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https://mathoverflow.net/questions/425569 | 5 | I am interested in the comparison between two different constructions which, as far as I can tell, are both supposed to produce rigorous constructions of Wess–Zumino-Witten conformal blocks.
More precisely, on the one hand, we have the construction of Reshetikhin–Turaev as well as Blanchet–Habegger–Masbaum–Vogel of "quantum representations" of mapping class groups of Riemann surfaces (possibly with marked points): see e.g. Section 3 of [Masbaum - Quantum representations of mapping class groups](https://webusers.imj-prg.fr/%7Egregor.masbaum/Masbaum.pdf) for a nice survey. On the other hand, the book of Bezrukavnikov–Finkelberg–Schechtman (BFS) [Factorizable sheaves and quantum groups](https://arxiv.org/abs/q-alg/9712001) produces sheaves over certain stacks $\mathcal{M}\_{A, \delta}$, which are, roughly, (a line bundle over) moduli stacks of curves with marked points **and** non-zero tangent vectors at these marked points; see for example Section 1.5 on page 9.
Question: how are these two constructions related?
I believe that they should be closely related, and this is perhaps well known to experts. I am confused by the appearance of tangent vectors in the moduli stacks $\mathcal{M}\_{A, \delta}$ above, though I have perhaps misunderstood what are supposed to be the conformal blocks in BFS. Any comments or references would be greatly appreciated!
| https://mathoverflow.net/users/484855 | Relation between TQFT representations and factorizable sheaves | To sum up: these tangent vectors are always present. In the literature on WZW one usually choose local formal coordinates at the marked point, ie an identification of the neighborhood of those with a chosen formal punctured disc, but this amount to essentially the same thing. The reason is compatibility with gluing: the moduli space of algebraic curves of genus $g$ with $n$ marked points with tangent vectors, is equivalent to the moduli space of genus $g$ surfaces with $n$ boundary components **together with a parametrization of the boundary** which is what you need to glue.
Btw, I wouldn't say BFS is a rigorous construction of the conformal blocks of the WZW model (this can already be made rigorous). Rather, my understanding is that it is a geometric construction of the modular functor associated with the modular tensor category of a quantum group at root of unity (hence, in particular, a direct geometric construction of that category as well). I think the point is to have a construction that looks like the construction of conformal blocks in WZW but on the quantum group side.
Once you know the modular tensor categories you get as the value of the circle for either of those constructions are the same, you can use e.g. Andersen-Ueno result (<https://arxiv.org/abs/math/0611087>) that a modular functor (in a somewhat restrictive sense) is determined by its genus 0 part to conclude these are equivalent.
| 1 | https://mathoverflow.net/users/13552 | 425676 | 172,813 |
https://mathoverflow.net/questions/425610 | 2 | Consider the following PDE on $\Omega\subset \mathbb{R}^n$ for $n\geq 2:$
\begin{align}
\Delta u - x\cdot \nabla u &= f(x),\text{ in } \Omega\\
u&=0 \text{ on }\partial \Omega
\end{align}
Are there any explicit expressions for a kernel $K$ such that,
$$u(x)=\int\_{\Omega} K(x,y)f(y)dy$$ when $\Omega=\mathbb{R}^n$ or $\Omega=B(0,1)$?
| https://mathoverflow.net/users/68232 | Kernel for an equation involving the Ornstein-Uhlenbeck operator | There is a trick that reduces the equation $u\_t=Lu$, $L=\Delta-x \nabla$ to the heat equation $u\_t-\Delta$. It is genuinely parabolic and gives the parabolic kernel in the whole space, from which the elliptic kernel can be deduced by integrating in time.
If $v\_t(t,x)=\Delta v(t,x)-x\nabla v (t,x)$ with $v(0,x)=f(x)$, then $u(t,x)=e^{-2t}v(t, e^tx)$ solves $u\_t(t,x)=\Delta u(t,x)-2u(t,x)$ with $u(0,x)=f(x)$.
It does not work in an a ball where probably an expansion in spherical harmonics can give the result for the elliptic case directly.
EDIT. Sorry for the mistake, but the equation for $u$ is wrong. Let us do it in two steps. First put $u(t,x)=v(t, e^t x)$. Then $u\_t(t,x)=e^{-2t}\Delta u(t,x)$ with $u(0,x)=f(x)$, which is a simple non-autonomous heat equation. Then, setting $$u(t,x)=w(\frac {1-e^{-2t}}{2}, x)$$ we have $w\_t=\Delta w$ with $w(0,x)=f(x)$.
Therefore the final transformation is $v(t,x)=w(\frac {1-e^{-2t}}{2}, e^{-t}x)$.
| 3 | https://mathoverflow.net/users/150653 | 425690 | 172,817 |
https://mathoverflow.net/questions/425652 | 8 | In his work, Albert Lautman thinks the genesis of some mathematical works as a dialectic that takes place between opposite notions, like between global and local. He argues that while those notions, in the natural language, are obviously linked by their opposition, the *souci* or concern that stems from it can only be studied in mathematics. In other words, the study of the crux of this opposition ends up being a mathematical study. He then suggests that the friction between such pairs of opposite notions, originally in the language, is the origin of many mathematical developments.
I would like to show that there are "lone" notions, that is notions not thought as part of a pair like one/many, intrinsic/extrinsic etc., that led through their study (or can be linked, maybe notwithstanding some historical reality of the study) to interesting mathematical developments. Hence my question :
>
> What are examples of stand-alone notions (in the language) that calls (by their importance and their wide meaning) for a formalisation which can be linked to new mathematical works ?
>
>
>
An example : The development of the theory of computability may be seen as an answer to the *souci* that comes from the informal concept of "computation". I can do a computation on a sheet of paper, Babbage's machine does computation, a sum is a computation... Then what unifies those ideas of computation ? Three attempts to capture the idea behind it have been made. Those are the theory of general recursive functions, Turing machines and the lambda calculus ; each of those embodies some aspect of the notion of computation, and it happens that those three definitions are equivalent. Hence, the Church-Turing thesis states, arguably, that those indeed capture the pre-conceptual notion of "computation".
This will be in a writing addressed to philosophers. So the simpler and the more fundamental, the better.
| https://mathoverflow.net/users/138089 | Formalisation of intuitive concepts in the language leading to mathematical progress | An example that is closely related to computation is proof. Mathematicians have been proving theorems ever since Euclid (and presumably even earlier). But it was not until the 20th century that the concept of a proof was formalized to the point where they could be studied as mathematical objects in their own right.
Though it does not have a well-known name like "Church–Turing Thesis," there is an analogous "formalization thesis" about proofs that I have mentioned [elsewhere on MathOverflow](https://mathoverflow.net/a/24919) as well as on the [Foundations of Mathematics mailing list](https://cs.nyu.edu/pipermail/fom/2010-May/014743.html):
>
> Given any precise mathematical statement, one can exhibit a formal
> sentence S in the first-order language of set theory with the property
> that any mathematically acceptable proof of the original mathematical
> statement can be mimicked to produce a formal proof of S from the axioms
> of ZFC.
>
>
>
| 4 | https://mathoverflow.net/users/3106 | 425692 | 172,818 |
https://mathoverflow.net/questions/425667 | 11 | [It is well known](https://math.stackexchange.com/questions/2965261/homotopy-type-of-the-diffeomorphism-group-of-the-sphere) that the diffeomorphism group of there sphere $\operatorname{Diff}(S^n)$ has the homotopy type of a product $X:=O(n+1)\times \operatorname{Diff}\_{\partial D^n}(D^n)$ of the orthogonal group and the diffeomorphism of the disk $D^n$ with are the identity in a neighborhood of the identity.
>
> Does the classifying space $B\operatorname{Diff}(S^n)$ have the homotopy type of $BX$? If not, do we at least have an isomorphism on the level of cohomology: $H^\ast(B\operatorname{Diff}(S^n);\mathbb{Z})\cong H^\ast(BX;\mathbb{Z})?$
>
>
>
Of course, if there was a homotopy equivalence $\phi:X\rightarrow \operatorname{Diff}(S^n)$ which is simultaneously a group homomorphism, the answer would be "yes". But the only homotopy equivalence I'm aware of (which comes from the link at the start of this post) is not obviously homotopic to a homomorphism.
The linked homotopy equivalence has the property that $\phi$ restricted to either factor is a homomorphism. Using this, it's easy to show that $H^\ast(B\operatorname{Diff}(S^n);\mathbb{Z})$ surjects onto the cohomlogy of each factor of $BX$, but I've been unable to piece this together to get what I want.
Motivation: I'm trying to understand characteristic classes for smooth sphere bundles whose structure group does *not* reduce to $O(n+1)$. It's clear that, except for a few small $n$, one should get "extra" characteristic classes, but I'm trying to understand if one also gets an analogue of Stiefel-Whitney, Euler, and Pontryagin classes satisfying all the "usual" properties, e.g., the mod $2$ reduction of the Euler class should be the top Stiefel-Whitney class; Stiefel-Whitney classes satisfy the Wu-formula, etc. I'd also be happy with pointers to the literature.
| https://mathoverflow.net/users/1708 | Characteristic classes of non-linear sphere bundles | For many values of $n$, the answer to both questions is no. Since the fundamental groups of $BX$ and $B\mathrm{Diff}(S^n)$ are finite for $n\ge5$ (This uses that $\pi\_0\mathrm{Diff}\_\partial(D^n)$ is isomorphic to the group of homotopy $(n+1)$-spheres which is finite by a famous result of Kervaire--Milnor.), their first cohomology groups vanishes, so the first degree in which one could hope to detect a cohomological difference is $2$. And there is indeed a difference in degree 2 in many cases.
To see this, assume $n\ge5$ and note first that since $BX$ and $B\mathrm{Diff}(S^n)$ have finite fundamental groups, if the second cohomology groups were isomorphic, then the same would hold for the first homology group (this uses that $H\_1$ and $H\_2$ are finitely generated and that $H\_1$ is torsion), so it suffices to show that
$$H\_1(B\mathrm{Diff}(S^n))\cong \pi\_0\mathrm{Diff}(S^n)^{\mathrm{ab}}\quad\text{and}\quad H\_1(BX)\cong \pi\_0X^{\mathrm{ab}}$$
are not isomorphic. Since $\pi\_0O(n+1)=\pm1$ and $\pi\_0\mathrm{Diff}\_\partial(D^n)$ are abelian, we have
$$\pi\_0X^{\mathrm{ab}}\cong \pi\_0\mathrm{Diff}\_\partial(D^n)\oplus \pm1.$$ Testing whether a diffeomorphism is orientation preserving gives an extension
$$0\rightarrow \pi\_0\mathrm{Diff}\_\partial(D^n)\rightarrow \pi\_0\mathrm{Diff}(S^n) \rightarrow\pm1\rightarrow 0$$ which admits a splitting induced by the standard $O(n+1)$-action on $S^n$, so we get
$$\pi\_1B\mathrm{Diff}(S^n)^{\mathrm{ab}}\cong\Big(\pi\_0\mathrm{Diff}\_\partial(D^n)\rtimes \pm1\Big)^{\mathrm{ab}}\cong \pi\_0\mathrm{Diff}\_\partial(D^n)\_{\pm 1}\oplus \pm 1$$ where $(-)\_{\pm 1}$ denotes taking coinvariants with respect to the action induced by the extension.
This shows that $H^2(BX)$ and $H^2(B\mathrm{Diff}(S^n))$ cannot be isomorphic as long as the $\pm 1$-action on the finite abelian group $\pi\_0\mathrm{Diff}\_\partial(D^n)$ is nontrivial. This action is given by conjugation with a reflection in one of the coordinates, which can be seen to induce multiplication with $-1$ on $\pi\_0\mathrm{Diff}\_\partial(D^n)$, so the action is nontrivial iff $\pi\_0\mathrm{Diff}\_\partial(D^n)$ is not annihilated by multiplication with $2$. As mentioned above, this group agrees with the group of homotopy $(n+1)$-spheres, which is known to not have this property in many cases (see e.g. the tables on [wikipedia](https://en.wikipedia.org/wiki/Exotic_sphere)).
| 11 | https://mathoverflow.net/users/32022 | 425693 | 172,819 |
https://mathoverflow.net/questions/425190 | 0 | Let $S,\Sigma$ in $\mathbb{R}^d$ be finite measure set. The Amrein-Berthier uncertainty principle states that there exists $C=C(S,\Sigma)>0$ such that for all $f\in L^2(\mathbb{R}^d)$, $\int\_{\mathbb{R}^d} |f|^2\leq C \left(\int\_{\mathbb{R}^d\setminus S} |f|^2+\int\_{\mathbb{R}^d\setminus \Sigma} |\widehat{f}|^2\right)$.
Suppose $C$ is the best constant. It is known that the constant $C$ satisfies $C(S,\Sigma)\leq A e^{A |S||\Sigma|}$ with $A\geq 1$.
Do we also have that $C(S,\Sigma)\to 1$ when $|S||\Sigma|\to 0$ ? In this case, is there a simple equivalent or an estimate of $C(S,\Sigma)-1$ when $|S||\Sigma|\to 0$ ?
| https://mathoverflow.net/users/107004 | Constant in Amrein-Berthier uncertainty principle | I remembered that this theorem is "essential" only when $|S||\Sigma|\geq1$. If $|S||\Sigma|<1$, one has an easier estimate (for $d=1$)
$$||f||^2\_{L^2}\leq\frac{1}{\sqrt{1-|S||\Sigma|}}\bigg(\int\_{S^c}|f|^2+\int\_{\Sigma^c}|\hat{f}|^2\bigg)$$
So you may see that $C\rightarrow1$ as $|S||\Sigma|$ goes to $0$.
I think this should be resaonable also for higher dimension. This inequality is very easy(only few lines) and due to F.Nazarov, you may find the proof in his paper **Local estimates for exponential polynomials and their applications to inequalities of the uncertainty principle type**, at the end of section 2.2.
It is conjectured that in $\mathbb{R}^d$, the sharp estimate of $C$ is of the form
$$C=Ce^{C(|S||\Sigma|)^{\frac{1}{d}}}$$
One can check the sharpness of $C$ by choosing $f$ Gaussian and $S$,$\Sigma$ balls centered at the origin. This sharp estimate has been proved when at least one of $S$ and $\Sigma$ is convex, but still open for general $S$ and $\Sigma$. You may see this in P.Jaming's paper **Nazarov's uncertainty principles in higher dimension**. This is one of my topic in research but sorry I do not know any other esimates on $C$.
| 1 | https://mathoverflow.net/users/484991 | 425703 | 172,821 |
https://mathoverflow.net/questions/425714 | 0 | Let $A \subset \mathbb{R}^{n}$ be a closed set. Does there exist an open set $O$ containing $A$, and a smooth function $f : O \to \mathbb{R} $ such that
$f(x) = 0$ for all $x \in A$,
$f(x) > 0$ and $\nabla f(x) \ne 0$ for all $x \in O \setminus A$,
and $f(x) \to \infty$ as $x \to \partial O$ ?
| https://mathoverflow.net/users/153602 | Existence of a particular positive definite and radially unbounded function | Not always possible.
Let $n = 1$. Take $A \subsetneq [0,1]$ be the usual ternary Cantor set.
Every point in $A$ is a limit point.
Let $O$ be any open set containing $A$, then $O$ must contain *some* interval of the form $(a,b)$ with $a,b\in A$. Since $f(a) = f(b) = 0$ is needed, if we want $f(x) > 0$ on $O\setminus A$, we must have that $f(x)$ attains an interior maximum where it has a critical point.
Similar examples can be built in higher dimensions.
| 3 | https://mathoverflow.net/users/3948 | 425715 | 172,824 |
https://mathoverflow.net/questions/425699 | 2 | Let $M\_t$ be a continuous time real valued martingale, and $\mathcal F\_t$ its natural filtration.
Suppose that $\mathcal F\_t \setminus \mathcal F\_s$ is nonempty for all $t > s$.
Let $\mathcal G$ be a sigma algebra, and define the filtration $\mathcal H\_t := F\_t \vee \mathcal G$.
**Question:** Is it true that $M$ is a $\mathcal H\_t$ martingale if and only if $\mathcal G$ is independent of $\mathcal F\_t$ for all $t$?
*Remark: The if direction follows from a monotone class argument.*
| https://mathoverflow.net/users/173490 | Enlargement of filtration | I think that I have a counterexample. Let $(X,Y)$ be a Brownian motion in $\mathbb{R}^2$. Then $M = \int\_0^\cdot X\_s \mathrm{d}Y\_s$ is a martingale, in the natural filtration of $(X,Y)$, in its own filtration $(\mathcal{F}\_t)\_{t \ge 0}$ and also in $(\mathcal{F}\_t \vee \sigma(X))\_{t \ge 0}$. Yet, $X$ is not independent of $M$ since $\langle M \rangle = \int\_0^\cdot X\_s^2\mathrm{d}s$ is not deterministic.
Remark: in this example, $\mathcal{F}$ is not immersed in $\mathcal{H}$ since $X$ is no more a martingale in $\mathcal{H}$.
| 4 | https://mathoverflow.net/users/169474 | 425723 | 172,826 |
https://mathoverflow.net/questions/425695 | 1 | Let $D\_+$ be the set of non-increasing functions $f: [0,T]\to [0,1]$ that are right-continuous. Let $(f\_n)\_{n\ge 1}\subset D\_+$ be a sequence of continuous functions s.t. $\lim\_{n\to\infty }f\_n(t)$ exits for each $t\in [0,T]$. Denote by $\hat f$ its pointwise limit and by $f$ the right-continuous modification of $\hat f$, i.e. $f(t):=\lim\_{s\searrow t}\hat f(s)$. Denote by $d$ the Skorokhod metric on $D\_+$. I have two questions:
1. Does $\lim\_{n\to\infty}d(f\_n,f)=0$ hold?
2. If not, does $\lim\_{n\to\infty}d(f\_n,f)=0$ hold by assuming additionally that $f\_n$ increases to $f$?
| https://mathoverflow.net/users/nan | Does pointwise convergence yield the convergence under Skorokhod topology? | The answer is no to both questions. E.g., suppose that $T=2$ and
$$f\_n(t)=1(t\le1-\tfrac1n)+n(1-t)1(1-\tfrac1n<t\le1).$$
Then $f\_n(t)$ increases in $n$ to $f(t)=1(t<1)$, and all the other conditions on $f\_n$ hold.
However, by the [definition of the Skorokhod metric](https://en.wikipedia.org/wiki/C%C3%A0dl%C3%A0g#Skorokhod_space), for each integer $n\ge1$, $t\_n:=1-\tfrac1{2n}$, and some strictly increasing continuous function $h\_n\colon[0,2]\to[0,2]$ we have
$$d(f\_n,f)+\tfrac1n\ge|f\_n(t\_n)-f(h\_n(t\_n))|=|\tfrac12-f(h(t\_n))|
=\tfrac12,$$
since $f(t)\in\{0,1\}$ for each $t\in[0,2]$. So, $d(f\_n,f)\not\to0$.
---
The idea of this example is simple: the range of each continuous decreasing function $f$ is connected, but the range of the right-continuous pointwise limit $f$ does not have to be connected, even if $f\_n$ increases to $f$.
It is easy to modify this example to make $f\_n(t)$ strictly decreasing in $n$ and in $t$, if so desired.
| 1 | https://mathoverflow.net/users/36721 | 425724 | 172,827 |
https://mathoverflow.net/questions/425585 | 9 | In the quadratic case, it does. Given an irreducible quadratic polynomial $f(x)=ax^2+bx+c$, the discriminant of the quadratic number field $\frac{\mathbb{Q}[x]}{f(x)}$ is $\operatorname{sqf}(d)$ or $4\cdot \operatorname{sqf}(d)$, depending on if $d \equiv 1 \mod 4$, where $d=b^2-4ac$ is the polynomial discriminant of $f(x)$, and $\operatorname{sqf}$ is the squarefree part.
One can reduce this to the case of local fields, in which one can ask a similar question. Fix a positive integer $n$. Given a separable polynomial $f \in \mathbb{Q}\_p[x]$ of degree $n$, does the discriminant of $f$ determine the valuation of the discriminant ideal of the associated extension? What about in the case where $p \nmid n$?
For the question over local fields, one should note that the dependence on the discriminant of $f$ does *not* factor through the valuation of the discriminant of $f$.
| https://mathoverflow.net/users/482554 | Does the discriminant of an irreducible polynomial of a fixed degree determine the discriminant of the number field it generates? | Another counterexample:
$f\_1(x)=x^3-9x-20, f\_2(x)=x^3-6x-18$, discriminant = $-4\*27\*73$ for both $f\_1$ and $f\_2$.
Both polynomials are irreducible over $\mathbb{Q}$ by the rational root test.
The 2-adic Newton polygons show that $f\_1$ has 2 roots of valuation=0 and 1 root of valuation=2, and $f\_2$ has 3 roots of valuation=1/3. So the ring of integers is ramified over 2 for $f\_2$ but not for $f\_1$, and their discriminants are not equal.
I found these polynomials by looking for integer solutions $(a=-3,b=-10)$ to $3a^2+3a+1=-(2b+1)$ which yield polynomials $f\_1=x^3+3ax+2b, f\_2=x^3+3(a+1)x+2(b+1)$ with same discriminant=$-4\*27\*(b^2+a^3)$.
Note: The initial version of this answer had mistakes as reported in comments below by KConrad.
| 3 | https://mathoverflow.net/users/59248 | 425731 | 172,828 |
https://mathoverflow.net/questions/425733 | 1 | Let $X$ be a Verre-threefold, which is by definition a $(2,2)$ hypersurface in $\mathbb{P}^2\times\mathbb{P}^2$, it is a Fano threefold. What is the semi-orthogonal decomposition of $D^b(X)$? It seems that there are very few materials in the literature. Is the residue category of $X$ an Enriques category in the sense that its Serre-functor is a involution up to shift by $2$?
| https://mathoverflow.net/users/41650 | Semi-orthogonal decomposition of Verra threefold | The projection of $X$ to each factor is a conic bundle, therefore there is a decomposition
$$
D^b(X) = \langle D^b(\mathbb{P}^2), D^b(\mathbb{P}^2,\mathcal{B}\_0) \rangle,
$$
where $\mathcal{B}\_0$ is the even part of the Clifford algebra. The second component is indeed an Enriques category.
| 3 | https://mathoverflow.net/users/4428 | 425739 | 172,832 |
https://mathoverflow.net/questions/425738 | 1 | Let $Y$ be a smooth algebraic variety and $i: X\hookrightarrow Y$ be a smooth divisor. We consider the derived functors $i^\*: D^b\_{coh}(Y)\to D^b\_{coh}(X)$ and $i\_\*: D^b\_{coh}(X)\to D^b\_{coh}(Y)$. By adjunction, for any $\mathcal{F}\in D^b\_{coh}(X)$ we have a canonical morphism $i^\*i\_\*\mathcal{F}\to \mathcal{F}$ in $D^b\_{coh}(X)$.
My question is: Can we complete $i^\*i\_\*\mathcal{F}\to \mathcal{F}$ to an exact triangle in $D^b\_{coh}(X)$? I am particularly interest in the case that $Y$ is Calabi-Yau and I guess that in this case the expected exact triangle should be
$$
i^\*i\_\*\mathcal{F}\to \mathcal{F}\to \mathcal{F}\otimes\omega\_X^{-1}
$$
but I cannot prove or disprove it.
| https://mathoverflow.net/users/24965 | How to complete $i^*i_*\mathcal{F}\to \mathcal{F}$ into an exact triangle for a smooth divisor $i: X\hookrightarrow Y$? | The cone of $i^\*i\_\*\mathcal{F} \to \mathcal{F}$ is isomorphic to $\mathcal{F} \otimes \mathcal{O}\_X(-X)[2]$.
EDIT. Let me write an argument for a sheaf $F$. Consider the distinguished triangle
$$
i^\*i\_\*F \to F \to F'.
$$
We need to identify $F'$.
Applying $i\_\*$ we obtain
$$
i\_\*i^\*i\_\*F \to i\_\*F \to i\_\*F'.\tag{\*}
$$
By projection formula and the Koszul resolution of $i\_\*\mathcal{O}\_X$, we have
$$
i\_\*i^\*i\_\*F \cong
i\_\*F \otimes i\_\*\mathcal{O}\_X \cong
\mathrm{Cone}(i\_\*F \otimes \mathcal{O}\_Y(-X) \to i\_\*F),
$$
and since the map in the right-hand side is obviously zero, we conclude that
$$
i\_\*i^\*i\_\*F \cong
i\_\*F \oplus i\_\*F \otimes \mathcal{O}\_Y(-X)[1].
$$
Now the first map in $(\*)$ is the projection to the direct summand, hence
$$
i\_\*F' \cong
i\_\*F \otimes \mathcal{O}\_Y(-X)[2] \cong
i\_\*(F \otimes \mathcal{O}\_X(-X)[2]).
$$
Since $i\_\*$ is fully faithful on the category of sheaves, it follows from this that
$$
F' \cong F \otimes \mathcal{O}\_X(-X)[2].
$$
| 6 | https://mathoverflow.net/users/4428 | 425740 | 172,833 |
https://mathoverflow.net/questions/425737 | 10 | Let $d \geq 1$ be an integer. Dirichlet's theorem on arithmetic progression implies that the arithmetic progression $a, a+d, a+2d, \ldots$ contains infinitely many primes if and only if $\gcd(a,d)=1$.
Suppose $K/\mathbb{Q}$ is a finite Galois extension. Cebotarev's density theorem implies that there are infinitely many primes that split completely in $K$.
Since there are finitely many $1 \leq a \leq d$ such that $\gcd(a,d)=1$, there exists at least some $1 \leq a\_0 \leq d$ with $\gcd(a\_0,d)=1$ such that the arithmetic progression $a\_0, a\_0+d, a\_0+2d, \ldots$ contains infinitely many primes that split completely in $K$.
My question is
>
> Is it true that for all $1 \leq a \leq d$ with $\gcd(a,d)=1$, the arithmetic progression $a, a+d, a+2d, \ldots$ contains infinitely many primes that split completely in $K$?
>
>
>
| https://mathoverflow.net/users/485009 | Infinitely many primes that split completely in an arithmetic progression | **Theorem**. *Let $K$ and $L$ be finite Galois extensions of $\mathbf Q$*. Set $F = K\cap L$.
(1) *If $F = \mathbf Q$, then for each conjugacy class $C$ in
${\rm Gal}(L/\mathbf Q)$ there are infinitely many primes that are unramified in $L$ with Frobebius conjugacy class $C$ and split completely in $K$*.
(2) *If $F \not= \mathbf Q$ then there is a conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$ such that no prime unramified in $L$ with Frobenius conjugacy class $C$ splits completely in $K$.*
(3) *A conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$ is the Frobenius conjugacy class of some prime unramified in $L$ that splits completely in $K$ if and only if $C \subset {\rm Gal}(L/F)$, in which case $C$ is the Frobenius conjugacy class of infinitely many primes unramified in $L$ that split completely in $K$.*
(Two sufficient conditions to have $F = \mathbf Q$ are (i) $[K:\mathbf Q]$ and $[L:\mathbf Q]$ are relatively prime and (ii) the discriminants of $K$ and $L$ are relatively prime. When $L = \mathbf Q(\zeta\_d)$, (ii) holds if $(d,{\rm disc}(K)) = 1$ since primes that ratify in $\mathbf Q(\zeta\_d)$ must divide $d$. Neither of these conditions is necessary.)
**Remark**. Using $L = \mathbf Q(\zeta\_d)$, we see the answer to the OP’s question if affirmative if and only if $K \cap \mathbf Q(\zeta\_d) = \mathbf Q$, and that even if $K \cap \mathbf Q(\zeta\_d) \not= \mathbf Q$ we can still describe exactly which elements of the group $(\mathbf Z/d\mathbf Z)^\times$, viewed as ${\rm Gal}(\mathbf Q(\zeta\_d)/\mathbf Q)$, contain a prime number that splits completely in $K$: it is the congruence classes mod $d$ that belong to ${\rm Gal}(\mathbf Q(\zeta\_d)/F)$, where $F = K \cap \mathbf Q(\zeta\_d)$.
*Proof*.
(1) We assume $F = \mathbf Q$. By Galois theory, the composite field $KL$ is Galois over $\mathbf Q$ and
${\rm Gal}(KL/\mathbf Q) \cong {\rm Gal}(K/\mathbf Q) \times {\rm Gal}(L/\mathbf Q)$.
Pick a conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$. Then $\{1\} \times C$ is a conjugacy class in ${\rm Gal}(KL/\mathbf Q)$. By Chebotarev, there are infinitely many primes $p$ unramified in $KL$ such that its Frobenius conjugacy class in ${\rm Gal}(KL/\mathbf Q)$ is $\{1\} \times C$, so such $p$ split completely in $K$ while having Frobenius conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$.
(2) We assume $F \not= \mathbf Q$. Now there is a restriction on the conjugacy classes $C$ in ${\rm Gal}(L/\mathbf Q)$ such that some prime number $p$ (not just infinitely many) unramified in $L$ can have Frobenius conjugacy class $C$ in ${\rm Gal}(L/\mathbf Q)$ while splitting completely in $K$. Such a prime $p$ splits completely in $F$, which implies $C \subset {\rm Gal}(L/F)$, and ${\rm Gal}(L/F)$ is a normal subgroup of ${\rm Gal}(L/\mathbf Q)$ since $F/\mathbf Q$ must be Galois. Since $F \not= \mathbf Q$, ${\rm Gal}(L/F)$ is a *proper* normal subgroup of ${\rm Gal}(L/\mathbf Q)$, so for $\sigma$ in ${\rm Gal}(L/\mathbf Q)$ that is not in ${\rm Gal}(L/F)$, there is *no* prime $p$ that is unramified in $L$, has Frobenius conjugacy class in ${\rm Gal}(L/\mathbf Q)$ equal to the conjugacy class of $\sigma$, and splits completely in $K$.
(3) We showed in the proof of (2) that if there is a prime unramified in $L$ that splits completely in $K$, then its Frobenius conjugacy class in ${\rm Gal}(L/\mathbf Q)$ lies in ${\rm Gal}(L/F)$. Conversely, let $C$ be a conjugacy class of ${\rm Gal}(L/\mathbf Q)$ that lies in the normal subgroup ${\rm Gal}(L/F)$. Pick $\sigma \in C$, so $\sigma \in {\rm Gal}(L/F)$. By Galois theory the restriction mapping ${\rm Gal}(KL/K) \to {\rm Gal}(L/F)$ is an isomorphism, so we can lift $\sigma$ to an automorphism $\sigma'$ in ${\rm Gal}(KL/K)$. By Chebotarev there are (infinitely many) primes $p$ unramified in $KL$ whose Frobenius conjugacy class in ${\rm Gal}(KL/\mathbf Q)$ is the conjugacy class of $\sigma'$. Let's show for such $p$ that (i) the Frobenius conjugacy class of $p$ in ${\rm Gal}(L/\mathbf Q)$ is $C$ and (ii) $p$ splits completely in $K$:
(i) since $\sigma'|\_{L} = \sigma$, the Frobenius conjugacy class of $p$ in ${\rm Gal}(L/\mathbf Q)$ is the conjugacy class of $\sigma$ in ${\rm Gal}(L/\mathbf Q)$, which is $C$,
(ii) since $\sigma'$ is trivial on $K$, the Frobenius conjugacy class of $p$ in ${\rm Gal}(K/\mathbf Q)$ is trivial, so $p$ splits completely in $K$.
QED
| 13 | https://mathoverflow.net/users/3272 | 425742 | 172,834 |
https://mathoverflow.net/questions/425743 | 1 | Is there a (closed) formula for the Alexander polynomial of the pretzel knot $P(2m+1,2n,2k+1)$, $m,k\ge 0 , n \ge 1$ ?
| https://mathoverflow.net/users/83515 | Alexander polynomial of the pretzel knot $P(2m+1,2n,2k+1)$ | It seems like the lower half of the Alexander polynomial of the pretzel knot $ P(2m+1,2n,2k+1)$ , up to multiplication by $\pm t^{\alpha}$ , is given by $$ \Delta\_{h}(t)= -nt + \sum\_{i=2}^{2m+1}(-1)^{i}( i+2n-1) t^{i} \\ +(2m+2n+1) \sum\_{i=2m+2}^{k+m+2} (-1)^{i} t^{i} $$
for $k\ge m\geq 0,n\ne 0.$ The subindex $h$ in $\Delta\_{h}(t)$ is to indicate that it is half of the Alexander polynomial. We can assume $k\ge m$ without loss of generality because we can exchange the parameters $2m+1$ and $2k+1$.
| 2 | https://mathoverflow.net/users/83515 | 425744 | 172,835 |
https://mathoverflow.net/questions/425481 | 3 | In many PDEs, I see the papers mention the energy of the PDE. And some papers and books mention Hamiltonians. I know that integrable systems have infinitely many conservation laws and these laws are usually written in books and papers but I do not know how to find them. I tried to see how these laws were found but I couldn't. In particular, I tried the KdV equation and some of its higher nonlinear hierarchy. I have read many books about this matter but I feel I do not fully understand the practical importance and some of its related mathematics. Could you please explain to me the general picture? I will handle the technicalities. Thanks in advance.
| https://mathoverflow.net/users/471464 | Hamiltonian, energy, and conservation laws of nonlinear PDEs | There is a great variety of methods to obtain conservation laws of nonlinear evolution equations. In a broad classification one can divide these in symmetry-based approaches ([Noether's theorem](https://en.wikipedia.org/wiki/Noether%27s_theorem) relates a symmetry to a conserved quantity) and direct approaches. From a practical perspective, if you wish to learn "how to find the conservation laws", the direct method using computer algebra tools seems the most efficient.
The direct method is more general than a method based on Noether's theorem, because it does not require that the linearization of the evolution equation be self-adjoint. If you wish to restrict yourself to self-adjoint linearizations the procedure is to identify the action functional which is minimized by the evolution. Each point symmetry of the action is then associated with a conservation law, and Noether's theorem gives an explicit formula to obtain it.
These two approaches are well explained in [Construction of conservation laws: how the direct method generalizes Noether’s theorem](https://personal.math.ubc.ca/~bluman/cyprus%20proceedings%20paper.pdf). The software to implement the direct method is described in [Direct Methods and Symbolic Software for Conservation Laws of Nonlinear Equations.](https://arxiv.org/abs/0803.0083)
| 1 | https://mathoverflow.net/users/11260 | 425747 | 172,837 |
https://mathoverflow.net/questions/425725 | 0 | [Edit, July 6, 2022: Removed erroneous characterization of Faber polynomials as an Appell sequence.]
[Dress and Siebeneicher](https://www.sciencedirect.com/science/article/pii/0001870889900273?via%3Dihub) in their tale of the Burnside family express an opinion (1.2) that, if I read it correctly, leads me to believe that the classic [Faber partition polynomials](https://oeis.org/A263916) which satisfy
$\ln[A(x)] = \ln[1 + a\_1 x + a\_2 x^2 + \cdots ] = \sum\_{\geq 1} -F\_n(a\_1,...,a\_n)\; \frac{x^n}{n}$
have the property
$[F\_n(a\_1,a\_2,...,a\_n)- (-a\_1)^n] \; mod(n) \; = 0$
for $n$ prime and integral indeterminates $a\_n$ .
I've been assured by a reputable authority that such is obvious. Can someone provide an 'obvious' proof or a least some other published hearsay on this?
Gessel and Ree in "Lattice paths and Faber polynomials" give (p. 4) a multinomial- coefficient type of expression for the coefficients of the Faber polynomials $FP\_n(u)$ for which $FP\_n(0)=F\_n[a\_1,a\_2,...,a\_n]$. (G & R use the notation $F\_n(u)$ for what I denote as $FP\_n(u)$.)
[Edit, July 6, 2022: Motivated by Peter's answer, I found two nice intros for the uninitiated to Kummer's theorem--"[Legendre’s and Kummer’s Theorems Again](https://www.ias.ac.in/article/fulltext/reso/015/12/1111-1121)" by Mihet and "[Revisiting Kummer's and Legendre's Formulae](https://www.ias.ac.in/article/fulltext/reso/010/02/0062-0071)" by Sury.]
---
On the 'ubiquity' of the Faber polynomials and Faber partition polynomials:
These Faber polynomials and their associated Faber partition polynomials crop up in multitude of discussions: in symmetric function theory in the Newton-Girard-Waring identities expressing the power symmetric polynomials in terms of the elementary symmetric polynomials; in operational calculus for a generic raising operator for Appell polynomials; in complex function theory in extending an analytic function defined on a closed curve to an analytic function within the curve (providing harmonic functions with prescribed boundary conditions); in an analog of Fourier series expansions of complex functions; in extracting the numerical values of the indeterminates of compositional partition polynomials given numerical evaluations of those polynomials; in relations between determinants and traces; in algebraic/geometric K-theory; in the combinatorial/analytic properties of random walks, lattice paths, noncrossing partitions, and associahedra; and in determining the compositional inverse of certain Laurent series--in fact, in an orgy with the family of reciprocal polynomials $R\_n$ birthed by $1/A(x) = \sum\_{n \geq 0} R\_n(a\_1,...,a\_n) x^n$, the compositional inverse $L^{(-1)}(z) = z + b\_1 +b\_2/z +b\_3/z^2 + \cdots $ of the formal Laurent series $L(z) =z + a\_1 +a\_2/z +a\_3/z^2 + \cdots$ is given by the umbral recursion formula $b\_n(a\_1,...,a\_n) = \frac{1}{n}[R\_n(b\_1-F.(a\_1,...),b\_2,b\_3,...,b\_{n-1},0] - R\_n(0,-b\_2,-2b\_3,...,-(n-2)b\_{n-1},0)].$
| https://mathoverflow.net/users/12178 | A Newton identity and the primes--the Faber partition polynomials and modular arithmetic | From $$\ln[A(x)] = \ln[1 + a\_1 x + a\_2 x^2 + \cdots ] = \sum\_{n \geq 1} -F\_n(a\_1,...,a\_n)\; \frac{x^n}{n}$$ and letting $A'(x) = A(x) - 1$ we have
$$\begin{eqnarray\*}
F\_n(a\_1,...,a\_n) &=& n [x^n] \sum\_{i=1}^{\infty} \frac{(-A'(x))^i}{i} \\
&=& (-a\_1)^n + n [x^n] \sum\_{i=1}^{n-1} \frac{(-A'(x))^i}{i} \\
%&=& n \sum\_{\lambda \, \vdash \, n} \frac{1}{\operatorname{len}(\lambda)} \binom{\operatorname{len}(\lambda)}{f\_1, \ldots, f\_n} \prod\_i (-a\_i)^{f\_i} \\
&=& (-a\_1)^n + n \sum\_{\lambda \, \vdash \, n,\; \lambda \neq 1^n} \frac{1}{\operatorname{len}(\lambda)} \binom{\operatorname{len}(\lambda)}{f\_1, \ldots, f\_n} \prod\_j (-a\_j)^{f\_j} \\
\end{eqnarray\*}$$
where the sum in the last line is over partitions $\lambda = 1^{f\_1} 2^{f\_2} \cdots n^{f\_n}$ with $\sum\_i if\_i = n$ and the length $\operatorname{len}(\lambda)$ defined as $\sum\_i f\_i$, and the partition $\lambda = 1^n$ giving $(-a\_1)^n$ is pulled out of the sum.
It remains to show that all of the multinomial coefficients in $[x^n](-A'(x))^i$ with $i < n$ must be divisible by $i$; they're certainly not divisible by any prime greater than $i$, which by hypothesis includes $n$.
Let $p$ be a prime factor of $\operatorname{len}(\lambda) < n$. There must be some $c$ for which $f\_c \neq 0 \pmod p$, since otherwise $p \mid n$ contradicting the primality of $n$. But then $$\binom{\operatorname{len}(\lambda)}{f\_1, \ldots, f\_n} = \binom{\operatorname{len}(\lambda)}{f\_c} \binom{\operatorname{len}(\lambda) - f\_c}{\{f\_j : j \neq c\}}$$ and by Kummer's theorem $\nu\_p \left( \binom{\operatorname{len}(\lambda)}{f\_c} \right) \ge \nu\_p \left(\operatorname{len}(\lambda)\right)$. Therefore the multinomial coefficient is indeed divisible by $\operatorname{len}(\lambda)$, giving the desired result.
| 1 | https://mathoverflow.net/users/46140 | 425753 | 172,840 |
https://mathoverflow.net/questions/425736 | -3 | Sorry for the not-perfect question. I am asking for a reference for the following relation:
$$\int f . g. h ...= \int\_{\xi\_1 +\xi\_2 +...=0} \hat{f}(\xi\_1) \hat{g}(\xi\_2)... d\xi\_1 d\xi\_2...$$
Could you please show me where to find details, statements and proofs about this relationship? Thanks in Advance.
| https://mathoverflow.net/users/471464 | Asking for reference about a relation related to Fourier transform | If $f\_1,\ldots,f\_n$ belong to the Schwarz space, Fourier inversion formula and Fourier-convolution properties yield
\begin{eqnarray\*}
\int\_{\xi\_1+\cdots+\xi\_n=0} \hat{f\_1}(\xi\_1) \cdots \hat{f\_n}(\xi\_n) d\xi\_1 \cdots d\xi\_n
&=& (\hat{f\_1} \* \cdots \* \hat{f\_n})(0) \\
&=& (2\pi)^{-1} \mathcal{F}(\overline{\mathcal{F}}(\hat{f\_1} \* \cdots \* \hat{f\_n}))(0) \\
&=& \mathcal{F}(f\_1 \cdots f\_n)(0) \\
&=& \int\_\mathbb{R} f\_1 \cdots f\_n(x) dx.
\end{eqnarray\*}
| 4 | https://mathoverflow.net/users/169474 | 425755 | 172,842 |
https://mathoverflow.net/questions/423877 | 6 | A Cartan-Hadamard 3-space $M$ is a complete simply connected 3-dimensional Riemannian manifold with nonpositive sectional curvature. A (smooth) convex surface $\Gamma\subset M$ is an embedded topological sphere with nonnegative second fundamental form $\mathrm{I\!I}$. The total (Gauss-Kronecker) curvature of $\Gamma$ is defined as
$$
\mathcal{G}(\Gamma):=\int\_\Gamma\det(\mathrm{I\!I}).
$$
It follows quickly from Gauss' equation and Gauss-Bonnet theorem that $\mathcal{G}(\Gamma)\geq 4\pi$. Suppose that $\mathcal{G}(\Gamma)= 4\pi$. Does it follow that the compact region of $M$ bounded by $\Gamma$ is Euclidean, i.e., all its sectional curvatures are zero?
**Note 1:** Schroeder and Strake showed in [this paper](https://www.jstor.org/stable/2048830?seq=1) (see Theorem 2) that the answer is yes, provided that $\Gamma$ is *strictly* convex, i.e., the second fundamental form is positive definite. Strict convexity appears to be an essential feature of the proof.
**Note 2:** On page 66 of [Lectures on Manifolds on Nonpositive Curvature](https://link.springer.com/book/10.1007/978-1-4684-9159-3) (see Exercise (b)), Gromov poses a more general question for total *absolute* curvature of closed surfaces in Cartan-Hadamard $3$-spaces (the term "absolute" is not explicitly mentioned).
| https://mathoverflow.net/users/68969 | Convex surfaces with minimal total curvature in Cartan-Hadamard 3-space | Note that curvature in sectional directions tangent to the surface, say $\Sigma$ vanish.
It seems sufficient to conclude that the usual Peterson–Codazzi equations hold for the surface.
It follows that there is a convex surface $\Sigma'$ in the Euclidian space that is isometric to $\Sigma$; plus they have identical second fundamental forms.
Now we want to use bow lemma (diffgeometric analog of arm lemma).
The bow lemma has a version for Hadamard spaces.
Applying it to plane arcs in $\Sigma'$, we get that the map $\Sigma'\to\Sigma$ is distance-noncontracting.
Applying Reshetnyak majorization theorem, to plane sections of $\Sigma'$, we get that the map $\Sigma'\to\Sigma$ is distance-preserving.
Finally, applying Kirszbraun, we get that the map $\Sigma'\to\Sigma$ extends to a distance-preserving map from the convex hull of $\Sigma'$.
| 2 | https://mathoverflow.net/users/1441 | 425764 | 172,844 |
https://mathoverflow.net/questions/425788 | 13 | I've been trying and failing to find a paper/article/blog post (I think it was a paper) on a particular algebraic structure. The paper describes a structure consisting of something like a constant $0$, an associative binary operation $+$ for which $0$ is the unit, and an $\omega$-ary operation $\Sigma$ that agrees with repeated addition if cofinitely many of the arguments are $0$. There were some other requirements I believe, because this is not enough for the following result. One of the points the paper made was that the set $[0,\infty]$ (the 'set of magnitudes') with its usual addition was the free object on one point for this algebraic structure, so this gave an elementary construction of the real numbers. This is also why the structure was called something like a 'magnitude algebra'.
I feel like this must be enough information to dig up the paper, but I have had no luck after a lot of trying, so I can only conclude I'm misremembering something. Does anyone know the text I'm looking for?
| https://mathoverflow.net/users/478205 | Looking for a half-remembered reference on 'magnitude algebras' | @Neil Strickland gave me all I needed in his comment (thank you!). Escardó and Simpson's paper is not what I was looking for, but they do cite it: *A Universal Characterization of $[0,\infty]$*, by Denis Higgs.
For those curious but not curious enough to find a text of the paper: a *magnitude module* is a variety of algebras with a constant $0$, a unary operation $h$, and an $\omega$-ary operation $\Sigma$ satisfying the identities
* $\Sigma(\Sigma(x\_{00},x\_{01},\dots),\Sigma(x\_{10},x\_{11},\dots),\dots) = \Sigma(\Sigma(x\_{00},x\_{10},\dots),\Sigma(x\_{01},x\_{11},\dots),\dots)$.
* $\Sigma(0,\dots,0,x,0,\dots) = x$.
* $h (\Sigma(x\_0,x\_1,\dots)) = \Sigma(h(x\_0),h(x\_1),\dots)$.
* $\Sigma(h(x),h^2(x),h^3(x),\dots) = x$.
Because $[0,\infty]$ is the free magnitude module on one point, every magnitude module $M$ admits a scalar multiplication-like action $[0,\infty] \times M \to M$, hence the name 'magnitude module'.
| 11 | https://mathoverflow.net/users/478205 | 425792 | 172,854 |
https://mathoverflow.net/questions/425727 | 21 | One of my graduate students asked me the following question, and I can't seem to answer it. Let $\Sigma\_g$ denote a compact oriented genus $g$ surface. For which $g$ does there exist an orientation-preserving diffeomorphism $f\colon \Sigma\_g \rightarrow \Sigma\_g$ with the following two properties:
1. $f$ has no fixed points.
2. The action of $f$ on $H\_1(\Sigma\_g)$ fixes no nonzero elements.
Since $f$ has no fixed points, you can use the Lefschetz fixed point theorem to deduce that the trace of the action of $f$ on $H\_1(\Sigma\_g)$ must be $2$. From this, you can easily see that no such $f$ can occur for $g=0$ and $g=1$. However, I can't figure out what is going on here for $g \geq 2$.
| https://mathoverflow.net/users/485005 | Fixed-point free diffeomorphisms of surfaces fixing no homology classes | Goodwillie's construction (in genus two) generalises to all higher genus as follows.
Let $P\_n$ be the regular $n$-gon in the plane with vertices at roots of unity. When $n$ is even, we can glue opposite (and thus parallel) sides to obtain an oriented surface $F\_n$. Suppose that $n = 4g + 2$. In this case $F\_n$ has genus $g$; also the rotation by $2\pi / (4g + 2)$ induces a homeomorphism $f\_n$ of $F\_n$ with exactly one fixed point, at the origin.
Now we take copies of $F\_{4g + 2}$ and $F\_{4h + 2}$, remove small disks about the origin of each, and glue along the so created boundaries. The resulting connect sum $F$ has genus $g + h$. In a neighbourhood of the gluing we interpolate between the homeomorphisms $f\_{4g + 2}$ and $f\_{4h + 2}$ (this is called a "fractional Dehn twist" in some places). The resulting homeomorphism $f \colon F \to F$ has the desired properties.
| 15 | https://mathoverflow.net/users/1650 | 425796 | 172,856 |
https://mathoverflow.net/questions/425801 | 2 | Define $N(F)$ to be the number of monomials of a multi-variable polynomial $F$. For example $N(x^2y+3xy-y^5)=3$.
If $\mathbf{x}=(x\_1,\dots,x\_n)$ and $F\_n(\mathbf{x})=\prod\_{k=1}^n(x\_1+\cdots+x\_k)$ then it's easy to verify that $N(F\_n)=C\_n$ where $C\_n=\frac1{n+1}\binom{2n}n$ are the Catalan numbers.
Consider now the polynomial
$$G\_n(\mathbf{x};q)=\prod\_{k=1}^n(x\_1+x\_2q+x\_3q^2+\cdots+x\_kq^{k-1}).$$
**Note.** $G\_n(\mathbf{x};1)=F\_n(\mathbf{x})$.
On the other hand, among the many variants of the $q$-Catalan polynomials there is one called the [*Carlitz's $q$-Catalan numbers*](https://www.sciencedirect.com/science/article/pii/0097316585900895?ref=cra_js_challenge&fr=RR-1) which can be presented by the recurrence $C\_0(q):=1$ and
$$C\_{n+1}(q)=\sum\_{k=0}^nC\_k(q)C\_{n-k}(q)\,q^{(k+1)(n-k)}.$$
I would like to ask:
>
> **QUESTION.** If $g\_{n,j}(\mathbf{x})$ denotes the coefficient of $q^j$ in $G\_n(\mathbf{x};q)$ (after expansion), then $N(g\_{n,j}(\mathbf{x}))$ equals the coefficient of $q^j$ in $C\_n(q)$. Is this true?
>
>
>
**Example.** Take $n=3$. Then, $G\_3(\mathbf{x};q)=x\_1^3 + 2x\_1^2x\_2q + (x\_1^2x\_3+x\_1x\_2^2)q^2+x\_1x\_2x\_3q^3$ and $C\_3(q)=1+q+2q^2+q^3$. Therefore,
$N(x\_1^3)=1, N(2x\_1^2x\_2)=1, N(x\_1^2x\_3+x\_1x\_2^2)=2$ and $N(x\_1x\_2x\_3)=1$.
| https://mathoverflow.net/users/66131 | Counting monomials and $q$-Catalan polynomials | Yes, bijectively.
Associate to a monomial $\prod\_{i=1}^n x\_i^{e\_i}$ with $\sum\_{i=1}^n e\_i =n$ the path where we walk one step to the right, then up $e\_n$ steps, then one step to the right, then up $e\_{n-1}$ steps, then $\dots$, and finally walk up $e\_1$ steps.
It's easy to see that this gives a bijection between monomials and paths with all steps up and right, and the path stays below the diagonal (a Dyck path) if and only if the monomial appears in $\prod\_{k=1}^n (x\_1 + \dots + x\_k)$.
Furthermore since each of our $e\_k$ steps up has $k-1$ squares to the right of it the number of squares below-right of our path is $\sum\_{k=1}^n e\_k (k-1)$, which is the power of $q$ next to which this monomial appears in $G\_n ( \mathbf x; q)$.
So $g\_{n,j}(\mathbf x)$ is a (weighted) sum of all the monomials corresponding to Dyck paths with $j$ squares below-right, and so the number of monomials appearing in $g\_{n,j}(\mathbf x)$ is the number of Dyck paths with $j$ squares below-right.
Since the generating function of this count of Dyck paths is the $n$th Carlitz $q$-Catalan number (if I understand correctly, this is one of the combinatorial descriptions in the paper you link), this proves your claim.
| 5 | https://mathoverflow.net/users/18060 | 425805 | 172,860 |
https://mathoverflow.net/questions/425799 | 1 | Let $X := \mathbb R^d$, $\lambda^d$ be the $d$-dimensional Lebesgue measure on $X$, and $f:X \to \mathbb R$ convex. Then [there is](https://math.stackexchange.com/questions/4483452/is-it-true-that-if-f-mathbb-rd-to-mathbb-r-is-continuous-then-the-differ) a Borel set $N \subset X$ such that $\lambda^d (N) = 0$ and $f$ is differentiable on $A := X \setminus N$. Clearly, $A$ is Borel measurable. It follows that the gradient $\nabla f: A \to X$ of $f$ is well-defined.
* If $f$ is differentiable on $X$, [then](https://math.stackexchange.com/questions/4482935/if-all-partial-derivatives-exist-at-each-point-of-a-then-f-in-mathcal-c1) $A = X$ and $f \in \mathcal C\_1 (X)$.
* Clearly, the Borel $\sigma$-algebra $\mathcal B(A)$ is a subset of the Borel $\sigma$-algebra $\mathcal B(X)$, i.e., $\mathcal B(A) \subset \mathcal B(X)$.
**My question:** Is the map $\nabla f$ measurable?
| https://mathoverflow.net/users/99469 | Let $f$ be convex and $A$ a Borel subset of $\mathbb R^d$ on which $f$ is differentiable. Is the gradient $\nabla f: A \to \mathbb R^d$ measurable? | Yes, since a convex function $f:\mathbb{R}^d\rightarrow\mathbb{R}$ is locally Lipschitz (see e.g. Theorem 6.3.1, pp. 236-239 of the book by L. C. Evans and R. F. Gariepy, *Measure Theory and Fine Properties of Functions*, CRC Press, 1992).
Recall that, as you wrote it yourself in the particular case of convex functions and more generally by Rademacher's theorem, if $g:\mathbb{R}^d\rightarrow\mathbb{R}$ is locally Lipschitz then $g$ is differentiable $\lambda^d$-almost everywhere. Moreover, in that case the distributional partial derivatives of $g$ are in $L^\infty\_{loc}(\mathbb{R}^d,\lambda^d)$ and can be identified $\lambda^d$-almost everywhere (say, on a Borel subset $A\subset\mathbb{R}^d$ with $\lambda^d(\mathbb{R}^d\smallsetminus A)=0$ as you defined it) with the classical partial derivatives of $g$ (for the details of all statements above for locally Lipschitz functions, see e.g. Theorem 5 in subsection 4.2.3, pp. 131-132 and Theorems 6.2.1, 6.2.2, pp. 235 of Evans-Gariepy), hence these partial derivatives are in $L^\infty\_{loc}(A,\lambda^d)$ and therefore are (Borel) measurable.
| 4 | https://mathoverflow.net/users/11211 | 425807 | 172,861 |
https://mathoverflow.net/questions/425557 | 6 | So I've read (for instance in the introduction to R.S de Jong's [thesis](https://www.math.leidenuniv.nl/%7Ejongrsde/publications/thesis.pdf) ) that the naive adaptation of the proof of the Mordell conjecture over function fields fails, even using Arakelov intersection theory. Most notably we lack a "good" canonical class inequality, for instance Bost, Mestre and Moret-Bailly showed in [this paper](http://www.numdam.org/item/AST_1990__183__69_0.pdf) that the analogue of Bogomolov-Miyao is false.
I was wondering if someone could explain the "proof" of Mordell which would rely on this inequality? I might well be explained in the Bost,Mestre and Moret-Bailly paper, but my french is not really up to the task....
| https://mathoverflow.net/users/152554 | Mordell conjecture over function fields | You can find a good explanation of the proof of the (effective) Mordell conjecture based on such an inequality in:
Parshin, A. N.: Application of ramified coverings in the theory of Diophantine equations. Math. USSR Sbornik 66 (1990), no. 1, 249–264.
The corresponding proof for function fields has been published by Parshin before in:
Parshin, A. N.: Algebraic curves over function fields. Izv. Akad. Nauk SSSR Ser. Mat. 32 (1968), 1191–1219.
| 2 | https://mathoverflow.net/users/61532 | 425824 | 172,865 |
https://mathoverflow.net/questions/423224 | 2 | It is well known and easy to see (modulo standard basic facts) that any compact 1-dimensional Alexandrov space with curvature bounded from below is isometric either to a circle or to a segment.
I am looking for a precise reference.
| https://mathoverflow.net/users/16183 | Isometric classification of 1-dimensional Alexandrov spaces | See 15.18 in our [Alexandrov geometry: foundations](https://arxiv.org/abs/1903.08539).
| 1 | https://mathoverflow.net/users/1441 | 425825 | 172,866 |
https://mathoverflow.net/questions/425794 | 9 | I would like to know which books I should read to understand the paper ["The local Langlands correspondence for $\mathrm{GL}\_n$ over $p$-adic fields"](https://www.math.uni-bonn.de/people/scholze/LocalLanglands.pdf) written by Peter Scholze.
I only know mathematics subjects at the master's degree level. I also learned about representation theory of finite groups.
Next year I will probably start my PhD in Lie theory but I'd like to read the above article as a personal project.
---
**EDIT:**
I found a good article that shows the necessary prerequisites to get a glimpse of the Langlands Program and I want to share it: [Prerequisites for the Langlands Program.](https://www.math.stonybrook.edu/%7Eaknapp/pdf-files/china1.pdf)
| https://mathoverflow.net/users/481822 | Roadmap to understand the Scholze's proof of the local Langlands correspondence for $\text{GL}_n$ over $p$-adic fields | @user860322 The local Langlands correspondence (LLC) for $\mathrm{GL}\_n$ can be understood (very sloppy) as:
There exists a group for $F$, called the Weil(-Deligne)-Group $W\_F$ (and not only for local fields), that encodes the (admissible) representation theory of $\mathrm{GL}\_n(F)$ for every $n$ in a *very nice* way (whatever it means). I don't think that Scholze's paper is the best way to learn about it. There is a nice survey of Clozel called '*Motives and automorphic representations*', which in my eyes would be much better for you.
But of course, there is no short way to go. If you really want it, I would suggest, you learn:
* Basics in algebraic number theory. You should at very least know, what a local ($p$-adic) field is, and how they are constructed. Neukirch covers it well, but you can also have a look in the first chapter of André Weil's Basic Number Theory - he treats it in its full generality, although it may not be appropiated to read in the beginning.
* (local) Class field Theory, since this is the special case $n=1$ for the LLC. I don't recommend you to read local CFT in Neukirch. Weil's Chapter 3 (or so) covers the character theory of local fields, but his approach is highly unintuitive in my eyes (but maybe you will find it great?). Ramakrishnan/Valenza's Fourier Analysis on Number Fields give a very brief and student-friendly introduction in Chapter 6 (if I am not mistaken).
* In Corvallis, Chapter 3.1., John Tate's Number Theoretical Background introduces the Weil Group $W\_F$, but for this you will need to know a little about cohomology theory (at least for the existence of $W\_F$). It may not be the easiest text to read as well.
* Representation Theory of $\mathrm{GL}(F)$ for $F$ local non-archimedean; the 'main paper' is well.. Bernstein/Zelevinsky, but Prasad/Raghuram have a very nice and readable survey. They also already mention the LLC and even give an ad-hoc definition of $W\_F$ (without cohomology theory).
But honestly, I would first learn about rep.theory of compact groups (maybe the first chapters in Bump's Lie Groups?); I think this is a nice intermediary step from finite to infinite groups. Since $\mathrm{GL}\_n(F)$ is locally compact with nice properties (sometimes also called an $l$-group), f.e. the identity $I\_n$ has a neighborhoud of open compact subgroups, in terms of representation theory, it might be seen as the 'next best case' after the compact one (but on this one I may be horribly wrong hehe).
* Probably, to scratch the surface of the bigger picture, it would be good if you learn what the adeles $\mathbb{A}\_K$ of a global field $K/\mathbb{Q}$ are. You can understand the adeles as an object that contains all the local information together and holds it in a very fashion way. Only thanks to the adeles, many things about classical $L$-functions that appeared to be very technical and come out of nowhere (for instance the Riemann zeta function) suddenly had a very natural explanation.
| 12 | https://mathoverflow.net/users/484997 | 425829 | 172,868 |
https://mathoverflow.net/questions/124616 | 26 | Recently I have been learning more about the h-principle and in particular the methods of "continuous sheaves". In many treatments of this I see people using "quasi-topological spaces" and I am trying to understand explicitly why we need to use these?
>
> **Background**
>
>
>
Very roughly, the idea is that one has a sheaf $\Phi$ of topological spaces on the site of smooth manifolds. A typical examples will be the sheaf of immersions. Often there is a related sheaf of spaces $\Psi$ which has a homotopy theoretic nature. For example the sheaf of sections of the jet bundles corresponding to jets of immersions. Now $\Psi$ is something which is easy to compute, and there is a comparison map
$$ \Phi \to \Psi$$
The h-principle will be satisfied if this is a weak equivalence for each manifold. By construction we usually know that this holds for $\mathbb{R}^n$, and we also know that $\Psi$ is not only a sheaf but a "homotopy sheaf". For example the value of $\Psi(U \cup V)$ will be equivalent to the homotopy fiber product
$$ \Psi(U) \times^h\_{\Psi(U \cap V)} \Psi(V)$$
(it is isomorphic to the ordinary fiber product by the sheaf condition).
So the h-principle would follow if we could show that $\Phi$ was also a "homotopy sheaf". One way to prove this is to show that the restriction maps $\Phi(U) \to \Phi(U\_0)$ are Serre fibrations.
This is usually too strong for open manifolds $U\_0$. For example for $\Psi$ the sheaf of smooth functions, this restriction map is not surjective and it is easy to show it is not a Serre fibration. This is related to the fact that the sheaf of smooth functions is not flabby.
However the sheaf of smooth functions is soft, and we might hope to prove something similar for these sheaves of spaces. Namely we might be able to show that the restriction map to closed submanifolds (with boundary) is a Serre fibration.
So at this stage we must do two things:
1. Extend the sheaf $\Phi$ to closed subsets in our manifolds
2. Show that $\Phi$ satisfies this a softness condition (namely that restriction to closed subsets is a Serre fibration).
After this the general argument goes on to show that indeed $\Phi$ satisfies the h-principle.
Quasitopological spaces enter the picture in step 1. The idea is to define the value of $\Phi$ on closed subsets in the usual way, as a colimit of the spaces assigned to open subsets containing the closed subset. In Gromov's text on the h-principle he has the following cryptic remark:
>
> There is no useful natural topology on this space; however there is a weaker structure, called a *quasi-topology*, which nicely behaves under direct limits.
>
>
>
I have seen similar things mentioned in various papers and texts where the h-principle comes up, but none I have come across have given an example of what goes wrong and why we actually need to use these more exotic spaces. Indeed in some other texts the issue is either ignored or skirted somehow, but I am left wondering when and why we would need to use these quasi-topological spaces.
**The Question**: Coming from algebraic topology, I know we now have several convenient categories of topological spaces which are cartesian closed and fairly well behaved categorically, for example the category of compactly generated Hausdorf spaces. Why can't we just take the direct limit in one of these categories? What goes wrong and how badly?
Do I really need to worry about this when $\Phi$ is, say, the space of smooth functions (with one of the Whitney $C^\infty$-topologies or some variant thereof)?
| https://mathoverflow.net/users/184 | Why are quasitopological spaces needed in sheaf theoretic approaches to the h-principle? | I recommend a very nice paper of Sander Kupers, "Three applications of delooping to $h$-principles". In an appendix he discusses and compares various convenient categories of spaces when working with continuous sheaves. However Kupers does not work with quasitopological spaces in the body of the paper. His preferred convenient category of spaces is instead the category of sheaves of sets on the site of manifolds and open embeddings (smooth, PL or topological). In the smooth case these are often called "differentiable spaces".
One reason this is a convenient category to work with is that in many applications one considers continuous sheaves $U \mapsto F(U)$ where $F(U)$ is a "moduli space" parametrizing some sort of "objects over $U$". In algebraic geometry it is a well established principle that moduli spaces are fruitfully described by specifying the functor they represent, i.e. as a sheaf of sets on some site. For example, the Grassmannian $G(k,n)$ represents the functor assigning to a space $S$ the set of all rank $k$ subbundles of the trivial rank $n$ vector bundle over $S$, and this is a more useful description of $G(k,n)$ than a specification of a topology on the set of $k$-dimensional subspaces of affine $n$-space. Moreover, in situations where the moduli functor is not representable, it is often more useful to work with the moduli functor itself rather than some coarse approximation of it. Importantly, for a continuous sheaf described in this way, the various notions of flexibility, microflexibility, softness etc get a very concrete meaning in terms of the moduli functor.
As a specific example, Kupers considers the case that $F(U)$ is the "space of foliations on $U$". In this case $F(U)$ can be defined in a very convenient way: it is taken to be the functor taking a manifold $M$ to the set of foliations on $M \times U$ of codimension $\dim(U)$ which are transverse to the projection onto $M$.
However one could consider equally well sheaves of sets on other geometrically natural sites, and have many of the same advantages. One could consider sheaves on topological spaces with respect to the open cover topology, as in the theory of topological stacks. Or, indeed, one could work with quasi-topological spaces: a quasi-topological space is a particular type of sheaf of sets on the category of compact Hausdorff spaces, with respect to the topology defined by finite families of jointly surjective maps. The reason it is natural to impose descent for such a fine topology is that any surjective map of compact Hausdorff spaces is a quotient map. I think this explains at least morally why quasi-topological spaces should arise naturally when studying continuous sheaves: we want $F(U)$ to be a moduli space, so it is natural to describe it as a sheaf of sets, and by retaining control of the moduli functor we can force also the value of the sheaf on a codimension zero submanifold with boundary, say, to have a modular interpretation.
---
Now let me also say that I believe that quasi-topological spaces are simply not very natural objects to work with and are probably never really the right tool for the job.
I want to compare them to a notion of ringed space introduced by T.A. Springer, in his book "Linear algebraic groups". In the first part of this book Springer introduces a "baby version" of scheme theory, adequate for describing the theory of varieties over a field. One way in which the theory is simplified is by the nonstandard definition of *ringed space* that he works with. He fixes a ground field $k$, and declares that a ringed space is a topological space $X$, together with a subsheaf of the sheaf of $k$-algebras $U \mapsto \mathrm{Funct}(U,k)$ on $X$. This simplifies life in some ways. For example, he can then define a morphism of ringed spaces to be a morphism of underlying topological spaces, satisfying the *property* of being compatible with the structure sheaves, rather than including the extra data of a map of sheaves. And he completely avoids talking about "locally" ringed spaces in this way, which would otherwise be necessary to have an equivalence of categories between $k$-algebras and affine schemes over $k$.
But in the end this definition is obviously a kludge - it is simply much more natural to work with an arbitrary sheaf of rings, without imposing this concreteness condition, and it is somehow an "accident" that it produces the "correct" theory when the scheme is of finite type over a field.
Now the definition of a quasi-topological space is strikingly similar. Recall that a quasi-topological space consists of a set $W$, and a subsheaf of the sheaf of sets $S \mapsto \mathrm{Funct}(S,W)$ on the category of compact Hausdorff spaces. Morphisms are natural transformations. It seems very natural to expect the category of *all* sheaves of sets on compact Hausdorff spaces to be a more flexible and convenient category than this one. When this question was asked, there was no established name for a sheaf of sets on compact Hausdorff spaces, but now we have one: this is precisely the notion of a **condensed set** of Clausen-Scholze. I would place a bet that any occurence of quasi-topological space in the literature is more conveniently described as a condensed set.
A final remark is that the algebro-geometric study of moduli spaces also teaches us that we should allow the parametrized objects to have automorphisms, which leads to replacing sheaves of sets with sheaves of groupoids and the theory of stacks. Once one has passed to 1-stacks it is natural to proceed further to higher stacks. I would therefore expect differentiable higher stacks or condensed anima to be useful categories of spaces for studying $h$-principles in general.
| 5 | https://mathoverflow.net/users/1310 | 425850 | 172,876 |
https://mathoverflow.net/questions/425855 | 1 | Let $D \subseteq \mathbb{R}^n$ be open, connected, bounded. Let $f : D \to \mathbb{R}$ be $C^1$ and assume that $f$ as well as $\partial\_i f$ extend as continuous functions to the closure $\overline{D}$ of $D$ for all $i \in \{1, \ldots, n\}$.
The paper [Functions Differentiable on the Boundaries of Regions](https://doi.org/10.2307/1968745) by Whitney shows in particular that if $D$ is quasi-convex (Property P in the paper) then one can extend $f$ as a $C^1$ function to all of $\mathbb{R}^n$.
I would guess this fails if one does not make a suitable assumption like quasi-convexity on the domain $D$ — does anyone know a good counterexample?
| https://mathoverflow.net/users/13338 | If function and derivative extend continuously to boundary of domain, can one extend as $C^1$ function? | The failure is essentially local, so the boundedness we can ignore for the purpose of this example.
* Let $D = \{ y < |x|^{1/4}: (x,y)\in \mathbb{R}^2\}$.
* Let $\phi:\mathbb{R}\to\mathbb{R}$ be such that $\phi(y) = 0$ for $y \leq 0$ and $\phi(y) = y^2$ for $y > 0$.
* Let $f:D\to \mathbb{R}$ be given by $f(x,y) = \mathrm{sign}(x) \phi(y)$.
We have $\partial\_x f \equiv 0$ and
$$
\partial\_y f =
\begin{cases}
0 & y \leq 0 \\
2\mathrm{sign}(x) y & y > 0
\end{cases}.
$$ It is not too hard to see that $f$ and its first derivatives have continuous extensions to $\bar{D}$.
Fixing $y >0$, applying the mean value theorem to $x\mapsto f(x,y)$, we see that any putative differentiable extension will require there having a point with $|x| \leq y^4$ that satisfy $\partial\_x f(x,y) = y^{-2}$. This shows it is impossible to have a $C^1$ extension to all $\mathbb{R}^2$ as the continuous extension of $\partial\_x f(0,0) = 0$.
| 3 | https://mathoverflow.net/users/3948 | 425857 | 172,878 |
https://mathoverflow.net/questions/425836 | 2 | I am looking to replicate, in the fiberwise setting, the result of Spivak/Wall that the fiber homotopy type of the normal sphere bundle of a manifold is preserved under homotopy equivalences.
A particularly slick way to prove this result, is to notice that the mapping cylinder of a homotopy equivalence $M \rightarrow M'$ is a relative Poincare duality space, and as such we can talk about its normal fibration (obtained from a relative embedding into $\mathbb{R}^N \times (0,1]$), which restricts to the normal fibrations of $M$ and $M'$ on either side. Thus, because a mapping cylinder of a homotopy equivalence has both end inclusions homotopy equivalences, we can conclude there is some fiberwise homotopy equivalence of normal spherical fibrations.
So here is my question: is there is enough fiberwise embedding theory in the literature to be able to conclude that the fiberwise mapping cylinder of a fiber homotopy equivalence of manifold bundles over $B$ embeds into $B \times (\mathbb{R}^{N} \times (0,1])$, such that $M,M'$ land in the boundary?
| https://mathoverflow.net/users/134512 | Preservation of fiberwise normal bundles under fiberwise homotopy equivalences | After looking at the actual construction of the fiberwise embedding of a bundle $M \rightarrow E \rightarrow B$ into $B \times \mathbb{R}^N$, it became clear how to adjust it to embed the mapping cylinder. For convenience, let's assume that $B$ is a finite CW complex.
To obtain a fiberwise embedding of the bundle $p:E \rightarrow B$ into $B \times \mathbb{R}^N$, simply pick an embedding $i$ of $E$ into $\mathbb{R}^N$ and define $E \rightarrow B \times \mathbb{R}^N$ by $x \rightarrow (p(x),i(x))$.
Suppose now we have a fiberwise homotopy equivalence between $E \rightarrow B$ and $E' \rightarrow B'$. Let $I$ be a relative embedding of the mapping cylinder into $\mathbb{R}^L \times (0,1]$. Now we can define a fiberwise embedding of the mapping cylinder of the bundle equivalence by $x \rightarrow (p(x),I(x))$.
| 1 | https://mathoverflow.net/users/134512 | 425869 | 172,886 |
https://mathoverflow.net/questions/425887 | 2 | Let $f:U \to V$ be a flat, quasi-finite, surjective morphism between two affine varieties defined over $\mathbb{C}$. Assume that every closed fiber is reduced. Consider the function $\eta$ that sends a closed point $v \in V$ to the cardinality of the fiber $f^{-1}(v)$ over $v$. Is this function semi-continuous? If so, is it upper or lower semi-continuous? Recall, if we add the assumption that $f$ is proper, then it is well-known that $\eta$ is [locally-constant](https://stacks.math.columbia.edu/tag/0E0N).
| https://mathoverflow.net/users/38832 | Semi-continuity in quasi-finite morphisms without properness | The function is lower semi-continuous, see EGA4, Prop. 15.5.1.
Note that since you assume the fibres reduced hence (because the characteristic is 0) geometrically reduced, the map $f$ is in fact étale; and closed points have residue field $\mathbb{C}$ which is algebraically closed; hence your $\eta(v)$ is indeed the function $n(v)$ from EGA4.
| 7 | https://mathoverflow.net/users/17988 | 425888 | 172,892 |
https://mathoverflow.net/questions/425647 | 1 | I want to compute a integral of a polynomial $f(x, y)$ over a polygon domain $D$ of $n$ sides.
$$
I(f) = \int\_{D} f(x, \ y) \ dx \ dy
$$
The vertex of this polygon are
$$\vec{p}\_{i} = (x\_i, \ y\_i) \ \ \ \ \ \ \ \ \forall \ i = 1, \ 2, \ \cdots , \ n$$
The main aproach is transform this integral over the domain in a integral over the boundary using Green's theorem.
$$
\int\_{D} \left(\dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y}\right)\ dA = \oint\_{C} P \ dx + Q \ dy
$$
Where the path $C$ is
$$
\partial D = C = C\_1 \cup C\_2 \cup \cdots \cup C\_{n} = \bigcup\_{i=1}^{n} C\_{i}
$$
And $C\_{i}$ is defined by the linear parametrization with $t \in \left[0, \ 1\right] $
$$
\vec{p}(t) = \left(1-t\right)\cdot \vec{p}\_{i} + t \cdot \vec{p}\_{i+1}
$$
$$
x(t) = \left(1-t\right) \cdot x\_{i} + t \cdot x\_{i+1}
$$
$$
y(t) = \left(1-t\right) \cdot y\_{i} + t \cdot y\_{i+1}
$$
And therefore the integral $I$ is just
$$
I = \sum\_{i = 1}^{n} \int\_{C\_i} P \ dx + Q \ dy = \sum\_{i = 1}^{n} \int\_{0}^{1} \left(P, \ Q\right) \cdot \dfrac{d\vec{p}}{dt} \ dt
$$
$$
\boxed{ I =
\sum\_{i = 1}^{n} \int\_{0}^{1} \left(P, \ Q\right) \cdot \left(\vec{p}\_{i+1}-\vec{p}\_{i}\right) \ dt}
$$
To compute this integral, I can choose whatever I want for $P$ and $Q$ such that
$$
f(x, \ y) = \dfrac{\partial}{\partial x} Q(x, \ y) - \dfrac{\partial}{\partial y} P(x, \ y)
$$
And then compute the sum.
**The question:** Why the expression is not unique?
Like, if I choose a pair $(P\_1, \ Q\_1)$, I get a function $g\_1(x\_1, \ y\_1, \ \cdots, \ x\_{n}, \ y\_{n})$.
But if I choose another pair $(P\_2, \ Q\_2)$, I get $g\_2(x\_1, \ y\_1, \ \cdots, \ x\_{n}, \ y\_{n})$ with
$$
g\_1 \ne g\_2
$$
**Example:** In the video [Michael Penn - Overkill - The area of a rectangle](https://youtu.be/Y2cBpBLJlfU) we have $f(x, y) = 1$.
Then he chose $P = 0$ and $Q = x$ to get
$$
I(1) = \sum\_{i = 1}^{n} \dfrac{(x\_{i+1}+x\_{i})(y\_{i+1}-y\_{i})}{2}
$$
But if I choose $P=-y$ and $Q = 0$ I get
$$
I(1) = \sum\_{i=1}^{n} \dfrac{(x\_{i+1}-x\_{i})(y\_{i+1}+y\_{i})}{2}
$$
| https://mathoverflow.net/users/173662 | Double integral in a polygon domain | In the example, I got that
$$
I(1) = \sum\_{i=1}^{n} \dfrac{(x\_{i+1}+x\_{i})(y\_{i+1}-y\_{i})}{2} = \sum\_{i=1}^{n} \dfrac{(x\_{i+1}-x\_{i})(y\_{i+1}+y\_{i})}{2}
$$
And it's indeed correct. If I expand around $j$, I get that
$$
I(1) = \sum\_{i=1}^{j-2} \square + x\_{j}\left(y\_{j+1}-y\_{j-1}\right) + y\_{j}\left(x\_{j-1}-x\_{j+1}\right)+ \sum\_{i=j+1}^{n} \square
$$
To say that
$$
\dfrac{\partial}{\partial x\_{j}} I(1) = y\_{j+1} - y\_{j-1}
$$
$$
\dfrac{\partial}{\partial y\_{j}} I(1) = x\_{j-1} - x\_{j+1}
$$
For other polynomials $f$, it's harder to get a closed formula to that.
| 1 | https://mathoverflow.net/users/173662 | 425903 | 172,894 |
https://mathoverflow.net/questions/425803 | 7 | Given a group $G$, one way to construct the classifying (topological) space $BG$ is to regard $G$ as a category with one object and morphisms $G$, take its nerve, and then apply the geometric realization functor.
Is an analogous statement true for *topological groups*? That is, given a topological group $G$, regarded as a topological category with one object, does the geometric realization of its homotopy coherent nerve model the classifying space $BG$?
---
**Edit**: I should have been clear about what I meant by the "homotopy coherent nerve" of a topological category; what I meant was the composition
$$\mathsf{Cat}\_{\mathsf{Top}}\xrightarrow{\operatorname{Sing}}\mathsf{Cat}\_{\mathsf{sSet}}\xrightarrow{N}\mathsf{sSet},$$
where $\mathsf{Cat}\_{\mathsf{Top}}$ and $\mathsf{Cat\_{sSet}}$ are the category of small topologically/simplicially enriched categories, $\operatorname{Sing}$ is the base change along the singular complex functor, and $N$ is the [homotopy coheret nerve functor](https://ncatlab.org/nlab/show/homotopy+coherent+nerve#the_homotopy_coherent_nerve).
| https://mathoverflow.net/users/144250 | Simplicial nerve of a topological group | This is an answer to the edited question.
First, observe that the composition of functors $\def\N{{\rm N}}\def\Sing{{\rm Sing}}\N∘\Sing$ in the main post computes
the homotopy colimit of the simplicial object $[n]↦(\Sing(G))^n$.
This can be seen as follows.
Observe that the functors $\Sing$ and the homotopy coherent nerve are Quillen equivalences.
After the composition $\N∘\Sing$ we also implicitly apply
the left Quillen functor given by the identity functor on the underlying categories from the Joyal model structure to the Kan–Quillen model structure on simplicial sets.
(Indeed, we need to compare the value of $\N∘\Sing$ to a *space* (like $\B G$), not an (∞,1)-category.)
Altogether the whole composition is a homotopy cocontinuous functor.
Such a functor can be uniquely specified (up to a contractible choice) by its values on generators and maps between them,
which in our case correspond to categories of the form $\{0→1→2→⋯→n\}$ for all $n≥0$.
As one can see from the definitions, in our case all these values are contractible spaces.
After we apply $\Sing$, we are free to apply a different Quillen equivalence, and then take the corresponding (unique up to a contractible choice) homotopy cocontinuous functor to simplicial sets.
For example, we can apply the right Quillen equivalence given by the usual enriched nerve functor, which goes from simplicial categories to Segal categories, and then apply the left Quillen equivalence given by the inclusion of Segal categories into complete Segal spaces, i.e., simplicial objects in simplicial sets equipped with a certain model structure.
(See Theorem 8.6 in Bergner's Three models for the homotopy theory of homotopy theories.)
The enriched nerve functor yields the simplicial object $[n]↦(\Sing(G))^n$.
The generators $\{0→1→2→⋯→n\}$ correspond to simplicial objects in simplicial sets given by representable presheaves of simplices.
The unique homotopy cocontinuous functor that sends these generators
to contractible space is simply the homotopy colimit functor.
Indeed, the homotopy colimit functor is homotopy cocontinuous
and it sends all representable presheaves to contractible spaces.
This proves the above claim.
Denote the above homotopy colimit by $\def\B{{\rm B}}\B(\Sing(G))$.
What is the easiest way to see that principal $G$-bundles over a topological group $G$ are classified by $\B(\Sing(G))$?
One way is provided by Theorem 8.5 in [arXiv:2203.03120](https://arxiv.org/abs/2203.03120) (see also Theorem 1.1 in [arXiv:1912.10544](https://arxiv.org/abs/1912.10544)),
which gives us an explicit formula for the classifying space of any homotopy coherent sheaf $F$ of spaces on the site of topological spaces equipped with numerable open covers.
(Recall that any open cover of a CW-complex is numerable, so this is a strictly more general setting than in the original post.)
The classifying space is given by the homotopy colimit of simplicial sets
$$\def\hocolim{\mathop{\rm hocolim}}\def\op{{\rm op}}\def\gs{{\bf Δ}}\hocolim\_{n∈Δ^\op}F(\gs^n),$$
where $\gs^n$ denotes the topological space given by the $n$-dimensional simplex.
In our case, $F(X)$ is the nerve of the category of principal $G$-bundles over $X$ and their isomorphisms.
Since any principal $G$-bundle over $\gs^n$ is trivial, we have $\def\C{{\rm C}}F(\gs^n)≃\B(\C(X,G))$, where $\C(X,G)$ denotes the group of continuous functions $X→G$.
Now
$$\hocolim\_{n∈Δ^\op}F(\gs^n)≃\hocolim\_{n∈Δ^\op}\B(\C(\gs^n,G)).$$
Since the functor $\B$ is homotopy cocontinuous, we have
$$\hocolim\_{n∈Δ^\op}\B(\C(\gs^n,G))≃\B(\hocolim\_{n∈Δ^\op}\C(\gs^n,G))≃\B(\Sing(G)),$$
as desired, proving that numerable principal $G$-bundles over an arbitrary topological space $X$ are classified by the simplicial set $\B(\Sing(G))$
(or the topological space $\B G$).
(This computation appears as Example 8.6 in the cited paper.)
| 3 | https://mathoverflow.net/users/402 | 425910 | 172,896 |
https://mathoverflow.net/questions/425906 | 1 | I am interested in the convergence of the following Euler product:
$$
\prod\_p \frac{1}{1-\chi(p)\cdot p^{-s}}.
$$
The product is over all primes (in increasing order), with $\chi(p)=+1$ if $p \bmod 4 =3$ and
$\chi(p)=-1$ if $p \bmod 4 =1$. Here $\chi(2)=\pm 1$, the sign does not matter. Also $s=\sigma+ it$ to use the standard notation.
Since the densities, for both set of primes, are identical, the signs will alternate nicely on average, and one would guess that the product converges not only for $\sigma>1$, but actually for $\sigma>0$. Is this the case, or is this still a conjecture?
Also, I'd love to see a plot of the orbit for a fixed value of $\sigma$, say $\sigma=0.75$. By orbit, I mean a scatterplot of the product in the complex plane (real vs. imaginary part), for a fixed value of $\sigma$, and $t$ varying between $0$ and (say) $1000$. See example [here](https://youtu.be/aPz7jy4jfEA) and [here](https://youtu.be/uMRp9FruOqE) for a truncated (finite) product when $\chi$ is constant equal to $1$.The hole in the orbit shrinks to a single point (the origin) if you add all the primes [if RH if true; if not it shrinks to an empty set].
I'm working on a number theory tutorial, and I'd like to introduce the students to a situation where the orbit not only never crosses the origin (as for the standard Riemann Hypothesis if $0.5<\sigma<1$), but in addition, stays away and never gets too close to the origin. If possible, not a conjectured result, but an established one.
On a side not, I am interested to know what the abscissa of convergence is, for the associated Dirichlet series.
**Update - Attempt with a different $\chi$**
Let $p\_k$ be the $k$-th prime ($p\_1=2$ and so on) and $\chi(p\_{2k})=-1$,
$\chi(p\_{2k+1})=+1$. Let $\chi$ be completely multiplicative. Could the abscissa of convergence be strictly less than $1$? If you look at the Euler product, say for $\sigma=0.75$, its convergence at $t=0$ is equivalent to the convergence (after taking the logarithm and standard asymptotic) of
$$
\sum\_{k=1}^\infty (-1)^{k+1}\frac{1}{p\_k^\sigma}.
$$
I believe that the Dirichlet test could positively answer the question. Actually, there was a question asked on MO eight years ago [see [here]](https://mathoverflow.net/questions/159534/does-this-alternating-euler-product-converge-for-all-res-0), about a similar convergence problem (proved to converge) so similar that this one must also converge if $\sigma=0.75$, and indeed, as long as $\sigma>0$.
Assuming this is correct, and since convergence for a specific $\sigma\_0=0.75$ at $t=0$ implies convergence for all $t$ for that particular $\sigma\_0$, it implies that the abscissa of convergence is well below $1$. Does this also hold for the Euler product? If yes (and I assume that the answer is yes), then can the product still be equal to zero? The norm of the product can be very close to zero. My guess is that for the product to be zero if $\sigma<1$ (that is, to diverge) you would need $t=\infty$. I could be wrong.
Now defining the concept of **hole**. In dynamical systems, it's called a repulsion basin, usually not circular. Let $L(z,\chi)$ be the Dirichlet function represented by the product or its associated series. Here the hole is centered at the origin $z\_0=(0,0)$ in the complex plane. It is the largest circle of radius $\rho$ such that
$|L(z,\chi)-z\_0|>\rho$ for all $z=\sigma+it$ in the complex plane, for a fixed value of $\sigma$, say $0.75$.
There is one thing where I am most certainly wrong: the hole is reduced to to a single point (the origin), contrarily to what I thought initially (a hole of radius $\rho>0$). But if you make a video of the orbit starting at $t=0$, no matter how fast your computer is, for a very, very long time the hole will be visible to the naked eye. It will shrink as $t$ increases, in the end to a single point (or worse, an empty set if there are some unexpected roots) but incredibly slowly. That's what it does for the standard RH case.
**Final update: non-trivial examples with a hole**
This does not contradict the universality property: this property is true if $\sigma<1$ and applies if $P$ is the set of all primes. Here $\sigma<1$ (typically but not necessarily) and $P$ can be a subset of primes, finite or infinite. A modified (generalized) universality property, in this context, would be: the orbit is dense between its internal boundary (the boundary of the hole), and its external boundary if bounded (e.g. if $\sigma>1$). Still have to prove it. Also, I assume that if there is a "thick" hole, there is only one.
I could not find "standard" examples (characters modulo $m$) with a hole if $P$ contains sufficiently many primes. I am pretty sure there are no such examples. So you need to look at cases where $P$ is infinite but sparse enough. It will work (still have to go through a round of double-checking to confirm this) if
$$\rho=\prod\_{p\in P} \frac{1}{1+p^{-s}}>0$$
regardless of $\chi$. In other words, the above product needs to converge. Then the radius of the hole is $\geq \rho$. Using my generalized version of the universality property, I guess the radius is $\rho$ (exactly). This is compatible with RH ($\rho=0$) and GRH (in that case $P$ is too big anyway, and $\rho=0$).
Examples with a hole with $\rho>0$:
* If $P$ is the set of all primes that are the sum of two cubes.
* If $P$ is the set of twin primes ?? (this case is borderline)
| https://mathoverflow.net/users/140356 | Abscissa of convergence for a very specific Dirichlet series / Euler product | What do you mean by the word "orbit"? Please define that term in the body of your question.
You ask where the product over $p$ converges. Although you did not specify it, I assume you mean *product with terms in the order of increasing $p$*. You need to specify the order of the terms if you're dealing with something not necessarily absolutely convergent.
Your $\chi$ is meant to be a totally multplicative function.
For odd primes $p$, your $\chi(p)$ is $\lambda(p)\chi\_4(p)$ where $\lambda$ is the Liouville function (the nonvanishing totally multiplicative cousin of the Moebius function that is $-1$ at all primes) and $\chi\_4$ is the nontrivial Dirichlet character mod $4$, so a calculation with Euler products shows for ${\rm Re}(s) > 1$ that (as GH from MO points out)
$$
\sum\_{n \geq 1} \frac{\chi(n)}{n^s} = \frac{1}{1-\chi(2)/2^s}\frac{L(2s,\chi\_4^2)}{L(s,\chi\_4)} = \left(1 + \frac{\chi(2)}{2^s}\right)\frac{\zeta(2s)}{L(s,\chi\_4)}.
$$
The Euler product for $\zeta(2s)$ converges absolutely for ${\rm Re}(s) > 1/2$ and $1 + \chi(2)/2^s$ is nonvanishing for ${\rm Re}(s) > 0$, so for ${\rm Re}(s) > 0$ your Euler product will converge exactly where
$\prod\_p (1 - \chi\_4(p)/p^s)$ converges.
The Generalized Riemann Hypothesis for $L(s,\chi\_4)$ implies for ${\rm Re}(s) > 1/2$ that $L(s,\chi\_4) \not= 0$ and that $L(s,\chi\_4) = \prod\_p 1/(1-\chi\_4(p)/p^s)$. Conversely, the convergence and nonvanishing
of $\prod\_p 1/(1-\chi\_4(p)/p^s)$ at a number $s\_0$ with ${\rm Re}(s\_0) > 1/2$ implies it equals $L(s\_0,\chi\_4)$ and that $L(s,\chi\_4) \not= 0$ for ${\rm Re}(s) > {\rm Re}(s\_0)$.
So it is not reasonable to expect the product to converge for ${\rm Re}(s) < 1/2$ and it is reasonable to expect the product to converge for ${\rm Re}(s) \geq 1/2$ excluding at zeros of $L(s,\chi\_4)$ on the critical line, but the only way you're going to have access to such results is through GRH.
It is utterly *hopeless* to expect a proof of convergence
of either $\prod\_p 1/(1-\chi\_4(p)/p^{s\_0})$ or $\sum \chi\_4(p)/p^{s\_0}$ at a single $s\_0$ where ${\rm Re}(s\_0) < 1$ unless you expect to be making dramatic progress towards GRH for $L(s,\chi\_4)$, and the same is true with $\chi\_4$ replaced by $\chi$. Concerning the Dirichlet series $\sum \chi(n)/n^s$, since $\chi(n) = \lambda(n)\chi\_4(n)$ for odd $n$ it's reasonable to expect $\sum\_{n \leq x} \chi(n) = O\_\varepsilon(x^{1/2+ \varepsilon})$ if you want to use GRH for $L(s,\chi\_4)$, so it's reasonable to expect that $\sum \chi(n)/n^s$ converges for ${\rm Re}(s) > 1/2$, but it's not reasonable to expect a proof of that without a major advance on GRH. For an analogous situation where a Dirichlet series was proved to converge on a half-plane bigger than its half-plane of absolute convergence by relying on a major result (modularity of elliptic curves over $\mathbf Q$), see my answer to an earlier MO question [here](https://mathoverflow.net/questions/25360/convergence-of-l-series).
Watch out: there is probably weird behavior for the Euler product on the critical line. We have $L(1/2,\chi\_4) \not= 0$ (not weird) and if $\prod\_p 1/(1-\chi\_4(p)/\sqrt{p})$ has a nonzero value, then that value is $\sqrt{2}L(1/2,\chi\_4)$ (weird!). See my paper "Partial Euler products on the critical line" or Kuo and Murty's paper "On a conjecture of Birch and Swinnerton-Dyer". Both appeared in Canadian J. Math. in 2005.
Since your purpose in working with the Euler product $\prod\_p 1/(1-\chi(p)/p^s)$ is purely to have a concrete example for some students, I suggest you work with $\chi\_4$, not $\chi$: the function $\chi\_4$ is a more commonly encountered object (a Dirichlet character), $\chi\_4$ doesn't have an artificial definition at $p = 2$, and the half-plane of convergence of the Dirichlet series $\sum \chi\_4(n)/n^s$ is already known to be precisely ${\rm Re}(s) > 0$.
| 7 | https://mathoverflow.net/users/3272 | 425913 | 172,898 |
https://mathoverflow.net/questions/425920 | 20 | Does there exist a pair of finite groups $G$ and $H$ satisfying both of the short exact sequences $1 \rightarrow G \rightarrow H \rightarrow A\_4 \rightarrow 1$ and $1 \rightarrow G \rightarrow H \rightarrow D\_6 \rightarrow 1$? Of course the homomorphisms $G \to H$ in these short exact sequences are not the same.
| https://mathoverflow.net/users/172799 | Pair of short exact sequences of groups | Call two finite groups $Q\_1$ and $Q\_2$ *compatible* if there exists a finite group $G$ with two isomorphic normal subgroups $N\_1$ and $N\_2$ such that $G/N\_i\cong Q\_i$.
One can show the following:
**Proposition:**
If two groups are compatible, then they have subnormal series of the same length with the same factor groups appearing in the same order.
*Proof:*
Let $Q\_1$ and $Q\_2$ be compatible with $G$ a witness of minimal order, and $N\_1$ and $N\_2$ the two corresponding isomorphic normal subgroups and let $\alpha$ be an isomorphism from $N\_1$ to $N\_2$. Let $M=N\_1\cap N\_2$. Note that $M$ and $\alpha(M)$ are isomorphic and normal in $N\_2$, so $N\_2/M$ and $N\_2/\alpha(M)$ are compatible, with $N\_2$ as a witness.
But $N\_2/M\cong N\_1N\_2/N\_1$ while $N\_2/\alpha(M)\cong N\_1/M\cong N\_1N\_2/N\_2$. Minimality of $G$ implies that $N\_1N\_2<G$, so that $Q\_1$ and $Q\_2$ have $G/N\_1N\_2$ as a non-trivial common quotient, but moreover the corresponding normal subgroups are compatible, so the result follows by induction.
$\square$
I've read somewhere that the above argument (which is in some sense a generalisation of the one by Robert) is due to Sims, but I'm not sure the argument itself is actually written anywhere.
In particular, it shows that $A\_4$ and $D\_6$ are not compatible, because they don't have such subnormal series. (Any series for $A\_4$ has a $C\_3$ "on top", and in $D\_6$, a $C\_2$ "on top".)
I've been interested in the question of determining which groups are compatible for a while. I think it's an interesting question and the answer is not known. See
Giudici, Glasby, Li, Verret, Arc-transitive digraphs with quasiprimitive local actions, Journal of Pure and Applied Algebra 223 (2019) 1217-1226
for some motivation and further results.
See also <https://math.stackexchange.com/questions/4295186/which-pairs-of-groups-are-quotients-of-some-group-by-isomorphic-subgroups/4296206#4296206>
| 26 | https://mathoverflow.net/users/22377 | 425924 | 172,901 |
https://mathoverflow.net/questions/425453 | 1 | Fix $\alpha \in (0,1)$ and $\psi\in C^{\infty}\_{c}(\mathbb{R}\to \mathbb{R})$. For a smooth function $\phi\geq 0$ define the integral $$J\_{\alpha}(\phi):=\int \frac{\psi}{\phi^{\alpha}}.$$
If $|\phi^{\prime}|$ is bounded below away from zero, then
$J\_{\alpha}$ exists for all $\alpha<1$. Indeed, one can simply integrate by parts to get
$$J\_{\alpha}=-\frac{1}{1-\alpha}\int \phi^{1-\alpha}\left(\frac{\psi}{\phi^{\prime}}\right)^{\prime}.$$
My question is about the case where $\phi^{\prime}$ changes sign finitely many times in the support of $\psi$. Using a smooth partition of unity, one may assume $\phi^{\prime}(x)=0$ at exactly one point $x\_{0}$ in the support of $\psi$.
For simplicity, let us further assume that $|\phi^{\prime\prime}|$ is bounded below away from zero near $x\_{0}$.
Obviously, we necessarily have $\alpha<1/2$. Just take $\phi$ to be a quadratic function.
I figured this out when $\phi^{\prime\prime}>C>0$: Write
$$\phi(x)=\phi(x\_{0})+(x-x\_{0})^{2}\int\_{0}^{1}(1-t) \phi^{\prime\prime}(x\_{0}+t(x-x\_{0}))dt\geq \frac{C}{2}(x-x\_{0})^{2}$$
In this case we have
$$|J\_{\alpha}(\phi)|\leq \frac{2}{C}\int \frac{|\psi|}{|x-x\_{0}|^{2\alpha}}$$
which exists for all $\alpha<1/2$.
What is left is the case $\phi^{\prime\prime}<c<0$.
**Summary :**
Fix $\alpha \in (0,1/2)$ and $\psi\in C^{\infty}\_{c}(\mathbb{R}\to \mathbb{R})$. Let $\phi\geq 0$ be a smooth function with a unique stationary point $x\_{0}$ in the support of $\psi$. Assume that $\phi^{\prime\prime}<c<0$.
Show that
$J\_{\alpha}(\phi)=\int \frac{\psi}{\phi^{\alpha}}$ is finite ?
| https://mathoverflow.net/users/116555 | Determine $\alpha \in (0,1)$ such that $J_{\alpha}(\phi):=\int \psi/\phi^{\alpha}$ exists? | $\newcommand\al\alpha\newcommand\R{\mathbb R}$Let us prove the following generalization of your desired statement:
>
> Let $\psi\colon\R\to\R$ be a continuous function with compact support $S$. Let $\phi\colon\R\to\R$ be a nonnegative twice differentiable function with the following property: for any $x\_0\in S$ such that $\phi'(x\_0)=0$ we have $\phi''(x\_0)<0$. Then $J\_\al(\phi):=\int\frac{\psi}{\phi^\al}$ is finite for any real $\al>0$.
>
>
>
The proof is based on a comment by Christian Remling. Indeed, take any $x\_0\in S$ such that $\phi(x\_0)=0$. Since $\phi$ is nonnegative, it follows that $\phi'(x\_0)=0$ and hence $\phi''(x\_0)<0$, which implies that $\phi<0$ in some punctured neighborhood of $x\_0$. This contradicts the condition that $\phi$ is nonnegative.
So, there is no $x\_0\in S$ such that $\phi(x\_0)=0$. So, $\phi>0$ on $S$. Since $\phi$ is continuous and $S$ is compact, we have $\phi\ge b$ on $S$ for some real $b>0$. Thus, for any real $\al>0$
$$|J\_\al(\phi)|\le\int\_S\frac{|\psi|}{b^\al}<\infty,$$
since $\psi$ is continuous and $S$ is compact.
We conclude that $J\_\al(\phi)$ is finite for any real $\al>0$, as claimed. $\quad\Box$
| 1 | https://mathoverflow.net/users/36721 | 425941 | 172,904 |
https://mathoverflow.net/questions/425895 | 1 | I have a question on subfactors of $B(H)$ (the von Neumann algebra of bounded operators on a complex Hilbert space).
Say I have a subfactor $M$ of $B(H)$. Is it true that another subfactor $N \subset B(H)$ which commutes with $M$ and is such that the vN algebra $(M\cup N)''$ generated by $M$ and $N$ is all of $B(H)$, is automatically the commutant of $M$?
I think that it automatically follows that $M \cap N=\mathbb C \,1\!\!1$ because the intersection is contained in the center of, say, $M$.
EDIT: Another way to put this question would be as follows. Consider pairs of commuting subfactors $M,N \subset B(H)$. A pair is trivial if $M=N'$ and it is generating if $(M\cup N)''=B(H)$. The question is now: Is there a non-trivial generating pair of subfactors $M,N \subset B(H)$?
| https://mathoverflow.net/users/485160 | Commuting and generating subfactors of $ B(H)$ | Let $N \subset M'$ be an *irreducible* inclusion of subfactors, meaning that $N \neq M'$ but the relative commutant is trivial, i.e., $N' \cap M' = \mathbb{C}\cdot I$. There are lots of examples of such things (Google "example of irreducible subfactor"). Then the commutant of the von Neumann algebra generated by $M \cup N$ is contained in both $M'$ (since it contains $M$) and $N'$ (since it contains $N$), so it must be trivial. Therefore this von Neumann algebra must be $B(H)$.
| 0 | https://mathoverflow.net/users/23141 | 425942 | 172,905 |
https://mathoverflow.net/questions/425943 | 0 | Let $\{ f\_n \}$ be a sequence of Schwartz functions on $\mathbb{R}^n$ converging to some Schwartz function $f$ in the $L^2(\mathbb{R}^n)$ norm.
Then, it is an elementary fact that we can extract a subsequence converging almost everywhere to $f$.
However, I wonder if we can extract "nicer" subsequences as well. For example, is it possible to extract a subsequence of $\{ f\_n \}$ that converges to $f$ in the Schwartz class topology(=the Frechet topology on the Schwartz space)?
Could anyone please provide any information?
| https://mathoverflow.net/users/56524 | $L^2$ convergence of Schwartz functions to a Schwartz function and possibility of extracting a "nicer" subsequence? | To get an example: consider non-negative smooth functions $\chi\_n$ with compact support in the intervals $[0, 1/n]$ with maximum $1$ at some point in this interval. Note that such functions indeed exist and belong to the Schwartz space (being smooth with compact support). Their $L^2$-norm is then at most $1/n$, thus this sequence converges to the zero function in the $L^2$-sense. But it does not converge to zero uniformly, let alone in the Schwartz sense.
| 5 | https://mathoverflow.net/users/12482 | 425946 | 172,906 |
https://mathoverflow.net/questions/425939 | 3 | Consider the following weak version of the Noether–Enriques theorem (field is $\mathbb{C}$):
>
> Let $\varphi:X\rightarrow Z$ be a morphism from a smooth projective surface onto a smooth curve with $F\_z:=\varphi^{-1}(z)\cong\mathbb{P}^1$ for every point $z$. Then there exists a Zariski open neighborhood $U$ of any point $z$ making the following diagram commute.
> $\require{AMScd}$
> \begin{CD}
> \varphi^{-1}(U) @>{\sim}>> U\times\mathbb{P}^1\\
> @V{\varphi}VV @VV{\text{projection}}V\\
> U @>{\sim}>> U
> \end{CD}
>
>
>
The critical point is to find a divisor $D$ on $X$ such that $D.F\_z=1$ or to find a section $s:Z\rightarrow X$ for $\varphi$. Beauville states in his book [Complex Algebraic Surfaces](https://doi.org/10.1017/CBO9780511623936) (c.f. Remark III.6) that this step can be done by using Tsen's lemma, and I want to know how exactly can we do this.
>
> In particular, I want to know how to make $X$ a conic?
>
>
>
Indeed, Beauville even claims that we can use Tsen's lemma to show this step for the usual Noether–Enriques theorem.
| https://mathoverflow.net/users/nan | Noether–Enriques using Tsen's lemma | I will assume that $X$ is proper. Then the generic fiber is a smooth projective curve of genus $0$ over the function field $K = k(Z)$ but any such curve can be embedded as a conic in $\mathbb{P}^2\_K$ using the anticanonical linear series. By Tsen's theorem, the conic has a $K$-point which we can spread out to a $U$-point of $X$ where $U$ is some open subset of $Z$. Finally by properness, this $U$-point extends to a section over $Z$.
| 3 | https://mathoverflow.net/users/12402 | 425949 | 172,907 |
https://mathoverflow.net/questions/425957 | 9 | I wonder whether $\sum\_{k=0}^n \exp(r\_k z)$ has a complex zero for any $n\in \mathbb{Z}\_n^\*,0=r\_0<r\_1<r\_2<\dotsb<r\_n$. I think the answer is affirmative.
| https://mathoverflow.net/users/481550 | Zeros of a complex function | An affirmative answer follows from (9) in [this paper by Ritt](https://www.ams.org/journals/tran/1929-031-04/S0002-9947-1929-1501506-6/S0002-9947-1929-1501506-6.pdf).
| 8 | https://mathoverflow.net/users/11919 | 425959 | 172,910 |
https://mathoverflow.net/questions/424902 | 7 | Let $B\_{0}\supseteq B\_{1}\supseteq\dots\supseteq B\_{\alpha}\supseteq\dots\,\,\left(\alpha<\kappa\right)$ be a descending sequence of complete Boolean algebras, $B\_{\kappa}:=\bigcap\_{\alpha<\kappa}B\_{\alpha}$, $G\_0$ a $V$-generic filter on $B\_0$ and for every $\alpha\leq\kappa$, $G\_{\alpha}:=G\cap B\_{\alpha}$. I want to understand the intersection $\bigcap\_{\alpha<\kappa}V\left[G\_{\alpha}\right]$, which clearly contains $V[G\_{\kappa}]$.
In *Iterating ordinal definability*, Zadrożny surveys without proof some results. First are general results:
>
> 1. (Grigorief) $\bigcap\_{\alpha<\kappa}V\left[G\_{\alpha}\right]\vDash ZF$
> 2. (Jech) If $B\_0$ is $\kappa$-distributive, then $\bigcap\_{\alpha<\kappa}V\left[G\_{\alpha}\right]=V[G\_\kappa]$ (so in particular satisfies $ZFC$)
>
>
>
[The latter appears as lemma 26.6 in the 1978 edition of Jech's Set Theory, but curiously I haven't found it in the 3rd Millennium edition].
Then he gives a more concrete characterization of $V\left[G\_{\kappa}\right]$, attributed to Sakarovitch:
>
> For $p,q\in B\_0^+$, let $p\sim q$ iff $\exists \alpha<\kappa$ such that $$ \inf\{d\in B\_\alpha \mid d\geq p\}=\inf\{d\in B\_\alpha \mid d\geq q\}$$
> Then the separative part of $B\_0/{\sim}$ is isomorphic to $B\_\kappa^+$.
> In particular, from $G\_0$ one can define a $B\_0/{\sim}$ generic $G\_0/{\sim}$, and if $B\_0$ is $\kappa$-distributive then $$\bigcap\_{\alpha<\kappa}V\left[G\_{\alpha}\right]=V[G\_\kappa]=V[G\_0/{\sim}]$$
>
>
>
I want to understand this result more, however Zadrożny references [Sakarovitch's PhD thesis](https://pascal-francis.inist.fr/vibad/index.php?action=getRecordDetail&idt=PASCAL7830080080), which is in French, not available online and as far as I can tell, has no adaptation to a paper. So my questions are:
1. Is there some other source where this result is presented?
2. Can someone provide a proof or at least a sketch?
3. What more can be said on this intersection, given the properties of the descending sequence?
A particular case I'm interested in, is when the sequence is given by "tails" of an iteration (or even product): assume $\langle P\_\alpha \mid\alpha\leq\kappa\rangle$ is an iteration, and set for every $\alpha$ $P\_\kappa=P\_\alpha \* \dot{P}^\alpha$ (so $\dot{P}^\alpha$ is the "tail" of the iteration), and "$B\_\alpha=ro(\dot{P}^\alpha)$" (I guess that for this to make perfect sense we should say something like $B\_\alpha=ro(P\_\kappa)/ro(P\_\alpha)$). Can something more concrete be said in this case?
Edit: note that if $P\_\kappa$ is a direct limit (and $\kappa$-distributive), we'd get that all elements are eventually equivalent so $B\_0/{\sim}$ is trivial. So on one hand this shows that the intersection is $V$. And on the other hand, the question is interesting only when non-direct limit is taken.
| https://mathoverflow.net/users/91680 | Understanding descending intersections of generic extensions | The theorem should state that the separative *quotient* of $B\_{0}/{\sim}$
is isomorphic to $B\_{\kappa}^+$.
*Proof.* Recall that the separative quotient of a poset $P$ is the
unique (up to isomorphism) separative $Q$ such that there is an order preserving $h:P\to Q$ such that $x$ is compatible with $y$ iff $h(x)$
is compatible with $h(y)$ (see Jech pg. 205). Since $B\_{\kappa}$
is separative as a Boolean algebra, we want to provide such $h:B/{\sim}\to B\_{\kappa}^{+}$.
Let $b\in B\_{0}^{+}$. For every $\alpha<\kappa$ let
$b\_{\alpha}=\inf \{ d\in B\_{\alpha}\mid d\geq b \} $,
and let $\bar{b}=\sup\{b\_{\alpha}\mid\alpha<\kappa\}$. Note that
$\{b\_{\alpha}\mid\alpha<\kappa\}$ is an ascending sequence, so in
fact for every $\beta$, $\bar{b}=\sup\{b\_{\alpha}\mid\beta\leq\alpha<\kappa\}$,
and this is an element of $B\_{\beta}$, so all-in-all $\bar{b}\in B\_{\kappa}$.
Now if $b'\sim b$ then for all large enough $\alpha$, $b\_{\alpha}=b\_{\alpha}'$
so $\bar{b}=\bar{b'}$. So the function $h([b])=\bar{b}$ is well
defined, and since in particular $\bar{b}\geq b>0$, it is into $B\_{\kappa}^{+}$. It is order preserving since if $[b]\leq[c]$ then for all large enough $\alpha$, we have $b\_\alpha \leq c\_\alpha $ so also $\bar{b}\leq\bar{c}$.
Let $[b],[c]\in B\_{0}/{\sim}$. We want to show they are compatible
iff $\bar{b}$ and $\bar{c}$ are compatible.
* If $[b],[c]$ are compatible, $[d]\leq[b],[c]$, then for all large
enough $\alpha$ $d\_{\alpha}\leq b\_{\alpha},c\_{\alpha}$ so $\bar{d}\leq\bar{b},\bar{c}$.
* If $[b],[c]$ are incompatible, it means that there is an unbounded
$I\subseteq\kappa$ such that for $\alpha\in I$, $b\_{\alpha}$ and
$c\_{\alpha}$ are incompatible. But this also implies that for *every*
$\alpha,\beta\in I$ $b\_{\alpha}$ and $c\_{\beta}$ are incompatible
(if $\alpha<\beta$ and there is e.g. $d\leq b\_{\alpha},c\_{\beta}$
then since $b\_{\alpha}\leq c\_{\beta}$ we get $d\leq b\_{\beta},c\_{\beta}$),
so, as we are in a complete Boolean algebra,
$$
\overline{b}\cdot\bar{c}=\sum\_{\alpha\in I}b\_{\alpha}\cdot\sum\_{\beta\in I}c\_{\beta}=\sum\_{\alpha,\beta}b\_{\alpha}\cdot c\_{\beta}=0
$$
i.e. $\bar{b},\overline{c}$ are incompatible. $\square$
Regarding the specific case, my advisor pointed out to me that if we are considering tails of a *full support product*, say of length $\kappa$, then the quotient poset will be $\kappa$-closed.
*Proof sketch.* Let $P=\prod\_{\alpha<\kappa}Q\_{\alpha}$
be a full support product of posets, and for $\alpha<\kappa$ let
$P\_{\alpha}=\prod\_{\alpha\leq\xi<\kappa}Q\_{\xi}$. Let $\left\langle p\_{i}\mid i<\kappa\right\rangle $ be sequence such
that $i<j$ implies $[p\_{i}]>[p\_{j}]$. Then to construct $q$ such that $[q]$ is a lower bound, we diagonalize - let $q(\xi)=p\_0(\xi)$ until the coordinate witnessing $[p\_0]>[p\_1]$, then $q(\xi)=p\_1(\xi)$ until $p\_2$ is smaller, and so on. I'll spare you the indexing monstrosity. $\square$
| 3 | https://mathoverflow.net/users/91680 | 425961 | 172,911 |
https://mathoverflow.net/questions/425960 | 6 | Last month @godelian, alias Christian Espíndola, in a FOM post has mentioned Joyal's proof of Godel's second incompleteness via the so-called **Arithmetic Universes**, introduced by Joyal around 1973, and further investigated by Vickers, Maietti and Taylor.
After perusing the relatively sparse literature on them, I decided that the last [reference](https://arxiv.org/abs/2004.10482) was the only one I could use as an entry door. Though the article makes a substantial effort to be self-contained, it is still to me a bit of a hard reading, especially in building the right intuition (though some great help comes from Taylor's answer here: [Categories of recursive functions](https://mathoverflow.net/questions/126513/categories-of-recursive-functions/126782#126782) and by Godelian [Has Goedel's Second Incompleteness Theorem been proven using Lawvere's Fixed Point Theorem?](https://mathoverflow.net/questions/424374/has-goedels-second-incompleteness-theorem-been-proven-using-lawveres-fixed-poi/) )
What I understand so far:
1 Joyal introduces Arithmetic Universes, ie pretopoi ( finite limits, effective quotients of equivalence
relations ) plus an axiom which essentially says: if you have an object **A** you also have
a list object **LIST[A]** which is to be intended as lists of elements of A (actually, LIST[-] is an endofunctor)
.
Note: an AU is in general NOT
an elementary topos, though of course all topoi with a natural number object NNO are trivially AUs. Topoi are universes of constructive set theory, they have too much (functions spaces, power object, all things that are indispensable for higher order logic, but not here).
First key point: one can inhabit an AU, and from its vantage point there is an internal typed language (developed by Maietti)
that makes it "roomy enough" to talk about primitive recursive functions, decidable predicates, etc. In other words, arithmetic universes are indeed universes of discourse which are paired with **Primitive Recursive Arithmetics (PRA)**. It goes without saying that there is a 2-category of AUs, let us call it AUS. The functors between them preserve the AU structure.
**FIRST QUESTION**: who are these AUs? if I take any ordinary model of PRA, the category of
all definable functions, predicates, and whatnot of M is an AU. But, is it true that all AUS come from that? In other words, is there a canonical way to complete/embed an AU to a set-based model of PRA?
2 the theory of the AUS is *essentially algebraic*, so there is a distinguished
initial object in AUS, let us call it **AU-0**. AU-0 is the freest AU of the multiverse AUS\*
3 Fact : AU-0 has an internal category, let us call it $au\_0$, which is ITSELF a AU
(in the internal sense).
$au\_0$ is quite intriguing: in a sense, it measures the degree of self-reflection of AU-0, ie how much AU-0 can code of itself. Moreover, as AU-0 is initial in AUS, ALL AUs have their own little au.
The construction of $au\_0$ from AU-0 leverages a number of things included the infamous list object above (see reference) .
4. Now we get to the real "main course", and also to the haziest area to me:
category theorists love internal cats, and have a technique called *EXTERNALIZATION*,
by which they can externalize internal cats. So, let us externalize little au-0:
$$
AUG = EXT(au\_0).
$$
AUG is a REAL object of the arithmetic multiverse, and therefore we can compare it with
AU-0.
The fundamental piece of magic is that by looking at the interplay between AU-G and AU-0
we can get Godel 1 and 2. How? Let me point out one thing:
AU-0 is INITIAL in AUS, so there is an AU functor R FROM AU-0 TO AU-G:
$$R\colon AU\text{-}0 \rightarrow AUG $$
There is something else, 'going
the other way around:
$$L\colon AUG \rightarrow AU\text{-}0 $$
Whereas T is an AU-functor, L is not. Nevertheless we can define $\Box= LR$ which is (correct me if I am wrong ) a *monad* on AU-0.
Now, the magic: $\Box$ is Joyal's categorical version of the famous Provability Operator in [Provability Logic](https://plato.stanford.edu/entries/logic-provability/)! It satisfies Lob's condition and can be used to do exactly what the provability operator does in algebraic logic, for instance G 1 and G2.
**SECOND QUESTION**: to me this last steps still sounds like magic: what is the proper way to think of the externalized GAU? If indeed the Box is monadic, what is its category of algebras?
| https://mathoverflow.net/users/15293 | Joyal arithmetic universes and the Box operator | As discussed in the comments, the AU's are all models of the initial one, which is equivalent to the pretopos completion of the syntactic category of PRA (in coherent logic). In particular a set model of PRA cannot be made initial because it could collapse two different arrows in the initial AU (i.e., there could be two primitive recursive functions that are extensionally equal but whose recursive definitions are different).
The externalization of the internal initial AU is precisely the Gödel numbering on each formula. More explicitly, each object of the externalization is an arrow $1 \to O$ where $O$ is the object of "objects" of the internal arithmetic universe, so it is an arrow from the terminal object to something that is a quotient by an equivalence relation of a certain primitive recursive predicate (i.e., a decidable subset of $\mathbb{N}$), and all the numbers in the chosen equivalence class are the Gödel codes of the corresponding formula (there are several depending on which coding we use).
Now the global section functor from the externalization to the initial AU is the object that represents the hom set $[1, -]$, and since the externalization is also a syntactic category, having an arrow from $1$ is equivalent to being provable. So this functor is indeed the correct way to code the provability of the formulas represented by Gödel numbers.
But $R$ and $L$ are not adjoints, so we cannot speak of the corresponding monad, for otherwise $L$ would have to be the right adjoint. To see that this is impossible, note that the construction of Joyal sentence $J$ (the self-referential sentence "I am provably false") is proven to be equivalent to $Incon(PRA)$. By construction, the application of the functor $R$ gives us the corresponding Joyal sentence $R(J)$ of the externalization, which is never $0$ if the externalization is consistent (this is Gödel's second incompleteness theorem). So, if $L$ and $R$ were an adjoint pair, we would have in particular that $[J, L(0)] \cong [R(J), 0]$. But $[R(J), 0]$ is empty when the externalization is consistent, while $[J, L(0)] \cong [J, \square \bot] \cong [J, J]$ which is nonempty.
| 6 | https://mathoverflow.net/users/12976 | 425965 | 172,913 |
https://mathoverflow.net/questions/425005 | 4 | Is it true that for every infinite group $G$ and every $m\in\mathbb{N}$ there are infinite subsets $A\_0,\dots,A\_{m-1}$ such that all the products $a\_0\cdot\dots\cdot a\_{m-1}$ with $a\_i\in A\_i$ are distinct?
(This is true at least when $G$ is either uncountable or contains an infinite abelian subgroup, and for $m=2$ this is true for all $G$. Also, one can always obtain arbitrarily large finite $A\_i$)
| https://mathoverflow.net/users/484402 | Existence of m infinite subsets in an arbitrary group such that all products of one element from each (in order) are distinct | Yes, such infinite subsets $A\_0, \dots, A\_{m-1}$ of an infinite group $G$ always exist. We can show this by wedging the problem into one of two polar opposite cases: abelian and ICC (i.e. every non-trivial element has an infinite conjugacy class).
**Fact.** An infinite group either has an infinite abelian subgroup or an infinite ICC quotient.
*Proof*. Let $FC(G) := \{ g \in G : [G : Z\_G(g)] < \infty \}$ denote the FC-centre of $G$ containing those elements whose centralizer is of finite index, or equivalently whose conjugacy class is finite. For $g, h \in FC(G)$ we have that $Z\_G(gh) \geq Z\_G(g) \cap Z\_G(h)$ and is also finite index, so $FC(G)$ is a subgroup of $G$ and in fact a characteristic one. There are two possibilities:
* If $FC(G)$ is infinite, then being an infinite FC-group (i.e. all conjugacy classes are finite) it contains an infinite abelian subgroup $\langle g\_0, g\_1, \dots \rangle$ constructed inductively: at each step, if the subgroup $\langle g\_0, g\_1, \dots, g\_i \rangle$ is finite, then its centralizer is a finite intersection of finite-index subgroups hence finite index itself and thus infinite so we can find some $g\_{i+1}$ commuting with and not contained in $\langle g\_0, g\_1, \dots, g\_i \rangle$.
* If $FC(G)$ is finite, then $G / FC(G)$ is infinite and ICC since the conjugacy class of any non-trivial $g FC(G) \in G / FC(G)$ will contain the infinitely many images of the conjugacy class of $g$ in $G$.
$\square$
So it suffices to prove the existence of such $A\_0, \dots, A\_{m-1}$ with the additional assumption that $G$ is abelian or ICC (since after finding such sets in a quotient of $G$, any choice of lifts to $G$ will still have the desired property). We now show that we can choose inductively $g\_0, g\_1, g\_2, \dots$ to put into the sets $A\_0, \dots, A\_{m-1}$ (say, cycling through the $A\_i$) so that at each step the desired uniqueness holds.
Start by putting $g\_0 = \dots = g\_{m-1} = 1$ into $A\_0, \dots, A\_{m-1}$. Suppose we are choosing $g\_n$ to add to $A\_i$. There are two cases to consider for distinct words $a\_0 a\_1 \dots a\_{m-1}$ and $b\_0 b\_1 \dots b\_{m-1}$ that possibly give equal group elements that involve $g\_n$ and otherwise only $g\_j$ with $j < n$: (1) $a\_i = g\_n$ and $b\_i \neq g\_n$, and (2) $a\_i = b\_i = g\_n$.
(1) simply rules out finitely many possibilities for $g\_n$.
(2) is equivalent to the equation $g\_n^{-1} (u^{-1} u') g\_n = v (v')^{-1}$ that we get by rewriting $u g\_n v = u' g\_n v'$ for $u = a\_0 \dots a\_{i-1}, v = a\_{i+1} \dots a\_{m-1}, u' = b\_0 \dots b\_{i-1}, v' = b\_{i+1} \dots b\_{m-1}$. Note that $u^{-1} u' \neq v (v')^{-1}$ as otherwise picking $a\_i = b\_i = 1$ would then give distinct words giving equal group elements, contrary to our inductive hypothesis. If $G$ is abelian then there is no solution to this conjugacy equation so we have no additional constraints on $g\_n$. If $G$ is ICC then the set of solutions is either empty (if $u^{-1} u'$ and $v (v')^{-1}$ are not conjugate) or is a coset of $Z\_G(u^{-1} u')$. In the latter case this is a coset of an infinite-index subgroup, by ICC, because $u^{-1} u' \neq v (v')^{-1}$ so their being conjugate rules out $u^{-1} u' = 1$.
Both (1) and (2) only rule out finitely many cosets of infinite-index subgroups (in (1) we have cosets of the trivial subgroup!), so since we cannot write an infinite group as a finite union of such cosets by a result of B.H. Neumann (see [MO question](https://mathoverflow.net/q/17396)), there is some choice of $g\_n$ for which neither (1) nor (2) can give a failure of the desired uniqueness, which completes the proof.
| 4 | https://mathoverflow.net/users/24447 | 425984 | 172,922 |
https://mathoverflow.net/questions/425352 | 1 | When I say that an $r$-sided simple (*i.e.*, not self-intersecting) polygon is inscribed into an $n$-sided regular polygon, I mean that every vertex of the simple $r$-gon is also a vertex of the regular $n$-gon. Let $M(r,n)$ stand for the number of different simple $r$-gons that can be inscribed into a regular $n$-gon. *What is the formula for $M(r,n)$?* I assume that, if $P$ and $Q$ are two $r$-sided simple polygons, and $Q$ can be obtained from $P$ by means of a rotation, then $P=Q$. (Without this assumption, $M(r,n)$ would obviously be equal to ${n \choose r}$.)
I solved some small cases with *Maple*, obtaining results like $M(4,6)=3$, $M(5,7)=3$ and $M(7,12)=66$. These results suggest that, if $r$ and $n$ are relatively prime (their gcd is $1$), then $M(r,n)=\frac{1}{n}{n \choose r}$. Is this hypothesis true? And what is the formula for $M(r,n)$ in the case $gcd(r,n)>1$? For $r=4$ and $n=6$, the formula $M(r,n)=\frac{1}{n}{n \choose r}$ gives the result $\frac{1}{6}\cdot 15=\frac{5}{2}$, which is of course wrong. The result $M(4,6)=3$ can be seen even without Maple. Suppose that, during a counterclockwise tour, the vertices of the regular hexagon appear in the order $A$, $B$, $C$, $D$, $E$, $F$. Then the three inscribable quadrilaterals are $ADEF$, $ACEF$ and $ACDF$.
| https://mathoverflow.net/users/481690 | Rotational invariance assumed, what is the number of $r$-sided simple polygons that can be inscribed into an $n$-sided regular polygon? | $$M(r, n) = \frac{1}{n} \sum\_{d \,\mid\, \gcd(r,n)} \varphi(d) \binom{n/d}{r/d}$$
See [OEIS A047996](http://oeis.org/A047996) for references.
| 2 | https://mathoverflow.net/users/46140 | 425988 | 172,924 |
https://mathoverflow.net/questions/425981 | 5 | In the description of the integral Adams spectral sequence, representations of the following quiver (with relations) arise naturally:
* We have two objects $A, B$,
* we have two arrows $\pi: A \rightarrow B$ and $\delta: B \rightarrow A$ and
* we have a single relation $\delta \circ \pi = 0$.
Does this quiver arise naturally in other contexts? What is known about its representation theory?
| https://mathoverflow.net/users/16981 | Quiver with two objects and two arrows composing to zero | This is a Nakayama algebra with Kupisch series [2,3] and the representation theory of Nakayama algebras (a Nakayama algebra is a quiver algebra with admissible relations whose quiver is a linear oriented line or a cycle) is completely understood in nearly all details. (see for example [Farnsteiner - Nakayama algebras: Kupisch series and Morita type](https://www.math.uni-bielefeld.de/%7Esek/select/Nakayama-alg1.pdf)).
It has 5 indecomposable modules and global dimension 2 and appears very often as a block of algebras in geometric representation theory for example for blocks of category $\mathcal{O}$ and it is also the simplest non-trivial representaiton-finite block of a Schur algebra. See also section 6.1 of [Chan and Marczinzik - On representation-finite gendo-symmetric biserial algebras](https://arxiv.org/abs/1607.05965) for this algebra and its "bigger brothers" and where they appear.
| 7 | https://mathoverflow.net/users/61949 | 425989 | 172,925 |
https://mathoverflow.net/questions/425987 | 2 | Consider the second-order ordinary differential system
$$
A y'' + i B y' - C y = 0,
$$
where $A$ and $C$ are real-valued $d\times d$ SPD matrices satisfying
$$
I \le A,C \le 2I,
$$
and $B$ is real-valued and symmetric, having operator 2-norm less than 2 (and $i = \sqrt{-1}$). I would like to show that
1. There are $2d$ independent solution of the form $y(t) = y\_0 e^{\lambda t}$ and
2. There are exactly $d$ eigenvalues $\lambda$ having positive real part.
| https://mathoverflow.net/users/98733 | A question about solutions to ordinary differential systems | EDIT 1: I think one needs to assume that $\|B\|<2$, since otherwise as @Iosif Pinelis example in the comments shows, the eigenvalues may be purely imaginary.
EDIT 2: I changed the notation so my answer become clearer. In the original answer, I had the following fact in mind: if $P(\xi)$ is the principle symbol of a linear $2$-differential operator satisfying $P(\xi)\geq c|\xi|^{2}$, then the eigenvalues of the $2$-order linear ODE $P(\xi+i\frac{d}{dt}\nu)y(t)=0$ must have the desired properties in the question. I basically reconstructed such a symbol from the given ODE, but then I reckoned this will not be needed in the actual answer.
---
Under the assumption that $\|B\|<2$, let us see that all eigenvalues have either positive or negative real part. Consider the matrix-valued polynomial of degree 2 in $\lambda$, $P:\mathbb{R}\rightarrow\mathrm{End}{\mathbb(\mathbb{R}^{d})}$, given by
$$P(\lambda)=\lambda^{2}A-\lambda B+C$$
Note how for every $\lambda\in\mathbb{R}$, by the assumptions in the question we find
$$c(1+\lambda^{2})I\leq P(\lambda)$$
For some constant $c>0$. In particular, $P(\lambda)$ is nonsingular for all $\lambda\in\mathbb{R}$.
By formally allowing $\lambda$ to be complex, we note how the eigenvalue problem for the ODE in the question becomes for $y(t)=y\_{0}e^{\lambda t}$,
$$P(i\lambda)y\_{0}=0$$
For $\lambda=i\eta$ with $\eta\in\mathbb{R}$, this reads,
$$P(-\eta)y\_{0}=0$$
which implies $y\_{0}=0$. Therefore, the eigenvalues of the ODE can't be purely imaginary, and must contain a real part.
This implies that the fundamental solutions exponentially decreases when either $t\rightarrow\infty$ or $t\rightarrow-\infty$. Let $M^{+}$ denote the space of $t\rightarrow\infty$ exponentially decreasing solutions and by $M^{-}$ the space of $t\rightarrow-\infty$ exponentially decreasing solutions. Let $I:M^{\pm}\rightarrow \mathbb{R}^{d}$ be the map $Iy=y(0)$. This map is clearly linear. We prove that this map is injective, hence $\operatorname{dim}{M^{\pm}}\leq d$. Since the space of solutions is $2d$ dimensional, this implies that $\operatorname{dim}{M^{\pm}}=d$, which is the required result. The proof also shows that all fundamental solutions must be of the form $y(t)=y\_{0}e^{\lambda t}$, because if $y(t)=y\_{0}t^{k}e^{\lambda t}$ for some $k\ne 0$, then $y(0)=0$, hence $y=0$ identically.
Without loss of generality, let $y\in M^{+}$ with $y(0)=0$. Denote by $\hat{y}(\eta)$ the fourier transform of $y$, albeit extneded to satisfy $y(t)=0$ for $t\leq 0$. Since $P(\lambda)$ is positive definite for all $\lambda\in\mathbb{R}$, we may write: $P(\eta)=M(\eta)M(\eta)$, where $M(\eta)$ is a matrix-valued polynomial of degree at most 1 in $\eta$.
By formally replacing $\eta$ with an ordinary derivative $i\frac{d}{dt}$,
the ODE can formally be written as,
$$P(i\frac{d}{dt})y(t)=0$$
Since $y(0)=0$ and the solution is rapidly decreasing, we can integrate by parts the following equation to obtain,
$$0=(P(i\frac{d}{dt})y,y)\_{L^{2}(\mathbb{R^+})}=(M(i\frac{d}{dt})y,M(i\frac{d}{dt})y)\_{L^{2}(\mathbb{R^+})}$$
By the Plancheral theorem, this quantity becomes by applying the Fourier transform on each flank,
$$\int\_{-\infty}^{\infty}(P(\eta)\hat{y}(\eta),\hat{y}(\eta))\_{\mathbb{R}^{d}}d\eta=0$$.
And since $P(\eta)\geq c(1+\eta^{2})I$ for all $\eta\in\mathbb{R}$, this yields $\hat{y}(\eta)=0$, which is to say $y=0$. This proves that $I$ was injective, and the solution space decomposes as described.
| 2 | https://mathoverflow.net/users/144247 | 426010 | 172,931 |
https://mathoverflow.net/questions/426003 | 1 | Let $f(x,y)=\frac{10xy-(x+y)+1}{8xy-2(x+y)+5}$ and $g(x,y)=\frac{1}{4}\left[1+\frac{1}{3}(4x-1)(4y-1)\right]$. I want to prove that for any $0.5\le a\le b\le 1$ and $0.7\le c\le d\le 1$, it holds that $g(f(a,b),f(c,d))>f(g(a,c),g(b,d))$. I have checked the result by programming a search algorithm that this seems to be true, but I am looking for a way to formally prove it.
| https://mathoverflow.net/users/52871 | An inequality in four variables | $\newcommand{\dif}{\text{dif}}\newcommand{\num}{\text{num}}\newcommand{\den}{\text{den}}\newcommand{\nnt}{\text{nn2}}$We want to show that
\begin{equation}
\dif:=g(f(a,b),f(c,d))-f(g(a,c),g(b,d))>0 \tag{?}\label{?}
\end{equation}
given conditions
\begin{equation}
\tfrac12\le a\le b\le 1,\quad \tfrac7{10}\le c\le d\le 1. \tag{C}\label{C}
\end{equation}
We have
\begin{equation}
\dif=\frac23\,\frac\num\den,
\end{equation}
where $\num$ is a certain polynomial of degree $\le2$ in $a$, in $b$, in $c$, and in $d$, whereas
\begin{equation}
\begin{aligned}
\den&:=a (8 b-2)-2 b+5) (c (8 d-2)-2 d+5) \\
&\times (2 a (4 b-1) (4 c-1) (4 d-1)-2 b (4 c-1) (4 d-1)+8 c d-2 c-2 d+41).
\end{aligned}
\end{equation}
Since each of factors in the latter expression is affine in $a$, in $b$, in $c$, and in $d$, it is straightforward to check that $\den>0$ given \eqref{C}.
So, it enough to show that
\begin{equation}
\num>0 \tag{??}\label{??}
\end{equation}
given \eqref{C}.
Recall that $\num$ is a polynomial of degree $\le2$ (in $a$, in $b$, in $c$, and) in $d$. The coefficient of $d^2$ in $\num$ is
\begin{equation}
2 (4 b-1) (4 c-1)\,\nnt,
\end{equation}
where $\nnt$ is a polynomial in $a,b,c$ of degree $\le1$ in $c$, and the coefficient of $c$ in $\nnt$ is $16 (4 a-1) (-4 + a + b + 2 a b)\le0$ (given \eqref{C}). So,
\begin{equation}
\nnt\le\nnt|\_{c=7/10}.
\end{equation}
Next, $\nnt|\_{c=7/10}$ is a polynomial in $a,b$ of degree $\le1$ in $b$, and the coefficient of $b$ in $\nnt|\_{c=7/10}$ is $\frac35\, ( 4 a-1) (64 a-13)>0$ (given \eqref{C}). So,
\begin{equation}
\nnt|\_{c=7/10}\le\nnt|\_{c=7/10,\,b=a}=-\frac35\,(1-a) (7 + 178 a + 256 a^2)\le0.
\end{equation}
So, $\nnt\le0$ and hence the coefficient of $d^2$ in $\num$ is $\le0$. So, $\num$ is concave in $d$. So, it is enough to prove \eqref{??} for $d\in\{1,c\}$.
This is done using similar concavity-cum-monotonicity reasoning, along with factoring polynomials. Here it will help to note that $\dfrac{\num|\_{d=1}}{9(1-c)}$ is a polynomial of degree $\le1$ in $c$, whereas $\dfrac{\num|\_{d=c}}{9(1-c)(4c-1)}$ is (a polynomial of degree $\le2$ in $b$ that is) concave in $b$.
| 2 | https://mathoverflow.net/users/36721 | 426016 | 172,932 |
https://mathoverflow.net/questions/426019 | 2 | Let $A$ be a non-unital $C^\*$-algebra. Let $S\subseteq A^\*$ be a set of continuous functionals that separates the points of $A$. Every element $\omega \in A^\*$ extends uniquely to a strictly continuous functional $\omega \in M(A)^\*$, so we can ask: does $S$ also separate the points of $M(A)$? Concretely, if $m \in M(A)$ and $\omega(m)=0$ for all $\omega \in S$, do we have $m=0$?
In some cases, this is automatically true. For instance, if $AS \subseteq S$ or $SA\subseteq S$ this is easily seen to be true. However, it is not clear to me if this is true in general. My intuition tells me that the answer is "no" but I have trouble finding a concrete counterexample. It's not even clear to me what happens if $A= C\_0(X)$ or $A=B\_0(H)$.
| https://mathoverflow.net/users/216007 | If $S\subseteq A^*$ is separating, does $S$ also separate $M(A)$? | You're right, it's not true in general. If $A = c\_0$ then the multiplier algebra is the same as the double dual, $l^\infty$. Then we want a linear subspace of $l^1$ which is weak\* dense (so it separates $c\_0$) but not norm dense (so it doesn't separate $l^\infty$). This can be achieved by taking any $\vec{a} \in l^\infty \setminus c\_0$ and letting $S$ be its kernel in $l^1$. For instance, taking $\vec{a} = 1$, we get $S = \{\vec{b} \in l^1: \sum b\_n = 0\}$. This separates $c\_0$ but not $l^\infty$.
| 5 | https://mathoverflow.net/users/23141 | 426025 | 172,935 |
https://mathoverflow.net/questions/426038 | 14 | For nice topological spaces (say Haudorff spaces) $X$ and $Y$, there is a bijection between continuous maps $X\to Y$ and isomorphism classes of geometric morphisms $\mathrm{Sh}(X)\to \mathrm{Sh}(Y)$.
Question: Is there a similar statement for "nice" schemes, i.e., that morphisms of schemes $X\to Y$ correspond bijectively to isomorphism classes of geometric morphisms from the étale topos of $X$ to the étale topos of $Y$?
| https://mathoverflow.net/users/485276 | Making the étale topos construction a fully faithful 2-functor from schemes to Grothendieck topoi | Below is a proof that the (pseudo)functor that sends a scheme to its petit étale topos is not fully faithful, for any category of schemes over an algebraically closed base field $k$, assuming that this category contains both $\mathrm{Spec}(k)$ and $\mathrm{Spec}(k[t])$.
*Proof:* The morphisms of schemes
$\mathrm{Spec}(k) \to \mathrm{Spec}(k[t])$ over $k$ correspond to maximal ideals of $k[t]$. At the level of toposes, these give geometric morphisms $\mathbf{Sets} \to \mathcal{E}$, with $\mathcal{E}$ the étale topos of $\mathrm{Spec}(k[t])$. However, not every geometric morphism $\mathbf{Sets} \to \mathcal{E}$ is of this form. There is another one induced by the morphism of schemes $\mathbf{Spec}(k(t)^\mathrm{sep}) \to \mathrm{Spec}(k[t])$, with $k(t)^\mathrm{sep}$ the separable closure of $k(t)$. $\square$
For other reasonable categories of schemes, I expect that the answer to your question will still be "no". However, in practice it might be enough that the functor is "almost fully faithful", see the positive results described in the [answer](https://mathoverflow.net/a/426051/) by D.-C. Cisinki.
| 12 | https://mathoverflow.net/users/37368 | 426046 | 172,938 |
https://mathoverflow.net/questions/414390 | 3 | For a knot $K$, let $J\_N(K)$ denote the $N$th Kashaev invariant of $K$. This [is the same as](https://arxiv.org/abs/math/9905075) the $N$th colored Jones polynomial evaluated at an $N$th root of unity (or $2N$th depending on your conventions for what the variable is).
In the literature on the volume conjecture it is common to use closed-form expressions for $J\_N(K)$. For example, [Kashev gives](https://arxiv.org/abs/q-alg/9601025) the formulas
$$
J\_N(4\_1) = \sum\_{k = 0}^{N-1} |(\omega)\_k|^2
$$
and
$$
J\_N(5\_2) = \sum\_{0 \le k \le l \le N-1} \frac{(\omega)\_k^2}{(\omega)\_l^\*} \omega^{-k(l+1)}
$$
where $\omega = e^{2\pi i/N}$ and $(q)\_k$ is the Pochammer symbol.
How can you compute these formulas? A number of authors mention that they follow from presenting the $4\_1$ and $5\_2$ knots in terms of surgery on the Borromean rings, but I can't find a precise derivation. More accurately, I once found a paper on the arXiv by Habiro and Le (?) that I think gives the computation, but I can't find it again. (I just checked again and their only joint paper is not about this, so I must be mistaken.)
Another possibility is to use the definition of the Kashaev invariant in terms of triangulations, not braid group representations. There are well-known ideal triangulations of the $4\_1$ complement with 2 tetrahedra and of the $5\_2$ complement with 3, so this seems like it might give the above expressions. However, I've never seen a detailed derivation of this approach either.
| https://mathoverflow.net/users/113402 | Closed-form expressions for the Kashaev invariant via surgery | I think the reference I was looking for was [H], which indeed uses presentations of various knots as surgery on the Borromean rings to derive closed forms for their Kashaev invariants.
Skein modules also show up, which I think is where the surgery formula comes from.
In [H] the author cites another paper "to appear" expanding on it, so now I'm curious if that paper was ever published.
[H] *Habiro, Kazuo*, On the colored Jones polynomials of some simple links, RIMS Kokyuroku 1172, 34-43 (2000). [ZBL0969.57503](https://zbmath.org/?q=an:0969.57503) [direct link](https://www.kurims.kyoto-u.ac.jp/%7Ekyodo/kokyuroku/contents/pdf/1172-3.pdf).
| 0 | https://mathoverflow.net/users/113402 | 426055 | 172,941 |
https://mathoverflow.net/questions/425849 | 7 | So this is probably blasphemist to ask and I've resisted asking this for a while. Essentially my question is why are locally symmetric spaces/Shimura varieties the "right" object to study from a number theoretic perspective?
The merit of them, say from a Langlandian perspective, is for sure beyond debate, and certainly not my question. Specifically they are spaces with a Hecke-action (and even a Galois one in the Shimura case), such that we can assign to Hecke-eigenspaces of its cohomology a Galois representation (or at least conjecturally).
However, it could be, at least naively, that there exist other spaces with similiar number-theoretic benefits.
| https://mathoverflow.net/users/152554 | Why are Shimura varieties the "right" objects? | This is a long comment. Number theorists have long been interested in modular functions and modular forms, which are functions on the complex upper half plane. Elliptic modular curves are the spaces on which these functions live. An important property of modular curves, which helps explain why they are of interest to number theorists, is that they have canonical models over number fields.
A modular curve is the quotient of a one-dimensional bounded symmetric domain (= hermitian symmetric domain) by a congruence subgroup. Remove the "one-dimensional", and you get a Shimura variety. Shimura varieties are known to have canonical models over number fields, and so have a reasonable claim to being the "right" generalization of modular curves.
From another perspective, a modular curve is a moduli variety for one-dimensional abelian varieties with additional structure. Remove the "one-dimensional" and you get a PEL Shimura variety. I suspect algebraic geometers would have considered these to be the "right" generalization if Shimura hadn't proved that Shimura curves (not of PEL-type) also have canonical models. Shimura varieties with integral weight are moduli variety for motives with additional structure, at least conjecturally.
Shimura varieties play two roles in the Langlands program, first as sources for Galois representations and second as a test of Langlands's idea that all L-functions arising from algebraic geometry are automorphic. For the first, PEL Shimura varieties suffice. Much is known about the zeta functions of Shimura varieties and almost nothing about the zeta functions of other varieties (except for elliptic curves over $\mathbb{Q}$ and some modest generalizations).
Thus, Shimura varieties are certainly good objects for number theorists to study (but not the only good objects).
| 3 | https://mathoverflow.net/users/nan | 426058 | 172,943 |
https://mathoverflow.net/questions/426052 | 2 | How to prove the following inequality $$\forall t>0,\quad\int^\infty\_0 \sin(rt)\frac{r^3}{\sinh^2(r)} dr\leq c \big(te^{-At}\big)$$
for some constants $A>0,c>0$
| https://mathoverflow.net/users/172078 | The inequality $\int^\infty_0 (\sin(rt)r^3/\sinh^2(r)) dr\leq cte^{-At}$ | This is to complement the answer by Carlo Beenakker by showing that
$$I(t)\le\pi^4 te^{-\pi t}\tag{1}\label{1}$$
for real $t\ge0$, where $I(t)$ is the integral in question.
Indeed, according to Carlo Beenakker,
$$I(t)=\frac{\pi ^3}{8} \frac{\pi t \cosh \pi t+2\pi t-3 \sinh \pi t}{\sinh^4(\pi t/2)}.$$
This expression for $I(t)$ can be obtained from [formula 3.986.4 of Gradshteyn and Ryzhik](http://fisica.ciens.ucv.ve/%7Esvincenz/TISPISGIMR.pdf) by using substitutions $\pi x=r$ and $\beta=\pi t/2$, to get
$$J(t):=\int\_0^\infty\frac{1-\cos rt}{\sinh^2 r}\,dr
=\frac{\pi t}2\,\coth\frac{\pi t}2-1,$$
and then noting that $I(t)=-J'''(t)$.
So, using the substitution $t=\dfrac{\ln(1+x)}\pi$, one rewrites
inequality \eqref{1} as
$$d(x):=\left(8 x^3+19 x^2+18 x+6\right) \ln(x+1)-3 x (x+1)^2 (x+2)\le0\tag{2}\label{2}$$
for real $x\ge0$.
In turn, inequality \eqref{2} follows immediately because $d(0)=d'(0)=d''(0)=d'''(0)=0$ and
$$d''''(x)=-\frac{2 x \left(36 x^3+120 x^2+139 x+58\right)}{(x+1)^4}\le0$$
for real $x\ge0$. $\quad\Box$
One may note that $I(t)\sim\pi^4 te^{-\pi t}$ as $t\to\infty$, so that the upper bound $\pi^4 te^{-\pi t}$ on $I(t)$ in \eqref{1} is exact.
| 5 | https://mathoverflow.net/users/36721 | 426061 | 172,945 |
https://mathoverflow.net/questions/426043 | 3 | Let $E$ be a nowhere dense subset of $\mathbb{R}\times \mathbb{R}$.
For $x\in \mathbb{R}$, define
$$E\_x=\{ y\in\mathbb{R}\mid (x,y)\in E\}.$$
Let $D$ denote the set of $x$ for which $E\_x$ is NOT nowhere dense in $\mathbb{R}$.
By the Kuratowski-Ulam Theorem, we know that $D$ is of first
cateogory in $\mathbb{R}$. My question is: can we strengthen
this conclusion by saying that $D$ is nowhere dense in $\mathbb{R}$? If not, is there a counterexample?
Thanks in advance!
| https://mathoverflow.net/users/122378 | Kuratowski-Ulam Theorem, nowhere dense set in product space | List the rational numbers as $\{q\_n : n = 1, 2, \cdots \}$ and for each $n$, let $X\_n = \{q\_n\} \times [n, n+1]$. Finally, let $X = \cup\_{n = 1}^\infty X\_n$. Then $X$ is nowhere dense in $\mathbb R\times\mathbb R$ and for $x$ a rational number, $E\_x$ is a non-degenerate closed interval, so the set $D$ is the set of rational numbers which, of course, fails to be nowhere dense in $\mathbb R$.
| 6 | https://mathoverflow.net/users/89233 | 426068 | 172,948 |
https://mathoverflow.net/questions/426049 | 1 | I would like to give the following object, $M=S^5 \setminus \sqcup\_{2 \text{ copies}} \text{int}(S^1\times D^4)$, a handle decomposition. It is then to be attached to another manifold. along the two copies of $S^1\times S^3$, so there is no need to start with a $0$-handle. My attempt, below, seems quite complicated (then I would like to generalize the construction to $S^5$ with more copies of $S^1\times D^4$ removed, under certain combinatorial conditions).
Is the following attempt a valid handle decomposition of $M$ ?
**Attempt to construct M:**
Start with two free floating $1$-handles $(D^1\times D^4, S^0\times D^4)$. They have co-attaching region $\text{co-att}=D^1\times S^3$. I believe that I am not allowed to decompose $D^4=D^1\times D^3$ and connect two $D^4$ of each handle in order to create a $S^1\times S^3$. In that case, take one $D^4=D^1\times D^3$ of each handle and connect the $D\_1$ of each two handles on one side to create a $S^1\times D^3$. Do the same on the other ends of the two 1-handles as well. The $D^1$'s of the co-attaching regions get connected as well to give a $S^1$. The result has boundary $(\sqcup\_{2 copies} S^1\times D^3)\cup S^1 \times S^3$. Take the result and an another copy of itself, and connect the four disjoint $S^1\times D^3$'s two by two, giving an object with two disjoint $S^1\times S^3$ as boundaries; these are the boundaries of the target manifold. Note that at this point we did not do any gluings of handles but only combination of disks along their boundaries. There are also two unwanted boundaries from the co- attaching regions $\text{co-att}$, which are $\#\_{2 copies} S^1\times S^3$. Fill part of this two spaces by attaching two $2$-handles $(D^2\times D^3,S^1\times D^3)$. The result are two copies of $D^4$, one for each "holes". Now the fundamental group is trivial, and we can close the remaining unwanted boundary with a 5-handle $(D^5\times D^0,S^4\times D^0)$.
| https://mathoverflow.net/users/477214 | An example of handle decomposition on modified $S^5$ | I didn't really follow the details of your description, but here is a handle decomposition along the lines you suggest. Start with $((S^1 \times S^3)\_a \coprod (S^1 \times S^3)\_b) \times I$ where $I= [0,1]$. Add a 1-handle along $((S^1 \times S^3)\_a \coprod (S^1 \times S^3)\_b) \times \{1\}$. Note that by choosing the attaching region of the $1$-handle, you can make it miss a copy of $S^1 \times B^3$ in each of $(S^1 \times S^3)\_a \times \{1\}$ and $(S^1 \times S^3)\_b\times \{1\}$
Now add two 2-handles, one along each of the aforementioned copies of $S^1 \times B^3$. If you glue copies of $S^1 \times B^4$ along $(S^1 \times S^3)\_a \times \{0\}$ and $(S^1 \times S^3)\_b\times \{0\}$ then all of the handles cancel and you've got $B^5$ with boundary $S^4$.
It follows that if you add a $5$-handle, you get $S^5$ minus a particular pair of copies of $S^1 \times B^4$. But since any pair of circles with a given in $S^5$ is isotopic to any other pair, this is a description of any pair. The same description would work for any number of circles.
| 2 | https://mathoverflow.net/users/3460 | 426086 | 172,951 |
https://mathoverflow.net/questions/426084 | 3 | Let $K$ be a cubic field and let $\mathcal{O}\_K$ be its ring of integers. Does there always exist elements $\alpha, \beta \in \mathcal{O}\_K$ with $\text{Tr}(\alpha) = \text{Tr}(\beta) = 0$ such that $\{1, \alpha, \beta\}$ forms a $\mathbb{Z}$-basis for $\mathcal{O}\_K$? If the answer is generally no, then what about the case when $K$ is a cyclic cubic field?
| https://mathoverflow.net/users/10898 | Trace-free basis for $\mathcal{O}_K$, $K$ a cubic field | No. There do not always exist such $\alpha$ and $\beta$. If $K$ is a cubic field and such $\alpha$ and $\beta$ exist, then for all $x \in \mathcal{O}\_{K}$, ${\rm Tr}\left(\frac{1}{3} \cdot x\right) \in \mathbb{Z}$ and this implies that $1/3$ is in the inverse different of $\mathcal{O}\_{K}$, and hence that the different of $\mathcal{O}\_{K}$ is contained in $(3)$. Since the norm of the different is the discriminant, this implies that the discriminant of $\mathcal{O}\_{K}$ is a multiple of $27$. This need not be true, even for cyclic cubic fields.
| 8 | https://mathoverflow.net/users/48142 | 426088 | 172,952 |
https://mathoverflow.net/questions/426075 | 5 | Let $\pi: M\to B$ be a fiber bundle of smooth manifolds with $B$ connected and each fiber of $\pi$ is a compact manifold. Let $G$ be a compact Lie group acting smoothly on $M$ such that
$\pi(g\cdot m)=\pi(m)$. It is clear that $G$ acts smoothly on each fiber $M\_b$ for $b\in B$.
Noe fix a $g\in G$. For each $b\in B$ we consider the fixed point submanifold $M\_b^g\subset M\_b$.
My question is: when $b$ varies, does the diffeomorphic type of $M\_b^g$ unchanged?
| https://mathoverflow.net/users/24965 | Does a compact Lie group action on a family of compact manifolds have diffeomorphic fixed point submanifolds? | I think the answer is yes.
Since $G$ is compact, there is a $G$-invariant Riemannian metric on $M$ (by averaging any metric). The orthogonal distribution to the fiber for this metric is a $G$-invariant Ehresmann connection, and the parallel transport for this connection thus commutes with the $G$-action.
This shows that the action of $G$ on any two fibers are conjugate. In particular, the fixed loci are diffeomorphic.
| 9 | https://mathoverflow.net/users/173096 | 426089 | 172,953 |
https://mathoverflow.net/questions/426096 | 1 | Can two increasing and differentiable convex functions agree exactly on a countable set of cardinality greater than two?
| https://mathoverflow.net/users/473751 | Intersecting points of increasing convex functions | Let
$$f\_\pm(x):=e^x(1\pm c\sin x)$$
for some $c\in(0,1/2]$ and all real $x$.
Then $f'\_\pm(x)=e^x(1\pm c(\cos x+\sin x))>0$ and $f''\_\pm(x)=e^x(1\pm 2c\cos x)\ge0$ for all real $x$.
So, $f\_+$ and $f\_-$ are increasing differentiable convex functions that agree exactly on the countable set $\pi\mathbb Z$.
---
Moreover, one can modify this example to make the derivatives of $f\_+$ and $f\_-$ agree on the set $\pi\mathbb Z$ as well. This can be done by the formula
$$f\_\pm(x):=e^x(1+c\sin x+\tfrac12\,(c\pm c)\sin^2x )$$
for some $c\in(0,1/4]$ and all real $x$.
| 1 | https://mathoverflow.net/users/36721 | 426098 | 172,955 |
https://mathoverflow.net/questions/426099 | 5 | The minimal model program attempts to classify algebraic varieties up to birational equivalence. For compact Riemann surfaces, Riemann's uniformization theorem tells us that the geometry of the curve is determined primarily by its genus. In particular, if $M$ is a compact Riemann surface of genus $g$, then $g=0 \implies M \cong \mathbb{P}^1$, $g =1 \implies M \cong \mathbb{T}^2$, and $g \geq 2 \implies M \cong \mathbb{H}/\Gamma$. Here, $\cong$ denotes biholomorphism (or isomorphism).
In higher dimensions, we forget the biholomorphic equivalence classes, even for smooth compact complex surfaces. I am convinced of the utility of this and understand that it can be very hard to identify algebraic varieties or complex manifolds up to biholomorphism. My question is:
>
> Is there a nice very illustrative example that shows that for complex surfaces, you are essentially forced to proceed with birational equivalence and not biholomorphic equivalence?
>
>
>
Note: This is an expository question, more than a mathematical question.
| https://mathoverflow.net/users/471309 | Example illustrating necessity of considering birational equivalence and not biholomorphic equivalence in MMP | This is a very long comment.
First I should point out that by GAGA, biholomorphic = isomorphic for smooth projective varieties so in this case, classification up to biholomorphism is the same as classification up to isomorphism. Note the non-compact case can be quite wild in higher dimensions (see [this example](https://mathoverflow.net/a/657/12402) or the fact that the unit ball and the unit polydisc in $\mathbb{C}^2$ are not biholomirphic) but I will only talk about the projective case here.
From the point of view of the minimal model program, the higher dimensional version of the classification of Riemann surfaces via uniformization is the classification of algebraic varieties into general type ($K\_X$ is big), Calabi-Yau ($K\_X \equiv 0$), and Fano ($-K\_X$ is ample). These correspond to negative, zero, and positive Ricci curvature in the same way that for Riemann surfaces, $g = 0,1$ and $\geq 2$ correspond to positively curved, flat, and hyperbolic respectively. General type varieties moreover have a unique birational representative, the *canonical model*.
One complication in higher dimensions is that not every variety fits into one of these categories. For example a variety can be positively curved in one direction and negatively in another, e.g. $\mathbb{P}^1 \times C$ where $g(C) \geq 2$ or $\mathrm{Bl}\_p(C \times D)$ where $g(C), g(D) \geq 2$.
The main goal of the minimal model program is to show that every algebraic variety can be decomposed (in an algorithmic way and involving well understood birational transformations) into building blocks which are of general type, Calabi-Yau, or Fano. This (conjecturally) reduces the classification of varieties to two parts: 1) understand the steps in the MMP which build up every variety from the basic building blocks, and 2) classify the basic building blocks.
The classification of the building blocks often involves producing moduli spaces (e.g. moduli spaces of canonical models which generalize $\mathcal{M}\_g$ to higher dimensions, and the $K$-moduli in the Fano case).
Finally there is the basic fact that in higher dimensions, there exist infinitely many non-isomorphic (and thus non-biholomorphic) but birational smooth projective varieties, e.g. a smooth projective surface $X$ can be blown up any number of times.
The MMP for surfaces then says that every surface is obtained as an iterated blowup (which has Fano fibers) of one of the following: 1) a minimal surface of general type, 2) a fibration by genus 1 curves (a Calabi-Yau fibration), 3) a Calabi-Yau surface, 5) a ruled surface (Fano fibration). Thus to classify smooth projective surfaces up to isomorphism = biholomorphism, we need to understand blowups, classify minimal surfaces of general type, and classify Fano and Calabi-Yau fibrations over a lower dimensional variety.
TLDR biholomorphism is the same as isomorphism for projective varieties by GAGA. The MMP posits that every smooth projective variety conjecturally can be decomposed, after a sequence of specific birational transformations, into the basic building blocks of general type, Calabi-Yau and Fano varieties. This is the higher dimensional generalization of the trichotomy for compact Riemann surfaces given by uniformization.
| 8 | https://mathoverflow.net/users/12402 | 426109 | 172,956 |
https://mathoverflow.net/questions/426103 | 9 | In classical mathematics, there exists only one Cauchy complete Archimedean ordered field, the Dedekind complete Archimedean ordered field. However, in constructive mathematics, there are multiple Cauchy complete Archimedean ordered fields, which are not provable to be equivalent to each other: one cannot in general prove that the Dedekind real numbers embeds into the initial Cauchy complete Archimedean ordered field. (By Cauchy complete we mean complete by Cauchy sequences, not Cauchy nets.)
The principal square root function is a function defined on the non-negative elements $[0, \infty)$ of a Cauchy complete Archimedean ordered field $\mathbb{R}$ such that it is a two-sided inverse of the square function $x^2$ when $x^2$ is restricted on the domain and codomain to $[0, \infty)$.
In classical mathematics, one could prove that the principal square root function exists by proving the fundamental theorem of algebra for any Cauchy complete Archimedean ordered field. However, in constructive mathematics, the fundamental theorem of algebra cannot in general be proven for Cauchy complete Archimedean ordered fields.
In classical mathematics, there is an alternative to proving that the principal square root function exists: by first proving that zero has a square root, and that there exists a two-sided inverse function of $x^2$ on the positive elements $(0, \infty)$ of $\mathbb{R}$. The square root of zero is zero in any integral domain, and because in any Cauchy complete Archimedean ordered field the square function $x^2$ is continuously differentiable on the open interval $(0, \infty)$ and its derivative $2x$ is always positive on the open interval $(0, \infty)$, by the inverse function theorem, there exists a two-sided inverse defined on $(0, \infty)$. (In constructive mathematics, the inverse function theorem still holds for $x^2$ because $x^2$ is uniformly differentiable on every closed subinterval of $(0, \infty)$. Then we use excluded middle to show that if the principal square root function is defined at $0$ and defined on the domain $(0, \infty)$, then it is defined on $[0, \infty)$. However, in constructive mathematics, by definition there is no excluded middle, so we cannot prove the last step.
The exponential function is defined by a particular Taylor series, and the logarithm function could be defined by an analytic continuation of a particular Taylor series. Thus, one could try to define the square root as the function
$$e^{\frac{1}{2} \ln(x)}$$
However, because $\ln(x)$ is undefined at $x = 0$, that function is only defined on the open interval $(0, \infty)$. Similarly to the previous attempt, in classical mathematics, one could use excluded middle to extend the function to $[0, \infty)$, but that isn't possible in constructive mathematics.
Is there a way of proving that the principal square root function on the non-negative elements of a Cauchy complete Archimedean ordered field actually exists? I might be missing something very obvious. Or is it not provable in constructive mathematics that such a function exists in all Cauchy complete Archimedean ordered fields?
This is fairly important because it is one of the functions used in defining the Euclidean metric in finite-dimensional vector spaces over Cauchy complete Archimedean ordered fields. If the latter is the case, then one might have to equip Cauchy complete Archimedean ordered fields with the additional structure of a principal square root function in order to do constructive Euclidean geometry.
| https://mathoverflow.net/users/483446 | Proof in constructive mathematics that the principal square root function exists in any Cauchy complete Archimedean ordered field | I think see how this is related to your other questions. These are not resolved last I checked, but the following will answer this question whichever way the other questions end up going.
I believe the following are constructively equivalent for any notion of $\mathbb R$ which is Cauchy complete assuming that unique choice is valid:
1. There are lattice functions $\min/\max:\mathbb{R}^2 \to \mathbb{R}$.
2. There is an an absolute value function $\mathbb{R} \to [0,\infty)$.
3. There is square root function $[0,\infty)\to[0,\infty)$.
The equivalence of 1 and 2 follows from these relations:
$$|x| = \max(x,-x), \qquad \max(x,y) = \frac{x + y + |y - x|}{2}, \qquad \min(x,y) = \frac{x + y - |y - x|}{2}.$$
That 3 implies 2 follows from $|x| = \sqrt{x^2}$.
The fun part is that 1 & 2 implies 3. As you have observed, there is a square root function $(0,\infty)\to(0,\infty)$ by the Inverse Function Theorem.
With this apparatus, we can define continuous functions $f\_n:[0,\infty)\to[0,\infty)$:
$$f\_n(x) = \begin{cases}
1/2^n & \text{when $0 \leq x \leq 1/4^n$} \\
\sqrt{x} & \text{when $x \geq 1/4^n$} \\
\end{cases}$$
As stated, that requires knowing whether $x \leq 1/4^n$ or $x \geq 1/4^n$, but we can work around this by patching two functions together:
* $f^{-}\_n:[0,1/4^n)\to[0,\infty)$ is the constant function with value $1/2^n$,
* $f^{+}\_n:(0,\infty)\to[0,\infty)$ is defined as $\max(1/2^n,\sqrt{x})$.
Since these functions agree on their overlap, and their domains comprise all of $[0,\infty)$ we do get a total function $f\_n:[0,\infty)\to[0,\infty)$ as a result. (This is where unique choice is used.)
Now the sequence of functions $(f\_n)\_{n=0}^\infty$ so defined converges uniformly on any closed bounded interval to a continuous function $f:[0,\infty)\to[0,\infty)$. It is easily seen that this is indeed the square-root function: $(f(x))^2 = x$ for all $x \geq 0$.
---
In the above, I discreetly used the fact that a uniformly Cauchy sequence $(f\_n )^\infty\_{n=0}$ of continuous functions on a closed bounded interval converges to a continuous function on that interval. This is a straightforward consequence of Cauchy completeness and unique choice, so nothing to worry about... but it just occurred to me that this is enough to prove the existence of the absolute value function. To do this without circular reasoning, we need to pick a sequence $(f\_n)\_{n=0}^\infty$ whose definition doesn't involve min/max, absolute values, square root. That takes some thought but I believe $f\_n(x) = x\tanh(n x)$ does the trick!
So it looks like all three statements above are equivalent because they are all true!
---
As mentioned in the comments, there are two common notions of "Cauchy complete" used in constructive mathematics. The "classical" notion, in symbolic form, is:
$$\forall \varepsilon > 0\,\exists N\,\forall m, n \geq N\,[-\varepsilon < x\_m - x\_n < \varepsilon]$$
(I avoided the customary absolute value for good form.) The stricter notion requires a modulus of convergence $\phi : \mathbb N\to\mathbb N$ such that: $$\forall m,n \geq \phi(N)\,\left[\frac{-1}{2^N} < x\_m - x\_n < \frac{1}{2^N}\right]$$
There's a bunch of equivalent variations but you get the idea...
The latter is usually preferred by constructivists because of things like the [Specker sequence](https://en.wikipedia.org/wiki/Specker_sequence).
The answer above is agnostic about this distinction since it is straightforward to supply the requisite modulus of convergence wherever Cauchy completeness is used.
| 9 | https://mathoverflow.net/users/2000 | 426110 | 172,957 |
https://mathoverflow.net/questions/426097 | 10 | What is the history of Eisenstein series? Did the mathematician Eisenstein actually encounter them?
There are, as far as I know, two major perspectives on what Eisenstein series are. The first is in the study of modular forms. By summing over lattice points, one obtains a holomorphic modular form
$$E\_{2k}(z) = \sum\limits\_{(m,n) \neq (0,0)} \frac{1}{(m+nz)^{2k}}$$
for $\operatorname{SL}\_2(\mathbb Z)$ which together with the cusp forms exhausts the space of modular forms of a given weight.
The other perspective is in considering a unitary representation $(\pi, V)$ of a Levi subgroup $M$ of a semisimple real Lie group $G$. The representation considered should be left $\Gamma \cap M$- invariant for a lattice $\Gamma$ of $G$. If $P$ is a parabolic subgroup of $G$ containing $M$ as a Levi subgroup, one forms the induced representation $\mathcal V = \operatorname{Ind}\_P^G \pi$ of $G$ and, wishing to embed $\mathcal V$ into $L^2(\Gamma \backslash G)$, one associates to each $\varphi\_{\pi} \in \mathcal V$ the Eisenstein series
$$E(\varphi\_{\pi},g) = \sum\limits\_{\gamma \in \Gamma \cap P \backslash \Gamma \cap G} \varphi\_{\pi}(\gamma g). \tag{1}$$
There is a bit more to it, as one needs to take continuous sums (integrals) of such Eisenstein series $E(\varphi\_{\pi\_s},g)$ for various unramified twists $\pi\_s$ of $\pi$, and analytically continue $E(\varphi\_{\pi\_s},g)$ so that the formula (1) no longer makes sense.
These perspectives can be connected by associating modular forms with automorphic forms on $\operatorname{SL}\_2(\mathbb Z) \backslash \operatorname{SL}\_2(\mathbb R)$.
Did these two points of view on Eisenstein series develop independently? Who were the first people to connect them?
| https://mathoverflow.net/users/38145 | History of points of view on Eisenstein series | **Q:** *What is the history of Eisenstein series? Did the mathematician Eisenstein actually encounter them?*
**A:** Yes, he did, see [Elliptic Functions According to
Eisenstein and Kronecker](https://ems.press/content/serial-article-files/8714).
The reference is [Beiträge zu Theorie der elliptischen Functionen. VI. Genaue Untersuchungen der unendlichen Doppelproducte, aus welchen die elliptischen Functionen als Quotienten zusammengesetzt sind, und der mit ihnen zusammenhangenden Doppelreihen](https://resolver.sub.uni-goettingen.de/purl?PPN243919689_0035) (1847).
Eisenstein writes the double sum (on page 223) as
where $w=\alpha m+\beta n$ and the sum is over the two integers $m,n$.
| 7 | https://mathoverflow.net/users/11260 | 426117 | 172,958 |
https://mathoverflow.net/questions/426119 | 1 | It is well known that if $\varphi$ is a Schwartz function on $\mathbb{R}$ (i.e. smooth and decaying at infinity faster than polynomials), then its Fourier transform decays faster than polynomials. More precisely, for any $M>0$ there exists a constant $C\_M>0$ such that
\begin{equation}\tag{1}\label{1}
|\widehat{\varphi}(\lambda)|=\left|\int\_{\mathbb{R}} \varphi(x)e^{-2\pi i x \lambda}\,dx\right|\le C\_M \lambda^{-M},
\end{equation}
for any $\lambda>0$. The proof of \eqref{1} uses integration by parts.
My questions regard potential relaxing of the smoothness assumption imposed on $\varphi$:
1. Is there a function which is non-differentiable at some point, but for which \eqref{1} still holds?
2. Take $\varphi(x)=|x|e^{-x^2}$. Does \eqref{1} hold for this function, and if not, what is the optimal rate of decay of $|\widehat{\varphi}(\lambda)|$?
I think the answer to the second question should be $\lambda^{-1}$ (same as in van der Corput estimate with a non-smooth cutoff function), but I am not able to work out the details. All hints will be appreciated.
| https://mathoverflow.net/users/157356 | Decay of the Fourier transform of a non-differentiable function | I answer question $(1)$, assuming only that $\varphi$ is integrable.
If $(1)$ holds, then $\hat{\varphi}$ is also integrable, so Fourier inversion formula applies. For almost every $x \in \mathbb{R}$,
$$\varphi(x) = \int\_{\mathbb{R}}\hat{\varphi}(\lambda)e^{i2\pi x\lambda}\mathrm{d}\lambda.$$
Since for every $n \in \mathbb{N}$, the functions $\lambda\mapsto\lambda^n\hat{\varphi}(\lambda)$ are integrable, the right-hand side defines a $\mathcal{C}^\infty$ function. Hence $\varphi$ coincides almost everywere with a $\mathcal{C}^\infty$ function.
Now, I answer question $(2)$. I hope that my computations are right. Set $\varphi(x) = |x|e^{-x^2}$ and $\psi(x) = e^{-|x|}$. For every non null $x$,
$$\varphi(x) + \psi(x) = |x|e^{-x^2}+e^{-|x|}.$$
$$\varphi'(x) + \psi'(x) = \mathrm{sign}(x)(e^{-x^2}-2x^2e^{-x^2}-e^{-|x|}).$$
$$\varphi''(x) + \psi''(x) = \mathrm{sign}(x)(-6xe^{-x^2}+4x^3e^{-x^2})+e^{-|x|}.$$
These quantities have the same limits at $0+$ as at $0-$.
Therefore, the function $\varphi+\psi$ is $\mathcal{C}^2$ and $(\varphi+\psi)^{(k)}$ for $k=0,1,2$ are integrable. Moreover, $(\varphi+\psi)''$ has bounded variation. Hence
$$(\hat{\varphi}+\hat{\psi})(\lambda) = o(\lambda^{-3})~\mathrm{as}~\lambda\to\pm\infty.$$
But
$$\hat{\psi}(\lambda) = \int\_0^\infty e^{-x}(e^{-i2\pi x\lambda}+e^{i2\pi x\lambda})\mathrm{d}x = \frac{1}{1+i2\pi\lambda}+\frac{1}{1-i2\pi\lambda} = \frac{2}{1+4\pi^2\lambda^2}.$$
Hence
$$\hat{\varphi}(\lambda) = \frac{-1}{2\pi^2\lambda^2} + o(\lambda^{-3})~\mathrm{as}~\lambda\to\pm\infty.$$
| 3 | https://mathoverflow.net/users/169474 | 426122 | 172,960 |
https://mathoverflow.net/questions/426123 | 6 | In *Locally Presentable and Accessible Categories*, page 12 (10),
>
> A topological space is finitely presentable in $\mathbf{Top}$, the category of topological spaces and continuous functions, iff it is finite and discrete.
>
>
>
But the explanation after this sentence makes little sense to me. In particular, I want a rigorous proof that every finitely presentable object in $\mathbf{Top}$ is finite and discrete.
| https://mathoverflow.net/users/239112 | Why finitely presentable objects in Top need to be discrete? | $\newcommand{\colim}{\operatorname{colim}}$I believe there is a mistake in the argument - as written, the obvious maps $A\to D\_n$ are continuous, so the map $A\to\colim D\_n$ certainly factors through the maps $D\_n\to\colim D\_n$. I also don't believe the claim that $\colim D\_n$ is indiscrete is accurate - all open subsets of $A$ are also open in that colimit.
Instead, the argument should work if instead you modify $D\_n$ so that a subset $U\subseteq A\sqcup\mathbb N$ is open iff $U$ is empty, the whole space or $U$ is *cofinite* and disjoint from $\{0,1,\dots,n-1\}$, with no condition on elemens of $A$.
Now, because $A$ is not discrete, but the subset of $D\_n$ on underlying set of $A$ *is* discrete, we see that the maps $A\to D\_n$ are not continuous. On the other hand, you can check the colimit in this case is indeed indiscrete - since no cofinite subset of $\mathbb N$ can be disjoint from all $\{0,\dots,n-1\}$, the only two sets which are open in each $D\_n$ are the empty set and all of $A\sqcup\mathbb N$. Thus we have a map $A\to\colim D\_n$ which doesn't factor through any of the $D\_n$.
Finiteness is an easier part - any infinite discrete space $X$ can be written as a colimit of its finite subspaces, $X=\colim X\_i$. But $X\to X$ certainly doesn't factor through any of the $X\_i$ unless $X$ itself is finite.
| 10 | https://mathoverflow.net/users/30186 | 426127 | 172,961 |
https://mathoverflow.net/questions/426136 | 7 | Question: If $X$ is a simplicial complex that's simply connected and $2$-dimensional, does there always exist a contractible subcomplex $Y$ satisfying $X^{(1)} \subseteq Y$?
The statement is true "down a dimension": If $X$ is connected and $1$-dimensional, then there exists a contractible subcomplex $Y$ satisfying $X^{(0)} \subseteq Y$, namely you can use a spanning tree. So, one could think of the desired $Y$ in the original problem as a $2$-dimensional analog of a spanning tree, with contractibility being the key desired property.
It seems there is a higher dimensional analog of "spanning tree" in the literature, at least for finite complexes, e.g., Definition 3.1 of [this](https://www.ams.org/journals/tran/2009-361-11/S0002-9947-09-04898-3/S0002-9947-09-04898-3.pdf). Note that Proposition 3.7 of that paper implies that (finite) simply connected $2$-complexes indeed have these sorts of $2$-dimensional "spanning trees". But these are not necessarily contractible, they're more of a homological analog of trees.
| https://mathoverflow.net/users/164670 | Contractible subcomplex containing 1-skeleton? | I think the answer is no.
In Hatcher's Algebraic Topology there is an example of an acyclic 2-dimensional complex with one 0-cell, two 1-cells and two 2-cells (Example 2.38). You start with a wedge of two circles, labeled $a$ and $b$, and attach two 2-dimensional cells by the words $a^5b^{-3}$ and $b^2(ab)^{-2}$. One can easily check that it is acyclic. The fundamental group has the presentation $<a, b\mid a^5b^{-3}, b^2(ab)^{-2}>$, and one can show that it is not trivial. Now form a new complex $X$ by attaching one more $2$-dimensional cell, along $b$. The fundamental group of $X$ is easily seen to be trivial. But no proper subcomplex of $X$ containing the $1$-skeleton is simply-connected, let alone contractible.
To make this example into a simplicial complex, choose a triangulation of $X$. This can be done because all the attaching maps are nice piece-wise linear maps. I believe it is still true that no simplicial subcomplex of $X$ that contains the $1$-skeleton is contractible. Removing a two-cell from the triangulation of $X$ is like punching a hole in a two cell in the original CW structure on $X$. We already know that punching a single hole in some of the cells does not make a contractible complex, and punching additional holes only makes the fundamental group bigger.
| 8 | https://mathoverflow.net/users/6668 | 426144 | 172,969 |
https://mathoverflow.net/questions/426126 | 19 | Suppose $X$ is an infinite set and $\mathcal{A}$ is a family of subsets of $X$.
(1) We say that $\mathcal{A}$ is **$k$-cover-free** if for every distinct $a\_0, a\_1, \cdots, a\_k \in \mathcal{A}$, $a\_0 \nsubseteq \bigcup\_{1 \leq i \leq k} a\_i$.
(2) We say that $\mathcal{A}$ is **$k$-good** if for every distinct $a\_1, \cdots, a\_k \in \mathcal{A}$, there exists $a\_0 \in \cal{A} \setminus \{a\_1, \cdots, a\_k\}$ such that $a\_0 \subseteq \bigcup\_{1 \leq i \leq k} a\_i$.
Note that the family $\mathcal{A}$ of all subsets of $X$ of size $2$ is both $1$-cover-free and $2$-good. So there are $1$-cover-free and $2$-good families of all possible infinite cardinalities. I would like to know if there are similar examples for $k \geq 2$.
**Question:** Let $k \geq 2$. Are there families of arbitrarily large cardinalities which are both $k$-cover-free and $(k+1)$-good? What if $k = 2$?
| https://mathoverflow.net/users/485348 | Family of sets with a covering property | Yes. Let $S$ be a set of cardinality $|S|\ge k+2$. Let $X=\binom Sk$, the set of all $k$-element subsets of $S$. For each $s\in S$ let $a\_s=\{x\in X:s\notin x\}$, and let $A=\{a\_s:s\in S\}$. It is easy to verify that $A$ is $k$-cover-free and $(k+1)$-good.
Given distinct elements $s\_0,s\_1,\dots,s\_k\in S$, we have $a\_{s\_0}\not\subseteq a\_{s\_1}\cup\cdots\cup a\_{s\_k}$ because $\{s\_1,\dots,s\_k\}\in a\_{s\_0}\setminus(a\_{s\_1}\cup\cdots\cup a\_{s\_k})$.
Given distinct elements $s\_1,\dots,s\_{k+1}\in S$, take any $s\_0\in S\setminus\{s\_1,\dots,s\_{k+1}\}$; we have $a\_{s\_0}\subseteq a\_{s\_1}\cup\cdots\cup a\_{s\_{k+1}}$ because $a\_{s\_1}\cup\cdots\cup a\_{s\_{k+1}}=X$.
**P.S.** In a comment the OP asked:
>
> Is possible to get such a family whose members are all finite?
>
>
>
The answer is negative; for $k\ge2$ a family which is $k$-cover-free and $(k+1)$-good can't contain uncountably many finite sets. In fact:
**Theorem.** For $k\ge2$ a family which is $k$-cover-free and $(k+1)$-good can't contain infinitely many finite sets of bounded size.
**Proof.** Assume for a contradiction that $k\ge2$ and that $A$ is a $k$-cover-free $(k+1)$-good family containing infinitely many finite sets of bounded size. Then $A$ contains an infinite *quasidisjoint* family of finite sets, i.e., there exist an infinite sequence of distinct finite sets $a\_i\in A$ and a set $d$ such that $a\_i\cap a\_j=d$ whenever $i\ne j$. Let $a'\_i=a\_i\setminus d$ and let $s=a\_1\cup\cdots\cup a\_k$.
Since $A$ is $(k+1)$-good, for each $i\gt k$ we can choose a set $b\_i\in A$ so that $b\_i\subseteq s\cup a\_i$ and $b\_i\notin\{a\_1,\dots,a\_k\}\cup\{a\_i\}$. Since $A$ is $k$-cover-free we have $b\_i\not\subseteq s$, whence $b\_i\cap a'\_i\ne\varnothing$, so the sets $b\_i$ ($i\gt k$) are distinct.
Since $s$ is finite, we can choose $i,j$ with $k\lt i\lt j$ and $b\_i\cap s=b\_j\cap s$.Now $b\_i,b\_j,a\_j$ are distinct elements of $A$, and $b\_j\subseteq b\_i\cup a\_j$, showing that $a$ is not $2$-cover-free. Since $k\ge2$, this contradicts our assumption that $A$ is $k$-cover-free.
| 13 | https://mathoverflow.net/users/43266 | 426152 | 172,974 |
https://mathoverflow.net/questions/426156 | 6 | For a positive integer $n$, let $\mathcal{P}$ be the power set of $[n]$. Consider the graph $G$ with $\mathcal{P}$ as its vertex set, and, for $S\_1,S\_2 \in \mathcal{P}$, the edge $(S\_1,S\_2)$ exists iff $S\_1 \subset S\_2$.
(That is, $G$ is the transitive closure of the Hasse diagram of the "$\subset$" relation.)
**Question:** Suppose we can color the vertices of $G$ using $k$ colors. How many vertices $v$ are there such that $v$ has at least one neighbor that has the same color as $v$?
Remarks:
* Note that the question asks for a lower bound on the number of such vertices that holds no matter how $G$ is colored with $k$ colors.
* Obviously, the bound will depend on $k$. I'm mostly interested in large $k$, e.g., $k \approx \epsilon|\mathcal{P}|$, for some small constant $\epsilon>0$.
| https://mathoverflow.net/users/156157 | Coloring of a graph representing the power set | For $k\ge n+1$ there is a proper coloring of $G$ where each set in $\mathcal{P}$ is colored by its cardinality. Then no vertex $v$ has a neighbor with the same color.
| 8 | https://mathoverflow.net/users/24076 | 426161 | 172,977 |
https://mathoverflow.net/questions/423434 | 8 | I recently encountered an interesting space. It is a compactification of the space of $ n$ points in $ \mathbb A^1 $ modulo translation, $ (\mathbb A^1)^n / \mathbb G\_a $.
Let $ n \in \mathbb N $ and let $ p([n]) = \{ (i,j) \in \{1, \dots, n\}^2 : i \ne j \} $. Let $ Y\_n $ be the subscheme of $ (\mathbb P^1)^{p([n])} $ defined by the equations
$$
\tau\_{ik} = \tau\_{ij} + \tau\_{jk} \quad \tau\_{ij} = -\tau\_{ji}
$$
If we restrict to the open set $ Y\_n^{\circ} $ where $ \tau\_{ij} \ne \infty $, then we have an isomorphism $ Y\_n^\circ \cong (\mathbb A^1)^n / \mathbb G\_a $ where $ \tau\_{ij} = x\_i - x\_j $.
In fact, $Y\_n $ has a stratification with strata
$$ (\mathbb A^1)^{S\_1} / \mathbb G\_a \times \cdots \times (\mathbb A^1)^{S\_m} / \mathbb G\_a$$
where $ \{1, \dots, n \} = S\_1 \sqcup \cdots \sqcup S\_m $.
Another interesting thing about $Y\_n $ is that it is the degeneration of the permutahedral toric variety (also known as the Losev-Manin space). This latter space can be realized as a subscheme of $ (\mathbb P^1)^{p([n])} $ by the equations
$$ \alpha\_{ij} \alpha\_{jk} = \alpha\_{ik} \quad \alpha\_{ij} = \alpha\_{ji}^{-1} $$
**Has anyone seen this space before?**
| https://mathoverflow.net/users/438 | A compactification of the space of points on the affine line | Given a linear space $L \subset \mathbb{A}^n$, the closure of $L$ inside of $(\mathbb{P}^1)^n$ is called the **Schubert variety** of $L$, or of the hyperplane arrangement obtained by intersecting the coordinate hyperplanes with $L$. Your variety is the Schubert variety of the Coxeter arrangement of type $A\_{n-1}$.
The Schubert variety was introduced by Ardila and Boocher <https://arxiv.org/pdf/1312.6874.pdf>. The stratification is not explicitly studied there, but it was observed both by Huh and Wang <https://arxiv.org/pdf/1609.05484.pdf> (who used it to prove the top-heavy conjecture for realizable matroids, which features prominently in Huh's Fileds Medal citation) and by Young, Xu, and myself <https://arxiv.org/pdf/1706.05575.pdf>. If you blow up first the point stratum, then the strict transforms of the 1-dimensional strata, and so on, you obtain the **augmented wonderful variety** <https://arxiv.org/pdf/2010.06088.pdf>, which is the topic of a lot of current research. The induced stratification of the affine patch in which all coordinates are nonzero (called the **reciprocal plane**, or the spectrum of the **Orlik--Terao algebra**) was first studied much earlier by Speyer and myself <https://arxiv.org/pdf/math/0410069.pdf>.
Sometimes the Schubert variety of a hyperplane arrangement is referred to as a **matroid Schubert variety**, since its cohomology ring is a matroid invariant (called the **graded Mobius algebra**), as is its intersection cohomology module. But that's kind of sloppy terminology, as the isomorphism type of the variety itself is not determined by the matroid.
In the case that you describe, $Y\_n$ can be regarded as the compactification of the configuration space of $n$ labeled points in $\mathbb{A}^1$ up to translation in which the distances between points are allowed to go either to zero or to infinity. The stratification that you describe comes from asking which collections of points are a finite distance away from each other, which gives a partition of the set $[n]$. This particular case was studied in detail by Young and myself <https://arxiv.org/pdf/1704.04510.pdf>.
| 6 | https://mathoverflow.net/users/10273 | 426165 | 172,978 |
https://mathoverflow.net/questions/426162 | 3 | **Background**
Recently I asked [a question](https://mathoverflow.net/questions/425803/simplicial-nerve-of-a-topological-group) on a particular construction ot the classifying space of a topological group. I got an answer, but it relied on nontrivial Quillen equivalences between various models of $(\infty,1)$-categories, and I was wondering whether there might be a more elementary approach. I then stumbled on [a paper by V.Hinich (arxiv 0704.2503)](https://arxiv.org/abs/0704.2503), which contained the line of argument I was thinking of.
**Question**
In the aformentioned paper, it is asserted (in Proposition 2.6.2) that for a simplicial group $G$, the $\overline{W}$-construction and the homotopy coheret nerve $\mathfrak{N}$ (applied to $G$ by regarding it as a simplicially enriched category with one object) evaluate to homotopy equivalent Kan complexes. The proof is rather short, but is unsatisfactory: The author constructs a comparison map $\overline{W}(G)\to\mathfrak{N}(G)$, computes the homotopy groups of $\overline{W}(G)$ and $\mathfrak{N}(G)$, observe that they coincide, and conclude that they have the same homotopy type without actually showing that the comparison map induces isomorphisms in the homotopy groups. Perhaps the last step might be obvious to the eyes of experienced, but it is not clear to a novice. Can someone explain why the cited proposition ought to be true?
| https://mathoverflow.net/users/144250 | Homotopy coherent nerve versus simplicial nerve | Both $\def\W{{\bar W}}\W$ and $\def\N{\mathfrak{N}}\N$ are right Quillen functors from the model category of simplicial groups to the model category of reduced simplicial sets (see the original paper by Dwyer–Kan, or Proposition V.6.3 in Goerss–Jardine).
Thus, to show that the natural transformation $\W→\N$ is a weak equivalence,
it suffices to show that the adjoint natural transformation $\def\L{{\bf L}}\L\_\N→\L\_\W$ of associated left adjoint functors (denoted by $\L\_{(-)}$) is a natural weak equivalence.
The natural transformation $\L\_\N→\L\_\W$ is a natural transformation of left Quillen functors from simplicial sets to simplicial groups.
Since reduced simplicial sets are generated by $\def\Z{{\bf Z}}\W\Z≃S^1$ under homotopy colimits,
to show that $\L\_\N→\L\_\W$ is a natural weak equivalence,
it suffices to show that $\L\_\N(S^1)→\L\_\W(S^1)$ is a weak equivalence of simplicial sets. This is done by a simple direct inspection.
| 3 | https://mathoverflow.net/users/402 | 426169 | 172,979 |
https://mathoverflow.net/questions/426160 | 3 | Let $T$ be a regular tournament, and $u \in V(T)$. Let $Out(u) \subset V(T)$ denote the set of vertices such that the edges between $u$ and them go out of $u$. Similarly define $In(u)$. Let two distinct vertices $u,v$ be called $\textbf{antipodal}$ if $Out(u) \setminus v = In(v) \setminus u$ (and thus also $In(u) \setminus v = Out(v) \setminus u$). Put more concretely, if $w$ is any other vertex, then either $u$ goes into $w$ AND $w$ goes into $v$, or vice versa.
Given $u$, let $v$ be called an $\textbf{antipode}$ of $u$ if $u$ and $v$ are antipodal. Note that this is poor notation as $u$ could have more than one antipode.
>
> Does every regular tournament have an antipodal pair of vertices? Furthermore, does every vertex have at least one antipode?
>
>
>
Note that the first problem is equivalent to showing that a regular tournament on $2n+1$ vertices contains an induced subgraph on $2n-1$ vertices that is a regular tournament.
The motivation for this problem is the following observation (if it's correct). $T$ has a size $2n-1$ subgraph that is a regular tournament if and only if, for any partition of $2n+1$ into odd numbers, $T$ can be partitioned into regular tournaments as subgraphs of sizes corresponding to the partition.
Perhaps this result is too strong to be true, I have only verified it for some very small examples.
| https://mathoverflow.net/users/143629 | About regular induced subgraphs of regular tournaments | There are regular tournaments [here](https://users.cecs.anu.edu.au/%7Ebdm/data/digraphs.html) and if you choose one at random except on the smallest sizes you will find a counterexample with high probability.
Enumeration gives a bound (which could be turned into an accurate estimate but I won't). Let $T(n)$ be the number of (labelled) regular tournaments with $n$ vertices. To make a regular tournament with $n+2$ vertices and an antipodal pair, take a regular tournament with $n$ vertices and add to it to new vertices $v\to w$. There are less than $2^n$ ways to add the edges between the new vertices and the old vertices such that $v,w$ are antipodal, then you can multiply by $n^2$ for the labels of the new vertices. So the number of regular tournaments on $n+2$ vertices that have an antipodal pair is at most $n^2 2^n T\_n$. However, $T\_{n+2}$ is approximately $2^{2n}n^{-1}T\_n$. So only an exponentially small fraction of regular tournaments have an antipodal pair.
Ref for enumeration: B. D. McKay, [The asymptotic numbers of regular tournaments, eulerian digraphs and eulerian oriented graphs](https://users.cecs.anu.edu.au/%7Ebdm/papers/rt.pdf), Combinatorica, 10 (1990) 367-377.
| 3 | https://mathoverflow.net/users/9025 | 426171 | 172,980 |
https://mathoverflow.net/questions/148862 | 5 | Consider the setting of categories enriched over a suitable monoidal category $\mathbb V$.
We define $$\mathrm{Dist}(X,Y):=\mathbb V−\mathrm{Cat}(X^ \mathrm{op}⊗Y,\mathbb V).$$
Recall the definition of [ends](http://ncatlab.org/nlab/show/end%E2%80%8E). Taking the end is an operation of signature
$$\mathrm{Dist}(X,X)→\mathbb V.$$
QUESTION: Is there an analogue for functors into V of higher arity. More explicitly: Is there a canonical operation of signature
$$\mathbb V−\mathrm{Cat}(X^?⊗X^?⊗X^?,V)→\mathbb V$$
where the ? are to be replaced by either op or nothing.
MOTIVATION: I like to think of the composition $\otimes$ of (2-ary)distributors and the right adjoints to $D\otimes−$ and $−\otimes E$ as "[horn-filling](http://ncatlab.org/nlab/show/horn)" (in the sense of viewing categories as simplicial sets).
I hope to find a similar situation for "3-ary distributors" - Whenever there is a (oriented?) tetrahedron of 3-ary functors with one side missing we should be able to find the missing side.
So given
$$U\in\mathrm{Dist}^3(X,Y,A)$$
$$V\in\mathrm{Dist}^3(Y,Z,A)$$
$$W\in\mathrm{Dist}^3(Z,X,A)$$
there should be
$$?(U,V,W)\in\mathrm{Dist}^3(X,Y,Z)$$
and related adjoints.
Of course $\mathrm{Dist}^3$ remains to be defined.
| https://mathoverflow.net/users/1261 | Ends and coends – analogues for higher arity – Horn Filling | This is exactly the subject of the paper [Coends of higher arity](https://arxiv.org/abs/2011.13881) by Loregian and de Oliveira Santos.
| 4 | https://mathoverflow.net/users/152679 | 426180 | 172,983 |
https://mathoverflow.net/questions/426066 | 6 | Let $k$ be a nonarchimedean local field and $G$ a reductive $k$-group,
which we assume to be semisimple and simply-connected. Recall that an abstract group $H$ is *perfect* if it is generated by commutators, that is, equals its derived subgroup.
**Question**: Is $G(k)$ perfect?
When $G$ is isotropic, $G(k)$ is known to be perfect. This is a consequence of the solution to the [Kneser–Tits problem](https://en.m.wikipedia.org/wiki/Kneser%E2%80%93Tits_conjecture) for $k$: the group $G(k)^+$ generated by (the rational points of) additive subgroups of $G$ is the same as the group generated by root subgroups of $G$ for a fixed maximal split torus, and the latter group is generated by commutators. It is known that $G(k)=G(k)^+$ when $G$ is simply-connected.
So the main thrust of my question is when $G$ is anistropic, or equivalently (hence the title), when $G(k)$ is compact in the analytic topology. In type $A$, Platonov and Jančevskiĭ have proved ([On a conjecture of Harder](https://mathscinet.ams.org/mathscinet-getitem?mr=379444)) by computations with division rings that $G(k)$ is perfect. (**Clarification**: This result is for $D^\times$, not $\text{SL}\_1(D)$.)
**Related question**: Are there any anisotropic simple $k$-groups not of type $A$?
I realize that if the answer to the related question is negative then this work of Platonov and Jančevskiĭ answers my first question. I once looked through the tables in Tits's Boulder notes [Classification of semisimple algebraic groups](https://mathscinet.ams.org/mathscinet-getitem?mr=224710) and it seemed like there were compact simple reductive groups only in type $A$. But this result was so surprising, and the tables so hard for me to follow, that I was not confident in the correctness of my understanding.
Bonus points for questions that work with a more general $k$, say, a complete discretely valued field with residue field of dimension $\leq1$.
**Clarification**: I was too blithe about what happens in type $A$. As krl pointed out in the [comments](https://mathoverflow.net/questions/426066/is-every-compact-simply-connected-reductive-p-adic-group-perfect/426080#comment1095360_426066), $\operatorname{SL}\_1(D)$ is not perfect, and in fact, its commutator subgroup is its pro-unipotent radical. Riehm's paper "[The norm 1 group of a $\mathfrak P$-adic division algebra](https://mathscinet.ams.org/mathscinet-getitem?mr=262250)" carefully analyzes the situation. So in the end, the answer to my question is "no" overall, but "yes" in many cases.
| https://mathoverflow.net/users/174855 | Is every compact simply-connected reductive p-adic group perfect? | It seems that the answer to the **Question** is "no" in general (in the anisotropic case): for example, if $D$ is a finite-dimensional central division algebra over $k$, then Theorem 1.9 in Platonov-Rapinchuk (based on the article of Riehm quoted above) implies that $$[\textrm{SL}\_1(D), \textrm{SL}\_1(D)] \neq \textrm{SL}\_1(D).$$
| 3 | https://mathoverflow.net/users/120578 | 426211 | 172,994 |
https://mathoverflow.net/questions/426219 | 1 | Let $A$ be a matrix of the Gaussian unitary ensemble (GUE) and $v\_1,v\_2$ be two orthonormal vectors.
I wonder if one can compute (or at least get a non-trivial lower bound on) the expectation value
$$\mathbf E\sqrt{4\langle Av\_1,v\_2 \rangle^2 + (\langle Av\_1,v\_1 \rangle-\langle Av\_2,v\_2 \rangle)^2}.$$
We can choose a unitary $U$ such that $Ue\_1=v\_1$ and $U e\_2 = v\_2$ then we have
$$\mathbf E\sqrt{4\langle U^\*AUe\_1,e\_2 \rangle^2 + (\langle U^\*AUe\_1,e\_1 \rangle-\langle U^\*AUe\_2,e\_2 \rangle)^2}.$$
Since the GUE is invariant under unitaries, this implies we have to compute (estimate from below)
$$\mathbf E\sqrt{4a\_{12}^2 + (a\_{11}-a\_{22})^2},$$
where $a\_{ij}$ are the matrix entries of $A.$
| https://mathoverflow.net/users/457901 | Expectation value of random GUE matrix | From the definion of the GUE for $n\times n$ Hermitian matrices $A$, one has the integral expression.
$$\mathbb{E}[f({\rm Re}\,a\_{12},{\rm Im}\,a\_{12},a\_{11},a\_{22})]=\frac{n^2}{2\pi^2}\iint\!\!\!\!\!\iint\_{-\infty}^{\infty} f(x\_1,x\_2,y\_1,y\_2)$$
$$\qquad\qquad\times\exp\left(-n(x\_1^2+x\_2^2+y\_1^2/2+y\_2^2/2)\right)\,dx\_1dx\_2dy\_1dy\_2.$$
The expression in the OP is cumbersome to evaluate, a simpler example is
$$\mathbb{E}[4|a\_{12}|^2+(a\_{11}-a\_{22})^2]=6/n.$$
A numerical evaluation gives
$$\mathbb{E}\left[\sqrt{4|a\_{12}|^2+(a\_{11}-a\_{22})^2}\right]=\frac{2.25676}{\sqrt{n}}.$$
| 2 | https://mathoverflow.net/users/11260 | 426221 | 172,998 |
https://mathoverflow.net/questions/426199 | 0 | Consider a set of N teams (N even number) that must make a
Round Robin Tournament. To each pair i; j, i ≠ j, of teams there is associated level
of interest si,j ∈ {1;2;3} of the match between them (1 = minimum interest, 2 = medium
interest, 3 = maximum interest). Define a calendar in such a way that:
1. in each day there is at least one game of maximum interest.
2. the minimum average level of interest between all days is maximized.
I have already written the solution for the first constraint, now i want to write the objective function relative to the second point. I have tried to write something like: Maximize the sum of the minimum average for each combination of teams in one matchday but it doesn't work
| https://mathoverflow.net/users/485367 | Round Robin volleyball Tournament | Assuming binary decision variable $x\_{ijd}$ indicates whether teams $i$ and $j$ play each other on day $d$, introduce a decision variable $z$, and maximize $z$ subject to linear constraints
$$z \le \frac{1}{N/2} \sum\_{i,j} s\_{ij} x\_{ijd} \quad \text{for all $d$}.$$
| 0 | https://mathoverflow.net/users/141766 | 426225 | 173,000 |
https://mathoverflow.net/questions/423925 | 2 | Suppose I have $n$ Boolean variables $x\_1,\dots,x\_n$, and an objective function of the form $f(x\_1,\dots,x\_n) = \sum\_{a\_1,\dots,a\_n}c\_{a\_1,\dots,a\_n} x\_1^{a\_1} \cdots x\_n^{a\_n}$ with $(a\_1,\dots,a\_n) \in \{0,1\}^n$. I know this is a bit vague, but suppose that there aren't that many nonvanishing coefficients and/or all the nonvanishing terms are of low degree. I want to maximize $f$ over $(x\_1,\dots,x\_n) \in \{+1,-1\}^n$. How hard is it? (If no coefficients vanish, then checking all $2^n$ vertices wouldn't take that much more time than writing down all $2^n$ coefficients.)
| https://mathoverflow.net/users/3621 | Optimizing a multilinear function over the vertices of the cube | (Making a CW answer to take this off the list of unanswered questions.)
As noted in comments by Jason Gaitonde, MAXCUT for a graph $G=(V,E)$ is equivalent to maximization of the quadratic multilinear polynomial
$$-\sum\_{(i,j)\in E}x\_ix\_j$$
over $\{-1,1\}^V$, hence the problem is NP-hard.
| 2 | https://mathoverflow.net/users/12705 | 426232 | 173,002 |
https://mathoverflow.net/questions/426092 | 4 | I'm looking for the name of what I suspect must be a standard property, and also for a possible statement about that property.
First the property: $W=a\_0\ldots a\_{n-1}$ has *this property* if for all $1\le k<n$,
$a\_0\ldots a\_{k-1}\ne a\_{n-k}\ldots a\_{n-1}$.
In particular, this implies that in any finite or infinite word, the blocks containing $W$ are disjoint. For the particular application that I have in mind, I start off with an infinite word $x$, and replace some subwords of $X$ of length $n$ spaced far apart by $W$'s. The consequence of the definition that is useful for me is that if $x$ initially contained no $W$'s, then the only $W$'s in the resulting sequence are those $W$'s that I "manually" inserted.
>
> Is there a name for this property?
>
>
> Is it the case that if $X$ is any mixing shift of finite type, then $X$ contains a word with this property?
>
(It's not hard to show that if $X$ is a full shift with an alphabet with two or more symbols, then $X$ contains words with this property.)
| https://mathoverflow.net/users/11054 | Word combinatorics terminology question | Yes you find these in all infinite mixing SFTs. More is true. As mentioned, these words are sometimes called **unbordered**, I'll use that word.
The following is Theorem 8.3.9 in [1].
>
> Theorem. Let $x \in A^\mathbb{N}$ for any alphabet $A$. If $x$ is not periodic, then for any $m$, there exists an unbordered word $w \sqsubset x$ with $|w| \geq m$.
>
>
>
Here **periodic** means $x = u^{\mathbb{N}} = uuuuu...$ for a finite-length word $u \in A^\*$; and $w \sqsubset x$ means that $w = x\_{[i, i+|w|-1]}$ for some $i$, i.e. it appears as a subword. So all you need is a non-periodic point in your subshift, in the weak sense that it's not literally of the form $u^{\mathbb{N}}$. To find these, we can apply the following result. The earliest written reference I know proving something like this is [2, Theorem 3.8], I'll just write a proof.
>
> Theorem. Let $X \subset A^\mathbb{N}$ be an infinite subshift (the shift need not be surjective). Then $X$ has a point which is not periodic.
>
>
>
Proof. We prove the contrapositive. Suppose $X$ has only periodic points. Let $Y \subset A^{\mathbb{Z}}$ be the $\mathbb{Z}$-subshift obtained as limit points of points in $X$. First suppose it is infinite. In this case, suppose that for some $n$, every word of length $n$ has a unique predecessor letter, i.e. $\forall u \in L: \exists! a \in A: au \in L$ where $L$ is the language of $Y$. This clearly implies the subshift $Y$ has at most $|A^n|$ points, contradicting infiniteness. So we can find arbitrarily long words $u$ which can be extended to the left by two distinct letters $a, b$. Taking a limit of such pairs, we obtain that there exists an infinite right tail $x \in A^{\mathbb{N}}$ which can be preceded by two distinct letters in points of $Y$. Thus the same is true in $X$, i.e. $ax, bx \in X$ for some distinct $a, b$. These cannot both be periodic. Next, suppose $Y$ is finite. Then in particular every point in $Y$ is periodic with some period $p$. This is a finite type condition, so because $Y$ is the limit points of $X$ we have that tails of points in $X$ actually have $p$-periodic tails after a bounded prefix (the subshifts $\sigma^n(Y)$ tend to $X$ in Hausdorff metric, so eventually you have to respect forbidden patterns of $X$). But then $X$ has only eventually periodic points, with a bound on the ``eventual'', so $X$ is finite. Square.
>
> Theorem. Let $X \subset A^\mathbb{N}$ or $X \subset A^\mathbb{Z}$ be an infinite subshift. Then there are unbordered words of unbounded length in the language of $X$.
>
>
>
Proof. For $\mathbb{N}$ just combine the above theorems. For $\mathbb{Z}$ cut off the left tail; this preserves infiniteness and gives you an ${\mathbb{N}}$-subshift. Square.
[1] *Lothaire, M.*, Algebraic combinatorics on words, Encyclopedia of Mathematics and Its Applications. 90. Cambridge: Cambridge University Press. xiii, 504 p. \textsterling 60.00/hbk (2002). [ZBL1001.68093](https://zbmath.org/?q=an:1001.68093).>
[2] *Ballier, Alexis; Durand, Bruno; Jeandal, Emmanuel*, [**Structural aspects of tilings**](http://dx.doi.org/10.4230/LIPIcs.STACS.2008.1334), Albers, Susanne (ed.) et al., STACS 2008. 25th international symposium on theoretical aspects of computer science, Bordeaux, France, February 21–23, 2008. Wadern: Schloss Dagstuhl – Leibniz Zentrum für Informatik (ISBN 978-3-939897-06-4). LIPIcs – Leibniz International Proceedings in Informatics 1, 61-72, electronic only (2008). [ZBL1258.05023](https://zbmath.org/?q=an:1258.05023).
| 6 | https://mathoverflow.net/users/123634 | 426233 | 173,003 |
https://mathoverflow.net/questions/426243 | 2 | Let $\mathbb{N}$ denote the set of positive integers, and consider the graph $(\mathbb{N}, E)$ where a set $\{a,b\}$ of two distinct positive integers belongs to $E$ if there is an integer $k>1$ such that $a = k\cdot b$ or $b = k \cdot a$.
**Questions.**
1. Is there a bijection $p:\mathbb{N}\to\mathbb{N}$ such that for all $k\in \mathbb{N}$ we have $\{p(k), p(k+1)\} \in E$?
2. Is there a bijection $p:\mathbb{Z}\to\mathbb{N}$ such that for all $k\in \mathbb{Z}$ we have $\{p(k), p(k+1)\} \in E$?
| https://mathoverflow.net/users/8628 | Hamiltonian path in divisibility graph | Yes, as for every countable graph on which any two vertices have infinitely many common neighbours. If you constructed a path $v\_1\ldots v\_m$, and $u$ is the first (with respect to a numeration chosen in advance) not visited vertex, then you may proceed with $\dots v\_{m+1}u$, where $v\_{m+1}$ is a not used yet common neighbour of $v\_m$ and $u$.
| 5 | https://mathoverflow.net/users/4312 | 426248 | 173,007 |
https://mathoverflow.net/questions/426241 | 1 | Let $\alpha, \beta:\mathbb R\_+\to [0,1]$ be continuous and decreasing functions s.t. $\alpha(0)=1=\beta(0)$ and $\alpha, \beta$ are continuously differentiable on $(0,\infty)$ satisfying for some $c>0$
$$-\frac{c}{\sqrt{t}}\le\alpha'(t), \beta'(t)\le 0,\quad \forall t>0.$$
Let $f,g :\mathbb R\_+\to [0,1]$ be continuous s.t. for all $t\ge 0$
\begin{eqnarray}
\|\alpha-\beta\|\_t &\le& C\sqrt{t} \big(\|\alpha-\beta\|\_t+\|f-g\|\_t\big) \\
\|f-g\|\_t &\le& C\sqrt{t}\left(\|\alpha-\beta\|\_t+\|f-g\|\_t + \int\_0^t|\alpha'(s)-\beta'(s)|ds\right),
\end{eqnarray}
where $C>0$ is some constant and $\|\cdot\|\_t$ denotes the uniform norm of functions on $[0,t]$. Can we prove the existence of $T>0$ s.t. $(\alpha,f)=(\beta,g)$ over the interval $[0,T]$?
| https://mathoverflow.net/users/261243 | Do these two pairs coincide at small time? | $\newcommand{\al}{\alpha}\newcommand{\be}{\beta}$The answer is no.
E.g., suppose that for all small enough $t>0$ we have
\begin{equation\*}
f(t)=t,\quad g(t)=0,\quad \al'(t)=-\frac1{\sqrt t}\,\Big(1+\sin\frac1t\Big),\quad \be'(t)=-\frac1{\sqrt t},
\end{equation\*}
with $\al(0)=1=\be(0)$.
Then all your conditions hold. In particular, for $t\downarrow0$,
\begin{equation\*}
|\al(t)-\be(t)|=\Big|\int\_{1/t}^\infty\frac{\sin z}{z^{3/2}}\,dz\Big|=O(t^{3/2})
\end{equation\*}
and
\begin{equation\*}
\int\_0^t|\al'(s)-\be'(s)|\,ds=\int\_{1/t}^\infty\frac{|\sin z|}{z^{3/2}}\,dz\asymp t^{1/2},
\end{equation\*}
so that
\begin{equation\*}
\begin{aligned}
\|\al-\be\|\_t &\le C\sqrt{t} \big(\|\al-\be\|\_t+\|f-g\|\_t\big) \\
\|f-g\|\_t &\le C\sqrt{t}\left(\|\al-\be\|\_t+\|f-g\|\_t + \int\_0^t|\al'(s)-\be'(s)|ds\right)
\end{aligned}
\tag{1}\label{1}
\end{equation\*}
for some real $C>0$ and all $t$ in some right neighborhood of $0$. So, inequalities \eqref{1} hold for some real $C>0$ and all real $t\ge0$.
Yet, $f\ne g$ in any right neighborhood of $0$. So, $(\al,f)\ne(\be,g)$ on $[0,T]$, for any real $T>0$.
| 1 | https://mathoverflow.net/users/36721 | 426253 | 173,008 |
https://mathoverflow.net/questions/426246 | 4 | Let $\mathfrak{g}$ be a complex simple Lie algebra. Let $S(\mathfrak{g})$ be the algebra of polynomial functions on $\mathfrak{g}$, viewed as a $\mathfrak{g}$-representation. Are the isotypic components of $S(\mathfrak{g})$ finite-dimensional?
This seems to be true for $\mathfrak{sl}\_n$, since this is the subject of "plethysm" and there are explicit formulae for these multiplicities. See e.g. [this](https://mathoverflow.net/questions/299900/symmetric-powers-for-lie-algebras) mathoverflow question and the reference therein.
| https://mathoverflow.net/users/385475 | Are isotypic components of $S(\mathfrak{g})$ finite-dimensional? | I think there must be some different ideas of what your question means, looking at the comments. I understood it to mean as in the decomposition of tensor powers of a $\mathfrak{g}$-representation into irreducible subrepresentations. In which case, the answer is a definite yes. For a start, the action of $\mathfrak{g}$ on any symmetric power of a representation preserves the degree. So in this way $S(\mathfrak{g}) = \bigoplus\_{i=0}^\infty S^{i}(\mathfrak{g})$. Then the isotypic components in this sense just refers to the decomposition of these finite dimensional pieces into irreducibles which are automatically finite dimensional themselves. The explicit decompositions themselves can be quite hard to work out although there are formulae for these things. This is the kind of thing that "plethysm" refers to.
The invariant polynomials in this view form trivial subrepresentations but they would not be regarded as a single isotypic component (indeed they appear at various degrees and there are more than $\mathrm{rank}\ \mathfrak{g}$ of them, those are just the generators)
Edit: having looked through the edits to the question I understand the confusion a little better. You originally asked about whether copies of each irreducible appeared finitely many times. I would hesitate to call these isotypic components but the answer here is no. As pointed out in the comments the collection of all trivial subreps form an infinite-dimensional component. This is, in essence, the span of all polynomials generated by the Casimir elements.
| 2 | https://mathoverflow.net/users/163024 | 426257 | 173,009 |
https://mathoverflow.net/questions/426268 | 1 | Let $S(\omega)$ denote the collection of "sparse" infinite subsets of $\omega$, that is, $X\subseteq \omega$ is a member of $S(\omega)$ if and only if both $X$ and $\omega\setminus X$ are infinite.
Is there ${\cal S}\subseteq S(\omega)$ such that for all $a\neq b \in \omega$ we have $|\{s\in {\cal S}: \{a,b\} \subseteq s\}| = 1$?
| https://mathoverflow.net/users/8628 | Steiner-like systems on $\omega$ | Sure, let $K$ be any countably infinite field and let $P$ be the projective plane (or a higher-dimensional projective space) over $K$. Let $S'$ be the set of lines in $P$ (where a line is regarded as a set of points), and transport the family $S'$ of subsets of $P$ to a family $S$ of subsets of $\omega$ via your favorite enumeration of $P$.
| 4 | https://mathoverflow.net/users/6794 | 426270 | 173,014 |
https://mathoverflow.net/questions/426261 | 2 | Let $a>0$. How to prove the following inequality $$\exists c>0,\exists A>0,\forall t>0:\quad\int^\infty\_0 \frac{\sin(rt)}{rt}\frac{r^4}{\sinh^2(r)} e^{-ar\coth(r)}dr\leq c \big(e^{-At}\big)$$
for some constants $A>0,c>0$
| https://mathoverflow.net/users/172078 | The inequality $\int^\infty_0 \frac{\sin(rt)}{rt}\frac{r^4}{\sinh^2(r)} e^{-ar\coth(r)}dr\leq c \big(e^{-At}\big)$ | $\newcommand{\R}{\mathbb R}\renewcommand{\S}{\mathcal S}$This follows from [Theorem IX.14](https://www.google.com/books/edition/II_Fourier_Analysis_Self_Adjointness/Kz7s7bgVe8gC?hl=en), which states the following:
>
> Let $T\in\S'(\R^n)$. Suppose that the Fourier transform $\hat T$ of $T$ can be continued analytically to the set $\{z\colon|\Im z|<A\}$ for some $A>0$. Suppose also that for each $B<A$ we have $\sup\{\|\hat T(\cdot+iy)\|\_1\colon y\in(-B,B)\}<\infty$. Then $T$ is a bounded continuous function and for each $B<A$ there is a real number $C\_B$ such that
> \begin{equation\*}
> |T(x)|\le C\_B e^{-B|x|}
> \end{equation\*}
> for $x\in\R^n$.
>
>
>
Indeed, for the integral in question,
\begin{equation\*}
I(t):=\int^\infty\_0 \frac{\sin rt}{rt}\frac{r^4}{\sinh^2r} e^{-ar\coth(r)}\,dr,
\end{equation\*}
and
\begin{equation\*}
T(t):=t I(t), \tag{1}\label{1}
\end{equation\*}
we have $T=\frac{\sqrt{2\pi}}{2i}\,\check f$, where
\begin{equation\*}
f(r):=\frac{r^3}{\sinh^2r} e^{-ar\coth(r)}
\end{equation\*}
and $\check f$ is the inverse Fourier transform of $f$ (defined by the formula $\check f(t):=\frac1{\sqrt{2\pi}}\,\int\_\R e^{irt}f(r)\,dr$), so that
\begin{equation\*}
\hat T=\frac{\sqrt{2\pi}}{2i}\,f.
\end{equation\*}
It is not hard to see that the function $T$ defined by \eqref{1} satisfies all the conditions of [Theorem IX.14](https://www.google.com/books/edition/II_Fourier_Analysis_Self_Adjointness/Kz7s7bgVe8gC?hl=en), cited above, with $A=\pi$. So, for each $B\in(0,\pi)$ there is a real number $C\_B$ such that $|T(t)|\le C\_B e^{-Bt}$ for all $t\ge0$ and hence
\begin{equation\*}
|I(t)|\le C\_B e^{-Bt}
\end{equation\*}
for $t\ge1$. On the other hand, for all $t\in[0,1]$,
\begin{equation\*}
|I(t)|\le c\_a:=\int^\infty\_0\frac{r^4}{\sinh^2r} e^{-ar\coth(r)}\,dr
\le c\_{a,B}\, e^{-Bt},
\end{equation\*}
where $c\_{a,B}:=c\_a e^B\in(0,\infty)$.
Thus, for each $B\in(0,\pi)$ and all $t\ge0$
\begin{equation\*}
|I(t)|\le\max(C\_B,c\_{a,B}) e^{-Bt},
\end{equation\*}
as desired.
| 3 | https://mathoverflow.net/users/36721 | 426273 | 173,015 |
https://mathoverflow.net/questions/426236 | 3 | Consider $k\geq 2$ biinfinite arithmetic progressions $\mathcal A\_i=a\_i+b\_i\mathbb Z$ (for $i=1,\ldots,k$) in $\mathbb Z$.
(We suppose that $a\_i$ and $b\_i\geq 2$ are strictly positive integers.
One can assume $a\_i$ and $b\_i$ coprime and $a\_i$ can be reduced modulo $b\_i$ without loss of generality.)
The set $\mathcal P=\prod\_{i=1}^k \mathcal A\_i$ of all possible products has strictly positive upper and lower densities $\overline{\delta}$ and $\underline{\delta}$ in $\mathbb Z$.
*Is it possible that $\overline{\delta}>\underline{\delta}$?
Can they be irrational? Are they easy to compute? Fixing $\epsilon>0$ and $k\geq 2$ are there only finitely many such products $\mathcal P$ with upper density at least $\epsilon$?*
Examples: $(1+2\mathbb Z)(1+3\mathbb Z)=\mathbb Z\setminus\{0\}$
(and both densities are $1$).
$(1+2\mathbb Z)(1+4\mathbb Z)=1+2\mathbb Z$ (and both densities are equal to $1/2$).
$(1+3\mathbb Z)(1+5\mathbb Z)$ seems to have densities above $.75$.
| https://mathoverflow.net/users/4556 | Density for products of arithmetic progressions | I think Yaakov Baruch is right in general.
Let $D$ be the l.c.m. of all the $b\_i$, and let $S$ be the set of all residue classes in $\mathbb Z/D\mathbb Z$ met by $\mathcal P$. I claim that the density is $|S|/D$.
The upper bound is trivial. In what follows, I assume $(a\_i,b\_i)=1$, otherwise pass to a (1-dim) sublattice.
Consider any residue class $x+D\mathbb Z\in S$; it is a product of some classes $c\_i+D\mathbb Z\subseteq a\_i+b\_i\mathbb Z$.
As far as I understand, most of the integers (in particular, most of those in $x+D\mathbb Z$) have (different) prime divisors $p\_i\equiv a\_i\pmod {b\_i}$. Each such integer $n\in x+D\mathbb Z$ is expanded as
$$
n=p\_1\dots p\_{k-1}\cdot \frac n{p\_1\dots p\_{k-1}}\in \mathcal P,
$$
as the last factor automatically belongs to $c\_k+D\mathbb Z$.
| 5 | https://mathoverflow.net/users/17581 | 426289 | 173,019 |
https://mathoverflow.net/questions/418976 | 3 | Say that a digraph $(V,E)$ is *reducible* if there exist $x,y\in V$ with $x\ne y$ and such that for all $z\in V$, $(x,z)\in E\leftrightarrow(y,z)\in E$ and $(z,x)\in E\leftrightarrow(z,y)\in E$. It is clear that a reducible digraph has a non-trivial automorphism (swapping $x$ and $y$ and fixing the rest), but the inverse is not necessarily true.
For some applications elsewhere, I'm interested in "irreducible" digraphs and especially "irreducible" posets. Now, (insert usual disclaimer about not being an expert), do these already have a (potentially better) name? and if yes, where have they been treated?
| https://mathoverflow.net/users/30392 | Digraph without "immediately isomorphic" vertices? | Just to provide a conclusion here (summing up the comments): *twin-free* seems to be the best and, dare I say, standard terminology; two points as in my question would then be called *twins*. Otherwise, *point-determining*, *R-thin*, or *mating-type* have been used to denote the same property.
| 0 | https://mathoverflow.net/users/30392 | 426291 | 173,020 |
https://mathoverflow.net/questions/370764 | 3 | It's all in the title: Is there really no name for categories in which all automorphisms are trivial? I've encountered some examples of these, the most prominent being the simplex category.
| https://mathoverflow.net/users/30392 | Name for "Category without nontrivial automorphisms"? | To sum up the comments & discussion: no established terminology seems to exist, but *essentially gaunt* would be a good name for categories without nontrivial automorphisms.
For a related notion, categories without notrivial *isomorphisms* are called *gaunt*, see <https://ncatlab.org/nlab/show/gaunt+category>
On the other side, categories without nontrial *endomorphisms* are called *one-way*, and those without *infinite descending chains of non-id morphisms* are called *direct*, see <https://ncatlab.org/nlab/show/direct+category>
| 0 | https://mathoverflow.net/users/30392 | 426299 | 173,021 |
https://mathoverflow.net/questions/426306 | 3 | If $K = \mathbb{Q}(\sqrt{d})$ is a real quadratic field, then any unit $u \in \mathcal{O}\_K^\times$ with $u > 1$ must not be too small: indeed, such a $u = u\_1 + u\_2 \sqrt{d}$ with $u\_1, u\_2 > 0$ must satisfy $u\_1^2 - d u\_2^2 = \pm 1$, so $u\_1 \gg \sqrt{d}$, say. Thus the gap between the smallest unit $u \in \mathcal{O}\_K^\times$ and $1$ must be quite large.
This seems a particular quirk of fields whose unit group is 1-generated. For real cubic fields, whose unit group is 2-generated, there seems to be substantial possibility for cancellation and even possible for the smallest unit $u > 1$ to be arbitrarily close to $1$ as the fields vary.
My question is this: let $K$ be a cyclic cubic field (in particular, necessarily totally real) having discriminant $d\_K = c\_K^2$. Let $u\_K \in \mathcal{O}\_K^\times$ be the smallest (in terms of usual archimedean valuation) element satisfying $u\_K > 1$. Write
$$\displaystyle u\_K = 1 + \kappa\_K.$$
Can one effectively bound $\kappa\_K$ in terms of $d\_K$ (or equivalently, $c\_K$)?
| https://mathoverflow.net/users/10898 | Size of a generator of the unit group in a cyclic cubic field | Asking about a smallest unit bigger than $1$ in a unit group of rank greater than $1$ feels like the wrong question, sort of like asking for a smallest algebraic integer of absolute value greater than $1$ in a number field (inside $\mathbf C$) of degree greater than $1$. The ring of integers is discrete when you use all Archimedean embeddings, but not fewer Archimedean embeddings (e.g., $\mathbf Z[\sqrt{2}]$ is dense in $\mathbf R$ but its image in $\mathbf R^2$ using the Euclidean embedding is a lattice). Likewise, the unit group of that ring is discrete using the logarithm mapping to a hyperplane in $\mathbf R^{r\_1+r\_2}$.
Use the regulator as a measure of the size of the unit group. For a unit group of rank $1$ with a real embedding, taking logarithms shows that comparing the size of the smallest unit greater than $1$ with a power of the absolute value of the discriminant is like comparing the regulator of that unit group with a multiple of the logarithm of the absolute value of discriminant. For cubic fields of rank $1$, Artin showed its regulator $R$ and (negative) discriminant $D$ satisfy $R > \log(|D|/4 - 6)$ except for the unique cubic field of discriminant $-23$. For totally real cubic fields, the unit rank is $2$ and Cusick showed the regulator $R$ and (positive) discriminant $D$ for such fields satisfy $R > (1/16)(\log(D/4))^2$. See Theorem 5.8 [here](https://kconrad.math.uconn.edu/blurbs/gradnumthy/unittheorem.pdf).
| 10 | https://mathoverflow.net/users/3272 | 426311 | 173,024 |
https://mathoverflow.net/questions/426307 | 4 | *Note: Here we consider the Lebesgue measure on $[0, 1]$.*
Let $f\_n: [0, 1] \to [0, 1]$ be a sequence of measurable functions.
We say a measurable subset $E$ of $[0, 1]$ is a *condensation set* of the sequence $f\_n$ if there exists a subsequence $f\_{n\_k}$ and a function $f: [0, 1] \to [0, 1]$ (both depending on $E$) such that for almost every $x \in E$, we have $f\_{n\_k} (x) \to f(x)$.
**Question:**
Let $f\_n: [0, 1] \to [0, 1]$ be a sequence of measurable functions.
Suppose there exists some $\varepsilon > 0$ such that any measurable subset of $[0, 1]$ of measure less than or equal to $\varepsilon$ is a condensation set of $f\_n$.
Does it follow that there exists some measurable function $F$, and a subsequence of $f\_n$ that converges pointwise a.e. to $F$?
| https://mathoverflow.net/users/173490 | A “compactness theorem” for measurable functions | Counterexample. For each $n$ let $k\_n$ be the characteristic function of $[0,\frac{1}{2^n}] \cup [\frac{2}{2^n},\frac{3}{2^n}] \cup \cdots$. Next observe that there are only countably many subsets of $[0,1]$ of the form: a finite union of open intervals with rational endpoints, whose total length is $\frac{1}{10}$. Enumerate these sets as $(A\_n)$, and for each $n$ let $f\_n$ be the function which is constantly $1$ on $A\_n$ and equals $k\_n$ on $[0,1]\setminus A\_n$.
For any subset $E$ of $[0,1]$ whose measure is at most $\frac{1}{10}$, we can find a subsequence $(A\_{n\_k})$ of $(A\_n)$ such that $m(E\setminus A\_{n\_k}) \to 0$. Then $(f\_{n\_k}) \to 1$ pointwise a.e. on $E$.
Now suppose that some subsequence $(f\_{n\_j})$ converges pointwise a.e. on $[0,1]$ to some function $F$. By Egoroff, we can assume that $(f\_{n\_j}) \to F$ uniformly on $A$, for some measurable $A \subseteq [0,1]$ with $m(A) > \frac{9}{10}$. But this is impossible, because for any $n\_0$ the set $B$ on which $f\_{n\_0}$ is constantly zero has measure at least $\frac{4}{10}$ (namely, $\frac{1}{2}$ on which $k\_{n\_0}$ is zero, minus at most $\frac{1}{10}$ on $A\_{n\_0}$), hence has positive measure intersection with $A$, and no subsequence of $(f\_n)$ converges to zero on a set of positive measure (thus showing that for sufficiently large $n$, $f\_n$ is not constantly zero, and hence has uniform distance $1$ to $f\_{n\_0}$, on this intersection).
| 4 | https://mathoverflow.net/users/23141 | 426312 | 173,025 |
https://mathoverflow.net/questions/426284 | 2 | Let $\psi \in C^{\infty}\_{c}(\Omega)$ where $\Omega$ is a bounded smooth domain, and $\phi$ the solution to
\begin{equation\*}
-\Delta \phi =\psi, ~\phi|\_{\partial \Omega}=0.
\end{equation\*}
My question is how to get the following estimate :
$$\|\phi\|\_{C^1( \overline{\Omega})} \leq C \|\psi\|\_{L^{\infty}(\Omega)}.$$
The usual elliptic gradient estimate using the Bernstein method shows
$$\|\phi\|\_{C^1( \overline{\Omega})} \leq C \sup|\psi|+ C\sup|\phi|+C\sup |\nabla \psi|.$$
The problem comes from [Stable Solutions of Elliptic Partial Differential Equations](https://i.stack.imgur.com/DqdAe.png).
Another question is how to prove:
* Let $u$ be a $L^1$ weak solution of $-\Delta u= f(u)$ with zero boundary condition, if $f \in C^{\alpha}$ for some $\alpha \in (0, 1)$ and $f(u)\in L^p$ for some $p>\frac{n}{2}$, then $u$ is $C^2$, hence a classic solution.
for the definition of $L^1$ weak solution, see
[definition](https://i.stack.imgur.com/IQ5A2.png).
| https://mathoverflow.net/users/166368 | Gradient estimate and $L^1$ theory for the Laplace operator | One way to get both is to use the estimates $
\|\phi\|\_{W^{2,p}(\Omega)} \leq C\|\psi\|\_{L^p(\Omega)}$
which hold when $1<p<\infty$ with a constant $C=C(p,\Omega,n)$. Taking $p>n$ by Sobolev embedding $$\|\phi\|\_{C^1(\Omega)} \leq c\_1 \|\phi\|\_{W^{2,p}(\Omega)} \leq c\_2 \|\psi\|\_{L^p(\Omega)} \leq c\_3 \|\psi\|\_{L^\infty (\Omega)},$$ since $\Omega$ is bounded. The last question is similar: since $f(u) \in L^p$, then $u \in W^{2,p}$ hence in $C^\theta$ with $\theta=1-\frac{2p}{n}$ and then $f(u)$ is Holder continuous. The Schauder theory now yields that $u \in C^2$.
| 2 | https://mathoverflow.net/users/150653 | 426316 | 173,028 |
https://mathoverflow.net/questions/426302 | 44 | A relation $R$ is *implicitly definable* in a structure $M$ if there is a formula $\varphi(\dot R)$ in the first-order language of $M$ expanded to include relation $R$, such that $M\models\varphi(\dot R)$ only when $\dot R$ is interpreted as $R$ and not as any other relation. In other words, the relation $R$ has a first-order expressible property that only it has.
(Model theorists please note that this is implicit definability *in a model*, which is not the same as the notion used in [Beth's implicit definability theorem](https://en.wikipedia.org/wiki/Beth_definability).)
Implicit definability is a very weak form of second-order definability, one which involves no second-order quantifiers. Said this way, an implicitly definable relation $R$ is one that is definable in the full second-order Henkin structure of the model, but using a formula with only first-order quantifiers.
**Examples.** Here are some examples of relations that are implicitly definable in a structure, but not definable.
* The predicate $E$ for being even is implicitly definable in the language of arithmetic with successor, $\langle\mathbb{N},S,0\rangle$. It is implicitly defined by the property that $0$ is even and evenness alternates with successor: $$E0\wedge \forall x\ (Ex\leftrightarrow\neg ESx).$$ Meanwhile, being even is not explicitly definable in $\langle\mathbb{N},S,0\rangle$, as that theory admits elimination of quantifiers, and all definable sets are either finite or cofinite.
* Addition also is implicitly definable in that model, by the usual recursion $a+0=a$ and $a+(Sb)=S(a+b)$. But addition is not explicitly definable, again because of the elimination of quantifiers argument.
* Multiplication is implicitly definable from addition in the standard model of Presburger arithmetic $\langle\mathbb{N},+,0,1\rangle$. This is again because of the usual recursion, $a\cdot 0=0$, $a\cdot(b+1)=a\cdot b+a$. But it is not explicitly definable, because this theory admits a relative QE result down to the language with congruence mod $n$ for every $n$.
* First-order truth is implicitly definable in the standard model of arithmetic $\langle\mathbb{N},+,\cdot,0,1,<\rangle$. The Tarski recursion expresses properties of the truth predicate that completely determine it in the standard model, but by Tarski's theorem on the nondefinability of truth, this is not a definable predicate.
My question concerns iterated applications of implicit definability. We saw that addition was implicitly definable over successor, and multiplication is implicitly definable over addition, but I don't see any way to show that multiplication is implicitly definable over successor.
**Question.** Is multiplication implicitly definable in $\langle\mathbb{N},S,0\rangle$?
In other words, can we express a property of multiplication $a\cdot b=c$ in its relation to successor, which completely determines it in the standard model?
I expect the answer is **No**, but I don't know how to prove this.
**Update.** I wanted to mention a promising idea of Clemens Grabmayer for a **Yes** answer (see [his tweet](https://twitter.com/clegra/status/1545771851586146304)). The idea is that evidently addition is definable from multiplication and successor (as first proved in Julia Robinson's [thesis](https://www.jstor.org/stable/2266510), and more conveniently available in Boolos/Jeffrey, Computability & Logic, Sect. 21). We might hope to use this to form an implicit definition of multiplication from successor. Namely, multiplication will be an operation that obeys the usual recursion over addition, but replacing the instances of $+$ in this definition with the notion of addition defined from multiplication in this unusual way. What would remain to be shown is that there can't be a fake version of multiplication that provides a fake addition, with respect to which it fulfills the recursive definition of multiplication over addition.
| https://mathoverflow.net/users/1946 | Is multiplication implicitly definable from successor? | Contrary to my initial expectation, the answer is Yes.
This answer is based on the idea of Clemens Grabmayer, which makes the observation that addition $+$ is definable from multiplication $\cdot$ and successor.
The idea generalizes to the following:
**Theorem.** Suppose that relation $R$ is implicitly definable in model $M$, that $S$ is implicitly definable in the expansion $\langle M,R\rangle$, and that $R$ is explicitly definable in $\langle M,S\rangle$. Then $S$ is implicitly definable in $M$.
**Proof.**
Suppose that $R$ is the unique relation fulfilling sentence $\varphi(\dot R)$ in $M$, in the language expanded with predicate $\dot R$. Suppose $S$ is the unique relation fulfilling sentence $\psi(R,\dot S)$ in $\langle M,R\rangle$. And suppose that $R$ is definable by formula $\theta(x,S)$ in $\langle M,S\rangle$, in that $Rx\leftrightarrow\theta(x,S)$.
Let $\Phi(\dot S)$ be the sentence asserting:
* $\varphi(\theta(x,\dot S))$, that is, the relation defined by $\theta(x,\dot S)$ fulfills property $\varphi$, and
* $\psi(\theta(x,\dot S),\dot S)$ holds, that is, the assertion $\psi(\dot R,\dot S)$ holds where $\dot R$ is interpreted by the relation defined by $\theta(x,\dot S)$.
I claim that this is an implicit definition of $S$ in $M$. The reason is that whatever relation interpretation is given to $\dot S$, it will have the property that the relation extracted from it via $\theta(x,\dot S)$ will have to be $R$, since it fulfills the implicit definition of $R$ given by $\varphi$. And further, since $\Phi$ asserts that $\psi$ is fulfilled by $\dot S$ relative to that relation, it follows that $\dot S$ must be $S$. $\Box$
The corollary is that:
**Corollary.** Multiplication is implicitly definable from successor.
**Proof.** Addition is implicitly definable in $\langle\mathbb{N},S,0\rangle$, and multiplication is implicictly definable over addition $\langle\mathbb{N},S,0,+\rangle$, and by the Boolos/Jeffrey observation, addition is explicitly definable from multiplication and successor. So we are in the case of the theorem. $\Box$
A more striking instance might be:
**Corollary.** First-order arithmetic truth for the standard model of arithmetic $\langle\mathbb{N},+,\cdot,0,1<\rangle$ is implicitly definable just from successor $\langle\mathbb{N},S,0\rangle$.
**Proof.** I intend to use the trinary truth predicate $\text{Tr}(\varphi,x,y,z)$, holding when $\mathbb{N}\models\varphi[x,y,z]$. This truth predicate is uniquely characterized on the standard model $\mathbb{N}$ by fulfilling the Tarski recursion, and so it is implicitly definable in $\langle\mathbb{N},+,\cdot\rangle$. But both addition and multiplication are definable from the truth predicate (this is why we use the trinary version, since with just successor we don't initially have any coding, but once we get $+$ and $\times$, then the usual coding kicks in), and they themselves are implicitly definable from successor. So by the theorem, truth is implicitly definable from successor. $\Box$
And one can of course iterate this by forming the predicate for truth-about-truth, and truth-about-truth-about-truth and so on, proceeding transfinitely up the hierarchy for quite some way.
But lastly, let me mention that the theorem falls short of proving that the property of being implicitly-definable-over is transitive. That seems to be false in light of counterexamples discussed in the comments.
| 25 | https://mathoverflow.net/users/1946 | 426322 | 173,029 |
https://mathoverflow.net/questions/426272 | 0 | Set $\pi=\frac{1}{4}(\delta\_{(1,0)}++\delta\_{(1,3)}+\delta\_{(1,1)}+\delta\_{(2,2)})$. Suppose that $\pi\in\Pi(\mu,\nu)$.
How to get the disintegration of $\pi$ with respect to $\mu$?
| https://mathoverflow.net/users/168083 | How to compute the unique disintegration w.r.t. the first coordinate? | By disintegration is meant that $\pi$ can by broken down as
$$
\pi(dx,dy) =\mu(dx)\pi\_{x}(dy),
$$
with $\pi\_x$ a Borel measurable family of sub-probability measures on $X$.
In your example:
$\pi\_{x\_1}={1\over 2}(\delta\_{0}+\delta\_{1})$ if $x\_1=1$; $=\delta\_{2}$ if $x\_1=2$; and $=0$ otherwise.
Then you have $W\_1(\pi\_{x\_1},\nu) = 1/2$ if $x\_1=1$, and $W\_1(\pi\_{x\_1},\nu)=1$ if $x\_1=2$.
| 1 | https://mathoverflow.net/users/42851 | 426324 | 173,030 |
https://mathoverflow.net/questions/426150 | 2 | This question has been bugging me for a while, I have an answer that is working sufficiently for the program I'm using, but it is a tad slow, and let's say imprecise. It is not an overtly difficult question, and for that I assume someone smarter than me has already found a better explanation.
To begin, we take a value $y \in \mathbb{C}$ such that $|\log(y)| < 1$. Then we look at the exponential:
$$
b^z = e^{\log(y) z/y}\\
$$
The value $b = y^{1/y}$ is inside what we call the Shell-Thron region, in the tetration circles. This is the area in which:
$$
\lim\_{n\to\infty}\exp\_b^{\circ n}(0) \,\,\text{converges}
$$
Or, where the iterated exponential converges.
Now, to highlight the question, we can start with $1 < b < e^{1/e}$. The exponential $b^z$ has two fixed points on the real line. There is an attracting fixed point $1 < y < e$ and a repelling fixed point $\mu > e$. Both of which satisfy:
$$
e^{\log(y)/y} = e^{\log(\mu)/\mu}\\
$$
This relationship carries over for all $|\log(y)| < 1$, not just when it is real valued. From here, my question is simple.
Can anyone describe the holomorphic function:
$$
\begin{align}
f(y): \{y \in \mathbb{C} : \, |\log(y)| < 1\} &\to \{\mu \in \mathbb{C} :\,|\log(\mu)| > 1\}\\
\frac{\log f(y)}{f(y)} &= \frac{\log(y)}{y}\\
\end{align}
$$
And when restricted to real values:
$$
f(y): (1,e) \to (e , \infty)\\
$$
The manner I am solving this currently is a tad involved. But if $h(z) = \frac{\log(z) y}{\log(y)} = \log\_b(z)$, and $h^{\circ n}$ are the iterates of $h$:
$$
f(y) = \lim\_{n\to\infty} h^{\circ n}(e)\\
$$
This formula works on the real line, and moderately well in the complex plane. But I'm not certain if it works for all $|\log(y)| < 1$, there may be issues. It essentially runs off the basis that $e$ is in the Julia set of $b^z$, and/or in the attracting basin of the fixed point $\mu$ for the logarithm $\log\_b(z)$. I don't know if this is true universally though, just for a large part of the domain in question.
This has been bugging me, and any help is appreciated. In essence, does anyone know any good ways of computing/describing/constructing $f$, other than the way I mentioned?
| https://mathoverflow.net/users/133882 | Finding the repelling fixed point of an exponential, knowing only its attracting one | Okay, so I found the answer to this problem! It's a tad different than I thought, so I'll give the run down. Special thanks to @RolandBacher for getting me to think about $e$ as a branching point. The answer was simpler than I thought once I dug through it.
Let
$$
g(y) = y^{1/y}\\
$$
Then, there exists two branches to the inverse of this function, call them $h\_1$ and $h\_2$. The first satisfies:
$$
h\_1 : (1,e^{1/e}) \to (1, e)\\
$$
And the second satisfies:
$$
h\_2 : (1,e^{1/e}) \to (e,\infty)\\
$$
This is just standard calculus when looking at the self-root function. Thereby, our function $f$ is just the analytic continuation of the expression:
$$
f(y) = h\_2(y^{1/y})\\
$$
This solves the problem entirely. I apologize for asking a question when the answer was right in front of me, lmao. Thanks for the help, Roland!
| 2 | https://mathoverflow.net/users/133882 | 426343 | 173,033 |
https://mathoverflow.net/questions/426326 | 3 | Consider a simple random walk $$\mathcal{X}\_t= \sum\_{n<t} X\_n,$$ where $P(X\_n=1)= P(X\_n=-1)= 1/2.$
If I put an extra condition that excludes cases with more than 5 consecutive +1, or -1 in the sum:
For every $n$, between 1 and t-4:
$$|X\_n+ X\_{n+1}+ X\_{n+2}+ X\_{n+3}+ X\_{n+4}|< 5.$$
Can I still expect that $\mathcal{X}\_t \ll \sqrt{t}$ almost surley?
If yes, how can we prove this?
| https://mathoverflow.net/users/422944 | Simple random walk with an extra condition | Let $Y\_n=(X\_n,X\_{n+1},X\_{n+2},X\_{n+3},X\_{n+4})$. The sequence $\{Y\_n\}$ is an aperiodic irreducible Markov chain on 30 states (vectors of $\pm 1$ that are not all $1$ or all $-1$). Its distribution is known as the Parry measure on a shift of finite type, See [4] or [5]. For the aperiodicity, it is important that the constraint is on sums of five-tuples and not on pairs. One could easily reduce the number of states to 16, since looking at 4-tuples also yields a Markov chain, but that takes a moments' thought and is not important, so let's stay with 30 states.
The invariance of the constraint to flipping the sign of all the variables implies that the unique stationary distribution $\pi$ is even:
$$\pi(x\_1,\ldots,x\_5)=\pi(-x\_1,\ldots,-x\_5)$$
This is not really necessary, but we note that the Markov chain $\{Y\_n\}$ is reversible, since the constraint is stable to flipping the direction of time.
The partial sums $$S\_t= \sum\_{n<t} X\_n$$ (note the change of notation)
form a mean zero additive functional in this Markov chain, so one fancy way to conclude and obtain a central limit theorem is to use the Kipnis-Varadhan Theorem [1], see [2] for a friendly exposition.
A more elementary approach is to observe that any finite irreducible aperiodic Markov chain exhibits
exponential decay of correlations (see e.g. [3]) so $|E(X\_m X\_{m+k})| \le C\alpha^k$
for some $\alpha<1$ and $C<\infty$.
Therefore
$$ES\_t^2\le 2\sum\_{m<t} \sum\_{k\ge 0} |E(X\_m X\_{m+k})| \le 2Ct/(1-\alpha)=C't.$$
Thus
$$E|S\_t| \le \sqrt{C't} \,.$$
[1] Kipnis, C., Varadhan, S. R. S. (1986) Central limit theorem for additive functionals of reversible markov processes. Commun. Math. Phys. 104 1-19.
[2] [https://www.math.arizona.edu/~sethuram/588/lecture8.pdf](https://www.math.arizona.edu/%7Esethuram/588/lecture8.pdf) Theorem 3.1
[3] <https://www.yuval-peres-books.com/markov-chains-and-mixing-times/> <https://pages.uoregon.edu/dlevin/MARKOV/mcmt2e.pdf> Theorem 4.9
[4] <https://dmg.tuwien.ac.at/aofa15/slides/Marcovici_AofA.pdf> Slides 12-33
[5] <https://personalpages.manchester.ac.uk/staff/Charles.Walkden/ergodic-theory/lecture15.pdf> Section 15.5
| 4 | https://mathoverflow.net/users/7691 | 426345 | 173,034 |
https://mathoverflow.net/questions/426047 | 13 | Recently Professor Peter Cameron posed a [number theory problem which is related to graphs of groups](https://cameroncounts.wordpress.com/2022/06/17/id-like-to-see-this-solved/). The problem is:
Problem:
>
> Let $n$ be a positive integer. Show that there exist subsets $A\_1, A\_2, …,A\_n$ of $\{1,2,…n\}$ with the properties
>
>
> 1. $|A\_i| = \varphi(i)$ for $i = 1,2,…,n$;
> 2. if $\mathrm{lcm}(i,j) ≤ n$, then $A\_i$ and $A\_j$ are disjoint, where $\mathrm{lcm}$ denotes the least common multiple.
>
>
>
Professor Cameron gave [some motivations for this question](https://cameroncounts.wordpress.com/2022/07/05/why-id-like-to-see-this-solved/) in his blog.
I checked it for a lot of numbers and the results seems to be true. Also, it seems that for some class of numbers there are patterns. For example if $n$ is prime and the statement is true, then it seems that it is true for the multiples of it.
**My question: Is it sufficient that we prove this problem for prime numbers?**
The first version of the paper published. Congratulate to Veronica Phan, a medical student who knows mathematics well. I think she is the first medical student which has common paper with professor Cameron.
[Paper](https://arxiv.org/abs/2207.07156)
| https://mathoverflow.net/users/19885 | A nice problem by Peter Cameron on subsets of $\{1,\dots,n\}$ | Let $F\_n$ be a set of all irreducible fraction $\frac{p}{q}$ such that $0<\frac{p}{q}\leq 1,1\leq p,q\leq n$ and for $i\in \{1,...,n\}$, $D\_i$ be subset of $F\_n$ which contain all irreducible fraction of the form $\frac{k}{i}$. We have $D\_i$ are pairwise disjoint and $|D\_i|=\varphi(i)$. So we want a function $f:F\_n\rightarrow \{1,...,n \}$ then we can take $A\_i=f(D\_i)$.
We construct $f$ as follow: if $\frac{m-1}{n}<\frac{p}{q}\leq \frac{m}{n}$ then $f(\frac{p}{q})=m$. If $f(\frac{p}{q})=f(\frac{p'}{q'})=m$ then $|\frac{p}{q}-\frac{p'}{q'}|<\frac{m}{n}-\frac{m-1}{n}=\frac{1}{n}$ (#).
-If $f(\frac{k}{i})=f(\frac{l}{i}), i\leq n$ then by (#) we have $|\frac{k-l}{i}|<\frac{1}{n}$, because $i\leq n$ we must have $k-l=0\Rightarrow k=l$. So the restriction of $f$ to $D\_i$ is injective for all $i\in \{1,...,n\}$, therefore $|A\_i|=|D\_i|=\varphi(i)$.
-If $m\in A\_i\cap A\_j,i\neq j$, then there exist irreducible fractions $\frac{k}{i}\in D\_i,\frac{l}{j}\in D\_j$ such that $f(\frac{k}{i})=f(\frac{l}{j})=m$, by (#) we have $\frac{P}{Q}=|\frac{k}{i}-\frac{l}{j}|<\frac{1}{n}$ with $\frac{P}{Q}$ is irreducible. Because $i\neq j$ so $\frac{k}{i}\neq \frac{l}{j}\Rightarrow \frac{P}{Q}>0\Rightarrow P\geq 1$.
We have $\frac{P}{Q}=\frac{|k\frac{lcm(i,j)}{i}-l\frac{lcm(i,j)}{j}|}{lcm(i,j)}>0\Rightarrow Q\leq lcm(i,j)$ (because $\frac{P}{Q}$ is irreducible) and $\frac{P}{Q}<\frac{1}{n}\Rightarrow Q>Pn\geq n$ so $lcm(i,j)>n$. So if $lcm(i,j)\leq n$, $A\_i,A\_j$ must be disjoint.
So $A\_1,...,A\_n$ are subsets of $\{1,...,n\}$ we want.
| 33 | https://mathoverflow.net/users/432274 | 426347 | 173,035 |
https://mathoverflow.net/questions/426341 | 17 | Recently, I encountered this problem:
*"Given a sequence of positive number $(x\_n)$ such that for all $n$,
$$x\_{n+1}=x\_n+\frac{n}{x\_1+x\_2+\cdots+x\_n}$$
Find the limit $\lim\_{n \rightarrow \infty} \frac{x\_n}{\sqrt{n}}.$"*
As pointed out in the comment section of the following topic [[1](https://math.stackexchange.com/questions/4488666/find-lim-n-to-infty-fracx-n-sqrtn-where-x-n1-x-n-fracnx-1x)], this problem is indeed open. Out of curiosity, I wonder if this problems has a research-level meaning? For example, does it stem from an interesting system dynamic question?
I'm aware this question might not suitable for Mathoverflow forum, but aside from this site, I don't think there is a more suitable place to ask this question
| https://mathoverflow.net/users/168269 | "Find $\lim_{n \to \infty}\frac{x_n}{\sqrt{n}}$ where $x_{n+1}=x_n+\frac{n}{x_1+x_2+\cdots+x_n}$" -where does this problem come from? | The OP asks where this recursion relation might appear in a research context. It appears as a discretization of the [Emden–Fowler nonlinear differential equation](http://eqworld.ipmnet.ru/en/solutions/ode/ode0302.pdf),
$$f''(t)=t^{p}[f(t)]^q,$$
for $p=1$, $q=-1$. A particular solution is
$$f(t)=\lambda t^{(p+2)/(1-q)},\;\;\lambda=\left[\frac{(p+2)(p+q+1)}{(q-1)^2}\right]^{1/(q-1)}.$$
One readily checks that the asymptotic limit $x\_n\rightarrow \sqrt{3n}$ for $x\_n=f'(n)$ is obtained for $p=1$, $q=-1$.
---
*This connection to the Emden-Fowler equation motivates the following*
**Conjecture:** The recursion relation
$$x\_{n+1}=x\_n+n^{p}(x\_1+x\_2+\cdots+x\_n)^q$$
has for $p\geq 0$, $q\leq 0$, $p+q>-1$, the limit
$$\lim\_{n\rightarrow\infty}n^{(1-q)/(1+p+q)}x\_n=\frac{p+2}{1-q}\left[\frac{(p+2)(p+q+1)}{(q-1)^2}\right]^{1/(q-1)}.$$
For $q=-1$ this is stated as an open problem on page 11 of [Sequences of Real Numbers](https://link.springer.com/chapter/10.1007/978-3-030-77139-3_1), by Sîntămărian & Furdui.
Here are two numerical tests (blue is the recursion for $x\_n$, gold the conjectured large-$n$ asymptotic):
| 12 | https://mathoverflow.net/users/11260 | 426351 | 173,037 |
https://mathoverflow.net/questions/426269 | 1 | This question is pretty much [this](https://math.stackexchange.com/questions/4477619/a-map-from-the-symmetric-algebra-generated-by-the-first-cohomology-to-the-cohomo) question stated in slightly different way. All projective spaces are complex ones. Let's assume $X$ is a connected CW complex. We have a natural map in the following form:
$$f: Sym^{\infty}(Hom^0(X, \mathbb{P}^{\infty}))\rightarrow Sym^{\infty}(Hom(X, \mathbb{P}^{\infty}))\rightarrow Hom(X, Sym^{\infty}(\mathbb{P}^{\infty}))$$
Superscript zero denotes the connected component corresponding to zero. Since $\mathbb{P}^{\infty}\cong K(\mathbb{Z}, 2)$ so $\pi\_0(Hom(X, \mathbb{P}^{\infty}))=H^2(X, \mathbb{Z})$.
By Dold-Thom $\pi\_i(Sym^{\infty}(Hom^0(X, \mathbb{P}^{\infty})))=H\_i(Hom^0(X, \mathbb{P}^{\infty})))$. There is a $H$-space structure on $Hom^0(X, \mathbb{P}^{\infty})$ induced from $\mathbb{P}^{\infty}$. This $H$-space structure turns $H\_i(Hom^0(X, \mathbb{P}^{\infty})))$ into a graded algebra. Consequently $\pi\_i(Sym^{\infty}(Hom^0(X, \mathbb{P}^{\infty})))$ is a graded algebra.
* Does $f$ map this algebra product to the cup product on the right side?
Note that homotopy groups of the right side is as the following:
We have $Sym^{\infty}(\mathbb{P}^{\infty})=\prod\_{i=1}^{\infty}K(2i, \mathbb{Z})$ so $\pi\_i(Hom(X, Sym^{\infty}(\mathbb{P}^{\infty})))=\bigoplus H^{\text{even}}(X, \mathbb{Z})$ if $i$ is even and $\bigoplus H^{\text{odd}}(X, \mathbb{Z})$ if $i$ is odd. So there is a natural cup product structure on the right.
Let's assume as Tom Goodwillie mentions in the comments that the homotopy equivalence between $Sym^{\infty}(\mathbb{P}^{\infty})$ and $\prod\_{i=1}^{\infty}K(2i, \mathbb{Z})$ is induced by mapping $\mathbb{P}^{\infty}$ to $\prod\_{i=1}^{\infty}K(2i, \mathbb{Z})$ where each map $\mathbb{P}^{\infty}\rightarrow K(2i, \mathbb{Z})$ is the map in homotopic to the one corresponding to the generator of $H^{2i}(\mathbb{P}^{\infty}, \mathbb{Z})$. Now the map from $\mathbb{P}^{\infty}$ to $\prod\_{i=1}^{\infty}K(2i, \mathbb{Z})$ is extended to a map from $Sym^{\infty}(\mathbb{P}^{\infty})$ to $\prod\_{i=1}^{\infty}K(2i, \mathbb{Z})$ by the $H$-space operation. So the final map is a map of $H$-spaces.
| https://mathoverflow.net/users/127776 | Dold-Thom and infinite symmetric power of an $H$-space | No, not quite.
Let's set the stage like this:
If $G$ is a connected $H$-space then $Sym^\infty G$ is a ring space, so that $Hom(X,Sym^\infty G)$ is also a ring space and $\pi\_\ast Hom(X,Sym^\infty G)$ is a graded ring.
Also, $Hom^0(X,G)$ is a connected $H$-space, so $Sym^\infty Hom^0(X,G)$ is a ring space and $\pi\_\ast Sym^\infty Hom^0(X,G)$ is a graded ring.
The canonical map $Sym^\infty Hom^0(X,G)\to Hom(X,Sym^\infty G)$ is a ring space map. Therefore $\pi\_\ast Sym^\infty Hom^0(X,G)\to \pi\_\ast Hom(X,Sym^\infty G)$ is a map of graded rings.
Now what does this have to do with cup products?
I would say that the graded ring $\pi\_\ast Hom(X,Sym^\infty G)$ is the cup product ring for cohomology of $X$ with coefficients in the graded ring $\pi\_\ast (Sym^\infty G)=H\_\ast (G)$.
That is, the multiplication
$$
H^i(X;H\_{2j}(\mathbb P^\infty))\times H^k(X;H\_{2\ell}(\mathbb P^\infty))\to H^{i+k}(X;H\_{2j+2\ell}(\mathbb P^\infty))
$$
is a cup product. But it's a cup product based on the multiplication
$$
(\*)\ \ \ H\_{2j} \mathbb P^\infty\times H\_{2\ell}\mathbb P^\infty\to H\_{2j+2\ell}\mathbb P^\infty
$$
(induced by the $H$-space structure). This cannot be identified with the cup product
$$
H^i(X;\mathbb Z)\times H^k(X;\mathbb Z)\to H^{i+k}(X;\mathbb Z),
$$
because the map $(\ast)$ above does not take generators to generators; there is a factor of ${j+\ell}\choose {j}$.
| 2 | https://mathoverflow.net/users/6666 | 426361 | 173,040 |
https://mathoverflow.net/questions/426350 | 1 | They introduce a new correlation. For $\pi\in \Pi(\mu,\nu)$ the set of coupling of two probability measures $\mu$ and $\nu$ on a Polish space $(X,d)$. The author introduces a plugin estimator.
Question: How to understand $\hat{\pi}$?
---
For the estimator of the denominator, I use the same notation as the answer.
\begin{equation\*}
D(\hat{\pi})=\sum\_{y,z}d(y,z)\hat{\pi}\_2(\{y\})\hat{\pi}\_2(\{z\})
=\sum\_{y,z}d(y,z)\frac{1}{N^2}\sum 1[\phi(Y\_n)\in\{y\}]\sum 1[\phi(Y\_m)\in\{z\}]
\end{equation\*}
Expand the product of these two summation,
$$
=\frac{1}{N^2}\sum\_{y,z}d(y,z)(\sum\_{i=1}^N 1[\phi(Y\_i)\in\{y\}]1[\phi(Y\_i)\in\{z\}]+\sum\_{i\neq j}1[\phi(Y\_i)\in\{y\}]1[\phi(Y\_j)\in\{z\}] )
$$
| https://mathoverflow.net/users/168083 | How to get the estimator? | $\newcommand{\de}{\delta}$We have
\begin{equation\*}
W(\pi):=\frac{N(\pi)}{D(\pi)}, \tag{1}\label{1}
\end{equation\*}
where
\begin{equation\*}
N(\pi):=\int W\_1(\pi\_{x\_1},\nu)\mu(dx\_1)=\int W\_1(\pi\_{x\_1},\pi\_2)\pi\_1(dx\_1),
\end{equation\*}
\begin{equation\*}
D(\pi):=\int d(y,z)\nu(dy)\nu(dz)=\int d(y,z)\pi\_2(dy)\pi\_2(dz),
\end{equation\*}
$\pi\_{x\_1}$ is the conditional distribution of $\eta$ given $\xi=x\_1$ if the joint distribution of a pair $(\xi,\eta)$ of random variables is the measure $\pi$ with marginals $\mu=\pi\_1$ and $\nu=\pi\_2$.
The standard notation $\de\_a$ is for the Dirac delta measure supported on the singleton set $\{a\}$. So,
\begin{equation\*}
\hat{\pi}:=\frac1N\,\sum\_{i=1}^N \de\_{(\phi(X\_i),\,\phi(Y\_i))}
\end{equation\*}
is the (random) probability measure such that
\begin{equation\*}
\hat{\pi}(A)=\frac1N\,\sum\_{i=1}^N 1((\phi(X\_i),\,\phi(Y\_i))\in A) \\
=\frac1N\,\sum\_{i=1}^N |\{i\in\{1,\dots,N\}\colon (\phi(X\_i),\,\phi(Y\_i))\in A\}|
\end{equation\*}
for any Borel set $A\in[0,1]^2$. Here $1(\mathcal A)$ is the indicator of an assertion $\mathcal A$, so that $1(\mathcal A)=1$ if $\mathcal A$ is true and $1(\mathcal A)=0$ if $\mathcal A$ is false.
Then for the first marginal $\hat\pi\_1$ of $\hat\pi$ and each $x\_1\in X=[0,1]$ we have
\begin{equation\*}
\hat\pi\_1(\{x\_1\})=\frac1N\,\sum\_{i=1}^N 1(\phi(X\_i)\in \{x\_1\}) \\
=\frac1N\,\sum\_{i=1}^N 1(X\_i\in\phi^{-1}(\{x\_1\}))
=\frac1N\,\sum\_{i=1}^N 1(X\_i\in G)
\end{equation\*}
for $G=\phi^{-1}(\{x\_1\})$,
and for the conditional distribution $\hat\pi\_{x\_1}$ of $\eta$ given $\xi=x\_1$ if the joint distribution of a pair $(\xi,\eta)$ is $\hat\pi$ we have
\begin{equation\*}
\hat\pi\_{x\_1}(B)=\frac{\hat\pi(\{x\_1\}\times B)}{\hat\pi(\{x\_1\}\times X)}, \tag{2}\label{2}
\end{equation\*}
where $B$ is any Borel subset of $X=[0,1]$; if the denominator of the ratio in \eqref{2} is $0$, let $\hat\pi\_{x\_1}$ be an arbitrary probability measure. Next,
\begin{equation\*}
\hat\pi(\{x\_1\}\times B)=\frac1N\,\sum\_{i=1}^N 1(\phi(X\_i)=x\_1,\,\phi(Y\_i)\in B) \\
=\frac1N\,\sum\_{i=1}^N 1(X\_i\in\phi^{-1}(\{x\_1\}),\,\phi(Y\_i)\in B).
\end{equation\*}
So, for $G=\phi^{-1}(\{x\_1\})$,
\begin{equation\*}
\hat\pi\_{x\_1}(B)=\hat\pi\_G(B):=\frac{\frac1N\,\sum\_{i=1}^N 1(X\_i\in G,\,\phi(Y\_i)\in B)}
{\frac1N\,\sum\_{i=1}^N 1(X\_i\in G)}. \tag{3}\label{3}
\end{equation\*}
So,
\begin{equation\*}
\begin{aligned}
N(\hat\pi)&=\int W\_1(\hat\pi\_{x\_1},\hat\pi\_2)\hat\pi\_1(dx\_1) \\
&=\sum\_{x\_1} W\_1(\hat\pi\_{x\_1},\hat\pi\_2)\hat\pi\_1(\{x\_1\}) \\
& =\sum\_{G\in \Phi} W\_1(\hat\pi\_G,\hat\pi\_2)\frac1N\,\sum\_{i=1}^N 1(X\_i\in G) \\
& =\sum\_{G\in \Phi} \frac{1}{N}|i\in\{1,\dots,N\}\ \text{s.t.}\ X\_i\in G|\,
W\_1(\hat{\pi}\_G,\hat{\pi}\_2).
\end{aligned}
\end{equation\*}
Also, if $d$ is the standard metric over $[0,1]$, then, as you have it in your "new edition",
\begin{equation}
\begin{aligned}
D(\hat\pi)&=\int|y-z|\hat\pi\_2(dz)\hat\pi\_2(dz) \\
&=\sum\_{y,z}|y-z|\hat\pi\_2(\{y\})\hat\pi\_2(\{z\}) \\
&=\sum\_{y,z}|y-z|\frac{1}{N^2}\sum\_{n,m=1}^N 1(\phi(Y\_n)=y,\phi(Y\_m)=z) \\
&=\frac{1}{N^2}\sum\_{n,m=1}^N \sum\_{y,z}|y-z|1(\phi(Y\_n)=y,\phi(Y\_m)=z) \\
&=\frac{1}{N^2}\sum\_{n,m=1}^N |\phi(Y\_n)-\phi(Y\_m)|.
\end{aligned}
\end{equation}
So,
\begin{equation\*}
W(\hat\pi):=\frac{N(\hat\pi)}{D(\hat\pi)}
=\frac{\sum\_{G\in \Phi} \frac{1}{N}|i\in\{1,\dots,N\}\ \text{s.t.}\ X\_i\in G|\, W(\hat{\pi}\_G, \hat{\pi}\_2)}{\frac{1}{N^2}\sum\_{n,m=1}^N|\phi(Y\_n)-\phi(Y\_m)|}
\end{equation\*}
is a plug-in estimator of $W(\pi)$.
| 1 | https://mathoverflow.net/users/36721 | 426367 | 173,043 |
https://mathoverflow.net/questions/425348 | 5 | Let $X$ be a compact metric space, $P\_X$ the set of Borel probability measures on $X$, and $K\_X$ the set of non-empty closed subsets of $X$. I will define the "topology of weak+Hausdorff convergence" to be the metrisable topology on $P\_X$ whose convergence is given by
\begin{align\*}
& \mu\_n \overset{\text{w+H}}\to \mu \\
&\ \ \ \, \Longleftrightarrow \\
(\mu\_n \to \mu \text{ weakly, and } \, \mathrm{supp}&\,\mu\_n \to \mathrm{supp}\,\mu \text{ in Hausdorff distance})\text{.}
\end{align\*}
This convergence is indeed metrisable, with a metric being given by
$$ d\_{\text{w+H}}(\mu,\nu) = d\_\text{w}(\mu,\nu) + d\_\text{H}(\mathrm{supp}\,\mu,\mathrm{supp}\,\nu) $$
where $d\_\text{w}$ is any metrisation of the topology of weak convergence, and $d\_\text{H}$ is Hausdorff distance.
[This is a fairly natural topology to consider when one is interested in probability measures simultaneously at the "probabilistic" level of describing the distribution of likelihoods and at the "deterministic" level of prescribing what is and is not possible.]
Although $d\_\text{w}$ and $d\_\text{H}$ each generate compact topologies, it is easy to see that $d\_{\text{w+H}}$ is not even complete: just take $\mu\_n$ to be a sequence of fully supported probability measures converging weakly to a Dirac mass; then $\mu\_n$ is $d\_{\text{w+H}}$-Cauchy but not (weak+Hausdorff)-convergent as $n \to \infty$. Still, it would be nice at least to be able to say that the topology of weak+Hausdorff is *Polish*, meaning that it is both separable and *admits* a complete metrisation. (So, for example, even though the Euclidean metric on $\mathbb{R} \setminus \mathbb{Q}$ is not complete, it is known that the topology on $\mathbb{R} \setminus \mathbb{Q}$ generated by the Euclidean metric *is* Polish.)
>
> Is the topology of weak+Hausdorff convergence Polish?
>
>
>
As in the answer to [Comparison of several topologies for probability measures](https://mathoverflow.net/questions/332006/), separability should be fairly easy: e.g. let $\mathcal{K}$ be a Hausdorff-dense countable subset of $K\_X$, and for each $C \in \mathcal{K}$ let $\mathcal{P}\_C$ be a weakly dense countable subset of the set of probability measures supported on $C$; then $\bigcup\_{C \in \mathcal{K}} \mathcal{P}\_C$ should be (weak+Hausdorff)-dense.
So the remaining question is that of complete metrisability: Is there a complete metric on $P\_X$ topologically equivalent to $d\_{\text{w+H}}$?
---
**My first guess:** The $\infty$-Wasserstein metric
$$ W\_\infty(\mu,\nu) = \inf\{\mathbb{P}\text{-}\mathrm{ess}\sup d(\boldsymbol{\cdot},\boldsymbol{\cdot}) \, : \, \mathbb{P} \text{ is a coupling of } \mu \text{ and } \nu \} $$
generates the topology of weak-Hausdorff convergence (perhaps generalising the answer to [The infinity Wasserstein distance $W\_\infty$ and the weak topology](https://mathoverflow.net/questions/404619/)) and is complete?
| https://mathoverflow.net/users/15570 | Is the topology of weak+Hausdorff convergence Polish? | The set $MS=\{(\mu,K)\in P\_X\times K\_X:\mathrm{supp}(\mu)=K\}$ is of type $G\_\delta$ in $P\_X\times K\_X$ and hence the weak+Hausdorff topology on $P\_X$ is Polish.
Indeed, fix any countable base $\{U\_n\}\_{n\in\mathbb N}$ of the topology of $X$ and observe that $$MS=\bigcap\_{n\in\mathbb N}\{(\mu, K)\in P\_X\times K\_X:\mu(U\_n)>0\;\Leftrightarrow\;U\_n\cap K\ne\emptyset\}.$$
So, it remains to show that for every $n\in\mathbb N$ the set
$$MS\_n:=\{(\mu, K)\in P\_X\times K\_X:\mu(U\_n)>0\;\Leftrightarrow\;U\_n\cap K\ne\emptyset\}$$is of type $G\_\delta$ in $P\_X\times K\_X$.
Observe that $(P\_X\times K\_X)\setminus MS\_n=MS\_n'\cup MS\_n''$ where
$$MS\_n':=\{(\mu,K)\in P\_X\times K\_X: \mu(U\_n)>0\;\wedge\;U\_n\cap K=\emptyset\}$$ and
$$MS\_n'':=\{(\mu,K)\in P\_X\times K\_X: \mu(U\_n)=0\;\wedge\;U\_n\cap K\ne\emptyset\}.$$
The known properties of the topologies on the spaces $P\_X$ and $K\_X$ ensure that the set $\{\mu\in P\_X:\mu(U\_n)>0\}$ is open in $P\_X$ and the set $\{K\in K\_X:K\cap U\_n\ne \emptyset\}$ is open in $K\_X$. Then $MS\_n'$ and $M\_s''$ are intersections of open and closed sets, so are of type $F\_\sigma$ in $P\_X\times K\_X$. Then $MS\_n$ is of type $G\_\delta$ in $P\_X\times K\_X$ and so is the set $MS=\bigcap\_{n\in\mathbb N}MS\_n$.
| 2 | https://mathoverflow.net/users/61536 | 426371 | 173,044 |
https://mathoverflow.net/questions/426356 | 2 | Given $C^1([a, b])$ functions $f\_n$ that converge to a continuous real-valued function $f\_n \to f$ on a closed interval $[a, b] \subset \mathbb R$, suppose
$$
\int\_a^b |f\_n'(x)|^{1 + \epsilon} dx < M
$$
for all $n > 0$ and fixed $M < \infty$. For a small $\epsilon$, does this implies
1. $f$ is absolutely continuous
2. $f$ is Lipschitz
Can we tell more about this function?
I have $\epsilon = 0$ does not hold for Cantor function and some approximation $f\_n$.
| https://mathoverflow.net/users/92099 | Does that exponent of (absolute value of derivative) is constrained implies Lipschitz continuity? | Q1: Yes. Since $f'\_n$ is bounded in $L^p$, a suitable subsequence will converge $f\_n'\to g$ weakly. Then also
$$
f(x) =\lim f\_n(x) = \lim \left( f\_n(a)+\int\_a^x f'\_n(t)\, dt \right) = f(a) + \int\_a^x g(t)\, dt ,
$$
so $f$ is absolutely continuous (with derivative in $L^p$).
Q2: No. For example $f(x)=|x|^{\alpha}$ has a derivative $f'\in L^p$ for $p<1/(1-\alpha)$, so can easily be approximated by functions with bounded $L^p$ derivatives.
| 4 | https://mathoverflow.net/users/48839 | 426372 | 173,045 |
https://mathoverflow.net/questions/426334 | 21 | An [earlier question by Joel David Hamkins](https://mathoverflow.net/questions/426302/is-multiplication-implicitly-definable-from-successor) asked whether multiplication is implicitly definable in the structure $(\mathbb{N},S)$ of the naturals with successor. Here $R$ is implicitly definable if there is a formula $\phi(\dot{R})$ that is satisfied only if $R = \dot{R}$. Joel's goal was to find a natural counterexample to implicit definability being transitive: $+$ is implicitly definable in terms of $S$, and $\cdot$ is implicitly definable in terms of $+$, and he hoped that $\cdot$ would not be implicitly definable in terms of $S$. [Alas, it is](https://mathoverflow.net/a/426322/22930), so we need an alternate example.
**Question:** Is the set of primes implicitly definable in $(\mathbb{N},S)$?
My guess is no, roughly because the set of primes has arbitrarily large gaps with pseudorandom behavior that cannot be captured by $S$ alone. In particular, I would conjecture that given any formula $\phi(P)$ on a predicate $P$ that is satisfied if $P = \textrm{Primes}$, there is some prime $p$ such that $\phi(P - \{p\})$ is true.
If primes is indeed not implicitly definable, it provides a nice counterexample to transitivity. From the [earlier question](https://mathoverflow.net/questions/426302/is-multiplication-implicitly-definable-from-successor) $\cdot$ is implicitly definable from $S$, and primes is explicitly definable from $\cdot$.
| https://mathoverflow.net/users/22930 | Is the set of primes implicitly definable from successor? | The set of primes is not implicitly definable in $(\mathbb{N},S)$. This is immediately implied by following:
**Theorem.** A unary predicate $P$ on $\mathbb{N}$ is implicitly definable in $(\mathbb{N},S)$ iff $P$ is a finally periodic set of naturals.
**Proof.** Suppose $P$ is finally periodic, i.e. there are naturals $n$ and $m>0$ such that $\forall x(x\ge n \to x\in P\mathrel{\leftrightarrow} x+m\in P)$. Then $P$ is implicitly definable by the following formula $\varphi(X)$:
$$\forall x\Big(X(x)\mathrel{\leftrightarrow}\exists y,z\big( S^{n+m}(z)=x\land S^m(y)=x\land X(y)\big) \lor \bigvee\limits\_{k<n+m\text{ and }k\in P}x=S^k(0)\Big)$$
On the other hand if $P$ is implicitly definable in $(\mathbb{N},S)$, then it is definable in $(\mathbb{N},S)$ by a monadic second-order formula. And by the classical result of Büchi a set of naturals has a monadic second-order definition in $(\mathbb{N},S)$ iff it is finally periodic (combination of results from [1] and [2]).**QED**
[1]Büchi, J. Richard. "Symposium on Decision Problems: On a Decision Method in Restricted Second Order Arithmetic." Studies in Logic and the Foundations of Mathematics. Vol. 44. Elsevier, 1966. 1-11, doi:[10.1016/S0049-237X(09)70564-6](https://doi.org/10.1016/S0049-237X%2809%2970564-6).
[2]Büchi, J. R. “Weak Second Order Arithmetic and Finite Automata”, Zeitschrift für Math. Log. und Grundl. der Math., 6 (1960), pp. 66–92, doi:[10.1002/malq.19600060105](https://doi.org/10.1002/malq.19600060105).
| 20 | https://mathoverflow.net/users/36385 | 426382 | 173,050 |
https://mathoverflow.net/questions/426374 | 1 | Take the $y$-axis and a set of $n$ curves starting from $y$-axis, labelled as $\mathcal{C}:=\{C\_1,C\_2,...,C\_n\}$. These curves fulfill the following conditions:
1. The curves all have a starting point somewhere on the $y$-axis and no two curves share the same starting point.
2. At least one curve is non-$x$-monotone.
3. Any curve intersects with any other curve at most once.
Now, I am thinking of a way to represent the set of curves $\mathcal{C}$ with a set of $x$-monotone curves, say $\mathcal{C}'=\{C\_{1}',C\_{2}',...,C\_{n}'\}$ where the following conditions are fulfilled:
1. All curves in $\mathcal{C}'$ have a starting point on the $y$-axis and no two curves share the same starting point.
2. All curves in $\mathcal{C}'$ are $x$-monotone.
3. Any curve intersects with any other curve at most once.
So I am looking for the following: would there exist a representation of $\mathcal{C}$ with $\mathcal{C}'$ in some way, or perhaps would there exist some example of $\mathcal{C}$ that is not representable by $\mathcal{C}'$? In particular, the order of the intersections of the curves do not matter, just simply the pairs of intersections in $\mathcal{C}$ and $\mathcal{C}'$ must remain the same.
Intuitively, I feel like there should exist some counter example for this, but I am struggling with coming up with such, but it also would seem difficult to prove the given statement.
It seems that maybe the proof of such could be argued with some type of a construction, for instance, labeling the curves from bottom to top as $C\_1,....,C\_n$ and have $C\_1',...,C\_n'$ correspond to these curves based on the number of intersection points on each curve. This construction seems unnecessarily complicated and I feel like there could be another way to go about this.
Also something I thought about was obtaining a general construction for this argument through something similar to a wiring diagram (see page 7 here: <https://arxiv.org/pdf/1410.2350.pdf>). However, I don't think this necessarily works since wiring diagrams are not necessarily $x$-monotone (although some of them can be).
| https://mathoverflow.net/users/485561 | Representation of $x$-non-monotone curves with one intersection each by $x$-monotone curves | **This answer, as pointed out by Jan in the comment, is incorrect, as the definitions slightly differ. I leave it here as it contains useful pointers.**
If I understood your definitions correctly, your $\mathcal{C}$ would be called an outerstring graph, and your $\mathcal{C}'$ a double outerstring graph. There are outerstring graphs that are not double outerstring graphs, in fact, even the order of the growth rates of these two families are different. For some literature and a nice summary of results, see the first two pages here: [Alexandre Rok and Bartosz Walczak: Outerstring graphs are χ-bounded](https://arxiv.org/abs/1312.1559).
| 2 | https://mathoverflow.net/users/955 | 426383 | 173,051 |
https://mathoverflow.net/questions/426332 | 0 | Let $r\in [0,1]\setminus\mathbb{Q}$, let $\mathbb{N}$ denote the set of non-negative integers and let $\mathbb{N}\_+=\mathbb{N}\setminus\{0\}$. For $n\in\mathbb{N}\_+$ let $$\alpha\_r(n)=\min\{\big|r-\frac{m}{n}\big|: m\in\mathbb{N}\}$$ be the best approximation of $r$ that can be obtained using $n$ as the denominator.
For $k\in\mathbb{N}\_+$ we say that $r\in[0,1]\setminus\mathbb{Q}$ is *$k$-good* if there is $c\in\mathbb{N}\_+$ as well as a "starting index" $N\_0\in \mathbb{N}\_+$ such that $$\alpha\_r(n) < c \cdot \frac{1}{n^k} \;\text{ for all } n\geq N\_0.$$
Note that for all irrational $r$ in the unit interval, we have $\alpha\_r(n) < \frac{1}{n}$ for all $n\in\mathbb{N}\_+$, so every such $r$ is $1$-good.
For which integers $k \geq 2$ is there a $k$-good irrational number $r\in[0,1]\setminus\mathbb{Q}$? Is there even such an $r$ that is simultaneously $k$-good for all integers $k\geq 2$? (Only the first question needs to be answered for acceptance.)
| https://mathoverflow.net/users/8628 | Measuring the quality of real approximation | This certainly can't happen for any $k > 1$. (There is no reason to assume that $k$ is an integer rather than real number). In fact, it can't even happen if you just restrict the $n$ to sufficiently large powers of two.
Suppose you know that for every sufficiently large integer $i$ there is an integer $m$ such that
$$\left|r - \frac{m\_i}{2^i}\right| < \frac{c}{2^{ik}}.$$
Compare this appoximation with the approximation for $n = 2^{i+1}$. We have
$$\left|r - \frac{m\_{i+1}}{2^{i+1}}\right| < \frac{c}{2^{(i+1)k}}$$
By the triangle inequality
$$\left| \frac{m\_i}{2^i} - \frac{m\_{i+1}}{2^{i+1}} \right| < \frac{c}{2^{ik}} + \frac{c}{2^{(i+1)k}}.$$
If $k > 1$, then for large enough $i$ (depending only on $c$) the RHS will be strictly less than $1/2^{i+1}$. On the other hand the left hand side is equal to
$$\left| \frac{2m\_{i} - m\_{i+1}}{2^{i+1}} \right|$$
and this is at least $1/2^{i+1}$ if it is non-zero. Hence the only way the inequalities can hold for all sufficiently large $i$ is if eventually we have $2 m\_i = m\_{i+1}$ for all $i \ge i\_0$. But then our sequence of rational approximations to $r$ are all equal to the (rational) number $m\_{i\_0}/2^{i\_0}$, which implies that $r$ is rational.
| 3 | https://mathoverflow.net/users/485571 | 426386 | 173,054 |
https://mathoverflow.net/questions/424400 | 0 | All,
Let $S(x,y,t)$ be a variable function in $x$, $y$, and $t$. After
performing Reynold averaging over area $\frac{1}{A}\int S(x,y,t) dA$, could $S$ still be a function in $x$, and $y$?
[Equations (1-18)](https://www.ecmwf.int/sites/default/files/elibrary/2017/Atmospheric_moist_convection.pdf) of dry statics energy ($s$) and specific humidity ($q$) find the gradient of area-averaged variables ($s$ and $q$) indicating that that Reynolds-averaged variables might be still a function in space. Any idea how that might be the case?
Thanks.
| https://mathoverflow.net/users/483883 | Could variable be still function in x and y after performing Reynolds averaging over area | Based on the comment by @CarloBeenakker, Coarse-grain averaging could be performed on a subset of the domain (spatial or temporal) so that the averaged value still varies with the independent dimension. I just wanted to document some resources (below) that might be helpful for others.
The classical Reynold average in which the average is performed on the whole data is based on the assumption that the data is stationary, yet if the data is not statistically stationary then the coarse-grain average might do the job by calculating the average at each subset of the data (recall running-average). I am not sure if the coarse-grain averaging is a subtype of the Reynold average.
The coarse-grain technique is discussed in the following textbooks
* Storm and cloud dynamics by Cotton and Anthes [Chapter 3](https://www.elsevier.com/books/storm-and-cloud-dynamics/cotton/978-0-12-088542-8)
* Mesoscale Meteorological modeling by Pielke Sr [Chapter 4](https://www.elsevier.com/books/mesoscale-meteorological-modeling/pielke-sr/978-0-12-385237-3)
| 0 | https://mathoverflow.net/users/483883 | 426387 | 173,055 |
https://mathoverflow.net/questions/426331 | 16 | Let $F\_n$ be a free group of rank $n$. Say that $w \in F\_n$ is *non-reversible* if there does not exist any $f \in \text{Aut}(F\_n)$ such that $f(w) = w^{-1}$.
**Original Question.** Intuitively, I expect that most elements of $F\_n$ are non-reversible. However, I have trouble coming up with examples. Does anyone know any ways to produce them? To avoid just focusing on low-rank situations (where there might be tricks), I'd like to find examples that are not contained in any proper free factor of $F\_n$.
A related question replaces $F\_n$ with the fundamental group of a surface and $\text{Aut}(F\_n)$ with the mapping class group. Here again it's not as easy as one would like to find examples, but in our paper [here](https://arxiv.org/abs/1807.00833) Margalit and I gave a geometric criterion that allowed us to find them. However, just like in the free group setting I expect that there are far more examples than our paper constructs, and that in some sense they should be "generic".
---
In this revision, I want to make a number of comments and pose some additional questions inspired in part by the comments on the original version.
1. Benjamin Steinberg pointed out that the answer to [this](https://math.stackexchange.com/questions/539898/group-element-not-taken-to-its-inverse-by-any-automorphism) math.stackexchange question claims to give examples for $n=2$. It is not surprising that this special case is more tractable than the case $n \geq 3$. Indeed, a classical theorem of Nielsen shows that $\text{Out}(F\_2) \cong \text{GL}(2,\mathbb{Z})$, so it is often fairly easy to understand automorphisms of $F\_2$ directly. Automorphisms of $F\_n$ for $n \geq 3$ are much more complicated.
2. It is natural to try to promote examples for $n=2$ to higher $n$ via the standard inclusion $\iota\colon F\_2 \hookrightarrow F\_n$. As YCor pointed out, if an element $w \in F\_2$ cannot be inverted by an endomorphism of $F\_2$ (much stronger than being non-reversible; let me call this property *non-endoreversible*), then $\iota(w)$ in $F\_n$ is also non-endoreversible (and hence non-reversible). To see this, assume that for some $w \in F\_2$ the endomorphism $\phi\colon F\_n \rightarrow F\_n$ takes $\iota(w)$ to $\iota(w)^{-1}$. Letting $r\colon F\_n \rightarrow F\_2$ be the retraction, we can then define $\phi'\colon F\_2 \rightarrow F\_2$ via the formula $\phi'(x)=r(\phi(\iota(x)))$. We then have $\phi'(w) = r(\phi(\iota(w))) = r(\iota(w)^{-1}) = \iota(w)^{-1}$.
3. Carl-Fredrik Nyberg Brodda pointed out that there are algorithms to determine whether or not a word $w' \in F\_n$ is the image of a word $w \in F\_n$ under an endomorphism of $F\_n$. This would give a finite procedure to verify that a specific element of $F\_n$ is non-endoreversible. In particular, if you apply this for $F\_2$ you can presumably get non-endoreversible elements of $F\_n$ for all $n$.
4. However, note that by design this will not be able to answer the harder question from the original post of finding such elements that do not lie in any proper free factor of $F\_n$.
5. Let me close by making a conjecture that makes precise the statement "Intuitively, I expect that most elements of $F\_n$ are non-reversible" from the original version of the question. This actually does not hold for the non-endoreversible elements (see the comments for a proof), so let’s focus on just the non-reversible elements. Fix some $n \geq 2$, and let $G\_k \subset F\_n$ be all elements of word length at most $k$ and $N\_k \subset F\_n$ be the set of all non-reversible elements of length at most $k$. I then conjecture that
$$\text{lim}\_{k \mapsto \infty} \frac{|N\_k|}{|G\_k|} = 1.$$
| https://mathoverflow.net/users/317 | Elements of a free group that can't be inverted by automorphisms | Proposition 8.7 of <https://arxiv.org/pdf/math/0303386.pdf#page107> says that the nonreversible elements are exponentially generic. This means you should get your limit in 5 with fast convergence. Note that they look only at cyclically reduced words which makes more sense since you can clearly reduce to conjugate elements for this problem. Here is the journal version [Generic properties of Whitehead's algorithm and isomorphism rigidity of random one-relator groups, Ilya Kapovich, Paul Schupp and Vladimir Shpilrain, Pacific J. Math](https://msp.org/pjm/2006/223-1/pjm-v223-n1-p06-p.pdf)
| 9 | https://mathoverflow.net/users/15934 | 426393 | 173,058 |
https://mathoverflow.net/questions/426264 | 5 | Let $\mathsf{RO}(X)$ stand for the collection of regular open subsets of a topological space $X$ and let $d(X)$ be its density. It is well-known (see Theorem~3.3 of Hodel's chapter in the Handbook) that every regular space satisfies the inequality $|\mathsf{RO}(X)|\leq 2^{d(X)}$. What is an example of an infinite $T\_3$ (regular + $T\_1$) space such that $|\mathsf{RO}(X)|<2^{d(X)}$?
I think a relevant observation regarding this question is the following. From Pierce's result on the cardinality of complete Boolean algebras (see [here](https://www.jstor.org/stable/2033324)), it can easily be deduced that if $X$ is infinite and $T\_2$, then $|\mathsf{RO}(X)|\geq \omega\_1$. Thus, $\textsf{CH}$ implies that if $X$ is infinite, $T\_3$ and separable, then $|\mathsf{RO}(X)| = 2^{d(X)}$. For this reason, if $X$ satisfies the desired characteristics of the question above, then necessarily $d(X)\geq \omega\_1$.
| https://mathoverflow.net/users/146942 | $|\mathsf{RO}(X)|$ vs. $2^{d(X)}$ for $T_3$ spaces | In fact, Joseph Van Name's argument gives a ZFC example. For $f \in \{0,1\}^{[0,1]}$ let $supp(f) = \{x \in [0,1] : f(x) = 1\}$. Let $X$ be the $\Sigma$-product given by $X = \{f \in \{0,1\}^{[0,1]}: |supp(f)| \leq \aleph\_0\}$ (with the subspace topology from the product). Then $X$ satisfies the countable chain condition because it is a dense subset of a separable space. Its weight is $\frak c$ so by Joseph Van Name's comment, $|RO(X)| \leq {\frak c}$. The density of $X$ is also $\frak c$ because given a collection of fewer than $\frak c$ elements of $X$, there is a subset of $[0,1]$ having cardinality smaller than $\frak c$ that contains all of their supports, and therefore is clearly not dense. Therefore, $2^{d(X)} = 2^{\frak c} > {\frak c} \geq |RO(X)|$.
One further comment is that if a space $Y$ has a collection $\mathcal U$ of $d(Y)$ pairwise disjoint non-empty open subsets, then $2^{d(Y)} = |RO(Y)|$ because in this case, for every ${\mathcal V} \subset {\mathcal U}$, the interior of the closure of $\cup {\mathcal V}$ gives a regular open set, and different choices of $\mathcal V$ give different regular open sets. (This is the reason that there is no separable example, even in ZFC.)
| 6 | https://mathoverflow.net/users/89233 | 426395 | 173,059 |
https://mathoverflow.net/questions/426407 | 10 | In a [previous question](https://mathoverflow.net/questions/425966/examples-of-locally-hyperbolic-groups), I asked about hyperbolic groups in which every finitely generated subgroup is hyperbolic. I am now curious about the reverse question: what are some examples of hyperbolic groups containing finitely generated non-hyperbolic groups?
Any non-coherent hyperbolic group will give an example (e.g. via the Rips construction). Do there exist examples of coherent hyperbolic groups containing f.g. non-hyperbolic subgroups?
| https://mathoverflow.net/users/371866 | Examples of hyperbolic groups with non-hyperbolic subgroups | It is an open problem to find a coherent hyperbolic group with a finitely generated, non-hyperbolic subgroup. See [Wise's survey article](https://www.degruyter.com/document/doi/10.1515/9780691185897-014/html?lang=en) for the state of the art on coherent groups.
Wise, Daniel T. (3-MGL)
**An invitation to coherent groups.**
*What's next?—the mathematical legacy of William P. Thurston*, 326–414,
Ann. of Math. Stud., 205, *Princeton Univ. Press, Princeton, NJ, 2020*.
I suspect this is listed as a question there, but if not, it's very much in that spirit.
The theorem of Gersten that I mentioned [in answer to your last question](https://mathoverflow.net/a/425992/1463) showed that there is no 2-dimensional example. All known coherent 3-dimensional groups are essentially 3-manifold groups. [Sam Nead's answer to that question](https://mathoverflow.net/a/425967/1463) explained why finitely generated subgroups of hyperbolic 3-manifold groups are also hyperbolic.
As Matt Zaremsky mentions in comments, the key examples of finitely presented non-hyperbolic subgroups of hyperbolic groups are given by [the original example of Noel Brady](https://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/S0024610799007644) and the recent work of [Italiano--Martelli--Migliorini](https://arxiv.org/abs/2105.14795). There are also related constructions by [Fujiwara](https://arxiv.org/abs/2106.08549), [Groves--Manning](https://arxiv.org/abs/2205.11290), [Kropholler](https://arxiv.org/abs/1808.09505), [Llosa Isenrich--Martelli--Py](https://arxiv.org/abs/2112.06531) and [Lodha](https://arxiv.org/abs/1403.6716).
It would be very surprising if any of these examples were to turn out to be coherent. In fact, it is reasonable to conjecture that these examples are all incoherent, but I believe it is unknown in each case. Proving that Brady's example (say) is incoherent wouldn't answer your question, but would be a nice thing to do.
| 14 | https://mathoverflow.net/users/1463 | 426413 | 173,064 |
https://mathoverflow.net/questions/426300 | 3 | Let $:\Sigma \to GL(d, \mathbb{R})$ be a continuous matrix cocycle over a topologically mixing subshift of finite type $(\Sigma, T)$. We denote by $\Sigma\_n$ the set of addmisible words with the length $n$. For any $n\in \mathbb{N},$ and $I \in \Sigma\_n$, we define
$$\|A(I)\|:=\sup\_{x\in [I]}\|A^{n}(x)\|.$$
We can also define the topological pressure for the potential $\log \|A\|$ as follows:
$$P(\log \|A\|):=\lim\_{n\to \infty} \frac{1}{n}\log \sum\_{|I|=n}\|A(I)\|.$$
Is it true that $P(\log \|A\|)$ can also be defined as follows:
$$P(\log \|A\|)=\lim\_{n\to \infty} \frac{1}{n}\log \sum\_{|I|\leq n}\|A(I)\|?$$
| https://mathoverflow.net/users/127839 | the definition of the topological pressure for matrices | Yes. This is a consequence of the following simple exercise about exponential growth: if $a\_n\geq 0$ is any sequence such that $P = \lim \frac 1n \log a\_n$ exists, then $\frac 1n \log \sum\_{k=1}^n a\_k \to P$ as well. (Apply this with $a\_n = \sum\_{|I|=n} \|A(I)\|$.)
To prove the exercise, observe that $\liminf \frac 1n \log \sum\_{k=0}^n a\_k \geq P$ is immediate, so it is enough to prove that $\limsup \frac 1n \log \sum\_{k=0}^n a\_k \leq P$. To do this, observe that for every $P' > P$ there is $C>0$ such that $a\_k \leq C e^{kP'}$ for all $k$, so $$\sum\_{k=0}^n a\_k \leq \sum\_{k=0}^n C e^{kP'} = C\frac{e^{(n+1)P'} - 1}{e^{P'} - 1}.$$
| 1 | https://mathoverflow.net/users/5701 | 426414 | 173,065 |
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