parent_url
stringlengths 37
41
| parent_score
stringlengths 1
3
| parent_body
stringlengths 19
30.2k
| parent_user
stringlengths 32
37
| parent_title
stringlengths 15
248
| body
stringlengths 8
29.9k
| score
stringlengths 1
3
| user
stringlengths 32
37
| answer_id
stringlengths 2
6
| __index_level_0__
int64 1
182k
|
|---|---|---|---|---|---|---|---|---|---|
https://mathoverflow.net/questions/429464 | 5 | $\DeclareMathOperator{\graph}{\operatorname{graph}}$
I would like to know if, given $f\in W^{1,2}(\mathbb{R}^n,\mathbb{R})$, it is true that we can always cover $\graph(f)\subset\mathbb{R}^{n+1}$ with the images of countably many Lipschitz maps $g\_k:\mathbb{R}^n\to\mathbb{R}^{n+1}$ and a set $E$ with $\mathscr{H}^n(E)=0$, i.e.
$$\graph(f)\subset \bigcup\_{k\in\mathbb{N}} g\_k(\mathbb{R}^n) \cup E$$
(which is the definition of $\graph(f)$ being countably rectifiable).
Any hint, reference or additional conditions for this to be true are very much appreciated.
| https://mathoverflow.net/users/351083 | Is the support of a Sobolev function a varifold? | Yes if you choose a suitable representative of a Sobolev function.
**Lemma.** *Let $f\in W^{1,p}(\mathbb{R}^n)$, $1\leq p<\infty$. Then for every $\epsilon>0$, there is a Lipschitz function $g:\mathbb{R}^n\to\mathbb{R}$ such that
$$
|\{ x\in\mathbb{R}^n:\, f(x)\neq g(x)\}|<\epsilon.
$$*
The proof follows from the pointwise inequality
$$
|f(x)-f(y)|\leq C|x-y|(M|\nabla f|(x)+M|\nabla f|(y)),
$$
where $M$ stands for the Hardy-Littlewood maximal function.
$\Box$
The lemma implies that you can find countably many Lipschitz functions $g\_i:\mathbb{R}^n\to\mathbb{R}$ such that such that the set
$$
E:=\mathbb{R}^n\setminus \bigcup\_{i=1}^\infty\{x\in\mathbb{R}^n:g\_i(x)\neq f(x)\}.
$$
has measure zero.
If you redefine $f$ on $E$ so that the function $f$ is constant on that set (that alters $f$ on a set of measure zero which is okay, because Sobolev functions are defined a.e.) it follows that graps of the functions $g\_i$ and a graph of a constant function cover the graph of $f$.
For more details see for example Lemma on p. 96 in:
**P. Hajłasz,** Change of variables formula under minimal assumptions. *Colloq. Math.* 64 (1993), no. 1, 93–101.
| 7 | https://mathoverflow.net/users/121665 | 429467 | 174,031 |
https://mathoverflow.net/questions/429452 | 6 | Say that $\{a\_n\}\_{n\geq 1}$, $|a\_n|\leq 1$, are such that $$\left|\sum\_{n\leq x} a\_n \log \frac{x}{n}\right|\leq \epsilon x\quad\text{for all $x\geq x\_0$.}$$ What sort of bound can we deduce on $S(x)=\sum\_{n\leq x} a\_n$?
---
*Naïve answer.*
It is easy to give a simple bound:
since, for $y>1$, $$\begin{aligned}
\left|\sum\_{n\leq x y} a\_n \log \frac{x y}{n}
- \sum\_{n\leq x/y} a\_n \log \frac{x/y}{n}
\right.&\left.
- 2 (\log y) \sum\_{n\leq x} a\_n\right| \\
&=
\left|\sum\_{x/y < n\leq x y} a\_n \log \frac{x y}{n}\right|\\
&= \left|\sum\_{x/y < n\leq x y} \log \frac{x y}{n}\right|\\
&\leq \int\_{x/y}^{x y} \log \frac{x y}{t} dt + \log y^2 \\
&= \left(y-\frac{1}{y}\right) x - \left(\frac{x}{y}-1\right) \log y^2,
\end{aligned}$$
we know that, for $x\geq x\_0/y$, assuming $y\leq x/e$,
$$\frac{1}{x}\left|\sum\_{n\leq x} a\_n\right|\leq \frac{\epsilon}{2 \log y} \left(y + \frac{1}{y}\right) + \frac{y - \frac{1}{y}}{\log y} - \frac{2}{ y} +\frac{2}{x}.$$
Writing $y = \exp(\delta)$, we easily see that the leading term of the right side is $\epsilon/\delta + 2 \delta$. Hence, setting $\delta = \sqrt{\epsilon/2}$ and thus obtain that $|S(x)|$ is at most roughly $\sqrt{8 \epsilon}\cdot x$.
---
Now, that is not just a simple bound, but a downright simple-minded one. Can one do better?
| https://mathoverflow.net/users/398 | (Explicit) Tauberian theorems: removing $(\log x/n)$ | $\newcommand\ep\epsilon$You cannot improve the upper bound $c\sqrt\ep\,x$ on $|S(x)|$ (where $c>0$ is a universal real constant factor) by more than a universal positive real constant factor.
Indeed, take any $\ep\in(0,1/4)$ and let
$$a\_n:=\sum\_{j\ge1}(-1)^j1(c^{j-1}<n<c^j),\quad c:=e^{\sqrt\ep}.$$
Let
\begin{equation}
T(x):=\sum\_{n\le x} a\_n\ln\frac{x}n.
\end{equation}
Let $h(u):=1-u+u\ln u\sim(1-u)^2/2$ as $u\to1$.
Then for natural $k\to\infty$
\begin{equation}
\begin{aligned}
T(c^k)
&= \sum\_{j\le k}(-1)^j\sum\_{c^{j-1}<n<c^j}\ln\frac{c^k}n \\
&= \sum\_{j\le k}(-1)^j\Big(o(c^k/k)+\int\_{c^{j-1}<n<c^j}\ln\frac{c^k}n\,dn\Big) \\
&=o(c^k)+ \sum\_{j\le k}(-1)^j[c^j h(c)+(c^j-c^{j-1})(k-j)\ln c] \\
&=o(c^k)+ \sum\_{j\le k}(-1)^j c^j\ep[1/2+(k-j)+O(\ep)] \\
&=O(c^k\ep).
\end{aligned}
\end{equation}
Note that, for $x$ between $c^{k-1}$ and $c^k$, we have $T(x)$ between $T(c^{k-1})$ and $T(c^k)$.
It follows that for $x\to\infty$
\begin{equation}
\sum\_{n\le x} a\_n\ln\frac{x}n=T(x)=O(x\ep),
\end{equation}
as required.
On the other hand, for natural $j$ and $x=c^{j-1}$,
$$|S(cx)-S(x)|=\sum\_{c^{j-1}<n<c^j}1
\asymp\sqrt\ep\,c^{j-1}=\sqrt\ep\,x,
$$
so that the upper bound $\asymp\sqrt\ep\,x$ on $|S(x)|$ cannot be improved,
as was claimed.
| 8 | https://mathoverflow.net/users/36721 | 429469 | 174,032 |
https://mathoverflow.net/questions/427693 | 4 | This is a question that has bothered myself and Gottfried Helms a fair amount of late. He has made his case for the following result, but a proof escapes both of us. The question is deceptively simple, but keeps eluding each of my attempts when we get into the finer details.
Let's start by calling:
$$
f(z) = e^{z}-1\\
$$
Then $f$ has a parabolic fixed point at $0$ with multiplier $1$. By which we have:
$$
f(z) = z + O(z^2)\\
$$
There exists an attracting petal, and a repelling petal; and conveniently they are separated by $i\mathbb{R}$. By which, we are only focused on the attracting petal, which is given as $\Re(z) < 0$. Using the standard Ecalle construction of an Abel function, which I'm mostly just following Milnor's *Dynamics in One Complex Variable*. We can construct an Abel function:
$$
\alpha(f(z)) = \alpha(z) + 1\\
$$
This function is $2 \pi i$ periodic, and is holomorphic for $\Re(z) < 0$. We are primarily focused near the boundary point $z = 0$--where the Abel function has a singularity.
Now we can invert this to the left of zero, so that we can construct a super function $\alpha^{-1}(z)$, and by this, we can get to our function in question:
$$
g(z) = \alpha^{-1}(\alpha(z) + 1/2)\\
$$
I will spare the details, but this function is also $2 \pi i$ periodic, and holomorphic for $\Re(z) < 0$. We don't need this though, just that it's holomorphic on a domain to the left of $0$, with $0$ on the boundary. This function satisfies the equation:
$$
g(g(z)) = f(z)\\
$$
---
Gottfried first came to me with this question, and so I am not the most versed in the construction of this series, but it relates to a question which he had investigated on this forum--of the $\sin$ half iterate. But, the essential statement, is that we can construct a formal series:
$$
h(z) = \sum\_{n=1}^\infty d\_n z^n\\
$$
Such that this series satisfies the formal manipulations:
$$
h(h(z)) = \sum\_{n=1}^\infty \frac{z^n}{n!}\\
$$
By which this series has a radius 0 convergence (it's a divergent series).
The question then becomes, is it possible to represent this divergent series, as Euler would represent the divergent series:
$$
\sum\_{n=0}^\infty (-1)^nn!z^n\\
$$
Using the Laplace transform:
$$
\int\_0^\infty \frac{e^{-zx}}{1+x}dx = \sum\_{n=0}^\infty (-1)^nn!z^n\\
$$
Or, rather, could we do something similar, in the same vein that Borel summation is done.
---
This means, we have the function $g(z)$, which is holomorphic, and a boundary value at $0$. We have a divergent/formal series at $z=0$. If we were to take:
$$
\mathcal{B}h(z) = \sum\_{n=0}^\infty d\_n\frac{z^n}{n!}\\
$$
Will this have a non-zero radius of convergence? By which we could Borel sum the divergent series. I know I'm missing something crucial, and every attempt I've made a using traditional arguments fails because $g(z)$ is a rather anomalous function.
Essentially, can we make the same treatment of Euler's series, of this series. By which we'd be able to pull out a bound $d\_n = O(c^n n!)$ for some $c > 0$.
| https://mathoverflow.net/users/133882 | Borel summation and the Abel function of $e^z-1$ | So, there's been some developments on the tetration forum, and I came across two papers:
*Introduction to 1-summability and the resurgence theory*, David Sauzin
<https://hal.archives-ouvertes.fr/hal-00860032v1/document#cite.LY>
And:
*The Fixed Point of the Parabolic Renormalization Operator*,
Oscar Lanford III, Michael Yampolsky
<https://arxiv.org/pdf/1108.2801.pdf>
Long story short, both of these papers provide our result. They specifically mention an expansion by Ecalle for the Abel function: $\alpha(f(z)) = \alpha(z) + 1$ that:
$$
\alpha(z) = \frac{1}{z} - A\log(z) + \sum\_{k=0}^\infty b\_k z^k
$$
Where in these papers they prove that $b\_k = O(c^kk!)$. This doesn't directly answer our question, but, through a bit of algebra, you can identify that:
$$
\alpha^{-1}(1/2 + \alpha(z)) = g(z) = \sum\_{k=0}^\infty g\_k z^k
$$
And that the values $g\_k$ have similar bounds to $b\_k$--this is just mule work as far as I'm concerned.
The word they use specifically, is that the coefficients are of Gevrey class 1. This is further identified by a conversation between Will Jagy and Ecalle, on the thread [Does the formal power series solution to $f(f(x))= \sin( x) $ converge?](https://mathoverflow.net/questions/45608/does-the-formal-power-series-solution-to-ffx-sin-x-converge).
This goes to show that we can express the half iterate $g(z)$ in Euler's manner. Numerical evidence has further confirmed this, as much of us at the tetration forum have been evaluating different and more general integrals using this basis.
This answers the question entirely to my satisfaction, and further answers it using work other than my own. I'm just answering it here for the benefit of anyone who comes across this.
| 2 | https://mathoverflow.net/users/133882 | 429477 | 174,034 |
https://mathoverflow.net/questions/429471 | 4 | **Question.** Let $\mathcal{C}$ be a small abelian category. Does the category $\mathrm{Ind}(\mathcal{C})$ of ind-objects of $\mathcal{C}$ have enough injectives?
I have seen many times that $\mathrm{Ind}(\mathcal{C})$ automatically has enough injectives. For example, Proposition 5 of [Akhil Mathew's note](https://amathew.wordpress.com/2012/07/25/ind-objects-in-an-abelian-category/) and the main theorem of
[this paper](https://www.ams.org/journals/tran/1972-170-00/S0002-9947-1972-0302738-0/S0002-9947-1972-0302738-0.pdf) claims such a result. The argument is something like this.
1. $\mathrm{Ind}(\mathcal{C})$ is an abelian category satisfying (AB5)
2. $``\displaystyle\bigoplus\_{X \in \mathrm{ob}\,\mathcal{C}}" X$ is a generator of $\mathrm{Ind}(\mathcal{C})$
3. Apply Theorem 1.10.1 of the Tôhoku paper
On the other hand, I found another result in [Kashiwara and Schapira's book](https://link.springer.com/chapter/10.1007/3-540-27950-4_16) which claims the exact opposite. Let $\mathrm{Mod}(k)$ be the category of $k$-vector spaces. Then Corollary 15.1.3 of this book implies that $\mathrm{Ind}(\mathrm{Mod}(k))$ does not have enough injectives.
I cannot spot any errors in both arguments. Which one is incorrect and why?
---
There is a weaker claim that if $\mathcal{C}$ is an artinian abelian category, then $\mathrm{Ind}(\mathcal{C})$ has enough injectives. For example, p.356, Proposition 7 of [Gabriel's paper](http://www.numdam.org/item/?id=BSMF_1962__90__323_0) implies this. This makes a lot more sense.
| https://mathoverflow.net/users/490435 | Does $\mathrm{Ind}(\mathcal{C})$ have enough injectives, if $\mathcal{C}$ is an abelian category? | As Dan Petersen said in the comments, $Mod(k)$ isn't small. Note that even in this case, $Ind(Mod(k))$ only has *small* direct sums, and in particular you cannot take the direct sum that appears in point 2. of your proof sketch (I'm assuming you meant direct sum, and not tensor product); and so you don't have that generator.
In fact it's not too hard to see that there simply is no generator, and Kashiwara-Schapira prove that it also does not have enough injectives. Their proof certainly uses vector spaces of increasing dimension/cardinality (I haven't read it).
About your comment : if you restrict to some small abelian subcategory $C\subset Mod(k)$, such as "vector spaces of cardinality/dimension less than some fixed infinite cardinal", then the theorem above will apply and show that $Ind(C)$ has enough injectives.
But note that the $Ind$ construction *depends* on the universe because it is defined by freely adding *small* filtered colimits. So if you say "ok, go to a universe where $Mod(k)$ is small, and apply $Ind$ there to get something with enough injectives", that will work, but the category you obtain is not the same as the one you obtain by defining $Ind(Mod(k))$ while taking all $k$-modules of your universe of reference. Keeping track of universes, this is like $Ind^U(Mod^U(k))$ while you would be taking $Ind^V(Mod^U(k))$ for some larger universe $V > U$.
| 7 | https://mathoverflow.net/users/102343 | 429492 | 174,036 |
https://mathoverflow.net/questions/429497 | 18 | For $n\geq 4$, let $V\_n$ be the maximum volume of the convex hull of $n$ points on the unit sphere (in $\mathbb{R}^3$, although information on higher dimensions is welcome as well). I'm sure the problem of computing $V\_n$ has been extensively studied and has a standard name: what is this name?
For which values of $n$ is the exact value of $V\_n$, and/or a configuration realizing it, known (at least combinatorially)? And, for small $n$, what are the best known configurations even if they are not proved to be optimal? Essentially, I'm looking for a list analogous to what [Wikipedia lists for the Thomson problem](https://en.wikipedia.org/wiki/Thomson_problem#Configurations_of_smallest_known_energy) but for the volume of the convex hull instead of electrostatic potential energy.
PS: the dual problem of finding the least possible volume cut by $n$ (hyper)planes all tangent to the unit sphere (i.e., the smallest convex polytope exscribed around a sphere, rather than the largest inscribed in one as described above) also seems interesting, so if it has a standard name I'm also curious to know.
| https://mathoverflow.net/users/17064 | Known configurations maximizing the volume of the convex hull of n points on the unit sphere | The problem is elementary for $n=5$.
We may regard that case as a combination of two triangular pyramids sharing a base $\triangle ABC$. Then the volume is always bounded by one-third the area of the common base times the distance between the two remaining points $D,E$ (the latter is always greater than or equal to the sum of the pyramid altitudes). Then this bound is saturated and both factors maximized by placing $D$ and $E$ at opposite points ("poles") and distributing $A,B,C$ uniformly along the "equator" (a maximally symmetric triangular bipyramid).
---
Let's look more broadly here, as other broader answers have yet to appear.
We begin with [this 2014 answer from Math Stack Exchange](https://math.stackexchange.com/questions/979660/largest-n-vertex-polyhedron-that-fits-into-a-unit-sphere#988708), which concentrates on four to eight points in three dimensions. Summarizing those results gives:
$n=4$ -- regular tetrahedron
$n=5$ -- triangular bipyramid, as above
$n=6$ -- regular octahedron
$n=7$ -- pentagonal bipyramid (this would seem to overcrowd the "equator" with five points, but plausible alternatives such as the capped octahedron would be worse)
$n=8$ -- not a cube, which isn't even close; rather a $D\_2$-symmetry object with triangular faces and vertices of order $4$ and $5$.
The linked answer cites Ref. [1] for proof.
Horváth and Lángi[2] give a lemma that all solutions with four or more points in three dimensions are polyhedra with triangular faces; and the faces are also simplices in higher dimensions. The above claims conform with this, as do well-known (but AFAIK not rigorously proved) arrangements for larger $n$ such as the laterally tricapped trigonal prism for $n=9$ and the regular icosahedron for $n=12$. The triangular face requirement excludes the regular dodecahedron, which in any event has long since been beaten for $20$ points.
Reference [2] also considers higher dimensionalities. For $n=d+1$ points on the surface of a $d$-dinensional ball the optimum is the regular simplex. With $n=d+2$ the optimum is described as combining regular-simplex cross-sections in two complementary subsets of the dimensions. Going back to the triangular bipyramid with $n=5$ and $d=3$, the relevant cross-sections are an equilateral triangle (the base of the bipyramid) in two of the dimensions, and a line segment (the axis) in the remaining dimension. With higher dimensionality the two orthogonal cross-sections should have as nearly equal dimensionalities as possible. Thus for $(n,d)=(6,4)$ we would select an equilateral triangle in two dimensions and another equilateral triangle in the remaining two dimensions, not a three-dimensional tetrahedron and a one-dimensional line segment; so the $(6,4)$ optimum is not a higher-dimensional bipyramid. For $n=d+3$ with $d$ odd the optimum is guaranteed by using three orthogonal sim-plectic cross-sections, such as the line segments which are diagonals of the regular octahedron for $n=6, d=3$.
**References**
1. Joel D. Berman, Kit Hanes (1970), "Volumes of polyhedra inscribed in the unit sphere in E³",
*Mathematische Annalen*, **188**, 1, 78-84.
2. Horváth, Ákos G.; Lángi, Zsolt (2016), "Maximum volume polytopes inscribed in the unit sphere", *Monatsh. Math.* **181**, No. 2, 341-354. ZBL1354.52016.
| 11 | https://mathoverflow.net/users/86625 | 429498 | 174,037 |
https://mathoverflow.net/questions/429506 | 3 | Let $G=GL\_n(\mathbb{F}\_q)$ be the (finite) group of all linear invertible transformations of the vector space $(\mathbb{F}\_q)^n$ over the finite field $\mathbb{F}\_q$.
$G$ acts naturally on the Grassmannian $Gr\_{k,n}$ of $k$-dimensional linear subspaces of $(\mathbb{F}\_q)^n$.
**What is the length of the corresponding representation of $G$ in the space of complex valued functions on $Gr\_{k,n}$?**
| https://mathoverflow.net/users/16183 | Length of representation of $GL_n(\mathbb{F}_q)$ in functions on Grassmannian | I believe that this is answered in Proposition 5.1 of [this](https://arxiv.org/abs/1811.08675) paper, which says that it is $\text{min}(k,n-k) + 1$.
| 5 | https://mathoverflow.net/users/317 | 429510 | 174,039 |
https://mathoverflow.net/questions/429473 | 0 | Computation model is defined as Hartmanis and Stearns [4](https://www.ams.org/journals/tran/1965-117-00/S0002-9947-1965-0170805-7/S0002-9947-1965-0170805-7.pdf), it is well known that Liouvilles constant
$$C\_L=\sum\_{i=1}^{\infty} 10^{-i!}$$ is computable in real time or linear time [1](https://www.ams.org/journals/tran/1965-117-00/S0002-9947-1965-0170805-7/S0002-9947-1965-0170805-7.pdf), [5](https://rjlipton.wpcomstaging.com/2012/06/15/why-the-hartmanis-stearns-conjecture-is-still-open/) especially Theorem 12 in [1](https://www.ams.org/journals/tran/1965-117-00/S0002-9947-1965-0170805-7/S0002-9947-1965-0170805-7.pdf).
Is there any example of transcendental number computational complexity of which is $\Theta(n \log n)$, that is, any first $n$ digits of the 10-base expansion of it can be outputed by Turing Machine (defined by Hartmanis, Stearns, Yamada, Robin) in $\Theta(n \log n)$?
Please see the following reference for real-time computation or linear time computation if there is any ambiguity:
[1](https://link.springer.com/article/10.1007/BF02759719),
[2](https://ieeexplore.ieee.org/document/5219459),
[3](https://www.cambridge.org/core/journals/journal-of-symbolic-logic/article/abs/hisao-yamada-realtime-computation-and-recursive-functions-not-realtime-computable-ire-transactions-on-electronic-computers-vol-ec11-1962-pp-753760/06282D96A6D1C5F481E9A8B2B5A55535),
[4](https://www.ams.org/journals/tran/1965-117-00/S0002-9947-1965-0170805-7/S0002-9947-1965-0170805-7.pdf)
[5](https://rjlipton.wpcomstaging.com/2012/06/15/why-the-hartmanis-stearns-conjecture-is-still-open/)
Hope concrete example, if one want to discuss computation model, please show the computing code by the model. If one think the theorem 12 in [1](https://www.ams.org/journals/tran/1965-117-00/S0002-9947-1965-0170805-7/S0002-9947-1965-0170805-7.pdf) is not correct, please refute it in hard code.
| https://mathoverflow.net/users/14024 | Examples of real-time transcendental number and superlinear-time trancsendental number | Too long for a comment, though not a complete answer, sorry.
I don't understand the exact influence of allowing multiple tapes on complexity for Turing machines, but let me offer the following anyway.
Any number whose binary expansion is a so-called Sturmian or Beatty sequence (definition below) is transcendental; see "Transcendence of Numbers with a Low Complexity Expansion" by Ferenczi and Mauduit.
A Sturmian sequence $s$ is defined via an irrational number $x$, and the nth bit is $s\_n = \lfloor (n+1) x \rfloor - \lfloor nx \rfloor$. Put another way, the nth bit is 1 iff there is an integer between $nx$ and $(n+1)x$.
Here's where I guess I can't give an explicit example, but I would think that by using numbers with various computability rates (since $x$ can be any irrational at all!), I would guess that you could achieve any superlinear runtime, including $n \ln n$. Perhaps someone else can be a little more explicit here.
| 2 | https://mathoverflow.net/users/116357 | 429516 | 174,040 |
https://mathoverflow.net/questions/429518 | 3 | By a theorem of Hagemann and Mitschke, a condition (A) that a variety $\mathcal{V}$ is congruence $n$-permutable, is equivalent to a condition (B) that there exist ternary terms $p\_1,\dots,p\_{n-1}$ such that $\mathcal{V}$ satisfies identities
$$x = p\_1(x,y,y),\ \ p\_i(x,x,y) = p\_{i+1}(x,y,y),\ \ p\_{n-1}(x,x,y) = y.$$
I know how to show that (B) implies (A), however I was not able to show the converse implication. I searched the literature, but it turns out that much of it is in German. I found no full proof in English of the implication (A) $\Rightarrow$ (B). The best I could find is the information that there is a third equivalent condition (C) that for every $A \in \mathcal{V}$ and every reflexive subalgebra $R$ of $A^2$, one has $R^{-1} \subseteq R^{n-1}$, where $R^{-1}$ is the converse relation and $R^{n-1}$ is the $(n-1)$-fold composition of $R$ with itself. I know how to show the equivalence of (C) and (B).
I would like to ask for a reference in English for the proof of the implication (A) $\Rightarrow$ (B) (or alternatively for sketching such a proof). In fact I am mostly interested in congruence $3$-permutability, so it would be enough for me if someone could provide a proof for the case $n = 3$. Note that a proof of the implication (A) $\Rightarrow$ (C) that does not make use of the equivalence of (A) and (B) would fulfill my request.
| https://mathoverflow.net/users/490477 | Reference request for a proof of the Mal'cev condition for congruence $n$-permutability | The original paper is in English:
Hagemann, Joachim; Mitschke, A.
On $n$-permutable congruences.
Algebra Universalis 3 (1973), 8-12.
The proof given there is only a partial proof which depends
on an earlier paper (in German) by Hagemann.
I did not find the argument in any of the standard textbooks by Burris and Sankpannavar (1981), McKenzie-McNulty-Taylor (1987), or the one by Cliff Bergman (2012).
I did find a complete proof in English of $(A)\Rightarrow(B)$ in Algebras, Lattices, Varieties
Volume II by
Ralph S. Freese,
Ralph N. McKenzie,
George F. McNulty.
Walter F. Taylor. It is their Theorem 6.11.
I will state Theorem 6.11 here and make a few comments on the proof.
**Theorem 6.11.**
For a variety $V$ and an integer $k \geq 2$ the following conditions are equivalent:
(i) $V$ has $k$-permutable congruences.
(ii) $F\_V(k + 1)$ has $k$-permutable congruences.
(ii') There exist terms $r\_i(x\_0 , x\_1 , \ldots, x\_k )$,
$i = 0, \ldots, k$ for $V$ such that the following are
identities of $V$:
$x\_0 \approx r(x\_0 , x\_1 , \ldots, x\_k)$,
$r\_k (x\_0 , x\_1 , \ldots, x\_k ) \approx x\_k$, and
$r\_{i−1}(x\_0 , x\_0 , x\_2 , x\_2 , \ldots) \approx r\_i (x\_0 , x\_0 , x\_2 , x\_2 ,\ldots )$ for $i$ even,
$r\_{i−1} (x\_0 , x\_1 , x\_1 , x\_3 , x\_3,\ldots) \approx r\_i (x\_0 , x\_1 , x\_1 , x\_3 , x\_3 \ldots)$ for $i$ odd.
(iii) There exist terms $p\_1 , \ldots, p\_{k−1}$ for $V$ such that the following are identities of $V$:
$x \approx p\_1 (x, z, z)$,
$p\_1 (x, x, z) \approx p\_2 (x, z, z)$
$\vdots$
$p\_{k−2} (x, x, z) \approx p\_{k−1} (x, z, z)$
$p\_{k−1} (x, x, z) \approx z$.
(iv) The subalgebra of $F\_V^2(x, y) $
generated by $\langle x, x\rangle$, $\langle x, y\rangle$
and $\langle y, y\rangle$ contains elements
$\langle a\_i , b\_i\rangle$, $i = 0, \ldots,k$, with $\langle y, y\rangle = \langle a\_0 , b\_0\rangle$, $\langle x, x\rangle = \langle a\_k , b\_k\rangle$, and $b\_i = a\_{i+1}$.
The steps you are interested in are (ii) $\Rightarrow$ (ii') $\Rightarrow$ (iii).
For (ii) $\Rightarrow$ (ii'), use the fact that the congruences
$\theta\_0 = \textrm{Cg}(\langle x\_0,x\_1\rangle, \langle x\_2,x\_3\rangle,\ldots)$ and $\theta\_1 = \textrm{Cg}(\langle x\_1,x\_2\rangle, \langle x\_3,x\_4\rangle,\ldots)$ $k$-permute and have join containing $\langle x\_0,x\_k\rangle$. The argument to get the $r\_i$'s follows a standard pattern.
For the implication (ii') $\Rightarrow$ (iii), define
$p\_i (x\_0 , x\_1 , x\_2 ) = r\_i (x\_0 , \ldots , x\_0 , x\_1 , x\_2 , \ldots , x\_2 )$
where there are $i$ instances of $x\_0$ followed by one instance of $x\_1$ followed by $k-i$ instances of $x\_2$.
| 1 | https://mathoverflow.net/users/75735 | 429525 | 174,041 |
https://mathoverflow.net/questions/429528 | 1 | Let $\mathcal{H}$ be a separable Hilbert space and let $x\_1,...,x\_n$ be points in $\mathcal{H}$. Let $\varepsilon >0 $ be given and consider the measures
$$
\mu := \frac1{n}\,\sum\_{i=1}^n\, \delta\_{x\_i} \mbox{ and }
\mu^{\varepsilon} := \frac1{n}\,\sum\_{i=1}^n\, G(x\_i,\varepsilon\, T).
$$
Here $T$ is any trace-class operator on $\mathcal{H}$ and $G(x,\Sigma) $ denotes the Gaussian measure on $\mathcal{H}$ with mean $x$ and covariance operator $\Sigma$.
Is there an upper bound on the 2-Wasserstein distance between $\mu$ and $\mu^{\varepsilon}$ that depends only on $\varepsilon$ and tends to 0 as $\varepsilon$ does?
| https://mathoverflow.net/users/36886 | Distance between empirical measures and thickened version | The claim follows by a synchronous coupling: $X = x\_I$ and $X^{\epsilon} = x\_I + \sqrt{\epsilon} \sum\_{j=1}^{\infty} \sqrt{\lambda\_j} \rho\_j e\_j$ where $I \sim \operatorname{Uniform}(\{1, \dots, n \})$; $e\_j$ are the eigenfunctions of $T$; $\lambda\_j$ are the corresponding eigenvalues; and $\{ \rho\_j \} \overset{i.i.d}{\sim} \mathcal{N}(0,1)$. Indeed, \begin{align\*}
\mathcal{W}\_2(\mu,\mu^{\epsilon})^2 \le E\left[ |X - X^{\epsilon} |^2 \right] = \epsilon E\left[| \sum\_{j=1}^{\infty} \sqrt{\lambda\_j} \rho\_j e\_j|^2 \right] = \epsilon \operatorname{trace}(T) \;.
\end{align\*}
| 1 | https://mathoverflow.net/users/64449 | 429532 | 174,043 |
https://mathoverflow.net/questions/428777 | 6 | $\newcommand\Sn{\mathit{Sn}}$A subset $A$ of a group $X$ is called *algebraic* if $A=\{x\in X: a\_0xa\_1x\dotsm xa\_n=1\}$ for some elements $a\_0,a\_1,\dotsc,a\_n\in X$.
Let $\mathcal A\_X$ be the family of all algebraic sets in $X$.
**Definition.** The *Steinhaus number* $\Sn(X)$ of an infinite group $X$ is the largest cardinal $\kappa$ such that for any cover $\mathcal C\subset\mathcal A\_X$ of $X$ with cardinality $\lvert\mathcal C\rvert<\kappa$ there exists a set $C\in \mathcal C$ such that $FCC^{-1}F=X$ for some finite set $F\subseteq X$.
By the famous Steinhaus–Weil Theorem, for any closed set $F$ of positive Haar measure in a compact topological group $X$ the set $FF^{-1}$ is a neighborhood of the identity in $X$.
This theorem implies that each compact topological group has $$\operatorname{cov}(\overline{\mathcal N}\_X)\le \Sn(X)\le \lvert X\rvert$$
where $\operatorname{cov}(\overline{\mathcal N}\_X)$ is the smallest cardinality of a cover of $X$ by closed sets of Haar measure zero.
Under Martin's Axiom, $\Sn(X)=\mathfrak c$ for each infinite compact Polish group.
>
> **Problem.** Is $\Sn(X)=\mathfrak c$ for any infinite compact Polish group in ZFC?
>
>
>
**Remark.** The answer is affirmative for commutative groups. This follows from the observation that each algebraic set in a commutative group is a coset of a subgroup $\{x\in X: x^n=1\}$.
| https://mathoverflow.net/users/61536 | Steinhaus number of a group | The answer to this problem is negative: *For the compact Polish group $X=S\_3^\omega$ we have $Sn(X)\le\mathfrak r$ where $$\mathfrak r=\min\{|\mathcal R|:\mathcal R\subseteq [\omega]^\omega\;\wedge\;\forall f\in 2^\omega\;\;\exists R\in\mathcal R\;\;|f[R]|=1\}.$$*
By induction it can be shown $\mathfrak r=\min\{|\mathcal R|:\mathcal R\subseteq[\omega]^\omega\;\wedge\;\forall f\in n^\omega\;\exists R\in\mathcal R\;\;|f[R]|=1\}$ for every $n\ge 2$. So, we can find a family $\mathcal R\subseteq[\omega]^\omega$ of cardinality $|\mathcal R|=\mathfrak r$ such that for every function $f:\omega\to S\_3$ there exists $R\in\mathcal R$ such that $|f[R]|=1$.
Then $$S\_3^\omega=\bigcup\_{R\in\mathcal R}\bigcup\_{g\in S\_3}A\_{R,g},\quad\mbox{where $A\_{R,g}=\{f\in S\_3^\omega:\{g\}=f[R]\}$}.$$
Observe that for every $R\in\mathcal R$ and $g\in S\_3$ the set $A\_{R,g}A\_{R,g}^{-1}=A\_{R,e}$ is nowhere dense in $S\_3^\omega$ and hence $FA\_{R,g}A\_{R,g}^{-1}F\ne S\_3^\omega$ for any finite set $F\subseteq S\_3^\omega$.
It remains to show that for every $R\in\mathcal R$ and $g\in S\_3$, the set $A\_{R,g}$ is algebraic in $S\_3^\omega$.
Let $d$ and $t$ be elements of order $2$ and $3$ in the symmetric group $S\_3$. It is easy to check that $S\_3=\{x\in S\_3:dxxdxx=e\}$ where $e$ is the identity of the group $S\_3$. On the other hand, $\{x\in S\_3:xtxxtx=t^2\}=\{e\}$ and hence $\{x\in S\_3:g^{-1}xtg^{-1}xg^{-1}xtg^{-1}x=t^2\}=\{g\}$.
Then $A\_{R,g}=\{x\in S\_3^\omega:a\_0xa\_1xa\_2xa\_3x=b\}$ where
$$a\_0(i)=a\_2(i)=\begin{cases}g^{-1}&\mbox{if $i\in R$};\\
d&\mbox{if $i\in\omega\setminus R$},
\end{cases}\quad
a\_1(i)=a\_3(i)=\begin{cases}tg^{-1}&\mbox{if $i\in R$};\\
e&\mbox{if $i\in\omega\setminus R$},
\end{cases}
$$
and
$$
b(i)=\begin{cases}t^2&\mbox{if $i\in R$};\\
e&\mbox{if $i\in\omega\setminus R$},
\end{cases}
$$
| 1 | https://mathoverflow.net/users/61536 | 429535 | 174,045 |
https://mathoverflow.net/questions/429079 | 7 | For $f: \mathbb R^n \to \mathbb R$ a locally integrable function, $\varepsilon \in (0, \infty)$, and $x \in \mathbb R^n$, define $I(f, \varepsilon, x)$ to be the averaged integral of $f$ over $B\_{\varepsilon} (x)$, the ball of radius $\varepsilon$ around $x$. That is,
$$I(f, \varepsilon, x) := \frac{1}{\mu(B\_\varepsilon (x))} \int\_{B\_\varepsilon (x)} f(y) \, dy$$
Define
$$
K(f, \varepsilon, x) :=
\begin{cases}
1 & \text{if }\; I(f, \varepsilon, x) > f(x),\\
-1 & \text{if }\; I(f, \varepsilon, x) < f(x),\\
0, &\text{if }\; I(f, \varepsilon, x) = f(x).\\
\end{cases}
$$
Finally, let
$$
H(f, \varepsilon, x) = \dfrac{1}{\varepsilon} \int\limits\_{(0, \varepsilon]} K(f, s, x) ds
$$
Intuitively, $H$ is the weighted average amount of time a function spends greater than (resp. less than) its value at a point, in an infinitesimal neighbourhood of said point.
**Question:** Let $f$ be a $C^2$ function. Is it true that $\lim\_{\varepsilon \to 0} H(f, \varepsilon, x)$ exists for almost every $x \in \mathbb R^n$?
**Remarks:**
1. A sample path of Brownian motion provides a counterexample in the $C^0$ case, as shown [here](https://mathoverflow.net/questions/358653/how-much-time-does-a-function-spend-above-or-below-its-average-value-around-a-po?rq=1). So $C^2$ is in some sense sharp.
2. The existence of the limit is immediate on the regions on which $f$ is super/sub-harmonic, as well as on the interior of the region on which $f$ is harmonic.
| https://mathoverflow.net/users/173490 | Does this "local time" type limit exist a.e. for $C^2$ functions? | Even $C^\infty$ isn't enough and even on $\mathbb R$. Take any nowhere dense compact set $A\subset\mathbb R$ of positive measure and set $f=0$ on $A$. Now let $I\_k$ be the complementary intervals to $A$ (ignore the two rays). Draw some positive bumps on the middle halves of $I\_1, I\_2,\dots I\_n$. If $n$ is large enough, then $K(f,s,x)$ will be $1$ on $[1/10,1]$ for all $x\in A$ and no sufficiently small in $C^0$ perturbation of $f$ will change that. However, the function is still identically $0$ in some small $\delta\_1$-neighborhood of $A$. Now draw some much smaller negative bumps still deep inside $I\_k$'s but closer to the endpoints on both sides. Then, using finitely many (but sufficiently many) intervals, you'll achieve that $K(f,s,x)=-1$ on $[\delta\_1/10,\delta\_1]$ for all $x\in A$ and, again, the property is stable and we still have $f=0$ in some $\delta\_2$-neighborhood of $A$. Then do it again and again alternating between positive and negative bumps.
| 3 | https://mathoverflow.net/users/1131 | 429544 | 174,050 |
https://mathoverflow.net/questions/429561 | 1 | I am looking for an example of a function $f:[0,1]\to\mathbb{R}$ which is in $L^p$ for some $p$ and whose graph is not a $1$-dimensional varifold in $\mathbb{R}^2$, that is such that it is not possible to write
$$
\operatorname{graph}(f)\subset \bigcup\_{n\in\mathbb{N}} g\_k(\mathbb{R}) \cup E
$$
for Lipschitz functions $g\_k:\mathbb{R}\to \mathbb{R}$ and a measurable set $E$ with null $\mathscr{H}^1$ measure.
This is a follow-up to these two questions [Is the support of a Sobolev function a varifold?](https://mathoverflow.net/questions/429464/is-the-support-of-a-sobolev-function-a-varifold) and [Hausdorff dimension of the graph of a BV function](https://mathoverflow.net/questions/327267/hausdorff-dimension-of-the-graph-of-a-bv-function) where it is proved that the graph of a (representative of a) Sobolev function, and more generally of a $BV$ function *is* indeed a varifold. Since varifolds support a notion of differentiability, my intuition suggests that the result must be false for functions that are only in $L^p$ (and that probably the counterexample is very easy to construct).
| https://mathoverflow.net/users/351083 | $L^p$ function whose graph is not a varifold | While writing this question down I realized that its answer is actually pretty trivial, but I am going to keep it here for the benefit of other users. It suffices to take any measurable function $\tilde{f}:[0,1]\to\mathbb{R}$ whose graph is not a varifold and compose it with a diffeomorphism $\phi$ that sends $\mathbb{R}$ to $(0,1)$ to get an $L^\infty$ function $f=\phi\circ \tilde{f}$ with the same property.
As shown in [If the Hausforff dimension of the graph of a function $u$ is $N$ and $\tilde u = u$ a.e. then $\dim\_H \mathrm{graph} \, \tilde u = N$ too](https://mathoverflow.net/questions/327331/if-the-hausforff-dimension-of-the-graph-of-a-function-u-is-n-and-tilde-u/327646#327646) we can in fact even take $f=0$ almost everywhere. By the way, this kind of construction is also exactly the reason why not all representatives of a $BV$ or Sobolev function have rectifiable graphs.
| 2 | https://mathoverflow.net/users/351083 | 429562 | 174,056 |
https://mathoverflow.net/questions/429547 | 2 | $(X, \tau\_X) $ and $(Y, \tau\_Y) $ be two topological spaces.
$\forall f\in Y^X$ with $\text{Gr}(f) $ is closed implies $f\in C(X, Y) $.
Question : Does this implies $(Y, \tau\_Y) $ is compact?
---
Notation:
$Y^X$: Set of all functions from $X$ to $Y$.
$C(X, Y) =\{f\in Y^X: f \text{ is continuous }\}$
$\text{Gr}(f) =\{(x,f(x)):x\in X\}\subset X×Y$
---
Question $1$:(prove/disprove)
$(\forall X$ and $\forall f\in Y^X \text{ with } \textrm{Gr}(f) \subset X\times Y$ closed $\implies f\in C(X, Y) ) \implies Y$ is compact.
Question $2$:(prove/disprove)
For a fixed non discrete space $(X, \tau) $ and $\forall f\in Y^X$ having closed graph is continuous then $Y$ is compact.
| https://mathoverflow.net/users/483536 | (Dis)prove : if every function with closed graph are continuous then the target space is compact | Let me try to answer these questions under the assumption that $Y$ is $T\_1$.
We start with a set $E$ along with a filter $\mathcal F\in\mathcal P(E)$. We can then cook up a topological space $X$ with underlying set $\{x\}\sqcup E$ with topology given by the discrete topology $E^\delta$ on $E$, and $\{x\}\sqcup U$ for every $U\in\mathcal F$. It follows from definition that, for every topological space $Y$, a map $f\colon X\to Y$ is continuous if and only if $f\rvert\_E\colon E\to Y$ *tends to* $f(x)\in Y$ along the filter $\mathscr F$, namely, for every neighborhood $V$ of $f(x)\in Y$, there exists a set $U\in\mathcal F$ such that $f(U)\subseteq V$. In short, the continuity captures the convergence along the filter.
What about the graph of a map $f\colon X\to Y$ being closed? Here we need the assumption of $Y$ being $T\_1$, so that we only have to test the existence of open neighborhoods at every point $(x,y)\in X\times Y$ where $y$ runs through all points of $Y\setminus f(x)$. Unwinding the definitions, it is equivalent to saying that, there exists a subset $U\in\mathcal F$ and an open neighborhood $V$ of $y\in Y$ such that
1. $y\not\in V$; and
2. for every $x\in U$, we have $f(x)\not\in V$.
Equivalently, it is saying that $y$ is not a cluster point of $f$ along $\mathcal F$.
---
A positive answer to Question 1. It is known that, a topological space is compact if and only if every filter on the underlying set admits a cluster point. We assume that $Y$ is non-compact, thus it is non-empty, and there exists a filter on $Y$ which does not admit any cluster point. We fix such a filter $\mathcal F$, an arbitrary point $y\in Y$, and let $E=Y$ be a set endowed with the filter $\mathcal F$. We now define $f\colon X\to Y$ by $f(x)=y$ and $f\rvert\_E=\operatorname{id}$. The previous analysis shows that $f$ is not continuous but its graph is closed.
---
A sketch of a negative answer to the second question. More precisely, we have
**Proposition.** For *every* topological space $X$, there exists a non-compact but Hausdorff topological space $Y$ such that every map $X\to Y$ with closed graph is continuous.
Roughly speaking, the compactness of $Y$ can be more complicated than what a fixed topological space $X$ can see. To see this, we introduce a slightly "quantitative" version of compactness:
**Definition.** Let $\kappa$ be a strong limit cardinal (i.e., for every $\lambda<\kappa$, we have $2^\lambda<\kappa$). We say that a topological space $X$ is *$\kappa$-compact* if, for every set $E$ with $\lvert E\rvert<\kappa$ and every filter $\mathcal F$ in $E$ (note: $\kappa$ being strong limit implies that $\lvert\mathcal F\rvert\le 2^{\lvert E\rvert}<\kappa$), every map $E\to X$ admits a cluster point along $\mathcal F$.
---
**Update:** This definition is equivalent, by the axiom of choice, to the following simpler one: every filter *base* $\mathcal F$ on $X$ with $\lvert\mathcal F\rvert<\kappa$ admits a cluster point in $X$. The usual proof shows that this $\kappa$-compactness is equivalent to the open-cover condition that every open cover of size $<\kappa$ contains a finite subcover.
---
**Lemma.** Let $\kappa$ be a strong limit cardinal and $Y$ a $\kappa$-compact topological space. Then for every topological space $X$ with $\lvert X\rvert<\kappa$, the projection $X\times Y\to X$ is closed.
The usual proof leads to
**Corollary.** Let $\kappa$ be a strong limit cardinal and $Y$ a $\kappa$-compact topological space. Then for every topological space $X$ with $\lvert X\rvert<\kappa$, a map $X\to Y$ is continuous if and only if its graph is closed.
Now in order to see the Proposition above, it suffices to construct a $\kappa$-compact non-compact topological space for a chosen strong limit cardinal $\kappa>\lvert X\rvert$. If I am not mistaken, we can take an ordinal $\lambda$ with cofinality greater than $\kappa$, with the order topology.
| 5 | https://mathoverflow.net/users/176381 | 429564 | 174,057 |
https://mathoverflow.net/questions/429579 | 8 | Let $M$ be a compact contractible manifold, $X\subset\partial M$ and $C\_X$ the cone over $X$.
>
> **Question:** Is it true that $C\_X$ embeds in $M$ with its boundary $\partial C\_X$ mapped to $X\subset \partial M$?
>
>
>
I am mostly interested in the piecewise linear case, that is, $M$ is a PL manifold, $X$ is a simplicial complex in $\partial M$, the embedding is a PL map, etc. I am also mostly interested in the case when $M$ is a 4-manifold, but a general answer is welcome too.
Note that the answer is "Yes" if $M$ is a ball.
| https://mathoverflow.net/users/108884 | If $M$ is contractible manifold and $X\subset \partial M$, does the cone over $X$ embed in $M$? | Not in the PL case - this follows from the results of ["Knot concordance in homology cobordisms"](https://arxiv.org/abs/1801.07770) by Hom, Levine, and Lidman.
They prove that for many pairs of a 3-manifold $Y$ and knot $K \subset Y$, any contractible 4-manifold with boundary $Y$ does not contain a PL embedded disc with boundary $K$ even though $Y$ does in fact bound contractible 4-manifolds.
A particular example given in the paper is taking your $M$ to be a contractible 4-manifold with boundary $-1/2$ surgery on the right handed trefoil, and $X = K$ to be the core of the surgery torus. Then $X$ does not bound a PL disc in $M$.
| 10 | https://mathoverflow.net/users/33041 | 429581 | 174,064 |
https://mathoverflow.net/questions/429199 | 1 | This might be related to [counting hamiltonian cycles](https://mathoverflow.net/questions/429025/counting-hamiltonian-cycles-in-graph-and-finding-a-coefficient-of-polynomial).
@**Peter Taylor** gave [negative result](https://mathoverflow.net/a/428266/12481) about the one dimensional case, but we believe his attack is
not directly applicable to this question.
Given positive integer $n$, find integer $m$ and $m \times n$
matrix $A = a\_{i,j}$ with positive integer entries.
Let $y\_1,...,y\_n$ be integer variables.
Consider the following integer program with constraints:
* $0 \le y\_i \le n$
* $\sum\_{j=1}^n y\_j = n$
* For $ 1 \le i \le m$: $\sum\_{j=1}^n y\_j a\_{i,j}= \sum\_{j=1}^n a\_{i,j}$
We require the integer program to have unique solution
of $y\_i$ all ones for chosen $(m,A)$. Let the solution be $(m\_0,A\_0)$.
>
> Q1: How small the unique solution can be in terms of $n$? Can we get
> $2^{m\_0} \max ( a \in A\_0)=\exp(o(n))$?
>
>
>
Getting $O(\exp(n))$ is easy by taking $m=1,a\_{1,i}=2^i$.
| https://mathoverflow.net/users/12481 | Only trivial solutions to system of linear diophantine equations possibly related to hamiltonian cycles in graphs | Yes, we can get $\exp(\omicron(n))$.
Assume for a moment that $n$ is a perfect square and $m=\sqrt{n}$.
The general case is essentially the same, just a bit more complicated.
The idea is to partition those $n=m^2$ variables $y\_1,\ldots\, y\_n$ in $m$ blocks of $m$ digits in base $b=n+1$. Define
$$a\_{ij} = \left\{
\begin{array}{ll}
b^{\,j-m(i-1)-1} & \text{if } \; m(i-1) \lt j \leq mi \\
0 & \text{otherwise}
\end{array}
\right.
$$
For example if $n=9$ we have $m=3$ and $b=10$ and
$$ A = \left(
\begin{array}{rrr|rrr|rrr}
1 & 10 & 100 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 10 & 100 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 10 & 100 \\
\end{array}
\right) $$
It is not hard to see that such $A$ only admits the trivial solution where all the $y\_j$ are equal to $1$, because $0 \le y\_j \lt n + 1 = b$.
Now $\max a\_{ij} = b^{m-1} \le b^m$ and thus
$$
2^m \cdot \max a\_{ij} \le 2^m b^m = (2b)^m
= \exp(m \, \ln(2b))
= \exp(\sqrt{n} \,\ln(2n+2))
$$
and $\sqrt{n} \,\ln(2n+2)$ is in $\omicron(n)$.
The question actually asked for *positive* integer entries $a\_{ij}$ rather than just non-negative entries.
But because $\sum\_{j=1}^n y\_j = n$, we can instead use $a'\_{ij} = a\_{ij} + 1$, and since
$\,\max a'\_{ij} = b^{m-1} + 1 \le b^m$
we arrive at the same bound.
**Edit** (based @joro's comment)
In summary, we have $\sqrt{n}$ equations with $\max a\_{ij} \sim n^{\sqrt{n}}$.
| 2 | https://mathoverflow.net/users/155625 | 429592 | 174,066 |
https://mathoverflow.net/questions/429396 | 2 | The suggested intuition behind mixed Hodge structures - developed
in particular to generalize Hodge decomposition of cohomology
groups from complex smooth complete varieties to more general algebraic varieties - is that one should think the cohomology groups $H^k(X)$ to be endowed with increasing filtrations whose successive quotients originate from cohomologies of appropriate smooth complete varieties,
hence admit (pure) Hodge structures, *but* of different weights.
Here is an motivating and so rather 'informal' example which I took from here: [Example of curves](https://en.wikipedia.org/wiki/Hodge_structure#Example_of_curves) where I'm missing this decisive feature that successive quotients of the associated filtration should come from the
cohomologies of smooth *complete* varieties.
The example works as follows:
>
> To motivate the definition, consider the case of a reducible complex
> algebraic curve $X$ consisting of two nonsingular components, $X\_1 $ and $X\_2$,
> which transversally intersect at the points $Q\_1$ and $Q\_2$.
> Further, assume that the components are not compact, but can be compactified by adding
> the points $P\_1 , ... , P\_n$. The first cohomology group of the curve $X$ (with compact support)
> is dual to the first homology group, which is easier to visualize.
> There are three types of one-cycles in this group. First, there are elements
> $ \alpha\_{i}, (i=1,..., n)$ representing small loops around the punctures
> $P\_{i}$. Then there are elements $ \beta\_{j} $ that are coming from the
> first homology of the compactification of each of the components.
> The one-cycle in $ X\_{k}\subset X$ ( $ k=1,2$ ) corresponding to a cycle in the
> compactification of this component, is not canonical: these elements are determined
> modulo the span of $\alpha\_{1} ,... , \alpha\_{n}$.
> Finally, modulo the first two types, the group is generated by a combinatorial cycle
> $\gamma $ which goes from $ Q\_{1}$ to $ Q\_{2}$ along a path in one component
> $X\_{1}$ and comes back along a path in the other component $X\_{2}$.
> This suggests that $ H\_{1}(X)$ admits an increasing filtration
>
>
>
$$ 0\subset W\_{0}\subset W\_{1}\subset W\_{2}=H\_{1}(X) $$
>
> whose successive quotients $W\_n/W\_{n−1}$ originate from the cohomology of smooth
> complete varieties, hence admit (pure) Hodge structures, albeit of different weights.
>
>
>
*Problem:* The last point I not understand. From cohomology of which concrete *smooth complete varieties* originate the cycles $ \alpha \_{i}$ generating $W\_0$ in this example?
The same question about the combinatorical cycle $\gamma $ generating $W\_2/W\_{1}$.
| https://mathoverflow.net/users/108274 | Example motivating mixed Hodge structures | We think of the cycle $\alpha\_i$ as coming from a point, specifically, the point we need to add to compactify the puncture $p\_i$.
Here "come from" refers to the excision exact sequence in compactly supported cohomology
When we have a variety $X$ obtained as the open subset of a variety $\overline{X}$ whose closed complement is $Z$, there is a long exact sequence
$$ H^{i-1}\_c(Z) \to H^i\_c( X) \to H^i\_c(\overline{X} ) \to H^i\_c(Z) $$
which in particular maps cycles in $H^0(Z)$ to cycles in $H^1(X)$. For $Z$ a finite union of points needed to compactify $X$, this produces the cycles $\alpha\_i$.
| 3 | https://mathoverflow.net/users/18060 | 429606 | 174,067 |
https://mathoverflow.net/questions/429262 | 2 | Let $\overline{X}$ be a smooth proper curve over $\mathbb{F}\_q$, for some $q$, $S$ a collection of $\mathbb{F}\_q$ points of $\overline{X}$, and set $X=\overline{X}-S$.
For a rank $n$ $\overline{\mathbb{Q}}\_{\ell}$-local system on $X$, it is known that the coefficients of the characteristic polynomials of Frobenii at all closed points generate a number field $E$, by the work of L. Lafforgue on function field Langlands.
**Question**: are there (known or conjectural) bounds on how many closed points are needed so that the coefficients of the Froebnii characteristic polynomials at these points generate the field $E$?
For example, it seems to me that in the case of $\overline{X}=\mathbb{P}^1\_{\mathbb{F}\_q}$, $S$ being four points, and $n=2$, the Frobenii at the $\mathbb{F}\_q$-points suffice to generate $E$, by the [computations of Kontsevich](https://www.math.nyu.edu/%7Etschinke/.manin/final/kontsevich/kontsevich.pdf) in Section 0.1 of the linked paper.
| https://mathoverflow.net/users/484855 | Generation of trace fields of Frobenii on local systems | Your proof in the case of four points assumes that the local monodromies at those four points are unipotent.
In general, one needs a bound not just on the set of ramification points but on the breaks/slopes of the sheaf at those points.
To see this is necessary, one can work already in the case of sheaves lisse of rank one on $\mathbb P^1\_{\mathbb F\_p}$ minus one point, where the Artin-Schreier sheaf $\mathcal L\_\psi ( x (x^{p^n}-x))$ has trace of Frobenius 1 at every point of degree dividing $n$ but has trace field $\mathbb Q(\mu\_p)$.
With this caveat, a bound is given by Deligne in Proposition 2.10 of the article [Finitude de l’extension de $\mathbb Q$ engendrée par des traces de Frobenius, en caractéristique finie](https://publications.ias.edu/sites/default/files/FrobTraces.pdf) which states that it suffices to take all closed points of degree at most
$$2n + 2 \log\_q^+ ( 2n^2 (b\_1 ( X) + \sum\_{s\in S} \alpha\_s \deg s))$$
where $\alpha\_s$ is the largest slope of the rank $n$ local system at $s$, $b\_1(X)$ is the first Betti number, and $\log\_q^+$ denotes the log base $q$, or $0$, whichever is larger.
| 2 | https://mathoverflow.net/users/18060 | 429607 | 174,068 |
https://mathoverflow.net/questions/429604 | 5 | Let $X$ be a smooth complex algebraic variety with $H^0(X,\mathcal{O}\_X) = \mathbb{C}$ and $V \subset X$ an open subvariety whose complement has codimension two. Now, let $L\_{\varepsilon}$ be a line bundle on $V\_{\varepsilon} = V \times Spec[\mathbb{C}[\varepsilon]/(\varepsilon ^2)$. If we denote by $j\_{\varepsilon}: V\_{\varepsilon} \hookrightarrow X\_{\varepsilon} = X \times Spec(\mathbb{C}[\varepsilon]/(\varepsilon ^2))$ the natural inclusion, then is it true that $\tilde{L}\_{\varepsilon} = (j\_{\varepsilon})\_\* (L\_{\varepsilon})$ defines a line bundle on $X\_{\varepsilon}$ ? If not, is there any other approach to extend $L\_{\varepsilon}$ to $X\_{\varepsilon}$ ?
| https://mathoverflow.net/users/45597 | Extension of first order deformations of a line bundle | Under some conditions on $X,V$, your line bundle can be extended to $X\_{\varepsilon}$. Indeed, let $\imath\_X:X\hookrightarrow X\_{\varepsilon}$ and $\imath\_V:V\hookrightarrow V\_{\varepsilon}$ be two closed immersions and $\mathcal{I}\_X,\mathcal{I}\_V$ be the ideal sheaves respectively. By the following exact sequence
$$ 0\to \mathcal{I}\_X \to \mathcal{O}^{\times}\_{X\_{\varepsilon}}\to \mathcal{O}^{\times}\_X\to 0,$$
if we consider universal $\delta$-functor theory and a natural transformation $H^0(X,-)\to H^0(U,-)$, we have a diagram of two exact sequences
$$
\begin{aligned}H^1(X,\mathcal{I}\_X)&\to \mathrm{Pic}(X\_{\varepsilon})\to \mathrm{Pic}(X)\to H^2(X,\mathcal{I}\_X)
\\ \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\downarrow \alpha&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\beta\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow \gamma\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\delta
\\ H^1(V,\mathcal{I}\_V)&\to \mathrm{Pic}(V\_{\varepsilon})\to \mathrm{Pic}(V)\to H^2(V,\mathcal{I}\_V)
\end{aligned}$$
(see Hartshorne Exercises 3.4.6). Note that in the case, $\mathcal{I}\_X\cong \mathcal{O}\_X$ as $\mathcal{O}\_{X\_{\varepsilon}}$-modules and $\mathcal{I}\_V\cong \mathcal{O}\_V$ respectively.
Since $X$ is smooth, $\gamma$ is an isomorphism. Hence, if $\alpha$ is a surjection and $\delta$ is an injection, then $\beta$ is a surjection by five lemma and any line bundle on $V\_{\varepsilon}$ can be extended to $X\_{\varepsilon}$. Note that considering the local cohomology exact sequence
$$ H^1\_Z(X,\mathcal{O}\_X)\to H^1(X,\mathcal{O}\_X)\to H^1(V,\mathcal{O}\_V)\to H^2\_Z(X,\mathcal{O}\_X)\to H^2(X,\mathcal{O}\_X)\to H^2(V,\mathcal{O}\_V),$$
where $Z:=X\setminus U$, if $H^2\_Z(X,\mathcal{O}\_X)=0$, then $\beta$ is a surjection (it is worth saying that if $\mathrm{codim}\_X Z\ge 3$, then $H^2\_Z(X,\mathcal{O}\_X)=0$ by SGA2, III, Proposition 3.3).
| 4 | https://mathoverflow.net/users/490443 | 429616 | 174,074 |
https://mathoverflow.net/questions/429588 | 9 | On a scheme, the coherent sheaves that are invertible objects for the tensor product (monoid) operation are precisely the coherent sheaves that are (Zariski) locally free of rank one. Is the same true for algebraic spaces? (I believe that this follows from a theorem of Nisnevich since there is simultaneously an etale local section and a Nisnevich local section, but hopefully somebody has a nice answer.)
This is motivated by my (incomplete!) answer to the following MO post: [Quotients of schemes by connected groups](https://mathoverflow.net/q/429489)
| https://mathoverflow.net/users/13265 | Are the tensor-invertible coherent sheaves on an algebraic space (Zariski) locally free of rank one? | There are counterexamples by Stefan Schröer. One of them is not locally separated (a *bug-eyed cover*, as Kollár calls it), another is a (non-normal) proper surface. See the paper [here](https://doi.org/10.1016/S0022-4049(02)00012-9).
About the link: some characters do not display properly, but one can click "View PDF" on top of the page, or use the [arXiv version](https://arxiv.org/abs/math/0110243). Thanks to Jason Starr and Sasha for the links.
| 8 | https://mathoverflow.net/users/7666 | 429622 | 174,075 |
https://mathoverflow.net/questions/429614 | 3 | Let $F\subset \Bbb{R}$ intersect every closed uncountable subsets of $\Bbb{R}$.
Does there exist $f:\Bbb{R}\to \Bbb{R}$ additive onto function such that $f(F) \subset \Bbb{R}$ has the property of Baire for every $F$ ?
I have explained my thoughts [here](https://math.stackexchange.com/q/4470490/977780) on MSE.
| https://mathoverflow.net/users/483536 | Does there exist $f:\Bbb{R}\to \Bbb{R}$ additive onto function such that $f(F) \subset \Bbb{R}$ has the property of Baire for every $F$? | Yes, there exists such a function: Consider the real line as a linear space over the field $\mathbb Q$ and find a linearly independent Cantor set $C\subseteq \mathbb R$
(using the Kuratowski-Mycielski Theorem 19.1 in Kechris' "Classical Descriptive Set Theory"). Identifying $C$ with $C\times C$, we can write $C$ and the union $\bigcup\_{\alpha\in\mathbb R}C\_\alpha$ of continuum many uncountable compact sets.
Since the set $C$ is linearly independent, there exists an additive function $f:\mathbb R\to\mathbb R$ such that $C\_\alpha\subseteq f^{-1}(\alpha)$ for every real number $\alpha\in \mathbb R$. This function $f$ has the required property: any Bernstein set $F\subset\mathbb R$ has non-empty intersection with each set $C\_\alpha$, $\alpha\in\mathbb R$, and hence $f[F]=\mathbb R$, so $f[F]$ has the Baire property in $\mathbb R$.
| 5 | https://mathoverflow.net/users/61536 | 429623 | 174,076 |
https://mathoverflow.net/questions/429619 | 25 | A function $f:X\to X$ on a group $X$ is called a *polynomial* if there exist $n\in\mathbb N=\{1,2,3,\dots\}$ and elements $a\_0,a\_1,\dots,a\_n\in X$ such that $f(x)=a\_0xa\_1x\cdots xa\_n$ for all $x\in X$. The smallest possible number $n$ in this representation is called the *degree* of the polynomial $f$ and is denoted by $\deg(f)$.
Let $\mathrm{Poly}(X)$ be the set of all polynomials on a group $X$.
In fact, $\mathrm{Poly}(X)$ is a submonoid of the monoid $X^X$ of all self-maps of $X$, endowed with the operation of composition of functions.
So, $|\mathrm{Poly}(X)|\le|X^X|=|X|^{|X|}$.
If the group $X$ is commutative, then each polynomial is of the form $f(x)=ax^n$ for some $a\in X$ and $n\in\mathbb N$. This implies that the number of semigroup polynomials on a finite Abelian group $X$ is equal to $|X|\cdot\exp(X)\le |X|^2$ where $\exp(X)=\min\{n\in\mathbb N:\forall x\in X\; (x^n=1)\}$.
>
> **Question 1.** *Is any reasonable upper bound on the number of polynomials on a finite group $X$?
> For example, is $|\mathrm{Poly}(X)|=o(|X|^{|X|})$?*
>
>
>
Each polynomial $f:X\to X$ on a finite Abelian group $X$ has degree $\deg(f)\le\exp(X)$.
**Question 2.** *Is $\deg(f)\le\exp(X)$ for any polynomial $f:X\to X$ on a finite group $X$?*
**Remark 2.** The affirmative answer to Question 2 would imply that $$|\mathrm{Poly}(X)|\le \sum\_{n=1}^{\exp(X)}|X|^{k+1}=\frac{|X|^{\exp(X)+2}-|X|^2}{|X|-1}.$$
**Remark 3.** Finite groups $X$ with $|\mathrm{Poly}(X)|=|X|\cdot\exp(X)$ are characterized in the following theorem.
**Theorem.** A finite group $X$ has $|\mathrm{Poly}(X)|=|X|\cdot\exp(X)$ if and only if $X$ is either commutative or is isomorphic to $Q\_8\times A$ for some nontrivial commutative group $A$ of odd order.
*Proof.* To prove the ``if'' part, assume that $X$ is either commutative or $X$ is isomorphic to $Q\_8\times A$ for some nontrivial commutative group $A$ of odd order. If $X$ is commutative, then the equality $|\mathrm{Poly}(X)|=|X|\cdot\exp(X)$ is clear.
Now assume that $X=Q\_8\times A$ for some nontrivial commutative group $A$ of odd order. [GAP-calculations of Peter Taylor](https://gist.github.com/pjt33/b9e1e62291110b164e091ef7c2f7ead5) show that the group $Q\_8$ has exactly 32 polynomials of each degree $k\in\{1,2,3,4\}$. This implies that
$$|\mathrm{Poly}(Q\_8\times A)|=32\cdot|\mathrm{Poly}(A)|=32\cdot |A|\cdot\exp(A)=4\cdot|X|\cdot\exp(A)=|X|\cdot\exp(X).$$
To prove the ``only if'' part, assume that $X$ is a finite non-commutative group with $|\mathrm{Poly}(X)|=|X|\cdot\exp(X)$.
For every $a\in X$ and $n\in\mathbb N$, consider the polynomial $p\_{a,n}(x)=ax^n$. The definition of $\exp(X)$ implies that the set $\mathrm{Pol}(X):=\{p\_{a,n}:a\in X,\;1\le n\le \exp(X)\}$ has cardinality $|X|\cdot\exp(X)$ and hence coincides with the set $\mathrm{Poly}(X)$. So, for any $a\in X$ there exists $n\le\exp(X)$ such that $axa^{-1}=x^n$ for all $x\in X$. This implies that every subgroup of $X$ is normal, so $X$ is a [Dedekind group](https://en.wikipedia.org/wiki/Dedekind_group). By the [classical Dedekind result](https://en.wikipedia.org/wiki/Dedekind_group), $X$ is isomorphic to the product $Q\_8\times A\times B$ where $A$ is a Abelian group of odd order and $B$ is a Boolean group, i.e., a group of exponent $\exp(B)\le 2$.
If the group $A$ and $B$ is trivial, then $|\mathrm{Poly}(X)|=|\mathrm{Poly}(Q\_8)|=128\ne |X|\cdot\exp(X)=32$.
Next, assume that the group $A$ is trivial and $B$ is not trivial. Then $|\mathrm{Poly}(B)|=|\{a,ax:a\in B\}|=2|B|$.
[GAP-calculations of Peter Taylor](https://gist.github.com/pjt33/b9e1e62291110b164e091ef7c2f7ead5) show that the group $Q\_8$ has exactly 32 polynomials of each degree $k\in\{1,2,3,4\}$. In particular, $Q\_8$ has exactly 64 polynomials of even degree and 64 polynomials of odd degree.
This implies that $|\mathrm{Poly}(X)|=64\cdot 2|B|=16|Q\_8\times B|=16|X|\ne 4|X|=|X|\cdot\exp(X)=|\mathrm{Poly}(X)|$. This contradiction shows that the group $A$ is nontrivial.
Taking into account that the group $Q\_8$ has exactly 32 polynomials of each degree $k\in\{1,2,3,4\}$, we conclude that $$|X|\cdot\exp(X)=|\mathrm{Poly}(X)|=|\mathrm{Poly}(Q\_8\times A\times B|=32\times|\mathrm{Poly}(A\times B)|=32\times |A\times B|\times \exp(A\times B)=4\times|Q\_8\times A\times B|\times \exp(A\times B)=4\cdot |X|\cdot\exp(A\times B)$$
and hence $\exp(Q\_8\times A\times B)=\exp(X)=4\exp(A\times B)$. Since $\exp(Q\_8\times A\times B)=4\exp(A),$ this implies that the Boolean group $B$ is trivial and hence $X=Q\_8\times A$. $\square$
| https://mathoverflow.net/users/61536 | The number of polynomials on a finite group | $\DeclareMathOperator\Poly{Poly}$**Proposition.** If $G$ is a simple non-abelian finite group, then $\Poly(G)=G^G$.
(Edit: this observation appears as the main therorem in [this paper](https://www.ams.org/journals/proc/1965-016-03/S0002-9939-1965-0175971-0/) by Maurer and Rhodes, Proc. AMS 1965. See also Theorem 2 [here](https://arxiv.org/abs/2206.11956) by Schneider-Thom. Thanks to Benjamin Steinberg for the reference.)
Here is the proof. It uses no machinery.
**Lemma.** There exists $f\in\Poly(G)$ whose support is a singleton.
[Here the support of $f$ means $f^{-1}(G\smallsetminus\{1\})$.]
Indeed, let $f$ have support $\{g\}$. Considering $x\mapsto hf(x)h^{-1}$ we see that all values in a single nontrivial conjugacy class are achieved by polynomials supported by $\{g\}$. By simplicity and taking products, we see that all maps supported by $g$ are definable as polynomials. Moreover after considering $x\mapsto f(gh^{-1}x)$ we obtain all functions supported by $\{h\}$. Since an arbitrary map is product of maps supported by singletons, we obtain the proposition.
Now let us prove the lemma. Let $X$ be a minimal subset among nonempty supports of elements of $\Poly(G)$ ($X$ exists because there exists a polynomial not constant $=1$). Say $X$ is the support of $f$. We have to show that $X$ is a singleton. Fix $g\in X$. So $u(x)=g^{-1}x$ is a polynomial. Also for each $h\in H$, the self-map $v$ defined $v(x)=hf(x)h^{-1}$ is a polynomial. Then $w\_h:x\mapsto [u(x),v(x)]$ is a polynomial as well. Its support is contained in $X\smallsetminus\{g\}$. So we obtain a contradiction (a strictly smaller nonempty support), unless $w\_h$ is constant equal to $1$ for each choice of $h$. The latter means that for each $x\in X\smallsetminus\{g\}$, the element $g^{-1}x$ commutes with $hf(x)h^{-1}$. That is, the nontrivial element $g^{-1}x$ commutes with a whole nontrivial conjugacy class. But the centralizer of a nontrivial conjugacy class is trivial (it is a normal subgroup, and can't be the whole group because the center is trivial). This is a contradiction unless $X\smallsetminus\{g\}$ is empty, which is precisely what we want. The proof is complete.
---
Remark (after Taras' comment, and also in the above Maurer-Rhodes reference): conversely, for a finite group $G$, the property $\Poly(G)=G^G$ implies that $G$ is simple non-abelian or $|G|\le 2$.
Indeed if $G$ is non-trivial and non-simple, then it has a non-trivial proper normal subgroup $N$, and polynomials have the nontrivial constraint $f(N)\subset f(1)N$.
Otherwise $G=\mathbf{Z}/p\mathbf{Z}$ for $p$ prime or $1$. For such a group, a "polynomial" has the form (using additive notation) $x\mapsto a+bx$ for some $a,b\in\mathbf{Z}/p\mathbf{Z}$ (i.e. is an affine self-map in this ring). There are thus $p^2$ such functions. And $p^2<p^p$ iff $p>2$.
| 22 | https://mathoverflow.net/users/14094 | 429635 | 174,078 |
https://mathoverflow.net/questions/429633 | 3 | We are given a convex shape $S$ lying inside the hypercube $[0,1]^d$ in the $d$-dimensional Euclidean space. Let the volume $V(S)$ of $S$ be $\tfrac12$ (*I guess nothing changes for any other fixed constant in $(0,1)$*).
---
**Question:** How can we prove or disprove that, for all $S$ and all $d\in\mathbb{N}$, there exists a set of points lying on its boundary such that their convex hull $C$ satisfy simultaneously the following properties?
* Its volume $V(C)$ is lower bounded by a constant independent of $d$.
* Its number of facets $\phi\_C$ grows at most polynomially in $d$.
---
---
***Note:*** *This problem is similar to [Approximation of a convex shape in the $d$-dimensional Euclidean space for $d\gg 1$](https://mathoverflow.net/q/429558/115803), but instead of bounding the number of vertices of the approximating convex polytope lying inside the given shape, we focus on bounding the number of its facets (the motivations are given below, at the end of the following paragraph).*
*The main intuition underlying this conjecture is the idea that (***very***) informally, for $d\gg 1$ if there are areas where the boundary of $S$ is far from being linear (e.g., portions of $(d-1)$-balls), the volume of the regions of $S$ bounded by such areas is disregardable compared to the volume of $S$. More generally, the same holds for areas formed by $\omega(\mathrm{poly}(d))$-many linear regions that can be viewed as good approximations of manifolds with significant curvature (in this context, such linear regions can be viewed as facets of an approximating polytope, which is conceptually linked to bounding $\phi\_C$ in this problem).*
| https://mathoverflow.net/users/115803 | Bounding the number of facets of a polytope to approximate a given convex shape in higher dimensions | I think ***if we assume the facets are simplexes***, the number of facets of such a polytope must grow more than exponentially, even in the easiest case where $S=[0,1]^d$. Fix a constant $\epsilon>0$.
Choose a finite subset of $\partial( [0,1]^d)$, spanning a polytope $C$ with $\phi\_C$ facets, and volume $V(C)\ge\epsilon$. We can partition $C$ into $\phi\_C$ simplexes $\{\Delta\_i\}\_{1\le i\le \phi\_C }$ with vertices in the center of $S$, so the volume of $C$ is not larger than $f$ times the largest volume of such simplexes. Parenthesis: the largest volume simplex included in a cube, say $[-1,+1]^d$, and with a vertex in the center $O$ of the cube, has wlog all its other vertices among the vertices of the cube: for we can move any vertex $v\neq O$ in the half-space whose boundary is the hyperplane for $v$, parallel to the facet of the simplex opposite to $v$, and disjoint to it, till we reach an extremal point of the cube (a vertex of the cube), and this does not decrease the volume. Thus the maximum volume among all simplexes included in $[-1,+1]^d$ with a vertex in the origin is reached by those with $d$ vertices in $\{-1,1\}^d$ plus $0$, and it is $1/d!$ times the maximum determinant of a $\{1,-1\}$ matrix of oder $d$: this is the [Hadamard's matrix problem](https://en.wikipedia.org/wiki/Hadamard%27s_maximal_determinant_problem#Hadamard%27s_bound_(for_all_n)), and the corresponding Hadamard determinant bound is $d^{d/2}$. Going back to the $\Delta\_i$ and to $[0,1]^d$, and normalizing over $V([-1,+1]^d=2^d$, we conclude that $V(\Delta\_i)\le {d^{d/2}}/({2^dd!})$ for all $i$, so $V(C)\le \phi\_C {d^{d/2}}/({2^dd!})$ so by Stirling formula $\phi\_C\ge \frac{2^dd!}{\epsilon d^{d/2}}\sim \sqrt{2\pi d}(4de^{-2})^{d/2} \epsilon^{-1}$ that grows more than exponentially.
| 3 | https://mathoverflow.net/users/6101 | 429639 | 174,079 |
https://mathoverflow.net/questions/429638 | 3 | The Manin-Mumford conjecture states that for an abelian variety A over a field F of characteristic 0 the torsion points are dense in an integral closed subvariety Z if and only if it is an abelian subvariety translated by a torsion element.
Both Raynaud's proof and the equidistribution proof prove this by reduction to the case of number fields. The hint as to why this works is "by a specialization argument". Let me now present my thoughts on this. I will now assume Manin-Mumford for number fields.
The abelian variety and its subscheme are certainly defined over some finite type Q-algebra S lying inside F. Denote the models of our given data by $\mathcal{A}$ and $\mathcal{Z}$. It suffices to show that the torsion points in $\mathcal{Z}$ aren't dense. Their specializations over any closed point are not dense by assumption. I don't know how to conclude from here. I want to study the failure of commutativity of intersecting the defining ideals and tensoring, ideally obtaining generic commutativity of tensoring with certain limits.
| https://mathoverflow.net/users/157701 | Why does the Manin-Mumford conjecture over number fields imply the conjecture over arbitrary fields of characteristic 0? | What you want to do is, by induction on the theorem in the number field $K$ case, prove that all torsion points in $\mathcal Z\_K$ lie in a finite union of torsion translates of abelian varieties (contained in $Z\_K$).
It follows that all torsion points in $\mathcal Z\_{\eta}$ specialize to points in a finite union of torsion translates of abelian varieties.
But since the specialization map is injective on torsion points, all torsion points in $\mathcal Z\_\eta$ lie in torsion translates of lifts of those abelian varieties. (We can ensure the abelian varieties list by choosing $K$ appropriately using Masser's specialization theorem, though this can't be quite what Raynaud did for timeline reasons.)
So all torsion points in $Z$ lie in finitely many torsion translates of abelian subvarieties, but these subvarieties may not be contained in $Z$. However, they have dimension at most the dimension of $Z$, so the only way the torsion points of $Z$ can be Zariski dense is if $Z$ is equal to one of these torsion translates, because otherwise their intersections have smaller dimension and are not dense.
Probably one can avoid Masser's theorem by thinking more carefully about what the lifts of torsion points on an abelian subvariety that doesn't lift look like.
| 5 | https://mathoverflow.net/users/18060 | 429640 | 174,080 |
https://mathoverflow.net/questions/429545 | 10 | Suppose $\langle L,\leq\rangle$ is a lattice with join $\sqcup$. Let $F\_1$ and $F\_2$ be principal filters on $L$. Thus, for $i\in I=\{1,2\}$ there are $x\_i\in L$ so that $F\_i=\{y\in L:x\_i\leq y\}$.
In this situation, $F\_1\cap F\_2$ is also a principal filter, because $F\_1\cap F\_2=\{y\in L:x\_1\sqcup x\_2\leq y\}$. This isn't too hard to see: clearly if $x\_1\sqcup x\_2\leq y$, then $x\_1\leq y$ (so $y\in F\_1$) and $x\_2\leq y$ (so $y\in F\_2$). Thus $\{y\in L: x\sqcup y\leq y\}\subseteq F\_1\cap F\_2$. On the other hand, if $y\in F\_1\cap F\_2$, then $x\_1\leq y$ (because $y\in F\_1$) and $x\_2\leq y$ (because $y\in F\_2$) and thus $x\_1\sqcup x\_2\leq y$. So $F\_1\cap F\_2\subseteq\{y\in L:x\_1\sqcup x\_2\leq y\}$.
Inductively, if $I$ is finite then $\bigcap\_{i\in I} F\_i$ is a principal filter. But what if $I$ is arbitrary? In general, of course, the intersection of an arbitrary family of principal filters might be empty (example: take the divisibility poset on $\mathbb{N}$, and let $[p]$ be the principal filter generated by a prime $p$. The intersection of all such filter is clearly empty, since no number is divisible by every prime). Ok, sure. But (now the actual question):
>
> provided each $F\_i$ is a principal filter and $\bigcap\_{i\in I} F\_i$ is nonempty, is it a principal filter?
>
>
>
In all the examples I can think of, ensuring the big intersection is nonempty essentially guarantees that the result is a principal filter. So e.g. in the lattice of subsets of $\mathbb{N}$, let $E$ be the set of all subsets of the even numbers, and let $F\_E$ be the set containing, for each $e\in E$, the principal filter generated by $e$. Then, lo and behold, the intersection of all members of $F\_E$ is the filter generated by $E$. And clearly if $L$ is complete, then the answer is an easy 'yes'. So I guess I'm really interested in what we can say *generally*, for any old lattice $L$.
| https://mathoverflow.net/users/25415 | Are arbitrary nonempty intersections of principal filters principal? | Nope. For a silly counterexample consider the lattice of nonzero real numbers with the usual order. Consider the principal filters $F\_i = \left\{y \ne 0: -\frac{1}{i} \le y \right\}$. Then $F=\bigcap\_i F\_i $ should be the principal filter generated by zero. But since zero is not in the lattice $F$ is the filter of positive elements. This is nonprincipal.
| 6 | https://mathoverflow.net/users/58082 | 429647 | 174,083 |
https://mathoverflow.net/questions/429620 | 0 | Let $K$ be a imaginary quadratic field, $R\_K$ be ring of integers of $K$, and $E/K$ be elliptic curve which has CM over $K$.
Let $\psi\_E$ be Hecke (Grössencharakter) character of $E/K$.
Let fix prime ideal $I＝(\pi)$ of $K$.
Then, why does $[I](P)＝0$ ($P\in E$) imply $[\psi(I)](P)＝0$?
If I could write $\psi(I)$ like $\psi(I)＝a\pi$ for some $a\in R\_K$,
$[\psi(I)](P)＝[a][\pi](P)＝0$, but I'm not confident.
| https://mathoverflow.net/users/144623 | Why does $[I](P)＝0$ ($P\in E$) imply $[\psi(I)](P)＝0$ ? ($\psi$ is Hecke character of elliptic curve) | This is essentially the same as your other recent CM-theory question, in a mild disguise; for both questions the point is that ***$\psi(I)$ is a generator of $I$.*** This follows easily from the fact that $\psi$ takes values in $K$, and for principal ideals $(\lambda)$ with generators sufficiently congruent to 1 we have $\psi(\ (\lambda)\ ) = \lambda$.
Perhaps an example would be helpful. Consider the elliptic curve of conductor 49, with Weierstrass equation $y^2 + xy = x^3 - x^2 - 2x - 1$. (This is 49a1 in Cremona's tables, and 49a4 in the [LMFDB](https://beta.lmfdb.org/EllipticCurve/Q/49/a/4); it also happens to be the modular curve $X\_0(49)$, but that isn't important here.)
This curve has CM by the ring of integers $R\_K = \mathbb{Z}[(1 + \sqrt{-7})/2]$ of $\mathbb{Q}(\sqrt{-7})$, and one can write down the corresponding Groessencharacter explicitly:
>
> For $\mathfrak{a}$ any integral ideal of $R\_K$ coprime to the ideal $\mathfrak{f} = \sqrt{-7} R\_K$, the value $\psi(\mathfrak{a})$ is the unique generator of $\mathfrak{a}$ whose image in $(R\_K / \mathfrak{f})^\times \cong (\mathbb{Z} / 7\mathbb{Z})^\times$ is a square.
>
>
>
Note this makes sense, because $R\_K$ has class number 1 and its only units are $\pm 1$; so every $\mathfrak{a}$ has exactly two generators, say $x$ and $-x$, and as $-1$ is not a square mod 7, precisely one of these is a square.
| 2 | https://mathoverflow.net/users/2481 | 429649 | 174,084 |
https://mathoverflow.net/questions/429598 | 3 | Let $V$ be a Riemann surface, $x\in V$, and $B:=B(x,r)$ some small ball (in a local chart). It is well known that there is a meromorphic function $f$ on $V$ with the only pole at $x$. What I’d like to ask of is if there is a meromorphic f on V that has a pole at x and additionally such that $|f|<1$ outside $B$?
| https://mathoverflow.net/users/138007 | Meromorphic function on the Riemann surfaces | For open surfaces, there are counterexamples. The first one was constructed by P. Myrberg:
Ueber die analytische Fortsetzung von beschrankten Funktionen, Ann. Acad. Sci. Fenn., Ser. A. I N:o 58 (1949)
Since this paper is difficult to obtain (and written in German), I refer to another paper
Heins, Maurice,
Riemann surfaces of infinite genus.
Ann. of Math. (2) 55 (1952), 296–317,
Which proves an even stronger result: there is an open Riemann surface, such that if you remove a disk from it, then on the remaining surface every non-constant meromorphic function takes all complex values, except at most two of them.
| 7 | https://mathoverflow.net/users/25510 | 429653 | 174,085 |
https://mathoverflow.net/questions/422885 | 1 | Let $\overline L= (L, h)$ be a hermitian $C^
\infty$ line bundle on an arithmetic variety $X\to\operatorname{Spec }\mathbb Z$ (I am reasoning in terms of higher Arakelov geometry, like in Gillet & Soule' papers).
$\overline L$ is said to be arithmetically ample if:
1. $L$ is relatively ample on $X$
2. $L\_{\mathbb C}$ is positive on $X\_{\mathbb C}$
3. A power of $L$ is generated by *small sections*.
Now assume that $L$ is relatively ample. It is not difficult to show that there exist a scaling $\alpha h$, with $\alpha\in\mathbb R\_{>0}$, such that $\overline L\_{\alpha}=(L,\alpha h)$ is arithmetically ample.
Then I ask the following:
>
> Are there some $X$(maybe different from $\mathbb P^n\_{\mathbb Z}$) and $L$ such that $\overline L\_{\alpha}$ is
> arithmetically ample for **any** $\alpha>0$.
>
>
> In other words, is it possible to find an arithmetically ample line
> bundle such that with **any** other scaling of the metric, it remains
> arithmetically ample?
>
>
>
It seems to be "just" a problem regarding successive minima in the theory of lattices... If I try to picture it in my mind, it seems impossible to find a linde bundle with such properties; there should be a lower bound $\alpha\_0>0$ such that below that value it is impossible to find generating small sections for the powers of $\overline L\_{\alpha}$.
| https://mathoverflow.net/users/65980 | Arithmetic ampleness and scalings of the metric | Let $\overline{L}$ be any arithmetically ample line bundle. In the way you have written it down, $\overline{L}\_{\alpha}$ is not arithmetically ample for $\alpha$ sufficiently large, more precisely for $\alpha\ge\alpha\_0=\exp\left(\frac{2\overline{L}^{\dim X}}{L\_{\mathbb{C}}^{\dim X-1}}\right)$:
Let $s\in H^0(X,L^{\otimes n})$ be any global section of any $n$-th tensor power of $L$. Then we have the following equality of arithmetic intersection numbers
$$n\overline{L}^{\dim X}=\overline{L}\cdot\ldots\cdot\overline{L}\cdot\overline{L}^{\otimes n}=\left(\overline{L}|\_{\mathrm{div}(s)}\right)^{\dim X-1}-\int\_{X(\mathbb{C})}\log |s|c\_1\left(\overline{L}\right)^{\dim X-1}.$$
By the arithmetic ampleness $\left(\overline{L}|\_{\mathrm{div}(s)}\right)^{\dim X-1}\ge 0$. Thus
$$\int\_{X(\mathbb{C})}\log |s|c\_1\left(\overline{L}\right)^{\dim X-1}\ge -n\overline{L}^{\dim X}.$$
Let us write $|s|\_{\alpha}$ for the norm of $s$ with respect to $\overline{L}\_{\alpha}^{\otimes n}$. As $|s|=\sqrt{h^{\otimes n}(s,s)}$, this means $|s|\_{\alpha}=\sqrt{\alpha^n h^{\otimes n}(s,s)}=\alpha^{n/2}|s|$. Thus
\begin{align\*}
\int\_{X(\mathbb{C})}\log |s|\_{\alpha}c\_1\left(\overline{L}\right)^{\dim X-1}&=\frac{n}{2}\log \alpha\int\_{X(\mathbb{C})}c\_1\left(\overline{L}\right)^{\dim X-1}+\int\_{X(\mathbb{C})}\log |s|c\_1\left(\overline{L}\right)^{\dim X-1}\\
&\ge \frac{n}{2}(\log \alpha) L\_{\mathbb{C}}^{\dim X-1}-n\overline{L}^{\dim X}\\
&=\frac{n}{2}\left((\log\alpha)L\_{\mathbb{C}}^{\dim X-1}-2\overline{L}^{\dim X}\right)
\end{align\*}
If $\alpha\ge\alpha\_0=\exp\left(\frac{2\overline{L}^{\dim X}}{L\_{\mathbb{C}}^{\dim X-1}}\right)$, then
$$\log\sup|s|\_{\alpha}\ge\frac{\int\_{X(\mathbb{C})}\log |s|\_{\alpha}c\_1\left(\overline{L}\right)^{\dim X-1}}{\int\_{X(\mathbb{C})}c\_1\left(\overline{L}\right)^{\dim X-1}}\ge0.$$
Thus, $\sup|s|\_{\alpha}\ge 1$ such that $s$ is not strictly small with respect to $\overline{L}\_{\alpha}^{\otimes n}$. As $s$ and $n$ were arbitrary, $\overline{L}\_{\alpha}^{\otimes n}$ does not have any strictly small global sections for any $n$. In particular, $\overline{L}\_{\alpha}$ is not arithmetically ample for every $\alpha\ge\alpha\_0$. Note, that by the ampleness of $L\_{\mathbb{C}}$ we always have $L\_{\mathbb{C}}^{\dim X-1}>0$ such that $\alpha\_0$ is always well-defined.
| 1 | https://mathoverflow.net/users/61532 | 429661 | 174,087 |
https://mathoverflow.net/questions/429648 | 6 | Let $ t>0 $, and we look at the random walk $S\_{n}=\sum\_{i=1}^{n}X\_{n}$ on $\mathbb{Z}$ with $S\_0=0$ where $$ \mathbb{P}\left(X\_{n}=1\right) =\frac{1}{2}\left(1+\frac{1}{n^{t}}\right)
$$ $$ \mathbb{P}\left(X\_{n}=-1\right) =1-\mathbb{P}\left(X\_{n}=1\right)=\frac{1}{2}\left(1-\frac{1}{n^{t}}\right)$$We would like to know when this walk is recurrent and when it is transient, depending on $t$. $S\_{n}$ is not a martingale, and we have $ \mathbb{E}\left(X\_{n}\right)=\frac{1}{n^{t}}$. so we have $ \mathbb{E}\left(S\_{n}\right)=\sum\_{i=1}^{n}\frac{1}{i^{t}}$ which might look look like something we can apply series convergence to, but I didn't achieve much with this approach.
I tried to bound $ \mathbb{P}\left(S\_{2n}=0\right)$: $$ \mathbb{P}\left(S\_{2n}=0\right) =\sum\_{s\in S}\left(\prod\_{i=1}^{2n}\mathbb{P}\left(X\_{i}=s\_{i}\right)\right)
\leq{2n \choose n}\left(\prod\_{i=1}^{n}\mathbb{P}\left(X\_{i}=1\right)\right)\left(\prod\_{i=n+1}^{2n}\mathbb{P}\left(X\_{i}=-1\right)\right)$$
(Where $ S $ is the set of all arrangements of $n$ $-1$s and $n$ $1$s) But this seems like a pretty bad bound, since the sum of it diverges to $\infty $. We also know that $ \mathbb{P}\left(X\_{1}=1\right)=1$, so we can change ${2n \choose n}$ to ${2n-1 \choose n}$, but this doesn't achieve much either. (and anyway, the above bound isn't that easy to work with)
My current attempt is to bound $ \mathbb{P}\left(S\_{2n}=0\right)$ from above with $ \mathbb{P}\left(S\_{2n}=0\right) \leq p\_n$, with $\sum\_{n=1}^{\infty}p\_{n}<\infty$ which implies $\sum\_{i=1}^{\infty}\mathbb{P}\left(S\_{2i}=0\right)<\infty$ and then using Borel–Cantelli lemma we will get $\mathbb{P}\left(S\_{2i}=0\quad\text{i.o.}\right)=0$. But all of my attempts to get a bound which is (1) convergent (2) easy to work with, have failed.
| https://mathoverflow.net/users/490586 | Parameterized simple asymmetric random walk | Recall that a random walk (or a Markov chain in general) is called recurrent if it almost surely (a.s.) returns to the initial state infinitely often.
We will show that in our case the walk is recurrent iff $t\ge1/2$.
The key here is the law of the iterated logarithm. Indeed, according to (say) [Theorem 1 of Chapter X](https://link.springer.com/book/10.1007/978-3-642-65809-9) (see the statement of this theorem reproduced at the end of this answer),
\begin{equation\*}
\begin{aligned}
&\limsup\_n\frac{S\_n-ES\_n}{\sqrt{2n\ln\ln n}}=1,\\
&\liminf\_n\frac{S\_n-ES\_n}{\sqrt{2n\ln\ln n}}=-1
\end{aligned}
\tag{1}\label{1}
\end{equation\*}
a.s.
Also, if $t\ge1/2$, then
\begin{equation\*}
ES\_n=\sum\_{i=1}^n EX\_i=\sum\_{i=1}^n 1/i^t=O(n^{1/2})=o(\sqrt{2n\ln\ln n}),
\end{equation\*}
(as $n\to\infty$), so that, by \eqref{1},
\begin{equation\*}
\begin{aligned}
&\limsup\_n\frac{S\_n}{\sqrt{2n\ln\ln n}}=1,\\
&\liminf\_n\frac{S\_n}{\sqrt{2n\ln\ln n}}=-1
\end{aligned}
\end{equation\*}
a.s., which implies that the walk is recurrent (because it cannot jump over $0$: If $S\_k<0<S\_m$ for some natural $k$ and $m$, then $S\_j=0$ for some natural $j$ between $k$ and $m$).
If now $t<1/2$, then
\begin{equation\*}
ES\_n=\sum\_{i=1}^n EX\_i=\sum\_{i=1}^n 1/i^t\asymp n^{1-t},
\end{equation\*}
so that, by \eqref{1}, $S\_n-ES\_n=o(ES\_n)$ a.s. and hence $S\_n\sim ES\_n\to\infty$ a.s., which implies that the walk is not recurrent.
Thus, the walk is recurrent iff $t\ge1/2$. (Basically, it all depends on how $n^{1-t}$ asymptotically compares with $\sqrt{n\ln\ln n}$)
---
**The formulation of the law of the iterated logarithm used in this answer:**
>
> Let $Y\_1,Y\_2,\dots$ be independent zero-mean random variables with $s\_k^2:=EY\_k^2<\infty$ for all $k$. Let $T\_n:=\sum\_{k=1}^n Y\_k$ and $B\_n:=\sum\_{k=1}^n s\_k^2$. Suppose that $B\_n\to\infty$ and for some sequence $(M\_n)$ of positive constants we have $|Y\_n|\le M\_n$ for all $n$ and
> \begin{equation\*}
> M\_n=o\Big(\sqrt{\frac{B\_n}{\ln\ln B\_n}}\Big).
> \end{equation\*}
> Then
> \begin{equation\*}
> \begin{aligned}
> &\limsup\_n\frac{T\_n}{\sqrt{2B\_n\ln\ln B\_n}}=1
> \end{aligned}
> \end{equation\*}
> a.s.
>
>
>
We used this result for $Y\_n:=X\_n-EX\_n$.
| 5 | https://mathoverflow.net/users/36721 | 429665 | 174,090 |
https://mathoverflow.net/questions/428994 | 8 | This is a soft question, I guess. $\Gamma$-convergence is a notion of convergence of functionals so that if $F\_n$ $\Gamma$-converges to $F$, then cluster points of $\arg\inf F\_n$ are minimizers of $F$. This is especially helpful if you want to minimize $F$ but find it easier to minimize $F\_n$.
However, if you look at the definition of [$\Gamma$-convergence](https://en.wikipedia.org/wiki/%CE%93-convergence) it becomes pretty untenable to prove in complicated settings. It requires you to perform 4 optimizations, and take two limits. In some simple cases one can prove $\Gamma$-convergence "by hand" but I was wondering - what is the state of the art for proving $\Gamma$-convergence in a complicated setting? I know that it is equivalent to convergence of the epi-graph, and also in first countable spaces there is an equivalent definition, that is still pretty hard to show.
>
> What techniques exist to prove $\Gamma$-convergence?
>
>
>
I know you can add and subtract continuous functions.
| https://mathoverflow.net/users/479223 | How do people prove $\Gamma$-convergence in more complicated settings? | I am mostly familiar with the simpler definition ("Definition in first-countable spaces" from the Wikipedia link):
>
> Given the functionals $F\_\varepsilon, F: X \to \overline{\Bbb{R}}$
> (indexed for $\varepsilon>0$, we say that $F\_\varepsilon$
> $\Gamma$-converges to $F$ if the two properties hold:
>
>
> (LI) For every $u \in X$ and every $u\_\varepsilon \to u$ we have
> $\liminf\_{\varepsilon \to 0} F\_\varepsilon(u\_\varepsilon) \geq F(u)$.
>
>
> (LS) For every $u \in X$ there exists $u\_\varepsilon \to u$ such that
> $\limsup\_{\varepsilon \to 0} F\_\varepsilon(u\_\varepsilon) \leq F(u)$.
>
>
>
The techniques used to prove $\Gamma$-convergence in this case vary with the problem studied.
* Generally (LI) comes from some lower-semicontinuity result (see some examples below)
* (LS) is easy if the constant sequence can be chosen. However, in some cases, $F\_\varepsilon$ and $F$ can have *different domains* in the sense that $F$ is $+\infty$ for any $u$ such that $F\_\varepsilon(u)$ is finite. In such cases, *recovery sequences* realizing the equality $\lim\_{\varepsilon \to 0}F\_\varepsilon(u\_\varepsilon)=F(u)$ need to be constructed by hand.
* the (LS) is usually proved on a dense subset of $X$ verifying additional regularity properties, facilitating the construction of the *recovery sequences* described above.
* the (LS) property is sometimes proved using a *slicing procedure*, i.e. the result is somehow reduced to the dimension one by considering the whole space as a product between a manifold of dimension $N-1$ and one dimensional spaces.
**Examples and references**
1. The first example I knew was the Modica-Mortola theorem. It gives an approximation of the perimeter using energies defined for smoother density functions. Consider $F, F\_\varepsilon : L^1(D) \to \overline{\Bbb{R}}$:
$$ F\_\varepsilon(u) = \begin{cases}
\int\_D (\varepsilon |\nabla u|^2 +\frac{1}{\varepsilon}u^2(1-u)^2) & \text{ if } u \in H^1(D) \\
+\infty & \text{ otherwise}
\end{cases}
$$
and
$$ F(u) = \begin{cases}
\frac{1}{3} \text{Per}(E\_1) & \text{ if } u \in BV(D,\{0,1\}), E\_1 = u^{-1}(\{1\}) \\
+\infty & \text{ otherwise}
\end{cases}
$$
Then $F\_\varepsilon$ $\Gamma$-converges to $F$. See [my blog](https://mathproblems123.wordpress.com/2012/03/31/modica-mortola-theorem/) for more details and a proof. This is an example where the constant sequence cannot be chosen as a recovery sequence in (LS). Characteristic functions of finite perimeter sets are not in $H^1$. The (LI) part comes from the lower semicontinuity of the total variation for $BV$ functions (abstract, but standard result).
2. A similar setting can be used for the total perimeter. Notice however that this is not a consequence of the Modica-Mortola theorem, since $\Gamma$-convergence is not stable for the sum. This kind of argument was used for approximating numerically partitions minimizing the total perimeter in the [following article](https://projecteuclid.org/journals/experimental-mathematics/volume-20/issue-3/Approximation-of-Partitions-of-Least-Perimeter-by-%CE%93-Convergence/em/1317924419.full). Similar approaches can be extended to Cheeger clusters, again with numerical implications. See [this link](http://www.cmap.polytechnique.fr/%7Ebeniamin.bogosel/cheeger.html). Reading these articles could given more insight about difficulties regarding proofs of properties (LI) or (LS) above. More references in this sense exist for finding Steiner trees, minimal surfaces and much more (take a look at publications related to $\Gamma$-convergence on E. Bretin's [webpage](http://math.univ-lyon1.fr/%7Eebretin/publication.html)).
3. More generally, the books of Andrea Brides on the subject are a must read. They go into more depth regarding the techniques involved. Consider for example: *Gamma-convergence for beginners* by A. Braides, *Approximation
of Free-Discontinuity Problems* again by A. Braides. And this is just scratching the surface.
I hope this answer helps giving some insights regarding $\Gamma$-convergence results in some basic cases.
| 4 | https://mathoverflow.net/users/13093 | 429666 | 174,091 |
https://mathoverflow.net/questions/429671 | 1 | Recall that a [preadditive category](https://en.wikipedia.org/wiki/Preadditive_category) is just a category $\mathcal{C}$ enriched in the category of abelian groups such that composition is linear with respect to the various group operations, so $$f\circ(g+h)=f\circ g+f\circ h,$$ $$(g+h)\circ f=g\circ f+h\circ f,$$ and a preadditive category with one object is a categorification of the notion of a noncommutative ring where we interpret the ring addition as the addition inherited from enrichment and multiplication as composition of morphisms.
>
> Is a preadditive groupoid with one object a categorification of the notion of a skew field, and if so where can I read more about it?
>
>
>
Naively it seems like we now have additive inverses and multiplicative inverses for each object; we only seem to be missing multiplicative commutativity to categorify fields. This leads to a followup question,
>
> Is there a notion of a category where composition of endoarrows is always commutative? That is, a category $\mathcal{C}$ such that for all endoarrows $f,g:X\rightrightarrows X$ we have $f\circ g=g\circ f$?
>
>
>
This seems like a bad definition since it involves equality, so perhaps a better request is a category $\mathcal{C}$ such that composition of endoarrows is commutative up to isomorphism in a (co?)slice category, with the above notion recovered as the 'strict' version.
| https://mathoverflow.net/users/92164 | One object preadditive groupoids as a categorification of skew fields | There aren't any nontrivial preadditive groupoids; a preadditive category always has zero morphisms, and if zero morphisms are invertible then every object is a zero object.
If you think of rings as one-object preadditive categories, then commutative rings can be thought of as one-object preadditive monoidal categories, by the Eckmann-Hilton argument. This is a more natural way to enforce commutativity since it's related to thinking of monoidal categories themselves as one-object $2$-categories and delooping etc.
| 3 | https://mathoverflow.net/users/290 | 429673 | 174,093 |
https://mathoverflow.net/questions/429374 | 2 | While looking at an analogue of Pontryagin duality for compact Discrete Valuation Rings (DVRs), I came about the observation that generally one *should* have an isomorphism of $A$-modules
$$\hom\_{\mathbb{Z}}\left(A,\mathbb{R}/\mathbb{Z}\right)\xrightarrow{\sim} K/A,$$
where $A$ is a DVR that is compact with respect to its canonical topology, and $K$ is its field of fractions. The left hand side consists of continuous $\mathbb{Z}$-module homomorphisms from $A$ with its canonical metric topology to $\mathbb{R}/\mathbb{Z}$ with its usual topology. The $A$-module structure on the left is given by $(a\cdot \varphi)(x)=\varphi(a\cdot x)$.
For DVRs, being compact is equivalent to being complete and having finite residue field. This means that if the characteristic of $A$ is equal to the characteristic of its residue field then $A\cong \mathbb{F}\_q[[t]]$ for some finite field $\mathbb{F}\_q$. In this case, we can choose an $\mathbb{F}\_q$-module isomorphism $i:\hom\_{\mathbb{Z}}(\mathbb{F}\_q,\mathbb{R}/\mathbb{Z})\xrightarrow{\sim}\mathbb{F}\_q$ and define
\begin{align\*}
\hom\_{\mathbb{Z}}(\mathbb{F}\_q[[t]],\mathbb{R}/\mathbb{Z})&\xrightarrow{} \mathbb{F}\_q((t))/\mathbb{F}\_q[[t]].\\
\varphi&\mapsto \sum\_{n=0}^{\infty}i\left(\left.\varphi\right|\_{\mathbb{F}\_q \cdot x^n}\right) x^{-(n+1)}
\end{align\*}
A tedious check of all the desired properties shows that this is an isomorphism of $A$-modules. One key step is the fact that $\varphi$ being continuous is equivalent to $\varphi$ acting by $0$ on $\mathbb{F}\_q\cdot x^{n}$ for all large enough $n$, and hence giving a well defined element of $\mathbb{F}\_q((t))/\mathbb{F}\_q[[t]]$.
I now want to extend this result to the mixed characteristic case. I'm guessing that the correct first step is to do this when $A\cong W(\mathbb{F}\_q)$ is the space of Witt vectors over its (finite) residue field. In particular, we can write $K/\mathbb{Q}\_p$ to be the unique absolutely unramified extension of degree $\log\_p(q)$ of $\mathbb{Q}\_p$. My guess for defining a good map would be to use the fact that there is a multiplicative section $\omega:\mathbb{F}\_q^{\times} \xrightarrow{} A^{\times}$ to the reduction map $A^{\times}\xrightarrow{} \mathbb{F}^{\times}\_q$ given by sending any element $g\in \mathbb{F}\_q^{\times}$ to $\lim\_{n\to\infty}\left(\tilde{g}\right)^{q^n}$, where $\tilde{g}$ is any lift of $g$ to $A$. Namely, we construct
\begin{align\*}
\hom\_{\mathbb{Z}}\left(A,\mathbb{R}/\mathbb{Z}\right)&\xrightarrow{\ell} K/A.\\
\varphi &\mapsto \sum\_{g\in \mathbb{F}\_q^{\times}}\varphi(g)\cdot g^{-1}
\end{align\*}
This map is well defined since $\varphi(g)$ must lie in the Prüfer $p$-group $\mathbb{Z}[1/p]/\mathbb{Z}$, since
$$\lim\_{n\to\infty} p^{n}\varphi(g)=\varphi\left(\lim\_{n\to\infty}p^n\cdot g\right)=0.$$
We can then naturally identify the groups $\mathbb{Z}[1/p]/\mathbb{Z}$ and $\mathbb{Q}\_p/\mathbb{Z}\_p$ and thus make sense of the expression $\sum\_{g\in \mathbb{F}\_q^{\times}}\varphi(g)\cdot g^{-1}$. Now, we check that $\ell$ is an $A$-module homomorphism. Since $p$ is a uniformizer, we have that $A$ is (topologically) generated by elements of the form $h\cdot p^n$ where $h\in \mathbb{F}\_q^{\times}$. On elements of this form we see that
\begin{align\*}
\ell\left((h\cdot p^n)\cdot \varphi\right)&=\sum\_{g\in \mathbb{F}\_q^{\times}}\varphi\left(\left(h\cdot p^n\right)\cdot g\right)\cdot g^{-1}\\
&=(h\cdot p^n)\cdot \sum\_{g\in \mathbb{F}\_q^{\times}}\varphi\left(hg\right)\cdot (hg)^{-1}\\
&=(h\cdot p^n)\cdot \ell(\varphi).
\end{align\*}
Since $\ell$ clearly preserves addition, a short continuity argument shows we are done. Thus, what remains is this:
>
> Is the above map $\ell$ bijective?
>
>
>
I have no idea why this map is injective. If $\ell(\varphi)\neq 0$ for some element $\varphi$ though, I can deduce that the map is surjective. This is done as follows. Given any $\mathbb{Z}$-module morphism $\varphi: A \xrightarrow{} \mathbb{R}/\mathbb{Z}$ I claim there is at least one $\mathbb{Z}$-module morphism $\varphi': A \xrightarrow{} \mathbb{R}/\mathbb{Z}$ such that $\varphi=p\cdot \varphi'$. From this subjectivity is obvious, since applying the claim to $\varphi$ repeatedly gives elements in the image of $\ell$ with arbitrarily low valuations and then multiplying by elements of $A$ and using the fact that $\ell$ is an $A$-module morphism gives the full codomain as the image.
To prove that such a map $\varphi'$ always exists, we do as follows. We define $\varphi'$ on $p\cdot A$ by $\varphi'(p\cdot x)=\varphi(x)$. By Baer's criterion, $\mathbb{R}/\mathbb{Z}$ is an injective object in the category of $\mathbb{Z}$-modules. This means we can lift the map $\varphi'$ to the full domain $A$. Carrying around slightly more data we can ensure that $\varphi'$ will be continuous, and hence we conclude the result.
Thank you!
| https://mathoverflow.net/users/159298 | Why is the natural map $\hom(A,\mathbb{R}/\mathbb{Z})\to K/A$ an isomorphism, $K/\mathbb{Q}_p$ unramified, $A=\mathcal{O}_K$? | This answer does not prove that the map described in the question is an isomorphism, but it does prove that an isomorphism between $\hom\_{\mathbb{Z}}(A,\mathbb{R}/\mathbb{Z})$ and and $K/A$ exists.
To begin, we note that $M \mapsto \hom\_{\mathbb{Z}}(A,M)$ is a right adjoint to the forgetful functor from the category of (locally compact) topological $A$-module to the category of (locally compact) topological groups. Hence, since $\mathbb{R}/\mathbb{Z}$ is a cogenerator in the target category we get that $\hom\_{\mathbb{Z}}(A,\mathbb{R}/\mathbb{Z})$ is a cogenerator in the category of locally compact topological groups and thus there exists a nonzero morphism $f:K/A\xrightarrow{} \hom\_{\mathbb{Z}}(A,\mathbb{R}/\mathbb{Z})$. Moreover, by factoring we get an injective map
$$f':(K/A)/\ker(f)\hookrightarrow{}\hom\_{\mathbb{Z}}(A,\mathbb{R}/\mathbb{Z}).$$
Submodules of $K/A$ are all of the form $\ker\_{n}=\left\{x\,\,\mathrm{s.t}\,\, \pi^n x=0\right\}$. multiplication by $\pi^n$ gives an isomorphism
$$K/A\xrightarrow{\sim} (K/A)/\ker\_n,$$
and hence we can consider $f'$ as a morphism $K/A\hookrightarrow{} \hom\_{\mathbb{Z}}(A,\mathbb{R}/\mathbb{Z})$. Both modules are clearly equal to their $p^{\infty}$-torsion. In particular, we can compare the size of their $p^n$-torsion for each $n$. Once we establish that they have the same number of elements for each $n$, then infectivity will imply surjectivity onto each $p^n$ and hence surjectivity onto the full module so we will be done.
Given any map $\varphi$ in $\hom\_{\mathbb{Z}}(A,\mathbb{R}/\mathbb{Z})$, every element $x$ in the image of $\varphi$ must have the property that $\lim\_{n\to\infty}p^n x$ goes to zero. The only such elements in $\mathbb{R}/\mathbb{Z}$ are those in the Prüfer $p$-group $\mathbb{Z}[1/p]/\mathbb{Z}$, and hence we get that
$$\hom\_{\mathbb{Z}}(A,\mathbb{R}/\mathbb{Z})=\hom\_{\mathbb{Z}}(A,\mathbb{Z}[1/p]/\mathbb{Z}).$$
Now, every module of a DVR is free and hence we have an (I believe topological) $\mathbb{Z}\_p$-module isomorphism $A\xrightarrow{} \bigoplus \mathbb{Z}\_p$ which commutes with the $\hom$. Pairing $\varphi: \mathbb{Z}\_p\xrightarrow{} \mathbb{Z}[1/p]/\mathbb{Z}$ with $\varphi(1)$ and checking the module structure, we get an isomorphism between $\hom(\mathbb{Z}\_p,\mathbb{Z}[1/p]/\mathbb{Z})$ and $\mathbb{Q}\_p/\mathbb{Z}\_p$. Counting $p$-torsion we get the correct number of elements and hence we conclude the result.
I am fairly sure the argument as stated here checks out, at least in the unramified case. I will be coming back in the next couple of days to flesh it out more.
| 0 | https://mathoverflow.net/users/159298 | 429675 | 174,094 |
https://mathoverflow.net/questions/291293 | 7 | Let $G$ be a finitely generated group and $S$ a finite generating set and consider the word metric associated to $S$.
If $g\in G$, define its stable translation length as $l(g)=\lim\_n \frac{d(e,g^n)}{n}$.
This number can actually be defined in a more general context: if $G$ acts by isometries on a set $X$, define $l(g)=\lim\_n \frac{d(x,g^n\cdot x)}{n}$ and this do not depend on the point $x$, but we restrict our attention to a word metric in the following.
If $G$ is hyperbolic, then there exists $C\in \mathbb{R}$, such that for every $g\in G$, $l(g)\in C\mathbb{Z}$.
My question is the following: are there examples of groups not satisfying this property for the word metric ? More precisely, fixig a word metric on a group $G$, can we find two elements $g,h\in G$ such that $l(g)$ and $l(h)$ are arbitrarily close ? (settled, see the comment of YCor below).
I am specially interested with hyperbolic elements in relatively hyperbolic groups, so another related question is the following: If $G$ is relatively hyperbolic, can one find two hyperbolic elements $g,h$ such that $l(g)$ and $l(h)$ are arbitrarily close ?
As noticed by YCor, it would be enough to find either a loxodromic element with irrational translation length, or to find a relatively hyperbolic group with loxodromic elements of rational translation length but arbitrarily large denominator.
| https://mathoverflow.net/users/111917 | Rational stable translation length | As already mentioned in the comments, there exist finitely generated groups having non-discrete translation spectra. (There seems to be only a few examples though, it would be interesting to have more.) About relatively hyperbolic groups, we have the following statement:
>
> **Proposition.** *Let $G$ be a relatively hyperbolic group. There exists some $\epsilon>0$ such that the translation number of every hyperbolic element is a rational $>\epsilon$.*
>
>
>
The fact that the translation number is rational is actually quite general. See in [my other answer here](https://mathoverflow.net/a/429689/122026) for the sketch of a proof that Morse elements in finitely generated groups have rational translation numbers. In order to bound below the translation number, we can consider the action of $G$ on the hyperbolic space $X$ obtained by coning-off $G$ over its parabolic subgroups, i.e. $X$ is the graph obtained from (the Cayley graph of) $G$ by adding an edge between any two vertices that belong to a common parabolic subgroup. The action of $G$ on $X$ is acylindrical and an element of $G$ is loxodromic in $X$ if and only if it is hyperbolic in $G$. Therefore, the translation length in $X$ of a hyperbolic $g \in G$ cannot be arbitrarily small. (See for instance Lemma 2.1 in Fujiwara's article *Subgroups generated by two pseudo-Anosov elements in a mapping class group I*.) But the obvious map $G \to X$ is $1$-Lipschitz, so the translation length of $g$ in $X$ is at most the translation number of $g$ in $G$. The desired conclusion follows.
| 1 | https://mathoverflow.net/users/122026 | 429696 | 174,099 |
https://mathoverflow.net/questions/429642 | 5 | I've only recently learned about Girard's theory of Dilators and Ptykes, and I find this theory very elegant, but it is not clear at all to me whether/how it can be used to produce ordinal notations for all the large recursive ordinals used in proof theory and ordinal analysis. The introduction of several papers on the topic seems to claim it is possible - but I can't find it done anywhere...
I think I can see how Dilators alone can be used to construct (something very similar to) the Veblen functions in (possibly infinitely) many variables and how to get ordinal notations up to maybe the large Veblen ordinal, or a little higher than this.
However - and I have no idea how to make this formal - these constructions feel very "predicative" and my intuition would be that there is some kind of limit to what we can build using these only, I would guess around the Bachmann-Howard ordinal... But maybe I'm wrong and there is a way to formalize something similar to the ordinal collapsing functions using this theory?
General Ptykes on the other hand, feel much more mysterious to me, and I'm not sure how they can be used - I find the literature on the topic doesn't provide many examples - but they do seem more powerful, so I wouldn't be surprised if they could go as high (and probably higher) than everything we get using ordinal collapsing functions... But I don't know how.
Basically, I'd be interested by any reference or answer that gives Dilator/Ptykes based description of Ordinals notations up to and above the Bachman-Howard ordinal. I'd be also interested in results that gives limitations to such methods...
| https://mathoverflow.net/users/22131 | How to build large recursive ordinals using Dillator and/or Ptykes? | There are at least two fundamentally different ways how one could reach Bachmann-Howard ordinal using dilator and ptykes.
One way is to allow recursion on ordinals for ptykes for all finite types. Then the supremum of all ordinals describable in this way will be exactly B-H ordinal. See outline of ptyx interpretation of modified Gödel's system $T$ in the draft of Girard's book about ptykes [Section 12.A;1]. I believe that the details of computations that this leads to B-H ordinal had been carried out by Päppinghaus, but I never looked into this details myself [2]. In a sense this approach is just a generalization of "predicative" ordinal notation systems to finite types.
The other way is via the recursion on dilators. The intuition is that we want to effectively define set-sized objects, e.g. dilators, by a recursion along a class-sized well-ordering $D(\mathsf{Ord})$, where $D$ is some dilator. This is basically what Girard's functor $\Lambda\colon \mathsf{Dil}\to \mathsf{Dil}$ [Section 9.6,1] does. It is possible to reach B-H ordinal by applying $\Lambda$ to a certain fairly tame dilator. Girard in [1] is proving that the fact that $\Lambda$ maps dilators to dilators is equivalent to $\Pi^1\_1\textsf{-}\mathsf{CA}\_0$. And he is using $\Lambda$-like functors to perform cut elimination in certain functorial proofs for systems of positive inductive definitions.
In our recent paper [3] Juan Aguilera, Andreas Weiermann and I defined a functor $B\colon \mathsf{Dil}\to\mathsf{Dil}$ that is very similar to $\Lambda$, but I think that our definition is considerably more compact. For $B$ we have two equivalent definitions. One definition basically is that $B(D)$ is a natural extension of a function $B\_{D(\omega)}\colon \omega\to \omega$ from a version of binary fast-growing hierarchy up to $D(\omega)$ to the type $\mathsf{Ord}\to\mathsf{Ord}$. Which demonstrates the functor $B$ to be a variant of recursion along dilators. The other definition of $B$ is in the terms of a term system for certain version of an ordinal collapsing function $\psi$. Thus making an explicit connection between Girard's idea of tame type 2 bar-recursion and ordinal collapsing. B-H ordinal is $B(\varepsilon^+,0)$, where $\varepsilon^+$ is a naturally defined dilator mapping an ordinal $\alpha$ to the smallest $\varepsilon$-number strictly above $\alpha$.
There is a number of recent works of Anton Freund on dilators and ptykes, see for example [4]. But the approach that he mostly pursuing is to apply a more traditional proof-theoretic techniques to dilators and ptykes rather than using them directly to define ordinal notation systems.
With regards to what are the limits of the approaches. The limit of the approaches based on recursion on ordinals is B-H ordinal (at least as long as we limit ourselves to finite types). Recursion on dilators seems to be closely connected to collapsing functions and although it have been studied much less than the latter, probably it will have similar limitations (the notation systems for systems that are not too much stronger than $\Pi^1\_1\textsf{-CA}\_0$ are fairly simple, but the extensions of the approach to $\Pi^1\_2\textsf{-CA}\_0$ and even weaker systems become quite complicated). I don't know about any works that extend ptyx-based approach beyond recursion on dilators. I have been working on this, but haven't yet published anything about it. There are some indications that it might be possible to reach the ordinal of full second-order arithmetic.
[1] J.-Y. Girard. Proof Theory and logical complexity, II. Book Draft. <https://girard.perso.math.cnrs.fr/ptlc2.pdf>
[2] Päppinghaus, Peter. "Ptykes in GödelsT und Definierbarkeit von Ordinalzahlen." Archive for Mathematical Logic 28.2 (1989): 119-141.
[3] Aguilera, J. P., F. Pakhomov, and A. Weiermann. "Functorial Fast-Growing Hierarchies." arXiv preprint arXiv:2201.04536 (2022).
[4] A. Freund. $\Pi^1\_1$-comprehension as a well-ordering principle. Adv. Math., 355, 2019
| 4 | https://mathoverflow.net/users/36385 | 429708 | 174,105 |
https://mathoverflow.net/questions/429711 | 2 | Let $A>0$ be fixed and consider $X\_1,\ldots$ i.i.d. nonnegative random variables such that $E[1/X\_1]<\infty$.
Is is true that $$\sup\_{a\in \big (0,\frac A{\sqrt n} \big]} \sum\_{i=1}^n 1\_{X\_i>a} \frac{a^3}{X\_i^2}$$
converges in probability to $0$ ?
With the crude bound $\sum\_{i=1}^n 1\_{X\_i>a} \frac{a^3}{X\_i^2} \leq \frac {A^2}n \sum\_{i=1}^n \frac{1}{X\_i}$, the supremum is clearly $O\_P(1)$. In my research I need it to be $o\_P(1)$, but I haven't been able to prove it.
| https://mathoverflow.net/users/490637 | Convergence in probability of a supremum | As suggested by Anthony Quas, the supremum in question can be rewritten as
$$s\_n:=\sup\_{b\ge\sqrt n/A}S\_n(b),$$
where
$$S\_n(b):=\frac1{b^3}\sum\_{i=1}^n Z\_i^2\,1(Z\_i<b)$$
and $Z\_i:=1/X\_i$, so that the $Z\_i$'s are iid positive random variables with $EZ\_i<\infty$.
Take now any real $c>0$. Then for all large enough $n$
$$S\_n(b)=T\_n(b,c)+U\_n(b,c),$$
where
$$T\_n(b,c):=\frac1{b^3}\sum\_{i=1}^n Z\_i^2\,1(Z\_i<c),$$
$$U\_n(b,c):=\frac1{b^3}\sum\_{i=1}^n Z\_i^2\,1(c\le Z\_i<b).$$
Next, for all large enough $n$, uniformly in $b\ge\sqrt n/A$,
$$0\le T\_n(b,c)\le\frac{nc}{b^3}\frac1n\sum\_{i=1}^n Z\_i
\sim\frac{nc}{b^3}\,EZ\_1\to0$$
and
$$0\le U\_n(b,c)\le\frac n{b^2}\frac1n\sum\_{i=1}^n Z\_i\,1(c\le Z\_i)\sim\frac{n}{b^2}\,EZ\_1\,1(c\le Z\_1)
=O(EZ\_1\,1(c\le Z\_1))\underset{c\to\infty}\longrightarrow0$$
almost surely (a.s.), by the strong law of large numbers.
So, $s\_n\to0$ a.s. and hence, indeed, in probability.
| 3 | https://mathoverflow.net/users/36721 | 429734 | 174,112 |
https://mathoverflow.net/questions/429688 | 2 | Fix a continuously differentiable but nowhere twice differentiable function $f$ on $\mathbb{R}$ supported on $[0,1]$. Is it true that for all $x\in[0,1]$ and all $\delta$ sufficiently small
\begin{align\*} & \sup\_{0<h\leq \delta}|f'(x+2h)-2f'(x+h)+f'(x)|\leq \\ & \quad C \delta^{-1}\sup\_{0<h\leq \delta}|f(x+2h)-2f(x+h)+f(x)|~~?
\end{align\*}
Here $C$ is allowed to depend on $f$, and does not depend on $\delta$ and $x$.
| https://mathoverflow.net/users/152618 | An inequality about the second-order difference | $\newcommand\de\delta\newcommand\lhs{\text{lhs}}\newcommand\rhs{\text{rhs}}$No. E.g., let
$$f(x):=x^3\sin\frac1x$$
for $x\in(0,1/2]$, with $f(0):=0$. Then $f$ can be obviously extended to a continuously differentiable function $f$ on $\mathbb R$ supported on $[0,1]$.
Moreover,
$$\rhs(\de):=\de^{-1}\sup\_{0<h\le\de}|f(2h)-2f(h)+f(0)| \\
=\de^{-1}\sup\_{0<h\le\de}\Big|-2 h^3 \Big(\sin\frac1h-4 \sin \frac{1}{2h}\Big)\Big|=O(\de^2)=o(\de),
$$
while for $\de=\dfrac1{2(2n+1)\pi}$ and natural $n\to\infty$
$$\lhs(\de):=\sup\_{0<h\le\de}|f'(2h)-2f'(h)+f'(0)| \\
\ge |f'(2\de)-2f'(\de)+f'(0)| \\
=4 \de \sin ^2\frac{1}{4 \de }\,
\Big|6 \de \sin\frac{1}{2 \de }-2 \cos \frac{1}{2 \de
}-1\Big| \\
\sim4 \de \sin ^2\frac{1}{4 \de }=4\de.
$$
So, $\lhs(\de)$ is not $O(\rhs(\de))$.
| 1 | https://mathoverflow.net/users/36721 | 429735 | 174,113 |
https://mathoverflow.net/questions/429744 | 3 | Denote by $f(n)$ the maximal number of distinct divisors of $k$ integer numbers $1\leq a\_1<a\_2<\ldots<a\_k\leq n$, where $k$ is not fixed and $a\_1+\ldots+a\_k\leq n$. I'm interested in the asymptotics of $f(n)$.
For example, $f(4)=3$ since 4 has 3 divisors, $f(10)=5$ since 10=6+4 and 1,2,3,4,6 divide 4 or 6.
Notice that if we take numbers $1,2,3,\ldots,t$ such that $1+\ldots+t=t(t+1)/2\leq n$ (so $t\sim\sqrt{2n}$), then we have exactly $t$ divisors, so $f(n)>\sqrt{n}$.
Question: is it possible to obtain a better asymptotics?
| https://mathoverflow.net/users/103116 | Maximal number of divisors of numbers whose sum does not exceed $n$ | No, if you do not care on multiplicative factor: namely, $f(n)\leqslant 2\sqrt{n}$. Use the following
**Lemma.** For any real $x>0$ and any positive integer $a$, $a$ has at most $a/x$ divisors which are not less than $x$.
**Proof.** Let $d\_1>d\_2>\ldots>d\_m\geqslant x$ be these divisors. Then $a/d\_1<a/d\_2<\ldots<a/d\_m$, and these are integers, thus $a/d\_m\geqslant m$ and $a\geqslant md\_m\geqslant mx$, so $m\leqslant a/x$ as needed.
Thus $a\_i$'s have at most $(a\_1+\ldots+a\_k)/x\leqslant n/x$ divisors which are at least $x$, and also at most $x$ divisors which are less than $x$, totally at most $x+n/x$ divisors. It remains to put $x=\sqrt{n}$ to get $f(n)\leqslant 2\sqrt{n}$.
| 6 | https://mathoverflow.net/users/4312 | 429746 | 174,116 |
https://mathoverflow.net/questions/429752 | 0 | Let $K$ be an imaginary quadratic field and $E/K$ be an elliptic curve which has complex multiplication on $K$.
Let $R\_K$ be ring of integers of $K$.
Let $ \hat{E}$ be its formal group of $E$.
Take $a \in{R\_K}$ and $[a] \in \operatorname{End}E$ . Then there is unique corresponding homomorphism of formal group, $[a](t) \in \operatorname{End}\hat{E}$.
My question is, how can I prove $[a](t)=at+\text{(term higher than degree $2$)}$ ?
| https://mathoverflow.net/users/144623 | Power series corresponding to $[a]\in \operatorname{End}(E)$ ($a \in R_K$) can be expressed as $[a](t)=at+\text{(term higher than degree $2$)}$? | Let $\omega\_E$ be an invaraint differential on $E$. Then $[a]$ satisfies $[a]^\*\omega\_E=a\omega\_E$. That's over $K$, so the same formula holds on the formal group, i.e., $\widehat{[a]}\omega\_{\hat E}=a\omega\_{\hat E}$. On the other hand, if you write $\widehat{[a]}(T)=cT+\text{h.o.t.}$, then $\widehat{[a]}\omega\_{\hat E}=c\omega\_{\hat E}$. Hence $c=a$.
| 5 | https://mathoverflow.net/users/11926 | 429755 | 174,117 |
https://mathoverflow.net/questions/429764 | 0 | I sent yesterday a paper to a journal of publisher Springer for consideration, in the same time I have got a new free distribution service and an open-access called [Research Square](https://www.researchsquare.com/). It looks like Arxiv; it is not peer-reviewed and makes research available in a fast way. Now I want to know: should I withdraw my paper from that research square service since I sent it to a journal for review, or should I leave it there for comments to improve my paper in the future? Is what I did ethical? More than that, I want to know the difference between that Research Square and [Arxiv](https://arxiv.org/).
| https://mathoverflow.net/users/51189 | Is it ethical to submit a paper to journal then to Research Square? And what is the difference between that research square and ArXiv? |
>
> I want to know the difference between that Research Square and Arxiv.
>
>
>
One of the differences is that Arxiv is a not-for-profit service and is not actively trying to sell you stuff.
| 7 | https://mathoverflow.net/users/1898 | 429771 | 174,121 |
https://mathoverflow.net/questions/429758 | 7 | $\newcommand{\Tors}{{\rm Tors}}
\newcommand{\tf}{{\rm\, t.f.}}
\newcommand{\Gt}{{\Gamma\!,\,\Tors}}
\newcommand{\Gtf}{{\Gamma\!,\tf}}
\newcommand{\Q}{{\mathbb Q}}
\newcommand{\Z}{{\mathbb Z}}
\newcommand{\HH}{{\mathbb H}}$Let $A$ be an abelian group. We denote by $A\_\Tors$ the torsion subgroup of $A$.
We set $A\_\tf=A/A\_\Tors\,$, which is a torsion free group.
Let $\Gamma$ be a finite group, and let $M$ be a $\Gamma$-module, that is, an abelian group on which $\Gamma$ acts.
We denote by $M\_\Gamma$ the group of coinvariants of $\Gamma$ in $M$, that is,
$$ M\_\Gamma=M\, \big/\bigg\{\sum\_{\gamma\in \Gamma}(\,{}^\gamma y\_\gamma-y\_\gamma\,)
\ \big|\ y\_\gamma\in M\bigg\}.$$
We write $M\_\Gt:= (M\_\Gamma)\_\Tors$ (which is the torsion subgroup of $M\_\Gamma$),
$\ M\_\Gtf=M\_\Gamma/ M\_\Gt\ $ (which is a torsion free group).
We consider the functors:
\begin{align\*}
&(\Gamma,M)\,\rightsquigarrow\, M\_\Gt\,,\\
&(\Gamma,M)\,\rightsquigarrow\, M\_\Gtf\otimes(\Q/\Z)=M\_\Gamma\otimes (\Q/\Z).
\end{align\*}
**Theorem.**
Let $\Gamma$ be a finite group, and let
$$0\to M\_1\xrightarrow{i} M\_2\xrightarrow{j} M\_3\to 0$$
be a short exact sequence of $\Gamma$-modules.
Then there exists a natural homomorphism
$$\delta\colon (M\_3)\_\Gt\to (M\_1)\_\Gamma\otimes (\Q/\Z)$$
such that the following sequence is exact:
\begin{multline\*}
(M\_1)\_\Gt \xrightarrow{i\_\*} (M\_2)\_\Gt \xrightarrow{j\_\*} (M\_3)\_\Gt\xrightarrow{\delta}\\
(M\_1)\_\Gamma\otimes(\Q/\Z)\xrightarrow{i\_\*} (M\_2)\_\Gamma\otimes(\Q/\Z)\xrightarrow{j\_\*} (M\_3)\_\Gamma\otimes(\Q/\Z)\to 0.
\end{multline\*}
>
> **Question.** Is this exact sequence known?
> If not, does this follow easily from some more general known exact sequence?
>
>
>
I do have a proof, I am asking for a reference!
**Special cases.**
$\DeclareMathOperator\Gal{Gal}$Let $\Gamma=\Gal(E/F)$ be the Galois group
of a finite Galois extension $E/F$ of nonarchimedean local fields.
Assume that our $\Gamma$-modules $M\_i$ for $i=1,2,3$ are finitely generated and torsion-free, and for each $i=1,2,3$, let $T\_i$ be the corresponding algebraic $F$-torus, which splits over $E$, with cocharacter group $M\_i$.
We have a short exact sequence of $F$-tori
$$ 1\to T\_1\to T\_2\to T\_3\to 1$$
and the corresponding Galois cohomology exact sequence
$$
H^1(F,T\_1)\to H^1(F,T\_2)\to H^1(F,T\_3)\xrightarrow{\delta}
H^2(F,T\_1)\to H^2(F,T\_2)\to H^2(F,T\_3)\to 1,
$$
where $H^n(F,T\_i):=H^n(\Gal(F\_s/F), M\_i\otimes F\_s^\times)$
is the cohomology of the *profinite* group $\Gal(F\_s/F)$.
This is our sequence in this special case.
More generally, let again $\Gamma=\Gal(E/F)$ be the Galois group
of a finite Galois extension $E/F$ of nonarchimedean local fields.
Assume that our $\Gamma$-modules $M\_i$ for $i=1,2,3$ are finitely generated, but now we do not assume that they are torsion-free.
For each $i=1,2,3$, we choose a torsion-free resolution
$$ 0\to M\_i^{-1}\to M\_i^0\to M\_i\to 0.$$
We can choose these resolutions compatibly, so that we obtain a short exact sequence of complexes of torsion-free $\Gamma$-modules
$$0\to (M\_1^{-1}\to M\_1^0)\to (M\_2^{-1}\to M\_2^0) \to (M\_3^{-1}\to M\_3^0)\to 0$$
and a short exact sequence of complexes of $F$-tori
$$0\to (T\_1^{-1}\to T\_1^0)\to (T\_2^{-1}\to T\_2^0) \to (T\_3^{-1}\to T\_3^0)\to 0.$$
Write $T\_i^\bullet$ for the complex of tori $(T\_i^{-1}\to T\_i^0)$.
Then we know the hypercohomology groups $\HH^n(F,T\_i^\bullet)$:
$$ \HH^1(F,T\_i^\bullet)\cong(M\_i)\_\Gt,\quad\ \HH^2(F,T\_i^\bullet)\cong(M\_i)\_\Gamma\otimes(\Q/\Z);$$
see M. Borovoi, [Abelian Galois cohomology of reductive groups](https://doi.org/10.1090/memo/0626), Memoirs AMS 132(626), 1998, Proposition 4.1.
We have a hypercohomology exact sequence
\begin{multline\*}
\HH^1(F,T\_1^\bullet)\to \HH^1(F,T\_2^\bullet)\to \HH^1(F,T\_3^\bullet)\xrightarrow{\delta}
\HH^2(F,T\_1^\bullet)\to \HH^2(F,T\_2^\bullet)\to \HH^2(F,T\_3^\bullet)\to 1.
\end{multline\*}
This is our sequence in this special case.
| https://mathoverflow.net/users/4149 | Is this exact sequence known? | $\newcommand{\bQ}{\mathbb{Q}}\newcommand{\bZ}{\mathbb{Z}}\DeclareMathOperator{\Tor}{Tor}\newcommand{\Tors}{\mathrm{Tors}}$I tried to write up the computation with some level of details, please let me know if anything looks dubious.
Consider the long exact sequence $$\ldots\to H\_1(\Gamma, M\_2)\to H\_1(\Gamma,M\_3)\to M\_{1,\Gamma}\to M\_{2,\Gamma}\to M\_{3,\Gamma}\to 0$$ which we will view as an (acyclic) complex and denote its terms by $C^n, n\leq 0$, for brevity, so that $C^0=M\_{3,\Gamma},C^{-1}=M\_{2,\Gamma}$ etc. Consider the spectral sequence associated with the derived functor of $-\otimes\_{\bZ}\bQ/\bZ$ and the 'bête' filtration on the complex $\ldots C^1\to C^0$. I'll use the cohomological grading conventions so that the first page of the spectral sequence looks like $E\_{1}^{i,j}=\Tor\_{-j}^{\bZ}(C^{-i},\bQ/\bZ)$. Note that for an abelian group $M$ we have $\Tor^{\bZ}\_0(M,\bQ/\bZ)=M\otimes\bQ/\bZ,\Tor^{\bZ}\_1(M,\bQ/\bZ)=M\_{\mathrm{Tors}}$, and $\Tor\_{>1}(M,\bQ/\bZ)=0$, hence the $1$st page looks like this, having only $2$ potentially non-zero rows:
$$\begin{matrix}\dots & \color{red}{H\_1(\Gamma,M\_2)\otimes\bQ/\bZ} & \color{red}{H\_1(\Gamma,M\_3)\otimes\bQ/\bZ} & M\_{1,\Gamma}\otimes\bQ/\bZ & M\_{2,\Gamma}\otimes\bQ/\bZ & M\_{3,\Gamma}\otimes\bQ/\bZ & \\ \dots & H\_1(\Gamma,M\_2)\_{\mathrm{Tors}} & H\_1(\Gamma,M\_3)\_{\Tors} & M\_{1,\Gamma,\Tors} & M\_{2,\Gamma,\Tors} & M\_{3,\Gamma,\Tors}\end{matrix}$$
The differentials $d\_{i,j}:E\_1^{i,j}\to E^{i+1,j}\_1$ are simply the maps induced by the maps in the above exact sequence. Since we started with an acyclic complex, the spectral sequence must converge to zero.
Now, since $H\_i(\Gamma, M)$ is annihilated by $|\Gamma|$ for all $\Gamma$-modules $M$, the two entries that are highlighted in red are in fact zero (and all the hidden entries in the top row are zero likewise). This implies that the only possibly non-trivial differential on the second page is $$E\_{2}^{-2,0}=\ker (E\_1^{-2,0}\to E\_1^{-1,0})=\ker (M\_{1,\Gamma}\otimes\bQ/\bZ \xrightarrow{i\_\*} M\_{2,\Gamma}\otimes\bQ/\bZ )\to E\_2^{0,-1}=\mathrm{coker} (M\_{2,\Gamma,\Tors} \xrightarrow{j\_\*} M\_{3,\Gamma,\Tors})$$
This differential must be an isomorphism, while all other $E\_2^{i,j}$ must already be zero as otherwise something would survive to the abutment $E\_3^{i,j}=E\_{\infty}^{i,j}$ of the spectral sequence. But this is exactly saying that the $6$-term sequence in question is exact, with $\delta$ being defined as the inverse to the differential $E\_2^{-2,0}\to E\_2^{0,-1}$.
| 5 | https://mathoverflow.net/users/39304 | 429772 | 174,122 |
https://mathoverflow.net/questions/429778 | 4 | Let $X$ be a finite $p$-local spectrum. For each $h \in \mathbb{N} \cup \{\infty\}$, let $K(h)$ be [Morava $K$-theory](https://en.wikipedia.org/wiki/Morava_K-theory) of height $h$. Recall that the coefficients $K(h)\_\ast$ are a graded field, and $K(h)\_\ast(X)$ is a finite-dimensional vector space over this graded field. Denote $|X|\_h := dim\_{K(h)\_\ast} K(h)\_\ast (X)$.
**Question:** If $h \leq h'$, then do we have $|X|\_h \leq |X|\_{h'}$?
In some special cases, the answer is *yes*:
* When $X = \mathbb S\_{(p)}$ the answer is *yes*: we have $|\mathbb S\_{(p)}|\_h = 1$ for all heights $h$.
* When $h' = \infty$, the answer is *yes*. Here, $K(\infty) = H\mathbb F\_p$ and $|X|\_\infty$ counts the number of cells of $X$. The "yes" answer is straightforward to show by induction on the number of cells.
* If $|X|\_{h'} = 0$, then the answer is *yes*. That is, $|X|\_{h'} = 0 \Rightarrow |X|\_{h} = 0$ for $h \leq h'$. This follows from the [thick subcategory theorem](https://ncatlab.org/nlab/show/thick+subcategory+theorem) (though I'm pretty sure this is in fact one of the easier observations which goes into the proof of that theorem).
* As $h \to \infty$ it's well-known (I think it's a simple connectivity argument?) that $K(h)\_\ast(X)$ is eventually just $(H\mathbb F\_p)\_\ast(X) \otimes\_{\mathbb F\_p} K(h)\_\ast(X)$. So the sequence $|X|\_h$ is eventually constant at the value $|X|\_\infty$. So for fixed $X$, there are at most finitely many exceptions to a "yes" answer.
So the simplest case I'm not sure about is when $h = 0$, $h' < \infty$, and $X$ is not a sum of shifts of spheres. For instance, probably the simplest 2-cell type 0 spectrum is $\Sigma^\infty \mathbb C \mathbb P^2$. In this case, we have $|\Sigma^\infty \mathbb{CP}^2|\_0 = |\Sigma^\infty \mathbb{CP}^2|\_\infty = 2$, so the question is whether $|\Sigma^\infty \mathbb{CP}^2|\_h$ ever drops to 1, i.e. whether $\Sigma^\infty \mathbb{CP}^2$ is ever $K(h)$-locally invertible...
| https://mathoverflow.net/users/2362 | Is $\operatorname{dim}_{K(h)_\ast} K(h)_\ast X$ increasing in $h$? | Yes, this is true, and appears in the early literature, although I do not immediately remember exactly where; I'd guess work of Wilson and/or Ravenel. I'll assume that $h>0$ for notational simplicity. There is a homology theory $E$ with $E\_\*=\mathbb{F}\_p[v\_h,v\_{h+1}^{\pm 1}]$, and this is a discrete valuation ring in the graded sense. It follows that when $X$ is finite, $E\_\*(X)$ is a finite sum of $n$ terms like $E\_\*$ and $m$ terms like $E\_\*/v\_h^k$, for some $n,m\geq 0$. It follows that $v\_h^{-1}E\_\*(X)$ is isomorphic to a direct sum of $n$ copies of $v\_h^{-1}E\_\*$. Using the fact that all formal groups of strict height $h$ become isomorphic after faithfully flat extension, we find that $|X|\_h=n$. On the other hand, we have a cofibre sequence $\Sigma^{|v\_h|}E\to E\to K(h+1)$, giving a short exact sequence relating $K(h+1)\_\*(X)$ to the cokernel and kernel of multiplication by $v\_h$ on $E\_\*(X)$. This gives $|X|\_{h+1}=n+2m$.
| 8 | https://mathoverflow.net/users/10366 | 429779 | 174,124 |
https://mathoverflow.net/questions/429776 | 4 | Consider a pointed compact Hausdorff space $(X,x\_0)$ and a closed pointed subspace $i:A\subset X$ such that there exists a continuous map $r:X\rightarrow A$ such that $r|\_A=\text{Id}\_A$. Set
$$q:(X,x\_0)\rightarrow(X/A,A/A)$$
be the collapsing map and let $\tilde{K}^0$ denote the reduced (complex) topological $K$-group of a space, then there exists an exact sequence
$$\tilde{K}^0(X/A)\xrightarrow{q^\*}\tilde{K}^0(X)\xrightarrow{i^\*}\tilde{K}^0(A)\rightarrow0$$
where the right hand side is surjective because of the retraction. By the construction of negative $K$-groups and the long exact sequence of $K^i$, we know that the map $q^\*:\tilde{K}^0(X/A)\rightarrow\tilde{K}^0(X)$ should be injective.
>
> Is there a direct way to see the injectivity only involving vector bundles but not $K^{-1}$ or analytic computation like in Karoubi or Park?
>
>
>
Pick $\alpha\in\tilde{K}^0(X/A)$, we know by our setting that there exists a vector bundle $E$ over $X/A$ such that $[E]=\alpha$. Suppose that $q^\*\alpha=0$, then $q^\*E$ is stably trivial i.e. $q^\*E\oplus H$ is trivial for some product bundle $H=X\times\mathbb{C}^n$.
>
> Now it suffices to show that $E$ is also stably trivial. How can one see it directly?
>
>
>
| https://mathoverflow.net/users/nan | Pullback of complex vector bundles along a retraction of compact Hausdorff spaces: a direct proof instead? | You may add $\mathbb C^n$ to $E$ so that in fact $q^\ast E$ has a trivialization $\phi$. Say that $E$ has rank $m$. The restriction of $q^\ast E$ to $A$ has also a trivialization $\psi$, simply because the restriction of $q$ to $A$ is a constant map. Comparing $\psi $ with the restriction of $\phi$ we get a map $u:A\to GL\_m(\mathbb C)$. The composed map $u\circ r:X\to GL\_m(\mathbb C)$ can now be used to modify $\phi$, making a trivialization of $q^\ast E$ that restricts to $\psi$ on $A$ and therefore comes from a trivialization of $E$ itself.
Alternatively, use classifying spaces. The problem is to show that if a map $i:X/A\to B$ is such that the composed map $i\circ q:X\to B$ is nullhomotopic then $i$ is nullhomotopic. (Here $B$ could $BU$ or any other based space.) A nullhomotopy of $i\circ q$ means an extension of $i$ to the mapping cone of $q$. The mapping cone of $q$ is homotopy equivalent to the suspension $\Sigma A$. A map $\Sigma A\to B$ extends to a map $\Sigma X\to B$ because of the retraction of $X$ to $A$. The composed map $X/A\to \Sigma A\to \Sigma X$ is nullhomotopic.
| 9 | https://mathoverflow.net/users/6666 | 429781 | 174,125 |
https://mathoverflow.net/questions/429780 | 4 | **Short version:** If I have a map $f:Y \to I$, and $\mu$ an ultrafilter on $Y$, under what condition can $\mu$ be written as a limit/sum/integral of ultrafilters on the fibers of $f$ along the ultrafilter $\eta = f\_\*(\mu)$ on $I$ ? Is it always possible? If not is there an explicit condition on $\mu$ and $f$ to detect if such a decomposition "along $f$" exists? check for this to be true? Is there a standard name for this type of decomposition or a keyword I should know related to this?
I apologize if this is a very basic question - but I couldn't find the correct keywords to find anything about this searching online.
**For more details:**
Let $I$ be set, $\eta$ an ultrafilter on $I$, and for each $i\in I$, let $Y\_i$ be a set and $\mu\_i$ an ultrafilter on $Y\_i$, then one can define an ultrafilter $\mu$ on $Y = \coprod\_{i \in I} Y\_i$ using the usual formula for limits of families of ultrafilter:
$$ A \in \mu \Leftrightarrow \{i | A \cap Y\_i \in \mu\_i \} \in \eta $$
this is this $\mu$ I referred to as the limit/sum/integral of the $\mu\_i$ above. If $f:Y \to I$ is the obvious map, then it should be noted that the direct image $f\_\*(\mu)$ of $\mu$ along $f$ gives back the ultrafilter $\eta$
Obviously one cannot completely recover the family $(\mu\_i)$ from $\mu$, but one can recover exactly its "germ" along $\eta$, in the sense that if one starts form another family $\mu'\_i$ of ultrafilter on $Y\_i$, and we build an ultrafilter $\mu'$ on $Y$ out of them, then one can show that
$$ \mu = \mu' \Leftrightarrow \exists J \in \eta, \forall j\in J, \mu\_j = \mu'\_j $$
But given $\mu$ and $f:Y \to I$ it seems hard to tell if there is such a familly $\mu\_i$ of ultrafilter on the fibers $Y\_i = f^{-1}\{i\}$ that will give $\mu$. I can see no way to get description of a $\mu\_i$, nor any reason for such a decomposition to exists ( though I don't really know how to show that it does not always exist either...)
So is there a condition on $\mu$ and $f$ to determine whether such a decomposition exists? Or does it exists for all $f$ and $\mu$?
| https://mathoverflow.net/users/22131 | Decomposition of an ultrafilter on the fibers of a map | First, let me dispose of the trivial cases where $f$ is constant or one-to-one on a set in $\mu$. In the case of constant $f$, say with value $i$, you can take $\eta$ principal at $i$ and let $\mu\_i$ be a suitable copy of $\mu$. In the one-to-one case, $\eta$ is a copy of $\mu$, and all the $\mu\_i$ can be principal.
So now let me confine attention to the non-trivial cases. There are plenty of non-principal ultrafilters on $\omega$ that do not admit such a decomposition. Indeed, those that do are precisely the limit points of countable discrete sets in the Stone-Cech remainder $\beta\omega-\omega$. Kunen proved the existence of weak P-points in $\beta\omega-\omega$, i.e., points that are not limit points of any countable subset (discrete or otherwise) of $\beta\omega-\omega$. But it was known earlier that some ultrafilters are not limit points of any countable discrete subset.
In the notation of your question, when such a decomposition exists, one says that $\mu$ is above $\eta$ in the Rudin-Frolik ordering. Googling "Rudin-Frolik ordering" should get you lots more information. (But note that the Rudin-Keisler ordering is a different, weaker ordering.)
| 5 | https://mathoverflow.net/users/6794 | 429785 | 174,126 |
https://mathoverflow.net/questions/429796 | 5 | This question is follow up of [this MO-post](https://mathoverflow.net/q/429619/61536).
First let us recall the necessary definitions.
A function $f:X\to X$ on a group $X$ is called a *polynomial* if there exists $n\in\mathbb N$ and elements $a\_0,\dots,a\_n\in X$ such that $f(x)=a\_0xa\_1x\cdots xa\_n$ for all $x\in X$.
Let $\mathrm{Poly(X)}$ be the set of all polynomials on a group $X$. It is a submonoid of the monoid $X^X$ of all self-maps of $X$.
Observe that each polynomial $f$ on a commutative group $X$ is of the form $f(x)=ax^n$ for some $a\in X$ and $n\in \mathbb N$, which implies that $|\mathrm{Poly}(X)|=|X|\cdot\exp(X)$ for each finite commutative group $X$.
Here $\exp(X):=\min\{n\ge 1:\forall x\in X\;\;(x^n=1)\}$ is the exponent of the group $X$.
Since any group $X$ acts effectively on $\mathrm{Poly}(X)$ by left (or right) shifts, the cardinal $|\mathrm{Poly}(X)|$ is divisible by $|X|$.
>
> **Problem 1.** Is $|\mathrm{Poly}(X)|$ divisible by $|X|\cdot\exp(X)$ for every finite group $X$?
>
>
>
The cardinality of the monoid $\mathrm{Poly}(X)$ was calculated by [Peter Taylor](https://gist.github.com/pjt33/b9e1e62291110b164e091ef7c2f7ead5) for all non-commutative groups $X$ of cardinality $|X|<24$.
>
> **Problem 2.** Is $|\mathrm{Poly}(X)|$ divisible by $|X|^2$ for every non-commutative finite group $X$?
>
>
>
Peter Taylow observed that the answer to Problem 2 is negative for the group $D\_{12}=GAP(12,4)$ with $|\mathrm{Poly}(D\_{12})|=648=2^3\times 3^4$. So, only Problems 1 and 3 remain open.
The calculations of Peter Taylor show that the following problem has affirmative answer for all finite groups of cardinality $<24$:
>
> **Problem 3.** Let $X$ be a finite group and $p$ be a prime number dividing $|\mathrm{Poly}(X)|$. Is $p$ a divisor of $|X|$?
>
>
>
**Remark 1.** The answer to Problems 2 and 3 are affirmative for finite simple groups since $\mathrm{Poly}(X)=X^X$ for any non-commutative simple finite group $X$, see the [answer of @YCor](https://mathoverflow.net/a/429635/61536) to [this MO-question](https://mathoverflow.net/q/429619/61536).
| https://mathoverflow.net/users/61536 | The number of polynomials on a finite group, II | This is an answer to problem 1 and problem 3:
As noted in the comments, $\operatorname{Poly}(G)$ with pointwise multiplication is a group for finite $G$. Consider the homomorphism $\operatorname{Poly}(G)\to G$ given by evaluating at $1$. It is surjective (as seen by constant polynomials), and the kernel contains an element of order $\operatorname{exp}(G)$, the identity polynomial. So the kernel has order divisible by $\operatorname{exp}(G)$, and thus $\operatorname{Poly}(G)$ has order divisible by $\operatorname{exp}(G)\cdot |G|$.
Let me also answer problem 3! If $|\operatorname{Poly}(G)|$ is divisible by $p$, then there is an element $f\in \operatorname{Poly}(G)$ of order $p$. This means all values of $f$ have order dividing $p$, and at least one is nontrivial. So $G$ must have an element of order $p$, thus $p$ divides $|G|$.
| 10 | https://mathoverflow.net/users/39747 | 429798 | 174,132 |
https://mathoverflow.net/questions/429757 | 1 | How to derive that after stereo-graphical projection, $\Delta u$ in $\mathbb{R}^n$ is transformed to
$$
\Delta\_{\mathbb{S}^n}u - \frac{n(n-2)}{4}u\ \text{in}\ \mathbb{S}^n.
$$
To be more precise, in this [paper](https://pdf.sciencedirectassets.com/272601/1-s2.0-S0022123610X00052/1-s2.0-S0022123609005114/main.pdf?X-Amz-Security-Token=IQoJb3JpZ2luX2VjEDYaCXVzLWVhc3QtMSJHMEUCIFbXMf27WHEDMDXJGUvxxDCMyxaXJ4xeMrZK%2BDRCzhsSAiEAjh42q5rClB4mJ8%2FRHiv9Ke%2BIKa%2BkbwxotqzQ8yjf6j8q1QQIvv%2F%2F%2F%2F%2F%2F%2F%2F%2F%2FARAFGgwwNTkwMDM1NDY4NjUiDIOTGR6hKf3HDGSdXiqpBMw5bPgD%2BWVLLv9a8PEZJsH%2BfwSTMx1Mzem2AXWpeMcZXy1yvUiBvqLDUFXJQn5KWtyo1AAV4EMkqeazJFtQL46RlRR7LdYUYmjGqOrB6q02wyYTqZVwh%2BlPpGF2Y7Ddr2YFzkcRCehhdCs7qAXfIn%2F8HIcD7P1w0pAVjsTcyPSx0BxctjlzYY%2FJTAE%2BMSjS65TAPyV7HpomETNJgpiwUTeWPQtteGhkExlU4xJyV6ERaRapeiW8JeYfZQUPR63l9wnjL4EBS6MeMXcVXHXnkQqUGVdNdRsUszNfksDz%2Bfe6Fx3OcBgLBbOpLE0zVS8vd9yNxDmWXyiUREhcELil2LcM9enRkpPvCr0k1qmtNgFA%2F1Sh3VlHO22FAHjcgLTtqb0dJtu%2BG%2FP0qWRdKdte44bOyN60CUKSZKl2qiRhX5A%2F7qGmaSYAF3GRgQdXz883O6xxa7NBQTO1DaOUrRNkHZ6LuwVvo0Lm7R96P5cxJIzw%2F3tqfRyRM3g28Gw09z7JX0EQPZG5cnCZr%2Fum14Oi83Es5HJf4PcPDaxC%2B3Cxl%2Ff8poNUFYhClYFm2Gf5hDdQUPgKanv13xfWqzMjLa21ruU4wiQ0GisDRuUCcmEpdBwwHSjYkmLcXUSreUkXhOyxPMD0GSmGQQzuEACTUCccWHB8wEweC52B569tJpx7mHxgAShcjWdsSIWsMN5B4yXc%2Bk2I9%2FfFjSFAoLAT6Z1MZ6QxqXNtQHfoyygwi8bSmAY6qQEe6sYbufDn8fQ3McnKb4rg1orNYcaILKZxUGjqP2KTZMbEXfffM32VjNHXXfEJlJmUnC1rdC7O%2FxUSAV8XAKH7un5KamQpZjoq8itiEbaZ%2FfriFWhYoXORsOeGvVJI4KVhcevTHmzl5pbmvIp69PQPNt5DZpWUXNpzjPUv0KrklyQy5SagKheWv8SDZRYnhqtIakiiby1d9DuOHeA4eFkbn7AB6hqd8iEO&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Date=20220904T135030Z&X-Amz-SignedHeaders=host&X-Amz-Expires=300&X-Amz-Credential=ASIAQ3PHCVTYXMPUFFST%2F20220904%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Signature=325680fa17a0b4af2bcae827e53cec8582bfc80afbadbc688fbfb3332beacaa4&hash=6b0e2422a57f2bbdbf93bdf9474dce3953b63973b546dec6bb3204d69a8edcac&host=68042c943591013ac2b2430a89b270f6af2c76d8dfd086a07176afe7c76c2c61&pii=S0022123609005114&tid=spdf-4aca3989-b294-4b32-9a25-89d860a8ba3c&sid=d209ff5679cc14414928ebe4e4a1dd0b985egxrqa&type=client&ua=5152545955500657510c&rr=745731d14c25afc3), it is asserted that the equation
$$-\Delta\_{\mathbb{S}^n}\,u+\frac{n(n-2)}{2}u-\tilde{K}u^{\frac{n+2}{n-2}}=0\quad\text{on}\ \mathbb{S}^n$$
can be reduced to
$$-\Delta u = K(y)u^{\frac{n+2}{n-2}}\quad\text{in}\ \mathbb{R}^n.$$
I need some tips or relevant references.
| https://mathoverflow.net/users/480661 | Laplacian on sphere after stereographic projection | the relation between the two Laplacians is a bit more complicated. I follow the survey [paper](https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society-new-series/volume-17/issue-1/The-Yamabe-problem/bams/1183553962.full) of Lee and Parker on the Yamabe problem.
The conformal Laplacian of a Riemannian metric $g$ is defined as
$$
L\_g = 4 \frac{n-1}{n-2} \Delta\_g + S\_g
$$
where $S\_g$ is the scalar curvature of the metric and $\Delta\_g = d\_g^\* d$ is the Hodge (positive) Laplacian. The main point is that if we do a conformal change of the metric $\tilde{g} = \varphi^2 g$ (where $\varphi>0$ is a smooth positive function) then
\begin{equation}\label{eq:conflap}
L\_g (u) = \varphi^{\frac{n+2}{2}} L\_{\tilde{g}} (\varphi^{\frac{2-n}{2}} u) .
\end{equation}
The metric of the unit sphere in stereographic projection coordinates is
\begin{equation\*}
g\_{S^n} = \varphi^2 g\_{\mathbb{R}^n} \hspace{2mm} \text{ with } \varphi = \frac{2}{(1+|x|^2)}.
\end{equation\*}
Using the equation for the conformal laplacian and that the scalar curvature of $S^n$ is $n(n-1)$ we obtain
\begin{equation}
\Delta\_{\mathbb{R}^n} u = \varphi^{\frac{n+2}{2}}
\left(
\Delta\_{S^n} + \frac{n(n-2)}{4}
\right)
(\varphi^{\frac{2-n}{2}} u) .
\end{equation}
On the other hand, the conformal Laplacian relates to prescribing scalar curvature in conformal classes as follows.
Let $K$ be a given function,
then the metric $\tilde{g} = u^{4/(n-2)} g$ has scalar curvature equal to $K$ if and only if $u$ solves the PDE
\begin{equation\*}
L\_g (u) = u^{\frac{n+2}{n-2}} K .
\end{equation\*}
In the case of the sphere, the metric $\tilde{g} = u^{4/(n-2)} g\_{S^n}$ has scalar curvature equal to $K$ if and only if
\begin{equation}\label{eq1}
\Delta\_{S^n} u + \frac{n(n-2)}{4} u = u^{\frac{n+2}{n-2}} \tilde{K}
\end{equation}
where $\tilde{K}= \frac{n-2}{4(n-1)} K$. At the same time, $\tilde{g}= \tilde{u}^{4/(n-2)} g\_{\mathbb{R}^n}$ where $\tilde{u} = \varphi^{\frac{n-2}{2}} u$ with $\varphi = 2 (1+|x|^2)^{-1}$. We conclude that $u$ solves the above equation if and only if $\tilde{u}$ solves
\begin{equation}
\Delta\_{\mathbb{R}^n} \tilde{u} = \tilde{u}^{\frac{n+2}{n-2}} \tilde{K} .
\end{equation}
| 3 | https://mathoverflow.net/users/127247 | 429805 | 174,137 |
https://mathoverflow.net/questions/429808 | 8 | Suppose $f:\mathbb{R}^n\to\mathbb{R}$ is integrable. Is it true that
$$
\lim\_{r\to 0}\frac{\displaystyle\int\_{B\_r(0)}f(y)~\mathrm dy}{r^{n-1}}=0 \quad ?
$$
This is obvious if $0$ is a Lebesgue point of $f$ or if $n=1$, but I would like to know if it's true in general.
| https://mathoverflow.net/users/351083 | How badly can the Lebesgue differentiation theorem fail? | Metafune has given an example of the limit failing to be $0$ at a particular point - namely for $n > 1$, the function $|x|^{-\alpha}$, with $1 \leq \alpha < n$ has that limit equal to $\infty$ at $0$.
However, you can still get some kind of affirmative result.
In general the limit in question is zero $\mathcal H^{n-1}$-a.e, where $\mathcal H^{n-1}$ denotes the $n-1$ dimensional Hausdorff measure.
This is Theorem 2.10 in *Measure Theory and Fine Properties of Functions* by Evans and Gariepy (2015 version).
| 17 | https://mathoverflow.net/users/173490 | 429810 | 174,138 |
https://mathoverflow.net/questions/429789 | 4 | Suppose that $\pi:X\to S$ is a *smooth projective* morphism of relative dimension 1. If $S$ is the spectrum of an algebraically closed field, then it is known that $X$ embeds into $\mathbb{P}^3\_S$.
**Question**: Can a similar statement be made for more general $S$?
I am primarily interested in Dedekind schemes $S$, but would also like to understand the general situation.
Let us assume that $X$ is geometrically integral, to try to account for the algebraically-closedness hypothesis. ([Edit]: In light of Jef's comment, I will strengthen this to "There is a section $S\to X$".)
Let us also assume any standard finiteness condition that could be helpful: Noetherian, quasi-compact, etc.
If the statement is not true in this generality, what if we add some cohomological niceness properties: $X/S$ is cohomologically flat, or maybe even $S$ affine.
**Question**: If we assume the fibers of $\pi$ are curves of genus $0$, can we embed $X$ into $\mathbb{P}^2\_S$?
I think this is true over Dedekind schemes, using the ample anti-canonical sheaf $\omega\_{X/S}^\vee$, since $\pi\_\*\omega\_{X/S}^\vee$ is locally free of rank 3, but I am not certain.
I would be completely content with an answer for Dedekind schemes $S$, but I would appreciate any help.
| https://mathoverflow.net/users/94086 | Does a smooth relative curve $X/S$ embed into $\mathbb{P}^3_S$? | This answer addresses the second question: "If we assume the fibers of $\pi$ are curves of genus $0$, can we embed $X$ into $\mathbb{P}^2\_S$?"
The answer to this is also **no** (providing there is a singular fibre).
Let $\pi: X \to \mathbb{P}^1$ be a conic bundle surface over an algebraically closed field $k$, i.e. $X$ is regular and every fibre is isomorphic to a plane conic (possibly singular).
Consider the relative anticanonical bundle $\omega\_{X/\mathbb{P}^1}^{-1}$. This is very ample when restricted to each fibre, and the pushforward to $\mathbb{P}^1$ is a vector bundle $V$ of rank $3$. We obtain an embedding
$$X \to \mathbb{P}(V)$$
which respects $\pi$, where $\mathbb{P}(V)$ denotes the corresponding $\mathbb{P}^2$-projective bundle over $\mathbb{P}^1$.
Now every vector bundle on $\mathbb{P}^1$ splits as a direct sum of line bundles, so we can write
$$V = \mathcal{O}(a\_1) \oplus \mathcal{O}(a\_2) \oplus \mathcal{O}(a\_3).$$
We obtain the trivial projective bundle if and only if $a\_1 = a\_2 = a\_3$. So we just need to give an example where this doesn't hold.
Firstly, if every fibre is smooth, then $X$ is a ruled surface. By the classification of ruled surfaces $X$ is a Hirzebruch surface $\mathbb{F}\_n$ for some $n$. But $\mathbb{F}\_n$ embeds into $\mathbb{P}^2 \times \mathbb{P}^1$ as
$$x\_0^n y\_0 = x\_1^n y\_1.$$
So to get a counter-example we need to consider a conic bundle with a singular fibre.
I take $X$ to be a smooth cubic surface in $\mathbb{P}^3$. For any line $L \subset X$, there is an associated conic bundle given by taking the residual intersection of the pencil of planes through $L$. This has exactly 5 singular fibres as $L$ meets exactly 10 other lines of $X$.
Now take $Y$ a conic bundle surface in $\mathbb{P}^2 \times \mathbb{P}^1$. This has bidegree $(2,d)$ for some $d$. Considering the discriminant of the associated quadratic form one sees that this has $3d \neq 5$ singular fibres, as required.
Incidently, one can show that any smooth cubic surface embeds into $\mathbb{P}(V)$ where $V = \mathcal{O}(0) \oplus \mathcal{O}(0) \oplus \mathcal{O}(1).$
| 4 | https://mathoverflow.net/users/5101 | 429813 | 174,139 |
https://mathoverflow.net/questions/429236 | 2 | I am interested in the following question:
Let $A,B\in\text{Mat}(2n\times2n;\mathbb{Z})$ be two integer matrices with the property that $\text{det}(A-A^T)=1=\text{det}(B-B^T)$. Are there known invariants that completely characterize congruency via unimodular integer matrices?
More precisely: Are there invariants under congruency of $A$ and $B$ that completely characterize whether or not some unimodular matrix $P\in\text{GL}\_{2n}(\mathbb{Z})$ exists such that $P\cdot A\cdot P^T = B ?$
We can assume $A$ and $B$ to be non-singular, if it makes an answer possible.
| https://mathoverflow.net/users/157865 | Classification of congruent integer matrices | [My Ph.D. thesis](http://www-personal.umich.edu/%7Ealimil/phdthesis.pdf) studied this and related questions from the perspective of arithmetic invariant theory. (There's also various other literature out there addressing related questions from a different number of directions; my contribution was mainly putting this material into the context of arithmetic invariant theory.)
One thing to note is that all skew-symmetric unimodular matrices are congruent, and so one can fix $A-A^T = B-B^T=J$, and change the problem to ${\rm Sp}\_{2n}(\Bbb Z)$-congruence of matrices $P$ with $P - P^T = J$. I observed in section 3.3 of the thesis that these congruence classes are in bijection with what I called "conjugate-self-balanced" (CSB) modules over the ring $\Bbb Z[t]/f(t)$ where $f(t) = det(t J - P)$ (this is the Alexander polynomial up to a change of variables).
(I came up with the terminology CSB modules based on analogy with prior work of Wood -- however, similar objects are also called "polarized modules" in the context of the classification of abelian varieties over finite fields. There's recently been a lot of computational work in that direction, and one interesting question is whether these tools could be adapted to work with Seifert pairings.)
So this is formally an answer to your question, but to actually do anything with this, you need to understand CSB modules over $\Bbb Z[t]/f(t)$ -- which may not be a Dedekind domain, or even a domain, or may even contain nilpotents -- meaning that even completely classifying modules over this ring may be messy even before bringing in the extra CSB structure.
| 7 | https://mathoverflow.net/users/422 | 429820 | 174,141 |
https://mathoverflow.net/questions/429827 | 5 | Let $C$ be a diagram. Consider a functor $F: C \to \mathbb{E}\_{\infty}(Sp)$ from the diagram to the category of $\mathbb{E}\_{\infty}$-rings in spectra. Let $R$ be the limit of this diagram.
Given the functor $F$, we can also construct $F^{\prime}: C \to Cat\_\infty$, $c \mapsto Mod\_{F(c)}$, where $Mod\_{F(c)}$ is the infinity category of modules over the ring $F(c)$. Then we have an adjunction $$Mod\_R\rightleftharpoons\lim\_C(c \mapsto Mod\_{R(c)}).$$ The map from $Mod\_R$ is simply given by ``tensoring up'' through the diagram. For the right adjoint,I have heard that it can be written explicitly as a homotopy limit of a diagram of modules. I was wondering if anyone had any reference for this. To explain what I mean by explicit description I have a small example below.
Example: For odd primes, we have an equalizer diagram for the $K(1)$- local sphere given by
$$L\_{K(1)}S \rightarrow K{{ \xrightarrow{\Psi^g}}\atop{\xrightarrow[i\_K ] {}}} K$$
where $g$ is a topological generator of $\mathbb{Z}\_p^{\times}$ and $\Psi^g$ denotes the Adams operation. This diagram gives rise to the adjunction
$$L\_{K(1)}Sp \rightleftharpoons \lim (Mod\_K \rightrightarrows Mod\_K)$$
In this case the objects of the limit category are given pairs $(M,\phi)$ where $\phi : M \wedge \_{\psi^g} K \to M$ is an isomorphism. The right adjoint is simply given by $lim(M \xrightarrow{\phi \circ \psi^g -1} M)$ ie the equaliser of $\phi \circ \psi^g$ and $id$. I am looking for a similar description for bigger diagrams with more simplices. Thank you in advance.
| https://mathoverflow.net/users/70889 | Explicit description of the right adjoint | See Theorem B in [On conjugates and adjoint descent](https://arxiv.org/abs/1705.04933), Horev-Yanovski, which states exactly this. The statement doesn't include this, but the body of the paper describes explicitly in what way the right adjoint is a limit of the right adjoints.
Note that your example is not a special case of your general setup because $Mod\_{L\_{K(1)}S}$ is not equivalent to $L\_{K(1)}Sp$.
| 6 | https://mathoverflow.net/users/102343 | 429831 | 174,145 |
https://mathoverflow.net/questions/429809 | 3 | I have a little technical question on Peter Scholze's [lectures on condensed mathematics](https://www.math.uni-bonn.de/people/scholze/Condensed.pdf).
On page 12, right above the Proof of Theorem 2.2, he says that for extremally disconnected sets the condition (ii) on page 7 is automatic. I can't see why. I understand that the map is injective under the presence of sections, but why is it surjective?
This assertion is then used in the proof of Theorem 2.2 and I guess, it was the reason for introducing extremally disconnected spaces here, so I believe that it is true, but I just can't see why.
| https://mathoverflow.net/users/473423 | Condensed mathematics | The key point is to use the fact that extremally disconnected sets are projective (in the category of compact Hausdorff spaces) to note the coequalizer diagram involved in the analogue of condition (ii) is a [*split* coequalizer.](https://ncatlab.org/nlab/show/split+coequalizer)
I'll explain some additional details in the next paragraphs, but first it's worthwhile to note that there is a small subtlety in formulating the analogue of condition (ii) because the pullback of extremally disconnected sets need not be extremally disconnected. Rather, you can always find an extremally disconnected presentation of the pullback.
So one way to formulate the analogue of condition (ii), for a presheaf $T$ on the category of extremally disconnecteds, is as follows:
For all surjective morphisms $X \to B$ of extremally disconnected sets, and all surjective morphisms $R \to X \times\_B X$ where $R$ is extremally disconnected (and the pullback is computed in $\mathrm{CompHaus}$, say), the corresponding diagram
$$T(B) \to T(X) \rightrightarrows T(R)$$
is an equalizer diagram.
This diagram is obtained by applying $T$ to the coequalizer diagram
$$R \rightrightarrows X \to B,$$
where the two maps $R \to X$ are the compositions of $R \to X \times\_B X$ with the two projections.
To see that the first diagram is a (split) equalizer, it suffices to show that the second is a split coequalizer.
We can construct such a splitting as follows.
First, since $\pi : X \to B$ is surjective, and $B$ is projective, we can find a splitting $\sigma : B \to X$, so that $\pi \circ \sigma = 1$.
Next, consider the map $X \to X \times\_B X$ whose composition with the first projection to $X$ is the identity and whose composition with the second projection is $\sigma \circ \pi$ (since $\sigma$ is a splitting of $\pi$, the two maps to $B$ do indeed coincide, so we get such a map from the universal property of the pullback).
Now the map $R \to X \times\_B X$ is surjective and $X$ is projective, so we can lift this map $X \to X \times\_B X$ to a map $\tau : X \to R$ with the property that the compositions $X \to R \to X\times\_B X \to X$ are $1$ and $\sigma\circ\pi$ respectively.
The two maps $\sigma$ and $\tau$ then give us the desired splitting of the coequalizer diagram.
Finally, to add to the comments on the original question, this was formalized in Lean (as part of the Liquid Tensor Experiment) as the following lemma:
<https://github.com/leanprover-community/lean-liquid/blob/a1f009de0f88731492e998e4fd8f27de3f6952af/src/condensed/extr/basic.lean#L262>
| 8 | https://mathoverflow.net/users/490721 | 429839 | 174,147 |
https://mathoverflow.net/questions/416510 | 4 | I'm reading the Atanas Atanasov's course notes of Joe Harris' course [Geometry of Algebraic Curves](https://staff.math.su.se/shapiro/UIUC/curvesHarris.pdf)
and have a question about a suggested modification of an dimension
countinging argument applying methods from deformation theory.
On page 22 one consideres a version of Hurwitz scheme
$$ V\_{d,g}:= \{(X, f: X \to \mathbb{P}^2) \ \vert \
X \text{ curve of genus } g, f \text{ has degree } d
\text{ and is birational } \\ \text{ onto a plane curve with }
\delta \text{ nodes } \} $$
together with two canonical canonical projection maps
$V\_{d,g} \to M\_g $ (to the 'naive' moduli set) and
$V\_{d,g} \to \mathbb{P}^{\delta} \backslash \Delta$.
Rather elementary considerations in the script show that $\dim V\_{d,g}=3+g−1$
if $d(d+3)/2 \ge 3 \delta$ but the **Remark 4.2** says:
>
> There is a serious problem with this argument if $3 \delta> d(d + 3)/2 $
> but this can be fixed using deformation theory.
>
>
>
Namely, the counting method in the script used as intermediate equalities $\dim V\_{d,g}=d(d+3)/2−3\Delta+2\Delta=3+g−1$. Of course for $3 \delta> d(d + 3)/2 $ these considerations make no any sense, but the final equality between left and right is known to be still true.
How to fix this gap using deformation theoretic arguments
as suggested in the remark 4.2 in detail?
| https://mathoverflow.net/users/108274 | Deformation theoretic argument on dimension counting of naive Hurwitz scheme | You can get a lower bound on the dimension of $V\_{d,g}$ using deformation theory as follows. The deformation obstruction theory of a map $f : X \to Y$ between smooth varieties (where $f$ and $X$ are allowed to deform by $Y$ is held fixed) is governed by the complex
$$
\mathbb{L}\_f = \left[f^\*\Omega\_Y \to \Omega\_X\right]
$$
concentrated in degrees $[-1,0]$. The infinitessimal automorphisms, first order deformations and obstructions respectively lie in the groups $\mathrm{Ext}^i(\mathbb{L}\_f, \mathcal{O}\_X)$ for $i = 0,1,2$. In particular, if $\mathrm{Ext}^0(\mathbb{L}\_f, \mathcal{O}\_X) = 0$, then we have the lower bound
$$
\dim \mathrm{Def}(f : X \to Y) \geq \mathrm{ext}^1(\mathbb{L}\_f, \mathcal{O}\_X) - \mathrm{ext}^2(\mathbb{L}\_f, \mathcal{O}\_X).
$$
In the example $f : X \to \mathbb{P}^2$ where $X$ is a smooth curve mapping birationally onto a nodal curve, the map $f$ is an immersion so $f^\*\Omega\_Y \to \Omega\_X$ is surjective. Let $K = \operatorname{ker} f$ so
$$
\mathbb{L}\_f = K[1]
$$
is $K$ in degree $-1$. Then $\mathrm{Ext}^i(\mathbb{L}\_f, \mathcal{O}\_X) = \mathrm{Ext}^{i - 1}(K, \mathcal{O}\_X)$ which is $0$ for $i = 0$ so we have the dimension inequality above. Moreover, by the exact sequence
$$
0 \to K \to f^\*\Omega\_{\mathrm{P}^2} \to \Omega\_X \to 0
$$
we see that $\mathcal{E}xt^i(K,\mathcal{O}\_X) = 0$ for $i > 0$. Denoting by $K^\vee = \mathcal{H}om(K, \mathcal{O}\_X)$, we have that
$$
H^i(X, K^\vee) = \mathrm{Ext}^i(K, \mathcal{O}\_X) = \mathrm{Ext}^{i+1}(\mathbb{L}\_f, \mathcal{O}\_X)
$$
by the local-to-global spectral sequence.
Putting this together, we have the lower bound
$$
\dim\_{[f : X \to \mathbb{P}^2]} V\_{d,g} \geq \chi(X, K^\vee).
$$
Using the exact sequence
$$
0 \to T\_X \to f^\*T\_{\mathbb{P}^2} \to K^\vee \to 0
$$
we get
$$
\chi(K^\vee) = \chi(f^\*T\_{\mathbb{P}^2}) - \chi(T\_X)
$$
and pulling back the Euler sequence
$$
0 \to O\_X \to O\_X(1)^{\oplus 3} \to f^\*T\_{\mathbb{P}^2} \to 0
$$
we have
$$
\chi(f^\*T\_{\mathbb{P}^2}) = 3\chi(\mathcal{O}\_X(1)) - \chi(O\_X).
$$
Putting this together and using Riemann-Roch, we conclude
$$
\dim\_{[f]}V\_{d,g} \geq \chi(K^\vee) = 3(d-1+g) - (1-g) - (3g-3) = 3d + g - 1
$$
To make this an equality, we would need to know that the obstructions $H^1(X, K^\vee)$ vanish, which follows if $H^1(X, f^\*T\_{\mathbb{P}^2}) = 0$ and more generally is equivalent to the surjectivity of
$$
H^1(X, T\_X) \to H^1(X, f^\*T\_{\mathbb{P}^2}).
$$
I don't see any reason why this has to hold in general but I'm not sure.
| 4 | https://mathoverflow.net/users/12402 | 429841 | 174,148 |
https://mathoverflow.net/questions/429739 | 15 | **Qeustion:**
Given a Lie algebra $\mathfrak{g}$ over $\mathbb{Q}\_\ell$ with an ideal $\mathfrak{g}^O$ and a subalgebra $\mathfrak{h}$,
such that $\mathfrak{g}=\mathfrak{g}^O+\mathfrak{h}$.
Now given a faithful representation
$$\varphi:\mathfrak{g}\hookrightarrow \mathfrak{gl}(V)$$
such that the restrictions on $\mathfrak{g}^O$ and $\mathfrak{h}$ are semisimple, Faltings claim that $\varphi$ is semisimple.
My question is: why this is true?
**Background:**
In Faltings' book "[Rational Points](https://link.springer.com/book/10.1007/978-3-663-06812-9)" Chapter VI "[Complements](https://link.springer.com/chapter/10.1007/978-3-663-06812-9_6)", he generalized his result about Tate conjecture for abelian varieties over number field into finitely generated field over $\mathbb{Q}$. The main idea is to combine the complex Hodge theory and Tate conjecture over number field.
For example, consider the case $K$ is the function field of the (smooth geometric irreducible) scheme $X$ over a number field $L$ with a rational point $p\in X(L)$, we have a split exact sequence
$$
e\rightarrow \widehat{\pi\_1(X\_{\mathbb{C}})}\rightarrow \pi\_1^{\text{ét}}(X)\rightarrow \text{Gal}(\overline{L}/L)\rightarrow e
$$
where $X\_\mathbb{C}$ is the base change from $L$ to $\mathbb{C}$, $V\_\ell(A)$ is the Tate module tensor with $\mathbb{Q}\_\ell$.
It terms out that the Galois representation $\rho:\text{Gal}(\overline{K}/K)\rightarrow \text{Aut}(V\_\ell(A))$ will factor through the étale fundamental group.
Hence to show the $\rho$ is semisimple, we reduce to show that $\rho\_1:\pi\_1^{\text{ét}}(X)\rightarrow \text{Aut}(V\_\ell(A))$ is semisimple.
Now we have:
1. $\rho\_1|\_{\widehat{\pi\_1(X\_{\mathbb{C}})}}$ is semisimple, from the complex Hodge theory by Deligne.
2. $\rho\_1|\_{\text{Gal}(\overline{L}/L)}$ is semisimple by Tate conjecture over number field by Faltings, here we taking the restriction via the splitting by the rational point $p\in X(L)$.
Faltings claim that, therefore the representation $\rho$ is semisimple. To do that, he taking $\mathfrak{g},\mathfrak{g}^O,\mathfrak{h}$ to be the Lie algebra of the complex $\ell$-adic group $\rho\_1(\pi\_1^{\text{ét}}(X)),\rho\_1(\widehat{\pi\_1(X\_{\mathbb{C}})})$ and $\rho\_1(\text{Gal}(\overline{L}/L))$. We want to show that $\mathfrak{g}$ is completely reducible in $V\_\ell(A)$.
We know that this already holds for $\mathfrak{g}^O$ and $\mathfrak{h}$, and $\mathfrak{g}^O$ is an ideal in $\mathfrak{g}$, and $\mathfrak{g}=\mathfrak{g}^O+\mathfrak{h}$, hence completely reducible for $\mathfrak{g}$.
**What I tried:**
1. From the definition, a faithful representation $\rho:\mathfrak{g}\rightarrow\mathfrak{gl}(V)$ is semisimple if $\mathfrak{g}=\mathfrak{c}\times\mathfrak{l}$ is reductive, and $\mathfrak{c}$ acting on $V$ is semisimple, where $\mathfrak{c}$ is the radical and is abelian, and $\mathfrak{l}$ is Levi factor.
We need at least show that $\mathfrak{g}$ is reductive.
But consider the standard Borel of $\mathfrak{sl}\_2$, it is the extension of trivial Lie algebra with a trivial Lie algebra, one is the Lie algebra of $\mathbb{G}\_m$ and another the Lie algebra of $\mathbb{G}\_a$, they are reductive, but the lie algebra of Borel is not since it is solvable and non-abelian.
The trouble here is that, we can't distinct Lie algebra of $\mathbb{G}\_a$ and $\mathbb{G}\_m$, we do need to use the fact that the radical acting on $V$ is semisimple to get reductive. I tried different attempts but failed. Even the case $\mathfrak{g}^O$ is semisimple and $\mathfrak{h}$ is abelian is still hard to prove.
2. I tried to apply the Hochschild-Serre spectral sequence to get the completely reducible: for an exact sequence
$$0\rightarrow \mathfrak{h}\rightarrow\mathfrak{g}\rightarrow\mathfrak{g}/\mathfrak{h}\rightarrow 0$$
we have
$$
H^p(\mathfrak{g}/\mathfrak{h},H^q(\mathfrak{h},V))\Rightarrow H^{p+q}(\mathfrak{g},V).
$$
We want to show that $H^1(\mathfrak{g},\text{Hom}\_k(V/W,W))=0$ for all sub representation $W$,
what we have is $H^1(\mathfrak{g}^O,\text{Hom}\_k(V/W,W))=0$, $H^1(\mathfrak{h},\text{Hom}\_k(V/W,W))=0$.
But to make spectral sequence works, we need
$H^1(\mathfrak{g}/\mathfrak{h},\text{Hom}\_k(V/W,W)^\mathfrak{h})=0$.
Again, I can't find a good way to fix it.
3. I tried to use universal enveloping algebra, and reduce to an algebra representation question. But we don't have a nice formula even for the universal enveloping algebra of semi-direct product of two Lie algebras.
4. I also tried to prove the result without using any Lie algebra. For example, taking the algebra generated by the image in $\text{End}(V\_\ell(A))$, or consider the representation of $\ell$-adic Lie groups, but does not help.
**Why I think it is a research level problem:**
1. In the note, Faltings used the terminology $\mathfrak{g}$ is reductive in $M$ to say that $M$ is a semisimple $\mathfrak{g}$-module.
His statements seems to be more natural if we have algebraic group in mind, and the claim is easy in the algebraic group setting. So I think what he really thought is the algebraic group.
But there are crucial difference between algebraic group and lie algebra: we can't distinct $\mathbb{G}\_m$ and $\mathbb{G}\_a$.
2. In Lei Fu's paper [On the semisimplicity of pure sheaves](https://www.ams.org/journals/proc/1999-127-09/S0002-9939-99-05414-3/S0002-9939-99-05414-3.pdf),
he uses more several pages to prove the same question over finite fields, and crucially using that $\text{Gal}(\overline{\mathbb{F}\_q}/\mathbb{F}\_q)\simeq \mathbb{Z}$ in the proof. If we adopt Faltings' argument, we can greatly simplify Fu's paper, by replacing complex Hodge theory by Weil conjecture.
| https://mathoverflow.net/users/486528 | A possible gap in Faltings note to prove the Tate conjecture for finitely generated field over $\mathbb{Q}$ | WLOG, we may assume that $\phi$ is injective and identify $\mathfrak{g}$ with its image in $\mathrm{End}(V)$. Our goal is to construct a reductive algebraic subgroup of $\mathrm{GL}(V)$, whose Lie algebra coincides with $\mathfrak{g}$.
We may assume, thanks to Deligne's results, that $\mathfrak{g}^0$ (which comes from Hodge theory) is a semisimple Lie algebra. In addition, we may also assume that $\mathfrak{h}$ is an algebraic Lie subalgebra of $\mathrm{End}(V)$; indeed, since $\mathfrak{h}$ is the Lie algebra attached to the Galois action on the Tate module, its algebraicity is a theorem of Bogomolov.
Let $G\_0 \subset \mathrm{GL}(V)$ be the connected semisimple algebraic group, whose Lie algebra coincides with $\mathfrak{g}^0$ and $H\subset \mathrm{GL}(V)$ be the connected reductive algebraic group, whose Lie algebra coincides with $\mathfrak{h}^0$. We know that $H$ normalizes $G\_0$.
Let us consider the Lie subalgebra $\overline{\mathfrak{g}^0}\subset \mathrm{End}(V)$ that is the normalizer of $\mathfrak{g}^0$ in $\mathrm{End}(V)$. Clearly, $\overline{\mathfrak{g}^0}$ is an algebraic Lie subalgebra of $\mathrm{End}(V)$;
we write $G$ for the connected algebraic subgroup of $\mathrm{GL}(V)$, whose Lie algebra coincides with $\overline{\mathfrak{g}^0}$.
We have
$$\mathfrak{g}^0, \mathfrak{h} \subset \overline{\mathfrak{g}^0}\subset \mathrm{End}(V).$$
By definition, $\mathfrak{g}^0$ is an ideal in $\overline{\mathfrak{g}^0}$.
This gives rise to a natural Lie algebra homomorphism from $\overline{\mathfrak{g}^0}$ to the Lie algebra $\mathrm{Der}(\mathfrak{g}^0)$ of derivations of $\mathfrak{g}^0$, which is just the restriction of the adjoint representation
$$\mathrm{Ad}: \overline{\mathfrak{g}^0} \to \mathrm{Der}(\mathfrak{g}^0)\subset \mathrm{End}(\mathfrak{g}^0).$$
Since $\mathfrak{g}^0$ is semisimple, its every derivation is inner one, i.e.,
$\mathrm{Der}(\mathfrak{g}^0)=\mathfrak{g}^0$. So, we get the Lie algebra homomorphism
$\rho: \overline{\mathfrak{g}^0} \to \mathfrak{g}^0$ that coincides with the identity map on $\mathfrak{g}^0$ (in particular, $\rho$ is surjective) and such that
$[\rho(x),y]=[x,y]$ for all $x \in \overline{\mathfrak{g}^0}, y\in \mathfrak{g}^{0}$.
Then $\ker(\rho)$ is an ideal of $\mathfrak{g} $ that meets $\mathfrak{g}^{0}$ precisely at $\{0\}$, because the center of $\mathfrak{g}^{0}$ is $\{0\}$. Hence,
$$\overline{\mathfrak{g}}=\mathfrak{g}^{0}\oplus \ker(\rho).$$ In other words, $\ker(\rho)$ is the centralizer $\mathrm{End}\_{\mathfrak{g}^{0}}(V)$ of $\mathfrak{g}^{0}$ in $\mathfrak{g}$ in $\mathrm{End}(V)$. Since $\mathfrak{g}^{0}$ is semisimple, the $\mathfrak{g}^{0}$-module $V$ is semisimple and the centralizer $\mathrm{End}\_{\mathfrak{g}^{0}}(V)$ is a semisimple associative subalgebra of $\mathrm{End}(V)$. Viewed as the Lie (sub)algebra, $\mathrm{End}\_{\mathfrak{g}^{0}}(V)$ is reductive algebraic and coincides with the Lie algebra of
the connected reductive algebraic subgroup $\mathrm{Aut}\_{\mathfrak{g}^{0}}(V)$
of $\mathrm{GL}(V)$. Clearly, both $G\_0$ and $G\_1$ are normal closed subgroups of $G$. They mutually commute, and their Lie algebras $\mathfrak{g}^{0}$ and $\mathfrak{g}^{1}=\mathrm{End}\_{\mathfrak{g}^{0}}(V)$ meet precisely at $\{0\}$. Hence, the intersection of $G^0$ and $G^1$ (in $\mathrm{GL}(V)$) is a finite central subgroup of both $G^0$ and $G^1$. In addition, $H$ is a closed reductive subgroup of $G$, because its Lie algebra $\mathfrak{h}$ lies in $\overline{\mathfrak{g}}$. So,
$$\mathfrak{h}\subset \overline{\mathfrak{g}}=\mathfrak{g}^{0}\oplus \mathfrak{g}^{1}.$$
If $\mathfrak{h}^{1}$ is the image of $\mathfrak{h}$ in $\mathfrak{g}^{1}$ under the projection map then
$$\mathfrak{g}=\mathfrak{g}^{0}+\mathfrak{h}=\mathfrak{g}^{0}\oplus \mathfrak{h}^1.$$
We are done if we can prove that $\mathfrak{h}^1$ is the Lie algebra of a reductive algebraic subgroup of $\mathrm{GL}(V)$.
The homomorphism of reductive algebraic group
$$\pi: G\_0 \times G\_1 \to G, (g\_0, g\_1)=g\_0 g\_1=g\_1 g\_0$$
is an isogeny, because its tangent map
$$d\pi: \mathfrak{g}^{0}\oplus \mathfrak{g}^{1} \to \overline{\mathfrak{g}}$$
is an isomorphism (the identity map). Let $\tilde{H}\subset G\_0 \times G\_1$ be the identity component of the preimage $\pi^{-1}(H)$ of $H$. Clearly, $\pi$ induces an isogeny $\tilde{H} \to H$, hence $\tilde{H}$ is a reductive algebraic group. Let us consider the composition of homomorphisms of reductive algebraic groups
$$\tilde{H} \stackrel{\pi}{\to} H \subset G\_0 \times G\_1\to G\_1$$ where the last map is the projection map. Let $H\_1 \subset G\_1$ be the image of $\tilde{H}$. Then the Lie algebra of $H\_1$ is precisely $\mathfrak{h}^1$! Since $H\_1$ is the image of reductive $\tilde{H}$, it is isomorphic to a quotient of $\tilde{H}$ and therefore is also reductive. Hence, $G\_0 \times H\_1$ is also reductive and therefore the image $\tilde{G}\subset \mathrm{GL}(V)$ of the homomorphism
$$G\_0 \times H\_1 \to \mathrm{GL}(V), (g\_0,h\_1)\mapsto g\_o h\_1=h\_1 g\_0$$
is a reductive algebraic subgroup, whose Lie algebra is
$$\mathfrak{g}^0\oplus \mathfrak{h}^1=\mathfrak{g}.$$
Therefore the $\mathfrak{g}$-module $V$ is semisimple.
| 11 | https://mathoverflow.net/users/9658 | 429849 | 174,149 |
https://mathoverflow.net/questions/429855 | 2 | Let $G$ be a (connected ?) algebraic group and $X$ a smooth, projective, and connected algebraic curve, both over an algebraically closed field $k$ of characteristic $0$.
My questions are then as follows:
1. By "local systems" in this context, *do we mean lisse $\ell$-adic étale sheaves* on $X$ (with coefficients in $\bar{\mathbb{Q}}\_{\ell}$ ?), for some prime $\ell$ ?
2. It is apparently well-known to experts that the dg stack $LocSys(X)^G$ of $G$-equivariant local systems on $X$ is quasi-smooth (cf. subsubsection 1.1.5 of Arinkin–Gaitsgory, *Singular support of coherent sheaves, and the geometric Langlands conjecture*, [arXiv:1201.6343](https://arxiv.org/abs/1201.6343)); recall that a dg algebraic stack $\mathscr{Z}$ is quasi-smooth iff for all points $z \in |\mathscr{Z}|$, $H^i(T\_z\mathscr{Z}) = 0$ whenever $i < 0$. *Where might I find a reference for this fact ?*
Thanks in advance!
**Edit**: Thanks to Will Samin for pointing out to me that it should've been $i < 0$ instead of $i \not = -1, 0$.
| https://mathoverflow.net/users/143390 | The stack of equivariant local system is quasi-smooth | These are both answered in section 10 of the linked paper but since this is done at a very high level I will try to provide a down-to-earth explanation.
1 No, absolutely not. We mean local systems in the D-module sense, i.e. $G$-torsors with flat connection. So for $G =G L\_n$ these are $D$-modules that, restricted to $\mathcal O\_X$, are rank $n$ vector bundles.
This is crucial as the base field for the moduli space of local systems must be the coefficient field of these local systems. For coherent sheaves on this space to be equivalent to $D$-modules on $Bun\_G$, the coefficient field of this space must match the coefficient field of $Bun\_G$, which is the base field $k$. The only sheaf theory with coefficient field naturally the base field $k$ is $D$-module theory.
2 This is proven in Proposition 10.4.5 of the linked paper. I think you have the definition wrong and they only demand $H^i$ vanish for $i<0$. The calculation they do shows that $H^i(T\_z \mathcal L)$ is $H^{i+1}$ of $X$ with coefficients in the adjoint representation composed with $\mathcal L$. This is believable as it's not too hard to see (in any sheaf theory - $\ell$-adic, Betti, $D$-module, whatever) that deformations of a local system are controlled by $H^1$ of the adjoint. Since curves have de Rham cohomology concentrated in degrees $0,1,2$, we have $H^i(T\_z \mathcal L) \neq 0$ only for $i=-1, 0,1$.
| 6 | https://mathoverflow.net/users/18060 | 429856 | 174,155 |
https://mathoverflow.net/questions/47009 | 19 | Let $\chi$ be an irreducible (complex) character of a finite group, $G$. The Schur index $m\_{K}(\chi)$ of $\chi$ over the field $K$ is the smallest positive integer $m$ such that $m\chi$ is afforded by a representation over the field $K(\chi)$. The most interesting case is $K=\mathbb{Q}$. Given the character table, or only the particular character one is interested in, one can usually derive bounds for $m(\chi)=m\_{\mathbb{Q}}(\chi)$. For example, $m(\chi)$ divides $\chi(1)$ and $n[\chi^n,1\_G]$ for all $n\in \mathbb{N}$ (Fein), and the Benard-Schacher Theorem tells us that $\mathbb{Q}(\chi)$ contains a primitive $m(\chi)$-th root of unity.
On the other hand, the example of the quaternion group $Q\_8$ and the dihedral group $D\_8$ shows that two groups might have identical character tables, but corresponding characters with different Schur indices. I am curious wether there are examples that are even worse than this.
**Notation:** To state this more precisely, I'll make the following assumptions: We are given two finite groups $G$ and $H$, such that there is a bijection $\tau\colon {\rm Cl}(G) \to {\rm Cl}(H)$ from the classes of $G$ to the classes of $H$, and such that $\psi \mapsto \psi \circ \tau$ is a bijection ${\rm Irr}(H)\to {\rm Irr}(G)$. Now:
>
> Is there an example with $m(\chi)/m(\chi\circ\tau)\notin \{1,2,1/2\}$ for some $\chi\in {\rm Irr}(H)$?
>
>
> Is there an example with $G$ of odd order and $m(\chi) / m(\chi\circ\tau)\neq 1$ for some $\chi \in {\rm Irr}(H)$?
>
>
>
Now let us assume that we know the power maps of the character table. These are the maps $\pi\_n^G\colon {\rm Cl}(G)\to {\rm Cl}(G)$ induced by $g\mapsto g^n$. (These maps are stored in the tables of the character table library of GAP.) Given these maps, one can compute $[\chi\_C, 1\_C]$ for cyclic subgroups $C\leq G$, for example. Also we can compute the Frobenius-Schur Indicator and thus the Schur index over $\mathbb{R}$.
Now assume that $\tau\circ \pi\_n^G = \pi\_n^H\circ \tau$ in the above situation (then $(G,H)$ is called a Brauer pair).
>
> Is there a Brauer pair $(G,H)$ such that $m(\chi)/m(\chi\circ\tau)\neq 1$ for some $\chi\in {\rm Irr}(H)$?
>
>
>
I would appreciate any examples or (pointers to) results that show the impossibility of such examples.
Thanks
| https://mathoverflow.net/users/10266 | What can be said about Schur indices, given only the character table? | The following is a theorem of K. Kronstein:
>
> **Theorem:** for $k$ a number field or a nonarchimedean completion of a number field, if it is possible to detect the Schur index $m\_k$ of all finite groups from their character table and power maps, then $m\_k(\chi) \leq 2$ for all characters $\chi$ of finite groups.
>
>
>
In particular, for a finite group $G$, the map $\tau: G \times G \to G \times G$ defined by $\tau(x,y) = (x,y^{-1})$ induces a bijection on conjugacy classes preserving the power maps $x \mapsto x^n$ and a bijection on characters, but does not necessarily preserve Schur indices larger than 2. Thus, the map $\tau$ provides a positive answer to the first and third questions. It also provides a positive answer to the second question once one produces an example of a group of odd order with a character with Schur index greater than 2.
I provide Kronstein's proof here:
Suppose $\chi$ is a character of $G$, $k$ is a number field or nonarchimedean completion of a number field, and the Schur index $m = m\_k(\chi)$ is at least $3$.
Let $K = k(\chi)$, let $V$ be an irreducible $KG$-module affording the character $m\chi$, and let $D = End\_{KG}(V)$. It is a division algebra with order $m$ in the Brauer group of $K$.
Consider the characters $\chi \boxtimes \chi$ and $\chi \boxtimes \chi^\vee$ on $G \times G$. Then
$$End\_{K(G\times G)}(V\boxtimes V) = D \otimes\_K D.$$
Since $m>2$, $D \otimes\_K D$ is not split, so $\chi\boxtimes \chi$ has Schur index greater than 1. However, $$End\_{K(G \times G)}(V \boxtimes V^\vee) = D \otimes\_K D^{op},$$
which splits over $K$, and thus $\chi \boxtimes \chi^\vee$ has Schur index 1.
---
*Reference:*
Karl Kronstein. Character tables and the Schur index. In *Representation Theory of Finite Groups and Related Topics*,
volume 21 of *Proceedings of Symposia in Pure Mathematics*, pages 97–98. American Mathematical Soc., 1971
| 2 | https://mathoverflow.net/users/125523 | 429861 | 174,156 |
https://mathoverflow.net/questions/429868 | 6 | In other words, does there exist a metric space $(E,\rho)$ with finite Hausdorff dimension but infinite packing dimension?
Here are my thoughts:
* I know that it is generally hard to relate Hausdorff and packing measures and dimensions (other than the fact that Hausdorff dimension is always less than or equal to packing dimension), and I was not able to find a direct proof using the definitions.
* Assume that $(E,\rho)$ is separable and complete. It is known that $n:=\dim\_C E \le \dim\_H E\le \dim\_P E$ (respectively denoting topological covering dimension, Hausdorff dimension, and packing dimension). In particular, if $(E,\rho)$ has finite Hausdorff dimension, then it has - as a topological space - finite covering dimension, and is thus homeomorphic to a subset of $\mathbb{R}^{2n+1}$ (see book by Hurewicz below; a similar embedding result is in the proof of Theorem 6.3.10 in Edgar's book). In particular, there exists a metric $\rho'$ on $E$ that induces the same topology as $\rho$ such that $(E,\rho')$ has packing dimension at most $2n+1$. Then the identity map $\textrm{id}\colon (E,\rho') \to (E,\rho)$ is bi-continuous (because the metrics induce the same topology), but this is not enough to relate their packing dimensions. If it was $\eta$-Hölder continuous for some $\eta \in (0,1)$, then we could deduce $\dim\_P (E,\rho) \le (2n+1) / \eta < \infty$, but this must a priori not be the case.
I am also interested in any results that require more specific assumptions on $E$, say separable, complete, compact, etc.
---
Edit: I refer to packing dimension as defined [here](https://en.wikipedia.org/wiki/Packing_dimension).
---
*Hurewicz, W.; Wallman, H.*, Dimension theory., 165 p. Princeton University Press (1941). [ZBL67.1092.03](https://zbmath.org/?q=an:67.1092.03).
*Edgar, Gerald A.*, Measure, topology, and fractal geometry, Undergraduate Texts in Mathematics. New York etc.: Springer-Verlag. ix, 230 p. DM 58.00/hbk (1990). [ZBL0727.28003](https://zbmath.org/?q=an:0727.28003).
| https://mathoverflow.net/users/322634 | Does finite Hausdorff dimension imply finite packing dimension? | A construction used (repeatedy) in the paper
*Edgar, G. A.*, [**Centered densities and fractal measures**](http://nyjm.albany.edu:8000/j/2007/13-4.html), New York J. Math. 13, 33-87 (2007). [ZBL1112.28004](https://zbmath.org/?q=an:1112.28004).
For more information, see that paper.
---
We construct a compact metric space $E$ for this purpose.
For $n=1,2,3,\dots$ let $k\_n \in \{2,3,4,5,\dots\}$ and $r\_n \in (0,1/2)$. More precise properties are to be specified later.
Let $T\_n$ be a set with $k\_n$ elements (with the discrete topology). Let
$$
K\_n = k\_1 k\_2\dots k\_n,\qquad R\_n = r\_1 r\_2\dots r\_n.
$$
so $K\_n \nearrow \infty$ and $R\_n \searrow 0.$ Also let $R\_0 = 1$.
Our space is
$E = \prod\_{n=1}^\infty T\_n$ with the product topology. So $E$ is separable, compact. Define metric $\rho$ on $E$ as follows:
Let $x = (x\_1,x\_2,\dots), y=(y\_1,y\_2,\dots) \in E$. For $x=y$, define $\rho(x,y) = 0$. For $x \ne y$, let $m \in \mathbb N$ be such that
$$
x\_j=y\_j\quad\text{for }1 \le j \le m,\quad\text{and }
x\_{m+1} \ne y\_{m+1}
$$
and define $\rho(x,y) = R\_m$.
Then
$$
E \text{ has diameter } R\_0 .
$$
$E$ is the disjoint union of $k\_1$ closed subsets $E[a\_1], a\_1 \in T\_1$,
$$
E[a\_1] = \{(x\_1,x\_2,\dots) \in E : x\_1 = a\_1\}
\text{ has diameter } R\_1 .
$$
Each set $E[a\_1]$ is the disjoint union of $k\_2$ closed subsets
$E[a\_1,a\_2], a\_2 \in T\_2$,
$$
E[a\_1,a\_2] = \{(x\_1,x\_2,\dots) \in E : x\_1 = a\_1, x\_2=a\_2\}
\text{ has diameter } R\_2
$$
Continue in this way, so that each $E[a\_1,\dots,a\_m]$ is the disjoint union
of $k\_{m+1}$ closed subsets $E[a\_1,\dots,a\_m,a\_{m+1}], a\_{m+1} \in T\_{m+1}$, and $E[a\_1,\dots,a\_m]$ has diameter $R\_{m}$.
For each $m$, the space $E$ is the disjoint union of $K\_m$ closed sets of diameter $R\_m$.
Define a "uniform" measure $\mu$ on $T$ so that
$$
\mu\big(E[a\_1,\dots,a\_m]\big) = \frac{1}{K\_m} .
$$
Now, let $s\in(0,\infty)$. The upper $s$-density of $\mu$ at a point $x \in E$ is
$$
\overline{D}\_\mu^s(x) =
\limsup\_{\eta\searrow 0} \frac{\mu(B\_\eta(x))}{(\operatorname{diam} B\_\eta(x))^s} =
\limsup\_{n \to \infty} \frac{1/K\_n}{R\_n^s}=
\limsup\_{n \to \infty} \frac{1}{K\_nR\_n^s} .
$$
The lower density is $$
\underline{D}\_\mu^s(x)
= \liminf\_{n \to \infty} \frac{1}{K\_nR\_n^s} .
$$
Now what we need to do is select $k\_n, r\_n$ at the beginning so that, for all $s \in (0,\infty)$ we have
$$
\limsup\_{n \to \infty} \frac{1}{K\_nR\_n^s} = +\infty,\qquad
\liminf\_{n \to \infty} \frac{1}{K\_nR\_n^s} = 0
$$
Then our metric space $(E,\rho)$ will have Hausdorff dimension $0$ and packing dimension $\infty$.
| 7 | https://mathoverflow.net/users/454 | 429872 | 174,158 |
https://mathoverflow.net/questions/429862 | 5 | Let $G$ be a finite group and let $p$ be a prime number such that $p\mid |G|$.
Let $\text{IBr}(G)$ denote the set of irreducible Brauer characters of $G$ for the prime $p$.
Assume $\mathbb{F}\_{q}$ is a splitting field for $G$ where $q=p^f$ for some positive integer $f$.
Set $r:=|\text{IBr}(G)|$.
Let $\{\rho\_1, ..., \rho\_r\}$ be the a set of representatives of all simple $\mathbb{F}\_{q}G$-modules up to isomorphism.
It is well-known that $\text{IBr}(G)$ is linearly independent.
Now, take the $\mathbb{F}\_{q}G$-traces of $\rho\_j$ at the $p'$-classes and write the results in a vector, for each $j$.
>
> Does the list of vectors obtained in that way always have the property that there are no repeated rows?
>
>
>
Example:
Doing the computations for $G=A\_5$, the alternating group acting on $5$ symbols, for the prime number $p=2$ yields the following:
$[ Z(2)^0, Z(2)^0, Z(2)^0, Z(2)^0 ]$
$[ 0\*Z(2), Z(2)^0, Z(2^2)^2, Z(2^2) ]$
$[ 0\*Z(2), Z(2)^0, Z(2^2), Z(2^2)^2 ]$
$[ 0\*Z(2), Z(2)^0, Z(2)^0, Z(2)^0 ]$
The list of vectors obtained in that way has the property that no two rows are identical.
>
> Is this always the case (for any finite group G and for any prime number p)?
>
>
>
A reference would be cool.
| https://mathoverflow.net/users/91107 | Do $F$-traces of simple modules at $p'$-classes uniquely determine the module? | It is still the case that the $\mathbb{F}\_{q}$-valued trace functions of the (say) $\ell$ non-isomorphic simple $\mathbb{F}\_{q}$-modules $V\_{1},V\_{2}, \ldots V\_{\ell}$, are linearly independent, where $G$ has $\ell$ $p$-regular classes.
This was known to Brauer and Nesbitt, and a proof may be found (for example) in the book of Curtis and Reiner (1962).
The usual proof is to note that for each $i$ we may take an $\mathbb{F}\_{q}$-linear combination of group elements which is represented by an idempotent of trace $1$ in ${\rm End}\_{\mathbb{F}\_{q}}(V\_{i})$ and represented by the zero matrix in ${\rm End}\_{\mathbb{F}\_{q}}(V\_{j})$ for all $j \neq i$.
Since (over $\mathbb{F}\_{q}$) the trace of $g$ in its action on any $V\_{k}$ is the trace of its $p^{\prime}$-part $g\_{p^{\prime}}$, the claimed result follows.
| 4 | https://mathoverflow.net/users/14450 | 429879 | 174,159 |
https://mathoverflow.net/questions/429886 | 1 | Let us define
$$
\mathbb{H}^{1} = H^{1}(-L,0) \times H^{1}(0,L) \ \ \text{and} \ \ \mathbb{L}^{2} = L^{2}(-L,0)\times L^{2}(0,L),
$$
where $H^{1}(I) = \big\lbrace u \in L^{2}(I) \ \text{and} \ u\_{x} \in L^{2}(I); I = (a,b) \big\rbrace$.
Besides these,
$$
\mathbb{M} = \big\lbrace (u,v) \in \mathbb{H}^{1}; u(-L) = v(L) = 0 \ \text{and} \ u(0) = v(0) \big\rbrace .
$$
Under the above conditions, we have that the phase space is given by
$$
\mathcal{H} = \mathbb{M} \times \mathbb{L}^{2}.
$$
Note that this space equipped with the inner product
$$
\langle (u\_{1},v\_{1},w\_{1},z\_{1}), (u\_{2},v\_{2},w\_{2},z\_{2}) \rangle = \int\_{-L}^{0}u\_{1\_{x}}\overline{u}\_{2\_{x}} + \int\_{0}^{L}v\_{1\_{x}}\overline{v}\_{2\_{x}} + \int\_{-L}^{0}w\_{1}\overline{w}\_{2} + \int\_{0}^{L}z\_{1}\overline{z}\_{2}
$$
is a Hilbert space.
| https://mathoverflow.net/users/481556 | Inequality involving Sobolev spaces | Just take a derivative, and you get
$$ i \lambda u\_x = f\_x - w\_x $$
So taking the $L^2(-L,0)$ norms on both sides, you get
$$ \lambda^2 \int |u\_x|^2 \leq \int |f\_x - w\_x|^2 $$
The RHS can be expanded and estimated using AM-GM to be
$$ \lambda^2 \int |u\_x|^2 \leq 2 \int |f\_x|^2 + |w\_x|^2 $$
The first term is bounded by $\|F\|^2$, and the second, by your assumption on the $L^2$ bound of $w\_x$, is bounded by $\|U\| \|F\|$.
| 2 | https://mathoverflow.net/users/3948 | 429890 | 174,163 |
https://mathoverflow.net/questions/426434 | 3 | *This question was originally posted last week in Math Stack Exchange (see [here](https://math.stackexchange.com/questions/4483240/compatibility-of-pullbacks-with-an-equivalence-relation)).*
I'm currently working on the proof of the existence of the sheafification in Angelo Vistoli’s 2007 [*Notes on Grothendieck topologies,
fibered categories
and descent theory*](https://arxiv.org/pdf/math/0412512.pdf), but i currently stuck on a statement in the proof of Theorem 2.64. $(ii)$. That is the following:
Note: In this context, a Grothendieck topology is a Singleton Grothendieck topology, such that every covering of an object $U$ in $C$ is a single map $\phi:T\rightarrow U$.
We have a site $C$, that is a category $C$ with a Grothendieck topology. Then we have a functor $$F:C^{op}\rightarrow Set$$
We define an equivalence relation $\sim$ for every object $U$ of $C$ on $F(U)$ as: $a\sim b$ if there exists a covering $\phi:T\rightarrow U$, such that the pullback $F(\phi)=\phi^\*$ of $a$ and $b$ coincide in $F(T)$. In other words $\phi^\*(a)=\phi^\*(b)$.
Now the statement is, that for every morphism $f:S\rightarrow U$ the pullback $F(f)=f^\*:F(U)\rightarrow F(S)$ is compatible with $\sim$. That means:
$$
a,b\in F(U):a\sim b\Rightarrow f^\*(a)\sim f^\*(b)
$$
**Question:** How do I prove this?
| https://mathoverflow.net/users/485069 | Compatibility of pullbacks with an equivalence relation | After I saw my mistake I got the proof really quick. We need the fibre product of $f$ and $\phi$, where $f$ is an arbitrarily morphism and $\phi$ the covering from $a\sim b$.
So we get the following commutative diagram:
$\require{AMScd}$
\begin{CD}
S\times\_UT @>{pr\_2}>> T\\
@V{pr\_1}VV @VV{\phi}V\\
S @>{f}>> U
\end{CD}
hence
$$
\phi\circ pr\_2=f\circ pr\_1.
$$
Using $F$ on both sides we get
$$
F(\phi\circ pr\_2)=F(f\circ pr\_1)\quad\text{ or }\quad pr\_2^\*\circ\phi^\*=pr\_1^\*\circ f^\*.
$$
Hence, $pr\_1^\*(f^\*(a))=(pr\_1^\*\circ f^\*)(a)=pr\_2^\*(\phi^\*(a))=pr\_2^\*(\phi^\*(b))=(pr\_1^\*\circ f^\*)(b)=pr\_1^\*(f^\*(b)).$
Since $pr\_1$ is a covering (Covering Axioms), we get $f^\*(a)\sim f^\*(b)$.
q.e.d.
| 0 | https://mathoverflow.net/users/485069 | 429901 | 174,166 |
https://mathoverflow.net/questions/429896 | 0 | How can we represent F(x,m) in the infinte polynominal of x,m?
(Note that F(x,m) is the incomplete elliptical integral of the first kind, and I used its representation in the wikipedia)
More specifically, what is the value of [F(x/2,(cos m)^-0.5)-F(m,(cos m)^-0.5)]\*(cos m)^-0.5 when terms of O(m^3),O(x^3) are ignored?
I searched about it for a while, but there was not so much information about the incomplete one actually.
| https://mathoverflow.net/users/489097 | Approximation of Incomplete elliptic integral of first kind | The series expansion in powers of $k$ of the incomplete elliptic integral of the first kind
$$F(\varphi,k)= \int\_0^\varphi \frac {d\theta}{\sqrt{1 - k^2 \sin^2 \theta}}$$
can be simply obtained by expanding the integrand,
$$F(\varphi,k)=\varphi+\frac{1}{4} k^2 (\varphi-\sin \varphi \cos \varphi)+\frac{3}{256} k^4 (12 \varphi-8 \sin 2 \varphi+\sin 4 \varphi)+{\cal O}(k^6),$$
and then if you wish you can further expand in powers of $\varphi$.
| 1 | https://mathoverflow.net/users/11260 | 429905 | 174,168 |
https://mathoverflow.net/questions/429893 | 3 | Given a first order elliptic operator $D:\Gamma(X; E)\to \Gamma(X; F)$ where $X$ is a closed manifold, and $E\to X, F\to X$ vector bundles, we know that $D$ induces a Fredholm operator
between the spaces of Sobolev sections $D: W^{k+1,2}(X;E)\to W^{k,2}(X;F) $.
Therefore we can compute its index which is a topological quantity given by the celebrated Atiyah-Singer index theorem.
Now suppose that $X$ is a compact manifold with boundary, and consider $D$ as an operator
$$D: W\_0^{k+1,2}(X;E)\to W\_0^{k,2}(X;F) $$
where $W\_0^{k,2}(X;E)$ is the space of Sobolev sections vanishing over $\partial X$: the completion of $C^\infty\_0(\mathrm {int} (X); E)$ in the Sobolev norm $\| \cdot \|\_{W^{k,2}} $.
$D$ is again a Fredholm operator.
>
> Is there a formula for its index?
>
>
>
Note: this is different from the case considered by Atiyah, Patodi and Singer; in my case I impose a boundary condition $f|\_{\partial X} =0$, in their case they impose $\Pi\_{-} f|\_{\partial X} = 0$ where $\Pi\_{-}$ is the $L^2$-projector to the negative eigenspace of the boundary operator $L$ (i.e. over a collar of the boundary $D = \sigma(\frac {\partial}{dt} + L$)).
| https://mathoverflow.net/users/99042 | Index formula for elliptic operators acting on Sobolev sections vanishing on the boundary (say $D: H_0^k(\Omega) \to H_0^{k-1}(\Omega)$) | Local boundary conditions such as the Dirichlet condition you mention were considered in Atiyah-Bott, The index problem for manifolds with boundary. 1964 Differential Analysis, Bombay Colloq., 1964 pp. 175–186 Oxford Univ. Press, London. (There is also a chapter by Atiyah in Palais's Seminar on the Atiyah-Singer Index Theorem with similar material.)
My understanding is that such boundary conditions do not typically lead to elliptic problems, eg for the signature operator. This was one of the motivations for the Atiyah-Patodi-Singer approach.
| 4 | https://mathoverflow.net/users/3460 | 429916 | 174,171 |
https://mathoverflow.net/questions/429915 | 2 | Assume we have an irreducible algebraic cycle $Z$ on $X\times Y$ where $X$ and $Y$ are projective varieties ($X$ is smooth) such that restriction of $Z$ to $U\times Y$ where $U\subset X$ is a Zariski open is a graph of a regular map from $U$ to $Y$. Is any cycle close enough to $Z$ on the Chow variety also graph of a morphism when restricted to $U\times Y$? In other words is the family of irreducible cycles with the aforementioned property a Zariski open subset of the Chow variety?
It seems that one can modify the proof of the fact $\text{Hom}$ scheme is an open sub-scheme of the Hilbert scheme to this setting, I just wanted to make sure this is correct and whether there is a simpler proof.
| https://mathoverflow.net/users/127776 | Cycles that are graphs of morphisms | Not as stated. Let $X$ be $\mathbb P^2$, $Y$ be the blowup of $\mathbb P^2$ on one point $P$, $Z$ the graph of the blowup map $Y \to X$, which is the graph of a regular map on the open set $U$ defined as the complement of $P$.
We can deform $Z$ in a family that moves the point $P$. The cycles in that family will each be regular on some open set, but it won't be the same open set $U$.
Allowing $U$ to vary, the answer is positive, since this occurs if and only if the intersection number of $Z$ with a general fiber of the projection to $X$ is $1$, and that's an open (and closed) condition in the Chow variety.
| 2 | https://mathoverflow.net/users/18060 | 429917 | 174,172 |
https://mathoverflow.net/questions/429914 | 0 | I am looking for an example of the following:
Find a bijective, differentiable function $f$ and continuous probability density functions $q\_1\ne q\_2$ such that $f\_\*q\_1=p=f\_\*q\_2$, where $f\_\*$ is the pushforward density and $p$ is continuous as well. What if continuity is strengthened to differentiability?
Edit: Intuitively this seems impossible, just by continuity considerations; e.g. pick a neighborhood where $q\_1$ and $q\_2$ differ, and invoke bijectivity of $f$.
| https://mathoverflow.net/users/486206 | Transforming two smooth densities to the same density | This is impossible if $f$ is injective, without further assumptions such as bijective, differentiable, etc. Let $Q\_1,Q\_2$ be probability measures on a measurable space $(\Omega, \mathcal{F})$, and assume $f\_\* Q\_1 = f\_\* Q\_2$ for some injective (bimeasurable) $f : (\Omega,\mathcal{F}) \to (\Xi,\mathcal{G})$. For any $A\in \mathcal{F}$, definitions give
$$
Q\_1(A) = f\_\* Q\_1 (f(A)) = f\_\* Q\_2 (f(A)) = Q\_2(A).
$$
Thus, $Q\_1 = Q\_2$.
| 3 | https://mathoverflow.net/users/99418 | 429919 | 174,173 |
https://mathoverflow.net/questions/429725 | 2 | There's been [some debate at the nLab](https://nforum.ncatlab.org/discussion/3306/identity-type/#Item_0) recently over the names of "identity type" and "path type" in certain dependent type theories.
One user wrote that
>
> Many cubical type theorists make the distinction between identity types like Martin-Löf’s identity types, Swan’s identity types, and the higher observational type theory identity types, where the rules for the identity types all imply the definitional J rule as a theorem (and they call all 3 types ’identity types’), vs the cubical path types where the rules only imply the J rule up to a path, and additional regularity rules are needed to make the J rule definitionally valid.
>
>
>
but Mike Shulman says that
>
> The identity types of higher observational type theory also do not satisfy definitional computation for J.
>
>
>
and a user named Nathan said that
>
> Cubical type theorists make the distinction between path types and identity types because path types are literally functions out of an interval, just like how paths in Euclidean space are functions out of the unit interval, not because of the definitional/propositional J distinction. Otherwise, why are the types in XTT called “path types” instead of “identity types”? They satisfy definitional J by coercion and regularity.
>
>
>
Mike Shulman also wrote the two comments earlier in the discussion:
>
> I would actually prefer that the phrase “identity type” be usable for whatever notion is appropriate for the particular type theory. This includes the Martin-Lof identity type (a.k.a. “jdentity type”), the path type of cubical type theroies, and the identity type of higher observational type theory, all of which satisfy different formal rules but are still “types of identifications”.
>
>
>
>
> Yes, I’m aware that cubical type theorists tend to use “identity type” to mean the Martin-Lof one. (Although in cubical type theory there is also a “Swan identity type” that’s different from both path types and ML identity types.) I’m saying I would rather advocate a different usage. Some cubical type theorists have taken to calling the ML identity type the “jdentity type”, since its defining feature is the J rule.
>
>
>
In dependent type theory (such as homotopy type theory) both "path type" and "identity type" generally refer to type families whose categorical semantics in $(\infty,1)$-categories are path space objects. However, in most definitions, the type family is called "identity type"; this includes Per Martin-Löf's definition, [Andrew Swan's definition](https://arxiv.org/abs/1808.00915), and Mike Shulman's definition in higher observational type theory. The sole exception to this seems to be in the various flavors of cubical type theory, where the corresponding type is called "path type". However, the induction principle for Martin-Löf's identity types is called "path induction", which along with the categorical semantics as path space objects indicates that "path type" may also be an appropriate name for the various other identity types.
Why are path types called path types in cubical type theory? And why aren't path types and identity types used interchangeably as synonyms elsewhere if all the types behave like path space objects?
| https://mathoverflow.net/users/483446 | Path types and identity types in dependent type theory | Prior to about a decade ago, no one used "path" terminology for identity types. The identification of the semantics of identity types with path objects dates to Awodey and Warren's [Homotopy theoretic models of identity types](https://arxiv.org/abs/0709.0248) and Voevodsky's [simplicial model of univalent foundations](https://arxiv.org/abs/1211.2851). After this, some people working in homotopy type theory started calling Martin-Lof identity types (which were the only ones being used at that time) "path types" and their elements "paths" to emphasize the homotopical intuition. This perspective was ascendant when the [HoTT Book](https://homotopytypetheory.org/book/) was written, and is the origin of the phrase "path induction" for identity elimination.
A little while after that, some of us started to back off from this terminology, partly because of rising interest in [topological/cohesive models and versions](https://arxiv.org/abs/1509.07584) of type theory, where in addition to the "homotopical" notion of "path" encoded by the identity types, there is a separate unrelated "topological" notion of "path" represented by actual functions out of an actual interval. (In fact, of course, topological paths also exist in Book HoTT, as soon as you define real numbers and topological spaces!) To reserve "path" for topological notions, we started to return to the terminology "identity type" and call their elements "identifications".
Separately and unrelatedly, cubical type theory arrived on the scene with two or even three notions of "identity type", which therefore needed different names to distinguish them. I was not one of the inventors of cubical type theories, so I can't say with certainty why its "path types" are called thusly, but I expect Nathan has it right. Unlike Martin-Lof identity types, cubical path types really are "paths" in the sense of "function out of an interval", although the interval is a primitive object (not even a type) rather than a topological set of real numbers. So inside cubical type theory, it makes some sense to reserve "path type" for those and "identity type" for the others.
| 7 | https://mathoverflow.net/users/49 | 429931 | 174,177 |
https://mathoverflow.net/questions/429925 | 4 | I have this problem at the moment which the strong topology $\beta (E;E^\* )$ is defined, when $E$ is a locally convex space. This topology is generated by the basic open sets:
$$U=\{x \in E : \sup\_{f \in B} |\langle f,x \rangle|<\varepsilon\},$$
where $B\subset E^\* $ is bounded. In this way, we say that $B$ is bounded if for all $x\in E, \ \sup\_{f\in B} |\langle f,x \rangle|<\infty,$ which is equivalent to the $\omega^\* -$boundless. If we consider the strong topology $\beta(E^\* ; E),$ now in $E^\* , $ the basic open sets are
$$V=\{f \in E^\* : \sup\_{x \in A} |\langle f,x \rangle|<\varepsilon\},$$
where $A\subset E$ is bounded. It is well known that a set $A$ is bounded if, and only if, is weakly bounded, and because of that, we have
$\sup\_{x \in A}|\langle f,x \rangle|<\infty, \ \forall f \in E^\* . $
So, my question is: when we say that $B\subset E^\* $ is bounded, do we mean that it is bounded in the strong topology or in the weak\* topology? Or are they equivalent?
| https://mathoverflow.net/users/489810 | Weak* bounded and strong bounded are the same? | In general, $\sigma(E^\*,E)$-bounded sets need not be $\beta(E^\*,E)$-bounded. For an example, let $E$ be the set of scalar sequences with only finitely many non-zero terms endowed with the norm $\|x\|\_\infty=\sup\{|x\_n|:n\in\mathbb N\}$. For the evaluations $\delta\_n(x)=x\_n$, the set $B=\{n\delta\_n:n\in\mathbb N\}$ is $\sigma(E^\*,E)$-bounded but not $\beta(E^\*,E)$-bounded.
A sufficient condition for the coincidence of weak$^\*$- and strongly bounded sets is *barrelledness* of the locally convex space $E$ since then $\sigma(E^\*,E)$-bounded sets are even equi-continuous.
I am not sure whether there is a standard what is meant by just boundedness and I would always specify to *pointwise boundedness* or *uniform boundedness on $E$-bounded sets*.
| 8 | https://mathoverflow.net/users/21051 | 429940 | 174,180 |
https://mathoverflow.net/questions/427437 | 2 | Let $n\geq 1$. Let $[n]=\{0<1\}^n$ equipped with the product order. Let $f:[n]\to [n]$ be a strictly increasing map. When $f$ is bijective, there exists a permutation $\sigma$ of $\{1,\dots,n\}$ such that $f(\epsilon\_1,\dots,\epsilon\_n)=(\epsilon\_{\sigma(1)},\dots,\epsilon\_{\sigma(n)})$.
>
> Is there such a representation theorem when $f$ is not bijective ?
>
>
>
For $n=2$, the only maps are $(\epsilon\_1,\epsilon\_2)\to (\epsilon\_1,\epsilon\_2)$, $(\epsilon\_1,\epsilon\_2)\to (\epsilon\_2,\epsilon\_1)$, $(\epsilon\_1,\epsilon\_2)\to (\min(\epsilon\_1,\epsilon\_2),\max(\epsilon\_1,\epsilon\_2))$ and $(\epsilon\_1,\epsilon\_2)\to (\max(\epsilon\_1,\epsilon\_2),\min(\epsilon\_1,\epsilon\_2))$. For $n\geq 3$, things become more complicated and I am not aware of any canonical representation, or at least of a way of listing all strictly increasing maps $f:[n]\to [n]$.
**Motivation**: When $f$ is bijective or for $n=2$, I can see $f$ as a continuous map $[0,1]^n\to [0,1]^n$ which moreover, for people interested in directed homotopy theory, takes a directed path of $[0,1]^n$ to another directed path of $[0,1]^n$ preserving the initial and final states of the $n$-cube. I would like to do the same thing for the other cases.
| https://mathoverflow.net/users/24563 | How to describe this set of maps of posets? | I have found a way published in a recent preprint (<https://doi.org/10.48550/arXiv.2209.02667>).
>
> **Theorem**: Let $n\geq 1$. Let $f=(f\_1,\dots,f\_n):[n]\to [n]$ be a stricly increasing map. Then there is the equality $
> f\_i(x\_1,\dots,x\_n) = \max\_{(\epsilon\_1,\dots,\epsilon\_n)\in
> f\_i^{-1}(1)} \min \{x\_k\mid \epsilon\_k=1\}$ for all $1\leq i\leq n$.
>
>
>
I explain the result with an example instead. Consider the map $f:[3]\to[3]$ depicted in the following picture (the top diagram is the source, the bottom diagram is the image):
Let $f=(f\_1,f\_2,f\_3)$. For boolean values, $\min$ means "and" and $\max$ means "or". If $x\_1=1$ and $x\_3=1$, or $x\_1=1$ and $x\_2=1$ and $x\_3=1$, then $f\_1(x\_1,x\_2,x\_3)=1$. Thus $f\_1(x\_1,x\_2,x\_3)=\max(\min(x\_1,x\_3),\min(x\_1,x\_2,x\_3))$. If $x\_1=1$ and $x\_2=1$, or $x\_2=1$ and $x\_3=1$, or $x\_1=1$ and $x\_2=1$ and $x\_3=1$, then $f\_2(x\_1,x\_2,x\_3)=1$. Thus $f\_2(x\_1,x\_2,x\_3)=\max(\min(x\_1,x\_2),\min(x\_2,x\_3),\min(x\_1,x\_2,x\_3))$. Finally, if $x\_1=1$ and $x\_2=1$, or $x\_1=1$ and $x\_3=1$, or $x\_2=1$ and $x\_3=1$, or $x\_1=1$ and $x\_2=1$ and $x\_3=1$, then $f\_3(x\_1,x\_2,x\_3)=1$. Thus $f\_3(x\_1,x\_2,x\_3)=\max(\min(x\_1,x\_2),\min(x\_1,x\_3),\min(x\_2,x\_3),\min(x\_1,x\_2,x\_3))$.
| 1 | https://mathoverflow.net/users/24563 | 429941 | 174,181 |
https://mathoverflow.net/questions/365603 | 0 | Let $f\_n:\mathbb{R}\rightarrow \mathbb{R}$ be a sequence of functions and define $F\_n:= f\_n\circ \dots\circ f\_1$. Then $F\_n$ is continuous. However, the pointwise limit need not be (consider Mateusz's example of
$$
f\_n = \frac{\sqrt{2} x}{\sqrt{1 + 4 x^2}} \qquad F\_n \to \operatorname{sign}(x)
$$
the sign function).
In general, what can be said about the limits of an iterated function system? Are such functions studied and if so what are they called?
All I figure at the moment is that they must necessary be of [Baire 1](https://en.wikipedia.org/wiki/Baire_function) type.
| https://mathoverflow.net/users/36886 | Infinite composition of continuous functions | Since you’re searching for names: there is a rich literature on *discrete non-autonomous dynamical systems* generated by an infinite family of continuous maps. Here "non-autonomous" means of course that the map you apply depends on the order of iteration, which is exactly your subject.
For instance:
1. Cánovas, J. S. (2006). On ω-limit sets of non-autonomous discrete systems. Journal of Difference Equations and Applications, 12(1), 95-100.
ISO 690
2. Shao, H., Chen, G., & Shi, Y. (2020). Some criteria of chaos in non-autonomous discrete dynamical systems. Journal of Difference Equations and Applications, 26(3), 295-308.
3. Sharma, P., & Raghav, M. (2015). Dynamics of nonautonomous discrete dynamical systems. arXiv preprint arXiv:1512.08868.
You can find a lot more, I suggested those just to make it clear which kind of name/keywords can be useful for you. You'll find lots of results on interval maps, real maps and on more general spaces. It all depends on which specific questions you want to address.
Good news for you: you don't have to consider Baire class 1 functions in their full glory, since given a sequence $\{F\_n\}$ of continuous functions which converges pointwise, the functions $f\_n$ solving the countable set of functional equations $$F\_{n}(x)=f\_n(F\_{n-1}(x)), \ n\ge 2,$$ even when they exist, aren't necessarily continuous. So you're exploring a (topologically small, I guess) portion of Baire class 1 maps.
| 1 | https://mathoverflow.net/users/167834 | 429955 | 174,185 |
https://mathoverflow.net/questions/429967 | 5 | Can anyone give an example of a projective, regular, geometrically reduced but non-smooth curve ?
Of course, the base field should be imperfect.
In Exercise 4.3.22 of Qing Liu's book *Algebraic Geometry and Arithmetic Curves*, a regular but non-smooth curve is given. But that curve is not geometrically reduced.
| https://mathoverflow.net/users/11599 | A regular, geometrically reduced but non-smooth curve | I believe a classic example is the curve define in $\mathbb P^2\_{\mathbb F\_p(t)}$, with coordinates $(x:y:z)$, by the equation
$$ t x^p + z^{p-1} y + y^p=0$$
for $p>2$.
Differentiating with respect to $y$, one can see that the curve is smooth wherever $z \neq 0$, and substituting in $z=0$, one can see the curve is smooth away from a single geometric point $(1:t^{1/p}:0)$. Since the derivatives of the defining equation with respect to $x,y,$ and $z$ all vanish at this point, using $p>2$ for the $z$ derivative, this curve is indeed non-smooth at that point. But since it has only one non-smooth point, it is geometrically reduced.
However, the curve is regular at this nonsmooth point, because the derivative with respect to $t$ of the defining equation is nonzero there.
| 14 | https://mathoverflow.net/users/18060 | 429970 | 174,189 |
https://mathoverflow.net/questions/425948 | 10 | Let $\{b\_n\}\_{n\geq0}$ be a sequence such that $b\_nb\_{n+1}=0$ and define
$$a\_n:=\sum\_{k=0}^n(-1)^{n-k}\binom{n}{k}b\_k.$$
If $\lim\_{n\to\infty}a\_n=0$, can we conclude that $b\_n=0$ for all $n$?
More generally, if $\{b\_n\}\_{n\geq0}$ is a sequence with infinitely many zeros and $\lim\_{n\to\infty}a\_n=0$, can we still conclude that $b\_n=0$ for all $n$?
Finally, what if we only assume $\{b\_n\}\_{n\geq0}$ contains at least ONE zero?
Till now, only the original question remains unsolved.
**Remark:** This question arised from the computation of the $K\_0$ groups of the smooth noncommutative $\mathbb{R}^{2n}$ which come from noncommutative field theory, see [enter link description here](https://arxiv.org/abs/2208.06253). I showed that the $K\_0$ groups of original noncommutative $\mathbb{R}^{2n}$ are all $\mathbb{Z}$. Then mimicking the smooth noncommutative tori, I construct the smooth noncommutative $\mathbb{R}^{2n}$ and I want to show that the $K\_0$ groups of the smooth cases are still all $\mathbb{Z}$. But this brings many new problems. First I consider a special class of the projectors of smooth noncommutative $\mathbb{R}^{2}$, and show that if this conjecture is true, then this class of projectors are 0 or 1. The other two questions I asked here are also related to the characterization of projectors of smooth noncommutative $\mathbb{R}^{2n}$.
| https://mathoverflow.net/users/168342 | A number sequence problem involving binomial transform | Unfortunately the argument that I originally posted contained a gap at the end.
The gap is explained at the end of the proof, where I also state 3 partial results.
Let
$$f(z)=\sum\_{n=0}^\infty b\_nz\_n/n!,\quad g(z)=\sum\_{n=0}^\infty a\_nz^n/n!.$$
Then your relation between $a\_n$ and $b\_n$ means
$$g(z)=e^{-z}f(z).$$ Suppose that $a\_n=o(1)$, then $|g(z)|=o(e^{|z|})$. Thus both $f$ and $g$ are of exponential type, and the indicator diagram $K$ of $f$ is contained in the
disk $D=\{ z:|z+1|\leq 1\}.$ This means that the Borel (=Laplace) transform
$$F(z)=\sum\_{n=0}^\infty b\_n z^{-n-1}$$
is analytic outside of this disk $D$. Thus the power series
$$F(1/z)=\sum\_{n=0}^\infty b\_nz^{n+1}$$
has radius of convergence at least $1/2$ and has an analytic continuation into the half-plane $\{ z:\Re z<1/2\}$. But this contradicts the Fabry Gap Theorem, since
your condition that $b\_nb\_{n+1}=0$ implies that the density
of non-zero coefficients is at most $1/2$, and Fabry's theorem says that under this condition $F(1/z)$ must have
a singularity on every arc of the circle of convergence which is bigger than semi-circle.
This argument does not work when $F(1/z)$ has infinite radius of convergence.
1. To exclude this case, one may assume that $|a\_n|$ tends to zero with geometric speed, that is $|a\_n|=O(\delta^n)$
for some $\delta\in(0,1)$. Then $0\not\in K$, and
$F(1/z)$ is not entire.
2. And of course, if the condition $b\_nb\_{n+1}=0$
is replaced by the condition that $f$ is even or odd, then the conclusion is true as well.
3. The argument above also shows that under the stated conditions one can conclude that $b\_n=O(\epsilon^n),\forall \epsilon>0$.
Thus my argument needs either a stronger assumption
(1) or (2), or leads to a weaker conclusion (3).
Reference.
L. Bieberbach, Analytische Fortsetzung, Springer 1955, Chap. 2
(Chap. 1 also contains all other facts used in this argument).
| 6 | https://mathoverflow.net/users/25510 | 429979 | 174,192 |
https://mathoverflow.net/questions/429977 | 11 | David Roberts wrote in the comment section of the blog post "[Convergence of an infinite sum in the rationals](https://thehighergeometer.wordpress.com/2022/07/29/convergence-of-an-infinite-sun-in-the-rationals/)" the following paragraph:
>
> Someone mentioned (I think on Twitter) that the Taylor series of rational functions should all be like this example (which is easy to see), but possibly also that this is the only class of power series that converges like this in the rationals, namely, if a power series converges on the rationals, then it is the Taylor series for a rational function. Not sure how one would show this.
>
>
>
Note that David Roberts is working inside of the rational numbers $\mathbb{Q}$, rather than the real numbers $\mathbb{R}$, in his blog post.
Is it true that in the rational numbers every convergent power series on the rational numbers is a Taylor series for a rational function on the rational numbers? If so, how would one go about proving this statement? If not, what counterexamples exist out there?
| https://mathoverflow.net/users/483446 | In the rational numbers, is every convergent power series a Taylor series for a rational function? | No. Enumerate the rational numbers $a\_1,a\_2,\dots$. Then for every sequence $c\_1, c\_2,\dots$ of rational numbers decreasing rapidly enough, the series
$$ \sum\_{n=1}^{\infty} c\_n x^n \prod\_{i=1}^{n-1} (x-a\_i ) $$
converges on each rational number. On $a\_m$ it takes the value $$ \sum\_{n=1}^{m} c\_n a\_m^n \prod\_{i=1}^{n-1} (a\_m-a\_i ) $$ which is rational.
By "decreasing rapidly enough", it suffices to have $$ \sum\_{n=1}^{\infty} c\_n |x|^n \prod\_{i=1}^{n-1} ( |x| + |a\_i|)< \infty $$ for each rational $x$, e.g. it suffices to have
$$|c\_n| < \frac{1}{ n^n \prod\_{i=1}^{n-1} ( n + |a\_i|)}$$ as then for $|x|<m$, for all $m \geq n$, the $n$'th term in the above sequence is bounded by $(x/m)^n$ and thus that sequence converges.
There are uncountably many series of this type, so they can't all come from rational functions.
| 21 | https://mathoverflow.net/users/18060 | 429981 | 174,193 |
https://mathoverflow.net/questions/429889 | 3 | Let $Y (N) $ be the moduli scheme of dimension two principally polarized Abelian schemes with level $N$. It is claimed in "[G.Laumon - Fonctions zeta des variétés de Siegel](http://www.numdam.org/article/AST_2005__302__1_0.pdf)" (Lemma 4.1) that to an algebraic representation $W$ of $\mathrm{GSp}\_{4}(\mathbb{Q})$ we can associate an $l$-adic smooth sheaf on $Y (N) [1/l] $.
Where can I find a proof of this please?
| https://mathoverflow.net/users/169282 | $l$-adic sheaf associated to an algebraic representation of $\mathrm{GSp}_{4}(\mathbb{Q})$ | This is a special case of Pink's "canonical construction" functor, which associates various kinds of coefficient sheaves on a Shimura variety (etale $\ell$-adic sheaves, vector bundles with connection, variations of Hodge structures, etc) to algebraic representations of the underlying group.
For more information see e.g. this paper by Torzewski: <http://link.springer.com/article/10.1007/s00229-019-01150-9#Sec10>.
| 2 | https://mathoverflow.net/users/2481 | 429982 | 174,194 |
https://mathoverflow.net/questions/429845 | 4 | I am looking into the practicalities of doing Math in FOL + PA with the FOL extended with equality and functions.
For a predicate you can easily extend the language such that a predicate is defined as a logical expression, which then can be comprehended or expanded in the proofs and theorems that follow.
However, if you want to do the same for a function, then it should be proven that the function is deterministic (giving only one result for a certain set of parameters) and total. If you define the function with other functions, then it that is obvious. However, if the function is defined using logical expressions then it is not obvious.
If I look in the literature I see often a function popping up (such as Gödel's β function) without this explicit proof. I assume that it has informally proven, but since I want to write a piece of text about this, my question is if there are any conventions how to deal with this and how you normally incorporate this in a logic language that can checked by a theorem checker?
This is not very theoretical, since you can get rid of function and rewrite the expressions as predicates.
Thanks in advance,
Lucas
| https://mathoverflow.net/users/5917 | Defining functions in FOL + PA | You might avoid the issue in first-order arithmetic by assuming that a function always selects the least option, or zero if there is no option. So a definition like
$$R(x,y,z)=z<y \wedge \exists v:x = vy+z$$
would actually be transformed into
$$R'(x,y,z)=(R(x,y,z) \wedge \forall w:R(x,y,w)\implies z\le w) \vee (z=0 \wedge \forall z: \neg R(x,y,z))$$
Alternatively, you might follow Solomon Feferman in using logical frameworks flexible enough to include symbols for partial functions and for [definedness](https://virtualmath1.stanford.edu/%7Efeferman/papers/definedness.pdf).
| 3 | https://mathoverflow.net/users/nan | 429983 | 174,195 |
https://mathoverflow.net/questions/429933 | 2 |
>
> Is there a discrete space Markov chain, starting from a fixed state, whose stationary distribution is a multimodal distribution and that mixes in polynomial time?
>
>
>
For example, Ising model on say a complete graph has a multimodal stationary distribution at low temperature. Critical $\beta$ (i.e. inverse temperature) is known to be $\beta\_c=1$. So $\beta>1$ is low temperature regime. Single-site Glauber dynamics for Ising model at low temperature on a complete graph is known to mix exponentially slow. Here is a [proof](https://link.springer.com/content/pdf/10.1007/s00440-008-0189-z.pdf). I believe faster mixing Swendsen-Wang dynamics which update multiple spins in one step also mix exponentially slow at low temperature.
So I am wondering if there is a fundamental barrier and no Markov chain can mix in polynomial time, starting from a fixed state, at low temperature for Ising model on say a complete graph? Or is it that no one has found such a Markov chain yet?
Coupling is often used to prove upper bounds on mixing time. If there exists a contractive coupling, then a chain mixes in polynomial time. Here is a [paper](https://arxiv.org/pdf/1805.00452.pdf) that proves that there exists a contractive coupling for a Markov chain on a continuous state space and a multimodal stationary distribution. For instance, see example 2.4 in this paper which talks about a Gaussian mixture.
| https://mathoverflow.net/users/479350 | Polynomial time mixing Markov chain for multimodal distribution | The OP seems specifically interested in ergodic Markov chains with a unique stationary distribution $\pi$. The following example is admittedly a bit contrived, but is helpful to illustrate some general principles discussed more below.
>
> **Claim:** There exists a Markov chain with a "bimodal" stationary distribution, such that the mixing time of the chain is logarithmic in $n$.
>
>
>
"Bimodal" just means there are two distinct states where $\pi$ is larger than in other states.
*Proof.* Let $\Delta\_i(t)$ denote the total variation distance between a Markov chain starting at the $i$th state and the stationary distribution $\pi$. For any accuracy parameter $\epsilon>0$, define the *mixing time*: $$
\tau(\epsilon) := \max\_i \min\{ t > 0 : \Delta\_i(t) < \epsilon \} \;.
$$ A chain that displays rapid mixing is given by the following lazy Markov chain with $n \times n$ transition matrix: $$
P\_{ij} = \begin{cases} 1/2 & \text{if $i=j$} \\
1/4 & \text{if $i \notin \{1,n \}$ and $j \in \{1, n \}$} \\ 1/(2 (n-1)) & \text{if $i \in \{1,n\}$ and $j\ne i$} \\
0 & \text{else}
\end{cases}
$$ The stationary distribution of this chain satisfies $$
\pi \propto \left( 1, \frac{2}{n-1}, \cdots, \frac{2}{n-1},1 \right)
$$ which is *bimodal* (concentrated at the endpoints), and with spectral gap $\gamma=1/2$, and since the chain is reversible, the mixing time satisfies $$
\tau(\epsilon) \le 2 \log\left(\frac{2 n-3}{\epsilon}\right)
$$ which is logarithmic in $n$. $\Box$
In general, energy and/or entropic barriers can cause chains that make only local moves to mix with exponential time. This is, e.g., the case with Glauber dynamics applied to the mean-field Ising (or Curie-Weiss) model at low temperature; see, e.g. Theorem 3 of [Levin/Luczak/Peres (2010)](https://link.springer.com/article/10.1007/s00440-008-0189-z) and references therein. In contrast, the above chain is contrived to make global moves, and doesn't really have any type of barriers, which is why it mixes so fast.
| 2 | https://mathoverflow.net/users/64449 | 429990 | 174,197 |
https://mathoverflow.net/questions/425726 | 8 | Let $f(x)$ be a polynomial of degree $d$ with integer coefficients. Let $G\_p^+$ be the Galois group of the polynomial $f(x)-y$ over $\overline{\mathbb{F}}\_p(y)$ and $G\_p$ be the Galois group of the same polynomial over $\mathbb{F}\_p(y)$. It is known (see B. Birch and H. Swinnerton-Dyer, "Note on a problem of Chowla", Lemma 1) that for large enough $p$ all the groups $G\_p^+$ are isomorphic and also isomorphic to the Galois group over $\mathbb C(y)$. My question is, what happens with $G\_p$ when $p$ varies?
More precisely, for all large enough $p$ fix the order of roots of $f(x)-y$. Then $G\_p^+$ and $G\_p$ are subgroups of $S\_d$. Let us fix the order in a compatible way: so that $G\_p^+$ is always the same group $G\subseteq S\_d$. Let $H$ be a subgroup in $S\_d$, denote by
$$
\pi\_f(x;H)
$$
the number of $p\leq x$ such that $G\_p=H$. Then I'm interested in asymptotic formulas for $\pi\_f(x;H)$ and their dependence on $f$. Is it true that the main term actually depends only on $G=G\_p^+$?
| https://mathoverflow.net/users/101078 | The distribution of certain Galois groups | Denote by $\Omega$ the splitting field of $f(x)-y$ over $\mathbb{Q}$, by $k/\mathbb{Q}$ the maximal constant extension inside $\Omega$, and set $G:=Gal(\Omega/\mathbb{Q}(y))$, $G^+ =Gal(\Omega/k(y))$. Up to avoiding finitely many primes $p$, one has "good constant reduction" and can obtain the picture over $\mathbb{F}\_p(y)$ by reducing given defining polynomials of $\Omega/k(y)$ (i.e., $f(x)-y$) and of $k/\mathbb{Q}$ individually modulo $p$; in other words, if $x\_p\in G/G^+$ is a representative of the Frobenius class of $p$ in $k/\mathbb{Q}$ (well-defined up to conjugation in $G/G^+$), then $G\_p$ is the subgroup of $G$ mapping onto $\langle x\_p\rangle$ under projection $G \to G/G^+$, and these subgroups occur with frequency described by Chebotarev's density theorem (applied to $k/\mathbb{Q}$); this should give the "main term" in the question.
So then, if I understand the last question correctly, it basically amounts to asking whether the normal subgroup $G^+$ uniquely determines $G$ (independently of the concrete shape of $f$). This would not be true for polynomials of arbitrary shape, but due to the form $f(x)-y$, it should be true; indeed, it is known (cf. Lemma 3.4 in <https://arxiv.org/pdf/math/0109071.pdf>) that $G\le S\_d$ is generated by $G^+$ together with the normalizer (in $G$) of a $d$-cycle $\sigma$ (the latter being the inertia group generator at infinity); on the other hand, the so-called branch cycle lemma implies that this inertia group generator must be conjugate in $G$ to all powers $\sigma^k$ ($k$ coprime to $d$), i.e., $G$ must contain the full symmetric normalizer (of order $d\cdot \varphi(d)$) of this $d$-cycle. So then, once $G^+$ is given **and the $d$-cycle has been fixed**, there is really no freedom anymore in the choice of $G$, meaning that the answer to the last question should be "yes".
**EDIT:** Shortly after confidently typing away the above, I'm noticing the following crucial **gap**: It is not clear (to me at least) that, after fixing $G^+$, the ``fixing the $d$-cycle" is something that can be done without loss of generality; i.e., there are groups $H\le S\_d$ that have several classes of $d$-cycles $\sigma$, such that $H\trianglelefteq \langle H, N\_{S\_d}(\sigma)\rangle$, but the groups on the right can have different orders! This is not yet sufficient for these $d$-cycles to actually belong to the monodromy of some polynomial $f(x)-y$ over $\mathbb{Q}$, but if they did, it would actually give different $G$ with the same $G^+$, i.e., it would be a counterexample to the above claim!
| 3 | https://mathoverflow.net/users/127660 | 430001 | 174,203 |
https://mathoverflow.net/questions/429999 | 1 | Geometry and combinatorics are two different branches of mathematics. Does there exist any connection between them? In many cases, mathematicians solve some geometric problems by reducing them to a combinatorial language. What are the general techniques to convert a geometrical problem to a combinatorial one? What are the known examples in literature?
What will be some good references to learn these techniques?
Thanks in advance.
| https://mathoverflow.net/users/490039 | Bridges between geometry and combinatorics | I would recommend the work of Adiprasito, Huh and Katz:
>
> K. Adiprasito, J. Huh, E. Katz, Hodge Theory for Combinatorial
> Geometries, Annals of Mathematics 188 (2018), 381–452. [[arXiv](https://arxiv.org/abs/1511.02888)]. [[Journal](https://web.math.princeton.edu/%7Ehuh/MatroidHodge.pdf)]
>
>
>
They actually use geometric intuition/techniques to solve problems in combinatorics (rather than using combinatorics to solve geometric problems).
| 0 | https://mathoverflow.net/users/7031 | 430002 | 174,204 |
https://mathoverflow.net/questions/430003 | 18 | I am a Masters student of math interested in physics. When I was an undergraduate, I took the introductory course of physics, but it is just slightly harder than high school physics course. To be precise, it just taught us how to use calculus in physics, without involving the higher knowledge of math such as manifold, PDE, abstract algebra and etc. By the way, the knowledge in that course is "discrete", the connections between fields of physics is omited.
My question is, what I should do to learn "real physics" by myself? What books or materials should I read?
| https://mathoverflow.net/users/490869 | How does a Masters student of math learn physics by self? | I can recommend Leonard Susskind's [Theoretical Minimum](https://theoreticalminimum.com):
>
> A number of years ago I became aware of the large number of physics
> enthusiasts out there who have no venue to learn modern physics and
> cosmology. Fat advanced textbooks are not suitable to people who have
> no teacher to ask questions of, and the popular literature does not go
> deeply enough to satisfy these curious people. So I started a series
> of courses on modern physics at Stanford University where I am a
> professor of physics. The courses are specifically aimed at people
> who know, or once knew, a bit of algebra and calculus, but are more or
> less beginners.
>
>
>
The name "theoretical minimum" is a reference to the notoriously rigorous [exam](https://arxiv.org/abs/hep-ph/0204295) a student needed to pass in order to study with Lev Landau. See also this [discussion.](https://physics.stackexchange.com/questions/13861/lev-landaus-theoretical-minimum)
| 13 | https://mathoverflow.net/users/11260 | 430004 | 174,205 |
https://mathoverflow.net/questions/429980 | 1 | Let $X\subseteq B(H)$ be an operator system and let $M\subseteq B(K)$ be a von Neumann algebra. We form the Fubini-tensor product
$$X \otimes\_\mathcal{F} M := \{z \in B(H\otimes K): (\sigma\otimes \iota)(z) \in M \text{ and } (\iota \otimes \tau)(z)\in X \text{ for all }\sigma\in B(H)\_\*, \tau \in B(K)\_\*\}.$$
We have a natural injective linear map
$$X \otimes\_\mathcal{F}M \to B(M\_\*,X): z \mapsto (\omega \mapsto (\iota \otimes \omega)(z)).$$
Is it true that the image of this map consists of all completely bounded maps $M\_\*\to X$?
Note that this is at least true if $X$ itself is a von Neumann algebra (However, this result is not very easy to find in the literature).
| https://mathoverflow.net/users/470427 | Which elements live in the image of the canonical map $X \otimes_\mathcal{F} M \to B(M_*, X)$? | I follow the book of Effros+Ruan (which is a book, so not viewable online, but really is the nicest source I think). For any operator spaces $X,Y$ we can consider the operator space projective tensor product $\newcommand{\proten}{\widehat\otimes}X\proten Y$ whose dual satisfies
$$ (X\proten Y)^\* = CB(X,Y^\*). $$
(To be precise, there is some completely isometric isomorphism here.) This is in Chapter 7, Corollary 7.1.5 to be precise.
The Fubini tensor product is also considered in Chapter 7, Theorem 7.2.3, which shows that $(X\proten Y)^\* \cong X^\* \bar\otimes\_{\mathcal F} Y^\*$ in general, so that $X^\* \bar\otimes\_{\mathcal F} Y^\* \cong CB(X,Y^\*)$.
So now apply this to a von Neumann algebra $M$ with predual $M\_\*$, and a dual operator space $X$ with predual $X\_\*$. Then
$$ CB(M\_\*, X) = (M\_\* \proten X\_\*)^\* = M \bar\otimes\_{\mathcal F} X. $$
Again there are (canonical) isomorphisms involved here, but chasing them down will show that they match the isomorphism given in the original question. In particular, this includes the isomorphism $M \bar\otimes\_{\mathcal F} X \cong X \bar\otimes\_{\mathcal F} M$.
Here I used that $X$ is a dual space, and in Chapter 7 of Effros and Ruan, we need this, and a "dual realisation" of $X$ as a weak$^\*$-closed subspace of $B(H)$. That is, the Original Question has a positive answer when $X \subseteq B(H)$ is weak$^\*$-closed.
When $X$ is only assumed norm closed, we can use the definition of the Fubini tensor product given in the original question, though I am not aware of much study of this. However, let $\overline{X}^{w^\*}$ be the weak$^\*$-closure of $X$ in $B(H)$. Then by definition(s),
$$ X \bar\otimes\_{\mathcal F} M \subseteq \overline{X}^{w^\*} \bar\otimes\_{\mathcal F} M \cong CB(M\_\*, \overline{X}^{w^\*}). $$
The isomorphism clearly identifies $X \bar\otimes\_{\mathcal F} M$ with those CB maps $T:M\_\*\rightarrow \overline{X}^{w^\*}$ which map into $X$, and by definition of what a CB map is, this is just the space $CB(M\_\*,X)$. Thus the original question is answered in the affirmative.
| 2 | https://mathoverflow.net/users/406 | 430008 | 174,207 |
https://mathoverflow.net/questions/430016 | 5 | Let $G=Gl\_n(\mathbb{C})$ and $\mathcal{N}$ be the nilpotent cone associated to it i.e nilpotent matrices inside $\mathfrak{g}=\mathfrak{gl}\_n(\mathbb{C})$.
We have the variety $\tilde{\mathcal{N}}$ with the Springer resolution $p:\tilde{\mathcal{N}} \to \mathcal{N}$ with the Springer sheaf $p\_{!}(\mathbb{Q}[\dim \mathcal{N}])$ with its action of the Weyl group $S\_n$. In this case we know it decomposes as $$\bigoplus\_{\lambda \in \mathcal{P}\_n}\operatorname{IC}(O\_{\lambda},\mathbb{Q})\otimes V\_{\lambda} $$ where $O\_{\lambda}$ is the nilpotent orbit associated to the partition $\lambda$ and $V\_{\lambda}$ is the $S\_n$ irreducible module indexed by $\lambda$.
I'm looking for an explicit description of $H\_c(\mathcal{N},p\_{!}(\mathbb{Q}[\dim \mathcal{N}]))^{S\_{\mu}}$ where $\mu$ is a partition of $n$ and $S\_{\mu}=S\_{\mu\_1} \times \dotsb \times S\_{\mu\_k}$. I think this should be related to the cohomology $H(T^\*(G/P\_{\mu}),\mathbb{Q})$ where $P\_{\mu}$ is the parabolic subgroup associated to $\mu$.
The only thing I was able to notice is that $$H\_c(\mathcal{N},p\_{!}(\mathbb{Q}[\dim \mathcal{N}]))^{S\_{\mu}}=\bigoplus\_{\lambda \in \mathcal{P}\_n}H\_{c}(\operatorname{IC}(O\_{\lambda},\mathbb{Q}))\otimes V\_{\lambda}^{S\_{\mu}} $$ and $V\_{\lambda}^{S\_{\mu}} \neq 0$ only if $\lambda \geq \mu$. This happens exactly when $O\_{\lambda} \subseteq \mathfrak{u}\_{\mu}$ where $\mathfrak{u}\_{\mu}$ is the Lie algebra of the unipotent radical of $P\_{\mu}$.
| https://mathoverflow.net/users/146464 | Invariants of cohomology of Springer sheaf | You want to look at the partial Grothendieck-Springer resolution, i.e. the variety of pairs $ (g \in G/ P\_\mu, v \in g \mathfrak p\_\mu g^{-1})$.
The partial Grothendieck-Springer resolution is smooth, so the shifted constant sheaf on it is perverse, and its pushforward to $\mathfrak g$ is pure.
The partial Grothendieck-Springer resolution is covered by the Grothendieck-Springer resolution, so the pushforward of the shifted constant sheaf is a summand of the pushforward of the shifted constant sheaf of the Grothendieck-Springer resolution.
In particular, the pushforward from the partial Grothendieck-Springer resolution is a middle extension from the subset of regular semisimple elements.
On a regular semisimple element, the fiber of the map from this variety to $\mathfrak g$ is $S\_n/S\_\mu$. It follows from this and the middle extension property that the pushforward from he partial Grothendieck-Springer resolution, restricted to the nilpotent cone, is the $S\_\mu$-invariants of the Springer sheaf.
So, by the Leray spectral sequence with compact support, the cohomology you want is the compactly supported cohomology of the space of pairs $g \in G/ P\_\mu, v \in g \mathfrak p\_\mu g^{-1}$ with $v$ nilpotent.
The fiber of the map to $G/P\_\mu$ is the product of the nilpotent cone of the associated Levi with a vector space. I think we know that the compactly supported cohomology of the nilpotent cone is supported in a single degree, as is the compactly supported cohomology of a vector space, so in the end you get a shift of the cohomology of $G/P\_\mu$.
| 9 | https://mathoverflow.net/users/18060 | 430029 | 174,214 |
https://mathoverflow.net/questions/429512 | 3 | $\DeclareMathOperator\ht{ht}$All rings are commutative Noetherian with identity.
Exercise 9.8 of Matsumura's book *Commutative ring theory*: Let $A$ be a ring and $A\subset B$ an integral extension. If $P$ is a prime ideal of $B $ with $\mathfrak p = P\cap A$ then $\ht P \leq \ht \mathfrak p$.
There are many examples that equality occurs; i.e. $\ht P = \ht \mathfrak p$. But the issue is different for the case of examples for which $\ht P \lt \ht \mathfrak p$. Can you give such examples, please?
Thank you.
| https://mathoverflow.net/users/47763 | Examples of integral ring extensions that $\operatorname{ht}P \lt \operatorname{ht}P\cap A$ | Take $A=\mathbb{Z}$ and $B=\mathbb{Z}[X]/(X^2+X,2X)$ $=$ $A[x]$, where $x$ denotes the residue class of $X$. Clearly, $x$ is integral over $A$. Every prime ideal of $B$ either contains $x$ or both $x+1$ and $2$. Hence $P$ = $(2,x+1)$ is a minimal prime ideal, so of height $0$. (Note that $P$ is also a maximal ideal of $B$). Since $2$ $\in$ $P$, it follows that $P\cap A$ $=$ $2\mathbb{Z}$, which is of height $1$.
| 5 | https://mathoverflow.net/users/31923 | 430039 | 174,216 |
https://mathoverflow.net/questions/430035 | 6 | For $n\geq 1$, $f\_n\in\mathcal{C}^1([0,1],\mathbb{R})$ such that $f\_n(x)\geq\sqrt{x}$ for $x\in[0,1]$, and
$$\lim\limits\_{n\to+\infty}\sup\_{x\in[0,1]}\big|f\_n(x)-\sqrt{x}\big|= 0.$$
Let $y\_n$ be the unique solution of
$$\begin{cases}
y\_n(0)=0 \\
y\_n'=f\_n(y\_n) \text{ on [0,1]}.
\end{cases}$$
**Question:** Is there a function $y\in\mathcal{C}^1([0,1],\mathbb{R})$ such that
$$\lim\limits\_{n\to+\infty}\sup\_{x\in[0,1]}\big|y\_n(x)-y(x)\big|= 0$$
which is solution of the system (which has itself an infinity of solutions)
$$\begin{cases}
y(0)=0 \\
y'=\sqrt{y} \text{ on [0,1]}
\end{cases}$$
and satisfies the condition: $y(x)>0$ for $x\in\,]0,1]$ (which, I hope, permits to characterize $y$).
| https://mathoverflow.net/users/159940 | Forcing the uniqueness of a solution of an ODE | $\newcommand\ep\varepsilon$First, the conditions that $f\_n\in\mathcal{C}^1([0,1],\mathbb{R})$ and $f\_n(x)\ge\sqrt{x}$ for $x\in[0,1]$ imply $f\_n(0)>0$. Since
\begin{equation\*}
\begin{cases}
y\_n(0)=0, \\
y\_n'=f\_n(y\_n) \text{ on [0,1]},
\end{cases}
\tag{2}\label{2}
\end{equation\*}
we see that $y\_n>0$ in a right neighborhood of $0$. Since $y'\_n=f\_n(y\_n)\ge0$, we see that $y\_n>0$ on $(0,1]$.
Letting then $u\_n:=\sqrt{y\_n}$, we get $2u\_n u'\_n=f\_n(u\_n^2)\ge u\_n$, whence $u'\_n\ge1/2$, $u\_n(x)\ge x/2$, and
\begin{equation\*}
y\_n(x)\ge x^2/4 \tag{3}\label{3}
\end{equation\*}
for all $x\in[0,1]$.
Next, for $\ep\in(0,1)$, let $z\_\ep$ be the unique solution of the ODE
\begin{equation\*}
z'\_\ep=\sqrt{z\_\ep}+\ep
\end{equation\*}
on $[0,1]$ with the initial condition $z\_\ep(0)=0$. It is not hard to see that
\begin{equation\*}
z\_\ep(x)\to x^2/4 \tag{4}\label{4}
\end{equation\*}
uniformly in $x\in[0,1]$ as $\ep\to0$. (See details on this at th end of this answer.)
Let now
\begin{equation\*}
\ep\_n:=\sup\_{x\in[0,1]}\big|f\_n(x)-\sqrt{x}\big|
=\sup\_{x\in[0,1]}\big(f\_n(x)-\sqrt{x}\big),
\end{equation\*}
so that $\ep\_n\to0$ (as $n\to\infty$), and then let
\begin{equation\*}
w\_n:=z\_{\ep\_n+1/n}.
\end{equation\*}
So, $y\_n(0)=0=w\_n(0)$,
\begin{equation\*}
y\_n'\le\sqrt{y\_n}+\ep\_n,\quad w\_n'=\sqrt{w\_n}+\ep\_n+1/n.
\end{equation\*}
Suppose that
\begin{equation\*}
x\_n:=\sup\{x\in[0,1]\colon y\_n\le w\_n\text{ on }[0,x]\}<1.
\end{equation\*}
Then $x\_n>0$ and $w\_n(x\_n)=y\_n(x\_n)$, and hence $y'\_n(x\_n)\ge w'\_n(x\_n)$, so that
\begin{equation\*}
\sqrt{w\_n(x\_n)}+\ep\_n=\sqrt{y\_n(x\_n)}+\ep\_n \\
\ge y'\_n(x\_n)\ge w'\_n(x\_n)=\sqrt{w\_n(x\_n)}+\ep\_n+1/n,
\end{equation\*}
a contradiction. So, $x\_n=1$ and hence, in view of \eqref{3},
\begin{equation\*}
x^2/4\le y\_n(x)\le w\_n(x)=z\_{\ep\_n+1/n}(x)\to x^2/4
\end{equation\*}
uniformly in $x\in[0,1]$, by \eqref{4}.
On the other hand, the only solution $y$ of the system
\begin{equation\*}
\begin{cases}
y(0)=0 \\
y'=\sqrt{y} \text{ on [0,1]}
\end{cases}
\end{equation\*}
such that $y>0$ on $(0,1]$ is given by the formula $y(x)=x^2/4$.
Thus, $y\_n\to y$ uniformly on $[0,1]$, as desired.
---
**Details on \eqref{4}:** Letting $t\_\ep:=\sqrt{z\_\ep}$, rewrite
$z'\_\ep=\sqrt{z\_\ep}+\ep$ as $2t\_\ep t'\_\ep=t\_\ep+\ep$. "Separating the variables", we find that $t\_\ep=g^{-1}\_\ep$, where
\begin{equation\*}
g\_\ep(t):=2t-2\ep\ln\frac{t+\ep}\ep.
\end{equation\*}
We have $g'\_\ep(t):=2-\frac{2\ep}{t+\ep}>0$ for $t>0$, so that the inverse function $g^{-1}\_\ep$ is well defined. Using the inequalities $\ln(t+\ep)\le t+\ep-1<t$, we get
\begin{equation\*}
g\_\ep(t)\ge (2-2\ep)t+2\ep\ln\ep
\end{equation\*}
for $t\ge0$, whence for $\ep\to0$ we have $t\_\ep(x)=g^{-1}\_\ep(x)\le\frac{x-2\ep\ln\ep}{2-2\ep}\to x/2$ and
\begin{equation\*}
z\_\ep(x)=t\_\ep(x)^2\le(1+o(1))x^2/4
\end{equation\*}
uniformly in $x\in[0,1]$. Also, similarly to \eqref{3}, $z\_\ep(x)\ge x^2/4$ for $x\in[0,1]$. Now \eqref{4} follows.
| 8 | https://mathoverflow.net/users/36721 | 430048 | 174,219 |
https://mathoverflow.net/questions/430068 | 0 | Consider $f \in L^{2}(0,1)$ and $g \in L^{\infty}(0,1)$ such that
1. $ \text{lim} ~g(x) = 0 \ \ \text{when} \ \ x \to 0^{+};$
2. $g(x) > 0 \ \forall x \in (0,1)$;
3. $\text{lim}~\dfrac{g(x)}{x^{\alpha}} = N > 0,$ when $x \to 0^{+}$, $0 < \alpha < 1$
Moreover, suppose
$$
\int\_{0}^{1}g(x)|f(x)|^{2} = M < \infty
$$
**Question:** Is it possible to get an estimate of the form
$$
\|f\|\_{L^{2}(0,1)}^{2} \leq C \|gf\|\_{L^{2}(0,1)}^{2} ?
$$
**My idea**
$$
\|gf\|\_{L^{2}(0,1)}^{2} \leq \|g\|\_{L^{\infty}(0,1)} \int\_{0}^{1}g(x)|f(x)|^{2} = M\|g\|\_{L^{\infty}(0,1)}
$$
It is correct to write
$$
\int\_{0}^{1}|f(x)|^{2} = \dfrac{1}{|g(s)|}\int\_{0}^{1}|g(s)||f(x)|^{2} dx
$$
But I can't do anything more than that :(
| https://mathoverflow.net/users/490936 | Get an estimate on $L^{2}(0,1)$ | No. Think of a sequence of $f\_n$ such that $\| f\_n\|\_2=1$ while $\|g f\|\_2\to 0$. And you might as well assume $g(x)= x^{\alpha}$.
It is equivalent to asking the same question with $f\_n\geq0$, $\| f\_n\|\_1 =1$, $\| x^{2\alpha}f\_n\|\_1\to0$ (just take $g\_n=\sqrt{f\_n}$ for the $L^2$ case).
Let us try $f\_n=\begin{cases} n & \textrm{when } x<\frac1n\\ 0 &\textrm{otherwise}\end{cases}.$ We check that it works:
$$\int f\_n \textrm{d} x =1,\quad \int\_0^1 x^{2\alpha} f\_n \textrm{d} x = \frac{1}{(2\alpha+1)n^{2\alpha}}\to0.$$
| 1 | https://mathoverflow.net/users/40120 | 430084 | 174,228 |
https://mathoverflow.net/questions/427590 | 0 | There is a great introduction by May, "[The Geometry of Iterated Loop Spaces](https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.170.2840&rep=rep1&type=pdf)". I really enjoy reading it, but it was written 50 years ago and contains outdated technical details related to the language of topological spaces. Now, as far as I understand, there is no doubt that this should be stated in the language of simplicial sets. What are the best references nowadays? I found, for example, [Simplicial and Operad Methods in Algebraic Topology](https://bookstore.ams.org/mmono-198). Is this a good introductory text?
In this question, I am interested in operads specifically in homotopy theory, in particular, the connection with iterated loop spaces (so [the question](https://mathoverflow.net/questions/139021/good-reference-for-studying-operads) about algebraic operads is not relevant to the current topic, as far as I understand, although the algebraic theory of operads also fascinates me).
| https://mathoverflow.net/users/148161 | How now to study operads in homotopy theory? | One of the most comprehensive references today is certainly:
>
> B. Fresse, *Homotopy of Operads and Grothendieck–Teichmüller Groups*, Mathematical Surveys and Monographs 217. <https://bookstore.ams.org/surv-217/>
>
>
>
However, the idea that topological spaces are obsolete and that we should only use simplicial sets is a bit misguided. How do you define a manifold using just simplicial sets? Or the little disks/cubes operads? Simplicial sets of course have their place, but there are times when you just can't avoid topological spaces.
| 8 | https://mathoverflow.net/users/36146 | 430092 | 174,231 |
https://mathoverflow.net/questions/430081 | 6 | I feel like this is maybe an incredibly trivial problem, and I'm just missing something. I may also be describing a well-known model that I cannot find the name for, so any comment/suggestion is appreciated.
But let's consider a percolation model on $\mathbb{Z}^d$ (or even $\mathbb{Z}^2$ if that's easier), with each open edge contributing weight $p \in (0, 1)$, with the condition that no loops can be formed by the open edges. More formally, let $E$ be the edges of the grid graph, and $\mathbf{\omega} \in \{ 0, 1 \}^{|E|}$ be an edge configuration, then define the measure of the edge configurations to be
$$ \phi\_p(\mathbf{\omega}) = \mathcal{Z}^{-1} p^{o(\mathbf{\omega})}(1-p)^{c(\mathbf{\omega})}\mathbf{1}\_A(\mathbf{\omega})$$
where $o(\mathbf{\omega})$ is the number of open edges of $\mathbf{\omega}$, $c(\mathbf{\omega})$ is the number of closed edges, and $A$ is the event that there is no cycle within the graph generated by the open bonds of $\mathbf{\omega}$. ($\mathcal{Z}$ is just some normalizing constant, which would not be 1 in this case.)
The motivation behind this model is that it serves as a highly simplified version of a certain [random cluster representation of Ising spin glasses](https://arxiv.org/abs/0707.0073). Basically it's just Bernoulli percolation with an extra "loopless" condition. Let's assume for now that the infinite-volume measure is actually well-defined, unless this poses more than a technical problem (?).
Note that this model clearly does not satisfy the [FKG inequality](https://en.wikipedia.org/wiki/FKG_inequality). Even worse, this model does not satisfy the [finite-energy property](https://sites.math.northwestern.edu/%7Eauffing/papers/BurtonKeane.pdf), meaning that conditional percolation ratio on an edge $e$ (given arbitrary bond configurations on other edges $E \setminus e$) cannot be bounded away from zero (due to the loopless condition). There are several natural questions about this model that I cannot seem to settle with classical techniques for standard percolation models.
* Is there a phase transition for this model, in the sense that there exists $p\_c \in (0, 1)$ where $p\_c = \inf\{ p: \text{there is at least one inifinite cluster with nonzero probability in } \phi\_p \}$. The Peierls contour argument doesn't seem to work due to the loopless condition. Neither does the classical self-dual argument in $\mathbb{Z}^2$ because I cannot find an obvious dual model. (Note $p\_c \geq 1/2$ trivially because the model is stochastically dominated by the standard Bernoulli model.)
* If a phase transition does exist, in the (super)critical phase, is the number of infinite clusters almost surely a constant? If so, how many infinite clusters are there? And does this number change as $p$ is varied? Note that the [Burton-Keane](https://sites.math.northwestern.edu/%7Eauffing/papers/BurtonKeane.pdf) argument would not work (or needs to be modified) due to the lack of apparent ergodicity and finite-energy property.
| https://mathoverflow.net/users/121745 | Infinite clusters for loopless percolation | These fascinating questions have been studied recently, e.g. by [Bauerschmidt, Crawford, Helmuth and Swan](https://arxiv.org/abs/1912.04854) (no percolation on $\mathbb{Z}^2$) and by [Bauerschmidt, Crawford and Helmuth](https://arxiv.org/abs/2107.01878) (percolation phase transition on $\mathbb{Z}^d$ for $d\geq 3$).
| 7 | https://mathoverflow.net/users/5784 | 430093 | 174,232 |
https://mathoverflow.net/questions/425711 | 3 | The [diamond lemma](https://www.cip.ifi.lmu.de/%7Egrinberg/algebra/diamond-talk.pdf) has recently come up in my teaching, and as always I've been looking for nice and simple applications. This has reminded me of the thesis
Kimmo Eriksson, *Strongly convergent games and Coxeter groups*, KTH Stockholm 1993,
which I have never been able to locate despite the existence of [a ProQuest page](https://www.proquest.com/docview/304095210) and many (Google Scholar says 60) citations in the literature.
Does anyone have a scan of this thesis? (I am aware of several papers by Eriksson, but I'm not sure how much of the thesis they cover.)
| https://mathoverflow.net/users/2530 | Eriksson's thesis "Strongly convergent games and Coxeter groups" | The thesis can now be found at <https://archive.org/details/eriksson-strongly-convergent-games-thesis> .
Thanks to Kimmo Eriksson for sending me a hard copy and allowing it to be shared!
| 2 | https://mathoverflow.net/users/2530 | 430106 | 174,235 |
https://mathoverflow.net/questions/430071 | 1 | Suppose you have Banach spaces $\mathcal B\_\alpha$ where $\alpha$ is in some index set $I$. Let $\mu\_\alpha$ be Gaussian measures on $\mathcal B\_\alpha$ with Cameron-Martin spaces $\mathcal H\_{\mu\_\alpha}$.
Is it then true that the product space of all the Banach spaces with the product measure (independent components), call it $\mathcal B$, has Cameron-Martin space $\mathcal H\_\mu=\{\mathbf v=\{v\_\alpha\}\_{\alpha\in I}:\sum\_{\alpha\in I}\|v\_\alpha\|\_{\mu\_\alpha}^2<\infty\}$?
I believe I read this somewhere but I can't find a reference.
| https://mathoverflow.net/users/479223 | Cameron-Martin space of product space | At least if $I$ is countable this should be true, but I do not have a reference:
Let $H$ be the CM space of the Frechet space $\mathcal{B} := \prod\_{\alpha \in I} \mathcal{B}\_{\alpha}$ with the product topology and the product measure. Recall that the Hilbert space reproducing kernel of $(\mathcal{B}, \mu)$ is isometrically isomorphic ($\Vert \cdot \Vert\_{L^2(\mathcal{B}^{\ast}, \mu)} = \Vert \cdot \Vert\_H$) to the CM space $H$ and is defined as the $L^2$ closure of $j(\mathcal{B}^{\ast})$ where $j$ is the canonical inclusion $\mathcal{B}^{\ast} \rightarrow L^2(\mathcal{B}, \mu)$.
In our case, the dual to $\mathcal{B}$ can be identified with the sequences in $\prod\_{\alpha \in I} \mathcal{B}\_{\alpha}^{\ast}$ with finite support and the pairing $\mathcal{B}^{\ast} \times \mathcal{B} \rightarrow \mathbb{R}$ given by
\begin{equation}
v(x) = \sum\_{\alpha \in I} v\_{\alpha} (x\_{\alpha}), ~~ v\_{\alpha} \in \mathcal{B}\_{\alpha}^{\ast}, x\_{\alpha} \in \mathcal{B}\_{\alpha} .
\end{equation}
where the sum is, of course, finite.
Restricted to $j(\mathcal{B}^{\ast})$, the $L^2$-norm is
\begin{equation}
\Vert v \Vert\_H = \Vert j(v) \Vert\_{L^2} = \int \sum\_{\alpha, \beta \in I} v\_{\alpha} (x\_{\alpha}) v\_{\beta} (x\_{\beta}) d \mu(x) = \sum\_{\alpha \in I} \int v\_{\alpha} (x\_{\alpha})^2 d \mu(x) = \sum\_{\alpha \in I} \Vert v\_{\alpha} \Vert\_{H\_{\alpha}}^2
\end{equation}
where we used the product structure of $\mu$. Hence indeed
\begin{equation}
H = \{\mathbf v=\{v\_\alpha\}\_{\alpha\in I}:\sum\_{\alpha\in I}\|v\_\alpha\|\_{\mu\_\alpha}^2<\infty\} .
\end{equation}
| 1 | https://mathoverflow.net/users/117692 | 430114 | 174,236 |
https://mathoverflow.net/questions/429812 | 15 | Let $f: \mathbb R^n \to \mathbb R$ be a locally integrable function. We say $x \in \mathbb R^n$ is a *strong Lebesgue point* of $f$ if
$$\lim\_{r \to 0} \frac{\int\_{B\_r (x)} |f(y) - f(x)| \, dy}{r^{n+1+\varepsilon}} = 0$$
for some $\varepsilon > 0$, potentially depending on $x$.
**Question:** Suppose every point in $\mathbb R^n$ is a strong Lebesgue point of $f$. Does it follow that $f$ is a constant function?
| https://mathoverflow.net/users/173490 | If a function $f$ is $(1+\varepsilon)$-times Lebesgue differentiable everywhere, is $f$ a constant function? | I realise I'm bumping into you again and already gave you an answer elsewhere after you posted this, but I thought I'd post my answer here for others to see. The answer is yes, $f$ has to be constant even if $\varepsilon = 0$. Here's the proof (where $\varepsilon = 0$):
---
Fix $x \in \mathbb{R}^n$. Then by Markov's inequality, we have:
$$\mu^n(\{y \in \mathbb{R}^n : |f(y) - f(x)| \geq s \} \cap B\_r(x)) \leq \frac{1}{s} \int\_{B\_{r}(x)} |f(y) - f(x)| dy$$
Then by setting $s = cd \space r$ (where $c$ and $d$ are arbitrary positive constants) and dividing both sides by $r^n$, we have:
$$\frac{\mu^n(\{y \in \mathbb{R}^n : |f(y) - f(x)| \geq cd \space r \} \cap B\_r(x))}{r^n} \leq \frac{1}{cd \space r^{n+1}} \int\_{B\_{r}(x)} |f(y) - f(x)| dy$$
The right-hand side tend to $0$ as $r \rightarrow 0^+$, therefore so does the left-hand side. In other words:
$$\lim\_{r \rightarrow 0^+} \frac{\mu^n(\{y \in \mathbb{R}^n : |f(y) - f(x)| \geq cd \space r \} \cap B\_r(x))}{r^n} = 0$$
---
Now (whenever $d < 1$) we can lower bound the above expression by:
$$\frac{\mu^n(\{y \in \mathbb{R}^n : |f(y) - f(x)| \geq c |y - x| \} \cap [B\_r(x) \setminus B\_{d \space r}(x)])}{r^n}$$
and note that $\frac{\mu^n(B\_{d \space r}(x))}{r^n} = V \space d^n$, where $V := \mu^n(B\_1(0))$. Therefore:
$$\limsup\_{r \rightarrow 0^+} \frac{\mu^n(\{y \in \mathbb{R}^n : |f(y) - f(x)| \geq c |y - x| \} \cap B\_r(x))}{r^n} \leq V \space d^n$$
However this limit holds for all $d \in \mathbb{R}^{>0}$, and so we can actually conclude that:
$$\lim\_{r \rightarrow 0^+} \frac{\mu^n(\{y \in \mathbb{R}^n : |f(y) - f(x)| \geq c |y - x| \} \cap B\_r(x))}{r^n} = 0$$
$$\lim\_{r \rightarrow 0^+} \frac{\mu^n(\{y \in \mathbb{R}^n : \frac{|f(y) - f(x)|}{|y - x|} \geq c \} \cap B\_r(x))}{r^n} = 0$$
---
Therefore $f$ is approximately differentiable at $x$ with $Df\_{ap}(x) = 0$ (the definition of approximately differentiable is given here <https://encyclopediaofmath.org/wiki/Approximate_differentiability>).
Now I want to quote the fact that $f$ having an everywhere zero approximate derivative implies that $f$ is constant. Unfortunately I couldn't find a good source for this, expect in the case of $n=1$ where it follows from theorem 14.18 in *The Integrals of Lebesgue, Denjoy, Perron, and Henstock* by Russell A. Gordon.
So for arbitrary dimensions, here is my own personal proof for completeness, modified from the $n=1$ case. Feel free to let me know if there are any mistakes.
>
> Assume, seeking a contradiction, that there exists $a, b \in \mathbb{R}^n$ such that $f(a) > f(b)$. Define $g(x) := f(x) + x \cdot \frac{f(a) - f(b)}{2|b - a|} \frac{b-a}{|b-a|}$.
>
>
> Then first of all $g$ is approximately differentiable with $D\_{ap}g(x)(h) = h \cdot \frac{f(a) - f(b)}{2|b - a|} \frac{b-a}{|b-a|}$ for all $x \in \mathbb{R}^n$.
>
>
> Now without loss of generality, we can assume that $a = 0$ and $b = (1,0,...,0)$ just so I can talk about moving from $a$ to $b$ as moving from left to right. Note that $D\_{ap}g(x)$ is positive in every direction which points rightward (by which I mean the direction has a positive dot product with $b$).
>
>
> In what follows let $x <\_1 y$ be defined to mean that the $1^{\text{st}}$ coordinate of $x$ is less than the $1^{\text{st}}$ coordinate of $y$. Same for similar relations like $=\_1$.
>
>
> Let $W \subseteq \mathbb{R}^n$ be the interior and boundary of some right-angled cone with vertex at $b$, and $a$ at the middle of its base.
>
>
> Now define $A := \{x \in W : g(x) \geq g(a)\}$. Let $\mathcal{F}$ be a family of subsets of $A$, such that $S \in \mathcal{F}$ whenever all the following hold:
>
>
> 1. If $x,y \in S$ are distinct, then $x \not =\_1 y$
> 2. If $x,y \in S$ with $x <\_1 y$, then $\frac{\mu^n(A \cap B\_{|x-y|}(y))}{\mu^n(B\_{|x-y|}(y))} \geq \frac{1}{2}$
>
>
> Note that $\mathcal{F}$ is non-empty as $\{a\} \in \mathcal{F}$, and is partially ordered by $\subseteq$ where every non-empty linearly ordered subset has an upper bound (the union). Therefore, by Zorn's lemma, $\mathcal{F}$ has a maximal element which we can denote by $Z$.
>
>
> Assume, seeking a contradiction, that $Z$ does not contain an element with maximal $1^{\text{st}}$ coordinate. Then there exists $p$ which is a limit point of $Z$ and has supremum $1^{\text{st}}$ coordinate. Let $(z\_k)\_{k=1}^\infty$ be a sequence in $Z$ which converges to $p$.
>
>
> Then for each $x \in Z$, we have:
> $$\mu^n(A \cap B\_{|x-p|}(p))$$
>
>
> $$= \lim\_{k \rightarrow \infty} \mu^n(A \cap B\_{|x-z\_k|}(z\_k))$$
>
>
> $$\geq \lim\_{k \rightarrow \infty} \frac{1}{2} \mu^n(B\_{|x-z\_k|}(z\_k))$$
>
>
> $$= \frac{1}{2} \mu^n(B\_{|x-p|}(p))$$
>
>
> (The justification for the first and last equalities comes from the fact that $\lim\_{k \rightarrow \infty} \mu^n(B\_{|x-z\_k|}(z\_k) \Delta B\_{|x-p|}(p)) = 0$, where $\Delta$ denotes the symmetric difference).
>
>
> So $\frac{\mu^n(A \cap B\_{|x-p|}(p))}{\mu^n(B\_{|x-p|}(p))} \geq \frac{1}{2}$ for all $x \in Z$. Also note that $g$ is approximately continuous at $p$, so we must have that $p \in A$.
>
>
> Then $Z \cup \{p\} \in \mathcal{F}$, which contradicts the maximality of $Z$. Therefore $Z$ contains an element with maximal $1^{\text{st}}$ coordinate. Denote the element as $q$.
>
>
>
>
> ---
>
>
> Now we must have that $q =\_1 b$, otherwise we could use the fact that $g$ is approximately differentiable at $q$ (with positive derivative in any rightward direction) to extend $Z$, again contradicting its maximality. But due to $q \in W$, we must have that $q = b$.
>
>
> So $b = q \in Z \subseteq A$. Hence $g(a) \leq g(b)$, which contradicts the assumption that $f(a) > f(b)$ from the beginning.
>
>
> Hence $f$ is constant, as $a$ and $b$ are arbitrary.
>
>
>
| 12 | https://mathoverflow.net/users/120665 | 430115 | 174,237 |
https://mathoverflow.net/questions/430089 | 2 | I have started reading about subgroup growth and, to my surprise, I haven't found a reference to whether direct products preserve subgroup growth.
Recall that, given a finitely generated group $G$, the function $s\_n(G)$ is given by
$$s\_n(G)=\#\{\text{subgroups of } G\text{ of index }\leq n\}.$$
We say that $G$ has *subgroup growth type* $f$ for some function $f$ if there are $a$, $b>0$ such that
\begin{align}
s\_n(G)&\leq f(n)^a \quad & &\text{for } n \text{ large enough},\\
f(n)&\leq s\_n(G)^b \quad & &\text{for infinitely many } n.
\end{align}
What I have found is the following: Let $G$ be a finitely generated group, let $N$ be a normal subgroup and let $Q=G/N$. Then Proposition 1.3.2 in *Subgroup growth* by A. Lubotzky and D. Segal states
\begin{align}
s\_n(G)&\leq s\_n(Q)s\_n(N)n^{\text{rk}(Q)},\\
s\_n(G)&\leq s\_n(Q)s\_n(N)c^n,\qquad \text{where}\ c=3^{d(Q)/3}.
\end{align}
Of course, these inequalities can be applied to a direct product $G\times G$ by taking $N=G\times 1$.
This seems to suggest that there should exist a group $G$ of intermediate growth and infinite rank such that $G\times G$ has strictly faster subgroup growth than $G$; more precisely, given any $a>0$,
$$s\_n(G\times G)>s\_n(G)^a \quad \text{for infinitely many } n.$$
Does such an example exist?
| https://mathoverflow.net/users/44172 | Subgroup growth of direct product | Subgroup growth of direct product is quite difficult in general. However, it is easier for pro-$p$ groups. You might like to see a recent paper: Y. Barnea and J.-C. Schlage-Puchta, [Branch groups, orbit growth, and subgroup growth types for pro-p group](https://www.cambridge.org/core/services/aop-cambridge-core/content/view/E142F008A7F6E25DD7A944CBA56F0692/S2050508620000086a.pdf/branch_groups_orbit_growth_and_subgroup_growth_types_for_pro_p_groups.pdf), Forum Math. Pi 8 (2020). We compute there the subgroup growth of the Grigorchuk group (and more complicated stuff). The big obstacle was exactly dealing with direct sums and we were able to solve it because we did think about it as a pro-$p$ group rather than in general.
| 2 | https://mathoverflow.net/users/5034 | 430122 | 174,240 |
https://mathoverflow.net/questions/430100 | 1 | I try to understand the Proof of Theorem 4.21 in [Carmona Delarue (2018)](https://link.springer.com/content/pdf/10.1007/978-3-319-58920-6.pdf). In the following, what I don't understand:
Processes are assumed to be defined on a complete filtered probability space $(\Omega, \mathcal F, \mathbb F=(\mathcal F\_t)\_{t \in [0,T]},\mathbb P)$ supporting a $d$-dimensional Wiener process $W=(W\_t)\_{t\in [0,T]}$ wrt. $\mathbb F$, the filtration $\mathbb{F}$ satisfying the usual conditions. We denote by $\mathbb{H}^{2,n}$ the Hilbert space $$\mathbb{H}^{2,n} = \{ Z \in \mathbb{H}^{0,n}: \mathbb{E} \int\_0^T |Z\_s|^2 dx < \infty \},$$ where $\mathbb{H}^{0,n}$ stands for the collection of all $\mathbb{R}^n$-valued progressively measurable processes on $[0,T]$.
Assume that, for each $(x,\mu) \in \mathbb{R}^d \times \mathcal{P}\_2(\mathbb{R}^d)$, the processes $B(\cdot,\cdot,x,\mu):[0,T]\times \Omega \to \mathbb{R}^d, (t,\omega) \mapsto B(t,\omega,x,\mu)$ and $\Sigma
(\cdot,\cdot,x,\mu):[0,T]\times \Omega \to \mathbb{R}^{d \times d}, (t,\omega) \mapsto \Sigma(t,\omega,x,\mu)$ are $\mathbb{F}$-progressively measurable and belong to $\mathbb{H}^{2,d}$ and $\mathbb{H}^{2,d\times d}$ respectively. Furthermore, assume that for any $t \in [0,T], \omega \in \Omega, x, x' \in \mathbb{R}^d$ and $\mu, \mu'\in \mathcal{P}\_2(\mathbb{R}^d)$, $$|B(t,x,\mu)-B(t,x',\mu')|+|\Sigma(t,x,\mu)-\Sigma(t,x',\mu')| \leq L (|x-x'|+W\_2(\mu,\mu')).$$
Temporarily fix some $\mu = (\mu\_t)\_{t \in [0,T]} \in \mathcal{C}([0,T],\mathcal{P}\_2(\mathbb{R}^d))$ and let $X\_0 \in L^2(\Omega,\mathcal{F}\_0,\mathbb{P},\mathbb{R}^d)$. Then "the classical existence result for Lipschitz SDE guarantees existence and uniqueness of a strong solution of the classical stochastic differential equation with random coefficients": $$dX\_t = B(t,X\_t,\mu\_t)dt + \Sigma(t,X\_t,\mu\_t)dW\_t$$
My problem: I don't find such a classical existence result. I think the $\mu$ part is mostly irrelevant here. It comes from McKean-Vlasov setting that is treated originally. I didn't want to omit it here in case it is relevant in some way. Crossing it out simplifies things a little. Still, when I look for the standard existence result (e.g. Karatzas Shreve, Chapter 5, Theorem 2.9) there is always some linear growth condition like $$ \|B(t,x)\|^2 + \|\Sigma(t,x)\|^2 \leq K^2(1+\|x\|^2).$$ I do not see, why this is implied by the conditions.
| https://mathoverflow.net/users/490964 | Finding an existence and uniqueness result of a strong solution of Lipschitz SDEs | Following the proof of [Theorem 4.21](https://link.springer.com/content/pdf/10.1007/978-3-319-58920-6.pdf), fix the environment $\mu$, and additionally suppress the dependence of the SDE coefficients on $\mu$, so that the nonlinear SDE reduces to a classical one.
>
> **Claim:** In the classical existence/uniqueness theorem for SDEs, the standard linear growth condition can be relaxed to: there exists $y\in \mathbb{R}^d$ such that for all $T>0$ $$
> E \int\_0^T ( \|B(t,y)\|^2 + \|\Sigma(t,y)\|^2 ) dt < \infty \;.\tag{$\star$} \label{A1}
> $$
>
>
>
Note that Asssumption (A1) in [Theorem 4.21](https://link.springer.com/content/pdf/10.1007/978-3-319-58920-6.pdf) is actually stronger than \eqref{A1}, since (A1) holds for all $y\in \mathbb{R}^d$.
*Proof.* Recall that the linear growth condition is used to prove that the SDE solution is a real-valued, progressively measurable process such that $E \int\_0^T \| X\_s \|^2 dt < \infty$, i.e., $X \in \mathbb{L}^2\_d(0,T)$; see, e.g., [Chapter 5](https://bookstore.ams.org/mbk-82). Here we show that \eqref{A1} can play the same role. Indeed, let $X^{k}$ denote the $k$th Picard iterate in the standard existence/uniqueness proof. Then \begin{align\*}
&E\|X\_t^{k+1}\|^2 \le 3 \left(E\|X\_0\|^2 + E \| \int\_0^T B(s,X\_s^k) ds \|^2 + E \| \int\_0^T \Sigma(s, X\_s^k) dW\_s \|^2 \right) \\
&~~\le 3 \left(E\|X\_0\|^2 + T E \int\_0^T \| B(s,X\_s^k) \|^2 ds + E \int\_0^T \| \Sigma(s, X\_s^k)\|^2 ds \right) \;. \tag{$1$} \label{1}
\end{align\*}
Since the SDE coefficients are uniformly Lipschitz continuous in space, for any $x \in \mathbb{R}^d$ and $s \in [0,T]$,
\begin{align\*}
& \| B(s,x) \|^2 + \| \Sigma(s, x)\|^2 \le 2 ( \| B(s,x) - B(s,y)\|^2 + \|B(s,y)\|^2) + 2 ( \| \Sigma(s,x)- \Sigma(s,y) \|^2 + \| \Sigma(s,y) \|^2 ) \\
&~~ \le 2 ( \|B(s,y)\|^2 + \| \Sigma(s,y) \|^2 ) + 4 L^2 |x-y|^2
\end{align\*}
Inserting this bound into \eqref{1} and invoking \eqref{A1} proves the claim.
$\Box$
| 2 | https://mathoverflow.net/users/64449 | 430124 | 174,242 |
https://mathoverflow.net/questions/430121 | 1 | I'm reading the book 'An Introduction to the Kahler-Ricci Flow' (Lecture Notes in Mathematics 2086). They discuss Bott-Chern cohomology on complex spaces:
Let $X$ be a complex space(i.e. analytic variety) with normal singularity.
>
> Lemma 4.6.1. Any pluriharmonic distribution on $X$ is locally the real part of a holomorphic function, i.e. the kernel of the $dd^c$ operator on the sheaf $\mathcal D'\_X$ of germs of distributions coincides with the sheaf $\mathfrak R\mathcal O\_X$ of real parts of holomorphic germs.
>
>
>
>
> Definition 4.6.2. A (1,1)-form (resp.(1,1)-current) with local potentials on $X$ is defined to be a section of the quotient sheaf $\mathcal C^\infty\_X/\mathfrak R\mathcal O\_X$(resp. $\mathcal D'\_X/\mathfrak R\mathcal O\_X$). We also introduce the Bott-Chern cohomology space $H^{1,1}\_{BC}(X):=H^1(X,\mathfrak R\mathcal O\_X)$.
>
>
>
My questions are:
1. The lemma seems suspicious to me. It implies if $dd^cf = 0$ for a smooth function $f$, then $f = Re(h)$ for some holomorphic function $h$. So, $f$ must be a real function? At least $if$ also satisfies the condition which is not real.
2. They say a (1,1)-form with local potentials can be described as a closed $(1,1)$-form $\theta$ on $X$ that is locally of the form $\theta = dd^cu$ for a smooth function $u$. I wnt to prove this. First, I need to use the short exact sequence
$$0\rightarrow\mathfrak R\mathcal O\_X\rightarrow \mathcal A^0\_X\rightarrow \mathcal A\_X^0/\mathfrak R\mathcal O\_X\rightarrow 0$$
but I have no idea explicitly what the first map is?
3. They mention that a closed $(1,1)$-forms and currents on $X$ are not necessary locally $dd^c$-exact in general. But in my knowledge(I may be wrong), it holds when $X$ is a manifold. What makes it different when $X$ is singular?
| https://mathoverflow.net/users/167083 | Bott-Chern cohomology for singular complex spaces |
>
> closed (1,1)-forms and currents on X
> are not necessary locally $dd^c$-exact in general
> What makes it different when X is singular?
>
>
>
The obstruction to local $dd^c$-lemma
is $R^1\pi\_\*(O\_{X'})$, where
$\pi:\; X' \to X$ is the resolution of
singularities. When $X$ is smooth
(more generally, when it
has rational singularities), this sheaf is
trivial, but if the singularities are bad
(say, when $X$ is the cone over a curve
of big genus), local $dd^c$-lemma might
fail.
>
> It implies if $dd^cf=0$ for a smooth function $f$, then
> $f=Re(h)$ for some holomorphic function $h$. So, $f$ must be a
> real function?
>
>
>
This is usually assumed. If $f$ is complex,
its real and imaginary parts are both real
parts of (a priori, different) holomorphic
functions.
| 3 | https://mathoverflow.net/users/3377 | 430128 | 174,244 |
https://mathoverflow.net/questions/430126 | 4 | I've asked two years ago a post on Mathematics Stack Exchange, were provided two excellent answers. I'm asking on MathOverflow in the hope that some professor can to expand/improve (if it is possible) these results answering my question. The post has the same title and identifier [*3757149*](https://math.stackexchange.com/questions/3757149/on-the-diophantine-equation-xm-1x1-yn-1y1-with-xy-over-integer) on Mathematics Stack Exchange.
I don't know if the following diophantine equation (problem) is in the literature. We consider the diophantine equation $$x^{m-1}(x+1)=y^{n-1}(y+1)\tag{1}$$
over integers $x\geq 2$ and $y\geq 2$ with $x>y$, and over integers $m\geq 2$ and $n\geq 2$. These are four integral variables $x,y,m$ and $n$. The solutions that I know for the problem $(1)$ are two, the solution $(x,y;m,n)=(3,2;2,3)$ and $(98,21;2,3)$.
>
> **Question 1.** Do you know if this problem is in the literature? Alternatively, if this problem isn't in the literature can you find more solutions? **Many thanks.**
>
>
>
If the equation or problem $(1)$ is in the literature please refer it answering this question as a reference request, and I try to search and read the statements for new solutions from the literature. In other case compute more solutions or add upto what uppers limits you got evidence that there aren't more solutions.
I would like to know **what work can be done** with the purpose to know if the problem $(1)$ have finitely many solutions $(x,y;m,n)$.
>
> **Question 2.** Are there finitely many solutions $(x,y;m,n)$ of stated problem $(1)$? I mean what relevant reasonings or heuristics you can to deduce with the purpose to study if the problem have finitely many solutions. **Many thanks**
>
>
>
If this second question is in the literature, please refer the literature answering this question as a reference request, and I try to search and read the statements from the literature.
| https://mathoverflow.net/users/142929 | On the diophantine equation $x^{m-1}(x+1)=y^{n-1}(y+1)$ with $x>y$, over integers greater or equal than two | You can make a lot of progress if you're willing to assume a deep conjecutre. The $N$-variable generalization of the $abc$-conjecture (<https://en.wikipedia.org/wiki/N_conjecture>) applied to your equation (with $N=4$) says that if $\gcd(x,y)=1$ and if no subsum of the $4$-term sum
$$ x^n + x^{n-1} - y^m - y^{m-1} $$
vanishes, then
$$ \max\{ x^n, y^m \} \le C\_\epsilon\cdot (xy)^{3+\epsilon}. \quad(\*) $$
Since you're assuming that $x>y\ge2$, this first implies that
$$ x^n \le C\_\epsilon\cdot x^{6+2\epsilon}, $$
so there are only finitely many solutions with $n\ge7$. If we assume that $n\le6$, then
$$ x^6 \le x^n \le x^n + x^{n-1} = y^m + y^{m-1} \le 2y^m, $$
so $x\le (2y^m)^{1/6}$. Then $(\*)$ gives
$$ y^m \le C\_\epsilon (xy)^{3+\epsilon} \le C\_\epsilon ((2y^m)^{1/6})^{3+\epsilon})y^{3+\epsilon} \le C'\_\epsilon y^{(m/2+3)(1+\epsilon)}. $$
This shows that there are only finitely many solutions with $m>m/2+3$, so only finitely many solutions with $m>6$.
So now you're reduced to a finite number of exponents, and since you've assumed that $x>y$, you need to handle exponents satisfying $2\le n<m\le 6$. Presumably for each choice of $(m,n)$, there are only finitely many solutions. In any case, since you want to avoid $x=y$, you're looking at integral points on the curve
$$ \frac{x^n + x^{n-1} - y^m - y^{m-1}}{x-y}=0. $$
For each $(m,n)$, there will only be infinitely many integer solutions if the polynomials has a linear or quadratic factor.
Finally, it's possible that the cases where some subsum vanishes can be treated directly, but in any case, the usual $abc$-conjecture will handle the three term vanishing sums, and the two term vanishing sums are easy.
The assumptions in this analysis include the assumption that $\gcd(x,y)=1$. If you don't want that assumption, you can write $x=ZX$ and $y=ZY$ with $Z=\gcd(x,y)$, so your equation becomes
$$ ZX^n + X^{n-1} = Z^{m-n+1} Y^m + Z^{m-n} Y^{m-1}. $$
Now you can try applying the $4$-variable $abc$ conjecture to this equation.
| 10 | https://mathoverflow.net/users/11926 | 430130 | 174,245 |
https://mathoverflow.net/questions/430082 | 4 | Let $n$ be a positive integer and $0 \leq i < n$. Define
$$
N(i) = \# \left\{ (x\_1,\dots, x\_s) \in [1, n]^s: x\_1^2 +\dots + x\_s^2 \equiv i \mod n \right\}.
$$
I am looking for a reference for the following result: if $s$ is large enough,
$$
n^{s-1} \ll N(i) \ll n^{s-1}
$$
is true for all $i$ and $n$. Could anyone provide a reference for this result? Thank you! Any input also appreciated.
| https://mathoverflow.net/users/48408 | Sum of many squares modulo $n$ | The estimate is false for $s\leq 4$, but it is true for $s\geq 5$. For example, if $s=3$ and $8\mid n$ and $i=7$, then $N(i)=0$. Or if $s=4$ and $n=4^k$ and $i=0$, then $N(i)\leq n^2$. For $s\geq 5$ the result follows easily from the fact that the number of ways to write a positive integer $m$ as a sum of $s$ squares is $\asymp\_s m^{s/2-1}$. For more details, see Vaughan's book "The Hardy-Littlewood method".
**Added.** Here are some additional details. From the mentioned bound
$$r\_s(m):=\#\{ x\_1,\ldots,x\_s \in \mathbb{Z}: x\_1^2+\ldots + x\_s^2 = m\} \asymp\_s m^{s/2 - 1},$$
it follows that
\begin{align\*}
N(i)&\leq\sum\_{k=1}^{sn+1} r\_s(kn-n+i)\ll\_s \sum\_{k=1}^{sn+1}(kn)^{s/2-1}\ll\_s n^{s-1},\\
N(i)&\geq\sum\_{k=1}^{n} r\_s(kn-n+i)\gg\_s \sum\_{k=1}^{n}(kn)^{s/2-1}\gg\_s n^{s-1}.
\end{align\*}
| 11 | https://mathoverflow.net/users/11919 | 430133 | 174,246 |
https://mathoverflow.net/questions/430129 | 7 | I've been trying to understand the Adams spectral sequence, and I've gotten myself confused about how derived descent is supposed to work, so I would like to understand a simple example.
Given a faithfully flat map of commutative rings $A \to B$, the usual rules of faithfully flat descent let us identify $A$-modules as $B$-modules equipped with descent data (one nice presentation of this is that a $B$-module with descent data is a comodule over the $B$-coalgebra $B \otimes\_A B$). If I understand correctly, derived descent is supposed to allow us to generalize this to more general maps of rings, and so in particular I'd like to understand derived descent along the morphism $\mathbb{Z} \to \mathbb{F}\_2$.
I've made a bit of progress, working in the derived category of abelian groups. I believe a cofibrant replacement for $\mathbb{F}\_2$ is
$$ \widetilde{\mathbb{F}}\_2 = \mathbb{Z}[\eta; d \eta = 2], $$
and I can compute the coalgebra structure on
$$ \widetilde{\mathbb{F}}\_2 \otimes\_\mathbb{Z} \widetilde{\mathbb{F}}\_2 = \mathbb{Z} [\eta\_1, \eta\_2; d \eta\_i = 2]. $$
But now I get stuck. Somehow, a (cofibrant?) comodule over $\widetilde{\mathbb{F}}\_2 \otimes\_\mathbb{Z} \widetilde{\mathbb{F}}\_2$ is supposed to be the same thing as a 2-completed abelian group, but I don't even see how we would distinguish, say, $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/4\mathbb{Z}$ from different comodule structures on $\widetilde{\mathbb{F}}\_2$. Maybe the answer is in a choice of coassociator for the comodule structure?
In any case, I would greatly appreciate help doing this calculation, or advice on a different way to think about derived descent along $\mathbb{Z} \to \mathbb{F}\_2$ that makes manifest the equivalence between 2-completed abelian groups and $\mathbb{F}\_2$-modules with derived descent data.
| https://mathoverflow.net/users/490054 | Basic example of derived descent | $\newcommand{\Z}{\mathbf{Z}}\newcommand{\FF}{\mathbf{F}}\newcommand{\H}{\mathrm{H}}$Let me do the universal case of your situation, which is understanding descent along the map $\Z[t] \to \Z$ sending $t\mapsto 0$. [To get your case, note that $\FF\_2 = \Z \otimes\_{\Z[t]} \Z$, where one of the maps $\Z[t] \to \Z$ sends $t\mapsto 0$, and the other map sends $t\mapsto 2$. In other words, replace $t$ everywhere below by $2$.]
How does one do descent along $\Z[t] \to \Z$? In other words, suppose I have a (derivedly) $t$-complete $\Z[t]$-module $M$; how do I encode in terms of a $\Z$-module and descent data?
Let $R\_1$ denote the complex $\Z \otimes\_{\Z[t]}^L \Z$, so that $\H\_\ast(R)$ is given by $\Z[\epsilon]/\epsilon^2$, where $\epsilon$ lives in $\H\_1$. If $M$ is a $\Z[t]$-module, then the associated $\Z$-module is the complex $M \otimes\_{\Z[t]}^L \Z$. I'll write this as $M\_0$.
What is the $R\_1 = \Z \otimes\_{\Z[t]}^L \Z$-comodule structure on $M\_0$? This is a map $M\_0 \to M\_0 \otimes\_\Z R\_1$, which on homology gives a map $M\_0 \to M\_0 \otimes\_\Z \Z[\epsilon]/\epsilon^2$. Projecting onto the second factor of $\Z[\epsilon]/\epsilon^2 \cong \Z \oplus \Z \cdot \epsilon$, we obtain a map $\delta\_1: M\_0 \to M\_0 [1]$. (The shift by $1$ is because $\epsilon$ lives in homological degree $1$.) Explicitly, $\delta\_1$ is the composite
$$M\_0 \to M\_0 \otimes\_\Z \Z[\epsilon]/\epsilon^2 \to M\_0 [1].$$
What is $\delta\_1$? Well, there is a cofiber sequence/distinguished triangle in chain complexes of $\Z[t]$-modules given by
$$\Z \cong (t)/(t^2) \to \Z[t]/t^2 \to \Z[t]/t = \Z.$$
Rotating this gives a cofiber sequence
$$\Z[t]/t^2 \to \Z \to \Z[1].$$
Applying $M \otimes\_{\Z[t]}^L -$, we get a cofiber sequence
$$M \otimes\_{\Z[t]}^L \Z[t]/t^2 \to M \otimes\_{\Z[t]}^L \Z = M\_0 \to M \otimes\_{\Z[t]}^L \Z[1] = M\_0 [1].$$
This second map $M\_0 \to M\_0[1]$ is precisely $\delta\_1$. If you want, this map $\Z \to \Z[1]$ is the "universal example" of a $\delta\_1$.
Summarizing, the $R\_1$-comodule structure on $M\_0$ encodes $\delta\_1$, and taking the fiber of $\delta\_1$ produces $M \otimes\_{\Z[t]}^L \Z[t]/t^2$. In other words, knowing the $R\_1$-comodule structure is the same as knowing $M \otimes\_{\Z[t]}^L \Z[t]/t^2$.
In the derived setting, one has not just the $R\_1$-comodule structure, but a whole bunch of other structure coming from $R\_n = \Z^{\otimes\_{\Z[t]}^L n+1}$. Note that $R\_n = R\_{n-1} \otimes\_{\Z[t]}^L \Z$. When one keeps track of the structure arising from $R\_n$, the same argument as above shows that one has a new $\delta$ showing up. This new $\delta$ will correspond to the cofiber sequence
$$\Z[t]/t^{n-1} \cong (t)/(t^{n-1}) \to \Z[t]/t^n \to \Z[t]/t = \Z,$$
which rotates to a cofiber sequence
$$\Z[t]/t^n \to \Z \to \Z[t]/t^{n-1} [1].$$
If you want, this map $\Z \to \Z[t]/t^{n-1} [1]$ is the "universal example" of a $\delta\_{n-1}$.
If one knows all this structure in a coherent way (which is exactly what the derived comodule structure is encoding), then the above discussion shows that you can recover $M \otimes\_{\Z[t]}^L \Z[t]/t^n$ for every $n\geq 1$. If $M$ is derived $t$-complete, then knowing $M \otimes\_{\Z[t]}^L \Z[t]/t^n$ for every $n\geq 1$ is the same as knowing $M$ itself. In other words, the derived comodule structure in all its coherent glory is the descent data for $\Z[t] \to \Z$. (One could think of the $\delta\_n$'s as being the differentials in the associated graded spectral sequence for the $t$-adic filtration on $M$, i.e., the $t$-Bockstein spectral sequence. This whole story can then be viewed as a version of Koszul duality between $\Z[t]^\wedge\_t = \Z[\![t]\!]$ and $\Z \otimes\_{\Z[t]}^L \Z = \Z[\epsilon]/\epsilon^2$.)
| 8 | https://mathoverflow.net/users/102390 | 430134 | 174,247 |
https://mathoverflow.net/questions/430136 | 21 | A well known equivalent of the Axiom of Choice is Krull's Maximal Ideal Theorem (1929): if $I$ is a proper ideal of a ring $R$ (with unity), then $R$ has a maximal ideal containing $I$. The proof is easy with Zorn's Lemma. The converse, Krull implies Zorn, is due to Hodges (1979).
A ring with a unique maximal ideal is said to be local. A standard homework problem is this: Show that if $R$ is a local ring with maximal ideal $M$, then every element outside of $M$ is a unit. The "usual" solution is as follows: Let $x$ be a nonunit and let $I$ be the ideal generated by $x$. By Krull, $I$ is contained in a maximal ideal which, by locality, must be $M$.
Since the solution relies on AC, a natural question arises:
Is the homework problem itself equivalent to AC? More precisely, assume that in every local ring, every element outside of the unique maximal ideal is a unit. Does this imply AC?
If the answer is no, is there a proof of the homework problem which does not rely on AC?
| https://mathoverflow.net/users/159728 | A Krull-like Theorem and its possible equivalence to AC | Nice question ! I believe the homework exercise implies AC.
Indeed, assume its conclusion holds, and let $R$ be a ring with no maximal ideal. I'm going to prove that $R$ is zero, thus proving Krull's theorem (apply this to $R/I$ for a proper ideal $I$).
Let $k$ be your favourite field. Then $k\times R$ is local : $0\times R$ is a maximal ideal, and any ideal of a product is of the form $I\times J$, so because $R$ has no maximal ideal, $0\times R$ is the only maximal ideal.
Thus, by the homework problem, anything outside is invertible - but $(1,0)$ is outside. This proves that $0$ is invertible in $R$, hence $R=0$.
| 33 | https://mathoverflow.net/users/102343 | 430138 | 174,249 |
https://mathoverflow.net/questions/429844 | 2 | *The sets are defined in $\mathbb{R}\_+^n$ $(n\geq 1)$. The relative interior of a convex set $C$ is denoted $\mathring C$.*
Let $S$ be the $n$-simplex:
$$S=\left\{x\in\mathbb{R}\_+^n,\,\sum\_{i=1}^n x\_i=1\right\}$$
and $E$ be a linear subspace such that $E\cap\mathring S\neq\varnothing$ (hence, $E$ contains a vector with positive coordinates) of codimension $2$ in $\mathbb{R}^n$. For $F$ a face of $E\cap S$, one can see that there exists a unique face $G$ of $S$ such that $\mathring F\subset\mathring G$. Hence, $E\cap G=F$.
**Question:** We assume that, in $\text{Aff}(G)$, $\text{Aff}(F)$ has codimension $2$. If a face $G'$ of $S$ satisfies $G'\cap E=F$, do $G'=G$?
| https://mathoverflow.net/users/159940 | Intersection of the simplex with a linear subspace of codimension $2$ | The answer is *Yes*.
By your assumption there exists $x\in\mathring F\subset \mathring G$.
We also have $x\in F\subset G'$, and the only way a face $G'$ can contain an interior point of a face $G$ is if $G\subseteq G'$.
Now, consider a small neighborhood of $x$. In this neighborhood, $F$, $G$ and $E$ look like linear subspaces, while $G'$ looks like a cone that contains the subspace $G$. We will keep this local perspective for the rest of this proof (we can assume that $x$ is the origin of our local linear space).
Since all subspaces contain $F$, we can factor by $F$. We find
$$G/F\cap E/F = (G\cap E)/F=F/F = \{0\}.$$
Moreover, since $\DeclareMathOperator{\codim}{codim}\codim\_G F=2$, we have $\dim G/F=2$; and since $\codim\_{\Bbb R^n} E=2$, we have $\codim\_{\Bbb R^n/F} E/F=2$. We thus find the direct decomposition
$$G/F \oplus E/F = \Bbb R^n/F.$$
This suffices to conclude that also $G'\subseteq G$, hence $G=G'$: fix $y\in G'/F$.
Then there is a unique decomposition $y=y'+y''$ with $y'\in G/F$ and $y''\in E/F$. Since $y'\in G/F$ and $G/F$ is a linear space, we have $-y'\in G/F\subset G'/F$.
Since $G'/F$ is a cone, we have $y+(-y')=y''\in G'/F$. Thus
$$y''\in E/F\cap G'/F = (E\cap G')/F=F/F=\{0\}$$
and we conclude $y=y'\in G/F$.
| 1 | https://mathoverflow.net/users/108884 | 430140 | 174,250 |
https://mathoverflow.net/questions/430158 | 2 | Under suitable large-cardinal assumptions, in the inner model $L(\mathbb R)$ one can have $\omega\_1$ and $\omega\_2$ measurable (this follows from determinacy).
I was wondering if it is possible to reconcile measurability of $\omega\_1$ (or other cardinals) with the model $L( \mathbb R^{\omega\_1} )$ or objects alike?
Here $\mathbb R^{\omega\_1}$ is the product space of $\omega\_1$-many copies of the real line.
| https://mathoverflow.net/users/491011 | Large cardinals and measurability in $L(A)$ | For each $\alpha<\omega\_1$, choose a real $r\_\alpha$ that codes the ordertype $\alpha$. This sequence $\langle r\_\alpha : \alpha < \omega\_1 \rangle$ then codes a sequence of surjections from $\omega$ to each countable ordinal. From this, you can then run the Ulam matrix construction and show that there is no countably complete ultrafilter on $\omega\_1$.
Let $F$ be a countably complete filter on $\omega\_1$ in $L(\mathbb R^{\omega\_1})$, and let $\sigma\_\alpha : \omega \to \alpha$ be the surjection coded by $r\_\alpha$. For $n<\omega$ and $\beta<\omega\_1$, let $S^\beta\_n = \{ \alpha > \beta : \sigma\_\alpha(n) = \beta \}$. Since $\bigcup\_{n<\omega} S\_n^\beta$ is the tail interval $(\beta,\omega\_1)$, we can take $n\_\beta$ to be the least $n$ such that $S^\beta\_n$ is $F$-positive, by countable completeness. Let $T\_n = \{ \beta : n\_\beta = n \}$. Since $\omega\_1$ is uncountable, there is $n$ such that $|T\_n| > 1$. So for such $T\_n$, if $\beta<\gamma$ are in $T\_n$ and $\alpha \in S\_n^\beta \cap S\_n^\gamma$, then $\sigma\_\alpha(n) = \beta = \gamma$, which is impossible. Thus $S\_n^\beta$ and $S\_n^\gamma$ are disjoint $F$-positive sets, and so $F$ is not an ultrafilter.
| 6 | https://mathoverflow.net/users/11145 | 430161 | 174,254 |
https://mathoverflow.net/questions/430143 | 4 | I'm studying the Meyer's book, "Wavelets and operators", and I'm confused about a proof of Bernstein's inequality at page 47, which is stated below:
"The function $\frac{\xi^\beta}{|\xi|^s}\hat\phi(\xi)$ is the Fourier transform of an integrable function." Here $\hat\phi(\xi)\in C^\infty\_0(\mathbb{R}^n)$ is a bump function, which is equal to 1 on $|\xi|\leq \frac{1}{2}$, 0 on $|\xi|\geq 1$, $|\beta|\geq 1\in\mathbb{N}$ and $|\beta|-1<s<|\beta|$.
Is it correct? How to prove it? Thanks in advance!
| https://mathoverflow.net/users/490996 | A proof of Bernstein's inequality | Well, I find that there could be another answer. We still only need prove the integrablity at inifity. Considering the homogeneous function $\frac{\xi^\beta}{|\xi|^s}$, by the proposition 2.4.8 in "Classical Fourier Analysis", its original function, denoted by $g(x)$, is smooth on $\mathbb{R}^n\backslash\{0\}$ and homogeneous of degree $-n-|\beta|+s$, that is, $g(x)\sim\frac{g(x')}{|x|^{n+|\beta|-s}}$. Then, we have
\begin{equation}
K(x)=\int \frac{\xi^\beta}{|\xi|^s}\hat\phi(\xi) e^{ix\cdot\xi}\,d\xi=\int\frac{\xi^\beta}{|\xi|^s}e^{ix\cdot\xi}\,d\xi-\int \frac{\xi^\beta}{|\xi|^s}[1-\hat\phi(\xi)] e^{ix\cdot\xi}\,d\xi
\end{equation}
For the second term, let $|\alpha|=m$ large enough, then
$$|x^\alpha K(x)|\lesssim \int \left|\sum\_{\alpha\_1+\alpha\_2=\alpha}D\_\xi^{\alpha\_1}\frac{\xi^\beta}{|\xi|^s}D\_\xi^{\alpha\_2}[1-\hat\phi(\xi)]\right|\,d\xi\lesssim 1,$$
since, for each $|\alpha\_2|\geq 1$, $D\_\xi^{\alpha\_2}[1-\hat\phi(\xi)]$ has compact support away from the origin and $[1-\hat\phi(\xi)]D\_\xi^{\alpha}\frac{\xi^\beta}{|\xi|^s}$ is integrable on $\mathbb{R}^n$. Therefore, choosing $m\geq n+2(|\beta|-s)$, we obtain
$$|K(x)|\lesssim \frac{|g(x')|}{|x|^{n+|\beta|-s}}+\frac{1}{|x|^{m-|\beta|+s}}\lesssim \frac{1}{|x|^{n+|\beta|-s}},\,\mathrm{when}\,|x|\geq 1$$
which conculdes the proof.
| 1 | https://mathoverflow.net/users/490996 | 430174 | 174,262 |
https://mathoverflow.net/questions/430176 | 8 | Let $E$ be a topos and let $\Omega\_E$ be its subobject classifier. We can consider $\Omega$ as an internal poset in $E$, and thus we may form the $E$-topos of internal presheaves $\hat{\Omega}\_E := [\Omega\_E^{\text{op}},E]$. This feels like a "fatter" version of the relative Sierpinski $E$-topos $\mathbb{S}\_E := [\cdot\!\to\!\cdot,E]$, but I don't know what it is geometrically. (Indeed, when $E$ is boolean, I suppose this is exactly the Sierpinski $E$-topos.)
I encountered this topos through essentially domain-theoretic considerations; in particular, when developing domain theory within a topos $E$, the partial map classifier monad of the topos lifts to a *lifting monad* $L$ on internal dcpos. When you consider $L(1) = \Omega\_E$, you have what should be considered a **Sierpinski space** in the sense of internal domain theory and denotational semantics. But taking $E$-valued sheaves on this "Sierpinski $E$-dcpo", I think we would not get the actual Sierpinski $E$-topos, but rather (I suppose), the rather strange topos that I opened the question with.
I am wondering whether anyone has some experience or insight on this topos, and am curious about the following questions:
1. Is there a description of the $E$-geometric theory that $\hat{\Omega}\_E$ classifies?
2. There is an essential geometric morphism of $E$-topoi $\mathbb{S}\to \hat{\Omega}\_E$ corresponding to the inclusion $2\hookrightarrow\Omega\_E$. What other properties does this morphism have?
3. Does this topos occur anywhere in the literature?
| https://mathoverflow.net/users/51336 | Internal presheaves on the subobject classifier | I'm not sure this constitute an answer to your question, but from what you are telling of your motivation it very well might be, and in any case it was too long to be a comment.
First - I don't think there is a good answer to your questions 1 and 2. The problem is that considering $\Omega$ as just a poset is a very "non-geometric" thing to do (in the sense of geometric logic) in fact this is somehow the typical non-geometric construction. So asking what its geometric properties are will probably not lead anywhere.
As such, I don't think there is any better description of this topos than the tautological one, that it classifies the ideals of $\Omega\_E$, in the sense that a morphism $T \to \widehat{\Omega\_E}$ classifies pairs of a morphism $f:T \to E$ and of an ideal of $f^\*(\Omega\_E)$. The non-geometric nature of $\Omega$ (when considered as a poset) means that $f^\*(\Omega\_E)$ can be pretty much anything, it has a comparison map to $\Omega\_T$ of course, but that just means it has a distinguished subobject - so that not really an interesting observation.
**However - there is another construction** you can make that seems to fit what you are talking about more closely and produce a much nicer result. The thing is $\Omega$ might not be geometric when seen as a "set" or "poset", but it is geometric when seen with more structure - for example, it is obviously geometric when considered as a frame - but what might be more interesting to you given what you are saying - it is geometric when considered as a DCPO. (what I mean here is that if $f^\sharp$ denotes the left adjoint to $f\_\*$ acting on the categories of DCPO, then $f^\sharp \Omega\_E = \Omega\_T$). This is because $\Omega$ is the ind-completion of $\{0 < 1\}$.
So, if you really treat $\Omega$ as a DCPO you'll get much better results. This means you can't look at general presheaves on $\Omega$ though. But there is another way to get a topos out of a DCPO that will be geometric: You can use what Ivan Di liberty and myself have called the "Scott topos" (see [here](https://arxiv.org/abs/1812.00652), and also Ivan Di Liberti's [thesis](https://arxiv.org/abs/2009.07320) and [paper](https://arxiv.org/abs/2009.14023)).
In Short, given a DCPO $P$, defines $S(P)$ as the category of (covariant) functor $P \to Set$ that preserves directed colimits. Then $S(P)$ is a (localic) topos, and you have a DCPO morphism from $P$ to the poset (DCPO) of points of $S(P)$. The Scott topos is a geometric construction on DCPO in the sense that the Scott topos $S(f^\sharp P)$ identifies with the pullback of the topos $S(P)$.
Moreover, in your present situation, as $\Omega$ is the ind-completion of {0<1} the Scott topos $S(\Omega)$ is exactly the Sierpinski topos.
| 10 | https://mathoverflow.net/users/22131 | 430179 | 174,263 |
https://mathoverflow.net/questions/430131 | 6 | $\newcommand{\GL}{\mathrm{GL}}\newcommand{\SO}{\mathrm{SO}}\newcommand{\SU}{\mathrm{SU}}\newcommand{\Spin}{\mathrm{Spin}}\renewcommand{\O}{\mathrm
O}\newcommand{\R}{\mathbb
R}\newcommand\Z{\mathbb Z}$If $\rho\colon \SU\_8\to\GL\_n(\R)$ is a representation, then we can lift $\rho$ to land in the spin group
$\Spin\_n$ in three steps:
* Since $\SU\_8$ is compact, $\GL\_n(\R)$ admits an $\SU\_8$-invariant product, and the image of
$\rho$ is contained in the subgroup of orthogonal matrices, which is isomorphic to $\O\_n$.
* Since $\SU\_8$ is connected, $\rho\colon\SU\_8\to\O\_n$ lands in the connected component of the
identity in $\O\_n$, which is $\SO\_n$.
* Now we want to lift $\rho$ across the double cover $\Spin\_n\to\SO\_n$. Such a lift exists because
$\SU\_8$ is simply connected.
I want to know whether this is always possible for the quotient $\SU\_8/\{\pm 1\}$. The first two steps are
unchanged, but this group is not simply connected, so a priori there could be a representation
$\rho\colon\SU\_8/\{\pm 1\}\to\SO\_n$ that does not lift to $\Spin\_n$. As additional evidence, the obstruction $w\_2(\rho)$ for lifting
from $\SO\_n$ to $\Spin\_n$ lives in $H^2(B(\SU\_8/\{\pm 1\});\Z/2)\cong\Z/2$, so it could be nonzero in
principle. But on all the examples I've calculated, the obstruction vanishes, and $\rho$ lifts to $\Spin\_n$.
This seems like the sort of thing that may have been worked out already, or that would follow from a clever
application of highest weight theory. But the fact that $\SU\_8/\{\pm 1\}$ isn't simply connected, and that I
want to look at real representations, makes the combinatorics more confusing for me. Is an example of a non-spin
representation of this group known, and if not, how would I go about working this out?
| https://mathoverflow.net/users/97265 | Is there a representation of $\mathrm{SU}_8/\{\pm 1\}$ that doesn't lift to a spin group? | $\def\ZZ{\mathbb{Z}}\def\RR{\mathbb{R}}\def\CC\mathbb{C}\def\SU{\text{SU}}\def\SO{\text{SO}}\def\Spin{\text{Spin}}\def\diag{\text{diag}}\def\Id{\text{Id}}$Such a lift is always possible. We will show:
**Theorem** Let $n \equiv 0 \bmod 8$ be a positive integer and let $\rho : \SU\_n/\{ \pm \Id\_n \} \to \SO\_m$ be a (continuous) representation. Then $\rho$ lifts to a map $\SU\_n/\{ \pm \Id\_n \} \to \Spin\_m$.
We have $\pi\_1(\SU\_n/ \{ \pm \Id\_n \}) \cong \pi\_1(\SO\_m) \cong \ZZ/ 2 \ZZ$ (for $m \geq 3$), and such a lift exists if and only if $\rho\_{\ast} : \pi\_1(\SU\_n/ \{ \pm 1 \}) \longrightarrow \pi\_1(\SO\_m)$ is $0$, so that is what we will be trying to show.
---
**Part one: Tori** Our first task will be to write down some useful tori. Let $S$ be the circle group $\{ z \in \CC : |z| = 1 \}$; we will also write $z$ as $e^{i \theta}$.
Map the interval $[0, 2 \pi]$ to $\SU\_n$ by $\theta \mapsto \diag(e^{i\theta/2}, e^{- i \theta/2}, e^{i\theta/2}, e^{- i \theta/2}, \dots, e^{i\theta/2}, e^{- i \theta/2})$. This is a path from $\Id\_n$ to $- \Id\_n$. In the quotient $\SU\_n/\{ \pm \Id\_n \}$, it becomes a closed loop and a group homorphism $\gamma: S \to \SU\_n/\{ \pm \Id\_n \}$. Moreover, $\gamma\_{\ast}: \pi\_1(S) \to \pi\_1(\SU\_n/\{ \pm \Id\_n \})$ is surjective. So it is enough to show that the induced map $\rho\_{\ast} \circ \gamma\_{\ast}$ from $\pi\_1(S)$ to $\pi\_1(\SO\_m)$ is $0$.
Put $Z(\theta) = \left[ \begin{smallmatrix} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \\ \end{smallmatrix} \right]$. If $m$ is even, let $T$ be the maximal torus in $\SO\_m$ of block diagonal matrices of the form $\diag(Z(\theta\_1), Z(\theta\_2), \ldots, Z(\theta\_{m/2}))$; if $m$ is odd, let $T$ be the block diagonal matrices of the form $\diag(Z(\theta\_1), Z(\theta\_2), \ldots, Z(\theta\_{m/2}), 1)$. We write $\delta$ for the inclusion $T \hookrightarrow \SO\_m$.
Choose coordinates on $\SO\_m$ so that the homomorphism $\rho \circ \gamma : S \to \SO\_m$ factors as $\delta \circ \beta$.
$$\begin{matrix}
S & \overset{\beta}{\longrightarrow} & T \\
\gamma \downarrow && \delta \downarrow \\
\SU\_n/{\pm \Id\_n} & \overset{\rho}{\longrightarrow} & \SO\_m.
\end{matrix}$$
It is well known that $\delta\_{\ast} : \pi\_1(T) \to \pi\_1(\SO\_m)$ is the map $\ZZ^{\lfloor m/2 \rfloor} \to \ZZ/2 \ZZ$ given by $(b\_1, b\_2, \ldots, b\_{\lfloor m/2 \rfloor}) \mapsto \left( \sum b\_i \bmod 2 \right)$. The map $\beta$, being a group homomorphism, must be $\theta \mapsto \diag(Z(b\_1 \theta), Z(b\_2 \theta), \ldots, Z(b\_{\lfloor m/2 \rfloor} \theta))$ (or the same thing with $1$ tagged on the end) for some integers $b\_1$, $b\_2$, ..., $b\_{\lfloor m/2 \rfloor}$. So our goal is to show that $\sum b\_j$ is even.
---
**Part two: Combinatorics** Consider the character of the group homomorphism $\rho \circ \gamma = \delta \circ \beta : S \to SO\_m$; it is a Laurent polynomial in the coordinate $z$, and we will write it $f(z)$.
Explicitly, we have
$$f(z) = \sum\_{j=1}^{\lfloor m/2} \left( z^{b\_j} + z^{-b\_j} \right) + \begin{cases} 0 & m \equiv 0 \bmod 2 \\ 1 & m \equiv 1 \bmod 2 \end{cases}.$$
**Lemma** We have $f(1) - f(-1) \equiv 4 \sum b\_j \bmod 8$.
**Proof** If $b\_j$ is even, then $1^{b\_j} + 1^{-b\_j} = (-1)^{b\_j} + (-1)^{-b\_j}$. If $b\_j$ is odd, then $1^{b\_j} + 1^{-b\_j} = (-1)^{b\_j} + (-1)^{-b\_j}+4$ $\square$
Therefore, our goal is to show that $f(1) \equiv f(-1) \bmod 8$.
Now, recall the representation $\rho : \SU\_n/\{ \pm 1 \} \to \SO\_m$ and restrict it to the torus $D := \{ \diag(z\_1, z\_2, \ldots, z\_n) : z\_j \in S,\ \prod z\_j = 1 \}$ inside $\SU\_n$. The character of $\rho$ is a symmetric polynomial in the $z\_j$ with integer coefficients, call it $g(z\_1, z\_2, \ldots, z\_n)$.
The point $-1$ in $S$ is mapped to $\diag(i, -i, i, -i, \ldots, i, -i)$ in $\SU\_n/(\pm \Id\_n)$. So $f(-1) = g(i, -i, i, -i, \ldots, i, -i)$. Of course, $1 \in S$ maps to $\diag(1,1,\ldots,1)$, so $f(1) = g(1,1,\ldots, 1)$. Thus, we are done if we prove the following lemma:
**Lemma** Let $n \equiv 0 \bmod 8$ and let $g(z\_1, z\_2, \ldots, z\_n)$ be a
symmetric polynomial with integer coefficients. Then
$$g(1,1,\ldots,1) \equiv g(i, -i, i, -i, \ldots, i, -i) \bmod 8.$$
**Proof** By the fundamental theorem of symmetric polynomials, $g$ is a polynomial in the elementary symmetric polynomials $e\_1, e\_2, \dotsc, e\_n$ with $\ZZ$-coefficients. So it is enough to show that
$$e\_k(1,1,\dotsc,1) \equiv e\_k(i, -i, i, -i, \dotsc, i, -i) \bmod 8$$
for $1 \leq k \leq n$. Organizing this into a single generating function, we want to show the polynomial congruence
$$(1+t^2)^{n/2} \equiv (1+t)^n \bmod{8 \ZZ[t]}.$$
To this end, it is enough to show that $(1+t^2)^{4} \equiv (1+t)^8 \bmod{8 \ZZ[t]}$ and then raise both sides to the $n/8$ power. This last condition can be checked by hand; we do indeed have
$$\begin{multline\*}
t^8 + 8 t^7 + 28 t^6 + 56 t^5 + 70 t^4 + 56 t^3 + 28 t^2 + 8 t + 1 \\ \equiv
t^8 + 4 t^6 + 6 t^4 + 4 t^1 + 1 \bmod{8 \ZZ[t]}. \qquad \square\end{multline\*}$$
---
**Remark:** The fact that $\rho$ is a representation of $\SU\_n/(\pm \Id\_n)$ means that $g$ has even degree. The condition that $\rho$ is a real (rather than complex) representation of $\SU\_n$ also imposes nontrivial conditions on $g$. But the lemma is right for all symmetric polynomials.
**Remark:** For the reader who doesn't like computing binomial coefficients, the formula $(1+t)^{2^k} \equiv (1+t^2)^{2^{k-1}} \bmod{2^k \ZZ[t]}$ can also be proved by induction on $k$.
| 6 | https://mathoverflow.net/users/297 | 430180 | 174,264 |
https://mathoverflow.net/questions/430185 | 3 | I am trying to understand the the following paper <https://arxiv.org/pdf/1810.10971.pdf>, in particular Example 2:
If $ Y \sim N(0,1)$, the standard normal on $\mathbb{R}$, then
$ \begin{align\*} \Big( \mathbb{E} \Big[ \frac{1}{m!} Y^{\otimes m }\Big]\Big)\_{m \geq 0 } = \exp\Big(\frac{1}{2} e\_1 \otimes e\_1 \Big) \in \prod\_{m \geq 0 } \mathbb{R}^{\otimes m} \end{align\*},$
where $e\_1$ is the unit basis vector of $\mathbb{R}$ and $\exp$ is defined in the following way:
\begin{align\*} \exp &: \prod\_{m \geq 0 } V^{\otimes m} \to \prod\_{m \geq 0 } V^{\otimes m},\\ exp(s) &:= \sum\_{m \geq 0} \frac{s^{\otimes m}}{m!} \end{align\*}
What does it mean to take a tensor product of random variables? In particular, what is $\mathbb{E} \Big[ Y^{ \otimes 3 }\Big]$, for example?
| https://mathoverflow.net/users/490927 | What is a tensor product of random variables? | In this context, I believe the tensor product on random variables is nothing other than the tensor product over the values of the RVs. (In other words, if $\Omega$ is a sample space and $X : \Omega \rightarrow V$ and $Y : \Omega \rightarrow W$ are RVs, then $X \otimes Y : \Omega \rightarrow V \otimes W$ is defined by $(X \otimes Y)(\omega) = X(\omega) \otimes Y(\omega)$.)
In Example 2, the RV $Y$ takes values in $\mathbb R$, which makes the tensor powers rather boring since $\mathbb R^{\otimes n} \simeq \mathbb R$ for any $n \geq 0$; we just have to take apart the notation to read out the usual (normalized) moments for the standard normal.
Thus we can expand (see Appendix A, for example)
$$
\exp \left ( \frac 12 e\_1 \otimes e\_1 \right ) = \left (1, 0, \frac 12, 0, \frac 1{2!} \left ( \frac12 \right )^2, 0, \ldots \right),
$$
noting that this notation gives the coefficients of $e\_1^{\otimes n}$ for $n \geq 0$.
Thus $\mathbb E [Y^{\otimes 3}] = 0 \in \mathbb R^{\otimes 3}$, since the standard normal has zero odd moments.
| 2 | https://mathoverflow.net/users/7961 | 430192 | 174,268 |
https://mathoverflow.net/questions/413965 | 8 | Let $Cat\_n$ denote the category of $n$-categories. Then [Joyal's category $\Theta\_n$](http://nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/Theta+category) is (1) a full [dense subcategory](http://nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/dense+subcategory) of $Cat\_n$ which (2) is also a [Reedy category](https://nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/Reedy+category).
**Question:** Is Joyal's category $\Theta\_n$ somehow uniquely characterized by (1) and (2)?
**Notes:**
* I've been ambiguous about what "category" and "$n$-category" mean -- the above statements are true whether "category" means "1-category" or "$(\infty,1)$-category". In the latter case, "$n$-category" can mean either "strict $n$-category" or "weak $(\infty,n)$-category" or various things in between.
* I'm interested in understanding the story for any $n \in \mathbb N \cup \{\infty\}$.
* One nice story which makes Joyal's category $\Theta\_n$ look very "canonical" is [the theory of generalized nerves](https://golem.ph.utexas.edu/category/2008/01/mark_weber_on_nerves_of_catego.html). I'd be interested, for example, if there were something to say in that framework about getting one's nerves to be indexed by Reedy categories.
* I'd also be interested if $\Theta\_n$ didn't turn out to be literally "unique" with respect to (1) and (2), but at least "minimal" with respect to (1) and (2), or something like that.
* When $n=1$, the question asks whether the simplex category $\Delta = \Theta\_1$ is the "unique" full subcategory of $Cat$ which is a Reedy category. It sounds like the answer is probably "no" in this case, but I don't know a good counterexample.
| https://mathoverflow.net/users/2362 | Is Joyal's category $\Theta_n$ the "only" Reedy category which is dense in $n$-categories? | I now believe what's really going on here isn't so much a question of Reedyness. Rather, we have the following:
**Claim:**
Let $\mathcal A \subseteq Cat\_n$ be a dense subcategory which is idempotent-complete. Then $\Theta\_n \subseteq \mathcal A$.
Here $Cat\_n$ means weak $(\infty,n)$-categories.
**Remarks:**
* This does have a "Reedy" upshot, since any Reedy category worth its salt (e.g. any elegant Reedy category) will be idempotent complete. And anyway, density is unaffected by passing to the idempotent completion, so asking for idempotent-completeness is a reasonable "normalization" condition.
* We deduce that $\Theta\_n$ is the unique minimal idempotent-complete dense subcategory of $Cat\_n$, or equivalently that $\Theta\_n$ is the intersection of all idempotent-complete dense subcategories of $Cat\_n$.
**Proof of Claim:**
For $A \in \mathcal A$ and $\theta \in \Theta$, let $A' \in Psh\_{Set}(\Theta\_n)$ be an [$n$-quasicategory](https://ncatlab.org/nlab/show/n-quasicategory) modeling $A$, and suppose that $\theta$ is not a retract of $A'$. Then any map $A' \to \theta$ factors through $\partial \theta$, i.e. $Hom(A', \partial \theta) \to Hom(A', \theta)$ an isomorphism. So, letting $D \theta$ be a fibrant replacement for $\partial \theta$, we have that $Hom(A', D\theta) \to Hom(A', \theta)$ is a split epimorphism.
Moreover, if $\theta$ is not a retract of $A'$ for any $A \in \mathcal A$, then the isomorphism $Hom(A', \partial \theta) \to Hom(A', \theta)$ is natural in $A'$, and so we get a splitting of $Hom(A', D\theta) \to Hom(A', \theta)$ which is natural in $A'$. That is, the map $\nu\_\mathcal{A}(D\theta) \to \nu\_\mathcal{A}(\theta)$ is a split epimorphism, where $\nu\_\mathcal{A} : Cat\_n \to Psh\_{Spaces}(\mathcal A)$ is the restricted Yoneda embedding. If $\mathcal A \subseteq Cat\_n$ is dense, so that $\nu\_\mathcal{A}$ is fully faithful, this implies that $D \theta \to \theta$ is a split epimorphism in $Cat\_n$.
That this is generally not the case follows from the following observations:
**Notation:** For $\theta \in \Theta$, let $H(\theta)$ denote the "long" $n$-fold hom-space of $\theta$, and let $H(D\theta)$ denote the "long" $n$-fold hom-space of $D\theta$ (where $n = dim(\theta)$). (So $H(\theta)$ is contractible -- its unique element is the "big" $n$-morphism of $\theta$ -- the one obtained by pasting together all the other $k$-morphisms of $\theta$ for $k \leq n$. $H(\partial \theta)$ is the boundary of this contractible space.)
Let $\hat \times$ denote the [pushout-product](https://ncatlab.org/nlab/show/pushout-product) of maps of spaces.
**Lemma:** For $\theta = [k \mid \theta\_1, \dots, \theta\_k] \in \Theta\_n$, we have $H(\theta) \simeq \ast$, $H(\partial [1]) = \emptyset$, and $H(\partial \theta) \simeq (H(\partial \theta\_1) \to \ast) \hat \times \cdots \hat \times (H(\partial \theta\_k) \to \ast)$.
**Proof:** That $H(\theta) = \ast$ is obvious and so also that $H(\partial [1]) = \emptyset$. We have $H(\theta) \cong H(\theta\_1) \times \cdots \times H(\theta\_k)$. The result then follows from the colimit decomposition of $\partial \theta$.
**Notation:** For $N \in \mathbb N$, let $\underline{N}^n \in \Theta\_n$ be defined inductively by $\underline{N}^1 = [N]$, and $\underline{N}^{n+1} = [N \mid \underline{N}^n,\dots,\underline{N}^n]$.
**Corollary:** For $N \geq 3$ and all $n$, $H(\underline{N}^n)$ is a sphere of finite dimension $\geq 1$.
**Proof:** When $n = 1$, $H(\underline{N}^1) \simeq S^{N-2}$, as can be computed using the homotopy coherent nerve. Then higher $H(\underline{N}^n)$'s are iterated joins of this space with itself (by the Lemma), so form connected spheres as claimed.
**Corollary:** For $N \geq 3$ and $\theta = \underline{N}^n$, the map $D\theta \to \theta$ is not a split epimorphism in $Cat\_n$.
**Proof:** Suppose there is a section. On the long hom-space, this is a map from a disk $D^m = H(\theta)$ to a sphere $S^{m-1} = H(D\theta)$. But any section must be the identity on all but the long hom-space, and therefore it must fix the boundary of the disk $D^m$. That is, the *section* actually induces a *retraction* from the disk $D^m$ to its boundary, which Brouwer showed is impossible.
---
Returning to the proof of the main claim, we see that for $N \geq 3$, we have that $\underline{N}^n$ must be a retract of an object of $\mathcal A$, and hence by idempotent-completeness, must lie in $\mathcal A$. As every $\theta \in \Theta\_n$ is a retract of some $\underline{N}^n$, it follows that $\theta \in \mathcal A$ as claimed.
| 2 | https://mathoverflow.net/users/2362 | 430196 | 174,269 |
https://mathoverflow.net/questions/430168 | 1 | (I asked this question [on MSE](https://math.stackexchange.com/questions/4521591) 10 days ago, but I got no answer.)
Let $X$ and $Y$ be two independent identically distributed binomial random variables with parameters $n \in \mathbb{N}$ and $p \in (0,1)$. Let $Z := XY$ be their product.
Is it true or false that $\tilde{Z} := (Z - \mathbf{E}[Z]) / \sqrt{\mathbf{Var}[Z]}$ converges in distribution to a standard normal random variable (as $n \to \infty$) ?
At a first glance, I would be tempted to write $X = \sum\_{i=1}^n A\_i$ and $Y = \sum\_{i=1}^n B\_i$, where $A\_i$ and $B\_i$ are Bernoulli random variables, and then to apply the central limit theorem to $Z = \sum\_{i=1}^n \sum\_{j = 1} A\_i B\_j$... but $A\_i B\_j$ are *not* independent...
Thanks for any help
**P.S.1** For $p=1/2$, it is easy to check that $\mathbf{E}[Z] = n^2 / 4$ and $\mathbf{Var}[Z] = n^3 / 8 + n^2 / 16$. Moreover, expanding $(XY - n^2/4)^k$ with the binomial theorem and using the formula for the moments of the binomial distribution, I got that
$$\mathbf{E}\left[\tilde{Z}^k\right] = \frac1{(n^3 / 8 + n^2 / 16)^{k/2}} \sum\_{j=0}^k \binom{k}{j} \left(\sum\_{i=0}^j {j \brace i} (n)\_{i} (1/2)^i\right)^2 (-n^2/4)^{k-j}$$
where ${j \brace i}$ are Stirling numbers of second kind and $(n)\_{i}$ is a falling factorial. With this formula, I verified that the first 20 moments of $\tilde{Z}$ tend to the moments of a standard normal variable. However, I still do not know how to prove this for all moments.
**P.S.2** I think that one cannot prove the claim only using the fact that the (normalized) $X$ and $Y$ converge in distribution to normal variables. In fact, [it is known](https://mathworld.wolfram.com/NormalProductDistribution.html) that the product of two independent normal variables is not a normal variable.
| https://mathoverflow.net/users/491015 | Does the (normalized) product of two independent binomial variables converge in distribution to a normal variable? | $\newcommand{\R}{\mathbb R}\newcommand\ep\epsilon\newcommand\tsi{\tilde\sigma}$Yes, of course. This follows by the multivariate (here, bivariate) so-called delta method.
Indeed, we may assume that
\begin{equation\*}
X=\sum\_{i=1}^n X\_i,\quad Y=\sum\_{i=1}^n Y\_i,
\end{equation\*}
where $X\_1,Y\_1,\dots,X\_n,Y\_n$ are independent identically distributed (iid) Bernoulli random variables (r.v.'s) with parameter $p\in(0,1)$.
For each $i\in[n]:=\{1,\dots,n\}$, let
\begin{equation\*}
V\_i:=(X\_i-p,Y\_i-p),
\end{equation\*}
so that $V\_1,\dots,V\_n$ are iid zero-mean random vectors in $\R^2$. Then
\begin{equation\*}
XY=n^2 f(\bar V)+n^2p^2, \tag{1}\label{1}
\end{equation\*}
where $\bar V:=\frac1n\,\sum\_{i=1}^n V\_i$ and for $(x,y)\in\R^2$
\begin{equation\*}
f((x,y)):=f(x,y):=(x+p)(y+p)-p^2,
\end{equation\*}
so that $f(0,0)=0$ and for $L:=f'(0,0)$ we have
\begin{equation\*}
L(x,y)=px+py
\end{equation\*}
and
\begin{equation\*}
|f(x,y)-L(x,y)|=|xy|\le\tfrac12\,\|(x,y)\|^2,
\end{equation\*}
where $\|(x,y)\|:=\sqrt{x^2+y^2}$.
So, condition (2.1) of [this paper](https://projecteuclid.org/journals/electronic-journal-of-statistics/volume-10/issue-1/Optimal-order-bounds-on-the-rate-of-convergence-to-normality/10.1214/16-EJS1133.full) holds (for any real $\ep>0$ and $M\_\ep=1$).
Moreover, $v\_3:=E\|V\_1\|^3<\infty$ and
\begin{equation\*}
\tsi:=\sqrt{EL(V\_1)^2}=\sqrt{p^2 E(X\_1-p+Y\_1-p)^2}=\sqrt{2p^3q}>0,
\end{equation\*}
where $q:=1-p$.
So, by \eqref{1} and Theorem 2.9 of the [same paper](https://projecteuclid.org/journals/electronic-journal-of-statistics/volume-10/issue-1/Optimal-order-bounds-on-the-rate-of-convergence-to-normality/10.1214/16-EJS1133.full),
\begin{equation\*}
\frac{XY-n^2p^2}{n^2\sqrt{2p^3q/n}}
=\frac{f(\bar V)}{\tsi/\sqrt n}\to Z\sim N(0,1) \tag{2}\label{2}
\end{equation\*}
(in distribution as $n\to\infty$).
Note also that $EXY=n^2p^2$ and
$$Var\,XY=2n^3p^3q+n^2p^2q^2=2n^3p^3q(1+O(1/n))\\
\sim2n^3p^3q=(n^2\sqrt{2p^3q/n})^2.$$
Thus, by \eqref{2},
\begin{equation\*}
\frac{XY-EXY}{\sqrt{Var\,XY}}\to Z,
\end{equation\*}
as desired.
Moreover, it follows from cited Theorem 2.9 that
\begin{equation\*}
\sup\_{z\in\R}\Big|P\Big(\frac{XY-EXY}{\sqrt{Var\,XY}}\le z\Big)-P(Z\le z)\Big|\le
\frac{C\_p}{\sqrt n}
\end{equation\*}
for some real $C\_p>0$ depending only on $p\in(0,1)$ and all natural $n$.
| 1 | https://mathoverflow.net/users/36721 | 430198 | 174,271 |
https://mathoverflow.net/questions/430193 | 4 | Question:
---------
Did Kolmogorov develop a set of axioms for probability theory motivated by Algorithmic Information Theory in the 1960s?
Context:
--------
In 1965, Andrey Kolmogorov considered three approaches to information theory(combinatorial, probabilistic, and algorithmic) [1] where in the algorithmic approach he introduces the Invariance theorem for the minimal description of datasets $X$ relative to Universal description languages $U$ and $U'$:
\begin{equation}
\lvert K\_U(X)-K\_{U'}(X) \rvert \leq \text{Cst} \tag{1}
\end{equation}
which allows us to formulate the Universal Distribution, an effective formalisation of Occam's razor [2].
In essence, he introduced the foundations for Algorithmic Information Theory and in the conclusion he clarifies that his motivation for doing so is to provide a clear definition of randomness as well as a robust foundation for probability theory where all probabilities have a deterministic and frequentist nature.
In fact, his PhD student Leonid Levin would rigorously define the algorithmic probability of observing a dataset $X$:
\begin{equation}
-\log\_2 P(X) = K\_U(X) - \mathcal{O}(1) \tag{2}
\end{equation}
which would later become known as Levin's Coding theorem [3].
References:
-----------
1. A. N. Kolmogorov Three approaches to the quantitative definition of information. Problems of Information and Transmission, 1(1):1--7, 1965
2. Peter Grünwald and Paul Vitanyí. Shannon Information and Kolmogorov Complexity. Arxiv. 2004.
3. L.A. Levin. Laws of information conservation (non-growth) and aspects of the foundation of probability theory. Problems Inform. Transmission, 10:206–210, 1974.
| https://mathoverflow.net/users/56328 | Kolmogorov's approach to probability theory | In 1970, Kolmogorov developed the 'Combinatorial foundations of information theory and the calculus of probabilities' in relation to a presentation at the International Congress of Mathematicians in Nice(1970). This text was eventually published in 1983:
* A.N. Kolmogorov. Combinatorial foundations of information theory and the calculus of probabilities. Russian Math. Surveys (1983).
While Kolmogorov admits that his treatment is incomplete, he presents strong arguments for the following conclusions:
1. Information theory must precede probability theory and not be based on it. By the very essence of this discipline, the foundations of information theory has a finite combinatorial character.
2. The applications of probability theory can be put on a uniform basis. It is always a matter of consequences of hypotheses about the impossibility of reducing in one way or another the complexity of the description of the objects in question.
In the last statement, Kolmogorov is implicitly referring to the Minimum Description Length or Kolmogorov Complexity of an object.
**Note:**
Since 1983, the important influence Kolmogorov's theory of Algorithmic Information has had on modern science may be partly determined by the scientific research direction of Deep Mind led by Shane Legg, whose Universal Intelligence Measure of an agent with policy $\pi$ is determined by the formula:
\begin{equation}
\Upsilon(\pi) := \sum\_{\mu \in E} 2^{-K(\mu)} V\_{\mu}^{\pi}
\end{equation}
where $E$ is the space of environments, $2^{-K(\mu)}$ is the algorithmic probability that an agent finds itself in this environment and $V\_{\mu}^{\pi}$ is the expected future discounted reward of an agent $\pi$ interacting with environment $\mu$. For the complete theory behind this formula, I may refer the reader to the PhD thesis of Shane Legg:
* Legg, Shane. Machine Super Intelligence (Thesis). Lugano, Switzerland: Università della Svizzera italiana. 2008.
This formula actually appears at 3:22 in the 2019 TED talk of Thore Graepel, a research scientist at Deep Mind, titled 'The human pursuit of artificial intelligence'.
| 4 | https://mathoverflow.net/users/56328 | 430209 | 174,276 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.