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https://mathoverflow.net/questions/404208
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1
|
Let $H=(H, (\cdot, \cdot))$ be a Hilbert space. Let $T\_1,T\_2:D \subset H \longrightarrow H$ be a self-adjoint operators (not necessarily bounded). It's well-know that the spectrum $\sigma(T\_i)$ of $T\_i$ satisfies $\sigma(T\_i) \subset \mathbb{R}$, for $i=1,2$ (see Theorem $29.2$ in $[3]$). Suppose that $T\_1$ and $T\_2$ are bounded below and has $N \in \mathbb{N}$ (real) eigenvalues arranged in the ascending order
$$
\lambda\_1(T\_i) \leq \lambda\_2(T\_i) \leq \lambda\_3(T\_i) \leq \cdots \lambda\_N(T\_i), \quad i \in \{1,2\}.
$$
As a consequence of the Min-Max Principle $($see $[2$, page $85]$ or $[1$, page $61])$, if
$$
(T\_1(u), u) \leq (T\_2(u), u),\; \forall \; u \in D \tag{1}
$$
then, for each $n \in \{1,\cdots, N\}$,
$$\lambda\_n(T\_1) \leq \lambda\_n(T\_2). \tag{2}$$
**Question.** If
$$
(T\_1(u), u) < (T\_2(u), u),\; \forall \; u \in D\setminus \{0\}
$$
and
then
$$\lambda\_n(T\_1) < \lambda\_n(T\_2) \tag{3}
$$ for each $n \in \{1,\cdots, N\}$?
I think so, because the Min-Max Principle establishes that, for $i=1,2$,
$$
\lambda\_n(T\_i)= \sup\_{u\_1, u\_2, \cdots u\_{n-1} \in H } \inf\_{v \in D\setminus \{0\} \atop v \in [u\_1, u\_2, \cdots u\_{n-1}]^{\perp} } \frac{(T\_i(v),v)}{\|v\|}.
$$
**Remark.** I did this question in [Math Stackexchange](https://math.stackexchange.com/questions/4246725/a-consequence-of-the-min-max-principle-for-self-adjoint-operators), but I don't received any comment or answer.
Any comment or reference are welcome.
$[1]$ Kato, T., Perturbation Theory for Linear Operators, $2$nd edition, Springer, Berlin, $1984$.
$[2]$ Reed, S. and Simon, B., Methods of Modern Mathematical Physics: Analysis of Operator,
Academic Press, Vol. IV, $1978$.
$[3]$ Bachman, G. and Narici, L. Functional Analysis. New York: Academic Press, $1966$.
|
https://mathoverflow.net/users/156344
|
A consequence of the Min-Max Principle for self-adjoint operators
|
I'm expanding my comment, in response to the OP's comment. Indeed, the case of just the lowest eigenvalue is perhaps not a good illustration of the full argument.
In general, let $u\_j$ be a normalized eigenvector for $\lambda\_j(T\_1)$, so $T\_1 u\_j=\lambda\_j(T\_1) u\_j$. Also make sure that the $u\_j$ are orthogonal (this is automatic, except in the case of degeneracies). Then
$$
\lambda\_n(T\_1)=\langle u\_n, T\_1 u\_n\rangle =\inf\_{v\perp u\_1,\ldots , u\_{n-1}} \langle v, T\_1 v\rangle < \inf \_{v\perp u\_1,\ldots , u\_{n-1}} \langle v, T\_2 v\rangle .
$$
The inequality is true because the infima are really minima: by the assumption on the existence of discrete spectrum in the range we're investigating and by the spectral theorem, the search for $v$ can be restricted to a suitable finite-dimensional subspace. We can then try the $v$ that minimizes $\langle v, T\_2 v\rangle$ in the other quadratic form.
We're done since obviously $\lambda\_n(T\_2)\ge \inf \_{v\perp u\_1,\ldots , u\_{n-1}} \langle v, T\_2 v\rangle$.
---
The argument can also be organized differently, maybe this version is more transparent: Let $M\subseteq H$, $\dim M=n$, be the ("an", in the case of degeneracies) space spanned by the eigenvectors of $T\_2$ with eigenvalues $\lambda\_1(T\_2),\ldots , \lambda\_n(T\_2)$. Then $\max\_{v\in M, \|v\|=1 }\langle v, T\_2 v\rangle = \lambda\_n(T\_2)$.
By assumption and since $M$ is finite-dimensional,
$$
t=\max\_{v\in M, \|v\|=1 } \langle v, T\_1 v\rangle < \lambda\_n(T\_2) .
$$
For any choice of $u\_1,\ldots , u\_{n-1}\in H$, there will be a $v\in M\ominus H$, $\|v\|=1$. Hence, by min-max, $\lambda\_n(T\_1)\le t<\lambda\_n(T\_2)$.
|
0
|
https://mathoverflow.net/users/48839
|
404254
| 165,802 |
https://mathoverflow.net/questions/404261
|
1
|
Let $G=p^{n+m}.Q$ be an extension group of the special $p$-group $p^{n+m}$ by a group $Q$. Now $p^{n+m}=p^n{{}^\cdot}p^m$. How does one show that $\frac{G}{p^n}\cong p^m.Q$? Or equivalently that $G \cong p^n{{}^\cdot}(p^m{.}Q)$ (non-split extension of $p^n$ by $p^m{.}Q$?
|
https://mathoverflow.net/users/148317
|
On a quotient of a finite extension group $G=p^{n+m}.Q$
|
I presume you mean that $G$ has a normal subgroup $P$ with $G/P \cong Q$, where $P$ is a special $p$-group with $Z(P) = [P,P] = \Phi(P)$ elementary abelian of order $p^n$, and $P/Z(P)$ elementary abelian of order $p^m$.
But $Z(P)$ is characteristic in $P$ and hence normal in $G$, so $G$ has the structure ${p^n}^. (p^m.Q)$, and $G/Z(P)$ has structure $p^m.Q$.
|
1
|
https://mathoverflow.net/users/35840
|
404264
| 165,805 |
https://mathoverflow.net/questions/397791
|
7
|
Given a function $f \in L^1 (\mathbb R)$, define the *roughness* $R\_f$ of $f$ at $x \in \mathbb R$ by
$$\DeclareMathOperator{\esssup}{\operatorname{esssup}}
R\_f (x) := \limsup\_{r \to 0+}\dfrac{r \esssup\_{y \in B\_r (x)} |f(y) - f(x)|}{\displaystyle\int\limits\_{B\_r (x)} |f(s) - f(x)| ds}
$$
where $\esssup$ denotes the essential supremum, and by convention we take $\frac{0}{0} = 1$.
**Question:** Let $f$ be continuous. Is it true that $f$ is differentiable almost everywhere if and only if $R\_f = 1$ almost everywhere?
*Remark: The “only if” direction is relatively straightforward, the “if” direction is the issue.*
|
https://mathoverflow.net/users/173490
|
An equivalent condition for differentiability almost everywhere?
|
The answer is **negative**: If $f'(x) = 0$ "too often", then $R\_f$ may fail to be equal to one almost everywhere.
---
Let $C$ be a [fat Cantor set](https://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set), let $I\_n = (a\_n, b\_n)$ ($n \geqslant 2$) be the sequence of all finite components of the complement of $C$, and let $f$ be a differentiable function with the following properties:
* $f(x) = 0$ for $x \in C$;
* on $I\_n$, $f$ is a smooth bump of a fixed shape, supported in $\tfrac1n I\_n$ (the middle $n$th part of $I\_n$), and with maximum equal to $|I\_n|^2$ (and hence the integral of $f$ over $I$ is equal to $\tfrac1n |I\_n|^3$).
Then $f'(x) = 0$ for $x \in C$ by the first property, so $f$ is everywhere differentiable. On the other hand, it is rather straigthforward to see that there is a constant $C$ such that for each $n$ and $t \in I\_n = (a\_n, b\_n)$ we have
$$ \int\_{a\_n}^t f(y) dy \leqslant \frac{C}{n} (t - a\_n) \sup\_{y \in [a\_n, t]} f(y) $$
and
$$ \int\_t^{b\_n} f(y) dy \leqslant \frac{C}{n} (b\_n - t) \sup\_{y \in [t, b\_n]} f(y) . $$
It follows that if $x \in C$ and $r$ is small enough, so that $B\_r(x)$ is disjoint with $\tfrac12 I\_2 \cup \tfrac13 I\_3 \cup \ldots \cup \tfrac1{n-1} I\_{n-1}$, then
$$ \int\_{B\_r(x)} f(y) dy \leqslant \frac{C}{n} |B\_r(x)| \sup\_{y \in B\_r(x)} f(y) . $$
This, in turn, implies that $R\_f(x) = \infty$ for every $x \in C$.
Thus, $R\_f(x) \ne 1$ on a set of positive Lebesgue measure.
|
2
|
https://mathoverflow.net/users/108637
|
404279
| 165,813 |
https://mathoverflow.net/questions/404218
|
2
|
Let $S$ be a set with $\lvert S\rvert=\lvert\mathbb{R}\rvert$. Suppose it has subsets $S\_x$ indexed by $x\in \mathbb{R}$ with $\lvert S\_x\rvert=\lvert\mathbb{R}\lvert$ for each $x\in \mathbb{R}$. Suppose that
* for any $s\in S$ we have $\lvert\{x\in\mathbb{R}\mathrel\vert s\in S\_x\}\rvert=2$
* for any $x\neq y\in \mathbb{R}$ we have $\lvert S\_x\cap S\_y\rvert=3$.
Consider functions $f:S\to \coprod\_{x\in\mathbb{R}}S\_x$ such that $f\circ \pi=\mathrm{id}$ where $\pi:\coprod\_{x\in\mathbb{R}}S\_x\to S$ is the projection. They are indexed by $2^{\mathbb{R}}$ because each $s\in S$ can go in $2$ subsets. Is there $f$ such that $\lvert f(S)\cap S\_x\rvert<\infty$ for each $x\in \mathbb{R}$?
|
https://mathoverflow.net/users/376005
|
Include each point of continuum in a subset so that each subset gets finitely many points
|
Let $M : = \mathbb R^2 \setminus\{(x, y): x^2 + y^2 \leq 1\}$, $\Delta := \{(x, x) \in \mathbb R^2\}$, and let $h$ be any bijection from $\mathbb R$ to the circle $\{(x, y) \in \mathbb R^2: x^2 + y^2 = 1\}$.
Define $S := (\mathbb{R}^2 \setminus \Delta) \sqcup M \sqcup (\mathbb R/{\sim})$, where $x \sim y$ if and only if $h(x)$ and $h(y)$ are diametrically opposite points. For $\alpha \in \mathbb R$, let $S\_\alpha$ consist of the points of the form $$L\_\alpha := \{(y, \alpha) : y \in \mathbb R \setminus \{\alpha\}\} \cup\{(\alpha, y) : y \in \mathbb R \setminus \{\alpha\}\} \subset \mathbb R^2 \setminus \Delta $$ (i.e. the vertical and horizontal lines going through $(\alpha, \alpha)$) as well as the tangent line to $h(\alpha)$ in $M$, as well as the image of $\alpha$ under the $\mathbb R \to \mathbb R/{\sim}$ map.
First of all, note that this satisfies the properties you described. For $s = (a, b) \in \mathbb R^2 \setminus \Delta$, $s$ lies in $S\_a$ and $S\_b$; for a point $s \in M$, $s$ lies in the two sets $S\_\alpha$ for which the line going through $s$ and $h(\alpha)$ is tangent to the circle, and of course every point in $\mathbb R /{\sim}$ also has two preimages.
It's also clear that any two $S\_\alpha$, $S\_\beta$ intersect at three points: two points in $\mathbb R^2 \setminus \Delta$ and one either in $\mathbb R/{\sim}$ or in $M$ depending on whether $h(\alpha)$, $h(\beta)$ are diametrically opposite points on the circle or not.
Lastly, for this example, the described function $f$ cannot exist. Indeed, suppose $f(S) \cap S\_x$ is finite for every $x$. Then in particular you have finitely many points mapping to each $L\_x$, and yet the union
$$\bigcup\_{x \in \mathbb R} L\_x \cap f(\mathbb R^2 \setminus \Delta)$$ has to be all of $\mathbb R^2\setminus \Delta$.
To paraphrase, you have a finite set of points on every vertical and horizontal line in $\mathbb R^2$ and their union is all of $\mathbb R^2$ — that cannot happen (indeed, let $V\_a$ be the set of $x$ coordinates of the finite set of points on the line $y = a$ and choose $s$ which isn't in $V\_n$ for any $n \in \mathbb Z$; then to fill all of $\mathbb R^2$, the horizontal line $x = s$ would have to contain all the points $(s, n)$ for all $n \in \mathbb Z$).
|
3
|
https://mathoverflow.net/users/177751
|
404281
| 165,815 |
https://mathoverflow.net/questions/404245
|
8
|
In the mod $p$ local Langlands correspondence for $\mathrm{GL}\_{2}(\mathbb{Q}\_{p})$, the irreducible supercuspidal representation $\left(\mathrm{ind}^{\mathrm{GL}\_{2}(\mathbb{Q}\_{p})}\_{\mathrm{GL}\_{2}(\mathbb{Z}\_{p})\mathbb{Q}\_{p}^{\times}}\mathrm{Sym}^{r}\overline{\mathbb{F}}\_{p}^{2}\right)/T$ of $\mathrm{GL}\_{2}(\mathbb{Q}\_{p})$ is mapped to the irreducible Galois representation $\mathrm{ind}(\omega\_{2}^{r+1})$ for $r\in\lbrace 0,\ldots,p-1\rbrace$. See Definition 1.1 in
[https://www.imo.universite-paris-saclay.fr/~breuil/PUBLICATIONS/GL2%28Qp%29II.pdf](https://www.imo.universite-paris-saclay.fr/%7Ebreuil/PUBLICATIONS/GL2%28Qp%29II.pdf)
(with $\eta=1$).
However, I notice that this is not compatible with the local class field theory: on $\mathbb{Z}\_{p}^{\times}$, the central character of the supercuspidal representation is the $r$-th power map after going modulo $p$, while the determinant character of the Galois representation is the $(r+1)$-th power map.
So is it defined this way to satisfy some other compatibilities (for example, reduction modulo $p$ of the $p$-adic local Langlands for $\mathrm{GL}\_{2}(\mathbb{Q}\_{p})$)?
|
https://mathoverflow.net/users/34414
|
A question about mod $p$ local Langlands for $\mathrm{GL}_{2}(\mathbb{Q}_{p})$
|
You seem to be expecting that mod $p$ local Langlands should satisfy the same compatibilities as "conventional" local Langlands (for smooth representations of $GL\_2(\mathbf{Q}\_p)$ and $WD(\mathbf{Q}\_p)$ with coefficients in $\mathbf{C}$).
However, before you can even talk about reduction mod $p$, you need to check that the coefficients can be descended from $\mathbf{C}$ to a number field. I.e., if $\pi$ is a smooth irred rep of $GL\_n(\mathbf{Q}\_p)$ on an $L$-vector space, where $L$ is some subfield of $\mathbf{C}$, then is its Langlands parameter $\phi\_{\pi}$ an $L$-valued Weil-Deligne rep? The answer, annoyingly, is "no": if $n$ is even, you have to add $\sqrt{p}$ to $L$ in order to get this to work. So it's common to re-normalise by twisting the correspondence for $GL\_n$ by $|\det|^{(n-1)/2}$; this makes it compatible with coefficient fields, and also works better for local-global compatibility.
It's this same shift which you are seeing in the mod $p$ theory, and here it's completely impossible to get rid of, even if you extend the coefficient fields as much as you like: the character $|\cdot|^{1/2}$ of $\mathbf{Q}\_p^\times$ is not $p$-adically unitary, so there is no way you can reduce it mod $p$.
|
6
|
https://mathoverflow.net/users/2481
|
404291
| 165,817 |
https://mathoverflow.net/questions/404257
|
2
|
I'm looking for an elegant way to show the following claim.
**Claim:** Let $m\_1, m\_2 \in \mathbb{R}^2$ be the two columns of matrix $M \in \mathbb{R}^{(2 \times 2)}$. The singular values of the matrix are $\sigma\_1 = \sqrt{\|m\_1\|\_2 + \left|\cos{\measuredangle \left( m\_1, m\_2 \right)}\right| \|m\_2\|\_2}$ and $\sigma\_2 = \sqrt{\sin{\measuredangle \left( m\_1, m\_2 \right)} \|m\_2\|\_2}$ so that $\sigma\_1 > \sigma\_2$.
What I have so far is only pretty messy first and second derivations of $max\_{a:\|a\|\_2=1} \|Ma\|\_2$ and $min\_{a:\|a\|\_2=1} \|Ma\|\_2$ w.r.t. the first coordinate of $a$, denoted by $a\_1$, after replacing the second coordinate of $a$, which we denote by $a\_2$, with $\sqrt{1-a\_1^2}$.
Another try that I had is to start by claiming that $\exists P$ s.t. $PM = \begin{bmatrix} \|m\_1\|\_2 & \|m\_2\|\_2 \cos a \\ 0 & \|m\_2\|\_2 \sin a\end{bmatrix}$, where $a := \angle(m\_1, m\_2) = \frac{m\_1^T m\_2}{\|m\_1\|\_2 \|m\_2\|\_2}$. Then, that $PM$ and $M$ have the same singular values, so we could just compute them for $PM$. Unfortunately, I'm getting an unexpected expression so I guess I've missed something...
Tnx!
|
https://mathoverflow.net/users/150065
|
Expressing the singular values of a 2-by-2 real-valued matrix by the norm of the two columns and the angle between them
|
Since $\sigma\_1^2=\lambda\_+$ and $\sigma\_2^2=\lambda\_-$ are the two eigenvalues of the symmetric matrix product $MM^t$, we have $\lambda\_++\lambda\_-={\rm tr}\,MM^t=\|m\_1\|^2+\|m\_2\|^2$. Hence we may write WLOG
$$\lambda\_\pm=\tfrac{1}{2}\left(\|m\_1\|^2+\|m\_2\|^2\right)\pm\Delta.$$
To determine $\Delta$ we equate $$\lambda\_+\lambda\_-={\rm det}\,MM^t=(m\_1\times m\_2)^2=\|m\_1\|^2\|m\_2\|^2\sin^2\measuredangle \left( m\_1, m\_2 \right),$$
hence
$$\lambda\_\pm=\tfrac{1}{2}\left(\|m\_1\|^2+\|m\_2\|^2\right)\pm\sqrt{\tfrac{1}{4}\left(\|m\_1\|^2+\|m\_2\|^2\right)^2-\|m\_1\|^2\|m\_2\|^2\sin^2\measuredangle \left( m\_1, m\_2 \right)}.$$
If the two vectors $m\_1$ and $m\_2$ have the same norm $\|m\|$, this simplifies to
$$\lambda\_\pm=\|m\|^2\bigl(1\pm\cos \measuredangle \left( m\_1, m\_2 \right)\bigr),\;\;\text{if}\;\;\|m\_1\|=\|m\_2\|\equiv\|m\|.$$
This differs from the result in the OP. Let me check, as an example,
$$M=\begin{pmatrix}
1&1\\
0&1
\end{pmatrix},\;\;\sigma\_1^2=\tfrac{3}{2}+\tfrac{1}{2}\sqrt 5,\;\;\sigma\_2^2=\tfrac{3}{2}-\tfrac{1}{2}\sqrt 5.$$
Since the angle between the vectors $m\_1={1\choose 0}$ and $m\_2={1\choose 1}$ is $\pi/4$, the formula in the OP would give $\sigma\_1^2=1+\sqrt 2$ and $\sigma\_2^2=\sqrt 2$, which is incorrect.
|
2
|
https://mathoverflow.net/users/11260
|
404301
| 165,822 |
https://mathoverflow.net/questions/404266
|
3
|
The group $\mathbb{Z}/2$ corepresents the functor $\mathrm{Inv}\colon\mathsf{Mon}\to\mathsf{Sets}$ sending a monoid $A$ to its set of involutory elements (those satisfying $a^2=1\_A$).
A similar story is true for $\mathbb{Z}$ and invertible elements, but let's instead tell it in the $\infty$-setting: namely, the $\infty$-category of $\mathbb{E}\_1$-monoidal functors $\mathbb{Z}\_\mathsf{disc}\to\mathcal{C}$ is just $\mathsf{Pic}(\mathcal{C})$, and thus $\mathbb{Z}\_\mathsf{disc}$ corepresents the functor
$$\mathsf{Pic}\colon\mathsf{Mon}\_{\mathbb{E}\_1}(\mathsf{Cats}\_{\infty})\to\mathcal{S}$$
However, replacing
* $\mathsf{Mon}\_{\mathbb{E}\_1}(\mathsf{Cats}\_\infty)$ by $\mathsf{Mon}\_{\mathbb{E}\_\infty}(\mathsf{Cats}\_\infty)$, the $\infty$-category of symmetric monoidal $\infty$-categories;
* $\mathcal{S}$ by $\mathsf{Grp}\_{\mathbb{E}\_\infty}(\mathcal{S})$;
changes the corepresenting object from $\mathbb{Z}\_{\mathsf{disc}}$ to the sphere spectrum $\mathbb{S}$. Similarly, if we pass to $\mathbb{E}\_k$ rather than $\mathbb{E}\_{\infty}$, we get $\Omega^kS^k$
instead of $\mathbb{S}$.
---
Now, define an **involutory object** of a monoidal $\infty$-category $\mathcal{C}$ to be a strong monoidal functor $(\mathbb{Z}/2)\_{\mathsf{disc}}\to\mathcal{C}$. By definition, $(\mathbb{Z}/2)\_{\mathsf{disc}}$ corepresents the functor
$$\mathsf{Inv}\colon\mathsf{Mon}\_{\mathbb{E}\_1}(\mathsf{Cats}\_{\infty})\to\mathcal{S}$$
sending $\mathcal{C}$ to $\mathsf{Inv}(\mathcal{C})\overset{\mathrm{def}}{=}\mathsf{Fun}^\otimes((\mathbb{Z}/2)\_{\mathsf{disc}},\mathcal{C})$.
**Question.** Is the functor
$$\mathsf{Inv}\colon\mathsf{Mon}\_{\mathbb{E}\_{k}}(\mathsf{Cats}\_\infty)\to\mathsf{Grp}\_{\mathbb{E}\_{k-1}}(\mathcal{S})$$
corepresentable by an $\mathbb{E}\_{k}$-monoidal category for $2\leq k\leq\infty$?
|
https://mathoverflow.net/users/130058
|
Corepresentability of involutory objects in monoidal $\infty$-categories
|
$$Fun^{\otimes}(\mathbb Z/2, C) \simeq map\_{E\_1}(\mathbb Z/2, C^\simeq) \simeq map\_{E\_k}(\mathrm{Ind}\_{E\_1}^{E\_k}\mathbb Z/2, C^\simeq)$$
where $\mathrm{Ind}\_{E\_1}^{E\_k}$ denotes the left adjoint to the forgetful functor.
So $Inv$ is representable, and the natural $E\_{k-1}$-structure (see my comments for why I wrote $E\_{k-1}$ and not $E\_k$ - it is possible that in this special case too we could get $E\_k$, but I don't see a reason why, and what I wrote works for any $E\_1$-space $X$) on this space gives $\mathrm{Ind}\_{E\_1}^{E\_k}\mathbb Z/2$ a natural co-$E\_{k-1}$-structure (in $E\_k$-spaces - with the coproduct monoidal structure).
Now does the space $\mathrm{Ind}\_{E\_1}^{E\_k}\mathbb Z/2$ have a reasonably concrete description ? I think it's something like a bar construction $Bar(E\_1,E\_k, \mathbb Z/2)$ so you can get an explicit description involving the space of little $k$-disks, but I'm not entirely sure you can get much better. I'd love to hear about a better description.
|
4
|
https://mathoverflow.net/users/102343
|
404304
| 165,824 |
https://mathoverflow.net/questions/404310
|
2
|
Let $\mathbb{F}\_{q^n}/\mathbb{F}\_q$ be an extension of finite fields.
Is a proper quotient of $\mathbb{F}\_{q^n}[x]$ considered as an $\mathbb{F}\_q$-algebra always a quotient of $\mathbb{F}\_q[x]$ (i.e. no extra generator is necessary)?
|
https://mathoverflow.net/users/378181
|
Is a proper quotient of $\mathbb{F}_{q^n}[x]$ considered as an $\mathbb{F}_q$-algebra always a quotient of $\mathbb{F}_q[x]$?
|
The answer is no. A counterexample: the quotient $\mathbb{F}\_4[x]/(x(x-1))$ is isomorphic to $\mathbb{F}\_4\times\mathbb{F}\_4$. If $\mathbb{F}\_2[x] / (f(x))$ were isomorphic to $\mathbb{F}\_4\times\mathbb{F}\_4$, $f(x)$ would need to be a product of two distinct irreducibles, each of degree two. But there is only one degree $2$ irreducible in $\mathbb{F}\_2[x]$.
|
5
|
https://mathoverflow.net/users/5263
|
404317
| 165,830 |
https://mathoverflow.net/questions/404055
|
1
|
Let $X$ is a strip between two different parallel lines $a$ and $b$ on a plane ($a,b\subset X$) and $h(x)=\min\limits\_{l\in \{a,b\}}\{d(x,l)\}$.
Let $(X,\*)$ be a topological group with the following property:
$$h(xy)\leq \max\{h(x),h(y) \}.$$
It is a locally compact, connected, simply connected, Hausdorff group. I think it might be a Lie group, but which one I don't know. This situation can be transferred to higher dimensions and there I am also interested in the next question.
Does such a group exist?
|
https://mathoverflow.net/users/175589
|
Group structure on the strip
|
Such group does not exist. To derive a contradiction, assume that the strip $X=\mathbb R\times(-1,1)$ admits a continuous group operation $X\times X\to X$, $(x,y)\mapsto xy$, such that $h(xy)\le\max\{h(x),h(y)\}$ for all $x,y\in X$.
Let $c$ be any point on the central line $L=\mathbb R\times\{0\}$ and $f:X\to X$ be the homeomorphism defined by $f(x)=x^{-1}c$. Observe that $xf(x)=xx^{-1}c=c\in L$ for every $x\in X$.
Since the line $L$ is nowhere dense in $X$, the set $L\cup f^{-1}[L]$ is nowhere dense in $X$ and hence we can choose an element $x\in X\setminus (L\cup f^{-1}[L])$. Then for the points $x\notin L$ and $y=f(x)=x^{-1}c\notin L$ we have
$$1=h(c)=h(xy)>\max\{h(x),h(x^{-1}c)\},$$
which contradicts the choice of $h$.
|
2
|
https://mathoverflow.net/users/61536
|
404328
| 165,832 |
https://mathoverflow.net/questions/403960
|
4
|
If I have a metric $d(\cdot,\cdot)$ on the set $\{1,\dots,n\}$, are there well-known necessary or sufficient conditions for the existence of a matrix norm $Q$ that induces that metric on the unit vectors $e\_1,\dots,e\_n$? That is, under what conditions can I find $Q\succeq0$ such that $$(e\_i-e\_j)^TQ(e\_i-e\_j) = d(i,j)^2$$ for all $i$ and $j$?
|
https://mathoverflow.net/users/70190
|
When does a finite metric induce a matrix norm?
|
Your quadratic form $Q$ is uniquely defined by $d$ on the hyperplane $H$ defined by $\sum x\_i=0$. Further, $Q|\_H\ge 0$ if and only if your metric space is isometric to a subset of a Eulcidean space.
|
2
|
https://mathoverflow.net/users/1441
|
404331
| 165,833 |
https://mathoverflow.net/questions/404325
|
2
|
In the [Young's lattice](https://en.wikipedia.org/wiki/Young%27s_lattice), the number of branches that connect the $N$'th layer to the $N+1$'th layer has the sequence:
$$
1,2, 4, 7, 12, 19, 30, 45, 67, 97, 139, \cdots
$$
Looking this up on OEIS, leads to [this result](https://oeis.org/A000070). This is nothing but the cumulative sum of integer partitions
$$ \sum\_{k=0}^N p(k)
$$
My question is: why do we expect this number to count the number of branches between two layers of the Young's lattice?
|
https://mathoverflow.net/users/140290
|
Number of branches between two layers of the Young's lattice
|
Contra what I originally thought, I’m not sure this is a general fact for all differential posets.
Nonetheless in the case of Young’s lattice it is easy to see this directly from the fact that if an element has $x$ edges coming in from below, it has $x+1$ going out above: we can always add a box in one more position than we can take away a box, in any Young diagram.
|
3
|
https://mathoverflow.net/users/25028
|
404332
| 165,834 |
https://mathoverflow.net/questions/404344
|
6
|
What would be the best book, article, or otherwise to reference for the specific construction of the classifying space for a discrete group $G$ which goes as follows?:
* Regard $G$ as a category with one object whose morphisms are the elements of $G$.
* Construct the simplicial sets $NG$ (i.e., the nerve of $G$) and $\mathcal{E}G$ (unsure if there's a standard notation; I mean that $\mathcal{E}G\_n$ should be $G^{n+1}$ with face maps given by deletion and degeneracy maps given by repetition).
* Take the geometric realizations $BG$ of $NG$ and $EG$ of $\mathcal{E}G$; then $BG$ is the classifying space with universal cover $EG$.
As far as I know, this is a fairly standard construction (although perhaps not *the* standard one). I'm just wondering about the best place to use as a reference for it.
|
https://mathoverflow.net/users/137445
|
Simplicial set construction of the classifying space
|
I believe that's called the Milgram bar construction:
* R.J. Milgram, *The bar construction and abelian $H$-spaces*, Illinois J. Math. 11 (1967), 242-250.
|
10
|
https://mathoverflow.net/users/2926
|
404346
| 165,841 |
https://mathoverflow.net/questions/403296
|
1
|
I encountered a sentence which says it is well known that problem
$$
\begin{cases}
-\Delta u =|u|^{p-1} u & in \,\, \Omega \\
u=0 & on \,\, \partial \Omega
\end{cases}
$$
has a solution for $1<p<\frac{N+2}{N-2}$ and doesn't have any solution for $p>\frac{N+2}{N-2}$.
The existence is okay by mountain pass theorem. But how about the non-existence case? Can some one give a reference or hint?
|
https://mathoverflow.net/users/76453
|
Non-existence result for $p>\frac{N+2}{N-2}$
|
It is not true that this equation always has no solutions in the supercritical case $p > \frac{N + 2}{N - 2}$.
The simplest counterexample is on an annulus, say $\Omega = B\_R \setminus B\_1$: in this case one may search for radial solutions by separating variables, which reduces to solving the second-order ODE
$$
-u'' - \frac{N - 1}{r} u' = u |u|^{p - 1}
$$
with boundary condition $0$ at $R$ and $1$. Here the mountain pass or constrained optimization approaches (applied in 1D) will give the existence of a nonnegative solution, similarly to the ball, so long as $p > 2$. The key point is that $r > 1$, so the coefficients are nonsingular.
The most well-known nonexistence result for this equation is that there are no nontrivial solutions on star-shaped domains. This based on the (Rellich-)Pohozaev identity. Here is a sketch assuming also that the domain is smooth (and, WLOG, star-shaped around the origin):
Set $T = x |\nabla u|^2 - 2 (\nabla u \cdot x) \nabla u - 2 x \frac{|u|^{p+1}}{p+1} $. This is a vector field, and a computation shows that
$$
\text{div} T = (N-2)|\nabla u|^2 - 2 N \frac{|u|^{p + 1}}{p+1}.
$$
At this point one integrates over $\Omega$ and applies the divergence theorem:
$$
\int\_\Omega (N-2)|\nabla u|^2 - 2 N \frac{|u|^{p + 1}}{p+1} = \int\_{\partial \Omega} T \cdot \nu,
$$
where $\nu$ is the outward unit normal vector to $\partial \Omega$. As $u = 0$ on $\partial \Omega$, we have $\nabla u = \pm |\nabla u| \nu$, so after some simplification
$$
\int\_{\partial \Omega} T \cdot \nu = - \int\_{\partial \Omega} |\nabla u|^2 x \cdot \nu.
$$
This is a nonpositive quantity as the domain is star-shaped (indeed, with more effort one can check that it is strictly negative for nontrivial solutions).
There is a second, simpler identity available: multiply the PDE by $u$ and integrate by parts:
$$
\int\_{\Omega} |\nabla u|^2 = \int\_{\Omega} |u|^{p+1}.
$$
Substituting this in,
$$
\int\_{\Omega} (N - 2 - \frac{2N}{p + 1})|\nabla u|^2 \leq 0.
$$
When $p > \frac{N + 2}{N - 2}$ the coefficient is positive and this implies that $u$ is constant $0$, a contradiction.
This general approach accounts for a large portion of nonexistence theorems about semilinear equations in the literature.
To answer the question more fully is difficult: generally one expects existence of positive solutions to be related to the topology of the domain. In the critical case ($p = \frac{N + 2}{N - 2}$) this is fairly well understood, see e.g. [Bahri, Coron](https://doi.org/10.1002/cpa.3160410302). In the supercritical case this is not understood nearly as well.
|
6
|
https://mathoverflow.net/users/378654
|
404354
| 165,842 |
https://mathoverflow.net/questions/404348
|
0
|
Suppose that $n$ is a natural number, $X$ is a set, and $S\subseteq X^{2}$ is a subset such that if $x,y\in X$, then there is a unique tuple $(x\_{0},\dots,x\_{n})$ where $x\_{0}=x,x\_{n}=y$ and $(x\_{i},x\_{i+1})\in S$ for $0\leq i<n$ (this condition is equivalent to saying that if $\chi\_{S}$ is the characteristic function of $S$, then each entry in the matrix $\chi\_{S}^{n}$ is $1$).
Then does there necessarily exist some $x\in X$ where $(x,x)\in S$? Is $|\{x\in X\mid(x,x)\in S\}|^{n}=|X|$?
We observe that by the Lowenheim-Skolem theorem, if there is an infinite set $S\subseteq X^{2}$ where every entry in $\chi\_{S}^{n}$ is $1$ but where there is no $x\in X$ with $(x,x)\in X$, then there is an example of every infinite cardinality.
Proposition: If $X$ is finite, then $|\{x\in X\mid(x,x)\in S\}|^{n}=|X|$.
Proof: Suppose that $X=\{1,\dots,r\}$. Suppose that the matrix $\chi\_{S}$ has characteristic polynomial $(x-\lambda\_{1})\dots(x-\lambda\_{r})$. Then each entry in $\chi\_{S}^{n}$ is $1$, so $\chi\_{S}^{n}$ has characteristic polynomial $$(x-\lambda\_{1}^{n})\dots(x-\lambda\_{r}^{n})=x^{n-1}(x-r).$$ Therefore, we conclude that there is some $i$ such that $|\lambda\_{i}|=\sqrt[n]{r}$ but where $\lambda\_{j}=0$ whenever $j\neq i$. However, $$\{x\in X\mid(x,x)\in S\}=\text{Tr}(\chi\_{A})=\lambda\_{1}+\dots+\lambda\_{r}=\lambda\_{i},$$ so $$\{x\in X\mid(x,x)\in S\}=\sqrt[n]{r}.$$
Example: Suppose that $X=K^{n}$. Let $S$ be the collection of all pairs
$$((x\_{1},\dots,x\_{n}),(a,g\_{x\_{2},\dots,x\_{n-1}}(x\_{1}),\dots,g\_{x\_{n-1}}(x\_{n-2}),g(x\_{n-1})))$$. Then every entry in the (possibly infinite) matrix $\chi\_{S}^{n}$ is $1$. However, I claim that $\{x\in X\mid(x,x)\in S\}=|K|$. More specifically, I claim that the mapping
$L:\{x\in X\mid(x,x)\in S\}\rightarrow K$ defined by $L(x\_{1},\dots,x\_{n})=x\_{n}$ is a bijection by constructing its inverse $M$. If $x\in X$, then we define a sequence $M(x)=(x\_{1},\dots,x\_{n})$ by reverse recursion; we set $x\_{n}=x$ and $x\_{i}=g\_{x\_{i+1},\dots,x\_{n-1}}^{-1}(x\_{i+1})$ for $i\in\{1,\dots,n-1\}$. It is not too hard to see that $M:K\rightarrow\{x\in X\mid(x,x)\in S\}$ and that $L$ and $M$ are inverses of each other. Therefore $|\{x\in X\mid(x,x)\in S\}|=|K|$ in this case.
|
https://mathoverflow.net/users/22277
|
Can this fixed point theorem generalize to infinite structures?
|
Here's one way to build an infinite counterexample with $n=2$ for simplicity:
We start with (say) $X\_0=\{0\}, S\_0=\emptyset$. Having defined $X\_m, S\_m$, we define $X\_{m+1}, S\_{m+1}$ as follows:
* To get $X\_{m+1}$, we add to $X\_m$ a fresh element $c\_{(x,y)}$ for each pair $(x,y)\in X\_m^2$ such that there is no $z\in X\_m$ with $(x,z),(z,y)\in S\_m$.
* To get $S\_{m+1}$, we add to $S\_m$ all pairs of the form $(x,c\_{(x,y)})$ and $(c\_{(x,y)},y)$ for $(x,y)\in X\_m^2$ such that no $z\in X\_m$ has $(x,z),(z,y)\in S\_m$.
Now let $X=\bigcup\_{m\in\mathbb{N}}X\_m$, $S=\bigcup\_{m\in\mathbb{N}}S\_m$.
This same idea works for any $n\ge 2$ and for any (infinite) cardinality.
|
2
|
https://mathoverflow.net/users/8133
|
404355
| 165,843 |
https://mathoverflow.net/questions/404340
|
5
|
The question is in the title: can a Landau-Siegel zero be the only zero off the critical line for a Dirichlet L-function or does its existence imply the existence of a complex non trivial zero in the critical strip off the critical line?
This question came to my mind considering the sequence of trivial zeros in decreasing order for zeta as a $2$-periodic signal whose Fourier transform would be a $1/2$-periodic signal made of Dirac peaks (the former future physicist in me is speaking, sorry), which if we compactify partially the critical strip by identifying the vertical lines of real parts $0$ and $1$ (this compactification process should be compatible with the functional equation as it relates the value at $s$ to the one at $1-s$) becomes a single Dirac peak which is supported on $1/2$, hence the real part of the non trivial zeros under RH. So my idea is that adding a Landau-Siegel zero would create a non periodic signal made by the decreasing sequence of real zeros, whose Fourier transform would not be periodic either, suggesting the existence of a complex non trivial zero in the critical strip off the critical line. In some sense, the Fourier transform is expected to *permute* zeros of L-functions, which seems consistent with the Deuring-Heilbronn phenomenon.
So would the existence of a Landau-Siegel zero create such a havoc that the analogue of RH for the considered Dirichlet L-function would fail completely?
|
https://mathoverflow.net/users/13625
|
Does the existence of a Landau-Siegel zero imply the existence of a complex zero off the critical line?
|
A result of Sarnak and Zaharescu, stated
in the contrapositive, implies the existence of a complex zero off the critical line for at least one Dirichlet L-function if one has a sufficiently strong Siegel zero: [ProjectEuclid link](https://www.projecteuclid.org/journals/duke-mathematical-journal/volume-111/issue-3/Some-remarks-on-Landau-Siegel-zeros/10.1215/S0012-7094-02-11134-X.short)
|
19
|
https://mathoverflow.net/users/766
|
404356
| 165,844 |
https://mathoverflow.net/questions/404350
|
1
|
$\DeclareMathOperator\PL{PL}$Consider the group $\PL\_{n,n-1}$ of orientation preserving PL self-homeomorphisms of $\mathbb R^n$ that also preserve $\mathbb R^{n-1}$ pointwise. It is usually understood as a simplicial group whose $k$-simplices are PL self-homeomorphisms $\mathbb R^n\times\Delta^k\to \mathbb R^n\times \Delta^k$ commuting with the projection on $\Delta^k$. Is this simplicial group contractible?
|
https://mathoverflow.net/users/9800
|
Is $\operatorname{PL}_{n,n-1}$ contractible?
|
The answer is yes, see the discussion above with [Connor Malin](https://mathoverflow.net/users/134512/connor-malin).
|
1
|
https://mathoverflow.net/users/9800
|
404359
| 165,845 |
https://mathoverflow.net/questions/404369
|
2
|
Is there an uncountable integral domain such its every countable subset is contained in a finitely generated $\mathbb{Z}$-algebra?
|
https://mathoverflow.net/users/378181
|
Uncountable integral domain such that every countable subset is contained in a finitely generated $\mathbb{Z}$-algebra
|
1- No. Indeed, let $A$ be an uncountable domain and $X$ a maximal algebraically independent subset (over the minimal subring $A\_0=\mathbf{Z}$ or $\mathbf{F}\_p$). Then $A$ is contained in an algebraic closure of the field of rational functions $\mathrm{Frac}(A\_0)(x:x\in X)$. In particular, it follows that $X$ is uncountable. Then, for every infinite countable subset $Y$ of $X$ there is no finitely generated subalgebra containing it.
---
2- Note: I'm more familiar with the analogous property for groups (every countable subgroup is contained in a finitely generated subgroup). There are uncountable groups with this property (e.g., symmetric groups — Galvin 1995), but no abelian ones. Similarly, there are many non-commutative (associative) algebras with this property, for instance, group algebras $FG$ of such groups $G$ for $F$ finitely generated field. I guess infinite-dimensional matrix algebras over finite fields are good candidates too.
---
3- Added: the conclusion still holds assuming that $A$ is an arbitrary scalar (= associative commutative unital) ring.
Indeed, let $A$ satisfy the condition. The first consequence is that $A$ is noetherian. Indeed, noetherian is equivalent to the condition that for every sequence $(a\_n)\_{n\ge 0}$, there exists $N$ such that for every $n\ge N$ there exists $b\_0,\dots,b\_{n-1}$ such that $a\_n=\sum\_{i<n}a\_ib\_i$. (If $A$ is not noetherian choose $J$ infinitely generated, $a\_0=0$ and choose by induction $a\_n$ not in the ideal $\sum\_{i<n}a\_iA$.) Hence, if $A$ is not noetherian, let $(a\_n)$ be such a sequence: it is contained in a finitely generated subring $B$, which is noetherian, which in turns yields a contradiction.
Now $A$ is noetherian. By the domain case, for every prime ideal $P$ of $A$, $A/P$ is countable. Since $A$ is noetherian, as $A$-module, $A$ is an iterated extension of various such $A/P$. It follows that $A$ is countable.
---
4- There is a property "uncountable cofinality" which means "cannot be written as union of an increasing sequence of proper subrings" (for a countable ring, it is equivalent to being finitely generated). The property "every countable subalgebra is contained in a finitely generated one" clearly implies uncountable cofinality. The argument in 1 above easily shows the stronger result that no uncountable domain has uncountable cofinality. (Similarly, every uncountable field has countable cofinality in the category of fields.)
However there exist uncountable scalar rings of uncountable cofinality. An example, due to Koppelberg and Tits, is the Boolean algebra $(\mathbf{Z}/2\mathbf{Z})^X$ for any infinite set $X$.
|
5
|
https://mathoverflow.net/users/14094
|
404373
| 165,850 |
https://mathoverflow.net/questions/402927
|
6
|
**The situation:**
Let $X$ be a 2 dimensional normal quasi-projective $\mathcal{O}\_K$-scheme, where $K$ is an algebraic number field. Assume the following conditions on $X$:
1. $X$ is integral.
2. $X\_K$ is geometrically integral.
3. $X \to \textrm{spec}(\mathcal{O}\_K)$ is surjective.
Let $X\to \bar{X}$ an open immersion into a projective scheme. (this exist since $X$ is quasi-projective).
In particular, 1-dimensional irreducible closed subschemes of $\bar{X}$ are either
1. Horizontal: spectra of finite flat extensions of $\mathcal{O}\_K$
2. Vertical: curves over $\mathcal{O}\_K/p\mathcal{O}\_K$, which is a finite field, for nonzero ideals $p\in \textrm{spec}(\mathcal{O}\_K)$.
**The claim I want to prove:**
Let $V$ be the vertical part of $\bar{X}\backslash X$. Theorem 2 of (Points entiers des variétés arithmétiques, Moret-Bailly) states (withouth proof) that there exists a "contraction" of $V$. i.e. a map $\bar{X}\to Y$ which is surjective, $\bar{X}\backslash V\to Y$ is an open immersion and the image in $V$ is a set of isolated points. I am looking for a proof of such a claim. If that helps, you may assume that $K=\mathbb{Q}$. You may assume that $\bar{X}$ is regular.
**What I have already found:**
The paper states states that this follows similarly from a [paper](http://%3Chttps://www.jstor.org/stable/2372985%3E) of Artin, but I couldn't understand how. I also found a [paper](http://www.numdam.org/item/10.24033/asens.1581.pdf) of Moret-Bailly which uses the existance of integral points on $X$ to prove what I am surching for. But I am looking for a proof which does not rely on this fact, since I am trying to write the proof of theorem 1 of (Points entiers des variétés arithmétiques, Moret-Bailly) using theorem 2 of the same paper. Conditions of the existance of such a contraction can also be found in theorem 27.1 of [this work](http://www.numdam.org/item/PMIHES_1969__36__195_0.pdf) , so you may just help me to find why these conditions apply to my case.
|
https://mathoverflow.net/users/352015
|
Contraction of some surfaces over a ring of algebraic integers
|
The paper [2] is a seminar talk announcing the results of [3]. In this talk, I explain how to deduce the existence of integral points (theorem 1) from the contraction theorem (theorem 2). In turn, the contraction theorem is due to Artin [1] in the geometric case (surfaces over finite fields), but Raynaud explained me the arithmetic analogue (see remark 4.5/2 in [2]). Thus, the arithmetic case is *not* in Artin's paper.
Then Szpiro and I found it more convenient to prove the existence of integral points directly (and then deduce the contraction theorem): this is what I did in [3], redirecting Raynaud's arguments for the contraction theorem, especially the crucial proposition 3.8 of [3], for which I must confess that Raynaud does not get proper credit.
[1] M. Artin, Some numerical criteria for contractability of curves on algebraic surfaces, Amer. J. Math. 84 (1962), 485-496. doi:10.2307/2372985
[2] L. Moret-Bailly, Points entiers des variétés arithmétiques, Séminaire de Théorie des Nombres, Paris 1985–86, p. 147-154, Progress in Mathematics vol. 71 (Birkhäuser)
[3] L. Moret-Bailly, Groupes de Picard et problèmes de Skolem I, Ann. scient. Ec. Norm. Sup. 22 (1989), 161–179. doi: 10.24033/asens.1581
|
6
|
https://mathoverflow.net/users/7666
|
404376
| 165,851 |
https://mathoverflow.net/questions/404296
|
4
|
Let $X$ be the solution to the one dimensional SDE
$dX\_t = \mu(t, X\_t)dt + \sigma(t, X\_t) dW\_t$, for $t \in [0, T]$.
with $X\_0= x\_0$ a.s. for some $x\_0 \in \mathbb R$.
Here $W\_t$ denotes a standard Brownian motion, and we assume $\mu$ and $\sigma$ are Lipschitz continuous and uniformly bounded.
For every $\varepsilon > 0$, denote by $\mathcal S\_{\varepsilon}$ the event $\sup\_{t \in [0, T]} |W\_t| \leq \varepsilon$.
**Question:** Considering $X$ as a $C[0, T]$-valued random variable, is it true that the conditioned random variables $X| \mathcal S\_\varepsilon$ converge in law to the deterministic solution $Y\_t$ of
$dY\_t = \mu(t, Y\_t) dt$, with $Y\_0 = x\_0$ a.s.?
|
https://mathoverflow.net/users/173490
|
Conditioning an SDE on the event that the driving noise is small
|
The answer is yes, provided that you write your equation in Stratonovich form, rather than Itô form (and assuming that $\mu$ and $\sigma$ are sufficiently smooth in their arguments). The reason is that in one dimension the solution to the Stratonovich equation is a continuous map of $W$ in the sup-norm topology, [as observed by Doss in 1977](http://www.numdam.org/item/AIHPB_1977__13_2_99_0/).
This breaks in higher dimensions, but the answer to your question remains the same although I don't know if anyone wrote it up in precisely this way. (Various proofs of the Stroock-Varadhan support theorem use closely related variants of this statement. Note that it is again the Stratonovich formulation which is relevant.)
|
3
|
https://mathoverflow.net/users/38566
|
404379
| 165,852 |
https://mathoverflow.net/questions/404357
|
2
|
Let $X$ be a Banach space. If $X^{\*\*}$ is linearly isometric to $L\_{1}(\mu)$ for some $\sigma$-finite measue $\mu$, we shall say that $X$ is an $L\_{1}$-pre-bidual.
Question 1. What are the examples of $L\_{1}$-pre-bidual ?
Question 2. Are there any characterizations or even references about $L\_{1}$-pre-biduals ?
Thank you !
|
https://mathoverflow.net/users/41619
|
Banach spaces whose biduals are $L_{1}$
|
If $X$ is an infinite dimensional Banach space such that $X^{\*\*}$ is isomorphic to $L^1(\mu)$ for some $\sigma$-finite measure, then $X^{\*\*}$ is non-reflexive, separable and has DPP (Dunford-Pettis property) since reflexivity, separability and DPP are isomorphic properties. This is not possible ([Banach spaces whose second conjugates are separable](https://mathoverflow.net/questions/404226/banach-spaces-whose-second-conjugates-are-separable/404306#404306))
**Edit:** $L^1(\mu)$ is separable when $\mu$ is a $\sigma$-finite measure and $L^1(\mu)$ is a dual Banach space.
a. If $\mu$ is a $\sigma$-finite measure, then $L^1(\mu)$ is isometrically isomorphic to $L^1(\nu)$ for some probability measure, e.g., see *Albiac & Kalton, "Topics in Banach Space Theory", Chapter 5.*
b. If $\mu$ is a probability measure, then $L^1(\mu)$ is a dual space if and only if $\mu$ is purely atomic.
*Proof.* $(\Leftarrow)$ If $\mu$ is purely atomic, then $L^1(\mu)$ is isomorphic to $\ell^1(supp\mu)$, which is a dual space.
$(\Rightarrow)$ $L^1(\mu)$ is weakly compactly generated. Every weakly compactly generated dual Banach space has RNP, and $L^1(\mu)$ has RNP iff $\mu$ is purely atomic; see *Diestel & Uhl, "Vector Measures", Section 7.7.7.*
c. If $\mu$ is a finite purely atomic measure, then $supp\mu$ is countable. Thus, $\ell^1(supp\mu)$ is separable.
|
3
|
https://mathoverflow.net/users/164350
|
404386
| 165,853 |
https://mathoverflow.net/questions/404372
|
2
|
We consider a sequence $u = (u\_k)\_{k\geq 1}$ such that $u\_k \geq 0$ for any $k \geq 1$. We assume that there exists a critical $p\_c \in \mathbb{R}$ such that, for any $q<p\_c <p$,
$$\sum\_{k=1}^\infty k^q u\_k < \infty \quad \text{and} \quad \sum\_{k=1}^\infty k^p u\_k = \infty.$$
I am interested in the asymptotic behavior of $\sum\_{k=1}^n k^p u\_k$ when $n\rightarrow \infty$ for $p > p\_c$.
More precisely, I expect this quantity to behaves roughly as $n^{p-p\_c}$ in the following sense.
**Conjecture:** For any $p > p\_c$ and $0 < \epsilon < p - p\_c$, there exists $0 < m , M < \infty$ such that, for any $n \geq 1$,
$$m n^{p - p\_c - \epsilon} \leq \sum\_{k=1}^n k^p u\_k \leq M n^{p - p\_c + \epsilon}.$$
The upper bound of the conjecture is easy to obtain. Indeed, we readily have that
$$\sum\_{k=1}^n k^p u\_k \leq \left( \sum\_{k=1}^n k^{p\_c - \epsilon} u\_k \right) n^{p-p\_c + \epsilon} \leq \left( \sum\_{k=1}^\infty k^{p\_c - \epsilon} u\_k \right) n^{p-p\_c + \epsilon},$$
hence the constant $M = \sum\_{k=1}^\infty k^{p\_c - \epsilon} u\_k$ works.
My question is: Is the conjecture true in the sense that the constants $m$ always exist to lower-bound $\sum\_{k=1}^n k^p u\_k$.
|
https://mathoverflow.net/users/39261
|
Asymptotic behavior of the moments of non-negative sequences
|
$\newcommand\ep\epsilon$The answer is no.
Indeed, let e.g.
$u\_k:=1$ if $k=k\_j:=2^{5^j}$ for some natural $j$, and $u\_k:=0$ otherwise. Then $p\_c=0$.
Take now any real $\ep>0$ and then take $p=2\ep$, so that condition $0< \epsilon <p-p\_c$ holds. Then for all large enough $j$ and $n=k\_{j+1}-1$ we have
$$n^{p-p\_c-\ep}=(k\_{j+1}-1)^\ep\ge2^{\ep 5^{j+1}/2}$$
and
$$\sum\_{k=1}^n k^pu\_k=\sum\_{i=1}^j k\_i^{2\ep}
\le2k\_j^{2\ep}=2\times2^{2\ep 5^j}
=o(2^{\ep 5^{j+1}/2})=o(n^{p-p\_c-\ep})$$
as $j\to\infty$.
So, there is no real $m>0$ such that $mn^{p-p\_c-\ep}\le\sum\_{k=1}^n k^pu\_k$ for all $n$.
|
2
|
https://mathoverflow.net/users/36721
|
404388
| 165,854 |
https://mathoverflow.net/questions/404391
|
3
|
Let $A$ be a commutative ring with $f,g\in A[x]$ monics. Consider the $A$-linear endomorphism $\mu\_g^{(f)}\in \mathrm{End}\_A\tfrac{A[x]}{\langle f\rangle}$ given by multiplication by $g$.
For monics $f\_1,f\_2\in A[x]$, how to directly prove that $\det \mu\_g^{(f\_1f\_2)}=\det\mu\_g^{(f\_1)}\det\mu\_g^{(f\_2)}$?
Writing down the matrix representation of $\mu\_g^{(f)}$ w.r.t the monomial basis $1,x,\dots ,x^{\deg f-1}$ is messy and I am unable to see anything through it. On the other hand, $\mu\_g^{(f)}=g(\mu\_x^{(f)})$, and the matrix representation of $\mu\_x^{(f)}$ w.r.t the monomial basis is the companion matrix of $f$, but again I see nothing smart to say about polynomial functions of a companion matrix.
|
https://mathoverflow.net/users/69037
|
Multiplicative identity of determinant of multiplicative action of a polynomial on a quotient ring (companion matrices)
|
You have an exact sequence, $0\to A[x]/f\_1\stackrel{f\_2}{\to} A[x]/f\_1f\_2\to A[x]/f\_2\to 0$. This splits as $A$-modules and then multiplication by $g$ in the middle is just the diagonal matrix of multiplication by $g$ in the two factors. So, determinant multiplies.
|
4
|
https://mathoverflow.net/users/9502
|
404394
| 165,855 |
https://mathoverflow.net/questions/404401
|
1
|
Let $X\sim\mathcal{N}(\boldsymbol{\mu}\_1,\mathrm{\Sigma}\_1)$ and $Y\sim\mathcal{N}(\boldsymbol{\mu}\_2,\mathrm{\Sigma}\_2)$. Then it is know that $\mathbb{P}(X>\boldsymbol{t})\leq\mathbb{P}(Y>\boldsymbol{t})$ implies $\mu\_i\leq \mu^{\prime}\_i$ and $\sigma\_{ii} = \sigma^{\prime}\_{ii}$ (Theorem 10 of Muller 2001, Ann. Inst. Stat. Math. 53(3) 567-575). Is there any similar results for absolute $|X|$ and $|Y|$?
|
https://mathoverflow.net/users/120111
|
Stochastic ordering of absolute multivariate normal random variables
|
$\newcommand\si\sigma$The condition $P(|X|>\boldsymbol{t})\le P(|Y|>\boldsymbol{t})$ implies $P(|X\_i|>t)\le P(|Y\_i|>t)$, for each $i$ and all real $t$; this follows by letting $t\_j=0$ for all $j\notin\{i\}$.
Fix any $i$ and, for brevity, let $m:=EY\_i$, $s:=\sqrt{Var\,Y\_i}$, $m\_1:=EX\_i$, and $s\_1:=\sqrt{Var\,X\_i}$, so that $X\_i\sim N(m\_1,s\_1^2)$ and $Y\_i\sim N(m,s^2)$.
Then
$$s\_1^2+m\_1^2=EX\_i^2=\int\_0^\infty 2t\,dt\,P(|X\_i|>t) \\
\le\int\_0^\infty 2t\,dt\,P(|Y\_i|>t)=s^2+m^2,\tag{0}$$
so that
$$s\_1^2+m\_1^2
\le s^2+m^2, \tag{1}$$
and, for $t\to\infty$,
$$P(|Y\_i|>t)\ge P(|X\_i|>t) \\
=\exp\Big\{-\frac{t^2}{(2+o(1))s\_1^2}\Big\},$$
whence
$$s\_1\le s.\tag{2}$$
Moreover, letting $t\downarrow0$ in $P(|X\_i|>t)\le P(|Y\_i|>t)$, we see that the density at $0$ of the distribution of $X\_i$ is no less that the density at $0$ of the distribution of $Y\_i$, which can rewritten as
$$s\_1^2 e^{m\_1^2/s\_1^2}\le s^2 e^{m^2/s^2}. \tag{3}$$
Thus, the condition $P(|X\_i|>t)\le P(|Y\_i|>t)$ implies (1),
(2), and (3).
One can show that, vice versa, if (2) and (3) hold, then $P(|X\_i|>t)\le P(|Y\_i|>t)$ for all real $t$.
---
**Details on the second equality in (0) in response to a comment:** For any random variable $Z$,
$$EZ^2=E\int\_0^{|Z|}2t\,dt
=E\int\_0^\infty 2t\,dt\,1(|Z|>t) \\
=\int\_0^\infty 2t\,dt\,E1(|Z|>t)
=\int\_0^\infty 2t\,dt\,P(|Z|>t).$$
|
1
|
https://mathoverflow.net/users/36721
|
404404
| 165,857 |
https://mathoverflow.net/questions/404390
|
0
|
In this post I present my variations of the problem involving Nagell-Ljunggren equation, that is explained in pages 10 and 11 of *Highlights in the Research Work of T. N. Shorey* by R. Tijdeman, from *Diophantine Equations*, (Editor) N. Saradha, Tata Institute of Fundamental Research, Narosa Publishing House (2008).
Consider for integers $x>1$, $1<y<z$ and integers $1<p$ and $2<n$ the equations $$y^p+z^p=(y^{p-1}+z^{p-1})\cdot\frac{x^n-1}{x-1},\tag{1}$$
and $$2yz=(y+z)\cdot\frac{x^n-1}{x-1}.\tag{2}$$
**Conjecture 1.** *Consider the equation* $(2)$ *over integers* $x>1$, $1<y<z$ *and* $2<n$, *then there exists a positive integer* $n\_0\geq 8$ *such that* $\forall n>n\_0$ *this equation has no solutions.*
**Conjecture 2.** *We consider the equation* $(1)$ *over integers* $x>1$, $1<y<z$, $1<p$, *and* $2<n$ *then there exists a positive constant* $C$ *such that the equation has no solutions whenever* $n\cdot p>C$.
Both equations ask for Lehmer means $L\_p(y,z)$ with all the digits equal to $1$ in the corresponding $x-$adic expansion
$$L\_p(y,z)=1+x+\ldots+x^{n-1}.$$
Wikipedia has an article for [Lehmer mean,](https://en.wikipedia.org/wiki/Lehmer_mean) and their relationship to the harmonic mean.
>
> **Question.** I would like to know if it is possible (what work can be done about it) to **prove or disprove** Conjecture 1 and/or Conjecture 2. **Many thanks.**
>
>
>
I don't know if preveious problems are in the literature, I hope that these problems have a good mathematical content and are interesting for you. I'm asking about what work can be done about the Conjecture 1 or well about the Conjecture 2. Finally I add two solutions from the following examples.
**Example 1.** One has that $(x,y,z;n,p)=(16,225,300;3,3)$ solves $(1)$ because $$\frac{225^3+300^3}{225^2+300^2}=272=1+1\cdot 16+1\cdot 16^2.$$
**Example 2.** Similarly $(x,y,z;n)=(11,91,247;3)$ solves $(2)$ because $$\frac{2}{\frac{1}{91}+\frac{1}{247}}=133=1+1\cdot 11+1\cdot 11^2.$$
|
https://mathoverflow.net/users/142929
|
Diophantine equations that involve Lehmer means with all digits equal to $1$ in their $x-$adic expansions
|
Conjecture 1 does not hold as for any even $n$, (2) has a solution:
$$(x,y,z)=(2,\frac23(2^n-1),2(2^n-1))$$
Likewise, Conjecture 2 fails as for $p=2$ and any even $n$, (1) has a solution:
$$(x,y,z)=(4,\frac2{15}(4^n-1),\frac25(4^n-1))$$
|
2
|
https://mathoverflow.net/users/7076
|
404410
| 165,862 |
https://mathoverflow.net/questions/404324
|
3
|
Let $A,B,C,D$ be the corners of a tetrahedron with positive volume and distinct sidelengths. Is there a positive $x$ and a planar straight-line embedding of a $K\_4$ graph with distinct vertices $A’,B’,C’,D’$ such that each edge of the tetrahedron is greater than the corresponding edge of the graph by exactly $x$?
|
https://mathoverflow.net/users/31310
|
Deflating a tetrahedron to a $K_4$ graph with equal changes to sidelengths
|
Yes, this is true. The main point is that the "first thing that goes wrong" cannot be two vertices coming together.
Let $a\_0$, $b\_0$, $c\_0$, $d\_0$, $e\_0$, $f\_0$ denote the edge lengths of the original tetrahedron, let $x$ be a variable, and let $a = a\_0 - x$, $b = b\_0 - x$, $c = c\_0 - x$, $d = d\_0 - x$, $e = e\_0 - x$, $f = f\_0 - x$. The set $U$ of possible 6-tuples of edge lengths of (nondegenerate) tetrahedra is cut out by some strict inequalities (described in the comment of Matt F.), and in particular it is an open subset of $\mathbb{R}^6$. So, there exists a smallest $x > 0$ such that $(a, b, c, d, e, f) \notin U$; fix this choice of $x$ and the corresponding quantities $a$, ..., $f$.
I claim that $a > 0$. Among the inequalities defining $U$ are the triangle inequalities for the faces. Since $(a, b, c, d, e, f)$ belongs to the closure of $U$, we must have $a + b - c \ge 0$, as well as $a + c - b \ge 0$. If $a = 0$, then these together imply $b = c$, but then $b\_0 = c\_0$, contradicting the assumption that the original edge lengths were distinct. (Without this assumption there are counterexamples, for example a tetrahedron with $AB = AC = BC < AD = BD = CD$.)
Now, we just have to show that $(a, b, c, d, e, f)$ are the pairwise distances between some 4 points $A$, $B$, $C$, $D$ of $\mathbb{R}^3$. Indeed, $A$, $B$, $C$, $D$ then lie in a plane because otherwise they would form a nondegenerate tetrahedron, contradicting $(a, b, c, d, e, f) \notin U$. On the other hand the points $A$, $B$, $C$, $D$ are distinct because $a > 0$ and likewise $b > 0$, ..., $f > 0$.
Presumably this follows easily from known facts about metric embeddings in $\mathbb{R}^n$, but here is a rather general proof using semialgebraic geometry/o-minimal geometry. The background theory can be found for example in van den Dries, Tame topology and o-minimal structures, chapter 6.
Fix the point $A$ at the origin of $\mathbb{R}^3$, and consider the function $\psi : \mathbb{R}^3 \times \mathbb{R}^3 \times \mathbb{R}^3 \to \mathbb{R}^6$ sending $(B, C, D)$ to the 6-tuple of distances formed by $A$, $B$, $C$, $D$. This function is semialgebraic, continuous, and proper: the preimage of a closed and bounded subset of $\mathbb{R}^6$ is bounded (because we fixed $A$, and using the triangle inequality). The set $U$ is contained in the image of $\psi$, by definition. Using the definable choice principle, there is a semialgebraic curve $\gamma : [0, x) \to \mathbb{R}^3 \times \mathbb{R}^3 \times \mathbb{R}^3$ which sends $t$ to a choice of tetrahedron realizing the distances $(a\_0 - t, \ldots, f\_0 - t)$. The curve $\gamma$ will be continuous after restriction to $[t\_0, x)$ for some $0 \le t\_0 < x$. By properness of $\psi$, since $\psi \circ \gamma$ converges (to $(a, \ldots, f)$) as $t$ approaches $x$, the curve $\gamma$ can be completed continuously to $[t\_0, x]$. Then the value of this extension at $x$ is the required configuration $(A = 0, B, C, D)$.
|
1
|
https://mathoverflow.net/users/126667
|
404412
| 165,864 |
https://mathoverflow.net/questions/404362
|
-1
|
I am considering a *Principle of Ubiquity*, expressed as follows - for a class theory where precisely the elements are sets - with the aid of set abstracts:
For $\alpha(y,z)$ a first order condition so that $\forall y(\exists w(y\in w)\to \exists w (\{z|\alpha(y,z)\}\in w))$:
$\forall v(\exists t (v\in t)\to\exists t(\{w|\forall x(v\in x\wedge \forall y(y\in x\to \{z|\alpha(y,z)\}\in x)\to w\in x)\}\in t))$
The set abstracts can be eliminated with the following Mendelsonian abstraction schema:
$\forall x(x\in \{x|\alpha\}\leftrightarrow\exists y(x\in y)\wedge\alpha)$
It is immediate that we get a theorem of infinity, as well as the least transitive closure of all sets; moreover, several further instances of replacement will hold, though with countable co-finality.
May Z with Ubiquity, instead of just the Axiom of Infinity, justify the Recursion Theorem?
|
https://mathoverflow.net/users/37385
|
Ubiquity beyond infinity, transitive closure and the recursion theorem?
|
The answer seems to be affirmative.
Let an adapted version of ubiquity be as follows:
For $\alpha(y,z)$ a functional first order condition so that
$\forall y(\exists w(y\in w)\to \exists w(\{z|\alpha(y,z)\}\in w))$:
$\forall v(\exists w(v\in w)\to\exists w(\{u|\forall x(v\in x\wedge \forall n \forall y(((n,y)\in x\to (n+1,\{z|\alpha(n,z)\})\in x))\to u\in x)\}\in w))$
Let $v=(0,X)$
Let $F=\{u|\forall x(v\in x\wedge \forall n \forall y(((n,y)\in x\to (n+1,\{z|\alpha(n,z)\})\in x))\to u\in x)\}$
We see that $(0,X)\in F$.
Moreover, for any $n$ and $y$, if $(n,y)\in F$ then $(n+1,\{z|\alpha(n,z)\}\in F)$
We see that function F is the least class with these properties, and from the assumption of Infinitude that F is a set.
As F is a set, Mendelsonian abstraction gives us that
$(0,X)\in F\wedge ((n+1,y)\in F\leftrightarrow\exists w((n,w)\in F\wedge \forall u(u\in w\leftrightarrow \alpha (n,u)))$
Am I done?
Edit: Unsurprisingly, Replacement entails Ubiquity.
|
0
|
https://mathoverflow.net/users/37385
|
404416
| 165,866 |
https://mathoverflow.net/questions/404421
|
8
|
By Dirchlet's hyperbola method, one can prove that the average number of divisors of integers $1 \leq n \leq X$ is $\log X$. This question concerns the number of integers $n \leq X$ such that the number of divisors, $d(n)$, is substantially larger than average. Indeed, what is known about the size of the set
$$\displaystyle \{1 \leq n \leq X : d(n) > (\log X)^A \}$$
where $A > 1$ is considered to be a large (but fixed) positive number?
|
https://mathoverflow.net/users/10898
|
Density of integers with many divisors
|
Theorem 1.11 and Theorem 1.22 of the paper by Norton, cited in the comment of Peter Humphries, show that for any fixed $A \ge \log 2$,
$$
\frac{X (\log\log X)^{O(1)}}{(\log X)^{B(A)}} \ll\_A
|\{1\le n\le X:d(n) \ge (\log X)^A\}| \ll\_A \frac{X}{(\log X)^{B(A)}},
$$
where
$$
B(A):=1+\frac{A}{\log 2}\left(\log\left(\frac{A}{\log 2}\right) -1 \right).
$$
Equation (1.37) of the same paper gives the correct order of magnitude: For every fixed $A>\log 2$,
$$
|\{1\le n\le X:d(n) \ge (\log X)^A\}| \asymp\_A \frac{X}{(\log X)^{B(A)} (\log\log X)^{1/2}}.
$$
|
12
|
https://mathoverflow.net/users/12947
|
404427
| 165,870 |
https://mathoverflow.net/questions/398543
|
0
|
Let $M$ be a closed manifold and assume that is given a family of elliptic operators $L\_t,~t\in [0,1]$ and a smooth function $F :[a,b] \to \mathbb{R}$ such that for each $t$ the elliptic problem $L\_tu = F(u)$ has a classical solution $u : M \to \mathbb{R}$.
I would like to know which kind of methods and techniques could be employed to ensure that we can get a continuous family of solution in the parameter $t$ provided if $L\_t$ is also continuous in this parameter.
If needed it can be assumed that the solutions $u$ are firstly obtained by variational methods (as some minimizers for a certain functional) and regularity is proved classicaly by standard procedures.
Can anyone give me some reference to this kind of analysis or, if possible, make some remarks on the general procedure?
|
https://mathoverflow.net/users/94097
|
Reference request and methods indication to the continuity of solutions to the problema $L_tu = F(u), ~t\in [0,1],$ and $L_t$ elliptic
|
This question is difficult to answer because the obstruction to continuous families of solutions is (usually) not technical but substantive, and has to do with the local uniqueness of solutions to the equation for a given $t$.
To see why, consider a simple example which illustrates the basic approach one might take: $F = 0$, $L\_t(u) = \text{div} (A\_t(x) \nabla u) - f(x)$, with $A\_t$ uniformly elliptic and smooth in both $x$ and $t$ variables and $f$ a smooth function with $\int\_M f = 0$. Then $L\_t(u) = 0$ admits a unique smooth solution with mean $0$ which may be characterized by the minimization of
$$
\int \frac{1}{2} A\_t \nabla u \cdot \nabla u + f u.
$$
Call this solution $u\_t$: our goal is to show $u\_t$ is continuous as a function of $t$.
The idea is that you plug $u\_t$ in to the equation for $u\_s$:
$$
L\_s(u\_t) = \text{div} A\_s \nabla u\_t - f= \text{div} [A\_s - A\_t] \nabla u\_t,
$$
using that $L\_t(u\_t) = 0$ at the end. Then we just estimate the right-hand side:
$$
|L\_s(u\_t)| \leq C \|A\_s - A\_t\|\_{C^1}\|\nabla u\_t\|\_{C^1} \leq C |t - s|,
$$
using that $A\_t$ is smooth and the regularity of $u\_t$. Then as $L\_s(u\_s) = 0$,
$$
|\text{div} A\_s \nabla (u\_s - u\_t)| \leq C |t - s|.
$$
Now multiplying by $u\_s - u\_t$ and integrating by parts:
$$
\int A\_s \nabla (u\_s - u\_t) \cdot \nabla (u\_s - u\_t) = - \int (u\_s - u\_t)\text{div} A\_s \nabla (u\_s - u\_t) \leq C | t - s|\int |u\_s - u\_t|.
$$
Uniform ellipticity and Holder's inequality give
$$
\|\nabla(u\_s - u\_t)\|\_{L^2}^2\leq C |t - s| \|u\_s - u\_t\|\_{L^2},
$$
so the Poincare inequality (recall $u\_s, u\_t$ have mean $0$) leads to
$$
\|u\_s - u\_t\|\_{W^{1,2}} \leq C |t - s|.
$$
This is basically the continuity estimate we want. If the fact that it is in $W^{1,2}$ topology is concerning, we only need to go back to the equation
$$
|\text{div} A\_s \nabla(u\_s - u\_t)|\leq C |t - s|
$$
and apply our elliptic regularity estimate of choice (De Giorgi-Nash, Schauder, whatever) to upgrade the norms. Many modifications and improvements are available, but this is the basic idea.
The most important ingredient in this argument was that for the solution $u\_s$ of $L\_s$, if $v$ has $|L\_s(v) - L\_s(u\_s)|$ small, then $v$ is close to $u\_s$. This was true here, as we checked, but this is likely to be the hardest step to generalize to nonlinear equations. Even here, note that if we did not assume that $u\_s$ all have mean value $0$, the argument fails (indeed, any family $u\_s + g(s)$ is still a solution to $L\_s(u\_s) = 0$).
If you have, e.g. unique minimizing solutions to the nonlinear problem, you can likely establish this kind of stability property for them and run a similar argument. If not, you can attempt instead to construct your family of solutions "all at once," for example viewing $L\_t$ as a perturbation of $L\_0$ and using fixed point arguments or implicit function theorems.
|
1
|
https://mathoverflow.net/users/378654
|
404429
| 165,871 |
https://mathoverflow.net/questions/404435
|
4
|
Let $U$ be a bounded domain in $\mathbb R^n$. Does there exist a smooth function $f$ with compact support in $U$ such that:
$$ \| f\|\_{W^{k,\infty}(U)} \leq (k!)^{2-\epsilon},$$
for some $\epsilon>0$?
Thanks,
|
https://mathoverflow.net/users/50438
|
Existence of a smooth compactly supported function
|
The answer is yes if $\epsilon<1$, and no when $\epsilon\geq 1$.
This follows from Carleman's quasianalyticity criterion, see for example, Hormander, Analysis of linear partial differential operators, Vol. I, Chap I, Section 1, Theorem 1.3.8.
(Carleman's original proof used Complex Analysis, and it was reproduced in the books on the subject. Hormander has an elementary proof, not using Complex Analysis. As far as I know, this proof was first published in the book I refer to).
|
15
|
https://mathoverflow.net/users/25510
|
404438
| 165,873 |
https://mathoverflow.net/questions/404431
|
5
|
We can characterise $\mathbb{Z}$ and $\mathbb{Z}/2$ as the corepresenting abelian groups of the functors
\begin{align\*}
\mathsf{Forget} &\colon \mathsf{Ab} \to \mathsf{Sets},\\
\mathrm{Inv} &\colon \mathsf{Ab} \to \mathsf{Sets}
\end{align\*}
given by $(A,\cdot\_A,1)\mapsto A$ and $A\mapsto\mathrm{Inv}(A)\overset{\mathrm{def}}{=}\left\{a\in A\ \middle|\ a^2=1\_A\right\}$.
[A similar approach](https://mathoverflow.net/a/404304) in the $\infty$-world gives the $\mathbb{E}\_\infty$-groups $QS^0$ and $\Omega Q\mathbb{RP}^\infty$. Passing to spectra via the [equivalence between $\mathbb{E}\_\infty$-groups and connective spectra](https://mathoverflow.net/questions/280980), we obtain the sphere spectrum $\mathbb{S}$ corresponding to $QS^0$ and a spectrum $E$ corresponding to $\Omega Q\mathbb{RP}^\infty$.
(One possible name for $E$ might be "$\mathbb{S}/2$" since it satisfies an analogous universal property to that of $\mathbb{Z}/2$, corepresenting "involutory objects". However, that notation already usually denotes the mod 2 Moore spectrum, so let's write $E$ for it instead.)
For comparison, their first $8$ homotopy groups are as follows:
$$
\begin{aligned}
\pi\_0(\mathbb{S}) &\cong \mathbb{Z},\\
\pi\_1(\mathbb{S}) &\cong \mathbb{Z}/2,\\
\pi\_2(\mathbb{S}) &\cong \mathbb{Z}/2,\\
\pi\_3(\mathbb{S}) &\cong \mathbb{Z}/24,\\
\pi\_4(\mathbb{S}) &\cong 0,\\
\pi\_5(\mathbb{S}) &\cong 0,\\
\pi\_6(\mathbb{S}) &\cong \mathbb{Z}/2,\\
\pi\_7(\mathbb{S}) &\cong \mathbb{Z}/16\times\mathbb{Z}/3\times\mathbb{Z}/5,
\end{aligned}
\quad\quad
\begin{aligned}
\pi\_0(E) &\cong \mathbb{Z}/2,\\
\pi\_1(E) &\cong \mathbb{Z}/2,\\
\pi\_2(E) &\cong \mathbb{Z}/8,\\
\pi\_3(E) &\cong \mathbb{Z}/2,\\
\pi\_4(E) &\cong 0,\\
\pi\_5(E) &\cong \mathbb{Z}/2,\\
\pi\_6(E) &\cong \mathbb{Z}/16\times\mathbb{Z}/2,\\
\pi\_7(E) &\cong \mathbb{Z}/2\times\mathbb{Z}/2\times\mathbb{Z}/2.
\end{aligned}
$$
(The ones for $E$ are taken from [Liulevicius](https://www.jstor.org/stable/1993859?seq=1#metadata_info_tab_contents); see also [MO 230790](https://mathoverflow.net/questions/230790).)
---
What (homotopy associative, homotopy commutative, $\mathbb{A}\_k$-, $\mathbb{E}\_k$-, or $\mathbb{E}\_\infty$-) ring spectra structures, if any, are there on $E$?
|
https://mathoverflow.net/users/130058
|
Ring spectra structures on a certain spectral analogue of $\mathbb{Z}/2$
|
There are no left-unital multiplications on E. If there were, then for any element $x$ in $\pi\_n(E)$, we would have $x+x = 1 \cdot x + 1 \cdot x = (1+1) \cdot x = 0$ because all elements in $ \pi\_0 E$ are 2-torsion. This is not satisfied by the homotopy groups in your table.
|
8
|
https://mathoverflow.net/users/360
|
404440
| 165,875 |
https://mathoverflow.net/questions/404449
|
1
|
This is a (probably basic) question about the generator of a Markov process.
Let $(E,d)$ be a locally compact metric space. We consider a Feller process $X=(\{X\_t\}\_{t \ge 0},\{P\_x\}\_{x \in E})$ on $E$.
That is, for any $t>0$, the semigroup $P\_t$ of $X$ maps $C\_{\infty}(E)$ into itself. Here, $C\_\infty(E)$ denotes the space of continuous functions on $E$ vanishing at infinity.
We assume that the generator $L$ of $X$ is described as follows: for $f \in C\_{\infty}(E)$ and $x \in E$,
\begin{align\*}
Lf(x)=\int\_{X \setminus \{x\}}(f(y)-f(x))c(x,y)\,\mu(dy).
\end{align\*}
Here, $\mu$ denotes a Radon measure on $E$, and $c(x,y)$ a nonnegative bounded function on $E \times E \setminus \text{diag}$ with compact support. Therefore, the Feller processs $X$ is a pure jump process.
For a bounded open set $U$, we define $\tau\_U=\inf\{t>0 \mid X\_t \notin U\}$. The part process $X^U$ of $X$ on $U$ is defined as
\begin{align\*}
X^U\_t=\begin{cases}
X\_t,&\quad t<\tau\_U,\\
\partial ,&\quad t \ge \tau\_U.
\end{cases}
\end{align\*}
We assume also that $X^U$ is also Feller process. **Then, can we describe the generator $L^U$ of $X^U$?**
We write $E\_x$ for the expectation under $P\_x$, the law of $X$ starting from $x$.
For $f \in C\_{\infty}(U)$ and $x \in U$, we have
\begin{align\*}
L^Uf(x)&=\lim\_{t \to 0}\frac{E\_{x}[f(X^U\_t)]-f(x)}{t}\\
&=\lim\_{t \to 0}\left(\frac{E\_{x}[f(X\_t)]-f(x)}{t}-\frac{E\_{x}[f(X\_t):t \ge \tau\_U]}{t} \right).
\end{align\*}
Can wa characterized the quantity $\lim\_{t \to 0}E\_{x}[f(X\_t):t \ge \tau\_U]/t$ in terms of $c(x,y)$ and $\mu$?
|
https://mathoverflow.net/users/68463
|
On the generator of a Markov process
|
My second advice for those who work with generators of Markov processes is: *Use Dynkin's characteristic operator!* [\*]
---
If $f$ is in the domain of the generator $L$ of $X$, then
$$ L f(x) = \lim\_{B \to \{x\}} \frac{E\_x f(X(\tau\_B)) - f(x)}{E\_x \tau\_B} \, , $$
where the expression under the limit means, for example, that $B = B(x, r)$ with $r \to 0^+$. The same formula holds for $f$ in the domain of the generator $L^U$ of $X^U$, as long as we restrict the class of sets $B$ to those that are contained in $U$.
So if $f$ is in the domain of both $L$ and $L^U$, then we get $L f(x) = L^U f(x)$. Therefore, if the intersection of the domains of $L$ and $L^U$ is sufficiently rich (for example, if for every $x \in U$ and every $f$ in the domain of $L^U$, there is $g$ in the domain of $L$ such that $f = g$ in a neighbourhood of $x$), then you get the result that you are looking for.
In the general case, things can get more complicated. However, your expression for the generator suggests that you know in advance that the domains of the generators are rich enough, so I will stop here.
---
*Edit:* For the record, let me add a short description of a peculiar example where there are too few functions in the intersection of the domains of $L$ and $L^U$.
Let $X\_t$ be the standard 1-D Brownian motion with added extra jumps from $2$ to $0$, according to the following rule. We let $X\_t$ run as the usual Brownian motion until the local time at $2$ hits an independently chosen, exponentially distributed threshold. At that time, we artificially move $X\_t$ to zero, and restart the above procedure. Lather, rinse, repeat.
A function $f$ belongs to the domain of $L$ if and only if it is $C^2\_\infty$ in $\mathbb R \setminus \{2\}$, with $f''\_+(2) = f''\_-(2)$ and
$$ (f'\_+(2) - f'\_-(2)) + (f(0) - f(2)) = 0 . $$
Suppose that $U = (-1, 1)$. If $f$ is in the domain of both $L$ and $L^U$, then the latter condition implies that $f = 0$ in the complement of $U$), so that the former condition leads to simply $f(0) = 0$.
It follows that every function $f$ in the intersection of the domains of $L$ and $L^U$ satisfies $f(0) = 0$. This is already bad, but still not *too* bad.
Consider, however, a similarly constructed process which jumps from $1 + n$ to $q\_n$, where $q\_n$ is the enumeration of rational numbers in $U = (-1, 1)$. By the same argument, every function in the intersection of the domains of $L$ and $L^U$ is zero at every $q\_n$, and hence identically zero. This *is* bad.
Of course these two processes are seemingly quite different from the ones mentioned in the question, but in fact I am rather sure one can cook up similar examples using only jumps governed by sufficiently irregular functions $c(x,y)$.
---
[\*] The first advice reads: *Don't do that unless you really have to.* Most of the time there is a way to avoid the use of the generator whatsoever.
|
2
|
https://mathoverflow.net/users/108637
|
404451
| 165,877 |
https://mathoverflow.net/questions/404230
|
5
|
Let $\mathrm{cov}\_H(C\_2^\omega)$ be the smallest cardinality of a cover of the Boolean group $C\_2^\omega=(\mathbb Z/2\mathbb Z)^\omega$ by closed subgroups of infinite index. It can be shown that
$$\max\{\mathrm{cov}(\mathcal M),\mathrm{cov}(\mathcal N)\}\le \mathrm{cov}(\mathcal E)\le\mathrm{cov}\_H(C\_2^\omega)\le \mathfrak r,$$
where $\mathcal E$ is the $\sigma$-ideal generated by closed subsets of Haar measure zero in $C\_2^\omega$ (the upper bound $\mathrm{cov}\_H(C\_2^\omega)\le\mathfrak r$ is proved, for example, [here](https://mathoverflow.net/q/378850/61536)). It is known that $\max\{\mathrm{cov}(\mathcal N),\mathrm{cov}(\mathcal M)\}$ can be strictly smaller than $\mathfrak b$ (this happens, for example, in the Laver and Mathias models).
>
> **Problem.** Is $\mathrm{cov}\_H(C\_2^\omega)<\mathfrak b$ consistent? What is the value of $\mathrm{cov}\_H(C\_2^\omega)$ in the Laver (or Mathias) model?
>
>
>
|
https://mathoverflow.net/users/61536
|
Can the Boolean group $C_2^\omega$ be covered by less than $\mathfrak b$ nowhere dense subgroups?
|
Lyubomyr Zdomskyy proved that in the Laver model $\mathrm{cov}\_H(2^\omega)=\omega\_1<\mathfrak b=\mathfrak c$.
His argument used the following known *Laver property* of the Laver model $V'$: for every
function $f:\omega\to\omega$ in $V'$ upper bounded by some function $h:\omega\to\omega$ in the ground model $V$, there exists $H:\omega\to [\omega]^{<\omega}$ in $V$ such that $|H(n)|\leq n+1$
and $f(n)\in H(n)$ for all $n\in\omega$.
Choose any increasing sequence $(k\_n)\_{n\in\omega}\in \omega^\omega\cap V$ such that $k\_0=0$ and $k\_{n+1}>k\_n+n+1$ for every $n\in\omega$. Consider the family $\mathcal H$ of closed nowhere
dense subgroups of $2^\omega$ of the form $ \prod\_{n\in\omega}H\_n, $ where
$H\_n$ is a proper subgroup of $2^{k\_{n+1}\setminus k\_n}$ and $(
H\_n)\_{n\in\omega}\in V$. We claim that for every $x\in 2^\omega$ there
exists $H\in\mathcal H$ with $x\in H$. Indeed, since $x{\restriction}
(k\_{n+1}\setminus k\_n)\in 2^{k\_{n+1}\setminus k\_n}$, the latter set
has size $2^{k\_{n+1}-k\_n}$, and the sequence $\langle
2^{k\_{n+1}-k\_n}\rangle\_{n\in\omega}$ lies in $V$, there exists a sequence
$\langle X\_n\rangle\_{n\in\omega}\in V$ such that
$$x{\restriction}(k\_{n+1}\setminus k\_n)\in
X\_n\in \big[ 2^{k\_{n+1}\setminus k\_n}\big]^{n+1}$$ for all $n\in\omega$. It
suffices to denote by $H\_n$ the subgroup of $2^{k\_{n+1}\setminus
k\_n}$ generated by $X\_n$ and note that $H\_n\neq 2^{k\_{n+1}\setminus
k\_n}$ because $|X\_n|=n+1$ and $k\_{n+1}-k\_n>n+1$.
Therefore in the model $V'$ we have $\mathrm{cov}\_H(2^\omega)\le|\mathcal H|\le|V\cap 2^\omega|=\omega\_1$.
|
3
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https://mathoverflow.net/users/61536
|
404455
| 165,879 |
https://mathoverflow.net/questions/404415
|
2
|
I am looking to find a solution, or even just prove the existence of one, to the following system of linear PDEs. They come up in a construction I am trying to work out in symplectic geometry. Here $(r\_1, \theta\_1, r\_2, \theta\_2)$ are the polar coordinates on $\mathbb{R}^4$, and $\beta$ and $\gamma$ are smooth functions in terms of those coordinates, that I would like to solve for:
$$2\gamma + r\_2\frac{\partial \gamma}{\partial r\_2} = 1$$
$$\frac{\partial \beta}{\partial \theta\_2}(r\_1^2-1) - r\_2^2\frac{\partial \gamma}{\partial \theta\_1} = 0$$
$$2\beta r\_1 + \frac{\partial \beta}{\partial r\_1}(r\_1^2-1) = r\_1$$,
and I need the conditions that $\beta = \gamma = 1$ when $r\_1^2 \geq \delta > 0$, and $\beta = r\_1$ when $r\_1 = 0$ and $\theta\_2 = 0, 0.5$. For certain reasons, I know that a solution (i.e. functions $\beta$ and $\gamma$) will not exist if I want $\beta = r\_1$ whenever $r\_1 = 0$, but I want to know if they might exist around two values of $\theta\_2$.
|
https://mathoverflow.net/users/379942
|
Existence of solution to a system of linear PDEs with boundary conditions
|
The general solution of your equations in a simply connected domain on which $r\_2\not=0$ and $r\_1\not=\pm1$ is
$$
\beta = \frac12 + \frac1{{(r\_1}^2{-}1)}\,
\left(\frac{\partial a}{\partial\theta\_1}+b(\theta\_1,r\_2)\right)
\quad\text{and}\quad
\gamma= \frac12 + \frac1{{r\_2}^2}\,
\left(\frac{\partial a}{\partial\theta\_2}+c(\theta\_2,r\_1)\right),
$$
where $a = a(\theta\_1,\theta\_2)$ is a function of $\theta\_1$ and $\theta\_2$ only, $b$ is a function of $\theta\_1$ and $r\_2$ only, and $c$ is a function of $\theta\_2$ and $r\_1$ only.
To see this, set $\bar\beta = ({r\_1}^2{-}1)\bigl(\beta-\tfrac12\bigr)$ and $\bar\gamma = {r\_2}^2\bigl(\gamma-\tfrac12\bigr)$ and note that the given equations imply the constant coefficient linear equations
$$
\frac{\partial\bar\beta}{\partial r\_1}
= \frac{\partial\bar\gamma}{\partial r\_2}
= \frac{\partial\bar\beta}{\partial\theta\_2}-\frac{\partial\bar\gamma}{\partial\theta\_1} = 0,
$$
which are easily solved.
I do not see how you can choose $a$, $b$, and $c$ so that your 'boundary conditions' are satisfied. Because the general solution depends on three functions of two variables, one would expect to be able to specify (initial) conditions along a surface in the domain, but not along a hypersurface (which is what boundaries generally are).
|
4
|
https://mathoverflow.net/users/13972
|
404456
| 165,880 |
https://mathoverflow.net/questions/404392
|
1
|
Let $K$ be a perfect field, and let $f\_1, \ldots, f\_m \in K[X\_1,\ldots,X\_n]$ be polynomials. Consider the affine scheme
$$X = \mathrm{Spec}(K[X\_1,\ldots, X\_n]/(f\_1,\ldots,f\_m))$$
and let $N = \dim(X)$. Given a closed point $\mathfrak{m} \in X$, we define the *multiplicity* of $\mathfrak{m}$ to be $N!$ times the leading coefficient of the Hilbert-Samuel polynomial of the local ring $\mathcal{O}\_{X,\mathfrak{m}}$. Since $X$ is an excellent scheme, it is known that the set of all multiplicities occuring among the (closed) points of $X$ is finite.
**My question**: Is it possible to find an upper bound for the maximal multiplicity of a closed point in $X$ in terms of (e.g. the degrees of) the polynomials $f\_1,\ldots, f\_m$ ?
Indeed, the special case $m=1$ should be easy: Write $f = f\_1$ and assume w.l.o.g. that $f(0,\ldots,0) = 0$. Then the multiplicity of $\mathfrak{m} = \langle X\_1, \ldots, X\_n \rangle$ equals the minimal degree of a monomial $X\_1^{i\_1} \cdots X\_n^{i\_n}$ appearing in $f$ with non-zero coefficient (which is certainly bounded from above by the degree of $f$). The case of general $\mathfrak{m}$ should follow by base change.
Any help is appreciated! In particular, I am not sure whether this is a difficult problem or not. (So if you have an opinion on that, then please let me know as well.)
**EDIT** Sept. 20th, 06:20 pm: Could it be the case that
$$\deg(f\_1)+\ldots + \deg(f\_m)$$
serves as an upper bound? To motivate this idea, suppose again that $f\_j(0, \ldots, 0) = 0$ for all $1 \leq j \leq n$. Then I tend to believe that the singularity of
$\mathrm{Spec}(K[X\_1,\ldots, X\_n]/(f\_1 \cdots f\_m))$ at $\mathfrak{m} = \langle X\_1, \ldots, X\_n \rangle$ (which should be bounded by $\deg(f\_1)+\ldots + \deg(f\_m)$ according to the special case $m=1$) is worse than the singularity of $X$ at $\mathfrak{m}$. But maybe my geometric intuition is mistaken here.
|
https://mathoverflow.net/users/125074
|
Bound for multiplicities of closed points on scheme
|
Edited: the proof below assumes $k$ is algebraically closed. The proof for the multiplicity inequality has been added.
Given $x \in X := V(f\_1, \ldots, f\_m)$, let $k$ be the *local dimension* of $X$ at $x$ (i.e. $k$ is the maximum of the dimension of all irreducible components of $X$ containing $x$).
**Claim:** $mult\_x(X) \leq $ the product of the largest $n-k$ elements of the sequence $\deg(f\_1), \dots, \deg(f\_n)$.
If $k = 0$, then the claim holds due to Bézout's theorem. This estimate is also sharp: given degrees $d\_1, \ldots, d\_n$, take generic homogeneous polynomials of the specified degrees in $n$-variables - they intersect only at the origin, and the multiplicity at the origin is precisely $\prod\_i d\_i$ due to Bézout's theorem.
In the case that $n > k \geq 1$, we will prove the following
**Sub-claim 1:** For each $j = 1, \ldots, n-k$, there are $g\_1, \ldots, g\_j$ such that each $g\_j$ is a linear combination of the $f\_i$, and $V(g\_1, \ldots, g\_j)$ is a complete intersection on a neighborhood of $x$.
**Proof**: Proceed by induction on $j$. For $j = 1$, set $g\_1 := f\_1$. If $k = n - 1$, then stop. Otherwise, for $2 \leq j \leq n - k$, we can assume by induction we found $g\_1, \ldots, g\_{j-1}$ such that $V(g\_1, \ldots, g\_{j-1})$ is a complete intersection on a neighborhood of $x$. We claim that there is a linear combination of the $f\_1, \ldots, f\_n$ which does not identically vanish on any irreducible component of $V(g\_1, \ldots, g\_{j-1})$. Indeed, otherwise since $k$ is infinite, we can find $m$ linearly independent linear combinations of $f\_1, \ldots, f\_m$ which vanish identically on one of the irreducible components of $V(g\_1, \ldots, g\_{j-1})$ containing $x$, which would mean that the local dimension of $X$ at $x$ is $n - j + 1 > k$, a contradiction. Now let $g\_j$ be that linear combination of the $f\_i$ which does not identically vanish on any irreducible component of $V(g\_1, \ldots, g\_{j-1})$, and repeat. $\square$
Once you find $g\_1, \ldots, g\_{n-k}$ as above, let $Y := V(g\_1, \ldots, g\_{n-k})$. The second observation is that $mult\_x(X) \leq mult\_x(Y)$, which follows from the following general fact:
**Sub-claim 2:** Let $X \subseteq Y \subseteq k^n$ be a chain of closed subschemes and $x$ be a closed point of $X$ such that $X$ and $Y$ have the same local dimension at $x$. Then $mult\_x(X) \leq mult\_x(Y)$.
**Proof**: For each $q \geq 0$, there is a surjection
$$O\_{x,Y}/m\_x^qO\_{x,Y} \to O\_{x,X}/m\_x^qO\_{x,X}$$
where $m\_x$ is the ideal of $x$. Treating this surjection as an $O\_{x,Y}$-module homomorphism, it follows that
$$l(O\_{x,Y}/m\_x^qO\_{x,Y}) \geq l(O\_{x,X}/m\_x^qO\_{x,X})$$
where $l$ is the "length" (see e.g. Atiyah-Macdonald, Proposition 6.9). By the assumption on dimension, for $q \gg 1$, both of these lengths are given by polynomials in $q$ of degree $d$, where $d := \dim\_x(X) = \dim\_x(Y)$. It follows that the coefficient of $q^d$ in the polynomial providing the values of $l(O\_{x,Y}/m\_x^qO\_{x,Y})$ can not be smaller than the corresponding coefficient of the polynomial for $l(O\_{x,X}/m\_x^qO\_{x,X})$. $\square$
The third observation is that if one of the $f\_i$ appears in linear combinations defining more than one $g\_j$, say $g\_{j\_1}, g\_{j\_2}$, then replacing $g\_{j\_2}$ by an element of the form $g\_{j\_2} + ag\_{j\_1}$ for an appropriate $a \in k$ we may ensure that $f\_i$ does not appear in the linear combination defining $g\_{j\_2}$. In this way we can find appropriate "invertible" linear combinations $g'\_1, \ldots, g'\_{n-k}$ of $g\_1, \ldots, g\_{n-k}$ such that
* $V(g\_1, \ldots, g\_{n-k}) = V(g'\_1, \ldots, g'\_{n-k})$, and
* there is a reordering of $f\_1, \ldots, f\_n$ such that $\deg(g'\_j) \leq \deg(f\_j)$, $j = 1, \ldots, n-k$.
The final observation is that $mult\_x(Y) = mult\_x(Y \cap H)$ for a generic affine subspace of dimension $k$ containing $x$. On the other hand, if $h\_1, \ldots, h\_k$ are linear polynomials such that $x$ is an isolated point of $V(g'\_1, \ldots, g'\_{n-k}, h\_1, \ldots, h\_k)$, then we are done by the $k = 0$ case.
|
1
|
https://mathoverflow.net/users/1508
|
404477
| 165,888 |
https://mathoverflow.net/questions/404472
|
4
|
It's well known that any oriented closed 3-manifold (topological or smooth) can be obtained by surgerizing along a (framed oriented) link $L$ in the 3-sphere $S^{3}$. Even better, Kirby found a complete set of relations that classify all 3-manifold in terms of links. In short, we have the map
$$\{\mbox{framed oriented link in }S^3\}/\{\mbox{kirby move}\} \xrightarrow{\sim} \{3\mbox{-manifold (oriented and closed)}\}.$$
A natural question arise.
1. Given two (oriented and framed) link presentations, is there an algorithm (that halts in finite time) determines if they give rise to the same $3$-manifold?
Or even better, one may ask
2. Given a link presentation $l$, is there any sort of normal form $[l]$ for it (in the sense that $l$ and $l'$ are related by Kirby moves if and only if $[l] = [l']$?
### Related
* A program [KLO](https://community.middlebury.edu/%7Emathanimations/klo/) lets you play with link presentations under Kirby moves. But it doesn't seem to determine if two link presentations produce the same $3$-manifold.
* The most popular programs **snappea** and **regina** present a $3$-manifold in terms of triangulation. It does not seem that they provide a way to determine from their Kirby diagram.
|
https://mathoverflow.net/users/124549
|
Normal form of framed links under Kirby moves
|
1. It is a folklore result that geometrisation solves the homeomorphism problem for (compact, connected, oriented) three-manifolds. [Kuperberg](https://arxiv.org/abs/1508.06720) discusses this, and improves the running time to elementary recursive. So, use Snappy to convert the given Kirby diagrams to triangulations and appeal to the above.
2. Sure. First place a reasonable complexity on diagrams. Say, count the number of crossings, add the bit-size of the framings, and break ties using some lexicographic complexity on encodings of graphs (as Burton does using isoSigs of triangulations). Now generate all diagrams up to the complexity of the given framed link. Use (1) to decide homeomorphism and take the smallest. This algorithm to compute normal forms is basically as fast as your solution to the homeomorphism problem (times the number of Kirby diagrams up to the given complexity, of course!).
And to repeat a remark from above - snappy can produce a triangulation from a framed link.
|
3
|
https://mathoverflow.net/users/1650
|
404479
| 165,889 |
https://mathoverflow.net/questions/404478
|
3
|
Let $G$ be a compact (Lie) group, and $H \subseteq H'$ two compact (Lie) subgroups. It is clear that we have an obvious surjective map of homogeneous spaces
$$
G/H \twoheadrightarrow G/H'.
$$
Will it be true in general that this fibration gives a fibre bundle? Will the fibre be isomorphic to the quotient space $H/H'$? Will it be true that the bundle is principal if and only if $H'$ is a normal subgroup of $G$?
P.S. Do such fibre bundles have a special name?
|
https://mathoverflow.net/users/326091
|
Principal bundles from a fibration of homogeneous spaces
|
I call such bundles "homogeneous bundles", but it's not a totally standard terminology.
It is true that the map $G/H\rightarrow G/H'$ is a fiber bundle map with fiber $H/H'$. One way to see this is to start with the principal bundle $H\rightarrow G \rightarrow G/H$. The group $H$ naturally acts on $H/H'$, and so we have an associated fiber bundle $H/H'\rightarrow G\times\_H (H/H')\rightarrow G/H$, where the notation $G\times\_H (H/H')$ refers to the quotient of $G\times (H/H')$ by the $H$-action $h\ast(g, h\_1 H') = (gh^{-1}, hh\_1 H')$.
The space $G\times\_H (H/H')$ is diffomeorphic to $G/H'$: the map which takes $[g, hH']$ to $ghH'$ is a diffeomorphism with inverse $gH'\rightarrow [g,eH']$. With respect to this diffeomorphism, the projection $G/H'\rightarrow G/H$ is the obvious one.
Now, if $H'$ is normal in $H$ (though it doesn't have to be normal in $G$), then the map $G/H'\rightarrow G/H$ is a principal $H/H'$ bundle. The $H/H'$ action on $G/H'$ is just the $H'/H$ action on $G\times\_H (H'/H)$ given by multiplication on the second coordinate: $(hH')\ast [g, h\_1 H'] = [g, h\_1 h H']$
Lastly, it is *not* necessary that $H'$ be normal in $H$ in order for the bundle to be principal. To see this, consider the chain of subgroups $$O(1)\subseteq O(2)\subseteq O(3)$$ with each $O(k)$ embedded in $O(k+1)$ as the top left $k\times k$ block. The corresponding bundle $O(2)/O(1)\rightarrow O(3)/O(1)\rightarrow O(3)/O(2)$ is the Hopf bundle, which is a principal $S^1$-bundle. However, $O(1)$ is not normal in $O(2)$ since conjugating by a matrix which swaps the standard basis of $\mathbb{R}^2$ doesn't preserve $O(1)$.
|
4
|
https://mathoverflow.net/users/1708
|
404483
| 165,890 |
https://mathoverflow.net/questions/404491
|
5
|
Let $G(k,n)$ denote the Grassmiannian of $k$-planes in $\mathbb C^n$. Let's define
$$ I\_j =\{ (\Lambda\_1,\Lambda\_2 ) \in G(k,n) \times G(l,n) \, | \, \dim(\Lambda\_1 \cap \Lambda\_2) \geq j \}. $$
These are an analytic subvarieties in $G(k,n) \times G(l,n)$. I would like to know if something is known about their homology class. It would be perfect if it was possible to express the Poincare duals of these homology classes in terms of the universal Chern classes (Chern classes of tautological bundles on the two factor Grassmannians, pulled back to the product). I am most interested in the case $k=l$, $n=2k$. The question seems to be closely related to Schubert calculus.
|
https://mathoverflow.net/users/94647
|
Intersection cycle in a product of Grassmannians
|
Let $V$ be the $n$-dimensional space such that $\Lambda\_i \subset V$. Then the condition $\dim(\Lambda\_1 \cap \Lambda\_2) \ge j$ is equivalent to the condition
$$
\mathrm{rank}(\Lambda\_1 \hookrightarrow V \to V/\Lambda\_2) \le k - j.
$$
This means that $I\_j$ is a degeneracy locus for the morphism
$$
\mathcal{U}\_1 \hookrightarrow V \otimes \mathcal{O} \to \mathcal{Q}\_2
$$
on $\mathrm{Gr}(k,V) \times \mathrm{Gr}(l,V)$ from the pullback of the tautological subbundle of the first factor to the tautological quotient bundle of the second factor. Therefore, its class is computed by **Porteous formula** in terms of Chern classes of $\mathcal{U}\_1$ and $\mathcal{Q}\_2$.
|
8
|
https://mathoverflow.net/users/4428
|
404493
| 165,896 |
https://mathoverflow.net/questions/404502
|
2
|
Given a fixed $p \in \{3,4,5,\ldots\}$, we define the strictly increasing sequence $\{a\_k\}\_{k\in \mathbb N}$ as follows. We set $a\_{p,1}=1$ and for each $k>1$, we set $a\_{p,k}$ to be the least integer strictly greater than $a\_{p,k-1}$ such that
$$a\_{p,k}=b\_1+b\_2+\ldots+b\_p$$
for some $\{b\_j\}\_{j=1}^{p} \subset \{a\_{p,1},\ldots,a\_{p,k-1}\}$.
For example, it is clear that $a\_{p,2}=p$ and $a\_{p,3}=2p-1$, $a\_{p,4}=3p-2$,...
My question is whether there is a strictly increasing sequence $\{p\_\ell\}\_{\ell\in \mathbb N}\subset \{3,4,\ldots\}$ such that the following statement holds: There are infinitely many $j\in \mathbb N$ that do not belong to any $\{a\_{p\_{\ell},k}\}\_{k=1}^{\infty}$ for all $\ell\in \mathbb N$.
|
https://mathoverflow.net/users/50438
|
On gaps in a sequence of integers
|
Unless I'm confused, this is true. The key fact is that your sequences are just infinite arithmetic progressions, i.e. $a\_{p,k} = 1 + (p-1)(k-1)$ for all $p,k$.
We can prove this formula by strong induction on $k$: clearly $a\_{p,1} = 1$ by definition. Assume that $a\_{p,k} = 1 + (p-1)(k-1)$ for $k < K$. Then all $a\_{p,k}$ for $k < K$ are equal to $1 \pmod{p-1}$, so by definition, $a\_{p,K}$ is the sum of $p$ integers equal to $1 \pmod{p-1}$ and must itself equal $1 \pmod{p-1}$. Therefore, $a\_{p,K} \geq 1 + (p-1)(K-1)$, since this is the smallest integer greater than $a\_{p,K-1} = 1 + (p-1)(K-2)$ equal to $1 \pmod{p-1}$.
But clearly $a\_{p,K} \leq 1 + (p-1)(K-1)$, since we can write
$1 + (p-1)(K-1) = 1 + \cdots + 1 + 1 + (p-1)(K-2) = a\_{p,1} + \cdots + a\_{p,1} + a\_{p,K-1}$. So, $a\_{p,K} = 1 + (p-1)(K-1)$, completing the proof by induction.
But with this formula, we can just give a density argument for your question. Take any sequence $\{p\_\ell\}$ for which $\sum\_{\ell} (p\_{\ell}-1)^{-1} < 1$. (For future reference, note that this implies $p\_{\ell}/\ell \rightarrow \infty$.)
Then, for every $N$, the number of integers in
$\{1, \ldots, N\} \cap \{a\_{p\_\ell,k}\}\_k$ is at most $\lceil N/(p\_{\ell}-1) \rceil \leq N/(p\_{\ell} - 1) + 1$. Therefore,
$\frac{|\{1, \ldots, N\} \cap \{a\_{p\_\ell,k}\}\_{\ell,k}|}{N} \leq N^{-1}
\sum\_{\ell \ s.t. \ p\_\ell \leq N} (N/(p\_{\ell} - 1) + 1) \leq
\frac{L}{N} + \sum\_{\ell} (p\_{\ell} - 1)^{-1} $,
where $L$ is the maximal $\ell$ so that $p\_{\ell} \leq N$. As $N$ approaches infinitely, the final term approaches $\sum\_{\ell} (p\_{\ell} - 1)^{-1} < 1$, since $L/N \rightarrow 0$ by the fact noted for future reference above.
The limiting density of positive integers in $\{a\_{p\_\ell,k}\}\_{\ell,k}$ is strictly less than $1$, so clearly there are infinitely many positive integers not in this set. (In fact, you can make the limiting proportion of positive integers in the set as small as desired, but not zero, since the set contains an infinite arithmetic progression.)
|
4
|
https://mathoverflow.net/users/116357
|
404510
| 165,901 |
https://mathoverflow.net/questions/404459
|
4
|
Let $G$ be a simple (i.e. *every proper normal subgroup is discrete*) simply connected compact Lie group. Define **the degree of $k$-nilpotence** of $G$ to be the Haar measure of the set
$$\{(x\_1,\dotsc,x\_{k+1}): [x\_1,\dotsc,x\_{k+1}]=1\}.$$
($[x,y]=x^{-1}y^{-1}xy$ and $[x\_1,\dotsc,x\_{k+1}]:=[[x\_1,\dotsc,x\_k],x\_{k+1}]$.)
The following question is raised in [Martino, Tointon, Valiunas, and Ventura - Probabilistic nilpotence in infinite groups](https://doi.org/10.1007/s11856-021-2168-3):
>
> Does a compact group with positive degree of $k$-nilpotence have an
> open $k$-step nilpotent subgroup?
>
>
>
Is it easy to answer this question when $G$ is a simple simply connected compact Lie group?
|
https://mathoverflow.net/users/84700
|
Can the degree of $k$-nilpotence of a simple simply connected compact Lie group be in $(0,1)$?
|
Here's a positive answer of the question for arbitrary compact Lie groups. For a group $G$, denote $W\_k(G)=\{x\in G^k:[x\_1,\dots,x\_k]=1\}$.
>
> Let $G$ be a compact Lie group. Then $W\_k(G)$ has nonzero Haar measure for some $k\ge 1$ if and only if $G^0$ is a torus.
>
>
>
One direction is obvious: if $G^0$ is a torus, then $(G^0)^k\subset W\_k^G$ has positive measure (for every $k\ge 2$).
Conversely, we assume $W\_k$ has nonzero Haar measure. If $k=1$, it follows that $G$ is finite and we are done. Assuming otherwise, $k\ge 2$.
The subset $W\_k(G)\subset G^k$ is Zariski-closed. Since it has nonzero Haar measure, it therefore contains $\prod\_{i=1}^k C\_i$ for cosets $C\_i$ of $G^0$. [If $G$ is connected, this is $G^k$, so $G$ is nilpotent, hence is a torus and the proof is finished.]
I assume the conventions $[x,y]=xyx^{-1}y^{-1}$ and that the iterated commutator is defined by $[x]=x$ and $[x\_1,\dots,x\_k]=[x\_1,[x\_2,\dots,x\_k]]$ for $k\ge 2$. Also, by $[X\_1,\dots,X\_k]$ I mean the set $\{[x\_1,\dots,x\_k]:x\_i\in X\_i\}$ (rather than the group it generates).
Now assume that $G^0$ is semisimple and nontrivial. I need an inductive argument. Namely, to reach a contradiction, choose $k\ge 2$ minimal for the property that there are cosets $C\_1,\dots,C\_k$ of $G^0$ such that $[C\_1,\dots,C\_k]$ is a singleton, say $w$. Necessarily $k\ge 2$.
Define $Y=\{[x\_2,\dots,x\_k]:x\_i\in C\_i,\forall i\ge 2\}$. Then $[x,y]=w$ for all $x\in C\_1$ and $y\in Y$. Then for all $s\in G^0$, $x\in C\_1,y\in Y$, we have $$w=[x,y]=[xs,y]=x[s,y]x^{-1}[x,y]= x[s,y]x^{-1}w,$$
and hence $[s,y]=1$. Thus, $Y$ is contained in the centralizer of $G^0$. Since $G^0$ is semisimple, its centralizer is finite. Hence $Y=[C\_2,\dots,C\_k]$ is finite, hence a singleton by connectedness. This contradicts the minimality of $k$.
Now the result is proved when $G^0$ is semisimple (i.e., when $G^0$ is semisimple, if $W\_k(G)$ has nonempty interior implies $G^0=1$). In general, let $Z$ be the center of $G^0$. If $W\_k(G)$ has nonempty interior, since the projection $G\to G/Z$ is open and so is $G^k\to (G/Z)^k$, the subset $W\_k(G/Z)$ (which contains the projection of $W\_k(G)$) has nonempty interior. So, by the semisimple case $(G/Z)^0$ is trivial. That is, $Z=G^0$.
[Edit: I fixed my previous argument]
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5
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https://mathoverflow.net/users/14094
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404514
| 165,903 |
https://mathoverflow.net/questions/404518
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1
|
Consider infinite matrices of the form
$$\left(
\begin{array}{ccccc}
a\_0 & a\_1 & a\_2 & a\_3 & . \\
0 & a\_0 & a\_1 & a\_2 & . \\
0 & 0 & a\_0 & a\_1 & . \\
0 & 0 & 0 & a\_0 & . \\
. & . & . & . & . \\
\end{array}
\right)$$
The elements on each diagonal coincide.
My questions are:
* Do they form a commutative ring?
* Can they be extended to form a field?
Now, let define an operation $\operatorname{reg} A=\sum\_{k=0}^\infty B\_k a\_k,$
where $B\_k$ are Bernoulli numbers.
What are the properties of this operation?
Let's define another operation $\det' A=\exp(\Re \operatorname{reg} \log A)$.
What are the properties of this operation?
**Motivation part.**
This is meant to be a matrix representation of divergent integrals and series.
For instance,
$\sum\_{k=1}^\infty 1=
\left(
\begin{array}{ccccc}
0 & 1 & 0 & 0 & . \\
0 & 0 & 1 & 0 & . \\
0 & 0 & 0 & 1 & . \\
0 & 0 & 0 & 0 & . \\
. & . & . & . & . \\
\end{array}
\right)$
$\sum\_{k=0}^\infty 1=
\left(
\begin{array}{ccccc}
1 & 1 & 0 & 0 & . \\
0 & 1 & 1 & 0 & . \\
0 & 0 & 1 & 1 & . \\
0 & 0 & 0 & 1 & . \\
. & . & . & . & . \\
\end{array}
\right)$
$\sum\_{k=0}^\infty k=
\left(
\begin{array}{ccccc}
1/12 & 1/2 & 1/2 & 0 & . \\
0 & 1/12 & 1/2 & 1/2 & . \\
0 & 0 & 1/12 & 1/2 & . \\
0 & 0 & 0 & 1/12 & . \\
. & . & . & . & . \\
\end{array}
\right)$
$\int\_0^\infty x dx=\int\_0^\infty \frac 2{x^3}=\left(
\begin{array}{ccccc}
1/6 & 1/2 & 1/2 & 0 & . \\
0 & 1/6 & 1/2 & 1/2 & . \\
0 & 0 & 1/6 & 1/2 & . \\
0 & 0 & 0 & 1/6 & . \\
. & . & . & . & . \\
\end{array}
\right)$
There are also some expressions that include divergent integrals that can be represented this way:
$(-1)^{\int\_0^\infty dx}=\left(
\begin{array}{ccccccc}
i & -\pi & -\frac{i \pi ^2}{2} & \frac{\pi ^3}{6} & \frac{i \pi ^4}{24} & -\frac{\pi
^5}{120} & . \\
0 & i & -\pi & -\frac{i \pi ^2}{2} & \frac{\pi ^3}{6} & \frac{i \pi ^4}{24} & . \\
0 & 0 & i & -\pi & -\frac{i \pi ^2}{2} & \frac{\pi ^3}{6} & . \\
0 & 0 & 0 & i & -\pi & -\frac{i \pi ^2}{2} & . \\
0 & 0 & 0 & 0 & i & -\pi & . \\
0 & 0 & 0 & 0 & 0 & i & . \\
. & . & . & . & . & . & . \\
\end{array}
\right)$
The $\operatorname{reg}$ operation gives the regularized value of the integral or series.
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https://mathoverflow.net/users/10059
|
What are the properties of this set of infinite matrices and operations on them?
|
If the matrices have entries from a (unital) ring $R$ then the set of such matrices is isomorphic to $R[[x]]$, the ring of formal power series over $R$. To see this, observe that the map sending the infinite matrix with $a\_0 = 0$, $a\_1 = 1$ and $a\_k = 0$ for $k \ge 2$ to $x$ is a ring isomorphism.
This also answers the second question: if $R$ is an integral domain then set of matrices embeds canonically in the field of fractions of $R[[x]]$ and this is the smallest field containing $R[[x]]$. In particular, if $R$ is a field then this field is $\{ \sum\_{k=-m}^\infty a\_k x^k : a\_k \in R, m \in \mathbb{N}\_0 \}$.
I'm uncertain how $\mathrm{reg}$ is (well)-defined, but certainly one can take $R$ to be the polynomial ring $\mathbb{C}[z]$ and then something like $\sum\_{k=0}^\infty B\_k(z) x^k$ is a well-defined element of $R[[x]] = \mathbb{C}[z][[x]]$. If, as in the correction then one wants Bernoulli numbers rather than the polynomials, just specialize to $\mathbb{C}[[x]]$ by evaluating at $z=0$.
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5
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https://mathoverflow.net/users/7709
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404519
| 165,904 |
https://mathoverflow.net/questions/404482
|
2
|
Let $\mathrm{sSet}^+ = \mathrm{sSet}^+\_{/ \Delta^0}$ be the model category of marked simplicial sets over the point. By Theorem 3.1.5.1 in Higher Topos Theory, this model category is Quillen equivalent to $\mathrm{sSet}$ with Joyal's model structure. The fibrant objects of $\mathrm{sSet}^+$ are the quasicategories in which precisely the equivalences are marked.
My question concerns the classification of fibrations of fibrant objects in $\mathrm{sSet}^+$. For $\mathrm{sSet}$ with Joyal's model structure this is understood (Corollary 2.4.6.5 in Higher Topos Theory): A map $f : X \rightarrow Y$ of quasicategories $X, Y$ is a fibration if and only if $f$ is an inner fibration and an isofibration (equivalences in $Y$ can be lifted along a preimage of the codomain). Since lifts for equivalences are already part of this classification, I suspect that the following holds:
A map of fibrant *marked* simplicial sets is a fibration in $\mathrm{sSet}^+$ if and only if its underlying map of simplicial sets is a fibration in $\mathrm{sSet}$ with Joyal's model structure.
Is this true?
|
https://mathoverflow.net/users/84063
|
Fibrations of fibrant marked simplicial sets
|
Yes, this is true. There are various ways to prove this. Here's the shortest argument I can think of. One direction is easy to prove, so let's prove the other direction.
Let $U \colon \mathbf{sSet}^+ \to \mathbf{sSet}$ denote the functor that forgets markings. We will use that the restriction of this functor to the full subcategory of fibrant objects in $\mathbf{sSet}^+$ (i.e. the naturally marked quasi-categories) is fully faithful and preserves cofibrations, fibrations, and weak equivalences. All of these properties are easy to prove.
Now, let $f \colon A \to B$ be a morphism of naturally marked quasi-categories, and suppose that $U(f) \colon U(A) \to U(B)$ is an isofibration of quasi-categories. We want to prove that $f$ is a fibration in the model structure on $\mathbf{sSet}^+$. Let $f = p\circ j$ be a factorisation of $f$ into a trivial cofibration $j \colon A \to C$ followed by a fibration $p \colon C \to B$ in $\mathbf{sSet}^+$. (Note that $C$ is fibrant, since $B$ is fibrant and $p$ is a fibration.) By the above preservation properties of $U$, $U(f) = U(p) \circ U(j)$ is a factorisation of $U(f)$ into a trivial cofibration followed by a fibration in the Joyal model structure on $\mathbf{sSet}$. But $U(f)$ is a fibration, so it is has the RLP wrt $U(j)$, and hence is a retract of $U(p)$ in $\mathbf{sSet}$. By the above fully faithfulness property of $U$ (restricted to the fibrant objects of $\mathbf{sSet}^+$), it follows that $f$ is a retract of $p$ in $\mathbf{sSet}^+$. Hence $f$ is a fibration in $\mathbf{sSet}^+$, since $p$ is.
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3
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https://mathoverflow.net/users/57405
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404540
| 165,909 |
https://mathoverflow.net/questions/254887
|
7
|
*In retrospect the original version of this question was impossibly bloated. Here's a better version:*
There are many results about when first-order sentences are preserved by algebraic operations on model classes; for example, in first-order logic the sentences preserved by taking substructures are those semantically equivalent to (= having the same models as) universal sentences.
I'm broadly interested in preservation results beyond first-order logic. In particular, I'd like to know if preservation under taking (arbitrary) Cartesian products nicely characterizable in *second-order logic* (preservation under taking substructures is actually simpler for $\mathsf{SOL}$ than $\mathsf{FOL}$ since the former can directly quantify over substructures).
To pose this precisely:
>
> Is there a computable set of $\mathsf{SOL}$-sentences $X$ such that the $\mathsf{SOL}$-sentences preserved by Cartesian products ("productive sentences") are exactly those semantically equivalent to elements of $X$?
>
>
>
Note that the productivity of a given $\mathsf{SOL}$-sentence is not set-theoretically absolute. However, neither is semantic equivalence between even individual $\mathsf{SOL}$-sentences, so this doesn't immediately give a negative answer to the question.
I'm also interested in what happens if we weaken "semantically equivalent to elements of $X$" to "semantically equivalent to possibly infinite conjunctions of elements of $X$."
|
https://mathoverflow.net/users/8133
|
Preservation results in abstract logics
|
EDIT: Now understanding "semantically equivalent", I change my answer (my previous answer, to a different question, is further below)...I'm working in ZFC.
Yes, there is such a computable set $X$; that is, $X$ is a set of productive second order sentences, and for every productive second order $\mathcal{L}$-sentence $\psi$, there is a semantically equivalent sentence $\varphi\in X$ (having the same models). Here we can take "productive" either to mean w.r.t finite products or set-sized products, and we fix countably many constant/function/relation symbols that we deal with.
Proof: The plan is to (i) give a recursive procedure to pass from sentences $\varphi$ to sentences $\psi\_\varphi$ such that $M\models\psi\_\varphi$ iff $M$ is a product $\Pi\_{i\in I}M\_i$ such that $I\neq\emptyset$ and each $M\_i\models\varphi$,
and (ii) then use $X=\{\psi\_\varphi\bigm|\varphi\text{ is an second order }\mathcal{L}\text{-sentence}\}$.
First let's consider the set-product version, so $I$ can be whatever set above.
Note that if we can do (i), then $X$ as in (ii) works: Each $\psi\_\varphi$
is easily productive, and if $\varphi$ is productive then $\varphi$ is semantically equivalent to $\psi\_\varphi\in X$. (If $M\models\varphi$ then $M\models\psi\_\varphi$ (take the product of only $M\_0=M$, with $I=\{0\}$), and if $M\models\psi\_\varphi$ then because $\varphi$ is productive, $M\models\varphi$.)
So it suffices to do (i). A slightly annoying thing to deal with here are empty models and models of cardinality 1, because they contribute less to the cardinality of a product. Let $M$ be the model in question. Then $\psi\_\varphi$ says: if $M\models\neg\varphi$ then $M$ is isomorphic to a product $\Pi\_{i\in I}M\_i\times\Pi\_{i\in J}M\_i$ where $I\cup J$ has cardinality $>1$ and each $M\_i\models\varphi$, and for each $i\in I$, $M\_i$ has cardinality 1, and for each $i\in J$, $M\_i$ has cardinality $\neq 1$. Note that we can now ignore the case that some $M\_i$ is empty, because this just makes the product empty (hence isomorphic to $M\_i$).
Note that we can identify (in SOL) whether $M$ is infinite.
*Case 1*: $M$ is infinite.
Note that in order to be a product as above, $M$ must have cardinality $\Pi\_{i\in J}\lambda\_i$ where $\lambda\_i=\mathrm{card}(M\_i)$. Since $M$ is infinite, and $\mathrm{card}(M\_i)>1$ for $i\in J$, we have $0<|J|<2^{|J|}\leq\mathrm{card}(M)$, so we can find a subset of $J'\subseteq M$ of cardinality $|J|$.
Similarly, each $\lambda\_i\leq\mathrm{card}(M)$, so we can find a subset $N'\_i\subseteq M$ with $\mathrm{card}(N'\_i)=\lambda\_i$, and a structure $M'\_i$ on $N'\_i$ such that $M'\_i\equiv M\_i$ (note these are all structures in the same finite language that $\varphi$ uses). Fixing bijections $\pi:J'\to J$ and $\sigma\_i:N'\_i\to M\_i$,
let $M'\_i$ be the structure on $N'\_i$ isomorphic to $M\_i$ via $\sigma\_i$.
We can find some (finitely many) relations describing these things; e.g.
we have a binary relation $R$ where $R(x,y)$ iff $x\in J'$ and $y\in N'\_{\pi(x)}$.
Suppose for the moment that $I=\emptyset$; we just want to assert that $M$
is isomorphic to the product $\Pi\_{i\in J}M\_i$. For this, we want to assert
(using SOL) that there is a binary relation $R$ (on the universe of $M$)
and some other relations $R^P$ etc for the symbols $P$ etc used in $\varphi$,
coding a set $J'$ and models $M'\_i$ as above, each $M'\_i$ models $\varphi$,
and $M$ is isomorphic to their product. So it just remains to assert the isomorphism. But this is just done by coding an isomorphism through another relation: Code an isomorphism
$\tau:M\to\Pi\_{i\in J}M\_i$ via the ternary relation $R\_\tau$ where
$R\_\tau(a,x,y)$ iff $a\in M$ and $x\in J'$ and $R(x,y)$ and $\tau(a)(\pi(x))=\sigma\_i(y)$.
(That is, $\tau(a)$ is an element of the product, i.e. a function
$\tau(a):J\to\bigcup\_{i\in J}M\_i$ with $\tau(a)(i)\in M\_i$ for each $i\in J$.
And $\pi:J'\to J$ is the bijection above. So $\tau(a)(\pi(x))=\tau(a)(i)\in M\_i$,
so this is of the form $\sigma\_i(y)$ for some $y\in N'\_i$.)
Thus, we just have to assert that there is a binary relation $R$ on $M$, and
various relations $R^{P\_1},\ldots,R^{P\_n}$ for the symbols $P\_1,\ldots,P\_n$ in $\varphi$,
such that $R,R^{P\_1},\ldots,R^{P\_n}$ code a collection of models, indexed
by $\mathrm{dom}(R)$, each modelling $\varphi$, and there is a ternary relation $R'$ on $M$, such that $R'$ codes an isomorphism between $M$ and the product
of the models just mentioned, as coded above.
This deals with the $I=\emptyset$ case. Now suppose $I\neq\emptyset$.
The slight irritation here is that these factors of the product don't add to the cardinality of $M$, so it might be that $I$ has cardinality bigger than $M$. However, note that this case is irrelevant, and in fact, we can assume that $I$ is finite. This is because there are only finitely many non-isomorphic models with 1 element and in the symbols used in $\varphi$, and the if $M\_i,M\_{i'}$ are isomorphic 1-element models, then note that their product is still isomorphic to the same model; likewise for infinite products of the same 1-element model. So we may assume $I$ is finite. That being the case, it is easy to modify the preceding paragraph to incorporate the full product $\Pi\_{i\in I}M\_i\times\Pi\_{i\in J}M\_i$.
*Case 2*: $M$ is finite.
We deal with $J$ just as before. For $I$, in case $M$ has too small cardinality,
we can just hard code all of the (finitely many) possible products
of 1-element models, in the language of $\varphi$, into a disjunction
of possibilities. That is, first compute a list $A\_1,\ldots,A\_k$ of all such products (up to isomorphism). Then when writing $\psi\_\varphi$, just say "There is $i\leq k$ such that $M$ is isomorphic to $A\_i\times\Pi\_{i\in J}M\_i$ (where $J,M\_i$ are as before)".
This completes the description of $\psi\_\varphi$, and note that $\varphi\mapsto\psi\_\varphi$ is recursive.
That was the case for arbitrary products. For finite products it is the same, except that we must also assert that $I,J$ are both finite. But since we have SOL, this is no problem.
---
REMARK: It's important above that "false" is productive, so that we can (and must) have sentences in $X$ which have no models. Suppose one modified the question and demanded that all the sentences in $X$ be true. Then (1) it is no longer possible. Similarly, (2) there is no computable set $X$ of true SOL sentences such that for each true SOL sentence $\varphi$, there is some $\psi\in X$ which is semantically equivalent to $\varphi$. For (2): Let $\delta$ be the least ordinal such that $V\_\delta\equiv\_{\Sigma\_2}V$. Then $V\_\delta$ is also the least set containing models of all consistent SOL sentences. But this means that using $X$, we can define $\delta$ in a $\Sigma\_2$ way, and since $X$ is computable, therefore $\{\delta\}$ is $\Sigma\_2$, which contradicts the definition of $\delta$. Note that this shows that in fact no such $X$ can be $\Sigma\_1$-definable. For (1) it is similar, because the true productive SOL sentences are still cofinal in the true SOL sentences. (Given $\varphi$ true productive, $\psi\_\varphi$ (as above) is "above" $\varphi$.)
---
Original answer (actually to another question): I think the answer to the first question is no, but I'm not totally sure what "semantically equivalent" should mean.
As a warm-up, first consider the version of the question for finite products (I'm not sure whether you meant finite or arbitrary products).
And avoiding the question of "semantically equivalent",
let's restrict attention to the second order language $\mathcal{L}=(0,1,+,\times)$ of arithmetic (without $<$), and show that the set $X$ of productive sentences of $\mathcal{L}$ is itself not computable.
(Also, I'm taking the product $P$ of two $\mathcal{L}$-models $M,N$
to have $c^P=(c^M,c^N)$ for constants $c=0,1$, and $+/\times$ are likewise defined component-wise. Likewise for arbitrary products.)
To show $X$ (the finite product version) is not computable, we just identify a computable sequence
$\left<\psi\_n\right>\_{n\in\mathbb{N}}$ of sentences
such that $\psi\_n$ is finitely productive iff $n\in 0$-jump.
Observe first that there is a finitely productive sentence $\psi$ of second order $\mathcal{L}$ which asserts "The model is a product of finitely many, but at least 1, copies of $\mathbb{N}$".
For this, say (i) The sub-model generated by $(0,1,+)$, extended with the restriction
of $\times$, is isomorphic to $\mathbb{N}$, and (ii) there is some $n\in\mathbb{N}$ and an isomorphism between the $n$-fold product of $\mathbb{N}$ and the full model.
Now for $n\in\mathbb{N}$ let $\psi\_n$ be the sentence $\psi$+"if the model is not isomorphic to $\mathbb{N}$ then $n\in 0$-jump",
where $0$-jump is defined using the copy of $\mathbb{N}$ found as in (i) above
(note that because it models $\psi$, we can indeed find a copy of $\mathbb{N}$ in this way to start with).
Then $\psi\_n$ is finitely productive iff $n\in 0$-jump: If $n\in 0$-jump
this is clear; if $n\notin 0$-jump then note that $\mathbb{N}\models\psi\_n$
but $\mathbb{N}^2\models\neg\psi\_n$ (it's easy to see that $\mathbb{N}$ and $\mathbb{N}^2$ are not isomorphic).
---
Edit: generalization for arbitrary products. The new thing here is to appropriately adapt $\psi$. But for this, I want to work instead
with the language $\mathcal{L}'=(0,1,+,\times,\leq)$.
I am taking products of relations to be defined using the "for all" quantifier, i.e. in our case, for the product $P$
of a sequence $\left<M\_i\right>\_{i\in I}$ of models,
$f\leq^Pg$ iff $f(i)\leq^{M\_i}g(i)$ for all $i\in I$.
I'll write $<^\*$ for the strict part of $\leq^P$ below
(note this need not be the same as the product of the strict parts of respective $\leq$s).
So we want a sentence $\psi$ which says "the model is the product
of set-many, but at least 1-many, copies of $\mathbb{N}$".
Let $P$ be the product of $I$-many copies of $\mathbb{N}$.
Given $j\in I$, let $f\_j\in P$ be the function $f\_j:I\to\mathbb{N}$
where $f\_j(j)=1$ and $f\_j(i)=0$ for $i\neq j$.
Note that the set $A=\{f\_j\bigm|j\in I\}$ is
definable over $P$: given $f\in P$, we have
$$ f\in A\iff [f\neq 0^P\text{ and for all }g\text{, if }g<^\*f
\text{ then }g=0^P],$$
where $<^\*$ is the strict part of $\leq^P$. Then from $A$,
we can uniformly identify the models $\mathbb{N}\_f$, for $f\in A$,
generated by $\{f\}$ by adjoining $0^P$ and closing under $+$, interpreting $1$ as $f$ and restricting $\times,\leq$ from $P$. And then $P$ is isomorphic to $\Pi\_{f\in A}\mathbb{N}\_f$.
So to assert that a given model $M$ (in the language above) is some product of set-many (at least 1) copies of $\mathbb{N}$, just define the set $A^M$ in the same manner, assert that every $f\in A^M$ generates (by adjoining $0$, closing under $+$, and interpreting $1$ as $f$) a model $\mathbb{N}\_f$ which together with the restriction of $\times^M,\leq^M$, is isomorphic to $\mathbb{N}$,
and $\mathbb{N}\_f\cap\mathbb{N}\_g=\{0^M\}$ when $f\neq g\in A$,
and that $M$ is isomorphic to the product $\Pi\_{f\in A}\mathbb{N}\_f$. To say the latter,
just say that there is a binary relation $R\subseteq M^2$ which codes an isomorphism $\pi:M\to\Pi\_{f\in A}\mathbb{N}\_f$, via $R(x,y)$ iff $y\in\mathrm{rg}(\pi(x))$.
It is easy enough to write down the properties that $R$ needs in order to code an isomorphism in this way.
This $\psi$ is (set-)productive, and has the right meaning. Now define $\psi\_n$ as before, and we get the same result as before.
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2
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https://mathoverflow.net/users/160347
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404543
| 165,910 |
https://mathoverflow.net/questions/404552
|
0
|
Let $s \in \mathbb{R}$ such that $0<s<1$. Consider the fractional Laplacian $(-\Delta)^s$ in the real line defined via [Fourier series](https://en.wikipedia.org/wiki/Fourier_series) as follows: if $f:[-\pi,\pi] \subset \mathbb{R} \longrightarrow \mathbb{C}$ is a periodic function and is written as
$$
f(x)=\sum\_{n \in \mathbb{Z}} f\_n e^{inx}
$$
then
$$
(-\Delta)^{s/2}f(x)=\sum\_{n \in \mathbb{Z}} |n|^{s} f\_n e^{inx}.
$$
**Question.** If we define $g: \mathbb{R} \longrightarrow \mathbb{R}$ by
$$
g(x):= |f(x)|,\; \forall \; x \in \mathbb{R}
$$
then is true that
$$
|(-\Delta)^{s/2}g(x)| \leq |(-\Delta)^{s/2}f(x)|? \tag{1}
$$
I didn't make any progress as I couldn't get any relation between the Fourier coefficients of $f$ and $g$ (it would be ideal if we had $g\_n=|f\_n|$ for each $n \in \mathbb{Z}$). There is some relation? The inequality in $(1)$ makes sense?
|
https://mathoverflow.net/users/156344
|
Inequality involving the fractional Laplacian
|
If true, this would follow from the integral expression for the fractional Laplacian:
$$(-\Delta)^{s/2} f(x) = \int\_{-\pi}^\pi (f(x) - f(y)) \nu(x - y) dy$$
for an appropriate kernel $\nu$. But, unfortunately, the claimed inequality is false: if, for example, $f$ is a non-zero odd function, then
$$(-\Delta)^{s/2} f(0) = 0$$
(by symmetry), while
$$(-\Delta)^{s/2} |f|(0) < 0$$
(by the integral expression given above, or by a version of the maximum principle).
---
On the positive side, we have the following inequality ("Markov property") for the corresponding quadratic (Dirichlet) forms:
$$ \langle |f|, (-\Delta)^{s/2} |f| \rangle \leqslant \langle f, (-\Delta)^{s/2} f \rangle . $$
|
3
|
https://mathoverflow.net/users/108637
|
404557
| 165,915 |
https://mathoverflow.net/questions/404547
|
0
|
Let $E$ be a separable $\mathbb R$-Banach space and $\lambda\_i$ be a finite symmetric measure on $\mathcal B(E)$ with $\lambda\_i(\{0\})=0$ and $$\int\_B1-\cos\langle x,x'\rangle\:\underbrace{(\lambda\_1-\lambda\_2)}\_{=:\:\sigma}({\rm d}x)=0\tag1$$ for all $B\in\mathcal B(E)$ and $x'\in E'$.
>
> How can we conclude that $\lambda\_1=\lambda\_2$?
>
>
>
The idea is the following: Let $$0\le f\_{x'}(x):=\sum\_{n\in\mathbb N}\frac{1-\cos\frac{\langle x,x'\rangle}n}{2^{n+1}}\le1\;\;\;\text{for }x\in E$$ for $x'\in E'$ and note that $$f\_{x'}(x)=0\Leftrightarrow\langle x,x'\rangle=0.\tag2$$ Let $(x\_n)\_{n\in\mathbb N}\subseteq E\setminus\{0\}$ be dense and $(x'\_n)\_{n\in\mathbb N}\subseteq E'$ with $\|x'\_n\|\_{E'}=1$ and $\langle x\_n,x'\_n\rangle=\left\|x\_n\right\|\_E$ for all $n\in\mathbb N$. Now let $$0\le g(x):=\sum\_{n\in\mathbb N}\frac{f\_{x\_n'}(x)}{2^n}\le1\;\;\;\text{for }x\in E$$ and note that $$g(x)=0\Leftrightarrow x=0\tag3.$$ By Lebesgue's dominated convergence theorem and $(1)$, $$\int\_Bg\:{\rm d}\sigma=0\tag4\;\;\;\text{for all }B\in\mathcal B(E).$$
>
> How can we conclude that $\sigma(B)=0$ for all $B\in\mathcal B(E)$ with $0\not\in B$?
>
>
>
|
https://mathoverflow.net/users/91890
|
If $\lambda_i$ is symmetric with $\lambda_i\{0\}=0$, why does $\int_B1-\cos\langle x,x'\rangle\:(λ_1-λ_2)({\rm d}x)=0$ imply $λ_1=λ_2$?
|
$\newcommand\B{\mathcal B}$Let $C:=E\setminus\{0\}$, so that $g>0$ on $C$. Let $\rho(B):=\sigma(B)$ for all $B\in\B(C)$, so that $\rho$ is a signed measure defined on $\B(C)$ such that
$$\int\_B g\,d\rho=0 \tag{1}$$
for all $B\in\B(C)$.
By the [Hahn decomposition theorem](https://en.wikipedia.org/wiki/Hahn_decomposition_theorem), $C$ can be partitioned into two sets, $P$ and $N$ in $\B(C)$ so that $P$ is a positive set for $\rho$ and $N$ is a negative set for $\rho$.
Substituting $P$ for $B$ in (1), we get $\rho(P)=0$. Similarly, $\rho(N)=0$.
So, $\rho=0$; that is, $\sigma(B)=0$ for all $B\in\B(E)$ with $0\notin B$, as desired.
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2
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https://mathoverflow.net/users/36721
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404563
| 165,916 |
https://mathoverflow.net/questions/404529
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4
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I am wondering how to prove the below Fourier transform is non-negative? I did much simulation and it seems to be non-negative.
$$\int\_0^\inf (be^{-at^p}-ae^{-bt^p})\cos(tx)dt, 0<a<b, \frac{1}{2}<p<1$$
|
https://mathoverflow.net/users/381188
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Fourier-positivity of a certain function
|
In the OP, it assumed that $1/2<p<1$. Let us first show that, actually, the desired conclusion holds for $p\in(0,1/2]$. Let
$$h(a):=\int\_0^\infty \frac{e^{-at^p}}a\,\cos(tx)\,dt.$$
As noted by Johannes Hahn, it suffices to show that $h$ is decreasing.
We have
$$-h'(a)=\int\_0^\infty g(t)\cos(tx)dt,$$
where
$$g(t):=\frac{e^{-a t^p} \left(a t^p+1\right)}{a^2},$$
and it is easy to see that $g$ is convex and decreasing to $0$ on $(0,\infty)$. So, by the Pólya criterion (see e.g. [page 2310](https://www.ams.org/journals/proc/2001-129-08/S0002-9939-01-05839-7/)), $-h'(a)\ge0$, so that $h$ is indeed decreasing.
---
Now for any $p\in(0,1]$, the above reasoning shows that it suffices to prove that the function $s\mapsto (1+|s|^p)e^{-|s|^p}$ is positive definite.
It appears that the function $0\le u\mapsto (1+u^{1/2})e^{-u^{1/2}}$ is completely monotone on and hence a mixture of exponential functions $0\le u\mapsto e^{-cu}$ with $c>0$. So, the function $s\mapsto (1+|s|^p)e^{-|s|^p}$ is a mixture of functions $s\mapsto e^{-c|s|^{2p}}$ with $c>0$, which are positive definite for any $p\in(0,1]$, as the characteristic functions of (symmetric) [stable\_distributions](https://en.wikipedia.org/wiki/Stable_distribution).
So, it remains to check that the function $0\le u\mapsto (1+u^{1/2})e^{-u^{1/2}}$ is completely monotone on $[0,\infty)$.
---
Actually, the function $0\le u\mapsto (1+u^{1/2})e^{-u^{1/2}}$ is the Laplace transform of the positive function $0<t\mapsto \dfrac{e^{-1/(4t)}}{4 \sqrt{\pi }\, t^{5/2}}$ and hence a mixture of exponential functions $0\le u\mapsto e^{-cu}$ with $c>0$; see details on this below. So, we are done.
---
*Proof of the statement on the Laplace transform:* Note that the second derivative of $(1+u^{1/2})e^{-u^{1/2}}$ in $u$ is $e^{-u^{1/2}}/(4u^{1/2})$.
So, after a simple rescaling, it is enough to show that
$$J(a):=\int\_0^\infty\exp\Big\{-\frac1t-at\Big\}\frac{dt}{2\sqrt t}=\frac{\sqrt\pi}2
\frac{e^{-2\sqrt a}}{\sqrt a}, \tag{1}$$
where $a>0$.
Using substitutions $t=u^2$ and then $u=1/(x\sqrt a)$, we get
$$J(a)=\int\_0^\infty\exp\Big\{-\frac1{u^2}-au^2\Big\}\,du
=K(a)/\sqrt a,$$
where
$$K(a):=\int\_0^\infty\exp\Big\{-ax^2-\frac1{x^2}\Big\}\,\frac{dx}{x^2}.$$
Note that $K'(a)=-J(a)$ and $K(a)=J(a)\sqrt a$. So, we get the differential equation
$$J'(a)=-\Big(\frac1{\sqrt a}+\frac1{2a}\Big)J(a),$$
whose general solution is given by
$$J(a)=\frac c{\sqrt a}\,e^{-2\sqrt a}$$
for a constant $c$.
To determine $c$, note that
$$K(a)=J(a)\sqrt a
=\int\_0^\infty\exp\Big\{-\frac1{u^2}-au^2\Big\}\,du\,\sqrt a
=\int\_0^\infty\exp\Big\{-\frac a{y^2}-y^2\Big\}\,dy$$
and
$$c=K(0+)=\int\_0^\infty\exp\{-y^2\}\,dy=\frac{\sqrt\pi}2.$$
So, (1) follows. $\quad\Box$
|
3
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https://mathoverflow.net/users/36721
|
404565
| 165,918 |
https://mathoverflow.net/questions/404523
|
4
|
Let $\Gamma$ be a torsion group (i.e. every element has finite order). I am interested in understanding central extensions of the form:
$\require{AMScd}$
\begin{CD}
0 @>>> \mathbb{R}^n @>\exp>> G @>\pi>> \Gamma @>>> 1\\
\end{CD}
Equivalently, I want examples of groups $\Gamma$ with non-trivial classes in $H^2(\Gamma,\mathbb{R}^n)$. When $\Gamma$ is finite I'm aware that $H^2(\Gamma,\mathbb{R}^n) = 0$ but I have little intuition or examples for infinite torsion groups.
Interestingly, any central extension such a $\Gamma$ by $\mathbb{R}^n$ has a canonical set-theoretic splitting $\sigma \colon \Gamma \to G$ with the property that:
$$ \sigma(\gamma) = g \quad \Leftrightarrow \quad \pi(g) = \gamma \text{ and } g \text{ has finite order}$$
where $\pi \colon G \to \Gamma$ is the projection. It is not too hard to show that if any group-theoretic splitting exists then it must be $\sigma$.
Using usual group cohomology arguments, this gives rise to a cocycle:
$$ \alpha \colon G \times G \to \mathbb{R}^n $$
I've managed to prove a few interesting properties of $\alpha$ (for example, $\alpha$ must be symmetric) but have not been able to show it is zero and I suspect a counterexample probably exists.
|
https://mathoverflow.net/users/381262
|
Central extensions of torsion groups by $\mathbb{R}^n$
|
The paper S. I. Adyan and V. S. Atabekyan, V. S.
*Central extensions of free periodic groups*,
Mat. Sb. 209 (2018), no. 12, 3–16; translation in
Sb. Math. 209 (2018), no. 12, 1677–1689
proves that if $n\geq 665$ is odd and $m\geq 2$, then the Schur multiplier $H^2(B(m,n),\mathbb Z)$ for the free Burnside group $B(m,n)$ of exponent $n$ on $m$-generators is free abelian of countable rank. In the [arXiv version](https://arxiv.org/abs/1811.07167)
this is Corollary 3 and so using the universal coefficients theorem, $H^2(B(m,n),\mathbb R^n)$ is quite huge. If you want an explicit class, then that is beyond my competency range.
|
3
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https://mathoverflow.net/users/15934
|
404591
| 165,925 |
https://mathoverflow.net/questions/400632
|
12
|
*(Below I'm thinking only about computably axiomatizable set theories extending $\mathsf{ZFC}$ which are arithmetically, or at least $\Sigma^0\_1$-, sound.)*
Say that a theory $T$ is omniscient iff $T$ **proves that** the following holds:
>
> For every formula $\varphi(x,y)$ there is some formula $\psi(z,y)$ such that $T$ proves: "For all $y$, if $\varphi(-,y)^V=\varphi(-,y)^{V[G]}$ for every set generic extension $V[G]$, then $\psi(-,y)$ is a truth predicate for $\varphi(-,y)^V$."
>
>
>
*(That's not a typo, I do want a "$\vdash\vdash$"-situation. Note that by the soundness assumption on $T$, there really does exist such a $\psi$ for every $\varphi$.)*
Two points about omniscience are worth noting:
* For each $\varphi,\psi$, the statement in quotes is indeed expressible in the language of set theory - in particular, although we can't talk about the full theory of a generic extension via a single sentence, for each $\varphi,\psi$ we only need to talk about a bounded amount of those theories. So it does in fact make sense.
* $\mathsf{ZFC}$ itself is not omniscient - since $\mathsf{ZFC}$ proves that $L$ is forcing-invariant, any omniscient theory must prove that $V$ is not a set forcing extension of $L$. Informally speaking, any omniscient theory must prove that $V$ is "much bigger" than any canonical inner model.
Omniscience strikes me as an implausibly strong property. However, I don't see an immediate reason why no consistent omniscient theory can exist. So my question is:
>
> **Is there a consistent omniscient theory at all?**
>
>
>
|
https://mathoverflow.net/users/8133
|
Is this definability principle consistent?
|
There is a consistent omniscient theory, at least assuming the consistency of a Woodin limit of Woodin cardinals.
The Maximality Principle (MP) asserts that if a sentence is forceable in $V$, it is forceable in every generic extension of $V$. In other words, if a sentence can be forced to be indestructible by set forcing, then the sentence was true all along. Variants of the principle were discovered independently by many people, including Stavi, Väänänen, Bagaria, Chalons, and Hamkins. The main reference is [Hamkins's paper](https://arxiv.org/abs/math/0009240). The Boldface MP (due to Hamkins) asserts the same but allowing hereditarily countable parameters. The Necessary MP (NMP, also due to Hamkins) asserts that the Boldface MP is true in all forcing extensions. MP and BMP are fairly weak, but Woodin [unpublished] showed the consistency strength of NMP lies between $\text{AD}$ and $\text{AD}\_\mathbb R + \Theta \text{ is regular}$. I claim ZFC + NMP is an omniscient theory. Since $\text{AD}\_\mathbb R + \Theta\text{ is regular}$ is consistent (by a theorem of Sargsyan its consistency follows from a Woodin limit of Woodin cardinals), so is ZFC + NMP.
The proof is essentially due to Hamkins, who showed that under NMP, if $W$ is a forcing invariant inner model, then for all cardinals $\lambda$, $H(\lambda)\cap W\preceq W$. His proof adapts to the case that $W$ is an *arbitrary* class with a forcing invariant definition using a parameter $x$, although of course one must assume $\lambda$ is above the hereditary cardinality of $x$. This implies that the truth predicate for $W$ is definable: $W\vDash \varphi(\overline p)$ if and only if $W\cap H(\lambda) \vDash \varphi(\overline p)$ for some/all $\lambda > |\text{tc}(p)|$.
I recite Hamkins's proof below since it's nice and I wanted to check that it works, but there are really no new ideas.
We want to show that the truth predicate for $(W,\in)$ is definable from $x$. By homogeneity, it suffices to show it is definable from $x$ over $V[G]$ where $G$ is generic for $\text{Col}(\omega,\text{tc}(x))$. We may therefore pass to $V[G]$ and assume without loss of generality that $x$ is hereditarily countable. (This is ok because $V[G]$ is also a model of NMP.)
Now fix a cardinal $\lambda$, and I will show $W\cap H(\lambda) \preceq W$. By induction on the complexity of $\varphi$, I'll show that $\varphi$ is absolute between $W\cap H(\lambda)$ and $W$. The only nontrivial step is to show that if $\varphi(\overline u) \equiv \exists t\, \psi(t,\overline u)$ and $\psi$ is absolute between $W\cap H(\lambda)$ and $W$, then $\varphi$ is absolute as well. So fix $\overline p\in W\cap H(\lambda)$ and suppose $W\vDash \varphi(\overline p)$.
Let $H$ be generic for $\text{Col}(\text{tc}(\overline p))$. In $V[H]$, one can force so that the minimum hereditary cardinality of a set $z\in W$ such that $W\vDash \psi(z,\overline p)$ is at most $\aleph\_0$. Once this is true, it is of course true in any outer model, so applying the Boldface MP in $V[H]$, $V[H]$ satisfies that the minimum hereditary cardinality of a set $z\in W$ such that $W\vDash \psi(z,\overline p)$ is at most $\aleph\_0$. Fix such a set $z\in W$. Then $z\in V$ (since $W\subseteq V$) and the hereditary cardinality of $z$ is at $|\text{tc}(\overline p)| < \lambda$. Thus there is some $z\in W\cap H(\lambda)$ such that $W\vDash \psi(z,\overline p)$, and so by our induction hypothesis, $W\cap H(\lambda)\vDash \psi(z,\overline p)$, and hence $W\cap H(\lambda)\vDash \varphi(\overline p)$ as desired.
|
8
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https://mathoverflow.net/users/102684
|
404594
| 165,927 |
https://mathoverflow.net/questions/404532
|
4
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Suppose that $f:\Bbb R^2\to\Bbb R$ is a *continuous non-linearity* and consider the following semi-linear elliptic PDE given by:
$$-\Delta u=f(x,u),\;\;x\in\Omega\subset\Bbb R^n,\tag{1}\label{1}$$
To avoid the mention of critical Sobolev exponents and to narrow down the scope of the answer, let us assume that $n=2$ and $\Omega$ is bounded. Assume that $u\in H^1\_0(\Omega)$ is a weak solution of \eqref{1}.
---
***Q1.*** *Under what assumptions on $f$ can we show that* (i) $u\in H^2(\Omega)$, (ii) *recover a classical solution of \eqref{1}*?
---
**Remark 1**. There are numerous conditions on $f$ that guarantee the existence of $u$. For $n=2$, see [this](https://mathoverflow.net/questions/391851/reference-request-for-semilinear-pdes-in-dimension-2/391865?noredirect=1#comment1000212_391865) question. For $n\geq 3$, see the the book [Semi-linear elliptic equations for beginners](https://www.springer.com/gp/book/9780857292261) by Badiale and Serra.
**Remark 2.** In Evan's [Partial Differential Equtions](http://home.ustc.edu.cn/%7Exushijie/pdf/textbooks/pde-evans.pdf) book, Chpater 6 exercise 4, if $f(x,u)=g(x)+h(u)$, where $g\in L^2(\mathbb{R}^n)$, $h(0)=0$, $h$ is smooth, and $h'\geq 0$, then a solution $u\in H^1(\Bbb R^n)$ to \eqref{1} is in fact in $H^2(\Bbb R^n)$.
**Remark 3.** From [this](https://people.kth.se/%7Ehenriksh/Henriks_page/research-papers/paper37.pdf) paper, where $\Omega$ is the unit ball in $\Bbb R^n$ with $n\geq 2$, and $f$ is Lipschitz with $\partial\_uf\geq -M$ for some $M\geq 0$, I quote the following sentence in the introduction:
>
> We remark that solutions to \eqref{1} are already in $W^{2,p}(\Omega)\cap C^{1,\alpha}(\Omega)$ for $p<\infty$ and $\alpha<1$ (since $f$ is bounded).
>
>
>
I think I can workout a proof as to why $u\in H^2(\Omega)$, but not $u\in C^{1,\alpha}(\Omega)$ for $\alpha<1$, which brings us to the second question.
---
***Question 2.*** *How does one obtain $u\in C^{1,\alpha}(\Omega)$*?
---
**Remark 4.** In the answer, assume *any* needed regularity on $\Omega$ to obtain regularity of $u$.
|
https://mathoverflow.net/users/105925
|
Regularity of weak solutions to semi-linear elliptic PDEs
|
There is a "standard" bootstrap argument which can be used to show regularity for semilinear equations. I sketch it here under the assumption that $|f(x, u)| \leq C(1 + |u|^p)$ for some $0 < p < \frac{2n}{n-2}$ in $n \geq 2$ (I know the question was posed in $n = 2$ but it is useful to see the role played by the critical exponent).
Assume we know that $u$ is in $L^q$ for some $p < q < \infty$, and that $\Omega$ is of class $C^{1,1}$. Then $\Delta u \in L^{q/p}$, and so by the Calderon-Zygmund theorem $u \in W^{2, q/p}$. The Calderon-Zygmund theorem is not always stated in this form (i.e. on domains), but this version may be found in Gilbarg-Trudinger (Section 9.6). Applying the Sobolev embedding gives that $u \in L^{q'}$, where $q' = \frac{nq}{np - 2q}$ (or $\infty$ if $2q > np$).
Notice that $q' > q$ if and only if $q > \frac{n}{2}(p - 1)$, and moreover $q'/q$ is an increasing function of $q$. This means that as long as we start with $u \in L^q$ for $q > \frac{n}{2}(p - 1)$, we may apply this procedure repeatedly to get that $u \in L^{q\_\*}$ for any $q\_\* < \infty$, with the number of steps needed always finite and depending only on $q, q\_\*$. For a sufficiently large value of $q\_\*$, $u \in W^{2, q\_\*}$ implies $\nabla u \in C^{0, \alpha}$ for $ \alpha \in (0, 1)$, from Sobolev embeddings.
The point, then, is to check that $u \in L^q$ for $q > \frac{n}{2}(p - 1)$. This is not guaranteed by the equation in general, but it does help that we know that $u \in W^{1, 2}$. This embeds into $L^{\frac{2n}{n-2}}$, so if $n = 2$ and $p$ is anything or $n > 2$ and $p < \frac{n + 2}{n - 2}$ we succeed.
This kind of argument is "sharp" in the sense that there are examples of solutions to semilinear equations which are not bounded. The most standard is $u(x) = |x|^{-\frac{2}{p - 1}}$ being a distributional solution of $-\Delta u = au^p$ for some $a \in \mathbb{R}$: this $u$ is not in $L^q$ for $ q \geq \frac{n}{2}(p - 1)$, and it is not in $W^{1, 2}$ unless $p > \frac{n + 2}{n - 2}$ (and $n \geq 3$). It is not sharp in other ways: it is not sharp at any of the endpoint cases, and the dependence on $x$ in the assumptions on $f(x, u)$ here is not sharp. Therefore if the question is, in $n = 2$ what are the optimal assumptions on $f$ to ensure regularity of $W^{1, 2}$ solutions, this does not give a complete answer.
|
4
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https://mathoverflow.net/users/378654
|
404596
| 165,929 |
https://mathoverflow.net/questions/404037
|
2
|
In a non weighted graph, the adjacency matrix ($A$) raised to the power $k$ will return the number of k-step paths between nodes $i$ and $j$ at the entry $a\_{ij}$. Is there an equivalent for weighted graphs? I.e. to obtain the *sum of the cumulative weights of all paths* between node pairs?
Applying the same approach as above, but to a weighted adjacency matrix, i.e. raising to the power $k$, returns the *sum of the product of the weights along each $k$ step path* between node pairs. This is useful when edge weight represents some probability (e.g. Markov models), however not when edge weight represents a length (e.g. geospatial network).
**Note** I want to avoid path search algorithms (e.g. Dijkstra's or Yen's) if possible, as this will be applied to very large networks so efficiency is important.
**Example**: Given a 4-node digraph (image linked below), with the weighted adjacency matrix (infinity represents no connection between nodes):
[Digraph described in the example](https://i.stack.imgur.com/fhxz1.png)
$$A =
\begin{matrix}
\infty & 2 & 3 & \infty \\
2 & \infty & 5 & 1 \\
\infty & \infty & \infty & 7 \\
\infty & 1 & \infty & \infty \\
\end{matrix}
$$
where e.g. the connection between nodes B & C is of length 5. I want to find the n-step length matrix, which for the 2 step case would be:
$$A\_2 =
\begin{matrix}
4 &\infty & 7 & 13 \\
\infty & 6 & 5 & 12 \\
\infty & 8 & \infty & \infty \\
3 & \infty & 6 & 2 \\
\end{matrix}
$$
Thanks in advance for any help.
|
https://mathoverflow.net/users/373251
|
Solution to the sum of k-step path lengths between node pairs in directed weighted graphs
|
Let $A^k\_{i,j} = (n^k\_{i,j}, s^k\_{i,j})$ where $n^k\_{i,j}$ is the number of paths of length $k$ between $i$ and $j$, and $s^k\_{i,j}$ is the sum of the lengths between these paths.
$A^1\_{i,j} = (1, w\_{i,j})$ if $ij$ is an edge, else $A^1\_{i,j}=(0,0)$
The idea of the recurrence is that a $(k+k')$-step path from $i$ to $j$ is composed of a $k$-step path from $i$ to an intermediate vertex $c$ followed by a $k'$-step path from $c$ to $j$.
Let fix $c$. Let $n\_1$ (resp $n\_2$) be the number of $k$-step paths from $i$ to $c$ (resp. $k'$-step paths from $c$ to $j$) and $s\_1$ (resp. $s\_2$) the sum of weigths on these paths.
We can easily see that the number of $(k+k')$-step paths from $i$ to $j$ passing through $c$ at step $k$ is $n\_1n\_2$. The sum of weights is $s\_1n\_2 + s\_2n\_1$ because each path from $i$ to $c$ will appear in exactly $n\_2$ paths, and conversely. To get the total number of $(k+k')$-step paths from $i$ to $j$, we just need to sum over the $c$. So the recurrence is given by :
$$A^{k+k'}\_{i,j} = (n^{k+k'}\_{i,j}, s^{k+k'}\_{i,j}) = \sum\_{c\in V}(n^k\_{i,c}n^{k'}\_{c,j}, n^k\_{i,c}s^{k'}\_{c,j}+s^k\_{i,c}n^{k'}\_{c,j})$$
We can adapt the definition of the matrix multiplication by replacing the multiplication by :
$(n\_1,s\_1)\otimes(n\_2,s\_2) = (n\_1n\_2, n\_1s\_2 + n\_2s\_1)$
and addition by :
$(n\_1,s\_1)\oplus(n\_2,s\_2) = (n\_1+n\_2, s\_1+s\_2)$
With it, we have :
$$A^{k+k'}\_{i,j} = \bigoplus\_{c\in V} A^{k}\_{i,c}\otimes A^{k'}\_{c,j}$$
and so $A^kA^{k'} = A^{k+k'}$
The new matrix multiplication is associative, so matrix exponentiation by squaring still work.
---
This kind of redefinition of the matrix multiplication is common when dealing with shortest paths, see the [min-plus matrix multiplication](https://en.wikipedia.org/wiki/Min-plus_matrix_multiplication)
|
1
|
https://mathoverflow.net/users/381833
|
404611
| 165,932 |
https://mathoverflow.net/questions/404573
|
2
|
Let $\mathcal{G} = G\_1 \rightrightarrows G\_0$ be a Lie Groupoid (although I am also interested in groupoids internal to other sites), the stack associated to $\mathcal{G}$, which is sometimes denoted $B \mathcal{G}$ is defined as the projection from the category of principal $\mathcal{G}$-bundles to $\mathsf{Man}$. However it is also well known that we can use the functor $r\_1: \mathsf{LieGpd} \hookrightarrow \mathsf{Pre}(\mathsf{Man}, \mathsf{Gpd})$, defined as
$$M \mapsto \left( C^\infty(M, G\_1) \rightrightarrows C^\infty(M, G\_0) \right)$$
and that the stackification of this presheaf of groupoids is equivalent to $B \mathcal{G}$.
However as Chris Schommer-Pries pointed out, $r\_1(G \rightrightarrows \*)$ is not a stack on $\mathsf{Man}$. It is however a stack on $\mathsf{Cart}$ the site of cartesian spaces. This can be seen by showing that if we let $\widehat{r}G = r\_1(G \rightrightarrows \*)$, then for $U \in \mathsf{Cart}$ equipped with a differentiably good open cover $\{ U\_i \to U \}$, the canonical map
$$\widehat{r}G(U) \to \text{holim} \left( \prod\_i \widehat{r}G(U\_i) \rightrightarrows \prod\_{i,j} \widehat{r}G(U\_{ij}) \mathrel{\substack{\textstyle\rightarrow\\[-0.6ex]
\textstyle\rightarrow \\[-0.6ex]
\textstyle\rightarrow}} \prod\_{i,j,k} \widehat{r}G(U\_{ijk}) \right)$$
is an equivalence of groupoids. One can see this by using a specific model of $\text{holim}$ from Hollander's Homotopy Theory of Stacks Corollary 2.11, and then recognizing the right hand side as the groupoid of Principal $G$-bundles on $U$. Since $U$ is a Cartesian space, this groupoid will be connected and this will prove that the map is an equivalence.
My question then is the following:
Is there a necessary and sufficient condition that one could put on a Lie groupoid $\mathcal{G}$ such that $r\_1(\mathcal{G})$ is already a stack on $\mathsf{Cart}$? I suspect that it should be any Lie groupoid, we would need to follow the same logic as above using principal groupoid bundles, i.e. building principal groupoid bundles from local data. Perhaps this has been done somewhere?
|
https://mathoverflow.net/users/124010
|
Necessary and sufficient conditions for a Lie groupoid to present a stack
|
Note that $\hat r\mathcal{G}$ is always a prestack. This is basically equivalent to the fact that $C^\infty(-,G\_1)$ is a sheaf, and it means that your functor
$$
\hat r\mathcal{G}(U) \to \mathrm{holim}(...)
$$
can only fail to be essentially surjective.
Essential surjectivity of this functor is equivalent to saying that every principal $\mathcal{G}$-bundle over a Cartesian space is trivializable.
A sufficient condition to garantee this is that your Lie groupoid $\mathcal{G}$ is an action groupoid (for an action of a Lie group $H$ on a smooth manifold $X$). In particular, this is true when your Lie groupoid is a Lie 2-group.
Namely, principal bundles for an action groupoid $X/\!/H$ are at the same time ordinary principal $H$-bundles (with additional structure). Any ordinary principal bundle has a global section over any Cartesian space. But a section is already enough to trivialize the $X/\!/H$-bundle.
You find these statements in Section 2.2. of:
*Nikolaus, Thomas; Waldorf, Konrad*, [**Four equivalent versions of nonabelian gerbes**](http://dx.doi.org/10.2140/pjm.2013.264.355), Pac. J. Math. 264, No. 2, 355-420 (2013). [ZBL1286.55006](https://zbmath.org/?q=an:1286.55006).
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2
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https://mathoverflow.net/users/3473
|
404613
| 165,934 |
https://mathoverflow.net/questions/404616
|
6
|
This problem is derived from [this post](https://mathoverflow.net/questions/403797/a-variation-of-the-ryll-nardzewski-fixed-point-theorem).
Let $G$ be a countable discrete group and $H\le G$ be a subgroup. Consider the $G$-action on $X=G/H$. Then the following amenability-like conditions are equivalent.
(i) Every $G$-invariant weak\*-compact convex subset $K$ of $\ell\_\infty(X)$ contains a constant function.
(ii) There is a sequence $\mu\_n$ in $\mathrm{Prob}(G)$ such that $\| \mu\_n x - \mu\_n y \|\_{\ell\_1(X)}\to0$ for every $x,y\in X$.
As in the case of other amenability-like conditions, one can easily give more equivalent conditions. However, none seems easy to check.
Is there an equivalent condition that is easy to check?
I would be happy if the criterion applies to any non-trivial case (if any...).
|
https://mathoverflow.net/users/7591
|
Trans-amenability of group actions
|
This is precisely what was called **$L$-amenability** by Kaimanovich and **lamenability** by Bartholdi (who were inspired by [Infinitely supported Liouville measures of Schreier graphs](https://mathscinet.ams.org/mathscinet-getitem?mr=3844999) of Juschenko and Zheng; "L" here stands for Liouville). In a sense, the aforementioned paper also provides a non-trivial example you are asking about. As you say, most of the usual definitions of group amenability can be adapted to this situation as well; in particular, for $\mu\_n$ one can take the sequence of convolution powers of a single measure, and one can reformulate condition (ii) in "Følner" terms. However, I don't think there is an "easy to check equivalent condition" - what is an easy to check condition for the usual group amenability?
|
12
|
https://mathoverflow.net/users/8588
|
404624
| 165,936 |
https://mathoverflow.net/questions/404384
|
0
|
For any set $X$, we let $[X]^2 = \big\{\{x,y\}:x\neq y \in X\big\}$.
If $G=(V,E)$ is a simple, undirected graph, and $v\in V$, let $N(v) = \{z\in V: \{v,z\}\in E\}$. Given any $v\in V$, we use the following notation:
1. Let $G\setminus\{v\} := (V\setminus \{v\}, E \cap [V\setminus\{v\}]^2)$, and
2. if $w\in V$ with $\{v,w\}\in E$, let $(G\setminus\{v\})^w := (V\setminus\{v\}, E\_{v,w})$ where $E\_{v,w}:= (E \cap [V\setminus\{v\}]^2) \cup \{\{w,z\}: z\in N(v)\}$.
A graph $G=(V,E)$ is said to be *vertex-critical* if $\chi(G\setminus\{v\}) < \chi(G)$ for every $v\in V$. We say $G$ is *collapse-critical* if $\chi\big((G\setminus\{v\})^w\big) < \chi(G)$ for all $v\in V$ and $w\in V$ such that $\{v,w\}\in E$. It is clear that $\chi(G\setminus\{v\})\leq \chi\big((G\setminus\{v\})^w\big)$, so collapse-criticality implies vertex-criticality.
**Question.** What is an example of a finite, connected graph that is vertex-critical, but not collapse-critical?
|
https://mathoverflow.net/users/8628
|
Two kinds of vertex-criticality
|
Unless I'm mistaken, your definition of $(G\setminus \{v\})^w$ is symmetrical, and generally referred as an edge-contraction, so collapse critical is the same as edge-contraction critical.
[this graph](https://math.stackexchange.com/questions/2430688/vertex-critical-graph-with-at-least-one-non-critical-edge) is vertex-critical, but not collapse-critical.
[duplicate](https://math.stackexchange.com/questions/2580899/distinguishing-vertex-edge-and-edge-contraction-critical-graphs)
|
1
|
https://mathoverflow.net/users/381833
|
404627
| 165,937 |
https://mathoverflow.net/questions/404587
|
1
|
Let $X\_0/\mathbb F\_q$ be a variety, and let $\mathcal F$ be a Weil sheaf on $X := (X\_0)\_{\overline{\mathbb{F}\_q}}$ that is pure of weight $n$. If $j < n$, does the weight $j$ piece of $H^i\_c(X,\mathcal F)$ necessarily vanish for all $i$?
I thought this was true, but I have made some computations that seem to contradict it.
|
https://mathoverflow.net/users/382874
|
Lowest weight of compactly supported cohomology with coefficients
|
This is not true.
If you take an open curve $U \subset C$ and a pure sheaf $\mathcal F$ on $U$ of weight $w$, with $j$ the open immersion $U \to C$ and $i$ the complementary closed immersion, then the exact sequence $$ 0 \to j\_! \mathcal F \to j\_\* \mathcal F \to i\_\* i^\* j\_\* \mathcal F \to 0$$ induces a long exact sequence on cohoomology
$$ H^\*\_c( U, \mathcal F) \to H^\* (C, j\_\* \mathcal F) \to H^\* (C- U, i^\* j\_\* \mathcal F ) $$
The first term is what we want to compute and the middle term is pure of weight $w+d$ in degree $d$ by Deligne. So low weight cohomology can only come from the third term.
However, come from the third term it does when $\mathcal F$ has unipotent local monodromy. For each Jordan block of size $n$ in the local monodromy around a point of $C - U$, $i^\* j\_\* \mathcal F$ has a single Frobenius eigenvalue of weight $w+ 1-n$, and that eigenvalue will show up in cohomology as $C-U$ is finite and thus cohomology is simply taking the sums of the stalks at its points.
The sheaves on $M\_{1,1}$ you get from $M\_{1,n}$ can be constructed from symmetric powers of the Tate module of the universal elliptic curve. Because the universal elliptic curve has semistable reduction at $\infty$, the Tate module has unipotent local monodromy there, and its symmetric powers do as well, explaining why the phenomenon occurs in this case.
|
3
|
https://mathoverflow.net/users/18060
|
404631
| 165,938 |
https://mathoverflow.net/questions/404546
|
40
|
As the "second-generation" proof of the Classification of Finite Simple Groups is being written up in the volumes by Gorenstein, Lyons, Aschbacher, Smith, Solomon, and others (see e.g. [this question](https://mathoverflow.net/questions/114943/where-are-the-second-and-third-generation-proofs-of-the-classification-of-fin)) certainly much of the work will go into fixing minor (and possibly major) issues and gaps in the proof, since the first announcement in 1983.
Here are two such gaps:
1. The classification of quasithin groups. G. Mason claimed a proof in an unpublished manuscript in 1981, but this was found to contain serious gaps. It would not be until 2004 that this gap would be fixed (see [this behemoth](http://homepages.math.uic.edu/%7Esmiths/papers/quasithin/quasithin.pdf), [1,2]).
2. In 2008, Harada and Solomon [3] filled a minor gap in the classification by describing groups with a standard component that is a cover of the Mathieu group $M\_{22}$, a case that was accidentally omitted from the proof of the classification due to an error in the calculation of the Schur multiplier of $M\_{22}$ (from the [Wikipedia page for CFSG](https://en.wikipedia.org/wiki/Classification_of_finite_simple_groups)).
I would like to see a longer such list! Thus:
**What other (major or minor) gaps have been discovered, and subsequently fixed, in the proof of the CFSG, since the announcement in 1983?**
Of course, if there are any gaps that are "known, but with a known fix" (but which have not yet made it into the aforementioned second-generation proof), then these would also be interesting to know.
${}$
References:
[1] *Aschbacher, Michael; Smith, Stephen D.*, The classification of quasithin groups. I: Structure of strongly quasithin $\mathcal K$-groups., Mathematical Surveys and Monographs 111. Providence, RI: American Mathematical Society (AMS) (ISBN 0-8218-3410-X/hbk). xiv, 477 p. (2004). [ZBL1065.20023](https://zbmath.org/?q=an:1065.20023).]
[2] *Aschbacher, Michael; Smith, Stephen D.*, The classification of quasithin groups. II: Main theorems: the classification of simple QTKE-groups., Mathematical Surveys and Monographs 112. Providence, RI: American Mathematical Society (AMS) (ISBN 0-8218-3411-8/hbk). xii, pp. 479-1221. (2004). [ZBL1065.20024](https://zbmath.org/?q=an:1065.20024).
[3] *Harada, Koichiro; Solomon, Ronald*, [**Finite groups having a standard component (L) of type $\widehat M\_{12}$ or $\widehat M\_{22}$.**](http://dx.doi.org/10.1016/j.jalgebra.2006.09.034), J. Algebra 319, No. 2, 621-628 (2008). [ZBL1135.20009](https://zbmath.org/?q=an:1135.20009).
|
https://mathoverflow.net/users/120914
|
Known and fixed gaps in the proof of the CFSG
|
Here is an answer from my point of view, immersed as I am -- Geoff
is right -- in the second generation project. First, a few general
comments. Our overriding purpose has been to expound a coherent proof
of CFSG that is supported completely by what we call ``Background
Results,'' an explicit and restricted list of published books and
papers, plus the assertion that every one of the $26$ sporadic groups
is determined up to isomorphism, as a finite simple group, by its
so-called centralizer-of-involution pattern. This list has changed
over the years. In our first volume it is explicitly listed as we
conceived it at the time (1990's). Further additions, mostly of
post-first-generation publications, are noted as they have been
adopted in subsequent volumes. (Some of these additions are
characterizations of some sporadic groups--for example, the Monster
and Baby Monster--by weaker data than centralizer-of-involution
pattern, so that they supplant the earlier Background Results
characterizing those groups.) The biggest additions, by far, are
Aschbacher and Smith's monumental books on the quasi-thin problem,
since we were hardly going to do it as well ourselves, let alone
better. Whatever errors may be in the second-generation proof,
therefore, are either in the Background Results or in our series.
Naturally, we have taken ideas and arguments from many papers and
books outside the Background Results in formulating our
proof. Occasionally in the course of understanding these results, or
adapting them for our purposes, we have uncovered gaps. None of these
is at all comparable in scope (by orders of magnitude) to the
well-known quasi-thin gap that Aschbacher and Smith bridged; in that
sense, they could be called ``minor.'' To deal with these gaps, when
they threatened our proof, we have either found alternative arguments
ourselves, or asked the authors for help. In every case, so far, the
gap has been closed in one of these two ways. However, and
unfortunately for the purposes of answering your question, we have not
kept a log of these incidents. Nor have we by any means intended to
examine every paper needed in the first-generation proof this way. We
are guided just by what we need in the second generation.
Here is an example of a minor gap that came to our attention in the
preparation of volume $9$. We needed a certain characterization of the
$7$- and $8$-dimensional orthogonal groups over the field of $3$
elements. We were guided by an important paper by Aschbacher that had
appeared relatively late and without much fanfare in the first
generation. There was an apparent gap -- very technical -- in the
paper, and Professor Aschbacher promptly supplied us with a
correction.
Another example that I know well, from before the CFSG, came in $1972$
in my paper pointing to the possible existence of the sporadic group
$Ly$. I asserted that if such a group existed, then every nonidentity
element of order a power of $5$ actually would have order
$5$. Koichiro Harada wrote me shortly thereafter that on the contrary,
there would be elements of order $25$. He was
right; I had miscalculated. Luckily, the miscalculation did not
affect the rest of the paper.
|
39
|
https://mathoverflow.net/users/99221
|
404644
| 165,941 |
https://mathoverflow.net/questions/404520
|
5
|
I am interested in proving an upper bound (expressed as a power of $N$, with $N\rightarrow\infty$ ) for the number of elements of the set
$$
A\_N=\{(k,l,m,n)\in([N,2N]\cap\mathbb{Z})^4: |k^2+l^2-m^2-n^2|\le|k+l-m-n|\}.
$$
My intuition is that the inequality defining this set cannot hold too often unless the quadruple exhibits some form of diagonal behavior, which would suggest a bound of the form $N^{2+\epsilon}$, with arbitrarily small $\epsilon>0$.
I can't quite formalize this idea, though.
|
https://mathoverflow.net/users/157356
|
A bound for the number of integer solutions to a simple inequality
|
$|A\_N|$ has order of magnitude $N^3$:
We use the change of variables $x=k-m$, $y=k+m$, $u=n-l$, $v=n+l$, so that the inequality becomes $|xy-uv|\le |x-u|$. If $x=0$ or $u=0$, there are clearly $\ll N^3$ solutions.
Given $x\ge u \ge 1$ and $y$, and assuming $xy\ge uv$, the
variable $v$ must satisfy
$$
v \in [2N+u,4N-u] \cap \left[ \frac{xy}{u}-\frac{x-u}{u}, \frac{xy}{u}\right].
$$
The other cases are similar.
For this intersection to be non-empty, we must have $x/u \asymp 1$, since $y \asymp N$. This implies that the width of the second interval is $O(1)$.
Thus there are $O(1)$ choices for $v$, which means that the number of solutions is $O(N^3)$.
To show that there are actually $\gg N^3$ solutions, choose
$$
x \in [0.5 N, 0.55 N], \ y\in [2.6N, 2.8N], \ y\equiv x \bmod 2, \ u\in [0.45N,0.5N], \ u\equiv 0 \bmod 2.
$$
Then the second interval for $v$ is contained in the first. We need to count the number of $v$ in the second interval such that $v\equiv u \bmod 2$.
The number of such $v$ is
$$
\ge \left\lfloor \frac{xy}{2u} \right\rfloor - \left\lfloor \frac{xy}{2u}-\frac{x-u}{2u} \right\rfloor
= \frac{x-u}{2u} -\psi\left( \frac{xy}{2u} \right) + \psi\left( \frac{xy}{2u} -\frac{x-u}{2u}\right) ,
$$
where $\psi(t)=t-\lfloor t \rfloor -1/2$. Summing the $\psi$'s over $u$ results in $o(N)$, so their total contribution
(after also summing over $x$ and $y$) is $o(N^3)$.
Summing the term $(x-u)/(2u)$ over $u,x,y$ contributes $\asymp N^3$.
|
9
|
https://mathoverflow.net/users/12947
|
404645
| 165,942 |
https://mathoverflow.net/questions/404625
|
12
|
A rational singularity is a singularity of a
complex variety $X$ such that for any
resolution $\pi:\; \tilde X\rightarrow X$ the
higher direct images $R^i\pi\_\*(O\_{\tilde X})$
vanish for all $i>0$. Suppose that $X$
has isolated rational singularity,
and $\tilde X\rightarrow X$ its resolution.
I expect that the fiber of $\pi$ over
the singular point is rationally
connected; I would be very grateful
for any reference to this. I need to apply
this to the local situation, so it would be
especially nice if the argument does not
use projectivity.
|
https://mathoverflow.net/users/3377
|
Does a resolution of a rational singularity have rationally connected fibers?
|
No. For instance the cone over an Enriques surface (with respect to any projective embedding) has rational singularity, but Enriques surface is not rationally connected.
|
15
|
https://mathoverflow.net/users/4428
|
404646
| 165,943 |
https://mathoverflow.net/questions/404647
|
6
|
$\DeclareMathOperator\cd{cd}$Recall that the **rational cohomological dimension** of a group $G$ is the supremum of the set of integers $k$ such that there exists a $\mathbb{Q}[G]$-module $M$ with $H^k(G;M) \neq 0$. Denote this by $\cd\_{\mathbb{Q}}(G)$.
If $G$ is a finite group, then it is easy to see that $\cd\_{\mathbb{Q}}(G) = 0$.
However, it is not clear to me what the rational cohomological dimension of a locally finite group is (where "locally finite" means that all finitely generated subgroups are finite). Since homology commutes with direct limits, the rational **homological** dimension of these group are $0$.
So my question is what can be said about the rational cohomological dimension of a locally finite group $G$. I mostly care about countable $G$. In fact, the easiest example where I don't know the answer is
$$G = \bigoplus\_{k=1}^{\infty} \mathbb{Z}/2\mathbb{Z}.$$
|
https://mathoverflow.net/users/384401
|
Rational cohomological dimension of a locally finite group
|
**Edited version.** If $G$ is a countable infinite locally finite group, then the rational cohomological dimension is exactly $1$. The rational cohomological dimension for an infinite group is never $0$ because if $\mathbb Q$ is a projective $\mathbb QG$-module, then you need an idempotent $e$ with $\mathbb QGe\cong \mathbb Q$ and that is impossible unless $G$ is finite and $e$ is the average of all elements of $G$.
Here are two proofs that the projective dimesion of $\mathbb QG$ is at most one for a countable locally finite group. This amounts to showing the augmentation ideal is projective. Then $\mathbb Q$ has projective dimension at most $1$ since $I\to \mathbb QG\to \mathbb Q\to 0$ is a projective resolution where $I$ is the augmentation ideal.
It follows from a result in Dicks book Groups, trees and projective modules (1980) characterizing projectivity of the augmentation ideal of $RG$ that $\mathbb QG$ has a projective augmentation ideal whenever $G$ is countable and locally finite. See [this question](https://mathoverflow.net/questions/297043/using-dunwoodys-results-on-cohomological-dimension-to-learn-about-a-von-neumann?rq=1) for more details.
The second argument is that by a result of Connell, $\mathbb QG$ is von Neumann regular if $G$ is locally finite. The augmentation ideal is countably generated as a left ideal if $G$ is countable and hence by a result of Kaplansky is projective (Kaplansky proved that a countably generated left ideal in a von Neumann regular ring is projective).
Thus for a countably infinite locally finite group the rational cohomological dimension is one.
It follows from Theorem 2 of [Derek Holt's 1981 paper (DOI link)](https://doi.org/10.1112/blms/13.6.557) that if $G$ is a locally finite group of cardinality $\aleph\_1$, then the rational cohomological dimension of $G$ is at least $2$.
|
4
|
https://mathoverflow.net/users/15934
|
404649
| 165,945 |
https://mathoverflow.net/questions/404607
|
10
|
A **state sum model** is a smooth invariant defined on smooth triangulated, or PL manifolds, by summing a local partition function over labels attached to the elements of the triangulation.
Typical examples in 4d are the [Crane-Yetter invariant](https://ncatlab.org/nlab/show/Crane-Yetter+model) on ribbon fusion categories and [Cui's invariant](https://www.ems-ph.org/journals/show_abstract.php?issn=1663-487X&vol=10&iss=4&rank=1) based on $G$-crossed braided spherical fusion categories, but there are also others like the Kashaev invariant.
A **chain mail invariant** is defined on [handle decompositions](https://en.wikipedia.org/wiki/Handle_decomposition), and assigns data to the handles and uses a [Kirby diagram](https://en.wikipedia.org/wiki/Kirby_calculus) or similar to calculate the invariant, typically by interpreting the diagram in a graphical calculus of a category.
State sum models are often chain mail invariants (see Roberts "Refined state-sum invariants of 3- and 4-manifolds", <https://arxiv.org/abs/1810.05833>, <https://doi.org/10.1007/s00220-017-3012-9>), and chain mail invariants are always state sum models since a triangulation gives rise to a handle decomposition.
An **extended TQFT** is a (higher) functor from the fully extended bordism category to some category of higher vector spaces. State sum models are expected to give extended TQFTs, and in low dimensions, this relationship can be made precise (<https://repositories.lib.utexas.edu/bitstream/handle/2152/ETD-UT-2011-05-3139/DAVIDOVICH-DISSERTATION.pdf>).
In 4d, there are TQFTs defined via path integrals such as the Seiberg-Witten and Donaldson invariants, which can detect exotic smooth structures. Since these theories are defined fully locally, one would expect them to be extended TQFTs, or even state sums, or even chain mails. But I'm not aware of an explicit, rigorous construction. (In fact, 4d state sum models are often homotopy invariants.)
*Are there rigorously defined ETQFTs, state sums or chain mail theories that detect smooth structures?*
|
https://mathoverflow.net/users/13767
|
Are there 4d state sum models, extended TQFTs or chain mail invariant that detect smooth structures?
|
[This MO answer](https://mathoverflow.net/a/362517/184) by Arun Debray gives an example in the unoriented case where two specific homeomorphic manifolds can be distinguished by a specific TFT of this kind.
In general all these constructions produce "semisimple" TFTs and it has been shown by David Reutter that in the oriented case such TFTs cannot detect smooth structures. See his paper [Semisimple 4-dimensional topological field theories cannot detect exotic smooth structure](https://arxiv.org/abs/2001.02288).
It is possible that some (extended) TFT with a more exotic target category could do the trick. (In fact the identity functor gives a trivial example that does detect all smooth structures, but obviously we want something less tautological).
|
7
|
https://mathoverflow.net/users/184
|
404652
| 165,946 |
https://mathoverflow.net/questions/404640
|
6
|
Given a probability $p \in (0,1)$ and parameter $\alpha \in (0,1)$, is there an entropy-based proof which yields a good upper bound for the sum
$$\sum\_{\ell = 0}^{\alpha n} \binom{n}{\ell}p^\ell(1-p)^{n-\ell}$$
when $n$ is large?
When $p = 1/2$, there is very simple proof (for example, see section 3.1 of [this paper](https://arxiv.org/pdf/1406.7872.pdf)) which upper bounds the above quantity by
$$2^{(H(\alpha) - 1)n}$$
when $H(\cdot)$ denotes the [binary entropy function](https://en.wikipedia.org/wiki/Binary_entropy_function).
Is there a proof using similar techniques which gives a bound for the more general sum above (which can be interpreted as the CDF of a binomial distribution with parameter $p$)?
I'd also be interested in other proofs for bounds on the above sum. The appropriate bound has already noted in [this answer](https://mathoverflow.net/a/93756/138628), but doesn't sketch out a proof establishing this result.
|
https://mathoverflow.net/users/138628
|
Is there an entropy proof for bounding a weighted sum of binomial coefficients?
|
Yes, if $\alpha<p$ (if $\alpha>p$, the sum is almost 1). To see this, write
$$
\sum\_{\ell = 0}^{\alpha n} \binom{n}{\ell}p^\ell(1-p)^{n-\ell}\leqslant t^{-\alpha n}(pt+(1-p))^n
$$
for every $t\in (0,1]$. Choose a positive $t=t\_0$ for which RHS is minimal possible, taking the logarithmic derivative equal to 0 we get $-\alpha/t\_0+p/(pt\_0+1-p)=0$, $-\alpha p t\_0-\alpha(1-p)+pt\_0=0$, $pt\_0(1-\alpha)=\alpha(1-p)$, $t\_0=\frac{\alpha(1-p)}{p(1-\alpha)}$. We see that if $\alpha\leqslant p$, this $t\_0$ is indeed in $(0,1]$, thus we get the upper bound $\theta^n$, where
$$
\theta=\frac{p^\alpha(1-p)^{1-\alpha}}{\alpha^\alpha (1-\alpha)^{1-\alpha}}.
$$
|
7
|
https://mathoverflow.net/users/4312
|
404654
| 165,947 |
https://mathoverflow.net/questions/404570
|
8
|
**Setup/Notation:**
Let $n,m\in \mathbb{N}$ and let $C(\mathbb{R}^n,\mathbb{R}^m)$ be the space of continuous functions from $\mathbb{R}^n$ to $\mathbb{R}^m$ equipped with the compact-open topology. Let $\mathcal{I}(\mathbb{R}^n,\mathbb{R}^m)$ be the subset of $C(\mathbb{R}^n,\mathbb{R}^m)$ consisting of injective functions.
*Observations - Effect of Dimension:*
* If $n\leq m$: The subset $\mathcal{I}(\mathbb{R}^n,\mathbb{R}^m)$ need not be closed. To see this note that the family
$$
f\_{n}(x):=\frac1{n}\cdot x,
$$
converges to $0$ (in the compact-open topology).
* If $n>m$: Then [Brouwer's Invariance Theorem](https://en.wikipedia.org/wiki/Invariance_of_domain) implies that $\mathcal{I}(\mathbb{R}^n,\mathbb{R}^m)=\emptyset$. In particular, $\overline{\mathcal{I}(\mathbb{R}^n,\mathbb{R}^m)}=\emptyset\neq C(\mathbb{R}^n,\mathbb{R}^m)$.
---
**Question:** Is there a critical $m^{\star}$ (depending on $n$) such that if:
$$
\begin{cases}
\overline{\mathcal{I}(\mathbb{R}^n,\mathbb{R}^m)}
=
C(\mathbb{R}^n,\mathbb{R}^m) &: m\geq m^{\star}\\
\overline{\mathcal{I}(\mathbb{R}^n,\mathbb{R}^m)}
\neq
C(\mathbb{R}^n,\mathbb{R}^m) & : m<m^{\star}
\end{cases}?
$$
If so, more precisely, what is $m^{\star}$ and how does it depend on $n$?
---
**Partial Thoughts/Observations''** - *(Edit)*: If we embed $C(\mathbb{R}^n,\mathbb{R}^m)$ into $C(\mathbb{R}^n,\mathbb{R}^{m+n})$ via $f\mapsto \tilde{f}\_{\infty}(x):=[x\mapsto (f(x),0)]$ then, for any $f\in C(\mathbb{R}^n,\mathbb{R}^{m})$ we can define the maps:
$$
\tilde{f}\_{n}(x):= (f(x),\frac1{n}\cdot x).
$$
Moreover, for each $n\in \mathbb{N}$, the map $f\mapsto \tilde{f}\_n$ is an embedding. So, then in this way... an affirmative pseudo-answer to the question below is ``kind of'' since:
$\lim\limits\_{n\to \infty}\,\tilde{f}\_n =\tilde{f}\_{\infty}$ and each $\tilde{f}\_n \in \mathcal{I}(\mathbb{R}^n,\mathbb{R}^{m+n})$ and is the image of some $C(\mathbb{R}^n,\mathbb{R}^m)$.
|
https://mathoverflow.net/users/36886
|
The closure of the set of injective continuous functions
|
Kuratowski proved in [Sur les théorèmes du „plongement" dans la théorie de la dimension. *Fundamenta Mathematicae* 28.1 (1937): 336-342](http://matwbn.icm.edu.pl/ksiazki/fm/fm28/fm28137.pdf) that the set of embeddings of an at most $n$-dimensional separable metrizable space into $\mathbb{R}^{2n+1}$ contains a dense $G\_\delta$-set.
Hence $m^\*\le 2m+1$.
|
5
|
https://mathoverflow.net/users/5903
|
404659
| 165,949 |
https://mathoverflow.net/questions/404656
|
2
|
A few procrastinal computations motivated by [Four infinite series involving Riemann zeta function](https://mathoverflow.net/questions/401468/four-infinite-series-involving-riemann-zeta-function) suggest the identity
$$\tan\left(\frac{\kappa-1}{\kappa+1}\frac{\pi}{2}\right)=\frac{1}{\pi}\sum\_{n=1}^\infty \frac{\kappa^n-1}{(\kappa+1)^n}\zeta(n+1)$$
for all $\kappa>0$. (Both sides change sign when replacing $\kappa$ by $1/\kappa$. It is thus enough to consider $\kappa\geq 1$.)
I checked it on $100$ random values up to 400 digits which convinced me that it must hold.
Is there an easy explanation?
(Remark: This identity, if true, provides of course easy answers to question M0401468 mentionned above.)
(Computational remark: For high precision computation, it is perhaps best to cut the sum at some not very large value $N$ (I worked with $N=400$) and to approximate the omitted terms by summing the first few terms contributing to $\zeta(N+1),\ldots$, by considering the associated geometric progressions.)
|
https://mathoverflow.net/users/4556
|
A expression for the tangent function involving $\zeta(n),n=2,3,\ldots$
|
When I was writing my answer, Dan Romik answer appeared. Mine is the same but with more detail.
We have
$$\log\Gamma(1-x)=\gamma x+\sum\_{n=2}^\infty\zeta(n)\frac{x^n}{n}$$
(this is known and is also an exercise in complex analysis).
Hence by differentiation
$$-\frac{\Gamma'(1-x)}{\Gamma(1-x)}-\gamma=\sum\_{n=2}^\infty \zeta(n) x^{n-1}$$
It follows that (I change $\kappa$ into $x$)
$$\sum\_{n=1}^\infty\frac{x^n-1}{(x+1)^n}\zeta(n+1)=
\frac{\Gamma'(1-\frac{1}{x+1})}{\Gamma(1-\frac{1}{x+1})}+\gamma
-\frac{\Gamma'(1-\frac{x}{x+1})}{\Gamma(1-\frac{x}{x+1})}-\gamma$$
$$=\frac{\Gamma'(\frac{x}{x+1})}{\Gamma(\frac{x}{x+1})}-\frac{\Gamma'(\frac{1}{x+1})}{\Gamma(\frac{1}{x+1})}=\frac{\Gamma'(y)}{\Gamma(y)}-\frac{\Gamma'(1-y)}{\Gamma(1-y)}=\frac{d}{dy}\log(\Gamma(y)\Gamma(1-y))=\frac{d}{dy}\log\frac{\pi}{\sin\pi y}$$
$$=-\frac{\pi\cos\pi y}{\sin\pi y}=-\pi\cot\pi y=\pi\cot\frac{\pi x}{x+1}$$
$$=-\pi\tan\Bigl(\frac{\pi}{2}-\frac{\pi x}{x+1}\Bigr)=-\pi\tan\Bigl(\frac{\pi}{2}\frac{1-x}{1+x}\Bigr)=\pi\tan\Bigl(\frac{\pi}{2}\frac{x-1}{x+1}\Bigr).$$
|
2
|
https://mathoverflow.net/users/7402
|
404660
| 165,950 |
https://mathoverflow.net/questions/404629
|
2
|
Let $r$ be the rank function of a matroid. If the matroid is representable (over a field), then $r$ must satisfy Ingleton's inequalities. On the other hand, there are matroids that satisfy Ingleton's inequalities but are still not representable (already on $8$ elements such examples exist). However, not satisfying Ingleton's inequalities implies actually that for each $n\in\mathbb{N}$ the polymatroid $n\cdot r$ is not representable as well. This leads to the following questions:
-Are there matroids that satisfy Ingleton's inequalities such that $n\cdot r$ is not representable for all $n\in\mathbb{N}$?
-If yes, what are the smallest such examples? Are there such examples on $8$ elements?
|
https://mathoverflow.net/users/36563
|
Non-representable matroids and Ingleton's inequality
|
I don't know the smallest such example, but here is an example with 14 elements.
* Let F be the Fano matroid: the 7-element matroid represented over any field of characteristic 2 by the set of non-zero vectors in $\{0,1\}^3$.
* Let N be the non-Fano matroid: the 7-element matroid represented over any field of characteristic not equal to 2 by the set of non-zero vectors in $\{0,1\}^3$.
* The direct sum of F and N has 14 elements, and its rank function satisfies Ingleton's inequality because it is a sum of two functions, each satisfying Ingleton's inequality.
In "[Lexicographic Products and the Power of Non-Linear Network Coding](https://arxiv.org/pdf/1108.2489.pdf)", Anna Blasiak, Eyal Lubetzky, and I present an inequality that is satisfied by any positive scalar multiple of the rank function of a matroid representable in characteristic 2, but is violated by the rank function of N. This is inequality (6.8) in the paper. We also present an inequality that is satisfied by any positive scalar multiple of the rank function of a matroid representable in characteristic not equal to 2, but is violated by the rank function of F. This is inequality (6.17) in the paper. Since the direct sum of N and F violates both inequalities, the polymatroid defined by any positive scalar multiple of its rank function is not representable over a field of any characteristic.
|
2
|
https://mathoverflow.net/users/8049
|
404663
| 165,952 |
https://mathoverflow.net/questions/404530
|
12
|
Let $a(n)=$ Number of ordered set partitions of $[n]$ such that the smallest element of each block is odd.
```
Example:
a(3) = 3: 123, 12|3, 3|12.
a(4) = 5: 1234, 124|3, 3|124, 12|34, 34|12.
```
See [A290384](http://oeis.org/A290384) for details.
Motivated by [Question 403336](https://mathoverflow.net/questions/403336/) and [Question 403386](https://mathoverflow.net/questions/403386), I consider the permanents $\operatorname{per}(A)$ where
$$A=\left[\left\lceil \frac{2j-k}{n+\cos^2(\frac{n\pi}{2})}\right\rceil \right]\_{1\le j,k\le n}$$ and $\lceil \cdot\rceil$ is the ceil function.
When $n=1,2,3,4,5,6$, $A=$
$$\left[ \begin {array}{c} 1\end {array} \right] ,$$
$$\left[ \begin {array}{cc} 1&0\\ 1&1\end {array} \right], $$
$$ \left[ \begin {array}{ccc} 1&0&0\\ 1&1&1
\\ 2&2&1\end {array} \right]
,$$
$$ \left[ \begin {array}{cccc} 1&0&0&0\\ 1&1&1&0
\\ 1&1&1&1\\ 2&2&1&1\end {array}
\right]
,$$
$$ \left[ \begin {array}{ccccc} 1&0&0&0&0\\ 1&1&1&0&0
\\ 1&1&1&1&1\\ 2&2&1&1&1
\\ 2&2&2&2&1\end {array} \right]
,$$
$$\left[ \begin {array}{cccccc} 1&0&0&0&0&0\\ 1&1&1&0
&0&0\\ 1&1&1&1&1&0\\ 1&1&1&1&1&1
\\ 2&2&1&1&1&1\\ 2&2&2&2&1&1
\end {array} \right]
$$
respectively.
Numerical computation indicates that
$$\operatorname{per}(A)=a(n)$$
for $1≤n≤21$.
```
n per(A)
1 1
2 1
3 3
4 5
5 23
6 57
7 355
8 1165
9 9135
10 37313
11 352667
12 1723605
13 19063207
14 108468169
15 1374019539
16 8920711325
17 127336119839
18 928899673425
19 14751357906571
20 119445766884325
21 2088674728868631
```
Thus, we obtain the following
**Conjecture 1.** For any positive integer $n$,$$\operatorname{per}(A) = a(n).$$
**Question 1.** Is this identity correct? How to prove it?
---
**EDIT**
Let
$$B=\left[\left\lceil \frac{j+k}{n+\cos^2(\frac{n\pi}{2})}\right\rceil \right]\_{1\le j,k\le n}$$
where $\lceil \cdot\rceil$ is the ceil function.
When $n=1,2,3,4,5,6$, $B=$
$$ \left[ \begin {array}{c} 2\end {array} \right], $$
$$\left[ \begin {array}{cc} 1&1\\ 1&2\end {array}
\right], $$
$$\left[ \begin {array}{ccc} 1&1&2\\ 1&2&2
\\ 2&2&2\end {array} \right] ,$$
$$ \left[ \begin {array}{cccc} 1&1&1&1\\ 1&1&1&2
\\ 1&1&2&2\\ 1&2&2&2\end {array}
\right] ,$$
$$ \left[ \begin {array}{ccccc} 1&1&1&1&2\\ 1&1&1&2&2
\\ 1&1&2&2&2\\ 1&2&2&2&2
\\ 2&2&2&2&2\end {array} \right] ,$$
$$ \left[ \begin {array}{cccccc} 1&1&1&1&1&1\\ 1&1&1&1
&1&2\\ 1&1&1&1&2&2\\ 1&1&1&2&2&2
\\ 1&1&2&2&2&2\\ 1&2&2&2&2&2
\end {array} \right]
$$
respectively.
We have the following
**Conjecture 2.** For any positive integer $n$,
$$\operatorname{per}(B)=\frac{3-(-1)^n}{2}F(n)$$
where $F(n)$ is the [Fubini numbers](http://oeis.org/A000670), i.e. the number of ordered partitions of $[n]$.
It is equivalent to
**Conjecture 3.** For any positive integer $n$,
$$\operatorname{per}(C)=F(n),$$
where $$C=\left[\left\lceil\frac{j+k}{n+1}\right\rceil\right]\_{1\le j,k\le n}.$$
Numerical computation indicates that it is true for $1≤n≤21$.
**Question 2.** Are these new results? How to prove them?
**A2**: By @Peter Taylor's comments, Conjectures 2 and 3 are known results.
---
**ADDED**
Today (2021-9-24), I discovered the following interesting arithmetic properties of $a(n)$
**Conjecture 4.** For any prime $p$,
$$a(n) \equiv a(n+p(p-1)) \pmod p.$$
Noting that $a(1)=a(2)=1$, we have
**Conjecture 5.** For any prime $p$ ,
$$a(p^2-p+1) \equiv 1 \pmod p,$$
$$a(p^2-p+2) \equiv 1 \pmod p.$$
Numerical computation indicates that it is true for $2≤p≤19$.
Moreover
**Conjecture 6.** For any prime $p$ and positive integer $n\geq h \geq 1$,
$$a(n) \equiv a(n+p^h(p-1)) \pmod{p^h}.$$
It is similar to [Daniel Barsky's congruence](http://www.mat.univie.ac.at/%7Eslc/opapers/s05barsky.html) for [Fubini numbers](http://oeis.org/A000670).
**Daniel Barsky's Congruence.** For any prime $p$ and positive integer $n\geq h \geq 1$,
$$F(n) \equiv F(n+p^{h-1}(p-1)) \pmod{p^h},$$
where $F(n)$ is the [Fubini numbers](http://oeis.org/A000670).
**Question 3.** Are these congruences correct? How to prove them?
---
**ADDED** (2021-9-27)
**Claim.** Conjecture 4 is true for $2 \leq p \leq 5$.
**Proof.** Obviously, we have $a(n)\equiv 1 \pmod{2}$.
For $3\leq p \leq 5$, we use the following generating function
$$\sum\_{n\geq 1} a(n)x^n = \sum\_{k\geq 1} \frac{(-1)^{k-1}}{\binom{-\frac1x-1}{k-1}\binom{\frac1x-1}k},$$
which is proved by Max Alekseyev.
When $p=3$,
\begin{align}
\sum\_{n\geq 1} a(n)x^n
&\equiv \sum\_{1 \leq k \leq 2} \frac{(-1)^{k-1}}{\binom{-\frac1x-1}{k-1}\binom{\frac1x-1}k}\\
&=-\frac{x}{x-1}+2\,{\frac {{x}^{3}}{ \left( x+1 \right) \left( 2\,{x}^{2}-3\,x+1
\right) }}\\
&= \sum \_{n=0}^{\infty } \left( -{\frac { \left( -1 \right) ^{n}}{3}}+{
\frac {{2}^{n}}{3}} \right) {x}^{n}
\pmod{3}
\end{align}
and $-{\frac { \left( -1 \right) ^{n}}{3}}+{\frac {{2}^{n}}{3}} \pmod{3}$ is a periodic sequence with least period $6$: repeat in the pattern $$1, 1, 0, 2, 2, 0.$$
This leads to $a(n) \equiv a(n+6) \pmod 3.$
Similarly, we have
\begin{align}
\sum\_{n\geq 1} a(n)x^n
&\equiv \sum\_{1 \leq k \leq 4} \frac{(-1)^{k-1}}{\binom{-\frac1x-1}{k-1}\binom{\frac1x-1}k}\\
&=\sum \_{n=0}^{\infty } \left( {\frac {13\,{2}^{n}}{30}}+{\frac {{4}^{n}
}{35}}+{\frac { \left( -2 \right) ^{n}}{10}}-{\frac { \left( -3
\right) ^{n}}{35}}-{\frac {13\, \left( -1 \right) ^{n}}{30}}-{\frac {
{3}^{n}}{10}} \right) {x}^{n}
\pmod{5},
\end{align}
so $a(n) \pmod{5}$ is a periodic sequence with least period $20$: repeat in the pattern $$1, 1, 3, 0, 3, 2, 0, 0, 0, 3, 2, 0, 2, 4, 4, 0, 4, 0, 1, 0.$$
This leads to $a(n) \equiv a(n+20) \pmod 5.$ **QED**
Moreover, we have the following
**Conjecture 7.** For any odd prime $p$,
\begin{equation}
\sum\_{n=1}^{p(p-1)}a(n) \equiv
\begin{cases}
p \pmod{p^2} &\mbox{if $p \equiv 1 \pmod 4$ }\\
0 \pmod{p^2} &\mbox{if $p \equiv 3 \pmod 4$ }
\end{cases}.
\end{equation}
It is true for $3 \leq p \leq 19$.
---
**ADDED** (2021-10-04)
**Conjecture 8.** For any prime $p$,
$$\sum\_{n=1}^{p(p-1)}\left(\frac{a(n)}{p}\right)\equiv 0 \pmod p,$$
where $\left(\frac{\cdot}p\right)$ is the Legendre symbol.
It is true for $2 \leq p \leq 19$.
|
https://mathoverflow.net/users/287674
|
Set partitions and permanents
|
Let me outline an approach for computing permanents in these conjectures. For the sake of concreteness, I will prove Conjecture 1 for an odd $n$. The matrix here is the sum of the following two 0-1 matrices (using [Iverson bracket](https://en.wikipedia.org/wiki/Iverson_bracket) notation):
$$A:=\big([2j-k \geq 1]\big)\_{j,k=1}^n$$
and
$$B:=\big([2j-k \geq n+1]\big)\_{j,k=1}^n$$
(notice that I intentionally redefine matrices $A$ and $B$). For example, for $n=5$, we have
$$A=\begin{bmatrix} 1&0&0&0&0\\ 1&1&1&0&0
\\ 1&1&1&1&1\\ 1&1&1&1&1
\\ 1&1&1&1&1\end{bmatrix}
\quad\text{and}\quad
B=\begin{bmatrix} 0&0&0&0&0\\ 0&0&0&0&0
\\ 0&0&0&0&0\\ 1&1&0&0&0
\\ 1&1&1&1&0\end{bmatrix}
$$
Our goal is to compute $\mathrm{per}(A+B)$ and show that it's equal to $a(n)$.
The crucial observation is that 0-1 matrices can be viewed as boards on which permanent enumerates [non-attacking rook placements](https://en.wikipedia.org/wiki/Rook_polynomial). Furthermore, our matrices have the shape of [Ferrers boards](https://en.wikipedia.org/wiki/Partition_(number_theory)#Ferrers_diagram), and the one for $B$ is a sub-board for that of $A$. From now on, I will not distinguish matrices $A$ and $B$ from the corresponding Ferrers boards.
I will use the notation and machinery from my [other answer](https://mathoverflow.net/q/386804), which computes the number of non-attacking rook placements (i.e., the permanent) for the *difference* of a Ferrers board with its sub-board. In the current problem, we need to compute the number of placements of $n$ non-attacking rooks in $A$, where each placement comes with multiplicity $2^t$, where $t$ in the number of rooks in $B\subset A$.
Board $A$ has row lengths
$$a:=(1,3,5,\dots,n-2,\underbrace{n,n,\dots,n}\_{(n+1)/2}),$$
while board $B$ has row lengths
$$b:=(\underbrace{0,0,\dots,0}\_{(n+1)/2},2,4,\dots,n-1).$$
By inclusion-exclusion here, we have
$$\mathrm{per}(A+B) = \sum\_{T\subseteq[n]} r\_n(A\_{\bar T}\| B\_T),$$
where $\bar T := [n] \setminus T$ is the complement of $T$. The analog of formula $(\star)$ here gives the following expression:
$$\mathrm{per}(A+B) = \sum\_{p\in\{0,1\}^n} \prod\_{i=1}^n \big(p\_i(a\_i-\sum\_{j=1}^{\tau\_A(i)-1} \delta\_j) + q\_i(b\_i-\sum\_{j=1}^{\tau\_B(i)-1} \delta\_j)\big),$$
where $q\_i:=1-p\_i$ and
$$\sigma:=\big(\underbrace{0,0,\dots,0}\_{(n+1)/2},1,2,\dots,n-1,\underbrace{n,n,\dots,n}\_{(n+1)/2}\big),$$
$$\delta:=\big(q\_1,q\_2,\dots,q\_{\frac{n+1}2},p\_1,q\_{\frac{n+1}2+1},p\_2,q\_{\frac{n+1}2+2},\dots,p\_{\frac{n-1}2},q\_n,p\_{\frac{n+1}2},p\_{\frac{n+1}2+1},\dots,p\_n\big),$$
$$\tau\_A:=\big( \frac{n+3}2,\frac{n+7}2, \dots, \frac{3n+1}2, \frac{3n+3}2,\frac{3n+5}2,\dots,2n\big),$$
$$\tau\_B:=\big(1,2,\dots,\frac{n+1}2,\frac{n+1}2+2,\frac{n+1}2+4,\dots,\frac{3n-1}2\big).$$
Correspondingly, we have
$$\sum\_{j=1}^{\tau\_A(i)-1} \delta\_j =
\begin{cases}
i-1 + \sum\_{j=i}^{\frac{n-1}2+i}q\_j, & \text{if}\ i\leq\frac{n-1}2;\\
n - \sum\_{j=i}^{n}p\_j, & \text{if}\ i\geq\frac{n+1}2.
\end{cases}$$
and
$$\sum\_{j=1}^{\tau\_B(i)-1} \delta\_j =
\begin{cases}
\sum\_{j=1}^{i-1} q\_j & \text{if}\ i\leq\frac{n-1}2;\\
i-1 - \sum\_{j=i-\frac{n-1}2}^{i-1}p\_j, & \text{if}\ i\geq\frac{n+1}2.
\end{cases}$$
The formula then becomes
$$\mathrm{per}(A+B) = \sum\_{p\in\{0,1\}^n} \prod\_{i=1}^{(n-1)/2} \big(p\_i(i - \sum\_{j=i}^{\frac{n-1}2+i}q\_j) - q\_i \sum\_{j=1}^{i-1} q\_j)\big)
\prod\_{i=(n+1)/2}^n \big(p\_i\sum\_{j=i}^{n}p\_j + q\_i(i-n + \sum\_{j=i-\frac{n-1}2}^{i-1}p\_j)\big).$$
We can see that if $\min\{i\,:\,q\_i=1\}\leq\frac{n-1}2$, then the corresponding summand is zero. Hence, we can restrict summation to $(p\_i,q\_i)=(1,0)$ for all $i\leq\frac{n-1}2$, and further the same holds for $i=\frac{n+1}2$. Shifting indices $i\to \frac{n+1}2+i$, we get the formula:
\begin{split}
&\mathrm{per}(A+B) = \sum\_{p\in\{0,1\}^{\frac{n-1}2}} (1 + \sum\_{j=1}^{\frac{n-1}2} p\_i) \prod\_{i=1}^{(n-1)/2} \big(p\_i\sum\_{j=i}^{\frac{n-1}2} p\_j + q\_i (1+\sum\_{j=1}^{i-1} p\_j)\big)(1+\sum\_{j=1}^{i-1} p\_j) \\
&=\sum\_{p\in\{0,1\}^{\frac{n-1}2}}
\bigg( (1 + \sum\_{j=1}^{\frac{n-1}2} p\_i) \prod\_{i=1\atop p\_i=1}^{(n-1)/2} \sum\_{j=i}^{\frac{n-1}2} p\_j\bigg) \cdot
\bigg( \prod\_{i=1\atop p\_i=0}^{(n-1)/2} (1+\sum\_{j=1}^{i-1} p\_j) \bigg) \cdot
\bigg( \prod\_{i=1}^{(n-1)/2} (1+\sum\_{j=1}^{i-1} p\_j) \bigg)
\end{split}
---
Now, let's show that this is exactly $a(n)$. More specifically, if we restrict summation to fixed $1 + \sum\_{j=1}^{\frac{n-1}2} p\_i =: k$, then the sum gives the number of ordered set partitions with $k$ parts.
Think of constructing a set partition by assigning elements $1,2,\dots,n$ in order to some part, and of $p\_i$ as the indicator for $2i+1$ being a smallest element in its part (element $1$ has to be the smallest in its part, and this where "$1+$ in the formula comes from). Then
* $(1 + \sum\limits\_{j=1}^{\frac{n-1}2} p\_i) \prod\limits\_{i=1\atop p\_i=1}^{(n-1)/2} \sum\limits\_{j=i}^{\frac{n-1}2} p\_j = k!$ accounts for the order of parts;
* $\prod\limits\_{i=1\atop p\_i=0}^{(n-1)/2} (1+\sum\limits\_{j=1}^{i-1} p\_j)$ accounts for assignments of $2i+1$ to one of $1+\sum\limits\_{j=1}^{i-1} p\_j$ parts, whose smallest elements are smaller than $2i+1$;
* $\prod\limits\_{i=1}^{(n-1)/2} (1+\sum\limits\_{j=1}^{i-1} p\_j)$ accounts for assignments of $2i$ to one of $1+\sum\limits\_{j=1}^{i-1} p\_j$ parts (whose smallest elements are smaller than $2i$).
Hence, $\mathrm{per}(A+B) = a(n)$. QED
|
8
|
https://mathoverflow.net/users/7076
|
404669
| 165,953 |
https://mathoverflow.net/questions/403336
|
3
|
Inspired by [Question 402572](https://mathoverflow.net/questions/402572), I consider the permanent of matrices
$$f(n)=\mathrm{per}(A)=\mathrm{per}\left[\operatorname{sgn} \left(\sin\pi\frac{j+2k}{n+1} \right)\right]\_{1\le j,k\le n},$$
where $n$ is a positive integer and $\operatorname{sgn}$ is the sign-function.
When $n=1,2,3,4,5,6,$ the matrices are
$$\left[ \begin {array}{c} -1\end {array} \right],$$
$$\left[ \begin {array}{cc} 0&-1\\ -1&0\end {array}\right],$$
$$\left[ \begin {array}{ccc} 1&-1&-1\\ 0&-1&0
\\ -1&-1&1\end {array} \right]
,$$
$$\left[ \begin {array}{cccc} 1&0&-1&-1\\ 1&-1&-1&0
\\ 0&-1&-1&1\\ -1&-1&0&1
\end {array} \right]
,$$
$$\left[ \begin {array}{ccccc} 1&1&-1&-1&-1\\ 1&0&-1&
-1&0\\ 1&-1&-1&-1&1\\ 0&-1&-1&0&1
\\ -1&-1&-1&1&1\end {array} \right]
,$$
$$\left[ \begin {array}{cccccc} 1&1&0&-1&-1&-1\\ 1&1&
-1&-1&-1&0\\ 1&0&-1&-1&-1&1\\ 1&-1
&-1&-1&0&1\\ 0&-1&-1&-1&1&1\\ -1&-
1&-1&0&1&1\end {array} \right]
.$$
Numerical computation indicates that
\begin{equation}
\mathrm{per}(A)=
\begin{cases}
-T\_n&\mbox{if $n$ is odd}\\
E\_n&\mbox{if $n$ is even}
\end{cases}
\end{equation}
for $3\leq n \leq 21$, where $T\_n$ and $E\_n$ are [tangent numbers](https://mathworld.wolfram.com/TangentNumber.html) and [secant numbers](https://mathworld.wolfram.com/SecantNumber.html), i.e.
$$\tan{x}=\sum\_{2\nmid n}T\_n\frac{x^n}{n!}=1\frac{x}{1!}+2\frac{x^3}{3!}+16\frac{x^5}{5!}+272\frac{x^7}{7!}+7936\frac{x^9}{9!}+\cdots$$
and
$$\sec{x}=\sum\_{2\mid n}E\_n\frac{x^n}{n!}=1+1\frac{x^2}{2!}+5\frac{x^4}{4!}+61\frac{x^6}{6!}+1385\frac{x^8}{8!}+50521\frac{x^{10}}{10!}+\cdots$$
respectively.
```
f(1) = -1
f(2) = 1
f(3) = -2
f(4) = 5
f(5) = -16
f(6) = 61
f(7) = -272
f(8) = 1385
f(9) = -7936
f(10) = 50521
f(11) = -353792
f(12) = 2702765
f(13) = -22368256
f(14) = 199360981
f(15) = -1903757312
f(16) = 19391512145
f(17) = -209865342976
f(18) = 2404879675441
f(19) = -29088885112832
f(20) = 370371188237525
f(21) = -4951498053124096
```
Thus, we obtain the following
**Conjecture.** For any positive integer $n$,
\begin{equation}
\mathrm{per}(A)=
\begin{cases}
-T\_n&\mbox{if $n$ is odd}\\
E\_n&\mbox{if $n$ is even}
\end{cases}.
\end{equation}
**Question.** Is this identity correct? How to prove it?
|
https://mathoverflow.net/users/287674
|
Tangent numbers, secant numbers and permanent of matrices
|
The conjecture has been proved! See the preprint *Proof of five conjectures relating permanents to combinatorial sequences* by Fu, Lin and me available from <http://arXiv.org/abs/2109.11506>.
|
4
|
https://mathoverflow.net/users/124654
|
404675
| 165,955 |
https://mathoverflow.net/questions/402572
|
21
|
Motivated by [Question 402249](https://mathoverflow.net/questions/402249) of [Zhi-Wei Sun](https://mathoverflow.net/users/124654/zhi-wei-sun), I consider the permanent of matrices
$$e(n)=\mathrm{per}\left[\operatorname{sgn} \left(\tan\pi\frac{j+k}n \right)\right]\_{1\le j,k\le n-1},$$
where $n$ is an odd integer greater than 1 and $ \operatorname{sgn} $ is the sign-function.
When $n=3,5,7,$ the matrices are
$$\left[ \begin {array}{cc} -1&0\\ 0&1\end {array}
\right] ,
$$
$$
\left[ \begin {array}{cccc} 1&-1&-1&0\\ -1&-1&0&1
\\ -1&0&1&1\\ 0&1&1&-1\end {array}
\right],
$$
$$
\left[ \begin {array}{cccccc} 1&1&-1&-1&-1&0\\ 1&-1
&-1&-1&0&1\\ -1&-1&-1&0&1&1\\ -1&-
1&0&1&1&1\\ -1&0&1&1&1&-1\\ 0&1&1&
1&-1&-1\end {array} \right].
$$
Numerical computation indicates that
$$e(n)=E(n-1)$$
for $3 \leq n \leq 23 $, where $E(n)$ are the [Euler numbers](https://encyclopediaofmath.org/wiki/Euler_numbers).
```
[e(3) = -1, E(2) = -1]
[e(5) = 5, E(4) = 5]
[e(7) = -61, E(6) = -61]
[e(9) = 1385, E(8) = 1385]
[e(11) = -50521, E(10) = -50521]
[e(13) = 2702765, E(12) = 2702765]
[e(15) = -199360981, E(14) = -199360981]
[e(17) = 19391512145, E(16) = 19391512145]
[e(19) = -2404879675441, E(18) = -2404879675441]
[e(21) = 370371188237525, E(20) = 370371188237525]
[e(23) = -69348874393137901, E(22) = -69348874393137901]
```
**Conjecture.** For any odd integer $n>1$, $e(n)=E(n-1)$.
**Question 1.** How to prove this conjecture?
---
**Edit.**
By [T. Amdeberhan's recent MO post](https://mathoverflow.net/questions/402596),
$$\mathrm{det}\left[\operatorname{sgn} \left(\tan\pi\frac{j+k}n \right)\right]\_{1\le j,k\le n-1}=(-1)^{\frac{n-1}{2}}=\sin(\frac{n\pi}{2})\neq 0,$$
we can consider the permanent of inverse matrices
$$e'(n)=\mathrm{per}(A^{-1}),$$
where $A=\left[\operatorname{sgn} \left(\tan\pi\frac{j+k}n \right)\right]\_{1\le j,k\le n-1}$ and $n$ is an odd integer greater than 1.
When $n=3,5,7,$ $A^{-1}$ equals
$$\left[ \begin {array}{cc} -1&0\\ 0&1\end {array}
\right],
$$
$$
\left[ \begin {array}{cccc} 1&-1&1&0\\ -1&-1&0&-1
\\ 1&0&1&1\\ 0&-1&1&-1\end {array}
\right],
$$
$$
\left[ \begin {array}{cccccc} 1&1&-1&1&1&0\\ 1&-1&-
1&1&0&-1\\ -1&-1&-1&0&-1&-1\\ 1&1&0
&1&1&1\\ 1&0&-1&1&1&-1\\ 0&-1&-1&1
&-1&-1\end {array} \right]
$$
respectively.
Numerical computation indicates that all elements of $A^{-1}$ are $-1, 0, 1 $ and
$$e'(n)=E(n-1)$$
for $3 \leq n \leq 23 $.
**Added Conjecture 1.** For any odd integer $n>1$, $e'(n)=E(n-1)$.
The following are some properties of the $A$ and $A^{-1}$ that may be easy to prove:
**Added Conjecture 2.** For any odd integer $n>1$, all elements of $A^{-1}$ are $-1, 0, 1 $.
**Added Conjecture 3.** For any odd integer $n>1$, the Characteristic Polynomials of $A$ and $A^{-1}$ are
$$
\mathrm{Char}(A)=\mathrm{Char}(A^{-1})={\frac { \left( \lambda+i \right) ^{n-1} + \left( \lambda-i \right) ^{n-1}}{2}},
$$
where $i^2=-1$ and $\mathrm{Char}(A)=\mathrm{det}(\lambda I-A)$.
**Question 2.** How to prove these added conjectures?
---
**Edit.**
**The connection between characteristic polynomials and recurrences of Euler numbers:**
Let
$$\mathrm{Char}(A)|\_{\lambda=iE}=\mathrm{det}(\lambda I-A)|\_{\lambda=iE}={\frac { \left( iE+i \right) ^{n-1} + \left( iE-i \right) ^{n-1}}{2}}=0,$$
we deduce
$$(E+1)^{n-1}+(E-1)^{n-1}=0,$$
which is [the recurrence formula for the Euler numbers](https://encyclopediaofmath.org/wiki/Euler_numbers) ($E^n≡E\_n≡E(n)$ in symbolic notation).
**Question 3.** Is this a coincidence?
---
**Edit.**
Let $B=\left[\operatorname{sgn} \left(\sin\pi\frac{j+k}n \right)\right]\_{1\le j,k\le n-1}$ where $n>1$ is an integer.
[Zhi-Wei Sun](https://mathoverflow.net/users/124654/zhi-wei-sun) and [Nemo](https://mathoverflow.net/users/82588/nemo) guessed in the comments that
$$\mathrm{per}(B)=E(n-1)$$
for any integer $n>1$.
Numerical calculations show that this is correct for $1<n\leq 23$.
It is easy to see that $\mathrm{det}(B)=0$ for any even number $n$ and it seems that
$$\mathrm{det}(B)=\mathrm{det}(A)=(-1)^{\frac{n-1}{2}}\neq 0$$
for any odd integer $n>1$.
When $n=3,5,7,$ $B$ and $B^{-1}$ are equal to
$$\left[ \begin {array}{cc} 1&0\\ 0&-1\end {array}
\right] , \left[ \begin {array}{cc} 1&0\\ 0&-1
\end {array} \right] ,$$
$$ \left[ \begin {array}{cccc} 1&1&1&0\\ 1&1&0&-1
\\ 1&0&-1&-1\\ 0&-1&-1&-1
\end {array} \right] , \left[ \begin {array}{cccc} 1&-1&1&0
\\ -1&1&0&-1\\ 1&0&-1&1
\\ 0&-1&1&-1\end {array} \right] ,$$
$$
\left[ \begin {array}{cccccc} 1&1&1&1&1&0\\ 1&1&1&
1&0&-1\\ 1&1&1&0&-1&-1\\ 1&1&0&-1&
-1&-1\\ 1&0&-1&-1&-1&-1\\ 0&-1&-1&
-1&-1&-1\end {array} \right] , \left[ \begin {array}{cccccc} 1&-1&1&-1
&1&0\\ -1&1&-1&1&0&-1\\ 1&-1&1&0&-1&1\\ -1&1&0&-1&1&-1\\ 1&0&-1&1&-1
&1\\ 0&-1&1&-1&1&-1\end {array} \right]
$$
respectively.
A natural question is: Does $B$ and $B^{-1}$ have the same properties as $A$ and $A^{-1}$ for any odd integer $n>1$.
**Added Conjecture 4.** For any odd integer $n>1$,
$$\mathrm{per}(B)=\mathrm{per}(B^{-1})=E(n-1)$$ and $$
\mathrm{Char}(B)=\mathrm{Char}(B^{-1})={\frac { \left( \lambda+i \right) ^{n-1} + \left( \lambda-i \right) ^{n-1}}{2}}.
$$
**Question 4.** If $$\mathrm{per}(C)=\mathrm{per}(C^{-1})=E(n-1)$$
and
$$\mathrm{Char}(C)=\mathrm{Char}(C^{-1})={\frac { \left( \lambda+i \right) ^{n-1} + \left( \lambda-i \right) ^{n-1}}{2}},$$
where $n$ is an odd number greater than $1,$ $C$ is an $(n-1)\times (n-1)$ matrix and all elements of $C$ are $-1, 0, 1 ,$ what can we say about $C$?
|
https://mathoverflow.net/users/287674
|
Euler numbers and permanent of matrices
|
The conjecture and the added Conjectures 1 and 2 have been proved.
See the preprint *Proof of five conjectures relating permanents to combinatorial sequences* by Fu, Lin and me available from <http://arXiv.org/abs/2109.11506>.
|
5
|
https://mathoverflow.net/users/124654
|
404676
| 165,956 |
https://mathoverflow.net/questions/404683
|
4
|
Can you provide a proof for the following claim:
>
> $$-\displaystyle\sum\_{n=1}^{\infty}\frac{J\_k(n)}{n} \cdot \ln\left(1-x^n\right)=\frac{x \cdot A\_{k-1}(x)}{(1-x)^k} \quad \text{for} \quad |x| < 1 \quad \text{and} \quad k>1$$
> where $J\_k(n)$ is the [Jordan's totient function](https://en.wikipedia.org/wiki/Jordan%27s_totient_function) and $A\_k(x)$ is the kth [Eulerian polynomial](https://en.wikipedia.org/wiki/Eulerian_number).
>
>
>
The first few Eulerian polynomials are: $A\_1(x)=1 , A\_2(x)=x+1 , A\_3(x)=x^2+4x+1$ , etc.
The SageMath cell that demonstrates this claim can be found [here](https://sagecell.sagemath.org/?z=eJw9jEEKgzAQRfc5xSxnQqKmLtNcoIcQFNMSEkcxjQild29KoZ-3-fD-L5MzXY0Vp9Nd0_dWRHex4oZRMblXuCPw1SjoFOSyzOFAVjBXhggSltVPoWRkaGEmIvsW2VUP2RlVJvW7aVmm9YFGnwNXR2x74CfqLE3zb6f0x5jQl-T3MPK2JozaELXfGQ2R7AfRYDEd&lang=gp&interacts=eJyLjgUAARUAuQ==).
|
https://mathoverflow.net/users/88804
|
An infinite series involving Jordan's totient function
|
The right hand side evaluates to
$$\sum\_{m=0}^{\infty} m^{k-1} x^m,$$
and so it remains to verify that the coefficient of $x^m$ in the l.h.s. is also $m^{k-1}$.
By differentiating the l.h.s., we have
\begin{split}
[x^m]\ \bigg(-\sum\_{n\geq 1} \frac{J\_k(n)}{n} \ln(1-x^n)\bigg) &= \frac1m [x^{m-1}]\ \sum\_{n\geq 1} J\_k(n)\frac{x^{n-1}}{1-x^n}\\
&=\frac1m \sum\_{n|m} J\_k(n) \\
&= m^{k-1}
\end{split}
as required.
|
6
|
https://mathoverflow.net/users/7076
|
404685
| 165,960 |
https://mathoverflow.net/questions/404619
|
4
|
Let $X$ be a compact metric space (without isolated points). The $\infty$-Wasserstein distance $W\_\infty$ on the space of Borel probability measures on $X$ can be described as $$W\_\infty(\mu,\nu) = \inf\{r>0 \mid \mu(U)\le\nu(U\_r)\,\forall \text{ open } U\subseteq X\},$$ where $U\_r=\{x\in X \mid d(x,U)\le r\}$. The topology induced by this distance is in general finer than the weak (or vague) topology. For example, if $X=[0,1]$ and $\delta\_x$ denotes the point mass at $x$, the measures $\mu\_n=\frac{n-1}{n}\delta\_0+\frac{1}{n}\delta\_1$ converge weakly to $\delta\_0$, but $W\_\infty(\mu\_n,\delta\_0)=1$ for every $n$. On the other hand, $W\_\infty$ dominates the Levy-Prokhorov distance, so $W\_\infty$-convergence does imply weak convergence.
**Question**: is $W\_\infty$-convergence equivalent to weak convergence if one restricts to *fully supported* measures?
(By *fully supported* I mean those measures $\mu$ such that $\mu(U)>0$ for every nonempty open set $U\subseteq X$.)
I believe this is the case at least for $X=[0,1]$, but I just want to make sure it is not obviously ridiculous!
This is a duplicate of a [question I asked](https://math.stackexchange.com/questions/4257317/the-infinity-wasserstein-distance-w-infty-and-the-weak-topology) on stackexchange, but perhaps it is too specialised for that site.
|
https://mathoverflow.net/users/384059
|
The infinity Wasserstein distance $W_\infty$ and the weak topology
|
First, I interpret your condition that "$X$ has no isolated points" in the following ways: First, every ball $B(x, \varepsilon)$ has non-empty interior. This means, in particular, that we can find arbitrarily fine partitions of $X$ with sets that each have non-empty interior. And second, if we have $X = \Omega\_1 \stackrel{.}{\cup} \Omega\_2$ for two non-empty sets $\Omega\_1, \Omega\_2$, then $\inf\_{x\_1 \in \Omega\_1, x\_2 \in \Omega\_2} d(x\_1, x\_2) = 0$. I am not entirely sure if these conditions can be stated in simpler terms, but they don't seem unreasonable to me.
If this is what you had in mind (it's certainly satisfied for $[0, 1]$ or convex and compact subsets of $\mathbb{R}^d$), then I believe the two notions of convergence are indeed equivalent, see the proposed proof below. It's rather lengthy, so I hope I didn't miss anything. Please let me know if the proof seems incomplete.
Let $\mu \in \mathcal{P}(X)$ be fully supported and $(\mu\_k)\_{k \in \mathbb{N}}$ be a sequence in $\mathcal{P}(X)$ such that $\mu\_k$ converges weakly to $\mu$. In particular, $d\_P(\mu\_k, \mu) \rightarrow 0$ for the Prokhorov metric $d\_P$.
Let $\mathcal{F}\_1, \mathcal{F}\_2, ...$ be a sequence of refinements of partitions of $X$, i.e., $\mathcal{F}\_N = \{\Omega\_{N, 1}, \dots, \Omega\_{N, N}\}$, $X = \stackrel{.}{\cup}\_{i=1}^N \Omega\_{N, i}$, such that $\delta(N) := \max\_{i=1, \dots, N} \max\_{x, y \in \Omega\_{N, i}}d(x, y) \rightarrow 0$ for $N \rightarrow 0$, each $\Omega\_{N, i}$ has non-empty interior and choose some fixed $x\_{N, i} \in \Omega\_{N, i}$. Note that by the "connectedness condition" on X, for any $J \subsetneq \{1, \dots, N\}$, it holds $\min\_{j \in J} \min\_{i \not\in J} d(x\_i, x\_j) \leq 2 \delta(N)$.
Define $\bar\mu^N := \sum\_{i=1}^N \delta\_{x\_{N, i}} \mu(\Omega\_{N, i})$. It is clear that $W\_\infty(\mu, \bar\mu^N) \leq \delta(N)$. The fact that the partitions are refinements implies, in particular, that $\bar\mu^{N\_2}(\Omega\_{N\_1, i}) = \bar\mu^{N\_1}(\Omega\_{N\_1, i})$ for $N\_2 \geq N\_1$.
Define $w(N) := \min\_{i=1, \dots, N} \mu(\Omega\_{N, i})$, which is positive by assumption on $\mu$.
Note that for any $N\_1, N\_2 \in \mathbb{N}$ with $N\_2 \geq N\_1$ and any $A \subset X$ open such that $\bar\mu^{N\_2}(A^{3 \delta(N\_1)}) \neq 1$, it holds
$$
\bar\mu^{N\_2}(A^{4 \delta(N\_1)}) \geq \bar\mu^{N\_2}(A^{2 \delta(N\_1)}) + w(N\_1),
$$
the reason being as follows: $\bar\mu^{N\_2}(A^{3 \delta(N\_1)}) \neq 1$ implies that there exists some $i \in \{1, \dots, N\_1\}$ such that $\Omega\_{N\_1, i} \cap A^{2 \delta(N\_1)} = \emptyset$ (since otherwise $A^{3 \delta(N\_1)} = X$). In particular, by connectedness of $X$ there exists such an $i$ such that $\Omega\_{N\_1, i} \subset A^{4 \delta(N\_1)}$. I.e.,
\begin{align}
\bar\mu^{N\_2}(A^{4 \delta(N\_1)}) &\geq \bar\mu^{N\_2}(A^{2 \delta(N\_1)}) + \bar\mu^{N\_2}(\Omega\_{N\_1, i})\\ &= \bar\mu^{N\_2}(A^{2 \delta(N\_1)}) + \bar\mu^{N\_1}(\Omega\_{N\_1, i})\\ &\geq \bar\mu^{N\_2}(A^{2 \delta(N\_1)}) + w(N\_1).
\end{align}
Now for $\varepsilon > 0$, we choose $N\_1$ large enough so that $\delta(N\_1) < \varepsilon$. Let $r(N\_1) := \min\{2 \delta(N\_1), w(N\_1)\}$. Choose $N\_2, k$ large enough so that $d\_P(\bar\mu^{N\_2}, \mu\_k) \leq r(N\_1)$.
Let $A \subset X$ open. Either $\bar\mu^{N\_2}(A^{3 \delta(N\_1)}) = 1$, and hence $\bar\mu^{N\_2}(A^{4 \delta(N\_1)}) \geq \mu\_k(A)$ holds trivially. Or, $\bar\mu^{N\_2}(A^{3 \delta(N\_1)}) < 1$ and hence by the above
$$
\bar\mu^{N\_2}(A^{4 \delta(N\_1)}) \geq \bar\mu^{N\_2}(A^{2 \delta(N\_1)}) + w(N\_1) \geq \bar\mu^{N\_2}(A^{r(N\_1)}) + r(N\_1) \geq \mu\_k(A),
$$
where the last step follows since $d\_P(\bar\mu^{N\_2}, \mu\_k) \leq r(N\_1)$. Therefore, $W\_\infty(\bar\mu^{N\_2}, \mu\_k) \leq 4 \delta(N\_1) \leq 4 \varepsilon$. By the triangle inequality, $W\_\infty(\mu, \mu\_k) \leq W\_\infty(\bar\mu^{N\_2}, \mu\_k) + W\_\infty(\bar\mu^{N\_2}, \mu) \leq 5 \varepsilon$, which yields the claim.
|
3
|
https://mathoverflow.net/users/106046
|
404689
| 165,961 |
https://mathoverflow.net/questions/404688
|
7
|
Let $(X\_i)\_{i\in \mathbb{N}}$ iid random variables such that there exists $\alpha>0$ where $\mathbb{P}\left(X\_1\in [x,x+1]\right)\leq \alpha$ for all $x\in \mathbb{R}$. Assume $\alpha$ small enough, does there exist a universal constant $C>0$ so that $$\mathbb{P}\left(\sum\_{i=1}^N X\_i\in [x,x+1]\right)\leq \frac{C\alpha}{\sqrt{N}}$$ for all $x\in\mathbb{R}$ and $N\in \mathbb{N}^\*$?
A few remarks :
* From the central limit theorem this is very natural to be expected. However here we do not assume $\mathbb{E}(X^2)<\infty$ and it is not so clear to me how to use the characteristic functions with the hypothesis.
* In the case of discrete variable $X\in \mathbb{Z}$, one should just do the computation. However I can't see why should it be the "worse case scenario" and how to compare it with the more general case.
|
https://mathoverflow.net/users/99045
|
Regularity for the sum of iid random variables
|
Yes, this is a special case of an inequality by Kesten (see [Theorem 2 and Corollary 1](https://www.mscand.dk/article/view/10950/8971)).
In particular, letting $L=\lambda=:t$ in that Corollary 1, we get the following:
>
> Let $X\_1,\dots,X\_n$ be independent identically distributed random variables, with $S\_n:=\sum\_1^n X\_k$. Then for any positive real $t$,
> $$Q(S\_n,t)\le\frac{CQ(X\_1,t)}{\sqrt{(1-Q(X\_1,t))\, n}},$$
> where $C$ is a universal positive real constant and $Q(X,u):=\sup\_{x\in\mathbb R}P(x\le X\le x+u)$ for positive real $u$.
>
>
>
|
7
|
https://mathoverflow.net/users/36721
|
404697
| 165,964 |
https://mathoverflow.net/questions/404699
|
1
|
I have a question about the proof of lemma 6.4.12 in the book Algebraic Operads (Loday-Vallette) which I do not seem to be able to fully complete on my own. Hopefully, somebody here can point out what I am not seeing.
Let me sketch the situation. Given a cooperad $(C,\Delta,\epsilon)$ and a operad $(P,\gamma,\eta)$, we consider its free right P-module $C \circ P$. Let $\alpha: C \longrightarrow P$ be a twisting morphism (of degree $-1$). The map $d^r\_\alpha$ is then defined as follows
$$ C \circ P \overset{\Delta\_{(1)} \circ P}{\longrightarrow} (C \circ\_{(1)} C) \circ P \overset{(C \circ\_{(1)} \alpha) \circ P}{\longrightarrow} (C \circ\_{(1)} P) \circ P \cong C \circ (P ; P\circ P) \overset{C \circ (P;\gamma)}{\longrightarrow} C \circ (P ; P) \cong C \circ P$$
In concrete terms,
$$d^r\_\alpha( c ; p\_1,\ldots,p\_n) = \sum\_i \pm (c\_{(1)}; p\_1,\ldots, \alpha c\_{(2)} \circ (p\_i,\ldots,p\_j), \ldots,p\_n) $$
where $\Delta\_{(1)}(c) = c\_{(1)} \otimes (i, c\_{(2)})$ denotes the sweedler notation.
The proof of the lemma then states the following
$$ [ d^r\_\alpha , d^r\_\alpha ] = d^r\_{[\alpha,\alpha]}$$
which reduces to
$$ (d^r\_\alpha)^2 = d^r\_{\alpha \* \alpha}$$
The book says the computation is similar to the case of twisted complex for associative (co)algebras. However, I think extra terms appear on the left-hand side, namely
$$ (c\_{(11)}; p\_1, \ldots, \alpha c\_{(12)} \circ (p\_i,\ldots,p\_j), \ldots ,\alpha c\_{(2)} \circ (p\_k,\ldots,p\_l), \ldots ,p\_n)$$
and
$$(c\_{(11)}; p\_1, \ldots, \alpha c\_{(2)} \circ (p\_i,\ldots,p\_j), \ldots ,\alpha c\_{(12)} \circ (p\_k,\ldots,p\_l), \ldots ,p\_n) $$
Even using coassociativity, I still do not see how these terms cancel as they are not of the right shape? Perhaps I am overlooking something simple?
Thanks in advance for taking your time to read and help!
|
https://mathoverflow.net/users/123496
|
Differential of the Twisted complex for algebraic operads
|
Since $\alpha$ is of degree $-1$, these terms come in pairs appearing with opposite signs that cancel each other. In other words, the (co)associativity for (co)operads has a sequential axiom and a parallel axiom, and these terms vanish because of the parallel axiom. A nice way to see that is to draw these as trees and remember that trees have levels helping you to remember the order of insertions, and that exchanging levels creates Koszul signs.
|
3
|
https://mathoverflow.net/users/1306
|
404701
| 165,965 |
https://mathoverflow.net/questions/404664
|
9
|
Let $\nu\_p(n)$ denote the $p$-adic valuation of $n$, i.e. the highest power of $p$ dividing $n$.
Consider the following two $q$-series formed by infinite products
$$\prod\_{n\geq1}\left(\frac{1+q^n}{1-q^n}\right)^2=\sum\_{k\geq0}a\_k\,q^k \qquad \text{and} \qquad
\prod\_{n\geq1}\left(\frac{1+q^n}{1-q^n}\right)^n=\sum\_{k\geq0}b\_k\,q^k.$$
Both $a\_k$ and $b\_k$ have combinatorial interpretations in the context of partitions. One reference for $b\_k$ would be: Corteel, S., Savelief, C., Vuletić, M.: Plane overpartitions and cylindric partitions. J. Combin. Theory Ser. A 118(4), 1239–1269 (2011). Some references for $a\_k$ include: Jeremy Lovejoy, Overpartition pairs, Annales de l'institut Fourier, vol.56, no.3, p.781-794, 2006.
I would like to ask:
>
> **QUESTION.** Is this true? If $k=j^2\geq1$ is a perfect square, then we have $\nu\_2(a\_k)=2=\nu\_2(2b\_k)$.
>
>
>
**ADDED.** I thought it might be proper to record the following extension: if
$$\prod\_{n\geq1}\left(\frac{1+q^n}{1-q^n}\right)^r=\sum\_{k\geq0}a\_k^{(r)}q^k$$
and (once again) $k=j^2$, then $\nu\_2(a\_k^{(r)})=\nu\_2(2r)$ independent of $j$.
|
https://mathoverflow.net/users/66131
|
$2$-adic valuations: a tale of two $q$-series
|
First notice that
$$\frac{1+q^n}{1-q^n} = 1 + 2\frac{q^n}{1-q^n}.$$
Computing modulo $8$, we have
$$\left(\frac{1+q^n}{1-q^n}\right)^2 \equiv 1 + 4\frac{q^n}{1-q^{2n}}\pmod8.$$
Correspondingly,
$$\prod\_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^2 \equiv 1+ 4\sum\_{n\geq 1} \frac{q^n}{1-q^{2n}}\pmod8.$$
For $k=2^s m$ with $s\geq 0$ and odd $m$, we have
$$a\_k = [q^k]\ \prod\_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^2 \equiv 4\tau(m)\pmod8,$$
where $\tau(m)$ is the number of divisors of $m$. If $k$ is a square, then $\tau(m)$ is odd, proving the result for $\nu\_2(a\_k)$.
---
Similarly, we have
$$\left(\frac{1+q^n}{1-q^n}\right)^n \equiv 1 + 2n\frac{q^n}{1-q^n}\pmod4$$
and
$$\prod\_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^n\equiv 1+ 2\sum\_{n\geq 1} n\frac{q^n}{1-q^{n}}\pmod4.$$
And again for $k=2^sm$, we have
$$b\_k = [q^k]\ \prod\_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^n \equiv 2\tau(m)\pmod{4},$$
proving the result for $\nu\_2(b\_k)$.
---
For the **ADDED** part, with the help of Lemma 4.7 for $p=2$ in [this paper](https://arxiv.org/abs/2004.14000) we have
$$\prod\_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^r\equiv 1+ 2r\sum\_{n\geq 1} \frac{q^n}{1-q^{(1+[t>0])n}}\pmod{2^{t+2}},$$
where $t:=\nu\_2(r)$. Then
$$a\_k^{(r)} = [q^k]\ \prod\_{n\geq 1}\left(\frac{1+q^n}{1-q^n}\right)^r \equiv 2r\cdot\tau(k)\pmod{2^{t+2}},$$
implying that $\nu\_2(a\_k^{(r)}) = \nu\_2(2r) = t+1$ when $k$ is a square.
|
14
|
https://mathoverflow.net/users/7076
|
404708
| 165,968 |
https://mathoverflow.net/questions/404713
|
2
|
[A question I remember from many years ago.]
Definition
>
> A **Riemann surface** is a connected complex manifold $X$ of complex dimension one. This means that $X$ is a connected Hausdorff space that is endowed with an atlas of charts to the open unit disk of the complex plane: for every point $x \in X$ there is a neighbourhood of $x$ that is homeomorphic to the open unit disk of the complex plane, and the transition maps between two overlapping charts are required to be holomorphic.
>
> [source](https://en.wikipedia.org/wiki/Riemann_surface)
>
>
>
When we define ordinary manifolds, we have to add the assumption "paracompact" if we want to rule out strange manifolds like the [long line](https://en.wikipedia.org/wiki/Long_line_(topology)). But that assumption is not included in the definition above. What is a reference for the (I think I remember) fact that a Riemann surface as defined above is necessarily paracompact?
|
https://mathoverflow.net/users/454
|
A Riemann surface is automatically paracompact
|
This is precisely the statement of [Radó's theorem](https://en.wikipedia.org/wiki/Rad%C3%B3%27s_theorem_(Riemann_surfaces)) (modulo the standard equivalence between paracompactness and second-countability for connected manifolds). I believe there are multiple proofs available, one of which is in section 1.3 of [Hubbard's book](http://matrixeditions.com/TeichmullerVol1.html) (referenced on Wikipedia.)
|
5
|
https://mathoverflow.net/users/30186
|
404714
| 165,969 |
https://mathoverflow.net/questions/404724
|
41
|
Linear algebra as we learn it as undergraduates usually holds for any field (even though we usually learn it for the complex, or real, numbers).
I am looking for a list of concepts, and results, in linear algebra that actually depend on the choice of field.
To start I propose the notion of an complex valued inner product. Here the anti-linear axiom requires an involution on the field.
|
https://mathoverflow.net/users/167165
|
Results in linear algebra that depend on the choice of field
|
The existence of Chevalley–Jordan decompositions depends on the perfectness of the field.
|
47
|
https://mathoverflow.net/users/1409
|
404737
| 165,976 |
https://mathoverflow.net/questions/404705
|
4
|
How can we from $\sum\_{n\leqslant x}\mu (n)=o(x)$ deduce $\sum\_{n\leqslant x}(-1)^{\omega(n)}=o(x)$, where $\omega(n)$ is the number of different primes dividing $n$ and
$\mu (n)$ is the Möbius function?
Heuristic: Up to x there must be roughly the same portion of numbers with odd and even number of prime divisors. First equality ( equivalent to prime number theorem ) prooves that for square free numbers, and there is no reason why should it not be valid for all numbers
|
https://mathoverflow.net/users/169583
|
Connection between Möbius and function
|
Yes, that deduction is possible, but somewhat indirectly.
The estimate $\sum\_{n \leq x} \mu(n) = o(x)$ is equivalent to the nonvanishing of $\zeta(s)$ on ${\rm Re}(s) = 1$ since both conditions are known to be equivalent to the prime number theorem. (Maybe one can directly deduce the estimate and the nonvanishing from each other without going *through* the prime number theorem, but that's a side issue here.) The function $\zeta(s)$ is known by elementary techniques to be meromorphic on ${\rm Re}(s) = 1$ with a simple pole at $s = 1$, so the nonvanishing of $\zeta(s)$ on that line is equivalent to the function $1/\zeta(s)$ being analytic on ${\rm Re}(s) = 1$. I will show $\sum\_{n \leq x} (-1)^{\omega(n)} = o(x)$ follows from analyticity of $1/\zeta(s)$ on ${\rm Re}(s) = 1$, so in a rather indirect way $\sum\_{n \leq x} (-1)^{\omega(n)} = o(x)$ follows from the estimate $\sum\_{n \leq x} \mu(n) = o(x)$.
Set $F(s) := \sum\_{n \geq 1} (-1)^{\omega(n)}/n^s$ for ${\rm Re}(s) > 1$.
As David Speyer observed in his answer, for ${\rm Re}(s) > 1$ we have
$$
F(s) =
\prod\_p \frac{1-2/p^s}{1-1/p^s} = \frac{1}{\zeta(s)}\prod\_p \frac{1-2/p^s}{(1-1/p^s)^2}
$$
and the $p$-th term on the right for ${\rm Re}(s) \geq 1$ is $1 + O(1/p^{2s})$ with $O$-constant $4$ since $|1/p^{s}| \leq 1/2^{{\rm Re}(s)} \leq 1/2$ and
$(1-2z)/(1-z)^2 = 1 + O(z^2)$ for $|z| \leq 1/2$ with $O$-constant $4$: $(1-2z)/(1-z)^2 - 1 = -z^2/(1-z)^2$ and $|1/(1-z)^2| \leq 4$. So the infinite product over $p$ on the right, which David calls $B(s)$, is holomorphic for ${\rm Re}(s) \geq 1$ (in fact, $B(s)$ converges and is holomorphic for ${\rm Re}(s) > 1/2$ since $\sum 1/p^{2s}$ converges for ${\rm Re}(s) > 1/2$). Since $\zeta(s)$ is meromorphic on ${\rm Re}(s) \geq 1$, $F(s)$ has a meromorphic continuation to ${\rm Re}(s) \geq 1$.
From $1/\zeta(s)$ being analytic on ${\rm Re}(s) \geq 1$, $F(s)$ has an analytic continuation to ${\rm Re}(s) \geq 1$. Now we will apply the following theorem, which is a consequence of Newman's Tauberian theorem (one of the approaches to proving the prime number theorem).
**Theorem**. *If the Dirichlet series* $\sum\_{n \geq 1} b\_n/n^s$ *converges for* ${\rm Re}(s) > 1$, $|b\_n| \leq 1$, *and the series has an analytic continuation to* ${\rm Re}(s) \geq 1$ *then* $(1/x)\sum\_{n \leq x} b\_n \to 0$ *as* $x \to \infty$.
In this theorem use $b\_n = (-1)^{\omega(n)}$ to see that analyticity of $1/\zeta(s)$ on ${\rm Re}(s) = 1$ implies $(1/x)\sum\_{n \leq x}(-1)^{\omega(n)} \to 0$ as $x \to \infty$.
What about a converse: does $(1/x)\sum\_{n \leq x}(-1)^{\omega(n)} \to 0$ as $x \to \infty$ imply $1/\zeta(s)$ is analytic on ${\rm Re}(s) =1$ (and thus imply $\sum\_{n \leq x} \mu(n) = o(x)$)? Set $A(x) = \sum\_{n \leq x} (-1)^{\omega(n)}$, so trivially $|A(x)/x|$ is bounded and we are assuming that in fact $A(x)/x \to 0$ as $x \to \infty$.
We already pointed out that the Dirichlet series $F(s) = \sum (-1)^{\omega(n)}/n^s$ has a meromorphic continuation to ${\rm Re}(s) \geq 1$ without using any information about possible zeros of $\zeta(s)$ on the line ${\rm Re}(s) = 1$.
**Theorem**. *Assume a Dirichlet series* $G(s) := \sum\_{n \geq 1} a\_n/n^s$ *converges for* ${\rm Re}(s) > 1$ *and* $G(s)$ *has a meromorphic continuation to* ${\rm Re}(s) \geq 1$. *If* $(1/x)\sum\_{n \leq x} a\_n \to 0$ *as* $x \to \infty$ *then* $G(s)$ *is holomorphic on the line* ${\rm Re}(s) = 1$.
Use this theorem with $a\_n = (-1)^{\omega(n)}$ to see that if $A(x)/x \to 0$ as $x \to \infty$, then $F(s)$ is holomorphic on ${\rm Re}(s) = 1$. What does analyticity of $F(s)$ on ${\rm Re}(s) = 1$ tell us about analyticity of $1/\zeta(s)$ on that line? In the formula for $F(s)$ as an infinite product over primes when ${\rm Re}(s) > 1$, let's extract the $p = 2$ term:
$$
F(s) =
\prod\_p \frac{1-2/p^s}{1-1/p^s} = \frac{1}{\zeta(s)}\frac{1-2/2^s}{(1-1/2^s)^2}\prod\_{p \geq 3} \frac{1-2/p^s}{(1-1/p^s)^2}.
$$
To reciprocate the product on the right over $p \geq 3$, let's see how quickly the reciprocals of the factors tend to $1$.
For $|z| \leq 1/3$ we have $(1-z)^2/(1-2z) = 1 + z^2/(1-2z)$ and $|1/(1-2z)| \leq 3$, so
$(1-z)^2/(1-2z) = 1 + O(z^2)$ with $O$-constant $3$.
Therefore $(1-1/p^s)^2/(1-2/p^s) = 1 + O(1/p^{2s})$ for ${\rm Re}(s) \geq 1$ and $p \geq 3$ with $O$-constant 3, so
$$
\frac{1}{\zeta(s)}\frac{1-2/2^s}{(1-1/2^s)^2} = F(s)\cdot\prod\_{p \geq 3} \frac{(1-1/p^s)^2}{1-2/p^s}
$$
where the product on the right converges and is analytic for ${\rm Re}(s) \geq 1$ (and in fact for ${\rm Re}(s) > 1/2$). Therefore analyticity of $F(s)$ on the line ${\rm Re}(s) = 1$ implies the expression on the left side of the displayed equation just above is analytic on ${\rm Re}(s) = 1$. The factor $1-1/2^s$ is nonvanishing on ${\rm Re}(s) > 0$, so
$(1-2/2^s)/\zeta(s)$ is analytic on ${\rm Re}(s) = 1$. Since $1-2/2^s$ is analytic on ${\rm Re}(s) = 1$ with zeros only at the points $s\_k = 1 + 2\pi ik/\log 2$ where $k \in \mathbf Z$, the meromorphic function $1/\zeta(s)$ is analytic on
the line ${\rm Re}(s) = 1$ except for possible poles at the points $s\_k$.
All that remains to be done to get the desired converse result is show $1/\zeta(s)$ has no pole at each $s\_k$, or equivalently that $\zeta(s)$ is nonvanishing at each $s\_k$. I don't see right now how to do that in the context of this problem. (While we *know* $\zeta(s) \not= 0$ on the whole line ${\rm Re}(s) = 1$ from methods used to prove the prime number theorem, it's not fair to use such arguments here.) I am reminded here of the trick of comparing the products $(1-2/2^s)\zeta(s)$ and $(1-3/3^s)\zeta(s)$ in one of the proofs that $\zeta(s)$ is analytic on ${\rm Re}(s) > 0$ except for a simple pole at $s = 1$, but I don't see how to modify that.
If someone has a useful suggestion to finish this argument, please put it in the comments.
|
9
|
https://mathoverflow.net/users/3272
|
404743
| 165,978 |
https://mathoverflow.net/questions/404706
|
7
|
For the sake of this question, we'll model a six functor formalism in the following way. Let $\mathsf{C}$ be a category *of spaces* (be it the category of schemes, or topological spaces) and consider a triangulated closed symmetric monoidal category $\mathsf{D}(X)$, with identity $\mathscr{O}\_X$, for each $X\in\mathsf{C}$.
Given a morphism $f:X\to Y$ in $\mathsf{C}$, we have functors $f\_\*,f\_!:\mathsf{D}(X)\leftrightarrows \mathsf{D}(Y):f^\*,f^!$ such that $f^\*\dashv f\_\*$, $f\_!\dashv f^!$, and $f^\*$ is strong symmetric monoidal.
If we also suppose that $\mathsf{C}$ has a final object $S$, it is natural to define cohomology and cohomology with compact support to be
$$H^i(X,M):=\hom\_{\mathsf{D}(S)}(\mathscr{O}\_S,p\_\*M[i])\qquad\text{and}\qquad H\_c^i(X,M):=\hom\_{\mathsf{D}(S)}(\mathscr{O}\_S,p\_!M[i]),$$
where $p:X\to S$ is the natural morphism. If $\mathsf{C}$ is the category of ringed spaces (or even of ringed sites, I think), this coincides with the usual definitions.
**I wonder then how could we obtain some sort of duality isomorphism similar to Poincaré and Serre duality.** Perhaps we could begin with
$$\operatorname{Ext}^i(M,p^!\mathscr{O}\_S)=\hom\_{\mathsf{D}(X)}(M,p^!\mathscr{O}\_S[i])=\hom\_{\mathsf{D}(S)}(p\_!M[-i],\mathscr{O}\_S)$$
and the latter should be something like $H^{-i}\_c(X,M)^\vee$, but I'm not sure how to make this precise.
|
https://mathoverflow.net/users/131975
|
How duality follows from a six functor formalism
|
Your description of the six functors does not mention any relations between the $!$-functors and the $\*$-functors or the tensor product, which is where these dualities are hiding.
Poincaré duality is a relation between the $\*$-functors and the $!$-functors. Typically there are canonical isomorphisms $f\_!=f\_\*$ when $f$ is proper and $f^!=f^\*[d]$ when $f$ is nice (e.g. smooth) of relative dimension $d$. Depending on how the $!$-functors are constructed, one of these isomorphisms will hold by definition but the other one will be nontrivial to prove. When $f:X\to S$ is nice, we get isomorphisms (where $1\_S$ is the unit object in $D(S)$)
$$
H^n(X,M) := \mathrm{hom}(1\_S, f\_\*f^\*(M)[n]) = \mathrm{hom}(1\_S,f\_\*f^!(M)[n-d])=:H\_{d-n}^\mathrm{BM}(X,M)
$$
and
$$
H^n\_c(X,M):= \mathrm{hom}(1\_S, f\_!f^\*(M)[n]) = \mathrm{hom}(1\_S,f\_!f^!(M)[n-d])=:H\_{d-n}(X,M),
$$
On the other hand, there are vertical identifications when $f$ is proper.
Serre duality uses a relation between the $!$-functors and the tensor product. Namely, for $f:X\to S$, the functor $f\_!$ is $D(S)$-linear; in particular we have the projection formula $f\_!(A\otimes f^\*(B))= f\_!(A)\otimes B$. By adjunction this implies the isomorphism
$$
f\_\*\mathrm{Hom}(A, f^!(B)) = \mathrm{Hom}(f\_!(A),B).
$$
One gets Serre duality when $f$ is proper (so $f\_!=f\_\*$) and $B=1\_S$ (so $f^!(B)$ is the "dualizing sheaf"). Combining the projection formula with Poincaré duality we can also deduce that $f\_\*$ preserves $\otimes$-dualizable objects when $f$ is both proper and nice, which gives a perfect pairing. But the usual formulation of Serre duality in terms of cohomology is specific to the derived category of quasi-coherent sheaves when $S=\mathrm{Spec}(k)$ is a field (note that in the quasi-coherent case the adjunction $(f\_!,f^!)$ only exists when $f$ is proper).
There are further duality statements, such as Verdier duality, which says that $\mathrm{Hom}(-,f^!(1\_S))$ is an equivalence of a certain subcategory $D\_c(X)\subset D(X)$ of "constructible"/"coherent" objects with its opposite. But unlike the previous two, this statement does not directly follow from the various relations between the six functors.
|
11
|
https://mathoverflow.net/users/20233
|
404750
| 165,980 |
https://mathoverflow.net/questions/404711
|
3
|
I am looking at the following Riemann surface (let's call it $M$),
\begin{equation}
y^n=\frac{(x-x\_1)(x-x\_3)}{(x-x\_2)(x-x\_4)}
\end{equation}
which is a Riemann surface of genus $n-1$. It can be thought of as a quotient of the complex plane by a Schottky group $\Sigma$,
\begin{equation}
M\cong\mathbb{C}'/\Sigma
\end{equation}
where $\mathbb{C}'$ is the domain of discontinuity of $\Sigma$. Then, we can look at the covering,
\begin{equation}
\pi\_{\Sigma}:\mathbb{C}'\rightarrow \mathbb{C}'/{\Sigma}
\end{equation}
which takes $z$ to $[z]$. The inverse map $w=\pi^{-1}\_{\Sigma}$ is multivalued on $\mathbb{C}'$,
\begin{equation}
w\sim\gamma w
\end{equation}
where $\gamma\in\Sigma$. However, the Schwarzian derivative of $w$ in some coordinate patch $z$ is single valued,
\begin{equation}
\{w,z\}=\frac{w'''}{w'}-\frac{3}{2}\left(\frac{w''}{w'}\right)^2.
\end{equation}
On overlapping coordinate patches, it transforms like,
\begin{equation}
\{w',z'\}=\left(\frac{dz}{dz'}\right)^2\{w,z\}+\{z',z\}
\end{equation}
which means it is a projective connection on $M$. Also, near the $z\_i$s, since we can use the coordinate, $y^n\sim(z-z\_i)$,
\begin{equation}
\{w',y\}\sim\{z,y\}=-\frac{1}{2}\frac{n^2-1}{y^2}+\cdots.
\end{equation}
According to [Faulkner - The Entanglement Renyi Entropies of Disjoint Intervals in AdS/CFT](https://arxiv.org/abs/1303.7221), these properties are enough to fix the Schwarzian derivative up to $3g-3$ unknown accessory parameters,
\begin{equation}
\{w,z\}=\Delta\left(\sum\_{i=1}^4\frac{1}{(z-z\_i)^2}+\frac{-z\_3+z\_1+z\_2+z\_4-2z}{(z-z\_1)(z-z\_2)(z-z\_4)}\right)+\sum\_{s=1}^{3(n-2)}p\_s\omega\_s
\end{equation}
where $\omega$s are a basis for the quadratic differentials on $M$ and $\Delta=1/2(1-1/n^2)$. I understand that we can add any linear combination of quadratic differntials on $M$ to the form of the Schwarzian because they transform uniformly, but the first term in the brackets is not clear to me. It should transform like a projective connection. Does anyone have a justification for including the first term multiplying $\Delta$?
|
https://mathoverflow.net/users/351553
|
Schwarzian derivative, accessory parameters, projective connections
|
First of all, you should take care about which coordinates you take. Namely, Faulkner uses the coordinate $x$ given by the defining equation $$y^n=\frac{(x-x\_1)(x-x\_2)}{(x-x\_3)(x-x\_4)}$$ as a branched projective structure (he denotes this by $z$ in his paper!). He then computes the Schwarzian derivative of $w$ - the Schottky uniformisation - with respect to $x$. As $x$ is a well-defined branched projective structure $${w,x}(dx)^2$$ is a well-defined meromorphic differential on $M$, having prescribed singularities at $x\_1,\dots,x\_4$. He derives in equation (4.10) in his paper that the quadratic residues must be $\Delta$. This is well-known since Mandelbaum.
The second term in the brackets is needed as ${w,x}(dx)^2$ has no pole over $x=\infty.$ It might look a bit unnatural as it has no pole at $z\_3$, but in fact $\frac{(dx)^2}{(x-x\_1)\dots(x-x\_4)}$ is a holomorphic quadratic differential.
|
1
|
https://mathoverflow.net/users/4572
|
404752
| 165,982 |
https://mathoverflow.net/questions/404761
|
2
|
To generate a uniform distribution on a sphere $S^n$ in $\mathbb R^{n+1}$, we can normalize a vector whose entries are $n+1$ i.i.d normal random variables. If $\rho$ is a correlation, $|\rho|<1$, how can we generate a uniform distribution on the manifold
$$
M = \left \{ x, y \in S^n: x^Ty = \rho \right\}\ ?
$$
|
https://mathoverflow.net/users/97209
|
Uniform distribution on a manifold
|
It can be done in pretty much the same way as for a single vector by using the fact that if you fix $x$, then the conditional distribution of $y-x$ is uniform on the sphere of radius $\sqrt{1-\gamma^2}$ in the hyperplane perpendicular to $x$. Therefore, first you generate $x$ (as you say, by normalizing a vector with i.i.d. standard normal coordinates), then you complement $x$ to an orthonormal basis $(x,e\_2,\dots, e\_n)$, generate, as above, a vector $y'$ uniformly distributed on the unit sphere of the space spanned by $e\_2,\dots, e\_n$, and finally put $y=\gamma x + \sqrt{1-\gamma^2} y'$.
PS While I was typing you replaced $\gamma$ with $\rho$ and $n$ with $n+1$.
|
2
|
https://mathoverflow.net/users/8588
|
404769
| 165,986 |
https://mathoverflow.net/questions/404760
|
20
|
**Example:** How can you guess a polynomial $p$ if you know that $p(2) = 11$? It is simple: just write 11 in binary format: 1011 and it gives the coefficients: $p(x) = x^3+x+1$. Well, of course, this polynomial is not unique, because $2x^k$ and $x^{k+1}$ give the same value at $p=2$, so for example $2x^2+x+1$, $4x+x+1$ also satisfy the condition, but their coefficients have greater absolute values!
**Question 1:** Assume we want to find $q(x)$ with integer coefficients, given its values at some set of primes $q(p\_i)=y\_i$ such that $q(x)$ has the least possible coefficients. How should we do it? Any suggestion/algorithm/software are welcome. (Least coefficients means: the least maximum of modulii of coefficients).
**Question 2:** Can one help to guess the polynomial $p$ such that $p(3) = 221157$, $p(5) = 31511625$ with the smallest possible integer coefficients? (Least maximum of modulii of coefficients). Does it exist?
(That example comes from the [question MO404817](https://mathoverflow.net/q/404817/10446) on count of 3x3 anticommuting matrices $AB+BA=0$ over $F\_p$.)
(The degree of the polynomial seems to be 10 or 11. It seems divisible by $x^3$, and I have run a brute force search bounding absolute values of the coefficients by 3, but no polynomial satisfying these conditions is found, so I will increase the bound on the coefficients and will run this search again, but the execution time grows too quickly as the bound increases and it might be that brute force is not a good choice).
**Question 3:** Do conditions like $q(p\_i)=y\_i$ imply some bounds on coefficients? E.g., can we estimate that the coefficients are higher than some bound?
|
https://mathoverflow.net/users/10446
|
How can you find an integer coefficient polynomial knowing its values only at a few points (but requiring the coefficients be small)?
|
You can certainly do better than brute force by considering modular constraints. If the solution is $p(x) = \sum\_i a\_i x^i$ then $p(x) - \sum\_{j=0}^{n-1} a\_j x^j$ is divisible by $x^n$ and $$\frac{p(x) - \sum\_{j=0}^{n-1} a\_j x^j}{x^n} = a\_n \pmod x$$
Solving for $a\_0$ in each of the given bases and using the Chinese remainder theorem gives an equivalence class for $a\_0$; for each possible value of $a\_0$ you can expand similarly for $a\_1$; and traversing this tree in order of increasing cost of the coefficients gives a directed search. This works in principle for any cost function which increases when any coefficient increases in absolute value.
[This Python code](https://gist.github.com/pjt33/c942040a9b8fa9bacc7f5d1bccc7ac2b) implements the idea and finds two polynomials with sum of absolute values of 29:
$$-2x^3 + 4x^4 + 3x^5 + 11x^6 + x^7 + 5x^8 + 3x^{10} \\
-2x^3 + 4x^4 + 3x^5 - 4x^6 + 9x^7 + 4x^8 + 3x^{10}$$
and one polynomial with maximum absolute value of 7:
$$-2x^3 + 4x^4 + 3x^5 - 4x^6 - 6x^7 - 3x^8 + 7x^9 + 2x^{10}$$
in a small fraction of a second.
---
Some follow-up questions in comments brought me to the realisation that if we're trying to interpolate $\{ (x\_i, y\_i) \}$ with the $x\_i$ coprime, there is at most one polynomial with coefficients in the range $[-\lfloor \tfrac{(\operatorname{lcm} x\_i) - 1}2 \rfloor, \lfloor \tfrac{\operatorname{lcm} x\_i}2 \rfloor]$, because the tree collapses to a chain. This gives the following algorithm for finding such a polynomial, if it exists:
```
M := lcm(x_i)
while any y_i is non-zero:
find a_0 by Chinese remainder theorem
if a_0 > floor(M / 2):
a_0 -= M
output a_0
update y_i := (y_i - a_0) / x_i
```
In the long term, the initial values of the $y\_i$ are reduced to negligibility by the repeated division by $x\_i$, so eventually each $y\_i$ will be reduced to a range which is bounded by $\frac{x\_i}{x\_i - 1} M$. This means that for a given set of $x\_i$ it's possible to compute a finite directed graph to see whether existence is guaranteed.
In the particular case that the $x\_i$ are $\{3, 5\}$ there are three cycles, all of them loops: $(0,0) \to (0,0)$ is the terminating loop which indicates that a solution exists, but there are also loops $(2, 1) \to (2, 1)$ and $(-2, -1) \to (-2, -1)$.
|
26
|
https://mathoverflow.net/users/46140
|
404772
| 165,988 |
https://mathoverflow.net/questions/404774
|
1
|
Consider the [Young's lattice](https://en.wikipedia.org/wiki/Young%27s_lattice). *What is the number of paths starting from the origin (0) to a specific Young diagram?*
For instance, the Young diagram corresponding to the integer partition 1+1+1 has 1 path leading to it, 2+1 has 2 paths leading to it and 2+2 has 2 paths leading to it.
Is there a generating function to get this combinatorics?
|
https://mathoverflow.net/users/140290
|
Number of paths to a specific vertex in the Young's lattice
|
These paths are the same thing as standard Young tableaux, which are enumerated by the famous [Hook Length Formula](https://en.wikipedia.org/wiki/Hook_length_formula).
|
3
|
https://mathoverflow.net/users/33089
|
404780
| 165,989 |
https://mathoverflow.net/questions/404786
|
2
|
Naturally given any $s\in (0,1)$, the fractional Laplacian, $(-\Delta\_g)^s u$ on a closed Riemannian manifold can be defined via spectral decomposition of $-\Delta\_g$. There is another formulation of the fractional Laplacian that I commonly see that is stated as follows:
$$(-\Delta\_g)^sf(x)=\int\_0^{\infty} (e^{-t\Delta\_g}f(x)-f(x))\,t^{-1-s}\,dt.$$
Are these two definitions equivalent? Is there any reference for this?
Thanks,
|
https://mathoverflow.net/users/50438
|
Fractional Laplacian on closed manifolds
|
Yes, they are equivalent. Up to a constant missing and a sign error in the displayed formula, it should read:
$$ (-\Delta\_g)^s f(x) = \frac{1}{\Gamma(-s)} \int\_0^\infty (e^{t \Delta\_g} f(x) - f(x)) t^{-1-s} dt . $$
In fact, this is true for quite general operators. If $L^2$ theory is what you are after, this is a direct consequence of the spectral theorem and the gamma-type integral
$$\lambda^s = \frac{1}{\Gamma(-s)} \int\_0^\infty (e^{-\lambda t} - 1) t^{-1-s} dt . $$
If you are interested in more general spaces of functions, then *Bochner's subordination* is a right keyword. A standard reference is, I think, Martínez–Sans:
* C. Martínez, M. Sanz, *The Theory of Fractional Powers of Operators*.
North-Holland Math. Studies 187, Elsevier, Amsterdam (2001)
|
4
|
https://mathoverflow.net/users/108637
|
404792
| 165,993 |
https://mathoverflow.net/questions/404805
|
3
|
Let $\lambda\vdash n$ denote the integer partition of $n$. Define the product $\mathcal{N}(\lambda)=\lambda\_1\lambda\_2\cdots\lambda\_r$ when $\lambda=(\lambda\_1\geq\lambda\_2\geq\cdots\geq\lambda\_r>0)$.
Let $\gamma$ be the Euler's constant. Lehmer proved that
$$\lim\_{n\rightarrow\infty}\frac1n\sum\_{\lambda\vdash n}\frac1{\mathcal{N}(\lambda)}=e^{-\gamma}.$$
I like to ask:
>
> **QUESTION.** Does the following limit exist? Or, what is the asymptotic growth of the sum?
> $$\lim\_{n\rightarrow\infty}\sum\_{\lambda\vdash n}\frac1{\mathcal{N}(\lambda)^2}.$$
>
>
>
**ADDED.** Glad to see that Fedor Petrov's answer would (potentially) work for
$$\lim\_{n\rightarrow\infty}\sum\_{\lambda\vdash n}\frac1{\mathcal{N}(\lambda)^r}.$$
|
https://mathoverflow.net/users/66131
|
asymptotic growth of a sum involving partitions
|
The limit equals 2. We have $$
\sum\_{\lambda\vdash n}\frac1{\mathcal{N}(\lambda)^2}=[x^n]\prod\_{k=1}^\infty\frac1{1-x^k/k^2}=\sum\_{m=0}^n[x^m]\prod\_{k=2}^\infty\frac1{1-x^k/k^2}\\
=b\_0+b\_1+\ldots+b\_n,$$
where
$$
\sum b\_i x^i=\prod\_{k=2}^\infty \frac1{1-x^k/k^2}=:f(x),
$$
the standard uniform convergence argument shows that $$\sum b\_i=\sup\_{0<x<1} f(x)=f(1)=\prod\_{k=2}^\infty\frac{k^2}{k^2-1}=2.$$
|
6
|
https://mathoverflow.net/users/4312
|
404810
| 166,001 |
https://mathoverflow.net/questions/404777
|
1
|
Let $B \ge 2$ be integer and $[x]$ denote the nearest integer
to real $x$.
For $2 \le B \le 10^5$ computations with mpmath suggest:
$$ [\zeta(1-1/B)]=-B+1 \qquad (1)$$
Is (1) true for all $B \ge 2$?
|
https://mathoverflow.net/users/12481
|
On the nearest integer to $\zeta(1-1/B),B \ge 2$
|
We can make the error mentioned by Wojowu in his comment explicit by using some results on the Laurent coefficients of the zeta function. There are a few results on this, but I'll just use Theorem 2 of [this paper](https://projecteuclid.org/journals/rocky-mountain-journal-of-mathematics/volume-2/issue-1/On-the-Hurwitz-zeta-function/10.1216/RMJ-1972-2-1-151.full) by Berndt (which gives a more general result for the Hurwitz zeta function). The relevant result is
**Theorem.** (Berndt) *Write*
$$
\zeta(s) = \frac{1}{s-1} + \sum\_{n=0}^\infty \gamma\_n (s-1)^n
$$
*for the Laurent series expansion of $\zeta$ around $s=1$. Then for $n\geq 1$, we have*
$$
\left|\gamma\_n\right| \leq \frac{4}{n\pi^n}.
$$
For $B\geq 2$ an integer, write
$$
\zeta\left(1-\frac{1}{B}\right) = -B + \gamma + R(B),
$$
where $\gamma=\gamma\_0\approx .577...$ is the usual Euler-Mascheroni constant and
$$
R(B) = \sum\_{n=1}^\infty \gamma\_n \left(-\frac{1}{B}\right)^n.
$$
Applying the triangle inequality and Berndt's result, we have
$$
\left|R(B)\right| \leq \sum\_{n=1}^\infty \frac{4}{n(B\pi)^n} \leq \frac{4}{B}\sum\_{n=1}^{\infty} \frac{1}{n\pi^n}.
$$
The sum on the right converges rapidly and is numerically $\approx 0.383...$. Thus
$$
\left| R(B) \right| \leq \frac{2}{B}.
$$
(By a more careful analysis, the constant 2 can be replaced with 1.) For $B \geq 26$, it follows that
$$
\frac{1}{2} < \gamma+ R(B) < 1.
$$
Therefore
$$
\left[ \zeta\left(1-\frac{1}{B} \right) \right] = -B+1
$$
for $B \geq 26$. The remaining cases can be verified via computer.
|
7
|
https://mathoverflow.net/users/307675
|
404816
| 166,003 |
https://mathoverflow.net/questions/404803
|
2
|
Let $D$ be a smooth domain of $\mathbb{R}^d$. Let $\partial D$ denote the boundary of $D$. We denote by $B(x,r)=\{y \in \mathbb{R}^d \mid |y-x|<r\}$ the Euclidean ball centered at $x$ with radius $r$
For $x \in \partial D$, we consider the following limit:
\begin{align\*}
\lim\_{r \to 0}\frac{1}{m(D \cap B(x,r))}\int\limits\_{D \cap B(x,r)}\frac{z-x}{r}\,m(dz)
\end{align\*}
Here, $m$ denotes the Lebesgue measure on $\mathbb{R}^d$.
Does this limit exists?
In my impression, this seems to converge to the inward (unit) normal vector at $x$.
|
https://mathoverflow.net/users/68463
|
On a characterization of inward unit normal vector
|
For small $r$ the curvature of the surface $\partial D$ can be neglected, so $D \cap B(x,r)$ is half the $d$-dimensional ball with radius $r$. Choosing the origin of the coordinate system at position $x$ and orienting the $x\_1$-axis along the inward normal, the integral is given by the vector $v$ with components
$$v\_p=\frac{2}{V\_{d}(r)}\int\cdots\int\_{-\infty}^\infty \theta\biggl(r-\sum\_{i=1}^d x\_i^2\biggr)\theta(x\_1)\frac{x\_p}{r}\,dx\_1dx\_2\ldots dx\_d$$
with $V\_d(r)$ the volume of the $d$-dimensional ball of radius $r$ and $\theta$ the unit step function.
Only the $p=1$ component is nonzero because of symmetry, and this component evaluates to
$$v\_1=\frac{\int\_0^r d\rho\int\_0^{\pi/2}d\theta\,\rho^{d} \cos \theta \sin^{d-2}\theta}{r\int\_0^r d\rho\int\_0^{\pi/2}d\theta\,\rho^{d-1} \sin^{d-2}\theta}=\frac{\Gamma \left(\frac{d}{2}+1\right)}{\sqrt{\pi }\, \Gamma \left(\frac{d+3}{2}\right)}.$$
So it's an inward normal vector, but not a unit vector.
|
3
|
https://mathoverflow.net/users/11260
|
404819
| 166,004 |
https://mathoverflow.net/questions/404806
|
2
|
$\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}$The compact simple Lie groups $\SO\_8(\mathbb{R}) $ and $\SO\_9(\mathbb{R}) $ both have rank 4. The group
$$
G=\SU\_3 \times \SU\_2 \times \operatorname{U}\_1
$$
also has rank 4. Does there exist a subgroup of $\SO\_8(\mathbb{R}) $ or $\SO\_9(\mathbb{R}) $ isomorphic to $ G $?
I already know that there is an inclusion of $ G $ into $\SU\_5 $ given by putting $ \SU\_3 $ in the top 3 by 3 block, $\SU\_2 $ in the bottom 2 by 2 block and then $ U\_1 $ corresponds to the block scalar subgroup $\exp(-\frac{2 \pi i t}{3})I\_3 \times \exp(\frac{2 \pi i t }{2}) I\_2 $ where $ I\_n $ are identity matrices of the appropriate size and $ t $ parameterizes the subgroup. Since $ \SU\_5 $ naturally embeds in $\SO\_{10}(\mathbb{R}) $ we can compose embeddings and get an embedding of $ G $ into $\SO\_{10}(\mathbb{R}) $.
|
https://mathoverflow.net/users/387190
|
Smallest dimension for faithful orthogonal representation
|
I made some mistakes in my first version of this answer (including giving the opposite reply …), so hopefully this is correct. Thanks to @ZoltanZimboras for help sorting it out (although, of course, if it is still wrong, then the fault is entirely mine).
I claim that there is no embedding of $G$ in $\operatorname{SO}\_9(\mathbb R)$ (and hence none in $\operatorname{SO}\_8(\mathbb R)$).
If $\{\alpha\_1, \alpha\_2, \alpha\_3, \alpha\_4\}$ is a system of simple roots for $\mathsf B\_4$, with $\alpha\_4$ short, then the subsystems generated by $\{\alpha\_1, \alpha\_2\}$ and $\{\alpha\_4\}$ are orthogonal and of types $\mathsf A\_2$ and $\mathsf A\_1$, respectively. I claimed that this gave an embedding of $\operatorname{SU}\_3 \times \operatorname{SU}\_2$ in $\operatorname{SO}\_9(\mathbb R)$, but @ZoltanZimboras pointed out that it actually only gives an embedding in $\operatorname{Spin}\_9(\mathbb R)$ (or whatever is the correct notation for the compact spin group). Note that, under this embedding, the image $\alpha\_4^\vee(-1)$ of $-1 \in \operatorname{SU}\_2$ in $\operatorname{Spin}\_9(\mathbb R)$ lies in the kernel of $\operatorname{Spin}\_9(\mathbb R) \to \operatorname{SO}\_9(\mathbb R)$, so we cannot simply compose with the projection to get an embedding in $\operatorname{SO}\_9(\mathbb R)$. The additional $\operatorname U\_1$ can be taken to be the image of $2\alpha\_1^\vee + 4\alpha\_2^\vee + 6\alpha\_3^\vee + 3\alpha\_4^\vee$.
We do not have an embedding of $G$ in $\operatorname{Spin}\_8(\mathbb R)$, since the absolute root system of $\operatorname{Spin}\_8(\mathbb R)$ is of type $\mathsf D\_4$, but $\mathsf D\_4$ contains no subsystem of type $\mathsf A\_2 + \mathsf A\_1$. (The orthogonal complement of a root in $\mathsf D\_4$ is of type $\mathsf A\_1 + \mathsf A\_1$.)
Suppose we had an embedding $G \to \operatorname{SO}\_9(\mathbb R)$. We claim that some absolute root of $G$ is short in the absolute root system of $\operatorname{SO}\_9(\mathbb R)$. Indeed, suppose not, and consider one of the roots of the $\operatorname{SU}\_2$ factor. Since it is long, its orthogonal complement in $\mathsf B\_4$ is of type $\mathsf B\_2 + \mathsf A\_1$, in Carter's notation (so the $\mathsf A\_1$ roots are long). The long roots in $\mathsf B\_2 + \mathsf A\_1$ form a system of type $3\mathsf A\_1$, which contains no subsystem of type $\mathsf A\_2$.
Let $\alpha$ be an absolute root of $G$ that is short in $\operatorname{SO}\_9(\mathbb R)$. Then the simple subgroup of $\operatorname{SO}\_9(\mathbb R)$ whose absolute root system is $\{\pm\alpha\}$ is adjoint (i.e., $\operatorname{SU}\_2/\langle-1\rangle$), but the simple subgroup of $G$ whose absolute root system is $\{\pm\alpha\}$ is simply connected (i.e., $\operatorname{SU}\_2$). (See [@nfdc's answer](https://mathoverflow.net/a/270280) to [Centralizers of subtori in reductive groups, derived subgroups](https://mathoverflow.net/questions/270205/centralizers-of-subtori-in-reductive-groups-derived-subgroups) .) Since these are the same group, we have a contradiction.
@ZoltanZimboras also asked, in a comment on the first version of the answer, how the resulting representations $\operatorname{SU}\_3 \to \operatorname{Spin}\_9(\mathbb R) \to \operatorname{SO}\_9(\mathbb R)$ and $\operatorname{SU}\_2 \to \operatorname{Spin}\_9(\mathbb R) \to \operatorname{SO}\_9(\mathbb R)$ decompose. I couldn't answer most questions about this representation, but, fortunately, since all computations have been in terms of what happens on tori, I can answer this one! We have that $(\alpha\_1^\vee, \alpha\_2^\vee) : (\operatorname S^1)^2 \to \operatorname{Spin}\_9(\mathbb R)$ and $\alpha\_4^\vee : \operatorname S^1 \to \operatorname{Spin}\_9(\mathbb R)$ factor through embeddings of maximal tori $(\operatorname S^1)^2 \to \operatorname{SU}\_3$ and $\operatorname S^1 \to \operatorname{SU}\_2$, which I will use as coördinates for those tori.
The weights of the (implicit) chosen maximal torus in $\operatorname{Spin}\_9(\mathbb R)$, in the natural representation *via* its projection to $\operatorname{SO}\_9(\mathbb R)$, are the trivial weight, together with $\alpha\_i + \dotsb + \alpha\_4$ and their opposites for $i \in \{1, 2, 3, 4\}$. Thus the weights of $\operatorname{SU}\_3 \times \operatorname{SU}\_2 \times \operatorname U\_1$ are the trivial weight $(0, 0; 0; 0)$, together with $(1, 0; 0; 2)$, $(-1, 1; 0; 2)$, $(0, -1; 0; 2)$, $(0, 0; 2; 0)$, and their opposites. Thus, if I haven't mixed up my computation, the restricted representation of $G$ is the sum of the trivial representation; the two fundamental representations of $\operatorname{SU}\_3$, on which $\operatorname{SU}\_2$ acts trivially, and on one of which the weight of $\operatorname U\_1$ is $2$ and on the other of which it is $-2$; and the representation of $\operatorname{SU}\_2$ with highest weight $2$, on which $\operatorname{SU}\_3 \times \operatorname U\_1$ acts trivially.
|
2
|
https://mathoverflow.net/users/2383
|
404831
| 166,008 |
https://mathoverflow.net/questions/404605
|
4
|
This question is based on the assumption that $V \ne L$ and we have $\omega\_1^L < \omega\_1$ (here $\omega\_1^L$ is equal to the supremum of ordinals accidentally writable by no-oracle Ordinal Turing Machines).
Consider Ordinal Turing Machines (called “$\omega\_{\alpha}$-machines”) with an oracle that provides access to all [transfinite initial ordinals](https://mathworld.wolfram.com/InitialOrdinal.html).
Any $\omega\_{\alpha}$-machine is an Ordinal Turing Machine equipped with an extra tape (the *oracle tape*). We may assume that this tape is read-only.
Let $t(\alpha)$ denote the symbol written on an $\alpha$-th cell of the oracle tape. Then $t(\alpha) = 1$ if and only if $\alpha$ is an initial ordinal.
If $\epsilon > 0$, then the $\epsilon$-stabilization time of a machine is the successor of $\gamma\_0$, where $\gamma\_0$ is the least ordinal such that the values of all symbols written on all cells of the initial segment of length $\epsilon$ of the output tape never change at any time $\gamma > \gamma\_0$. If $\epsilon = 0$, then the $\epsilon$-stabilization time of a machine is the successor of $\gamma\_0$, where $\gamma\_0$ is the least ordinal such that the values of all symbols written on *all cells* (i.e. cells indexed by any element of the class of all ordinals) of the entire output tape never change at any time $\gamma > \gamma\_0$. If a machine halts, then $\gamma\_0$ is not greater than the halting time. If an ordinal $\gamma\_0$ does not exist, then the corresponding $\epsilon$-stabilization time is $0$.
Let $F\_{\epsilon}(i)$ denote the $\epsilon$-stabilization time of an $i$-th $\omega\_{\alpha}$-machine, assuming that all computations start with no ordinal parameters (i.e. empty input).
Assuming that we have fixed a particular way to encode a countable ordinal by an infinite binary sequence of length $\omega\_0=\omega$, the ordinal $\tau\_0$ is defined as the supremum of ordinals *eventually writable* by $\omega\_{\alpha}$-machines with empty input on the initial segment of length $\omega\_0=\omega$ of the output tape. The reasoning behind this definition of $\tau\_0$ is that there may be $\omega\_{\alpha}$-machines whose initial output segments of length $\omega$ stabilize at a time $\ge \omega\_1$ (i.e. $F\_{\omega}(i) \ge \omega\_1$), yet all other output segments are irrelevant: they stabilize at an arbitrarily large time or *even diverge*. That is, I suppose that there may exist a machine $M\_n$ such that, for example, $F\_0(n) = 0$, yet $F\_{\omega}(n) \ge \omega\_1$. In this case, if the eventually stable content on the initial $\omega$-segment of the output tape encodes an ordinal, this countable ordinal is eventually writable by $M\_n$.
The ordinal $\tau\_1$ is defined as follows: $$\tau\_1 = \sup \{F\_0(i) : i \in \mathbb{N}\}.$$
Is it possible to estimate how large are $\tau\_0$ and $\tau\_1$ (at least, give a “reasonably accurate” estimate for the lower/upper bounds)? In particular, is $\tau\_0$ larger than the least ordinal $\delta$ such that $L\_{\delta} \prec\_{\Sigma\_3} L\_{\omega\_1}$ (the latter is mentioned in [this comment](https://mathoverflow.net/questions/259100/memorable-ordinals#comment639579_259105) and [this answer](https://mathoverflow.net/a/259500) on Mathoverflow)?
|
https://mathoverflow.net/users/122796
|
How large are the stabilization times of Ordinal Turing Machines with an oracle for the transfinite initial ordinals?
|
For ordinal Turing machines with an oracle $S⊂Ord$,
- the set/class of output locations that are written at some point is $Σ^{L[S]}\_{1,S}$ (i.e. $Σ\_1$ in $L[S]$ with a predicate for $S$) and can be arbitrary such,
- the set/class of output locations that are eventually 1 is $Σ^{L[S]}\_{2,S}$ and can be arbitrary such,
- any set in $L[S]$ (but no set not in $L[S]$) can be 'accidentally' written at some point.
The supremum of eventually writable (as subsets of $ω$) ordinals $τ\_0$ is the least ordinal that is not $Δ^{L[S]}\_{2,S}$ definable. The supremum of stabilization times $τ\_1$ is the supremum of $Δ^{L[S]}\_{2,S}$ definable ordinals (possibly uncountable). Similarly, the supremum of writable (as subsets of $ω$) ordinals is the least ordinal not $Δ^{L[S]}\_{1,S}$ definable, and the supremum of halting times is the supremum of $Δ^{L[S]}\_{1,S}$ definable ordinals.
Your question corresponds to $S=\mathrm{Card}$ (cardinals). I actually considered $S=\mathrm{Card}$ in the question [Cardinal Register Machines](https://mathoverflow.net/questions/280638/cardinal-register-machines), but it is worth it to elaborate it further.
*Case 1:* There is no sharp for a proper class of measurable cardinals. Then:
- $K^{L[\mathrm{Card}]}$ is an iterate of the core model $K$. ($K^{L[\mathrm{Card}]}$ does not have a sharp since otherwise enough cardinals will be indiscernible to generate a proper of class of measurables $K^{L[\mathrm{Card}]}$.)
- If $V=K$ (which holds if $V=L$), then for sets below the least measurable cardinal (or arbitrarily if there are none), $Δ^{L[\mathrm{Card}]}\_{1,\mathrm{Card}}=Δ\_2$ and $Δ^{L[\mathrm{Card}]}\_{2,\mathrm{Card}}=Δ\_3$. I think the only difference between $K$ and $L[\mathrm{Card}]$ (if $V=K$) is that every measurable cardinal $κ$ (and its measure) in $K$ is converted into a Prikry sequence $κ,κ^{+},κ^{++},...$ for $κ^{+ω}$.
- Non-absoluteness: There is an ordinal Turing machine $M$ such that if $V$ is a (set) generic extension of $K$ (not sure if this condition is needed), then for every set of ordinals $s$, there is a generic extension $V[G]$ such that in $V[G]$ (and using $\mathrm{Card}^{V[G]}$), $M$ outputs $s$ and halts. (For example, $M$ can check which cardinal successors are computed correctly for the least $L\_α[\mathrm{Card}]$ that correctly computes enough cardinals.) I think such $M$ (if it works for all such $s$ and $V$) cannot halt if the above sharp exists: In every case, $\mathrm{Card}$ should in a sense give indiscernibles (and thus be unavailable for coding) until $L[\mathrm{Card}∩α]$ contains all reals in $K$, and if the sharp exists, $M$ should be stuck in the pre-decoding phase.
*Case 2:* There is a sharp for a proper class of measurable cardinals. Then, $K^{L[\mathrm{Card}]}$ is an iterate of the minimal inner model $K'$ with a proper class of measurable cardinals. Writable (as subsets of $ω$) countable ordinals are exactly $Δ\_2^{K'}∩ω\_1^{K'}$, and eventually writable — $Δ\_3^{K'}∩ω\_1^{K'}$. Note that these are $Δ^1\_3$ (in $V$) and independent of forcing. As an aside, with stronger large cardinal axioms, more structure becomes forcing-independent; for example, see my questions [Complexity of L[cf]](https://mathoverflow.net/questions/279724/complexity-of-l-mathrmcf) and [Inner models with all sets generic](https://mathoverflow.net/questions/356597/inner-models-with-all-sets-generic).
|
3
|
https://mathoverflow.net/users/113213
|
404838
| 166,014 |
https://mathoverflow.net/questions/404840
|
-1
|
Does there exist a non constant almost surely continuous stochastic process $X$ on $[0, \infty)$ with $X\_t$ independent of $X\_{t+1}$ for all $t \geq 0$?
|
https://mathoverflow.net/users/173490
|
A periodically independent stochastic process
|
Stupid answer: the trivial process $X\_t=0$.
Less stupid answer: for every half-integer $n/2$, choose $X\_{n/2}$ independently according to your favorite probability distribution and then interpolate linearly for other values of $t$.
|
2
|
https://mathoverflow.net/users/47135
|
404842
| 166,015 |
https://mathoverflow.net/questions/404070
|
2
|
Let $G$ be a finite group, $\tau$ a group automorphism of $G$ of period two and $m$ a natural number. Following [1, Definition 2.1], a complex fusion category $\mathcal{C}$ is called a *quadratic category* with $(G,\tau,m)$ if its Grothendieck ring has basis $\{X\_g, Y\_g \ | \ g \in G\}$ and fusion rules (let $e$ be the neutral element of $G$):
* $X\_gX\_h = X\_{gh}$,
* $X\_gY\_e = Y\_g = Y\_eX\_{g^{\tau}}$,
* $Y\_e^2 = X\_e + m\sum\_{g \in G} Y\_g$.
Note that *quadratic* just requires the action of $G$ on the non-group part of the basis to be transitive, as stated in the second line above, in particular, we can have $Y\_{g\_1}=Y\_{g\_2}$ with $g\_1 \neq g\_2$.
Let $\mathcal{C}$ be a spherical quadratic category with $(G,\tau,m)$. By [1, Theorem 2.2], if $G$ is an odd group and $m$ is an odd number, then $G$ is abelian and $g^{\tau} = g^{-1}$ for any $g \in G$.
The Haagerup category is a quadratic category with $(C\_3, -1, 1)$.
**Question**: Is there a complex fusion category, quadratic with $(C\_3, 1, 1)$?
According to above statements, such a fusion category would be a *twisted* Haagerup category without spherical structure, in fact without pivotal structure by applying [2, Corollary 2.14].
Recall that the existence of a pivotal structure on every fusion category is a well-known conjecture [3, Conjecture 2.8]. If this conjecture is true then above question has a negative answer, but if this conjecture is false then above category could be a first counter-example. Anyway, it should be possible to answer above question independently of this conjecture.
---
*References*
[1]: Izumi, Masaki The classification of $3^n$ subfactors and related fusion categories. Quantum Topol. 9 (2018), no. 3, 473–562.
[2]: Ostrik, Victor. Pivotal fusion categories of rank 3. Mosc. Math. J. 15 (2015), no. 2, 373-396, 405.
[3]: Etingof, Pavel; Nikshych, Dmitri; Ostrik, Viktor. On fusion categories. Ann. of Math. (2) 162 (2005), no. 2, 581-642.
|
https://mathoverflow.net/users/34538
|
A twisted Haagerup category without pivotal structure
|
**Quick answer**: no.
Andrew Schopieray pointed out to me the PhD thesis of Josiah E. Thornton [2].
---
Let $\mathcal{F}$ be a fusion ring with basis $B=\{ b\_1, \dots, b\_r \}$. Let $G$ be the group of invertible elements $b\_i$ of $B$ (i.e. $\textrm{FPdim}(b\_i)=1$). The fusion ring $\mathcal{F}$ is called *quadratic* if the action of $G$ on $B \setminus G$ is transitive. If $G=B$ such a fusion ring is called *pointed*, and if $|B \setminus G|=1$ it is called *near-group*. A fusion category with a quadratic Grothendieck ring is called (in the literature) a *quadratic category* or a *generalized near-group category*.
**Theorem.**
A categorification $\mathcal{C}$ of a quadratic fusion ring must admit a spherical (so pivotal) structure.
*Proof.* By [2, Theorem IV.3.6.], $\mathcal{C}$ must be $\varphi$-pseudo-unitary (i.e. pseudo-unitary up to Galois automorphism), and then spherical (so pivotal) by [1, Proposition 2.16].
---
*References*
[1] V. Drinfeld, S. Gelaki, D. Nikshych, V. Ostrik, *On braided fusion categories. I.*, Selecta Math. (N.S.) 16 (2010), no. 1, 1--119.
[2] J.E. Thornton, *Generalized near-group categories*, Thesis (Ph.D.)–University of Oregon (2012) 72 pp.
|
1
|
https://mathoverflow.net/users/34538
|
404843
| 166,016 |
https://mathoverflow.net/questions/404775
|
0
|
Consider the function $\_2F\_1(5.5, 1, 5;-|x|^2]$ for $x\in \mathbb{R}^n.$ I want to show that this function is positive. I checked that it does not have any roots so can I conclude the inequality by using continuity in $x$ of the function $\_2F\_1(5.5, 1, 5;-|x|^2]$?
|
https://mathoverflow.net/users/68232
|
How to show the following inequality $_2F_1(5.5, 1, 5;-|x|^2]>0$?
|
You have the integral representation
$${}\_2F\_1\left(\begin{matrix}11/2,1\\5\end{matrix};-x\right)=\frac 4{x^4}\int\_0^x\frac{(x-t)^3}{(1+t)^{11/2}}\,dt,$$
which follows by expanding $1/(1+t)^{11/2}$ using the binomial theorem and integrating termwise. This should prove the positivity.
Note that this integral is a remainder term in the Taylor expansion of $1/(1+x)^{3/2}$. So your series can be summed explicitly as $1/(1+x)^{3/2}$ minus a Taylor polynomial. This is also easy to see by writing
$$\frac{(11/2)\_k}{(5)\_k}=\frac{4!}{(3/2)\_4}\frac{(3/2)\_{k+4}}{(k+4)!}.$$
Your series is essentially just the binomial series for $1/(1+x)^{3/2}$ minus the first four terms. More explicitly,
$${}\_2F\_1\left(\begin{matrix}11/2,1\\5\end{matrix};-x\right)=\frac{128}{315x^4}\left(\frac 1{(1+x)^{3/2}}-1+\frac 32\,x-\frac{15}8\,x^2+\frac{35}{16}\,x^3\right).$$
Perhaps this fact is also useful to you.
|
2
|
https://mathoverflow.net/users/10846
|
404845
| 166,017 |
https://mathoverflow.net/questions/404782
|
2
|
We will be working over an algebraically closed field of characteristic 0. We say that a projective variety $X\subset \mathbb{P}^n$ has projectively isomorphic plane sections if there is an open set $U\in(\mathbb{P}^n)^\vee$ such that the hyperplane sections $H\cap X,\ H\in U$ are all projectively isomorphic, i.e. for any two $H\_1,H\_2\in U$ there is some automorphism $f$ of $\mathbb{P}^n$ such that $f(H\_1)=H\_2, f(X\cap H\_1)=X\cap H\_2$.
Consider a smooth cubic hypersurface $X\subset \mathbb{P}^3.$ According to Theorem 2.12 of the paper "[Some Remarks on Varieties with Projectively
Isomorphic Hyperplane Sections](https://link.springer.com/article/10.1007/BF01263521)" by Rita Pardini, $X$ has projectively isomorphic hyperplane sections if and only if it's a projection of a normal rational ruled surface in $\mathbb{P}^4.$ I suspect that this is not the case in this situation.
Is there a way to see that $X$ does not have projectively isomorphic hyperplane sections? More generally, what are the ways to identify such projections of normal rational ruled surfaces of degree $d$ from $\mathbb{P}^{d+1}$ to $\mathbb{P}^d$?
|
https://mathoverflow.net/users/131868
|
Does a smooth cubic in $P^3$ have projectively isomorphic sections?
|
Try this: Consider a "Lefschetz pencil" of planes through a general line. I believe these cut the smooth cubic surface in a pencil of irreducible plane cubic curves that are generally smooth, and the special ones have only one node, in particular are stable curves. Hence the smooth ones converge to the nodal ones in the moduli space of stable curves of genus one. That space is hausdorff, so the smooth ones cannot all be isomorphic, much less projectively equivalent. Does this work?
|
3
|
https://mathoverflow.net/users/9449
|
404847
| 166,018 |
https://mathoverflow.net/questions/404817
|
17
|
There are rare algebraic varieties such that the number of points over finite fields $\mathbb{F}\_p$ is given by a polynomial in $p$. One notable series of examples is the commuting variety: $[A,B]=0$ of $n\times n$ matrices $A,B$ over finite field. The computation was obtained by [Feit and Fine in 1960](https://projecteuclid.org/journals/duke-mathematical-journal/volume-27/issue-1/Pairs-of-commuting-matrices-over-a-finite-field/10.1215/S0012-7094-60-02709-5.short) and many generalizations have been obtained recently. But it seems results in natural generality are not yet achieved (see below).
**Question 1:** Consider pairs of $3\times 3$ anticommuting $AB+BA = 0 $ over finite fields $\mathbb{F}\_p$, is true that their number is polynomial in $p$, for $p>2$ ? May be one needs to exclude some other primes, not only $p=2$ or $p=2$ is the only exception ?
**Question 2:** If the number of points is indeed given by a polynomial ($p>2$), then it is given by the polynomial found by Roland Bacher and Peter Taylor in [MO 404760](https://mathoverflow.net/q/404760/10446):
$$2p^{10}+7p^9-3p^8-6p^7-4p^6+3p^5+4p^4-2p^3$$
([My direct calculation](https://www.kaggle.com/alexandervc/count-anticommuting-matrices-over-f-p-attempt-3) yields 221157 and 31511625 matrices for $p=3,5$ respectively,
and colleagues found that it is the only polynomial of degree 10 which satisfies these conditions and has minimal possible coefficients. Heuristic to search for polynomials with the smallest possible coefficients works quite fine in my experience for such questions.)
**Question 3:** Bonus question. It might be count is polynomial for any $n$ and $n$x$n$ anticommuting $[A,B]=0$, and there is nice generating function over $n$ for such polynomials - similar to Feit,Fine result for commuting matrices ? (Well, it might be better to leave it for separate question).
======================
PS
For 2x2 case the polynomial seems to be given by $+p^5+3p^4-2p^3-2p^2+p$ for $p>2$, checked till $p=19$.
Anticommuting variety has been studied recently e.g.: [Anti-commuting varieties](https://arxiv.org/abs/1805.00378).
======================
**Context**
We might expect similar results not only for pairs, but triples, n-tuples of commuting/anticommuting matrices: [MO271752](https://mathoverflow.net/q/271752/10446), but commuting/anticommuting is just an example, it should be true for much wider class of algebras - "categorical exponential formula" might be the right context for such questions [MO272045](https://mathoverflow.net/q/272045/10446), [MO275524](https://mathoverflow.net/q/275524/10446), however presently known forms seems does not cover even Feit,Fine case. See also very nice results and connections with Hasse-Weil zeta function in [Yifeng Huang 2021](https://yifeng-huang-math.github.io/files/talk_gocc.pdf). The general question about varieties which are polynomial count seems also not so simple as [disproof of Kontsevich conjecture](https://arxiv.org/abs/math/0012198) indicates.
Such varieties thought to be defined over the mysterious ["field with one element"](https://en.wikipedia.org/wiki/Field_with_one_element).
|
https://mathoverflow.net/users/10446
|
Number of $3\times 3$ anticommuting matrices over finite fields $\mathbb{F}_p$ is (or is not?) polynomial in $p$?
|
Let's work over a field $K$, which when finite is supposed to have $q$ elements. I'll assume the characteristic to be $\neq 2$, since in characteristic 2 we get the commuting variety which is well-known. Let $Q=Q(K)$ be the set of anticommuting pairs $(A,B)$.
If $A$ has eigenvalues $(x,y,z)$ (possibly on an extension), the eigenvalues of $B\mapsto AB+BA$ are, $2x,2y,2z$, and, with multiplicity 2, $x+y$, $x+z$, $y+z$. Write $W\_A$ as the kernel of this operator, i.e., the set of $B$ such that $(A,B)\in Q$.
We essentially need to discuss in terms of the conjugacy class of $A$, and more precisely in terms of the cardinal of $W\_A$ and of the conjugacy class of $A$.
Write $Q=Q\_1\sqcup Q'\sqcup Q''$, where $Q\_1$ is the set of pairs $(A,0)$ with $A$ invertible; $Q'$ is the set of pairs $(A,B)$ with $A$ invertible, $B\neq 0$, $Q''$ is the set of pairs $(A,B)$ with $A$ not invertible. Define $V'$ and $V''$ as the first projection of $Q'$ and $Q''$.
Write $u\_q=(q^3-1)(q^3-q)(q^3-q^2)$ for the cardinal of $\mathrm{GL}\_3(\mathbf{F}\_q)$. Then $V\_1$ has cardinal $u\_q$.
From the eigenvalues fact above, the elements of $V'$ have eigenvalues $(x,-x,y)$ for some nonzero $x,y$. Write $V'=V\_2\sqcup V\_3\sqcup V\_4\sqcup V\_5$ with
* $V\_2$ (resp. $V\_3$) the set of $K$-diagonalizable (resp. non-$K$-diagonalizable) invertible matrices with eigenvalues of the form $(x,-x,y)$ with $y\neq\pm x$; (note that $y$ is the trace).
* $V\_4$ (resp. $V\_5$) the set of diagonalizable (resp. non-diagonalizable) invertible matrices with eigenvalues $(x,x,-x)$ (note that $x$ is the trace)
Write $V''=V\_6\sqcup \dots\sqcup V\_{16}$ with
* $V\_6$ (resp. $V\_7$) the set of $K$-diagonalizable (resp. non-$K$-diagonalizable) matrices with eigenvalues of the form $(0,x,y)$, $x,y$ nonzero and $x\neq\pm y$
* $V\_8$ (resp. $V\_9$) the set of diagonalizable (resp. non-diagonalizable) matrices with eigenvalues of the form $(0,x,x)$, $x\neq 0$ (note that $x$ is half the trace)
* $V\_{10}$ (resp. $V\_{11}$) the set of $K$-diagonalizable (resp. non-$K$-diagonalizable) matrices with eigenvalues of the form $(0,x,-x)$, $x$ nonzero
* $V\_{12}$ (resp. $V\_{13}$) the set of diagonalizable (resp. non-diagonalizable) matrices with eigenvalues of the form $(0,0,x)$, $x\neq 0$
* $V\_{14}$, $V\_{15}$, $V\_{16}$ the set of nilpotent matrices of rank 2, 1, 0 respectively.
For each $i\in\{2,\dots,16\}$, all matrices $A$ in $V\_i$ have a centralizer in $\mathrm{GL}\_3(\mathbf{F}\_q)$ of the same size, say $c\_i$ (and hence a conjugacy class of cardinal $u\_q/c\_i$), and have $W\_A$ of the same size, say $q^{d\_i}$ (we will compute them below). Also denote by $e\_i$ the number of conjugacy classes in $V\_i$.
Let $Q\_i$ be the inverse image of $V\_i$ in $Q'$ (for $2\le i\le 5$) and in $Q''$ for $i\ge 6$.
Then for $i\ge 6$ the cardinal of $Q\_i$ is $u\_qe\_iq^{d\_i}/c\_i$, and for $2\le i\le 5$ it is $u\_qe\_i(q^{d\_i}-1)/c\_i$, the difference being only because we excluded $B=0$ in those cases. We now compute case by case.
* $d\_2=d\_3=2$, $d\_4=4$, $d\_5=2$, $d\_6=d\_7=d\_8=d\_9=1$, $d\_{10}=d\_{11}=3$, $d\_{12}=4$, $d\_{13}=2$, $d\_{14}=3$, $d\_{15}=5$, $d\_{16}=9$.
* $c\_2=c\_6=c\_{10}=(q-1)^3$, $c\_3=c\_7=c\_{11}=(q+1)(q-1)^2$, $c\_4=c\_8=c\_{12}=(q-1)(q^2-1)(q^2-q)$, $c\_5=c\_9=c\_{13}=q(q-1)^2$, $c\_{14}=(q-1)q^2$, $c\_{15}=(q-1)^2q^3$, $c\_{16}=u\_q$.
* $e\_2=(q-1)(q-3)/2$ (choose $y\neq 0$, then $x\notin\{0,y,-y\}$, up to $x\to -x$), $e\_3=(q-1)^2/2$ (choose $y\neq 0$ and a non-square $x^2\neq 0$), $e\_4=e\_5=e\_8=e\_9=e\_{12}=e\_{13}=q-1$, $e\_6=(q-1)(q-3)/2$ (choose $y\neq 0$, then then $x\notin\{0,y,-y\}$, up to $x\leftrightarrow y$), $e\_7=(q-1)^2/2$ (choose nonzero $y$ and non-square $x^2$), $e\_{10}=e\_{11}=(q-1)/2$, $e\_{14}=e\_{15}=e\_{16}=1$.
So the desired cardinal is $|Q\_1|+\sum\_{i=2}^5|Q\_i|+\sum\_{i=6}^{16}|Q\_i|$, namely
$$|Q(\mathbf{F}\_q)|=u\_q+u\_q\sum\_{i=2}^5 e\_i(q^{d\_i}-1)/c\_i+u\_q\sum\_{i=6}^{16} e\_iq^{d\_i}/c\_i.$$
Computation yields
$$|Q(\mathbf{F}\_q)|=2q^{10} + 7q^9 - 3q^8 - 6q^7 - 4q^6 + 3q^5 + 4q^4 - 2q^3,$$
which confirms your expectation.
|
18
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https://mathoverflow.net/users/14094
|
404860
| 166,022 |
https://mathoverflow.net/questions/404830
|
5
|
If $(a\_{n})$ is a conditionally convergent series in real field, then for any real number $\alpha$, there exists a rearrangement $(a\_{k\_{n}})$ of $(a\_{n})$ such that for all even $n$, $a\_{k\_{n}} \geq 0$, for all odd $n$, $a\_{k\_{n}} \leq 0$, and $\sum a\_{k\_{n}} = \alpha$.
This problem boils down to the following problem: can every conditional real series be rearranged to an alternative convergent series. If this is solved, then apply the extensions of Riemann's theorem by [Sierpiński](https://mathoverflow.net/questions/47589/how-to-rearrange-only-negative-part-of-a-conditionally-convergent-series-to-get), the original problem is done.
I appreciate any suggestion about this problem. Thanks in advance.
|
https://mathoverflow.net/users/138147
|
A restricted version of Riemann series theorem: rearrangements with alternating signs
|
Here you prescribe in addition the sequence of signs of the rearranged series in the Riemann-Dini theorem to be alternating, but note that any non-stationary binary sequence of signs does as well. More precisely:
>
> Let $(a\_k)\_{k\in\mathbb N} $ be an infinitesimal sequence of non-zero real
> numbers such that $\sum\_{k\ge0}a^+\_k=\sum\_{k\ge0}a^-\_k=+\infty$.
>
>
> Let $\epsilon\in \{-1,1\}^\mathbb{N}$ be a non-eventually constant
> sequence.
>
>
> Let $\alpha\in\mathbb R \cup\{ \pm\infty\}$.
>
>
> Then there exists a permutation $\sigma$ of $\mathbb N$ such that
> $$\sum\_{k=0}^\infty a\_{\sigma(k)}=\alpha$$
> $$\text{sgn}\,a\_{\sigma(k)}=\epsilon\_k.$$
>
>
>
To this end: extract a subset $S\subset\mathbb N$ such that $\sum\_{k\in S}a\_k$ is absolutely summable and $a\_k$ are positive resp. negative for infinitely many $k\in S$ (therefore for infinitely many $k\in \mathbb N\setminus S$ as well, because of the assumption $\sum\_{k\ge0}a^+\_k=\sum\_{k\ge0}a^-\_k=+\infty$).
Then do the Riemann-Dini bijection $\tau:\mathbb N\to \mathbb N \setminus S$ relatively to the sequence $(a\_k)\_{k\in \mathbb N\setminus S }$ and the number $\alpha-\sum\_{k\in S}a\_k$, namely $$\sum\_{k=0}^\infty a\_{\tau(k)}=\alpha-\sum\_{k\in S}a\_k.$$
Finally, insert the coefficients $\{a\_k\}\_{k\in S}$ in some order into the series $\sum\_{k=0}^\infty a\_{\tau(k)}$, so as to get a rearrangement with the prescribed final sequence of signs $(\dagger)$. Since $\sum\_{k\in S}a\_k$ is absolutely summable, the order of the insertion does not affect the convergence and the value of the sum, which is $\alpha$ as wanted.
$(\dagger)$ This can easily be done, for the reason that any non eventually constant binary sequence contains any other non eventually constant binary sequence as a subsequence, in such a way that the complement is also non eventually constant.
|
4
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https://mathoverflow.net/users/6101
|
404861
| 166,023 |
https://mathoverflow.net/questions/404844
|
2
|
I wonder whether one can exactly calculate the following integral in terms of $d$ and $p\geq 1$ or not, or a better bound(than the trivial one I am going to give) in terms of $d,p$:
$$\left(\int\_{[0,1)^d}\|x\|\_2^p\,dx\right)^{1/p},$$
where $\|x\|\_2$ is the Euclidean distance from $x$ to 0. Trivially, one has $\|x\|\_2\leq\sqrt{d}$, but this ignores the effect of $p$. Can we have a better explicit bound which also relies on $p$, or even an explicit formula?
|
https://mathoverflow.net/users/174600
|
Exact formula or non-trivial upper bound on p-norm of $f(x)=\|x\|_2$ in $[0,1)^d$
|
$\DeclareMathOperator\E{E}\DeclareMathOperator\Var{Var}\DeclareMathOperator\P{P}$Note that
$$\int\_{[0,1)^d}\|x\|\_2^p\,dx
=\E S\_d^{p/2},\tag{1}\label{1}$$
where $S\_d:=\sum\_1^d U\_j^2$ and the $U\_j$'s are iid random variables uniformly distributed on the interval $[0,1]$.
Note next that $\E S\_d=d/3$ and $\Var S\_d=4d/45<d/10$. So, by [Cantelli's inequality](https://en.wikipedia.org/wiki/Cantelli%27s_inequality),
$$\P(S\_d\ge d/6)\ge1-\frac{\Var S\_d}{\Var S\_d+(d/3-d/6)^2} \\
\ge 1-\frac{d/10}{d/10+(1/3-1/6)^2 d} \\
=1-\frac{1/10}{1/10+(1/3-1/6)^2}=:c\in(0,1).$$
So,
$$\E S\_d^{p/2}\ge(d/6)^{p/2} \P(S\_d\ge d/6)
\ge c(d/6)^{p/2}$$
and hence
$$\Bigl(\int\_{[0,1)^d}\|x\|\_2^p\,dx\Bigr)^{1/p}
=(\E S\_d^{p/2})^{1/p}
\ge c^{1/p}\sqrt{d/6}
\ge c\sqrt{d/6}$$
for $p\ge1$.
So, the trivial upper bound $\sqrt d$ on $\bigl(\int\_{[0,1)^d}\|x\|\_2^p\,dx\bigr)^{1/p}$ is optimal up to a universal constant factor.
---
For $p\ge2$, one can can do with a much simpler reasoning: by Jensen's inequality,
$$\Bigl(\int\_{[0,1)^d}\|x\|\_2^p\,dx\Bigr)^{1/p}
\ge\Bigl(\int\_{[0,1)^d}\|x\|\_2^2\,dx\Bigr)^{1/2}
=\sqrt{d/3}.$$
---
One may also note that, for any real $p>0$, by the Fatou lemma,
$$\liminf\_{d\to\infty}\Bigl(\int\_{[0,1)^d}\|x\|\_2^p\,dx\Bigr)^{1/p}
\Big/\sqrt{d/3}\ge1.$$
In view of \eqref{1}, this follows because, by the law of large numbers, $S\_d/d\to \E U\_1^2=1/3$ in probability (as $d\to\infty$).
|
3
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https://mathoverflow.net/users/36721
|
404864
| 166,025 |
https://mathoverflow.net/questions/404863
|
1
|
The expectation values of the [1D simple random walk](https://en.wikipedia.org/wiki/Random_walk#One-dimensional_random_walk) $S\_n$ can be [shown](https://mathworld.wolfram.com/RandomWalk1-Dimensional.html) to have the asymptotic behavior of
$$ \lim\_{n\to\infty} \frac{a\_n}{n^{1/2}} = \sqrt{\frac{2}{\pi}}, \tag{1}\label{1}$$
with $a\_n = S\_n$.
On the other hand, the Riemann Hypothesis is famously equivalent to the statement that
$$ \lim\_{n\to\infty} \frac{a\_n}{n^{1/2 +\varepsilon}} = 0, \tag{2}\label{2}$$
for any fixed $\varepsilon>0$ and setting $a\_n=L(n)$, the [cumulative (or summatory) Liouville function](https://mathworld.wolfram.com/LiouvilleFunction.html) $L(n)=\sum\_n \lambda(n)$.
I am interested in about what we know on the difference between the two asymptotic behaviours \eqref{1} and \eqref{2}. Clearly the random walk asymptotics \eqref{1} implies \eqref{2} but not vice versa.
How can it be shown that $a\_n=L(n)$ violates \eqref{1}, would its fulfillment have interesting implications? Are there other/simple examples of $a\_n$ that fulfill \eqref{2}?
|
https://mathoverflow.net/users/83999
|
Asymptotics of cumulative Liouville function under RH versus simple random walk
|
I figured the remarks I gave in the comments deserve to be gathered up into a more coherent form as an answer.
One thing I will start with is that comparing $L(n)$ (or $M(n)$, the Mertens function) to the values $S\_n$ is arguably not the right heuristic. Indeed, $S\_n$ averages over all random walks, which can have their peaks and troughs at different places, and so they cancel out. Analysing a single random walk gives a different asymptotic: this is governed by [the law of the iterated logarithm](https://en.wikipedia.org/wiki/Law_of_the_iterated_logarithm), which asserts that the random walk will oscillate with extremal values on the order of $\pm\sqrt{2n\log\log n}$.
However it turns out that for subtle arithmetic reasons we do not expect the arithmetic functions of interest to reach the same bounds. Most literature (see e.g. [this paper](https://arxiv.org/abs/math/0310381v1)) concerns the Mertens function $M(n)$ so I will discuss this one, but all the relevant heuristics should also hold for $L(n)$. An unpublished conjecture of Gonek is that the maximal order of $|M(n)|$ is not $\sqrt{n\log\log n}$, but rather $\sqrt{n}(\log\log\log n)^{5/4}$, which in particular implies that the ratio $|M(n)|/\sqrt{n}$ is unbounded.
While Gonek's conjecture is still open, as is unboundedness of $|M(n)|/\sqrt{n}$, we do have unconditional results which imply that this sequence does not have a limit. Specifically, we know that $M(n)/\sqrt{n}$ has strictly negative liminf and strictly positive limsup (indeed, they are at least $\pm1.8$ respectively, see [Wikipedia](https://en.wikipedia.org/wiki/Mertens_conjecture#Disproof_of_the_conjecture) for more details and references.) This shows that $|M(n)|/\sqrt{n}$ crosses zero infintiely often, and infinitely often it takes values greater than $1$, so it cannot have a limit. (Same is known for Liouville though with different bounds on liminf and limsup, see [Wikipedia](https://en.wikipedia.org/wiki/Liouville_function#Conjectures_on_weighted_summatory_functions).)
|
3
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https://mathoverflow.net/users/30186
|
404868
| 166,029 |
https://mathoverflow.net/questions/404853
|
2
|
Let $R$ be noncommutative unital ring and $M$ a projective (right) $M$-module. Assume that $R$ embedds into $M$ as a right -module.
A) If $R$ is a semisimple ring, then of course $R$ admits an $R$-module complement. But this is a very strong assumption. What are weaker but sufficient criteria for an $R$-module complement to exist.
B) What is a non-free example where $M$ does not admit an $R$-module complement in $M.
C) What is a non-free module example where $M$ does admit a complement?
|
https://mathoverflow.net/users/167165
|
Module complements to rings embedded in a projective module
|
For (A) a ring $R$ is selfinjective if it is injective as an $R$-module (on e should say left or right). By definition of injective this means $R$ has a complement in any module. Examples include Frobenius and quasi-frobenius rings.
If $R$ is von Neumann regular, then any finitely generated submodule of a projective module has a complement. So again if $R$ embeds in a projective module there is a complement.
|
2
|
https://mathoverflow.net/users/15934
|
404871
| 166,031 |
https://mathoverflow.net/questions/404880
|
2
|
Let $f$ be a function of $\geq 2$ real variables defined on a convex cone $\mathcal{C}$ in the upper half plane, with $f(0) = 0$. Suppose $f$ is subadditive, i.e. $f(x\_1+y\_1, \dots,
x\_n+y\_n) \leq f(x\_1, \dots, x\_n) + f(y\_1, \dots, y\_n)$ in its domain, $f \geq 0$, and $f$ is nondecreasing in $x$ and nonincreasing in $y$.
I'd like to know whether $f$ is necessarily $\left(0,1\right]$-superhomogeneous, i.e., whether $f(\lambda x) \geq \lambda f(x)$ for $\lambda \in \left(0,1\right]$.
I can show that $f$ is superhomogeneous for $\lambda$ as above in $\mathbf{N}^{-1}$, I'd like to know if that's true on the interval $\left(0,1\right]$.
|
https://mathoverflow.net/users/43628
|
Superhomogeneity of subadditive functions
|
The answer is no. E.g., let $f(0):=0$ and $f(x):=1+m\_+$ if $|x|\in[2^{m-1},2^m)$ for $x\in\mathcal C:=[0,\infty)^n$ and an integer $m$; that is, $f(x)=1+(1+\lfloor\log\_2|x|\rfloor)\_+$ for all $x\in\mathcal C\setminus\{0\}$. Here, $m\_+:=\max(0,m)$ and $|x|$ is the Eucludean norm of $x$.
**Details:** Clearly, $f$ is nonnegative and nondecreasing in each of the coordinates of the argument. Take now any $x$ and $y$ in $\mathcal C$ such that $|x|\in[2^{k-1},2^k)$ and $|y|\in[2^{m-1},2^m)$ for integers $k$ and $m$ such that $k\le m$. Then $f(x)+f(y)=1+k\_+ +1+m\_+\ge2+m\_+$ and $|x+y|\le|x|+|y|<2^k+2^m\le2^{m+1}$, whence
$f(x+y)\le1+(m+1)\_+\le2+m\_+\le f(x)+f(y)$, so that $f$ is subadditive.
However, the condition $f(tx)\ge tf(x)$ will fail to hold if e.g. $x\in\mathcal C$, $|x|=1$, and $t\in(1/2,1)$ -- because then $f(x)=2$ whereas $f(tx)=1<tf(x)$.
|
2
|
https://mathoverflow.net/users/36721
|
404883
| 166,035 |
https://mathoverflow.net/questions/404876
|
1
|
Consider a separable Hilbert space $\mathcal H$ and the bounded linear operators $B(\mathcal H)$.
Consider a function $T: [0, \infty) \to B(\mathcal H)$, under what assumptions on $T(t)$ is it true that
$$\big(\int\_0^c T(t) \, dt \big) (x) = \int\_0^c T(t)x \, dt \ , \ \ \ \forall x \in \mathcal H , c \in (0, \infty)$$
for the Bochner integral? I am aware of a similar question for semigroups of BLO on Banach spaces,[The Bochner integral about a semigroup of bounded linear operators on a Banach space](https://mathoverflow.net/q/271377/161092), but I am interested in general operator-valued functions.
|
https://mathoverflow.net/users/161092
|
Action of Bochner integral of operator-valued functions on vectors
|
I understand you assume that $T:[0,c]\to\mathcal{B(H)}$ is Bochner integrable in order to write $\int\_0^c T(t)dt$ as Bochner integral. Then for any $x\in\mathcal H$ the map $ [0,c]\ni t\mapsto \mathcal H$ is also Bochner integrable and the identity you wrote holds. More generally: for a measure situation $(X,\mathcal S,\mu)$, a couple of B-spaces $\mathbb E$ and $\mathbb F$, a bounded linear operator $L:\mathbb E\to \mathbb F$, and a Bochner integrable map $f:X\to \mathbb E$, the composition $L\circ f:X\to \mathbb F$ is Bochner integrable and $\int\_X L f(u) d\mu(u)=L\int\_X f(u))d\mu(u)$ (in your case $L$ is the evaluation map $\mathcal{B(H)}\ni A\mapsto Ax\in\mathcal H$).
(The proof is immediate if $f:=v\chi\_S$ with $v\in\mathbb E$ and $S\in\mathcal S$; by linearity it generalizes immediately to integrable simple functions $f:X\to \mathbb E$; it further generalizes immediately to $f\in \mathcal L^1(\mu,\mathbb E)$, by the very definition of Bochner integrable function and integral).
|
4
|
https://mathoverflow.net/users/6101
|
404889
| 166,037 |
https://mathoverflow.net/questions/404626
|
3
|
It is a well-known fact that for every compact oriented odd-dimensional manifold $\mathcal{M}$ with boundary it holds that
$$\chi(\mathcal{M})=\frac{1}{2}\chi(\partial\mathcal{M}).$$
In particular, if you take a $3$-dimensional manifold with boundary given by a genus $g$ surface, then its Euler characteristic is $\chi=1-g$.
Is there any relation known between the Euler characteristic of a pseudomanifold with boundary and its boundary? Maybe some explicit formula as above, or at least some inequality relating the two Euler characteristics? I am mainly interested in the $3$-dimensional case. Maybe pseudomanifolds are too general and something like the statement above is only true in the case of "normal" pseudomanifolds, in which all links are themselve pseudomanifolds.
---
Let me briefly define, what I mean when taking about "pseudomanifolds with boundary":
**Simplicial Complex:**
>
> Let $\mathcal{V}$ be a finite set. Then a collection of non-empty
> finite subsets of $\mathcal{V}$, denoted by
> $\Delta\subset\mathcal{P}(\mathcal{V})$, is called "(abstract)
> simplicial complex", if it satisfies the following two properties:
>
>
> 1. $\Delta$ contains all singletons, i.e. $\{v\}\in\Delta$ for all $v\in\mathcal{V}$.
> 2. For any non-empty $\tau\subset\sigma$ for some $\sigma\in\Delta$ it holds that $\tau\in\Delta$.
>
>
>
**Pseudomanifolds:**
>
> Let $\Delta$ be a finite abstract $d$-dimensional simplicial complex.
> We call the corresponding geometric realization $\vert\Delta\vert$ a
> "$d$-dimensional pseudomanifold", if the following conditions are
> fulfilled:
>
>
> 1. $\Delta$ is "pure", i.e. every simplex $\sigma\in\Delta$
> is the face of some $d$-simplex.
> 2. $\Delta$ is "non-branching", i.e. every $(d-1)$-simplex
> is face of exactly one or two $d$-simplices.
> 3. $\Delta$ is "strongly-connected", i.e. for every pair of
> $d$-simplices $\sigma,\tau\in\Delta\_{d}$, there is a sequence of
> $d$-simplices $\sigma=\sigma\_{1},\sigma\_{2},\dots,\sigma\_{k}=\tau$
> such that the intersection $\sigma\_{l}\cap\sigma\_{l+1}$ is a
> $(d-1)$-simplex for every $l\in\{1,\dots,k-1\}$.
>
>
> The $(d-1)$-simplices from condition (2), which are the face of only
> one $d$-simplices are called "boundary simplices". The (geometrical
> realization of the) subcomplex of all these simplices is called
> "boundary of the pseudomanifold" and we denote this subcomplex by
> $\partial\Delta$. If $\partial\Delta\neq\emptyset$, then we call
> $\vert\Delta\vert$ "pseudomanifold with boundary", otherwise just "pseudomanifold."
>
>
>
|
https://mathoverflow.net/users/259525
|
Euler characteristic of pseudomanifolds with boundary
|
Ok, to convert my comment to an answer. Let $S$ be a closed orientable triangulated surface of genus $\ge 1$. Let $M$ be the cone over $S$. Then $M$ has a natural orientable pseudomanifold structure. However, $\chi(M)=1$, while $\chi(S)$ can be any nonpositive even number.
The moral is that there are way too many pseudomanifolds (even "normal" ones, in your sense). If you want to have the standard relation $\chi(\partial M)=\chi(M)/2$, consider working with (say, rational) "homology manifolds" (with boundary). But in the 3-dimensional setting, all homology manifolds are manifolds.
|
4
|
https://mathoverflow.net/users/39654
|
404892
| 166,039 |
https://mathoverflow.net/questions/404852
|
3
|
A proper edge $k$-coloring of a graph is an assignment of $k$ colors to the edges of the graph so that no two adjacent edges have the same color. The smallest integer $k$ such that $G$ has a proper edge $k$-coloring is the chromatic index of $G$.
Giving a simple graph $G$, the well-known Vizing's theorem tells us that the chromatic index of any simple graph $G$ is either $\Delta(G)$ or $\Delta(G)+1$. In particular, if the chromatic index of a graph $G$ is $\Delta(G)+1$, then we say that $G$ is of class two.
Suppose that $G$ is a graph of class two and $\varphi$ is a proper edge coloring using $\Delta(G)+1$ colors. I wonder whether there always exists an edge $uv$ such that
$\varphi(uv)\cup \{\varphi(e)~|~e~is~an~edge~adjacent~to~uv\}$ covers all of the $\Delta(G)+1$ colors.
I guess the answer is yes, however, cannot find any source supporting this. If anyone knows some relative references or can prove or disprove this, please reply me. Thanks in advance.
|
https://mathoverflow.net/users/375270
|
An edge coloring problem for class two graphs
|
Yes!
Let $uv$ be an edge colored by the last color, say $\Delta+1$. If $uv$ is incidence with all colors, then it is the required edge. So the only case is that every edge colored with $\Delta+1$ is not incident with some color in $[\Delta]$, and thus it can be recolored by that color. This results an edge $\Delta$-coloring, a contradiciton.
|
5
|
https://mathoverflow.net/users/148974
|
404928
| 166,060 |
https://mathoverflow.net/questions/404778
|
4
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Which classes of (scalar or systems of) linear first or second order hyperbolic PDEs $Lf=g$ in $n$ variables and spaces $S$ of functions have the property that there is a retarded Green's function $G:S\to S$ with $LG=1$. Retarded means that for every $g\in S$ whose support is disjoint from the past causal cone of $x$, the support of the retarded solution $g=Gf$ of $Lf=g$ is disjoint from the past causal cone of $x$.
When can one choose $S$ as a space of smooth (i.e., $C^\infty$) functions?
Where can I read about the mathematical tools for studying this and similar questions?
A particular case I am interested in is hyperbolic $L$ of the form
$$L=\pmatrix{\alpha & (a-d)^\* \cr b-d & B},$$
where $d$ is the exterior derivative, ${}^\*$ is the Minkowski adjoint, $\alpha$ is a scalar field on Minkowski space, $a,b$ are covector fields, and $B$ is a matrix field mapping covectors to covectors. Hyperbolicity is defined as in [Wikipedia](https://en.wikipedia.org/wiki/Hyperbolic_partial_differential_equation#Hyperbolic_system_of_partial_differential_equations).
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https://mathoverflow.net/users/56920
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spaces of smooth functions for linear hyperbolic PDE
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For normally hyperbolic operators (those whose principal symbol is the same as for the wave operator, but possibly acting on a vector bundle, the theory of fundamental solutions/Green functions (as distributions that would be acting on smooth functions) is very well developed in
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> *Bär, Christian; Ginoux, Nicolas; Pfäffle, Frank*, [**Wave equations on Lorentzian manifolds and quantization.**](http://dx.doi.org/10.4171/037), ESI Lectures in Mathematics and Physics. Zürich: European Mathematical Society Publishing House (ISBN 978-3-03719-037-1/pbk). viii, 194 p. (2007). [ZBL1118.58016](https://zbmath.org/?q=an:1118.58016). [arXiv:0806.1036](https://arxiv.org/abs/0806.1036)
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Of course, the condition of global hyperbolicity on the (Lorentzian) metric associated to the wave-like principal symbol is crucial there. The simplest space of smooth functions that is closed under the action of the retarded Green function is $C\_+^\infty$, the space of smooth functions with retarded support, consisting of functions whose support is contained in $J^+(K)$, the future causal influence set of some compact subset $K$. But there is a also a larger space $C\_{pc}^\infty$ with the same property, the space of smooth functions with past compact support, consisting of functions whose support has compact intersection with $J^-(K)$ for any compact $K$. These and other natural support restrictions are conveniently described in
>
> *Sanders, Ko*, [**A note on spacelike and timelike compactness**](http://dx.doi.org/10.1088/0264-9381/30/11/115014), Classical Quantum Gravity 30, No. 11, Article ID 115014, 10 p. (2013). [ZBL1272.83015](https://zbmath.org/?q=an:1272.83015). [arXiv:1211.2469](https://arxiv.org/abs/1211.2469)
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Going beyond normally hyperbolic equations, the following reference shows that the basic properties of Green functions that are needed for the discussion above are also shared by symmetric hyperbolic systems (or those that can be reduced to them)
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> *Bär, Christian*, [**Green-hyperbolic operators on globally hyperbolic spacetimes**](http://dx.doi.org/10.1007/s00220-014-2097-7), Commun. Math. Phys. 333, No. 3, 1585-1615 (2015). [ZBL1316.58027](https://zbmath.org/?q=an:1316.58027). [arXiv:1310.0738](https://arxiv.org/abs/1310.0738)
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I don't know as much about the case of sub-$C^\infty$ regularity. If that's what you are interested in, perhaps others can give more information.
**UPDATE:** The example system given in the updated question is just the first order form of a normally hyperbolic equation whose principal symbol is the same as for $d^\* B^{-1} d$ (so normal hyperbolicity here is with respect to a metric that is composed of the background metric and the operator $B$). I'm assuming here that the inverse $B^{-1}$ exists and that the equation does not lead to any integrability conditions (e.g., no new equations of order 1 or lower appear after applying the exterior derivative $d$ to the second row of $L$). Otherwise, it's not even clear that the system is indeed hyperbolic.
Speaking at a higher level of generality, you might say that $L$ belongs to the class of *generalized normally hyperbolic* operators. At least that's the terminology that I used in a recent paper, for lack of a better one in the literature (AFAIK). See Lemma 3 and the definition preceding it in
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> *García-Parrado, Alfonso; Khavkine, Igor*, [**Conformal Killing initial data**](http://dx.doi.org/10.1063/1.5126683), J. Math. Phys. 60, No. 12, 122502, 13 p. (2019). [ZBL1435.83020](https://zbmath.org/?q=an:1435.83020). [arXiv:1905.01231](https://arxiv.org/abs/1905.01231)
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More specifically, if we let $\square\_B = d^\* B^{-1} d$, there should exist
$$
M = \pmatrix{
1 & -d^\* \\
B^{-1} d & B^{-1} (\square\_B - d d^\*)} + \text{l.o.t},
$$
such that
$$
L M = \pmatrix{ \square\_B & 0 \\ 0 & \square\_B }
+ \text{l.o.t} .
$$
You see that $L M$ is normally hyperbolic in the usual way and hence has a retarded Green function $G\_{LM}$. The retarded Green function for $L$ is then just $G\_L = M G\_{LM}$. Hence $L$ is Green hyperbolic in the sense of Bär.
*N.B.:* Naively, the upper right corner of $L M$ might actually be $O(\partial^2)$, instead of $O(\partial)$, and hence contribute to the principal symbol. The point of the lower order terms in $M$ is to try to cancel that contribution to the principal symbol. If such a cancellation is impossible (basically when the adjoint operator $L^\*$ has non-trivial integrability conditions), then the principal symbol of $LM$ will only be upper triangular, with $\square\_B$ on the diagonal. It should still be possible to construct $G\_{LM}$ then, by exploiting the upper triangular structure.
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5
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https://mathoverflow.net/users/2622
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404934
| 166,063 |
https://mathoverflow.net/questions/404875
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2
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This is a follow-up question to [Positive integer solutions to the diophantine equation $(xz+1)(yz+1)=z^4+z^3 +z^2 +z+1$](https://mathoverflow.net/questions/403542/positive-integer-solutions-to-the-diophantine-equation-xz1yz1-z4z3-z/404278#404278)
Let \begin{equation}
P(x,n)= 1+x+x^2+ \cdots + x^n, \end{equation}
\begin{equation}
h(m,n)= \frac{m^n-1}{m+1} \end{equation} if $n$ is even and \begin{equation} h(m,n)=\frac{m^{n}-m}{m^2-1}
\end{equation} if $n$ is odd,
where $x, n > 0, m>1$ are non negative integers.
Then;
**Claim 1**: For all $n>0, m>1$, $m\cdot h(m,n)+1$ divides $P(h(m,n), n)$
**Claim 2:** Furthermore, for a fixed pair $(m,n), h(m,n)$ is the largest integer $x$ such that $mx+1 \ | \ P(x,n)$ i.e $mx+1 \ \nmid \ P(x,n)$ for all $x>h(m,n)$.
These claims are verified for all $n < 10^4, m < 10^2$, a high level of certainty! To prove these claims, my idea would be to use proof by induction but it appears tricky to proceed with induction. How does one go about proving these results? (An elementary proof is preferred or a counterexample).
Claim 2 is particularly interesting because, if its true, then, by just looking at the sizes of $m, n$, one will be able to rule out divisors of $P(x,n)$ of the form $mx+1$.
There’s nothing particularly unique about polynomial $P(x, n)$ and $mx+1$ to limit Claim 2 to $P(x, n) $ or $mx+1$; Only that $P(x,n)$ has no factor $mx+1$ with $m>1$ in its factorization which is why $m$ is set greater than $1$. If $m=1$ is allowed, claim 2 is false as $x+1$ divides $P(x,n)$ for all $x$ when $n+1$ is composite. If Claim 2 is true, the following conjecture is reasonable enough;
**Conjecture 1**
Let $f(x)$ and $g(x)$ be two polynomials of a non negative integer $x$ with integer coefficients, g(x) does not factor into $ (a/b)h(x)f(x) $ for some integers $a, b \not = 0 $ and polynomial $h(x) $ with integer coefficients. Then, $f(x)$ divides $g(x)$ for **finitely** many $x$.
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https://mathoverflow.net/users/166404
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Positive divisors of $P(x,n)=1+x+x^2+ \cdots + x^n$ that are congruent to $1$ modulo $x$
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Conjecture 1 follows from a (multiple of) [Bezout identity](https://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity) for polynomials:
$$mu(x)f(x)+mv(x)g(x)=md(x),$$
where $d(x) :=\gcd(f(x),g(x))$ and a positive integer $m$ is chosen such that polynomials $mu(x)$, $mv(x)$, $md(x)$ have integer coefficients. Then $f(x)\mid g(x)$ for some $x$ implies that $\frac{f(x)}{d(x)}\mid m$, which has a finite number of solutions as soon as $d(x)$ has degree smaller than $f(x)$.
In fact, I used this identity to design an [algorithm](https://mathoverflow.net/q/30204) for finding all $x$ with $f(x)|g(x)$.
Also, when $d(x)\equiv 1$ (i.e., $f(x)$ and $g(x)$ are coprime), the smallest possible $m$ is called the *reduced resultant*, which can be computed as explained in <https://mathoverflow.net/q/333279>
---
**ADDED.** As for the claims, they follow from the formula for the absolute value of resultant of $P(x,n)$ and $mx+1$ (see [A062160](https://oeis.org/A062160)) given by
$$P(m,n)=\frac{m^{n+1} - (-1)^{n+1}}{m+1}.$$
It follows that if $(mx+1)\mid P(x,n)$, then $(mx+1)\mid P(m,n)$.
And as soon as $mx+1 > P(m,n)$, i.e., $x > \frac{m^n - 1}{m+1}$, no divisibility can happen.
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6
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https://mathoverflow.net/users/7076
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404937
| 166,064 |
https://mathoverflow.net/questions/404942
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1
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I asked the following question in a forum more suitable for statistics, but I didn't get any answer; I hope, someone could shed light on my question:
I have three random variables, $X\_1$, $X\_2$, and $X\_3$, which they are distributed normally. If we consider an estimator which reads as:
$$\hat{\theta} = a\_1 X\_1 + a\_2 X\_2 +a\_3X\_3,$$
with the constraint: $a\_1 + a\_2 + a\_3 = 1$, we know that for $a\_i \propto 1/\sigma\_i^2$, where $\sigma\_i$'s are the standard deviations of $X\_i$'s, the estimator has the smallest variance (for proof, see Theorem 3.2., [here](https://arxiv.org/pdf/1710.04055.pdf)).
Now, my question is: What is the significance of choosing the coefficients in the above sum as Jeffreys' priors, that is, $a\_i \propto 1 / \sigma\_i$?
Does $a\_i \propto 1 / \sigma\_i$ result, for example, in the largest acceptable variance of the estimator? Perhaps, one can say, in the case of complete ignorance, i.e., uninformative priors, the variance of $\hat{\theta}$ should be the largest acceptable variance. Is this argument valid?
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https://mathoverflow.net/users/nan
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Jeffreys' priors as coefficients of a linear estimator
|
The coefficients $a\_i$ of the random variables $X\_i$ are not any prior probabilities at all -- because prior probabilities are coefficients, not of random variables, but of probability distributions.
The choice $a\_i\propto 1/\sigma\_i$ in your setting equalizes the variances of the random variables $a\_iX\_i$, and that is all it does.
---
Even though priors have actually nothing to do with your question, for criticisms of Jeffreys' "noninformative" priors one may want to see e.g. [Section 4.1](https://projecteuclid.org/journals/statistical-science/volume-24/issue-2/Harold-Jeffreyss-Theory-of-Probability-Revisited/10.1214/09-STS284.full).
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1
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https://mathoverflow.net/users/36721
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404943
| 166,066 |
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