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https://mathoverflow.net/questions/440311 | 10 | I know that the topic is classical and even "folklore", but many treatments make use of local coordinates and such treatments are rather messy. Could somewhere maybe provide some reference(s) to a modern treatment, using the language of modern differential geometry (with complex line bundles, connections and so on), of magnetic monopoles including a complete description, which is as explicit as possible without being too messy, of the basic Dirac monopole? If the holomorphic structure of $S^2$, which turns it into $\mathbb{C}P^1$, plays a role, I would also like that to be mentioned and made clear, if possible. Of course a combination of references would be ok too. If someone would like to describe it in an answer, that would be great of course!
I suspect I may find that somewhere in the various Cambridge lecture notes online (maybe in one of D. Tong's lecture notes), or in some articles/books by N. Manton for example. I remember it was very briefly mentioned in the Atiyah and Hitchin book on magnetic monopoles, but I kind of would like more details please. I will dig in the literature, but I suspect such references would be generally useful to others too, which is why I thought about writing this post. I will edit and report on what I find too.
Edit 1: first, there is a lot of information in Tong's lecture notes (for example, his gauge theory notes) on Dirac monopoles. What follows is a very short description which I found in the article by Gibbons and Manton called "The Moduli Space Metric for Well-separated
BPS Monopoles".
A Dirac vector potential, which really means a connection on a complex line bundle, in this case defined on $\mathbb{R}^3 \setminus \{ \mathbf{0} \}$, is one which satisfies:
$$ \nabla \times \mathbf{w} = - \frac{\mathbf{r}}{r^3} $$
and
$$ \mathbf{w}(-\mathbf{r}) = \mathbf{w}(\mathbf{r}). $$
I understand the first equation, which is really the Bogomolny equation. Indeed, we can think of it as $F = \* d \phi$, where $F$ is the curvature of the connection $\mathbf{w}$, $\*$ denotes the Hodge star with respect to the (flat) Euclidean metric on $\mathbb{R}^3 \setminus \{ \mathbf{0} \}$ and $\phi$ is a scalar potential (physically a Higgs field). In this case, $\phi = 1 / r$.
Could someone perhaps comment on the second equation please? How can you compare the values of a connection at $2$ different points? Is that using the fact that we are working inside a domain of $\mathbb{R}^3$, whose tangent bundle is naturally trivial? I guess this is what they mean please?
Edit 2: I should add that I have just stumbled on a book by Yakov Shnir entitled "Magnetic Monopoles" which seems to contain a lot of relevant material (including on multimonopoles). It definitely seems "modern", judging from the preview provided by Google.
I also found a nice detailed treatment of Dirac monopoles in the book by N. Manton and P. Sutcliffe entitled "Topological Solitons", in section 8.1. I think I will definitely benefit from reading that section. I do find the work of P. Dirac on one hand beautiful mathematically, yet difficult to interpret physically, and often his formulas have many possible interpretations. I guess I am mostly thinking about the Dirac equation, but his monopoles also have their own subtleties too!
Update: I found section 8.1 in the book by N. Manton and P. Sutcliffe called "Topological Solitons" to be exactly what I wanted. Not only is the language "modern", but they also answer many questions I had too. For example, why can't we just modify Maxwell's equations so that we have that the divergence of $B$ is (up to some factor) the magnetic charge density? Well, such a naive way of introducing magnetic monopoles has several issues which they discuss.
Dirac's work was much more subtle. He essentially notices that he could introduce a magnetic monopole mathematically using a connection on a $U(1)$ bundle over $\mathbb{R}^3 \setminus \{ \mathbf{0} \}$. One may use 2 patches to cover that region, say, using spherical coordinates, with $\theta \neq 0$ being one region and $\theta \neq \pi$ being another. In these coordinates, it looks as though the magnetic potential has a singular ray in each of the two patches (the so called Dirac strings), but these are just fake coordinate singularities. The connection only has a singularity at the origin of $\mathbb{R}^3$. The integrality of the first Chern class of the $U(1)$ bundle leads to a quantization condition on the electric/magnetic charges. Dirac (or "Monopoleon", as W. Pauli used to call him, I think) was brilliant at finding subtle loopholes in the equations which we use to describe nature.
Whether they exist or not in our universe remains to be seen (as of today's date). But their existence does not lead to contradictions with Maxwell's equations.
Finally, I wonder, purely speculatively (and I am not a physicist, so it is most probably false), whether there are black hole solutions of Einstein's equations, which are also magnetic monopoles, such that the singularity of the monopole coincides with the singularity of the black hole. Do those exist (at least mathematically)? If not, is there any fundamental reason why they cannot exist, please (mathematically and/or physically)?
| https://mathoverflow.net/users/81645 | Modern treatment of Dirac monopoles and related topics | It is from 1993 but I find it definitely worth mentioning for this question.
*Brylinski, Jean-Luc*, [**Loop spaces, characteristic classes and geometric quantization**](https://dx.doi.org/10.1007/978-0-8176-4731-5), Modern Birkhäuser Classics. Basel: Birkhäuser. xvi, 300 p. (2008).
Chapter VII (pages 257-277) is called "The Dirac Monopole" and includes treatment of Dirac monopoles in terms of a 3-cohomology class assigned to an explicitly constructed stack on the 3-sphere.
The idea is to interpret Dirac's vision of a magnetic field having singularity at the origin by viewing wave functions of particles moving in such magnetic field as sections of some (complex) line bundle $L$ on $\mathbb R^3\setminus\{0\}$.
Brylinski attributes the idea to Deligne; another reference, for general constructions of this kind, is
*Dixmier, Jacques; Douady, Adrien*, [*Champs continus d’espaces hilbertiens et de $C^\*$-algèbres*](https://doi.org/10.24033/bsmf.1596), Bull. Soc. Math. Fr. 91, 227-284 (1963).
We identify the above $\mathbb R^3\setminus\{0\}$ with $X:=S^3\setminus\{0,\infty\}$, where $0$ and $\infty$ are some two opposite points on $S^3$ ("north pole" and "south pole"), so that $X$ is homotopy equivalent to the "equator" $S^2\subset S^3$. Brylinski picks a line bundle $L$ on $X$ which realizes $2\pi i$ times the canonical generator of $H^2(X;\mathbb Z)$. For an open set $U\subseteq S^3$ he then defines the groupoid $\mathscr G(U)$ to have objects of the form $(L\_0,L\_\infty,\phi)$ where $L\_0$ is a line bundle on $U\_0:=U\setminus\{0\}$, $L\_\infty$ is a line bundle on $U\_\infty:=U\setminus\{\infty\}$, and $\phi$ is an isomorphism of line bundles on $X\cap U=U\_0\cap U\_\infty$,
$$
\phi:\left((L\_0|\_{U\_0\cap U\_\infty})^\*\otimes(L\_\infty|\_{U\_0\cap U\_\infty})\right)\cong L|\_{X\cap U}.
$$
He then assigns to this stack certain 3-cohomology class, shows that it is nonzero, so proportional to the fundamental class of $S^3$, and interprets it in several ways: as the obstruction to the existence of a global object, i. e. to nonemptiness of $\mathscr G(S^3)$; as the obstruction to $\operatorname{SU}(2)$-equivariance of $\mathscr G$; and as something he calls 3-curvature of $\mathscr G$ (central notion in the book).
| 2 | https://mathoverflow.net/users/41291 | 442758 | 178,640 |
https://mathoverflow.net/questions/442720 | 2 | Landau in the first equation of [Über die Gitterpunkte in einem Kreise](https://doi.org/10.1007/BF01203524) uses the following formula for the Bessel function of the first kind:
$$\frac{1}{2\pi i } \int\_{1-\infty}^{1+i\infty} \frac{\mathrm e^{As-B/s}}{s^4} \, \mathrm d s= (A/B)^{3/2} J\_3(2 \sqrt{AB}))
$$ valid for all $A,B>0$.
Does anyone know of a reference or a proof of that? Since Landau gives no reference or hint I imagine it is either well-known or completely trivial.
| https://mathoverflow.net/users/9232 | infinite integral defining Bessel functions | First, change the variable in the integral $s=kz$, where $k=\sqrt{B/A}$.
You obtain
$$\left(A/B\right)^{3/2}
\int \frac{\exp{\sqrt{AB}\left(z-1/z\right)}}{z^4}dz.$$
Then deform the contour of integration from the vertical line
to $|z|=1$, this is possible since the real part of
$(z-1/z)$ is bounded in a simply connected region which contains both contours.
And finally refer to the well known generating function
for Bessel functions
$$\sum\_{-\infty}^\infty J\_n(x)z^n=\exp\left(\frac{x}{2}\left(z-1/z\right)\right).$$
Combined with Laurent's theorem this implies
that our integral is $J\_3(2\sqrt{AB})$.
| 2 | https://mathoverflow.net/users/25510 | 442769 | 178,644 |
https://mathoverflow.net/questions/442469 | 7 | Given a covering space $p \colon X \to Y$, we get an injection $p^\* \colon \pi\_1(X) \to \pi\_1(Y)$, and we know that the image $p^\*(\pi\_1(X))$ is normal in $\pi\_1(Y)$ if an only if $p$ is *regular*, that is if the deck group of the covering $p$ acts transitively on preimages of points in $Y$. This is a satisfyingly topological criterion for a group-theoretic property.
Is a similarly topological criterion possible for the case where $p^\*(\pi\_1(X))$ is not just normal in $\pi\_1(Y)$ but characteristic?
Edit: If it helps, $X$ can be assumed to be aspherical. The initial context in which this came up was surfaces.
| https://mathoverflow.net/users/14257 | When do covering spaces correspond to characteristic subgroups? | Suppose $X$ is a $K(\pi,1)$. Then by definition every endomorphism, hence every automorphism $f\_\star\colon \pi\_1(X,x) \to \pi\_1(X,x)$ may be realized by a continuous map $f\colon (X,x) \to (X,x)$.
If $p\colon (\tilde X,\tilde x) \to (X,x)$ is a covering map such that $p\_\star(\pi\_1(\tilde X, \tilde x))$ is characteristic in $\pi\_1(X,x)$, then every homotopy equivalence $f\colon (X,x) \to (X,x)$ preserves the image $p\_\star(\pi\_1(\tilde X, \tilde x))$ so by the lifting criterion (see Hatcher chapter 1, for example), there exists a lift $\tilde f\colon (\tilde X,\tilde x) \to (\tilde X,\tilde x)$ such that $p\tilde f = fp$. In fact, if every homotopy equivalence lifts to $\tilde X$, then the converse of my use of the lifting criterion implies that $p\_\star(\pi\_1(\tilde X,\tilde x))$ is characteristic in $\pi\_1(X,x)$.
| 6 | https://mathoverflow.net/users/135175 | 442771 | 178,645 |
https://mathoverflow.net/questions/442762 | 3 | Given an $m\times n$ chess board, we place $p$ $2\times 1$ dominoes on the board so that they don't overlap. How many ways can we place them?
When each square of the board is covered by a domino this is the well-known tiling problem. Is there any research on the case where $p$ is small such that we have to leave some empty squares on the board?
| https://mathoverflow.net/users/115114 | An "incomplete" tiling? | If we sum over all $p$, then we obtain the number of monomer-dimer tilings of an $m\times n$ chessboard. This number is known to be #P-complete and hence very likely computationally intractable. For a reference, see citation [7] of the paper by Yong Kung [here](https://arxiv.org/abs/cond-mat/0610690).
| 9 | https://mathoverflow.net/users/2807 | 442778 | 178,647 |
https://mathoverflow.net/questions/442788 | 1 | Suppose that $X,Y$ are independent random $d$-dimensional vectors each uniformly distributed on the unit sphere, and let $Z=Y\cdot\sqrt{1-\alpha}$ be a uniformly selected vector on a slightly smaller sphere.
I'm interesting in calculating (or lower-bounding) the probability that $X$ and $Z$ are far from each other, namely, what is:
$$
\Pr[\lVert X-Z \rVert^2\le \alpha]
$$
as a function of $d,\alpha$?
This quantity appears to play a key role in a vector quantization paper I'm trying to understand (<https://arxiv.org/pdf/2010.03246.pdf>, Section 4.3), which they unfortunately didn't analyze.
| https://mathoverflow.net/users/501044 | Let $\alpha\in(0,1),d\in\mathbb N^+$ and $X,Y\in\mathbb S^d$ be uniform, what is $\Pr[\lVert X-Y\cdot\sqrt{1-\alpha} \rVert^2\le \alpha]$? | $\newcommand\al\alpha$By spherical symmetry, the conditional distribution of $\|X-Y\,\sqrt{1-\al}\|$ given $Y$ does not depend on $Y$. So, letting $e\_1:=(1,0,\dots,0)$ and writing $X=(X\_1,\dots,X\_d)$, we have
$$\begin{aligned}
&P(\|X-Y\,\sqrt{1-\al}\|^2\le\al) \\ &=P(\|X-e\_1\,\sqrt{1-\al}\|^2\le\al) \\
&=P((X\_1-\sqrt{1-\al})^2+X\_2^2+\cdots+X\_d^2\le\al) \\
&=P((X\_1-\sqrt{1-\al})^2+1-X\_1^2\le\al) \\
&=P(X\_1\ge\sqrt{1-\al}) \\
&=\tfrac12\,P(X\_1^2\ge1-\al).
\end{aligned}$$
Noting that $X\_1^2$ has the beta distribution with parameters $1/2,(d-1)/2$, we conclude that
$$\begin{aligned}
P(\|X-Y\,\sqrt{1-\al}\|^2\le\al)
&=\tfrac12\,(1-F\_{1/2,\,(d-1)/2}(1-\al)) \\
&=\tfrac12\,F\_{(d-1)/2,\,1/2}(\al),
\end{aligned}$$
where $F\_{a,b}$ is the cdf of the beta distribution with parameters $a,b$.
| 1 | https://mathoverflow.net/users/36721 | 442790 | 178,651 |
https://mathoverflow.net/questions/442776 | 1 | I have two questions.
Let $(X\_t)\_{t\geq 0}$ be a Lévy process with Lévy measure $\nu$. The jump process $\Delta X=\left(\Delta X\_t\right)\_{t\geq 0}$ is defined by
$\Delta X\_t=X\_t-X\_{t-}$, for every $t\geq0$, with $X\_{t-}$ left limit in $t$.
For every $0\leq t <\infty$ e $A \in \mathcal{B}(\mathcal{R}-\{0\})$, let
$N(t,A)(w)= \#\left\{ 0 \leq s \leq t: \Delta X\_t(w) \in A \right\}$
if $w \in \Omega\_0$
and
$N(t,A)(w)=0$ if $w \in \Omega\_0^c$, with $\Omega\_0$ a measurable set with probability $1$ such that $t\longrightarrow X\_t(w)$ is cadlag for every $w \in \Omega\_0$.
Let $\nu(\cdot)=\mathbb{E}\left[N(1,\cdot)\right]$ be the intensity measure. We say that $A$ is bounded below if $0 \notin \overline{A}$.
Theorem
1.If $A$ is bounded below, then $\left(N(t,A)\right)\_{t\geq 0}$ is a Poisson process of intensity $\nu(A)$.
2.If $A\_1, \dots , A\_m \in \mathcal{B}(\mathcal{R}-\{0\})$ are disjoint e bounded below then the random variable $N(t,A\_1) \dots N(t,A\_m)$ are indpendent.
So my questions are:
i) How can I prove the second statement?
ii) How can I deduce from this theorem that $\nu(A)<\infty$ for all $A$ bounded below?
| https://mathoverflow.net/users/501039 | The Lévy process jumps | Your questions are covered by, say, Theorem 3 in Section 4 of [Lalley's notes](http://galton.uchicago.edu/%7Elalley/Courses/385/LevyProcesses10-23-2017.pdf). The proof (given there) is somewhat involved.
| 0 | https://mathoverflow.net/users/36721 | 442800 | 178,654 |
https://mathoverflow.net/questions/442791 | 2 | Let $A$ be a closed (densely defined) operator on a Hilbert space $H$.
We define for a natural number $k$, the operator $A^k$ with its natural domain.
Is $A^k$ closed?
| https://mathoverflow.net/users/216007 | If $A$ is a closed operator, is $A^k$ closed? | Here's a counterexample (subject perhaps to what you consider "natural").
Take a separable Hilbert space with orthonormal basis $\{u\_n : n = 1, 2, \ldots\}$ and the operator $A$ defined by
$$ A u\_n = \cases{ n u\_{n+1} & if $n$ is odd\cr 0 & if $n$ is even}$$
with domain $D = \{x = \sum\_n c\_n x\_n : \sum\_n |c\_n|^2 < \infty, \sum\_{n \text{ odd}} n^2 |c\_n|^2 < \infty\}$. Then $A$ is closed and densely defined. For integers $k > 1$ I would expect the "natural" domain of $A^k$ to be contained in $D$. However, $A^k x = 0$ where it is defined, so that $A^k$ is not closed.
| 7 | https://mathoverflow.net/users/13650 | 442808 | 178,656 |
https://mathoverflow.net/questions/442816 | 19 | I’m currently writing my first paper, a little paper, and I’d like to have some nice graphics in it. Much of the proof work is tedious analysis and I’d like to give any potential reader visual references to help.
To give an idea of what I’m looking for in particular, I’d like to have some pictures of the plane with square lattice points, perhaps some of the points labeled with their coordinates. I’d also like to be able to create images of complex integrals (Argand diagram with labeled coordinate axes, little arrow along the path to orient the path).
Anyone have experience generating these graphics who can direct me to online resources/libraries/software (preferably free)?
EDIT: Thank you very much everyone for all the effort into your responses. There are multiple great answers here. Exploring as many as I can will be my goal!
| https://mathoverflow.net/users/138669 | Where can I create nice looking graphics for a paper? | Ipe is a fantastic tool. There's also a bit of a learning curve, but it's not too bad. I've seen people who are proficient at it generate amazing figures in a matter of minutes.
<https://ipe.otfried.org/>
Edit: I wish I had known about it a few years ago, it would've saved me many hours of TikZ coding.
| 19 | https://mathoverflow.net/users/103164 | 442822 | 178,661 |
https://mathoverflow.net/questions/442818 | 7 | $\DeclareMathOperator\SL{SL}$By the celebrated results of Culler and Shalen, a closed $3$-manifold contains an incompressible surface if its $\SL\_2(\mathbb{C})$ character variety is infinite.
Now, for manifolds with positive $b\_1,$ there is a much easier argument for existence of incompressible surfaces. In that regard, Culler-Shalen theory looks more striking when applied to rational homology spheres. Also maybe it is more interesting to look at hyperbolic homology spheres, since it is really surfaces of higher genera that we would like to detect.
Now my question is that, looking at the litterature, I was not able to find an example of a rational hyperbolic homology sphere with infinite $\SL\_2(\mathbb{C})$ character variety. I am sure they exist though, so I would like to see if anyone has a nice example.
I would guess that the surgery on a knot along a boundary slope corresponding to a genus $g\geq 2$ surface would be a good candidate, but I have no idea whether those would typically have infinite $\SL\_2(\mathbb{C})$ character variety.
| https://mathoverflow.net/users/62201 | Hyperbolic homology spheres with infinite $\mathrm{SL}_2(\mathbb{C})$ character variety | Take two knot complements (say of $K,K'$) and glue them together, interchanging meridians and longitudes. This is called splicing and produces a homology sphere $S(K,K')$; if both knots are non-trivial then the boundary torus where you glued is incompressible. It's standard that this produces a higher dimensional component of the character variety. See eg Boden-Curtis, *Splicing and the SL2(C) Casson invariant*,
Proc. Amer. Math. Soc. **136** (2008), no. 7, 2615–2623. You don't need any Culler-Shalen theory for this (see Theorem 3.1).
Because of the incompressible torus, $S(K,K')$ is not hyperbolic. But there is always a hyperbolic homology sphere $Y$ with a degree one map $Y \to S(K,K')$. Then the character variety of $Y$ will have $\dim > 0$ as well. There are variations/elaborations on this theme, eg you could assume that there is an invertible homology cobordism from $S(K,K')$ to $Y$ which induces the degree one map.
| 8 | https://mathoverflow.net/users/3460 | 442823 | 178,662 |
https://mathoverflow.net/questions/442055 | 16 | $\DeclareMathOperator\Ind{Ind}$Let $C$ be a (small) category, and $I$ a finite category, is it true that the natural functor $\Ind(C^I) \to \Ind(C)^I$ is an equivalence of categories? where $\Ind$ denotes the category of ind-objects (so the free completion under filtered colimits) and the exponentials are for categories of functors.
This is proved by Lurie in Higher topos theory (proposition 5.3.5.15) when $C$ is an infinity category and $I$ is a finite poset. He gives an example to show that this cannot be generalized to the case of $I$ a finite simplicial set — but finite category is much more restrictive and his example doesn't rule this out at all.
I'm mostly interested in the case where $C$ is a 1-category and already has finite colimits, but I'm curious of any interesting things that can be said more generally.
| https://mathoverflow.net/users/22131 | $\operatorname{Ind}(C^I) = \operatorname{Ind}(C)^I$? | So, the question appeared more subtle than I initially thoughts so I have written a short paper with more references and the details of what I'm going to say below. It is available [here](https://arxiv.org/abs/2307.06664).
Here is the summary:
First, Makkai's theorem cited by Ivan is indeed false. Building on Ben Wieland's comment (thank you very much for this!), one get the following counter-example.
Let $I=B(\mathbb{Z}/2\mathbb{Z})$ the category with one object $\*$ and one non-trivial automorphism satisfying $f^2=1$ and let $C$ be the subcategory of $I \times \omega$ containing all the objects, and all the arrows $(\*,n) \to (\*,m)$ when $n < m$ but only the identity when $n = m$.
In particular, functors $I \to C$ are all constant, as $C$ has no non-trivial endomorphisms, but there is a non-trivial functor $I \to Ind(C)$: indeed the colimits of the chain of the $(\*,n)$ is in the Ind completion, and as a presheaf, it is the pullback along the projection $C \to I$ of the unique representable presheaf, so it comes with a $\mathbb{Z}\_2$-action and hence is a non-trivial functor $ I \to Ind(C)$, which doesn't belong to $Ind(C^I)$.
As $C$ has no non-trivial endomorphisms it is Cauchy complete, hence $C$ identifies with the full subcategory of $\omega$-presentable objects of $Ind(C)$ and hence the exponential $Ind(C)^I$ is a counter-example to Makkai's theorem.
Now this construction can be generalized to any category and any ordinal, and this leads to the following theorem:
**Theorem:** Let $I$ be a category and $\kappa$ be a regular cardinal. The following condition are equivalent.
1. $I$ is $\kappa$-small, has no non-identity endomorphisms, and its posetal relfection is well-founded.
2. for all category $C$, the functor $Ind\_\kappa(C^I) \to Ind\_\kappa(C)^I$ is an equivalence.
3. for all $\kappa$-accessible category $A$, the category $A^I$ is $\kappa$-accessible with its $\kappa$-presentable objects being the functors from $I$ to $\kappa$-presentable objects of $A$.
In particular, one recovers Lurie's version of the theorem (I mean for 1-categories), as well as Meyer's result mentioned by Peter Haine. Also Makkai theorem is false even for uncountable $\kappa$. But this isn't the end of the story. If we are interested in category with colimits, then everything works more nicely and we get a positive answer to my questions:
**Theorem:** Let $I$ be a category and $\kappa$ be a regular cardinal. The following conditions are equivalent.
1. $I$ is $\kappa$-small.
2. for all category $C$ with $\kappa$-small colimits, the functor $Ind\_\kappa(C^I) \to Ind\_\kappa(C)^I$ is an equivalence.
3. for all locally $\kappa$-presentable category $A$, the category $A^I$ is locally $\kappa$-presentable, with its $\kappa$-presentable objects being the functors from $I$ to $\kappa$-presentable objects of $A$.
The proofs of these and additional details are in the paper linked above.
It seems Makkai's theorem was used in a few places in the literature - but from what I can tell, all uses I've seen are covered by these two theorems.
| 17 | https://mathoverflow.net/users/22131 | 442827 | 178,663 |
https://mathoverflow.net/questions/405276 | 2 | When the tensor algebra is presented, it is usually constructed as the direct sum of all tensor powers of the space. By this construction, the graded structure of the tensor algebra is easy to prove. However, I have not been able to find any proofs of the graded structure of the tensor algebra using just the universal property. Here's my question: **Has anybody come up with a proof that the tensor algebra is graded using just the universal property?** I will admit that I have a hard time using universal properties in general to prove things about objects, so I've had a hard time trying to come up with an argument myself.
The reason I'm interested in this is that I've started studying category theory recently, and I've heard about the adjoint functor theorem, which seems to provide a way to define the tensor algebra without using the traditional construction in the first place (I may be wrong about this since I haven't learned enough category theory yet). I was wondering how easy it would be to prove that the tensor algebra is graded from this kind of definition.
Another reason I'm interested in this is that from studying geometric algebra, I've seen that there is an alternative grading structure for Clifford algebras that is similar to the grading of the tensor algebra (that only respects anti-commuting products, not arbitrary products like is usually implied by the use of the word "grade"), and I was wondering if an argument for the grading of the tensor algebra could be adapted to proving this as well. I'm not expecting an answer to this problem though; the question is about tensor algebras.
| https://mathoverflow.net/users/394901 | Proving the graded structure of the tensor algebra from only the universal property | After all of this time I found a simple and direct solution. I can't believe I didn't think of this before:
Let $V$ be an $R$-Module and let $T(V)$ be its tensor algebra. Define $T^n(V)$ to be the span of all products of $n$ vectors in $T(V)$, and consider the (for now possibly distinct) space $\bigoplus\_{n = 0}^\infty T^n(V)$. Multiplying two homogeneous tensors in this direct sum produces another homogeneous tensor, and we can extend this by linearity to show that $\bigoplus\_{n = 0}^\infty T^n(V)$ is a graded algebra.
Now let $A$ be an arbitrary algebra with a module homomorphism $f : V \to A$. By the universal property of the tensor algebra, this extends to an algebra homomorphism $g : T(V) \to A$. We can define the restriction of $g$ onto each $T^n(V)$ and then extend by linearity to produce a map $h : \bigoplus\_{n = 0}^\infty T^n(V) \to A$. This map is an algebra homomorphism that is an extension of $f$ (and it's trivial to show that this extension is unique), showing that $\bigoplus\_{n = 0}^\infty T^n(V)$ satisfies the universal property that uniquely defines the tensor algebra. Thus, $T(V) \cong \bigoplus\_{n = 0}^\infty T^n(V)$, showing that $T(V)$ itself is graded.
| 0 | https://mathoverflow.net/users/394901 | 442828 | 178,664 |
https://mathoverflow.net/questions/442856 | 10 | $\DeclareMathOperator{\Hom}{Hom}$
Dear all,
The question is for teaching purposes and rather basic, so I hope that it also allows (relatively) easy answer.
By abstract homotopy theory we know that if the object $A$ in a model category $\mathcal{C}$ is cofibrant, then for any object $X$ the left homotopy relation on $\Hom(A,X)$ is equivalence relation. The only thing that can go wrong if $A$ is not cofibrant, then the relation might not be transitive.
So the question is: in the Quillen model structure on topological spaces, what is an example of two spaces $A,X$ with $A$ non-cofibrant and three maps $f,g,h$ such that the transitivity of left homotopy is violated?
| https://mathoverflow.net/users/123432 | Example of non-transitive homotopy relation | This is not possible in the Quillen model structure on topological spaces. (Here I am interpreting in the weak sense: "$f$ and $g$ are left homotopic if there exists *some* cylinder object $A \amalg A \to C \xrightarrow{\sim} A$ and a map $C \to Y$ restricting to $f$ and $g$". If you restrict to using one fixed cylinder object, like $A \times [0,1]$ or $A$, then the answer depends on exactly which.)
Suppose we have three maps $f,g,h: X \to Y$, $C$ and $D$ are cylinder objects for $A$, $H: C \to Y$ is a left homotopy from $f$ to $g$, and $K: D \to Y$ is a left homotopy from $g$ to $h$. Without loss of generality we may assume that $C$ is a *good* cylinder object (the map $A \amalg A \to C$ is a cofibration) by factoring $A \amalg A \rightarrowtail C' \xrightarrow{\sim} C$ . We get a map $C \amalg\_A D \to Y$ which restricts to $f$ and $h$, and so it is sufficient to prove that $C \amalg\_A D$ is a cylinder object: equivalently, that the map $C \amalg\_A D \to A$ is an equivalence.
Consider the following diagram of pushouts:
$$
\require{AMScd}
\begin{CD}
\emptyset \amalg A \amalg \emptyset @>>> \emptyset \amalg A \amalg A @>>> \emptyset \amalg D\\
@VVV @VVV @VVV\\
A \amalg A \amalg \emptyset @>>> A \amalg A \amalg A @>>> A \amalg D\\
@VVV @VVV @VVV\\
C \amalg \emptyset @>>> C \amalg A @>>> C \amalg\_A D\\
\end{CD}
$$
The bottom-left vertical map is a cofibration, hence so are all the bottom vertical maps. The middle composite is a weak equivalence (disjoint unions of spaces preserve weak equivalences, cofibrant or not), and hence the bottom composite is a weak equivalence (topological spaces are a left proper model category -- pushouts along cofibrations preserve weak equivalences). However, this means that in the composite $C \to C \amalg\_A D \to A$, two out of the three maps are equivalences, and hence so is the third: $C \amalg\_A D$ is a cylinder object for $A$.
| 11 | https://mathoverflow.net/users/360 | 442868 | 178,673 |
https://mathoverflow.net/questions/442861 | 5 | Let $E$ be a Banach space. A set $C \subseteq E$ is called *ideally convex* if for every bounded sequence $(x\_n)$ in $C$ and for every sequence $(\lambda\_n)$ in $[0,1]$ that sums up to $1$ the vector $\sum\_n \lambda\_n x\_n$ is also in $C$.
(So a bounded set is ideally convex if and only if it is *$\sigma$-convex* in [this](https://mathoverflow.net/q/56161/102946) sense. But in the following question, $C$ is automatically unbounded.)
**Question.** Let $C \subseteq E$ be ideally convex and dense in $E$. Does it follows that $C = E$?
**Remark.**
Every open convex set is ideally convex (by the Hahn-Banach separation theorem). In this special case it is not particularly difficult to prove that the question has a positive answer: an open, convex, and dense set $C \subseteq E$ is automatically equal to $E$.
| https://mathoverflow.net/users/102946 | Large ideally convex sets | Just take your favorite decreasing sequence $\lambda\_k$ of positive numbers with sum $1$ and inductively construct the vectors $x\_k\in C$ such that $\left \|\sum\_{k=1}^n\lambda\_k x\_k-x\right\|\le\lambda\_{n+1}$ (then automatically $\|x\_n\|\le \frac{\lambda\_{n}+\lambda\_{n+1}}{\lambda\_n}\le 2$ for $n\ge 2$) to get a series converging to $x$. I use that the density of $C$ implies the density of $\lambda C$ for any $\lambda>0$, of course.
| 7 | https://mathoverflow.net/users/1131 | 442870 | 178,674 |
https://mathoverflow.net/questions/442772 | 47 | Is the following metric characterization of the real line true (and known)?
>
> A nonempty complete metric space $(X,d)$ is isometric to the real line if and only if for every $c\in X$ and positive real number $r$ there exist two points $a,b\in X$ such that $d(a,b)=2r$ and $\{a,b\}=\{x\in X:d(x,c)=r\}$.
>
>
>
**Added in Edit:** Without the completeness of $(X,d)$ this hypothetical characterization of the real line is not true: using the idea from the answer of @PietroMajer, by transfinite induction of length $\mathfrak c$, one can construct a dense $\mathbb Q$-linear subspace $L$ of the Euclidean plane $\mathbb R^2$ such that $|\{x\in L:\|x\|=r\}|=2$ for every positive real number $r$. More details can be found in [this preprint](https://arxiv.org/abs/2305.07354).
| https://mathoverflow.net/users/61536 | A metric characterization of the real line | Yes: in the new version of the question, with the word "complete" added, this is indeed a characterization of the real line.
In order to generate as much confusion as possible, but also for convenience, let's give the name "Banakh space" to any metric space satisfying the condition in your question with the word "complete'' deleted.
**Theorem:** Every complete Banakh space is isometric to $\mathbb R$ (with the usual metric).
This follows fairly easily from:
**Lemma:** Every Banakh space (whether complete or not) contains an isometric copy of a dense subset of $\mathbb R$.
To prove the theorem from the lemma, suppose $(X,d)$ is a complete Banakh space. Assuming the lemma, there is an isometric copy of some dense $Q \subseteq \mathbb R$ in $X$. Because $(X,d)$ is complete, the closure of this copy of $Q$ in $X$ is an isometric copy of the Cauchy completion of $Q$, which is $\mathbb R$. Let the map $r \mapsto \bar r$ be an isometric embedding of $\mathbb R$ in $X$. We are done if we can show that $X = \{ \bar r :\, r \in \mathbb R \}$. Aiming for a contradiction, suppose there is some $x \in X \setminus \{ \bar r :\, r \in \mathbb R \}$, and let $a = d(x,\bar 0)$. But then there are (at least) three points in $X$ at distance $a$ from $\bar 0$: $x$, $\bar a$, and $- \bar a$, contradicting that $X$ is a Banakh space.
Now to prove the lemma. We'll build up an embedding of a dense subset of $\mathbb R$ into $X$, one piece at a time. The first step is to find an isometric copy of $A = \{0\} \cup \{2z+1 :\, z \in \mathbb Z\}$ (the odd integers plus $0$) inside $X$.
To start, let $\bar 0$ be any point of $X$. There are exactly two points at distance $1$ from $\bar 0$: let's call these $\bar 1$ and $-\bar 1$ (it doesn't matter which point gets which label). By the Banakh property, $d(-\bar 1,\bar 1) = 2$.
Next, observe that there are exactly two points in $X$ with distance $2$ from $\bar 1$. We already have a name for one of these points, $-\bar 1$; let's call the other one $\bar 3$. By the Banakh property, $d(-\bar 1,\bar 3) = 4$. So we have:
$$d(-\bar 1,0) = d(\bar 0,\bar 1) = 1,\ d(-\bar 1,\bar 1) = d(\bar 1,\bar 3) = 2,\ d(-\bar 1,\bar 3) = 4.$$
To compute $d(\bar 0,\bar 3)$, we use the triangle inequality twice:
$$d(\bar 0,\bar 3) \leq d(\bar 0,\bar 1)+d(\bar 1,\bar 3) = 3,$$
$$d(\bar 0,\bar 3) \geq d(-\bar 1,\bar 3)-d(-\bar 1,\bar 0) = 3.$$
Hence $d(\bar 0,\bar 3) = 3$, and we have an isometric embedding of $\{-1,0,1,3\}$ into $X$.
For the next step, we can add in $-3$ the same way we did $3$. That is, observe that there are exactly two points in $X$ with distance $2$ from $-\bar 1$. One of these points is $\bar 1$, and we call the other one $-\bar 3$. By the Banakh property, $d(-\bar 3,\bar 1) = 4$. Using the triangle inequality as above, we can get $d(-\bar 3,\bar 0) = 3$. Lastly, because $d(-\bar 3,\bar 1) = 4 \neq 2 = d(\bar 3,\bar 1)$, we have $-\bar 3 \neq \bar 3$, but both of these are at distance three from $\bar 0$. Thus by the Banakh property, $d(-\bar 3,\bar 3) = 6$.
Thus we obtain an isometric embedding of $\{-3,-1,0,1,3\}$ into $X$.
Next we add in $5$ and $-5$ in a similar fashion. (I'll go through the details rather than just leaving it at "similar" though.) There are two points at distance $2$ from $\bar 3$. One of them is $\bar 1$; let's call the other $\bar 5$. The Banakh property gives $d(\bar 1,\bar 5) = 4$. But we also know already that $d(-\bar 3,\bar 1) = 4$, so another use of the Banakh property gives $d(-\bar 3,\bar 5) = 8$. Once we know both $d(-\bar 3,\bar 5)$ and $d(\bar 3,\bar 5)$ (i.e., the distance from $\bar 5$ to the least and greatest of our previously constructed points) the triangle inequality fills in all the other distances from previously constructed points to $\bar 5$. For example,
$$d(\bar 0,\bar 5) \leq d(\bar 0,\bar 3)+d(\bar 3,\bar 5) = 5,$$
$$d(\bar 0,\bar 5) \geq d(-\bar 3,\bar 5)-d(-\bar 3,\bar 0) = 5.$$
Similar computations give $d(-\bar 1,\bar 5) = 6$, and so we have an isometric embedding of $\{-3,-1,0,1,3,5\}$ into $X$.
Next add in $-5$ the same way we did $5$. That is, observe that there are exactly two points in $X$ with distance $2$ from $-\bar 3$. One of these points is $-\bar 1$, and we call the other one $-\bar 5$. The distances from $-\bar 5$ to $-\bar 1, \bar 0, \bar 1, \bar 3$ are computed just as they were for $\bar 5$. Then we observe that $-\bar 5 \neq \bar 5$ (because their distances to $\bar 1$ are different) but they are both distance $5$ from $\bar 0$, and this implies $d(-\bar 5,\bar 5) = 10$.
Thus we obtain an isometric embedding of $\{-5,-3,-1,0,1,3,5\}$ into $X$.
Continuing in this way, we can, two points at a time, build up an isometric embedding of $A$ into $X$, denoted by the map $z \mapsto \bar z$.
Once this is done . . . do it again! By the same exact method, we can find an isometric embedding of $\frac{1}{3} A$ (all odd integer multiples of $\frac{1}{3}$, plus $0$) in $X$, beginning with the same base point $\bar 0$ as before. Let's denote this new embedding by $z \mapsto \bar{\bar z}$ (so $\bar{\bar 0} = \bar 0$). But then notice that $A \subseteq \frac{1}{3}A$ and, by the Banakh property, for each $a \in A \setminus \{0\}$ there can only be two points of $X$ at distance $|a|$ from $\bar 0$. Thus $\{-\bar a,\bar a\} = \{-\bar{\bar a},\bar{\bar a}\}$. In this way we see that $\{ \bar z :\, z \in A \}$ is naturally included in $\{ \bar{\bar z} :\, z \in \frac{1}{3}A \}$.
The same works with $\frac{1}{3^k}$ in place of $\frac{1}{3}$ for any $k$. Doing this for every $k$, and gluing things together in the obvious way, we get our isometric embedding of a dense subset of $\mathbb R$ in $X$. (Namely, the set of all fractions with a power of $3$ in the denominator.)
| 27 | https://mathoverflow.net/users/70618 | 442872 | 178,675 |
https://mathoverflow.net/questions/442850 | 0 | Given $0<\alpha, \beta<1$, $a,b>0$, $a^2+b^2<1$.
I am trying to determine the asymptotic behaviour of
$$F(a,b):=\int\_{\substack{a/2<x<2a\\\\b/\sqrt{2}<\sqrt{1-x^2}<\sqrt{2}b}}\frac{dx}{|x-a|^{\alpha}|\sqrt{1-x^2}-b|^{\beta}}$$
as $a^2+b^2\to 1$ and $a\rightarrow 0$ simultaneously.
Notation: Given $A,B>0$, as customary we use $ A\approx B$ to denote the fact that there exist two constants $c\_1, c\_2$ such that $c\_1 A\leq B \leq c\_2 A$.
We can estimate
$$|\sqrt{1-x^2}-b|=\frac{|x^2+b^2-1|}{\sqrt{1-x^2}+b}\approx b^{-1}|x^2+b^2-1|\approx |x^2+b^2-1|\qquad (1)$$
since $b\approx 1$ when $a$ is small enough and $a^2+b^2$ is close enogh to $1$.
Now, use the estimate $(1)$ then translate $x-a\to y$ to get
$$F(a,b)\approx \int\_{\substack{-a/2<y<a\\\\ -b^2 <y^2+2a-(1-(a^2+b^2))<b^2/2}}\frac{dy}{|y|^{\alpha}|y^2+2a-(1-(a^2+b^2))|^{\beta}}$$
Let us look then at
$$G(a,b):= \int\_{\substack{0<y<a\\\\ 0<y^2+2a-(1-(a^2+b^2))<b^2/2}}\frac{dy}{|y|^{\alpha}|y^2+2a-(1-(a^2+b^2))|^{\beta}}.$$
When $0<y<a$ we have $y^2+2a\approx a$. Indeed, when $0<y<a<1$ we have
$2a\leq y^2+2a \leq a^2+2a<3a$.
It is easy now to estimate $G$ both when $a>>1-(a^2+b^2)$ and when $a<<1-(a^2+b^2)$.
What I need help with is finding the asymptotic behaviour of $F$ (equivalently $G$)
when $a\approx 1-(a^2+b^2)$ which is the most interesting part of course.
| https://mathoverflow.net/users/116555 | The asymptotic behaviour of a singular integral | $\newcommand{\al}{\alpha}\newcommand{\be}{\beta} $As discussed in comments, given
\begin{equation\*}
a/2<x<2a \tag{00}\label{00}
\end{equation\*}
and $a\to0$ (so that $b\to1$), the restriction $b/\sqrt{2}<\sqrt{1-x^2}<\sqrt{2}b$ will automatically hold (eventually). So,
\begin{equation\*}
F(a,b)=\int\_{a/2}^{2a}\frac{dx}{|x-a|^\al|\sqrt{1-x^2}-b|^\be}. \tag{10}\label{10}
\end{equation\*}
Refining your formula (1) slightly, we get
\begin{equation\*}
|\sqrt{1-x^2}-b|=\tfrac{1+o(1)}2\,|x^2-c^2|, \tag{20}\label{20}
\end{equation\*}
where
\begin{equation\*}
c:=\sqrt{1-b^2}, \tag{30}\label{30}
\end{equation\*}
so that $0<a<c\to0$.
Suppose now that the limit
\begin{equation\*}
r:=\lim\frac ca \tag{35}\label{35}
\end{equation\*}
exists. Then $r\in[1,\infty]$.
If $r=\infty$, then, by \eqref{20} and \eqref{00}, $|\sqrt{1-x^2}-b|\sim\tfrac12\,c^2$ and hence, by \eqref{10},
\begin{equation\*}
F(a,b)\sim(2/c^2)^\be\int\_{a/2}^{2a}\frac{dx}{|x-a|^\al}
=(2/c^2)^\be\frac{2^{1-\al}+2^{\al-1}}{1-\al}\,a^{1-\al}. \tag{40}\label{40}
\end{equation\*}
If $r\in(1,\infty)$ (or $r=1$ and $\al+\be<1$), then, by \eqref{10} and \eqref{20},
\begin{equation\*}
F(a,b)\sim2^\be\int\_{a/2}^{2a}\frac{dx}{|x-a|^\al |x^2-r^2 a^2|^\be}
=2^\be a^{1-\al-2\be} C\_{\al,\be;r}, \tag{50}\label{50}
\end{equation\*}
where
\begin{equation}
C\_{\al,\be;r}:=\int\_{1/2}^{2}\frac{du}{|u-1|^\al |u^2-r^2|^\be}.
\end{equation}
The constant $C\_{\al,\be;r}$ admits a very complicated expression in terms of the gamma function and certain hypergeometric functions; this expression is hardly of any interest.
It is also clear from these considerations that $F(a,b)$ does not have any definite asymptotic if the limit $r$ in \eqref{35} does not exist. The asymptotic of $F(a,b)$ in the case when $r=1$ and $\al+\be\ge1$ (whenever such an asymptotic exists) will depend on the way $\frac ca$ goes to $1$.
| 3 | https://mathoverflow.net/users/36721 | 442892 | 178,681 |
https://mathoverflow.net/questions/442886 | 2 | Let $X$ be a set and $\widehat{F}(X)$ the restricted free profinite group on $X$. To get $\widehat{F}(X)$ we define a profinite topology on $F$ (the free abstract group on $X$) and take $\widehat{F}(X)$ as the completion relative to this topology. If $X$ is a finite set, then $\widehat{F}(X)$ is the profinite completion of $F$. Also, we can write
$$\widehat{F}(X) = \varprojlim \widehat{F}(X\_i)$$
where $X\_i$ varies on the set of the finite subsets of $X$.
>
> I would like to know if $\widehat{F}(X)$ is torsion-free. This problem can be reduced to finite $X\_i$.
>
>
>
In principle, I thought it was a trivial thing, but it seems that the profinite completion (even in residually finite case) of a torsion-free group need not to be torsion-free.
I don't know if $\widehat{F}(X)$ has a non-trivial torsion element. Does anyone know if this is true?
Also, if it is false in the profinite case, what about the pro-$p$ case?
| https://mathoverflow.net/users/123172 | Is the free profinite group (or pro-$p$) torsion-free? | Both the free profinite and free pro-$p$ groups are torsion-free (also free pro-solvable groups or free pro-$\mathcal C$ groups for any collection $\mathcal C$ of groups closed under extensions, subgroups and homomorphic images). The reason is any closed subgroup of such a group is a projective profinite group; this can be proved using Shapiro's lemma and cohomology or can be deduced by reducing first to the case of an open subgroup and then using Nielsen-Schreier and Marshall Hall. You can find a proof in the book of Ribes and Zalesskii.
If a profinite group has nontrivial torsion, then it has a finite cyclic subgroup of prime order and this subgroup is closed. Since a finite cyclic group of prime order is not projective (because the quotient map $\mathbb Z/p^2\mathbb Z\to \mathbb Z/p\mathbb Z$ does not split), we deduce a free profinite or pro-$p$ group does not have torsion.
| 7 | https://mathoverflow.net/users/15934 | 442893 | 178,682 |
https://mathoverflow.net/questions/442902 | 2 | Let $B$ be an invertible real matrix and let $Q=\{A \text{ real}\mid AB^{T} \text{ is symmetric}\}$. Is the subset $S=\{ A \in Q\mid A+A(B^{-1}A)^{2} \text{ is symmetric}\}$ of measure zero in $Q$? I need this for the final step for my proof that I've been working on for 4 years straight; I think that it might intuitively be true, since e.g. the condition for the product of two symmetric matrices to be nonsymmetric is quite strict (that the two matrices be noncommutative) and therefore the conditions for the product of a nonsymmetric matrix and a symmetric matrix to be symmetric should be equally strict as well.
EDIT: It suffices that this holds for all $B$ save a different set of measure zero.
| https://mathoverflow.net/users/113020 | Question on density of certain set of matrices | It suffices to check whether $B^{-1}S=:S'$ has measure zero in $B^{-1}Q=:Q'$. We have
$$Q'={\bf Sym}\_n(\mathbb R),\qquad S'=\{{\bf Sym}\_n(\mathbb R)|B(\Sigma+\Sigma^3)\in{\bf Sym}\_n(\mathbb R)\}.$$
The subset $S'$ is algebraic in $Q'$. Either it has measure zero, or $S'=Q'$.
Let us consider the latter case: then the equation $B(\Sigma+\Sigma^3)=(\Sigma+\Sigma^3)B^T$ is satisfied identically in ${\bf Sym}\_n(\mathbb R)$. By homogeneity, this amounts to saying that $B\Sigma\equiv\Sigma B^T$ in ${\bf Sym}\_n(\mathbb R)$. One checks easily that $B=\beta I\_n$ for some $\beta\in\mathbb R$.
In conclusion, if $B\in{\bf GL}\_n(\mathbb R)$ is not a homothety, then $S$ has measure zero in $Q$.
| 4 | https://mathoverflow.net/users/8799 | 442904 | 178,685 |
https://mathoverflow.net/questions/442890 | 0 | Let $X,Y$ be two independent random variables.
The Kac-Bernstein theorem states that if $X+Y,X-Y$ are also independent, then $X,Y$ are Normal.
Are there analogues of this theorem for non-normal, continuous distributions? For instance, what functions $f\_i(X,Y)$ whose mutual independences guarantee that $X,Y$ are exponentially, or Beta distributed?
| https://mathoverflow.net/users/113397 | Analogues of Kac-Bernstein characterisation theorem for non-normal distributions | Such problems were systematically considered in the book *Characterization Problems in Mathematical Statistics* by Kagan, Linnik, and Rao; see e.g. [this review](https://projecteuclid.org/journals/annals-of-statistics/volume-5/issue-3/Review--A-M-Kagan-Yu-V-Linnik-C-Radhakrishna/10.1214/aos/1176343861.full). In particular, see Ch. 8 there on characterization of families of distributions admitting sufficient statistics, and also Section 6.2 on characterization of the gamma distribution. The list of families of distributions characterized in the book includes the normal (of course), exponential, gamma, beta, Cauchy, Wishart, and other distributions.
See also many papers citing this book.
| 1 | https://mathoverflow.net/users/36721 | 442920 | 178,690 |
https://mathoverflow.net/questions/442927 | 1 | Let $A \subset [0,1]^n$ with $A$ measurable and such that $\mathcal{L}^n (A)= \delta >0$, and consider a partition of $[0,1]^n$ in $\epsilon$-cubes (i.e. cubes of side $\epsilon)$.
For $\epsilon \to 0$ (say, $\epsilon= 1/n$ with $n \in \mathbb{N}$ so that $1/\epsilon$ is always an integer) we should have that, at the limit, it holds the Lebesgue differentiation theorem, so you have that points in $A$ have density 1 and those in $A^c$ have density 0. What I mean is that given every (in fact $\mathcal{L}^n$ almost every, but whatever) point $x$ in $[0,1]^n$ I think that if you take the sequence of the $\epsilon$ -cubes $Q\_\epsilon (x)$ such that they contain $x$, one should have that
$$
\frac{| A \cap Q\_\epsilon (x)|}{\epsilon ^n} \to I\_{A} (x),
$$ with $I\_A$ the indicator function of $A$.
In fact the Lebesgue theorem is about shrinking balls centered at the point $x$ you consider, but at the limit I think this shouldn't matter if you consider finer and finer partitions.
What I would like to know is if there is a sort of quantitative version of this result for fixed $\epsilon$. What I mean is, given a partition $\{Q\_\epsilon (i)\}\_{i=1, \dots, \epsilon^{-n}}$ , having some result which gives you a sort of balance between how big $f\_{A, \epsilon } (i) =\frac{| A \cap Q\_\epsilon (i)|}{\epsilon ^n}$ is and in how many cubes you have this.
So for example, if I fix $\epsilon$ we can't have $ f\_{A, \epsilon} (i)= 1$ for more than $\delta/ \epsilon ^n$ indices, but it's easy to give an example of an $A$ such that $f\_{A, \epsilon} (i)= \delta$ for every $i$.
I was thinking that since the convergence from the Lebesgue theorem is pointwise I can play a bit with Egoroff's theorem or something like that, but maybe these kind of estimates are already considered and have a name in the literature (although I couldn't find anything).
| https://mathoverflow.net/users/109382 | Quantitative version of Lebesgue points theorem | $\newcommand\ep\epsilon\newcommand\de\delta$Let $\ep=1/m$ for a natural $m$, so that
$$N\_\ep:=\ep^{-n},$$
the number of the $\ep$-cubes, is an integer.
Take any $c\in(0,1]$ and let
$$N\_{A,\ep}(c):=\#\{i\in[N\_\ep]\colon f\_{A,\ep}(i)\ge c\},$$
the number of the $\ep$-cubes with relative $A$-content $\ge c$; as usual, $[k]:=\{1,\dots,k\}$. Then, by Markov's inequality,
$$N\_{A,\ep}(c):=\sum\_{i\in[N\_\ep]}\,1(f\_{A,\ep}(i)\ge c)
\le\sum\_{i\in[N\_\ep]}\,\frac{f\_{A,\ep}(i)}c \\
=\frac1c\,\sum\_{i\in[N\_\ep]}\,|A\cap Q\_\ep(i)|\,N\_\ep
=\frac1c\,|A|\,N\_\ep=\frac\de c\,N\_\ep=:M\_{\de,\ep}(c).$$
So, $M\_{\de,\ep}(c)$ is an upper bound on the number $N\_{A,\ep}(c)$ of the $\ep$-cubes with relative $A$-content $\ge c$. This bound is nontrivial only if $c>\de$.
Also, the bound $M\_{\de,\ep}(c)$ is exact in (say) the following sense:
Suppose that $c\in[\de,1]$ is such that $M\_{\de,\ep}(c)$ is an integer (which will be $\le N\_\ep$, since $c\ge\de$). Clearly, then there is a measurable subset $A$ of $[0,1]^n$ such that $f\_{A,\ep}(i)=c\,1(i\le M\_{\de,\ep}(c))$ for all $i\in[N\_\ep]$. Then $N\_{A,\ep}(c)=M\_{\de,\ep}(c)$, so that the upper bound $M\_{\de,\ep}(c)$ on $N\_{A,\ep}(c)$ is attained.
On the other hand, whenever the exact upper bound $M\_{\de,\ep}(c)$ on $N\_{A,\ep}(c)$ is nontrivial (that is, for any $c\in(\de,1]$, the only lower bound on $N\_{A,\ep}(c)$ is the trivial bound $0$ -- which will be attained if e.g. $f\_{A,\ep}(i)=\de$ for all $i\in[N\_\ep]$.
| 1 | https://mathoverflow.net/users/36721 | 442931 | 178,693 |
https://mathoverflow.net/questions/442928 | 2 | Let $C$ be the square $[-1,1]^2$. Let $a\_1,\dots,a\_m$ be points chosen independently and uniformly at random from $C$. Let $d\_m$ (dispersion) be the random variable $\max\_{x \in C}{\min\_{j \in [m]}{\|x-a\_j\|\_2}}$. I would like to find a function $\epsilon(m)$ such that $\Pr[d\_m \leq \epsilon(m)] \geq $ constant (no dependence on m).
In computational geometry terminology, this means, find an $\epsilon(m)$ such that a "random" $m$-subset of $C$ is an $\epsilon(m)$-net for $(C,\mathcal{B})$ (where $\mathcal{B}$ is the class of all $L\_2$-balls in $\mathbb{R}^2$) with constant probability.
In the one-dimensional case (here $C = [-1,1]$), we can take $\epsilon(m)$ to be $K\frac{\log m}{m}$ for a large enough constant $K$. This follows from Theorem 3.1 of "On the lengths of the pieces of a stick broken at random (Holst, 1980)".
I found [this](https://mathoverflow.net/questions/317410/mean-maximum-distance-for-n-random-points-on-a-unit-square?rq=1) similar question on mathoverflow, about the expected value of maximum distance between any two points in $\{a\_1,\dots,a\_m\}$. I don't see how the bound there can be used to get an $\epsilon(m)$ though.
On page 2 of the paper "Quasi-Monte-Carlo methods and the dispersion of point sequences (Rote-Tichy, 1996)" it says that "If the range space $R$ has finite Vapnik-Chervonenkis dimension $d$ then a random subset of $X$ [to be thought of as $C$ in our question] of size $(d/\epsilon)\log (1/\epsilon)$ is an $\epsilon$-net with high probability." This would be great for us as $\mathcal{B}$ (the class of all $L\_2$-balls in $\mathbb{R}^2$) has finite VC dimension ($= d+1$ for the $d$-dimensional case). Rote-Tichy says that this is proved in the paper "Epsilon-nets and simplex range queries (Haussler-Welzl, 1987)". But on this paper, I could only find such theorems for random subsets of a *finite* $A \subset X$ (Theorem 3.3, Corollary 3.7, etc.).
| https://mathoverflow.net/users/316923 | Dispersion of a "random" subset of $[-1,1]^2$ | For every constant dimension $d$, (I.e. when $C = [0,1]^d$), the answer is asymptotically $\varepsilon\_d(m) = \Theta\left(\frac{\log m}{m}\right)^{1/d}$. In the $\Theta$ notation I’m hiding constant factors that depend on $d$. This follows from the [coupon collectors problem](https://en.wikipedia.org/wiki/Coupon_collector%27s_problem): let us partition a $[0,1]^d$ cube into $\varepsilon^{-d}$ sub-cubes of side-length $\varepsilon$, I.e. $[0,1]^d = \left( \bigcup\_i [i \varepsilon, (i + 1) \varepsilon]\right)^d$.
Now, let us draw $m$ points uniformly at random from $[0, 1]^d$, and let’s keep track of which sub-cubes are being hit.
By coupon collectors, as soon as $m \gg \varepsilon^{-d} \log(\varepsilon^{-d})$, with high probability we will hit each sub-cube, and when $m \ll \varepsilon^{-d} \log(\varepsilon^{-d})$ with high probability we will miss at least one sub-cube.
Now it is enough to show the following two elementary geometric fact:
1. Any subset $S\subset [0, 1]^d$ which misses at least one sub-cube, has dispersion at least $\varepsilon/2$ (take the center of a missed cube as a witness $x$).
2. Any subset $S \subset [0,1]^d$ which hits every sub-cube has dispersion at most $\sqrt{d}\varepsilon$ (the diameter of a cube of side length $\varepsilon$).
Related: I recommend the [great answer](https://mathoverflow.net/a/442259/468679) by Iosif Pinelis with much more precise calculation of the exact constant for the case where $d=1$.
| 3 | https://mathoverflow.net/users/468679 | 442938 | 178,695 |
https://mathoverflow.net/questions/442921 | 0 | I am going through Terence Tao's "*Nonlinear Dispersive Equations (Local & Global Analysis)*" and trying to work through some of his exercises. However, I find myself being stumped by Exercise A.21. The question goes as follows:
>
> Let $0<\alpha<1$ and $1\le p\le\infty$. If $f\in S\_x(\mathbb{R}^d)$, we define the *Hölder norm* $\|f\|\_{\Lambda\_\alpha^p(\mathbb{R}^d)}$by the formula
> $$
> \|f\|\_{\Lambda\_\alpha^p(\mathbb{R}^d)}:=\|f\|\_{L^p\_x(\mathbb{R}^d)} + \sup\_{h\in\mathbb{R}^d; 0<|h|\le1}\frac{\|f^h-f\|\_{L^p\_x(\mathbb{R}^d)}}{|h|^\alpha}
> $$
> where $f^h(x)=f(x+h)$ is the translate of $f$ by $h$. Show that
> $$
> \|f\|\_{\Lambda\_\alpha^p(\mathbb{R}^d)} \sim\_{p,\alpha,d} \|f\|\_{L^p\_x(\mathbb{R}^d)} + \sup\_{N\ge 1}N^\alpha\|P\_{N}f\|\_{L^p\_x(\mathbb{R}^d)}
> $$
> (Hint: to control the former by the latter, obtain two bounds for the $L^p\_x(\mathbb{R}^d)$ norm of $P\_Nf^h-P\_Nf$, using triangle inequality for the high frequency case $N\gtrsim |h|^{-1}$ and the fundamental theorem of calculus in the low frequency case $N\lesssim |h|^{-1}$.)
>
>
>
So when attempting to control the former with the latter in the high frequency case, I took his hint and applied triangle inequality, which landed me at
$$
\|P\_Nf^h-P\_Nf\|\_{L^p\_x(\mathbb{R}^d)} \le 2\|P\_Nf\|\_{L^p\_x(\mathbb{R}^d)}
$$
Thereafter,
$$
\frac{\|P\_Nf^h-P\_Nf\|\_{L^p\_x(\mathbb{R}^d)}}{|h|^\alpha} \lesssim N^\alpha\|P\_Nf\|\_{L^p\_x(\mathbb{R}^d)}
$$
From here, I found myself in a rather unpleasant dead-end since moving forward will require me to take the infinite sum over $N$'s, which in this present form will not converge.
May I know how should I side-step this problem? Or am I misunderstanding his hint? Thanks in advance!
| https://mathoverflow.net/users/501150 | Littlewood-Paley characterisation of Hölder regularity | You need to remember that $N$ is a **dyadic number**, so that $\sum\_{N \geq c} N^{-\alpha}$ converges for positive $\alpha$.
So you have
$$ \sum\_{N \geq |h|^{-1}} \frac{\|P\_Nf^h-P\_Nf\|\_{L^p\_x(\mathbb{R}^d)}}{|h|^\alpha} = \sum\_{N \geq |h|^{-1}} N^\alpha\|P\_Nf^h-P\_Nf\|\_{L^p\_x(\mathbb{R}^d)}\cdot \frac{1}{N^\alpha|h|^\alpha} $$
Split the sum using Holder and put the first factor in $\ell^\infty$ and sum the second factor to get a constant (which is $\approx 1$, independently of $h$.)
| 3 | https://mathoverflow.net/users/3948 | 442939 | 178,696 |
https://mathoverflow.net/questions/442733 | 2 | Are there incompatible Turing degrees $a,b$ s.t any degree computable in $a$ either computes $a$ or is computed by $b$?
Obviously, if $a$ was above $b$ then $a$ would be a strong minimal cover of $b$. But do incomparable such degrees exist and is there a name for them?
Or am I missing some obvious fact that implies such degrees can't exist? I feel like I should already know the answer but i don't seem to.
Edit: I meant to rule out the case where this is trivially possible so I should have added: and this relation holds for no $b$ below $a$.
| https://mathoverflow.net/users/23648 | Incompatible degrees $a,b$ s.t. $x < a$ implies $x \leq b$ | This type of algebraic property is observed in the local structure of the enumeration degrees: there we call them Ahmad pairs. Ahmad (a student of Lachlan) showed that there are incomparable $\Sigma^0\_2$ enumeration degrees $a$ and $b$ such that for all enumeration degrees $x$ if $x<a$ then $x<b$. She also showed that there are no symmetric Ahmad pairs. This property plays a key role in our attempts to analyze the 2-quantifier theory of the structure.
In the c.e. Turing degrees there are no Ahmad pairs because of Sacks splitting theorem (in an Ahmad pair the first element cannot be the join of two lower ones). In the $\Delta^0\_2$ Turing degrees there are symmetric Ahmad pairs: any pair of distinct minimal degrees.
| 6 | https://mathoverflow.net/users/501163 | 442941 | 178,698 |
https://mathoverflow.net/questions/442926 | 1 | Let $[a, b]$ be a nonempty interval, $o \in C^1([a, b])$ be such that $o>0$ and $o'<0$ and assume we found some $v \in L^\infty(\mathbb{R})$ such that
\begin{equation}\tag{1}\label{1}
\int\_a^b \varphi' v o ~\mathrm{d}x \leq 0 \quad \forall \varphi \in C\_0^\infty([a, b], [0, \infty))=:U, \quad \frac{o(b)}{o(a)} \leq v \leq 1 \text{ a.e. }
\end{equation}
and
$$\tag{2}\label{2}
\int^b\_a \varphi'vo~\mathrm{d}x < 0
$$
for at least one $\varphi \in U$. My question is whether we can find $\zeta \in L^\infty(\mathbb{R})$ with $\int\_a^b \zeta ~\mathrm{d}x >0$ such that $v+\delta \zeta$ still fulfils $(1)$ for $\delta$ arbitrarily small. I also don't know whether \eqref{1} and \eqref{2} can actually be satisfied at once.
I thought about $\zeta$ adding a very small constant where $v<1$ (on sets with measure greater than $0$) which should lead to the integral inequality not being fulfiled. If one could exclude the case of $v=1$ on a set with measure greater than zero, then we could take $v = \zeta$. I also tried to choose $\zeta$ such that $\zeta o$ is constant which would render the integral condition zero. But then we would very likely violate the box constraints. Of course, $\zeta$ can also be negative, but I would not know how to construct such example.
| https://mathoverflow.net/users/500621 | How much "room" in inequality $\displaystyle \int_a^b \varphi' ov ~\mathrm{d}x \leq 0$ | Let's upgrade Iosif's [comment](https://mathoverflow.net/questions/442926/how-much-room-in-inequality-displaystyle-int-ab-varphi-ov-mathrmdx#comment1143194_442926) to an answer.
Let $\chi$ be a smooth bump function supported in $[-\epsilon,\epsilon]$. For any $\varphi\in U$ with support within $[a+\epsilon,a-\epsilon]$, then $\chi\*\varphi$ is also in $U$. So
$$ \int (\chi\*\varphi)(x) v(x) o(x) ~dx = 0 \implies \int\_{a+\epsilon}^{b-\epsilon} \int\_{-\epsilon}^{\epsilon} \varphi'(x) \chi(y) v(x+y) o(x+y) ~dy ~dx = 0$$
The output of the $y$ integral is a smooth function on $[a+\epsilon,b-\epsilon]$, and hence Iosif's argument shows that it must be constant.
Taking a limit as $\epsilon\to 0$ using $\chi$ an approximation to identity, we find that $vo$ must be almost everywhere constant.
| 3 | https://mathoverflow.net/users/3948 | 442942 | 178,699 |
https://mathoverflow.net/questions/442953 | 5 | I have been using the notion of "semisimple linear category" for a while now, but I never bothered to write down a definition. I've always just waved my hands and said "Schur's lemma holds and every object is the direct sum of simple objects". I find myself writing some notes on the subject, and it occurred to me that I had some problems coming up with the correct definition.
The natural reference is nLab. Semisimple linear categories are definition 2.2 of their article ["semisimple category"](https://ncatlab.org/nlab/show/semisimple+category). However, I have an issue with their (and by proxy, Müger's) approach. Namely, they require that the category has finite biproducts (direct sums). Next, they require that for any pair of objects $V$, $W$ the natural map
$$\sum\_{i\in I}\mathrm{Hom}(V,X\_i)\otimes \mathrm{Hom}(X\_i,W)\to\mathrm{Hom}(V,W)$$
is an isomorphism, where $\{X\_i,\,\, i\in I\}$ is a collection of simple objects chosen to make this isomorphism work. My understanding is that these $X\_i$ are supposed to be representatives of the different isomorphism classes of simple objects in $\mathscr{C}$. However, in general there might be infinitely many different such isomorphism classes. This would mean that the direct sum is infinite. This is a problem, since we only required *finite* biproducts. How do we fix this?
My guess is that if every object is the direct sum of (finitely many) simple objects, then this sum will always be finite, and hence we don't have any problems. If we replace this hom-space-symmetry axiom with the simpler "Every object is the direct sum of finitely many simple objects" do we recover the correct definition?
| https://mathoverflow.net/users/159298 | What is the correct definition of semisimple linear category? | If I were to try to define "semisimple linear category", and this is indeed something I do try to define, I would say: it is a category, linear over your base commutative ring, which has finite direct sums and splittings of idempotents, and such that for every object $X$, the ring $\mathrm{End}(X)$ is semisimple.
I would then remark that "has finite direct sums and splittings of idempotents" is an extremely natural condition for a linear category, as it is precisely asking that the linear category be Cauchy complete (i.e. it should contain [(co)weighted] (co)limits for all [linear] diagrams such that (co)limits of those shapes are preserved by all linear functors).
Then I would point out that these conditions imply that the category is abelian and that every object is a finite direct sum of simple objects. Also the converse: an abelian category comes with a notion of "simple object", and if every object is a finite direct sum of simples, then it is semisimple in the above sense.
Credit where it's due: I learned this definition from David Reutter, and a version is in his paper with Chris Douglas on semisimple 2-categories.
| 8 | https://mathoverflow.net/users/78 | 442957 | 178,704 |
https://mathoverflow.net/questions/442981 | 3 | Let $n\geq 2$ and denote by $B\subset \mathbb{R}^n$ the closed unit ball.
Does there exist a closed subset $A\subset B$ containing $0\in \mathbb{R}^n$ with the following properties i,ii,iii?
i) $\{0\}$ is a connected component of $A$.
ii) $0\in \overline{A\setminus\{0\}}$ and
iii) each connected component $C\neq \{0\}$ of $A$ intersects $\partial B$: $C\cap \partial B\neq \emptyset$.
A comment: $A$ needs to have infinitely many connected components by i) and ii). The first two conditions i) and ii) are easily satisfied for a set $A$ consisting of the elements of a sequence in $B\setminus \{0\}$ converging to $0$ together with its limit point. However I do not know how to construct an example of a closed set $A$ where in addition iii) holds.
| https://mathoverflow.net/users/66777 | Closed subset of unit ball with peculiar connected components | No.
Note that for any integer $n>0$ there is a component $C\_n\subset A$ that contains two points $x\_n$ and $y\_n$ such that $|x\_n|=\tfrac1n$ and $|y\_n|=1$.
Recall that $C\_n$ is a closed set.
Pass to a subsequence of $C\_n$ that converges in the sense of Hausdorff; denote its limit by $C\_\infty$.
We may assume that $y\_n\to y\_\infty$.
Observe that $C\_\infty$ is a closed connected subset of $A$ that contains $0$ and $y\_\infty$ --- a contradiction.
| 8 | https://mathoverflow.net/users/1441 | 442983 | 178,711 |
https://mathoverflow.net/questions/442511 | 5 | Apparently the Galois group $G$ of a Galois extension $E/F$ can be viewed as an “étale sheaf” on the set $X$ of intermediate Galois extensions equipped with an appropriate Grothendieck topology (see short discussion at [Galois Group as a Sheaf](https://mathoverflow.net/questions/30030/galois-group-as-a-sheaf)). For now I will assume $G$ is abelian. In this setting an open cover of $X$ will be a set of intermediate extensions $\{E/E\_i/F\}$ which together generate $E$.
My question is about how much this viewpoint transfers the properties of sheaves on topological spaces, e. g. coherent sheaves on schemes. I think the correct analogy is that $G$ is the structure sheaf, and discrete $G$-modules $M$ are coherent sheaves on $X$. If this is correct, is the correct notion of restriction to a subfield $L$ the quotient of $M$ by the invariants of $\operatorname{Gal}(E/L)$?
If the answer is yes, it is clear how to define cochains and coboundary operator via restriction maps, as in the construction of Čech cohomology. Does this give a cochain complex? If so we obtain an associated cohomology $\grave{H}^\bullet(G, M)$. As with Čech cohomology, a priori this is defined with respect to a choice of cover. Is there some choice of cover which will give a canonical result? What is the relation, if any, between this and the typical group cohomology? What is the correct analogy of a short exact sequence of sheaves? Is there a better definition of cohomology in terms of derived functors?
To what extent do these constructions “globalize”? For instance, can we somehow patch together these objects, in the same way affine schemes patch into general schemes? Are there interesting analogues of moduli spaces of Galois groups? In general I am interested in whatever geometric aspects of Galois theory (or class field theory more specifically) this perspective may elucidate.
If anyone can shed light themselves or suggest a reference I would greatly appreciate it. Please excuse me if these questions are not research level.
| https://mathoverflow.net/users/351164 | “Sheaf cohomology” of Galois groups | The usual geometric point of view on the Galois group of $F$ is that the Galois group is the etale fundamental group of $\operatorname{Spec} F$. It's possible that you're already aware of this point of view and are looking for something different, but let me say a few words about it anyway.
This analogy comes not from thinking of the fundamental group as loops, but as the automorphism group of the universal cover. The starting point is the observation that if $X$ is any reasonable topological space, say connected for simplicity, then the fundamental group $\pi\_1(X,x)$ is isomorphic to the group of deck transformations of the universal cover $\tilde{X} \to X$, via monodromy. Working in the etale site over $\operatorname{Spec} F$ (which is now our $X$), the "universal cover" is $\operatorname{Spec}F^{\text{sep}}$ and the Galois group $G\_F = \operatorname{Gal}(F^{\text{sep}} / F)$ of $F$ becomes the group of deck transformations. Strictly speaking, $\operatorname{Spec} F^{\text{sep}}$ isn't a *finite* etale cover of $\operatorname{Spec} F$, which is usually dealt with by viewing $\operatorname{Spec} F^{\text{sep}}$ as a limit of connected etale covers $\operatorname{Spec} E \to \operatorname{Spec} F$, where $E$ is a finite (Galois) extension of $F$. So the etale fundamental group of $\operatorname{Spec} F$ is identified with $G\_F$ as a profinite group.
If $\mathcal{F}$ is an abelian sheaf on the etale site of $\operatorname{Spec} F$, then $M := \mathcal{F}(\operatorname{Spec} F^{\text{sep}})$ has a natural $G\_F$-action. This construction gives an equivalence between abelian sheaves on the etale site and discrete $G\_F$-modules. Moreover, $M^{G\_F} \cong \mathcal{F}(\operatorname{Spec} F)$, which implies that $H^\*(G\_F,M) \cong H^\*\_{et}(\operatorname{Spec} F, \mathcal{F})$.
In sum, I think that any reference about algebraic geometry containing the word "etale" may be relevant. As for globalizing these constructions, the keyword you're looking for may be "descent". For example, the Stacks Project contains a lot of material about these topics.
| 3 | https://mathoverflow.net/users/501202 | 442992 | 178,714 |
https://mathoverflow.net/questions/442851 | 1 | Let $f:[0, \infty)\to [0, \infty)$ be non-increasing (and not necessarily differentiable nor continuous) and satisfy $$f(t)\leq f(0)-C\int\_{0}^{t}f(s)^{1/2}ds,$$ where $C>0$. How can one show that then $$f(t)\leq g(t)\quad \text{for all}~t\leq t\_{\*},$$ where $t\_{\*}>0$ and $g$ is a differentiable function on the interval $[0,t\_\*)$ such that $$g'(t)=-Cg(t)^{1/2}\quad \text{for all}~t\leq t\_{\*}, \quad g(0)=f(0)?$$
Obviously if we would have equality in the above integral inequality and differentiability of $f$, we would have $f=g$, but is it still true that $f\leq g$ under this weaker assumptions?
Thanks in advance!
| https://mathoverflow.net/users/163368 | Integral inequality implies majorization by solution of ODE | The problem with this question, compared to [this one of yours](https://math.stackexchange.com/questions/4658708/integral-inequality-implies-majorization-by-solution-of-ode), is that the vector field on the right hand side of the ODE is not a non-decreasing function of $g$. If you try to make the example of @fedja rigorous, you can see that you can manage to build even smooth counterexamples. To make everything work, you need an ODE of the form $g’(t)=v(g(t))$, where $v$ is non-decreasing.
---
Another example, this time with a positive vector field:
$$ g’(t)=g(t)(2-g(t)), $$
with initial data $f(0)=g(0)=1$. Then
$$ g(t)=\frac{2e^{2t}}{e^{2t}+1}. $$
The corresponding integral inequality
$$ f(t)\leq 1+\int\_0^t f(s)(2-f(s))ds $$
admits as a solution, for instance,
$$ f(t)=\frac{1+2e^{-(x-10)^2}}{1+2e^{-(10)^2}}. $$
Clearly, $f(10)=3>2>g(10)$. The problem is that the vector field $v(x)=x(2-x)$ fails to be non-decreasing in its argument.
---
Heuristic explanation. The second term of your integral equation is like a ‘bag’ that saves you a quantity of energy $v(f(s))$ per second. The inequality you want to prove is disproved as soon as you make $f$ gain more total energy than $g$.
If $v$ is a decreasing function, then as soon as $f(t)<g(t)$, you have $v(f(t))>v(g(t))$. That is, $f$ collects more energy than $g$ per second. So to make a large $f$, it is convenient to keep $f$ small for a suitable amount of time (so that it satisfies the inequality) and at the same time make it stay in a favorable range where you can collect more energy (in my previous example, $v$ is maximized at $x=1$), and use that energy later to grow past $g$ as soon as you have collected enough energy (while $g$ is forced to grow by the ODE and as soon as it gets large it will start collecting energy with a smaller rate with respect to $f$).
| 2 | https://mathoverflow.net/users/272040 | 442995 | 178,715 |
https://mathoverflow.net/questions/442994 | 15 | In [his 1926 paper](https://ipparco.roma1.infn.it/pagine/deposito/2011/TF.pdf) Fermi states without further explanation that it follows from the Thomas-Fermi equation
$$\frac{d^2\psi(x)}{dx^2}=\frac{\psi(x)^{3/2}}{\sqrt{x}},\label{1} \tag{1}$$
and boundary conditions
$$\psi(0)=1,\quad\psi(\infty)=0\label{2}\tag{2},$$
that
$$\int\limits\_0^\infty{\frac{\psi^{5/2}(x)}{\sqrt{x}}}dx=-\frac{5}{7}\psi^\prime(0),\label{3} \tag{3}$$ where $\psi^\prime(x)=\frac{d\psi(x)}{dx}$. It seems Fermi considered this inference trivial. However, the proof given below and inspired by [Kleinert, p. 429](https://hagenkleinert.de/documents/pi/HagenKleinert_PathIntegrals.pdf), although simple, is not at all trivial.
The Thomas-Fermi equation \eqref{1} can be considered as a Euler-Lagrange equation coresponding to the action principle
$$
\delta S=0,\;\; S=\int\limits\_0^\infty\left [\frac{1}{2}\left (\frac{d\psi}{dx}\right)^2+\frac{2}{5}\frac{\psi^{5/2}(x)}{\sqrt{x}}\right ]dx.\label{4}\tag{4}
$$
The following infinitesimal deformation of the "coordinate" function
$$\psi(x)\to \bar\psi(x)=\psi(\lambda x),\;\lambda=1+\epsilon,\;\epsilon\ll 1,$$
respects the boundary conditions \eqref{2} and changes the action functional \eqref{4} to
$$\bar S=\int\limits\_0^\infty\left [\frac{1}{2}\left (\frac{d\bar\psi(x)}{dx}\right)^2+\frac{2}{5}\frac{\bar\psi^{5/2}(x)}{\sqrt{x}}\right ]dx=\int\limits\_0^\infty\left [\frac{\lambda}{2}\left (\frac{d\psi(y)}{dy}\right)^2+\frac{2}{5}\frac{\psi^{5/2}(y)}{\sqrt{\lambda y}}\right ]dy,$$ where $y=\lambda x$. Therefore, using $\lambda =1+\epsilon,\;\lambda^{-1/2}\approx 1-\epsilon/2$, we get
$$\bar S =S+\epsilon \int\limits\_0^\infty\left [\frac{1}{2}\left (\frac{d\psi(y)}{dy}\right)^2-\frac{1}{5}\frac{\psi^{5/2}(y)}{\sqrt{y}}\right ]dy,$$
and since we must have $\delta S=\bar S-S=0$ for any $\epsilon\ll 1$, we conclude
$$\int\limits\_0^\infty \left (\frac{d\psi(x)}{dx}\right)^2 dx=\frac{2}{5}\int\limits\_0^\infty \frac{\psi^{5/2}(x)}{\sqrt{x}} dx.\label{5} \tag{5}$$
On the other hand, using \eqref{1}, \eqref{2} and integration by parts, we have
$$
\begin{split}
\int\limits\_0^\infty \frac{\psi^{5/2}(x)}{\sqrt{x}} dx &=\int\limits\_0^\infty \psi(x)\frac{d^2\psi(x)}{dx^2}dx\\
&=\left .\psi(x)\frac{d\psi(x)}{dx}\right |\_0^\infty-\int\limits\_0^\infty \left (\frac{d\psi(x)}{dx}\right)^2 dx\\
& =-\left .\frac{d\psi(x)}{dx}\right |\_{x=0}-\int\limits\_0^\infty \left (\frac{d\psi(x)}{dx}\right)^2 dx,
\end{split}$$ and the Fermi result \eqref{3} immediately follows from \eqref{5}.
Is there a simpler proof of \eqref{3} that Fermi might have had in mind?
| https://mathoverflow.net/users/32389 | How did Fermi calculate this integral? | The way this is taught in text books,$^\ast$ which is likely the way Fermi reasoned, is to compare two alternative integrations by parts. One the one hand,
$$\int\_0^\infty \frac{\psi^{5/2}(x)}{\sqrt{x}}\,dx=-5\int\_0^\infty \psi'(x)\psi^{3/2}(x)\sqrt{x}\,dx$$
$$\qquad\qquad=-\frac{5}{2}\int\_0^\infty x\frac{d}{dx}[\psi'(x)]^2\,dx=\frac{5}{2}\int\_0^\infty [\psi'(x)]^2\,dx.$$
On the other hand,
$$\int\_0^\infty \frac{\psi^{5/2}(x)}{\sqrt{x}}\,dx=\int\_0^\infty \psi(x)\psi''(x)\,dx=-\psi'(0)-\int\_0^\infty[\psi'(x)]^2\,dx.$$
Hence
$$\int\_0^\infty \frac{\psi^{5/2}(x)}{\sqrt{x}}\,dx=-\frac{5}{7}\psi'(0).$$
---
$^\ast$ See, for, example, page 17 of [Roi Baer's notes](http://vintage.fh.huji.ac.il/~roib/LectureNotes/DFT/dft02.pdf)
| 19 | https://mathoverflow.net/users/11260 | 443000 | 178,716 |
https://mathoverflow.net/questions/442765 | 1 | Let $W$ be a standard one dimensional Brownian motion, $\mathcal F\_t$ its natural filtration, and $\mathbb P$ be the induced Wiener measure on $\Omega := C[0, 1]$.
Given a $C[0, 1] $ valued random variable $F$, define the translation map $T\_F: \Omega \to \Omega$ by $T\_F (\omega) = \omega + F(\omega)$, and denote the pushforward of $\mathbb P$ under this map by $T\_F^{\ast} \, \mathbb P$.
We view $F$ as a stochastic process and say that $F$ is adapted to $W$ if as a stochastic process, $F\_t$ is $\mathcal F\_t$ adapted. Denote by $W^{1, 2}$ the space of absolutely continuous functions on $[0, 1]$ with derivative in $L^2$.
>
> **Question:** Is it true that $\mathbb Q$ is equivalent to $\mathbb P$ if and only if $\mathbb Q = T\_F ^\ast \, \mathbb P$ for some adapted $F$ such that $F(0) = 0$ and $F \in W^{1, 2}$ almost surely?
>
>
>
| https://mathoverflow.net/users/173490 | Converse Cameron-Martin theorem for shifts by adapted processes | If $F\_t(\omega)=\int\_0^t h\_s(\omega)\,ds$, with $h$ progressive and such that $\int\_0^1 h\_s^2(\omega)\,ds\le C$ for some finite constant $C$, then indeed $T^\*\_F\Bbb P$ is equivalent to $\Bbb P$.
Let $X\_t$ denote the coordinate process on $\Omega$, a Brownian motion under $\Bbb P$. Now define the martingale $M\_t:=\int\_0^t h\_s\,dX\_s$ and the associated exponential martingale $L\_t:=\exp(-M\_t-{1\over 2}\langle M\rangle\_t)$. (This is a martingale by Novikov's criterion.)
Finally define $\Bbb Q$ to be the probability measure on $\Omega$ with density $L\_1$ with respect to $\Bbb P$. By Girsanov's theorem, the process $Y\_t:=X\_t+F\_t$ is a $\Bbb Q$-Brownian motion. Then for any bounded measurable $f:\Omega\to\Bbb R$,
$$
\int\_\Omega f\,d\Bbb P = \int\_\Omega f(Y)\,d \Bbb Q=
\int\_\Omega L\_1 f(Y)\,d \Bbb P.
$$
Thus, $\int\_\Omega f\,d\Bbb P=0$ iff $\int\_\Omega L\_1 f(Y)\,d \Bbb P=0$, and in turn iff $\int\_\Omega f\,dT^\*\_F\Bbb P=\int\_\Omega f(Y)\,d \Bbb P=0$ because the density $L\_1$ is strictly positive and finite a.s. $\Bbb P$.
The boundedness condition on $h$ can be partially relaxed by localization. Suppose $\int\_0^1 h^2\_s\,ds<\infty$, $\Bbb P$-a.s. Define, for each positve integer $n$, $T\_n:=\inf\{t: \int\_0^t h\_s^2\,ds>n\}$, and then $F^{(n)}\_t:=\int\_0^{t\wedge T\_n} h\_s\,ds$. By the preceding paragraph, $\Bbb P$ is equivalent to $T^\*\_{F^{(n)}}\Bbb P$ for each $n$. But $F^{(n)}=F$ on the event $G\_n:=\{\int\_0^1 h^2\,ds\le n\}$. Because $\Bbb P(\cup\_nG\_n)=1$, it follows that $T^\*\_F\Bbb P \ll\Bbb P$.
| 2 | https://mathoverflow.net/users/42851 | 443006 | 178,719 |
https://mathoverflow.net/questions/442445 | 2 | Let $(\Omega, \mathcal{F})$ be a measurable space. Let $E: \mathcal{F}\to B(H)$ be a regular resolution of the identity on the Hilbert space $H$, see e.g. Rudin's functional analysis book.
Suppose that $g: \Omega \to \mathbb{C}$ is Borel-measurable and that $\xi \in \mathscr{D}\left(\int\_\Omega g dE\right)=: \mathscr{D}\_g$. We can then define
$$\eta:= \left(\int\_\Omega g dE\right)\xi\in H.$$
Is it true that
$$dE\_{\eta, \eta}= |g|^2 dE\_{\xi, \xi}?$$
I.e., if $A\in \mathcal{F}$, do we have
$$E\_{\eta, \eta}(A)= \int\_{A}|g|^2 dE\_{\xi, \xi}.$$
I believe I can show this if $g$ is bounded. But I am mostly interested in the case where $g$ is unbounded.
Thanks in advance for your help!
| https://mathoverflow.net/users/216007 | Identity for spectral resolution: $dE_{\xi, \xi}= |g|^2 dE_{\eta, \eta}$ | If $g=\sum\_k\beta\_k\,1\_{X\_k}$ is simple,
$$
\Big\langle\Big(\int\_\Omega g\,dE\Big)\xi,\xi\Big\rangle=\int\_\Omega g\,dE\_{\xi,\xi}=\sum\_k\beta\_k\,E\_{\xi,\xi}(X\_k)=\Big\langle\sum\_k\beta\_k\,E(X\_k)\xi,\xi\Big\rangle.
$$
Since $g$ is measurable, then there exist simple functions $g\_n$ such that $g\_n\to g$ and $|g\_n|^2\nearrow |g|^2$ (this is easily achievable by finding first simple functions that increase to $(\operatorname{Re}g)^+$, $(\operatorname{Re}g)^-$, $(\operatorname{Im}g)^+$, $(\operatorname{Im}g)^-$, respectively).
For any $\nu$ we have, using monotone convergence in each of the sixteen integrals coming from decomposing $g$ and $dE\_{\xi,\nu}$ as a linear combinations of positive functions/measures,
$$
\langle \eta,\nu\rangle=\int\_\Omega g\,dE\_{\xi,\nu}=\lim\_n\int\_\Omega g\_n\,dE\_{\xi,\nu}=\lim\_n\Big\langle\Big(\int\_\Omega g\_n\,dE\Big)\xi,\nu\Big\rangle
$$
So, writing $g\_n=\sum\_k\beta\_{n,k}\,1\_{X\_{n,k}}$,
\begin{align}
E\_{\eta,\eta}(A)
&=\langle E(A)\eta,\eta\rangle
=\Big\langle E(A)\Big(\int\_\Omega g\,dE\Big)\xi,\eta\rangle\\[0.3cm]
&=\lim\_n\Big\langle E(A)\Big(\int\_\Omega g\,dE\Big)\xi,\Big(\int\_\Omega g\_n\,dE\Big)\xi\Big\rangle\\[0.3cm]
&=\lim\_n\sum\_k\overline{\beta\_{n,k}}\Big\langle E(A)\Big(\int\_\Omega g\,dE\Big)\xi,E(X\_{n,k})\xi\Big\rangle\\[0.3cm]
&=\lim\_n\sum\_k\overline{\beta\_{n,k}}\Big\langle \Big(\int\_\Omega g\,dE\Big)\xi,E(A)E(X\_{n,k})\xi\Big\rangle\\[0.3cm]
&=\lim\_n\sum\_k\overline{\beta\_{n,k}}\Big\langle \Big(\int\_\Omega g\,dE\Big)\xi,E(A\cap X\_{n,k})\xi\Big\rangle\\[0.3cm]
&=\lim\_n\lim\_m\sum\_k\sum\_j\overline{\beta\_{n,k}}\beta\_{m,j}\Big\langle E(X\_{m,j})\xi,E(A\cap X\_{n,k})\xi\Big\rangle\\[0.3cm]
&=\lim\_n\lim\_m\sum\_k\sum\_j\overline{\beta\_{n,k}}\beta\_{m,j}\Big\langle E(A\cap X\_{n,k})E(X\_{m,j})\xi,\xi\Big\rangle\\[0.3cm]
&=\lim\_n\lim\_m\Big\langle \Big(\int\_A \overline{g\_n}g\_m\,dE\Big)\xi,\xi\Big\rangle\\[0.3cm]
&=\lim\_n\Big\langle \Big(\int\_A \overline{g\_n}g\,dE\Big)\xi,\xi\Big\rangle\\[0.3cm]
&=\Big\langle \Big(\int\_A \overline{g}g\,dE\Big)\xi,\xi\Big\rangle\\[0.3cm]
&=\int\_A|g|^2\,dE\_{\xi,\xi}.
\end{align}
| 1 | https://mathoverflow.net/users/3698 | 443020 | 178,720 |
https://mathoverflow.net/questions/443026 | 0 | I'd like to know the estimate of the following sum
$$\sum\_{n\leq x}\sum\_{d|n}\Lambda(d)\frac{\phi(d)}{d} $$
where $\Lambda(d)$ is the von mangoldt function and $\phi(d)$ is the Euler totient function. Do you know how to compute this?
| https://mathoverflow.net/users/159935 | Estimating a sum involving the von Mangoldt function | Following Joshua Stucky's remark, the sum can be rewritten as the following sum over prime numbers:
$$\sum\_{p\leq x}(\log p)\left(1-\frac{1}{p}\right)\sum\_{k=1}^\infty\left\lfloor\frac{x}{p^k}\right\rfloor.$$
Hence the sum is upper bounded by
$$\sum\_{p\leq x}(\log p)\left(1-\frac{1}{p}\right)\frac{x}{p-1}=x\sum\_{p\leq x}\frac{\log p}{p}=x\log x+O(x),$$
and it is lower bounded by
$$\sum\_{p\leq x}(\log p)\left(1-\frac{1}{p}\right)\left(\frac{x}{p}-1\right)>x\sum\_{p\leq x}\frac{\log p}{p}-x\sum\_{p\leq x}\frac{\log p}{p^2}-\sum\_{p\leq x}\log p=x\log x+O(x).$$
Therefore, the sum is
$$\sum\_{n\leq x}\sum\_{d|n}\Lambda(d)\frac{\phi(d)}{d}=x\log x+O(x).$$
| 5 | https://mathoverflow.net/users/11919 | 443029 | 178,722 |
https://mathoverflow.net/questions/442888 | 3 | I have non-negative $d\times d$ matrices $A$, $B$ and need a tractable way to compute the sum of all entries of $\exp(-t(A-B))$ where $A$ is diagonal and $B$ symmetric rank-$1$. IE
$$f(t)=\langle\exp(-t(A-B))\rangle$$
Where $t>0$ and $\langle M\rangle$ represents sum of all entries of $M$.
**What expansion would be useful to estimate $f(t)$?**
I know that norms of $A$, $B$ and $A-B$ are less than one, $A-B$ is positive semi-definite and norm of $B$ is relatively small compared to norm of $A$.
I can only afford to perform $c\times d$ operations for small values of $c$.
Golden-Thompson [inequality](https://terrytao.wordpress.com/2010/07/15/the-golden-thompson-inequality/) gives efficient to compute and tight bound for $\operatorname{Tr}$, but similar factoring is loose for $\langle\cdot\rangle$. Computing $f(t)$ would extend Velikanov's [approach](https://arxiv.org/abs/2206.11124) for computing SGD error after $t$ steps to small values of $t$.
[Notebook](https://www.wolframcloud.com/obj/yaroslavvb/nn-linear/forum-bounding-rank1.nb)
| https://mathoverflow.net/users/7655 | Approximating sum of entries of $\exp(A-B)$ for diagonal $A$ and rank-$1$ $B$? | This Asymptote code seems to work perfectly and for any $t$ in your range the estimate uses $Cd$ operations and is a guaranteed upper bound though I am not sure whether $C$ is small enough for you (I hope it is).
If you try to run it online on <http://asymptote.ualberta.ca/> , the final pause() command should be removed. The output gives the value of $t$ (I run it over powers of 2 to make the comparison quick), the truth and the quick upper bound after it. All explanations tomorrow.
```
int d=300;
srand(seconds());
real p=0.6+2*unitrand();
real[] h,A;
real[][] C;
for(int k=0;k<d;++k) h[k]=exp(-p*log(k+1))*(1+2*unitrand());
h=reverse(sort(h));
real a=1/(2*max(h)+sum(h));
for(int k=0;k<d;++k)
{
C[k]=new real[];
for(int l=0;l<d;++l) {C[k][l]=a^2*h[k]*h[l];}
real u=a*h[k];
A[k]=2*(u-u^2); C[k][k]+=1-2*(u-u^2);
}
real S(real[][] A)
{
real s=0; for(int k=0;k<d;++k) for(int l=0;l<d;++l) s+=A[k][l];
return s;
}
real[][] U=copy(C);
for(int kk=0;kk<19;++kk)
{
int t=2^kk;
write(t);
real la=A[d-1]-a^2*h[d-1]^2;
for(int p=0;p<4;++p)
{
real s=-1,ss=0;
for(int k=0;k<d;++k) {s+=a^2*h[k]^2/(A[k]-la); ss+=a^2*h[k]^2/(A[k]-la)^2;}
la-=s/ss;
}
pair r1=(-3.45888846213541,1.77794875224373), r2=(-0.541111537864589,5.00955180872487),
d1=(0.0960769769474388,0.281952388994648), d2=(-0.762743643614106,-0.350175117369014);
write(S(U));
real s=0; pair ss=(0,0),sss=(0,0);
pair r=-1/r1;
for(int k=0;k<d;++k) {ss+=a*h[k]/(1+t*r*(A[k]-la)); sss+=(a*h[k])^2/(1+t*r*(A[k]-la));}
pair al=ss/(1-t*r*sss);
pair[] Q;
for(int k=0;k<d;++k) {Q[k]=-(2/d1/r1)*((1+t*r*al*a*h[k])/(1+t*r*(A[k]-la)));}
r=-1/r2;
ss=(0,0); sss=(0,0);
for(int k=0;k<d;++k) {ss+=a*h[k]/(1+t*r*(A[k]-la)); sss+=(a*h[k])^2/(1+t*r*(A[k]-la));}
pair al=ss/(1-t*r*sss);
for(int k=0;k<d;++k) {Q[k]-=(2/d2/r2)*((1+t*r*al*a*h[k])/(1+t*r*(A[k]-la)));}
for(int k=0;k<d;++k) s+=Q[k].x^2;
s*=(1-la)^t;
write(s);
write("************");
U*=U;
}
pause();
```
Now the explanations.
To the best of my understanding, you are aiming at finding $\langle(1-V)^t w,w\rangle$ with low *relative* error where $w$ is the vector of all $1$'s and $V=A-h\otimes h$ where $A=\text{diag}(a\_j)$ and $h=(h\_j)$ with $a\_j,h\_j>0$ ($j=1,\dots,d$).
The first observation is that we can factor out $(1-\lambda)$ where $\lambda$ is the least eigenvalue of $V$. It is the $\lambda$ that makes possible to have an equality in the positive definiteness Cauchy-Schwarz inequality $\sum\_j{a\_j-\lambda}x\_j^2\ge\left(\sum\_j h\_jx\_j\right)^2$ which is the least root of the equation
$$
\sum\_j\frac{h\_j^2}{a\_j-\lambda}=1\,.
$$
I'm finding it by Newton iterations (I used 4 in the above program but if you want to go really high in $t$, 6 would be better). Note that it is crucial to get the correct root, so one should start with the initial approximation $a\_{j\_0}-h\_{j\_0}^2$ where $a\_{j\_0}=\min\_j a\_j$. It is tempting to start with $0$, but that will often throw you in between $a\_j$ and result in a wrong root, which will ruin everything.
Now we have $V=\lambda I+V'$ where $V'=A'-h\otimes h$, $A'=\text{diag\,}(a\_j-\lambda)$ is a SPD matrix with the least eigenvalue $0$ and the corresponding unit eigenvector $e\_0$ that has positive entries (proportional to $\frac{h\_j}{a\_j-\lambda}$). This
$$
(I-V)^t=(1-\lambda)^t(1-\tfrac 1{1-\lambda}V')^t\approx (1-\lambda)^t\exp(-\frac t{1-\lambda}V')
$$
and $\approx$ is an upper bound.
So, from now on, I'll remove the primes, denote $a\_j-\lambda$ by $a\_j$, $V'$ by $V$, and $\frac{t}{1-\lambda}$ by $t$ (I didn't divide $t$ by $(1-\lambda)$ in the code above because $\lambda$ was very small in your setting and you do not need it when finding $e^{-tV}$ but if you are after the discrete case, it helps a bit).
Theoretically, $\langle e^{-tV}w,w\rangle$ (BTW, I recommend computing it after this renormalization because the geometric progression $(1-\lambda)^t$ doesn't affect the relative error but can easily result in mantissa overflow) is just $\langle w,e\_0\rangle^2+\sum\_{j\ge 1}e^{-\gamma\_j t}\langle w,e\_j\rangle^2$ where $\gamma\_j$ and $e\_j$ are the remaining eigenvalues and eigenvectors. The idea is to replace it with $\langle F(tV)w,w\rangle$ where $F(t)\ge e^{-t}$ is a good approximation to $e^{-t}$ (we will discuss below how good we want it) and $\langle F(tV) w, w\rangle$ is reasonably quick to compute.
All I really know how to compute quickly is $w(z)=(I+zV)^{-1}w$. This calls for solving the system of equations
$$
w(z)\_i (1+za\_i)-z\langle w(z),h\rangle h\_i=1\,.
$$
Let $\alpha=\langle w(z),h\rangle$. Then $w(z)\_i=\frac{1+z\alpha h\_i}{1+z a\_i}$ and $\alpha$ can be found from
$$
\sum\_i \frac{1+z\alpha h\_i}{1+z a\_i}h\_i=\alpha\,,
$$
i.e., $\alpha=\frac{s\_1}{1-zs\_2}$ where $s\_q=\sum\_i \frac{h\_i^q}{1+z a\_i}$, $q=1,2$.
Thus we can afford *rational functions of not too high degree*. This is the key. We can now use $F(t)=P(t/2)^{-2}$ where $P$ is the partial sum of the Taylor series for $e^t$ (I used $P(t)=1+t+t^2/2+t^3/6+t^4/24$ in the above program because I was aiming at the range of hundreds (Asymptote will just run out of memory if you try to make a $d\times d$ matrix with $d=10^4$ and it is also about 100 times slower than C++, because it is an interpreter, not compiler; for $d\asymp 10^4$ I would rather recommend degree 8).
If $r\_m$ are the (complex!) roots of $P$ (which have to be found once with high precision; I actually found the roots of $P(t/2)$ in the upper half-plane, those are the mysterious $r1$ and $r\_2$ in the code) and $d\_m=P'(r\_m)$, then we have the representation
$$
\frac 1{P(t)}=\sum\_m\frac {d\_m}{t-r\_m}=-\sum\_m \frac 1{r\_md\_m}\frac 1{1+(-1/r\_m)t}=-2\Re {\sum\_m}' \frac 1{r\_md\_m}\frac 1{1+(-1/r\_m)t}
$$
where ${\sum}'$ is taken over the roots in the upper half-plane (so you have to run the above computation just for 2 roots for degree 4, 4 roots for degree 8, etc.).
Thus we get
$$
P(tV/2)^{-1}w=-2\sum'\_m \Re[(r\_md\_m)^{-1}w(-r\_m^{-1})t/2]
$$
which is readily computable as above, and
$\langle P(tV/2)^{-2}w,w\rangle=\|P(tV/2)^{-1}w\|^2$ (one more summation of squares of entries).
Now it is time to discuss precision.
Since $\sum\_{j\ge 0}\langle w,e\_j\rangle^2=\|w\|^2=d$, we have
$$
\langle e^{-tV}w,w\rangle=\sum\_{j\ge 1}\langle w,e\_j\rangle^2\left[\tfrac {\langle w,e\_0\rangle^2}{d-\langle w,e\_0\rangle^2}+e^{-t\gamma\_j}\right]
$$
and
$$
\langle F(t)w,w\rangle-\langle e^{-tV}w,w\rangle=
\sum\_{j\ge 1}(F(t\gamma\_j)-e^{-t\gamma\_j})\langle w,e\_j\rangle^2\,.
$$
Note that all terms are positive, so the relative error in the sum is at most the maximum of the relative errors in the individual terms. Note also that, since $e\_0$ has positive entries, we have $\langle w,e\_0\rangle\ge 1$. Thus we are interested in
$$
E=\max\_{t>0}\frac{F(t)-e^{-t}}{\frac 1{d-1}+e^{-t}}=
max\_{t>0}\frac{P(t/2)^{-2}-e^{-t}}{\frac 1{d-1}+e^{-t}}\,.
$$
This can be investigated theoretically, but I'll just make a small table (the first number is the degree of the polynomial and the second is $d$) so that you can see that for $n=4, d=100$ I have only 10% error *in the worst case scenario* (your data can never go that high) while for $d=10^4$, with $n=4$ I can theoretically go 150% above the truth while $n=8$ gives you guaranteed 5% precision at the expense of 2 extra roots. If you can afford $n=14$, you can go to ten millions range and declare your computations "precise".
*Edit:* The $\lambda$-story.
The estimate $E=E(d,n)$ for the relative error above is *uniform* in all $t>0$, but it is based of the assumption that the lowest eigenvalue $\gamma\_0$ is exactly $0$, so not only the term $\langle w,e\_0\rangle^2e^{-\gamma\_0 t}$ is evaluated exactly by $\langle w,e\_0\rangle^2 F(t)$, but it also stays fixed independently of $t>0$. The relative error bound is based *not* on the idea that $F(t)$ approximates $e^{-t}$ with small relative error on the whole semi-axis (that is impossible and you have a blow-up beyond $t=40$ or so for $n=16$) but on the idea that both $F(t)$ and $e^{-t}$ become negligibly small compared to $\frac 1{d-1}$ when the relative error between them becomes large.
In actual computation, you'll not be able to keep $\gamma\_0$ exactly at $0$ (the computation of $\lambda$ has some error), so it is important to understand what happens if it is small. The answer is that as long as $|\gamma\_0|t<0.01$, say, nothing changes in the bounds; just $1$ gets replaced by $0.99$ in the formula for $E$. However, if this product is large, the whole thing falls apart dramatically because you accumulate a huge relative error in the term that was presumed to be evaluated exactly and the numerator in $\frac 1{d-1}$ will also become $e^{-\gamma\_0 t}$. So, you are theoretically guaranteed to have the relative error as claimed *only* if you use the $\lambda I$ reduction *and* compute $\lambda$ with the error at most $0.01 t^{-1}$. Otherwise you'll be outputting junk for large $t$. On the other hand, the theoretical guarantee (once you just check that its *assumptions* hold) is a sure one, so you do not need to check that its *conclusion* holds directly.
*Moral*: make sure that the error in $\lambda$ is much smaller than $1/t$ and you will not need to check anything else after that, provided that there is no stupid error in the code, which would reveal itself on small data.
```
2 10 0.17245166347402
2 100 0.961455640371778
2 1000 4.28015125191342
2 10000 19.7816887991447
2 100000 101.301563701775
2 1000000 570.64054529331
2 10000000 3464.50750307103
*********
4 10 0.0162756483432251
4 100 0.106015462775259
4 1000 0.429607071204345
4 10000 1.4053481903916
4 100000 4.41442731700713
4 1000000 14.7171384159884
4 10000000 54.2101332254995
*********
6 10 0.00166712645387837
6 100 0.0139026251293181
6 1000 0.0705786452201884
6 10000 0.2484898912299
6 100000 0.708166083384971
6 1000000 1.85420759122078
6 10000000 4.87739303482548
*********
8 10 0.00016984164765101
8 100 0.00166787956056493
8 1000 0.0110935381765316
8 10000 0.0492442082940087
8 100000 0.159219723419999
8 1000000 0.420853472933991
8 10000000 1.00146598253247
*********
10 10 1.73531792069359e-05
10 100 0.000183806352715792
10 1000 0.00151002712819166
10 10000 0.00873278982007405
10 100000 0.0352404205756958
10 1000000 0.107570103457829
10 10000000 0.271655462239004
*********
12 10 1.78880885047298e-06
12 100 1.94620128830683e-05
12 1000 0.000180991851874676
12 10000 0.00131944915308362
12 100000 0.00681706198345839
12 1000000 0.0256053500194781
12 10000000 0.0750096474609068
*********
14 10 1.86072838564969e-07
14 100 2.04069133751507e-06
14 1000 2.00507453637578e-05
14 10000 0.000172432808234548
14 100000 0.00112204219089126
14 1000000 0.0052891118636613
14 10000000 0.0187925172629428
```
| 5 | https://mathoverflow.net/users/1131 | 443030 | 178,723 |
https://mathoverflow.net/questions/443032 | 6 | I am looking for deformations of the 4-sphere with 8-dimensional isometry group, like a 4-dimensional Berger sphere.
| https://mathoverflow.net/users/142151 | Deformations of the 4-sphere with 8-dimensional isometry groups | There cannot be an 8-dimensional group $G$ acting effectively on $S^4$ by Riemannian isometries. The following argument may not be the best, but it explains why this is true. (I will assume that $G$ acts effectively, since, otherwise, we can quotient by the closed subgroup $K\subset G$ that acts trivially, and work with $G/K$, which does act effectively, instead.)
First, such a $G$ cannot act transitively on $S^4$, since the effective transitive Lie group actions on spheres were classified by Borel, and the only effective transitive compact connected group action on $S^4$ is the standard one with $G = \mathrm{SO}(5)$, a 10-dimensional Lie group. [If you don't like relying on Borel's classification, here is a 'low-tech' argument to rule out the transitive case: If a connected compact $d$-dimensional group $G$ acts transitively and effectively on $S^4$, then we can write $S^4 = G/K$, where $K$ is the $G$-stabilizer of a point. This $K$ must be a compact connected subgroup of $\mathrm{SO}(4)$ of dimension $d{-}4$. All the proper connected subgroups of $\mathrm{SO}(4)$ are conjugate to a subgroup of $\mathrm{U}(2)$. So, if $K$ were of dimension less than $6$, then $S^4$ would have a $G$-invariant, almost complex structure. However, $S^4$ does not support an almost complex structure.]
Now look at a generic $G$-orbit $\Sigma$, which will be a smooth submanifold. If it is codimension $1$, then $G$ acts by isometries on $\Sigma$ and the normal subgroup $H\subset G$ that fixes $\Sigma$ pointwise is discrete. However, the dimension of $G/H$ can be at most $6$ (since it acts effectively on a compact Riemannian $3$-manifold), so $G$ has dimension at most $6$. Similarly, if $G$ fixes a point of $S^4$, then $G$ has dimension at most $6$.
If all of the $G$-orbits were of dimension $1$ or of dimension $2$ then this would define a nontrivial splitting of the tangent bundle of $S^4$, which does not exist.
Thus, the $G$-orbits of an $8$-dimensional Lie group acting by isometries on $S^4$ endowed with some Riemannian metric would have dimension either $1$ or $2$ and there would have to be some of each dimension. A 'generic' $G$-orbit $\Sigma$ would be of dimension $2$. If $H\subset G$ is the normal subgroup that fixes $\Sigma$ pointwise, then $H$ has dimension at most $1$ and $G/H$ has dimension at most $3$, so $G$ has dimension at most $4$.
| 11 | https://mathoverflow.net/users/13972 | 443036 | 178,724 |
https://mathoverflow.net/questions/442859 | 3 | Consider a smooth map $f:M\rightarrow N$ between smooth manifolds.
Ehresmann's theorem states that if $f$ is a proper submersion, then it is locally trivializable; in particular, this implies that nearby fibers are homeomorphic. If we weaken the condition of properness, we can still say something of the sort near a compact component of a fiber. For any $x\in M$, let $F\_x$ denote the connected component of $f^{-1}\!\big(f(x)\big)$ containing $x$. Then the proof of Ehresmann's theorem can be adapted to show the following: If $f$ is a submersion and $F\_x$ is compact, then there is an open set $V\ni f(x)$ and a local trivialization $V\times F\_x\rightarrow M$. In particular, this implies that there is an open set $U\ni x$ such that $F\_y$ is compact for all $y\in U$. This can be phrased succinctly by saying that "compactness of fiber components is an open condition" for any submersion $f$. The proof that I have in mind very much relies on $f$ being a submersion, but I have had trouble thinking of an example of any $f$ lacking this property (namely, that compactness of fiber components is an open condition). As such, I am seeking an example of such an $f$ (or if I am wrong in my guess that one might exist, an explanation of non-existence).
| https://mathoverflow.net/users/147463 | When is compactness of fiber components an open condition? | Let $f:X\rightarrow Y$ be a continuous map between locally compact Hausdorff spaces. For $x\in X$ denote by $A\_x:=f^{-1}(f(x))$ the fiber over $f(x)$ and by $C\_x$ the connected component of $A\_x$ containing $x$.
Assume that $C\_{x}$ is compact for some given $x\in X$.
**Claim**: there is an open neighborhood $V\subset X$ of $x$ such that $C\_z$ is compact for all $z\in V$.
Proof of Claim: By applying the Lemma below with $C:=C\_x$ and $A:=A\_x$, we find a compact open neighborhood $N$ of $C\_x$ in $A\_x$. Since $N$ is compact and $A\_x\setminus N$ is closed, there is a relatively compact, open neighborhood $U\subset X$ of $N$ whose closure does not meet $A\_x\setminus N$.
Its boundary $\partial U$ is mapped by $f$ to a compact set in the complement of $f(x)$. Pick $W\subset Y$ open containing $f(x)$ and disjoint from $f(\partial U)$. Let $V:=U\cap f^{-1}(W)$.
If $z\in V$, then $z\in U$ but its connected component $C\_z$ of the fiber $A\_z$ never touches $\partial U$, since $\partial U$ maps to the complement of $W$, while $f(z)\in W$. Thus $C\_z$ is compact as closed subset of $\overline{U}$.
**Lemma**: Let $C$ be a compact connected component of a locally compact Hausdorff topological space $A$. Then $C$ has an open neighborhood $N$ which is compact.
Proof of Lemma: $C$ admits a relatively compact open neighborhood $B\subset A$. Then $C$ is also a connected component of the compact Hausdorff space $\overline{B}$; thus $C$ is also a quasicomponent of $\overline{B}$.
Therefore, for each $b\in \partial B$, there exists a closed and open subset $U\_b$ of $\overline{B}$ in the complement of $b$, which contains $C$. The complements of the $U\_b$'s form an open cover of the compact set $\partial B$.
We can pick a finite subcover corresponding to $b\_1,\ldots b\_k$ and conclude that $N:=U\_{b\_1}\cap\ldots\cap U\_{b\_k}$ is an open and closed subset of $\overline{B}$ with $N\cap \partial B=\emptyset$ (if $\partial B=\emptyset$, then we set $N:=B$). It follows that $N$ is open in $B$ (and thus in $A$) and also compact.
| 3 | https://mathoverflow.net/users/66777 | 443037 | 178,725 |
https://mathoverflow.net/questions/443013 | 2 | Suppose $X \in \mathbb{R}^{n \times d}$ is a random matrix where $n > d$. Given a matrix $A \in \mathbb{R}^{n \times n}$ such that $AX$ is a zero matrix in expectation, i.e., $\mathbb{E}\_{X}[AX] = 0$. Let $\sigma^2$ be the variance of the norm of $AX$, i.e., $\sigma^2:=\mathbb{V}[\lVert AX \rVert^2\_F]$.
Now I would like to study the property of the random matrix $B=X(X^\top X)^{-1}X^\top AX$. Do we also have $\mathbb{E}[B] = 0$ and can we bound $\mathbb{V}[\lVert B \rVert^2\_F]$ by $\sigma^2$?
| https://mathoverflow.net/users/478836 | expectation and variance of the norm of a random matrix | $\newcommand\si\sigma\newcommand\bm[1]{\begin{bmatrix}#1\end{bmatrix}}$No and no: In general, (i) $EB\ne0$ and (ii) we cannot bound $Var\,\|B\|\_F^2$ by $\si^2$.
E.g., let $n=3$, $d=1$,
$$y\_1:=\bm{1\\ -1\\ -1},\quad y\_2:=\bm{-1\\ 1\\ -1},\quad
y\_3:=\bm{-1\\ -1\\ 1},\quad y\_4:=\bm{1\\ 1\\ 1},
$$
$$A:=\bm{1&-1&0\\0&1&0\\0&0&1}.$$
Let $Y$ be a random matrix such that $P(Y=y\_j)=1/4$ for $j=1,2,3,4$. Let
$$X:=A^{-1}Y.$$
Then $EAX=EY=0$ and $\|AX\|\_F^2=\|Y\|\_F^2=3$ almost surely (a.s.), so that
$$\si^2=Var\,\|AX\|\_F^2=0.$$
However,
$$EB=\bm{0\\ 0\\ -1/6}\ne0.$$
Also, the values of $\|B\|\_F^2$ at $Y=y\_1$ and at $Y=y\_4$ are the non-equal numbers $2$ and $8/3$, respectively, so that $Var\,\|B\|\_F^2$ is strictly greater than $0=\si^2$. (In fact, $Var\,\|B\|\_F^2=1/9$.)
---
One might feel some affinity for the OP's conjectures. Indeed,
(i) we have $B=P\_X AX$, where $P\_X$ is the matrix of the orthoprojector onto the column space of the matrix $X$. So, if $P\_X$ did not depend on $X$, the conjectured conclusion $EB=0$ would follow from $EAX=0$ and the linearity of any orthoprojector. Also,
(ii) since $P\_X$ is the matrix of an orthoprojector, the equality $B=P\_X AX$ does imply $\|B\|\_F^2\le\|AX\|\_F^2$. However, as shown above, this will not in general imply that $Var\,\|B\|\_F^2\le Var\,\|AX\|\_F^2$.
| 1 | https://mathoverflow.net/users/36721 | 443045 | 178,727 |
https://mathoverflow.net/questions/443047 | 2 | Let $X, Y$ be smooth projective connected complex varieties of the same pure dimension $d$ and $f : X\to Y$ a finite flat surjective morphism.
Let $\Delta\_X$ be the closed subscheme of $X\times X$ that is the scheme-theoretic image of the diagonal morphism $X\to X\times X$, and $\Delta\_Y$ the scheme-theoretic image of the diagonal morphism $Y\to Y\times Y$.
>
> Do we have $f^{-1}\Delta\_Y=\Delta\_X$ scheme-theoretically? (i.e. $\Delta\_Y\times\_{Y\times Y}(X\times X)=\Delta\_X$)
>
>
>
I'd expect $\Delta\_X$ to be one of the irreducible components of $f^{-1}\Delta\_Y$, but was wondering about an explicit example and whether one could say something about the discrepancy between $f^{-1}\Delta\_Y$ and $\Delta\_X$.
| https://mathoverflow.net/users/497064 | Finite flat pullback of the diagonal | If $f$ is finite flat of degree $d$, then $f \times f \colon X \times X \to Y \times Y$ has degree $d^2$, but $\Delta\_f \colon \Delta\_X \to \Delta\_Y$ has degree $d$. So equality cannot hold scheme-theoretically unless $f$ is an isomorphism.
A fairly explicit case is the finite étale Galois case, where $(f \times f)^{-1}(\Delta\_Y)$ is the disjoint union of the graphs $\Gamma\_\sigma$ of deck transformations $\sigma \colon X \to X$.
If $f$ is only étale but not Galois, then $(f\times f)^{-1}(\Delta\_Y)$ will be smooth and $\Delta\_X$ is still a connected component, but the other components will not map isomorphically onto $X$ under their projections.
If $f$ is generically étale (i.e. separable), then there is a dense open $U \subseteq Y$ above which the above holds, so $(f \times f)^{-1}(\Delta\_Y)$ is still generically smooth (i.e. geometrically reduced). But for each component of the branch divisor $D \subseteq Y$, there will be another irreducible component of $(f \times f)^{-1}(\Delta\_Y)$ intersecting $\Delta\_X$ above $D$.
In the inseparable case, you get situations where $\Delta\_X$ occurs with multiplicity $>1$, but that only happens in positive characteristic.
| 5 | https://mathoverflow.net/users/82179 | 443049 | 178,728 |
https://mathoverflow.net/questions/443016 | 6 | I'm looking to compute normalizers of finite subgroups of $\mathrm{GL}(n, \mathbb{Z})$ and its possible that they are infinite but they are always finitely presented. For $\mathrm{GL}(n, \mathbb{Z})$ the solution is actually described here:
<https://www.ams.org/journals/mcom/1973-27-121/S0025-5718-1973-0333025-7/S0025-5718-1973-0333025-7.pdf>
My question is whether there's a general method for computing finitely presented normalizers when the group and subgroup presentations are known? The subgroup is finite.
| https://mathoverflow.net/users/173855 | Is there a general method for computing finitely generated normalizers? | Here’s a strategy that sometimes works but is likely hopeless in general. Suppose $G$ acts on a complex $K$ with the property that each finite subgroup of $G$ fixes a point. For example, if the complex $K$ happens to enjoy the CAT(0) property, this will be satisfied.
In this situation, suppose $H$ is a finite subgroup. It has a fixed subspace $X \subset K$. Each $g \in G$ sends the fixed-point set $X$ for $H$ to the fixed-point set for $gHg^{-1}$, so if $g$ normalizes $H$, then $g$ actually preserves $X$, and if $X$ has the property that $H$ is its pointwise stabilizer, the converse holds; namely, if $g$ preserves $X$, then $g$ normalizes $H$.
Anyway, there are standard techniques to attempt to compute a presentation of a group from its action on a simply connected complex, so if $X$ enjoys this property, then you might attempt to compute a presentation for $N\_G(H)$ this way.
| 3 | https://mathoverflow.net/users/135175 | 443053 | 178,730 |
https://mathoverflow.net/questions/443024 | 5 |
>
> Let $f\in C^1(\mathbb T)=C^1(\mathbb R/\mathbb Z)$ be a function such that
> $$\hat f(k):=\int\_{\mathbb T}f(x)e^{-2\pi ikx}\,dx=0,\qquad \forall k\in\{-N+1,\cdots,-1,0,1,\cdots, N-1\}.$$
> Do we have $\|f\|\_{L^\infty}\leq \frac CN \|f'\|\_{L^1}$ for some $C>0$ indpendent of $f$ and $N$?
>
>
>
We have $\|f\|\_{L^\infty}\leq \frac1{4N} \|f'\|\_{L^\infty}$, see [this MSE post](https://math.stackexchange.com/questions/4389189/if-the-fourier-coefficient-hatfk-of-f-in-c1-mathbb-t-is-zero-for-all/4395619#4395619) for a clever proof. I just wonder, if we change $\|f'\|\_{L^\infty}$ to $\|f'\|\_{L^1}$, do we have the same inequality?
Maybe the decay $\frac1N$ is too fast to be true. If this is not ture, can we find a $C\_N$ such that
$$\|f\|\_{L^\infty}\leq C\_N \|f'\|\_{L^1} \qquad \text{and }\ \ \ \ \lim\_{N\to\infty}C\_N=0?$$
Note that by the fundamental theorem of calculus we obtain $\|f\|\_{L^\infty}\leq \|f'\|\_{L^1}$. Also, by Cauchy inequality and the Plancherel identity,
\begin{align\*}
\|f\|\_{L^\infty}&\leq\sum\_{|k|\geq N}|\hat f(k)|=\sum\_{|k|\geq N}\frac{\left|\widehat{f'}(k)\right|}{2\pi |k|}\\
&\leq \frac1{2\pi}\left(\sum\_{|k|\geq N}\left|\widehat{f'}(k)\right|^2\right)^{1/2}\left(\sum\_{|k|\geq N}\frac1{|k|^2}\right)^{1/2}\\
&\leq \frac1{2\pi}\left\|f'\right\|\_{L^2}\left(\frac2N\right)^{1/2}=\frac C{\sqrt N}\left\|f'\right\|\_{L^2}.
\end{align\*}
Any help would be appreciated!
| https://mathoverflow.net/users/141451 | If the Fourier coefficient $\hat{f}(k)$ of $f\in C^1(\mathbb T)$ is zero for all $|k|<N$, then $\|f\|_{L^\infty}\leq \frac CN \|f'\|_{L^1}$? | There is no chance. If you could do it, the linear functional $g\mapsto (\int g)(0)$ would have small norm on the corresponding subspace of $L^1$ and, thus, extend to a functional of small norm in the whole $L^1$. Thus, we would have a small in $L^\infty$ function $g$ whose Fourier coefficients are $\frac 1n$ for $|n|>N$, but it would differ only by a trigonometric polynomial from the full series $G(t)=2i\sum\_{n\ge 1}\frac {\sin nt}n$, which has a jump discontinuity at $0$. So $g$ cannot eliminate it, and "Whoosh, fly all our hopes away"...
| 10 | https://mathoverflow.net/users/1131 | 443057 | 178,732 |
https://mathoverflow.net/questions/443060 | 2 | Let $A\subset B$ be an inclusion of $\mathbb{Z}/n\mathbb{Z}$-graded noetherian domains. Is the integral closure of $A$ in $B$ also $\mathbb{Z}/n\mathbb{Z}$-graded?
This is true for the $G$-graded case with $G$ a nice torsion-free integral commutative monoid, see Theorem 2.3.2 of [Huneke-Swanson, Integral Closure of Ideals, Rings, and Modules] for $G=\mathbb{Z}^r\times\mathbb{N}^s$ (even without noetherian condition).
| https://mathoverflow.net/users/3848 | Is the integral closure of a $\mathbb{Z}/n\mathbb{Z}$-graded noetherian domain in a bigger $\mathbb{Z}/n\mathbb{Z}$-graded domain also graded? | Example 2.3.3 in Huneke and Swanson answers this in the negative.
| 2 | https://mathoverflow.net/users/460592 | 443066 | 178,736 |
https://mathoverflow.net/questions/443052 | 10 | Let
$r(B)$ be the number of integers $1 \leq n \leq B$ such that $n = x^2 + y^2$ for some $x, y \in \mathbb{Z}.$
Then it is a known theorem of Landau that
$$
r(B) \sim C \frac{B}{\sqrt{\log B}}
$$
for some $C > 0$. I was wondering if there is a generalization of this result for
slightly more general function $ax^2 + b y^2$ where $a, b$ are positive integers.
Thank you!
| https://mathoverflow.net/users/84272 | A generalisation of theorem of Landau on sum of two squares? | The general result you seek is due to Paul Bernays, who studied under E. Landau in Göttingen and proved in his dissertation (1912) that $$\tag{$\star$}\sum\_{n \le x} b\_Q(n) \sim C\_Q \frac{x}{\sqrt{\log x}}$$
as $x \to \infty$ where $Q(x,y)=ax^2+bxy+cy^2$ is a primitive positive-definite binary quadratic form ($b^2−4ac<0$, $a>0$, $a,b,c \in \mathbb{Z}$) and $b\_Q$ is the indicator function of integers representable as $Q(x,y)$ ($x,y \in \mathbb{Z}$). The constant $C\_Q$ depends only on the discriminant of $Q$.
You want to apply it with $b=0$ (in the above notation); if $a,b$ (in your notation) are not coprime you need to first reduce to the primitive case before applying Bernays’ result.
(E.g. if $Q\_1$ and $Q\_2$ are two binary quadratic form with the same discriminant, it does not necessarily follow that $\sum\_{n \le x} b\_{Q\_1}(n) \sim \sum\_{n \le x} b\_{Q\_2}(n)$ unless you first assume $Q\_i$ are primitive.)
The dissertation is titled “Über die Darstellung von positiven, ganzen Zahlen durch die primitiven, binären quadratischen Formen einer nicht-quadratischen Diskriminante”, and is available [here](https://eudml.org/doc/203258).
Since the dissertation is in German, let me mention that there are two alternative proofs, giving rate of decay for the error term in $(\star)$.
R. W. K. Odoni in “On norms of integers in a full module of an algebraic number field and the distribution of values of binary integral quadratic forms” (Mathematika, Lond. 22, 108–111 (1975)) proved that $(\star)$ holds with a relative saving of $(\log x)^{c\_Q}$ for some $c\_Q>0$.
O. M. Fomenko, in “Distribution of values of Fourier coefficients for modular forms of weight 1” (J. Math. Sci., New York 89, No. 1, 1051–1071 (1998)) gave a third proof, that leads to formulas for $C\_Q$ and $c\_Q$. See section 2 of this paper for a useful overview of other results.
| 25 | https://mathoverflow.net/users/31469 | 443080 | 178,740 |
https://mathoverflow.net/questions/443046 | 6 | $\DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\Sing}{Sing} \DeclareMathOperator{\Top}{Top}
\DeclareMathOperator\sSets{sSets}$Is there a category of “sufficiently nice” topological spaces such that the singular simplicial complex functor $\Sing\_\bullet:\Top \to \sSets$ is fully faithful ?
>
> Is $\Hom\_{\Top}(X,Y)=\Hom\_{\sSets}(\Sing\_\bullet X, \Sing\_\bullet Y)$ for $X$ and $Y$ topological spaces nice enough ?
>
>
>
Is there a good reference for facts like this ?
Note that $X:=\Bbb{Q}$ and $Y:=\Bbb Z$ form a counterexample, as the singular simplicial complex of a totally disconnected space is constant. Thus totally disconnected spaces are not nice enough. See details at [Is the singular simplicial functor full](https://mathoverflow.net/questions/261925/is-the-singular-simplicial-functor-full).
| https://mathoverflow.net/users/494312 | Is the singular simplicial complex functor $\operatorname{Sing}_\bullet:\operatorname{Top} \to \operatorname{sSets}$ fully faithful for nice spaces? | Here is a simple counterexample with $X = Y = \mathbb{R}$:
Send a simplex $\sigma : |\Delta^n| \to \mathbb{R}$ to the affine function $F(\sigma) : |\Delta^n| \to \mathbb{R}$ with the same values at the vertices of $|\Delta^n|$. This is the identity on 0-simplices, so it could only come from the identity map of $\mathbb{R}$, but it's not the identity for $n > 0$.
Your question is implicitly about the counit $|\mathrm{Sing}\,X| \to X$. In general, $|\mathrm{Sing}\,X|$ is much bigger than $X$. The "problem" is that there are not enough maps in the simplex category (i.e. $|{-}| : \Delta \to \mathrm{Top}$ is not fully faithful). That is why, above, the value of $F(\sigma)$ at a non-vertex point $x \in |\Delta^n|$ did not have to have any particular relation to the value of $\sigma$ at $x$.
| 15 | https://mathoverflow.net/users/126667 | 443092 | 178,744 |
https://mathoverflow.net/questions/443093 | 7 | I do not know whether this question (in history of math) is proper for MathOverflow, but I know no other places where it can be asked with a hope to obtain an answer.
Reading the biography of Henry Maurice Sheffer (a famous logician whose name is attributed to the "Sheffer stroke", one of two binary operations that generate all other logical operations) in [Wikipedia](https://en.wikipedia.org/wiki/Henry_M._Sheffer) I learned that ``Sheffer was a Polish Jew born in the western Ukraine''. The same information is repeated in [JewAge](https://www.jewage.org/wiki/en/Article:Henry_M._Sheffer_-_Biography). On the other hand, [MacTutor](https://mathshistory.st-andrews.ac.uk/Biographies/Sheffer/) claims that Sheffer was born 1 September 1882 in Odessa, that time it was in Russian Empire, and not in western Ukraine at all (those times the western Ukraine was under Austrian Empire). The paper "[The known and unknown H.M. Sheffer](https://www.jstor.org/stable/27795017)" of M.Scanland writes simply that Sheffer was born in Ukraine in 1883.
So, what is the truth? Sheffer was born in Odessa (the south of Ukraine) or in the western Ukraine. If in the western Ukraine, then at which town or village? And what is the year of his birth? 1882 or 1883? Who can know this information?
| https://mathoverflow.net/users/61536 | The place and year of birth of Henry Maurice Sheffer | The confusion on the year of birth can be traced back to official records. The 1918 draft registration card of Henry Maurice Sheffer states September 1, 1883 as date of birth. However, the 1904 naturalization record states September 1, 1882. In one more document, a passport application from 1910, Henry Sheffer "solemnly swears" that he was born on September 1, 1882. I would consider that the most authoritative official source (even though it is at odds with the inscription on the tomb stone).
| 11 | https://mathoverflow.net/users/11260 | 443095 | 178,745 |
https://mathoverflow.net/questions/442789 | 5 | Suppose $G$ is a $\mathbb{Q}$-algebraic group (I am interested in the semisimple case) acting rationally on a vector space $V\_\mathbb{Q}$. Let $x \in V\_\mathbb{Q}$ be a non-zero rational vector. Consider the stabilizer group $G\_{x}$ of $x$. This is a closed subgroup in $G$.
Consider a compactly supported continuous $f:V\_{\mathbb{R}}\rightarrow \mathbb{R}$ and then consider the following integral
\begin{equation}
\int\_{G(\mathbb{R}) / G\_{x}(\mathbb{R})} f(gx )dg .
\end{equation}
1. When is this integral well defined for all rational points? That is, when can I perform an integration on the homogeneous space $G(\mathbb{R})/G\_x(\mathbb{R})$ for all rational points. In the semisimple case, this is the same as asking if there any general conditions to guarantee that $G\_x$ will be unimodular for any rational point $x \in V\_\mathbb{Q}$.
2. Once it is well-defined, when is it finite for any $f$?
For example, some very generous conditions are when $G(\mathbb{R})$ acts transitively on non-zero points, or when $G(\mathbb{R}) x$ forms the non-zero points of a subspace in $V\_\mathbb{R}$.
My guess is that such questions must have been considered in representation theory of algebraic groups but I don't really know where to start looking.
| https://mathoverflow.net/users/94546 | Integrating on orbits of algebraic groups | One condition that I came across is that it is sufficient to have $G(\mathbb{C}) \cdot x $ a closed subvariety of $V\_\mathbb{C}$. Then $G(\mathbb{C}) \cdot x$ is a closed affine variety. This guarantees in particular that $G\_x(\mathbb{C})$ must be reductive from Matsushima's criterion which says if $G$ is a connected reductive $\mathbb{Q}$-group and $H \subseteq G$ is a $\mathbb{Q}$-subgroup then $G/H$ is affine if and only if $H$ is reductive.
This also applies if $G(\mathbb{C}) \cdot x$ is just affine but not closed but I don't know if there is a nice way to check this for some $x \in V\_\mathbb{Q}$.
| 1 | https://mathoverflow.net/users/94546 | 443099 | 178,747 |
https://mathoverflow.net/questions/443097 | 6 | Let $X$ be a compact Hausdorff space and $p\in X$ be a non-isolated point. Is it always possible to find a net $(x\_\alpha)\_{\alpha\in (I,\leq)}$ in $X\setminus\{p\}$ converging to $p$ such that $(I,\leq)$ is *totally ordered* (or a regular cardinal, which gives you the same thing)?
I would expect this to be false, but I do not know how to construct a counter-example.
I tried some standard spaces like the $1$-point compactification of an uncountable discrete space (because here the infinite point has no countable basis of neighborhoods) or a product of continuum many copies of $\{0,1\}$, but none of the worked. Probably the Stone-Cech-Compactification of the integers could be a counter-example but I really have no exprerience in working with that space ... Any ideas?
| https://mathoverflow.net/users/153400 | When can we find a net, defined on a totally ordered index set, converging to a non-isolated point in a compact Hausdorff space? | Yes. For every non-isolated $p\in X$, there is a well-ordered net in $X\setminus\{p\}$ converging to $p$ as long as $X$ is compact and Hausdorff. This result was observed in the 1992 paper Convergent free sequences in compact spaces by I. Juhász and Z. Szentmiklóssy, so let me paraphrase their argument. Suppose that $\lambda$ is a limit ordinal and $(A\_\alpha)\_{\alpha<\lambda}$ is a descending sequence of closed subsets of a compact Hausdorff space $X$ with $\bigcap\_{\alpha<\lambda}A\_\alpha=\{p\}$ but where $|A\_\alpha|>1$ for $\alpha<\lambda$. Let $x\_\alpha\in A\_\alpha\setminus\{p\}$ for $\alpha<\lambda$. Then whenever $U$ is a neighborhood of $p$, by compactness, there is some $\alpha<\lambda$ where $x\_\beta\in A\_\beta\subseteq U$ whenever $\lambda>\beta\geq\alpha$. Therefore, $x\_\alpha\rightarrow p$.
We still need to construct the sequence $(A\_\alpha)\_{\alpha<\lambda}$. Let $\lambda$ be the smallest cardinal such that there exists a collection of closed sets $(C\_\alpha)\_{\alpha<\lambda}$ where $\{p\}=\bigcap\_{\alpha<\lambda}C\_\alpha$ but where $p\in (C\_\alpha)^\circ$ for each $\alpha<\lambda$. We can therefore set $A\_\alpha=\bigcap\_{\beta<\alpha}C\_\beta$.
We also observe that some form of compactness is necessary for this argument to work; if $p\in\beta\mathbb{N}\setminus\mathbb{N}$, then there is no well-ordered net in $\mathbb{N}$ that converges to $p$.
**Unique accumulation makes the proof work**
Compact Hausdorff spaces can be characterized as the Hausdorff spaces where each net that accumulates at a unique point $p$ actually converges to the point $p$. This characterization of compact Hausdorff spaces allows us to break our result up into a couple of lemmas.
We say that a filter $\mathcal{F}$ on a topological space $X$ accumulates at the point $p\in X$ if $p\in\bigcap\_{R\in\mathcal{F}}\overline{R}$. We say that a net $(x\_{d})\_{d\in D}$ on the same space $X$ accumulates at the point $p\in X$ the filter generated by $\{\{x\_{e}\mid e\geq d\}\mid d\in D\}$ accumulates at the point $p$. Equivalently, the net $(x\_d)\_{d\in D}$ accumulates at $p$ precisely when $U\cap\{x\_e\mid e\geq d\}\neq\emptyset$ whenever $U$ is a neighborhood of $p$.
Lemma: Let $X$ be a Hausdorff space. Then the following are equivalent:
1. $X$ is compact.
2. If $(x\_d)\_{d\in D}$ is a net that accumulates at the unique point $p\in X$, then $(x\_d)\_{d\in D}$ converges to the point $p.$
Lemma: Let $X$ be a topological space. Let $p\in X$ be a non-isolated point. Suppose that $\{p\}=\bigcap\_{U\in\mathcal{N}(p)}\overline{U}$ where $\mathcal{N}(p)$ is the filter of neighborhoods of $p$ (this is a weakening of regularity). Then there is a cardinal $\lambda$ and well-ordered net $(x\_\alpha)\_{\alpha<\lambda}$ where if $(x\_\alpha)\_{\alpha<\lambda}$ accumulates at a point $q$, then $q=p$.
Proof: Let $\lambda$ be the smallest cardinal such that there exists a transfinite sequence $(U\_\alpha)\_{\alpha<\lambda}$ of neighborhoods of $p$ with $\bigcap\_{\alpha<\lambda}\overline{U\_\alpha}=\{p\}$. Then since $\lambda$ is minimized, the set $\bigcap\_{\beta\leq\alpha}\overline{U\_\beta}\setminus\{p\}$ is non-empty for each $\alpha<\lambda$. Then set $x\_\alpha\in\bigcap\_{\beta\leq\alpha}\overline{U\_\beta}\setminus\{p\}$ for $\alpha<\lambda$. Therefore, $\bigcap\_{\beta<\lambda}\overline{\{x\_\alpha\mid \alpha\geq\beta\}}\subseteq\bigcap\_{\beta<\lambda}\overline{U\_\beta}=\{p\}.$ $\square$
By combining the two above lemmas, we can conclude that for each point $p$ in a compact Hausdorff space $X$, there is a well-ordered net in $X\setminus\{p\}$ converging to $p$.
| 4 | https://mathoverflow.net/users/22277 | 443108 | 178,751 |
https://mathoverflow.net/questions/442370 | 3 | The following post builds on [this post](https://mathoverflow.net/posts/442315/edit); I'll begin by quoting the setting.
---
**Background from Previous Question:**
$\newcommand\SS{P}\newcommand\TT{Q}$Call a Gaussian probability measure $\SS$ on $\mathbb{R}^d$ *isotropic* if its covariance matrix is diagonal with non-vanishing determinant; i.e. $\Sigma\_{i,i}>0$ for $i=1,\dots,d$ and $\Sigma\_{i,j}=0$ whenever $i\neq j$ for each $i,j=1,\dots,d$.
*Note: My definition of "isotropic" includes "the usual isotropic Gaussian measures," which, from my limited understanding, are assumed to have a covariance of the form $\sigma I\_d$ for some $\sigma>0$.*
Let $\mathcal{P}$ the set of *isotropic* Gaussian probability measures on $\mathbb{R}^d$ and let $\mathcal{Q}$ be the set of probability measures on $\mathbb{R}^d$ with Lebesgue density equipped with TV distance.
Consider the *information projection (of I-projection)* defined by
\begin{align}
\pi:Q &\rightarrow \mathcal{P}
\\
\pi(\TT) &:= \operatorname\*{argmin}\_{\SS\in \mathcal{P}}\, D(\SS\parallel\TT)
\end{align}
---
**Definitions:**
Let $\mathcal{B}(\mathbb{R}^d)$ denotes the Borel $\sigma$-algebra on the $d$-dimensional Euclidean space $\mathbb{R}^d$, $\mu$ denote the $d$-dimensional Lebesgue measure, and $\nu$ the standard Gaussian probability measure on $\mathbb{R}^d$. Let $\mathcal{W}\_2$ denote the $2$-Wasserstein distance.
Fix a parameter $R>0$ and let $\mathcal{Q}\_R$ denote the set of Borel probability measures $\mathbb{Q}$ on $\mathbb{R}^d$ which:
* $\mathbb{Q}\ll \mu$,
* $ \frac{d\mathbb{Q}}{d\mu} \in L^2(\mathcal{B}(\mathbb{R}^d),\mu)$,
* $\mathcal{W}\_2(\mathbb{Q},\nu)\le R$.
Is the map $\mathbb{Q}\mapsto \pi(\mathbb{Q})$ is locally-Lipschitz when considered on $\mathcal{Q}\_R$?
| https://mathoverflow.net/users/491352 | Conditions for: (local) lipschitz stability of I-projection | The answer is still no -- for any, however small $R>0$.
Indeed, let $d=1$. Let $g$ be the pdf of $\nu$; that is, $g$ is the standard normal pdf. Let $Q\_h$ be the probability measure with pdf $(1-p)g+pq\_h$, where $q\_h$ is as in [the previous answer](https://mathoverflow.net/a/442343/36721) and $p\in(0,1)$ is a fixed number which is close enough to $0$ so that your condition involving $R$ hold. Then the arguments provided in the previous answer will hold here just as well:
Since $a$ is rather small, $Q\_h$ is somewhat close to the mixture of (i) $\nu$ and (ii) the mixture of the rather narrow normal distributions $N(1,a^2)$ and $N(-1,a^2)$ with slightly unequal weights, $c\_h\,(1+h)$ and $c\_h\,(1-h)$ respectively. So, a minimizer $P\_h$ of the Kullback–Leibler divergence $D(P\parallel Q\_h)$ in $P$ should be sufficiently close to $(1-p)\nu+pN(1,a^2)$ or $(1-p)\nu+pN(-1,a^2)$ depending on whether the small perturbation $h$ is $>0$ or $<0$, respectively. Thus, an infinitesimally small change from, say, $h>0$ to $-h<0$ will result in quite a nonnegligible change from $P\_h\approx (1-p)\nu+pN(1,a^2)$ to $P\_{-h}\approx (1-p)\nu+pN(-1,a^2)$. (If $h=0$, then there will be two minimizers.)
| 2 | https://mathoverflow.net/users/36721 | 443118 | 178,753 |
https://mathoverflow.net/questions/442998 | 16 | I am interested in the low-degree stable homotopy group $\pi\_2^{s}(X)$ of a path-connected space $X$. Using the Atiyah-Hirzebruch spectral sequence, we have the short exact sequence $0\to H\_1(X,\mathbb{Z}/2)\to \pi\_2^{s}(X)\to H\_2(X,\mathbb{Z})\to 0$.
**Question:** Does it always split?
---
**Edit:** I've been working on a solution to my question, and I believe I've made progress. However, I've hit a roadblock in the proof and I'm hoping for some assistance. I will award the best answer to whoever can help me finish my proof or provide insight into my original question.
I think my exact sequence splits canonically! My idea is to show the following dual exact sequence splits canonically:
$$
0\to H^2(X,\mathbb{Q}/\mathbb{Z})\to\mathrm{Hom}\_{\mathbb{Z}}(\pi\_2^s(X),\mathbb{Q}/\mathbb{Z})\stackrel{h}{\to} H^1(X,\mathbb{Z}/2)\to 0,
$$
where I use the canonical isomorphism $\mathrm{Hom}\_{\mathbb{Z}}(H\_1(X,\mathbb{Z}/2),\mathbb{Q}/\mathbb{Z})\cong H^1(X,\mathbb{Z}/2)$.
Next, I define $\phi$ as the generator of $\mathrm{Hom}\_{\mathbb{Z}}(\pi^s\_2(K(\mathbb{Z}/2,1)),\mathbb{Q}/\mathbb{Z})\simeq\mathbb{Z}/2$. For each $f\in H^1(X,\mathbb{Z}/2)\cong[X,K(\mathbb{Z}/2,1)]$, I consider the pullback $f^\*\phi\in\mathrm{Hom}\_{\mathbb{Z}}(\pi^s\_2(X),\mathbb{Q}/\mathbb{Z})$. This defines a map $[\phi]:H^1(X,\mathbb{Z}/2)\mapsto\mathrm{Hom}\_{\mathbb{Z}}(\pi^s\_2(X),\mathbb{Q}/\mathbb{Z})$ that sends $f$ to $f^\*\phi$.
I'm tempted to believe that $[\phi]$ is a homomorphism and $[\phi]\circ h=\mathrm{id}\_{H^1(X,\mathbb{Z}/2)}$, but I cannot come up with a proof. If you have any insights into this, I would greatly appreciate it. Thank you!
---
**Edit2** I’m really happy to see some of you enjoyed this question! Then what’s left for myself is to figure out how my argument fails, especially the naturality part.
| https://mathoverflow.net/users/472749 | The second stable homotopy group | I love this question! I've enjoyed thinking of it. Below, I show why the sequence splits always.
Let $X\to Y=K(H\_1(X,\mathbb{Z}/2),1)$ be the map inducing the identity in $H\_1(-,\mathbb{Z}/2)$. By naturality, we have have a commutative diagram
$$
\begin{array}{cccccccc}
0&\to& H\_1(X,\mathbb{Z}/2)&\to& \pi\_2^{st}(X)&\to&H\_2(X,\mathbb{Z})&\to&0
\\
\downarrow&&\cong\downarrow&&\downarrow&&\downarrow&&\downarrow&&\\
0&\to& H\_1(Y,\mathbb{Z}/2)&\to& \pi\_2^{st}(Y)&\to&H\_2(Y,\mathbb{Z})&\to&0
\end{array}
$$
Hence, if the bottom sequence splits, the upper one too.
Let $A=H\_1(X,\mathbb{Z}/2)$. We have another commutative diagram where all vertical maps are isomorphisms:
$$
\begin{array}{cccccccc}
0&\to& A\otimes \mathbb{Z}/2&\to& A\hat{\otimes} A&\to&A\wedge A&\to&0\\
\downarrow&&\cong\downarrow&&\cong\downarrow&&\cong\downarrow&&\downarrow&&\\
0&\to& H\_1(Y,\mathbb{Z}/2)&\to& \pi\_2^{st}(Y)&\to&H\_2(Y,\mathbb{Z})&\to&0
\end{array}
$$
Here $A\hat{\otimes}A$ and $A\wedge A$ are the quotients of $A\otimes A$ by the relations $a\otimes b+b\otimes a=0$, $a,b\in A$, in the first case, and $a\otimes a=0$, $a\in A$, in the second case. The morphism $A\otimes \mathbb{Z}/2\to A\hat{\otimes} A$ is given by $a\otimes 1\mapsto [a\otimes a]$. This actually holds for any $Y=K(A,1)$ with $A$ abelian. See:
Brown, Ronald; Loday, Jean-Louis Van Kampen theorems for diagrams of spaces. With an appendix by M. Zisman. Topology 26 (1987), no. 3, 311–335. <https://www.sciencedirect.com/science/article/pii/0040938387900048?via%3Dihub>
The top sequence in the second commutative diagram splits because $A=H\_1(X,\mathbb{Z}/2)$, so it is a short exact sequence of $\mathbb{Z}/2$-vector spaces.
We can also proceed without using the Brown-Loday paper, as hinted below by Tom Goodwillie in a comment. It suffices to show that $\pi\_2^{st}(Y)$ is a $\mathbb{Z}/2$-vector space. We have $\pi\_2^{st}(Y)=\pi\_4(\Sigma^2Y)$ since $Y$ is connected. The space $Y$ is a product of copies of $\mathbb{R}P^\infty$. By the splitting of the suspension of a product, $\Sigma^2 Y$ is a wedge of copies of $\Sigma^2(\mathbb{R}P^\infty\wedge\stackrel{n}\cdots\wedge\mathbb{R}P^\infty)$. The latter space is $4$-connected for $n>2$, hence $\pi\_2^{st}(Y)$ is a direct sum of copies of $\pi\_4\Sigma^2\mathbb{R}P^\infty=\mathbb{Z}/2$ and $\pi\_4\Sigma^2(\mathbb{R}P^\infty\wedge\mathbb{R}P^\infty)=\mathbb{Z}/2$.
**P.S.** A previous version of this answer contained a partial proof. The argument was similar, but this final version is even simpler.
| 11 | https://mathoverflow.net/users/12166 | 443120 | 178,754 |
https://mathoverflow.net/questions/443127 | 5 | Let $\mathcal{F}$ be a saturated coherent subsheaf of $T\mathbb{CP}^n$. In particular, $\mathcal{F} \subset T\mathbb{CP}^n$ is a holomorphic vector subbundle outside a subset $Z \subset \mathbb{CP}^n$ of complex codimension $2$. Assume that $c\_1(\mathcal{F}) = 0$.
**Question:** Does there exists a non-zero holomorphic vector field $X \in H^0(T\mathbb{CP}^n)$ which is tangent to $\mathcal{F}$ at every point?
By being tangent I mean that the sheaf generated by the vector field $\mathcal{O}\_{\mathbb{CP}^n} \cdot X$ is contained in $\mathcal{F}$. Another equivalent formulation of the Question (as I believe it follows from Hartog's) is the following.
Does the vector bundle $\mathcal{F}|\_{\mathbb{CP}^n \setminus Z}$ admits a non-zero holomorphic section?
| https://mathoverflow.net/users/195890 | Vector fields tangent to distributions with zero first Chern class | Not in general.
Let $F \in H^0(\mathbb P^3, \mathcal O\_{\mathbb P^3}(3))$ be a general cubic form and let $H \in H^0(\mathbb P^3, \mathcal O\_{\mathbb P^3}(1))$ be a linear form. The kernel of
$\omega = 3FdH - HdF \in H^0(\mathbb P^3, \Omega^1\_{\mathbb P^3}(4))$ is a saturated subsheaf $T\_{\mathcal F}$ of $T\_{\mathbb P^3}$ with trivial determinant and $H^0(\mathbb P^3, T\_{\mathcal F})=0$. Indeed, a non-zero section of $T\_{\mathcal F}$ would determine a non-zero vector field on the cubic surface $\{F=0\}$, but we know that smooth cubic surfaces carry no holomorphic vector field.
| 4 | https://mathoverflow.net/users/605 | 443129 | 178,756 |
https://mathoverflow.net/questions/442777 | 0 | I am looking for existing constructions of [vertex-symmetric graphs](https://en.wikipedia.org/wiki/Vertex-transitive_graph) on $n$ nodes that have a girth at least $g$ and are dense, i.e., have at least $n^{1 + \epsilon}$ edges, where $\epsilon>0$ may depend on $g$: in particular, if $g = O(1)$, then $\epsilon$ must be a constant bounded away from $0$ (w.r.t. $n$).
I am aware of [Cayley graphs](https://en.wikipedia.org/wiki/Cayley_graph), which are vertex-transitive, but I did not find any high-girth constructions that give *dense* graphs.
| https://mathoverflow.net/users/501040 | Dense vertex-symmetric graphs with high girth | If you want an explicit construction (that does not involve randomness), you can use classical constructions of Ramanujan graphs, such as the one from [this paper of Morgenstern](https://doi.org/10.1006/jctb.1994.1054).
Given an odd prime power $q$ and an even integer $d$, Theorem 4.13 in that paper gives you a Cayley graph on $n\sim q^{3d}/2$ vertices, with degree $q+1$, and girth $g$ at least (roughly) $2d$. So the number of edges is (roughly) $n^{1+3/2g}$.
| 1 | https://mathoverflow.net/users/45855 | 443130 | 178,757 |
https://mathoverflow.net/questions/443117 | 21 | *(I asked this question on [Math.SE earlier](https://math.stackexchange.com/questions/4660421/why-do-we-need-canonical-well-orders) but received no response and am therefore moving it here, please note that I realise this question is probably incredibly naïve for the experienced set-theorist, but for an outsider it seems like an important question to ask, and am therefore asking it. )*
Von-Neumann ordinals can be thought of as "canonical" well-orders, indeed every well-order $(W,<)$ has a unique ordinal that is its "order type".
This raises the question of *why* a canonical order is needed, it seems to me that every application of ordinals can be done by using a "large enough" well-ordered set instead that is guaranteed by Hartogs' lemma$^{\*}$, for example, instead of performing a transfinite process on an ordinal, we perform it on the "large enough" well ordered set $(X, <)$ whose existence is guaranteed by Hartogs' lemma. Using this method we can prove the first basic applications of ordinals such as Zorn's Lemma$^{\dagger}$ (see for example [Asaf Karagila's answer](https://math.stackexchange.com/a/4132884/789929) to [Zermelo set theory and Zorn's lemma](https://math.stackexchange.com/questions/4132858/zermelo-set-theory-and-zorns-lemma)).
$^{\*}$ For the purposes of this question let Hartogs' Lemma state: For every set $S$, there exists a well-ordered set $(X, <)$, such that there is no injection from $X\to S$.
$^{\dagger}$ Interestingly popular set-theory books give the exact same argument using ordinals, which are totally superfluous (and need not exist without replacement)!
Remarks/Notes:
* The above observations seem to imply, that the "working mathematician" can totally ignore ordinals, but I am more interested in why they are so important to the working *set*-theorist/logician (given that they literally are a set-theorist's "bread and butter").
* This is not an entirely useless question that does not "affect things" in any way, since ordinals $\ge \omega+\omega$ need not exist in $\mathsf{ZFC}-\mathsf{Replacement}$, and indeed the above method gives a proof of Zorn's lemma in
$\mathsf{ZFC}-\mathsf{Replacement}$. Given that many find replacement dubious, this seems like a strong argument for not using ordinals. (OK, without replacement sets of size larger than $\aleph\_{\omega}$ need not exist, but assuming replacement the above method can easily construct such large sets without the need for ordinals.)
* I suppose one can ask a similar question about cardinal *numbers*: Why do we need cardinal numbers, when we can reason about cardinalities using simply injections and bijections on sets?
* Ordinals seem to give us a "uniform definability" but is that actually useful?
One answer that I have received is "convenience", but if convenience is *the* answer why do we need a formal notion that takes hours to develop when an informal notion seems to suffice (formally)?
| https://mathoverflow.net/users/328267 | Why do we need "canonical" well orders? | This isn't about just any choice for a 'canonical' well-order, but the von Neumann ordinals in particular have nice properties that you don't get just from well-orders. They admit a logically simple definition which enables certain arguments about complexity of definitions to go through.
A relation $R$ being well-founded is a $\Pi\_1$ property—it can be expressed with a single unbounded universal quantifier. (Namely, when saying that *every* nonempty subset of the domain has an $R$-minimal element, that "every" is an unbounded quantifier.) This implies that being well-founded is downward absolute: if you thin down your universe to have fewer sets then $R$ will still be well-founded in the thinner universe. After all, if every nonempty subset of the domain has an $R$-minimal element, that's still true if you throw out some of those subsets.
Is being well-founded upward absolute? If $R$ is well-founded, does it stay well-founded even if we expand our universe by adding new sets? For this, we would like a $\Sigma\_1$ way of characterizing well-foundedness—expressed using a single unbounded *existential* quantifier. Then, if the witnessing object exists it continue to exists in a larger universe, so $R$ stays well-founded.
One attempt at this is: $R$ is well-founded if and only if there's a ranking function $\rho$ from the domain of $R$ to an ordinal. (That is, $x \mathbin{R} y$ iff $\rho(x) < \rho(y)$.) We'd like $\rho$ to still be a ranking function in the larger universe. But the problem is, if being an ordinal is only $\Pi\_1$, then how do we know the codomain of $\rho$ is still an ordinal in the larger universe?
This is where von Neumann ordinals have an advantage over just well-orders. You can express that $\alpha$ is a von Neumann ordinal just by quantifying over the elements of $\alpha$. This means that $\alpha$ is still a von Neumann ordinal if you add new sets. So this gives a $\Sigma\_1$ way to characterize well-foundedness, whence it is upward absolute. Being also downward absolute, we simply say it's absolute.
[Technical caveat here: I'm only talking about so-called transitive extensions/submodels, where both universes are transitive sets or classes. One can talk about other, nonstandard models, where you can add new elements to old sets, but let me set those aside.]
Why care about absoluteness? If you're a set theorist this is clear. A lot of the daily work of the set theorist involves moving up and down between universes, and you want to know what properties carry up or down. But even if you're not a set theorist this can be relevant. A question mathematicians sometimes want to know is whether the axiom of choice was necessary for such and such theorem. For example, maybe you know of the Galvin–Glazer proof of Hindman's theorem using idempotent ultrafilters and you want to know whether you really need choiceful objects like ultrafilters to prove it. An instance of [Shoenfield's absoluteness theorem](https://en.wikipedia.org/wiki/Absoluteness#Shoenfield%27s_absoluteness_theorem) says that Hindman's theorem is actually provable in ZF, so you didn't need AC all along. While a non-logician may be happy to treat the theorem as a blackbox, if one digs into why it works one'll see that the absoluteness of well-foundedness is key in the proof.
Finally, are von Neumann ordinals (or some other nice canonical choice for well-orders) really necessary for this sort of absoluteness result? If you look at some contexts where von Neumann ordinals aren't available then you don't have that well-foundedness is upward absolute. For instance, [this mathoverflow answer](https://mathoverflow.net/a/432772/64676) addresses the case with ZFC - Replacement.
| 25 | https://mathoverflow.net/users/64676 | 443135 | 178,759 |
https://mathoverflow.net/questions/443133 | 2 | For the proof of a certain combinatorial statement on subsets of ${\mathbb F}\_2^n$
in a paper I and several other people are working on, the following statement was helpful.
**Proposition.**
Let $V$ be a vector space over a field $F$, let $I$ be a finite set, and let
$U\_i \subseteq V$ for $i \in I$ be linear subspaces such that
$\dim \sum\_{j \in J} U\_j \ge \#J$ for each subset $J \subseteq I$.
Then there exists a linearly independent family $(v\_i) \in \prod\_{i \in I} U\_i$.
I have a proof for this, but I also have the feeling that this should be known.
So my question is: Has somebody seen this somewhere, and if so, can you give
a reference?
---
**EDIT:** Following up on Fedor Petrov's answer, I found the original
reference:
R. Rado, *A theorem on independence relations*, [Quart. J. Math. Oxford Ser. **13** (1942), 83-89](https://doi.org/10.1093/qmath/os-13.1.83).
The statement I was looking for is Theorem 1 there.
| https://mathoverflow.net/users/21146 | Linearly independent vectors from a family of subspaces | This is exactly Rado theorem on independent transversal (for the corresponding linear matroid).
Strictly speaking, to use it you should deal with finite sets, but a dimension of a vector subspace always equals to a rank of its appropriate finite subset, so there is no problem with it.
| 2 | https://mathoverflow.net/users/4312 | 443138 | 178,761 |
https://mathoverflow.net/questions/443144 | 9 | In a now [classical paper](https://zbmath.org/?q=an:0438.10030), Iwaniec proved the following theorem.
**Theorem.** Let $T \geq 2$, $T^{1/2} < T\_0 \leq T$ and $T \leq t\_1 < t\_2 < \cdots < t\_R \leq 2T$, $t\_{r+1} - t\_r \geq T\_0$. Then
$$
\tag{1}
\sum\_{r=1}^R \int\_{t\_r}^{t\_r+T\_0} \left|\zeta\left(\tfrac{1}{2}+it\right) \right|^4 dt \ll \left( R T\_0 + R^{1/2}T\_0^{-1/2} T\right) T^\varepsilon.
$$
Immediately after the statement, Iwaniec says that this leads to the estimate
$$
\tag{2}
\int\_{T}^{2T} \left|\zeta\left(\tfrac{1}{2}+it\right) \right|^{12} dt \ll T^{2+\varepsilon},
$$
but gives no details of this calculation. Thus my question: **how does (2) follow from (1)?**
| https://mathoverflow.net/users/307675 | Large values of $\zeta(1/2+it)$ from sums of short moments | I wrestled with this question for an annoyingly long time. Now understanding things much better, I'm a bit embarrassed that this took me so long to figure out (such is the process of learning though). I was quite frustrated that I couldn't find the details of this anywhere, aside from some helpful comments in [this paper](https://arxiv.org/abs/math/0512016) of Ivić. To spare others my frustration, I'm posting the answer in a Q&A style, as well as for my own reference.
It suffices to show that if $V > 0$ and $t\_1,\ldots, t\_L \in [T,2T]$ with $t\_{l+1}-t\_l \geq 2$ and $|\zeta(\frac{1}{2}+it\_l)| \geq V$, then
$$
L \ll T^{2+\varepsilon}V^{-12},
$$
for then the estimate for the twelfth moment follows by the argument originally given by Heath-Brown in his paper first proving the estimate (2). Note that we may also assume that $V > T^{1/8+\varepsilon/2}$, since the contribution to the integral (2) of the $t$ with $|\zeta(\frac{1}{2}+it)| \leq T^{1/8+\varepsilon/2}$ is
$$
\leq T^{1+4\varepsilon} \int\_{T}^{2T} \left|\zeta\left(\tfrac{1}{2}+it\right) \right|^{4} dt \ll T^{2+5\varepsilon}
$$
by the standard estimate for the fourth moment of $\zeta$. In particular, we may assume $V^4 > T^{1/2+2\varepsilon}$.
Cover the interval $[T,2T]$ by $R$ disjoint intervals $I\_1,\ldots,I\_{R}$ of length $V^4T^{-2\varepsilon}$ ($> T^{1/2}$) such that each point $t\_l$ lies in some interval $I\_r$. Then
$$
R \leq L \leq \sum\_{r=1}^R \sum\_{t\_l \in I\_r} \frac{\left|\zeta\left(\tfrac{1}{2}+it\_l\right) \right|^4}{V^4}.
$$
At this point, we need that for $t \in [T,2T]$,
$$
\tag{3}
\left|\zeta\left(\tfrac{1}{2}+it\right) \right|^{2k} \ll \log T \int\_{t-1/3}^{t+1/3} \left|\zeta\left(\tfrac{1}{2}+iu\right) \right|^{2k} du,
$$
which can be found in the paper of Ivić linked above. Since the points $t\_l$ are spaced by at least 2,
$$
\sum\_{t\_l \in I\_r} \left|\zeta\left(\tfrac{1}{2}+it\_l\right) \right|^4 \ll \log T \int\_{I\_r'} \left|\zeta\left(\tfrac{1}{2}+it\right) \right|^4 dt,
$$
where $I\_r'$ is the interval $I\_r$ enlarged by $1/3$ on either end.
Applying the estimate (1), we find that
\begin{align\*}
\sum\_{r=1}^R \log T \int\_{I\_r'} \left|\zeta\left(\tfrac{1}{2}+it\right) \right|^4 dt &\ll \left(RV^4 T^{-2\varepsilon} + R^{1/2}V^{-2}T^{1+\varepsilon} \right)T^{\varepsilon} \\
&\ll R T^{-\varepsilon} V^4 + R^{1/2}T^{1+2\varepsilon}V^{-2}.
\end{align\*}
and so in particular
$$
\tag{4}
R \leq L \ll RT^{-\varepsilon} + R^{1/2}T^{1+2\varepsilon}V^{-6}
$$
Since $RT^{-\varepsilon} < cR$ for any constant $c > 0$ and $T$ sufficiently large, we must have
$$
R\ll R^{1/2}T^{1+2\varepsilon}V^{-6}.
$$
Thus
$$
R \ll T^{2+4\varepsilon} V^{-12},
$$
and the required bound for $L$ now follows from (4).
| 10 | https://mathoverflow.net/users/307675 | 443145 | 178,763 |
https://mathoverflow.net/questions/443124 | 2 | I apologize in advance for how vague this request is.
A few weeks ago, I came upon a paper that (if I recall correctly) proves that **the hull of a cut-and-project tiling is a fiber bundle over a torus**. This is *not* the paper by Sadun cited below, but rather it specifically deals with cut-and-project tilings. If I recall correctly, it was published in a journal of physics, or the author was a physicist. I don't recall the name being one of the "big names" in tiling spaces (like Bellisard, Forrest, Hunton, Kellendonk, Putnam, etc.). I am interested mainly in the methods that they used in the paper, and not the result itself. Sadly, the computer on which I found the paper died shortly after I found it and I lost everything related to that paper. I have no author name, no paper title, and very vague search terms to go off of in my hunt.
Edit: it might be that the result I'm thinking of is actually related to something with bounded displacement rather than the hull being a fiber bundle.
Does anyone happen to have any idea what paper this might be?
***References***
*Sadun, Lorenzo; Williams, R. F.*, [**Tiling spaces are Cantor set fiber bundles.**](https://doi.org/10.1017/S0143385702000949), Ergodic Theory Dyn. Syst. 23, No. 1, 307-316 (2003). [ZBL1038.37014](https://zbmath.org/?q=an:1038.37014).
| https://mathoverflow.net/users/165348 | Reference request: Cut-and-project method gives rise to a fiber bundle over the torus | I have found the sought-after paper. It was *Duneau, M., and Christophe O.* "Displacive transformations and quasicrystalline symmetries." Journal de Physique 51.1 (1990): 5-19.
In section 3, they show that a cut and project pattern can be mapped onto a lattice by what they call (but never define) a "modulation".
| 3 | https://mathoverflow.net/users/165348 | 443147 | 178,765 |
https://mathoverflow.net/questions/442538 | 5 | I was asked to post a different question following a wording error on [my previous question](https://mathoverflow.net/questions/442533/jigsaw-puzzle-on-set-family), so here it is.
>
> Given a set family $\mathcal{F}$ of $[n]$ (with certain additional properties), such that every element of $[n]$ lies in exactly $K$ sets in $\mathcal{F}$, what properties of $\mathcal{F}$ are sufficient to guarantee that we can partition $\mathcal{F}$ into $K$ subfamilies $\mathcal{F}\_i$ such that each subfamily is a partition of $[n]$?
>
>
>
What I am looking for is a nontrivial sufficient condition that guarantees the existence of such a partition of $\mathcal{F}$(e.g. in the style of Hall's theorem), and whether any papers have studied these kinds of questions before.
I feel this kind of question is very natural, and greatly appreciate any reference on this type of question.
| https://mathoverflow.net/users/475875 | "JigSaw Puzzle" on Set Family II | This problem is NP-complete, a nice reference is this answer here by András Salamon:
<https://cstheory.stackexchange.com/a/356/419>
If you are interested in results about the complete family, see [Baranyai's theorem](https://en.wikipedia.org/wiki/Baranyai%27s_theorem).
| 2 | https://mathoverflow.net/users/955 | 443163 | 178,769 |
https://mathoverflow.net/questions/443078 | 2 | Suppose $Y \to X$ is a finite morphism of varieties over $\mathbb C$, with $Y$ and $X$ both smooth. Is $Y$ always birational to a smooth hypersurface $Y' \subset X \times \mathbb A^1$, such that the projection $Y' \to X$ induces $\mathbb C(X) \subset \mathbb C(Y)$?
| https://mathoverflow.net/users/5279 | Does every finite map of smooth varieties birationally embed as a smooth hypersurface of 1d trivial bundle over the base? | This is an answer to the question in the original post. The answer is positive in many cases, in particular whenever $X$ is affine. There is nothing to prove if the degree $n$ of the finite morphism equals $1$. Thus, assume that the degree $n$ is at least $2$.
By the Primitive Element Theorem, $Y$ is birational over $X$ to a hypersurface $Y'''$ in $X\times \mathbb{A}^1$ with its first projection to $X$, i.e., $Y'''$ is the zero scheme of a global section $g'''$ of the structure sheaf $$g\_n(a\_n,\dots,a\_m,\dots,a\_1,a\_0;t) = a\_nt^n + \dots + a\_mt^m + \dots + a\_1t+a\_0$$ for $(a\_n,\dots,a\_m,\dots,a\_1,a\_0) \in \mathcal{O}\_X(X)^{\oplus(n+1)}$ with $a\_n$ and $a\_0$ both nonzero, and where $\mathbb{A}^1=\text{Spec}\ \mathbb{Z}[t]$. Of course $Y'''$ is birational over $X$ to the zero scheme $Y''$ of the following *monic* section $g''$, $$g\_n(1,\dots,b\_m,\dots,b\_1,b\_0;t) = a\_n^{n-1}g\_n(a\_n,\dots,a\_m,\dots,a\_1,a\_0;t/a\_n) = $$ $$t^n + \dots + b\_mt^m + \dots + b\_1t+b\_0,$$ where $b\_m$ equals $a\_ma\_n^{n-m-1}$ for $m=0,\dots,n$. The discriminant $D$ of this monic polynomial is a nonzero element of $\mathcal{O}\_X(X)$. Finally, $Y''$ is birational over $X$ to the zero scheme $Y'$ of the following section $g'$, $$g\_n(D^n,\dots,c\_m,\dots,c\_1,c\_0;t) = D^ng\_n(1,\dots,b\_m,\dots,b\_1,b\_0;t+(1/D)),$$ where $c\_m$ is the section $$c\_m=\sum\_{\ell=0}^{n-m}\binom{m+\ell}{\ell}D^{n-\ell}b\_{m+\ell}.$$ By construction, the zero scheme $Y'$ of $g'$ is étale over $X$.
| 4 | https://mathoverflow.net/users/13265 | 443181 | 178,774 |
https://mathoverflow.net/questions/443112 | 3 | Let $R$ be a regular local ring of dimension $n$. Let $i:\text{Flat}\to\text{Fin}$ be the fully faithful inclusion of the category of flat finitely generated type $R$-modules into all finitely generated modules.
* If $n=1$ then $M$ is flat iff $M$ is torsionfree, and the functor $M\mapsto M/(torsion)$ is left adjoint to $i$.
* If $n=2$ then $M$ is flat iff $M$ is reflexive, and the functor $M\mapsto M^{\vee\vee}$ is left adjoint to $i$.
If $n=3$ (or in fact $n\ge 3$), is there a nice characterization of flat modules? Is there a left adjoint to $i$, providing a flatification functor? Or are there reasons why such a left adjoint can't exist?
| https://mathoverflow.net/users/17988 | Flatness over regular local rings of dimension 3 | Fix $R$ (not necessarily local) and $M$. Let us call a map $u:M\to \overline{M}$ a *free hull* if $\overline{M}$ is finite free and for every finite free $F$ the induced map $\mathrm{Hom}(\overline{M},F)\to \mathrm{Hom}({M},F)$ is bijective.
>
> **Claim**: $M$ has a free hull if and only if $M^\vee$ is free, and in that case $\overline{M}=M^{\vee\vee}$ (with the obvious map $v\_M:M\to M^{\vee\vee})$.
>
>
>
Namely, assume $u:M\to \overline{M}$ is a free hull. Then (taking $F=R$) we conclude that $\overline{M}^\vee\cong M^{\vee}$, hence $M^{\vee}$ is free since $\overline{M}$ is.
Conversely, for any $M$, any map $\varphi:M\to F$ with $F$ free (or just reflexive) factors as
$$M\xrightarrow{v\_M} M^{\vee\vee} \xrightarrow{\varphi^{\vee\vee}} F^{\vee\vee}
\xrightarrow{v\_F^{-1}}F$$
and if $M^\vee$ is free so is $M^{\vee\vee}$, whence the claim.
**EDIT:** As Matthieu observes, everything also works with "free" replaced by "locally free".
| 1 | https://mathoverflow.net/users/7666 | 443182 | 178,775 |
https://mathoverflow.net/questions/182440 | 13 | The aperiodic monotile problem asks whether there exists a single tile that every tiling of the plane made with it results non-periodic. What is known about this problem? If this tile exists, how can it be/not be? how could it (not) be constructed?
EDIT: I'm thinking about the most restrictive case where the single tile should be homeomorphic to an euclidean closed disk.
EDIT:
The kind of answer I'm looking for is, for example, [this result](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.84.5262&rep=rep1&type=pdf) implies that if the monotile is to exist, it cannot be convex.
| https://mathoverflow.net/users/49443 | (non-)existence of the aperiodic monotile | This recent preprint claims to find such a tile.
[David Smith, Joseph Samuel Myers, Craig S. Kaplan, Chaim Goodman-Strauss, “An aperiodic monotile”,](https://arxiv.org/abs/2303.10798) (2023-03-20) arXiv:2303.10798
>
> A longstanding open problem asks for an aperiodic monotile, also known as an "einstein": a shape that admits tilings of the plane, but never periodic tilings. We answer this problem for topological disk tiles by exhibiting a continuum of combinatorially equivalent aperiodic polygons. We first show that a representative example, the "hat" polykite, can form clusters called "metatiles", for which substitution rules can be defined. Because the metatiles admit tilings of the plane, so too does the hat. We then prove that generic members of our continuum of polygons are aperiodic, through a new kind of geometric incommensurability argument. Separately, we give a combinatorial, computer-assisted proof that the hat must form hierarchical -- and hence aperiodic -- tilings.
>
>
>
Chapter 1 of this paper gives a historical overview of previous results.
[Akiva Weinberger](https://mathoverflow.net/q/182440/5340) points to this preprint in a comment to this question, though I actually first heard about it from another source and then searched MathOverflow for the right question.
| 22 | https://mathoverflow.net/users/5340 | 443193 | 178,778 |
https://mathoverflow.net/questions/443169 | 28 | I would like to ask the following question:
>
> Is it possible to find two non-constant polynomials $p(x), q(x)$ with integer
> coefficients, such that $\gcd(p(n), q(m))=1$ for every $(n, m)\in
> \mathbb{N}^2$?
>
>
>
If such $p(x), q(x)$ exist, we will call them "completely" coprime, since all of their values will be coprime.
Obviously $p(x), q(x)$ must not have the same root, but this does not seem to help. The problem seems to be quite simple and I suspect that the answer is no, but I was unable to prove this.
| https://mathoverflow.net/users/38851 | Two polynomials which are "completely" coprime | Gro-Tsen's solution is elegant. Here's a more elementary solution that doesn't directly use any algebraic number theory. I'll change the polynomials to $f(x)$ and $g(x)$. We'll assume that
$$\gcd(f(n),g(m))=1 \quad\hbox{for all $m,n\in\mathbb N$ } \qquad(\*)
$$
and derive a contradiction. We may assume, WLOG, that $f(x)$ and $g(x)$ have no common factors in $\mathbb Q[x]$. We first note that if there is a prime $p$ that divides the resultant $\operatorname{Res}\_x(f(x),g(x))$, then $f(x)$ and $g(x)$ have a common root mod $p$, and lifting that root to an integer $n\in\mathbb N$, we find that $p\mid\gcd(f(n),g(n))$, which contradicts $(\*)$. More generally, for every integer $m\in\mathbb N$, if there is a prime $p$ dividing $\operatorname{Res}\_x\bigl(f(x),g(x+m)\bigr)$, then we can repeat the argument to find an $n\in\mathbb N$ such that $p\mid\gcd(f(n),g(n+m))$, again contradicting $(\*)$. To summarize, we've proven that
$$
(\*)\quad\Longrightarrow\quad\operatorname{Res}\_x\bigl(f(x),g(x+m)\bigr)=\pm1\quad\hbox{for every $m\in\mathbb N$.}$$
This means that the polynomial
$$\operatorname{Res}\_x\bigl(f(x),g(x+y)\bigr)^2\in\mathbb Q[y]$$
is identically equal to $1$ for every $y\in\mathbb N$, which means that it's identically equal to $1$ as a polynomial. Taking square roots, for one of the choices $\epsilon\in\{\pm1\}$, we have proven that
$$
\operatorname{Res}\_x\bigl(f(x),g(x+y)\bigr) = \epsilon.
$$
Using standard formal properties of the resultant over any ring (in our case, the ring $\mathbb Q[y]$), there are polynomials $A(x,y),B(x,y)\in\mathbb Q[x,y]$ satisfying
$$
A(x,y)f(x) + B(x,y)g(x+y) = \epsilon \quad\text{in the polynomial ring $\mathbb Q[x,y]$.}
$$
We are assuming that $f(x)$ is not a constant polynomial, so it has a root $x\_0$ living in some finite extension $K$ of $\mathbb Q$. (This admittedly uses a little bit of field theory.) Substituting $x=x\_0$, we find that
$$
B(x\_0,y)g(x\_0+y) = 1 \quad\text{in the polynomial ring $K[y]$.}
$$
It follows that $\deg\_yg(x\_0+y)=1$ in $K[y]$, and thus that $\deg\_xg(x)=1$ in $\mathbb Q[x]$. Reversing the roles of $f$ and $g$ gives $\deg\_xf(x)=1$. We have thus reduced to the case that $f$ and $g$ are non-constant linear polynomials, in which case $f$ and $g$ have roots modulo $p$ for all but finitely many primes $p$.
| 21 | https://mathoverflow.net/users/11926 | 443203 | 178,781 |
https://mathoverflow.net/questions/442937 | 3 | Weighted $K\_5$ have the unique property that their edge set can be interpreted as the disjoint union of their shortest and their longest Hamilton cycle.
That makes $K\_5$ attractive for designing new TSP heuristics; especially extremal $K\_5$ in TSP instances seem interesting for generating starting tours (e.g. the $K\_5$ with the longest shortest tour) for tour expansion heuristics.
Now as there are $\binom{n}{5}$ $K\_5$ in a TSP instance with $n$ vertices, even the slightest speedup in calculating the optimal Hamilton cycle in a weighted $K\_5$ would be desirable, leading to the
>
> **Question:**
>
>
> what is the minimal number of additions and comparisons that are necessary to determine the edge-set of a shortest Hamilton cycle in a weighted $K\_5$?
>
>
>
| https://mathoverflow.net/users/31310 | Fastest algorithm for calculating optimal tours in weighted $K_5$ | OK, I guess I have exhausted my ideas, so I'll make a claim on 14 additions and 8 (actually 7 and a half in a certain sense) comparisons.
Let A,B,C,D,E be the vertices of the pentagon. Consider all cycles that contain the edge AB (there are 6 of them and the other 6 are their complements, so if I find *both* the shortest and the longest of these six cycles, I'll have just to compute the weight of the shortest and the weight of the complement of the longest one and to compare them to reach the final conclusion.
Now compute X=BC+DE, Y=BD+CE, Z=BE+CD and x=DE+AC, y=CE+AD, z=CD+AE (6 additions). Note that to make a cycle, we can combine any small letter with any capital one *provided that they are different*. Thus we have to find the minimum and the maximum of
x+Z, x+Y; y+X, y+Z; z+Y, z+X; Note that we can make the comparison in each pair between ; with the common small letter without knowing the sums to select 3 candidates for the minima and 3 for the maxima and it cannot happen that the splitting will be into the set of the first elements in each pair and the set of the second ones (because the sums are equal). Thus, after 3 comparisons (actually 2 and a half because the first two determine the outcome of third one with probability 1/2 on random data), we will have something like x+Z, y+X, z+X for the minimum candidates and the complementary set for the maximum ones. Now let's find the minimum. Note that y+X and z+X can still be compared without computation, so we make this comparison, then compute the value of the smaller sum, compute the value of the remaining sum and compare the two (2 additions, 2 comparisons). Double that for finding the maximum and we see that we have spent 10 additions and 7 comparisons by the moment. Now the total weight of the minimal cycle can be found by adding the weight of AB (one addition) but the weight of the complement of the maximal one requires 3 additions (we have one of the sums X,Y,Z in it and need to add 3 extra edges. After we compute all that, we make the final comparison and we are done in (14,8) in the worst case scenario, as promised.
I would love to see someone to beat that (especially as far as the comparisons are concerned because the execution of the if(...)... ;else...; construct takes more time than the execution of a=b+c; command), so don't hurry to accept this answer!
| 3 | https://mathoverflow.net/users/1131 | 443209 | 178,784 |
https://mathoverflow.net/questions/443191 | 2 | $\newcommand{\Ex}{\mathbb E}\newcommand{\diff}{\ \mathrm d}$Let
* $(\Omega, \mathcal F, \mathbb P)$ be a probability space.
* $B=(B^1, \ldots, B^N)$ independent one-dimensional Brownian motions.
* $X=(X\_0^1, \ldots, X\_0^N)$ independent real-valued random variables.
* $X$ independent of $B$
* $(P\_t, t\ge0)$ a Markov semi-group.
* $V, F:\mathbb R \to \mathbb R$ smooth functions with compact supports.
* $\*$ the convolution operation.
We consider a particle system
$$
X\_t^i = X\_0^i + \sigma B\_t^i - \int\_0^t V (X\_s^i) \diff s - \int\_0^t F \* \eta\_s (X\_s^i) \diff s,
$$
where
$$
\eta\_s := \bigg( \frac{1}{N} \sum\_{j=1}^N \delta\_{X\_0^j} \bigg ) P\_s.
$$
It is menitoned at page $14$ of [this](https://www.tandfonline.com/doi/abs/10.1080/07362994.2019.1622426) paper that
>
> The particles $X^r, 1 \leq r \leq N$, are not independent but they are independent conditionally to the knowledge of the initial random variables $X\_0^1, \ldots, X\_0^i, \ldots, X\_0^N$.
>
>
>
This statement is very intuitive to me because the dependence of $X^r, 1 \leq r \leq N$ comes from the random measure $\eta\_s$. After conditioning, this measure becomes "non-random". However, I could not see how to establish the above statement rigorously.
>
> Could you elaborate on how to obtain above claim?
>
>
>
My definition of [conditional independence](https://math.stackexchange.com/questions/26042/understanding-conditional-independence-of-two-random-variables-given-a-third-one#comment56059_26055) is
>
> $X,Y$ are conditionally independent given $Z$ if and only if
> $$
> \mathbb P [X \in A, Y \in B | Z] = \mathbb P [X \in A | Z] \cdot \mathbb P [Y \in B | Z]
> \quad \text{a.s.} \quad \forall A,B \in \mathcal B (\mathbb R).
> $$
>
>
>
| https://mathoverflow.net/users/99469 | Interacting particle system: how are the particles independent conditionally to the knowledge of their initial positions? | Let $\{\nu\_x\}\_{x \in \mathbb R^n}$ be the regular conditional probability measures on $\Omega$ associated with $X$, and $\mu\_X$ the law of $X$ on $\mathbb R^n$.
Denote by $E$ the event
$$\left \{ \nu\_X ( \bigcap\_i \, \{X^i \in A\_i\} ) =
\prod\_i \nu\_X (X^i \in A\_i) \, , \, \forall A\_i \in \mathcal B(C[0, T])\right \}.$$
By definition of conditional independence, we need to show that $\mathbb P(E) = 1.$
But for $\mu\_X$-a.e. $x$, the $X^i\_0$ are deterministic under $\nu\_x$, and hence also the process $\eta\_s$. As such, for $\mu\_X$ a.e. $x$, under $\nu\_x$ each $X^i$ is a standard diffusion SDE driven by $B^i$ with non-random coefficients and deterministic initial condition, for which it is known there is a strong solution.
Here independence of $B$ from $X$ guarantees that $B$ is still an independent collection of Brownian motions under each $\nu\_x$.
Thus there exist deterministic maps $\Phi\_{i, x}$ such that $X^i = \Phi\_{i, x} (B\_i)$ for all $i$ almost surely under $\nu\_x$ for $\mu\_X$-a.e. $x$. Independence of the $X^i$ under $\nu\_x$ for $\mu\_X$-a.e $x$ thus follows from that of the $B\_i$.
In other words, denoting by $S$ the set
$$\{ x \in \mathbb R^n \, | \, \nu\_x ( \bigcap\_i \, \{X^i \in A\_i\} ) =
\prod\_i \nu\_x (X^i \in A\_i) \, , \, \forall A\_i \in \mathcal B(C[0, T]) \}$$
we have $\mu\_X (S) = 1$, and so
$$\mathbb P (E) = \int\_{\mathbb R^n} \mathbf 1\_S (X(\omega)) \, d\mathbb P (\omega) = \int\_{\mathbb R^n} \mathbf 1\_S (x) \, d\mu\_X (x) = 1.$$
Thus we conclude conditional independence of the processes $X^i$ as desired.
| 1 | https://mathoverflow.net/users/173490 | 443213 | 178,785 |
https://mathoverflow.net/questions/442626 | 7 | I am struggling to understand lemma 7.20 of the paper *Stack Semantics and the Comparison of Material and Structural Set Theories* by Mike Shulman ([arXiv:1004.3802](https://arxiv.org/abs/1004.3802)). It contains formal sequents of the form $$U \Vdash \ulcorner V\Vdash \phi\urcorner$$ and I do not understand how I can remove the Quine-Corners in a systematic way. I would love to see an explicit example.
So let let $p: V\to U$ be a morphism in the base category $\mathscr S$, let $e:X\to Y$ be a morphism in the slice category $\mathscr S/V$ and let $\phi$ be the statement that $e$ is epi. How do I remove the Quine corner in the expression below systematically?
$$U\Vdash \ulcorner V \Vdash \forall Z:Ob. \forall f,g:Y\to Z. fe =ge \to f = g\urcorner$$
**Edit.** I have moved the question from the proofassistant stackexchange site to this site because I was told to do so. I hope it is welcome here.
I believe that I can do the specific example above by using dependent types. I should get the following statement:
$$U\Vdash \forall v:V. \forall y:Y(v).\exists x:X(v). e(v,x) = y$$
This is just a guess by me. I believe that $\ulcorner V\Vdash \phi\urcorner$ should mean that $\phi$ holds fiberwise, so I tried to express that in the internal language of the base category $\mathscr S$.
When I apply the Kripke-Joyal semantic to the above sequent then I see that it holds if and only if the pullback of $e$ along each morphism into the interpretation of $U,v:V$ is epi. That should be the same thing as
$$U,v:V\Vdash \forall Z:Set.\forall f,g: Y(v)\to Z. f\circ e(v) =g\circ e(v) \to f = g$$
and hence the idempotence works out in that example. Have I done that correctly? If yes, how do I do it systematically?
| https://mathoverflow.net/users/219922 | The idempotence of Mike Shulman's "Stack semantics" | The stack semantics as described in that paper doesn't involve any type theory, only the first-order language of categories. Writing $\ulcorner V \Vdash \phi \urcorner$ means to write out the definition of $\Vdash$, which is by induction over the structure of $\phi$. In your example, to say that
$$V \Vdash \forall Z. \forall f,g:Y\to Z. fe=ge \to f=g$$
means, by definition of $\Vdash$, to say that
>
> For any $p:W\to V$, any $Z \in \mathcal{S}/W$, and any $f,g : p^\*Y \to Z$ over $W$, if $f\circ p^\*e = g \circ p^\*e$ then $f=g$.
>
>
>
(Strictly speaking, we should pass to a new $W,p$ with each universal quantifier or implication, but we can combine them all.) Thus,
$$U \Vdash \ulcorner V \Vdash \forall Z. \forall f,g:Y\to Z. fe=ge \to f=g\urcorner$$
means
$$U \Vdash \forall W.\forall p:W\to V. \forall Z. \forall q:Z\to W. \forall f,g: p^\*Y\to Z. f\circ p^\*e = g\circ p^\*e \to f = g.
$$
only even more verbose, because we have to write out the meaning of "$p^\*$" and the fact that $f,g$ are morphisms over $W$.
| 4 | https://mathoverflow.net/users/49 | 443217 | 178,787 |
https://mathoverflow.net/questions/443224 | 2 | Let $C$ be an algebraic curve over $\mathbb{C}$ and $\omega\_C$ be its canonical bundle. We may assume that $C$ has genus $g\geq2$. Let $x\in C$ be an arbitrary point.
>
> **Question:** What is the image of $H^0(C,\omega\_C-x)$ in the Grassmannian $G(g-1,\, H^0(C,\omega\_C))$ as $x$ varies along $C$?
>
>
>
| https://mathoverflow.net/users/148748 | Image of $H^0(C,\omega_C-x)$ in $G(g-1,H^0(C,\omega_C))$ | First, note that projectification provides a natural isomorphism $$G(g-1, \, H^0(C, \omega\_C)) \simeq \mathbb{G}(g-2, \, \mathbb{P}H^0(C, \omega\_C))=\mathbb{P}H^0(C, \omega\_C)^\*.$$
Next, observe that the canonical map $$\varphi \colon C \to \mathbb{P}H^0(C, \omega\_C)^\*$$ associates to $x \in C$ the hyperplane $\mathbb{P}H^0(C, \, \omega\_C - x) \subset \mathbb{P}H^0(C, \omega\_C)$.
Thus, the locus you are interested in is just the canonical image of $C$.
| 2 | https://mathoverflow.net/users/7460 | 443226 | 178,788 |
https://mathoverflow.net/questions/443212 | 16 | $\DeclareMathOperator{\ap}{ap}$ $\DeclareMathOperator{\rad}{rad}$ The average power of an integer $m$ is given by
$$
\ap(m):=\log\_{\rad(m)}(m)=\frac{\log(m)}{\log(\rad(m))},
$$ where $\rad(m)=\prod\_{p|m}p$.
$\ap(m) \ge 2$ if and only if $\rad(m)^2 \le m$ , which is obviously satisfied by the set of powerful integers $P\_2$ defined by the relation "$p|m$ implies $p^2|m$".
Clearly
$$
\sum\_{P\_2} \frac{1}{m}=\prod\_p(1+\frac{1}{p^2}+\frac{1}{p^3}+ \ldots
) =\prod\_p\left(1+\frac{1}{p(p-1)}\right)=1.9436\ldots
$$ converges. From numerical computation, it seems that we may have $$
\sum\_{ap(m) \ge 2} \frac{1}{m} =2.11\ldots\;.
$$ Does it actually converge? Is there a similar expression as for $P\_2$? Note $\ap(48) \ge 2$ but $48=2^4.3 \notin P\_2$
| https://mathoverflow.net/users/112259 | Does the sum of reciprocal of integers with average power at least two converge? | $\DeclareMathOperator{\ap}{ap}$ $\DeclareMathOperator{\rad}{rad}$
I think so. Fix $\rad(m)=p\_1\ldots p\_k=:P$. Denote by $\Omega$ the set of positive integers with all prime divisors in $\{p\_1,\ldots,p\_k\}$. Then $m=PQ$ where $Q\in \Omega$ and $Q\geqslant P$.
For $s\in (0,1)$ we have
$$
\prod\_{p|m}(1-p^{s-1})^{-1}=\sum\_{n\in \Omega} n^{s-1}\geqslant P^s\sum\_{n\in \Omega,n\geqslant P}n^{-1}.
$$
Thus
$$
\sum\_{n\in \Omega,n\geqslant P}n^{-1}\leqslant \prod\_{p|m}p^{-s}(1-p^{s-1})^{-1}=\prod\_{p|m}(p^s-p^{2s-1})^{-1}.
$$
For $s=1/2$ the right hand side reads as $\prod(p^{1/2}-1)^{-1}$. Thus
$$
\sum\_{m:\ap(m)\geqslant 2} m^{-1}\leqslant \sum\_{p\_1,\ldots,p\_k}\prod\_{i=1}^k\frac1{p\_i(\sqrt{p\_i}-1)}=\prod\_p\left(1+\frac1{p(\sqrt{p}-1)}\right)<\infty.
$$
| 19 | https://mathoverflow.net/users/4312 | 443228 | 178,790 |
https://mathoverflow.net/questions/442971 | 3 | Does positive set theory [$\sf GPK^+\_\infty$](https://en.wikipedia.org/wiki/Positive_set_theory#Axioms) prove the existence of a set $K= \{x \mid x \text { is von Neumann ordinal } \lor x=K\}$
| https://mathoverflow.net/users/95347 | Does positive set theory prove the existence of a set of all ordinals and itself? | I came to know that this question had been answered to the positive by the founder of $\sf GPK^+\_\infty$ himself at an article of him titled: [Inconsistency of GPK + AFA](https://onlinelibrary.wiley.com/doi/10.1002/malq.19960420109). At Journal Math.Log.Quart 1996: vol. 42, issue 1, page 107
| 3 | https://mathoverflow.net/users/95347 | 443234 | 178,791 |
https://mathoverflow.net/questions/443031 | 14 | Question is as mentioned in the title:
Are there any introductory notes on deformation theory that are easier to read for differential geometers?
I am learning about differential graded Lie algebras (and $L\_\infty$-algebras). I am aware of Marco Manetti's notes [Deformation theory via differential graded Lie algebras](https://arxiv.org/abs/math/0507284).
Are there any other notes that are easier to read for people who are trained in differential geometry?
I know some commutative algebra, for example definition (and few properties) of Noetherian, Artinian rings. I can recall (or read) some more commutative algebra if required.
I read from the paper [From Lie Theory to Deformation Theory and Quantization](https://arxiv.org/abs/0704.2213) that
>
> "Deformation Theory is a natural generalization of Lie Theory, from Lie groups and their linearization, Lie algebras, to differential graded Lie algebras and their higher order deformations, quantum groups."
>
>
>
So, this gave some hope that there would be some notes from the point of view of Lie theory/differential geometry.
| https://mathoverflow.net/users/118688 | (An introduction to) deformation theory (written) for differential geometers | A classical subject in deformation theory with a differential-geometric flavour is that of deformation quantization. There are several good introductory references I can recommend, all of which introduce the Maurer–Cartan formalism for DG Lie algebras in deformation theory:
* *Cattaneo, Alberto S.*, Formality and star products (lecture notes taken by D. Indelicato), Gutt, Simone (ed.) et al., Poisson geometry, deformation quantisation and group representations. Cambridge: Cambridge University Press (ISBN 0-521-61505-4/pbk). London Mathematical Society Lecture Note Series 323, 79-144 (2005). [ZBL1077.53074](https://zbmath.org/?q=an:1077.53074).
* [some later chapters in French] *Cattaneo, Alberto; Keller, Bernhard; Torossian, Charles; Bruguières, Alain*, Déformation, quantification, théorie de Lie, Panoramas et Synthèses 20. Paris: Société Mathématique de France (ISBN 2-85629-183-X/pbk). vii, 186 p. (2005). [ZBL1093.53095](https://zbmath.org/?q=an:1093.53095).
* [in German] *Waldmann, Stefan*, [**Poisson-Geometrie und Deformationsquantisierung**](https://doi.org/10.1007/978-3-540-72518-3), Berlin: Springer (ISBN 978-3-540-72517-6/pbk). xii, 612 p. (2007). [ZBL1139.53001](https://zbmath.org/?q=an:1139.53001).
* *Esposito, Chiara*, [**Formality theory. From Poisson structures to deformation quantization**](https://doi.org/10.1007/978-3-319-09290-4), SpringerBriefs in Mathematical Physics 2. Cham: Springer (ISBN 978-3-319-09289-8/pbk; 978-3-319-09290-4/ebook). xii, 90 p. (2015). [ZBL1301.81003](https://zbmath.org/?q=an:1301.81003).
Another reference for deformation quantization, taking a Lie-theoretic point of view (but whose prerequisites are only "[basic linear algebra and differential geometry](https://ems.press/books/elm/93)"), is
* *Calaque, Damien; Rossi, Carlo A.*, [**Lectures on Duflo isomorphisms in Lie algebra and complex geometry**](https://doi.org/10.4171/096), EMS Series of Lectures in Mathematics. Zürich: European Mathematical Society (EMS) (ISBN 978-3-03719-096-8/pbk). viii, 106 p. (2011). [ZBL1220.53006](https://zbmath.org/?q=an:1220.53006).
Besides Manetti's new book
* *Manetti, Marco*, [**Lie methods in deformation theory**](https://doi.org/10.1007/978-981-19-1185-9), Springer Monographs in Mathematics. Singapore: Springer (ISBN 978-981-19-1184-2). xii, 574 p. (2022). [ZBL07529473](https://zbmath.org/?q=an:07529473).
which centers around deformations of compact complex manifolds, there is also a freely available shorter "book" (a 183-page article) by Manetti
* *Manetti, Marco*, [**Lectures on deformations of complex manifolds. Deformations from differential graded viewpoint**](https://arxiv.org/abs/math/0507286), Rend. Mat. Appl., VII. Ser. 24, No. 1, 1-183 (2004). [ZBL1066.58010](https://zbmath.org/?q=an:1066.58010) [PDF](https://www1.mat.uniroma1.it/ricerca/rendiconti/visualizza-volume.cgi?anno=2004&volume=24&issue=1).
where you can also find differential forms in the main examples.
For the passage from the DG Lie algebra / L$\_\infty$ algebra viewpoint to derived deformation theory see [this question](https://mathoverflow.net/q/392885/163420).
| 6 | https://mathoverflow.net/users/163420 | 443237 | 178,792 |
https://mathoverflow.net/questions/443231 | 7 | Let $G$ be a finitely presented simple group. By Kuznetsov (1958), $G$ has decidable word problem. However, by Scott [1], $G$ may have undecidable conjugacy problem. Is anything known about other decision problems for $G$, in generality? I am particularly interested in the question of the subgroup membership problem, i.e. the problem of, given a set of words $w\_1, \dots, w\_k$ and a word $w$, deciding whether or not $w$ belongs to the subgroup $\langle w\_1, \dots, w\_k \rangle$ of $G$. This is undecidable already for the direct product $H \times H$ of any group $H$ admitting a finitely presented quotient with undecidable word problem (in particular, we can take $H$ to be a free group of rank $2$), but such direct products are of course very far from being simple.
${}$
[1] *Scott, Elizabeth A.*, [**A finitely presented simple group with unsolvable conjugacy problem**](https://doi.org/10.1016/0021-8693(84)90174-1), J. Algebra 90, 333-353 (1984). [ZBL0544.20029](https://zbmath.org/?q=an:0544.20029).
| https://mathoverflow.net/users/120914 | Subgroup membership problem in simple groups | As another example, the problem of computing the order of an element of the finitely presented simple Brin–Thompson group $2V$ is undecidable by
*Belk, James; Bleak, Collin*, [**Some undecidability results for asynchronous transducers and the Brin-Thompson group (2V)**](https://doi.org/10.1090/tran/6963), Trans. Am. Math. Soc. 369, No. 5, 3157-3172 (2017). [ZBL1364.20015](https://zbmath.org/?q=an:1364.20015).
| 9 | https://mathoverflow.net/users/24447 | 443241 | 178,794 |
https://mathoverflow.net/questions/443235 | 2 | Let $\mu,\nu$ be finite Borel measures on $\mathbb R$.
Assume that there is an open interval $(a,b)$ on which the Laplace transforms exist and coincide:
$$
\int\_{-\infty}^\infty e^{-tx}\,d\mu(x) = \int\_{-\infty}^\infty e^{-tx}\,d\nu(x) <\infty
$$
for all $t\in(a,b)$.
Does this imply that $\mu = \nu$?
| https://mathoverflow.net/users/485160 | Injectivity of two sided Laplace transform | I suppose that $\mu$ and $\nu$ are real measures and "exist"
means absolute convergence. Then both transforms can be extended to bounded analytic functions in the strip
$\{ t+i\sigma:a<t<b\}$ and coincide in this trip by uniqueness theorem for analytic functions. But then the restrictions on the line $t=t\_0$ for some $t\_0\in(a,b)$
can be considered as Fourier transforms of $e^{-t\_0x}d\mu(x)$
and $e^{-t\_0x}\nu(x)$, therefore by the uniqueness theorem of Fourier transform, these measures coincide, and thus $\mu=\nu$.
| 2 | https://mathoverflow.net/users/25510 | 443243 | 178,795 |
https://mathoverflow.net/questions/443247 | -1 | In a German science journal I read that PH Schoute asked this in the Educational Times around 1900:
What is the sum of
$$S = \frac{1}{1}+\frac{1}{2+3}+\frac{1}{4+5+6}+\frac{1}{7+8+9+10} +\cdots\text{?}$$
I don't know if an answer has been found. Anyone?
| https://mathoverflow.net/users/501420 | What's this infinite sum? | You can have a very good approximation of the partial sums writing first
$$\frac2{n(n^2+1)}=\frac 2{n(n+i)(n-i)}=\frac{2}{n}-\frac{1}{n-i}-\frac{1}{n+i}.$$
Using generalized harmonic numbers
$$S\_p=\sum\_{n=1}^p\frac2{n(n^2+1)}=2 H\_p-H\_{p-i}-H\_{p+i}+H\_i+H\_{-i}.$$
Using asymptotics
$$S\_p=\left(H\_{-i}+H\_i\right)-\frac{1}{p^2}+\frac{1}
{p^3}-\frac{1}{p^5}+\frac{2}{3
p^6}+\frac{1}{p^7}+O\left(\frac{1}{p^8}\right).$$ As @Robert Israel wrote in [comments](https://mathoverflow.net/questions/443247/whats-this-infinite-sum#comment1144169_443247)
$$\left(H\_{-i}+H\_i\right)=2 \gamma + 2 \text{Re}(\Psi(i)).$$
| 1 | https://mathoverflow.net/users/42185 | 443249 | 178,797 |
https://mathoverflow.net/questions/443245 | 7 | On p. 49 in Gould's book Combinatorial Identities, the author states that the sum $$\sum\_{k=0}^{n-1}(-1)^k\binom{n}{k}\binom{2n}{2k}^{-1}$$ "... arises naturally in a statistical problem; it amounts to the evaluation of the moments of a certain distribution". Does someone know what exactly Gould is referring to? I am looking for references...
| https://mathoverflow.net/users/159851 | A reference for a sum found in Gould's Combinatorial Identities book | $\renewcommand{\b}{\binom}\renewcommand{\B}{\text{B}}$As noted in [Carlo Beenakker's comment](https://mathoverflow.net/questions/443245/a-reference-for-a-sum-found-in-goulds-combinatorial-identities-book#comment1144176_443245),
\begin{equation\*}
s\_n:=\sum\_{k=0}^{n-1}(-1)^k\b nk\b{2n}{2k}^{-1}=\frac{2 n+1-(-1)^n}{2 (n+1)}. \tag{1}\label{1}
\end{equation\*}
The immediately related formula
\begin{equation\*}
\sum\_{k=0}^n(-1)^k\b nk\b{2n}{2k}^{-1}=\frac{[1+(-1)^n](2n+1)}{2 (n+1)}
\end{equation\*}
(with the upper summation limit $n$)
is formula 4.2.8.5 in Vol.1 of the handbook by Prudnikov, Brychkov, and Marichev (in Russian).
It is immediate from \eqref{1} that $(s\_n)$ is not the sequence of moments of any probability distribution. Indeed, otherwise the sequence $(s\_{2n})$ would be convex, which it is not. So, the statement "it amounts to the evaluation of the moments of a certain distribution" cannot be true, unless "amounts to" is understood as something like "has something to do with".
However, it may be of help to present a (quite standard) derivation of \eqref{1}. It is based on the following representation of the reciprocals of the binomial coefficients in terms of the beta function (which of course has to do with the beta distribution):
\begin{equation\*}
\b sr^{-1}=r\B(r,s-r+1) \tag{2}\label{2}
\end{equation\*}
for integers $r$ and $s$ such that $0<r\le s$,
where $\B(a,b):=\int\_0^1 dx\, x^{a-1}(1-x)^{b-1}$.
By the definition of $s\_n$ in \eqref{1} and identity \eqref{2}, we get
\begin{equation\*}
\begin{aligned}
s\_n&=1+\sum\_{k=1}^{n-1}(-1)^k\b nk\b{2n}{2k}^{-1} \\
&=1+\int\_0^1 dx\, \sum\_{k=1}^n(-1)^k\b nk 2k x^{2k-1}(1-x)^{2n-2k} \\
&=1-2n\int\_0^1 dx\,[(-1)^n x^{2n-1}+x(1-2x)^{n-1}] \\
&=\frac{2 n+1-(-1)^n}{2 (n+1)}.\quad\Box
\end{aligned}
\end{equation\*}
One may note that the latter integral looks something like an expression for the $n$th moment of a probability distribution.
| 9 | https://mathoverflow.net/users/36721 | 443255 | 178,799 |
https://mathoverflow.net/questions/443262 | 3 | I don't know how to solve part (a) of exercise 1.5.8 in Robin Hartshorne's book Deformation Theory 1.5.8 (page 42):
>
> Consider the Hilbert scheme of zero-dimensional closed subschemes
> of $\mathbb{P}^4\_k$ of length $8$, the field $k$ is algebraically closed. There is one component of dimension $32$
> that has a nonsingular open subset corresponding to sets of eight
> distinct points. We will exhibit another component containing a nonsingular open
> subset of dimension $25$.
>
>
>
>
> (a) Let $R=k[x,y,z,w]$, let $\mathfrak{m}$ be a maximal ideal in
> this ring, and let $I=V+\mathfrak{m}^3$, where
> $V$ is a $7$-dimensional subvector space of
> $\mathfrak{m}^2/\mathfrak{m}^3$. Let $B=R/I$, and
> let $Z= \operatorname{Spec}(B)$ be the associated closed subscheme of
> $\mathbb{A}^4 \subset \mathbb{P}^4$.
> Show that the set of all such $Z$, as the point of its support
> ranges over $\mathbb{P}^4$, forms an irreducible $25$-dimensional
> subset of the
> Hilbert scheme $H=\operatorname{Hilb}^8(\mathbb{P}^4)$.
>
>
>
How can I show that this subscheme of the Hilbert scheme is irreducible? I have no idea even how to start. Is there a general strategy how to deal with that kind of questions to show that certain subscheme of a moduli space is irreducible? Clearly it suffice to construct an irreducible open dense subscheme sitting inside it, but I not see how to manage it in this exercise.
| https://mathoverflow.net/users/501436 | Exercise 1.5.8 from Robin Hartshorne's Deformation Theory | As the comment also indicates, consider the Grassmannian $ Gr\_3(\operatorname{Sym}^2 \Omega\_{\mathbb{P}^4} ) $ where the vector bundle $ \operatorname{Sym}^2 \Omega\_{\mathbb{P}^4} $ has fiber $ m\_p^2/m\_p^3 $ over a point $ p \in \mathbb{P}^4 $. Then there is a morphism $ Gr\_3(\operatorname{Sym}^2 \Omega\_{\mathbb{P}^4} ) \rightarrow Hilb^8(\mathbb{P}^4) $ and it suffices to give it by specifying it on $ \mathbb{C} $-valued points. By the universal property, a point in $ Gr\_3(\operatorname{Sym}^2 \Omega\_{\mathbb{P}^4} ) ( \mathbb{C} ) $ is given by a point in $ \mathbb{P}^4 $ along with a $ 3 $-dimensional quotient $ \operatorname{Sym}^2 \Omega\_{\mathbb{P}^4}|\_p = m\_p^2 / m\_p^3 \rightarrow Q \rightarrow 0 $, or equivalently a $ 7 $-dimensional subspace, which in turn gives the appropriate length $ 8 $ subscheme of $ \mathbb{P}^4 $.
It only remains to see that $Gr\_3(\operatorname{Sym}^2 \Omega\_{\mathbb{P}^4} ) $ is $ 25 $-dimensional and irreducible.
| 2 | https://mathoverflow.net/users/152391 | 443270 | 178,803 |
https://mathoverflow.net/questions/443264 | 2 | As a term of a Serre spectral sequence, I would like to compute the cohomology group with compact support $H^\*\_c(\mathcal{M}\_{1,[2]},\mathbb{V}\_1)$ of the moduli space of genus $1$ curves with $2$ unordered marked points, with values in $\mathbb{V}\_1$, the standard representation of $SL\_2(\mathbb{Z})$, coming from the cohomology group $H^1(E,\mathbb{Q})$, where $E$ is an elliptic curve.
I know that $H^\*\_c(\mathcal{M}\_{1,1},\mathbb{V}\_1)=0$, and I would like to try to use this result to obtain a similar one for the case of $2$ unordered marked points.
My attempt so far consists in using mapping class groups, the Birman exact sequence
\begin{equation\*}
1\to \pi\_1(S\_{1,1})\to \text{PMod}(S\_{1,2})\to \text{PMod}(S\_{1,1})\to 1
\end{equation\*}
and the Hochschild-Serre spectral sequence of group cohomology, to obtain $H^\*(\mathcal{M}\_{1,2},\mathbb{V}\_1)$ (which I believe to be identically $0$), using the fact that the action of the fibre $\pi\_1(S\_{1,1})\cong \mathbb{Z}\*\mathbb{Z}$ on the local system is trivial, as $\mathbb{V}\_1$ is a representation of $\text{PMod}(S\_{1,1})\cong SL\_2(\mathbb{Z})$.
Since I am not used to mapping class groups and group cohomology, I have, now, two questions:
1. Is the reasoning so far correct?
2. Assuming it is, this takes care only of the moduli stack $\mathcal{M}\_{1,2}$ with ordered points. How do I obtain the cohomology of the unordered one? I think that would consist in getting rid of the action of the elliptic involution, exchanging the marked points, but it is not clear to me how to effectively do it.
To be coherent with other results I have, I would hope this cohomology groups not to be identically $0$.
I would appreciate any comment on my problem, or any helpful reference for the computation of $H^\*\_c(\mathcal{M}\_{1,[2]},\mathbb{V}\_1)$, also with other methods.
| https://mathoverflow.net/users/477884 | Computation of $H^*_c(\mathcal{M}_{1,[2]},\mathbb{V}_1)$ | Your approach is good but you must have messed something up with the Hochschild-Serre spectral sequence.
I prefer the sheafy approach over thinking about the Hochschild-Serre spectral sequence, so let me explain it in these terms. So we have the forgetful map $\pi : M\_{1,2} \to M\_{1,1}$, and we may write
$$ H^\bullet\_c(M\_{1,2},V) = H^\bullet\_c(M\_{1,1},R\pi\_!V ).$$
Now $V=\pi^\ast V$ is actually pulled back from $M\_{1,1}$ and we have $$R\pi\_! \pi^\ast V \cong (R\pi\_!\mathbf Q) \otimes V \cong (V \otimes V)[-1] \oplus V(-1)[-2]$$
where in the first step we used the projection formula and in the second we used a section to split $R\pi\_!\mathbf Q$ in the derived category. Representation theory of $\mathrm{SL}(2)$ shows that $V \otimes V \cong V\_2 \oplus \mathbf Q(-1)$, where $V\_2$ is the second symmetric power of $V$ and $\mathbf Q(-1)$ is a Tate twist of the trivial representation. From the Eichler-Shimura isomorphism we know $H^\bullet\_c(M\_{1,1},-)$ with coefficients taken in all of these summands: $H^2\_c(M\_{1,1},\mathbf Q(-1)) \cong \mathbf Q(-2)$, $H^1\_c(M\_{1,1},V\_2)\cong \mathbf Q(0)$, all other cohomologies vanish. So $H^\bullet\_c(M\_{1,2},V)$ is $\mathbf Q(-2)$ in degree $3$, $\mathbf Q(0)$ in degree $2$, and vanishes otherwise.
To deal with unordered points we use that $H^\bullet\_c(M\_{1,[2]},V) = H^\bullet\_c(M\_{1,2},V)^{\Sigma\_2}$, where $\Sigma\_2$ acts by permuting the markings. Unfortunately the computation in the previous paragraph isn't immediately $\Sigma\_2$-equivariant. But in fact $\Sigma\_2$ acts on $M\_{1,2}$ preserving the map $\pi$; it acts by multiplication by $-1$ on the universal elliptic curve. To see this we consider a curve with two marked points $(x,y)$. We can translate to make the first coordinate the origin, and our markings are then $(0,y-x)$. Clearly if we had swapped $x$ and $y$ we would have gotten $x-y$ instead. Inversion in $V$ acts as the identity on $V \otimes V$, so $H^\bullet\_c(M\_{1,[2]},V) = H^\bullet\_c(M\_{1,2},V)$.
Hope this is coherent with other results you have!
| 7 | https://mathoverflow.net/users/1310 | 443274 | 178,804 |
https://mathoverflow.net/questions/443280 | 1 | [Can move to cs or tcs stackexchange if thats a better home]
I remember back around 2016 DIMACS used to host a list of problem instances of various famous problems in the NP-complete class and harder classes. Usually "problem instance data", "solution" and sorted by difficulty for testing new algorithms/heuristics against.
After quite a bit of googling I couldn't find that page so I was curious where one can find known "hard" instances to play around with.
Do any research groups still host such files? (The DIMACS data was nice because it included very easy to very difficult instances and the entire range between).
| https://mathoverflow.net/users/46536 | Where to find hard instances of subsetsum and other famous np-complete problems for testing heuristics against? | Here's the DIMACS Implementation Challenges page:
<http://dimacs.rutgers.edu/programs/challenge/>
| 3 | https://mathoverflow.net/users/141766 | 443281 | 178,806 |
https://mathoverflow.net/questions/443251 | 4 | To be more precise, suppose that $M$ is a model of ZF, for simplicity (or tactility) a set in some larger model $V$ of ZFC+Con(ZF), and suppose that $M\vDash``\alpha$ is an ordinal$"$. Must there be a model $N$ of ZFC (appearing in $V$) such that $\alpha\in N$ and an isomorphism of the partial structures given by the transitive closures of $\alpha$ in $M$ and $N$?
If yes, then can we also assume that $M$ and $N$ agree on cardinality for ordinals ${\leq}\alpha$?
| https://mathoverflow.net/users/478588 | Does every ordinal appearing in a model of ZF appear in a model of ZFC? | As Noah mentioned in the comments, the answer to the first question is "yes".
The answer to the second question is consistently "no", assuming a little consistency: if there is a transitive model of ZF in $V$, then there will be transitive models $M$ of ZF in $V$ and ordinals $\alpha$ of $M$ for which the extra requirement fails. For let $P$ be a countable transitive model of ZF of minimal height. Let $Q=L^P$, so $Q$ models ZFC and has the same ordinals as does $P$ (and $Q$ is also countable). Let $\lambda=\aleph\_\omega^Q$. So $Q\models$ "$\lambda$ is singular". Let $G$ be $(Q,\mathrm{Coll}(\omega,{<\lambda}))$-generic. Let $\mathbb{R}^\*$ be the resulting symmetric set of reals, i.e. $\mathbb{R}^\*=\bigcup\_{\alpha<\lambda}\mathbb{R}\cap Q[G\upharpoonright\alpha]$. Let $M=(L(\mathbb{R}^\*))^{Q[G]}$. Then $M\models$ ZF and $\mathbb{R}^\*=\mathbb{R}\cap M$ and $M\models$ "$\lambda=\omega\_1$", so $M\models$ "$\omega\_1$ is singular", since $Q=L^M\subseteq M$.
Now suppose there is a model $N$ of ZFC such that the ordinals of $N$ which are ${\leq\lambda}$ are (externally) isomorphic to those of $M$ which are ${\leq\lambda}$, and $M,N$ have the same cardinals $\leq\lambda$. Thus, $\lambda=\omega\_1^N$. But $L\_\lambda^N=L\_\lambda^M$, and the sequence $\left<\aleph\_n^{L^M}\right>\_{n<\omega}=\left<\aleph\_n^{L\_\lambda^M}\right>\_{n<\omega}$ is definable over $L\_\lambda^M$, so it is in $N$, so $N\models$ "$\lambda$ is singular", a contradiction.
(Without the assumption that $\lambda+1$ is wellfounded, it doesn't seem clear that we must get $L\_\lambda^N=L\_\lambda^M$, so the last paragraph seems to become problematic.)
| 8 | https://mathoverflow.net/users/160347 | 443286 | 178,808 |
https://mathoverflow.net/questions/443268 | 3 | Firs of all I ask my question, then I explain how this question arises in my mind and lastly what I tried to solve it.
QUESTION:
Let $P\_{n,N}(k)$ be the number of composition of an integer $k$ in $n$ parts bounded by $N.$ Prove that, if $n$ and $N$ are sufficiently large, $$P\_{n,N}(k)\sim N^{n-1}\sqrt{\dfrac{6}{\pi n}}\exp{\left[-\frac{3}{2n}\left(\frac{nN+n-2k}{N}\right)^{2}\right]}.$$
PROLOGUE:
Following a suggestion given by Daniele Tampieri in [his answer](https://mathoverflow.net/questions/102597/autobiographies-of-mathematicians/376423#376423) to [this question](https://mathoverflow.net/questions/102597/autobiographies-of-mathematicians/376423%7Bdb005102-8faa-494e-b99a-769c1b99bc26%7D376423) (I acknowledge Daniele for this inspiring advice), I read the autobiographical work of F. Tricomi [1] where the author describes, together with some autobiographical facts, some of his works. In particular he speaks about his paper [2]. Tricomi says that, in this paper, he considers the number $P\_{n,N}(k)$ of composition of an integer $k$ in $n$ parts bounded above by $N$ (I retained the same notations used by Tricomi) and that he proves that, asymptotically, if $n$ and $N$ are sufficiently large, the above estimate holds. That formula reminds me the Hardy-Ramanujan-Rademacher formula for the number of partitions of an integer.
I found the description of this paper a little bit surprising because Tricomi is best known for his work on ordinary differential equations and special function, not for works on number theory/combinatorics.
Since I am not able to find a copy of the original paper of Tricomi, my question is how to prove his asymptotic formula.
WHAT I TRIED TO DO:
I write down the generating function $$\sum\_k P\_{n,N}(k)x^k$$ which is trivially $$\left(\frac{x-x^{N+1}}{1-x}\right)^{n}$$ from which one can find the explicit expression $$P\_{n,N}(k)=\sum\_{j}(-1)^j\binom{n}{j}\binom{k-1-Nj}{n-1}.$$
Then I tried to apply the classical result for the asymptotic estimates in analytic combinatorics (see e.g. [3]) without success.
Perhaps some asymptotic for large binomial coefficient of which I am not aware is necessary?
REFERENCES
[1] Tricomi, Francesco; La mia vita di matematico attraverso la cronistoria dei miei lavori. (Bibliografia commentata 1916–1967), Padova: CEDAM – Casa Editrice Dottor Antonio Milani, pp. XII+172 (1967), MR0274255, Zbl 0199.28603.
[2] Tricomi, Francesco; Su di una variabile casuale connessa con un notevole tipo di partizioni di un intero. Giornale dell'Istituto Italiano degli Attuari, 2, 455-468 (1931), [JFM 57.0614.02](https://zbmath.org/57.0614.02), [Zbl 0003.01703](https://zbmath.org/0003.01703).
[3] Flajolet, Philippe; Sedgewick, Robert; *Analytic combinatorics.* Cambridge University Press, Cambridge, 2009. xiv+810 pp., MR2483235, Zbl 1165.05001.
| https://mathoverflow.net/users/165036 | Proof of an asymptotic formula by Tricomi | $\newcommand{\si}{\sigma}
\newcommand{\Z}{\mathbb Z}$As noted in [Pietro Majer's](https://mathoverflow.net/questions/443268/proof-of-an-asymptotic-formula-by-tricomi#comment1144269_443268) comment, the meaning of $\sim$ in the claim that, if $n$ and $N$ are sufficiently large, then
\begin{equation\*}
P\_{n,N}(k)\sim N^{n-1}\sqrt{\frac6{\pi n}}
\exp\Big(-\frac3{2n}\Big(\frac{nN+n-2k}N\Big)^2\Big) \tag{1}\label{1}
\end{equation\*}
should be clarified.
Looking at the constant factor $\sqrt{\frac6\pi}$ on the right-hand side of \eqref{1}, one should apparently assume that $A\sim B$ is supposed to mean here that $A/B\to1$ (as it is common). So, then the claim is that \eqref{1} holds uniformly in $k$ (or at least for each feasible value of $k$) as $n,N\to\infty$.
Then \eqref{1} will of course be false. For instance, if $k=n$, then $P\_{n,N}(k)=1$, whereas the right-hand side of \eqref{1} goes to $\infty$ (super-exponentially fast).
Below we will see for what values of $k$ the asymptotic relation \eqref{1} holds. We will also see that a slight modification of \eqref{1} holds for finite $N\ge2$ and such values of $k$.
---
The proper setting for statements like \eqref{1} is the so-called local (central) limit theorems (LLT's) of probability theory. Indeed, let
\begin{equation\*}
S\_{n,N}:=X\_{1,N}+\dots+X\_{n,N},
\end{equation\*}
where for each $N$ we have that $X\_{1,N},\dots,X\_{n,N}$ are independent random variables (r.v.'s) each uniformly distributed on the set $[N]:=\{1,\dots,N\}$. Then
\begin{equation\*}
P\_{n,N}(k)=N^n P(S\_{n,N}=k). \tag{2}\label{2}
\end{equation\*}
Moreover, one has the following LLT:
\begin{equation\*}
P(S\_{n,N}=k)=\frac1{\si\_N\sqrt{2\pi n}} e^{-z\_k^2/2}
+o\Big(\frac1{\si\_N\sqrt n}\Big) \tag{3}\label{3}
\end{equation\*}
uniformly in $k\in\Z$, where $z\_k:=z\_{k;n,N}:=\dfrac{k-n\mu\_N}{\si\_N\,\sqrt n}$,
$\mu\_N:=EX\_{1,N}=(N+1)/2$, and $\si\_N:=\sqrt{Var\,X\_{1,N}}=\sqrt{(N^2-1)/12}$. (The condition $N\to\infty$ is not needed. One may also note that the case $N=2$ of \eqref{3} is also the case $p=q$ of the de Moivre–Laplace theorem.)
>
> It follows immediately from \eqref{2} and \eqref{3} that \eqref{1} holds as $n,N\to\infty$ uniformly in all values $k$ that are in the normal deviation zone from the mean $n\mu\_N$ of $S\_{n,N}$ -- that is, uniformly in $k$ such that $|z\_k|=O(1)$. Moreover, the slight modification of \eqref{1} for finite $N\ge2$ -- with $N^{n-1}$ replaced by $N^n/\sqrt{N^2-1}$ and $\dfrac{nN+n-2k}N$ replaced by $\dfrac{nN+n-2k}{\sqrt{N^2-1}}$ --
> holds, again uniformly in $k$ such that $|z\_k|=O(1)$.
>
>
>
The proof of \eqref{3} is similar to the proof of, say, Theorem 1 in Ch. VII of [Petrov's book](https://link.springer.com/book/10.1007/978-3-642-65809-9). That theorem is stated for a sequence $(X\_j)$ of (lattice-valued) r.v.'s, whereas in our case we have to deal with a double-index array $(X\_{j,N})$. However, the $X\_{j,N}$'s have the specific and easy to deal with uniform distribution on the set $[N]$, which makes the proof of \eqref{3} significantly easier overall than the proof of the mentioned theorem. Indeed, in view of (say) Lemma 2 of [Gamkrelidze](https://epubs.siam.org/doi/10.1137/1133053), we can deal with the most difficult integral $I\_3$ in the proof of the mentioned theorem the same way as with the easy integrals $I\_2$ and $I\_4$ (the four integrals $I\_1,\dots,I\_4$ are defined at the bottom of p. 190 of Petrov's book).
| 5 | https://mathoverflow.net/users/36721 | 443290 | 178,809 |
https://mathoverflow.net/questions/443272 | 5 | Suppose $B$ and $C$ are commutative unital $C^{\ast}$-algebras with $B \subseteq C$ (unital). Let $c$ be an element of $C$ such that $c \ast 1 = 1 \ast c$ in the pushout (in the category of commutative unital $C^{\ast}$-algebras) $C \ast\_B C$. Is it valid to conclude that $c$ belongs to $B$?
| https://mathoverflow.net/users/130868 | Elements that commute with $1$ in the pushout of a $C^{\ast}$-algebra | The answer is yes: $1 \ast c = c \ast 1$ implies $c \in B$.
---
To see this, let me first translate the statement into fully categorical language. If we restrict to $\|c\| \le 1$ wlog, then the elements $c \in C$ are in bijection with unital $\*$-homomorphisms $\mathscr{C}(D) \to C$, where $D \subseteq \mathbb{C}$ is the unit disk; this is precisely the statement of functional calculus.
Thus, we can formulate the original question as follows: if a morphism $\mathscr{C}(D) \to C$ equalizes the two inclusion maps $C \rightrightarrows C \ast\_B C$, then does itt factor across $B$? If this property even holds for all objects $A$ in place of $\mathscr{C}(D)$, then this is known as the inclusion $B \hookrightarrow C$ being an [effective monomorphism](https://ncatlab.org/nlab/show/effective+monomorphism). It is thus enough to show that this is the case.
Equivalently, under Gelfand duality it is enough to show that every epimorphism in the category of compact Hausdorff spaces is effective. Since this category has pullbacks, it is enough to show that every epimorphism is [regular](https://ncatlab.org/nlab/show/regular+epimorphism). This is well-known to be the case, see e.g. the paper [Algebra $\cap$ Topology = compactness](https://www.sciencedirect.com/science/article/pii/0016660X71900018?via%3Dihub) of Herrlich and Strecker.
| 3 | https://mathoverflow.net/users/27013 | 443302 | 178,814 |
https://mathoverflow.net/questions/443310 | 2 | Suppose $x$ is a **unknown** sequence of non-negative integers with $n$ elements and $\sum\_{i=1}^n x\_i = c$ for a constant $c$.
What is the minimum possible value of $\sum\_{i=1}^n x\_i(x\_i-1)(x\_i-2)$?
I know the answer when we want a lower bound for $\sum\_{i=1}^n x\_i(x\_i-1)$. We can expand it and use the *Cauchy-Schwarz Inequality*. But I don't see anything for solving it in this form.
PS: Actually I want a good lower bound over this expression; by good I mean non-trivial ones.
| https://mathoverflow.net/users/501463 | Minimum possible value of $\sum_{i=1}^n x_i(x_i-1)(x_i-2)$ for fixed sum | (Updated as per suggestions in the comments.)
I will use
$$
S[x]\equiv \sum\_{i=1}^n x\_i(x\_i-1)(x\_i-2).
$$
Obviously, $S[x]\geq 0$.
First, let $n=2$, and set $x\_1=c/2+y$ and $x\_2=c/2-y$. Then
$$
\sum\_{i=1}^n x\_i(x\_i-1)(x\_i-2)=\frac{(c-2)(12y^2+c(c-4))}{4}.
$$
This means that for $c>2$ the minimum is achieved for the minimal possible value of $|y|$, which is either $0$ or $\pm \tfrac{1}{2}$ depending on whether $c$ is even or odd, which corresponds to $x\_1=x\_2$ or $x\_1=x\_2\pm 1$.
Now, consider a general $n>1$. Suppose that there is a pair $i,j$ such that $|x\_i-x\_j|>1$. Suppose that $x\_i+x\_j>2$. Then by the above argument, we can alter the values of $x\_i,x\_j$ while preserving $x\_i+x\_j$ to make sure that $|x\_i-x\_j|\leq 1$ and decrease the sum $S[x]$ by doing so. If instead $x\_i+x\_j\leq 2$, then we must have that $x\_i=0$ and $x\_j=2$ or vice versa. Therefore, if $x$ achieves the minimum of $S[x]$, then for every pair $i,j$ we have $|x\_i-x\_j|\leq 1$ or $\{x\_i,x\_j\}=\{0,2\}$. If the latter option holds for at least one pair $i,j$ then, by the former property, the only other value allowed in the sequence is $1$.
The above argument implies that if a sequence $x$ achieves the minimum of $S[x]$, then one of two options hold:
1. $x\_i=a$ or $x\_i=a+1$, $\forall i\in \{1,\cdots,n\}$, for some integer $a$. (This includes the case when all $x\_i$ are identical.)
2. $x\_i\in \{0,1,2\}$, $\forall i\in \{1,\cdots,n\}$.
The second option can only be realized for $c\leq 2n$ in which case the minimum of $S[x]=0$ can be achieved. For $c>2n$ we need to solve the following problem: let $a\geq 0$ and $0\leq q<n$ be integers such that $(n-q)a+q(a+1)=c$. What is the minimal value of $S(q,a)=(n-q)a(a-1)(a-2)+q(a+1)a(a-1)$? (We can assume that $q<n$ since $q=n$ is equivalent to $q=0$ with $a$ replaced by $a+1$.)
The condition $(n-q)a+q(a+1)=c$ is equivalent to $na+q=c$, which uniquely determines $a=\lfloor c/n\rfloor$ and $q=c-na$, which in turn uniquely fixes the value of $S(q,a)$ which gives the tight lower bound.
| 6 | https://mathoverflow.net/users/32985 | 443313 | 178,818 |
https://mathoverflow.net/questions/443306 | 2 | Consider $L^p[ 0,1]$ for $1\leq p < \infty$ or, if you prefer, $L^p(\mu)$ where $\mu$ is a finite Borel measure with compact support. Let $(\phi)\_{i\in I}$ be a subset of measurable functions that is contained in $L^p$ for every $1\leq p < \infty$ and assume that it is total for $L^{p\_0}$ for some $1\leq p\_0< \infty. $
Could you deduce that $(\phi\_i)\_{i\in I}$ is total for $L^p$ for every $1\leq p < \infty$?
If $1\leq p\_1 \leq p\_0$ it is enough to recall that the set of linear combinations of step functions is dense in every $L^p$, and argue as it follows. Let $f \in L^{p\_1}$ and $\varepsilon > 0.$ Then, there exists $g \in L^{p\_1}\cap L^{p\_0}$, which is a linear combination of step functions, such that $\|f-g\|\_{p\_1} < \varepsilon.$ Moreover, since $(\phi)\_{i\in I}$ is total in $L^{p\_0},$ there exists $h$ in its linear span such that $\|g-h\|\_{p\_0} < \varepsilon.$ Since the measure is finite, it follows that $\|g-h\|\_{p\_1} \leq C \|g-h\|\_{p\_0},$ with $C > 0.$ Finally, $$\|f-h\|\_{p\_1} \leq \|f-g\|\_{p\_1} + \|g\_h\|\_{p\_1} < \varepsilon (1+C),$$ as we wanted. What does it happen with the rest of exponents $p > p\_0$ ?
| https://mathoverflow.net/users/145367 | Total sets for $L^p$ for every $1\leq p < \infty$ | No, this is not true. The set $S$ of all simple function orthogonal to $x^{-\frac 13}$ is not dense $L^2(0,1)$. To see that it is dense in $L^1(0,1)$, take $s$ simple, let $A=\int\_0^1 s(x)x^{-\frac 13}$ and choose $t=\frac 23 A \delta^{-\frac 23}\chi\_{(0, \delta)}$. Then $s-t \in S$ and $\|t\|\_1 \leq C \delta^{\frac 13} \to 0$ as $\delta \to 0$.
| 4 | https://mathoverflow.net/users/150653 | 443317 | 178,819 |
https://mathoverflow.net/questions/443305 | 1 | I want a reference of the following statement which I think is true. Let $X$ and $Y$ be Banach spaces with $X$ finite dimensional. Then $(X\otimes\_\epsilon Y)^\*$ is isometrically isomorphic to $(X^\*\otimes\_\pi Y^\*)$ and $(X\otimes\_\pi Y)^\*$ is isometrically isomorphic to $(X\otimes\_\epsilon Y)^\*$ where $\otimes\_\epsilon$ and $\otimes\_\pi $ denote the injective and projective tensor product of Banach spaces respectively. For tensor product of Banach spaces see the book <https://www.amazon.in/Raymond-A-Ryan/e/B001K88BH0/ref=dp_byline_cont_pop_ebooks_1>.
| https://mathoverflow.net/users/136860 | Duality of projective and injective tensor product | Since Ryan's book is mentioned in the original question, here's some pointers for an answer based on this source:
* See Section 3.4 for what the dual space of $X\otimes\_\epsilon Y$ is. If you combine Proposition 3.14 (and the comments after) with Proposition 3.22 (see the comments after it) we get $(X\otimes\_\epsilon Y)^\* = I(X,Y^\*) = I(Y,X^\*)$ the space of integral operators.
* Now look at Theorem 5.33 which shows that when $X^\*$ has the Radon-Nikodym Property and the approximation property (which is true when $X$ is finite-dimensional) then $I(Y,X^\*) = X^\* \hat\otimes\_\pi Y^\*$. See also the proof of Proposition 5.52.
For the dual of $X\otimes\_\pi Y$ see Section 2.2 where it's shown that $(X\otimes\_\pi Y)^\* = B(X,Y^\*)$ the space of all bounded linear maps. We always have that $X^\* \hat\otimes\_\epsilon Y^\*$ is a subspace of $B(X,Y^\*)$, see Section 3.1, and it's easy to see that when $X$ is finite-dimensional, then we have equality here.
| 2 | https://mathoverflow.net/users/406 | 443321 | 178,820 |
https://mathoverflow.net/questions/443323 | 2 | Let $p$ be an odd prime, and consider the group
$$\{U\in \operatorname{SL}\_n(\mathbb{Z}/p\mathbb{Z}) : U^{t}U=I \bmod p \}\subseteq \operatorname{SL}\_n(\mathbb{Z}/p\mathbb{Z}).$$
I wonder what is the structure of this subgroup? Can the generators of this subgroup be written or calculated ?
We know that if $U^{t}U=I$ with $U\in \operatorname{GL}\_n(\mathbb{Z})$, then $U$ must be a signed permutation. So I came up with the above subgroup.
| https://mathoverflow.net/users/482299 | A subgroup of $\mathrm{SL}_n(\mathbb{Z}/p\mathbb{Z})$ | This is the [special orthogonal group](https://groupprops.subwiki.org/w/index.php?title=Orthogonal_group_for_the_standard_dot_product) over the field $\mathbb{Z}/p \mathbb{Z}$. Or, rather, one of the "special orthogonal groups". For any invertible symmetric matrix $Q$, one can consider the group of matrices $U$ in $\text{SL}\_n$ such that $U Q U^T = Q$, which forms a subgroup $\text{SO}(Q)$. If one allows matrices in $\text{GL}\_n$ instead, this is $O(Q)$.
I'll limit myself to $p$ odd throughout.
**"Split" versus "non-split"** If $n=2m+1$ is odd, then any two nondegenerate quadratic forms define the same orthogonal group.
If $n=2m$ is even, then there are two classes of quadratic forms, called the "[split](https://groupprops.subwiki.org/wiki/Split_orthogonal_group)" and the "non-split" class. The case you have described is split if $p^{m} \equiv 1 \bmod 4$ and is non-split if $p^{m} \equiv 3 \bmod 4$. I'll denote the split and non-split cases by $SO\_{2m}^+(p)$ and $SO\_{2m}^-(p)$.
**Order of the group** The order of $SO\_{2m+1}$ is
$$p^{m^2} \prod\_{i=1}^m (p^{2i}-1).$$
The orders of $SO\_{2m}^+(p)$ and $SO\_{2m}^-(p)$ are
$$p^{m(m-1)} (p^m-1) \prod\_{i=1}^{m-1} (p^{2i}-1) \ \text{and} \ p^{m(m-1)} (p^m+1) \prod\_{i=1}^{m-1} (p^{2i}-1)$$
respectively. Note, in particular, that they can't be isomorphic since they have different sizes.
**Generators** The orthogonal group $O\_n$ is generated by reflections; for a vector $x$ with $x \cdot x \neq 0$, the reflection over $x$ is the linear map $r\_x(v) := v - \tfrac{x \cdot v}{x \cdot x} x$. Reflections have determinant $-1$, so the special orthogonal group $SO\_n$ is generated by pairs of reflections; the product $r\_x r\_y$ fixes the $(n-2)$-plane $x^{\perp} \cap y^{\perp}$, so we can also say that $SO\_n$ is generated by rotations fixing planes of dimension $n-2$. A shorter list of generators can be found in [Matrix Generators for the Orthogonal Groups](https://www.maths.usyd.edu.au/u/ResearchReports/Algebra/RylTay/1997-7.html), Rylands and Taylor , 1998.
**Structure** The orthogonal group $O\_n$ has two characters to $\{ \pm 1 \}$: The determinant map, which sends every reflection to $-1$, and the [spinor norm](https://groupprops.subwiki.org/wiki/Spinor_norm), which sends the reflection $r\_x$ to $\left( \tfrac{x \cdot x}{p} \right)$ (this is the quadratic residue symbol). I'll write $K\_{2m+1}$, $K\_{2m}^+$ or $K\_{2m}^-$ for the common kernel of these characters. So $K$ is an indexed $2$ subgroup of the corresponding $SO$.
For $n=2m+1$ odd, the group $K\_n$ is the [Chevalley group](https://groupprops.subwiki.org/wiki/Chevalley_group_of_type_B) $B\_m(p)$. The group $B\_1(p)$ is isomorphic to $\text{PSL}\_2(p)$, which is the alternating group $A\_4$ if $p =3$ and is otherwise simple. I believe that $B\_m(p)$ is simple for all $m \geq 2$ and $p \geq 3$, but I couldn't find a reference for this.
For $n$ even in the case you care about, where $Q = \text{Id}\_n$, the group $K\_n$ contains $- \text{Id}\_n$, which generates the center. For the other quadratic form that you didn't use, the spinor norm of $- \text{Id}\_n$ is $-1$, and the center of $K\_{n}$ is trivial.
The quotients of $K\_{2m}^+(p)$ and $K\_{2m}^-(p)$ by their centers are the [Chevalley group](https://groupprops.subwiki.org/wiki/Chevalley_group_of_type_D) $D\_m(p)$ and the twisted Chevalley group ${}^2 D\_m(p^2)$ respectively. We have $D\_2(p) \cong \text{PSL}\_2(p) \times \text{PSL}\_2(p)$. Again, $\text{PSL}\_2(p)$ is simple for any $p \geq 5$, whereas $\text{PSL}\_2(3) \cong A\_4$. I believe that $D\_m(p)$ and ${}^2 D\_m(p^2)$ are simple for $m \geq 3$ (and $p$ odd), but again I couldn't find a reference
I'm using [Wikipedia](https://en.wikipedia.org/wiki/List_of_finite_simple_groups#Groups_of_Lie_type) as my source for group theoretic notation.
| 7 | https://mathoverflow.net/users/297 | 443328 | 178,824 |
https://mathoverflow.net/questions/443312 | 1 | Let $\mathbb{F}$ be a field of characteristic $2$ and define $S$ to be the set of all triples $(i,j,k)\in\lbrace 1,\dotsc,n\rbrace^3$ with $\left|i-j\right|=1$, $\left|i-k\right|>1$, and $\left|j-k\right|>1$. Define $V=\bigoplus\limits\_{(i,j)\in\lbrace 1,\dots,n\rbrace^2}\mathbb{F}\sigma\_{i,j}$ to be the vector space with basis the quotient of $\lbrace \sigma\_{i,j}\rbrace\_{(i,j)\in\lbrace 1,\dots,n\rbrace^2}$ by the relation $\sigma\_{i,j}=\sigma\_{j,i}~\forall~(i,j)\in\lbrace 1,\dots,n\rbrace^2$. We can consider the vector subspace $V\_S=\sum\_{(i,j,k)\in S}\mathbb{F}(\sigma\_{i,k}+\sigma\_{j,k})$ (I don't know if this is an standard notation, but I want to represent the sum of subspaces). It is obvious that the above sum is not direct, since for every tuple $(i-1,i,j,j+1)$ with $i<j$ and $\left|j-i\right|>1$ we have
$$(\sigma\_{i-1,j}+\sigma\_{i-1,j+1})+(\sigma\_{i,j}+\sigma\_{i,j+1})+(\sigma\_{i-1,j}+\sigma\_{i,j})+(\sigma\_{i-1,j+1}+\sigma\_{i,j+1})=0.$$
My question is if we can find the number of summands needed in $V\_S$ such that the vector space is represented as a direct sum, which is equivalent to finding $\dim(V\_S)$. It is easy to prove that $\left|S\right|=(n-2)(n-3)$ and with a bit of extra work one can see that the problem mentioned above can be solved by deleting $\frac{(n-3)^2}{4}$ summands in case $n$ is odd and $\frac{(n-2)(n-4)}{4}$ summands in case $n$ is even. However, this is not enough, for the case $n=6$ there is one extra summand to delete which does not correspond to the above situation. I guess there must be many more extra summands to delete for bigger $n$.
This problem appeared when trying to compute the homology of a certain family of groups, and I am not an expert on combinatorics, so I'm quite lost.
Any hints or help will be appreciated.
| https://mathoverflow.net/users/482329 | Counting the number of summands in a vector space over characteristic $2$ to get a direct sum | The dimension is $n(n-3)/2$. The argument is as follows. The set of pairs $(i,j)$ with $i$ and $j$ between $1$ and $n$, and $j-i \geqslant 2$ has cardinality $\binom{n-2}{2}=\frac{(n-1)(n-2)}{2}$. The $\sigma\_{i,j}$ with these indices are the only ones involved. So let $W$ be the vector space spanned by these. For each element of $S$, we get a vector in $W$ such that the sum of the coordinates is zero. This gives us a subspace of $W$ of codimension one, and dimension $n(n-3)/2$. I claim that the images of elements of $S$ span this subspace. To see this, order the pairs $(i,j)$ lexicographically. Then for each $(i,j)$ apart from the first (which is $(1,3)$), there is an element of $S$ giving a vector whose last non-zero coordinate is $\sigma\_{i,j}$. These span the codimension one subspace.
| 1 | https://mathoverflow.net/users/460592 | 443351 | 178,833 |
https://mathoverflow.net/questions/443340 | 1 | I am trying to recover the result given by equation 10 in the article [here](http://proceedings.mlr.press/v119/lu20b/lu20b.pdf). I am unable to get rid of the integral, any help would be much appreciated. To keep the description as self contained as possible, I will describe the relevant notations etc., a more detailed reference is the article [1](http://proceedings.mlr.press/v119/lu20b/lu20b.pdf) itself. Here is my attempt:
Suppose, the overparametrized deep ResNet is modeled via a mean-field ODE:
$$
\dot{X\_\rho}(x,t)=\int\_{\theta}f(X\_{\rho}(x,t),\theta)\rho(\theta,t)d\theta.
$$
Here $x$ denotes the input at the layer $t=0$ and $X\_{\rho}(x,1)$ is the output at the final layer $t=1$.
In this model, every residual block $f(\cdot,\theta\_i)$ is considered as a particle and optimization (training) will be done over the distribution of the particles $\rho(\theta,t)$ where $\theta$ denotes the parameters of the Residual block and $t$ denotes the $t-th$ layer of the block. We will further represent $\int\_{\theta}f(X\_{\rho}(x,t),\theta)\rho(\theta, t)d\theta= F(X\_{\rho}(x,t);\rho)$. We also know that $\rho(\theta, t)$ is a density for every $t$. Thus the ODE equation above is is reduced to:
$\dot{X\_\rho}(x,t)=F(X\_{\rho}(x,t);\rho).$
Let $E(x,\rho)$ be the cost function that depends on the mismatch of the output of the neural net, $X\_{\rho}(x,1)$, and the true output $y(x)$. For emphasis we note that $X\_{\rho}(x,1)$ is the final output of the neural net at layer $t=1$ corresponding to the true input $x$. Now we calculate the sensitivity of the cost function $E(x,\rho)$ with respect to the parameter $\rho(\theta,t)$ that describes the distribution of the weight parameters $\theta$ at every $t-$th layer of the neural net. We will sometimes supress the explicit dependence of $E(x,\rho)$ on its arguments and simply write $E:=E(x,\rho)$ for convenience.
$\frac{dE(x,\rho)}{d\rho}=\frac{\partial E}{\partial X\_{\rho}(x,1)}\frac{d X\_{\rho}(x,1)}{d \rho}.$
To calculate $\frac{d X\_{\rho}(x,1)}{d \rho}$ we will use the adjoint sensitivity method.
Recall:
$ X\_{\rho}(x,1)=x+\int\_{0}^1 F(X\_{\rho}(x,t);\rho) dt$
$ \frac{d X\_{\rho}(x,1)}{d\rho}=\int\_{0}^1 \bigg[\frac{\partial {F(X\_{\rho}(x,t);\rho)}}{\partial X\_{\rho}(x,t)}\frac{d{X\_{\rho}(x,t)}}{d \rho}+\frac{\partial F(X\_{\rho}(x,t);\rho)}{\partial \rho}\bigg]dt-\int\_{0}^1 \lambda(t) \bigg[\frac{d \dot{X\_\rho}(x,t)}{d\rho}-\frac{\partial F}{\partial X\_{\rho}(x,t) }\frac{{d X\_{\rho}(x,t) }}{d \rho}-\frac{\partial F}{\partial \rho}\bigg]$.
Note that the second integral is zero due to the ODE equation above. More specifically, $\frac{d}{d\rho}\bigg(\dot{X\_\rho}(x,t)-F(X\_{\rho}(x,t);\rho)\bigg)=0$.
Consider the term
$-\int\_{0}^1 \lambda (t)\frac{d \dot{X\_\rho}(x,t)}{d\rho} dt=-\int\_{0}^1 \lambda (t)\frac{d}{dt}\frac{d {X\_\rho}(x,t)}{d\rho} dt$
Evaluating using integration by parts
$-\int\_{0}^1 \lambda (t)\frac{d \dot{X\_\rho}(x,t)}{d\rho} dt=-\lambda(t) \frac{d X\_{\rho}(x,t)}{d\rho}|\_{t=0}^{t=1}
+\int\_0^1 \frac{d\lambda (t)}{dt} \frac{d X\rho}{d\rho} dt $
We will choose $\lambda$ such that $\lambda(1)=0$. We also note that $\frac{dX\_{\rho}}{d\rho}|\_{t=0}=0$.
Using these we can rewrite,
$ \frac{d X\_{\rho}(x,1)}{d\rho}=\int\_{0}^1 \bigg[(\lambda(t)+Id)\frac{\partial F}{\partial X\_{\rho}(x,t)}+\frac{d\lambda(t)}{dt}\bigg]\frac{d X\_{\rho}(x,t)}{d\rho} dt+\int\_{0}^1 (\lambda+Id)\frac{\partial F}{\partial \rho} $
Now we choose $\lambda(t)$ to satisfy the ODE equation:
$(\lambda(t)+Id)\frac{\partial F}{\partial X\_{\rho}(x,t)}+\frac{d\lambda(t)}{dt}=0$
along with the condition $\lambda(1)=0$. This is equivalent to the system for $\tilde{\lambda}=\lambda+Id$,
$-\tilde{\lambda}(t)\frac{\partial F}{\partial X\_{\rho}(x,t)}=\frac{d\tilde{\lambda}(t)}{dt} $
and $\tilde{\lambda}(1)=Id$.
It can be independently verified that $\tilde{\lambda}(t)=J\_{\rho}(x,t)$ where $J\_{\rho}(x,t)=\frac{d X\_{\rho}(x,1)}{d X\_{\rho}(x,t)}$ satisfies the system along with final value at $t=1$, see also [1](http://proceedings.mlr.press/v119/lu20b/lu20b.pdf)(eqn 9).
Thus we get,
$\frac{d X\_{\rho}(x,1)}{d\rho}=\int\_{0}^1 J\_{\rho}(x,t)\frac{\partial F}{\partial \rho} dt$ and so,
$\frac{dE(x,\rho)}{d\rho}=\frac{\partial E}{\partial X\_{\rho}(x,1)} \int\_{0}^1 J\_{\rho}(x,t)\frac{\partial F}{\partial \rho} dt$
whereas in the article, [1](http://proceedings.mlr.press/v119/lu20b/lu20b.pdf)(eqn 10), it is evaluated:
$\frac{dE(x,\rho)}{d\rho}=\frac{\partial E}{\partial X\_{\rho}(x,1)} J\_{\rho}(x,t)\frac{\partial F}{\partial \rho}$
| https://mathoverflow.net/users/21422 | Adjoint sensitivity analysis for a cost functional under an ODE constraint | Ah, that is just about the meaning of the expression $\frac{\partial E(x,\rho)}{\partial\rho}$. Since $\rho$ is *a function* of $t$, it really means "a function $D(t)$ such that
$$
E(x,\rho+\Delta\rho)-E(x,\rho)\approx \int\_0^1 D(t)\Delta\rho(t)\,dt
$$
for all small perturbations $\Delta\rho(t)$".
What you did was to compute the derivative treating $\rho$ like a constant, i.e., your computation is formally valid assuming $\Delta\rho(t)=h$ throughout the whole interval, in which case your formula is just a partial case of their formula, i.e.,
$$
E(x,\rho+h)-E(x,\rho)\approx h\int\_0^1 D(t)\,dt.
$$
However you derivation is incomplete because you need to find the linearization for *all* $\Delta\rho$, not just constants. Fortunately, you hardly need to change anything in it: almost mechanical insertion of $\Delta\rho$ where it belongs should do the trick.
| 3 | https://mathoverflow.net/users/1131 | 443356 | 178,834 |
https://mathoverflow.net/questions/443367 | 4 | Inspired by [The set of all limits of sub-series of an absolute convergent series](https://math.stackexchange.com/questions/2062357/the-set-of-all-limits-of-sub-series-of-an-absolute-convergent-series) is the following true?:
>
> Let $a\_n$ be a strictly decreasing sequence and $\sum\_1^\infty a\_n=\ell<\infty$ is a convergent series. Is it true to say that the set of all possible value of all subseries $\sum a\_{n\_i}$ of $\sum a\_n$ is whole $[0,\ell]$?
>
>
>
| https://mathoverflow.net/users/36688 | The set of all possible values of subseries of a convergent positive term series | For convenience define $S\_n = \sum\_{j\le n} a\_j$ and $T\_n = \sum\_{j > n} a\_j$.
Suppose there is $n$ such that $a\_n > T\_n$. Then for $S\_{n-1} + T\_n < x < S\_n$,
$x$ is not the sum of a subseries.
On the other hand, if $a\_n \le T\_n$ for all $n$, then every $x \in [0,\ell]$ is the sum of a subseries. This can be obtained "greedily": include $a\_n$ iff the sum of $a\_n$ and already-included terms $\le x$.
| 9 | https://mathoverflow.net/users/13650 | 443370 | 178,838 |
https://mathoverflow.net/questions/443353 | 1 | Let $C$ be a connected curve of arithmetic genus $g$
over algebraically closed field $k$ of characteristic zero having only nodes
as singularities together with finite morphism
$f: C \to \mathbb{P}^1$.
In [3264 and All That](https://www.cambridge.org/core/books/3264-and-all-that/DC062983CC5F8B7CDD37CFEBCCA5FEA4) by Eisenbud and Harris is claimed on page 313 that the the
automorphism group of the map $f$ — that is, automorphisms $\phi$ of $C$ such that
$f \circ \phi =f$ — is finite.
**Q1**: Why and how to prove it? If $C$ is smooth, then it follows immediately
from Hurwitz's automorphisms theorem. Clearly it suffice to check
that for every $q \in f^{-1}(p)$ the stabilizer of $q$ is finite. Why is
it true?
**Q2**: Is the automorphism group of $f$ also finite if we weaken the assumptions
on base field $k$? For example if we allow positive cahracteristic?
| https://mathoverflow.net/users/501436 | Finiteness of automorphism group of finite map $f: C \to \mathbb{P}^1$ | Q1: since maps of curves are determined by their action on the function field, we get an injection $\mathrm{Aut}(C/\mathbb{P}^1) \to \mathrm{Aut}(k(C)/k(t))$ but the latter is just a Galois group (or automorphism group of a field extension if you don't like this terminology for a non-Galois extension) of a finite extension of fields and hence is finite.
In particular, this shows that $|\mathrm{Aut}(C/\mathbb{P}^1)| \le \deg{f}$ where equality is achieved for so-called Galois covers (since for $C$ smooth it is equivalent to $k(C) / k(\mathbb{P}^1)$ Galois). This argument does not depend on the base curve being $\mathbb{P}^1$, just that you have a nonconstant map of (possibly singular) curves.
Q2: the above proof works over any field.
| 3 | https://mathoverflow.net/users/154157 | 443375 | 178,840 |
https://mathoverflow.net/questions/443364 | 2 | I read the following on Wikipedia's page on [Monadic Second-Order Logic of Two Successors (MS2S)](https://en.wikipedia.org/wiki/S2S_(mathematics)):
>
> Weak S2S (WS2S) requires all sets to be finite (note that finiteness
> is expressible in S2S using Kőnig's lemma).
>
>
>
Is this statement an error? I would think that only [Weak Kőnig's Lemma(WKL)](https://en.wikipedia.org/wiki/Reverse_mathematics#Weak_K%C5%91nig%27s_lemma_WKL0) would be expressible in MS2S since it is restricted to binary trees whereas [Kőnig's Lemma](https://en.wikipedia.org/wiki/K%C5%91nig%27s_lemma) is not limited to finite tree width. If it is possible to express the full Kőnig's Lemma, how would one do so? If not, how would one express WKL in MS2S? I looked at the [axiomatization of S2S](https://mathoverflow.net/questions/434659/axiomatization-of-s2s) but wasn't sure how to go about expressing WKL using it or another representation such as an infinite tree automata.
| https://mathoverflow.net/users/38049 | How can Kőnig's Lemma be expressed in Monadic Second-Order Logic of 2 Successors? |
>
> I would think that only Weak Kőnig's Lemma(WKL) would be expressible in MS2S
>
>
>
You're slightly misreading the passage - the point is that Konig's Lemma can be used to show that finiteness is definable in MS2S. (That said, you are right that even WKL would be enough.)
Here's the idea. First, note that in MS2S we can define the ordering on $2^{<\omega}$: we have $\sigma\preccurlyeq\tau$ iff every predecessor-closed set containing $\tau$ also contains $\sigma$. From this **and (weak) Konig's Lemma** we can define infiniteness: $X$ is infinite iff every downwards-closed set $Y\supseteq X$ contains a chain with no greatest element.
The point is not that KL, or WKL, is somehow being expressed in MS2S; rather, we are using (W)KL in the metatheory so to speak in order to prove something about the strength of MS2S. This is in fact nontrivial: over general structures, monadic second-order logic is *incomparable* with weak monadic second-order logic despite the name of the latter. (To see why this is plausible, consider an equivalence relation with infinitely many infinite classes *and* infinitely many classes of each finite cardinality.)
| 3 | https://mathoverflow.net/users/8133 | 443380 | 178,842 |
https://mathoverflow.net/questions/440126 | 9 | First I would like to apologize if this post breaks any rule regarding career advice or opinion-based questions. Given that construct QFT (CQFT) is a rather small community, I found this is the only site where some current/past practitioners can chime in.
For an idea of my mathematical level, in terms of analysis I have finished Folland's book on real analysis (which is mainly about measure/integration theory and basic parts of functional analysis) and I am currently going through Reed & Simon's book on functional analysis. I am also comfortable with general topology and I have gone through most of Tu's book on (smooth) manifolds. I also know a little bit of PDE theory. On the physics side of things, I am largely self taught and I have gone through most of Folland's book on QFT, Hall's book on quantum mechanics, and a small smattering of Peskin & Schroeder. The major physics topic I am not very familiar with is E&M, which makes learning from physics books difficult as they often teach QFT through QED.
I am interesting in going into CQFT, but I am a little puzzled on the best way to go about it. I read a few other questions on here that provide some very nice resources. The problem I am having is some of the math-oriented books assume too much math and the physicists assume too much physics (and speak in a completely different language than what I am used to as a mathematics student). Should I be learning QFT from only CQFT texts, or should I use traditional/heuristic physics books as well (e.g. Peskin & Schroeder) to get an intuition about the subject? Is there any other approach that is recommended?
| https://mathoverflow.net/users/498931 | Approach to learning constructive QFT | CQFT is very much still an open research subject. I don't think it is known what the best approach is. So all I can do is share my own opinion. (And a warning: I'm just an interested observer!)
First: **Learn you some classical electrodynamics!** Quantum field theory is after all *a theory of fields*, and classical electrodynamics is the main example of a field theory! Skipping it is a bit like skipping spheres to study homotopy categories.
Electrodynamics books tend to be a little cluttered, since they must also serve the needs of future engineers and experimentalists. You could skip around in Griffiths or Jackson, but I think the most efficient & logical approach to the core material of electrodynamics is [Landau & Lifshiftz, Volume 2](https://ia903206.us.archive.org/4/items/landau-and-lifshitz-physics-textbooks-series/Vol%202%20-%20Landau%2C%20Lifshitz%20-%20The%20classical%20theory%20of%20fields%20%284th%2C%201994%29.pdf). Landau takes a straight route to Maxwell's equations, and then treats electro- and magnetostatics, EM waves, scattering and so forth as applications.
---
To your larger question: I think you should not attempt to read the CQFT literature without reading some of the physics literature as well. The CQFT literature doesn't exist in isolation, so studying it alone is at best only hearing part of a conversation. For another, most of the existing textbooks on CQFT are basically research monographs. You're expected to bring some intuition for the problem domain. (The exception to this is Dimock's [Quantum Mechanics & Quantum Field Theory: A Mathematical Primer](https://www.cambridge.org/core/books/quantum-mechanics-and-quantum-field-theory/06B6B13B344880D492F4A09C9C1E713D), which could only be better if it were longer.)
But the physics literature is very large! And written by people who either enjoy trolling mathematicians or are genuinely unable to distinguish between the $SO(3)$ and $\mathfrak{so}(3)$. So, yes, you have to be choosy. Since how to be choosy is a matter of taste, this is where I get opinionated.
A few suggestions, specifically about which physics books to read to gain intuition and context. (I'll let others try to make a map to topics of contemporary interest in CQFT.)
**Avoid particle physics**.
So many people have gone down this road, and it doesn't go anywhere! You just end up with mathematical translations of Peskin & Schroeder. Lots of calculational techniques, very few quantum fields. Frankly, I think this approach is basically doomed. Particle physics is:
* too complicated: the details of the calculations obscures the basic structures.
* not well understood: e.g., the infrared structure of even *trivial* theories like the quantum Maxwell theory is one of the [major research themes](https://arxiv.org/abs/1703.05448) of the past decade, never mind the problem of confinement.
* probably broken at a foundational level: It's highly likely that neither quantum electrodynamics nor the Higgs field nor the Standard Model itself actually exists as a continuum model.
So the easy examples don't exist, in the same sense that the translation invariant probability measure on the integers doesn't exist, and the harder examples are so hard you might win a Millenium prize. I'm not saying don't study particle physics, but don't make it your only focus.
**Learn QFT via statistical physics**
Take the other path: Learn QFT from the statistical physicists. It's easier! They have non-trivial solvable examples! In low dimensions! Also: the core organizing principle of QFT -- renormalization as a means of isolating the low-frequency behavior -- was discovered by people doing statistical physics (Kadanoff & Wilson), so you might as well follow along in their footsteps.
Also, there's so much good mathematics here, from basic analysis and combinatorics in the classical theory of gases up to the recent work of Fields medalists like Hairer and Duminil-Copin.
If I were doing it over again, I'd probably read Goodstein's [States of Matter](https://store.doverpublications.com/048664927x.html) together with Ruelle's more rigorous [Statistical Mechanics: Rigorous Results](https://www.worldscientific.com/worldscibooks/10.1142/4090#t=aboutBook). Then progress to Shankar's [Quantum Field Theory and Condensed Matter: An Introduction](https://www.cambridge.org/core/books/quantum-field-theory-and-condensed-matter/2CA9970800C3D31D6E641736186B3FBD). Also, Baxter's [Exactly Solved Models in Statistical Mechanics](https://physics.anu.edu.au/research/ftp/mpg/baxter_book.php).
**Spend some time learning about QFT on curved spacetimes**.
You learned all that differential geometry, so you might as well use it! And thinking about what QFT on curved spacetime should be is a good thing to do, since it forces you to focus on the fundamentals. You can't just Fourier analyze everything in sight.
If you're a mathematician, the best thing to do here is to read Robert Wald's writings, starting with the little red book [Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics](https://press.uchicago.edu/ucp/books/book/chicago/Q/bo3684008.html). Wald is a very careful thinker and writer and he works with mathematics in very tasteful ways. (And I just learned, he's also written a [grad level textbook on electromagnetism](https://press.princeton.edu/books/hardcover/9780691220390/advanced-classical-electromagnetism), which looks like a nice alternative to Landau.)
Even if you don't want to think about black holes and the like, you should make an effort to understand the Unruh Effect: Observers in relatively accelerated rest frames *in flat Minkowski space* will agree on the value of a quantum field at a point, but they will not agree on how many particles are in a system. The reality of particles depends on the representation of the algebra of observables.
**Read lattice QFT papers**
The physics literature on lattice QFT tends to have very clear definitions. They have to, because they're usually oriented towards numerical simulations, and you can't just hand-wave the details to a computer. It's also usually the starting point for rigorous CQFT constructions.
Start off with LePage's [Lattice QCD for Novices](https://arxiv.org/pdf/hep-lat/0506036.pdf). Then (if you still want particle physics), pick up Montvay & Munster's [Quantum Fields on a Lattice](https://www.cambridge.org/core/books/quantum-fields-on-a-lattice/4401A88CD232B0AEF1409BF6B260883A#).
| 11 | https://mathoverflow.net/users/35508 | 443387 | 178,843 |
https://mathoverflow.net/questions/443297 | 3 | I'm interested in computing eta invariants of Dirac operators (on spinor bundles tensored with some vector bundles) on the total space of $S^{2n}$ bundles over odd-dimensional manifolds. I found the papers of [Bismut-Cheeger](https://www.jstor.org/stable/1990912) and [Dai](https://www.jstor.org/stable/2939276) where a formula is given in terms of Bismut-Cheeger eta forms, but I'm having a hard time actually computing them.
Are there places where they are computed explicitly for some $S^{2n}$-bundles? [This MO answer](https://mathoverflow.net/a/237960/) gives various explicit computations of eta invariants but I didn't find $S^{2n}$ bundles there.
Edit:
To add a bit more context, I’m interested in the following restricted situation.
Let $Spin(2n)$ act on $S^{2n}$ fixing the north and the south poles. Pick a metric compatible with this action. Now pick an $Spin(2n)$ bundle $P$ with a connection over an odd dimensional base $B$.
From this data we have a fibration $S^{2n}\to X\to B$. Then I should be able to compute the eta form of the Dirac operator on $X$.
What I’m confused is that, as the eta form is local on $B$, it should be a local expression of the metric on $B$ and the $Spin(2n)$ connection on $P$. But the eta form is an odd-degree form, and I can’t think of any way to write a local expression in terms of the metric and the $Spin(2n)$ connection! Does it mean that the eta form in this class of examples is simply zero?
I would also like to consider the case when we have an $Spin(2n)$ equivariant vector bundle $E$ over $X$ and compute the eta form for the Dirac operator tensored with $E$. I’m still at a loss what nonzero term I can write in this situation.
| https://mathoverflow.net/users/5420 | Explicit computations of Bismut-Cheeger eta form for $S^{2n}$ bundles | If you consider a fibre bundle with compact structure group that acts by isometries on the typical fibre and preserves a given Dirac operator, then the $\eta$-form can be read of from an infinitesimally equivariant $\eta$-invariant on the typical fibre, the fibre bundle curvature, and its action on the typical fibre. This is explained [in this article](https://doi.org/10.1515/crll.2000.073) and much better in an [article](https://doi.org/10.1016/j.aim.2021.108163) by Liu and Ma. If the fibre is even-dimensional, the infinitesimally equivariant $\eta$-invariant vanishes, so your suspicion is correct.
In your special case, there is an orientation reversing, fibre preserving isometry that swaps north and south poles. If this respects the superconnection in question, then you get another argument for the vanishing of the $\eta$-form.
| 2 | https://mathoverflow.net/users/70808 | 443391 | 178,845 |
https://mathoverflow.net/questions/443376 | 3 | I apologize if this question is basic in some sense. I was looking for an example of a non-proper HNN-extension and I found [this](https://mathoverflow.net/questions/405536/examples-of-non-proper-profinite-hnn-extensions).
In the comments, markvs mentioned the Baumslag-Solitar group $B(2,3)$. We can show that this is non-Hopfian using Britton's lemma so that is not residually finite. So, its natural try to use this to get a non-proper example. We know also that $B(2,3)$ is an HNN-extension of $\Bbb{Z}$ with respect to the natural $f: \langle a^2 \rangle \to \langle a^3 \rangle$. Now, how to proceed now? I mean, there is a way to decompose $\widehat{B(2,3)}$ as an HNN-extension of $\widehat{\Bbb{Z}}$ or something?
Intuitively, makes sense to me to do $B(2,3) = \operatorname{HNN}(\Bbb{Z})$ then $\widehat{B(2,3)} = \operatorname{HNN}(\Bbb{\widehat{Z}})$, but I don't know if it works.
| https://mathoverflow.net/users/123172 | Profinite completion of Baumslag-Solitar group as a profinite HNN-extension | The largest residually finite quotient of $\mathrm{BS}(2,3)$ is the semidirect product $\mathbf{Z}\ltimes\_{2/3}\mathbf{Z}[1/6]$.
So, if I'm correct, its profinite completion is
$$\widehat{\mathbf{Z}}\ltimes\_{2/3}\widehat{\mathbf{Z}[1/6]}.$$
Each time, the $2/3$ means that the generator $1$ acts by multiplication by $2/3$. Denoting by $P$ the set of primes, note that $\widehat{\mathbf{Z}}=\prod\_{p\in P}\mathbf{Z}\_p$ and $\widehat{\mathbf{Z}[1/6]}=\prod\_{p\in P-\{2,3\}}\mathbf{Z}\_p$. The group of automorphisms of the latter is also profinite, namely $\prod\_{p\in P-\{2,3\}}\mathbf{Z}\_p^\times$, so indeed the homomorphism $\mathbf{Z}\to \widehat{\mathbf{Z}[1/6]}^\times$ mapping $1$ to multiplication by $2/3$ extends uniquely to a continuous homomorphism $\widehat{\mathbf{Z}}\to \widehat{\mathbf{Z}[1/6]}^\times$.
A similar description should hold for all $\mathrm{BS}(m,n)$ whenever $\gcd(m,n)=1$. (Otherwise it's quite different since then the profinite completion is not solvable.)
| 2 | https://mathoverflow.net/users/14094 | 443400 | 178,846 |
https://mathoverflow.net/questions/443198 | 2 | Given a word $w \in X^{\pm 1}$ representing an element of the free group $F(X)$ there is a (usually non-unique) sequence $w=w\_0 \to w\_1 \to \cdots \to w\_r$ with $|w\_i|>|w\_{i+1}|$ where $w\_r$ is the unique reduced form of $w$, i.e. there are no subwords of the form $xx^{-1}, x \in X^{\pm 1}$. I know of this fact, even have a proof of it in some notes I wrote. **Does anybody know a reference for this?**
Similarly let $\mathbb X$ be a graph of groups and let $w$ be some word in some symbols representing an element of the fundamental group $\pi\_1(\mathbb X,v)$. Britton's lemma famously, and conveniently, describes what normal forms are in the case of HNN extensions. **Does anybody know of a reference** for the fact that given $w$ as above, there is a sequence $w=w\_0 \to w\_1 \to \cdots \to w\_r$ of decreasing *syllable length* such that $w\_r$ is reduced (i.e. has minimal syllable length among all words representing the same group element)?
Again, I can prove this myself, but a reference would be convenient.
| https://mathoverflow.net/users/38698 | Passing to normal forms in graphs of groups | In comments, the OP indicates that what they really want is a *uniqueness* result for reduced words in arbitrary graphs of groups. (Indeed, what the question actually asks for, that any word can be transformed into a reduced word by successively cancelling, is obvious by induction on length.)
The desired uniqueness result exists, but people don't usually write it down in full generality, because it's a bit of a mess to state. Instead, they usually write down the following morally equivalent result, which can be thought of as uniqueness for the trivial element.
**Theorem:** Any word in the fundamental group of a graph of groups that represents the trivial element is not reduced.
For instance, this is Theorem 11 of Serre's classic *Trees*.
The appropriate uniqueness result follows immediately: given two reduced words $u,v$ both representing the same element $g$, we have that $uv^{-1}$ represents the trivial element, and so is not reduced. Since $u$ and $v$ are themselves reduced, the only possible cancellation occurs at the concatenation point of the words. Now apply this cancellation and induct on length. Following convention, I will leave it as an exercise to state the uniqueness result that this proves. ;)
Finally, I'll add an important philosophical remark, of which I'm sure the OP is well aware. The existence and uniqueness of normal forms for graphs of groups is equivalent to the statement that the Bass--Serre tree is a tree -- existence shows that the graph is connected, and uniqueness shows that it has no loops. So if you're content to prove that it's a tree via another technique (e.g. the topological approach of Scott--Wall) then you can also deduce existence and uniqueness.
| 6 | https://mathoverflow.net/users/1463 | 443408 | 178,847 |
https://mathoverflow.net/questions/443068 | 1 | In [Juven Wang, Zheng-Cheng Gu, and Xiao-Gang Wen - Field theory representation of gauge-gravity symmetry-protected topological invariants, group cohomology and beyond](https://arxiv.org/abs/1405.7689), the authors calculated many group cocycles in explicit form. See [n-cocycles of finite abelian groups from cohomology group](https://mathoverflow.net/questions/154885/n-cocycles-of-finite-abelian-groups-from-cohomology-group) for more discussions. In all of those examples, the group cocycle is always additive with respect to at least 1 argument, e.g.
$$\omega(a,b,c)=\exp\Big(\frac{2\pi ip}{N^{2}}a(b+c-[b+c]\_{N})\Big)$$
(for some 3-cocyles in $H^{3}(\mathbb{Z}\_{N};U(1))$) where
$$[b+c]\_{N}=\begin{cases}b+c, & 0\leq b+c<N\\
b+c-N, & b+c\geq N.
\end{cases}$$
Similarly, for some 4-cocycles in $H^{4}(\mathbb{Z}\_{N\_{1}}\times\mathbb{Z}\_{N\_{2}};U(1))$
$$\omega((a\_{1},a\_{2}),\dotsc,(d\_{1},d\_{2}))=\exp\Big(\frac{2\pi ip\_{II(12)}}{N\_{12}N\_{2}}a\_{1}b\_{2}(c\_{2}+d\_{2}-[c\_{2}+d\_{2}]\_{N\_{2}}\Big)$$
where $N\_{12}:=\gcd(N\_{1},N\_{2})$, see the first ref above for more examples.
The additivity is not so obvious in the first place (since group cochains are not homomorphism in general).
Is there a simple reason for cocycles to be additive (w.r.t. say, the first argument)? I didn't find an explanation in my homological algebra textbook (and I am sorry if I missed something obvious because I am a physicist).
(I am also wondering how general the additivity is.)
| https://mathoverflow.net/users/500508 | Additivity of group cocycles? | There are plenty of cocycles which are not additive in any variable, and plenty of cohomology classes that do not admit additive representatives. For example, if $G$ is finite, then there are no [nontrivial] additive functions $G \to \mathbb{Z}$, and so no [nontrivial] $\mathbb{Z}$-valued cohomology class has an additive (in any variable) representative.
What's happening in your examples is that the class in question is a cup product. In the first case, you have the group $G = \mathbb{Z}\_N$, and recall that as a ring $\operatorname{H}^\bullet(G; \mathbb{Z}) = \mathbb{Z}[\phi]/N\phi$, where $\phi \in \operatorname{H}^2$ is the generator. Recall also that the Bockstein map $\beta : \operatorname{H}^{\bullet-1}(G; U(1)) \to \operatorname{H}^\bullet(G; \mathbb{Z})$ is an isomorphism, and that $\beta^{-1}\phi \in \operatorname{H}^{1}(G; U(1)) = \hom(G, U(1))$ is the standard inclusion $a \mapsto \exp(2\pi i a / N)$. But the Bockstein map is $\operatorname{H}^\bullet(G; \mathbb{Z})$-linear, and so $\omega = \beta^{-1}(\phi^2) = \phi \cup \beta^{-1}(\phi)$. The additivity that you observe follows from the additivity of $\beta^{-1}(\phi)$.
The second case is similar: the class you care about is a cup product of a class in $\operatorname{H}^2(G; \mathbb{Z})$ with a class in $\operatorname{H}^2(G; U(1))$. This latter class comes from a bihomomorphism.
In general, multihomomorphisms do supply cocycles. Usually the map from multihomomorphisms to cohomology classes is neither injective nor surjective, but it sometimes is, especially in low degree and with specific coefficients. For example, $\operatorname{H}^1(G;A) = \hom(G,A)$ is always a space of homomorphisms. In degree 2, the Künneth formula shows that for $G$ finite abelian, $\operatorname{H}^2(G; U(1))$ is surjected (but not injected) by the set of bihomomorphisms. If $G$ is an elementary abelian $2$-group, then $\operatorname{H}^\bullet(G; \mathbb{F}\_2) = \operatorname{Sym}^\bullet(\hom(G, \mathbb{F}\_2))$ manifestly consists entirely of multihomomorphisms. Furthermore, in this case the standard inclusion $\mathbb{F}\_2 \hookrightarrow U(1)$ induces a surjection $\operatorname{H}^\bullet(G; \mathbb{F}\_2) \to \operatorname{H}^\bullet(G; U(1))$, so again you find that every class is represented by a multihomomorphism. If $G$ is an elementary abelian $p$-group with $p$ odd, then $\operatorname{H}^\bullet(G; \mathbb{F}\_p) = \operatorname{Alt}^\bullet(\hom(G, \mathbb{F}\_p)) \otimes \operatorname{Sym}^\bullet(\hom(G, \mathbb{F}\_p))$, with the $\operatorname{Alt}$ part the subalgebra generated in degree $1$; that $\operatorname{Alt}$ part manifestly consists of multihomomorphisms. This $\operatorname{Alt}$ part injects (via the standard $\mathbb{F}\_p \hookrightarrow U(1)$) into the $U(1)$-cohomology, giving you some classes you can easily realize multihomomorphicly. Further cup products then supply cocycles which are partially-multihomomorphic, i.e. additive in some but not all variables.
| 3 | https://mathoverflow.net/users/78 | 443409 | 178,848 |
https://mathoverflow.net/questions/443406 | 5 | Let f be a polynomial with integer coefficients.
Let B(f) be the set of all values of f on positive integers.
B(f) = {f(n)| n is a positive integer} = {f(1), f(2), ...}
A positive integer k is called "good" if it is a sum of distinct members of B(f).
Otherwise we say k is "bad".
For instance, if f(x)=x^3, then it is known that there are finitely many bad numbers. In other words, all but finitely many natural numbers can be written as sum of distinct cubes.
If f(x)=2x^2 + 2, then there are infinitely many bad numbers since every odd number is bad.
My question : if gcd of coefficients of f is 1, are there only finitely many bad numbers?
| https://mathoverflow.net/users/30650 | Covering all but finitely many integers via some given polynomials | With the right adaptations, the answer should be yes. In particular, the trivially necessary assumption is that the gcd of the values (not just of the coefficients) is 1. Then it is shown in
K. F. Roth, G. Szekeres, Some asymptotic formulae in the theory of partitions (1954) that every sufficiently large integer is a sum of distinct values of $f$.
| 9 | https://mathoverflow.net/users/127660 | 443410 | 178,849 |
https://mathoverflow.net/questions/427785 | 3 | Let $X$ be an affine variety and $G$ an affine algebraic group (for example $\operatorname{PGL}\_n$). How do I compute the Selmer set
$$ \operatorname{Sel}\_\zeta(\mathbb{Q},G) = \{\tau \in H^1(\mathbb{Q},G) \ | \ \tau\_\nu \in \zeta(X(\mathbb{Q}\_\nu)) \ \text{for all places} \ \nu\} $$
where $(\zeta \mapsto \zeta(x))$,
$H^1(X,G) \to H^1(\mathbb{Q}\_\nu,G)$ comes from a point $x: \operatorname{Spec}(\mathbb{Q}\_\nu) \to X$ and $(\tau \mapsto \tau\_\nu)$ comes from the map $\mathbb{Q} \to \mathbb{Q}\_\nu$?
| https://mathoverflow.net/users/489009 | How to compute Selmer set? | Assume that $X(\mathbb{Q}\_\nu)$ is nonempty for every place $\nu$. Since $X$ is affine, $H^1(X,G)$ is trivial. This means that $\zeta(x) = e$ for every $x \in X(\mathbb{Q}\_\nu)$, which means that $e \in \operatorname{Sel}\_\zeta(\mathbb{Q},G)$. So the Selmer set is nonempty.
| -1 | https://mathoverflow.net/users/489009 | 443420 | 178,851 |
https://mathoverflow.net/questions/443392 | 3 | Suppose $\mathcal{M} = (M, +, 0)$ is a cancelable commutative monoid. Let $G$ be the maximal subgroup of $M$, i.e.
$$G = \{ a \in M \colon (\exists b \in M)\, a + b = 0 \}.$$
For $a, b \in M$ say $a \preceq b$ if $(\exists c \in M)\, a + c = b$ and say $a \equiv b$ if $a \preceq b$ and $b \preceq a$. Note for all $a$, $\{b \in M \colon a \equiv b\} = \{a + g \colon g \in G\}$. In particular, if $a \equiv b$ and $c \equiv d$ then $a + b \equiv c + d$ and so there is a monoid $\mathcal{M}/\equiv$.
Is it always the case that $\mathcal{M}\cong (\mathcal{M}/ \equiv) \times G$ as monoids?
What if $G$ is finite?
What if for all $a \in M$, $\{b \in M\colon b\preceq a\}$ is finite?
| https://mathoverflow.net/users/8106 | Cancelable commutative monoids with finite maximal subgroups | Here is a general construction that encompasses @R. van Dobben de Bruyn's example but the idea is taken from his answer. I'll write commutative monoids additively. I'll use $K(M)$ for the Grothendieck group of a commutative monoid $M$ and $M^\times$ for the group of units.
Let $0\to A\to B\xrightarrow{f} C\to 0$ be any exact sequence of finitely generated abelian groups with $C\ncong A\times B$, e.g., $0\to2\mathbb Z/4\mathbb Z\to \mathbb Z/4\mathbb Z\to \mathbb Z/2\mathbb Z\to 0$. Let $Y$ be a finite generating set for $B$.
Let $N$ be any nontrivial finitely generated cancellative monoid with trivial group of units, say with finite set of generators $X$ with $0\notin X$. Let $M$ be the submonoid of $N\times B$ generated by $N\times \{0\}$, $\{0\}\times A$ and $X\times Y$. Notice that $K(M)\cong K(N)\times B$ by construction since $(0,y)=(x,y)-(x,0)$ for $x\in X$ and $y\in Y$. By construction, $M^\times = \{0\}\times A$ and $M/M^{\times}$ is isomorphic to the submonoid of $N\times C$ generated by $N\times \{0\}$ and $X\times f(Y)$. In particular $K(M/M^\times)\cong K(N)\times C$ by the same argument as before. Thus if $M\cong (M/M^\times) \times M^\times$, then $K(N)\times C\cong K(N)\times A\times B$. But then $C\cong A\times B$ by the structure theorem for finitely generated abelian groups, a contradiction.
| 4 | https://mathoverflow.net/users/15934 | 443433 | 178,856 |
https://mathoverflow.net/questions/443432 | 2 | Suppose $(Z\_1, Z\_2)$ is the zero-mean bivariate normal distribution with covariance $\left( \begin{matrix} 1 & \rho; \\ \rho & 1\end{matrix} \right)$ with positive $\rho > 0$. What I want to know a valuable tighy lower bound of the probability
$$\mathrm{P} \left( Z\_1 > z\_1, Z\_2 < z\_2\right)$$
where $z\_1, z\_2$ are some positive constants.
PS:
There are various valuable lower bound for the tail probability $\mathrm{P} \left( Z\_1 > z\_1, Z\_2 > z\_2\right)$ with positive $\rho > 0$, see p495-p499 in [Continuous Bivariate Distributions](https://link.springer.com/book/10.1007/b101765). Can we transform $\mathrm{P} \left( Z\_1 > z\_1, - Z\_2 > -z\_2\right)$ to $\mathrm{P} \left( Z\_1 > z\_1, Z\_2 > z\_2\right)$ keeping the correlation coefficient positive?
| https://mathoverflow.net/users/153595 | The lower bound of bivariate normal distribution | Let $h:=z\_1$, $k:=z\_2$, and $r:=\rho\ge0$. We want to lower-bound $P(Z\_1>h,Z\_2<k)$. Formula (2.11) in the [paper by Willink](https://www.tandfonline.com/doi/abs/10.1081/STA-200031505) (cited the book you linked) gives the following upper bound on $P(Z\_1>h,Z\_2>k)$:
$$P(Z\_1>h,Z\_2>k) \\
\le\Phi(-h)\Big[\Phi\Big(\frac{rh-k}{\sqrt{1-r^2}}\Big)
+re^{(h^2-k^2)/2}\,\Phi\Big(\frac{rk-h}{\sqrt{1-r^2}}\Big)\Big] \tag{1}\label{1}$$
for $h>0$ and states certain optimality properties of this bound; here, as usual, $\Phi$ denotes the standard normal cdf.
Since $P(Z\_1>h,Z\_2<k)=1-\Phi(h)-P(Z\_1>h,Z\_2>k)=\Phi(-h)-P(Z\_1>h,Z\_2>k)$, immediately from \eqref{1} we get the following lower bound on $P(Z\_1>h,Z\_2<k)$:
$$P(Z\_1>h,Z\_2<k) \\
\ge\Phi(-h)\Big[\Phi\Big(\frac{k-rh}{\sqrt{1-r^2}}\Big)
-re^{(h^2-k^2)/2}\,\Phi\Big(\frac{rk-h}{\sqrt{1-r^2}}\Big)\Big]. \tag{2}\label{2}$$
Inequality \eqref{2} will turn into the equality when $r=0$.
| 2 | https://mathoverflow.net/users/36721 | 443435 | 178,857 |
https://mathoverflow.net/questions/443444 | 11 | Let $k\in\mathbb{Z}\_{>0}$, and $s\in\mathbb{N}$, and for $m\_1,\ldots,m\_k$ some nonnegative integers, consider the problem of maximizing the product
$$
(1+m\_1)(1+m\_2)\cdots(1+m\_k)
$$
under the constraint $m\_1+\cdots+m\_k=s$.
I would like to know:
1. The exact formula $M(k,s)$ for the maximal value of the product.
2. A complete description of the tuples $(m\_1,\ldots,m\_k)$ which achieve the maximum.
Of course with the $m\_i$ taking continuous real values, this is just the equality case of the arithmetic mean-geometric mean inequality and maximization calls for $m\_i$'s that are "as equal to each other as possible", but forcing integer values makes this rather messy.
I can try to work my way through this, but it would feel like reinventing the wheel.
A solution to 1. and 2. or pointer to the relevant literature would be appreciated.
| https://mathoverflow.net/users/7410 | A discrete optimization problem related to the AM-GM inequality | This was essentially answered by Nate in the comments, but here are some more details. As Nate argues, $|m\_i - m\_j| \leq 1$ for all distinct $i,j$. Thus, if $s=ak+r$, where $a,r \in \mathbb{N}$ and $r < k$, then there is a unique choice (up to permuting variables) which maximizes the product. Namely, set $r$ of the variables to $\lceil s/k \rceil$ and the rest to $\lfloor s/k \rfloor$.
This problem is related to [Turán's Theorem](https://en.wikipedia.org/wiki/Tur%C3%A1n%27s_theorem), which concerns the maximum possible number of edges in a graph on $s$ vertices with no $K\_{k+1}$ subgraph. The answer is given by the Turán graph, which is unique. More generally, Zykov proved that among $K\_{k+1}$-free graphs, the Turán graph also has the most number of complete graphs $K\_t$ for all $t \leq k$. The case $t=k$ leads to your optimization problem (once you have already established that such a graph must be $k$-partite).
| 10 | https://mathoverflow.net/users/2233 | 443448 | 178,861 |
https://mathoverflow.net/questions/443438 | 1 | Consider the (inhomogeneous) minimal surface equation for functions $u,f:D\to \mathbb{R}$ for some smooth domain $D\subset \mathbb{R}^n$
$$Lu:=\operatorname{div} \frac{\nabla u}{\sqrt{1+|\nabla u|^2}}=f.$$
Then is it true that $u\_1\leq u\_2$ on $\mathbb{R}^n\setminus D$ and $Lu\_1\geq Lu\_2$ implies $u\_1\leq u\_2$ on $D$?
| https://mathoverflow.net/users/68232 | comparison principle for the minimal surface equation | For the case that $D$ is bounded, as Leo Moos has noted, more details can be found in a textbook of minimal surface, cf. Lemma 1.26 in T. Colding, W. Minicozzi, *A Course in Minimal Surfaces*.
That is, let $F(X)=\frac{X}{\sqrt{1+|X|^2}},$ then
\begin{equation}
F(\nabla u\_1)-F(\nabla u\_2)=\left(\int\_0^1dF\big(\nabla u\_2+t(\nabla u\_1-\nabla u\_2)\big)
dt\right)(\nabla u\_1-\nabla u\_2).
\end{equation}
From this, one can conclude that $v=u\_1-u\_2$ satisfies an equation of the form
$$\operatorname{div}(a\_{i,j}\nabla v)\leq 0,$$
where the matrix is defined as $(a\_{i,j})=\int\_0^1dF(\nabla u\_2+t(\nabla u\_1-\nabla u\_2))
dt.$
In particular, for a unit vector $V$ and a vector $X$, we have
$$dF(x)V=\frac{V}{\sqrt{1+|X|^2}}-\frac{\langle V,X\rangle}{(1+|X|^2)^{\frac{3}{2}}}X.$$
Thus,
$$(1+|X|^2)^{\frac{3}{2}}\langle V, dF(X)V\rangle=(1+|X|^2)-\langle V,X\rangle^2\geq 1, $$
which means $(a\_{i,j})$ is positive, therefore, the usual comparison principle gives the claim.
| 3 | https://mathoverflow.net/users/166368 | 443452 | 178,863 |
https://mathoverflow.net/questions/443450 | -1 | [This answer](https://mathoverflow.net/a/295742/501568) says that if $X$ is a random variable and $X\_+ = \mathrm{max}(0, X)$, then $X\_+ = \int\_0^\infty I\_{\{X > x\}}\mathrm{d}x$. I'd like to know how to derive this starting with $A \in \mathcal{S} \implies \int\_S 1\_A\mathrm{d}\mu = \mu(A)$ (from ["Desired Properties"](https://stats.libretexts.org/Bookshelves/Probability_Theory/Probability_Mathematical_Statistics_and_Stochastic_Processes_(Siegrist)/03:_Distributions/3.10:_The_Integral_With_Respect_to_a_Measure)).
My thinking is that for a probability space $(\Omega, F, \mu)$, $\{X > t\} \in F \implies \int\_\Omega I\_{\{X > t\}}\mathrm{d}\mu = \mu(\{X > t\})$ but it's the wrong variable of integration. Am I on the right track?
| https://mathoverflow.net/users/501568 | Random variable as an integral of an indicator function | The layer cake representation of a non-negative measurable function, $X$, is applied in the proof of proposition 2.1 [here](https://stat.uiowa.edu/sites/stat.uiowa.edu/files/cae/Lo_Expectation.pdf).
| 0 | https://mathoverflow.net/users/501568 | 443470 | 178,869 |
https://mathoverflow.net/questions/442086 | 7 | In Yau and Schoen's differential geometry,in Ch5 before Thm 3.5,the author says
When $R$(scalar curvature of a manifold M)$>0$,there exists a unique Green's function $G$ to the operator $L=-\Delta+aR$ and $LG\_{P}=\delta\_{P}$.Here $P$ is an arbitrary point of $M$.
In the normal coordinate of $P$,we have
$$
G\_P(x)=\frac{1}{(n-2) \omega\_{n-1}} r^{2-n}(1+o(1)).
$$
I wonder where can I find this existence result and how to get the expansion at point $P$.
| https://mathoverflow.net/users/148247 | Existence and estimates of Green's function on Riemannian manifold | For further details on the existence argument, see Chapter 4 of Aubin's book, "Nonlinear Analysis on Manifolds: Monge-Ampère Equations" [1] for the harmonic case, and [2] for the metaharmonic case. The asymptotic expansion formula near $P$ is derived from the first term of the Green function, which is obtained by multiplying the fundamental solution in $\mathbb{R}^n$ by a function that equals 1 near $P$.
[1] T. Aubin, "Nonlinear Analysis on Manifolds: Monge-Ampère Equations," Springer-Verlag New York, 1982.
[2] Yanir A. Rubinstein, "Lecture 16: Metaharmonic Functions," University of Maryland, 2019. Available at [http://www.math.umd.edu/~yanir/742/742-16-19.pdf](http://www.math.umd.edu/%7Eyanir/742/742-16-19.pdf).
| 3 | https://mathoverflow.net/users/166368 | 443480 | 178,876 |
https://mathoverflow.net/questions/443488 | 7 | The Lie operad $\text{Lie}$ is generated by a binary operator $[\ ,\ ]$, modulo a degree two relation (skew commutativity $[x,y]=-[y,x]$) and a degree three relation (Jacobi $[x,[y,z]]+[y,[z,x]]+[z,[x,y]]$). See e.g. the paper ["The operad Lie is free"](https://arxiv.org/pdf/0802.3010.pdf).
**Question:** do "non(skew)commutative Lie algebras" ever show up in nature? i.e. is there an operad $\text{Lie}^{\mathrm{nc}}$ generated by a binary operator $[\ ,\ ]$ subject to Jacobi *and some subset of the relations generated by Jacobi and skew-commutativity* (so there is a map $\text{Lie}^{\mathrm{nc}}\to \text{Lie}$) *but not including skew-commutativity*, and so that its algebras $\text{Alg}\_{\text{Lie}^{\mathrm{nc}}}$ show up in e.g. algebra or geometry?
| https://mathoverflow.net/users/119012 | Non(skew)commutative Lie algebras? | Such objects are known as *Leibniz algebras*.
A Leibniz algebra is a module $M$ together with a bilinear pairing $$[-,-]\colon M⊗M→M$$ that satisfies the Leibniz identity: $$[a,[b,c]]=[[a,b],c]+[b,[a,c]].$$
Leibniz algebras for which $[a,b]=-[b,a]$ are precisely Lie algebras.
The concept was introduced and studied by A. Blokh: [A generalization of the concept of a Lie algebra](https://www.mathnet.ru/eng/dan/v165/i3/p471).
Later, it was studied by Loday, who also coined the name “Lebniz algebra”: [Une version non commutative des algèbres de Lie: les algèbres de Leibniz](http://www.numdam.org/item/RCP25_1993__44__127_0/).
The article [Leibniz algebra](https://ncatlab.org/nlab/show/Leibniz+algebra) contains many more additional references.
| 9 | https://mathoverflow.net/users/402 | 443494 | 178,879 |
https://mathoverflow.net/questions/443239 | 6 | I am trying to understand the following situation for $G=GL(2)$, when going from the compact trace formula to the non-compact case.
The integral over $G(\mathbb{A})^1\_\gamma \backslash G(\mathbb{A})^1$ may diverge for a given conjugacy class $\gamma$. Say $\gamma$ is the matrix of the translation by $1$. Then $G(\mathbb{A})\_\gamma$ is the set of upper triangular matrices with same diagonal entries. I wan to compute the integral
$$\int\_{G(\mathbb{A})^1\_\gamma \backslash G(\mathbb{A})^1} f(x^{-1}\gamma x)dx$$
to see why it diverges. In Arthur's notes, it reduces to
$$\int\_{G\_\gamma(\mathbb{A}) \backslash P(\mathbb{A}) } \int\_{P(\mathbb{A}) \backslash G(\mathbb{A})} f((pk)^{-1} \gamma pk) dpdk$$
and this reduces apparently to a constant times
$$\prod\_p (1-p^{-1})^{-1}$$
which is infinite (except when the constant is zero). I do not understand this computation (that I would also like to see for more general groups).
| https://mathoverflow.net/users/128718 | Divergence of integrals in the trace formula | I take it that the question is "Why does this iterated integral diverge?," but correct me if that is not the question.
The issue here is the outer integral: when $\gamma$ is as you describe, then if $P$ is the group of upper triangular matrices, the quotient $G\_\gamma\backslash P$ is a rank $1$ split torus, the adelic integral over which is infinite. Things get more complicated for regular unipotent elements in higher rank, but the problem is always that the orbit is not closed in $G$. In this case, the matrix $I\_2$ lies in the closure for example.
This non-closed orbit issue is morally why the adelic integral diverges. This is visible in the calculation, so I'll quickly sketch it. The idea is that if this converges, it converges to the product of all the local integrals (which we will see do converge). That is, we can assume that $f= \prod\_p f\_p$ is a pure tensor in the space of test functions, and consider the local integrals
$$
\int\_{G\_\gamma(\mathbb{\mathbb{Q}\_p}) \backslash P(\mathbb{Q}\_p) } \int\_{P(\mathbb{Q}\_p) \backslash G(\mathbb{Q}\_p)} f\_p((bk)^{-1} \gamma bk) dkdb
$$ for each $p\leq \infty$. For this to make sense, we need $f\_p = \mathbb{1}\_{GL\_2(\mathbb{Z}\_p)}$ to be the characteristic function of the integral points for almost all $p$.
The inner integral is compact since $P\backslash GL\_2$ is just a projective line, so won't impact whether the iterated integral diverges or not. In particular, there is a smooth function $f'\_p$ (just shoving the compact integral into the notation) such that
$$
\int\_{G\_\gamma(\mathbb{Q}\_p) \backslash P(\mathbb{Q}\_p) } \int\_{P(\mathbb{Q}\_p) \backslash G(\mathbb{Q}\_p)} f\_p((bk)^{-1} \gamma bk) dkdb =\int\_{G\_\gamma(\mathbb{Q}\_p) \backslash P(\mathbb{Q}\_p) } f'\_p(b^{-1} \gamma b) db.
$$
Note that we still have $f'\_p = \mathbf{1}\_{GL\_2(\mathbb{Z}\_p)}$ for almost every prime $p$. At such primes, the integral is
$$
\int\_{\mathbb{Q}\_p^\times } \mathbb{1}\_{GL\_2(\mathbb{Z}\_p)}\left(\begin{pmatrix}t& \\ &1\end{pmatrix} \begin{pmatrix}1&1 \\ &1\end{pmatrix}\begin{pmatrix}t^{-1}& \\ &1\end{pmatrix}\right) dt =\int\_{\mathbb{Q}\_p^\times } \mathbb{1}\_{GL\_2(\mathbb{Z}\_p)}\left(\begin{pmatrix}1&t \\ &1\end{pmatrix}\right) dt .
$$
If we normalize the Haar measure on $\mathbb{Q}\_p^\times$ so that $vol(\mathbb{Z}\_p^\times)=1$, this last integral is $\sum\_{i=0}^\infty p^{-i}=(1-p^{-1})^{-1}$, converging to the local zeta value $\zeta\_p(1)$. Notice that this integral ``sees'' the limit point of $G\_\gamma\backslash G$, since as $t\to 0$,
$$\begin{pmatrix}1&t \\ &1\end{pmatrix} \longmapsto \begin{pmatrix}1& \\ &1\end{pmatrix}. $$
Formally taking the product over all $p$, you see the divergent product $\prod\_p(1-p^{-1})^{-1}$ times some factor coming from finitely many places where $f'\_p\neq\mathbb{1}\_{GL\_2(\mathbb{Z}\_p)}$. This factor could vanish for a given $f$, but it doesn't always so that the integral generally diverges.
In general, orbital integrals for non-semisimple elements will be integrals over non-closed orbits, so even though the local integrals make sense, the global need some notion of regularization to give a well-defined distribution.
| 4 | https://mathoverflow.net/users/62154 | 443508 | 178,881 |
https://mathoverflow.net/questions/443443 | 6 | In [Generalisation of the quantum exterior algebra](https://mathoverflow.net/questions/258512/generalisation-of-the-quantum-exterior-algebra) the quantum exterior algebra is discussed:
$$
K\langle x\_1,\dotsc x\_n\rangle/(x\_i^2,x\_i x\_j + q\_{i,j}x\_j x\_i),
$$
with nonzero field elements $q\_{i,j}$ for $i<j$.
Where was this algebra first introduced and is there a standard text about such things? Is the standard name really the *quantum exterior algebra*? I have searched for it but not had any luck finding a good reference.
Moreover, where is the dimension of such algebras calculated? It must surely be the same as the normal exterior algebra . . .
| https://mathoverflow.net/users/499575 | Quantum exterior algebra | Quantum affine space has been studied since the nineties. The coordinate ring of quantum affine space is a quantum polynomial algebra, whose definition I think you can imagine, and the Koszul dual of a quantum polynomial algebra is a quantum exterior algebra. There are conditions on the $q\_{i,j}$ in order for the algebras constructed this way to be associative (or, if you like, to have the dimension you imagine them to have), and I'm sure you can work these out yourself. Quantum complete intersections started appearing soon afterwards, and have been studied by many people including Lucho Avramov. They started becoming relevant to the study of block theory of finite groups due to an example of Jon Alperin, that I further studied with Ed Green, and later with Radha Kessar. Hochschild cohomology and support varieties have also been studied, see the work of Petter Bergh, of Karin Erdmann, and of Steffen Oppermann.
| 7 | https://mathoverflow.net/users/460592 | 443512 | 178,882 |
https://mathoverflow.net/questions/443491 | 1 | Let $c>0$ be a very small constant and $N \in \mathbb N$ very large.
Assume we have a function $f(x)$ for $x \in S^1$ defined as
$$
f(x) = \sum\_{k=\lfloor N/(1+c) \rfloor}^{N} c\_k \sin(kx+b\_k)
$$
for some coefficients $b\_k, c\_k \in \mathbb R$, that is all the sines appearing have frequency similar to $N$.
I know that $|f(x)|\leq 1$ on $[-\pi/4, \pi/4]$. Can I get a good bound for arbitrary $x \in S^1$?
I'd love to get a bound independent of $N$ but I am not sure whether I can expect that.
I'd also love references where the authors treat these kinds of objects, but **I do not know what this kind of functions are called in the literature**.
| https://mathoverflow.net/users/173610 | Inequality for sums of sines with similar frequency | In other words, in particular, you want to create an entire function $f$ of exponential type $a=\pi/4$ bounded on the real axis by $C$ and satisfying $f(n)=(-1)^n$ near $N$ on an interval of length $cN$. Note that by symmetrizing ($F(z)=\frac 12(f(z)+\bar f(\bar z))$), you can make it real-valued on $\mathbb R$, and then it will have a zero between any two integers in that interval. Now Jensen in the disk of comparable radius kills all your hopes: the best bound you can get is still exponential in $N$.
| 2 | https://mathoverflow.net/users/1131 | 443514 | 178,883 |
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