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https://mathoverflow.net/questions/427057 | 1 | Let $C$ be a closed convex set of $\mathbb{R}^n$ $(n\geq 1)$, with asymptotic cone $C^{as}$ having for interior $\text{Int}\big(C^{as}\big)$. Let $u\in\mathbb{R}^n\setminus\{0\}$ such that
\begin{align}
u \notin C^{as} \text{ and }(-u) & \in \text{Int}\big(C^{as}\big).
\end{align}
For $x\in C$, let $p(x)$ be the projection of $x$ on $\partial C$ toward the direction $u$:
$$p(x)=x+\inf\{\lambda\geq 0;\,x+\lambda u\notin C\}\,u.$$
**Question:** is $p$ Lipschtiz?
This question is motivated by the answers given for the close question: [Is this projection on the boundary of a convex Lipschitz?](https://mathoverflow.net/questions/426915/is-this-projection-on-the-boundary-of-a-convex-lipschitz/426918?noredirect=1#comment1097752_426918)
| https://mathoverflow.net/users/159940 | Lipschitz aspect of a projection on the boundary of a convex | $\newcommand{\ep}{\varepsilon}\newcommand{\R}{\mathbb R}\newcommand{\epi}{\operatorname{epi}}$Yes, $p$ is Lipschitz.
Indeed,
\begin{equation\*}
p(x)=x+f(x)u
\end{equation\*}
for $x\in C$, where
\begin{equation\*}
f(x):=\inf E\_x,\quad E\_x:=\{t\ge0\colon x+tu\notin C\}.
\end{equation\*}
So, it suffices to show that the function $f$ is Lipschitz on $C$.
Let us use the [Rockafellar, p. 62](https://www.google.com/books/edition/Convex_Analysis/1TiOka9bx3sC?hl=en&gbpv=1) notation $0^+C$ for $C^{as}$ and then $(0^+C)^\circ$ for the interior of $0^+C$.
For each $x\in C$, we have
\begin{equation\*}
f(x)=\infty\iff E\_x=\emptyset\iff (\forall t\ge0\ x+tu\in C)\iff u\in0^+C;
\end{equation\*}
the latter $\iff$ follows by [Rockafellar, Theorem 8.3, p. 63](https://www.google.com/books/edition/Convex_Analysis/1TiOka9bx3sC?hl=en&gbpv=1). So, in view of the condition $u\notin0^+C$, we see that
\begin{equation}
\text{$0\le f<\infty$ on $C$. } \tag{0}\label{0}
\end{equation}
Without loss of generality (wlog), the vector $u$ is of unit length.
Since
\begin{equation\*}
v:=-u\in(0^+C)^\circ,
\end{equation\*}
there is some real $\ep>0$ such that
\begin{equation\*}
v+\ep B\subseteq 0^+C, \tag{1}\label{1}
\end{equation\*}
where $B$ is the closed unit ball centered at the origin.
It is enough to show that the restriction of $f$ to the intersection of $C$ with any 2-D affine plane parallel to the vector $u$ is $L$-Lipschtiz, with the same Lipschitz constant
\begin{equation\*}
L:=1+l,\quad l:=1/\ep \tag{2}\label{2}
\end{equation\*}
for all such 2-D affine planes.
Therefore, wlog the dimension $n$ is $2$. Also then, wlog $-u=v=(0,1)$. It then follows that
\begin{equation\*}
f(a,b):=f((a,b))=b-g(a) \tag{3}\label{3}
\end{equation\*}
for $(a,b)\in C\subseteq\R^2$, where
\begin{equation\*}
g(a):=\inf\{y\in\R\colon(a,y)\in C\}
\end{equation\*}
for all real $a$, so that $C$ is the epigraph $\epi(g):=\{(a,y)\in\R^2\colon y\ge g(a)\}$ of the (necessarily convex) function $g$. Here, the standard convention $\inf\emptyset=\infty$ is used (even though we will see in a moment that in our case in fact we have $g(a)<\infty$ for all real $a$).
By \eqref{1} and \eqref{2}, $\epi(g)$ contains $\epi(h)$ for a function $h$ given by a formula of the form $h(a)=B+l|a-A|$ for some real $A,B$ and all real $a$. Then $g\le h$; in particular, $g<\infty$ on $\R$. Also, by \eqref{3} and \eqref{0}, $g(a)>-\infty$ for some real $a$. So, for the convex function $g$ we have $g>-\infty$ on $\R$. So, $-\infty<g<\infty$ on $\R$.
Now it follows from $g\le h$ that the limit $g'(\infty)$ of the (say) right derivative $g'(a)$ of the (finite convex) function $g$ as $a\to\infty$ is $\le h'(\infty)=l$. Similarly, the limit $g'(-\infty)$ of the right derivative $g'(a)$ of the convex function $g$ as $a\to-\infty$ is $\ge h'(-\infty)=-l$. Since the right derivative $g$ of the convex function $g$ is nondecreasing, we see that $-l\le g'\le l$. So, $g$ is $l$-Lipschitz.
Thus, by \eqref{3} and \eqref{2}, $f$ is indeed $L$-Lipschtiz. $\quad\Box$
| 2 | https://mathoverflow.net/users/36721 | 427913 | 173,530 |
https://mathoverflow.net/questions/427916 | 0 | Spanier mentions that locally contractible implies homologically locally connected but I'm wondering whether contractible implies homologically locally connected?
Definition of homologically locally connected:
"A space $X$ is said to be homologically locally connected in dimension $n$ if for every $x \in X$ and neighborhood $U$ of $x$ there exists a neighborhood $V$ of $x$ in $U$ such that $\tilde{H}\_q(V) \to \tilde{H}\_q(U)$ is trivial for $q \leq n$. It is said to be homologically locally connected if it is homologically locally connected in dimension $n$ for all $n$."
| https://mathoverflow.net/users/489082 | Does contractible imply homologically locally connected? | Recall that $E$, the *earring space*, is the union in the plane of a countable collection of shrinking circles, all tangent to the $y$ axis at the origin. To obtain the desired counterexample, form the cone on $E$.
---
**Edit:** My example just above is very similar to that of Henrik Rüping (in the comments) and works for much the same reason (also mentioned by Henrik Rüping in the comments). Seeing the further questions asked by the original poster prompted me to give the example below.
---
Let $S$ be the following subset of the plane:
$$S = \{(0, 0)\} \cup \{ (1/n, 0) \mid \mbox{$n > 0$ a natural number}\}$$
Let $A = (0, 1)$ be the given point in the plane, called the apex.
Let $I\_n$ be the closed line segment in the plane connecting the apex $A$ to the point $(1/n, 0)$. Finally, let $I\_\infty$ be the closed line segment connecting the apex to the origin $(0, 0)$. We define the cone $C(S)$ to be the set
$$C(S) = I\_\infty \cup \left( \cup\_{n > 0} I\_n \right)$$
equipped with the subspace topology. The space $C(S)$ is sometimes called the *broom space*. (I think it looks a bit more like a "brush" than a broom, but anyway.)
We make two claims.
**Claim 1:** Straight-line homotopy gives a deformation retraction of the broom $C(S)$ to its apex $A$.
Thus $C(S)$ is contractible.
**Claim 2:** The broom $C(S)$ is not (homologically) locally connected at the origin $(0, 0)$.
| 2 | https://mathoverflow.net/users/1650 | 427926 | 173,535 |
https://mathoverflow.net/questions/427899 | 1 | Let $\{ X\_n(\omega,x)\}\_{n \ge 0}$ be a Markov chain with and underlying probability space $(\Omega,\Sigma,\mathbb{P})$ and state space $X= \mathbb{S}^1$. Suppose this markov chain admits unique ergodic measure which is full supported.
I would like to estimate $\mu(C)$, where $C \subset \mathbb{S}^1$, is a particular subset with positive measure.
I would like to know if in any way $\mu(C)$ is related to the following quantity:
$$\sup\_{x \in \mathbb{S}^1}\mathbb{E}\_x[\tau^C(\omega,x)] $$
where
$$\tau^C(\omega,x):= \min\{ n \ge 1 \colon X\_n(\omega,x) \in C\} $$
is defined for $x \in C$. Here $ X\_n(\omega,x)$ is the position of the markov chain at time $n$ with initial condition $x$ and randomness $\omega$.
In particular, I would like to know if some inequality of the sort
$$\mu(C) \ge\sup\_{x \in \mathbb{S}^1} \frac{1}{\mathbb{E}\_x[\tau^C(\omega,x)]} $$
holds or not.
The reason for this question is because it is well known that for a Markov Chain in a discrete state space it is possible to recover the invariant measure via the positive recurrent states. I was wandering if something like this was possible too in this case.
Thanks in advance.
| https://mathoverflow.net/users/480325 | Connection between invariant measure and positive recurrence for continuum state space markov chain | Yes, let's assume that sup (which I think should be in the denominator in the last expression) is finite, as otherwise it is always true. Suppose it is bounded by A, the each return time is also quite finite, as by chebyshev inequality $P(T > 2A) < \frac 1 2$, and using the markov property this makes each return time subgeometric, compared with the same geometric distribution. Now using the ergodic theorem calculate $\mu(C)$ by counting the number of time the process is in C. Let $a > \mu(C)$. If $T\_1, T\_1 + T\_2, $ etc are those times, then $ {\sum^n 1\_C(X\_i)} < na = P(T\_1 + ... + T\_{na} > n) $ and so both are about 1. Therefore you expect that $n < E(T\_1 + ... + T\_{na}) < na \sup(E(T) $ and therefore $a > \frac 1 {\sup(E(T)}$. I had started out arguing that they returns times were subgeometric, which I though I was going to use, but seem not to have, but I'm going to leave it there in case it occurs to me why I wanted it.
| 0 | https://mathoverflow.net/users/143907 | 427929 | 173,536 |
https://mathoverflow.net/questions/427924 | 1 | It is well known that the (real analytic) Eisenstein series is defined, in the slash notation, as follows
$$E\_{s}(\tau) = \sum\limits\_{\gamma\in\Gamma\_{\infty}\backslash\text{SL}(2,\mathbb{Z})}\left.y^{s}\right\vert\_{\gamma},$$
where $s\in\mathbb{C}$ and $\Gamma\_{\infty}$ is the stabilizer of $\text{SL}(2,\mathbb{Z})$. This sum converges when $\text{Re}(s)>1$ and when continued analytically to all of the $s$-plane, it reads
$$E\_{s}(\tau) = y^{s} + \frac{\Lambda(1-s)}{\Lambda(s)}\frac{1}{y^{s-1}} + \sum\limits\_{j=1}^{\infty}\frac{4\sigma\_{2s-1}(j)\sqrt{y}\ K\_{s-\tfrac{1}{2}}(2\pi jy)}{\Lambda(s)j^{s - \tfrac{1}{2}}}\cos(2\pi jx),$$
where $\sigma\_{2s-1}(j)$ is the divisor sigma function, $K\_{\nu}(z)$ is the modified Bessel function of the second kind, and $\Lambda(s)$ is the completed zeta function as a function of $2s$, $\Lambda(s) = \pi^{-s}\Gamma(s)\zeta(2s)$ ,that obeys the functional equation $\Lambda\left(\tfrac{1}{2}-s\right) = \Lambda(s)$. This real analytic Eisenstein series is an eigenfunction of the Laplacian defined on the space $\text{SL}(2,\mathbb{Z})\backslash\mathbb{H}$, where $\mathbb{H}$ is the upper half-plane. I have the following questions:
1. Does there exist explicit formulae in literature for the analytic continuation of $E\_{s}(\tau)$ defined on the space with Hecke and Fricke subgroup quotients, i.e. $\Gamma\_{0}(N)\backslash\mathbb{H}$ and $\Gamma\_{0}^{+}(N)\backslash\mathbb{H}$?
2. I suspect the analytic continuation formulae to contain certain L-functions $L^{\*}(s)$ akin to how $\Lambda(s)$ shows up in the above definition. If they don't then is it possible to reexpress these formulae in terms of some nice L-functions?
Here, the Hecke group $\Gamma\_{0}(N)$ is a discrete subgroup of $\text{SL}(2,\mathbb{Z})$ and is a subgroup of index $2$ in the Fricke group $\Gamma\_{0}^{+}(N)$.
| https://mathoverflow.net/users/99716 | Analytic continuation of the Eisenstein series defined over Hecke and Fricke subgroups | One natural generalization of $E\_s$ to $\Gamma\_0(N) \backslash \mathbb{H}$ is the family of Eisenstein series, $$E\_{s; \chi\_1, \chi\_2}(\tau) = \sum\_{\gamma \in \Gamma\_{\infty} \backslash \mathrm{SL}\_2(\mathbb{Z})} \chi\_1(c) \chi\_2(d) (q\_2 y)^s \Big| \gamma, \quad \gamma = \begin{pmatrix} a & b \\ c & d \end{pmatrix},$$ where $\chi\_1$ and $\chi\_2$ are Dirichlet characters modulo $q\_1$, $q\_2$ satisfying $\chi\_1(-1) = \chi\_2(-1)$. These are automorphic on $\Gamma\_0(q\_1 q\_2)$ with Nebentypus character $\chi\_1 \overline{\chi\_2}$.
The series $E\_{s; \chi\_1, \chi\_2}$ have an analytic continuation and have natural completions $E^\*$, with the completed $L$-function of $\chi\_1 \chi\_2$ playing the role of $\Lambda$. The functional equation should now relate Eisenstein series attached to different characters, taking the form $$E\_{s; \chi\_1, \chi\_2}^\* = E\_{1-s, \overline{\chi\_2}, \overline{\chi\_1}}^\*.$$ For a detailed reference (also including weight other than zero) see the paper [*Explicit calculations with Eisenstein series* by M. Young](https://arxiv.org/abs/1710.03624), especially section 4.
| 4 | https://mathoverflow.net/users/489142 | 427930 | 173,537 |
https://mathoverflow.net/questions/427934 | 4 | Below is a simple determinant. I need to show that it is not 0, so that the corresponding matrix is invertible.
$$
D = \begin{vmatrix}
0! & 1! & 2! & \ldots & x!\\
1! & 2! & 3! & \ldots & (x+1)! \\
2! & 3! & 4! & \ldots & (x+2)! \\
\vdots & \vdots & \vdots & \ldots & \vdots \\
y! & (y+1)! & (y+2)! & \ldots & (x+y)!
\end{vmatrix}
$$
Remark: As pointed out in the comments, obviously we must have $y=x$ in order to have a square matrix.
Obviously, we can factor out $0!1!\ldots y!$ and get entries which are falling factorials, but I do not see how to continue.
The determinant of a similar 3X3 matrix was considered [here](https://math.stackexchange.com/questions/2053416/determinants-question-with-factorials) and a stronger statement was proved on the remainder that the determinant has module 4 (after division by the obvious factors).
| https://mathoverflow.net/users/85939 | Determinant with factorials is not 0? | This is the [Hankel determinant](https://en.wikipedia.org/wiki/Hankel_matrix) associated to the sequence $m\_n = \mathbb{E}(X^n) = n!$ of moments of an exponential distribution with mean $1$. Some general results can be used to show that the sequence of Hankel determinants associated to the moments of a random variable are always positive iff the induced measure on $\mathbb{R}$ has infinite support, and some more general results can be used to exactly calculate the Hankel determinants as in the comments by calculating an appropriate sequence of orthogonal polynomials. The relevant orthogonal polynomials for the exponential distribution are the [Laguerre polynomials](https://en.wikipedia.org/wiki/Laguerre_polynomials).
I gather this is very classical material but I don't know a reference; you can see a writeup in slightly unusual language [here](https://qchu.wordpress.com/2012/09/18/moments-hankel-determinants-orthogonal-polynomials-motzkin-paths-and-continued-fractions/).
**Edit:** Ah, here's a reference: Section 2.7 of Krattenthaler's *[Advanced Determinant Calculus](https://arxiv.org/abs/math/9902004)*.
| 10 | https://mathoverflow.net/users/290 | 427939 | 173,541 |
https://mathoverflow.net/questions/427561 | 3 | Say that a logic $\mathcal{L}$ is **nowhere-negative** iff for every $\mathcal{L}$-theory $T$ there is a structure $\mathfrak{A}$ such that $$\mathit{Th}\_\mathcal{L}(\mathfrak{A})=\mathit{Ded}\_\mathcal{L}(T),$$ where $\mathit{Th}\_\mathcal{L}(\mathfrak{A})=\{\varphi\in\mathcal{L}:\mathfrak{A}\models\_\mathcal{L}\varphi\}$ is the $\mathcal{L}$-theory of the structure $\mathfrak{A}$ and $\mathit{Ded}\_\mathcal{L}(T)=\{\varphi\in\mathcal{L}: T\models\_\mathcal{L}\varphi\}$ is the $\mathcal{L}$-deductive-closure of $T$. Basically, $\mathcal{L}$ is nowhere-negative iff every deductively-closed theory is the full theory of some structure.
For example, first-order logic is unsurprisingly non-nowhere-negative: consider any incomplete but deductively closed theory. On the other hand, equational logic **is** nowhere-negative by the usual proof of the HSP theorem: given a class of structures $\mathbb{K}$ we construct a single structure $\mathcal{A}$ satisfying precisely those equations true in every element of $\mathbb{K}$, and if we apply this to $\mathbb{K}$ = the class of models of some equational theory $T$ then the resulting $\mathcal{A}$ will satisfy all and only the equational sentences provable from $T$.
Re: the name, note that if $\mathcal{L}$ contains sentences $\varphi,\psi$ with nonempty and exactly complementary model classes then $\mathcal{L}$ cannot be nowhere-negative: this is because any structure must satisfy exactly one of $\varphi$ and $\psi$ and neither $\varphi$ nor $\psi$ is in the deductive closure of $\emptyset$, so $\mathit{Ded}\_\mathcal{L}(\emptyset)\not=\mathit{Th}\_\mathcal{L}(\mathfrak{A})$ for any structure $\mathfrak{A}$.
My question is about the coarse structure of the nowhere-negative fragments of first-order logic:
>
> **Question**: Is there a ("reasonably natural") maximal nowhere-negative sublogic of first-order logic?
>
>
>
(Actually, the question I'm really interested in is the broader "Is every nowhere-negative sublogic $\mathcal{L}\subseteq\mathsf{FOL}$ contained in a *maximal* nowhere-negative sublogic of $\mathsf{FOL}$?," but I think that's a bit ambitious.) Note that this is not immediately solved via Zorn's lemma, since the union of a chain of nowhere-negative logics could fail to be nowhere-negative. I'm separately interested in what nowhere-negativity is *actually* called, since I vaguely remember seeing the same notion with a different name before but I can't track it down at the moment.
| https://mathoverflow.net/users/8133 | Is there a maximal fragment of FOL with "no negation at all?" | There is a simple characterization that $\def\LL{\mathcal L}\LL$ is nowhere negative iff all $\LL$-theories have the *disjunction property*:
>
> **Proposition.** For any $\LL\subseteq\mathrm{FO}$, the following are equivalent:
>
>
> 1. $\LL$ is nowhere negative.
> 2. For every $\LL$-theory $T$ and every finite set of $\LL$-formulas $\{\phi\_i:i\in I\}$,
> $$T\models\bigvee\_{i\in I}\phi\_i\implies\exists i\in I\:T\models\phi\_i.$$
> Note that this is equivalent to the special cases where $|I|$ is either $0$ (which amount to the consistency of $T$) or $2$.
> 3. The same for finite theories $T$.
>
>
>
**Proof.**
$1\to2$: Let $A$ be a model whose theory is $T$. Then $A\models\bigvee\_{i\in I}\phi\_i$ implies $A\models\phi\_i$ for some $i\in I$ by the definition of satisfaction of disjunctions.
$2\to3$ is trivial.
$3\to1$: For any $\LL$-theory $T$, the $\mathrm{FO}$-theory
$$T^\*=T+\{\neg\phi:\phi\in\LL,T\not\models\phi\}$$
is consistent: if it were not, there would be a finite $T'\subseteq T$ and a finite set $\{\phi\_i:i\in I\}$ of $\LL$-sentences such that $T\not\models\phi\_i$ (hence $T'\not\models\phi\_i$) for each $i\in I$, and $T'\cup\{\neg\phi\_i:i\in I\}$ is inconsistent. But then $T'\models\bigvee\_{i\in I}\phi\_i$, contradicting the disjunction property of $T'$.
Then any model of $T^\*$ has the property that its $\LL$-theory consists exactly of the $\LL$-consequences of $T$. QED
Note that the only properties of FO we used here is that it is closed under Boolean connectives, and obeys the compactness theorem.
It follows immediately from the Proposition that the property of being nowhere negative *is* stable under unions of chains after all, hence by Zorn’s lemma, maximal nowhere negative sublogics of FO exist (as long as we fix the language so that the class of all FO sentences is a set).
While it is neither here nor there, let me mention another fact from the comments: the equational logic example generalizes to the logic $\LL\subseteq\mathrm{FO}$ of sentences built from atomic formulas by means of $\exists$, $\forall$, and $\land$. The fact that all $\LL$-theories have the disjunction property, hence $\LL$ is nowhere negative, is a consequence of the property that for every family of models $\{A\_i:i\in I\}$ (where $I$ may be empty) and every $\LL$-sentence $\phi$,
$$\prod\_{i\in I}A\_i\models\phi\iff\forall i\in I\:A\_i\models\phi.$$
The left-to-right implication follows from the fact that all $\LL$-sentences are positive (monotone), hence preserved under surjective homomorphisms (incuding projections from products); the right-to-left implication follows from the fact that $\LL$-sentences are strict Horn sentences. I do not know whether $\LL$ is maximal.
| 6 | https://mathoverflow.net/users/12705 | 427946 | 173,543 |
https://mathoverflow.net/questions/427965 | 2 | Let $M$ a non-well founded model of Finite $\sf ZF$, which is $\sf ZF$ with axiom of infinity replaced by the axiom stating that all sets are finite. So there must be a set $\zeta$ that $M$ thinks it's a natural yet it possess's an infinite descending membership chain as seen from the outside of $M$, i.e. we have $\{ \zeta-1, \zeta-2, \zeta-3, ...\} \subset \zeta $, where each $\zeta -n = \zeta -n-1 \cup \{\zeta-n-1\} $ for all $n \in \mathbb Z$.
Since $M$ is a model of $\sf ZF$, so there must be stages: $$....,V\_{\zeta-2}, V\_{\zeta-1}, V\_\zeta, V\_{\zeta + 1}, V\_{\zeta+2},...$$
Now we know that some models of $\sf ZF$ can have external downward rank-shifting automorphisms $j$ on them, i.e. $j(V\_\alpha) =V\_\beta$ where $\beta < \alpha$ for some non-standard ordinal $\alpha$.
>
> Can there be a non-well founded model $M$ of Finite $\sf ZF$, that admits an external automorphism $j$ such that: $j(V\_{\zeta+n})= V\_{\zeta +n -1}$ ?
>
>
>
| https://mathoverflow.net/users/95347 | About external automorphism on non-well founded model of Finite ZF? | The answer is no, there cannot be even a single instance of this for an automorphism $j:M\to M$. The reason is that if $j(V\_{n+1})=V\_n$ for some (possibly nonstandard $n$), then we would have $j(n+1)=n$, and so $j$ does not respect that one of these numbers is even and the other odd. So the map couldn't be truth-preserving on $M$.
| 3 | https://mathoverflow.net/users/1946 | 427967 | 173,551 |
https://mathoverflow.net/questions/427970 | 1 | Let $X$ be a proper, geometrically connected, geometrically integral variety over $\mathbf{F}\_q$. There exists a finite field extension $k/\mathbf{F}\_q$ of degree $d$ and an alteration $X'\to X\_k$ defined over $k$, where $X'$ is smooth and projective over $k$.
For $\ell$ a prime not dividing $q$, we have an injective map
$$H^i\_{ét}(X\_{\overline{k}},\overline{\mathbf{Q}}\_{\ell})\subset H^i\_{ét}(X'\_{\overline{k}},\overline{\mathbf{Q}}\_{\ell})$$
such that geometric Frobenius $F$ on $H^i\_{ét}(X'\_{\overline{k}},\overline{\mathbf{Q}}\_{\ell})$ restricts to the $d$-th power of geometric Frobenius on $H^i\_{ét}(X\_{\overline{k}},\overline{\mathbf{Q}}\_{\ell})\simeq H^i\_{ét}(X\_{\overline{\mathbf{F}}\_q},\overline{\mathbf{Q}}\_{\ell})$.
Fix a field isomorphism $\iota: \overline{\mathbf{Q}}\_{\ell}\simeq\mathbf{C}$.
By the Weil conjectures applied to $X'$, $F$ has eigenvalues $\lambda$ with complex absolute value $$|\iota(\lambda)|=q^{di/2}$$ It looks like this implies that this is also satisfied by geometric Frobenius (to the power $d$) on $H^i\_{ét}(X\_{\overline{\mathbf{F}}\_q},\overline{\mathbf{Q}}\_{\ell})$.
This also seems to imply that geometric Frobenius on $H^i\_{ét}(X\_{\overline{\mathbf{F}}\_q},\overline{\mathbf{Q}}\_{\ell})$ has eigenvalues with complex absolute value $q^{i/2}$.
This feels wrong: I would expect the eigenvalues to have complex absolute value **bounded above** by $q^{i/2}$ (resp. $q^{di/2}$). If $X$ was itself smooth then Poincaré duality would give the reverse inequality, and so purity, but without smoothness I would expect this to not be true.
>
> What's being missed?
>
>
>
(maybe an incorrect use/statement of de Jong's theorem? I guess $X'$ needs not be projective, but only open in smooth projective)
| https://mathoverflow.net/users/nan | Purity for proper varieties | The claim that alteration induces an injection on cohomology is wrong, as the example of the resolution of a nodal cubic curve by $\mathbb P^1$ shows (resolutions being a special case of alterations).
The induced map from $H^1$ of the nodal cubic to $H^1$ of $\mathbb P^1$ is not an injection, because $H^1$ of $\mathbb P^1$ vanishes but $H^1$ of the nodal cubic does not.
| 3 | https://mathoverflow.net/users/18060 | 427971 | 173,552 |
https://mathoverflow.net/questions/427942 | 15 | For an integer $n$, let $\ell(n)$ denote the maximal number of consecutive $1$s in the binary expansion of $n$. For instance,
$$ \ell(71\_{10}) = \ell(1000111\_2) = 3. $$
Consider the set $E$ of all integers $n \in \mathbb{N}$ such that $\ell(n)$ is even.
It seems intuitively obvious that $E$ should have natural density $1/2$:
$$ d(E) = \lim\_{N\to \infty} \frac{|E \cap [0,N)|}{N} = 1/2.$$
Can one prove that this is actually the case?
Less ambitiously, can one show that
$$ \bar{d}(E) = \limsup\_{N\to \infty} \frac{|E \cap [0,N)|}{N} > 0?$$
---
Edit to add: The following sketch of an argument seeems to show that $$1/3 \leq \underline{d}(E) \leq \bar{d}(E) \leq 2/3.$$
Note that the binary expansion of any integer $n$ can be written as
$(n)\_2 = u 1^{\ell(n)}v$, where $u,v \in \{0,1\}^\*$, $u$ is either empty or ends with a $0$, $v$ is either empty or begins with a zero, and the longest block of consecutive $1$s in $u$ has length strictly less than $\ell(n)$.
Divide $E$ into three sets, depending on $v$ in the decomposition above: $E\_{bad}$ consists of $n\in E$ for which $|v| \leq 1$, $E\_{0}$ consists of $n\in E$ for which $v = 00v'$, and $E\_1$ consists of $n \in E$ for which $v = 01v'$. The set $E\_{bad}$ is small enough that we don't need to worry about it.
Now, define the map $\phi\_0 \colon\ E\_0 \to \mathbb{N} \setminus E$ by (using the expansion above):
$$ (\phi\_0(n))\_2 = u1^{\ell(n)+1}0v'.$$
Similarly, define $\phi\_1 \colon\ E\_1 \to \mathbb{N} \setminus E$ by
$$ (\phi\_1(n))\_2 = u1^{\ell(n)+1}0v'.$$
It is not hard to show that these maps are both injective. Additionally, for almost all $n$, $\phi\_0(n)/n$ is close to $1$, and likewise for $\phi\_1$. From here, one can infer that
$$ \bar d(E) \leq 2(1 -\bar d(E)), $$
and consequently $\bar d(E) \leq 2/3$. A symmetric argument shows that $\underline d(E) \geq 1/3$.
| https://mathoverflow.net/users/14988 | The parity of the maximal number of consecutive 1s in the binary expansion of an integer | Perhaps surprisingly, the random variable $\ell(n)$ (with $n$ drawn uniformly from $[0,N)$) concentrates too much around $\log\_2\log\_2 N$ (where $\log\_2$ denotes the logarithm to base $2$) to have a limiting parity probability - the variance stays bounded as $N \to \infty$, as opposed to growing to infinity. One only recovers a limiting law when the fractional part $\{\frac{1}{2} \log\_2 \log\_2 N \}$ of half the double logarithm of $N$ converges to a limit, and when one does so the parity probability will usually converge to a limit that deviates slightly from $1/2$.
To simplify the calculations a little let us assume that $N$ is of the form $N = 2^{2^{2k+1}}$ (I'll leave it as an exercise to the reader to handle the general case) in the asymptotic regime $k \to \infty$. Then the binary expansion of a randomly chosen element $n$ of $[0,N)$ consists of $2^{2k+1}$ independent Bernoulli variables (each taking $0$ and $1$ with values $1/2$). We think of this as the initial segment of an infinite sequence of Bernoulli variables. Now we perform the standard trick of viewing this sequence as a [renewal process](https://en.wikipedia.org/wiki/Renewal_theory). After each $0$, the number of $1$s one encounters before one reaches the next $0$ is $a-1$ where $a$ is distributed according to a [geometric distribution](https://en.wikipedia.org/wiki/Geometric_distribution) of expectation $2$. One can thus interpret this sequence as $a\_1-1$ zeroes followed by a one, then $a\_2-1$ zeroes followed by a one, and so forth ad infinitum, where $a\_1,a\_2,\dots$ are iid geometric distributions of expectation $2$. By the law of large numbers, we see with probability $1-o(1)$ that the first $t$ for which $a\_1+\dots+a\_t$ exceeds $2^{2k+1}$ will lie in the range $[2^{2k}-2^{4k/3},2^{2k}+2^{4k/3}]$ (say). Also, by symmetry we see that with probability $1-o(1)$, the maximum value of the $a\_i$ for $i \leq 2^{2k}+2^{4k/3}$ will already be attained for $i \leq 2^{2k}-2^{4k/3}$. Putting these two together, we see that with probability $1-o(1)$, $\ell(n)$ will equal $\sup\_{1 \leq i \leq 2^{2k}} a\_i-1$. So asymptotically we just need to understand the distribution of $\sup\_{1 \leq i \leq 2^{2k}} a\_i-1$. We have the exact formula
$$ {\bf P}( \sup\_{1 \leq i \leq 2^{2k}} a\_i-1 < t ) = \prod\_{i=1}^{2^{2k}} {\bf P}(a\_i-1 < t)$$
$$ = (1-2^{-t})^{2^{2k}}$$
for any positive integer $t$, so in particular
$$ {\bf P}( \sup\_{1 \leq i \leq 2^{2k}} a\_i-1 - 2k < s ) = \exp( - 2^{-s} ) + o(1)$$
for any fixed $s$. Thus in the limit $k \to \infty$, $\ell(n) - 2k$ converges in distribution to a discrete random variable $X$ with distribution function
$$ {\bf P}( X < s ) = \exp( - 2^{-s} ).$$
(Is there a name for this sort of random variable? EDIT: it is a discrete Gumbel distribution, see update below.) The quantity $\frac{|E \cap [0,N)|}{N}$ then converges to the probability that $X$ is even, which is
$$ \sum\_{j \in {\bf Z}} \exp(-2^{-2j-1}) - \exp(-2^{-2j}) = 0.4998402\dots$$
which is very slightly less than $1/2$. (If one picked a different subsequence of $N$ one would obtain a different limit; for instance if $N = 2^{2^{2k}}$ then the same analysis would ultimately give the complementary limiting probability of $0.500157\dots$.)
UPDATE: after a tip in the comments, I'll remark that a refinement of the above analysis will eventually show that the distribution of $\ell(n)$ is asymptotic to the integer part $\lfloor \mathrm{Gumbel}(\log\_2 \log\_2 N, \log\_2 e)\rfloor$ of a [Gumbel distribution](https://en.wikipedia.org/wiki/Gumbel_distribution), in the sense that the [Levy metric](https://en.wikipedia.org/wiki/L%C3%A9vy_metric) (for instance) between the two distributions goes to zero as $N \to \infty$ (without any further restriction on the natural number $N$). In retrospect this sort of answer was a natural guess, given the usual role of the Gumbel distribution in [extreme value theory](https://en.wikipedia.org/wiki/Extreme_value_theory).
Some references for further reading (gathered from following links in the comments):
*Gordon, Louis; Schilling, Mark F.; Waterman, Michael S.*, [**An extreme value theory for long head runs**](http://dx.doi.org/10.1007/BF00699107), Probab. Theory Relat. Fields 72, 279-287 (1986). [ZBL0587.60031](https://zbmath.org/?q=an:0587.60031).
*Chakraborty, Subrata; Chakravarty, Dhrubajyoti; Mazucheli, Josmar; Bertoli, Wesley*, A discrete analog of Gumbel distribution: properties, parameter estimation and applications, [ZBL07482747](https://zbmath.org/?q=an:07482747).
| 25 | https://mathoverflow.net/users/766 | 427983 | 173,556 |
https://mathoverflow.net/questions/427984 | 9 | Classically, given a compact Lie group $G$, there is a topological space $BG$ which classifies principal $G$-bundles. This means that there is an equality of sets {principal $G$-bundles up to isomorphism} = {maps $X \to BG$ up to homotopy}.
**Question:** is there a "infinity categorical" refinement of this statement? (And if so, what are good references?)
For example, is there a statement along the lines of " every principal $G$-bundle over a space $X$ is "coherently isomorphic" (whatever this means...) to the pullback of the universal bundle over $BG$ under some map $X \to BG$ which is unique "up to coherent homotopy"?
| https://mathoverflow.net/users/396191 | Homotopy coherent generalization of classifying space theory | $\newcommand{\cS}{\mathcal{S}}\newcommand{\Fun}{\mathrm{Fun}}\newcommand{\LMod}{\mathrm{LMod}}\newcommand{\Sp}{\mathrm{Sp}}$Hey Laurent :) Let $X$ be a space (which I'll view as a Kan complex), and let $\cS$ denote the $\infty$-category of spaces. You could think of the homotopy-coherent categorification of "maps $X \to BG$ up to homotopy" as $\cS\_{/BG}$. Then Theorem 2.2.1.2 of Higher Topos Theory (see Section 2.2 of <https://arxiv.org/pdf/1403.4325.pdf> for a bite-sized summary) says that there's an equivalence $\Fun(X, \cS) \simeq \cS\_{/X}$. So, taking $X = BG$ (i.e., the nerve of the one-object topological category whose morphism space is $G$), we see that $\Fun(BG, \cS) \simeq \cS\_{/BG}$. But a functor from $BG \to \cS$ is exactly the data of a homotopy-coherent $G$-action on a space, as desired. (One other useful fact is that if $X$ is assumed to be connected, and $\Sp$ denotes the $\infty$-category of spectra, then there's a Koszul duality equivalence $\Fun(X, \Sp) \simeq \LMod\_{\Sigma^\infty\_+ \Omega X}(\Sp)$, where $\Sigma^\infty\_+ \Omega X$ is viewed as an $\mathbf{E}\_1$-algebra in $\Sp$. Taking $X = BG$, we see that $\Fun(BG, \Sp) \simeq \LMod\_{\Sigma^\infty\_+ G}(\Sp)$; you can interpret this as saying that the $\mathbf{E}\_1$-algebra $\Sigma^\infty\_+ G$ plays the role of a spherical group ring of $G$.)
| 10 | https://mathoverflow.net/users/102390 | 427985 | 173,557 |
https://mathoverflow.net/questions/427849 | 2 | The following lemma is from Fisher, Marsden, and Moncrief's paper: the structure of the space of solutions of Einstein's equations:1
**1.1. Lemma.**
If Ein( $\left.{ }^{(4)} g\right)=0$, and ${ }^{(4)} h$ is any symmetric two tensor, then
$$
\delta\left[\operatorname{DEin}\left({ }^{(4)} g\right) \cdot ^{4} h\right]=0
$$
where $\delta=\delta\_{(4) g}$ is the divergence with respect to ${ }^{(4)} g$.
They gave the following proof:
The contracted Bianchi identities assert that $\delta \operatorname{Ein}\left({ }^{(4)} g\right)=0$. Differentiation gives the identity
$$
\*\*\left[\text { D } \delta\left({ }^{(4)} g\right) \cdot ^{4}h\right] \cdot \operatorname{Ein}\left({ }^{(4)} g\right)+\delta\left[D \operatorname{Ein}\left({ }^{(4)} g\right) \cdot^{4} h\right]=0\*\*
$$
where $\delta\left({ }^{(4)} g\right)=\delta\_{(4)}$ indicates the functional dependence of $\delta$ on ${ }^{(4)} g$, and $\left[\mathrm{D} \delta\left({ }^{(4)} g\right) \cdot ^{4} \mathrm{h}\right]$ is the linearized divergence operator acting on Ein $\left({ }^{(4)} g\right)$. The lemma follows since $\operatorname{Ein}\left({ }^{(4)} g\right)=0$.
Sadly, I cannot see how they found the identity in the second sentence of the proof. Could you give me some hints/suggestions on how to derive the identity in the second sentence of Lemma 1.1?
Any help/suggestions would be highly appreciated. Thanks so much.
1*Fischer, Arthur E.; Marsden, Jerrold E.; Moncrief, Vincent*, [**The structure of the space of solutions of Einstein’s equations. I: One Killing field**](http://www.numdam.org/item?id=AIHPA_1980__33_2_147_0), Ann. Inst. Henri Poincaré, Nouv. Sér., Sect. A 33, 147-194 (1980); [eudml](https://eudml.org/doc/76090). [ZBL0454.53044](https://zbmath.org/?q=an:0454.53044), [MR605194](https://mathscinet.ams.org/mathscinet-getitem?mr=605194).
| https://mathoverflow.net/users/298774 | Understanding the proof of lemma 1.1 from Fisher, Marsden, and Moncrief's paper | I presume the formula you are asking about is the long one highlighed by $\*$ $\*$ in your question, while the standard "contracted Bianchi identity" $\delta \operatorname{Ein}\left({ }^{(4)} g\right)=0$ poses no mystery to you. The Longer equation is simply obtained by using the Leibniz rule while applying the functional derivative $\mathrm{D}$ that last identity. I would consider that a pretty clear explanation, but possibly that is not the point you are confused about.
Perhaps you are wondering why the Leinbniz rule applies to $\delta$, which is an operator. A quick way to see why is just to write the corresponding expression in coordinates. There is no need to be very explicit, the following schematic form suffices:
$$
\delta \mathrm{Ein} = \partial \cdot \mathrm{Ein} + \Gamma \cdot \mathrm{Ein} .
$$
That is, the covariant divergence operator $\delta(-)$ consists of a coordinate derivative part $\partial\cdot(-)$ and of the part containing multiplication by Christoffel symbols $\Gamma\cdot(-)$. Unsurprisingly, the coordinate derivative part has no dependence on the metric, hence formally $\mathrm{D}\partial = 0$. So $\mathrm{D}\delta$ comes entirely from the Christoffel part, which quite obviously obeys the Leibniz rule.
If that's still not where your confusion lies, please clarify your question further.
| 1 | https://mathoverflow.net/users/2622 | 427991 | 173,560 |
https://mathoverflow.net/questions/427989 | 7 | Freedman’s theorem shows that all 3-dimensional homology spheres bound topologically a contractible 4-manifold. It is well known that the Poincaré homology sphere does not bound a smooth contractible 4-manifold. Is there other explicit examples or is there an invariant of homology sphere characterizing this property(whether or not smoothly bound a contractible 4-manifold)?
| https://mathoverflow.net/users/1190 | Examples of homology sphere that bound a nonsmoothable contractible 4-manifold | There are certainly lots of obstructions coming from gauge theory, for instance Frøyshov's $h$-invariant or Ozsváth and Szabó's $d$-invariant. If an integer homology sphere $P$ has $d(P) \neq 0$, then $P$ does not even bound a *rational* homology ball. There are $\mathbb{Z}/2\mathbb{Z}$ refinements (more precisely, obstructions to bounding *spin* rational homology balls).
In addition to this, and closer to your question: using periodic ends and Seiberg–Witten theory, Taubes proved that if $Y$ is a homology 3-sphere that bounds a negative definite 4-manifold with non-diagonal intersection form (e.g. $Y$ is the Poincaré sphere, which bounds the $E\_8$-plumbing of spheres), then $Y \# {-Y}$ does not bound a *contractible* 4-manifold. (It does bound an integer homology 4-ball, namely $(Y\setminus B^3)\times [0,1]$.) As far as I know, there is no single invariant capturing this information. Certainly there is no single invariant capturing whether a 3-manifold bounds an integer homology 4-ball.
Some keywords you might care about are: (integer or rational) homology cobordism group, homology cobordism invariants, Heegaard Floer correction terms (or variants thereof, like involutive Floer homology).
| 5 | https://mathoverflow.net/users/13119 | 428014 | 173,566 |
https://mathoverflow.net/questions/428020 | -3 | Let be a positive integer and
= () be the digit sum of such that
+ 1 ≡ 0 (mod 2).
Is it that if is prime then is also prime?
e.g. =47(prime)-> =4+7=11 (prime)
| https://mathoverflow.net/users/489235 | Digit sum of a prime number | Not necessarily, but it takes a while to see this, because
you're not allowing $q$ to be a multiple of $2$, and it cannot be
a multiple of $3$ (other than $3$ itself) because then the same
would be true of $p$. So the smallest candidate for $q$ is $25$,
which happens for the first time at $p=997$.
| 13 | https://mathoverflow.net/users/14830 | 428021 | 173,568 |
https://mathoverflow.net/questions/427997 | 5 | Let $d \geq 2$, and let $f \in W^{1, 1} (\mathbb R^d)$ be a Sobolev function.
**Question:** For any $a, b \in \mathbb R$ such that $\text{essinf } f \leq a < b \leq \text{esssup } f$, is it true that $\mu\left(f^{-1} ([a, b])\right) > 0$?
*Note:* Here $\mu$ denotes the Lebesgue measure.
| https://mathoverflow.net/users/173490 | Intermediate value property for Sobolev functions | the one dimensional case is clear since $W^{1,1}$ functions have representatives that are absolutely continuous, see [1] Sec. 4.9.1. The general case can be reduced to that case:
Let $a<a\_1<b\_1<b$. Since $A:=\{x \in \mathbb R^d: f(x)<a\_1\}$ has $\mu(A)>0$,
it has a point of density; WLOG that point is the origin. Since $D:=\{x \in \mathbb R^d: f(x)>b\_1\}$ has $\mu(D)>0$,
it has a point of density; WLOG that point is $z=(0,0,\ldots,0,c)$ where $c>0$.
Find $0<r<c/3$ so that $A$ occupies at least $3/4$ of $B(0,r)$ and $D$ occupies at least $3/4$ of $B(z,r)$. Thus
$A\_1:=A \cap (D-z)$ satisfies $\mu(A\_1) \ge \mu B(0,r)/2$. By Theorem 2, Sec. 4.9.2 in [1], $f$ has a representative such that for a.e. $x=(x\_1,\ldots,x\_d)$ in $A\_1$, the function
$t \mapsto f(x\_1,\ldots,x\_d+t)$ is absolutely continuous, and the claim follows by Fubini.
Evans, Lawrence C., and Ronald F. Gariepy. "Measure theory and fine properties of functions", CRC Press. Studies in Advances Mathematics (1992).
| 5 | https://mathoverflow.net/users/7691 | 428035 | 173,572 |
https://mathoverflow.net/questions/428004 | -1 | In the 3-dimensional hyperbolic space there are given a plane $\mathcal{P}$ and four distinct lines $a\_1, a\_2, r\_1, r\_2$ in such positions that $a\_1$ and $a\_2$ are perpendicular to $\mathcal{P}$, $r\_1$ is coplanar with $a\_1, r\_2$ is coplanar with $a\_2$, finally $r\_1$ and $r\_2$ intersect $\mathcal{P}$ at the same angle. Rotate $r\_1$ around $a\_1$ and rotate $r\_2$ around $a\_2$; denote by $\mathcal{S}\_1$ and $\mathcal{S}\_2$ the two surfaces of revolution they sweep out. Show that the common points of $\mathcal{S}\_1$ and $\mathcal{S}\_2$ lie in a plane.
| https://mathoverflow.net/users/489216 | 3-dimensional hyperbolic space | Let $L$ be the geodesic segment in $P$ with one endpoint at $a\_1 \cap P$ and the other endpoint at $a\_2 \cap P$. Let $b$ be the midpoint of $L$. We define a plane $Q$ by requiring it to contain $b$ and by requiring it to be perpendicular to $L$ (and thus to $P$). Thus reflection in $Q$ exchanges $a\_1$ and $a\_2$. It is now an exercise to prove that reflection in $Q$ exchanges the “cones” $S\_1$ and $S\_2$. Thus $S\_1 \cap S\_2$ lies in the plane $Q$.
---
Here I am assuming that the lines $r\_1$ and $r\_2$ actually meet $P$ — the question is slightly unclear on this detail.
| 0 | https://mathoverflow.net/users/1650 | 428038 | 173,573 |
https://mathoverflow.net/questions/428041 | 3 | The classic Hahn-Mazurkiewicz theorem has the following consequence: Let $X$ be a compact, connected topological manifold. Then there is a continuous surjective map $f: [0,1] \rightarrow X$.
It is also true that such a $f$ cannot be injective unless $X=[0,1]$ (since then $f$ would be a homeomorphism). I was wondering if some weaker notions of this is possible, where injectivity holds (this would be very useful for my research). Specifically, suppose we have a compact, connected Riemannian manifold $(M,g)$, without any boundary. Let $d$ be the metric induced on $M$ by $g$. Then I want to know the following:
For every $\epsilon > 0$, does there exist a map $f\_{\epsilon}: [0,1] \rightarrow M$ that is continuous and injective, such that for any $x \in M$ there exists $y \in [0,1]$ such that $d(x,f\_{\epsilon}(y)) < \epsilon$?
Does anything change if $M$ is non-compact?
| https://mathoverflow.net/users/151406 | Space filling curves | I would call such $f\_\epsilon$ *coarsely $\epsilon$-dense*, or more simply *$\epsilon$-dense*. These exist in the compact case as follows. Tile $M$ by $n$-balls (these may overlap, but only finitely). Each ball is homeomorphic to $[0, 1]^n$, so admits an $\epsilon/2$-dense smooth Hilbert arc. A small perturbation of these makes them pairwise disjoint. Choose an ordering of the $n$-balls. Now connect the Hilbert arcs together, in order. Tubular neighbourhoods allow us to do this smoothly and injectively.
In the complete, non-compact case this is not possible. This is because the continuous image of a compact set is compact, and thus contained in a metric ball.
| 4 | https://mathoverflow.net/users/1650 | 428042 | 173,575 |
https://mathoverflow.net/questions/428043 | 2 | The last problem in 2022 IMC Day 1 strongly correlates with graph theory. In its official solution, the fundamental approach can be rephrased as follows.
>
> Give a digraph $G=(V,E)$. We call a subset of $E$ *admissible* such that it doesn't contain any directed paths of length $2$. Let $b(G)$ denote the minimal number of admissible sets that cover $E$.
>
>
>
The comment stated that $b(G) = \left( 1 + o \left( 1 \right) \right) \log\_2 \chi(G)$, where $\chi(G)$ is the chromatic number of $G$.
The lower bound is easy. Given an admissible cover $\mathcal{E}$, for each vertex, we convert the information on whether $E\_i$ in $\mathcal{E}$ contains an edge incident to $v$ into a binary string. Now color $V$ according to their corresponding string. Hence we conclude that $2^{b(G)} \geqslant \chi(G)$.
However, the upper bound is intractable. In its solution, it only states that $b(G) \leqslant 2\left \lceil \log\_2 \chi(G) \right \rceil$. I searched a lot but failed to find any context about $b(G)$. Talking about the chromatic number of digraphs, I can reflect no more than the Gallai-Roy theorem, which states that there must be a path of length larger than $\chi(G)$. But it seems rare to consider covers splitting path into pieces and I don't know enough tools to solve this sort of problem.
So my question is, how can I prove the upper bound of $b(G)$ is also $\left( 1 + o \left( 1 \right) \right) \log\_2 \chi(G)$? Thanks.
| https://mathoverflow.net/users/151440 | Minimal digraph covering with no 2-path edge sets is of size $\left( 1 + o \left( 1 \right) \right) \log_2 \chi(G)$ | This is beautiful application of Sperner theorem on antichains. Consider a bijection between the vertices and the family $\binom{[n]}{n/2}$, which realizes the maximum in the mentioned theorem.
Element $i$ from 1 to $n$ are the palette. One may color a vertex in any color containing in the corresponding set. The antichain property implies that the coloring is proper.
P. S. In a folklore it is related with a Nero Wolfe, who have deal with $m$ people, there are a murderer and a witness among them. Each day Nero asks Archi to collect some of these people and Nero succeeds iff he has the witness and does not have the murderer in th set. What is the smallest number of days, Nero needs to win?
| 2 | https://mathoverflow.net/users/479618 | 428050 | 173,577 |
https://mathoverflow.net/questions/428045 | 2 | Let ${\cal P}(\omega)$ denote the power-set of $\omega$. We order it by set inclusion $\subseteq$ and say that ${\cal C}\subseteq {\cal P}(\omega)$ is a *chain* if for all $A, B\in {\cal C}$ we have $A\subseteq B$ or $B \subseteq A$.
Interestingly, [chains in ${\cal P}(\omega)$ can have length $2^{\aleph\_0}$](https://mathoverflow.net/questions/395002/chains-of-length-2-kappa-in-cal-p-kappa). Zorn's Lemma implies that every chain is contained in a maximal chain with respect to set inclusion.
**Question.** What is the cardinality of the collection of maximal chains in ${\cal P}(\omega)$ having uncountable cardinality?
| https://mathoverflow.net/users/8628 | Maximal uncountable chains in ${\cal P}(\omega)$ | Any maximal chain in $\mathcal{P}(\omega)$ is determined by (that is, is the unique maximal chain containing) a *countable* subchain; this is basically just a restatement of the fact that $\omega\_1$ doesn't embed into $\mathcal{P}(\omega)$. This gives an upper bound of $2^{\aleph\_0}$ for the number of distinct maximal chains.
On the other hand, it's easy to show that there are at least $2^{\aleph\_0}$ distinct size-continuum maximal chains in $\mathcal{P}(\omega)$ - each bijection $\mathbb{N}\rightarrow\mathbb{Q}$ gives rise to such a chain via Dedekind cuts, and different bijections give rise to incompatible chains in this way - so this bound is sharp.
| 7 | https://mathoverflow.net/users/8133 | 428051 | 173,578 |
https://mathoverflow.net/questions/427972 | 4 | $\newcommand\norm[1]{\lVert#1\rVert}$For any $p \in [1,2]$, $r \ge 0$, and integer $d \ge 1$, define a mixed-norm $\eta:\mathbb R^d \to \mathbb R$ by $\eta(x) := \norm x\_2 + r\norm x\_p$, for any $x \in \mathbb R^d$.
Do there exist scalars $a = a(d,p,r) \ge 0$ and $q=q(p,r) \in [1,\infty]$ such that $c\_1 \le \dfrac{\eta(x)}{a\norm x\_{q}} \le c\_2$ for every nonzero $x \in \mathbb R^d$, where $c\_1$ and $c\_2$ are absolute constants?
I'm particularly interested in the case $p=1$.
| https://mathoverflow.net/users/78539 | Is a mixture of $\ell_p$-norms $\eta(x):=\lVert x\rVert_2 + r\lVert x\rVert_p$ always dimensionlessly equivalent to some $\ell_q$-norm? | This is impossible unless the following holds:
>
> $q=2$ and either $p=2$ or $r=0$.
>
>
>
The above exception is due to the fact that it implies $\frac{1}{1+r}\eta(\cdot)=\|\cdot\|\_q=\|\cdot\|\_2$, for which the desired inequality trivially exists.
We argue ad absurdum; hence, suppose the desired inequality exists. If $r=0$, then $p$ is redundant in your definition of $\eta$; hence, in that case, we appropriate $p$ for the $2$ in the $2$-norm $\|\cdot\|\_2$. A *fortiori*, assume in the proof below that $p=2$ when $r=0$. (Note that this assumption implies that the exception above simplifies to $p=q=2$).
First, observe that the constant $a(d,p,r)$ and the absolute constants $c\_1,c\_2$ are essentially positive constants; hence, necessarily we must have that $a\_1:=\liminf\_{d\to\infty}\frac{1}{a(d,p,r)}>0$ and $a\_2:=\limsup\_{d\to\infty}\frac{1}{a(d,p,r)}<\infty$ (otherwise, either $c\_1$ will be forced to be $0$ or $c\_2$ will be forced to be $\infty$ in your desired inequality because $x\in\mathbb{R}^d$ implies $x\in\mathbb{R}^{d’}$ for all $d’>d$ via the canonical imbedding of $\mathbb{R}^d$ in $\mathbb{R}^{d’}$).
It follows from the above that for all non-zero $x\in\mathbb{R}^d$ and all $d\ge1$, we must have
\begin{array}
\label{(1)}
&a\_1c\_1\le\frac{\eta(x)}{\|x\|\_q}\le a\_2c\_2\,.
\end{array}
In other words, the norms $\eta$ and $\|\cdot\|\_q$ are equivalent on the real vector space $c\_{00}$ of finitely supported sequences (i.e. real sequences with only finitely many nonzero terms). However, this is impossible because of the following well-known (and in any case easily proven) result:
>
> $c\_{00}$ is dense in the Banach space $(X\_q,\|\cdot\|\_q)$, where $X\_q=\ell^q$ for any $q\in[1,\infty)$ and $X\_q=c\_0$ for $q=\infty$.
>
>
>
Recall that whenever $1\le q\le q’\le\infty$, then $X\_q\subseteq X\_{q’}$; thus, because $p\le2$ and $\eta(\cdot)=\|\cdot\|\_2+r\|\cdot\|\_p$, it follows that $(\ell^p,\eta)$ is a well-defined Banach space as the completion of $(c\_{00},\eta)$—note that this is true when $r=0$ since we assume $p=2$ in that case.
Now, let $x= (k^{-1/p})\_{k\ge1}$ and $y=(k^{-1/q})\_{k\ge1}$, and consider their truncated sequences in $c\_{00}$,that is
\begin{align}
&x\_n=(1,2^{-1/p},\ldots,n^{-1/p},0,0,0,\ldots)\,,\\
&y\_n= (1,2^{-1/q},\ldots,n^{-1/q},0,0,0,\ldots)\,.
\end{align}
Thanks to the divergence of the harmonic series, we know that $\|x\_n\|\_p\to\infty$ and $\|y\_n\|\_q\to\infty$; however, observe that $p<q$ implies $x\_n\to x$ in $(X\_q,\|\cdot\|)$ and $p>q$ implies $y\_n\to y$ in $(\ell^p,\eta)$, either of which contradicts Inequality (1). We must therefore necessarily have that $p=q$. However, this implies that $q=2$, because otherwise $p=q<2$ and this time considering $z=(k^{-\frac{2}{2+q}})\_{k\ge1}$ and its truncated sequences in $c\_{00}$,
$$z\_n=(1,2^{-\frac{2}{2+q}},\ldots,n^ {-\frac{2}{2+q}},0,0,0,\ldots)\,,$$
then Inequality (1) implies that
$$(a\_1c\_1-r)\|z\_n\|\_q\le\|z\_n\|\_2\le(a\_2c\_2-r)\|z\_n\|\_q\,,$$
which leads to a contradiction as $n\to\infty$ (because $\|z\_n\|\_2\to\|z\|\_2\ne0$ whereas $\|z\_n\|\_q$ is unbounded).
Q.E.D
| 2 | https://mathoverflow.net/users/166628 | 428054 | 173,579 |
https://mathoverflow.net/questions/427950 | 1 | I am trying to use the coarea formula to get estimates on the measure of an epsilon-neighbourhood of a set. Specificly, given a compact 'nice' set $A\subseteq \mathbb{R^d}$, possibly with more than one connected components which are not convex, I'm hoping to get an upper estimate on $$\lambda(A^\epsilon)-\lambda(A),$$
where $\lambda$ is the Lebesgue measure on $\mathbb{R}^d$ and $A^\epsilon= \cup\_{a\in A} B\_\epsilon(a)$. The notion of niceness is something that should be not just convex. I'm pretty sure it should be possible to write
$$ \lambda(A^\epsilon)-\lambda(A)= \int\_0^\epsilon \nu(A^{(t)})dt $$
by a Fubini like argument, where $A^{(t)}=\{x: d(x,A)=t \}$ and $\nu$ is some measure. I found [this thread](https://mathoverflow.net/questions/291561/relative-volume-increase-of-delta-fattening-of-a-compact-set) and related threads, which seem to rely on the Minkowski-Steiner formula. Is there perhaps some results of this nature using the co-formula area for non convex sets?
| https://mathoverflow.net/users/143153 | Coarea formula for measure of epsilon neighbourhood | Your formula is true (up to appropriate constants) if you take $\nu$ to be the Minkowski content, assuming that the volume of the tubular neighborhood is locally Lipschitz as a function of the tube radius. This is because the Minkowski content is defined as the derivative of the volume of the tubular neighborhood.
One condition ensuring that the tube volume is locally Lipschitz is that the distance function to $A$ has Clarke gradient bounded below in the $\epsilon$-neighborhood of $A$. It also ensures that Minkowski contents coincide with the Hausdorff measures of the boundaries.
Finally, although that's not the question, it is possible to bound the volume of the tubular neighborhood without any regularity conditions, using the $\epsilon$-covering numbers of $A$.
| 2 | https://mathoverflow.net/users/112954 | 428061 | 173,582 |
https://mathoverflow.net/questions/428029 | 2 | I asked this question on [Mathematics Stackexchange](https://math.stackexchange.com/questions/4505957/characterization-of-extendible-distributions), but got no answer.
I found the following [question](http://www.cmls.polytechnique.fr/perso/golse/MAT431-10/pc10.pdf) which characterize the extension of a distribution in $\mathbb{R}$:
>
> Let $f \in L\_{\text{loc}}^{1}(\mathbb{R}\_{>0})$ such that $f \geq 0$ a.e. Show that $f$ extends to a distribution $F$ in $\mathbb{R}$ if and only if there exists $k \geq 0$ $$\int\_{\varepsilon}^{1}f(x)dx=O(\varepsilon^{-k}),$$
> for $\varepsilon \rightarrow 0^{+}$.
>
>
>
I would like to know if someone could give me some hint to solve this problem. The only thing I could think was evaluate the distribution $T\_f$ generated by $f$ in functions like $\varphi\_\varepsilon(x)$ satisfying $\operatorname{supp}(\varphi\_\varepsilon) \subset (\varepsilon,1)$, but without success.
Another question I have is:
>
> Is there any generalization of this result to higher dimensions?
>
>
>
| https://mathoverflow.net/users/115618 | Characterization of extendible distributions | $\newcommand{\R}{\mathbb R}$Without loss of generality, $k$ is a positive integer.
First, the "if" part. For $t\in(0,1]$, let
\begin{equation\*}
(Jf)(t):=\int\_t^1 dx\,f(x).
\end{equation\*}
Here it is assumed that
\begin{equation\*}
(Jf)(t)=O(t^{-k})
\end{equation\*}
for $t\in(0,1]$.
As suggested in the comment by paul garrett, this implies
\begin{equation\*}
h(t):=(J^kf)(t)=O(\ln\tfrac1t)
\end{equation\*}
for $t\in(0,1]$. So, $h$ extends to a distribution $H$ (by the formula $H\phi:=\int\_{(0,1]} h\phi$ for smooth $\phi$ with compact support $S\_\phi$) and hence $f$ extends to the distribution $(-1)^k H^{(k)}$, because $f=(-1)^k h^{(k)}$ on $(0,1]$.
For the "only if" part, we can use the characterization of distributions as generalized multiple derivatives of continuous functions.
Suppose that $f$ extends to a distribution $F$. Then, according to [Theorem 6.26](https://rads.stackoverflow.com/amzn/click/com/0070542368), there exist a continuous function $g$ and a nonnegative integer $k$ such that for all smooth functions $\phi$ with support $S\_\phi\subseteq(0,2]$
\begin{equation\*}
F\phi=(-1)^k\int\_{\R} g\phi^{(k)}
\end{equation\*}
and hence
\begin{equation\*}
\int\_{(0,\infty)}f\phi=(-1)^k\int\_{\R} g\phi^{(k)}. \tag{1}\label{1}
\end{equation\*}
Let $\Phi$ be any nonnegative smooth function such that
\begin{equation\*}
\text{$\Phi=1$ on $[1,\infty)$ and $\Phi=0$ on $(-\infty,1/2]$} \tag{2}\label{2}
\end{equation\*}
and let $\psi$ be any nonnegative smooth function with $S\_\psi\subset[-1,2]$ such that
\begin{equation\*}
\text{$\psi=1$ on $[0,1]$.}
\end{equation\*}
For $t\in(0,1]$ and real $x$, let
\begin{equation\*}
\phi\_t(x):=\Phi(x/t)\psi(x),
\end{equation\*}
so that $\phi\_t$ is a nonnegative smooth function with $S\_{\phi\_t}\subseteq[t/2,2]\subset(0,2]$ such that $\phi\_t=1$ on $[t,1]$.
Then for $t\in(0,1]$ we have $\phi\_t^{(k)}=O(t^{-k})$ on $\R$ (see Detail at the end of this answer) and hence,
by the condition $f\ge0$ a.e., \eqref{1}, and the continuity of $g$
\begin{equation\*}
0\le\int\_t^1 f\le\int\_{(0,\infty)}f\phi\_t=(-1)^k\int\_{\R} g\phi\_t^{(k)}
=(-1)^k\int\_0^2 g\phi\_t^{(k)}=O(t^{-k}),
\end{equation\*}
as desired. $\quad\Box$
**Detail:** Because $\Phi$ is smooth and in view of \eqref{2}, we see that $\Phi$ and all its derivatives, of any order, are bounded. Because $\psi$ is smooth and has a compact support, we see that $\psi$ and all its derivatives, of any order, are bounded. So, by the Leibniz and chain rules of differentiation,
\begin{equation\*}
\phi\_t^{(k)}(x)=\sum\_{j=0}^k\binom kj \Phi^{(j)}(x/t)t^{-j}\psi^{(k-j)}(x) =O(t^{-k})
\end{equation\*}
for $t\in(0,1]$.
| 2 | https://mathoverflow.net/users/36721 | 428065 | 173,583 |
https://mathoverflow.net/questions/427918 | 13 | It's well known that there are a shocking number of identities for the usual Jacobi theta function $$ \theta\_3(x) = \sum\_{n=-\infty}^{\infty} x^{n^2}. $$
So I wanted to turn my attention to slowly decreasing exponents. If I make my $n$ decay too slow such as $$f(x) = \sum\_{n=1}^{\infty} x^{\log n}, $$
we basically have the Riemann zeta function and end up in well explored territory via the relation
$$f(x) = \sum\_{n=1}^{\infty} x^{\log n} = \zeta( - \log x). $$
So then I started to consider some slowly growing exponents that don't grow TOO slowly and the obvious first candidate is
$$ G(x) = \sum\_{n=0}^{\infty} x^{\sqrt{n}}. $$
Has anyone looked into this object/similar algebraic exponents? Is it known to obey any interesting identities? It seems like a natural object to consider even without an application/motivation.
| https://mathoverflow.net/users/46536 | Is anything known about the series $\sum_{n=0}^{\infty} x^{\sqrt{n}} $? | The series
$$\tag{0}\label{0}
G\_a(x)=\sum\_{n=0}^\infty x^{\sqrt[a]{n}}
$$
can be calculated for $0<x<1$ and $a\geq1$ using Cauchy's residue theorem. We use the fact that $\pi \cot(\pi n)$ has residues 1 at $n\in \mathbb Z$, and deform the integration contour to the imaginary axis in the usual way. After the substitution $n\to (-\nu\,/\log x)^a$ we get
\begin{align}\tag{1}\label{1}
G\_a(x)
%&=
%\frac{\Gamma(1+a)}{(-\log x)^a} + \frac 1 2
%+ \frac{2a}{(-\log x)^a} \times\\
%&\quad\int\_0^\infty \nu^{a-1}
%\exp\left[-\nu \cos\left(\frac{\pi}{2a}\right)\right]
%\sin\left[ \nu \sin\left(\frac{\pi}{2a}\right)\right]
%\left(\coth\left[\pi \left(\frac{\nu}{-\log %x}\right)^a\right]-1\right)\mathrm{d}\nu \\
&=\frac{\Gamma(1+a)}{(-\log x)^a} + \frac 1 2
+ \frac{2a}{(-\log x)^a} \int\_0^\infty \frac{\nu^{a-1}
\exp\left[-\nu \cos\left(\frac{\pi}{2a}\right)\right]
\sin\left[ \nu \sin\left(\frac{\pi}{2a}\right)\right]}
{\exp\left[2\pi \left(\frac{\nu}{-\log x}\right)^a\right]-1} \mathrm{d}\nu,
\end{align}
or, after back-substitution,
$$\tag{2}\label{2}
G\_a(x)= \frac{\Gamma(1+a)}{(-\log x)^a} + \frac 1 2
+ \int\_0^{\mathrm i\infty} \frac{
x^{\sqrt[a]{n}} - x^{\sqrt[a]{-n}}}
{\mathrm e^{-2\pi \mathrm i n}-1} \mathrm{d} n.
$$
The $1/2$ can be seen as the first Euler–Maclaurin correction. The integral \eqref{2} converges exponentially fast. This result is identically obtained using the [Abel–Plana formula](https://en.wikipedia.org/wiki/Abel%E2%80%93Plana_formula).
**Edit 08.08.22,23:20 CEST**
As pointed out by @Joseph, we better consider the series
$$\tag{3}\label{3}
H\_a(z) = G\_a(\mathrm e^z)-1 = \sum\_{n=1}^\infty \mathrm e^{\sqrt[a]{n} z},
$$
with $\mathrm{Re}(z) < 0$ and again $a\geq1$.
An even simpler evaluation is the following: expanding the exponential into a Taylor series,
$$\tag{4}\label{4}
\mathrm e^{\sqrt[a]{n} z} = 1 + \sum\_{k=1}^\infty n^{k/a} \frac{z^k}{k!}
$$
and interchanging sums, we directly get a representation involving the Riemann zeta function,
$$\tag{5}\label{5}
H\_a(z) = \frac{\Gamma(1+a)}{(-z)^{a}} + \sum\_{k=0}^\infty \zeta\left(-\frac{k}{a}\right)\frac{z^k}{k!},
$$
where the first term is regularized as in \eqref{2}.
The term $-1/2$ is identified with the zeta-regularized sum for $k=0$ and is moved back into the sum, which therefore starts at $k=0$.
A Mathematica function that checks \eqref{3} against \eqref{5} reads
```
r[a_,z_] := Gamma[1+a]/(-z)^a + Sum[Zeta[-k/a] z^k/k!, {k,0,100}]
- NSum[Exp[n^(1/a) z], {n,1,∞}, NSumTerms->200000,
WorkingPrecision->50, Method->"EulerMaclaurin"]
```
I guess that for the original case $a=2$, Eq. \eqref{5} can be related to square-free numbers from @Joseph's answer through their generating function $\zeta(s)/\zeta(2s)$, cf. <https://en.wikipedia.org/wiki/Square-free_integer>.
| 10 | https://mathoverflow.net/users/90413 | 428077 | 173,587 |
https://mathoverflow.net/questions/428026 | 17 | While playing around with the MO question [Determinant with factorials is not 0?](https://mathoverflow.net/questions/427934/determinant-with-factorials-is-not-0) about a determinant of the Hankel matrix of entries $(i+j-2)!$, having the value $\prod\_{k=0}^{n-1}k!^2$, I stumbled on the following.
A permutation $\pi\in\mathfrak{S}\_n$ is called a [*derangement*](https://en.wikipedia.org/wiki/Derangement) if it has no fixed points. Let $d\_n$ be the number of derangement permutations in $\mathfrak{S}\_n$ which may be presented by the formula
$$d\_n=n!\sum\_{k=0}^n\frac{(-1)^k}{k!}.$$
>
> **QUESTION.** It appears that we have $\det((i+j-2)!)=\det(d\_{i+j-2})$. Why? Why not?
>
>
>
| https://mathoverflow.net/users/66131 | A coincidence or a fact: determinants of two matrices | Orangeskid's guess is correct: a more general fact holds that the binomial transform preserves Hankel determinants.
For a matrix $(a\_{ij})$ (it is convenient to enumerate rows and columns from 0,not from 1) denote $$b\_{ij}=\sum\_{k, s}{i\choose k}{j\choose s}a\_{ks}.$$
This matrix transform corresponds to a left and right multiplicaton by unitriangular matrices, thus it preserves determinant. Now if $a\_{ij}=f(i+j)$ is a Hankel matrix, then $$b\_{ij}=\sum\_t f(t)\sum\_{k+s=t} {i\choose k}{j\choose s}=\sum\_tf(t){i+j\choose t}$$
is a Hankel matrix corresponding to the binomial transform of $f$.
It remains to recall that $n! =\sum\_k {n\choose k} d\_k$ (combinatorially ${n\choose k} d\_k$ counts the number of permutations of $\{1,\ldots, n\}$ with exactly $n-k$ fixed points, thus this formula), that means that the sequence of factorials is the binomial transform of the sequence of dearrangements.
| 21 | https://mathoverflow.net/users/4312 | 428093 | 173,592 |
https://mathoverflow.net/questions/428072 | 6 | Let $G$ be a group of order $2^n$. Does $G$ have a normal abelian subgroup of order at least $2^{n/2}$?
(This is true, via computations in GAP, for $n \le 8$.
The question is similar to one posed here: <https://math.stackexchange.com/questions/44275/abelian-subgroups-of-p-groups/44283#44283>
However, that question, and answer, involves groups of order $p^n$, for odd primes $p$, and I need $p$ to be even!)
| https://mathoverflow.net/users/11124 | Is the largest normal abelian subgroup of a finite 2-group $P$ of order at least the square root of the order of $P$? | In
*Alperin, J. L.*, [**Large abelian subgroups of p-groups**](http://dx.doi.org/10.2307/1994193), Trans. Am. Math. Soc. 117, 10-20 (1965). [ZBL0132.27204](https://zbmath.org/?q=an:0132.27204),
the second part of Theorem 1 gives a group of order $2^{50}$ with no abelian subgroups of order greater than $2^{24}$.
| 14 | https://mathoverflow.net/users/22989 | 428096 | 173,593 |
https://mathoverflow.net/questions/427969 | 6 | By a egg-box **diagram** I will simply mean a (possibly infinite) rectangular array of holes, with some of the holes containing an egg (denoted by a filled-in circle) and the rest of the holes are empty (denoted with an empty circle). For instance
$$
\begin{array}{|c|c|c|c|}
\hline
\bullet & \circ & \circ &\circ \\ \hline
\bullet & \circ & \bullet &\bullet \\\hline
\circ & \bullet & \bullet & \bullet\\\hline
\end{array}
$$
is a $3\times 4$ egg-box diagram.
When I say that two eggs are **connected** in a diagram, I mean that we can pass from one egg to the other egg via a series of rook **moves** (i.e., moving within a given column or row) without ever stopping on an empty hole. All the eggs in the previous diagram are connected; it takes at most 4 moves to get from one egg to another. For added simplicity, we consider "staying put" as a row (and as a column) move. (Consequently, eggs connected in $n$ moves are also connected in $n+1$ moves.) In the diagram above, note that the egg in the top left corner is connected to the egg in the bottom right corner in three moves, by a column-row-column movement, but not a row-column-row movement.
By a **permutation** I will just mean a permutation of the rows and columns. For instance, if we consider the permutation where we switch the two middle columns and cyclically shift the rows downward, and we apply that permutation to the diagram above, then
$$
\begin{array}{|c|c|c|c|}
\hline
\circ & \bullet & \bullet & \bullet\\\hline
\bullet & \circ & \circ &\circ \\ \hline
\bullet & \bullet & \circ&\bullet \\\hline
\end{array}
$$
is the diagram we obtain.
By an **automorphism of a diagram** I will mean a permutation that results in the same diagram. For example, the only automorphisms of the first diagram are the identity permutation and the permutation that only switches the last two columns.
The types of diagrams I care about satisfy an extremely strong **switching property**: If there are two eggs that lie in the same row (or column), then there is an automorphism that switches the eggs. (Note: Such an automorphism may need to permute many other rows and columns too.)
**Question**: Is there a diagram satisfying the switching property, where all the eggs are connected in three moves, but (at least) two eggs are not connected in three moves if we start with a column movement?
The motivation for my questions comes from the fact that I can translate some of my work in ring theory into this combinatorial situation. I'm wondering how much of what we can prove algebraically is really just a consequence of combinatorial ideas. In particular, if we change "three" to "two" in the previous question, the answer is no, and this gives a much simpler proof for something I had previously proven using some deep algebra.
| https://mathoverflow.net/users/3199 | Automorphisms of special egg-box diagrams | It turns out (surprisingly) that the answer to my question is yes, and there are even finite examples. I came up with the following diagram
$$
\begin{array}{|c|c|c|c|c|c|}
\hline
\circ & \circ & \bullet & \circ & \bullet & \bullet\\\hline
\circ & \bullet & \circ & \bullet & \circ & \bullet\\\hline
\bullet & \circ & \circ & \bullet & \bullet & \circ\\\hline
\bullet & \bullet & \bullet & \circ & \circ & \circ\\\hline
\end{array}
$$
where for any pair of rows, there is exactly one column with eggs in just those two rows.
| 1 | https://mathoverflow.net/users/3199 | 428112 | 173,597 |
https://mathoverflow.net/questions/428081 | 6 | Let $L$ be a non-splittable link in $S^3$. Non-splittable means that there is no smooth embedding $s:S^2\to S^3\setminus L$ which splits $L$, i. e. such that both connected components of $S^3\setminus\operatorname{im}s$ intersect $L$.
The [Hopf link](https://en.wikipedia.org/wiki/Hopf_link) is an example.
>
> Proof. Let $l\_1, l\_2:S^1 \to S^3$ be two smooth embeddings with images $L\_1$ and $L\_2$, forming the Hopf link $L=L\_1\sqcup L\_2$, $s:S^2\to S^3\setminus L$ a smooth embedding which splits $L$, $B$ a connected component of $S^3\setminus \operatorname{im}s$. Suppose that $L\_1\subset B$ and $L\_2\cap B=\varnothing$. By Alexander's theorem, $B$ is homeomorphic to the 3-ball, so $l\_1$ is contractible as a map to $B=B\setminus L\_2$, hence contractible as a map to $S^3\setminus L\_2$. This is a contradiction (for example, one can construct a retraction of $S^3\setminus L\_2$ onto $L\_1$, which is homeomorphic to $S^1$, so non-contractible).
>
>
>
Can there still be a *continuous* embedding $s:S^2\to S^3\setminus L$ which splits $L$? The analogue of Alexander's theorem is false for continuous embeddings, so the above proof doesn't work.
| https://mathoverflow.net/users/485324 | Can a non-splittable link be split by a wild sphere? | Let $L\_1 \cup L\_2$ be a nonsplittable link in $S^3$, and let $\phi : S^2 \to S^3 \setminus (L\_1 \cup L\_2)$ be an embedding of a $2$-sphere. We want to show that the $S^2$ does not separate $L\_1$ from $L\_2$.
Choose a points $p\_1$ and $p\_2$ on $L\_1$ and $L\_2$, so $S^3 \setminus \{ p\_1, p\_2 \} \cong S^2 \times \mathbb{R}$ and $H\_2(S^3 \setminus \{ p\_1, p\_2 \} ) \cong \mathbb{Z}$. The sphere $\phi(S^2)$ will separate $L\_1$ from $L\_2$ if and only if $\phi\_{\ast}[S^2]$ is nontrivial in $H\_2(S^3 \setminus \{ p\_1, p\_2 \} )$.
The OP points me to a [paper of Papakyriakopoulos](https://www.maths.ed.ac.uk/%7Ev1ranick/papers/papa.pdf) which proves that $\pi\_2(S^3 \setminus (L\_1 \cup L\_2))$ is trivial. Therefore, there is a homotopy between $\phi$ and a constant map $S^2 \to S^3 \setminus (L\_1 \cup L\_2)$, with the homotopy staying within $S^3 \setminus (L\_1 \cup L\_2)$. But then, in particular, the homotopy stays within $S^3 \setminus \{ p\_1, p\_2 \}$, so the two homotopic maps induce that same map $H\_2(S^2) \to H\_2(S^3 \setminus \{ p\_1, p\_2 \} )$. So $\phi\_{\ast}$ is zero on $H\_2$, and thus $\phi(S^2)$ does not separate $p\_1$ from $p\_2$.
| 2 | https://mathoverflow.net/users/297 | 428114 | 173,598 |
https://mathoverflow.net/questions/428018 | 4 | This is a second part of my [previous question](https://mathoverflow.net/questions/428017/what-is-a-large-field-problem), which I decided to split into two parts not to mix up different topics at one giant question.
Again, according to V. Rivasseau (section 1.5 of [Constructive Renormalization Theory](https://arxiv.org/pdf/math-ph/9902023.pdf)), the solution for the large field problem is to consider a single scale analysis of the RG. The text explains that the main tool of constructive RG is to do a cluster expansion and then a Mayer expansion. The cluster expansion is related to the partition function of the theory and the Mayer expansion is related to its logarithm, which is the pressure.
My first question concerns the terminology: why is it called *constructive*? And if one *expands* the partition function and the pressure in (what I understand is the) power series of the coupling parameter $\lambda$, why isn't it considered a *perturbation* theory?
My second question concerns the objective of the theory. If I understood it correctly, the point is to expand the pressure in power series of $\lambda$ and prove it is uniformly bounded, so it has nonzero radius of convergence when the cutoffs are removed. What information can we get from that? Does it imply that perturbation theory converges? And if so, why is that? Isn't the whole idea to obtain the Schwinger (or $n$ point functions)? Can one obtain such functions from the pressure?
| https://mathoverflow.net/users/152094 | Cluster expansion, Mayer expansion and perturbative renormalization group | I think there is a misunderstanding here on what is the expansion parameter.
Perturbative renormalization expands in a power series of the interaction strength (the coupling parameter $\lambda$); this expansion typically has zero radius of convergence.
Constructive renormalization, instead, does not expand in powers of $\lambda$, but in the number of particles that interact. The interaction strength is not expanded, it is retained to all orders in $\lambda$.
The objective of the theory is to show that this expansion (known as a cluster or Mayer expansion) has a finite radius of convergence. This is typically the case if the interaction decays sufficiently rapidly at large distances.
And yes, convergence of the pressure implies convergence of correlators, see for example <https://doi.org/10.1063/1.523040>
| 2 | https://mathoverflow.net/users/11260 | 428127 | 173,602 |
https://mathoverflow.net/questions/428134 | 5 | Here is what I observed :
Let $8p+1 = 256a^2+(2b-1)^2$ with $a$ and $b$ be a positive integers, $p$ and $8p+1$ both prime numbers.
Then $8p+1$ divides $(2^p+1)/3$ only if you can write $8p+1$ as $256a^2+(2b-1)^2$.
For example :
* $1049 = 256 \cdot 2^2+(2 \cdot 3 - 1)^2$ and $1049 = 131 \cdot 8+1$ so $1049$ divides $(2^{131}+1)/3$
* $12569 = 256 \cdot 7^2+(2 \cdot 3 - 1)^2$ and $12569 = 1571 \cdot 8+1$ so $12569$ divides $(2^{1571}+1)/3$
* $2120057 = 256 \cdot 91^2+(2 \cdot 6 - 1)^2$ and $2120057 = 265007 \cdot 8+1$ so $2120057$ divides $(2^{265007}+1)/3$
* $137 = (8 \cdot 17 + 1)$ but you can't write $137$ as $256a^2+(2b-1)^2$ so $137$ does not divide $(2^{17}+1)/3$
For the moment, I didn't find a counterexample with this condition.
I need help for proving it but I don't know how to start.
I thought about Wagstaff numbers and Sophie Germain primes that say if $p \equiv 1 \pmod{4}$ and $2p+1$ is prime then $2p+1$ divides $(2^p+1)/3$ but for $8p+1$ it doesn't work.
This question looks like the one I asked on [Math stack exchange](https://math.stackexchange.com/q/4507944/932679) (for $2^p-1$ and the divisibility by $8p+1$) but I didn't get an answer for the question.
If you found a counterexample please tell me.
| https://mathoverflow.net/users/264367 | Condition for $8p+1$ divides $(2^p+1)/3$? | This follows easily from the octic reciprocity law for $2$. More precisely, it states that if $q$ is a prime of the form $q=8n+1$ and $n$ is odd, then $q=a^2+256b^2$ for some integers $a,b$ (note that $a$ is necessarily odd) if and only if $2$ is a biquadratic residue but not an octic residue, i.e. there is $x \in \mathbb F\_q$ with $x^4=2$ but no $y\in \mathbb F\_q$ with $y^8=2$.
This condition is equivalent to $2^{2n}\equiv 1\pmod q$ and $2^{n}\not\equiv 1\pmod q$, which is the same as $q\mid \frac{2^n+1}{3}$ since $2^{2n}-1=3(2^n-1)\frac{2^n+1}{3}$. For instance, your observation works for non-prime (but odd) values of $p$.
For the reference on octic reciprocity, see, for example, the first paragraph of [Whiteman AL. The Sixteenth Power Residue Character of 2](https://www.cambridge.org/core/journals/canadian-journal-of-mathematics/article/sixteenth-power-residue-character-of-2/E9DF1AA4086D7764A9A21C9FFAED5A36)
| 9 | https://mathoverflow.net/users/101078 | 428139 | 173,605 |
https://mathoverflow.net/questions/428084 | 3 | This question might be below the level of MO, so apologies in advance. I posted the [same question](https://math.stackexchange.com/questions/4503962/does-the-set-of-ideals-whose-jacobson-radical-nilradicals-coincide-form-a-la) in MS about a week ago without an answer so far.
Let $R$ be a unital commutative ring and $L(R)$ denote the [lattice of ideals](https://planetmath.org/latticeofideals) of $R$. Let $\mathcal{S}\subseteq L(R)$ be the set of all ideals whose Jacobson radicals and nilradicals coincide.
**Q1:** For which classes of rings $\mathcal{S}$ is a sublattice (respectively, $\sigma$-complete or complete sublattice) of $L(R)$?
**Q2:** For which classes of rings $\mathcal{S}$ is a lattice? Specifically, we could define $I\vee\_S J$ as the *smallest* ideal in $\mathcal{S}$ containing $I+J$ for a given pair $I,J\in\mathcal{S}$. For which classes of rings $I\vee\_S J$ exists for any $I,J\in\mathcal{S}$?
For example, $\mathcal{S}=L(R)$ iff $R$ is a [Hilbert-Jacobson ring](https://en.wikipedia.org/wiki/Jacobson_ring), so $\mathcal{S}$ is trivially a complete lattice.
---
**edit (Aug 9, 2022):** Professor Keith Kearnes' reply below indeed answers the question in the title by providing a counterexample. I appreciate his detailed response. I'm carrying the two questions in the body of the post to a subsequent [separate question](https://mathoverflow.net/questions/428175/set-of-ideals-whose-jacobson-radical-nilradicals-coincide-ii) that look for an affirmative answer.
| https://mathoverflow.net/users/164350 | Does the set of ideals, whose Jacobson radical & nilradicals coincide, form a sublattice? | Let's call an ideal $I\lhd R$ *Jacobson* if $J(I)=\sqrt{I}$.
I will answer the question in the title by constructing, in stages, an example of a unital ring $R$ where the set of Jacobson ideals
is not a sublattice of the ideal lattice of $R$. It will turn out that more is true about this example: its ordered set of Jacobson ideals does not form a lattice at all.
Since the formation of Jacobson radical or nilradical commutes with intersection, it follows that if $A, B\lhd R$ are Jacobson, then $J(A\cap B)=J(A)\cap J(B)=\sqrt{A}\cap \sqrt{B}=\sqrt{A\cap B}$, so $A\cap B$ is also Jacobson. This means that the example I want to construct should have Jacobson ideals $A$ and $B$ where $A+B$ is not Jacobson.
**Stage 1.** Let $\mathbb Q$ be the field of rational numbers and let $L$ be the subring of $\mathbb Q$ consisting of fractions $m/n$ with odd denominator.
The key fact here is that $L$ is a local integral domain with field of fractions equal to $\mathbb Q$.
**Stage 2.**
Let $S$ be the subring of
$\mathbb Q^{\omega}$ consisting of those tuples
$\textbf{q}=(q\_0,q\_1,\ldots)\in \mathbb Q^{\omega}$
which are eventually constant and which satisfy the condition that the limit
$q\_{\infty}:=\lim\_{n\to\infty} q\_n$ belongs in the subring $L\leq \mathbb Q$. Saying that $\textbf{q}\in S$
means the same thing as saying that
$\textbf{q}\in \mathbb Q^{\omega}$ and
all but finitely many entries of $\textbf{q}$ are equal to
some fixed $q\_{\infty}\in L$.
The key facts here are that
* $S$ has trivial Jacobson radical. This is because the $n$th
coordinate projection $\pi\_n\colon S\to \mathbb Q\colon \textbf{q}\mapsto q\_n$
is surjective. Since the image of $\pi\_n$ is a field we get that $\ker(\pi\_n)\lhd S$ is a maximal ideal of $S$.
This ideal contains exactly those
elements of $S$ that vanish in the $n$th coordinate.
If one intersects the maximal ideals of this type, one is left with
$\{\textbf{0}\}=\{(0,0,\ldots)\}$ only. Thus, if one intersects all maximal ideals of $S$,
one must get the zero ideal of $S$.
* $S$ has trivial nilradical. This is because the nilradical is contained
in the Jacobson radical.
* $S$ has a retraction onto a
subring $L'$ that is isomorphic to the $L$ from Stage 1.
Here, the retraction is the map $(q\_0,q\_1,q\_2, \ldots)\mapsto (q\_{\infty},q\_{\infty},q\_{\infty},\ldots)$ where $q\_{\infty}$ is $\lim\_{n\to\infty} q\_n$.
**Stage 3.** The ring that I have been aiming for is $R:=\{(\textbf{u},\textbf{v})\in S\times S\;|\;u\_{\infty}=v\_{\infty}\}$. To make this clear, let me repeat the definition using more words than symbols: $R$ is the subring of $S\times S$ consisting of those pairs of tuples that have the same limit.
The key facts here are that
* $R\leq S\times S$ is a subdirect embedding. That is, $R$ projects onto $S$ in each of the two factors.
This means that the composition of the embedding $R\leq S\times S$
with either projection
$\pi\_i\colon S\times S\to S$, $i=1,2$, is surjective.
Let the kernels of these two projections be $A, B\lhd R$, so that
$A=\ker(\pi\_1)=\{\{(\textbf{0},\textbf{v})\in S\times S\;|\;v\_{\infty}=0\}$
and
$B=\ker(\pi\_2)=\{\{(\textbf{u},\textbf{0})\in S\times S\;|\;u\_{\infty}=0\}$.
Notice that any pair of tuples in $A$ or $B$ is zero in all but finitely
many coordinates.
* $A$ and $B$ are Jacobson ideals of $R$.
This is because $R/A\cong S$, $R/B\cong S$,
and $S$ has trivial Jacobson radical.
* Let $L''$ be the subring of $R$ consisting of all pairs
$(\textbf{u},\textbf{v})=((\ell,\ell,\ell,\ldots),(\ell,\ell,\ell,\ldots))$ for some $\ell\in L$.
I will refer to $L''$ as the 'diagonal copy' of $L$ inside $R$.
It is clear that $L''\cong L$.
* Let $C=A+B$. This is the ideal of $R$ consisting of exactly those pairs of tuples in $R$ that are zero almost everywhere.
I claim that $C$ is the kernel of a retraction of $R$
onto $L''$. To see this, first consider the intersection $C\cap L''$:
since pairs of tuples in $C$ are zero almost everywhere and $L''$
is the diagonal copy of $L$ in $R$, the only common pair
must be the diagonal pair with all coordinates zero. Thus $C\cap L'' = \{(\textbf{0},\textbf{0})\}$.
Now I want to show that $R=C+L''$. Choose any
$x=(\textbf{u},\textbf{v})\in R$
and define $\ell:=u\_{\infty}=v\_{\infty}$. The pair
$d:=((\ell,\ell,\ldots),(\ell,\ell,\ldots))$ belongs to $L''$
while the difference $x-d\in R$ is zero almost everywhere,
hence belongs to $C$. Thus, $x=(x-d)+d\in C+L''$. This completes the proof that
$C$ is the kernel of a retraction of $R$ onto $L''\cong L$.
(That is, $C$ is an ideal of $R$, $L''$ is a subring of $R$, and the conditions $C\cap L'' = \{(\textbf{0},\textbf{0})\}, C+L''=R$ guarantee that $L''$ is a complete irredundant set of coset representatives for $C$ in $R$. The map that sends an element of $R$ to its $C$-coset representative in $L''$ is the retraction.)
We explained above why $A, B\lhd R$ are Jacobson ideals.
Now I argue that $C:=A+B$ is not Jacobson,
i.e., that $J(C)\neq \sqrt{C}$. For this we work in
$R/C\cong L''\cong L$. It suffices to show that
the zero ideal of this quotient is not Jacobson.
Equivalently, we must show that
the $J(0)\neq \sqrt{0}$ in $L$.
But, $L$ is a local domain that is not a field,
so $J(0)\neq 0=\sqrt{0}$.
The final observation I want to make is that
$A$ and $B$ do not have a join in the ordered set of Jacobson ideals
of $R$, so the Jacobson ideals of $R$ do not form a lattice.
For this it suffices to note that the ideals of $L$ are
$$
L\supsetneq (2) \supsetneq (4) \supsetneq (8) \supsetneq \cdots \supsetneq (0),
$$
that $J(L)=\sqrt{L}=L$, and $J((2^k))=\sqrt{(2^k)}=(2)$ for $k>0$.
This shows that every ideal of $L$ is Jacobson except the zero ideal.
In particular, since there is no least nonzero ideal of $L$, there is no least Jacobson ideal of $L$. Now $L\cong R/C$ for $C=A+B$. We can translate this information back to $R$ to conclude that every ideal above $C$
is Jacobson except $C$ itself, and there is no least ideal properly above $C$. Writing this in terms of $A$ and $B$, this means that every ideal of $R$ containing both $A$ and $B$ is Jacobson except $A+B$, and there is no least element in the set of ideals properly above $A+B$. Hence there is no join of $A$ and $B$ in the ordered set of Jacobson ideals of $R$.
\\\
| 2 | https://mathoverflow.net/users/75735 | 428144 | 173,608 |
https://mathoverflow.net/questions/427811 | 4 | A metric space $M$ is called *two-point homogeneous* if for any pair of points $(p,q)$ in $M$ any distance preserving map $f\colon\{p,q\}\to M$ can be extended to an isometry $\bar f\colon M\to M$.
The following statement seems to be an easy corollary of [Gleason--Yamabe theorem](https://terrytao.wordpress.com/2011/06/17/hilberts-fifth-problem-and-gleason-metrics/):
>
> Any two-point homogeneous locally compact length-metric space is a Riemannian manifold.
>
>
>
Is there a reference for this statement?
| https://mathoverflow.net/users/1441 | Locally compact + two-point homogeneous => Riemannian | Tits proves in [Tits, J. Sur certaines classes d'espaces homogènes de groupes de Lie, Acad. Roy. Belg. Cl. Sci. Mém. Coll. in 8$^\circ$ 29 (1955), no. 3, 268 pp. MR0076286], page 220, the following.
*If $M$ is a locally compact and connected metric space which is 2-point homogeneous (in your sense) then $M$ is isometric to euclidean space or to a rank 1 Riemannian symmetric space.*
I don't know any newer references. Helgason mentions in his book some related results.
| 5 | https://mathoverflow.net/users/37911 | 428148 | 173,609 |
https://mathoverflow.net/questions/428149 | 0 | Let be given positive integers $m,n,r$, with $r \leq \min(m, n)$, and a finite field of $q$ elements $\mathbb{F}\_q$.
I'm looking for an efficient algorithm to enumerate (i.e., generate one by one) all the $m \times n$ matrices over $\mathbb{F}\_q$ that have rank equal to $r$.
One obvious solution is to enumerate the matrices over $\mathbb{F}\_q$ and to test if each of them has rank equal to $r$, so the complexity of each enumeration step is $O(mn^{\omega-1})$, where $\omega$ is the exponent of matrix multiplication (i.e., multiplying two $n \times n$ matrices has complexity $O(n^{\omega})$).
However, I think that more efficient algorithm should exist. For example, if the matrices are enumerated using some kind of Gray code, then each matrix $M$ and the next matrix $M^\prime$ differ only by one entry. Therefore, it might be possible to compute the rank of $M^\prime$ in a more efficient way, by somehow using the previous computation of the rank of $M$.
Thanks for any help/reference.
| https://mathoverflow.net/users/489345 | Enumerating (i.e. generating one by one) matrices of given rank over a finite field | You may fix an ordering of $\mathbb{F}\_q^n$ (like lexicographic or which you prefer), then start the following: choose the first row, then choose the second etc. Denote by $r(i)$ the dimension of the span of the first chosen $i$ rows; also let $B(i)$ be (some) set of linearly independent rows between the first $i$ rows. If $r(i)<r$, choose the next row arbitrarily and modify $r(i)$, $B(i)$ accordingly.
If $r(i)=r$ for the first time, proceed with linear combinations of the rows from $B(i)$ (they are in the natural bijection with $\mathbb{F}\_q^r$, so you have some order pre-chosen there).
This enumerates all matrices of rank at most $r$. If you need rank exactly $r$, you must also assure that $r(i)+n-i\geqslant r$ on all steps.
| 1 | https://mathoverflow.net/users/4312 | 428150 | 173,610 |
https://mathoverflow.net/questions/428137 | 3 | Let $\Gamma \subset \mathrm{SL}(2,\mathbb{R})$ be a lattice. If $N\_1, N\_2$ are a pair of independent parabolic subgroups contained in $\Gamma$, why must $\Gamma$ contain a hyperbolic element? By parabolic subgroup, I mean "subgroup containing only parabolic elements, other than the identity." This is used in the proof of Theorem 10.1, here:
[https://people.math.harvard.edu/~ctm/papers/home/text/papers/abel/abel.pdf](https://people.math.harvard.edu/%7Ectm/papers/home/text/papers/abel/abel.pdf)
If it helps, $\Gamma$ here is the stabilizer of a 1-form defining a Teichmuller curve.
| https://mathoverflow.net/users/123002 | Parabolic elements and hyperbolic elements in SL(2,R) | This very special case follows from "general principles" (namely a version of the ping-pong lemma) but it is also possible to give a direct proof, as follows.
---
Suppose that $a$ and $b$ are the distinct points at infinity fixed by the two parabolic subgroups $A$ and $B$ (note change of notation). Since $\mathrm{SL}(2, \mathbb{R})$ is three-transitive, we can conjugate $\Gamma$ and so assume that $a = \infty$ and $b = 0$. We deduce that elements of $A$ now have the form
$$
\begin{pmatrix}
1 & r\\
0 & 1
\end{pmatrix}
$$
while elements of $B$ have the form
$$
\begin{pmatrix}
1 & 0\\
s & 1
\end{pmatrix}
$$
Taking an inverse if needed, we obtain such elements of $A$ and $B$ where $r$ and $s$ are non-zero and have the same sign. We now multiply to find
$$
\begin{pmatrix}
1 & r\\
0 & 1
\end{pmatrix}
\begin{pmatrix}
1 & 0\\
s & 1
\end{pmatrix}
=
\begin{pmatrix}
1 + rs & r\\
s & 1
\end{pmatrix}
$$
This has trace $2 + rs > 2$ so is hyperbolic, as desired.
---
Hmm. Now that I write this, I realise that there is a third proof using the classification of fixed points of isometries, and the intermediate value theorem. It helps to draw a picture and think dynamically.
| 4 | https://mathoverflow.net/users/1650 | 428154 | 173,611 |
https://mathoverflow.net/questions/427956 | 2 | Let $T\_{p,q}$ be line joining $(0,0)$ and $(p,q).$ Now let us define the set
$$L= \bigcup\_{p\in[0,1]\cap \mathbb{Q}}T\_{p,1} \bigcup\_{q\in[0,1]\cap \mathbb{Q}}T\_{1,q} $$
and consider $P=[0,1]\times[0,1]\setminus L.$ $P$ should be a set of positive Lebesgue measure.
Question: Does there exist set of positive Lebesgue measure, $A,B \subset \mathbb{R}$ such that $A\times B\subset P?$
Addition after 1st comment:
The above property also not holds for the set $$D=\{(x,y)\in [0,1]\times [0,1]:x-y\notin \mathbb{Q}\}.$$ This is classic example and can be proved using Steinhaus theorem. But a suitable translation and rotation of $D$ contains $A\times B,$ where $A,B$ are sets of positive Lebesgue measure. This leads me to ask the above question since if $P$ doesn't have the above property, then any translation and rotation of $P$ would not also have this property.
(I believe that $P$ wouldn't contain $A\times B,$ where $A,B$ are sets of positive Lebesgue measure but I can't prove it as it doesn't have a nice definition, unlike $D$, to apply Steinhaus theorem.)
| https://mathoverflow.net/users/483450 | Problem regarding set of positive Lebesgue measure in $\mathbb{R}^2$ | There are no such $A,B$. If $A\subset[0,1]$ is a Lebesgue measurable set of positive measure, the set $\mathbb Q\_+ A:=\{qa: q\in\mathbb Q\_+,\, a\in A\}$ has full measure in $\mathbb R\_+$, for it has at least one point of density $1$ and it is invariant by multiplication by positive rationals. Therefore if $B\subset[0,1]$ is a Lebesgue measurable set of positive measure, $\mathbb Q\_+ A\cap B\neq\emptyset$, that is $qa=b$ for some $q\in\mathbb Q\_+, \ a\in A ,\ b\in B$, that is $(a,b)\in (A\times B)\cap L$.
| 2 | https://mathoverflow.net/users/6101 | 428156 | 173,613 |
https://mathoverflow.net/questions/427995 | 2 | Is it true that, in the category $\mathbf{Top}$ of topological spaces and continuous maps, the compactly generated weakly Hausdorff spaces are precisely the spaces arising as filtered colimits of compact Hausdorff spaces?
Note added in the light of @Tyrone's comment. The answer to this question is NO
as can be seen by consider the directed colimit of maps $S^1\to S^1\to S^1\to\dots$ in which the $n$th map is defined $z\mapsto x^{n!}$. The colimit contains a point (namely the image of $1\in S^1$)
whose preimage is dense and so the colimit fails the weaker separation axiom $T\_1$.
In the light of this I should broaden the question: what I am really interested in is finding attractive ways of drawing attention to the weakly Hausdorff property and how it sits in relation to the world of compact Hausdorff spaces.
| https://mathoverflow.net/users/124943 | Possible characterisation of compactly generated weakly Hausdorff spaces | Peter Scholze's [comment](https://mathoverflow.net/questions/427995/possible-characterisation-of-compactly-generated-weakly-hausdorff-spaces#comment1101190_427995)
>
> The correct characterization is that they are the topological spaces that can be written as filtered colimits of compact Hausdorff spaces along injective transition maps. It is pretty clear that all CGWH spaces are of this form (being CG, it is the filtered colimit of the images of maps from CH spaces; and those images are themselves CH by WH); the other direction is e.g. Proposition A.14 in Schwede's "[Global homotopy theory](https://arxiv.org/abs/1802.09382)".
>
>
>
is the correct answer to my question: thank you for this!
| 0 | https://mathoverflow.net/users/124943 | 428161 | 173,614 |
https://mathoverflow.net/questions/428172 | 3 | Montesinos proved that the double branched cover $\Sigma(T)$ of an algebraic tangle $T$ in a $3$-ball is a graph manifold. I wonder if the converse true: Is $T$ algebraic if $\Sigma(T)$ is a graph manifold?
If unknown, how about this simpler question:
Is $T$ algebraic if $\Sigma(T)$ is a Seifert manifold?
| https://mathoverflow.net/users/23935 | Characterizing algebraic tangle by their double branched covers | Yes, this is true. Suppose that $T$ is the given tangle in the three-ball, and the branched double cover is a graph manifold. By the uniqueness of the [JSJ decomposition](https://en.wikipedia.org/wiki/JSJ_decomposition), the decomposition of the graph manifold into Seifert manifolds descends to give a tangle decomposition of $T$. This reduces full case to your simpler case where the branched double cover is a Seifert fibered space with boundary.
Applying an [equivariant torus theorem](https://www.ams.org/journals/tran/1991-326-02/S0002-9947-1991-1034664-8/S0002-9947-1991-1034664-8.pdf), we can further reduce to the case where the branched double cover is Seifert fibered over the disk with two cone points. That is, to torus knot complements, where the result holds.
| 5 | https://mathoverflow.net/users/1650 | 428178 | 173,618 |
https://mathoverflow.net/questions/428145 | 2 | Is there a well known result that states that as $t \to \infty$, 'almost all' zeros of any Dirichlet L function $L(s,\chi)$ lie in the region $R= \{\sigma+i t\mid |\sigma -\frac{1}{2}| \leq \Phi(t) \}$ for a positive function $\Phi$ "slowly going to zero" as $t\to \infty$, in the sense that the limit of the fraction of the # of zeros of the $L$ function, as $t \to \infty$, within $R\cap \{[0,1] \times[-t,t]\}$, is $1$?
In this case, precise quantifiers above would also be needed. I would be grateful for any possible reference (possibly there is a place in Iwaniec-Kowalski's book where this is mentioned precisely).
| https://mathoverflow.net/users/111052 | 'Almost all' zeros of the Dirichlet L function lies 'near' the critical line? | One way to address this question is via a zero density result, of which there are a great variety. One can find a nice survey of this technology in Chapter 10 of Iwaniec and Kowalski. For instance, their Theorem 10.4 (which they call the Grand Density Theorem) gives a pretty general result that certainly works for an individual Dirichlet $L$-function with some control on the conductor. They also remark that work of Montgomery (*Topics in Multiplicative Number Theory*) can give some improvements which appear to be desirable based on the statement of the question.
| 6 | https://mathoverflow.net/users/2627 | 428182 | 173,621 |
https://mathoverflow.net/questions/428183 | 5 | Let $p \equiv 1 \pmod 4$ be a prime and $E\_n$ denote the $n$-th Euler number. While investigating $E\_{p-1} \pmod{p^2}$ I have encountered this summation (modulo $p$)
\begin{align\*}
\sum\_{k =1}^{\frac{p-3}{2}} \frac{a\_k}{2k+1} \pmod p.
\end{align\*}
where $a\_k = \frac{1\cdot 3\cdots (2k-1)}{2\cdot 4 \cdots 2k}$.
Another way to express this sum (this is how I first saw it) is
\begin{align}
\sum\_{k = 1}^{\frac{p-3}{2}} {2k \choose k} \frac{1}{4^k(2k + 1)}.
\end{align}
Like the questions says, I am wondering if a simple expression is known for this sum, or if sums like it have been studied before. Any references or insights would be appreciated!
**Some observations:** Using Wilson's theorem, we have
\begin{align}
\frac{1\cdot 3\cdots (p-2)}{2\cdot 4 \cdots (p-1)} \equiv \frac{-1}{(2\cdot 4 \cdots (p-1))^2} \equiv \frac{-1}{2^{p-1}\left(\frac{p-1}{2}\right)^2!} \equiv 1 \pmod p
\end{align}
($p \equiv 1 \pmod 4 \implies \left(\frac{p-1}{2}\right)^2!\equiv -1 \pmod p$). Hence
\begin{align}
\frac{1\cdot 3\cdots (2k-1)}{2\cdot 4 \cdots 2k} \equiv \frac{1\cdot 3\cdots (p-2(k+1))}{2\cdot 4 \cdots (p-(2(k+1)-1))} \pmod p
\end{align}
then $a\_k \equiv a\_{\frac{p-1}{2}-k} \pmod p$
and the sum collapses a bit,
\begin{align\*}
\sum\_{k =1}^{\frac{p-3}{2}} \frac{a\_k}{2k+1} \equiv \sum\_{k =1}^{\frac{p-3}{4}} \frac{a\_k}{2k+1} - \sum\_{k =1}^{\frac{p-3}{4}} \frac{a\_{\frac{p-1}{2}-k}}{2k} \equiv -\sum\_{k =1}^{\frac{p-3}{4}} \left(\frac{a\_k}{(2k)(2k+1)} \right) \pmod p
\end{align\*}
but this doesn't seem to help. The symmetry of $a\_k$ described above gives $\sum\_{k =1}^{\frac{p-3}{2}}(-1)^k a\_k \equiv 0 \pmod p$, which may be of interest to others.
| https://mathoverflow.net/users/171396 | Is there a simple expression for $\sum_{k =1}^{(p-3)/2} \frac{1\cdot 3\cdots (2k-1)}{2\cdot 4 \cdots 2k\cdot(2k+1)} \bmod p$? | It is known that $$\sum\_{k=0}^\infty\frac{\binom{2k}k}{(2k+1)4^k}=\frac{\pi}2.$$ Motivated by this, I proved in my paper [*On congruences related to central binomial coefficients*](https://doi.org/10.1016/j.jnt.2011.04.004) [J. Number Theory 131(2011), 2219-2238] the following result (as part (i) of Theorem 1.1 in the paper):
Let $p$ be any odd prime. Then
$$\sum\_{k=0}^{(p-3)/2}\frac{\binom{2k}k}{(2k+1)4^k}\equiv(-1)^{(p+1)/2}\frac{2^{p-1}-1}p\pmod{p^2}$$
and
$$\sum\_{k=(p+1)/2}^{p-1}\frac{\binom{2k}k}{(2k+1)4^k}\equiv pE\_{p-3}\pmod{p^2},$$
where $E\_0,E\_1,E\_2,\ldots$ are the Euler numbers.
| 8 | https://mathoverflow.net/users/124654 | 428190 | 173,623 |
https://mathoverflow.net/questions/423336 | 3 | Let $V$ be a finite dimensional complex Hilbert space. Let $L(V)$ denote the collection of all linear operators from $V$ to $V$. An operator $\mathcal{E}:L(V)\rightarrow L(V)$ is said to be positive if whenever $A\geq 0$, we have $\mathcal{E}(A)\geq 0$ as well. We say that $\mathcal{E}$ is completely positive if $\mathcal{E}\otimes 1\_{L(W)}:L(V\otimes W)\rightarrow L(V\otimes W)$ is positive for each finite dimensional complex Hilbert space $W$. A linear mapping $\mathcal{E}:L(V)\rightarrow L(V)$ is said to be trace preserving if $\text{Tr}(\mathcal{E}(A))=\text{Tr}(A)$ whenever $A\in L(V)$. A channel is a completely positive trace preserving map $\mathcal{E}:L(V)\rightarrow L(V)$. A unital channel is a channel where $\mathcal{E}(1\_V)=1\_V$.
If $A\_1,\dots,A\_r\in L(V)$, then define a mapping $\Phi(A\_1,\dots,A\_r):L(V)\rightarrow L(V)$ by $\Phi(A\_1,\dots,A\_r)(X)=A\_1XA\_1^\*+\dots A\_rXA\_r^\*$. Then the mapping $\Phi(A\_1,\dots,A\_r)$ is a completely positive mapping, and every completely positive mapping $\mathcal{E}:L(V)\rightarrow L(V)$ is of this form.
Define $$\rho\_{2}(A\_1,\dots,A\_r)=\rho(\Phi(A\_1,\dots,A\_r))^{1/2}.$$
The Cauchy-Schwarz inequality holds for $\rho\_{2}$: $$\rho(A\_1\otimes B\_1+\dots+A\_r\otimes B\_r)\leq\rho\_2(A\_1,\dots,A\_r)\rho\_2(B\_1,\dots,B\_r).$$
Define $$\rho\_{2,d}(A\_1,\dots,A\_r)$$
$$=\sup\{\frac{\rho(A\_1\otimes X\_1+\dots+A\_r\otimes X\_r)}{\rho\_{2}(X\_1,\dots,X\_r)}\mid X\_1,\dots,X\_r\in M\_d(\mathbb{C}),\rho\_{2}(X\_1,\dots,X\_r)\neq 0\}.$$ Observe that $\rho\_{2,d}(A\_1,\dots,A\_r)\leq\rho\_{2,g}(A\_1,\dots,A\_r)\leq\rho\_2(A\_1,\dots,A\_r)$ whenever $d\leq g$, and $\rho\_{2,d}(A\_1,\dots,A\_r)=\rho\_2(A\_1,\dots,A\_r)$ whenever $d\geq\dim(V).$
Theorem: $\Phi(A\_1,\dots,A\_r)=\Phi(B\_1,\dots,B\_r)$ if and only if there is an $r\times r$ unitary matrix $(u\_{i,j})\_{i,j}$ where $B\_{i}=\sum\_{j=1}^{r}u\_{i,j}A\_{j}$ for $1\leq i\leq r$.
The $\leftarrow$ direction is easy to prove, and a proof of the direction $\rightarrow$ can be found in the book The Theory of Quantum Information by John Watrous.
Lemma: Suppose that $A\_1,\dots,A\_r,B\_1,\dots,B\_r,X\_1,\dots,X\_r,Y\_1,\dots,Y\_r$ are matrices over the same field and whose dimensions are proper so that $A\_1\otimes X\_1+\dots A\_r\otimes X\_r,B\_1\otimes Y\_1+\dots B\_r\otimes Y\_r$ both make sense and have the same dimension. Suppose that $(u\_{i,j})\_{i,j},(v\_{i,j})\_{i,j}$ are $r\times r$-matrices over the field $K$ and
$(u\_{i,j})\_{i,j}^{-1}=(v\_{i,j})\_{i,j}^{T}$. Furthermore, suppose that
$A\_i=\sum\_{j=1}^{r}u\_{i,j}B\_j,X\_i=\sum\_{j=1}^{r}v\_{i,j}Y\_j$ for $1\leq i\leq r$. Then
$$A\_1\otimes X\_1+\dots A\_r\otimes X\_r=B\_1\otimes Y\_1+\dots B\_r\otimes Y\_r.$$
I was able to prove the following fact (it is not too hard to verify that this fact is correct using computer calculations).
Theorem: If $\Phi(A\_1,\dots,A\_r)=\Phi(B\_1,\dots,B\_r)$, then $\rho\_{2,d}(A\_1,\dots,A\_r)=\rho\_{2,d}(B\_1,\dots,B\_r)$.
Proof: If $\Phi(A\_1,\dots,A\_r)=\Phi(B\_1,\dots,B\_r)$, then there is a unitary map $(u\_{i,j})\_{i,j}$ where $A\_i=\sum\_{j=1}^{r}u\_{i,j}\cdot B\_j$ for $1\leq i\leq r$. Therefore, set $(v\_{i,j})\_{i,j}=((u\_{i,j})\_{i,j}^{-1})^{T}$. Then whenever $X\_1,\dots,X\_r\in M\_n(\mathbb{C})$, and $Y\_i=\sum\_{j=1}^{r}v\_{i,j}X\_{j}$, we have $\rho\_{2}(X\_1,\dots,X\_r)=\rho\_{2}(Y\_1,\dots,Y\_r)$, and $\rho(A\_1\otimes X\_1+\dots+A\_r\otimes X\_r)=\rho(B\_1\otimes Y\_1+\dots+B\_r\otimes Y\_r)$. Therefore, $$\frac{\rho(A\_1\otimes X\_1+\dots+A\_r\otimes X\_r)}{\rho\_2(X\_1,\dots,X\_r)}=\frac{\rho(B\_1\otimes Y\_1+\dots+B\_r\otimes Y\_r)}{\rho\_2(Y\_1,\dots,Y\_r)}$$
whenever $\rho\_2(Y\_1,\dots,Y\_r)\neq 0$, so $\rho\_{2,d}(B\_1,\dots,B\_r)\leq\rho\_{2,d}(A\_1,\dots,A\_r)$. The reverse inequality is established in a similar manner. Q.E.D.
Therefore, if $\mathcal{E}:L(V)\rightarrow L(V)$ is a completely positive mapping, then we can define $\rho\_{2,d}(\mathcal{E})$ by letting $\rho\_{2,d}(\mathcal{E})=\rho\_{2,d}(A\_1,\dots,A\_r)^{2}$ where $\mathcal{E}=\Phi(A\_1,\dots,A\_r)$.
If $d\geq\dim(V)$, then $\rho\_{2,d}(\mathcal{E})=\rho(\mathcal{E})$.
If $1\leq d<\dim(V)$, then is there a characterization of $\rho\_{2,d}(\mathcal{E})$ that does not require us to decompose $\mathcal{E}$ as $\Phi(A\_1,\dots,A\_r)$? Is there such a characterization of $\rho\_{2,d}(\mathcal{E})$ in the special case when $\mathcal{E}$ is a channel? What about when $\mathcal{E}$ is a unital channel or a mixed unitary channel? Can $\rho\_{2,d}(\mathcal{E})$ be generalized to the case when $\mathcal{E}$ is no longer necessarily completely positive?
It would be great if there were a quantum algorithm that often efficiently computes $\rho\_{2,d}(\mathcal{E})$ when there is a quantum computer that sends the mixed state $D$ to the mixed state $\mathcal{E}(D)$, but perhaps this is too much to ask for.
**Added 5/27/2022**
Suppose that $(e\_{a}\mid a\in\Sigma)$ is an orthonormal basis for $W$. Let
$A:V\rightarrow V\otimes W$ be a linear operator. Suppose that
$A=\sum\_{a\in\Sigma}A\_a\otimes e\_a$. Then the mapping $\mathcal{E}\_A:L(V)\rightarrow L(V)$ defined by letting $\mathcal{E}\_A(X)=\text{Tr}\_W(AXA^\*)$ ($\text{Tr}\_W$ denotes the partial trace) is a completely positive mapping, and
$\text{Tr}\_W(AXA^\*)=\sum\_{a\in\Sigma}A\_aXA\_a^\*$, so every completely positive mapping is of the form $\mathcal{E}\_A$ for some $A$.
Furthermore, if $B\in L(U,W\otimes U)$, and $B=\sum\_{b\in\Sigma}e\_b\otimes B\_b$, then $$\text{Tr}\_W(A\otimes B^\*)=\sum\_{a\in\Sigma}A\_a\otimes B\_b^\*.$$
Therefore, $$\rho\_{2,d}(\mathcal{E})^{1/2}=
\sup\{\frac{\rho(\text{Tr}\_W(A\otimes B^\*))}{\rho(\mathcal{E}\_{B})^{1/2}}\mid B\in L(U,W\otimes U)\}$$
whenever $\dim(U)=d$. This characterization of $\rho\_{2,d}(\mathcal{E})^{1/2}$ depends on the choice of $A$ and is not much different than the other definition of $\rho\_{2,d}(\mathcal{E})^{1/2}$. It is known that
if $\mathcal{E}\_{A\_1}=\mathcal{E}\_{A\_2}$, then
$A\_1=(1\_{V}\otimes O)A\_2$ for some unitary map $O\in L(W)$.
| https://mathoverflow.net/users/22277 | Approximations of the spectral radii of completely positive superoperators | Yes. We can characterize $\rho\_{2,d}(\mathcal{E})$ whenever $\mathcal{E}$ is completely positive without needing to first decompose $\mathcal{E}$ as $\Phi(A)$ or $\Phi(A\_1,\dots,A\_r).$ As a consequence, we can define $\rho\_{2,d}(\mathcal{E})$ for all linear operators $\mathcal{E}:L(V)\rightarrow L(V)$, but if $\mathcal{E}$ is not completely positive, then $\rho\_{2,d}(\mathcal{E})$ is usually infinite, so $\rho\_{2,d}(\mathcal{E})$ it not well behaved when we do not assume that $\mathcal{E}$ is completely positive.
Let $U\_1,U\_2,V\_1,V\_2,U\_1^\sharp,U\_2^\sharp,V\_1^\sharp,V\_2^\sharp,U,V$ be finite dimensional complex inner product spaces.
If $A\_1,\dots,A\_r:U\_2\rightarrow V\_2,B\_1,\dots,B\_r:U\_1\rightarrow V\_1$ are linear, then define a mapping $\Gamma(A\_1,\dots,A\_r;B\_1,\dots,B\_r):L(U\_1,U\_2)\rightarrow L(V\_1,V\_2)$ by letting $$\Gamma(A\_1,\dots,A\_r;B\_1,\dots,B\_r)(X)=\sum\_{k=1}^rA\_kXB\_k^\*.$$
Suppose now that $A\_1,\dots,A\_r:U\_2\rightarrow V\_2,B\_1,\dots,B\_r:U\_1\rightarrow V\_1$. Let $S:L(U\_2,V\_2)\rightarrow L(U\_2^\sharp,V\_2^\sharp)$ be linear and let
$T:L(U\_1,V\_1)\rightarrow L(U\_1^\sharp,V\_1^\sharp)$ also be linear. Then define
$\mho(\Gamma(A\_1,\dots,A\_r;B\_1,\dots,B\_r),S,T):L(U\_1^\sharp,U\_2^\sharp)\rightarrow L(V\_1^\sharp,V\_2^\sharp)$ by letting
$$\mho(\Gamma(A\_1,\dots,A\_r;B\_1,\dots,B\_r),S,T)=\Gamma(S(A\_1),\dots,S(A\_r);T(B\_1),\dots,T(B\_r)).$$ The mapping $\mho$ is well-defined; i.e. it depends on $\Gamma(A\_1,\dots,A\_r;B\_1,\dots,B\_r)$ rather than the particular choice of $A\_1,\dots,A\_r;B\_1,\dots,B\_r$.
$\rho\_{2,d}(\mathcal{E})$ is the maximum value of $$\frac{\rho(\mho(\mathcal{E},1\_{L(V)},T))^2}{\rho(\mho(\mathcal{E},T,T))}$$
where $T:L(V)\rightarrow L(U)$ is a linear operator and $\dim(U)=d$, and this definition generalizes to all linear operators $\mathcal{E}:L(V)\rightarrow L(V)$ when we replace the word 'maximum' with 'supremum'.
Suppose now that $\mathcal{E}=\Gamma(A\_1,\dots,A\_r;B\_1,\dots,B\_r)$ where $A\_1,\dots,A\_r,B\_1,\dots,B\_r$ are linearly independent and $\Phi(A\_1,\dots,A\_r)$ is not nilpotent. Let $T:L(V)\rightarrow L(V)$ be a linear mapping where $T(A\_j)=A\_j,T(B\_j)=\alpha\cdot A\_j$ for $1\leq j\leq r$.
Then
$\mho(\mathcal{E},1\_{L(V)},T)=\mho(\mathcal{E},T,T)=\alpha\cdot\Phi(A\_1,\dots,A\_r)$.
Therefore, $$\rho\_{2,d}(\mathcal{E})\geq\frac{\rho(\alpha\cdot\Phi(A\_1,\dots,A\_r))^2}{\rho(\alpha\cdot\Phi(A\_1,\dots,A\_r))}=\alpha\cdot\rho(\Phi(A\_1,\dots,A\_r)).$$
Since $\alpha$ can be made arbitrarily large, we have $\rho\_{2,d}(\mathcal{E})=\infty$ in this case.
**A modest generalization: added 8/15/2022**
There is another way to generalize $\rho\_{2,d}(\mathcal{E})$ to some operators that are not completely positive, but this generalization is based on a conjecture.
Conjecture: Suppose that $\mathcal{E}:L(V)\rightarrow L(V)$ is completely positive and $\alpha\geq 0$. Then $\rho\_{2,d}(\mathcal{E}+\alpha\cdot 1\_{L(V)})=\rho\_{2,d}(\mathcal{E})+\alpha$.
From this conjecture, we can define $\rho\_{2,d}(\mathcal{E})$ whenever there is some $\alpha\geq 0$ where $\mathcal{E}+\alpha\cdot 1\_{L(V)}$ is completely positive by letting $\rho\_{2,d}(\mathcal{E})=\rho\_{2,d}(\mathcal{E}+\alpha\cdot 1\_{L(V)})-\alpha$. Here, we can have negative values of $\rho\_{2,d}(\mathcal{E})$, so $\rho\_{2,d}$ more closely resembles the maximum value of a Hermitian operator than the spectral radius.
**Projective mappings: added 3/21/2023**
Define $\rho\_{2,d}^P(A\_1,\dots,A\_r)$ to be the supremum of all values of the form $$\frac{\rho(A\_1\otimes\overline{RA\_1S}+\dots+A\_r\otimes\overline{RA\_rS})}{\rho(RA\_1S\otimes\overline{RA\_1S}+\dots+RA\_rS\otimes\overline{RA\_rS})^{1/2}}$$
where $R\in M\_{d,n}(\mathbb{C}),S\in M\_{n,d}(\mathbb{C})$. I conjecture that $\rho\_{2,d}^P(A\_1,\dots,A\_r)=\rho\_{2,d}(A\_1,\dots,A\_r)$ (computer calculations support this conjecture at least sometimes), but it seems like $\rho\_{2,d}^P(A\_1,\dots,A\_r)$ is easier to work with than $\rho\_{2,d}(A\_1,\dots,A\_r)$.
If $S:U\rightarrow V,R:V\rightarrow U$ are linear maps, then define a mapping $T\_{R,S}:L(V)\rightarrow L(U)$ by setting $T\_{R,S}(A)=RAS$ for $A\in L(V)$.
$$\mho(\Gamma(A\_1,\dots,A\_r;B\_1,\dots,B\_r),T\_{R,S},T\_{R,S})(X)$$
$$=\Gamma(T\_{R,S}(A\_1),\dots,T\_{R,S}(A\_r);T\_{R,S}(B\_1),\dots,T\_{R,S}(B\_r))(X)$$
$$=\Gamma(RA\_1S,\dots,RA\_rS;RB\_1S,\dots,RB\_rS)(X)=\sum\_{k=1}^{r}RA\_kSX(RB\_kS)^\*$$
$$=\sum\_{k=1}^rRA\_kSXS^\*B^\*\_kR^\*
=R\big(\Gamma(A\_1,\dots,A\_r;B\_1,\dots,B\_r)(SXS^\*)\big)R^\*.$$
Similarly,
$$\mho(\Gamma(A\_1,\dots,A\_r;B\_1,\dots,B\_r),1\_{L(V)},T\_{R,S})(X)
=\Gamma(A\_1,\dots,A\_r;RB\_1S,\dots,RB\_rS)(X)$$
$$=\sum\_{k=1}^rA\_kX(RB\_kS)^\*=\sum\_{k=1}^rA\_kXS^\*B\_k^\*R^\*
=\big(\Gamma(A\_1,\dots,A\_r;B\_1,\dots,B\_r)(XS^\*)\big)R^\*.$$
Therefore, $\mho(\mathcal{E},T\_{R,S},T\_{R,S})(X)=R\big(\mathcal{E}(SXS^\*)\big)R^\*$ and $\mho(\mathcal{E},1\_{L(V)},T\_{R,S})(X)=\big(\mathcal{E}(XS^\*)\big)R^\*$ whenever $\mathcal{E}:L(V)\rightarrow L(V)$ and $R,S,X$ are suitable linear operators. Therefore, if $\mathcal{E}:L(V)\rightarrow L(V)$, then define $\mho\_{R,S}^-(\mathcal{E})(X)=\big(\mathcal{E}(XS^\*)\big)R^\*$ and
$\mho\_{R,S}^+(\mathcal{E})(X)=R\big(\mathcal{E}(SXS^\*)\big)R^\*$.
$\rho\_{2,d}^P(A\_1,\dots,A\_r)$ is the supremum of all values of the form
$$\frac{\rho(\mho\_{R,S}^-(\Phi(A\_1,\dots,A\_r)))}{\rho(\mho\_{R,S}^+(\Phi(A\_1,\dots,A\_r))^{1/2}}.$$ Therefore, we may define $\rho\_{2,d}^P(\mathcal{E})$ for all completely positive operators $\mathcal{E}:L(V)\rightarrow L(V)$ by setting
$\rho\_{2,d}^P(\mathcal{E})=\rho\_{2,d}^P(A\_1,\dots,A\_r)^2$ whenever $\mathcal{E}=\Phi(A\_1,\dots,A\_r)$. Then $\rho\_{2,d}^P(\mathcal{E})$ is the supremum of all values of the form $\frac{\rho(\mho\_{R,S}^-(\mathcal{E}))^2}{\rho(\mho\_{R,S}^+(\mathcal{E}))}$ where $R\in L(U,V),S\in L(V,U)$ and $U$ is a complex inner product space $\dim(U)=d$. This definition of $\rho\_{2,d}^P(\mathcal{E})$ makes sense even when $\mathcal{E}:L(V)\rightarrow L(V)$ and $V$ is an infinite dimensional complex Hilbert space (we will probably need to use Stinespring's dilation theorem to generalize $\rho\_{2,d}$ to infinite dimensional spaces).
Example: $\rho\_{2,d}^P(\mathcal{E})$ can be infinite for superoperators which are positive but not completely positive. For example, the transpose map $T:M\_n(\mathbb{C})\rightarrow M\_n(\mathbb{C})$ defined by $T(X)=X^T$ is positive but not completely positive whenever $n>1$, and we shall prove that $T$ is not completely positive by showing that $\rho\_{2,d}^P(T)=\infty$. A straightforward calculation yields $\rho\_{2,d}^P(T)=\sup\_{R\in M\_{d,n}(\mathbb{C}),S\in M\_{n,d}(\mathbb{C})}\frac{\rho(SR)^2}{\rho(R\overline{S}\overline{R}S)}.$
If $R,S$ are rank-1 matrices, then we can set $R=uv^\*,S=wx^\*$ for vectors $u,v,w,x$. In this case,
$\frac{\rho(SR)^2}{\rho(R\overline{S}\overline{R}S)}=|\frac{v^\*w}{v^Tw}|^2$. If $v=w=[1,i]^T$, then $\frac{\rho(SR)^2}{\rho(R\overline{S}\overline{R}S)}=(2/0)^2=+\infty.$
It seems like a good way of showing that a superoperator $\mathcal{E}:L(V)\rightarrow L(V)$ is not completely positive is to show that $\rho\_{2,d}^P(\mathcal{E})>\rho(\mathcal{E})$ by finding matrices $R,S$ where $\rho(\mho\_{R,S}^-(\mathcal{E}))^2>\rho(\mho\_{R,S}^+(\mathcal{E}))\cdot\rho(\mathcal{E})$.
| 1 | https://mathoverflow.net/users/22277 | 428197 | 173,625 |
https://mathoverflow.net/questions/428217 | 11 | Let $X ⊆ \mathbb{P}^n$ be a smooth projective variety (over $\mathbb{C}$). I think we can find a chain of irreducible varieties $X = X\_0 ⊆ X\_1 ⊆ X\_2 ⊆ \cdots ⊆ X\_k = \mathbb{P}^n$ whose dimension increases by one at every step by writing $X = \mathcal{V}(f\_1, \dots, f\_n)$ and dropping some of the $f\_i$ until the dimension of the irreducible component containing $X$ increases, and then proceeding by induction on $k = \operatorname{codim}(X)$.
Is it possible to a chain where all of the $X\_i$ are smooth?
| https://mathoverflow.net/users/123448 | Is every smooth projective variety contained in a chain of smooth projective varieties of increasing dimension? | Suppose that $\operatorname{dim}(X)>1$ and that such a chain exists. Since $\operatorname{Pic}(\mathbf{P}^n)\simeq \mathbf{Z}$, the variety $X\_{k-1}$ is an ample divisor in $\mathbf{P}^n$, and hence by the Lefschetz hyperplane theorem we have $\operatorname{Pic}(X\_{k-1})\simeq \mathbf{Z}$. So $X\_{k-2}$ is an ample divisor on $X\_{k-1}$, and again its Picard group is $\mathbf{Z}$. By induction, we obtain that each $X\_{i-1}$ is an ample divisor on $X\_{i}$, and then by hyperplane Lefschetz for $\pi\_1$ we obtain that $\pi\_1(X)\simeq \pi\_1(\mathbf{P}^n)$ is the trivial group. So to conclude, an abelian variety of dimension at least two embedded in $\mathbf{P}^n$ gives a counterexample. (N.B. There exist abelian surfaces in $\mathbf{P}^4$, constructed by Horrocks and Mumford).
**Edit.** Of course this contradicts Sasha's answer posted roughly at the same time. I am puzzled as to where the mistake is.
| 18 | https://mathoverflow.net/users/3847 | 428220 | 173,632 |
https://mathoverflow.net/questions/428214 | 5 | I am reading David Geraghty's paper, 'Modularity lifting theorems for ordinary Galois representations'(<https://link.springer.com/article/10.1007/s00208-018-1742-4>) and I have a related question, which, despite some search in the literature, remains puzzling to me (I should say that I am not quite familiar with this field of research and my question may be too naive to the experts): Geraghty's paper is for unitary groups that are compact at infinity. Are there any modularity lifting theorems for non-compact unitary groups? It seems many results on modularity liftings are for compact groups (except perhaps GL(2), GSp(4)?) What are the difficulties for non-compact ones? I am especially interested in the case of split unitary groups GU(n,n). Thanks in advance!
| https://mathoverflow.net/users/478677 | modularity lifting theorems for non-compact unitary groups | You might like to read the introduction of Harris' 2013 Crelle paper "The Taylor-Wiles method for coherent cohomology" (see [link](https://webusers.imj-prg.fr/%7Emichael.harris/coherentTW.pdf)). Here is an excerpt:
>
> In practice, all the higher-dimensional results, with the exception of
> [GT] and [Pi], have been based on topological cohomology of
> zero-dimensional Shimura varieties. This is because the Taylor–Wiles
> method does not work well in the presence of torsion, and there are no
> general methods for comparing torsion in the cohomology of locally
> symmetric spaces to automorphic forms.
>
>
>
(The references [GT] and [Pi] refer to works of Genestier–Tilouine and Pilloni, both focussing on $GSp(4)$.) Harris goes on to write (emphasis mine):
>
> We obtain no new results about Galois representations, and in fact **I believe that practically everything one wants to say about automorphy of Galois representations can be obtained from the zero-dimensional case**, as in [CHT], using Langlands functoriality for classical groups (see [A] and, in special cases, [CHLN]). Our purpose is rather to prove [freeness results for coherent cohomology as a Hecke module].
>
>
>
This was written a decade ago, and in the meantime the technology has got much better. So torsion in the cohomology of symmetric spaces is not quite so mysterious now as it was in 2013, and this has been a crucial input in recent dramatic advances in modularity over non-totally-real fields (e.g. the 10-author paper by Allen et al) or for non-regular weights (e.g. Boxer–Calegari–Gee–Pilloni). However, it definitely still remains true that the 0-dimensional case is much easier than the positive-dimensional one.
| 8 | https://mathoverflow.net/users/2481 | 428233 | 173,634 |
https://mathoverflow.net/questions/428254 | 0 | Working in a suitable extension of $\sf ZF-Reg.$, can we have a transitive model $M$ of $\sf ZF-Reg.$ such that for every set in $M$ there is a partition on it in $M$ all compartments of which are non-singleton finite sets. And such that any subset $X$ of $M$ that is a family of pairwise disjoint larger than singleton finite sets then $X \in M$.
| https://mathoverflow.net/users/95347 | Can we have all classes that can be partitioned into non-singleton finite sets, be sets? | No, this is not possible: the ordinals of $M$ will always provide a counterexample.
Even if regularity fails, the class $\mathsf{Ord}$ of ordinals cannot be a set. But we can partition $\mathsf{Ord}$ into (say) two-element sets via the equivalence relation $$\alpha\sim\beta\quad\iff\quad \sup\{\lambda: 2\cdot\lambda<\alpha\}=\sup\{\lambda: 2\cdot\lambda<\beta\}.$$ The classes of this relation are $\{0,1\},\{2,3\},...,\{\omega,\omega+1\}, ...$ etc.
(Perhaps more elegantly, let $\approx$ be any equivalence relation on $\omega$ all of whose classes are finite and have at least two elements. Then consider the relation $\approx'$ defined by $\alpha\approx'\beta$ iff there is a limit ordinal $\lambda$ and finite ordinals $m,n$ such that $\lambda+m=\alpha$, $\lambda+n=\beta$, and $m\approx n$.)
| 7 | https://mathoverflow.net/users/8133 | 428257 | 173,639 |
https://mathoverflow.net/questions/428231 | 5 | Let $F$ be a finite field extension of the $p$-adic numbers $\mathbb{Q}\_p$, whose residue field has $q$ elements. Let $\mathfrak{p}$ be the prime ideal of $F$. Given a finite field extension $K/F$, write $\operatorname{Disc}\_\mathfrak{p}(K)$ for $q^{v\_\mathfrak{p}(\operatorname{Disc}(K/F))}$. Then **Serre's mass formula** states that for all $n\geq 1$, we have
$$
\sum\_{[K:F] = n \text{ totally ramified}} \frac{1}{\#\operatorname{Aut}(K/F)}\cdot\frac{1}{\operatorname{\operatorname{Disc}\_{\mathfrak{p}}(K)}} = \frac{1}{q^{n-1}}.
$$
This result first appeared in Serre's 1978 paper **Une “ formule de masse ” pour les extensions totalement ramifiees de degre donne d’un corps local**. As far as I can tell, this paper is only available in the book **Oeuvres/Collected Papers. 3 volumes.** Given that I am unable to obtain this book, I am looking for a reference on the internet that contains either Serre's original paper or an exposition of his proof (ideally both proofs from the paper).
I would also (delightedly) accept an answer that directly explains the proof (i.e. no reference); such an answer would in my opinion be a great addition to the learning resource that is MO, seeing as the proof is so hard to find on the internet!
| https://mathoverflow.net/users/169035 | Looking for proof of Serre's mass formula | The article is not only available in Serre's Collected Papers: it first appeared in a journal, after all. Here's a scan from the Comptes Rendus archives: <https://gallica.bnf.fr/ark:/12148/bpt6k6234149b/f323.item>.
| 7 | https://mathoverflow.net/users/3272 | 428260 | 173,640 |
https://mathoverflow.net/questions/428232 | 5 | In the page 482 of his [article](https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.449.3854&rep=rep1&type=pdf), Fickle wrote the following argument:
Let $Y$ be a homology $3$-sphere. Next
* Add a $2$-handle to $Y \times [0,1]$ and produce a $4$-manifold with
boundary $S^1 \times S^2$,
* Cap off the boundary with a $3$-handle and a $4$-manifold,
* Turn the resulting $4$-manifold upside down.
Then we may conclude that $Y$ bounds a Mazur manifold. Why this is true?
| https://mathoverflow.net/users/475366 | Fickle's argument for Mazur manifolds | Call the manifold built in this way $W$. If you turn the handlebody for $W$ relative to $Y$ upside down, it becomes a handlebody with a single handle in indices $0$, $1$, and $2$. This is almost the same as saying it's a Mazur manifold; you need to check that it's contractible. Because $\pi\_1(W)$ is cyclic (it's got just the one generator from the $1$-handle), this is the same as saying that $H\_1(W) = 0$. This in turn follows from the fact that $Y$ is a homology sphere, as follows.
Suppose the $2$-handle (in the upside-down handlebody) goes over the 1-handle $n$ times, and has framing $k$. Then $Y$ has a surgery description where the linking matrix is
$$
\pmatrix{0 & n\\n & k}.
$$
In order for $Y$ to be a homology sphere, this must have determinant $1$, so $n= \pm 1$ and hence $H\_1(W) = 0$.
By the way, this argument predates Fickle's paper. See for instance Casson-Harer, *Some homology lens spaces which bound rational homology balls*, Pacific J. Math. 96 (1981), no. 1, 23–36.
| 7 | https://mathoverflow.net/users/3460 | 428261 | 173,641 |
https://mathoverflow.net/questions/428153 | 2 | Given positive integer $n$, we are looking for a set
of $n$ positive integers $a\_i$.
The following linear integer program must have only
the trivial integer solution of all ones.
* $0 \le x\_i \le \frac{n}{2}$
* $\sum x\_i = n$
* $\sum a\_i x\_i = \sum a\_i$
One exponential example is to take $a\_i=C^i$.
We experimented with an integer programming solver and
couldn't find small solutions, possibly because of
the law of small numbers.
To our surprise taking $n=30$ and random numbers in the range
$[2^{29},2^{30}]$ gave non-trivial solutions.
>
> Q1 How small can $\max a\_i$ be in terms of $n$?
> Can we get $\exp(o(n))$?
>
>
>
>
> Q2 If $\max a\_i$ is polynomial in $n$, what bound
> can we get for the number of solutions of the integer program?
>
>
>
| https://mathoverflow.net/users/12481 | Only trivial solution to a pair of constrained linear diophantine equations | The answer to the first question is negative.
Let $A$ denote the set of weights $\{a\_i\}$. Strengthen the first constraint to $0 \le x\_i \le 2$.
If we have two different subsets $S\_1, S\_2 \subset A$ of the same cardinality, their sums must be different, since otherwise we can assign $$x\_i = \begin{cases} 2 & \textrm{if } a\_i \in S\_1 \setminus S\_2 \\
0 & \textrm{if } a\_i \in S\_2 \setminus S\_1 \\
1 & \textrm{otherwise}\end{cases}$$ to get a second solution to the program.
Therefore the $\binom{n}{\lfloor n/2 \rfloor}$ subsets of half of the weights (rounded down) must all have different (positive integer) sums, so that one of those subsets must have weight at least the number of subsets, and the sum of all of the weights must be at least $\binom{n}{\lfloor n/2 \rfloor} + \binom{\lceil n/2 \rceil + 1}{2} = \Theta(n^{-1/2} 2^n)$.
Thus $\max a\_i \in \Omega(n^{-3/2} 2^n)$.
| 1 | https://mathoverflow.net/users/46140 | 428266 | 173,642 |
https://mathoverflow.net/questions/426170 | 6 | Suppose $A,B \in M\_{n}(\Bbb{R})$ such that $A = \left[C\_{1}\middle|\frac{I}{0\dots0}\right], B= \left[C\_{2}\middle|\frac{I}{0\dots0}\right]$ , where $A$ and $B$ have different first columns (represented as $C\_{1}, C\_{2}$).
Thus we have $B = A+ \xi e\_{1}^T$, where $\xi$ is a $n \times 1$ column vector and $e\_{1}^{T} = [1, 0,\ldots,0]$
$\textbf{Assumptions}:$
Let $\lambda\_{i}, i=1, \ldots, n$ denote the eigenvalues of $AB^2$. Suppose we have the condition that $|\lambda\_{i}|<1 \, \forall i$.
Let $\beta\_{i}, i=1, \ldots, n$ denote the eigenvalues of $A^2B$. Suppose we have the condition that $|\beta\_{i}|<1 \, \forall i$.
$\textbf{Claim}:$
Then I have an intuition that $\textbf{ $\det(AB+A+I) < 0$ and $\det(BA+B+I)<0$ is not possible.}$
That is both of the determinants cannot be negative. I am not sure how to prove it?
$\textit{Some thoughts:}$
$(a) \textbf{Using $A = B+ \xi e\_{1}^T$, we have }$
\begin{align}
AB+A+I &= A^2 + A + I + A \xi e\_{1}^{T},\\
BA+B+I &= A^2 + A + I + \xi e\_{1}^{T}(I+A)
\end{align}
Since the absolute values of the eigenvalues of $AB^2,$ and $A^2B$ are less than one, that means we have $|\det(AB^2)| < 1$ and $|\det(A^2B)|<1$.
\begin{align}
A^2B &= A^3 + A^2 \xi e\_{1}^{T},\\
AB^2 &= A^3 + (\xi e\_{1}^{T})^2 + 2A^{2} \xi e\_{1}^{T}
\end{align}
$\textbf{(b)}$ If we proceed via the method of contradiction. if $\det(AB+A+I)<0$ and $\det(BA+B+I)<0$, then some of the eigenvalues of $AB^2$ or $A^2B$ will be greater than one in absolute value. THis would then violate the assumption of $|\lambda\_{i}|<1 \forall i$,, $|\sigma\_{i}|<1, \forall i$ .
$\textbf{(c)}$ Another thought is to use the perturbation argument: Fix $A, B$. Define $B(\epsilon):= A + \epsilon (B-A)$. For $\epsilon = 0$, we get $\det(A^2+A+I) \geq 0$ and hence the statement holds. For $\epsilon=1$, we have $B(1) = B$. If the statement fails in this case then there should be a minimal $\epsilon$ for which the statement is false. There might be a contradiction for $\epsilon < 1$?
| https://mathoverflow.net/users/105018 | Is it impossible for determinants of these matrices to both be negative? | This is a partial answer, now with added material.
Assume that $n$ is a multiple of 3, the other two cases should be similar. I'll denote $C\_1\mapsto a$ and $C\_2\mapsto b$, such that for, e.g., $n=6$,
$$\tag{1}\label{1}
A = \begin{pmatrix}
a\_1 & 1 & 0 & 0 & 0 & 0 \\
a\_2 & 0 & 1 & 0 & 0 & 0 \\
a\_3 & 0 & 0 & 1 & 0 & 0 \\
a\_4 & 0 & 0 & 0 & 1 & 0 \\
a\_5 & 0 & 0 & 0 & 0 & 1 \\
a\_n & 0 & 0 & 0 & 0 & 0
\end{pmatrix},\quad
A^{-1} = \begin{pmatrix}
0 & 0 & 0 & 0 & 0 & 1/a\_n \\
1 & 0 & 0 & 0 & 0 & -a\_1/a\_n \\
0 & 1 & 0 & 0 & 0 & -a\_2/a\_n \\
0 & 0 & 1 & 0 & 0 & -a\_3/a\_n \\
0 & 0 & 0 & 1 & 0 & -a\_4/a\_n \\
0 & 0 & 0 & 0 & 1 & -a\_5/a\_n
\end{pmatrix}.
$$
Note that $\det A = (-1)^{n-1}a\_n$. The inverse of $A$ is easily calculated, see \eqref{1}, such that the determinant of $AB+A+I$ (and also of the other case, exchange $a\_i$ and $b\_i$) can be expressed via a Schur complement of the matrix $A^{-1}(AB+A+I) = B+I+A^{-1}$:
We take the Schur complement w.r.t. the first and last row/column and get
$$\tag{2}\label{2}
\det(AB+A+I)=\det
\begin{pmatrix}
\alpha\_1-\alpha\_3 & \beta\_1-\beta\_2\\
\alpha\_2-\alpha\_3 & \beta\_1-\beta\_3
\end{pmatrix}
=\det
\begin{pmatrix}
\alpha\_1 &1& \beta\_2\\
\alpha\_2 &1& \beta\_3\\
\alpha\_3 &1& \beta\_1
\end{pmatrix},
$$
where in the last step we used basic determinant rules.
The $\alpha\_j$ and $\beta\_j$, with $j\in\{1,2,3\}$, are simply given by
$$\tag{3}\label{3}
\alpha\_j = -\delta\_{3,j} + \sum\_{k=0}^{n/3-1} a\_{3k+j},
\qquad
\beta\_j = -\delta\_{3,j} + \sum\_{k=0}^{n/3-1} b\_{3k+j},
$$
with Kronecker's $\delta$. I’ve renamed OPs $\beta\mapsto\nu$.
For the eigenvalue assumption $|\lambda\_i|, |\nu\_i| <1$ it should be sufficient to consider the characteristic polynomials
\begin{align}\tag{4a}\label{4a}
P\_a(\lambda)&=\det(A^2 B - \lambda I),\\
P\_b(\nu )&=\det(A B^2 - \nu I),\tag{4b}\label{4b}
\end{align}
which can be calculated in a similar fashion: now we use $A^{-1}(A^2 B - \lambda I) = A B - \lambda A^{-1}$ and build the Schur complement w.r.t. the first, second and last row/column.
If the eigenvalues fulfill $|\lambda\_i|<1$, the (real) zeroes of $P\_a(\lambda)$ are between $\lambda=\pm 1$. At this point, I am not sure how to handle the complex zeroes (but see edit below). However, as $P\_a(\lambda) \sim (-\lambda)^n$ for large $|\lambda|$, it should fulfill $P(-1)>0$ and $(-1)^{n}P(1)>0$. Let's evaluate $P\_a(1)$ as one example,
$$\tag{5}\label{5}
P\_a(1)=\det(A^2B-I)=
\det
\begin{pmatrix}
\alpha\_1 & \alpha\_3 & \beta\_2\\
\alpha\_2 & \alpha\_1 & \beta\_3\\
\alpha\_3 & \alpha\_2 & \beta\_1
\end{pmatrix}.
$$
Note the similarity of \eqref{2} and \eqref{5}, and that \eqref{5} can be generalised to $P\_a(\lambda)$ using polynomials $\alpha\_j(\lambda),\beta\_j(\lambda)$.
So, the problem for arbitrary $n$ can be reduced to a discussion of the sign of the determinants of related $3\times3$ matrices.
Disclaimer: there might be sign errors due to even/odd $n$ and row/column permutations, please check.
**Edit 12.08.22, 09:00 CEST:**
From now on, we only consider even $n$, such that $n\mod 6 \equiv 0$, to get rid of the $(-1)^n$ terms.
As noted in my comments, $P\_a(1)>0$ and $P\_a(-1)>0$ are necessary conditions for $|\lambda\_i|<1$, because complex $\lambda\_i$ appear in complex conjugate pairs $\lambda\_{i'}=\lambda\_i^\*$, such that
$$\tag{6}\label{6}
|\lambda\_i|<1 \Rightarrow (\lambda\_i \pm 1)(\lambda\_i^\* \pm 1)>0,
$$
and all factors in
$$\tag{7}\label{7}
P\_a(\lambda) = \prod\_{i=1}^{n/2} (\lambda\_i - \lambda)(\lambda\_{i'} - \lambda)
$$
are positive for $\lambda=\pm 1$. Here, we grouped the real eigenvalues in arbitrary pairs $(i,i')$.
As shown above, $P\_{a,b}(1)$ have the simple representation \eqref{5}.
Hence, we consider the matrix
$$\tag{8}\label{8}
D(\alpha,\gamma,\beta)=\begin{pmatrix}
\alpha\_1 &\gamma\_3& \beta\_2\\
\alpha\_2 &\gamma\_1& \beta\_3\\
\alpha\_3 &\gamma\_2& \beta\_1
\end{pmatrix}
$$
with 3d vectors $\alpha,\beta,\gamma$, and formulate a geometric version of the problem. Define $\delta=(1,1,1)^T$, then the OPs conjecture holds, if
$$\tag{9}\label{9}
D(\alpha,\alpha,\beta)>0 \land D(\alpha,\beta,\beta)>0
\Rightarrow
D(\alpha,\delta,\beta)>0 \lor D(\beta,\delta,\alpha)>0.
$$
Note (a) that the determinant in 3D is known as triple product,
$$\tag{10}\label{10}
\det(a,b,c) = a \cdot (b \times c) = b \cdot (c \times a) = c \cdot (a \times b),
$$ and (b) that the cyclic index permutations of in \eqref{2}, \eqref{5} and \eqref{8} are rotations by $120^\circ$ around $\delta$. I guess that the OP question can now be answered through a discussion of the (rotated) directions of $\alpha$, $\beta$ and $\delta$ in 3D.
| 1 | https://mathoverflow.net/users/90413 | 428268 | 173,643 |
https://mathoverflow.net/questions/428128 | 4 | Consider two measurable spaces $X\_1 = (\mathbb{R}^m,\mathcal{B}(\mathbb{R}^m),\mu\_1)$ and $X\_2 = (\mathbb{R}^m,\mathcal{B}(\mathbb{R}^m),\mu\_2)$ and the product spaces
$$X\_1^{q} = (\times\_{i=1}^q\mathbb{R}^m,\otimes\_{i=1}^q\mathcal{B}(\mathbb{R}^{m}),\mu\_1^{\otimes q})\ \ \ \text{ and }\ \ \ X\_2^{q} = (\times\_{i=1}^q\mathbb{R}^m,\otimes\_{i=1}^q\mathcal{B}(\mathbb{R}^{m}),\mu\_2^{\otimes q})$$
where
$\mu\_1^{\otimes q} = \underbrace{\mu\_1\otimes \cdots \otimes\mu\_1}\_{q}$ and $\mu\_2^{\otimes q} = \underbrace{\mu\_2\otimes \cdots \otimes\mu\_2}\_{q}$, respectively.
Let
$$
D(\mu,\nu) = \sup\_{\substack{\text{all closed balls } B \text{ of }\mathbb{R}^m\\ \text{in the Euclidean norm} }} \left|\mu(B)-\nu(B)\right|
$$
be the discrepancy metric between probability measures $\mu$ and $\nu$ (see e.g. [this paper](https://arxiv.org/pdf/math/0209021.pdf) for more details on this metric).
>
> **My question.** For probability measures $\mu\_1$ and $\mu\_2$, does there exist $k>0$ independent of $\mu\_1$ and $\mu\_2$
> such that
> $$
> D(\mu\_1^{\otimes q},\mu\_2^{\otimes q})\le k D(\mu\_1,\mu\_2)\ \ \ ?
> $$
>
>
>
**Note.** For other probability metrics the answer is in the affirmative (e.g., for the total variation metric, [see Eq. (4.5) of this paper](https://www.jstor.org/stable/2237258?casa_token=k0z7kEIC5zQAAAAA%3AGUNJrIaK7tlNuZMXk0GGZgURhDXke5QdrH4Klrv1EUFHJrx38434gnUID43_RuUUBU2mFSOMbACAUyjvcyb1u_rAdoLs9aITyTW3aH6GWyHZe3v9zgb8)) but I couldn't find anything about the discrepancy metric above. I suspect that, if true, this should be a rather known result. Any suggestion or comment is very welcome.
| https://mathoverflow.net/users/62673 | Bounds on discrepancy metric of product measures | Analogous to the TV metric, the requested upper bound holds for the discrepancy metric with $k=q$. The result given below can also be easily extended to general product probability measures $\mu=\otimes\_{i=1}^q \mu\_{i}$ and $\nu=\otimes\_{i=1}^q \nu\_i$ to obtain $$
D(\mu,\nu)\le \sum\_{i=1}^q D(\mu\_i, \nu\_i)\;.
$$
---
Suppose that $\mu\_1$ and $\mu\_2$ are probability measures. Let $B = \{ (x\_1, \dots, x\_q) : \sum\_i (x\_i - c\_i)^2 \le R^2 \}$ be a closed ball in $\mathbb{R}^q$ with radius $R>0$ centered at $(c\_1, \dots, c\_q) \in \mathbb{R}^q$. Let $\mathbf{1}\_A$ denote the indicator function of the set $A$.
By telescoping, and invoking Tonelli’s theorem to write $(\mu\_1^{\otimes q}-\mu\_2^{\otimes q})(B)$ as an iterated integral, note that
\begin{align\*} & |(\mu\_1^{\otimes q}-\mu\_2^{\otimes q})(B)|=
\left| \int\_{\mathbb{R}^q} \mathbf{1}\_B(x\_1, \dots, x\_q) \biggl( d\mu\_1(x\_1) \cdots d\mu\_1(x\_q) - d\mu\_2(x\_1) \cdots d\mu\_2(x\_q) \biggr) \right| \\
&= \left| \int\_{\mathbb{R}^q} \mathbf{1}\_B(x\_1, \dots, x\_q) \sum\_{ i=1}^q d\mu\_1(x\_1) \cdots d\mu\_1(x\_{i-1}) \biggl(d\mu\_1(x\_{ i}) - d\mu\_2(x\_{ i})\biggr) d\mu\_2(x\_{i+1}) \dots d\mu\_2(x\_q) \right| \\
&= \left| \sum\_{ i=1}^q \int\_{\mathbb{R}^{q-1}} \left(\int\_{\mathbb{R}} \mathbf{1}\_B(x\_1, \dots, x\_q)\biggl(d\mu\_1(x\_{ i}) - d\mu\_2(x\_{ i})\biggr) \right) d\mu\_1(x\_1) \cdots d\mu\_1(x\_{i-1}) d\mu\_2(x\_{i+1}) \dots d\mu\_2(x\_q) \right| \\
&\le \sum\_{ i=1}^q \int\_{\mathbb{R}^{q-1}} \left| \int\_{\mathbb{R}} \mathbf{1}\_B(x\_1, \dots, x\_q)\biggl(d\mu\_1(x\_{ i}) - d\mu\_2(x\_{ i})\biggr) \right| d\mu\_1(x\_1) \cdots d\mu\_1(x\_{i-1}) d\mu\_2(x\_{i+1}) \dots d\mu\_2(x\_q) \\
&\le \sum\_{i=1}^q \sup\_{\gamma\_i, \rho\_i} \left| \int\_{\mathbb{R}}\mathbf{1}\_{(x\_i-\gamma\_i)^2 \le \rho\_i^2} \biggl(d\mu\_1(x\_{ i}) - d\mu\_2(x\_{ i}) \biggr) \right| \\
& \qquad \int\_{\mathbb{R}^{q-1}} d\mu\_1(x\_1) \cdots d\mu\_1(x\_{i-1}) d\mu\_2(x\_{i+1}) \cdots d\mu\_2(x\_q) \\
&\le q D(\mu\_1, \mu\_2) \;.
\end{align\*}
Since $B$ is arbitrary, the conjectured upper bound holds with $k=q$.
---
**Remark.** One can do much better than this linear upper bound in some special cases. For instance, suppose that $\mu\_1 = \mathcal{N}(0,1)$ and $\mu\_2 = \mathcal{N}(0, \sigma^2)$. Then, by moving to hyperspherical coordinates, it's not too hard to show that $$
D(\mu\_1^{\otimes q},\mu\_2^{\otimes q}) = \left| \frac{\Gamma(\frac{q}{2}, \frac{q \log{\sigma}}{\sigma^2-1}) - \Gamma(\frac{q}{2}, \frac{q \sigma^2 \log(\sigma)}{\sigma^2-1})}{ \Gamma(\frac{q}{2}) } \right| \; $$ which converges to one with $q$ and $k(q):= D(\mu\_1^{\otimes q},\mu\_2^{\otimes q}) /D(\mu\_1,\mu\_2) $ grows sublinearly. This Gaussian case is rather exceptional because the corresponding product measure can be directly written in terms of the Euclidean distance.
| 2 | https://mathoverflow.net/users/64449 | 428269 | 173,644 |
https://mathoverflow.net/questions/428064 | 3 | Let $S$ be a scheme and $X\to Y$ be a morphism over $S$. Then we have an induced homomorphism of sheaves $h\_X=\operatorname{Hom}\_S({-}, X)\to h\_Y=\operatorname{Hom}\_S({-}, Y)$ over the small étale site $S\_\text{étale}$.
Question: When is $h\_X\to h\_Y$ a surjective homomorphism between sheaves of sets over $S\_\text{étale}$?
I find this result in the [comment](https://math.stackexchange.com/questions/22008/%C3%89tale-local-sections-of-a-smooth-surjective-morphism#comment633866_22008) of Ehsan M. Kermani under [Étale Local Sections of a Smooth Surjective Morphism](https://math.stackexchange.com/questions/22008/%C3%89tale-local-sections-of-a-smooth-surjective-morphism). Is there any proof or reference of this claim?
| https://mathoverflow.net/users/153842 | Surjective sheaf homomorphisms induced by morphisms of schemes | **Proposition.** Let $f\colon X\to Y$ be a surjective morphism such that $\forall y\in Y$ there exists a point $x\in f^{-1}(y)$ where $f$ is smooth (i.e. every fiber has a non-reduced point). Then $h\_X\to h\_Y$ is surjective on the (small or big) étale site.
**Proof:** Take $f\in h\_Y(S)$. The property of having a smooth point in every fiber is stable under base change so we may assume $S=Y$. Consider the open subset $X^\circ$ of $X$ where $f$ is smooth. The restriction $f^\circ$ of $f$ to $X^\circ$ is smooth and surjective. Apply EGA IV 4 Cor. 17.16.3 (ii) to $f^\circ$. We find an étale covering $Y'\to Y$ such that $Y'\times\_Y X$ admits a section $s\colon Y'\to Y'\times\_Y X$ which is what we need.
**Remark 1.** By EGA IV 4 Cor. 17.16.2 if $f\colon X\to Y$ is locally of finite presentation and faithfully flat then it admits a section after a faithfully flat base change. Therefore, in this case $h\_X$ surjects over $h\_Y$ on the (small or big) fppf site.
**Remark 2.** The property in the proposition is equivalent to the surjectivity of $h\_X\to h\_Y$. Indeed, the surjectivity implies that $f$ admits a section étale locally on $Y$.
| 1 | https://mathoverflow.net/users/88385 | 428270 | 173,645 |
https://mathoverflow.net/questions/428259 | 11 | Let $H\_1,\ldots,H\_n$ be hyperplanes in $\Bbb R^d$. Denote $\mathcal{H} :=\{H\_1,\ldots,H\_n\}$ and let $c(\mathcal{H})$ be the number of regions in the complement: $\Bbb R^d\setminus \bigcup H\_i$.
**Question:** What is the complexity of computing $c(\mathcal{H})$?
Here we are assuming that $H\_i$ are defined explicitly over $\Bbb Q$, and that the dimension $d$ is NOT bounded. For a fixed $d$ there is plenty of literature, see e.g. Halperin-Sharir [Arrangements](http://www.csun.edu/%7Ectoth/Handbook/chap28.pdf) survey. As far as I can tell, none of that literature is applicable.
Note that for *graphical arrangements* $\{x\_i-x\_j=0 : (ij)\in E\}$ corresponding to the graph $G=([n],E)$, the number of regions $c(\mathcal H)$ is an evaluation of the chromatic (and therefore Tutte) polynomial, and thus #P-hard. See e.g. Welsh's ``[Complexity: knots, colouring and counting](https://www.cambridge.org/core/books/complexity-knots-colourings-and-countings/84DC92FA83A7B5A39231D8396321D45A)'' book, Chapter 6.
**Comment:** It feels like this should be well known, so maybe this is a reference request. The problem is in PSPACE and feels similar to $\exists \Bbb R$ (see [Wikipedia page](https://en.wikipedia.org/wiki/Existential_theory_of_the_reals)), except it's a counting problem. Is it $\exists \Bbb R$-hard, for example?
| https://mathoverflow.net/users/4040 | Complexity of counting regions in hyperplane arrangements | The problem is $\#\mathsf{P}$-complete. As you already noted, the problem is $\#\mathsf{P}$-hard even when we restrict to graphical arrangements, so it remains to show that the problem is in $\#\mathsf{P}$. Directly from the definition of $\#\mathsf{P}$, we see that it suffices to show that we can give a short description of a region of a given arrangement whose regionhood can be verified in polynomial time. Every region has a set of facets, so we can describe a region by a linear program that lists the facets and specifies on which side of the facet the region lies. To verify that an alleged region really is a region, we need to check that (a) the linear program has a solution; (b) the solution set is full-dimensional; (c) for every hyperplane $H$ in the arrangement not listed as a facet, one of the two constraints "above $H$" and "below $H$" would be redundant if added to the linear program. The theory of linear programming tells us that all these checks can be performed in polynomial time. For example, for part (b), we can check, for each coordinate direction $x\_i$, that maximizing $x\_i$ and minimizing $x\_i$ yield different answers.
| 9 | https://mathoverflow.net/users/3106 | 428272 | 173,647 |
https://mathoverflow.net/questions/428224 | 2 | Let $\alpha>1$ be a constant and define $B\_n$ as the number of (labeled) *balanced* graphs with $n$ vertices and $\left\lceil \alpha n\right\rceil $ edges. The paper [Strongly Balanced Graphs
and Random Graphs](https://people.clas.ufl.edu/avince/files/StronglyBalanced.pdf "Ruciński, A. and Vince, A. (1986). J. Graph Theory, 10: 251-264. doi:10.1002/jgt.3190100214. zbMATH review at https://zbmath.org/?q=an:0651.05057") by A. Ruciński and A. Vince implies that $B\_n\geq1$. What better lower bounds are known? In particular, is it known to be bounded from below by $n^{(1+o(1))\alpha n}$?
Edit: adding a definition of balanced graphs. The *density* of a graph $G$ with $v$ vertices and $e$ edges is $$\mathrm{den}(G)=\frac{e}{v}$$
A graph $G$ is *balanced* if $\mathrm{den}(G)\geq \mathrm{den}(H)$ for every subgraph $H$ of $G$. Balanced graphs appear naturally when counting copies of a given graph in the random graph, as explained in the paper I referred to.
| https://mathoverflow.net/users/163685 | Lower bound on the number of balanced graphs | The bound of Ruciński and Vince is for *strongly balanced*, which is a more strict condition. If only *balanced* is required, the example of connected regular graphs provides a bound much greater than $n^{\Omega(n)}$.
The total number of regular graphs with $n$ vertices is
$$\alpha(n) \frac{2^{n^2/2}\sqrt{2e}}{\pi^{n/2} n^{n/2}},$$
where $\alpha(n)$ is a constant depending on $n\pmod{4}$.
B. D. McKay and N. C. Wormald, [Asymptotic enumeration by degree sequence of graphs of high degree](https://www.sciencedirect.com/science/article/pii/S019566981380042X), European J. Combin., 11 (1990) 565-580.
Note that the formula is stated for all regular graphs, not necessarily connected, but the vast majority of regular graphs are connected. (All degrees apart from 0,1,2 guarantee almost sure connectivity.)
| 2 | https://mathoverflow.net/users/9025 | 428273 | 173,648 |
https://mathoverflow.net/questions/384146 | 4 | The question is inspired by G. Rzadkowski and M. Urlinska's examples in their paper [A Generalization of the Eulerian Numbers](https://arxiv.org/abs/1612.06635). They refer to the discussion
[Expressions involving Eulerian numbers of the second kind](http://mathoverflow.net/questions/45756/), with Pietro Majer's
[answer](https://mathoverflow.net/q/46126) being particularly relevant.
There are slightly different definitions of the Eulerian numbers, we follow the definitions in GKP, Concrete Mathematics, (6.35) and (6.41).
In the OEIS the numbers are listed as [A173018](https://oeis.org/A173018) and [A201637](https://oeis.org/A201637).
The first-order Eulerian numbers are defined as
$$ \left\langle n\atop k \right\rangle = (k+1) \left\langle n-1\atop k \right\rangle + (n-k) \left\langle n-1\atop k-1 \right\rangle, $$
with boundary conditions $\left\langle 0\atop 0 \right\rangle=1$, $\left\langle n\atop k \right\rangle =0$ for $k<0$ or $k > n$.
The second-order Eulerian numbers are defined by the recurrence
$$ \left\langle\!\!\left\langle n\atop k\right\rangle\!\!\right\rangle = (k+1) \left\langle\!\!\left\langle n-1\atop k\right\rangle\!\!\right\rangle + (2n-k-1) \left\langle\!\!\left\langle n-1\atop k-1\right\rangle\!\!\right\rangle . $$
The same boundary conditions as for the first-order numbers apply.
Based on a hypothesis about the representation of the Bernoulli numbers (formula 29), one can derive from Rzadkowski and Urlinska's formula 20 and the last formula in their paper:
$$ \frac{1}{n+1}\, \sum\_{k=0}^{n-1} (-1)^{k} \frac{ \left\langle n\atop k \right\rangle }{ \binom{n}{k} } = \frac{1}{2}\sum\limits\_{k=0}^{n-1}(-1)^k \frac{\left\langle\!\!\left\langle n-1\atop k \right\rangle\!\!\right\rangle} { \binom {2n-1}{k+1}} \quad(n \ge 0). $$
Apparently, both sides have for $n \ge 1$ the value $B\_{n}(1)$, where $B\_{n}(x)$ denotes the Bernoulli polynomials. Can this equation be proved, or can the relation between the two kinds of Eulerian numbers be expressed more succinctly?
**Edit**: Definitions adapted to GKP, Concrete Mathematics and adjusted the conclusion to them.
| https://mathoverflow.net/users/174295 | How are the Eulerian numbers of the first-order related to the Eulerian numbers of the second-order? | The identity is valid. This is a corollary to a proof of Amy M. Fu, [Some Identities Related to the Second-Order Eulerian Numbers](https://arxiv.org/abs/2104.09316).
A second proof follows from recent work of Cormac O'Sullivan, [Stirling's approximation and a hidden link between two of Ramanujan's approximations](https://arxiv.org/abs/2208.02898).
**Postscript**: Following Donald Knuth's definition of the Bernoulli numbers, (note that Knuth switched to $B\_1 = \frac12$ in TAOCP, vol. 1 since 47-th printing, Oct. 2021, and in Concrete Mathematics, since 34-th printing, Jan. 2022), we can sum up:
Both sides of the identity represent the Bernoulli numbers, except $B\_0$.
| 0 | https://mathoverflow.net/users/174295 | 428286 | 173,652 |
https://mathoverflow.net/questions/428288 | 3 | Serre's criterion says that for a scheme to be normal is equivalent to it being $R\_1$ (i.e. regular in codimension $1$) and $S\_2$ (i.e. regular functions on $X-Y$ extend to $Y$ if $Y$ has codimension at least $2$).
What would be examples of:
1. a scheme which is $R\_1$, but not $S\_2$ (i.e. not normal)?
2. a scheme which is $S\_2$, but not $R\_1$ (i.e. not normal)?
| https://mathoverflow.net/users/198139 | Normal schemes and Serre's criterion | 1. Glue two planes at a point, i.e., take take $\mathrm{Spec}(A)$, where
$$
A = \{(f,g) \in k[x\_1,x\_2] \oplus k[y\_1,y\_2] \mid f(0,0) = g(0,0) \}.
$$
2. Take any singular curve, e.g., $\mathrm{Spec}(k[x,y]/xy)$.
| 7 | https://mathoverflow.net/users/4428 | 428291 | 173,654 |
https://mathoverflow.net/questions/428256 | 3 | Let $\Sigma\_g$ denote a Riemann surface and let $X$ denote the complex surface $\Sigma\_g \times \Sigma\_g$. Then can there exist holomorphic embeddings of $\Sigma\_l$ into $X$ for $l < g$?
What about in the symplectic category i.e
if $\omega$ denotes the area 1 form on $\Sigma\_g$ and we equip $X = \Sigma\_g \times \Sigma\_g$ with the form $\omega \oplus \omega$. Then does there exist a symplectic embedding $\Sigma\_l$ into $X$ for $l < g$?
| https://mathoverflow.net/users/92483 | Holomorphic/Symplectic embedding of Riemann surfaces | I am just posting my comment as one answer.
**Lemma.** For integers $\ell < g$, for every continuous map from $\Sigma\_\ell$ to $\Sigma\_g$, the pullback map on $H^2$ is the zero map.
**Proof.** The pullback map on $H^1$ has rank no greater than $2\ell$, since that is the rank of $H^1(\Sigma\_\ell)$. Since $H^1(\Sigma\_g)$ has rank $2g>2\ell$, there exists a nonzero element $\alpha$ in the kernel. Since the cup product pairing on $H^1(\Sigma\_g)$ is nondegenerate, there is an element $\beta$ in $H^1(\Sigma\_g)$ such that $\alpha\cup \beta$ is nonzero in $H^2(\Sigma\_g)$. Since the pullback map on cohomology is a ring homomorphism, the pullback of $\alpha\cup \beta$ is zero. Since $H^2(\Sigma\_\ell)$ is torsion-free, and since $\alpha\cup \beta$ is a nonzero multiple of the generator of $H^2(\Sigma\_g)$, the pullback map on $H^2$ is zero. **QED**
Thus, for a generator $\omega$ of $H^2(\Sigma\_g,\mathbb{R})$, for every continuous function from $\Sigma\_\ell$ to $\Sigma\_g\times \Sigma\_g$, the pullback of $\omega\oplus \omega$ is zero in $H^2(\Sigma\_\ell,\mathbb{R})$. Since a symplectic form on $\Sigma\_\ell$ has nonzero cohomology class, there is no differentiable map from $\Sigma\_\ell$ to $\Sigma\_g\times \Sigma\_g$ that pulls back $\omega\oplus \omega$ to a symplectic form on $\Sigma\_\ell$.
As noted by @abx, there is no nonconstant holomorphic map from $\Sigma\_\ell$ to $\Sigma\_g$. The lemma shows that there is not even a map that pulls back $\omega$ to a differential form with nonzero cohomology class (every nonconstant holomorphic map pulls back $\omega$ to a differential form with nonzero cohomology class).
| 3 | https://mathoverflow.net/users/13265 | 428294 | 173,656 |
https://mathoverflow.net/questions/428278 | 4 | Let $G$ be a group. I have two questions about the homology of $G$:
1. Consider a finite exact sequence
$$0 \rightarrow M\_1 \rightarrow \cdots \rightarrow M\_m \rightarrow 0$$
of $G$-modules. How are the homology groups $H\_k(G;M\_i)$ related?
2. Consider a filtration of $G$-modules
$$0 = N\_0 \subset N\_1 \subset \cdots \subset N\_n = N.$$
How can I relate $H\_k(G;N)$ to the homology groups $H\_k(G;N\_{i+1}/N\_i)$?
These feel related since the simplest cases of both of them reduce to the long exact sequence in group homology associated to a short exact sequence of coefficients. Maybe there is some kind of spectral sequence?
| https://mathoverflow.net/users/489452 | Groups homology with coefficients fitting into filtration or exact sequence | 1. Think of your exact sequence as a resolution of $M\_m$. It's not necessarily a resolution by free $G$-modules, or by projective $G$-modules; it's just a resolution by $G$-modules. You get a spectral sequence whose $E\_1$-term is the direct sum of the $H\_\*(G; M\_i)$ for all $i<m$, and which converges to $H\_\*(G; M\_m)$.
When the $G$-modules $M\_i$ *are* projective for all $i<m$, then the $E\_1$-page degenerates on to a single line, and then running the spectral sequence reduces to the usual process for calculating $Tor$: take a projective resolution, apply the tensor product (at this point you have recovered the one nonzero line in the $E\_1$-page), then take homology (at this point you have both the $E\_2$-page and the final answer).
This "resolution spectral sequence" is a standard homological tool: see application 5.9.8 in Weibel's homological algebra textbook for its construction.
2. Having a filtration on the coefficient module $N$ yields a filtration on the bar complex, when you tensor $N$ with the bar resolution of $\mathbb{Z}$ by $G$-modules. The resulting filtered chain complex yields a spectral sequence whose input is homology of $G$ with coefficients in the associated graded $G$-module of your filtration of $N$, and whose output is the homology of $G$ with coefficients in $N$. This was suggested in the comments as well. The resulting spectral sequence is also a standard tool, and although I don't know a textbook that discusses it in exactly this level of generality, any treatment of the spectral sequence of a filtered chain complex will be applicable to this one.
| 2 | https://mathoverflow.net/users/nan | 428309 | 173,660 |
https://mathoverflow.net/questions/428311 | 2 | I am wondering what would be the value of $C(f)$ for the following inequality to hold? E.g., $C(f)$ could be some quantity related to the Lipschitz constant or the size of the domain.
$$\left(\int f(x,x) dx\right)^2 \leq C(f)\cdot \iint f(x,y)^2 dxdy$$
| https://mathoverflow.net/users/118287 | For which value of $C(f)$ would the following inequality hold? | The question is not well posed, as it is not quite clear in what terms you want $C(f)$ to be expressed.
If one only uses the terms you did mention -- the Lipschitz constant and the size of the domain, then no finite $C(f)$ exists. Indeed, suppose that the domain, of "size" $a\in(0,\infty)$, is $[0,a]^2$. Take any real $c>0$ and let
\begin{equation\*}
f(x,y):=\max(0,c-|y-x|)
\end{equation\*}
for real $x$ and $y$. Then $f$ is $1$-Lipschitz, the left-hand side of your inequality is
\begin{equation\*}
\text{lhs}:=\Big(\int f(x,x) dx\Big)^2=c^2a^2,
\end{equation\*}
and the double integral on the right-hand side of your inequality is
\begin{equation\*}
\begin{aligned}
\text{rhs}&:=\iint f(x,y)^2\,dx\,dy \\
&\le\int\_0^a dx\,\int\_{-\infty}^\infty dy\,\max(0,c-|y-x|)^2 \\
&=\int\_0^a dx\,\int\_{x-c}^{{x+c}} dy\,(c-|y-x|)^2 \\
&=2\int\_0^a dx\,\int\_0^c dz\,(c-z)^2\le c^3a.
\end{aligned}
\end{equation\*}
So, if your inequality holds, then
\begin{equation\*}
C(f)\ge\frac{\text{lhs}}{\text{rhs}}\ge\frac{c^2a^2}{c^3a}=\frac ac.
\end{equation\*}
Since the "height" $c$ of the function $f$ can be arbitrarily small, there is no finite $C(f)$ expressed in terms of the Lipschitz constant and the size of the domain such that your inequality holds.
---
On a somewhat positive note, suppose that $f$ is $L$-Lipschitz for some $L\in(0,\infty)$, so that $|f(x,y)|\ge\max(0,|f(x,x)|-L|y-x|)$ for all $x,y$. Then
\begin{equation\*}
\begin{aligned}
\iint f(x,y)^2\,dx\,dy &\ge\int dx\,\int dy\,\max(0,|f(x,x)|-L|y-x|)^2 \\
&=\frac23\,\int dx\,\frac{|f(x,x)|^3}L \\
&=\frac2{3L}\,\int dx\,|f(x,x)|^3.
\end{aligned}
\tag{5}\label{5}
\end{equation\*}
**Remark:** We see that, perhaps unexpectedly, the third power of $f$ appears here. However, the "physical" dimensions of $\iint f(x,y)^2\,dx\,dy$ and $\frac2{3L}\,\int dx\,|f(x,x)|^3$ are the same: If, say, we measure $f$ in grams (g) and $x,y$ in centimeters (cm), then $\iint f(x,y)^2\,dx\,dy$ and $\frac2{3L}\,\int dx\,|f(x,x)|^3$ will both be measured in g$^2\,$cm$^2$, because $L$ will be measured in g/cm. This confirms that the third power of $f$ is the right one in $\int dx\,|f(x,x)|^3$.
(I believe that the [centimeter-gram-second system of units (CGS)](https://en.wikipedia.org/wiki/Centimetre%E2%80%93gram%E2%80%93second_system_of_units#:%7E:text=The%20centimetre%E2%80%93gram%E2%80%93second%20system,as%20the%20unit%20of%20time.) -- with its three base units -- is much better than the (unfortunately) universally accepted [International System of Units (Si)](https://en.wikipedia.org/wiki/International_System_of_Units) with its seven base units and a myriad derived units. :-))
Assuming that the "size" of the domain understood as the Lebesgue measure $|D|$ of the set $D:=\{x\colon f(x,x)\ne0\}$ is finite, by the Hölder inequality we get
\begin{equation\*}
\int dx\,|f(x,x)|^3=\int\_D dx\,|f(x,x)|^3\ge\frac1{|D|^2}\,\Big(\int\_D dx\,|f(x,x)|\Big)^3.
\end{equation\*}
So,
\begin{equation\*}
\Big(\int dx\,f(x,x)\Big)^2\le\Big(\int dx\,|f(x,x)|\Big)^2 \\
\le\Big(\frac{3L|D|^2}2\Big)^{2/3}
\Big(\iint f(x,y)^2\,dx\,dy\Big)^{2/3}. \tag{10}\label{10}
\end{equation\*}
We saw in the first, "negative" part of this answer that the reason why there is no finite $C(f)$ expressed in terms of the Lipschitz constant and the size of the domain such that your inequality holds is that the "height" of the function $f$ can be arbitrarily small. Now, if we try to avoid this by requiring that, say,
\begin{equation\*}
\iint f(x,y)^2\,dx\,dy\ge b^2>0,
\end{equation\*}
then \eqref{10} will yield
\begin{equation\*}
\Big(\int dx\,f(x,x)\Big)^2
\le\Big(\frac{3L|D|^2}{2b^2}\Big)^{2/3}\,
\iint f(x,y)^2\,dx\,dy.
\end{equation\*}
Alternatively, if we try to avoid small $f$'s by requiring that
\begin{equation\*}
\int dx\,|f(x,x)|\ge h>0,
\end{equation\*}
then \eqref{10} will yield
\begin{equation\*}
\Big(\int dx\,f(x,x)\Big)^2
\le\frac{3L|D|^2}{2h}\,
\iint f(x,y)^2\,dx\,dy.
\end{equation\*}
| 1 | https://mathoverflow.net/users/36721 | 428313 | 173,661 |
https://mathoverflow.net/questions/428315 | 3 | Consider the following Sylvester equation, where each of the known coefficient matrices ($A$, $B$, $C$) is symmetric positive definite and has dimensions $n \times n$
\begin{align\*}
C = A^TXA + B^TXB.
\end{align\*}
In my case, the coefficient matrices are such that the equation has a unique real solution which is also symmetric.
I have been trying to prove that the solution must also be positive definite and I am struggling. The struggle makes me think that this is actually not true.
I would be grateful for either a pointer to related literature, a way to prove my hypothesis or a counterexample. If it is not true in general, are there any known conditions on $A$ and $B$ which make this true?
| https://mathoverflow.net/users/168454 | Solution to a Sylvester equation with positive definite coefficients | This is not true. For example, if
$$
A = \begin{bmatrix}
1 & 0 \\ 0 & 4
\end{bmatrix}, B = \begin{bmatrix}
4 & 0 \\ 0 & 1
\end{bmatrix}, C = \begin{bmatrix}
17 & 16 \\ 16 & 17
\end{bmatrix}
$$
then each of $A$, $B$, and $C$ is symmetric and positive definite. However, it is straightforward to check that the unique solution to the Sylvester equation is
$$
X = \begin{bmatrix}
1 & 2 \\ 2 & 1
\end{bmatrix},
$$
which is not positive (semi)definite.
| 7 | https://mathoverflow.net/users/11236 | 428317 | 173,664 |
https://mathoverflow.net/questions/428240 | 3 | I'm currently learning about sheaf theory with Angelo Vistoli’s 2007 [Notes on Grothendieck topologies,
fibered categories and descent theory](http://homepage.sns.it/vistoli/descent.pdf). And in page 35, there is the following definition of a refinement and a subordinate grothendieck topology:
**Refinement:**
Let $C$ be a category, $\{U\_i\xrightarrow{\phi\_i} U\}\_{i\in I}$ a set of arrows. A refinement $\{Va\xrightarrow{\psi\_a} U\}\_{a\in A}$ is a set of arrows such that for each index $a\in A$ there is some index $i\in I$ such that $V\_a\xrightarrow{\psi\_a} U$ factors through $U\_i\xrightarrow{\phi\_i} U$.
**Subordinate Grothendieck Topology:**
Let $C$ be a category, $\mathcal{T}$ and $\mathcal{T'}$ two topologies on $C$. We say that $\mathcal{T}$ is subordinate to $\mathcal{T'}$, and write $\mathcal{T}\prec\mathcal{T'}$, if every covering in $\mathcal{T}$ has a
refinement that is a covering in $\mathcal{T'}$. If $\mathcal{T}\prec\mathcal{T'}$ and $\mathcal{T'}\prec\mathcal{T}$, we say that $\mathcal{T}$ and $\mathcal{T'}$ are equivalent, and write $\mathcal{T}\equiv\mathcal{T'}$.
Now we have the following **main-proposition**: Let $\mathcal{T}$ and $\mathcal{T'}$ be two Grothendieck topologies on the same category $C$. If $\mathcal{T}$ is subordinate to $\mathcal{T'}$, then every sheaf in $\mathcal{T'}$ is also a sheaf in $\mathcal{T}$ .
In particular, two equivalent topologies have the same sheaves.
Vistoli proved this proposition with sieves and I questioned myself: Can it be proven 'easier' without sieves?
What do I mean with 'easier'? The prove with sieves in Vistolis paper uses several statements (Cor. 2.40, Prop. 2.42, Lemma 2.43,Prop. 2.46, Prop. 2.48) and with 'easier' I mean with less theory.
>
> **This part can be ignored, because it is using a false statement.**
> My first idea was using the following statement: Let $F:C^{op}\rightarrow Set$ a presheaf. Then $F$ is a sheaf if and only if the following diagram is an equalizer for all coverings $\{U\_i\xrightarrow{\phi\_i}U\}\_{i\in I}$ in $C$:
> $$ F(U)\rightarrow \prod\_{i}F(U\_i) \rightrightarrows \prod\_{i,j}F(U\_i\times\_UU\_j) $$ where the function $F(U) → \prod\_i F(U\_i)$ is induced by the restrictions $F(U)\xrightarrow{\phi\_i^\*} F(U\_i)$ and $pr\_1^\*:\prod F(U\_i) \rightarrow \prod\_{i,j}F(U\_i\times\_UU\_j)$ and $pr\_2^\*:\prod F(U\_i) \rightarrow \prod\_{i,j}F(U\_i\times\_UU\_j)$. So using that statement I want to proof the following **main-lemma:** Let $C$ be a category, $\mathcal{T}$ and $\mathcal{T'}$ two topologies on $C$ with $\mathcal{T}\prec\mathcal{T'}$ and $F:C^{op}\rightarrow Set$ a sheaf in $\mathcal{T}$. Let $\{U\_i\xrightarrow{\phi\_i} U\}\_{i\in I}$ be a covering in $\mathcal{T}$ and $\{Va\xrightarrow{\psi\_a} U\}\_{a\in A}$ the refinement of $\{U\_i\xrightarrow{\phi\_i} U\}\_{i\in I}$, then the diagram $$ F(U)\rightarrow \prod\_{a}F(V\_a) \rightrightarrows \prod\_{a,b}F(V\_a\times\_UV\_b) $$ is an equalizer. Unfortunately, I don't know a way to prove this lemma.
>
>
>
**Question 1:** Is the way of proving the **main-proposition** without sieves even 'easier' (pretopology)?
**Answer:** No, see Answer of Marc Hoyois.
**Question 2:** How do I prove the **main-lemma**?
**Answer:** The **main-lemma** is wrong, because 'refinements' are not well-defined.
| https://mathoverflow.net/users/485069 | Proof without sieves: Equivalent grothendieck topologies have the same sheaves | Let me break down the statement you are trying to prove into two independent facts. This answer is not really in the spirit of the question since I will make maximal use of sieves, but for such foundational matters I think it is much more efficient to embrace sieves rather than to avoid them (so my subjective answer to Question 1 is no: it is not easier without sieves). Let me also point out that the statement of your "main lemma" does not make sense since "the refinement" is not well-defined, and if you mean "any" or "some" refinement then the statement is false (it goes in the wrong direction!).
First, let $\mathfrak U=\{U\_i\to U\}\_i$ be a family of maps in a category $C$. It generates a sieve $\langle\mathfrak U\rangle\subset \operatorname{Hom}(-,U)$, consisting of all maps to $U$ that factor through one of the $U\_i$.
**Fact 1.** For a presheaf $F$ on $C$, $F$ satisfies descent with respect to the family $\mathfrak U$ iff it satisfies descent with respect to the sieve $\langle\mathfrak U\rangle$.
**Proof.** Descent wrt the sieve means that
$$
F(U) \stackrel{\sim}{\to} \operatorname{Hom}(\langle\mathfrak U\rangle, F).
$$
Descent wrt the family means that
$$
F(U) \stackrel{\sim}{\to} \operatorname{lim}\_{n\in\Delta}\operatorname{Hom}(\check C\_n(\mathfrak U),F),
$$
where $\check C\_\bullet(\mathfrak U)\to \operatorname{Hom}(-,U)$ is the Čech nerve of the morphism of presheaves $\coprod\_i \operatorname{Hom}(-,U\_i) \to \operatorname{Hom}(-,U)$. But the sieve $\langle\mathfrak U\rangle \subset \operatorname{Hom}(-,U)$ is by definition the image of this morphism, hence the colimit of its Čech nerve. So then
$$
\operatorname{Hom}(\langle\mathfrak U\rangle, F) = \operatorname{Hom}(\operatorname{colim}\_{n\in\Delta^{\operatorname{op}}}\check C\_n(\mathfrak U),F) = \operatorname{lim}\_{n\in\Delta}\operatorname{Hom}(\check C\_n(\mathfrak U),F),
$$
which proves the claim.
Second, define a *quasi-topology* on a category $C$ to be a collection of sieves which is stable under pullbacks. For example, the sieves generated by the covering families of a pretopology form a quasi-topology.
**Fact 2.** Let $J$ be a quasi-topology on $C$ and $F$ a presheaf on $C$. Then $F$ is a sheaf for $J$ (i.e., satisfies descent wrt all sieves in $J$) iff it is a sheaf for the Grothendieck topology $\bar J$ generated by $J$.
**Proof.** Let $K$ be the finest quasi-topology on $C$ such that every $J$-sheaf is a $K$-sheaf: a sieve $R\subset \operatorname{Hom}(-,U)$ is in $K$ iff for every morphism $f\colon V\to U$ in $C$, every $J$-sheaf satisfies descent wrt $f^\*(R)$. Tautologically $J\subset K$. It will suffice to show that $K$ is a topology, so that $\bar J\subset K$, so that every $K$-sheaf (in particular every $J$-sheaf) is a $\bar J$-sheaf. Only the "local character" of $K$ is not given: if $R,S$ are sieves on $U$ such that $S$ is in $K$ and $f^\*(R)$ is in $K$ for every $f$ in $S$, then $R$ must be in $K$. Since $K$ is stable under pullbacks, it suffices to show that every $J$-sheaf $F$ satisfies descent wrt $R$. To see this consider the intersection $R\cap S$. Since $S$ is the colimit of the representable presheaves $\operatorname{Hom}(-,V)$ with $f\colon V\to U$ in $S$, $R\cap S$ is correspondingly the colimit of $f^\*(R)$ for $f$ in $S$. By assumption each map $f^\*(R)\hookrightarrow \operatorname{Hom}(-,V)$ becomes an isomorphism after applying $\operatorname{Hom}(-,F)$, hence also the map $R\cap S\hookrightarrow S$. Similarly, $R$ is the colimit of $\operatorname{Hom}(-,V)$ for $f\colon V\to U$ in $R$, so $R\cap S$ is the colimit of $f^\*(S)$ for $f$ in $R$. Since $K$ is a quasi-topology, the sieve $f^\*(S)$ on $V$ is in $K$, so we deduce as before that the inclusion $R\cap S\hookrightarrow R$ induces an isomorphism after applying $\operatorname{Hom}(-,F)$. Since both inclusions $R\cap S\hookrightarrow S$ and $R\cap S\hookrightarrow R$ as well as $S\hookrightarrow \operatorname{Hom}(-,U)$ become isomorphisms, so does the inclusion $R\hookrightarrow \operatorname{Hom}(-,U)$, as desired.
**Remark.** Fact 1 is Lemma C2.1.3 in Johnstone's Elephant and Fact 2 is a special case of Proposition C2.1.9. The statement of Fact 2 (and its proof) is from SGA4 Corollary II.2.3. I wrote both proofs in a way that makes sense for presheaves of animas on an ∞-category $C$. For presheaves of sets we can improve Fact 2 by weakening the notion of quasi-topology to Johnstone's notion of sifted coverage, which is a collection of sieves with the only requirement that the pullback of a covering sieve contains a covering sieve. However, I do not know if (and do not expect that) this refined statement holds for presheaves of animas.
| 5 | https://mathoverflow.net/users/20233 | 428343 | 173,673 |
https://mathoverflow.net/questions/428346 | 7 | Consider a set $X\subseteq \mathbb{R}$ such that
1. $X$ is *not* separable wrt its subspace topology
2. For all $r\in\mathbb{R}$ there exists a sequence $(x\_n)\_{n\in\omega} \subset X$ converging to $r$
In a model containing such a set $\text{AC}\_\omega(X)$ (choice for countable families of non-empty subsets of $X$) would of course fail, but not that critically.
For example, the unique way I've seen to prove the consistency of the existence of a non-separable set of reals is the one that shows the consistency of an infinite, Dedekind-finite set of reals, but our set, though being non-separable, is well-behaved enough to witness density in a sequencial manner.
My questions are:
* Is its existence consistent relative to $\text{ZF}$? Has it been proved somehwere?
* In case the answer above is "no", does this remind you similar results (besides the most known ones that can be found in Jech' *Axiom of Choice*)?
Thanks!
| https://mathoverflow.net/users/141146 | Consistency of a strange (choice-wise) set of reals | The existence of such a set follows from $``\mathbb{R}$ is a countable union of countable sets.$"$ Let $\mathbb{R} = \bigcup\_{n<\omega} S\_n,$ each $S\_n$ countable. Let $T\_n = \{x \in \mathbb{R}: \exists m \le n \exists y \in S\_m (x \le\_T y)\}$ and $X\_n = (2^{-n-1}, 2^{-n}) \setminus T\_n.$
We will show $X = \bigcup\_{n<\omega} X\_n$ is a subset of $(0,1)$ with the desired properties. Condition (2) follows from $X$ having cocountable intersection with each $(2^{-n}, 1).$ Suppose condition (1) fails. Let $r \in S\_n$ encode a dense sequence $\langle r\_i: i<\omega \rangle \subset X.$ Then $r\_i \in T\_n$ for all $i,$ so $2^{-n}<r\_i,$ contradiction.
Edit:
It turns out the nonexistence of such a set is equivalent to $\text{AC}\_{\omega}(\mathbb{R}).$ First, $\text{AC}\_{\omega}(\mathbb{R})$ implies every set of reals is separable. For the other direction, suppose $\langle S\_n: n<\omega \rangle$ is a sequence of nonempty sets of reals without a choice function. Let $T\_n = \{x \in \mathbb{R}: \forall m \le n \exists y \in S\_m (y \le\_T x)\}$ and $X\_n = (2^{-n-1}, 2^{-n}) \cap T\_n.$
We will show $X = \bigcup\_{n<\omega} X\_n$ is a subset of $(0,1)$ with the desired properties. Condition (2) follows from the fact that each $T\_n$ is nonempty and closed under addition by rational numbers. Suppose condition (1) fails. Let $r$ encode a dense sequence $\langle r\_i: i<\omega \rangle \subset X.$ Then $\{x \in \mathbb{R}: x \le\_T r\}$ is a countable set which meets each $S\_n,$ which contradicts the fact that $\langle S\_n \rangle$ has no choice function.
| 13 | https://mathoverflow.net/users/109573 | 428354 | 173,674 |
https://mathoverflow.net/questions/428308 | 2 | I have a sparse square matrix and want to see if it is full rank (so that I can apply the implicit function theorem).
$$\left[\begin{array}{cccccccccc}
0 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0\\
x\_{1}^{2} & Nx\_{1} & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0\\
0 & c & 0 & 0 & 0 & 0 & 0 & -x^2\_{1} & 0 & 0\\
x\_{2}^{2} & 0 & Nx\_{2} & 0 & 0 & -1 & 0 & 0 & 0 & 0\\
0 & 0 & c & 0 & 0 & 0 & 0 & 0 & -x^2\_{2} & 0\\
0 & 0 & 0 & 0 & z\_1 & 0 & 0 & -1 & 1 & 0\\
x\_{3}^{2} & 0 & 0 & Nx\_{3} & 0 & 0 & -1 & 0 & 0 & 0\\
0 & 0 & 0 & c & 0 & 0 & 0 & 0 & 0 & -x^2\_{3}\\
0 & 0 & 0 & 0 & z\_2 & z\_2 & 0 & 0 & -1 & 1
\end{array}\right]$$
where all variables are strictly positive, and $\sum x\_i=1$.
Given that it is sparse, one approach that we considered is to do row-reductions and rearranging to reduce it to a block matrix. This is possible and yields:
$$\left[\begin{array}{ccc}
A & B & 0 \\
0 & C & D \\
E & 0 & F\end{array}\right]=\left[\begin{array}{ccc|ccc|ccc}
N & 0 & 0 & x\_{1}-1 & x\_{1} & x\_{1} & 0 & 0 & 0\\
0 & N & 0 & x\_{2} &x\_2-1 & x\_{2} & 0 & 0 & 0\\
0 & 0 & N & x\_{3} & x\_{3} & x\_3-1 & 0 & 0 & 0\\ \hline
0 & 0 & 0 & x\_1z\_1 & 0 & 0 & -1 & 1 & 0\\
0 & 0 & 0 & x\_1z\_2 & x\_2z\_2 & 0 & 0 & -1 & 1\\
0 & 0 & 0 & x\_1 & x\_2 & x\_3 & 0 & 0 & 0\\ \hline
c & 0 & 0 & 0 & 0 & 0 & -x\_{1}^2 & 0 & 0\\
0 & c & 0 & 0 & 0 & 0 & 0 & -x\_{2}^2 & 0\\
0 & 0 & c & 0 & 0 & 0 & 0 & 0 & -x\_{3}^2
\end{array}\right]$$
| https://mathoverflow.net/users/119725 | Full-rank matrix | OK, let's call the block matrix above $M$. First eliminate $N$ by a substitution $c\mapsto d N$. Then substitute $z\_i \mapsto d y\_i$ to eliminate $d$. Then you can construct the Schur complement w.r.t. the first and last rows/columns of $M$ to get $\det(M) = -c^2 N x\_1^2 x\_2^2 x\_3^2 \det P$, with
$$
P =
\begin{pmatrix}
x\_1 y\_1 + x\_1^{-1}-x\_2^{-1}-{x\_1^{-2}} & x\_1^{-1}-x\_2^{-1}+x\_2^{-2} &
x\_1^{-1}-x\_2^{-1} \\
x\_1 y\_2+x\_2^{-1}-x\_3^{-1} & x\_2 y\_2+x\_2^{-1}-x\_3^{-1}-x\_2^{-2} &
x\_2^{-1}-x\_3^{-1}+x\_3^{-2} \\
x\_1 & x\_2 & x\_3 \\
\end{pmatrix}
$$
It might be easier to discuss this determinant.
| 5 | https://mathoverflow.net/users/90413 | 428358 | 173,675 |
https://mathoverflow.net/questions/427876 | 5 | Given any topological space $(X,\tau)$, it seems to me that the uniformity generated by all continuous pseudometrics $d:(X,\tau)\times (X,\tau)\to [0,\infty)$ is the *fine uniformity* of the associated completely regular topology of $\tau$. This would yield an easy desciption of the left adjoint of the functor assigning to a uniform space the generated topological space (a quick search in the web only yields such a left adjoint as the composition of the adjoints of CompReg $\to$ Top and Unif $\to$ CompReg).
**Is this correct and do you have a reference?**
| https://mathoverflow.net/users/21051 | Is the fine uniformity generated by all continuous pseudometrics? | It is true, and I learned of many of the details in the somewhat obscure paper
>
> D. Thampuran, [*On Completely Regular Spaces*](https://purl.pt/2750/1/j-5293-b-vol33-fasc4-art4_PDF/j-5293-b-vol33-fasc4-art4_PDF_01-B-R0300/j-5293-b-vol33-fasc4-art4_0000_capa1-208_t01-B-R0300.pdf), Portugaliae Mathematica \**33*, (1974).
>
>
>
Here is the basic idea of how things work.
The full subcategory $CReg$ on the completely regular spaces is bireflective in $Top$. The reflector sends a space $(X,\tau)$ to its completely regular modification $cr(X)$, which is the underlying set $X$ given the topology generated by all cozero sets of $(X,\tau)$.
On the other hand, there is a descrition of $cr(X)$ in terms of continuous pseudometrics. First observe the following.
>
> *Lemma:* A pseudometric $d:X\times X\rightarrow[0,\infty)$ is $\tau$-continuous if and only if the topology generated by $d$ is weaker than $\tau$.
>
>
>
Thus under $\tau$ in the lattice of topologies on $X$ is a family of pseudometric topologies. The supremum of these topologies need not be pseudometric, but will of course be completely regular. This is true because each pseudometric topology is completely regular, and the supremum of any family of completely regular topologies is completely regular.
>
> *Lemma:* A set $U\subseteq X$ is a cozero set if and only if it is an open subset of some weaker topology on $X$ which is generated by a continuous pseudometric.
>
>
>
Putting these observations together yields the following.
>
> *Corollary:* Let $\mathcal{D}$ be the set of all continuous pseudometrics on $(X,\tau)$. Then the completely regular modification $cr(X)$ is the coarsets topology making each member of $\mathcal{D}$ continuous.
>
>
>
So now the affirmiative answer of Jochen's query is clear. The *fine uniformity* on a completely regular space $(X,\tau)$ is the supremum (in the lattice of uniformities) of all uniformities on $X$ which are compatible with $\tau$. This uniformity is generated by the family of all continuous pseudometics on $(X,\tau)$, a fact which appears as exercise 8.1.C.c in Engelking's *General Topology* (pg.437 of my edition). (Engelking works only with separated uniformities, but none of the details of the exercise require this.)
On the other hand, by the above discussion, if $(X,\tau)$ is not completely regular, then its family of continuous pseudometrics describes its completely regular modification. Thus the composite functor
$$Top\rightarrow CReg\rightarrow Unif$$
is indeed described in the single process given in Jochen's question.
| 3 | https://mathoverflow.net/users/54788 | 428368 | 173,677 |
https://mathoverflow.net/questions/428375 | 5 | This line
>
> The symbol may naturally be thought of as an element in the K-theory
> of X
>
>
>
appears in the [nLab page on principal symbols](https://ncatlab.org/nlab/show/symbol+of+a+differential+operator) for differential operators. What does this mean? Are they talking about K-theory or K-homology? How does one produce a class from the symbol of the operator?
| https://mathoverflow.net/users/153228 | The principal symbol as an element in the K-theory | It's a bit easier to see this using a slightly non-standard definition of topological K-theory. Given a locally compact Hausdorff space $X$, let $\bf{E}$ be a complex of vector bundles, i.e. a sequence
$$0 \to E\_0 \xrightarrow{\alpha\_0} E\_1 \xrightarrow{\alpha\_1} \ldots \xrightarrow{\alpha\_{n-1}} E\_n \to 0$$
where the $\alpha\_i$'s are bundle maps and $\alpha\_{i+1} \circ \alpha\_i = 0$. The *support* of $\bf{E}$ is by definition the set of all $x \in X$ such that the fiber of $\bf{E}$ over $x$ is not exact. A *homotopy* between complexes $\bf{E}$ and $\bf{F}$ is a complex over $X \times [0,1]$ whose restriction to $X \times 0$ is isomorphic to $\bf{E}$ and whose restriction to $X \times 1$ is isomorphic to $\bf{F}$. Finally, declare that two compactly supported complexes of vector bundles over $X$ are equivalent if there is a compactly supported homotopy between them, and let $C(X)$ denote the set of all equivalence classes.
$C(X)$ is an abelian group under Whitney sum of complexes, and it has a subgroup $C\_0(X)$ consisting of complexes with empty support, i.e. the complex is exact over every point in $X$.
**Proposition:** $K(X) \cong C(X) / C\_0(X)$
There is a proof in Atiyah's book on K-theory, for example.
With that in hand, let $D$ be an elliptic operator mapping smooth sections of a vector bundle $E$ to smooth sections of a vector bundle $F$ over the same compact base manifold $M$. Let $\pi \colon TM \to M$ denote the tangent bundle of $M$. For each $x \in M$ and $V \in T\_x M$, the symbol of $D$ is a linear map
$$\sigma(x, V) \colon E\_x \to F\_x$$
which varies smoothly in $x$ and $V$. Thus it defines a complex of vector bundles over TM:
$$0 \to \pi^\* E \xrightarrow{\sigma} \pi^\* F \to 0$$
The condition that $D$ is elliptic means that $\sigma$ is invertible - and hence the complex above is exact - outside of the zero section of $TM$, which is compact since $M$ is compact. So by the definition of K-theory above we get a class $[\sigma] \in K(TM)$.
| 14 | https://mathoverflow.net/users/4362 | 428382 | 173,682 |
https://mathoverflow.net/questions/427601 | 7 | Let $\epsilon <1/2$. Let $X$ be a random variable in $\mathbb Z$ such that $\mathbb P (X=x)\le \epsilon $ for any $x\in \mathbb Z$ (you may add any moment or regularity conditions on $X$ if needed). Let $S\_n$ be a sum of $n$ independent copies of $X$. Show that for any $x\in \mathbb Z$
$$\mathbb P (S\_n=x) \le C\cdot \epsilon \cdot n^{-1/2}$$
for some universal constant $C$.
I'm also looking for high dimensional generalizations of this claim in which the factor $n^{-1/2}$ is replaced with $n^{-d/2}$ where $d$ is the dimension (I'm thinking of $d$ as fixed). Clearly, for such a statement to hold we need an assumption saying that $X$ is "truly $d$ dimensional". I wonder if the following statement is correct:
Let $M$ large and let $X$ be a random variable in $\mathbb Z ^d$ such that $\mathbb P (X=x)\le M^{-d/2}$ for any $x\in \mathbb Z ^d$ and $\mathbb P(X\cdot v=y) \le 1/M$ for any vector $v\in \mathbb R ^d$ and $y>0$. Then, for any $x\in \mathbb Z ^d$ we have
$$\mathbb P (S\_n=x) \le C\_d (Mn)^{-d/2}$$
for some constant $C\_d$ depending only on $d$.
| https://mathoverflow.net/users/161778 | Local probabilities for lattice random walk | For the one dimensional case, a quite nice bound is in Theorem 4.2 of [1]. See also [2]. The dependence on $\epsilon$ that you seek was first shown by Kesten[3].
The combinatorial approach was revived in [4]. The sharpest result is quite recent, see [5], which also identifies the worst case. Look there first.
[1] Esseen, Carl-Gustav. "On the concentration function of a sum of independent random variables." Zeitschrift für Wahrscheinlichkeitstheorie und Verwandte Gebiete 9, no. 4 (1968): 290-308. <https://link.springer.com/content/pdf/10.1007/BF00531753.pdf>
[2] <https://link.springer.com/article/10.1023/A:1022654631571>
[3] KESTEN, H. (1969). A sharper form of the Doeblin-Lévy-Kolmogorov-Rogozin inequality for concentration functions. Math. Scand. 25 133–144.
[4] LEADER, I. and RADCLIFFE, A. J. (1994). Littlewood-Offord inequalities for random variables. SIAM J.
Discrete Math. 7 90–101.
[5] <https://arxiv.org/pdf/2201.09861.pdf>
| 4 | https://mathoverflow.net/users/7691 | 428389 | 173,685 |
https://mathoverflow.net/questions/427528 | -1 | This posting is a follow up of [this](https://mathoverflow.net/questions/427351/can-we-write-tangled-type-theory-without-reference-to-type-sequences)
Language multi-sorted FOL, with sorts (types) indexed by the naturals, equality symbol restricted to same type, while membership symbol restricted from lower to higher types, i.e. $x^n\_i=x^m\_j$ is well formed formula only when $n=m$, and $x^n\_i \in x^m\_j$ is well formed only when $n<m$
A type-graph on formula $\phi$ is a graph on the type indices used in $\phi$, this is precisely defined (meta-theoretically) here as a set of singletons and unordered pairs where each pair in it is the Boolean union of two distinct singletons in it, the nodes are singletons of types, the edges are pairs of types. An acyclic graph here can be defined as a graph having no subgraph of it having as many edges as nodes. A linear graph is an acyclic graph in which no node has more than two edges springing from it, i.e. no singleton in it is a proper subset of more than two distinct pairs in it!
**Axioms:** Multi-sorted ID axioms +
**Extensionality:**
$i=1,2,3,\dotsc; j=0,1,2,\dotsc, j<i \\ \forall x^i \forall y^i: \forall z^j (z^j \in x^i \iff z^j \in y^i) \to x^i=y^i$
**Comprehension:**
$i=1,2,3,\dotsc; j=0,1,2,\dotsc, j< i \\ \exists x^i \forall y^j ( y^j \in x^i \iff \phi ) $
Where $\phi$ is a well formed formula with acyclic type-graph.
Now $\sf TTT$ of Randall Holmes ([see: [Holmes - NF is consistent, p:11](https://arxiv.org/abs/1503.01406), [Holmes - The equivalence of NF-style set theories with “tangled” type theories; the construction of $\omega$-models of predicative NF (and more), p:4-5](https://randall-holmes.github.io/Papers/tangled.pdf)]) seems to be equivalent to restricting $\phi$ in comprehension to well formed formulas having *linear* type-graphs, which is a special case of comprehension here, so this theory is an extension of $\sf TTT$. On the other hand, it is known that if we allow $\phi$ to be any well formed formula in this language then this leads to [inconsistency](https://mathoverflow.net/a/427355/95347).
>
> Is there a clear inconsistency with this theory?
>
>
>
| https://mathoverflow.net/users/95347 | Is there an obvious inconsistency with this extension of Tangled Type Theory? | I do not know whether acyclicity avoids paradox here.
| 2 | https://mathoverflow.net/users/485780 | 428390 | 173,686 |
https://mathoverflow.net/questions/428384 | 9 | This is a cross-post! For the original post on SE (9 upvotes, no answer) see:
<https://math.stackexchange.com/questions/4475853/is-a-complex-algebraic-set-with-a-zariski-dense-subset-of-algebraic-points-alrea>
Let $X$ be a complex algebraic set, i.e. the (not necessarily irreducible) vanishing set of some polynomials in $\mathbb{C}[X\_1,\ldots,X\_n]$. If $X$ contains a Zariski dense subset of points with coordinates in the algebraric numbers $\bar{\mathbb{Q}}$, is it true that $X$ is defined over $\bar{\mathbb{Q}}$, i.e. that $X$ can be defined as the vanishing set of some polynomials in $\bar{\mathbb{Q}}[X\_1,\ldots,X\_n]$?
It seems to be the case that the converse of this stament holds, as discussed [here](https://math.stackexchange.com/questions/504766/are-the-algebraic-valued-points-of-a-variety-dense-in-complex-valued-points?noredirect=1&lq=1). I read that the implication in my question follows from Lagrange interpolation, but it is not clear to me how Lagrange interpolation works in the present multivariable case and how it can be used to answer my question.
For context: I try to understand a proof of Mann's theorem in [this](https://math.berkeley.edu/%7Escanlon/papers/csp.pdf) survey on applications of o-minimality (from mathematical logic) to diophantine geometry. We want to prove the following:
**Theorem.** Let $Y\subseteq\mathbb{C}^n$ be an algebraic set and consider $\mathbb{G}=(\mathbb{C}^\times)^n$, the $n$th power of the multiplicative group of units in $\mathbb{C}$. Then the set $Y\cap\mathbb{G}^\mathrm{tor}$ of points on $Y$ such that each coordinate is a root of unity is a finite union of cosets of subgroups of $\mathbb{G}$.
At some point on page 15, we want to use bounds from Galois theory and need to reduce to the case that $Y$ is defined over a number field. In the paragraph in paranthesis on p. 15, the following argument is sketched: From basic properties of the Zariski topology using that $\mathbb{G}$ is open, we can see that $Y\cap\mathbb{G}^\mathrm{tor}=\overline{Y\cap\mathbb{G}^\mathrm{tor}}\cap\mathbb{G}^\mathrm{tor}$, and may thus assume that $Y=\overline{Y\cap\mathbb{G}^\mathrm{tor}}$. Then $Y$ contains a dense set of points with algebraic coordinates, since the elements of $\mathbb{G}^\mathrm{tor}$ have algebraic coordinates. Now the author Scanlon claims that it follows from Lagrange interpolation that $Y$ is defined over $\bar{\mathbb{Q}}$. The fact that $Y$ is defined over a number field then just follows from Noetherianess of $\bar{\mathbb{Q}}[X\_1,\ldots,X\_n]$.
A similar reduction also seems to work in the case of torsion points on abelian varieties defined over $\mathbb{C}$. I would also be grateful for details or references regarding this case.
Thanks for your help!
| https://mathoverflow.net/users/489043 | Is a complex algebraic set with a Zariski dense subset of algebraic points already defined over the algebraic numbers? | More is true. Let $K/F$ be a field extension. Let $X$ be the vanishing set of some polynomials in $K[X\_1,\dots, X\_n]$. If $X$ contains a Zariski dense set of points with coordinates in $F$, then $X$ is the vanishing set of some polynomials in $F$.
Proof: Choose a basis $\alpha\_i$ for $K/F$. Every polynomial in $K[X\_1,\dots, X\_n]$ vanishing on $X$ can be written as a (finite) combination $\sum\_i \alpha\_i f\_i$ with $f\_i \in F[X\_1,\dots, X\_n]$, by decomposing every coefficient in this basis.
For $x\_1,\dots, x\_n\in F$, if $ \sum\_i \alpha\_i f\_i(x\_1,\dots, x\_n)=0$ then $f\_i (x\_1,\dots, x\_n)=0$ for each $i$, by the definition of basis. So the $f\_i$ are polynomials vanishing on a Zariski dense set of points in $X$ and thus vanish on $X$.
Thus the vanishing locus of the $f\_i$, for all the polynomials $\sum\_i \alpha\_i f\_i$ in a set of polynomials defining $X$, is $X$, so $X$ is defined over $F$.
| 14 | https://mathoverflow.net/users/18060 | 428394 | 173,688 |
https://mathoverflow.net/questions/428360 | 4 | There exists an explicit bijection (due to Cayley, that has built up a very nice table to describe this) between the positive roots of the lattice $E\_7$ and $\mathbb{F}\_2^6 \setminus \{0\}$ (where $\mathbb{F}\_2$ is the field with two elements. Btw, this also preserves orthogonality. There is also a relation between $E\_8$ and $\mathbb{F}\_2^8 \setminus \{0\}$. Is there an explicit description of the features of this relation in the literature?
| https://mathoverflow.net/users/4096 | Relation between positive roots of $E_8$ and $\mathbb{F}_2^8 \setminus \{0\}$ | Let $L$ denote the root lattice of $E\_8$. The group $W(E\_8)$ acts (linearly) on $L/2L \cong \mathbb{F}\_2^8$, and hence as a permutation of $2^8=256$ elements. This permutation action has the following three orbits:
1. $\{0\}$, an orbit of size 1.
2. The classes represented by roots, an orbit of size 120. There are 240 roots in $L$, but each root is equivalent to its negative mod $2L$. In particular, each element of this orbit is represented by a unique positive root.
3. The *frames*, an orbit of size 135. The $E\_8$ lattice has $240\cdot9$ vectors of length$^2=4$, and each one is congruent to 16 such vectors (including itself). For example, in a rather standard basis (in which $E\_8 \supset D\_8 \subset \mathbb{Z}^8$), the vector $(2,0^7)$ is congruent to the 16 vectors $(0^a, \pm 2, 0^{7-a})$.
Incidentally, the stabilizer of a root is $W(E\_7)$, whereas the stabilizer of a frame is $W(D\_8)$. You do indeed find that $\#W(E\_8) = 120\cdot \#W(E\_7) = 135 \cdot \#W(D\_8)$.
For further details, see for example Section 8.1.2 of [Berkeley lectures on Lie and quantum groups](http://categorified.net/LieQuantumGroups.pdf). (That section is based on lectures by Richard Borcherds, an expert in lattices.)
| 8 | https://mathoverflow.net/users/78 | 428401 | 173,690 |
https://mathoverflow.net/questions/428406 | 4 | An element $s$ of a group $G$ is a *logical generator* of $G$ iff every element of $G$ can be defined in the first order language of groups with $s$ as a parameter. In this case we may call $G$ a *logically cyclic* group. This is equivalent to say that for any elementary extension $G^{\ast}$ of $G$ and any automorphism $\alpha:G^{\ast}\to G^{\ast}$, the condition $\alpha(s)=s$ implies $\alpha\_{|\_G}=\mathrm{id}$.
As a result, the set $\{ s\}$ is an $\mathrm{Aut}$-basis of $G$ in terms of [1] (i.e., every automorphism of $G$ is uniquely determined by its value on $s$).
I started to study logically cyclic groups in [2]. Recently, I realized that in fact I don't have any example of a group which is not logically cyclic but it has a singleton $\mathrm{Aut}$-basis. So, here is my question:
**Is it true that if $\{ s\}$ is an $\mathrm{Aut}$-basis of $G$, then $s$ is a logical generator of $G$?**
[1] G. Cutolo, C. Nicotera: Subgroups defining automorphisms in locally nilpotent groups, Forum Mathematicum, **15** (2003).
[2] M. Shahryari: On logically cyclic groups, J. Group Theory, **18** (2015).
| https://mathoverflow.net/users/44949 | Logical generators of groups and $\mathrm{Aut}$-bases | $\{1\}$ is an Aut-basis of the group of profinite integers $\hat{\mathbb Z}$, but since this group has cardinality $2^\omega$, it has no logical generator. However, every element of $\hat{\mathbb Z}$ is *type-definable* from $1$ (i.e., any pair of distict elements are distinguishable by a first-order formula with parameter $1$); using Ali Enayat’s terminology, the structure $\langle\hat{\mathbb Z},+,1\rangle$ is *Leibnizian*. (Conversely, if $G$ is the group reduct of a $\mathbb Z$-group such that $\{1\}$ is an Aut-basis of $G$, then all elements of $G$ are type-definable from $1$, and $G$ embeds in $\hat{\mathbb Z}$; see my paper [Rigid models of Presburger arithmetic](https://arxiv.org/abs/1803.05797).)
Similarly, if $p$ is any prime, the group of $p$-adic integers $\mathbb Z\_p$ has no logical generator (being uncountable), but all its elements are type-definable from $1$, thus $\{1\}$ is an Aut-basis.
Note that “every element of $G$ is type-definable from $s$” gives you a general property to consider that is intermediate between “$\{s\}$ is an Aut-basis of $G$”, and “$s$ is a logical generator of $G$”.
| 6 | https://mathoverflow.net/users/12705 | 428413 | 173,692 |
https://mathoverflow.net/questions/428396 | 3 | **1. Question**
How to get from the formulas
$$ \left| \frac{\log 2}{\log 3} - \frac{p}{q} \right| < c\_a\frac{1}{q^{B\_a}} \ \ \ \ \ \ \ \ \ \ \ (1.1)$$
and / or
$$ \left| \frac{\log 2}{\log 3} - \frac{p}{q} \right| \geq c\_b\frac{1}{q^{B\_b}} \ \ \ \ \ \ \ \ \ \ \ (1.2)$$
to the formula
$$ \left| 3^p - 2^q \right| \geq c\_r\frac{1}{q^{B\_r}}3^p\ \ \ \ \ \ \ \ \ \ \ \ (1.3)$$
with constants $c\_a, c\_b, B\_a, B\_b \in \mathbb{R}$ given and $\frac{p}{q} \in \mathbb{Q}$?
It is also important for me how exactly $c\_r, B\_r \in \mathbb{R}$ depend on the other constants $c\_a, c\_b, B\_a$, and $B\_b$. All 6 constants are > 0 and $B\_a, B\_b \in [2, 8)$.
The lower bound in (1.3) is claimed by <https://terrytao.wordpress.com/2011/08/21/hilberts-seventh-problem-and-powers-of-2-and-3/> (proposition 3 and corollary 4) . Terry Tao uses (1.2) with $\geq$ instead of (1.1) with $<$. Maybe both inequalities are required to prove (1.3) for the two cases $3^p > 2^q$ and $3^p < 2^q$.
Something very similar is [the answer by user Lierre to the MO question distance between powers of 2 and powers of 3](https://mathoverflow.net/a/116844/163727), but I would like to see more detailed steps and how to calculate the constants in (1.3).
**2. Ansatz A**
\begin{align}
3^p - 2^q &= 3^p - (3^{\log\_3(2)})^q\\
&= 3^p - 3^{q\cdot \log\_3(2)}\\
&= 3^p - 3^{q\cdot \log\_3(2) - p + p}\\
&= 3^p - 3^p\cdot3^{q\cdot \log\_3(2) - p}\\
&= 3^p - 3^p\cdot3^{q\cdot(\log\_3(2) - \frac{p}{q})}\\
&= 3^p(1 - 3^{q\cdot(\log\_3(2) - \frac{p}{q})}) \ \ \ \ \ \ \ \ \ \ (2.1)\\
\end{align}
Note that
\begin{align}
3^p > 2^q &\iff \log\_3(3^p) > \log\_3(2^q)\\
&\iff p > q\log\_3(2)\\
&\iff \frac{p}{q} > \log\_3(2)\\
&\iff 0 > \log\_3(2) - \frac{p}{q}\\
&\iff \log\_3(2) - \frac{p}{q} < 0 \ \ \ \ \ \ \ \ \ \ (2.2)\\
\end{align}
and analogously
$$2^q > 3^p \iff \log\_3(2) - \frac{p}{q} > 0 \ \ \ \ \ \ \ \ \ \ (2.3)$$.
**3. Ansatz B**
\begin{align}
3^p - 2^q &= 3^p(1 - \frac{2^q}{3^p})\\
&= 3^p(1 - \frac{(e^{\log2})^q}{(e^{\log3})^p})\\
&= 3^p(1 - \frac{e^{q\log2}}{e^{p\log3}})\\
&= 3^p(1 - e^{q\log2 - p\log3}) \ \ \ \ \ \ \ \ \ \ (3.1)\\
\end{align}
| https://mathoverflow.net/users/163727 | Simple estimation of difference of powers of 2 and powers of 3? | This is about bounding $e^x$ such as $e^x \geq 1 + x$.
For example, when $2^q>3^p$ (ie. $\log\_3 2>\frac{p}q$), from (1.2) we have
$$q\log 2 \geq c\_b \frac{\log 3}{q^{B\_b-1}} + p\log 3,$$
which under exponentiation translates to
$$2^q \geq e^{c\_b \frac{\log 3}{q^{B\_b-1}}}3^p.$$
Then
$$2^q - 3^p \geq \big(e^{c\_b \frac{\log 3}{q^{B\_b-1}}}-1\big)3^p \geq c\_b \frac{\log 3}{q^{B\_b-1}}3^p.$$
The other cases are considered similarly.
| 2 | https://mathoverflow.net/users/7076 | 428425 | 173,693 |
https://mathoverflow.net/questions/258394 | 12 | Is it possible to unite two $n$-vertex trees such that the resulting graph has bounded tree-width?
Formally, does there exists a constant $k$ such that given two $n$-vertex trees $T\_1$ and $T\_2$ there exist labeled trees $T\_1'=([n], E\_1)$ and $T\_2' =([n],E\_2)$ with $T\_1' \simeq T\_1$, $T\_2' \simeq T\_2$ and
$$
\texttt{treewidth} (T\_1' \cup T\_2') \leq k,
$$
where $T\_1' \cup T\_2' = ([n], E\_1 \cup E\_2)$.
**Special cases**
1. $T\_1 \simeq P\_n$ and $T\_2 \simeq P\_n$. While one can 'construct' a grid out of two large paths (and a grid has 'large' tree-width), $T\_1$ and $T\_2$ can be united into a path $P\_n$ (which has treewidth 1).
2. $T\_1$ is a star, $T\_2$ is an arbitrary tree. In this case, no matter how we unite the trees we get a chordal graph with maximum clique size 3, and therefore of tree-width 2.
3. It is also possible to unite a path with an arbitrary tree into a graph of bounded tree-width. But it is not clear if this is always possible for a pair of arbitrary trees.
| https://mathoverflow.net/users/83519 | Tree-width of a union of two trees | It is shown in [Bogdan Alecu, Vadim Lozin, Daniel A. Quiroz, Roman Rabinovich, Igor Razgon, Viktor Zamaraev: "The treewidth and pathwidth of graph unions"](https://arxiv.org/abs/2202.07752) that the general question has a negative answer.
| 2 | https://mathoverflow.net/users/151145 | 428426 | 173,694 |
https://mathoverflow.net/questions/428408 | 4 | Consider a norm on $\mathbb C^2$ as $\|(z\_1,z\_2)\|:=\max\{|z\_1|,|z\_2|,\frac{1}{\sqrt{2}}|z\_1+iz\_2|\}.$
*Question.* Is $(\mathbb C^2,\|.\|)$ linearly isometric to $(\mathbb C^2,\|.\|\_{\infty})$ where $\|(z\_1,z\_2)\|\_\infty:=\max\{|z\_1|,|z\_2|\}?$
| https://mathoverflow.net/users/136860 | A space isometric to $\ell_\infty^2$ | There is no such map $f$. Let's try to map from the second space (with the funny norm, which I'll denote simply by $\|\cdot\|$) back to $(\mathbb C^2, \|\cdot \|\_{\infty})$. Let $u=f(e\_1)$, $v=f(e\_2)$, so $\|u\|\_{\infty}=\|v\|\_{\infty}=1$. Since $|1\pm i|^2=2$, so $\|(1,\pm 1)\|=1$, we also have $\|u\pm v\|\_{\infty}=1$. However, if $|z|=1$ and $w\not= 0$, then $|z\pm w|>1$ for one choice of sign. So if (say) $|u\_1|=1$, then $v\_1=0$. It follows that $u=e^{i\alpha}e\_1$, $v=e^{i\beta}e\_2$, or the other way around.
But then $\|f((1,-i)/\sqrt{2})\|\_{\infty}=1/\sqrt{2}$ even though $\|(1,-i)/\sqrt{2}\|=1$.
| 3 | https://mathoverflow.net/users/48839 | 428432 | 173,698 |
https://mathoverflow.net/questions/428438 | 0 | I'm looking at the <https://arxiv.org/abs/1904.09193> paper (version 2, from 2021) and think it has a few errors. I think I found three small places where the paper needs to be corrected (in the sense that the corrected version is valid - none of the key conclusions are undermined by this).
1. On page 5 it says "Lemma 3.5. Let Q ∈ $2^{N\_∞}$ be given. If Q(ω) = 1 and ∀n ∈ N. Q(n) = 1, then ∀p ∈ $2^{N\_∞}$ . Q(p) = 1." I believe the second $2^{N\_∞}$ should be $N\_∞$ (because Q ∈ $2^{N\_∞}$, you'd want to evaluate Q at an element of $N\_∞$)
2. A few lines down is the statement "Theorem 3.6 ([Esc13, Theorem 3.15]). There is a function ε : $2^{N\_∞}$ → $N\_∞$ such that for every
Q ∈ $2^{N\_∞}$ , if Q(ε(Q)) = 1, then ∀p ∈ $2^{N\_∞}$ . Q(p) = 1.". The last $2^{N\_∞}$ should be $N\_∞$ for the same reason.
3. At the end of the proof of that theorem, in "Hence
Q(p) = 1 for every p ∈ $2^N$", $2^N$ should be $N\_∞$ again for the same reason.
Perhaps this is a social question - how do I contact an author? - but I also wanted to post here to see whether what I am proposing is right and in case other people are hitting this when trying to read this paper.
P.S. I formalized these proofs at <https://us.metamath.org/ileuni/nninfall.html> and <https://us.metamath.org/ileuni/nninfsel.html> in case that helps.
| https://mathoverflow.net/users/489586 | Did I find a few (small) errors in the Pradic and Brown 2021 paper that Schroeder-Bernstein implies excluded middle? | It may interest you that Martín Escardó also formalized the proofs, see [`Cantor-Schröder-Bernstein-gives-EM`](https://www.cs.bham.ac.uk/%7Emhe/TypeTopology/CantorSchroederBernstein.CSB.html#6327) in his [TypeTopology](https://www.cs.bham.ac.uk/%7Emhe/TypeTopology/index.html) library.
Regarding contacting the authors, I am pretty sure this [Pierre Pradic](https://www.swansea.ac.uk/staff/p.r.a.pradic/) is the right one. Chad has always made it hard to be tracked down. If you search the [coq-club](https://sympa.inria.fr/sympa/info/coq-club) mailing list archives from 2011 or so, you'll find his email address.
| 1 | https://mathoverflow.net/users/1176 | 428445 | 173,700 |
https://mathoverflow.net/questions/428447 | 10 | I've seen the following sentence come up a few times in papers:
>
> Let $E$ be the universal elliptic curve over the modular curve $Y\_1(N)$. Then the localization of $E$ at any choice of cusp is isomorphic to the Tate curve with some suitable level structure.
>
>
>
Could somebody explain what exactly this sentence means? What does it mean to localize $E$ at a cusp? (Here $Y\_1(N)$ is the open modular curve so it has no cusps.) And in what sense is this localization of $E$ at a cusp given by the Tate curve?
My only exposure to Tate curves has been from Silverman's Advanced Topics book, where he explains how the Tate Curve $E\_q$ over $\mathbf{Q}\_p$ can be $p$-adically uniformized.
But I'm not so comfortable with how the Tate curve shows up when dealing with universal elliptic curves. Could someone shed some light on the connection between Tate curves and universal elliptic curves?
| https://mathoverflow.net/users/394740 | Universal elliptic curve and the Tate curve | Let $E\_1(N)$ denote the universal elliptic curve you are referring to. Note that we must assume $N\ge 4$, or else there is no such universal elliptic curve. Localizing $E\_1(N)$ at a cusp means, roughly, to look at how it behaves near a cusp. As you rightly point out, $Y\_1(N)$ has no cusps. In this case, the analytic picture that the sentence suggests you consider is the restriction of $E\_1(N)$ to a small punctured neighborhood at a cusp.
The algebraic analog of a small neighborhood of a point $x$ is the spectrum of the completion of the local ring at $x$. In the case of a curve $X$ over a field $k$, the complete local ring at a smooth $k$-rational point is isomorphic to $k[[t]]$. Thus the algebraic analog of a small punctured neighborhood at $x$ is $\text{Spec }k((t))$. To be precise, for $x\in X$, there is a map $l\_x : \text{Spec }k[[t]]\cong\text{Spec }\widehat{\mathcal{O}\_{X,x}}\rightarrow \text{Spec }\mathcal{O}\_{X,x}\rightarrow X$. If $Y\subset X$ is an open not containing $x$, then the pullback of $l\_x$ by the inclusion $Y\hookrightarrow X$ gives a map $\text{Spec }k((t))\rightarrow Y$. The quoted sentence presumably asks you to consider the the restriction of $E\_1(N)$ to $\text{Spec }k((t))$ via this map.
As for the relation with the Tate curve, the canonical reference is probably chapter 8 of Katz-Mazur. Below I give a rough sketch of the relationship:
In Silverman's advanced topics book, he gives an explicit equation for the Tate curve, where the coefficients are power series in $q$. In his book $q$ is taken to be a nonzero element of a $p$-adic field, but it's not hard to see that the same equations in fact defines an elliptic curve over $\mathbb{Z}((q)) := \mathbb{Z}[[q]][q^{-1}]$ (in fact it defines a stable 1-pointed curve of genus 1 over $\mathbb{Z}[[q]]$, whose fibers above $q = 0$ give "nodal cubics"). If you consider $Y\_1(N)$ over a field $k$, then these equations define the Tate curve as an elliptic curve over $k((q))$, with $j$-invariant
$$j(q) = \frac{1}{q} + 744 + 196884q + \cdots\in k((q)).$$
At this point, one can "simply" say that the Tate curve, viewed as a stable pointed curve over $k[[q]]$, is the universal deformation of its central fiber, and $k[[q]]$ is the universal deformation ring of a nodal cubic. This also implies that if $\overline{\mathcal{M}\_{1,1}}$ is the moduli stack of stable pointed curves of genus 1 and $\text{Spec }R\rightarrow\overline{\mathcal{M}\_{1,1}}$ is the completed etale local ring at the cusp, then $R$ is isomorphic to $k[[q]]$, and via this isomorphism the associated map $\text{Spec }k[[q]]\rightarrow \overline{\mathcal{M}\_{1,1}}$ corresponds to the Tate curve. This covers the case $N = 1$. For general $N\ge 4$, one should note that the universal elliptic curve over $Y\_1(N)$ is just the pullback of the universal elliptic curve over the moduli stack of elliptic curves. Thus, the complete local rings near the cusps are given by the Tate curve with scalars extended to $k[[q^{1/n}]]$, where $n$ denotes the width of the cusp.
Here is a more verbose but elementary illustration of the previous paragraph: For $N = 1$, $Y(1) = Y\_1(N)$ is the coarse moduli scheme of elliptic curves. Explicitly, $Y(1)$ is an affine line, with coordinate given by the $j$-invariant; we will write $Y(1) = \text{Spec }k[j]$ (think of the case $k = \mathbb{C}$ and $Y(1) = \mathcal{H}/\text{SL}\_2(\mathbb{Z})$). In this case a punctured neighborhood at $j = \infty$ is given by $\text{Spec }k((j^{-1}))$. Taking $q$-expansions formally (i.e., map $j^{-1}$ to $j(q)^{-1}$), we obtain an isomorphism $\text{Spec }k((j^{-1}))\cong\text{Spec }k((q))$, and hence have a map
$$t : \text{Spec }k((q))\longrightarrow Y(1)$$
where we view $\text{Spec }k((q))$ as a punctured neighborhood of the cusp "$j = \infty$" of $Y(1)$. Note that when $k = \mathbb{C}$, and $Y(1)$ is the spectrum of the ring of modular functions of level 1 on the upper half plane which are meromorphic at the cusps, then the map $t$ induces precisely the analytic $q$-expansion map at the level of rings.
The map $t$ defines a $k((q))$-valued point of $Y(1)$. It is a defining property of coarse schemes that $t$ uniquely defines an equivalence class of elliptic curves over $k((q))$, where two elliptic curves are equivalent if and only if they are isomorphic over an algebraic closure (i.e., they are twists of each other). This equivalence class is exactly the data of a $j$-invariant, and the $j$-invariant corresponding to $t$ is by definition the image of $j\in\Gamma(Y(1),\mathcal{O}\_{Y(1)})$ in $k((q))$. Since the image of $j$ is $j(q)$, $t$ corresponds to the twist-equivalence class of an elliptic curve with $j$-invariant $j(q)$. Since the Tate curve has the right $j$-invariant, $t$ must correspond to the class of the Tate curve.
This gives an algebraic picture of the Tate curve. There is also an analytic picture, starting from the universal elliptic curve over the upper half plane. I've worked out some of these details in section 5 of my notes [here](https://static1.squarespace.com/static/60d92f4398f4bb6a7b5a99c6/t/60d936a12b37c143c12bd42a/1624848034730/Katz+Modular+Forms.pdf), though I hesitate to link them since I wrote them a long time ago for myself, and to my knowledge nobody has ever checked it for correctness.
| 7 | https://mathoverflow.net/users/15242 | 428459 | 173,703 |
https://mathoverflow.net/questions/428355 | 1 | Suppose that $V\_1,V\_2$ are two commutative von Neumann algebras and $V\_1 \subset V\_2$. Being in particular commutative $C^\*$-algebras we have that $V\_1 \cong C(X\_1), V\_2 \cong C(X\_2)$ for some topological compact spaces $X\_1,X\_2$. Our inclusion $V\_1 \subset V\_2$ yields a continuous surjection $\pi:X\_2 \to X\_1$ going in the opposite direction. Since $V\_1,V\_2$ are von Neumann algebras then $X\_1,X\_2$ are in fact very special (the so called hyperstonean spaces). I wonder whether in this case:
>
> **Question** Is it true that $\pi$ maps clopen sets into clopen sets?
>
>
>
| https://mathoverflow.net/users/24078 | Continuous surjection between spectra of commutative von Neumann algebras |
>
> Is it true that π maps clopen sets into clopen sets?
>
>
>
This is true if and only if the inclusion $V\_1→V\_2$ is a morphism of von Neumann algebras, i.e., its image is closed in the ultraweak topology.
See Proposition 2.48 in [arXiv:2005.05284](https://arxiv.org/abs/2005.05284).
| 2 | https://mathoverflow.net/users/402 | 428477 | 173,707 |
https://mathoverflow.net/questions/428352 | 6 | I have posted a few questions on MSE, ~~most notably this one~~, which revolve around the same issue and have received no answers, so I decided to ask the same here.
In the following, $K(A, n)$ is the minimal Eilenberg-MacLane Kan complex given by $$K(A, n)\_q=\{\text{normalized $n$-cocycles } \Delta^q \to A\}.$$ This is constructed via the Dold-Kan correspondence, which applies to show that $\mathsf{sAb}(K(A, n), K(A', n))
\cong \mathsf{Ab}(A, A')$. In light of proving the existence of an isomorphism $\mathsf{sSet}(K(A, n), K(A', n))\cong \mathsf{Ab}(A, A')$, the following issue arises.
Is every simplicial map $\Phi:K(A, n) \to K(A', n)$ a simplicial homomorphism of groups? May in his book Simplicial objects in Algebraic Topology, Lemma 25.1, claims that the answer is positive and I know no other reference for this result. The proof is by induction: in degrees $q<n$, $K(A, n)\_q=\{\ast\}=K(A', n)\_q$ and there is nothing to prove. By minimality, $K(A, n)\_n=\pi\_n(K(A, n))$ and $\pi\_n(\Phi)=\Phi\_n$, hence $\Phi\_n$ is a homomorphism. Now come the problems: suppose for $q\ge n+1$ that we have proven that $\Phi\_{q-1}$ is homomorphism. Consider $\Phi\_q(x+y)$ and $\Phi\_q(x)+ \Phi\_q(y)$. These two elements of $K(A', n)$ have the same boundary by the induction hypothesis. May concludes that they are homotopic. Why this? We know that $\pi\_q(K(A', n)\cong 0$, but $\Phi\_q(x+y)$ and $\Phi\_q(x)+ \Phi\_q(y)$ need not determine elements of $\pi\_q(Y)$, since there is no reason why their boundary should be constant at the base-point. If we accept that they are homotopic relative their boundary, then they are equal by minimality and the argument is concluded. I don't want to sound skeptical about May's proof, but I am not convinced, and I even don't know if the result itself is true.
Why I doubt this is true:
* It is well-known that $[K(A, n), K(A', n)]\cong H^n(K(A, n), A')\cong \mathsf{Ab}(A, A')$. If the isomorphism above holds true, then we get an isomorphism $[K(A, n), K(A', n)] \cong \mathsf{sSet}(K(A, n), K(A', n))$. (And this groups can be finite.) Using the explicit form of the involved isomorphisms, this essentially says that two homotopic maps $K(A, n) \to K(A', n)$ are equal, which I highly doubt to be true, even using minimality of the codomain.
* The only other place I know in literature where something similar is
treated is a book (in German) by Lamotke, Semisimpliziale
Algebraische Topologie. In Theorem (Satz) VIII.3.11, he proves that
every simplicial map $K(A, n) \to K(A', n)$ is homotopic to a
homomorphism and every two homotopic homomorphisms are equal. Although this
does not rule out the possibility that every simplicial map $K(A, n)
\to K(A', n)$ be a homomorphism on the nose, if this is true, why not
saying it?
So, my question boils down to: is this fact true, and if yes, why?
[Edit: I have now removed the question on MSE.]
| https://mathoverflow.net/users/482564 | Is every simplicial map $\Phi:K(A, n) \to K(A', n)$ a simplicial homomorphism of groups? | Denote by $$\def\U{{\sf U}}\def\sSet{{\sf sSet}}\def\sAb{{\sf sAb}}\def\Ch{{\sf Ch}}\def\N{{\sf N}}\U\colon \sAb→\sSet$$ the forgetful functor,
which is the right adjoint of $$\def\Z{{\bf Z}}\Z\colon\sSet→\sAb$$
and by $$Γ\colon \Ch→\sAb$$ the Dold–Kan functor given by the right adjoint of the normalized chains functor $$\N\colon\sAb→\Ch.$$
Thus $$\def\K{{\sf K}}\K(A,n)=\U Γ(A[n]).$$
We have a chain of isomorphisms of sets
$$\sSet(\K(A,n),\K(A',n))≅\sSet(\U Γ A[n], \U Γ A'[n])≅\Ch(\N\Z\U Γ A[n],A'[n])$$
(by adjunctions $\Z⊣\U$, $\N⊣Γ$),
$$\def\Hom{\mathop{\sf Hom}}\def\H{{\sf H}}\Ch(\N\Z\U Γ A[n],A'[n])≅\Hom(\H\_n(\N\Z\U Γ A[n]),A')$$
(since both chain complexes vanish below degree $n$),
and if $n>0$, we have
$$\Hom(\H\_n(\N\Z\U Γ A[n]),A')≅\Hom(A,A')$$
(by the Hurewicz theorem).
This proves the desired isomorphism
$$\sSet(\K(A,n),\K(A',n))≅\Hom(A,A').$$
(If $n=0$, then $\Hom$ refers to maps of sets, not groups.)
The same argument also proves that
$$\sSet(\K(A,m),\K(A',n))≅0$$
(i.e., the only simplicial map factors through the basepoint) if $m>n>0$.
If $m<n$, then the Hurewicz theorem is not applicable, and in this case we have nontrivial cohomology operations.
| 3 | https://mathoverflow.net/users/402 | 428479 | 173,708 |
https://mathoverflow.net/questions/428472 | 3 | I am trying to show that
$$\lim\_{n\to\infty}\frac{1}{n}\sum\_{i=1}^{n}\prod\_{j=0}^{i}\frac{kn-j-k}{kn-j}=\frac{k^{k+1}-(k-1)^{k+1}}{(k+1)k^{k}}$$
for all $k\in\mathbb{N}$, $k\geq 4$.
I could verify the statement with Mathematica, but I could not find a self-standing proof. The product is telescopic, but I could only derive lower bounds instead of the exact value of the limit.
| https://mathoverflow.net/users/489614 | Limit of the average of telescopic products | Start by re-expressing the product term
\begin{align}\frac{(kn-k)(kn-k-1)\cdots(kn-k-i)}{(kn)(kn-1)\cdots(kn-i)}
&=\frac{(kn-i-1)\cdots(kn-i-k)}{(kn)\cdots(kn-k+1)}=\frac{\binom{kn-1-i}k}{\binom{kn}k}.
\end{align}
So, the given sum $S\_n(k):=\sum\_{i=1}^n\prod\_{j=0}^i\frac{kn-j-k}{kn-j}$ reads
\begin{align}
S\_n(k)&=\frac1{\binom{kn}k}\sum\_{i=1}^n\binom{kn-1-i}k
=\frac1{\binom{kn}k}\sum\_{j=0}^{n-1} \binom{kn-n-1+j}{k} \\
&=\frac1{\binom{kn}k}\left[\sum\_{\ell=0}^{kn-2} \binom{\ell}k
-\sum\_{\ell=0}^{kn-n-1} \binom{\ell}k\right] \\
&=\frac1{\binom{kn}k}\left[\binom{kn-1}{k+1}-\binom{kn-n}k\right] \\
&=\frac{kn-1}{k+1}\cdot\prod\_{j=1}^k\frac{kn-1-j}{kn-j}-
\frac{kn-n}{k+1}\cdot\prod\_{j=1}^k\frac{kn-n-1-j}{kn-j} \end{align}
where we have utilized the identity $\sum\_{\ell=0}^N\binom{\ell}k=\binom{N+1}{k+1}$. Therefore, going back to the required limit
\begin{align}
\lim\_{n\rightarrow\infty}\frac1nS\_n(k)
&=\lim\_{n\rightarrow\infty}
\frac{kn-1}{(k+1)n}\cdot\prod\_{j=1}^k\frac{kn-1-j}{kn-j}-
\frac{(k-1)n}{(k+1)n}\cdot\prod\_{j=1}^k\frac{(k-1)n-1-j}{kn-j} \\
&=\frac{k}{k+1}-\frac{k-1}{k+1}\cdot\frac{(k-1)^k}{k^k}
=\frac{k^{k+1}-(k-1)^{k+1}}{(k+1)\,k^k}.
\end{align}
Your claim has been verified.
| 8 | https://mathoverflow.net/users/66131 | 428486 | 173,709 |
https://mathoverflow.net/questions/428501 | 12 | A *corner* of an integer partition is a location at where a box can be added to its Ferrers diagram to give a new partition.
E.g. the partition $\{1,1,1\}$ has two corners, and $\{1,2\}$ has three corners.
Let $p(n,r)$ denote the number of integer partitions of $n$ with $r$ corners.
Let $P(x,t) = \sum\_{n,r\geq0} p(n,r)x^nt^r = t + t^2x + 2t^2x^2 + (2t^2+t^3)x^3 + \ldots$ be the generating function for these numbers.
1. Is there an elementary (**EDIT:** i.e. probabilistic) proof that $P(x,1-x) = 1$?
Thus for $x \in [0,1)$, the map $\lambda \to x^{|\lambda|}(1-x)^{\#corners (\lambda)}$ defines a probability distribution on the set of partitions of all integers.
2. Is it well known that $P(x,t) = \prod\_{k=1}^{\infty} \frac{1-x^{k-1}(1-t)}{1-x^k} = \frac{(1-t,x)\_\infty}{(x,x)\_\infty}$?
Note: In [Aubrey Blecher, Charlotte Brennan, Arnold Knopfmacher, and Toufik Mansour, "Counting corners in partitions"](https://www.researchgate.net/publication/273286302_Counting_corners_in_partitions), the following formula (Theorem 2.2) is given
$P(x,t) = t + t^2 \sum\_{j\geq 1} \frac{x^j}{1-x^j} \prod\_{i=1}^{j-1} \frac{1-(1-t)x^i}{1-x^i}$
| https://mathoverflow.net/users/489627 | Generating function for counting partitions with corners | Maybe this is not the kind of answer you are looking for, but it is easy to see that
$$ P(x,t) = \prod\_{k=1}^{\infty} \frac{1-x^{k-1}(1-t)}{1-x^k}$$
because the number of corners of a partition is one plus its number of distinct parts (you always have a corner at the "top" of a series of repeated parts, plus one at the very bottom of the partition). So the term of $\frac{1-x^k(1-t)}{1-x^k} = 1 + t \frac{x^k}{1-x^k}= 1 + t(x^k + x^{2k} + \cdots)$ correspond to choosing how many parts equal to $k$ in your partition you want. And then there is an extra factor of $1-x^0(1-t)=t$ for the extra corner at the bottom every partition has.
I guess this answers part 1 as well, right?
**EDIT**: Okay, in response to the comment below, here is a probabilistic proof that weighting each partition $\lambda$ by $x^{|\lambda|}(1-x)^{\#\mathrm{corners}(\lambda)}$ gives a probability distribution on the set of all partitions. Imagine constructing a partition as follows. We start with the empty partition. Then we focus on its unique corner. We flip a coin that is heads with probability $x$ and tails with probability $1-x$. If we get heads, we add a box in that corner, and then move on to consider the "next" corner of the partition we've built so far, moving left to right and top to bottom. (I am always using "English" notation for partitions/Young diagrams, by the way.) When we flip a tails at a corner, we leave that box empty, but we still move on to the next corner. Unless we flipped tails at the bottom corner (i.e., in a row with no boxes in it), in which case we stop and output the partition we've made. It is not hard to see that we produce each $\lambda$ with probability $x^{|\lambda|}(1-x)^{\#\mathrm{corners}(\lambda)}$. Maybe a small additional argument is needed to explain why we terminate with probability $1$.
| 13 | https://mathoverflow.net/users/25028 | 428502 | 173,713 |
https://mathoverflow.net/questions/428505 | 0 | Let M be the generator matrix of a $N\times N$ lattice, and $\tilde{N}$ the set of Voronoi relevant vectors. The Voronoi cell for the origin can be written as $\text{Vor}\_{\bf 0}(M)=\left\{{\bf x}: |{\bf x}|\leq|{\bf x-c}|\text{ for } \forall {\bf c}\in \tilde{N}\right\}$. My question is given $\tilde{N}$, how to find the closest lattice vector for a given point ${\bf x}\in R^N$? Equivalently, given $\tilde{N}$, how to translate the given point ${\bf x}$ back to $\text{Vor}\_{\bf 0}$?
If $\tilde{N}$ is not given, the closest point can be found with the algorithm introduced [here](https://publications.lib.chalmers.se/records/fulltext/14990/local_14990.pdf), and the same algorithm can be modified to find $\tilde{N}$. The motivation for my question is that I would like to find the closest points for a large number of ${\bf x}$'s, for the same lattice. Running the closest point search algorithm is very expensive, hence I am hoping to use relevant vectors to find the closest points.
This is my attempt to translate a given point ${\bf x}$ back to $\text{Vor}\_{\bf 0}$ given the relevant vectors $\tilde{N}$
```
for c in $\tilde{N}$
overlap = x^T * c/norm(c)^2
if abs(overlap) > 1/2
x = x - round(overlap) * c
end
end
```
which seems working for 2D lattices, but not for 3D lattices. Hence I am not sure if the algorithm is correct, or if it is possible to do that for general lattices.
| https://mathoverflow.net/users/476103 | How to find the closest point given the Voronoi relevant vectors? | I believe state of the art for this is [Short Paths on the Voronoi Graph and Closest Vector
Problem with Preprocessing](https://arxiv.org/abs/1412.6168) by Bonifas and Dadush.
For instance, see the following quote from the abstract
>
> Our main technical contribution is a randomized procedure that given the Voronoi relevant vectors of a lattice - the lattice vectors inducing facets of the Voronoi cell - as preprocessing and any "close enough" lattice point to the target, computes a path to a closest lattice vector of expected polynomial size. This improves on the $\tilde{O}(4^n)$ path length given by the MV algorithm. Furthermore, as in MV, each edge of the path can be computed using a single iteration over the Voronoi relevant vectors.
>
>
>
This is to say that Bonifas+Dadush should imply a (randomized) algorithm of complexity $O(\mathsf{poly}(N)|\tilde{N}|)$ (as a side note, using $N$ for a dimension and $\tilde{N}$ for a set is perhaps not the best choice of notation).
Note that this may not mean *efficient*, as $|\tilde{N}|$ can grow exponentially in $N$.
But, if you have a lattice for which you know that $|\tilde{N}|$ is small, it should efficiently solve your problem.
Note that one can often solve CVP efficiently even in lattices that have exponentially many Voronoi relevant vectors.
This is true if they have small *extension complexity*, see [Lifts for Voronoi Cells of Lattices](https://arxiv.org/abs/2106.04432) by Schymura, Seidel, and Weltge.
It's also worth mentioning that there have been improvements for finding $\tilde{N}$.
In particular, see [A Deterministic Single Exponential Time Algorithm for Most Lattice Problems](http://cse.ucsd.edu/sites/cse/files/cse/assets/research/theory/MicciancioVoulgaris_2010_stoc.pdf) by Micciancio+Voulgaris.
| 0 | https://mathoverflow.net/users/101207 | 428506 | 173,714 |
https://mathoverflow.net/questions/428410 | 8 | Does the parallelogram law for vectors of equal length imply the full parallelogram law? That is,
if for all norm one vectors $x$ and $y$ in a Banach space $X$ it holds that $\lVert x-y\rVert^2+\lVert x+y\rVert^2=4$, does it follow that $X$ is isometric to a Hilbert space?
I suspect the answer is "no", but I cannot come up with an example. Of course, it is enough to consider the question in two dimensions.
| https://mathoverflow.net/users/69275 | Parallelogram law for vectors of equal length | I will give it a try, based on Day's idea. Let $X$ be a two-dimensional Banach space with the given property and denote by $B\_X$ its unit ball. Consider the ellipsoid of maximal volume (denoted by $B\_2$) contained in $B\_X$ (the John's ellipsoid) and denote by $\|\cdot\|\_2$ the induced Euclidean norm. The goal is to show that $B\_X=B\_2$.
From John's theorem concerning the ellipsoid of maximal volume (or Loewner's Lemma for two dimensions in Day's paper), it follows that $B\_X$ and $B\_2$ have at least four contact points. Unless $B\_X=B\_2$, the contact points cannot form a dense subset of $B\_X$. Assuming $B\_X\neq B\_2$, we can find contact points $x$ and $y$ such that $\displaystyle\frac{x+y}{\|x+y\|}\in B\_X$ is not a contact point. Hence $\|x+y\|<\|x+y\|\_2$. Therefore:
$$
4=\|x+y\|^2+\|x-y\|^2<\|x+y\|\_2^2+\|x-y\|\_2^2=4
$$
This is a contradiction, hence $B\_X=B\_2$.
| 4 | https://mathoverflow.net/users/69275 | 428510 | 173,717 |
https://mathoverflow.net/questions/428519 | 11 | Consider, for instance, the categories of $C^k$-manifolds, where $k=0,1,2,...,\infty,\omega$. ($C^\omega$ means real analytic.) Are these categories pairwise non-equivalent?
Of course, the obviuos forgetful functor is not an equivalence, because it is not full: for $k<l$, there are $C^k$ functions which are not $C^l$.
I was just thinking about possible ways of proving non-equivalence of categories and came up with this question. For example, one can prove non-equivalence by showing that limits or colimits of certain shape exist in one category but not in the other. I don't know whether one can deal with manifolds in this way.
What I can prove is that the category of *complex*-analytic manifolds is not equivalent to any of the above.
>
> This follows from the fact that any real manifold other than the point (which is the terminal object) has non-trivial automorphisms. For $C^k$-manifolds with $k=0,1,...,\infty$ this is very easy to show, as there is a $C^\infty$-diffeomorphism of the ball $|x|<1$ which is the identity near its boundary sphere $|x|=1$. (One can construct such a diffeomorphism uning bump functions.) This does not work for the real analytic case. However, by a theorem of Grauert, for any real analytic manifold $M$ there is a closed analytic embedding $i$ of $M$ into some $\mathbb R^n$. Then one can choose a non-zero vector tangent to $i(M)$ at some point, extend it to a constant vector field along $i(M)\subset\mathbb R^n$ and project this vector field orthogonally to get a bounded vector field on $M$ which is not identically zero. As $M$ is complete in the induced metric (because $\mathbb R^n$ is complete and $i$ is closed), the flow of this vector field will be globally defined and thus it will contain a non-trivial analytic self-diffeomorphism of $M$.
>
>
> On the other hand, there are lots of complex analytic manifolds with no non-trivial automorphisms. For example, a generic compact Riemann surface of genus $\geq 3$ is known to have this property.
>
>
>
| https://mathoverflow.net/users/485324 | Are different categories of manifolds non-equivalent (as abstract categories)? | Your method of looking for automorphisms of objects works for $C^k$ manifolds as well, $0 < k \le \infty$. This follows from [a result of Filipkiewicz](https://www.cambridge.org/core/journals/ergodic-theory-and-dynamical-systems/article/isomorphisms-between-diffeomorphism-groups/BE5563E03AD7DA6A94D08F92CFA4AF90) (which I found in Kathryn Mann's [excellent survey](https://www.ams.org/journals/notices/202104/rnoti-p482.pdf)):
>
> If $M, N$ are smooth manifolds without boundary and $\varphi: \text{Diff}^p(M) \to \text{Diff}^q(N)$ is an isomorphism of groups (with $1 \le p, q \le \infty$), then $p = q$ and $\varphi$ is induced by a $C^p$ diffeomorphism $w: M \to N$.
>
>
>
(Recall that every $C^p$ manifold for $p > 0$ carries a compatible smooth structure, so the condition that $M,N$ are smooth manifolds is harmless --- cf. Hirsch, Differential Topology, chapter 2).
If $F: \mathsf{Man}\_p \to \mathsf{Man}\_q$ is an equivalence of categories, then $F$ induces an isomorphism $\text{Diff}^p(M) \cong \text{Diff}^q(F(M))$, and hence $p = q$ and $M \cong F(M)$. Not only do we have $p = q$, we also see that any autoequivalence of $\mathsf{Man}\_p$ is the identity on isomorphism classes. Probably one can say something stronger (maybe they're all naturally isomorphic to the identity).
This handles all but the case where one of $p,q$ is zero or $\omega$. It seems plausible to me that Filipkiewicz' proof goes through when $p = 0$ (though I haven't read carefully enough to say this confidently). I don't know about the case of $\omega$, and Filipkiewicz asks about this as the final question in his paper.
| 17 | https://mathoverflow.net/users/40804 | 428521 | 173,721 |
https://mathoverflow.net/questions/428531 | 2 | Let $X, Y, Z$ be discrete random variables with $X$ and $Y$ independent of $Z$, while $X$ and $Y$ can be dependent. For the mutual information, we have $I(X; Y,Z) = I(X;Y)$. Now consider $I(X; f(Y,Z))$ for some deterministic function $f$. Does $I(X; f(Y,Z))$ depend on $Z$? If not, is there a way to express $I(X; f(Y,Z))$ in terms of $X$ and $Y$ only?
| https://mathoverflow.net/users/101100 | Mutual information and bivariate function of independent variables | $I(X;f(Y, Z))$ can depend on the $Z$ (or more specifically the distribution of $Z$). Consider the following example, $Z \sim B(p)$, $X \sim B(0.5)$, $Y=X$ satisfying $X,Y \perp \!\!\! \perp Z$. Let $F=\max(Y, Z)$ be a variable from the output of a deterministic function of $Y, Z$. $X, F$ have the following joint distribution,
1. $P(X=0,F=0) = (1-p)/2$.
2. $P(X=0,F=1) = p/2$.
3. $P(X=1,F=0) = 0$.
4. $P(X=1,F=1) = 1/2$.
$I(X;F) = \log(2) + \frac{p}{2}\log(p) - \frac{1+p}{2}\log(1+p)$ which is dependent on $p$.
| 1 | https://mathoverflow.net/users/478836 | 428538 | 173,728 |
https://mathoverflow.net/questions/426330 | 2 | Let $X$, $Y$, $Z$ be discrete random variables, with $Y$ and $Z$ independent. Does the following equality hold?
$$
\max\_{f\_{Y,Z}} \big\{ \ I(X; f\_{Y,Z}(Y,Z)) \ \big\} \le \max\_{f\_X, f\_Y} \big \{ \ I(X; f\_Y(Y), f\_Z(Z)) \ \big \}
$$
where the maximization is taken over all non-injective deterministic functions.
| https://mathoverflow.net/users/101100 | Maximization of information over set of non-injective functions | It seems the inequality should be held in the opposite way due to the fact that the function family on the left that you are taking maximum over is a superset of the function family on the right. More specifically, let $F\_1:=\{(y,z) \to f\_1(y,z)\}$ and $F\_2:= \{(y,z) \to (f\_y(y),f\_z(z))\}$ where $f\_1,f\_y,f\_z$ are arbitrary non-injective functions. We have $F\_1 \supset F\_2$.
| 0 | https://mathoverflow.net/users/478836 | 428550 | 173,731 |
https://mathoverflow.net/questions/428493 | 1 | Let $X\neq \emptyset$ be a set. We say $E\subseteq {\cal P}(X)$ is a *linear set system* if for all $a\neq b\in X$ there is exactly one $e\in E$ with $\{a,b\}\subseteq e$.
Is there an infinite cardinal $\kappa$ and a linear set system $E\subseteq {\cal P}(\kappa)$ with $\kappa \notin E$ such that for all $x\in \kappa$ we have $|\{e\in E: x\in e\}| < \kappa$?
**EDIT.** Had to exclude cases like $E = \{\kappa\}$ which Noah Schweber made me aware of (thanks!).
| https://mathoverflow.net/users/8628 | Flat linear set systems | Denote $\deg x =|\{e\in E\colon x\in e\}|$.
Take any $e\in E$; choose $x\notin e$. Then $e$ meets any $e'\ni x$ by at most one element, whence $|e|\leq \deg x<\kappa$.
Now, any $x$ is contained in $\deg x<\kappa$ sets of cardinality $<\kappa$ each; moreover (if this is needed), any of them except one has cardinality at most $\deg y<\kappa$ (for an arbitrary $y\neq x$). All this implies that their union is of cardinality $<\kappa$; but their union **is** $\kappa$.
| 2 | https://mathoverflow.net/users/17581 | 428561 | 173,736 |
https://mathoverflow.net/questions/428530 | 1 | This post records a little bit more on this question: [Partitioning convex polygons into triangles of equal area and perimeter](https://mathoverflow.net/questions/428243/partitioning-convex-polygons-into-triangles-of-equal-area-and-perimeter).
The basic question of the above linked post was about this claim: ""For any convex polygon, there is some finite value of a positive integer n such that the polygon allows partition into n triangles all of which are of same area." The claim may not be true (please see above post).
**Question:** What happens if we replace 'triangles' in above claim with 'convex quadrilaterals'? The new claim thus generated appears more likely to be true.
**Note:** In the new claim, one can replace "area" with "perimeter" or diameter or combinations such as "area and perimeter", " area and diameter",... and generate further questions.
| https://mathoverflow.net/users/142600 | Partitioning convex polygons into quadrilaterals of equal area and perimeter | Convex quadrilateral decompositions of equal area are always possible. Given a convex polygon $P$ with at least $5$ sides, draw a line from one of the vertices of $P$ to a point $Q$ on one of the edges furthest from $P$ (in the graph sense). As we vary the location of $Q$ continuously, we'll find some decomposition into two convex pieces with strictly fewer sides whose areas are in a rational ratio.
Repeating as necessary, we obtain a collection of quadrilaterals whose areas are rational multiples of one another; from there, we just need to subdivide the quadrilaterals into smaller pieces the size of their least common divisor $d$, which we can do by cutting off quadrilateral-shaped chunks of area $d$ from each piece by connecting two points on opposite sides. (As we slide the two points freely, we can obtain any fraction of the area between $0$ and $1$.)
This only leaves the case where we start with a triangle, which is easy: since decompositions are affinely equivalent, we can assume our triangle is equilateral and draw three line segments from the center to the midpoints of the side to obtain our decomposition.
| 2 | https://mathoverflow.net/users/89672 | 428563 | 173,737 |
https://mathoverflow.net/questions/428433 | 2 | My apologies if this question is below the level of MO. I posted the [same question](https://math.stackexchange.com/questions/4507690/dcc-on-the-powers-of-ideals) in MS about a week ago without an answer so far.
Let $R$ be a unital commutative ring. $R$ is called *strongly $\pi$-regular* if it satisfies the DCC on the powers of principal ideals, that is, the chain $\hspace{3mm}I\supseteq I^2 \supseteq I^3 \supseteq\dots\supseteq I^n\supseteq\dots\hspace{3mm}$ stops for each $x\in R$, where $I=xR$.
>
> Do we have a term coined for those rings that satisfy DCC on the *powers* of all maximal ideals (respectively, prime ideals, semiprime ideals, semiprimitive ideals, all ideals) ?
>
>
>
| https://mathoverflow.net/users/164350 | DCC on the powers of ideals | I am not aware of a term for rings with this property, and I have rarely seen it in the literature. There is a small result about such rings in Proposition 3.22 of the following paper:
*Lam, T. Y.; Reyes, Manuel L.*, [**A prime ideal principle in commutative algebra**](http://dx.doi.org/10.1016/j.jalgebra.2007.07.016), J. Algebra 319, No. 7, 3006-3027 (2008). [ZBL1168.13002](https://zbmath.org/?q=an:1168.13002).
Part of that result says that a commutative noetherian ring satisfying this DCC property on powers of maximal is already artinian.
| 3 | https://mathoverflow.net/users/778 | 428569 | 173,741 |
https://mathoverflow.net/questions/428562 | 2 | Given a complete smooth Toric surface (over $\mathbb C$), is its intersection form well-known? Or is there an algorithm to calculate it? Thanks in advance.
| https://mathoverflow.net/users/88180 | Intersecton form of complete smooth Toric surface | The Picard group of a toric surface is generated by the simple toric divisors $D\_1,D\_2,\dots,D\_n$, and if $v\_1,v\_2,\dots,v\_n$ are the generators of the rays of the corresponding fan, there are relations
$$
\sum f(v\_i)D\_i = 0
$$
for all linear functions $f$. The intersection product is determined by these linear relations and simple intersection formulas
$$
D\_i \cdot D\_j =
\begin{cases}
1, & \text{if $|i-j| = 1$},\\
0, & \text{if $|i-j| > 1$},
\end{cases}
$$
where the indicies are assumed to by ordered cyclically (and to compute $D\_i \cdot D\_i$ one should use the linear relations).
| 6 | https://mathoverflow.net/users/4428 | 428583 | 173,746 |
https://mathoverflow.net/questions/428577 | 5 | So I was interested in formally assigning values to the completely divergent series $G(x) = \sum\_{n=0}^{\infty} n!x^n $. I guess the question COULD end here if you already have an idea of how to tackle this but feel free to continue reading for a strategy i think MIGHT work.
We start by consider a different totally divergent series $$F(s) = \sum\_{n=1}^{\infty} \log(n)^s $$
This doesn't converge for any choice of $s \in \mathbb{C}$. Now its worth observing that the series
$$ \zeta(-s) = 1 + 2^z + 3^z + 4^z ... $$
Has the property that for positive integers $k$ one "formally" has
$$ \frac{d^k}{ds^k} \left[ \zeta(-s) \right]\_{@(s = 0)} = \sum\_{n=1}^{\infty} \log(n)^k $$
So its natural then to define our diverging logarithmic series everywhere by writing
$$F(z) = \sum\_{n=1}^{\infty} \log(n)^z = \frac{d^z}{ds^z} \left[ \zeta(-s) \right]\_{@(s = 0)}$$
Where $z$ is taken to be an arbitrary complex number and we use the standard cauchy definition of the fractional derivative. Let's call this operator $Q$. To be explicit
$$ Q[f] = \frac{d^{\alpha}}{dx^{\alpha}} \left[ f(x) \right]\_{@(x=0)} $$
So your domain begins with $x$ and ends with $\alpha$ after applying our "Q-transform".
From here one can see easily that
$$ Q\left[ \sum\_{n=0}^{\infty} n! x^n \right] = \Gamma(\alpha+1)^2 $$
So it might be fruitful to consider then the expression
$$ Q^{-1} \left[ \Gamma(\alpha+1)^2 \right] $$
Unfortunately I don't know how to define the inverse Q-transform and perhaps that is as hard (or harder) than summing this series in the first place but I think its worth a shot.
| https://mathoverflow.net/users/46536 | Evaluating the series $\sum_{n=0}^{\infty} n! x^n$ and inverse variable-fractional-derivatives | $$
G(x) = \sum\_{n=0}^\infty n!x^n
\tag1$$
Another approach is to observe that the series $G(x)$ formally satisfies the differential equation
$$
x^2 G'(x) + (x-1) G(x) + 1 = 0 .
\tag2$$
The unique solution of $(2)$ with $\lim\_{x\to 0}G(x) = 1$ is
$$
\widetilde{G}(x) = -\frac{1}{x}\;e^{-1/x}\;\operatorname{Ei}\_1\left(-\frac{1}{x}\right) ,
\tag3$$
where $\operatorname{Ei}\_1$ is the [exponential integral function](https://en.wikipedia.org/wiki/Exponential_integral).
For $x<0$, the series $(1)$ is [Borel summable](https://en.wikipedia.org/wiki/Borel_summation) to $(3)$.
| 8 | https://mathoverflow.net/users/454 | 428595 | 173,752 |
https://mathoverflow.net/questions/428223 | 4 | $\DeclareMathOperator\BS{BS}$The linearity of the Baumslag-Solitar groups $\BS(m, n)=\langle a, t\mid t^{-1}a^mt=a^n\rangle$ is completely understood, and it may be phrased as: $\BS(m, n)$ is linear if and only if it is residually finite\*.
A *generalised Baumslag-Solitar (GBS) group* is a group which may be realised as the fundamental group of a graph of groups with all vertex and edge groups infinite cyclic. For example, Baumslag-Solitar groups themselves are GBS groups, with a single vertex and a single loop edge, while one-relator groups with center form another class of examples.
I am wondering if the above characterisation of linear Baumslag-Solitar groups extends to GBS groups:
>
> Do there exist non-linear, residually-finite GBS groups?
>
>
>
---
\*See either [this](https://mathoverflow.net/q/119719/6503) old MO question, or combine the papers:
* [Classifying residual finiteness] S. Meskin *Nonresidually finite one-relator groups* Trans. Amer. Math. Soc. 164 (1972), 105–114
* [Classifying linearity, in Russian] R.T. Vol’vachev *Linear representation of certain groups with one relation* Vestsi Akad. Navuk BSSR Ser. Fi z.-Mat. Navuk 1985, no. 6, 3–11, 124
| https://mathoverflow.net/users/6503 | When is a generalised Baumslag-Solitar group linear? | (1) I've been looking a little more. My belief is now:
**Conjecture.** *Let $G$ be a generalized Baumslag-Solitar group (i.e. the Bass-Serre fundamental group of a nonempty finite graph of groups in which every vertex and edge group is infinite cyclic). Let $\langle x\rangle$ be one vertex group. Then we have one of the following:*
(a) for some $n\ge 1$, $\langle x^n\rangle$ is a normal subgroup [this can be read on the graph, see below]. Then $G$ is virtually direct product of $\mathbf{Z}$ and a free group (and hence linear).
(b) $G$ is a solvable, Baumslag-Solitar group, and in particular is residually finite [assuming the graph is reduced, this means that there is a single vertex, at most a single self-loop with one inclusion being surjective]
(c) Otherwise, $G$ is not residually finite.
In (a) this can be read as follows: the partial isomorphism defined by any loop (including self-loops) is $\pm$ a partial identity. See also (4) below.
Reduced means that for every non-self loop, both inclusions are proper. If the (finite) graph of groups is not reduced, one can collapse non-reduced edges until one gets a reduced graph, without changing the Bass-Serre fundamental group.
---
(2) Now let me prove (sketch) the conjecture when the graph is a tree (with self-loops allowed).
Let me start with a graph with a single vertex (and finitely many self-loops). Such a group is thus defined by a family of pairs $(n\_i,m\_i)\_{i\in I}$ of nonzero integers with $I$ finite.
Let us specify (a),(b),(c) to this case:
*(a) for some $n\ge 1$, $\langle x^n\rangle$ is a normal subgroup: this means that $|m\_i|=|n\_i|$ for all $i$;*
(b) $G$ is a solvable, Baumslag-Solitar group: this means that $|I|=1$ (say $I=\{1\}$) and $\min(|m\_1|,|n\_1|)=1$;
(c) other cases.
Since the case $|I|\le 1$ is already settled, we deal with $|I|\ge 2$. Assume $G$ is residually finite. By the case $|I|=1$, we know that for each $i$, either $|n\_i|=|m\_i|$ or $\min(|n\_i|,|m\_i|)=1$. We can assume that whenever $\min(|n\_i|,|m\_i|)=1$, we have $1\in\{n\_i,m\_i\}$.
Just to illustrate, a typical example would then be $|I|=3$, with the pairs $(1,3),(-2,1),(4,-4)$. The corresponding presentation is
$$\langle t\_1,t\_2,t\_3,x\mid t\_1xt\_1^{-1}=x^3,t\_2x^{-2}t\_2^{-1}=x,t\_3x^4t\_3^{-1}=x^{-4}\rangle.$$
Suppose that $(1,m)$ and $(n,1)$ both occur among the $(n\_i,m\_i)$, for some $|n|,|m|\ge 2$. So we have the subgroup with presentation
$$\langle t,u,x\mid txt^{-1}=x^m,u^{-1}xu=x^n\rangle$$
In every finite quotient of this group, $g\mapsto tgt^{-1}$ defines an injective endomorphism of $\langle x\rangle$, hence bijective. Similarly for $u$. Hence $\langle x\rangle$ is normal in every finite quotient. Hence $[t^{-1}xt,u^{-1}xu]$ is trivial in every finite quotient. But it is not trivial in the group itself, by standard facts on HNN extensions.
A similar argument holds if $(1,n\_i)$, $(1,n\_j)$ both occur among the pairs with $i\neq j$ and $|n\_i|,|n\_j|\ge 2$: then $[t\_i^{-1}xt\_i,t\_j^{-1}xt\_j]$ is not trivial but vanishes in every finite quotient. (Same with $(n\_i,1)$ and $(n\_j,1)$.)
Next, if $(1,n)$ and $(m,m)$ both occur with $|n|\ge 2$: this corresponds to relators $txt^{-1}=x^n$, $[u,x^m]=1$: then $[u,t^{-1}x^mt]$ is nontrivial but vanishes in every proper quotient. Finally if $(1,n)$ and $(m,-m)$ both occur with $|n|,|m|\ge 2$, then $[u,t^{-1}x^{2m}t]$ is nontrivial but vanishes in every proper quotient (to check that it is nontrivial one has to check separately the case $|n|=2$).
If all these are excluded and $|I|\ge 2$, this means that we are in case (a).
---
(3) Here is the simplest case with a non-self loop: two vertices joined with two edges. This corresponds to a presentation
$$\langle t,x,y\mid x^k=y^\ell,x^m=ty^nt^{-1}\rangle.$$
We get $x^{kn}=y^{\ell n}=t^{-1}x^{\ell m}t$. Indeed the discussion is:
If $|kn|=|\ell m|$, then $\langle x^{kn}\rangle$ is normal cyclic and we are in case (a). Otherwise, an argument similar to the Baumslag-Solitar case (and to the ones above) implies that the group is not residually finite.
---
(4) In general, let me clarify "this can be read on the graph" in (a): given a sequence of oriented edges $e\_1,\dots,e\_p$ forming a loop ($p\ge 1$). Let the left and right inclusion of $e\_j$ be given by multiplication by $n\_j$, $m\_j$ respectively. Say that the loop is unimodular if $\prod m\_j=\prod n\_j$. Say the GBS graph of groups is unimodular if every loop is unimodular. Then (a) means that the graph is unimodular. (It is enough to check loops in a generating subset of the (ordinary, i.e; Poincaré) fundamental group of the graph.)
~~I haven't checked, but believe it's easy~~, [Edit: one easily checks] that if the graph is unimodular then $G$ is virtually direct product of $\mathbf{Z}$ and a free group. Indeed, after modding out by a normal cyclic subgroup contained in every edge group, the resulting group is Bass-Serre fundamental group of a finite graph of finite (cyclic) groups, and hence is virtually free.
And if the graph is not unimodular (and reduced with at least 2 edges) then one has to check the failure of residual finiteness. I did it above if the lack of unimodularity is witnessed by a self loop, and more quickly with it is witnessed by a loop of size 2. But I think there is no serious difficulty in general.
Eventually I believe the conjecture follows using routine arguments. Thus the linear examples are "obviously linear" (virtually $\mathbf{Z}\times$free or solvable BS groups) and the other ones are non-residually finite for a reason similar to the Baumslag-Solitar case, while the solvable examples appear as an exception.
| 2 | https://mathoverflow.net/users/14094 | 428598 | 173,754 |
https://mathoverflow.net/questions/428446 | 6 | When I started studying the basics of $\phi^{4}\_{d}$, I looked for papers or lecture notes which would give me some general ideas about the topic and which would construct and/or prove the basic results of the theory. One of the main targets of this theory is, for example, the rigorous construction of the formal object:
$$d\mu = e^{-S(\phi)}e^{\lambda\int \phi^{4}(x)dx}d\phi$$
I think I understood the problems one encounter when trying to define such measure, i.e. the large field problem and the need of the introduction of a cutoff to regularize the theory.
However, trying to learn something about $\phi^{4}\_{d}$ alone, in my experience, has been a really difficult task. There is a huge number of papers and each one seem to do something different. And by something different, I don't mean just the technique used in the paper, or the realization of the renormalization group, but each paper seem to deal with particular problems. For example, some use statistical mechanics terminology, study the pressure of the system (the logarithm of the partition function normalized by the volume). Some, use Schwinger functions and effective actions. And so on.
Also, some start studying the $\phi^{4}\_{d}$ alone, some include some addition terms, say $g \phi^{2}(x)$ or some derivative. It is not clear if one needs it or not, or *when* does one need it or not.
In summary, I got to the conclusion that I need to start from somewhere safe and start learning from the most basic to get to the most advanced.
I am interested in the construction of $\phi^{4}\_{d}$ in terms of constructive/non-perturbative QFT using renormalization group techniques (scale decomposition and so on).
**Question:** Can you rank from the easiest to the most advanced cases of $\phi^{4}\_{d}$ and point me the papers in which the constructions are given? For example, I think the case $d=1$ is probably the most trivial one, but I have never seen anyone treating this case. What next? Maybe $d=2$? Maybe $d=3$? Do we need extra terms there? What papers should I get to learn the most constructions of the easiest cases?
| https://mathoverflow.net/users/152094 | Reference request for $\phi^{4}_{d}$ theory - where to begin? | For an introduction to the basics of quantum field theory you could look into [Introduction to Quantum Field Theory for Mathematicians](https://souravchatterjee.su.domains//qft-lectures-combined.pdf). Lectures 13 and 18-22 introduce the $\phi^4$ model in 3+1 dimensions and the perturbative calculation of transition probabilities (from momenta $p\_1,p\_2$ to $p\_3,p\_4$). The first order term in the coupling constant is finite, but the second-order term diverges. Renormalization is then introduced to obtain a finite answer.
This perturbative approach to QFT is not mathematically rigorous, but you will obtain answers to some of the questions stated in the post (like "why add a $\phi^2$ term?" --- it gives the particles a mass).
For more rigour you then want to turn to the constructive approach to QFT. I understand your interest is in bosonic fields. For a broad overview you could take a look at [A Perspective on Constructive Quantum Field Theory](https://arxiv.org/abs/1203.3991), to see that there exists a great variety of approaches in this category.
One rather recent development that I think requires the least amount of background is the [Tree Quantum Field Theory](https://arxiv.org/abs/0807.4122) of Gurau, Magnen, and Rivasseau. This a reformulation of the combinatorial core of constructive quantum field theory, which is neither based on functional integrals, nor on Feynman graphs, but on marked trees. For an application specifically to the $\phi^4$ model, see [Constructive $\phi^4$ field theory without tears](https://arxiv.org/abs/0706.2457), by Magnen and Rivasseau.
One advantage of focusing on this modern approach is that it might be an entry point for original research (which may or may not be your objective).
| 6 | https://mathoverflow.net/users/11260 | 428604 | 173,755 |
https://mathoverflow.net/questions/428581 | 0 | I have an integral of the form
$$
I = \int\limits^{1}\_{0} \exp\left(\dfrac{vt}{(v+1)^2 + v^2} - vt\right) dv
$$
and I want to prove that $I\leq c t^{-1}$ for the sufficiently large $t$, where $c$ is a constant independent of $t$.
Can anyone give me some hints or references to prove this expansion?
| https://mathoverflow.net/users/489455 | Asymptotic estimation of an integral | (This question should be on math.stackexchage.com.)
Substitute $v=t^{-1/2}u$, then it becomes
$$ t^{-1/2} \int\_0^{t^{1/2}} e^{-2u^2}\bigl(1 + O(u^3/t^{1/2})\bigr)\,du
= \sqrt{\frac{\pi}{8t}} + O(t^{-1}).$$
| 2 | https://mathoverflow.net/users/9025 | 428605 | 173,756 |
https://mathoverflow.net/questions/428555 | 24 |
>
> **Question:** Can I have an arbitrarily large finite family of lines $\ell\_1,\dotsc,\ell\_n\subset\Bbb R^3$ so that the average angle between two (distinct) lines is $\ge \pi/3$?
>
>
>
We can assume that all lines pass through the origin, and by angle I mean the smaller one of the two angles defined by two intersecting lines.
If my integral calculus hasn't failed me, then the expected angle between two random lines with uniformly chosen directions is exactly $1\, \mathrm{rad}$, wheras $\pi/3\approx 1.0472 >1$.
So generic arrangements won't do it. Can I be more clever? Or is there perhaps some standard measure theory argument that tells me that in the limit, uniformly chosen lines is the best I can do? If so, can we put a bound on $n$ for where this stops being possible?
I know from random experiments that $n=17$ is possible, but already for $n=18$ I have no examples so far.
---
**Update**
The [comment](https://mathoverflow.net/questions/428555/are-there-arbitrarily-large-families-of-lines-in-bbb-r3-with-average-angle#comment1102298_428555) by alesia answers the question as posed (and I am looking forward to accept their [answer](https://mathoverflow.net/a/428607)): there are arbitrarily large families of lines with average angle $> \pi/3$.
Though one should note that the bound $\pi/3$ still seems kind of substantial: the more lines I have, the close the average will be to $\pi/3$.
I won't change the question (I might post a separate one), but the following seems to be the more fundamental question:
>
> Is there a probability distribution on lines in $\Bbb R^3$ so that the expected value for the angle between two (independently sampled) lines is $>\pi/3$?
>
>
>
I think *No* and I am looking forward to ideas. The value $\pi/3$ is attained by alesia's construction (uniformly
choosing directions from an orthogonal basis).
| https://mathoverflow.net/users/108884 | Are there arbitrarily large families of lines in $\Bbb R^3$ with average angle $\ge \pi/3$? | Start with $n$ times each coordinate axis, so $3n$ lines in total. The average angle between non necessarily distinct lines is $1/3\*0 + 2/3\*\pi/2 = \pi/3$. So the average angle between "distinct" lines (here, distinct lines may coincide geometrically due to multiplicities) is greater than $\pi/3$.
You can then perturb the lines to make them geometrically distinct while keeping an average angle larger than $\pi/3$.
Although that's not the question, I'm not sure what happens if one wants an infinite number of lines.
| 15 | https://mathoverflow.net/users/112954 | 428607 | 173,757 |
https://mathoverflow.net/questions/428336 | 6 | Let $a,b\in\mathbb C$ be suc that $\max\{|a+b|,|a-b|\}\leq 1$ but $|a|+|b|>1.$ According to this paper by Arias, Figiel, Johnson and Schechtman <https://www.jstor.org/stable/2155206?origin=crossref#metadata_info_tab_contents>, the two dimensional complex subspace spanned by $(1,0,a)$ and $(0,1,b)$ in complex $\ell\_\infty^3$ is not isometric to complex $\ell\_\infty^2$. Can anyone prove that?
| https://mathoverflow.net/users/136860 | Subspaces of $\ell_\infty^3$ | A funny isometry invariant to distinguish these normed spaces is: *The space of spheres of radius $2$ in the unit sphere of $(X,\|\cdot\|\_X)$ which are maximal by inclusion*, as described below. It turns out that for $\ell\_\infty^2(\mathbb C)$ it is a torus $\mathbb S^1\times \mathbb S^1$, and for the space $Y:=\text{span}\big((1,0,a),(0,1,b)\big)\subset \mathbb C^3$ as defined above it is a disjoint union of two circles, $\mathbb S^1\times \mathbb S^0$.
Another one is: *The length of the poset of all spheres of radius $2$ in the unit sphere of $(X,\|\cdot\|\_X)$ ordered by inclusion.* One gets respectively $2$ and $3$.
Consider the unit sphere $S\_X:=\{x\in X:\|x\|\_X=1\}$ of the normed space $(X,\|\cdot\|\_X)$ as a metric space, and the set $\Sigma\_2(S\_X)$ of all spheres of radius equal to the diameter $2$ of $S\_X$. This is both a metric space with the Hausdorff distance, and a POS with the inclusion relation, and these structures are clearly an isometry invariant of $X$. We can further consider the subspace $\Sigma^\*\_2(S\_X)$ of all maximal elements of $\Sigma\_2(S\_X)$ w.r.to inclusion. For instance, in the case of $X:=\ell^2\_\infty(\mathbb C)$, denoting by $\Delta$ the closed unit disk of $\mathbb C$, the sphere of radius $2$ and center $(x,y)\in S\_X$ is either $\{-x\}\times\Delta$, or $\Delta\times\{-y\}$, or $\big(\{-x\}\times\Delta\big)\cup\big(\Delta\times\{-y\}\big)$ according whether $|y|<|x|=1$, or $|x|<|y|=1$, or respectively $|x|=|y|=1$. In particular, maximal spheres of radius $2$ are exactly those of the third type, that is with center $(x,y)\in\partial\Delta\times\partial\Delta$. Since these spheres are homeomorphically parametrised by their centers, we conclude that in this case the space $\Sigma^\*\_2(S\_X)$ is topologically a torus $\mathbb S^1\times\mathbb S^1$. Note that this easily generalises for all $\ell\_\infty^d(\mathbb C)$, in particular for $d=3$: now the maximal spheres of radius $2$ in $S\_{\ell\_\infty^3(\mathbb C)}$ are exactly those of center $(x,y,z)\in\partial\Delta\times\partial\Delta\times\partial\Delta$.
With some more computations, yet by elementary arguments, it is also easy to compute this object for the given complex space $Y:=\text{span}\big((1,0,a),(0,1,b)\big)\subset \mathbb C^3$ as defined above. We just have to look at the traces on $Y$ of the inclusion of spheres of radius $2$ in the unit sphere of $S\_{\ell\_\infty^d(\mathbb C)}$: it follows from the assumptions on $a$ and $b$ that the maximal spheres of radius $2$ of $S\_Y$ are exactly those whose center $(x,y,z)\in Y$ has $|x|=|y|=|z|=1$. Moreover, for any $x\in\partial\Delta$ there are exactly $2$ distinct values of $y$ for which $(x,y,ax+by)\in Y$, which are of the form $y=\theta\_1 x$ and $y=\theta\_2 x$, for some complex numbers $\theta\_1\neq\theta\_2$ of unit modulus (reason: by the assumptions one has $|a+\theta b|\le 1$ for $\theta=1$ and $|a+\theta b|=|a|+|b|> 1$ for $\theta=\text{sgn}b /\text{sgn}a$, so the circle of center $a$ and radius $|b|$ has exactly $2$ intersections with the unit circle $\partial \Delta$, namely $|a+\theta\_1b|=|a+\theta\_2b|=1$ for $\theta\_1|=|\theta\_2|=1$, so that also $|ax+(\theta\_1x)b|=|a+(\theta\_2x)b|=1$ for any $|x|=1$). By similar arguments one checks that a strict inclusion of spheres of radius $2$ in $S\_{\ell\_\infty^3(\mathbb C)}$ gives a strict inclusion on the trace on the space $Y$, so that there are no other maximal spheres in $S\_Y$.
So the set of the maximal spheres now is a space homeomorphic to the union of the graphs of $\partial\Delta\ni x\mapsto \big(\theta\_j x, (a +b\theta\_j)x\big)$, ($j=1,2$),thus to $\mathbb S^1\times \mathbb S^0$.
**Rmk** As a variant, we may consider the length of $\Sigma\_2(S\_X)$ as a partially ordered set (the maximum cardinality of its chains). Then, by the above computations, one gets length $d$ for $\ell^d\_\infty(\mathbb C)$, and length $3$ for the space $Y$.
| 5 | https://mathoverflow.net/users/6101 | 428609 | 173,758 |
https://mathoverflow.net/questions/428135 | 6 | I have encountered with the term "Morse index" multiple times while reading papers in PDEs (e.g. [1] and [2]). However the definition differs for each context. As far as I know this came from Morse theory in differential topology. But I am not sure how the definition of "Morse index" in [2] can be related to the definition in the differential topology. The definition in [2] is:
For a given (fractional) linear differential operator $L\_+$, **Morse index of $L\_+$** in the sector of even functions is defined by
$$\mathcal{N}\_{-,\text{even}}(L\_+):=\#\{e<0:e\text{ is eigenvalue of }L\_+\text{ restricted to }L\_\text{even}^2(\mathbb{R})\}.$$
As this seems functional analysis terminology, how does this relate to the definition in the differential topology, which essentially is the number of critical points in a given manifold? Negative eigenvalue of a linear differential operator somewhat resembles critical points in a manifold?
Sorry for my ignorance, I am new to this topic, and thank you in advance.
**References**
[1] Kelei Wang and Juncheng Wei, "[Finite Morse index implies finite ends](https://doi.org/10.1002/cpa.21812)" Communications on Pure and Applied Mathematics 72.5 (2019): 1044-1119, [MR3935478](https://mathscinet.ams.org/mathscinet-getitem?mr=MR3935478), [Zbl 1418.35190](https://zbmath.org/?q=an%3A1418.35190).
[2] Rupert L. Frank and Enno Lenzmann, "[Uniqueness of non-linear ground states for fractional Laplacians in $\Bbb R$](https://doi.org/10.1007/s11511-013-0095-9)" Acta mathematica 210.2 (2013): 261-318, [MR3070568](https://mathscinet.ams.org/mathscinet-getitem?mr=MR3070568), [Zbl 1307.35315](https://zbmath.org/?q=an%3A1307.35315).
| https://mathoverflow.net/users/151368 | Morse index in PDEs | In finite dimensional Morse theory, you study a function $f:M\to\mathbb{R}$ and look for it's critical points, i.e., where $(df)\_p = 0$. Then, Morse theory says that a count of these critical points is related to the underlying topology of $M$, as long as you count correctly. To do so, you should define the **Hessian** of $f$, $(D^2f)\_p$. (This is a well-defined symmetric bilinear form at a critical point, see e.g. <https://www.asc.ohio-state.edu/terekcouto.1/texts/hessian.pdf>). The Morse index is the number of *negative* eigenvalues of the Hessian. In Morse theory, you usually assume that the Hessian has no zero eigenvalues (and then call $f$ a **Morse function**). Now, Morse theory says, for example that
$$
\sum\_{(df)\_p =0} (-1)^{\textrm{index}(f,p)} = \chi(M)
$$
(where $\chi(M)$ is the Euler characteristic).
---
Now, when analyzing certain PDE's, some of this story holds up and some of it breaks down. A common PDE arises as the critical point equation for some "energy" which is a function $E : M \to \mathbb{R}$, where $M$ is an (infinite dimensional) Hilbert/Banach/etc space/manifold. For simplicity, let's assume that $M=\mathcal{H}$ is a vector space (e.g. Hilbert space). Now, you can define
$$
(dE)\_u(v) = \frac{d}{dt}\Big|\_{t=0} E(u+tv).
$$
If $E$ is sufficiently nice, this will exist and for $E$ with certain structure, the equation $(dE)\_u = 0$ will be a PDE. (I will give an example below).
Now, if $(dE)\_u = 0$, we can define the Hessian of $E$ at $u$ by
$$
(D^2E)\_u(\varphi,\varphi) : = \frac{d^2}{dt^2}\Big|\_{t=0} E(u+t\varphi).
$$
(this just defines the quadratic form $\mathcal{H}\to\mathbb{R}$ associated to the Hessian which is a symmetric bilinear form $\mathcal{H}\times \mathcal{H}\to\mathbb{R}$, but it's easy to recover the full Hessian from the quadratic form).
Now, we run into a potential problem. $\mathcal{H}$ is infinite dimensional, so even if it is possible to diagonalize $(D^2 E)\_u$, there should be an infinite number of eigenvalues (or continuous spectrum, etc.). However, in certain cases, the *negative* eigenvalues will be finite. (This is not always the case, there's entire fields devoted to studying the case where the Hessian has infinite positive and negative eigenvalues, you can find a survey on Floer theory to read.)
The reason you might expect the *negative* eigenvalues to be finite, is that one often studies functionals with some good "coercivity" property. By this, I just mean loosely that it's often possible to *minimize* the functional in some class. If a minimizer exists, it will have no negative eigenvalues (this is just the second derivative test from calculus). So you may dream that there are some other non-minimizing solutions with only finitely many negative eigenvalues.
---
To give a concrete example, I will use the PDE from (1) since I am familiar with this. The functional is $E:H^1(\mathbb{R}^n) \cap L^4(\mathbb{R}^n)\to\mathbb{R}$ (one can get very confused about exactly what spaces we are using, it's probably best to think of this as a formal discussion at first go, but of course most/all things can be made rigorous) defined by
$$
E(u) = \int\_{\mathbb{R}^n} \frac 12 |\nabla f|^2 + W(f)
$$
for $W(t) = \frac 1 4 (1-t^2)^2$ a "double well potential." Now, it's an exercise to show that $(dE)\_u=0$ if and only if $u$ is smooth and solves the PDE
$$\tag{\*}
\Delta u = W'(u).
$$
Moreover, you can check that the Hessian becomes
$$
(D^2E)\_u(\varphi,\varphi) = \int\_{\mathbb{R}^n} |\nabla \varphi|^2 + W''(u) \varphi^2.
$$
It *turns out* that there exist solutions $u$ so that the Hessian is non-negative. These solutions actually minimize $E$ in an appropruiate sense, so this is not suprising. Moreover (and this takes much more work to prove) there exist solutions so that $(D^2E)\_u$ has finitely many negative eigenvalues, i.e., *finite Morse index solutions*. There *also* exist solutions with infinite Morse index.
The finite Morse index assumption could be seen as an interesting class of solutions to study on its own right. Another motivation would be that usually one first studies solutions that minimize the functional. Then, after that theory is well developed, people study the stable case (i.e., no negative eigenvalues of the Hessian, but no global comparison arguments are allowed). Then, finite Morse index is a natural generalization (and in these cases admits a rich class of examples whose behavior is interesting to characterize).
There's also a deeper story that well-motivates the finite Morse index assumption in this (and other cases). Consider $(N,g)$ a closed Riemannian manifold. Suppose we want to study the same PDE $\Delta\_g u = W'(u)$ on $N$ (this makes sense since the Laplacian can be defined via the Riemannian metric). It turns out that you can use an infinite dimensional version of Morse theory to *find* solutions to this PDE (this is done [here](https://arxiv.org/abs/1608.06575)). This is closely related to the Morse equality in the finite dimensional case, namely since the functional $E: H^1(N) \to \mathbb{R}$ is even, you can consider it as a functional on the quotient of $H^1(N)\setminus\{0\}/\sim$ by $u\sim-u$. This space has a lot of topology, leading to many critical points (this idea goes back to a beautiful [paper](http://www.math.uchicago.edu/%7Eshmuel/QuantCourse%20/Gromov,%20Dimension,%20non-linear%20spectra%20and%20width.pdf) of Gromov).
It turns out to be natural to add a scaling parameter $\varepsilon^2\Delta\_g u = W'(u)$. As $\varepsilon\to0$, there exists more and more solutions, with higher and higher Morse index.
For a fixed Morse index solution, we can prove that the level set ${u=0}$ will collapse to a minimal hypersurface. This is a complicated story, but the limit as $\varepsilon\to 0$ can be studied in one way by dilating by $\varepsilon^{-1}$. Appropriately doing so, you will always be able to pass to a limit to find a solution to the PDE (\*) with finite Morse index. Understanding their structure is then important. The Wang--Wei paper you refer to accomplishes certain very important properties in the $n=2$ dimensional case (later some generalizations to higher dimensions have been done).
In case you are interested in this specific PDE further, you can see these lecture notes: [here](http://www.cmls.polytechnique.fr/perso/pacard/Publications/70.pdf), [here](http://math.uchicago.edu/%7Eguaraco/princeton2019.pdf), or [here](http://web.stanford.edu/%7Eochodosh/AllenCahnSummerSchool2019.pdf).
---
As an addendum: the second paper you linked (the ground state paper) considers only even functions for various reasons. This is why there is this restriction. I am less familiar with this topic, but I think they're interested in PDE's of the form $(\Delta)^s u = F(u)$ and specifically *radial* solutions. In dimension $n=1$, a radial solution should be the same thing as an even solution.
| 8 | https://mathoverflow.net/users/1540 | 428613 | 173,759 |
https://mathoverflow.net/questions/428616 | 3 | I've edited this post two years ago on Mathematics Stack Exchange, with identifier **3590406** and same title [*On conjectures about the arithmetic function that counts the number of Sophie Germain primes*](https://math.stackexchange.com/questions/3590406/on-conjectures-about-the-arithmetic-function-that-counts-the-number-of-sophie-ge), there were some comments from professors (about the veracity of these consejectures). I ask what work can be done to find out a counterexample or if you can to refute some of these conjectures.
In this post we denote (for a fixed positive integer or real number $x$) as $$\operatorname{Germain}(x)=\#\{\text{ primes }p\leq x\,|\,2p+1\text{ is also prime}\}$$
the arithmetic function that counts the number of primes $p$ less than a given positive real $x$ satisfying that $2p+1$ is also prime. As general reference I add the article from Wikipedia [*Sophie Germain prime*](https://en.wikipedia.org/wiki/Sophie_Germain_prime) that refers that is unproven the existence of infinitely many Sophie Germain primes, and the articles also from Wikipedia [*Second Hardy–Littlewood conjecture*](https://en.wikipedia.org/wiki/Second_Hardy%E2%80%93Littlewood_conjecture) and for [*Legendre's constant*](https://en.wikipedia.org/wiki/Legendre%27s_constant). I was inspired in these articles and a few experiments using Pari/GP scripts to state the following conjectures.
**Conjecture 1.** *One has* $$\operatorname{Germain}(x+y)\leq \operatorname{Germain}(x)+\operatorname{Germain}(y)$$
*for all integer* $x\geq 2$ *and all integer* $y\geq 2$.
**Conjecture 2.** *One has that* $\forall x>87$
$$\operatorname{Germain}(x)<2C\frac{x}{\log^2 x-20}$$
*where* $C$ *denotes the Hardy–Littlewood's twin prime constant.*
>
> **Question.** Are these known, or is it possible to prove or refute any of previous conjectures? Can you find counterexamples for these or add heuristics to know what about the veracity of this kind of conjectures? **Many thanks.**
>
>
>
I've tested the first conjecture only for the segments of integers $2\leq x,y\leq 500$. The second conjecture is true for the segment $88\leq x\leq 25000$, but my belief is that is false, I've tested different constants $\mu$ in the expression $2C\frac{x}{\log^2 x-\mu}$ that are approximately $\mu\approx 20$. I can add in a comment the scripts written in Pari/GP that I've used to check it.
References:
-----------
[1] G. H. Hardy and J. E. Littlewood, *Some Problems of 'Partitio Numerorum.' III. On the Expression of a Number as a Sum of Primes*, Acta Math. (44), J. E. (1923) pp. 1–70.
| https://mathoverflow.net/users/142929 | On conjectures about the arithmetic function that counts the number of Sophie Germain primes | A reasonable conjecture is that
$$ \operatorname{Germain}(x) = 2 C \int\_{2^x} \frac{dx}{\log^2 x} + O( x^{1-\delta} )$$ for some $\delta >0$. This is the Hardy-Littlewood conjecture with power-saving error term.
A function-field analogue of this conjecture follows from the methods of my paper [On the Chowla and twin primes conjectures over $\mathbb F\_q[t]$](https://arxiv.org/abs/1808.04001) with Mark Shusterman.
A good way to test the validity of these conjectures is to see how they relate to this asymptotic. For example, Conjecture 1 follows as long as $x$ and $y$ are not too far apart.
The integral $\int\_{2^x} \frac{dx}{\log^2 x}$ has an asymptotic expansion $$x /(\log x)^2 + 2x / (\log x)^3 + 6 x / (\log x)^4+ \dots= x / (\log^2 x - 2 \log x + O(1))$$ and the power savings error term is, once we put it on the denominator, smaller than that $O(1)$, thus it is reasonable to conjecture that
$$\operatorname{Germain}(x) = 2 C \frac{x}{ \log^2 x - 2 \log x + O(1) } $$
and one could even look for numerical evidence about how large this $O(1)$ should be.
So indeed Conjecture 2 should be wrong but a modified version with the $-2 \log x$ could well be correct.
| 6 | https://mathoverflow.net/users/18060 | 428621 | 173,760 |
https://mathoverflow.net/questions/428542 | 5 | Let $G$ be a connected reductive group defined over a finite field $F$ of characteristic $> 3$. Is it true that the commutator group of $G(F)$ is perfect? This is true if $G$ is assumed to be semisimple.
| https://mathoverflow.net/users/4690 | Is the derived group of the G(F) perfect | $\newcommand{\ssc}{\text{sc}}
\newcommand{\der}{\text{der}}
\DeclareMathOperator{\im}{im}$**Yes,** this is true for a reductive $F$-group $G$, if
for the *simply connected semisimple* $F$-group $G^\ssc$, see below,
the group of $F$-points $G^\ssc(F)$ is perfect,
that is, the derived group $G^\ssc(F)^\der$ coincides with $G^\ssc(F)$.
>
> **Theorem.** Let $G$ be a (connected) reductive group over a field $F$.
> Let $G^\ssc$ denote the universal cover of the commutator subgroup $[G,G]$ of $G$.
> Consider Deligne's homomorphism
> $$\rho\colon\, G^\ssc\twoheadrightarrow [G,G]\hookrightarrow G.$$
> If the group $G^\ssc(F)$ is perfect, then $G(F)^\der=\rho\bigl(G^\ssc(F)\bigr)$, and hence $G(F)^\der$ is perfect as well.
>
>
>
*Proof.*
Consider the regular maps of $F$-varieties
\begin{align\*}
\varkappa&\colon G\times G\to G,\qquad (x,y)\mapsto xyx^{-1}y^{-1}\\
\varkappa^\ssc&\colon G^\ssc\times G^\ssc\to G^\ssc,\quad (x,y)\mapsto xyx^{-1}y^{-1}.
\end{align\*}
By definition,
\begin{align\*}
G(F)^\der&=\bigl\langle\varkappa\bigl(G(F)\times G(F)\bigr)\bigr\rangle,\\
G^\ssc(F)^\der&=\bigl\langle\varkappa^\ssc\bigl(G^\ssc(F)\times G^\ssc(F)\bigr)\bigr\rangle,
\end{align\*}
where $\langle S \rangle$ denotes the subgroup generated by a set $S$.
Deligne observed that $\varkappa$ factors via $G^\ssc$,
that is, there exists a regular map
$$\widetilde\varkappa\colon G\times G\to G^\ssc$$
such that
$$\varkappa=\rho\circ\widetilde\varkappa\colon\, G\times G\,\to\, G^\ssc\,\to\, G.$$
It follows that
$\im\varkappa\_F\subseteq \im\rho\_F$, where we use the subscript $\_F$ to denote
maps on $F$-points, for instance, $\rho\_F\colon G^\ssc(F)\to G(F)$.
Therefore,
\begin{align}\tag{1}\label{1}
G(F)^\der=\langle\im \varkappa\_F\rangle\subseteq\langle\im\rho\_F\rangle=\rho\bigl(G^\ssc(F)\bigr).
\end{align}
Moreover, we have
$$\widetilde\varkappa\circ(\rho\times\rho)=\varkappa^\ssc\colon\ G^\ssc\times G^\ssc\ \to\ G\times G\ \to\ G^\ssc$$
and
$$\rho\circ\widetilde\varkappa\circ(\rho\times \rho)=\rho\circ\varkappa^\ssc\colon\, G^\ssc\times G^\ssc\to G\times G\to G^\ssc\to G.$$
We see that
\begin{align\*}
\im\varkappa\_F=\im(\rho\circ\widetilde\varkappa)\_F\supseteq\im&\bigl(\rho\circ\widetilde\varkappa\circ(\rho\times \rho)\bigr)\_F\\
&=\im(\rho\circ\varkappa^\ssc)\_F=\rho(\im \varkappa^\ssc\_F),
\end{align\*}
whence
\begin{align}\tag{2}\label{2}
G(F)^\der=\langle\im\varkappa\_F\rangle\supseteq\rho\langle\im\varkappa^\ssc\_F\rangle= \rho \bigl(G^\ssc(F)^\der\bigr).
\end{align}
From \eqref{1} and \eqref{2} we obtain that
\begin{align}\tag{3}\label{3}
\rho\bigl(G^\ssc(F)^\der\bigr)\subseteq G(F)^\der\subseteq \rho\bigl(G^\ssc(F)\bigr).
\end{align}
Now assume that $G^\ssc(F)$ is perfect. Thus means that
$$G^\ssc(F)^\der=G^\ssc(F),\quad\text{whence}\quad \rho\bigl(G^\ssc(F)^\der\bigr)=\rho\bigl(G^\ssc(F)\bigr),$$
and we obtain from \eqref{3} that $G(F)^\der=\rho\bigl(G^\ssc(F)\bigr)$.
Since $G^\ssc(F)$ is perfect, so is $\rho\bigl(G^\ssc(F)\bigr)$. Thus $G(F)^\der$ is perfect, as required.
| 6 | https://mathoverflow.net/users/4149 | 428626 | 173,763 |
https://mathoverflow.net/questions/428631 | 2 | For any set $X$ and cardinal $\kappa$, we denote by
$\ [X]^\kappa :=\ \binom X\kappa\ $ the collection of subsets of $X$ having cardinality $\kappa$.
If $X$ is a set, we call a set system $E\subseteq {\cal P}(X)$ *linear* if for all $a\neq b\in X$ there is exactly one $e\in E$ with $\{a,b\}\subseteq e$. For example, for every set $X$, the set system $[X]^2$ is linear.
For which integers $k\in \omega$ with $k > 2$ is there a linear set system $E\subseteq [\omega]^k$?
| https://mathoverflow.net/users/8628 | $k$-regular linear set systems | I think they exist for all $k \geq 2$. Here is a proof. Fix an enumeration of $[\omega]^2$. For the first step in the process add an arbitrary $k$-set $q\_1=q(p\_1)$ containing the first pair $p\_1$ of the enumeration. At step $i$, if $\ p\_i\subseteq q(p\_j)\ $ for some $\ j<i\ $ then let $\ q(p\_i):=q(p\_j).\ $ Otherwise, select any $(k-2)$-set $\ r\subseteq \omega\setminus p\_i\ $ that is not contained in
$\, \bigcup\_{t=1}^{i-1}\,q(p\_t),\ $ and let
$$ q(p\_i)\ :=\ p\_i\cup r $$
for every $\ i>1.\ $ Then $\ E\ :=\ \{q(p\_i):\ i\in\mathbb N\}\ $
is the solution.
| 4 | https://mathoverflow.net/users/2233 | 428639 | 173,768 |
https://mathoverflow.net/questions/427603 | 6 | In a compact connected Lie group $G$, each element is conjugate to an element of a maximal torus $T$. For a classical group, one can pick a basis of the tautological representation such that $T$ is represented by diagonal matrices. The conjugacy classes in $G$ are in bijective correspondence with the orbits of the Weyl group $W$ on $T$. Thus, say, for $\operatorname{SU}(n)$ the conjugacy class of a unitary matrix $g$ is determined by its (unordered) eigenvalues, and the same holds for $\operatorname{SO}(2n+1)$.
For $\operatorname{SO}(2n)$, the torus $T$ can be represented by matrices $\operatorname{diag}(\gamma\_1,\dotsc,\gamma\_n,\gamma\_n^{-1},\ldots,\gamma\_1^{-1})$, and $W$ acts by permutations on $\gamma\_1,\dotsc,\gamma\_n$ and an *even* number of changes $\gamma\_i\leftrightarrow\gamma\_i^{-1}$ (contrary to $\operatorname{SO}(2n+1)$, where an arbitrary number of switches can be done, by virtue of an extra eigenvalue $1$). Thus the eigenvalues alone determine a pair of conjugacy classes which differ by one switch.
>
> **Q:** How, given a matrix $g\in\operatorname{SO}(2n)$, to determine the corresponding element of the torus $T$?
>
>
>
It can be done by diagonalizing $g$ by a matrix $C$ and checking whether $\det(C)=1$ or $-1$, but I was hoping for a more efficient solution (calculating the eigenvalues of $g$ is faster then finding $C$). Are there any other invariants which can, given the eigenvalues $\gamma\_1,\dotsc,\gamma\_n,\gamma\_n^{-1},\dotsc,\gamma\_1^{-1}$ of $g$, tell whether $g$ is conjugate to $\operatorname{diag}(\gamma\_1,\dotsc,\gamma\_n,\gamma\_n^{-1},\dotsc,\gamma\_1^{-1})$ or to $\operatorname{diag}(\gamma\_1^{-1},\gamma\_2,\dotsc,\gamma\_n,\gamma\_n^{-1},\dotsc,\gamma\_2^{-1},\gamma\_1)$?
| https://mathoverflow.net/users/5018 | How to distinguish conjugacy classes in SO(2n) efficiently? | One method, I don't know whether it is the most efficient, is to compute the following polynomial quantity $Q(g) = \mathrm{Pf}(g-g^t)$ for $g\in\mathrm{SO}(2n)$, where $g^t$ means the transpose of $g$ and $\mathrm{Pf}$, known as the *Pfaffian*, is a polynomial of degree $n$ on the skew-symmetric $2n$-by-$2n$ matrices that satisfies $\det(S) = \mathrm{Pf}(S)^2$ for skew-symmetric $2n$-by-$2n$ matrices.
This polynomial satisfies $Q(CgC^{-1}) = \det(C)\,Q(g)$ for $C$ in $\mathrm{O}(2n)$, so it takes different values on two elements of $\mathrm{SO}(2n)$ that are conjugate in $\mathrm{O}(2n)$ but not in $\mathrm{SO}(2n)$. (Also, if $Q(g)=0$, then the conjugacy class of $g$ in $\mathrm{SO}(2n)$ is the same as its conjugacy class in $\mathrm{O}(2n)$.)
| 7 | https://mathoverflow.net/users/13972 | 428641 | 173,770 |
https://mathoverflow.net/questions/428610 | 0 | A functional Hilbert space $\mathscr H=\mathscr H(\Omega)$ is a Hilbert space of complex valued functions on a (nonempty) set $\Omega$, which has the property that point evaluations are continuous i.e. for each $\lambda\in \Omega$ the map $f\mapsto f(\lambda)$ is a continuous linear functional on $\mathscr H$. The Riesz representation theorem ensure that for each $\lambda\in \Omega$ there is a unique element $k\_{\lambda}\in \mathscr H$ such that $f(\lambda)=\langle f,k\_{\lambda}\rangle$ for all $f\in \mathscr H$. The collection $\{k\_{\lambda} : \lambda\in \Omega\}$ is called the reproducing kernel of $\mathscr H$. For $\lambda\in \Omega$, let $\hat{k\_{\lambda}}=\frac{k\_{\lambda}}{\|k\_{\lambda}\|}$ be the normalized reproducing kernel of $\mathscr H$.
For a bounded linear operator $A$ on $\mathscr H$, we define the following norms:
\begin{align\*}
N\_1(A):=\sup\{\big|\langle T\widehat{k}\_{\lambda},\widehat{k}\_{\mu}\rangle\big|: \lambda,\mu\in\Omega\} \qquad \textrm{and} \qquad N\_2(A):=\sup\{\|T\widehat{k}\_{\lambda}\|: \lambda\in\Omega\}.
\end{align\*}
>
> Is $N\_1=N\_2$? Clearly $N\_1(A)\leq N\_2(A)$.
>
>
>
| https://mathoverflow.net/users/113054 | A reproducing kernel Hilbert space | A useful test case for RKHS (which is not like the interesting examples, but does satisfy the definitions) is $\Omega={\mathbb N}$ and $H=\ell^2({\mathbb N})$. Note that $\hat{k\_n}$ is just the usual unit basis vector that is $1$ in position $n$ and $0$ everywhere else.
Viewing $T$ as an ${\mathbb N}\times {\mathbb N}$ matrix, $N\_1(T)$ is the maximum absolute value of all matrix entries, and $N\_2(T)$ is the maximum $\ell^2$-norm of all columns in the matrix.
It is then easy to find examples where $N\_1(T)$ is strictly less than $N\_2(T)$, because this is basically asking for vectors in $\ell^2$ whose sup norm is strictly smaller than their $\ell^2$ norm.
In fact, we could have built a counterexample with $\Omega$ being a 2-element set; then the RKHS is just ${\mathbb C}^2$ and you could take $T$ to be the $2\times 2$ matrix with all entries equal to $1$.
| 2 | https://mathoverflow.net/users/763 | 428643 | 173,771 |
https://mathoverflow.net/questions/428638 | 1 | Let $N$ be a Riemannian manifold, and $\widetilde{N}\subset N$ a closed submanifold. If we look at the total spaces of the tangent bundles, we get that $T\widetilde{N}$ is a submanifold of $TN$.
If we now use the metric to identify $\widetilde{M}=T^\*\widetilde{N}$ with $T\widetilde{N}$, and $M=T^\*N$ with $TN$, this gives an embedding of $\widetilde{M}$ into $M$.
Both of $\widetilde{M}$ and $M$ are of course symplectic manifolds, with canonical symplectic forms. Is there a coordinate-free way of seeing that this is a symplectic embedding, i.e. that the canonical symplectic form on $M$ restricts to the canonical symplectic form on $\widetilde{M}$?
| https://mathoverflow.net/users/940 | Cotangent bundles to Riemannian manifolds and submanifolds | May I please call the submanifold $M$ instead of $\tilde N$?
For a smooth manifold $N$, $T^\ast N$ does not just have a canonical symplectic form $\omega$. It also has a canonical $1$-form $\alpha$ such that $d\alpha=\omega$. For $w\in T^\ast N$, the element $\alpha\_w\in T^\ast\_w(T^\ast N)$ can be defined in a coordinate-free way by giving its value at $v\in T\_w(T^\ast N)$:
$$
\langle \alpha\_w,v\rangle=\langle w,\pi\_\ast v\rangle
$$
where $\pi\_\ast :T\_w(T^\ast N)\to T\_{\pi(w)} N$ is induced by the bundle projection $\pi:T^\ast N\to N$.
In the presence of a Riemannian structure on $N$ this becomes:
For $w\in T N$, the element $\alpha\_w\in T^\ast\_w(T N)$ is defined by giving its value at $v\in T\_w(T N)$:
$$
\langle \alpha\_w,v\rangle=w\cdot \pi\_\ast v
$$
Here "$\cdot$" means Riemannian inner product, and $\pi:T N\to N$ is the bundle projection.
If the $\alpha $ for $N$ restricts to the $\alpha$ for $M$ then also the $\omega$ restricts to the $\omega$. And surely this is the case; for $w\in TM$ and $v\in T\_w(TM)$ the inner product $w\cdot \pi\_\ast v$ is the same whether calculated in $M$ or in $N$.
| 3 | https://mathoverflow.net/users/6666 | 428652 | 173,776 |
https://mathoverflow.net/questions/428651 | 7 | Jech exercise 13.3 states:
>
> If $M$ is closed under Gödel operations and extensional, and $\pi$ is the transitive collapse of $M$, then $\pi(G\_i(X,Y))=G\_i(\pi X,\pi Y)$ for all $i=1,\ldots,10$ and all $X$, $Y\in M$.
>
>
>
I'm able to show this for the first 5 Gödel operations ($\{X,Y\}$, $X\times Y$, $\varepsilon(X,Y)$, $X-Y$, and $X\cap Y$), but I'm a little stuck on $G\_6(X)=\bigcup X$. I can see that the problem is essentially equivalent to showing that
$$\left(\bigcup X\right)^M=\bigcup X,$$
as then $\pi(\bigcup X)=\pi((\bigcup X)^M)=(\bigcup\pi X)^{\pi M}=\bigcup\pi X$ since $\bigcup$ is a $\Delta\_0$ operation and $\pi X$ is transitive.
The problem is that $M$ is *not* generally transitive, so the $\Delta\_0$ observation doesn't help on that side, and I don't see any other reason why the equation above should be true. I don't even see a reason to believe that $(\bigcup X)^M$ exists!
Since $M$ is extensional, it is enough to show that $M\cap(\bigcup X)^M=M\cap\bigcup(M\cap X)=M\cap\bigcup X$, but again, why should that be true?
I've already checked Frédéric Wang's solutions and Monk's notes. The former don't address this case at all, and the latter seems to implicitly assume that $M$ is transitive, though I may be missing something.
I anticipate problems with the remaining operations for similar reasons, so I hope that clarity on this one resolves the rest.
(Even if this turns out to be false, I doubt it affects anything of note. In most cases of interest (or so it seems), $M$ is an elementary submodel of $L\_\delta$ (or similar), and in those cases $\bigcup$ is a defined notion, hence absolute between $M$ and $L\_\delta$.)
| https://mathoverflow.net/users/4133 | Why do $\pi$ and $\bigcup$ commute for Gödel-closed extensional classes? | Here is what I wrote when I solved this problem a few years ago:
As $M$ is extensional, the transitive collapse is an isomorphism. The statement $C=G\_i(A,B)$ can be expressed by a $\Delta\_0$ formula $\phi\_i(A,B,C)$. If we assume that these operations are absolute for extensional classes, then $Z=G\_i(X,Y)$ if and only if $V\models \phi\_i[X,Y,Z]$, if and only if $M\models \phi\_i[X,Y,Z]$, if and only if $\pi(M)\models \phi\_i[\pi(X),\pi(Y),\pi(Z)]$, if and only if $V\models \phi\_i[\pi(X),\pi(Y),\pi(Z)]$, if and only if $\pi(Z)=G\_i(\pi(X),\pi(Y))$.
So it suffices to show that these $\Delta\_0$ formulas are absolute not only for transitive classes, but extensional classes closed under the Gödel operations. Note that once we show that the given sets satisfy the conditions, extensionality proves that they are the unique set satisfying the condition.
$\phi\_6[X]$ is
$$
(\forall z\in Z, \exists x\in X,\, z\in x)\land (\forall x\in X,\, \forall u\in x,\, u\in Z).
$$
The set $\bigcup X$ clearly satisfies the second conjunct in $M$, and we will show the other. Let $z\in \left(\bigcup X\right)\cap M$. As $M$ is closed under the operations, it contains the two distinct sets $G\_3(\bigcup X-\{z\},X)$ and $G\_3(\bigcup X,X)$. By extensionality, there exists some ordered pair $(z,x)\in \left(\left(\bigcup X\right)\times X\right)\cap M$ with $z\in x$. Hence, by the previous exercise $x\in M$.
| 7 | https://mathoverflow.net/users/3199 | 428656 | 173,777 |
https://mathoverflow.net/questions/428653 | 0 | Consider a random scalar variable $X$ with arbitrary measure.
I'm after a basis of polynomial functions $\{p\_k\}\_{k=0}^\infty$ which are orthonormal with respect to $X$ in the sense that
\begin{equation}
\mathbb{E}\_X [p\_k(X)p\_{k'}(X)] = \delta\_{kk'}.
\end{equation}
When discussing orthogonal polynomial bases, the measure of integration is usually assumed.
For example, if $X \sim \mathcal{N}(0,1)$, then $\{p\_k\}\_{k=0}^\infty$ are the Hermite polynomials.
However, it seems there ought to exist generic expressions for such orthogonal polynomials, with the coefficients given in terms of moments of $X$. For example, applying the typical Gram-Schmidt procedure, one can quickly find that
\begin{align}
p\_0(X) &= 1 \\
p\_1(X) &= \frac{X - \mathbb{E}[X]}{\sqrt{\text{Var}[x]}} \\
p\_2(X) &= \ \ \ ...
\end{align}
Are there known expressions for the rest of this polynomial basis (or even just the next few elements)?
In light of the expression for $p\_1$, perhaps centered moments or cumulants are involved.
| https://mathoverflow.net/users/60511 | Orthogonal polynomials w.r.t. an arbitrary measure | For example, you can write orthogonal polynomials as determinants
$p\_n(x) = c\_n \, \det \begin{bmatrix}
m\_0 & m\_1 & m\_2 &\cdots & m\_n \\
m\_1 & m\_2 & m\_3 &\cdots & m\_{n+1} \\
\vdots&\vdots&\vdots&\ddots& \vdots \\
m\_{n-1} &m\_n& m\_{n+1} &\cdots &m\_{2n-1}\\
1 & x & x^2 & \cdots & x^n
\end{bmatrix}$,
where $c\_n$ is some constant for normalization and $m\_k$ is the k-th moment.
A good book concerning orthogonal polynomials is Akhiezer, The Classical Moment problem
| 1 | https://mathoverflow.net/users/483414 | 428666 | 173,778 |
https://mathoverflow.net/questions/428669 | 11 | The character table of a finite group will be called *integral* if all its entries are integers. There are $11$ such groups up to order $16$, namely $C\_1$, $C\_2$, $C\_2^2$, $S\_3$, $D\_8$, $Q\_8$, $C\_2^3$, $D\_{12}$, $C\_2 \times D\_8$, $C\_2 \times Q\_8$ and $C\_2^3$. There are $76$ such groups up to order $120$ (see Appendix), the last one of the list, namely $S\_5$, is of specific interest as it is the first non-solvable one. In fact, every symmetric group has an integral character table. It can be proved using the following result (Lemma 7.15 in [this note](https://eugeniomaths.files.wordpress.com/2020/07/ghsvrevised12july.pdf) by [Sam Raskin](https://web.ma.utexas.edu/users/sraskin/)):
>
> **Lemma**: Let $G$ be a finite group such that for all $g \in G$ of order $n$
> and for all $m$ coprime to $n$, $g^m$ is conjugate to $g$. Then $G$
> has an integral character table.
>
>
>
**Question**: What is known about the finite groups with integral character table? Are they classified?
I mainly ask for references reviewing what is known about such groups. Otherwise here is a list of questions:
* Is the converse of above lemma true?
* I guess that $C\_2$ is the only such group which is simple. Is it true?
* I guess that $C\_1$ is the only such group of odd order. Is it true?
* Is there such a group with a non-abelian and non-alternating simple normal subgroup?
* Is every finite group a normal subgroup of such a group?
---
**Appendix**: List of all the finite groups with integral character table up to order $120$.
```
gap> for o in [1..120] do
> if o=1 then Print("\n","|G| ","Nr ","G ","\n","\n");fi;
> n:=NrSmallGroups(o);;
> for i in [1..n] do
> G:=SmallGroup(o,i);;
> irr:=Irr(G);;
> s:=Size(irr);;
> c:=0;;
> for j in [1..s] do
> if Conductor(irr[j])<>1 then
> c:=1;;
> break;
> fi;
> od;
> if c=0 then
> Print(o," ",i," ",StructureDescription(G),"\n");
> fi;
> od;
> od;
|G| Nr G
1 1 1
2 1 C2
4 2 C2 x C2
6 1 S3
8 3 D8
8 4 Q8
8 5 C2 x C2 x C2
12 4 D12
16 11 C2 x D8
16 12 C2 x Q8
16 14 C2 x C2 x C2 x C2
18 4 (C3 x C3) : C2
24 12 S4
24 14 C2 x C2 x S3
32 27 (C2 x C2 x C2 x C2) : C2
32 34 (C4 x C4) : C2
32 35 C4 : Q8
32 43 C8 : (C2 x C2)
32 44 (C2 x Q8) : C2
32 46 C2 x C2 x D8
32 47 C2 x C2 x Q8
32 49 (C2 x C2 x C2) : (C2 x C2)
32 50 (C2 x Q8) : C2
32 51 C2 x C2 x C2 x C2 x C2
36 10 S3 x S3
36 13 C2 x ((C3 x C3) : C2)
48 38 D8 x S3
48 40 Q8 x S3
48 48 C2 x S4
48 51 C2 x C2 x C2 x S3
54 14 (C3 x C3 x C3) : C2
64 134 ((C4 x C4) : C2) : C2
64 137 (C4 : Q8) : C2
64 138 ((C2 x C2 x C2 x C2) : C2) : C2
64 177 ((C4 x C4) : C2) : C2
64 178 (C4 : Q8) : C2
64 182 C8 : Q8
64 202 C2 x ((C2 x C2 x C2 x C2) : C2)
64 211 C2 x ((C4 x C4) : C2)
64 212 C2 x (C4 : Q8)
64 215 (C2 x C2 x D8) : C2
64 216 (C2 x ((C4 x C2) : C2)) : C2
64 217 ((C4 x C4) : C2) : C2
64 218 (C2 x ((C4 x C2) : C2)) : C2
64 224 ((C2 x Q8) : C2) : C2
64 225 (C4 : Q8) : C2
64 226 D8 x D8
64 230 Q8 x D8
64 239 Q8 x Q8
64 241 ((C4 x C2 x C2) : C2) : C2
64 242 ((C4 x C4) : C2) : C2
64 243 ((C4 x C2 x C2) : C2) : C2
64 244 (C4 : Q8) : C2
64 245 (C2 x C2) . (C2 x C2 x C2 x C2)
64 254 C2 x (C8 : (C2 x C2))
64 255 C2 x ((C2 x Q8) : C2)
64 261 C2 x C2 x C2 x D8
64 262 C2 x C2 x C2 x Q8
64 264 C2 x ((C2 x C2 x C2) : (C2 x C2))
64 265 C2 x ((C2 x Q8) : C2)
64 267 C2 x C2 x C2 x C2 x C2 x C2
72 40 (S3 x S3) : C2
72 41 (C3 x C3) : Q8
72 43 (C3 x A4) : C2
72 46 C2 x S3 x S3
72 49 C2 x C2 x ((C3 x C3) : C2)
96 209 C2 x D8 x S3
96 212 C2 x Q8 x S3
96 226 C2 x C2 x S4
96 227 ((C2 x C2 x C2 x C2) : C3) : C2
96 230 C2 x C2 x C2 x C2 x S3
108 17 ((C3 x C3) : C3) : (C2 x C2)
108 39 ((C3 x C3) : C2) x S3
108 44 C2 x ((C3 x C3 x C3) : C2)
120 34 S5
```
| https://mathoverflow.net/users/34538 | Finite groups with integral character table | There is no complete classification, but some structural results are known. To give you something to search for: such groups are called $\mathbb{Q}$-groups. There is a whole book devoted to their structure: [Structure and Representations of $\mathbb{Q}$-Groups](https://link.springer.com/book/10.1007%2FBFb0103426) by D. Kletzing. You will find there answers to many, if not all, of your questions.
In particular, the converse of the lemma is indeed true. It follows from two facts: firstly, characters "separate" conjugacy classes, i.e. if two elements are not conjugate, then there exists an irreducible character that takes different values on them; and secondly, if $m$ is coprime to the order of $g$, then for every irreducible character $\chi$ the values $\chi(g)$ and $\chi(g^m)$ are Galois conjugates.
| 16 | https://mathoverflow.net/users/35416 | 428671 | 173,780 |
https://mathoverflow.net/questions/428685 | 0 | We call an $n\times n$-matrix ${\bf A}\in \text{Mat}(n\times n, \mathbb{R})$ a *metric matrix* if
1. ${\bf A}\_{ii} = 0$ for all $i\in \{1,\ldots,n\}$,
2. ${\bf A}\_{ij} = {\bf A}\_{ji}$ for all $i,j \in \{1,\ldots,n\}$ (that is, ${\bf A}$ is symmetric), and
3. ${\bf A}\_{ik} \leq {\bf A}\_{ij} + {\bf A}\_{jk}$ for all $i,j,k \in \{1,\ldots, n\}$ (that is, the triangle inequality holds).
We say that ${\bf A}\in \text{Mat}(n\times n, \mathbb{R})$ is $\mathbb{R}^k$-*realizable* for some positive integer $k$ if there is a subset $S\subseteq \mathbb{R}^k$ having $n$ elements, and a bijection $\varphi:\{1,\ldots,n\}\to S$ such that
$$\|\varphi(i) - \varphi(j)\| = {\bf A}\_{ij} \text{ for all } i,j\in \{1,\ldots n\}.$$
(Note that $\|\cdot\|$ denotes the Euclidean norm.) So for instance, the following metric matrix ${\bf A}\in \text{Mat}(4\times 4, \mathbb{R})$ is not $\mathbb{R}^2$-realizable, but it is $\mathbb{R}^3$-realizable:
$${\bf A} = \begin{pmatrix}
0 & 1 & 1 & 1\\
1 & 0 & 1 & 1\\
1 & 1 & 0 & 1\\
1 & 1 & 1 & 0\\
\end{pmatrix}$$
**Questions.** (Only the first question needs to be answered for acceptance.)
1. Given an integer $n>1$, is every metric $n\times n$-matrix $\mathbb{R}^{n-1}$-realisable?
2. For every metric matrix ${\bf A}\in \text{Mat}(n\times n, \mathbb{N})$ let its *metric dimension* $\text{mdim}({\bf A})$ denote the smallest positive integer $k$ such that ${\bf A}$ is $\mathbb{R}^k$-realizable. Is the problem of finding $\text{mdim}({\bf A})$ given ${\bf A}\in \text{Mat}(n\times n, \mathbb{N})$ a polynomial-time problem with respect to $n$?
| https://mathoverflow.net/users/8628 | Realizability of metric matrices | In other words, you ask whether every finite metric space may be isometrically embedded to Euclidean space $\mathbb{R}^n$. Not every. A necessary and sufficient condition is given by the non-negativity of the so called [Cayley--Menger determinants](https://en.m.wikipedia.org/wiki/Cayley%E2%80%93Menger_determinant).
| 5 | https://mathoverflow.net/users/4312 | 428692 | 173,783 |
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