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https://mathoverflow.net/questions/445795 | 7 | These are fairly standard terms, but for the sake of completeness: An ultrafilter $\mathcal{U}$ on $\omega$ is a *p-point* if whenever $(A\_n)\_{n<\omega}$ is a partition of $\omega$ such that $A\_n \notin \mathcal{U}$ for all $n$, there is an $X \in \mathcal{U}$ such that $X \cap A\_n$ is finite for all $n$. $\mathcal{U}$ is *Ramsey* if the same holds but with $|X\cap A\_n| = 1$ for all $n$. Clearly Ramsey ultrafilters are p-point ultrafilters. The name Ramsey comes from the fact that if $\mathcal{U}$ is a Ramsey ultrafilter, then any graph on $\omega$ has a homogeneous set in $\mathcal{U}$.
Fix an edge relation $E$ on $\omega$ making it into a copy of the random graph. (For concreteness, we could say that $nEm$ if and only if the $\min(n,m)$th binary bit of $\max(n,m)$ is $1$.) A remarkable property of the random graph is that for any finite partition $X\_0,X\_1,\dots,X\_{n-1}$ of $\omega$, there is an $i<n$ such that $(X\_i,E)$ is isomorphic to $(\omega,E)$. (See Proposition 3 [here](https://arxiv.org/abs/1301.7544).) This implies the following.
>
> **Proposition.** There is an ultrafilter $\mathcal{U}$ on $\omega$ with the property that for any $X \in \mathcal{U}$, there is a $Y \subseteq X$ such that $(Y,E)$ is isomorphic to $(\omega,E)$.
>
>
>
*Proof.* For each finite partition $P$ of $\omega$, let $F\_P$ be the (clopen) set of ultrafilters $\mathcal{U}$ in $\beta \omega \setminus \omega$ satisfying that for the unique $X$ in $\mathcal{U} \cap P$, there is a $Y \subseteq X$ such that $(Y,E)$ is isomorphic to $(\omega,E)$. By the above fact, each $F\_P$ is non-empty. The family $\\{F\_P:P\text{ a finite partition of }\omega\\}$ has the finite intersection property. To see this, note that if $P\_0,P\_1,\dots,P\_{n-1}$ are finite partitions of $\omega$, then for any mutual refinement $Q$ of the $P\_i$'s, $F\_Q \subseteq \bigcap\_{i<n} F\_{P\_i}$. Therefore, by compactness, there is an ultrafilter $\mathcal{U}$ such that for every $P$, $\mathcal{U} \in F\_P$. For any set $X \in \mathcal{U}$, we have that $\mathcal{U} \in F\_{ \\{X,\omega\setminus X\\} }$, so $X$ has the required property. $\square$
Clearly, any such ultrafilter cannot be Ramsey, since an $E$-homogeneous set will contain no subset isomorphic to the random graph. Call such an ultrafilter *aggressively non-Ramsey*.
>
> **Question.** Is it consistent with $\mathsf{ZFC}$ that there is an aggressively non-Ramsey p-point?
>
>
>
| https://mathoverflow.net/users/83901 | Can there be a p-point ultrafilter that is 'aggressively non-Ramsey'? | No, such a filter cannot exist. Suppose $\mathcal U$ is a $p$-point. For $s\in 2^{<\omega}$, let $A\_s$ consist of all $m<\omega$ such that
$$\forall n\in\mathrm{dom}(s)\ (nEm\Leftrightarrow s(n)=1).$$
Note that there is a unique $x\in 2^\omega$ so that $A\_{x\upharpoonright n}\in \mathcal U$ for all $n<\omega$. As $\mathcal U$ is a $p$-point, there is $B\in\mathcal U$ with $B\subseteq^\ast A\_{x\upharpoonright n}$ for all $n<\omega$. The graph $(B, E)$ has the following property: For all $n\in B$, all but maybe finitely many $m\in B$ satisfy
$$n E m\Leftrightarrow x(n)=1.$$
This easily implies that there is no $C\subseteq B$ so that $(C,E)$ is (isomorphic to) the random graph, so $\mathcal U$ is not aggresively non-Ramsey.
| 3 | https://mathoverflow.net/users/125703 | 445817 | 179,654 |
https://mathoverflow.net/questions/445809 | 5 | I am studying Paley-Littlewood theorem in Harmonic analysis, and I met an exercise. I would like to construct a function $f$ as a counterexample to show that the inequality
\begin{equation}
\| f \|^2\_p \leq C \sum\_{j \in \mathbb{Z}} \| P\_j f\|^2\_p,
\end{equation}
is not true when $p<2,$ and here $C$ is a constant number only related to $f.$ Here,$P\_j$ is the Paley-Littlewood operator. More explicitly, for $\beta(x) \in C\_0^{\infty}(\mathbf{R}^n)$ and ${\rm supp}\,{
\beta} \subset \{1/2<|x|<2 \},$ and satisfying $1=\sum\_{j \in \mathbb{Z}} \beta(2^{-j}x), x \neq 0.$ We define the Fourier transform of $P\_j f$ is $\beta(2^{-j}\xi)\hat{f}(\xi).$
My attempt is, the function $f$ has the following form: $f(x)=e^{2 \pi imx }\phi(x),$ here $m$ is a constant number to be determined, and $\phi(x)$ is a Schwartz function. If $m$ is properly selected, then for some $P\_j$, we have $P\_j f=0.$ But I am stuck here.
Can anybody help me? Some suggestions or reference books are welcomed!
| https://mathoverflow.net/users/503783 | How to give a counterexample of this estimate related to Paley-Littlewood theorem? | This question is really about non-coincidence of different function spaces: the right hand side in your inequality is equal to the the square of the norm in (homogeneous) Besov space $\dot{B}\_{p}^{0,2}$. And by Littlewood--Paley theorem, the left-hand side is equivalent to the (square of) norm in Tribel--Lizorkin space $\dot{F}\_{p}^{0,2}$ which is defined as
\begin{equation}
\Big(\int \Big( \sum\_j |P\_j f|^2 \Big)^{p/2}\Big)^{2/p},
\end{equation}
which is bigger than
$$
\sum\_j \|P\_j f\|^2\_p
$$
if $p<2$ by Minkowski's integral inequality. So, if such inequality was true then we would have $\|f\|\_{\dot{B}\_{p}^{0,2}}\asymp \|f\|\_{\dot{F}\_p^{0,2}}$.
The fact that different function spaces do not coincide is standard, different counterexamples can be found probably in many different books about function spaces, for example in the book by Sawano "Theory of Besov Spaces" (see Theorem 2.16 there).
| 7 | https://mathoverflow.net/users/69086 | 445818 | 179,655 |
https://mathoverflow.net/questions/445822 | 2 | Let $X\_0$ be a smooth projective variety over a finite field $\mathbb{F}\_q$. Let $X$ be the corresponding variety over the algebraic closure $\bar{\mathbb{F}}\_q$. Let $Fr\_q\colon X\to X$ be the geometric Frobenius.
**Is it true that the eigenvalues of $Fr\_q$ on $H^i(X,\mathbb{Q}\_l)$ ($l\ne char(\mathbb{F}\_q)$) are algebraic integers independent of $l$? Is the same true if $X\_0$ is not necessarily smooth and projective?**
A reference would be helpful.
**Remark.** I know that in the smooth projective case the eigenvalues are algebraic numbers (I do not know if they are integers).
Sorry if this question is well known to experts, I am not from this field.
| https://mathoverflow.net/users/16183 | Eigenvalues of Frobenius in $l$-adic cohomology | This was proven by Deligne in the smooth projective case [Weil I, Thm. I.6], and later in the smooth proper case [Weil II, Cor. 3.3.9].
In general (already for smooth *quasi*-projective varieties), we don't even know whether $\dim H^i(X\_{\text{ét}},\mathbf Q\_\ell)$ is independent of $\ell$. See for instance [Katz, p. 28, 2(a)] or [Illusie, 3.5(c)]. I also wrote a paper on this [vDdB] that may contain some more references.
---
**References.**
[Weil I] P. Deligne, [*La conjecture de Weil. I*](http://www.numdam.org/item/PMIHES_1974__43__273_0/). Publ. Math. Inst. Hautes Étud. Sci. **43**, p. 273-307 (1973). [ZBL0287.14001](https://zbmath.org/?q=an:0287.14001).
[Weil II] P. Deligne, [*La conjecture de Weil. II*](http://www.numdam.org/item/PMIHES_1980__52__137_0/), Publ. Math. Inst. Hautes Étud. Sci. **52**, p. 137-252 (1980). [ZBL0456.14014](https://zbmath.org/?q=an:0456.14014).
[Katz] N. M. Katz, *Review of $\ell$-adic cohomology*. In: [Motives](http://dx.doi.org/10.1090/pspum/055.1). Proc. Symp. Pure Math. **55**.1, p. 21-30 (1994). [ZBL0817.14006](https://zbmath.org/?q=an:0817.14006).
[Illusie] L. Illusie, [*Miscellany on traces in $\ell$-adic cohomology: a survey*](https://doi.org/10.1007/s11537-006-0504-3). Jpn. J. Math. (3) **1**.1, p. 107-136 (2006). [ZBL1156.14309](https://zbmath.org/?q=an:1156.14309).
[vDdB] R. van Dobben de Bruyn, [*The equivalence of several conjectures on independence of $\ell$*](https://doi.org/10.46298/epiga.2020.volume4.5570). Épijournal de Géom. Algébr. **4**, Article No. 16 (2020). [ZBL1460.14053](https://zbmath.org/?q=an:1460.14053).
| 7 | https://mathoverflow.net/users/82179 | 445825 | 179,658 |
https://mathoverflow.net/questions/445735 | 6 | Consider the [Vandermonde's determinant](https://en.wikipedia.org/wiki/Vandermonde_matrix) computed by
$$V(x\_1,\dots,x\_m):=\det(x\_j^{i-1})\_{i,j=1}^m=\prod\_{1\leq i<j\leq m}(x\_i-x\_j).$$
The number of [plane partitions](https://en.wikipedia.org/wiki/Plane_partition) in an $n\times m\times m$ box (MacMahon) is given by the neat product formula
$$\prod\_{i=1}^n\prod\_{j=1}^m\prod\_{k=1}^m\frac{i+j+k-1}{i+j+k-2}.$$
**Notation.** Let $\mathcal{F}\_m$ be the set of all $m$-element subsets of $[n+m]=\{1,2,\dots,n+m\}$. Write $\mathbf{J}=\{j\_1,\dots,j\_m\}$ for $\mathbf{J}\in\mathcal{F}\_m$. The special element $\{1,2,\dots,m\}\in\mathcal{F}\_m$ is denoted by $\mathbf{I}$, in which case $V(\mathbf{I})=(m-1)!!=1!\cdot2!\cdots(m-1)!$.
I would like to ask:
**QUESTION.** Is this expansion true? A combinatorial proof is desired, if possible.
$$\prod\_{i=1}^n\prod\_{j=1}^m\prod\_{k=1}^m\frac{i+j+k-1}{i+j+k-2}
=\sum\_{\mathbf{J}\in\mathcal{F}\_m}\left(\frac{V(\mathbf{J})}{V(\mathbf{I})}\right)^2.$$
**Remark.** Observe the cute fact $\frac{V(\mathbf{J})}{V(\mathbf{I})}$ is always an integer (let's make Fedor's comment explicit).
**Proof.** Replacing $\mathbf{x}=(x\_1,\dots,x\_m)$ by $\mathbf{J}=(j\_1,\dots,j\_m)$ into
$$\det\left[\binom{x\_j}i\right]
=\frac1{\prod\_{i<j}(j-i)}\det\left[\prod\_{k=0}^{i-1}(x\_j-k)\right]
=\frac{\det\left[x\_j^{i-1}\right]}{\prod\_{i<j}(j-i)}
=\prod\_{1\leq i \lt j\leq m} \frac{x\_j-x\_i}{j-i}.$$
results in integer entries, and thus integer determinant $\frac{V(\mathbf{J})}{V(\mathbf{I})}$. $\qquad\square$
| https://mathoverflow.net/users/66131 | Plane partitions as sums of determinants | I haven't worked out the details, but $V(\mathbf{J})/V(\mathbf{I})$ is
the principal specialization of a Schur function. Then $\left(
\frac{V(\mathbf{J})}{V(\mathbf{I})}\right)^2$ corresponds to a pair of
SSYT (semistandard Young tableaux), which can be merged into a plane
partition as in EC2, proof of Theorem 7.20.1. Summing over
$\mathbf{J}$ should give all plane partitions fitting in an $n\times
m\times m$ box.
| 7 | https://mathoverflow.net/users/2807 | 445834 | 179,661 |
https://mathoverflow.net/questions/445841 | 2 | [This answer](https://mathoverflow.net/a/55356/148161) states that the category of finitary monads is locally presentable and monadic over the category $\mathrm{Set}^{\mathbb{N}}$. Where can I find proof of this claim?
More generally: I've looked through the literature in nlab ([finitary monad](https://ncatlab.org/nlab/show/finitary+monad), [monad](https://ncatlab.org/nlab/show/monad#References), [Lawvere theory](https://ncatlab.org/nlab/show/Lawvere+theory#references)) and found nothing about this category other than a proof of its equivalence with the category of Lawvere theories (written in several places). I would welcome links to any literature that discusses this category in more detail.
| https://mathoverflow.net/users/148161 | Literature about the category of finitary monads | These claims are proven more generally for the category $\mathrm{Mnd}\_f(\mathscr A)$ of finitary monads on a locally presentable category $\mathscr A$ in Lack's [On the monadicity of finitary monads](https://www.sciencedirect.com/science/article/pii/S0022404999000195). (This is the same Lack as in the linked answer.)
| 5 | https://mathoverflow.net/users/152679 | 445844 | 179,663 |
https://mathoverflow.net/questions/445849 | 2 | **Motivation.** Today my sons played a card game, in which a fixed number $n$ of cards was lying on the table. A move consists of adding an unused card to the cards on the table, and removing a card from the table, so that after the move, there are again $n$ cards on the table. Which made me ponder the following question.
**Formalization.** For any integer $n>1$ let $[\mathbb{N}]^n$ be the collection of subsets of $\mathbb{N}$ consisting of $n$ elements. For which integers $n>1$ is there a bijection $\varphi:\mathbb{N}\to [\mathbb{N}]^n$ such that for all $k\in\mathbb{N}$ we have $|\varphi(k) \cap \varphi(k+1)| = n-1$?
| https://mathoverflow.net/users/8628 | Inspired by a card game: finding a path through $[\mathbb{N}]^n$ | $[\mathbb{N}]^n$, with edges between $a,b\in[\mathbb{N}]^n$ if $\#(a\cap b)=n-1$, is an infinite graph in which all vertices have infinite degree. Moreover, for any two vertices $a,b$ in $[\mathbb{N}]^n$ and any finite subset $E\subseteq[\mathbb{N}]^n$ not containing $a,b$, there is a path from $a$ to $b$ which does not pass through any vertices of $E$: to find it, one can consider some $N\in\mathbb{N}$ bigger than all natural numbers appearing in elements of $E$, and then you go from $a$ to $\{N+1,\dots,N+n\}$ and from $\{N+1,\dots,N+n\}$ to $b$ in $\leq 2n$ steps.
So one can create a Hamiltonian path in the following way:
* Number the vertices $u\_1,u\_2,\dots$ in $[\mathbb{N}]^n$.
* Start the path in the vertex $u\_1$.
* At each step, take the first vertex that you have not visited and visit it using a path that does not pass through any previously visited vertices.
| 7 | https://mathoverflow.net/users/172802 | 445852 | 179,665 |
https://mathoverflow.net/questions/445833 | 1 | In a previous post [Lift chain complex from $\mathbb{F}\_2$ to $\mathbb{Z}$](https://mathoverflow.net/questions/163346/lift-chain-complex-from-mathbbf-2-to-mathbbz) the body of the question mentions that this (lifting a chain complex from $\mathbb Z/2\mathbb Z$ to $\mathbb Z$) is always possible :
"Now, this can always be done (for example, by using the structure theorem for chain complexes over $\mathbb F\_2$)."
What would be a constructive way of doing this?
(Unlike the original post, I'm removing the other two restrictions.)
| https://mathoverflow.net/users/16739 | How to lift a chain complex from $\mathbb{Z}/2\mathbb{Z}$ to $\mathbb{Z}$ | I presume “constructive” means a computational algorithm is desired.
Assuming the chain complex $F$ over $\def\Z{{\bf Z}}\Z/2$ is bounded from below, we are going to construct by induction on $n$ a basis $E\_n=A\_n⊔B\_n⊔C\_n$ of $F\_n$ with the following properties:
* $A\_n$ is a basis of exact elements in $F\_n$;
* $A\_n⊔B\_n$ is a basis of closed elements in $F\_n$.
For sufficiently small $n$ we have $F\_n=0$, which gives $A\_n=B\_n=C\_n=∅$.
If $A\_{n-1}$, $B\_{n-1}$, $C\_{n-1}$ have already been constructed,
take $A\_n$ to be the image of $C\_{n-1}$ under the differential.
Then extend $A\_n$ to a basis $A\_n⊔B\_n$ of closed elements in $F\_n$ using Gaussian elimination over $\Z/2$. Finally, extend $A\_n⊔B\_n⊔C\_n$ to a basis of $F\_n$ using Gaussian elimination again.
Now it is easy to lift $F$ to a chain complex over $\Z$: take the same basis elements and construct the corresponding differentials over $\Z$ as matrices with respect to these bases, where all entries are zero except for the entries corresponding to elements of $C\_{n-1}$ mapping to elements of $A\_n$, which we take to be 1.
| 4 | https://mathoverflow.net/users/402 | 445854 | 179,666 |
https://mathoverflow.net/questions/445861 | 2 | When I look at the count of distinct least prime factors for a range of consecutive integers, I am seeing the same minimum number appear again and again. I am wondering if this number represents the true minimum.
Consider a range of consecutive integers defined by $R(x+1,x+c) = x+1, x+2, x+3, \dots, x+c$ with $C(x+1,x+c)$ = the count of distinct least prime factors for $R(x+1,x+c)$
Let $p\_k$ be the the $k$th prime which is the greatest prime less than or equal to $\left\lfloor\frac{c}{2}\right\rfloor$
I am finding that for $x \ge 1$, $c \ge 4$, the mimimum $C(x+1,x+c)$ that I can find is $k+2$.
Consider $c=61$, with $\left\lfloor\frac{61}{2}\right\rfloor = 30$ and the greatest prime less than $30$ is $p\_{10}=29$. Using the Chinese Remainder Theorem, I can find $x$ where the count of distinct least prime factors is $10+2=12$. In this case, $x=210504408624479$
Here is another example. consider $c=225$, with $\left\lfloor\frac{225}{2}\right\rfloor = 112$ and the greatest prime less than $112$ is $p\_{29}=109$. In this case, the minimum that I find is $29+2=31$ with $x=2082073176015230647073633038337038148767197882834487$.
I have a simple java application that allows me to find $x$ for a given $c$ with $k+2$ distinct least prime factors.
Is there a counter example that shows an $x,c$ where $C(x+1,x+c) < k+2$?
---
Edit: Added $x \ge 1$ and $c \ge 4$ to clarify bounds.
| https://mathoverflow.net/users/15915 | Estimating the minimum number of distinct least prime factors found in range of $c$ consecutive integers | I think that the statement "I am finding that for any $x$, any $c$, the mimimum $C(x+1,x+c)$" should be reformulated, clarifying that if we set the value of $x$, then $c$ is free to run on its domain and vice versa, by specifying also which is the aforementioned domain of the pair $(x,c)$ (since the closed interval of positive integers $C$ depends on both $x$ and $c$), otherwise we could simply take $c:=2$ and conclude that it does not exist any pair of consecutive integers containing $k+2$ prime numbers, since $\min(k) : p\_k \in \mathbb{P} = 1$ and then $\min\_k(p\_k)+2 \geq 3$ if $k \in \mathbb{Z}^+$ by hypothesis.
Basically, I have constructed my counterexample here by simply taking $\overline{c} := 2$ so that $\left\lfloor\frac{\overline{c}}{2}\right\rfloor = \left\lfloor\frac{2}{2}\right\rfloor = 1$ and it follows that $[x+1, x+2]$ cannot contain more primes than the cardinality of the set {x+1, x+2} itself (i.e., two), but it is trivial to point out that $\nexists p\_k \in \mathbb{P} : p\_k \leq \left\lfloor\frac{\overline{c}}{2}\right\rfloor$, while for $c:=4$ we can find $k+2=3$ primes if $x=1$ (i.e., by considering the peculiar set {2, 3, 4, 5} such that $p\_k=p\_1=2$ is (least or) equal to $\left\lfloor\frac{4}{2}\right\rfloor$).
Anyway, I am starting to think about Rosser's Theorem (or its further improvements as Dusart's bound), maybe it would just be enough to give you a proper answer (since $p\_k > k \cdot k(\log(k))$ holds for any $k$ as above).
| 1 | https://mathoverflow.net/users/481829 | 445873 | 179,673 |
https://mathoverflow.net/questions/445874 | 0 | Let $n$ be a positive integer, and $r:=\frac{p}{q}<1$ where $\mathrm{gcd}(p,q)=1$.
I am interested in the product $n\cdot r$
Whenever $n$ is a multiple of $q$, a property of rational numbers is that
$$
\{n\cdot r\}=0
$$
where $\{.\}$ represents the fractional part of a number.
I am interested in studying the local and global minima of the fractional part of $n\cdot r$ as $n$ increases. If we were to plot $\{n\cdot r\}$ vs $n$, we would observe a periodic pattern.
My question is as follow: For what values of $r$ or what is the condition on $r$ such that all the local minima of $\{n\cdot r\}$ are global? Has this been studied before?
For example $r=0.75$ and $n$ is an increasing positive integer. We notice that all the local minima in the fractional part are zero, where as for $r=0.55$, we see that there are local minima that are not global.
| https://mathoverflow.net/users/393675 | On the existence of locals that are global in $np\;\mathrm{mod}\;q$ | Local minima which are not global minima will exist for all rational numbers $r=\frac{p}{q}$ as long as $p\neq 1,q-1$. Indeed, by [Bézout's identity](https://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity), there is some $n$ such that $\{nr\}=\frac{1}{q}$, and except in those two cases above, both $(n-1)r$ and $(n+1)r$ will have greater fractional part, as it will be of the form $\frac{k}{q}$ for $k$ which cannot be $0$ or $1$.
Though this is beyond the scope of your question, let me mention for completeness that such non-global local minima will also occur whenever $r$ is irrational. Indeed, for irrational $r$ we have that $\{nr\}$ is dense in $(0,1)$, so in particular it has no global minima, and it will take value between $0$ and $\min(r,1-r)$, which you can check will necessarily be a local minimum.
| 2 | https://mathoverflow.net/users/30186 | 445876 | 179,674 |
https://mathoverflow.net/questions/445789 | 2 | Let $f:X\rightarrow Y$ be a morphism of smooth projective varieties, and $NE(X),NE(Y)$ the Mori cones of curves of $X$ and $Y$. Assume that $NE(X)$ is finitely generated. Then is $NE(Y)$ finitely generated as well?
If $f$ is birational I think the answer is positive. Let $C\subset Y$ be an irreducible curve and $\Gamma$ its strict transform in $X$. Then $\Gamma\sim a\_1\Gamma\_1+\dots + a\_r\Gamma\_r$ with $a\_1,...,a\_r\geq 0$ and where $\Gamma\_1,\dots,\Gamma\_r$ are the generators of $NE(X)$. So $C\sim a\_1f\_{\*}\Gamma\_1 + \dots + a\_rf\_{\*}\Gamma\_r$ and hence $NE(Y)$ is generated by $f\_{\*}\Gamma\_1,\dots,f\_{\*}\Gamma\_r$. Is this argument correct?
If $f$ is not birational could it happen that $NE(Y)$ is not finitely generated even if $NE(X)$ is?
Thank you very much.
| https://mathoverflow.net/users/14514 | Mori cones and projective morphisms | It suffices to assume that $f$ is surjective (equivalently, dominant). Then for $C \subset Y$ any irreducible curve there exists an irreducible curve $D \subset X$ such that $f(D) = C$. (A schemy proof: let $D$ be the closure in $X$ of any closed point in $f^{-1}(c)$, where $c$ is the generic point of $C$.) It follows that $f$ induces a surjection from $NE(X)$ to $NE(Y)$, so if $NE(X)$ is finitely generated then so is $NE(Y)$.
The same argument also implies that if $\overline{NE(X)}$ is finitely generated then so is $\overline{NE(Y)}$.
| 3 | https://mathoverflow.net/users/519 | 445878 | 179,675 |
https://mathoverflow.net/questions/445891 | 1 | Let $\mathcal H(n)$ be the set of $n\times n$ Hermitian matrices, and $\mathcal S(n) \subset \mathcal H(n)$ be the subset of density matrices, i.e., $A \in \mathcal S(n)$ iff $A$ is Hermitian, positive semidefinite and of trace one. Does there exist $c>0$ such that the following inequality
$$\big| \operatorname{tr} (ALBL) - \operatorname{tr}\left(AL^2B\right) \big| \leq c \|L\|^2 \operatorname{tr} \big((I-A)B\big)$$
holds for all $A, B \in \mathcal S(n)$ and $L \in \mathcal H(n)$?
PS: By choosing a suitable basis, we may assume without loss of generality that $A$ is diagnosable, i.e.
$$A=\sum\_{k=1}^n a\_{kk}P^{kk},$$
where $a\_{ij}:=(A)\_{ij}$ denote the elements of $A$ and $P^{ij}$ is the matrix whose only non-zero element is $(P^{ij})\_{ij}=1$. Note further the above inequality is linear with respect to $A$, it suffices to deal with the case that $a\_{11}=1$ and $a\_{kk}=0$ for $k\neq 1$. Hence, the l.r.s. and r.h.s. become respectively
$$\big| (LBL)\_{11} - (L^2B)\_{11} \big| \leq c \|L\|^2 \big(1-(B)\_{11}\big)?$$
Is the above inequality true for some suitable $c$? I'm unable to carry out the computation.
| https://mathoverflow.net/users/493556 | On a matrix trace inequality | The answer is **false**.
Counterexample: $L$ is the 2x2 all-$1$ matrix, $\epsilon ↘ 0$,
$B=\left[\begin{array}{cc}
1-\epsilon & \sqrt{\epsilon (1-\epsilon)} \\ \sqrt{\epsilon (1-\epsilon)} & \epsilon
\end{array}\right]$.
Then the LHS converges to $1$ and $1-B\_{11}$ converges to $0$, so there can't be such $c$.
| 1 | https://mathoverflow.net/users/125498 | 445901 | 179,682 |
https://mathoverflow.net/questions/423178 | 12 | Consider the following two player game based on the Euclidean algorithm: Positions are given by $(a,b)$ in $\mathbb N^2\setminus\{(0,0)\}$ (where $\mathbb N=\{0,1,2,\ldots\}$) defining a greatest common divisor in $\mathbb N$. Moves are given by
$(a,b)\longrightarrow (\min(a,b),\max(a,b)-k\min(a,b))$ for
$k$ in $\{1,\ldots,\lfloor \max(a,b)/\min(a,b)\rfloor\}$ where we assume $\min(a,b)>0$.
Two players move in turn until arriving at $(d,0)$ or $(0,d)$.
At this point, the player who can no longer move announces the
greatest common divisor $d$ of $a$ and $b$ defining the initial position $(a,b)$ and wins.
Winning positions are easy to describe: $(a,b)$ is
winning if and only if $\min(a,b)/\max(a,b)$
is smaller than the inverse $(-1+\sqrt{5})/2$ of the golden number $(1+\sqrt{5})/2$.
*Has this game be described somewhere?*
The variation where the last player with a move (i.e. playing a position $(a,b)$ with $\min(a,b)$ a divisor of $\max(a,b)$) wins
is also easy to describe: A position $(a,b)$ is winning if either $(a=b)$ or if $ab>0$ and $\max(a,b)/\min(a,b)>(1+\sqrt{5})/2$.
| https://mathoverflow.net/users/4556 | Euclid's algorithm as a combinatorial game | Following up on Peter Taylor's answer, here's a fairly complete annotated bibliography.
What you describe as the variant came first:
A. J. Cole and A. J. T. Davie, A game based on the Euclidean algorithm and a winning strategy for it, Math. Gaz. 53 (1969) 354-357. <https://doi.org/10.2307/3612461>
The analysis of your initial game was posed and answered in *Mathematics Magazine*:
J. W. Grossman, A Nim-type game, Problem #1537, Math. Mag. 70 (1997) 382. <https://doi.org/10.1080/0025570X.1997.11996580>
P. D. Straffin, Solution to Problem #1537, Math. Mag. 71 (1998) 394-395. <https://doi.org/10.1080/0025570X.1998.11996686>
Using continued fractions in the analysis:
T. Lengyel, A Nim-type game and continued fractions, Fibonacci Quart. 41 (2003) 310-320. <https://www.fq.math.ca/Scanned/41-4/lengyel.pdf>
Refining the N-positions of your initial game\*:
G. Nivasch, The Sprague-Grundy function of the game Euclid, Discrete Math. 306 (2006) 2798-2800. <https://doi.org/10.1016/j.disc.2006.04.020>
Using the structure of the Calkin-Wilf tree in the analysis:
S. Hofmann, G. Shuster, J. Steuding, Euclid, Calkin & Wilf--playing with rationals, Elem. Math. 63 (2008) 109-117. <https://doi.org/10.4171/EM/95>
Refining the N-positions of your variant game\*:
G. Cairns, N. B. Bao, T. Lengyel, The Sprague-Grundy function of the real game Euclid, Discrete Math. 311 (2011) 457-462. <https://doi.org/10.1016/j.disc.2010.12.011>
\*The N- and P-position analysis of the two versions of the game is the same, but refining the N-positions by their Sprague-Grundy values differs.
| 8 | https://mathoverflow.net/users/14807 | 445905 | 179,683 |
https://mathoverflow.net/questions/445900 | 4 | Recall that a Boolean algebra is a complemented distributive lattice. The set of subspaces of a vector space comes very close to being a boolean algebra. It satisfies all the required properties, except being distributive. Even then it is a [modular lattice](https://ncatlab.org/nlab/show/modular+lattice), which is almost the distributivity property. This raises the question of constructing a Boolean algebra from this complemented modular lattice. So,
* Does there exist a left (or right) adjoint to the forgetful functor $F$ from the category of Boolean algebras (and bounded-lattice homomorphisms) to that of complemented modular lattices?
* If so, what does it look like?
| https://mathoverflow.net/users/54507 | Boolean algebra of the lattice of subspaces of a vector space? | I'm assuming "bounded-lattice homomorphism" means what I would call "0,1-homomorphism", i.e., it takes top to top and bottom to bottom? Then there is no adjoint, left or right.
Let $L$ be the lattice of subspaces of a two-dimensional vector space $V$ (over any field). I claim that there is no 0,1-homomorphism from $L$ into any Boolean algebra. This is because we can find three one-dimensional subspaces of $V$ any two of which are complements in $L$. But then the same is true of their images under any 0,1-homomorphism, and you cannot find three elements of any Boolean algebra with this property.
So, no left adjoint to the forgetful functor because every Boolean algebra has at least one 0,1-homomorphism into the two-element Boolean algebra $\{0,1\}$ and thus there is no Boolean algebra $B$ with the property that ${\rm Hom}(B, \{0,1\}) \cong {\rm Hom}(L, \{0,1\})$.
Now let $M$ be the lattice of subspaces of the vector space $\mathbb{F}\_2^3$, the three-dimensional vector space over the two-element field. Letting $B\_4$ be the four-element Boolean algebra, if I have calculated correctly\* there are exactly fifty-eight 0,1-homomorphisms from $B\_4$ into $M$. But the number of 0,1-homomorphisms from $B\_4$ into any Boolean algebra is either a power of two or infinite. So there is no Boolean algebra $B$ with ${\rm Hom}(B\_4, B) \cong {\rm Hom}(B\_4, M)$ and thus no right adjoint.
---
${}^\*$Because $M$ has sixteen elements, two of which (top and bottom) have exactly one complement, and fourteen of which have exactly four complements. So there are $2 + 14\cdot 4 = 58$ 0,1-homomorphisms from $B\_4$ into $M$.
| 6 | https://mathoverflow.net/users/23141 | 445914 | 179,688 |
https://mathoverflow.net/questions/445906 | 2 | Suppose $X$ and $Y$ are two non-negative, independent random variables such that $X \succcurlyeq\_{st} Y$. That is, $X$ first-order stochastically dominates $Y$. Suppose that $X$ and $Y$ have smooth CDFs that admit a density.
Let $a, b > 0$ be two constants such that $a > b$. Is the following true?
$$a X + b Y \succcurlyeq\_{st} a Y + b X$$
I suspect that the answer is no because various results in Shaked and Shanthikumar to this effect talk about random variables ranked under the reversed-hazard-rate order, and not FOSD (Theorem 4.A.36,4.A.37 e.g.).
| https://mathoverflow.net/users/78761 | Weighted sum of two random variables ranked by first order stochastic dominance | $\newcommand\de\delta$Indeed, a counterexample is as follows: $a=2$, $b=1$,
$$X\sim\frac12(\de\_0+\de\_2),\quad Y\sim\frac12(\de\_0+\de\_1),$$
where $\de\_x$ is the Dirac measure supported on the singleton set $\{x\}$.
Then $X\succcurlyeq\_{st}Y$, but
$$a X + b Y \not\succcurlyeq\_{st} a Y + b X,$$
because $P(aX+bY\ge 2)=\frac12\not\ge\frac34=P(aY+bX\ge 2)$.
(Your smoothness condition is inessential, since any distribution can be appropriately approximated by a smooth distribution.)
| 4 | https://mathoverflow.net/users/36721 | 445916 | 179,689 |
https://mathoverflow.net/questions/445884 | 3 | Let $M = (\mathbb{Z}\_p)^2$ be a Galois representation, with Galois action given by $\rho: G\longrightarrow SL\_2(\mathbb{Z}\_p)$. I am trying to understand how sensitive the Galois cohomology group $H^1(G, M)$ (or more interesting to me, $H^1\_f(G, M)$) is to perturbations in $\rho$.
For example, say I replace $\rho$ by $\rho'$, with $\overline{\rho} = \overline{\rho'}$, i.e. the residual representations are isomorphic.
Denote by $M'$ the Galois module $(\mathbb{Z}\_p)^2$ with the action $\rho'$. Is it true that
$H^1(G, M)\cong H^1(G, M')$?
Thanks in advance!
| https://mathoverflow.net/users/174655 | Deformations of Galois cohomology | The answer to your example question is "No".
Let $G$ be isomorphic to $\widehat{\mathbf{Z}}$, and $\phi$ the topological generator of $G$ (corresponding to $1 \in \widehat{\mathbf{Z}}$). Then one computes $H^0(G, M) = M^{\phi = 1}$ and $H^1(G, M) = M / (\phi - 1)M$, where $\phi$ is the generator of $G$.
Now $M \cong \mathbf{Z}\_p$ be the trivial one-dimensional representation, and let $M'$ be the unramified representation with Frobenius acting as $1 + p$. Then $H^1(G, M) = M = \mathbf{Z}\_p$, whereas $H^1(G, M') = \mathbf{F}\_p$.
(In this case, we do at least have $H^1(G, M) \otimes \mathbf{F}\_p = H^1(G, M') \otimes \mathbf{F}\_p$. But even this is just a small-number coincidence, coming from the fact that the $H^2$'s vanish, and will fail if you take more complicated groups.)
What *is* true is that if $M \cong M' \bmod p$ (and $G$ has reasonable finiteness properties), then the cohomology *complexes* $R\Gamma(G, M)$ and $R\Gamma(G, M')$ are "congruent mod $p$ in the derived category", in the sense that $R\Gamma(G, M) \otimes^{\mathbf{L}} \mathbf{F}\_p = R\Gamma(G, M') \otimes^{\mathbf{L}} \mathbf{F}\_p = R\Gamma(G, \overline{M})$ (where $\overline{M}$ is the common residual representation). But much can go wrong in translating that into a concrete statement about cohomology groups.
The situation for $H^1\_f$ is worse – far, far worse – because in general there is no sensible intrinsic definition of $H^1\_f$ of a mod $p$ representation. (A large part of the machinery of Iwasawa theory is designed to measure, control, and in some cases cleverly exploit, the discrepancy between $H^1\_f$'s of congruent representations).
(EDIT: You might find the [Greenberg-Vatsal paper](https://arxiv.org/abs/math/9906215) interesting in this line -- it shows a result which roughly amounts to "Selmer groups of congruent modular forms are congruent", but there is a *huge* amount of work involved in making this precise and avoiding all the technical pitfalls.)
| 4 | https://mathoverflow.net/users/2481 | 445918 | 179,690 |
https://mathoverflow.net/questions/445932 | 2 | Let $ G $ be a quasisimple finite group. Let $ d\_{min} $ be the minimum dimension of a nontrivial irrep of $ G $. Must it be the case that the image of all (nontrivial) dimension $ d\_{min} $ irreps of $ G $ are conjugate in $ SU(d\_{min}) $?
For example all $ SL(2,5) $ subgroups of $ SU(2) $ are conjugate. This is especially interesting given that there are actually two distinct irreps of $ SL(2,5) $ into $ SU(2) $, they just happen to have conjugate image.
And I'm pretty sure that all $ A\_5,GL(3,2) $ and $ 3.A\_6 $ subgroups of $ SU(3) $ are conjugate.
And I'm pretty sure that all $ SL(2,9) \cong 2.A\_6 , 2.A\_7 $ and $ Sp(4,3) $ subgroups of $ SU(4) $ are also unique up to conjugacy
| https://mathoverflow.net/users/387190 | Image of minimal degree representation of quasisimple group unique up to conjugacy | Looking at the Atlas, I'd say that the smallest counterexample is probably the Mathieu group $M\_{11}$. There are three $10$-dimensional irreducible characters, not conjugate in $SU(10)$.
Edit: Actually, $L\_2(8)$ is smaller, with four 7 dimensional irreducibles. Three of them are conjugate in $SU(7)$ but the fourth isn't. I somehow missed this, first time through. Similarly for $L\_2(16)$.
| 4 | https://mathoverflow.net/users/460592 | 445933 | 179,695 |
https://mathoverflow.net/questions/445843 | 2 | I am looking for a proof (or references) for the following result:
>
> If $P\in \mathbb{Z}[x]$ then there exists a nonzero polynomial $Q\in \mathbb{Z}[x]$
> such that
> $$H(PQ)\leq M(P)$$
> where
> $H(R)=\max\{|a\_0|,|a\_1|,\dots,|a\_n|\}$ and $
> M(R)=|a\_n|\prod\_{k=1}^n\max\{1,|\alpha\_k|\}$ are respectively the height and the Mahler measure of a polynomial $R(x)=\sum\_{k=o}^na\_kx^k=a\_n\prod\_{k=1}^n (x-\alpha\_k).$
>
>
>
Comments are also welcome.
| https://mathoverflow.net/users/94262 | Given $P\in \mathbb{Z}[x]$ there is a nonzero $Q\in \mathbb{Z}[x]$ such that $H(PQ)\leq M(P)$ | In Chapter 3 of Jonas Jankauskas's dissertation thesis, entitled ["Heights of Polynomials"](https://epublications.vu.lt/object/elaba:1917005/1917005.pdf), we learn that the inequality
$$\min\_{Q \in \mathbb{Z}[x] \setminus \{0\}} H(PQ) \le \lfloor M(P)\rfloor$$
where $P \in \mathbb{Z}[x]$ and $\lfloor r \rfloor$ denotes the integral part of $r \in \mathbb{R}$, follows from [Siegel's lemma](https://en.wikipedia.org/wiki/Siegel%27s_lemma) [3].
The references given by Jonas Jankauskas are
* [[2]](https://zbmath.org/0405.12021), written in French, which gives a simplified proof of an enhanced [1, Théorème], also written in French,
* [3], written in English.
I can confirm that the original theorem of Martine Pathiaux [1, Théorème] yields the desired result.
Taking more time to parse [3], I might be able to update this answer with the precise statement of [3] which proves the above inequality.
**Key words:** *Reduced height, Siegel's lemma*.
---
**Addendum.** After reading further [3], but still superficially, I didn't find any direct means to derive the desired result from Siegel's lemma. (If $P$ is monic, we can indeed consider a linear homogenous system of equations with coefficients in $\mathbb{Z}$. As I haven't been able to conclude from there, I'll ask around if such an approach can be successful).
The papers [1] and [2] both establish the result for $P$ an **irreducible** polynomial. In her final remark [1, page 13-08], Martin Pathiaux asserts that the reducible case can be addressed by simply "adding more inequations" to her system. This is accurate, but it requires a good understanding of her proof.
For the sake of seeing **a complete proof of the fact referred to by the OP**, I have written below a proof along the lines of [2].
This proof doesn't add anything original. It consists only of the details which quickly convince me that the reducible case holds also true.
**Generalizing [1, Théorème 1] to reducible polynomials.**
Let $P(x) = a\_0 + a\_1 x + \cdots + a\_d x^d \in \mathbb{C}[x]$. The *height* of $P$ is the real number $H(P) = \max\_{i} \vert a\_i \vert$.
The *Mahler measure* of $P$ is the real number
$M(P) = \vert a\_d \cdot \lambda\_1 \cdots \lambda\_r \vert$ where $\lambda\_1, \dots, \lambda\_r \in \mathbb{C}$ are the roots of $P$ in $\mathbb{C}$ whose complex modulus is at least $1$.
We set ourselves to prove:
>
> **Claim.** [1, Final remark]
> Let $P \in \mathbb{Z}[x]$. Then there is $Q \in \mathbb{Z}[x] \setminus \{0\}$ such that $H(PQ) \le M(P)$.
>
>
>
We first mention the two results which have allowed Maurice Mignotte to significantly simplify the original proof of [1, Théorème 1] by Martine Pathiaux.
>
> **Proposition.** [2, Corollaire 1]
> Let $a\_{i,j} \in \mathbb{C}$, with
> $i \in \{1, \dots, m\}, j \in \{1, \dots, n\}$ and
> $L\_{i}(\mathbf{x}) = \sum\_{i = 1}^n a\_{i,j} x\_j$ for
> $\mathbf{x} = (x\_1, \dots, x\_n) \in \mathbb{C}$. Let $A\_i = \max\_j \vert a\_{i,j} \vert$.
> Let $M \in \mathbb{N}\_{> 0}$.
> Then we can find $\mathbf{x} \in \mathbb{Z}^n \setminus \{0\}$ such that $\vert x\_j \vert \le M$ for every $j \in \{1, \dots, n\}$
> and
> $$\vert L\_i(\mathbf{x}) \vert < \sqrt{2n \log(18m)} A\_i \frac{M + 1}{(M + 1)^{n/m} - 2}$$
> for every $i \in \{1, \dots, m\}$, provided that $(M + 1)^{n/m} > 2$.
>
>
>
The following lemma is a quantitative version of the [fundamental theorem for symmetric polynomials](https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial#Fundamental_theorem_of_symmetric_polynomials).
>
> **Lemma.** [2, Section III, see remark at the bottom of page 3] Let $R$ be a commutative ring with identity. Let $Q \in R[x\_1, \dots, x\_d]$ by a symmetric polynomial.
> Then there is a polynomial $\tilde{Q} \in R[x\_1, \dots, x\_d]$, with total degree $\deg(\tilde{Q}) \le \max\_i \deg\_{x\_i}(Q)$ and such that
> $$Q(x\_1, \dots, x\_d) = \tilde{Q}(e\_{1, d}, e\_{2, d}, \dots, e\_{d,d})$$ where
> $e\_{1, d} = \sum\_{i = 1}^d x\_i, e\_{2, d} = \sum\_{1 \le i < j \le d} x\_i x\_j, \dots, e\_{d, d} = x\_1 \cdots x\_d$
> are [the elementary symmetric polynomials](https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial) on $x\_1, \dots, x\_d$.
>
>
>
>
> *Proof of the lemma.* Revisit one of the classical proofs of the [fundamental theorem for symmetric polynomials](https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial#Fundamental_theorem_of_symmetric_polynomials) and
> check in the induction step that $\deg(\tilde{Q}) \le \max\_i\deg\_{x\_i}(Q)$ additionally holds for every $i$.
>
>
>
We are now in position to prove the claim.
>
> *Proof of the claim.* We can assume that $P$ is not constant, since otherwise the result is trivial. (The result is also obvious if $\deg(P) = 1$, but we won't need this fact).
> We can also assume, without loss of generality, that $P$ is primitive, i.e., the coefficients of $P$ are coprime integers,
> and that the leading coefficient of $P$ is positive.
> As $\mathbb{Z}[x]$ is a [unique factorization domain](https://en.wikipedia.org/wiki/Unique_factorization_domain#:%7E:text=7%20References-,Definition,n%20with%20n%20%E2%89%A5%200), we can write
> $P(x) = P\_1^{\mu\_1}(x) \cdots P\_k^{\mu\_k}(x)$ where $\mu\_l \in \mathbb{N}\_{> 0}$ and
> $P\_i \in \mathbb{Z}[x]$ is an irreducible primitive polynomial of degree $d\_i > 0$, with positive leading coefficient, for
> every $i \in \{1, \dots, k\}$.
> For $i \in \{1, \dots, k\}$, we denote by $\theta\_{i, 1}, \dots, \theta\_{i, d\_i}$ the roots of $P\_i$ in $\mathbb{C}$.
> We shall prove that there exists, for all $n \in \mathbb{N}\_{> 0}$ sufficiently large, a polynomial
> $D(x) = d\_0 + d\_1x + \cdots + d\_{n - 1}x^{n - 1} \in \mathbb{Z}[x]$
> which satisfies the inequality $\vert d\_j \vert \le M(P)$ for every $j \in \{0, \dots, n - 1\}$ together with the equation
> $D^{(\mu)}(\theta\_{i, j}) = 0$ for every
> $i \in \{1, \dots, k\}, j \in \{1, \dots, d\_i\}$ and $\mu \in \{0, \dots, \mu\_i - 1\}$
> where $D^{(\mu)}$ denotes the $\mu$-th derivative of $D$.
> Let us observe that such a polynomial $D$ obviously satisfies $H(D) \le M(P)$. In addition, this polynomial $D$ is divisible by $P$ in $\mathbb{Q}[x]$.
> The fact that $P$ divides $D$ in $\mathbb{Z}[x]$ follows immediately from [Gauss's lemma](https://en.wikipedia.org/wiki/Gauss%27s_lemma_(polynomials)).
> The proof of the existence of $D$ is a straightforward application of the method developed in [2].
> We proceed by showing, for all $n$ sufficiently large, the existence of a polynomial
> $D \in \mathbb{Z}[x] \setminus \{0\}$ satisfying the inequations
> $\vert D^{(\mu)}(\theta\_{i, j}) \vert < \epsilon\_{i, j, \mu}(n)$ for some upper bound $\epsilon\_{i, j, \mu}(n)$
> which can be rendered sufficiently small to enforce $D^{(\mu)}(\theta\_{i, j}) = 0$
> for every $i, j$ and $\mu$.
> To this end, we set $M = \lfloor M(P) \rfloor$ and apply the above proposition to the linear forms
> $L\_{i, j, \mu}(\mathbf{x}) = \sum\_{l = 0}^n a\_{i, j, \mu, l} x\_l$ with
> $a\_{i, j, \mu, l} = \max(l(l - 1) \cdots (l - \mu + 1), 0) \cdot \theta\_{i, j}^{l - \mu}$ and $n > 0$. (There are $m = \deg(P)$ such forms.)
> Since we have
> $A\_{i, j, \mu} := \max\_l \vert a\_{i, j, \mu, l} \vert \le \mu\_i ! \max(\vert\theta\_{i, j}\vert, 1)^{n - 1}$, the above
> proposition provides us with some
> $\mathbf{d} = (d\_0, \dots, d\_{n - 1}) \in \mathbb{Z}^n \setminus \{0\}$ such that $\vert L\_{i, j, \mu}(\mathbf{d}) \vert < \epsilon\_{i, j, \mu}(n)$ for every $i, j$ and $\mu$, where
>
> $$\epsilon\_{i, j, \mu}(n) = \sqrt{2n \log(18 \deg(P))}\mu\_i ! \frac{\max(\vert\theta\_{i, j}\vert^{n - 1}, 1)(M + 1)}{(M + 1)^{n/\deg(P)} - 2}.$$
> Let us set $D(x) := d\_0 + d\_1x + \cdots + d\_{n - 1}x^{n - 1}$, so that $L\_{i, j, \mu}(\mathbf{d}) = D^{(\mu)}(\theta\_{i, j})$ by the definitions of $D$ and $L\_{i, j, \mu}$.
> Let $r\_{i, \mu}(n) = \vert D^{(\mu)}(\theta\_{i, 1}) \cdots D^{(\mu)}(\theta\_{i, d\_i}) \vert$ and let $q\_i > 0$ be the leading coefficient of $P\_i$.
> We have thus
> $$q\_i^{n - 1} r\_{i, \mu}(n) < q\_i^{n - 1} \epsilon\_{i, 1, \mu}(n) \cdots \epsilon\_{i, d\_i, \mu}(n) = O\left(\frac{n^{d\_i/2} M(P\_i)^{n - 1}}{(M + 1)^{nd\_i/\deg(P)}}\right).$$
> It follows that
> $$r\_i (n):= q\_i^{n - 1} r\_{i, 0}(n) \cdots q\_i^{n - 1} r\_{i, \mu\_i - 1}(n) = O\left(\frac{n^{\mu\_i d\_i/2} M(P\_i^{\mu\_i})^{n - 1}}{(M + 1)^{n \mu\_i d\_i/\deg(P)}}\right)$$
> for every $i \in \{1, \dots, k\}$ and hence
> $$r\_1(n) \cdots r\_k(n) = O\left( n^{\deg(P)/2} (\frac{M}{M + 1})^n\right) = o(1).$$
> Since $\theta\_{i, 1}, \dots, \theta\_{i, d\_i}$ are the roots of $P\_i$, we have
>
> $e\_{j, d\_i}(\theta\_{i, 1}, \dots, \theta\_{i, d\_i}) \in \frac{1}{q\_i} \mathbb{Z}$ for every elementary symmetric polynomial $e\_{j, d\_i}$.
> By the above lemma, we know that $q\_i^{n - 1} r\_{i, \mu}(n) \in \mathbb{N}$ for every $i \in \{1, \dots, d\_i\}$ and every $\mu \in \{0, \dots, \mu\_i - 1\}$.
> Thus we have $r\_1(n) \cdots r\_k(n) \in \mathbb{N}$ and $r\_1(n) \cdots r\_k(n) = o(1)$,
> which entails that $r\_i(n) = r\_{i, \mu}(n) = D^{(\mu)}(\theta\_{i, j}) = 0$ for all $n$ large enough and for every $i, j$ and $\mu$, as desired.
>
>
>
---
[1] M. Pathiaux, "Sur les multiples de polynômes irréductibles associés à certains nombres algébriques", 1972.
[2] M. Mignotte, "Sur les multiples des polynômes irréductibles", 1975.
[3] E. Bombieri and J. Vaaler, "On Siegel's Lemma", 1984.
| 7 | https://mathoverflow.net/users/84349 | 445935 | 179,696 |
https://mathoverflow.net/questions/445944 | 6 | Let $f\in C^{\infty}([-1,0])$ be real-valued and suppose that
$$ \left| \int\_{-1}^{0} f(t)\,e^{\lambda t} \,{\rm d} t \right| \leq e^{-\sqrt{\lambda}},$$
for all $\lambda > 0$. Does it follow that $f(t)$ is zero on $-\epsilon \leq t\leq 0$, for some $\epsilon>0$?
My hunch is that this is actually not true but fail to come up with a counter example.
| https://mathoverflow.net/users/50438 | On an asymptotic integral decay | I'll change $[-1,0]$ to $[0,1]$ (so $t\to -t$), which seems easier on the brain.
$f(t)=e^{-1/t}$ is a counterexample: For $0<t<1/\sqrt{\lambda}$, we have $f\le e^{-\sqrt{\lambda}}$, so this part of the integral satisfies the desired bound, and if $t>1/\sqrt{\lambda}$, then $e^{-\lambda t}\le e^{-\sqrt{\lambda}}$.
| 11 | https://mathoverflow.net/users/48839 | 445948 | 179,700 |
https://mathoverflow.net/questions/445949 | 2 | In Cox, Little and Schenck's book **Toric Varieties** they show that a toric variety $ X\_{\Sigma} $ over a field of characteristic zero is complete if and only if the support is all of $ N\_{\mathbb{R}} $. This proof was very specific to varieties over fields of characteristic zero. When I looked at Oda and Fulton's books they seemed to work over characteristic zero as well.
Is a toric variety $ X\_{\Sigma} $ over an algebraically closed field $ k $ of characteristic $ p>0 $ complete if and only if the support of $ \Sigma $ is all of $ N\_{\mathbb{R}} $? I believe the answer should be yes for the following reason, but if anyone could give me a reference, then I would appreciate it.
Let $ R $ be a DVR of mixed characteristic with special fibre $ k $ and generic fibre $ L $. Also, let us define a scheme $ \mathcal{X}\_{\Sigma} $ over $ \operatorname{Spec}(R) $ to be
\begin{equation\*}
\mathcal{X}\_{\Sigma} =\left(\coprod\_{\sigma \in \Sigma} \operatorname{Spec}(R[\sigma^{\vee} \cap M])\right)/\sim.
\end{equation\*}
We claim that the scheme $ \mathcal{X}\_{\Sigma} $ is faithfully flat over $ \operatorname{Spec}(R) $. The claim is local, so it suffices to prove that $ \operatorname{Spec}(R[\sigma^{\vee} \cap M]) $ is faithfully flat over $ \operatorname{Spec}(R) $. Since $ R[\sigma^{\vee} \cap M] $ is an $ R $-algebra, the claim follows.
If $ A $ is a DVR over $ k $, then let $ \operatorname{Spec}(B) $ be a faithfully flat, family of DVRs over $ \operatorname{Spec}(R) $, such that the special fibre is $ \operatorname{Spec}(A) $.
A normal, toric variety $ X\_{\Sigma} $ over a field $ L $ of characteristic zero is complete if and only if the support of $ \Sigma $ is all of $ N\_{\mathbb{R}} $. So since $ \operatorname{Spec}(B) $ is faithfully flat over $ \operatorname{Spec}(R) $, the valuative criterion of properness will hold for the morphism $ \mathcal{X}\_{\Sigma} \times\_{\operatorname{Spec}(R)} \operatorname{Spec}(L) \to \operatorname{Spec}(L) $ if and only if it holds for the morphism $ X\_{\Sigma} \to \operatorname{Spec}(k) $.
| https://mathoverflow.net/users/470753 | Is a toric variety over a field of positive characteristic complete if and only if the support is all of $ N_{\mathbb{R}} $? | The answer is yes. Fulton's proof is in fact valid positive characteristic as well. In the proof of "full support $\Rightarrow$ completeness" he uses the valuative criterion which, as you noticed, is true for all characteristic. For the opposite direction you only need to replace the zero characteristic language from Fulton's proof: if there is $v \in N$ which is not in any cone of $\Sigma$, then consider the corresponding one parameter subgroup $\lambda\_v$. If you choose a system of coordinates on $N$ so that $v = (v\_1, \ldots, v\_n)$, then $\lambda\_v$ is simply the map $k^\* \to X$ given by $t \mapsto (t^{v\_1}, \ldots, t^{v\_n})$. If $X$ were complete, then $\lambda\_v$ would extend to a map from $k \to X$. However, for every cone $\sigma \in \Sigma$, since $v \not\in \sigma$, there is $u\in \check{\sigma}$ such that $\langle u, v \rangle < 0$. Then $\chi^u$ is not defined at $0$, so that $0$ can not be mapped to anywhere on the open affine subset of $X$ corresponding to $\sigma$. Since $X$ is a union of these subsets, it follows that $X$ is not complete, as required.
| 4 | https://mathoverflow.net/users/1508 | 445956 | 179,703 |
https://mathoverflow.net/questions/445816 | 3 | Let $G$ be a second countable profinite group, $g\in G$ and $g^G:=\{hgh^{-1}~|~h\in G\}$ the conjugacy class of $g$ in $G$. Theorem 3.2 in Wesolek's [Conjugacy class conditions in locally compact second countable groups](https://www.ams.org/journals/proc/2016-144-01/S0002-9939-2015-12645-7/) gives a characterization of a profinite group having an open conjugacy class. For example, $g^G$ is open if and only if the Haar measure on $g^G$ is greater than $0$. In particular, if $g^G$ is open, then $g$ must has finite order in $G$, see Lemma 3.1 of the paper. I'm interested in the following case:
**Question:** Let $G$ be a second countable infinite profinite group, $g\in G$ of finite order such that the conjugacy class $g^G$ of $g$ generates $G$ as an abstract group. Is $g^G$ open in $G$?
Since infinite profinite group is uncountable, the set $g^G$ must be infinite. Moreover, $g^G$ and $G$ have the same cardinality by the question [Does a countable set generate a countable group?](https://math.stackexchange.com/questions/346244/does-a-countable-set-generate-a-countable-group?stw=2). In this sense, one can expect that the Haar measure on $g^G$ should be greater than $0$, and hence it should be true.
| https://mathoverflow.net/users/492970 | Open conjugacy classes in a second countable profinite group | Yes, there exist profinite groups $G$ with a conjugacy class of empty interior and consisting of elements of finite order, generating $G$ as an abstract group.
Let $H$ be a nonabelian finite simple group, and $S$ a nontrivial conjugacy class in $H$. Consider $G=H^\mathbf{N}$ and $T=S^\mathbf{N}$. Then $T$ is a single conjugacy class in $G$ and has empty interior. To be done, it is enough to check that $T$ generates $G$.
This is equivalent to asking that $S^n=H$ for some $n$ (where $S^n$ is the set of products of *exactly* $n$ elements of $S$). I don't know if it's always true. However it is enough to show that it's true in a single case.
Indeed, using that for a prime $p$ in $4\mathbf{N}+1$, every nonzero element is a sum of two nonzero squares, we see that every upper or lower unipotent matrix in $\mathrm{SL}\_2(\mathbf{F}\_p)$ is product of two conjugates of the matrix $u=\begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}$, and hence that every element of $\mathrm{SL}\_2(\mathbf{F}\_p)$ is product of exactly 6 $\mathrm{SL}\_2(\mathbf{F}\_p)$-conjugates of $u$. (For other odd primes it still holds that every element is sum of three nonzero squares and hence every element of $\mathrm{SL}\_2(\mathbf{F}\_p)$ is product of exactly 9 $\mathrm{SL}\_2(\mathbf{F}\_p)$-conjugates of $u$.) Hence $H=\mathrm{PSL}\_2(\mathbf{F}\_p)$ works.
[NB The proof also works, in the same way, for the product $\prod\_n\mathrm{PSL}\_2(\mathbf{F}\_{p^n})$, which has the additional requirement of being topologically finitely generated. Note that in this case the element $u$ still has order $p$.]
---
Edit (given follow-up question in comments): for an odd prime $p$, $\mathrm{SL}\_3(\mathbf{Z}\_p)$ has finite Prüfer rank and is generated by the conjugacy class of the diagonal matrix $d$ given by the diagonal $(1,-1,-1)$.
Indeed, first of all, $\mathrm{SL}\_n$ over a Euclidean domain is generated by elementary matrices. So it is enough to show that the matrix $\begin{pmatrix}1 & x & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$ is a product of conjugates of $d$. It is indeed the product $dd'$ where $d'=\begin{pmatrix}1 & x & 0\\ 0 & -1 & 0 \\ 0 & 0 & -1\end{pmatrix}$. Now $d'$ is conjugate to $d$, namely $d'=udu^{-1}$ with $u=\begin{pmatrix}1 & -x/2 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$.
| 2 | https://mathoverflow.net/users/14094 | 445975 | 179,711 |
https://mathoverflow.net/questions/445951 | 3 | It seems well-known that any smooth plane quartic can be written as the vanishing of $Q\_0Q\_2 -Q\_1^2$. Is there a good way to work out these quadratic factors $Q\_0,Q\_1,Q\_2$? For example, given the Klein quartic $X^3Y + Y^3Z + Z^3X = 0$ what would these quadratics be?
If not, is there at least some computational package for this purpose? Thanks for your help!
| https://mathoverflow.net/users/160814 | Writing a smooth plane quartic as the vanishing of $Q_0Q_2 - Q_1^2$ for quadratic $Q_0,Q_1,Q_2$ | The condition on your coefficients is an underdetermined system of quadratic equations. So specializing some of them (for instance setting them $0$), you can try to get a system with a finite number of (complex) solutions, and the chances are good that they are actually rational. Such systems can be handled with SageMath (which uses Singular, msolve or other systems as backend). In the specific case, a possible solution is
\begin{align}
Q\_0 &= X^2 - XY + Y^2 + XZ\\\\
Q\_1 &= X^2 - XY + YZ\\\\
Q\_2 &= X^2 - XZ + YZ + Z^2.
\end{align}
**Added later upon request in the comment:** Here is the rather naive and straightforward Sage code which runs 160 seconds on my machine. It shows that a rational solution requires at least $11$ nonzero coefficients.
```
e2 = [(i, j) for i in range(3) for j in range(3-i)]
n2 = len(e2)
R = PolynomialRing(QQ, 'a', 3*n2)
Rxy.<x, y> = R[]
T0, T1, T2 = [sum(R.gen(m+k*n2)*x^i*y^j for m, (i, j) in enumerate(e2))
for k in range(3)]
f = x^3*y + y^3 + x
l = (T0*T2-T1^2-f).coefficients()
N = R.ngens()
for m in range(1, N):
print('m = ', m)
for s in Subsets(range(N), m):
i = min(s)
if i >= n2:
continue
l0 = l + [R.gen(i)-1] + [R.gen(j) for j in range(N) if
j != i and not j in s]
I = ideal(l0)
if I.dimension() == 0:
V = I.variety()
if len(V) > 0:
break
else:
continue
break
v = V[0]
Q0, Q1, Q2 = [sum(c.subs(v)*xy for c, xy in Q) for Q in [T0, T1, T2]]
print(f'Q0 = {Q0}\nQ1 = {Q1}\nQ2 = {Q2}\n{Q0*Q2-Q1^2 == f}')
```
| 7 | https://mathoverflow.net/users/18739 | 445982 | 179,715 |
https://mathoverflow.net/questions/445927 | 1 | Given two von-Neumann algebra factors $\mathcal M,\mathcal N$, is $\mathcal M\cap\mathcal N$ a factor?
And how about the intersection of infinitely many factors?
**Notes:**
* I know that the intersection is a von-Neumann algebra. (This is immediate from the definition of a von-Neumann algebra as a SOT-closed algebra with adjoints and 1.)
* I know that this does not hold for *Type I* factors. (See comments [here](https://mathoverflow.net/questions/442906/intersection-of-finitely-many-type-i-von-neumann-algebra-factors).)
| https://mathoverflow.net/users/101775 | Intersection of von-Neumann algebra factors | The answer is no. There are subfactors $N\subset M$ with finite Jones index $[M:N]$ with $N^{\prime}\cap M=\mathbb{C}\oplus \mathbb{C}$. For example, consider a type $II\_1$ factor $P$ and let $\alpha$ be an outer automorphism of $P$. Put $M= P\otimes M\_2(C)$ and $N$ be the algebra of all diagonal matrices $(x,\alpha(x)),$ for $x\in P$.
| 2 | https://mathoverflow.net/users/164194 | 445985 | 179,716 |
https://mathoverflow.net/questions/445643 | 22 | Ever since Simpson's paper [Sim], it was observed that many different cohomology theories arise in the following way: we begin with our space $X$, we associate to it a stack $X\_\text{stk}$ (which depends on the chosen cohomology theory), and then the cohomology of $X$ coincides with the quasi-coherent cohomology of $X\_\text{stk}$.
We have even more! Very often cohomology theories come from a six-functor formalism $D(X)$ and we often have that $D(X)=D\_\text{qc}(X\_\text{stk})$.
As far as I know, this is known for: (Often there are many references discussing this, I put the first one that came to mind just to help the reader find something about it.)
* de Rham cohomology; [GR] (We can also recover the underlying Hodge filtration. [Bh §2.3])
* Crystalline cohomology; [RG]
* Prismatic cohomology; [Bh]
* Syntomic cohomology; [Bh]
* Dolbeault cohomology; [Sim2]
* Betti cohomology; [PS]
* Deligne cohomology (I think); [PS]
The most notable absences from this list seem to be étale / $\ell$-adic cohomology and rigid cohomology. **Is there a stacky approach to these cohomology theories as well?**
References:
[Sim] C. Simpson - [Homotopy over the complex numbers and generalized de Rham cohomology](https://www.tib.eu/en/search/id/BLCP%3ACN013382849/Homotopy-Over-the-Complex-Numbers-and-Generalized/)
[Sim2] C. Simpson - [The Hodge filtration on nonabelian cohomology](https://arxiv.org/pdf/alg-geom/9604005.pdf)
[Bh] B. Bhatt - [Prismatic $F$-Gauges](https://www.math.ias.edu/%7Ebhatt/teaching/mat549f22/lectures.pdf)
[GR] D. Gaitsgory, N. Rozenblyum - [Crystals and D-modules](https://arxiv.org/pdf/1111.2087.pdf)
[RG] R. Gregoric - [The Crystalline Space and Divided Power Completion](https://web.ma.utexas.edu/users/gregoric/CrystalMath.pdf)
[PS] M. Porta, F. Sala - [Simpson’s shapes of schemes and stacks](https://people.dm.unipi.it/sala/assets/pdf/porta_sala_shapes.pdf)
| https://mathoverflow.net/users/131975 | Is there a ring stacky approach to $\ell$-adic or rigid cohomology? | This is an interesting question.
First, I think the [PS] reference does not give the "correct" Betti stack. In my notes on 6 functors, I define a different stack $X\_B$ such that $D\_{\mathrm{qc}}(X\_B)$ is equivalent to (the left-completed version of) $D(X,\mathbb Z)$, for any locally compact Hausdorff space $X$. It represents the functor taking any scheme $S$ to the continuous functions from $|S|$ to $X$. The [PS] reference would only be able to see the locally constant sheaves on $X$ instead, and is effectively passing to the homotopy type instead.
Roughly speaking, a prerequisite for a stacky approach to some cohomology theory is that this cohomology theory satisfies a categorical Künneth formula: $D(X)\otimes\_{D(\ast)} D(Y)\cong D(X\times Y)$. At least, $X\mapsto X\_B,X\_{\mathrm{dR}}$ etc. all commute with finite limits (and I think this should always be true for such stacks), and the functor $X\mapsto D\_{\mathrm{qc}}(X)$ often takes fibre products to tensor products (of presentable stable $\infty$-categories). The latter may fail a little bit in general, but at least it should be "close" to being true. For example, for locally compact Hausdorff spaces it is true at least for finite-dimensional ones.
For $\ell$-adic sheaves, the categorical Künneth formula fails badly. Thus, one cannot really expect a stacky approach. However, one can (and I do) hope for a stack $X\_{\ell,n}$ such that $D\_{\mathrm{et}}(X,\mathbb Z/\ell^n)$ embeds fully faithfully into $D\_{\mathrm{qc}}(X\_{\ell,n})$.
In fact, if one assumes that $X$ lives over $\mathbb Q\_\ell$, and one is allowing stacks in almost schemes, then such a thing has been defined (implicitly) through the work of Lucas Mann on $p$(=$\ell$)-adic 6 functors.
So over $\mathbb Q\_\ell$, it exists, but I don't know how to descend it to $\mathbb Q$.
Edit: And for rigid cohomology, I explained a construction of such a stack in my course, I hope I will one day update the notes to include it.
| 13 | https://mathoverflow.net/users/6074 | 445987 | 179,717 |
https://mathoverflow.net/questions/445947 | 7 | I am interested in studying fluid dynamics and am searching for a good introductory textbook. I know just the very basics of fluids on the physics side. For mathematical prerequisites, I have completed a course on integration theory, and have a basic understanding of functional analysis and PDE, though by no means am I an expert.
I have done some searching online and most of the books I have come across lean more towards physics, numerical methods, and/or mathematical modeling. What I am instead looking for is something that is more mathematical and analysis/PDE oriented (and ideally one that does not assume too much of this material and introduces it as necessary), targeted towards early graduate students in mathematics. A good example of what I would like are Professor Tao's [lecture notes](https://terrytao.wordpress.com/category/teaching/254a-incompressible-fluid-equations/).
Are there any textbooks in this direction?
| https://mathoverflow.net/users/498931 | Textbook suggestions for rigorous fluid dynamics | A few possibilities:
* JC Robinson, JL Rodrigo, & W Sadowski (2016) Classical theory of the three-dimensional Navier-Stokes equations.
* OA Ladyzhenskaya (1963) The mathematical theory of viscous incompressible flow.
| 4 | https://mathoverflow.net/users/119114 | 445988 | 179,718 |
https://mathoverflow.net/questions/445973 | 1 | This question is a related question see this post [Vague convergence VS Laplace transform convergence](https://mathoverflow.net/questions/445665/vague-convergence-vs-laplace-transform-convergence). But now we assume that
\begin{equation}
\int\_0^\infty e^{-sx}\mu\_n(dx)\to \int\_0^\infty e^{-sx}\mu(dx),
\end{equation}
for all $s\in (0, \infty)$, where $\mu\_n$ is a finite measure. Please check this book for [vague convergence](https://www.google.com/books/edition/Foundations_of_Modern_Probability/L6fhXh13OyMC?hl=en&gbpv=0&bsq). Can we get what from the above Laplace transform convergence? From the above post, my understanding is that vague convergence is weaker than weakly convergence (please correct me, if I am wrong). Is it possible to get vague convergence and why or why not?
| https://mathoverflow.net/users/147009 | Which kind of convergence can we get from Laplace transform convergence? | $\newcommand{\thh}{\theta}$In general, you cannot get the vague convergence here. E.g., suppose that $\mu=0$ and $\mu\_n(dx)=x^2\,1(n<x<n+1)\,dx$. Then for each real $s>0$
\begin{equation\*}
0\le\int\_0^\infty e^{-sx}\mu\_n(dx)=\int\_n^{n+1} e^{-sx}x^2\,dx
\le e^{-sn}(n+1)^2\to0
\end{equation\*}
(as $n\to\infty$), so that
\begin{equation\*}
\int\_0^\infty e^{-sx}\mu\_n(dx)\to\int\_0^\infty e^{-sx}\mu(dx).
\end{equation\*}
However, letting $f(x):=\frac1{1+x}$ for real $x\ge0$, we have $f\in C\_0([0,\infty))$ and
\begin{equation\*}
\int\_0^\infty f(x)\mu\_n(dx)=\int\_n^{n+1} f(x)x^2\,dx
\ge \frac1{1+n+1}n^2\to\infty,
\end{equation\*}
so that
\begin{equation\*}
\int\_0^\infty f(x)\mu\_n(dx)\to\infty\ne0=\int\_0^\infty f(x)\mu(dx).
\end{equation\*}
---
However, assuming that the measure $\mu$ is finite, you can get the following:
\begin{equation\*}
\int\_0^\infty f(x)\mu\_n(dx)\to\int\_0^\infty f(x)\mu(dx) \tag{$\*$}\label{1}
\end{equation\*}
for all $f\in C\_c([0,\infty))$, that is, for all continuous functions $f$ on $[0,\infty)$ with a compact support.
Indeed, take any such $f$, so that $f=0$ on $[a,\infty)$ for some real $a>0$. Consider first the case when $f\ge0$.
Take any real $h>0$ and let
\begin{equation\*}
\nu\_{n,h}(dx):=e^{-hx}\mu\_n(dx),\quad \nu\_h(dx):=e^{-hx}\mu(dx).
\end{equation\*}
Then for any real $s\ge0$
\begin{equation\*}
\int\_0^\infty e^{-sx}\nu\_{n,h}(dx)=\int\_0^\infty e^{-(s+h)x}\mu\_n(dx)
\to\int\_0^\infty e^{-(s+h)x}\mu(dx)=\int\_0^\infty e^{-sx}\nu\_h(dx).
\end{equation\*}
So, by the [previous answer](https://mathoverflow.net/a/445676/36721), $\nu\_{n,h}\to\nu\_h$ weakly. So,
\begin{equation\*}
\int\_0^\infty f(x)e^{-hx}\mu\_n(dx)=\int\_0^\infty f(x)\nu\_{n,h}(dx)
\to\int\_0^\infty f(x)\nu\_h(dx)=\int\_0^\infty f(x)e^{-hx}\mu(dx)
\end{equation\*}
for each real $h>0$.
Next,
\begin{equation\*}
\int\_0^\infty f(x)\mu\_n(dx)\ge\int\_0^\infty f(x)e^{-hx}\mu\_n(dx)
=\int\_0^a f(x)e^{-hx}\mu\_n(dx) \\
\ge e^{-ha}\int\_0^a f(x)\mu\_n(dx)
=e^{-ha}\int\_0^\infty f(x)\mu\_n(dx),
\end{equation\*}
so that
\begin{equation\*}
\int\_0^\infty f(x)\mu\_n(dx)=e^{\thh\_{n,h,f}ha}\int\_0^\infty f(x)e^{-hx}\mu\_n(dx)
\end{equation\*}
for some $\thh\_{n,h,f}\in[0,1]$ depending on $n,h,f$. Similarly,
\begin{equation\*}
\int\_0^\infty f(x)\mu(dx)=e^{\thh\_{h,f}ha}\int\_0^\infty f(x)e^{-hx}\mu(dx)
\end{equation\*}
for some $\thh\_{h,f}\in[0,1]$ depending on $h,f$. So, for each real $h>0$,
\begin{equation\*}
\begin{aligned}
\limsup\_n\int\_0^\infty f(x)\mu\_n(dx)&\le e^{ha}\limsup\_n\int\_0^\infty f(x)e^{-hx}\mu\_n(dx) \\
& =e^{ha}\int\_0^\infty f(x)e^{-hx}\mu(dx) \\
& \le e^{ha}\int\_0^\infty f(x)\mu(dx) \\
& \underset{h\downarrow0}\longrightarrow
\int\_0^\infty f(x)\mu(dx)
\end{aligned}
\end{equation\*}
and
\begin{equation\*}
\begin{aligned}
\liminf\_n\int\_0^\infty f(x)\mu\_n(dx)&\ge\liminf\_n\int\_0^\infty f(x)e^{-hx}\mu\_n(dx) \\
&=\int\_0^\infty f(x)e^{-hx}\mu(dx) \\
&\ge e^{-ha}\int\_0^\infty f(x)\mu(dx) \\
& \underset{h\downarrow0}\longrightarrow
\int\_0^\infty f(x)\mu(dx).
\end{aligned}
\end{equation\*}
So, $\limsup\_n\int\_0^\infty f(x)\mu\_n(dx)=\liminf\_n\int\_0^\infty f(x)\mu\_n(dx)=\int\_0^\infty f(x)\mu(dx)$ and hence \eqref{1} holds
for any nonnegative $f\in C\_c([0,\infty))$.
Finally, writing $f=f\_+-f\_-$, where $f\_\pm:=\max(0,\pm f)$, we get \eqref{1} for all $f\in C\_c([0,\infty))$. $\quad\Box$
| 1 | https://mathoverflow.net/users/36721 | 445994 | 179,719 |
https://mathoverflow.net/questions/445934 | 6 | Let $\mathcal A$ be an additive 1-category, equipped with some class of weak equivalences $\mathcal W$. Let $\mathcal A[\mathcal W^{-1}]$ be the localization of $\mathcal A$ at $\mathcal W$ (so $\mathcal A[\mathcal W^{-1}]$ is an $\infty$-category). Let's assume that $\mathcal W$ is stable under finite direct sums, so that $\mathcal A[\mathcal W^{-1}]$ is an [additive $\infty$-category](https://ncatlab.org/nlab/show/additive+%28infinity%2C1%29-category), and hence admits a (unique) enrichment in the $\infty$-category $Sp\_{\geq 0}$ of connective spectra.
**Question:** Under what conditions does the $Sp\_{\geq 0}$-enrichment of $\mathcal A[\mathcal W^{-1}]$ lift to an enrichment in connective $H\mathbb Z$-modules (= topological abelian groups = connective chain complexes)?
I want to say the answer is "always", but probably there is some technical condition which must be assumed, since I don't have a great handle on how to think of $H\mathbb Z$-modules in a "model-independent" way... The idea should be that the strictness of the additive structure on $\mathcal A$ itself translates to strictness on the additive structure of $\mathcal A[\mathcal W^{-1}]$.
I'd be happy to understand things under more restrictive hypotheses, e.g. hypotheses implying that $\mathcal A[\mathcal W^{-1}]$ is not just additive but in fact stable.
| https://mathoverflow.net/users/2362 | When is an $\infty$-categorical localization of an additive 1-category enriched in topological abelian groups? | The answer is indeed always.
The fact that $\mathcal A$ is an additive 1-category makes it canonically a module over $Proj\_\mathbb Z$, the 1-category of finitely generated projective $\mathbb Z$-modules.
Your assumption on $W$ makes this action compatible with $W$, and because localization is a product-preserving procedure, $\mathcal A[W^{-1}]$ acquires the structure of a module over $Proj\_\mathbb Z$.
Now for an additive $\infty$-category $\mathcal B$, a module structure over $Proj\_\mathbb Z$ induces a lift to $D\_{\geq 0}(\mathbb Z)$ of the mapping connective spectra of $\mathcal B$ (in fact, an actual enrichment but this is more complicated to construct). This follows from the equivalence $D\_{\geq 0}(\mathbb Z) \simeq Fun^\times((Proj\_\mathbb Z)^{op}, Spaces)$ induced by the restricted Yoneda embedding.
| 3 | https://mathoverflow.net/users/102343 | 446007 | 179,722 |
https://mathoverflow.net/questions/446014 | 6 | Is there a known example of a set $S$ of Diophantine equations such that
1. $S$ is computable;
2. it is a theorem that every equation in $S$ has (at most) finitely many solutions;
3. the function that maps an element of $S$ to its set of solutions is uncomputable?
There are some famous finiteness theorems in number theory whose proofs are ineffective; e.g., Faltings's theorem that a curve of genus at least 2 has at most finitely many rational points, but in the examples I can think of, there is no proof that an algorithm does not exist—we just don't know of one.
By the way, although I said "Diophantine equations"—implying integer solutions—I'd be satisfied with an example where we're considering solutions in some other ring with a countably infinite number of elements (e.g., $\mathbb{Q}$).
| https://mathoverflow.net/users/3106 | Hilbert's tenth problem for equations with finitely many solutions | Yes (assuming that you're representing finite sets in an appropriately canonical manner). The proof of the MRDP theorem gives a stronger result: there is a computable function $f$ such that, for every $e$, the c.e. set $W\_e$ is equal to the $f(e)$th Diophantine set $D\_{f(e)}$ (in some fixed standard enumerations of each). This lets us transfer all the usual undecidability results around properties of c.e. sets to properties of Diophantine sets.
In particular, the fact that there is a computable sequence of indices for finite c.e. sets such that the corresponding sequence of *canonical* indices for *finite* sets is not computable gives the result you want.
| 8 | https://mathoverflow.net/users/8133 | 446016 | 179,725 |
https://mathoverflow.net/questions/446011 | 3 | Consider an $\left(\infty, 1\right)$-category $\mathcal{C}$ with a terminal object $1$. (I'm particularly interested in the case where $\mathcal{C}$ is a topos.) It is known that the forgetful functor $U$ from the under-category ${1}/{\mathcal{C}}$ to $\mathcal{C}$ creates colimits of diagrams with weakly contractible index categories. Is there an example of a diagram $F : \mathcal{I} \to {1}/{\mathcal{C}}$ such that
* $\mathcal{I}$ is *not* weakly contractible
* $U$ still creates the colimit of $F$?
| https://mathoverflow.net/users/503911 | Does the forgetful functor from a pointed $\left(\infty, 1\right)$-category only create weakly contractible colimits? | A short answer is that if $C$ is the terminal category, then $1/C \to C$ is an equivalence, and hence creates all colimits.
Less flippantly, we can take $C$ to be the ordinary topos of sets. Then the forgetful functor creates $I$-shaped colimits if $I$ is connected.
If $C$ is cocomplete, then the colimit of $F$ in $1/C$ is the pushout of a diagram
$$
1 \leftarrow colim\_I 1 \to colim\_I F
$$
and so the critical question is whether the colimit of the constant diagram with value $1$ is equivalent to $1$.
| 3 | https://mathoverflow.net/users/360 | 446018 | 179,726 |
https://mathoverflow.net/questions/446009 | 16 | Reading about ultraproducts in model theory and in Banach spaces leads to two distinct definitions. E.g., for an ultrapower given by an ultrafilter $\mu$ on $\mathbb{N}$, both notions of ultrapower are quotients of the space of sequences $\sigma: \mathbb{N} \to \mathbb{R}$.
**In model theory.** Two sequences $\sigma,\tau$ are equivalent in the case that $\{ i \in \mathbb{N} : \sigma\_i = \tau\_i \} \in \mu$. For example, the hyperreals are a model-theoretic ultrapower $\lim\_\mu \mathbb{R}$.
**For Banach spaces** (noting we also restrict to bounded sequences). Two sequences $\sigma,\tau$ are equivalent in the case that $\lim\_\mu \| \sigma\_i - \tau\_i \| = 0$. For example, any ultrapower $(\mathbb{R})\_\mu$ is isomorphic to $\mathbb{R}$.
Putting aside the difference that in Banach spaces we only look at bounded sequences, these two equivalence relations are not the same. Therefore, it seems confusing to call both constructions an *ultraproduct*. This leads to my question:
*Is there any argument that shows that the difference between both of these constructions is negligible?* Perhaps there are a few possible answers:
* There is almost a quotient map $\lim\_\mu \mathbb{R} \to (\mathbb{R})\_\mu$ for which one could lift any statement about the Banach space ultrapower to the model-theoretic one.
* The additional “non-standard” elements in the model-theoretic object are not useful for Banach spaces. *Though in that case, why call the Banach space construction an ultraproduct?!*
* There isn’t any formal result relating the two constructions, but they are at least similar-looking.
Are there any perspectives that fit into the first two points — or at least does not fit into the last point?
| https://mathoverflow.net/users/153883 | Ultraproducts of Banach spaces versus model theoretic ultraproduct | The ultraproduct of Banach spaces is the ultraproduct in the sense of metric structures in continuous logic. For a nice survey on this topic, see [Model theory for metric structures](https://faculty.math.illinois.edu/%7Ehenson/cfo/mtfms.pdf) by Ben Yaacov, Berenstein, Henson, and Usvyatsov. The ultraproduct of metric structures is defined in Section 5, pp. 23-27.
The restriction to bounded sequences in the Banach space ultraproduct is connected to the fact that in the definition of a metric structure, each sort carries the structure of a bounded metric space. One way to view a Banach space as a metric structure is to have a sort $B\_n$ for each unit ball of radius $n$, together with the inclusion maps $B\_n\to B\_m$ when $n<m$. Then in the ultraproduct construction, we only ever consider sequences from a fixed sort.
Every structure in the sense of ordinary model theory is a metric structure with the discrete metric, so indeed these two notions of ultraproduct are instances of the same construction.
| 18 | https://mathoverflow.net/users/2126 | 446021 | 179,728 |
https://mathoverflow.net/questions/446025 | 13 | I know not all groups can be realized as the automorphism group of a group. For example, it is well-known that no group can have $\mathbb Z/n\mathbb Z$, with $n > 1$ odd, as automorphism group. Now I'm wondering the same question about automorphism groups of rings, is there any result about this?
Since the inverse Galois problem (EDIT: over $\mathbb Q$) is widely believed to have an affirmative answer, I guess there is no obstruction for a finite group to be the automorphism group of a field, in particular of a ring, but what about infinite groups?
| https://mathoverflow.net/users/133679 | Is every group the automorphism group of a ring? | **EDIT:** Apologies for the delay. I was too tired yesterday for anything more than a few comments. So let me turn this into a proper answer for everyone's sake, with a few more details added, and recap the discussion from yesterday.
**1. Case: $G$ finite**
It follows from Artin's theorem that every finite group $G$ is the Galois group of *some* finite Galois extension $L/L^G$, but we have no control over $L^G$ (this would be the inverse Galois problem). As Keith Conrad points out in the comments, however, this is a non-problem in the context of OP's question, because every finite group is the automorphism group of some finite, not necessarily Galois extension $L/\mathbb{Q}$ as shown in **[Fr80]**. A somewhat simpler proof of this is also given in **[Ge83]** (in German). This settles OP's question in the finite case.
**2. Case: $G$ infinite**
An infinite Galois group in the sense of infinite Galois theory is necessarily profinite. Conversely, every profinite group is the Galois group (in the sense of infinite Galois theory) of *some* extension $L/L^G$ as shown in **[Wat74]**, alas once again we have no control over $L^G$.
But once again this is a non-problem in the context of OP's question. Even more generally it is proved in **[DugGöb87]** that for every prescribed infinite (not just profinite) group $G$ and every prescribed base field $K$ there exists an extension $L = L(G,K) / K$ (in their notation $R(K,G)$) such that $\operatorname{Aut}\_K(L) \cong G$. Taking $K:=\mathbb{Q}$, we get $G \cong \operatorname{Aut}(L)$ (the full automorphism group of $L$), which settles OP's question in the infinite case.
Note that, as pointed out by Keith Conrad in the comments, the class of infinite groups is much wider than that of profinite groups, for example because profinite groups are either finite or uncountably infinite. I implicitly (and **incorrectly**) assumed from the paper's title that in the profinite case their construction would in fact produce a Galois extension of $K$. But firstly, this was overly optimistic, given that we don't even know the full answer for a general finite $G$ and $K:=\mathbb{Q}$, and secondly, we can simply take $K \cong \bar{K}$ algebraically closed (which in turn of course does not admit proper algebraic extensions). So, beware that their usage of the term "Galois group" is very **non-standard** as far as infinite Galois theory is concerned, as pointed out by Keith Conrad in the comments.
I hope I haven't forgotten anything.
**References:**
**[DugGöb87]** [Manfred Dugas and Rüdiger Göbel. “All Infinite Groups Are Galois Groups Over Any Field.” Transactions of the American Mathematical Society, vol. 304, no. 1, 1987, pp. 355–84](https://www.jstor.org/stable/2000718)
**[Fr80]** [M. Fried."A note on automorphism groups of algebraic number fields".Proc. Amer. Math. Soc. 80 (1980), 386-388](https://www.ams.org/journals/proc/1980-080-03/S0002-9939-1980-0580989-8/S0002-9939-1980-0580989-8.pdf)
**[Ge83]** [Geyer, WD. Jede endliche Gruppe ist Automorphismengruppe einer endlichen Erweiterung $K|\mathbb{Q}$. Arch. Math 41, 139–142 (1983)](https://link.springer.com/article/10.1007/BF01196869)
**[Wat74]** [William C. Waterhouse."Profinite groups are Galois groups".Proc. Amer. Math. Soc. 42 (1974), 639-640](https://www.ams.org/journals/proc/1974-042-02/S0002-9939-1974-0325587-3/S0002-9939-1974-0325587-3.pdf)
| 11 | https://mathoverflow.net/users/1849 | 446028 | 179,731 |
https://mathoverflow.net/questions/418993 | 3 | **THE QUESTION**
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> > Let $(X,\mathcal{X})$ be a standard Borel space, $T \colon X \to X$ a measurable map, and $\mu$ a $T$-mixing probability measure.
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> > $$ \#\{n \in \mathbb{N} : |\mu(A \cap T^{-n}(B))-\mu(A)\mu(B)| \geq \varepsilon \} \leq N\_{A,\varepsilon} \text{?} $$
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Below, I will discuss the meaning and significance of this question in much more detail.
---
**-- BUT FIRST, A PROGRESS UPDATE --**
*By adapting Ronnie Pavlov's very nice comments, I can give a result in the direction towards my question, plus a consequent possible strategy for obtaining an affirmative answer to my question.*
Let $(X,\mathcal{X})$ be a standard Borel space, $T \colon X \to X$ a measurable map, and $\mu$ a $T$-invariant probability measure. For each $A,B \in \mathcal{X}$ write
$$ \rho\_n(A,B) = |\mu(A \cap T^{-n}(B))-\mu(A)\mu(B)|. $$
**Basic Proposition.** *To show that $(X,\mathcal{X},\mu,T)$ has the property described in my question, it is sufficient just to consider $A$ belonging to a set $\mathcal{Y} \subset \mathcal{X}$ where $\mathcal{Y}$ is closed under pairwise intersections and $\sigma(\mathcal{Y})=\mathcal{X}$.*
**Proposition A.** *If $(X,\mathcal{X},\mu,T)$ is mixing then for every $A \in \mathcal{X}$, every sequence $(B\_k)\_{k \in \mathbb{N}}$ in $\mathcal{X}$ and every $\varepsilon>0$,*
$$ \#\{ n \in \mathbb{N} : \#\{k \in \mathbb{N} : \rho\_n(A,B\_k) < \varepsilon \} < \infty \} < \infty\text{.} $$
(Note that taking constant sequences $B\_k=B$ precisely recovers the usual definition of mixing. So this result says that mixing automatically implies a seemingly slightly stronger version of mixing.)
*With this, we have the following potential strategy for proving an affirmative answer to the question.*
**Definition.** I will say that *$(X,\mathcal{X},\mu,T)$ has re-orderable correlations* if there exists a set $\mathcal{Y} \subset \mathcal{X}$ closed under pairwise intersections and with $\sigma(\mathcal{Y})=\mathcal{X}$, such that for every $A \in \mathcal{Y}$ there exists an unbounded sequence $(r\_n)\_{n \in \mathbb{N}}$ in $\mathbb{N}$, a sequence $(M\_N)\_{N \in \mathbb{N}}$ in $\mathbb{N}$ and a function $\delta \colon (0,\infty) \to (0,\infty)$ such that for every $\varepsilon>0$ and $N \in \mathbb{N}$, if there exists $B \in \mathcal{X}$ with
$$ \#\{n \in \mathbb{N} : \rho\_n(A,B) \geq \varepsilon\} \geq M\_N $$
then there exists $\tilde{B} \in \mathcal{X}$ with
$$ \rho\_{r\_n}(A,\tilde{B}) \geq \delta(\varepsilon) \quad \forall \, n \in \{1,\ldots,N\}\text{.} $$
**Proposition B.** *If $(X,\mathcal{X},\mu,T)$ is mixing and has re-orderable correlations, then $(X,\mathcal{X},\mu,T)$ has the property described in my question.*
So a possible strategy towards proving an affirmative answer is to show that mixing (or perhaps something weaker than mixing) implies re-orderable correlations.
For example, Ronnie Pavlov has shown in the comments that a two-sided Bernoulli shift has re-orderable correlations (with $\mathcal{Y}$ the set of cylinder sets and with $\delta(\varepsilon)=\varepsilon$), and so a two-sided Bernoulli shift fulfils the property described in my question.
Proofs of the above results will be given in the "PROOFS" section at the end of this post.
**-- END OF PROGRESS UPDATE --**
Now I will return to discussing the significance of the question in more detail.
---
**Introduction**
My question is about whether a mixing measure-preserving dynamical system necessarily exhibits a certain (very weak) kind of "uniformity" of mixing.
The question is particularly motivated by consideration of random dynamical systems (RDS): For a RDS, one can define notions of "almost-sure decay of random correlations" for observables defined over the product of the state space and the underlying probability space; but the result of such a definition is that when we reduce to the "deterministic" case in which the action of the RDS is independent of the noise, since the observables still incorporate the noise, the property of "almost-sure decay of random correlations" does not clearly reduce to classical "decay of correlations" (i.e. mixing): rather, it needs to incorporate at least some level of "extra uniformity" to take account of the randomness in the observable. But perhaps this "extra uniformity" is automatically guaranteed, in which case there is no problem - this is the motivation behind my question.
**The structure of the rest of this post is as follows:**
* First, I will introduce a relevant notion of convergence of functions that lies strictly between pointwise convergence and uniform convergence.
* Then, I will define (for classical measure-preserving dynamical systems) both classical mixing and the kind of "uniformity" of mixing that I am asking about in my question. (I will then re-pose my question in terms of the terminology introduced.)
* Then, I will describe how my stronger version of mixing is precisely equivalent to saying that the system is guaranteed to have "almost-sure decay of random correlations" when considered as a trivial RDS over a noise space.
* Finally, I will give all the proofs of results stated along the way.
---
**"Uniform convergence modulo re-ordering"**
Suppose we have a set $S$, a function $f \colon S \to \mathbb{R}$ and a sequence $(f\_n)\_{n \in \mathbb{N}}$ of functions $f\_n \colon S \to \mathbb{R}$.
**Definition.** We say that *$f\_n \to f$ uniformly modulo re-ordering as $n \to \infty$* if for all $\varepsilon > 0$ there exists $N \in \mathbb{N}$ such that every $x \in S$ has
$$ \#\{n \in \mathbb{N} : |f\_n(x)-f(x)| \geq \varepsilon \} \leq N\text{.} $$
[The name "uniformly modulo re-ordering" is based on characterisation 2 in Proposition 1 below.]
Note that uniform convergence modulo re-ordering is indeed an asymptotic property: if $f\_n \to f$ uniformly modulo re-ordering and we have a sequence of functions $\tilde{f}\_{\!\!n} \colon S \to \mathbb{R}$ such that $\tilde{f}\_{\!\!n}=f\_n$ for all sufficiently large $n$, then $\tilde{f}\_{\!\!n} \to f$ uniformly modulo re-ordering. (Taking $\tilde{N}$ such that $\tilde{f}\_{\!\!n}=f\_n$ for all $n \geq \tilde{N}$, just replace $N$ with $N+\tilde{N}$.)
**Other characterisations of uniform convergence modulo re-ordering**
We now give a couple of alternative characterisations of uniform convergence modulo re-ordering (one of which assumes additional measurable structure).
Given any $\mathcal{N} \subset \mathbb{N}$, we will say that a function $\pi \colon \mathbb{N} \to \mathbb{N}$ is *$\mathcal{N}\!$-almost bijective* if $\pi$ is injective and $\mathbb{N} \setminus \mathcal{N} \subset \pi(\mathbb{N})$.
For each $x \in S$, let $\mathcal{N}\_x:=\{n \in \mathbb{N} : f\_n(x)=f(x)\}$.
**Proposition 1.** (A) *The following two statements are equivalent:*
1. *$f\_n \to f$ uniformly modulo re-ordering as $n \to \infty$.*
2. *There exists an $S$-indexed family $(\pi\_x)\_{x \in S}$ of functions $\pi\_x \colon \mathbb{N} \to \mathbb{N}$ with $\pi\_x$ being $\mathcal{N}\_x\!$-almost bijective for all $x \in S$, such that defining $g\_n(x):=f\_{\pi\_x(n)}(x)$, we have $g\_n \to f$ uniformly as $n \to \infty$.*
(B) *Furthermore, if $S$ is equipped with a $\sigma$-algebra $\mathcal{S}$ such that $f\_n$ and $f$ are $\mathcal{S}$-measurable, then statements 1 and 2 above are equivalent to:*
3. *For every probability measure $\mathbb{P}$ on $(S,\mathcal{S})$ and every $\varepsilon>0$,*
$$ \sum\_{n \in \mathbb{N}} \mathbb{P}(x \in S : |f\_n(x)-f(x)| \geq \varepsilon ) < \infty\text{.} $$
---
**Mixing, and my question.**
Let $(X,\mathcal{X},\mu,T)$ be a measure-preserving dynamical system. (I assume in this terminology that $\mu$ is a probability measure.) For each $n \in \mathbb{N}\_0$, we define the function $\rho\_n \colon \mathcal{X} \times \mathcal{X} \to [0,1]$ by
$$ \rho\_n(A,B) = |\mu(A \cap T^{-n}(B)) - \mu(A)\mu(B)|. $$
**Definition.** We say that $(X,\mathcal{X},\mu,T)$ is *mixing* if $\rho\_n(A,B) \to 0$ as $n \to \infty$ for all $A,B \in \mathcal{X}$.
**Remark.** If $(X,\mathcal{X},\mu,T)$ is mixing then we have that for all bounded measurable $g\_1,g\_2 \colon X \to \mathbb{R}$,
$$ \int\_X g\_1 . (g\_2 \circ T^n) \, d\mu \to \int\_X g\_1 \, d\mu \int\_X g\_2 \, d\mu \ \text{ as } n \to \infty. $$
Now for each $A \in \mathcal{X}$ and $n \in \mathbb{N}$, we write $\rho\_n(A,\,\boldsymbol{\cdot}\,) \colon \mathcal{X} \to [0,1]$ for the map $B \mapsto \rho\_n(A,B)$. So then, $(X,\mathcal{X},\mu,T)$ is mixing if and only if for each $A \in \mathcal{X}$, $\rho\_n(A,\,\boldsymbol{\cdot}\,) \to 0$ pointwise as $n \to \infty$. Hence we can define the following potentially stronger notion of mixing.
**Definition.** I will say that $(X,\mathcal{X},\mu,T)$ is *nicely mixing* if for each $A \in \mathcal{X}$, $\rho\_n(A,\,\boldsymbol{\cdot}\,) \to 0$ uniformly modulo re-ordering as $n \to \infty$.
Equivalently, $(X,\mathcal{X},\mu,T)$ is nicely mixing if for each $A \in \mathcal{X}$, as $n \to \infty$ the function $B \mapsto \mu(A \cap T^{-n}(B))$ converges to the function $B \mapsto \mu(A)\mu(B)$ uniformly modulo re-ordering.
This notion of "nicely mixing" may at first seem like a somewhat strange definition, but the Theorem further below gives another characterisation of "nicely mixing".
Now my question again:
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> > Assume that $(X,\mathcal{X})$ is a standard Borel space. Is it necessarily the case that if $(X,\mathcal{X},\mu,T)$ is mixing then $(X,\mathcal{X},\mu,T)$ is nicely mixing?
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**Remark.** Regarding the same question with actual uniform convergence in place of the weaker "uniform convergence modulo re-ordering", the answer is fairly clear: If $T$ is invertible and $\mu(A) \in (0,1)$, then we do *not* have that $\rho\_n(A,B) \to 0$ uniformly across $B \in \mathcal{X}$ as $n \to \infty$, since for each $n$ we can just set $B=T^n(A)$, in which case $\rho\_n(A,B)=\mu(A)-\mu(A)^2$ for all $n$.
---
**Motivation: Sample-pathwise mixing in random dynamical systems**
For a measurable skew-product map $\Theta \colon \Omega \times X \to \Omega \times X$ whose state space is a standard Borel space $(X,\mathcal{X})$ and whose base is an invertible measure-preserving dynamical system $(\Omega,\mathcal{F},\mathbb{P},\theta)$, given a $\Theta$-invariant measure $\mu$ on $\Omega \times X$ that projects onto $\mathbb{P}$ with disintegration $(\mu\_\omega)\_{\omega \in \Omega}$, one can define a "random correlation function" $\rho\_{g\_1,g\_2}(n,\omega)$ for a pair of bounded measurable functions $g\_1,g\_2 \colon \Omega \times X \to \mathbb{R}$, by
\begin{align\*}
&\rho\_{g\_1,g\_2}(n,\omega) = \\
&\ \left| \int\_X g\_1\!(\omega,x) \, (g\_2 \circ \Theta^n)(\omega,x) \, \mu\_\omega(dx) - \int\_X g\_1(\omega,x) \, \mu\_\omega(dx) \int\_X g\_2(\theta^n\omega,y) \, \mu\_{\theta^n\omega}(dy) \right|\text{.}
\end{align\*}
For each $g\_1,g\_2$, this is well-defined up to $\mathbb{P}$-a.s. equality.
(Such "random correlation functions" appear in e.g. the question [Two mixing rates of random dynamical system](https://mathoverflow.net/questions/327938/) and the paper [Quenched decay of correlations for slowly mixing systems](https://doi.org/10.1090/tran/7811).)
It might seem natural to define "almost-sure mixing" as follows: for every pair of bounded measurable functions $g\_1,g\_2 \colon \Omega \times X \to \mathbb{R}$, we have that for $\mathbb{P}$-almost all $\omega \in \Omega$, $\rho\_{g\_1,g\_2}(n,\omega) \to 0$ as $n \to \infty$.
However, if the answer to my question is *no*, then this definition does not "reduce to classical mixing" when the skew-product structure is a direct product structure. To be precise, we have the following theorem (which, for simplicity, I will formulate just in terms of indicator-function observables rather than general bounded measurable observables, although I expect the theorem to generalise to bounded measurable observables).
**Theorem.** *A measure-preserving dynamical system $(X,\mathcal{X},\mu\_X,T)$ is nicely mixing if and only if for every invertible measure-preserving dynamical system $(\Omega,\mathcal{F},\mathbb{P},\theta)$, taking $\Theta(\omega,x):=(\theta\omega,T(x))$ and $\mu:=\mathbb{P} \otimes \mu\_X$, every $A,B \in \mathcal{F} \otimes \mathcal{X}$ has $\rho\_{\mathbf{1}\_A,\mathbf{1}\_B}(n,\,\boldsymbol{\cdot}\,) \overset{\mathbb{P}\text{-a.s.}}{\to} 0$ as $n \to \infty$.*
The "if" direction can, more specifically, take the following form:
**Proposition 2.** *Suppose $(\Omega,\mathcal{F},\mathbb{P},\theta)$ is the Bernoulli shift on the sequence space $\Omega=[0,1]^\mathbb{Z}$, where $[0,1]$ is equipped with the Lebesgue measure. Suppose $\Theta = \theta \times T$ and $\mu=\mathbb{P} \otimes \mu\_X$, where $(X,\mathcal{X},\mu\_X,T)$ is not nicely mixing. Then one can find $A,B \in \mathcal{F} \otimes \mathcal{X}$ such that for $\mathbb{P}$-almost every $\omega \in \Omega$, $\rho\_{\mathbf{1}\_A,\mathbf{1}\_B}(n,\omega) \not\to 0$ as $n \to \infty$.*
So the theorem holds even if "every invertible measure-preserving dynamical system" is replaced by "every Bernoulli automorphism".
The proof of the Theorem is based on the Borel-Cantelli Lemmas (used in a similar manner to the answer to my question [Does a sequence of coin-tosses a.s. have a subsequence on which the remainder of the sequence can be identified with the position in the sequence?](https://mathoverflow.net/questions/418835/)).
**Remark.** One other approach that I've seen to studying mixing in RDS is simply to consider $\omega$-independent bounded observables $g\_1,g\_2 \colon X \to \mathbb{R}$; but a potential problem with this as an approach towards defining mixing is that it respects the Cartesian-product structure of $\Omega \times X$ (as opposed to treating $\Omega \times X$ just as a space equipped with a projection onto $\Omega$), which goes somewhat contrary to the general philosophy of random dynamical systems theory.
---
**PROOFS.**
*Proof of Proposition 1(A).* First assume 1. It is clear that $f\_n \to f$ pointwise as $n \to \infty$, and so in particular, for every $x \in X$ and every non-empty subset $A$ of $\mathbb{N}$, the set $\{|f\_n(x)-f(x)| : n \in A\}$ has a maximum. Therefore, for each $x \in X$, by recursion we can construct an injective function $\pi\_x \colon \mathbb{N} \to \mathbb{N}$ such that for each $n \in \mathbb{N}$,
$$ |f\_{\pi\_x(n)}(x)-f(x)| = \max \{|f\_m(x)-f(x)| : m \in \mathbb{N} \!\setminus\! \pi\_x(\mathbb{N}\_{<n}) \}\text{.} $$
It is clear that $n \mapsto |f\_{\pi\_x(n)}(x)-f(x)|$ is decreasing and hence convergent; and since $\pi\_x$ is injective and $|f\_n(x)-f(x)| \to 0$ as $n \to \infty$, it is then clear that the limit $\lim\_{n \to \infty} |f\_{\pi\_x(n)}(x)-f(x)|$ is $0$. Since, again, $n \mapsto |f\_{\pi\_x(n)}(x)-f(x)|$ is decreasing, it then clearly follows that every element of $\mathbb{N} \setminus \mathcal{N}\_x$ is in the range of $\pi\_x$. So $\pi\_x$ is $\mathcal{N}\_x\!$-almost bijective. It remains to show that $g\_n \to f$ uniformly. Fix $\varepsilon>0$, and let $N$ be as in the definition of uniform convergence modulo re-ordering. Then for all $x \in S$ and $n > N$, since every $m < n$ has $|f\_{\pi\_x(m)}(x)-f(x)| \geq |f\_{\pi\_x(n)}(x)-f(x)|$ and we also have (by injectivity of $\pi\_x$) that
$$ \#\{m \in \mathbb{N} : |f\_{\pi\_x(m)}(x)-f(x)| \geq \varepsilon \} \leq N\text{,} $$
it follows that $|f\_{\pi\_x(n)}(x)-f(x)|<\varepsilon$, as required.
Now suppose 2. Fix $\varepsilon>0$, and let $N$ be such that every $n > N$ and $x \in S$ has $|g\_n(x)-f(x)|<\varepsilon$. Then for each $x \in S$, every $n \in \mathbb{N}$ with $|f\_n(x)-f(x)| \geq \varepsilon$ has $n \in \pi\_x(\mathbb{N})$ and $\pi^{-1}(n) \leq N$. $\quad\square$
*Proof of Proposition 1(B).* For convenience, write $S\_{n,\varepsilon}:=\{x \in S : |f\_n(x)-f(x)| \geq \varepsilon\}$.
First assume 1. Fix $\varepsilon>0$, and let $N$ be as in the definition of uniform convergence modulo re-ordering. For any $R \subset \mathbb{N}$ with $\#R > N$, we have $\bigcap\_{n \in R} S\_{n,\varepsilon} = \emptyset$. So
$$ \sum\_{n \in \mathbb{N}} \mathbb{P}(S\_{n,\varepsilon}) = \sum\_{i=1}^N \mathbb{P}\!\left( \bigcup\_{R \subset \mathbb{N}, \, \#R=i} \ \bigcap\_{n \in R} S\_{n,\varepsilon} \right) \leq N\text{.} $$
Now assume that 1 fails (i.e. $f\_n$ does not converge to $f$ uniformly modulo re-ordering), and take a counterexemplary $\varepsilon>0$. For each $N \in \mathbb{N}$, choose $x\_N \in S$ such that
$$ \#\{n \in \mathbb{N} : x \in S\_{n,\varepsilon} \} \geq 3^N\text{.} $$
Take
$$ \mathbb{P} = \sum\_{N \in \mathbb{N}} 2^{-N}\delta\_{x\_N}. $$
Then $\sum\_{n \in \mathbb{N}} \mathbb{P}(S\_{n,\varepsilon})$ is at least $(\frac{3}{2})^N$ for all $N$, and hence is infinite. $\quad\square$
*Proof of "only if" direction in Theorem.* Suppose that $(X,\mathcal{X},\mu\_X,T)$ is nicely mixing, and take any invertible measure-preserving dynamical system $(\Omega,\mathcal{F},\mathbb{P},\theta)$. Let $\mathcal{A}$ be the set of all $A \in \mathcal{F} \otimes \mathcal{X}$ with the property that every $B \in \mathcal{F} \otimes \mathcal{X}$ has $\rho\_{\mathbf{1}\_A,\mathbf{1}\_B}(n,\,\boldsymbol{\cdot}\,) \overset{\mathbb{P}\text{-a.s.}}{\to} 0$ as $n \to \infty$. We will show that
* $\mathcal{A}$ is closed under complements and countable disjoint unions;
* $\mathcal{A}$ includes all rectangles $A=E \times Y$ with $E \in \mathcal{F}$ and $Y \in \mathcal{X}$.
Dynkin's $\pi$-$\lambda$ theorem then gives the desired result.
Given a set $A \subset \Omega \times X$ and a point $\omega \in \Omega$, we write $A\_\omega \subset X$ for the $\omega$-section of $A$. Note that
$$ \rho\_{\mathbf{1}\_A,\mathbf{1}\_B}(n,\omega) = |\mu\_X(A\_\omega \cap T^{-n}(B\_{\theta^n\omega}))-\mu\_X(A\_\omega)\mu\_X(B\_{\theta^n\omega})|. $$
It is easy to check that $\rho\_{\mathbf{1}\_{(\Omega \times X) \setminus A},\mathbf{1}\_B}(n,\omega)=\rho\_{\mathbf{1}\_A,\mathbf{1}\_B}(n,\omega)$, and so $\mathcal{A}$ is closed under complements. Now let $(A(k))\_{k \in \mathbb{N}}$ be a mutually disjoint sequence in $\mathcal{A}$, take any $B \in \mathcal{F} \otimes \mathcal{X}$, and let $\Omega' \subset \Omega$ be a $\mathbb{P}$-full measure set such that for all $\omega \in \Omega$ and $k \in \mathbb{N}$, $\rho\_{\mathbf{1}\_{A(k)},\mathbf{1}\_B}(n,\omega) \to 0$ as $n \to \infty$. Letting $A=\bigcup\_{k \in \mathbb{N}} A(k)$, the triangle inequality gives
$$ \rho\_{\mathbf{1}\_A,\mathbf{1}\_B}(n,\omega) \leq \sum\_{k \in \mathbb{N}} \rho\_{\mathbf{1}\_{A(k)},\mathbf{1}\_B}(n,\omega)\text{.} $$
Now for each $\omega$, $\sum\_{k \in \mathbb{N}} \mu\_X(A(k)\_\omega) \leq 1$ and $\rho\_{\mathbf{1}\_{A(k)},\mathbf{1}\_B}(n,\omega) \leq \mu\_X(A(k)\_\omega)$. Hence the dominated convergence theorem for discrete sums can be applied to give that for all $\omega \in \Omega'$,
$$ \sum\_{k \in \mathbb{N}} \rho\_{\mathbf{1}\_{A(k)},\mathbf{1}\_B}(n,\omega) \to 0 \ \text{ as } n \to \infty $$
and so $\rho\_{\mathbf{1}\_A,\mathbf{1}\_B}(n,\omega) \to 0$ as $n \to \infty$. Thus $\mathcal{A}$ is closed under countable disjoint unions. It now remains to show that $\mathcal{A}$ includes all measurable rectangles. First note that for a measurable rectangle $A=E \times Y$, for any $B \in \mathcal{F} \otimes \mathcal{X}$, we have
$$ \rho\_{\mathbf{1}\_A,\mathbf{1}\_B}(n,\omega) = \mathbf{1}\_E(\omega)\rho\_{\mathbf{1}\_{\Omega \times Y},\mathbf{1}\_B}(n,\omega). $$
Hence it is sufficient just to consider rectangles of the form $A=\Omega \times Y$ with $Y \in \mathcal{X}$. Let us define a $\sigma$-algebra $\Sigma\_{\mu\_X,T}$ on the $\sigma$-algebra $\mathcal{X}$: namely, let $\Sigma\_{\mu\_X,T}$ be the smallest $\sigma$-algebra on $\mathcal{X}$ with respect to which the map $Z \mapsto \mu\_X(W \cap T^{-n}(Z))$ is measurable for every $W \in \mathcal{X}$ and $n \in \mathbb{N}\_0$. Now fix $Y \in \mathcal{X}$ and $B \in \mathcal{F} \otimes \mathcal{X}$, and let $A=\Omega \times Y$. The map
$$ \omega \ \ \mapsto \ \ \mu\_X(W \cap T^{-n}(B\_\omega)) = \! \int\_X \mathbf{1}\_W(x)\mathbf{1}\_B(\omega,T^n(x)) \, \mu\_X(dx) $$
is $\mathcal{F}$-measurable for all $W \in \mathcal{X}$ and $n \in \mathbb{N}\_0$, and so the map $\omega \mapsto B\_\omega$ is $(\mathcal{F},\Sigma\_{\mu\_X,T})$-measurable. So define the probability measure $\tilde{\mathbb{P}}$ on the measurable space $(\mathcal{X},\Sigma\_{\mu\_X,T})$ to be the pushforward measure of $\mathbb{P}$ under the map $\omega \mapsto B\_\omega$. For each $\varepsilon>0$ and $n \in \mathbb{N}$, let
$$ \mathcal{E}\_{\varepsilon,n} = \{\tilde{B} \in \mathcal{X} : |\mu\_X(Y \cap T^{-n}(\tilde{B}))-\mu\_X(Y)\mu\_X(\tilde{B})| \geq \varepsilon \}. $$
Note that $\mathcal{E}\_{\varepsilon,n} \in \Sigma\_{\mu\_X,T}$ and, since $\mathbb{P}$ is $\theta$-invariant,
$$ \mathbb{P}(\omega : \rho\_{\mathbf{1}\_A,\mathbf{1}\_B}(n,\omega) \geq \varepsilon ) = \tilde{\mathbb{P}}(\mathcal{E}\_{\varepsilon,n}). $$
Since $(X,\mathcal{X},\mu\_X,T)$ is nicely mixing, Proposition 1(B) gives that for all $\varepsilon>0$, $\sum\_{n \in \mathbb{N}} \tilde{\mathbb{P}}(\mathcal{E}\_{\varepsilon,n}) < \infty$, and so
$$ \sum\_{n \in \mathbb{N}} \mathbb{P}(\omega : \rho\_{\mathbf{1}\_A,\mathbf{1}\_B}(n,\omega) \geq \varepsilon ) < \infty. $$
Hence, by the First Borel-Cantelli Lemma, for each $\varepsilon>0$, $\mathbb{P}$-almost every $\omega \in \Omega$ has that $\rho\_{\mathbf{1}\_A,\mathbf{1}\_B}(n,\omega)<\varepsilon$ for all sufficiently large $n$; and taking a sequence of $\varepsilon$-values tending to $0$ then gives that for $\mathbb{P}$-almost every $\omega \in \Omega$, $\rho\_{\mathbf{1}\_A,\mathbf{1}\_B}(n,\omega) \to 0$ as $n \to \infty$. So $A \in \mathcal{A}$. $\quad\square$
*Proof of Proposition 2.* Choose an $\tilde{A} \in \mathcal{X}$ and $\varepsilon>0$ such that for all $N \in \mathbb{N}$ one can find $\tilde{B} \in \mathcal{X}$ with
$$ \#\{n \in \mathbb{N} : |\mu\_X(\tilde{A} \cap T^{-n}(\tilde{B}))-\mu\_X(\tilde{A})\mu\_X(\tilde{B})| \geq \varepsilon\} > N\text{.} $$
Set $A=\Omega \times \tilde{A}$. In analogy to the proof of 3$\Rightarrow$1 in Proposition 1(B), for each $N \in \mathbb{N}$ let $B\_N \in \mathcal{X}$ be such that the set
$$ R\_N := \{n \in \mathbb{N} : |\mu\_X(\tilde{A} \cap T^{-n}(B\_N))-\mu\_X(\tilde{A})\mu\_X(B\_N)| \geq \varepsilon\} $$
is of cardinality at least $3^N$. Let
$$ O = \bigcup\_{N=1}^\infty ([2^{-N},2^{1-N}] \times B\_N) \, \subset [0,1] \times X \text{,} $$
and let $B$ be the pre-image of $O$ under the projection $((\omega\_i)\_{i \in \mathbb{Z}},x) \mapsto (\omega\_0,x)$ from $\Omega \times X$ to $[0,1] \times X$. So
$$ \rho\_{\mathbf{1}\_A,\mathbf{1}\_B}(n,(\omega\_i)\_{i \in \mathbb{Z}}) = \sum\_{N=1}^\infty \mathbf{1}\_{[2^{-N},2^{1-N}]}(\omega\_n)|\mu\_X(\tilde{A} \cap T^{-n}(B\_N))-\mu\_X(\tilde{A})\mu\_X(B\_N)|. $$
Note in particular that the random variables $\rho\_{\mathbf{1}\_A,\mathbf{1}\_B}(n,\,\boldsymbol{\cdot}\,)$ are mutually $\mathbb{P}$-independent. Furthermore, writing $Q\_n:=\{N \in \mathbb{N} : n \in R\_N \}$, we have
$$ \mathbb{P}(\omega : \rho\_{\mathbf{1}\_A,\mathbf{1}\_B}(n,\omega) \geq \varepsilon ) = \sum\_{N \in Q\_n} 2^{-N} $$
for each $n$, and so
$$ \sum\_{n \in \mathbb{N}} \mathbb{P}(\omega : \rho\_{\mathbf{1}\_A,\mathbf{1}\_B}(n,\omega) \geq \varepsilon ) = \sum\_{N,n \text{ with } N \in Q\_n} 2^{-N} = \sum\_{N \in \mathbb{N}} 2^{-N}\#R\_N = \infty. $$
Hence the Second Borel-Cantelli Lemma gives that for $\mathbb{P}$-almost all $\omega$, there are infinitely many $n$ for which $\rho\_{\mathbf{1}\_A,\mathbf{1}\_B}(n,\omega) \geq \varepsilon$. $\quad\square$
**Proofs of results in progress update**
*Proof of Basic Proposition.* Let $\mathcal{A}$ be the collection of all sets $A \in \mathcal{X}$ with the property described in the question. By Dynkin's $\pi$-$\lambda$ theorem, it is sufficient to show that $\mathcal{A}$ is closed under complements and countable disjoint unions. Firstly, $\rho\_n(A,B)=\rho\_n(X \setminus A,B)$, and so $\mathcal{A}$ is closed under complements. Now take a sequence $(A\_i)\_{i \geq 1}$ in $\mathcal{A}$ that is mutually disjoint, and let $A=\bigcup\_{i \geq 1} A\_i$. For each $\varepsilon>0$, let $k(\varepsilon) \in \mathbb{N}$ be such that $\mu\!\left( \bigcup\_{i > k(\varepsilon)} A\_i \right)<\frac{\varepsilon}{2}$. For all $B \in \mathcal{X}$ and $n \in \mathbb{N}$, the triangle inequality gives
$$ \rho\_n(A,B) \leq \sum\_{i=1}^\infty \rho\_n(A\_i,B) < \tfrac{\varepsilon}{2} + \sum\_{i=1}^{k(\varepsilon)} \rho\_n(A\_i,B)\text{.} $$
Hence we may take
$$ N\_{A,\varepsilon} = \sum\_{i=1}^{k(\varepsilon)} N\_{A\_i,\frac{\varepsilon}{2k(\varepsilon)}}\text{.} \quad\square $$
To prove Proposition A, we first need the following lemma (whose proof is adapted from a comment in the question <https://math.stackexchange.com/questions/4005728/.>)
**Lemma.** *For any countable collection $\{A\_n\}\_{n \in \mathbb{N}} \subset \mathcal{X}$ there exists a compact metrisable topology on $X$ generating $\mathcal{X}$ such that for every $n$, $A\_n$ is clopen.*
*Proof.* For each $n$, let $\mathcal{P}\_n$ be the coarsest partition of $X$ for which $A\_1,\ldots,A\_n$ can all be expressed as unions of members of $\mathcal{P}\_n$. Let $(\tau\_n)\_{n \in \mathbb{N}}$ be an increasingly fine sequence of compact metrisable topologies on $X$ such that for each $n$, $\mathcal{X}=\sigma(\tau\_n)$ and every member of $\mathcal{P}\_n$ is $\tau\_n$-clopen. The increasing fineness implies that for integers $n \geq m$, if $x\_i \overset{\tau\_n}{\to} x$ as $i \to \infty$ then $x\_i \overset{\tau\_m}{\to} x$ as $i \to \infty$. Equip $X^\mathbb{N}$ with the product topology, i.e. the compact metrisable topology whose convergence corresponds to saying that for each $n \in \mathbb{N}$, the $n$-th coordinate converges in $\tau\_n$. It is then easy to see that the diagonal $\{(x,x,x,\ldots) : x \in X \}$ is closed, and so compact. Let $\tau$ be the topology on $X$ that is identified by the bijection $x \mapsto (x,x,\ldots)$ with the induced topology on $\{(x,x,\ldots) : x \in X \}$
from our topology on $X^\mathbb{N}$. Given $n \in \mathbb{N}$ and $\tilde{A} \in \mathcal{P}\_n$, if $(x\_i)\_{i \in \mathbb{N}}$ is a sequence in $\tilde{A}$ such that $(x\_i,x\_i,\ldots)$ converges to a point $(x,x,\ldots) \in X^\mathbb{N}$ as $i \to \infty$, then in particular $x\_i \overset{\tau\_n}{\to} x$, and so $x \in \tilde{A}$. Hence for all $n$, every member of $\mathcal{P}\_n$ is $\tau$-closed, and so in particular, $A\_n$ is $\tau$-clopen. $\quad\square$
*Proof of Proposition A if $T$ is invertible.* Arguing by contrapositive, suppose we have a counterexemplary $A$, $(B\_k)$ and $\varepsilon$, and write
$$ R = \{ n \in \mathbb{N} : \#\{k \in \mathbb{N} : \rho\_n(A,B\_k) < \varepsilon \} < \infty \}\text{.} $$
Fix a compact metrisable topology on $X$ whose Borel $\sigma$-algebra is precisely $\mathcal{X}$, such that the sets $T^n(A)$, $n \in \mathbb{N}\_0$, are all clopen. Since $X$ is compact, any bounded subset of $L^\infty(\mu)$ is relatively compact in the metrisable topology on $L^\infty(\mu)$ generated by the set of mappings $\{g \mapsto \int\_X f.g \, d\mu : f \in C(X) \}$. Hence we can find $g \in L^\infty(\mu)$ and a subsequence $(B\_{m\_k})$ of $(B\_k)$ such that $\int\_{B\_{m\_k}} f \, d\mu \to \int\_X f.g \, d\mu$ for all $f \in C(X)$. For each $n \in \mathbb{N}$, we have
\begin{align\*} \rho\_n(A,B\_{m\_k}) = & \ |\mu(T^n(A) \cap B\_{m\_k})-\mu(A)\mu(B\_{m\_k})| \\ \overset{k \to \infty}{\to} & \ \left| \int\_{T^n(A)} g \, d\mu - \mu(A) \int\_X g \, d\mu \right| \\ =& \ \left| \int\_A g \circ T^n \, d\mu - \mu(A) \int\_X g \, d\mu \right| =: \rho\_n(A,g)\text{,}
\end{align\*}
and so for each $n$ in the unbounded set $R$, we have $\rho\_n(A,g) \geq \varepsilon$, and so $\mu$ is not $T$-mixing. $\quad\square$
*Proof of Proposition A when $T$ is non-invertible.* We use the "natural extension" to reduce to the invertible case. It is well-known that, since $(X,\mathcal{X})$ is a standard Borel space and $\mu$ is a $T$-invariant probability measure, the set $X\_{\text{ext}} := \{(x\_i)\_{i \in \mathbb{Z}} \in X^\mathbb{Z} : T(x\_i)=x\_{i+1} \forall \, i \in \mathbb{Z} \}$ is non-empty and on $X\_{\text{ext}}$ (equipped with the induced $\sigma$-algebra from $\mathcal{X}^{\otimes \mathbb{Z}}$) there is a unique probability measure $\mu\_{\text{ext}}$ all of whose coordinate-projections coincide with $\mu$. Write $T\_{\text{ext}} \colon X\_{\text{ext}} \to X\_{\text{ext}}$ for the left-shift map
$$ T\_{\text{ext}}((x\_i)\_{i \in \mathbb{Z}}) = (x\_{i+1})\_{i \in \mathbb{Z}} = (T(x\_i))\_{i \in \mathbb{Z}}\text{.} $$
Obviously $T\_{\text{ext}}$ is invertible, and one can show that if $\mu$ is $T$-mixing then $\mu\_{\text{ext}}$ is $T\_{\text{ext}}$-mixing, in which case Proposition A applies to $(\mu\_{\text{ext}},T\_{\text{ext}})$. We then recover Proposition A for $(\mu,T)$ by taking sets $\pi\_0^{-1}(A)$ and $\pi\_0^{-1}(B\_k)$. $\quad\square$
*Proof of Proposition B.* Assume that $(X,\mathcal{X},\mu,T)$ has re-orderable correlations but fails the property described in my question. Take $\mathcal{Y}$ as in the definition of re-orderable correlations, and (on the basis of the Basic Proposition) take a counterexemplary $A \in \mathcal{Y}$ and $\varepsilon>0$ for the property described in my question. Let $(r\_n)$ be the sequence corresponding to $A$ as in the definition of re-orderable correlations. So we can find $\delta>0$ and a sequence $(\tilde{B}\_N)\_{N \in \mathbb{N}}$ in $\mathcal{X}$ such that for all $n,N \in \mathbb{N}$ with $N \geq n$ we have $\rho\_{r\_n}(A,\tilde{B}\_N) \geq \delta$. Hence in particular, since the set $\{r\_n\}\_{n \in \mathbb{N}}$ is obviously an infinite set, the conclusion of Proposition A fails by taking $\delta$ in place of $\varepsilon$. So $(X,\mathcal{X},\mu,T)$ is not mixing. $\quad\square$
| https://mathoverflow.net/users/15570 | Does mixing automatically imply this seemingly stronger "uniform modulo re-ordering" version of mixing? | The answer to my question is **yes!!** [The assumption that $(X,\mathcal{X})$ is a standard Borel space is not needed.]
A beautiful proof of a modified version of the statement has been provided by user65023, who says that it is "inspired by results in Blum-Hanson (1960): [On the mean ergodic theorem for subsequences](https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society-new-series/volume-66/issue-4/On-the-mean-ergodic-theorem-for-subsequences/bams/1183523588.full)." I now give the proof of the answer to my question, in my own words.
---
First some generalities:
Fix a set $S$, a sequence $(f\_n)\_{n \in \mathbb{N}}$ of functions $f\_n \colon S \to \mathbb{R}$, and a function $f \colon S \to \mathbb{R}$.
**Definition.** We say that $f\_n$ **converges uniformly modulo re-ordering to** $f$ if for all $\varepsilon>0$ there exists $N \in \mathbb{N}$ such that for all $x \in S$,
$$ \#\{n \in \mathbb{N} : |f\_n(x)-f(x)|\geq\varepsilon \} \leq N. $$
Now for any finite $J \subset \mathbb{N}$ with $\#J \geq 2$, let $\Delta J$ denote the smallest distance between distinct elements of $J$.
**Proposition.** *Suppose that, over the set of all finite subsets $J$ of $\mathbb{N}$, we have the following convergence*:
$$ \sup\_{x \in S} \left| f(x) - \tfrac{1}{\#J}\!\sum\_{n \in J} f\_n(x) \right| \to 0 \ \ \text{ as } \, (\#J,\Delta J) \to (\infty,\infty)\text{.} $$
*Then $f\_n$ converges uniformly modulo re-ordering to $f$.*
*Proof.* Arguing by contrapositive, suppose we have $\varepsilon>0$ and a sequence $(x\_k)\_{k \in \mathbb{N}}$ in $S$ such that $\#\{n : |f\_n(x\_k)-f(x\_k)| \geq \varepsilon\} \to \infty$ as $k \to \infty$. Without loss of generality, suppose there is a subsequence $(x\_{m\_k})$ of $(x\_k)$ such that, writing $I\_k=\{n : f(x\_{m\_k}) \geq f\_n(x\_{m\_k}) + \varepsilon\}$, we have $\#I\_k \to \infty$ as $k \to \infty$. Take a finite $J\_k \subset I\_k$ for each $k$, such that $\#J\_k$ and $\Delta J\_k$ both tend to $\infty$ as $k \to \infty$. We have that $ f(x\_{m\_k}) - \frac{1}{\#J\_k}\!\sum\_{n \in J\_k} f\_n(x\_{m\_k}) \geq \varepsilon$ for all $k$. $\ \square$
---
Now a re-statement and proof of the result:
Let $(X,\mathcal{X},\mu,T)$ be a mixing probability-preserving transformation. Fixing $A \in \mathcal{X}$, define $f\_n \colon \mathcal{X} \to \mathbb{R}$ and $f \colon \mathcal{X} \to \mathbb{R}$ by
\begin{align\*}
f\_n(B) &= \mu(A \cap T^{-n}(B)) \\
f(B) &= \mu(A)\mu(B).
\end{align\*}
**Theorem.** *$f\_n$ converges uniformly modulo re-ordering to $f$.*
We now prove this, using the above Proposition. We will equip the space of $[0,1]$-valued functions on $X$ identified up to $\mu$-almost sure equality with its natural topology, namely the topology of $L^p$ convergence for any $p \in [1,\infty)$ or equivalently the topology of convergence in probability.
Since $\mu$ is $T$-invariant, we have that for all $B \in \mathcal{X}$ and $n,N$ with $0 \leq n \leq N$,
\begin{align\*} f\_n(B) &= \int\_{T^{-N}(B)} \mathbf{1}\_{T^{n-N}(A)} \, d\mu \\
f(B) &= \int\_{T^{-N}(B)} \mu(A) \, d\mu.
\end{align\*}
So for any non-empty finite $J \subset \mathbb{N}$ and any $B \in \mathcal{X}$, applying the above with $N=\max J$ gives
$$ f(B) - \tfrac{1}{\#J}\!\sum\_{n \in J} f\_n(B) \ = \ \int\_{T^{-\max J}(B)} \mu(A) - \tfrac{1}{\#J}\!\sum\_{n \in J} \mathbf{1}\_{T^{n-\max J}(A)} \, d\mu. $$
Therefore, to be able to apply the Proposition, it will be sufficient to show the following.
**Lemma.** *Over the set of all finite subsets $J$ of $\mathbb{N}$, we have the following convergence*:
$$ \tfrac{1}{\#J}\!\sum\_{n \in J} \mathbf{1}\_{T^{n-\max J}(A)} \to \mu(A) \ \ \text{ as } \, (\#J,\Delta J) \to (\infty,\infty)\text{.} $$
*Proof.* We have
\begin{align\*}
\| \mathrm{RHS}-\mathrm{LHS} \|\_{L^2(\mu)}^2 &= \left\| \tfrac{1}{\#J}\!\sum\_{n \in J} (\mu(A) - \mathbf{1}\_{T^{n-\max J}(A)}) \right\|\_{L^2(\mu)}^2 \\
&= \left( \tfrac{1}{\#J} \right)^{\!2} \sum\_{n\_1,n\_2 \in J} \int\_X (\mu(A) - \mathbf{1}\_{T^{n\_1-\max J}(A)})(\mu(A) - \mathbf{1}\_{T^{n\_2-\max J}(A)}) \, d\mu \\
&= \left( \tfrac{1}{\#J} \right)^{\!2} \sum\_{n\_1,n\_2 \in J} \int\_X \mu(A)^2 - \mu(A)\mathbf{1}\_{T^{n\_2-\max J}(A)} - \mu(A)\mathbf{1}\_{T^{n\_1-\max J}(A)} + \mathbf{1}\_{T^{n\_1-\max J}(A)}\mathbf{1}\_{T^{n\_2-\max J}(A)} \, d\mu \\
&= \left( \tfrac{1}{\#J} \right)^{\!2} \sum\_{n\_1,n\_2 \in J} \left( \int\_X \mathbf{1}\_{T^{n\_1-\max J}(A)}\mathbf{1}\_{T^{n\_2-\max J}(A)} \, d\mu - \mu(A)^2 \right) \\
&= \left( \tfrac{1}{\#J} \right)^{\!2} \sum\_{n\_1,n\_2 \in J} \big( \mu(A \cap T^{-|n\_1-n\_2|}(A)) - \mu(A)^2 \big) \\
&= \left( \tfrac{1}{\#J} \right)^{\!2} \sum\_{n \in J} \big( \mu(A) - \mu(A)^2 \big) \ + \ \left( \tfrac{1}{\#J} \right)^{\!2} \!\!\! \sum\_{\substack{\text{distinct} \\ n\_1,n\_2 \in J}} \big( \mu(A \cap T^{-|n\_1-n\_2|}(A)) - \mu(A)^2 \big) \\
&\leq \underbrace{\tfrac{1}{\#J}}\_{\to 0} \ + \ \underbrace{\sup\_{n \geq \Delta J} \big| \mu(A \cap T^{-n}(A)) - \mu(A)^2 \big|}\_{\to 0}. \quad \square
\end{align\*}
[NB: The fact that $T$ is mixing only entered at the very final step.]
| 0 | https://mathoverflow.net/users/15570 | 446034 | 179,732 |
https://mathoverflow.net/questions/446033 | 3 | Consider the matrix $$A:=\left(
\begin{array}{cccc}
0 & a & 0 & 0 \\
f & 0 & b & 0 \\
0 & e & 0 & c \\
0 & 0 & d & 0 \\
\end{array}
\right)$$
I noticed that if I square this matrix then the eigenvalues of $A^2$ are two-fold degenerate. Does anyone see how this follows? I don't want an explicit computation but rather an argument that generalizes to arbitrary matrix sizes.
The same phenomenon follows (aside from an eigenvalue 0, since the matrix size is odd) if the matrix is continued analogously, i.e. if I consider
$$ A:=\left(
\begin{array}{ccccc}
0 & a & 0 & 0 & 0 \\
f & 0 & b & 0 & 0 \\
0 & e & 0 & c & 0 \\
0 & 0 & d & 0 & r \\
0 & 0 & 0 & g & 0 \\
\end{array}
\right).$$
| https://mathoverflow.net/users/496243 | Eigenvalues two-fold degenerate | For the eigenvalues of the matrix powers the following identity holds:
>
> If $A$ is a square ($d \times d$) matrix with associated eigenvalues $\lambda\_1,\dots,\lambda\_d$, then the eigenvalues of $A^n$ are
> $$\lambda\_1^n,\dots,\lambda\_d^n$$.
>
>
>
This can be shown by considering the eigenvalue/eigenvector equality for $A$
$$A\;\mathbf{e}\_i=\lambda\_i \mathbf{e}\_i,$$
where $\mathbf{e}\_1, \mathbf{e}\_2, \dots, \mathbf{e}\_d$ are the eigenvectors, and take the product of the power of $A$ with an eigenvector and progressively replace products of $A$ with the eigenvector with $\lambda\_i \mathbf{e}\_i$
\begin{align}
A^d \mathbf{e}\_i&=A^{d-1} A\mathbf{e}\_i \\&=A^{d-1}\lambda\_i \mathbf{e}\_i\\&=\lambda\_i A^{d-2} A\mathbf{e}\_i \\&= \dots\\&=\lambda\_{i}^{d}\mathbf{e}\_i.\end{align}
In your case the eigenvalues of $A$ are
$\lambda\_1 \approx -0.707107 \sqrt{-\sqrt{(-a f - b e - c d)^2 - 4 a c f d} + a f + b e + c d} $
$\lambda\_2 \approx 0.707107 \sqrt{-\sqrt{(-a f - b e - c d)^2 - 4 a c f d} + a f + b e + c d} $
$\lambda\_3 \approx-0.707107 \sqrt{\sqrt{(-a f - b e - c d)^2 - 4 a c f d} + a f + b e + c d} $
$\lambda\_4 \approx 0.707107 \sqrt{\sqrt{(-a f - b e - c d)^2 - 4 a c f d} + a f + b e + c d} $
so the pairs $(\lambda\_1, \lambda\_2 )$ and $(\lambda\_3, \lambda\_4 )$ have same magnitudes and opposite signs, which results in them being equal pairwise once you square them.
| 1 | https://mathoverflow.net/users/483817 | 446035 | 179,733 |
https://mathoverflow.net/questions/427090 | 2 | Let $(X, T, \mathcal F, \mu)$ be a nonatomic standard probability space equipped with a measure preserving transformation $T$. We say $T$ is *uniformly weak mixing* if for every $\varepsilon > 0$, there exists some $N > 0$ such that for all measurable sets $A, B \in \mathcal F$
$$| \frac{1}{n} \sum\_{k=1}^n \mu(T^{-k} A \cap B) - \mu(A) \mu(B)| < \varepsilon$$
for all $n > N$.
**Question:** If $T$ is uniformly weak mixing, then does it hold that for all $f \in L^\infty (X)$, there exists some measurable set $E$ of full measure such that
$$\frac{1}{n} \sum\_{k=1}^n T^k f \to \int f d\mu$$
uniformly on $E$?
| https://mathoverflow.net/users/173490 | Uniformly weak mixing transformations | @John Griesmer answered this question: "I don't think there is a system that satisfies your definition of uniformly weak mixing. Such a system must be ergodic, and therefore admit Rohlin towers. Given $>0$ and a Rohlin tower $\{,,…,^{−1}\}$ with $=()$, you can set $==\cup \cup \ldots \cup ^{/2}$ and find that $\frac{1}{+1}\sum^{+1}\_{=1}\mu(^{−}\cap )\approx \mu()\approx 1/2$. In other words, given an initial segment of integers, every ergodic MPS on a nonatomic probability space admits a subset which is approximately invariant under that segment."
| 1 | https://mathoverflow.net/users/503770 | 446036 | 179,734 |
https://mathoverflow.net/questions/445810 | 0 | Let $\mathcal{T}$ be a triangulated category that is generated by one object, say $A$ in the sense that the smallest triangulated subcategory containing $A$ and closed under coproducts and isomorphisms is $\mathcal{T}$ itself. Denote by $K\_0(\mathcal{T})$ the Grothendieck group of $\mathcal{T}$. Is $K\_0(\mathcal{T})$ generated by the equivalence class of $A$?
| https://mathoverflow.net/users/45397 | Generators of triangulated category and Grothendieck groups | Fernando Muro has already answered this in the comments, but perhaps a reference would help. This is all spelled out in Neeman's book on triangulated categories. In particular, see Definition 4.5.8 and Proposition 4.5.11, and their proofs.
| 3 | https://mathoverflow.net/users/11540 | 446039 | 179,736 |
https://mathoverflow.net/questions/446040 | 6 | I have the following recurrence relation and boundary condition?
$$
f(n,m) = \frac{\alpha n}{n+m} f(n-1,m) + \frac{\beta m}{n+m} f(n,m-1) + 1
$$
$$
f(n,0) = \frac{1-\alpha^{n+1}}{1-\alpha}, f(0,m) = \frac{1-\beta^{m+1}}{1-\beta}
$$
Is it possible to get a exact solution for this recurrence relation?
| https://mathoverflow.net/users/503932 | How to solve recurrence relation with 2 variables？ | Let
$$g(n,m)=\sum\limits\_{i=0}^{n}\sum\limits\_{j=0}^{m}\binom{i+j}{j}\binom{n-i+m-j}{m-j}\alpha^i\beta^j$$
I conjecture that
$$g(n,m)=\binom{n+m}{m}f(n,m)$$
Here is the PARI prog to verify this conjecture:
```
f(n, m) = if(n==0 || m==0, (m==0)*(1 - a^(n+1))/(1-a)+(n==0)*(1 - b^(m+1))/(1-b)-(n==0 && m==0), a*n/(n+m)*f(n-1, m)+b*m/(n+m)*f(n, m-1)+1)
g(n, m) = sum(i=0, n, sum(j=0, m, binomial(i+j,j)*binomial(n-i+m-j,m-j)*a^i*b^j))
test(n, m) = g(n, m)==binomial(n+m, m)*f(n, m)
n=6; x=sum(i=0, n, sum(j=0, n, test(i, j)))
```
Here is the proof of a user with the nickname **svv** on a scientific forum [dxdy.ru](https://dxdy.ru/) (here is the [link to the topic](https://dxdy.ru/topic154023.html)):
Let
$$g(n,m)=\binom{n+m}{m}f(n,m)$$
Then we have
$$g(n,m)=\alpha\,g(n-1,m)+\beta\,g(n,m-1)+\binom{n+m}{m}$$
If we make a little change
$$g(n,m)=[n\geqslant 0]\cdot[m\geqslant 0]\cdot\left(\alpha\,g(n-1,m)+\beta\,g(n,m-1)+\binom{n+m}{m}\right)$$
where square brackets denote Iverson brackets, then the basic conditions given for $f(n,0)$ and $f(0,m)$ holds.
From the recurrence relation and $g(0,0)=1$ it follows that $g(n,m)$ as a polynomial of $\alpha,\beta$ for $n,m\geqslant 0$ has degree $n$ by $\alpha$ and degree $m$ by $\beta$:
$$g(n,m)=\sum\limits\_{i=0}^n \sum\limits\_{j=0}^m c\_{n,m}^{i,j} \alpha^i \beta^j$$
Substitute this into a recurrence relation and equate the coefficients at the same degrees of $\alpha$ and $\beta$. Then we have
$$c\_{n,m}^{i,j}=\begin{cases}\binom{n+m}{m}&\text{if}\;i=j=0\\c\_{n-1,m}^{i-1,j}+c\_{n,m-1}^{i,j-1}&\text{otherwise}\end{cases}$$
Note that the differences between the first lower and first upper indices are the same in all terms: $n-i=(n-1)-(i-1)$. Similarly, the differences between the second lower and second upper indices are the same. This suggests using (maybe temporarily) other coefficients, more invariant, so to speak, instead of $c$:
$$d\_{p,q}^{i,j}=c\_{p+i,q+j}^{i,j},\quad c\_{n,m}^{i,j}=d\_{n-i,m-j}^{i,j}$$
Then we have
$$d\_{p,q}^{i,j}=\begin{cases}\binom{p+q}{q}&\text{if}\;i=j=0\\d\_{p,q}^{i-1,j}+d\_{p,q}^{i,j-1}&\text{otherwise}\end{cases}$$
If the conditions had the form
$$d\_{p,q}^{i,j}=\begin{cases}1&\text{if}\;i=j=0\\d\_{p,q}^{i-1,j}+d\_{p,q}^{i,j-1}&\text{otherwise}\end{cases}$$
then the solution would be
$$d\_{p,q}^{i,j}=\binom{i+j}{j}$$
For the proof, see [this answer](https://math.stackexchange.com/a/698182/929945).
Due to the linearity of the recurrence relation, if now $d\_{p,q}^{0,0}$ is multiplied by the coefficient $\binom{p+q}{q}$, that is, take $d\_{p,q}^{0,0}=\binom{p+q}{q}$, all other $d\_{p,q}^{i,j}$ will be multiplied by the same coefficient, so we get
$$d\_{p,q}^{i,j}=\binom{i+j}{j}\binom{p+q}{q}$$
Returning to $c\_{n,m}^{i,j}$, we get the explicit form
$$c\_{n,m}^{i,j}=\binom{i+j}{j}\binom{n-i+m-j}{m-j}$$
and the desired formula.
| 3 | https://mathoverflow.net/users/231922 | 446045 | 179,738 |
https://mathoverflow.net/questions/446048 | 5 | This question is a more precise version of [this question.](https://mathoverflow.net/questions/446033/eigenvalues-two-fold-degenerate)
Let's assume we have the matrix
$$\left(
\begin{array}{ccccc}
0 & a & 0 & 0 & 0 \\
f & 0 & b & 0 & 0 \\
0 & e & 0 & c & 0 \\
0 & 0 & d & 0 & r \\
0 & 0 & 0 & g & 0 \\
\end{array}
\right)$$
If we square it, we get the matrix
$$\left(
\begin{array}{ccccc}
a f & 0 & a b & 0 & 0 \\
0 & a f+b e & 0 & b c & 0 \\
e f & 0 & b e+c d & 0 & c r \\
0 & d e & 0 & c d+g r & 0 \\
0 & 0 & d g & 0 & g r \\
\end{array}
\right)$$
We see that this matrix decomposes into two submatrices
$$C\_1:=\left(\begin{array}{cccc}
a f & a b & 0 \\
e f & b e+c d & c r \\
0 & d g & g r \\
\end{array}\right)$$
and
$$C\_2:=\left(\begin{array}{cc}
a f+b e & b c \\
d e & c d+g r \\
\end{array}\right)$$
Now, one can check explicitly that the two submatrices are isospectral apart from one eigenvalue zero. I wonder if there is an abstract argument why this is so?
It would for instance follow if we can write the matrices as $C\_1 = AB$ and $C\_2= BA$, but I don't see how such a decomposition could work. In particular, this does not seem to be restricted to 5x5 matrices but holds for arbitrary matrices of the above form.
| https://mathoverflow.net/users/496243 | Matrices with same eigenvalues | To answer say a previous question, call such tridiagonal matrix $T$; the matrices $T$ and $-T$ (of dimension $n$) are unitarily congruent (they have the same spectra), that is there is an 'alternating' $\pm 1$ diagonal matrix $U$ such that $U^\*TU=-T$.
As you noticed when you square $T$ and rearrange it by a permutation $P$ the diagonal blocks have the same eigenvalues. This is true as: for $P\_i$ the permutation matrix exchanging $2i+1\leftrightarrow i+1$, set $P=\prod\_{i=1}^{\lfloor \frac{n-1}{2}\rfloor}P\_i=P\_{\lfloor \frac{n-1}{2}\rfloor}\cdots P\_1$; we get $PTP^\*=\begin{pmatrix}0\_{\lceil\frac{n}{2}\rceil}&A\\B&0\_{\lfloor \frac{n}{2}\rfloor}\end{pmatrix}.$ The square block $0\_m$ is a submatrix with all zeros of dimension $m$. Squaring the last identity gives your matrix $\begin{pmatrix}AB=C\_1&0\\0&BA=C\_2\end{pmatrix}.$
| 6 | https://mathoverflow.net/users/121643 | 446055 | 179,740 |
https://mathoverflow.net/questions/445895 | 2 | **I. First Set**
Before going to Elkies' nonic, we start with something a bit simpler. There is a list of j-function formulas in this [MSE post](https://math.stackexchange.com/a/4580319/4781). For example, for prime levels $p = 5,7,13,$ we have,
$$j=\frac{(x^2+10x+5)^3}x$$
$$j=\frac{(x^2+5x+1)^3(x^2+13x+49)}x$$
$$j=\frac{(x^4+7x^3+20x^2+19x+1)^3(x^2+5x+13)}x$$
Assume $j$ as **any** number, hence a free parameter. Then these equations have groups $\text{PGL}(2,5), \text{PGL}(2,7), \text{PGL}(2,13),$ respectively, so generally is not solvable in radicals. However, $x$ is solvable if $j$ is the $j$-function **OR** if we do this,
$$\frac{(x^2+10x+5)^3}x =\frac{(n^2+10n+5)^3}n$$
for **any** non-zero $n$. For example, let $n=2$,
$$\frac{(x^2+10x+5)^3}x =\frac{29^3}2$$
Its irreducible $5$th-deg factor has Frobenius group $F(5)=4\times5=20$ hence is now solvable. Likewise for its higher siblings which have $F(7)=6\times7=42$ and $F(13)=12\times13=156$.
**Edit** (May 3): For an example where the result is **non-solvable**, let
$$j = \frac{(x + 432)(x^2 + 80x - 3888)^3}{7^7 x^3}$$
which has familiar discriminant (without its numerical factors) as $D = -(j-1728)^3\, j^4$. Then,
$$\frac{(x + 432)(x^2 + 80x - 3888)^3}{x^3} = \frac{(n + 432)(n^2 + 80n - 3888)^3}{n^3}$$
does **not** yield a solvable sextic factor.
**Q:** So the procedure doesn't work on just any rational function. What condition should be satisfied such that the procedure yields a solvable factor?
---
**II. Second Set**
Searching the literature for similar "formulas", the paper *[Galois Number Fields with Small Root Discriminant](http://cda.morris.umn.edu/%7Eroberts/dpr/Research_files/lowgrd.pdf)* by J. Jones and D. Roberts was a gold mine. They had degrees for $p \leq 10$ and even for $p=17$ which I mentioned in an old [MO post](https://mathoverflow.net/questions/226449/whats-so-special-about-these-17th-deg-equations). But this equating of,
$$P(j,x) = P(j,n)$$
was not there. A small sample for deg 9,
$$j = \frac{(x^3 + 4x^2 + 10x + 6)^3}{(4x^2 + 13x + 32)}$$
Again, assume $j$ as a free parameter. This equation in Magma notation is $\text{9T32}$ and has order $9\times12\times14 = 1512$ (unsolvable). However,
$$\frac{(x^3 + 4x^2 + 10x + 6)^3}{(4x^2 + 13x + 32)} = \frac{(n^3 + 4n^2 + 10n + 6)^3}{(4n^2 + 13n + 32)}$$
(after removing the linear factor) is $\text{8T36}$ and has order $12\times14 = 168$ (solvable). Elkies' nonic is also $\text{9T32}$ but a bit trickier,
$$j^2+(9x^4 - 42x^3 - 675x^2 - 1485x - 441)j - (x^3 - 9x^2 - 69x - 123)^3 =0$$
$$j^2+(9n^4 - 42n^3 - 675n^2 - 1485n - 441)j - (n^3 - 9n^2 - 69n - 123)^3 =0$$
Eliminating $j$ between the two using resultants and getting rid of two linear factors yields a $16$-deg equation with order $12\times14\times16 = 2688$ (and solvable according to Magma).
**Note:** For consistency, I still used the variable $j$ for the second set, but they are not to be understood as level-9 formulas for the j-function.
---
**III. Question**
So why is it for these two sets (and some others) that eliminating the common variable $j$ between $P(j,x) = P(j,n) = 0$ suddenly yields a parametric equation with a solvable Galois group?
| https://mathoverflow.net/users/12905 | On Elkies' $\text{9T32}$ nonic and a shared property with j-function formulas | This has relatively little to do with the j-invariant itself. If you take any rational function $f(x)\in k(x)$ and let $G:=Gal(f(x) - t/k(t))$ (also referred to as the monodromy group of $f$), then by elementary Galois theory, $Gal(f(x)-f(y) / k(y))$ is a point stabilizer in $G$ (in the usual action on the roots). In your cases $G$ is $P\Gamma L\_2(q)$ acting on $q+1$ points ($q=5,7,8,13$) which has solvable point stabilizer.
(Edit: Of course the reason why this observation "wasn't there" in the linked Jones/Roberts paper is that this is comparatively the "trivial case", working for every $f$. More interesting are the (exceptional) cases where a group is a monodromy group of two essentially *different* rational functions in the same Galois closure, which then gives shrinking of $Gal(f(x)-g(y)/k(y))$ in the analogous way - the sextic above together with the right quintic (relates to the $j$-invariant) is one such pair.)
| 1 | https://mathoverflow.net/users/127660 | 446059 | 179,742 |
https://mathoverflow.net/questions/446053 | 8 | Let $\Omega$ be an open subset of $\mathbb R^n$ for $n \geq 2$, and $p \in \Omega$.
Let $k$ be a positive integer. Suppose that $f: \Omega \setminus \{p\} \to \mathbb R$ is in $C^k$, and $\lim\_{x \to p} D^k f (x)$ exists.
**Question:** Is it true that $f$ admits an extension to $\Omega$ that is in $C^k (\Omega)$?
| https://mathoverflow.net/users/173490 | An extension problem | $\newcommand\R{\mathbb R}\newcommand{\Om}{\Omega}$This is to detail the [comment by Fedor Petrov](https://mathoverflow.net/questions/446053/an-extension-problem#comment1152071_446053).
Without loss of generality, $p=0$ and $\Om$ is an open ball in $\R^n$ centered at $p=0$. Assume, slightly more generally, that $f\colon\Om\_0\to\R^d$ for some natural $d$, where $\Om\_0:=\Om\setminus\{0\}$. (Actually, the original problem is clearly equivalent to the "$d$-dimensional" one.) Let $g\colon\Om\_0\to\R$ denote any coordinate-function of the vector function $f$.
Let $L^kf:=\lim\_{x\to0}(D^kf)(x)\in((\R^n)')^d$ and $L^kg:=\lim\_{x\to0}(D^kg)(x)\in(\R^n)'$, where $(\R^n)'$ is the space of all linear functionals on $\R^n$; let us identify $(\R^n)'$ with $\R^n$ in the standard manner, and then $((\R^n)')^d$ with $\R^{nd}$.
Consider first the case $k=1$. Take any sequence $(x\_r)$ in $\Om\_0$ such that $x\_r\to0$ and $0$ is not on the straight line segment $[x\_m,x\_r]$ from $x\_m$ to $x\_r$ for any natural $m,r$; let us call such a sequence $(x\_r)$ *good*. Then, by the mean value theorem, for any (coordinate-function) $g$, any natural $m$ and $r$, and some $t\_{m,r}\in(0,1)$ we have
\begin{equation}
g(x\_r)-g(x\_m)=(D^1g)(x\_m+t\_{m,r}(x\_r-x\_m))(x\_r-x\_m)\to0
\end{equation}
as $m,r\to\infty$, since $x\_r-x\_m\to0$, $x\_m+t\_{m,r}(x\_r-x\_m)\to0$, and hence $(D^1g)(x\_m+t\_{m,r}(x\_r-x\_m))\to L^1g\in(\R^n)'=\R^n$. So, by the Cauchy convergence test, the sequence $(g(x\_r))$ converges to a limit in $\R$ and hence the sequence $(f(x\_r))$ converges to a limit in $\R^d$.
Now take any sequence $(x\_r)$ in $\Om\_0$ such that $x\_r\to0$, now allowing $0$ to be on the straight line segment $[x\_m,x\_r]$ for some natural $m,r$. We can consecutively approximate the sequence $(x\_r)$ by a good sequence $(y\_r)$ such that $|x\_r-y\_r|+|f(x\_r)-f(y\_r)|<1/r$ for all natural $r$ -- **it is only here that we use the condition $n\ge2$** (here we also use the continuity of $f$ on $\Om\_0$). So, for any sequence $(x\_r)$ in $\Om\_0$ such that $x\_r\to0$, the sequence $(f(x\_r))$ converges to a limit in $\R^d$. Clearly, this limit, which let us denote by $f(0)$, does not depend on the choice of a sequence $(x\_r)$ in $\Om\_0$ such that $x\_r\to0$, so that $f(0)=\lim\_{x\to0}f(x)$. Let then $g(0):=\lim\_{x\to0}g(x)$.
Moreover, using the mean value theorem again (as well as the condition $L^kg:=\lim\_{x\to0}(D^kg)(x)$), for all $x\in\Om\_0$ and some $t\_x\in(0,1)$,
\begin{equation}
g(x)-g(0)-(L^1g)(x)=(D^1g)(t\_x x)(x)-(L^1g)(x)=((D^1g)(t\_x x)-L^1g)(x)=o(|x|)
\end{equation}
as $x\to0$, where $|x|$ denotes the Euclidean norm of $x$. Thus, $(D^1g)(0)=L^1g$ and hence $(D^1f)(0)=L^1f$.
So, the case $k=1$ is done. The general case of any natural $k$ now follows by backward induction, utilizing the vector extension of the original problem and identifying the arising finite-dimensional linear spaces of linear functionals with spaces $\R^N$ of appropriate dimensions $N$.
| 4 | https://mathoverflow.net/users/36721 | 446086 | 179,747 |
https://mathoverflow.net/questions/221232 | 9 | Let $d$ be a fundamental discriminant and let $\chi$ be the associated primitive real character of modulus $\vert d \vert$. Assuming GRH, Littlewood proved that as $\vert d \vert$ grows large,
$$L(1, \chi) \leq (2 + o(1)) e^\gamma \log \log(\vert d \vert).$$
Granville and Soundararajan provide a treasure-trove of information about $L(1,\chi)$ in their GAFA paper (Vol. 13, 2003).
But I've never seen a bound as above in which the $o(1)$ is made explicit. (I admit that when logarithms are inside of logarithms, my brain tries to jump out of my ear and run away, so I've only looked for an hour or two.) Is there any statement proven (assuming GRH) along the lines of the following -- for some explicit constant $K$ and explicit decaying function $F$?
Desired form: If $\vert d \vert > K$ then $L(1, \chi) \leq (2 + F(\vert d \vert)) e^\gamma \log \log(\vert d \vert).$
If not, is there a well-known obstruction to proving such statements?
| https://mathoverflow.net/users/3545 | Effective bound of $L(1,\chi)$ | Explicit upper and lower bounds for $L(1,\chi)$, conditional on the generalised Riemann hypothesis, are given in Theorem 1.5 of the paper ["Conditional bounds for the least quadratic non-residue and related problems"](https://doi.org/10.1090/S0025-5718-2015-02925-1) by Youness Lamzouri, Xiannan Li and Kannan Soundararajan. In particular, for $q \geq 10^{10}$ and $\chi$ a primitive Dirichlet character modulo $q$, we have the explicit upper and lower bounds
$$|L(1,\chi)| \leq 2e^{\gamma}\left(\log \log q - \log 2 + \frac{1}{2} + \frac{1}{\log \log q}\right)$$
and
$$\frac{1}{|L(1,\chi)|} \leq \frac{12e^{\gamma}}{\pi^2} \left(\log \log q - \log 2 + \frac{1}{2} + \frac{1}{\log \log q} + \frac{14 \log \log q}{\log q}\right).$$
(There is a corrigendum to this paper, but Theorem 1.5 remains unchanged.)
This method generalises quite nicely to other $L$-functions, as explained in the paper ["Explicit bounds for $L$-functions on the edge of the critical strip"](https://doi.org/10.1016/j.jnt.2018.01.001) by Allysa Lumley.
| 4 | https://mathoverflow.net/users/3803 | 446087 | 179,748 |
https://mathoverflow.net/questions/446052 | 3 | There are two versions of Cauchy identity for Schur functions, namely
$$
\sum\_{\lambda}s\_\lambda(\underline x)s\_\lambda(\underline y) = \prod\_{i,j=1}^n\frac 1{1-x\_iy\_j}\ ,\qquad {\rm (1)}
$$
and
$$
\sum\_{\lambda}s\_\lambda(\underline x)s\_{\lambda'}(\underline y) = \prod\_{i,j=1}^n (1+x\_iy\_j)\ .\qquad {\rm (2)}
$$
(Usual notations employed here: $\underline x=(x\_1,\dots,x\_n)$, $\lambda$ runs over all partitions (of length $\leq n$), $\lambda'$ is the conjugate partition.)
I found in the literature (Stanley's 1989 paper, Macdonald's book) the following generalization of (1) to Jack symmetric poylnomials:
$$
\sum\_{\lambda} \frac{J^{(a)}\_ \lambda(\underline x)J^{(a)} \_ \lambda(\underline y) }{\langle J^{(a)}\_ \lambda,J^{(a)}\_ \lambda\rangle\_ a}= \prod\_{i,j=1}^n\biggl(\frac 1{1-x\_iy\_j}\biggr)^{1/a}\ .
$$
(Here, $J^{(a)}\_ \lambda$ are the Jack polynomials (in the $J$-normalization) and $\langle,\rangle\_a$ the deformed Hall inner product, namely $\langle p\_\lambda,p\_\mu\rangle\_a=a^{\ell(\lambda)}z\_\lambda \delta\_{\lambda\mu}$.)
QUESTION: does there exist a similar generalization of (2) for Jack polynomials?
| https://mathoverflow.net/users/178792 | Cauchy identity for Jack functions | The dual Cauchy identity for Jack polynomials also exists, and is better expressed in terms of the $P$-normalized Jack polynomials:
$$
\sum\_{\lambda} P^{(a)}\_ \lambda(\underline x)P^{(1/a)} \_ {\lambda'}(\underline y) = \prod\_{i,j=1}^n\bigl(1+x\_iy\_j\bigr)\ .
$$
(Cf., e.g., formula (2.6) in [these notes by I.G. Macdonald](https://kurims.kyoto-u.ac.jp/EMIS/journals/SLC/opapers/s28macdonald.pdf).)
| 5 | https://mathoverflow.net/users/178792 | 446099 | 179,750 |
https://mathoverflow.net/questions/446043 | 1 | [*Bring's curve*](https://en.wikipedia.org/wiki/Bring%27s_curve) or *Bring's surface* with genus 4 and $5!=120$ automorphisms can be given by the homogeneous equations,
$$x\_1+x\_2+x\_3+x\_4+x\_5 = x\_1^2+x\_2^2+x\_3^2+x\_4^2+x\_5^2 = \\x\_1^3+x\_2^3+x\_3^3+x\_4^3+x\_5^3 = 0$$
This is also a property of the Bring quintic $x^5+ax+b = 0$ since its roots $x\_i$ obey,
$$\sum\_{i=1}^5 x\_i^k = 0,\quad k = 1,2,3$$
But there is also a surface with **six** coordinates such that,
$$y\_1 + y\_2 + y\_3 + y\_4 + y\_5 + y\_6 = y\_1^2 + y\_2^2 + y\_3^2 + y\_4^2 + y\_5^2 + y\_6^2 = \\ y\_1^4 + y\_2^4 + y\_3^4 + y\_4^4 + y\_5^4 + y\_6^4 = y\_1^7 + y\_2^7 + y\_3^7 + y\_4^7 + y\_5^7 + y\_6^7 = 0$$
This seems to be a property of the sextic $y^6+ay^3+by+c = 0$ since its roots $y\_i$ obey,
$$\quad\sum\_{i=1}^6 y\_i^k = 0,\quad k = 1,2,4,\color{red}7$$
**Questions:**
1. The general quintic and sextic can be transformed in radicals into $x^5+ax+b = 0$ and $y^6+ay^3+by+c = 0$, respectively. So does it follow if the first surface has $5!=120$ automorphisms, then the second has $6!=720$? (In [this table](https://en.wikipedia.org/wiki/Hurwitz%27s_automorphisms_theorem#Automorphism_groups_in_low_genus), no genus $< 12$ has more than 720.)
2. If the surface in $x$ has genus 4, what is the genus of the surface in $y$?
3. Any other special property, such as *why* is it valid for the high exponent $k=7$?
| https://mathoverflow.net/users/12905 | Bring's curve $\sum_{i=1}^5 x_i^k = 0$ for $k = 1,2,3$ and an analogue $\sum_{i=1}^6 y_i^k = 0$ for $k = 1,2,4,7$ | The equations $\sum\_{i=1}^6 y\_i^k = 0$ for $k=1,2,4,7$ do not cut out a curve because the $k=1,2,4$ equations imply the one for $k=7$. (The 7th power sum is a polynomial in the power sums of degree 1 through 6, and there is no way to get 7 as a sum of numbers in $\{3,5,6\}$.) So you have some surface with 720 automorphisms.
If some family of sextics $y^6 + a y^3 + b y + c = 0$ is a sextic resolvent of $x^5 + Ax + B = 0$, then the corresponding curve -- call it $C$ -- is isomorphic to the Bring curve, and thus also has genus 4 and automorphisms by $S\_5$, acting by permutations of the $y\_i$ by a transitive subgroup of $S\_6$. If you find enough $S\_6$-invariant equations in $y\_1,\ldots,y\_6$ to cut out a 1-dimensional variety containing $C$, then that variety is the union of 6 curves each of which is isomorphic with the Bring curve and has automorphism group one of the six transitive subgroups of $S\_6$ that's isomorphic with $S\_5$.
| 1 | https://mathoverflow.net/users/14830 | 446104 | 179,751 |
https://mathoverflow.net/questions/445931 | 6 | $\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\O{O}$I’m studying the group $\O(5,5,\mathbb{Z})$, the indefinite orthogonal matrices with integer entries. In particular, I want to know about its homology/cohomology. It doesn’t seem like much is known about this, so I’m starting with the first integral homology, which is its abelianization. To do this, I'm trying to determine the commutator subgroup of $\O(5,5,\mathbb{Z})$. From the Cartan-Dieudonné theorem it is pretty easy to show that
$$[\O(5),\O(5)] = \SO(5)$$
I'm pretty sure we can use this same theorem to show that
$$[\O(5,5),\O(5,5)] = \SO(5,5)$$
Now, I'm interested in $[\O(5,5,\mathbb{Z}),\O(5,5,\mathbb{Z})]$. Given the above, my guess is that it is equal to $\SO(5,5,\mathbb{Z})$ (we certainly have that $[\O(5,5,\mathbb{Z}),\O(5,5,\mathbb{Z})]\subseteq \SO(5,5,\mathbb{Z})$). However, we cannot use the Cartan-Dieudonné theorem in this case since we are not working over a field. Furthermore, the only calculation of a commutator subgroup of integral matrices that I have seen is that for $\SL(n,\mathbb{Z})$, but this relied on knowing generators for $\SL(5,\mathbb{Z})$. I don't know any generators for $\O(5,5,\mathbb{Z})$.
Update: I misspoke in the comments, the quadratic form I am considering is $\sum\_{n=1}^5 x\_ix\_{i+5}$. This is the one that is relevant to string theory. That being said, the generators for $O(5,5,\mathbb{Z})$ are given in Becker, Becker, Schwarz “String Theory Aand M-Theory”, equation (7.76) and (7.77):
$$\begin{pmatrix}
0 & I\_5 \\
I\_5 & 0
\end{pmatrix} \text{ and } \begin{pmatrix} I\_5 & 0\\
N\_{IJ} & I\_5
\end{pmatrix}
$$
where $N\_{IJ}$ is an anti-symmetric matrix of integers.
| https://mathoverflow.net/users/503849 | Computing a Commutator Subgroup | With the latest update, indicating that the quadratic form in question is, up to reordering, $\sum x\_i x\_{-i}$ (here we number the basis elements as $e\_1,\ldots,e\_5,e\_{-5},\ldots,e\_{-1}$), the answer is affirmative.
This can be seen as follows:
1. Note that the group in question is the group of $\mathbb{Z}$-points of a Chevalley—Demazure group scheme of type $\mathsf{D}\_5$ (for an intermediate weight lattice). For this see, for example, "Structure of Chevalley groups over commutative rings" by N. Vavilov (<https://mahalex.net/misc/Structure.djvu>, p. 255). Below I denote the split group of this type by $\operatorname{SO}(10,\mathbb{Z})$.
2. Introduce the elementary subgroup $\operatorname{EO}(10,R)\leqslant \operatorname{SO}(10,R)$, generated by elementary orthogonal transvections (here $R$ is a commutative ring). The detail can be found in the same paper.
3. Use Corollary 4.4 of ["Generators, relations and coverings of Chevalley groups over commutative rings"](https://doi.org/10.2307/2373742) by M. Stein to see that $\operatorname{EO}(10,R)$ is perfect.
4. Use the stability results of Stein from ["Stability theorems for K1, K2 and related functors modeled on Chevalley groups"](https://doi.org/10.4099/math1924.4.77) or the fact that $\mathbb{Z}$ is Euclidean to show that $\operatorname{EO}(10,\mathbb{Z})=\operatorname{SO}(10,\mathbb{Z})$.
5. Conclude that $[\operatorname{O}(10,\mathbb{Z}),\operatorname{O}(10,\mathbb{Z})] = [\operatorname{SO}(10,\mathbb{Z}), \operatorname{SO}(10,\mathbb{Z})] = \operatorname{SO}(10,\mathbb{Z})$.
| 8 | https://mathoverflow.net/users/5018 | 446109 | 179,753 |
https://mathoverflow.net/questions/446107 | 5 | How many digits of $\sqrt{2}$ are known to date, in base 10 and in base 2? I am trying to produce the largest sequence known to date, and would like to sense if I can do it either alone or with hiring someone. I should have no problems producing 1 trillion, but almost sure I can't produce 1 quadrillion in any reasonable amount of time.
The goal is for marketing purposes and not to impress the mathematical community, but instead to attract clients to buy services that I offer. I'd like to also know about the digits of $\pi$, since producing a longer list (albeit for $\sqrt{2}$ or the golden ratio) would have a stronger impact.
**Update on 5/4/2023**
I am aware of what is in Wikipedia on this subject, and about the integer square root and how to compute it efficiently (for instance using the gmpy2 library). I develop PRNGs based on digits of millions of quadratic irrationals using new fast formulas, starting at arbitrary large locations, see [here](https://mltechniques.com/2022/12/13/military-grade-fast-random-number-generator-based-on-quadratic-irrationals/). If I claim that I can do better than what is in Wikipedia because I don't know the most recent computations, I could be accused of false advertising when stating that I have the longest sequence.
I want to avoid this, thus my question. The target customers will understand the random character of these digits, especially if offering a competition featuring 50k previous digits of one of these numbers starting at some location, and offer a large award (that I know no one will win) for correctly predicting the next 20k digits (with participants not knowing which starting location and which quadratic irrational I use).
I've made quite a bit of research on this topic, for instance - among others - proving that the digits of $\sqrt{2}$ and $\sqrt{3}$ are uncorrelated. The definition of correlation is in the comments.
| https://mathoverflow.net/users/140356 | How many digits of $\sqrt{2}$ are known to date? | In one sense, *all* of the base-2 digits (or, I guess, "bits") of $\sqrt{2}$ are known because we have a closed-form formula, according to the [OEIS](https://oeis.org/A004539): $$\begin{align} a(n) &= \frac{1}{2} - \frac{2\arctan(\cot(2^{-\frac{3}{2}+n}\pi))}{\pi} + \frac{\arctan(\cot(2^{-\frac{1}{2}+n}\cdot\pi))}{\pi} \\[10pt] &= \left\lfloor 2^{-\frac{1}{2}+n}\right\rfloor -2\left\lfloor 2^{-\frac{3}{2}+n}\right\rfloor \end{align}$$
**ETA**:
According to [Wikipedia](https://en.wikipedia.org/wiki/Square_root_of_2#Records_in_computation), the record number of digits for $\sqrt{2}$ as of this writing is $10^{13}+1000$. Further, [this website](http://www.numberworld.org/y-cruncher/records.html) purports to contain information regarding world records for the computation of various constants. I am unsure of how credible that site is but it came up in the citations on the Wikipedia page.
| 17 | https://mathoverflow.net/users/9840 | 446118 | 179,755 |
https://mathoverflow.net/questions/446114 | 10 | I'm specifically assuming that we have replacement instead of collection; collection breaks things (because then there is a set that contains a map from $n$ to $C$ for every $n\in\mathbb N$, and you can look within that set to get an injection from an infinite subset of $C$ to some infinite set, so $\aleph(C)$ couldn't have been $\aleph\_0$.
The construction that seems to work in ZFCA is to take a model of ZFA with infinitely many atoms and then take the direct limit of $L(A\_n)$, where $A\_n$ is the set of the first $n$ atoms. This seems to be a model of ZFCA, because every set belongs to $L(A\_n)$ for some $n$ and therefore inherits a well-ordering from that (as $L(A\_n)$ still has a global choice function definable from parameters).
I can't see how replacement fails where collection did, because I can't think any $\varphi$ such that $\forall n\exists!x\varphi(n,x)$ and $\varphi(n,x)$ implies that $x$ is an injection from $n$ into the class of atoms, or anything else that would cause the union of all such $x$s to not fit into any $L(A\_n)$.
This of course cannot be extended to a model of GBCA, because the existence of a proper class that is smaller than the universe prevents the existence of a global choice function.
If the construction works and can be extended to ZFC, it does prove that the Hartogs number of classes is interesting in NBG even with the axiom of choice (NBG being conservative over ZFC).
| https://mathoverflow.net/users/152182 | Is it consistent with ZFC(A) for the Hartogs number of a proper class to be $\aleph_0$? | Yes. If you start with infinitely many urelements and then take the sets whose kernel is finite (the kernel of a set is the set of urelements in its transitive closure), the resultant inner model will satisfy Replacement. A more generalized argument is included in my dissertation (<https://arxiv.org/abs/2303.14274>, Theorem 26).
Let ZFU$\_\text{R}$ denote ZF set theory with urelements axiomatized with Replacement. In general, given a model $U$ of ZFU$\_R$ an ideal $\mathcal{I}$ of the class of all urelements that contains every urelement singleton, the inner model $U^\mathcal{I}$, which contains all urelements and sets whose kernel is in $\mathcal{I}$, is a model of ZFU$\_R$ also. And $U^\mathcal{I}$ satisfies AC if $U$ does. So in the case of ZFCU$\_\text{R}$, the Hartogs number of a proper class can be any infinite cardinal.
For ZFCU$\_\text{R}$ + Collection, the Hartogs number of a proper class cannot be any limit cardinal by the argument you have given. But this is the only constraint: for every successor infinite cardinal $\kappa^+$, there can be models of ZFCU$\_\text{R}$ + Collection where the Hartogs number of the class of urelements is $\kappa^+$. This follows from Lemma 21 and Theorem 26 in my dissertation.
The result for ZFCU$\_\text{R}$ **can** be extended to GBU$\_\text{R}$ (in fact, KMU$\_\text{R}$) with a global choice function. This is due to Felgner (see Theorem 108 in my dissertation). In class theory with urelements, it is important to distinguish different versions of second-order AC. For example, the existence of a global choice function doesn’t imply the universe can be well-ordered, as shown in Felgner’s model.
However, it is not known, for example, if KMU$\_\text{R}$ + Collection + Global Choice (not Global Well-Ordering) is consistent with a proper class with Hartogs number, say, $\aleph\_1$.
| 10 | https://mathoverflow.net/users/504023 | 446122 | 179,759 |
https://mathoverflow.net/questions/446126 | 7 | Given two finite presentations of torsion-free groups, is there an algorithm to determine whether the given groups are isomorphic or not?
I have found results for narrower classes (for example, they are briefly reviewed in [The isomorphism problem for finitely generated fully residually free groups](https://www.sciencedirect.com/science/article/pii/S002240490600079X)).
| https://mathoverflow.net/users/148161 | Is the isomorphism problem solvable for torsion-free groups? | Novikov's centrally-symmetric group $\mathfrak{A}\_P$ is a torsion-free group with undecidable word problem, constructed in [1]. Novikov did not prove it is torsion-free but, as Adian points out in [Adi1957b] (p. 76 of [my translation](https://arxiv.org/abs/2208.08560), and referenced as in there; he calls the group $F\_0$ in that article), it is not difficult to prove this fact using Novikov's work.
The statement of the Adian-Rabin theorem is today done via Markov properties (as defined by Markov when he proved the corresponding theorem for semigroups a few years earlier). And this is how Adian does it in [Adi1957b], too, but in his main article [Adi1957a], which contains all the detailed proofs, there are no Markov properties, only a slightly smaller family of properties (I call them "pseudo-Markov"). Instead, what Adian does in [Adi1957a] is the following:
Let $A, B$ be any two words in $\mathfrak{A}\_P$. From these two words and $\mathfrak{A}\_P$ he constructs a new finitely presented group, which he calls $\mathfrak{A}\_{q, A, B}$, and which has the following property: $\mathfrak{A}\_{q, A, B}$ is trivial if and only if $A = B$ in $\mathfrak{A}\_P$. Furthermore, the groups $\mathfrak{A}\_{q, A, B}$, for all $A, B$, are all torsion-free (he does not state it directly but it follows from the same type of argument as in his behemoth Main Lemma).
This yields the weaker Adian-Rabin theorem, but it also yields the undecidability you want; the family $\mathfrak{A}\_{q, A, B}$, together with the trivial group, is recursively enumerable (indeed the way one constructs the groups is given explicitly, and in modern terminology it is just taking some HNN-extensions of $\mathfrak{A}\_P$), and so if we could decide the isomorphism problem for torsion-free groups, then we could decide the isomorphism in this class, so we could also decide the word problem for $\mathfrak{A}\_P$, a contradiction.
(Footnote: note that there are Markov properties $\mathcal{P}$ such that the isomorphism problem for all groups with $\mathcal{P}$ is decidable (trivially, we may take the property "being trivial"; but less trivially we may take "being hyperbolic" or "being abelian"). But I suspect that given any Markov property $\mathcal{P}$ such that there are groups with undecidable word problem with $\mathcal{P}$, the isomorphism problem for the class of groups with $\mathcal{P}$ is undecidable; I think I am missing some simple argument for why this is the case.)
[1] *Novikov, P. S.*, [**On the algorithmic insolvability of the word problem in group theory**](https://doi.org/10.1090/trans2/009/01), Am. Math. Soc., Transl., II. Ser. 9, 1-122 (1958). [ZBL0093.01304](https://zbmath.org/?q=an:0093.01304).
| 9 | https://mathoverflow.net/users/120914 | 446131 | 179,761 |
https://mathoverflow.net/questions/445953 | 4 | Say that a function $f \in \omega^\omega$ witnesses that an $\omega$-REA set $A = \bigoplus\_{i \in \omega} A^{[i]}$ is non-arithmetic if $A^{[\leq f(n)]} \not\leq\_T 0^n$. Say that $A$ is effectively non-arithmetic if there is such an $f \leq\_a A$ (i.e. arithmetic in $A$)?
Is every non-arithmetic $\omega$-REA set effectively non-arithmetic?
---
Note that, we can't hope to always have $f$ arithmetic. To see this, note that we can build $0^{\omega}$ with the pieces spread out. So, if we want to ensure that $f^{k}\_i(n) = \phi\_i(0^{k}, n)$ isn't such a witness we can wait until after we've set $A^{[\leq m]} = 0^{k+1}$ at which point we can produce indexes for $A^{[m + n]}$ that yields $0^{k+2}$ if $f^{k}\_i(k+2)$ diverges or $m + n > f^{k}\_i(k+2)$ and otherwise is $\emptyset$. It is straightforward to extend this approach to ensure that no arithmetic $f^{k}\_i$ witnesses $A$ is non-arithmetic.
However, as every non-arithmetic $\omega$-REA set has a witness that's computable in $0^{\omega}$ this approach doesn't obviously extend to showing that $f$ can't be arithmetic in $A$.
---
Also note that if the claim is false it can't be false in a reletivizeable fashion since (in some sense) since given an $\omega$-REA operator A(X) that's never arithmetic in X we can take G to be $\omega$ generic below $0^\omega$ with A(G) of degree $0^\omega$ and there will be a witness to the fact that A(G) isn't arithmetic in G computable in $0^\omega$.
| https://mathoverflow.net/users/23648 | Effectively non-arithmetic $\omega$-REA degrees | Unfortunately, I believe I can show there are non-arithmetic $\omega$-REA degrees that aren't effectively $\omega$-REA.
To this end we need to build $A$ to be $\omega$-REA with $A >\_a 0$ to satisfy
$$R\_{n,i}: \phi\_i(A^{(n)})\downarrow \implies (\exists m,k)\left(\phi\_i(A^{(n)};m) =k \implies A^{[\leq k]} \leq\_T 0^{m} \right)$$
where $\phi\_i(A^{(n)})\downarrow$ means the function is total.
Meeting requirements $R\_{0,i}$ is relatively straightforward. Given a construction of a $\omega$-REA $A >\_a 0$ we start building $\hat{A}$ just like $A$ can simply look at approximations to the construction [1] and freeze $\hat{A}\_s$ if we see $\phi\_0(A;0)\downarrow = k$. Then simply let $\hat{A}^{[\leq k]}$ be computable and restart the original construction of $A$ at the $k+1$-st component, e.g., $\hat{A}^{[k+1]} = A^{[1]}$. If $\hat{A}^{[k\_i + 1]} = A^{[i]}$ then we don't allow $R\_{0,i+1}$ to affect $\hat{A}^{[\leq k\_i + 1]}$ but if we can make $\phi\_{i+1}(A;i+1)$ converge to $k\_{i+1}$ by freezing the approximation to $\hat{A}^{[> k\_i + 1]}$ we do so and let sufficiently many more columns be computable to ensure that $\hat{A}^{[\leq k\_{i+1} + 1]} \leq 0^{i+1}$.
However, to do this for $n > 0$ we need some way to control both the behavior of $A^{n}$ at the same time as the columns of $A$. We leverage Harrington's construction of a low $\omega$-REA set.
The short version of this is that since $A$ is just built by taking a low set and then copying over columns by jump inversion we can use the same approach with the $n$-th jump of $A$ (where we can compute $\phi\_i(A^{n})$ and the same spacing out trick on the $n$-th jump will entail spacing out in $A$. The much longer version is below.
Construction of low $\omega$-REA set
------------------------------------
Per Odifreddi vol 2 we build an $\omega$-REA operator $A(X)$ with the following properties
$$A(X)' \equiv\_T A(X) \oplus X' \equiv\_T A(X')$$
$$A(X) >\_T X$$
This ensures that $A(X)^{n}\equiv\_T A(X^{n}) \equiv\_T A(X) \oplus X^{n} \not\leq\_T X^{n}$ making $A(\emptyset)$ a non-arithmetic low (for the arithmetic jump) $\omega$-REA set.
To build this we actually build $A(X)' \equiv\_T A(X) \oplus X' \equiv\_T J\_e(X')$ and use the fixed point theorem to find $e$ with $A(X) = J\_e(X)$. The construction proceeds by first building $A^{[1]}(X) = L(X)$ where $L$ is a low non-computable r.e. set. Then, assuming that
$$A^{[\leq 2n +1]}(X)' \equiv\_T A^{[\leq 2n +1]}(X) \oplus X' \equiv J\_e^{[\leq n]}(X')$$
(via known reductions) we use the ZBC lemma (extension of Sacks jump inversion) relativized to $A^{[\leq 2n +1]}(X)$ to construct $A^{[2n +2]}(X), A^{[2n +3]}(X)$ to ensure that above equation holds for $n+1$. Finally, we allow $A^{[\geq i]}(X)$ to be frozen on a finite initial segment if we have the opportunity to place $i \in A(X)'$ thus ensuring that $A(X)' \equiv\_T \oplus\_{n \in \omega} A^{[\leq 2n +1]}(X)'$
Modifying the Construction
--------------------------
We modify the above construction by inserting extra computable columns as above. In particular, we build $A(X)$ so that the first $k\_0$ columns are computable then the next column is the low set $L(X)$ followed by more computable columns followed by the two columns from the ZBC lemma to jump to $A^{[\leq 1]}(X)$ and so on. In general we ensure that (note that the value of $k\_m$ will depend on $X$ and we mean the value relative to $X$ in these equations).
$$A^{[\leq k\_m]}(X) \leq\_T 0^m$$
$$A^{[\leq k\_m + 2]}(X)' \equiv\_T A^{[\leq m]}(X')$$
Now, we ensure that if $m = \langle n,i\rangle$ and we can cause $\phi\_i(X;m + n)\downarrow = z$ to converge by freezing some finite part of the columns extending $A^{[\leq k\_{m-1} + 2]}(X)$ we do so and ensure that $k\_m > z > k\_{m-1} + 2$. We again build $A = A(\emptyset)$ using a fixed point as above.
Now, suppose that $f$ witnessed the fact that $A$ was effectively non-arithmetic. Then, for some $n$ $f$ would be computable in $A^{n}(\emptyset) \equiv\_T A(0^{n})$ so for some $i$ we have $f = \phi\_i(A(0^{n}))$. Let $m = \langle n,i\rangle$ and $z = \phi\_i(A(0^{n});m + n)$.
The first $k^{0^n}\_{m}$ columns of $A(0^{n})$ are computable in $0^{m + n}$ hence, by induction on the second equation above using the fact that $k\_{m+1} > k\_m + 2$, so too are the first $k^{0^n}\_{m} > z$ columns of $A(\emptyset)$. Thus, showing that $R\_{n,i}$ is satisfied.
[1]: We can view the construction of an $\omega$-REA set as an enumeration of axioms and using this gives us a stagewise approximation $A\_s$ (finite binary valued partial function) with the property that if $\sigma$ is an initial segment of $A$ then infinitely often $\sigma \prec A\_s$.
| 1 | https://mathoverflow.net/users/23648 | 446137 | 179,764 |
https://mathoverflow.net/questions/445642 | -4 | Newtons law for gravity states that:
$$F\_{12} = \frac{G m\_1 m\_2} {|x\_1-x\_2|^2}$$
The function :
$$k(x,y):=\exp(-| x-y|^2)$$
is known to be a positive definite function, called [the RBF-kernel](https://en.wikipedia.org/wiki/Radial_basis_function_kernel).
It follows that
$$k(x\_1,x\_2) := \exp( -|x\_1-x\_2|^2) = \exp(-\frac{G m\_1 m\_2}{F\_{12}})$$
is a positive definite , symmetric function. My question, is if you know of any other situations, where terms like
$$\exp(-\frac{m\_1 m\_2 G}{F\_{12}})$$
are shown to be positive definite kernels and what is the explanation for this?
(By this I mean: Do you know of situations where one proves that functions like:
$$f(a,b) = \exp(-\frac{m\_a m\_b G}{F\_{ab}})$$
are positive definite - and how is it proved? Here $m\_x$ denotes numbers $>0$ and $F\_{xy}$ is a function of $x,y$.)
Thanks for your help.
| https://mathoverflow.net/users/165920 | Inverse square-law as a positive definite kernel? | Please look pages 141-147 of the book:
S. Saitoh and Y. Sawano, Theory of Reproducing Kernels and Applications,
Developments in Mathematics 44, Springer (2016).
2023.5.4.20:36
| 2 | https://mathoverflow.net/users/504063 | 446143 | 179,766 |
https://mathoverflow.net/questions/446147 | 3 | Is the following statement true?
For all $n$ large enough, there exists an $M\_n \in [-1,1]^{n \times n}$ such that the smallest singular value of $M\_n$, $\sigma\_n(M\_n) \gtrsim \sqrt{n}$.
If $n$ is such that there exists a Hadamard matrix of order $n$, we can take $M\_n$ to be that.
On Terry Tao's [blog](https://terrytao.wordpress.com/2009/03/04/random-matrices-the-distribution-of-the-smallest-singular-values), it says that:
>
> To oversimplify somewhat, the conclusion of this work is that the least singular value $\sigma\_n(M\_n)$ has size comparable to $1/\sqrt{n}$ with high probability.
>
>
>
This does not rule out the possibility that for all $n$ large enough, the random signed matrix $M\_n$ satisfies $\sigma\_n(M\_n) \gtrsim \sqrt{n}$ with a tiny non-zero probability.
| https://mathoverflow.net/users/316923 | Existence of a matrix with bounded entries and large smallest singular value | The OP wrote $[-1, 1]$ but refers to Hadamard matrices, whose entries are more strongly in $\{ -1, 1 \}$. Assuming that the former is meant, we can do this with the [sine transform matrix](https://en.wikipedia.org/wiki/Discrete_sine_transform). Take
$$M\_{ab} = \sin \frac{\pi a b}{n+1} \ \text{for} 1 \leq a,b \leq n.$$
Then
$$M M^T = \frac{n+1}{2} \text{Id}\_n$$
so the singular values of $M$ are $\sqrt{\tfrac{n+1}{2}}$.
I imagine that other variants on the discrete Fourier transform would work simmilarly.
---
If you do want entries in $\{ -1, 1 \}$, this is also possible; see Theorem 1.2 in [Approximately Hadamard matrices and Riesz bases in random frames](https://arxiv.org/abs/2207.07523). Thanks to Aleksei Kulikov for making me aware of this paper.
| 7 | https://mathoverflow.net/users/297 | 446159 | 179,767 |
https://mathoverflow.net/questions/445980 | 1 | I am interested in a concrete case of a representation in de Finetti's Theorem for expressing the distribution of exchangeable binary sequence in terms of product measures. Namely, let
$$X\_1,\ldots,X\_n$$
be a random $0-1$ string with exactly $k$ $1$'s. How could one obtain a concrete respresentation of the measure in this particular case? This could be very standard and a reference might do. Thanks.
| https://mathoverflow.net/users/24494 | Concrete representation in de Finetti's Theorem | The distribution over $(X\_1, \ldots, X\_n)$ you are considering is extremally exchangeable, i.e. it is a vertex of the simplex of $n$-exchangeable distributions. It can not be extended to an $(n+1)$-exchangeable distribution $(X\_1, \ldots, X\_{n+1})$ of which it is the marginal. One consequence of De Finetti's theorem is the mixtures of iid measures are precisely those sequences which can be exchangeably extended to arbitrary length. So your original question cannot have a solution.
It has been mentioned in the comments that there are fragments of De Finetti's theorem for finite exchangeable sequences. **But these involve negative probabilities.** Geometrically, the IID distributions do not generate the space of all $n$-exchangeable distributions via convex combinations, but they still do so via affine combinations (this is where the negative coefficients come from). Finite representations are not unique, so for any given length $n$ you can find many different De Finetti representations by inverting certain matrices (given in the comments).
Maybe the most canonical one is this: In an iid sequence with bias $p$, the number $K$ of ones is $Bin(n,p)$ distributed. If we want to write your target distribution (with $P(K=n)=1$) as a continuous mixture with density $d(p)$, it is necessary and sufficient that
$$\int\_0^1 \binom{n}{k} p^k(1-p)^{n-k} d(p)\mathrm dp = \delta\_{nk}$$
that is $d(p)$ is orthogonal to the Bernstein basis polynomials. Such $d(p)$ must integrate to 1, thus is a signed probability density. The polynomial basis that is *defined* by this orthogonality relation is known as [dual Bernstein basis](http://www.ag.jku.at/pubs/tdb98.pdf) and seems to have some relevance in Computer Graphics, given the ubiquity of Bernstein polynomials. Here you have a (signed) probabilistic reading :)
| 2 | https://mathoverflow.net/users/4080 | 446162 | 179,769 |
https://mathoverflow.net/questions/446139 | 8 | In his famous list of [Problems in Low-Dimensional Topology](https://math.berkeley.edu/%7Ekirby/problems.ps.gz), Kirby states the following as Problem 3.14 (B), which is attributed to Thurston:
>
> Conjecture: Suppose $G$ (an arbitrary group I suppose) acts properly
> and discontinuously on a contractible 3-manifold with compact
> quotient. Suppose also that $G$ has no subgroup isomorphic to $Z\oplus Z$. Then $G$ is conjugate to a discrete group of isometries of
> hyperbolic 3-space.
>
>
>
Has this been settled, and if not would you still consider it interesting after the Geometrization theorem?
Going through Kirby's references I was not able to spot this question in one of Thurston's works; perhaps it is formulated differently there. A concrete reference would be appreciated.
| https://mathoverflow.net/users/69681 | Problem 3.14 from Kirby's list | This problem is answered in the literature, with a caveat.
As indicated in the comments, it follows from the orbifold theorem + geometrization conjecture (to handle the case of orbifolds without fixed points).
The caveat is that the action needs to be assumed smooth (or PL). Otherwise, there exists wild involutions such as the [Bing involution](https://arxiv.org/abs/2209.07597). One could incorporate this into an action on $\mathbb{R}^3$ by taking an orbifold quotient whose underlying space is $S^3$ and admits a reflection symmetry, then take a Bing involution preserving the orbifold locus by inserting into a part of the sphere being reflected. The resulting group action would have quotient which is not an orbifold. I think this was an underlying assumption of the problem as stated, but I thought I would clarify.
With the smoothness assumption, now this follows from the [orbifold theorem](https://en.wikipedia.org/wiki/Orbifold#3-dimensional_orbifolds) (see Problem 3.46 of Kirby’s list posed by Geoff Mess). I won’t go through the history of the proofs of cases of this (see the link), but it follows for orientable orbifolds modulo previous results by the proof of the [geometrization](https://en.wikipedia.org/wiki/Geometrization_conjecture?wprov=sfti1) theorem by Perelman.
For the non-orientable case, take an index-two subgroup $G’<G$ which is orientation-preserving. By the orientable case (and no $Z+Z$ subgroup assumption), $\mathbb{R}^3/G’$ is an orientable hyperbolic 3-orbifold. By Selberg’s lemma, there is a finite-index subgroup $G’’ < G’$ for which $M’’=\mathbb{R}^3/G’’$ is a manifold. By passing to a further subgroup (the *core*), we may assume that $G’’\lhd G$, so $G/G’’$ acts as a finite group of transformations of $M’’$. Now one may apply a result of [Dinkelbach-Leeb](https://msp.org/gt/2009/13-2/p15.xhtml) (Theorem H) to see that the quotient orbifold $M’’/(G/G’’)= \mathbb{R}^3/G$ is a hyperbolic orbifold.
| 11 | https://mathoverflow.net/users/1345 | 446165 | 179,771 |
https://mathoverflow.net/questions/446169 | 2 | Given (possibly non-Abelian) groups $H,G$ with $H \subseteq G$ and $f \in G$, I write $\langle H, f \rangle$ for the subgroup of $G$ generated by $H \cup \{f\}$.
Write $T(H)$ for the free product of $H$ with a multiplicative copy $y^{\mathbb{Z}}$ of the group of integers. We can represent a generic element $t(y)$ of $T(H)$ as a "term" $t(y) = h\_0 y^{\alpha\_0} \cdot \cdot \cdot h\_n y^{\alpha\_n}$ where $n \in \mathbb{N}$, $h\_0,...,h\_n \in H$ and $\alpha\_0,...,\alpha\_n \in \mathbb{Z}$. We set $t(f) := h\_0 f^{\alpha\_0} \cdot \cdot \cdot h\_n f^{\alpha\_n} \in G$ for this term.
Say that $f$ *cancels a term over $H$* if there is a $t(y) \in T(H) \setminus \{1\}$ with $t(f)=1$. This is the case for instance if $f$ commutes with a non-trivial element of $H$.
**Question**: Is it the case that if $f$ cancels a term over $H$, then each element of $\langle H, f\rangle$ cancels a term over $H$?
I suspect not, but I don't have a way to produce a counterexample.
| https://mathoverflow.net/users/45005 | Elements in group extensions which cancel unary terms in the language of groups | No. Let $G = \langle x, y \rangle$ be a $2$-generated non-free group with $x$ of infinite order such that $F = \langle x, g\rangle \cong F\_2$ for some $g \in G$. Then we get a counterexample by taking $H = \langle x \rangle$ and $f = y$. Indeed, since $G$ is not free, there is some nontrivial word $w \in F\_2$ such that $w(x, y) = 1$, so "$y$ cancels a term over $\langle x \rangle$", but since $\langle x, g\rangle$ is free the element $g$ does not cancel a term over $\langle x \rangle$.
There are probably many constructions of such a $G$. Here is one. Start with $F = F\_2 = \langle x, g\rangle$. Let $y$ be the automorphism of $F$ that swaps $x$ and $g$ and let $G = F\langle y \rangle \cong F \rtimes C\_2$. Observe that $G = \langle x, y\rangle$, since $g = x^y$. (In this case $y$ cancels a term over $\langle x \rangle$ for the boring reason that $y^2 = 1$, but $\langle x, x^y\rangle$ is free.)
| 6 | https://mathoverflow.net/users/20598 | 446173 | 179,774 |
https://mathoverflow.net/questions/446172 | 3 | I am trying to understand the asymptotics of Haar Moments on general compact Lie groups (in particular, subgroups of $\mathrm{SU}(n)$). I have learned that closed form formulae for these moments are only available for some classical groups, namely the orthogonal, unitary and symplectic groups. However, my understanding of the derivation of these formulae leads me to conjecture that the asymptotics of the moments for *any* compact Lie group $\mathcal{G}$ are inversely polynomial in the dimension of the largest irreducible sub-representation of $\mathcal{G}$. I have however been unable to find any results to this effect.
Does any one know of such results, or any counterexamples to the conjecture above? I am interested only in the asymptotic dependence on the dimension parameters and not in the explicit form of the integrals.
| https://mathoverflow.net/users/57449 | Asymptotics of Haar moments on general Lie groups | The generalization of Weingarten calculus to compact Lie groups is studied in [Expectation values of polynomials and moments on general compact Lie groups](https://arxiv.org/abs/2203.11607), see section 4.
| 2 | https://mathoverflow.net/users/11260 | 446174 | 179,775 |
https://mathoverflow.net/questions/446142 | 2 | Let $E,E'$ be a pair of elliptic curves defined over $\mathbb{Z}$. Let $T\_p[E], T\_p[E']$ be their associated ($p$-adic) Tate modules. These are Galois representations for the absolute Galois group of $\mathbb{Q}$, which we denote by $G$, denote them by $\rho,\rho'$, respectively.
Let $\overline{\rho},\overline{\rho'}$ be their associated residual representations and assume that $\overline{\rho}\cong\overline{\rho'}$. Is it true that $H^1(G,T\_p[E]) \cong H^1(G,T\_p[E'])$?
Thanks in advance!
| https://mathoverflow.net/users/174655 | Galois cohomology of Tate modules | Let $S$ be a finite set containing all places where an elliptic curve $E$ has bad reduction as well as $p$ and $\infty$. Write $T$ for $T\_pE$ and $G\_S$ for the Galois group of the maximal extension of $\mathbb{Q}$ unramified outside $S$. Then, for any $k>0$ the exact sequence $0\to T \to^{[p^k]} T \to E[p^k]\to 0$ is exact. Since $T^{G\_S}=0$, we get an isomorphism $H^1(G\_S,T)[p^k] \cong E(\mathbb{Q})[p^k]$ and a short exact sequence
$$ 0\to H^1(G\_S,T)/p \to H^1(G\_S, E[p]) \to H^2(G\_S,T)[p] \to 0.$$
If $p>3$, then $E(\mathbb{Q})[p^{\infty}]\cong H^1(G\_S,T)\_{\mathrm{tors}} = H^1(\mathbb{Q}, T)\_{\mathrm{tors}}$ only depends on the $G$-fixed part of $E[p]$. Therefore the torsion subgroup are isomorphic groups for $E$ and $E'$ in this case. For $p=2$ or $p=3$ or for larger number fields that fails.
The short exact sequence shows that the $\mathbb{Z}\_p$-rank of $H^1(G\_S,T)$ may also depend on more than just the Galois module $E[p]$, since there is no reason that the dimension of $H^2(G\_S,T)[p]$ as an $\mathbb{F}\_p$-vector space does not depend on this alone.
Here is an explicit counter example. Take $p=3$ and the curves $E$ <https://www.lmfdb.org/EllipticCurve/Q/20449g1/> and $E'$ <https://www.lmfdb.org/EllipticCurve/Q/20449d2/> . They have isomorphic $3$-torsion as they are the twist by $D=-143$ of a well known example of such a pair.
Now $E$ has rank $2$ and, if you believe BSD or if you are willing to calculate more than on that page, the $3$-primary part of the Tate-Shafarevich group has order $1$. Instead $E'$ has rank $0$ but its Tate Shafarevich group has order $9$.
Consider the Cassels sequence
$$ 0\to \operatorname{Sel}(T) \to H^1(G\_S,T) \to \bigl(E(\mathbb{Q}\_3)\otimes \mathbb{Q}\_3/\mathbb{Z}\_3\bigr)^{\vee} \to \operatorname{Sel}(E[p^{\infty}])^{\vee}, $$
where ${}^{\vee}$ stands for the Pontryagin dual, the Selmer groups are the projective and the direct limit of the usual $3^k$-Selmer groups.
For the curve $E$, we have $\operatorname{Sel}(T) \cong \mathbb{Z}\_3^2$, $\operatorname{Sel}(E[3^{\infty}])^{\vee}\cong \mathbb{Z}\_3^2$, while the local term is free of rank $1$ with the map to the right being injective since the points of infinite order in $E(\mathbb{Q})$ will not reduce to torsion point locally. We conclude that $H^1(G\_S,T)$ is free of rank $2$ for $E$.
For $E'$ instead, the group $\operatorname{Sel}(T')$ is trivial, while $\operatorname{Sel}(E'[3^{\infty}])^{\vee}$ is isomorphic to $\bigl(\mathbb{Z}/3\mathbb{Z}\bigr)^2$. The local term is still free of rank $1$. Hence, no matter whether the right hand map is injective or not, we have that $H^1(G\_S, T')$ is free of rank $1$ for $E'$.
Unsurprisingly, the $3$-Selmer groups $\operatorname{Sel}(E[3])$ are isomorphic of dimension $2$ over $\mathbb{F}\_p$, just for one of the curves this comes from global points while for the other is it the non-trivial Tate-Shafarevich group.
| 5 | https://mathoverflow.net/users/5015 | 446180 | 179,779 |
https://mathoverflow.net/questions/446182 | 17 | Any finitely presented simple group has solvable word problem, and hence recursive Dehn function. I'm curious though how wild these recursive functions could possibly be.
One concrete question is whether there are even any examples known of a finitely presented simple group whose Dehn function is exponential. The only examples I can think of at the moment of finitely presented simple groups where something is known about their Dehn functions are Burger-Mozes groups (quadratic Dehn function) and Thompson's groups $T$ and $V$ (polynomial Dehn function, conjecturally quadratic).
Another (vaguer) question is whether there is some stronger bound on what sorts of functions can be Dehn functions of finitely presented simple groups (stronger than just being recursive, I mean). E.g., maybe there's some reason they can't be super-exponential?
| https://mathoverflow.net/users/164670 | Dehn functions of finitely presented simple groups | To answer the vaguer question: I think there is no known bound on the Dehn functions of finitely presented simple groups. Recall:
**Boone–Higman Embedding Theorem.**
A finitely presented group has solvable word problem if and only if it can be embedded in a recursively presented simple group.
In their paper, they asked about strengthening their theorem:
**Question.** Is every finitely generated group G with soluble word problem embeddable in some finitely presented simple group?
This is still open (as I learnt from a talk by Jim Belk last year) and if the answer were yes, then we could embed groups with "arbitrarily bad" solvable word problem in finitely presented simple groups, whereas a bound on the Dehn function gives a bound on the complexity of the word problem.
*Boone, William W.; Higman, Graham*, [**An algebraic characterization of groups with soluble word problem**](https://doi.org/10.1017/S1446788700019108), J. Aust. Math. Soc. 18, 41-53 (1974). [ZBL0303.20028](https://zbmath.org/?q=an:0303.20028) [MR0357625](https://mathscinet.ams.org/mathscinet-getitem?mr=357625).
| 13 | https://mathoverflow.net/users/24447 | 446193 | 179,780 |
https://mathoverflow.net/questions/316474 | 31 | It is well-known that any positive rational number can be written as the sum of finitely many distinct unit fractions. This is easy since
$$\frac1n=\frac1{n+1}+\frac1{n(n+1)}\quad\text{for all}\ n=1,2,3,\ldots.$$
In 2015 I thought that this easy fact should have a further refinement which is somewhat sophisticated. Note that the series $\sum\_p\frac1{p-1}$ and $\sum\_p\frac1{p+1}$ (with $p$ prime) diverge just like the harmonic series $\sum\_{n=1}^\infty\frac1n$. Also, for any positive integer m there are infinitely many primes $p$ congruent to $1$ (or $-1$) modulo $m$ (by Dirichlet's theorem). Motivated by this, I made the following conjecture in Sept. 2015.
**Conjecture**. For any rational number $r>0$, there are finite sets $P\_r^-$ and $P\_r^+$ of primes such that
$$r=\sum\_{p\in P\_r^-}\frac1{p-1}=\sum\_{p\in P\_r^+}\frac1{p+1}.$$
This appeared as Conjecture 4.1 of [this published paper](https://doi.org/10.1007/978-3-319-68032-3_20) of mine.
For example,
$$2=\frac1{2-1}+\frac1{3-1}+\frac1{5-1}+\frac1{7-1}+\frac1{13-1}$$
with $2,3,5,7,13$ all prime, and
$$1=\frac1{2+1}+\frac1{3+1}+\frac1{5+1}+\frac1{7+1}+\frac1{11+1}+\frac1{23+1}$$
with $2,3,5,7,11,23$ all prime. Also,
\begin{align\*}\frac{10}{11}=&\frac1{3-1}+\frac1{5-1}+\frac1{13-1}+\frac1{19-1}+\frac1{67-1}+\frac1{199-1}
\\=&\frac1{2+1}+\frac1{3+1}+\frac1{5+1}+\frac1{7+1}+\frac1{43+1}+\frac1{131+1}+\frac1{263+1}
\end{align\*}
with $2,3,5,7,13,19,43,67,131,199,263$ all prime. The reader may see more numerical data in [my detailed introduction](http://maths.nju.edu.cn/~zwsun/UnitFraction.pdf) to this conjecture.
After learning this conjecture from me, Prof. Qing-Hu Hou and Guo-Niu Han checked my above conjecture seriously and their computational results support my conjecture. For example, in 2018 Prof. Han found 2065 distinct primes $p\_1<\ldots<p\_{2065}$ with $p\_{2065}\approx 4.7\times10^{218}$ such that $$\frac1{p\_1+1}+\ldots+\frac1{p\_{2065}+1}=2.$$
My question is whether the above conjecture is true. I would like to offer 500 US dollars as the prize for the first correct solution.
**Remark**. Let $r$ be any positive rational number, and let $\varepsilon\in\{\pm1\}$. As the series $\sum\_p\frac1{p+\varepsilon}$ (with $p$ prime) diverges, there is a unique prime $q$ such that $$\sum\_{p<q}\frac{1}{p+\varepsilon}\le r<\sum\_{p\le q}\frac1{p+\varepsilon}.$$ Thus
$$0\le r\_0:=r-\sum\_{p<q}\frac1{p+\varepsilon}<\frac1{q+\varepsilon}\le1.$$
If $r\_0=\sum\_{j=1}^k\frac1{p\_j+\varepsilon}$ with $p\_1,\ldots,p\_k$ distinct primes, then $p\_1,\ldots,p\_k$ are all greater than $q$, and
$$r=\sum\_{p<q}\frac1{p+\varepsilon}+\sum\_{j=1}^k\frac1{p\_j+\varepsilon}.$$ Therefore it suffices to consider the conjecture only for $r<1$.
| https://mathoverflow.net/users/124654 | Can we write each positive rational number as $\frac1{p_1-1}+\ldots+\frac1{p_k-1}$ with $p_1,\ldots,p_k$ distinct primes? | This conjecture is true (as is the version for $p-h$ for any $h\neq 0$).
The proof is too long to reproduce here, but the preprint is at <https://arxiv.org/abs/2305.02689>
EDIT:
Quick summary as requested: My previous paper (<https://arxiv.org/abs/2112.03726>) essentially shows that for any set $A$ which is large enough and satisfies some mild number theoretic conditions we can write $1$ as the sum of distinct unit fractions with denominators from $A$. (The most important of these conditions is 'smoothness', that no $n\in A$ is divisible by a large prime power.)
Such a result was previously proved by Croot in 2003, but with a stronger smoothness condition. The proof is a strengthening of Croot's method, using the circle method and some combinatorial analysis.
What's new is observing that we can show that the set $\{p-h\}$ satisfies these mild number theoretic conditions - which is to be expected, since these conditions are all 'generic', and we expect $p-h$ to look like a typical integer of the same size from a multiplicative point of view. In fact very classical techniques suffice here (e.g. the Siegel-Walfisz theorem, Selberg's sieve, and an Erdos-Kac type theorem due to Halberstam from the 1950s).
This means we can write $1$ as the sum of $\frac{1}{p-h}$. But this is robust, in that we can remove any finite set of primes and still find such a sum. Using the identity $n=1+\cdots+1$ we can therefore write any integer $n$ as the sum of distinct $\frac{1}{p-h}$. Now to write any $\frac{n}{m}$ do the above with primes $p\equiv h\pmod{m}$ and the set $A=\{\frac{p-h}{m}\}$.
| 16 | https://mathoverflow.net/users/385 | 446198 | 179,781 |
https://mathoverflow.net/questions/8247 | 122 | I'm collecting advanced exercises in geometry. Ideally, each exercise should be solved by one trick and this trick should be useful elsewhere (say it gives an essential idea in some theory).
If you have a problem like this please post it here.
**Remarks:**
* I have been collecting such problems for many years. The current collection is at [arXiv](https://arxiv.org/abs/0906.0290); the paper version is available at [amazon](https://rads.stackoverflow.com/amzn/click/com/B08QS38Z23).
* At the moment, I have just a few problems in topology and in geometric group theory and only one in algebraic geometry.
* Thank you all for nice problems --- I decided to add bounty once in a while and choose the best problem (among new or old).
| https://mathoverflow.net/users/1441 | One-step problems in geometry | Let $A$ be a set of intially labelled points in $\mathbb{R}^d$. We may take any line containing at least $k$ labelled points and label any point on this line. For which minimal size $|A|$ (as a function of $d, k$) it may occur that we can label (by performing finitely many such operations) every point of $\mathbb{R}^d$?
| 1 | https://mathoverflow.net/users/4312 | 446218 | 179,788 |
https://mathoverflow.net/questions/446157 | 21 | The canonical inclusion $\beta\mathbb{Q}\setminus \mathbb{Q} \hookrightarrow \beta\mathbb{Q}$ is not the Stone-Čech compactification of $\beta\mathbb{Q}\setminus \mathbb{Q}$. Even so, this doesn't necessarily mean that $\beta(\beta\mathbb{Q}\setminus\mathbb{Q})$ and $\beta\mathbb{Q}$ are not homeomorphic, just that this particular map doesn't work for this compactification, so that they might be homeomorphic less "canonically". For example, $\beta(\mathbb{Q}\setminus\{0\})\cong \beta\mathbb{Q}$ but the map $\mathbb{Q}\setminus\{0\}\hookrightarrow \beta\mathbb{Q}$ is not the Stone-Čech compactification of $\mathbb{Q}\setminus\{0\}$, even though it's a dense embedding into a compact space homeomorphic to $\beta(\mathbb{Q}\setminus\{0\})$.
I've tried looking online for some properties of $\beta\mathbb{Q}\setminus\mathbb{Q}$ that would exclude a homeomorphism, though while looking I could only find properties it has in common with $\mathbb{Q}$, with exception of homogeneity (though I don't think this particular property amounts to much). For example, $\beta\mathbb{Q}\setminus \mathbb{Q}$ is zero-dimensional, not extremally disconnected.
As in the title, is $\beta\mathbb{Q}\cong \beta(\beta\mathbb{Q}\setminus \mathbb{Q})$?
**Edit:** By suggestion of @R. van Dobben de Bruyn's [comment](https://mathoverflow.net/questions/446157/are-beta-mathbbq-and-beta-beta-mathbbq-setminus-mathbbq-homeomorp#comment1152375_446157), I've checked cardinalities of both spaces. In the article "[The Stone-Čech compactification of the rational world](https://doi.org/10.1017/S0017089500007205)" by M. P. Stannett it's shown that $\beta\mathbb{Q}\setminus\mathbb{Q}$ is separable, so that $$\lvert\beta(\beta\mathbb{Q}\setminus\mathbb{Q})\rvert \leq 2^{2^{d(\beta(\beta\mathbb{Q}\setminus\mathbb{Q}))}} \leq 2^{2^{d(\beta\mathbb{Q}\setminus\mathbb{Q})}} = 2^\mathfrak{c}$$ while the inclusion $\beta\mathbb{Q}\setminus \mathbb{Q}\hookrightarrow \beta\mathbb{Q}$ induces a surjection $\beta(\beta\mathbb{Q}\setminus\mathbb{Q})\to \beta\mathbb{Q}$, thus $$\lvert\beta\mathbb{Q}\rvert = \lvert\beta(\beta\mathbb{Q}\setminus\mathbb{Q})\rvert = 2^\mathfrak{c}.$$ This shows in particular that there's no issue with cardinality. I think trying to approach it with weight would provide similar results, though I haven't checked that.
**Edit2:** Here's my proof of my claim that $\beta\mathbb{Q}\setminus\mathbb{Q}\hookrightarrow \beta\mathbb{Q}$ is not the Stone-Čech compactification of the space $\beta\mathbb{Q}\setminus\mathbb{Q}$. Note that $\beta\mathbb{Q}\setminus\mathbb{Q}$ is dense in $\beta\mathbb{Q}$ since $\mathbb{Q}$ is nowhere locally compact. The space $\beta\mathbb{Q}\setminus \mathbb{Q}$ is not $C^\*$-embedded in $\beta\mathbb{Q}$: the decomposition $$\beta\mathbb{Q}\setminus \{0\} = (\overline{\mathbb{Q}}\_+\setminus\{0\}) \cup (\overline{\mathbb{Q}}\_-\setminus \{0\})$$ of $\beta\mathbb{Q}\setminus\{0\}$ where $\mathbb{Q}\_+ = (0, \infty)\cap\mathbb{Q}$ and $\mathbb{Q}\_- = (-\infty, 0)\cap \mathbb{Q}$ into two disjoint closed sets in $\beta\mathbb{Q}\setminus\{0\}$ shows that $\DeclareMathOperator\sgn{sgn}\sgn:\mathbb{Q}\setminus\{0\}\to \mathbb{R}$ can be continuously extended to $\sgn^\*:\beta\mathbb{Q}\setminus\{0\}\to\mathbb{R}$ but clearly not to whole of $\beta\mathbb{Q}$. If the function $\sgn^\*\restriction\_{\beta\mathbb{Q}\setminus\mathbb{Q}}$ were to continuously extend to $\beta\mathbb{Q}$, the extension would have to be equal to $\sgn^\*$ on $\beta\mathbb{Q}\setminus\{0\}$, which is impossible since $\sgn^\*$ doesn't extend to $\beta\mathbb{Q}$. Since for $A\subseteq \beta\mathbb{Q}$, we have that $\DeclareMathOperator\cl{cl}\cl\_{\beta\mathbb{Q}}A =\beta A$ (that is $A\hookrightarrow \cl\_{\beta\mathbb{Q}} A$ is the Stone-Čech compactification of $A$) iff $A$ is $C^\*$-embedded in $\beta\mathbb{Q}$, the inclusion $\beta\mathbb{Q}\setminus \mathbb{Q} \hookrightarrow \beta\mathbb{Q}$ isn't the Stone-Cech compactification of $\beta\mathbb{Q}\setminus \mathbb{Q}$.
| https://mathoverflow.net/users/150060 | Are $\beta \mathbb{Q}$ and $\beta(\beta\mathbb{Q}\setminus\mathbb{Q})$ homeomorphic? | Let me summarize the discussion in the comments as an answer. Let $\chi(x, Y)$ be the character of $x$ in $Y$ i.e. the least cardinality of a local basis of the point $x$ in space $Y$.
**Proposition 1.** If $p\in \beta X\setminus X$ then $\chi(p, \beta X)$ is uncountable.
*Proof:* If it werre $\chi(p, \beta X) = \aleph\_0$ we would find a sequence $(a\_n)\subseteq X$ with $a\_n\to p$ and $a\_n\neq a\_m$ for $n\neq m$. Since $\beta X\setminus \{p\}$ is $\sigma$-compact, it's Lindelöf regular, so normal. Moreover, $A = \{a\_n : n\in\mathbb{N}\}$ is a closed discrete subset of $\beta X\setminus \{p\}$. Thus the function $f:A\to [-1, 1]$ given by $f(a\_n) = (-1)^n$ has a continuous extension $\tilde f$ to $\beta X\setminus \{p\}$, but no extension to $\beta X$. This is a contradiction, because $\tilde f\restriction\_X$ needs to continuously extend to $\beta X$, and thus the extension needs to be equal to $\tilde f$ on $\beta X\setminus \{p\}$.
**Proposition 2.** If $S\subseteq X$ is dense, $X$ regular, $p\in S$, then $\chi(p, S) = \chi(p, X)$.
This can be found in the Handbook of Set-theoretic Topology.
**Proposition 3.** If $p\in \beta(\beta\mathbb{Q}\setminus \mathbb{Q})$ then $\chi(p, \beta(\beta\mathbb{Q}\setminus \mathbb{Q}))$ is uncountable.
*Proof:* If $p\in \beta(\beta\mathbb{Q}\setminus\mathbb{Q})\setminus (\beta\mathbb{Q}\setminus\mathbb{Q})$ then $\chi(p, \beta(\beta\mathbb{Q}\setminus\mathbb{Q}))$ is uncountable by proposition 1. If $p\in \beta\mathbb{Q}\setminus\mathbb{Q}$, then $\chi(p, \beta(\beta\mathbb{Q}\setminus\mathbb{Q})) = \chi(p, \beta\mathbb{Q}\setminus\mathbb{Q}) = \chi(p, \beta\mathbb{Q})$ by proposition 2, which is uncountable, again by proposition 1.
Thus $\beta\mathbb{Q}$ has elements of countable character while $\beta(\beta\mathbb{Q}\setminus \mathbb{Q})$ has no such points, so the two spaces are not homeomorphic.
| 7 | https://mathoverflow.net/users/150060 | 446233 | 179,794 |
https://mathoverflow.net/questions/446015 | 11 | Apologies for a naive question (especially for Iwasawa theorists): it is well-known
and trivial to prove that the usual (elementary) construction of $p$-adic L functions
attached to odd Dirichlet characters leads to a function which is identically zero
(experts even have a highbrow explanation for this) for a "silly reason" as one
expert says (essentially $\chi(-1)+1=0$). But still, isn't there any kind
of $p$-adic "object" which interpolates the Euler numbers for instance, essentially
$2L(\chi\_{-4},-2k)$, or $3L(\chi\_{-3},-2k)$, similar to Kubota--Leopoldt interpolating $\zeta(1-2k)$ after removing some $p$-Euler factor ?
| https://mathoverflow.net/users/81776 | $p$-adic L function of an odd Dirichlet character | **Theorem**. Let $p > 2$ be prime, and let $\chi$ be a Dirichlet character (non-trivial, and of prime-to-$p$ conductor, for
simplicity). Choose $t \in \mathbb{Z} / (p-1)\mathbb{Z}$ such that
$(-1)^t = \chi(-1)$. Then there exists a continuous function $L\_{p,
t}(\chi, -) : \mathbb{Z}\_p \to \mathbb{C}\_p$ such that for all
integers $k \ge 0$ with $k = t \pmod {p-1}$, we have $L\_{p, t}(\chi, k) =
(1 - \chi(p) p^{k-1}) \cdot L(\chi, 1 - k)$.
(*Edited to add*: We can say much more than continuity; in fact if $\mathcal{O}$ is the ring of integers of the extension of $\mathbb{Q}\_p$ generated by the values of $\chi$, we have $L\_{p, t}(\chi, s) = F( (1 + p) ^ s - 1)$ for some power series $F \in \mathcal{O}[[T]]$.)
Taking a more highbrow view, one should think of $L\_p(\chi)$ as an element of the Iwasawa algebra $\mathcal{O}[[\mathbb{Z}\_p^\times]]$. This ring is a direct product of $(p-1)$ subrings, indexed by $\mathbb{Z} / (p-1)$, each of them isomorphic to $\mathcal{O}[[T]]$. For any $\chi$, the element $L\_p(\chi)$ will project to zero in half of those components, but which half will depend on the parity of $\chi$.)
(My go-to reference for a modern account of $p$-adic Dirichlet $L$-functions is Dasgupta, [Factorization of p-adic Rankin L-series](https://services.math.duke.edu/%7Edasgupta/papers/Factorization.pdf), section 3.1.)
**EDIT, 9.5.23**. There seems to be some confusion here about the role of the Teichmüller char $\omega$. Note that there were no $\omega$'s at all in what I wrote! The interpolation formula I gave,
$$L\_{p, t}(\chi, k) = \star \cdot L(\chi, 1 - k) \ \text{for $k \ge 0$ st $k = t \bmod {p-1}$},$$
(where $\star$ is a local correction factor at $p$ which I want to ignore for now), is a special case of the more general statement
$$L\_{p, t}(\chi, k) = \star \cdot L(\chi \omega^{t - k}, 1 - k) \ \text{for all $k \ge 0$}.$$
If we set $t = 0$, then this gives the interpolation formula from Henri's remark. But the theorem is true for all $t$ with $(-1)^t = \chi(-1)$; so if $\chi$ is odd, you *cannot* set $t = 0$; but there are still lots of other interesting choices of $t$ you can use.
| 8 | https://mathoverflow.net/users/2481 | 446243 | 179,797 |
https://mathoverflow.net/questions/446222 | 1 | Is there a commutative subalgebra $A\subset B(H)$ containing the 1 dimensional scalars with the following property:
>
> The algebra $A$ has trivial intersection with the set of commutator elements $xy-yx$
>
>
>
| https://mathoverflow.net/users/36688 | A subalgebra of $B(H)$ which does not contain a commutator element | From the results of
*Brown, Arlen; Pearcy, Carl*, [**Structure of commutators of operators**](https://doi.org/10.2307/1970564), Ann. Math. (2) 82, 112-127 (1965). [ZBL0131.12302](https://zbmath.org/?q=an:0131.12302).
an element of $B(H)$ is a commutator $xy-yx$ if and only if it is not of the form $\lambda I + C$ for some non-zero scalar $\lambda$ and compact $C$. So your algebra would have to contain an element of the form $\lambda I + C$ but not $C$ (unless $C=0$), and so cannot contain the scalars unless it consisted solely of the scalars, which are indeed not commutators by the classical results of [Wintner](https://mathscinet.ams.org/mathscinet-getitem?mr=20724) and [Wielandt](https://mathscinet.ams.org/mathscinet-getitem?mr=30701).
| 10 | https://mathoverflow.net/users/766 | 446252 | 179,800 |
https://mathoverflow.net/questions/446257 | 7 | Let $W(\alpha)$ denote the set of all (countable) ordinals *writable* by [Ordinal Turing Machines](https://arxiv.org/abs/math/0502264) with a single (countable) parameter $\alpha$, i.e. each computation starts with a single ($\alpha$-th) cell on the input tape marked with a non-zero symbol (all other cells are marked with a zero symbol).
Question: does there exist an ordinal $\beta$ such that there exists at least one ordinal $\gamma < \beta$ such that $\gamma \notin W(\beta)?$ If no, why? If yes, how large is the smallest such $\beta$?
| https://mathoverflow.net/users/122796 | Gaps in the ordinals writable by Ordinal Turing Machines with a single countable parameter | The answer is yes. For example, take $\beta=\omega\_1$, the first uncountable ordinal. Since there are only countably many programs, there can be only countably many writable ordinals relative to $\beta$ as input. But there are uncountably many ordinals $\gamma$ below $\beta$, and so most of them are not writable from $\beta$.
The smallest such ordinal will be exactly $\omega\_1^L$. To see this, consider any $\beta<\omega\_1^L$. Thus, $\beta$ is countable in $L$, and this will be revealed in some countable stage of the constructibility hierarchy $L\_\eta$. Consider the algorithm that on input $\beta$ begins to construct copies of the constructible hierarchy until it finds a stage $L\_\eta$ that can see that $\beta$ is countable. Thus, this structure provides an $\omega$-enumeration of the ordinals $\gamma<\beta$. Therefore they will all be writable from $\beta$, since there will be a program that searches for this $L\_\eta$ and then writes the $n$th such $\gamma$ in that enumeration.
| 8 | https://mathoverflow.net/users/1946 | 446274 | 179,804 |
https://mathoverflow.net/questions/445815 | 4 | I'm reading *"BGG category $\mathcal{O}$"* by Humphreys.
In section 3.12 we look into the projective modules over $\mathfrak{sl}(2,\mathbb{C})$. If $\lambda\in\mathbb{Z}$ is a weight and $\mu=-\lambda-2$ is the other weight linked to $\lambda$ then from calculation and the BGG reciprocity we get that the projective cover $P(\mu)$ of $L(\mu)$ is in the non-split extension $$0\to M(\lambda)\to P(\mu)\to M(\mu)\to0$$
where $M(\lambda),M(\mu)$ are the Verma's modules.
After that, there is an exercise to calculate the action of the Casimir element $z:=h^2+2h+4yx$ on $P(\mu)$. I know how to prove that $(z-c)^2=0$ when $c:=\chi\_\lambda(z)=\lambda^2+2\lambda$. I also know how to prove that $z$ doesn't act as a scalar, but I don't know how exactly it acts.
Also, I would like to know more details about the structure of $P(\mu)$, and how $x,y$ acts on the nontrivial/interesting weight spaces (coming from $M(\mu)$). Is there a general way to describe this? I try to compute but it seems not a lot of fun.
| https://mathoverflow.net/users/496537 | Structure of projective indecomposable modules for $\mathfrak{sl}_2$ | Denoting the highest weight vectors of
$M(\lambda)$ and $M(\mu=-\lambda-2),\, \lambda\in\mathbb{Z}\_{>0}$ as $v$ and $v\_\mu$, we choose a preimage $w\in P(\mu)$ of $v\_\mu$ with $h\cdot w =\mu w$. Then for nontrivial extension the vector $x\cdot w$ does not vanish, hence $x\cdot w=A\_1 y^{\lambda}\cdot v$.
The action of $x$ can be calculated directly,
\begin{align}
& x\cdot v\_{n}=n(\lambda-n+1)v\_{n-1}\\
& x\cdot w\_{n}=A\_1 v\_{\lambda+n}-n (1 + \lambda + n)w\_{n-1}
\end{align}
where $v\_{n}=y^{n}\cdot v$ and $w\_{n}=y^{n}\cdot w$. Hence $x$ acts with Jordan blocks, and contributes nontrivially to the action of the Casimir.
This calculation relies on the assumptions of category $\mathcal{O}$. There are other extensions with non-semisimple action of $h$, which are more interesting in physics and are related to the discussion of logarithmic CFTs.
| 0 | https://mathoverflow.net/users/102775 | 446277 | 179,805 |
https://mathoverflow.net/questions/446246 | 9 | The Division Paradox is the fact that there are models of ${\sf ZF \neg C}$ in which a set can be partitioned into a set that is bigger than it — equivalently, in which there are sets $X$ and $Y$ such that $|X| < |Y|$ yet there is a surjection from $X$ onto $Y$. For example, there are models of ${\sf ZF \neg C}$ in which $\mathbb{R}$ can be partitioned into a set that is bigger than it. Taylor & Wagon's ["A Paradox Arising from the Elimination of a Paradox"](https://www.tandfonline.com/doi/full/10.1080/00029890.2019.1559416) describe a number of other examples as well.
My question is: can an analogous phenomenon arise for proper classes in standard class theories without class-theoretic analogues of choice — for example, ${\sf NBG}$ or ${\sf MK}$ without global choice or limitation of size?
While there's not a direct class-theoretic analogue of the notion of a "partition" for proper classes — after all, it's not sensible to speak of a class *of* proper classes — we can nonetheless "encode" cardinality comparisons between proper classes by understanding injections, surjections, and bijections simply as classes of ordered pairs — instead of sets of ordered pairs — with the relevant properties. So we can ask whether there are models of (say) ${\sf NBG}$ without global choice in which there are proper classes $X$ and $Y$ such that: (1) there is an injection from $X$ to $Y$, but not vice versa, and (2) there is a surjection from $X$ to $Y$. If there are such models, then these would be class-theoretic instances of the Division Paradox.
I was looking at Theorems 5 and 6 of Wagon & Taylor's paper in particular and wondering if the examples they describe (provided by Asaf Karagila) admit of a class-theoretic analogue, but I couldn't work this out for myself. One thing that is shown there is that ${\sf ZF}$, in conjunction with dependent choice and the claim that all sets of reals are Lebesgue measurable, proves that $|\mathbb{R} \cup \omega\_1| < |\mathbb{R} \times \omega\_1|$, yet there is a surjection from $\mathbb{R} \cup \omega\_1$ to $\mathbb{R} \times \omega\_1$. Can we prove a class-theoretic analogue of this result (or something similar) in a choiceless setting — e.g., if we replaced $\omega\_1$ with the class of all ordinals?
| https://mathoverflow.net/users/504195 | Class-theoretic division paradox | The answer is yes.
First, let us observe the following lemma. Let us work in GBc, that is, Gödel-Bernays set theory with the axiom of choice, but only choice for sets, and not global choice.
**Lemma.** If global choice fails, then there is a proper class $A$ with no class injection $F:\newcommand\Ord{\text{Ord}}\Ord\hookrightarrow A$.
**Proof.** Suppose global choice fails. Let $A$ be the class of all well-orderings of any rank-initial segment $V\_\alpha$ of the universe. Any injection of $\newcommand\Ord{\text{Ord}}\Ord$ into $A$ would have to hit arbitrarily large $V\_\alpha$, and from it we could therefore define a well ordering of the universe. Namely, $x<y$ when this is true for the first order in the injection that puts them in an order. $\Box$
Meanwhile, every proper class maps surjectively onto $\Ord$, since it must contain sets of arbitrarily large ranks, and so we can map surjectively to an unbounded subclass of $\Ord$, but by omitting the missing elements, we get a copy of $\Ord$.
**Theorem.** If global choice fails, then there are classes $X$ and $Y$ such that
1. $X$ injects into $Y$, but not conversely. i.e. $X<Y$.
2. $X$ surjects onto $Y$.
**Proof.** Let $X=\omega\times A$ and $Y=X\sqcup\Ord$. It is clear that $X$ injects into $Y$ since it is a subclass. The class $X$ also surjects onto $Y$, because we can use $\{0\}\times X$ to map onto the ordinals, since $X$ has elements of unbounded rank, and then shift $\{n+1\}\times X$ to $\{n\}\times X$. But there is no injection of $Y$ to $X$, since I claim indeed that there is no injection of $\Ord$ to $X$. Any such injection would have to pick elements of the form $(n,a)$ for $a\in A$, and from it we could define an injection $\Ord\hookrightarrow A$ by looking at the first ordinal which hits any $(n,a)$ and deleting other uses of $a$ (and then reindexing the domain). $\Box$
Another way to state the example is as follows.
**Theorem.** If global choice fails, then there is a class $X$ with an equivalence relation $\sim$, such $X$ has strictly more equivalence classes than elements. That is,
1. There is a class function $X$ to $X$ such that distinct elements map to $\sim$-inequivalent elements, but
2. There is no class function $F:X\to X$ that is constant on $\sim$-classes with a different value on different classes.
**Proof.** Use the same $X$, and define $\sim$ to make all $(0,a)$ of the same rank equivalent, but all others inequivalent. That is, in the first section only, we identify the elements of $A$ by rank. This amounts to putting a copy of $\Ord$ into the quotient, as in $Y$. We can map $X$ injectively to the quotient by shifting sections, but there is no function $F$ from the quotient to $X$ since that would provide a map from Ord to $X$, which is impossible as we have argued. $\Box$
Since global choice implies that all proper classes are equinumerous, what the theorem shows in consequence is that the class partitition paradox phenomenon is simply equivalent to the failure of global choice.
Finally, let me pull the analysis back to ZFC.
**Theorem.** The following are equivalent in ZFC.
1. There is no definable global well order (with parameters).
2. The partition paradox holds for some definable classes. That is, there is a definable class $X$ having a equivalence relation $\sim$ so that $X$ has strictly more $\sim$-equivalence classes than elements.
**Proof.** Note first that by equipping any ZFC model with its definable classes, we get a model of GBc. If there is such a definable global well order, and this implies that all proper classes are definably equinumerous with $\Ord$. In this case, the partition paradox does not occur.
If there is no such definable global well order, then we can run the counterexample above, which involves definable classes only. $\Box$
Let me clarify the formalism of the statement of the theorem, since these statements are not directly first-order assertions in set theory. The situation is that if there is a definable global well order, then one can prove in each case separately that no definable class is an instance of the partition paradox. And if there is no definable global well order, then we get the specific definable instance of the proof. In other words, if that instance is not a paradoxical case, then we can deduce the scheme for all other definitions that they are also are not paradoxical.
| 11 | https://mathoverflow.net/users/1946 | 446286 | 179,807 |
https://mathoverflow.net/questions/446263 | 4 | **I. Degree 8**
Assume the $j\_i$ to be free parameters. The following octics in $x$ belong to $8T43,$ have group $\text{PGL}(2,7)$, and order $2\times168 = 336.$
\begin{align}
{j\_1}\; &=\frac{(x^2 + 5x + 1)^3(x^2 + 13x + 49)}x\\
{j\_2}^2 &=\frac{j\_2\,(-7x^4 - 196x^3 - 1666x^2 - 3860x + 49)+(x^2 + 14 x + 21)^4}x\\
{j\_3}\; &=\frac{(x + 1)^6(x^2 + x + 7)}x\\
{j\_4}^2 &=\frac{7 j\_4\,(x + 1)^4+(x + 1)^7 (x - 7)}x
\end{align}
Expanded out, some of the coefficients are quadratic in $j\_i$. They also have nice discriminants $D\_i$. (Note: $D\_2$ has another square factor I've suppressed for brevity. I have a feeling the second octic may have a simpler version.)
\begin{align}
D\_1 &= -7^7(j\_1-1728)^4\,{j\_1}^4\\
D\_2 &= -7^7(j\_2-256)^4\,{j\_2}^6 \\
D\_3 &= -7^7(j\_3-108)^3\,{j\_3}^5\\
D\_4 &= -7^7(j\_4-64)^4\,{j\_4}^{12}
\end{align}
For general $j\_i$, these octics are not solvable in radicals. However, if we substitute the following integers,
\begin{align}
j\_1 &= -640320^3\quad \\
j\_2 &= -63^2,\, 396^4\quad \\
j\_3 &= -300^3\quad \\
j\_4 &= -2^9,\, 2^{12}\quad
\end{align}
then they *become solvable*. The first two $j\_i$ are famous being in the Chudnovsky and Ramanujan pi formulas, and the last two can be used in [Ramanujan-Sato pi formulas](https://en.wikipedia.org/wiki/Ramanujan%E2%80%93Sato_series) as well.
---
**II. Eta quotients**
There are infinitely many such radical $j\_i$ given by,
\begin{align}
\quad j\_1 &=\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{8}+2^8 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{16}\right)^3 \\
\quad j\_{2} &=\left(\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^{12}+2^6 \left(\frac{\eta(2\tau)}{\eta(\tau)}\right)^{12}\right)^2 \\
\quad j\_{3} &=\left(\left(\frac{\eta(\tau)}{\eta(3\tau)}\right)^{6}+3^3 \left(\frac{\eta(3\tau)}{\eta(\tau)}\right)^{6}\right)^2 \\
\quad j\_{4} &=\left(\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^{4} + 4^2 \left(\frac{\eta(4\tau)}{\eta(\tau)}\right)^{4}\right)^2 = \left(\frac{\eta^2(2\tau)}{\eta(\tau)\,\eta(4\tau)} \right)^{24}
\end{align}
where $\eta(\tau)$ is the [*Dedekind eta function*](https://en.wikipedia.org/wiki/Dedekind_eta_function#Eta_quotients) and $j\_1$ is just the j-function. (For examples of $\tau$, click on the link.)
Thus, the octics in $x$ are also formulas for the $j\_i$ and where $x$ involve eta quotients (though I didn't include them to prevent clutter).
---
**III. Degree 6**
So far, I've only found two with group $\text{PGL}(2,5)$ hence $6T14$ and order $2\times60 = 120$. (***Edit***: For instant comparison, we add the other two $6T14$ from Joachim König's answer, in red.)
\begin{align}
j\_1 &=\frac{(x^2 + 10x + 5)^3}x\\
\color{red}{j\_2} &=\frac{(x+1)^4(x^2+6x+25)}{x}\\
\color{red}{{j\_3}^2} &= \frac{j\_3\,(2 y^6 + 29y^5 + 85y^4 + 50y^3) + 5^4 y^6}{(2y - 1)} \\
j\_4 &=\frac{(x + 1)^5 (x + 5)}x
\end{align}
with discriminants,
\begin{align}
D\_1 &= 5^5\,(j\_1-1728)^2\,{j\_1}^4\\
D\_2 &= 5^5\,(j\_2-256)^3\,{j\_2}^3 \\
D\_3 &= 5^5(j\_3-108)^3\,{j\_3}^{11}\\
D\_4 &= 5^5\,(j\_4-64)^2\,{j\_4}^4
\end{align}
It is preferred the constant factor is only $5^5$. *Note*: $D\_3$ actually has the extra square factor $(32 j\_3+84375)^2$, but I truncated it for aesthetics.
---
**IV. Question:**
So are there analogous sextic $6T14$ formulas in $x$ for $j\_2, j\_3$, where the $j\_i$ are linear or quadratic, and discriminants similar to the others? (Note: Has been answered in the affirmative.)
---
**V. Addendum:**
*(The addendum has been moved as an "answer" to prevent clutter and loading issues.)*
| https://mathoverflow.net/users/12905 | Finding a sextic analogue to the solvable octic $\frac{(x + 1)^6(x^2 + x + 7)}x = -k^3$ where $e^{(\pi/3)\sqrt{d}}\approx k^3+41.999999999999\dots$ | Didn't think about eta quotients, but if the suggested analogies of ramification types are anything to go by, then
$$j\_2 = \frac{(x+1)^4(x^2+6x+25)}{x}$$
(which matches the prescribed discriminant exactly), and
$$j\_3^2 = \frac{2(x+1)^3(x^3+32x^2+231x+400) j\_3 + 5^4(x+1)^6}{64x}$$
(or possibly some reparametrization of it). The last equation has additional divisors a $2$-power as well as a square of a linear factor (however not corresponding to a branch point of the function field extension); on the other hand, the quintic subextension of the Galois closure could be parameterized by $j\_3 =y^3(y-5)^2$, which does give exactly matching discriminant. Not sure if there's any direct reason why reparametrization should allow for extra factors to vanish, just like for general number fields it's not always possible to find a polynomial whose discriminant matches the field discriminant.
| 4 | https://mathoverflow.net/users/127660 | 446287 | 179,808 |
https://mathoverflow.net/questions/445495 | 4 | Let $S\_1,S\_2,S\_3$ be three simple closed curves on the 2-sphere $\mathbb{S}^2$. (With no smoothness or rectifiability assumption)
For each $i$, let $M\_i$ denote the minimal surface (i.e. disc) bounded by $S\_i$, as provided by Douglas. Note that $M\_i$ is contained in the interior of $\mathbb{S}^2$ in $\mathbb{R}^3$.
Suppose that the intersections of the $S\_i$ follow the same pattern as the intersections of the equator of the earth with two distinct Meridian circles; that is, $S\_i \cap S\_j$ is a pair of points for every $i \neq j$, and $S\_i$ separates the two points of $S\_{i+1} \cap S\_{i+2}$ for every $i$, where addition is modulo 3.
>
> Question: Must $M\_1 \cap M\_2 \cap M\_3$ be a single point?
>
>
>
| https://mathoverflow.net/users/69681 | Is the intersection of such a triple of minimal surfaces in the 3-ball a single point? | ~~This isn't an answer, but is too long for a comment.~~
**Your question has a negative answer in general -- see below.**
One initial comment: There is no reason in general for there to be a unique Douglas-Rado disk (i.e. an area minimizer in the class of disks) so you can't really speak of "the the minimal surface (i.e. disc) bounded by $S\_i$, as provided by Douglas."
That being said, there are, in general, also many other minimal disks spanning a curve that are not Douglas-Rado solutions (i.e. are not area minimizing disks but just minimal).
If you consider solutions of the latter type, your question is definitely false.
As an example: Consider $S\_1$ to be the curve given by starting with the three parallel circles $\mathbb{S}^2\cap \{x\_3=0, \pm \epsilon\}$ (for $\epsilon>0$ small) which we denote by $C\_0$ and $C\_{\pm \epsilon}$. Form $S\_1$ by cutting out small pieces from the three circles around the intersection with the $x\_1=0$ plane and then gluing in two "necks" connecting the three circles. The one connecting $C\_\epsilon$ and $C\_0$ is on the $x\_2>0$ side and the connecting $C\_{0}$ and $C\_{-\epsilon}$ is on the $x\_2<0$ side. This gives the curve $S\_1$
By the [bridge principle](https://link.springer.com/article/10.1007/BF01192091) for stable minimal surfaces as long as the necks are small enough there is a stable minimal disk, $M\_1$, spanning $S\_1$. It looks like three stacked disks joined by small bridges.
For $S\_2$ and $S\_3$ take two small rotations around the $x\_3$-axis of the circle $\{x\_1=0\}\cap \mathbb{S}^2$. If you choose the rotations properly (in particular small enough compared to the size of the necks used in constructing $S\_1$) then the three curves satisfy your hypothesis. In this case there are unique minimal disks $M\_2$ and $M\_3$ spanning $S\_2$ and $S\_3$ which are flat. It is not hard to see that $M\_1\cap M\_2\cap M\_3$ consists of three points.
This doesn't quite mean you can't solve your problem, but it does mean you have to use the area minimizing property. Indeed, in this picture there should be at least two area-minizing disks spanning $S\_1$ and for both the triple intersection should be a disk
**Edit:**
In fact, as pointed out by @LeoMoos, this construction can be turned into a counterexample to the original question. The needed modification is as follows:
Pick $\epsilon\in (0,1)$ sufficiently large so any minimal surface spanning $C\_0\cup C\_{-\epsilon}\cup C\_\epsilon$ has three components and hence is three disks. Such an $\epsilon$ can be seen to exist by using catenoid barriers. In particular, the area minimizer in the GMT sense (more formally the $\mathbb{Z}\_2$-minimizing integral current) with this boundary is the union of the three flat disks.
It follows that in this case, if the necks are thin enough, then the Douglas-Rado solution with boudary $S\_1$ looks like the three disks joined by thin bridges. To see this just send the neck size to zero in this case $S\_1$ converges as a $\mathbb{Z}\_2$ current to the three circles which means $M\_1$ converges as a $\mathbb{Z}\_2$ current to the three disks.
The rest of the argument is the same.
| 2 | https://mathoverflow.net/users/127803 | 446290 | 179,809 |
https://mathoverflow.net/questions/446299 | 2 | Suppose $R$ is a discrete valuation ring with fraction field $K$. Let $X\subset \mathbf{P}^n\_{C\_K}$ be a closed subscheme, flat over $C\_K$, a smooth projective curve over $K$.
Let $C\_R$ be a flat regular proper model for $C$ over $R$ and take $\mathcal{X}$ the scheme-theoretic closure of $X$ in $\mathbf{P}^n\_{C\_R}$.
>
> Is $\mathcal{X}$ necessarily flat over $C\_R$? What if $C\_R = \mathbf{P}^1\_R$?
>
>
>
I'd expect "no" because the scheme-theoretic closure is flat when $C\_K$ is replaced by the open complement of a point in a regular integral scheme of dimension $1$, whereas $C\_R$ is of dimension $2$.
| https://mathoverflow.net/users/501361 | Flat scheme-theoretic closure | Explicitly, lets let $R = \mathbb{C}[x]\_{(x)}$ so $K = \mathbb{C}(x)$. You can then let $C\_K = {\bf P}^1\_K$ and $C\_R = {\bf P}^1\_R$. $C\_R$ has a chart that looks like $\mathbb{C}[x]\_(x)[y]$. This is just a localization of $\mathbb{C}[x,y]$. Blowup $(x,y)$ in the latter, and localize (equivalently, blowup $(x,y)$ in the appropriate chart of $C\_R$). You get $\mathcal{X} \to C\_R$ which is not flat.
Now, $\mathcal{X} \subseteq \mathbb{P}^1\_{C\_R}$, so base change to $C\_K$ and you get $X \subseteq \mathbb{P}^1\_{C\_K}$. It's obviously flat over $C\_K$ (in fact $X \to C\_K$ is an isomorphism). When you close it up, you just recover $\mathcal{X}$ though (and you can't recover anything else, since you have the generic point of $\mathcal{X}$ in $X$).
This has nothing to do with $\mathbb{C}$, you can do the same with any DVR.
| 4 | https://mathoverflow.net/users/3521 | 446301 | 179,811 |
https://mathoverflow.net/questions/446297 | 1 | Let $\Omega\subseteq\mathbb{R}^n$ open and convex. It is elementary that if $u\in C^2(\Omega)$ then
$$u \text{ is convex}\iff D^2u\geq0 \ \text{ in } \Omega$$
I was looking for a similar characterization for $u\in C(\Omega)$. If I remember correctly it can be formalized using distribution theory, but I was wondering if it can also be done somehow using viscosity theory. I don't know exactly how to formalize it and prove it(if it's true, of course).
Can you give me some hint or reference?
| https://mathoverflow.net/users/490711 | Viscosity characterization of convex functions | You can always consider the second (distribution) derivative of a continuous function and require that it is non-negative, which means that for all $T\in \mathbb R^n$ and all $\phi\in \mathscr D(\Omega;\mathbb R\_+)$
\begin{multline}
\sum\_{1\le j,k\le n}\langle \frac{\partial^2u}{\partial x\_j\partial x\_k}(x)T\_k T\_j, \phi(x)\rangle\_{\mathscr D'(\Omega), \mathscr D(\Omega)}
=
\sum\_{1\le j,k\le n}T\_k T\_j\langle u(x),\frac{\partial^2\phi} {\partial x\_j\partial x\_k}(x)\rangle\_{\mathscr D'(\Omega), \mathscr D(\Omega)}
\\
=\sum\_{1\le j,k\le n}T\_k T\_j\int u(x)\frac{\partial^2\phi} {\partial x\_j\partial x\_k}(x)dx\ge 0.
\end{multline}
Note also that non-negative distributions are actually Radon measures, so that, for all $T\in \mathbb R^n$,
$$
\sum\_{1\le j,k\le n}\frac{\partial^2u}{\partial x\_j\partial x\_k}(x)T\_k T\_j
\quad\text{ is a non-negative Radon measure.}
$$
Now, if $\rho$ is a non-negative compactly supported smooth function with integral 1 and $
\rho\_\epsilon(x)=\rho(x/\epsilon)\epsilon^{-n},
$
you may require that
$$
u\ast \langle (D^2\rho\_\epsilon) T\cdot T\rangle\ge 0,
$$
which is a pointwise condition when $u$ is continuous.
| 1 | https://mathoverflow.net/users/21907 | 446304 | 179,813 |
https://mathoverflow.net/questions/446307 | 10 | $$C\_{n} = \sum\_{i=1}^n (-1)^{i-1} \binom{n-i+1}{i} C\_{n-i}$$
Are there any good combinatorial proofs or algebraic proofs of this?
| https://mathoverflow.net/users/504256 | Proving an identity about Catalan numbers | $C\_n$ is the number of *Catalan sequences* $(x\_1,\ldots,x\_{2n})$ of $\pm 1$ with zero sum and non-negative prefix sums $x\_1+\ldots+x\_k$, for $k=1,\ldots,2n$. Note that any such sequence contains an index $j\in \{1,2,\ldots,n\}$ for which $x\_j=1$, $x\_{j+1}=-1$. Call $(j,j+1)$ a special pair. Then ${n-i+1\choose i}$ is the number of ways to choose $i$ special pairs (they are of course disjoint), and $C\_{n-i}$ is the number of Catalan sequences with these $i$ pairs being special. Then your identity is just inclusion-exclusion. Note that the same argument shows that $$C\_n=\sum\_i (-1)^{i-1}{n+1+t-i\choose i}C\_{n-i}$$
for every $t=0,1,\ldots,n-1$, if we consider special pairs with $j\in \{1,2,\ldots,n+t\}$.
| 18 | https://mathoverflow.net/users/4312 | 446311 | 179,815 |
https://mathoverflow.net/questions/446298 | 2 | Let $p$ be a prime and ${\mathbb F}\_p$ the finite field with $p$ elements. There is a canonical ring map ${\mathbb Z} \to {\mathbb F}\_p \cong {\mathbb Z}/ p {\mathbb Z}$. Denote the image of $n$ by $[n]\_p$.
Now consider the set of algebraic integers $\overline {\mathbb Z} \subset {\mathbb C}$, which is the set of roots of monic polynomials with integer coeffiecients. Let $\overline{\mathbb F}\_p$ be the algebraic closure of ${\mathbb F}\_p$.
**Question:** Is there a ring map $\overline{\mathbb Z} \to \overline{\mathbb F}\_p$ which extends the natural map ${\mathbb Z} \to {\mathbb F}\_p$? For example, such a map should send $a + b \sqrt{2}$ to $[a]\_p + [b]\_p \sqrt{2}$ (if this ever makes sense).
| https://mathoverflow.net/users/46433 | A ring map from algebraic integers to algebraic closure of $\mathbb F_p$ | This is basic ramification theory that you can find in any textbook on algebraic number theory; for instance [Neukirch]. As in my comments, I will show slightly more than nonemptiness of $\operatorname{Hom}(\overline{\mathbf Z},\overline{\mathbf F}\_p)$, namely determine this set via Galois theory.
Fix an algebraic closure $\overline{\mathbf F}\_p$ of $\mathbf F\_p$. For a finite Galois extension $\mathbf Q \to K$ with ring of integers $\mathcal O\_K$, consider the left action by precomposition
$$\operatorname{Gal}(K/\mathbf Q) \times \operatorname{Hom}\_{\text{Ring}}(\mathcal O\_K,\overline{\mathbf F}\_p) \to \operatorname{Hom}\_{\text{Ring}}(\mathcal O\_K,\overline{\mathbf F}\_p).$$
Now [Neukirch, Prop. I.9.1] says that the action on the set of primes $\mathfrak p \subseteq \mathcal O\_K$ above $p$ is transitive (and this set is nonempty). Given such a prime $\mathfrak p$, [Neukirch, Prop. I.9.4] says that $\operatorname{Stab}\_{\operatorname{Gal}(K/\mathbf Q)}(\mathfrak p) \to \operatorname{Gal}((\mathcal O\_K/\mathfrak p)/\mathbf F\_p)$ is surjective. But $\operatorname{Hom}(\mathcal O\_K,\overline{\mathbf F}\_p)$ is a torsor under $\operatorname{Gal}((\mathcal O\_k/\mathfrak p)/\mathbf F\_p)$ by Galois theory, so the action above is transitive. The stabiliser of $\phi \colon \mathcal O\_K \to \overline{\mathbf F}\_p$ is the inertia subgroup of $\ker \phi$, more or less by definition [Neukirch, Prop. I.9.6].
Now we just need a straightforward limit argument. Recall that $\operatorname{Gal}(\overline{\mathbf Q}/\mathbf Q)$ is the limit of $\operatorname{Gal}(K/\mathbf Q)$ for all finite Galois extensions $\mathbf Q \to K$ (with its natural profinite topology), and likewise $\operatorname{Hom}(\overline{\mathbf Z},\overline{\mathbf F}\_p)$ is the limit of $\operatorname{Gal}(\mathcal O\_K,\overline{\mathbf F}\_p)$ over all finite Galois extensions $\mathbf Q \to K$. Since a cofiltered limit of finite nonempty sets is nonempty [Tag [0A2R](https://stacks.math.columbia.edu/tag/0A2R)] (or a countable cofiltered limit of surjective maps is nonempty), we see $\operatorname{Hom}(\overline{\mathbf Z},\overline{\mathbf F}\_p) \neq \varnothing$, and a limit of torsors under finite groups is a torsor under the limit group.
---
**References.**
[Neukirch] J. Neukirch, [*Algebraic number theory*](https://doi.org/10.1007/978-3-662-03983-0). Grundlehren der Mathematischen Wissenschaften **322**. Springer, 1999. [ZBL0956.11021](https://zbmath.org/?q=an:0956.11021).
| 6 | https://mathoverflow.net/users/82179 | 446312 | 179,816 |
https://mathoverflow.net/questions/439348 | 1 | This question is a copy of one I asked in the [Math StackExchange forum](https://math.stackexchange.com/questions/4623074/what-lattices-beyond-the-laminated-lattices-particularly-in-%E2%89%A4-24d-belong-to-a) a few days ago. I don't know if it qualifies as a research-level question, but it may be something beyond most people on the Math StackExchange forum.
Back in 2016, in his proposed answer to the question asked at <https://math.stackexchange.com/questions/1790672/packing-of-n-balls/1795044#1795044> , achille hui ( <https://math.stackexchange.com/users/59379/achille-hui> ) gave the following definition of laminated lattices, which I found to be very clearly worded and useful for what I will be asking further down.
"Laminated lattices $\Lambda\_n$ can be defined/constructed recursively.
* For $n = 1$, $\Lambda\_1$ is "the" lattice of even integers.
* For $n > 1$, $\Lambda\_n$ is "a" $n$-dim lattice [which] satisfies
1. the minimal spacing among lattice points is $2$.
2. contains "a" $\Lambda\_{n-1}$ as sub-lattice.
3. subject to 1) and 2), the volume of its fundamental cell is minimal."
He goes on to give the following textualization and note, which will also seem to work for my proposed generalization further down (although the last sentence of the second paragraph would no longer be correct):
"Geometrically, one can think of $\Lambda\_n$ as stacking copies of a lower dimensional laminated lattice $\Lambda\_{n-1}$ as tightly as possible without reducing the minimum lattice spacing.
With this definition, there is no guarantee $\Lambda\_n$ is unique for a given $n$. Indeed, it isn't unique in general. However, $\Lambda\_n$ is unique for $n \le 10$ and $14 \le n \le 24$."
I am envisioning a category of lattices (perhaps call them "psuedo-laminated lattices", if that isn't already defined to mean something else) where the definition is the same as achille hui's above apart from his #3 in the recursive definition:
"3. for at least one of the $\Upsilon\_{n-1}$'s it contains as a sub-lattice, among the $n$-dim lattices which satisfy 1) and which contain *that* $\Upsilon\_{n-1}$ as a sub-lattice, the volume of its fundamental cell is minimal."
(The Greek letter Upsilon doesn't seem to be used much in mathematics, and I like the way it looks, so I'm using $\Upsilon\_n$ here to denote a "pseudo-laminated" lattice in dimension $n$. I didn't think it necessary to repost the part of achille hui's definition of laminated lattices other than his condition #3 (where the one substantive change is) just to replace the symbols used, but each $\Lambda$ there would become "a" $\Upsilon$ (or maybe "an" $\Upsilon$) in my definition.)
This recursive definition would follow what I think of as an "any possible resolution of ties beginning with the earliest one" or "share of inheritance" (with inheritance shared in the case of any tie) model. If $\Lambda\_{13}^\mathsf{mid}$, the $\Lambda\_{13}$ with the mid-ranked kissing number of the three $\Lambda\_{13}$'s, was only infinitesimally more dense than $\Lambda\_{13}^\mathsf{max}$ or $\Lambda\_{13}^\mathsf{min}$, it wouldn't matter that the lattice known in actuality as $\Lambda\_{14}$ (which contains $\Lambda\_{13}^\mathsf{max}$ and $\Lambda\_{13}^\mathsf{min}$ as sub-lattices, but apparently not $\Lambda\_{13}^\mathsf{mid}$)... it wouldn't matter that that lattice is denser than any lattice formed by layers of $\Lambda\_{13}^\mathsf{mid}$ (which would in this case be the sole $\Lambda\_{13}$) and respecting the same minimum distance of $2$ between lattice points (or alternatively the same minimal norm of $4$). The densest lattice(s) with the same minimal norm which contained $\Lambda\_{13}^\mathsf{mid}$ (which would again be the sole $\Lambda\_{13}$ in this scenario) would be the laminated lattice(s) in dimension $14$. So why shouldn't that/those lattice(s) be considered as laminated lattices in actuality, given that tie in dimension $13$ and what the result of a resolution in favor of $\Lambda\_{13}^\mathsf{mid}$ would be?
Of course, a simple answer to that question is, "That's not how laminated lattices are defined." So, I'm basically asking, "But, what if they were defined that way?" Or rather I'm proposing a slightly generalized category of lattices using that definition as one worthy of note.
For all I know, the set of $\Upsilon\_n$'s could remain strictly larger than the set of $\Lambda\_n$'s for all $n \ge 14$ ($n = 14$ is the lowest $n$ for which there are any $\Upsilon\_n$'s that are not $\Lambda\_n$'s). Or it could be for all I know that the $\Upsilon\_{14}$'s other than $\Lambda\_{14}$ are sublattices of $\Lambda\_{15}$, and that $\Lambda\_{15}$ is the densest minimal norm $4$ lattice in dimension $15$ containing any of the $\Upsilon\_{14}$'s and is thus the sole $\Upsilon\_{15}$. And thus the set of pseudo-laminated lattices becomes equivalent to the set of laminated lattices from dimensions 15 through at least $25$ (if $\Lambda\_{24}$ (the Leech lattice) is the sole $\Upsilon\_{24}$, then the sets of $\Upsilon\_{25}$'s and $\Lambda\_{25}$'s must be equivalent), although the set of pseudo-laminated lattices could become larger again in higher dimensions. Or there could be (for all I know) a "collapsing" of the number of pseudo-laminated lattices to $1$ somewhere after dimension $15$ but before dimension $25$ where the number of laminated lattices itself expands again, seemingly not to collapse back to $1$ for a long while, if at all.
So my question is basically which of the scenarios I mentioned in the last paragraph is the case, and what additional pseudo-laminated lattices there are (or are presently known) in dimensions $14$ through $24$ (and to a lesser extent in dimensions beyond $24$). I don't know if this category of lattices has been studied at all, although I doubt I'm the first person who's thought of this.
Thanks to anyone who can help in answering my question here. Feel free to also comment on the utility and noteworthiness of this proposed category as compared to the category of laminated lattices as they are actually defined.
| https://mathoverflow.net/users/155650 | What lattices beyond the laminated lattices (particularly in $\le 24D$) belong to a slightly expanded category that includes "descendants" of Λ13_mid? | I finally found the answer to this question in a 1993 paper by W. Plesken and M. Pohst, *Constructing integral lattices with prescribed minimum. II* ( <https://www.ams.org/journals/mcom/1993-60-202/S0025-5718-1993-1176715-1/S0025-5718-1993-1176715-1.pdf> ). They are the lattices in Figure 1 of that article that are to the left or right of the main path of (strongly) laminated lattices, but to the right of the dashed dividing line, in $14$ through $18$ dimensions (where that "main path" forks beginning in dimension $11$ before merging at dimension $14$ are where the classic set of (strongly) laminated lattices itself has multiple non-isometric cases in a single dimension, with the number after the decimal point ascending with descending kissing number). To be specific, the lattices on the "lambda-lattices" side of that figure marked $14.2$, $14.4$, $14.3$, $15.2$, $15.3$, $15.4$, $16.2$, $16.3$, $16.4$, $17.2$, $17.3$, $17.4$, $18.2$ and $18.3$ are what I was thinking of as the "pseudo-laminated lattices" other than the laminated ones. Jacques Martinet in Chapter 3 of *Perfect lattices in Euclidean spaces*, which I bought earlier this year in search of an answer to this question, calls this group of lattices the *weakly laminated lattices above* $\Lambda0$ (or above $\Lambda1$ if I were to start the definition as with the one I quoted in the question), with what I had learned as being the laminated lattices being the *strongly laminated lattices*, or simply the *laminated lattices* (above $\Lambda0$ or $\Lambda1$). It was on the page after this definition in Martinet's work that I was directed to the two papers by Plesken and Pohst on the subject, where I finally found what I was looking for in the second of those two papers. The Leech lattice is indeed the sole weakly laminated lattice above $\Lambda0$ or $\Lambda1$ in $24$ dimensions, but it's not until dimension $19$ that all paths merge, and not until dimension $22$ (rather than $18$) that the Kappa lattices fully merge back into that path if you define allow for all weakly laminated lattices above $K7$, $K8$ aka $K8.1$ and $K9$ aka $K9.1$ are included (you have to use points other than deep holes of prior layers, presumably shallow holes but I don't know that for a fact, three times in a row in the Kappa family to avoid it quickly merging back into the Lambda family).
| 0 | https://mathoverflow.net/users/155650 | 446318 | 179,818 |
https://mathoverflow.net/questions/446281 | 2 | I have to prove this:
Let $\alpha\in(0,1)$ and $f\in L^q(a,b)$, $1\leq q<\frac 1\alpha$, and $\mathcal{I}\_{a+}^\alpha f=0$. Then $f(x)=0$ for almost all $x\in (a,b)$.
Where $(\mathcal{I}\_{a+}^\alpha f)(x):=\frac{1}{\Gamma(\alpha)} \int\_{a}^{x} f(y)(x-y)^{\alpha-1}$.
Can Somebody help me?
Thanks
| https://mathoverflow.net/users/501039 | If the Riemann-Liouville fractional integral of $f$ is zero then $f=0$ a.e | $\newcommand{\al}{\alpha}\newcommand{\Ga}{\Gamma}$Let $g:=\Ga(\al)\mathcal{I}\_{a+}^\al f$, so that $g=0$ on $(a,b)$. Then for any $z\in(a,b)$
\begin{equation}
\begin{aligned}
0&=\int\_a^z dx\,(z-x)^{-\al}g(x) \\
&=\int\_a^z dx\,(z-x)^{-\al}\int\_a^x dy\,f(y)(x-y)^{\al-1} \\
&=\int\_a^z dy\,f(y)\int\_y^z dx\,(z-x)^{-\al}(x-y)^{\al-1} \\
&=\int\_a^z dy\,f(y)\int\_0^1 du\,(1-u)^{-\al}u^{\al-1} \quad \Big[u:=\frac{x-y}{z-y}\Big] \\
&=\int\_a^z dy\,f(y)\,\Ga(1-\al)\Ga(\al);
\end{aligned}
\end{equation}
the interchange of the order integration is possible because $f\in L^q(a,b)$ for some $q\ge1$ and hence $f\in L^1(a,b)$.
So, $\int\_a^z f=0$ for all $z\in(a,b)$ and thus $f=0$ almost everywhere (a.e.) on $(a,b)$. $\quad\Box$
---
*Details on the last sentence in the proof above:* Let $\mu$ be the measure over $(a,b)$ defined by the formula $\mu(A):=\int\_A f$ for all Lebesgue-measurable $A\subseteq(a,b)$. Then the condition that $\int\_a^z f=0$ for all $z\in(a,b)$ implies that $\mu=0$ on the algebra over $(a,b)$ generated by all intervals of the form $(a,z)$ for $z\in(a,b)$. By the uniqueness of the extension of measure, $\mu=0$ on all Borel subsets of $(a,b)$. Also, clearly $\mu=0$ on the sets of Lebesgue measure $0$. So, $\mu=0$ on all Lebesgue subsets of $(a,b)$. So, for each real $t>0$ and $A\_t:=f^{-1}((t,\infty))$ we have $0=\mu(A\_t)\ge t|A\_t|$, where $|\cdot|$ is the Lebesgue measure. So, $|A\_t|=0$ for all real $t>0$.
So, $f\le0$ a.e. on $(a,b)$. Similarly, $f\ge0$ a.e. on $(a,b)$. Thus, $f=0$ a.e. on $(a,b)$.
---
The condition $q<1/\al$ was not needed or used in the proof above.
| 2 | https://mathoverflow.net/users/36721 | 446320 | 179,819 |
https://mathoverflow.net/questions/446276 | 4 | Let $U\_1\subset \mathbb R^3$ be a simply connected bounded open set with a smooth boundary and let $U\_2$ be a neighborhood of $U\_1$. Does there exist a real-analytic diffeomorphism $\psi: U\_2 \to W\_2$ for some $W\_2\subset \mathbb R^3$ such that
$$ \psi(U\_1) \subset B \subset \psi(U\_2)$$
where $B$ is the unit ball centered at the origin?
| https://mathoverflow.net/users/50438 | Finding a real-analytic diffeomorphism | The answer is positive. First, the fact that $U\_1$ is simply connected implies that every connected component $S\_i$ ($1\le i\le n$) of its smooth boundary $\partial U\_1$ is diffeomorphic to the $2$-sphere (not completely obvious, but true). Second, after the Schönfliess theorem in dimension $3$, each of these $2$-spheres bounds a domain $D\_i$ smoothly ($C^\infty$) diffeomorphic with the compact $3$-ball. Since $U\_1$ is bounded and connected, it is easy to conclude that one of these ball domains, say $D\_1$, contains $U\_1$. So, $U\_1$ is the complement of $D\_2\cup\dots\cup D\_n$ in the interior of $D\_1$. Now, any smooth diffeomorphism $D\_1\to B$ is isotopic to the identity (by "Alexander's trick") and hence extends to a global smooth diffeomorphism $\phi:R^3\to R^3$. This provides a positive answer to your question but with instead of $\psi$ a diffeomorphism $(1-\epsilon)\phi$ which is only smooth. I guess that $U\_2$ is bounded. Hence, the Whitney approximation theorem gives on some neighborhood of $\bar U\_2$ a real analytic $\psi$ arbitrarily close to $(1-\epsilon)\phi$
in the $C^1$ topology (i.e. at every point $x\in\bar U\_2$, the value $\psi(x)$ and the three partial derivates $\partial \psi/\partial x(x)$, $\partial \psi/\partial y(x)$, $\partial \psi/\partial z(x)$ are close to those of $(1-\epsilon)\phi$.) If close enough, then $\psi$ still restricts to an embedding $U\_2\to R^3$ (as one sees using the local inversion theorem) and satisfies your demand.
| 3 | https://mathoverflow.net/users/105095 | 446322 | 179,821 |
https://mathoverflow.net/questions/446334 | 4 | If $A$ is an abelian variety over a finite field $\mathbf{F}\_q$, then $A(\mathbf{F}\_q)$ (resp. $A(\overline{\mathbf{F}}\_q)$) is a finite (resp. infinite torsion) group, but $A(\mathbf{F}\_q(t))$ is a finitely generated and possibly infinite abelian group.
This question concerns a similar situation for étale group schemes over a field, and it came up while thinking about Serre's method of specialization for Mordell-Weil groups via Hilbert irreducibility.
Suppose $G$ is a commutative *étale* group scheme over $\mathbf{F}\_q$ (I am not assuming $G$ is of finite type. Only *locally* of finite type).
>
> Is there an example of such $G$ such that $G(\mathbf{F}\_q)$ is finite, or at least torsion, and $G(\mathbf{F}\_q(t))$ is finitely generated and with a point of infinite order?
>
>
>
Suppose now that $f: G\_1\to G\_2$ is a surjective map of commutative étale group schemes over $\overline{\mathbf{F}}\_q$ (not necessarily of finite type). Assume that $G\_2(\overline{\mathbf{F}}\_q)$ is a *finitely generated free* abelian group.
>
> Is there a homomorphism of commutative étale group schemes over $\overline{\mathbf{F}}\_q$, $g : G\_2\to G\_1$, such that $f\circ g$ is the identity on $G\_2$?
>
>
>
In other words, a section to $f$ that is a homomorphism.
| https://mathoverflow.net/users/501361 | Étale group schemes and specialization | If $k \to \ell$ is any regular field extension (i.e. $k$ is algebraically closed in $\ell$), then $X(k) \to X(\ell)$ is a bijection when $X \to \operatorname{Spec} k$ is étale. Indeed, it suffices to show this when $X = \operatorname{Spec} K$ for some finite separable field extension $k \to K$, and then the statement is that the natural map $\operatorname{Hom}\_k(K,k) \to \operatorname{Hom}\_k(K,\ell)$ is a bijection. If $k \to K$ is an isomorphism, then both sides are a singleton, and otherwise both sides are empty by assumption.
Note that $k \to \ell$ is regular if and only if $\ell$ is a geometrically integral $k$-algebra [Tags [037Q](https://stacks.math.columbia.edu/tag/037Q) and [030W](https://stacks.math.columbia.edu/tag/030W)]. In particular, this holds when $\ell = k(X)$ is the function field of a geometrically integral $k$-variety $X$. Notably, $k \to k(t)$ (in this case regularity is easily checked by hand).
(A previous version of this answer contained a more geometric argument when $\ell = k(C)$ for a smooth geometrically integral curve with a rational point, but neither dimension $1$ nor the rational point is needed, and the current proof is actually easier.)
For the second question, recall that the category of finite étale $k$-schemes is equivalent to the category of finite sets with a continuous $\Gamma=\operatorname{Gal}(\bar k/k)$-action. Taking disjoint unions gives an equivalence between étale $k$-schemes and discrete sets with a continuous $\Gamma$-action. The group objects are therefore discrete groups with a continuous $\Gamma$-action, i.e. groups $A$ with a group homomorphism $\phi \colon \Gamma \to \operatorname{Aut}(A)$ such that for every $a \in A$, the stabiliser $\Gamma\_a$ is open (equivalently, closed of finite index).
This gives the required counterexample: let $U \subseteq \Gamma$ be a subgroup of index $2$ (when $k = \mathbf F\_q$ there is a unique such subgroup), and write $H$ for the quotient $\Gamma/U$, with nontrivial element $\sigma$. Now consider the augmentation $\mathbf Z[H] \twoheadrightarrow \mathbf Z$ of $H$-modules. The quotient is the constant étale scheme $\mathbf Z$ corresponding to the trivial $\Gamma$-action. But the surjection is not split as the only invariant elements in $\mathbf Z[H]$ are multiples of $\sigma + 1$, which maps to $2$ under the augmentation.
| 8 | https://mathoverflow.net/users/82179 | 446346 | 179,829 |
https://mathoverflow.net/questions/446341 | 10 | Define $P(x)$ to be positive if $P(x)>0$ for $x>0$.
I can prove that a quadratic positive polynomial is the ratio of 2 polynomials with non negative coefficients, for example $\displaystyle x^2-x+1/3=\frac{x^6+1/27}{x^4+x^3+2/3 x^2+1/3 x+1/9}$, and similarly for every $x^2-x+c$ where $c>1/4$. The full proof is not hard and involves some recursive polynomials related to Chebyshev polynomials of the second kind.
This leads me to wonder
**QUESTION**: in general is it true that $P$ positive $\Leftrightarrow$ $\exists Q,R$ with non-negative coefficients, such that $\displaystyle P=Q/R$?
| https://mathoverflow.net/users/2480 | Is every positive polynomial the ratio of 2 positive coefficient polynomials? | It is well-known. It is even known that you may take $R=(1+x)^m$ for large enough $m$. See, for example, [John Scholes's solution to Problem 11 of the 38th IMO 1997 shortlist](https://prase.cz/kalva/short/soln/sh9711.html). Note that by the real fundamental theorem of algebra, the general case reduces to the case $\deg P\leqslant 2$.
| 17 | https://mathoverflow.net/users/4312 | 446347 | 179,830 |
https://mathoverflow.net/questions/445656 | 2 | Let $(a\_k)\_{k \geq 1}$ be random variables taking values on a finite subset $B$. Assume that
$$
(1) \quad \Pr\Big (\lim\_{n\rightarrow +\infty}d(\frac{1}{n}\sum\_{k=1}^n 1\_{[a\_k = b]}, [v\_\ell(b,\theta\_0),v\_u(b,\theta\_0)])= 0\Big)=1 \quad \forall b\in B
$$
where $d$ stays for distance (i.e., absolute value difference); $v\_\ell$ and $v\_u$ are real-valued functions of $b\in B$ and $\theta \in \mathbb{R}$; $\theta\_0\in \mathbb{R}$ is a *specific* value of $\theta$.
Define
$$
\Theta\_n\equiv \Big\{\theta \in \mathbb{R}: \frac{1}{n}\sum\_{k=1}^n 1\_{[a\_k = b]}\in [v\_\ell(b,\theta),v\_u(b,\theta)] \forall b \in B\Big\}
$$
**Question**: Is it possible to find an upper for $\Pr(\theta\_0\in \Theta\_n)$?
**My attempt so far:**
We have that $E\Big(\frac{1}{n}\sum\_{k=1}^n 1\_{[a\_k = b]}\Big)\leq v\_u(b,\theta\_0)$ (see [here](https://mathoverflow.net/questions/445629/bound-the-expectation-of-an-average#445629)). By Markov inequality
$$
\Pr \Big(\frac{1}{n}\sum\_{k=1}^n 1\_{[a\_k = b]} \geq r\Big)\leq \frac{v\_u(b,\theta\_0)}{r}
$$
for each $r>0$. From here?
---
As advised in the comment below, the problem as posed only provides trivial bounds. Therefore, in what follows, I add some additional structure. In particular, I show that
$$
(2) \quad \Pr(a\_k=b | a\_{1},\dots, a\_{k-1}) \in \big[\nu\_{\ell}(b, \theta\_0), \nu\_u(b,\theta\_0)\big] \quad \text{ for all } k, b
$$
is a sufficient condition for (1), in the hope that this can suggest additional relevant details.
Proof of (1):
From (2), we have that
$$
\frac{1}{n}\sum\_{k=1}^n \Pr(a\_k=b | a\_{1},\dots, a\_{k-1})\in \big[\nu\_{\ell}(b, \theta\_0), \nu\_u(b,\theta\_0)\big] \text{ for all } n
$$
Let
$$
S\_n(b)\equiv\sum\_{k=1}^n \frac{1}{k} \left({1}\_{[a\_k=b]}- \Pr(a\_k=b| a\_1, \dots, a\_{k-1})\right).
$$
Observe that the sequence $(S\_n(b))\_{n\geq 1}$ is an $L^2$-bounded
martingale. Therefore, $S\_n(b)$ converges almost surely as $n\to +\infty$. Further, $\frac{1}{n}\sum\_{k=1}^{n-1} S\_k(b)$ converges to the same limit as $S\_n(b)$.
Let
$$
Q\_n(b)\equiv \sum\_{k=1}^n \left(1\_{[a\_k=b]}- \Pr(a\_k=b| a\_1, \dots, a\_{k-1}\right).
$$
Observe that
$$
Q\_n(b) = n S\_n(b)-\sum\_{k=1}^{n-1} S\_{k}(b).
$$
By the convergence of $S\_n(b)$ and $\frac{1}{n}\sum\_{k=1}^{n-1} S\_k(b)$ to the same limit, $\frac{1}{n}{Q\_n(b)}$ converges almost surely to 0 as $n\to +\infty$.
Hence,
$$
\frac{1}{n}\sum\_{k=1}^n 1\_{[a\_k=b]}-\frac{1}{n}\sum\_{k=1}^n \Pr(a\_k=b| a\_1, \dots, a\_{k-1})\overset{a.s.}{\to} 0 \text{ as $n\to +\infty$.}
$$
Therefore, $\frac{1}{n}\sum\_{k=1}^n 1\_{[a\_k=b]}$ and $\frac{1}{n}\sum\_{k=1}^n \Pr(a\_k=b| a\_1, \dots, a\_{k-1})$ have the same limit points which belong to $ \big[\nu\_{\ell}(b, \theta\_0), \nu\_u(b,\theta\_0)\big] $ and (1) holds.
| https://mathoverflow.net/users/42412 | Bound the probability that a point belongs to a set | As I wrote before in a comment, the only upper bound on $\Pr(\theta\_0\in\Theta\_n)$ under these very general conditions is the trivial bound $1$.
Indeed, suppose that for some $b\_\*\in B$
\begin{equation}
\Pr(a\_1=a\_2=\cdots=b\_\*)=1
\end{equation}
and for all $b\in B$
\begin{equation}
v\_\ell(b,\theta\_0)=1(b=b\_\*)=v\_u(b,\theta\_0).
\end{equation}
Then $\Pr(a\_k=b)=v\_\ell(b,\theta\_0)=v\_u(b,\theta\_0)$ for all natural $k$ and all $b\in B$. So, condition (1) holds and $\Pr(\theta\_0\in\Theta\_n)=1$ for all natural $n$.
| 2 | https://mathoverflow.net/users/36721 | 446349 | 179,831 |
https://mathoverflow.net/questions/446324 | 1 | Let $I(Y;X)$ denote the mutual information between $Y$ and $X$. If we have $I(Y;X\_{i}) < B$ for all $i \quad (1 \leq i \leq N)$, could we also get the upper-bound of $I(Y; X\_{1}, X\_{2}, ..., X\_{N})$?
| https://mathoverflow.net/users/478341 | Upper bound of $I(Y; X_{1}, ..., X_{N})$ when we have $I(Y;X_{i}) < B$ for all $i$ $(1 \leq i \leq N)$ | Fix any integer $N\ge2$. Let $X:=(X\_1,\dots,X\_N)$. There is no upper bound $U\_N(B)$ on $I(Y;X)$ such that $U\_N(B)\to0$ whenever $\max\_{i=1}^N I(Y;X\_i)< B\downarrow0$. That is, each of the $I(Y;X\_i)$'s may be arbitrarily small while $I(Y;X)$ is not small. Informally, each $X\_i$ may contain little or no information about the distribution of $Y$, but the information about the distribution of $Y$ contained in all the $X\_i$'s together may be essential, even for $N$ as small as $2$.
Indeed, suppose that $X\_1,\dots,X\_N$ are independent Rademacher random variables (r.v.'s), so that $P(X\_i=1)=P(X\_i=-1)=1/2$ for each $i$. Then $Y:=X\_1\cdots X\_N$ is a Rademacher r.v. independent of each of the $X\_i$'s. So, $\max\_{i=1}^N I(Y;X\_i)=0<B$ for any real $B>0$. However (assuming the natural logarithm in the definition of the mutual information), $I(Y;X)=\ln2\not\to0$ as $B\downarrow0$.
**Details on the equality $I(Y;X)=\ln2$:** Assuming the usual convention $0\ln0:=0$, we have
\begin{equation}
\begin{aligned}
&I(Y;X) \\
&=\sum\_{(y,x\_1,\dots,x\_N)\in\{-1,1\}^{N+1}}
P(Y=y,X\_1=x\_1,\dots,X\_N=x\_N) \\
&\qquad\qquad\qquad\times\ln\frac{P(Y=y,X\_1=x\_1,\dots,X\_N=x\_N)}{P(Y=y)P(X\_1=x\_1,\dots,X\_N=x\_N)} \\
&=\sum\_{(y,x\_1,\dots,x\_N)\in\{-1,1\}^{N+1}}
P(X\_1=x\_1,\dots,X\_N=x\_N) \\
&\qquad\qquad\qquad\times1(y=x\_1\cdots x\_N)\,
\ln\frac{P(X\_1=x\_1,\dots,X\_N=x\_N)}{\tfrac12\,P(X\_1=x\_1,\dots,X\_N=x\_N)} \\
&=\sum\_{(x\_1,\dots,x\_N)\in\{-1,1\}^N}
P(X\_1=x\_1,\dots,X\_N=x\_N) \\
&\qquad\qquad\qquad\times
\ln\frac{P(X\_1=x\_1,\dots,X\_N=x\_N)}{\tfrac12\,P(X\_1=x\_1,\dots,X\_N=x\_N)}=\ln2.
\end{aligned}
\end{equation}
| 3 | https://mathoverflow.net/users/36721 | 446352 | 179,832 |
https://mathoverflow.net/questions/446344 | 4 | Recently I learned from the Stacks project that for every abelian category ${\mathcal A}$, there is a natural isomorphism $K\_0({\mathcal A})\cong K\_0(D^{b}(\mathcal A))$.
When we set $\mathcal A$ to be the abelian category $M(X)$ of coherent sheaves on a Noetherian scheme $X$,we get that $G\_0(X)\cong K\_0(D^{b}(M(X)))$.
Is it true that $K\_i({\mathcal A})\cong K\_i(D^{b}({\mathcal A}))$ for every abelian category ${\mathcal A}$ and $i\geq 0$?
I saw in Amnon Neeman’s 1997 paper K-theory for Triangulated Categories I(A):Homological Functors that my question was answered in the affirmative on page 341,as Theorem 7.1.
But I am new to the K-theory of triangulated categories.
So I would like to confirm that this is true.
Another question is, assuming $K\_i({\mathcal A})\cong K\_i(D^{b}({\mathcal A}))$ for every abelian category ${\mathcal A}$ and $i\geq 0$, does this isomorphism provide a practical method of computing the higher $G$-theory of a Noetherian scheme?
| https://mathoverflow.net/users/477848 | Can higher G-theory of Noetherian schemes be computed by derived categories? | Yeah this is true, as Neeman explains this follows from his more general theorem of the heart for the bounded derived category with the standard t-structure.
You can see a modern treatment ( with a shorter proof in the language of $\infty$-categories ) in Barwick's paper
*Barwick, Clark*, [**On exact (\infty)-categories and the theorem of the heart**](https://doi.org/10.1112/S0010437X15007447), Compos. Math. 151, No. 11, 2160-2186 (2015). [ZBL1333.19003](https://zbmath.org/?q=an:1333.19003).
There, in the first appendix, Barwick discusses applications of this theorem to calculating the G-theory of noetherian schemes, namely one of the claims is that the G-theory of such scheme coincides with the K-theory of an abelian category of Cohen–Macaulay complexes.
| 7 | https://mathoverflow.net/users/44499 | 446353 | 179,833 |
https://mathoverflow.net/questions/446338 | 2 | Given a surface $X:f(x,y,z)=0\subset \mathbb{A}^{3}\_{\mathbb{C}}$ with only ADE singularities, how does one determine the correct singularity type of $X$ by computing the normal forms?
Does a similar procedure also exist in positive characteristic?
| https://mathoverflow.net/users/211978 | Normal forms of ADE singularities | This may not be exactly what you are asking about, but what follows has some references. I'm guessing you are already aware of:
* V. I. Arnolʹd, *Critical points of smooth functions, and their normal forms*. Uspehi Mat. Nauk 30 (1975), no. 5(185), 3–65.
There are a couple different ways. I think you are indicating you can just do changes of coordinates to get to the form you want. That works, and it's not too bad. However, I think for many cases, doing resolutions of singularities and computing the dual graph is frequently faster.
See for instance:
* M. Artin, *On isolated rational singularities of surfaces*, American Journal of Mathematics, Vol. 88, No. 1 (Jan., 1966), pp. 129-136.
Indeed, this procedure works even in positive characteristic. However, in characteristics 2,3,5 there are different types of the "same" ADE singularity (ie, ADE singularities with the same looking resolutions but different analytic isomorphism types).
* Artin, M. *Coverings of the rational double points in characteristic p*. Complex analysis and algebraic geometry, pp. 11–22. Iwanami Shoten, Tokyo, 1977.
* Greuel, G.-M.; Kröning, H. *Simple singularities in positive characteristic.* Math. Z. 203 (1990), no. 2, 339–354.
Such characterizations are also typically possible in mixed characteristic (ie, if you have a 2-dimensional Gorenstein local ring of multiplicity 2 whose residue field is char. $p > 0$ but whose fraction field is char. $0$). As before, in many cases you can just do change of coordinates to figure out the form, and indeed that's what Artin was doing above to write down the forms. Some other references which do some of these explicit changes in coordinates in mixed characteristic include the following (Lipman's paper predates Artin's characteristic $p > 0$ paper, and he did the E8 case quite explicitly if I recall correctly).:
* J. Lipman, *Rational singularities*, Publ. Math. IHES 36 (1969), 195-280.
* Carvajal-Rojas, L. Ma, T. Polstra, -, K. Tucker. *Covers of rational double points in mixed characteristic*. J. Singul. 23 (2021), 127–150.
Note, in mixed characteristic, there's no way $\mathbb{Z}\_p[x,y]/(p^2 + x^2 + y^3)$ is isomorphic to $\mathbb{Z}\_p[x,y]/(x^2 + y^2 + p^3)$. The ring element $p$ is special.
| 5 | https://mathoverflow.net/users/3521 | 446354 | 179,834 |
https://mathoverflow.net/questions/445853 | 3 | **One sided Shift**
Let be $M$ separable metric space. Consider $X=M^{\mathbb{N}}$ the sequence space equipped with the product metric $d(x,y)=\sum\_{i=1}^\infty |x\_i-y\_i|/2^i$ . Define the shift map $\sigma:X\to X$ by putting in each sequence $x=(x\_i)$, $\sigma(x)\_i=x\_{i+1}$. Then we have that the shift has the following propery: given $x\in X$ and $\epsilon>0$ there exists a uniform nonnegative integer $K$ such that for each $z\in X$ there exist $y\in \sigma^{-K}(z)\cap B(x,\epsilon)$.
**Weighted Shift**
Now $X=\ell^p({\mathbb{N}})$ with $1 \leq p < \infty$
endowed with it standard norm.
We fix values $0 < c < c'$ and a sequence $(\alpha\_n)\_{n \geqslant 1}$ satisfying $\alpha\_n \in (c, c')$ for each $n \in \mathbb{N}$. The weighted shift associated to the sequence $(\alpha\_n)\_{n \geqslant 1}$
is defined as the linear map $L : X \to X$ given by
$$
L((x\_n)\_{n \geqslant 1}) = (\alpha\_n x\_{n+1})\_{n \geqslant 1} \,.
$$
For each $k,n\in\mathbb{N}$ we define $\beta\_k^n \equiv \alpha\_k \ldots \alpha\_{k+n-1}$, if
\begin{eqnarray}
\label{dnn}
\sum\_n (\beta\_1^n)^{-p}<\infty
\end{eqnarray}
then is a known result that $L$ is **frequently hypercyclic** in $\ell^p(\mathbb{N})$, which means that,there exists $x$ such that for any open set $V$ the set $N(x, V)=\{n: T^nx\in V\}$ has positive lower density, i.e
$$
\liminf\_{n\to \infty} \dfrac{1}{n}\# N(x,V)\cap\{1,\ldots,n \}>0.
$$
**My Question:**
I'am looking for to know if there exist any condition over the weight sequence $(\alpha\_n)$ in order to have a property a a bit more stronger that frequently hypercyclicity for the the weighted shift.
More precisely I would like to know any condition over the weight sequence $(\alpha\_n)$ in order to have the following property(silimar to the ordinary one-sided shift): given $x\in X$ and $\epsilon>0$ a uniform nonnegative integer $K$ such that for each $z\in X$ there exists some $y\in L^{-K}(z)\cap B(x,\epsilon)$.
| https://mathoverflow.net/users/98969 | Is the weighted shift strong frequently hypercyclic? | The answer to the question is no. If the sequence $\alpha = (\alpha\_{n})\_{\geqslant 1}$ satisfies the condition $c<\alpha\_{n}<c^\prime$ there is no other condition on a sequence $\alpha$ for the answer to be affirmative. Let's assume that the answer to the question is affirmative. That is, there exists $\alpha = (\alpha\_{n})\_{n\geqslant 1}\in [c,c^{\prime}]^{\mathbb{N}-{0}}$ such that for every open ball $B(x,\epsilon)\subset \ell^p({\mathbb{N}})$ there exists $K\in \mathbb{N}$ such that for every $z\in \ell^p({\mathbb{N}})$ there exists $y=(y\_{n})\_{n\in \mathbb{N}} = (y\_0,y\_1,y\_3,\ldots,y\_n,\ldots)$ satisfying the condition
$$
{\rm (i)}\; y\in\ell^{p}(\mathbb{N}),
\qquad
{\rm (ii)}\;y\in L^{-K}\_{\alpha}(z)
\qquad \mbox{ and } \qquad
{\rm (iii)}\; y\in B(x,\epsilon)
$$
We observe that for every integer $K\geqslant 1$ and every $({\bf u}\_{n})\_{n\in\mathbb{N}}\in \ell^p(\mathbb{N})$ we have, in a susceptible way,
$$
L^{K}\_{\alpha}\Big( ({\bf u}\_{n})\_{n\in\mathbb{N}} \Big) = \left( \Big( \prod\_{\ell=1}^{K}\alpha\_{\ell+n}\Big)\cdot {\bf u}\_{K+n} \right)\_{n\in\mathbb{N}}.
$$
From the observation we made above about the $L\_{\alpha}$ operator the first condition below is an immediate consequence of the assumption we made. Furthermore, the second condition is a consequence of the first condition and the third condition is a consequence of the second condition.
**Condition 1.** there exists $\alpha = (\alpha\_{n})\_{n\geqslant 1}\in [c,c^{\prime}]^{\mathbb{N}-{0}}$ such that for any open ball $B(x, \epsilon)\subset \ell^p({ \mathbb{N}})$ there exists $K\in \mathbb{N}$ such that for every $z=(z\_{n})\_{n\in \mathbb{N}}$ there exists $ y=(y\_{n})\_{n\in \mathbb{N}}$ satisfying
$$
{\rm (i)}\; \|y\|\_{p}<\infty,
\qquad
{\rm (ii)}\;\Big( \prod\_{\ell=1}^{K}\alpha\_{\ell+n}\Big)\cdot y\_{K+n}=z\_{n} \quad n\in\mathbb{N}
\quad \mbox { and } \quad
{\rm (iii)}\;\| y-x\|\_p<\epsilon
$$
**Condition 2.** there exists $\alpha = (\alpha\_{n})\_{n\geqslant 1}\in [c,c^{\prime}]^{\mathbb{N}-{0}}$ such that for any open ball $B(x, \epsilon)\subset \ell^p( {\mathbb{N}})$ there exists $K\in \mathbb{N}$ such that for every $z=(z\_{n})\_{n\in \mathbb{N}}$ there exists $ y=(y\_{n})\_{n\in \mathbb{N}}$ satisfying
$$
{\rm (i)}\; \sum\_{n=0}^{\infty}|z\_n|^{p}<\infty,
\qquad
{\rm (ii)}\; y\_{K+n}= \Big( \prod\_{\ell=1}^{K}\alpha\_{\ell+n}^{-1}\Big)\cdot z\_{n}\quad n\in\mathbb{N}
\quad \mbox { and } \quad
{\rm (iii)}\;\sum\_{n=0}^{K-1}\left|y\_{n} - x\_n\right|^{p}
+
\sum\_{n=K}^{\infty}\left|\Big( \prod\_{\ell=1}^{K}\alpha\_{\ell+n}^{-1}\Big)\cdot z\_{n} - x\_n\right|^{p} <\epsilon
$$
**Condition 3.** there exists $\alpha = (\alpha\_{n})\_{n\geqslant 1}\in [c,c^{\prime}]^{\mathbb{N}-\{0\}}$
such that for any open ball $B(x, \epsilon)\subset \ell^p( \mathbb{N})$
there exists a $K\in \mathbb{N}$ such that for every pair $z=(z\_{n})\_{n\in \mathbb{N}}$
and $\tilde{z}=(\tilde{z}\_{n})\_{n\in \mathbb{N}}$ in $\ell^p(\mathbb{N})$
there exists $ y=(y\_{n})\_{n\in \mathbb{N}}$ which fits both $z$ and $\tilde{z}$ satisfying
$$
{\rm (i)}\; \sum\_{n=0}^{\infty}|z\_n|^{p}<\infty,
\qquad
{\rm (ii)}\; y\_{K+n}= \Big( \prod\_{\ell=1}^{K}\alpha\_{\ell+n}^{-1}\Big)\cdot z\_{n}\quad n\in\mathbb{N}
\quad \mbox { and } \quad
{\rm (iii)}\sum\_{n=0}^{K-1}\left|y\_{n} - x\_n\right|^{p}
+
\sum\_{n=K}^{\infty}\left|\Big( \prod\_{\ell=1}^{K}\alpha\_{\ell+n}^{-1}\Big)\cdot z\_{n} - x\_n\right|^{p} <\epsilon
$$
$$
{\rm (iv)}\; \sum\_{n=0}^{\infty}|\tilde{z}\_n|^{p}<\infty,
\qquad
{\rm (v)}\; y\_{K+n}= \Big( \prod\_{\ell=1}^{K}\alpha\_{\ell+n}^{-1}\Big)\cdot \tilde{z}\_{n}\quad n\in\mathbb{N}
\quad \mbox { and } \quad
{\rm (vi)}\; \sum\_{n=0}^{K-1}\left|y\_{n} - x\_n\right|^{p}
+
\sum\_{n=K}^{\infty}\left|\Big( \prod\_{\ell=1}^{K}\alpha\_{\ell+n}^{-1}\Big)\cdot \tilde{z}\_{n} - x\_n\right|^{p} <\epsilon
$$
Since $z=(z\_{n})\_{n\in \mathbb{N}}$ and $\tilde{z}=(\tilde{z}\_{n})\_{n\in \mathbb{N}}$ are arbitrary we can choose their $K$ first coordinates equal and the others different such that $\|\tilde{z}-z \|\_p\geqslant 3\epsilon $. Adding member by member the equalities {\rm (ii)} and {\rm (v)} will result in $z=(z\_{n})\_{n\in \mathbb{N}}=\tilde{z}=(\tilde{z}\_{n})\_{n\in \mathbb{N}}$ since $0<c<\alpha\_{n}$. However, this is a contradiction.
| 2 | https://mathoverflow.net/users/36917 | 446356 | 179,835 |
https://mathoverflow.net/questions/446357 | 3 | When I was reading an article by CHUN-GANG JI (A SIMPLE PROOF OF A CURIOUS CONGRUENCE BY ZHAO), he mentioned in the acknowledgement
the following identity
$$\sum\_{i+j+k=p,\text{ } i,j,k\gt 0}{p\choose i}{p\choose j}{p\choose k}={3p\choose p}-3{2p\choose p}+3$$
Can anybody give me a lead to a combinatorial or algebraic proof of this identity ?
| https://mathoverflow.net/users/481754 | A combinatorial identity involving binomial coefficients | Without the inequality on the $i, j, k$, the sum on the left would be $\binom{3p}{p}$. This is an immediate consequence of the Chu-Vandermonde convolution identity
$$\sum\_{i+j = k} \binom{x}{i}\binom{y}{j} = \binom{x+y}{k}$$
which is treated in many places, for example in Concrete Mathematics by Graham, Knuth, and Patashnik.
By an inclusion-exclusion argument, the adjustment one must make to account for the inequality in the sum is to subtract the sum where at least one of the $i, j, k$ is $0$, then add back in the sum where at least two of them are $0$. For example, the sum over all cases where $k = 0$ is
$$\sum\_{i+j=p} \binom{p}{i}\binom{p}{j} = \binom{2p}{p}$$
where again we use Chu-Vandermonde. This leads to the answer you want.
| 6 | https://mathoverflow.net/users/2926 | 446359 | 179,836 |
https://mathoverflow.net/questions/446215 | 7 | The question, which came up in a conversation with my advisor Ola Kwiatkowska, is pretty much in the title:
Let $Z\subseteq[0,1]^3$ be zero-dimensional. Is it possible for $[0,1]^3\setminus Z$ to be embeddable in $\Bbb R^2$?
| https://mathoverflow.net/users/49381 | Can you remove a zero dimensional subspace from a cube and obtain a planar space? | I think the answer is negative in any dimension $n\geq 2$.
**Theorem:**
$\mathbb R^n$ can not be covered by a zero-dimensional set and a set homeomorphic to a subspace of $\mathbb R^{n-1}$.
**Proof:**
Suppose for contradiction that $\mathbb R^n=A\cup B$ where A is zero-dimensional and $i:B\to\mathbb R^{n-1}$ is an embedding. Let $I$ be the interior of $i(B)$ in $\mathbb R^{n-1}$ and let $R=i(B)\setminus I$.
Then $R$ has empty interior in $\mathbb R^{n-1}$ and thus it is known to be of dimension at most $n-2$. Also $\mathbb R^n=A\cup i^{-1}(I)\cup i^{-1}(R)$.
However, $I$ is an open subset of $\mathbb R^{n-1}$, hence its homeomorphic image $i^{-1}(I)$ is not dense in $\mathbb R^n$.
Consequently $A\cup i^{-1}(R)$ is the union of a zero-dimensional space and at most $(n-2)$-dimensional space and covers a non-empty open subset of $\mathbb R^n$, which is a contradiction.
| 6 | https://mathoverflow.net/users/128723 | 446364 | 179,839 |
https://mathoverflow.net/questions/443751 | 5 | Let $V$ be a Banach space and $B(V)$ be the bounded operators on $V$. Now, the space $B(V)$ has a norm, the strong topology (which is the topology of pointwise convergence) and it has a stronger topology called ultrastrong or $\sigma$-strong topology. It is known that these two topologies coincide on norm-bounded sets. I want to know if the $\sigma$-strong topology is the strongest topology which coincides with the strong topology (respectively itself) on bounded sets.
Edit:
The $\sigma$-strong topology is a topology generated by certain seminorms. Namely for all sequences $b\_n$ with $\sum\_n \|b\_n\|^2<\infty$, the seminorm
$B(V)\times B(V) \to \mathbf{R}$
given by the square root of $(f,g)\mapsto \sum\_n \|f(b\_n)-g(b\_n)\|^2$ should be continuous and these seminorms generate the topology.
Edit:
The answer seems to be "no" for this general case. I have no example though.
To simplify this question, I removed a second part. Thank you for the helpful comments on both parts.
| https://mathoverflow.net/users/501466 | Is the sigma-strong topology generated by bounded sets? | The finest topology that coincides with $\tau\*$ ($\sigma$-strong in this case) on $\tau$-bounded (norm-bounded in this case) subsets is the mixed topology $\gamma(\tau,\tau^\*)$, introduced by [A. Wiweger, Linear spaces with mixed topology. Studia Mathematica 20 (1961), 47--68](https://eudml.org/doc/217004); see 2.2.2. For the Hilbert space $\ell\_2$ (or perhaps any separable infinite dimensional Banach space?), the inequalities
$$\mbox{$\sigma$-strong} \le \mbox{uniform convergence on compact subsets} \le \gamma(\mbox{norm},\mbox{$\sigma$-strong})$$
are strict (Addendum: I'm no longer so sure if the second inequality is strict).
---
A note for posterity: I was noticed that there is some confusion in the literature as to the difference of the above topologies. E.g., Proposition I.8.6.3 in [Blackadar's book](https://mathscinet.ams.org/mathscinet-getitem?mr=2188261) erroneously claims that the $\sigma$-strong topology on $B(H)$ coincides with the strict topology arising as the multiplier of $K(H)$. The essentially same claim is seen in Chapter 8 in [Lance's book](https://mathscinet.ams.org/mathscinet-getitem?mr=1325694). In fact, these topologies are different (although they coincide on norm bounded subsets, and that's what's needed in daily life), as the following example shows.
Take a sequence of mutually orthogonal rank one orthogonal projections $p\_n$ on a separable Hilbert space $H$.
Then, the $\sigma$-strong closure of $\{ \sqrt{n} p\_n \}$ contains $0$. Indeed, for any positive linear functional $f$ on $B(H)$, one has $\liminf\_n f(n p\_n)=0$, for otherwise $f(p\_n)>\frac{\epsilon}{n}$ for all $n$ and $\lim\_m f(\sum\_{n=1}^m p\_n)=\infty$, a contradiction. However, $T:=\sum\_{n=1}^\infty \frac{1}{\sqrt{n}}p\_n \in K(H)$ (or the compact subset $\{ Tv : \|v\|\le1\})$ separates $0$ from $\{ \sqrt{n} p\_n \}$ in the strict topology (or the compact open topology).
| 6 | https://mathoverflow.net/users/7591 | 446375 | 179,843 |
https://mathoverflow.net/questions/446374 | 8 | The broad theme that underlies this question is: to what extent can the study of finite groups be reduced to the study of $p$-groups?
I imagine that it is possible for a pair of nonisomorphic finite groups $G$ and $H$ to have isomorphic Sylow subgroups. That is to say, for every prime $p$, $G$ and $H$ have isomorphic Sylow $p$-subgroups. In particular, $G$ and $H$ must have the same number of elements.
Let us call $G$ and $H$ **Sylow-isomorphic** if they have isomorphic Sylow subgroups.
1. For a given positive integer $n$ (that we may obviously assume *isn't* a prime power), how many Sylow-isomorphic finite groups of order $n$ can there be?
2. Can a pair of non-isomorphic finite simple groups be Sylow-isomorphic?
**Some results:**
1. Every finite nilpotent group is isomorphic to the external direct product of its Sylow subgroups, so a pair of finite nilpotent groups are Sylow-isomorphic if and only if they are actually isomorphic.
2. Let $C\_n$ denote the cyclic group of order $n$. If $p>2$ is an **odd** prime, $\text{Aut}(C\_p)\cong (\mathbb{Z}/p\mathbb{Z})^\times\cong C\_{p-1}$. We know that $C\_p\rtimes C\_{p-1}$ and $C\_p\times C\_{p-1}$ are non-isomorphic (the former is non-abelian while the latter is abelian), but clearly they are Sylow isomorphic. This shows that it is possible for a pair of non-isomorphic solvable groups to be Sylow-isomorphic.
**Additional question:**
I've heard that it is often useful to study the **normalizers** of $p$-groups in finite groups (this method is sometimes called **local analysis**, and normalizers of $p$-groups are called $p$-local subgroups).
Since the Sylow $p$-subgroups of a finite group $G$ are conjugate, their normalizers must also be conjugate, so we can unambiguously talk about the isomorphism type of a Sylow $p$-normalizer of a finite group. Suppose $G$ and $H$ have isomorphic Sylow $p$-normalizers for every prime $p$. This also means, by the way, that $G$ and $H$ must be Sylow-isomorphic. In this situation, let's call $G$ and $H$ **locally isomorphic**. Then what can we say about how $G$ and $H$ are related?
**Edit:** on finding a pair of non-isomorphic, but locally isomorphic groups
If $G$ has a normal Sylow $p$-subgroup for any prime $p$, then any group that is locally isomorphic to $G$ must be isomorphic to $G$, so we must restrict our attention to groups without any normal Sylow subgroups. This means we must ignore supersolvable groups altogether, since if $G$ is supersolvable and $p$ is the largest prime divisor of $|G|$, then a Sylow $p$-subgroup of $G$ is normal.
Also, one might consider a group in which the notion of "locally-isomorphic" coincides with "Sylow-isomorphic": that is, groups in which every Sylow subgroup is self-normalizing. Such a group (assuming it isn't of prime power order) cannot be solvable, because of the following theorem:
**Theorem:** (R. Carter) Let $G$ be a finite solvable group. Then $G$ has a self-normalizing nilpotent subgroup, and all such subgroups are conjugate (and thus isomorphic)
So if $G$ is solvable, then there is at most **one** prime $p$ such that a Sylow $p$-subgroup of $G$ is self-normalizing.
| https://mathoverflow.net/users/502468 | Nonisomorphic finite groups with isomorphic Sylow subgroups | To answer your question 2, there are very few pairs of finite simple groups with the same order. There's $PSL\_3(4)$ and $PSL(4,2)$, and there's $P\Omega\_{2n+1}(q)$ and $PSp\_{2n}(q)$ for $q$ odd. None of these are Sylow isomorphic in your sense.
As a contribution to your question 1, finite groups can be not only Sylow isomorphic but they can even be indistinguishable from the point of view of fusion at any prime. This is equivalent to the $\mathbb{Z}$-completion of the classifying spaces being homotopy equivalent. Probably the smallest example is $(\mathbb{Z}/3\rtimes\mathbb{Z}/4) \times (\mathbb{Z}/5\rtimes\mathbb{Z}/2)$ and $(\mathbb{Z}/3\rtimes\mathbb{Z}/2) \times (\mathbb{Z}/5\rtimes\mathbb{Z}/4)$, with the action of $\mathbb{Z}/4$ having the subgroup of index two in its kernel in both cases.
| 11 | https://mathoverflow.net/users/460592 | 446381 | 179,846 |
https://mathoverflow.net/questions/446116 | 8 | $\DeclareMathOperator\SO{SO}$I asked this initially in math stack exchange, but thought to ask it here since it is more advanced and related to my research topic. I study optimization on Lie groups from a Riemannian geometry viewpoint. Naturally, one of the Lie groups I use the most is $\SO(3) \subset \mathbb{R}^{3 \times 3}$ equipped with the bi-invariant metric $\langle \xi, \eta\rangle = \frac{1}{2} \operatorname{tr}(\xi^T\eta )$, where $\xi,\eta \in \mathfrak{so}(3)$ are $3 \times 3$ skew-symmetric matrices. The way I parameterize $\SO(3)$ and $\mathfrak{so}(3)$ is not important ofc. I could just as easily use unit quaternions.
Lately in my studies, I've been interested in *commutation error*, and ways to bound it.
I was especially curious if there existed $K > 0$ such that $$\left\|\log(R^{-1}S)\right\| \leq K \left\|\log(S) - \log(R)\right\|$$ for all $R,S \in \SO(3)$ such that $\log(S), \log(R), \log(R^{-1}S)$ exists. Here, $\log(\cdot)$ is the Lie logarithm. Under my parameterization, it coincides with the matrix logarithm.
After generating random rotation matrices $R,S$ in a script, to my surprise I found $K=1$ *every single time*. I have yet to randomly generate or analytically derive a counterexample. Once again, I am only focusing on pairs $(R,S)$ such that each term above exists.
If true, this is a very powerful inequality! I'm surprised to see very little discussion about online. Surely, someone must've noticed this. Any idea how I could prove this?
| https://mathoverflow.net/users/141449 | Is it true $\left\|\log(RS)\right\|≤\left\|\log(R)+\log(S)\right\|$ for all $R,S \in \mathrm{SO}(3)$, where $\|\cdot\|$ is the Frobenius norm? | Now that I have written out the completely elementary proof above for the quaternions and for $\mathrm{SO}(3)$, I feel that I should point out that the statement $|\log(ab)|\le |\log(a) + \log(b)|$, suitably interpreted, is true for any compact connected Lie group $G$ endowed with a biïnvariant Riemannian metric. The proof is not completely elementary; but it only relies on facts that are covered in any first course on Riemannian geometry.
Let $G$ be a compact, connected Lie group with Lie algebra ${\frak{g}}=T\_eG$, and let $<,>:\frak{g}\times\frak{g}\to\mathbb{R}$ be an $\mathrm{Ad}(G)$-invariant positive definite inner product on $\frak{g}$. Let $|v| = <v,v>^{1/2}$ for $v\in\frak{g}$, as usual. There is a unique biïinvariant Riemannian metric $g$ on $G$ that equals $g\_0=<,>$ on ${\frak{g}} = T\_eG$. Let $d:G\times G\to \mathbb{R}$ be the associated distance function of $g$. It satisfies $d(ac,bc)=d(ca,cb)=d(a,b)$ for all $a,b,c\in G$.
Because the sectional curvature of $g$ is non-negative, the Lie group exponential map (which is equal to the exponential map of $g$ at $e\in G)$, i.e., $\exp:{\frak{g}}\to G$, satisfies $\exp^\*(g)\le g\_0$ (as smooth quadratic forms on $\frak{g}$). (For a proof, see, for example, Helgason's *Differential Geometry, Lie Groups, and Symmetric Spaces*.)
Now, Let $U\subset G$ be the connected, dense open set of points $a\in G$ for which there is a unique $g$-length-minimizing geodesic from $e$ to $a$. (The complement of $U$ is a closed algebraic set in $G$ called the *cut locus* of $G$.) Then for each $a$ in $U$, there is a unique element $\log a\in\frak{g}$, such that the curve $\gamma(t) = exp\bigl(t\log(a)\bigr)$ for $0\le t\le 1$ is that unique $g$-length-minimizing geodesic. Then $\log:U\to\frak{g}$ is a smooth inverse to $\exp$ in the sense that $\exp(\log a) = a$.
By construction, $|\log(a)| = d(a,e)$, so, in particular, the function $a\mapsto|\log(a)|$ extends continuously to all of $G$, even though $\log$ does not.
Now, for $a,b\in U$, we have the following estimate:
$$
|\log(ab^{-1})| = d(e,ab^{-1}) = d(b,a) = d(a,b)
$$
and the righthand number is less than or equal to the length of the curve
$\gamma(t) = \exp\bigl((1{-}t)\log(a)+t\log(b)\bigr)$ for $0\le t\le 1$. However, because $\exp^\*(g)\le g\_0$, the $g$-length of $\gamma$ is less than or equal to the $g\_0$-length of the line segment $\alpha(t) = (1{-}t)\log(a)+t\log(b)$ in $\frak{g}$, which is $|\log(a)-\log(b)|$. (Note that $\alpha(t)$ does not have to lie in the image of $\log$ for all $0<t<1$ in order for this statement to hold.)
Thus, $|\log(ab^{-1})| \le |\log(a)-\log(b)|$ for all $a,b\in U$, i.e., for all $a$ and $b$ for which $\log$ is defined. Replacing $b$ by $b^{-1}$ and using the fact that $\log(b^{-1}) = -\log b$ yields
$$
|\log(ab)| \le |\log(a)+\log(b)|
$$
for all $a,b$ in the domain of $\log:U\to \frak{g}$. (It doesn't matter whether $ab$ is in the domain of $\log$.)
| 7 | https://mathoverflow.net/users/13972 | 446385 | 179,848 |
https://mathoverflow.net/questions/446384 | 6 | This might seems like a bit of philosophical question and so maybe if I keep reading a bit more, I might get my answer. But, I ask nonetheless.
In my attempt to study Shimura varieties, I came across the definition of a locally symmetric space associated to connected reductive group $G/\mathbb{Q}$ as follows:
>
> Let $A\_G\subset G$ denote the maximal $\mathbb{Q}$-split torus in the center of $G$ and $A\_\infty=A\_G(\mathbb{R})$. Let $K\_{\infty}\subset G(\mathbb{R})$ denote a maximal compact subgroup. If, for any topological group $\mathfrak{G}$, the connected component of the identity is denoted by $\mathfrak{G}^\circ$, then we define $$X=G(\mathbb{R})/A^\circ\_\infty K\_\infty^\circ.$$ Now, let $\Gamma\subset G(\mathbb{Q})$ be a subgroup such that $\Gamma\cap G(\mathbb{Q})\cap \operatorname{GL}\_n(\mathbb{Z})$ has fintie index in both $\Gamma$ and $G(\mathbb{Q})\cap \operatorname{GL}\_n(\mathbb{Z})$ for some faithful representation $G\hookrightarrow \operatorname{GL}\_n$. The locally symmetric space is defined as $$ X(\Gamma)=\Gamma\backslash X. $$
>
>
>
My question is basically, **why?**
Intuitively what is this trying to achieve? As in, why are we quotienting out by $K\_\infty^\circ$ and $A\_\infty^\circ$ (maybe this is the least we need to do to ensure some kind of compactness-like result, or better cohomology groups)?
I do have some familiarity with moduli spaces of elliptic curves with level structures and I know that we can get examples using this construction (which helps construct Galois representations using certain modular forms), but when I see this construction, it just looks like generalising without any end goal in mind (which is obviously not true and why I said I should probably keep reading rather than ask this question — but I cannot help myself).
I am sure the question sounds vague, but I just don't have an image in mind when I think of these locally symmetric spaces and it would greatly help it someone could help fix that.
| https://mathoverflow.net/users/157428 | Definition of locally symmetric space of reductive groups | There is a very natural, intrinsic definition of a "symmetric space", as a manifold (Riemannian or Hermitian) with an extra symmetry of a certain prescribed type. It is then a *theorem*, not a definition, that all such objects have the form $G(\mathbb{R}) / A\_\infty^\circ K\_\infty^\circ$ for a reductive group $G$. You can find this perspective, for instance, in Milne's "[Shimura Varieties and Moduli](https://www.jmilne.org/math/xnotes/svi.pdf)".
From this perspective, *defining* a symmetric space as $G(\mathbb{R}) / A\_\infty^\circ K\_\infty^\circ$ is rather misleading. It's common to define it in this way in number-theory texts, because it avoids wading through large amounts of quite intricate (aka: beautiful) differential geometry before getting to the number-theoretic parts of the theory; but it makes the motivation highly non-obvious.
(*Edit*: Just to clarify, this is about *symmetric spaces*, not *locally symmetric spaces*. A locally symmetric space is the quotient of a symmetric space by a discrete subgroup of its automorphism group $G$, so once you understand the motivation for symmetric spaces, the motivation for locally symmetric spaces should hopefully be clear.)
| 12 | https://mathoverflow.net/users/2481 | 446402 | 179,852 |
https://mathoverflow.net/questions/446407 | 0 | I'm working on a problem that involves vectors and scalar values, and I'm looking for a closed-form solution. I hope someone can help me with this or provide insights into how to approach it. Here's the problem:
Given vectors $\mathbf{W}$, $\mathbf{S}$, and $\mathbf{T}$, and a scalars $q$.
The goal is to find a scalar $x$ so that the following constraint is satisfied:
$$
\sum \left(\frac{\mathbf{W} \cdot (q + x)}{\sum \mathbf{W} \cdot \mathbf{S}} \cdot \mathbf{T}\right) = q
$$
I have tried using optimization techniques, which work for specific instances but do not provide a general closed-form solution. I'm wondering if there's a closed-form solution to this problem, or if there are any alternative methods or insights to find the scalar $x$ for arbitrary vectors and scalar $q$.
For clarity, by $A\cdot B$ I mean the element-wise product
$$A\cdot B=[a\_1b\_1,a\_2b\_2,a\_3b\_3]$$
If helpful, here's the same definition and constraints in pseudocode:
```
find x such that sum((W * (q + x) / sum(W * S)) * T) == q
```
I've searched:
1. "vector equation closed-form"
2. "scalar equation with vectors"
3. "sum of products of vectors"
4. "linear algebra vector optimization"
But I'm not seeing anything that catches my eye.
Any help or guidance would be greatly appreciated!
| https://mathoverflow.net/users/113891 | Seeking closed-form solution for vector equation | The sum of the entrywise product of two vectors is just their dot product. So this question is asking how to find $x$ so that
$$
(q+x)(\mathbf{W} \cdot \mathbf{T}) = q\mathbf{W} \cdot\mathbf{S}.
$$
This is just a scalar equation; rearranging and solving gives
$$
x = \frac{q\mathbf{W} \cdot(\mathbf{S} - \mathbf{T})}{\mathbf{W} \cdot \mathbf{T}}
$$
as the unique solution.
| 2 | https://mathoverflow.net/users/11236 | 446410 | 179,854 |
https://mathoverflow.net/questions/446235 | 7 | It seems that Professor Lennart Carleson gave a series of Lectures at UCLA in 1985. For example, one could find several mentions about these lectures in the book by Garnett & Marshal (see for example, notes at the end of chapters II and III). I would like to know whether is aware of theses lectures; preferably available in electronic form?
I'm not sure about the context of these lectures; But, it seems that it was about some topics in complex analysis; potential theory; (quasi-)conformal mappings, harmonic measures and so on.
*Garnett, John B.; Marshall, Donald E.*, Harmonic measure, New Mathematical Monographs 2. Cambridge: Cambridge University Press (ISBN 978-0-521-72060-1/pbk). xv, 571 p. (2008). [ZBL1139.31001](https://zbmath.org/?q=an:1139.31001).
| https://mathoverflow.net/users/62739 | Carleson's lectures at UCLA | I have contacted the mathematics department at UCLA, and then received the following information from Professor Garnett (emphasis mine):
>
> For nine or ten winter quarters beginning 1985 Lennart Carleson gave a lecture course at UCLA. The topics included Potential Theory, Harmonic Measure, Complex Dynamics, Iteration Theory, Conformal Mappings and some others I can't recall. **There are no lecture notes published or recorded in the UCLA Mathematics Reading Room.** However, the Complex Dynamics became the very nice book Complex Dynamics by Carleson and Gamelin, and parts of the lectures on conformal mappings and harmonic measure are in my book Harmonic Measure written with Donald Marshall.
>
>
> The most elementary course was second term graduate real analysis using the book by Wheeden and Zymund.
>
>
>
>
> **[The 1985] course was about harmonic measure. Almost everything from it is in Garnett-Marshall Chapters I - IV and then VI, VIII and IX.**
>
>
>
| 8 | https://mathoverflow.net/users/36721 | 446411 | 179,855 |
https://mathoverflow.net/questions/446387 | 7 | In the article "Automorphism groups of simple algebras and group algebras" (1978), Janusz conjectures the following:
>
> The group algebra $\mathbb{Q} G$ for a non-trivial finite group has an outer automorphism (=non-inner automorphism)?
>
>
>
>
> Question: Is this conjecture solved ?
>
>
>
| https://mathoverflow.net/users/61949 | Do rational group algebras have an outer automorphism? | This conjecture was proved in Feit, Walter; Seitz, Gary M.
On finite rational groups and related topics.
Illinois J. Math. 33 (1989), no. 1, 103–131. The proof relies on the classification since this easily reduces to the case of simple groups.
| 10 | https://mathoverflow.net/users/15934 | 446413 | 179,856 |
https://mathoverflow.net/questions/446383 | 8 | Let $\Omega$ be a bounded open set in $\mathbb{R}^n$ with $C^\infty$ boundary, $n\ge 3$. Define
$$V(z)=\int\_\Omega \frac{1}{|z-y|^{n-2}}dy$$
Is it true that $V(z) \in C^{\infty}(\partial \Omega)$?
---
Motivation: Classical Holder estimate says that if $f \in C\_c^\alpha(\mathbb{R}^n)$, then the Newtonian potential given by
$$V(z)=\int\_{\mathbb{R}^n} \frac{f(y)}{|z-y|^{n-2}}dy$$
is $C^2$ with respect to $z$ variable. If $f$ is merely bounded, then $V$ is $C^1$ and we cannot expect better regulartiy. This can be seen from the classical example that $f=\chi\_{B\_1}$, where $\chi$ is the characteristic function and $B\_1$ is the unit ball.
However, for this example, one can see that $V\in C^{\infty}(\partial B\_1)$. From this example, it is natural to study the smoothness of $V$ along smooth boundaries. This motivates the original question. Of course, one can consider similar problems for more general Riesz potentials.
**Further Update**:
The question actually was motivated from a talk yesterday given by Jian Lu from South China Normal University. The talk is on chord log Minkowski problem. The speaker actually have already proved the smoothness of $V(z)$ along boundaries of any smooth convex bodies, and the details can be seen in <http://arxiv.org/abs/2304.14220> (2nd version). Hence I was led to ask about the regularity on generic smooth sets.
| https://mathoverflow.net/users/51546 | Regularity of Newtonian potential along smooth boundary | Sure. By a smooth dyadic decomposition it suffices to show that convolutions of the form
$$ \varepsilon^{-n} \int\_\Omega \varphi\left(\frac{y-z}{\varepsilon}\right)\ dy$$
for $0 < \varepsilon \lesssim 1$ and $\varphi$ a fixed bump function are smooth on $\partial \Omega$ uniformly in $\varepsilon$ (multiply by $\varepsilon^2$ and sum over dyadic $\varepsilon>0$ for a suitably chosen $\varphi$ to recover the Newton potential). This is trivial for large $\varepsilon$, so we may assume $\varepsilon$ small.
The strategy here is to transform this expression to eliminate all negative powers of $\varepsilon$, as this is the only obstruction to non-uniformity.
Locally we may parameterize $\Omega$ as a half-space $\{ (y', y\_n): y\_n \geq f(y') \}$ for some smooth function $f$, and then for $z = (z',f(z'))$ and $\varepsilon$ small enough the above expression becomes
$$ \varepsilon^{-n} \int\_{{\bf R}^{n-1}} \int\_0^\infty \varphi\left(\frac{(y'-z', f(y')-f(z')+t)}{\varepsilon}\right)\ dy' dt$$
which after a change of variables $y' = z'+\varepsilon w$, $t = \varepsilon s$ becomes
$$ \int\_{{\bf R}^{n-1}} \int\_0^\infty \varphi\left(\left(w, \frac{f(z'+\varepsilon w)-f(z')}{\varepsilon}+s\right)\right)\ dw ds. \quad (1) $$
By the fundamental theorem of calculus we have
$$ \frac{f(z'+\varepsilon w)-f(z')}{\varepsilon} = \int\_0^1 w \cdot \nabla f(z' + \varepsilon \theta w)\ d\theta$$
which can then be seen to (locally) be a smooth function of $z'$ and $w$ uniformly in $\varepsilon$. From this it follows from repeated differentiation under the integral sign and the chain rule that the expression in (1) is a smooth function of $z'$ uniformly in $\varepsilon$, giving the claim.
| 12 | https://mathoverflow.net/users/766 | 446421 | 179,859 |
https://mathoverflow.net/questions/446420 | 1 | Diagonally dominant matrices are required in many linear algebra algorithms such as the Gauss-Seidel algorithm.
Some matrices can be made diagonally dominant by permuting its rows and others cannot.
Given an $n$ by $n$ integer matrix $A$, what is the complexity of deciding the existence of rows permutation that results in diagonally dominant matrix? Is it NP-complete?
| https://mathoverflow.net/users/8784 | Diagonally dominant matrix via rows permutation | Create a bipartite graph $G$ with $2n$ vertices, say $u\_1,\ldots,u\_n, v\_1,\ldots,v\_n$, where the edge set of $G$ represents matrix entries that could potentially appear on the diagonal, if the rows of $A$ were permuted to make it diagonally dominant. In other words, the bipartite graph has an edge from $u\_i$ to $v\_j$ if and only if $|a\_{ij}| \geq \sum\_{k \neq j} |a\_{ik}|$. Observe that $G$ contains a perfect matching if and only if $A$ can be made diagonally dominant by permuting its rows. Thus, there is a polynomial-time algorithm to decide the existence of a row permutation that results in a diagonally dominant matrix. In fact, constructing the graph $G$ from the $n \times n$ matrix $A$ takes $O(n^2)$ time, and using the observation that every vertex in the set $\{u\_1,\ldots,u\_n\}$ belongs to at most two edges of $G$, it is possible to come up with an algorithm that decides in $O(n)$ time whether $G$ contains a perfect matching.
| 2 | https://mathoverflow.net/users/8049 | 446422 | 179,860 |
https://mathoverflow.net/questions/446433 | -2 | Let $X$ be a finite CW-complex of $n$.
1. For $i\geq 2$, $\pi\_i (X)$ is a $\mathbb{Z}\pi\_1 (X)$-module.
2. for $i\geq 2$, $H\_i (\tilde{X})$ is a finitely generated $\mathbb{Z}\pi\_1 (X)$-module, where $\tilde{X}$ is the universal covering of $X$.
I know that the connection between homotopy groups and homology groups is given by the Hurewicz-Theorem. But my question is that:
**My question**: Can we express $\pi\_i (X)$ (as a $\mathbb{Z}\pi\_1 (X)$-module) in terms of $H\_i (X)$'s (as $\mathbb{Z}$-module) or $H\_i (\tilde{X})$ (as $\mathbb{Z}\pi\_1 (X)$-module) for $2\leq i\leq n$? (particularly when they are free).
| https://mathoverflow.net/users/114476 | Relation between $\mathbb{Z}\pi_1 (X)$-module $\pi_n (X)$ and $\mathbb{Z}$-module $H_n (X)$ or $\mathbb{Z}\pi_1 (X)$-module $H_n (\tilde{X})$ | Not sure if you're already aware of an exact sequence which implies Hurewicz's theorem, but if $X$ is a space with $\pi\_i(X)=0$ for $1<i<n$ then we have
$$H\_{n+1}(X)\to H\_{n+1}(\pi)\to \pi\_n(X)\_{\pi}\to H\_n(X)\stackrel{\psi}{\to} H\_n(\pi)\to0$$ where $\pi=\pi\_1(X)$ is the fundamental group and $H\_\*(\pi)$ is discrete group cohomology, and $\pi\_n(X)\_{\pi}$ denotes the coinvariants. See e.g. Ken Brown "Cohomology of Groups" exercise VII.7.6.
Generally, without assumptions on the homotopy groups, that map $\psi$ exists (in all degrees) and is canonical: You take a projective resolution of $\mathbb Z$ over $\mathbb Z\pi$, and you take the complex of free $\mathbb Z\pi$-modules $C\_\*(\text{universal cover of }X)$, to build a canonical chain map between those resolutions that induces $\psi$. This is also explained in the above reference somewhere in Chapter II.
| 4 | https://mathoverflow.net/users/12310 | 446435 | 179,865 |
https://mathoverflow.net/questions/446442 | 11 | Let $B$ be some compact, path connected $n$-manifold without boundary such that its cobordism class is trivial, so that there exists some other $n+1$ manifold $M$ with $\partial M= B$. While there is not a unique choice for $M$ (for example $\mathbb{S}^1$ can be seen as the boundary of the connected sum of $n\geq 0$ torus $\mathbb{T}^2$) I was wondering if given $B$, some lower bounds could be obtained for the Betti numbers of $M$.
My intuition tells me that complicated manifolds cannot be the boundary of very simple manifolds, but I don't know if there are any results regarding my question. I have tried using the Mayer–Vietoris sequence but I have not get to anything menaningful.
On principle $B$ is not endowed with any particular structure $G$, but I could be very interested if such bounds were to depend also on $G$ apart from the topology of $B$.
| https://mathoverflow.net/users/504402 | Lower bounds for Betti numbers of a manifold given its boundary? | Let me assume that both $M$ and $B$ are orientable.
From the long exact sequence of the pair $(M,B)$, Poincaré–Lefschetz duality, and the universal coefficient theorem, for every $k$ we get an exact sequence (I'll use rational or real coefficients throughout):
$$
H\_{n-k}(M)^{\vee} \cong H^{n-k}(M)\cong H\_{k+1}(M,B) \to H\_k(B) \to H\_k(M),
$$
so $b\_{n-k}(M) + b\_k(M) \ge b\_k(B)$.
You will get a similar statement for $\mathbb{F}\_2$-Betti numbers if you drop the assumption of orientability.
If you look at 3-manifolds, you can get something better by looking at the torsion in $H\_1$ and at the linking form. You can show that a rational homology 3-sphere (i.e. $b\_1 = b\_2 = 0$) cannot bound a rational homology 4-ball (i.e. $b\_1 = b\_2 = b\_3 = 0$) unless the order of its first homology is a square. (In fact, the linking form on it has to be metabolic, which goes in the direction of the half lives/half dies principle mentioned in Fernando Muro's comment.) There are even finer obstructions coming from gauge theory telling you that certain integer homology 3-spheres (i.e. $H\_1 = 0$) cannot bound *smooth* rational homology 4-balls.
I don't know enough surgery theory to be able to say that in higher dimensions you "frequently" get equality in the inequality $b\_{n-k}(M) + b\_k(M) \ge b\_k(B)$. Kervaire proved that you can always get equality if $B$ is an integer homology sphere and $n \ge 4$. Anyone wants to pitch in with more information?
| 9 | https://mathoverflow.net/users/13119 | 446446 | 179,868 |
https://mathoverflow.net/questions/446428 | 4 | The fundamental theorem of surfaces states that if symmetric matrices $g\_{ij}$, $l\_{ij}\colon U\subset R^2\to R$, where $U$ is open and $g\_{ij}$ is positive definite satisfy the Gauss and Codazzi equations, then there exists a surface $X\colon U\to R^3$ with $g\_{ij}$, $l\_{ij}$ as the first and second fundamental forms.
I wonder whether an approximate version of this result has been investigated, namely suppose that $g\_{ij}$, $l\_{ij}$ do not necessarily satisfy the Gauss-Codazzi equations exactly, but the right hand side of these equations are some small functions. Does there exist then a surface $X$ whose first and second fundamental forms are close to $g\_{ij}$, $l\_{ij}$?
In other words, if $g\_{ij}$, $l\_{ij}$ satisfy the Gauss-Codazzi equations approximately, then do there exist perturbations of $g\_{ij}$, $l\_{ij}$ which satisfy the Gauss-Codazzi equations exactly? In the case I am interested in, the surfaces are convex, so we may assume that $l\_{ij}$ is positive semi-definite if that helps.
| https://mathoverflow.net/users/68969 | Approximate isometric embeddings of surfaces | I think that the answer is 'yes' if $U$ is simply-connected, because there is a way to construct a candidate 'approximate surface' from 'approximate solutions' of Gauss and Codazzi, but a more useful answer would be one that gave you estimates of how close the metric and second fundamental form of the approximate surface are to the given data in terms of the failure of the Gauss and Codazzi equations of the original data.
Here is what I mean by constructing an approximate solution: There's no loss of generality in assuming that $U$ is the set in the $uv$-plane given by $|u|,|v|\le 1$ and that $g$ and $l$ are continuous or even smooth on the boundary. We can choose $1$-forms $\omega\_1$ and $\omega\_2$ on $U$ such that $g = \omega\_1^2+\omega\_2^2$ and we can write $l = l\_{ij}\omega\_i\omega\_j$ where $l\_{ij}=l\_{ji}$ and set $\omega\_{3i} = l\_{ij}\omega\_j$. Also, let $\omega\_{12}=-\omega\_{21}$ satisfy $\mathrm{d}\omega\_i = -\omega\_{ij}\wedge\omega\_j$.
Now consider the matrix valued $2$-form
$$
\eta = \begin{pmatrix}
0 & 0 & 0 & 0\\
\omega\_1 & 0 &\omega\_{12} & -\omega\_{31}\\
\omega\_2 & -\omega\_{12} & 0 & -\omega\_{32}\\
0 & \omega\_{31} & \omega\_{32} & 0
\end{pmatrix} = A\,\mathrm{d}u + B\,\mathrm{d} v.
$$
The Gauss and Codazzi equations would hold if and only if
$\mathrm{d}\eta + \eta\wedge\eta = 0$, i.e.
$$
B\_u-A\_v + [A,B]=0.
$$
In general, we'll have
$$
\mathrm{d}\eta + \eta\wedge\eta
=\begin{pmatrix}
0 & 0 & 0 & 0\\
0 & 0 &H & -C\_1\\
0 & -H & 0 & -C\_2\\
0 & C\_1 & C\_2 & 0
\end{pmatrix}\,\mathrm{d}u\wedge\mathrm{d}v
= F\,\mathrm{d}u\wedge\mathrm{d}v \,,
$$
where $C\_1$, $C\_2$, and $H$ are 'small' in some appropriate sense (which it will be important to make precise if you want this to be useful).
Now, consider the solution of the matrix ODE with initial condition
$$
L'(u) = L(u)A(u,0), \qquad L(0)=I\_4\,.
$$
and use this to solve the matrix ODE with initial condition
$$
M\_v(u,v) = M(u,v)B(u,v),\qquad M(u,0) = L(u).
$$
Of course, $M$ won't satisfy $M^{-1}\mathrm{d}M = \eta$ unless $C\_1=C\_2=H=0$.
But, by construction, $M^{-1}\mathrm{d}M-\eta = E\,\mathrm{d}u$, where it should be possible to estimate the size of $E$ in terms of the given data $g$, $l$ and $C\_1$, $C\_2$, $H$, and this estimate should be boundable in such a way that when $C\_1$, $C\_2$, and $H$ are small in the appropriate sense, then $E$ will be as well. The point is that $E(u,v)$
satisfies the matrix ODE with initial condition
$$
E\_v = [E,B] + F,\qquad E(u,0) = 0,
$$
so $|E|$ can be bounded in terms of $|B|$ and $|F|$. (Clearly, if $F\equiv0$, then $E\equiv0$.)
The first column of $M$ will look like
$$
\begin{pmatrix}1\\x(u,v)\\y(u,v)\\z(u,v)\end{pmatrix},
$$
and $(x,y,z):U\to\mathbb{R}^3$ will give an 'approximate surface'. With the right estimates on $C\_1$, $C\_2$, and $H$ (which may involve coefficients that depend on $g$ and $l$), you should be able to say how close its first and second fundamental form are to the originally given data.
Of course, obtaining good estimates would be an interesting project. What one would really want would be estimates that were expressed in terms of the $g$-geometry of the open set $U$ and the $g$-geometry of the second fundamental form, without reference to the local coordinates $(u,v)$. This should be possible, and I think that the estimates would be interesting. I suspect something like this has been done in the literature, maybe in the computer science literature where it might be used in surface reconstruction from local data (which is only approximately known). I wouldn't be surprised to find that there is something about this in the discrete differential geometry literature, but I'm not familiar with much of that enormous corpus.
| 5 | https://mathoverflow.net/users/13972 | 446447 | 179,869 |
https://mathoverflow.net/questions/446444 | 6 | I am interested to use the mapping degree for simplicial maps between (oriented) abstract simplicial complexes. What I mean by "elementary": My preference would be to use a definition of mapping degree that is *not* based on homology and *not* based on the geometric realization of the complexes.
As I am new to this area, I have consulted several sources to find a comprehensive definition. In Allen Hatcher's and in Sergey Matveev's Algebraic Topology books it is defined through homology. In Jiri Matousek's book about the Borsuk-Ulam Theorem it is defined for simplicial complexes without orientation, so it only defines degree mod 2. In Wikipedia, nCatLab, and Wolfram MathWorld I could not find it either. In Brouwer's original paper from 1912 the definition of mapping degree is for simplicial complexes, but I find it extremely hard to follow. And if I understand it correctly, it is based on the geometric realization of the simplicial complex.
Hence my question, are you aware of a good source and reference for an "elementary" yet rigorous definition?
EDIT: Based on good comment from Achim Krause below, restricting my request to oriented n-dimensional simplicial complexes all of whose (n−1)-dimensional faces are contained in exactly two n-dimensional ones, with appropriate orientations.
| https://mathoverflow.net/users/156936 | Abstract simplicial complexes - Reference for an elementary definition of mapping degree for simplicial maps? | I think the right generality to restrict to is the following:
Let $n$ be a positive integer, and let $X$ and $Y$ be $n$-dimensional oriented simplicial complexes, with the following properties:
1. Every $n-1$-face is contained in exactly two $n$-faces (with opposite orientations relative to the $n-1$-face, i.e. if $\sigma$ is an $n-1$-face, there should be exactly two elements $x$ such that $\{x\}\cup \sigma$ is an $n$-face, with opposite orientations. Here and below I think of an orientation as assigning to each ordering of the vertices of each $n$-simplex a number $+1$ or $-1$ which flips if the ordering is changed by an odd permutation).
2. In $Y$, it is possible to go from any $n$-face to any other $n$-face by a path of neighbouring $n$-faces (where two $n$-faces are neighbouring if they share an $n-1$-face). I thought at first this would follow from 1. and connectedness, but something like "Two copies of $\partial\Delta^3$ glued together along a single vertex" shows that this is an independent condition.
Then:
**Definition.** The mapping degree of a map $f: X\to Y$ is computed as follows: Pick an $n$-simplex $\sigma$ of $Y$, and let
$$
\operatorname{deg}(f) := \sum\_{\widetilde{\sigma}} \varepsilon\_{\widetilde{\sigma}}
$$
where the sum is over all $n$-simplices $\widetilde{\sigma}$ in $X$ which are preimages of $\sigma$, and $\varepsilon\_{\widetilde{\sigma}} \in \{+1,-1\}$ depending on whether the orientations of $\widetilde{\sigma}$ and $\sigma$ agree or disagree under $f$.
**Lemma.** The mapping degree is well-defined, i.e. doesn't depend on the choice of $\sigma$ in $Y$.
*Proof.* Since by assumption we can pass from any $n$-face to any other $n$-face through a sequence of neighbouring $n$-faces, we only need to discuss the case of neighbouring $n$-faces. Let $\sigma$, $\sigma'$ be two neighbouring $n$-faces in $Y$, and let $\tau = \sigma\cap \sigma'$ be their shared $n-1$-simplex. Then every preimage $\widetilde{\tau}$ of $\tau$ in $X$ satisfies exactly one of the following:
1. It bounds two $n$-faces, both of which lie over $\sigma$, contributing opposite signs to the sum defining $\deg(f)$ via $\sigma$.
2. It bounds two $n$-faces, both of which lie over $\sigma'$, contributing opposite signs to the sum defining $\deg(f)$ via $\sigma'$.
3. It bounds two $n$-faces, one lying over $\sigma$, one over $\sigma'$, contributing equal signs to each of the sums defining $\deg(f)$ via $\sigma$ or $\sigma'$.
In all cases we see that the contributions to $\deg(f)$ defined via either $\sigma$ or $\sigma'$ agree. Since every $n$-simplex lying over $\sigma$ or $\sigma'$ appears exactly once in this enumeration, this proves that $\deg(f)$ defined via $\sigma$ or $\sigma'$ agree. $\square$
Note that this is an unwinding of the definition of mapping degree via *cohomology* (which agrees, by the universal coefficient theorem, with the mapping degree defined via homology). Indeed, under the assumptions discussed above, $H^n(Y)\cong \mathbb{Z}$ is generated by a cocycle taking a fixed positively oriented $n$-simplex to $1$, and all other $n$-simplices to $0$ (and any two such cocycles are cohomologous). To determine the map $f^\*: H^n(Y)\to H^n(X)$, we precompose one such cocycle, and decompose the resulting cocycle on $X$ again as sum of such cocycles (which gives the signed sum over preimages).
| 9 | https://mathoverflow.net/users/39747 | 446450 | 179,871 |
https://mathoverflow.net/questions/446443 | 0 | I am struggling to model my problem correctly since multiple days.
Maybe someone can give me a hint.
I have two levels, both with a fixed number of slots (i.e. 200 each).
The items I want to put on the slots are ordered and earlier items have to be positioned on earlier slots.
Imagine a fixed queue of vehicles driving on one of two levels of a train. For every vehicle one can decide whether to drive on the lower or the upper level. Once a vehicle is put in position the following vehicle can either stand behind its predecessor or drive on the opposite level.
Certain slots can remain empty due to capacity constraints. I.e. the first 5 slots on the upper level have a weight constraint of 8 tons, so if all the vehicles weigh 2 tons, one of the 5 slots has to remain empty.
The general model I have developed is working but I can not figure out how to model these precedence constraints.
My current idea is the following:
$p \in P^{top}$ are the slots on the top level and $p \in P^{bot}$ the slots on the bottom level
$v \in V$ are the vehicles. For every vehicle $v$ we build a set of pairs that models the precedence requirements for every vehicle according to the order v < u: $(v, u) \in A$.
Led $x\_{pv}$ be a binary decision variable that is 1 if $v$ is on $p$ and 0 otherwise.
Precedence rules on the bottom level:
$ \sum\limits\_{p\in P^{bot}}px\_{pv} \sum\limits\_{p\in P^{bot}}x\_{pu} \leq \sum\limits\_{p\in P^{bot}}px\_{pu} \forall (v, u)\in A$
Precedence rules on the top level:
$ \sum\limits\_{p\in P^{top}}px\_{pv} \sum\limits\_{p\in P^{top}}x\_{pu} \leq \sum\limits\_{p\in P^{top}}px\_{pu} \forall (v, u)\in A$
Order of the vehicles:
$\sum\limits\_{p\in P}x\_{p(v+1)} \leq \sum\limits\_{p\in P}x\_{pv} \forall v\in V$ \ {last element}
By themselves these three constraints seem to be working. Together they do not do what is intended . I know I am missing a (some) key elements here.
| https://mathoverflow.net/users/477317 | Precedence constraints in assignment problem | Here are linear constraints for the bottom:
$$\sum\_{p\in P^{bot}}px\_{pv} \le \sum\_{p\in P^{bot}}(p-1)x\_{pu} +|P^{bot}|\sum\_{p\in P^{top}}x\_{pu}\quad \forall (v, u)\in A$$
Equivalently,
$$\sum\_{p\in P^{bot}}px\_{pv} \le \sum\_{p\in P^{bot}}(p-1-|P^{bot}|)x\_{pu}+|P^{bot}|\quad \forall (v, u)\in A$$
The constraints for the top are similar.
| 0 | https://mathoverflow.net/users/141766 | 446456 | 179,872 |
https://mathoverflow.net/questions/446461 | 2 | For a prime $p$, let $\varphi\_p\colon\mathbb Z\to\mathbb Z/p\mathbb Z$ denote the canonical homomorphism from the integers onto the group of order $p$.
>
> Given an integer $n\ge 3$, what is the smallest $\varepsilon=\varepsilon(n)>0$ such that for any subset $A\subset\mathbb Z$ with $|A|=n$, there exists a prime $p$ satisfying $(1-\varepsilon(n))n<|\varphi\_p(A)|<n$?
>
>
>
| https://mathoverflow.net/users/9924 | Prime divisors of $\prod(a_i-a_j)$ | Maybe I misunderstand the question, but doesn't the set $A=\{i\cdot n!\,|\,1\le i\le n\}$ have $\lvert\varphi\_p(A)\rvert=1$ for all $p$ for which $\varphi\_p$ is not injective on $A$?
| 4 | https://mathoverflow.net/users/18739 | 446470 | 179,876 |
https://mathoverflow.net/questions/446445 | 7 | Let $\Omega \subset \mathbb{R}^n$ be a compact domain and for given functions $g: \partial \Omega \times [0,T] \to \mathbb{R}$ and $h: \Omega \to \mathbb{R}$ consider the heat equation
$$
\begin{cases}
\frac{\partial u}{\partial t} - \Delta u &= 0 \qquad \text{on } \Omega \times (0,T]\\
u &= g \qquad \text{on } \partial \Omega \times [0,T]\\
u &= h \qquad \text{on } \Omega \times \{t = 0\}
\end{cases}
$$
The existence of a weak solution to this equation has been proven Theorem 6.1 in *Non-Homogeneous Boundary Value Problems and Applications II* (Lions and Magenes, 1972). Throughout this book it is assumed that $\Omega$ is a $C^{\infty}$-domain.
I was wondering if this condition can be relaxed for certain domains, in particular for a hypercube $\Omega = [0,1]^n \subset \mathbb{R}^n$? Texts other than Lions and Magenes only ever seem to the consider homogeneous Dirichlet condition $g \equiv 0$. Furthermore, a book considering parabolic equations that is analogous to *Elliptic Problems in Nonsmooth Domains* (Grisvard, 2011) doesn't seem to exist.
Is there anything known about the existence of solutions to such PDEs?
| https://mathoverflow.net/users/295304 | Existence of solutions to the heat equation on nonsmooth domains | $\newcommand\R{\mathbb R}\newcommand{\Om}{\Omega}$It is convenient to reverse the time by the substitution $t\leftrightarrow T-t$ and also rescale it by a factor of $2$. So, the boundary value problem becomes the following:
\begin{align}
\frac{\partial u}{\partial t} +\frac12 \Delta u &= 0 \quad \text{on } D, \tag{1}\label{1}
\\
u &= f \quad \text{on } C, \tag{2}\label{2}
\end{align}
where $D:=\Om\times(0,T)$; $C:=\{(x,t)\in\partial D\colon t>0\}
=(\partial\Om\times(0,T])\cup(\Om\times\{T\})$ (which corresponds to the "lower boundary" of $D$, according to [Doob](https://www.ams.org/tran/1955-080-01/S0002-9947-1955-0079376-0/S0002-9947-1955-0079376-0.pdf), p. 218, who chose to not reverse the time); $f(x,t):=g(x,t)$ if $(x,t)\in\partial\Om\times[0,T]$, and $f(x,t):=h(x)$ if $(x,t)\in\Om\times\{T\}$. For the just given definition of $f$ to be consistent, we need the condition $g(x,T)=h(x)$ for all $x\in\partial\Om$; of course, the corresponding consistency condition ($g(x,0)=h(x)$ for all $x\in\partial\Om$) was needed in the original setting, before the time reversal.
Let $(B\_t)\_{t\ge0}$ be a standard Brownian motion in $\R^n$. For any $z\in D\cup C$, let $\tau\_z:=\inf\{t\ge0\colon z+(B\_t,t)\notin D\cup C\}$, the exit time of the Brownian motion from $D\cup C$ starting at the point $z$. As in Doob's paper, let $Z(z,C):=B\_{\tau\_z}$.
Assume that $f$ is continuous. Then, according to Doob's Theorem 2.1, the function $u$ on $D$ defined by the formula
\begin{equation\*}
u(z):=Ef(Z(z,C)) \tag{3}\label{3}
\end{equation\*}
for $z\in D$ will be parabolic, that is, $u$ will satisfy condition \eqref{1}.
Suppose also that for some $y\in\partial\Om$ the following Poincaré condition is satisfied: there exist a nonempty open cone $K\_y\subset\R^n$ with the vertex at $y$ and a neighborhood $V\_y\subset\R^n$ of $y$ such that $K\_y\cap V\_y\cap\Om=\emptyset$. Then, using the Kolmogorov 0-1 law, it is easy to see that for each $t\in[0,T]$ we have $\tau\_{(y,t)}=0$ almost surely, which in turn implies that $u(z)\to f(y,t)$ as $D\ni z\to(y,t)$.
Let us say that $\partial\Om$ satisfies the Poincaré condition if the Poincaré condition is satisfied for every point in $\partial\Om$.
(Of course, if $\partial\Om$ is $C^1$ or if $\Om$ is the interior of a polytope with a nonempty interior, then $\partial\Om$ satisfies the Poincaré condition.)
We conclude that, if $f$ is continuous and $\partial\Om$ satisfies the Poincaré condition, then formula \eqref{3} gives a solution to problem \eqref{1}--\eqref{2}, with condition \eqref{2} satisfied in the following sense: for each $z\_0\in C$ we have $u(z)\to f(z\_0)$ as $D\ni z\to z\_0$.
---
More generally (cf. the first full paragraph on p. 230 of Doob's paper), if $f$ is a bounded Baire function and $u$ is given by \eqref{3}, then \eqref{1} will hold and, moreover, we will have $u(z)\to f(z\_0)$ as $D\ni z\to z\_0$ for each $z\_0=(y\_0,t\_0)\in C$ such that (i) $f$ is continuous at $z\_0$ and (ii) either $t\_0=T$ or the Poincaré condition is satisfied for $y\_0$.
| 7 | https://mathoverflow.net/users/36721 | 446471 | 179,877 |
https://mathoverflow.net/questions/446482 | 6 | For a positive integer $N$, we define $$\Gamma(N)=\big \{ \begin{bmatrix} a & b \newline c & d\end{bmatrix}\in \operatorname{SL}\_2(\mathbb{Z}): \begin{bmatrix} a & b \newline c & d\end{bmatrix}\equiv \begin{bmatrix} 1 & 0 \newline 0 & 1\end{bmatrix}(\operatorname{mod}N)\big\}$$
Then, [Diamond, Shurman] defines the set of equivalence classes $S(N)$ of enhanced elliptic curves associated to $\Gamma(N)$ to be the set of equivalence classes of all triples $(E,P,Q)$, where $E/\mathbb{C}$ is an elliptic curve, and $P,Q\in E(\mathbb{C})[N]$ such that the Weil pairing $e\_N(P,Q)=e^{2\pi i/N}$ .
Defining a Weil pairing:
>
> Let $P\_1,P\_2\in E(\mathbb{C})[N]$ (possibly equal). Firstly, fix $E(\mathbb{C})=\mathbb{C}/\Lambda$ for some lattice $\Lambda=\mathbb{Z}\omega\_1+\mathbb{Z}\omega\_2\subset \mathbb{C}$. Now, there is a matrix $\gamma\in M\_{2\times 2}(\mathbb{Z}/N\mathbb{Z})$ such that $\begin{bmatrix} P\_1 \\ P\_2\end{bmatrix}=\gamma \begin{bmatrix} \omega\_1/N+\Lambda \\ \omega\_2/N+\Lambda\end{bmatrix}$. We define $e\_N(P\_1,P\_2)=e^{2\pi i\operatorname{det}(\gamma)/N}$.
>
>
>
Now, I know that there exists a smooth affine curve $X\_{\widetilde{\Gamma}(N)}/\operatorname{Spec}\mathbb{Z}[1/N]$ which represents the moduli problem (assume $N$ is large enough): $$X\_{\widetilde{\Gamma}(N)}(S)=\{(E,\alpha):E/S \operatorname{\ is\ an\ elliptic\ curve\ }, \alpha:\underline{(\mathbb{Z}/N\mathbb{Z})^{\oplus 2}}\_S\xrightarrow{\cong}E[N]\}.$$
At the face of it, it seems like $X\_{\widetilde{\Gamma}(N)}(\mathbb{C})=S(N)$. But, the problem I am having is that an isomorphism $\alpha:(\mathbb{Z}/N\mathbb{Z})^{\oplus 2}\xrightarrow{\cong}E(\mathbb{C})[N]$ gives two points $P,Q\in E(\mathbb{C})[N]$ such that $e\_N(P,Q)=e^{2\pi i x/N}$ for some $x\in (\mathbb{Z}/N\mathbb{Z})^\times$, which is not the same as a point in $S(N)$
Am I missing something? Are these two the same?
Reference:
*Diamond, Fred; Shurman, Jerry*, A first course in modular forms, Graduate Texts in Mathematics 228. Berlin: Springer (ISBN 0-387-23229-X/hbk). xv, 436 p. (2005). [ZBL1062.11022](https://zbmath.org/?q=an:1062.11022).
| https://mathoverflow.net/users/157428 | Definition of modular curve associated to $\Gamma(N)$ | This is a subtle issue (which has come up before on this site several times, see e.g. [is the modular curve X(N) defined over Q?](https://mathoverflow.net/questions/192156/is-the-modular-curve-xn-defined-over-q?rq=1) for a related question).
Your $S(N)$ is naturally a scheme over $\mathbb{Z}[1/N, \zeta\_N]$. Your $X\_{\widetilde{\Gamma}(N)}$ is, as you have defined it, a scheme over $\mathbb{Z}[1/N]$. But there is actually a natural map $\mathbb{Z}[1/N, \zeta\_N] \hookrightarrow \mathcal{O}(X\_{\widetilde{\Gamma}(N)})$ which sends $\zeta\_N$ to $e\_N(P, Q)$; and if you regard $X\_{\widetilde{\Gamma}(N)}$ as a $\mathbb{Z}[1/N, \zeta\_N]$-scheme *via this morphism*, it coincides with $S(N)$.
| 10 | https://mathoverflow.net/users/2481 | 446485 | 179,882 |
https://mathoverflow.net/questions/446462 | 4 | Let $k$ be a perfect field. I am looking for a $k$-algebra $R$ with the following properties.
* $R$ is of finite type over $k$ and is a domain;
* for all ${\mathfrak p}\in{\rm Spec}(R)$, the local ring $R\_{\mathfrak p}$ is Cohen-Macaulay and ${\rm emb.\, dim.}(R\_{\mathfrak p})-{\rm dim}(R\_{\mathfrak p})\leq1$;
* for at least one ${\mathfrak p}\in{\rm Spec}(R)$, the local ring $R\_{\mathfrak p}$ is not Gorenstein.
I would also be happy with the weaker condition "not a complete intersection" in place of "not Gorenstein".
A related question: is there an example of reduced local noetherian ring $T$, which is Cohen-Macaulay, has ${\rm emb.\, dim.}(T)-{\rm dim}(T)=1$, but is not Gorenstein (resp. not a complete intersection)?
Any help would be appreciated.
| https://mathoverflow.net/users/17308 | Example of a certain type of Cohen-Macaulay ring | There is no such example. A Cohen-Macaulay local ring with $\operatorname{embdim}(R)-\dim(R) \le 1$ is a hypersurface, which is in particular a complete intersection. Indeed, we may pass to the completion to suppose $R$ is complete. Then by Cohen's structure theorem, $R \cong S/I$ where $(S,\mathfrak{n})$ is a regular local ring and $I \subseteq \mathfrak{n}^2$. In this situation, $\dim S=\operatorname{embdim}(R)$, and as $R$ is Cohen-Macaulay, the Auslander-Buchsbaum formula implies $\operatorname{pd}\_S(R) \le 1$. If it is $0$, then $R$ is regular and we're done. If it is $1$, then from the exact sequence $0 \to I \to S \to R \to 0$ we see that $I$ is a free $S$ module. As $I$ is an ideal, it must be principal, so $R \cong S/(f)$ for some $f$.
| 6 | https://mathoverflow.net/users/155965 | 446488 | 179,884 |
https://mathoverflow.net/questions/446453 | 6 | If a real analytic variety $V$ in $\mathbb{R}^n$ is both bounded and contractible, is it true that $V$ must be a single point?
Here a real analytic variety is the set of zeros of a real analytic function $f: \mathbb{R}^n \rightarrow \mathbb{R}$.
This is certainly true if $V$ is a compact real analytic manifold (without boundary). But what about varieties that are not manifolds? Answers in other settings (complex analytic varieties or real algebraic varieties) would be interesting too.
| https://mathoverflow.net/users/54756 | Contractible real analytic varieties | It is a consequence of Sullivan's work
*Sullivan, D.*, Combinatorial invariants of analytic spaces, Proc. Liverpool Singularities-Sympos. I, Dept. Pure Math. Univ. Liverpool 1969-1970, Lect. Notes Math. 192, 165-168 (1971). [ZBL0227.32005](https://zbmath.org/?q=an:0227.32005).
that every compact $k$-dimensional real-analytic subset $V\subset {\mathbb R}^n$ is a mod 2 pseudo-manifold: Every $k-1$-dimensional simplex in a triangulation of $V$ is contained in an even number of $k$-simplices. (This is immediate from the main result of Sullivan's paper about local structure of $V$ as a cone over a base of even Euler characteristic.) Now, take the sum of all $k$-simplices in the given triangulation of $V$. This will be a mod 2 cycle. Since there are no simplices of dimension $k+1$, this cycle is not a boundary. Hence, $H\_k(V, {\mathbb Z}\_2)\ne 0$. In particular, $V$ cannot be contractible. Note that this argument is pretty much the same as in the smooth case, once you have Sullivan's local result.
Sullivan's paper is freely available [here](https://www.math.stonybrook.edu/%7Edennis/publications/PDF/DS-pub-0007.pdf).
| 4 | https://mathoverflow.net/users/39654 | 446497 | 179,888 |
https://mathoverflow.net/questions/446493 | 0 | Let $c>0$ and let $Y$ be the space of all distributions of compact support in $(-1,1)$ with singular support at $\{0\}$. Let $X$ be **subspace** of $Y$ such that for any $\phi \in X$ there holds:
$$ \int\_{t^2>r^2} \phi'(t)^2\,dt \leq c\,\int\_{t^2>r^2-r^4} \phi(t)^2\,dt \quad \forall\, r\in (0,1).$$
Is it true that $X$ must be finite dimensional?
| https://mathoverflow.net/users/50438 | Finite dimensionality of a subspace | $\newcommand\de\delta$The answer is no. E.g., let $X$ be the linear span of the set $\{\de,\de',\de'',\dots\}$, where $\de$ is the Dirac delta distribution supported on $\{0\}$. Then $X$ satisfies your condition but is infinite dimensional.
| 1 | https://mathoverflow.net/users/36721 | 446502 | 179,890 |
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