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https://mathoverflow.net/questions/423417 | 6 | A fusion category is called *noncommutative* if its Grothendieck ring is noncommutative. Let us call a fusion category *strongly noncommutative* if every fusion category Morita equivalent to it (i.e. same Drinfeld center up to equiv.) is noncommutative.
**Question**: Is there a strongly noncommutative fusion category (say over $\mathbb{C}$)?
If so, what are the known examples?
Note that if $G$ is a finite group then $Vec(G)$ is not strongly noncommutative (even if $G$ is noncommutative) because it is Morita equivalent to $Rep(G)$ which has a commutative Grothendieck ring. Moreover, the Extended Haagerup fusion categories are also not strongly noncommutative because they form a Morita equivalent class and one of them is commutative.
This post is in the same spirit than [this one](https://mathoverflow.net/q/422308/34538) about strongly simple fusion categories. The main difference is that I know examples of strongly simple fusion categories whereas I do not know a single strongly noncommutative fusion category.
The next step would be about fusion categories which are both strongly simple and strongly noncommutative.
| https://mathoverflow.net/users/34538 | Is there a strongly noncommutative fusion category? | Consider the symmetric group group $G = S\_3$ of order $6$. Then $\mathrm{H}^3\_{\mathrm{gp}}(G;\mathrm{U}(1)) \cong \mathbb Z/6\mathbb Z$. Choose a generator $\omega \in \mathrm{H}^3\_{\mathrm{gp}}(G;\mathrm{U}(1))$. Then $\omega$ restricts nontrivially to every nontrivial subgroup of $G$.
It follows that $\mathbf{Vec}^\omega[G]$ is not Morita-equivalent to any other fusion category. (Recall that, for any $G,\omega$, fusion categories equivalent to $\mathbf{Vec}^\omega[G]$ are indexed by pairs consisting of a subgroup $H \subset G$ together with a 2-cochain $\psi$ on $H$ solving $\mathrm{d}\psi = \omega|\_H$.)
But $G$ is noncommutative.
| 11 | https://mathoverflow.net/users/78 | 423436 | 172,105 |
https://mathoverflow.net/questions/423428 | 2 | Consider two continuous functions $f,g : [0,1]\rightarrow\mathbb{R}$ of bounded variation, and let $\mu\_f, \mu\_g : \mathcal{B}([0,1])\rightarrow\mathbb{R}$ be their associated Lebesgue-Stieltjes (signed) measures.
Let further $h:= f\star g : [0,1]\rightarrow\mathbb{R}$ be the [$x=1/2$ centered] concatenation of $f$ and $g$ (that is: $h(x) = f(2x)$ for $x\in[0,1/2]$, and $h(x)= g(2x-1) + (f(1)-g(0))$ for $x\in(1/2,1]$).
Is there an algebraic (or otherwise explicit or insightful) relation between the measures $\mu\_h$ and $\mu\_f, \mu\_g$?
| https://mathoverflow.net/users/472548 | Does the map $f \mapsto \mu_f$ (BV to Lebesgue-Stieltjes measure) behave nicely under function concatenation? | $\newcommand{\B}{\mathcal B}\renewcommand{\S}{\mathcal S}$Note that for any $a$ and $b$ such that $0\le a\le b\le1$
\begin{equation}
\mu\_h((a,b])=h(b)-h(a)=
\left\{
\begin{alignedat}{2}
& \mu\_f(2(a,b])&&\text{ if }(a,b]\subseteq[0,1/2], \\
& \mu\_g(2(a,b]-1)&&\text{ if }(a,b]\subseteq(1/2,1],
\end{alignedat}
\right.
\end{equation}
where $2A:=\{2x\colon x\in A\}$ and $2A-1:=\{2x-1\colon x\in A\}$ for any $A\subseteq\mathbb R$.
So, for any left-open subinterval $(a,b]$ of the interval $[0,1]$,
\begin{equation}
\begin{aligned}
\mu\_h((a,b])&= \mu\_h((a,b]\cap[0,1/2])+\mu\_h((a,b]\cap(1/2,1]) \\
&=\tilde\mu\_h((a,b]) \\
&:=\mu\_f(2((a,b]\cap[0,1/2]))+\mu\_g(2((a,b]\cap(1/2,1])-1).
\end{aligned}
\end{equation}
The function $\tilde\mu\_h$ is a finite signed measure on the semiring (say $\S$) of all left-open subintervals of the interval $[0,1]$, and $\sigma(\S)=\B([0,1])$. So, by the uniqueness of the measure extension (cf. e.g. [Proposition 13 of Kisil - Introduction to functional analysis](http://www1.maths.leeds.ac.uk/%7Ekisilv/courses/math3263m014.html)),
\begin{equation}
\begin{aligned}
\mu\_h(B)&=\mu\_f(2(B\cap[0,1/2]))+\mu\_g(2(B\cap(1/2,1])-1) \\
&=\mu\_f(2B\cap[0,1])+\mu\_g((2B-1)\cap(0,1])
\end{aligned}
\end{equation}
for all $B\in\B([0,1])$.
| 1 | https://mathoverflow.net/users/36721 | 423438 | 172,106 |
https://mathoverflow.net/questions/423384 | 1 | I'm reading a paper which has this line:
A direct computation shows that $P\Omega\_8$($\mathbb K$) has an elementary abelian subgroup $X = 2^2$ such that $C\_{P\Omega\_8(\mathbb K)}(X) = T\_4.2^{1+4}\_+$. Now the action of $2^{1+4}\_+$ on $T\_4$ is also explicitly determined.
$\mathbb K$ is algebraically closed with char not equal 2 and $P\Omega\_8$($\mathbb K$) is the adjoint algebraic group of type D.
What is the action exactly here? I suppose the four 2 invert the four $T\_1$? What about the center? Not sure...
| https://mathoverflow.net/users/477707 | action of the extra-special group | Take $T\_4$ as the diagonal torus, which is composed by 4-tuples of 2x2 rotation matrices $a,b,c,d$: $$\begin{bmatrix}a & & & \\ & b & &\\ & & c &\\ & & & d\end{bmatrix}$$.
Let $T$ denote the matrix $\begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix}$ and $I$ the $2\times2$ identity matrix.
Then the extraspecial group $E=2^{1+4}\_+$ acts on $T\_4$ by conjugation of the following matrices:
$\begin{bmatrix}X & & & \\ & X & &\\ & & X &\\ & & & X\end{bmatrix}$, $\begin{bmatrix}& X & & \\ X & & &\\ & & & X\\ & & X & \end{bmatrix}$, $\begin{bmatrix}& & X & \\ & & & X\\ X & & &\\ & X & & \end{bmatrix}$ and$\begin{bmatrix}& & & X \\ & & X &\\ & X & &\\ X & & &\end{bmatrix}$.
where $X$ represents an element that is either $T$ or $I$, and the number of $T$s is always even.
This group has $32$ elements, because every matrix above represent $8$ elements. There is a matrix representation of $Q\_8 \circ Q\_8$ on [GroupNames](https://people.maths.bris.ac.uk/%7Ematyd/GroupNames/1/ES+(2,2).html). By checking the matrices on the GroupNames page and exchanging $T$ for $-1$ and $I$ for $1$, it is evident that $Q\_8 \circ Q\_8$ is a subgroup of $E$. Since $E$ and $Q\_8 \circ Q\_8$ both have $32$ elements, we have $E=Q\_8 \circ Q\_8$.
Bonus: The group $X$ is generated by the following elements of the maximal torus.
$\begin{bmatrix}-I & & & \\ & -I & &\\ & & I &\\ & & & I\end{bmatrix}$ and $\begin{bmatrix}-I & & & \\ & I & &\\ & & -I &\\ & & & I\end{bmatrix}$
Another way of seeing this group is via the projections from the universal cover to the orthogonal group and then the projective orthogonal group, i.e. $Spin\_8^+(K) \rightarrow \Omega\_8^+(K) \rightarrow P\Omega\_8^+(K)$.
Let $\hat{T}\_4$ be the Cartan subgroup of $Spin\_8^+(K)$ and $T'\_4$ its image in $\Omega\_8^+(K)$. Then the image of the exponent-2 subgroup of $\hat{T}\_4$ in $\Omega\_8^+(K)$ are composed of the elements in $T'\_4$ that have order $2$ and reversing an even number of indicies, i.e. the elements $(a, b, c, d) \rightarrow (s\_1 a,s\_2 b,s\_3 c, s\_4 d)$ where the $s\_n$ are either $+1$ or $-1$ and $s\_1s\_2s\_3s\_4=1$. There are $8$ such elements.
In the map $\Omega\_8^+(K) \rightarrow P\Omega\_8^+(K)$, the element $(a, b, c, d) \rightarrow (-a,-b,-c,-d)$ is identified with the identity element, so the image of the subgroup in $T'\_4$ in $T\_4$ has only 4 elements.
| 1 | https://mathoverflow.net/users/125498 | 423440 | 172,107 |
https://mathoverflow.net/questions/422654 | 5 | Let $C\_2$ be the cyclic group of order $2$ and $\mathbb{F}\_2$ the field with $2$ elements. Consider the group algebra $A:= \mathbb{F}\_2 (C\_2\times C\_2)$. It is well-known that $A$ has infinite representation type. Is there a classification of the finite dimensional indecomposable $A$-modules (and the Auslander-Reiten quiver of $\text{mod}\,A$) in this case?
| https://mathoverflow.net/users/145920 | What are the indecomposable modules over $\mathbb{F}_2(C_2\times C_2)$? | A complete description of the indecomposable modules for $C\_2\times C\_2$, including the Auslander-Reiten quiver is available in David Benson's book *Modular Representation Theory: New Trends and Methods*, Springer 1984, pp.176-181. This is over $\bar{\mathbb{F}}\_2$; the result over arbitrary fields is given in the following few pages.
| 9 | https://mathoverflow.net/users/152674 | 423462 | 172,111 |
https://mathoverflow.net/questions/423456 | 3 | Let $\gamma:[0,1]\to \mathbb{R}^2$ be a rectifiable curve and $\Gamma=\gamma[0,1]$ be its image.
Is it possible to cover $\Gamma$ by a countable collection of sets $N,R\_1,R\_2,\dots$ such that $N$ has vanishing 1-dimensional Hausdorff measure and each $R\_i$ admits a bi-Lipschitz embedding into $\mathbb R$?
If yes, is there an analogue of this statement for $k$-rectifiable sets in $\mathbb{R}^n$?
| https://mathoverflow.net/users/13441 | Decomposition of rectifiable curves in $\mathbb R^2$ | Regardless of its dimension, a $k$-rectifiable set $M \subset \mathbf{R}^n$ can be covered by a collection $(N\_j \mid j \geq 0 )$ of sets, with $\mathcal{H}^k(N\_0) = 0$ and where the $N\_j$, $j \geq 1$ are $k$-dimensional embedded $C^1$ submanifolds of $\mathbf{R}^n$. Here by 'covered' I mean that $M \subset \cup\_{j \geq 0} N\_j$.
This is Lemma 11.1 in Leon Simon's lecture notes on geometric measure theory.
| 7 | https://mathoverflow.net/users/103792 | 423464 | 172,112 |
https://mathoverflow.net/questions/423463 | 2 | Let $\Sigma$ be a non-compact orientable connected two-manifold without boundary. Let $f,g\colon \Sigma\to \Sigma$ be two homeomorphisms. Suppose there is a homotopy $H\colon \Sigma\times [0,1]\to \Sigma$ from $f$ to $g$ such that $H(-,t)\colon \Sigma\to \Sigma$ is a homeomorphism for each $t\in [0,1]$.
Now, $f,g$ induce maps $\mathcal E(f),\mathcal E(g)\colon \mathcal E(\Sigma)\to \mathcal E(\Sigma)$, where $\mathcal E(\Sigma)$ denotes the space of ends of $\Sigma$.
>
> Is it true that $\mathcal E(f)=\mathcal E(g)$?
>
>
> If the above question has a false answer, then do I need to consider
> some restriction, like $\pi\_1(\Sigma)$ is (in)finitely-generated?
>
>
>
Of course, if $H$ itself is a proper map, then $\mathcal E(f)=\mathcal E(g)$, for any $\Sigma$.
| https://mathoverflow.net/users/363264 | Isotopic homeomorphisms of surface induces same map on the space of ends | Yes, the induced maps $\mathcal{E}(f)$ and $\mathcal{E}(g)$ are equal. This is because two isotopic transversely oriented separating circles in $\Sigma$ determine the same subset of $\mathcal{E}(\Sigma)$ and because such subsets give a basis for the topology of $\mathcal{E}(\Sigma)$.
| 3 | https://mathoverflow.net/users/1650 | 423467 | 172,113 |
https://mathoverflow.net/questions/423453 | 7 | A topological space $X$ is concentrated on a set $D$ iff for any open set $G$ if $D\subseteq G$, then $X\setminus G$ is countable.
What is an example of a separable metrizable (uncountable) meager (meaning a countable union of nowhere dense subsets, also known as a set of first category) space $X$ such that $X$ is concentrated on countable dense set?
| https://mathoverflow.net/users/112417 | What is an example of a meager space X such that X is concentrated on countable dense set? | **ADDED LATER**
The answer to your question is that there is such a space $X$ if and only if $\mathfrak{b} = \aleph\_1$.
*If $\mathfrak{b} = \aleph\_1$, then there is such a space.*
To see this, first note that $\mathfrak{b} = \aleph\_1$ if and only if there is an uncountable subset of the irrationals concentrated on $\mathbb Q$. This is proved by van Douwen in section 10 of his article in the *Handbook of Set Theoretic Topology*. By modifying his argument slightly, we can show that if $\mathfrak{b} = \aleph\_1$ then there is an uncountable subset of the Cantor space concentrated on a countable subset of the Cantor space.
[The proof of this goes as follows. If $\mathfrak{b} = \aleph\_1$, then one can construct via transfinite recursion a length-$\omega\_1$ sequence $\langle f\_\alpha :\, \alpha < \omega\_1 \rangle$ of functions that is unbounded with respect to $\leq^\*$, but also with the property that $\alpha < \beta$ implies $f\_\alpha \leq^\* f\_\beta$. This means that a subset of this sequence is $\leq^\*$-bounded if and only if it is countable. It's a well-known fact that the space $\omega^\omega$, endowed with the usual product topology, is homeomorphic to the space of irrational numbers. It is also fairly well known that the irrationals are homeomorphic to $C \setminus D$, where $C$ is the Cantor space and $D$ is some countable relatively dense subset of $C$. Let $Y$ be the image of your sequence of functions under some homeomorphism $\omega^\omega \rightarrow C \setminus D$. If $U$ is an open set containing $D$, then $C \setminus U$ is a compact subset of $C$. It's not too difficult to show that every compact subset of $\omega^\omega$ is $\leq^\*$-bounded above by some function (or even more -- it is $\leq$-bounded), and this means $C \setminus U$ contains only countably many points of $Y$.]
Let $Y$ be, as in the previous paragraph, an $\aleph\_1$-sized subset of the Cantor space $C$ that concentrates on a countable $D \subseteq C$. Let $X = Y \cup \mathbb Q$. As a subspace of the reals, this set $X$ meets all your requirements.
*A proof that if there is a space $X$ as described in your post then $\mathfrak{b} = \aleph\_1$.*
Suppose there is an uncountable separable metrizable space $X$ that is concentrated on a countable set $D \subseteq X$. Let $\{ d\_0, d\_1, d\_2, \dots \}$ be an enumeration of $D$. For each $x \in X \setminus D$, define a function $f\_x : \omega \rightarrow \omega$ by setting $f\_x(n) = \min\{ k :\, \mathrm{dist}(x,d\_n) > \frac{1}{k} \}$.
Let $Y$ be a subset of $X \setminus D$ with $|Y| = \aleph\_1$. The set of functions $\{ f\_x :\, x \in Y \}$ is an $\aleph\_1$-sized subset of $\omega^\omega$, and I claim it is unbounded with respect to $\leq^\*$. In other words, I claim this set of functions witnesses $\mathfrak{b} = \aleph\_1$.
To see this, suppose instead there is some $g \in \omega^\omega$ such that $f\_x \leq^\* g$ for all $x \in Y$. By a pigeonhole argument, there is some function $h \in \omega^\omega$, differing from $g$ in only finitely many places, such that $f\_x \leq h$ for uncountably many $x \in Y$. Let $U = \bigcup\_{n \in \omega} B\_{1/h(n)}(d\_n)$. This is an open set containing $D$, and our definition of the $f\_x$'s ensures that $x \notin U$ if and only if $f\_x \leq h$. Thus there are uncountable many $x$'s with $x \notin U$, contradicting the fact that $Y$ concentrates on $D$.
Let me point out that this argument is essentially contained in an earlier MO post by Taras Banakh found [here](https://mathoverflow.net/questions/416331/example-of-an-uncountable-scattered-space-with-some-properties). (A version of the argument seems to be in van Douwen's article too, and I don't know whether it really originated there either, but anyway I first learned it from Taras' post.)
$$$$
**ORIGINAL POST**
Here is a consistent example of a subspace of $\mathbb R$ with these properties.
First, there is an uncountable subset $X$ of $\mathbb R$ concentrating on $\mathbb Q$ if and only if $\mathfrak{b} = \aleph\_1$. This is proved in chapter 10 of van Douwen's article in the *Handbook of Set-Theoretic Topology*. Note that $X$ concentrates on $\mathbb Q$ if and only if every subset of $X$ does, so if $\mathfrak{b} = \aleph\_1$ then there is an $\aleph\_1$-sized set of reals concentrating on $\mathbb Q$.
Now suppose $\aleph\_1 = \mathfrak{b} < \mathrm{non}(\mathcal M)$ (the least size of a non-meager subset of $\mathbb R$). This situation is consistent -- it happens in the random real model, for example. In such a model, let $X$ be an $\aleph\_1$-sized set concentrating on $\mathbb Q$. Adding countably many points to $X$ if necessary, we may (and do) assume $X$ is dense in $\mathbb R$. Then (viewed as a subspace of $\mathbb R$) it meets all your requirements.
$$$$
*A sketch of an argument as to why $\mathfrak{b} = \aleph\_1$ implies there is an $\aleph\_1$-sized set concentrating on $\mathbb Q$*:
Since you may not have access to van Douwen's article, here is a sketch of the idea I quoted above. If $\mathfrak{b} = \aleph\_1$, then one can construct via transfinite recursion a length-$\omega\_1$ sequence $\langle f\_\alpha :\, \alpha < \omega\_1 \rangle$ of functions that is unbounded with respect to $\leq^\*$, but also has the property that $\alpha < \beta$ implies $f\_\alpha \leq^\* f\_\beta$. It's a well-known fact that the space $\omega^\omega$, endowed with the usual product topology, is homeomorphic to the space of irrational numbers. Let $X$ be the image of your sequence of functions under any such homeomorphism. If $U$ is an open set containing $\mathbb Q$, then $\mathbb R \setminus U$ is a $\sigma$-compact subset of the irrationals. It's not too difficult to show that every $\sigma$-compact subset of $\omega^\omega$ is $\leq^\*$-bounded above by some function, and this means $\mathbb R \setminus U$ can only contain countably many points of $X$.
| 11 | https://mathoverflow.net/users/70618 | 423469 | 172,115 |
https://mathoverflow.net/questions/423493 | 1 | If $\boldsymbol{A}\in\mathbb{R}^{n\times n}$ is the cost-matrix of an assignment problem, then the usual statement of the problem of finding an optimal assignment is to identify $n$ elements $a\_{i,\,\pi(i)},\ i=1\cdots n$ of least cost-sum, i.e. to *directly* determine the solution set from $\boldsymbol{A}$ by modifying its entries e.g. according to the Hungarian algorithm.
>
> **Question:**
>
>
> can the following interpretation of the assignment problem fail to report the optimal solution:
>
>
> `determine the sequence of line-exchanges that renders the sum of the diagonal-elements optimal`?
>
>
>
If the permutation-formulation also generates the optimal solution to the assignment problem, that would yield a greedy algorithm:
Exchanging two lines also exchanges two on-diagonal elements with two off-diagonal elements with known effect on the cost-sum of the elements on the diagonal.
**Addendum:**
as the counter example of Brendan McKay shows, swapping pairs of rows or columns may not suffice to find the optimal assignment.
A preprocessing step that can deal with the counter example would be to first rotate the columns until the diagonal-elements have optimal cost-sum and only then strive for further improvements via swapping pairs of rows or columns.
It may even be necessary to interleave line-swapping with column-rotations frequently so that the elements on the principal diagonal always have better cost-sum than the elements on other diagonals.
| https://mathoverflow.net/users/31310 | Interpreting optimal matchings as permutations | $$\pmatrix{ 2&3&0&0\\0&2&3&0\\0&0&2&3\\3&0&0&2}$$
Every swap of two columns or swap of two rows decreases the trace. However, there is a permutation putting all the 3s on the diagonal.
| 4 | https://mathoverflow.net/users/9025 | 423501 | 172,122 |
https://mathoverflow.net/questions/423498 | 0 | Let $A$ be a Banach algebra with a bounded approximate identity, and let $E$ be a Banach left $A$-module. Suppose neither $A$ nor $E$ has the Schur property.
**Question:** Given a weakly null sequence $(w\_n)$ in $E$, does there exist
* a weakly null sequence $(a\_n)$ in $A$
* a bounded sequence $(x\_n)$ in $E$
such that $w\_n = a\_nx\_n$ for all $n\in\mathbb{N}$?
| https://mathoverflow.net/users/164350 | Do the weakly null sequences in a Banach module factor? | No; here's a counter-example. Let $E$ be some Banach space without the [Schur Property](https://en.wikipedia.org/wiki/Schur%27s_property), give $E$ the zero product, and let $A=E\oplus\mathbb C$ be the unitisation, with natural character $\epsilon:A\rightarrow\mathbb C; a=(x,\lambda)\mapsto\lambda$. Use the character to turn $E$ into a left $A$-module:
$$ a\cdot x = \epsilon(a) x \qquad (a\in A, x\in E). $$
If $(a\_n)$ is weakly-null in $A$ then $\epsilon(a\_n)\rightarrow 0$ and so $(a\_n\cdot x\_n)$ is norm-null. Taking $(w\_n)$ to be any weakly-null sequence in $E$ which is not norm-null (which exists as $E$ fails to have the Schur property) yields the counter-example.
| 1 | https://mathoverflow.net/users/406 | 423506 | 172,123 |
https://mathoverflow.net/questions/423392 | 1 | This question is a modification of the one asked [here](https://mathoverflow.net/q/423367/160416), which turned out to ask for something too strong to be true.
Given $k>0$ and a positive integer $n$, let $X, Y$ be two vertex sets of size $n$ and define a random bipartite graph $G(k,n)$ on $X \sqcup Y$ in an Erdos-Renyi fashion by putting an edge between each pair $x, y$ with $x\in X$, $y\in Y$ with probability $\frac{k}{n}$.
Define the discrepancy $\text{Disc}(G)$ of the resulting bipartite graph as the maximum of
$$\left|\frac{E(A,B)}{kn}-\frac{|A||B|}{n^2}\right|$$
over all the subsets $A \subset X$, $B\subset Y$, where $E(A, B)$ is the number of edges between vertices in $A$ and vertices in $B$.
>
> Given $\varepsilon>0$, does there exist $K(\varepsilon)>0$ such that, for every $k>K(\varepsilon)$, the probability that $\text{Disc}(G(k,n))>\varepsilon$ goes to zero as $n \to \infty$.
>
>
>
In other words, do all pairs of subsets have roughly the expected number of edges between them with high probability? Note that the naive approach of bounding $\mathbb{P}\left(\left|\frac{E(A,B)}{kn}-\frac{|A||B|}{n^2}\right|>\varepsilon\right)$ independently for each pair $A,B$ and then use the union bound on all possible pairs $A, B$ cannot work, since one can prove that, for any fixed $k, \varepsilon$,
$$\sum\_{A,B} \mathbb{P}\left(\left|\frac{E(A,B)}{kn}-\frac{|A||B|}{n^2}\right|>\epsilon\right) \to \infty.$$
Moreover, the constant $K(\varepsilon)$ is important since, as shown by James Martin in the previous version of that question, the statement is false if $k$ is too small with respect to $\varepsilon$.
In case the statement is true, I'm also interested in the natural generalization to $r$-partite $r$-uniform hypergraphs. That is, fix vertex sets $X\_1, \ldots, X\_r$ and put an edge between $x\_1, \ldots, x\_r$ for each $x\_1 \in X\_1, \ldots x\_r\in X\_r$ with probability $\frac{k}{n^{r-1}}$. Define the discrepancy as the maximum of
$$\left|\frac{E(A\_1,\ldots, A\_r)}{kn}-\frac{|A\_1|\cdots|A\_r|}{n^r}\right|$$
over all the $r$-tuples of subsets $(A\_i \subset X\_i)$, where $E(A\_1, \ldots, A\_r)$ is the number of edges between $A\_1, \ldots, A\_r$. Given $\varepsilon>0$, does there exist $K\_r(\varepsilon)>0$ such that, for every $k>K\_r(\varepsilon)$, the probability that the discrepancy is greater than $\varepsilon$ goes to zero as $n \to \infty$?
| https://mathoverflow.net/users/160416 | Discrepancy of random bipartite graphs (2) | The answer is yes, and the union bound actually does work.
By applying the [multiplicative Chernoff bound found here](https://en.wikipedia.org/wiki/Chernoff_bound#Multiplicative_form_(relative_error)) with $\delta=\frac{\varepsilon n^r}{N}$ and $\mu=\frac{kN}{n^{r-1}}$ where $N=|A\_1|\cdots|A\_r|$ we find that
$$\mathbb{P}\left(\left|\frac{E(A\_1,\ldots, A\_r)}{kn}-\frac{|A\_1|\cdots|A\_r|}{n^r}\right|>\varepsilon\right)\le 2\exp\left(-\frac{\delta^2\mu}{2+\delta}\right)$$
$$=2\exp\left(-\frac{\varepsilon^2 k n^{r+1}}{2N+\varepsilon n^r}\right)$$
$$\le 2\exp\left(-\frac{\varepsilon^2 k n}{2+\varepsilon}\right)$$
Therefore if $\frac{\varepsilon^2 k}{2+\varepsilon}>r\log 2$, the discrepancy is at most $\varepsilon$ almost surely as $n\to \infty$.
| 1 | https://mathoverflow.net/users/160416 | 423508 | 172,124 |
https://mathoverflow.net/questions/423507 | 13 |
>
> Has there ever been a set theory without an empty set? Is this possible?
>
>
>
I ask because we usually take the empty set to exist axiomatically or obtain it through separation and a nonempty set together with the standard parameter-free predicate $X\neq X$, but it seems possible to have a 'set theory' without an axiom asserting the existence of an empty set or an axiom of separation.
I put 'set theory' in quotations because such a nonstandard axiomatization might not really deserve to be called a set theory per-se (it wouldn't prove the existence of intersections of disjoint sets), but more formally I mean
>
> Has a theory in the language of set theory whose axioms do not prove the existence of an empty set ever been explored?
>
>
>
| https://mathoverflow.net/users/92164 | Set theory without the empty set | For a good discussion of this matter, see: Kanamori, Akihiro *The empty set, the singleton, and the ordered pair*. Bull. Symbolic Logic 9 (2003), no. 3, 273–298.
| 22 | https://mathoverflow.net/users/18939 | 423509 | 172,125 |
https://mathoverflow.net/questions/423505 | 4 | A *triangulation* of a convex polytope $P\subset\Bbb R^n$ is a partition of $P$ into $n$-simplices $\{\Delta\_1,...,\Delta\_m\}$ each of which has all its vertices among the vertices of $P$. A polytope may have many different triangulations.
>
> **Question I:** do all combinatorially equivalent polytopes have the same triangulations?
>
>
>
More precisely, let $P,Q\subset\Bbb R^n$ be combinatorially equivalent, $\phi:\mathcal F(P)\to\mathcal F(Q)$ be a face-lattice isomorphism, and $\{\Delta\_1,...,\Delta\_m\}$ a triangulation of $P$.
If $\Delta\_i$ has vertices $p\_1,...,p\_{n+1}\in\mathcal F\_0(P)$, then let $\phi(\Delta\_i)\subset Q$ be the simplex with vertices $\phi(p\_1),...,\phi(p\_{n+1})$. Do the simplices $\phi(\Delta\_1),...,\phi(\Delta\_m)$ form a triangulation of $Q$? That is, do they have disjoint interiors and cover all of $Q$?
>
> **Question II:** if not, is there always a **universal triangulation**? That is, a special triangulation for $P$ that carries over to every combinatorially equivalent polytope in the way described above?
>
>
>
| https://mathoverflow.net/users/108884 | Do combinatorially equivalent polytopes have the same triangulations? | Just to mark this question as answered, the comment by Tobias Fritz is spot on. In [this answer](https://math.stackexchange.com/a/2122958/79593) to a Math Stack Exchange question, Francisco Santos completely resolves your questions. On the one hand, the answer to Question I is no: combinatorially equivalent polytopes can have different triangulations. On the other hand, the answer to Question II is: yes, there is at least one triangulation they all share (the so-called "pulling triangulations").
| 4 | https://mathoverflow.net/users/25028 | 423514 | 172,126 |
https://mathoverflow.net/questions/423379 | 4 | The $l$-th Catalan number ${2l\choose l}\frac{1}{l+1}$ is equal
to the number of sequences $s\_0,\ldots,s\_{l+1}$ of length $l+2$ with the following
properties:
(1) $s\_0=s\_{l+1}=1$ and $s\_1,\ldots,s\_l$ are integers $\geq 2$,
(2) $s\_i$ divides $s\_{i-1}+s\_i+s\_{i+1}$ for $i=1,\ldots,l$.
(The proof is easy: consecutive terms $s\_is\_{i+1}$ in such sequences encode leaves of finite binary trees with $l+1$ leaves. This encoding of binary trees has nice properties: the integer $(s\_{i-1}+s\_i+s\_{i+1})/s\_i$ is the distance between the consecutive leaves
encoded by $s\_{i-1},s\_i$ and $s\_i,s\_{i+1}$. Leaves correspond always to coprime integers $s\_i,s\_{i+1}$ and heights of leaves can be computed by applying Euclid's algorithm (essentially by summing all quotients occuring during the computation) to the coprime pair
$(s\_i,s\_{i+1})$. Interior vertices of the binary tree encoded by $s\_0,\ldots,s\_{l+1}$ correspond to segments $s\_i,s\_{i+1},\ldots,s\_{i+k}$ such that $s\_{i+1},\ldots,s\_{i+k-1}$
are all strictly larger than $\max(s\_i,s\_{i+k})$ (and the height of such an interior vertex is given as for leaves by Euclid's algorithm applied to $s\_i,s\_{i+k}$ which are always coprime). In particular the
binary tree of the left child of the root vertex (for $l\geq 1$) is
encoded by the initial terms $s\_0,\ldots,s\_i=2$. The subtree for the right child corresponds to $s\_i=2,\ldots,s\_{l+1}$.)
Examples (with integers concatenated):
$l=0$: $11$,
$l=1$: $121$,
$l=2$: $1231,1321$,
$l=3$: $12341,12531,13231,13521,14321$.
(Recipe for recursive constructions: All such sequences of length $l+3$ are obtained from such sequences $l+2$ by inserting in all possible ways between two adjacent terms their sum.)
*Is there a reference for this description?* (I could not find it in Stanley's list but a mistake on my behalf is likely.)
| https://mathoverflow.net/users/4556 | Reference for a definition of Catalan numbers | I'm converting my comments to an answer. This interpretation of the Catalan numbers is indeed well-known. It appears for instance as Exercise 92 in Chapter 2 of Stanley's book on [Catalan numbers](https://doi.org/10.1017/CBO9781139871495), as well as [Exercise 6.19(iii)](https://math.mit.edu/~rstan/ec/catalan.pdf) in his *Enumerative Combinatorics, Vol. 2*. It is also discussed, for instance, in the paper [Sum-Difference Sequences and Catalan Numbers](https://doi.org/10.1007/s006050050025) by Aigner & Schulze, who attribute this interpretation to Van Lint.
(By the way, it is a little unclear why you write "$s\_i$ divides $s\_{i-1}+s\_i+s\_{i+1}$" instead of "$s\_i$ divides $s\_{i-1}+s\_{i+1}$", but of course that amounts to the same thing.)
| 2 | https://mathoverflow.net/users/25028 | 423516 | 172,127 |
https://mathoverflow.net/questions/422764 | 5 | The paper "Regular Functions on Certain Infinite-dimensional Groups" by Kac and Peterson describes the construction of a group associated to the datum of a Kac-Moody algebra in a way I haven't seen before. Let me outline the construction.
Let $A$ be a symmetrizable generalized Cartan matrix. Define $\frak{g}$ by the usual, Serre-type relations. Denote by $\frak{g}'$ the derived algebra of $\frak{g}$. Let $F$ denote a field of characteristic 0.
The following is a quote from the above-mentioned paper, section 1C.
>
> Let $G^\*$ be the free product of the additive groups $\frak{g}\_\alpha$, $\alpha \in \Delta^{\mathrm{re}}$ [the real roots], with canonical inclusions $i\_\alpha: \frak{g}\_\alpha$$ \to G^\*$. For any integrable $\frak{g}'$-module $(V, \pi)$, define a homomorphism $\pi^\*: G^\* \to \operatorname{Aut}\_F(V)$ by $\pi^\*(i\_\alpha(e)) = \operatorname{exp} \pi(e)$. Let $N^\*$ be the intersection of all $\operatorname{Ker}\pi^\*$, put $G = G^\* / N^\*$[...].
>
>
>
There are two examples given below the definition:
>
> a) Let $A$ be the Cartan matrix of a split simple finite-dimensional Lie algebra $\frak{g}$ over $F$. Then the group $G$ associated to $\frak{g} := \frak{g}'$$(A)$ is the group $\underline{G}(F)$ of $F$-valued points of the connected simple-connected algebraic group $\underline{G}$ associated to $\frak{g}$ [...]
>
>
>
>
> b) Let $\frak{g}$ be as in a), and let $\tilde{A}$ denote the extended Cartan matrix of $\frak{g}$. Then the group $G$ associated to $\frak{g}'$$(\tilde{A})$ is a central extension by $F^\*$ of $\underline{G}(F[z, z^{-1}])$ [...]
>
>
>
I have managed to verify a) over the complex numbers for certain classical groups using basic Lie theory, though ideally I would like to see a proof using purely algebraic methods. I have no clue how to tackle the second one. I am thankful for references hinting to any of the following:
* Example calculations using this definition
* Other papers which make use of this definition (I am aware of the predecessor paper "Infinite flag varieties and conjugacy theorems" by the same authors.)
as well as for any advice on how to retrace the examples, especially in cases where one cannot read "integrable" as "comes from a representation of $G$".
| https://mathoverflow.net/users/175824 | Constructing a Kac-Moody group as a quotient of the free product of its root subgroups | Note first that the derived Kac-Moody algebra $\mathfrak g'$ is the Kac-Moody algebra $\mathfrak g\_{\mathcal D}$ associated to the Kac-Moody root datum $\mathcal D$ of simply connected type (see [1, Example 7.11]), in the sense of [1, Definition 7.13]. Let $\mathfrak G\_{\mathcal D}$ denote the constructive Tits functor of type $\mathcal D$ (see [1, Definition 7.47]), restricted to the category of fields of characteristic zero.
Let $F$ be a field of characteristic zero, and let $G$ be the group constructed by Kac-Peterson over $F$. In other words, if $\mathrm{exp}\_V: G^\*\to \mathrm{GL}(V)$ is the representation of $G^\*$ over the integrable $\mathfrak g\_{\mathcal D}$-module $V$, and if $\exp=\bigoplus\_V\exp\_V$ is the direct sum of all these representations of $G^\*$, then $G=\exp(G^\*)$.
The group $\mathfrak G\_{\mathcal D}(F)$ also acts on the direct sum of all the integrable $\mathfrak g\_{\mathcal D}$-modules (with similar formulas), and this action factors through $G$ (see [1, Section 7.4.3]). In other words, there is a surjective morphism $\pi:\mathfrak G\_{\mathcal D}(F)\to G$. Its kernel lies in the standard torus of $\mathfrak G\_{\mathcal D}(F)$ by the Recognition Theorem [1, Theorem 7.71], and since it is easy to check that $\pi$ is injective on this torus (see e.g. [1, Definition 7.31]), $\pi$ is actually an isomorphism. In other words, $G$ is just $\mathfrak G\_{\mathcal D}(F)$.
The fact that $\mathfrak G\_{\mathcal D}(F)$ satisfies the properties (a) and (b) in the question is then well-known: see for instance [1, Exercice 7.50] for (a) and [1, Section 7.6] for (b).
Note that, modulo some technical details that need some ironing out (whence the reference to the book [1](https://www.ems-ph.org/books/book.php?proj_nr=232) rather than Tits' papers), everything above follows from the work of Jacques Tits (see in particular [2](https://www.sciencedirect.com/science/article/pii/0021869387902146)).
[1](https://www.ems-ph.org/books/book.php?proj_nr=232) T. Marquis, [An introduction to Kac-Moody groups over fields](https://www.ems-ph.org/books/book.php?proj_nr=232), EMS Textbooks in Mathematics, European Mathematical Society (EMS), Zürich, 2018
[2](https://www.sciencedirect.com/science/article/pii/0021869387902146) J. Tits, [Uniqueness and presentation of Kac–Moody groups over fields](https://www.sciencedirect.com/science/article/pii/0021869387902146), J. Algebra 105 (1987), no. 2, 542–573.
| 3 | https://mathoverflow.net/users/106751 | 423524 | 172,131 |
https://mathoverflow.net/questions/423496 | 1 | Let $H=(V,E)$ be a [hypergraph](https://en.wikipedia.org/wiki/Hypergraph) such that $\emptyset\notin E$. We say that $C\subseteq V$ is a *(vertex) cover* if for all $e \in E$ we have $C\cap e\neq \emptyset$. The minimum size that a cover can have is denoted by $\nu(H)$.
A set $M\subseteq E$ is said to be a *matching* if $M$ consists of pairwise disjoint members of $E$. Clearly, for any matching $M\subseteq E$ we have $|M|\leq \nu(H)$. We say that $H$ is *balanced* if there is a matching $M$ with $|M| = \nu(H)$.
**Question.** Let $H=(V,E)$ be a hypergraph such that for all finite $E\_0\subseteq E$ the hypergraph $(V, E\_0)$ is balanced. Does this imply that $H$ itself is balanced?
| https://mathoverflow.net/users/8628 | Hypergraphs with finite matching / covering balance | If infinite edges are allowed, there are trivial counterexamples. Let $H=(V,E)$ where $V$ is an infinite set and $E$ is the set of all cofinite subsets of $V$; then $\nu(H)=\aleph\_0$ while $\nu((V,E\_0))=1$ for any nonempty finite subset $E\_0$ of $E$.
If the edges of the hypergraph $H=(V,E)$ are nonempty **finite** sets, then the answer to your question is yes. Let $M\subseteq E$ be a maximal matching, so that $\bigcup M$ is a vertex cover and $|M|\le\nu(H)\le|\bigcup M|$. If $M$ is infinite then $\nu(H)=|M|$ and we're done, so suppose $M$ is finite. Then $\nu(H)$ is finite, and a simple compactness argument shows that there must be a finite set $E\_0\subseteq E$ with $\nu((V,E\_0))=\nu(H)$. Since $(V,E\_0)$ is balanced, there is a matching $M\_0\subseteq E\_0$ with $|M\_0|=\nu(H)$.
| 2 | https://mathoverflow.net/users/43266 | 423527 | 172,133 |
https://mathoverflow.net/questions/423502 | 2 | **Note:** I am reposting this [question](https://math.stackexchange.com/q/4436917/369800) from Math Stack Exchange, which failed to receive an answer after several weeks and a bounty. Also, I believe it fits the requirements for this website, as it relates to a research paper.
**Question**
Let $X$ be a random variable with distribution function $F$ on a probability space $(\Omega, \mathcal F, P)$.
Suppose that there exist $\alpha \in (0,2)$ and a slowly varying function $\ell(\cdot)$ such that
$$
\bar F(x) := 1 - F(x) = \frac{C\_1(x)}{x^\alpha} \ell(x) \quad \text{and} \quad F(-x) = \frac{C\_2(x)}{x^\alpha} \ell(x) \quad \text{for $x > 0$,}
$$
where $C\_1(\cdot), C\_2(\cdot)$ are non-negative functions with $C\_i := \lim\_{x \to \infty} C\_i(x)$, and $C\_1 + C\_2 > 0$.
Why then does the following hold?
>
> There exist $C, \tilde C > 0$ such that
> $$ \sum\_{n=1}^\infty P\big( |X| > a\_n \big) \leq \sum\_{n=1}^\infty \frac{C}{nf(n)} \leq \tilde C \int\_1^\infty \frac{dt}{t f(t)},
> $$
> where we define $a\_n := [n f(n) \ell(n) ]^{1/\alpha}$ for an arbitrary positive function $f$ with the properties
> $$ \limsup\_{t \to \infty} \sup\_{0 \leq t \leq x} \frac{f(t)}{f(x)} < \infty \quad \text{and} \quad \int\_1^\infty \frac{dt}{tf(t)}< \infty.
> $$
>
>
>
If not, what if we also required that $f$ be slowly varying, too?
**Background and Thoughts**
This comes from Cai's 2006 paper "[Chover-Type Laws of the Iterated Logarithm for Weighted sums of $\rho^\*$-Mixing Sequences](https://www.hindawi.com/journals/ijsa/2006/065023/)". Cai writes on page 5 that this is "easily seen" based on the representation of $F$ above. I don't see why. What follows below is my attempt so far.
$$
\begin{aligned}
\sum\_{n=1}^\infty P\big( |X| > a\_n \big) &= \sum\_{n=1}^\infty \Big[ P\big( X > a\_n \big) + P\big( X < - a\_n \big) \Big] \\
&\leq \sum\_{n=1}^\infty \Big[ \bar F(a\_n) + F(-a\_n) \Big] \\
&= \sum\_{n=1}^\infty \frac{C\_1(a\_n) + C\_2(a\_n)}{a\_n^\alpha} \ell(a\_n) \\
&\leq \sum\_{n=1}^\infty \frac{C}{a\_n^\alpha} \ell(a\_n) = C \sum\_{n=1}^\infty \frac{1}{nf(n)} \cdot \frac{\ell(a\_n)}{\ell(n)},
\end{aligned}
$$
where the inequality on the last line holds for some $C>0$, since $C\_1(\cdot)$ and $C\_2(\cdot)$ are convergent.
If $\frac{\ell(a\_n)}{\ell(n)} = \ell\Big( \big[ n f(n) \ell(n) \big]^{1/\alpha} \Big) \Big/ \ell(n)$ were bounded, then the first desired inequality would follow. However, it's not clear why this would have to hold.
***Updates***
* Indeed, I *think* the author implicitly uses the fact (?) that $\left\{ \frac{\ell(a\_n)}{\ell(n)} \right\}$ is a bounded sequence several times in the paper. But I still don't know why that's the case.
* Although not stated in the paper, maybe we need some more restrictions on $f$, such that it is also a slowly varying function. If $f$ were slowly varying, then $u(x) := x^{1/\alpha}\cdot[f(x) \ell(x)]^{1/\alpha}$ would be regularly varying with coefficient $1/\alpha$. And $\frac{\ell(a\_n)}{\ell(n)} = \frac{\ell\big(u(n)\big)}{\ell(n)}$. Since $u(x) \to \infty$ and is regularly varying, $\ell \circ u$ is slowly varying. Hence, $\frac{\ell(a\_n)}{\ell(n)} = \frac{ \ell \big(u(n) \big)}{\ell(n)}$ is slowly varying. But, of course, slowly varying functions aren't necessarily bounded.
| https://mathoverflow.net/users/104268 | Inequality with slowly varying functions | $\newcommand{\al}{\alpha}$This inequality is false, for any $\al\in(0,2)$.
Indeed, consider first the case $\al\ne1$. Suppose that $C\_1(x)+C\_2(x)=1$ for all $x>1$,
\begin{equation}
\ell(x)=e^{b\sqrt{\ln x}} \tag{1}\label{1}
\end{equation}
for some real $b$ and all $x>1$, and $f(t)=\ln^2 t$ for all $t>2$.
Then, reasoning as in your post, for all large enough $n$ we have
\begin{equation}
P(|X|>a\_n)\asymp\frac1{nf(n)}\frac{\ell(a\_n)}{\ell(n)}
=\frac1{n\ln^2 n}\frac{\ell(a\_n)}{\ell(n)},
\end{equation}
$a\_n=n^{1/\al+o(1)}$,
\begin{equation}
\frac{\ell(a\_n)}{\ell(n)}=\exp\{b(\sqrt{1/\al}-1+o(1))\sqrt{\ln n}\}.
\end{equation}
Letting now $b=1$ if $\al\in(0,1)$ and $b=-1$ if $\al\in(1,2)$, we see that for all large enough $n$
\begin{equation}
\frac{\ell(a\_n)}{\ell(n)}>\ln n
\end{equation}
and hence
\begin{equation}
\sum\_n P(|X|>a\_n)=\infty,
\end{equation}
whereas $\int\_1^\infty \frac{dt}{t f(t)}<\infty$, so that the inequality
\begin{equation}
\sum\_{n=1}^\infty P\big( |X| > a\_n \big) \le \tilde C \int\_1^\infty \frac{dt}{t f(t)}
\end{equation}
cannot hold for any real $\tilde C$.
In the case $\al=1$, instead of \eqref{1} similarly consider
\begin{equation}
\ell(x)=\exp\frac{\ln x}{\ln\ln x}
\end{equation}
for $x>e$.
| 2 | https://mathoverflow.net/users/36721 | 423533 | 172,136 |
https://mathoverflow.net/questions/423510 | 3 | Let $d,m$ be positive integers. Suppose that $A\_{i,j}$ is a $d\times d$-matrix with real entries whenever $i,j\in\{1,\dots m\}$.
Let $A$ be the $dm\times dm$ matrix that can be written as a block matrix as
$$A=\begin{bmatrix}
A\_{1,1} & \cdots & A\_{1,m} \\
\vdots & \ddots & \vdots \\
A\_{m,1} & \dots & A\_{m,m}
\end{bmatrix}.$$
1. Is $$\rho(A)^2\leq\min(d,m)\cdot \rho\left(\sum\_{i,j}A\_{i,j}\otimes A\_{i,j}\right)?$$
2. For all $d,m$, can we select real matrices $A\_{i,j}$ so that
$$\rho(A)^2=\min(d,m)\cdot \rho\left(\sum\_{i,j}A\_{i,j}\otimes A\_{i,j}\right)>0?$$
3. If $$\rho(A)^2=\min(d,m)\cdot \rho\left(\sum\_{i,j}A\_{i,j}\otimes A\_{i,j}\right)>0,$$
and $d\leq m$ then is $\text{Rank}(A\_{i,j})\leq 1$ for each $i,j$?
My computer calculations suggest that the answer to these questions is 'yes'. This is a follow up of [this question](https://mathoverflow.net/questions/422925/is-rhox-1-dots-x-r2-r-leq-fracdr-cdot-rhox-1-otimes-x-1-dotsx-r-o). This question can be generalized to the case when each $A\_{i,j}$ is a complex matrix. One can rephrase the complex version of this question in terms of the completely depolarizing channel in quantum information theory.
| https://mathoverflow.net/users/22277 | An inequality for the spectral radius of block matrices | I'll answer #1 only, leaving the rest to you or someone else to figure out. The answer is affirmative.
Dropping the trivial case $m=1$, we may assume by density that we are in the generic position, i.e., that $A\_{ij}^T\ne 0$ and together have no non-trivial invariant subspace (this is just to avoid considering degenerate cases, which can also be done using dimension reduction but I prefer not to bother with this extra technicality).
Consider the mapping $X\mapsto F(X)=\sum\_{i,j}A\_{ij}XA\_{ij}^T$ on the cone $K$ of symmetric non-negative definite $d\times d$ matrices $X$. Notice that if $X\ne 0$ and $F(X)=0$, then all $A\_{ij}^T$ map the entire space to the kernel of $X$, which contradicts our generic position assumption. Thus the mapping $X\mapsto \frac 1{\operatorname{Tr}F(X)}F(X)$ is well-defined on the convex compact $Q=K\cap\{X:\operatorname{Tr}X=1\}$ and maps $Q$ to itself. So, by the Brouwer fixed point theorem, the linear mapping $F$ has an eigenvector in the cone with positive eigenvalue, which we can normalize to $1$ by scaling $A$, i.e., we have
$$
\sum\_{i,j}A\_{ij}XA\_{ij}^T=X\,,
$$
which, clearly, implies $\rho(\sum\_{i,j}A\_{ij}\otimes A\_{ij})\ge 1$.
I now claim that $X$ is non-degenerate. Indeed, otherwise the kernel of $X$ would be non-trivial and invariant under all $A\_{ij}^T$.
Now for a vector $y=(y\_1,\dots,y\_m)\in\mathbb R^{md}$ define its norm by
$$
\|y\|^2=\sum\_j|X^{-1/2}y\_j|^2
$$
where $|\cdot|$ stands for the usual Euclidean norm in $\mathbb R^d$. Let $y$ be a unit vector (in this norm) on which the induced operator norm of $A$ is attained.
Then $\|Ay\|^2=\sum\_{i} \langle X^{-1/2}(Ay)\_i,\xi\_i\rangle^2$ for some unit vectors $\xi\_i\in\mathbb R^d$. Now use Cauchy-Schwarz:
$$
\langle X^{-1/2}(Ay)\_i,\xi\_i\rangle^2= \left[\sum\_j \langle X^{-1/2}A\_{ij}y\_j,\xi\_i\rangle\right]^2
\\
=
\left[\sum\_j \langle X^{-1/2}y\_j,X^{1/2}A\_{ij}^TX^{-1/2}\xi\_i\rangle\right]^2
\\
\le \left[\sum\_j |X^{-1/2}y\_j|^2\right]\left[\sum\_j |X^{1/2}A\_{ij}^TX^{-1/2}\xi\_i|^2\right]
$$
This allows one to estimate the induced operator norm of $A$ and, thereby $\rho(A)$ by
$$
\rho(A)^2\le\|A\|^2\le
\sum\_{i,j}|X^{1/2}A\_{ij}^TX^{-1/2}\xi\_i|^2
\\
=\sum\_{i,j}\langle X^{-1/2}A\_{ij}XA\_{ij}^TX^{-1/2}\xi\_i,\xi\_i\rangle\,.
$$
Now let $L$ be the linear subspace of $\mathbb R^d$ spanned by all vectors $\xi\_i$. Obviously, $\operatorname{dim}L\le\min(d,m)$. Let $P$ be the orthogonal projection to $L$. Then for every non-negative definite operator $S$ in $\mathbb R^d$, we have $\langle S\xi\_i,\xi\_i\rangle\le\operatorname{Tr} (PSP)$, which yields the bound
$$
\sum\_{i,j} \operatorname{Tr}(PX^{-1/2}A\_{ij}XA\_{ij}^TX^{-1/2}P)
= \operatorname{Tr}\left(\sum\_{i,j} PX^{-1/2}A\_{ij}XA\_{ij}^TX^{-1/2}P\right)
\\
=
\operatorname{Tr}\left(PX^{-1/2}\left[\sum\_{i,j} A\_{ij}XA\_{ij}^T\right]X^{-1/2}P\right)=
\operatorname{Tr}(PX^{-1/2}XX^{-1/2}P)
\\
=\operatorname{Tr} P=\operatorname{dim}L\le\min(d,m)
$$
as required.
| 3 | https://mathoverflow.net/users/1131 | 423534 | 172,137 |
https://mathoverflow.net/questions/423550 | 2 | A prime $q$ such that $2q+1$ is also a prime is a Sophie Germain prime.
Cramer's conjecture tells gap between consecutive primes is bound by $O(\log^2p)$.
Is there a similar conjecture for Sophie Germain primes?
| https://mathoverflow.net/users/10035 | Is there a Cramer's conjecture for Sophie Germain primes? | Heuristics says that $n$ is Sophie Germain prime with probability roughly $1/\log^2n$. Thus the probability that $C\log^an$×consecutive numbers starting from $n$×are not Sophie Germain primes is about $(1−1/\log^2n)^{C\log^an} \sim e^{−C\log^{a−2} n}$ which is too large when $a<3$ and equals $n^{-C} $ when $a=3$. So, it is natural to predict $O(\log^3p)$ bound.
| 9 | https://mathoverflow.net/users/4312 | 423553 | 172,139 |
https://mathoverflow.net/questions/423471 | 5 | I was reading [this post](https://physics.stackexchange.com/questions/264274/is-the-usage-of-the-fock-space-a-postulate-in-qft) from PSE and it reminded me an [old question of mine](https://mathoverflow.net/questions/369253/creation-and-annihilation-operators-in-qft), in which the use of creation and annihilation operators were discussed. Both questions got answers which agreed on the fact that in QFT one is not always assuming a Fock space as the underlying Hilbert space of the theory. That made me think about the role of Fock spaces to justify path integrals in some contexts. Let me elaborate.
To keep the ideas simple and keep the notations from my previous linked post, I will deal with fermionic models. Suppose our Hilbert space $\mathscr{H}$ has dimension $n < +\infty$. In many-body quantum theory, one considers its associated Fock space $\mathcal{F}^{-}(\mathscr{H})$ which has dimension $2^{n}$. There is a vast literature about rigorous treatments of many-body systems. This Fock space identifies, in a natural way, with a Grassmann algebra with finitely many generators $\psi\_{1},...,\psi\_{2^{n}}$ by simply identifying each creation operators on the Fock space with a generator of the Grassmann algebra and the vacuum vector $\Omega$ with $1$. In this setting, one can give mathematical meaning to the path integral representation by using:
(i) The fact that the trace of an operator $A$ on $\mathcal{F}^{-}(\mathscr{H})$ has a natural "translation" in terms of Grassmann integrals and
(ii) The Lie-Trotter formula for bounded operators.
At least from my perspective, every reference I know that discusses path integrals for fermions does it in the above setting: finite-dimensional Hilbert spaces of one-particle system, finite dimensional Fock spaces and so on. However, as stressed before, in a more general setting, we couldn't assume a Fock space as the underlying Hilbert space. The fact that (at least to my knowledge) no reference discusses path integral formulations in more abstract setting makes me think this approach either only works for many-body systems, in which Fock spaces can be used, or it is just easier/more intuitive to do this way.
My point is: I don't see why Fock spaces are so much necessary apart from being more intuitive. So, can't we justify the path integral method in a more abstract way, under some additional hypothesis? Suppose $\mathcal{F}^{-}(\mathscr{H})$ is replaced by an arbitrary finite-dimensional complex Hilbert space $\mathscr{H}$, with creation and annihilation operators $\varphi^{\dagger}(x)$ and $\varphi(x)$ and some Hamiltonian $H$ on $\mathscr{H}$. If we assume:
(1) These creation and annihilation operators satisfy CAR relations
(2) The existence of a unique vacuum vector $\Omega$, which is the ground state for $H$
(3) $\mathscr{H}$ is spanned by successive applications of creation operators to $\Omega$
Then:
**Q1:** Doesn't the above hypothesis fully define a fermionic system?
**Q2:** Aren't (1) to (3) enough to establish a isomorphism between $\mathscr{H}$ and a finitely generated Grassmann algebra?
**Q3:** Doesn't the path integral formulation, using the Lie-Trotter formula hold in the very same way as before?
In summary: Are the hypothesis (1) to (3) enough to replace the Fock space by an arbitrary Hilbert space and still get the same analysis?
The hypothesis (1) to (3) seems exactly the kinds of hypothesis physicists are assuming when studying their fermionic models, but the lack of discussions about path integrals on these more abstract settings (i.e. outside Fock spaces and many-body quantum mechanics) makes me question whether there is something wrong with the above.
| https://mathoverflow.net/users/150264 | Can Fock spaces be replaced by arbitrary Hilbert spaces under some hypothesis to justify path integrals? | As soon as you assume the structure of a CAR algebra, then you are automatically dealing with a Fock space. To define a CAR algebra structure, it must be generated by something, and that something is the single-particle Hilbert space $\mathscr{H}$.
Let $\mathscr{H}$ be a complex Hilbert space, and let $\mathscr{A}$ be the CAR algebra it generates. There is a unique state $\Omega$ that is a $0$-eigenstate of all annihilation operators. Then the GNS representation based on $\Omega$ is unitarily equivalent to the Fock space $\mathcal{F}^-(\mathscr{H})$. Now, this is all quite independent of the details of the Hamiltonian $H$. A free field Hamiltonian is compatible with and preserves this structure. We could, if we wanted, consider a more general Hamiltonian, and it would generate an automorphism of the algebra $\mathscr{A}$, but it would not preserve the CAR structure, in the sense that it would not preserve $\mathscr{H}$ or the mapping from $\mathscr{H}$ to creation and annihilation operators.
In infinite-dimensional systems, the situation is even worse. Nothing prevents us from considering the CAR algebra of the field on a particular time-slice. But the interacting Hamiltonians we would want to write down are symmetric but not self-adjoint on the Fock space, so they do not generate a unitary time evolution. No one knows (or at least there is no general agreement) how to write down relations, such as the canonical anticommutation relations, that would define the algebra of interacting fields. So we are forced to approximate with free fields and do perturbation theory.
| 5 | https://mathoverflow.net/users/226696 | 423569 | 172,141 |
https://mathoverflow.net/questions/423492 | 4 | Let us recall that a basis $(x\_{n})\_{n}$ for a Banach space $X$ is boundedly complete if for every scalar sequence $(a\_{n})\_{n}$ with $\sup\limits\_{n}\|\sum\limits\_{i=1}^{n}a\_{i}x\_{i}\|<\infty$, the series $\sum\limits\_{n=1}^{\infty}a\_{n}x\_{n}$ converges in norm.
Let $(x\_{n})\_{n}$ be a bounded sequence in a Banach space $X$. We set
$$ \textrm{ca}((x\_{n})\_{n})=\inf\_{n}\sup\_{k,l\geq n}\|x\_{k}-x\_{l}\|.$$
Then $(x\_{n})\_{n}$ is norm-Cauchy if and only if $\textrm{ca}((x\_{n})\_{n})=0$.
Let $(x\_{n})\_{n}$ be a basis for a Banach space $X$. We introduce a quantity measuring non-bounded completeness as follows:
$$\textrm{bc}((x\_{n})\_{n})=\sup\Bigg\{\textrm{ca}\Big(\big(\sum\_{i=1}^{n}a\_{i}x\_{i}\big)\_{n}\Big)\colon
\big(\sum\_{i=1}^{n}a\_{i}x\_{i}\big)\_{n}\subseteq B\_{X}\Bigg\},$$ where $B\_{X}$ is the closed unit ball of $X$. Clearly, $(x\_{n})\_{n}$ is boundedly complete if and only if $\textrm{bc}((x\_{n})\_{n})=0$.
We think about the $\textrm{bc}$-values of some known bases and obtain the following:
1. $\operatorname{bc}((e\_{n})\_{n})=1$, where $(e\_{n})\_{n}$ is the unit vector basis of $c\_{0}$.
2. $\operatorname{bc}((s\_{n})\_{n})=1$, where $(s\_{n})\_{n}$ is the summing basis of $c\_{0}$.
3. $\operatorname{bc}((e\_{n})\_{n=0}^{\infty})=2$, where $(e\_{n})\_{n=0}^{\infty}$ is the unit vector basis of $c$ ($e\_{0}=(1,1,1,\ldots)$).
4. $\operatorname{bc}((e\_{n})\_{n})=1$, where $(e\_{n})\_{n}$ is the unit vector basis of the James space $\mathcal{J}$.
5. Let $x\_{1}=e\_{1}$ and $x\_{n}=-x\_{n-1}+(n-1)e\_{n}$ for $n\geq 2$, where $(e\_{n})\_{n}$ is the unit vector basis of $c\_{0}$. Then $\operatorname{bc}((x\_{n})\_{n})=1$.
6. $\operatorname{bc}((f\_{n})\_{n})=2$, where $(f\_{n})\_{n}$ is the Haar basis of $L\_{1}[0,1]$.
7. $\operatorname{bc}((f\_{n})\_{n})=2$, where $(f\_{n})\_{n}$ is the Faber-Schauder basis of $C[0,1]$.
The examples above yield naturally the following question:
Question. $\textrm{bc}((x\_{n})\_{n})=1$ or $2$ for every basis $(x\_{n})\_{n}$ that is not boundedly complete ?
Prof. William B. Johnson has the following guess with respect to Question:
Guess. Let $(x\_{n})\_{n}$ be a basis for a Banach space $X$ that is not boundedly complete. If $(x\_{n})\_{n}$ is monotone and shrinking, then $\textrm{bc}((x\_{n})\_{n})=1$ or $2$.
I can not prove Guess. Thank you !
| https://mathoverflow.net/users/41619 | Boundedly complete bases | As to the question, the answer is "no". The simplest counterexample is the closure of finite support sequences in the norm $\sup\_{j\ge 1}|a\_j|+\sum\_{j\ge 1}{|a\_j-a\_{j+1}|}$ (decaying to $0$ sequences of finite total variation) with the standard basis $x\_n(k)=\delta\_{nk}$. Then for any partial sum sequence in the unit ball, the total variation component of the norm gets exhausted eventually and the differences become essentially $(0,0,\dots,0,a\dots, a,0,0,\dots)$ with norm $3|a|\le 3/2$ (every entry of a sequence in the unit ball should be at most $1/2$ in absolute value). On the other hand, the coefficient sequence $1/2,1/2,1/2,\dots$ realizes this bound giving $\mathbf{bc}((x\_n)\_n)=3/2$.
I cannot tell anything about the Guess because I don't know what the adjectives "monotone" and "shrinking" mean applied to a basis in a Banach space and you didn't bother to explain ;-)
*Edit:* OK, since Bill explained the words, I'll attempt to refute the guess as well. It will be essentially the same construction but using the quadratic variance instead of the usual one.
Define $X$ to be the space of decaying to $0$ sequences $a=(a\_j)\_{j\ge 1}$ with
$$
\|a\|\_0=\sup\_{j\ge 1}|a\_j|+\left[\sum\_{j\ge 1}|a\_j-a\_{j+1}|^2\right]^{\frac 12}<+\infty\,.
$$
To make the standard basis monotone, we'll just change the norm to the equivalent one given by
$\|a\|=\sup\_m\|a^m\|\_0$ where $a^m=(a\_1,\dots,a\_m,0,0,\dots)$.
Now, if $\|a\|\le 1$, then we still have all $|a\_j|\le\frac 12$, so, since the variation component of the norm will eventually exhaust itself except for the endpoint jump, we still can say that $\mathbb{bc}\le \frac12(1+\sqrt 2)$ with the sequence of all $1/2$ giving exactly that value.
The only thing that remains to show is that our standard basis is shrinking. One can, probably, compute the dual space directly, but we'll take a shortcut.
One key property of our space is that if we have finitely many elements $a\_q\in X$ of norm $1$ with *separated* supports and numbers $b\_q$ with $\sum\_q|b\_q|^2\le 1$, then $\|\sum\_q b\_qa\_q\|\le 1$.
Now let $\psi$ be a linear functional on $X$ corresponding to the sequence $(\psi\_1,\psi\_2,\dots)$ with the usual pairing
$$
\psi(a)=\sum\_{j\ge 1}\psi\_ja\_j\,.
$$
If an arbitrarily faraway tail of $\psi$ has norm at least $\varepsilon$, then for every $Q$, we can inductively construct $Q$ sequences $a\_q\in X$ of norm $1$ with finite separated supports such that $\psi(a\_q)\ge\varepsilon/4$. But then $\psi(Q^{-1/2}\sum\_qa\_q)\ge \varepsilon\sqrt Q/4$, which contradicts the boundedness of $\psi$. Thus the tails must tend to $0$ in the norm and we are done.
If you look at this construction closely, you'll realize that it can yield any $\mathbb bc\in(1,2)$ but not one in $(0,1)$. So, the interesting question really is
**Question:** Can we construct a basis (any or monotone and shrinking) in a Banach space with $\mathbb{bc}\in (0,1)$?
| 2 | https://mathoverflow.net/users/1131 | 423570 | 172,142 |
https://mathoverflow.net/questions/423559 | 4 | Fix a faithful functor $\Gamma: \mathsf C\longrightarrow \mathsf{Set}$ and think of it as the "underlying points". When it exists, a left adjoint $\mathrm{disc}\dashv \Gamma$ can be thought of as the "discrete objects" functor. When it exists, a further left adjoint $\pi\_0\dashv \mathrm{disc}\dashv \Gamma$ can be thought of as "connected components.
For $\mathsf C$ the category of graphs and graph morphisms, $\pi\_0$ takes a graph to the set of its connected components, where vertices lie in the same component if they are connected by a path.
For $\mathsf C$ the category of directed graphs, the same assertion holds.
**Question.** How to modify the adjunction above to obtain a "strongly connected components" functor? Vertices of a directed graph are strongly connected if there is a path between them in each direction.
| https://mathoverflow.net/users/69037 | Strongly connected components as adjoint functor? | The existence of such an adjoint triple in particular requires the strongly connected components functor to be a left adjoint. But in fact this fact is not a left adjoint, since it doesn't preserve colimits.
To see this, consider the span formed by the discrete graph on two vertices, included in the two two-vertex graphs which have one edge each (in opposite directions). Their pushout is the two-vertex graph with an edge in each direction, which is strongly connected. At the level of strongly connected components, you get the identity span between two-element sets, so their pushout still has two elements. Thus the pushout is not preserved.
**Conclusion:** There is no adjoint triple with the strongly connected components functor on the left.
There's some ambiguity about which category of directed graphs you actually have in mind, but the above argument works in all obvious candidates. If it doesn't work in yours, then you should be more explicit about what your category is.
| 8 | https://mathoverflow.net/users/27013 | 423572 | 172,143 |
https://mathoverflow.net/questions/423532 | 6 | For $\mathcal{M}$ a (countable) nonstandard model of $\mathsf{PA}$, let $\mathsf{SS}(\mathcal{M})$ be the set of sets of natural numbers coded by elements of $\mathcal{M}$. There are various ways to define this, for example $$\{X\subseteq\mathbb{N}:\exists a\in\mathcal{M}\;\forall k\in\mathbb{N}\;(k\in X\iff p\_k\vert a)\}$$ (where $p\_k$ is the $k$th prime and we conflate $\mathbb{N}$ with its canonical image in $\mathcal{M}$).
I'm curious about how this could differ from the following analogue: let $$\mathsf{SS}^-(\mathcal{M})=\{X\subseteq\mathbb{N}:\exists a\in\mathcal{M}\;\forall k\in\mathbb{N}\;[k\in X\iff \exists n\in\mathbb{N}\;(p\_k^n\not\vert a)]\}.$$
Intuitively, elements of $\mathbb{N}$ are **prevented** from entering $X$ by corresponding primes dividing $a$ "too much." (This version has come up in a separate problem I'm playing with, and I'd like to understand it better.)
It's easy to show that $\mathsf{SS}(\mathcal{M})$ is the set of elements of $\mathsf{SS}^-(\mathcal{M})$ whose complements are also in $\mathsf{SS}^-(\mathcal{M})$, and it's not hard to show that every $\mathcal{M}$ has an elementary extension $\mathcal{N}$ such that $\mathsf{SS}^-(\mathcal{N})=\mathsf{SS}(\mathcal{N})$. However, this still leaves a lot open. In particular:
>
> Is there an $\mathcal{M}$ with $\mathsf{SS}(\mathcal{M})\not=\mathsf{SS}^-(\mathcal{M})$?
>
>
>
Equivalently per the above, is there an $\mathcal{M}$ whose $\mathsf{SS}^-$ is not closed under complementation?
| https://mathoverflow.net/users/8133 | A "negative" standard system | If $\mathcal{M}$ is a nonstandard model of PA then for any set $X \in \mathsf{SS}(\mathcal{M})$, $X' \in \mathsf{SS}^-(\mathcal{M})$. Thus if $\mathsf{SS}(\mathcal{M}) = \mathsf{SS}^{-}(\mathcal{M})$ then $\mathsf{SS}(\mathcal{M})$ is closed under the Turing jump. Since not every Scott set is closed under the jump, $\mathsf{SS}(\mathcal{M})$ is not always equal to $\mathsf{SS}^-(\mathcal{M})$.
**Claim.** If $X \in \mathsf{SS}(\mathcal{M})$ then $X' \in \mathsf{SS}^-(\mathcal{M})$.
*Proof.* Suppose $X$ is in $\mathsf{SS}(\mathcal{M})$. Thus there is some $a$ such that
$$
\forall k \in \mathbb{N}\, (k \in X \iff p\_k \mid a).
$$
Now let $b$ be a fixed nonstandard number in $\mathcal{M}$ and let $c$ be a nonstandard number such that
$$
\forall n, k < b\, (p\_k^n \mid c \iff \varphi^a\_k(k) \text{ does not converge in $\leq n$ steps}).
$$
Here, using $a$ as an oracle for $\varphi\_k(k)$, means that when $\varphi\_k(k)$ asks a question about $m$ to the oracle, we check if $p\_m \mid a$ to determine the answer. Note that the existence of such a $c$ can be proved in PA.
Now let $Y = \{k \in \mathbb{N} \mid \exists n \in \mathbb{N} \, (p\_k^n \nmid c)\}$. I claim that $Y = X'$. To show this it is enough to show that for all $n, k \in \mathbb{N}$, $p\_k^n \mid c$ if and only if $\varphi^X\_k(k)$ does not converge in $\leq n$ steps.
Note that for $n$ and $k$ standard natural numbers, the convergence or nonconvergence of $\varphi^X\_k(k)$ within $n$ steps is witnessed by a standard natural number (encoding the transcript of the computation). And since running $\varphi^X\_k(k)$ for $n$ steps never requires asking questions of the oracle at nonstandard numbers, there is no difference between using $X$ and $a$. Thus $\mathcal{M}$ can check that the witness to convergence or divergence of $\varphi^X\_k(k)$ within $n$ steps is also a witness to the convergence or divergence of $\varphi^a\_k(k)$ within $n$ steps and therefore by definition of $c$, $p\_k^n \mid c$ if and only if $\varphi^X\_k(k)$ does not converge in $\leq n$ steps.
| 4 | https://mathoverflow.net/users/147530 | 423575 | 172,145 |
https://mathoverflow.net/questions/423576 | 5 | Reading about the calculation of the Brauer group of rational numbers, the calculations of the group are extremely lengthy and technical. First of all, it will be very helpful to me if someone can explicitly give me a class which corresponds to a nontrivial member of the Brauer group of rational numbers. Secondly, I appreciate it if someone can sketch the calculation if we assume we know the Brauer group of every local field is $\mathbb{Q}/\mathbb{Z}$.
| https://mathoverflow.net/users/483209 | Brauer group of rational numbers | What reference are you reading?
For a field $K$, every finite-dimensional central simple $K$-algebra $A$ is isomorphic to ${\rm M}\_n(D)$ where $n$ is a positive integer and $D$ is a division ring with center $K$ where $\dim\_K(D)$ is finite. The number $n$ is unique and $D$ is unique up to $K$-algeba isomorphism. The equivalence classes defining ${\rm Br}(K)$ just involve declaring ${\rm M}\_n(D) \sim D$. That is, we identify two $A$'s if they have isomorphic $D$'s. Thus representatives for ${\rm Br}(K)$ are the finite-dimensional $K$-central *division algebras*, up to isomorphism.
The group law on ${\rm Br}(K)$ uses the tensor product of representatives:
$[A][A'] = [A \otimes\_K A']$ (isomorphic central simple $K$-algebras have isomorphic tensor products over $K$). Since we can represent each element of ${\rm Br}(K)$ by a unique finite-dimensional $K$-central division algebra up to isomorphism, in terms of $K$-central division algebras you could say the group law in ${\rm Br}(K)$ is $[D][D'] = [D'']$ where
$D \otimes\_K D' \cong {\rm M}\_n(D'')$ for some $n$ and $D''$. (**Note**: a tensor product of two $K$-central division rings need not be a division ring, *e.g.*, ${\mathbf H} \otimes\_{\mathbf R} {\mathbf H} \cong {\rm M}\_{4}(\mathbf R)$, so in ${\rm Br}(\mathbf R) = \{[\mathbf R],[\mathbf H]\} =\{1,[\mathbf H]\}$ we have $[\mathbf H]^2 = 1$.)
To ask for a nontrivial element of ${\rm Br}(K)$ is to ask for a finite-dimensional $K$-central division algebra other than $K$ itself. The simplest examples besides $K$ are the quaternion algebras over $K$, which are defined to be the $4$-dimensional central simple $K$-algebras. Every quaternion algebra over $K$ that's not isomorphic to ${\rm M}\_2(K)$ will be a division ring. (For some $K$ there are no quaternion algebras over $K$ except for ${\rm M}\_2(K)$, such as when $K$ is algebraically closed or finite, in fact in these cases ${\rm Br}(K)$ is trivial: all finite-dimensional $K$-central simple algebras are just the rings ${\rm M}\_n(K)$ up to isomorphism.) A concrete account of quaternion algebras, which avoids working in characteristic $2$ (even though the general theory can handle this case if you develop it in a more abstract way) is [here](https://kconrad.math.uconn.edu/blurbs/ringtheory/quaternionalg.pdf).
In ${\rm Br}(K)$, the quaternion algebras (including the trivial or "split" case ${\rm M}\_2(K)$) are representatives of the $2$-torsion subgroup: every element of ${\rm Br}(K)$ with order dividing $2$ is represented by a unique quaternion algebra over $K$, so the tensor product operation is a group law on quaternion algebras over $K$.
It's very helpful to be able to understand quaternion algebras well before trying to deal with the full Brauer group of a field, just as quadratic fields are a good testing ground to understand theorems about general number fields.
The construction called *cyclic algebras* gives you an $n^2$-dimensional $K$-central simple algebra from a cyclic Galois extension $L/K$ of degree $n$. (The Hamilton quaternions come in this way from the quadratic extension $\mathbf C/\mathbf R$: $\mathbf H = \mathbf C + \mathbf Cj$ where $j^2 = -1$ and $jz = \overline{z}j$ for all $z \in \mathbf C$.) Number fields and local fields have lots of cyclic Galois extensions (think of the unramified extensions of $\mathbf Q\_p$ or suitable subfields of cyclotomic extensions of $\mathbf Q$), so you can get lots of central simple algebras of number fields and local fields this way. An account of cyclic algebras is in Section 14 of Chapter 5 of Lam's *A First Course in Noncommutative Rings*, and in that section there is an example of a $9$-dimensional division ring over $\mathbf Q$ built from a cyclic cubic extension of $\mathbf Q$ (the cubic subfield of the $7$th cyclotomic field).
Also see Definition 7.10 and Theorem 7.13 [here](https://kconrad.math.uconn.edu/blurbs/gradnumthy/cfthistory.pdf).
There is a more general way of building $n^2$-dimensional central simple algebras from degree-$n$ Galois extensions (not necessarily having a cyclic Galois group) called a *crossed product algebra*. It is not always clear when these constructions are nontrivial, meaning they are not isomorphic to ${\rm M}\_n(K)$
The general concepts I have mentioned here can be applied to an arbitrary base field $K$: $\mathbf Q$, $\mathbf R$, $\mathbf Q\_p$, finite fields, and so on. The fact that ${\rm Br}(\mathbf Q\_p) \cong \mathbf Q/\mathbf Z$ amounts to showing every $n^2$-dimensional division ring over $\mathbf Q\_p$ can be built as a cyclic algebra using the degree-$n$ unramified extension of $\mathbf Q\_p$ (which is a cyclic extension). See Theorem 7.14 [here](https://kconrad.math.uconn.edu/blurbs/gradnumthy/cfthistory.pdf). Figuring out when two such constructions can be isomorphic, in terms of the parameters that show up in the construction, turns out to be the same as asking what the elements of order $n$ look like in $\mathbf Q/\mathbf Z$. A detailed proof of the calculation of ${\rm Br}(K)$ when $K$ is a local field is in Jacobson's *Basic Algebra II*: see the sections on division rings and the Brauer group near the end of Chapter $9$.
| 11 | https://mathoverflow.net/users/3272 | 423579 | 172,147 |
https://mathoverflow.net/questions/423577 | 2 | This question is motivated by a vague analogy between **true paths** in priority arguments and **realizers** - relative to an oracle - in the sense of intuitionistic logic. Intuitively, I'm looking for a precise way to ask the following: "can downward density be proved without infinite injury?" *(Incidentally, there is a vague similarity-of-purpose with [this other question of mine](https://mathoverflow.net/questions/422758/how-much-downward-density-can-we-get-without-injury).)*
Precisely, for $X, A$ noncomputable c.e. sets, say that $A$ is *efficiently below $X$* iff $\emptyset<\_TA<\_TX$ and there is a pair $(F,G)$ of ${\bf 0'}$-computable total functions with the following properties:
* For each $a\in\omega$, $F(a)$ is a pair $\langle u,i\rangle\in \omega\times 2$ such that if $i=0$ then $\Phi\_a^{A}(u)\uparrow$ and if $i=1$ then $\Phi\_a^{A}(u)=1-X(u)$; and
* For each $b\in\omega$, $G(b)$ is a pair $\langle v,j\rangle\in\omega\times 2$ such that if $j=0$ then $\Phi\_b^\emptyset(v)\uparrow$ and if $j=1$ then $\Phi\_b^\emptyset(v)=1-A(v)$.
My question, then, is the following:
>
> Is there a noncomputable c.e. set with no noncomputable c.e. sets efficiently below it?
>
>
>
Note that if we were to replace ${\bf 0'}$ with ${\bf 0''}$ we would get a **uniform-in-$X$** negative answer since the true path in the usual proof of Sacks density is ${\bf 0''}$-computable. Meanwhile, we would get a *positive* answer if we looked at general density instead of specificially downward density, but for boring reasons since any analogous "efficient" witness to an interval $(X,Y)$ being nonempty would a fortiori compute $X'$; this also shows that we get a trivial positive answer if we replace ${\bf 0'}$ by any smaller degree.
| https://mathoverflow.net/users/8133 | Hard-to-"realize" instances of downward density | Every noncomputable c.e. set has a noncomputable c.e. set efficiently below it. The reason is that "$A$ is efficiently below $X$" is equivalent to "$A$ is low" and it is possible to show that every noncomputable c.e. set has a noncomputable low c.e. set below it.
To prove this let me first restate your definitions a bit. Let's say that $X$ is efficiently not above $Y$ if there is a $0'$ computable function $F \colon \omega \to \omega\times 2$ such that for all $e$, one of the two following options holds.
* For some $n \in \omega$ we have $F(e) = (n, 0)$ and $\Phi^X\_e(n)\uparrow$
* For some $n \in \omega$ we have $F(e) = (n, 1)$ and $\Phi^X\_e(n)\downarrow \neq Y(n)$.
Now "$X$ is efficiently below $Y$" in your sense means that $X$ is efficiently not above $Y$, $0$ is efficiently not above $X$ and $X \leq\_T Y$.
**Claim 1.** If $X$ and $Y$ are c.e. then $X$ is efficiently not above $Y$ if and only if $Y \nleq\_T X$ and $X$ is low.
*Proof.* First suppose that $Y \nleq\_T X$ and $X$ is low. Use $0'$ to compute $F$ witnessing that $X$ is efficiently not above $Y$ as follows. Given any $e$, search for an $n$ such that either $\Phi^X\_e(n)\uparrow$ or $\Phi^X\_e(n) = 0$ and $n$ eventually enters $Y$ or $\Phi^X\_e(n) = 1$ and $n$ never enters $Y$. Note that for a fixed $n$ we can figure out if one of these three situations holds (and if so, which one) using $X'$ and hence $0'$ since $X$ is low. Also we are guaranteed to find at least one such $n$ since $X$ does not compute $Y$. Use this $n$ to define $F(e)$.
Now suppose $X$ is efficiently not above $Y$. Clearly this implies $Y \nleq\_T X$. To see why it implies $X$ is low, let $F$ be a witness to $X$ being efficiently not above $Y$. Now use $F$ to compute $X'$ as follows. Given $e$, produce $\tilde{e}$ such that if $\Phi^X\_e(e)$ diverges then $\Phi\_{\tilde{e}}^X$ diverges on every input and if $\Phi^X\_e(e)$ converges then $\Phi\_{\tilde{e}}^X$ converges on every input. Note that $e \mapsto \tilde{e}$ is computable and the second bit of $F(\tilde{e})$ tells you whether $\Phi^X\_e(e)$ converges or not.
**Claim 2.** If $0 <\_T A <\_T X$ are c.e. and $A$ is low then $A$ is efficiently below $X$.
*Proof.* Since $0$ is low and $X \nleq\_T 0$, $0$ is efficiently not above $A$. Since $A$ is low and $A \nleq\_T X$, $A$ is efficiently not above $X$. And since $A \leq\_T X$ this implies $A$ is efficiently below $X$.
**Claim 3.** For any noncomputable c.e. set $X$ there is a low noncomputable c.e. set computable from $X$.
*Proof.* If $X$ is low then any set computable from it is also low so the usual density theorem can be applied directly. If $X$ is not low then any noncomputable low c.e. set below it must be strictly below it. But given a noncomputable c.e. set $X$ it is easy to construct a noncomputable low c.e. set $A \leq\_T X$ using a finite injury construction (to make it low we simply want to preserve instances of programs converging, to make it noncomputable we occasionally want to put an element in and to make it below $X$ we can use permitting).
| 3 | https://mathoverflow.net/users/147530 | 423589 | 172,151 |
https://mathoverflow.net/questions/423584 | 4 | Suppose $X\subset \mathbb S^1$ is a finite subset of the group $\mathbb S^1=\mathbb R/\mathbb Z$. We say that $t\in \mathbb S^1$ is a half symmetry of $X$ if $|(X+t)\cap X|>|X|/2$.
**Question.** Can the number of half symmetries of $X$ be larger than $|X|$? If so, is there some reasonable (like $c|X|$) upper bound on the number of half symmetries of $X$?
| https://mathoverflow.net/users/13441 | Number of "half symmetries" of a finite subset of $\mathbb S^1$ | **Simple construction.** First, let me answer the question as posed. Let $X$ be any subset of size of $2m$ of $\mathbb{Z}/(3m-1)\mathbb{Z}$. Then every shift of $X$ is a half-symmetry of $X$. Since $\mathbb{Z}$ embeds into $\mathbb{S}^1$, this gives an example with about $(3/2)|X|$ half-symmetries.
**Better construction (slightly informal)**. Let $m$ be an arbitrary integer. Pick each element of $\mathbb{Z}/m\mathbb{Z}$
independent with probability $\tfrac{1}{2}+2\sqrt{\frac{\log m}{m}}$. Then $\mathbb{E}[X]=m/2+\sqrt{m\log m}$. Let $t\neq 0$, and consider $I\_t=|(X+t)\cap X|$. It is distributed approximately as a binomial random variable with mean $(m/2+\sqrt{m\log m})^2/m\approx m/4+\sqrt{m\log m}$. By Chernoff's bound, $P[I\_t\leq m/4]<1/m$. Taking the union bound over all $t\in\mathbb{Z}/m\mathbb{Z}$, we obtain a set $X$ with $(2-o(1))|X|$ half-symmetries.
**Better construction (more formal).** I said *approximately a binomial random variable* because different elements of $(X+t)\cap X$ are not completely independent. However, it is easy to fix: Choose $m$ to be prime, with $|X|$ being chosen by picking elements independent as above. By multiplying everything by $t^{-1}$, we see that the random variables $I\_t$ and $I\_1$ are identically distributed. So, consider $I\_1$. Let $R\_x$ be the characteristic random variable of the event event $x\in X\cap(X+1)$. The variables $R\_0,R\_2,R\_4,\dotsc,R\_{p-3}$ are independent because they depend on independent choices for elements of $X$. Namely, $R\_0$ depends only on whether $0\in X$ and $1\in X$, whereas $R\_2$ depends only on whether $2\in X$ and $3\in X$, etc. For the similar reason, the random variables $R\_1,R\_3,R\_5,\dotsc,R\_{p-2}$ are independent. Let $A=R\_0+R\_2+\dotsb+R\_{p-3}$ and $B=R\_1+R\_3+\dotsb+R\_{p-2}$. So, $I\_t=A+B+R\_{p-1}$ is a sum of two binomial random variables and a little correction. We can bound the deviations of each of these binomial random variables by Chernoff to show that both events "$A\leq m/8$" and "$B\leq m/8$" are unlikely, and hence $I\_1$ is very likely to be greater than $m/4$. Then taking the union bound over different $t\neq 0$ completes the formal proof.
**Simple almost construction.** Let $X$ be the set of quadratic residues in $\mathbb{Z}/p\mathbb{Z}$. Then $X\cap (X+t)$ has size almost $|X|/2$ for every $t$. So, it fails to be a example of a set with $2|X|$ half-symmetries only barely.
| 5 | https://mathoverflow.net/users/806 | 423590 | 172,152 |
https://mathoverflow.net/questions/422930 | 19 | Let $M$ be the *saddle surface* in $\mathbb R^3$ defined by $x^2-y^2+z=0$. For any $r\geq 0$ and $(x\_0,y\_0,z\_0)\in\mathbb R^3$, let $rM+(x\_0,y\_0,z\_0)$ denotes the surface obtained by scaling $M$ by $r$ and then *translating* by $(x\_0,y\_0,z\_0)$. (Note that by $rM$, with $r=0$, we mean $\lim\_{r\to 0} rM$, which is two perpendicularly intersecting plane.) And let $B$ be the unit ball.
Is it true that
***$$\mathrm{area}\left[(rM+(x\_0,y\_0,z\_0))\cap B\right]\leq 2\pi,$$***
with equality holds ***if and only if*** $r=0$ and $(x\_0,y\_0,z\_0)=(0,0,0)$ (which gives two intersecting equatorial disks)?
**Edit:** Using the first variation formula, I can show some partial results:
* For any fixed $z\_0$, the area of $(r(M+(0,0,z\_0)))\cap B$ decreases as $r$ increases, for all $r\in(0,+\infty)$.
* $(r,x\_0,y\_0,z\_0)=(0,0,0,0)$ is a strict local maximum.
| https://mathoverflow.net/users/174092 | All saddles in the unit ball have area $<2\pi$? | It is actually next to trivial if you choose the right parameterization (and rather puzzling if you don't, so it can make a decent take-home exam problem in multivariate calculus).
I'll use the line cover $x(s,t)=(s+t,s-t,4st)$. The area element is then
$2\sqrt{1+8s^2+8t^2}\,ds\,dt< 2(\sqrt{1+8s^2}+\sqrt{1+8t^2})\,ds\,dt$.
Now for a ball $B$ of radius $R$ centered at $(u,v,w)$ and for fixed $s$, we have the line in $t$ whose moving speed is $\sqrt{2+16s^2}=\sqrt 2\sqrt{1+8s^2}$ and whose square distance from the center of the ball is at least
$$
\min\_t[(s+t-u)^2+(s-t-v)^2]=2(s-\tfrac{u+v}2)^2=2(s-s\_0)^2.
$$
Thus, integrating in $t$ first, we have
$$
\int\_{s,t:x(s,t)\in B}2\sqrt{1+8s^2}\,dt\,ds\le \int\_{s\in\mathbb R}2\sqrt 2\sqrt{[R^2-2(s-s\_0)^2]\_+}\,ds=\pi R^2
$$
(time = line length/speed; length = $2\sqrt{R^2-\text{(line distance to the center)}^2}$).
The other integral is done in exactly the same way, only you need to integrate with respect to $s$ first. Hence, the area is $<2\pi R^2$, which is equivalent to the requested bound after scaling.
| 12 | https://mathoverflow.net/users/1131 | 423592 | 172,153 |
https://mathoverflow.net/questions/423612 | 16 | Let $ L \subseteq \mathbb{R}^2 $ be a smooth real algebraic curve. Let's fix some parameter $ \delta \in \mathbb{R} $ and for every point $ (x,y) \in L $ define $$ L\_{\delta}(x,y) = (x,y) + \delta n(x,y), $$ where $ n(x,y) $ is a normal vector to $L$ at $(x,y)$. The equidistant curve is then $$ L\_\delta = \{L\_\delta (x,y) : (x,y) \in L\}. $$
The idea is very simple: we take a normal vector with length $ \delta $ and "roll" it along the curve. Note that this is not the same as the set of points at the same distance $\delta$ from $L$.
I am trying to find the equation of $L$. But I do not even understand if it is algebraic. For example, let's consider the parabola $ L$ given by $y - x^2 = 0 $. It is easy to find a parametrization of $ L\_\delta $:
$$ x = t - \delta \frac{2t}{\sqrt{4t^2+1}}, $$
$$ y = t^2 + \delta \frac{1}{\sqrt{4t^2+1}}. $$
But how do I find the equation oh $x,y$ from this one? I looked up some literature about computational algebraic geometry. As I understood, there are only theorems for the case when the parametrization is rational in the variable $t$, which is not the case here. $ \mathbb{R} $ is not algebraicaly closed, maybe this is also an issue.
For me the question is interesting when $L$ is an algebraic surface in $ \mathbb{R}^n $, but the case of curves on the plane looks so simple that I am sure somebody has already solved it. Any help would be appreciated!
| https://mathoverflow.net/users/483248 | Is the "equidistant curve" to an algebraic curve algebraic? | Yes, $L\_\delta$ is algebraic. You can find its equations by elimination theory as follows: Let $L$ be defined by the polynomial equation $F(x,y) = 0$. Now consider the polynomial equations
$$
F(x,y)=(u{-}x)-aF\_x(x,y)= (v{-}y)-aF\_y(x,y)= a^2(F\_x(x,y)^2+F\_y(x,y)^2)-\delta^2=0.
$$
for (x,y,a,u,v). This is 4 equations for 5 unknowns. You can now use elimination theory to find a polynomial equation $G(u,v)=0$ in the ideal of the above system of equations that does not involve $x$, $y$, or $a$. This will be the algebraic equation of the parallel curve you want.
In your example of $y-x^2$, you find, for example, that the curve has degree $6$ and has the equation
$$
\begin{aligned}
0=&16\,{u}^{6}+16\,{u}^{4}{v}^{2}-40\,{u}^{4}v + \left( -48\,{b}^{2}+1
\right) {u}^{4}\\
&\quad-32\,{u}^{2}{v}^{3}+ \left( -32\,{b}^{2}+32 \right) {u
}^{2}{v}^{2}+ \left( 8\,{b}^{2}-2 \right) {u}^{2}v+ \left( 48\,{b}^{4}
-20\,{b}^{2} \right) {u}^{2}\\
&\quad+16\,{v}^{4}+ \left( -32\,{b}^{2}-8
\right) {v}^{3}+ \left( 16\,{b}^{4}-8\,{b}^{2}+1 \right) {v}^{2}+
\left( 32\,{b}^{4}+8\,{b}^{2} \right) v\\
&\quad-16\,{b}^{6}-8\,{b}^{4}-{b}^{2
}
\end{aligned}
$$
where, to save typing, I have writteen $b$ instead of $\delta$.
**Remark:** If you apply this method to Willie Wong's 'counterexample', you get the equation
$$
0 = \bigl((u-v)^2-2b^2\bigr)\bigl((u+v)^2-2b^2\bigr),
$$
i.e., the equation of the 4 lines $u\pm v = \pm b\sqrt2$, as expected.
| 25 | https://mathoverflow.net/users/13972 | 423618 | 172,158 |
https://mathoverflow.net/questions/423627 | 5 | Apologizes if this is a basic question, but I am new to the area of finite dimensional algebras. I am reading the paper "Unbounded derived categories and the finitistic dimension conjecture" by Jeremy Rickard (<https://arxiv.org/abs/1804.09801>) and have a question about the proof of Theorem 4.3.
In the proof, $P\_i$ is a projective resolution of the module $M\_i$, and has length $d\_i$.
The author then concludes that $Tor\_{d\_i}^A(M\_i,A/rad(A))$ is non-zero.
My question is: why is this the case? does this follow from some property of finite dimensional algebras? or this a more general fact?
| https://mathoverflow.net/users/483271 | About a recent paper of Rickard on finitistic dimension | I use $M$ instead of $M\_i$ and $d$ instead of $d\_i$.
For a finite dimensional algebra $A$ over a field $K$ we have in general
$D Ext\_A^i(Y,DZ)=Tor\_i^{A}(Y,Z)$ using the duality $D=Hom\_K(-,K)$.
Thus $Tor\_d^{A}(M,A/radA)=D Ext\_A^d(A/rad A, D(M))$ is non-zero since $D(M)$ has injective dimension at least $d$ ($D$ is a duality so this is equivalent to $M$ having projective dimension at least $d$).
Here I used that in general for a simple module $S$ and a module $N$ with minimal injective coresolution $0 \rightarrow N \rightarrow I^0 \rightarrow I^1 \rightarrow \cdots $ we have $Ext\_A^n(S,N)$ being non-zero if and only if the injective envelope $I(S)$ of $S$ is a direct summand of $I^n$. Note that $A/rad A$ is simply the direct sum of all simple $A$-modules.
| 5 | https://mathoverflow.net/users/61949 | 423629 | 172,162 |
https://mathoverflow.net/questions/423630 | 8 | Let us recall that a *symmetric* on a set $X$ is any function $d:X\times X\to[0,\infty)$ such that
for every $x,y\in X$ the following two conditions are satisfied:
$\bullet$ $d(x,y)=0$ if and only if $x=y$;
$\bullet$ $d(x,y)=d(y,x)$.
A topological space $X$ is called *symmetrizable* if there exists a symmetric $d$ on $X$ such that a subset $U\subseteq X$ is open if and only if for every $x\in X$ there exists $\varepsilon>0$ such that $B\_d(x,y)\subseteq U$, where $B\_d(x,\varepsilon)=\{y\in X:d(x,y)<\varepsilon\}$.
By the Urysohn Metrization Theorem, every regular second-countable space is metrizable and hence symmetrizable.
**Question 1.** *Is each Hausdorff second-countable space symmetrizable?*
A weaker version of Question 1 is also of interest:
**Question 2.** *Is each countable first-countable Hausdorff space symmetrizable?*
---
**Added in Edit.** I have just realized that Question 2 has a simple affirmative answer (which I present below), so only Question 1 remains open.
**Added in Edit 31.05.2022.** Ups: Question 1 has a negative answer!
| https://mathoverflow.net/users/61536 | Is every second-countable Hausdorff space symmetrizable? | I have just realized that the first question has a simple affirmative answer.
**Theorem 1.** *Every countable first-countable $T\_1$ space $X$ is symmetrizable.*
*Proof.* For every $x\in X$, fix a neighborhood base $(U\_n(x))\_{n\in\omega}$ such that $U\_{n+1}(x)\subseteq U\_n(x)$ for all $n\in\omega$. Since $X$ is countable, there exists a linear order $\le$ on $X$ such that for every $x\in X$ the initial interval $\{y\in X:y\le x\}$ is finite.
It is easy to see that the topology of $X$ is generated by the symmetric
$$d(x,y)=\inf\big\{2^{-n}:\max\{x,y\}\in U\_n(\min\{x,y\})\big\}.\quad\square$$
---
**Important Remark.** After looking at a reference Dave L. Renfro suggested in a comment as possibly being relevant, I discovered that affirmative answers to both my questions follow from
**[Theorem 2.9](http://www.mathnet.ru/links/dbf92f01dafb7bfdaa4d7d96dd652ad7/rm5901.pdf) (Arhangelski, 1966):** Every first-countable $T\_1$ space with a $\sigma$-discrete (closed) network is symmetrizable.
But reading the proof of this theorem I discovered that it works only for regular spaces. Arhangelski writes that one loses no generality assuming that the $\sigma$-discrete network is closed, i.e., consists of closed sets. But after taking the closure of elements of a network in a $T\_1$-space (even in a Hausdorff space), the network property can be destroyed. So, Question 2 seems to stay open and not answered even for Hausdorff (not mention $T\_1$) spaces.
---
**Added in Edit 31.05.2022.** To my big surprise (and contrary to what was claimed by Arhangelski in his Theorem 2.9), I have just discovered that Question 1 has negative answer!
First let us prove that the symmetrizability of second-countable spaces is equivalent to the perfectness. Recall that a topological space $X$ is *perfect* if every closed subset of $X$ is of type $G\_\delta$.
**Theorem 2.** *A second-countable (Hausdorff) $T\_1$-space $X$ is symmetrizable if (and only if) it is perfect.*
*Proof.* If $X$ is symmetrizable and Hausdorff, then the first-countability of $X$ implies that $X$ is semi-metrizable and perfect, see the paragraph before Theorem 9.8 in [Gruenhage's "Generalized metric spaces"](https://www.sciencedirect.com/science/article/pii/B9780444865809500136).
Conversely, if a second-countable $T\_1$-space $X$ is perfect, then each open subset of $X$ is an $F\_\sigma$ in $X$, which implies that $X$ has a countable closed network. Now we can apply Arhangelski's Theorem 2.9 to conclude that $X$ is symmetrizable. $\square$
Let $\mathrm{non}(\mathcal M)$ denote the smallest cardinality of a nonmeager set in the real line.
**Example 1.** *There exists a second-countable Hausdorff space of cardinality $\mathrm{non}(\mathcal M)$ which is not perfect and hence not symmetrizable.*
*Proof.* Take any nonmeager linear subspace $L$ of cardinality $\mathrm{non}(\mathcal M)$ in $\mathbb R^\omega$ such that for every $n\in\omega$ the intersection $L\_n=L\cap(\{0\}^n\times\mathbb R^{\omega\setminus n})$ is dense in $\{0\}^n\times\mathbb R^{\omega\setminus n}$. Consider the quotient space $X=L\_\circ/\_\sim$ of $L\_\circ=L\setminus\{0\}$ by the equivalence relation $\sim$ defined by $x\sim y$ iff $\mathbb R x=\mathbb Ry$. Since the space $L\_\circ$ is Baire and the quotient map $q:L\_\circ\to X$ is open, the space $X$ is second-countable and Baire. It is easy to check that the closure of every nonempty set in $X$ contains the set $q[L\_n\setminus\{0\}]$ for some $n\in\omega$. This implies that the space $X$ is *superconnected* in the sense that for every nonempty open sets $U\_1,\dots,U\_n$ in $X$ the intersection of their closures $\overline U\_1\cap\dots\cap\overline U\_n$ is not empty.
Now take any disjoint nonempty open sets $U,V$ in $X$. Assuming that $V$ is of type $F\_\sigma$, we can apply the Baire Theorem and find a nonempty open set $W\subseteq V$ whose closure in $X$ is contained in $V$. Then $\overline{U}\cap\overline{W}=\emptyset$, which contradicts the superconnectedness of $X$. $\square$
The cardinality $\mathrm{non}(\mathcal M)$ is the above example can be lowered to $\mathfrak q\_0$, where $\mathfrak q\_0$ is the smallest cardinality of a second-countable metrizable space which is not a $Q$-space (= contains a subset which is not of type $G\_\delta$).
A topological space is *submetrizable* it it admits a continuous metric.
Each submetrizable space is *functionally Hausdorff* in the sense that for any distinct elements $x,y\in X$ there exists a continuous function $f:X\to\mathbb R$ such that $f(x)\ne f(y)$.
**Example 2.** *There exists a submetrizable second-countable space $X$ of cardinality $\mathfrak q\_0$, which is not symmetrizable.*
*Proof.* By the definition of the cardinal $\mathfrak q\_0$, there exists a second-countable metrizable space $Y$, which is not a $Q$-space and hence contains a subset $A$ which is not of type $G\_\delta$ in $X$. Let $\tau'$ be the topology on $X$, generated by the subbase $\tau\cup\{X\setminus A\}$ where $\tau$ is the topology of the metrizable space $Y$. It is clear that $X=(Y,\tau')$ is a second-countable space containing $A$ as a closed subset. Since $\tau\subseteq\tau'$, the space $X$ is submetrizable. Assuming that $X$ is symmetrizable and applying Theorem 2, we conclude that $X$ is perfect and hence the closed set $A$ is equal to the intersection $\bigcap\_{n\in\omega}W\_n$ of some open sets $W\_n\in\tau'$. By the choice of the topology $\tau'$, for every $n\in\omega$ there exists open sets $U\_n,V\_n\in \tau$ such that $W\_n=U\_n\cup(V\_n\setminus A)$. It follows from $A\subseteq W\_n=U\_n\cup(V\_n\setminus A)$ that $A=A\cap W\_n=A\cap U\_n\subseteq U\_n$.
$$A=\bigcap\_{n\in\omega}W\_n=A\cap\bigcap\_{n\in\omega}W\_n=\bigcap\_{n\in\omega}(A\cap W\_n)=\bigcap\_{n\in\omega}(A\cap U\_n)\subseteq \bigcap\_{n\in\omega}U\_n\subseteq \bigcap\_{n\in\omega}W\_n=A$$
and hence $A=\bigcap\_{n\in\omega}U\_n$ is a $G\_\delta$-set in $X$, which contradicts the choice of $A$. This contradiction shows that the submetrizable second-countable space $X$ is not symmetrizable. $\square$
On the other hand we have the following partial affirmative answer to Question 1.
**Theorem 3.** Martin's Axiom implies that every second-countable $T\_1$ space of cardinality $<\mathfrak c$ is perfect and hence symmetrizable.
*Proof.* [It is known](https://www.ams.org/journals/proc/1980-078-02/S0002-9939-1980-0550513-4/S0002-9939-1980-0550513-4.pdf) that Martin's Axiom implies that every second-countable $T\_1$-space $X$ of cardinality $\mathfrak c$ is a $Q$-space, which means that every subset of $X$ is of type $G\_\delta$. In particular, $X$ is perfect and by Theorem 2 is symmetrizable. $\square$
| 13 | https://mathoverflow.net/users/61536 | 423632 | 172,163 |
https://mathoverflow.net/questions/423597 | 4 | **Definition.** A topological space $X$ is a *$Q$-space* if every subset of $X$ is of type $G\_\delta$.
It is clear that every $Q$-space has countable pseudocharacter (= all singletons are $G\_\delta$) and is perfect (= all closed subsets are $G\_\delta$).
It is well-known that Martin's Axiom (MA) implies that every second-countable space of cardinality $<\mathfrak c$ is a $Q$-space.
[A standard proof](https://www.ams.org/journals/proc/1980-078-02/S0002-9939-1980-0550513-4/S0002-9939-1980-0550513-4.pdf) of this fact actually shows more, namely, that every topological space $X$ with countable separating weight and $|X|<\mathfrak c$ is a $Q$-space.
Let us recall that a topological space $X$ has *countable separating weight* if it admits a countable open cover $\mathcal U$ such that for any distinct points $x,y\in X$ there exists a set $U\in\mathcal U$ that contains $x$ but not $y$.
It is easy to see that every topological space with countable separating weight has countable pseudocharacter. The space $\omega\_1$ of ordinals with the order topology is a first-countable but not a $Q$-space.
**Question.** *Let $X$ be a first-countable Lindelof Hausdorff space of cardinality $|X|<\mathfrak c$. Is $X$ a $Q$-space under MA? Is $X$ perfect under MA?*
**Remark.** The "Lindelof" requirement can not be removed from this question as the ordinal $\omega\_1=[0,\omega\_1)$ with the standard order topology is first-countable but not perfect (because the closed set of countable limit ordinals is not of type $G\_\delta$ in $\omega\_1$ by Fodor's Lemma) and hence $\omega\_1$ is not a $Q$-space.
| https://mathoverflow.net/users/61536 | Is every first-countable Lindelof space of cardinality $<\mathfrak c$ a $Q$-space under MA? | I think that the answer to both of your questions is negative.
Let $X$ be a subspace of the real line with its usual topology of size $\omega\_1$. Then, clearly, $X$ is first countable, Lindelöf and Hausdorff. Thus, its Alexandroff duplicate, $A(X)$, is also first countable, Lindelöf and Hausdorff (see here [The Alexandroff Duplicate and its subspaces](https://polipapers.upv.es/index.php/AGT/article/view/1880) ). Now, since every point in $X\times \{1\}$ is isolated, it is clear that $X\times \{1\}$ is a uncountable and discrete subspace of $A(X)$; in particular, $X\times \{1\}$ is not Lindelöf. Since every Lindelöf perfect space is hereditarily Lindelöf, it follows that $A(X)$ is not perfect.
Hence, $A(X)$ is a first countable, Lindelöf and Hausdorff space of cardinality $\omega\_1$ which is not perfect (thus, not a $Q$-space). Finally, since $\mathsf{MA}+\neg \mathsf{CH}$ is consistent, this procedure would in turn give an example of a first countable, Lindelöf and Hausdorff space of cardinality $<\mathfrak{c}$ which is not perfect (thus, not a $Q$-space).
| 4 | https://mathoverflow.net/users/146942 | 423635 | 172,164 |
https://mathoverflow.net/questions/423631 | 1 | Let $X$ be a connected, smooth affine algebraic variety over an algebraically closed field $K$ of characteristic zero. Assume we have a finite group $G$ acting on $X$ by morphisms of $K$-schemes. Choose a closed point $p\in X$. Is there a $G$-invariant affine open $U$, such that $p\in U$ and $\Omega\_{X/K}(U)$ is free over $O\_{X}(U)$? Notice that, in this case, the quotient space $X/G$ is also an affine $K$-variety, and that the quotient map $\pi: X\rightarrow X/G$ is finite. Furthermore, the quotient $\pi$ is étale at a point $p\in X$ if and only if $p$ is not a fixed point of any element $g\in G$. As this locus is open in $X$, my statement holds for every point $p$ in a dense open in $X$. If $p$ is fixed by every element in $G$, then choose an affine open $U$ containing $p$, which trivializes the sheaf of Kahler differentials. Then the open $V=\cap\_{g\in G}g(U)$ contains $p$ and satisfies the property. I have not been able to show that this holds for points that are fixed by a proper, non-trivial subgroup of $G$.
Notice that this statement is equivalent to showing that $\pi\_{\*}\Omega\_{X}$ is a finite locally free sheaf over $\pi\_{\*}O\_{X}$.
| https://mathoverflow.net/users/476832 | On the trivialization of the sheaf of kahler differentials on the G-invariant topology | Just to make the discussion in the comments a little more concrete: given the orbit $A=G\cdot p$, we can choose an affine open containing these points, and find functions $f\_a$ for $a\in A$ on this open such that $f\_a(a')=\delta\_{a,a'}$ for $a,a'\in A$ (by [Sun-tzu's remainder theorem](https://en.wikipedia.org/wiki/Chinese_remainder_theorem)). For any locally free sheaf $\mathcal{F}$, we can choose a set of sections $\sigma\_{1,a},\dots, \sigma\_{n,a}$ on $U$ that give a basis of the residue $\mathcal{F}\_a$. The sums $\sum\_{a\in A}f\_a\sigma\_{k,a}$ are then sections that give a basis of the fiber on a possibly smaller open subset $U'$ containing $A$. If you're careful about how you choose everything with respect to $G$, I believe you can get that $U'$ is manifestly $G$-invariant, but you can also just intersect it with all its images under $G$.
| 3 | https://mathoverflow.net/users/66 | 423637 | 172,166 |
https://mathoverflow.net/questions/423614 | 3 | I am mainly thinking about the group $\Gamma(N)$. A weakly modular form of weight one is a holomorphic function $f: \mathfrak{H} \to \mathbb{C}$ satisfying
$$
f(\gamma \tau) = (c\tau+d)f(\tau), \qquad \gamma \in \Gamma(N),
$$
and which is also "meromorphic at the cusps". It corresponds to an algebraic section of the Hodge line bundle $\underline{\omega}$ over $Y(N) = \mathfrak{H}/\Gamma(N)$.
I believe that, at least for small values of $N$, the Hodge line bundle is actually trivial (in other words, the universal elliptic curve over $Y(N)$ admits a Weierstrass equation). In particular, the trivialization of the $\underline{\omega}$ should correspond to some explicit weakly modular form of weight 1 with no zeros.
Where can I find explicit formulas for such modular forms? For instance, can they be defined by a series on $\mathfrak{H}$ (much like Eisenstein or Poincaré series)?
| https://mathoverflow.net/users/483252 | Explicit expressions for "weakly holomorphic" modular forms of weight 1 | Completing David's answer, it is clear that any eta quotient of level $N$ and weight 1 satisfies the required condition. If you ask in addition that the
form be holomorphic, it is not difficult to write a program giving all holomorphic eta quotients of a given level. For instance there are none in prime level
$N\equiv1\pmod4$ (and more generally 50 levels out of the first 100 have none), but otherwise some levels have hundreds. The smallest level is of course level $4$ with
the eta quotient $\eta(\tau)^{-4}\eta(2\tau)^{10}\eta(4\tau)^{-4}=\theta(\tau)^2$.
This begs for instance the additional question: do there exist weakly modular forms of weight $1$ and level $N\equiv1\pmod4$ prime? I assume this is known.
| 3 | https://mathoverflow.net/users/81776 | 423646 | 172,169 |
https://mathoverflow.net/questions/423642 | 5 | Are there examples of compact complex manifolds $X$ with $K\_X$ nef, but $X$ is not Kähler? Perhaps even non-Moishezon examples?
Here, nef can be defined as follows: For any $\varepsilon>0$ there is a Hermitian metric $h\_{\varepsilon}$ on $K\_X$ with curvature $\Theta\_{h\_{\varepsilon}} \geq - \varepsilon \omega$, where $\omega$ is a positive-definite real $(1,1)$--form on $X$.
| https://mathoverflow.net/users/471309 | Compact complex non-Kähler manifolds with nef canonical bundle | Let $X$ and $Y$ be compact complex manifolds. Note that $K\_{X\times Y} \cong \pi\_1^\*K\_X\otimes \pi\_2^\*K\_Y$. If $Y$ has trivial canonical bundle, then $K\_{X\times Y} \cong \pi\_1^\*K\_X$. Now the pullback of a nef line bundle is again nef, see Proposition 1.8 (i) of [*Compact Complex Manifolds with Numerically Effective Tangent Bundles*](https://www-fourier.ujf-grenoble.fr/%7Edemailly/manuscripts/dps1.pdf) by Demailly, Peternell, and Schneider. So one can construct many examples by choosing $X$ with $K\_X$ nef and $Y$ non-Kähler with $K\_Y$ trivial.
**Example:** Let $X$ be a curve of genus $g > 1$ and $Y$ be a primary Kodaira surface. Then $X\times Y$ is a non-Kähler threefold with $K\_{X\times Y}$ nef. Note that $X\times Y$ is also not Moishezon as it contains $Y$ as a complex submanifold and $Y$ is not Moishezon.
For more examples of non-Kähler manifolds with $K\_Y$ trivial, see the introduction of [*Non-Kähler Calabi-Yau Manifolds*](https://arxiv.org/abs/1401.4797) by Tosatti. As is pointed out in Proposition 1.1, if $K\_Y$ is trivial (or even just torsion), then it admits a metric $h$ with curvature $\Theta\_h = 0$ and hence $K\_Y$ is nef, so we don't even need to take a product with $X$ in the above construction.
| 8 | https://mathoverflow.net/users/21564 | 423647 | 172,170 |
https://mathoverflow.net/questions/423571 | 15 | I would like an example of the following situation, or a proof that no such example exists.
$\textbf{Situation}$: Two connective (EDIT: I'm fine with dropping this condition) spectra $X$ and $Y$ such that the spaces $\Omega^\infty\Sigma^nX$ and $\Omega^\infty\Sigma^nY$ are equivalent (as spaces) for all $n$, but as spectra, $X$ and $Y$ are $\textit{not}$ equivalent. It was pointed out to me by Maxime that another way to phrase things is that $X$ and $Y$ are two (grouplike) $E\_\infty$ spaces that are equivalent as $E\_n$-spaces for all finite $n$ but not as $E\_\infty$-spaces.
I think (as Charles pointed out in the comments, this is wrong) it would be sufficient to find an example of the following: An infinite loop space $A$ (with terms $A\_0=\Omega A\_1,A\_1,...$ and structure maps $a\_i:A\_i\rightarrow \Omega A\_{i+1}$ such that $a\_0$ is the identity) and an equivalence $f:A\_0\rightarrow A\_0$ (of spaces) such that for any (infinite loop) equivalence $\phi:A\rightarrow A$ with 0th component $\phi\_0:A\_0\rightarrow A\_0$, the composite $\phi\_0\circ f$ is not homotopic to the component of any infinite loop map. Because then we can consider the infinite loop space $B$ with the same terms $B\_i=A\_i$ and the same structure maps $b\_i=a\_i$ for $i>0$ but with the $b\_0$ changed to $f$. The condition on $f$ implies that there is no way to fill in the 0th component of any attempted equivalence between $A$ and $B$.
The problem is that due to my insanely low supply of brain cells the only infinite loop spaces I can really come to grips with are Eilenberg-Maclane, and among Eilenberg-Maclane spectra such an example can't exist since those determined by their homotopy groups...
| https://mathoverflow.net/users/163893 | "Phantom" non-equivalences of spectra? | For this we can use a swindle-type technique.
Let $B = \bigoplus\_{n=2}^\infty H\Bbb Z/2$, and $A = \bigoplus\_{n=2}^\infty \Sigma^{-n} H\Bbb Z/2$. We can construct maps $A \to B$ by specifying their effect on each summand of $A$.
* The first map, $f: A \to B$, is the sum of the composites $$\Sigma^{-n} H\Bbb Z/2 \xrightarrow{Sq^n} H\Bbb Z/2 \to \bigoplus\_{n=2}^\infty H\Bbb Z/2,$$ where the second map includes as the $n$'th summand.
* The second map, $g: A \to B$, is the same, except the we map the $n$'th summand of $A$ to the $(n+1)$'st summand of $B$. (In particular, the summand with $n=2$ is completely missed by $g$.)
Define $X = cofib(f)$ and $Y = cofib(g)$. Since these sums start at $n=2$, $B$ is the connective cover of both $X$ and $Y$ by the long exact sequence in homotopy. I claim that $X$ and $Y$ have equivalent connective covers $\tau\_{\geq m} X$ and $\tau\_{\geq m} Y$ for all $m < 0$, and that they are inequivalent spectra. This suffices to give an example for your question, since taking the associated infinite loop spaces factors through some connective cover.
---
First let's show that they have equivalent covers. Note that $\tau\_{\geq m+1} X$ is the cofiber of a map
$$
\bigoplus\_{2 \leq n \leq -m} \Sigma^{-n} H\Bbb Z/2 \xrightarrow{\tau\_{\geq m} f} \bigoplus\_{n=2}^\infty H\Bbb Z/2
$$
and similarly for $\tau\_{\geq m+1} Y$. However, the maps $\tau\_{\geq m} f$ and $\tau\_{\geq m} g$ are equivalent maps in the homotopy category: both are the the same finite list of Steenrod squares summed with countably many extra summands of $H\Bbb Z/2$ on the target. Thus, these connective covers are equivalent.
---
Next, let's show that $X$ and $Y$ are not equivalent. To see this, first note that $Y$ has $H\Bbb Z/2$ as a summand: the map $g: A \to B$ completely misses the summand $H\Bbb Z/2$ and allows us to split it off. However, the cofiber sequence defining $X$ gives an exact sequence
$$
[X,H\Bbb Z/2] \to [B, H\Bbb Z/2] \to [A, H\Bbb Z/2]
$$
which is more explicitly
$$
[X,H\Bbb Z/2] \to \prod\_{n=2}^\infty H^0(H\Bbb Z/2) \xrightarrow{\prod Sq^n} \prod\_{n=2}^\infty H^n(H\Bbb Z/2).
$$
This second map is injective, and so by exactness the restriction
$$
[X,H\Bbb Z/2] \to [B, H\Bbb Z/2] \cong Hom(\pi\_0 X, \Bbb Z/2)
$$
must be trivial. Therefore, $X$ can't have $H\Bbb Z/2$ as a summand and thus can't be equivalent to $Y$.
| 14 | https://mathoverflow.net/users/360 | 423649 | 172,172 |
https://mathoverflow.net/questions/423669 | 3 | A theorem of Edmonds (see Theorem 3.1. of ["Deformation of Maps to Branched Coverings in Dimension Two"](https://www.jstor.org/stable/1971246?seq=1)) says that
**Theorem 1**: *A degree-one map between closed orientable surfaces is homotopic to a pinch map (quotient map obtained from identifying a connected compact bordered sub-surface with one boundary component to a point).*
By Theorem 1, we have Theorem 2 below.
**Theorem 2**: *A degree-one map between closed orientable surfaces, when it induces an injective map between the fundamental groups, is homotopic to a homeomorphism.*
Edmonds' proof of Theorem 1 is based on the induction of the genus. Also, there is a proof of Theorem 2 **without** using induction; for example, see the first proof of Theorem 8.9. of "A Primer on Mapping Class Groups."
>
> **Question:** Is there any proof of Theorem 1 **without** using induction on the genus?
>
>
>
I am looking for a reference that has a proof of Theorem 1 in the flavor of the proof of Theorem 8.9. like as given in "A Primer on Mapping Class Groups. *My Question can be a little vague or weird. Sorry for this.*
| https://mathoverflow.net/users/363264 | Classification of degree one map between two closed orientable surfaces without using induction on the genus | The first proof of 8.9 in the Primer uses the genus (to prove that all pants decompositions of $S$ have the same number of curves). Proving that the genus of a surface is well-defined is, at some point, an induction.
It is possible (in my mind at least) that the second proof of Theorem 8.9, based on harmonic maps, gives a proof which more carefully hides the use of induction. (I mean, at some point there will be a proof that diffeomorphisms preserve dimension, which will rely on the pigeonhole principle, which is more-or-less an induction.)
---
There is an area of foundations called [reverse mathematics](https://en.wikipedia.org/wiki/Reverse_mathematics). I have not heard of an analysis of low-dimensional topology (or more specifically the mapping class group) from this viewpoint. Perhaps a first step would be to formalise some of the most important tools (the Alexander trick, the Dehn-Nielsen-Baer theorem) in [Lean](https://en.wikipedia.org/wiki/Lean_(proof_assistant)) (for example).
I imagine that this would be a huge undertaking. But I am very old, and have none of the courage of youth.
| 3 | https://mathoverflow.net/users/1650 | 423674 | 172,176 |
https://mathoverflow.net/questions/423640 | 3 | **Edited:** Due to work of [Raymond and Scott](https://doi.org/10.1007/BF01220468), there exist diffemorphisms (of certain three-dimensional nil-manifolds) whose $n$th power is diffeotopic to the identity, but which are not themselves homotopic to finite order homeomorphisms.
Do they thus provide counterexamples to the claim that that actions by homeomorphisms (on an aspherical three manifold) are conjugate (by a homeomorphism) to isometric actions?
| https://mathoverflow.net/users/21985 | Conjugacy of topological actions on aspherical three manifolds to isometric actions |
>
> Do they thus provide counterexamples to the claim that that actions by
> homeomorphisms (on an aspherical three manifold) are conjugate (by a
> homeomorphism) to isometric actions?
>
>
>
Here is a counterexample which I find a bit simpler.
Suppose that $M$ is a closed, connected, oriented hyperbolic three-manifold. Thus $M$ is aspherical, as desired. Also, $\mathrm{Isom}(M)$ is finite. Let $f$ be a homeomorphism of $M$ which is the identity outside of a small ball $B$ and is not the identity on $B$. Then (difficult exercise) $f$ has infinite order. Now suppose that $g$ is any homeomorphism. Then (easy exercise) $gfg^{-1}$ again has infinite order. Thus $f$ is not conjugate to an isometry.
| 1 | https://mathoverflow.net/users/1650 | 423677 | 172,177 |
https://mathoverflow.net/questions/423676 | 6 | At the end of Section 1.1 of [3-manifold groups](https://arxiv.org/abs/1205.0202v3) it is written that *"the decomposing spheres are not unique up to isotopy, but two different sets of decomposing spheres are related by ‘slide homeomorphisms’"* and I am trying to understand what the author meant.
The given references do not treat the topic (maybe Theorem 3 of [The homotopy type of homeomorphisms of 3-manifolds](https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society-new-series/volume-1/issue-1/The-homotopy-type-of-homeomorphisms-of-3-manifolds/bams/1183542348.full), but if it does, then it is written in a quite mysterious way for my knowledge).
I guess that *"two different sets of decomposing spheres are related by ‘slide homeomorphisms’"* means that for every two sets of decomposing spheres there exists a slide homemomorphism sending one set of spheres in the other one.
However, it is clear that two sets of decomposing spheres containing a different number of spheres cannot be related by slide homeomorphism.
Maybe the author meant that the two sets of decomposing spheres have to be minimal (but in this case, notice that the decomposing system of spheres treated in the [mysterious article](https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society-new-series/volume-1/issue-1/The-homotopy-type-of-homeomorphisms-of-3-manifolds/bams/1183542348.full) that I cited before are not minimal).
Can someone cite some article stating the exact result?
(In the case the result I am looking for is Theorem 3 in the [mysterious article](https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society-new-series/volume-1/issue-1/The-homotopy-type-of-homeomorphisms-of-3-manifolds/bams/1183542348.full), then can someone explain me what this theorem is saying exactly?)
| https://mathoverflow.net/users/483301 | Uniqueness of the set of decomposing spheres in prime decomposition of a 3-manifold | Here is a simple example.
Suppose that $A$, $B$, and $C$ are closed, connected, oriented, prime three-manifolds, so that no two are homeomorphic and none are the three-sphere. (For example, lens spaces with fundamental groups of distinct sizes.) We define a manifold $M = A \ \# \ B \ \#\ C$. Implicit in this a pair of spheres: say $S$ separating $A$ from $B \ \#\ C$ and $S'$ separating $A \ \# \ B$ from $C$.
"Clearly" $M$ is homeomorphic to $B \ \# \ C \ \#\ A$. However, here we find a different pair of spheres: say $T$ separating $B$ from $C \ \#\ A$ and $T'$ separating $A$ from $B \ \# \ C$. "Clearly" there is no homeomorphism of $M$ sending $S \cup S'$ to $T \cup T'$ because the two systems have different separation properties.
---
So reading "slide homeomorphisms" as "homeomorphism of the ambient space" is wrong. I believe that the offending sentence would be clearer if written as follows.
>
> The decomposing spheres are not unique up to isotopy, but two
> different sets of decomposing spheres are related by ‘sphere
> slides’.
>
>
>
The proof is somewhat complicated, and relies on finding the correct "sphere surgery sequence". My suggested references are Hatcher's *Notes on basic three-manifold topology* and the second would be Casson's *Three-dimensional topology*.
| 7 | https://mathoverflow.net/users/1650 | 423678 | 172,178 |
https://mathoverflow.net/questions/423673 | 10 | The Schwarzian derivative of an entire holomorphic function $f$ is defined as
$$Sf:=\left(\frac{f^{''}}{f'}\right)'-\frac{1}{2}\left(\frac{f^{''}}{f'}\right)^2.$$
In the following, we only consider entire holomorphic functions.
If $Sf=0$, then it is well known that $f$ must be a linear function.
If $Sf$ is a constant, then it is also well known that $f$ takes the form $f(z)=a+be^{cz}$, where $a,b,c$ are complex numbers.
**My question** is the following: Let $f$ be a locally injective entire function (the injectiveness is to guarantee that $Sf$ is also entire), if $Sf$ is a polynomial of degree bigger than or equal to $1$, what does $f$ look like? Can anyone give an example?
| https://mathoverflow.net/users/51546 | On entire functions with polynomial Schwarzian derivative | The answer is this:
$$f(z)=\int\_{z\_0}^z e^{Q(\zeta)}d\zeta,$$
where $Q$ is a polynomial, and this is the general form of
an entire function whose Schwarzian is a polynomial. The crucial fact that $f$ has finite order.
Then $f'$, a function of finite order without zeros must be of the form $e^Q$.
This is a special case of the theorem of R. Nevanlinna which characterizes
solutions of the Schwarz equation
$Sf=P$, where $P$ is a polynomial. The Schwarz differential equation is equivalent to the linear differential equation
$$w''+(P/2)w=0,$$
namely, the general solution $f=w\_1/w\_2$, where $f\_1,f\_2$ are two
linearly independent solutions of the linear differential equation.
And the fact that all solution of the linear differential equation with
polynomial coefficient are entire functions of finite order, is classical and well-known. It follows from asymptotic expansions of these solutions for large values of the independent variable, or in a simpler way, from the Wiman-Valiron theory.
Ref. R. Nevanlinna, Analytic functions, Chap.XI,3.
Remark. For generic $P$, the general solution $f$ of $Sf=P$ is
meromorphic but not entire. The condition that it has an entre solution is somewhat complicated:
$$Q''-(1/2){Q'}^2=P,$$
so this is a condition for a Riccati equation to have a polynomial
solution $Q'$.
| 17 | https://mathoverflow.net/users/25510 | 423680 | 172,179 |
https://mathoverflow.net/questions/423683 | 0 | The starting point of this question is the following true statement for graphs:
A simple, undirected graph $G = (V,E)$ is [bipartite](https://en.wikipedia.org/wiki/Bipartite_graph) if and only if for all $E\_0\subseteq E$ the graph $(V, E\_0)$ is bipartite.
Note that any graph is bipartite if it does not have any odd cycles. Using this, it is not hard to prove the above statement.
A [hypergraph](https://en.wikipedia.org/wiki/Hypergraph) $H=(V,E)$ is *bipartite* if there is $D\subseteq V$ such that whenever $e\in E$ and $|e|> 1$, we have that $D$ intersects $e$, and also $V\setminus D$ intersects $e$.
One might hope that if $H = (V, E)$ is a hypergraph such that for all finite $E\_0\subseteq E$ the hypergraph $(V, E\_0)$ is bipartite, we get that $H$ itself is bipartite. But if $[\omega]^\omega$ is the collection of infinite subsets of $\omega$, then the hypergraph $(\omega, [\omega]^\omega)$ shows that is not true.
**Question.** Let $H = (V, E)$ be a hypergraph such that there is $n\in \omega$ with $|e|\leq n$ for all $e\in E$ and such that for all finite $E\_0\subseteq E$ the hypergraph $(V, E\_0)$ is bipartite. Does it necessarily follow that $H$ itself is bipartite?
| https://mathoverflow.net/users/8628 | Hypergraphs such that all finite subhypergraphs are bipartite | Even if all edges are finite (not necessarily of uniformly bounded size), this is correct. It follows from the compactness argument: the set $2^V$ of all maps $V\to \{0,1\}$ is a Tychonoff compact set, for each $e$ the set $F(e)$ of all functions which are constant on $e$ is open. So, if such sets cover $2^V$, then finitely many of them cover $2^V$. In other words, if the hypergraph is not bipartite, a certain finite subhypergraph is not bipartite.
| 1 | https://mathoverflow.net/users/4312 | 423685 | 172,180 |
https://mathoverflow.net/questions/423616 | 2 | Consider the one-dimensional stochastic differential equation:
$$dX\_t = {\bf 1}\_{\{X\_t>0\}}\big(b(t,X\_t)dt + a(t,X\_t)dW\_t\big),\quad \forall t>0,$$
or equivalently
$$dX\_t = b(t,X\_t)dt + a(t,X\_t)dW\_t,\quad \forall t\le \tau,\quad \mbox{with } \tau:=\inf\{t\ge 0: X\_t\le 0\},$$
where $(W\_t)\_{t\ge 0}$ is a standard Brownian motion and $\mathbb P(X\_0>0)=1$. Under which conditions on the coefficients $b, a$ does pathwise uniqueness or even uniqueness in law hold?
| https://mathoverflow.net/users/nan | Uniqueness of the solution to some degenerate SDE | The solution of the SDE in question is an example of a time-inhomogeneous diffusion process that is stopped/killed/terminated when it first hits the origin. For pathwise uniqueness until time $t \wedge \tau$, Theorem 3.7 in Chapter 5 of the following classic should do the trick.
*Ethier, Stewart N.; Kurtz, Thomas G.*, Markov processes. Characterization and convergence, Wiley Series in Probability and Mathematical Statistics. New York etc.: John Wiley & Sons. X, 534 p. \textsterling 49.10 (1986). [ZBL0592.60049](https://zbmath.org/?q=an:0592.60049).
In this setting the theorem precisely states the following.
>
> **Theorem** (Pathwise Uniqueness). Let $U \subset \mathbb{R}$ be an open set, let $T>0$, and suppose that there exists a constant $L$ such that $$
> |b(t,x)-b(t,y)| \vee |a(t,x)-a(t,y)| \le L |x-y| \qquad 0 \le t \le T \quad x,y \in U \;.$$
> Given two solutions $X\_t, Y\_t$ of the SDE, let $$
> \tau = \inf\{ t \ge 0 \mid X\_t \notin U~~ \text{or}~~ Y\_t \notin U \} \;.
> $$ Then $P[X\_0 = Y\_0] = 1$ implies that $P[X\_{t \wedge \tau} = Y\_{t \wedge \tau}~~\text{for} ~~0 \le t \le T] = 1$.
>
>
>
To be sure, choose the open set to be $U = \{ x > 0 \}$. The proof of this theorem is standard, and involves computing the mean-squared difference between $X\_{t \wedge \tau}$ and $Y\_{t \wedge \tau}$, applying Lipschitz continuity of the coefficients, and then invoking Grönwall's inequality.
| 0 | https://mathoverflow.net/users/64449 | 423689 | 172,182 |
https://mathoverflow.net/questions/423670 | 1 | Given $r$ r.v.s $x\_k$: $x\_k=1$ with probability $p\_k$ and $0$ otherwise. Let $s\_i=\sum\_{j=1}^r c\_{ij}p\_j$ where $c\_{ij}\in[0,1]$, $1\le i\le n$. Let $E\_i$ denote the event $\sum\_{j=1}^r c\_{ij}x\_j>(1+\delta)s\_i$. I want to compute an upper-bound of $\Pr(\bigcup\_{i=1}^n E\_i)$. A loose bound is $\sum\_{i=1}^n E\_i$. Another bound is $1-\Pi\_{i} [1-\Pr(E\_i)]$. However, since $E\_i$ are mutually dependent, how to derive a tighter bound?
| https://mathoverflow.net/users/52871 | Tight upper-bound on dependent events | In general, really nothing can be said here. Very much will depend on how close the rows of the matrix $(c\_{ij})$ are to one another.
Even in the very special case when $p\_k=1/2$ for all $k$, this is a very difficult problem, considered in detail in Chapter 5 "Bernoulli Processes" of Talagrand's book [*Upper and Lower Bounds for Stochastic Processes*](https://link.springer.com/book/10.1007/978-3-642-54075-2).
Arguably the main idea there is that of generic chaining, going back to Kolmogorov. If you have enough information on how close the rows of the matrix $(c\_{ij})$ are to one another, you can try a custom adaptation of the generic chaining method.
| 1 | https://mathoverflow.net/users/36721 | 423690 | 172,183 |
https://mathoverflow.net/questions/423707 | 2 | It seems well known in modal logic society that $\Box(\Box p\to p) \to \Box p$ in Kripke semantics of $GL$ implies well-foundedness of the relation i.e. no infinite ascending chains are allowed.
And validity of $GL$ additionally implies irreflexivity and transitivity of the relation.
But I am wondering what is the initial-like reference for the facts mentioned.
**P.S.** This question may not make clear anything interesting, this is just reference request to experts.
**UPDATE** It seems that Segerberg's thesis is tha earliest reference for the fist fact.
Concerning the second fact I reckon Dick de Jongh has proved this one, that said, unfortunately I could not find any reference.
| https://mathoverflow.net/users/73577 | Initial reference on Gödel-Löb axiom in Kripke semantic of $GL$ | The paper [*Provability: the emergence of a mathematical modality*](https://www.jstor.org/stable/20015550) by Boolos and Sambin says (bottom of page $9$) that the first fact was independently gotten by Kripke and Segerberg, and gives the latter's $1971$ thesis as a reference. However, I can't find a copy of Segerberg's thesis online; the only promising link I found, [http://lpcs.math.msu.su/~zolin/ml/pdf/Segerberg\_Essay\_1971.pdf](http://lpcs.math.msu.su/%7Ezolin/ml/pdf/Segerberg_Essay_1971.pdf), seems broken.
The second fact you mention has trivial first conjunct and false second conjunct (non-transitive frames may still be well-founded), and so I doubt there is a citation for it.
| 4 | https://mathoverflow.net/users/8133 | 423711 | 172,186 |
https://mathoverflow.net/questions/423665 | 3 | Let $M$ be a Riemannian manifold; if $d$ is the distance on $M$, we can consider the distance $D$ between any two continuous curves given by $D(c, \gamma) = \max \_{t \in [0,1]} d(c(t), \gamma(t))$.
>
> Let $c:[0,1] \to M$ a continuous curve. Is it true that for every $\varepsilon > 0$ there exists a differentiable (or even smooth) curve $\gamma$ such that $D(c, \gamma) < \varepsilon$?
>
>
>
I believe that this is true, and I also believe that I have proven it (the case of Peano-like curves make me doubt my reasoning, though). The idea is to consider sufficiently many points on $c$, join each two consecutive of them by a unique minimizing geodesic, join all these geodesic segments in a single zig-zag line and show that this line is at distance at most $\varepsilon$ from $c$. The problem is that its proof is 1.5 pages long (a lot of playing around with the injectivity radius), technical and boring. It would help me if, instead of including it in an article that I am writing, I could replace it with just a reference.
>
> Assuming the result above is true, can you please provide a citable reference?
>
>
>
| https://mathoverflow.net/users/54780 | There exists differentiable curves arbitrarily close to the continuous ones | It turns out that something much more general is true and can be found in the literature.
>
> **Theorem [Thm 3.3, Hirsch, *Differential Topology*]** Let $M$ and $N$ be $C^s$-manifolds (with boundary), $1\le s\le\infty$. Then $C^s(M,N)$ is dense in $C\_S^r(M,N)$ for $0\le r<s$, where $C\_S^r(M,N)$ is equipped with the "strong" topology (when $r=0$, it becomes the compact-open topology).
>
>
>
In particular, when $s=\infty$, $r=0$, $M$ is compact and $N$ is equipped with a (Riemannian) metric, it leads to the result that you want.
| 12 | https://mathoverflow.net/users/176381 | 423713 | 172,187 |
https://mathoverflow.net/questions/423695 | 0 | These days I'm trying to research relations between prime numbers and the notion of chirality in the $xy$-plane. Wikipedia has the article [*Chirality*](https://en.wikipedia.org/wiki/Chirality).
I don't know if this relation or the problem for which I ask here is in the literature; please add a comment in such case.
**Definition.** In this post, I consider a definition ad hoc, defining as **admissible polygons** $P$ the connected regions in the $xy$-lattice that are arrangements of unit squares with $\text{area}(P)=p>3$ a prime number, and also by definition I consider that these polygons must have a chiral shape, that is these pieces are chirals. A tessellation of a connected region $R$ by means of admissible polygons is called an **admissible tessellation of** $R$ (in our problem $R=S$ will be a square $S$ in the $xy$-lattice).
(The rationale for previous definitions is that any polygonal representation of the integers $1,2$ or $3$ aren't chirals, and on the other hand, again by definition one can consider that a given composite integer can be represented as a rectangle or square $A\times B$, that isn't chiral polygon, of sides $1<A\leq B$ in the $xy$-lattice as an arrangement of $n=A\cdot B$ unit squares. I emphasize with these brackets that these pieces will not be used in the tessellations in this post).
**Conjecture.** *There exists a constant* $N\_0$ *such that every square* $S=N\times N$ *of side* $N$ *with* $N>N\_0$ *can be represented by admissible polygons.*
>
> **Question.** I would like to know what work can be done to prove the veracity of the previous conjecture. **Many thanks.**
>
>
>
With admissible polygons it is easy get admissible tessellations for the square $4\times 4$, $5\times 5$, $6\times 6$ and $7\times 7$. We have the following obvious statement.
**Proposition.** *Given an admissible tessellation for the square* $S=N\times N$ *of* $N^2$ *unit squares, it is immediate to get an admissible tessellation for the square* $\hat{S}$ *of side* $2N$, *since this second square* $\hat{S}$ *admits the obvious decomposition into four squares* $S$.
Thus we must deal with squares $S=N\times N$ with $N>7$ an odd integer. Exploiting a specialization of a well-known identity $1+3+5=3^2$ I can tessellate the square $9\times 9$, because this square can be split as two copies of the square $(1+3)\times(1+3)$, plus the tessellated region corresponding to the square $5\times 5$ plus a region that can be tessellated with two admissible polygons with areas $7$ and $17$, being these chirals (I just emphasize it to provide details).
I think that the conjecture isn't obvious.
If you have problems getting it, and it is required, I can add in a figure some of the mentioned tessellations.
| https://mathoverflow.net/users/142929 | Primes and chirality: a definition and question in the context of tessellations for squares | I claim that there is an $N$ so that any rectangle with both sides at least $N$ can be decomposed into squares of sides $4,5,6$ and $7.$ If we show that, then the same applies squares of sides at least $N.$ I will show this for $N=1178.$ With a little more work that number could be decreased to $90.$ Although that is probably not optimal.
Let $a,b>0$ be relatively prime integers.
It is a result of [Frobenious](https://en.wikipedia.org/wiki/Coin_problem) that any integer $m\geq f(a,b)=ab-a-b+1$ can be written in the form $m=as+bt$ with $s,t$ non-negative.
Using just $a\times a$ squares we can make a rectangle $a\times ab$ and using just $b \times b$ squares we can make a rectangle $b \times ab.$ Using those blocks we can make a rectangle $m \times ab$ for any $m\geq f(a,b)=ab-a-b+1$
Hence
* Using $4 \times 4$ and $5 \times 5$ squares we can make a rectangle $20 \times m$ for $m \geq f(4,5)=12.$
-Using $7 \times 7$ and $9 \times 9$ squares we can make a rectangle $63 \times m$ for $m \geq f(7,9)=48$
* Using all four sizes of squares we can make any rectangle $m \times n$ provided that $m \geq 48$ and $n \geq f(20,63)=1178.$
Sticking to squares and rectangles which can be decomposed into vertical rows each of which itself can be decomposed into horizontal columns of widths $4,5,6,7$ or $9$ (with the further restriction that each row uses only two widths) , is probably far from optimal. However, by the method above we can get
* Any row $2x \times 2y$ with $\min(2x,2y) \geq 12$ using $(a,b)=(4,6).$
* Any row $3x \times 3y$ with $\min(3x,3y) \geq 18$ using $(a,b)=(6,9).$
* Any row $ab \times m$ with $m \geq f(a,b)=ab-a-b+1$ for any of the eight relatively prime pairs drawn from $\{4,5,6,7,9\}$
Ignoring the first two types, we can get any rectangle $p \times q$ with with $p \in \{20,28,30,35,36,42,45,63\}$ and $q \geq 48.$
Combining those we can get any rectangle $r \times q$ with
$r \in \tiny{\{ 20, 28, 30, 35, 36, 40, 42, 45, 48, 50, 55, 56, 58, 60, 62, 63, 64, 65, 66, 68, 70, 71, 72, 73, 75, 76, 77, 78, 80, 81, 82, 83, 84, 85, 86, 87, 88\}}$ or $r \geq 90$ and $q \geq 48.$
| 4 | https://mathoverflow.net/users/8008 | 423720 | 172,189 |
https://mathoverflow.net/questions/423443 | 2 | Let $X$ be a toric variety, and let $\pi: \mathbb A^n-V(B) \to X = (\mathbb A^n-V(B))/(\mathbb C^\*)^\rho$ be the quotient map defining $X$ in the Cox construction. A subvariety $Y\subset X$ is called *quasismooth* if $\pi^{-1}(Y)$ is smooth. Clearly, $Y$ smooth implies $Y$ quasismooth.
If we assume now that $X$ is smooth, is $Y$ quasismooth if and only if it is smooth?
| https://mathoverflow.net/users/122729 | Quasismooth vs smooth in a smooth toric variety | Yes. Morphism $\pi$ is smooth if $X$ is smooth. Smoothness is stable under base change, so $\pi$ is smooth implies $\pi^{-1}(Y) \to Y$ is smooth. By [Lemma 35.17.4 in "Stack Project"](https://stacks.math.columbia.edu/tag/034D) "being regular" (which is equivalent to "being smooth" when the base field is $\mathbb C$) is a property local in the smooth topology, hence, $\pi^{-1}(Y)$ smooth, $\pi^{-1}(Y) \to Y$ smooth and surjective implies $Y$ is smooth.
Why $\pi$ is smooth if $X$ is smooth. $X$ is smooth iff defining fan $\Sigma$ is smooth. Consider fan $\tilde{\Sigma}$ in $\mathbb R^{\Sigma(1)}$ corresponding to toric variety $\mathbb C^{\Sigma(1)} - Z(\Sigma)$ which for every cone $\sigma \in \Sigma$ have cone $\tilde{\sigma} = \operatorname{cone}(e\_{\rho} | \rho \in \sigma(1)) \in \tilde{\Sigma}$ and all corresponding faces (see Cox Little Schenck "Toric varieties" §5.1 for more accurate construction of $\tilde{\Sigma}$). We have to show that morphism $\pi : X\_{\tilde{\Sigma}} \to X\_{\Sigma}$ induced by canonical morphism of fans $\tilde{\Sigma} \to \Sigma$ is smooth. Since smoothness is a property local on source, it is enough to check smoothness of restriction $\pi|\_{U\_{\tilde{\sigma}}}$ for every affine open set $U\_{\tilde{\sigma}}, \sigma \in \Sigma$. Computation in coordinates shows that morphism $\pi|\_{U\_{\tilde{\sigma}}} : U\_{\tilde{\sigma}} \to U\_{\sigma}$ is isomorphic to standard projection $\pi : \mathbb C^{k} \times (\mathbb C^\*)^{\dim X-k} \times (\mathbb C^\*)^{r} \to \mathbb C^{k} \times (\mathbb C^\*)^{\dim X-k}$ where $k=\dim \sigma$, which is definitely smooth.
| 2 | https://mathoverflow.net/users/54337 | 423722 | 172,190 |
https://mathoverflow.net/questions/423700 | 5 | The classifying topos of a geometric theory $\mathbb T$ is a topos $\mathcal E\_\mathbb T$ such that for any other Grothendieck topos $\mathcal E$, the category of geometric morphisms from $\mathcal E\_\mathbb T$ to $\mathcal E$ (or the other way round -- I tend to confuse the direction of geometric morphisms) is equivalent to the $\mathbb T$-models in $\mathcal E$.
I noticed that in a [document of Caramello](https://www.oliviacaramello.com/Unification/ToposTheoreticPreliminariesOliviaCaramello.pdf) (which I think is part of her book *Theories, Sites, Toposes*) a different notation is used: the classifying topos of $\mathbb T$ is denoted by $\mathbf{Set}[\mathbb T]$ instead of $\mathcal E\_\mathbb T$.
**Question:** In which sense can the classifying topos of $\mathbb T$ be viewed as *adjoining* $\mathbb T$ to the category $\mathbf{Set}$ of *sets*?
| https://mathoverflow.net/users/483320 | Notation classifying topos | There is indeed a strong analogy between this situation and adjoining a polynomial variable to a ring, except that the direction of the arrows is somewhat messed up:
A ring homomorphism $\mathbb{Z}[X] \to R$ is the same thing as a ring homomorphism $\mathbb{Z} \to R$ (of which there is exactly one) together with one arbitrarily chosen element of $R$. (More precisely, we have a bijection, natural in the ring $R$.)
A geometric morphism $\mathcal{E} \to \mathrm{Set}[\mathbb{T}]$ is the same thing as a geometric morphism $\mathcal{E} \to \mathrm{Set}$ (of which there is exactly one, up to unique isomorphism) together with one arbitrarily chosen $\mathbb{T}$-model in $\mathcal{E}$. (More precisely, we have an equivalence of categories, natural in the topos $\mathcal{E}$.)
In case you wonder how to adjoin a $\mathbb{T}$-model to another topos than $\mathrm{Set}$, you can simply take the product $\mathcal{E}[\mathbb{T}] = \mathcal{E} \times \mathrm{Set}[\mathbb{T}]$, just like we have $R[X] = R \otimes \mathbb{Z}[X]$. (But be aware that the product of toposes is *not* given by the product of the underlying categories.) (And one could generalize to the case where $\mathbb{T}$ is not an ordinary geometric theory but instead a geometric theory internal to $\mathcal{E}$.)
In case you wonder, if you can adjoin a model of an arbitrary geometric theory to a topos, then what else can you adjoin to a ring apart from just a new ring element: you could for example "adjoin" two elements $X$ and $Y$ with the property that $X^2 = 5Y$; this would yield $R[X, Y]/(X^2 - 5Y)$. But the analogy arguably starts breaking down here.
The fact that the geometric morphisms go in the opposite direction from the ring homomorphisms (and that we use the product of toposes where we used the tensor product (coproduct) of rings) can be explained by saying that toposes are "geometric" objects and rings are "algebraic" objects. Or alternatively by the observation that while an element of a ring can be "pushed forward" along a ring homomorphism, a $\mathbb{T}$-model in a topos can be pulled back along a geometric morphism.
| 11 | https://mathoverflow.net/users/166281 | 423723 | 172,191 |
https://mathoverflow.net/questions/322965 | 33 | $\zeta(3)$ has at least two well-known representations of the form $$\zeta(3)=\cfrac{k}{p(1) - \cfrac{1^6}{p(2)- \cfrac{2^6}{ p(3)- \cfrac{3^6}{p(4)-\ddots } }}},$$
where $k\in\mathbb Q$ and $p$ is a cubic polynomial with integer coefficients. Indeed, [we can take $k=1$ and$$ p(n) =n^3+(n-1)^3=(2n-1)(n^2-n+1)=1,9,35,91,\dots \qquad $$](https://tpiezas.wordpress.com/2012/05/04/continued-fractions-for-zeta2-and-zeta3/) (this one generalizes in the obvious way to the odd zeta values $\zeta(5),\zeta(7),...$) or, as shown by Apéry, [$k=6$ and
$$ p(n) = (2n-1)(17n^2-17n+5)= 5,117,535,1463,\dots
. $$](https://mathoverflow.net/questions/265688/about-a-ramanujan-sato-formula-of-level-10-a-recurrence-and-zeta5) **Numerically, I have found that $k=\dfrac87$ and $p(n) = (2n-1)(3n^2-3n+1)$ also works.** (Is that known? Maybe Ramanujan obtained that as some by-product?)
The question:
>
> * Are there other values of $k$ where such a polynomial exists?
> * Must all those polynomials have a zero at $\dfrac12$ for some deeper reason?
>
>
>
| https://mathoverflow.net/users/29783 | Representations of $\zeta(3)$ as continued fractions involving cubic polynomials | See **NOTE** below.
This MO inquiry is over 3 yrs old now.
By the date **the question about the $\zeta(3)$ CF with $k=8/7$** was made (Feb, 2019), it can be answered in the negative nowadays, since it was '*(re)-discovered*' (and tagged as a *new* conjectured CF for Apéry's constant) by a team from Technion - Institute of Technology (Israel) using a highly specialized CAS named **The Ramanujan Machine**. [See here](http://www.ramanujanmachine.com/). These findings for $\zeta(3)$ and several other constants were published in [Arxiv](https://arxiv.org/pdf/1907.00205v4.pdf) (Table 5. pg. 16) in May, 2020 and also in [Science Journal Nature](https://www.nature.com/articles/s41586-021-03229-4) (Feb, 2021). See Ref. below.
So, I think it deserves to be called Wolfgang's $\zeta(3)$ CF. It provides about 1.5 decimal digits per iteration.
Are there **other values of k** where such a polynomial exists?.
Answer is yes.
In addition to known $k=1,\,8/7,\,6$, ($k=6$ CF is equivalent to Apéry's recursion employed to prove $\zeta(3)$'s irrationality), $k=5/2$ is currently also known (June, 2019) and $k=12/7$ is conjectured (2020).
This [youtube video shows at 26:10](https://youtu.be/h0FW7l7z-C4?t=1578) $\zeta(3)$ Continued Fractions for $k=8/7$ (Wolfgang's) and $k=12/7$.
Must all those polynomials have a **zero at 1/2** for some deeper reason?
$k=5/2$ CF does not have a zero at 1/2, but $k=1,6,8/7,12/7$ do (all have companion numerator sequence polynomials $q(n)=-n^6$). There are some conjectures in [this paper](https://arxiv.org/ftp/arxiv/papers/2111/2111.04468.pdf) to look for candidates to prove (or improve) the irrationality (measure) of some constants based on the type of factors and roots that $q(n)$ must have.
---
**NOTE**.
As Wolfgang has pointed out in his answer, there was a "Séminaire de Théorie des Nombres" held in Bordeaux University on March 21th, 1980. In the [Exposé N°23 by Christian Batut and Michael Olivier](https://gdz.sub.uni-goettingen.de/download/pdf/PPN320141322_0009/LOG_0026.pdf) "Sur l'accéléracion de la convergence de certaines fractions continues" (in French), the $\zeta(3)$ CF with $k=8/7$ is found [on page 23-20 3.2.4](https://gdz.sub.uni-goettingen.de/id/PPN320141322_0009?tify=%7B%22pages%22:%5B246%5D,%22panX%22:0.482,%22panY%22:0.73,%22view%22:%22export%22,%22zoom%22:0.349%7D). In this presentation, Apéry's equivalent $\zeta(3)$ CF with $k=6$ is also found (23-19 3.2.2), together with some CFs for Catalan's and other constants. Other $\zeta(3)$ CFs like $k=5/2$ or $k=12/7$ are not shown.
To preserve the spirit of the original response, I will leave this answer at this point.
---
**Ref:** Raayoni, G., Gottlieb, S., Manor, Y. et al. Generating conjectures on fundamental constants with the Ramanujan Machine. Nature 590, 67–73 (2021). <https://doi.org/10.1038/s41586-021-03229-4>
| 24 | https://mathoverflow.net/users/141375 | 423736 | 172,195 |
https://mathoverflow.net/questions/423728 | 2 | Given an ergodic and non-singular dynamic system (definition provided [here](https://web.williams.edu/Mathematics/csilva/NonsingularET_Apr.pdf)) $(X, \mathcal{B}, \mu\_1, T)$ where $(X, \mathcal{B}, \mu\_1)$ is a measure space and $T$ is a fixed transformation, we then will have $\mu\_1$ equivalent to $T\mu$ where $T\mu\_1$ is defined by $T\mu\_1(A)=\mu\_1[T^{-1}(A)]$ for each $A\in \mathcal{B}$. Now suppose $\mu\_2$ is another measure that is equivalent to $\mu\_1$ and makes $(X, \mathcal{B}, \mu\_2, T)$ ergodic and non-singular. Hence, we will have $T\mu\_2\sim\mu\_2\sim\mu\_1\sim T\mu\_1$. Do we have:
$$
\frac{d\,\mu\_1}{d\,\mu\_2}\circ T = \frac{d\,T\mu\_1}{d\,T\mu\_2}
$$
$\mu\_1$-almost everywhere? Here the "$\mu\_1$-almost everywhere" can be replaced by "$\mu\_2$-almost everywhere" or any one of those four measures. By definition of $T\mu\_2$, for each $A\in\mathcal{B}$, we have:
$$
\int\_X\chi\_A(x)d\,T\mu\_2 = \int\_X\chi\_A(Tx)d\,\mu\_2
$$
I tried to show the following equation holds for each $A\in\mathcal{B}$ but cannot:
$$
\int\_A\frac{d\,\mu\_1}{d\,\mu\_2}(x)d\,T\mu\_2 = \int\_A\frac{d\,\mu\_1}{d\,\mu\_2}\circ T(x)d\,\mu\_2 = \int\_A\frac{d\,T\mu\_1}{d\,T\mu\_2}(x)d\,\mu\_2
$$
Any hints or counter-examples will be appreciated
| https://mathoverflow.net/users/151332 | In general is $\frac{d\,\mu_1}{d\,\mu_2}\circ T = \frac{d\,T\mu_1}{d\,T\mu_2}$? | I am fairly sure this is not the case (unless I am missing something in the definition of $T$)
Consider the special case where $T$ is invertible with inverse $T\_{\rm inv}$. Then, it holds for any test function $f$
\begin{aligned}
\int f \,dT\mu\_1 &= \int f \circ T \,d\mu\_1 \\
&= \int f \circ T \,\frac{d\mu\_1}{d\mu\_2} \,d\mu\_2 \\
&= \int f \circ T \, \frac{d\mu\_1}{d\mu\_2} \,d(T\_{\rm inv} (T \mu\_2))\\
&= \int f \,\frac{d\mu\_1}{d\mu\_2} \circ T\_{\rm inv} \,dT\mu\_2
\end{aligned}
and thus it actually holds
$$
\frac{dT\mu\_1}{dT\mu\_2} = \frac{d\mu\_1}{d\mu\_2} \circ T\_{\rm inv}.
$$
If I understand correctly, the following simple example satisfies your assumptions and shows that your claim does not hold: Take $X = \{1, 2, 3\}$, $\mu\_1 = (1/6, 2/6, 3/6)$ (meaning $\mu\_1 = \frac{1}{6}\delta\_1 + \frac{2}{6} \delta\_2 + \frac{3}{6} \delta\_3$), $\mu\_2 = (2/6, 3/6, 1/6)$ and $T(1) = 2, T(2) = 3, T(3) = 1$, and one quickly finds that $T$ is invertible but also that $\frac{dT\mu\_1}{dT\mu\_2} = \frac{d\mu\_1}{d\mu\_2} \circ T\_{\rm inv} \neq \frac{d\mu\_1}{d\mu\_2} \circ T$.
| 2 | https://mathoverflow.net/users/106046 | 423741 | 172,196 |
https://mathoverflow.net/questions/423744 | 7 | I encountered a problem where I need to compute:
$$\mathbb{E}(U) = \mathbb{E}(\min(X\_1, .. , X\_6))$$
The problem is that I have little information on the $X\_i$. Basically I know $\mathbb{E}(X\_i)$ and $\operatorname{var}(X\_i)$ which are identical for all $i$; say $\mu$ and $\nu$ respectively. One may assume for simplicity that the $X\_i$ are independant.
I found [this question](https://math.stackexchange.com/questions/308230/expectation-of-the-min-of-two-independent-random-variables) that answers the problem as long as we know the distributions; the thing is that I don't know these.
Question: *Can we figure out an upper bound on $\mathbb{E}(U)$ which is not $\mathbb{E}(U) \le \mu$ with little information?*
Edit from comments: $X\_i$ are non-negative, finite, integer.
| https://mathoverflow.net/users/342793 | Expected value of min of variables - what informations do I need? | One has an upper bound of $\lfloor \mu \rfloor + (\mu - \lfloor \mu \rfloor)^6 $, and this is best possible - i.e. one can obtain $\mathbb E(U)$ arbitrarily close to this.
To see this, let's first consider the case where we know the $X\_i$ are integer-valued with mean $\mu$ but the variance is unrestricted. Then we can obtain $\mathbb E(U) =\lfloor \mu \rfloor + (\mu - \lfloor \mu \rfloor)^6 $ by a distribution with probability $1- (\mu - \lfloor \mu \rfloor)$ on $\lfloor \mu \rfloor $ and $(\mu - \lfloor \mu \rfloor)$ on $\lfloor \mu \rfloor +1$.
This is optimal since we are trying to maximize $\sum\_{n=1}^\infty \mathbb P(X \geq n)^6$ given $\sum\_{n=1}^\infty \mathbb P(X \geq n)=\mu$, for which increasing the larger values of $\mathbb P(X \geq n)$ and decreasing the smaller values always gives an improvement, so an optimum is obtained when the larger values are all $1$ and can't be increased and the smaller values are all $0$ and can't be decreased, i.e. when the probability distribution is supported on at most two values.
Let's now see that we can achieve $\mathbb E(U)$ arbitrarily close to $\lfloor \mu \rfloor + (\mu - \lfloor \mu \rfloor)^6 $ with a given variance $\nu$. Our previous construction works for $\nu= (\mu - \lfloor \mu \rfloor) (1- (\mu - \lfloor \mu \rfloor))$, and it is not possible to have variance smaller than this, so it suffices to handle the case when $\nu$ is larger.
If we shift $\epsilon$ probability mass from $\lfloor \mu \rfloor$ to $\lfloor \mu \rfloor+1$ and $\epsilon/m$ mass from $\lfloor \mu \rfloor$ to $\lfloor \mu \rfloor+m$, we have not changed the mean but have raised the variance by $\epsilon (m+1)$. Taking $$\epsilon = \frac{\nu- (\mu - \lfloor \mu \rfloor) (1- (\mu - \lfloor \mu \rfloor)) }{ m+1} $$
we see that for $m$ sufficiently large, the distribution is still well-defined after the mass shift, and taking $m$ sufficiently large we may take $\mathbb E(U)$ arbitrarily close to its initial value.
| 7 | https://mathoverflow.net/users/18060 | 423773 | 172,202 |
https://mathoverflow.net/questions/423771 | 6 | Given a $n \times n$ matrix $A = (a\_{ij})$, I was wondering if there was any theory or research interest relevant to the term
$$ \prod\_{i,j} a\_{ij}$$
the product of all the entries of the matrix.
| https://mathoverflow.net/users/71233 | Product of the entries of a matrix | By using the arithmetic-geometric mean inequality, if each entry $a\_{i,j}$ in $A$ is positive, we can bound several quantities related to $A$ below by the product $\prod\_{i,j}a\_{i,j}$. The geometric-arithmetic mean inequality states that $(x\_1\dots x\_n)^{1/n}\leq\frac{1}{n}(x\_1+\dots+x\_n)$ whenever $x\_1,\dots,x\_n$ are non-negative. Therefore, $n(x\_1\dots x\_n)^{1/n}\leq x\_1+\dots+x\_n$ whenever $x\_1,\dots,x\_n$ are non-negative.
Suppose $A$ is a matrix with non-negative entries. Then we obtain the following bound for the permanent of $A$:
$$\text{per}(A)=\sum\_{\sigma\in S\_{n}}\prod\_{k=1}^{n}a\_{k,\sigma(k)}\geq n!\cdot(\prod\_{\sigma\in S\_{n}}\prod\_{k=1}^{n}a\_{k,\sigma(k)})^{1/n!}=n!\cdot\big((\prod\_{i,j}a\_{i,j})^{(n-1)!}\big)^{1/n!}$$
$$=n!\cdot(\prod\_{i,j}a\_{i,j})^{1/n}.$$
Here, equality is reached if and only if every entry in $A$ is the same. While this inequality is easy to prove, the Van der Waerden's conjecture is a result that was proven in 1980 that strengthens this inequality whenever $A$ is doubly stochastic.
If $A$ is doubly stochastic, then by again applying the geometric-arithmetric mean inequality, we obtain
$\prod\_{i,j}a\_{i,j}\leq n^{-n^2}.$
Van der Waerden's conjecture states that
$$n!\cdot(\prod\_{i,j}a\_{i,j})^{1/n}\leq\frac{n!}{n^n}\leq \text{per}(A)$$
whenever $A$ is doubly stochastic.
For stochastic matrices, the product of all entries can be interpreted in terms of Markov chains.
Observation: Suppose that $(X\_r)\_r$ is an irreducible aperiodic Markov chain with underlying set $\{1,\dots,n\}$ and with transition matrix $A$. Furthermore, suppose that every entry in $B$ is $1/n$.
For almost all tuples $(y\_r)\_r\in\{1,\dots,r\}^{\omega}$, we have
$$\lim\_{N\rightarrow\infty}P(X\_0=y\_0,\dots,X\_N=y\_N)^{1/N}=\prod\_{i,j}a\_{i,j}^{n^{-2}}\leq 1/n.$$
If each entry in $A$ is positive, then the spectral radius $\rho(A)$ of $A$ is an eigenvalue of $A$.
The $i,j$-th entry in $A^{N}$ is the sum of all products of the form
$a\_{i,i\_{1}}\dots a\_{i\_{N-1},j}$. However, the geometric mean value of
$a\_{i,i\_{1}},\dots,a\_{i\_{N-1},j}$ is about $(\prod\_{i,j}a\_{i,j})^{1/n^2}$, so the geometric mean value of the product $a\_{i,i\_{1}}\dots a\_{i\_{N-1},j}$ is about $(\prod\_{i,j}a\_{i,j})^{N/n^2}$. And since the $i,j$-th entry is the sum of $n^{N-1}$ many factors, we estimate that the $i,j$-th entry in $A^N$ is about $n^{N-1}(\prod\_{i,j}a\_{i,j})^{N/n^2}$ which is about
$[n\cdot(\prod\_{i,j}a\_{i,j})^{1/n^2}]^{N}$. Therefore, we have
$$n\cdot(\prod\_{i,j}a\_{i,j})^{1/n^2}\leq\rho(A).$$
| 10 | https://mathoverflow.net/users/22277 | 423780 | 172,206 |
https://mathoverflow.net/questions/423789 | 8 | Let $A,B$ be two homeomorphic topological subspaces of $\mathbb{R}^3$ such that their complements $\mathbb{R}^3 - A, \mathbb{R}^3 - B$ are not homeomorphic to each other. Must $A \cong B$ contain a homeomorphic image of the Cantor set?
(It is known that there are homeomorphic images $A,B$ of the Cantor set such that $\mathbb{R}^3 - A, \mathbb{R}^3 - B$ are not homeomorphic, see e.g. <https://www.sciencedirect.com/science/article/pii/016686418690060X>)
---
UPDATE 1:
The answer is no, as Wojowu's answer shows. This leads to
**Question 2:** Must $A \cong B$ contain a homeomorphic image of $\mathbb{Q}$?
---
UPDATE 2:
After Wojowu's answer, the interesting question remaining is
**Question 3:** Let $A,B$ be two closed, countable, topological subspaces of $\mathbb{R}^3$ homeomorphic to each other. Must their complements $\mathbb{R}^3 - A, \mathbb{R}^3 - B$ be homeomorphic to each other?
| https://mathoverflow.net/users/69681 | On homeomorphic subsets of $\mathbb{R}^3$ with non-homeomorphic complements | Let $A=\mathbb Q^3$ and $B=\{0\}\times\mathbb Q^2$. It is a [classical result](https://mathoverflow.net/q/26001/30186) that they are homeomorphic (both homeomorphic to $\mathbb Q$), and their complements are not homeomorphic as $\mathbb R^3-B$ contains a subset homeomorphic to an open ball, while $\mathbb R^3-A$ doesn't (e.g. by the invariance of domain theorem).
However, since $A,B$ are countable, they don't contain a copy of the Cantor set.
---
The answer to the new question is also negative. Indeed let $A=\{(0,0,n)\mid n\in\mathbb N\}$ and $B=\{(0,0,1/n)\mid n\in\mathbb N\}$. Both of these are discrete and countable, so are homeomorphic. On the other hand, the complement of $A$ is a manifold, while $(0,0,0)$ in $\mathbb R^3-B$ has no Euclidean neighbourhood.
| 15 | https://mathoverflow.net/users/30186 | 423790 | 172,209 |
https://mathoverflow.net/questions/423725 | 5 | For metric spaces $(M\_1, d\_1)$ and $(M\_2, d\_2)$, it is [an exercise](https://math.stackexchange.com/questions/781734/proving-a-metric-induces-the-product-topology) that the product topology on $M\_1\times M\_2$ is induced by the metric $d((x\_1, y\_1), (x\_2, y\_2)) =d\_1(x\_1, x\_2) + d\_2(y\_1, y\_2)$.
Do you know if this statement generalises to premetric spaces?
Here, we call $(M,\tilde{d})$ a *premetric space* if $M$ is a set and $\tilde{d}:M\times M\rightarrow[0,\infty)$ is such that $\tilde{d}(x,x)=0$ for all $x\in M$.
| https://mathoverflow.net/users/472548 | Product topology from two premetric spaces induced by sum of premetrics? | The answer to this question is negative.
Consider the subspace $M\_1=\{0\}\cup\{\frac 1n+\tfrac{i}{nm}:n,m\in\mathbb N\}$ of the complex plane and the space $M\_2=M\_1\cup\{\frac1n:n\in\mathbb N\}$ endowed with the symmetric
$$d\_2(x,y)=\begin{cases}|x-y| &\mbox{if $0\notin \{x,y\}$ or $x,y\in\mathbb R$ or $x=y$};\\
1&\mbox{otherwise}
\end{cases}
$$
It can be shown that the product $M\_1\times M\_2$ is not sequential, so its topology cannot be generated by a premetric, in particular, it is not generated by the symmetric $d\_1+d\_2$.
| 5 | https://mathoverflow.net/users/61536 | 423792 | 172,210 |
https://mathoverflow.net/questions/423787 | 4 | The ordinary cohomology ring $H^\*(X)$ of a smooth projective toric variety $X$ has a combinatorial description: the (quotient) Stanley-Reisner ring of its fan. This ring is generated by $T$-invariant divisors of $X$, where $T$ is the torus acting on $X$. (With these assumptions, $H^\*(X)$ is also $A^\*(X)$, the Chow ring.) Can one describe hard Lefschetz in a combinatorial way in this context? Namely, how does $sl\_2$ act on $H^\*(X)$?
| https://mathoverflow.net/users/138150 | Combinatorial description of Hard Lefschetz for toric varieties | Hard Lefschetz depends on the choice of integral ample Cartier class. In combinatorial terms, this is the choice of integral strictly convex support function $\phi : N\_{\mathbb R} \to \mathbb R$. If you have such function then Lefshetz operator is just multiplication by $\sum\_{i=1}^m -\phi(u\_i) \tau\_i$, in terms of your SE question, where $u\_i$ is a ray generators corresponding to toric invariant divisor $D\_i$. In general, no canonical choice is available.
If $X=X\_{\Sigma}$ is a Gorenstein Fano complete toric variety then anticanonical class $-K\_{\Sigma} = \sum\_{\rho \in \Sigma(1)} D\_{\rho}$ is a natural choice. In this case the Lefshetz operator (or the action of $e \in \mathfrak{sl}\_2$, if you prefer) is a multiplication by $\sum\_{i=1}^m \tau\_i$, in terms of your SE question.
If toric variety $X=X\_P$ comes from the full dimensional lattice polytope $P \subset M\_{\mathbb R}$ then $\phi\_P(n) = \min\_{m \in P \cap M} (n,m)$ is a natural choice of integral strictly convex support function. Hence, the Lefshetz operator is a multiplication by $\sum\_{i=1}^m (-\min\_{m \in P \cap M} (m,u\_{i})) \tau\_i$.
| 5 | https://mathoverflow.net/users/54337 | 423798 | 172,214 |
https://mathoverflow.net/questions/423800 | 7 | For $\alpha,\beta\in\mathbb{C}$ and $\gamma\in\mathbb{C}\setminus\{0,-1,-2,\dotsc\}$, Gauss' hypergeometric function ${}\_2F\_1(\alpha,\beta;\gamma;z)$ can be defined by the series
\begin{equation}\label{Gauss-HF-dfn}
{}\_2F\_1(\alpha,\beta;\gamma;z)=\sum\_{n=0}^{\infty}\frac{(\alpha)\_n(\beta)\_n}{(\gamma)\_n}\frac{z^n}{n!},\quad |z|<1.
\end{equation}
The following special cases are well-known:
\begin{align\*}
{}\_2F\_1(a,b;b;z)&=\frac{1}{(1-z)^a},\\
{}\_2F\_1(1,1;2;z)&=-\frac{\ln(1-z)}{z},\\
{}\_2F\_1\biggl(\frac12,1;\frac32;z^2\biggr)&=\frac1{2z}\ln\frac{1+z}{1-z},\\
{}\_2F\_1\biggl(\frac12,1;\frac32;-z^2\biggr)&=\frac{\arctan z}{z},\\
{}\_2F\_1\biggl(\frac12,\frac12;\frac32;z^2\biggr)&=\frac{\arcsin z}{z},\\
{}\_2F\_1\biggl(\frac12,\frac12;\frac32;-z^2\biggr)&=\frac{\ln\bigl(z+\sqrt{1+z^2}\bigr)}{z}.
\end{align\*}
See Chapter 5 and page 109 in the book [1] below.
Lemma 2.6 in the paper [2] below reads that, for $0\ne|t|<1$ and $n=1,2,\dotsc$,
\begin{equation\*}
{}\_2F\_1\biggl(\frac{1-n}{2}, \frac{2-n}{2};1-n;\frac1{t^2}\biggr)
=\frac{t}{2^n\sqrt{t^2-1}\,} \biggl[\biggl(1+\frac{\sqrt{t^2-1}\,}{t}\biggr)^n -\biggl(1-\frac{\sqrt{t^2-1}\,}{t}\biggr)^n\biggr].
\end{equation\*}
Corollary 4.1 in the paper [3] below states that, for $n=0,1,2,\dotsc$,
\begin{multline}\label{Gauss-HF-Spec-Value}
{}\_2F\_1\biggl(n+\frac{1}{2},n+1;n+\frac{3}{2};-1\biggr)
=\frac{(2n+1)!!}{(2n)!!}\frac{\pi}{4}\\
+\frac{2n+1}{2^{2n}}\sum\_{k=1}^{n} (-1)^{k} \binom{2n-k}{n} \frac{2^{k/2}}{k}\sin\frac{3k\pi}{4}.
\end{multline}
**My question is:** can one find an elementary function $f(t)$ such that
\begin{equation\*}
{}\_2F\_1\biggl(\frac{1}{2},\frac{1}{2};2;t\biggr)=f(t), \quad |t|\le1?
\end{equation\*}
In other words, is the Gauss hypergeometric series $\_2F\_1\bigl(\frac{1}{2},\frac{1}{2};2;t\bigr)$ an elementary function?
References
1. N. M. Temme, *Special Functions: An Introduction to Classical Functions of Mathematical Physics*, A Wiley-Interscience Publication, John Wiley & Sons, Inc., New York, 1996; available online at <http://dx.doi.org/10.1002/9781118032572>.
2. Feng Qi, Qing Zou, and Bai-Ni Guo, *The inverse of a triangular matrix and several identities of the Catalan numbers*, Applicable Analysis and Discrete Mathematics **13** (2019), no. 2, 518--541; available online at <https://doi.org/10.2298/AADM190118018Q>.
3. Feng Qi and Mark Daniel Ward, *Closed-form formulas and properties of coefficients in Maclaurin's series expansion of Wilf's function composited by inverse tangent, square root, and exponential functions*, arXiv (2022), available online at <https://arxiv.org/abs/2110.08576v2>.
| https://mathoverflow.net/users/147732 | Is the Gauss hypergeometric series ${}_2F_1\bigl(\frac{1}{2},\frac{1}{2};2;t\bigr)$ an elementary function? | Maple does it in terms of complete elliptic integrals $\rm{K}$ and $\rm{E}$ ...
$$
{\mbox{$\_2$F$\_1$}\left(\frac12,\frac12;\,2;\,t\right)}={\frac {4\left( t-1 \right){\rm K} \left(
\sqrt {t} \right) +4{\rm E} \left(
\sqrt {t} \right) }{t\pi}}
\tag1$$
But that does not show it is **elementary**. In fact, I suspect it is not elementary.
---
Recall the known formulas:
$$
\rm{K}\big(\sqrt{t}\big) = \frac{\pi}{2}\;
{}\_2F\_1\left(\frac12 , \frac12 ; 1 ; t\right) ,
\\
\rm{E}\big(\sqrt{t}\big) = \frac{\pi}{2}\;
{}\_2F\_1\left(-\frac12 , \frac12 ; 1 ; t\right) .
$$
By themselves, they are not elementary. $(1)$ should follow from these two and a
contiguous formula for the hypergeometrics.
$$
{c\;\mbox{$\_2$F$\_1$}(a-1,b;\,c;\,x)}
+ \left( x-1 \right) c\;{\mbox{$\_2$F$\_1$}(a,b;\,c;\,x)}
+ \left( b-c \right) x\;{\mbox{$\_2$F$\_1$}(a,b;\,c+1;\,x)}
=0
$$
| 12 | https://mathoverflow.net/users/454 | 423802 | 172,216 |
https://mathoverflow.net/questions/421690 | 2 | Let $V$ be a finite dimensional complex inner product space. 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 letting $\Phi(A\_1,\dots,A\_r)(X)=A\_1XA\_1^\*+\dots+A\_rXA\_r^\*$ for all operators $X\in L(V)$.
We have $$\rho(A\_1\otimes B\_1+\dots+A\_r\otimes B\_r)\leq\rho(\Phi(A\_1,\dots,A\_r))^{1/2}\rho(\Phi(B\_1,\dots,B\_r))^{1/2}$$
whenever $A\_1,\dots,A\_r,B\_1,\dots,B\_r$ are matrices.
In particular,
$$\frac{\rho(A\_1\otimes X\_1+\dots+A\_r\otimes X\_r)}{\rho(\Phi(X\_1,\dots,X\_r))^{1/2}}\leq \rho(\Phi(A\_1,\dots,A\_r))^{1/2}$$
whenever $\rho(\Phi(X\_1,\dots,X\_r))\neq 0.$
Define a quantity $\rho\_{2,d}$ by letting
$$\rho\_{2,d}(A\_1,\dots,A\_r)$$
$$=\sup\{\frac{\rho(A\_1\otimes X\_1+\dots+A\_r\otimes X\_r)}{\rho(\Phi(X\_1,\dots,X\_r))^{1/2}}\mid \rho(\Phi(X\_1,\dots,X\_r))\neq 0,X\_1,\dots,X\_r\in M\_{d}(\mathbb{C})\}.$$
Given matrices $A\_1,\dots,A\_r$, do there necessarily exist $d\times d$-complex matrices $X\_1,\dots,X\_r$ where
$$\rho\_{2,d}(A\_1,\dots,A\_r)
=\frac{\rho(A\_1\otimes X\_1+\dots+A\_r\otimes X\_r)}{\rho(\Phi(X\_1,\dots,X\_r))^{1/2}}?$$
If $d=1$, then the supremum can be reached since
$$\rho\_{2,1}(A\_1,\dots,A\_r)=\sup\{\frac{\rho(\alpha\_1X\_1+\dots+\alpha\_rX\_r)}{(|\alpha|\_1^2+\dots+|\alpha\_r|\_1^2)^{1/2}}\mid(\alpha\_1,\dots,\alpha\_r)\neq\mathbf{0}\}$$
$$=\sup\{\rho(\alpha\_1X\_1+\dots+\alpha\_rX\_r):|\alpha\_1|^2+\dots+|\alpha\_r|^2=1\}.$$
The sumpremum may also be reached when $d\geq\text{Dim}(V)$.
The value $$\frac{\rho\_{2,d}(A\_{1},\dots,A\_r)}{\rho(\Phi(A\_1,\dots,A\_r))^{1/2}}$$ seems to be a number in the interval $[0,1]$ that is a sort of maximum value of a kind of correlation coefficient between $(A\_1,\dots,A\_n)$ and a collection of $d\times d$-matrices $(X\_1,\dots,X\_n)$. Said differently, for $d<\text{Dim}(V)$ this value seems to be a coefficient that tells one how random $(A\_1,\dots,A\_n)$ is.
| https://mathoverflow.net/users/22277 | Can the supremum of this quotient of spectral radii be reached? | **Yes.** The equality can in fact be reached. Our strategy will be to produce a compact set $K\_{d,r}\subseteq M\_{d}(\mathbb{C})^{r}$ such that if $(X\_1,\dots,X\_r)\in K\_{d,r}$, then $\rho(\Phi(X\_1,\dots,X\_r))=1$, and where
$$\rho\_{2,d}(A\_1,\dots,A\_r)=\max\{\rho(A\_1\otimes X\_1+\dots+A\_r\otimes X\_r)\mid(X\_1,\dots,X\_r)\in K\_{d,r}\}.$$
Let us go over a few definitions and facts to give some context to our construction. These facts can easily be found in John Watrous's 2018 book called [The Theory of Quantum Information](https://cs.uwaterloo.ca/%7Ewatrous/TQI/TQI.pdf).
A mapping $\mathcal{E}:L(V)\rightarrow L(W)$ is said to be positive if whenever $X$ is positive semidefinite, $\mathcal{E}(X)$ is also positive semidefinite. A mapping $\mathcal{E}:L(V)\rightarrow L(W)$ is said to be completely positive if the mapping $\mathcal{E}\otimes 1\_{U}:L(V\otimes U)\rightarrow L(W\otimes U)$ is positive for each finite dimensional complex Hilbert space $U$.
A mapping $\mathcal{E}:L(V)\rightarrow L(W)$ is said to be trace preserving if $\text{Tr}(\mathcal{E}(X))=\text{Tr}(X)$ for each $X\in L(V)$.
Proposition: Let $\mathcal{E}:L(V)\rightarrow L(W)$ be a linear operator.
1. $\mathcal{E}$ is completely positive if and only if there are $A\_1,\dots,A\_r$ where $\mathcal{E}=\Phi(X\_1,\dots,X\_r)$.
2. If $\mathcal{E}$ is defined by letting $\mathcal{E}(X)=A\_1XB\_1^\*+\dots+A\_rXB\_r^\*$, then $\mathcal{E}$ is trace preserving if and only if $A\_1^\*B\_1+\dots+A\_r^\*B\_r=1\_V$.
A quantum channel is a completely positive trace preserving operator $\mathcal{E}:L(V)\rightarrow L(W)$. In quantum information theory, the quantum channels are the main morphisms between quantum states.
Proposition: If $\mathcal{E}$ is a quantum channel, then $\rho(\mathcal{E})=1$.
Let $K\_{d,r}$ be the collection of all tuples $(X\_1,\dots,X\_r)\in M\_{d}(\mathbb{C})^r$ such that $X\_1^\*X\_1+\dots+X\_r^\*X\_r=1\_d$. Said differently, $K\_{d,r}$ is the collection of all tuples $(X\_1,\dots,X\_r)\in M\_{d}(\mathbb{C})^r$ such that
$\Phi(X\_1,\dots,X\_r)$ is a quantum channel. The set $K\_{d,r}$ is clearly a closed set, and $K\_{d,r}$ is bounded since $d=\text{Tr}(X\_1^\*X\_1+\dots+X\_r^\*X\_r)=\|X\_1\|\_2^2+\dots+\|X\_r\|\_2^2$ where $\|\cdot\|\_2$ denotes the Frobenius norm, so $K\_{d,r}$ is compact.
Lemma: Let $A\_1,\dots,A\_r\in L(V)$. Suppose that there is no $x\in V\setminus\{0\}$ with $A\_1x=\dots=A\_rx=0$. Furthermore, suppose that there is no subspace $W\subseteq V$ with $W\neq\{0\},W\neq V$, and $W=A\_1[W]+\dots+A\_r[W]$. Then there is a $\lambda>0$ along with a positive definite $P$ with $\Phi(A\_1,\dots,A\_r)(P)=\lambda P$.
Proof: Now, let $\mathcal{Q}$ be the collection of all positive semidefinite matrices in $L(V)$ with trace $1$, and let $F:\mathcal{Q}\rightarrow\mathcal{Q}$ be the mapping defined by letting $$F(P)=\frac{\Phi(A\_1,\dots,A\_r)(P)}{\text{Tr}\big(\Phi(A\_1,\dots,A\_r)(P)\big)}.$$ Then $\mathcal{Q}$ is convex, and $F$ is a continuous bijection, so by the Brouwer fixed point theorem, there is some $P\in\mathcal{Q}$ with
$F(P)=P$. Therefore, we have $\Phi(A\_1,\dots,A\_r)(P)=\lambda P$ for some positive $\lambda$.
Now,
$$\text{Im}(P)=\text{Im}(\lambda P)=\text{Im}(A\_1PA\_1^\*+\dots+A\_rPA\_r^\*)$$
$$=\text{Im}(A\_1PA\_1^\*)+\dots+\text{Im}(A\_rPA\_r^\*)=\text{Im}(A\_1P)+\dots+\text{Im}(A\_rP)$$
$$=A\_1[\text{Im}(P)]+\dots+A\_r[\text{Im}(P)].$$
This is only possible if $\text{Im}(P)=V$. $\square$
Let $O\_{d,r}$ be the collection of all $(X\_1,\dots,X\_r)\in M\_{d}(\mathbb{C})^{r}$ where $\Phi(X\_1,\dots,X\_r)$ is not nilpotent. Let $E\_{d,r}$ be the collection of all $(X\_1,\dots,X\_r)\in M\_d(\mathbb{C})^r$ where there is no subspace $W\subseteq\mathbb{C}^d$ with
$W=X\_1^\*[W]+\dots+X\_r^\*[W]$ and $W\neq\{0\},W\neq\mathbb{C}^d$ and where there is no $x$ with $X\_j^\*x=0$ for $1\leq j\leq r$.
The sets $O\_{d,r},E\_{d,r}$ are dense subsets of $M\_{d}(\mathbb{C})^{r}$ with $E\_{d,r}\subseteq O\_{d,r}$.
Proposition: Suppose that $(X\_1,\dots,X\_r)\in E\_{d,r}$. Then there is an invertible matrix $B$ and some positive number $\lambda$ with $(\lambda BX\_1B^{-1},\dots,\lambda BX\_rB^{-1})\in K\_{d,r}$.
Proof: Suppose that $B$ is invertible and $\lambda$ is positive. Then $(\lambda BX\_1B^{-1},\dots,\lambda BX\_rB^{-1})\in K\_{d,r}$ if and only if
$$\sum\_{k=1}^{r}\lambda^2 (B^{-1})^\*X\_k^\*B^\*BX\_k^\*B^{-1}=I$$ if and only if
$$\sum\_{k=1}^{r}\lambda^2 X\_k^\*B^\*BX\_k=B^\*B$$.
Therefore, there are $\lambda,B$ with $(\lambda BX\_1B^{-1},\dots,\lambda BX\_rB^{-1})\in K\_{d,r}$ if and only if there is a positive $\mu$ and a positive definite $P$ with $$\mu\Phi(X\_1^\*,\dots,X\_k^\*)P=P.$$
On the other hand, the existence of such a positive definite $P$ and positive $\mu$ is guaranteed by the above lemma. $\square$
Now, define a mapping $G\_{A\_1,\dots,A\_r}:O\_{d,r}\rightarrow\mathbb{R}$ by letting
$$G\_{A\_1,\dots,A\_r}(X\_1,\dots,X\_r)=\frac{\rho(A\_1\otimes X\_1+\dots+A\_r\otimes X\_r)}{\rho(\Phi(X\_1,\dots,X\_r))^{1/2}}.$$ Since the mapping $G$ is continuous, we have
$$\rho\_{2,d}(A\_1,\dots,A\_r)=\sup\{G\_{A\_1,\dots,A\_r}(X\_1,\dots,X\_r)\mid (X\_1,\dots,X\_r)\in E\_{d,r}\}.$$
Since $$G\_{A\_1,\dots,A\_r}(X\_1,\dots,X\_r)=G\_{A\_1,\dots,A\_r}(\lambda BX\_1B^{-1},\dots,\lambda BX\_rB^{-1})$$ whenever $B$ is invertible and $\lambda$ is a non-zero complex number, by the above proposition, we know that
$$\{G\_{A\_1,\dots,A\_r}(X\_1,\dots,X\_r)\mid (X\_1,\dots,X\_r)\in E\_{d,r}\}$$
$$=\{G\_{A\_1,\dots,A\_r}(X\_1,\dots,X\_r)\mid (X\_1,\dots,X\_r)\in K\_{d,r}\}.$$
Therefore, since $K\_{d,r}$ is compact, there is some $(Z\_1,\dots,Z\_r)\in K\_{d,r}$ with
$$G\_{A\_1,\dots,A\_r}(Z\_1,\dots,Z\_r)=\max\{G\_{A\_1,\dots,A\_r}(X\_1,\dots,X\_r)\mid (X\_1,\dots,X\_r)\in K\_{d,r}\}$$
$$=\rho\_{2,d}(A\_1,\dots,A\_r).$$
I have ran computations that maximize $G\_{A\_1,\dots,A\_r}$, and in these computations the maximum seems to actually be reached.
| 0 | https://mathoverflow.net/users/22277 | 423803 | 172,217 |
https://mathoverflow.net/questions/328216 | 4 | Any compactly generated presentable stable $\infty$-category $C$ is known to be dualizable (with respect to Lurie's tensor product), so there is a coevaluation map:
$$Sp \to C \otimes C^{dual}.$$
Can one describe this map (or the image of the sphere spectrum) more concretely in terms of the compact generators of $C$?
| https://mathoverflow.net/users/18116 | Counit map for compactly generated categories | Recall that, for $A,B$ stable presentable $\infty$-categories, we can compute the tensor and internal hom using the equivalences $A\otimes B \simeq \operatorname{Fun}^\mathrm{lim}(A^\mathrm{op},B)$ and $\operatorname{hom}(A,B) \simeq \operatorname{Fun}^\mathrm{L}(A,B)$. If $A$ is moreover compactly generated, we also have $\operatorname{Fun}^\mathrm{L}(A,B) \simeq \operatorname{Fun}^\mathrm{ex}(A^\omega, B)$.
Thus, for $C$ a stable compactly-generated $\infty$-category, a colimit-preserving functor $\mathrm{Spt} \to C \otimes C^\vee$ is the same thing as an object of $C \otimes C^\vee \simeq \operatorname{Fun}^\mathrm{lim}(C^\mathrm{op}, \operatorname{Fun}^\mathrm{ex}(C^\omega,\mathrm{Spt}))$. So, what are some good functors $C^\mathrm{op} \times C^\omega \to\mathrm{Spt}$? One reliable choice is the (restricted) mapping spectrum functor $(x,y)\mapsto \operatorname{map}(x,y)$. This is a sensible choice, since the functor $C^\mathrm{op} \to \operatorname{Fun}(C^\omega,\mathrm{Spt})$ given by $x\mapsto \operatorname{map}(x,{-})$ preserves limits and factors through the full subcategory $\operatorname{Fun}^\mathrm{ex}(C^\omega,\mathrm{Spt})$ of the target. And indeed, one can check that this, together with the evaluation functor $C^\vee \otimes C \to \mathrm{Spt}$ defined by $(F, x) \mapsto F(x)$, satisfies the triangle identities for dual objects.
Alternatively, the object of $C\otimes C^\vee$ that we are interested in can be described as the coend $\int^{x\in C^\omega} x\otimes \operatorname{map}(x,{-})$, which is might align better with one's linear-algebraic intuitions.
| 4 | https://mathoverflow.net/users/94624 | 423807 | 172,218 |
https://mathoverflow.net/questions/423810 | 10 | The [Davenport constant](https://en.wikipedia.org/wiki/Davenport_constant) $D(G)$ of a finite abelian group $(G,+)$ is the least positive integer $k$ such that every sequence in $G$ of length $k$ has a zero-sum (nonempty) subsequence.
It seems that the Davenport constant is explicitly known only in a handful of cases (such as when $G$ is cyclic, or is a $p$-group, or has rank $2$). Is anything known about how hard it is to compute the Davenport constant?
| https://mathoverflow.net/users/183188 | How hard is it to compute the Davenport constant? | The Davenport constants for all groups of order less than thirty-two is computed in [arXiv.1702.02997](https://doi.org/10.48550/arXiv.1702.02997), using GAP. The algorithm is presented and the complexity is discussed in section 6. The run-time varies by orders of magnitude from one group to another, dependent on the number of equivalence classes that need to be considered.
| 9 | https://mathoverflow.net/users/11260 | 423812 | 172,219 |
https://mathoverflow.net/questions/423811 | 5 | Let
$$
F=\sum\_{i\ge0}\frac1{(T+2)^i}\left(\frac T{T+1}\right)^{3^i}\in\mathbb F\_3\left(\!\!\left(\frac1T\right)\!\!\right).$$
Does $F$ belong to $\mathbb F\_3(T)$?
Here, truncations of the series do not give a good approximation of $F$.
| https://mathoverflow.net/users/33128 | Is this function rational? | It is not rational.
First of all we denote $1/T=x$, then $$F=\sum\_i \left(\frac{x}{1+2x}\right)^i(1+x)^{-3^i}\in \mathbb{F}\_3((x)).$$
Now denote $x/(1+2x)=y$. Note that rationality in $x$ is equivalent to rationality in $y$ (and also that $\mathbb{F}\_3((x))=\mathbb{F}\_3((y))$). We have $x=y/(1-2y)$; $1+x=(1-y)/(1-2y)$; $$(1+x)^{-1}=\frac{1-2y}{1-y}=\frac{1+y}{1-y}=1-(y+y^2+y^3+\ldots).$$
Thus $$F=\sum\_i y^i(1-y^{3^i}-y^{2\cdot 3^i}-\ldots)=\frac1{1-y}-\sum\_{i\geqslant 0, k\geqslant 1}y^{i+k\cdot 3^i}.$$We should prove that the sequence of coefficients $a\_n$ of the power series $$\sum\_{i\geqslant 0, k\geqslant 1}y^{i+k\cdot 3^i}=\sum a\_ny^n$$
is not eventually periodic modulo 3 (since rationality of a power series with coefficients in a finite field is equivalent to eventual periodicity of its coefficients).
We have $a\_n=|A(n)|$, where $A(n)$ is the set of all non-negative integers $i<n$, for which $3^i$ divides $n-i$. It is straightforward that $A(k)=A(k+3^{k-1})$ for all $k>1$ and $A(k+3^{n-1})=A(k)\sqcup \{k\}$ for all $n>k>1$. This yields $a\_{k+3^{k-1}}=a\_k$ and $a\_{k+3^{n-1}}=a\_k+1$ for all $n>k>1$.
Now assume that, on the contrary, $a\_{n+T}\equiv a\_{n} \pmod 3$ for certain integer $T>0$ and all $n\geqslant n\_0$. Consider the powers of 3 of the form $3^{m-1}$, $m>n\_0$. By pigeonhole principle, there exist two of them which are congruent modulo $T$, say, $T$ divides $3^{n-1}-3^{k-1}$ for certain $n>k>n\_0$. Then $T$ divides $(k+3^{n-1})-(k+3^{k-1})$ but from above we have $a\_{k+3^{n-1}}=a\_k+1=a\_{k+3^{k-1}} +1$. This contradicts to our periodicity assumptions.
| 11 | https://mathoverflow.net/users/4312 | 423855 | 172,233 |
https://mathoverflow.net/questions/423851 | 1 | My questions come from the paper [Logarithmic Sobolev inequalities for some
nonlinear PDE’s](https://www.idpoisson.fr/malrieu/wp-content/uploads/sites/3/2018/01/malrieu-spa2001.pdf) written by F. Malrieu (May 2001) where author omitted a good amount of details to be filled. Suppose that $W$ is convex, even, with polynomial growth and $U$ is uniformly convex. Let $\mu\_N$ to be the probability measure with density $$\mu\_N = \frac{1}{Z\_N}\,\exp\left(-\sum\_{i=1}^N U(x\_i) - \frac{1}{2N}\sum\_{i,j=1}^N W(x\_i-x\_j)\right)$$ with $Z\_N$ being a normalization constant rendering $\mu\_N$ to be a probability density function, also let $\bar{u}$ be the unique minimizer (stationary measure) of the free energy functional defined by $$\eta(f) = \int f(x)\,\log f(x)\,\mathrm{d} x + \int U(x)\,f(x)\,\mathrm{d} x + \frac 12 \iint W(x-y)\,f(x)\,f(y)\,\mathrm{d}x\,\mathrm{d}y,$$ i.e., $\bar{u}$ is the unique solution of
$$ \bar{u}(x) = \frac{1}{Z}\,\exp\left(U(x) - W\*\bar{u}(x)\right) $$ with $Z = \int \exp\left(U(x) - W\*\bar{u}(x)\right)\,\mathrm{d}x$. It is implicitly used in pp. 15 and pp.16 of the aforementioned paper that we expect to have $$\|\mu\_{1,N} - \bar{u}\|\_1 \leq \frac{K}{\sqrt{N}} \quad \textrm{and} \quad \mathrm{W}\_2(\mu\_{1,N},\bar{u}) \leq \frac{K}{\sqrt{N}}, \tag{a}$$ where $\mu\_{1,N}$ represents the first marginal of $\mu\_N$ and $K > 0$ is some fixed constant. Can anyone help me to figure out the claimed bounds on the $L^1$ distance and the $2$-Wasserstein distance ?
---
Remark: I personally do not think the proof of the first bound in **(a)** automatically yields the second inequality in **(a)**
| https://mathoverflow.net/users/163454 | Bounding $2$-Wasserstein distance and the $L^1$ distance | Referring to the proof of Prop 3.21 in Malrieu 2001, after the triangle inequality is applied twice, two of the terms are bounded via the upper bound $$
W\_2(u\_t,u\_t^{(1,N)}) \vee W\_2(\mu\_{1,N},\bar{u}) \le \sup\_{s \ge 0} \sqrt{E|X\_s^{1,N}-\bar{X}\_s^1|^2} \tag{$\star$}$$ which accounts for the factor $2$. The reason this bound holds is because of the supremum over $s$, which indicates that it holds for all $s\ge0$ including $s=t$ and $s=\infty$ which gives (via the coupling characterization of the 2 Wasserstein distance) the upper bounds in ($\star$) on $W\_2(u\_t,u\_t^{(1,N)})$ and $W\_2(\mu\_{1,N},\bar{u}) $, respectively. To finish, Theorem 3.3 is invoked.
For the related $L^1$ bound between the densities, one uses the analogous upper bound $$
\|u\_t-u\_t^{(1,N)}\|\_1 \vee \|\mu\_{1,N}-\bar{u}\|\_1 \le \sup\_{s \ge 0} \|u\_s-u\_s^{(1,N)}\|\_1 \;.
$$ To finish, Prop. 3.13 is invoked with $k=1$.
| 2 | https://mathoverflow.net/users/64449 | 423856 | 172,234 |
https://mathoverflow.net/questions/423862 | 0 | Let $D$ be an operator defined by $D(c\_n)\_{n\in\mathbb{N}} = (a\_n c\_n)\_{n\in\mathbb{N}}$. It's a well known fact that $D$ is well defined as an operator from $l^2(\mathbb{N})$ to $l^2(\mathbb{N})$ if $(a\_n)\_{n\in\mathbb{N}} \in l^\infty(\mathbb{N})$.
Does the converse hold, i.e. if $D\in L(l^2(\mathbb{N}))$ then $(a\_n)\_{n\in\mathbb{N}} \in l^\infty(\mathbb{N})$?
For me this seems like an easy to answer question, but I did not find any proofs or statements in the literature. I think I found a proof by myself but I want to be sure about that and I have the feeling that sequences like $(\log n)\_{n\in\mathbb{N}}$ make problems.
| https://mathoverflow.net/users/483443 | Diagonaloperators on $l^2(N)$ | If $ (a\_n) \notin l^{\infty} $, we can find subsequence $ (a\_{n\_i}) $ such that $ |a\_{n\_i}| > 2^i $. Then we can take $ (c\_n) \in l^2 $ like this:
$$ c\_k = \frac{1}{i}, \text{ if } k = n\_i \text{ for some } i, $$
$$ c\_k = 0 \text{ otherwise.} $$
Then I hope it's clear that $ (a\_n c\_n) \notin l^2 $.
| 4 | https://mathoverflow.net/users/483248 | 423864 | 172,237 |
https://mathoverflow.net/questions/423838 | 1 | Given a $\sigma$-algebra $\scr F$ on $\Omega$, say that an accuracy scoring rule for $\scr F$ is a function $s$ from the set of all (countably additive) probabilities on $\scr F$ to the $\scr F$-measurable functions on $\Omega$ with values in $[-\infty,M]$ (for some fixed real $M$). A scoring rule $s$ is proper iff $\int\_\Omega s(p) \, dp \ge \int\_\Omega s(q) \, dp$ for all pairs of probabilities $p$ and $q$, and strictly proper if additionally equality only holds when $p=q$.
If $\scr F$ is countably generated, it has a strictly proper scoring rule (see my answer [here](https://mathoverflow.net/questions/374094/example-of-a-strictly-proper-scoring-rule-on-a-general-measurable-space)). The same is true for any measure space that has only atomic probability measures, like the one [here](https://www.ams.org/proc/1983-089-04/S0002-9939-1983-0718982-7/S0002-9939-1983-0718982-7.pdf).
Question:
1. Are there any strictly proper scoring rules for a $\scr F$ that is not countably generated and that has a nonatomic probability measure?
2. If yes, is this true for all $\scr F$?
**Note:** The answer to (2) is negative if there are cases where there are more than ${\mathfrak c}^{|\Omega|}$ probability measures on $\scr F$. By a result of [Paris and Koonen](https://www.sciencedirect.com/science/article/pii/0003484371900015), it is relatively consistent with the existence of a measurable cardinal that there be a measurable cardinal $\kappa$ that has $2^{2^\kappa}$ normal measures, so it is relatively consistent with the existence of a measurable cardinal that the answer to (2) is negative.
| https://mathoverflow.net/users/26809 | Existence of a strictly proper scoring rule on a $\sigma$-algebra that is not countably generated | There is a somewhat boring positive answer to the first question. It is shown in [K. P. S. Bhaskara Rao, and B. V. Rao. [Borel spaces.](https://eudml.org/doc/268562) Warszawa: Instytut Matematyczny Polskiej Akademi Nauk, 1981.] on page 15 that the $\sigma$-algebra on $[0,1]$ generated by analytic sets is not countably generated. However, since analytic sets are universally measurable, every Borel probability measure has a unique extension to this $\sigma$-algebra. It follows that every proper scoring rule for the probability measures on $[0,1]$ with the Borel sets works also for this larger space that is not countably generated.
| 2 | https://mathoverflow.net/users/35357 | 423910 | 172,250 |
https://mathoverflow.net/questions/423911 | 7 | What is the asymptotic growth rate of $$f(x) = \int\_e^\infty e^{ - x t / \log t} dt$$ as $x \to 0$?
As an example of what is meant by "growth rate" consider $$g(x) = \int\_e^\infty e^{-x t} dt = \frac {e^{-ex}} x \approx x^{-1}.$$
By comparing to $g$, it is easy to see that $f(x) \approx x^{-1 - o(1) }$, but we would like to know what the lower order correction is.
We are guessing that $f(x) \approx \log^p(1/x)/x$ for some $p >0$, but unsure how to show this or find $p$.
| https://mathoverflow.net/users/52896 | Asymptotics for $\int\exp( -x t / \log t)dt$ | Denote $t/\log t=y$. Then $y$ increases from $e$ to $\infty$ when $x$ goes from $e$ to $\infty$, and $dt(1/\log t-1/\log^2 t)=dy$, thus $dt\sim \log t\cdot dy\sim \log y\cdot dy$ for large $t$ (since $\log y=\log t-\log\log t\sim \log t$). Over any finite interval, the integral is uniformly bounded, thus we get $$f(x)\sim \int\_e^\infty e^{-xy}\log y dy.$$
Here denote $xy=\tau$ to get
$$
\int\_e^\infty e^{-xy}\log y dy=x^{-1}\int\_{ex}^\infty e^{-\tau}\log (\tau/x) d\tau=x^{-1}\left(\int\_{ex}^\infty e^{-\tau}\log \tau d\tau-\log x \int\_{ex}^\infty e^{-\tau}d\tau\right)\\
=-\frac{\log x+O(1)}x.
$$
Therefore, $f(x)\sim -\frac{\log x}x$.
| 12 | https://mathoverflow.net/users/4312 | 423913 | 172,251 |
https://mathoverflow.net/questions/423909 | 6 | I suppose the following result follows
from Ambrose-Singer theorem, but I cannot
find a reference, and the arguments I found
in the literature are usually weaker. The idea
is that holonomy over a null-homotopic loop is bounded
by the supremum of the curvature times the area of the
2-dimensional surface segment bounded by the loop.
THEOREM-CONJECTURE
Let $D$ be a unit disk, and $(B, \nabla)$ a trivial
vector bundle on $D$ with connection (not necessarily
orthogonal). Assume that the curvature
$R$ of $\nabla$ is uniformly bounded,
that is, $R(x, y)$ belongs to a compact subset $K$
in $End(B\_m)$ for all $x, y \in T\_m D$ of length 1.
Then the holonomy of $\nabla$ around the boundary
of $D$ is bounded by a uniform constant which
depends on $K$ only.
I think I can prove this, but there are
some segments of the proof which are tricky
and take too much effort.
Can someone please point me to a reference,
or to some relevant papers. Many thanks in advance.
| https://mathoverflow.net/users/3377 | Holonomy bounded in terms of area and the curvature | There are in fact more precise versions, expressing the parallel translation around a loop as the identity map plus a curvature integral over a homotopy. References:
Section 3.1 of Werner Ballmann's lecture notes on vector bundles:
<http://people.mpim-bonn.mpg.de/hwbllmnn/archiv/conncurv1999.pdf>
Deane Yang's notes "Holonomy equals curvature" on his web page
[https://cims.nyu.edu/~yangd/papers/holonomy.pdf](https://cims.nyu.edu/%7Eyangd/papers/holonomy.pdf)
Buser-Karcher, Gromov's almost flat manifolds, page 92.
| 7 | https://mathoverflow.net/users/127309 | 423915 | 172,252 |
https://mathoverflow.net/questions/396671 | 1 | Let $X$ be a non-discrete Euclidean building. Let $x \in X$, $\Delta\_x$ be the germ of a Weyl-chamber based at $x$ and $\xi$ be a point at infinity. Choose $y \in \Delta\_x$.
Is there an apartment containing $x$, $y$ and $\xi$?
Addendum: is there an apartment containing $x$, $\Delta\_x$ and $\xi$?
| https://mathoverflow.net/users/127739 | Apartment in non-discrete Euclidean building with prescribed properties | See Proposition 1.8 in Parreau's paper on non-discrete Euclidean buildings, which proves several foundational results of this nature.
[https://www-fourier.ujf-grenoble.fr/~parreau/Recherche/Parreau\_1999\_immeubles.pdf](https://www-fourier.ujf-grenoble.fr/%7Eparreau/Recherche/Parreau_1999_immeubles.pdf)
| 1 | https://mathoverflow.net/users/130882 | 423943 | 172,257 |
https://mathoverflow.net/questions/423944 | 7 | What is the difference (if any) between "fourier transform" and "SO(3) fourier transform"?
I searched on Google but couldn't find a satisfiable answer.
Thanks in advance :)
| https://mathoverflow.net/users/483509 | What is the difference (if any) between "fourier transform" and "SO(3) fourier transform"? | The usual (discrete) Fourier transform expands a function in the basis set $e^{i n\phi}$, $\phi\in(0,2\pi)$, $n\in\mathbb{Z}$. The $\text{SO}(3)$ Fourier transform uses as basis the [Wigner D-functions](https://en.wikipedia.org/wiki/Wigner_D-matrix) $D\_{\ell}^{m,n}$, which are an orthogonal basis for the rotation group,
$$f=\sum\_{\ell=1}^L\sum\_{m,n=-\ell}^\ell f\_{\ell,m,n}D\_\ell^{m,n}.$$
The integer $L$ is the degree of the transform. It is possible to rewrite this as a usual Fourier transform by expanding the Wigner-D function into a Fourier sum and then one can apply the Fast Fourier Transform algorithm, see [A Fast Fourier Algorithm on the Rotation Group](https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.122.6676&rep=rep1&type=pdf). The complexity of this operation scales as $L^3\log L$.
This is an example of [noncommutative harmonic analysis](https://en.wikipedia.org/wiki/Noncommutative_harmonic_analysis).
| 14 | https://mathoverflow.net/users/11260 | 423946 | 172,258 |
https://mathoverflow.net/questions/423938 | 5 | I present you a family of posets here. I don't say the posets themselves have a conventional name. However I'm sure the general construction of this kind has received some terminology, related to Grothendieck construction, comma categories, category of elements, etc. I can't exactly nail it, so it would be very helpful if you could do it for me.
As usual, write $\Delta$ for the simplex category: the category of inhabited finite linearly ordered set and order-preserving sets. Let $\Delta\_-$ and $\Delta\_+$ denote the wide subcategory of degeneracy maps and face maps, respectively, of $\Delta$.
Let $\sigma$ be a simplex of the simplicial nerve $N(\Delta\_-)$:
$$
\sigma\colon [n\_0]\twoheadrightarrow [n\_1]
\twoheadrightarrow \dotsb \twoheadrightarrow [n\_k],
$$
where each $\twoheadrightarrow$ lies in $\Delta\_-$.
I would like to define a poset $P(\sigma)$ as:
$$
P(\sigma) := \left\{ \tau \overset{d}{\rightarrowtail} \tau' \overset{u}{\subset} \sigma \right\}.
$$
Before we define an order on this set, we need to clearify the meaning of the symbols here. Firstly $\tau' \overset{u}{\subset} \sigma$ denotes a face $\tau'$ of $\sigma$ in the nerve $N(\Delta\_-)$. It is determined by a face map $u\colon [l] \rightarrowtail [k]$ in $\Delta\_+$, and we have
$$
\tau' = u^\*(\sigma)\colon [n\_{u(0)}]\twoheadrightarrow [n\_{u(1)}]
\twoheadrightarrow \dotsb \twoheadrightarrow [n\_{u(l)}].
$$
Secondly, $\tau \overset{d}{\rightarrowtail} \tau'$ denotes the vertex-wise family of face maps, i.e.
$$
\tau\colon [m\_0]\twoheadrightarrow [m\_1]
\twoheadrightarrow \dotsb \twoheadrightarrow [m\_l]
$$
is a diagram in $\Delta\_-$ and $d\_i\colon [m\_i] \rightarrowtail [n\_{u(i)}]$, for $i=0,1,\dotsc, l$, are face maps in $\Delta\_+$ which, in $\Delta$, commutes all the squares.
We need to define an order on $P(\sigma)$. Given $\tau \overset{d}{\rightarrowtail} \tau' \overset{u}{\subset} \sigma$ and $\omega \overset{e}{\rightarrowtail} \omega' \overset{v}{\subset} \sigma$,
we say
$$
\left(\omega \overset{e}{\rightarrowtail} \omega' \overset{v}{\subset} \sigma\right) \le \left(\tau \overset{d}{\rightarrowtail} \tau' \overset{u}{\subset} \sigma\right),
$$
iff we have $\omega \overset{\exists f}{\rightarrowtail} \exists\omega'' \overset{\exists w}{\subset} \tau$ in the commutative way, i.e. $v=u\circ w$ in $\Delta\_-$ and
$e\_i = d\_{w(i)}\circ f\_i$.
I feel a strong déjà-vu looking at this, but I can't write it down into a conventional categorical construction. This clearly looks like a slice category, so if we can name the category with its objects simplices of the simplicial nerve $N(\Delta\_-)$ and its morphisms $\bullet \rightarrowtail \bullet \subset \bullet$, we are done. However I can't go beyond that point, so your help would be very helpful.
| https://mathoverflow.net/users/84247 | What is the name for the construction of this poset related to coherence of degeneracies of the simplex category? | Here’s one way to see it, if I’m not misunderstanding your definition.
* For a small category $\newcommand{\C}{\mathbf{C}}\C$, take its *categorical nerve* $\newcommand{\N}{\mathbf{N}}\N\C$ to be the functor $\newcommand{\op}{\mathrm{op}} \Delta^{\op} \to \mathbf{Cat}$ defined by $(\N\C)\_k = \C^{[k]}$; and take its *semi-nerve* $\N\_{+}\C$ to be the restriction of this to to $\Delta\_{+}$.
* The Grothendieck construction $\int\_{\Delta\_{+}} \N\_{+}\C$ is a split fibration over $\Delta\_{+}$. Its objects are strings $\sigma\_0 \to \cdots \to \sigma\_n$ in $\C$; its morphisms are $(\tau \overset{g}{\to} \tau' \overset{u}{\subseteq} \sigma)$, where $u$ is a face map and $\tau'$ is the restriction of $\sigma$ along $u$, and $g$ is a ladder in $\C$ from $\tau$ to $\tau'$.
Taking $\C := \Delta$, this is very nearly the category you describe in your last para (and so its slices are nearly the category you want overall), but it’s a bit more general: its objects are arbitrary strings in $\Delta$ (not just degeneracies) and its vertical maps are arbitrary ladders, not necessarily of face maps.
So we cut down to a subcategory. This can be done already at the level of the functor $\N\_{+}\C : \Delta\_{+}^{\op} \to \mathbf{Cat}$, before taking the Grothendieck construction. Suppose $\C$ has two distinguished wide subcategories of maps; call them $a$, $b$. Then write $\N^{a,b}\C : \Delta^\op \to \mathbf{Cat}$ (and $\N\_{+}^{a,b}$ similarly) for the functor where $(\N^{a,b}\C)\_k$ is the subcategory of $\C^[k]$ whose objects are strings of $a$-maps, and whose arrows are ladders of $b$-maps.
Then $\int\_{\Delta\_{+}} \N\_{+}^{-,+}\Delta$ is the category you describe in the last paragraph; and its slices are the posets you want overall.
In particular, $\Delta$ is playing two different roles here, which can be generalised separately:
* the *base* of the categorical semi-nerve, $\Delta\_{+}$ along with its inclusion into $\mathbf{Cat}$;
* the *target* of the categorical semi-nerve, $\Delta$ with its two distinguished wide subcategories.
(This categorical (semi-)nerve is an instance of a well-established construction, the [generalised nerve/realisation](https://ncatlab.org/nlab/show/nerve+and+realization); and the Grothendieck construction is of course very standard. Cutting down to a subcategory of the nerve in this particular way is not something I’ve seen before.)
| 2 | https://mathoverflow.net/users/2273 | 423963 | 172,263 |
https://mathoverflow.net/questions/423008 | 9 | A [theorem of Alon and Füredi](http://www.cs.tau.ac.il/%7Enogaa/PDFS/Publications/Covering%20the%20cube%20by%20affine%20hyperplanes.pdf) says that if $A$ and $B$ are finite, nonempty subsets of the field $\mathbb F$, and if a polynomial $P(x,y)\in\mathbb F[x,y]$ vanishes on all, but exactly one point of the grid $A\times B$, then
$\deg P\ge |A|+|B|-2$.
Can one improve this bound given that $P$ is homogeneous?
>
> What is the smallest possible degree of a *homogeneous* polynomial $P(x,y)$ vanishing on all, but exactly one point of the grid $A\times B$?
>
>
>
The underlying field $\mathbb F$ can be assumed to have characteristic $2$.
| https://mathoverflow.net/users/9924 | Alon-Füredi for homogeneous polynomials | In general homogeneity does not improve the bound. Take $A=\{1,q,q^2,\ldots,q^{a-1}\}$, $B=\{1,q,q^2,\ldots,q^{b-1}\}$, $f(x,y)=\prod\_{i=-(b-1)}^{a-2}(x-q^iy)$.
| 3 | https://mathoverflow.net/users/4312 | 423970 | 172,267 |
https://mathoverflow.net/questions/423978 | 1 | $A\subset \Bbb{R}$ is meager if $A$ can be expressed as a countable union of nowhere dense sets.
Let $f:[a, b]\to \Bbb{R}$ is absolutely continuous, i.e., for every $\epsilon>0$, there exists $\delta>0 $ such that whenever a finite sequence of pairwise disjoint sub-intervals $(a\_n, b\_n) $ of $[a, b]$ satisfies $\sum\_{n} b\_n-a\_n<\delta$, then $\sum\_{n} |f(b\_n) -f(a\_n)| <\epsilon$.
Given an absolutely continuous $f$, is it always true that $f(M) $ is meager for every meager set $M$?
The Cantor function is an example of a function that maps the Cantor set (meager) to all of $[0, 1]$ (non meager). But I found that the Cantor function is not absolutely continuous.
One more question, if a function maps meager sets to meager sets, does it satisfy the Lusin-N property? And what about the converse?
I asked this question on MSE. <https://math.stackexchange.com/q/4463264/977780>
This is my first question on MO.
| https://mathoverflow.net/users/483536 | Topological analog of the Lusin-N property | Let $K$ be a nowhere dense closed subset of $[0,1]$ of positive Lebesgue measure $\delta>0$. Such a set $K$ can be obtained using the standard technique for constructing a nowhere dense closed set of positive measure.
Define a function $f:\mathbb{R}\rightarrow\mathbb{R}$ by letting
$f(x)=m(K\cap(-\infty,x))$ where $m$ denotes the Lebesgue measure. Then $f(0)=0,f(1)=\delta$. Then the function $f$ satisfies $|f(y)-f(x)|\leq|y-x|$. Therefore, $f$ is Lipschitz continuous and therefore absolutely continuous as well.
I now claim that $f[K]=[0,\delta]$. Suppose therefore that $0\leq y\leq\delta$. Then let $x\_0$ be the least element with $f(x\_0)=\delta$. Therefore, whenever $x<x\_0$, we have $f(x)<\delta$. This implies that $m((x,x\_0)\cap K)>0$ which means that $x\_0$ is a limit point of $K$. Since $K$ is a closed set, we have $x\_0\in K$ as well. Therefore, $y=f(x\_0)\in K$. We conclude that $f[K]=[0,\delta]$ which has positive measure.
Now, every continuous bijection will automatically map meager sets to meager sets. On the other hand, suppose $f:[0,1]\rightarrow[0,1]$ is the Cantor function, and $I:[0,1]\rightarrow[0,1]$ is the identity function, then $f+I:[0,1]\rightarrow[0,2]$ will be a continuous bijection, so $f+I$ maps meager sets to meager sets. The function $f+I$ is clearly not absolutely continuous since $f$ is not absolutely continuous, so does not satisfy the Luzin N property.
| 5 | https://mathoverflow.net/users/22277 | 423980 | 172,272 |
https://mathoverflow.net/questions/423954 | 3 | Let $f\_n: [0, 1] \to \mathbb R$ be a sequence of functions.
Given a measurable subset $E$ of $[0, 1]$, we say that the sequence $f\_n$ is *equicontinuous on $E$* if for every $x \in E$, and $\varepsilon > 0$ there exists a $\delta > 0$ and all $n \in \mathbb N$ we have $|f\_n(x) - f\_n(y)| < \varepsilon$ for all $y \in [0, 1]$ such that $|y - x| < \delta$.
We say a sequence of functions $f\_n: [0, 1] \to \mathbb R$ has uniformly bounded variation if there exists some $M > 0$ such that each $f\_n$ has total variation less than or equal to $M$.
>
> **Question:** Given any sequence of functions $f\_n: [0, 1] \to \mathbb R$ of uniformly bounded variation, is it true that there exists a measurable subset $E$ of $[0, 1]$ with Lebesgue measure $1$ and a subsequence $f\_{n\_k}$ of lower density $1$ such that $f\_{n\_k}$ is equicontinuous on $E$?
>
>
>
*Note:* We define the *lower density* of a subsequence $f\_{n\_k}$ to be the number
$$\liminf\_n\frac {\#\{k \, | \, k \in \mathbb N, \, n\_k < n\}}{n}$$
where $\#$ denotes the cardinality of a finite set.
| https://mathoverflow.net/users/173490 | Is every sequence of functions with uniformly bounded variation almost equicontinuous? | Consider the pairs $(n,k)\in\mathbb{N}^2$ with $0\leq k<2^n$, they form a sequence in lexicographic order. Consider now the functions $f\_{n,k}$ which are defined as $0$ in $[0,\frac{k}{2^n}]$, $1$ in $[\frac{k+1}{2^n},1]$, and $2^nx-k$ in $[\frac{k}{2^n},\frac{k+1}{2^n}]$, and suppose there is a set $E\subseteq [0,1]$ and a subsequence of $f\_{n,k}$ as in the question.
If we define $A\_n=\bigcup\{(\frac{k}{2^n},\frac{k+1}{2^n});(n,k)\text{ is in the subsequence}\}$, then the fact that the subsequence has density $1$ implies that when $n$ tends to $\infty$, the measure of $A\_n$ tends to $1$. This implies that there is a point $x\in E$ which is in infinitely many of the $A\_n$: for example, take a subsequence $A\_{n\_i}$ such that $\sum\_i(1-m(A\_{n\_i}))<1$; then $m(\cap\_k A\_{n\_i})>0$), so $E\cap(\cap\_k A\_{n\_i})\neq\emptyset$.
Suppose then that $x\in E\cap(\cap\_i A\_{n\_i})$ and we are given $\varepsilon<\frac{1}{2}$; there is no $\delta$ that satisfies the definition of equicontinuity at $x$: indeed, take $i$ such that $2^{-n\_i}<\delta$, then there is some interval $(\frac{k}{2^{n\_i}},\frac{k+1}{2^{n\_i}})$ contained in $(x-\delta,x+\delta)$ and such that $(n\_i,k)$ is in the subsequence. So the image of $(x-\delta,x+\delta)$ under the function $f\_{n\_i,k}$ is all $[0,1]$, thus it contains values at distance $>\varepsilon$ of $f\_{n\_i,k}(x)$.
| 2 | https://mathoverflow.net/users/172802 | 423982 | 172,273 |
https://mathoverflow.net/questions/423955 | 5 | Let $f\_n: [0, 1] \to \mathbb R$ be a sequence of continuously differentiable functions. We say that the sequence $f\_n$ is *equidifferentiable* if for every $x \in [0, 1]$ and every $\varepsilon > 0$, there exists a $\delta > 0$ such that for all $n \in \mathbb N$,
$$\frac{|f\_n (x) - f\_n (y) - f\_n’(x)(x-y)|}{|x-y|} < \varepsilon$$
for all $y$ with $|x - y| < \delta$.
**Question:** Given a sequence $f\_n$ of continuously differentiable functions, is it true that $f\_n$ are equidifferentiable if and only if the sequence $f’\_n$ is equicontinuous?
| https://mathoverflow.net/users/173490 | Equidifferentiable functions | "If" part follows from Lagrange theorem: $f\_n(x)−f\_n(y)=f\_n'(\theta)(x−y)$ for certain $θ$ between $x$ and $y$, and $|f\_n'(x)−f\_n'(\theta)|<\varepsilon$ provided that $x$ and $\theta$ are close enough.
"Only if" part does not hold in general. Let $f\_n'$ be supported on $[1/n,1/n+1/n^2]$ and vary on this segment from 0 to 1. Then for all $x\ne 0$ the claim is obvious, since $f\_n$ are locally constant at $x$ for all large enough $n$. For $x=0$ the inequality reads as $|f\_n(y)-f\_n(0)|<\varepsilon y$, this also holds for large enough $n$, since $f\_n(y)-f\_n(0)=0$ for $y\leqslant 1/n$ and $0\leqslant f\_n(y)-f\_n(0)\leqslant 1/n^2\leqslant y/n$ for $y>1/n$.
| 10 | https://mathoverflow.net/users/4312 | 423984 | 172,274 |
https://mathoverflow.net/questions/423967 | 10 | Let $X$ be an integral scheme with function field $K$. If $U\subset X$ is an open subscheme, we may consider the restriction functor
$$\textsf{QCoh}(X) \to \textsf{QCoh}(U).$$
I don't know much about 2-categories (I'm probably thinking about (2,1)-categories, to be more precise), but I wonder if we may consider a 2-colimit of this and obtain an equivalence of categories
$$\operatorname{2-colim}\_U\textsf{QCoh}(U) \xrightarrow{\sim} \textsf{Vect}(K).$$
If this is true, I wonder moreover if the natural functor
$$\textsf{QCoh}(X) \to \operatorname{2-colim}\_U\textsf{QCoh}(U)\cong \textsf{Vect}(K)$$
is the one which sends a quasi-coherent module over $X$ to its stalk on the generic point.
| https://mathoverflow.net/users/131975 | Is it true that $\operatorname{2-colim}_U \textsf{QCoh}(U) = \textsf{Vect}(K_X)$, as $U$ shrinks to the generic point? | Let $x$ be a point in a scheme $X$. There are two posets, namely the poset of affine opens containing $x$, $A(x)$, and the poset of opens containing $x$, $O(x)$.
The inclusion $A(x)^{op} \to O(x)^{op}$ is (homotopy) cofinal - by Quillen's theorem A, it suffices to show that for any open $U$, $A(x)\_{/U}$ is weakly contractible. But it is a co-directed poset : for any $V,W \to U$, $V\cap W$ is an open and therefore contains some affine open $x\in O\subset V\cap W\subset U$; therefore it is weakly contractible.
It follows that, in the $(2,1)$-category of *presentable* $1$-categories, $colim\_{U\in O(x)^{op}}QCoh(U) \simeq colim\_{U\in A(x)^{op}} QCoh(U)$.
Now note that the latter diagram in fact lives in the category of $E\_0$ *presentable* categories, that is, presentable categories with a chosen basepoint (namely $O\_U$) that are equivalent to something of the form $(Mod\_R,R)$ for some (commutative) ring $R$.
By a $(2,1)$-categorical analogue of theorem 4.8.5.11. in Lurie's *Higher algebra* (which says that $R\mapsto (Mod\_R,R)$, as a functor from rings to $E\_0$ *presentable* categories, is fully faithful and colimit preserving) (it's actually not an analogue, it follows strictly from this theorem by specializing to $Set$-modules in presentable $(\infty,1)$-categories), and using the fact that $A(x)$ is weakly contractible, so colimits over it in $E\_0$-objects are computed underlying, it follows that $colim\_{U\in A(x)^{op}} QCoh(U)\simeq colim\_{U\in A(x)^{op}} Mod\_{O\_U(U)}$ is equivalent to modules over $colim\_{U \in A(x)^{op}} O\_U(U)= colim\_{U\in A(x)^{op}} O\_X(U)= O\_{X,x}$.
Note that here the colimit is taken in ordinary associative rings, but it's a filtered colimit, so it can also be taken in commutative rings.
In conclusion, for any point $x$, $colim\_{U\in O(x)^{op}} QCoh(U)$ is equivalent to modules over $O\_{X,x}$.
If $x$ is a generic point, then $O(x)$ is the category of all opens, and so you get the corresponding statement for $colim\_U QCoh(U)$.
Note that here I've used the following version of "$2-\mathrm{colim}$": homotopy colimits in the $(2,1)$-category (in the sense of "$(\infty,1)$-categories with $1$-truncated mapping spaces") of *presentable* $1$-categories. I'm assuming that this shouldn't matter too much
| 15 | https://mathoverflow.net/users/102343 | 423987 | 172,275 |
https://mathoverflow.net/questions/423969 | 4 | Let $\{U\_\lambda\}\_{\lambda\in\Lambda}$ be an open covering of $\mathbb{R}^n$.
Given a family of functions $f\_\lambda:U\_\lambda\rightarrow \mathbb{R}\,(\lambda\in\Lambda)$ such that $f\_\lambda-f\_\mu: U\_\lambda\cap U\_\mu\rightarrow\mathbb{R}$ is constant for any $\lambda,\mu\in\Lambda$, is there a function $f:\mathbb{R}^n\rightarrow \mathbb{R}$ such that for each $\lambda\in\Lambda$, the function $f-f\_\lambda$ is constant on $U\_\lambda$?
It seems like that this assertion is true, but I don't have any proof.
| https://mathoverflow.net/users/483533 | Existence of a function on the Euclidean space which differs by constants from locally defined functions | This doesn't work if the $U\_\lambda$ are not connected, for example we can take $U\_1=(-\infty,0)\cup(1,\infty)$, $U\_2=(-1,1)$ and $U\_3=(0,2)$ and the functions $f\_1$ defined by $f\_1(x)=x$ if $x<0$ and $x+1$ if $x>0$, and $f\_2(x)=f\_3(x)=x$.
But if the $U\_\lambda$ are connected, it is true. The relationship of this with homology can be seen using homology of small simplices (Hatcher proposition 2.21).
As in Hatcher's book let $C\_n(\mathbb{R}^n)$ be the group of $n$-chains in $\mathbb{R}^n$ generated by the singular $n$-simplices. Now we can take the cover $\mathcal{U}=\{U\_\lambda\}\_\lambda$ and let $C\_n^\mathcal{U}(\mathbb{R}^n)$ be the subgroup of $C\_n(\mathbb{R}^n)$ generated by simplices which are contained in some element of $\mathcal{U}$.
We can define a homomorphism $\alpha:C\_1^\mathcal{U}(\mathbb{R}^n)\to\mathbb{R}$ which assigns to each $1$-simplex $\gamma:[0,1]\to\mathbb{R}^n$ the value $f\_\lambda(\gamma(1))-f\_\lambda(\gamma(0))$, for some $\lambda$ such that $\gamma([0,1])\subseteq U\_\lambda$.
This is well defined and it is $0$ in boundaries (as the $2$-simplices are also contained in some $U\_\lambda$), so it gives you a homomorphism $\overline{\alpha}:H\_1^\mathcal{U}(\mathbb{R}^n)\to\mathbb{R}$. But by Hatcher proposition 2.21, $H\_1^\mathcal{U}(\mathbb{R}^n)\cong H\_1(\mathbb{R}^n)=0$, so $\alpha$ is $0$ in cycles.
So you can define your function $f$ by $f(0)=0$, and for any other $x\in\mathbb{R}^n$, $f(x)=\alpha(\gamma)$, where $\gamma$ is a path (sum of small segments) from $0$ to $x$. By the previous paragraph, $f$ is well defined, and $f-f\_\lambda$ is constant because for $x,y\in U\_\lambda$, we can take a sequence of segments $[x\_0,x\_1],[x\_1,x\_2],\dots,[x\_{n-1},x\_n]$ contained in $U\_\lambda$ going from $x=x\_0$ to $y=x\_n$, so that $f\_\lambda(y)-f\_\lambda(x)=\sum\_{i=1}^n(f\_\lambda(x\_i)-f\_\lambda(x\_{i-1}))$, which also coincides with $f(y)-f(x)$.
| 3 | https://mathoverflow.net/users/172802 | 423990 | 172,277 |
https://mathoverflow.net/questions/423905 | 2 | Let $\pi: E \to D$ be an exact Lefschetz fibration with corners (fibers with boundary)over the disk. Fix a point $\theta \in \partial D$ and consider the fiber $F\_\theta = \pi^{-1}(\theta)$ over that point. Then the vanishing cycles in $F\_\theta$ are exact Lagrangian spheres. By vanishing cycle we mean a sphere embedded in the fiber that degenerates to a point along a path that joins $\theta$ with a critical point of $\pi$.
As pointed out by Paul Seidel in his book, the fact that vanishing cycles are exact Lagrangian spheres is trivial except when $\dim E = 4$ (see 16b, page 221 of "Fukaya categories and Picard-Lefschetz theory"). A bit later in that same chapter is pointed out that any exact Lagrangian sphere in an exact Symplectic manifold $F$ appears as the vanishing cycle in some Lefschetz fibration that has $F$ as one of its fibers.
Now my question: assume $\dim E = 4$ (I'm not sure if this is relevant but maybe it simplifies it in terms of the framings of the Lagrangian spheres). Assume also that we fix the Lefschetz fibration $\pi: E \to D$. Suppose that the set of vanishing cycles generates the homology $H\_1(F\_\theta)$. Let $V \subset F\_\theta$ be an exact Lagrangian sphere (in this case, a simple closed curve).
>
> Is $V$ a vanishing cycle? That is, does necessarily exist a path in $D$ joining $\theta$ with some critical point that has $V$ as its vanishing cycle?
>
>
>
| https://mathoverflow.net/users/43097 | Are all exact Lagrangian spheres, vanishing cycles? | No, this is not necessarily true. In fact, this is *never* the case if $E = B^4$, and $F$ has connected boundary and genus $g \ge 2$.
Since $H\_1(E) = 0$, the vanishing cycles span $H\_1(F)$.
An Euler characteristic computation tells you that the Lefschetz fibration has $2g$ critical points.
Choosing a set of paths in $D$ from a basepoint to the critical points, we get a set $C$ of $2g$ vanishing cycles, and Dehn twists along them generate a subgroup $\Gamma$ of the mapping class group of $F$. Every other vanishing cycle is in the orbit $\Gamma\cdot C$, so that in particular the subgroup generated by all vanishing cycles is $\Gamma$ itself. But Humphries proved that the mapping class group of a surface of genus $g \ge 2$ and with one boundary component cannot be generated by fewer than $2g+1$ Dehn twists, so $\Gamma$ is a proper subgroup of the mapping class group of $\Gamma$, so there has to be a simple closed curve which is not a vanishing cycle. (I believe that the reference for Humphries' result is: Humphries, *Generators for the mapping class group*, 1979.)
The torus knots $T(2,2g+1)$ display this phenomenon very concretely: there is a Lefschetz fibration on $B^4$ whose fibres have genus $g$ and whose boundary has an open book with binding $T(2,2g+1)$. (This comes from Morsifying the function $f\colon \mathbb{C}^2 \to \mathbb{C}$ mapping $(x,y)$ to $x^2+y^{2g+1}$.) The surface of genus $g$ is a plumbing of $2g$ bands, and the vanishing cycles can be taken to be the cores of these bands. But the subgroup they generate is a proper subgroup of the mapping class group: the proof by authority is that there is a well-known presentation that includes these generators but also has an extra Dehn twist.
| 2 | https://mathoverflow.net/users/13119 | 423994 | 172,278 |
https://mathoverflow.net/questions/423996 | 2 | Let $\mathcal{B}$ be a bicategory. Section 4.10 of Gordon, Power and Street's paper "Coherence for Tricategories" states that there is a bicategory $\textbf{st}\mathcal{B}$ and a biequivalence $\eta\_{\mathcal{B}}: \mathcal{B} \rightarrow \textbf{st}\mathcal{B}$ such that, for any $2$-category $\mathcal{C}$, precomposition with $\eta\_{\mathcal{B}}$ induces a bijection between $2$-functors from $\textbf{st}\mathcal{B}$ to $\mathcal{C}$ and pseudofunctors from $\mathcal{B}$ to $\mathcal{C}$.
Further, Corollary 3.6 in Alexander Campbell's article "How strict is strictification?" states that this bijection extends to an isomorphism $\textbf{Gray}(\textbf{st}\mathcal{B},\mathcal{C}) \simeq \textbf{Bicat}(\mathcal{B},\mathcal{C})$ of $2$-categories. The left-hand side is the $\operatorname{Hom}$-$2$-category in the $\textbf{Gray}$-category $\textbf{Gray}$, and thus consists of $2$-functors, pseudonatural transformations and modifications; the right-hand side is the $\operatorname{Hom}$-$2$-category in the tricategory of bicategories, pseudofunctors, pseudonatural transformations and modifications.
I am interested in the generalization of the above $2$-isomorphism to the setting of $\mathcal{V}$-bicategorical structures, where $\mathcal{V}$ is a closed, symmetric, (co)complete monoidal category.
More precisely, let $\mathcal{V}$-$\textbf{Cat}$ be the closed monoidal $2$-category of (small) $\mathcal{V}$-categories, $\mathcal{V}$-functors and $\mathcal{V}$-transformations. Following the definitions given by Garner and Shulman in "Enriched categories as a free cocompletion", we obtain the notions of bicategories, pseudofunctors, pseudonatural transformations and modifications enriched in $\mathcal{V}$-$\textbf{Cat}$ (as well as their strict variants), assembling into a tricategory $\mathcal{V}$-$\mathbf{Bicat}$.
We may now ask whether the above coherence results for bicategories also hold for $\mathcal{V}$-bicategories: given a $\mathcal{V}$-bicategory $\mathcal{B}$, is there a $\mathcal{V}$-$2$-category $\textbf{st}\mathcal{B}$ together with a $\mathcal{V}$-biequivalence $\eta\_{\mathcal{B}}: \mathcal{B} \rightarrow \textbf{st}\mathcal{B}$ giving rise to $\mathcal{V}$-$2$-isomorphisms between $\mathcal{V}$-$\textbf{Gray}(\textbf{st}\mathcal{B},\mathcal{C})$ and $\mathcal{V}$-$\textbf{Bicat}(\mathcal{B,C})$, for any $\mathcal{V}$-$2$-category $\mathcal{C}$? If this is the case, is there a reference for the statement (preferably with some sketch of proof)? Or is it perhaps generally accepted as folklore?
I would also be interested in less general statements. For instance, if I understand correctly, in the bicategorical setting this statement is somewhat simpler if we assume $\mathcal{C} = \textbf{Cat}$. Is the above statement true in the case $\mathcal{C} = \mathcal{V}$-$\textbf{Cat}$? (The $\mathcal{V}$-pseudofunctors we consider then become a special case of modules studied in "Enriched categories as a free cocompletion")
Restricting further, is either of the statements (general $\mathcal{C}$ or $\mathcal{C} = \mathcal{V}$-$\textbf{Cat}$) true if $\mathcal{B}$ is a $\mathcal{V}$-monoidal category, perhaps by some general results on pseudomonoids in monoidal $2$-categories? Or even further, is any of the above statements true in the case when $\mathcal{V}$ is locally presentable?
| https://mathoverflow.net/users/146218 | Strictification of $\mathcal{V}$-pseudofunctors | In section 4 of my paper [Not every pseudoalgebra is equivalent to a strict one](https://arxiv.org/abs/1005.1520), I sketched a proof that for any monoidal 2-category $\mathcal{W}$ with small sums preserved on both sides by its tensor product (which includes $\mathcal{V}\text{-Cat}$ for nice enough $\mathcal{V}$), there is a 2-monad on the 2-category of $\mathcal{W}$-graphs whose strict algebras and pseudo algebras are, respectively, strict $\mathcal{W}$-categories and (unbiased) $\mathcal{W}$-bicategories.
Moreover, under appropriate hypotheses on $\mathcal{W}$ (one of which, the existence of an appropriate factorization system, holds for $\mathcal{V}\text{-Cat}$), the general 2-monadic coherence theorem holds for this 2-monad. This means that the inclusion of the 2-category of pseudoalgebras into that of strict algebras has a strict left 2-adjoint, and the components of the adjunction unit are equivalences (thus every $\mathcal{W}$-bicategory is equivalent to its strictification). This universal property lies in between the two mentioned in your first two paragraphs: instead of an isomorphism of 2-categories of strict/pseudo functors, pseudonatural transformations, and modifications, we have an isomorphism of 1-categories of strict/pseudo functors and [icons](https://ncatlab.org/nlab/show/icon), which in particular entails a bijection between sets of strict/pseudo functors.
One could probably start from this and then work to extend it to pseudonatural transformations and modifications. I don't know offhand a reference that does this.
| 3 | https://mathoverflow.net/users/49 | 423997 | 172,279 |
https://mathoverflow.net/questions/424002 | 3 | The question is in the title:
>
> Is it possible to define the category of sets using the 'one hom-class' definition without relying on tuples as arrows or another 'typing trick'?
>
>
>
I used to view the 'triples trick' (where we say that an arrow $f$ in a category is actually a triplet $(f,X,Y)$ consisting of the arrow paired with its intended domain/codomain) as an 'artificial typing', used to present 'naturally typed' categories using the one hom-class definition.
A recent [embarrasing question](https://mathoverflow.net/questions/418804/category-with-domain-codomain-relations) of mine, however, made it clear that we can't rely on entire relations in place of functions to specify domains/codomains in the 'one hom-class' definition of a category. This makes me unsure if it's even possible to give an 'untyped' definition of the category of sets, since saying 'the objects are sets and the arrows are functions' leaves us at a loss for how to select codomains using a function unless we 'artificially type' the arrows by viewing them as ordered pairs with the codomain prespecified.
Needless to say, if it isn't possible to define categories like ${\bf Set}$ without somehow 'typifying' the arrows, constructions like the 'triples trick' immediately seem less artificial and more like essential tools for defining categories in a 'one hom-class' fashion.
| https://mathoverflow.net/users/92164 | Is there a 'one hom-class' definition of the category of sets not relying on tuples for arrows? | It is certainly possible to define the class of functions as something else than ordered pairs: for example, we can define a function to be a set $Z$, whose individual elements must be ordered pairs $(a,b)$, and the first components of these pairs must be disjoint sets $a$. Then the union of all $a$ is the domain of function and the set of all $b$'s is the codomain. If $(a,b)∈Z$, the associated function sends all elements of $a$ to $b$.
| 8 | https://mathoverflow.net/users/402 | 424009 | 172,281 |
https://mathoverflow.net/questions/423953 | 1 | Let $T$ be a set, $R$ be a ring with $1$ and $B, S\_t$ be $R$-modules $\forall t \in T$
My task is to state and prove the dual to the following statement:
Given momomorphisms $j\_t: S\_t \rightarrow B$. Then there are equivalences:
1. There is an isomorphism $B \rightarrow \coprod\_{t\in T} S\_t$ (coproduct, i.e. direct sum)
2. $B=\sum\_{t\in T}S\_t$ and $S\_{t\_0} \cap (\sum\_{t \not= t\_0}S\_t) = 0$ $\forall t\_0 \in T$.
(Here $\sum$ means a sum of modules, not necessarily direct sum).
To state the dual, we first say that we have epimorphisms $\pi\_t :B \rightarrow B/S\_t$.
The dual to 1 is easy:
1') There is an isomorphism $B \rightarrow \prod\_{t\in T} B/S\_t$
To dualize 2, I try to rewrite it in "language of arrows":
2.a) $B=\sum\_{t\in T}S\_t\iff$ the natural map $\coprod\_{t\in T} S\_t \rightarrow B$ is surjective.
2.b) $S\_{t\_0} \cap (\sum\_{t \not= t\_0}S\_t) = 0$ $\forall t\_0 \in T \iff$ the composite $S\_{t\_0} \rightarrow B \rightarrow B/ \sum\_{t\not=t\_0}S\_t$ is injective.
Now it's easy to dualize 2.a:
2'.a) the natural map $B \rightarrow \prod\_{t\in T} B/S\_t$ is surjective (i.e. $\cap\_{t\in T} S\_t = 0$)
I have no idea how to dualize 2.b. What should be the dual object to $B/ \sum\_{t\not=t\_0}S\_t$?
The intuition tells me to simply write $S\_{t\_0} + (\bigcap\_{t\not=t\_0} S\_t) = B, \forall t\_0\in T$ as a dual to 2.b, but then the statement would be true only for finite set T.
| https://mathoverflow.net/users/474573 | Product-coproduct duality | The brief answer is the following:
>
> the dual of a category of modules (i.e., any category equivalent to $(\mathrm{Mod}(R))^{\mathrm{op}}$ for some ring $R$) is not itself a category of modules, so you have to be careful when you try to dualize statements about modules.
>
>
>
To be more precise, it is true that, given a ring $R$, the category $(\mathrm{Mod}(R))^{\mathrm{op}}$ is Abelian, so categorical statements that just involve *finite* limits and colimits (e.g., kernels, cokernels, finite products=coproducts) tend to dualize smoothly. On the other hand, problems may occur when you try to dualize statements that involve infinite co/limits.
The fact is that categories of modules are (Ab.5) but not (Ab.5$^\*$) Abelian categories, that is, in $\mathrm{Mod}(R)$ directed colimits are exact but inverse limits may fail to be exact.
This easy observation has the following consequence on lattices of submodules: given a right $R$-module $M$, consider a directed family of submodules $\{M\_i\}\_I$ of $M$ and a submodule $K\leq M$, then:
$$
K\cap\sum\_IM\_i=\sum\_I(K\cap M\_i).
$$
On the other hand, given an inverse system of submodules $\{M\_j\}\_J$ of $M$ and a submodule $H\leq M$, it is not difficult to find examples where
$$
H+\bigcap\_{J}M\_j\neq \bigcap\_{J}(H+M\_j).
$$
The failure of this dual equality makes it impossible to characterize products in $\mathrm{Mod}(R)$ in the way you want. Let me try to make you understand why: given your family of morphisms $\{j\_t\colon S\_t\to B\}\_{T}$, by the universal property of the coproduct, there is a canonical morphism $j\colon \coprod\_TS\_t\to B$ and this is an isomorphism if, and only if:
1. *$j$ is surjective*, that is, $\mathrm{Im}(j)=\sum\_TS\_t=B$;
2. *$j$ is injective*, that is, $\mathrm{Ker}(j)=0$. Now note that:
$$
\mathrm{Ker}(j)=\mathrm{Ker}(j)\cap \coprod\_TS\_t=\mathrm{Ker}(j)\cap \sum\_{\text{$T'\subseteq T$ finite}}\left(\coprod\_{T'}S\_t\right)=\sum\_{\text{$T'\subseteq T$ finite}}
\left(\mathrm{Ker}(j)\cap \coprod\_{T'}S\_t\right),$$
where, for the last equality, we have used condition (Ab.5), that is, the exactness of directed colimits. Hence, $\mathrm{Ker}(j)=0$ if, and only if, $\mathrm{Ker}(j)\cap \coprod\_{T'}S\_t=0$ for each finite subset $T'\subseteq T$ (and from this stronger characterization it is not difficult to recover the one you propose considering, instead of finite subsets of $T$, the bigger subsets of the form $T\setminus\{t\_0\}$ with $t\_0$ varying in $T$).
Let us now try to do the same (actually, the dual) with products: the family of morphisms $\{\pi\_t\colon B\to B/S\_t\}\_{T}$ gives you, by the universal property of products, a canonical morphism $\pi\colon B\to \prod\_TB/S\_t$. This $\pi$ is an isomorphism if, and only if:
1. *$\pi$ is injective*, that is, $\mathrm{Ker}(\pi)=0$;
2. *$\pi$ is surjective*, that is, $\mathrm{Im}(\pi)=\prod\_{T}B/S\_t$. Now note that we still have:
$$
\mathrm{Im}(\pi)=\mathrm{Im}(\pi)+ 0=\mathrm{Im}(\pi)+\bigcap\_{\text{$T'\subseteq T$ finite}}\left(\prod\_{T\setminus T'}B/S\_t\right)
$$
but, to go further, one would need to use the (Ab.5$^\*$) condition which, in general, does not hold in $\mathrm{Mod}(R)$.
---
Some final observations:
1. The unique hope I see to find some kind of criterion like the one you propose would be to take into account the derived functor of inverse limits of modules.
2. There are very nice Abelian categories that happen to be (Ab.5$^\*$) but not (Ab.5), for example, consider the category of compact Hausdorff topological Abelian groups, categories of linearly compact vector spaces or linearly compact modules over some linearly topologized ring (e.g., a commutative complete local Noetherian ring, with the unique topology for which the powers of the maximal ideal are a base of neighborhoods of $0$).
3. In the categories described in part 2 you have criteria for "internal product decomposition" along the lines you suggest in your question but, on the other hand, in such categories the criterion for "internal coproduct decomposition" fails.
4. Finally, the unique bicomplete Abelian category that is both (Ab.5) and (Ab.5$^\*$) is the trivial Abelian category $0$. In particular, in an Abelian category with products and coproducts you can hope for at most one of the two "internal decomposition criteria" (but never both if your category is not $0$): either it holds for coproducts (like for categories of modules) or it holds for products (like for compact Hausdorff topological groups).
| 1 | https://mathoverflow.net/users/24891 | 424017 | 172,283 |
https://mathoverflow.net/questions/424006 | 3 | Is completion of infinite degree extension of perfectoid fields perfectoid ?
It is known that finite extension of perfectoid fields is also perftoid from tilting correspondence, but what about infinite cases ?
Infinite degree extension of perfectoid is not perfectoid because it is not always complete, but completion of it possibly be perfectoid.
If you could find some counter example(infinite degree extension of perfectoid field $K$ whose completion is not perfectoid),I would be appreciated.
| https://mathoverflow.net/users/144623 | Completion of infinite degree extension of perfectoid fields is perfectoid? | I'm assuming you mean infinite *algebraic* extensions, as otherwise there is no standard way of completing them.
Let $K$ be a perfectoid field, let $L$ be an infinite algebraic extension. Then $L$ admits a unique valuation extending that of $K$, and hence we can take the completion $\widehat L$. It is clearly complete and the valuation is nondiscrete (since the one on $K$ was). It remains to check the Frobenius is surjective on $O\_{\widehat L}/p$. We have $O\_{\widehat L}/p=O\_L/p$. Write $L$ as a direct limit of finite extensions $L\_i/K$. Then $O\_L$ is the direct limit of $O\_{L\_i}$, and $O\_L/p$ is the direct limit of $O\_{L\_i}/p$. Since each $L\_i$ is perfectoid, the Frobenius is surjective on each $O\_{L\_i}/p$, and hence it is also so on the direct limit. We conclude $\widehat L$ is indeed perfectoid.
| 6 | https://mathoverflow.net/users/30186 | 424019 | 172,284 |
https://mathoverflow.net/questions/423999 | 14 | Marcus du Sautoy, in the section Riemann's Final Twist (pp. 278-80) in his book The Music of the Primes, discusses a discovery of Jon Keating of a connection in Riemann's Nachlass between Riemann's simultaneous investigations on the hydrodynamics of a ball/ellipsoid of fluid and the non-trivial zeros of the Riemann zeta function--both involve critical lines on which important imaginary numbers are distributed.
Does anyone know of a reference which discusses more thoroughly the critical line appearing in the hydrodynamics problem? (I suppose it has something to do with Hermite functions.)
Excerpts from The Music of the Primes:
>
> The physicists believe that the reason Riemann's zeros will be in a straight line is that they will turn out to be frequencies of some mathematical drum. A zero off the line would correspond to an imaginary frequency which was prohibited by the theory. It was not the first time that such an argument had been used to answer a problem. Keating, Berry and
> other physicists all learnt as students about a classical problem in hydrodynamics whose solution depends on similar reasoning. The problem concerns a spinning ball of fluid held together by the mutual gravitational interactions of the particles inside it. For example, a star is a ball of
> spinning gas kept together by its own gravity. The question is, what happens to the spinning ball of fluid if you give it a small kick? Will the fluid wobble briefly and remain intact, or will the small kick destroy the ball completely? The answer depends on showing why certain imaginary numbers lie in a straight line. If they do, the spinning ball of fluid will
> remain intact. The reason why these imaginary numbers do indeed line up is related very closely to the quantum physicists' ideas about proving the Riemann Hypothesis. Who discovered this solution? Who used the mathematics of vibrations to force these imaginary numbers onto a straight line? None other than Bernhard Riemann.
>
>
>
>
> Before Keating set off for Gottingen, one of his colleagues in the mathematics department, Philip Drazin, recommended looking at the part of the Nachlass in which Riemann tackles the classical problem of hydrodynamics.
>
>
>
>
> At the library in Gottingen, Keating ordered the two different parts of the Nachlass that he wanted to consult: one on Riemann's ideas about the zeros in his zeta landscape, and the second on his work on hydrodynamics. When only one pile of papers appeared from the vaults, Keating mentioned that he had asked to see two parts. Both 'parts' were on the same sheets of paper, the librarian told him. As Keating explored the pages, he found to
> his amazement that Riemann had been concocting his proof about rotating balls of fluid at the very same time that he'd been thinking about the points at sea level in his zeta landscape. The very method by which modern-day physicists were proposing to force Riemann's zeros to line up had been used by Riemann to answer the hydrodynamics problem.
>
>
>
>
> There, in front of Keating on the same pieces of paper, were Riemann's thoughts on both problems.
>
>
>
>
> Yet again, the Nachlass had revealed how far Riemann was ahead of his time. He could not have failed to recognise the significance of his solution of the problem in fluid dynamics. His method had shown why certain imaginary numbers that emerged from his analysis of the ball of fluid were all in a straight line. Yet at the same time, and on the same paper, he was trying to prove why the zeros in his zeta landscape all lay on a straight line. In the year following his discoveries about primes and hydrodynamics, he was recording his new ideas in the little black book which, infuriatingly, disappeared from the archives. With it have disappeared Riemann's thoughts on uniting these two themes from number theory and physics.
>
>
>
| https://mathoverflow.net/users/12178 | Riemann, fluid dynamics, and critical lines | **Q:** *Does anyone know of a reference which discusses more thoroughly the critical line appearing in Riemann's hydrodynamics problem?*
**A:** A recent reference is [Elliptical instability in hot Jupiter systems](https://arxiv.org/abs/1309.1624) by Cébron et al. (2013). The stability analysis of Riemann was not quite correct, it turns out.
Stability diagram of ellipsoids with a (resp. b) the longest (resp. shortest) equatorial axis and c the polar axis. For perturbations that are linear in the spatial coordinates, Riemann obtained unstable ellipsoids between the solid blue line and the black solid uppermost line, but Chandrasekhar showed in 1965 that the correct unstable zone is the blue one. The green zone corresponds to unstable ellipsoids for quadratic perturbations.
---
For a hydrodynamic approach to the Riemann zeta function (unrelated to the above), see [The Riemann hypothesis illuminated by the
Newton flow of $\zeta$](https://iopscience.iop.org/article/10.1088/0031-8949/90/10/108015/pdf) by Neuberger, Feilers, Maier, and Schleich.
>
> We analyze the Newton flow of the Riemann zeta function $\zeta$ and
> rederive in an elementary way the Riemann-von Mangoldt estimate of the
> number of non-trivial zeros below a given imaginary part. The
> representation of the flow on the Riemann sphere highlights the
> importance of the North pole as the starting and turning point of the
> separatrices, that is of the continental divides of the Newton flow.
> We argue that the resulting patterns may lead to deeper insight into
> the Riemann hypothesis. For this purpose we also compare and contrast
> the Newton flow of $\zeta$ with that of a function which in many ways
> is similar to $\zeta$, but violates the Riemann hypothesis.
>
>
>
---
*Update, 9 June 2022: professor Keating's answer to my email inquiry*
>
> It is now many years since I saw the papers in the Riemann Nachlass
> and since I had this conversation, so I cannot be certain what I said,
> but essentially Riemann considered the stability of a rotating gas of
> particles held together by gravity. He computed the linear stability
> in the usual way, by considering small perturbations and computing the
> frequencies. Complex frequencies correspond to unstable modes (as you
> know) and real frequencies to stable vibrations. He showed that all of
> the frequencies are real by demonstrating that they are the
> eigenvalues of a self-adjoint operator (eg as one shows that the zeros
> of Bessel functions are real). So the critical line in this case is
> the real line (of frequency).
>
>
> Experts tell me that Riemann made some mistakes in the calculations,
> but I have not looked into this carefully. Chandrasekhar wrote a book about the problem.
>
>
> Juan Marin at Harvard has been going through the Nachlass carefully.
> He wrote to me that
>
>
> "It also includes a version of Riemann's letter to Betti on how to
> find the attraction due to any homogeneous right ellipsoidal cylinder,
> including why "for the roots of the given equation F = 0 ... the
> singularities of the integrand are all real."
>
>
> Indeed in a letter to Prym, Betti writes that Riemann was working on
> number theory on his last days. I found evidence in another notebook
> suggesting he worked specifically on the twin prime conjecture while
> finishing his article on the "hydraulics" of sound vibrations and the
> anatomy of the ear."
>
>
> So it seems that Riemann worked on vibrations and number theory
> simultaneously on several occasions.
>
>
> Jon Keating
>
>
>
| 12 | https://mathoverflow.net/users/11260 | 424021 | 172,285 |
https://mathoverflow.net/questions/424033 | 0 | I am attempting to factor an $N^{\text{th}}$ degree polynomial with coefficients strictly equal to $1$ given by the equation
$$\sum\_{n=1}^{N} x^n$$
Although the Galois group for anything beyond a quartic is not generally soluble, I had hoped that an existing result had been established for this particular case. If not, I was curious if generalizing Tchirhausen transforms to the $N^{\text{th}}$ order and employing the [Lagrange inversion theorem](https://en.wikipedia.org/wiki/Lagrange_inversion_theorem) would allow me to examine series solutions with a special case being the Bring ultraradicals.
| https://mathoverflow.net/users/170939 | Factor $\sum_{n=1}^{N} x^n$ | This is not a research level question, nevertheless I feel like answering it. If you divide the polynomial by $x$, and multiply by $x-1$, you get $x^N-1$. The latter factors over $\mathbb{Q}$ as the product of the cyclotomic polynomials $\Phi\_d(x)$ with $d$ dividing $N$ (the factor corresponding to $d=1$ is $x-1$ that we added artificially). These factors are irreducible over $\mathbb{Q}$. Factoring over $\mathbb{R}$ is even easier, you just need to group the nonreal roots of $x^N-1$ (i.e. the $N$-th roots of unity except $\pm 1$) into complex conjugate pairs. Hope this helps!
| 7 | https://mathoverflow.net/users/11919 | 424034 | 172,287 |
https://mathoverflow.net/questions/424001 | 2 | *Note: Throughout, we denote by $\mathcal L$ the Lebesgue measure on $\mathbb R$.*
Let $g: [0, 1] \to \mathbb R$ be a continuous function of bounded variation. Denote by $\mu\_g$ its associated Lebesgue–Stieltjes measure, and $\lvert\mu\_g\rvert$ its total variation measure.
We say that a function $f: [0, 1] \to \mathbb R$ is *absolutely continuous* with respect to $g$ if $f$ is continuous, and for every $\varepsilon > 0$, there exists a $\delta > 0$ such that whenever $I\_k$, ($k = 1, \dotsc, n$) are disjoint open intervals with $\sum\_{i = 1}^n \lvert\mu\_g\rvert (I\_n) <\delta$, we have $\sum \mathcal L(f(I\_n)) < \varepsilon$.
For fixed $x \in [0, 1]$, denote by $E\_x$ the set $\{y \in [0, 1] \, \mid \, g(x) - g(y) \neq 0\}$, and consider the limit
$$\lim\_{y \to x\, ,\, y \in E\_x} \frac{f(x) - f(y)}{g(x) - g(y)}.$$
We shall say that the above limit exists, and denote it by $\frac{df}{dg}(x)$ if for every $r > 0$, the set $E\_x \cap B\_r (x)$ is nonempty, and the limit along $E\_x$ exists in the usual sense.
Note that for $\lvert\mu\_g\rvert$ a.e. $x \in [0, 1]$, $E\_x \cap B\_r (x)$ is nonempty for every $r > 0$.
**Question:** Let $g$ be a continuous function of bounded variation as above, and $f$ a function absolutely continuous with respect to $g$. Is it true that the following two statements hold?
1. $\frac{df}{dg}$ exists $\lvert\mu\_g\rvert$-a.e.
2. For every $x \in [0, 1]$, we have the following fundamental theorem of calculus style formula:
$$\int\_0^x \frac{df}{dg} \, dg = f(x) - f(0).$$
**Remark:**
The "Riemann FTC" version of the above is true, and not overly difficult to prove — if the limit $\frac{df}{dg}$ exists everywhere, then the integral formula holds.
It is thus left to see if the "Lebesgue FTC" version holds — if $f$ is absolutely continuous with respect to $g$, then $\frac{df}{dg}$ exists $
\lvert\mu\_g\rvert$-a.e., and the integral formula holds.
| https://mathoverflow.net/users/173490 | Fundamental theorem of calculus for Lebesgue–Stieltjes integrals? | This works, because: (1) Your definition of absolute continuity is equivalent to the other standard definition, namely, the condition that $|\mu\_g|(A)=0$ implies $|\mu\_f|(A)=0$ (your condition implies that $f\in BV$, so $\mu\_f$ is well defined).
(2) By a sufficiently general version of the (Lebesgue) differentiation theorem, $\lim\_{h\to 0} \mu\_h(x,x+h)/|\mu\_g|(x,x+h)$ exists for $|\mu\_g|$-a.e. $x$ and if $\mu\_h\ll |\mu\_g|$, then the limit computes the Radon-Nikodym derivative $d\mu\_h/d|\mu\_g|$. Thus your quotient converges to $(d\mu\_f/d|\mu\_g| )/(d\mu\_g/d|\mu\_g|)$; the denominator takes the values $\pm 1$, so the division doesn't make any trouble.
In general, a Radon-Nikodym derivative satisfies $\int f (d\mu/d\nu)\, d\nu = \int f\, d\mu$. This gives the formula from part (2) of your question.
| 5 | https://mathoverflow.net/users/48839 | 424039 | 172,288 |
https://mathoverflow.net/questions/423238 | 17 | Given $n$, what is the smallest value $\delta\_n$ satisfying the following:
>
> For any group of $n$ runners with constant but distinct speeds,
> starting from the same point and running clockwise along the unit
> circle, there exists a moment, at which the length of each of $n$
> empty circle segments between two consecutive runners does not exceed
> $\delta\_n$.
>
>
>
More formally, for $n \in \mathbb{N}$, $M=\{m\_1,\dots,m\_n\} \subset \mathbb{N}$, and $t \in [0,1]$, we sort the fractional parts of the values $tm\_i$ in ascending order: $0 \le \{tm\_{i\_1}\} \le\dots\le \{tm\_{i\_n}\} \le 1$. Then we consider the longest empty segment between two consecutive 'runners':
$$ \text{LongestArc}(M,t) := \max\Big\{ \{tm\_{i\_2}\}-\{tm\_{i\_1}\}, \dots, \{tm\_{i\_n}\}-\{tm\_{i\_{n-1}}\}, 1+\{tm\_{i\_1}\}-\{tm\_{i\_n}\} \Big\}.$$
Finally, we put
$$ \delta\_n := \mathop{\smash{\mathrm{sup}}}\_{|M|=n} \inf\_{t \in [0,1]} \text{LongestArc}(M,t).$$
Has the problem of finding these values (or, perhaps, an equivalent one) been studied earlier? I have looked through some papers related to the [Lonely Runner Conjecture](https://en.wikipedia.org/wiki/Lonely_runner_conjecture) (including [this one](https://arxiv.org/abs/1508.07289)) but found nothing.
It is clear that $\delta\_2=\frac12$. With a little more effort goes $\delta\_3=\frac25$, one of the 'extremal' sets here is $M=\{0,1,3\}$. One may check that $\delta\_4 \ge \frac13$ by taking $M=\{0,1,2,4\}$. I suspect this bound to be tight. Maybe someone sees a counterexample or a short argument, why is that so?
For large $n$, I believe I can show that $\delta\_n \ge \frac{\log n -O(\log \log n)}{n}$. However, I don't have any (decent) upper bound on $\delta\_n$ apart from the fact that $\delta\_n=o(1)$ as $n \to \infty$.
Finally, I found that the 'dual' problem concerning the values
$$ \epsilon\_n := \inf\_{|M|=n} \mathop{\smash{\mathrm{sup}}}\_{t \in [0,1]} \text{LongestArc}(M,t)$$
has earlier been [considered](https://mathoverflow.net/questions/176822/not-lonely-runners) on MathOverflow (not in the literature, though). One may also state two more 'dual' extremal problems related to the shortest arc (instead of the longest one), but I haven't found anything published about them as well.
I would appreciate any related reference!
| https://mathoverflow.net/users/482790 | Has the following problem, resembling the lonely runner conjecture, been studied? | I think the answer to the question as posed is "yes", see Konyagin-Ruzsa-Schlag (2011)
[https://gauss.math.yale.edu/~ws442/papers/DILATES.pdf](https://gauss.math.yale.edu/%7Ews442/papers/DILATES.pdf)
It looks like nothing beyond the bounds discovered in the comments and Fedja's answer above is known.
| 7 | https://mathoverflow.net/users/5575 | 424047 | 172,291 |
https://mathoverflow.net/questions/416279 | 6 | Let $f : A \to B$ be a monomorphism of strict $\omega$-categories, and let $d : \partial \mathbb G\_n \to A$ be an attaching map. There is an induced map $g : A \cup\_{\partial \mathbb G\_n} \mathbb G\_n \to B \cup\_{\partial \mathbb G\_n} \mathbb G\_n$. Here the pushout is taken in strict $\omega$-categories.
**Question:** Is $g$ a monomorphism?
Since folk cofibrations are monomorphisms, this is the case when $f$ (and hence $g$) is a folk cofibration, at least. I suspect the answer is *yes* in general though.
Here a *folk cofibration* is a map which is a retract of transfinite composites of cobase-changes of maps of the form $\partial \mathbb G\_n \to \mathbb G\_n$, i.e. a cofibration in the [folk model structure](https://arxiv.org/abs/0712.0617) on strict $\omega$-categories. $\mathbb G\_n$ denotes the $n$-globe, and $\partial \mathbb G\_n$ denotes its boundary. The title question is equivalent to the question as I've stated it above because folk cofibrations are generated by maps of the form $\partial \mathbb G\_n \to \mathbb G\_n$ under transfinite composition and retracts, and monomorphisms are stable under transfinite composition and retracts.
**EDIT:** Monomorphisms of strict $\omega$-categories (or even of strict 1-categories) are not stable under pushout along an arbitrary map. For example, push out the monomorphism $B\mathbb N \to B\mathbb Z$ along the map $B\mathbb N \to B(\mathbb N / (2=1))$, and you get the map $B \mathbb Z \to \ast$ (in other words, an invertible idempotent is automatically an identity), which is not a monomorphism.
According to [this MO question](https://mathoverflow.net/questions/249589/pushouts-of-injective-monoid-homomorphisms), the pushout of one injective monoid homomorphism along another injective monoid homomorphism need not be injective. Since monoids are 1-object categories (and the inclusion preserves pushouts), this means that the pushout of a monomorphism of 1-categories along a monomorphism of 1-categories need not be a monomorphism. The inclusion of 1-categories into $\omega$-categories preserves pushouts, so this is also the case in $\omega$-categories. Thus the retreat I've made in my question to pushing out a monomorphism along a folk cofibration seems warranted. But it would be nice to have an *explicit* example illustrating that the pushout of a monomorphism along a monomorphism need not be a monomorphism in $Cat\_\omega$.
| https://mathoverflow.net/users/2362 | Are monomorphisms of strict $\omega$-categories stable under pushout along folk cofibrations? | No.
Let $B = \partial \mathbb G\_2 \vee \mathbb G\_2$ be obtained by gluing together the boundary of a 2-globe with a 2-globe, in such a way that the 1-morphisms are composable. Let $D = \mathbb G\_2 \vee \mathbb G\_2 = B \amalg\_{\partial \mathbb G\_2} \mathbb G\_2$ be the result of freely filling in that globe boundary. Let $x$ denote the first atomic 2-cell (the one which was glued on) and let $y$ denote the second atomic 2-cell (the one which is in $B$). Then $y \circ\_0 x$ is the unique 2-cell with its given boundary in $D$.
Let $A \subset B$ be obtained by deleting the atomic 2-cell. Then $A$ still has two nondegenerate 2-cells (in $B$ these were obtained by whiskering $y$ with the two nondegnerate 1-cells composable with it). Now $C := A \amalg\_{\partial \mathbb G\_2} \mathbb G\_2$ has two distinct 2-cells mapping to $y \circ\_0 x$ in $D$. That is, $C \to D$ is not a monomorphism, even though $A \to B$ is.
| 3 | https://mathoverflow.net/users/2362 | 424050 | 172,292 |
https://mathoverflow.net/questions/423934 | 4 | Let $S(n)$ be the (unitary) matrix group of $n\times n$ permutation matrices. This is clearly a finite group of order $n!$. It is well known that we can add diagonal unitary matrices with any finite root of unity entries to this group and the new group is also finite. For any finite group of permutations and diagonal matrices we can always add diagonal matrices with higher root of unity entries to the group and the group will remain finite. There is, therefore, no maximal finite group in this case. Here we are interested in a slightly different notion of maximal finite group (see below).
Let $U\_n\in U(n)$ be a unitary matrix which is not a monomial matrix (also known as generalized symmetric matrices). In other words, $U\_n$ is not a product of a permutation and a diagonal matrix.
A permutation group, $S(n)$, is maximal if the addition of any non-monomial matrix, $U\_n$, generates a group $G=\langle S(n), U\_n \rangle$ where the order of $G$ is infinite. When this is the case, I will refer to the group $S(n)$ as a maximal finite group.
With this definition of a maximal finite group I'm interested in the following questions:
Can we prove the existence of maximal groups $S(n)$ for some $n$?
Can we prove that there exists an integer $k$ such that for all $n>k$ and any non-monomial matrix $U\_n$, that $\langle S(n), U\_n \rangle$ necessarily generates an infinite order group?
| https://mathoverflow.net/users/482648 | Existence of 'maximal' finite permutation groups? | The standard representation of Sn+1 is faithful and n-dimensional. We may also assume it preserves a Hermitian inner product. When restricted to a standard copy of Sn, it becomes isomorphic to the permutation representation. Use this isomorphism to choose a basis, and hence realise Sn+1 as living between your S(n) and U(n), hence providing a negative answer to the last question. (Sn+1 is not generated by a monomial matrix as there is no homomorphism from Sn+1 to Sn for n at least 4).
| 8 | https://mathoverflow.net/users/425 | 424058 | 172,296 |
https://mathoverflow.net/questions/424056 | -1 | I am having the following integral:
$$I = \int u\, J^s(\partial\_x \overline{u})- \overline{u}\, J^s(\partial\_x u))dxdy$$
where $J^S= (I-\Delta)^\frac{2}{2}$, $\mathbb{R} \ni s \geq 1$ and $u=u(x,y)$,
$u:\mathbb{R}^2 \to \mathbb{C}$.
My question: Is it allowed to do the integration by parts for the fractional differentiation so I can get zero? That is:
\begin{align}
I &= \int u\, J^s(\partial\_x \overline{u})- \overline{u}\, J^s(\partial\_x u))dxdy\\
&= \int u\, J^s(\partial\_x \overline{u})+ \partial\_x\overline{u}\, J^s( u))dxdy\\
&= \int u\, J^s(\partial\_x \overline{u})-\int u\, J^s(\partial\_x \overline{u})\\
&=0
\end{align}
Are the above calculations right? If so, how to prove it?
Thanks in Advance.
| https://mathoverflow.net/users/471464 | Is integration by parts allowed for the $J^s$ derivative, where $s \in \mathbb R$ | Your claim is false, since $J^s$ is self-adjoint, moving $J^s$ from one term to the other does not incur a minus sign. So instead of the two terms cancelling, they actually double.
If $u$ is a Schwartz function (or more generally belonging to a suitable Sobolev space) you can compute by using Plancherel:
You have
$$ \int u J^s (\nabla \bar{u}) = \int \hat{u} (1 + |\xi|^2)^{s/2} (-i) \xi \bar{\hat{u}} = - \int (1 + |\xi|^2)^{s/2} i\xi \hat{u} \bar{\hat{u}} = - \int J^s(\nabla u) \bar{u} $$
| 1 | https://mathoverflow.net/users/3948 | 424059 | 172,297 |
https://mathoverflow.net/questions/201680 | 7 | Consider the one-dimensional Schrodinger operator on the real line $\mathbb{R}$ given by
$$ L = - \partial\_x^2 + V $$
where $V$ is a potential with the following properties:
* $V$ is non-negative, and infinitely differentiable
* $|V| = \frac{1}{|x|^2}$ for $|x| \gg 1$ (so in particular it is in $L^p$ for any $p\geq 1$).
**Question**: Are there general theorems concerning the (pointwise or local energy) decay of the Schrodinger ($i\partial\_t \Phi = L\Phi$) equations for such operators? I am particularly interested in the case where $V$ has more than one local maximum.
---
**Motivation**:
First let us consider the case where $V$ has exactly one local (and hence global) maximum. In the classical particle picture we see that the global maximum corresponds to an unstable fixed point of the dynamics, and for most energies a particle either has large energy so it flies over the hump, or has small energy so it comes in, turns around, and bounces off. Only at a very specific energy can you achieve the balancing act of sending in a particle that comes eventually to rest on the top of the hump.
This classical picture is essentially sufficient for analyzing the quantum picture. If we assume that the local maximum is non-degenerate, then we can prove relatively simply an integrated local energy decay estimate for solutions. One version of which states that, assuming the local maximum is at the origin, for every positive $\delta$ there exists some $b$ such that
$$ \int\_0^\infty \int\_{\mathbb{R}} \frac{1}{(1 + (bx)^2)^{\frac12 + \delta}} \Phi\_x^2 ~\mathrm{d}x ~\mathrm{d}t \tag{\*}$$
is bounded by some universal constant times some quantity depending only on the initial data.
In the case where $V$ has more than one local maximum, the classical picture is drastically different. We see that between two consecutive local maxima of the potential, we expect classically there to be stable trapped particles, since classical particles cannot jump over the hump. So the classical picture would contradict local energy decay, since a purely classical picture would admit spatially localised solutions over a non-trivial range of energies.
On the other hand, when dealing with quantum phenomenon there should be tunneling where particles can escape through finite barriers. So I expect the intuition to be that in the quantum case some (possibly weaker) energy decay is still available.
What is known about this problem? Is my intuition okay?
**Related question**:
At the present I just don't have a starting point to search from. So if someone can give me a few names and/or papers to start looking, or some keywords to search for, that would be appreciated.
---
**Edit**: As Christian Remling pointed out, there is the standard RAGE theorem which in particular implies that energy will escape from any compact set. What I seek is something a bit stronger. The RAGE theorem (as far as I know) does not give explicit rates, and I am hoping for some sort of result giving either the localised energy has an explicit rate of decay $\leq t^{-\alpha}$ for some $\alpha > 0$ or that the decay can be made explicit in the integral sense (something like equation (\*) above but with possibly the $L^1$ integration in $t$ replaced by some higher $L^p$ for $p < \infty$).
| https://mathoverflow.net/users/3948 | Decay of solutions to Schrodinger equation with local minimum in potential | There are several decay results in the 1D case, but probably they are not enough for you. Goldberg and Schlag (Comm. Math. Phys. 251 (2004) 157–178) proved pointwise decay of the $L^\infty$ norm as $t^{-1/2}$ like in the free case, for any potential such that $(1+|x|)V\in L^1$, plus some spectral assumptions. Of course this rules out the case $V\sim|x|^{-2}$ that is of interest to you. Anyway, in these results the local structure of $V$ is irrelevant. Maybe this suggests that either localization is so strong to produce a bound state, or you have dispersion like in the free case.
| 2 | https://mathoverflow.net/users/7294 | 424078 | 172,300 |
https://mathoverflow.net/questions/424082 | 3 | Let $f\colon X\to Y$ be a continuous map between topological spaces, which you can assume to be Hausdorff if you like. Say that $f$ has property $P$ if for every compact subset $L\subseteq Y$, there exists a compact subset $K\subseteq X$ with $f(K)=L$. Is there a more standard name for this?
Clearly a map with property $P$ must be surjective. Any proper surjective map has property $P$. A product projection $Y\times Z\to Y$ has property $P$ iff $Z\neq\emptyset$, but is only proper if $Z$ is compact. More generally, I think that any surjective fibre bundle projection has property $P$. Any map that has a continuous section has property $P$.
| https://mathoverflow.net/users/10366 | What is this property of surjective continuous maps called? | A map $f\colon X\to Y$ is a *compact-covering* map if every compact $K\subseteq Y$ is the image of some compact $C\subseteq X$.
| 3 | https://mathoverflow.net/users/112417 | 424084 | 172,303 |
https://mathoverflow.net/questions/424065 | -1 | I am studying the use of the commutator for finding the estimate of energy. During my looking through many papers I found that [this paper](https://www.aimsciences.org/article/doi/10.3934/dcds.2019145) contains a possible typo. [Here is the archive version](https://arxiv.org/abs/1809.02027) which has the same prospective typo!
Another question about the commutator. When I espand the follawing commutator $[J^s,u]\partial\_x u$ the last two terms are cancel eachohter! Are my calculations right?
\begin{align}
[J^s,u]\partial\_x u &= J^s(u\, \partial\_x u)- u J^s(\partial\_x u)\\
&=J^s(u) \partial\_x u + u J^s(\partial\_x u) - u J^s(\partial\_x u)\\
&=J^s(u) \partial\_x u
\end{align}
Is the above right?
| https://mathoverflow.net/users/471464 | A question about the commutator $[J^s,u]\partial_x u$ | The computation is not right in general. $J^s$ does not satisfy the Leibniz rule. The idea is that at least one of the derivatives hits the function $w$ in the paper, So the authors used the commutator to get rid of it.
| 1 | https://mathoverflow.net/users/471464 | 424090 | 172,305 |
https://mathoverflow.net/questions/423691 | 0 | Let
* $(E,\mathcal E)$ be a measurable space
* $\mathcal E\_b:=\left\{f:E\to\mathbb R\mid f\text{ is bounded and }\mathcal E\text{-measurable}\right\}$
* $(\kappa\_t)\_{t\ge0}$ be a Markov semigroup on $(E,\mathcal E)$
* $Q$ denote the weak generator of $(\kappa\_t)\_{t\ge0}$; i.e. $$\mathcal D(Q):=\left\{f\in\mathcal E\_b\mid\forall x\in E:[0,\infty)\ni t\mapsto(\kappa\_tf)(x)\text{ is right-differentiable at }0\right\}$$ and $$(Qf)(x):=\left.\frac{\rm d}{{\rm d}t}(\kappa\_tf)(x)\right|\_{t=0+}\;\;\;\text{for }x\in E\text{ and }f\in\mathcal D(Q)$$
* $(\Omega,\mathcal A,\operatorname P)$ be a probability space
* $(Y\_t)\_{t\ge0}$ be an $(E,\mathcal E)$-valued time-homogeneous Markov process on $(\Omega,\mathcal A,\operatorname P)$ with transition semigroup $(\kappa\_t)\_{t\ge0}$
* $\alpha$ be a transition kernel on $(E,\mathcal E)$ and $$Af(x):=\int\_E ( f(y)-f(x)) \:\alpha(x,{\rm d}y)\;\;\;\text{for all }x\in E\text{ and }f\in\mathcal D(A):=\mathcal E\_b$$
>
> **Question**: How can we construct an $(E,\mathcal E)$-valued time-homogeneous Markov process $(X\_t)\_{t\ge0}$ on $(\Omega,\mathcal A,\operatorname P)$ with weak generator $$Lf=Qf+Af\;\;\;\text{for all }f\in\mathcal D(L)\subseteq\mathcal D(Q)\cap\mathcal D(A)?\tag1$$
>
>
>
The idea is that the local behavior between jumps of $(X\_t)\_{t\ge0}$ is described by $(Y\_t)\_{t\ge0}$ and, assuming that $\alpha(x,B)=c(x)\mu(x,B)$ for all $x\in E$ for some $\mathcal E$-measurable $c:E\to[0,\infty)$ and a Markov kernel $\mu$ on $(E,\mathcal E)$, the jumps occur at a state-dependent rate $c$ and are performed according to the state-depedendent distribution $\mu$.
The process should be described by something like $$X\_t=\sum\_{n\in\mathbb N\_0}1\_{[\tau\_n,\:\tau\_{n+1})}(t)Y^{(n)}\_{t-\tau\_n}\;\;\;\text{for all }t\ge0\tag1,$$ where $\tau\_n$ is the time of the $n$th-jump and the $Y^{(n)}$ are independent copies of $Y$.
>
> However, how do we need to define the $\tau\_n$ precisely and how do we see that the weak generator of $(1)$ is actually equal to $L$?
>
>
>
I'm aware of the following simpler result: If $(W\_n)\_{n\in\mathbb N\_0}$ is a time-homogeneous Markov chain on $(\Omega,\mathcal A,\operatorname P)$ with transition kernel $\kappa$ and $(N\_t)\_{t\ge0}$ is a Poisson process on $(\Omega,\mathcal A,\operatorname P)$ with intensity $r>0$ and $W$ is independent of $N$, then $$Z\_t:=W\_{N\_t}\;\;\;\text{for }t\ge0$$ is a time-homogeneous Markov process with transition semigroup $\left(e^{t(\kappa-r)}\right)\_{t\ge0}$ and generator $r\left(\kappa-\operatorname{id}\_{\mathcal E\_b}\right)$.
In particular, if $W$ is a random walk with step distribution $\alpha^{-1}\nu$; i.e. $W\_n=\sum\_{i=1}^n\xi\_i$ for all $n\in\mathbb N$ for some independent identically $\alpha^{-1}\nu$-distributed process $(Z\_n)\_{n\in\mathbb N}$ on $(\Omega,\mathcal A,\operatorname P)$, then the generator of $Z$ is given by $$\mathcal E\_b\ni g\mapsto\int g(\;\cdot\;+y)-g\:\nu({\rm d}y).$$
Maybe a similar construction and hence an expression different from $(1)$ from which it is easier to derive the desired result is possible in the setting of this question.
| https://mathoverflow.net/users/91890 | Construction of a Markov process with prescribed local behavior and state-dependent jump distribution | The construction given by the OP is almost correct. Here is a slight correction: $$
X\_t = \sum\_{n=0}^{\infty} 1\_{[\tau\_n,\tau\_{n+1})}(t) Y\_{t - \tau\_n}^{n} \;, \tag{1}
$$ where we have introduced
* $\{\tau\_i\}$ are a sequence of jump times defined via $\tau\_{i+1}=\tau\_i+\xi\_i$, $\tau\_0=0$, and $\{\xi\_i \} \overset{i.i.d.}{\sim} \operatorname{Exp}(1)$ ; and,
* $\{Y^{i}\}$ are independent realizations of $Y$ with $Y\_0^i=x$ if $i=0$ and else sample $Y\_0^i \mid (Y^0, \dots, Y^{i-1}, \xi\_0, \dots, \xi\_i) \sim \alpha(Y\_{\tau\_i - \tau\_{i-1}}^{i-1}, \cdot)$ .
In other words,
$$
X\_t = \begin{cases}
Y^0\_t & t < \tau\_1 \;, \\
Y^1\_{t-\tau\_1} & \tau\_1 \le t < \tau\_2 \;, \\
Y^2\_{t-\tau\_2} & \tau\_2 \le t < \tau\_3 \;, \\
\vdots
\end{cases}
$$
To see that the weak generator of (1) is indeed $L=Q+A$, write $f (X\_t) - f(x) = \rm{I} + \rm{II} + \rm{III}$ where \begin{align\*}
\rm{I} &:= (f(X\_t) - f(x)) 1\_{\{t < \tau\_1 \}} \;, \\
\rm{II} &:= (f(Y\_0^1) - f(Y\_{\tau\_1}^0)) 1\_{\{t \ge \tau\_1 \}} \;, \\
\rm{III} &:= (f(Y\_{\tau\_1}^0) - f(x) + f(X\_t) - f(Y\_0^1)) 1\_{\{t \ge \tau\_1 \}} \;.
\end{align\*}
Then \begin{align\*}
E[\rm{I}] &= e^{-t} ( \kappa\_t f(x) - f(x) ) = e^{-t} E \int\_0^t Qf (Y\_s^0) ds \;, \\
E[{\rm II} \mid \tau\_1 = s] &= E[f(Y\_0^1) - f(Y\_{s}^0)] 1\_{\{ t \ge s \}} = E[ A f(Y\_s^0) ] 1\_{\{ t \ge s \}} \;, \\
E[ \rm{II} ] &= E \int\_0^{\infty} E[ {\rm II} \mid \tau\_1 = s] e^{-s} ds = E \int\_0^t e^{-s} A f(Y\_s^0) ds \;.
\end{align\*} One can similarly show that $E( \rm{III} )$ is $O(t^2)$ for $t \in [0,1]$. Therefore, combining the above and using $(e^{-s} - e^{-t}) \le (t-s)$ for $t \ge s$, one obtains that for all $ t \in [0,1]$ $$
E[f(X\_t)] - f(x) = E\int\_0^t (A f(Y\_s^0) + Q f(Y\_s^0)) ds + O(t^2) \;.
$$
While this construction/analysis covers the case of constant jump rates, the case of state-dependent jump rates can be treated similarly as discussed in the comments below.
| 1 | https://mathoverflow.net/users/64449 | 424091 | 172,306 |
https://mathoverflow.net/questions/64407 | 7 | Consider the complex $n$-by-$n$ matrices $M\_n$.
Suppose that $A\_i$, for $i=1,\ldots,n^2$, satisfy $\mathrm{Tr}(A\_i^\*
A\_j)=\delta\_{ij}$, so that together they form an orthonormal basis for
$M\_n$. Define a linear map $T \colon M\_n \to M\_n \otimes M\_n$ by $T(A\_i) = A\_i \otimes A\_i$.
>
> Question: when is $T$ completely positive?
>
>
>
For example, if $A\_i$ are the matrices with a single entry one and the rest zeroes in some fixed basis of $\mathbb{C}^n$, then $T$ is completely positive. In fact, I think these might be the only examples. If $T$ is completely positive, then the following are equivalent to $A\_i$ being matrix units as in the above example:
* each $A\_i$ has rank one;
* each positive semidefinite $A\_i$ has trace one;
* the set $\{0,A\_1,\ldots,A\_{n^2}\}$ is closed under multiplication;
* $T(1)$ is idempotent;
* $T^\*(1) \leq 1$;
* $T$ preserves trace.
These are sufficient conditions, but proving they are sufficient doesn't use $\mathrm{Tr}(A\_i^\* A\_j)=\delta\_{ij}$ at all. Are they necessary?
| https://mathoverflow.net/users/10368 | When is this map completely positive? | Further to the linear map $T: M\_n \rightarrow M\_n \otimes M\_n$ defined above by setting $T(A\_i):=A\_i \otimes A\_i$, consider the linear map $E: M\_n \rightarrow \mathbb{C}$ defined by setting $E(A\_i) := 1$ for all $i=1,...,n^2$.
Let $\eta: \mathbb{C} \rightarrow M\_n \otimes M\_n^\ast$ and $\epsilon: M\_n^\ast \otimes M\_n \rightarrow \mathbb{C}$ be the cups and caps of CP maps:
$$
\eta(1) := \sum\_{i} A\_i \otimes A\_i^\ast
\hspace{1cm}
\epsilon(A\_i^\ast \otimes A\_j) := \delta\_{ij}
$$
Also, let $\sigma := M\_n \otimes M\_n^\ast \rightarrow M\_n^\ast \otimes M\_n$ be the swap of CP maps:
$$
\sigma(A\_i \otimes A\_j^\ast) := A\_j^\ast \otimes A\_i
$$
If $T$ is a CP map, then so is $E$, because the latter can be obtained from the former by composition and tensor product with the CP maps $id\_{M\_n}$, $\eta$, $\epsilon$ and $\sigma$:
$$
E = \left(id\_{M\_n}\otimes (\epsilon \circ \sigma)\right) \circ (T \otimes id\_{M\_n}) \circ \eta
$$
By the way $T$ and $E$ are defined, $(T, E, T^\dagger, E^\dagger)$ is a special commutative $\dagger$-Frobenius algebra of linear maps. If $T$ is CP, then $(T, E, T^\dagger, E^\dagger)$ is a special commutative $\dagger$-Frobenius algebra of CP maps.
By Corollary 7 of <https://arxiv.org/abs/2110.07074v2>, such an algebra necessarily arises by "doubling" of a special commutative $\dagger$-Frobenius algebra $(t, e, t^\dagger, e^\dagger)$ of linear maps. The latter are defined as follows, for some choice of OBN $|\phi\_i\rangle$ on $\mathbb{C}^n$:
$$
t(|\phi\_i\rangle) := |\phi\_i\rangle\otimes|\phi\_i\rangle
\hspace{1cm}
e(|\phi\_i\rangle) := 1
$$
Hence, $T$ takes the form $T(|\phi\_i\rangle\langle\phi\_j|) := |\phi\_i\rangle\langle\phi\_j| \otimes |\phi\_i\rangle\langle\phi\_j|$, as originally conjectured.
| 2 | https://mathoverflow.net/users/128347 | 424092 | 172,307 |
https://mathoverflow.net/questions/424097 | 3 | The title may be inappropriate, and I apologize for that.
I'm writing a reading report on my harmonic analysis course. My topic is the Littlewood-Paley inequality for arbitrary intervals, which was proved in [Rubio de Francia, 1985](https://www.ems-ph.org/journals/show_abstract.php?issn=0213-2230&vol=1&iss=2&rank=1). I found it hard to understand the proof of Lemma 2.3 in that paper, beacuse I've learnt nothing about $A\_p$-weights. However, I found that page 8 of [Lacey’s paper](https://arxiv.org/abs/math/0306417v3) proves the same lemma using a vector-valued Littlewood–Paley inequality, in a very short space (only two lines are related). So I'm asking about the details of the proof using vector-valued Littlewood–Paley inequality.
Now I restate the main problem here.
For any set $I\in\mathbb R$, we define the operator $\widehat{S\_If}=\chi\_I\hat f$. For every interval $I$ and $c>0$, we denote by $cI$ the interval with the same center as $I$ and length: $|cI|=c|I|$.
Given disjoint intervals $\{I\_k=(a\_k,b\_k)\}\_{k\in\mathbb Z}$, we define that the Whitney decomposition of each $I\_k$ consists of $\{I\_{k,j}\}\_{j\in\mathbb Z}$, where
$$I\_{k,j}=\begin{cases}
\left[a\_k+(b\_k-a\_k)\frac{2^{-j}}3, a\_k+(b\_k-a\_k)\frac{2^{-j+1}}3\right], & j\geq 0,\\
\left[b\_k-(b\_k-a\_k)\frac{2^{j+1}}3, b\_k-(b\_k-a\_k)\frac{2^{j}}3\right], & j\leq -1.
\end{cases}$$
Note that $2I\_{k,j}\subset I\_k$ and $\sum\_{k,j}\chi\_{2I\_{k,j}}(x)\leq 5$ for all $x\in\mathbb R$. We are going to prove that
>
> **Theorem (Lemma 2.3 in [Rubio de Francia, 1985](https://www.ems-ph.org/journals/show_abstract.php?issn=0213-2230&vol=1&iss=2&rank=1)).** Given disjoint intervals $\{I\_k\}\_{k\in\mathbb Z}$ and its Whitney decomposition as above, then for all $1<p<\infty$ we have
>
> $$\left\|\left(\sum\_{k,j\in\mathbb Z}\left|S\_{I\_{k,j}}f\right|^2\right)^{\frac12}\right\|\_p\sim\_p\left\|\left(\sum\_{k\in\mathbb Z}|S\_{I\_k}f|^2\right)^{\frac12}\right\|\_p,\qquad f\in L^p(\mathbb R).$$
>
>
>
Page 8 of Lacey’s paper sketches the proof of this Theorem, as he said, “by a vector-valued Littlewood-Paley inequality”. But he didn’t say which one the vector-valued LP inequality is.
I have known a kind of vector-valued Littlewood–Paley inequality, which states that
$$\left\|\left(\sum\_{k,j\in\mathbb Z}\left|S\_{\Delta\_j}f\_k\right|^2\right)^{\frac12}\right\|\_p\lesssim\_p\left\|\left(\sum\_{k\in\mathbb Z}|f\_k|^2\right)^{\frac12}\right\|\_p,\qquad f\in L^p(\mathbb R),$$
where $\Delta\_j=(-2^{j+1},-2^j]\cup[2^j,2^{j+1})$ is the Littlewood–Paley dyadic intervals.
But I don't know how to use this vector-valued Littlewood–Paley inequality to prove the Theorem. I tried to mimic the proof of the above inequality, but I failed since it became very messy.
Maybe this is not the one used in Lacey’s proof. I want to know which one does Lacey use, and how it is used to prove the Theorem.
Any help would be appreciated!
| https://mathoverflow.net/users/141451 | A kind of vector-valued Littlewood–Paley inequality for arbitrary intervals | Here you need to apply slightly non-standard Littlewood--Paley inequality. It is well known (however, an exact reference does not come to my mind immediately but I believe any proof of standard L.--P. inequality works equally well in this case) that the Littlewood--Paley inequality holds not only for the intervals $[2^j, 2^{j+1})$ but also for arbitrary lacunary intervals in $\mathbb{R}$, that is, you may take the intervals $[\lambda\_k,\lambda\_{k+1})$ as long as $\lambda\_{k+1}/\lambda\_k\ge c > 1$ (or $\lambda\_{k+1}/\lambda\_k\le c < 1$; and you can also shift and dilate such collection). And the collection $\mathrm{Well}(\Omega)$ is obviously lacunary in such sence (or at least a union of two lacunary collections).
Now, we apply this observation to get:
\begin{equation}
\|S^\Omega f\|\_{L^p} = \Big\| \Big( \sum\_{\omega\in\Omega} |S\_\omega f|^2 \Big)^{1/2} \Big\|\_{L^p}\asymp \Big\| \Big( \sum\_{\omega\in\Omega} |S^{\mathrm{Well}(\omega)}S\_\omega f|^2 \Big)^{1/2} \Big\|\_{L^p} = \Big\| S^{\mathrm{Well}(\Omega)} f \Big\|\_{L^p} \ \text{(1)}.
\end{equation}
The inequality we used here is
$$
\Big\| \Big( \sum\_{\omega\in\Omega} |S^{\mathrm{Well(\omega)}}f\_\omega|^2 \Big)^{1/2} \Big\|\_{L^p} \lesssim \Big\| \Big( \sum\_{\omega\in\Omega} |f\_\omega|^2 \Big)^{1/2} \Big\|\_{L^p}
$$
for arbitrary functions $f\_\omega$ --- this is the kind of vector-valued Littlewood--Paley inequality Lacey (and Rubio de Francia) used. I am not sure that it can be proved easier than by following the path Rubio de Francia mentioned, that is, using the uniform boundedness of operators $S^{\mathrm{Well}(\omega)}$ in $L^2(w)$ for $w\in A\_2$. The key observation here is that the collections $\mathrm{Well}(\omega)$ are obtained from one another by some affine transformation so this uniform boundedness is not surprising once you know the boundedness of just one operator, say $S^{\mathrm{Well}([-1/2. 1/2])}$ on $L^2(w)$. The reverse inequality in the middle of formula (1) follows by duality.
Rubio de Francia mentioned the weighted Littlewood--Paley inequality because in general weighted norm inequalities imply vector-valued ones. One of possible references for these things is the book by Rubio de Francia and Garcia-Cuerva "Weighted norm inequalities and related topics" (here you may apply Theorem 6.4 from page 519).
| 4 | https://mathoverflow.net/users/69086 | 424107 | 172,311 |
https://mathoverflow.net/questions/423998 | 2 | Let $X$ be a smooth projective variety over a field $k$. Let $L$ be an invertible sheaf on $X$. Suppose that $L$ is big and globally generated. Can one conclude that the associated morphism $\phi\_L : X \to \mathbb{P}^n$ is generically finite? Thanks in advance.
| https://mathoverflow.net/users/483548 | Morphism attached to a big and globally generated line bundle | I don't see it in Lazarsfeld's Positivity, but one way to see it is this, let $Z=\phi\_L(X)$, let $\eta$ be the generic point of $Z$ and let $f:X\_{\eta}→\eta$ be the base change. Note $L|\_{X\_{\eta}}$ is still big and globally generated. Furthermore $X\_\eta$ is a projective variety over the field $k(\eta)$ and $$L|\_{X\_{\eta}}=\phi^\* \mathcal{O}(1)|\_{X\_{\eta}}=f^\*k(η)=\mathcal{O}\_{X\_η}.$$ Now the only time the structure sheaf of a projective variety can be big is if the variety is 0-dimensional, so $X\_{\eta}$ is zero dimensional, thus $X \to Z$ is generically finite.
| 0 | https://mathoverflow.net/users/3521 | 424117 | 172,313 |
https://mathoverflow.net/questions/424067 | 3 | When regarding a ternary quadratic form $Q(x,y,z)$, is a classic question to consider which integers $n$ can be represented by $Q$. It is also classic to wonder how "well distributed" are the lattice points $(x,y,z)\in\mathbb{Z}^3$ that satisfies $Q(x,y,z)=n$ when $n\rightarrow\infty$ along a certain sequence.
When $Q$ is just the modulus square, I have found a lot of literature but I am interested in more general $Q$. On chapter 11 of Topics in Classical
Automorphic Forms by Henryk Iwaniec, we find conditions to $n$ so that this condition holds. However, Iwaniec said that the result is due to Duke and that he asked a few less on $n$.
I have tried to read Duke papers:
Representation of integers
by positive ternary quadratic forms and
equidistribution of lattice points on ellipsoids
And
On ternary quadratic forms,
But I couldn't find the conditions he imposed over $n$ so that we have equidistribution. I am interested in asking the less possible to $n$ and knowing the difference between the hypothesis of Duke and those of Iwaniec.
Thank you to everyone.
| https://mathoverflow.net/users/411616 | Duke and Schulze-Pillot condition for equidistribution | I recommend that you study the paper by Duke and Schulze-Pillot, Representation of integers by positive ternary quadratic forms and equidistribution of lattice points on ellipsoids (Inventiones, 1990).
Theorem 1 shows that if the number of representations $r(n,Q)$ exceeds $n^{1/2-1/176}$, then the representations become equidistributed on the relevant ellipsoid as $n\to\infty$. So the question is how to guarantee that $r(n,Q)$ is large. By Theorem 3 (see also the subsequent Corollary), it suffices that $n$ is primitively represented by some form in the spinor genus of $Q$. By Theorem 2, if we exclude finitely many explicitly computable square classes for $n$, the condition simplifies to: $n$ is primitively represented by some form in the genus of $Q$. The latter is equivalent to: $n$ is primitively represented by $Q$ modulo $4\det(a\_{ij})$, where $(a\_{ij})\in\mathrm{M}\_3(\mathbb{Z})$ is the symmetric matrix such that
$$Q(x\_1,x\_2,x\_3)=\frac{1}{2}\sum\_{i,j}a\_{ij}x\_ix\_j.$$
It remains to determine the exceptional square classes. Regarding that, see the proof of Theorem 2 and the references given. In particular, if $n$ is coprime to $\det(a\_{ij})$, then the only exceptional square class is the set of squares (see Footnote 5 in [arXiv:1402.1332](https://arxiv.org/abs/1402.1332)).
| 5 | https://mathoverflow.net/users/11919 | 424127 | 172,316 |
https://mathoverflow.net/questions/423675 | 3 | Let $a,b$ be positive integers with $x^2-ay^2-bz^2+abv^2=0$ having only the zero solution over $\mathbb Z$ and consider the Fuchsian group
\begin{equation\*}
\Gamma=\left\{\begin{bmatrix}
k+\sqrt{a}l & m+\sqrt{a}n \\
b(m-\sqrt{a}n) & k-\sqrt{a}l \\
\end{bmatrix}
\colon k,l,m,n\in\mathbb Z, k^2-al^2-bm^2+abn^2=1\right\}.
\end{equation\*}
I would like to know a reference where the abelianization of $\Gamma$ is described. Can one find an explicit non-zero group homomorphism $\Gamma\to\mathbb Z$?
| https://mathoverflow.net/users/50457 | Abelianizations of arithmetic Fuchsian groups | Let $\mathbb H^2$ be the hyperbolic plane, on which $\Gamma$ acts properly discontinuously and cocompactly. Thus the quotient space $O = \Gamma \backslash \mathbb H^2$ is a closed hyperbolic surface with conical singularities. If these have angles $2\pi/q\_1, \ldots, 2\pi/q\_s$ then a presentation for $\Gamma$ is given by
$$ \Gamma \cong \langle a\_1,b\_1 \ldots, a\_g,b\_g, c\_1, \ldots, c\_s | [a\_1, b\_1] \cdots [a\_g, b\_g] \cdot c\_1 \cdots c\_s,\, c\_1^{q\_1}, \ldots, c\_s^{q\_s} \rangle$$
from which the abelianisation is immediate to read.
There is no closed formula for the invariants $g, q\_1, \ldots, q\_s$. They can be computed in a given example by using software developed by John Voight (see the answer to this related question : [How do you find the genus of a Fuchsian group derived from a quaternion algebra?](https://mathoverflow.net/questions/108517/how-do-you-find-the-genus-of-a-fuchsian-group-derived-from-a-quaternion-algebra?rq=1)).
There is also an asymptotic relation between $g$ and the covolume of $\Gamma$. Namely, if $\mathrm{Vol}(O)$ is the hyperbolic volume of $O$ then it holds that
$$ g \sim \frac{\mathrm{Vol}(O)}{4\pi},\: s, q\_i = o(\mathrm{Vol}(O)$$
as $\mathrm{Vol}(O) \to +\infty$ (this follows from work of Mikołaj Frączyk, see <https://arxiv.org/abs/1810.10515> ; you could even get effective bounds by being more careful). In particular this recovers the result of Long--Maclachlan--Reid in Ian Agol's comment that the abelianisation is infinite for almost all $\Gamma$, since the volume is a proper function of $(a, b)$ (there is a closed formula but you need first to compute the discriminant of the quaternion algebra from the Hilbert symbol, which at this moment i'm not exactly sure how to do in general).
I'm not aware of any general construction of a non-trivial morphism from $\Gamma$ to $\mathbb Z$. Maybe the Jacquet--Langlands correspondence could be used for that, though i'm not sure how explicit this would be.
| 2 | https://mathoverflow.net/users/32210 | 424143 | 172,319 |
https://mathoverflow.net/questions/424100 | 3 | This is my first overflow question, so let me apologize in advance if this question is too low level.
I was asking the same question in stackexchange, but didn't get an answer; check [here](https://math.stackexchange.com/questions/4465381/why-does-the-fourier-algebra-ag-consits-precisely-with-the-set-of-matrix-coe) for details.
Short: The Fourier algebra $A(G)$ of a locally compact group $G$ is the closed linear span of matrix coefficients of the left regular representation $\lambda$ (w.r.t. some certain norm). As one can show, $A(G)$ is isometric isomorphic the predual (=normal functionals =ultraweakly continuous functionals) of the group vn-algebra $VN(G)$.
Now Thm 2.4.3. in Kanituh-Lau (check link for details) asserts that $A(G)$ consists precisely of the matrix coefficients of $\lambda$. This doesn't seem to clear for me and I don't really get the proof, especially the reference to Dixmier.
EDIT:
For the proof Kanituh-Lau start with a second countable lc $G$, conclude that $L^2(G)$ is separable [clear] and that $VN(G)$ is countably generated [also clear]. Then they assert that every normal positive functional of $VN(G)$ is of the form $T \mapsto\langle T\xi,\xi \rangle$ and conclude that then any normal functional is of the form $T\mapsto \langle T\xi,\eta \rangle$ [why??] with a reference to Dixmier, C\*-algebras, which most certainly is a typo. So you will already help me out with a reliable reference, why
(a) Any normal positive functional on $VN(G)$ is of the form $T \mapsto\langle T\xi,\xi \rangle$
and why
(b) This is enough to show that any normal functional on $VN(G)$ is of the form $T\mapsto \langle T\xi,\eta \rangle$.
I guess (a) implies (b) is not that hard and is based on some polarization/ Jordan decomposition type argument. But proving (a) in the first place is where I'm stuck.
If both (a) and (b) are done, then we have $T\mapsto \langle T\xi,\eta \rangle=\varphi\_T(\overline{\eta}\*\breve{\xi})$, where $\varphi\_T$ is the corresponding functional on $A(G)$ induced by the isomorphism of $A(G)^\*$ and $VN(G)$, and $\overline{\eta}\*\breve{\xi}$ is the desired element of $A(G)$.
| https://mathoverflow.net/users/483532 | Why does the Fourier algebra $A(G)$ consist precisely of the set of matrix coefficients of the LRR? | The key concept you need is of a von Neumann algebra being in "standard form". The classical proof, as given by Eymard, and which Kanituh-Lau seem to follow, is to first use that for a second countable $G$, $VN(G)$ has a separating vector in $L^2(G)$, from which it follows that all normal positive functionals have the form $x\mapsto (x\xi|\xi)$. Then some locally compact group theory can boost this up to the general case. As far as I can tell, the Kanituh-Lau presentation is basically identical, excepting their erroneous reference. In any case, the heavy lifting is performed by von Neumann algebra theory, and most textbooks carry the material you need, if you look hard enough.
A more modern approach is perhaps to follow the 2nd volume of Takesaki, and use the general theory of left Hilbert Algebras. In particular, Chapter VII, Section 3 treats the Fourier algebra and group von Neumann algebra. Lemma 3.7 then gives you what you want. However, even here, the key result is delegated to Chapter V, Theorem 3.15, which follows the "separating vector" ideas. However, I think the idea of Chapter IX, Section 1, about "standard forms" put things in a nice context.
For a more gentle approach, I really like the thesis of Zwarich, [Von Neumann Algebras for Abstract Harmonic Analysis](https://uwspace.uwaterloo.ca/handle/10012/3920). See in particular section 4.3 (but the rest of the thesis develops most of the machinery you need in a self-contained way). Note however that Zwarich uses "quasi-Hilbert algebras" (see Defn 4.3.3) which are not the same a left Hilbert Algebras (I *think*).
| 4 | https://mathoverflow.net/users/406 | 424150 | 172,322 |
https://mathoverflow.net/questions/424070 | 0 | Let $H=(V,E)$ be a [hypergraph](https://en.wikipedia.org/wiki/Hypergraph). We say that $C\subseteq V$ is a *choice set* if $|C\cap e| = 1$ for all $e\in E$.
**Question.** Let $H=(V,E)$ be a hypergraph with $e$ finite for all $e\in E$, and suppose that for all finite sets $E\_0\subseteq E$ the hypergraph $(V, E\_0)$ has a choice set. Does $H$ itself necessarily have a choice set?
| https://mathoverflow.net/users/8628 | Choice sets in hypergraphs with finite edges | For finite edges this follows indeed from compactness. Consider the Cartesian product $K=\prod e$ of all edges in Tychonoff topology. This $K$ may be understood as the set of simultaneous choices $v\_e\in e$ for all edges $e$. For every pair $e\ne f$ of edges consider the open set $U(e,f)\subset K$ determined as follows: a vertex chosen in $e$ belongs to $f$. If these open sets do not cover $K$, then a choice set exists. If they do cover $K$, then finitely many of them cover $K$, that means that the choice set does not exist for finitely many edges.
| 3 | https://mathoverflow.net/users/4312 | 424153 | 172,323 |
https://mathoverflow.net/questions/424158 | 3 | By $[\omega]^\omega$ we denote the collection of infinite subsets of $\omega$. Two sets $A,B\in[\omega]^\omega$ are said to be *almost disjoint* if $A\cap B$ is finite. An *almost disjoint family* is a set ${\cal A}\subseteq [\omega]^\omega$ in which every two distinct members are almost disjoint. A standard application of Zorn's Lemma shows that any almost disjoint family is contained in a maximal almost disjoint (MAD) family (maximal with respect to $\subseteq$).
A "pathological" MAD family is $\{E, \omega\setminus E\}$ where $E = \{2n:n\in \omega\}$. We will consider infinite MAD families only. (A diagonalisation argument shows that every infinite MAD family is uncountable.)
**Question.** Is there an infinite MAD family ${\cal M}\subseteq [\omega]^\omega$ with $\bigcap {\cal M} = \emptyset$ and a set $R\subseteq \omega$ such that $|R\cap M| = 1$ for all $M\in {\cal M}$?
| https://mathoverflow.net/users/8628 | MAD family with the choosability property | This answer only deals with the case that **$R$ is infinite.** I thought that I would be able to modify it to the finite case - thanks to Ilya Bogdanov for spotting the mistake in my argument. (His answer shows that for finite $R$ such family indeed exists.) And thanks to bof for explaining in a comment that my original argument was unnecessarily complicated.
---
$\newcommand{\mc}[1]{\mathcal{#1}}$Let us assume that $R$ and $\mc M$ fulfill the conditions given in the question. Moreover, let $R$ be an infinite set.
Then the system $\{R\}\cup\mc M$ is an almost disjoint family. (For every $M\in \mc M$, the intersection $R\cap M$ is finite.) Thus from maximality we get $R\in\mc M$.
But now $|R\cap R|=|R|\ne 1$, contradicting "choosability".
---
The above argument can be shortly summarized as follows: If an infinite sets has a finite intersection with each element of a MAD family $\mc M$, then this set belongs to $\mc M$.
| 6 | https://mathoverflow.net/users/8250 | 424162 | 172,327 |
https://mathoverflow.net/questions/422192 | 2 | Define the Jaccard distance between two continuous vectors $a, b\in [0,1]^p$ as
\begin{equation}
J(a,b) = 1 - \frac{\|a\odot b\|\_1}{\|a\odot b\|\_1+\|a-b\|\_1}
\end{equation}
where $\odot$ is the Hadamard product (element-wise product).
Is it a metric? Note that $a,b \in [0,1]^p$ rather than $\{0,1\}^p$.
I've tried with the naive approach. After some messy algebra, I need to prove the following
\begin{equation}
\|a-b\|\_1\|a\odot c\|\_1\|b\odot c\|\_1 + \|a-b\|\_1\|b-c\|\_1\|a\odot c\|\_1 + \|a-b\|\_1 \|b-c\|\_1 \|a-c\|\_1 + \|b-c\|\_1\|a\odot b\|\_1 \|a\odot c\|\_1 + \|a-b\|\_1\|b-c\|\_1\|a\odot c\|\_1 \geq \|a-c\|\_1\|a\odot b\|\_1 \|b\odot c\|\_1.
\end{equation}
| https://mathoverflow.net/users/482033 | Is the Jaccard distance between continuous vectors a metric? | All norms are supposed to be $1$-norms. Rewrite $J(a,b)$ as
$$
J(a,b)=\frac{\|a-b\|}{\|a-b\|+\|a\odot b\|}.
$$
Notice that
$$
\|b\odot c\|=\sum\_i b\_ic\_i
=\sum\_i\bigl((b\_i-a\_i)c\_i+a\_ic\_i\bigr)
\leq\sum\_i|b\_i-a\_i|+\sum\_ia\_ic\_i=\|a-b\|+\|a\odot c\|.
$$
Similarly,
$$
\|a\odot b\|\leq \|b-c\|+\|a\odot c\|.
$$
Hence we have
$$
J(a,b)+J(b,c)
=\frac{\|a-b\|}{\|a-b\|+\|a\odot b\|}+\frac{\|b-c\|}{\|b-c\|+\|b\odot c\|}\\
\geq \frac{\|a-b\|}{\|a-b\|+\|b-c\|+\|a\odot c\|}
+\frac{\|b-c\|}{\|b-c\|+\|a-b\|+\|a\odot c\|}\\
=\frac{\|a-b\|+\|b-c\|}{\|a-b\|+\|b-c\|+\|a\odot c\|}
\geq \frac{\|a-c\|}{\|a-c\|+\|a\odot c\|}=J(a,c),
$$
where the last inequality follows from $\|a-b\|+\|b-c\|\geq \|a-c\|$.
| 2 | https://mathoverflow.net/users/17581 | 424164 | 172,328 |
https://mathoverflow.net/questions/424159 | 4 | Given any $\epsilon > 0$, are there infinitely many $(a,b) \in \mathbb{Z}^2$ with $(a,b) = 1$ such that $$\left|\pi - \frac{a}{b}\right| < \frac{\epsilon}{b^2}?$$
According to [this document](https://math.osu.edu/sites/math.osu.edu/files/What_is_2018_Markov_Lagrange_Spectra.pdf), if we prove that $\pi$ has unbounded coefficients in its continued fraction expansion, then the answer to the above question is affirmative.
| https://mathoverflow.net/users/134295 | Markov constant of $\pi$ | I believe that this is an open problem. See e.g. the bottom of page 202 of [this article](https://www.maa.org/sites/default/files/pdf/pubs/BaileyBorweinPiDay.pdf) by Bailey and Borwein. BTW the question was asked here [before](https://mathoverflow.net/questions/99300/is-pi-well-approximable).
| 4 | https://mathoverflow.net/users/11919 | 424166 | 172,329 |
https://mathoverflow.net/questions/424140 | 3 | A linear process $(X\_t)\_{t \in \mathbb{Z}}$ is usually written as a moving-average process with infinity order:
\begin{equation}\label{linear\_process}\tag{Eq. 1.1}
X\_{t} = \sum\_{j =0 }^\infty \psi\_{j} \varepsilon\_{t-j}, \forall t \in \mathbb{Z}.
\end{equation}
where $\varepsilon\_{t}$ is a i.i.d. white noise ($E(\varepsilon\_{t})=0,\,\, E |\varepsilon\_{t}|^2< \infty$) and $\sum\_{j =0 }^\infty \psi\_{j}^2 < \infty$.
According to this [paper](https://www.pnas.org/doi/pdf/10.1073/pnas.93.22.12128), page 12130, the author says:
Mallows (12) argues that a linear process such as in (\ref{linear\_process}) is close to a Gaussian process if $\max\_{j\geq 0}|\psi\_j|$ is small.
I would like to know if you have any relatively simple examples for this statement. I tried to think of a simple example, but I couldn't. I don't want to go to the Mallows paper before I go through here.
| https://mathoverflow.net/users/479236 | Linear process close to a Gaussian process | $\newcommand{\R}{\mathbb R}\newcommand{\Z}{\mathbb Z}\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}$Let $\psi\_j:=0$ for $j=-1,-2,\dots$. Then
\begin{equation\*}
X\_t=\sum\_{j\in\Z}X\_{t,j}
\end{equation\*}
for $t\in\Z$, where
\begin{equation\*}
X\_{t,j}:=\psi\_{t-j}\ep\_j.
\end{equation\*}
Let
\begin{equation\*}
B:=\sqrt{\sum\_{j\in\Z} \psi\_i^2},\quad m:=\max\_{j\ge0}|\psi\_j|=\max\_{j\in\Z}|\psi\_j|.
\end{equation\*}
Suppose that $B>0$ and $m$ vary in any manner such that
\begin{equation\*}
m/B\to0. \tag{1}\label{1}
\end{equation\*}
Let us show that then $X\_t/B$ converges in distribution to a standard normal random variable, for each $t\in\Z$.
---
For each real $\de>0$,
\begin{equation\*}
\begin{aligned}
L&:=\frac1{B^2}\sum\_{j\in\Z}EX\_{t,j}^2\,1(|X\_{t,j}|\ge\de B) \\
&=\frac1{B^2}\sum\_{j\in\Z}E(\psi\_{t-j}\ep\_j)^2\,1(|\psi\_{t-j}\ep\_j|\ge\de B) \\
&\le\frac1{B^2}\sum\_{j\in\Z}\psi\_{t-j}^2E\ep\_j^2\,1(|\ep\_j|\ge\de B/m) \\
&=\frac1{B^2}\sum\_{j\in\Z}\psi\_{t-j}^2E\ep\_0^2\,1(|\ep\_0|\ge\de B/m) \\
&=E\ep\_0^2\,1(|\ep\_0|\ge\de B/m)\to0.
\end{aligned}
\end{equation\*}
Hence,
\begin{equation\*}
\frac1{B^2}\sum\_{j\in\Z}EX\_{t,j}^2\,1(|X\_{t,j}|<\de B)=1-L\to1,
\end{equation\*}
\begin{equation\*}
\begin{aligned}
&\frac1{B^2}\sum\_{j\in\Z}(EX\_{t,j}\,1(|X\_{t,j}|<\de B))^2 \\
&=\frac1{B^2}\sum\_{j\in\Z}(EX\_{t,j}\,1(|X\_{t,j}|\ge\de B))^2
\le L\to0,
\end{aligned}
\end{equation\*}
\begin{equation\*}
\begin{aligned}
&\Big|\frac1B\sum\_{j\in\Z}EX\_{t,j}\,1(|X\_{t,j}|<\de B)\Big| \\
&=\Big|\frac1B\sum\_{j\in\Z}EX\_{t,j}\,1(|X\_{t,j}|\ge\de B)\Big| \\
&\le\frac1B\sum\_{j\in\Z}E|X\_{t,j}|\,1(|X\_{t,j}|\ge\de B)\
\le \frac L\de\to0,
\end{aligned}
\end{equation\*}
\begin{equation\*}
\sum\_{j\in\Z}P(|X\_{t,j}|\ge\de B)\le L\to0.
\end{equation\*}
So, by [Theorem 18 in Chapter IV](https://link.springer.com/book/10.1007/978-3-642-65809-9), $X\_t/B$ converges in distribution to a standard normal random variable, for each $t\in\Z$.
Thus, under condition \eqref{1}, all the one-dimensional distributions of the process $(X\_t)$ are asymptotically normal.
---
Similarly considered are all the finite-dimensional distributions of the process $(X\_t)$ -- that is, all the joint distributions of $(X\_{t\_1},\dots,X\_{t\_p})$ for integers $t\_1<\cdots<t\_p$. This is done by writing
\begin{equation\*}
\sum\_{i=1}^p c\_i X\_{t\_i}=\sum\_{j\in\Z}Y\_j
\end{equation\*}
for any real $c\_1,\dots,c\_p$, where
\begin{equation\*}
Y\_j:=\phi\_j\ep\_j,\quad\phi\_j:=\sum\_{i=1}^p c\_i \psi\_{t\_i-j},
\end{equation\*}
so that $\sum\_{j\in\Z}\phi\_j^2<\infty$ and $\max\_{j\in\Z}|\phi\_j|\le m\sum\_{i=1}^p |c\_i|$.
| 3 | https://mathoverflow.net/users/36721 | 424180 | 172,335 |
https://mathoverflow.net/questions/424173 | 9 | **Introduction:** We have a question of how to calculate the number of $n$-variables Laurent monomials of degree at most $d$.
**For example:** If $n=2$, $d=2$ then we have 19 monomials, which are:
$x^{-2}$, $x^{-2}y$, $x^{-2}y^2$,
$x^{-1}y^{-1}$, $x^{-1}$, $x^{-1}y$, $x^{-1}y^2$,
$y^{-2}$, $y^{-1}$, 1, $y$, $y^{2}$,
$xy^{-2}$, $xy^{-1}$, $x$, $xy$,
$x^2y^{-2}$, $x^2y^{-1}$, $x^2$. They correspond to the following 19 pairs of powers: $(-2,0)$, $(-2,1)$, $(-2,2)$,
$(-1,-1)$, $(-1,0)$, $(-1,1)$, $(-1,2)$,
$(0,-2)$, $(0,-1)$, $(0,0)$, $(0,1)$, $(0,2)$,
$(1,-2)$, $(1,-1)$, $(1,0)$, $(1,1)$,
$(2,-2)$, $(2,-1)$, $(2,0)$.
So how about, e.g., $n=4$, $d=4$ ?
**Formulation:**
Let $n,d\in\mathbb N$. Let
\begin{align\*}
\Omega(n,d) &= \{\alpha\in\mathbb N^n: \alpha\_1+\dotsb+\alpha\_n \leq d\},\\
\Gamma(n,d) &= \Omega(n,d) + \Omega(n,d),\\
\Delta(n,d) &= \Omega(n,d) - \Omega(n,d).
\end{align\*}
Then, the elements of $\Delta(n,d)$ correspond to the tuples of powers of the monomials.
The cardinality of $\Omega(n,d)$ is well-known to be $\binom{n+d}{n}$. The cardinality of $\Gamma(n,d)$ can be shown to be $\binom{n+2d}{n}$. However, is there a formula to compute the cardinality of $\Delta(n,d)$ ?
**Some results:** We did some attempts and got that
\begin{align\*}
|\Delta(n,1)| &= n(n+1)+1,\\
|\Delta(1,d)| &= 2d+1,\\
|\Delta(2,d)| &= (2d+1)^2 - d(d+1),\\
|\Delta(3,d)| &= (2d+1)^3 - \frac{7}{3}d(d+1)(2d+1),\\
|\Delta(4,d)| &= (2d+1)^4 - \frac{1}{12}d(d+1)(157d^2+157d+46).
\end{align\*}
We still haven't found a way to calculate $|\Delta(n,d)|$ in general. By definition of $\Delta(n,d)$ as above, I think we may have a geometric approach. An orientable suggestion would be nice.
| https://mathoverflow.net/users/483716 | Number of Laurent monomials of n variables with degree at most d | One such formula is
$$\sum\_{p=0}^n \binom{n}{p} \binom{d}{p} \binom{d+n-p}{n-p}.$$
To derive this, let $P \subseteq [n]$ be the set of variables with positive exponents and let $p = |P|$. There are $\binom{n}{p}$ ways to choose $P$. After choosing $P$, we must choose a monomial in $\{ x\_i \}\_{i \in P}$ of degree $\leq d$ where each variable has degree $\geq 1$; there are $\binom{d}{p}$ ways to do this. And we must choose a monomial in $\{ x\_i \}\_{i \in [n] \setminus P}$ of degree $\leq d$ where each variable has degree $\geq 0$; there are $\binom{d+n-p}{n-p}$ ways to do this. I think this is equivalent to Ira Gessel's formula in the comment below.
My previous answer is preserved below the line:
---
One such formula is
$$\sum\_{\substack{0 \leq p,q \leq d\\ p+q \leq n}} \binom{n}{p,q} \binom{d}{p} \binom{d}{q}$$
where $\binom{n}{p,q}$ is the trinomial coefficient $\tfrac{n!}{p! q! (n-p-q)!}$.
To see this, let $P$ and $Q \subset [n]$ be the sets of variables whose exponents are positive and negative. To choose a monomial in your set, first choose the index sets $P$ and $Q$, which we can do in $\binom{n}{|P|,|Q|}$ ways. Then choose two monomials, one in the variables $\{ x\_i \}\_{i \in P}$, and one in the variables $\{ x\_j \}\_{j \in Q}$, where each variable occurs to degree $\geq 1$ and the total degree is $\leq d$. The number of monomials in $p$ variables where each variable occurs to degree $\geq 1$ and the total degree is $\leq d$ is $\binom{d}{p}$. (Note that this is even correct when $d<p$.)
This isn't a particularly nice formula, but I don't see how to simplify it more. (EDIT: But Ira Gessel did; see his comment below.) Note that it is manifestly a polynomial in $d$, for fixed $n$.
| 15 | https://mathoverflow.net/users/297 | 424183 | 172,337 |
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