$, $e\_1=(1,0,\dots,0) \in \mathbb R^n$, $\mu\_2 := w^\top \mu$, $\alpha := \dfrac{e\_1^\top \Sigma w}{w^\top \Sigma w}$.
Of course, this can be generalized by replacing $e\_1$ with any unit-vector $v$.
| https://mathoverflow.net/users/78539 | Radon transform of the function $h(x_1,\ldots,x_n) = x_1 g(x_1,\ldots,x_n)$, where $g$ is the density of multivariate Gaussian $N(\mu,\Sigma)$ | $\newcommand{\si}{\sigma}\newcommand{\Si}{\Sigma}\newcommand{\ep}{\varepsilon}\newcommand{\vpi}{\varphi}\newcommand{\R}{\mathbb R}$In this "Gaussian" setting especially, it is convenient to approximate the delta function by the normal distribution $N(0,\ep^2)$ with $\ep\downarrow0$, so that
\begin{equation\*}
R[f](w,b)=\lim\_{\ep\downarrow0}R\_\ep[f](w,b), \tag{1}\label{1}
\end{equation\*}
where
\begin{equation\*}
R\_\ep[f](w,b):=\int\_{\R^n}dx\,\vpi\_\ep(w^\top x-b)f(x),
\end{equation\*}
\begin{equation\*}
\vpi\_\ep(t):=\frac1\ep
\vpi\Big(\frac t\ep\Big),
\end{equation\*}
and $\vpi$ is the standard normal density.
Now, for $h(x)\equiv x\_1 g(x)$ and $g$ the density of $N(\mu,\Si)$, we can write
\begin{equation\*}
R\_\ep[h](w,b)
=\int\_{\R^n}dx\,g(x) x\_1 \vpi\_\ep(w^\top x-b)
=E e\_1^\top X\, \vpi\_\ep(w^\top X-b),
\end{equation\*}
where $e\_1:=[1,0,\dots,0]^\top\in\R^{n\times1}=\R^n$ and $X\sim N(\mu,\Si)$ .
Note that the joint distribution of $e\_1^\top X$ and $w^\top X$ is bivariate normal with respective means
\begin{equation\*}
\mu\_1:=e\_1^\top\mu\quad\text{and}\quad\mu\_2:=w^\top\mu, \tag{2}\label{2}
\end{equation\*}
respective standard deviations
\begin{equation\*}
\si\_1:=\sqrt{e\_1^\top\Si e\_1} \quad\text{and}\quad \si\_2:=\sqrt{w^\top\Si w}, \tag{3}\label{3}
\end{equation\*}
and correlation
\begin{equation\*}
\rho:=\frac{e\_1^\top\Si w}{\si\_1\si\_2}. \tag{4}\label{4}
\end{equation\*}
So, straightforward calculations yield
\begin{equation\*}
R\_\ep[h](w,b)
=
\frac{ \rho \si \_1 \si \_2 (b-\mu \_2)+\mu \_1 (\si
\_2^2+\ep ^2)}{\sqrt{2 \pi } (\si \_2^2+\ep ^2){}^{3/2}}\,
\exp\Big\{-\frac{(b-\mu \_2){}^2}{2 (\si \_2^2+\ep ^2)}\Big\}.
\end{equation\*}
Finally, by \eqref{1},
\begin{equation\*}
R[h](w,b)
=
\frac{\rho\si\_1(b-\mu\_2)+\mu\_1\si\_2}{\sqrt{2\pi}\,\si \_2^2}\,
\exp\Big\{-\frac{(b-\mu\_2)^2}{2 \si \_2^2}\Big\},
\end{equation\*}
with $\mu\_1,\mu\_2,\si\_1,\si\_2,\rho$ given by \eqref{2}--\eqref{4}.
| 2 | https://mathoverflow.net/users/36721 | 418914 | 170,551 |
https://mathoverflow.net/questions/418905 | 2 | When comparing two sub-$\sigma$-algebras on a probability space $(\Omega,\Sigma,\pi)$, say $\mathcal{X}$ and $\mathcal{Y}$, say that $\mathcal{X}$ is strictly coarser than $\mathcal{Y}$ if the completion of $\mathcal{X}$ does not contain $\mathcal{Y}$. Here completion always refers to the restriction of $\pi$. Do there exist probability spaces $(\Omega,\Sigma,\pi)$ satisfying the following property?
* For any countably-generated sub-$\sigma$-algebra $\mathcal{X}$ strictly coarser than $\Sigma$, and containing a set of interior measure (strictly between $0$ and $1$), there exists an atomless sub-$\sigma$-algebra $\mathcal{U}\subseteq\Sigma$, independent of $\mathcal{X}$.
Furthermore, does this imply the following property?
* For every $\mathcal{X}$ as above, there exists an independent sub-$\sigma$-algebra $\mathcal{U}\subseteq\Sigma$ independent of $\mathcal{X}$ such that the completion of $\mathcal{X}\vee\mathcal{U}$ contains $\Sigma$.
I know the first can't hold in $[0,1]$ equipped with the Borel measure from [Ramachandran (1979)](https://projecteuclid.org/journals/annals-of-probability/volume-7/issue-3/Existence-of-Independent-Complements-in-Regular-Conditional-Probability-Spaces/10.1214/aop/1176995044.full), although it does hold in that case when $X$ is restricted to be generated by a countable partition of $[0,1]$.
| https://mathoverflow.net/users/479356 | Spaces with atomless independent $\sigma$-sub-algebras | The answer to the first question is yes. There is a class of probability spaces known under various names such as superatomless, saturated, nowhere countably-generated, $\aleph\_1$-atomless, and a couple of other names that have exactly this property. Note that the restriction to sub-$\sigma$-algebras admitting sets of interior measure is superfluous. Otherwise, the $\sigma$-algebra $\Sigma$ would be trivially independent of it. [This paper](https://doi.org/10.1016/j.jmateco.2007.03.003) might be a good entry to the topic.
A typical example of such a probability space would be the independent product measure on uncountably many copies of the unit interval endowed with the uniform distribution. It is the canonical example in a special sense. If you take a probability space $(\Omega,\Sigma,\pi)$ and identify two measurable sets $A$ and $B$ with $\pi(A\Delta B)=0$, you obtain the so-called *measure algebra*. There is a fundamental theorem due to Dorothy Maharam that for every probability space, the measure algebra is a countable weighted sum of product measures obtained from $[0,1].$ I think one might also be able to use this to prove that the second part of the question holds true.
| 3 | https://mathoverflow.net/users/35357 | 418916 | 170,553 |
https://mathoverflow.net/questions/255246 | 5 | In [this](https://mathoverflow.net/questions/250926/characterization-of-exact-groups-via-the-existence-of-amenable-actions-on-unital) recent MOF question I asked whether exact groups could be characterized via the existence of amenable actions on unital C\*-algebras. The answer, provided by Caleb Eckhardt in a comment, was that an amenable action on a unital C\*-algebra leads to an amenable action on the compact spectrum of the center of said algebra, and hence exactness follows from a well known result.
The present question is pretty similar to my previous question, but I am now employing a different amenability condition, as follows:
**Definition**. An action $\theta$ of a discrete group $G$ on a C\*-algebra $A$ is said to satisfy the *approximation property* if there exists a net
$\{ a\_i \}\_{i\in I}$
of finitely supported functions
$$
a\_i: G \to A,
$$
which is bounded in the sense that
$$
\sup \_{i\in I} \Big\Vert {\sum\_{g\in G } a\_i(g)^\* a\_i(g)} \Big\Vert < \infty ,
$$
and such that
$$
\lim \_{i \rightarrow \infty } \sum\_{h\in G } a\_i(gh)^\* b\,\theta\_ g( a\_i(h)) = b,\quad \forall b\in A .
$$
See the discussion before [**1**, Definition 20.11].
This condition is weaker than Anantharaman-Delaroche's usual condition of amenability (definition 4.3.1 in Brown and Ozawa) in that it does not require the $a\_i$ to take values in the center of $A$. Nevertheless it is enough for most purposes and in particular it implies that the crossed product is nuclear provided $A$ is nuclear [**1**, Proposition 25.10].
Thus, here is the new version of my question:
>
> **Question**: Suppose that a discrete group $G$ admits an action on a unital
> C\*-algebra $A$, satisfying the above approximation property. Is $G$
> necessarily exact?
>
>
>
Notice that, precisely because the $a\_i$ are allowed to take values outside the center of $A$, it is not immediately clear that the above condition passes to the center, and hence Caleb's answer might no longer work in this case.
Reference:
[1]
[Partial Dynamical Systems Fell Bundles and
Applications](https://arxiv.org/abs/1511.04565)
| https://mathoverflow.net/users/97532 | Characterization of exact groups via the existence of amenable actions on unital C*-algebras, part 2 | The answer to the question is positive: see Remark 6.6 in the paper
<https://arxiv.org/pdf/1904.06771.pdf>
The approximation property implies amenability in the sense of Claire Anantharaman Delaroche, this has been proved in
<https://arxiv.org/pdf/1907.03803.pdf>
for discrete groups, see also
<https://arxiv.org/pdf/1904.06771.pdf>
Indeed, it has been stablished recently that the approximation property is equivalent to amenability even in the more general context of locally compact groups by Ozawa-Suzuki in
<https://arxiv.org/abs/2011.03420>
and that the above question in this context also has a positive answer (see Corollary 3.6).
| 3 | https://mathoverflow.net/users/75215 | 418918 | 170,554 |
https://mathoverflow.net/questions/384312 | 5 | $\newcommand{\Cc}{\mathcal{C}}$
$\newcommand{\Dd}{\mathcal{D}}$
$\newcommand{\Z}{\mathbb{Z}}$
$\newcommand{\Q}{\mathbb{Q}}$
$\newcommand{\tensor}{\otimes}$
$\newcommand{\colim}{\rm colim}$
$\newcommand{\Sp}{Sp}$
$\newcommand{\iHom}{\underline{\rm Hom}}$
(This is the follow-up of [this question](https://mathoverflow.net/questions/383783/). I have repeated the motivation for the question.)
The following situation frequently arises in abstract category theory: we have symmetric closed monoidal categories $\Cc$ and $\Dd$ and a closed symmetric monoidal functor $f^\*: \Dd \to \Cc$ that admits both a left adjoint $f\_!: \Cc \to \Dd$ and a right adjoint $f\_\*: \Cc \to \Dd$, and we are interested in the relation between $f\_!$ and $f\_\*$. This is often called a [Wirthmüller context](https://ncatlab.org/nlab/show/Wirthm%C3%BCller+context), named after the inspiring example of the Wirthmüller isomorphism in equivariant homotopy theory. (The symbol $f^\*$ is just notation: there is not necessarily a morphism $f$ around.)
To compare $f\_!$ and $f\_\*$, Fausk-Hu-May [FHM] assume in the article *isomorphisms between left and right adjoints* the existence of an object $C \in \Cc$ satisfying $f\_!(C) \simeq f\_\*(1\_{\Cc})$, where $1\_{\Cc}$ denotes the monoidal unit of $\Cc$. From this data, a natural transformation $f\_\*(-) \to f\_!(- \tensor C)$ is constructed and conditions are given for when this is an isomorphism (the 'formal Wirthmüller isomorphism').
As remarked by Balmer-Dell'Ambrogio-Sanders [BDS] in the article *Grothendieck-Neeman duality and the Wirthmüller isomorphism*, the object $C$ is *a priori* not unique and it doesn't come with a `characterizing description'. My main question is:
**Question (1): Do we have counterexamples for essentially uniqueness of an object $C$ satisfying $f\_!(C) = f\_\*(1\_{\mathcal{C}})$?**
I actually want to follow [BDS] and stay in a more restricted setting. We assume that $\Cc$ and $\Dd$ are tensor-triangulated (or, if you prefer, stably symmetric monoidal stable $\infty$-categories) and that they are generated by a set of compact objects and admit internal homs (or equivalently: the tensor products commute with coproducts in both variables). Moreover, we assume that $\Cc$ and $\Dd$ are *rigid*, i.e. the compact objects are the same as the strongly dualizable objects. Finally, we assume that $f^\*: \Dd \to \Cc$ is exact and strong monoidal and that it admits both a left adjoint $f\_!: \Cc \to \Dd$ and a right adjoint $f\_\*: \Cc \to \Dd$.
Examples of rigid tensor triangulated categories are spectra $\operatorname{Sp}$, genuine $G$-spectra $\operatorname{Sp}^G$ for a compact Lie group $G$ and the derived category $D(R)$ for a ring $R$. Several more advanced examples of such categories and functors are in examples 4.6 - 4.8 in the article *Grothendieck-Neeman duality and the Wirthmüller isomorphism* by Balmer-Dell'Ambrogio-Sanders, based on 'finite group schemes', 'motivic homotopy theory' and 'cohomology rings of classifying spaces'. However, I don't really understand them and in particular I don't know if they give counterexamples to question (1) above. Hence:
**Question (2): What are simple examples of functors $f^\*: \Dd \to \Cc$ satisfying the above conditions?**
**Attempts:**
I expect that it shouldn't be hard to come up with counterexamples for (1), but somehow I cannot make them work. For example:
* Consider a finite group $G$ and let $f: H \hookrightarrow G$ be a subgroup inclusion. Then the restriction functor $f^\*: \Sp^{G} \to \Sp^H$ between (genuine) equivariant spectra is symmetric monoidal with left adjoint $f\_!(X) = G \ltimes\_H X$ and right adjoint $f\_\*(X) = F(G/H\_+,X)$. In this case we actually have $f\_! \simeq f\_\*$ by the Wirthmüller isomorphism. And I think that if $C \in \Sp^H$ satisfies $f\_!(C) \simeq f\_\*(\mathbb{S}\_H) \simeq f\_!(\mathbb{S}\_H)$, then we already have $C \simeq \mathbb{S}\_H$, so this won't give a counterexample.
* Consider a finite group $G$ and a normal subgroup $N < G$, and let $f: G \to G/N$ be the projection. Then $f^\*: \Sp^{G/N} \to \Sp^G$ between (genuine) equivariant spectra is symmetric monoidal with left adjoint $f\_!(X) = X/N$ and right adjoint $f\_\*(X) = X^N$. Now an object $C$ with $f\_!(C) \simeq f\_!(\mathbb{S}\_G)$ is no longer unique since $f\_!(\Sigma^{\infty}\_+G) \simeq f\_!(\Sigma^{\infty}\_+ G/N)$. But we don't have the Wirthmüller isomorphism $f\_! \simeq f\_\*$ so it doesn't immediately produce a counterexample.
I also tried a little outside the range of equivariant homotopy theory, but there I fail to even come up with functors satisfying the required criteria:
* Let $f: \Z \hookrightarrow \Q$ and let $f^\*: \Dd = D(\Q) \to \Cc = D(\Z)$ be the forgetful functor. This is symmetric monoidal (since $\Q$ is a (derived) localization of $\Z$). However, it does not preserve compact objects: $\Q$ is not finitely generated over $\Z$. **Edit:** as remarked in the comments, it is actually only lax symmetric monoidal, since the monoidal unit is not sent to the monoidal unit.
* Let $f^\*: \Dd = D(\Z/2\Z) \to \Cc = D(\Z)$ be the forgetful functor. This preserves compact objects, but it is not symmetric monoidal: we have $\Z/2Z \tensor^L\_{\Z/2\Z} \Z/2\Z \simeq \Z/2\Z$, but $\Z/2\Z \tensor^L\_{\Z} \Z/2\Z$ has higher Tor-groups so cannot be $\Z/2$.
* If $\Cc$ is cartesian closed and pointed, and $\Dd = 1$ is the trivial category, then the functor $f^\*: 1 \to \Cc: \* \mapsto \*$ is symmetric monoidal, has both adjoints $f\_!$ and $f\_\*$, and any object $C$ satisfies $f\_!(C) \simeq f\_\*(1\_{\Cc})$. But $\Cc$ cannot be tensor-triangulated (as $- \times -$ doesn't preserve coproducts in either variable.)
| https://mathoverflow.net/users/144100 | Non-uniqueness of $C$ with $f_!(C) = f_*(1_{\mathcal{C}})$ | Consider the projective line $\mathbb{P}^1\_k$ over a field $k$. The structure morphism $f:\mathbb{P}^1\_k\rightarrow \mathrm{Spec}(k)$ induces a functor $f^\*:D(k) \rightarrow D\_{qc}(\mathbb{P}^1\_k)$ which is fully faithful (equivalently, $f\_\*(1) \simeq 1$) and which satisfies Grothendieck--Neeman duality. Since $f\_\*(1)\simeq 1$ is (trivially) a rigid separable commutative algebra, there is a canonical map $\theta:1 \to \omega\_f$ such that $f\_\*(\theta)$ an isomorphism (by Lemma 4.5 of my paper "[A characterization of finite étale morphisms in tensor triangular geometry](https://arxiv.org/abs/2106.14066)"). Thus, $f\_\*(1) \simeq f\_\*(\omega\_f) \simeq f\_!(1)$ where the last isomorphism is the Wirthmüller isomorphism provided by [BDS]. However, $\omega\_f = \Sigma \mathcal O\_{\mathbb{P}^1\_k}(-2)$ is not isomorphic to $1=\mathcal O\_{\mathbb{P}^1\_k}$.
Thus $C=\omega\_f^{-1}$ and $C=1$ both satisfy $f\_!(C)\simeq f\_\*(1)$ showing that $C$ need not be unique.
| 4 | https://mathoverflow.net/users/1148 | 418934 | 170,559 |
https://mathoverflow.net/questions/418658 | 9 | Consider the sequence defined by
\begin{align}
c\_0 &{}= 1 \\
c\_n &{}= 2\,n\,c\_{n-1}-\frac{1}{2}\sum\_{m=1}^{n-1}c\_m\,c\_{n-m}.
\end{align}
How can you prove that it has the following asymptotics (strongly suggested by numerics)
$$
c\_n\sim \frac{2}{\pi}\Gamma(n)\,2^n\ \ \ ?
$$
To make the question more motivated: the associated divergent series is the asymptotic expansion of a simple ratio of modified Bessel functions
$$
R(z) = \frac{K\_0(-\frac{1}{4z})}{K\_1(-\frac{1}{4z})}\sim \sum\_{n\ge 1}c\_n z^n = 1+2z+6z^2+24z^3+\dotsb.
$$
Indeed, as pointed out in the [comment](https://mathoverflow.net/questions/418658/asymptotics-of-a-quadratic-recursion#comment1075111_418658) by Richard Stanley, one has
$$
4z^2 R'+4z R+1-R^2 = 0.
$$
So the question boils down to what can be said about the large order behaviour of the terms of the asymptotic expansion of a particular known function. If $R$ were analytic one could have used Darboux theorems to relate the answer to the type of nearest singularity. For a divergent non-Borel summable function I don't know whether there are general results. By the way, having the Borel transform in closed form could help because then one could deform the integration contour off the positive real axis to get information, in the spirit of Nevanlinna theorems.
| https://mathoverflow.net/users/174308 | Asymptotics of a quadratic recursion | **TL;DR:** I have a proof of your conjectured asymptotic formula, modulo the correctness of a certain alternative description of your $c\_n$ sequence.
---
I tried to complete Iosif Pinelis's elegant analysis by finding a way to derive the value of the constant $2/\pi$ (denoted $a$ in Iosif's answer) from the quadratic recurrence relation. The problem with that relation is that the behavior it implies for $c\_n$ for large $n$ seems to depend in a very sensitive way on the initial values of the sequence, so I concluded that this approach has little chance of working.
Fortunately, I've now discovered another, *linear* recurrence relation for the same sequence $c\_n$ that has better behavior and gives your claimed asymptotics without much effort. The relation is:
$$
c\_0=1, \qquad c\_n = g\_n + \sum\_{k=1}^n h\_k c\_{n-k} \quad (n\ge 1), \qquad (\*)
$$
where I define
\begin{align}
g\_n &= \frac{((2n)!)^2}{2^{3n}(n!)^3},
\\
h\_n &= \frac{((2n)!)^2}{2^{3n}(n!)^3}\cdot \frac{2n-1}{2n+1}.
\end{align}
(**Edit:** this corrects a small typo from the earlier version. As you pointed out in a comment, $g\_n$ can also be expressed as $\frac{2^n \Gamma(n+1/2)^2}{\pi \Gamma(n)}$.)
I haven't verified rigorously that this relation is equivalent to your quadratic relation, but numerically it gives the correct sequence 1, 2, 6, 24, 126, 864, 7596, ..., and I believe this should be straightforward to prove. The reasoning that led me to it involves your description of the sequence as coming from an asymptotic expansion for a ratio of two Bessel functions. I started with the relation
$$
K\_0(-1/4z) = K\_1(-1/4z) \times \sum\_{n} c\_n z^n,
$$
and, expanding both Bessel functions in a power series (actually not quite a traditional power series because of some nasty-looking transcendental terms, but those can be factored out), massaged this into a linear system of equations satisfied by the $c\_n$'s, which gave me the recurrence after a bit of additional guesswork.
To rigorously prove the relation, one can work with this Bessel function picture and do the analysis more carefully, or one can try to prove directly that the linear recurrence is equivalent to the quadratic recurrence without any reference to Bessel functions. I suspect this is doable through an inductive argument, probably involving formulating and proving some auxiliary hypergeometric summation identities.
Finally, if we assume that $(\*)$ is correct, we can prove your claim that
$$
c\_n \sim \frac{2}{\pi} 2^n (n-1)!.
$$
Observe that
$$
\frac{c\_n}{2^n (n-1)!} = \frac{g\_n}{2^n (n-1)!} + \frac{h\_n}{2^n (n-1)!}
+ \frac{1}{2^n (n-1)!} \sum\_{k=1}^{n-1} h\_k c\_{n-k}
$$
Using Stirling's formula, you can check that each of the first two terms in this expression converges to $1/\pi$. The third term (the normalized sum) can be easily shown to be $O(1/n)$ (the first summand is $O(1/n)$, and the remaining summands are $O(1/n^2)$ and there are $n-2$ of them). Here I am using the fact that the sequence $c\_n/2^n (n-1)!$ is bounded, as shown in Iosif Pinelis's answer (and as can probably also be shown from the linear recurrence without much effort).
| 7 | https://mathoverflow.net/users/78525 | 418935 | 170,560 |
https://mathoverflow.net/questions/418835 | 3 | Let $(X\_n)\_{n \geq 0}$ be an i.i.d. sequence of $\{0,1\}$-valued random variables $X\_n \sim \mathrm{Bernoulli}(\frac{1}{2})$, i.e. a sequence of independent tosses of a fair coin.
>
> Does there exist a (non-random) Borel-measurable function $h \colon \{0,1\}^{\mathbb{Z}\_{\geq 0}} \to \mathbb{Z}\_{\geq 0}$ such that
> $$ \mathbb{P}(\exists\,\text{infinitely many } k \geq 0 \text{ s.t. }h((X\_{k+n})\_{n \geq 0})=k ) = 1 \, \text{?} $$
>
>
>
| https://mathoverflow.net/users/15570 | Does a sequence of coin-tosses a.s. have a subsequence on which the remainder of the sequence can be identified with the position in the sequence? | Such a function $h$ does not exist.
Indeed, given a Borel-measurable function $h \colon \{0,1\}^{\mathbb{Z}\_{\geq 0}} \to \mathbb{Z}\_{\geq 0},$ define for each $k \ge 0$ the events
$$A\_k=\Bigl\{h \Bigl((X\_{n})\_{n \geq 0}\Bigr)=k \Bigr\}$$
and
$$B\_k=\Bigl\{h \Bigl((X\_{k+n})\_{n \geq 0}\Bigr)=k \Bigr\} \,.$$
Clearly,
$$\mathbb{P}(A\_k)=\mathbb{P}(B\_k) \;\; \mbox{for every} \; k \ge 0.
$$
The $A\_k$ are disjoint events that partition the probabiility space, so
$$ 1=\sum\_{k \ge 0} \mathbb{P}(A\_k)=\sum\_{k \ge 0}\mathbb{P}(B\_k) \,.
$$
Therefore, by the (first) Borel-Cantelli lemma,
$$ \mathbb{P}(\exists\,\text{infinitely many } k \geq 0 \text{ s.t. }h((X\_{k+n})\_{n \geq 0})=k ) = 0 \,.
$$
<https://en.wikipedia.org/wiki/Borel%E2%80%93Cantelli_lemma>
| 2 | https://mathoverflow.net/users/7691 | 418938 | 170,561 |
https://mathoverflow.net/questions/418932 | 12 | My question is about Simpson's motivicity conjecture, that is the conjecture that for any (cohomogically) rigid irreducible connection $(M,\nabla)$ on a smooth complex scheme $X$ is of geometric origin in the sense that there exists $Y\overset{f}{\to}X$ such that $(M,\nabla)$ is a subquotient of $R^nf\_\*\mathcal{O}\_Y$ with the Gauss-Manin connection.
My main question can be put over the top as "why should we believe it?" I'm aware that predictions of this conjecture have been proved, for instance by H.Ésnault and her colaborators. I'd be more interested in a "plausability" sort of criterion. For instance, there are some deformation theoretic arguments explaining why the Fontaine-Mazur conjecture should hold (I'm choosing F-M because it is of a very similar "shape"). Considering this came from Simpson's work on non-abelian Hodge theory, presumably there is a non-abelian Hodge theoretic reason to expect the validity of this conjecture...
Also, how "sharp" do we expect this conjecture to be? Do we know that subquotients of Gauss-Manin connections are rigid?
| https://mathoverflow.net/users/152554 | Simpson's motivicity conjecture | At the time Simpson formulated his conjecture, he had proved that rigid local systems correspond to rational (in a suitable sense) variations of Hodge structure. And, as you point out, we now know that they are integral (Esnault-Groechenig). So rigid local systems have many of the earmarks of geometric local systems.
As for your other questions, geometric local systems, or their subquotients (which would be rationally summands), need not be rigid.
| 11 | https://mathoverflow.net/users/4144 | 418940 | 170,562 |
https://mathoverflow.net/questions/418941 | 1 | The paper spatial interaction and the statistical analysis of lattice systems by Besag (1974) presents an alternative proof for the Hammersley-Clifford theorem.
In order to prove the HM theorem, Besag says that that under the following assumptions
1. The number of possible values at each site/vertex is finite.
2. $0$ is a possible value at every site.
the following expansion holds for every cdf $\mathbb{P}(\mathbf{x})$ and is unique on the sample space $\Omega$,
$$
\begin{align}
Q(\textbf{x}) &= \mathbb{P}(\mathbf{x})/\mathbb{P}(\mathbf{0})
\\
&= \sum\_{1 \leq i \leq n} x\_iG\_i(x\_i)
+\sum\_{1 \leq i \leq j \leq n}x\_ix\_jG\_{i,j}(x\_i,x\_j)\\ &+ \sum\_{1 \leq i \leq j \leq k \leq n} x\_ix\_jx\_kG\_{i,j,k}(x\_i,x\_j,x\_k)+\ldots + x\_1x\_2 \ldots x\_nG\_{1,2,\ldots,n}(x\_1,x\_2,\ldots,x\_n)
\end{align}
$$
This equality looks slightly related to the inclusion-exclusion principle and in fact Besag says that we can use it to prove a HM for some restricted classes of lattices. How did Besag arrived at this equality? I'm really clueless.
My question is related to this [question](https://mathoverflow.net/questions/406500/canonical-representation-of-the-a-probability-distribution-for-hammersley-cliffo) that remains unanswered.
| https://mathoverflow.net/users/137267 | Hammersley-Clifford theorem | $\newcommand{\x}{\mathbf x}\newcommand{\tx}{\tilde{\mathbf x}}\newcommand{\0}{\mathbf0}\newcommand{\R}{\mathbb R}$You are citing Besag's paper ([formulas (3.1) and (3.3) there](http://links.jstor.org/sici?sici=0035-9246%281974%2936%3A2%3C192%3ASIATSA%3E2.0.CO%3B2-3)) incorrectly. What the paper actually has is this:
\begin{equation\*}
\begin{aligned}
Q(\x)&:=\ln\frac{P(\x)}{P(\0)}
\\
&= \sum\_{1\le i\le n} x\_iG\_i(x\_i)
+\sum\_{1\le i<j\le n}x\_ix\_jG\_{i,j}(x\_i,x\_j)\\
&+ \sum\_{1\le i<j<k\le n} x\_ix\_jx\_kG\_{i,j,k}(x\_i,x\_j,x\_k) +\cdots \\
&+x\_1\cdots x\_nG\_{1,\dots,n}(x\_1,\ldots,x\_n).
\end{aligned}
\tag{1}\label{1}
\end{equation\*}
This is indeed a version of the inclusion-exclusion formula. To show this, identify the vector $\x=(x\_1,\dots,x\_n)\in\R^n$ with the function $[n]\ni i\mapsto x\_i$, which will also be denoted by $\x$; here, as usual, $[n]:=\{1,\dots,n\}$. For each $J\subseteq[n]$, let $\tx\_J$ be the vector/function on $[n]$ such that $(\tx\_J)(i):=x\_i$ for $i\in J$ and $(\tx\_J)(i):=0$ for $i\notin J$. Let $\x\_J:=\x|\_J=(x\_i\colon i\in J)$, the restriction of the function $\x$ to $J$, so that $\tx\_J$ is the extension by zeroes of the the function $\x\_J$ from $J$ to $[n]$. Let $|J|$ denote the cardinality of $J$.
For any $K\subseteq[n]$, let
\begin{equation\*}
H\_K(\x\_K):=\sum\_{J\subseteq K}(-1)^{|K|-|J|}Q(\tx\_J);
\end{equation\*}
note that the last expression depends on $\x$ only through $\x\_K$, so that $H\_K(\x\_K)$ is well defined.
Then
\begin{equation\*}
\begin{aligned}
&\sum\_{K\subseteq[n]}H\_K(\x\_K) \\
&=\sum\_{K\subseteq[n]}\sum\_{J\subseteq K}(-1)^{|K|-|J|}Q(\tx\_J) \\
&=\sum\_{J\subseteq[n]}Q(\tx\_J)
\sum\_{K: J\subseteq K\subseteq[n]}(-1)^{|K|-|J|} \\
&=\sum\_{J\subseteq[n]}Q(\tx\_J)
\sum\_{k=|J|}^n(-1)^{k-|J|}\binom{n-|J|}{k-|J|} \\
&=\sum\_{J\subseteq[n]}Q(\tx\_J)
\sum\_{m=0}^{n-|J|}(-1)^m\binom{n-|J|}m \\
&=\sum\_{J\subseteq[n]}Q(\tx\_J)
\,1(|J|=n) \\
&=Q(\tx\_{[n]})=Q(\x).
\end{aligned}
\tag{2}\label{2}
\end{equation\*}
Note also that $H\_K(\x\_K)=0$ if $x\_j=0$ for some $j\in K$; this follows in view of the natural bijection between the **(**set of all subsets $J$ of the -- necessarily nonempty -- set $K$ such that $J\ni j$**)** and **(**the set of all subsets $J$ of $K$ such that $J\not\ni j$**)**.
So, $H\_K(\x\_K)=0$ if $p\_K(\x)=0$, where
\begin{equation\*}
p\_K(\x):=\prod\_{j\in K}x\_j.
\end{equation\*}
So, letting $G\_K(\x\_K):=H\_K(\x\_K)/p\_K(\x)$ if $p\_K(\x)\ne0$ and $G\_K(\x\_K):=0$ if $p\_K(\x)=0$, we get $H\_K(\x\_K)=p\_K(\x)G\_K(\x\_K)$ for all $\x$ and all $K$. Thus, by \eqref{2},
\begin{equation\*}
Q(\x)=\sum\_{K\subseteq[n]}p\_K(\x)G\_K(\x\_K), \tag{3}\label{3}
\end{equation\*}
which is just another, more compact way of writing \eqref{1}.
Of course, one may refer to $G\_K(\x\_K)$ as a partial divided difference (of the function $Q$) of order $|K|$ with respect to $\x\_K=(x\_i\colon i\in K)$. In particular,
\begin{equation\*}
G\_{\{1\}}(x\_1)=\frac{Q(x\_1,0,\dots,0)-Q(0,0,\dots,0)}{x\_1}
\Big(=\frac{Q(x\_1,0,\dots,0)}{x\_1}\Big)
\end{equation\*}
if $x\_1\ne0$ and
\begin{equation\*}
G\_{\{1,2\}}(x\_1,x\_2)=\frac{Q(x\_1,x\_2,0,\dots,0)-Q(x\_1,0,0,\dots,0)
-Q(0,x\_2,0,\dots,0)+Q(0,0,\dots,0)}{x\_1x\_2}
\end{equation\*}
$x\_1x\_2\ne0$.
So, \eqref{1} and, equivalently, \eqref{3} may be considered difference analogues of the Maclaurin expansion for functions of $n$ variables.
Representations \eqref{1} and, equivalently, \eqref{3} are essentially unique: the values of $G\_K(\x\_K)$ are uniquely determined for all $\x$ and $K$ such that $p\_K(\x)\ne0$. This can be verified by reasoning quite similar to \eqref{2}; this is left as an exercise for now.
| 2 | https://mathoverflow.net/users/36721 | 418945 | 170,563 |
https://mathoverflow.net/questions/418844 | 7 | Define "$\alpha$ starts a gap of order $n+1$ and length $\beta$" iff $\mathcal P^n(\omega)\cap (L\_{\alpha+\beta}\setminus L\_\alpha)=\emptyset\land\forall\gamma\in\alpha: L\_\alpha\setminus L\_\gamma\neq\emptyset$ where $\mathcal P^n$ is the powerset operation iterated $n$ times.
Define "$\alpha$ is in a gap of order $n$" as $\forall m<n:\mathcal P^m(\omega)\cap (L\_{\alpha+1}\setminus L\_\alpha)=\emptyset$
Define $\text{ZFC}^-$ as $\text{ZFC}-$(Powerset axiom). This theory is equiconsistent with $Z\_2$, second order arithmetic.
I believe gaps in the constructible universe were first talked about by Putnam in 1963, or at least that's the oldest source I've read, but that's besides the point. [According to Marek and Srebrny](https://pdf.sciencedirectassets.com/272681/1-s2.0-S0003484300X0006X/1-s2.0-0003484374900059/main.pdf?X-Amz-Security-Token=IQoJb3JpZ2luX2VjENj%2F%2F%2F%2F%2F%2F%2F%2F%2F%2FwEaCXVzLWVhc3QtMSJIMEYCIQCxo9GqSQOYZSIIuUy2DSaSS8MTj%2BKHq50tH95EUPNm0AIhAL5fp%2B2gW6Yq05kNaTP4PZoHTM3UhTT9CSA7L8COg%2BrwKvoDCGEQBBoMMDU5MDAzNTQ2ODY1Igx4J6ktYLU7RXb%2BsmYq1wOcQf%2F44N0cNoOBXXPgsLlTVcp4c78i1oFBjZSbP3wuwq6AA2VrkN21XS0EwQGSRmNXfu8kv0RrtnmfbOiAek7fYbLQVMK0rXANYj8SBoSj6LufLnuJrbxf6SvUHmmaZddRgfmK2mQaB7bU%2FWNtkRyTK6021D3LhyJpRo1DNkSRHmoHhSX5QI0s8%2Fv8SZoJS3gtrqrxQ2%2FQc82TYNR6AJS6ZuO6f9Yqs3UfhVItG4H85ZFyLlW9kP4fxwK1j8jZ%2FiTUZKI29HqUuT%2FRlvvw2mP%2FEU%2BQvhbByTJg2xWs6S%2FOrBvfygk7M8sFFGbHC%2FQ6fInUjqrL17jZbiGCUtn90syE1tgqoSc4L7Ao1GmjYiwFyBeMzUFXqkgE%2FTwoh4su%2F%2FsmnygvXoFqQZ5HbywGgtruC8ZdpklpYgmSTJybPcBB%2FklduYZmEo3Oy1QMAqCIbc3WhDgeb9%2BdUVWRwU4lT4Ih1zFCcs4WBxuvOVbMelGl5UStESBgLrI792DK%2FDemC1EyHptDhng3rZFrgz3wOCRtD8v8hxRtEzsCf2Nxi1NhybARf9Ti%2BFyAiUteaXsJA0DSbZa%2Fmrn5TgPitudIIqjVnlzbkYrrAOIhV6DmaA7%2FrQfSgKFz5pIwxKzykQY6pAFbFAKhgacu9s38EN4BA6dk%2BhEUSKV51S6cIi8WURUb0mqKK%2Bkld1ITcJo3%2Fi2ASeCPX1bBrD%2B6BAgT9fEOY9LsJmjXcTD2Boa%2Fv81ZTcY3Hzo411hd6ZsrVDykqwg7pN8FAoFsdXvV7roOMxMIP3K5%2F918%2BNG0k6RSiiv75cHcT4YCkKKJZ%2F%2FK7S%2FyO4c5gdBr14EjqnS9lUjDuwQm8D4ClcBdww%3D%3D&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Date=20220324T164913Z&X-Amz-SignedHeaders=host&X-Amz-Expires=300&X-Amz-Credential=ASIAQ3PHCVTYYYD5USHB%2F20220324%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Signature=29e8ea0c118a303fabdbdd97302ae500b418eacc5ad81ea4382c4ec57ea02b53&hash=a680b28c2b25f6f51d588c3dae657f44b0461067414a3e6e8b15000831d8178b&host=68042c943591013ac2b2430a89b270f6af2c76d8dfd086a07176afe7c76c2c61&pii=0003484374900059&tid=spdf-202fd20b-5df8-4322-af50-0a130923fc8c&sid=056b19b77010d543ae1a4e94453ba94f2960gxrqb&type=client&ua=530152570058545e5d5d&rr=6f10e618297b9be2), the least ordinal that starts a gap of second order is the minimal model height of $\text{ZFC}^-$ and [on p. 372 theorem 3.7] the first ordinal to start a gap of third order is also the minimal model height of $\text{ZFC}^-+\exists\omega\_1$.
I have four questions associated with this property:
1. Does the pattern hold that the least ordinal to start a gap of order $n+2$ is also the minimal model height of $\text{ZFC}^-+\exists\omega\_n$ or $Z\_{n+2}$?
2. If so, are all the ordinals $\alpha$ that start gaps of order $n+2$ exactly those for which
$L\_\alpha\models\text{ZFC}^-+\exists\omega\_n$ and $L\_\alpha\cap\mathcal P^{n+1}(\omega)\models Z\_{n+2}$? If the answer to (1) is unknown then how about (2) when $n=0$?
3. Does the ordinal $\sup\{\min\{\alpha~|~\mathcal P^n(\omega)\cap (L\_{\alpha+1}\setminus L\_\alpha)=\emptyset\}: n\in\mathbb N\}$ itself start any gaps?
4. Is there a countable ordinal that starts a gap of every order (of order $n\forall n\in\mathbb N$) simultaneously?
EDIT: As pointed out by Farmer S, question (4) does not make sense as it is contradicted by page 371 of the very same paper.
Thus, I will change it to the following:
Can an ordinal **be in a gap** of every order (order $n~\forall n\in\mathbb N$) simulteneously?
| https://mathoverflow.net/users/120848 | Can countable ordinals start gaps of every order in the constructible universe? | I presume that by "starts a gap (of order $n$)" you mean "starts a gap of positive length (of order $n$)", since length $0$ is trivial.
Note that the answers given by Monroe Eskew and Fedor Pakhomov were written when there was a different definition of "starts a gap" than what is now there.
1. Yes (by $L\_\alpha\models\mathrm{ZFC}^-$, there are no bounded subsets of $\alpha$ in $L\_{\alpha+1}\backslash L\_\alpha$; and if $L\_\alpha\models$"$\omega\_n$ is the largest cardinal" but $L\_\alpha\not\models\mathrm{ZFC}^-$, then $L\_{\alpha+1}\backslash L\_\alpha$ contains some bounded subset of $\alpha$, hence some subset of its $\omega\_n$).
2. No; you need to add the condition that $L\_\alpha\models$"$\omega\_n$ is the largest cardinal", and then you get the characterization.
3. No (again assuming that the length of a gap has to be $>0$);
let $\alpha$ be that sup.
Then the sequence $\left<\alpha\_n\right>\_{n<\omega}$, where $\alpha\_n$ is the first start of a gap of order $n$, is $\Sigma\_1$-definable over $L\_\alpha$ without parameters, and in $L\_\alpha$, every set is countable. But then it follows that $L\_\alpha$ is the $\Sigma\_1$-hull of the empty set in $L\_\alpha$ (that is, every element of $L\_\alpha$ is $\Sigma\_1$-definable over $L\_\alpha$ from no parameters), which implies that $\alpha$ does not start a gap.
4. No (ignoring the fact that "starts a gap of order $0$" was not defined);
$\alpha$ can not simultaneously start a gap of two distinct orders (this is alluded to in the case of orders 2 and 3 on p. 371 of the paper, near the bottom of the page). E.g. for orders 2 and 3, if cofinally many $\gamma<\alpha$ project to $\omega$, then $\omega$ is the largest cardinal in $L\_\alpha$. (Note Monroe's comment was written when the definition of "starts a gap" was different.)
| 7 | https://mathoverflow.net/users/160347 | 418951 | 170,564 |
https://mathoverflow.net/questions/418944 | 2 | Let $K$ be compact Hausdorff, let $U\subset K$ be open and dense, and let $x\in K\backslash U$. Can we find a disjoint collection $\{U\_i,~ i\in I\}$ of open subsets of $U$ and a collection $\{K\_i,~ i\in I\}$ of compact sets such that $K\_i\subset U\_i$, for every $i$, and $x\in \overline{\bigcup K\_i}$?
| https://mathoverflow.net/users/53155 | Can a point of a compact set be approximated by a disjoint union? | Yes. This is possible. Recall that a subset of a topological space $X$ of the form $f^{-1}[\{0\}]^{c}$ for some continuous function $f:X\rightarrow \mathbb{R}$ is known as a cozero set.
Let $\mathcal{V}$ be a maximal collection of disjoint cozero subsets of $K$ subject to the condition that $\bigcup\mathcal{V}\subseteq U$. Such a set $\mathcal{V}$ is guaranteed to exist by Zorn's lemma, and $\bigcup\mathcal{V}$ is necessarily dense in $K$.
Then for each $V\in\mathcal{V}$, let $f\_{V}:K\rightarrow[0,1/3]$ be a continuous function with
$f\_{V}^{-1}[(0,1/3]]=V$. Then for each $n\geq 1$, let
1. $A\_{V,n}=f\_{V}^{-1}[[\frac{1}{3n+2},\frac{1}{3n}]]$,
2. $B\_{V,n}=f\_{V}^{-1}[[\frac{1}{3n+3},\frac{1}{3n+1}]]$,
3. $A\_{V,n}^{+}=f\_{V}^{-1}[(\frac{1}{3n+2.5},\frac{1}{3n-0.5})]$, and
4. $B\_{V,n}^{+}=f\_{V}^{-1}[(\frac{1}{3n+3.5},\frac{1}{3n+0.5})]$.
Observe that each set $A\_{V,n},B\_{V,n}$ is necessarily compact. The sets
$A\_{V,n}^{+},B\_{V,n}^{+}$ are open in $X$ with $A\_{V,n}\subseteq A\_{V,n}^{+},B\_{V,n}\subseteq B\_{V,n}^{+}$. Furthermore, the sets
$(A\_{V,n}^{+})\_{n=1}^{\infty}$ are disjoint subsets of $V$, and the sets
$(B\_{V,n}^{+})\_{n=1}^{\infty}$ are also disjoint subsets of $V$. Therefore,
$\{A\_{V,n}^{+}\mid V\in\mathcal{V},n\geq 1\}$ is a collection of disjoint open sets, and $\{B\_{V,n}^{+}\mid V\in\mathcal{V},n\geq 1\}$ is another collection of disjoint open sets. However, we have $V=\bigcup\_{n=1}^{\infty}A\_{V,n}\cup B\_{V,n}$, so
$$\text{Cl}\_{K}\big(\bigcup\_{V\in\mathcal{V}}\bigcup\_{n=1}^{\infty}A\_{V,n}\big)\cup\text{Cl}\_{K}\big(\bigcup\_{V\in\mathcal{V}}\bigcup\_{n=1}^{\infty}B\_{V,n}\big)=\text{Cl}\_{K}\big(\bigcup\_{V\in\mathcal{V}}\bigcup\_{n=1}^{\infty}A\_{V,n}\cup B\_{V,n}\big)=K.$$ Therefore, if $x\in X$, then
$x\in\text{Cl}\_{K}\big(\bigcup\_{V\in\mathcal{V}}\bigcup\_{n=1}^{\infty}A\_{V,n}\big)$ or
$x\in\text{Cl}\_{K}\big(\bigcup\_{V\in\mathcal{V}}\bigcup\_{n=1}^{\infty}B\_{V,n}\big).$
| 3 | https://mathoverflow.net/users/22277 | 418952 | 170,565 |
https://mathoverflow.net/questions/318890 | 2 | The method of steepest descent provides an asymptotic approximation for integrals of the form:
$$I = \int\_C \exp(M f(z))\mathrm dz$$
for large positive $M$, where $f(z)$ is analytic in the region of interest, $C$ a contour and $f(z)$ goes to zero at the endpoints of the contour. The asymptotic approximation is:
$$I \sim \sqrt{\frac{2\pi}{f''(z\_0)}}\exp(M f(z\_0)+i\theta)$$
where $z\_0$ is a saddle-point of $f$ ($f'(z\_0)=0$) such that the original contour $C$ can be deformed (fixing the endpoints) to pass through $z\_0$ in a steepest descent direction (which defines the angle $\theta$) and assuming that it doesn't go through any other saddle-points for simplicity.
Is there a generalization for higher-dimensional integrals? For example,
$$J = \int\_{C\_1}\mathrm dz\_1\int\_{C\_2}\mathrm dz\_2 \exp(M g(z\_1,z\_2))$$
I haven't studied much of multi-dimensional complex analysis, so I'm not sure what conditions should be imposed on $g(z\_1,z\_2)$. But as suggested in (<https://en.wikipedia.org/wiki/Several_complex_variables>), say that $g(z\_1,z\_2$ can be represented as a convergent power series in the region of interest.
| https://mathoverflow.net/users/16615 | Steepest descent integration in several dimensions | You can look for **Multivariable Morse Lemma** to get an extension of Steepest Descent into multiple complex variables $z=[z\_1, z\_2, ...,z\_n]$. Higher dimensional asymptotics as $M\rightarrow\infty$ for this multiple integral with $f:\mathbb{C}^n\rightarrow\mathbb{C}$ and $C=C\_1\times C\_2\times ...\times C\_n$ a multiple complex contour domain, $$J(M)=\int\_C e^{M f(z)}dz$$ is obtained taking $z\_0=[z\_{10}, z\_{20}, ...,z\_{n0}]$, the point where $f'(z\_0)=\nabla f(z)|\_{z\_0}=0$ then, under regularity conditions (single non-degenerate saddle point), $$J(M)=\frac{(2\pi/M)^{\frac{n}{2}}e^{Mf(z\_0)}}{\sqrt{\det{[-f''(z\_0)]}}}[1+O(1/M)]$$ here $f''(z)$ is the Hessian matrix having eigenvalues $\lambda=[\lambda\_1,\lambda\_2,...,\lambda\_n]$ with $|\arg(-\lambda\_k)|<\frac{\pi}{2}$ $$\sqrt{\det{[-f''(z\_0)]}}=e^{\frac{i}{2}\phi}\prod\_{k=1}^n|\lambda\_k|^{1/2}\ne0$$ where $\phi=\sum\_{k=1}^n \arg(-\lambda\_k)$. Note that if $z\_0\in\mathbb{R}^n$ and $\Im[f(z)]=0\ \ \forall z\in\mathbb{R}^n\ $ then $\ \phi=0$.
Also if $z\_0\in\mathbb{R}^n,\ \Re[f(z)]=0\ \ \forall z\in\mathbb{R}^n\ $ and $|\arg\sqrt{-\lambda\_k}|\le\frac{\pi}{4}$ then $\ \phi=m\cdot\frac{\pi}{2}\ $ where $m$ is the number of negative eigenvalues minus the number of positive ones (stationary phase method).
The case of multiple saddle points $z\_0^{(\ell)}$ s.t. $f'(z\_0^{(\ell)})=0$ and $\det{[-f''(z\_0^{(\ell)})]}\ne0,\ \ \ell=1,2,...L$ is worked the same way. This gives $$J(M)=\sum\_{\ell=1}^L\frac{(2\pi/M)^{\frac{n}{2}}e^{Mf(z\_0^{(\ell)})}}{\sqrt{\det{[-f''(z\_0^{(\ell)})]}}}[1+O(1/M)]$$
Chapters VIII-IX in
*Wong, R.*, [**Asymptotic approximations of integrals**](http://dx.doi.org/10.1137/1.9780898719260), Classics in Applied Mathematics 34. Philadelphia, PA: SIAM (ISBN 0-89871-497-4/pbk). xvii, 543 p. (2001). [ZBL1078.41001](https://zbmath.org/?q=an:1078.41001).
provide asymptotic methods for multidimensional integrals. Cases of degenerate saddles where Hessian vanishes $\det[-f''(z\_0^{(\ell)})]=0$ for some $\ell$ is more complex. They are found in Ch VIII sect. 5 pg. 435 and Ch IX sect. 4 pg 491.
I hope this helps.
| 2 | https://mathoverflow.net/users/141375 | 418958 | 170,567 |
https://mathoverflow.net/questions/418947 | 2 | I asked this question on MSE some time ago but didn't get a response.
Everything can be assumed in $ \mathbb{C} $, or atleast in characteristic $ 0 $.
Consider degree $ d $ hypersurfaces in projective space $ \mathbb{P}^n $. These correspond, upto nonzero scalars, to homogeneous polynomials $ h $ of degree $ d $ in $ \mathbb{C} [ x\_0, ... , x\_n ] $. If $ M\_d = \{ x^{\alpha} \} $, ($ \alpha $ a vector of $n+1 $ non-negative integers summing to $ d $) is the set of monomials of degree $ d $, the set $ M\_d $ forms a basis for $ \mathbb{C} [ x\_0, ... , x\_n ]\_d $, so write $ h $ as $ \sum c\_{\alpha} x^{\alpha} $. Then there is a homogeneous polynomial $D(d,n) $ in the coefficients $ c\_{\alpha} $, called the discriminant which satisfies the property that the hypersurface $ V(h) $ is singular iff $ D(d,n) $ vanishes. So $ D(d,n) $ itself defines a hypersurface in $ \mathbb{P}^{ \binom{d+n}{n} - 1 } $.
Question (1) : Is the hypersurface $ V (D(d,n) )$ singular always?
Here is an example, taking $ d=n=2 $, so we're looking at conics in $ \mathbb{P}^2 $. Write $$ h = ax^2 + by^2 + cz^2 + dxy + eyz + fzx $$ Then one can compute $$ D(2,2) = 8abc + 2def - 2ae^2 - 2bf^2 - 2cd^2 $$ which defines a singular cubic fourfold in $ \mathbb{P}^5 $.
The singular locus is (a slight modification of) the $ 2 $-uple Veronese embedding given by $$ [u,v,w] \rightarrow [u^2, v^2, w^2, 2uv, 2vw, 2wu] $$
We see that the singular locus corresponds precisely to the non-reduced conics, of the type $ h=(ux+vy+wz)^2 $.(As every conic, upto an invertible change of coordinates is either $ xy - z^2 $ which is isomorphic to a $ \mathbb{P}^1 $ or $ xy $ which is two intersecting lines or $ x^2 $ which is a nonreduced line.) This leads to my next question-
Question (2) : Does the singular locus of $ V(D(d,n)) $ correspond to nonreduced degree $ d $ hypersurfaces always?
| https://mathoverflow.net/users/152391 | Singular locus of the discriminant variety | (Details of what follows can be found in any exposition of dual varieties such as [Lamotke's paper](https://www.sciencedirect.com/science/article/pii/0040938381900136).)
Given a smooth projective variety $X\subset\mathbb{P}^M$ we can look at the subvariety $D(X)\subset\mathbb{P}^M\times\mathbb{P}^{M\*}$ which is the locus of pairs $(p,H)$ where $H\in\mathbb{P}^{M\*}$ is a hyperplane in $\mathbb{P}^M$ containing $p$ such that $T\_pX$ and $T\_pH$ are *not* transversal at $p$. One can show that $D(X)$ is the projective bundle over $X$ of a suitable vector bundle on $X$ of rank $M-\dim(X)$. It follows that $D(X)$ is a smooth variety of dimension $M-1$. The image $X^{\*}$ of $D(X)$ in $\mathbb{P}^{M\*}$ is called the dual variety of $X$ and the morphism $D(X)\to X^{\*}$ is usually birational. Moreover, one can show that the hyperplane $L(p)$ in $\mathbb{P}^{M\*}$ corresponding to $p$ is tangent to $X^{\*}$ at a smooth point of $X^{\*}$.
The relevance of the above to your questions is that one can take $X$ to be the $d$-tuple Veronese embedding of $\mathbb{P}^n$ in $\mathbb{P}^M$ for $M=\binom{n+d}{d}-1$. In that case, the map $D(X)\to X^{\*}$ *is* birational and $X^{\*}$ is what you have called $D(d,n)$. Note that a hypersurface $Y$ in $\mathbb{P}^n$ of degree $d$ is of the form $H\cap X$ for a suitable hyperplane $H$ in $\mathbb{P}^M$.
When $d>1$ there is a hypersurface $Y$ in $\mathbb{P}^n$ which is singular at (at least) $2$ distinct points. The point in $X^{\*}$ corresponding to $Y$ is a singular point. The converse is also true. This answers both your questions. (The answer to the second question is in the negative when $d>2$ or $n\geq 3$, when one can find reduced hypersurfaces which have more than one singular point.)
| 6 | https://mathoverflow.net/users/124862 | 418961 | 170,568 |
https://mathoverflow.net/questions/418899 | 3 | Let $G$ be a connected simple graph. For any spanning tree $T$ of $G$, let $l(T)$ be the number of leaves of the graph $T$. Consider $\ell=\min\_Tl(T)$, can I find a spanning tree $T$ with $l(T)=\ell$, such that the set of leaves $A$ of $T$ is very close to an independent set.
For example, I guess that there exists a vertex $a\in A$ such that $A\setminus\{a\}$ is an independent set in the graph $G$.
My idea is that maybe we can justifying the tree $T$ step by step to get a extreme tree with some properties.
Is there any results or references for this question?
| https://mathoverflow.net/users/160959 | Property of the spanning tree with minimal leaves | If I am not mistaken, and if I understand you correctly, it seems to me that you are right.
The following statement is true.
>
> Let $G$ be a connected graph and $T$ be the spanning tree with the
> smallest number of leaves and $\ell=l(T)$. Let $A$ be the set of all
> leaves of tree $T$. If $|A|=\ell>2$, then $A$ is an independent set of
> graph $G$.
>
>
>
Here is a brief proof. Let $x$ and $y$ be two leaves and $e=xy$ be an edge of graph $G$.
Then the graph $H=T+e$ has a cycle. Denote this cycle by $C$. If all vertices of the cycle $C$ have degree $2$ in $H$, then $C=H$ and our graph $G$ is Hamiltonian, and this contradicts the condition $\ell>2$.
Hence there exists a vertex $a$ of cycle $C$ of degree $3$ or more in $H$.
Let $e'=ab$ be an edge of $C$. The graph $T'=H-e'$ is a spanning tree of graph $G$ and $l(T')<\ell$. Contradiction.
| 3 | https://mathoverflow.net/users/173068 | 418968 | 170,569 |
https://mathoverflow.net/questions/418799 | 1 | The New York Times, [reporting on Dennis Sullivan's Abel prize](https://www.nytimes.com/2022/03/23/science/abel-prize-mathematics.html), recounts the incident that lured Sullivan from chemical engineering to mathematics:
>
> One day during an advanced calculus lecture, the professor drew two shapes on the blackboard — one a circle, the other more blobby, like a kidney. He then said you could stretch either one to fit on the other.
>
>
> That was not particularly surprising. But then the professor said there was a way — and essentially just one way — to do the stretching such that the stretching was the same in all directions.
>
>
>
>
> “This blew my mind,” Dr. Sullivan recalled. “This was not like mathematics I’d learned up to that point. It was much deeper.”
>
1. In the first quoted paragraph, what is the meaning of "fit on"?
2. In the second quoted paragraph, what is the meaning of the professor's statement?
| https://mathoverflow.net/users/10503 | Uniqueness of "stretching" (subject to constraints) for a two-dimensional figure |
>
> In the first quoted paragraph, what is the meaning of "fit on"?
>
>
>
There is a [homeomorphism](https://en.wikipedia.org/wiki/Homeomorphism) between any pair of (open, Riemannian) disks.
>
> In the second quoted paragraph, what is the meaning of the professor's statement?
>
>
>
There is an (essentially) unique [conformal map](https://en.wikipedia.org/wiki/Conformal_map) from any (open) disk in the plane to the unit disk in the plane. This is (part of) the [uniformisation theorem](https://en.wikipedia.org/wiki/Uniformization_theorem).
| 3 | https://mathoverflow.net/users/1650 | 418973 | 170,571 |
https://mathoverflow.net/questions/418977 | 5 | I've copied over this question from [what I asked on StackExchange](https://math.stackexchange.com/questions/4412970/if-k-rtimes-mathbbz-is-a-finitely-generated-group-but-k-isnt-must-the-f), in the hope that an expert here can readily answer the question.
Is there an example of a group $G=K\rtimes \mathbb{Z}$ satisfying the following three conditions?
* $G$ is finitely generated;
* $K$ is **not** finitely generated;
* the fixed points of $\phi(1)$, which is the automorphism on $K$ corresponding to $1\_\mathbb{Z}$, are **not** a finitely generated group.
I suspect there is an example, but I don't have enough experience with infinite groups to come up with one right away.
$\\\\$
One idea, which may or may not work:
finding matrices $M\_1,\ldots,M\_k\in\text{GL}\_n(\mathbb{C})$ and $g\in\text{GL}\_n(\mathbb{C})$
such that $S=\langle g^{-n}\,M\_j\,g^n\rangle\_{1\leq j\leq k,\;n\in\mathbb{Z}} $ is not finitely generated
but such that $g$ commutes with surprisingly many matrices in $S$.
| https://mathoverflow.net/users/105730 | If $K\rtimes \mathbb{Z}$ is a finitely generated group but $K$ isn't, must the fixed points of $1_\mathbb{Z}$ be a finitely generated group? | No. Fix $p\ge 2$. Take the group
$$G=\{M(x,y,z;n):(x,y,z)\in\mathbf{Z}[1/p],n\in\mathbf{Z}\}$$where $$M(x,y,z;n)=\begin{pmatrix}1 & x & z \\ 0 & p^n & y\\ 0 & 0 & 1\end{pmatrix}$$
and $K$ the set of such $M(x,y,z;n)$ for $n=0$, and identify $\mathbf{Z}$ to powers of $M(0,0,0,1)$.
Then $G$ is finitely generated (namely by $\{M(0,0,0;1),M(1,0,0;0),M(0,1,0;0)\}$), $K$ is not finitely generated, and indeed the centralizer of $M(0,0,0;1)$ in $K$ is not finitely generated (isomorphic to the abelian group $\mathbf{Z}[1/p]$).
| 11 | https://mathoverflow.net/users/14094 | 418978 | 170,572 |
https://mathoverflow.net/questions/418975 | 1 | A similar post on [MSE](https://math.stackexchange.com/questions/4407255/the-nearness-of-transversal-pre-images-of-slices-of-a-trivial-tubular-neighborho) without answer.
Let $f\colon M'\to M$ be a smooth map between two orientable closed smooth manifolds and $S$ be a smoothly embedded closed orientable submanifold of $M$ of co-dimension one. Under these assumptions, we have a trivial tubular neighborhood, i.e., we have an embedding $\varphi\colon [-1,1]\times S\hookrightarrow M$ with $\varphi(0\times S)=S$. Write $S\_t:=\varphi(t\times S)$ for all $t\in [-1,1]$.
Suppose, $f\pitchfork \varphi(\varepsilon\times S)$ for each $-1<\varepsilon <1$. Let $S'$ be a component of the co-dimension one closed orietable submanifold $f^{-1}(S)$ of $M'$.
Define a property $\mathscr P(\varepsilon)$ for $\varepsilon\in (0,1)$ as follows: There exists a component $S\_\varepsilon'$ of $f^{-1}(S\_\varepsilon)$ and an embedding $\varphi\_\varepsilon'\colon [0,\varepsilon]\times S'\hookrightarrow M'$ with $\varphi\_\varepsilon'(0,S')=S'$, $\varphi'\_\varepsilon(\varepsilon,S')=S'\_\varepsilon$ such that $f\big(\text{im }\varphi\_\varepsilon'\big)\subseteq \varphi\big([0,\varepsilon]\times S\_\varepsilon\big)$. Similarly, define $\mathscr P(\varepsilon)$ for $\varepsilon\in (-1,0)$.
>
> Which of the following facts is true is/are if any?
>
>
> $(1)$ $\mathscr P(\varepsilon)$ does hold for all sufficiently small
> $\varepsilon$, $(2)$ There is a sequence $\varepsilon\_n\to 0$ such
> that $\mathscr P(\varepsilon\_n)$ does hold for each $n$.
>
>
>
>
> In case, both $(1)$ and $(2)$ are false, what about if we further
> assume the pull-back $f^\*\colon \text{Vect}(M)\to \text{Vect}(M')$ is a
> bijection?
>
>
>
| https://mathoverflow.net/users/363264 | Transversal pre-image of a small enough trivial tubular neighborhood contains a trivial tubular neighborhood | The condition $\mathcal P(\epsilon)$ is needlessly complicated. It's always true without passing to components.
Let $U$ be a tubular neighborhood of $S$ diffeo to $[-1,1]\times S$. Let $\pi: U\to \mathbb R$ be the projection onto the first factor. Let $V=f^{-1}(U)$. Then $V$ is compact and the transversality assumption is equivalent to $h=\pi\circ f$ not having critical points in $V$. Therefore $V=h^{-1}([-1,1])$ is a product by standard Morse theory. The same works for $h^{-1}([0,\epsilon])$ for any $\epsilon$.
| 3 | https://mathoverflow.net/users/18050 | 418992 | 170,578 |
https://mathoverflow.net/questions/418962 | 3 | Pierre-Gilles Lemarie-Rieusset, *The Navier-Stokes Problem in the 21st Century* treats the heat equation on $\mathbb{R}^3$ for time $t\geq 0$, and proves uniqueness of suitably smooth solutions by a kind of energy argument. For a solution $u(t, x)$ with $u(0,x)=0$ he looks at the integral over all $x\in \mathbb{R}^3$
\begin{equation}\int |u(t,x)|^2 e^{-|x|}\,dx.\end{equation}
Using the equation $u\_t=\Delta u$, calculations very like other energy proofs show
\begin{equation}
\frac{d}{dt}\int |u(t,x)|^2 e^{-|x|}\,dx = -2\int |\vec{\nabla} u|^2 e^{-|x|}\,dx\ + 2\int u e^{-|x|} \sum\_{j=1}^{3}\frac{x\_j}{|x|}\partial\_j u\,dx.
\end{equation}
Those calculations begin with integration by parts and proceed by pretty straightforward calculus. Lemarie-Rieusset immediately concludes
\begin{equation}
\frac{d}{dt}\int |u(t,x)|^2 e^{-|x|}\,dx \leq \frac{1}{2}\int |u(t,x)|^2 e^{-|x|}\,dx.
\end{equation}
But I do not see how.
| https://mathoverflow.net/users/38783 | Unique solutions to the heat equation on $\mathbb{R}^3$ | Sorry, maybe my previous comment was not clear enough. You have $$\frac{d}{dt} \int |u|^2 e^{-|x|}\, dx=-2\int |\nabla u|^2 e^{-|x|}\, dx +2 \int u \nabla u \cdot \frac{x}{|x|}e^{-|x|}\, dx.$$
Now use $2|u \nabla u| \leq \frac 12 |u|^2+2|\nabla u|^2$
to estimate the last integral.
| 8 | https://mathoverflow.net/users/150653 | 419003 | 170,581 |
https://mathoverflow.net/questions/418996 | 4 | There are $n$ men, standing one at each vertex of a convex $n$-gon. If they are allowed to move together along sides or diagonals of the polygon to reach another vertex, how many different ways are there to do so without meeting another one?
See OEIS [A350599](http://oeis.org/search?q=A350599) for the first few numerical values.
| https://mathoverflow.net/users/479476 | Combinatorics related plane geometry | Assume that the paths may not cross and each man must move.
Label the vertices $1,2,\dots,n$ in clockwise order. Let the man at
vertex $i$ move to vertex $\pi(i)$, so $\pi$ is a permutation of
$1,2,\dots,n$. If we draw an arrow from vertex $i$ to $\pi(i)$, then
we get a disjoint union of noncrossing cycles of length $\geq
3$. We can obtain such cycles by choosing a noncrossing partition of
the vertices with no blocks of size 1 and 2, and then orienting the
boundary of the convex hull of each block in two ways. Thus in
Exercise 5.35(b) of *Enumerative Combinatorics*, vol. 2, we should set
$f(i)=2$ for $i\geq 3$ and $f(1)=f(2)=0$. If the desired answer is
$h(n)$, then by this exercise we have
$$ x+\sum\_{n\geq 1}h(n)x^{n+1} = \left( \frac{x}{1+2\sum\_{n\geq 3}
x^n}\right)^{\langle -1\rangle} $$
$$ = \left( \frac{x(1-x)}{(1+x)(1-2x+2x^2)}\right)^{\langle
-1\rangle}, $$
where $\langle -1\rangle$ denotes compositional inverse.
If a man is allowed to stand still, replace $1+2\sum\_{n\geq 3}x^n$ by
$1+x+2\sum\_{n\geq 3}x^n$.
Possibly you can get some kind of explicit formula for $f(n)$ out of
this, but it will be messy.
| 6 | https://mathoverflow.net/users/2807 | 419016 | 170,583 |
https://mathoverflow.net/questions/418972 | 7 | $\DeclareMathOperator\RRe{Re}\DeclareMathOperator\Spin{Spin}\DeclareMathOperator\Sym{Sym}$Let $\mathcal{H}\_2(\mathbb{O})$ denote the (10-dimensional) real vector space of octonionic Hermitian matrices of size 2. Recall that a matrix $(a\_{ij})$ with octonionic entries is called Hermitian if $a\_{ji}=\bar a\_{ij}$.
One has a linear imbedding $j\colon \mathcal{H}\_2(\mathbb{O})\to \Sym^2(\mathbb{R}^{16})$ to the space of real quadratic forms on $\mathbb{R}^{16}=\mathbb{O}^2$ given by
$$(j(A))(\xi)=\sum\_{i,j=1}^2\RRe((\bar\xi\_iA\_{ij})\xi\_j),$$
where $\xi=(\xi\_1,\xi\_2)\in \mathbb{O}^2$.
**QUESTION. I am looking for a characterization of the image of $j$.**
Ideally I need a description analogous to the complex case as follows. The space $\mathcal{H}\_n(\mathbb{C})$ of complex Hermitial matrices is imbedded into the space of real quadratic forms on $\mathbb{R}^{2n}=\mathbb{C}^n$ via the similar map
$$j'(A)(\xi)=\sum\_{i,j=1}^n \bar\xi\_iA\_{ij}\xi\_j.$$
It is known that its image consists precisely of real quadratic forms invariant under the multiplication $\xi\mapsto z\cdot\xi$ for any complex number with $|z|=1$.
However, as far as I can see, this description does not seem to generalize to the octonionic situation.
**REMARK.** I am aware of a representation theoretical characterization of the image of $j$. There is an action of the group $\Spin(1,9)$ on $\mathbb{R}^{16}=\mathbb{O}^2$ (see e.g. p.29 here <https://arxiv.org/pdf/math/0105155v4.pdf> ). The symmetric square representation in $\Sym^2(\mathbb{R}^{16})$ is a sum of exactly two non-isomorphic irreducible subspaces (see the answer to this post [A representation of Spin(9,1)](https://mathoverflow.net/questions/243889/a-representation-of-spin9-1)). One of them is the image of $j$.
| https://mathoverflow.net/users/16183 | Quadratic forms on $\mathbb{R}^{16}$ coming from octonions | I'm revising my answer because whether the image of the map $j$ that the OP defines is equal to the $10$-dimensional subspace $H$ of $\mathrm{Sym}^2(\mathbb{R}^{16})$ that is invariant under $\mathrm{Spin}(9,1)$ depends on which particular action of $\mathrm{Spin}(9,1)$ on $\mathbb{R}^{16}$ one chooses.
First, independent of the map $j$ there is a characterization of $H$ in terms of the $\mathrm{Spin}(9,1)$-action on $\mathbb{R}^{16}=\mathbb{O}^2$ in other terms that may be what the OP wants. This goes as follows: $\mathrm{Spin}(9,1)$ preserves an $8$-sphere $\Sigma$ of $8$-dimensional subspaces $L\subset\mathbb{O}^2$ such that $\mathbb{O}^2\setminus\{0\}$ is foliated smoothly by the punctured planes $L\setminus\{0\}$ for $L\in\Sigma$. Each of these $8$-planes $L$ carries a 'natural' definite quadratic form defined up to a positive multiple such that, for each $g\in \mathrm{Spin}(9,1)$ and $L\in\Sigma$, the induced isomorphism $g:L\to g(L)$ identifies the two quadratic forms up to a multiple.
Then $H$ consists of the quadratic forms on $\mathbb{O}^2$ that restrict to each $L\in\Sigma$ to be a multiple of its 'natural' quadratic form.
All of this follows from the description of $\mathrm{Spin}(9,1)\subset\mathrm{SL}(\mathbb{R}^{16})=\mathrm{SL}(\mathbb{O}^{2})$ given at the end of my [Notes on spinors in low dimension](https://arxiv.org/abs/2011.05568), where the Lie algebra of $\mathrm{Spin}(9,1)$ is described as
$$
{\frak{spin}}(9,1) =
\left\{\pmatrix{
a\_1 +x\,I\_8 & C\,R\_{\bf w} \cr
C\,L\_{\bf x} & a\_3 -x\,I\_8 }\
:\
\matrix{ x\in\mathbb{R},\cr {\bf w},{\bf x}\in\mathbb{O},\cr
\ a\in{\frak{spin}}(8)}\ \right\}
\subset{\frak{sl}}(16,\mathbb{R})\,,
$$
where $C$ denotes conjugations in the octonions, $R\_{\bf w}$ denotes right multiplication by ${\bf w}$ in the octonions, $L\_{\bf x}$ denotes left multiplication by ${\bf x}$ in the octonions, and the assignments $a\mapsto a\_k$ for $k=1,3$ are Lie algebra isomorphisms $\mathfrak{spin}(8)\to \mathfrak{so}(8)$ described in the Notes referenced above.
For example, the subspaces $L\in\Sigma\simeq S^8 = \mathbb{OP}^1$ are of the form either $L\_\infty = \{\ (q,0)\ | \ q\in\mathbb{O\ }\}$ or of the form $L\_w = \{\ (wq, \bar q)\ |\ q\in\mathbb{O}\ \}$ for some $w\in\mathbb{O}$. Alternatively, they can be described as either of the form $L'\_\infty = \{\ (0,q)\ | \ q\in\mathbb{O\ }\}$ or of the form $L'\_x = \{\ (\bar q, qx)\ |\ q\in\mathbb{O}\ \}$ for some $x\in\mathbb{O}$. (Note that $L\_w = L'\_x$ when $w\bar x = 1$ while $L\_0 = L'\_\infty$ and $L\_\infty = L'\_0$.) [Here, I am writing elements of $\mathbb{O}^2$ as pairs $(p,q)$ to save space, but they really should be written in the form $\textstyle\begin{pmatrix}p\\ q\end{pmatrix}$ to be consistent with the matrix notation used above for ${\frak{spin}}(9,1)$.]
Now, with this identification of $\mathbb{R}^{16}$ with $\mathbb{O}^2$, $H$ is not equal to the image of $j$, as is easy to check.
On the other hand, if we take the conjugate of $\mathrm{Spin}(9,1)$ under the matrix
$$
\pmatrix{I\_8 & 0 \cr 0 & C }
$$
(which does not belong to $\mathrm{Spin}(9,1)$), then, for the $H$ associated to this conjugate subgroup, the image of $j$ *is* equal to $H$.
| 6 | https://mathoverflow.net/users/13972 | 419038 | 170,589 |
https://mathoverflow.net/questions/418662 | 0 | Consider roots $f = 0$ of a nicely-behaved real function $f(x, t)$ of two (real) variables.
Namely, points $(x, t)$ on which $f$ vanishes, $f(x, t) = 0$.
Suppose that $x$ can be written as function of time $t$, $x = x(t)$. By the multivariate chain rule,
$$
\partial\_x f \cdot x'(t) + \partial\_t f = 0 \;.
$$
This yields an *explicit* formula for $x'(t)$, if $\partial\_x f \neq 0$ at the point of evaluation.
A similar argument works also in the multivariate case, when $\mathbf{x} \in \mathbb{R}^n$. For any time $t$, think of $F(\cdot, t)$ as an operator on $\mathbb{R}^n$, and the argument follows.
I'm interested in derivatives $\frac{d^l\mathbf{x}}{dt^l}$ of higher orders $l > 0$, in the multivariate case. Based on the above intuition, one might expect a similar formula to exist, allowing to calculate $\frac{d^l\mathbf{x}}{dt^l}$ in terms of the derivative tensors of $F$. The purpose being, to calculate high-order time derivatives at a given operator root.
**Question:**
Before setting off to derive such a formula (if exists), is there such a result in the literature? Perhaps something similar?
**Edit:**
I am aware of the Faà di Bruno's formula, and of its multivariate counterparts.
**Edit 2:**
While an expression for the first-order derivative $\frac{d\mathbf{x}}{dt}$ is commonly found in the literature when discussing the implicit function theorem as @RyanBudney notes, this question is about derivatives of *higher* orders.
| https://mathoverflow.net/users/17083 | Calculating derivatives of arbitrary-order at an operator's root | Although this question sounds quite innocent, a systematic treatment of higher order derivatives of implicit functions is quite involved. On a second thought, this is no surprise if you think about how complicated higher derivatives of concatenations $f\circ g$ get (see [Faà di Bruno's formula](https://en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno%27s_formula)). For higher order derivatives of implicit functions, you can have a look at [The Combinatorics of Higher Derivatives of Implicit Functions](https://arxiv.org/abs/1711.07501).
| 2 | https://mathoverflow.net/users/9652 | 419039 | 170,590 |
https://mathoverflow.net/questions/419031 | 7 | Suppose $n$ is a positive integer greater than 2, and $F$ is an arbitrary field with at least 4 elements.
Denote $\text{GL}\_n(F)$ the general linear group in the usual sense and $U\_n(F)$ the unipotent group in the sense that it consists of $n\times n$ upper-triangular matrices with 1's on the diagonal.
The inclusion $U\_n(F)\hookrightarrow\text{GL}\_n(F)$ induces a (unstable) homology homomorphism $f\_k: H\_k(U\_n(F);\mathbb{Z})\to H\_k(\text{GL}\_n(F);\mathbb{Z})$ for any positive integer $k$.
And there is another map, a stable homology map: $g\_k: H\_k(U\_n(F);\mathbb{Z})\to H\_k(\text{GL}(F);\mathbb{Z})$.
How can I detect the image of $f\_k$ (or $g\_k$)? Is $\text{im}(f\_k)$ or $\text{im}(g\_k)$ always zero?
Background: In Suslin's paper [1] (Sublemma 4.4.2), he asserts that when $k=n=\text{char}(F)=3$ then $\text{im}(f\_k)=\text{im}(g\_k)=0$ with a reference directed to [2]. But from [2] it's quite unclear to me how to deduce the assertion made in [1].
[1] A. Suslin, K3 of a field, and the Bloch group (Russian), Trudy Mat. Inst. Steklov
183 (1990), 180-199. English transl. in Proc. Steklov Inst. Math. (1991), 217-239.
[2] A. Suslin, [Stability in algebraic K-theory](https://link.springer.com/chapter/10.1007/BFb0062181), pp. 304-333 in Lecture Notes in Math.
966, Springer-Verlag, 1982.
| https://mathoverflow.net/users/471160 | How can I detect the homology image of a unipotent group in the general linear group? | Suppose first that $F$ is a finite field of characteristic $p$. Then $U\_n(F)$ is a Sylow $p$-subgroup of $GL\_n(F)$, and so using the transfer in group homology one sees that the image of $f\_k$ (for $k>0$) is precisely the $p$-torsion in $H\_k(GL\_n(F);\mathbb{Z})$. Unstably, this is not well understood, but it cannot be trivial by general principles in group cohomology (see the top answer to [this MO question](https://mathoverflow.net/questions/64688/non-vanishing-of-group-cohomology-in-sufficiently-high-degree)). But it follows that the image of $g\_k$ is contain in the $p$-torsion of $H\_k(GL(F);\mathbb{Z})$, and this is completely understood by Quillen's calculations: it is trivial.
Suppose now that the field $F$ is infinite; it is then a "ring with many units" in the sense of Nesterenko--Suslin (*Homology of the full linear group over a local ring, and Milnor's K-theory*), and we may apply the results of that paper. There is a group $UT\_n(F)$ of all upper triangular matrices (not necessarily having 1's on the diagonal), and a split extension
$$1 \to U\_n(F) \overset{i}\to UT\_n(F) \overset{q}\to D\_n(F) \to 1$$
where $D\_n(F)$ are the diagonal matrices. If follows by iteratedly applying Theorem 1.11 of Nesterenko--Suslin that the map $q$ induces an isomorphism in integral homology, but then $i$ induces the trivial map in integral homology. As the inclusion of $U\_n(F)$ into $GL\_n(F)$ factors over $UT\_n(F)$, it follows that $f\_k$ (and hence $g\_k$) is trivial.
| 10 | https://mathoverflow.net/users/318 | 419040 | 170,591 |
https://mathoverflow.net/questions/419030 | 4 | Let $κ$ be an infinite cardinal, $S$ a set of cardinality $κ$, and let
$I = [0, 1]$ be the closed unit interval. Define an equivalence
relation $E$ on $I × S$ by $(x,α) E (y,β)$ if either $x = 0 = y$
or $(x,α) = (y,β)$. Let $H(κ)$ be the set of all equivalence
classes of $E$; in other words, $H(κ)$ is the quotient set obtained from $I ×S$ by collapsing the subset ${0}×S$ to a point.
For each $x ∈ I$ and each $α ∈ S$, $(x,α)$ denotes the element
of $H(κ)$ corresponding to $(x,α) ∈ I × S$. The topology induced from the metric $d$ on $H(κ)$ defined by
$d((x,α),(y,β))=|x − y|$ if $α = β$, and $d((x,α),(y,β))=
x + y$ if $α\not=β$.
The set $H(κ)$ with this topology is called the hedgehog
of spininess $κ$ and is often denoted by $J(κ)$.
The space is a complete, non-compact, metric space
of weight $κ$.
A topological space $X$ is an absolute retract for metrizable spaces $(M)$
provided that it is in $(M)$ and is a retract of each space $Y$ in
$(M)$ of which it is embedded as a closed subset.
${\bf Question.}$ Is the hedgehog $J(κ)$ of spininess $κ$ an absolute retract ?
| https://mathoverflow.net/users/112417 | Hedgehog of spininess $κ$ is an absolute retract? | The answer is positive because the hedgehog $J(κ)$ is $AE$.
| 1 | https://mathoverflow.net/users/112417 | 419049 | 170,595 |
https://mathoverflow.net/questions/419052 | 12 | For which algebraic numbers $\alpha$ is there a valuation on the number field ${\mathbb {Q}}(\alpha)$ for which the infinite series $\sum\_{n=0}^\infty \alpha^n$ converges to $1/(1-\alpha)$?
| https://mathoverflow.net/users/3621 | Geometric series in algebraic number fields | This holds precisely for elements which aren't roots of unity. Indeed, by the product formula we have $\prod\_v|\alpha|\_v=1$, where the product runs over all (finite and infinite) primes. This shows that either all $|\alpha|\_v$ are equal to $1$, or at least one of them is smaller than $1$. If all $|\alpha|\_v$ are equal to $1$, then $\alpha^n$ doesn't tend to zero in any valuation, so $\sum\alpha^n$ cannot converge. This happens precisely when $\alpha$ is an algebraic integer (that property follows from having $|\alpha|\_v\leq 1$ for all the finite places $v$) with all conjugates of absolute value $\leq 1$ (infinite places), which by a [theorem of Kronecker](https://mathoverflow.net/q/10911/30186) characterizes roots of unity.
If $|\alpha|\_v<1$ for some place $v$, then $\sum\alpha^n=\frac{1}{1-\alpha}$ in the valuation induced by $v$, regardless of whether it is a finite or infinite place.
The above answer is assuming you consider valuations/absolute values coming from all places in your question. If you only consider finite places, then there isn't such a complete characterization. Rather, you just want $\alpha$ to have positive valuation at some finite place, which you can characterize by saying $1/\alpha$ is *not* an algebraic integer.
| 18 | https://mathoverflow.net/users/30186 | 419055 | 170,597 |
https://mathoverflow.net/questions/419064 | 8 | **Edit:** In the comments, Tyrone points out that West's positive answer to Borsuk's conjecture implies that every compact ENR is homotopy equivalent to a finite CW complex. It follows that the only finitely dominated spaces which are homotopy equivalent to compact ENRs are those finitely dominated spaces whose Wall finiteness obstruction is trivial. This completely settles the question.
---
A space $X$ is said to be *finitely dominated* if there is a finite CW complex $K$ and maps $r: X \to K$, $s: K \to X$ such that $r \circ s: K \to K$ is homotopic to the identity map, i.e., $X$ is a homotopy retract of $K$.
Equivalently, $X$ is finitely dominated if there is a space $K'$ homotopy equivalent to a finite CW complex and maps $r: X \to K'$, $s: K' \to X$ such that $r' \circ s': K' \to K'$ is the identity, i.e., $X$ is a (strict) retract of $K'$.
A space $X$ is said to be an *ENR* (Euclidean neighborhood retract) if there is an embedding $i: X\to \Bbb R^n$ such that
$i(X)$ is a retract of some open neighborhood $U \subset \Bbb R^n$.
It is known that
$X$ is a compact ENR if and only if $X$ is a (strict) retract of a finite CW complex (cf. Hatcher's book App. A).
The above leads to the following question:
**Question** *Is every finitely dominated space homotopy equivalent to a **compact** ENR? If not, what are the obstructions?*
**Notes:** (1) The question asks whether or not the property of being a homotopy retract of a finite complex is the same as that of being homotopy equivalent to a strict retract of a finite complex.
(2) If $X$ is simply connected and finitely dominated, then Wall shows that X is homotopy equivalent to a finite CW complex. It follows that X is homotopy equivalent to a compact ENR. So if there a counterexample, if it exists, is necessarily not 1-connected.
| https://mathoverflow.net/users/8032 | Finite domination and compact ENRs | It was originally conjectured by Borsuk that every compact ANR should be homotopy equivalent to a finite CW complex. While it was known that every separable ANR has the homotopy type of a countable CW complex, the finer statement was an open problem for some time in the '60s and '70s.
It was J. West
>
> *[Mapping Hilbert Cube Manifolds to ANR's: A Solution of a Conjecture of Borsuk](https://www.jstor.org/stable/1971155)*, Ann. Math., **106**, (1977), 1-18.
>
>
>
who finally arrived at a positive solution to Borsuk's conjecture (see Corollary 5.2).
T. Chapman had previously shown that every compact Hilbert Cube manifold has the homotopy type of a finite CW complex. In the paper linked above, West showed that every compact ANR has the homotopy type of compact Hilbert cube manifold, thus concluding;
>
> *Theorem*: [West] Every compact ANR has the homotopy type of a finite CW complex.
>
>
>
Since Wall has shown that there are finitely dominated spaces not of the homotopy type of a finite complex, it must be that there are fintiely dominated spaces not of the homotopy type of any compact ANR.
| 11 | https://mathoverflow.net/users/54788 | 419065 | 170,598 |
https://mathoverflow.net/questions/419088 | 2 | Can such a set $A=$ {$a\_1,.. a\_k$} exist, such that:
1. $\sum\_i a\_i = 1$ and $a\_i $ are rational positive numbers
2. $k$ is and odd number, and is at least $3$.
3. We can partition $A$ in two parts of value $ \frac{1}{2}$ each
4. $\forall a\_j \in A$, let $B\_j := A - a\_j$. We can partition $B\_j$ into two groups of value $\frac{1-a\_j}{2}$.
**Examples**:
{$\frac{1}{3} , \frac{1}{3}, \frac{1}{9}, \frac{1}{9}, \frac{1}{9}$} respect property 1, 2 4 but not 3.
{$\frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{8}, \frac{1}{8}$} respect 1, 2, 3 but not 4: removing any element yields an infeasible problem.
{$\frac{1}{3}, \frac{1}{3}, \frac{1}{6} ,\frac{1}{6} $} respect properties 1, 3, 4, but not 2.
I suppose it should be impossible that such a partition exist. What are your thoughts?
By the way, I don't think it is important but $\forall a\_j, \quad a\_j \le \frac{1}{3}$
| https://mathoverflow.net/users/342793 | Odd partition with extra properties | Multiply by the least common multiple $M$ of the denominators to get the equivalent problem:
1. The $a\_i$ are positive integers.
2. $k$ is an odd number.
3. We can partition $A$ into two parts of equal sum.
4. If we remove any element of $A$, we can partition the remaining elements into two parts of equal sum.
In the rescaled problem, at least one of the $a\_i$ is odd, since otherwise we could have rescaled by $\tfrac M2$ to get integers, contradicting $M$ being the least common multiple of the denominators.
Property 3 tells us that the sum of the elements of $A$ is even. Therefore if we remove an odd $a\_i$ in step 4, each of the parts needs to sum to a non-integral total, despite being made of integers.
Therefore by contradiction there is no such set $A$.
| 7 | https://mathoverflow.net/users/46140 | 419099 | 170,608 |
https://mathoverflow.net/questions/419097 | 5 | Let $k=\mathbb{F}\_q$ be a finite field with $q$ elements and let $X$ be a quasi-projective $k$-scheme. I saw somewhere claims the following results (without explanation):
1. Let $N$ be a positive integer and let $i: Z\hookrightarrow X$ be the closed immersion of the (finite) disjoint union of ${\rm Spec}(\kappa(x))$ for all the closed points $x\in X$ of degree $[\kappa(x):k]<N$. For $n< N$, we have a bijection
$${\rm Sym}^ni: ({\rm Sym}^nZ)(k)\xrightarrow{\cong}({\rm Sym}^nX)(k).$$
2. Suppose $X={\rm Spec}(L)$ for a finite extension $L/k$ of degree $d$. Then $({\rm Sym}^nX)(k)$ is empty if $d\nmid n$, and is a singleton if $d\mid n$.
I don't have any idea with them. Can anyone help me to prove these results? Thanks also for any idea/hint/reference!
For the construction of symmetric products of varieties, one can see Section 3.1 in [Hilbert and Chow Schemes of Points,
Symmetric Products and Divided Powers](https://people.kth.se/%7Edary/hilbchowsymdiv-20080416.pdf).
| https://mathoverflow.net/users/42571 | About closed points in symmetric product schemes over a finite field | There are probably many ways to answer your question depending on what your preferred point of view on schemes and symmetric products are. Let me offer the following approach.
Forget that $k$ is a finite field, let it just be a perfect field, call $\newcommand{\alg}{\operatorname{alg}}k^{\alg}$ a fixed chosen algebraic closure and $\Gamma := \newcommand{\Gal}{\operatorname{Gal}}\Gal(k^{\alg}/k)$ its absolute Galois group. We see a $k$-scheme $X$ of finite type through its set $X(k^{\alg})$ of geometric points endowed with a (continuous) action of $\Gamma$; for $k \subseteq k' \subseteq k^{\alg}$, the set $X(k')$ is the set of $\Gamma\_{k'}$-fixed points of $X(k^{\alg})$ where $\Gamma\_{k'} := \Gal(k^{\alg}/k')$. We note that $\newcommand{\Sym}{\operatorname{Sym}}\Sym^n(X)$ has the set $\Sym^n(X(k^{\alg}))$ (of $n$-element multisubsets of $X(k^{\alg})$) as geometric points, with the obvious Galois action. In particular, $(\Sym^n(X))(k)$ is the set of $\Gamma$-stable $n$-element multisubsets of $X(k^{\alg})$. Also note that $\newcommand{\Spec}{\operatorname{Spec}}\Spec(L)$ is seen as the homogeneous $\Gamma$-set $\Gamma/\Gamma\_L$ of left cosets of $\Gamma\_L$ under $\Gamma$.
Now remember that $k$ is a finite field. Then $\Gamma$ is procyclic with progenerator given by the Frobenius $\sigma \colon x \mapsto x^q$ (i.e., a $\Gamma$-set is just a set with a finite-order permutation $\sigma$). When $L$ is the extension of degree $d$, the $\Gamma$-set $\Spec L$ is given by a $d$-cycle; a closed point in $X$ with residue field $L$ is such a $d$-cycle in the Galois action on the geometric points of $X$.
With this in mind, claim (1) essentially says that if $\sigma$ is cyclic permutation on an $n$-element multiset (a $\sigma$-stable $n$-element multisubset of $X(k^{\alg})$), we can see it as a multiset sum of $d$-cycles for $d\leq n$, which we then see as a single $n$-element $\sigma$-stable multisubset of the disjoint union of “archetype” $d$-cycles (one for each $d\leq n$).
As for (2), it says that a $d$-cycle has a single $\sigma$-stable $n$-element multiset when $d|n$ (namely, take every element of the cycle $n/d$ times), and none if $d$ does not divide $n$.
| 5 | https://mathoverflow.net/users/17064 | 419104 | 170,610 |
https://mathoverflow.net/questions/410625 | 10 | Let $P\subset \Bbb R^n$ be an inscribed convex polytope, that is, all its vertices are on a common sphere of radius $r$.
Let $G$ be the edge-graph of $P$. For convenience, assume $V(G)=\{1,\dotsc,s\}$. Let $\ell\_{ij}$ denote the length of the edge of $P$ corresponding to $ij\in E(G)$.
>
> **Question.** Let $p\_1,\dotsc,p\_s\in\Bbb R^{m}$ be points so that
>
>
> * the points are on a common sphere $S$,
> * $\lVert p\_i-p\_j\rVert\le\ell\_{ij}$ for all $ij\in E(G)$,
> * the center of $S$ lies in the convex hull $\operatorname{conv}\{p\_1,\dotsc,p\_s\}$.
>
>
> Is it then true that the radius of $S$ is at most $r$? If no, does this change if $n=m$?
>
>
>
In other words, are the skeleta of inscribed polytopes "as expanded as possible" for the given edge-lengths?
Note that the condition on the convex hull is necessary. Without this we could choose an arbitrarily large sphere $S$, and place all the $p\_1,\dotsc,p\_s$ in an arbitrarily small patch of $S$, so that $S$ is their circumsphere.
---
**The case $n=m=2$**
I will demonstrate my general ideal on the case $n=m=2$, which I hope to somehow generalize to all cases with $n=m$. I do not yet have an idea for $m>n$.
Let $P\subset\Bbb R^2$ be an inscribed polygon with circumradius $r$ and vertices $v\_1,...,v\_s$ in circular order. Let $\alpha\_i$ be the angle between $v\_i$ and $v\_{i+1}$ (indices mod $s$) as seen from the circumcenter. Then $\alpha\_1+\dots+\alpha\_s=2\pi$.
Suppose now that we have such a set of points $p\_1,...,p\_s$ with circumradius $r'>r$. Let $\beta\_i$ be the angle between $p\_i$ and $p\_{i+1}$ as seen from the circumcenter. Since $\|p\_i-p\_{i+1}\|\le \|v\_i-v\_{i+1}\|$ but also $\|p\_i\|>\|v\_i\|$ it is easy to see that $\beta\_i<\alpha\_i$. In particular, $\beta\_1+\cdots+\beta\_s<2\pi$, and the closed polyline with vertices $p\_1,...,p\_s$ must have zero winding number around the circumcenter. But since the convex hull of the $p\_i$ contains the origin, there are three points $p\_{i\_1},p\_{i\_2},p\_{i\_3}$ with $i\_1<i\_2<i\_3$ so that already the convex hull of these contains the circumcenter. It is then easy to see that $$\beta\_{i\_1}+\cdots+\beta\_{i\_2}+\cdots+\beta\_{i\_3}\ge \pi$$
and
$$\beta\_{i\_3+1}+\beta\_{i\_3+2}+\cdots+\beta\_{i\_1-1}\ge \pi,$$
in contradiction to $\beta\_1+\cdots+\beta\_s<2\pi$. $\;\square$
To generalize this, one would need to find a suitable notion of "winding number" (mapping degree) for point arrangements with $n>3$, somehow using the face structure provided by the polytope. One might then be able to construct a similar argument using space angles (aka fractions of the sphere, as described by Matt F. in the comments).
| https://mathoverflow.net/users/108884 | Given the skeleton of an inscribed polytope. If I move the vertices so that no edge increases in length, can the circumradius still get larger? | The answer to the question is unfortunately no, in the case of $n=m=3$. There is a simple example to illustrate this.
Let $P$ be the cube with vertices $(\pm \frac{1}{2},\pm \frac{1}{2},\pm \frac{1}{2})$. Very obviously, every edge is of length $1$, and a quick calculation shows this is inscribed upon a sphere of radius $\frac{\sqrt{3}}{2}$.
We now describe the locations of 8 vertices, one for each sign-string of length 3. Vertex $V\_{sign}$ with $s=\#$positive entries of $sign$ is at the point $(sin(\frac{s\pi}{3}),cos(\frac{s\pi}{3}),0)$. It is easy to check that the lengths of the edges is exactly preserved. However, these vertices lie on a sphere of radius $1 > \frac{\sqrt{3}}{2}$.
There are three things wrong with this example, two of which are immediately fixable. First, $\vec{0}$ is not interior. A small reduction of the radius can fix this. Secondly, the convex hull of these points is not full dimensional. Again a small reduction of the radius combined with some small perturbation in the third coordinate fixes this.
Finally, the images of the faces of $P$ form a complex which does not contain the origin. This is both what allows the example to exist, and makes it a not satisfying example. Perhaps an additional condition which could contribute to a winding number argument could be: the cone generated by all edges incident to a vertex contains the origin. This could replace that the origin is in the convex hull of the points.
| 3 | https://mathoverflow.net/users/445044 | 419107 | 170,611 |
https://mathoverflow.net/questions/419059 | 4 | The following is exercise 5 on p. 176 in Hirsch's "Differential Topology" (corrected 6th printing):
>
> Let $\eta = (p,E,B)$ be a fixed vector bundle over a compact manifold $B$, $\partial B = \varnothing$. An *$\eta$-submanifold* $(M,f)\subset V$ of a manifold $V$ is a pair $(M,f)$ where $M\subset V$ is a compact submanifold and $f$ is a bundle map from the normal bundle of $M$ to $\eta$ (this requires $\dim \eta = \dim V - \dim M$). Two $\eta$-submanifolds $(M\_i,f\_i)\subset V$ are $\eta$-*cobordant* if there is an $\eta$-submanifold $(W,f)\subset V\times I$ such that $\partial (W,f) = (M\_0,f\_0)\times 0 \cup (M\_1,f\_1)\times 1$ (using an obvious notation). The set of $\eta$-cobordism classes corresponds bijectively to the homotopy set $[V,E^\*]$.
>
>
>
Notes: here $I = [0,1]$, $E^\*$ is the Thom space (or one-point compactification) of $E$, and "bundle map" means linear isomorphism on each fiber (see p. 88 of Hirsch).
**Questions**: are there references where this type of cobordism is introduced and/or studied in further detail? Is "$\eta$-cobordism" standard terminology for this type of cobordism?
| https://mathoverflow.net/users/89166 | A cobordism theory from Hirsch's "Differential Topology" (reference request) | As noted in the comments this is in Tom Diecks book (section 21.2), and in Wall's differential topology (Section 8.1).
| 4 | https://mathoverflow.net/users/12156 | 419108 | 170,612 |
https://mathoverflow.net/questions/419106 | 1 | Let $G$ be a $d$-generated group. Then my first question is how to see free reduced group $FV(G)$ in the variety containing $G$. What I understood is: "Let $W \subset F\_d$ ($d$-generated free group) be the collection of words which are laws for $G$, i.e. for every $w \in W$ image of word map $w$ on $G$ is identity. Let $F\_m$ be m-generated free group and $S$ be the union of images of words $w \in W$ in $F\_m$. If $FV(G)$ is the $m$-generated reduced free group then $FV(G) = \frac{F\_m}{\langle S \rangle}$."
My second question is when $G$ is finite group then is it true that finitely generated free reduced group $FV(G)$ will also be finite? Where can I read more about free reduced group?
Thank you!
| https://mathoverflow.net/users/117282 | Reduced free group | I'll give a more detailed answer. My comment above about your first question was wrong because I read it too quickly and didn't catch the difference between $m$ and $d$. Your description of the free object in question 1 is off.
Take the cyclic group $C\_n$ of order $n\geq 3$. Then $C\_n$ is $1$-generated and the laws it satisfies in one variable are all consequences of $x^n=1$. But it also satisfies $xy=yx$, which is not a consequence of this law when $n\geq 3$.
If $A$ is a universal algebra, then the variety generated by $A$ is defined by all laws satisfied by $A$. So you should factor the free group out by the verbal subgroup generated by all laws satisfied by $G$.
For the second question, the answer is yes as I said in the comment. If $A$ is a universal algebra, then the free object in the variety generated by $A$ on a generating set $X$ embeds in $A^{A^X}$ via the map $\psi\colon X\to A^{A^X}$ with $\psi(x)(f)=f(x)$. This $X$ generates a relatively free object because any mapping $f\colon X\to A$ factors through $\psi$ by projecting the the $f$-coordinate and so every map from the free object on $X$ in the signature of $A$ to $A$ factors through the map induced by $\psi$. Now it is easy to verify that $\psi(X)$ generates a relatively free algebra in the variety generated by $A$ using the description as quotients of subalgebras of direct products of copies of $A$.
| 5 | https://mathoverflow.net/users/15934 | 419110 | 170,613 |
https://mathoverflow.net/questions/418900 | 8 | I am curious about the reverse math status of the below statement. Note that we work in second-order RM, i.e. 'closed set' is interpreted as in Simpson's excellent SOSOA.
*For any closed $E\subset [0,1]$ and $\epsilon >0$, there are $x\_0, \dots, x\_k \in E$ such that $\cup\_{i\leq k} B(x\_i, \epsilon)$ covers $E$.*
It seems that enough induction suffices to prove this theorem, but I cannot seem to formulate a nice minimal/small fragment.
Note that the above statement can be interpreted as asking for a finite sub-covering of the *uncountable* covering $\cup\_{x\in E} B(x, \epsilon)$ of $E$.
| https://mathoverflow.net/users/33505 | Reverse Mathematics strength of fixed radius covering theorem | The statement in provable in $\mathrm{WKL}\_0$.
Consider the following proof. Fix $k>1/\epsilon$, let $I=\{i<k:E\cap[i/k,(i+1)/k]\ne\varnothing\}$, and let $\{x\_i:i\in I\}$ be such that $x\_i\in E\cap[i/k,(i+1)/k]$. Then for any $x\in E$, we have $x\in[i/k,(i+1)/k]$ for some $i\in I$, hence $|x-x\_i|\le1/k<\epsilon$.
Now, let us formalize this. By Simpson’s definition, $E=[0,1]\smallsetminus\bigcup\_n(a\_n,b\_n)$, where $\{a\_n,b\_n:n\in\mathbb N\}$ is a sequence of rationals. By the Heine–Borel lemma, provable in $\mathrm{WKL}\_0$, we have
$$E\cap[i/k,(i+1)/k]=\varnothing\iff\exists n\:[i/k,(i+1)/k]\subseteq\bigcup\_{i<n}(a\_i,b\_i),$$
which is a $\Sigma^0\_1$ property. Thus, $I=\{i<k:E\cap[i/k,(i+1)/k]\ne\varnothing\}$ exists by bounded $\Sigma^0\_1$-comprehension, provable in $\mathrm{RCA}\_0$. Let $T$ be the set of sequences $\langle\vec x^0,\dots,\vec x^n\rangle$ such that
* $\vec x^j$ is a vector $\langle x^j\_i:i\in I\rangle$,
* $x^j\_i$ is a dyadic rational with denominator (at most) $2^j$,
* $|x^{j+1}\_i-x^j\_i|\le2^{-(j+1)}$,
* $[x^j\_i-2^{-j},x^j\_i+2^{-j}]\cap[i/k,(i+1)/k]\smallsetminus\bigcup\_{l<j}(a\_l,b\_l)\ne\varnothing$.
Then $T$ is a $3^k$-ary (or so) tree with elements of arbitrarily large height, hence it has an infinite branch by $\mathrm{WKL}\_0$. It is easy to see that such a branch determines a vector $\vec x=\langle x\_i:i\in I\rangle$ of reals such that $x\_i\in[i/k,(i+1)/k]\cap E$.
| 11 | https://mathoverflow.net/users/12705 | 419111 | 170,614 |
https://mathoverflow.net/questions/418451 | 4 | Is there a Mayer–Vietoris-type sequence for the homology of a coproduct of two Hopf algebras over an ideal? The definition of the coproduct can be found in [Agore - Categorical Constructions for Hopf Algebras](https://arxiv.org/abs/0905.2613v3). Maybe the question should be: what does the spectral sequence that computes this homology look like?
Edit: As was requested I specify what kind of homology I am talking about. I was mainly thinking about homology of Lie algebras and discrete groups. So, I am interested in $\operatorname{Tor}^H(k,M)$ for an $H$-module $M.$
| https://mathoverflow.net/users/143549 | Mayer–Vietoris sequence for coproduct of Hopf algebras | In the case of discrete groups, the MV sequence follows from the following argument: if you have $G=H\*K$ then you have a short exact sequence of the form:
$$0\to \mathbb{Z}G\stackrel{i}{\to} \mathbb{Z}G/H\oplus \mathbb{Z}G/K\stackrel{p}{\to} \mathbb{Z}\to 0,$$ where $i$ sends $g$ to $(gH,gK)$ and $p$ sends $gH$ and $gK$ to 1.
The fact that this sequence is exact can be proven combinatorially using the fact that if you take the bases $\{h\}\_{h\in h}$ and $\{k\}\_{k\in K}$ for $\mathbb{Z}H$ and $\mathbb{Z}K$ then a basis for $\mathbb{Z}G$ is given by the set of all alternating $\textit{words}$ in these bases. This argument can be generalized also the the definition of coproduct of two Hopf algberas given by Agore. If you have Hopf algebras $H\_1$ and $H\_2$ and $H=H\_1\*H2$, then choose bases $\{a\}\_{a\in A}$ for $H\_1$ and $\{b\}\_{b\in B}$ for $H\_2$ that contain 1. You then get a basis for $H$ by words in these bases:
Write $A' = A\backslash \{1\}$ and $B'=B\backslash\{1\}$. We have the following basis for $H$:
$$\{1\} \cup \{a\}\_{a\in A'}\cup \{b\}\_{b\in B'}\cup \{ab\}\_{a\in A',b\in B'}\cup\cdots $$
and you can prove in the same way that you get a short exact sequence of $H$-modules given by
$$0\to H\to \text{Ind}\_{H\_1}^{H}k\oplus \text{Ind}\_{H\_2}^H k\to k.$$ Applying now the relevant functors will give you a long exact sequence, which is the Mayer-Vietoris sequence you need here, using the fact that for a right module $M$ it holds that $\text{Tor}^H\_\*(\text{Ind}\_{H\_1}^H k,M)\cong \text{Tor}^{H\_1}\_\*(k,M)$.
I assume here that $k$ is a field for this construction to work.
| 2 | https://mathoverflow.net/users/41644 | 419122 | 170,619 |
https://mathoverflow.net/questions/419117 | 5 | I am interested in the asymptotics of the integral
$$I(a):=\int\_0^\infty \sqrt{x}\operatorname{Erfc}(x+a)\,\mathrm{d}x$$
for $a>0$. I think that $I(a)$ should be decaying exponentially as $I(a)\lesssim e^{-a^2}$ for large $a$. Numeric integration indicates that
$$ \lim\_{a\to\infty} I(a) e^{a^2} a^{5/2} =C \in(0,1)$$
for some constant $C$.
Are there integral tables where I could obtain explicit expressions for $I(a)$ for finite $a$? If not, are the asymptotics above correct, and what is $C$?
So far I could only find results when the error function is integrated against integer powers of $x$ e.g. in Section 2.14 of [https://intra.ece.ucr.edu/~korotkov/papers/Korotkov-book-integrals.pdf](https://intra.ece.ucr.edu/%7Ekorotkov/papers/Korotkov-book-integrals.pdf)
| https://mathoverflow.net/users/89934 | Asymptotics of error function integral with square root | $$\int\_0^\infty \sqrt{x}\,\mathrm{Erfc}(x+a)\,\mathrm{d}x\rightarrow \frac{e^{-a^2}}{(2a)^{5/2}},\;\;\text{for}\;\;a\rightarrow \infty,$$
so $C=2^{-5/2}.$ Corrections are smaller by a factor $1/a$.
I obtained this asymptotics by integrating the large-$a$ expansion of the error-function,
$$\text{Erfc}\,(x+a)\rightarrow \frac{e^{-(a+x)^2}}{\sqrt{\pi } a},\;\;\text{for}\;\;a\rightarrow \infty.$$
| 7 | https://mathoverflow.net/users/11260 | 419123 | 170,620 |
https://mathoverflow.net/questions/419127 | 2 | The question:
(1) Is every compact Lie group $G$ isomorphic (as a topological group) to some quotient $H/N$ where $H$ is a torsion-free compact metrizable group?
Or equivalently:
(2) Is every compact metrizable group $G$ isomorphic (as a topological group) to some quotient $H/N$ where $H$ is a torsion-free compact metrizable group?
((1)$\implies$(2) follows from the fact that every compact metrizable group is isomorphic to an inverse limit of compact Lie groups)
A related question that came up in our research:
(3) Is there a compact *connected* metrizable non-abelian torsion-free group?
(I asked the latter question without the assumption on connectivity [here](https://math.stackexchange.com/questions/4409106/is-there-a-compact-metrizable-non-abelian-torsion-free-group?noredirect=1#comment9226292_4409106), and YCor pointed out that the Heisenberg group over the $p$-adic integers is a suitable example.
| https://mathoverflow.net/users/479121 | Compact Lie groups as quotients of torsion-free compact metrizable groups | The answer to all questions is "no" (and even without the metrizability requirements).
a) A negative answer to (3) implies a negative answer to (1).
Indeed, if $f:G\to H$ is a surjective continuous homomorphism between compact groups, then $G^\circ\to H^\circ$ is surjective as well. So a non-abelian connected compact group (such as $\mathrm{SO}(3)$) is not quotient of any torsion-free compact group.
b) The answer to (3) is negative: **every connected torsion-free compact group $G$ is abelian**. Indeed, let $G$ be a connected compact group. Then (see Bourbaki Lie, Chap. IX Appendix I) there exists a family $(S\_i)\_{i\in I}$ of connected simple Lie groups (with finite center) and a surjective homomorphism $S=\prod S\_i\to [G,G]$ whose kernel is central in $S$. Since each $S\_i$ contains non-central elements of finite order, it follows that $[G,G]$ is not torsion-free, unless $I$ is empty, in which case $G$ is abelian.
| 3 | https://mathoverflow.net/users/14094 | 419130 | 170,621 |
https://mathoverflow.net/questions/419094 | 3 | This question just came to my mind and I have no idea as to how to approach it. Let $z\_1,z\_2,\dots,z\_n$ be $n$ be any complex numbers in the unit disc $|z| \leq 1.$ Consider a complex function on the unit disc wiith real values
$$ f(z)=\sum\_{i=1}^n \frac{|z-z\_i|}{n}.
$$
My questions:
* Does there exist a $z \in |w| \leq 1 $ so that $f(z)=|z|$?
* If not can we make $f(z)$ arbitrarily close to $|z|$ for some $ z \in |w| \leq 1 $?
* What about the maximum and the minimum value of $f(z)$?
Lots of thanks for any responces\hints\suggestions
| https://mathoverflow.net/users/158175 | A question about average deviation of given $n$ complex numbers | The answer is no. E.g., let $n=3$ and $z\_j=e^{i(j-1)2\pi/3}$ for $j=1,2,3$. Then $f(z)>|z|+15/100>|z|$ if $|z|\le1$.
---
This counterexample generalizes to any $n\ge3$. Indeed, take any $n\ge3$ and let $z\_j=e^{i(j-1)2\pi/n}$ for $j=1,\dots,n$. Then
$$f(z)=\frac1n\,\sum\_{j=1}^n f\_j(z),$$
where $f\_j(z):=|z-z\_j|$.
Note that for each $j$
the function $f\_j$ is convex and, moreover, $f\_j$ is strictly convex on any straight line not through the point $z\_j$. Since $n\ge3$, there is no straight line passing through all the points $z\_1,\dots,z\_n$. So, the function $f=\frac1n\,\sum\_{j=1}^n f\_j$ is strictly convex.
Also, for any real $a$ and $b$, the derivative of $f$ at $0$ along the vector $(a,b)=a+ib$ is
$$\begin{aligned}&\frac d{ds}\,f(s(a+ib))\Big|\_{s=0} \\
&=-\frac1n\,(a,b)\cdot\sum\_{j=1}^n
\Big(\cos\frac{(j-1)2\pi}n,\sin\frac{(j-1)2\pi}n\Big) \\
&=-\frac1n\,(a,b)\cdot(0,0)=0,
\end{aligned}$$
where $\cdot$ denotes the dot product.
Thus, the strictly convex function $f$ has a unique minimum at $0$, and the minimum value of $f$ is $1$. That is, $f(0)=1<f(z)$ for all $z$ such that $0<|z|\le1$.
So, for all $z$ such that $0<|z|\le1$, we have
$f(z)>1\ge|z|$ and hence $f(z)\ne|z|$. Also, $f(0)=1\ne0$. We conclude that there is no $z$ such that $|z|\le1$ and $f(z)=|z|$, which proves the claim.
---
On the other hand, the answer is yes for $n=1$ and $n=2$.
| 1 | https://mathoverflow.net/users/36721 | 419132 | 170,622 |
https://mathoverflow.net/questions/419075 | 5 | I am studying some papers in the analysis of nonlinear PDEs and I am encountering the $U^p$ and $V^p$ spaces for the first time. Where can I find references more detailed than papers?
**Edited**
The spaces I mentioned above are defined in section two of this paper:
<https://arxiv.org/pdf/0708.2011.pdf>.
| https://mathoverflow.net/users/471464 | Looking for references to study $U^p$ and $V^p$ spaces | You can take a look at Herbert Koch's contribution in
*Koch, Herbert; Tataru, Daniel; Vişan, Monica*, [**Dispersive equations and nonlinear waves. Generalized Korteweg-de Vries, nonlinear Schrödinger, wave and Schrödinger maps**](http://dx.doi.org/10.1007/978-3-0348-0736-4), Oberwolfach Seminars 45. Basel: Birkhäuser/Springer (ISBN 978-3-0348-0735-7/pbk; 978-3-0348-0736-4/ebook). xii, 312 p. (2014). [ZBL1304.35003](https://zbmath.org/?q=an:1304.35003).
Chapter 4 of Koch's notes there is specifically concerning the $U^p$ and $V^p$ spaces.
You should note that these notes are from 2014, and so there are a few newer developments covered in the PhD thesis mentioned in [Raffaele Scandone's comment](https://mathoverflow.net/questions/419075/looking-for-references-to-study-up-and-vp-spaces#comment1076145_419075) that are not included in this discussion. Though for reaching the paper you linked to certainly the material in the book should be more than sufficient.
| 8 | https://mathoverflow.net/users/3948 | 419134 | 170,623 |
https://mathoverflow.net/questions/418295 | 3 | Let $X$ be an affinoid variety over a discretely valued non-archimedean field $k$ with valuation ring $\mathcal{R}$. Fix a uniformizer $\omega$. On the section 3.2 of the paper <https://arxiv.org/abs/1501.02215> it is stated that, for every flat $O(X)^{\circ}$-module $Q$, the functor
$$A\mapsto \widehat{A\otimes\_{O(X)^{\circ}}Q}\otimes\_{\mathcal{R}}k$$
where the completion is with respect to the $\omega$-adic topology, is left exact. Showing this would involve showing that the kernel of the compled tensor is torsion, however, I do not understand why this is the case.
| https://mathoverflow.net/users/476832 | On the exactness of some completed tensor products | This proposition has been removed from the paper in the published paper, available in the author's webpage.
| 1 | https://mathoverflow.net/users/476832 | 419135 | 170,624 |
https://mathoverflow.net/questions/419131 | 10 | A [**geometric theory**](https://ncatlab.org/nlab/show/geometric%20theory) is made up of sequents of restricted form: It may only be of the form $$\phi \vdash \psi$$
possibly with free variables (which are implicitly taken universal closure). $\phi, \psi$ are *geometric formulae* which can only involve $\top, \bot, =, \exists, \wedge$ (binary conjunction) and $\bigvee$ (arbitrary disjunction). Geometric theories enjoy some good properties. In particular, it plays well with *geometric morphisms*, hence the name.
My question is: **is there a (complete, sound) deductive system that involves *only* geometric sequents**? I think we can do this just by writing down the obvious rules. But I'm not familiar with infinitary disjunction. I'm also not quite sure once it gets to the predicative logic part.
Is there any existing work on this? An answer or a pointer to relevant material is welcome.
| https://mathoverflow.net/users/136535 | Deductive system involving only geometric sequents | For geometric theories in a countable fragment, (axiomatized by countably many sequents where the disjunctions are also countable), there exists a sound and complete system that appears already in Makkai and Reyes "First-order categorical logic". In page 159, it is described in the section "Completeness of a one-sided system for coherent logic". One should be aware that what Makkai and Reyes called there "coherent" is what we call now "geometric". Also, although they use a Gentzen style sequent calculus, their system is easily seen to be equivalent to the system of geometric logic through sequents as described in the question, which appears, e.g. in Johnstone's "Sketches of an Elephant" section D1.3
If one removes the countability condition, then the system is no longer complete for usual set models but there is still a completeness theorem in terms of models in toposes (and, in fact, in terms of Boolean-valued models). In the same book Makkai and Reyes also treat these cases.
If one insists in having a complete system in terms of set models even when the theory has $\kappa$ sequents for some uncountable $\kappa$, the workaround is to relax the notion of geometric formula to that of a $\kappa$-geometric formula, in which conjunctions are not necessarily finitary, but of size less than $\kappa$. These are preserved by the $\kappa$-geometric morphisms, which are those geometric morphisms whose inverse image additionally preserves all $\kappa$-small limits. The deductive system in this case features a new infinitary rule and is described in my paper "Infinitary generalizations of Deligne's completeness theorem", The Journal of Symbolic Logic, volume 85, Issue 3, pp. 1147 - 1162.
| 12 | https://mathoverflow.net/users/12976 | 419137 | 170,626 |
https://mathoverflow.net/questions/419133 | 1 | Fix a half-space $H = \{x\_1 \geq 0: ~ (x\_1,\dots,x\_n) \in \mathbb{R}^n\}$. Let $p$ be a distribution with support in $\mathbb{R}^n$. I am interested in the following way of estimating the weight $p(H) = \Pr\_{\mathbf{x} \sim p}\left[\mathbf{x} \in H\right]$ using random Voronoi partitions.
Take $m$ i.i.d. samples $\mathbf{c}^{(1)},\dots,\mathbf{c}^{(m)} \sim p$. Let $\mathcal{V}$ be the Voronoi partition induced by the centers $\mathbf{c}^{(1)},\dots,\mathbf{c}^{(m)}$. Namely, $\mathcal{V} = (V\_1,\dots,V\_m)$, where $V\_i \subseteq \mathbb{R}^n$ is the set of points that are closer in $\ell\_2$ distance to $\mathbf{c}^{(i)}$ than to any $\mathbf{c}^{(j)}$ for $j \neq i$ (with ties broken arbitrarily).
Consider the estimate $\hat{p}(H) = p\left(\bigcup\{V\_i: ~ \mathbf{c}^{(i)} \in H\} \right)$. In words, $\hat{p}(H)$ is the probability according to $p$ of all the Voronoi neighborhoods with centers in $H$.
**My Question:** is $\hat{p}(H)$ an unbiased estimator of $p(H)$? Namely, is it true that:
$$
\mathbb{E}\_{\mathbf{c}^{(1)},\dots,\mathbf{c}^{(m)} \sim p}\left[ \hat{p}(H) \right] = p(H)
$$
| https://mathoverflow.net/users/158769 | Approximating the probability of a half-space using random Voronoi diagrams | No. E.g., suppose that $n=1$, $m=2$, and the pdf $f$ of each of the $m=2$ iid sample points $X\_i:=\mathbf c^{(i)}$ ($i=1,\dots,m$) is given by the formula $f(x)=e^{-x-1}1(x>-1)$ for real $x$.
Then, by straightforward calculations,
$$p(H)=e^{-1}= 0.367\ldots\ne0.506\ldots=\frac{-2+6 e^2-8 e^3+6 e^4+e^6}{3 e^6}=E\hat p\_H.$$
---
Intuitively, the reason for this is clear: the Voronoi diagram is all about distances, whereas a distribution does not have to care about distances at all.
| 2 | https://mathoverflow.net/users/36721 | 419143 | 170,627 |
https://mathoverflow.net/questions/419074 | 4 | Let $n<d$ be positive integers, and let $0< \epsilon\leq 1-\frac{n}{d}$ be a real number. I would like a succinct description of the following quantity:
$$
f\_{\epsilon}(n,d):=\min\_{\substack{x\_1\geq \dots \geq x\_d\geq 0\\ x\_1+\dots+x\_n =1-\epsilon\\ x\_{n+1}+\dots+x\_d=\epsilon}} \left(\sum\_{1 \leq i\_1 < \dots < i\_{n+1} \leq d} x\_{i\_1}\cdots x\_{i\_{n+1}}\right).
$$
In the absence of such a description, I would be interested in good lower/upper bounds on this quantity.
(I wasn't sure what tags to use.)
The assumption $\epsilon\leq 1-\frac{n}{d}$ ensures that the function is defined.
| https://mathoverflow.net/users/150898 | Bound on sum of products of positive real numbers with prescribed subset sums | Here is a succinct description of $f\_{\epsilon}(n,d)$ (I only have time to sketch the argument-- hopefully it is right).
Let
$$
h(x\_1,\dots, x\_d)=\left(\sum\_{1 \leq i\_1 < \dots < i\_{n+1} \leq d} x\_{i\_1}\cdots x\_{i\_{n+1}}\right),
$$
and let $x \in \mathbb{R}^d$ be a feasible element of the minimization (i.e., $x\_1\geq \dots \geq x\_d\geq 0, x\_1+\dots+x\_n=1-\epsilon$, and $x\_{n+1}+\dots+x\_d=\epsilon$).
For any $i<j \in \{n+1,\dots, d\}$, let
$$g(t)=h(x\_1,\dots, x\_{i-1},x\_i+t,x\_{i+1},\dots,x\_{j-1},x\_j-t, x\_{j+1},\dots, x\_d).$$
It is straightforward to verify that $g(t)$ is a quadratic function with negative second derivative. So $g(t)$ is minimized at an extreme point. It follows that $h(x)$ is minimized when $x\_{r+1},\dots, x\_n$ are as ``unbalanced" as possible. This implies that the minimum is attained for some $x \in \mathbb{R}^d$ for which
$$
(x\_{n+1},\dots, x\_d)=\left(\frac{1-\epsilon}{n},\dots, \frac{1-\epsilon}{n},\epsilon-\frac{k(1-\epsilon)}{n},0,\dots,0\right),
$$
where $\frac{1-\epsilon}{n}$ is repeated $k$ times in the tuple, and
$$
k=\lfloor \frac{\epsilon n}{1-\epsilon} \rfloor.
$$
We can do the same trick for the first $n$ coordinates of $x$, and prove that the minimum is attained for $x \in \mathbb{R}^d$ that satisfies
$$
(x\_1,\dots, x\_n)=\begin{cases} (\frac{1-\epsilon}{n},\dots,\frac{1-\epsilon}{n}) ,& \epsilon \geq \frac{1-\epsilon}{n}\\
(1-n\epsilon,\epsilon,\dots, \epsilon) ,& \epsilon\leq \frac{1-\epsilon}{n}.
\end{cases}
$$
It follows that
$$
f\_\epsilon(n,d)=\begin{cases} \binom{n+k}{n+1} \left(\frac{1-\epsilon}{n}\right)^{n+1}+\binom{n+k}{n} \left(\frac{1-\epsilon}{n}\right)^n \left[\epsilon-k\left(\frac{1-\epsilon}{n}\right)\right] ,& \epsilon \geq \frac{1-\epsilon}{n}\\
(1-n\epsilon)\epsilon^n,& \epsilon\leq \frac{1-\epsilon}{n}.
\end{cases}
$$
| 3 | https://mathoverflow.net/users/150898 | 419153 | 170,631 |
https://mathoverflow.net/questions/418576 | 9 | For any set $X$ and cardinal $\mu \neq \emptyset$, we denote by $[X]^\mu$ the collection of subsets of cardinality $\mu$. If $\kappa, \mu \neq \emptyset$ are cardinals and $f: [X]^\mu\to \kappa$ is a map, we say that $H\subseteq X$ is *homogeneous with respect to $f$* if the restriction $f|\_{[H]^\mu}: [H]^\mu \to \kappa$ is constant.
For cardinals $\lambda, \mu, \kappa\neq \emptyset$ and any set $X\neq \emptyset$ we write $$X \to (\lambda)^\mu\_\kappa$$ if for every map $f: [X]^\mu\to\kappa$ there is $H\subseteq X$ such that $H$ is homogeneous with respect to $f$ and $|H|=\lambda$.
With the help of the Axiom of Choice ${\sf (AC)}$ one can prove that $X \not\to (\omega)^\omega\_2$ for every infinite $X$ (see Theorem 7, p. 5 of [this recommended introduction to infinite combinatorics](https://users.metu.edu.tr/burakk/lecturenotes/village2017lecturenotes.pdf), thank you to Burak for writing it!).
**Question.** Does the statement "$X \not\to (\omega)^\omega\_2$ for every infinite set $X$" imply ${\sf (AC)}$?
| https://mathoverflow.net/users/8628 | Does "$X \not\to (\omega)^\omega_2$ for every infinite $X$" imply ${\sf AC}$? | The answer is no, the statement that for every set $X$ we have $$X\not\to(\omega)^\omega\_2$$ does not imply the axiom of choice.
This was shown by Kleinberg and Seiferas in 1973, see
>
> [MR0340025](https://mathscinet.ams.org/mathscinet-getitem?mr=340025) (49 #4782)
> Kleinberg, E. M.; Seiferas, J. I.
> Infinite exponent partition relations and well-ordered choice.
> J. Symbolic Logic 38 (1973), 299–308.
> <https://doi.org/10.2307/2272066>
>
>
>
For $\kappa$ a (well-ordered) infinite cardinal, $\kappa$-well-ordered choice, $\mathsf{AC}\_\kappa$, is the statement that every $\kappa$-sequence of nonempty sets admits a choice function.
The axiom of well-ordered choice $\mathsf{WOC}$ is the statement that $\mathsf{AC}\_\kappa$ holds for all infinite well-ordered $\kappa$.
This statement is strictly weaker than the axiom of choice: it does not imply that $\mathbb R$ is well-orderable, and even if we add this assumption, the result is still weaker than choice. See for instance theorem 5.1 in
>
> [MR1351415](https://mathscinet.ams.org/mathscinet-getitem?mr=1351415) (96h:03087)
> Higasikawa, Masasi
> Partition principles and infinite sums of cardinal numbers.
> Notre Dame J. Formal Logic 36 (1995), no. 3, 425–434. <https://doi.org/10.1305/ndjfl/1040149358>
>
>
>
However, as shown in the paper by Kleinberg and Seiferas, $\mathsf{WOC}$ plus the existence of a well-ordering of $[\omega]^\omega$ rules out all infinite exponent partition relations. It is still open (as far as I know) whether $\mathsf{WOC}$ suffices for this result. What Kleinberg and Seiferas show is that, under $\mathsf{WOC}$, either all infinite exponent partition relations fail, or else $\omega\to(\omega)^\omega\_2$. (And the latter fails if $[\omega]^\omega$ is well-orderable.)
| 11 | https://mathoverflow.net/users/6085 | 419160 | 170,634 |
https://mathoverflow.net/questions/419162 | 7 | Theorem. Let $\phi:X\rightarrow Y$ be a quasi-isometry between two (Gromov) hyperbolic spaces $X$ and $Y$. If $X$ and $Y$ are proper, then ϕ induces a homeomorphism between their boundaries.
The proof of the above statement is well-written in Bridson and Haefliger's book.
My question is that `can we drop the condition that $X$ and $Y$ are proper?'. In some papers about boundaries of hyperbolic spaces, the authors usually say that the above theorem is true without mentioning that $X$ and $Y$ are proper. If you know the answer or any references, then let me know.
| https://mathoverflow.net/users/173504 | Induced homeomorphism from a quasi-isometry between hyperbolic spaces | Properness is already needed to have a well-defined boundary at infinity, i.e., with a topology not depending on the chosen base point. This is Proposition III.3.7 in Bridson-Haefliger, which builds on some previous lemmata that are applications of the Arzela-Ascoli theorem. To apply the Arzela-Ascoli theorem one needs bounded sets to be compact.
| 8 | https://mathoverflow.net/users/39082 | 419166 | 170,635 |
https://mathoverflow.net/questions/419090 | 1 | Let $X,Y$ be metric space and suppose that $f:X\rightarrow Y$ is a *uniform embedding*; i.e.:
$$
\omega(d\_X(x,z))\leq d\_Y(f(x),f(z)) \leq \Omega(d\_X(x,z)),
$$
where $\omega\leq \Omega$ are both strictly increasing continuous functions mapping $[0,\infty)$ to itself and which fix $0$.
Is $f$ a quasisymmetry? I.e.: does there exist a monotone function $\eta:[0,\infty)\rightarrow [0,\infty)$ satisfying
$$
\frac{d\_Y(f(x),f(y))}{d\_{Y}(f(x),f(z))} \leq \eta\left(\frac{d\_X(x,y)}{d\_X(x,z)}\right)
.
$$
**Note on Edit:** *I have reduced my previous question down to this more general one; since quasisymmetries preserve the doubling property.*
| https://mathoverflow.net/users/469470 | When are uniform embeddings quasisymetric | This is false. For example take $f: [0,\infty)\to [0,\infty)$ given by
$$
f(x)=\begin{cases}0& \text{ if } x=0\\
e^{-1/x}& \text{ if } x>0
\end{cases}$$
Then $\lim\_{x\to 0}\frac{f(2x)}{f(x)}=\infty$ so $f$ is not quasisymmetric. But $f$ is continuous and monotone hence gives a homeomorphism onto the image when restricted to $[0,1]$. Since $[0,1]$ is compact $f$ is uniformly continuous on it. Same for $f^{-1}$.
Let me now answer the original question (which the OP erased) which asked for a weaker conclusion which is still false.
The question was the following.
Let $f: X\to Y$ be a bijection such that both $f$ and $f^{-1}$ are uniformly continuous. Suppose $X$ is doubling i.e. there exists a constant $C>$ such any ball of radius $r$ can be covered by at most $C$ balls of radius $r/2$.
Question: Does this imply that $Y$ is also doubling?
The answer is NO.
The answer is yes if $f$ is bi-Lipschitz in which case the statement is trivial but any weaker conditions on the modulus of continuity should not be sufficient.
Here is an example of a metric of the closed $2$-disk which is not doubling.
Take a 2-disk $\bar D=\{x^2+y^2\le 1, z=0\}$ in the $xy$ plane in $\mathbb R^3$ and attach 10 vertical intervals of length 2 at different points of the interior of the disk. The doubling constant of the resulting space $\tilde X\_1$ on scale 1 will be about 10.
The space $\tilde X\_1$ is no longer a disk but we can make it one by replacing the vertical intervals by very thin "fingers" (graphs of bump functions on tiny disks) of the same height 2 so that the the new space $X\_1$ is Hausdorff close to $\tilde X\_1$ and is a graph of a function from $\bar D$ to $\mathbb R$.
Now take a very small $\epsilon$, take small disk $D\_\epsilon$ in $D$ if radius $\epsilon$ away from the fingers and repeat the procedure on $D\_\epsilon$ on scale $\epsilon$ by attaching 100 vertical intervals of length $2\epsilon$. Replace the smaller intervals by fingers. then the doubling constant on scale $\epsilon$ will be 100. Iterate. The limit space $X\_\infty$ will be a graph of a function on $\bar D$ but will not be doubling. The projection map $f: X\_\infty\to \bar D$ is a homeomorphism. It's 1-Lipschitz. The inverse is uniformly continuous because the spaces involved are compact.
| 1 | https://mathoverflow.net/users/18050 | 419190 | 170,641 |
https://mathoverflow.net/questions/419158 | 4 | I was trying to solve the following problem:
Let $f: D \longrightarrow D$ be proper holomorphic (so that means it is a Blaschke product with finitely many factors). Suppose $\{ a\_1, ..., a\_n \} \subset D$ be the set of branch points of $f$ and the ramification degree of $a\_j$ is $m\_j \in [2, \infty) \cap \mathbb{N}$ (by this I mean there exists $\alpha\_j \in D$ such that $f( \alpha\_j) = a\_j$ and $\alpha\_j$ is a ramification point of degree $m\_j$).
The only progress I have made so far is that I can extend this map to be a holomorphic map from the Riemann sphere to itself. Then I can get a lower bound on the degree of the rational map using the Riemann-Hurwitz formula.
I would like to know how to construct such a function and if this problem has connections to any other phenomenon in (algebraic) geometry. Thanks in advance.
| https://mathoverflow.net/users/143311 | Constructing proper holomorphic self-mappings of the unit disk with a given set of branch points and corresponding ramification degrees | This (existence and uniqueness) is proved in the paper in much more general setting (in fact, the result is due to E. Picard):
M. Heins, ‘On a class of conformal metrics’, Nagoya Math. J. 21 (1962) 1–50.
For a more recent and simpler proof of the special case that you ask, see
MR1479037 (99d:30009)
Zakeri, Saeed
On critical points of proper holomorphic maps on the unit disk.
Bull. London Math. Soc. 30 (1998), no. 1, 62–66.
| 7 | https://mathoverflow.net/users/25510 | 419194 | 170,642 |
https://mathoverflow.net/questions/419195 | 2 | Consider two dg-algebras $A,B$ and their respective derived categories $D(A),D(B)$. A natural way to give a covariant functor is to take an $(A,B)$-bimodule $X$ and to tensor with it, that is
$$D(A)\to D(B), \quad M\mapsto M\otimes^{\mathbb{L}}\_A X.$$
However it is "well known" (according to Toen) that there exist triangulated functos which are not given by the tensor product with an $(A,B)$-bimodule. However, in [this paper](https://arxiv.org/abs/math/0408337v7), he shows that if we take into account the dg-category structure on $D(A), D(B)$, then the mapping space between the two of them admits a similar description: consider the full subcategory of $Ho((A,B)\text{-modules})$ of right quasi-representable objects, which he calls $\mathcal{F}(D(A),D(B))$. Then we have a weak equivalence of simplicial sets
$$Map(D(A),D(B))\cong N\left(\mathcal{F}(D(A),D(B)\right)$$
which is also true by replacing $D(A),D(B)$ by more general dg-categories. This seems like a really cool result to me, but I don't think I quite understand it. If I understand it correctly, then by looking at $\pi\_0$ of these simplicial sets, what I get is that covariant functors $D(A)\to D(B)$ up to natural equivalence are in bijection with $\pi\_0\left[N\left(\mathcal{F}(D(A),D(B)\right)\right]$. Naively you'd hope that this latter set corresponds to "nice" objects in the derived category of $(A,B)$-modules, but that seems a bit to nice.... I guess my first question is therefore how can I use this in "practice", i.e. if I run into a functor in the "wild" that "looks like" it could be given by a tensor product of a complex of bimodules, how well can I "approximate" it by tensor product of a complex of bimodules?
My other question is how this translates to the contravariant functor case. Can I just equally approximate contravariant functors by $\text{Hom}(-,X)$?
| https://mathoverflow.net/users/152554 | "Approximating" functors by Hom/Tensor product | I'm not sure how to phrase it in this language, so this answer is more of a comment (but too long for comments)
Note that it is related to your question via the slogan "dg-categories are the same as $\mathbb Z$-linear stable $\infty$-categories".
Taking this slogan for granted, I will answer your question in the setting of stable $\infty$-categories.
Given $A,B$ ring spectra (I'm thinking here of dg algebras as ring spectra, by viewing them as generalized Eilenberg-MacLane spectra), then the $\infty$-category of (homotopy) colimit preserving functors $Mod\_A\to Mod\_B$ (\*) is equivalent to the $\infty$-category of $(A,B)$-bimodules. This is a stable version of the Eilenberg-Watts theorem (or yet again a stable version of Morita theory)
(\*) $Mod\_A$ denotes the $\infty$-category of $A$-module spectra; by a theorem of Shipley, if $A$ is a dga then this is equivalent to the $\infty$-category of dg-$A$-modules with quasi-isomorphisms inverted
The key observation here is that $(A,B)$-bimodules are the same (with your convention that "modules" means "right modules") as $B\otimes\_\mathbb S A^{op}$-modules : note that the tensor product is taken over the sphere spectrum, i.e. it's really a smash product, even if $A,B$ were initially dgas. This is why you're missing some functors if you only look at $B\otimes^L\_\mathbb Z A^{op}$-modules.
If you want something that is classified by "ordinary derived" $(A,B)$-bimodules, i.e. $B\otimes^L\_\mathbb Z A^{op}$-modules (more generally $B\otimes^L\_k A^{op}$ if $A,B$ are dg-$k$-algebras), you have to look at $D(\mathbb Z)(\simeq Mod\_\mathbb Z$)-linear colimit-preserving functors $D(A)\to D(B)$, which, by the above slogan, should be the same as dg-functors.
Note that in any case, you do not get all functors, because $-\otimes\_A^L X$ preserves colimits, so this is the best you can get. If you run into an additive ( $\infty$-)functor $F: D(A)\to D(B)$ in the wild, you can universally approximate it by a colimit-preserving functor $- \otimes\_A F(A) \to F$, where $F(A)$ is an $(A,B)$-bimodule using the functoriality of $F$ and the right $A$-linear maps $a\cdot - : A\to A$ (you can be more precise in this description - this is where the additivity is needed).
This morphism will be an equivalence if and only if $F$ preserves colimits. If it preserves finite (homotopy) colimits, then this morphism is an equivalence on all perfect $A$-modules.
For contravariant functors, the story is indeed similar, except that now if you want something that has to do with $\hom(-,X)$, you will want functors $D(A)^{op} \to D(B)$ which preserve limits (i.e. send colimits in $D(A)$ to limits in $D(B)$). Let me sketch it below, with details missing.
With this in mind, such a functor is the same as a finite limit preserving functor $Perf(A)^{op} \to D(B)$, and $Perf(A)^{op} \simeq Perf(A^{op})$, so this is the same thing as a colimit preserving functor $D(A^{op}) \to D(B)$ (here I'm using that $Perf(A)$ and $D(B)$ are stable so preserving finite limits is equivalent to preserving finite colimits).
This is now an $(A^{op},B)$-bimodule, and the corresponds works as follows : given such a module $M$, restrict $-\otimes\_{A^{op}}M$ to perfect modules, use the equivalence $\hom\_A(-,A)$ between $A$ and $A^{op}$-modules to obtain that the corresponding functor $Perf(A)^{op}\to D(B)$ is $\hom\_A(-,M)$. But this is clearly the restriction of the limit preserving functor $\hom\_A(-,M) : D(A)^{op} \to D(B)$, so this is the corresponding functor ($\hom\_A$ here, as everything in this answer, is derived). Again, if you want an ordinary (derived) bimodule, you'll need to impose some form of $\mathbb Z$-(or $k$-)linearity.
As before, you can also approximate a functor that you meet "in the wild" by such a thing, this time in the other direction though.
| 3 | https://mathoverflow.net/users/102343 | 419198 | 170,644 |
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