parent_url
stringlengths 37
41
| parent_score
stringlengths 1
3
| parent_body
stringlengths 19
30.2k
| parent_user
stringlengths 32
37
| parent_title
stringlengths 15
248
| body
stringlengths 8
29.9k
| score
stringlengths 1
3
| user
stringlengths 32
37
| answer_id
stringlengths 2
6
| __index_level_0__
int64 1
182k
|
|---|---|---|---|---|---|---|---|---|---|
https://mathoverflow.net/questions/430864 | 2 | At the Wikipedia there are the differential formulation for [Euler-Bernoulli Beam](https://en.wikipedia.org/wiki/Euler%E2%80%93Bernoulli_beam_theory) \eqref{1} and [Timoshenko Beam](https://en.wikipedia.org/wiki/Timoshenko%E2%80%93Ehrenfest_beam_theory) \eqref{2}
$$
\begin{align}
&\dfrac{d^2}{dx^2}\left(EI\dfrac{d^2w}{dx^2}\right) = q(x) \label{1}\tag{1}\\
&\begin{cases}\dfrac{d^2}{dx^2}\left(EI\dfrac{d\varphi}{dx}\right) = q(x) \\
\dfrac{dw}{dx} = \varphi - \dfrac{1}{\kappa AG} \cdot \dfrac{d}{dx}\left(EI\dfrac{d\varphi}{dx}\right)
\end{cases} \label{2}\tag{2}
\end{align}
$$
Both of formulations supposes that the undeformed beam is at $(\vec{e}\_x)$ direction, the distributed charge $q$ is at $(\vec{e}\_z)$ direction and $w$ is the displacement at $(\vec{e}\_z)$ direction.
The deduction uses, for example that
$$
Q = \dfrac{dM}{dx}
$$
But the shear force $\vec{Q}$ and the momentum $\vec{M}$ are not colinear:
$$
\vec{Q} = Q \cdot \vec{e}\_{z} = \dfrac{dM}{dx} \cdot \vec{e}\_z \ne \dfrac{dM}{dx} \cdot \vec{e}\_{y} = \dfrac{d}{dx}\left(M \cdot \vec{e}\_{y}\right) = \dfrac{d}{dx} \vec{M}
$$
**Question:** Then, is there a formulation $(1)$ and $(2)$ using vectorial notation? Like for example
$$
\vec{Q} = \nabla \times \vec{M}
$$
Cause
$$
\vec{Q} =
\begin{bmatrix}
0 \\ 0 \\ Q
\end{bmatrix}=
\begin{bmatrix}
\dfrac{-dM}{dz} \\ 0 \\ \dfrac{dM}{dx}
\end{bmatrix} =
\det \begin{bmatrix}
\vec{e}\_x & \vec{e}\_y & \vec{e}\_z \\
\dfrac{d}{dx} & \dfrac{d}{dy} & \dfrac{d}{dz} \\
0 & M & 0
\end{bmatrix}
=
\begin{bmatrix}
\dfrac{d}{dx} \\ \dfrac{d}{dy} \\ \dfrac{d}{dz}
\end{bmatrix} \times
\begin{bmatrix}
0 \\ M \\ 0
\end{bmatrix}
$$
**Motivation:** I want an analytic model for a 3D beam which neutral line follows an arbitrary path $p(t) \in \mathbb{R}^{3}$. When I tried to get it, I could not use the scalar rotations cause the vectors' directions were not the same and I should use rotations.
| https://mathoverflow.net/users/173662 | Mechanics: Model beam using differential vectorial formulation | The beam equations \eqref{1} and \eqref{2} "admit a vector formulation" in the sense that they can be rigorously deduced from the 3D theory of (nonlinear) elasticity. I'm not aware of a precise reference dealing specifically with those two examples, but the *Encyclopedia of Physics* entry [1] by Antman (which also alludes briefly to the deduction of \eqref{1} in §26, example $\delta$, page 698) presents two different classes of methods for solving the problem, namely projection methods ([1], §11, pp. 660-663) and the asymptotic method ([1], §13, pp. 664-665). Antman gives a fairly complete (even if somewhat dated) survey of the theories of beams, rods and all continua with 1D behavior, considering also the history of the subject. Well, my two cents.
**Edit after the downvote**. After I saw the OP and my answer downvoted, I decided to add a few words of explanation. According to Stuart Antman ([1], §1, p. 641),
>
> A *theory of rods*2 or, equivalently a *one-dimensional theory of solids* is a characterization of the behavior of slender three-dimensional solid bodies by a set of equations having the parameter of a curve and the time as the only independent variables.
>
> 2We use "rod" as a generic name for "arch", "bar", "beam", "column", "ring", "shaft", etc. We employ rod both in the intuitive sense of a slender body and in several precise mathematical senses. The meaning will be clearer from the context.
>
>
>
Said that, by carefully reading the question and its *motivation*, one sees that the techniques shown in [1] exactly suit the needs of the Asker: his search for a *particular* 3D equation for the models \eqref{1} and \eqref{2} is probably useless since it would be unsuitable for the study of a non-rectilinear beam, while the general procedures described in [1] where the equations describing a 1D medium are deduced from the standard, general 3D equations by applying them to *particular* constitutive equations, are applicable for beams described by a general spatial curve.
**Reference**
[1] Stuart S. Antman, "The Theory of Rods", in Flügge, S. & Truesdell, C. (Eds.) *Festkörpermechanik/Mechanic of Solids*, *Handbuch der Physik/Encyclopedia of Physics*, Vol. VI part 2, Springer-Verlag, 1972, 641-703.
| 3 | https://mathoverflow.net/users/113756 | 430955 | 174,521 |
https://mathoverflow.net/questions/429953 | 5 | We are given a convex shape $S$ in the $d$-dimensional Euclidean space, whose boundary is formed by portions of $2d$ different spheres, one portion per sphere. The radius of each sphere is the same, $a+1$, and the spheres centers are $\pm a\textbf{e}\_j$ for $1\le j\le d$, where $a\ge 0$. Hence, for $a>0$ each sphere passes through one and only one of the $2d$ points in the set $P:=\{\pm\textbf{e}\_j | 1\le j\le d\}$.
Finally, for each point $\mathbf{p}\in P$, the portion of the sphere $\mathcal{S}\_{\mathbf{p}}$ passing through $\mathbf{p}$ consists of all points of $\mathcal{S}\_{\mathbf{p}}$ whose distance from $\mathbf{p}$ is smaller or equal to the distance between $\mathbf{p}$ and any point of the other $2d-1$ spheres.
---
**Questions:** How can we provide a tight lower bound of $a$ in terms of $d$ to have that $\frac{V(S)}{2^d}$ is a constant bounded away from $0$ when $d\to\infty$?
| https://mathoverflow.net/users/115803 | Volume of a shape whose boundary consists of portions of spheres symmetrically placed about the origin in $d\gg 1$ dimensions | If we set $a = \beta d^2$ for $\beta > 0$ then $$\frac{V(S)}{2^d} \to \exp(-1/(6\beta)).$$
The idea is to choose $x \in [-1,1]^d$ uniformly at random. Then $\frac{V(S)}{2^d}$ is just the probability that $x \in S$.
Let $S\_j^+$ (respectively $S\_j^-$) denote the ball centered at $e\_j a$ (respectively $-e\_j a$) of radius $a+1$. For a point $x \in [-1,1]^d$, note that $x \in S\_j^+$ if and only if $$(x\_j - a)^2 + \sum\_{i \neq j} x\_i^2 \leq (a + 1)^2$$
i.e. if and only if $$-2x\_j a + \|x \|\_2^2 \leq 2a + 1\,.$$
Arguing the same for $S\_j^{-}$ we have that $x \in S\_j^+ \cap S\_j^-$ if and only if $$ \|x\|\_2^2 \leq 2a(1 - |x\_j|) + 1\,.$$
This implies that $x \in S = \bigcap\_j (S\_j^+ \cap S\_j^-)$ if and only if $$\|x \|\_2^2 \leq 2a (1 - \|x \|\_\infty) + 1$$
i.e. if $$\frac{\|x\|\_2^2}{d} \leq 2\frac{a}{d^2} \left (d (1 - \|x \|\_{\infty}) \right) + \frac{1}{d}.$$
As $d \to \infty$ we have $\|x\|\_2^2/d \to 1/3$, while $d(1 - \|x \|\_\infty)$ converges to an exponential random variable $Y$ of mean $1$. Thus we have $$\frac{V(S)}{2^d} \to P(1/3 \leq 2 \beta Y) = \exp(-1/(6\beta)).$$
| 4 | https://mathoverflow.net/users/69870 | 430958 | 174,522 |
https://mathoverflow.net/questions/430474 | 0 | Let $R$ be a non commutative ring. We will say that an element of $R$ is isolated if it is zero divisor and nothing nonzero annihilates it at the same time on both sides.
Note that there are many classes of rings that do not contain isolated elements.
$\bullet$ If $R$ is an integral ring, it does not contain isolated elements.
$\bullet$ If $R$ is a nilpotent ring, it does not contain isolated elements.
$\bullet$ If $R$ is a reversible ring, it does not contain isolated elements.
$\bullet$ If $R$ is a finite unital ring, it does not contain isolated elements.
Do you know of any indecomposable ring that has no isolated elements and is neither reversible, nor integral, nor nilpotent, nor finite unital?
| https://mathoverflow.net/users/168671 | Do you know of any indecomposable ring that has no isolated elements and is neither reversible, nor integral, nor nilpotent, nor unitary? | Let $F$ be a field, and let
$$
R:=F\langle x,y\, :\, x^2=xy=y^2=0\rangle.
$$
Notice that $\{1,x,y,yx\}$ is an $F$-basis for $R$ as an $F$-vector space.
This ring is not reversible since $xy=0$ but $yx\neq 0$.
If by "integral" you mean a domain, then this ring is clearly not a domain (since it isn't reversible).
It is not nilpotent, since it contains $1\neq 0$.
It is not finite if $F$ is an infinite field.
This ring is indecomposable, as the only idempotents are the trivial ones. (To see this, first note that the constant term of an idempotent $e$ must be idempotent, as this is a graded ring, graded by degree. After replacing $e$ by $1-e$ if necessary, we may assume the constant term is $0$. But then $e$ is nilpotent, and the only nilpotent idempotent is $0$.)
Every element in this ring without $1$ in its support is annihilated, from both the left and the right, by $yx$. As no element with $1$ in its support is a zero-divisor, there are no isolated elements.
| 4 | https://mathoverflow.net/users/3199 | 430959 | 174,523 |
https://mathoverflow.net/questions/430961 | 2 | Let $X := \mathbb{P}^1$, $S\subset X$ a finite set of points, $U := X - S$, and $j : U\rightarrow X$ the inclusion.
Let $F$ be a complex local system on $U$ of rank $r$, and let $F\_0$ be a typical fiber, so $F\_0$ is a complex vector space. On p14 of Katz's book *Rigid local systems*, he says that
$$\chi(X,j\_\*F) = r\cdot\chi(U,\mathbb{C}) + \sum\_{s\in S}\text{dim}\_{\mathbb{C}}F\_0^{I(s)}$$
where $I(s)\cong\mathbb{Z}$ denotes the local monodromy group at $s$ (the fundamental group of a punctured neighborhood of $s$).
Why is this true? This is a very naive question, and I'm clearly just missing some basic points about the cohomology of local systems. I'm hoping someone here can help fill me in.
In the simple case $S = \{s\}$, let $D$ be a small neighborhood of $s$, $D^\* := D - s$, then Mayer Vietoris would give:
$$\chi(X,j\_\*F) = \chi(U,F) + \chi(D,j\_\*F) - \chi(D^\*,F)$$
Since $h^0(D,j\_\*F) = h^0(D^\*,F) = F\_0^{I(s)}$, we have
$$\chi(X,j\_\*F) = \chi(U,F) - h^1(D,j\_\*F) + h^1(D^\*,F) + h^2(D,j\_\*F) - h^2(D^\*,F)$$
1. I think we should have $\chi(U,F) = r\cdot \chi(U,\mathbb{C})$ (why?).
2. Assuming that local system cohomology is homotopy invariant, we should have $h^2(D^\*,F) = 0$.
3. By comparison with group cohomology, we have $h^1(D^\*,F) = r$.
Thus the desired result would follow in this simple case as long as $\dim\_{\mathbb{C}}F\_0^{I(s)} = h^2(D,j\_\*F) - h^1(D,j\_\*F)$. Is this true?
| https://mathoverflow.net/users/88840 | Formula for the Euler characteristic of a local system on $\mathbb{P}^1$ | The answer to your question at the end is negative. In fact, $h^2(D, j\_\*F)= h^2(D, j\_\* F)=0$. In fact, the cohomology of a sufficiently small disc around a point in any complex variety, with coefficients in any fixed constructible sheaf, vanishes in all positive degrees.
This is because taking global sections on a small disc around a point is the same as taking the stalk at that point and is an exact functor.
The error is in your step 3. In fact group cohomology shows $ h^1(D\*, F)= h^0(D^\*,F) = \dim\_{\mathbb C} F\_0^{I\_S}$, since the group cohomology $H^1(\mathbb Z, M)$ is the coinvariants of $M$ and thus has the same dimension as the invariants of $M$ for finite-dimensional $M$.
| 5 | https://mathoverflow.net/users/18060 | 430962 | 174,524 |
https://mathoverflow.net/questions/430609 | 7 | Let $a\_n$ is a *binomial sum*, for example
$$
a\_n := \sum\_{k} \binom{n-k}{k} \quad \text{or} \quad \sum\_{k=0}^n\binom{n+k}{n-k}\binom{2k}{k}
\quad \text{or} \quad \sum\_{k=0}^n\sum\_{\ell=0}^k\binom{n}{k}\binom{n+k}{k}\binom{k}{\ell}^3
$$
These are [Fibonacci](https://oeis.org/A000045), [Delannoy](https://oeis.org/A001850) and [Apéry numbers](https://oeis.org/A005259). More generally, we can consider
the sequence $\{a\_n\}$ obtained as a *diagonal* of a rational function over $\Bbb Z$ (so in particular it is *P-recursive*), see e.g. Stanley, "Enumerative Combinatorics", Chapter 6.
**Question:** Given $m\in \Bbb N$ and a binomial sum $\{a\_n\}$, is it decidable whether $m$ divides $a\_n$ for all $n\in \Bbb N$?
This might seem a bit naive, and I supect this should be true, but couldn't find the reference. Note the [Skolem's Problem](https://en.wikipedia.org/wiki/Skolem_problem) which is different but has a similar flavor (and not known to be decidable).
| https://mathoverflow.net/users/4040 | Congruences of binomial sums | I found the reference: Theorem 9.1 in Boris Adamczewski and Jason P. Bell, [Diagonalization and Rationalization of algebraic Laurent series](https://arxiv.org/abs/1205.4090), *Ann. Sci. Éc. Norm. Supér.* **46** (2013), 963–1004.
The authors even discuss the *Apéry sequence* modulo 5, right after the theorem.
It's sort of amazing when the reference request works out so well. I thank Boris Adamczewski and Jason Bell for telling me about this remarkable work.
| 4 | https://mathoverflow.net/users/4040 | 430966 | 174,527 |
https://mathoverflow.net/questions/430971 | -1 | If $0<\beta<1$ and $0<x<1,$ how to prove that $$h(x)-2x+(4-2^{1+\beta})x^{1+\beta}<0,$$ where $$h(x)=(1+x)^{1+\beta}-x^{1+\beta}-1.$$The numerical simulation shows that it is true.
| https://mathoverflow.net/users/484832 | A proof of an interesting inequality | $\newcommand\be\beta$We have to show that
$$g(x):=(1+x)^{1+\be}+(3-2^{1+\be}) x^{1+\be}-2 x-1<0; \tag{1}\label{1}$$
here and in what follows, $\be$ and $x$ are in $(0,1)$.
Let $r(x):=\dfrac x{1+x}$, so that $r$ is increasing on $(0,1)$.
Hence, $$g''(x)=(1+\be)\be x^{\be-1}(r(x)^{1-\be}+3-2^{1+\be})$$
can only switch sign from $-$ to $+$ (if at all) as $x$ increases from $0$ to $1$.
Noting also that $g(0)=g(1)=0$ and $g'(0)=\be-1<0$, we get \eqref{1}.
| 1 | https://mathoverflow.net/users/36721 | 430973 | 174,529 |
https://mathoverflow.net/questions/430997 | 5 | If $G$ is a finite group acting on a finite set $X$, we have Burnside's formula that counts the number of orbits $|X/G|$ as:
$$ |X/G| = \frac1{|G|} \sum\_{g\in G} |X^g|,
$$
with $X^g$ being the set of elements in $X$ fixed by $g$.
Now, consider the finite set $X$ of $\mathbb F\_p$-points of $\mathbb P^1\_{\mathbb F\_p}$, the one dimensional projective space over the finite field $\mathbb F\_p$, with $p$ elements, and the finite set $Y:=X\times X$ (Cartesian product). The 2-element group $G:=\{e,g\}$ acts by permuting the factors on $Y$ ($G$ is the symmetric group on 2 elements, and $e$ its identity), and it is well-known that (it follows, for example, from the Zeta function computation in <https://math.stackexchange.com/q/799101>):
$$ Y/G = Sym^2(\mathbb P^1\_{\mathbb F\_p}) = \mathbb P^2\_{\mathbb F\_p}.
$$
As such, we have $|Y/G|=|\mathbb P^2\_{\mathbb F\_p}|=p^2+p+1$.
On the other hand, one would say that $|Y^e|=|Y|=|X|^2=(p+1)^2$, and $|Y^g|=|X|=p+1$, since the elements fixed under the permutation are in the diagonal of $Y=X\times X$, isomorphic to $X$, so Burnside's formula gives, apparently, the wrong answer:
$$ |Y/G|=\frac12((p+1)^2+(p+1))=\frac{p^2+3p+2}{2}.$$
Maybe I am using the wrong notion of $\mathbb F\_p$-points of a variety (/scheme?), but I believe there should be a simple explanation of why the computations do not match. Any help is appreciated.
| https://mathoverflow.net/users/409881 | How to make Burnside's formula compatible with point counting for varieties over finite fields? | The issue is that $X(\mathbb{F}\_p)/G$ is not the same thing as $(X/G)(\mathbb{F}\_p)$. A simpler example is to take $p$ odd, $X = \mathbb{A}^1$ and let $S\_2$ act by $\pm 1$. There are $\tfrac{p+1}{2}$ orbits, but the quotient space is $\mathbb{A}^1$ with the quotient map $x \mapsto x^2$. The quotient space has $p$ $\mathbb{F}\_p$-points, but only $\tfrac{p+1}{2}$ of them (the squares) are in the image of $X(\mathbb{F}\_p)$.
Points of $(X/G)(\mathbb{F}\_p)$ index Frobenius stable $G$-orbits, so the other $\tfrac{p-1}{2}$ points correspond to the orbits of the form $\{ \pm x \}$ where $x^p = -x$ (other than the point $x=0$).
If you want to count $(X/G)(\mathbb{F}\_p)$, there is a combined Burnside/Lefschetz formula. Choose $\ell$ relatively prime to $p$ and $|G|$, then a formula of Grothendieck tells us that $H^j(X/G, \mathbb{Q}\_{\ell}) \cong H^j(X, \mathbb{Q}\_{\ell})^G$, and this isomorphism is Frobenius equivariant. So the trace of Frobenius on $H^j(X/G, \mathbb{Q}\_{\ell})$ is the same as the trace of Frobenius restricted to the subspace $H^j(X, \mathbb{Q}\_{\ell})^G$. Now, the linear operator $\tfrac{1}{|G|} \sum\_{g \in G} g$ on $H^j(X, \mathbb{Q}\_{\ell})$ is an idempotent whose image is $H^j(X, \mathbb{Q}\_{\ell})^G$. So the trace of Frobenius restricted to $H^j(X, \mathbb{Q}\_{\ell})^G$ is the same as the trace of $\tfrac{1}{|G|} \sum\_{g \in G} \text{Frob} \circ g$. So we get
$$\#((X/G)(\mathbb{F}\_p)) = \tfrac{1}{|G|} \sum\_{g \in G} \sum\_j (-1)^j \text{Tr}{\big(}\text{Frob} \circ g : H^j(X, \mathbb{Q}\_{\ell}) \longrightarrow H^j(X, \mathbb{Q}\_{\ell}){\big)}.$$
I am not sure that there is a Burnside/Lefschetz style formula for $\#X(\mathbb{F}\_p)/G$. Here is a troubling example: Take $p$ odd, $X = \mathbb{P}^1$ and let $S\_2$ act by $[x:y] \mapsto [cy : x]$. Then $X(\mathbb{F}\_p)$ has $p+1$ points. If $c$ is a quadratic residue, then there are two fixed points for $S\_2$, namely $[\pm \sqrt{c}:1]$, otherwise there are none. So there are either $\tfrac{p+1}{2}$ or $\tfrac{p-1}{2}$ orbits for $S\_2$ on $X(\mathbb{F}\_p)$ depending on whether or not $c$ is a quadratic residue. I find it hard to imagine a Burnside/Lefschetz style formula which could take this information into account.
---
It might help to say that there is a purely topological version of this question. Let $G$ be a finite group and let $X$ be a compact topological space with an action of $G$ and also with an endomorphism $\phi$ that commutes with the $G$-action. (More generally, we could imagine that there is an automorphism $\sigma$ of $G$ with $\phi \circ g = \sigma(g) \circ \phi$.)
Then the two questions are how to count $(X/G)^{\phi}$ and how to count $(X^{\phi})/G$. These aren't the same thing: If there is a $G$-orbit which $\phi$ permutes nontrivially, then it will contribute to the first count but not the second.
| 8 | https://mathoverflow.net/users/297 | 430999 | 174,538 |
https://mathoverflow.net/questions/430724 | 3 | $\DeclareMathOperator\ex{ex}$We write $K\_{2,\dots,2}^{(r)}$ to denote the $r$-uniform hypergraph with vertex set $\{1,2\}\times\{1,\dots,r\}$ and hyperedge set $\{(1,1),(1,2)\}\times \{(2,1),(2,2)\} \dots \times \{(r,1),(r,2)\}$.
For integer $n$, the Turan number $\ex(n,K\_{2,\dots,2}^{(r)})$ denotes the maximum number of hyperedges in a $n$-vertex $r$-uniform hypergraph $H$ which does not have $K\_{2,\dots,2}^{(r)}$ as a subhypergraph.
Reading some papers from the 90's, I gather that for $r\ge 3$, the bounds
$$n^{r-r/(2^r-1)}\ll \ex(n,K\_{2,\dots,2}^{(r)}) \le n^{r-1/2^{r-1}}$$were known. I believe the lower bound is by the probabilistic deletion method, while the upper bound is by induction.
Does this remain the state of the art? I'm particularly curious about $r=3$.
| https://mathoverflow.net/users/130484 | Bounds for $\mathrm{ex}(n,K_{2,\dots,2}^{(r)})$ | A recent paper by Conlon, Pohoata and Zakharov provides new lower bounds and a survey on the history of the problem.
It seems that the upper bound is the state of the art.
Now we know that for every $r \geq 2$ there are lower bounds better than the one obtained from probabilistic deletion.
In particular for $r =3$ it holds that $ex(n, K^{(3)}\_{2,2,2}) = \Omega(n^{8/3})$ (which was in fact already proven by Katz, Krop and Maggioni before).
*Conlon, David; Pohoata, Cosmin; Zakharov, Dmitriy*, Random multilinear maps and the Erdős box problem, Discrete Anal. 2021, Paper No. 17, 8 p. (2021). [ZBL1482.05167](https://zbmath.org/?q=an:1482.05167).
*Katz, Nets Hawk; Krop, Elliot; Maggioni, Mauro*, [**Remarks on the box problem**](http://dx.doi.org/10.4310/MRL.2002.v9.n4.a11), Math. Res. Lett. 9, No. 4, 515-519 (2002). [ZBL1031.42018](https://zbmath.org/?q=an:1031.42018).
| 2 | https://mathoverflow.net/users/45545 | 431001 | 174,539 |
https://mathoverflow.net/questions/430741 | 16 | Let $F\_i$ denote the $i$th Fibonacci number (with $F\_1=F\_2=1$). Define
$$ P\_n(x) = \prod\_{i=1}^n (1+x^{F\_{i+1}}). $$
Let $\nu\_k(n)$ denote the number of coefficients of the polynomial $P\_n(x)$
that are equal to the positive integer $k$. Evidence suggests that for
sufficiently large $n$ (depending on $k$), $\nu\_k(n)$ is a linear
polynomial in $n$. These polynomials for $1\leq k\leq 12$ are
empirically given by $2n$, $4n-8$, $8n-32$, $12n-68$, $16n-112$,
$24n-192$, $24n-224$, $36n-352$, $40n-432$, $48n-544$, $40n-512$, and
$88n-1056$. How can one prove these observations and generalize to any
$k$? Similar conjectures can be made for a wide class of more
general polynomials. See my papers <https://arxiv.org/abs/1901.04647>
and <https://arxiv.org/abs/2101.02131> for some related results. The
linear algebraic techniques in these papers might be applicable to
the present problem.
**Addendum.** The case $k=1$ follows easily from properties of the Fibonacci triangle poset discussed in Section 3 of <https://arxiv.org/abs/2101.02131>.
| https://mathoverflow.net/users/2807 | Number of coefficients equal to $k$ in certain "Fibonacci polynomials" | I will use the set up of [my answer](https://mathoverflow.net/q/266215) to a previous question about "Fibonacci polynomials".
The key observation is that the coefficient of $x^m$ in $P(x)$ equals the number of "unrollings" of the Zeckendorf representation $Z\_m$ (viewed as a $01$-string) of $m$, where any substring $001$ can be unrolled into $110$. Let $U(Z\_m)$ denote the number of such unrollings. It is easy to see that
$$U(0^p1v) = U(v) + \left\lfloor\frac{p}2\right\rfloor U(0v).$$
Then one can notice that
$$\nu\_k(n) = \nu'\_k(n) + \nu'\_k(n-1) - [k=1],$$ where $\nu'\_k(n)$ is the number of coefficients of $x^m$ in $P(x)$ equal $k$ with $m<F\_{n+2}$ (ie. $|Z\_m|\leq n$). Here the [Iverson bracket](https://en.wikipedia.org/wiki/Iverson_bracket) $[k=1]$ eliminates double counting of $v=0^n$ and $v=0^{n-1}$, which both correspond to $m=0$ .
The aforementioned recurrence for $U$ allows us to obtain one for $\nu'\_k(n)$ as follows.
Let $a\_1(n,k)$ be the number of strings $v$ of length $n$ starting with 0 such that $U(v)=k$, and let $a\_2(n,k\_1,k\_2)$ be the number of strings $v$ of length $n$ starting with 0 such that $U(v)=k\_1$ and $U(0v)=k\_2$. Clearly, $a\_2(n,k\_1,k\_2)=0$ whenever $k\_1>k\_2$ and we find it convenient to assume that both $a\_1$ and $a\_2$ are zero for $n<0$. Then we have
$$\nu'\_k(n) = a\_1(n,k) + a\_1(n-1,k),$$
where the terms enumerate $v$ starting with 0 and 1, respectively. The count $a\_1$ satisfies the recurrence:
\begin{split}
a\_1(n,k) &= [k=1] + a\_1(n-2,k) \\
&+ \sum\_{p=2}^{2k-1} \sum\_{k\_2 = \lceil k/(1+c\_p)\rceil}^{\lfloor (k-1)/c\_p\rfloor} a\_2(n-1-p,k-c\_pk\_2,k\_2),
\end{split}
where $p$ runs over the possible lengths of the initial run of 0's in $v$, the first term corresponds to $v=0^n$ and the second term corresponds to $p=1$, and $c\_p:=\left\lfloor\frac{p}2\right\rfloor$. Finally, the count $a\_2$ satisfies the recurrence:
\begin{split}
&a\_2(n,k\_1,k\_2) = [(k\_1,k\_2)=(1,1)] \\
&+ [k\_1=k\_2]\cdot\sum\_{c=1}^{k\_1-1} \sum\_{t = \lceil k\_1/(1+c)\rceil}^{\lfloor (k\_1-1)/c\rfloor} a\_2(n-1-2c,k\_1-ct,t) \\
& + [k\_1<k\_2]\cdot a\_2(n-1-(2\lfloor \tfrac{k\_1-1}{k\_2-k\_1}\rfloor+1), (k\_1-1) \bmod (k\_2-k\_1) + 1,k\_2-k\_1),
\end{split}
where the terms correspond to $v=0^n$, even $p>0$, and odd $p>0$, respectively.
---
**Example $k=1$.** In this case we have $a(n,1) = 1 + a\_1(n-2,1)$, implying that $a\_1(n,1)=1+\left\lfloor\frac{n}2\right\rfloor$. Then $a\_2(n,0,1) = 0$ and $a\_2(n,1,1) = 1$. It follows that $\nu'\_1(n) = n+1$ and $\nu\_1(n) = 2n$.
**ADDED.** It can be seen that $a\_2(n,k\_1,k\_2)$ stabilizes as $n$ grows and its limit $\bar a\_2(k\_1,k\_2)$ satisfies the recurrence:
\begin{split}
&\bar a\_2(k\_1,k\_2) = [(k\_1,k\_2)=(1,1)] \\
&+ [k\_1=k\_2]\cdot \sum\_{c=1}^{k\_1-1} \sum\_{t = \lceil k\_1/(1+c)\rceil}^{\lfloor (k\_1-1)/c\rfloor} \bar a\_2(k\_1-ct,t) \\
& +
[k\_1<k\_2]\cdot \bar a\_2( (k\_1-1) \bmod (k\_2-k\_1) + 1, k\_2-k\_1).
\end{split}
Then it is easy to find the coefficient $q\_k$ of $n$ in the linear formula for $\nu\_k(n)$ (which is the quadruple of such coefficient in $a\_1(n,k)$) as
$$q\_k = 2[k=1] + 4\sum\_{c=1}^{k-1} \sum\_{k\_2 = \lceil k/(1+c)\rceil}^{\lfloor (k-1)/c\rfloor} \bar a\_2(k-ck\_2,k\_2).$$
Here are the values of $q\_k$ for $k\leq 30$:
```
2, 4, 8, 12, 16, 24, 24, 36, 40, 48, 40, 88, 48, 72, 96, 108, 64, 152, 72, 176, 144, 120, 88, 312, 144, 144, 200, 264, 112, 416
```
There seems to be no easier way to get the free term of $\nu\_k(n)$ than computing it as $\nu\_k(n) - q\_kn$ for large enough $n$.
| 5 | https://mathoverflow.net/users/7076 | 431002 | 174,540 |
https://mathoverflow.net/questions/430929 | 3 | Let $X$ be a connected and compact $d$-dimensional smooth manifold; where $d$ is a positive integer. Does *(or rather, when does)* there exist a metric $\rho$ on $X$ generating $X$'s topology and a countable number of sets $\{X\_n\}\_{n}$ such that
* $\bigcup\_n\,X\_n = X$,
* Each $X\_n$ is $\rho$-geodesically convex,
* Each $X\_n$ is contractable,
* $\mu(X\_n\cap X\_m)=0$ whenever $n\neq m$ *(almost disjoint)*,
* There is a $d$-dimensional Ahfors regular measure on $(X,\rho)$ such that
$$
n\neq m \Rightarrow \rho(X\_n\cap X\_m)=0.
$$
**Comment:** *Ideally, I would like $\{X\_n\}\_n$ to be a finite set; but I'm open to countably infinite also..*
---
**Definition**:
* A measure $\mu$ is Ahfors regular if there are $q,c,C>0$ such that
$$
c\,r^q \le \mu(Ball\_{\rho}(x,r)) \le C\,r^q.
$$
| https://mathoverflow.net/users/36886 | Partitioning a smooth manifold into geodesically convex sets | Choose a triangulation of $X$.
Let us equip $X$ with a length metric such that each simplex is standard.
We may think that $X$ subcomplex of a standard simplex $S$ of large dimension.
Since each face of $S$ is convex; it follows that each simplex is a convex set in $X$.
Therefore the covering of $X$ by the simplexes of maximal dimension meets your requirement.
Let me also mention that from [our result](https://arxiv.org/abs/2103.15189) with Alexander Lytchak it follows that there is no finite covering for generic Riemannian metric on $X$.
Namely, for any covering $\{X\_i\}$ of $X$, we have that $X\_i\cap X\_j$ has nonempty interior for some $i\ne j$;
hence $\mu(X\_i\cap X\_j)>0$.
| 4 | https://mathoverflow.net/users/1441 | 431007 | 174,541 |
https://mathoverflow.net/questions/430953 | 1 | Let $M$ be an $n-m$ dimensional sub-manifold of $\mathbf R^n$ defined by the following set of equations:
\begin{equation}
f\_1(\vec x)=0, \\
\vdots \\
f\_m(\vec x)=0,
\end{equation}
where $\vec x$ are coordinates in $\mathbf R^n$.
Given a point $\vec x\_0$ in $M$, how can the Ricci scalar (calculated from the induced metric on $M$) at $\vec x\_0$ be expressed in terms of the functions $f\_1,...,f\_m$?
| https://mathoverflow.net/users/485792 | Ricci scalar of submanifold of $\mathbf R^n$ | *This is an extended comment answering [Ricci scalar of sub-manifold of $\mathbf R^n$](https://mathoverflow.net/questions/430953/ricci-scalar-of-sub-manifold-of-mathbf-rn?noredirect=1#comment1109048_430953)*
Assuming the $f\_i$ are independent, at every point $x$ their gradients span the orthogonal complement to the tangent space of the manifold.
The second fundamental form is given by the normal projection of the ambient derivative. So if $V,W$ are tangent vector fields to $M$, then at the point $p$ you have that $II\_p(V,W)$ is the projection to the span of $\{\nabla f\_i\}$ of $\partial\_V W$. Using that the $f\_i$ are independent, this means $II\_p(V,W)$ is uniquely determined by the $m$ values
$$ \langle \partial\_V W, \nabla f\_i(p) \rangle $$
(The reconstitution formula, when $\nabla f\_i$ are orthogonal, is just to multiply this against $|\nabla f\_i(p)|^{-2} \nabla f\_i(p)$ and sum. When they are not, there's a matrix inversion involved which makes the formula messy.)
The above inner product, using that $\langle W,\nabla f\_i(p)\rangle = 0$ as $W$ is a tangent and $\nabla f\_i(p)$ is a normal, can be rewritten as
$ \langle W, \partial\_v \nabla f\_i(p)\rangle$ which is nothing more than the Hessian of $f\_i$ evaluated in the direction $V\otimes W$.
---
More about the matrix inversion: the idea is that given any vector in the normal space you can write it as
$$ \sum \alpha\_i \nabla f\_i $$
You wish to reconstruct $\alpha\_i$ from the measuraments
$$ \beta\_j = \langle\sum\_i \alpha\_i \nabla f\_i, \nabla f\_j\rangle = \sum\_i \underbrace{\langle \nabla f\_j , \nabla f\_i \rangle}\_{= A\_{ji}} \alpha\_i $$
Linear independence guarantees that the matrix $A\_{ji}$ is invertible; in fact you have explicit formulas of this in terms of cofactors. These are highly multilinear expressions in terms of $\nabla f$, but do not involve Hessians.
For now, denote by $B\_{ji}$ the inverse matrix to $A\_{ji}$ above. So you can write
$$ II\_p(V,W) = \sum\_{i,j} B\_{ji} \nabla f\_j \nabla^2\_{V,W} f\_i $$
Note finally for the expression that shows up in the Gauss equations
$$ \langle II\_p(V,W), II\_p(X,Y)\rangle = \sum\_{i,j,k,l} \langle B\_{ji} \nabla f\_j \nabla^2\_{V,W} f\_i , B\_{kl} \nabla f\_k \nabla^2\_{X,Y} f\_l\rangle $$
the RHS can be reorganized as
$$ \sum\_{i,j,k,l} B\_{ji} B\_{kl} A\_{jk} \nabla^2\_{V,W} f\_i \nabla^2\_{X,Y} f\_l = \sum\_{i,l} B\_{li} \nabla^2\_{V,W} f\_i \nabla^2\_{X,Y} f\_l$$
So the formula for the Riemann curvature (with all indices lowered) is actually not too bad.
For the Ricci scalar you also need to take the trace, which requires computing the tangential projection using $\nabla f\_i$: you can do this just by subtracting the normal projection from the identity. This can complicate the expression quite a bit.
| 2 | https://mathoverflow.net/users/3948 | 431013 | 174,542 |
https://mathoverflow.net/questions/431015 | 3 | Let $Z$ be an object in a stable (or triangulated/whatever) category $\mathcal C$. I believe it follows from Thomason's theorem (see [The classification of triangulated subcategories](https://doi.org/10.1023/A:1017932514274)) that the triangulated categories generated by
$$Y = Z \oplus \Sigma Z$$
and
$$X = Y \oplus \Sigma Y = Z \oplus \Sigma Z \oplus \Sigma Z \oplus \Sigma^2 Z$$
are the same. This means that $X$ and $Y$ can each be constructed from one one another in finitely many steps, where each step consists of taking the fiber or cofiber of a map between previously-constructed objects. One direction of this is obvious — clearly $X$ can be constructed from $Y$ in this way. The other direction is not so clear to me.
**Question:** Let $Z$ be an object in a stable category $\mathcal C$, and let $Y,X$ be as above. How can one explicitly construct $Y$ from $X$ in finitely many steps, using just fibers and cofibers?
**Notes:**
1. It takes finitely many steps to construct a de/suspension, extension, or direct sum from fibers and cofibers, so one is free to use these as steps as well.
2. The argument from Thomason's theorem doesn't actually rely on $Y$ being of the form $X \oplus \Sigma X$; it's only important that $Y$ represent the zero class in the $K$-theory $G = K\_0(\langle Y \rangle\_{thick})$ of the thick subcategory it generates. The argument then is that $Y$ and $Y \oplus \Sigma Y$ both generate the zero subgroup of $G$, and both generate $\langle Y \rangle\_{thick}$ as thick subcategories (they are "dense" in Thomason's sense) and so by Thomason's theorem, the triangulated categories they generate are the same.
3. In light of (2), it may be better simply to assume that $Y$ represents the zero class in $K$-theory. On the other hand, it's possible that the construction I'm looking for will be more explicit in the case where $Y = X \oplus \Sigma X$, and if so, then I'm interested in having this extra explicit-ness.
| https://mathoverflow.net/users/2362 | How to construct $X \oplus \Sigma X$ from $X \oplus \Sigma X \oplus \Sigma X \oplus \Sigma^2 X$ without splitting an idempotent? | The cofibre of $0\oplus 1\oplus 1\oplus 1$ on $X=Z\oplus \Sigma Z\oplus\Sigma Z\oplus\Sigma^2Z$ is $Z\oplus\Sigma Z=Y$.
| 10 | https://mathoverflow.net/users/10366 | 431018 | 174,544 |
https://mathoverflow.net/questions/410332 | 8 | Let $r(n):=r\_3(\mathbb{F}\_3^n)=\max\{|A|: A \subset \mathbb{F}\_3^n, \ A \text{ is 3-AP-free}\}$.
[Edel](https://link.springer.com/article/10.1023%2FA%3A1027365901231) proved that $r(n)\geq 2.217^n$ for sufficiently large $n$. His proof is by giving a construction of a cap-set $A$ in $\mathbb{F}\_3^{480}$. Then observing that $A^k \subset \mathbb{F}\_3^{480k}$ is also a cap set, that is,
$$r(480k)\geq |A^k|=|A|^k.$$
Is this the best known lower bound? Are there other known approaches to this problem other than construction in low dimension and then using this product argument?
Is this product argument expected to be the best we could expect? That is, do we hope to construct an $A$ such that this argument is tight?
I'd appreciate any references or answers to some of these questions.
Any help on the tags or on how to better ask these questions would be nice also.
| https://mathoverflow.net/users/225950 | Known approaches for the lower bound on cap-set problem | I've just proved a new lower bound of $2.218^n$, in my paper 'New lower bounds for cap sets': <https://arxiv.org/abs/2209.10045>.
My new bound comes from extending Edel's ideas, with better computational methods (including a SAT solver) and introducing a new theoretical construction. I also conjecture that a lower bound of $2.233^n$ is possible, which is explained in section 4 of my paper.
My lower bound comes from a cap set in $\mathbb{F}\_3^{56232}$, although I do also give cap sets in 396, 420 and 462 dimensions which beat Edel (slight correction to your post: Edel's bound comes from a cap set in $\mathbb{F}\_3^{480}$).
| 8 | https://mathoverflow.net/users/490000 | 431023 | 174,547 |
https://mathoverflow.net/questions/430957 | 1 | $\DeclareMathOperator\Ext{Ext}$Let $(R,\mathfrak m)$ be a Noetherian local ring. Let $F,G$ be finitely generated free $R$-modules and $f:F\to G$ be an $R$-linear map such that $f(F)\subseteq \mathfrak m G$. Let $X$ be a finitely generated $R$-module, and let $\alpha : 0\to F \to A\_{\alpha} \to X \to 0$ be a short exact sequence i.e. $[\alpha]\in \Ext^1\_R(X,F)$. We have a following push-out diagram with $[\beta] \in \Ext^1\_R(X,G)$.
$$\require{AMScd}\begin{CD}
\sigma : 0 @>>> F @>>> A\_\alpha @>>> X @>>> 0 \\
@. @VVfV @VVV @| \\
\beta : 0 @>>> G @>>> A\_\beta @>>> X @>>> 0.
\end{CD}$$
My question is: must it be necessarily true that $[\beta] \in \mathfrak m \Ext^1\_R(X,G)$?
Some thoughts: The answer is affirmative if $F\cong G\cong R$. Indeed, in this case, $f:R\to R$ must be given by multiplication by some $x\in R$. Since $f(F)\subseteq \mathfrak m G$, so $x\in \mathfrak m$. Then in $[\beta]=x[\alpha]\in \Ext^1\_R(X,R)$, hence $[\beta]\in x\Ext^1\_R(X,R)\subseteq \mathfrak m \Ext^1\_R(X,G) $.
| https://mathoverflow.net/users/135389 | On image of map $\text{Ext}^1_R(X,F)\to \text{Ext}^1_R(X,G)$ induced by $R$-linear map of free modules $F\to G$ with entries in the maximal ideal | This is true for any $\mathrm{Ext}$-degree (and, in fact, without many hypotheses except that $\mathfrak m$ is finitely generated and $F$ is free).
Let $(x\_1,\dots,x\_n)$ be generators of the maximal ideal $\mathfrak m$. Then there is a surjection $G^n \to \mathfrak m G$ given by
$$(g\_1,\dots,g\_n) \mapsto \sum x\_i g\_i.$$
Because $F$ is free, the map $F \to \mathfrak m G$ lifts to a map $F \to G^n$.
Now $\mathrm{Ext}(X,-)$ respects sums and takes the map multiplication-by-$x$ map $M \to M$ to the multiplication-by-$x$ map on $\mathrm{Ext}(X,M)$ (I'm assuming $R$ is commutative here, because otherwise $\mathrm{Ext}$ doesn't necessarily take values in $R$-modules). Therefore, the composite
$$\mathrm{Ext}(X,G)^n \cong \mathrm{Ext}(X,G^n) \to \mathrm{Ext}(X,\mathfrak m G) \to \mathrm{Ext}(X,G)$$
is given by $(w\_1,\dots,w\_n) \mapsto \sum x\_i w\_i$, taking values in $\mathfrak m \mathrm{Ext}(X,G)$.
As a result, because the map $\mathrm{Ext}(X,F) \to \mathrm{Ext}(X,G)$ factors through this map $\mathrm{Ext}(X,G^n)$, it also factors through $\mathfrak m \mathrm{Ext}(X,G)$.
| 1 | https://mathoverflow.net/users/360 | 431026 | 174,549 |
https://mathoverflow.net/questions/430807 | 3 | Let $G=(V,E)$ be a finite undirected graph which we equip with its usual graph *geodesic distance* $d\_G$ making $(G,d\_G)$ into a metric space; let $1<\#V<\infty$. For a given $1<N< \#V$ *what conditions do I need* on $G$ so that does there exist *disjoint* subsets $V\_1,\dots,V\_N\subseteq V$ such that
* $\biguplus\_{n=1}^N\, V\_n = V$,
* $d\_{(V\_n,E\_n)}(x,y)=d\_{G}(x,y)$ for every $x,y\in V\_n$,
Here $E\_n:=\{(v,w):\,v,w\in V\_n\}$ *denotes the collection of edges connecting any two vertices in the "part" $V\_n$* and where $d\_{(V\_n,E\_n)}$ denotes the graph geodesic defined on the graph $(V\_n,E\_n)$ *(note, for arbitrary choices of $\{V\_n\}\_{n=1}^N$ we always have $d\_{(V\_n,E\_n)}\ge d\_G$)*.
| https://mathoverflow.net/users/36886 | "Geodesic coherent" partition of a graph | [Pilipczuk and Siebertz](https://arxiv.org/abs/1807.03683) proved that every planar graph has such a partition with an even stronger property. Namely, each part $V\_i$ is a geodesic path, and the graph obtained by contracting each part has [treewidth](https://en.wikipedia.org/wiki/Treewidth) at most 8. This result was strengthened by [Dujmović, Joret, Micek, Morin, Ueckerdt, and Wood](https://arxiv.org/abs/1904.04791), who proved that every planar graph is a subgraph of the [strong product](https://en.wikipedia.org/wiki/Strong_product_of_graphs) of a graph of treewidth at most 8 and a path. This theorem is now known as the Planar Graph Product Structure Theorem and has been the key tool in settling several long standing open problems on planar graphs. Similar partitions exist for other graph classes (beyond planar). Determining which graph classes admit a product structure theorem is now a very active research area. As a start, see this
[survey](https://arxiv.org/abs/2001.08860) and the references therein for more information.
**Disclaimer.** I am one of the authors of the above survey.
| 4 | https://mathoverflow.net/users/2233 | 431028 | 174,551 |
https://mathoverflow.net/questions/431031 | 3 |
>
> Does the process of 'constructing the category of presheaves' always/never stabilize? Does it stabilize for some special class of categories?
>
>
>
That is, work in a foundation that allows for multiple levels of 'categorical largeness' and let $\mathcal{C}$ be a category. Recursively define
1. $\mathcal{C}\_0=\mathcal{C}$,
2. $\mathcal{C}\_{\alpha+1}={\bf Set}^{\mathcal{C}\_\alpha^{op}}$,
3. $\mathcal{C}\_\lambda=\varinjlim\_{\alpha<\lambda}\mathcal{C}\_\alpha$.
>
> Does there always/never exist some $\alpha\in O\_n$ such that $$\alpha\leq\beta\implies \mathcal{C}\_\alpha\simeq\mathcal{C}\_\beta?$$
>
>
>
I tried playing around with the case $\mathcal{C}={\bf Set}$ and it seems like we don't have the above property for finite $\alpha$, since ${\bf Set}$ is locally small and ${\bf Set}^{{\bf Set}^{op}}$ is not locally small (natural transformations are proper-class sized collections of sets), and higher finite iterations begin sending proper classes to sets etc. -- I don't know if this process stabilizes at $\omega$ or some higher limit ordinal, however.
Further, if we begin with the terminal category ${\bf 1}$ then ${\bf Set}^{{\bf 1}^{op}}\cong{\bf Set}$ so the above sequence repeats itself shifted back one step, and if we begin with the initial category ${\bf 0}$ then ${\bf Set}^{{\bf 0}^{op}}\cong{\bf 1}$, so the above sequence repeats itself shifted back two steps.
| https://mathoverflow.net/users/92164 | Stabilization of taking presheaf categories | The question is extremely dependent on how size issues are handled and many choice that can be made, so that it is very hard to give a general answer.
If you really work with general presheaves categories then I cannot think of any interesting functors $C\_i \to C\_{i+1}$ that would define an interesting value at limit ordinal. For example the Yoneda embedding is not defined as if $C\_0 \neq \varnothing$, $C\_1$ will be a large category, so $C\_{2}$ will have large hom-sets, and so the Yoneda embeddings $C\_2 \to C\_3$ will not be defined unless "Set" also become bigger as the process go, in which case the process will obviously never stabilize.
Similarly, the covariant functoriality of the presheaf construction as suggested by Peter Lumsdain in the comment is not available either as it is no longer defined when working with large categories so there are no way to use it. The contravariant functoriality of the presheaf construction is always available, but I do not see how to use it to construct a tower.
Taking the colimit to be a disjoint union ( colimit over a discrete category) is not a viable solution either: the presheaf category are always connected categories (they have initial objects and terminal objects for example) while the coproduct are always disconnected, so the value at a limit ordinal and at a successor ordinal will never agree.
The only way I can think of to make sense of this construction, is to replace the category $Set^{C^{op}}$ by the construction $\widehat{C}$ the free co-completion of $C$. When $C$ is small then $\widehat{C} \simeq Set^{C^{op}}$, when $C$ is locally small then $\widehat{C}$ the full subcategory of $Set^{C^{op}}$ of so called "[small presheaves](https://ncatlab.org/nlab/show/small+presheaf)", that is the presheaves that are small colimits of representable. When $C$ has a large hom-set then $\widehat{C}$ is no longer related to $Set^{C^{op}}$, but the advantage of iterating $\widehat{C}$ is that one never gets out of locally small categories.
Here To fix what "small" means, I chose an inaccessible cardinal $\kappa$, and "small" means $\kappa$-small. Interestingly, $\kappa$ doesn't need to be inaccessible, regular is enough.
Once we restrict to small presheaf, then there are at least two interesting way to build such tower. One can use the Yoneda embeddings, or one can as suggested by Peter Lumsdaine, or one can start with some functor $C\_0 \to C\_1$ and then use the co-variant functionality of $\widehat{C}$ to build functor $C\_1 \to C\_2$ etc...
I claim that the process "stops" at the ordinal $\kappa$, in the sense that $C\_\kappa \simeq C\_{\kappa+1}$ and it never stops before. (so if you only induce on small ordinal it never stops).
To be clear, the Yoneda embedding $C \to \widehat{C}$ is never an equivalence (for example, the initial object is never in the essential image), I'm only saying that there will be an equivalence of categories $C\_\kappa \simeq \widehat{C\_\kappa}$ other than the Yoneda embeddings. It is not completely clear to me at this point that all subsequent values of the sequence are also equivalent to $C\_\kappa$ (it is clear for the $\kappa+n$ of course, but I don't quite know what happen at $\kappa+\omega$ yet).
If one use Peter Lumsdaine's version then the process also clearly stops at the ordinal $\kappa$ (and this time really stay at the value it had for $\kappa$). I haven't thought to this specific example enough, but it is likely that one can show that it never stabilize before $\kappa$.
If one really wants to keep the construction $Fun(C^{op}, Set)$ without restricting to small presheaves, and one somehow find a way to make it into an interesting tower (which I don't think is possible), then it will very probably never converge:
Indeed, it is for example easy to show that there is no category $C$ such that $Fun(C,Set) \simeq C$ using (a 2-categorical version of) [Lawvere fix-point theorem](https://ncatlab.org/nlab/show/Lawvere%27s+fixed+point+theorem) and the fact that $Set$ has some endofunctors without no fix-point (for example the covariant power-set endofunctor).
I suspect one can adjust this argument to show that there is no category $C$ such that $C \simeq Fun(C^{op},Set)$ as well, but I can't quite find the argument right now. This would clearly show that no variant of the construction with $Fun(C^{op},Set)$ would ever converge.
| 11 | https://mathoverflow.net/users/22131 | 431033 | 174,553 |
https://mathoverflow.net/questions/431030 | 6 | The proof that $AC$ is independent of $\sf ZF$ axioms is done by forcing and constructibility, and these don't beg any consistency strength more than that of $\sf ZF$.
>
> Is there a known similar proof of independence of $AC$ from $\sf Z$ that is done at the consistency level of $\sf Z$ itself?
>
>
>
>
> More generally what is the smallest consistency level we need to prove that $AC$ is independent from axioms of $\sf Z$?
>
>
>
| https://mathoverflow.net/users/95347 | Is there a proof of independence of AC from Z that is done in Z? | Towards a partial answer:
I have not read it myself yet, but it seems that the relative consistency of $\mathsf{AC}$ with $\mathsf{Z}$ was proved by Mathias in his paper [*The strength of MacLane set theory*](https://www.dpmms.cam.ac.uk/%7Eardm/maclane.pdf). To quote Mathias' summary (available here):
>
> The paper shows that Z + AC is indeed consistent relative to Zermelo's system Z, but the inadequacy, demonstrated in [Slim Models](https://www.dpmms.cam.ac.uk/%7Eardm/slim.pdf), of Z for recursive constructions necessitates an oblique approach.
>
>
>
Of course this leaves open the question of whether $\neg\mathsf{AC}$ is also relatively consistent with $\mathsf{Z}$. In the absence of replacement, forcing becomes quite tricky, so the "usual" approach seems fraught to say the least. Mathias has intensely studied forcing over weak set theories (leading up to his analysis of and advocacy of "provident set theory" $\mathsf{Provi}$ in this context), but at a quick glance I don't see any results that suffice here. I suspect, though, that the answer is again affirmative.
| 9 | https://mathoverflow.net/users/8133 | 431035 | 174,554 |
https://mathoverflow.net/questions/431038 | 4 | An element $g\in G$ in a group $G$ is called **infinitely divisible** if $b=y^n$ for infinitely many different $n\in {\Bbb Z}$. It is not hard to find a finite CW-complex (or even a compact manifold) with a fundamental group containing an infinitely divisible element. For example, consider a group generated by $x$ and $b$ with a relation $xbx^{-1}=b^2$. Then $b$ is infinitely divisible. However, in a hyperbolic manifold every element can be represented by a unique shortest geodesic, which implies that infinite divisibility does not occur. Now, suppose that $G$ is a finitely generated Gromov hyperbolic group (a posteriori, it is finitely presented, as Gromov proved). It seems that it cannot contain infinitely divisible elements of infinite order. I would be very grateful for any reference to this statement.
| https://mathoverflow.net/users/3377 | Infinitely divisible elements in Gromov hyperbolic groups | YCor has answered your question in the comments.
But here is another proof anyway. The "asymptotic translation lengths" of elements (in a fixed word hyperbolic group) are uniformly rational. Also, the asymptotic translation length of an element is zero if and only if the element has finite order. Thus if an element is infinitely divisable, so is its asymptotic translation length, which thus must be zero, hence the element is torsion.
Bowditch's paper *Tight geodesics in the curve complex* says that this result is contained in Gromov's long paper *Hyperbolic groups*, and "an elegant proof can be found in [D]" (which is Delzant's paper *Sous-groupes distingues et quotients des groupes hyperoliques*). I could not find the result in either. But Bowditch is very reliable, so I probably missed it...
| 6 | https://mathoverflow.net/users/1650 | 431043 | 174,557 |
https://mathoverflow.net/questions/420989 | 4 | As you may know, there has been very recently a big breakthrough concerning upper bounds for the capset problem over $\mathbb{F}\_3^n$ (and further generalizations to $\mathbb{F}\_q^n$). I was wondering which other configurations have been studied so far in this context. For instance, the corner problem has also been studied, but it does not arise from a linear equation.
Also, on the other direction, lower bounds for the capset problem come from a work by Edel, so I would like to know what is known for other configurations in the finite field setting.
| https://mathoverflow.net/users/46573 | Extremal problems in additive combinatorics (over finite fields) | The methods of Edel's lower bound, which I have improved in [this paper](https://arxiv.org/abs/2209.10045), are not specific to the setting of $\mathbb{F}\_3^n$ as far as I can see. My result and Edel's both come from considering 3 cap sets $A\_0, A\_1, A\_2 \subseteq \mathbb{F}\_3^6$ and then extending them, but these ideas can be used for any $\mathbb{F}\_p^n$, I believe. Taking direct products and their unions, which is the construction of Edel, should be possible in general characteristic.
It does depend what you mean by the problem in other finite fields, since cap set is a definition specific to $\mathbb{F}\_3^n$. Do you mean no non trivial solutions to $x+y+z=0$, no non trivial solutions to $x+y=2z$, no 3APs or no lines? In characteristic 3, these are equivalent, but they are not in general!
The no lines definition is often called a 'cap' in the design theory and finite geometry literature, and is the object of study in Edel's paper. He does state and prove everything for general finite fields, but the only examples of affine caps he discusses are in characteristic 3. By contrast, my paper only studies the setting of $\mathbb{F}\_3^n$.
The construction relies on 2 things:
1. A collection of cap sets, which satisfy certain structural conditions (I call this 'extendable', Edel calls it 'property $E\_L$')
2. Indexing sets (which I call 'admissible sets', but Edel calls 'capsets')
Depending on what you mean by 'other configurations in the finite field setting', I think a version of the extended product of Edel should be possible, but it requires good examples from low dimensions, which are mostly (I think) only known for cap sets (that is, in $\mathbb{F}\_3^n$).
| 6 | https://mathoverflow.net/users/490000 | 431049 | 174,559 |
https://mathoverflow.net/questions/431045 | 1 | Let $f:X\to Y$ be a birational morphism of smooth projective variety. We assume that $f(V)\simeq U$ isomorphism induced by $f$, where $V\subset X$ and $U\subset Y$ are two Zariski open sets. Let $x\in V$, $C$ be a curve passing through $x$ in $X$ and $L$ be a line bundle over $Y$. Then is the following true?
$$f^{\*}L\cdot C=L\cdot\overline{f(C\cap V)}$$
where $f^{\*}L$ denotes the pullback of the line bundle $L$ and $\overline{f(C\cap V)}$ is the Zariski closure of $f(C\cap V)$ in $Y$.
Any suggestions/comments are welcome!
| https://mathoverflow.net/users/211682 | Intersection pairing and birational morphisms | First note $\overline{f(C \cap V)} = f(C)$, since $f$ is closed and $C$ (and hence also $f(C)$) is irreducible. Also $f$ induces a birational map $C \to f(C)$, so $f\_\* [C] = [f(C)]$ where $[\cdot]$ denotes rational equivalence classes.
Then by the projection formula¹
$$f\_\* (f^\* L \cdot [C]) = L \cdot f\_\* [C] = L \cdot [f(C)].$$
In a more general situation you will get $f\_\*(f^\*L \cdot [C]) = \deg(f|\_C) L \cdot [f(C)]$, if $f$ is not a birational map $C \to f(C)$.
---
¹ see e.g. Fulton's *Intersection Theory*, Proposition 2.3.
| 2 | https://mathoverflow.net/users/111897 | 431061 | 174,564 |
https://mathoverflow.net/questions/430655 | 3 | I ask the same question on MathStackExchange but receive no answer.
I'm reading Kollár Mori Chapter 2.3. And they state the following lemma:
>
> Lemma 2.29. Let $X$ be a smooth variety and $\Delta = \sum a\_iD\_i$ a sum of distinct prime divisors. Let $Z\subseteq X$ be a closed subvariety of codimension $k$. Let $p:B\_Z\rightarrow X$ be the blow up of $Z$ and $E\subseteq B\_ZX$ the irreducible component of the exceptional divisor which dominates $Z$.(If $Z$ is smooth, then this is the only component.) Then
> $$a(E,X,\Delta) = k-1-\sum\_i a\_i\cdot \mathrm{mult}\_Z(D\_i)$$
>
>
>
I have a question about this sentence:
>
> Let $p:B\_Z\rightarrow X$ be the blow up of $Z$ and $E\subseteq B\_ZX$ the irreducible component of the exceptional divisor which dominates $Z$.(If $Z$ is smooth, then this is the only component.)
>
>
>
Does that mean the exceptional divisor of the blowing up of a singular irreducible subvariety can be reducible? Are there any examples of this phenomenon? In general, what does a blowing up of singular subvariety look like?(what contributes to nondominate component?)
| https://mathoverflow.net/users/167083 | Blowing up of a singular subvariety | Possibly the simplest example is to consider the blowup of a reduced and irreducible curve $C$ in a smooth 3-fold $X$ with a point $P\in C\subset X$ which is locally analytically isomorphic to
$$ 0\in \Gamma=\mathbb{V}(xy,yz,zx)\subset \mathbb{A}^3\_{x,y,z} $$
(i.e. in a small neighbourhood of $P\in X$, the curve $C$ looks like the union of the three coordinate lines $\Gamma\subset \mathbb{A}^3$). Thus we can understand what happens in a neighbourhood above $P\in X$ by considering the blowup of $\Gamma\subset\mathbb{A}^3$, even though $\Gamma$ is not irreducible.
The ideal $I=(xy,yz,zx)$ defining $\Gamma$ is generated by three equations $f=xy$, $g=yz$, $h=zx$ and these three equations satisfy two syzygies $zf=xg$, $zf=yh$. Thus the blowup is isomorphic to the complete intersection of codimension 2
$$ \sigma : \operatorname{Bl}\_\Gamma \mathbb{A}^3 = \mathbb{V}(zf-xg, \; zf-yh) \subset \mathbb{A}^3\_{x,y,z}\times \mathbb{P}^2\_{f,g,h} \to \mathbb{A}^3\_{x,y,z}. $$
There are four exceptional divisors. Three of them are isomorphic to $\mathbb{A}^1\times\mathbb{P}^1$ and they each dominate one of the three irreducible components of $\Gamma$. The last one isomorphic to $\mathbb{P}^2$ and it dominates the origin.
As this example shows, once you start blowing up subvarieties $V\subset X$ with non-lci singularities I believe this phenomenon (of having extra exceptional divisors which dominate the singularities of $V$) is essentially pretty typical.
| 3 | https://mathoverflow.net/users/104695 | 431068 | 174,566 |
https://mathoverflow.net/questions/431062 | 3 | Let $h \in C^2\_{\mathrm{ub}}(\mathbb{R}^{2n})$, where $C\_{\mathrm{ub}}^k$ consists of $C^k$-functions that are bounded and uniformly continuous along with their derivatives up to $k$th-order.
It is clear that the Hamiltonian vector field $X$ is $C^1$ and globally Lipschitz, hence the Hamiltonian flow $\Phi\_t$ exists for all times.
Since $X$ is $C^1$, so is the flow. Can one say something similar regarding the uniform continuity? Is the flow of a function $h$ as described above uniformly continuous?
Any help would be much appreciated.
| https://mathoverflow.net/users/485160 | Uniform continuity of Hamiltonian flow | Denote by $L$ the Lipschitz constant of the Hamiltonian vector field $\mathfrak{X}$ and by $\varphi\_t$ the flow generated by $\mathfrak{X}$. Then for any $x, y \in \mathbb{R}^{2n}$, by the chain rule, \begin{align\*}
& |\varphi\_t(x) - \varphi\_t(y)|^2 = \\
& \quad |x - y|^2 + 2 \int\_0^t \langle \mathfrak{X}(\varphi\_s(x)) - \mathfrak{X}(\varphi\_s(y)), \varphi\_s(x) - \varphi\_s(y) \rangle ds \\
& \le |x - y|^2 + 2 L \int\_0^t | \varphi\_s(x) - \varphi\_s(y) |^2 ds
\end{align\*}
By [Grönwall's inequality](https://en.wikipedia.org/wiki/Gr%C3%B6nwall%27s_inequality#Integral_form_for_continuous_functions),
$$
|\varphi\_t(x) - \varphi\_t(y)|^2 \le \exp(2 L t) |x-y|^2 \;.
$$
Thus, the Hamiltonian flow is Lipschitz continuous and hence uniformly continuous. $\Box$
Note that all we really need in this proof is the [one-sided Lipschitz property](https://en.wikipedia.org/wiki/Lipschitz_continuity#One-sided_Lipschitz) of the Hamiltonian vector field $\mathfrak{X}$.
| 3 | https://mathoverflow.net/users/64449 | 431069 | 174,567 |
https://mathoverflow.net/questions/431091 | 4 | The Wikipedia article on Bass-Serre theory claims that graphs of groups (in the context of Bass-Serre theory) "can be viewed as one dimensional versions of orbifolds." I hazily see a connection between a graph of groups and the notion of an orbifold, but I have no concrete sense of what the connection is nor how it is a 1 dimensional analogue. This is my first post on mathoverflow as a graduate student so apologies if my question is ill-posed or too basic.
| https://mathoverflow.net/users/163850 | In what sense is Bass-Serre theory the one-dimensional version of orbifold theory | A graph is a $1$-dimensional manifold with singularities and a graph of groups is, roughly, a $1$-dimensional orbifold with singularities. Every graph of groups has a [Bass-Serre covering tree](https://en.wikipedia.org/wiki/Bass%E2%80%93Serre_theory#Bass%E2%80%93Serre_covering_trees) which is its universal cover as an orbifold, and it can be written as the quotient of this tree by its fundamental group $\pi\_1$. This is exactly analogous to presenting an orbifold as a quotient of a manifold by a group action (when such a presentation exists).
Essentially the only example I understand in any detail is the graph of groups given by a copy of $C\_2$ attached to a copy of $C\_3$ by an edge, which, thought of as an orbifold, is the quotient of a certain tree (which I think is pictured [here](https://en.wikipedia.org/wiki/Modular_group#Tessellation_of_the_hyperbolic_plane)) in the hyperbolic plane $\mathbb{H}$ by the action of the [modular group](https://en.wikipedia.org/wiki/Modular_group) $C\_2 \ast C\_3 \cong PSL\_2(\mathbb{Z})$. You can see some pictures of this orbifold and some of its finite covers (corresponding to finite index subgroups of the modular group) in [this blog post](https://qchu.wordpress.com/2015/11/29/drawing-subgroups-of-the-modular-group/).
| 4 | https://mathoverflow.net/users/290 | 431095 | 174,573 |
https://mathoverflow.net/questions/431047 | 10 | Let $\mathcal P\_n$ be the set of trigonometric polynomials of degree less than or equal to $n$ and let $\lVert\cdot\rVert\_\infty$ be the supremum norm. The error of the best approximation of $f$ of degree $n$ is defined as
$$e\_n(f)=\inf\_{p\in\mathcal P\_n}\lVert f-p\rVert\_\infty.$$
A theorem of Bernstein says that if
$$\sum\_{n=1}^\infty \frac{e\_n(f)}n n^\alpha<\infty$$
for $\alpha>0$ then $f$ is $\alpha$-Hölder. I conjecture that this extends to the $\alpha=0$ case with the statement that $f$ is continuous and of bounded variation instead, and also that the reciprocal holds.
There's a result of Newman and Rivlin in [Approximation of monomials by lower degree polynomials](https://doi.org/10.1007/bf01835995) (the lemma starting on the bottom of the second page, p. 452) that you can use to prove that the function
$$F\_k(x)=\sum\_{h=0}^\infty f\_k(h)\cos(hx)$$
where
$$f\_k(x)=-\frac{d}{dx}\frac 1{\log^k(x)}$$
and the exponents are composition, satisfies
$$e\_n\gg\frac{1}{\log(n)\log^2(n)\dotsm \log^{k-1}(n)(\log^k(n))^2}.$$
And it *really* looks like $\sum\frac{e\_n}n$ is on the verge of diverging but doesn't quite get there.
Is the conjecture true?
| https://mathoverflow.net/users/210944 | A function is of bounded variation if and only if the errors of its best approximation by trigonometric polynomials satisfy $\sum\frac{e_n}n<\infty$? | This is not true. If $f(x)=x \sin( \frac 1x)$ near the origin, then $f$ is $1/2$ Holder continuos and $e\_n(f) \approx \frac{1}{\sqrt n}$ by Jackson theorem, so that the series $\sum\_n \frac{e\_n(f)}{n}$ converges. But is not of bounded variation, since $f' \not \in L^1$.
EDIT
Also the converse does not hold. Take $f(x)=-\frac{1}{\log x}$ for $x$ close to 0, which is continuous and increasing. If $P\_n$ realizes the best approximation then
$$
|f(0)-f(\delta)| \leq |f(0)-P\_n(0)|+|P\_n(0)-P\_n(\delta)|+|P\_n(\delta)-f(\delta)| \leq 2e\_n(f)+Cn \delta,$$
where $C \geq \|P\_n\|\_\infty$ for every $n$, using Bernstein inequality $\|P'\_n\|\_\infty \leq n\|P\_n\|\_\infty$. Taking $\delta=n^{-2}$ we get
$$\frac{1}{2\log n} \leq 2e\_n(f)+\frac Cn$$ and then $e\_n(f) \geq \frac{1}{8\log n}$ for large n and the series $\sum\_n \frac{e\_n(f)}{n}$ diverges.
| 5 | https://mathoverflow.net/users/150653 | 431098 | 174,574 |
https://mathoverflow.net/questions/431083 | 31 | Given $\ell\ge 1$, we say a graph $G$ is $\ell$-good if for each $u,v\in G$ (not necessarily distinct), the number of walks of length $\ell$ from $u$ to $v$ is odd. We say a graph $G$ is good if it is $\ell$-good for some $\ell\ge 1$.
Do good graphs exist? For clarity, I am only talking about simple graphs (which lack loops and multiple edges).
**Context:** In Stanley’s book on Algebraic combinatorics, Exercise 1.13 is about proving an interesting property held for all good graphs. A friend of mine told me that after solving the exercise, he realized he didn’t know of any example of such graphs. I too am stumped about whether such graphs can exist.
A computer search revealed that none exist with $7$ or fewer vertices. I am unclear about the specifics of the search, they were done by my friend.
| https://mathoverflow.net/users/130484 | Do graphs with an odd number of walks of length $\ell$ between any two vertices exist? | A graph without loops cannot be good.
Assume the contrary, let $G$ have $n$ vertices and be good.
Let $A$ be the adjacency matrix of $G$, let $\lambda\_1,\ldots,\lambda\_n$ be its eigenvalues over some extension of $\mathbb{F}\_2$. We have $\sum\_{i=1}^n \lambda\_i=\mathrm{tr} A=0$.
That $A$ is good means that $A^\ell$ is an all-1 matrix over $\mathbb{F}\_2$. It has rank 1, thus at least $n-1$ eigenvalues of $A^\ell$ are 0. On the other hand, the eigenvalues of $A^\ell$ are $\lambda\_1^\ell,\ldots,\lambda\_n^\ell$. Therefore, at least $n-1$ $\lambda\_i$'s are zero, and, since $\sum \lambda\_i =0$, all $\lambda\_i$'s are 0. Thus $A$ is nilpotent. Since $A^\ell$ has rank 1, we get $A^{\ell+1}=0$ (indeed, denote $\mathrm{im} A^{\ell}:=X$, then $\dim X=1$. We have $\mathrm{im} A^{\ell+1}\subset \mathrm{im} A^{\ell}=X$, and also $\mathrm{im} A^{\ell+1}=AX$. Since $\dim X=1$, either $\mathrm{im} A^{\ell+1}=\{0\}$, or $AX=\mathrm{im} A^{\ell+1}=X$; in the latter case $A$ is not nilpotent since $A^kX=X\ne \{0\}$ for all $k=0,1,2,\ldots$). So, $A\cdot A^\ell=0$, that means that the sum of entries in every row of $A$ is even, i.e., every vertex in $G$ must have even degree.
Now pick a vertex $v$ and let $W$ be the set of all walks of length $\ell$ from $v$ to $v$. The cardinality of $W$ is odd by hypothesis. The operation $\rho$ of reversing a walk is an involution on $W$, so the number of fixed points of $\rho$ is odd; these fixed points consists of walks of the form "take any walk of length $\ell/2$ starting at $v$ and then retrace your steps back to $v$" (so in particular, $\ell$ must be even). But because every vertex has even degree, in particular there is an even number of choices for the last step of the walk of length $\ell/2$, so the total number of walks of length $\ell/2$ must be even. This is a contradiction.
| 37 | https://mathoverflow.net/users/3106 | 431102 | 174,576 |
https://mathoverflow.net/questions/428023 | 3 | My question concerns the proof of Proposition 4.2 in Bhatt-Mathew’s paper on the arc-topology, but my confusion is completely general and anyone familiar with limits in $\infty$-categories would know what to do. The situation is that they have a map of $R$-algebras $V\to \tilde V:=V\_{\mathfrak p}\times V/\mathfrak p$ which is a covering map for their Grothendieck topology, and they consider an $\infty$-category $\mathcal C$ which is compactly generated by cotruncated objects and a functor $\mathcal F:R\mathrm{-alg}\to\mathcal C$ which preserves finite products ($V$ is a valuation ring with prime ideal $\mathfrak p$ and residue field $\kappa(\mathfrak p)$, but this is not important). They then claim that the totalisation $\lim\mathcal F(\tilde V^{\otimes\bullet+1})$ coincides with the pullback $\mathcal F(V\_{\mathfrak p})\times\_{\mathcal F(\kappa(\mathfrak p))}\mathcal F(V/\mathfrak p)$. Here, $\tilde V^{\otimes\bullet+1}$ means the cosimplicial $R$-algebra $\tilde V\rightrightarrows\tilde V\otimes\_V\tilde V\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow}\ldots$. It’s important that in this situation $V\_{\mathfrak p}\otimes\_V V\_{\mathfrak p}=V\_{\mathfrak p}$, $V/\mathfrak p\times\_VV/\mathfrak p=V/\mathfrak p$, $V\_{\mathfrak p}\otimes\_VV/\mathfrak p=\kappa(\mathfrak p)$, and $\kappa(\mathfrak p)\otimes\_VV\_{\mathfrak p}=\kappa(\mathfrak p)=\kappa(\mathfrak p)\otimes\_VV/\mathfrak p$. In fact, once you make these identifications, you can forget what $V$ and $\mathfrak p$ are and use the fact $\mathcal F$ preserves products to write down $\mathcal F(\tilde V^{\otimes\bullet+1})$ very concretely; the first few terms are
$$\mathcal F(V\_{\mathfrak p})\times \mathcal F(V/\mathfrak p)\rightrightarrows\mathcal F(V\_{\mathfrak p})\times \mathcal F(V\_{\mathfrak p}\otimes\_V V/\mathfrak p)\times\mathcal F(V/\mathfrak p\otimes\_VV\_{\mathfrak p})\times\mathcal F(V/\mathfrak p)\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow}\ldots$$
where both of the terms with a tensor product are canonically equivalent to $\mathcal F(\kappa(\mathfrak p))$ and all the maps are the canonical ones.
I’m having trouble seeing the claim that this limit coincides with the pullback. (If it were a diagram in a 1-category it would be completely obvious.) The claim that $\mathcal C$ is compactly generated by cotruncated objects means that it suffices to show the result after applying $\operatorname{Map}\_{\mathcal C}(x,-)$ and in this way replace $\mathcal C$ by $\mathcal S\_{\leq n}$, the full subcategory of $\mathcal S$ on $n$-truncated spaces. Then we can truncate the above diagram so it is indexed by $\Delta\_{\leq n+2}$ and deal with a finite diagram. The two approaches I’ve tried are to compute the above (infinite) totalization as a homotopy limit in $\mathcal S$ and to break up the products and show some finite sub-diagram of the resulting diagram is coinitial (left cofinal) in this specific situation. More specifically, I’ve tried to show the diagram $D$ is coinitial, where $D$ has four objects $\mathcal F(V\_{\mathfrak p})$, $\mathcal F(V/\mathfrak p)$, and two copies of $\mathcal F(\kappa(\mathfrak p))$ and four non-identity arrows: one arrow each from the first two objects to each copy of the second. However, this doesn’t appear to be true. On the other hand, I haven’t found a nice enough way to express the homotopy limit in $\mathcal S$ to allow me to make the desired identification. Any tips would be a big help, thank you.
**Edit** A simplified model: objects $X,Y,Z$ with maps $X\to Y$, $Z\to Y$ in an $\infty$-category with limits and a tensor product $\otimes$ which commutes with $\times$; we have $X\otimes Y=Y=Z\otimes Y$, $X\otimes Z=Y$, $X\otimes X=X$, $Y\otimes Y=Y$, $Z\otimes Z=Z$, and we consider the cosimplicial object $(X\times Z)^{\otimes\bullet+1}$ with first three terms
$$X\times Z\rightrightarrows X\times (X\otimes Z)\times (Z\otimes X)\times Z\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow}X\times (X\otimes X\otimes Z)\times(X\otimes Z\otimes X)\times(Z\otimes X\otimes X)\times (Z\otimes Z\otimes X)\times(Z\otimes X\otimes Z)\times(X\otimes Z\otimes Z)\times Z,$$
i.e.
$$X\times Z\rightrightarrows X\times Y\times Y\times Z\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow}X\times Y^{\times 6}\times Z.$$
Breaking apart the product, the three maps above all send $X\to X$, and then send $X=X\otimes X\to X\otimes X\otimes Z$, $X\otimes Z\otimes X$, or $Z\otimes X\otimes X$.
The claim is that the limit of the above diagram is $X\times\_YZ$.
| https://mathoverflow.net/users/37110 | How to simplify this homotopy totalization coming from an arc-cover into a pullback? | Let $X$, $Y$, and $Z$ be Kan complexes. We wish to show that
$X\times\_YZ$ in $\mathrm{Spc}$ can be computed as the limit of the diagram
$$(\*)\qquad X\times Z\rightrightarrows X\times Y\times Y\times Z\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow}X\times Y^{\times 6}\times Z\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow \\[-1em] \rightarrow}\ldots$$
in $\mathrm{Spc}$. This will follow from claim that in $\mathrm{Fun}(\Lambda\_2^2,\mathrm{Spc})$, the geometric realization (i.e. colimit) of the diagram
$$(\dagger)\qquad\ldots\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow \\[-1em] \rightarrow}h\_0\sqcup(\sqcup^6h\_2)\sqcup h\_1\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow}
h\_0\sqcup h\_2\sqcup h\_2\sqcup h\_1\rightrightarrows h\_0\sqcup h\_1$$
is $\mathrm{const}\_\*$, the constant functor $\Lambda\_2^2\to\Delta^0$.
Here, $h:(\Lambda\_2^2)^{\mathrm{op}}\to\mathrm{Spc}$ denotes the Yoneda embedding and 0,1,2 are vertices of $\Lambda\_2^2$. This claim implies the desired conclusion since we can write (with notation as in the book by Land and $F:\Lambda\_2^2\to\mathrm{Spc}$ given by $X\to Y\leftarrow Z$)
$$\operatorname{Map}\_{\operatorname{Fun(\Lambda\_2^2,\mathrm{Spc})}}(\mathrm{const}\_\*,F)\simeq\operatorname{Map}\_{\mathrm{Spc}}(\Delta^0,F)\simeq\operatorname{Map}\_{\mathrm{Spc}}(\Delta^0,\lim F)\simeq\operatorname{Fun}(\Delta^0,\lim F)=\lim F$$
on the one hand and
$\operatorname{Map}\_{\operatorname{Fun(\Lambda\_2^2,\mathrm{Spc})}}(\mathrm{const}\_\*,F)\simeq\lim\_\Delta(\*)$
on the other.
To prove that the colimit of $(\dagger)$ is $\mathrm{const}\_\*$, we can check vertex-by-vertex by HTT.5.1.2.3. For vertices 0 and 1 it suffices to know that the colimit $K\_1$ of the diagram
$$\qquad\ldots\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow \\[-1em] \rightarrow}\Delta^0\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow}\Delta^0\rightrightarrows\Delta^0$$
in $\mathrm{Spc}$ (with every map the identity) is weakly contractible, while for vertex 2 we must show that the colimit $K\_2$ of the diagram
$$\qquad\ldots\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow \\[-1em] \rightarrow}\sqcup^8\Delta^0\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow}\Delta^0\sqcup \Delta^0\sqcup\Delta^0\sqcup\Delta^0\rightrightarrows\Delta^0\sqcup\Delta^0$$
is weakly contractible. We use Corollary 5.5 of [this nLab page](https://ncatlab.org/nlab/show/Reedy+model+structure#OverDeltaWithValuesInSimplicialSets), which allows us to identify $K\_1$ with $\Delta^0$ up to weak equivalence and $K\_2$ with $J$ (Joyal interval/nerve of the walking isomorphism) up to weak equivalence. Both simplicial sets are weakly contractible, so we’re done.
The same method allows us to show that we can compute the pullback $X\times\_YZ$ via the totalization
$$X\times Z\rightrightarrows X\times Y\times Z\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow} X\times Y\times Y\times Z\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow \\[-1em] \rightarrow}\ldots.$$
It boils down to the same check for the geometric realization
$$\qquad\ldots\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow \\[-1em] \rightarrow}h\_0\sqcup h\_2\sqcup h\_2\sqcup h\_1\substack{\rightarrow\\[-1em] \rightarrow \\[-1em] \rightarrow}
h\_0\sqcup h\_2\sqcup h\_1\rightrightarrows h\_0\sqcup h\_1.$$
Here the diagrams corresponding to vertices 0 and 1 are the same, while the geometric realization for the vertex 2 corresponds to $\Delta^1$, which is weakly contractible.
| 0 | https://mathoverflow.net/users/37110 | 431107 | 174,578 |
https://mathoverflow.net/questions/430220 | 2 | A Markov Chain $M$ has only one stationary distribution $q$.
For distribution $p$, with $D\_{TV}(p,Mp)=x$, can we bound $D\_{TV}(p,q)$?
Clearly, $x=0$ implies $D\_{TV}(p,q)=0$. Does general bound hold?
We may write $q=\lim\_{n\mapsto \infty} \frac{p+Mp+M^2p+\cdots}{n}$.
| https://mathoverflow.net/users/4987 | On the distance to the stationary distribution | It is impossible to bound $D\_{TV}(p,q)$ in terms of $x=D\_{TV}(p,Mp)$ without further assumptions on the chain, like expansion. This is due to the phenomenon known as **metastability**.
Rich examples are discussed in [1], [2] and [3], for instance.
The simplest example is a chain on two states $a,b$ with transition probabilities
$M(a,b)=M(b,a)=\epsilon$ and $M(a,a)= M(b,b)=1-\epsilon$.
Then $q$ such that $q(a)=q(b)=1/2$ is the unique stationary measure, but $p$ with $p(a)=1$ satisfies
$D\_{TV}(p,Mp)=\epsilon$ yet $D\_{TV}(p,q)=1/2.$
[1] Olivieri, Enzo, and Maria Eulália Vares. Large deviations and metastability. No. 100. Cambridge University Press, 2005.
[2] Capocaccia, D., Cassandro, M. and Olivieri, E., 1974. A study of metastability in the Ising model. Communications in Mathematical Physics, 39(3), pp.185-205.
[3] Levin, David A., Malwina J. Luczak, and Yuval Peres. "Glauber dynamics for the mean-field Ising model: cut-off, critical power law, and metastability." Probability Theory and Related Fields 146, no. 1 (2010): 223-265.
| 4 | https://mathoverflow.net/users/7691 | 431108 | 174,579 |
https://mathoverflow.net/questions/431113 | 3 | I'm looking into matrix representations of Hurwitz groups, beginning with 2x2 matrices. There are many representations with finite characteristic, namely the ${\rm PSL}(2,p^n)$ groups, where $n=1$ if $p=7$ or $p=1(\operatorname{mod} 7)$, and $n=3$ otherwise. However, what about in characteristic zero?
If we define matrices $A=\begin{pmatrix}0&-1 \\ 1&0 \\\end{pmatrix}, B=\begin{pmatrix}0&-r^{-1} \\ r&1 \\\end{pmatrix}$, where $r^7=1$, then one can verify that $A^2=B^3=(AB)^7=I$ (working in the projective special linear group). The group $H=\langle A,B\rangle$ clearly has all the finite characteristic 2x2 matrix representations as quotients, but I doubt it has the other finite Hurwitz groups as quotients. Therefore, it would seem that $H$ must have some hidden extra relations.
My question is: What are those hidden relations? Or am I mistaken and the relations: $A^2=B^3=(AB)^7=I$ are all there is?
| https://mathoverflow.net/users/38744 | Relators of the "most general" 2x2 matrix Hurwitz group | I think the answer to your question, is yes, this is a faithful projective representation of the triangle group $G = \langle a,b \mid a^2,b^3,(ab)^7 \rangle$, although I have only a hazy understanding of the details.
In Corollary 3.2 of [this paper by Plesken](https://www.sciencedirect.com/science/article/pii/S0021869306001189) it is proved that, up to Galois conjugacy, there is a unique complex irreducible $2$-dimensional projective representation of the group, and he points out before the proof that there can be no faithful linear $2$-dimensional representation, because $G$ is perfect and $-I\_2$ is the only element of order $2$ with determinant $1$. The representation that he writes down gives looks similar to yours.
Then, near the beginning of the proof of Theorem 4.2 of the paper, Plesken remarks that Hurwitz studied a projective representation of $G$ into ${\rm PSL}\_2({\mathbb R})$, which is faithful, so that appears to imply that your representation is faithful.
| 3 | https://mathoverflow.net/users/35840 | 431125 | 174,584 |
https://mathoverflow.net/questions/428243 | 7 | This post is based on <https://math.stackexchange.com/questions/2822589/dissect-square-into-triangles-of-same-perimeter>, [On a possible variant of Monsky's theorem](https://mathoverflow.net/questions/394721/on-a-possible-variant-of-monskys-theorem) and [Cutting convex polygons into triangles of same diameter](https://mathoverflow.net/questions/392596/cutting-convex-polygons-into-triangles-of-same-diameter).
**Question 1:** Is this statement true: "For any convex polygon, there is some finite value of a positive integer *n* such that the polygon allows partition into *n* triangles all of which are of same area."?
**Note:** The above claim, if true, will automatically hold for infinitely many values of *n* - one only needs to subdivide the equal area triangular pieces further into equal area triangles. Further, replacing 'area' with 'perimeter' or 'diameter' above generates further questions.
**Question 2:** Is this claim true: "For any given convex polygon, there is at least one finite value of *n* such that the polygon allows partition into *n* triangles all of *equal area and equal perimeter*"?
**Note:** If question 1 has a negative answer, that would invalidate question 2. And there could be convexity-relaxed versions of both above questions.
**An extension added on 15th August, 2022:** If claim 1 is proved invalid (answer below makes progress in that direction), one could ask a further question: If a convex *m*-gon can be cut into some number *n* of triangles, what is the *lower bound* on the number of distinct areas/perimeters/diameters these n triangles can have - as a function of *m* and *n*?
| https://mathoverflow.net/users/142600 | Partitioning convex polygons into triangles of equal area and perimeter | Question 1 indeed has a negative answer, as previous responders have speculated. See e.g. the 2008 paper by Monsky and Jepsen:
[Constructing Equidissections for Certain Classes of Trapezoids"](https://doi.org/10.1016/j.disc.2007.10.031), Discrete Mathematics, 308 (23): 5672–5681, doi:10.1016/j.disc.2007.10.031, Zbl 1156.51304
(I copied this reference from the Wikipedia page on "equidissections".)
I think the simplest examples they prove have no equidissections are trapezoids $(0,0), (1,0), (0,1), (a,1)$ where $a$ is transcendental.
Interestingly there are still (algebraic) values of $a$ for which it is not known whether or not any equidissections exist.
| 2 | https://mathoverflow.net/users/167341 | 431126 | 174,585 |
https://mathoverflow.net/questions/431104 | 2 | Recall a set of integers $S$ is said to be an additive basis for the natural numbers if there is a $k$ such that every positive integer is expressible as a sum of at most $k$ elements of $S$. Similarly, a set $S$ is said to be an asymptotic additive basis for the natural numbers if there is a $k$ such that every sufficiently large integer is the sum of at most $k$ (not necessarily distinct) elements of $S$. Lagrange's four-square theorem can be thought of as a statement that the squares are an additive basis with $k=4$.
Given a set $S$, we will write $S^2= \{s^2: s \in S\}$.
**Question:** Is there an example of a set $S$ which is not an additive basis but where $S \cup S^2$ is an additive basis? (The same question then for asymptotic additive basis but I will not focus on that here.)
Note that any set with positive [Schnirelmann density](https://en.wikipedia.org/wiki/Schnirelmann_density) is an additive basis, so one naive way of solving this would be to exhibit a set $S$ which is not an additive basis but where $S \cup S^2$ has positive Schnirelmann density but this does not work; if $S$ has Schnirelmann density density zero then so will $S^2$.
| https://mathoverflow.net/users/127690 | Additive basis of a set union the square of the set | In the paper "On additive bases. II", Deshouillers and Fouvry prove a conjecture (made in part I, by a different set of authors) that for each sequence $K$ of positive integers, there is a set $A$ such that $A^k$ is a basis precisely when $k$ belongs to $K$. See
J. London Math. Soc. (2) 14 (1976), no. 3, 413–422.
In particular, one can have $A$ not a basis but $A^2$ is a basis (so that $A\cup A^2$ is also a basis). This case in fact follows from what is done in Part I.
| 6 | https://mathoverflow.net/users/16510 | 431145 | 174,590 |
https://mathoverflow.net/questions/431155 | 3 | Let $A\subset \mathbb R$. Is it true that
$$
\dim(A+A)\le 2\dim A
$$
for some dimensions – say, lower box for the LHS and upper box for the RHS.
| https://mathoverflow.net/users/8131 | Dimension of sumset vs sum of dimensions | $A+A$ is a Lipschitz image of the set $A\times A\subset \mathbb{R}^2$ under the map $(x,y)\to x+y$. If $A$ is covered by $N$ balls of radius $\varepsilon$, then $A\times A$ is covered by $N^2$ balls of radius, say, $\sqrt{2}\varepsilon$, thus the box dimension (lower or upper) of $A\times A$ does not exceed twice that of $A$, and the Lipschitz map does not increase the dimension.
| 5 | https://mathoverflow.net/users/4312 | 431156 | 174,591 |
https://mathoverflow.net/questions/431065 | 2 | Let $A\_n$ be a sequence of $d \times d$ matrices converging to a matrix $A$, all invertible and diagonalizable. We can define the Lyapunov spectrum of the corresponding dynamical system:
$$ \chi = \big\{ \lim\_{n \to \infty} \frac1n \log\|A^{(n)}v\| :\ v \in \Bbb R^d \big\}$$
where $A^{(n)} = A\_n A\_{n-1} A\_{n-2} \cdots A\_1$. Is it the case, in general, that $\chi$ equals the Lyapunov spectrum of the matrix $A$? If not, can we make this true by assuming that the sequence converges fast enough (for some notion of "fast enough")?
| https://mathoverflow.net/users/380456 | Lyapunov exponents of convergent sequence of matrices | So the answer is yes. A brief version is that it suffices to check this for the top exponent and then to use exterior powers to deduce the result for subsequent exponents.
(**Slightly modified to make $\lambda$ the leading Lyapunov exponent**)
Next, if $E$ is the sum of the generalized eigenspaces of all the eigenvalues of maximal modulus, and $F$ is the sum of the remaining generalized eigenspaces, for any $\epsilon>0$, there is a power of $A$ such that $e^{n(\lambda-\epsilon)}\|e\|\le \|A^ne\|\le e^{n(\lambda+\epsilon)}\|e\|$ for all $e\in E$ and $\|A^nf\|\le e^{n\mu}\|f\|$ where $\lambda$ is the logarithm of the absolute value of the dominant eigenvalue(s) of $A$ and $\mu<\lambda$. Now you can build a “cone” of points whose $F$ component is at most $\epsilon$ their $E$ component. This cone is invariant under small perturbations of $A^n$. Now a bit of triangle inequality gives you the result you’re looking for.
| 1 | https://mathoverflow.net/users/11054 | 431161 | 174,592 |
https://mathoverflow.net/questions/431167 | 10 | Let $K$ be a number field. Let $\mathcal{O}\_K^{\*}$ be the units in the ring of integers of this field.
I am interested in knowing how many units $u,v \in \mathcal{O}\_K^{\*}$ exist such that $u + v$ is also a unit. This is equivalent to knowing units $u$ such that $1+u$ is also a unit.
In particular, I'm interested when we have cyclotomic number fields as $K$. Some easy examples can be constructed here by using elements of the form $1+\zeta\_n^m$.
Are there any results of this type, perhaps about density of such units $u$? Where is the starting point for this?
| https://mathoverflow.net/users/94546 | When are sums of units in a number field also units? | I only want to add to Gerry's comment that it is a well-known result by Siegel that the number of exceptional units for a fixed number field is finite. There are even bounds on this number by Evertse, which only depend on the degree of the field extension. There also exist examples of number fields (by Triantafillou) where there does not exist a single exceptional unit.
(This would have been a comment, but I am currently not able to make them)
| 11 | https://mathoverflow.net/users/155336 | 431169 | 174,593 |
https://mathoverflow.net/questions/431172 | 2 | Could anyone please recommend a known website where I can find a database/library that has systems of polynomial equations with $n$ variables and $m$ parameters?
I need some real examples to test my elimination algorithm.
Systems with more variables than equations will also do.
Thanks
| https://mathoverflow.net/users/491824 | Library/Database of parametric polynomial systems | There is a [database of polynomial systems](http://homepages.math.uic.edu/%7Ejan/demo.html), which comes with [PHCpack](http://homepages.math.uic.edu/%7Ejan/PHCpack/phcpack.html) by Jan Verschelde.
| 3 | https://mathoverflow.net/users/7076 | 431187 | 174,599 |
https://mathoverflow.net/questions/431188 | 6 | Let
$$
H\_\lambda=-\frac{d^2}{dx^2}+\lambda^2 x^2,\quad\lambda>0.
$$ It is known that the spectrum of $H\_\lambda$ is the set $\{(2n-1)\lambda,n\in \Bbb N^\*\}$. Now put
$$
(U\_\mu \phi)(x)= e^{\mu\over 2}\phi (e^{\mu}x)\mu \in \Bbb R.
$$
It is easy to check that $\{U\_\mu,\mu\in\Bbb R\}$ forms a one-parameter unitary group and that
$$
U\_\mu H\_1 U^{-1}\_\mu = e^{-2\mu}\bigl( -\frac{d^2}{dx^2}+ e^{4\mu}x^2 \bigr) ,\quad\mu\in\Bbb R
$$
and can be analytically continued into regions of complex $\mu$. Hence for $\lambda,\mu\in\Bbb C$, we have
$$U\_\mu (H\_1-\lambda) U^{-1}\_\mu = e^{-2\mu}\biggl( -\frac{d^2}{dx^2}+ e^{4\mu}x^2-\lambda e^{2\mu}\biggr).$$
This seems to imply that the spectrum of the operator $-\frac{d^2}{dx^2}+ e^{4\mu}x^2$, $\mu\in \Bbb C$ is the set $\{(2n-1)e^{2\mu},n\in \Bbb N^\*\}$: is this result true? And if the answer is affirmative, how can we rigorously prove it?
@AlexandreEremenko. Here is the definition of the spectrum Let $T$ be a closed linear operator from a complex Banach space $X$
into $X$ with dense domain $D(T)$. Then the resolvent set $\rho(T)$ of $T$ is defined to be the set of all complex numbers $\lambda$ for which $T-\lambda I: D(T)\to X$ is bijective and $(T-\lambda I)^{-1}:X\to D(T)$ is a bounded operator, where $I$ is the identity operator on $X$. The spectrum $\sigma (T)$ is simply the complement of $\rho(T)$ in $\Bbb C$.
the point spectrum $\sigma\_p (T)$ of $T$ is the set of all complex numbers
$\lambda$ such that $T-\lambda I$ is not injective. The continuous spectrum $\sigma\_c (T)$ of $T$ is the set of all
complex numbers $\lambda$ such that the range $R(T-\lambda I) $ of $(T-\lambda I)$ is dense in $X, (T-\lambda I)^{-1}$ exists, but is unbounded. The residual spectrum $\sigma\_r (T)$ of $T$ is the set of all complex numbers $\lambda$ such that $(T-\lambda I)^{-1}$ is bounded, but the range $ R(T-\lambda I)$ is not dense in $X$. It is easy to
see that $\sigma\_p (T), \sigma\_c (T)$ and $\sigma\_r (T)$ are mutually disjoints and
$$ \sigma(T)=\sigma\_p (T)\sqcup \sigma\_c (T)\sqcup \sigma\_r (T) .$$
| https://mathoverflow.net/users/172078 | Spectrum of the complex harmonic oscilllator | Indeed, this is the result of [Davies - Pseudo-Spectra, the Harmonic Oscillator and Complex Resonances](https://www.jstor.org/stable/53393) (1982): The resolvent operator $(H-zI)^{-1}$ of
$$H=-d^2/dx^2+cx^2,\;\;\operatorname{Re}c>0,\;\; \operatorname{Im}c>0,$$
is compact for all $z$ not in the spectrum consisting of the set $\{(2n-1)\sqrt c,\;\;n=1,2,3,\dotsc\}$. The spectrum is referred to as a "pseudo-spectrum", because the associated eigenfunctions do not form a basis of the Hilbert space.
| 8 | https://mathoverflow.net/users/11260 | 431191 | 174,600 |
https://mathoverflow.net/questions/431117 | 19 | Is there a good generalisation of Laurent series for several complex variables?
I am interested in generalised power series that have some terms with negative powers, but not too many. In single variable complex analysis, "not too many" means that the (Laurent) series has only a finite number of terms with a negative power of the variable. In several variables, I want that at least something like $\frac1{1- z/w} = \sum\_{n\geq 0} z^n w^{-n}$ counts as generalized Laurent series: While the exponent of $w$ may become arbitrarily small, it is at least bounded in terms of the exponent of $z$.
Has someone thought about this kind of series?
| https://mathoverflow.net/users/175280 | Laurent series in several complex variables | The easiest way to deal with series like $\sum\_{n=0}^\infty z^n w^{-n}$ is with iterated Laurent series. This series is an element of the ring $\mathbb{Z}((w))[[z]]$: power series in $z$ whose coefficients are Laurent series in $w$. (In this case Laurent polynomials in $w$ would suffice.)
A much more general, though more complicated, approach is through [Hahn series](https://en.wikipedia.org/wiki/Hahn_series) (also called Mal'cev–Neumann series)
in which we have an indeterminate with exponents from an ordered group, with the condition that the exponents corresponding to nonzero terms are well ordered. (The well-ordered condition implies that multiplication of these series is well-defined.) To represent $\sum\_{n=0}^\infty z^n w^{-n}$ in this way, we take as our exponent group the additive group $\mathbb{Z}\times\mathbb{Z}$ ordered lexicographically. With $x$ as the indeterminate, the series under consideration are of the form $\sum\_{(i,j)\in \mathbb{Z}\times\mathbb{Z}} x^{(i,j)}$. We multiply monomials by $x^{(i\_1,j\_1)} x^{(i\_2,j\_2)}=x^{(i\_1+i\_2, j\_1+j\_2)}$. We may identify $x^{(i,j)}$ with $z^iw^j$. Then
\begin{equation\*}
\sum\_{n=0}^\infty z^n w^{-n}=\sum\_{n=0}^\infty x^{(n,-n)}
\end{equation\*}
is allowable since the exponent set $\{(0,0), (1,-1), (2,-2),\dots\}$ contains no infinite decreasing sequence. On the other hand
\begin{equation\*}
\sum\_{n=0}^\infty z^{-n} w^{n}=\sum\_{n=0}^\infty x^{(-n,n)}
\end{equation\*}
is not allowed since the exponent set contains the infinite decreasing sequence $(0,0)>(-1,1)>(-2,2)>\dots$.
| 16 | https://mathoverflow.net/users/10744 | 431192 | 174,601 |
https://mathoverflow.net/questions/431197 | 2 | Suppose $x\in SG(\sigma^2)$ is a sub-Gaussian random vector, i.e.
$\left<u,x\right>\quad \forall u\in \mathbb{S}^{n-1}$ is a sub-Gaussian random variable.
My question is : what condition on the random matrix $A$ can guarantee that $Ax$ is again a sub-Gaussian random vector?
I know that $\|A\|\in L^{\infty}$ is one of the conditions. But this one is too strong. Is there any weaker condition?
| https://mathoverflow.net/users/491840 | What condition on random matrix can preserve sub-Gaussian property? | If you only have the hypothesis of sub-Gaussianity, this is the best you can do. Work in dimension $n=1$ for simplicity, let $X\sim N(0,1)$, and let $A$ be independent of $X$. If $AX$ is to be sub-Gaussian, the [Laplace transform condition](https://en.wikipedia.org/wiki/Sub-Gaussian_distribution) will demand
$$
\mathbb{E}[e^{\lambda A X}]= \mathbb{E}[e^{\lambda^2 A^2 /2}] \leq B e^{\lambda^2 b}
$$
for some constants $B,b>0$ and all $\lambda$. But, this tells us something about the distribution of $A$. Namely, by a Chernoff bound, we have the concentration estimate
$$
P\{|A|^2/2 > t^2\} \leq e^{-\lambda^2 t^2} \mathbb{E}[e^{\lambda^2 A^2 /2}] \leq B e^{-\lambda^2 t^2} e^{\lambda^2 b}, ~~~\lambda\in \mathbb{R}.
$$
Hence, $P\{|A|^2 > 2 b\} = 0$, and $A$ is bounded a.s.
| 1 | https://mathoverflow.net/users/99418 | 431203 | 174,605 |
https://mathoverflow.net/questions/431202 | 13 | As far as I understand, the cobordism hypothesis provides a construction of all (appropriately defined) fully-extended TQFTs. In particular, given a fully-dualizable object in a certain category, one can in principle construct the entire TQFT (e.g. the partition function on an arbitrary closed manifold of full dimensionality).
I recently came across [the work of Grady and Pavlov](https://arxiv.org/pdf/2111.01095.pdf), where they claim a proof of the geometric cobordism hypothesis, which should be an analogous statement for arbitrary quantum field theories (i.e. not necessarily topological ones).
My question is: does this lead to a more-or-less explicit construction of any non-trivial quantum field theories? If so, this would be extremely interesting since only a handful of interacting quantum field theories have been constructed in more than two dimensions, and none in more than three dimensions.
A related question is: Are there any tools in the work of Grady and Pavlov (or elsewhere in the literature) that allow one to *compute* any interesting quantities in non-trivial quantum field theories in more than two dimensions? An example of such a quantity would be the spectrum of the Hamiltonian on a given codimension-1 manifold.
I would be happy to hear any relevant comments, and dispelling of confusions I might have about what has been achieved in this line of developments.
| https://mathoverflow.net/users/491848 | Practical consequences of the geometric cobordism hypothesis |
>
> My question is: does this lead to a more-or-less explicit construction of any non-trivial quantum field theories? If so, this would be extremely interesting since only a handful of interacting quantum field theories have been constructed in more than two dimensions, and none in more than three dimensions.
>
>
>
One explicit example can be found on page 112 of the slides <http://dmitripavlov.org/lecture-1.pdf>.
Given a Lie group $G$, a level for $G$, and an invariant polynomial of degree 2 on the Lie algebra of $G$,
it constructs the prequantum Chern–Simons field theory
as a fully extended 3-dimensional $G$-gauged functorial field theory.
A paper with complete proofs is forthcoming.
Concerning quantization of functorial field theories,
it was one of the motivations behind writing the two papers.
The geometric cobordism hypothesis reduces the problem
to the much more tractable computation of the right side, which amounts to computing certain mapping spaces of simplicial presheaves.
In principle, we know what to do to compute the right side,
and constructing a single point (as opposed to computing the entire space of quantum field theories) is even easier.
Again, details are forthcoming. The only example that is written up so far is the case of quantum mechanics. See Section 6.1 in the paper, and also the introduction.
| 8 | https://mathoverflow.net/users/402 | 431206 | 174,606 |
https://mathoverflow.net/questions/431205 | 2 | Let $T$ be a ring with involution $s:T\rightarrow T$. And let
$$h:T\otimes T^\text{op} \rightarrow T\otimes T^\text{op}$$ be the ring automorphism given by $h(a\otimes b)=s(b)\otimes s(a)$.
**suppose that $$ K\_{0}(h): K\_{0}(T\otimes T^\text{op})\rightarrow K\_{0}(T\otimes T^\text{op}) $$ is the identity map.**
I was wondering if the induced homorphism
$$ K\_{\ast}(h): K\_{\ast}(T\otimes T^\text{op})\rightarrow K\_{\ast}(T\otimes T^\text{op}) $$ in K-theory, is the identity map?
Notice that "op" is used for the opposite multiplication.
| https://mathoverflow.net/users/165456 | Induced map in k-theory by an involution | I don't think so: Let us take $T=\mathbb{Z}[x^{\pm 1}]$, and let us take $s$ the identity. Then $T\otimes T^{op}= \mathbb{Z}[x^{\pm 1},y^{\pm 1}]$ has $K\_0 = \mathbb{Z}$ detected by rank (by Grothendieck-Serre), so $K\_0(h)$ is the identity. But I think in $K\_1$ the elements given by the units $x$ and $y$ are different, and are interchanged by $K\_1(h)$.
| 1 | https://mathoverflow.net/users/39747 | 431210 | 174,607 |
https://mathoverflow.net/questions/430761 | 2 | $\DeclareMathOperator\SL{SL}$Fix an integer $p\geq 1$ and a cocompact lattice $\Gamma\subset \SL(p+1,\mathbb{R})$. Consider the manifold
$$
M\_{\Gamma}:=\SL(p+1,\mathbb{R})/\Gamma.
$$
Let $A\subset \SL(p+1,\mathbb{R})$ be the subgroup of diagonal matrices with positive entries. The action
$$
A\curvearrowright M\_{\Gamma};\ a\cdot x\Gamma=(ax)\Gamma
$$
is called the **Weyl chamber flow**. Denote by $\mathcal{F}$ the orbit foliation on $M\_{\Gamma}$.
Is $\mathcal{F}$ Riemannian?
| https://mathoverflow.net/users/150945 | Is the orbit foliation of the Weyl chamber flow Riemannian? | These foliations are very far from being Riemannian.
Consider the case $p=1$, and take $PSL(2,\mathbb R)$ instead of $SL(2,\mathbb R)$ (just to simplify a bit). I'll explain how to construct an example where one non-compact leaf converges to a compact one (a circle). Such behavior implies that the foliation is not Riemannian.
Let $\Gamma\_g\subset PSL(2,\mathbb R)$ be the fundamental group of a genus $g\ge 2$ surface. You can think of $PSL(2,\mathbb R)$ the unit tangent bundle to $\mathbb H^2$ (the hyperbolic plane). In fact $SO(2)\setminus PSL(2,\mathbb R)$ can be identified with $\mathbb H^2$. Then $\mathbb H^2/\Gamma\_g$ is a compact hyperbolic surface $S\_g$. The action of the diagonal group on $PSL(2,\mathbb R)/\Gamma\_g$ is the geodesic flow on the unit tangent bundle of $S\_g$. Now, this flow has closed orbits - corresponding to closed geodesic on $S\_g$. Take such a closed geodesic $\gamma$, then it's not hard to construct a different geodesic that converges to $\gamma$. To see how this looks like, consider, for example Fig 3, in <https://arxiv.org/pdf/1610.07409.pdf> .
| 2 | https://mathoverflow.net/users/943 | 431213 | 174,610 |
https://mathoverflow.net/questions/431209 | 3 | Let $V$ be an elementary abelian $p$-group of size $p^n$. Let $G$ be a finite group with $V\unlhd G$ such that $G/V=H$ is simple (like $\operatorname{PSL}(m,q)$ with $q$ a power of $p$ or any other finite simple group of Lie type in characteristic $p$). Moreover, $V$ is a faithful irreducible $H$-module (equivalently, $V$ is a minimal normal subgroup of $G$ and $C\_G(V)=V$).
Ideally, I would like to show that $G$ must be the semidirect product $V\rtimes H$, or $|V|$ must be somewhat large compared to $|H|$. **A quick example:** when $p=2$ and $H=\operatorname{PSL}(n,2)$, so that $V$ is the natural $n$-dimemsional $\mathbb{F}\_2$-module for $H$, according to "U. Dempwolff, On the second cohomology of $\operatorname{GL}(n,2)$", $G$ must be the mentioned semidirect product.
Any suggestions and references are appreciated.
| https://mathoverflow.net/users/64643 | Extensions of a simple group by an elementary abelian $p$-group | I think this can often fail even for the non-trivial representation $V$ of smallest possible dimension.
For a reductive group $G$ over $\mathbf{Z}\_p$, the group $G(\mathbf{Z}/p^2 \mathbf{Z})$ will typically be a non-split extension of $G(\mathbf{F}\_p)$ by the adjoint representation $V$, and for various $G$ the adjoint representation $V$ will be the smallest non-trivial representation. For example, take $G = E\_8$.
An even simpler example is the group $\Gamma = \mathrm{PSL}\_2(\mathbf{F}\_p)$ for $p > 3$. Then the smallest non-trivial representation over $\mathbf{F}\_p$ will be the adjoint representation $V$ of dimension $3$, but $\mathrm{PSL}\_2(\mathbf{Z}/p^2 \mathbf{Z})$ gives a non-split extension of $\Gamma$ by $V$, since the lift of any unipotent element in $\Gamma$ will have order $p^2$.
| 7 | https://mathoverflow.net/users/491858 | 431217 | 174,611 |
https://mathoverflow.net/questions/425117 | 4 | $\DeclareMathOperator\SU{SU}\DeclareMathOperator\SL{SL}$What are the maximal closed subgroups of $ \SU\_3 $?
This question is inspired by [Lie subgroups of SU(3)](https://mathoverflow.net/questions/65522/lie-subgroups-of-su3). Interesting partial answers to that question, treating only the case of connected subgroups, are given by [José Figueroa-O'Farrill](https://mathoverflow.net/a/65530) and [Neil Strickland](https://mathoverflow.net/a/65834).
Easier example: The three maximal closed subgroups of $ \SU\_2 $ are the binary octahedral group, the binary icosahedral group and the normalizer of the maximal torus. Details given here: [What are the finite subgroups of $SU\_2(C)$?](https://math.stackexchange.com/q/40351)
My attempt:
The maximal closed subgroups of $ \SU\_3 $ are
$$
U\_2\cong S(U\_2\times U\_1)
$$
and
$$
\operatorname{SO}\_3(\mathbb{R}) \times C\_3 \cong \langle\operatorname{SO}\_3(\mathbb{R}),\zeta\_3 I\rangle
$$
(I think this is the full normalizer, maybe there is some extra finite order stuff? Weirdly this type of subgroup doesn't seem to show up in table 5 of [Antoneli, Forger, and Gaviria - Maximal Subgroups of Compact Lie Groups](https://arxiv.org/abs/math/0605784)) and the normalizer of the maximal torus
$$
N(T)=T^2 \rtimes S\_3
$$
where here the symmetric group $ S\_3=W $ is the Weyl group of $ \SU\_3 $. The finite maximal closed subgroups are
$$
3 \times \Sigma\_{168} \;, \; 3.\Sigma\_{216},\;, \; 3.\Sigma\_{360}
$$
where $ \Sigma\_{168} \cong \SL\_3(2) $ is a the simple group of order $ 168 $, $ \Sigma\_{360} \cong A\_6 $ is the simple group of order $ 360 $ and $ \Sigma\_{216} \cong \mathbb{F}\_3^2 \rtimes \SL\_2(3) $ is an affine transformation group, sometimes called the Hessian group. I believe that $ 3.\Sigma\_{216} $ is the full automorphism group of an extra special $ 3 $-group of order $ 27 $ of $ + $ type, $ \operatorname{Aut}(3^{2+1}\_+) $.
| https://mathoverflow.net/users/387190 | What are the maximal closed subgroups of $ \operatorname{SU}_3 $? | $\DeclareMathOperator\SU{SU}\DeclareMathOperator\GL{GL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\Cl{Cl}$Yes the the above is the correct list of maximal closed subgroups of $ \SU\_3 $.
[Antoneli, Forger, and Gaviria - Maximal Subgroups of Compact Lie Groups](https://arxiv.org/abs/math/0605784)
classifies all maximal closed subgroups of $ \SU\_n $ whose identity component is not simple (here trivial counts as simple). According to this paper, pages 1013–1018, the maximal closed subgroups of $ \SU\_3 $ of this type are the normalizer of the maximal torus
$$
N(T)=S(U\_1 \times U\_1 \times U\_1): S\_3
$$
as well as
$$
S(U\_2 \times U\_1 )\cong U\_2.
$$
The maximal closed subgroups with trivial identity component are the finite groups:
$$
3.A\_6
$$
of order $ 3(360)=1080 $ (known as the [Valentiner group](https://en.wikipedia.org/wiki/Valentiner_group))
and
$$
\langle \zeta\_3I\rangle \times \GL\_3(\mathbb{F}\_2)
$$
of order $ 3(168)=504 $. Both these two groups are central extensions by $\langle\zeta\_3 I\rangle$ of a finite simple group. But the first is perfect central extension, indeed a Schur cover. While the second is just a direct product. A third finite maximal closed subgroup is the complex reflection group with Shephard–Todd number 25 (see [Complex reflection group](https://en.wikipedia.org/wiki/Complex_reflection_group)) and order $ 3(216)=648 $ which happens to be a central extension again by $\langle\zeta\_3 I\rangle$ of the [Hessian group](https://en.wikipedia.org/wiki/Hessian_group) of order 216. (Since you are from quantum computing this last subgroup would be known to you as the (determinant-1 subgroup of the) qutrit Clifford group.)
The only maximal closed subgroup of $ \SU\_3 $ with nontrivial simple identity component is the direct product
$$
\langle\zeta\_3I\rangle \times \SO\_3(\mathbb{R}).
$$
To summarize, the maximal closed subgroups of $ \SU\_3 $ are
\begin{align\*}
&N(T)\\
& S(U\_2\times U\_1)\cong U\_2\\
& \langle\zeta\_3 I\rangle \times \SO\_3(\mathbb{R}) \\
& \langle\zeta\_3 I\rangle \times \GL\_3(\mathbb{F}\_2) = 3 \times PerfectGroup(168) \\
& 3.A\_6=PerfectGroup(1080)\\
& S(\Cl\_1(3))=SmallGroup(648,532)
\end{align\*}
where $ S(\Cl\_1(3)) $ is the determinant-1 subgroup of the single qutrit Clifford group. Every closed subgroup of $ \SU\_3 $ is contained in one of these $ 6 $ maximal groups.
Also since unitary $ t $ designs are popular in quantum computing it may be of interest to you that $ 3.A\_6 $ is a unitary 3-design and $ \zeta\_3 \times \GL\_3(\mathbb{F}\_2) $ and $ S(\Cl\_1(3)) $ are both unitary 2-designs.
This is consistent with claim 3 of [my answer](https://math.stackexchange.com/a/4477296/758507) to [Finite maximal closed subgroup of connected Lie group](https://math.stackexchange.com/questions/4272017/finite-maximal-closed-subgroup-of-connected-lie-group)
that all maximal $ 2 $-design subgroups of $ \SU\_n $ (all $ 3 $ designs are also $ 2 $ designs) are finite maximal closed subgroups of $ \SU\_n $.
The only other closed subgroups of $ \SU\_3 $ which are unitary $ t $-designs for $ t \geq 2 $ are the $ \GL\_3(\mathbb{F}\_2) $ subgroup of $ \langle\zeta\_3 I\rangle \times \GL\_3(\mathbb{F}\_2) $ and the commutator subgroup of the qutrit Clifford group, SmallGroup(216,88), which has size $ 3(72)=216 $. These are again unitary $ t $-designs for $ t=2 $.
In addition to all 5 of these 2-designs there is one other Lie primitive subgroup (i.e. not contained in any proper positive dimensional closed subgroup): it is a subgroup of the qutrit Clifford group of size $ 6(36)=216 $.
| 5 | https://mathoverflow.net/users/387190 | 431221 | 174,613 |
https://mathoverflow.net/questions/431177 | 5 | Consider a complete smooth Riemannian manifold $(M,g)$.
I think that it is not difficult to prove that the $k$ Hodge Laplacian is essentially selfadjoint in the relevant $L^2$ space of $k$ forms, when defining it on the smooth compactly supported $k$ forms.
The standard argument by Chernoff should apply to this case, too.
I am interested in the validity of identities like this one (maybe on some specific domain) $$\delta\_{k+1}(\Delta\_{k+1})^{\alpha} = (\Delta\_k)^{\alpha} \delta\_{k+1}\tag{1}$$
where $\delta$ is the *closure* of the usual formal adjoint of $d$ (both defined on the smooth compactly supported forms) and $\Delta$ is the analogous closure (i.e., the unique selfadjoint extension) of the Hodge Laplacian (defined again on smooth compactly supported forms). $\alpha$ is any real. The powers are defined by spectral calculus using the fact that the spectrum of the operator is positive.
For reasons arising from physics, I am quite confident that (1) holds at least when applying both sides to smooth $k+1$-forms with compact support. However I did not manage to find books or papers on these issues, barring some quite old articles. I am not an expert so I suppose that, simply, I have not searched thoroughly.
A modern reference on these issues would be very appreciated.
| https://mathoverflow.net/users/45539 | Properties of non-integer powers of the Hodge Laplacian | This is a nice question. I think there is a trick that makes it obvious enough not to need a special reference. First, if both sides are to be applied to $k+1$-forms, to be consistent your desired identity needs to be
$$
\delta\_{k+1} (\Delta\_{k+1})^{\alpha} = (\Delta\_k)^{\alpha} \delta\_{k+1} . \tag{1}
$$
The trick consists of starting with the operator $D = d + \delta$, aka the *Hodge-Dirac operator*, which maps the space of all forms to itself. $D$ is obviously symmetric on smooth compactly supported forms. The Hodge-de Rham Laplacian is then $\Delta = D^2$. By block decomposition, $d\_k$ is the block of $D$ that maps $k$-forms to $k+1$-forms, $\delta\_k$ is the block of $D$ that maps $k$ forms to $k-1$-forms, while $\Delta\_k$ is the block of $\Delta$ that maps $k$-forms to $k$-forms. If you can show that $D$ is essentially self-adjoint on the core of smooth forms with compact support, then the desired identity (1) falls out of the block decomposition of the automatic functional-calculus identity
$$
D |D|^{2\alpha} = |D|^{2\alpha} D .
$$
But $D$ is essentially self-adjoint because $\Delta$ is, since the resolvent of one can be used to construct the resultant of the other by the formula
$$
\frac{1}{\lambda - D} = \frac{\lambda + D}{\lambda^2 - D^2} .
$$
To make the argument precise, note that the identity $(\lambda - D) \frac{\lambda + D}{\lambda^2 - D^2} = \mathrm{id}$ on smooth compactly supported forms shows that for any non-real $\lambda$ the range of $\lambda - D$ is dense in $L^2$ forms. Hence, by Corollary to Thm.VIII.3(c) in Reed-Simon v.1, $D$ is essentially self-adjoint.
I hope I didn't make a mistake in the last argument. In any case, the essential self-adjointness of $D$ can probably be shown in a more classical way, as it was done for the usual spinor Dirac operator, for instance in
>
> *Wolf, Joseph A.*, [**Essential self adjointness for the Dirac operator and its square**](http://dx.doi.org/10.1512/iumj.1973.22.22051), Indiana Univ. Math. J. 22, 611-640 (1973). [ZBL0263.58013](https://zbmath.org/?q=an:0263.58013).
>
>
>
| 2 | https://mathoverflow.net/users/2622 | 431222 | 174,614 |
https://mathoverflow.net/questions/430745 | 7 | What is the origin of the abacus bijection (aka the rim hook bijection, aka the Stanton-White bijection, aka James's bijection)?
Igor Pak, in his 2000 article "Ribbon tile invariants" (Transactions of the American Mathematical Society, volume 352 (2000), pages 5525-5561), summarizes the situation thus: "The theorem goes back to Nakayama and Robinson (see [R], [JK]). In modern times it was rediscovered by Stanton and White (see [SW], [FS]) and is sometimes attributed to them." But I have found no other attribution to Nakayama and Robinson (perhaps because I have no access to [R]). I have however looked at [JK] which contains a statement of the result, attributed to G. James (the book's first author).
Pak is a scrupulous mathematical historian (see for instance his excellent essay [https://www.math.ucla.edu/~pak/papers/cathist4.pdf](https://www.math.ucla.edu/%7Epak/papers/cathist4.pdf), included as an appendix in Richard Stanley's book on the Catalan numbers), so I'm inclined to take him at his word regarding Nakayama and Robinson. Still, I'd feel better knowing more details.
[FS] S. Fomin, D. Stanton, Rim hook lattices, St. Petersburg Math. J. 9 (1998), 1007–1016.
[JK] G. James, A. Kerber, The Representation Theory of the Symmetric Group, Addison-Wesley, Reading, MA, 1981.
[R] G. de B. Robinson, Representation Theory of the Symmetric Group, Edinburgh University Press and Univ. of Toronto Press, 1961.
[SW] D. Stanton, D. White, A Schensted algorithm for rim hook tableaux, J. Comb. Theory, Ser. A 40 (1985), 211–247.
| https://mathoverflow.net/users/3621 | Origin of the abacus bijection | After looking at Robinson's book, my coauthors and I settled upon the following wording:
The version of the bijection that we use is due to Gordon James (see \cite{JamesKerber}) but different forms of it seem to have been discovered independently by various people working in the field of modular representation theory around 1950; this community includes H.\ Farahat, J.\ S.\ Frame, D.\ E.\ Littlewood, T.\ Nakayama, M.\ Osima, G.\ de B.\ Robinson, R.\ A.\ Staal, and R.\ M.\ Thrall. The interested reader may find more details in the book \cite{Robinson} and the references it contains.
| 4 | https://mathoverflow.net/users/3621 | 431225 | 174,615 |
https://mathoverflow.net/questions/431219 | 7 | Let $\*$ be a binary operation on a set $M$, with an identity element $e\in M$.
A *monoid representation* of $(M,\*,e)$ is a map $\delta:M\to (S\to S)$ for some set $S$, such that $\delta(e)=\mathrm{id}\_S$, and $\delta(a\*b)=\delta(a)\circ\delta(b)$ for all $a,b\in M$. (A representation could also be called an *action*, I suppose?)
* $\delta$ is *faithful* if $\delta$ is injective as a function from $M$ to $S^S$.
* $\delta$ is *irreducible* if there is no subset $\emptyset\subsetneq T\subsetneq S$ with $\delta(m)(t)\in T$ for all $t\in T$ and $m\in M$.
Which monoids $(M,\*,e)$ have faithful irreducible representations?
For example, all groups do have such representations, but the monoid $\{e,a\}$ with $a^2=a\ne e$ does not since we can take $T=\{\delta(a)(t)\}$ for a fixed $t\in S$.
>
> Is there a characterization, or a name for such monoids?
>
>
>
Example: let $M$ be generated by $f,g:\{0,1,2,3\}\to\{0,1,2,3\}$
where $f(0)=1$, $g(0)=2$, $f(1)=g(1)=1$, and $f(2)=3$, $f(3)=2$, $g(2)=3$, $g(3)=3$. The monoid is
$$\begin{matrix}
& && e &&\\
&&f & &g\\
&f^2 & gf&&fg & g^2\\
& & fgf&&&fg^2\\
\end{matrix}$$
which has the ideals:
$$M(gf)=\{fgf,gf\}, M(g^2)=\{g^2,fg^2\}\quad\text{(minimal)}$$
$$ M(f),M(g) \quad\text{(not minimal)}$$
| https://mathoverflow.net/users/4600 | Which monoids have a faithful irreducible representation? | **Cleaner rewrite:**
I have a bit more time, so here is a cleaner rewrite. This notion is usually called transitive rather than irreducible, although the terms irreducible and minimal are both used.
If $M$ has a minimal left ideal $L$,then $L$ is a transitive $M$-set and every transitive action is a quotient of $L$ and hence $M$ has a faithful transitive action iff it acts faithfully on $L$. In particular, if $M$ is finite then all its minimal left ideals are isomorphic as $M$-sets (since they are quotients of each other and finite) and $M$ has faithful transitive action iff it acts faithfully on one (equals all) of its minimal left ideals.
The proof is trivial. If $S$ is a transitive $M$-set, then $Ls$ is invariant and hence $Ls=S$ for all $s\in S$. Thus, if we fix $s\in S$, then $m\mapsto ms$ is a surjective $M$-set map $L\to S$.
There is some information on Clifford and Preston Volume 2, Chapter 11.5 on the infinite case. For instance they show noncommutative free monoids have a faithful and transitive action and nontrivial free products of semigroups. See also Tully, E. J. (1961). Representation of a Semigroup by Transformations Acting Transitively on a Set. American Journal of Mathematics, 83(3), 533.
**Further Remark** It is more common to consider transitive actions by partial mappings because many more semigroups have faithful transitive actions in this setting.
| 6 | https://mathoverflow.net/users/15934 | 431227 | 174,616 |
https://mathoverflow.net/questions/431224 | -3 | I am reading
* imenez, J., Echevarria, J.I., Sousa, T. and Gutierrez, D. (2012), *SMAA: Enhanced Subpixel Morphological Antialiasing* Computer Graphics Forum, 31: 355-364. <https://doi.org/10.1111/j.1467-8659.2012.03014.x>
where I have encountered these three equations
\begin{gather\*}
e\_l=\lvert L-L\_l\rvert>T \\
c\_\text{max} = \max(c\_t, c\_r,c\_l,c\_r,c\_{2l}) \\
e\_{l'}=e\_l \land c\_l > 0.5 c\_\text{max}.
\end{gather\*}
Because $e\_l$ is a number being either 0 or 1, and $c\_l$ is always going to be a fraction number, I am wondering what is the AND result of these two? Would it be either 1 or 0 or it is going to be the value of $c\_l$ when $e\_l$ is 1?
| https://mathoverflow.net/users/491860 | Can you do boolean and of 1 and a number less than 1? | The paragraph before says
>
> We calculate the maximum contrast $c\_{max}$ for all these edges and compare it with the contrast for the left edge [this is $c\_l$]. If the latter is above a threshold of $0.5\cdot c\_{max}$ the edge is preserved; otherwise, it is ignored.
>
>
>
So this reads to me like a condition $c\_l \gt 0.5\cdot c\_{max}$.
We also have the definition of $e\_l$:
>
> a straightforward algorithm would calculate $e\_l=|L−L\_l|\gt T$, where $e\_l$ is the **boolean value** that codes whether the edge is active, [emphasis added]
>
>
>
so we should think of $e\_l$ not as $0$ or $1$, but a *boolean*, $T$ or $F$, giving the truth value of the condition "$|L−L\_l|\gt T$", which may or may not be true.
Moreover, after he displayed formulas, we have
>
> where $c\_t$, $c\_r$, $c\_b$, $c\_l$, $c\_{2l}$ are the contrast deltas for the edges shown in Figure 5, and $e'\_l$ represents the final boolean value (active or not) for the left edge boundary.
>
>
>
so again, we have that $e'\_l$ is a boolean, and so the condition $e\_l = e'\_l$ is checking if two booleans are equal (i.e. both true or both false).
The paper is using a symbol, $\wedge$, where a word would do, in a false economy, or else insisting on avoiding a word in an equation environment.
| 2 | https://mathoverflow.net/users/4177 | 431230 | 174,618 |
https://mathoverflow.net/questions/431215 | 3 | Let $d \ge 2$ be a positive integer. For $x=(x\_1,\dotsc,x\_{d-1},x\_d)$, we write $x'=(x\_1,\dotsc,x\_{d-1})$. Let $\mathbb{H}^d=\{x=(x',x\_d) \mid x\_d>0\}$ denote the $d$-dimensional upper half-space.
Then, we obtain the following result.
>
> Let $u \in C^2\_c(\mathbb{R}^{d-1})$, $v \in C\_c^1(\mathbb{R}^{d-1})$.
> Then, there exists $w \in C\_b^2(\mathbb{H}^d\cup \partial \mathbb{H}^d)$ such that
> $w(x',0)=u(x')$ and $\partial\_d w(x',0)=v(x')$ for every $x' \in \mathbb{R}^{d-1}$.
>
>
>
This assertion in proved in Lemma B.1 in [Baur and Grothaus - Construction and Strong Feller Property of Distorted Elliptic Diffusion with Reflecting Boundary](https://doi.org/10.1007/s11118-013-9355-8).
In the proof, it is important to introduce a bounded linear operator.
Let $\varphi \in C\_c^2(\mathbb{R}^{d-1})$ such that $\int\_{\mathbb{R}^{d-1}}\varphi(z)\,dz=1.$ We define a linear operator $P\_1\colon C\_c^2(\mathbb{R}^{d-1}) \to C\_b^2(\mathbb{H}^d\cup \partial \mathbb{H}^d)$ by
\begin{align\*}
P\_1h(x)=x\_d \int\_{\mathbb{R}^{d-1}}\varphi(\zeta)h(x'+x\_d\zeta)\,d\zeta,\quad x=(x',x\_d).
\end{align\*}
The authors of the above paper states that the operator $P\_1$ is extended to a bounded linear operator from $C\_c^1(\mathbb{R}^{d-1})$ to $C\_b^2(\mathbb{R}^d \cap \{x\_d \ge 0\})$. This means that there exists $C>0$ such that for every $h \in C\_c^2(\mathbb{R}^{d-1})$,
\begin{align}
&\sup\_{x \in \mathbb{R}^d \cap \{x\_d \ge 0\}} \{|P\_1h(x)|+|\nabla P\_1h(x)|+\|\nabla^2 P\_1h(x)\|\} \\
&\le C \sup\_{z \in \mathbb{R}^{d-1}}\{|h(z)|+|\nabla h(z)|\} \tag{A}
\end{align}
Here, $\|\nabla^2 P\_1h(x)\|$ denotes the Hilbert—Schmidt norm of the Hessian matrix $\nabla^2P\_1h$.
However, I could not understand why the estimate (A) holds. if you know the proof of this estimate, please let me know.
| https://mathoverflow.net/users/68463 | Boundedness of an extension operator | Let me give an estimate for the first derivatives of $v:=P\_1 h$ in terms of the sup-norm of $h$.
Setting $z=x'+x\_d \xi$ we get
$$v(x)=\int\_{\mathbb {R}^{d-1}}\phi \left (\frac{z-x'}{x\_d}\right )h(z) x\_d^{2-d} \, dz.
$$
Then
$$v\_{x\_d}(x)=(2-d)\int\_{\mathbb {R}^{d-1}}\phi \left (\frac{z-x'}{x\_d}\right )h(z) x\_d^{1-d} \, dz+\int\_{\mathbb {R}^{d-1}}\psi \left (\frac{z-x'}{x\_d}\right )h(z) x\_d^{1-d} \, dz
$$
with $\psi(y)=-y\cdot \nabla \phi (y),\ y \in \mathbb R^{d-1}$. Now the estimate follows since the scaling $x\_d^{1-d}$ makes constant the $L^1$ norms of the mollifiers.
The estimates for the tangential derivatives are similar. For second order derivatives, one first differentiate $h$ and then uses similar arguments. Note however that $v$ is not bounded for large $x\_d$ (take formally $h=1$).
| 2 | https://mathoverflow.net/users/150653 | 431266 | 174,625 |
https://mathoverflow.net/questions/431251 | 1 | Let $G$ be a profinite topological group with two closed subgroup $G\_1$ and $G\_2$. Suppose $G\_1$ is normal in $G$ and $G=G\_1G\_2$. Let $H\_i$ be an open subgroup in $G\_i$ for $i=1,2$.
**Question:** Is $ H\_1H\_2:=\{h\_1h\_2\mid h\_1\in H\_1 \text{ and } h\_2\in H\_2\}$ also open in $G$?
| https://mathoverflow.net/users/11750 | Openness of product of two open subgroups | Yes (even without assuming $G\_1$ normal).
Indeed, first, since $G\_1G\_2=G$, for some finite subset $F$ we have $FH\_1H\_2F=G$, i.e., finitely many left-right translates of the compact subset $H\_1H\_2$ cover $G$. Hence (Baire) $H\_1H\_2$ has nonempty interior.
Next, $H\_1\times H\_2$ left-right acts on $G$ and $H\_1H\_2$ is a single orbit (the orbit of 1). Hence by homogeneity, every point is an interior point. This means that $H\_1H\_2$ is open.
| 5 | https://mathoverflow.net/users/14094 | 431267 | 174,626 |
https://mathoverflow.net/questions/431275 | 6 | Consider the surface $S\_{\epsilon}$ defined as:
\begin{align}
%S&=\{\vec x \in \mathbf{R}^3: x=0\}, \\
S\_{\epsilon}&=\{\vec x \in \mathbf{R}^3:f\_{\epsilon}(\vec x)\equiv\epsilon ((x^2 + y^2 - 4)^2 + z^2 - 1) + x=0\}.
\end{align}
The topology of $S\_{\epsilon}$ is $\mathbf S^2$ when $\epsilon$ is small but non-zero, and the topology of $S\_0$ is $\mathbf{R}^2$. But I find this puzzling because I thought that the topology changes only when the constraining function, $f\_{\epsilon}(\vec x)$, has a critical point (i.e. $\vec\nabla f\_\epsilon=0$ at some point on the surface).
However as $\epsilon\to0$, $f\_\epsilon$ never has a critical point, yet the topology is changed. Can someone explain this?
| https://mathoverflow.net/users/485792 | Topology change induced by small perturbation | The formal statement you are thinking of when you assert "The topology changes only when..." is Ehresmann's theorem: a proper smooth submersion is a fiber bundle, and hence all fibers are diffeomorphic. Here "proper" means that the inverse image of any compact set is compact. It is a useful fact that proper maps are closed; because the set of critical points is closed, it follows that under these assumptions the set of critical values are also closed.
Thus if $M$ is some noncompact manifold and $f: M \to N$ is a **proper** smooth function for which $n\_0$ is not a critical value, then neither is any $n \in U$ in a neighborhood around $n\_0$, and $f^{-1}(U) \to U$ defines a proper smooth submersion. Ehresmann's theorem now asserts that $f^{-1}(n\_0)$ is diffeomorphic to any $f^{-1}(n)$ with $t \in U$.
If I were to attempt to apply Ehressman's theorem here, I would define a function $F: \Bbb R^4 \to \Bbb R^2$ by $F(\vec x, \epsilon) = (f\_\epsilon(\vec x), \epsilon)$ and consider $F^{-1}(0,\epsilon)$ as $\epsilon$ changes. As you say $F$ is a submersion. But the function $F$ is not proper, because $F^{-1}(0,0)$ is not compact. So I cannot apply Ehressman.
I do not know any particularly powerful theorem which asserts something like you want without the properness assumption, so it is not surprising that you see different preimages have different topological type.
| 10 | https://mathoverflow.net/users/40804 | 431276 | 174,627 |
https://mathoverflow.net/questions/431248 | 3 | Consider a long cylinder $C = D \times (-L,L) \subset \mathbf{R}^3$, with heat applied to its horizontal boundary according to $\varphi$ and perfectly insulated ends. The steady state $u: C \to \mathbf{R}$ representing the temperature is the solution of the system
\begin{equation}
\begin{cases}
\Delta u = 0 \quad \text{in $C$} \\
u = \varphi \quad \text{on $\partial D \times (-L,L)$} \\
\frac{\partial u}{\partial \nu} = 0 \quad \text{ on $D \times \{-L,L\}$}.
\end{cases}
\end{equation}
In heuristic terms, 'most important' for the values of $u$ on $D \times \{ 0 \}$ are the boundary values with similar height. Heat applied further away, 'nearer to the ends' of the cylinder is still 'felt' by $u$ on $D \times \{ 0 \}$, but 'much less'.
**Question.** How does one establish estimates that quantify the 'waning influence' of $\varphi$ on $u$ with the height? Maybe this would be in terms of a weight function $w: (-L,L) \to \mathbf{R}$, say $w(t) = \lvert L - t \rvert$ or something stronger?
*Edit.* Apparently there are estimates with exponentially decaying weight $w(t) = \mathrm{e}^{-C \lvert t \rvert}$, perhaps something like
\begin{equation}
\lvert u(\cdot,0) \rvert\_{L^2(D)}
\leq C \lvert \mathrm{e}^{-C \lvert t \rvert} \varphi\rvert\_{L^2(\partial D \times (-L,L))}.
\end{equation}
Although I am inclined to believe these bounds by fiat, I am ultimately most interested in the arguments used to derive them (or indeed a reference to such arguments). This wasn't explicit in the wording of the earlier version of the question, so I've amended it to clarify this.
| https://mathoverflow.net/users/103792 | Heating a long cylinder: steady states | $\newcommand{\R}{\mathbb R}$Sorry for being too sketchy in the following answer, time permitting, I'll try to expand.
---
*Step 0.* Some more-or-less classical potential theory. Let $D$ be an open set in $\R^d$ (with $d \geqslant 3$ for simplicity), and assume that $D$ is sufficiently regular (for example, Lipschitz). Let $p\_t^D(x, \xi)$ be the *heat kernel* in $D$: the fundamental solution of the heat equation $$\frac{\partial u}{\partial t}(t, x) = \Delta u(t, x) $$ with $u(t, x) = 0$ when $x \in \partial D$. In other words, $$u(t, x) = \int\_D p\_t^D(x, \xi) f(\xi) d\xi$$ is the (unique) solution of the heat equation with initial condition $u(0, x) = f(x)$.
The *Green function* in $D$ can be defined by the time integral of the heat kernel:
$$G\_D(x, \xi) = \int\_0^\infty p\_t^D(x, \xi) dt .$$
Note that this is always finite (except when $x = \xi$, of course), because $p\_t^D(x, \xi) \leqslant p\_t^{\R^d}(x, \xi)$, and the integral of $p\_t^{\R^d}(x, \xi)$ is the Newtonian potential kernel $c\_d |x - \xi|^{2 - d}$. Furthermore, $G\_D(x, \xi)$ is zero on the boundary (this is not quite obvious, though), and — formally — we have $$ \Delta\_x G\_D(x, \xi) = \int\_0^\infty \Delta\_x p\_t^D(x, \xi) dt = \int\_0^\infty \frac{\partial p\_t}{\partial t}(x, \xi) dt = 0 - \delta\_\xi(x) ,$$ so that $G\_D(x, \xi)$ is the fundamental solution for the Poisson problem $$ \Delta u(x) = -f(x) $$ in $D$, with $u(x) = 0$ for $x \in \partial D$. And indeed, one can rigorously prove that if $$u(x) = \int\_D G\_D(x, \xi) f(\xi) d\xi$$ for, say, continuous and bounded $f$, then indeed $\Delta u = -f$ in $D$ and $u = 0$ on $\partial D$.
Finally, if $D$ is regular enough ($C^{1,1}$ is the usual condition), then the Dirichlet problem in $D$: $$\Delta u(x) = 0$$ with Dirichlet boundary condition $u(x) = f(x)$ for $x \in \partial D$ can be solved using the *Poisson kernel*: $$u(x) = \int\_{\partial D} f(\xi) P\_D(x, \xi) \sigma(d\xi),$$ where $\sigma$ is the surface measure on $\partial D$ and the Poisson kernel $P\_D(x, \xi)$ is the boundary derivative of the Green function: $$P\_D(x, \xi) = \frac{\partial G\_D(x, \cdot)}{\partial \nu}(\xi) = \lim\_{s \to 0^+} \frac{G\_D(x, \xi + s \nu)}{s} \, ,$$ where $\nu$ is the inward normal vector at $\xi$. This follows relatively easily from the divergence theorem (or Green's identities).
By the way, for a general open set $D$, the solution of the Dirichlet problem is given in terms of the *harmonic measure*: $$ u(x) = \int\_{\partial D} f(\xi) P\_D(x, d\xi) ,$$
which is again closely related to the Green function, but this is a completely different story.
---
*Step 1.* First, consider the Poisson problem in $D \times \R$, with boundary data given by $f : \partial D \times \R \to \R$ (let us denote the boundary data by $f$ rather than $\varphi$, which we will need for the eigenfunctions; here $f$ is an arbitrary bounded and continuous function). The solution is given by the harmonic measure, which, due to translation invariance of the problem, is translation invariant itself:
$$ \begin{aligned} u(x, y) & = \int\_{\partial D \times \R} f(\xi, \eta) P\_{D \times \R}(x, y, d\xi d\eta) \\ & = \int\_{\partial D \times \R} f(\xi, y + \eta) P\_{D \times \R}(x, 0, d\xi d\eta) \end{aligned} $$
for an appropriate measure $P\_{D \times \R}(x, y, d\xi d\eta)$. If $D$ is nice enough — say $C^{1,1}$ — then $P\_{D \times \R}(x, y, d\xi d\eta)$ has a density function $P\_{D \times \R}(x, y, \xi, \eta)$ with respect to the surface measure $\sigma(d\xi) d\eta$, and this density is called the *Poisson kernel*. Thus,
$$ u(x, y) = \int\_{\R} \int\_{\partial D} f(\xi, \eta) P\_{D \times \R}(x, y, \xi, \eta) \sigma(d\xi) d\eta . $$
Translation invariance means that $P\_{D \times \R}(x, y, \xi, \eta) = P\_{D \times \R}(x, 0, \xi, \eta - y)$.
---
*Step 2.* How fast does $P\_{D \times \R}(x, 0, \xi, \eta)$ decay with $|\eta|$? In my comment to the question, I sketched a probabilistic argument which shows exponential decay. Here is a more analytic (but still potential-theoretic) version of the same argument.
Let $p\_t^D(x, \xi)$ be the heat kernel in $D$, and $p\_t^{D \times \R}(x, y, \xi, \eta)$ be the heat kernel in $D \times \R$. Thus,
$$ p\_t^{D \times \R}(x, y, \xi, \eta) = p\_t^D(x, \xi) (4 \pi t)^{-1/2} e^{-(\eta - y)^2 / (4t)} . $$
Again if $D$ is nice enough ($C^{1,1}$ is more than enough, Lipschitz is already fine), than $p\_t^D$ is known to be *intrinsically ultracontractive*. In particular,
$$ p\_t^D(x, \xi) \approx C e^{-\lambda\_1 t} \varphi\_1(x) \varphi\_1(\xi) $$
for $t > 1$. Here $\approx$ means that the ratio is bounded from above and below by positive constants.
Using an estimate $0 \leqslant p\_t^D(x, \xi) \leqslant p\_t^{\R^2}(x, \xi)$ (where $p\_t^{\R^2}$ is the usual Gauss–Weierstrass kernel) for $t < 1$ and intrinsic ultracontractiviety for $t > 1$, by direct integration, we find the following estimate of the Green function, valid when $|\eta|$ is large enough (here I omit the details):
$$\begin{aligned} G\_{D \times \R}(x, 0, \xi, \eta) & = \int\_0^\infty p\_t^{D \times \R}(x, 0, \xi, \eta) dt \\ & \approx \varphi\_1(x) \varphi\_1(\xi) \int\_0^\infty e^{-\lambda\_1 t} t^{-1/2} e^{-\eta^2 / (4 t)} dt \\ & \approx \varphi\_1(x) \varphi\_1(\xi) e^{-\sqrt{\lambda\_1} |\eta|} . \end{aligned} $$
Now the Poisson kernel is the normal derivative of the Green function. Thus, if $D$ is a $C^{1,1}$ set,
$$ P\_{D \times \R}(x, 0, \xi, \eta) \approx \varphi\_1(x) e^{-\sqrt{\lambda\_1} |\eta|} . $$
This may look as if we "differentiate both sides of an inequality", but it is not the case: since the Green function is zero on the boundary, the normal derivative reduces to a simple limit of $G\_D(x, 0, \xi + s \nu, \eta) / s$, where $\nu$ is the inward normal vector at $\xi$.
---
*Step 3.* Now it remains to translate this into a result on $D \times (-L, L)$ with zero Neumann boundary condition on the bases. This, however, is pretty standard: if $u$ is the solution of the problem on $D \times (-L, L)$, then the function $v$ given by
$$ v(x, y + 4 n L) = u(x, y) , \qquad v(x, y + 2 n L) = u(x, -y) $$
whenever $x \in D$, $y \in (-L, L)$ and $n \in \mathbb Z$, is a solution of the corresponding Poisson problem in $D \times \R$. Using the Poisson representation for $v$, we find that
$$ u(x, y) = \int\_{(-L, L)} \int\_{\partial D} f(\xi, \eta) \sum\_{n = -\infty}^\infty (P\_{D \times \R}(x, y, \xi, \eta + 4 n L) + P\_{D \times \R}(x, y, \xi, -\eta + 2 n L)) \sigma(d\xi) d\eta . $$
By using the estimate for the Poisson kernel found above, we easily see that again
$$ u(x, 0) \approx \varphi\_1(x) \int\_{(-L, L)} \int\_{\partial D} f(\xi, \eta) e^{-\sqrt{\lambda\_1} \eta} \sigma(d\xi) d\eta $$
uniformly in $L$ large enough and $f$.
| 5 | https://mathoverflow.net/users/108637 | 431285 | 174,630 |
https://mathoverflow.net/questions/429291 | 8 | *Throughout, I work in $\mathsf{MK}$ in order to be able to conveniently quantify over logics; if one prefers, we can restrict attention to (say) $\Sigma\_{17}$-definable logics and work in $\mathsf{ZFC}$. By "logic" I mean "regular logic containing $\mathsf{FOL}$ and having countably many formulas in a finite language" (for example, $\mathsf{SOL}$).*
---
Given a logic $\mathcal{L}$, consider the following two $\mathcal{L}$-theories $\mathscr{ZFC}(\mathcal{L})$ and $\mathscr{M}(\mathcal{L})$ defined as follows. We let $\mathscr{ZFC}(\mathcal{L})$ be the $\mathcal{L}$-theory consisting of
* the "boring" $\mathsf{ZFC}$-axioms Pairing, Union, Infinity, Choice, Regularity, and Extensionality, and
* the Separation and Replacement schemes modified to allow formulas coming from $\mathcal{L}$.
Note that $\mathscr{ZFC}(\mathsf{SOL})$ is [not quite](https://mathoverflow.net/questions/385904/analogues-of-worldly-cardinals-for-an-unusual-version-of-second-order-mathsf) the same thing as "second-order $\mathsf{ZFC}$." Meanwhile, $\mathscr{M}(\mathcal{L})$ is $\mathscr{ZFC}(\mathcal{L})$ **plus** for each $\mathcal{L}$-formula $\varphi(x)$ the reflection instance $$\forall x[\varphi(x)\rightarrow\exists \alpha(x\in V\_\alpha\wedge \varphi(x)^{V\_\alpha})].$$
It's easy to state and prove in (first-order!) $\mathsf{MK}$ that $V\_\alpha\models\mathscr{ZFC}(\mathcal{L})$ for every logic $\mathcal{L}$ whenever $\alpha$ is inaccessible; somewhat conversely, assuming $\mathsf{V=L}$ this is optimal already for $\mathcal{L}=\mathsf{SOL}$ since $V\_\alpha\models\_\mathsf{SOL}\mathscr{ZFC}(\mathsf{SOL})$ only if $\alpha$ is $L$-inaccessible.
The situation for $\mathscr{M}$ is more complicated. For example, if $\alpha$ is the least inaccessible then $V\_\alpha\not\models\_\mathsf{SOL}\mathscr{M}(\mathsf{SOL})$ since the sentence "$\mathsf{Ord}$ is inaccessible" is second-order expressible and holds in $V\_\alpha$ but not in any smaller $V\_\beta$. Instead, we need to go a bit higher. Say that a cardinal $\kappa$ is *chromatic* iff $\kappa$ is an inaccessible limit of inaccessibles and the following holds (letting $I$ be the set of inaccessibles $\le\kappa$):
>
> For every family $C=(c\_i)\_{i\in\omega}$ of $2$-colorings $c\_i: [I]^2\rightarrow 2$, there is some $\lambda\le\kappa$ such that for all inaccessible $\alpha<\lambda$ and all $i\in\omega$ with $c\_i(\{\alpha,\lambda\})=1$ there is some $\beta\in(\alpha,\lambda)$ with $c\_i(\{\alpha,\beta\})=1$.
>
>
>
If $\kappa$ is a chromatic cardinal, then for every countable logic $\mathcal{L}$ there is a $\lambda<\kappa$ such that $V\_\lambda\models\_\mathcal{L}\mathscr{M}(\mathcal{L})$. Basically, for a given $\mathcal{L}$ with formulas $(\varphi\_i)\_{i\in\omega}$ and inaccessible cardinals $\alpha<\beta<\kappa$ we let $c\_i(\{\alpha,\beta\})=1$ iff $V\_\beta\models\varphi(\alpha)$. And essentially trivially, this is optimal.
My first question is about chromaticity itself:
>
> **Question 1**: What are chromatic cardinals in more familiar language?
>
>
>
I suspect chromaticity is *much* weaker than Mahlo-ness, but I don't immediately see how to prove that.
My second question is about the specific strength of the second-order analogue of $\mathsf{ZFC}$ + reflection as indicated above:
>
> **Question 2**: What is the consistency strength of "There is an inaccessible $\alpha$ such that $V\_\alpha\models\_\mathsf{SOL}\mathscr{M}(\mathsf{SOL})$"?
>
>
>
It's easy to see that this is weaker than the existence of a chromatic cardinal; I'm interested in getting a better sense of how much weaker it is.
| https://mathoverflow.net/users/8133 | On the strength of higher-logic analogues of $\mathsf{ZFC}$ + Montague's Reflection Principle | The first chromatic cardinal is the first Mahlo cardinal. (Per the connection with the reflection in the question, I assume that in the definition, $α$ and the first argument of $c\_i$ need not be inaccessible.) If we allowed $c\_i$ for $i≤δ$, then the first chromatic cardinal would be the first Mahlo above $δ$.
If $κ$ is Mahlo, then it is chromatic. Define club $C⊆κ$ with $λ∈C ⇔ ∀α<λ \, ∀i \, \min(\{β>α:c\_i(α,β)\})<λ$ where $\min(S)$ is modified to return 0 if $S$ is empty. Then any inaccessible $λ∈C$ satisfies the desired property.
In the other direction, if $κ$ is below the first Mahlo, then there is a one-to-one function that for every inaccessible below $κ$ returns a smaller ordinal. For example, pick a club $D$ that excludes inaccessibles. If $β$ is the least inaccessible above some element of $D$, then set $f(β)=\max(β∩D)$. Let $λ=\min(D\setminus β)$, and if there are inaccessibles in $(β,λ)$, let $λ'$ be largest inaccessible or limit of inaccessibles $<λ$. Set (for example) $f(λ')=β$, and analogously to $D$ pick a club $E⊂λ'$ above $β$ that excludes inaccessibles. Analogously proceed by recursion until all inaccessibles are taken care of.
For your second question, despite the power of second order logic, the cardinals in question are precisely inaccessible $α$ with $V\_α ≺\_{Σ\_{1,V}} L\_1(V\_{α+1})$, where $Δ\_{0,V}$ formulas allow querying about $x$ whether $∃α \, x=V\_α$ (for the true $V$), and $L$ is the constructible hierarchy. $Σ\_{1,V}$ formulas are upwards absolute, and $L\_1$ corresponds to definability, so the above reduces to the desired reflection relation. If there is an inaccessible $λ$ with $V\_λ ≺\_{Σ\_2} V$, then $α$ for the question exists for every parameter-free definable $Δ^V\_2$ logic (even if there are no Mahlos in $V$; also, we can allow parameters in $V\_λ$).
| 3 | https://mathoverflow.net/users/113213 | 431287 | 174,632 |
https://mathoverflow.net/questions/431277 | 5 | I need a general method for solving systems of logical equations like:
$$
\begin{equation\*}
\begin{cases}
c\_{0} = a\_{0} \land b\_{0}\\\\
c\_{1} = a\_{0} \land b\_{1} ⊕ a\_{1} \land b\_{0}\\\\
c\_{2} = a\_{0} \land b\_{2} ⊕ a\_{1} \land b\_{1} ⊕ a\_{2} \land b\_{0}\\\\
c\_{3} = a\_{1} \land b\_{2} ⊕ a\_{2} \land b\_{1}\\\\
c\_{4} = a\_{2} \land b\_{2}
\end{cases}
\end{equation\*}
$$
Where c is known and a and b are unknown variables.
This system is a system of logical nonlinear equations, I want to know if it is possible to find a general solution for such a system. The number of unknowns is 1 more than the number of equations. Solutions will be symmetric (a and b can be swapped).
The challenge is not unsolvable and there is an example of a [solution](https://macton.github.io/codingame/nintendo/00000001.html) . However, it seems to me that there is a simpler solution.
| https://mathoverflow.net/users/491919 | Solve system of logical equations | Let $A=a\_0+a\_1x+a\_2x^2\cdots+a\_nx^n$, $B=b\_0+b\_1x+\cdots+b\_mx^m$, and $C=AB=c\_0+\cdots+c\_{n+m}x^{n+m}$, where arithmetic occurs over $\mathbb{F}\_2$. Then your problem is exactly equivalent to recovering $\{A,B\}$ from $C$. This is the problem of [factorization of polynomials over finite fields](https://en.wikipedia.org/wiki/Factorization_of_polynomials_over_finite_fields). Although, remarkably, the factors can be computed in polynomial time, I don't there there are any particularly simple algorithms for doing so.
| 9 | https://mathoverflow.net/users/8938 | 431289 | 174,633 |
https://mathoverflow.net/questions/431299 | 0 | Consider a metric space $(X,d)$ and let $\kappa$ be a cardinal. We say that $(X,d)$ is $\kappa$-**homogenous**, if every (surjective) isometry $h:X\_1 \to X\_2$ between subspaces of $(X,d)$ of size $< \kappa$ extends into an automorphism of $(X,d)$, i.e., into an isometry $f:(X,d) \to (X,d)$ such that $f\upharpoonright X\_1 = h$. For instance, the plane $\mathbb{R}^2$ is (at least) $\omega$-homogenous when equipped with the standard Euclidean metric.
**Question:** Does there exists a *compact* metric space $(X,d)$ which is $\omega$-homogenous but not $\omega\_1$-homogenous? In other words, does there exists a $\omega$-homogenous compact metric space $(X,d)$ and countably infinite sets $X\_1,X\_2 \subseteq X$ which are isometric via $h$ but $h$ does not extend into an automorphism of $(X,d)$?
My guess is that the answer should be yes, but I'm having a hard time coming up with a concrete example. Part of the problem is that I can't think of any homogenous compact metric spaces besides $S^{n-1}$.
| https://mathoverflow.net/users/322046 | $\omega$-homogenous space which is not $\omega_1$-homogenous | No: if $X$ compact metric is $\omega$-homogeneous then it is $\omega\_1$-homogeneous.
Indeed let $f:X\_1\to X\_2$ be a bijective isometry, with $X\_1$ countable. Write $X\_1$ as increasing union of finite subsets $F\_n$. So $f\_{|F\_n}$ extends to a bijective isometry $g\_n$ of $X$. Since the isometry group of $X$ is compact metrizable under uniform convergence, up to extract, we can suppose that $g\_n$ uniformly converges to some bijective isometry $g$. Then on each $F\_n$, $g\_n$ eventually equals $f\_{|F\_n}$, so $g$ extends $f\_{|F\_n}$ for each $n$. Thus $g$ extends $f$.
| 1 | https://mathoverflow.net/users/14094 | 431309 | 174,638 |
https://mathoverflow.net/questions/431318 | 6 | I've been trying to understand the (4 line!) proof of Lemma 2.3 of [Limits of small functors](https://arxiv.org/abs/math/0610439), on small functors into copresheaf categories $\mathbf{Set}^\mathcal C$. To me it seems to be using that the functor $\mathbf{Set}^\mathcal C \to \mathbf{Set}^{\text{ob } \mathcal C}$, defined by precomposing with the inclusion $\text{ob } \mathcal C \hookrightarrow \mathcal C$ with $\text{ob } \mathcal C$ the discrete category of objects, creates isomorphisms, which is false of course. So I'm wondering if I'm missing something.
Let $S\colon \mathcal K \to \mathcal M$ be a functor with $\mathcal M$ small cocomplete. $S$ is called *small* on p653 of [1](https://arxiv.org/abs/math/0610439) if there exists a small full subcategory $\mathcal B \subseteq \mathcal K$ and a natural isomorphism $S \cong \text{Lan}\_J (S \circ J)$, where $J \colon \mathcal B \hookrightarrow \mathcal K$ is the inclusion and $\text{Lan}\_J (S \circ J)$ is the left Kan extension of $S \circ J$ along $J$ (which exists by the assumption on $\mathcal M$).
Now if $\mathcal M = \mathbf{Set}^\mathcal C$ is a copresheaf category then there is a weaker notion as well: $S \colon \mathcal K \to \mathbf{Set}^\mathcal C$ is *pointwise small* if for each $C \in \mathcal C$ the composite $\text{ev}\_C \circ S \colon \mathcal K \to \mathbf{Set}$ is small, where $\text{ev}\_C \colon \mathbf{Set}^\mathcal C \to \mathbf{Set}$ is evaluation at $C$. The lemma asserts
**Lemma.** If $\mathcal C$ is small and $S \colon \mathcal K \to \mathbf{Set}^\mathcal C$ is pointwise small then $S$ is small.
Following the proof and using the pasting lemma for left Kan extensions (e.g. Theorem 4.47 of [Kelly's book](http://www.tac.mta.ca/tac/reprints/articles/10/tr10abs.html)) it is straightforward to see that the hypotheses imply the existence of a small full subcategory $J \colon \mathcal B \hookrightarrow \mathcal K$ such that $\text{ev}\_C \circ S \cong \text{Lan}\_J (\text{ev}\_C \circ S \circ J) \colon \mathcal K \to \mathbf{Set}$ simultaneously for all $C$. Now left Kan extensions into $\mathbf{Set}$ are defined pointwise, that is $\text{Lan}\_J (\text{ev}\_C \circ S \circ J)(Z)$ is a colimit for each $Z \in \mathcal K$ and, fixing $Z$, there is a unique way in which these colimits combine into a colimit that defines $\text{Lan}\_J(S \circ J)(Z) \colon \mathcal C \to \mathbf{Set}$ (See sections X.3 and V.3 of "Categories work"). The uniqueness here is such that the family of universal cocones combines into the universal cocone defining the latter colimit and such that
$$\text{Lan}\_J(S \circ J)(Z)(C) = \text{Lan}\_J(\text{ev}\_C \circ S \circ J)(Z).$$
We conclude that the left Kan extension $\text{Lan}\_J(S \circ J)\colon \mathcal K \to \mathbf{Set}^\mathcal C$ satisfies
$$\text{Lan}\_J(S \circ J)(Z)(C) \cong (\text{ev}\_C \circ S)(Z) = S(Z)(C)$$
and we need these isomorphisms to combine into a natural isomorphism $S \cong \text{Lan}\_J(S \circ J)$. But I don't see how they are natural in the variable $C$: there does not seem to be a reason for the unique functoriality of $\text{Lan}\_J(S \circ J)(Z) \colon \mathcal C \to \mathbf{Set}$ to coincide with that of $S(Z)$. In particular there seems to be no reason for the family of natural isomorphisms $\text{ev}\_C \circ S \cong \text{Lan}\_{J\_C} (\text{ev}\_C \circ S \circ J\_C) \colon \mathcal K \to \mathbf{Set}$, exhibiting the smallness of the $\text{ev}\_C \circ S$, to be natural in $C$ in whatever sense.
**Question 1.** What am I missing here?
It can also be that I misunderstood the notion of smallness as follows. If the universal transformations defining the relevant left Kan extensions are canonical in the following way then the lemma does hold when the last two mentions of "small" are replaced by "nicely small":
**Definition** $S\colon \mathcal K \to \mathcal M$ is *nicely small* if there exists a small full subcategory $J\colon \mathcal B \to \mathcal K$ such that for all $Z \in \mathcal K$ the cowedge
$$ \mathcal K(Z, JB) \times (S \circ J)(B) \to SZ, $$
given by the action of $S$ on morphisms, defines $SZ$ as the coend $\int^{B \in \mathcal B} K(Z, JB) \times (S \circ J)(B)$.
It is straightforward to prove that nice smallness implies smallness. I've tried to prove the opposite but I seem to run into a "naturality issue" similar to the above one. With nice smallness instead of smallness one can prove the lemma: roughly, this is because the universal cowedges defining the $\text{ev}\_C \circ S$ as left Kan extensions of their own restrictions are now given by the action of $S$ on morphisms in $Z$, which commutes with the action of $S$ on morphisms in $C$.
**Question 2.** I would like to hear any thoughts on the relation between nice smallness and smallness. Surely it is not the case that Day and Lack are implicitly using the notion of nice smallness instead of that of smallness?
| https://mathoverflow.net/users/133974 | Day and Lack's "Limits of small functors": Lemma 2.3 | Small and nicely small are indeed equivalent. I would consider this a fairly classical observation in the topic of small functor, at least when $M = Set$, so I wouldn't be surprised that Day and Lack are using it implicitly - maybe without realizing it - but I haven't had time to go re-read their paper in details to answer that part of the question.
Of course in the case $M = Set$ the question of functionality in $c \in C$ doesn't appear, but my point is that the argument is that "nicely small objects are closed under small colimits" and is a purely "levelwise" thing when we had dependency in $c \in C$.
The idea is as follows:
I'm starting with the case of $M = Set$ first for simplicity, and also because this is the case I would consider a standard fact and on which I'm confident I won't say something false.
If $J$ is a small full subcategory of $K$, then we have an adjunction $L : Fun(J,Set) \leftrightarrows Fun(K,Set): R$ given by $L$ the Kan extension and $R$ the restriction. As Kan extensions are computed pointwise (and the coend version of the Yoneda Lemma I guess) the unit $X \to RL X$ is an isomorphism. So $L$ is a fully faithful inclusion. The condition of "nice smallness" you are expressing (with $J$ fixed) is expressing that the counit is an isomorphism, i.e. that the object $S$ is in the essential image of $L$.
This has two consequences:
1. If $S$ is nicely small for a given $J$, then it is also nicely small for any $J'$ such that $J \subset J'$. (because $L\_J$ will factor through $L\_{J'}$).
2. The full subcategory of nicely small objects with respect to $J$ is closed under small colimits (because $L$ is fully faithful and preserves colimits).
Put together this implies that nicely small objects are closed under any small colimits as given a small diagram of such one can always find a "small but big enough" $J$ that contains all the full subcategories of $K$ needed to see that all the objects appearing in the diagram are nicely small. It follows that any small colimit of representable is "nicely small", and hence any small functor is "nicely small".
Now, unless I'm completely missing the problem you are talking about, in the discussion above, everything seems to be nicely functorial if we add a parameter $c \in C$, and carries over immediately to the case of $M = Set^C$ hence solving your problem.
| 8 | https://mathoverflow.net/users/22131 | 431323 | 174,642 |
https://mathoverflow.net/questions/431320 | 4 | Suppose we are in the following situation: $(X,d)$ is a metric space and $Y$ is a subspace of $X$. Furthermore we have a different metric $\delta$ defined on $Y$ such that $\delta$ is bi Lipschitz equivalent to $d|\_Y$. Is it possible to extend $\delta$ to e metric $\bar{\delta}$ on the whole $X$ such that $\bar{\delta}$ is bi Lipschitz equivalent to $d$?
I suspect the answer to be no in such generality, but I would also be interested in particular cases of metric spaces for which the answer is positive.
| https://mathoverflow.net/users/153260 | Extending a metric in a bi-Lipschitz way | Yes. Up to multiply $d$ with a scalar, we can suppose that for some $0<c\le 1$ we have $cd \le\delta\le d$ on $Y\times Y$.
Define $$d'(x,x')=\min(d(x,x'),D(x,x'));\quad \text{where}$$ $$D(x,x')=\inf\_{y,y'\in Y} d(x,y)+\delta(y,y')+d(y',x').$$
Clearly $cd\le d'\le d$ on $X\times X$. It remains to prove the triangle inequality for $d'$: $d'(x,x'')\le d'(x,x')+d'(x',x'')$.
There are four cases to consider.
1. If $d'(x,x')=d(x,x')$ and $d'(x',x'')=d(x',x'')$, then $d'(x,x'')\le d(x,x'')\le d(x,x')+d(x',x'')=d'(x,x')+d'(x',x'')$.
2. Suppose $d'(x,x')=D(x,x')$ and $d'(x',x'')=d(x',x'')$. Fix $\varepsilon>0$. Choose $y,y'\in Y$ such that $d'(x,x')\ge d(x,y)+\delta(y,y')+d(y',x')-\varepsilon$. Then
$$d'(x,x')+d(x',x'')\ge d(x,y)+\delta(y,y')+d(y',x')+d(x',x'')-\varepsilon$$
$$\ge d(x,y)+\delta(y,y')+d(y',x'')-\varepsilon\ge d'(x,x'')-\varepsilon.$$
Since $\varepsilon$ is arbitrary, the inequality follows.
3. Case $d'(x,x')=d'(x,x')$ and $d'(x',x'')=D(x',x'')$: reduces to the previous case by switching $x$ and $x''$.
4. Suppose $d'(x,x')=D(x,x')$ and $d'(x',x'')=D(x',x'')$. Fix $\varepsilon>0$. Fix $y,y'\_1,y'\_2,y''\in Y$ such that $d'(x,x')\ge d(x,y)+\delta(y,y'\_1)+d(y'\_1,x')-\varepsilon$ and $d'(x',x'')\ge d(x',y'\_2)+\delta(y'\_2,y'')+d(y'',x'')-\varepsilon$.
So $$d'(x,x')+d'(x',x'')\ge $$
$$d(x,y)+\delta(y,y'\_1)+d(y'\_1,x')+d(x',y'\_2)+\delta(y'\_2,y'')+d(y'',x'')-2\varepsilon$$
$$\ge d(x,y)+\delta(y,y'\_1)+d(y'\_1,y'\_2)+\delta(y'\_2,y'')+d(y'',x'')-2\varepsilon$$
$$\ge d(x,y)+\delta(y,y'\_1)+\delta(y'\_1,y'\_2)+\delta(y'\_2,y'')+d(y'',x'')-2\varepsilon$$
$$\ge d(x,y)+\delta(y,y'')+d(y'',x'')-2\varepsilon$$
$$\ge d'(x,x'')-2\varepsilon.$$
Since $\varepsilon$ is arbitrary, we deduce the triangle inequality.
| 4 | https://mathoverflow.net/users/14094 | 431324 | 174,643 |
https://mathoverflow.net/questions/431185 | 0 | **Setup:$\quad$**
Suppose that $(X\_n)$ is a stationary ergodic process with $E|X\_1|<\infty$.
Given $X^{(n)}=(X\_1, \dots, X\_n)$, select a standard Efron bootstrap subsample $(X\_{n,1}^\*, \dots, X\_{n,m(n)}^\*)$ by pulling $m(n)$ times with replacement from a uniform distribution $U(\{X\_1, \dots, X\_n\})$, i.e.
$$
X\_{n,i}^\* = X\_{Z\_{n,i}}, \quad Z\_{n,i} \overset{\text{iid}}{\sim} U(\{1,\dots,n\}), \quad i=1,\dots,m(n),
$$
where the $Z\_{n,i}$ are independent of $(X\_n)$ and form a triangular array with independent rows.
Let the ***bootstrap mean*** $\mu\_{m(n)}^\*$ be the sample mean of the bootstrap subsample, i.e.
$$
\mu\_{m(n)}^\* = \frac{1}{m(n)} \sum\_{i=1}^{m(n)} X\_{n,i}^\*.
$$
**Question:**
In the case that $(X\_n)$ is a stationary ergodic process, are there any known results about when the following WLLN holds?
$$
\mu\_{m(n)}^\* \overset{P}{\longrightarrow} E[X]
$$
as $n \to \infty$.
**What I've found:**
* In the case that the $X\_i$ are i.i.d. and $m(n) \to \infty$, it is known that the WLLN above holds for any $m(n) \to \infty$ (e.g. see p.2848 of this 2003 [survey](https://www.emis.de/journals/HOA/IJMMS/Volume2003_45/825942.pdf) by Csörgő and Rosalsky).
* Additionally, Einmal and Rosalsky later [proved](https://www.tandfonline.com/doi/full/10.1081/SAP-200064490?scroll=top&needAccess=false) that
$$
\mu\_{m(n)}^\* - \frac{1}{n} \sum\_{i=1}^n X\_i \overset{P}{\longrightarrow} 0
$$
holds for any $(X\_n)$ (not necessarily independentent or identically distributed), provided $m(n) \uparrow \infty$ and
$$
\frac{X\_n}{\sqrt{m(n)}} \overset{\text{a.s.}}{\longrightarrow} 0.
$$
This, however, doesn't cover all stationary ergodic processes with $E|X\_1|<\infty$.
| https://mathoverflow.net/users/104268 | WLLN for bootstrap means of stationary ergodic processes? | **Answered in comments above**
It seems as though the answer should be yes. I would suggest writing $X\_n$ as $Y\_n+Z\_n$ where $Y\_n$ is $X\_n$ if $|X\_n|\le m(n)^{1/3}$ and 0 otherwise; similarly $Z\_n$ is $X\_n$ if $|X\_n|>m(n)^{1/3}$ and 0 otherwise. Then the Einmal and Rosalsky result applies to the Bootstrap averages of the $Y\_n$, so all that remains is to check that the Bootstrap averages of the $Z\_n$ approach 0 in probability. I believe that follows from Markov's inequality once you know that $\mathbb EZ\_n\to 0$.
| 0 | https://mathoverflow.net/users/11054 | 431335 | 174,644 |
https://mathoverflow.net/questions/431313 | 2 | Let $S$ be a hyperbolic surface of genus $g \geq 2$.
A *discrete* geodesic lamination on $S$ is a set of disjoint, simple, closed geodesics.
Let $L\_{1}$ and $L\_{2}$ be two discrete geodesic laminations on $S$. We say that $L\_1$ and $L\_2$ fill $S$, if $S \setminus (L\_{1} \cup L\_{2})$ is a disjoint union of open topological disks.
Or equivalently, let $i(\alpha,L\_{i})$ be the intersection number of the closed curve $\alpha$ with the leaves of the lamination $L\_{i}$, then for any closed curve $\alpha$, $i(\alpha,L\_{1}) + i(\alpha,L\_{2}) > 0$ (including the closed curves that belong to $L\_1$ and $L\_2$, and recall that if $\alpha \in L\_{i}$, then $i(\alpha,L\_{i}) = 0$).
**My question is**: If we take two discrete geodesic laminations $L\_1$ and $L\_2$ on $S$, such that $L\_{1}$ and $L\_{2}$ have no common leaf. Can we add leaves to $L\_1$ and leaves to $L\_2$, to obtain two laminations $L^{'}\_{1}$ and $L^{'}\_{2}$ (then $L\_{i} \subset L^{'}\_{i}$) such that $L^{'}\_{1}$ and $L^{'}\_{2}$ fill $S$ ?
Since every geodesic lamination can be completed to $3g-3$ closed curve (then if $\alpha \notin L$, $i(\alpha,L) > 0$), **the question remains to the following:** Let $\alpha$,$\beta\_{1}$,...,$\beta\_{k}$ be disjoint, simple, closed geodesics in $S$. Can we find a simple closed geodesic $\gamma$ such that $i(\gamma,\alpha) >0$ and $\forall j$,$i(\gamma,\beta\_{j}) = 0$.
| https://mathoverflow.net/users/169634 | Pair of laminations that fill on a closed surface | The answer is "no". For consider the case where $L\_1 = L\_2$ is a single simple closed geodesic.
---
Now you've added the hypothesis that $L\_1$ and $L\_2$ have no common leaf, the answer becomes "yes". Here is a sketch.
Let $X\_i = S - L\_i$. If $X\_1$ (say) is a union of pants then $L\_1$ is a pants decomposition and so is maximal. Suppose not. Then we pick a lamination $\lambda\_1 \subset X\_1$ which restricts in each non-pants component $X' \subset X\_1$ to a *filling lamination*. That is, a closed subset of $X'$ which is a union of simple geodesics, none of which are loops, so that $X' - \lambda\_1$ is a union of disks and peripheral annuli. Furthermore $\lambda\_1$ admits a transverse measure (so has no leaves that spiral about a component of $L\_1$). Filling laminations exist by Thurston's theory of pseudo-Anosov maps.
We do the same for $L\_2$: that is, we pick a lamination $\lambda\_2$ that lives in and fills all non-pants components of $X\_2$. Now define $\Lambda\_i = L\_i \cup \lambda\_i$
**Exercise**: Since $L\_1$ and $L\_2$ share no leaves the same holds of $\Lambda\_1$ and $\Lambda\_2$.
We deduce that there is a definite angle $\epsilon$ so that, at every point of $\Lambda\_1 \cap \Lambda\_2$, the two laminations intersect with an angle of at least $\epsilon$. In particular $\Lambda\_1 \cup \Lambda\_2$ fills $S$.
We now choose sufficiently close Hausdorff approximations of $\Lambda\_i$ by pants decompositions $L\_i'$. We deduce that $L\_i'$ contains $L\_i$. Also $L\_1'$ and $L\_2'$ fill, as desired.
| 2 | https://mathoverflow.net/users/1650 | 431340 | 174,645 |
https://mathoverflow.net/questions/405935 | 3 | The paper 'Trisections, intersection forms and the Torelli group' by Peter Lambert-Cole quotes the following formula for the Casson invariant of a knot $K$ in a homology $3$-sphere in terms of the linking form $l$ on a Seifert surface $\Sigma$ for $K$ of genus $g$:
$$
\lambda'(K)=\sum\_{i=1}^g \big(l(a\_i,a\_i)l(b\_i,b\_i)-l(a\_i,b\_i)l(a\_i,b\_i)\big)+\sum\_{1 \le i<j \le g} \big(l(a\_i,a\_j)l(b\_i,b\_j)-l(a\_i,b\_j)l(a\_j,b\_i)\big).
$$
Here $\{a\_i,b\_i\}$ is a geometric symplectic basis for $\Sigma$.
Does anyone have a reference for this formula? I have looked for it in both Saveliev's and Akbulut & McCarthy's books, but I was not able to find anything similar.
| https://mathoverflow.net/users/156392 | Formula for the Casson invariant in terms of the linking form | Maybe you've already found a source or this is irrelevant by now, but I believe I took it from Morita's paper "Casson's invariant for homology 3-spheres and characteristic classes of surface bundles I". Specifically Proposition 3.2.
| 3 | https://mathoverflow.net/users/126853 | 431345 | 174,647 |
https://mathoverflow.net/questions/430869 | 2 | Who first used the corner quotes, $\ulcorner$ and $\urcorner$, for the notion of Gödel number? They can also be written as`\Godelnum` with Sam Buss's macro.
They were used by Joseph R. Shoenfield, in *Mathematical Logic*, 1967, as from page 122.
The corner quotes are used prevalently in provability logic, and in other areas of logic. Two important citations are Craig Smorynski, “The Incompleteness Theorems”, in *Handbook of Mathematical Logic*, as from 1st edition 1979; and Paanu Raatikainen, “Gödel's incompleteness theorems”, in [Stanford Encyclopedia of Philosophy](https://plato.stanford.edu/entries/goedel-incompleteness/).
Edit: The corner quotes $\ulcorner\urcorner$ were indeed first used by Quine, and long before Shoenfield, but not for the notion of Gödel number.
Edit 2: I do not find the corner quotes in Martin Davis, *The Undecidable*, or in the books by Moztowski or Kleene.
| https://mathoverflow.net/users/37385 | First use of corner quotes for Gödel numbers | While this is not a full answer, I hope the following observations can still be of some use.
Kreisel and Lévy were using corner quotes, explicitly for Gödel numbers, around the same time as Schoenfield (G. Kreisel and A. Lévy, [Reflection Principles and their Use for Establishing the Complexity of Axiomatic Systems](https://doi.org/10.1002/malq.19680140702), Zeitschr. f. math. Logik und Grundlagen d. Math. 14(7-12):97-142, 1968; but note that the paper was submitted on 12 December 1966).
The same notation is used in a related way in [three earlier abstracts of Kreisel's](https://www.jstor.org/stable/2271325) (The subformula property and reflection principles, JSL 28(4):305-306, 1963; Reflection principle for Heyting's arithmetic, JSL 28(4):306-307, 1963; Reflection principles and $\omega$-consistency, JSL 28(4):307-8, 1963). There, a distinction is made between free variables $n$ ranging over proofs and $\ulcorner A \urcorner$ ranging over formulae in expressions such as $\mathrm{Prov}(n,\ulcorner A\urcorner)$.
| 4 | https://mathoverflow.net/users/103227 | 431347 | 174,648 |
https://mathoverflow.net/questions/431331 | 5 | Let $(P,\leq)$ be a directed set with uncountable cofinality. For every element $p\in P$, we are given a finite set $c\_p\subset P\smallsetminus \{p\}$ of "incompatible elements". We say that a subset $Q\subseteq P$ is *compatible* if for every $q\in Q$ we have $c\_q\cap Q = \emptyset$.
**Question:** is it always possible to find a *cofinal* compatible subset?
*Remarks:*
* I can show that the answer is yes if there is a finite uniform upper bound on the cardinalities of the sets $c\_p$.
* It was pointed out to me that the assumption that the cofinality be uncountable is certainly necessary, as seen by considering $(\mathbb{N},\leq)$ with $c\_n = \{1,2,...,n-1\}$.
* I suspect that a (positive) solution would require some argument in infinite combinatorics.
* I am very happy to assume the axiom of choice (or any other useful set of axioms).
| https://mathoverflow.net/users/54309 | Searching for cofinal subsets of directed sets subject to finite constraints | No, it is not always possible. For a counterexample, let $P$ be the set of all finite subsets of some uncountable set $X$, ordered by inclusion (that is, $a \leq b \Leftrightarrow a \subseteq b$).
In this poset, a subset $D$ of $P$ is cofinal if and only if for every finite subset $a$ of $X$, there is some $b \in D$ with $b \supseteq a$.
For each $a \in P$, let $c\_a$ be the (finite) set of all proper subsets of $a$. Suppose $D$ is cofinal in $P$, and fix some $a \in D$. There is some $x \in X \setminus a$, and (because $D$ is cofinal) there is some $b \in D$ with $b \supseteq a \cup \{x\}$. But now we have $b \in D$ and $a \in D$ and $a \in c\_b$.
| 4 | https://mathoverflow.net/users/70618 | 431353 | 174,649 |
https://mathoverflow.net/questions/431152 | 6 | A characteristic feature of [Berkovich spaces](https://en.wikipedia.org/wiki/Berkovich_space) is that they are locally connected (in fact, locally contractible). I'd like to understand the proof. The key ingredient seems to be Corollary 2.2.8 in Berkovich's [book](https://www.google.com/books/edition/Spectral_Theory_and_Analytic_Geometry_Ov/6x5PIl2-DkkC?hl=en). A key step in the proof is to take a connected component $W\_n$ of a certain set $V\_n$, but there is no explanation of why this connected component should be open (as I think it needs to be). I'd like to understand why.
**Question:** Why are Berkovich spaces locally connected?
Ideally, I'd appreciate a clarification of Berkovich's proof, or else a pointer to some other proof in the literature. Also, I'd be happy to understand the case of the Berkovich spectrum of a ring.
| https://mathoverflow.net/users/2362 | Why are Berkovich spaces locally connected? | Let $X = \mathcal{M}(A)$ be an affinoid space. There is a map $\mathcal{M}(A) \to \operatorname{Spec}(A)$ sending a seminorm to its kernel. It is continuous, surjective and induces a bijection between the connected components of $\mathcal{M}(A)$ and $\operatorname{Spec}(A)$. Let me prove the last point.
First assume that $\operatorname{Spec}(A)$ is connected. Then $\mathcal{M}(A)$ is connected too. Indeed, consider a clopen subset $U$ of $\mathcal{M}(A)$. We can construct an analytic function $e$ on $\mathcal{M}(A)$ that is 0 on $U$ and 1 on the complement. But the global sections on $\mathcal{M}(A)$ are nothing but $A$ itself (Tate's theorem), so $e$ defines an element of $A$, which is idempotent. Since $\operatorname{Spec}(A)$ is connected, $A$ has only the trivial idempotents, so $e$ equals 0 or 1, hence $U$ is either the whole $\mathcal{M}(A)$ or the empty set.
For the general case, pick a connected component of $\operatorname{Spec}(A)$. It is a Zariski-closed subset of $\operatorname{Spec}(A)$, hence of the form $\operatorname{Spec}(B)$, with $B$ a quotient of $A$, so in particular an affinoid algebra. The previous argument shows that its preimage $\mathcal{M}(B)$ is connected. This finishes the proof of the bijection.
Finally, recall that the affinoid algebra $A$ is noetherian, so $\operatorname{Spec}(A)$ has finitely many connected components, and similarly for $\mathcal{A}$. As a consequence, those connected components are open.
I think this is what you need to complete the proof of local connectedness.
I am not sure what you mean about the Berkovich spectrum of a ring at the end of your post, so I will not comment on that.
| 2 | https://mathoverflow.net/users/4069 | 431354 | 174,650 |
https://mathoverflow.net/questions/431338 | 5 | Consider, $M$, a smooth $m$ dimensional submanifold of $\mathbf R^n$. Does there exist a smooth map $X: \mathbf{R}^m\to\mathbf R^n$ such that $M=X(\mathbf R^m)$?
$X$ may have points at which the Jacobian is singular, which means that $M$ doesn't have to be diffeomorphic to $\mathbf{R}^m$. Furthermore, the stereographic map shows that there exists a smooth map between manifolds of different topology, so that $M$ and $\mathbf{R}^m$ apparently don't even have to be homoeomorphic. This then raises the question whether any $M$ can be represented thus.
| https://mathoverflow.net/users/485792 | Can a smooth manifold be realised as the image of a smooth function? | My comment turned answer:
Any smooth $m$-manifold $M$ admits a complete Riemannian metric (for example, as [this answer](https://mathoverflow.net/a/18851/172802) says, any manifold embeds into some Euclidean space as a closed subset by Whitney embedding theorem).
So if we endow any connected smooth manifold $M$ with a complete metric and choose any point $p\in M$, then the exponential map $exp\_p:T\_pM\to M$ gives a smooth map (Prop 5.7c of [1]) which is surjective, because any two points in $M$ are joined by a geodesic (Cor 6.15 of [1]).
[1]: John M. Lee. *Riemannian Manifolds. An introduction to curvature*. Springer, 1997.
| 13 | https://mathoverflow.net/users/172802 | 431358 | 174,651 |
https://mathoverflow.net/questions/431343 | 2 | Suppose $\mathfrak{g}$ is a real form of a semisimple Lie algebra $\mathfrak{g}\_\mathbb{C} = \mathfrak{g} \otimes\_\mathbb{R} \mathbb{C}$. Then we have the following:
* There is an equivalence of monoidal categories between the category of finite-dimensional *complex* representations of $\mathfrak{g}$ and the category of finite-dimensional *complex* representations of $\mathfrak{g}\_\mathbb{C}$.
* One can classify the *irreducible* real representations of $\mathfrak{g}$. These are either restrictions to $\mathfrak{g}$ of irreducible complex representations of $\mathfrak{g}\_\mathbb{C}$ (which remain irreducible over $\mathfrak{g}$) or real forms of irreducible complex representations of $\mathfrak{g}\_\mathbb{C}$ (in this case there is a real structure on the underlying space of the representation that commutes with the action of $\mathfrak{g}$). See, for example, Theorem 1 on page 65 of Onishchik, *Lectures on Real Semisimple Lie Algebras*.
What I'd like to know is if the category of finite-dimensional *real* representations of $\mathfrak{g}$ is semisimple. In other words, is every finite-dimensional *real* representation of $\mathfrak{g}$ completely reducible? Despite spending quite some time searching through the literature, I can't seem to find the answer to this question.
| https://mathoverflow.net/users/29738 | Are finite-dimensional real representations of semisimple real Lie algebras completely reducible? | As pointed out by YCor in the comments, the answer is Yes. A reference has been pointed out to me: Chapter III, Section 7, Theorem 8 in Jacobson's book *Lie algebras*.
| 1 | https://mathoverflow.net/users/29738 | 431359 | 174,652 |
https://mathoverflow.net/questions/431363 | 2 | I have seen an equivalence claimed in a few places, but I do not know of a reference that actually proves it with details and it has been a while since I took graduate courses on all this. Apologies if this is something standard. A reference would be appreciated, particularly if it illuminates the general picture.
The simplest concrete version of my question is the following. Let
$$\mathcal{H} = -\Delta + |x|^2$$
be the quantum harmonic oscillator Hamiltonian in $\mathbb{R}^d$ and for $k\in\mathbb{N}$ define a Hilbert space by taking the closure of the Schwarz class functions $\mathcal{S}(\mathbb{R}^d)$ in the norm defined by
$$
\|f\|\_k^2 = \int\_{\mathbb{R}^d}|\mathcal{H}^k f(x)|^2dx
$$
I have seen it claimed in a few places (for example Proposition 2.3 in <https://aip.scitation.org/doi/abs/10.1063/1.5048726?journalCode=jmp>) that this norm is equivalent to the norm given by a combination of a weighted $L^2$ norm and an ordinary Bessel-Sobolev norm
$$
|||f|||\_k^2 = \|\mathcal{F}^{-1}(1+|\xi|^2)^{k}\mathcal{F}f\|^2\_{L^2(\mathbb{R}^d)} + \| |x|^{2k}f \|^2\_{L^2(\mathbb{R}^d)}
$$
I think I see that $\|\cdot\|\_k$ is controlled by $|||\cdot|||\_k$, but I do not know how to prove the reverse inequality.
Based on how I have seen this talked about (for example, the proof of that proposition in that reference) my understanding is that there is a simple proof of this that begins by viewing $\mathcal{H}$ as the operator with Fourier symbol given by $a(x,\xi) = 4\pi^2|\xi|^2+|x|^2$.
Does anyone here know how such a proof would go? Or of any proof of this claim?
| https://mathoverflow.net/users/43134 | Equivalence of Hilbert space norm associated to the harmonic oscillator and a sum of Sobolev and weighted $L^2$ norms | Chapter 9 of my book [https://www-users.cse.umn.edu/~garrett/m/v/current\_version.pdf](https://www-users.cse.umn.edu/%7Egarrett/m/v/current_version.pdf)
(that is mostly aimed at applications to automorphic forms) considers this. See section 9.8, in particular.
I really don't know of other references for this sort of computation/comparison, but, while it's a bit annoying, it's not toooo subtle. :)
| 3 | https://mathoverflow.net/users/15629 | 431364 | 174,654 |
https://mathoverflow.net/questions/431368 | 4 | $\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\PSp{PSp}\DeclareMathOperator\USp{USp}\DeclareMathOperator\BSp{BSp}\DeclareMathOperator\BUSp{BUSp}\DeclareMathOperator\BPSp{BPSp}$Let $\USp(n,\mathbb{C})$ and $\Sp(n,\mathbb{C})$ be the compact symplectic group and the symplectic group, respectively. This is,
$$
\Sp(n,\mathbb{C})=\left\{A\in M(2n,\mathbb{C}): A^{tr}J\_{2n}A=J\_{2n}\right\}
$$
where $M(n,\mathbb{C})$ is the set of $n\times n$ matrices with entries in $\mathbb{C}$,
$$
J\_{2n}=\begin{pmatrix}
0 & I\_{n}\\
-I\_{n} & 0\\
\end{pmatrix},
$$
and $\USp(n,\mathbb{C})$ is isomorphic to $\Sp(n,\mathbb{C})\cap \mathrm{U}(2n,\mathbb{C})$.
Let $\PSp(n,\mathbb{C})$ be the projective symplectic group, i.e. $\Sp(n,\mathbb{C})/\{\pm I\_{2n}\}$.
It can be shown that the inclusion $\USp(n,\mathbb{C}) \to \Sp(n,\mathbb{C})$ is a homotopy equivalence.
[Hatcher, AT, pg 271 and 381] says that the quaternionic Grassmannian $G\_{m}(\mathbb{H}^{\infty})$ is a model for $\BUSp(n,\mathbb{C})$, and that the integral cohomology $H^{\*}(G\_{m}(\mathbb{H}^{\infty}))\cong \mathbb{Z}[\alpha\_{1},\alpha\_{2},\dots,\alpha\_{m}]$ with $|\alpha\_{i}|=4i$.
This result makes you think that $\BSp(n,\mathbb{C})$ has a CW structure with cells only in degrees $4i$. I would like to find such structure for $\BSp(n,\mathbb{C})$ and also, know whether this structure induces one of the same kind in $\BPSp(n,\mathbb{C})$.
The closest thing I know about their CW structures is that $G\_{m}(\mathbb{H}^{\infty})$ has a CW structure with each $G\_{m}(\mathbb{H}^{k})$ a finite subcomplex, [Hatcher, VB and Kthy, pg 31-34].
Thank you for your ideas.
| https://mathoverflow.net/users/121001 | CW structure for $\mathrm{BSp}(n,\mathbb{C})$ and $\mathrm{BPSp}(n,\mathbb{C})$ in degrees $4i$ | I'll assume that by $BG$ you mean ``any space of the form $E/G$, where $E$ is a contractible space with free $G$-action''. (The alternative would be to define $BG$ as the geometric realisation of a specific simplicial space, but then $BU(n)$ would not have even cells.) But then you can take $G=Sp(n,\mathbb{C})=\text{Aut}\_{\mathbb{H}}(\mathbb{H}^n)$ and $E=\text{Inj}\_\mathbb{H}(\mathbb{H}^n,\mathbb{H}^\infty)$ and we get $BSp(n,\mathbb{C})=G^{\mathbb{H}}\_n(\mathbb{H}^\infty)$. This has a CW structure based on quaternionic Schubert cells (which have real dimension divisible by $4$) just as in the real and complex cases. We can also describe $G^{\mathbb{H}}\_n(\mathbb{H}^\infty)$ as the space of isometric $\mathbb{H}$-linear embeddings $\mathbb{H}^n\to\mathbb{H}^\infty$ mod the action of isometric $\mathbb{H}$-linear automorphisms of $\mathbb{H}^n$, so $G^{\mathbb{H}}\_n(\mathbb{H}^\infty)$ qualifies as $BUSp(n,\mathbb{C})$ as well as $BSp(n,\mathbb{C})$.
There is an inclusion $SO(3)\simeq PSU(2)\to PSp(1,\mathbb{C})$, which is a homotopy equivalence and so gives a homotopy equivalence of classifying spaces. It is known that $H^\*(BSO(3);\mathbb{Z}/2)\simeq\mathbb{Z}/2[w\_2,w\_3]$ with $|w\_2|=2$ and $|w\_3|=3$. This implies that any CW complex homotopy equivalent to $PSp(1,\mathbb{C})$ must have odd-dimensional cells.
| 4 | https://mathoverflow.net/users/10366 | 431385 | 174,659 |
https://mathoverflow.net/questions/430915 | 4 | Let $\Sigma$ be a compact orientable connected $2$-manifold with a non-empty boundary. Let $\widehat \pi(\Sigma)$ denote the set of free homotopy classes of
curves in $\Sigma$. We say $x\in \widehat \pi(\Sigma)$ is peripheral, if there is representative $\alpha\colon \Bbb S^1\to \Sigma$ of $x$ with $\text{im}(\alpha)\subseteq \partial \Sigma$.
>
> **Question:** Let $x\in\widehat \pi(\Sigma)$ be a *non-trivial non-peripheral* element. Does there exist a simple closed curve
> $\beta\subseteq \Sigma$ such that the geometric intersection of
> $\alpha$ and $\beta$ is non-zero, where $\alpha$ is a representative
> of $x$.
>
>
>
| https://mathoverflow.net/users/363264 | Characterization of a non-trivial non-peripheral element of the free homotopy classes of a compact bordered surface | This is true for all (compact, connected, oriented) surfaces that admit essential non-peripheral simple closed curves.
The sphere, disk, annulus and pants do not admit essential non-peripheral curves. The sphere and disk have trivial fundamental group. So the answer is "yes" vacuously for the sphere and disk, and "no" for the annulus and pants (as suggested by Johannes’).
Suppose that $S$ is any other compact, connected, oriented surface.
Now $S$ admits *filling laminations*. Any sufficiently good (Hausdorff close) simple closed curve approximation to a filling lamination does what you want. The existence of such laminations is a consequence of Thurston’s theory of pseudo-Anosov maps.
| 3 | https://mathoverflow.net/users/1650 | 431391 | 174,661 |
https://mathoverflow.net/questions/431375 | 1 | I'm reading through the paper [Poincaré type and spectral gap inequalities with fractional Laplacians on Hamming cube](https://arxiv.org/abs/1802.04411).
However, I'm having a difficult time understanding the following proof: [Lemma 2.1 page 3](https://i.stack.imgur.com/epHpF.png)).
I understand the general goal of the proof, but the second half in which we establish a lower bound for the integral contains several steps that I find puzzling. Where (and how) do we make use of the fact that $\eta=c\_1^{0.1}$, and why was this specific value chosen? Where does the $\sqrt{3/4}$ come from? Why exactly does $0<\beta\leq 2$ imply that the smallness of $c\_1$ is independent of $\beta$? Any help would be greatly appreciated.
| https://mathoverflow.net/users/491992 | Proof of lower bound on variance | It looks to me like the authors just wanted to put in some ridiculous values which "are clearly sufficient" so that they don't have to work out the details.
The choice $\eta = c\_1^{1/10}$ should ensure that the term $\int\_{|g-1|\leq \eta} |g|^\beta \operatorname{sgn}(g) \, d\mu$ converges to 1 for $c\_1 \rightarrow 0$, uniformly across all $g$ satisfying $\mathbb{E}[g^2] = 1$ and $\mathbb{E}[(g-\mathbb{E}[g])^2 \leq c\_1$ and uniformly across all $\beta$ (this is where you need an upper bound on $\beta$). The term $\sqrt{3/4}$ is then just some value that is sufficiently close to 1 for the remainder of the proof.
To see this convergence, it suffices to show that $\mu(|g-1|\leq \eta)$ goes to 1 using, e.g., Chebyshev's inequality. To this end, we need to use the vague statement "$|1-\mathbb{E}[g]| \ll 1$", which should be made rigorous by establishing a bound on $|\mathbb{E}[g] - 1|$ only depending on $c\_1$. I haven't worked out the details, but I guess something like $|\mathbb{E}[g] - 1| \leq f(c\_1)$, where $f(x) \approx x^{1/2}$ seems reasonable. The choice of $\eta$ is now simply large enough such that \begin{align}
\mu(|g-1| > \eta) \leq \mu(|g-\mathbb{E}[g]| > \eta - f(c\_1)) \leq \mu(|g-\mathbb{E}[g]| > c\_1^{1/3}) \rightarrow 0
\end{align}
for $c\_1 \rightarrow 0$, where of course the $1/3$ exponent is again a rather arbitrary choice by myself, as anything below $1/2$ should work.
| 0 | https://mathoverflow.net/users/106046 | 431409 | 174,665 |
https://mathoverflow.net/questions/431407 | 2 | Let $f: \mathbb R \to \mathbb R$ be a nonnegative measurable function, and $\{q\_n\}$ some enumeration of the rational numbers. Suppose for every $0 < r < 1$ it holds that
$$\sum\_{n = 0}^\infty r^n f(x + q\_n)$$
converges for almost every $x \in \mathbb R$.
**Question:** Does this imply that for every compact set $K$ of $\mathbb R$, that $\text{esssup}\_K f < \infty$?
| https://mathoverflow.net/users/173490 | Does this condition on $f$ imply essential boundedness on compacts? | No, this is not true. Let $f \in L^1(\mathbb R)$ and $g\_r(x)=\sum\_{n=0}^\infty r^n f(x+q\_n) \leq \infty$. Then $\|g\_r\|\_1=\frac{\|f\|\_1}{1-r}$ and hence $g\_r(x)<\infty$ a.e. If $E\_r$ is the null set where $g\_r$ may not converge then the series converge for all $x$ outside $\cup\_n E\_{1-\frac 1n}$, for every $r<1$, since $g\_r$ is increasing in $r$. Now it is sufficient to choose $f \in L^1$ but not locally bounded.
| 6 | https://mathoverflow.net/users/150653 | 431416 | 174,667 |
https://mathoverflow.net/questions/430194 | 4 | Let $M$ be a countable transitive model of $\sf ZF$, Let $V\_\alpha, V\_{\alpha+1}$ be two stages of the cumulative hierarchy in $M$, let $f: V\_\alpha \to V\_{\alpha+1}$ be a bijection such that for any set $S \in V\_{\alpha+2}$ we have both $f[S] \in M ; f^{-1}[S] \in M$ . Let $F: V\_{\alpha+1} \to V\_{\alpha+2}; F(x)=f[x]=\{f(y) \mid y \in x\}$.
>
> Is there a proof that we may have a set $S \in V\_{\alpha+3}$ such that we don't have both $F[S] \in M; F^{-1}[S] \in M$?
>
>
>
| https://mathoverflow.net/users/95347 | Is it possible to have such nesting of functions? | I think I have an answer to this question of mine. if $M$ in the question satisfy $AC$, then there must exist some set $S$ in $V\_{\alpha+3}$ that doesn't have both $F[S]; F^{-1}[S]$ being elements of $M$, otherwise we can interpret $\sf NF+AC$ which is inconsistent. However, the proof of that result is long.
| 0 | https://mathoverflow.net/users/95347 | 431417 | 174,668 |
https://mathoverflow.net/questions/431383 | 3 | Let $f:X\to S$ be a morphism between algebraic varieties which are smooth over a field of characteristic zero. We define the (derived) direct image functor $f\_+:\mathsf{D}^b(\mathcal{D}\_X)\to \mathsf{D}^b(\mathcal{D}\_S)$ as in basically every textbook. (That's what [HTT] denotes by $\int\_f$ on page 40, for example.) This functor also restricts to $f\_+:\mathsf{D}^b\_\text{qc}(\mathcal{D}\_X)\to \mathsf{D}^b\_\text{qc}(\mathcal{D}\_S)$ and $f\_+:\mathsf{D}^b\_\text{h}(\mathcal{D}\_X)\to \mathsf{D}^b\_\text{h}(\mathcal{D}\_S)$.
It is true that if $f$ is a closed embedding and $M\in \mathsf{D}^b(\mathcal{D}\_X)$ is concentrated in degree 0, then $\mathscr{H}^i(f\_+ M)=0$ for all $i\neq 0$. [HTT, Proposition 1.5.24.]
Given that this is very close to Artin vanishing on the context of perverse sheaves, I would expect the same result as above to be true more generally for affine + quasi-finite morphisms (at least when $M$ is holonomic).
**More precisely, is it true that the functor $f\_+:\mathsf{D}^b\_\text{h}(\mathcal{D}\_X)\to \mathsf{D}^b\_\text{h}(\mathcal{D}\_S)$ is right t-exact (with respect to the usual t-structures) when $f$ is affine and t-exact when $f$ is also quasi-finite?**
Reference: [HTT] is R. Hotta, K. Takeuchi, T. Tanisaki - D-modules, Perverse Sheaves, and Representation Theory.
| https://mathoverflow.net/users/131975 | Artin vanishing for D-modules (i.e., when is $f_+$ t-exact?) | The functor $f\_+$ by definition is the composite of a right t-exact functor (tensor product with the transfer bimodule) and a left t-exact functor (the derived pushforward of sheaves). In the case $f$ is affine, the latter functor is t-exact, so $f\_+$ is indeed right-exact.
On the other hand, D-module pushforward under a quasi-finite morphism is not necessarily t-exact. For example, consider the (non-affine) open embedding $f:\mathbb A^2 - \{(0,0)\} \hookrightarrow \mathbb A^2$. I claim the pushforward $f\_+(\mathcal O)$ is not concentrated in degree 0.
EDIT: I didn't catch the ``also'' in the question. It is true that when $f$ is quasi-finite, then $f\_+$ is left t-exact, though I don't have an entirely straightforward argument for this.
By Zariski's main theorem, it is enough to check the cases:
(i) $f$ is an open embedding, and (ii) $f$ is finite. In case (i), $f\_+$ is easily seen to be left t-exact (as the transfer bimodule is a localization, or alternatively, as the left adjoint to $f\_+$ is t-exact). In case (ii), the result then follows from the main result of [Raskin](https://arxiv.org/abs/1611.04940), noting that when $f$ is finite $f\_+=f\_!$ (so in particular, the pro-category is unnecessary). For holonomic modules, the finite case can be dealt with by using that $f\_+$ is right exact (as $f$ is affine) and $f\_+ = \mathbb D f\_+ \mathbb D$. (I think the general proof follows similar lines, but requires a bit more care.)
| 3 | https://mathoverflow.net/users/7762 | 431424 | 174,669 |
https://mathoverflow.net/questions/431425 | 1 | ### I would like to know if this equation is solvable for $a$ and $\alpha$:
\begin{equation}
\Sigma = \Gamma + a \left( \alpha 1^\top + 1\alpha^\top \right) +a^2 b
\end{equation}
---
1. $\Sigma$ & $\Gamma$ are known. Both are $D\times D$ matrices.
2. $\Sigma$ is symmetric positive definite.
3. $\Gamma$ is symmetric positive semi definite.
4. $\alpha$ is a $D$ vector.
5. $\sum\_{i=0}^D \alpha\_i =0$.
6. $a$ is a scalar.
7. $b$ is a $1$s matrix of $D\times D$.
8. The $1$ is a $D$ vector of $1$s .
If it is solvable, can you explain how?
| https://mathoverflow.net/users/492060 | Two unknowns: one vector, one scalar, one equation | A small rearrangement yields
$$
\Sigma-\Gamma = a(\alpha + a 1 )1^T + a1 \alpha^T.
$$
So for solvability the rank of $\Sigma-\Gamma$ can at most be $2$.
If $\mathrm{rank}\ \Sigma-\Gamma = 0$, then $\Sigma = \Gamma$. Then a solution is $a=0$ and any $\alpha$ with $\alpha^T1=0$. If we assume $a\neq 0$ we equivalently need to solve
$$
0 = \alpha 1^T + 1 \alpha^T + a 11^T
$$
This implies $\alpha$ is a nonzero multiple of $1$ contradicting the requirement $\alpha^T1=0$. So if $\Sigma = \Gamma$ the solution set is
$$
M = \{ (a,\alpha) | a=0, \alpha^T1=0\}.
$$
For the other cases first note that $\mathrm{rank}\ \Sigma-\Gamma >0 $ implies $a\neq 0$. Furthermore, if you multiply the original equation from left and right by $1^T$ and $1$ you obtain that any solution would have to verify the scalar equation
$$
1^T (\Sigma-\Gamma) 1 = a^2 D^2. \tag{$\*$}\label{star}
$$
The other terms vanish because $\alpha^T 1=0$ by assumption. This gives you two potential values for $a$. Also we obtain the next necessary condition that $1^T (\Sigma-\Gamma) 1 \neq 0$ (or even $>0$ if you are looking for real solutions).
When you plug these in the equation is linear in $\alpha$. So for the two potential values of $a$ we need to solve (using $a\neq 0$)
$$
\frac{1}{a}(\Sigma-\Gamma) - a 1 1^T= \alpha 1^T + 1 \alpha^T.
$$
Using $\alpha^T 1 =0$, we get from muliplication by $1$ and division by $D$ that
$$
\frac{1}{aD}(\Sigma-\Gamma)1 - a 1= \alpha.\tag{$\*\*$}\label{starstar}
$$
This at least satisfies $\alpha^T 1 =0$ as from \eqref{star} we get
$$
\alpha^T 1 = \frac{1}{aD}1^T(\Sigma-\Gamma)1 - a D = 0.
$$
So if there is a solution for the two possible $a$'s then it is given by the $\alpha$ from \eqref{starstar}. But note that this does not have to be a solution. This would still depend on $\Sigma - \Gamma$.
| 3 | https://mathoverflow.net/users/85570 | 431437 | 174,671 |
https://mathoverflow.net/questions/431448 | 3 | What is the consistency strength of the following prinicple?
Every set is a member of a [supertransitive](https://en.wikipedia.org/wiki/Supertransitive_class) model of $\sf ZFC$.
Formally this means adding the following sentence to axioms of $\sf ZFC$:
$\forall x \exists M :( M \models {\sf ZFC}) \land \operatorname {supertransitive}(M) \land x \in M$
Is it equivalent to having a proper class of worldly cardinals?
| https://mathoverflow.net/users/95347 | What is the strength of having all sets being elements of supertransitive models of ZFC? | Every level of the cumulative hierarchy is supertransitive, so "There is a proper class of worldly cardinals" clearly implies the principle in question.
Conversely, if $M\models\mathsf{ZFC}$ is supertransitive then $M$ computes powersets correctly: if $x\in M$ then $\mathcal{P}(x)\subseteq M$ so $\mathcal{P}(x)^M=\mathcal{P}(x)$. Consequently, every supertransitive model of $\mathsf{ZFC}$ is some $V\_\alpha$.
So the answer to your question is **yes**, by virtue of the fact that the supertransitive models of $\mathsf{ZFC}$ are essentially-by-definition the worldly levels of the cumulative hierarchy.
| 6 | https://mathoverflow.net/users/8133 | 431468 | 174,678 |
https://mathoverflow.net/questions/431461 | 2 | Let $M$ be a smooth (or just topological) closed manifold. Let $C(M)$ denote the cone over $M$, i.e.
$C(M)$ equals to $M\times [0,\infty)$ with $M\times \{0\}$ contracted to a point. The image of $M\times \{0\}$ in $C(M)$ is called the origin.
**What is the dualizing complex of $C(M)$? In particular what is its stalk at the origin?**
| https://mathoverflow.net/users/16183 | Dualizing complex of the cone over a manifold | The stalk of the dualizing complex at a point is the shift of reduced homology of the link at that point. In this case, the link is $M$ and so the homology of the stalk in degree $i$ is $\tilde H\_{i-1}(M)$.
| 6 | https://mathoverflow.net/users/52918 | 431469 | 174,679 |
https://mathoverflow.net/questions/431398 | 1 | On the nLab, given a local $S$-topos $E$, a concrete sheaf is defined as an object that is separated with respect to the local isomorphisms (the morphisms that are inverted by the global sections functor $\Gamma:E\rightarrow S$): <https://ncatlab.org/nlab/show/concrete+sheaf#in_a_local_topos>
However, separated means that these morphisms are merely send to monos and not to isos. So I cannot see why a concrete sheaf would, in particular, be a sheaf.
I'm pretty sure I made a mistake in my reasoning, so I was wondering if somebody had a simple explanation.
| https://mathoverflow.net/users/122987 | Concrete sheaves | To summarize the discussion in the comments:
* There are two nontrivial Grothendieck topologies used in the definition of a concrete sheaf: the Grothendieck topology T used to define sheaves, and the Grothendieck topology C used to define concrete presheaves.
* The Grothendieck topology T (and its site S) are given to us.
* The Grothendieck topology C is defined by specifying the generating covering family of an object X∈S as the family of all maps 1→X, where 1 is the terminal object in the site S.
* C-separated presheaves are precisely concrete presheaves on S in the usual sense.
* In particular, C-separated T-local presheaves on S (i.e., C-separated T-sheaves) are precisely concrete sheaves on S.
* A C-local presheaf F (i.e., a C-sheaf) has a very simple form: it sends an object Y∈S to the set of maps Y(\*)→F(\*), where F(\*) denotes the set of maps 1→F, where 1 is the terminal object of S.
| 2 | https://mathoverflow.net/users/402 | 431475 | 174,681 |
https://mathoverflow.net/questions/431474 | 4 | **Question**: How to calculate this summation $S=\sum\_{k=0}^m a^k b^{m-k} {n\_1\choose k} {n\_2\choose m-k} $? Where $m<n\_1,m<n\_2$
**Remark1**: When $a=b$, I know the above summation $S=a^m\sum\_{k=0}^m {n\_1\choose k} {n\_2\choose m-k} =a^m {n\_1+n\_2\choose m} $.
**Remark2**: This summation looks somewhat similar to the usual binomial formula $\sum\_{k=0}^m a^k b^{m-k} {m\choose k} =(a+b)^m$. So is there also a similar formula for $S$?
| https://mathoverflow.net/users/492126 | How to calculate this summation $\sum_{k=0}^m a^k b^{m-k} {n_1\choose k} {n_2\choose m-k} $? | In terms of a hypergeometric function you would have
$$S=\sum\_{k=0}^m a^k b^{m-k} {n\_1\choose k} {n\_2\choose m-k}=b^m \binom{n\_2}{m} \, \_2F\_1\left(-m,-n\_1;-m+n\_2+1;a/b\right).$$
I don't see a simpler closed-form expression for arbitrary parameters, but if you fix $n\_1$ this does simplify to a simple rational function of $m,n\_2,a/b$.
| 5 | https://mathoverflow.net/users/11260 | 431476 | 174,682 |
https://mathoverflow.net/questions/431128 | 3 | Let $\Sigma$ be a compact oriented connected bordered surface other than the pair of pants. Let $\Gamma:=\{\gamma\_i\}$ be a finite collection of simple closed curves on $\Sigma$ such that each component of $\Sigma\setminus\cup\_i\gamma\_i$ is homeomorphic to either $\Bbb S^1\times (0,1)$ or $\{z\in \Bbb R^2:|z|<1\}$. Notice that $\gamma\_i$ may intersect with $\gamma\_j$.
>
> Suppose $c$ is a closed curve (*not necessarily simple*) on $\Sigma$
> such that $\text{GI}(c,\gamma\_i)=0$ for each $i$, where $\text{GI}$
> denotes the geometric intersection number. Is it true that $c$ can be
> freely homotoped so that $c\cap \gamma\_i=\varnothing$ for each $i$?
>
>
>
| https://mathoverflow.net/users/363264 | A closed curve can be homotopic to remove all intersections with a filling $\Gamma$ if it has zero geometric intersection numbers with $\Gamma$ | This is true, here's a proof, by a kind of "Whitney trick".
Perturb the set of curves $\{\gamma\_i\} \cup \{c\}$ to put it into general position, so they are pairwise transverse and there is no triple point. Let $|c| = \sum\_i |c \cap \gamma\_i|$.
For each $i$ such that $|c \cap \gamma\_i| > 1$, since $GI(c,\gamma\_i)=1$ it follows that $\Sigma - (c \cup \gamma\_i)$ contains a component whose closure $B$ is a *bigon of $c$ and $\gamma\_i$*, meaning a closed disc having the property that $\partial B = \alpha \cup \beta$ where $\alpha = B \cap c = \partial B \cap c$ is a subarc of $\partial B$ and $\beta = B \cap \gamma\_i = \partial B \cap \gamma\_i$ is a subarc of $\partial B$.
If $c$ is not already disjoint from the $\gamma\_i$'s then, as $i$ varies and $B$ varies over all bigons of $c$ and $\gamma\_i$, there exists a bigon $B$ that is innermost with respect to inclusion. Letting $B$ be a bigon of $c$ and $\gamma\_i$, it follows that $B \cap \bigcup\_{j \ne i} \gamma\_j$ is a union of arcs in the $\gamma\_j$'s that cross from $\alpha$ to $\beta$, each such arc having one endpoint on $\alpha$ and the other endpoint on $\beta$. Now isotope $c$ to push $\alpha$ across $B$ and out past $\beta$ on the other side of $B$. This reduces $|c|$ by $2$.
By induction, $|c|$ can be reduced to $0$, at which point $c$ is disjoint from each of the $\gamma\_i$'s.
| 3 | https://mathoverflow.net/users/20787 | 431494 | 174,687 |
https://mathoverflow.net/questions/431466 | 1 | $\DeclareMathOperator\SL{SL}$Let $ G $ be a noncompact simple Lie group. For example $ \SL\_n $. Let $ \Gamma $ be a lattice in $ G $. Consider the action of $ \Gamma $ on the Lie algebra of $ G $ by conjugation. Is this representation of $ \Gamma $ always irreducible?
For example, I think it is true that all lattices in $ \SL\_2(\mathbb{R}) $ and $ \SL\_2(\mathbb{C}) $ are Ad-irreducible.
| https://mathoverflow.net/users/387190 | A lattice in $ \operatorname{SL}_n $ is Ad-irreducible | Per the request to post it as an answer.
Notice that the Ad representation is a polynomial representation into $\operatorname{GL}(\operatorname{Lie}(G))$.
We do know that $\operatorname{Ad}(G)$ acts irreducibly, and $\Gamma$ is Zariski dense by Borel's density theorem. Hence $\operatorname{Ad}\rvert\_{\Gamma}$ is also irreducible.
| 4 | https://mathoverflow.net/users/8857 | 431496 | 174,688 |
https://mathoverflow.net/questions/430944 | 14 | I need to find the determinant of matrix defined by
\begin{align\*}
& a\_{i,1}=a\_{1,j}=1,\quad \forall 1\leq i,j\leq n,\\ & a\_{i,j}=a\_{i-1,j}+a\_{i,j-1}+i-j, \quad \forall 1< i,j\leq n.
\end{align\*}
Numerically, for $n=1$ to $12$; I found that $\det(A)=F\_n$ where $F\_n$ is the Fibonacci sequence.
How to prove it if true?
Addition: edit I correct the formula
I found
$$a\_{i,j}=\displaystyle{\binom{i+j-1}{j}}-{\binom{i+j-2}{i}}+j-i\quad \forall 1\leq i,j\leq n.$$
Let $\Delta\_n=\det(A)$, it's clear that $\Delta\_1=\Delta\_2=1$.
I must prove that $\Delta\_n=\Delta\_{n-1}+\Delta\_{n-2}, \forall n\ge 3$.
**Addition 2**
**My friend jandri gave me this proof**: It proves that $\det(A\_n)=F\_n$ with $A\_n$ is the matrix of order $n$ defined by:
$a\_{i,1}=a\_{1,j}=1$ for $1\leq i,j\leq n$ and $a\_{i,j}=a\_{i-1,j}+a\_{i,j-1}+i-j$ for $1< i,j\leq n$.
We introduce the matrix $B\_n $ whose general term is defined by $b\_{i,j}=a\_{i,j}+i-j$ for $1\leq i,j\leq n$.
We have $b\_{i,1}=i$, $b\_{1,j}=2-j$ and for $i,j\geq2$ : $b\_{i,j}=b\_{i-1,j}+b\_{i,j-1}$.
We perform on the matrix $A\_n$ the operations $C\_j\leftarrow C\_j-C\_{j-1}$ for $j$ from $n$ to $2$ and after the operations $L\_i\leftarrow L\_i-L\_{i-1}$ for $i$ from $n$ to $2$. This gives a matrix that has $0$ in row $1$ and column $1$ (except for the first term which is équal to $1$) and the rest is the matrix $B\_{n-1}$, so $\det( A\_n)=\det(B\_{n-1})$.
Then we perform the same operations on the matrix $B\_n+xJ\_n$ where $J\_n$ is the matrix of order $n$ such that all the terms of which are $1$.
We obtain the matrix written in 4 blocks: $\left (\begin{array}{c|c} 1+x & L \\ \hline C& B\_{n-1} \end{array}\right)$ where $ L$ is the row in which all the terms are $-1$ and $C$ is the column in which all the terms are $1$. We deduce $\det(B\_n+xJ\_n)=(1+x)\det(B\_{n-1}+\frac1{x+1}J\_{n-1})$.
We then obtain by induction on $n$ that $\det(B\_n+xJ\_n)=F\_{n+1}+xF\_n$ whence $\det(B\_n)=F\_{n+1}$ then $\det( A\_n)=F\_n$.
| https://mathoverflow.net/users/126827 | Determinant equal to Fibonacci sequence | The idea of Jandri can also be applied to a slightly more general case:
Let $$ a(i,j,x,y,z,t)= x\binom{i+j-1}{j}-y \binom{i+j-2}{i}+z(j-i)+t,$$
$A\_n(x,y,z,t)$ be the matrix with entries $a(i,j,x,y,z,t)$ and $D\_n(x,y,z,t)=\det(A\_n(x,y,z,t).$
We apply the following operations on the matrix $A\_n(x,y,z,t):$ We change column $C\_j$ to $C\_j-C\_ {j-1}$ from $j=n$ to $j=2.$ Then we change row $R\_i$ to $R\_i-R\_{i-1}$ from $i=n$ to $i=2.$ We then get the block matrix
$$\begin{pmatrix}x+t&L \\ C & A\_{n-1}(x,y,0,0)\end{pmatrix}.$$
Here $L$ is a row vector all of whose entries are $z-y$ and $C$ is a column vector with entries $x-z$.
Observing that
$$\begin{pmatrix}x+t&L \\ C & A\_{n-1}(x,y,0,0)\end{pmatrix}=\begin{pmatrix}1&0 \\{\frac{C}{x+t}} & I\end{pmatrix}\begin{pmatrix}x+t&0 \\ 0& A\_{n-1}(x,y,0,0)-\frac{C}{x+t}L \end{pmatrix}\begin{pmatrix}1& \frac{L}{x+t} \\ 0 & I\end{pmatrix} $$
and that $CL=(x-z)(z-y)J\_{n-1}$ we get taking determinants
$$D\_n(x,y,z,t)=(x+t)D\_{n-1}(x,y,0,\frac{(x-z)(y-z)}{x+t}).$$
Let now $G\_n(x,y)=\sum\_{i=0}^{\lfloor{n/2}\rfloor}{n-j\choose j}y^jx^{n-j},$
which satisfies $G\_n(x,y)=xG\_{n-1}(x,y)+xy G\_{n-2}(x,y)$ and $G\_n(1,1)=F\_{n+1}$
and define
$$d\_n(x,y,z,t)=(x+t)G\_{n-1}(x,y)+(x-z)(y-z) G\_{n-2}(x,y).$$
These polynomials satisfy $d\_n(x,y,z,t)=(x+t)d\_{n-1}(x,y,0,\frac{(x-z)(y-z)}{x+t})$ and the same initial values as $D\_n(x,y,z,t).$
Therefore we get $$D\_n(x,y,z,t)=d\_n(x,y,z,t).$$
For example $D\_n(1,1,z,0)=F\_{n}+(1-z)^2F\_{n-1},$
| 2 | https://mathoverflow.net/users/5585 | 431503 | 174,691 |
https://mathoverflow.net/questions/431411 | 5 | $\newcommand{\H}{\mathcal{H}}$
$\newcommand{\A}{\mathcal{A}}$
Recall that a coideal $\H$ over $\omega$ is **selective** if for every $\{A\_n : n < \omega\} \subseteq \H$, where $i < j \implies A\_i \supseteq A\_j$, there exists some $B \in \H$ such that $B/n := \{m \in B : m > n\} \subseteq A\_n$ for all $n \in B$. There are two canonical examples of selective coideals, both of which can be constructed in $\mathsf{ZFC}$:
* The entire $[\omega]^\omega$ is of course a selective coideal.
* Let $\A$ be a mad (maximal almost disjoint) family of subsets of $\omega$. Let $\H$ be the set of all infinite subsets of $\omega$ which cannot be covered up to a finite set by finitely many members of $\A$. Then $\H$ is a selective coideal.
Are there any more examples of selective coideals that $\mathsf{ZFC}$ can prove to exist? Note that I'm not looking for selective coideals that exist with further assumptions (e.g. under $\mathsf{CH}$, there are $2^{2^{\aleph\_0}}$ many Ramsey ultrafilters, but I'm not asking for such examples).
| https://mathoverflow.net/users/146831 | Non-trivial examples of selective coideals of $\omega$ | See Section 12 in [S. Todorcevic, Topics in Topology, Lecture notes in Mathematics Vol. 1652, Springer-Verlag Berlin Heidelberg 1997](https://doi.org/10.1007/BFb0096295)
| 3 | https://mathoverflow.net/users/492151 | 431506 | 174,693 |
https://mathoverflow.net/questions/431509 | 0 | Fix $m\in \mathbb{N}.$
For each $n\in \mathbb{N},$ let $A\_n\in \mathbb{M}\_{m}(\mathbb{C}),$ $X\_n\in \mathbb{C}^m,$ and $B\_n\in \mathbb{C}^m.$ Suppose that
$$X\_{n+1}=A\_n X\_n+B\_n,$$
$$\lim\_{n\rightarrow \infty} A\_n=A\_0,$$
$$|\det(A\_0)|>1,$$
the moduli of all entries of $A\_0$ are greater than 1,
and
$$\limsup\_{n\rightarrow \infty}\|B\_n\|^{1/n}<1,$$
where $\|\cdot\|$ is the Euclidean norm.
Is it possible that
$$\limsup\_{n\rightarrow \infty}\|X\_n\|^{1/n}=1?$$
I have the feeling that the answer should be "No." But I do not know how to prove it.
For $m=1,$ it is not difficult to show that the answer is no. Could somebody suggest how to prove or disprove it? Any references (books or papers) are very welcome? Thank you.
| https://mathoverflow.net/users/492150 | Vector recurrences (asymptotic property) | $\newcommand\1{\mathbf1}\newcommand\R{\mathbb R}$Yes, this is of course possible. E.g., this will be so if for all $n$ the matrix $A\_n$ is the diagonal matrix with $1,2\dots,2$ on the diagonal, $B\_n=0$, and $X\_n=[1,0,\dots,0]^\top$.
---
The OP has changed the question, thereby invalidating the above answer. After the change, the answer is still the same: yes, this is possible. E.g., let $m=2$ and for all $n$ let $A\_n:=\begin{bmatrix}3&2\\ 2&3\end{bmatrix}=:A\_0$ and $B\_n:=0$.
Then all your conditions on $A\_n$ and $B\_n$ hold. However, $1$ is an eigenvalue of $A\_n=A\_0$ for all $n$.
It remains to let $X\_1$ be an eigenvector corresponding to the eigenvalue $1$ of $A\_n$; say, we can let $X\_1:=[-1,1]^\top$.
More generally, for any $m\ge2$ and for all $n$ we can let $A\_n=A\_0$ have (say) $3$ everywhere on the diagonal and $2$ everywhere off the diagonal, so that the eigenvalues of $A\_n=A\_0$ are $1$ (of multiplicity $m-1$) and $1+2m$ (so that $\det A\_0=1+2m>1$). Indeed, the vector $\1:=[1,\dots,1]^\top\in\R^m$ is an eigenvector of $A\_n=A\_0$ corresponding to the eigenvalue $1+2m$, and all nonzero vectors in $\R^m$ orthogonal to $\1$ are eigenvectors of $A\_n=A\_0$ corresponding to the eigenvalue $1$.
| 2 | https://mathoverflow.net/users/36721 | 431516 | 174,694 |
https://mathoverflow.net/questions/431422 | 16 | Context
=======
In a [recent paper](https://arxiv.org/abs/2209.06838) involving entanglement in linear optics, we came across some summations involving [Catalan numbers](https://en.wikipedia.org/wiki/Catalan_number) and permutations. In particular, these sums arise when doing integration over the unitary group $\mathrm U(n)$ in the limit that $n\to\infty$. When doing this, we encounter the asymptotic form of the [Weingarten function](https://en.wikipedia.org/wiki/Weingarten_function), which is the origin of the sums over the Catalan numbers. One of these sums that is of most interest to us seems to surprisingly simplify dramatically, which makes us curious to understand why.
---
Definitions
===========
Let $\delta$ be the [Kronecker delta](https://en.wikipedia.org/wiki/Kronecker_delta) defined as $\delta\_{a,b} = 1$ if $a=b$ and $0$ otherwise. Let $C\_n$ be the $n^{\rm th}$ [Catalan number](https://en.wikipedia.org/wiki/Catalan_number) $C\_n := \frac{1}{n+1}\binom{2n}{n}$.
Let $S\_{2\ell}$ denote the permutation group on $2\ell$ elements. For $\tau \in S\_{2\ell}$, define $\#(\tau)$ to be the number of cycles in the disjoint cycle decomposition of $\tau$. Indeed let $\{c\_i^{(\tau)} \mid i\in \{1,\dots, \#(\tau)\}\}$ be the cycle decomposition, and define $|c\_i^{(\tau)}|$ to be the length of the $i^{\rm th}$ cycle. Let $|\tau|$ be the minimum number of transpositions needed to generate the permutation $\tau$, and [note that](https://math.stackexchange.com/questions/616889/worst-case-number-of-transpositions-needed-to-generate-a-permutation) $|\tau| = 2\ell - \#(\tau)$.
Define $\xi^{(\ell)}\colon S\_{2\ell} \to \{1,2,\dots, \ell \}$ as
$$
\xi^{(\ell)}(\tau)= \log\_2\sum\_{j\_1,\dots,j\_\ell=1}^2 \delta\_{j\_{\lceil\tau(2\ell)/2\rceil},j\_{\lceil\tau(1)/2\rceil} }\delta\_{j\_{\lceil\tau(2)/2\rceil},j\_{\lceil\tau(3)/2\rceil}} \dots \delta\_{j\_{\lceil\tau(2\ell-2)/2\rceil},j\_{\lceil\tau(2\ell-1)/2\rceil}}.
$$
For $\ell \in \mathbb N$ and $\kappa \in \mathbb Z$, define $a^{(\ell)}\_{\kappa} \in \mathbb Z$ as
$$
a^{(\ell)}\_\kappa := \sum\_{\substack{\tau\in S\_{2\ell} \text{ s.t.}\\\xi^{(\ell)}(\tau) = |\tau|+\kappa}} (-1)^{\#(\tau)} \prod\_{i=1}^{\#(\tau)} C\_{|c\_i^{(\tau)}| - 1}.
$$
---
Example: $\ell = 1$ and $\kappa = 0$
====================================
* When $\ell = 1$, $\xi^{(1)}(\tau) = 1$ for both permutations (the identity and the swap) because $\lceil1/2\rceil = \lceil 2/2\rceil = 1$.
* When $\tau$ is the identity permutation, $|\tau| = 0$, so it does not contribute to the sum in $a^{(\ell=1)}\_{\kappa=0}$.
* When $\tau$ is the swap, $|\tau| = 1 = \xi^{(1)}(\tau)$, so it does contribute. The swap permutation has one cycle of length 2. Therefore,
$$
a\_{\kappa=0}^{(\ell=1)} = (-1)^{2-1}C\_{2-1} = -C\_1 = -1.
$$
---
Theorem
=======
In a *very* roundabout way, we proved in [our paper](https://arxiv.org/abs/2209.06838) that
$$
a\_{\kappa=1}^{(\ell)} = \frac{(-1)^{\ell +1}}{2 \ell -1} \binom{2 \ell -1}{\ell -1}.
$$
---
Conjecture
==========
For all $\ell \in \mathbb N$, $a\_{\kappa=0}^{(\ell)} = (-1)^\ell 4^{\ell-1}$.
**Evidence for the conjecture**: the conjecture is true for $\ell \leq 5$. We generate the following table for $\kappa = 0$ in Mathematica by iterating through all permutations.
| $\ell$ | $a\_{\kappa=0}^{(\ell)}$ | |
| --- | --- | --- |
| $1$ | $-C\_1$ | $= -1$ |
| $2$ | $-4 C\_0^2 C\_1+4 C\_0 C\_2$ | $=4$ |
| $3$ | $-9 C\_0^4 C\_1+15 C\_0^2 C\_1^2-C\_1^3+18 C\_0^3 C\_2-6 C\_0 C\_1 C\_2-9 C\_0^2 C\_3$ | $=-16$ |
| $4$ | $-16 C\_0^6 C\_1+80 C\_0^4 C\_1^2-40 C\_0^2 C\_1^3+48 C\_0^5 C\_2-112 C\_0^3 C\_1 C\_2 +16 C\_0 C\_1^2 C\_2+8 C\_0^2 C\_2^2-48 C\_0^4 C\_3+24 C\_0^2 C\_1 C\_3+16 C\_0^3 C\_4$ | $=64$ |
| $5$ | $-25 C\_0^8 C\_1+250 C\_0^6 C\_1^2-380 C\_0^4 C\_1^3+80 C\_0^2 C\_1^4-C\_1^5+100 C\_0^7 C\_2 -600 C\_0^5 C\_1 C\_2+440 C\_0^3 C\_1^2 C\_2-20 C\_0 C\_1^3 C\_2 + 130 C\_0^4 C\_2^2 -50 C\_0^2 C\_1 C\_2^2 -150 C\_0^6 C\_3+320 C\_0^4 C\_1 C\_3-60 C\_0^2 C\_1^2 C\_3-40 C\_0^3 C\_2 C\_3 +100 C\_0^5 C\_4-60 C\_0^3 C\_1 C\_4-25 C\_0^4 C\_5$ | $=-256$ |
---
Question
========
Can one find a closed form for $a\_{\kappa}^{(\ell)}$ for all $\kappa$ and $\ell$? Of particular interest for us is when $\kappa = 0$; how does one prove or disprove our conjecture that $a\_{\kappa=0}^{(\ell)} = (-1)^\ell 4^{\ell-1}$? If our conjecture is indeed true, is there any insight as to *why* the seemingly complicated expression simplifies so dramatically?
Unfortunately, it does not seem that the roundabout way which helped us solve the $\kappa = 1$ case works for the $\kappa = 0$ case. We imagine that a first step to answering these questions is to simplify $\xi^{(\ell)}$, probably using some properties of permutations. Is there any nice interpretation of what $\xi^{(\ell)}(\tau)$ is representing about the permutation $\tau$? We spent some time and came up with a few graphical and combinatorial ways of understanding $\xi^{(\ell)}$, but nothing we’ve been able to use to make any real progress.
---
Thanks!
=======
| https://mathoverflow.net/users/492058 | Conjecture on sum over permutations of products of Catalan numbers | The problem naturally fits in the framework of breakpoint graphs (per Peter Taylor's observation), which makes it possible to obtain a differential equation for the generating function
$$H(x,u,s\_1,s\_2,\dots) := \sum\_{\ell\geq 0} x^{2\ell} \sum\_{\tau\in S\_{2\ell}} u^{\xi^{(\ell)}(\tau)} \prod\_{i=1}^{\#(\tau)} s\_{|c\_i^{(\tau)}|}$$
by modification of the proof of Lemma 3.2 in my paper <https://arxiv.org/abs/1503.05285> similarly how it was done for function $G(x,u,s\_1,s\_2,\dots)$ in Theorem 4.1. In fact, $H$ represents an analog of function $G$, with just a minor complication that it runs only over even $n=2\ell$.
Then it may be possible to obtain an explicit expression for $H$ and/or derive the anticipated identity directly from the differential equation. This approach can potentially be extended to other values of $\kappa$. Please let me know if you are interested to collaborate on working out details in this direction.
| 6 | https://mathoverflow.net/users/7076 | 431527 | 174,699 |
https://mathoverflow.net/questions/431510 | 12 | In differential geometry it is often natural to speak of infinite-dimensional manifolds (e.g., the manifold of mappings between two smooth manifolds). Different versions of [generalized smooth spaces](https://ncatlab.org/nlab/show/generalized+smooth+space) are proposed for this purpose. I find the most natural and preferable approach: categories of sheaves on suitable sites (among other things, such categories are automatically Grothendieck topoi with all the properties that follow). Ideally, I'm looking for a textbook on differential geometry from scratch that actively uses, wherever appropriate, generalized smooth spaces, which are defined as the category of sheaves on some site (this is my only requirement). If such textbooks do not exist, then any literature on generalized smooth spaces of this kind, where some definitions are given and some theorems are proved, would be useful to me.
| https://mathoverflow.net/users/148161 | Are there textbooks on differential geometry in the language of smooth sets or smooth derived stacks? | “[Diffeology](http://math.huji.ac.il/%7Epiz/Site/TheBook-rep.html)” by Patrick Iglesias-Zemmour is probably the closest match.
He develops differential forms and de Rham cohomology, fiber bundles, connections, and symplectic geometry in the language of diffeological spaces, i.e., concrete sheaves of sets on the site of smooth manifolds. This book is closest in style to a conventional differential geometry textbook.
Another book is “[Synthetic geometry of manifolds](https://users-math.au.dk/kock/SGM-final.pdf)” by Anders Kock,
which treats differential forms, Lie groups and algebras, principal bundles with connections, jets and differential operators. It has a somewhat different focus (e.g., infinitesimals and the internal language of toposes) than the previous book.
In relation to this one can also mention “Models for smooth infinitesimal analysis” by Ieke Moerdijk and Gonzalo Reyes, which covers some foundational topics in differential geometry, like differential forms.
| 12 | https://mathoverflow.net/users/402 | 431529 | 174,700 |
https://mathoverflow.net/questions/431292 | 4 | A space $X$ is said to be Menger if for each sequence $(\mathcal{U}\_n)$ of open covers of $X$, there is a sequence $(\mathcal{V}\_n)$ such that $\mathcal{V}\_n$ is a finite subcollection of $\mathcal{U}\_n$, $n\in\omega$, and $\{\bigcup\mathcal{V}\_n:n\in\omega\}$ is an open cover of $X$. A space $X$ is Lindelöf if each open cover has a countable subcover.
My question was motivated from the fact that classical examples of Lindelöf spaces that are not Menger, have dense Menger subspaces.
| https://mathoverflow.net/users/476373 | Is there any example of a Lindelöf space that has no Menger dense subspaces? | Every space $X$ with a dense Menger subspace must be *weakly Menger*, that is, for each sequence $\{\mathcal{U}\_n: n<\omega \}$ of open covers there is a finite sub-collection $\mathcal{V}\_n \subset \mathcal{U}\_n$ such that $\bigcup \{\mathcal{V}\_n: n <\omega \}$ is dense in $X$. Therefore it's enough to find a Lindelöf non-weakly Menger space.
Given a topological space $(Y, \tau)$ let us denote by $\mathcal{K}[Y]$ the space of all compact nowhere dense subsets of $Y$ endowed with the Pixley–Roy topology, that is, the topology generated by the family $\{[F,U]: F \in \mathcal{K}[Y], U \in \tau, F \subset U\}$ where $[F,U]=\{G \in \mathcal{K}[Y]: F \subset G \subset U\}$.
Let $\mathbb{P}$ be the space of irrationals and consider the space $Z=\mathcal{K}[\mathbb{P}]$. van Douwen, Tall and Weiss proved that $Z$ is a ccc non-separable first-countable zero-dimensional Baire space without isolated points (see Theorem 3 of *van Douwen, Eric K.; Tall, Franklin D.; Weiss, William A. R.*, [Non-metrizable hereditarily Lindelöf from CH](https://www.ams.org/journals/proc/1977-064-01/S0002-9939-1977-0514998-1), Proc. Am. Math. Soc. (to appear) [ZBL0345.54016](https://zbmath.org/?q=an:0345.54016).) and therefore, by the Corollary from page 140 of the same paper, under CH, $Z$ contains a dense Luzin subspace $X$. Luzin means that every nowhere dense subset of $X$ is countable and it is easy to see that this feature, along with the ccc of $X$ implies that $X$ is hereditarily Lindelof. It remains to prove that $X$ is not weakly Menger, but since $X$ is dense in $Z$ it suffices to prove that $Z$ is not weakly Menger.
Indeed, since $\mathbb{P}$ is not Menger, there is a countable sequence $\{\mathcal{U}\_n: n < \omega \}$ of open covers of $\mathbb{P}$ which witnesses that. For every $K \in Z$ let $\mathcal{U}^K\_n$ be a finite subcollection of $\mathcal{U}\_n$ which covers $K$. Then $\mathcal{O}\_n=\{[K, \bigcup \mathcal{U}^K\_n]: K \in Z \}$ is an open cover of $Z$ for every $n<\omega$. Let $\mathcal{G}\_n$ be a finite subcollection of $\mathcal{O}\_n$ and let $\mathcal{F}\_n$ be the finite subset of $Z$ such that $K \in \mathcal{F}\_n$ if and only if $[K, \bigcup \mathcal{U}^K\_n] \in \mathcal{G}\_n$. Then $\bigcup \{\mathcal{U}^K\_n: K \in \mathcal{F}\_n \}$ is a finite subcollection of $\mathcal{U}\_n$, for every $n<\omega$, and therefore there is a point $y \in \mathbb{P} \setminus (\bigcup \{ \bigcup \{\mathcal{U}^K\_n: K \in \mathcal{F}\_n \}: n < \omega\})$. It follows that $[\{y\}, \mathbb{P}]$ is a non-empty open subset of $Z$ which is disjoint from $\bigcup \{\mathcal{G}\_n : n < \omega \}$, thus showing that $Z$ is not weakly Menger.
| 5 | https://mathoverflow.net/users/11647 | 431530 | 174,701 |
https://mathoverflow.net/questions/430032 | 3 |
>
> Suppose we have $N$ lines in general position (any two lines, but no three lines, meet at a point) ($N\geq 3$). Let the smallest bounded region have area $1$. Determine the minimum (or possibly infimum) of the total bounded area.
>
>
>
For example, $3$ lines create one triangular region while $4$ create one quadrangular region and two triangular regions (the complete region thus bounded is called a complete quadrilateral). I know that $N$ lines in general position will create ${N+1\choose 2}+1$ regions, ${N+1\choose 2}+1-2N=\frac{(N-1)(N-2)}{2}$ of those bounded on all sides by line segments, but I wanted to figure out how to make the regions maximally “equal” and how close it’s possible to get to the ideal minimum of $\frac{(N-1)(N-2)}{2}$ for a given $N$ (where all bounded regions have area $1$). I have little to no experience in the field of maximizing or minimizing geometrical quantities (especially in such a large mathematical space in which to minimize/maximize as this problem gives), so solving this is entirely beyond my experience.
It is easy to show that there is no maximal bounded area for $N\geq 4$ by drawing some three lines arbitrarily close to meeting at a point. In this case, the resulting triangular region becomes arbitrarily close to a point while maintaining an area of $1$ in the problem's scaling, and the area of the remaining regions grows arbitrarily large by comparison. But I really don't know how to do any of this when it comes to minimizing the total bounded area.
For $N=3$, with only one region, the minimum area is trivially $1$. For $N=4$, some fiddling around on Desmos has convinced me that the minimum area is indeed the ideal of $3$. For $N=5$, I believe that the minimum area still remains the ideal of $6$ (although here my fiddling around on Desmos gets much more nonrigorous and guesswork-y). For $N\geq 6$, however, any proofs or even good guesses of any sort elude me entirely.
If finding the exact minimum area (or infimum in case there’s somehow a minimum that can be approached but not actually reached, which I would not expect) is in fact too difficult to do well, I would also be interested in the growth rate of the actual minimal area relative to the ideal as $N\to\infty$. Also, I'm new to MO (my typical haven is MSE), so please feel free to correct me if I'm unclear/incorrect in any way or messed up/missed any tags or anything like that.
I originally posted this [problem](https://math.stackexchange.com/q/4417420/664819) on Math Stack Exchange, but never received a solution. I then posed it to one of my math professors, but he dismissed it as seeming too hard and suggested I email it to a particular nother professor. I did, but haven't received a response yet and am pessimistic about the probability of a response at all. Incomplete ideas are welcome too. Thank you for your help!
Edit 1: The other professor did write back. He said this seemed like a problem in discrepancy theory, it does seem very hard or even impossible to do explicitly, and the next best thing would be to determine upper and lower bounds for the minimal area.
Edit 2: I wound up contacting Dr. János Pach (an expert in relevant fields) regarding the problem on the recommendation of my first professor. Apparently the problem (originally in a slightly different form) is a minor unsolved problem first posed by Fejes Tóth in 1987:
>
> L. Fejes Tóth, On Spherical Tilings Generated by Great Circles, Geometriae Dedicata, 23 (1987), 67–71
>
>
>
And the only notable result on it since its posing comes from Dan Ismailescu in 2003:
>
> D. Ismailescu, Slicing the pie. U.S.-Hungarian Workshops on Discrete Geometry and Convexity (Budapest, 1999/Auburn, AL, 2000). Discrete Comput. Geom. 30 (2003), no. 2, 263–276.
>
>
>
So certainly pretty darn hard. :)
| https://mathoverflow.net/users/490865 | Minimize total area bounded by $N$ lines in general position | I wound up contacting Dr. János Pach (an expert in relevant fields) regarding the problem on the recommendation of my first professor. Apparently the problem (originally in a slightly different form) is a minor unsolved problem first posed by Fejes Tóth in 1987:
>
> L. Fejes Tóth, On Spherical Tilings Generated by Great Circles, Geometriae Dedicata, 23 (1987), 67–71
>
>
>
And the only notable result on it since its posing comes from Dan Ismailescu in 2003:
>
> D. Ismailescu, Slicing the pie. U.S.-Hungarian Workshops on Discrete Geometry and Convexity (Budapest, 1999/Auburn, AL, 2000). Discrete Comput. Geom. 30 (2003), no. 2, 263–276.
>
>
>
So certainly pretty darn hard. :)
| 1 | https://mathoverflow.net/users/490865 | 431537 | 174,704 |
https://mathoverflow.net/questions/431460 | 2 | Let $I$ be the Dynkin diagram vertex set and $K$ be a proper nonempty subset of it. Let $w\_0^K$ be the longest word of the Dynkin subdiagram $K$, which might be a disjoint union of connected Dynkin diagrams.
I saw in the paper "[Preprojective algebras and partial flag varieties](http://aif.cedram.org/item?id=AIF_2008__58_3_825_0)" by Geiss, Leclerc and Schroer the following statement:
“We can write $w\_0 = w^K\_0 v\_K$ with $\ell(w^K\_0) + \ell(v\_K) = ℓ(w\_0).$ Therefore there exist reduced words $i$ for $w\_0$ starting with a factor $(i\_1,\dotsc,i\_{r\_K})$ which is a reduced word for $w^K\_0$.”
I could run this through examples, like writing the longest element $w\_0=s\_3s\_2s\_3s\_1s\_2s\_3s\_1s\_2s\_1$ of type $B$ as $w\_0=\boldsymbol{s\_1s\_2s\_1} {s\_3s\_2s\_1s\_3s\_2s\_3}$, if $K=\{1،2\}$.
I was wondering if there is an algorithm showing how to do this in general, at least if the size of $K$ is known, like $|K|=|I|-1$.
| https://mathoverflow.net/users/169192 | Particular reduced expression of the longest element of Weyl group | Whenever $\ell(wv)=\ell(w)+\ell(v)$, you can construct a reduced word for $wv$ by producing one for $w$ and one for $v$ and then concatenating them. So, if you know an algorithm for producing reduced words, you can just do that.
On the other hand, there’s actually a particularly nice interpretation in this case. Reduced words for $w\_0$ are in bijection with convex orders on roots (orders such that if $\alpha,\beta,\alpha+\beta$ are all roots, then $\alpha+\beta$ is always between $\alpha$ and $\beta$). The bijection is obtained by sending $(i\_1,\dots, i\_m)$ to the order $\alpha\_{i\_1} < s\_{i\_{1}}\alpha\_{i\_{2}} < s\_{i\_1}s\_{i\_{2}}\alpha\_{i\_{3}}<\cdots $; the lowest $k$ roots in this order are those sent to negative roots by $s\_{i\_k}\cdots s\_{i\_1}$. So, the reduced word you want comes from choosing this order so that the roots in the span of $K$ are below all other roots.
There are various ways of doing this. For example, if we choose vectors $\mathbf{x}=(x\_i,y\_i) \in \mathbb{R}^2$ for each simple root such that $y\_i>0$, we can extend linearly to assign a vector to every root $\mathbf{x}\_{\alpha}$. If these are generic, every root will have a different slope, and we can order roots by slope from lowest to highest. If we assume that $x\_i\leq 0$ for all $i\in K$, and $x\_i\gg 0$ for $i\notin K$, then this slope ordering will give us what we want once the positive $x\_i$’s are big enough.
In your $B\_3$ example, we would choose, say $(x\_1,y\_1)=(-1,1), (x\_2,y\_2)=(0,1), (x\_3,y\_2)=(2,1)$. We would then get the order on positive roots:
$$\alpha\_1 < \alpha\_1+ \alpha\_2 < \alpha\_2 < \alpha\_1+ \alpha\_2+\alpha\_3 <\alpha\_2+\alpha\_3 <\alpha\_1+ \alpha\_2+2\alpha\_3 < \alpha\_2+2\alpha\_3 <\alpha\_3$$
This corresponds to the reduced word $s\_1s\_2s\_1s\_3s\_2 s\_3s\_1s\_2s\_3$ (Assuming I did my computations right).
| 5 | https://mathoverflow.net/users/66 | 431540 | 174,705 |
https://mathoverflow.net/questions/431512 | 1 | I have just asked the calculation of the following summation [see here](https://mathoverflow.net/questions/431474/how-to-calculate-this-summation-sum-k-0m-ak-bm-k-n-1-choose-k-n-2-ch) $$S(a,b,m,n\_1,n\_2)=\sum\_{k=0}^m a^k b^{m-k} {n\_1\choose k} {n\_2\choose m-k}, $$
which is motivated by the calculation of the following limit (if exist)
$$L\_0=\lim\_{m \to \infty}\frac{\sqrt{ab}S(a,b,m-1,n\_1-1,n\_2-1)}{S(a,b,m,n\_1,n\_2)}$$
where $a>0, b>0, \frac{m}{n\_1},\frac{m}{n\_2}$ are kept fixed.
**My question is**: Whether the above limit exists ($a\neq b$)? If it exists, then how to calculate it? When $a=b$, the result is simple $\lim\_{m \to \infty}\frac{a\cdot a^{m-1} {n\_1+n\_2-2\choose m-1}}{a^{m} {n\_1+n\_2\choose m}}=\lim\_{m \to \infty}\frac{m(n\_1+n\_2-m)}{(n\_1+n\_2)(n\_1+n\_2-1)}$ and this limit obviously exists, where we have used $S(a,b,m,n\_1,n\_2)=a^m {n\_1+n\_2\choose m}$ for $a=b$.
Despite the nice comments and answers in the [previous question](https://mathoverflow.net/questions/431474/how-to-calculate-this-summation-sum-k-0m-ak-bm-k-n-1-choose-k-n-2-ch), I still don't know how to calculate the limit.
| https://mathoverflow.net/users/492126 | How to calculate this limit (if exist)? | $\newcommand\ka\kappa\newcommand{\de}{\delta}\newcommand\ep\varepsilon$The problem obviously reduces to this one: find
\begin{equation}
L:=\lim\_{m\to\infty}\frac{S(c,m,n\_1,n\_2)}{S(c,m-1,n\_1-1,n\_2-1)},
\end{equation}
where $c:=a/b\ne1$ and
\begin{equation}
S(c,m,n\_1,n\_2):=\sum\_{k=0}^m c^k A\_k(m,n\_1,n\_2),
\end{equation}
where
\begin{equation}
A\_k(m,n\_1,n\_2):=\binom{n\_1}k \binom{n\_2}{m-k}.
\end{equation}
In accordance with the linked post, assume that $m<n\_1$ and $m<n\_2$, so that $m\to\infty$,
\begin{equation}
n\_1/m\to\nu\_1\in[1,\infty),\quad n\_2/m\to\nu\_2\in[1,\infty).
\end{equation}
By considering the ratios $r\_k:=c^{k+1}A\_{k+1}(m,n\_1,n\_2)/(c^k A\_k(m,n\_1,n\_2))$, one sees that
\begin{equation}
S(c,m,n\_1,n\_2)\sim\sum\_{k\sim \ka m} c^k A\_k(m,n\_1,n\_2),
\end{equation}
where
\begin{equation}
\ka:=\frac{c \nu \_1-\sqrt{\left(c \nu \_1+c+\nu \_2-1\right){}^2-4 (c-1) c \nu \_1}+c+\nu \_2-1}{2 (c-1)}\in(0,1).
\end{equation}
(More specifically, note that $r\_k\ge1$ for $k\le k\_\*$ and $r\_k\le1$ for $k\ge k\_\*$, for a certain integer $k\_\*\sim\ka m$. Next, note that for each real $\ep>0$ there is some $\de>0$ such that for all large enough $m$ and all $k=0,\dots,m-1$ we have the implication
if $k/m-\ka\ge\de\implies r\_k<1-\ep$. So, for integers $k\ge(\ka+3\de)m$ we will have $c^k A\_k(m,n\_1,n\_2)\le(1-\ep)^{\de m}c^{k\_\*+1} A\_{k\_\*+1}(m,n\_1,n\_2)$ and hence
$\sum\_{k\ge (\ka+3\de)m} c^k A\_k(m,n\_1,n\_2)\le m (1-\ep)^{\de m} c^{k\_\*+1} A\_{k\_\*+1}(m,n\_1,n\_2)
=o(\sum\_{|k-\ka m|\le3\de m} c^k A\_k(m,n\_1,n\_2))$.
Similarly, $\sum\_{k\le (\ka-3\de)m} c^k A\_k(m,n\_1,n\_2)
=o(\sum\_{|k-\ka m|\le3\de m} c^k A\_k(m,n\_1,n\_2))$.
Finally here, note that we can choose $\de>0$ to be however small.)
Also, for $k\sim \ka m$,
\begin{equation}
\frac{A\_k(m-1,n\_1-1,n\_2-1)}{A\_k(m,n\_1,n\_2)}
=\frac{(m-k)(n\_1-k)}{n\_1 n\_2}\sim\frac{(1-\ka )(\nu\_1-\ka )}{\nu\_1\nu\_2}.
\end{equation}
So,
\begin{equation}
L=\frac{(1-\ka )(\nu\_1-\ka )}{\nu\_1\nu\_2}.
\end{equation}
| 4 | https://mathoverflow.net/users/36721 | 431551 | 174,712 |
https://mathoverflow.net/questions/431565 | 6 | The density theorem in the ordinary category theory asserts that every presheaf on a small category is a colimit of representables in a canonical way. In Lemma 5.1.5.3 of *Higher Topos Theory*, Lurie proves an $\infty$-categorical version of this. There is a part of the proof which is not clear to me, and I hope someone kindly gives me an elaboration of the proof. (This question has already been asked previously on MSE [(see here)](https://math.stackexchange.com/questions/3974923/proof-of-lemma-5-1-5-3-in-jacob-luries-htt), but the question has not received an accepted answer yet. Given that it was asked about a year and a half ago, I decided that it wouldn't be inappropriate to ask the question again here.)
The lemma asserts the following:
>
> If $S$ is a simplicial set, then $\operatorname{id}\_{\mathcal{P}(S)}$ is the left Kan extension of the Yoneda embedding $j:S\to\mathcal{P}(S)=\operatorname{Fun}(S^{\mathrm{op}},\mathcal{S})$ along itself.
>
>
>
In the proof, he reduces the claim to the following assertion:
>
> Let $\mathcal{C}\subset \mathcal{P}(S)$ be the essential image of the Yoneda embedding. Let $X\in\mathcal{P}(S)$ be an arbitrary object and $s$ an arbitrary object. Set $\mathcal{E}=(\mathcal{C}\_{/X})^\triangleright\times\_{\mathcal{P}(S)}(\mathcal{P}(S)\_{j(s)/})$ and $\mathcal{E}^0=(\mathcal{C}\_{/X})\times \_{(\mathcal{C}\_{/X})^\triangleright}\mathcal{E}$. Then the inclusion $\mathcal{E}^0\subset \mathcal{E}$ is a weak homotopy equivalence.
>
>
>
Lurie then finishes the proof by claiming that both $\mathcal{E}$ and $\mathcal{E}^0$ deformation retracts onto $\mathcal{E}^1= \mathcal{C}\_{/X}\times \_{\mathcal{C}}\{\operatorname{id}\_{j(s)}\}$. **It is this very last step I am having trouble understanding.** Why do $\mathcal{E}$ and $\mathcal{E}^0$ deformation retract onto $\mathcal{E}^1$? It is easy to see that they both have the homotopy type of $\operatorname{Map}\_{\mathcal{P}(S)}(j(s),X),$ so I know that the claim isn't unreasonable. Also, the claim is easy to see if we are working with ordinary categories (and set-valued presheaves); but the proof (or my proof) does not seem to carry on direclty to the $\infty$-categorical case, because it relies heavily on the strict composition of 1-categories.
Any help (including alternative ways to see that $\mathcal{E}^0\subset \mathcal{E}$ is a weak homotopy equivalence, or even alternative proofs of Lemma 5.1.5.3) is appreciated. Thanks in advance.
| https://mathoverflow.net/users/144250 | Density Theorem for $\infty$-Categories (HTT, Lemma 5.1.5.3) | I'm going to deal with the case of $\mathcal E^1\subset \mathcal E^0$. Note that by composition of pullbacks, $\mathcal E^0 = \mathcal C\_{/X}\times\_{\mathcal P(S)}\mathcal P(S)\_{j(s)/}$, while $\mathcal E^1= \mathcal C\_{/X}\times\_\mathcal C\{j(s)\}$.
Further, because $\mathcal C\to \mathcal P(S)$ is a full subcategory inclusion, you can rewrite $\mathcal E^0 = \mathcal C\_{/X}\times\_\mathcal C \mathcal C\_{j(s)/}$
Now we write the obvious thing : the homotopy $\mathcal E^0\times\Delta^1\to \mathcal E^0$ should be given by the following thing:
* on the $\mathcal C\_{j(s)/}$ factor, go $\mathcal E^0\times\Delta^1\to \mathcal C\_{j(s)/}\times\Delta^1\to \mathcal C\_{j(s)/}$ where the second map is the canonical homotopy witnessing that $\mathcal C\_{j(s)/}$ deformation retracts onto $\{j(s)\}$;
* on the $\mathcal C$ factor, same as above but postcompose further with $\mathcal C\_{j(s)/}\to \mathcal C$
* On the $\mathcal C\_{/X}$ factor, you'll have to observe that there *is* a composition map $\mathcal C\_{/X}\times\_\mathcal C \mathcal C\_{j(s)/}\to map(X,j(s))$. Indeed this is like $\{j(s)\}\times\_{\hom(\{0\},\mathcal C)} \hom(\Lambda^2\_1,\mathcal C)\times\_{\hom(\{2\}, \mathcal C)} \{X\}$, and the forgetful map $\hom(\Delta^2,\mathcal C)\to \hom(\Lambda^2\_1,\mathcal C)$ is an acyclic fibration, so in particular has a section. So you can use that section, and then evaluate on $\Delta^{\{0,2\}}$, and forget about the $\{j(s)\}\times\_{\hom(\{0\},\mathcal C)}$ part to land in $\hom(\Delta^1,\mathcal C)\times\_{\hom(\{2\},\mathcal C)}\{X\}= \mathcal C\_{/X}$.
Here I'm being a bit sloppy between $\mathcal C\_{/X}$ and $\hom(\Delta^1,\mathcal C)\times\_{\hom(\{1\},\mathcal C)}\{X\}$ (I always forget whether they are literally isomorphic or just equivalent) but they are equivalent so it does not really matter.
So now, you have an equivalence ${}\_{j(s)}\hom(\Lambda^2\_1,\mathcal C)\_X\to {}\_{j(s)}\hom(\Delta^2,\mathcal C)\_X$, and the point is now that there is a homotopy $\Delta^1\times\hom(\Delta^2,\mathcal C)\to \hom(\Delta^1,\mathcal C)$ that more or less witnesses the fact that there is a map from $0\to 2$ to $1\to 2$ in the triangle $\Delta^2$, and so if you string these together, you will get a homotopy $\Delta^1\times\mathcal E^0\to \mathcal C\_{/X}$ that looks exactly like the natural transformation $(j(s)\to y\to X, j(s) = j(s))\to (y\to X, j(s)\to y)$ that you would write down $1$-categorically.
Now these three homotopies are compatible and so they do assemble as a map from $\mathcal E^0\times\Delta^1$ to the pullback, which is precisely $\mathcal E^0$. It is clear from the construction that it starts at the identity, and ends in $\mathcal E^1$.
Basically the key point here was the third homotopy: given what you wrote in your question about "composition", I'm guessing that this is what was missing. In particular, I would suggest trying to see if you can do the same kind of the thing for the case of $\mathcal E$, and replace the words "compose" in your $1$-categorical proof with something along the lines of what I did here. If that doesn't work, let me know and I'll try to modify my answer to incorporate that part as well.
Also note that, as Zhouhang pointed out in the comments, Kerodon has a different proof of this fact altogether, or rather it's arranged differently : it observes that the yoneda embedding is *dense*, and earlier proved that if $f : C\to D$ is dense, then the identity of $D$ was the left Kan extension of $f$ along itself (and also relates this to the condition that the "canonical diagram" be a colimit diagram). It feels like a better proof, so I would suggest looking at that too.
| 4 | https://mathoverflow.net/users/102343 | 431572 | 174,717 |
https://mathoverflow.net/questions/431564 | 12 | As the title suggest it seems standard conjectures mean different things depending on the context. I had the impression that in characteristic 0 they are a list of conjectures about varieties over an arbitrary field of characteristic zero and any Weil cohomology theory (not necessarily the usual suspects) just like the way it is explained in the [Wikipedia](https://en.wikipedia.org/wiki/Standard_conjectures_on_algebraic_cycles) page.
After reading [this](https://arxiv.org/pdf/1006.1116.pdf) paper that claims the standard conjecture over any field of char $0$ follows from Suslin's Lawson homology conjecture I am confused.
* First, the base field $k$ admits embeddings into $\mathbb{C}$ as mentioned in the first page third paragraph, so $k$ cannot be any field of char $0$.
* Second, in the proof of the Proposition $2.2$ page $5$ the cohomology theory that is used is the singular cohomology and it is not just any Weil cohomology theory.
Are the conjectures stated in the generality of the Wikipedia page obviously not correct or is it possible somehow to go from singular cohomology and $\mathbb{C}$ to any Weil cohomology theory and field of characteristic $0$?
| https://mathoverflow.net/users/127776 | What exactly do the standard conjectures in characteristic zero refer to? | To prove that the standard conjectures are true for any Weil cohomology over a given field $k$, it suffices to prove them for an arbitrary chosen Weil cohomology over each subfield of finite type of $k$.
Here is a way to summarize the principle of the proof.
There is, for each field $k$ the category $M(k)$ of pure motives over $k$, where the Hom' s are Chow groups up to numerical equivalence. The standard conjectures for a given Weil cohomology theory over $k$ imply that one can modify the tensor structure on $M(k)$ and turn it into a tannakian category. This also implies the standard conjectures are true for any other Weil cohomology: any such cohomology will define a fiber functor of $M(k)$ and thus will be isomorphic to the fiber functor defined by the original Weil cohomology, at least after a change scalar to a bigger field of coefficients.
Now, if a field $k$ is a filtered union of subfields $k\_i$, such that, for each $i$, there is a Weil cohomology over $k\_i$ for which the standard conjectures hold, then the category $M(k)$ is the filtered 2-colimit of the tannakian categories $M(k\_i)$: this comes from the fact this is true for the version of motives where the Hom's are Chow groups up to rational equivalence, and from the preceding discussion. This implies that $M(k)$ is tannakian itself and that any Weil cohomology on $k$ defines a fiber functor on $M(k)$, from which one can deduce the standard conjectures for any Weil cohomology on $k$ rather formally.
| 17 | https://mathoverflow.net/users/1017 | 431573 | 174,718 |
https://mathoverflow.net/questions/431539 | 2 | Meusnier's theorem states that all curves on a surface $S$ embedded in $\Bbb R^3$ passing through a given point $P$ and having the same tangent $v\in T\_PS$ also have the same normal curvature.
I think the same is true for [geodesics torsion.](https://encyclopediaofmath.org/wiki/Geodesic_torsion) See [here](https://en.wikipedia.org/wiki/Darboux_frame#Geodesic_curvature,_normal_curvature,_and_relative_torsion) for my definition of geodesics torsion.
My proof goes as: Take a curve $\gamma : (-\varepsilon, \varepsilon)\to S$ with $\gamma(0)=P,\,\gamma'(0)=v$ and let $u=n\times\gamma',$ where $n$ is the Gauss map. Then
$$\langle dn(v),n\times v\rangle =\langle (n\circ\gamma)',u \rangle(0)=-\langle n\circ\gamma, u' \rangle (0)=-\langle n\circ\gamma,-\kappa\_g v-\tau\_g n \rangle(0) =\tau\_g.$$
However, the inputs of this formula are independent of the curve. Therefore geodesic curvature depends only on $P$ and $V.$
But this statement and almost one-line proof cannot be found anywhere in the literature. Have I done something wrong?
| https://mathoverflow.net/users/54507 | Is the Meusnier's theorem true for geodesics torsion? | I believe that your $\tau\_g$ is what É. Cartan calls the 'geodesic torsion' in his 1945 book *Les systèmes différentiels extérieurs et leurs applications géométriques*. He denotes his geodesic torsion by $1/T\_g$. He gives your formula for $1/T\_g$ in Chapter 7 as part of equation $(17)$.
The point is that what the OP calls $\tau\_g$ is simply the value of what Cartan calls the *third fundamental form* $\Psi$ of the surface $S$ evaluated on the tangent vector to the curve, which is why it depends only on the tangent vector to the curve and not on any higher derivatives.
Note: Cartan's third fundamental form $\Psi$ (see equation $(14)$ in Chapter 7) is *not* what is nowadays called the 'third fundamental form' and usually denoted $I\!I\!I$. (The modern $I\!I\!I$ is just the pullback via the Gauss map of the standard metric on the $2$-sphere.) Instead, if, in an orthonormal coframe field, the first fundamental form of $S$ is $I = {\omega\_1}^2 + {\omega\_2}^2$
and the (usual) second fundamental form is $I\!I = h\_{11}\,{\omega\_1}^2 + 2h\_{12}\,\omega\_1\omega\_2 + h\_{22}\,{\omega\_2}^2$, Cartan's third fundamental form is
$$
\Psi = h\_{12}\,{\omega\_1}^2 + (h\_{22}{-} h\_{11})\,\omega\_1\omega\_2 - h\_{12}\,{\omega\_2}^2.
$$
| 5 | https://mathoverflow.net/users/13972 | 431574 | 174,719 |
https://mathoverflow.net/questions/431521 | 5 | Let $\mathcal{F}$ denote a family of countable subsets of $\mathbb{R}$, such that for each $U, V\in\mathcal{F}$ we have that $U\subseteq V$, or $V\subseteq U$. Let $(\mathcal{F}, \preceq)$ denote the inclusion partial order of $\mathcal{F}$.
1. Is it true that $(\mathcal{F}, \preceq)$ is isomorphic to $(S, \leq)$ where $S$ is a subset of $\mathbb{R}$?
2. Is it true that $\bigcup\_{U\in\mathcal{F}}U$ is a countable set?
| https://mathoverflow.net/users/100140 | Question about a family of nested countable subsets of $\mathbb{R}$ | $\let\sset\subseteq\def\cF{\mathcal F}\def\R{\mathbb R}\def\Q{\mathbb Q}$The answer to both questions is negative: let $\preceq$ be a well order of type $\omega\_1$ on a subset of $\R$, and let $\cF$ consist of all proper initial segments of $\preceq$. Then $\cF$ is a family of countable sets totally ordered by $\subseteq$, but $\bigcup\cF$ is uncountable, and $(\cF,{\sset})$ does not embed in $(\R,{\le})$ as $(\omega\_1,{\le})$ does not embed there.
In fact, for any $\cF$ as in the question, the following are equivalent:
1. $(\cF,{\sset})$ embeds in $(\R,{\le})$.
2. $\bigcup\cF$ is countable.
3. $(\cF,{\sset})$ has countable cofinality.
1 → 3: Any subset of $\R$ has a countable cofinal subset.
3 → 2: If $C\sset\cF$ is a cofinal countable subset, then $\bigcup\cF=\bigcup C$ is a countable union of countable sets, hence countable.
2 → 1: Let $I=\bigcup\cF$. Then $\{(x,y):\forall U\in\cF\,(y\in U\to x\in U)\}$ is a total preorder on $I$, hence it includes a total order $\preceq$ on $I$, and every $U\in\cF$ is an initial segment of $\preceq$. Since the lexicographic product $(I,{\preceq})\times2$ embeds in $(\Q,{\le})$, we can find an embedding $f\colon(I,{\preceq})\to(\Q,{\le})$ such that $f(x)>\sup\{f(y):y\prec x\}$ for all $x\in I$. Then $U\mapsto\sup f[U]$ is an embedding of $(\cF,{\sset})$ in $(\R,{\le})$.
Note that the example in the beginning is, in a sense, the worst that can happen: as any total order, $(\cF,{\sset})$ has a well-ordered cofinal subset $C$. If $C$ is not countable, it can only have order type $\omega\_1$, as otherwise some element of $\cF$ is uncountable. Thus, $(\cF,{\sset})$ has cofinality $\omega\_1$, $|\bigcup\cF|=\aleph\_1$, and using a similar argument as above, $(\cF,{\sset})$ embeds in the long line $\omega\_1\times[0,1)$.
| 7 | https://mathoverflow.net/users/12705 | 431575 | 174,720 |
https://mathoverflow.net/questions/431568 | 14 | In his "Théorie de chaleur" Fourier proves that the zeros of Bessel function $J\_0(x)$ are all real.
I want to ask if there is a modern version of this proof exist in literature?
If someone can provide me with an elegant compact proof or reference for it , it would be helpful.
| https://mathoverflow.net/users/145223 | Fourier's proof of reality of all roots of Bessel function $J_0(x)$ | Fourier proof was incomplete. Fourier used the following
**Statement.** A real entire function has only real zeros if
its derivatives have the following property:
If $x$ is a real root of $f^{(n)}$ then $f^{(n-1)}(x)f^{(n+1)}(x)<0.$
But he verified this only for polynomials. Fourier's proof was criticized by Cauchy, and Fourier defended his arguments, but the statement above is not correct for unrestricted entire functions. The full story is described in the paper of
Pólya, [Some problems connected with Fourier's work on transcendental functions](https://doi.org/10.1093/qmath/os-1.1.21), *Quart. J. Math.* 1 (1930) 21-34.
In this paper he stated what is called since then the “Fourier–Pólya Conjecture”: If $f$ is a real entire function of genus $0$, then the number of points where $f^{(n)}(x)=0$ but
$f^{(n-1)}(x)f^{(n+1)}(x)>0$ is equal to the number of pairs of non-real conjugate zeros.
This conjecture was proved only in 2000:
H. Ki and Y. Kim, [On the number of non-real zeros of real entire functions
and the Fourier–Pólya conjecture](https://web.archive.org/web/20190308101528id_/http://pdfs.semanticscholar.org/f133/581f3b56582852908c6013c5d0fea0b183c6.pdf), *Duke Math. J.* 104 (2000) 45–73.
(The earlier paper of Kim
Kim, Young-One,
[Critical points of real entire functions whose zeros are distributed in an infinite strip](https://web.archive.org/web/20190321094420/https://core.ac.uk/download/pdf/82785929.pdf),
*J. Math. Anal. Appl.* 204 (1996), no. 2, 472–481.
already contained what is needed to justify Fourier's argument).
Thus Fourier's method has been justified, after 170+ years of research, but it is incomparably more complicated than the "modern" proof based on the observation that zeros of Bessel's function are eigenvalues of a self-adjoint operator, and these eigenvalues are real by a two-line linear algebra argument. If I remember correctly, this observation is due to Poisson.
See also my lectures with an exposition of other related conjectures of Pólya: [https://www.math.purdue.edu/~eremenko/dvi/kent.pdf](https://www.math.purdue.edu/%7Eeremenko/dvi/kent.pdf)
| 24 | https://mathoverflow.net/users/25510 | 431579 | 174,721 |
https://mathoverflow.net/questions/431543 | 1 | A functional Hilbert space $\mathscr H=\mathscr H(\Omega)$ is a Hilbert space of complex valued functions on a (nonempty) set $\Omega$, which has the property that point evaluations are continuous i.e. for each $\lambda\in \Omega$ the map $f\mapsto f(\lambda)$ is a continuous linear functional on $\mathscr H$. The Riesz representation theorem ensure that for each $\lambda\in \Omega$ there is a unique element $k\_{\lambda}\in \mathscr H$ such that $f(\lambda)=\langle f,k\_{\lambda}\rangle$ for all $f\in \mathscr H$. The collection $\{k\_{\lambda} : \lambda\in \Omega\}$ is called the reproducing kernel of $\mathscr H$. For $\lambda\in \Omega$, let $\hat{k\_{\lambda}}=\frac{k\_{\lambda}}{\|k\_{\lambda}\|}$ be the normalized reproducing kernel of $\mathscr H$.
For a bounded linear operator $A$ on $\mathscr H$, we define the following norms:
\begin{align\*}
N(A):=\sup\{\|A\widehat{k}\_{\lambda}\|: \lambda\in\Omega\}.
\end{align\*}
>
> Is $ N(A)= N(A^\*)$?
>
>
>
| https://mathoverflow.net/users/113054 | Is $\sup\{\|A\widehat{k}_{\lambda}\|: \lambda\in\Omega\}=\sup\{\|A^*\widehat{k}_{\lambda}\|: \lambda\in\Omega\}$? where $A$ is an operator on RKHS | No, this fails already when $\Omega=\{ 1, 2\}$, so $H=\mathbb C^2$ and $\widehat{k}\_j=e\_j$. We can take
$$
A= \begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix} .
$$
Then $N(A)=\sqrt{2}$, $N(A^\*)=1$.
| 2 | https://mathoverflow.net/users/48839 | 431593 | 174,724 |
https://mathoverflow.net/questions/431592 | 1 | Suppose that $f: X \to Y$ is a smooth proper map between two smooth manifolds. Is it always possible to represent $f$ as a composition of a closed embedding $g: X \to Z$ with a proper submersion $h: Z \to Y$?
**Motivation:**
Firstly, closed embeddings and proper submersions are in a sense the simplest kinds of proper maps, so in some cases we could hope reduce proofs about proper maps to those two cases.
Secondly, projective morphisms in algebraic geometry are defined precisely as morphisms that admit such decomposition for $Z = \mathbb{P}^n\_Y$, and so far all examples of proper maps that I have come up with are in a sense similar.
Finally, if $X$ is compact, we always have such decomposition as $X \to X \times Y \to Y$, where the first map is the closed embedding of the graph of $f$ and the second map is the projection onto $Y$.
| https://mathoverflow.net/users/157863 | Decomposing proper map into closed embedding and proper submersion | Yes. Choose a smooth (but not necessarily closed) embedding $i:X\to W$ where the manifold $W$ is compact, for example a sphere. Together $i$ and $f$ give a smooth map $X\to W\times Y$ that is both proper and an embedding.
| 3 | https://mathoverflow.net/users/6666 | 431595 | 174,726 |
https://mathoverflow.net/questions/431585 | 3 | Suppose that $\tau \in \mathbf{H}$ belongs to the complex upper half plane. The quotient $\mathbf{C}/(\mathbf{Z}+\mathbf{Z}\tau)$ gives an elliptic curve over $\mathbf{C}$. Write this elliptic curve as $E\_{\tau}$. We can write $E\_{\tau}$ as follows:
$$E\_{\tau}: y^2 = 4x^3 - \left( \dfrac{27 j(\tau)}{j(\tau)-1728} \right)x - \left( \dfrac{27 j(\tau)}{j(\tau)-1728} \right),$$
where $j(\tau)$ is the $j$-invariant function. This is a model for the universal elliptic curve $E \to Y\_1(N)$ over a modular curve $Y\_1(N)$.
**In very vague terms**, my question is: suppose we have an eigen-cuspform $f \in S\_1(\Gamma\_1(N))$. Given the differential $\omega = dx/y$ on $E$, how do we compute the "$f$-part" of $\omega$? That is, how do we compute the direct summand of $\omega$ where the Hecke operators act via the Hecke eigenvalues of $f$?
**More rigorously:**
Let $H^0(E, \Omega\_{E/Y\_1(N)})$ be the space of holomorphic one-forms on $E$ over $Y\_1(N)$ and let $f \in S\_1(\Gamma\_1(N))$ be an eigen-cuspform. Given the differential $\omega = dx/y \in H^0(E, \Omega\_{E/Y\_1(N)})$, how do we explicitly compute the $f$-isotypical component of $\omega$ under the action of the Hecke operators?
| https://mathoverflow.net/users/394740 | Hecke operators on universal elliptic curves | The universal elliptic curve can be written $y'^2 = 4x'^3- g\_2 x - g\_3$ where $g\_2$ and $g\_3$ are Eisenstein series. Given any differential on $E$, simply change coordinates to this family, then divide by $dx'/y'$ to obtain a modular form of weight $1$.
The point is that $g\_2$ is modular of weight $4$ and $g\_3$ is modular of weight $6$, so if $y$ and $x$ are coordinates in a $\Gamma\_1(N)$-invariant presentation of the universal family, then $x/x'$ is a modular function of weight $2$ and $y/y'$ is a modular function of weight $3$. Thus $dx/y $ is equal to $ dx'/y'$ times a modular function of weight $2-3=-1$. Dividing the $\Gamma\_1(N)$-invariant equation for your differential form by a modular function of weight $-1$ produces a modular form of weight $1$.
I'm not sure the presentation you gave necessarily works for this, because I think it is only the correct one up to possible quadratic twist. The correct model is the one that actually has a section of order $N$.
Modular forms of weight $1$ certainly don't correspond to differentials on the modular curve. Rather these are modular forms of weight $2$. The relation between differentials on $E$ and differentials on modular curves is simply that they satisfy similar transformation laws under $SL\_2(\mathbb Z)$, or, algebro-geometrically, that the line bundle of relative differentials squares to the line bundle of differentials of the base.
| 4 | https://mathoverflow.net/users/18060 | 431605 | 174,729 |
https://mathoverflow.net/questions/431602 | 2 | Let $X$ be a compact complex manifold with canonical bundle $K\_X$. Assume the Kodaira dimension $\kappa(X)$ is positive (but *not* maximal, i.e., $X$ is *not* of general type). Let $\varphi\_m : X \dashrightarrow Y\_m \subset \mathbb{P}^{N\_m}$ denote the $m$th pluricanonical map given by the sections of the $m$th tensor power $K\_M^{\otimes m}$.
>
> Suppose $X$ does not contain any rational curves. Does $Y\_m$ contain any rational curves?
>
>
>
In more detail, if $K\_X$ is semi-ample, the base of the Iitaka map $\varphi : X \to X\_{\text{can}}$ given by the linear system $|K\_X^{\otimes \ell} |$ for $\ell>0$ sufficiently large has ample canonical bundle (I've seen this stated, but am not certain about $K\_{X\_{\text{can}}}$ being ample, for general $X$ with $K\_X$ semi-ample). In any case, if $K\_{X\_{\text{can}}}$ is ample, then Mori's newness result says that $X\_{\text{can}}$ has no rational curves.
I'm wondering if this type of phenomenon occurs for the pluricanonical maps, in general, *eventually*. That is,
>
> Suppose $X$ does not contain any rational curves. For $m>0$ sufficiently large, do the base spaces $Y\_m$ of the pluricanonical maps have rational curves?
>
>
>
| https://mathoverflow.net/users/105103 | Rational curves on the image of the pluricanonical maps | Not only could $Y\_m$ contain a rational curve for all $m$, $Y\_m$ could *be* a rational curve for all $m$.
Take $C$ a hyperelliptic curve, $E$ an elliptic curve, $\tau$ the hyperelliptic involution on $C$, $\sigma$ the translation on $E$ by a point of order $2$.
Let $X = (E \times C)/ (\sigma \times \tau)$.
Since $\sigma$ has no fixed points, $\sigma \times \tau$ has no fixed points, so $K\_X$ pulls back to $K\_E\otimes K\_C= K\_C$, so $H^0 (X, K\_X^{\otimes m})$ is a subspace of $$H^0(E \times C, K\_C^{\otimes m})= H^0(C, K\_C^{\otimes m}),$$ specifically, the $(\sigma \times \tau)$-invariant part.
The action of $\sigma$ is trivial so this is just the $\tau$-invariant part. For $g$ the genus of $C$, $K\_C$ is the pullback of $\mathcal O\_{\mathbb P^1}(g-1)$ along $\mathbb P^1 = C/\tau$, so the $\tau$-invariant part of $H^0(C, K\_C^{\otimes m})$ is simply $H^0(\mathbb P^1, \mathcal O\_{\mathbb P^1} ( m(g-1)))$ and thus $Y\_m = \mathbb P^1$.
| 6 | https://mathoverflow.net/users/18060 | 431607 | 174,731 |
https://mathoverflow.net/questions/431450 | 1 | Quick context: This is a transposition of exercise 6.3 of Remco Van der Hofstad (2016) and it is relevant to some problem i encoutered in my research.
For each $n \in \mathbb{N}$, define a series of positive numbers $w\_i^n$ with $1\leq i\leq n$. Also denote an uniformly chosen numbers by $w\_U = W\_n$ where $U$ is picked uniformly in $\{1,...,n\}$ (meaning $\mathbb{P}(W\_n = w\_i^n) = 1/n$ for any $i\leq n$). We also suppose that :
**a)**$W\_n$ converges in distribution to a positive variable in the sense that $\lim\_{n\rightarrow\infty}\mathbb{P}(W\_n < t) = \mathbb{P}(W < t)$
**b)** $\lim\_{n\rightarrow\infty} \mathbb{E}W\_n = \mathbb{E}W$
**c)** $\lim\_{n\rightarrow\infty} \mathbb{E}W\_n^2 = \mathbb{E}W^2$
The question is to show that **A.** if a) and b) hold then $\max\_{i\leq n} w\_i^n = o(n)$ and that **B.** if a), b) and c) then $\max\_{i\leq n} w\_i^n = o(\sqrt{n})$.
As for my approach, I have noticed **A.** and **B.** are proven the same way. To prove **A.** I wanted to use the Borel Cantelli Lemma to show that $w\_i^n/n > \epsilon$ only a finite amount of times. We have :
\begin{equation}
\sum\_{ i\leq n } 1\_{ w\_i^n/n > \epsilon } = \sum\_{ i\leq n} 1\_{w\_i^n > n\epsilon}
\end{equation}
which can be rewritten as
\begin{equation}
\sum\_{ i\leq n } 1\_{ w\_i^n/n > \epsilon } = \sum\_{ d = \lceil n\epsilon \rceil } d\mathbb{P}( W^n = d)
\end{equation}
The rightmost part can be majored by something close to the expectation of $W$ regardless of $n$. My intuition tells me that taking $n$ to $\infty$ feel like it should be enough to finish the proof but i feel like i did not use all the hypotheses... I'm open to any ideas about the proof.
PS: I asked an adjacent [question](https://math.stackexchange.com/questions/4540856/proof-that-max-id-in-sqrtn-tends-to-0-when-sum-i-d-in2) yesterday which might be relevant.
| https://mathoverflow.net/users/164762 | Maximum of a sequence is $o(\sqrt{n})$ | First suppose that hypotheses (a) and (b) hold. I will write $w\_{i,n}$ instead of $w\_i^n$ for clarity.
Given $\epsilon>0$, the Monotone convergence theorem implies that there exists $M$ such that $E(W)-E(W \wedge M) <\epsilon$, where $\wedge$ indicates minimum. By (a) and the bounded convergence theorem,
$$E(W\wedge M)=\int\_0^M P(W \ge t) \, dt =\lim\_{n \to \infty} \int\_0^M P(W\_n \ge t) \, dt
=\lim\_{n \to \infty} E(W\_n\wedge M)\,.$$
Invoking (b), we infer that
$$\lim\_{n \to \infty} \frac1n \sum\_{\{i:\, w\_{i,n}>M\} } (w\_{i,n}-M): = \lim\_{n \to \infty}E[W\_n-(W\_n \wedge M)]= \tag{\*} $$ $$ = E(W)-E(W\wedge M) <\epsilon \,. $$
If, for some $n$, $$\max\_i w\_{i,n}>2\epsilon n +M \tag{\*\*}\,,$$ then the sum on the LHS of $(\*)$ is greater than $2\epsilon n$, so $(\*\*)$ can only occur for finitely many $n$. Therefore,
$$\limsup\_{n \to \infty} \frac1n \max\_i w\_{i,n} \le 2\epsilon \,.$$
Since $\epsilon>0$ is arbitrary, we conclude that this limsup must equal $0$.
---
---
If we assume (a) and (c) hold (hypothesis (b) is not needed here) , then applying the above result to $w\_{i,n}^2$ instead of $w\_{i,n}$, yields that
$$\limsup\_{n \to \infty} \frac1n \max\_i w\_{i,n}^2 =0\,,$$
as required.
| 2 | https://mathoverflow.net/users/7691 | 431608 | 174,732 |
https://mathoverflow.net/questions/302516 | 19 | Classically, we can explicitly construct the free Abelian group $\newcommand{\Z}{\mathbb{Z}}\Z[X]$ on a set $X$ as the set of finitely-supported functions $X \to \Z$, and so easily see that the unit map $X \to \Z[X]$ (inclusion of generators) is injective.
Constructively, this construction of $\Z[X]$ doesn’t work for arbitrary $X$. The unit map $X \to \Z[X]$ should send $x \in X$ to the function $1\_x : X \to \Z$, sending $y \in X$ to $1$ if $y = x$ and $0$ otherwise; but this definition-by-cases requires that $X$ has decidable equality (i.e. for every $x,y \in X$, either $y = x$ or $y \neq x$).
Similarly, most other classical explicit constructions of $\Z[X]$ rely at some point on decidable equality of $X$. The abstract-nonsense syntactic construction of $\Z[X]$ as a free model of an algebraic theory works fine constructively; but from this construction, it’s not clear that the unit map $X \to \Z[X]$ is injective.
**Is there some constructive proof that the unit map $X \to \Z[X]$ is injective for arbitrary $X$, or can it fail?**
[Note this is a self-answered question. It came up since the fact was required [here](https://mathoverflow.net/a/272617/2273); asking several experts, the answer seems to be not well-known. After a couple of days, I worked out an answer via an explicit construction, and then also got a reply pointing to a slightly less explicit answer in the literature; so I’m writing it up here to make the answer and construction more prominently available.]
| https://mathoverflow.net/users/2273 | Constructively, is the unit of the “free abelian group” monad on sets injective? | I'd like to present a more modular proof. The key lemma (which I learnt from Christian Sattler) is that any set $X$ is a filtered colimit of finite sets. Indeed, let $I$ be the category of pairs $(n, f)$ with $n \in \mathbb N$ and $f : [n] \to X$, where a morphism $(n, f) \to (m, g)$ is a function $p : [n] \to [m]$ such that $f = g \circ p$. Since the category of finite sets has finite colimits, so does $I$, and so in particular $I$ is filtered. Now $X$ is the colimit $\mathop{\mathrm{colim}}\_{(n,f) \in I} [n]$ in the obvious way.
The free abelian group functor $F$ commutes with colimits, being a left adjoint, so $FX \cong \mathop{\mathrm{colim}}\_{(n,f) \in I} F[n]$. Moreover, the functor $U$ sending an abelian group to its underlying set commutes with *filtered* colimits, so $UFX \cong \mathop{\mathrm{colim}}\_{(n,f) \in I} UF[n]$.
The unit maps $[n] \to UF[n]$ are injective since $[n]$ has decidable equality. Filtered colimits commute with pullbacks in sets, so they preserve monomorphisms. This means precisely that the unit map $X \to UFX$ is also injective, as needed.
If we look more closely at the characterisation of equality in a sequential colimit of sets, we obtain an explicit description of $UFX$ as in previous answers.
| 4 | https://mathoverflow.net/users/113846 | 431619 | 174,735 |
https://mathoverflow.net/questions/431627 | 0 | Let $$A\_n=\{(x\_1,x\_2,x\_3,\cdots,x\_n):x\_i \in [q] \text{ for } i \in [n], x\_1 < x\_2, x\_2 > x\_3, x\_3 < x\_4, \cdots , (-1)^{n}x\_{n-1} < (-1)^{n} x\_n\}.$$ What is the cardinality of $A\_n$?
I have 2 observations. First, let $$B\_n=\{(x\_1,x\_2,x\_3,\cdots,x\_n):x\_i \in [q] \text{ for } i \in [n], x\_1 \leq \frac{q}{2}, x\_2 > \frac{q}{2}, x\_3 \leq \frac{q}{2}, \cdots\}.$$ We have $B\_n \subset A\_n$, so $|A\_n| \geq |B\_n| \geq (\frac{q}{2}-1)^n$.
Second, let $\boldsymbol{x}=(x\_1,x\_2,x\_3,\cdots,x\_n)$ be a uniform random variable in $\mathbb{F}\_q^n$, and event $E\_i$ represents $(-1)^{i+1}x\_i<(-1)^{i+1}x\_{i+1}$. We have $$Pr(E\_i)=\frac{1}{2}, Pr(\boldsymbol{x}\in A\_n)=Pr(E\_1 E\_2 \cdots E\_{n-1}).$$ If $E\_i$'s were independent, then $Pr(\boldsymbol{x}\in A\_n)=(\frac{1}{2})^{n-1}$, i.e. $|A\_n| = (\frac{1}{2})^{n-1}q^{n}$. However, $E\_i$'s are not independent.
**My question is**, can we find an upper bound of $|A\_n|$? Say, can we prove $|A\_n| \leq (\frac{3}{4})^{n-1}q^{n}$?
| https://mathoverflow.net/users/492228 | Upper bound of the number of oscillatory sequences | In the terminology of Chapter 3 of *Enumerative Combinatorics*,
vol. 1, second ed., you are asking for the value of the strict order
polynomial of the zigzag poset $Z\_n$ at $q$ (see Exercise
3.66). Equivalently, it is equal to $(-1)^n\Omega\_{Z\_n}(-q)$, where
$\Omega\_{Z\_n}$ is the ordinary order polynomial. The number of linear
extensions of $Z\_n$ is the Euler number $E\_n$. It follows that for
fixed $n$, $A\_n \sim \frac{E\_n}{n!} q^n$. Moreover, it is well known
that $\frac{E\_n}{n!} \sim 2\left( \frac{2}{\pi}\right)^{n+1}$.
| 7 | https://mathoverflow.net/users/2807 | 431629 | 174,737 |
https://mathoverflow.net/questions/417965 | 3 | Suppose $X,Y,Z$ are real-valued random variables on some probability space. Suppose I know everything there is to know about the moments
$$
M\_{p,q,r}=\mathbb{E}(X^pY^qZ^r)
$$
for $p,q,r\in\mathbb{N}\_0$. Suppose also that these moments do not grow too fast so the moment problem is determinate, i.e., the moments completely determine the joint distribution of $X,Y,Z$.
How can I characterize $X\perp Y\mid Z$, namely the conditional independence of $X$ and $Y$ given $Z$,
entirely in terms of the moments $M\_{p,q,r}$?
I tried to search the internet for such conditional independence criteria but what I could find was not helpful to me.
| https://mathoverflow.net/users/7410 | Moment criterion for conditional independence | If the moment problem is determinate and in addition $Z$ is bounded, then $X\perp Y\mid Z$ if and only if
$$\mathbb E\left[X^pY^qZ^r\right]=\lim\_{n\to\infty}\mathbb E\left[X^p\mathbf Z\_n^{\mathsf T}\right]\mathbb E\left[\mathbf Z\_n\mathbf Z\_n^{\mathsf T}\right]^+\mathbb E\left[\mathbf Z\_nZ^r\mathbf Z\_n^{\mathsf T}\right]\mathbb E\left[\mathbf Z\_n\mathbf Z\_n^{\mathsf T}\right]^+\mathbb E\left[\mathbf Z\_nY^q\right]$$
where $\mathbf Z\_n$ is the column vector $\left(1,Z,\dots,Z^n\right)$ and $^+$ denotes the Moore–Penrose pseudoinverse. I suspect that the assumption that $Z$ is bounded can be removed, but use it in my argument here.
The key to this result is the interpretation of conditional expectation as an orthogonal projection. Just as $\mathbb E\left[X\right]$ is the $\mu$ that minimizes $\mathbb E\left[\left(X-\mu\right)^2\right]$, the conditional expectation is $\mathbb E\left[X\mid Z\right]$ is $g\left(Z\right)$ for the measurable function $g$ that minimizes $\mathbb E\left[\left(X-g\left(Z\right)\right)^2\right]$, assuming $X$ and $Z$ have finite variance. In the Hilbert space of real-valued random variables with finite variance up to almost sure equality, where the inner product is defined by $\left\langle X,Y\right\rangle:=\mathbb E\left[XY\right],$ this is just the orthogonal projection of $X$ onto the subspace of measurable functions of $Z$.
If $Z$ is bounded, then by the Stone–Weierstrass theorem, the subspace of measurable functions of $Z$ is the closed linear span of $1,Z,Z^2,\dots$. Hence the orthogonal projection of $X$ onto this subspace is the limit of the orthogonal projection of $X$ onto the linear span of $1,Z,\dots,Z^n$ as $n\to\infty$. In other words, assuming $X$ has finite variance and $Z$ is bounded,
$$\mathbb E\left[X\mid Z\right]=\lim\_{n\to\infty}\mathbb E\left[X\mathbf Z\_n^{\mathsf T}\right]\mathbb E\left[\mathbf Z\_n\mathbf Z\_n^{\mathsf T}\right]^+\mathbf Z\_n$$
in mean square, where $\mathbf Z\_n$ is as above. The expression inside the limit on the right-hand side is the aforementioned orthogonal projection, which should be familiar from OLS regression.
To deduce the result, if $X\perp Y\mid Z$, then
$$\mathbb E\left[X^pY^qZ^r\right]=\mathbb E\left[\mathbb E\left[X^pY^q\mid Z\right]Z^r\right]=\mathbb E\left[\mathbb E\left[X^p\mid Z\right]\mathbb E\left[Y^q\mid Z\right]Z^r\right],$$
and the converse follows from the fact that the right-hand expression depends only on the joint distributions of $\left(X,Z\right)$ and of $\left(Y,Z\right)$, together with the assumption that the moment problem is determinate. The final step deducing convergence uses the fact that $Z$ is bounded again.
As an example, suppose $\left(X,Z\right)$ and $\left(Y,Z\right)$ are both bivariate Gaussian and $X\perp Y\mid Z$. Even though $Z$ is not bounded, the expression on the right hand-side of the equation for $\mathbb E\left[X\mid Z\right]$ has already converged once $n=1$, since uncorrelated bivariate Gaussians are independent. Hence the original equation still holds in the case $p=q=1$. Assuming $\mathrm{Var}\left(Z\right)>0$, this reduces in the case $r=0$ to $$\mathrm{Cov}\left(X,Y\right)=\frac{\mathrm{Cov}\left(X,Z\right)\mathrm{Cov}\left(Z,Y\right)}{\mathrm{Var}\left(Z\right)},$$
which can be shown to coincide with the solution obtained via maximum entropy.
| 2 | https://mathoverflow.net/users/492208 | 431637 | 174,741 |
https://mathoverflow.net/questions/431449 | 2 | $\DeclareMathOperator\Tr{Tr}$I have a problem with understanding the proof of Proposition 6.8 in the book ,,Elements of Noncommutative Geometry''. One can find the formulation of this proposition [here](https://ibb.co/V2BHGkY) and the proof [here](https://ibb.co/dDhDHxS). The context is the following: $V$ is a *real* infinite dimensional vector space equipped with a *bilinear* form $g$: let us assume that the complex orthogonal structure $J$ is chosen, namely $J$ is an operator on $V$ such that $J^2=-I$ and $g(Jx,Jy)=g(x,y)$ for $x,y \in V$. This allows us to view $V$ as a *complex* vector space with the action $ix:=Jx$. Once $J$ is chosen, there is a scalar product $\langle x,y \rangle:=g(x,y)+ig(Jx,y)$. The space $V$ with this scalar product will be denoted by $V\_J$. On the other hand, one can consider the complexification $V^{\mathbb{C}}$ of $V$ and equip it with the inner product $\langle \langle x,y \rangle \rangle:=2g^{\mathbb{C}}(\overline{x},y)$ where $g^{\mathbb{C}}$ is the extension of $g$ to the *bilinear* form on $V^{\mathbb{C}}$. One has then the projection $P\_J:=\frac{1}{2}(I-iJ)$ in $V^{\mathbb{C}}$. If we put $W\_J$ to be the range of this projection it turns out that viewing $P\_J$ as an operator $V\_J \to W\_J$ yields a unitary $P\_J$ (this is the reason to choose this constant $2$ in the definition of the inner product on $V^{\mathbb{C}}$). The Schiwger term is defined as $\alpha(A,B):=-\frac{1}{2} \Tr[A\_-,B\_-]$ where $A\_-:=\frac{1}{2}(A+JAJ)$ is the *antilinear* part of $A$ (similarly for $B$) where $A$ is such that $[J,A]$ is Hilbert-Schimdt. Note that we are computing the trace of the commutator so the first impression is that it should be trivially zero-however our operators are antilinear-therefore it is nontrivial.
After explaining the general context let me explain where exactly my problems pop out: I don't see the last two equalities.
>
> The first one is as follows:
> $$\Tr(iP\_J[A\_-,B\_-]P\_J)+\Tr(-iP\_{-J}[A\_-,B\_-]P\_{-J})=2i\Tr(P\_JA\_-P\_{-J}B\_-P\_J-P\_JB\_-P\_{-J}A\_-P\_J)$$
> Here I don't see how the projection $P\_{-J}$ appears in the middle of these two terms: $P\_{-J}=I-P\_J$ so one can have $P\_J$ in the middle instead of $P\_{-J}$ but still-there is no appearance of $P\_J$ in the middle on the left hand side. I would like to see why this equality is valid
>
>
>
>
> For the second one, namely:
> $$2i\Tr(P\_JA\_-P\_{-J}B\_-P\_J-P\_JB\_-P\_{-J}A\_-P\_J)=-4i\alpha(A,B)$$
> authors claim that this follows since $P\_J:V\_J \to W\_J$ is a unitary operator. As far as I understood-according to the formulation of the Proposition-$\alpha(A,B)$ is computed as a trace in $V^{\mathbb{C}}$ while the other trace probably takes place in $V\_J$. But somehow these projections $P\_J$ and $P\_{-J}$ go the other way around so I'm confused what is really going on.
>
>
>
I will be very grateful for the explanation!
| https://mathoverflow.net/users/24078 | Another formula for the Schwinger term — problems with a calculation | I combine the two questions in the OP in a single question: find a proof of proposition 6.8,
$$
\alpha(A,B)=\tfrac{1}{8}i\,{\rm Tr}\,J[J,A][J,B].\qquad\qquad(6.8)
$$
---
I note that the Schwinger term $\alpha(A,B)$ is defined as the trace of antilinear operators on a complex vector space, while the trace in (6.8) involves linear operators on a real vector space. The first step is then to rewrite the Schwinger term as a trace of linear operators on a real vector space,
\begin{equation}
\alpha(A,B)=\tfrac{1}{2}i\,{\rm Tr}\,J[A\_-,B\_-]=\tfrac{1}{8}i\,{\rm Tr}\,J[(A+JAJ),(B+JBJ)].\qquad\qquad(1)
\end{equation}
I have substituted the definitions
\begin{equation}
A\_-=\tfrac{1}{2}(A+JAJ),\;\;B\_-=\tfrac{1}{2}(B+JBJ),\qquad\qquad(2)
\end{equation}
where $J^2=-1$.
Since these are traces of linear operators, we can permute them cyclically, ${\rm Tr}\,XY={\rm Tr}\,YX$. I work out the trace in equation 1,
\begin{equation}
{\rm Tr}\,J[(A+JAJ),(B+JBJ)]=4\,{\rm Tr}\,J[A,B],\qquad\qquad(3)
\end{equation}
and compare with
\begin{equation}
{\rm Tr}\,J[J,A][J,B]=2\,{\rm Tr}\,J[A,B].\qquad\qquad(4)
\end{equation}
I thus arrive at
\begin{equation}
\alpha(A,B)=\tfrac{1}{4}i\,{\rm Tr}\,J[J,A][J,B].\qquad\qquad(5)
\end{equation}
This is almost [proposition 6.8](https://ibb.co/V2BHGkY) from the book cited by the OP, except that there the coefficient is $1/8$ instead of $1/4$. I have searched for an [alternative source](https://arxiv.org/abs/hep-th/9402098) by the same authors, where the coefficient (on page 15) is given as $1/4$, like I found, so this is likely a typo in their book.
| 2 | https://mathoverflow.net/users/11260 | 431640 | 174,743 |
Subsets and Splits