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https://mathoverflow.net/questions/431642 | 43 | Let $M$, $N$ be connected nondiscrete compact smooth manifolds. Can the ring of continuous functions on $M$ be isomorphic to the ring of smooth functions on $N$?
| https://mathoverflow.net/users/148161 | Do rings of smooth functions differ from rings of continuous functions? | No. In both the smooth function ring and the continuous function ring a maximal ideal $\frak m$ consists of the functions vanishing at some point. In the smooth case $\frak m/\frak m^2$ is the cotangent space of the manifold at that point, while in the continuous case $\frak m^2=\frak m$.
| 60 | https://mathoverflow.net/users/6666 | 431644 | 174,744 |
https://mathoverflow.net/questions/431625 | 10 | A finitely generated group $G$ *algebraically fibers* if there is an epimorphism $G\to\mathbb{Z}$ with finitely generated kernel. Since this kernel is finitely generated, we can ask whether \*it\* algebraically fibers. If it does, then we can keep going, etc etc etc. In any examples I can think of, this process always terminates, either by hitting a group that does not map onto $\mathbb{Z}$ at all (for example a perfect group), or that does but never with finitely generated kernel (for example a free group). But I don't see any reason in general that it should necessarily terminate, and I'm curious to find an example where it doesn't.
So, here is the concrete question: Does there exist an infinite sequence $G\_1,G\_2,G\_3,\dots$ of finitely generated groups such that for each $i$ there exists a short exact sequence $1\to G\_{i+1} \to G\_i \to \mathbb{Z} \to 1$?
It would suffice for example to find a residually solvable, non-solvable group whose derived series consists entirely of finitely generated groups. Another thing that would suffice is finding a finitely generated group $G$ and an automorphism $\phi$ of $G$ such that $G\cong G\rtimes\_\phi \mathbb{Z}$.
One reason to be interested in such an example would be a sort of domino effect of BNS-invariants: in such a sequence we would conclude that each $\Sigma^1(G\_i)$ is non-empty, as "revealed" by $G\_{i+1}$.
| https://mathoverflow.net/users/164670 | Iterated algebraic fibering | Let $G\_1$ be the wreath product of $\mathbb{Z}$ with $\mathbb{Z}$. There is a surjective homomorphism $G\_1\rightarrow \mathbb{Z}$ whose kernel $G\_2$ is isomorphic to $G\_1$. This gives a sequence of groups in which each $G\_i$ is isomorphic to $G\_1$.
If we write $G\_1:=\langle a,b: [a,a^{b^i}]\,\forall i\rangle$, the homomorphism is defined by $a\mapsto 1$ and $b\mapsto 0$. The elements $a':=aba^{-1}b^{-1}$ and $b$ generate the kernel and satisfy the same relations as $a$ and $b$.
| 5 | https://mathoverflow.net/users/124004 | 431660 | 174,749 |
https://mathoverflow.net/questions/430897 | 9 | A $\mathrm{C}^\*$-algebra $\mathcal{A}\subset B(\mathsf{H})$ is a norm-closed, self-adjoint subalgebra of bounded operators on a Hilbert space. If we then take a unital self-adjoint (possibly closed) subspace (not subalgebra) we have an *operator system* $\mathcal{O}\subset \mathcal{A}$.
(I am aware there are abstract definitions of both $\mathrm{C}^\*$-algebras and operator systems).
My question is:
>
> *Why are operator systems studied?*
>
>
>
I presume that they are not studied for their own sake. I have tried to find the answer to this question by skimming papers, but to no avail. All I can distill is that perhaps they can help study $\mathrm{C}^\*$-algebras, either their $\mathrm{C}^\*$-envelopes, or perhaps their ambient $\mathrm{C}^\*$-algebras $\mathcal{A}\supset \mathcal{O}$. Perhaps they are studied to learn about completely positive maps, the appropriate morphisms of operator systems.
I would have thought they are not coming from a physics motivation: from my point of view the algebra structure for observables makes more sense than the vector structure. I have asked this question in that context [here](https://physics.stackexchange.com/questions/728432/operator-systems-vs-operator-algebras). If the motivation is well addressed in some paper, I would be happy to be pointed to such a reference.
There is one MO user who I think I could email this question to directly but that would be rather presumptive on my part.
| https://mathoverflow.net/users/35482 | Why operator systems? | Collating some comments, it appears that one reason why operator systems are studied is because they are *useful*:
* Uri Bader mentions [a paper of Kalantar & Kennedy](https://arxiv.org/pdf/1405.4359.pdf) which makes extensive use of Hamana's theory for injective envelopes for operator systems equipped with a group action.
* მამუკა ჯიბლაძე explains that conditional complete C\*-algebras may be recovered from their underling injective operator system, and so the underlying operator system may be used to study the algebra.
* t.c. points to [a paper of Connes & van Suijlekom](https://arxiv.org/pdf/2004.14115.pdf), where an operator systems approach to noncommutative geometry is explored, and t.c. suggests that this approach has physical applications in mind.
Of a slightly different bent, Narutaka OZAWA alludes to the question "*what are the appropriate morphisms for a category of C$^\*$-algebras?*", and suggests that a usual choice is too rigid, something which is not the case for the category of operator systems.
| 4 | https://mathoverflow.net/users/35482 | 431662 | 174,750 |
https://mathoverflow.net/questions/431665 | 5 | The well-known *matrix determinant lemma* states that for an invertible square matrix $A$ and column vectors $u,v$ one has
$$
\det(A + uv^T) = \det(A)(1 + v^T A^{-1} u).
$$
Is there any analogous formula in case we are adding to $A$ a matrix which is not rank-one, but still has some special structure? The case I have in mind is when $A,B$ are square matrices, $I$ is the identity matrix, $v$ is a vector and we are dealing with tensor (or Kronecker, if you wish) products:
$$
\det(A \otimes I + I \otimes B + (v^T v) \otimes I) = ?
$$
Eventually I'd like to work with more components of the tensor product, but for simplicity let me just state the question for two.
| https://mathoverflow.net/users/2192 | Matrix determinant lemma for non-rank-one updates | If $A,B$ are square matrices, then $A\otimes I+I\otimes B$ is commonly known as the Kronecker sum of $A$ and $B$. If $A$ has eigenvalues $\mu\_1,\dots,\mu\_m$ and $B$ has eigenvalues $\nu\_1,\dots,\nu\_n$, then $A\otimes I-I\otimes B$ has eigenvalues
$\mu\_i-\nu\_j$ for $1\leq i\leq m,1\leq j\leq n$. Therefore, $\det(A\otimes I-I\otimes B)=\text{res}(\chi(A),\chi(B))$ (which denotes the resultant of the characteristic polynomial of $A$ and $B$).
Therefore (Peter Taylor originally mentioned that we can do something like this), by the matrix determinant lemma, we have
$$\text{Det}(A\otimes I+I\otimes B+pq^T\otimes I+I\otimes uv^T)$$
$$=\text{res}\big(\chi(A+pq^T),\chi(B+uv^T))=\text{res}(\det(xI-A-pq^T),\det(xI-B-uv^T)\big)$$
$$=\text{res}\big(\det(xI-A)\cdot(1-q^T(xI-A)^{-1}p),\det(xI-B)\cdot(1-v^T(xI-B)^{-1}u)\big)$$
$$=\text{res}\big(\chi(A)\cdot(1-q^T(xI-A)^{-1}p),\chi(B)\cdot(1-v^T(xI-B)^{-1}u)\big).$$
| 4 | https://mathoverflow.net/users/22277 | 431679 | 174,753 |
https://mathoverflow.net/questions/431676 | 11 | Let $X=(X\_1,\ldots,X\_n)$ be an iid sequence of random variables, and let $\nu$ be a *uniformly random* integer in the range $1,\ldots,n$. Then $\xi\_\nu$ is a random entry of $X$. Is it always true that after deleting such a random entry from $X$, the remaining sequence is still iid?
Due to the uniform randomness of the index $\nu$, it is tempting to answer that the leftover sequence always remains iid. This is correct, if we also assume that $\nu$ is chosen *independently* of $X$.
What happens, however, if no such independence is assumed? For example, consider a 0-1 valued iid sequence. Remove a random 1, chosen uniformly at random among all 1s. If there is no 1, then remove a random 0. In this situation, the position of the removed entry is still uniformly random, but not independent of the original sequence. Can such a removal destroy the iid property in the remaining sequence?
| https://mathoverflow.net/users/101180 | Can deleting a random entry from an iid sequence destroy the iid property? | The independence will be then in general lost. E.g., let $X\_1,\dots,X\_n$ be independent random variables each uniformly distributed on $[0,1]$. Let $M:=\max(X\_1,\dots,X\_n)=X\_\nu$, so that $\nu$ is uniformly distributed on $[n]:=\{1,\dots,n\}$. Let $(Y\_1,\dots,Y\_{n-1})$ be the leftover sequence, after the removal of $X\_\nu$. Then, conditionally on $M$, the $Y\_i$'s are iid uniformly distributed on $[0,M]$.
So, for $n\ge2$ and $i\in[2]$ we have
$E(Y\_i|M)=M/2$ and $E(Y\_1Y\_2|M)=E(Y\_1|M)E(Y\_2|M)=(M/2)^2$. So,
$$EY\_1Y\_2=E(M/2)^2>(EM/2)^2=EY\_1\,EY\_2,$$
with the inequality taking place because $Var\,M>0$.
Thus, $Y\_1$ and $Y\_2$ are not independent.
---
Intuitively, it can be expected that the $Y\_i$'s are positively dependent. Indeed, if $M$ is small, then all $Y\_i$'s will be small. So, if $Y\_1$ turns out to be small, a reason for that may be that $M$ is small, and then $Y\_2$ will be small. Thus, the smallness of $Y\_1$ seems to make $Y\_2$ tend to be small.
| 19 | https://mathoverflow.net/users/36721 | 431681 | 174,754 |
https://mathoverflow.net/questions/431680 | 3 | Let $X$ be a nodal maximally non-factorial Fano threefold. If there is $1$-node and no other singularities, they by the work of Kuznetsov-Shinder <https://arxiv.org/pdf/2207.06477.pdf> Lemma 6.18, $D^b(X)$ contains a categorical ordinary double point subcategory $\mathcal{P}\subset D^b(X)$, which is generated by a $\mathbb{P}^\infty$ object $P$, such that the category $\mathcal{P}^{\perp}$ is a smooth category.
I have a question
Let $X$ be a nodal maximally non-factorial Fano threefold with no other singularities, if $D^b(X)$ contains two categorical ordinary double point subcategories $\mathcal{P}$ and $\mathcal{P}'$ such that the $\mathbb{P}^{\infty}$ objects generating $\mathcal{P}$ and $\mathcal{P}'$ are $P$ and $P'$ respectively and $P\not\cong P'$, can we say that $X$ has at least two node?
Assume not, then $X$ has only one node, then by Lemma above, $D^b(X)=\langle\mathcal{P}^{\perp},\mathcal{P}\rangle$ such that $\mathcal{P}^{\perp}$ is a smooth category, if we can further show that $\mathcal{P}'\subset\mathcal{P}^{\perp}$, then this is a contradiction, then $X$ has at least two nodes.
But I am not sure if $\mathcal{P}'\subset\mathcal{P}^{\perp}$ is true.
| https://mathoverflow.net/users/41650 | Semi-orthogonal decomposition for maximally non-factorial Fano threefolds | If you assume that $\mathcal{P}$ and $\mathcal{P'}$ are semiorthogonal, this is true. The easiest way to see this is by looking at the singularity category. If $X$ has one node, (the idempotent completion of) the singularity category of $X$ is the category of $\mathbb{Z}/2$-graded vector spaces, and if
$$
D^b(X) = \langle \mathcal{P}, \mathcal{P}', \dots \rangle,
$$
the singularity category contains a direct sum of two copies of the category of $\mathbb{Z}/2$-graded vector spaces, a contradiction.
| 6 | https://mathoverflow.net/users/4428 | 431692 | 174,757 |
https://mathoverflow.net/questions/431616 | 3 | In the context of Banach space theory, what is the correct terminology: *extremally* disconnected or *extremely* disconnected. Looking through the internet I have met using both [extremely](https://math.stackexchange.com/questions/1808481/proof-that-ck-is-a-grothendieck-space-for-k-an-extremely-disconnected-comp) and [extremally](https://math.stackexchange.com/q/128388) disconnected. Or is there a difference between those two notions?
| https://mathoverflow.net/users/61536 | Extremely disconnected or extremally disconnected? | The correct word is **extremally** (which is not in standard English) and is indeed often mistakenly corrected to the more common "extremely" — possibly, in recent case, with the unfortunate help of automatic correctors. Extremely would mean "very very", while "extremally" purports to mean, as far as I understand "in an extremal way".
An MSN search *anywhere* = "extremally disconnected" vs "extremely disconnected" with "year $<$ 1963" yields 24 papers vs 1 paper (note that this mostly detects the spelling of the MSN review, although when misspelled it fitted the spelling of the paper in a few cases I checked). Without the year restriction (say, with year $<2021$, to fix things) it yields 600 papers vs 64 papers. In Google Scholar, the only few occurrences before 1950 are with "extremally disconnected" (I tried to open [a 1943 paper by Hewitt](https://doi.org/10.1215/S0012-7094-43-01029-4) but instead noticed along the way that Duke Math J. still likes to get our money out of 80 years old papers).
| 4 | https://mathoverflow.net/users/14094 | 431701 | 174,759 |
https://mathoverflow.net/questions/431556 | 20 | I can prove the following result.
>
> **Theorem 1.** Let $f\_n:\mathbb{R}^n\to \mathbb{R}$ be a sequence of convex functions
> that converges almost everywhere to a function $f:\mathbb{R}^n\to\mathbb{R}$.
> Then $f$ is convex in the
> sense that there is a convex function $F:\mathbb{R}^n\to\mathbb{R}$
> such that $F=f$ a.e.
>
>
>
I am sure, it must be known.
>
> **Question 1.** Where can I find a proof?
>
>
>
**Edit.** I added one more question (see below) and the answers published back in October 2022 apply to Question 1.
Theorem 1 implies the following fact that I could not find in any book:
>
> **Theorem 2.** If $f\in W^{2,1}\_{\rm loc}(\mathbb{R}^n)$ (Sobolev space) satisfies
> $v^T D^2f(x)v\geq 0$ for almost all $x$ and all $v$, then $f$ is convex.
>
>
>
Indeed, if $f\_\epsilon$ is an approximation by convolution (with a positive mollifier), then $v^T D^2f\_\epsilon(x)v\geq 0$ and hence $f\_\epsilon$ is convex since it is smooth. Now $f\_\epsilon\to f$ a.e. along with Theorem 1 implies that $f$ is convex.
>
> **Question 2.** Has Theorem 2 been written somewhere?
>
>
>
This is a simple result, but I am quite surprised I could not find it in any textbook.
| https://mathoverflow.net/users/121665 | Convergence of convex functions | It follows from Theorem 10.8 in
R. Tyrrell Rockafellar. Convex analysis. Princeton Mathematical Series, No.
28. Princeton University Press, Princeton, N.J., 1970.
This theorem essentially says the following: if $f\_n$ are convex functions on an open domain $\Omega\subset\mathbb{R}^n$ that converge pointwise (to a finite value) on a *dense* subset of $\Omega$, then the limit exists for every point of $\Omega$, this limit is a convex function, and the convergence is uniform on every compact set inside $\Omega$.
| 9 | https://mathoverflow.net/users/140505 | 431703 | 174,760 |
https://mathoverflow.net/questions/431672 | 7 | Let $X(t)$ be a diffusion process on $\mathbb{R}^d$ generated by
\begin{align}
\mathcal{D} = \nabla^2 + \sum\_{i=1}^d b\_i(x) \frac{\partial}{\partial x\_i},
\end{align}
where $b\_i(x) \in \mathcal{C}\_b^2(\mathbb{R}^d)$, $i=1,2,\ldots,d$, and initial condition $x \in \mathbb{R}^d$. Ikeda and Watanabe prove the following theorem in [$\ast$, Chapter VI, $\S$9]
**Theorem 1:** Let $\varphi: [0,T] \rightarrow \mathbb{R}^d$ be a $\mathcal{C}^2$ curve such that $\varphi(0) = x$. Furthermore let $\lambda\_1$ be the first eigenvalue of the boundary value problem
\begin{align}
\begin{cases}\big(\frac{1}{2}\nabla^2 + \lambda\big)\psi = 0\\ \psi |\_{\partial D} = 0, \end{cases}
\end{align}
with associated eigenfunction $c$. Here $D = \{x: |x| < 1\}$. Then we have
\begin{align}
P\_x(w: \|w - \varphi\|\_T < \epsilon) \sim c \exp\left(\int\_0^T L(\varphi, \dot\varphi) \mathrm{d}t\right)\exp\left(-\frac{\lambda\_1T}{\epsilon^2}\right), \quad \text{as} \quad \epsilon \downarrow 0,
\end{align}
where $L(x,\dot x)$ is the Onsager-Machlup (OM) function of $X(t)$, given by
\begin{align}
L(x,\dot x)= -\frac{1}{2} |\dot x - b(x) | ^2 - \frac{1}{2} (\nabla \cdot b)(x).
\end{align}
In the above $\| \cdot \|\_T = \sup\_{t \in [0,T]} |w(t)| $ and $w \in \mathcal{C}^0([0,T] \rightarrow \mathbb{R}^d)$.
I am interested in an extension of the above result that describes the OM function for diffusion processes where $b = b(t,x)$ is a not a time-homogeneous drift. Ideally, the result should cover cases where $b$ is *not* a continuous function of $t$ (but it can be bounded). In particular, I am interested in the case where $b(t,x) = b(t)$ is a sample path for a [telegraph process](https://en.wikipedia.org/wiki/Telegraph_process). The most relevant result I have found is in a paper by Bardina et al. [$\ast\ast$] where they extend the result for diffusion process $X(t)$ on a real separable Hilbert space $H$, given by
\begin{align}
\begin{cases}
dX(t) = (AX(t) + F(t,X(t))dt + B dW(t), \quad t \in [0,1] \\
X(0) = x \in H.
\end{cases}
\end{align}
For sufficiently nice $A$ and $B$, this admits a unique solution. Let $W^A(t)$ be the solution to the above when $F \equiv 0, x = 0$. Then the authors show that, given suitable assumptions on $A$,$B$, and $F$,
\begin{align}
\lim\_{\epsilon \downarrow 0} \frac{P(\|X-\varphi\| < \epsilon)}{P(\|W^A\| < \epsilon)} = \exp \left(\hat{L}(\varphi,\dot \varphi)\right)
\end{align}
where
\begin{align}\hat{L}(\varphi,\dot\varphi) = -\frac{1}{2}\int\_0^1 \left \lvert B^{-1} \left(A \varphi (t) + F(t, \varphi(t)) - \dot\varphi (t) \right)\right\lvert\_H \mathrm{d}t - Tr(\mathcal{S}\_{PR^\ast})
\end{align}
with $\mathcal{S}\_{PR^\ast}$ a certain bounded linear operator determined by $\nabla\_x F$ and $\varphi$. This seems to be very close to what I need, however, two questions remain for me:
1. In the paper, the authors are not clear on the meaning of $\nabla\_x$ when $F$ is not differentiable. I presume it then refers to the distributional derivative but it is not clear.
2. The authors assume in their derivation that $F$ is Lipschitz, but the sample paths of the telegraph process are discontinuous. Are there any similar results that allow for discontinuities in the drift?
**References**
$\ast$*Ikeda, Nobuyuki; Watanabe, Shinzo*, *Stochastic differential equations and diffusion processes*, North-Holland Mathematical Library.
$\ast\ast$*Bardina, Xavier; Rovira, Carles; Tindel, Samy*, [**Onsager-Machlup functional for stochastic evolution equations**](http://dx.doi.org/10.1016/S0246-0203(02)00009-2), Ann. Inst. Henri Poincaré, Probab. Stat. 39, No. 1, 69-93 (2003). [ZBL1016.60064](https://zbmath.org/?q=an:1016.60064).
| https://mathoverflow.net/users/474675 | Onsager-Machlup functional when drift is time-dependent | Using the argument of [http://users.sussex.ac.uk/~md326/MAP.pdf](http://users.sussex.ac.uk/%7Emd326/MAP.pdf) or <https://arxiv.org/abs/2209.04523>
We have that if $\mu\_0$ is a centered Gaussian measure then its Onsager-Machlup function is $\operatorname{OM}\_{\mu\_0}(z)=\frac{1}{2}\|z\|\_{\mu\_0}^2$. If $\mu$ is equivalent to $\mu\_0$ with density $\frac{d\mu}{d\mu\_0}\propto e^{-\Phi}$, then
\begin{align\*}
\frac{\mu(B\_\varepsilon(z\_1))}{\mu(B\_\varepsilon(z\_2))}&=\frac{\int\_{B\_\varepsilon(z\_1)}\mu(du)}{\int\_{B\_\varepsilon(z\_2)}\mu(du)}\\
&=\frac{\int\_{B\_\varepsilon(z\_1)}e^{-\Phi(u)}\mu\_0(du)}{\int\_{B\_\varepsilon(z\_2)}e^{-\Phi(u)}\mu\_0(du)}\\
&=\frac{\int\_{B\_\varepsilon(z\_1)}e^{-\Phi(u)+\Phi(z\_1)-\Phi(z\_1)}\mu\_0(du)}{\int\_{B\_\varepsilon(z\_2)}e^{-\Phi(u)+\Phi(z\_2)-\Phi(z\_2)}\mu\_0(du)}\\
&=\frac{e^{-\Phi(z\_1)}}{e^{-\Phi(z\_2)}}\frac{\int\_{B\_\varepsilon(z\_1)}e^{-\Phi(u)+\Phi(z\_1)}\mu\_0(du)}{\int\_{B\_\varepsilon(z\_2)}e^{-\Phi(u)+\Phi(z\_2)}\mu\_0(du)}
\end{align\*}
If $\Phi$ is say, locally Lipschitz, in $z$ then you can bound $|\Phi(z\_i)-\Phi(u)|\leq L |z\_i-u|\lt L\varepsilon$. Therefore the Onsager-Machlup function is
\begin{equation}
\operatorname{OM}\_\mu(z)=
\begin{cases}
\Phi(z)+\frac{1}{2}\|z\|\_{\mu\_0}^2 &\text{ if }z\in \mathcal H\_{\mu\_0}\\
\infty &\text{ else}.
\end{cases}
\end{equation}
By Girsanov, the law of $$dX(t)=b(t,X(t))dt+c dB(t),$$
$\mu^c$, has a density with respect to the law of $$cB(t),$$
$\mu\_0^c$, given by
$$\frac{d\mu^c}{d\mu\_0^c}=\exp\left(\frac{1}{c^2}\left(\int\_0^T b(t,B(t))dB(t)-\frac{1}{2}\int\_0^T b^2(t,B(t))dt\right)\right).$$
We must convert the Ito integral to Stratonovich which is defined pathwise. Therefore we have that the density is $\mu\_0^c$-a.s. equal to
$$\frac{d\mu^c}{d\mu\_0^c}=\exp\left(\frac{1}{c^2}\left(\int\_0^T b(t,B(t))\circ dB(t)-\frac{c^2}{2}\int\_0^T b\_x(t,B(t)) dt-\frac{1}{2}\int\_0^T b^2(t,B(t))dt\right)\right),$$
where $\circ dB(t)$ represents then Stratonovich integral. So long as the this exponent is locally Lipschitz in $B$ then we have that
\begin{equation}
\operatorname{OM}\_\mu(z)=
\begin{cases}
\frac{1}{2c^2}\int\_0^T((z'(t)-b(t,z(t)))^2+c^2b\_x(t,z(t)))dt &\text{ if }z\in \mathcal H\_{\mu\_0}\\
\infty &\text{ else}.
\end{cases}
\end{equation}
Also, in the case where $b$ is independent of $X$, then the process $X(t)=\int\_0^t b(s) ds+B(t)$ is just a mean shifted Brownian motion which is a Gaussian process. In this case, the Onsager-Machlup function is just $\frac{1}{2}\|z-\int\_0^\cdot b(s)ds\|\_{W\_0^{1,2}}^2$ for $z\in W\_0^{1,2}$.
| 3 | https://mathoverflow.net/users/479223 | 431704 | 174,761 |
https://mathoverflow.net/questions/431709 | 0 | I would like to know if there is a method to solve the Problem.
**Problem**:
Maximize the following function: $$f(p\_{1,i},p\_{2,i},\dotsc,p\_{m,i})=\sum\_{i=1}^{n}\begin{bmatrix}p\_{1,i} & p\_{2,i} & \cdots & p\_{m,i} \end{bmatrix}\*\begin{bmatrix}e\_1 \\ e\_2 \\ \vdots \\ e\_m \end{bmatrix} \* c\_i$$
where we know the values of each $e\_j$ ($j \in \{1, \dotsc, m\}$) and $c\_i$.
The values of the $p\_{j,i}$ should be either 1 or 0 and we have the following constraints:
1. For all $j \in \{1,\dotsc,m\}$: $\sum\_{i=1}^n p\_{j,i} \le 1$.
2. For all $i \in \{1,\dotsc,n\}$: $\sum\_{j=1}^m p\_{j,i} = t\_i$, where $t\_i$ is a known value.
I would be grateful for every hint, method or solution.
| https://mathoverflow.net/users/492293 | Method for (binary) optimization under constraints | This is the [transportation problem](https://en.wikipedia.org/wiki/Transportation_theory_(mathematics)) in a bipartite network, with a supply of $1$ at each $j$ node and a demand of $t\_i$ at demand node $i$. The problem can be solved via linear programming, a minimum-cost network flow algorithm, or a specialized algorithm. Because of [total unimodularity](https://en.wikipedia.org/wiki/Unimodular_matrix#Total_unimodularity), you can relax the binary $\{0,1\}$ restriction to $\ge 0$ and still obtain an optimal binary solution.
| 4 | https://mathoverflow.net/users/141766 | 431710 | 174,764 |
https://mathoverflow.net/questions/430844 | 3 | Consider the following NLS:
$$i u\_t + \Delta u- 2 \operatorname{Re} u = F(u),$$
where $F(u):=(u + \bar{u} + |u|^2)u.$
In [Scattering for the Gross–Pitaevskii equation](https://arxiv.org/abs/math/0510080), the authors S. Gustafson, K. Nakanishi, and T.-P. Tsai used a change of variables to get the free evolution as a unitary group:
$$u \mapsto v:=V^{-1}u:= U^{-1} \operatorname{Re} u + i \operatorname{Im}u,$$
where $U:=\sqrt{- \Delta (2-\Delta)^{-1}}$.
Then $v$ satisfies the equation
$$i v\_t - \sqrt{- \Delta (2-\Delta)} v = - i V^{-1} i F(V v).$$
I have been trying to understand the motivation of this change of variable but no result! The main obstacles for me is the squire root, I can not see where the square root came from? Any help is appreciated.
**Updates**
I tried to write the equation as a dynamical system of two PDEs, real and imaginary parts, and tried to diagonalized the operator but I couldn't get the same result.
| https://mathoverflow.net/users/471464 | Change of variables for obtaining a unitary group | Let $u=u\_1+iu\_2$, $v=v\_1+iv\_2$ with $u\_1,u\_2,v\_1,v\_2$ being real valued.
What one would like to do is find a $2\times 2$ matrix of real symmetric operators
$$
\begin{pmatrix}
A & B\\
C & D
\end{pmatrix}
$$
such that if one lets
$$
\begin{pmatrix}v\_1\\ v\_2\end{pmatrix}:=\begin{pmatrix}
A & B\\
C & D
\end{pmatrix}
\begin{pmatrix}u\_1\\ u\_2\end{pmatrix}\ ,
$$
the free equation $i\partial\_t u=-\Delta u+2\Re u$ is equivalent to an evolution for $v$ which is of the form $i\partial\_t v=H v$ where $H$ is a real symmetric operator, or rather its complexification.
One could start with general $A,B,C,D$ and see what happens, but with the hindsight provided by the authors of the mentioned article, we know we will be able to arrange for $B,C=0$, and $D={\rm Id}$. Let's go ahead and rename $A=U^{-1}$, for some $U$ TBA.
In terms of real and imaginary components, the equation for $u$ is equivalent to the system
$$
\left\{
\begin{array}{ccc}
\partial\_t u\_1 &=& -\Delta u\_2\\
\partial\_t u\_2 &=& \Delta u\_1 -2u\_1
\end{array}
\right.
$$
while for the components of $v$ we must have
$$
\left\{
\begin{array}{ccc}
\partial\_t v\_1 &=& H u\_2\\
\partial\_t v\_2 &=& -H u\_1
\end{array}
\right.
$$
Since we assumed the relations $v\_1=U^{-1}u\_1$, $v\_2=u\_2$, we immediately get
$$
\partial\_t v\_1=-U^{-1}\Delta v\_2
$$
which suggests $H=-U^{-1}\Delta$.
But we also get
$$
\partial\_t v\_2=\partial\_t u\_2=(\Delta-2)u\_1=(\Delta-2)Uv\_1
$$
which requires also $H=-(\Delta-2)U$. So consistency for the choice of $H$ needs the equation
$$
-U^{-1}\Delta=(2-\Delta)U
$$
to hold. Assuming $U$ commutes with $\Delta$, we obtain the desired square root formula.
| 2 | https://mathoverflow.net/users/7410 | 431713 | 174,765 |
https://mathoverflow.net/questions/429875 | 6 | A very fluffy question in which I'm ignorant of homology/cohomology, grading etc:
The open-closed and closed-open string maps relating the symplectic (co)homology and Hochschild (co)homology of the wrapped Fukaya category of some symplectic manifold, are proven to be (ring, not sure about BV-algebra?) isomorphisms in certain cases (see e.g. [Ganatra](https://sheelganatra.com/wrapcy.pdf), [Ritter-Smith](https://arxiv.org/pdf/1201.5880v7.pdf)). It is also known ([Abbondandolo-Schwarz](https://onlinelibrary.wiley.com/doi/abs/10.1002/cpa.20090), [Abouzaid](https://arxiv.org/pdf/1312.3354.pdf)) that the symplectic (co)homology of a cotangent bundle is (BV-algebra) isomorphic to the homology of the free loop space of the base manifold, which is (BV-algebra) isomorphic to the Hochschild (co)homology of the singular cochains of the manifold (see e.g. [Abbaspour](https://hal.archives-ouvertes.fr/hal-00794525v1/document)).
My superficial understanding of Hochschild (co)homology is that it measures deformations, but I'm not very familiar with Hochschild (co)homology constructions mentioned above, nor with the constructions of Hochschild (co)homology for categories.
**Question:** Can I, and if so in what (geometric or algebraic) sense, interpret symplectic (co)homology as encoding deformations? I guess I can write down a Maurer-Cartan equation in symplectic (co)homology; what do solutions encode/measure?
| https://mathoverflow.net/users/490756 | Relation between symplectic (co)homology and Hochschild (co)homology and deformations | First, the closed-open string map is expected to be an isomorphism between BV algebras. In the case when $X$ is a Weinstein manifold, it is known to be an isomorphism of BV algebras, just make sure that it is the wrapped Fukaya category $\mathcal{W}(X)$ that is under consideration.
Assume that $X$ is a smooth affine variety and admits a compactification $Y$ by a normal crossing divisor $D$. Then one can define the so-called *relative Fukaya category* $\mathcal{F}(Y,D)$, whose objects are objects in the compact Fukaya category $\mathcal{F}(X)$, and whose morphisms are defined by counting holomorphic polygons passing through $D$. In many known situations, $\mathcal{F}(Y,D)\subset\mathcal{F}(Y)$ is a generating full subcategory. It is expected that $\mathcal{F}(Y,D)$ is a deformation of $\mathcal{F}(X)$ with respect to the image of a cochain $s\in\mathit{SC}^\ast(X)$ under the closed open string map
$\mathit{CO}:\mathit{SC}^\ast(X)\rightarrow\mathit{CC}^\ast(\mathcal{F}(X))$
where $\mathit{SC}^\ast$ is the complex computing symplectic cohomology, and $\mathit{CC}^\ast$ is the Hochschild cochain complex. Note that $\mathit{CO}$ is in fact an $L\_\infty$-homomorphism, and both $s$ and $\mathit{CO}(s)$ are Maurer-Cartan elements with respect to the corresponding $L\_\infty$-structures, so one can use it to deform $\mathcal{F}(X)$.
The cochain $s$ is constructed by counting holomorphic thimbles in $Y$ passing through the divisor $D$ and asymptotic to Reeb orbits on the ideal contact boundary $\partial\_\infty X$. In nice senarios, e.g. when $Y$ has Kodaira dimension $-\infty$ and $D$ is anticanonical, one can show that the cochain $s\in\mathit{SC}^\ast(X)$ is actually a cocycle, therefore defines a class $[s]\in\mathit{SH}^0(M)$ in the (in general even degree) symplectic cohomology. This class is called a *Borman-Sheridan class*. One can use it to deform the wrapped Fukaya category of $X$ as well, since the closed-open string map can be defined for $\mathcal{W}(X)$.
Elements in the odd degree symplectic cohomology (or the underlying cochain complex), on the other hand, can be regarded as the non-commutative analogues of algebraic vector fields, so one can deform objects in the Fukaya category along these vector fields. This is explored in detail by Seidel in his works on categorical dynamics, see for example <https://arxiv.org/abs/1108.0394>.
| 3 | https://mathoverflow.net/users/43423 | 431719 | 174,767 |
https://mathoverflow.net/questions/431567 | 7 | Let $N(\sigma,T)$ denote the number of zeros $\rho=\beta+\gamma i$ of the Riemann zeta function satisfying $\beta\ge \sigma$ and $0<\gamma\le T$, counted with multiplicity. Then the "Density Hypothesis" usually means one of the two following unproven conjectures.
**1."Weak" Density Hypothesis:**
Let $\varepsilon>0$ be fixed. Then
$$N(\sigma,T)=O(T^{2(1-\sigma)+\varepsilon})$$
uniformly in $1/2\le \sigma\le 1$ as $T\to +\infty$.
**2. "Strong" Density Hypothesis:** There exists a real number $\alpha\ge 1$, such that
$$N(\sigma,T)=O(T^{2(1-\sigma)}(\log T)^{\alpha})$$
uniformly in $1/2\le \sigma\le 1$ as $T\to +\infty$.
From the work of Hoheisel, it is known that if the (Strong) density hypothesis is true, then
\begin{align}
\psi(x+x^{\vartheta})-\psi(x)\sim&\;x^{\vartheta} &&(1)\\[1mm]
\pi(x+x^{\vartheta})-\pi(x)\sim&\; \min(1,\vartheta^{-1})\frac{x^{\vartheta}}{\log x} &&(2)\\[1mm]
p\_{n+1}-p\_{n}=&\;O(p\_n^{\vartheta}) &&(3)
\end{align}
all hold for any fixed $\vartheta>1/2$.
**Question:** Are there any other (at least remotely significant) consequences that would follow if one of the above density hypotheses could be proven (without assuming the Lindelöf or Riemann Hypotheses)? $(1)-(3)$ surely cannot be an exhaustive list.
**Some background:** The density hypothesis arose from A.E. Ingham's paper *On the difference between consecutive primes* (1937), where he showed that, if $\zeta(\tfrac{1}{2}+ti)=O(t^{c})$ as $t\to +\infty$, for some fixed $c>0$, then
$$N(\sigma,T)=O(T^{2(1+2c)(1-\sigma)}(\log T)^{5})$$
uniformly in $1/2\le \sigma \le 1$ as $T\to +\infty$.
Thus, if the Lindelöf hypothesis is true, then we may take $c=\varepsilon$ and conclude that the weak density hypothesis is true. As explained in [this post](https://mathoverflow.net/questions/52576/zeta-function-zero-density-theorems), the linear term $2(1-\sigma)$ is the best possible (or, at the very least, naturally occuring). As argued in the same post, the density hypothesis has been proven unconditionally for $\sigma \ge 25/32$ by Bourgain, although the conclusion over this range is somewhat stronger than the density hypothesis itself.
Supposing for the moment that we could in fact prove that the density hypothesis is true, then what would we gain from it?
In Ivić's book, it sais on p. 47 that "Estimates of $N(\sigma,T)$" play an important role in many applications of zeta-function theory, and for some applications concerning primes the reader is referred to chapter 12." Now $(1)-(3)$ are basically the results he is referring to. So where else are estimates of $N(\sigma,T)$ important to zeta-function theory, especially if the density hypothesis is true?
| https://mathoverflow.net/users/153908 | What consequences would follow from the density hypothesis? | The "strong" density hypothesis implies that if $\epsilon>0$ fixed and $X$ is large, then the Lebesgue measure of the $x\in[0,X]$ such that the interval $[x,x+x^{\epsilon}]$ does not contain a prime is $o(X)$. (See the bottom of page 131 in Montgomery's "Topics in Multiplicative Number Theory".) For comparison, RH implies that one may replace $x^{\epsilon}$ with $f(x)(\log x)^2$, where $f(x)$ monotonically increases to infinity as $x\to\infty$ (proved in Selberg's paper "On the normal density of primes in small intervals, and the difference between consecutive primes", which can be found in his collected works). The threshold $x^{\epsilon}$ is interesting because [Maier's matrix method](https://projecteuclid.org/journals/michigan-mathematical-journal/volume-32/issue-2/Primes-in-short-intervals/10.1307/mmj/1029003189.full) shows that an asymptotic prime number theorem for $\pi(x+(\log x)^A)-\pi(x)$ cannot hold for any fixed $A>2$, so such almost-all results are the best for which one could hope.
But the density hypothesis is not the only meaningful bound on $N(\sigma,T)$. There is also Selberg's unconditional estimate
$$N(\sigma,T)\ll T^{1-\frac{1}{4}(\sigma-\frac{1}{2})}\log T,$$
which is a key tool in the first proofs of Selberg's central limit theorems for $\log|\zeta(\frac{1}{2}+it)|$ and $\arg\zeta(\frac{1}{2}+it)$. (See [here](https://projecteuclid.org/journals/duke-mathematical-journal/volume-80/issue-3/On-the-distribution-of-zeros-of-linear-combinations-of-Euler/10.1215/S0012-7094-95-08028-4.short).) These central limit theorems provide inspiration for conjectures on how the moments of $\zeta(s)$ grow, and they even suggest how one might possibly go about proving such conjectures. Perhaps the definitive example of this is due to [Soundararajan](https://arxiv.org/abs/math/0612106). Even though Soundararajan's results rely on RH, the inspiration from Selberg is clear in the proofs.
[Ford and Zaharescu](https://faculty.math.illinois.edu/%7Eford/wwwpapers/fz.pdf) used Selberg's estimate to study the limiting distribution of the fractional parts of the imaginary parts of the nontrivial zeros of $\zeta(s)$. As [Ford, Soundararajan, and Zaharescu](https://faculty.math.illinois.edu/%7Eford/wwwpapers/fsz.pdf) show (perhaps unexpectedly), this distribution is closely connected with the pair correlation of zeros of $\zeta(s)$ and the distribution of primes in short intervals (this is interesting because Selberg's density estimate is sharp near $\mathrm{Re}(s) = \frac{1}{2}$, but not near $\mathrm{Re}(s) =1$, which is the purview of the density hypothesis).
Selberg's estimate has other nice applications, but these come to mind easily.
| 8 | https://mathoverflow.net/users/111215 | 431727 | 174,770 |
https://mathoverflow.net/questions/431714 | 1 | Is it possible to find an example of a family $\mathcal{F}$ of $n$ finite distinct non-empty sets, a universe of maximum size $n/4$, with at least $\lfloor \frac{2}{3}{n \choose 2} \rfloor$ unordered couples of sets with at least one element in common between the two sets, and no element belonging to at least $n/2$ sets of the family?
| https://mathoverflow.net/users/136218 | Existence of a family of sets with some properties | Take a finite projective plane: for the sake of concreteness, the Fano plane $PG(2, 2)$. It has seven points $P$ and seven lines $L$, where each line goes through three points and each pair of lines intersects.
Take a second set $S$ which is disjoint from $P$ and an element $x$ which is not in either. Consider $$\mathcal{F} = L \times 2^S \cup \{\{x\}\}$$ and let $n = |\mathcal{F}| = 7 \cdot 2^{|S|} + 1$. Every element of $P$ is in $\frac{3}{7}(n-1)$ sets; every element of $S$ is in $\frac{1}{2}(n-1)$ sets; $x$ is in one set. Every pair of sets which does not include $\{x\}$ has a non-empty intersection. The size of the universe is $|S| + 8$, so this family satisfies all of the constraints provided that $4|S| + 32 \le 7 \cdot 2^{|S|} + 1$, which is true if $|S| \ge 3$.
---
For more general values of $n$, let $a$ be the largest integer such that $7 \cdot 2^a < n$. If we take two disjoint sets $S\_1, S\_2$ with $|S\_1| = |S\_2| = a$ then $L \times \left(2^{S\_1} \cup 2^{S\_2} \right)$ has more than $n$ sets, but each element of $S\_1 \cup S\_2$ occurs exactly $7 \cdot 2^{a-1} < \frac n2$ times. Each element of $P$ occurs exactly $3 \cdot 2^{a+1}$ times, so when choosing $\mathcal{F} \subset L \times \left(2^{S\_1} \cup 2^{S\_2} \right)$ we should be careful to keep the projection onto $L$ balanced, but if we do this then each element of $P$ will occur no more than $\frac{3}{7}n + 3$ times.
Then we simply require that the universe be sufficiently small. Now the size of the universe is $7 + 2a$, and we find that the construction works for $n \ge 44$.
| 0 | https://mathoverflow.net/users/46140 | 431743 | 174,774 |
https://mathoverflow.net/questions/431686 | 0 | Suppose we have a 5 tuple of positive real numbers $(l\_1,l\_2,m\_1,m\_2,m\_3)$, with $m\_i \in (0,\pi)$ for all $i$. Now fix a point $v\_1$ in the hyperbolic plane. Then consider a geodesic of length $l\_1$ starting at $v\_1$. suppose that ends at $v\_2$. At $v\_2$ draw another geodesic of length $l\_2$ which makes an angle $m\_1$ at $v\_2$ with first line. Suppose the second geodesic ends at the point $v\_3$. Then draw another geodesic of length $l\_1$ at $v\_3$ making an angle $m\_2$ with the second geodesic. Suppose end point of the third geodesic is $v\_4$. Now draw another geodesic of length $l\_2$ making an angle $m\_3$ at $v\_4$ with the third geodesic. Let the end point of the last geodesic is $v\_5$.
Then can we put some condition on the tuple to get $v\_5=v\_1$? If yes what that condition should be?
Thanks in advance.
| https://mathoverflow.net/users/490039 | When a polygonal line become a loop in hyperbolic plane? | Your broken line closes iff the triangles $(v\_1,v\_2,v\_3)$
and $(v\_3,v\_4,v\_5)$ are equal, and $[v\_1,v\_3]$ is their common side. So there are two conditions: one is that $m\_1=m\_3$, by the hyperbolic law of cosines.
To state the second one,
Apply the hyperbolic law of cosines to find the length $x$ of the side $[v\_1,v\_3]$, then the hyperbolic law of sines to find the angles of the two triangles at $v\_3$, and write the condition that the sum of these angles at $v\_3$ is equal to your $m\_2$.
| 1 | https://mathoverflow.net/users/25510 | 431749 | 174,777 |
https://mathoverflow.net/questions/431484 | 5 | Fix $\epsilon>0$.
For all large $N$, does there exist $A\subset [N]:=\{1,\dots,N\}$ such that both $A+A$ and $A^c:=[N]\setminus A$ lack arithmetic progressions of length $N^\epsilon$?
I am aware Rusza used “niveau sets” to create dense subsets of $[N]$ whose sumsets lack progressions of length $\exp(\log^{2/3+o(1)}N)$ (see <https://eudml.org/doc/206433>). But I don’t understand the construction well enough to know if it lacks long progressions in the complement (and thus satisfies the above).
My motivation is that I read that it was an open problem to determine for $r\ge 2$ if there exists $\epsilon\_r> 0$, so any $r$-coloring $c:[N]\to [r]$ has some color class $A$ where $A+A$ contains a progression of length $N^{\epsilon\_r}$. If one could find some $\epsilon>0$ where my question is false, then we could inductively take $\epsilon\_r =(r-1)\epsilon$. So I am hoping to learn if this approach is viable.
| https://mathoverflow.net/users/130484 | Progressions in sumset or complement | Ah, I realized how to construct such $A$.
The idea is to just crudely mimic the construction for the off-diagonal van der Waerden numbers $w(3,k)$.
We fix parameters $D,M,\rho$ which will be optimized later.
We shall sample $M$ points $x\_1,\dots,x\_M$ from the $D$-dimensional torus $\Bbb{T}^D := \Bbb{R}^D/\Bbb{Z}^D$. We then take $A\_0\subset \Bbb{T}^D$ to be $\{x\_1,\dots,x\_M\}+B\_0$ where $B\_0$ is the box $[0,\rho]^D+\Bbb{Z}^D$.
One then chooses $\theta \in \Bbb{T}^D$ uniformly at random and takes $A=\{n\in[N]: n\theta \in A\_0\}$. I claim that we can let $D=D(N)\to \infty$ as $N\to \infty$ while ensuring $A+A$ and $A^c$ both lack progressions of length $N^{O(1/D)}$.
In short, one just modifies several calculations from my paper (<https://arxiv.org/abs/2111.01099>). I might update with further details later, but if I’m not mistaken, this blocks progressions of length $\exp(O(\sqrt{\log N \log\log N}))$.
The intuition is that this looks like a union of $M$ generalized arithmetic progressions (GAP's) of rank $D$, $G\_1,...,G\_M$ where each $G\_i$ "looks generic" and thus doesn't have any arithmetic progressions much longer than $N^{1/D}$. The $M$ GAP'S are sufficiently randomly distributed in $[N]$ that they block progressions in the complement, as was shown in the aforementioned paper.
Meanwhile, $A+A$ looks like a union of $M^2$ GAP's of rank $D$, which again all look generic. Taking $M$ much smaller than $N^{1/D}$, these $M^2$ GAP's in $A+A$ shouldn't really interact, resulting in there not being progressions dramatically longer than $N^{1/D}$.
| 2 | https://mathoverflow.net/users/130484 | 431752 | 174,778 |
https://mathoverflow.net/questions/431683 | 12 | This is an obvious continuation of [an MO question](https://mathoverflow.net/q/431642). Let $r,s\in\mathbb N\cup\{\infty\}$ with $r\neq s$, and $M,N$ two connected manifolds of positive dimension (which roots out the trivial case of a single point). I wonder whether the rings $C^r(M)$ and $C^s(N)$ of real-valued continuously $r$-th (resp. $s$-th) differentiable functions are non-isomorphic?
More precisely, we have the following (combinations of) variants of conditions on $M$ and $N$, although they might not affect the result nontrivially.
1. Smoothness:
* weakest: $M$ is a $C^r$-manifold and $N$ is a $C^s$-manifold;
* strongest: both $M$ and $N$ are $C^\infty$.
2. Compactness:
* weakest: both are paracompact Hausdorff (usually present in the definition of manifolds);
* strongest: both are compact Hausdorff.
There is also a variant of manifolds with boundaries, but I am not looking at this generality in this question.
It seems that, in this case, Morse's lemma might not hold, thus the structure of vanishing ideals $\mathfrak m\_x=\ker(C^r(M)\to\mathbb R)$ at a point $x\in M$ might be more complicated (say, maybe not regular). However, it seems possible that [Tom Goodwillie's answer](https://mathoverflow.net/a/431644/) applies to the situation that one of $r$ and $s$ is $\infty$.
| https://mathoverflow.net/users/176381 | Is $C^r(M)$ non-isomorphic to $C^s(N)$ for $r\neq s$ and nontrivial manifolds $M,N$? | If $r\neq s$, then the rings $C\_r(M),C\_s(N)$ cannot be isomorphic nor can they be elementarily equivalent to each other. I claim that for all $r<\infty$, there is a first order formula $\phi$ where $C\_r(M)\models\phi$ if and only if $s=r$.
If $p/q$ is a reduced rational number where $q$ is odd, and $f$ is a real valued function (or real number), then we shall write $f^{p/q}$ for the unique function such that $(f^{p/q})^q=f^p$. While the function $f^{p/q}$ is unique, the function $f^{p/q}$ may no longer be contained in a ring of $r$-times continuously differentiable functions.
If $\alpha=p/q$ is a reduced rational number, then let $\phi\_\alpha$ denote the first statement "For each $f$, the object $f^\alpha$ exists and is unique." More precisely, $\phi\_\alpha$ is the statement "For each $f$, there is a unique $g$ with $g^q=f^p$."
Now, suppose that $\alpha=p/q$ is a reduced rational number with $q$ odd and that $\alpha$ is not an integer. Then $\frac{d^r}{dx^r}x^{\alpha}=\frac{\alpha!}{(\alpha-r)!}x^{\alpha-r}$ which is continuous and defined for every real number $x$ if and only if $\alpha>r$.
For non-logicians, the following proposition says that for $\alpha=p/q$ reduced, $q>1$, $q$ odd, the ring $C\_r(M)$ is closed under the function $f\mapsto f^\alpha$ if and only if $\alpha>r$.
Proposition: Suppose that $\alpha=p/q$ is a reduced rational number with $q$ odd and where $\alpha$ is not an integer. Then $C\_r(M)\models\phi\_\alpha$ if and only if $\alpha>r$.
Proof: Suppose $\alpha<r$. Let $U\subseteq M$ and let $V\subseteq\mathbb{R}^n$ where both $U$ and $V$ are open. Let $\phi:U\rightarrow V$ be a chart. We can assume that $\mathbf{0}\in V$ since we may translate the set $V$ if necessarily. Now suppose that $g:V\rightarrow\mathbb{R}$ be a smooth function with compact support where $g|\_W=1$ for some neighborhood $W$ of $\mathbf{0}$. Let $h:V\rightarrow\mathbb{R}$ be the function defined by letting $h(x\_1,\dots,x\_n)=x\_1\cdot g(x\_1,\dots,x\_n).$ Let $f:M\rightarrow\mathbb{R}$ be the function defined by letting
$f(\mathbf{x})=0$ for $\mathbf{x}\not\in U$ and
$f(\mathbf{x})=h(\phi(\mathbf{x}))$ for $\mathbf{x}\in U$. Then $f\in C^r(M)$. On the other hand, $h(x\_1,\dots,x\_n)^\alpha=x\_1^\alpha$ in a neighborhood of $\mathbf{0}$ which is not a $C^r$ function. Therefore, since $f(\mathbf{x})^\alpha=h(\phi(\mathbf{x}))^\alpha$ for $\mathbf{x}\in U$, the function $f^\alpha$ is not in $C^r(M).$
Now assume that $r<\alpha$ and $f\in C^r(M)$. Then $f^\alpha\in C^r(M)$ since $f^\alpha$ is the composition of the two $C^r$ functions, namely
$f$ and $x\mapsto x^\alpha.$
$\square$
| 6 | https://mathoverflow.net/users/22277 | 431757 | 174,780 |
https://mathoverflow.net/questions/431765 | 4 | The determinant line bundle of a coherent sheaf $\mathcal{F}$ on an $n$-dimensional (smooth) analytic space is defined as
\begin{equation}
\det \mathcal{F} := \bigotimes\_i^n (\det \mathcal{E}\_i)^{⊗ (-1)^i}
\end{equation}
where $\mathcal{E}\_\bullet \to F$ is a locally free resolution of $\mathcal{F}$ (which we can take to have length at most $n$). It can be shown that this is independent of the resolution taken, and that if $\mathcal{F}$ is torsion-free then
\begin{equation}
\det \mathcal{F} \cong \left(\bigwedge^{\operatorname{rk} \mathcal{F}} \mathcal{F}\right)^{\*\*}
\end{equation}
. Since $\bigwedge^k$ and double duals are both functorial we see that a morphis $\mathcal{F} \to \mathcal{F}$ of \**torsion-free* shaves **of the same rank** induces a morphism between their determinant line bundles. Can we say the same thing about any two coherent sheaves of the same rank?
**My thoughts so far:** It seems like the obvious way to do this would be to take free resolutions $\mathcal{E}\_\bullet, \mathcal{E}'\_\bullet$ of $\mathcal{F}, \mathcal{F}'$, which gives a map $f\_\bullet: \mathcal{E}\_\bullet \to \mathcal{E}'\_\bullet$ and to just take the alternating tensor product of the maps induced by $f\_i$ from $\det \mathcal{E}\_i \to \det \mathcal{E}'\_i$. However those maps don't exist because $\det$ is only functorial on vector bundles of the same rank?
| https://mathoverflow.net/users/123448 | Is the determinant line bundle of a coherent sheaf functorial (between sheaves of the same rank)? | No.
Working on projective space, consider a composition $$ \mathcal O \to \mathcal O \oplus \mathcal O/\mathcal O(-1) \to \mathcal O $$ where $\mathcal O/\mathcal O(-1)$ is the constant sheaf on a hyperplane, and the maps are just defined to be the identity on $\mathcal O$ and $0$ on $\mathcal O/\mathcal O(-1)$.
Since the composition is the identity, the induced map on determinants of the composition is the identity. By functoriality, the composition of the induced map on determinants must be the identity.
But taking determinants, we get $$\mathcal O \to \mathcal O(1) \to \mathcal O$$ and since every map $\mathcal O(1) \to \mathcal O$ vanishes, there is no composition of maps $\mathcal O \to \mathcal O(1)$ and $\mathcal O(1) \to \mathcal O$ that gives the identity.
| 10 | https://mathoverflow.net/users/18060 | 431767 | 174,783 |
https://mathoverflow.net/questions/431723 | 6 | In Martin-Löf type theory, a weakly Tarski universe is a type $\mathcal{U}$ with a type family $\mathcal{T}(A)$ indexed by terms $A:\mathcal{U}$, which is closed under identity types, dependent products, and dependent sums: there are functions
$$\Sigma\_\mathcal{U}:\prod\_{A:\mathcal{U}} (\mathcal{T}(A) \to \mathcal{U}) \to \mathcal{U}$$
$$\Pi\_\mathcal{U}:\prod\_{A:\mathcal{U}} (\mathcal{T}(A) \to \mathcal{U}) \to \mathcal{U}$$
$$\mathrm{Id}\_\mathcal{U}:\prod\_{A:\mathcal{U}} \mathcal{T}(A) \to (\mathcal{T}(A) \to \mathcal{U})$$
with equivalences
$$\mathrm{equiv}\_{\Sigma}(A, B):\mathcal{T}(\Sigma\_\mathcal{U}(A, B)) \simeq \sum\_{x:\mathcal{T}(A)} \mathcal{T}(B(x))$$
$$\mathrm{equiv}\_{\Pi}(A, B):\mathcal{T}(\Pi\_\mathcal{U}(A, B)) \simeq \prod\_{x:\mathcal{T}(A)} \mathcal{T}(B(x))$$
$$\mathrm{equiv}\_{\mathrm{Id}}(A, a, b):\mathcal{T}(\mathrm{Id}\_\mathcal{U}(A, a, b)) \simeq a =\_{\mathcal{T}(A)} b$$
A weakly Tarski universe is a strict Tarski universes if the equivalences are promoted to definitional equalities:
$$\mathcal{T}(\Sigma\_\mathcal{U}(A, B)) \equiv \sum\_{x:\mathcal{T}(A)} \mathcal{T}(B(x))$$
$$\mathcal{T}(\Pi\_\mathcal{U}(A, B)) \equiv \prod\_{x:\mathcal{T}(A)} \mathcal{T}(B(x))$$
$$\mathcal{T}(\mathrm{Id}\_\mathcal{U}(A, a, b)) \equiv a =\_{\mathcal{T}(A)} b$$
In both cases, we can define the internal equivalence types $A \simeq\_\mathcal{U} B:\mathcal{U}$ for $A:\mathcal{U}$ and $B:\mathcal{U}$.
There are two definitions of univalent Tarski universes possible: that the canonical transport function
$$\mathrm{transport}^\mathcal{T}(A, B):A =\_\mathcal{U} B \to \mathcal{T}(A) \simeq \mathcal{T}(B)$$
is an equivalence of types, and that the canonical function
$$\mathrm{idtoequiv}(A, B):A =\_\mathcal{U} B \to \mathcal{T}(A \simeq\_\mathcal{U} B)$$
is an equivalence of types. In strict Tarski universes, these two definitions of univalent universes are the same, because the two types $\mathcal{T}(A \simeq\_\mathcal{U} B)$ and $\mathcal{T}(A) \simeq \mathcal{T}(B)$ are definitionally equal to each other
$$\mathcal{T}(A \simeq\_\mathcal{U} B) \equiv \mathcal{T}(A) \simeq \mathcal{T}(B)$$
However, in weakly Tarski universes, the two types are only equivalent to each other by the function
$$\mathrm{equiv}\_{\simeq}:\mathcal{T}(A \simeq\_\mathcal{U} B) \simeq (\mathcal{T}(A) \simeq \mathcal{T}(B))$$
derived from $\mathrm{equiv}\_{\Sigma}$, $\mathrm{equiv}\_{\Pi}$, and $\mathrm{equiv}\_{\mathrm{Id}}$.
Is it still true that every weakly Tarski universe which is univalent by $\mathrm{transport}^\mathcal{T}$ is also univalent by $\mathrm{idtoequiv}$, and is it still true that every weakly Tarski universe which is univalent by $\mathrm{idtoequiv}$ is also univalent by $\mathrm{transport}^\mathcal{T}$? If not, under what conditions do the two notions of univalence for weakly Tarski universes coincide?
| https://mathoverflow.net/users/483446 | Univalence for weakly Tarski universes | Yes, $\mathrm{transport}^\mathcal{T}$ is an equivalence if and only if $\mathrm{idtoequiv}$ is an equivalence. To show this, it is enough to prove that the underlying functions of $\mathrm{transport}^\mathcal{T}(p)$ and $\mathrm{equiv}\_\simeq (\mathrm{idtoequiv}(p))$ are equal for all $p : A =\_\mathcal{U} B$. By path induction, it is enough to show that $\mathrm{equiv}\_\simeq$ maps the identity function in $\mathcal{T}(A \to\_\mathcal{U} A)$ to the identity function in $\mathcal{T}(A) \to \mathcal{T}(A)$, but the identity function in $\mathcal{T}(A \to\_\mathcal{U} A)$ is defined as $\mathrm{equiv}\_\simeq^{-1}(\lambda x.x)$, so this is true.
| 2 | https://mathoverflow.net/users/62782 | 431772 | 174,784 |
https://mathoverflow.net/questions/431776 | 4 | Let $1\leq p <\infty$ and let $p^{\prime}$ denote its conjugate exponent. Consider the following operator on Schwartz functions:
$$Tf(x)=\int\_{0}^{\infty}t^{\frac{n}{2 p^{\prime}}-1}e^{-t}
\int\_{|x-y|^2\leq t}\frac{f(y)}{|x-y|^{\frac{n}{p^{\prime}}}}dy dt,\qquad x\in \mathbb{R}^{n}.$$
I have tried to prove that $T$ is bounded from $L^{p}$ to $L^{\infty}$ but failed so far.
Young's inequality for convolution is not useful with the $y$-integral as $|\cdot|^{\frac{n}{p^{\prime}}}$ is not in $L^{p^{\prime}}(B(t))$ with $B(t)$ the standard ball centered at the origin with radius $t>0$.
Hardy-Little-wood-Sobolev inequality is not useful for obtaining $L^{\infty}$ boundedness.
One could look at the Hardy-Littlewood maximal operator $\displaystyle Mf(x)=\sup\_{r>0} \frac{1}{B(x,r)} \int\_{B(x,r)}\frac{f(y)}{|x-y|^{\frac{n}{p^{\prime}}}}dy$ since
$$\frac{1}{t^{\frac{n}{2}}}\int\_{|x-y|^2\leq t}\frac{f(y)}{|x-y|^{\frac{n}{p^{\prime}}}}dy\leq Mf(x).$$
The maximal operator is known to be bounded from $L^{p}$ to $L^{p}$ for all $1<p\leq \infty$ and from $L^1$ to weak $L^{1}$. I have no idea about the boundedness of $M$ from $L^p$ to $L^{\infty}$ when $p<\infty$.
Is it true that
$$\|Tf\|\_{L^{\infty}}\leq C \|f\|\_{L^{p}}$$
for any $1\leq p<\infty$ or is there a counterexample ?
| https://mathoverflow.net/users/116555 | Is this an $L^p-L^{\infty}$ operator? | No this is not true. Take $n=1$, $p=2$ and $$f(y)=\frac{1}{\sqrt {|y|} (|\log|y||)^\alpha}\chi\_{(-1,1)}(y)$$ with $\frac 12 < \alpha <1$. Then $f \in L^2(\mathbb R)$ but the innermost integral diverges at $x=0$ for every $t>0$.
EDIT The same counterexample can be done for general $n$ and $1<p<\infty$ (the case $p=1$ is in the answer by @Willie Wong). Take $$f(y)=\frac{1}{{|y|^{\frac np}} (|\log|y||)^\alpha}\chi\_{B\_r}(y)$$ with $r<1$ and $\frac 1p <\alpha <1$. Then $Tf(0)=\infty$ and, by Fatou, $\lim\_{x \to 0}Tf(x)=\infty$.
| 5 | https://mathoverflow.net/users/150653 | 431781 | 174,785 |
https://mathoverflow.net/questions/431775 | 3 | Let $\mathcal{O}$ be a Dedekind domain and $K = \mathrm{Frac}(\mathcal{O})$ its field of fractions. Let $E / K$ be an elliptic curve and $\mathcal{E} / \mathcal{O}$ its Neron model and $\mathcal{E}^\circ$ the connected component of the identity (fiberwise).
Then the natural map $H^1\_{\mathrm{fppf}}(\mathcal{O}, \mathcal{E}^\circ[n]) \to H^1(K, E[n])$ lands inside $\mathrm{Sel}\_n(E,K) \subset H^1(K, E[n])$. Is there a straightforward description of the kernel and cokernel?
| https://mathoverflow.net/users/154157 | Selmer groups and fppf cohomology | The following paper of Kestutis Cesnavicius answer these and related questions completely: <https://arxiv.org/abs/1301.4724>
| 3 | https://mathoverflow.net/users/110362 | 431791 | 174,789 |
https://mathoverflow.net/questions/431300 | 0 |
>
> Let $ \{a\_k\}\_{i=1}^n $ is a positive sequence. For $ 0<p<\infty $, space $ L^{p,\infty} $ is defined by
> $$
> \left\{f:\|f\|\_{p,\infty}=\inf\left\{C>0:a\_f(\lambda)\leq C/\lambda^p\right\}\right\}
> $$
> where
> $$
> a\_f(\lambda)=|\{x\in\mathbb{R}^n:|f(x)|>\lambda\}|.
> $$
> Then
> $$
> \left\|\sum\_{k=1}^{n}f\_k\right\|\_{1,\infty}\leq \left(1+\sum\_{k=1}^{n}a\_k\right)\sum\_{k=1}^{n}\left(\ln(1+a\_k^{-1})\|f\_k\|\_{1,\infty}\right)
> $$
>
>
>
I do not know why there are $ a\_k $ in the inequality. Can you give me some hints or references?
| https://mathoverflow.net/users/241460 | How to prove this inequality for the norm $ \|\cdot\|_{1,\infty} $? | For any $ \lambda>0 $, we have
\begin{align}
\left|\left\{x:\left|\sum\_{i=1}^nf\_k\right|>\left(1+\sum\_{i=1}^na\_k\right)\lambda\right\}\right|&\leq\left|\left\{x:\sum\_{i=1}^n|f\_k|>\left(1+\sum\_{i=1}^na\_k\right)\lambda\right\}\right|\\
&\leq\left|\left\{x:\sum\_{i=1}^n(|f\_k|-a\_k\lambda)>\lambda\right\}\right|\\
&\leq\left|\left\{x:\sum\_{i=1}^n(|f\_k|-a\_k\lambda)\_+>\lambda\right\}\right|.
\end{align}
For any $ \varepsilon>0 $, set
$$
g\_k(x)=\min\{(|f\_k|-a\_k\lambda)\_+,\lambda(1+\varepsilon)\}.
$$
Then
\begin{align}
\left|\left\{x:\left|\sum\_{i=1}^nf\_k\right|>\left(1+\sum\_{i=1}^na\_k\right)\lambda\right\}\right|&\leq\left|\left\{x:\sum\_{i=1}^ng\_k>\lambda\right\}\right|\\
&\leq\frac{\left\|\sum\_{i=1}^ng\_k\right\|\_{1,\infty}}{\lambda}\\
&\leq\frac{\sum\_{i=1}^n\|g\_k\|\_{1}}{\lambda}.
\end{align}
Finally, note that
\begin{align}
\|g\_k\|\_1&=\int\_{0}^{\lambda(1+\varepsilon)}|\{x:|g\_k|>\mu\}|d\mu\\
&\leq\int\_{0}^{\lambda(1+\varepsilon)}|\{x:|f\_k|>\mu+a\_k\lambda\}|d\mu\\
&\leq\int\_{0}^{\lambda(1+\varepsilon)}\frac{\|f\_k\|\_{1,\infty}}{\mu+a\_k\lambda}d\mu=\ln\left(\frac{a\_k\lambda+\lambda(1+\varepsilon)}{a\_k\lambda}\right).
\end{align}
Combining all above and letting $ \varepsilon\to 0 $, we can get the proof.
| 1 | https://mathoverflow.net/users/241460 | 431798 | 174,790 |
https://mathoverflow.net/questions/431777 | 5 | I was reading [this post about the Bell Numbers](https://math.stackexchange.com/a/1527678) where users Lucian and Vladimir Reshetnikov give us Dobiński's formula for the Bell numbers
$$ B(x) = \frac{1}{e} \sum\_{k=1}^{\infty} \frac{k^x}{k!}. $$
Now I was trying to reason about this function on the complex plane. It's easy to reason that for no value of $x$ such that $|x| < \infty$ will this expression have a singularity so this function definitely doesn't have any poles.
Now the other question I wanted to turn my attention to are, can we characterize the zeroes of this function? Are there only finitely many or countably many? Is there some natural characterization of the zeroes (like they all live on some well known curve(s))?
There also appears to be some Riemann surface structure attached to this wherein $k^x$ is multivalued for complex $x$ as it is $e^{\ln(k)x}$ where the choice of branch of $\ln(k)$ needs to be made.
| https://mathoverflow.net/users/46536 | Has anyone characterized the zeroes of the Bell numbers? | The function $B(z)$ is an example of an almost periodic function. The zeroes of an almost periodic function that is holomorphic on some strip are also almost periodic, so such a function either has no zeroes at all or infinitely many zeros.
Let $f(z)=\sum\_{k=0}^{\infty}a\_ke^{b\_kz}$ where each $b\_k$ is real and each $a\_k$ is complex. Suppose that $\alpha,\beta\in[-\infty,\infty]$ and that the sum converges uniformly and absolutely on compact subsets of the strip $\{z:\alpha<\text{Re}(z)<\beta\}$.
Now, there is a sequence of real numbers $r\_n$ with $r\_n\rightarrow\infty$ where
$$\lim\_{n\rightarrow\infty}(e^{ir\_nb\_k})\_{k=0}^{\infty}=(1)\_{k=0}^{\infty}.$$
This ensures that $f(z+ir\_n)\rightarrow f(z)$ and $f(z-ir\_n)\rightarrow f(z)$ uniformly on compact subsets of $\{z:\alpha<\text{Re}(z)<\beta\}$. Therefore, by Hurwitz' theorem, the locations of the zeroes of $f(z+ir\_n)$ will converge to the locations of the zeros of $f(z)$.
| 3 | https://mathoverflow.net/users/22277 | 431803 | 174,792 |
https://mathoverflow.net/questions/431802 | 3 | $\DeclareMathOperator\GL{GL}\DeclareMathOperator\SL{SL}\DeclareMathOperator\SU{SU}$I'm a PhD student in physics working in the broad area of photonic quantum computing. My current project looks at the equivalence of any two $n$-photon $m$-mode Fock states under linear optical evolution.
These two Fock states can be rewritten as symmetric and homogenous polynomials of degree $n$ in $m$ variables. The linear optical evolution implies a linear $\SU(m)$ i.e. special unitary matrices of dimension $m$, change of basis for these variables. So basically, given any two Fock states, I'm interested in calculating the polynomial invariants of representations of the compact Lie group $\SU(m)$.
I've been going through the book [*‘Algorithms in invariant theory’* by Bernd Sturmfels](https://doi.org/10.1007/978-3-7091-4368-1) but it only mentions the first fundamental theorem for $\GL(m,\mathbb{C})$ and $\SL(m,\mathbb{C})$ Lie groups. Does there then exist a relation between the polynomial invariants of $\SU(m)$ and its complexification $\SL(m,\mathbb{C})$?
The only related result I could find was in the article [‘Lifting smooth homotopies of orbit spaces’ by Gerald Schwarz](http://www.numdam.org/item?id=PMIHES_1980__51__37_0). The proposition (5.8) in it mentions the following algebra isomorphism:
\begin{equation}
\mathbb{R}[W]^{\rho (K)} \otimes \mathbb{C} \simeq \mathbb{C}[V]^{\sigma (G)},
\end{equation}
where $K$ is a compact Lie group and $G = K\_{\mathbb{C}}$ is its complexification. Their representations are also defined as $\rho: K \rightarrow \GL(W)$ and $\sigma: G \rightarrow \GL(V)$, where W is a real vector space and $V = W\_{\mathbb{C}}=W\otimes\_{\mathbb{R}}\mathbb{C}$.
I was hoping to get an easier description or examples of this algebra isomorphism if possible or any resources that point to the same. Also, is it possible to calculate the non-polynomial invariants that are related to the $L^{2}$ norm of the Fock state polynomials since they are also invariants under the $\SU(m)$ action?
| https://mathoverflow.net/users/490056 | Generators of polynomial invariant ring of compact Lie groups | The method/result you are looking for is commonly known under the name "unitary trick" (of Hurwitz and Weyl), - this keyword should bring you a great deal of accessible explanations of all possible levels (since your post does not give a clear idea of your level, it is better if you look through several different sources and choose a suitable one).
| 3 | https://mathoverflow.net/users/1306 | 431805 | 174,793 |
https://mathoverflow.net/questions/431769 | 7 | I am only a beginner in the field of type theory, and I'm wondering if the community could point me out a few open problems in the field. I have a good background in logic, in particular, proof theory and I have a reasonable knowledge on category theory. While I am interested in homotopy type theory, I'd like to receive answers that point to flavors of type theory other than HoTT because open problems in HoTT are not difficult to find on the web (problems in HoTT are still welcome though). This is my very first question in this site so please feel free to edit, point mistakes and criticize my question.
Thank you for your attention!
| https://mathoverflow.net/users/492345 | Open problems in type theory | Thierry Coquand has a list of open problems, based on Vladimir Voevodsky's research on Dependent Type Theory. Scroll to the end of this [presentation.](https://web.archive.org/web/20190518164032/https://eutypes.cs.ru.nl/eutypes_pmwiki/uploads/Meetings/Coquand.pdf)
And I presume you know the HoTT open problem list at [nLab](https://ncatlab.org/nlab/show/open+problems+in+homotopy+type+theory).
| 6 | https://mathoverflow.net/users/11260 | 431807 | 174,794 |
https://mathoverflow.net/questions/431804 | 0 | The ratio between the number of unordered couples of sets, with empty intersection between the two sets, and the total number of unordered couples of sets, for a powerset on $n$ elements without the empty set, $\mathcal{P}([n]) \setminus \emptyset$, is:
$$\frac{{n+1 \brace 3}}{{2^n-1 \choose 2}}=\frac{(1 + 3^n - 2^{n+1})}{(2^n-1)(2^n-2)}$$
Where ${n+1 \brace 3}$ denotes a Stirling number of the second kind. Is it possible to find a finite separating union closed family $\mathcal{F}$, $\emptyset \notin \mathcal{F}$, with size of the universe $|U(\mathcal{F})| = n$, with a biconnected Hasse diagram graph (without articulation vertices), and with the ratio defined as above higher than the value for $\mathcal{P}([n]) \setminus \emptyset$?
| https://mathoverflow.net/users/136218 | Number of couples of sets with empty intersection in a separating union-closed family of sets | The numerator grows faster than the denominator, so we can do better by making a minimal extension of a previous powerset: $2^{[k]} \cup \{[m] : k+1 \le m \le n \} \setminus \emptyset$ gives $$\frac{(1 + 3^k - 2^{k+1})}{(2^k+n-k-1)(2^n+n-k-2)}$$ and for $n \ge 5$ the optimal value of $k$ is less than $n$ and grows logarithmically.
```
k First n for which k is optimal
2 2
3 3
4 4
5 7
6 12
7 24
8 46
9 91
10 182
11 366
```
By brute force enumeration, $2^{[n]} \setminus \emptyset$ is optimal for $n \le 4$.
| 1 | https://mathoverflow.net/users/46140 | 431809 | 174,795 |
https://mathoverflow.net/questions/431748 | 8 | Here are some definitions:
A space is *homotopy finite* if it is homotopy equivalent to a finite CW complex.
A space *finitely dominated* if it is a retract of a homotopy finite space.
A space $X$ is a *Poincaré duality space* of dimension $d$ if there exists a pair
$$
({\mathscr L},[X])
$$
consisting of a rank one local system $\mathscr L$ on $X$ (i.e., a local coefficient system on $X$ which is locally isomorphic to $\Bbb Z$) and $[X] \in H\_d(X;{\mathscr L})$ is a twisted homology class such that the cap product homomorphism
$$
\cap [X]: H^\ast(X;{\mathscr E}) \to H\_{d-\ast}(X;{\mathscr E} \otimes {\mathscr L})
$$
is an isomorphism in all degrees, where $\mathscr E$ runs over all local systems on $X$.
My question is this:
**Question:** *Are there finitely dominated Poincaré duality spaces which are **not** homotopy finite?*
(Note: I am a bit embarrassed about not knowing the answer to this question.)
**Remarks:**
(1). If $X$ is a finitely dominated space with finitely presented fundamental group $\pi$, then Wall's finiteness obstruction $w(X) \in K\_0({\Bbb Z}[\pi])$ is defined. It is known that $X$ is homotopy finite if and only if $w(X)$ lies in the summand
${\Bbb Z} \cong K\_0({\Bbb Z}) \subset K\_0({\Bbb Z}[\pi])$.
(2). When $X= B\pi$ where $\pi$ is a discrete, finitely presented group and $X$ satisfies Poincaré duality, then $\pi$ is called a *Poincaré duality group.* In this case it automatically follows that $X$ is finitely dominated and $\pi$ is torsion free. It remains an open question as to whether $X$ is homotopy finite. However, it is known by work of Ian Leary that $w(X)$ is always a $2$-torsion element. Consequently, if $K\_0({\Bbb Z}[\pi])$ contains no 2-torsion, it follows that $X$ is homotopy finite.
(3). It has been conjectured that for any torsion free group $\pi$, the class group $K\_0(\Bbb Z[\pi])$ is torsion-free. If this conjecture holds, then every finitely presented Poincaré duality group $\pi$ will have the property that $B\pi$ is homotopy finite.
| https://mathoverflow.net/users/8032 | Finite domination and Poincaré duality spaces | Corollary 5.4.2 of Wall's article `Poincaré complexes I', *Ann. Math.* **86** (1967) 213-245 gives examples of 4-dimensional Poincaré complexes $X$ with fundamental group of prime order $p\geq 23$ for which the Wall finiteness obstruction $\chi(X)$ is non-zero.
Incidentally, Theorem 1.3 of the same article is very close to the result that you attributed to me, except that I put in extra hypotheses that imply that (in Wall's notation) $\sigma(X)^\*=\sigma(X)$. The thing I thought of as new in my article was considering PD groups over other rings such as $\mathbb{Q}$.
I think that Wall's $\sigma(X)$ is your $w(X)$ and Wall's $\chi(X)$ is the image of your $w(X)$ in the quotient $K\_0(\mathbb{Z}[\pi])/K\_0(\mathbb{Z})$.
| 9 | https://mathoverflow.net/users/124004 | 431812 | 174,797 |
https://mathoverflow.net/questions/431794 | 4 | Let $u$ be a solution of the heat equation
$$u\_t - \Delta u = 0, \qquad t >0, \ x \in \mathbb T^d$$
and $v$ be a solution of the bi-harmonic heat equation
$$v\_t +\Delta^2 v = 0, \qquad t >0, \ x \in \mathbb T^d$$
with the same initial data $f$.
Is it true that, for every fixed time $T >0$,
$$\|v(T,\cdot)\|\_{L^2} \le \|u(T,\cdot)\|\_{L^2}$$
holds? That is, in some sense, $v$ is "more dissipated"?
| https://mathoverflow.net/users/110835 | $L^2$ norm for solutions of evolution equations driven by different elliptic operators | Not necessarily. I mean, it depends upon the torus you consider. Notice that in the case of the standard one ${\mathbb T}^d={\mathbb R}^d/{\mathbb Z}^d$, the answer is positive. But if you torus is ${\mathbb R}^d/a{\mathbb Z}^d$, then it is positive if $a\le2\pi$ and negative otherwise. The reason is that both semi-groups $H\_t$ and $B\_t$ are co-diagonal with an orthogonal eigenbasis, but they have eigenvalues $e^{-t\mu}$ and $e^{-t\mu^2}$, where $\mu\_0=0<\mu\_1,\ldots$ are the eigenvalues of $-\Delta$. To have the required inequality is equivalent to having $\mu^2\ge\mu$ for every eigenvalue, that is $\mu\_1\ge1$. Whence the condition on the size of the torus.
More generally, if ${\mathbb T}^d={\mathbb R}^d/\Lambda$ where $\Lambda$ is a lattice, the inequality amounts to the fact that every non-zero element of the dual lattice $\Lambda^\*$ has norm $\ge\frac1{2\pi}$. Here $\Lambda^\*$ is the set of points $\alpha\in{\mathbb R}^d$ such that $\langle \alpha, p\rangle\in{\mathbb Z}$ for every $p\in\Lambda$.
**Edit**. Here are the details. Let me consider a general torus $T={\mathbb R}^d/\Lambda$ where $\Lambda$ is a lattice. Both the Laplacian and its square are diagonal in the Fourier basis of exponentials
$$\phi\_\alpha(x):=c\_\alpha\exp(2i\pi\alpha\cdot x),\qquad \alpha\in\Lambda^\*,$$
where the constant $c\_\alpha$ normalizes: $\|\phi\_\alpha\|\_{L^2}=1$.
The corresponding eigenvalues are $4\pi^2|\alpha|^2$, respectively $(4\pi^2|\alpha|^2)^2$. Decomposing an arbitrary data
$$u(0,\cdot)=\sum\_{\alpha\in\Lambda^\*}a\_\alpha\phi\_\alpha,$$
we have
$$u(T,\cdot)=\sum\_{\alpha\in\Lambda^\*}e^{4\pi^2|\alpha|^2T}a\_\alpha\phi\_\alpha,\qquad v(T,\cdot)=\sum\_{\alpha\in\Lambda^\*}e^{(4\pi^2|\alpha|^2)^2T}a\_\alpha\phi\_\alpha.$$
Using the fact that the Fourier basis is $L^2$-orthogonal, we find
$$\|u(T)\|\_{L^2}^2=\sum\_{\alpha\in\Lambda^\*}e^{8\pi^2|\alpha|^2T}|a\_\alpha|^2,\qquad \|v(T)\|\_{L^2}^2=\sum\_{\alpha\in\Lambda^\*}e^{32(\pi^2|\alpha|^2)^2T}|a\_\alpha|^2.$$
The required inequality is equivalent to
$$e^{32(\pi^2|\alpha|^2)^2T}\le e^{8\pi^2|\alpha|^2T}$$
for $T>0$ and every $\alpha$, that is to $2\pi|\alpha|\ge1$ for every non-zero element of $\Lambda^\*$.
| 4 | https://mathoverflow.net/users/8799 | 431815 | 174,798 |
https://mathoverflow.net/questions/431800 | 3 | For the elliptic equation with non-divergence form
$$
\sum\_{i,j=1}^na\_{ij}(x)\partial\_{ij}^2u=f\text{ in }B(0,1)\quad\text{and}\quad u=g\text{ on }\partial B(0,1),
$$
where $ \{a\_{ij}(x)\} $ is a matrix-valued function such that for any $ \xi\in\mathbb{R}^n $ and $ x\in B(0,1) $
$$
\mu|\xi|^2\leq\sum\_{i,j=1}^na\_{ij}(x)\xi\_i\xi\_j\leq\mu^{-1}|\xi|^2,\text{ with }\mu>0,
$$
and $ a\_{ij}(x)\in C^{0,\alpha}(\overline{B(0,1)}) $ with $ \|a\|\_{C^{0,\alpha}(B(0,1))}\leq M $, $ M>0 $. $ g\in C^{2,\alpha}(\overline{B(0,1)}) $ and $ f\in C^{0,\alpha}(\overline{B(0,1)}) $. Then Schauder estimates imply that
$$
\|u\|\_{C^{2,\alpha}(B(0,1))}\leq C\left\{\|f\|\_{C^{0,\alpha}(B(0,1))}+\|g\|\_{C^{2,\alpha}(B(0,1))}\right\}.
$$
I know that to show it, we can consider $ u-g $ and assume that $ g=0 $. I want to ask that if $ g\in C^{0,\alpha}(\overline{B(0,1)}) $ and $ f=0 $, what regularity of $ u $ can I get? I wonder if $ u\in C^{0,\alpha}(\overline{B(0,1)}) $ and satisfies
$$
\|u\|\_{C^{0,\alpha}(B(0,1))}\leq C\|g\|\_{C^{0,\alpha}(B(0,1))}.
$$
Can you give me some hints or references?
| https://mathoverflow.net/users/241460 | Schauder estimates with boundary conditions | The result is true. Let $L=\sum\_{ij}a\_{ij}D\_{ij}$ and consider $$L^{-1}: C^{2+\alpha}(\partial \Omega) \mapsto C^{2+\alpha}(\bar \Omega)$$ with $L^{-1}f=u$ is $Lu=0$ and $u=f$ at the boundary.
$L^{-1}$ is bounded from $C^{2+\alpha}(\partial \Omega)$ to $ C^{2+\alpha}(\bar \Omega)$ by the Schauder theory and from $C(\partial \Omega)$ to $C(\bar \Omega)$, by the maximum principle. By interpolation it is bounded from $C^\alpha (\partial \Omega)$ to $C^\alpha (\bar \Omega)$ (see for example Chapter 1 in the book "Analytic semigroups and optimal regularity in parabolic problems", by A. Lunardi).
| 2 | https://mathoverflow.net/users/150653 | 431835 | 174,804 |
https://mathoverflow.net/questions/431852 | 4 | Lurie's $\infty$-categorical Dold-Kan Correspondence relates simplicial objects and sequential diagrams in a stable $\infty$-category. Is there any reference for an equivalence to a category of homotopy chain complexes? That this equivalence holds is well-known, and I feel comfortable with the argument, but I'd like to save myself the trouble of writing it down and to give credit to whoever first did so. Actually I'm interested in cochain complexes, but that isn't important.
Specifically, I'm thinking of a homotopy coherent cochain complex as follows: let $\mathcal{P}\_{fin}\mathbb{Z}\_{>0}$ be the poset of finite subsets of the positive integers. Call $S\in\mathcal{P}\_{fin}\mathbb{Z}\_{>0}$ *orderly* if for all positive integers $n$, $n\leq \max(S)\implies n\in S$. A cochain complex in a pointed $\infty$-category $\mathcal{D}$ is a functor $$C:\mathcal{P}\_{fin}\mathbb{Z}\_{>0}\to \mathcal{D}$$ such that $C(S)=0$ whenever $S$ is not orderly.
| https://mathoverflow.net/users/28033 | Reference for the equivalence between chain complexes and sequential diagrams in a stable $\infty$-category | I believe <https://arxiv.org/abs/2109.01017> does what you want! The description of coherent chain complexes used there is a bit different than what you suggest, but they look equivalent at first glance.
| 4 | https://mathoverflow.net/users/39747 | 431856 | 174,812 |
https://mathoverflow.net/questions/431842 | 6 | Let $(M, \omega)$ be a holomorphic symplectic manifold of (complex) dimension $2n$. Let $x$ be a point in $M$. My understanding from the discussion and answers to [this MO question](https://mathoverflow.net/q/322447/184) is that there exists a neighborhood $U \subseteq M$ of $x \in M$, and a neighborhood $V \subseteq T^\* \mathbb{C}^n$, such that $V$ is symplecto-bi-holomorphic to $U$, where $T^\*\mathbb{C}^n$ is given its standard homomorphic symplectic form. In other words there is a holomorphic symplectic version of Darboux's theorem. If I have miss-understood that discussion please correct me!
Now given a holomorphic function $f: \mathbb{C}^n \to \mathbb{C}$, the graph of $df$ in $T^\*\mathbb{C}^n$ is a holomorphic Lagrangian submanifold.
**Question:** Given any holomorphic Lagrangian submanifold $L \subseteq M$, is it locally isomorphic to a holomorphic Lagrangian of this form?
More specifically, given any $x \in L$, do there exist a neighborhood $U \subseteq M$ of $x$ and a neighborhood $V \subseteq T^\*\mathbb{C}^n$ and a symplecto-bi-holomorphism $U \cong V$, as above, such that under this isomorphism $L \cap U$ coincides with the graph of $df$ for some holomorphic function $f$ (where $f$ is defined on, say, $V \cap \mathbb{C}^n$)?
Remark: We know (see [here](https://mathoverflow.net/a/350111/184)) that there is no holomorphic version of the Weinstein Lagrangian Neighborhood theorem. This question is morally asking if there is a much weaker local version that still holds holomorphically.
| https://mathoverflow.net/users/184 | Are holomorphic Lagrangians locally graphs? | The answer is 'yes'. Specifically, the holomorphic version of the Darboux-Weinstein theorem holds, just as it does in the smooth category. In particular, if $L\subset M$ is a holomorphic Lagrangian submanifold where $M$ is a holomorphic symplectic complex manifold and $x\in L$ is specified, then there is an open $x$-neighborhood $U\subset M$ on which there exist holomorphic coordinates $z^1,\ldots,z^{2n}$ such that $L\cap U$ is defined by $z^{n+1} = z^{n+2} = \cdots = z^{2n}=0$ and the symplectic form $\omega$ on $U$ has the form
$$
\omega = \mathrm{d}z^1\wedge\mathrm{d}z^{n+1} + \cdots +\mathrm{d}z^n\wedge\mathrm{d}z^{2n}.
$$
The proof is the same as in the classical case, because one only uses ODE existence and uniqueness, which is true in the holomorphic category just as it is in the smooth category.
| 9 | https://mathoverflow.net/users/13972 | 431857 | 174,813 |
https://mathoverflow.net/questions/431859 | 2 | Let $\wp(u) = \frac{1}{u^2} + \sum\limits\_{\omega \in L, \omega \neq 0} \left(\frac{1}{(u-\omega)^2} - \frac{1}{\omega^2}\right)$ be a Weierstrass pe function.
My question is, how can I calculate $\wp(αu), α\in \Bbb{C}$, $αL⊆L$
If $αL=L$, the answer is easy, we can replace sum up over $L$ to sum up over $αL$, $\wp(αu)=1/α^2\wp(u)$.
So the problem is in the case of $αL\subsetneq L$. Are there some known results ?
| https://mathoverflow.net/users/144623 | How can I calculate $\wp(αu), α\in \Bbb{C}$, $αL⊆L$ | $\alpha L$ is of finite index in $L$ so you can write $L = \cup\_{i=1}^n (\alpha L + \lambda\_i)$ for some $\lambda\_i, n$.
If you rewrite the sum defining $\wp(\alpha u)$ as a sum over $\cup\_{i=1}^n (\alpha L + \lambda\_i)$ and rearrange terms (without worrying too much about convergence) you should get something like
$$\wp(\alpha u) = (1/\alpha)^2 \sum\_{i=1}^n \wp(u - \lambda\_i/\alpha) + c$$
for some constant $c$. As Gro-Tsen suggested, you should think in terms of isogenies of elliptic curves and this boils down to Velú's formulas.
| 3 | https://mathoverflow.net/users/2290 | 431879 | 174,819 |
https://mathoverflow.net/questions/431867 | 1 | I'm reading Escobar's The Yamabe Problem On Manifolds With Boundary.
He says
>
> Let $(y\_{1},\cdots,y\_{n})$ be normal coordinates around $0\in \partial M$, such that $\eta(0)=-\frac{\partial}{\partial y\_{n}}$, and second fundamental form of $\partial M$ at 0 has a diagonal form.
>
>
>
Here $M$ is a Riemannian manifold with boundary and $0$ is a nonumbilic point on $\partial M$.$\eta$ represents the outward normal.
I wonder how can we take such a normal coordinates. As I know,normal coordinates can be taken at boundary point only when the boundary is locally totally geodesic. And in this case the second fundamental form must be 0.
In fact if we take a normal coordinates at $0$, which satisfies $g\_{ij}(0)=\delta\_{ij},\Gamma\_{ij}^{k}(0)=0.$
Then we compute
$$h\_{ij}(0)=g(\nabla\_{\partial y^{i}}\partial y^{n},\partial\_{y^{j}})=\Gamma\_{in}^{k}g\_{kj}=0.$$
So how can we take such a normal coordinate? Any help will be thanked.
| https://mathoverflow.net/users/148247 | Local geometry of nonumbilic points | You can—at least locally—find a normal coordinate system adapted to any submanifold $N \subset M$. (You can extend $M$ past its boundary to have $N = \partial M$ lie in the interior, but really that is not necessary here.)
Now suppose you have normal coordinates $(y^1,\dots,y^n)$ near a point $0 \in \partial M$, with $\partial y^n = -\eta$ in a neighbourhood of said point.
The formula you give for the second fundamental form does not look right; it should be in terms of the induced connection $\nabla'$ on $\partial M$:
\begin{equation}
h\_{ij}
= g(\nabla'\_{\partial y^i}\partial y^j,\eta)
= g(\partial y^j,\nabla'\_{\partial y^i} \partial y^n),
\end{equation}
where the sign is switched around because $\partial y^n = -\eta$.
As $h\_{ij}(0)$ is a symmetric $(n-1) \times (n-1)$ matrix, we can diagonalise it via an orthonormal change of basis. You leave $y^n$ unchanged, and modify the rest of the coordinates $(y^1,\dots,y^{n-1})$ through this same change of basis to obtain the desired coordinate system.
| 1 | https://mathoverflow.net/users/103792 | 431882 | 174,821 |
https://mathoverflow.net/questions/431883 | 2 | $\newcommand{\complex}{\mathbb{C}}\newcommand{\real}{\mathbb{R}}\newcommand{\proj}{\mathbb{P}}\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Seg{Seg}$I apologize in advance for my naïve understanding of real algebraic geometry. I define a *real projective variety* to be a subset of $\mathbb{P}(\real^n)$ that is the zero locus of some finite collection of real homogeneous polynomials $f\_1,...,f\_p \in \real[x\_1,...,x\_n]$.
Let $X, Y \subseteq \mathbb{P}(\real^n)$ be real projective varieties, and let
$$\Seg: \proj(\real^n)\times \proj(\real^n) \rightarrow \proj(\real^{n}\otimes \real^n)$$
be the Segre embedding $(v,w)\mapsto v\otimes w$. Is it true that $\Seg(X \times Y) \subseteq \proj(\real^n \otimes \real^n)$ is a real projective variety?
Assuming that $\Seg(X \times Y)$ is a real projective variety, I have a follow-up question: Let $\Pi\in \Hom\_{\real}(\real^n \otimes \real^n)$ be the linear map that acts as $\Pi(v \otimes w)=\frac{1}{2}(v\otimes w + w \otimes v)$. Is it true that $\Pi(\Seg(X \times Y)) \subseteq \proj(\real^n \otimes \real^n)$ is a real projective variety?
| https://mathoverflow.net/users/150898 | Is the Segre embedding of two real varieties a real variety? | $\mathrm{Seg}(X\times Y)$ is a real projective variety since the full Segre map is an isomorphism of real algebraic varieties onto its image.
As for your second question, I think the answer is "no". If you take $X=Y=\mathbf{P}(\mathbf R^2)$, the composition $\Pi\circ\mathrm{Seg}$ is nothing but a quotient map for the action of the symmetric group $S\_2$ on $\mathbf{P}(\mathbf R^2)\times\mathbf{P}(\mathbf R^2)$. The diagonal is the set of fixed points of this action. In general, if a finite group acts on a smooth real algebraic variety and if there are isotropy groups of even order, the quotient is only semi-algebraic and not real algebraic.
Here, this can be seen easily: if you denote the basis of $\mathbf{R}^2$ by $X,Y$, the image of $\Pi\circ\mathrm{Seg}$ is contained in the projectivization $\mathbf P(\mathrm{Sym}^2)$ of the symmetric square $\mathrm{Sym}^2$ of $\mathbf{R}^2$ whose basis is $X^2,XY,Y^2$. The image is the subset of $\mathbf P(\mathrm{Sym}^2)$ of nonzero real quadratic forms in $X,Y$ having positive discriminant, which is clearly a semi-algebraic and non-algebraic subset of $\mathbf P(\mathrm{Sym}^2)$.
| 4 | https://mathoverflow.net/users/85592 | 431889 | 174,822 |
https://mathoverflow.net/questions/431847 | 2 | I would like to know if there is a way of finding the inverse function of $f(x)=e^{-\varepsilon x}\sinh(x)$ with $-1<\varepsilon<0$.
It seems there is no simple way even if we consider Lambert or Gudermann-like functions.
| https://mathoverflow.net/users/492387 | Expression of the inverse function of $f(x)=e^{-\varepsilon x}\sinh(x)$ | $$y=e^{- \epsilon x} \sinh (x)=\sum\_{n=0}^\infty \frac{(1-\epsilon )^n+(-1)^{n+1} (1+\epsilon )^n}{2 n!}\,x^n$$ Now use [series reversion](https://mathworld.wolfram.com/SeriesReversion.html) using the explicit formula for the $n^{\text{th}}$ term as given by Morse and Feshbach.
This will give
$$x=y+\sum\_{n=2}^\infty a\_n y^n$$ the first coefficients being
$$\left(
\begin{array}{cc}
2 & \epsilon \\
3 & \frac{1}{6} (3 \epsilon -1) (3 \epsilon +1) \\
4 & \frac{2}{3} \epsilon (2 \epsilon -1) (2 \epsilon +1) \\
5 & \frac{1}{120} (5 \epsilon -3) (5 \epsilon -1) (5 \epsilon
+1) (5 \epsilon +3) \\
6 & \frac{2}{15} \epsilon (3 \epsilon -2) (3 \epsilon -1) (3
\epsilon +1) (3 \epsilon +2) \\
\end{array}
\right)$$
| 2 | https://mathoverflow.net/users/42185 | 431892 | 174,824 |
https://mathoverflow.net/questions/431888 | 4 | **Context**
Given a finitary monad $T:\operatorname{gSet}\_n\to\operatorname{gSet}\_n$ we can define categories $\operatorname{Comp}\_k^T$ of $k$-computads for $T$, for any $k=0,\cdots,n+1$. This is nicely explained in Schommer-Pries' [thesis](https://arxiv.org/abs/1112.1000), for example and the original source is this [paper](https://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.53.2118) of Batanin. Essentially a $k$-computad $C$ is defined inductively as a tuple $(C\_k,C\_{\leq k -1},s,t)$ where $C\_k$ is a set (the set of $k$-cells), $C\_{\leq k-1}$ is a $(k-1)$-computad and $s,t$ are maps $C\_k\to [F\_{k-1}(C\_{\leq k-1})]\_{k-1}$ where $F\_{k-1}$ is the functor that constructs the $T$-algebra generated by a $(k-1)$-computad. These are required to satisfy the usual globularity conditions $ss=st$ and $ts=tt$. Then one completes the induction by defining $F\_k$ by a certain pushout diagram in $\operatorname{Alg\_T}$ (using the fact that $\operatorname{Alg\_T}$ is cocomplete, because $T$ is finitary).
**Questions**
1. Is the category $\operatorname{Comp}\_k^T$ always cocomplete?
2. Does the functor $\operatorname{Comp}\_k^T\to\operatorname{Set}$ that takes a computad to its set of $i$-cells for some $i\leq k$ preserve colimits?
3. Is there a reference for these facts?
If there is no reference, then
4. Is the sketch of proof below correct, or am I missing something?
**Sketch of proof of 1. and 2.**
One constructs the colimit of some diagram $C(i)$ of $k$-computads by induction on $k$: one constructs a $k$-computad whose set of $k$-cells is the colimit of the diagram on $k$-cells and whose underlying $(k-1)$-computad is the colimit of the diagram of underlying $(k-1)$-computads. Source and target maps $\operatorname{colim}\_i C(i)\_k\to [F\_{k-1}(\operatorname{colim}\_i C(i)\_{\leq k-1})]\_{k-1}$ can be defined by the composite $$C(i)\_k\to [F\_{k-1}(C(i)\_{\leq k-1})]\_{k-1}\to\operatorname{colim}\_i [F\_{k-1}(C(i)\_{\leq k-1})]\_{k-1}\to[\operatorname{colim}\_iF\_{k-1}(C(i)\_{\leq k-1})]\_{k-1}=[F\_{k-1}(\operatorname{colim}\_i C(i)\_{\leq k-1})]\_{k-1}$$ where the equality comes from the fact that $F\_{k-1}$ is left adjoint and the final arrow is induced by the maps $[F\_{k-1}(C(i)\_{\leq k-1})]\_{k-1}\to[\operatorname{colim}\_iF\_{k-1}(C(i)\_{\leq k-1})]\_{k-1}$ which are the maps of underlying $(k-1)$-morphisms associated to the canonical maps of $T$-algebras $F\_{k-1}(C(i)\_{\leq k-1})\to\operatorname{colim}\_iF\_{k-1}(C(i)\_{\leq k-1})$.
Now one needs to check that $s,t$ satisfy globularity and that the construction has the right universal property. This also seems completely straightforward.
**What I have found in the literature**
In the above cited paper of Batanin, it is proved that when the monad is *truncable* and preserves finite pullbacks, the category $\operatorname{Comp}\_n^T$ is an elementary topos, so it particular has finite colimits. It is also proved that if additionally $T$ preserves wide pullbacks and the unit is cocartesian then $\operatorname{Comp}\_n^T$ is a presheaf topos, so in particular is cocomplete. This appears as Theorem 4.1.
I am hoping, however, that if one only needs cocompleteness then the hypotheses are not necessary and there is the above direct proof.
| https://mathoverflow.net/users/14078 | Is the category of computads for a finitary monad on $n$-globular sets cocomplete? | The answer to $1$ and $2$ are both yes. I don't know if this appears in the literature. The argument you give seems reasonable - I don't completely follow your notation but the general idea is that in the "inductive" definition of computads, you can show that colimits of "k-computads" are computed by taking the colimit of the (k-1)-computads and the sets of k-cells separately, which seems to be exactly what you are saying. Colimits of general computads are then computed by taking the colimit of their "underlying k-computads" for all k.
I would consider this as a folklore results in the area, and I wouldn't be surprised if it is written somewhere, at least for the special case of the strict $\infty$-category monads, but I couldn't find a reference to give you.
Note however that the results of Batanin you quote is unfortunately false - the category of computads for the free strict $\infty$-category (which satisfies all the assumption of Batamin's theorem) is not a topos. In fact it is [not cartesian closed](https://arxiv.org/abs/0710.5202) (or [here](https://arxiv.org/abs/1209.0414)).
A [latter paper](https://arxiv.org/abs/math/0209035) of Batanin gives a finer criterion for such categories of computads to indeed be presheaf categories which I think\* is correct. This second paper applies for example to the computads for Batanin's definition of weak $\infty$-categories.
\*: To be clear, I have to admit I have never been able to fully understand the argument given by Batanin in this second paper, and the paper has never been published, so I can't tell if the proof given there is correct (though I have no reasons to doubt it is). But, this being said I have very good reason to believe the result itself is definitely correct due to my [own](https://arxiv.org/abs/1711.00744) work on the topic which involves similar assumptions.
| 6 | https://mathoverflow.net/users/22131 | 431904 | 174,826 |
https://mathoverflow.net/questions/431731 | 1 | Let $M$ be some non-well-founded model of $\sf ZF$, can we have a sequence $(S\_n)\_{n \in \mathbb N}$ of nonempty sets in $M$, where each $S\_n \subset \mathcal P(S\_{n+1})$; and such that there exists a sequence of bijective functions $(f\_n)\_{n \in \mathbb N}: S\_{n+1} \to S\_n$, having : $f\_n(x) = f\_{n+1}[x]$?
This can work in $\sf ZF-Reg.$ like singleton maps "$x \mapsto \{x\}$" between sets of iterated singletons. But here $M$ satisfies Foundation.
| https://mathoverflow.net/users/95347 | Can we have such an infinite descending sequence of functions with prior ones inside their successors? | (Note that I am assuming the axiom of choice in the metatheory. If you want to ask if this holds without choice, clarify and add the appropriate tag. I suspect that this particular compactness instance should be true even without choice, but I am not an expert, nor did I think much about it.)
This is true by a standard compactness argument (assuming ZF is consistent). You can even demand that each $S\_n$ be singleton.
Namely, fixing an arbitrary model $M$ of ZF, the set $\{x\_n=\{x\_{n+1}\}\mid n\in \mathbf N\}^\dagger$ is a type in $M$ in variables $(x\_n)\_{n\in\mathbf N}$ (clearly, every finite part is realised in $M$), so there is an elementary extension $N\succeq M$ in which it is realised. In particular, $N$ is a model of ZF (in fact, of the theory of $M$) with the sequences you wanted.
$\dagger$: naturally, here, $x\_n=\{x\_{n+1}\}$ is short for $x\_{n+1}\in x\_n\land\forall y(y\in x\_n\rightarrow y=x\_{n+1})$
| 3 | https://mathoverflow.net/users/54415 | 431911 | 174,828 |
https://mathoverflow.net/questions/431900 | 6 | Let $\xi$ be a random variable valued in the space of Schwartz distributions $\mathcal{S}'(\mathbb{R}^d)$.
For any open set $R\subset\mathbb{R}^d$ let $\Sigma(R)$ be the $\sigma$-algebra generated by $\{\xi(f)\,|\,f\in\mathcal{S}(\mathbb{R}^d), \,\textrm{supp}\,f\subset R\}$.
For any $\lambda\in(0,\infty)$ the random variable $\xi\_\lambda$ valued in $\mathcal{S}'(\mathbb{R}^d)$ is defined by $\xi\_\lambda(f) = \xi(f\_\lambda)$ for all $f\in\mathcal{S}(\mathbb{R}^d)$, where $f\_\lambda(x) = \lambda^{d/2}\,f(\lambda x)$.
Assume that:
(1) the $\sigma$-algebras $\Sigma(R\_1)$ and $\Sigma(R\_2)$ are independent for any disjoint open sets $R\_1,R\_2\subset\mathbb{R}^d$.
(2) for any $\lambda\in(0,\infty)$ the law of $\xi\_\lambda$ coincides with the law of $\xi$.
Is it true that under the above assumptions $\xi$ must be the white noise?
| https://mathoverflow.net/users/47256 | Abstract characterization of white noise | These assumptions are not sufficient. Take $d=1$ and for simplicity let us work with the antiderivative of $\xi$. You are then asking if a process with independent increments and Brownian scaling is necessarily a Brownian motion. Any $\alpha$-stable Levy process $L$ has independent increments and scaling $(L\_t)\_{t\geq 0}\sim\lambda^{1/\alpha}(L\_{\lambda t})\_{t\geq0}$, so the process $X\_t=L\_{t^{\alpha/2}}$ is a counterexample.
My guess is that translation adding translation invariance and nontriviality (note that actually $\xi\equiv 0$ is also a counterexample) to the assumptions should be sufficient.
| 6 | https://mathoverflow.net/users/100941 | 431914 | 174,829 |
https://mathoverflow.net/questions/431795 | 0 | While reading a preprint [Eldan, Lehec, and Shenfeld - Stability of the logarithmic Sobolev inequality via the Föllmer Process](https://arxiv.org/abs/1903.04522) I came across the following SDE in Section 3:
$$d X\_t=d B\_t+\nabla \log P\_{1-t} f\left(X\_t\right) d t$$
where $B\_t$ is the Brownian motion and $P\_t$ denotes the heat semigroup.
Given a probability measure $\mu(dx)=f(x)\gamma(dx)$ where $\gamma$ is the gaussian density, one can consider $X\_t$ to be the process that is close to a Brownian motion while having the law $\mu$ at time $t=1$.
I would like to understand why the following change of measure formula holds for every test function $u$,
$$\mathbb{E}\left[u\left(X\_1\right) \mid \mathcal{F}\_t\right]=\frac{P\_{1-t}(u f)\left(X\_t\right)}{P\_{1-t} f\left(X\_t\right)}$$
where $\mathcal{F}\_t$ denotes the natural filtration associated with the process $X\_t$. I looked into Girsanov's theorem in this regard but I don't see any exponential martingales in the above expression, so I am not sure what is going on here.
| https://mathoverflow.net/users/68232 | Change of measure formula for the Föllmer process | One can show it using the Feynman-Kac formula. In particular, from Feynman-Kac, we know that $h(x, t) := \mathbb{E}[u(X\_1) \mid \mathcal{F}\_t ]$ solves the following PDE:
$ \frac{\partial}{\partial t} h(x, t) + \nabla \log P\_{1-t} f(X\_t)^T \nabla\_x h(x, t) + \frac{1}{2} \Delta\_x h(x, t) = 0.$
with the terminal condition $h(x, 1) = u(x)$.
Now, note that your proposed function for $h(x, t)$, namely $\frac{P\_{1-t} (uf)}{P\_{1-t}(f) }$ satisfies the terminal condition, so you just need to plug your proposed function to the above equation to show that it satisfies the PDE above.
| 2 | https://mathoverflow.net/users/116327 | 431925 | 174,832 |
https://mathoverflow.net/questions/431814 | 2 | We have an algebraic number $a$ and a real number $b$. Can the following inequality have infinitely many solutions for $n \in \mathbb{N}$?
$$ \{an\} \in [b - \frac{1}{2^n}, b + \frac{1}{2^n}] $$
Here $\{x\}$ denotes the fractional part of $x$, $\{x\} = x - [x]$.
**Background**. I encountered this problem when I was dealing with some kronecker approximations, I wanted to show that for a given point $(b\_1t \mod 2\pi, b\_2t \mod 2\pi)$ where $b\_1,b\_2$ are algebraic numbers who are linearly independent over rational numbers, one can not get exponentially close to a given fixed point $(\phi\_1,\phi\_2) \in [-1,1]^2$. Some special cases of this problem could be solved by Baker but I was unable to solve it in the general case.
| https://mathoverflow.net/users/492366 | Almost Diophantine approximation | Such $b$ exists for every real $a$.
Define an increasiung sequence of positive integers $n\_1,n\_2,\dots$ as follows. Picj $n\_1$ so that $\{an\_1\}<1/2$. If $n\_i$ has been already defined, choose $n\_{i+1}>n\_i$ such that
$$
\{an\_{i+1}\}\in\left[\{an\_i\},\{an\_i\}+\frac1{2^{n\_i+1}}\right];
$$
such $n\_{i+1}$ clearly exists (notice here that $\{an\_i\}\leq 1-1/2^{n\_i}$). Then the sequence of segments
$$
\left[\{an\_i\},\{an\_i\}+\frac1{2^{n\_i}}\right]
$$
is nested, hence its intersection is a desired point $b$.
| 1 | https://mathoverflow.net/users/17581 | 431933 | 174,834 |
https://mathoverflow.net/questions/431936 | 7 | Faculty members are encouraged to highlight the connection between the courses we teach and climate change, and raise awareness of the issue in our lectures, across subjects in my university. I am wondering what is usually done in that respect in mathematics. To formulate a question (answers to any of the three variants are welcomed) :
* What are the mathematics of planet earth initiatives that can be retold within undergraduate and graduate classes?
* Outside of optimization / numerical analysis, are there other areas where net impact on energy via mathematics can be highlighted and has been investigated (cryptography perhaps at a graduate level)?
* Has anyone tried to convert (negative) powers of $n$ in error estimates into KW, or given examples of the impact of optimization / numerical analysis in terms of reduction of energy usage?
I first asked this question on Math SE, but it received a -3 vote in a few minutes, so I removed it. Maybe this is a more adequate place to ask.
---
To clarify : more than (sets of) lectures on the mathematics of climate change, or awareness lectures, I was wondering whether some examples can be found that could be embedded in run of the mill Calculus classes, Algebra classes, or other classes taught in many universities.
---
To clarify even further. What we are encouraged to do is to include in our curriculum elements concerning adaptation to climate change. It can take many forms, such as adding additional lectures to the common core modules, possibly coming from other disciplines; it can also be of the form described above. Since this is the part relevant to MO, that is the focus of this question. The lively debate in the comments about the motivations of such a question and /or politics within mathematics wasn't intended. I should have worded it perhaps in terms of adapting traditional "real world" examples that appear in many classes to climate change.
| https://mathoverflow.net/users/40120 | Mathematics of sustainable development and energy sobriety in the classroom | My former colleague David Mond at Warwick [has developed some materials on this issue](https://homepages.warwick.ac.uk/~masbm/climate.html). He has some talks (at school and u/g level) on Climate change and game theory. He also lists some links there, to [a page by John Baez](https://math.ucr.edu/home/baez/what/hong_kong.html), and an [MSRI document](http://library.msri.org/msri/MathClimate.pdf), even if both of these are somewhat old now.
| 13 | https://mathoverflow.net/users/6107 | 431938 | 174,835 |
https://mathoverflow.net/questions/431924 | 1 | Let $G$ be a finite group, $p$ a prime number, and $k$ an algebraically closed field of characteristic $p$. Then we can consider the cohomological variety of $G$, namely the maximal spectrum $V\_G$ of the graded algebra $H^\bullet(G,k)=\operatorname{Ext}^\bullet\_{k[G]}(k,k)$. By the Quillen stratification, this space has a stratification by quotients of the cohomological varieties of elementary abelian subgroups $E$ of $G$ (where an elementary abelian subgroup is one of the form $(\mathbb{Z}/p\mathbb{Z})^n$). To be precise, for an elementary abelian subgroup $E$, one has a map $V\_E/W\_E\to V\_G$, where $W\_E=N\_G(E)/C\_G(E)$. These maps are inseparable isogenies away from the locus cutting out smaller subgroups $E'$ of $E$, and together they give our stratification.
Now my $\textbf{question}$ is: if the Sylow $p$-subgroup $P$ of $G$ is itself elementary abelian, then do we obtain that $V\_G$ is simply $V\_P/W\_P$?
Clearly this will be true on an dense open locus; further it holds in the case when $P$ is normal by the Lyndon-Hochschild-Serre spectral sequence. But I'm not clear about the general case.
| https://mathoverflow.net/users/97652 | Cohomological variety in case that Sylow subgroup is elementary abelian | Yes. There is a stronger result too, which predates Quillen's theorem: if $G$ is a finite group whose Sylow $p$-subgroup $P$ is abelian, then the restriction map $H^\*(G;\mathbb{F}\_p)\rightarrow H^\*(P;\mathbb{F}\_p)$ has image equal to the fixed points for the action of the normalizer $N\_G(P)$ on $H^\*(P;\mathbb{F}\_p)$. Here I'm using $\mathbb{F}\_p$ to denote the field of $p$ elements; the same theorem for other fields of characteristic $p$ follows. This theorem is due to Swan in "The $p$-period of a finite group", *Ill. J. Math* **4** (1960) 341-346. Or see Theorem II.6.8 of the book "Cohomology of Finite Groups" by Adem and Milgram.
| 3 | https://mathoverflow.net/users/124004 | 431939 | 174,836 |
https://mathoverflow.net/questions/431901 | 4 | Let $D \subset \mathbf{R}^2$ be the unit disc, and $L > 0$. Let $u: D \times (-L,L) \to \mathbf{R}$ satisfy
\begin{equation}
\begin{cases}
\Delta u = 0 \quad \text{ on $D \times (-L,L)$ } \\
\frac{\partial u}{\partial z} = 0 \quad \text{ on $D \times \{ -L , L \}$}.
\end{cases}
\end{equation}
**Question.** Is it possible that the singular set $S(u) = \{ u = 0 \} \cap \{ \lvert Du \rvert = 0 \}$ is exactly $\{(0,0,0)\}$? Does the answer depend on $L$?
* Trying separation of variables is a natural reflex, but this produces $u$ with either no singularities or a much larger singular set.
* Without the Neumann-type boundary conditions on the ends of the cylinder there are examples: the harmonic polynomial $p(x,y,z) = 2x^3 - 3xy^2 - 3xz^2$ is one.
| https://mathoverflow.net/users/103792 | Is there a harmonic function with just one singular point? | **Yes**, this is possible. An explicit example is
$$u(x, y, z) = 1 - I\_0\left(\sqrt{x^2 + y^2}\right) \, \cos z$$
when $L = \pi$, and $u\big(\frac{\pi x}{L}, \frac{\pi x}{L}, \frac{\pi x}{L}\big)$ for a general $L$. Here $I\_0$ is the Bessel $I$ function.
| 7 | https://mathoverflow.net/users/108637 | 431940 | 174,837 |
https://mathoverflow.net/questions/431793 | 7 | Consider the Fréchet spaces $C^\infty(\mathbb{R},\mathbb{R})$ and $\mathbb{R}^\infty$, and the continuous linear map
$$
J\colon C^\infty(\mathbb{R},\mathbb{R}) \to \mathbb{R}^\infty
$$
returning the infinite jet at 0, which is a surjection by [Borel's lemma](https://ncatlab.org/nlab/show/Borel%27s+theorem). Here $\mathbb{R}^\infty$ is the set of all sequences of real numbers, with the family of seminorms induced by the truncations to the first $n$-coordinates. The map $J$ does not have a continuous linear section, but it has a continuous *non-linear* section. What I'd like to know is if $J$ has a **smooth** nonlinear section, even if just in a neighbourhood of $0$. Here smoothness is taken in the sense of [Michal–Bastiani](https://ncatlab.org/nlab/show/Michal-Bastiani+smooth+map).
I recently learned that smooth maps $\mathbb{R}^n \to \mathbb{R}^\infty$ lift smoothly to $C^\infty(\mathbb{R},\mathbb{R})$ (via Enxin Wu's paper *[Homological Algebra for Diffeological Vector Spaces](https://arxiv.org/abs/1406.6717)*), which makes this projection maps a *subduction* of the [associated diffeological spaces](https://ncatlab.org/nlab/show/diffeological+space#RelationBetweenDeffeologicalAndFrechetStructure), making our friend $J$ above a diffeological principal bundle (which, I remind you, are not assumed locally trivial!). But I do wonder if it's a bundle in the traditional sense, in the category of Fréchet manifolds. Perhaps a continuous section could be smoothed, but I really am grasping at straws.
| https://mathoverflow.net/users/4177 | Is the Borel lemma projection a smooth principal bundle? | **No**, there is not even any $C^1$ (in the Michal−Bastiani, i.e. Keller $C\_c$ sence) map $\mathbb R^\infty=G\sqsupseteq{\rm dom}\,f\to E=C^\infty(\mathbb R)=C^\infty(\mathbb R,\mathbb R)$ with ${\rm dom}\,f$ a zero neighbourhood and $J\circ f={\rm id}$ on ${\rm dom}\,f$. The argument goes as follows. Supposing there is, for $j={\rm D}\,f(0)$ we have $J\circ j={\rm id}$ on $G$. Then for $F$ the subspace of $E$ formed by functions that are infinitely flat at $0$, and for $\rho:E\to F$ given by $x\mapsto x-j\circ J\,x$ we have $\rho$ the identity on $F$ which is shown to be impossible in Corollary 7.1.3 on page 206 in the Frölicher−Kriegl book *Linear Spaces and Differentiation Theory*. Note that since all spaces here are Fréchet, there is only one reasonable concept of smoothness that (in this restricted case) also equals that in the FK book. A $\underline{\rm Con}$−morphism (there and here; in this case) just means a continuous linear map.
| 6 | https://mathoverflow.net/users/12643 | 431944 | 174,839 |
https://mathoverflow.net/questions/431958 | 3 | Let $f: [0, 1] \to \mathbb R$ be a measurable function. A function $g: [0, 1] \to \mathbb R$ is said to be a *condensation limit* of $f$ if $g$ is continuous and agrees with $f$ on a dense subset of $[0, 1]$.
Let $k \geq 1$ be an integer, and $f: [0, 1] \to \mathbb R$ be a measurable function whose graph is dense in $[0, 1] \times \mathbb R$.
**Question:** Is the set of $k$ times continuously differentiable condensation limits of $f$ dense in $C^k$?
*Note: Here $C^k$ denotes the space of $k$ times continuously differentiable functions under the norm $\| g \|\_{C^k} := \sum\_{i = 0}^k \sup \_{x \in [0, 1]} \lvert g^{(i)} (x) \rvert$, where $g^{(i)}$ is the $i$’th derivative of $g$.*
| https://mathoverflow.net/users/173490 | Restriction to dense subset of functions whose graph is dense | **Yes.** This is true.
Proposition: Let $A\subseteq[0,1]\times\mathbb{R}$ be dense. Then for each $k\geq 0$ and $\epsilon>0$ and $g\in C^k([0,1])$, there is some $C^k$ function $f:[0,1]\rightarrow\mathbb{R}$ where $\{x:(x,f(x))\in A\}$ is dense in $[0,1]$ and where $\|f-g\|<\epsilon$.
Proof: We shall construct a sequence of functions by recursion. Let $f\_0=g$. If $f\_n$ has already been constructed, then let $U\_n=(0,1)\setminus\overline{\{x:(x,f\_n(x))\in A\}}$. If $U\_n=\emptyset$, then we have already constructed our required function $f$ and we do not need to continue this construction. Otherwise, $U\_n$ is a non-empty open set, so $U\_n$ is the union of open intervals. Choose the interval $(a\_n,b\_n)$ in $U\_n$ with the maximum length, and then let $f\_{n+1}:[0,1]\rightarrow\mathbb{R}$ be a smooth function where $\|f\_{n+1}-f\_n\|\_{C^k}<\epsilon 2^{-(n+1)}$ and where $f\_{n+1}-f\_n$ is supported on $(a\_n,b\_n)$ but where
$(c,f\_{n+1}(c))\in A$ for some $c\in(a\_n+(b\_n-a\_n)/3,a\_n+2(b\_n-a\_n)/3)$. Then $f\_n$ is Cauchy, so $f\_n\rightarrow f$ for some $f$ in the norm, (and therefore $f\_n\rightarrow f$ pointwise as well). We observe that $(U\_n)\_n$ is a decreasing sequence of open sets with $b\_n-a\_n\rightarrow 0$. We also observe that $|g-f\|<\epsilon$. Q.E.D.
One can also generalize this question in many ways (such as if we replace $[0,1]$ and $\mathbb{R}$ with differentiable manifolds, complex manifolds, topological spaces and if we require $f$ to be smooth, holomorphic, harmonic, continuous, etc.), and I suspect that in many of those generalized questions the answer is also **yes**.
| 4 | https://mathoverflow.net/users/22277 | 431961 | 174,844 |
https://mathoverflow.net/questions/431970 | 1 | Let $d$ be a large positive integer and fix $r \ge 0$. Set $S := B\_2^n \cap [-r,r]^d$, where $B\_2^d$ is the euclidean unit-ball in $\mathbb R^d$. Finally, let $\omega(S)$ be the *Gaussian width* of $S$, defined by
$$
\omega(S) := \mathbb E \sup\_{x \in S} x^\top z,
$$
where the expectation is over $z \sim N(0,I\_d)$.
>
> **Question.** What is a good upper-bound for $\omega(T)$, valid for large $d$ ?
>
>
>
**Note.** Using [Proposition 1 of this manuscript](https://arxiv.org/pdf/1705.10696.pdf) with $T = [-1/\sqrt{d},1/\sqrt{d}]^d$ (the convex hull of $m=2^d$ points in $B\_2^d$), and $s=1/(r\sqrt{d})$, I'm able to obtain the following upper-bound
$$
\omega(S) = s\cdot\omega(s B\_2^d \cap T) \lesssim r\sqrt{d\log(em)} \land \sqrt{d} = rd \land \sqrt{d}.
$$
Unfortunately, the above bound is not very good for my purposes.
| https://mathoverflow.net/users/78539 | Gaussian width of intersection of cube and ball in high-dimensional euclidean space | The answer is, up to a constant factor $\omega(S) = \Theta(\min(\sqrt{d}, rd))$.
To see the upper bound, we can use the fact that if $S \subset Q$, then $\omega(S) \leq \omega(Q)$, therefore
$$
\omega(B\_2^n \cap r B\_\infty^n) \leq \min(\omega(B\_2^n), r \omega(B\_\infty^n)) = \min( \mathbb{E}\_{z \sim \mathcal{N}(0, I)} \|z\|\_2, r\, \mathbb{E}\_{z \sim \mathcal{N}(0, I)} \|z\|\_1) = \mathcal{O}(\min(\sqrt{d}, rd)).
$$
For the lower bound, note that for any given vector $z$, we can chose $\tilde{z} \in S$ to be $\tilde{z}\_i := \min(1/\sqrt{d}, r) \mathrm{sgn}(z\_i)$. With this choice, we have for any fixed $z$
$$\sup\_{x \in S} x^T z \geq \tilde{z}^T z = \min(1/\sqrt{d}, r) \sum\_i |z\_i| = \min(1/\sqrt{d}, r) \|z\|\_1,$$
and therefore $\omega(S) \geq \min(1/\sqrt{d}, r) \mathbb{E}\_{z\sim\mathcal{N}(0, I)} \|z\|\_1 = \Omega(\min(\sqrt{d}, rd)$.
Note that this is not inconsistent with the upper bound given by Proposition 1 in the cited paper --- the hypercube $[-r, r]^d$ has $2^d$ vertices, not $2d$.
| 2 | https://mathoverflow.net/users/468679 | 431975 | 174,845 |
https://mathoverflow.net/questions/431953 | 6 | Let $u\_n : \mathbb{R}^n \to \mathbb{R}$ be a sequence of harmonic functions which converge uniformly on compact subsets. The limit function $u$ (which we assume to be not identically $0$) is clearly harmonic (via mean value property). Suppose we denote by $Z\_{f}$ the set of zeros of a function $f$.
My question is, is there a convergence of $Z\_{u\_n}$ to $Z\_u$ in any reasonable sense (maybe when restricted to compact regions)? This is probably well-known, and in that case, this is mainly a reference request. Thanks in advance!
| https://mathoverflow.net/users/492517 | Limit of zero sets of harmonic functions | **Yes.**
With an appropriate topology, the function mapping a non-zero harmonic function to its zero set is continuous. The result actually applies to open mappings from a locally compact locally connected space $X$ to $\mathbb{R}$, and harmonic functions are just a particular example of such open mappings.
If $X$ is a topological space, then let $H(X)$ denote the collection of all closed subsets of $X$. The set $H(X)$
If $X$ is a compact Hausdorff space and $U,V\_1,\dots,V\_n$ are non-empty open subsets of $X$, then let $[U;V\_1,\dots,V\_n]$ be the collection of all closed subsets $C\subseteq X$ where $C\subseteq U$ and where $C\cap V\_1\neq\emptyset,\dots,C\cap V\_n\neq\emptyset$. Then the Vietoris topology on $H(X)$ is the topology with basis consisting of all sets of the form $[U;V\_1,\dots,V\_n]$.
Proposition: Let $X$ be a locally compact locally connected Hausdorff space. Let $H$ be a collection of continuous open mappings $f:X\rightarrow\mathbb{R}$. Give $H$ the topology of uniform convergence on compact sets. Define a mapping $T:H\rightarrow X\cup\{\infty\}$ by letting $T(h)=Z(h)\cup\{\infty\}$. Then the function $T$ is continuous.
Proof: Suppose that $h\in H$. Let $z\in Z(h)$. Let $U$ be a neighborhood of $z$. Then there is some connected open set $V$ with $z\in V\subseteq\overline{V}\subseteq U$ and where $V$ is compact. In this case, we have $h(r)>0,h(s)<0$ for some $r,s\in V$. Let $\delta=\min(h(r),-h(s))$. Therefore, if $g\in H$ and $\|g|\_U-h|\_U\|<\delta$, then $g(r)>0,g(s)<0$. Since $V$ is connected and $r,s\in V,g(r)>0,g(s)<0$, there is some $t\in V$ with $g(t)=0$ (this is the reasoning that Giorgio Metafune has made in the comments).
We conclude that if $Z(h)\cap U\neq\emptyset$, then there is some $\delta>0$ where if $\|(g-h)|\_U\|<\delta$, then $Z(g)\cap U\neq\emptyset$ as well.
Suppose now that $h\in H$, and $T(h)\in[U;V\_1,\dots,V\_n]$ for open subsets $U,V\_1,\dots,V\_n\subseteq X\cup\{\infty\}$. Then $Z(h)\cup\{\infty\}\subseteq U$ and $Z(h)\cup\{\infty\}\cap V\_j\neq\emptyset$ for $1\leq j\leq n$. Without loss of generality, assume that $1\leq m\leq n$ and $\infty\not\in V\_j$ for $1\leq j\leq m$ where $\infty\in V\_j$ for $m<j\leq n$. Then $V\_j$ is an open subset of $X$ with $Z(h)\cap V\_j\neq\emptyset$ for $1\leq j\leq m$.
Therefore, for $1\leq j\leq m$ there is a relatively compact open $W\_j\subseteq V\_j$ and some $\delta\_j$ where if $\|(g-h)|\_{W\_j}\|<\delta\_j$, then $Z(g)\cap V\_j\neq\emptyset$. Now, set $W=W\_1\cup\dots\cup W\_m$ and $\delta=\min(\delta\_1,\dots,\delta\_m)$. Then whenever $\|(g-h)|\_W\|<\delta$, we have $Z(g)\cap V\_j\neq\emptyset$ for $1\leq j\leq m$.
Now let $C=X\setminus U$. Then $C$ is a compact subset of $X$ with $0\not\in g[C]$. Therefore, there is some $\epsilon>0$ where $|g(c)|\geq\epsilon$ for each $c\in C$. Therefore, if $\|(h-g)|\_C\|<\epsilon$, then $g(c)\neq 0$ whenever $c\in C$, so $Z(g)\subseteq U$.
Therefore if $g\in H$ and $\|(h-g)|\_{C\cup W}\|<\min(\delta,\epsilon)$, then
$g\in[U;V\_1,\dots,V\_n]$ as well. We therefore conclude that the function $T$ is continuous at the point $h$.
Q.E.D.
| 4 | https://mathoverflow.net/users/22277 | 431979 | 174,847 |
https://mathoverflow.net/questions/431973 | 0 | $H$ is an $n\times m$ matrix with non-negative coefficients and $n < m$. $H'$ is the transpose of $H$.
Are the following statements true?
1. If $\det(HH’) > 0$, the rows of $H$ define the edges of an $n$-dimensional conic polyhedron. No row of $H$ is a linear combination of the other rows of $H$ using non-negative coefficients.
2. If $\det(HH’) = 0$, a subset of the rows of $H$ define the edges of an $n$-dimensional conic polyhedron, but at least one row of $H$ either duplicates another row up to some scalar, lies on a face of the conic polyhedron or lies inside of the conic polyhedron. That is, at least one row is a linear combination of the other rows using non negative coefficients.
If these are true, how can I find one row (or enumerate each row) of $H$ which is a linear combination of the other rows of $H$ with non-negative coefficients when $\det(HH’) = 0$?
| https://mathoverflow.net/users/170274 | $\det(HH’) = 0$ for nonnegative $H$ | 1. Yes, because $H$ must be of rank $n$ (for $n=\operatorname{rk}(HH^\top)\leq\operatorname{rk}(H)$), so none of its rows is a linear combination of the others.
2. No, the matrix $$H = \begin{pmatrix} 1 & 1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 & 1 \end{pmatrix}$$ provides a counter-example. The simplest linear relation between its rows with non-negative coefficients is $H\_{1,\*}+H\_{2,\*}=H\_{3,\*}+H\_{4,\*}$.
| 0 | https://mathoverflow.net/users/5018 | 431981 | 174,849 |
https://mathoverflow.net/questions/431982 | 2 | *Question:* If $M$ is a compact smooth finite-dimensional manifold with boundary, is the inclusion of a closed subspace $A \subseteq M$ a cofibration? (I'm specifically interested in the case when $A$ is a smooth submanifold with boundary).
Does the following sketch proof work? *Sketch:* $M$ is homotopy equivalent to its interior which is a manifold with empty boundary, and since manifolds (I guess without boundary?) are absolute neighbourhood retracts (ANR's) and ANR's have the property that the inclusion of closed subspaces are cofibrations, the inclusion of $A \to M$ will be a cofibration.
I've seen similar statements for manifolds without boundary, for example when $A$ is a submanifold and both $A$ and $M$ have empty boundary, then it follows via Morse theory that $(X, A)$ have the homotopy type of a CW-pair and so the inclusion turns out to be a cofibration.
There is a potential answer to my question given on this site (<https://mathoverflow.net/a/16636/83360>), though I'm not sure if this applies to manifolds with boundary. Further, I haven't seen anything in the literature which covers the case when we consider manifolds with nonempty boundary (which is the reason for asking this question). If this has been shown somewhere in the literature, a reference would be very helpful.
| https://mathoverflow.net/users/83360 | If $M$ is a compact smooth finite-dimensional manifold with boundary, is the inclusion of a closed subspace $A \subseteq M$ a cofibration? | If the closed subset is locally compact and locally contractible then yes, the inclusion is a cofibration. This is surprisingly not very well known, but it follows from the classification of finite dimensional ANR's and the following fact from the answer of Tyrone [here](https://mathoverflow.net/questions/374160/is-the-alexander-horned-sphere-a-cofibration):
>
> If $X$ is an ANR and $j:A\subseteq X$ is a closed subspace, then $A$ is an ANR if and only if the inclusion $j$ is a cofibration.
>
>
>
| 6 | https://mathoverflow.net/users/134512 | 431988 | 174,852 |
https://mathoverflow.net/questions/46239 | 0 | I'm unable to access the following paper: J. Dénes, The representation of a permutation as the product of a minimal number of transpositions and its connection with the theory of graphs, Publ. Math. Inst. Hung. Acad. Sci. 4 (1959) 63-70. Can anyone help me?
| https://mathoverflow.net/users/3356 | Request for a copy of a paper of J. Dénes on permutation factorisations | The *PUBLICATION OF THE MATHEMATICAL INSTITUTE OF THE HUNGARIAN ACADEMY OF SCIENCES* is available online.
<http://real-j.mtak.hu/view/journal/A_Magyar_Tudom=E1nyos_Akad=E9mia_Matematikai_Kutat=F3_Int=E9zet=E9nek_k=F6zlem=E9nyei.html>
You can download this specific paper here: <http://real-j.mtak.hu/510/>
| 3 | https://mathoverflow.net/users/37580 | 432008 | 174,856 |
https://mathoverflow.net/questions/431999 | 2 | Let $a,b \in \mathbb R$, $R \ge 0$, and $c > 0$. Define $C := \{(x,y) \in \mathbb R^2 \mid x^2 + y^2 \le 1,\,x^2 + c y^2 \le R^2\}$, and set
$$
\alpha := \sup\_{(x,y) \in C} ax + b y.
$$
>
> **Question.** In terms of $a,b,c,R$, is there an analytic formula for $\alpha$ ?
>
>
>
Some special cases
------------------
* If $c=1$, then $C = \{z \in \mathbb R^2 \mid \|z\|\_2 \le R'\}$, where $R' := \min(1,R)$. Thus, $\alpha = R'\sqrt{a^2+b^2}$.
* If $a=0$, then we are maximizing $yb$ over $[-1,1] \cap [-R/\sqrt c,R/\sqrt c]$, and so $\alpha = |b|\min(1,R/\sqrt c)$.
* $\lim\_{c \to \infty} \alpha = |b|R'$. This is because the domain $C$ is shrunk to the interval $[-R',R'] \times \{0\}$.
| https://mathoverflow.net/users/78539 | Analytic value of $\alpha := \sup_{(x,y) \in C} ax+by$, where $C := \{(x,y) \in \mathbb R^2 \mid x^2 + y^2 \le 1,\,x^2 + c y^2 \le R^2\}$ | We have
$$\alpha = \left\{\begin{array}{ll}
\sqrt{a^2 + b^2} & \mathrm{if} ~ (a^2 + b^2)R^2 \ge a^2 + cb^2, \\[6pt]
R\sqrt{a^2 + b^2/c}& \mathrm{if} ~ a^2c^2 + b^2c \ge (a^2c^2 + b^2)R^2, \\[6pt]
|a|\sqrt{\frac{c-R^2}{c-1}}
+ |b|\sqrt{\frac{R^2 - 1}{c-1}} & \mathrm{otherwise}.
\end{array}
\right.$$
**Proof**:
If $a^2 + b^2 = 0$, it is easy. In the following, assume that $a^2 + b^2 > 0$.
We split into three cases:
**Case 1**: $(a^2 + b^2)R^2 \ge a^2 + cb^2$
By Cauchy-Bunyakovsky-Schwarz inequality, we have
$(ax + by)^2 \le (a^2 + b^2)(x^2 + y^2) \le a^2 + b^2$.
On the other hand, letting
$x = \frac{a}{\sqrt{a^2+b^2}}$
and $y = \frac{b}{\sqrt{a^2+b^2}}$, we have
$x^2 + y^2 = 1$ and $x^2 + cy^2 \le R^2$ and
$ax + by = \sqrt{a^2 + b^2}$.
The desired result follows.
**Case 2**: $a^2c^2 + b^2c \ge (a^2c^2 + b^2)R^2$,
By Cauchy-Bunyakovsky-Schwarz inequality, we have
$(ax + by)^2 \le (a^2 + b^2/c)(x^2 + cy^2) \le R^2(a^2 + b^2/c)$.
On the other hand, letting
$x = \frac{acR}{\sqrt{a^2c^2 + b^2c}}$ and $y = \frac{bR}{\sqrt{a^2c^2 + b^2c}}$, we have
$x^2 + y^2 \le 1$ and $x^2 + cy^2 = R^2$ and
$ax + by = R\sqrt{a^2 + b^2/c}$.
The desired result follows.
**Case 3**: $(a^2 + b^2)R^2 < a^2 + cb^2$
and $a^2c^2 + b^2c < (a^2c^2 + b^2)R^2$
Clearly, $c \ne 1$. It is easy to prove that
$$\alpha = \sup\_{0 \le y \le \min(1, R/\sqrt c)} |a|\min(\sqrt{1-y^2}, ~ \sqrt{R^2 - cy^2}) + |b| y.$$
(i) If $c > 1$, from $(a^2 + b^2)R^2 < a^2 + cb^2$
and $a^2c^2 + b^2c < (a^2c^2 + b^2)R^2$, we have $R > 1$ and $c > R^2$ and
$$\frac{|b|R}{\sqrt{a^2c^2 + b^2c}} < \sqrt{\frac{R^2-1}{c-1}} < \frac{|b|}{\sqrt{a^2+b^2}} . \tag{1}$$
We have
$$\alpha = \sup\_{0 \le y \le R/\sqrt c} |a|\min(\sqrt{1-y^2}, ~ \sqrt{R^2 - cy^2}) + |b| y
= \max(\alpha\_1, ~ \alpha\_2)$$
where
$$\alpha\_1 = \sup\_{0 \le y \le \sqrt{(R^2-1)/(c-1)}} |a|\sqrt{1-y^2} + |b| y$$
and
$$\alpha\_2 = \sup\_{ \sqrt{(R^2-1)/(c-1)} \le y \le R/\sqrt c} |a|\sqrt{R^2 - cy^2} + |b| y.$$
Using (1), it is not difficult to prove that
$$\alpha\_1 = \alpha\_2 = |a|\sqrt{\frac{c-R^2}{c-1}}
+ |b|\sqrt{\frac{R^2 - 1}{c-1}}.$$
(*Note*: For example, consider $\alpha\_1$. Let $f(y) = |a|\sqrt{1-y^2} + |b| y$.
Using $\frac{|b|}{\sqrt{a^2 + b^2}} > \sqrt{(R^2-1)/(c-1)}$, we have $f'(y) = - \frac{|a| y}{\sqrt{1 - y^2}} + |b| > 0$
on $[0, \sqrt{(R^2-1)/(c-1)}]$.
Thus, $\alpha\_1 = f(\sqrt{(R^2-1)/(c-1)}) = |a|\sqrt{\frac{c-R^2}{c-1}}
+ |b|\sqrt{\frac{R^2 - 1}{c-1}}$.)
The desired result follows.
(ii) If $0 < c < 1$, from $(a^2 + b^2)R^2 < a^2 + cb^2$
and $a^2c^2 + b^2c < (a^2c^2 + b^2)R^2$, we have $R < 1$ and $R^2 > c$ and
$$\frac{|b|R}{\sqrt{a^2c^2 + b^2c}} > \sqrt{\frac{R^2-1}{c-1}} > \frac{|b|}{\sqrt{a^2+b^2}} . \tag{2}$$
We have
$$\alpha = \sup\_{0 \le y \le 1} |a|\min(\sqrt{1-y^2}, ~ \sqrt{R^2 - cy^2}) + |b| y = \max(\alpha\_3, ~ \alpha\_4)$$
where
$$\alpha\_3 = \sup\_{0 \le y \le \sqrt{(1-R^2)/(1-c)}} |a|\sqrt{R^2 - cy^2} + |b| y,$$
and
$$\alpha\_4 = \sup\_{\sqrt{(1-R^2)/(1-c)} \le y \le 1} |a|\sqrt{1-y^2} + |b| y$$
Using (2), it is not difficult to prove that
$$\alpha\_3 = \alpha\_4 = |a|\sqrt{\frac{c-R^2}{c-1}}
+ |b|\sqrt{\frac{R^2 - 1}{c-1}}.$$
The desired result follows.
$\phantom{2}$
We are done.
| 2 | https://mathoverflow.net/users/141801 | 432010 | 174,857 |
https://mathoverflow.net/questions/432000 | 0 | The following interesting problem was asked at [Aops](https://artofproblemsolving.com/community/u738252h2922192p26122750) and I wonder if it was based on some research paper:
>
> Let $K$ be a convex body in $\mathbb R^2$, such that the diameter of $K$ is less than $\sqrt2$.
>
> Prove that there is a lattice-point-free translation of $K$.
>
>
>
Is the 'convex condition' necessary?
Can this problem be generalized to $\mathbb R^n$?
| https://mathoverflow.net/users/70464 | Lattice-point-free body diameter | For higher dimension, this does not hold, see the answer by Sergei Ivanov [here](https://mathoverflow.net/q/42774/4312).
For dimension 2, in the closed set $K$ without lattice-point free translate (in other words, such that $K$ has a point in every translate $\mathbb{Z}^2+c$ of the lattice) we may even find two points for which both coordinates differ at least by 1. This may be proved as follows. Fix real $a\in [0,1]$ and consider the union of vertical lines $V\_a:=\{(x, y) \colon x-a\in \mathbb{Z}\}$. In this set $V\_a$, the ordinates take all values modulo 1, thus there exist two points $p\_a, q\_a\in V\_a$ with ordinates differing at least by 1. If they have different abscissas, that's what we need. So, these $p\_a, q\_a$ have equal abscissa $v\_a$, congruent to $a$ modulo 1. Therefore there exist two values $a, b$ for which $|v\_a-v\_b|$ is more than 1 or arbitrarily close to 1. Between points $p\_a, q\_a, p\_b, q\_b$ let $p\_a$ be the lowest, then $p\_a$ and the highest of $q\_a, q\_b$ is (almost) necessary pair of points.
| 2 | https://mathoverflow.net/users/4312 | 432029 | 174,863 |
https://mathoverflow.net/questions/432026 | 8 | In an old paper of Glaisher, I find the following formulas:
$$\dfrac{\sin(\pi x)}{\pi x}=1-\dfrac{x^2}{1^2}-\dfrac{x^2(1^2-x^2)}{(1.2)^2}-\dfrac{x^2(1^2-x^2)(2^2-x^2)}{(1.2.3)^2}-\cdots$$
$$\cos(\pi x/2)=1-x^2-\dfrac{x^2(1^2-x^2)}{(1.3)^2}-\dfrac{x^2(1^2-x^2)(3^2-x^2)}{(1.3.5)^2}-\cdots$$
These are trivial since the $n$th partial sum is equal to the $n$th partial product of
the product formulas for the sine and cosine
But he also gives
$$\sin(\pi x/2)=x-\dfrac{x(x^2-1^2)}{3!}+\dfrac{x(x^2-1^2)(x^2-3^2)}{5!}-\cdots$$
$$\cos(\pi x/2)=1-\dfrac{x^2}{2!}+\dfrac{x^2(x^2-2^2)}{4!}-\dfrac{x^2(x^2-2^2)(x^2-4^2)}{6!}+\cdots$$
$$\dfrac{\sin(\pi x/3)}{\sqrt{3}/2}=x-\dfrac{x(x^2-1^2)}{3!}+\dfrac{x(x^2-1^2)(x^2-2^2)}{5!}-\cdots$$
$$\cos(\pi x/3)=1-\dfrac{x^2}{2!}+\dfrac{x^2(x^2-1^2)}{4!}-\dfrac{x^2(x^2-1^2)(x^2-2^2)}{6!}+\cdots$$
Apparently he considers them trivial. If they are, please explain and feel free to
downvote me.
| https://mathoverflow.net/users/81776 | Trivial (?) product/series expansions for sine and cosine | You can derive at least some of these formulas from expansions of $\sin x\theta$ and $\cos x\theta$ as Taylor series in $\sin\theta$,
$$\sin x\theta=x\sin \theta-\frac{x(x^2-1)}{3!}\sin^3\theta+\frac{x(x^2-1)(x^2-3^2)}{5!}\sin^5\theta+-\cdots$$
$$\cos x\theta=1-\frac{x^2}{2!}\sin^2 \theta+\frac{x^2(x^2-2^2)}{4!}\sin^4\theta-\frac{x^2(x^2-2^2)(x^2-4^2)}{6!}\sin^6\theta+-\cdots$$
(For a worked out proof, see this [MSE post.](https://math.stackexchange.com/a/4380218/87355))
Substituting $\theta=\pi/2$ gives the $\sin(\pi x/2)$ and $\cos(\pi x/2)$ series. Substituting $\theta=\pi/3$ gives slightly different series than in the OP...
| 10 | https://mathoverflow.net/users/11260 | 432030 | 174,864 |
https://mathoverflow.net/questions/431833 | 1 | This is a 2 part question:
1). I am looking for a (hopefully accessible to beginning grad student who knows matrix perturbation theory) reference for doing concrete calculations of perturbed discrete spectra for operators that also have continuous spectrum. We can assume perturbation is compact and hence continuous spectrum is unchanged. What would be your recommendations ? Are there any toolboxes that already exist to do such computations ?
2). As I understand, the main issue is that now one has to project onto the subspace corresponding to continuous spectra.
However, in computations, continuous spectra is approximated by additional discrete spectra. Can I expect to find reliable answers if I just a very high resolution and apply the algorithms used for purely discrete case to the resulting data ?
Is there any tutorial available anywhere ? I assume physicists are doing these types of calculations all the time, so something digestible must be out there ?
| https://mathoverflow.net/users/30684 | Spectral perturbation theory of discrete spectra in presence of continuous spectrum | For this purpose ("beginning grad student") it would make sense to focus on the case that the discrete eigenvalues appear in an energy range that does not overlap with the continuous spectrum (say, $E>E\_0$). One can then simply use the [formulas](https://en.wikipedia.org/wiki/Perturbation_theory_(quantum_mechanics)) for perturbation theory of discrete spectra, and replace the $\sum\_n$ over energies $E>E\_0$ in the continuous spectrum by an integral $\int\_{E\_0}^\infty dE\,\rho(E)$ weighted by the density of states $\rho(E)$.
To second order in $\lambda$ one thus has
$$E\_n(\lambda) = E\_n^{(0)} + \lambda \left \langle n^{(0)} \right |V\left |n^{(0)} \right \rangle + \lambda^2\sum\_{k \ne n} \frac{\left |\left \langle k^{(0)} \right |V\left |n^{(0)} \right \rangle \right |^2} {E\_n^{(0)} - E\_k^{(0)}}$$
$$\qquad\qquad+ \lambda^2\int\_{E\_0}^\infty dE\,\rho(E) \frac{\left |\left \langle E \right |V\left |n^{(0)} \right \rangle \right |^2} {E\_n^{(0)} - E}+{\cal O}(\lambda^3).$$
Here $|n^{(0)}\rangle$ and $|E\rangle$ are the unperturbed eigenstates in the discrete and continuous spectrum, respectively. I have assumed the eigenvalue $E\_n^{(0)}$ has multiplicity one, otherwise the usual approach of [degenerate perturbation theory](https://en.wikipedia.org/wiki/Perturbation_theory_(quantum_mechanics)#Effects_of_degeneracy) applies.
Concerning the question, "how to approximate the continuous spectrum" in a numerical calculation, indeed one would discretize this. For example, if the continuous spectrum consists of plane waves $e^{ikx}$ one could impose periodic boundary conditions over a length $L$, and discretize $k\_m=2\pi m/L$, $m\in\mathbb{Z}$. The integral $\int dE\,\rho(E)$ would then be a sum over $m$ and $E$ would be replaced by the dispersion relation $E(k\_m)$.
---
The case of non-isolated discrete levels, embedded in the continuum, is more complicated. The coupling to the continuum turns the bound state into a resonance, a quasi-bound state with a finite life-time. This can be calculated in perturbation theory by a formula known as [Fermi's golden rule.](https://en.wikipedia.org/wiki/Fermi%27s_golden_rule) I have the impression from the way the question is formulated that this is not the problem at hand, correct me if I'm wrong.
| 1 | https://mathoverflow.net/users/11260 | 432049 | 174,868 |
https://mathoverflow.net/questions/432053 | 2 | I am relatively new to the world of braids/knots so really sorry if this question is simple. However, I am not able to find if there is any theorem/procedure that determines if a closed braid, given its representation in the Artin braid group, is a link or an unlink. Or, any theorem that says this cannot be determined? I have read the textbook A Study of Braids by Kunio Murasugi to get familiar with the concepts of Alexander's Theorem.
Any suggestion or recommendation of literature is really appreciated.
Edit: To be more specific, I am looking for some specific papers/algorithms (that can be efficiently implemented in computers), using which, I can determine if the closure of a braid (in the form of generators $\sigma\_1$, $\sigma\_2$, ...) gives a trivial knot/link or non-tirvial knot/link.
| https://mathoverflow.net/users/492606 | Determine if a closed braid is a link/unlink | A braid gives a braid closure. This can be drawn as a knot (or link) diagram. There are then various approaches to solve the unknot (or unlink) recognition problem given a diagram. This begins with work of Haken, and then work of Hass, Lagarias, and Pippenger, and then work of Lackenby.
There is also another line of research, which is perhaps more in line with your stated interests (in braids). Namely, Birman with various co-authors and then work of Dynnikov work more directly with the presentation of the knot or link as a braid.
You can find a discussion and references at the Wikipedia page on the [unknotting problem](https://en.wikipedia.org/wiki/Unknotting_problem).
| 4 | https://mathoverflow.net/users/1650 | 432062 | 174,872 |
https://mathoverflow.net/questions/432071 | 4 | $\newcommand\Set{\mathbf{Set}}\newcommand\Ob{\mathbf{Ob}}\newcommand\Hom{\mathbf{Hom}}$Work in a foundation that admits a countable hierarchy of notions of ‘set’, and say that a category is *$n$-small* iff its object and arrow collections are $n$-sets. Denote the category of $n$-sets by $\Set\_n$.
For each $n$-small category $\mathcal{C}$ and each object $X\in\Ob\_\mathcal{C}$, define
\begin{gather\*}
\Hom\_\mathcal{C}(\ \ ,X)=\operatorname{cod}^{-1}(X)\subseteq\Hom\_\mathcal{C} \\
\Bigl({}=\bigcup\_{W\in\Ob\_\mathcal{C}}\Hom\_\mathcal{C}(W,X)\Bigr).
\end{gather\*}
For each $n$-small category $\mathcal{C}$, define a functor
\begin{gather\*}
\Hom\_\mathcal{C}(\ \ ,-):\mathcal{C}\to\Set\_{n+1} \\
X\mapsto\Hom\_\mathcal{C}(\ \ ,X) \\
f:X\to Y\longmapsto {f\circ{}}:\Hom\_\mathcal{C}(\ \ ,X)\to\Hom\_\mathcal{C}(\ \ ,Y).
\end{gather\*}
$\Hom\_\mathcal{C}(\ \ ,-)$ is always faithful, so $n$-small categories are never abstract. Working in an appropriate foundation (see [An axiomatic approach to higher order set theory](https://arxiv.org/abs/2206.10060) (disclaimer: I am the author of this paper)) all classically considered ‘abstract’ categories like Freyd's homotopy category are just $0$-large $1$-small categories, admitting canonical faithful functors into $\Set\_2$.
Taking this view, it seems like categories are only ‘abstract’ if we work in a set-theoretical foundation that is ‘too small’ to see the larger categories of sets that all categories embed into. This makes me wonder,
>
> Are there any categories we care about that remain abstract in set theories admitting a countable hierarchy of notions of ‘set’? What if we extend the hierarchy of notions of set to arbitrary ordinal heights?
>
>
>
| https://mathoverflow.net/users/92164 | There are no abstract categories | The argument you give in your original post is essentially the proof that *every small category is concrete.*
So if you work in a setting that have enough universes/inaccessible cardinal/notion of smallness so that it is reasonable to only consider "small" categories, then of course all categories are concrete.
However, as soon as you allow yourself to form something like the category of *all* sets, you will get examples of non-concrete categories:
The typical examples of non-concrete categories are the category of all groupoids with isomorphism classes of functors between them. Or the homotopy category of all spaces/simplicial sets.
So unless you insist on only talking about small categories (which definitely make sense in some foundation), you'll always get examples of non-concrete categories.
But I would add that even if you consider that "all categories are small", a better point of view on concreteness is that, (like for smallness) one shouldn't talk about "concrete categories" in the abstract but of $n$-concrete categories exactly as you talked about $n$-small sets. Even in settings where everything is small in some sense, the notion of smallness is still relevant: for e.g. the category of $n$-small sets isn't closed under all $m$-small colimits for $m>n$. So even in a setting where every category is concrete in some sense, the notion of concreteness could still be relevant.
| 20 | https://mathoverflow.net/users/22131 | 432073 | 174,876 |
https://mathoverflow.net/questions/431974 | 4 | There is a descent spectral sequence computing $\pi\_\*L\_{K(n)}S^0$ with $E\_2$-term
$$E\_2^{s,t}\cong H^s\_c(\mathbb{G}\_n,(E\_n)\_t)$$
It is mentioned in [Barthel-Beaudry](https://arxiv.org/pdf/1901.09004.pdf) (in the description of Figure 3.30) that if $p>2$ and $2(p-1)>n^2$, then there are only nonzero entries when $t$ is a multiple of $2(p-1)$.
>
> **Question.** Why does it have this property?
>
>
>
I’m trying to figure out the reason for this, but I could not read directly from the group cohomology, nor can I find a reference for that.
| https://mathoverflow.net/users/492531 | On the sparsity of the descent spectral sequence computing homotopy groups of the K(n)-local sphere | This sparsity holds even without the assumptions that $p>2$ and $2(p-1)>n^2$. (Those are used to get a horizontal vanishing line). As in the comments, it comes down to the fact that you can use $\widehat{E(n)}\simeq L\_{K(n)}E(n)$ in place of $E\_n$, and $\widehat{E(n)}$ has homotopy groups concentrated in degrees that are a multiple of $2(p-1)$.
One way to get sparsity directly from the group cohomology is to use that there's a copy of $\mathbb{Z}\_p^\times$ sitting inside the center of $\mathbb{G}\_n$, and further sitting inside this is the subgroup $\mathbb{F}\_p^\times\subset\mathbb{Z}\_p^\times$ of roots of unity (the Teichmüller lifts). Because the order of $\mathbb{F}\_p^\times$ is invertible in $\pi\_\ast E\_n$, you get an isomorphism
$$
H^\ast\_c(\mathbb{G}\_n;\pi\_\ast E\_n)\cong H^\ast\_c(\mathbb{G}\_n/\mathbb{F}\_p^\times;(\pi\_\ast E\_n)^{\mathbb{F}\_p^\times}).
$$
An element $\ell\in\mathbb{Z}\_p^\times$ corresponds to multiplication by $\ell$ on the formal group of $E\_n$, so acts on $\pi\_{2t}E\_n$ as multiplication by $\ell^t$. This implies
$$
(\pi\_t E\_n)^{\mathbb{F}\_p^\times} =
\begin{cases}
\pi\_t E\_n &2(p-1)\mid n\\0&\text{otherwise},
\end{cases}
$$
and thus $H^\ast\_c(\mathbb{G}\_n;\pi\_t E\_n) = 0$ unless $2(p-1)\mid n$.
| 10 | https://mathoverflow.net/users/111541 | 432074 | 174,877 |
https://mathoverflow.net/questions/432088 | 10 |
>
> What are your favorite models of the KL-axioms?
>
>
>
The motivation is having some basic models to understand the axiom scheme as presented e.g. in [Synthetic Geometry of Manifolds](https://users-math.au.dk/kock/SGM-final.pdf) by Kock.
In that text he references his other text, [Synthetic Differential Geometry](https://users-math.au.dk/kock/sdg99.pdf), as well as two other sources for models -- the examples in the above text are forgetful functors from categories of rings to the category of sets, and these are cool, but I was wondering if any other 'standard' models have cropped up in the past decade. Any pointers are appreciated.
| https://mathoverflow.net/users/92164 | Models of the Kock-Lawvere axioms | Here is what I once worked out; I think I remember seeing this published somewhere too but I forget where.
*Most of the interesting Kock-Lawvere algebra can be captured in a ring*:
$$R\_1= \mathbb{R}[t\_1,t\_2,\dots]/(t\_1^2,t\_2^2,\dots)$$
For example, this satisfies the principle that only zero can annihilate all infinitesimals:
$$(\forall c\in R)\big[(\forall d \in R)(d^2=0\to cd=0)\to c=0\big]$$
*Most of the interesting Kock-Lawvere logic can be captured in a Kripke model based on this*, in the sense of intuitionist logic. The Kripke model’s first stage is $R\_1$ as above, the second stage is $R\_2$ as the image of $R\_1$ under the map $t\_1\to 0$, the third stage is $R\_3$ as the image of $R\_2$ under the map $t\_2\to 0$, etc.
For example, this satisfies various principles that infinitesimals are in a neighborhood of zero:
$$(\forall d \in R)(d^2=0\to \neg\neg d=0)$$
$$\neg(\forall d \in R)(d^2=0\to d=0)$$
$$(\forall x \in R)\neg(\forall y \in R)(\neg \neg x=y \to x=y)$$
The last of these, which I call the [fuzziness of identity](https://mathoverflow.net/a/391800/44143), is interesting as a logical statement which follows from the Kock-Lawvere axioms and contradicts classical logic.
*Most of the interesting Kock-Lawvere
analysis can be captured in a second-order Henkin model based on this*, where the functions from $R$ to $R$ are identified with the smooth functions from $\mathbb{R}$ to $\mathbb{R}$, and $f(r)$ is interpreted as the result of taking the standard part of $r$ (the image of $r$ in $\mathbb{R}$ under the map sending all $t\_i$ to 0), taking the power series of $f$ around that standard part, and applying that power series to $r$. E.g., under this interpretation, $\cos(t\_1+t\_2)= 1-(t\_1+t\_2)^2/2$, and no more terms are needed since they would all vanish.
For example, this satisfies the central axiom of microlinearity:
$$(\forall f\in R^R)(\exists c\in R)(\forall d \in R)(d^2=0\to f(d)=f(0)+cd)$$
The basic model can also be extended to interpret the order $x<y$ as equivalent to the standard part of $x$ being less than the standard part of $y$. For instance, this falsifies trichotomy but satisfies $(x>0)\vee (x<1)$. The axioms for the other order $x\le y$ are more awkward to fit with these models.
These models are limited, but they are simpler than the topos models, they are accurate enough that they helped me find and correct a minor [error](https://mathoverflow.net/q/432151/44143) in Moerdijk and Reyes’s admirably clear presentation of the axioms, and they may help you understand the Kock-Lawvere axioms better, as they did for me.
| 8 | https://mathoverflow.net/users/nan | 432100 | 174,886 |
https://mathoverflow.net/questions/432106 | 16 | As the title says, for every Lebesgue integrable function $f:\mathbb{R}\to\mathbb{R}$ is there a Riemann integrable function $g:\mathbb{R}\to\mathbb{R}$ such that $f=g$ almost everywhere?
For example, $\chi\_{\mathbb{Q}\cap [0, 1]}$ is a well known example of lebesgue integrable function that is not riemann integrable, but $\chi\_{\mathbb{Q}\cap [0, 1]}=0$ almost everywhere and 0 is a riemman integrable function.
| https://mathoverflow.net/users/481551 | Equivalence between Lebesgue integrable and Riemann integrable functions | Let $A$ be a measurable subset of $[0,1]$ such that both it and its complement have positive measure in every open interval in $[0,1]$ (see [here](https://math.stackexchange.com/questions/57317/construction-of-a-borel-set-with-positive-but-not-full-measure-in-each-interval) for example). Its characteristic function is dominated by $1\_{[0,1]}$, so it is Lebesgue integrable, but it is discontinuous everywhere on $[0,1]$ even after being modified on a null set, so it is not a.e. equal to a Riemann integrable function.
| 29 | https://mathoverflow.net/users/23141 | 432110 | 174,888 |
https://mathoverflow.net/questions/432104 | 3 | Let
$$ n\ := \sum\_{k=0}^L\ a\_k\cdot10^k $$
where $\ L\ $ is a non-negative integer, and $\ a\_k\ $ are decimal digits, and $\ a\_L>0.$
Let $\ n\ $ be called ***mixed*** $\ \Leftarrow:\Rightarrow$
$$ n\ =\ \sum\_{k=0}^L a\_k\cdot\prod\_{k=0}^L a\_k $$
For instance, $\,\ n:=1\,\ $ and $\,\ n:=144\ $ are mixed.
**Problem 1:** describe all mixed numbers.
============
============
One may also consider mixed numbers $\,\ n:=\sum\_{k=0}^L\, a\_k\cdot\beta^k\,\ $
base $\ \beta,\ $ where $\ 1<\beta\in\mathbb Z.\ $
The only mixed number base $\ 2\ $ is $\ n:=1.$
In general,
$\quad$
$\qquad\qquad$ for arbitrary base $\ \beta\ $ there are only
$\qquad\qquad$ finitely many mixed numbers base $\ \beta.$
$\quad$
**Problem 2:** describe all mixed numbers for arbitrary base $\ \beta$.
---
---
**Remark:** The case of odd base $\ \beta\ $ is distinctly different from the even case.
---
---
| https://mathoverflow.net/users/110389 | Diophantine entertainment -- mixed natural numbers | <https://oeis.org/A038369> "Numbers $k$ such that $k$ = (product of digits of $k$) times (sum of digits of $k$)."
"$0, 1, 135, 144$. The list is complete. Proof: One shows that the number of digits is at most $84$ and then it is only necessary to consider numbers of the forms $2^i3^j7^k$ and $3^i5^j7^k$. - David W. Wilson, May 16 2003."
Many references to the literature are given at that page. In particular, Ezra Bussmann, S·P Numbers in bases other than 10, The Mathematical Gazette, 85(503)(2001) 245–248. <https://doi.org/10.2307/3622010> finds all the examples in all bases $b$, $2\le b\le12$. $1$ is an example in every base; the others are
Base $4$: $12$
$5$: $341$
$7$: $22; 242; 1254; 2343; 116655; 324236; 424644$
$9$: $13; 281876; 724856; 7487248$
$10$: $135; 144$
$11$: $253; 419; 2189; 7634; 82974$
$12$: $128; 173; 353$
| 10 | https://mathoverflow.net/users/3684 | 432111 | 174,889 |
https://mathoverflow.net/questions/432091 | 11 | The question is in the title. Here is a short motivation. The general quadratic Diophantine equation is
$$
x^TAx+bx+c=0,
$$
where $x$ is a vector of $n$ variables, $A$ is $n \times n$ matrix with integer entries, $b$ vector, $c$ integer. We can try simplify it by linear substitution $x=Hy$, where $y$ are new variables and $H$ is $n\times n$ matrix with integer entries and determinant $\pm 1$. The equation becomes
$$
y^T(H^TAH)y + bHy + c = 0.
$$
The first entry in matrix $H^TAH$ is $h^TAh$, where $h$ is the first column of $H$. If there is a non-zero vector $h$ such that $h^TAh=0$, then in the new equation the $y\_1^2$ term vanishes, and equation becomes linear in $y\_1$. In many cases, this can be used to solve the equation.
The equation $x\_1^2+x\_2^2=3x\_3^2+1$ is the simplest quadratic equation for which this method does not work. Here $h^TAh=0$ reduces to $h\_1^2+h\_2^2-3h\_3^2=0$, which does not have non-trivial integer solutions. Hence, no linear substitution makes the equation linear in any of the variables. How to solve such equations?
There is a general method outlined by by Grunewald and Segal that should work for all quadratic equations: the idea is that (i) the integral orthogonal group of A is finitely generated, and there is an algorithm for listing the generators, and (ii) there is a finite set of solutions to the equation such that all other solutions can be constructed from this finite set by actions of this integral orthogonal group. However, how to implement this idea for this specific equation? Also, maybe there is an easier method?
I remark that I am looking for more or less explicit description of all integer solutions. The answer "Try $z=0,\pm 1,\pm 2,\dots$ and for each $z$ list all representations of $3z^2+1$ as the sum of squares" does not count.
| https://mathoverflow.net/users/89064 | How to describe all integer solutions to $x^2+y^2=3z^2+1$? | Ok, I now was able to solve the equation myself.
If $(x,y,z)$ is any solution, then $(y,x,z)$, $(x,-y,z)$, $(x,y,-z)$, and $(x,3z-2y,2z-y)$ are also solutions. To check the last one, observe that
$$
x^2 + (3z-2y)^2 - 3(2z-y)^2 = x^2 + y^2 - 3z^2 = 1.
$$
All these transformations are invertible: if we apply any of them twice, we go back to the solution we have started with. To check this for the last one, observe that
$$
3(2z-y)-2(3z-2y) = y, \quad 2(2z-y)-(3z-2y)=z.
$$
Let us prove that all integer solutions to our equation can be produced from $(x,y,z)=(1,0,0)$ by a sequence of this transformations. This gives a complete description of the solution set.
Let us say that two solutions are in the same orbit if they can be obtained from each other by a sequence of these transformations. We will prove that all solutions are in the same orbit as $(1,0,0)$. In any orbit, choose a solution $(x,y,z)$ with $|z|$ minimal. By swapping $x$ and $y$ and changing signs of $y$ and $z$ if needed, we may assume that $0\leq |x|\leq y$ and $0\leq z$. Because this solution has minimal $|z|$ in the orbit, we must have $|2z-y|\geq z$. If $2z-y\geq 0$, this implies that $2z-y \geq z$, or $z \geq y$. But then $2z^2\geq 2y^2 \geq x^2+y^2 = 3z^2+1$, which is a contradiction. If $2z-y<0$, then $|2z-y|\geq z$ reduces to $y-2z\geq z$, or $y\geq 3z$. But then $9z^2 \leq y^2 \leq x^2+y^2 = 3z^2+1$, which is possible only if $z=0$. Hence, $(x,y,z)=(0,1,0)$, and, by applying the operation $(x,y,z)\to (y,x,z)$ for the last time, we arrive at $(x,y,z)=(1,0,0)$. Hence, all solutions are at the same orbit.
| 14 | https://mathoverflow.net/users/89064 | 432118 | 174,890 |
https://mathoverflow.net/questions/432114 | 5 |
>
> Suppose that $\{X\_{ij}\}\_{1\leqslant i,j\leqslant n}$ are iid random variables with $\mathbb{E}(X\_{11})=0$ and $\mathrm{Var}(X\_{11})=1$, does the following convergence hold:
> $$
> \max\_{1\leqslant j\leqslant n}\biggl\{\frac{1}{n^2}\sum\_{1\leqslant i\neq i'\leqslant n}(X\_{ij}X\_{i'j})\biggr\} \to 0 \qquad \text{almost surely}?
> $$
>
>
>
Comment: I first posted this question on [MSE](https://math.stackexchange.com/questions/4545745/maximal-inequality-of-iid-random-variables-x-ij-1-leqslant-i-j-leqslant), someone suggested me to post it on mathoverflow since it is an open problem.
---
Background
----------
I am reading the AoP paper "*Limit of the smallest eigenvalue of a large dimensional sample covariance matrix*" by Z. Bai and Y. Yin (1993). Their Lemma 2 states a generalization of the well-known Marcinkiewicz-Zygmund strong law of large numbers to the case of multiple arrays of iid random variables.
>
> **[Lemma 2 in [Bai and Yin (1993)](https://projecteuclid.org/journals/annals-of-probability/volume-21/issue-3/Limit-of-the-Smallest-Eigenvalue-of-a-Large-Dimensional-Sample/10.1214/aop/1176989118.full)]** Let $\{\xi\_{ij},i,j=1,2,\ldots\}$ be a double array of iid random variables and let $\alpha>1/2,\beta\geqslant 0$ and $M$>0 be constants. Then as $n\to\infty$,
> $$
> \max\_{j\leqslant Mn^{\beta}} \biggl|n^{-\alpha}\sum\_{i=1}^n (\xi\_{ij}-c)\biggr|\to0\quad \text{almost surely},
> $$
> if and only if
> $$
> (i)\quad \mathbb{E}|\xi\_{11}|^{(1+\beta)/\alpha}<\infty
> $$
> $$
> (ii)\quad c = \left\{
> \begin{array}{ll}
> \mathbb{E} \,\xi\_{11},& \text{if }\alpha\leqslant 1, \\
> \text{any number}, &\text{if }\alpha>1.
> \end{array}
> \right.
> $$
>
>
>
By our assumptions and taking $\alpha=\beta=M=1$, $\xi\_i=X\_{ij}^2$ in this lemma, we have
$$
\max\_{j\leqslant n}\biggl|\frac{1}{n}\sum\_{i,j}X\_{ij}^2-1\biggr|\to0\quad \text{almost surely}.
$$
This result is for square terms. I wonder if there is a similar result for the **cross terms**
$$
\max\_{1\leqslant j\leqslant n}\biggl\{\frac{1}{n^2}\sum\_{i\neq i'}(X\_{ij}X\_{i'j})\biggr\} \to 0 \qquad \text{almost surely}?
$$
---
Attempt
-------
I can prove that $(1/n^2)\sum\_{i\neq i'}(X\_{ij}X\_{i'j})\to 0\; a.s.$ for any fixed $j$. But I do not know how to deal with the problem with "$\max$".
For fixed $j$,
$$
\mathrm{Var}\Bigl(\sum\_{i\neq i'}X\_{ij}X\_{i'j}\Bigr)=2\sum\_{i\neq i'}\mathrm{E}\bigl(X\_{ij}^2\bigr)\cdot\mathrm{E}\bigl(X\_{i'j}^2\bigr)=2(n^2-n),
$$
then by Chebyshev's inequality, for any $\varepsilon>0$,
$$
\Pr\biggl(\frac{1}{n^2}\sum\_{i\neq i'}(X\_{ij}X\_{i'j})>\varepsilon\biggr) =O\Bigl(\frac{1}{n^2}\Bigr),
$$
which is summable. Hence, by using the Borel-Cantelli lemma, we have
$$
\frac{1}{n^2}\sum\_{i\neq i'}(X\_{ij}X\_{i'j})\to 0\qquad
\text{almost surely}.$$
If we consider $\max\_{1\leqslant j\leqslant n}$, and use the trivial inequality to bound it, we have
$$
\Pr\biggl(\max\_{1\leqslant j\leqslant n}\biggl\{\frac{1}{n^2}\sum\_{i\neq i'}(X\_{ij}X\_{i'j})\biggr\}> \varepsilon\biggr)
\leqslant n\cdot \Pr\biggl(\frac{1}{n^2}\sum\_{i\neq i'}(X\_{i1}X\_{i'1})>\varepsilon\biggr)=O\Bigl(\frac{1}{n}\Bigr),$$
which means
$$
\max\_{1\leqslant j\leqslant n}\biggl\{\frac{1}{n^2}\sum\_{i\neq i'}(X\_{ij}X\_{i'j})\biggr\}\to 0\qquad
\text{in probability}.\tag{\*}
$$
How can we improve the result (\*) to "almost surely"?
| https://mathoverflow.net/users/481055 | Maximal inequality of iid random variables $\{X_{ij}\}_{1\leqslant i,j \leqslant n}$ | In the Math Stack Exchange post, I gave a proof based on Lemma 2 in Bai and Yin (1993). I will give an alternative proof.
Expressing $\sum\_{1\leqslant i\neq i'\leqslant n}X\_{i,j}X\_{i',j}$ as
$\left(\sum\_{i=1}^n X\_{i,j}\right)^2-\sum\_{i=1}^nX\_{i,j}^2$ and considering dyadic numbers, we are reduced to show that
$$
\tag{1}\max\_{1\leqslant j\leqslant 2^n}\frac{1}{2^{n}}\max\_{1\leqslant \ell\leqslant 2^n}\left\lvert \sum\_{i=1}^\ell X\_{i,j}\right\rvert\to 0\mbox{ almost surely}
$$
$$
\tag{2}\max\_{1\leqslant j\leqslant 2^n}\frac{1}{2^{2n}} \sum\_{i=1}^{2^n} X\_{i,j}^2\to 0\mbox{ almost surely}.
$$
By the Borel-Cantelli lemma, in order to prove (1), it suffices to show that
$$
\tag{1'}\forall\varepsilon>0, \sum\_{n\geqslant 1}\mathbb P\left(\max\_{1\leqslant j\leqslant 2^n}\frac{1}{2^{n}}\max\_{1\leqslant \ell\leqslant 2^n}\left\lvert \sum\_{i=1}^\ell X\_{i,j}\right\rvert>\varepsilon\right) <\infty$$
which reduces, by a union bound, to prove that
$$
\tag{1''}\forall\varepsilon>0, \sum\_{n\geqslant 1}2^n\mathbb P\left( \frac{1}{2^{n}}\max\_{1\leqslant \ell\leqslant 2^n}\left\lvert \sum\_{i=1}^\ell X\_{i,0}\right\rvert>\varepsilon\right) <\infty$$
Let us prove (1''). Using Corollary 1.5 in [1](https://www.esaim-ps.org/articles/ps/pdf/2019/01/ps180007.pdf) (applied with $r'=r=2$ and $B=\mathbb R$, hence $C\_{r',B}=1$, we know that for each $q>0$, there exist constants $A(q)$ and $B(q)$ such that for each $x>0$ and each independent sequence $(Y\_i)$,
$$
\mathbb P\left(\max\_{1\leqslant \ell\leqslant N}\left\lvert \sum\_{i=1}^\ell Y\_i\right\rvert >x\right)\leqslant A(q)\int\_0^1u^{q-1}\mathbb P\left(\max\_{1\leqslant i\leqslant N}\lvert Y\_i\rvert>xB(q)u\right)du+A(q)x^{-q}\left(\mathbb E\left[Y\_i^2\right]\right)^{q/2}.
$$
Applying this inequality to $Y\_i=X\_{i,0}$, $x=2^n\varepsilon$ and $q=3$ shows that
$$
\sum\_{n\geqslant 1}2^n\mathbb P\left( \frac{1}{2^{n}}\max\_{1\leqslant \ell\leqslant 2^n}\left\lvert \sum\_{i=1}^\ell X\_{i,0}\right\rvert>\varepsilon\right)\leqslant A(3)\sum\_{n\geqslant 1}2^{2n}\int\_0^1u^{2}\mathbb P\left( \lvert X\_{i,0}\rvert>\varepsilon B(3)2^nu\right)du + A(3)\sum\_{n\geqslant 1}2^{n}2^{-3n/2}\varepsilon^{-3/2}
$$
and finiteness of the first series follows from the elementary fact that $\sum\_{n\geqslant 1}2^{2n}\mathbb P\left(Y>2^n\right)\leqslant 4\mathbb E\left[Y^2\right]$.
(2'') follows from an application of the similar argument as the usual strong law of large number, but the truncation level changes. Let
$Y\_{i,j}=X\_{i,j}^2$ and for a fixed $\varepsilon$ and $n$, let
$$
Z\_{n,i,j}=Y\_{i,j}\mathbf{1} \{Y\_{i,j}\leqslant\varepsilon 2^{2n} \}-\mathbb E\left[Y\_{i,j}\mathbf{1}\{Y\_{i,j}\leqslant\varepsilon 2^{2n}\}\right],
$$
$$
W\_{n,i,j}=Y\_{i,j}\mathbf{1}\{Y\_{i,j}>\varepsilon 2^{2n}\}-\mathbb E\left[Y\_{i,j}\mathbf{1}\{Y\_{i,j}>\varepsilon 2^{2n}\}\right],
$$
We thus have to show that
$$
\tag{2'}\max\_{1\leqslant j\leqslant 2^n}\frac{1}{2^{2n}} \sum\_{i=1}^{2^n} Z\_{n,i,j}\to 0\mbox{ almost surely}.
$$
$$
\tag{2''}\max\_{1\leqslant j\leqslant 2^n}\frac{1}{2^{2n}} \sum\_{i=1}^{2^n} Z\_{n,i,j}\to 0\mbox{ almost surely}.
$$
For (2'), it suffices to show (by the Borel-Cantelli lemma) that
$\sum\_m\mathbb E\left[\left(\max\_{1\leqslant j\leqslant 2^n}\frac{1}{2^{2n}} \sum\_{i=1}^{2^n} Z\_{n,i,j}\right)^2\right]<\infty$.
For (2''), note that $$ \left(\{\max\_{1\leqslant j\leqslant 2^n}\frac{1}{2^{2n}} \sum\_{i=1}^{2^n} Z\_{n,i,j}\neq 0 \}\right)\subset\bigcup\_{i,j=1}^{2^n}\left(\{Y\_{i,j}>\varepsilon 2^{2n}\}\right).$$
---
[1](https://www.esaim-ps.org/articles/ps/pdf/2019/01/ps180007.pdf) Deviation inequalities for Banach space valued martingales differences sequences and random fields Davide Giraudo ESAIM: PS 23 922-946 (2019)
| 3 | https://mathoverflow.net/users/17118 | 432135 | 174,895 |
https://mathoverflow.net/questions/432109 | 1 | Suppose that $f$ is a continuous, nonconstant function on $[0,1]$. Fix some $0<a<1$. Is it possible to establish the following inequality
$$ |f(x+h)-f(x)| \leq C \left[ |h|^a + |2f(x)-f(x+h)-f(x-h)| \right] ~~~~~\forall x,x-h,x+h\in I, $$
where $C$ only depends on $f$ and $a$?
$|h|^a$ in the RHS of the inequality is added to restrict the scenario when $f$ is differentiable.
| https://mathoverflow.net/users/152618 | Can the second-order difference control the first-order difference for nowhere differentiable functions? | No. E.g., take any $b\in(0,1)$ such that $1-b<a$. Take any strictly decreasing sequence $(x\_n)$ in $[0,1]$ converging to $0$.
Then there clearly exists a continuous function $f$ on $[0,1]$ such that $f(0)=0$ and for all natural $k$ and all $x\in[x\_{2k},x\_{2k-1}]$ we have
$$f(x)=2h\_{2k}^{-b}(x-x\_{2k}),$$
where $h\_n:=x\_{n-1}-x\_n>0$; for instance, for each natural $k$, you can define $f$ on $[x\_{2k+1},x\_{2k}]$ by linear interpolation: $f(x):=2h\_{2k+1}^{-b}(x\_{2k}-x)$ for $x\in[x\_{2k+1},x\_{2k}]$. Note that $f(x)\to0$ for $x\in[x\_{2k},x\_{2k-1}]$ and $k\to\infty$.
For each natural $k$, let now $x=x\_{2k}$ and $h=h\_{2k}/2$. Then $|f(x+h)-f(x)|=h\_{2k}^{1-b}$ and $|h|^a + |2f(x)-f(x+h)-f(x-h)|=h^a=h\_{2k}^a/2^a=o(|f(x+h)-f(x)|)$ as $k\to\infty$. So, there is no real $C$ such that $|f(x+h)-f(x)|\le C[|h|^a + |2f(x)-f(x+h)-f(x-h)|]$ for all such $x$ and $h$.
| 2 | https://mathoverflow.net/users/36721 | 432144 | 174,898 |
https://mathoverflow.net/questions/428693 | 6 | Let $V$ be the real vector space of finitely supported functions $f: \Omega\to \mathbf{R}$ such that $\sum\_\omega f(\omega)=0$, where $\Omega$ is a given uncountable set.
Endow $V$ with the weak topology $\sigma:=\sigma(V,\mathbf{R}^\Omega)$, so that, more explicitly, a net $(f\_i)$ in $V$ is $\sigma$-convergent to $f \in V$ iff $\sum\_\omega f\_i(\omega)g(\omega)\to \sum\_\omega f(\omega)g(\omega)$ for all $g: \Omega\to \mathbf{R}$.
>
> **Question 1.** Does there exist a $\sigma$-sequentially-closed convex cone in $V$ which is not $\sigma$-closed?
>
>
>
Of course, any real vector space $E$ with uncountable dimension can be seen, up to isomorphism, as the vector space of finitely supported functions $f:\Omega\to \mathbf{R}$ with $\mathrm{dim}(E)=|\Omega|$, and $\mathbf{R}^\Omega$ can be seen as the algebraic dual $E^\star$.
Hence, ignoring the additional constraint $\sum\_\omega f(\omega)=0$, we may ask:
>
> **Question 2.** Let $E$ be a real vector space with uncountable dimension. Does there exist a $\sigma(E,E^\star)$-sequentially-closed convex cone in $V$ which is not $\sigma(E,E^\star)$-closed?
>
>
>
Ps. I know that the answer to Question 1 is negative if $\Omega$ is countable, but such examples may exist if $\Omega$ is uncountable. A related MSE question with a beautiful answer can be found [here](https://math.stackexchange.com/questions/2134762/is-there-any-example-of-a-sequentially-closed-convex-cone-which-is-not-closed); cf. also this [MO question](https://mathoverflow.net/questions/234537/every-convex-sequentially-closed-set-is-closed) for subsets which are not necessarily cones.
Ps2. A related result (Treves, "Topological Vector Spaces, Distributions and Kernels" (1967), p.201, Exercise 19.1): "Let $E$ be a locally convex Hausdorff tvs over $\mathbf{C}$. Then every linear subspace of $E$ is $\sigma(E,E^\star)$-closed." In our case, let $E\_0$ be a vector subspace of a real vector space $E$, let $\mathscr{B}\_0$ be a basis of $E\_0$ and extend it to a basis $\mathscr{B}$ of $E$. Since every linear functional is $\sigma(E,E^\prime)$-continuous, then $E\_0$ can be written as intersection of $\sigma(E,E^\prime)$-closed sets $\{x=\sum\_{b \in \mathscr{B}}\lambda\_{x,b}b \in E: \lambda\_{x,b\_0}=0\}$ for each $b\_0 \in \mathscr{B}\setminus \mathscr{B}\_0$. This proves that, if an example in Question 2 exists, then the $\sigma(E,E^\prime)$-sequentially closed convex cone is not a subspace.
| https://mathoverflow.net/users/32898 | Weakly sequentially closed convex cone which is not weakly closed | The answer is **negative**, for both questions.
Let $\{Z\_1,Z\_2\}$ be a partition of $\Omega$ such that $|Z\_1|=|Z\_2|$.
Let also $h: Z\_1\to Z\_2$ be a bijection and fix $a \in Z\_1$.
For each nonempty finite subset $B\subseteq Z\_1$, define
$$
\tilde{e}(B):=\frac{1}{|B|^2}\sum\_{b \in B}(e\_b-e\_{h(b)}),
$$
where $e\_z\in V$ stands for the function such that $e\_z(t)=1$ if $t=z$ and $e\_z(t)=0$ otherwise.
Note that $\tilde{e}(B)$ belongs to $$\Sigma:=\{f \in V: f\ge 0 \text{ and }\sum\nolimits\_\omega f(\omega)=1\}$$ for each finite nonempty subset of $Z\_1$. Now, let $\mathscr{B}$ be the family of nonempty finite subsets of $Z\_1\setminus \{a\}$ and $C$ be *the sequential weak-closure of the convex cone* $C\_0$, where
$$
C\_0:=\mathrm{co}\left(\mathrm{cone}\left(\left\{\tilde{e}(\{a\})+\tilde{e}(B): B \in \mathscr{B}\right\}\right)\right).
$$
(Here, $\mathrm{co}$ is the convex hull and $\mathrm{cone}(S):=\{\lambda x: \lambda>0, x \in S\}$.)
It is routine to check that $C$ is, indeed, a convex cone. To complete the proof, it will be enough to show that $\tilde{e}(\{a\}) \notin C$ and, on the other hand, $\tilde{e}(\{a\})$ belongs to the weak-closure of $C\_0$, so that $C$ is a sequentially weak-closed convex cone in $\Sigma$ which is not weak-closed.
$\text{ }$
$\bullet$ First, let us show that $\tilde{e}(\{a\}) \notin C$. For the sake of contradiction, suppose that, for each $n\ge 1$, there exist a positive integer $k\_n$, positive scalars $\lambda\_{n,1},\ldots,\lambda\_{n,k\_n} \in \mathbf{R}$, and
nonempty finite pairwise disjoint
sets $B\_{n,1},\ldots,B\_{n,k\_n}\in \mathscr{B}$ such that the sequence $(p\_n)\_{n\ge 1}$ is weak-convergent to $\tilde{e}(\{a\})$, where
$$
p\_n:=\sum\_{i=1}^{k\_n}\lambda\_{n,i} \left(\tilde{e}(\{a\})+\tilde{e}(B\_{n,i})\right).
$$
Set also $B\_0:=\bigcup\_{n\ge 1}\bigcup\_{i=1}^{k\_n}B\_{n,i}$.
Recalling that $\lim\_n \langle p\_n,u\rangle =\langle \tilde{e}(\{a\}),u\rangle$ for each $u \in V$ and choosing $u=e\_a$, we obtain $\lim\_n \sum\_{i=1}^{k\_n}\lambda\_{n,i}=1$; here, $\langle f,g\rangle:=\sum\nolimits\_\omega f(\omega)g(\omega)$ for each $f \in C$ and $g \in \mathbf{R}^\Omega$. First, suppose that $|B\_0|<\infty$ and write each $p\_n$ as $$\sum\_{\emptyset\neq B\subseteq B\_0}\lambda\_{n,B} \left(\tilde{e}(\{a\})+\tilde{e}(B)\right),$$ where each $\lambda\_{n,B}$ is possibly $0$ and at least one of them is nonzero. Letting $u$ be the characteristic function of $Z\_1$, we obtain
$$
1=\langle \tilde{e}(\{a\}), u\rangle
=\lim\_{n\to \infty}\langle \sum\_{\emptyset \neq B\subseteq B\_0}\lambda\_{n,B}\left(\tilde{e}(\{a\})+\tilde{e}(B)\right),u\rangle
=\lim\_{n\to \infty} \sum\_{\emptyset \neq B\subseteq B\_0}\lambda\_{n,B}\left(1+\frac{1}{|B|}\right).
$$
Since $\lim\_n\sum\_{\emptyset \neq B\subseteq B\_0}\lambda\_{n,B}=1$, we obtain that
$$
\lim\_{n\to \infty} \sum\_{\emptyset \neq B\subseteq B\_0}\frac{\lambda\_{n,B}}{|B|}=0.
$$
However, this is impossible since
$$
\liminf\_{n\to \infty} \sum\_{\emptyset \neq B\subseteq B\_0}\frac{\lambda\_{n,B}}{|B|} \ge
\liminf\_{n\to \infty} \sum\_{\emptyset \neq B\subseteq B\_0}\frac{\lambda\_{n,B}}{|B\_0|}=\frac{1}{|B\_0|}>0.
$$
This contradiction proves that $B\_0$ is an infinite set. Hence, setting $z\_n:=\max \bigcup\_{i=1}^{k\_n}B\_{n,i}$ for each $n\ge 1$, we obtain that there exists a strictly increasing subsequence $(z\_{n\_m})\_{m\ge 1}$ with the property that $z\_{n\_m}\notin \bigcup\_{1\le t<n\_m}\bigcup\_{i=1}^{k\_t}B\_{t,i}$ for each $m\ge 1$. Finally, let $u\_0 \in \mathbf{R}^\Omega$ be the function supported on $\{z\_{n\_t}: t\ge 1\}$ defined by
$$
u\_0(z\_{n\_m}):=m \cdot \frac{\max\{|B\_{n\_m,i}|^2: i \in [1,k\_{n\_m}]\}}{\min\{\lambda\_{n\_m,i}: i \in [1,k\_{n\_m}]\}}
$$
for all $m\ge 1$. It follows by construction that
$$
\langle p\_{n\_m},u\_0\rangle\ge p\_{n\_m}(z\_{n\_m})u\_0(z\_{n\_m})
= u\_0(z\_{n\_m}) \sum\_{i=1}^{k\_{n\_m}}\lambda\_{n\_m,i} \left(\tilde{e}(\{a\})+\tilde{e}(B\_{n\_m,i})\right)(z\_{n\_m})
\ge u\_0(z\_{n\_m}) \sum\_{i=1}^{k\_{n\_m}}\lambda\_{n\_m,i} \tilde{e}(B\_{n\_m,i})(z\_{n\_m})
\ge u\_0(z\_{n\_m}) \cdot \frac{\min\{\lambda\_{n\_m,i}: i \in [1,k\_{n\_m}]\}}{\max\{|B\_{n\_m,i}|^2: i \in [1,k\_{n\_m}]\}}=m,
$$
for all $m\ge 1$. This shows that the subsequence $\left(\langle p\_{n\_m}, u\_0\rangle \right)\_{m\ge 1}$ cannot be convergent to $\langle \tilde{e}(\{a\}), u\_0\rangle$, contradicting the hypothesis that $(p\_n)\_{n\ge 1}$ is weak-convergent to $\tilde{e}(\{a\})$. Therefore $\tilde{e}(\{a\})\notin C$.
$\text{ }$
$\bullet$ Lastly, let us show that $\tilde{e}(\{a\})$ belongs to the weak-closure $\overline{C\_0}$ of $C\_0$. To this aim, suppose for the sake of contradiction that $\tilde{e}(\{a\})\notin \overline{C\_0}$. Thanks to the
Strong Separating Hyperplane Theorem, there exists a linear functional $f \in V$ such that
$$
f(\tilde{e}(\{a\}))=-1
\quad \text{ and }\quad
f(p) \ge 0 \text{ for all }p \in \overline{C\_0}.
$$
Now, since $\Omega$ is uncountable, there exists a positive integer $k\_0$ and an uncountable subset $\tilde{Z}\_1\subseteq Z\_1\setminus \{a\}$ such that
$$
\forall b \in \tilde{Z}\_1, \quad f(\tilde{e}(\{a\})+\tilde{e}(\{b\})) \le k\_0,
$$
Note that, since $\tilde{e}(\{a\})+\tilde{e}(B)$ belongs to $C\_0$ for each $B \in \mathscr{B}$, then $f(\tilde{e}(B)) \ge
1$.
To conclude, let $B$ be a finite subset of $\tilde{Z}\_1$ such that $|B|=k\_0+2$. It follows that
$$
0\le f(\tilde{e}(\{a\})+\tilde{e}(B))
$$
$$
=f\left(\frac{1}{|B|}\sum\_{b \in B}(\tilde{e}(\{a\})+\tilde{e}(\{b\}))+\tilde{e}(B)-\frac{1}{|B|}\sum\_{b \in B}\tilde{e}(\{b\})\right)
$$
$$
=\frac{1}{|B|}\sum\_{b \in B}f\left(\tilde{e}(\{a\})+\tilde{e}(\{b\})\right)+\left(\frac{1}{|B|^2}-\frac{1}{|B|}\right)\sum\_{b \in B}f\left(\tilde{e}(\{b\})\right)
$$
$$
\le \frac{k\_0}{|B|}+|B|\left(\frac{1}{|B|^2}-\frac{1}{|B|}\right)
<0.
$$
This contradiction proves that $\tilde{e}(\{a\})\in \overline{C\_0}$, completing the proof.
| 0 | https://mathoverflow.net/users/32898 | 432162 | 174,902 |
https://mathoverflow.net/questions/432080 | 6 | Let $C\subseteq \mathbb R^n$ be non-empty, convex and compact. For $v\in S^{n-1}$, let $H\_v$ be the supporting hyperplane in the direction of $v$ (i.e., $H\_v$ is the boundary of the smallest closed half-space with outward normal $v$ that contains $C$). Let $U\subseteq S^{n-1}$ be the set of directions $v$ such that $H\_v$ meets $C$ at exactly one point.
**Main question:** Does $S^{n-1}\backslash U$ have measure zero?
If not, then I have a second question: Is $U$ dense in $S^{n-1}$?
For $n=2$, it's easy to see that $S^{n-1}\backslash U$ is countable (otherwise there are uncountably many nondegenerate line segments in the boundary of $C$, and hence $C$ has infinite perimeter). But a cylinder shows that in general $S^{n-1}\backslash U$ need not be countable.
| https://mathoverflow.net/users/26809 | For most directions does the supporting hyperplane meeting a bounded convex set meet it in one point? | The support function of $C$, restricted to the unit sphere, is differentiable exactly at directions such that (the relevant) hyperplane normal to that direction has a single contact point with $C$.
Because $C$ is bounded, its support function is Lipschitz. Rademacher's theorem then says it is differentiable almost everywhere, giving a positive answer to the question.
| 5 | https://mathoverflow.net/users/112954 | 432165 | 174,903 |
https://mathoverflow.net/questions/430285 | 2 | Some time ago, I asked a [question](https://mathoverflow.net/questions/424067/duke-and-schulze-pillot-condition-for-equidistribution) about equidistribution on a paper of Duke and Schulze-Pillot that was usefully answered.
However, on the answer there was a statement that was unimportant for me back then but catch my attention now. One of the steps of the answer was to use the following equivalence property:
Let $Q$ be a ternary quadratic form and $n$ a non-negative integer. "$n$ is primitively represented by some form in the genus of $Q$" is equivalent to "$n$ is primitively represented by $Q$ modulo $4\det(a\_{ij})$, where $(a\_{ij})\in\mathrm{M}\_3(\mathbb{Z})$ is the symmetric matrix such that
$$Q(x\_1,x\_2,x\_3)=\frac{1}{2}\sum\_{i,j}a\_{ij}x\_ix\_j."$$
I believe that this statement is true but have failed to found a reference for it, although I have read some other papers that use the same or similar results (e.g. this [one](https://arxiv.org/pdf/1402.1332.pdf) , page 5, line 10). I wonder if somebody could give me a reference or a simple proof for this property in case it's true. Otherwise, I would really appreciate a similar congruence property that's correct (if exists).
| https://mathoverflow.net/users/411616 | Primitive representation of integers by some form on the genus of a quadratic form | The quoted text was written by me. As I corrected myself recently in a comment below the original post: I secretly assumed that $n$ was coprime to $\det(a\_{ij})$. In that case, I believe that modulo $4\det(a\_{ij})$ is sufficient. See Hilfssatz 13 in Siegel: Über die analytische Theorie der quadratischen Formen, Ann. of Math. 36 (1935), 527-606. Note that when applying this theorem, $b=0$ for $p>2$, and $b=1$ for $p=2$ (because $p\mid S$ implies $p\nmid T$ by our coprimality assumption).
| 3 | https://mathoverflow.net/users/11919 | 432174 | 174,906 |
https://mathoverflow.net/questions/432179 | 3 | For $A \in \mathbb R^{m \times n}$ and the induced norms:
$$
\| A \|\_1 = \max\_{x \ne 0} \frac{\|Ax\|\_1}{\|x\|\_1}
$$
$$
\| A \|\_2 = \max\_{x \ne 0} \frac{\|Ax\|\_2}{\|x\|\_2}
$$
... where:
$$
\|x\|\_1 = \sum\_{k=1}^n |x\_k|
$$
$$
\|x\|\_2 = \sqrt{\sum\_{k=1}^n |x\_k|^2}
$$
... does the following inequality hold in general?
$$
\|A\|\_2 \le \|A\|\_1
$$
| https://mathoverflow.net/users/146993 | Is the matrix induced L1-norm greater than the induced L2-norm? | To avoid ambiguity I will write $\lVert\cdot\rVert\_{p\to r}$ for the $\ell\_p$-to-$\ell\_r$-norm. Note that in general, $\lVert A\rVert\_{1\to r} = \max\_{1\leq j\leq n} \lVert (Ae\_j)\rVert\_r$.
Let $A$ be the $n\times n$ matrix whose top row has $1$ in every entry, and all other entries of the matrix are $0$. Then by the remark above, $\lVert A\rVert\_{1\to 1} =1$. On the other hand,
$$ \lVert A\rVert\_{2\to 2} \geq \lVert A^\* \rVert\_{2\to 2} \geq \lVert A^\*e\_1\rVert\_2 = \sqrt{n} $$
giving a counterexample to your question. (In fact this lower bound is an equality, although this is not needed to answer the question.)
| 7 | https://mathoverflow.net/users/763 | 432181 | 174,909 |
https://mathoverflow.net/questions/432168 | 2 | For any positive number $t$, nonnegative integer $n$, and nonzero vector $a \in \mathbb R^n$, define
$$
\begin{split}
K\_n(a,t) &:= \inf\_{x \in \mathbb R^n} \|x-a\|\_2 + t\|x\|\_1,\\
M\_n(a,t) &:= \min(\|a\|\_2,t\|a\|\_1),\\
R\_n(a,t) &:= K\_n(a,t)/M\_n(a,t).
\end{split}
$$
**Note.** $K\_n$ defines the Peetre's K-functional between $(\mathbb R^n,\ell\_2)$ and $(\mathbb R^n,\ell\_1)$, where the $\ell\_p$ norm of a vector $x=(x^1,\ldots,x^n)$ is defined by $\|x\|\_p := (\sum\_i |x^i|^p)^{1/p}$.
**Question.** Is it possible to construct $a\_n \in \mathbb R^n$ for each $n$, such that $\lim\_{n \to \infty} R\_n(a\_n,t) = 0$ ?
Motivation
----------
Clearly, one always has $K\_n(a,t) \le M\_n(a,t)$ with equality when $n=1$. What is not clear is whether one can construct $a$ with growing dimension $n$ such that $K\_n(a,t) \ll M\_n(a,t)$ eventually.
| https://mathoverflow.net/users/78539 | Fix positive $t$. Construct $a_n \in \mathbb R^n$ such that $(\inf_x \|x-a_n\|_2 + t\|x\|_1 )/\min(\|a_n\|_2,t\|a_n\|_1) \to 0$ | This is to extend Christian Remling's [comment](https://mathoverflow.net/questions/432168/fix-positive-t-construct-a-n-in-mathbb-rn-such-that-inf-x-x-a-n-2#comment1112378_432168) to all real $t>0$, with an explicit lower bound on $K/M$, where $K:=K\_n(a,t)$ and $M:=M\_n(a,t)$.
$\newcommand\norm[1]{\lVert#1\rVert}$The key here, as in Christian Remling's [comment](https://mathoverflow.net/questions/432168/fix-positive-t-construct-a-n-in-mathbb-rn-such-that-inf-x-x-a-n-2#comment1112378_432168), is the observation that $\norm x\_1\ge\norm x\_2$ for all $x\in\mathbb R^n$. Indeed, this observation implies
$$K
\ge\inf\_{x\in\mathbb R^n}\bigl(\lvert\norm x\_2-\norm a\_2\rvert+t\norm x\_2\bigr)
=\inf\_{u\ge0}\bigl(\lvert u-\norm a\_2|+tu\bigr)
=\min(1,t)\norm a\_2$$
and
$$M\le\norm a\_2,$$
whence
$$\frac KM\ge\min(1,t).$$
(In particular, $K\ge M$ if $t\ge1$.)
| 3 | https://mathoverflow.net/users/36721 | 432184 | 174,910 |
https://mathoverflow.net/questions/432186 | 0 | Let $R$ be a henselian DVR with fraction field $K$ and residue field $k$ of characteristic $p>0$. Let $\overline K$ be an algebraic closure of $K$, $\overline R$ the normalization of $R$ in $\overline K$ and $\overline k$ the residue field of $\overline R$. Let $G=\mathrm{Gal}(\overline K/K)$ be the absolute Galois group of $K$. Write $S, \eta, \overline{\eta}, s, \overline s$ for the $\mathrm{Spec}$ respectively of $R, K, \overline K, k, \overline k$. Let $\Lambda =\mathbb Z\_{\ell}$ with $\ell \not = p$ (or $\mathbb Q\_{\ell}$ or any torsion sheaf where $p$ is invertible). Let $X$ be a proper scheme over $S$. We consider the nearby cycle sheaf $\mathrm R\Psi\Lambda$ on the geometric special fiber $X\_{\overline s}$.
Assume now that $X$ has semi-stable reduction, implying that the special fiber $X\_s = \bigcup\_{1\leq i \leq r} Y\_i$ is a normal crossing divisor. For $E\subset \{1,\ldots,r\}$ let $Y\_E := \bigcap\_{i\in E} Y\_i$ and for $1\leq m \leq r$ define $Y^{(m)} = \bigsqcup\_{\#E=m} Y\_E$.
In his paper "Exposé 1: autour du théorème de monodromie locale", Illusie explains how to compute the nearby cycle sheaf $\mathrm R\Psi\Lambda$. Théorème 3.2.(c) states that there are isomorphisms of $G-\Lambda-$sheaves on $X\_{\overline s}$
* $\mathrm R^0\Psi\Lambda \simeq \Lambda,$
* $\mathrm R^1\Psi\Lambda \simeq \left(\bigoplus\_{i} \Lambda\_{Y\_i}/\Lambda\right)(-1)$ where $\Lambda$ injects diagonally in the sum,
* $\mathrm R^q\Psi\Lambda \simeq \bigwedge^q \mathrm R^1\Psi\Lambda.$
I have a doubt regarding this statement that I'd like to clear up. It is my geometric intuition that $\mathrm R^q\Psi\Lambda$ should be supported on $Y^{(q)}\_{\overline s}$, but according to the statement is looks like it is a constant sheaf on the whole of $X\_{\overline s}$. May I ask which is true?
---
(Edit) In regards to Will Sawin's reply, I understood the origin of my confusion. I thought that the notation $\Lambda\_{Y\_i}$ merely meant a copy of the constant sheaf $\Lambda$ (on the whole special fiber) for each $Y\_i$, but it actually denotes the constant sheaf $\Lambda$ supported on $Y\_i$.
| https://mathoverflow.net/users/125617 | Nearby cycles for schemes with semi-stable reduction | According to this statement, $\mathrm R^q\Psi\Lambda$ is supported on points where $\mathrm R^1\Psi\Lambda $ has rank at least $q$ (by a property of wedge powers) and thus supported on points where $\bigoplus\_{i} \Lambda\_{Y\_i}$ has rank at least $q+1$ (by a property of quotients) and thus on $Y^{(q+1)}$ (since each summand has rank $1$ at $x$ if $x\in Y\_i$ and rank $0$ at $x$ otherwise).
| 3 | https://mathoverflow.net/users/18060 | 432187 | 174,912 |
https://mathoverflow.net/questions/432202 | 6 | When is $\mathbb{Q}(\sqrt{p+\sqrt{p}})$ a Galois extension of $\mathbb{Q}$?
I was motivated by the question that $\mathbb{Q}(\sqrt{5+\sqrt{5}})$ is a Galois extension of $\mathbb{Q}$. Here is a rough sketch of the proof.
First, note that the polynomial
$$f(x)=(x-\sqrt{5+\sqrt{5}}) (x+\sqrt{5+\sqrt{5}})(x-\sqrt{5-\sqrt{5}})(x+\sqrt{5-\sqrt{5}})$$
is irreducible by the Eisenstein irreducibility criterion. So we claim that $\mathbb{Q}(\sqrt{5+\sqrt{5}})$ is the splitting field of the separable polynomial $f(x)$. It suffices to show that $\sqrt{5-\sqrt{5}}\in\mathbb{Q}(\sqrt{5+\sqrt{5}})$. To do this, we first note that $5+\sqrt{5}$ and hence $\sqrt{5}$ is in $\mathbb{Q}(\sqrt{5+\sqrt{5}})$.
Then, by noticing that
$$\sqrt{5-\sqrt{5}}=\frac{\color{red}{\sqrt{5-1}}\sqrt{5}}{\sqrt{5+\sqrt{5}}}= \frac{\color{red}2\sqrt{5}}{\sqrt{5+\sqrt{5}}},\tag{1}$$
we see that $\sqrt{5-\sqrt{5}}$ is in $\mathbb{Q}(\sqrt{5+\sqrt{5}})$ as desired.
By the above argument, we see that $\mathbb{Q}(\sqrt{p+\sqrt{p}})$ is a Galois extension of $\mathbb{Q}$ when $p-1$ is a square number. (The importance of $p-1$ being a square number can be seen from equation (1)-See the part highlighted in red)
I am wondering if this is both a necessary and sufficient criterion for $\mathbb{Q}(\sqrt{p+\sqrt{p}})$ to be a Galois extension of $\mathbb{Q}$. In other words, $\mathbb{Q}(\sqrt{p+\sqrt{p}})$ is a Galois extension of $\mathbb{Q}$ if and only if $p$ is prime and $p-1$ is a square number.
Some things that I have noticed regarding this problem:
1. $p=3$ doesn’t satisfy the suggested criterion and isn’t a Galois extension. See (<https://math.stackexchange.com/questions/4052247/showing-that-mathbbq-sqrt3-sqrt3-is-not-galois>)
2. For general a $p$, $f(x)=x^4-2px^2+(p^2-p)$.
3. I am not sure if we need $p$ to be prime. The only place where we used $p$ is prime is when we used the Eisenstein irreducibility criterion. But do we really need $f(x)$ to be irreducible?
4. This is where I am probably getting a bit ahead of myself. But when we say that $p-1$ is a square number, most of us naturally assume that $p$ is in $\mathbb{Z}^+$ and $p-1$ is square in $\mathbb{Z}^+$. But can we further generalise the result by allowing $p$ to be in $\mathbb{Q}$ and $p-1$ be square in $\mathbb{Q}$?
| https://mathoverflow.net/users/369754 | When is $\mathbb{Q}(\sqrt{p+\sqrt{p}})$ a Galois extension of $\mathbb{Q}$? | Note that $f$ cannot have an irreducible factor of degree $3$ since it is even, so if $f$ is reducible, it means that the minimal polynomial of $\sqrt{p+\sqrt{p}}$ has degree at most $2$, so your extension is automatically Galois in this case.
Hence we may assume without loss of generality that $f$ is irreducible.
Your question then may be rephrased as follows: when is the Galois group of $f=X^4-2pX^2+p^2-p$ is the Klein group or the cyclic group of order $4$ ?
The answer (and much more!) is given in Corollary 4.5. of [this paper by Keith Conrad](https://kconrad.math.uconn.edu/blurbs/galoistheory/cubicquartic.pdf)
Applied to your specific polynomial, the answer is : exactly when $p^2-p$ is a square (Klein group case) or when $p-1$ is a square (cyclic case).
Another link with a direct proof [may be found here](https://math.stackexchange.com/questions/649466/galois-group-of-a-biquadratic-quartic)
| 11 | https://mathoverflow.net/users/36683 | 432209 | 174,916 |
https://mathoverflow.net/questions/432208 | 13 | I want to grasp the moving frames method but I find some obstacles. I don't know if this question is suitable for MO, if it is not the case please let me know and I will move it.
I am aware there are other related questions here like [this one](https://mathoverflow.net/questions/337294/moving-frames-method-for-non-matrix-lie-group) or [this one](https://mathoverflow.net/questions/336149/about-the-cartans-moving-frame-method?rq=1), but they don't answer my doubts.
Given a Lie group $G$ and a homogeneous space $X\equiv G/H$, the goal of the moving frames method is to study submanifolds $M$ of $X$. In particular we want to know if two given submanifolds $M$ and $\tilde{M}$ are "congruent", in the sense that there is a "movement" $g\in G$ such that $g(M)=\tilde{M}$.
I know I have a $\mathfrak{g}$-valued differential form on $G$ which is left invariant, the Maurer-Cartan form $\theta$. The left-invariance of $\theta$ allows us to show (see [griffiths1974cartan](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&cad=rja&uact=8&ved=2ahUKEwjxmv-g3tf6AhUCcxoKHQC6DLcQFnoECAsQAQ&url=https%3A%2F%2Fprojecteuclid.org%2Fjournals%2Fduke-mathematical-journal%2Fvolume-41%2Fissue-4%2FOn-Cartans-method-of-Lie-groups-and-moving-frames-as%2F10.1215%2FS0012-7094-74-04180-5.full&usg=AOvVaw3bbvtFV6C_skrb20r37Tg9), lemma (1.3)) that given two maps $f,\tilde{f}$ from, let's say, $\mathbb{R}^n$ to $G$ then $\tilde{f}(x)=g\cdot f(x)$ if and only if $\tilde{f}^\*(\theta)=f^\*(\theta)$. This way $f^\*(\theta)$ provide a set of invariants to characterize submanifolds of $G$ (not of $X$!).
Suppose our submanifolds $M$ and $\tilde{M}$ are parametrized respectively by maps $\alpha:\mathbb{R}^n \to X$ and $\tilde{\alpha}:\mathbb{R}^n \to X$. If we are in a particular case, for example $G=E(2)$, $X=\mathbb{E}^2$ and $M,\tilde{M}$ curves, we have a "canonical" way to lift $\alpha$ and $\tilde{\alpha}$ to $f\_{\alpha}, f\_{\tilde{\alpha}}:\mathbb{R} \to G$ (the unitary tangent vector and its orthogonal, together with the curve point itself). This way, if
$$
f\_{\alpha}^\*(\theta)={f}\_{\tilde{\alpha}}^\*(\theta)
$$
we conclude that the "curves of frames" $f\_{\alpha}$ and $f\_{\tilde{\alpha}}$ are congruent, and therefore $\alpha$ and $\tilde{\alpha}$ are congruent.
The key fact here is, I think, that the assignment
$$
\alpha \mapsto f\_{\alpha}
$$
is $G$-invariant, in the sense that $\tilde{\alpha}=g\alpha$ if and only if $f\_{\tilde{\alpha}}=g f\_{\alpha}$. Otherwise we could have congruent curves in $E(2)$ which couldn't be detected by the invariants (because of the "bad assignment" of frames to the curves).
**Question 1**
Back to the general case: can we always find such a "canonical lift"? Is there a method to find it? Or is the moving frames method restricted to a bunch of particular cases?
$\blacksquare$
**Question 2**
Can you provide at least a brief list of examples of these assignments? For example:
* Curves in $\mathbb{R}^3$ with Euclidean movements: the Frenet frame.
* Surfaces in $\mathbb{R}^3$ with Euclidean movements: a frame made with the point, the normal vector and two ortogonal vectors aligned with the principal directions of the surface (or is not this necessary?).
* ...
$\blacksquare$
I have read the article of Griffiths, the corresponding chapter of *Cartan for begginers* and *From Frenet to Cartan*, but I am still blocked with these doubts.
| https://mathoverflow.net/users/129995 | Moving frames method | I think you might want to read a couple of articles on the moving frame that carefully discuss this issue (and show that it is more subtle than most people realize).
The first is a paper by Mark Green, *The moving frame, differential invariants and rigidity theorems for curves in homogeneous spaces* (Duke Math. J. 45 (1978), no. 4, 735–779). Mark discusses the 1-dimensional case in great detail and shows, by example, that, in fact, there are cases in which a `canonical' lifting by Cartan's method does not exist, and one must extend the method by introducing a family of liftings in order to find all of the invariants.
The second is a set of lecture notes by Gary Jensen,
*Higher order contact of submanifolds of homogeneous spaces*,
(Lecture Notes in Mathematics, Vol. 610. Springer-Verlag, Berlin-New York, 1977.) He discusses the foundations of the subject in great detail and gives a number of illuminating examples.
Of course, there are lots of more recent papers and articles, but I think that these two illustrate a lot of the issues and point out some of the pitfalls that aren't apparent at first sight.
| 16 | https://mathoverflow.net/users/13972 | 432215 | 174,918 |
https://mathoverflow.net/questions/432155 | 1 | We introduce the following functional to study Yamabe problem with boundary.
$$
Q\_g(\varphi)=\frac{\int\_M\left(|\nabla \varphi|\_g^2+\frac{n-2}{4(n-1)} R\_g \varphi^2\right) d v+\frac{n-2}{2} \int\_{\partial M} h\_g \varphi^2 d \sigma}{\left(\int\_M|\varphi|^{2 n /(n-2)} d v\right)^{(n-2) / 2}}
$$
$$
Q(M)=\inf \left\{Q\_g(\varphi): \varphi \in C^1(\bar{M}), \varphi \neq 0\right\}
$$
In Escobar's The Yamabe Problem On Manifolds With Boundary,he says if $Q(M)<Q(S^{+}\_{n})$,the functional has a critical point which is the solution of Yamabe Problem.
He also gives the formula of $Q(S\_{n}^{+})$
$$
Q\left(S\_{+}^n\right)=\frac{\int\_{\mathbb{R}\_{+}^n}\left|\nabla u\_{\varepsilon}\right|^2}{\left(\int\_{\mathbb{R}\_{+}^n} u\_{\varepsilon}^{2 n /(n-2)}\right)^{(n-2) / n}}=n(n-2)\left(\int\_{\mathbb{R}\_{+}^n} u\_{\varepsilon}^{\frac{2 n}{n-2}}\right)^{2 / n},
$$
$$
u\_{\varepsilon}\left(x, x\_n\right)=\left(\frac{\varepsilon}{\varepsilon^2+|x|^2+x\_n^2}\right)^{(n-2) / 2}
$$
I wonder how this is derived.
When studying the Yamabe problem on compact manifolds,we use conformal change to calculate $Q(S\_{n})$.Since the functional is conformally invariant, we use conformal change to transfer the problem to $R^{n}$,which becomes the problem of best Sobolev constant.
How can we do this when studying the problem on manifolds with boundary?Any help will be thanked.
| https://mathoverflow.net/users/148247 | How to calculate the infimum of Yamabe functional on upper hemisphere | The first observation is that $Q\_g(\phi)$ is conformally invariant.
Define
\begin{align\*}
L\_gu & := -\Delta u + \frac{n-2}{4(n-1)}Ru , \\
B\_gu & := \partial\_\nu u + \frac{n-2}{2}Hu ,
\end{align\*}
where $\nu$ is the outward-pointing unit normal and everything is defined with respect to the metric $g$.
On the one hand, integration by parts gives
\begin{equation\*}
Q\_g(u) = \frac{\int\_M u\,Lu\,dv\_g + \oint\_{\partial M} u\,Bu\,d\sigma\_g}{\left( \int\_M \lvert u \rvert^{\frac{2n}{n-2}} \, dv\_g \right)^{\frac{n-2}{n}}} .
\end{equation\*}
On the other hand, if $\hat g = w^{\frac{4}{n-2}}g$, then
\begin{align\*}
L\_{\hat g}u & = w^{-\frac{n+2}{n-2}}L\_g(wu) , \\
B\_{\hat g}u & = w^{-\frac{n}{n-2}}B\_g(wu) .
\end{align\*}
Therefore
\begin{equation\*}
Q\_{\hat g}(u) = Q\_g(wu) .
\end{equation\*}
It follows from stereographic projection that $Q(S\_+^n) = Q(\mathbb{R}\_+^n)$, where $\mathbb{R}\_+^n$ is the upper half plane $\{ x\_n \geq 0 \}$.
The next observation is that $u\_\varepsilon$ really minimizes $Q(\mathbb{R}\_+^n)$.
One argument goes as follows:
By the [Concentration Compactness Principle](https://doi.org/10.4171/RMI/12), a minimizer $u$ exists.
Moreover, the usual argument using elliptic regularity and the maximum principle implies that a minimizer is smooth.
By an [adaptation of the Obata argument](https://doi.org/10.1002/cpa.3160430703), if $u$ is a minimizer, then $u^{\frac{4n}{n-2}}dx^2$ is an Einstein metric on $S\_+^n$ with respect to which the boundary is minimal.
This requirement implies that $u$ takes the form $u\_\varepsilon$.
Note that the latter argument proceeds by using stereographic projection again to study the minimizer $u$ on $S\_+^n$.
| 1 | https://mathoverflow.net/users/121820 | 432227 | 174,922 |
https://mathoverflow.net/questions/432223 | 7 | Let $X \neq \emptyset$ be a set. We say that ${\cal F} \subseteq {\cal P}(X)$ is a *down-set* if ${\cal F}$ is closed under taking subsets. Whenever $a \in X$, we let ${\cal F}\_a = \{ S \in F : a \in S\}$.
We say ${\cal G} \subseteq {\cal P}(X)$ is *intersecting* if $S \cap T$ is non-empty whenever $S, T \in {\cal G}$.
Consider the following statement:
>
> (C): Whenever $X$ is a non-empty set, ${\cal F} \subseteq {\cal P}(X)$ is a down-set, and ${\cal G} \subseteq {\cal F}$ is an intersecting family, then there is $a \in X$, and an injection $\iota: {\cal G} \to {\cal F}\_a$.
>
>
>
It is not known whether (C) is true for finite non-empty sets, and this open problem is called Chvatal's conjecture. I was thinking that for $X = \omega$ maybe a family with the finite intersection property could be used to show that (C) is false for infinite ground sets $X$, but I failed to come up with an argument.
**Question.** What is a counterexample to (C) in the infinite case?
| https://mathoverflow.net/users/8628 | Counterexample for Chvatal's conjecture in an infinite set | Let $\kappa$ be a cardinal of
countable cofinality that is strictly larger than the continuum.
I will construct a counterexample to (C) on
$X=(\kappa\times \omega)\cup \omega$.
Since $\kappa$ has countable cofinality, there exists
a countable increasing sequence of cardinals
$(\kappa\_i)\_{i\in\omega}$ with limit $\kappa$.
For each ordinal $\mu<\kappa$, let $n\_{\mu}$
be the least $n\in\omega$ such that $\mu\in \kappa\_n$.
For each $\mu<\kappa$ let
$A\_{\mu}\subseteq X$ be the countable set
$$
A\_{\mu} = (\{\mu\}\times \omega)\cup \{2^c\cdot 3^d\;|\; c\leq n\_{\mu} \leq d\}.
$$
Let $\mathcal G = \{A\_{\mu}\;|\;\mu<\kappa\}$.
Let $\mathcal F$ be the down-closure of ${\mathcal G}$ in ${\mathcal P}(X)$.
Necessarily $\mathcal F$ is a down-set in ${\mathcal P}(X)$ containing $\mathcal G$.
I claim that
* $\mathcal G$ is intersecting.
* $\mathcal G$ has cardinality $\kappa$.
* Each ${\mathcal F}\_a$ has size $<\kappa$.
The last two items show that there is no injection from $\mathcal G$
to any ${\mathcal F}\_a$.
To show that $\mathcal G$ is intersecting, note that
if $\mu \leq \nu$, then $2^{n\_{\mu}}3^{n\_{\nu}}\in A\_{\mu}\cap A\_{\nu}$.
To see that $|{\mathcal G}| = \kappa$, it is enough to
observe that the $A\_{\mu}$'s are indexed by $\kappa$
and that they are distinct from each other since
$(\mu,0)\in A\_{\mu}-A\_{\nu}$
when $\mu\neq \nu$.
It remains to show that
each ${\mathcal F}\_a$ has size $<\kappa$.
Here there are cases depending on whether
(i) $a \in \kappa\times \omega$,
(ii) $a$ has the form $2^c\cdot 3^d$ for some $c\leq d$,
or (iii)
$a$ does not belong to any of the sets $A\_{\mu}$.
In case (i) $a=(\lambda,k)\in \kappa\times \omega$.
There is exactly one set $A\_{\mu}$
containing $a$, namely $A\_{\lambda}$.
Since ${\mathcal F}$ is the order ideal
generated by $\mathcal G$, the set
${\mathcal F}\_a$ is ${\mathcal P}(A\_{\lambda})\_a$.
Since $A\_{\lambda}$ is countably infinite,
${\mathcal P}(A\_{\lambda})\_a$ has size continuum, which is $<\kappa$.
In case (ii) we have $a=2^c\cdot 3^d$ for some $c\leq d$.
If $a\in A\_{\mu}$ for some $\mu$, then
$c\leq n\_{\mu}\leq d$. There are at most finitely
many $n\_{\mu}$'s that can satisfy this, let $N$ be the largest such one.
The elements of ${\mathcal F}\_a$ are subsets
of those $A\_{\mu}$'s where $c\leq n\_{\mu}\leq d$.
Hence
${\mathcal F}\_a\subseteq \bigcup\_{\mu\leq \kappa\_N} {\mathcal P}(A\_{\mu})\_a$.
Since each $A\_{\mu}$ is countably infinite
${\mathcal P}(A\_{\mu})\_a$ has size $2^{\aleph\_0}$.
Therefore $|{\mathcal F}\_a|\leq 2^{\aleph\_0}\cdot \kappa\_{N} = \max(2^{\aleph\_0},\kappa\_N)<\kappa$.
In case (iii) ${\mathcal F}\_a$ is empty, so has size $<\kappa$. \\\
| 3 | https://mathoverflow.net/users/75735 | 432245 | 174,927 |
https://mathoverflow.net/questions/432195 | 3 | This problem has been asked in [MSE](https://math.stackexchange.com/questions/4548511/an-inequality-equivalent-to-h%c3%b6rmanders-condition-sup-y-in-mathbb-rn-int), but got no answers. I guess that this exam problem may be a small lemma in some research papers, so I post it here on MathOverflow.
>
> Let $K\in L\_{\text{loc}}^1(\mathbb R^n\setminus\{0\})$. Prove that
> $$\sup\_{y\in\mathbb R^n}\int\_{\{x: |x|>2|y|\}}|K(x-y)-K(x)|\,dx<\infty\label{1}\tag{1}$$
> if and only if
> $$\sup\_{r>0}\frac1{r^n}\int\_{B(0,r)}\int\_{\{x: |x|>2r\}}|K(x-y)-K(x)|\,dx\,dy<\infty.\label{2}\tag{2}$$
>
>
>
This is an old exam problem on Harmonic Analysis. Formula \eqref{1} is called the Hörmander's condition for singular integrals. The proof of \eqref{1}$\Rightarrow$\eqref{2} is quite easy: assume
$$\int\_{\{x: |x|>2|y|\}}|K(x-y)-K(x)|\,dx\leq M,\qquad \forall y\in\mathbb R^n,$$
then for $r>0$ and $y\in B(0,r)$ we have $\{x: |x|>2r\}\subset \{x: |x|>2|y|\}$, so
$$\int\_{\{x: |x|>2r\}}|K(x-y)-K(x)|\,dx\leq \int\_{\{x: |x|>2|y|\}}|K(x-y)-K(x)|\,dx\leq M,$$
hence
$$\frac1{r^n}\int\_{B(0,r)}\int\_{\{x: |x|>2r\}}|K(x-y)-K(x)|\,dx\,dy\leq \frac1{r^n}\int\_{B(0,r)}M\,dy=M|B(0,1)|,\ \ \ \forall r>0.$$
This completes the proof of \eqref{1}$\Rightarrow$\eqref{2}.
However, for \eqref{2}$\Rightarrow$\eqref{1}, I don't know how to start.
Any help would be appreciated!
| https://mathoverflow.net/users/141451 | An inequality equivalent to Hörmander's condition $\sup_{y\in\mathbb R^n}\int_{\{x: |x|>2|y|\}}|K(x-y)-K(x)|\,dx<\infty$ | Condition \eqref{2} implies \eqref{1} with $5|y|$ instead of $2|y|$. Fix $y$, let $r=|y|$ and $I=\int\_{|x| >5|y| }|K(x-y)-K(x)|\, dx$. Then
$$I \leq \int\_{|x| >5r }|K(x-y)-K(x-z)|\, dx+\int\_{|x| >5r }|K(x-z)-K(x)|\, dx:=I\_1(z)+I\_2(z)
$$
for every $|z| \leq r$. If $K$ is the supremum in \eqref{2}, then $r^{-n} \int\_{B(0,r)} I\_2(z)\, dz \leq K$. In $I\_1$ we set $\xi=x-z$ so that $|\xi| \geq 4r$ and
$$I\_1(z) \leq \int\_{|\xi| \geq 4r} |K(\xi-(y-z))-K(\xi)|\, d\xi.$$ Since $|y-z| \leq 2r$, then
$$r^{-n}\int\_{B(0,r)} I\_1(z)\, dz \leq r^{-n} \int\_{B(0,2r)}|K(\xi-w)-K(\xi)|\, dw
\leq 2^n K.$$ The estimate of $I$ in terms of $K$ now follows by averaging the inequality $I \leq I\_1(z)+I\_2(z)$ over $B(0,r)$.
| 3 | https://mathoverflow.net/users/150653 | 432260 | 174,931 |
https://mathoverflow.net/questions/431969 | 3 | Suppose $\{a\_n(t)\}\_{n \geq 0}$ is a collection of differentiable (or simply smooth) functions such that i) $0 \leq a\_n(0) \leq 1$ for all $n\in \mathbb N$ (ii) $a\_n(t) \approx 1 - \mu2^{-n}$ uniformly in $t$ for $n \gg 1$ (iii) $a'\_n = a^2\_{n+1} - a\_n.$ My goal is to show that $$a\_n(t) \xrightarrow{t \to \infty} \mathrm{e}^{-\mu 2^{-n}} \quad \text{for all $n \in \mathbb N$.}$$ Any help (or hints) will be greatly appreciated!
---
Remark: The condition (ii) is a bit unclear, I should emphasize here that $a\_n(t) < 1$ for all (fixed) $n \in \mathbb N$ and for all $t \geq 0$, this is the reason that I did not write condition (ii) as
condition (ii') $a\_n(t) \approx 1$ uniformly in $t$ for $n \gg 1$.
---
Remark: As discussed in the comments, I need to impose an initial datum ${a\_n(0)}\_{n \geq 0}$ such that $\sup\_{n \in \mathbb N} a\_n(0) = 1$ but there does not exists an $N \in \mathbb N$ for which $a\_N(0) = 1$. Also, one can savely assume that $a\_n(0) < a^2\_{n+1}(0)$ for all $n \in \mathbb N$.
| https://mathoverflow.net/users/163454 | Showing convergence of an infinite ODE system | Theorem 2 below offers a sufficient condition for convergence to the nontrivial equilibrium that you are referring.
The results are contingent on the uniqueness of solutions to this infinite-dimensional ODE -- I will include the proofs when possible, as needed, but, and unless I am missing something, they are quite simple.
**Question.** Let $\left(a\_n(t)\right)\_{n\in\mathbb{N}}$ be the solution to the infinite-dimensional ODE
$$a\_n'(t)=a\_{n+1}^2(t)-a\_n(t) \,\,\,\,(\star)$$
for $n\in\mathbb{N}$, with initial condition $\left(a\_n(0)\right)\_{n\in\mathbb{N}}$. For which initial conditions, the solution $\left(a\_n(t)\right)\_{n\in\mathbb{N}}$ converges to an equilibrium distinct from the origin and the all 1's equilibrium?
**Remark 1 [Domain]**. The cube $\left[0,1\right]^{\mathbb{N}}$ is invariant to this dynamics. That is, if we initialize the system within this infinite-dimensional cube, then the solutions will remain there. You cannot impose condition i) as in your question. But, if you assume $a\_n(0)\in\left[0,1\right]$ for all $n$, then $a\_n(t)\in\left[0,1\right]$ for all $n,t$.
**Remark 2 [Equilibria parametrization].** The equilibria is given by $\mathcal{E}=\left\{\left(a\_{n}\right)\_{n\in\mathbb{N}}\,:\,a\_{n+1}^2=a\_n\mbox{ for all }n\right\}$ which can be rewritten as $\mathcal{E}=\left\{\left(b^{2^{-(n-1)}}\right)\_{n\in\mathbb{N}}\,:\,\mbox{ for all }b\in\left[0,1\right]\right\}$. Using your parametrization, we can further write it as $\mathcal{E}=\left\{\left(e^{-\mu 2^{-(n-1)}}\right)\_{n\in\mathbb{N}}\,:\,\mbox{ for all }\mu\geq0\right\}\cup \left\{\mathbf{0}\right\}$. In other words, given any $\mu\geq 0$, then the sequence $\left(e^{-\mu 2^{-(n-1)}}\right)\_{n\in\mathbb{N}}$ is an equilibrium for $(\star)$. That is, this provides a one-dimensional parametrization for the equilibria of the system. In particular, let us define for simplicity
$${\sf eq}\_{\mu}\overset{\Delta}=\left(e^{-\mu 2^{-(n-1)}}\right)\_{n\in\mathbb{N}}$$
the equilibrium associated with the parameter $\mu\geq 0$.
**Remark 3 [Causal structure].** Observe that the evolution of $\left(a\_m(t)\right)\_{t\geq 0}$, for $m\geq N$, does not depend on the evolution of $\left(a\_n(t)\right)\_{t\geq 0}$ for any $n<N$. The state variable $a\_{n+1}$ impacts $a\_n$ but not the other way around. This implies the following: the *tail* of the initial condition is what determines the asymptotic behavior of this dynamical system. In other words, for any $N$, the sub-sequence of the initial condition $\left(a\_n(0)\right)\_{n\leq N}$ is irrelevant for the asymptotic behavior of the system.
The next result reveals an important property of this ODE.
**Lemma 1[Monotonicity].** If $a\_n(0)\leq \overline{a}\_n(0)$ for all $n\in\mathbb{N}$, then $a\_n(t)\leq \overline{a}\_n(t)$ for all $n\in\mathbb{N}$ and $t\geq 0$.
We have an immediate corollary to Lemma 1.
**Corollary 1 [Invariant sub-regions].** Let $\mu\_1<\mu\_2$. If there exists $N\in\mathbb{N}$ so that $e^{-\mu\_2 2^{-(n-1)}}\leq a\_{n}(0)\leq e^{-\mu\_1 2^{-(n-1)}}$ for all $n\geq N$, then $e^{-\mu\_2 2^{-(n-1)}}\leq \lim\inf\_{t\rightarrow \infty}a\_{n}(t)\leq \lim\sup\_{t\rightarrow \infty}a\_{n}(t)\leq e^{-\mu\_1 2^{-(n-1)}}$ for all $n\in\mathbb{N}$.
As a consequence to Corollary 1, if the initial condition $\left(a\_n(0)\right)\_{n\in\mathbb{N}}$ is bounded as $e^{-\mu\_2 2^{-(n-1)}}\leq a\_{n}(0)\leq e^{-\mu\_1 2^{-(n-1)}}$ eventually, i.e., for $n\geq N$ for some $N$, then the dynamical system $\left(a\_n(t)\right)\_{n\in\mathbb{N}}$ cannot converge to ${\sf eq}\_{\mu}$ for any $\mu\notin \left[\mu\_1,\mu\_2\right]$.
**Theorem 1 [Monotonicity 2].** If $a\_n(0)<a\_{n+1}(0)^2$ for all $n$, then $a\_n(t)$ is increasing in $t$ for all $n$, i.e., $a\_{n}(t)>a\_{n}(t')$ for any $t>t'$ and for any $n$.
As a corollary to Theorem 1, and to the fact that $a\_n(t)\in\left[0,1\right]$ for all $t$, we have convergence to an equilibrium whenever the initial condition is given by $a\_n(0)<a\_{n+1}(0)^2$ for all $n$. In particular, via combining Corollary 1 and Theorem 1, we have the following sufficient condition for convergence to a nontrivial equilibrium.
**Theorem 2 [Sufficient condition].** If $a\_n(0)<a\_{n+1}(0)^2$ for all $n$, and further $e^{-\mu\_2 2^{-(n-1)}}<a\_n(0)<e^{-\mu\_1 2^{-(n-1)}}$ eventually for some $\mu\_1,\mu\_2\in\left(0,\infty\right)$, then there exists $\mu\in\left[\mu\_1,\mu\_2\right]$ so that $a\_n(t)\overset{t\rightarrow \infty}\longrightarrow e^{-\mu2^{-(n-1)}}$ for all $n$.
--------------------------- **Other results** --------------------------------------
**Lemma 2[Invariance of the diagonal].** If $a\_n(0)=b\in \left[0,1\right]$ for all $n$, then $a\_n(t)=a\_m(t)$ for all $n,m$. In particular, $a\_n'(t)=a\_n(t)^2-a\_n(t)$ for all $n$.
The proof is trivial, but relies on the uniqueness of solutions to the ODE $(\star)$.
**Theorem 3 [Origin's attraction].** If $\sup\_{n\in\mathbb{N}} a\_n(0)<1$, then, $a\_n(t)\overset{t\rightarrow \infty}\longrightarrow 0$ for all $n\in\mathbb{N}$.
>
> *Proof.* Consider the sequence $\left(\overline{a}\_n(0)\right)\_n$ so that $\overline{a}\_n(0)=\sup\_{n\in\mathbb{N}}a\_n(0)=:c\in\left(0,1\right)$ for all $n$. Then, from Lemma 1, we have that $\overline{a}\_n(t)\geq a\_n(t)$ for all $n$ and $t\geq 0$. From Lemma 2, we have that $\overline{a}'\_n(t)=\overline{a}\_n(t)^2-\overline{a}\_n(t)$. Therefore, $\overline{a}\_n(t)\overset{t\rightarrow \infty}\longrightarrow 0$ for all $n$. Thus, $a\_n(t)\overset{t\rightarrow \infty}\longrightarrow 0$ from the boundedness and since $\left[0,1\right]^{\mathbb{N}}$ is invariant to the infinite-dimensional dynamical system $(\star)$.
>
>
>
**Theorem 4 [1's attraction].** If $a\_n(0)=1$, for all $n>N$ for some $N\in\mathbb{N}$ large enough, then $a\_n(t)\overset{t\rightarrow \infty}\longrightarrow 1$ for all $n\in\mathbb{N}$.
>
> *Proof.* It is trivial to check that $a\_n(t)=1$ for all $t\geq 0$ for all $n>N$. we have $a\_n'(t)=1-a\_n(t)$ for $n=N$ and hence $a\_n(t)\overset{t\rightarrow \infty}\longrightarrow 1$. From here, it is trivial to check that $a\_n(t)\longrightarrow 1$ for all $n$.
>
>
>
| 3 | https://mathoverflow.net/users/138242 | 432268 | 174,932 |
https://mathoverflow.net/questions/432262 | 4 | Let $x\_0=1$ and
$$x\_{k+1} = (1-a\_k)\left(\frac{3}{2}-\frac{1}{2}\frac{1}{x\_k}\right)$$
where $a\_n$ is a known sequence satisfying that $a\_k\in(0,1)$ for all $k$ and $a\_k\to 0$ as $k\to\infty$. How to prove that $x\_k\to 1$ as $k\to\infty$?
---
The difficulty here is that
1. It is not known how fast $a\_k$ converges to zero, and I don't know how it affect the convergence of $x\_k$;
2. $x\_k$ may change sign and is not monotone, so I don't know how to prove $x\_k$ even converges;
3. Furthermore, if we assume $x\_k$ do converge to some limit $x^\*$, then by taking the limit,
$$x^\*=(1-0)\left(\frac{3}{2}-\frac{1}{2}\frac{1}{x^\*}\right)$$
I find there are two possible solution $x^\*=1/2$ or $x^\*=1$. How to exclude the case that $x^\*=1/2$?
| https://mathoverflow.net/users/490600 | Convergence of a sequence | If, as you say, $a\_k<0.1$ for all $k$, then we can prove by induction that $x\_k>\frac{3}{4}$ for all $k$, with induction step $x\_{k+1}>0.9\left(\frac{3}{2}-\frac{1}{2\cdot\frac{3}{4}}\right)=\frac{3}{4}$. By a similar induction we get $x\_k\in\big[\frac{3}{4},1\big]\;\forall k$.
So $1-x\_k\in\big[0,\frac{1}{4}\big]$ for all $k$. Now note that
$$
\begin{split}
1-x\_{k+1} &= 1-(1-a\_k)\left(\frac{3}{2}-\frac{1}{2x\_k}\right)\\
&=-\frac{1}{2}+\frac{1}{2x\_k}+a\_k\left(\frac{3}{2}-\frac{1}{2}x\_k\right)\\
&\leq\frac{1-x\_k}{2x\_k}+3a\_k\leq\frac{2}{3}(1-x\_k)+3a\_k.
\end{split}
$$
So as $a\_k\to0$, we also have $1-x\_k\to 0$.
| 7 | https://mathoverflow.net/users/172802 | 432270 | 174,933 |
https://mathoverflow.net/questions/432250 | 1 | Fix a Gaussian random matrix $A$ with $E[A\_{ij}]=0$ for $i, j=1,\dots n$ and $E[A\_{ij}^2]=\frac{1}{n}$. Let $v\_1$ be the leading eigenvector of $A$. What is the non-asymptotic upper bound for $v\_1$, that is something like
$$
P(v\_1\cdot u\ge t)\le e^{-\alpha t}
$$
where $u$ is distributed uniformly on the unit sphere.
Is there any reference for this tail probability? Thank you!
---
Let $\{v\_1,v\_2,\dots, v\_n\}$ be the eigenvectors corresponding to the eigenvalues $\lambda\_1,\dots, \lambda\_n$ of a matrix $A$ from GOE. Each of the eigenvectors $v\_1,\dots, v\_n$ is distributed uniformly on
\begin{equation}
S\_+^{n-1}:=\{x=(x\_1,\dots, x\_n): x\_i\in R, \|x\|\_2=1, x\_1>0\}.
\end{equation}
| https://mathoverflow.net/users/168083 | What is the non-asymptotic upper bound for the leading eigenvector of the random matrix? | We know the distribution of $x=v\_1\cdot u$, with $v\_1$ of length $n$ uniformly distributed on the unit $n$-sphere and $u$ an arbitrary unit vector. This distribution is given by
$$P(x)=\frac{\Gamma(n/2)}{\sqrt{\pi}\,\Gamma(n/2-1/2)}(1-x^2)^{n/2-3/2}\,\theta(1-x^2),\;\;n>1,$$
with $\theta(x)$ the unit step function. (Here is one [derivation.](https://mathoverflow.net/a/359646/11260)) So for $n\gg 1$ this becomes a Gaussian, $P(x)\propto e^{-nx^2/2}$, with mean zero and variance $1/n$.
Plot of $P(x)$ for $n=2,3,5,10,20$ (larger $n$ gives more peaked distribution).
| 1 | https://mathoverflow.net/users/11260 | 432277 | 174,936 |
https://mathoverflow.net/questions/432258 | 2 | For any sequence $\omega\in[1, \infty)^{\mathbb{N}}$, define the weighted $\ell^p\_\omega$-norm of the sequence $v$ by
$$\Vert v\Vert\_{\ell^p\_\omega} := \left(\sum\_{k=1}^\infty \omega\_k^p |v|\_k^p\right)^{1/p} .$$
**My question is:** If $v\in\ell^q$ and $p>q$, under which conditions on $\omega$ will $v\in\ell^p\_\omega$?
---
Obviously, since $p>q$, we can use Hölder‘s inequality to bound
$$
\Vert v\Vert\_{\ell^p\_\omega}^p
= \Vert \omega^p v^{p-q}v^q\Vert\_{\ell^1}
\le \Vert \omega^p v^{p-q}\Vert\_{\ell^\infty}\Vert v^q\Vert\_{\ell^1}
= \Vert \omega^{p/(p-q)} v\Vert\_{\ell^\infty}^{p-q} \Vert v\Vert\_{\ell^q}^q .
$$
This gives the condition $\Vert\omega^{p/(p-q)}v\Vert\_{\ell^\infty} <\infty$.
(Here, all operations on the sequences have to be understood element-wise.)
* But is this condition necessary?
* Moreover, can we find an unbounded increasing sequence $\omega$ (preferably algebraically increasing like $\omega\_k=k^s$ for $s=\frac1q-\frac1p$) such that $\ell^q\subseteq\ell^p\_\omega$?
| https://mathoverflow.net/users/75500 | Can a weighted $\ell^p$ norm be bounded by an unweighted $\ell^q$ norm? | The answer to both questions is no.
* Taking $q=1$, $p=2$, $v\_k = k^{-(1+\varepsilon)}$ and $\omega\_k = k^{(1+3/2\varepsilon)/2}$ for some $\varepsilon>0$ provides a counter-example to the first assertion.
* [This answer](https://math.stackexchange.com/a/2296826/303111) provides a counter-example for the second assertion when $\omega$ increases algebraically.
It can be generalized to any explicitly given sequence $\omega$.
| 2 | https://mathoverflow.net/users/75500 | 432289 | 174,940 |
https://mathoverflow.net/questions/432201 | 5 | Here is what I know. Assume $M\cong G/K$ is an irreducible hermitian symmetric space. Denote the Lie-algebra of $K$ by $\mathfrak{t}$. Proposition 1.2. chapter 3 in
[Wienhard - Bounded cohomology and geometry](https://arxiv.org/abs/math/0501258)
says that for a symmetric space, being hermitian is equivalent to the existence of a non-zero element $Z$ in the center of $\mathfrak{t}$. In particular $K\subset \operatorname{Stab}\_G(Z)$ is a subset of the stabilizer of $Z$ under the coadjoint action. If one could show that $K=\operatorname{Stab}\_G(Z)$, then $M=G/K$ can be identified with the coadjoint orbit $\mathcal{O}\_Z\cong G/\operatorname{Stab}\_G(Z)$.
| https://mathoverflow.net/users/492740 | Can all hermitian symmetric spaces be realised as coadjoint orbits? | This is true. One can use a few facts from Helgason's *Differential Geometry, Lie Groups, and Symmetric Spaces* to show that, indeed, $K = \mathrm{Stab}\_G(Z)$.
Since $M=G/K$ is an irreducible Hermitian symmetric space, one can apply Theorem 6.1 from Chapter VIII of Helgason. The main point is that one can regard $G$ as a subgroup of $\mathrm{Aut}(\frak{g})$ with $K\subset G$ connected, and the circle $\mathrm{exp}(\mathbb{R}\cdot Z)\simeq S^1$ is the center of $K$. (See Helgason's Proposition 6.2.) Moreover, as Helgason explains, there will be a real number $t$ such that the element $z = \exp(tZ)$ is an element of order $2$ with the property that the $G$-centralizer of $z\in K$ is equal to $K$. Since the $G$-stabilizer of $Z$ is clearly contained in the $G$-centralizer of $z$, it follows that the $G$-stabilizer of $Z$ must equal $K$ as well.
| 8 | https://mathoverflow.net/users/13972 | 432297 | 174,943 |
https://mathoverflow.net/questions/432263 | 3 | We begin by considering the usual general first order linear equation of the form
$$ a\_0 y' + a\_1 y + a\_2 = 0 $$
Where $a\_i,y \in \mathbb{C} \rightarrow \mathbb{C}$. Now it's well known from everyone's favorite undergrad ODE class that this has a general solution in terms of integration factors.
$$ y = - e^{-\int \frac{a\_1}{a\_0}} \int \left[ e^{\int \frac{a\_1}{a\_0}} \frac{a\_2}{a\_0} \right] dx$$
Now theres at least two integrals going on here and for almost all choices of $a\_0, a\_1, a\_2$ these integrals CANNOT be expressed in terms of elementary functions. BUT we can declare that this ENTIRE abstract solution is an "elementary" general solution, i.e. it involves only finitely many elementary functions, and "elementary" operators which for now just includes integrals, derivatives, and function inverses.
So the question is kind of natural, what if we look at second order equations?
$$ a\_0 y'' + a\_1 y' + a\_2 y + a\_3 = 0 $$
I expect that this has NO such elementary general solution (since we would've found it by now) BUT as far as I recall I have never seen an explicit proof of such.
So to be clear I want to know, at the OPERATOR-level why is there no elementary general solution to that equation? And more generally, what machinery/theory is required to settle such problems?
Some hints of what this theory could encompass
We have the trick of matrix exponentials i.e. if the matrix
$$ \begin{bmatrix} 0 & -1 \\ \frac{a\_1}{a\_0} & \frac{a\_2}{a\_0} \end{bmatrix} $$
commutes with its own derivative then the matrix exponential trick must work and so this equation is "solvable". So our gadget for gauging solvability must somehow be connected to these matrices (but it's not even clear how to define that object for non linear equations).
| https://mathoverflow.net/users/46536 | Is there a theory of "elementary closed form solution" at the operator level for differential equations? | There are several theories which deal with this question.
One is the differential algebra, see, for example the little book
I. Kaplansky, Introduction to differential algebra,
Publ. de l'Institut de Mathématique de l'Université de Nancago, No. V. Actualités Scientifiques et Industrielles No. 1251. Hermann, Paris, 1976.
Another is a generalization of Galois theory which is due to Picard and Vessiot:
MR1199112
Kuga, Michio
Galois' dream: group theory and differential equations.
Birkhäuser Boston, Inc., Boston, MA, 1993,
and
MR3289210 Khovanskii, Askold Topological Galois theory. Solvability and unsolvability of equations in finite terms. Springer, Heidelberg, 2014.
| 5 | https://mathoverflow.net/users/25510 | 432300 | 174,946 |
https://mathoverflow.net/questions/428933 | 3 | Let $F$ be a number field, $G$ an $F$-simple affine algebraic group.
Then is the Weil restriction $\operatorname{Res}\_{F/\mathbb{Q}} G$ $\mathbb{Q}$-simple?
I couldn’t find any references.
| https://mathoverflow.net/users/159361 | The Weil restriction of a simple algebraic group | As noted, the answer is yes. Let $G$ be a simple algebraic group (meaning $F$-simple) over $F$.
Assume first that $F$ is Galois over $\mathbb{Q}$. Then $(\mathrm{Res}\_{F/\mathbb{Q}}G)\_{F}\simeq G\_{1}\times\cdots\times G\_{r}$, where the $G\_{i}$ are the conjugates of $G$ under $\mathrm{Gal}(F/\mathbb{Q})$. If $H$ is a smooth connected normal algebraic subgroup of $\mathrm{Res}\_{F/\mathbb{Q}}G$, then $H\_{F}$ is a product of a certain number of the $G\_{i}$ (Milne, Algebraic Groups, 21.51), which is stable under the Galois action. But $\mathrm{Gal}% (F/\mathbb{Q})$ acts transitively on the set of $G\_{i}$, and so $H=\mathrm{Res}\_{F/\mathbb{Q}}G$.
In the general case, the same argument works if $G$ is geometrically simple. If not, then we may suppose that $G=\mathrm{Res}\_{F^{\prime}/F}G^{\prime}$ with $G^{\prime}$ geometrically simple (ibid. 24.3) and apply the argument to $G^{\prime}$.
This argument fails for algebraic groups like $\mathbb{G}\_{a}^{n}$ because for them the decomposition into simple objects is not unique.
Note that the proof only uses standard properties of semisimple groups, for which I gave a modern reference. I don't understand the comment of LSpice.
| 3 | https://mathoverflow.net/users/492821 | 432303 | 174,947 |
https://mathoverflow.net/questions/432281 | 5 | Suppose $R$ is a semisimple ring and if $L$ is a minimal left ideal. Let $B$ be the direct sum of all minimal left ideals isomorphic to $L$ ($B$ is called a simple component corresponding to $L$). It is a standard result that $B$ is a two-sided ideal. The argument critically uses the fact that $R$ is semisimple.
I am searching for an example of a ring $R$ and a simple component that is not a two-sided ideal.
Since, we need a minimal left ideal, we may consider the ring to be Artinian as they have always have a minimal left ideal.
I first tried considering the endomorphism ring of a vector space with a countable basis, but later I learned that it is neither semisimple nor Artinian.
I figured out that we cannot consider a simple Artinian ring, for it is semisimple. Also, we cannot consider any Artinian ring $R$ whose Jacobson radical $J(R)=0$ for the same reason.
Does anyone know any example?
| https://mathoverflow.net/users/165646 | Simple component that is not a two-sided ideal | I'm not sure why you believe it depends on $R$ being semisimple. The usual argument goes like this and makes no reference to semisimplicity:
If $r\in R$ and $L'\cong L$ where $L'$ is a left ideal of $R$, then by right multiplication, $r$ defines a homomorphism of left $R$ modules $L'\to R$. Because $L'$ is simple, either $L'r=\{0\}$ or $L'r\cong L'$. Either way, this shows right multiplication sends simple left ideals isomorphic to $L$ into the simple component for $L$, i.e. the component is an ideal of $R$.
Now, one thing that might not happen in general is for the component to be a summand. For example, in the full linear ring of transformations of a countable dimensional vector space, there is certainly a nonzero component there (in fact it is the only nontrivial ideal), but the ring is also prime, and so it cannot be decomposed into pieces.
Certainly the name "simple component" is meant to reflect the fact that in semisimple Artinian rings, the simple components are actually simple rings in their own right, and their product is the semisimple Artinian ring. But the existence of the ideal itself is not confined to semisimple rings.
| 12 | https://mathoverflow.net/users/19965 | 432304 | 174,948 |
https://mathoverflow.net/questions/432291 | 5 | This question is related to [Homotopy type theory : how to disprove that $0=\mathrm{succ}(0)$
in the type $\mathbb N$](https://mathoverflow.net/questions/421450/homotopy-type-theory-how-to-disprove-that-0-operatornamesucc0-in-the-ty).
Section 2.13 in [The HoTT Book](https://homotopytypetheory.org/book/) uses "the encode-decode method to characterize the path space of the natural numbers" by defining a type family:
$$\mathrm{code} : \mathbb N \to \mathbb N \to \cal U$$
with
$$\begin{array}{rcl}
\mathrm{code}(0,0) & :\equiv & \mathbf 1\newline
\mathrm{code}(\mathrm{succ} (m),0) & :\equiv & \mathbf 0\newline
\dots & :\equiv & \dots\newline
\dots & :\equiv & \dots
\end{array}$$
To my understanding, $\mathrm{code}$ is only well-defined, if (in particular) $0:\mathbb N$ and $\mathrm{succ}(0):\mathbb N$ are **not** judgementally equal. How can we be sure that this is the case?
| https://mathoverflow.net/users/492810 | Homotopy type theory: why are $0:\mathbb N$ and $\mathrm{succ}(0):\mathbb N$ not judgementally equal? | Daniel's answer is correct that the judgmental distinctness of $0$ and $\mathsf{succ}(m)$ is not what justifies a definition by pattern-matching. However, it is still a meaningful question of how to prove that $0$ and $\mathsf{succ}(m)$ are not judgmentally equal. (Stretching terminology a bit, Daniel answered your [X](https://xyproblem.info/) and I'm going to answer your Y.)
Note that this is a metatheoretic question, in contrast to the internal proof that $0\neq \mathsf{succ}(m)$. Thus, even though judgmental equality implies typal equality, the answer to that question does not answer this one.
It's also worth noting that this is not a trivial question, even though $0$ and $\mathsf{succ}$ are distinct constructors. From an "algebraic" perspective on type theory, judgmental equality is just the smallest congruence generated by certain rules. Thus, two terms that don't "look" like they could possibly be equal might turn out to be equal by passing through some chain of forwards and backwards equalities to other terms.
For the same reason, it is a "global" question about the type theory, not one that can be answered by referring only to $\mathbb{N}$. For instance, in a higher inductive type such as the interval $\mathsf{I}$, with constructors $\mathsf{zero},\mathsf{one} : \mathsf{I}$ and $\mathsf{seg}:\mathsf{zero}=\mathsf{one}$, there is no "local" reason for $\mathsf{zero}$ and $\mathsf{one}$ to be judgmentally equal; but if the type theory also includes the equality reflection rule, then $\mathsf{seg}$ implies that $\mathsf{zero}\equiv\mathsf{one}$.
One answer to this question involves giving an algorithm for checking whether two terms are judgmentally equal. Probably the simplest such algorithm involves giving a "rewriting system" under which every term can be reduced to a "normal form", and then two terms are judgmentally equal if they reduce to the same normal form (up to $\alpha$-equivalence). Roughly speaking, this rewriting system consists of what the HoTT Book calls "computation rules". One then has to prove that this rewriting system is terminating, and that the relation of "reducing to the same normal form" is a congruence; thus it coincides with judgmental equality. Then one can simply observe that $0$ and $\mathsf{succ}(m)$ are (when $m$ is a variable) distinct normal forms.
In practice, nowadays more complicated algorithms are used. Among other reasons, this is in order to also include what the HoTT Book calls "uniqueness rules". These algorithms often go by names like "normalization by evaluation". The crucial idea is that there are two "phases" of the algorithm, one which applies the computational rewrites, and another which applies the uniqueness rules in the *expansionary* direction ($f \mapsto (\lambda x. f(x))$ or $u \mapsto (\pi\_1(u),\pi\_2(u))$) by inspecting their types. But the basic idea is the same: after proving that the algorithm is terminating and complete, we can run the algorithm on two terms to verify that they are unequal (or equal).
With that said, I don't know whether anyone has *actually* done this for Book HoTT. It's known for MLTT, on which Book HoTT is based, but Book HoTT adds not just axioms (which don't disrupt such an algorithm) to MLTT but new judgmental equalities (the computation rules for point-constructors of HITs), which then have to be incorporated in the algorithm. So (even laying aside the point that "Book HoTT" is not precisely specified because it is open-ended with respect to what HITs are definable) there may technically be an open question here. However, I think no type theorist who's familiar with such proofs would have much doubt that standard techniques ought to apply to Book HoTT (and it's possible that someone has already done it).
**Edit:** Simon has pointed out in the comments what I should clearly have realized myself, namely that another way to deduce that two terms are not judgmentally equal is to give a model in which they are interpreted by distinct things. Thus, for instance, the set-theoretic model proves this for MLTT, and the simplicial set model proves it for Book HoTT (modulo details in the construction of that model that aren't yet in the literature, such as showing that the univalent universes are closed under a sufficiently large class of higher inductive types).
| 10 | https://mathoverflow.net/users/49 | 432308 | 174,949 |
https://mathoverflow.net/questions/380787 | 0 | It is well known that if a family of meromorphic functions is not normal (a family is said to be normal if each sequence of functions in the family has a subsequence which converges locally uniformly to a limit function which is either meromorphic or identically $\infty$) on some domain, then the corresponding family of derivatives may or may not be normal on that domain.
For example, $\mathcal{F}:=\{f\_n= nz, n\in\mathbb{N}\}$ is not normal on $|z|<1.$ However, the corresponding family of derivatives $\mathcal{F'}=\{n\}$ is normal on $|z|<1.$
Furthermore, the family $\mathcal{G}:=\{nz^2\}$ and its derivative $\mathcal{G'}=\{2nz\}$ are not normal on $|z|<1.$
Observe that the family $\mathcal{G}$ has a zero of order $2$ at $z=0$ on $|z|<1$ and its corresponding family of derivatives is not normal.
With the above observation in mind, I am curious to know the following:
Does there exist a family of meromorphic functions whose each zero is of multiplicity $2$ and which is not normal on $|z|<1,$ but the corresponding family of derivatives is normal?
Any help shall be largely appreciated.
| https://mathoverflow.net/users/143655 | Is there any non-normal family $\mathcal{F}$ of meromorphic functions on $|z|<1$ whose each zero has multiplicity $2$ but $\mathcal{F'}$ is normal | The answer to this question is negative.
This follows easily from the following result of Chen and Lappan (Adv. Math.,vol. 24, 1996, 517-524):
Let $\mathcal{F}$ be a family of meromorphic functions in a domain $D$ such that each $f\in\mathcal{F}$ has zeros of multiplicity at least $k+1,$ where $k$ is a positive integer. If the family $\mathcal{G}=\left\{f^{(k)}:f\in\mathcal{F}\right\}$ is normal in $D,$ then $\mathcal{F}$ is also normal in $D.$
| 1 | https://mathoverflow.net/users/143655 | 432319 | 174,952 |
https://mathoverflow.net/questions/432317 | 8 | Let $K$ be a number field. $O\_K$ be its ring of integers, so $O\_K^\*$ are the units.
We have sequence
$1 \rightarrow O\_K^\* \rightarrow K^\* \rightarrow K^\*/O\_K^\* \rightarrow 1$
Note that $K^\*/O\_K^\*$ is essentially the group of principal fractional ideals.
Does this sequence split for all number fields? It seems true for $K = Q$ and
$Q[i]$. But I doubt that it is always true for every number field.
| https://mathoverflow.net/users/3208 | Does this exact sequence split? | The exact sequence in the original post splits for every number field $K$. To see this, let $P\_K$ be the multiplicative group of nonzero principal fractional ideals of $K$, and let $I\_K$ be the multiplicative group of all nonzero fractional ideals of $K$. Clearly, $P\_K$ is a subgroup of $I\_K$, and it is isomorphic to $K^\*/O\_K^\*$. As $I\_K$ is free abelian (the nonzero prime ideals of $O\_K$ form a free generating set), $P\_K$ is also free abelian (see [here](https://en.wikipedia.org/w/index.php?title=Free_abelian_group&oldid=317358888#Subgroup_closure)). Let $X$ be a set of free generators of $P\_K$, and write each $x\in X$ as $(f(x))$ with a suitable $f(x)\in K^\*$. The map $f:X\to K^\*$ extends uniquely to a homomorphism $P\_K\to K^\*$, and it splits the exact sequence in the original post.
| 14 | https://mathoverflow.net/users/11919 | 432321 | 174,953 |
https://mathoverflow.net/questions/432294 | 4 | Let $\alpha(x) : \mathbb{R} \to (0,\infty)$ have bounded variation (BV) and suppose $\inf\_{\mathbb{R}} \alpha > 0$. Consider the second order differential operator
$$H : =-\partial\_x (\alpha(x) \partial\_x) : L^2(\mathbb{R}) \to L^2(\mathbb{R}).$$
It's not too hard to show that
$$ \{u \in L^2(\mathbb{R}) : u, u' \in L^2(\mathbb{R}) \cap L^\infty(\mathbb{R}) \text{ and } Hu \in L^2(\mathbb{R}) \} \\
= \{u \in L^2(\mathbb{R}) : u, \, \alpha u' \text{ are locally absolutely continuous and } Hu \in L^2(\mathbb{R})\},
$$
and that $H$ is self-adjoint with respect to this domain $\mathcal{D}(H)$, which is a dense subset of $L^2(\mathbb{R})$.
>
> I would like to know whether the Sobolev space $H^1(\mathbb{R})$ is contained in the domain of the square root operator $H^{1/2}$ (defined via the spectral theorem).
>
>
>
Recall the well-known fact that $\mathcal{D}(H^{1/2})$ is equal to the form domain associated to $H$ (i.e., the completion of $\mathcal{D}(H)$ with respect to the norm $\| u\|^2\_{+1} := \langle Hu,u \rangle\_{L^2} + \langle u,u \rangle\_{L^2} \approx \|u\|^2\_{H^1}$.)
If $\alpha$ has some more regularity, for instance if $\alpha$ is $W^{1, \infty}$ instead of merely having BV, then it is immediate that $C\_0^\infty \subseteq \mathcal{D}(H) \subseteq \mathcal{D}(H^{1/2})$. So in that case, for any $u \in H^1(\mathbb{R})$, if we approximate $u$ in $H^1(\mathbb{R})$ by a sequence $u\_j \in C^\infty\_0(\mathbb{R})$, these $u\_j$ are also Cauchy with respect to $\| \cdot \|\_{+1}$, and hence $u \in \mathcal{D}(H^{1/2})$.
We don't necessarily have the inclusion $C\_0^\infty \subseteq \mathcal{D}(H)$ in the BV case. But can we come up with a different subspace of functions contained in $\mathcal{D}(H)$, that has a sequence converging to $u$ in $L^2(\mathbb{R})$ and Cauchy with respect to $\| \cdot \|\_{+1}$? This is where I am stuck.
| https://mathoverflow.net/users/87862 | Is the Sobolev space $H^1(\mathbb{R})$ contained in the domain of $(-\partial_x \alpha(x) \partial_x)^{1/2}$? | We can actually do this directly and my comment above is not that relevant. Let $u\in H^1$ and also assume that $u$ is compactly supported, so $\int u'=0$. Approximate $\alpha u'$ in $L^2$ by $v\_n\in C\_0^{\infty}$. Here we can also insist that $\int v\_n/\alpha=0$. Then also $v\_n/\alpha\to u'$ in $L^2$, so $u\_n(x)=\int\_{-\infty}^x v\_n/\alpha\, dt\to u$ locally uniformly and thus $u\_n\to u$ in $H^1$.
Clearly, $u\_n\in D(H)\subseteq D(H^{1/2})$, and as you already explained yourself, this implies that $H^1\subseteq D(H^{1/2})$.
| 1 | https://mathoverflow.net/users/48839 | 432325 | 174,955 |
https://mathoverflow.net/questions/432312 | 6 | By an (intuitionistic) **propositional formula** $\varphi(x\_1,\ldots,x\_n)$ I mean a formula built up from a (finite) number of variables $x\_1,\ldots,x\_n$ using connectors $\top, \bot, \land, \lor, \Rightarrow$. Given such a formula $\varphi(x\_1,\ldots,x\_n)$, given a Heyting algebra $H$ and elements $u\_1,\ldots,u\_n \in H$ we can evaluate $\varphi(u\_1,\ldots,u\_n)$ in the obvious way (resulting in an element of $H$).
Given a propositional formula $\varphi(t,x\_1,\ldots,x\_n)$, I would like to know if $\bigwedge\_t \varphi(t,x\_1,\ldots,x\_n)$ and $\bigvee\_t \varphi(t,x\_1,\ldots,x\_n)$, which are evaluated in a complete Heyting algebra $H$ as the inf, resp. sup, of all values of $\varphi(t,x\_1,\ldots,x\_n)$ where $t$ ranges over $H$, i.e., by quantifying over truth values, can be rewritten (by eliminating the $\bigwedge$ or $\bigvee$ quantified variable $t$) as a propositional formula in $x\_1,\ldots,x\_n$ and possibly other variables $p\_1,\ldots,p\_m$ depending on $H$ (but on nothing else). More precisely:
**Question:** is it true that for any propositional formula $\varphi(t,x\_1,\ldots,x\_n)$ there exist propositional formulae $\varphi^\wedge(x\_1,\ldots,x\_n,p\_1,\ldots,p\_m)$ and $\varphi^\vee(x\_1,\ldots,x\_n,p\_1,\ldots,p\_m)$ (for some $m$) such that, for any complete Heyting algebra $H$ there exist $p\_1^H,\ldots,p\_m^H \in H$ such that, for all $u\_1,\ldots,u\_n \in H$, the following hold?
* $\bigwedge\_{v\in H} \varphi(v,u\_1,\ldots,u\_n) = \varphi^\wedge(u\_1,\ldots,u\_n,p\_1^H,\ldots,p\_m^H)$
* $\bigvee\_{v\in H} \varphi(v,u\_1,\ldots,u\_n) = \varphi^\vee(u\_1,\ldots,u\_n,p\_1^H,\ldots,p\_m^H)$
**Comments:**
* Unless I am mistaken, the analogous question for Boolean algebras is easily seen to have a positive answer (rewrite $\varphi(t,\underline{x})$ as $(a(\underline{x})\land t) \oplus b(\underline{x})$ for propositional formulas $a(\underline{x}), b(\underline{x})$, and then $\bigwedge\_t\varphi(t,\underline{x})$ is $b(\underline{x}) \land \neg a(\underline{x})$ while $\bigvee\_t\varphi(t,\underline{x})$ is $a(\underline{x}) \lor b(\underline{x})$).
* As an example, if $\varphi(t,x) := (t\Rightarrow x)$, then $\varphi^\wedge(x) = x$ and $\varphi^\vee(x) = \top$. On the other hand, if $\varphi(t,x) := (x\Rightarrow t)$, then $\varphi^\wedge(x) = \neg x$ and $\varphi^\vee(x) = \top$. (More generally, for any $\varphi$ that is order-preserving in $t$, we get $\varphi^\wedge$ and $\varphi^\vee$ by substituting $\bot$ and $\top$ respectively for $t$, and for any $\varphi$ that is order-reversing, $\top$ and $\bot$.)
* The possible need for extra parameters $p\_1,\ldots,p\_m$ is illustrated by taking $\varphi(t) := t\lor\neg t$, in which case $\bigwedge\_t \varphi(t)$ is the truth value of LEM in $H$, which certainly depends on $H$ (so we can't write it as a propositional formula of zero variable).
* If the answer to my question is negative, I would appreciate a pointer to literature, if there is any, on the class of formula obtained by closing the propositional variables by the propositional connectors and the quantifiers $\bigwedge,\bigvee$ ranging over propositional variables.
| https://mathoverflow.net/users/17064 | Variable elimination for propositional formulas in Heyting algebras | The answer for $\bigwedge\_t$ is no. Perhaps the idea here can be adapted to $\bigvee\_t$.
Consider the propositional formula $t \vee (t \to x)$ in the complete Heyting algebra of open subsets of $\mathbb{R}$.
*Claim.* For any open set $U \subseteq \mathbb{R}$, $\bigwedge\_t t \vee (t \to U)$ is the set $U^\bullet := \{r \in \mathbb{R} : r \in U\text{ or }r\text{ is isolated in }\mathbb{R} \setminus U\}$.
*Proof of claim.* Clearly if $r \in U$, then $r \in (t \vee (t \to U))$ for any open set $t$. If $r$ is isolated in $\mathbb{R}\setminus U$, then for any open set $t$, we either have that $r \in t$ or $(\mathbb{R} \setminus t) \cup U$ contains a neigbhorhood of $r$. Conversely, if $r\notin U$ but $r$ is also not isolated in $\mathbb{R} \setminus U$, then $r \notin ((\mathbb{R} \setminus \{r\}) \vee ((\mathbb{R} \setminus \{r\}) \to U))$.
So now consider some propositional formula $\varphi(x,\bar{p})$. We need to show that this fails to be equal to $\bigwedge\_{t} t \vee(t \to x)$ for all open sets $x$. Since there are only finitely many open sets in the tuple $\bar{p}$, we can find a non-empty open set $U \subseteq \mathbb{R}$ such that for each $i < |\bar{p}|$, either $U\wedge p\_i = U$ or $U \wedge p\_i = \bot$. For each $i < |\bar{p}|$, let $q\_i = \top$ if $U \wedge p\_i = U$ and let $q\_i = \bot$ if $U \wedge p\_i = \bot$. We now have that for any open $V$, $U \wedge \varphi(V,\bar{p}) = U \wedge \varphi(V,\bar{q})$.
Let $F$ be a closed subset of $U$ that is homeomorphic to Cantor space plus a single isolated point. Let $r$ be the single isolated point of $F$. Let $V = \mathbb{R} \setminus F$. Clearly we have that $V^\bullet = V \cup \{r\}$.
*Claim.* $U\wedge \varphi(V,\bar{q})$ is either $U$, $U \wedge V$, or $\bot$.
*Proof of claim.* We prove this by induction on propositional formulas in the single variable $x$ (i.e., we prove that for any propositional formula $\psi(x)$, $U \wedge \psi(V) \in \{U,U \wedge V,\bot\}$). Clearly we have that the statement is true for $U \wedge \bot = \bot$, $U \wedge \top = U$, and $U \wedge x$ (which is $U \wedge V$ when $x=V$).
If the statement is true for two propositional formulas $\psi(x)$ and $\chi(x)$, then it's easy to check that the statement is true for $\psi(x) \wedge \chi(x)$ and $\psi(x) \vee \chi(x)$. This just leaves $\psi(x) \to \chi(x)$. If $U \wedge \psi(V) = \bot$ or $U \wedge \chi(V) = U$, then $U \wedge (\psi(V) \to \chi(V)) = U \in \{U,U \wedge V,\bot\}$. If $U \wedge \psi(V) = U$, then $U \wedge (\psi(V) \to \chi(V)) = \chi(V) \in \{U,U \wedge V,\bot\}$. Finally, if $U \wedge \psi(V) = U \wedge V$ and $U \wedge \chi(V) $ is $U$ or $U \wedge V$, then $U \wedge ( \psi(V) \to \chi(V)) = U \wedge V \in \{U,U \wedge V,\bot\}$. So in every case, we have that $U \wedge (\psi(V) \to \chi(V)) \in \{U,U \wedge V, \bot\}$. Therefore, by induction, the same is true for $U \wedge \varphi(V,\bar{q})$.
Finally, note that $U \wedge V^\bullet \notin \{U,U\wedge V, \bot\}$, whence $\varphi(V,\bar{p}) \neq V^\bullet = \bigwedge\_t t \vee (t \to V)$.
Since we can do this for any formula $\varphi(x,\bar{p})$, we have that $\bigwedge\_t t \vee (t \to V)$ is not equal to any propositional formula with parameters.
| 6 | https://mathoverflow.net/users/83901 | 432328 | 174,957 |
https://mathoverflow.net/questions/421918 | 3 | Given a polytope $P$, what do the points of the secondary polytope correspond to?
I know that the vertices of the secondary polytope correspond to regular triangulations of $P$.
But what do the interior points of the secondary polytope correspond to?
| https://mathoverflow.net/users/5690 | Secondary polytope | As pointed out by Sam Hopkins in a comment above, secondary polytopes can be seen as a particular case of the fiber polytopes of Billera and Sturmfels (<https://doi.org/10.2307/2946575>).
This fiber polytope view provides the answer (or, at least, one answer) to what *each point* in the secondary polytope represents:
Consider a point configuration $A=\{a\_1,\dots,a\_n\}\subset \mathbb R^d$ (For the secondary polytope of a polytope $P$, take as $A$ the set of vertices of $P$, and substitute $P$ for $\operatorname{conv(A)}$ in all the description below). There is a canonical projection
\begin{align}
\pi: \Delta^{n-1} &\to \operatorname{conv(A)}\\
e\_i &\mapsto a\_i,
\end{align}
where $e\_1,\dots,e\_n$ are the standard basis in $\mathbb R^n$ and $\Delta^{n-1}$ is their convex hull, that is, the standard $n-1$-simplex.
Now, consider the space $\mathcal S$ of all *sections* of $\pi$. That is, elements of $\mathcal S$ are maps $s:\operatorname{conv(A)} \to \Delta^{n-1}$ such that $\phi\circ s$ is the identity map. Put differently, they are maps that choose a point in the fiber $\pi^{-1}(x)$ for each point $x\in \operatorname{conv(A)}$. (You can restrict to continuous sections, or to piecewise-linear sections, but you don't need to; what I say below holds for arbitrary sections too, as long as you can integrate them to compute their average).
Finally, consider the map that computes the average (or barycenter) of each such section. This is a map from $\mathcal S$ to the fiber $\pi^{-1}(b)$, where $b$ is the barycenter of $\operatorname{conv(A)}$. The secondary polytope of $\operatorname{conv(A)}$ is nothing but the image of this last map.
That is, each point in the secondary polytope corresponds, not uniquely, to a section of $\pi$; or, more precisely, it corresponds to the set of all sections with that average.
Vertices of the secondary polytope are special in that each of them is the image of a unique, and extremal, section, obtained by ``coherently'' picking the vertex of each fiber in a fixed direction. The section obtained in this coherent way is a union of faces of $\Delta^{n-1}$ forming (via $\pi$) the regular triangulation of $A$ corresponding to that vertex. This is why regular triangulations are also sometimes called *coherent*.
| 4 | https://mathoverflow.net/users/22608 | 432332 | 174,960 |
https://mathoverflow.net/questions/432333 | 3 | This question arose through reading "Interactions between homotopy theory and algebra" ([the first chapter by Goerss and Schemmerhorn](https://arxiv.org/abs/math/0609537)). In particular, I am struggling with the proof of Proposition 4.32, the cofiber sequence of cotangent complexes associated to two composable ring maps.
We fix a ring $R$ and take a map $A \to B$ in the category of $R$-algebras. We now take a cofibrant replacement $X \to A$ in the category of simplicial $R$-algebras and factorise $X \to B$ as $X \to Y \to B$, where $X \to Y$ is a cofibration and $Y \to B$ is an acyclic fibration.
There is a clear map $A \otimes \_X Y \to B$ by the universal property of the colimit and then they claim that this must be a weak equivalence. I cannot figure out why this is the case. Of course, it is enough to show that $Y \to A \otimes\_X Y$ is a weak equivalence, and this intuitively makes sense since we are pushing out along a weak equivalence, but I can't seem to produce a proof.
Would this be true in a general model category?
NOTE: wherever I wrote "cofibration", the original text says "free", and perhaps this is used, but I thought this might be true in greater generality.
| https://mathoverflow.net/users/170467 | Pushout along weak equivalence gives weakly equivalent object | This is true in general in any *left proper* model category. To be left proper means the pushout of a weak equivalence (for you, $X\to A$) along a cofibration (for you, $X\to Y$) is again a weak equivalence (for you, $Y \to A\otimes\_X Y$).
It is not true that any pushout of a weak equivalence is a weak equivalence (there are many easy counterexamples, e.g., in the category of topological spaces, because pushouts are not homotopy pushouts), but the point of left properness is that the pushout square formed by a span, where one leg is a cofibration and one leg is a weak equivalence, is a homotopy pushout square.
The other property Goerss and Schemmerhorn need in that proof you referenced, that $A\otimes\_X Y$ is cofibrant as an $A$-algebra, is true because the map $A\to A\otimes\_X Y$ is a cofibration (since it's a pushout of the cofibration $X\to Y$), and $A$ is the initial $A$-algebra.
| 7 | https://mathoverflow.net/users/11540 | 432334 | 174,961 |
https://mathoverflow.net/questions/385116 | 9 | Where can one find a Vinogradov-Korobov zero-free region for Dirichlet L-functions? It has to be in a standard reference, but I'm having a non-trivial time finding it.
| https://mathoverflow.net/users/398 | Vinogradov-Korobov for Dirichlet L-functions? | My paper <http://arxiv.org/abs/2210.06457> establishes several explicit Vinogradov--Korobov type zero-free regions for Dirichlet $L$-functions. In particular, Theorem 1.1 states the following:
Let $q \geq 3$, and let $\chi\pmod{q}$ be a Dirichlet character. The Dirichlet $L$-function $L(\sigma+it,\chi)$ does not vanish in the region
\begin{equation}\label{smallt}
\sigma \geq 1-\frac{1}{10.5 \log q+61.5(\log |t|)^{2 / 3}(\log \log |t|)^{1 / 3}}, \quad|t| \geq 10.
\end{equation}
Also, there exists an absolute and effectively computable constant $Y > 0$ such that $L(\sigma+it,\chi)$ does not vanish in the region
\begin{equation}\label{larget}
\sigma \geq 1- \frac{1}{ 10.1\log q + 49.13(\log |t|)^{2/3}(\log\log|t|)^{1/3}},\qquad |t| \geq Y.
\end{equation}
Additionally, a short, self-contained proof of the Vinogradov--Korobov zero-free region for Dirichlet $L$-functions (assuming an explicit upper bound for the Hurwitz zeta function $\zeta(s,u)$ proven by Ford, and a convenient form of Jensen's formula from complex analysis) is provided in the appendix of the paper (albeit with worse constants than Theorem 1.1 above).
| 7 | https://mathoverflow.net/users/167279 | 432336 | 174,962 |
https://mathoverflow.net/questions/432338 | 4 | So there's an elementary (but in my opinion quite interesting!) result which is that the Laurent series expansion of
$$\frac{1}{1-e^x} = -\frac{1}{x} + \frac{1}{2} - \frac{1}{12}x - \frac{1}{720}x^3 \cdots$$
Now the reason that is interesting is because each of those coefficients are equal to $\frac{1}{n!}B\_n$ where $B\_n$ is $n$th Bernoulli number and so more deeply one has the result
$$ \Gamma(\alpha + 1) \frac{d^\alpha}{dx^\alpha} \left[ \frac{1}{1-e^x} - 1 \right] = (-1)^{\alpha-1} \alpha \zeta(1-\alpha) $$
If you choose your fractional derivative carefully. ^^ (there might be an off by 1 error there, I need to double check)
So naturally I was inspired to go hunting for other elementary functions and to look at their coefficients/fractional derivatives in the hopes of finding something cool.
So I decided to look at $\frac{1}{1-e^{e^x-1}}$ the laurent series expansion of this is
$$\frac{1}{1-e^{e^x-1}} = -\frac{1}{x} + 1 - \frac{x}{6} - \frac{x^2}{24} - \frac{x^3}{90} - \frac{x^4}{720} + \frac{59}{60480}x^5 + \frac{17}{20160}x^6 + \frac{169}{453600} x^7 + \frac{47}{483840}x^8 - \frac{181}{119750400}x^9 \cdots$$
This was generated using this python code:
```
import sympy as sym
import math
x = sym.symbols('x')
w = 1/(1 - sym.exp(sym.exp(x)-1))
result = w.series(x, 0, 10).removeO()
print(result)
```
So I went over to OEIS and tried looking for those denominators and found nothing, I also looked at the denominators scaled by n! and similarly came up empty handed.
So this has me rather surprised since this is a pretty simple series to write down. My first question then that I want to ask is: is there some recurrence or functional equation that these coefficients/some scaled version of these coefficients obey?
If no recurrence then at least some kind of interpretation of their combinatorial significance would be cool.
| https://mathoverflow.net/users/46536 | Is there a recurrence for the coefficients of the Laurent series expansion of $\frac{1}{1-e^{e^x - 1}}$? | $e^{e^x-1}$ is the exponential generating function for [Bell numbers](https://en.wikipedia.org/wiki/Bell_number) ${\cal B}\_n$:
$$e^{e^x-1} = \sum\_{n\geq 0} {\cal B}\_n \frac{x^n}{n!}.$$
Then
$$g(x) := \frac{e^{e^x-1}-1}{x} = \sum\_{n\geq 0} {\cal B}\_{n+1} \frac{x^n}{(n+1)!}.$$
Correspondingly, the coefficient of $x^n$ in $g(x)^{-1}$ can be computed via [Faà di Bruno's formula](https://en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno%27s_formula) as
\begin{split}
\frac1{n!}\left.\left(\frac{d}{dx}\right)^n g(x)^{-1}\right|\_{x=0} &= \frac1{n!}\sum\_{k=1}^n (-1)^k k! B\_{n,k}( \frac{{\cal B}\_2}2, \frac{{\cal B}\_3}3, \dots) \\
&=\sum\_{k=1}^n (-1)^k \hat B\_{n,k}( \frac{{\cal B}\_2}{2!}, \frac{{\cal B}\_3}{3!}, \dots),
\end{split}
where $B\_{n,k}$ and $\hat B\_{n,k}$ are the exponential and ordinary [Bell polynomials](https://en.wikipedia.org/wiki/Bell_polynomials), respectively. Up to a sign this gives the coefficient of $x^{n-1}$ in $\frac1{1-e^{e^x-1}} = -\frac{g(x)^{-1}}x$.
PS. [This Sage code](https://sagecell.sagemath.org/?z=eJxNj80KgzAQhO-C77A3d7X-pEehb1IqahMJMRtJ46GI715_KDiHhfkY2Jm3VDDI0PROKoWW6jiCTQwPsJk4jZdh9gw5fGaLgLmgl4EUVNsH53U7oqHNdnIcm8mNX3Z2h3wzhCkemGfbSY86u1N5XFDOgwbN4FseJHJuMkF0crNzFEXBGygvf5jiKI72iD0iVVGI6l958poDqmSxa_0My3XUmtAPp-1DxA==&lang=sage&interacts=eJyLjgUAARUAuQ==) computes the coefficients using the derived formula.
| 11 | https://mathoverflow.net/users/7076 | 432341 | 174,963 |
https://mathoverflow.net/questions/432342 | 5 | Let $E \to \mathbb{C}P^\infty$ be any topological complex vector bundle over the infinite complex projective space. I'm wondering if it makes sense to possibly define a "holomorphic structure" on $E$. This a priori requires a complex structure on $\mathbb{C}P^\infty$, which is also something I don't know whether it exists or is well-defined, given that we're working with an infinite dimensional manifold. But it feels natural that there should be at least some notion of holomorphicity on the tautological line bundle over $\mathbb{C}P^\infty$. Is there anything in the literature about this?
| https://mathoverflow.net/users/143629 | Does it make sense to define a holomorphic structure on $\mathbb{C}P^\infty$ and vector bundles over it? | Yes, there is lots of literature on this subject.
However, Tyurin proved that all vector bundles on $CP^\infty$ are
direct sum of line bundles. There are several more recent papers by
Penkov and Tikhomirov about vector bundles on $C P^\infty$ (they treat other infinite-dimensional
manifolds, too).
* A. N. Tjurin (Tyurin), *Vector bundles of finite rank over infinite varieties*, Izv. Akad. Nauk SSSR Ser. Mat. 40:6 (1976), 1248–1268; English transl. in Math. USSR-Izv. 10:6 (1976), 1187–1204. <https://doi.org/10.1070/IM1976v010n06ABEH001832>
* Penkov, I. B. (D-JACOB); Tikhomirov, A. S. (RS-HSEM)
*On the Barth–Van de Ven–Tyurin-Sato theorem*, (Russian)
Mat. Sb. 206 (2015), no. 6, 49–84; translation in
Sb. Math. 206 (2015), no. 5-6, 814–848 <https://doi.org/10.1070/SM2015v206n06ABEH004480>, <https://arxiv.org/abs/1405.3897>
* Penkov, I. B.; Tikhomirov, A. S. Triviality of vector bundles on twisted ind-Grassmannians. (Russian) Mat. Sb. 202 (2011), no. 1, 65–104; translation in Sb. Math. 202 (2011), no. 1-2, 61–99, <https://doi.org/10.1070/SM2011v202n01ABEH004138>, <https://arxiv.org/abs/0706.3912>
| 8 | https://mathoverflow.net/users/3377 | 432346 | 174,964 |
https://mathoverflow.net/questions/432355 | 0 | Fix $a \in \mathbb R^n$ and let $\|\cdot\|$ be any norm on $\mathbb R$ (e.g $\ell\_1$ norm). For any $a \in \mathbb R^n$, it is clear that the function $f\_a(x) := \|x-a\|\_2 + \|x\|$ is strictly convex and has a unique minimizer $x(a)$.
**Question.** Given $a,b \in \mathbb R^n$, can $\|x(a)-x(b)\|\_2$ be bounded in terms of some norm of $a-b$ ?
| https://mathoverflow.net/users/78539 | Is the mapping $F(a):= \arg\min_{x \in \mathbb R^n} \|x-a\|_2 + \|x\|_1$ non-expansive? | First of all, the function $f\_a$ is not strictly convex, and hence, you should expect multiple minimizers. As such, non-expansiveness (even in some generalized sense) does not seem likely. Consider the one-dimensional case where
$$f\_a(x) = |x-a| + |x|$$
for which $\operatorname{argmin} f\_a = [0,a]$. So you can select minimizers which depend continuously on $a$, but this may not be what you want…
| 4 | https://mathoverflow.net/users/9652 | 432356 | 174,967 |
https://mathoverflow.net/questions/432266 | 4 | Suppose $x\_0=x\_1=1$, define $y\_k=x\_k+\frac{1}{2}(x\_k-x\_{k-1})$ and $x\_{k+1}=y\_k-\eta y\_k^3$ where $\eta\in(0,1/8)$. If we know $x\_k\to 0$ as $k\to\infty$. How to show that $x\_k=\Theta(1/\sqrt{k})$?
---
It suffices to show that $x\_k^{-2}=\Theta(k)$. By Taylor expansion, we have
$$x\_{k+1}^{-2} = (y\_k-\eta y\_k^3)^{-2} = y\_k^{-2}(1-\eta y\_k^2)^{-2} = y\_k^{-2}(1+2\eta y\_k^2+o(y\_k^2)) = y\_k^{-2}+2\eta+o(1)$$
It seems that $x\_{k+1}^{-2}$ and $y\_k^{-2}$ form a linearly increases sequence. If here $y\_k^{-2}$ is replaced by $x\_k^{-2}$, then we obtain exactly $x\_k^{-2}=\Theta(k)$. However, $y\_k^{-2}\ne x\_k^{-2}$, and I got stuck at this step. Can someone give a hint?
| https://mathoverflow.net/users/490600 | Convergence rate of a sequence | Suppose that we already know that $x\_k>0\forall k$. Clearly $(x\_k)$ and $(y\_k)$ are decreasing sequences which converge to $0$. Then we can prove that $b\_k:=\frac{x\_{k}}{x\_{k-1}}\to 1$. To do it note first that $b\_k\in[0,1]\forall k$ and
$$b\_{k+1}=\frac{y\_k-\eta y\_k^3}{x\_k}=\frac{x\_k+\frac{1}{2}(x\_k-x\_{k-1})-\eta y\_k^3}{b\_kx\_{k-1}}=\frac{3}{2}\frac{x\_k}{b\_kx\_{k-1}}-\frac{1}{2b\_k}-\eta\frac{y\_k^3}{b\_kx\_{k-1}}=$$
$$=\frac{3}{2}-\frac{1}{2b\_k}-\eta\frac{y\_k^3}{b\_kx\_{k-1}}>\frac{3}{2}-\frac{1}{2b\_k}-\eta\frac{x\_k^3}{b\_kx\_{k-1}}=\frac{3}{2}-\frac{1}{2b\_k}-\eta x\_k^2.$$
So $b\_{k+1}>\frac{3}{2}-\frac{1}{2b\_k}-\eta x\_k^2$. Now note that $x\_k<1-\frac{3}{2}\eta$ for all $k\geq3$, so that $b\_{k+1}>\frac{3}{2}-\frac{1}{2b\_k}-\eta(1-\frac{3}{2}\eta)^2>\frac{3}{2}-\frac{1}{2b\_k}-\frac{1}{8}(1-\frac{3}{16})^2=\frac{2903}{2048}-\frac{1}{2b\_k}\sim1.407-\frac{1}{2b\_k}$.
Using this you can prove by induction that $b\_k>0.7$ for all $k$ (as base case you need to check $b\_2,b\_3>0.7$). Using that fact and that:
* $b\_{k+1}>\frac{3}{2}-\frac{1}{2b\_k}-x\_k^2$
* $x\_k\to0$
it's not difficult to prove that $b\_k\to 1$.
Now let's prove that $x\_k=\Theta(\frac{1}{\sqrt{k}})$. To do it we will compare it with the sequence $a\_k=\frac{1}{10\sqrt{k}}$. Note that we have $a\_{k+1}<a\_k-10a\_k^3$. So if we prove that for big enough $k$ we have $x\_{k+1}>x\_k-10x\_k^3$, then our series will decrease slower than $a\_k$ and we will be done.
This is equivalent to proving $\frac{x\_k-x\_{k+1}}{x\_k^3}<10$ for big enough $k$. So let's study $c\_k:=\frac{x\_k-x\_{k+1}}{x\_k^3}$. Of course $c\_k$ is always positive, and we have
$$c\_{k}=\frac{x\_k-x\_{k+1}}{x\_k^3}=\frac{x\_k-y\_k+\eta y\_k^3}{b\_k^3x\_{k-1}^3}=\frac{\frac{1}{2}(x\_{k-1}-x\_k)+\eta y\_k^3}{b\_k^3x\_{k-1}^3}=\frac{1}{2b\_k^3}c\_{k-1}+\frac{\eta}{b\_k^3}\left(\frac{y\_k}{x\_{k-1}}\right)^3.$$
Now using that $b\_k\to 1$ and $\frac{y\_k}{x\_{k-1}}\to1$, the inequality implies that for big $k$ we have $c\_k<0.7c\_{k-1}+1$. So for big enough $k$ we have $c\_k<10$, as we wanted.
| 1 | https://mathoverflow.net/users/172802 | 432369 | 174,971 |
https://mathoverflow.net/questions/432365 | 1 | In Proposition 5.7 on page 34 in [lectures on condensed mathematics](https://www.math.uni-bonn.de/people/scholze/Condensed.pdf) Peter Scholze shows that $\mathbb{Z}[S]^\blacksquare$ is solid. He shows that the two relevant expressions are isomorphic, however, in the definition of solidity it says that the isomorphism has to come from the natural map $\mathbb{Z}[S]\to\mathbb{Z}[S]^\blacksquare$. Can somebody tell me where this is proved?
| https://mathoverflow.net/users/473423 | Isomorphism of RHoms in condensed mathematics | I guess that this basically follows from keeping track of isomorphisms. Seemingly it is a bit harder to trace the following isomorphisms, due to the occurrence of the shift $\require{AMScd}\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\iHom{\underline{Hom}}\DeclareMathOperator\fin{fin}\DeclareMathOperator\colim{colim}\mathbb Z[-1]$:
\begin{align\*}
R\Hom(\prod\_J\mathbb R/\mathbb Z,\mathbb Z)&=\bigoplus\_J\mathbb Z[-1]\\
R\Hom(\prod\_J\mathbb Z,\mathbb Z)&=\bigoplus\_J\mathbb Z
\end{align\*}
I am not sure what the author of these notes had in mind then, but one can cook up an argument by adapting the proof of Thm 4.3 as follows.
**Claim.** The canonical map $\bigoplus\_J\mathbb Z\to R\iHom(\prod\_J\mathbb Z,\mathbb Z)$ is an isomorphism. Namely, $\bigoplus\_J\mathbb Z$ is (derived condensed) reflexive.
**Proof.** We examine the morphism
\begin{CD}
\colim\_{I\subseteq\_{\fin}J}R\Hom(\prod\_I\mathbb R/\mathbb Z,\mathbb Z)@>>>\colim\_{I\subseteq\_{\fin}J}R\Hom(\prod\_I\mathbb R,\mathbb Z)@>>>\colim\_{I\subseteq\_{\fin}J}R\Hom(\prod\_I\mathbb Z,\mathbb Z)\\
@VVV@VVV@VVV\\
R\Hom(\prod\_J\mathbb R/\mathbb Z,\mathbb Z)@>>>R\Hom(\prod\_J\mathbb R,\mathbb Z)@>>>R\Hom(\prod\_J\mathbb Z,\mathbb Z)
\end{CD}
of fiber sequences. By the proof Thm 4.3, the middle terms are all zeros, and the left-most vertical map is an isomorphism, thus so is the right-most map.
In Prop 5.7, note that, for profinite sets $T$, the map $\mathbb Z[T]\to\mathbb Z[T]^\blacksquare$ can be rewritten as the canonoical double dual map $\mathbb Z[T]\to \mathbb Z[T]^{\vee\vee}$, where $(-)^\vee:=R\iHom(-,\mathbb Z)$, and it suffices to check that this induces an equivalence $\mathbb Z[T]^{\vee\vee\vee}\to\mathbb Z[T]^\vee$ after taking a further dual $(-)^\vee$, and this should follow formally from the reflexivity of $\mathbb Z[T]^\vee$.
| 2 | https://mathoverflow.net/users/176381 | 432384 | 174,972 |
https://mathoverflow.net/questions/431872 | 4 | I am looking at (the limitation of) the extension of the singular value decomposition to tensors. I would like to show that there is a tensor $A\_{i,j,k}$ that **cannot** be decomposed in the following singular value decomposition fashion
$$A\_{i,j,k}=\sum\_n \lambda\_n u\_{i,n}v\_{j,n}w\_{k,n} \tag 1\label{1}$$
where $\sum\_iu\_{i,m}^\*u\_{i,n}=\sum\_iv\_{i,m}^\*v\_{i,n}=\sum\_iw\_{i,m}^\*w\_{i,n}=\delta\_{m,n}$, $x^\*$ denotes the complex conjugate of $x$ and $\lambda\_n\ge 0$. Consider Equation \eqref{1} and confine ourselves to the complex number field. We can see the degree of freedom on the left-hand side is greater than that of the right-hand side whether in the complex number field or the real number field. For example, for a tensor $A$ of order $3$ and dimension $n$ in each component, the number of dimensions in $\mathbf R$ on the left hand side is $n^3$, whereas the number of dimensions in $\mathbf R$ on the right hand side is $3\bigl(n^2-{n+1\choose2}\bigr)+n=\frac12n(3n-1)$. The polynomial in $n$ of the left hand side has one higher degree than the right hand side. The critical value is $n=1$. Obviously there are $A$'s that cannot be decomposed as the right hand side. Is there a quick, intuitive way of finding one or even a systematic way of finding all the $A$'s that violates Equation \eqref{1}?
| https://mathoverflow.net/users/32660 | Singular value decomposition for tensor | [Asher Peres: *Higher Order Schmidt Decomposition*](https://arxiv.org/abs/quant-ph/9504006),
Physics Letters, A 202, No. 1, 16-17 (1995), [MR1337627](https://mathscinet.ams.org/mathscinet-getitem?mr=MR1337627), [Zbl 1020.81540](https://zbmath.org/?q=an%3A1020.81540), gives the necessary and sufficient condition.
| 2 | https://mathoverflow.net/users/32660 | 432386 | 174,973 |
https://mathoverflow.net/questions/432343 | 0 | In [the LPS paper "Ramanujan graphs"](http://www.ma.huji.ac.il/%7Ealexlub/PAPERS/ramanujan%20graphs/ramanujanGraphs.pdf) the adjacency matrix of $X^{p,q}$, for simplicity say that $p,q\equiv1\mod{4}$ and $\left(\frac{p}{q}\right)=1$ (so, nonbipartite) and $n=\lvert X^{p,q}\rvert$, is considered with spectrum $p+1=\lambda\_0>\lambda\_1\ge\cdots\ge\lambda\_{n-1}>-(p+1)$. Then, the "angle" $\theta\_j$ is introduced as $\lambda\_j=2\sqrt{p}\cos\theta\_j$, where
$$\theta\in\begin{cases}[0,\pi] & \text{$\lambda$ is in the bulk of the spectrum} \\\ i\mathbb{R}\_+ & \lambda>2\sqrt{p} \\\ \pi+i\mathbb{R} & \lambda<-2\sqrt{p}.\end{cases}$$
I think it is implicitly used that "$\cos$" is really $\cos(x)=\frac{1}{2}(e^{ix}+e^{-ix})$ and $\sin(x)=\frac{1}{2i}(e^{ix}-e^{-ix})$, in order to allow for $\lambda$ outside of the bulk.
I am looking at Case ii. in the proof of Theorem 4.1, on pgs. 13/14 of the linked PDF version. The crucial expression ends up being (for even $k$)
$$\sum\limits\_{0\le j<n}\frac{\sin((k+1)\theta\_j)}{\sin\theta\_j}.$$
It is claimed without explanation that this sum equals
$$\frac{2}{p^{\frac{k}{2}}}\frac{p^{k+1}-1}{p-1}+o(p^\frac{k}{2}).$$
My first question is: **Where does this asymptotic evaluation come from?**
Then, the paper reaches the conclusion that
$$\sum\limits\_{0<j<n}\frac{\sin((k+1)\theta\_j)}{\sin\theta\_j}\in O(p^{k\epsilon})$$
(having already resolved the trivial eigenvalue) and claims then that $\theta\_j\in\mathbb{R}$ as a consequence. My second question is: **Why does this consequence hold?** My best guess for this is that for $\lambda$ not in the bulk, the corresponding $\theta$ has $t:=e^{i\theta}=\frac{\lambda}{2\sqrt{p}}\pm\sqrt{\frac{\lambda^2}{4p}-1}$ (the sign agreeing with $\lambda$'s) satisfying $\lvert t\rvert>1$ and we can compute that
$$\frac{\sin((k+1)\theta)}{\sin\theta}=\frac{1}{t^k}(1+t^2+t^4+\cdots+t^{2k})$$
so that for very large (still even) $k$ and small $\epsilon$, this positive term would dominate. However I am not sure how correct this is.
| https://mathoverflow.net/users/159965 | Series analyzed in Lubotzky–Phillips–Sarnak "Ramanujan Graphs" | **1.** Regarding your first question, we have $e^{i\theta\_0}=p^{-1/2}$ as emphasized two lines below (4.11). Therefore,
$$\frac{\sin((k+1)\theta\_0)}{\sin\theta\_0}=
\frac{e^{-i(k+1)\theta\_0}-e^{i(k+1)\theta\_0}}{e^{-i\theta\_0}-e^{i\theta\_0}}
=\frac{p^{(k+1)/2}-p^{-(k+1)/2}}{p^{1/2}-p^{-1/2}},$$
so that
$$\frac{2p^{k/2}}{n}\cdot\frac{\sin((k+1)\theta\_0)}{\sin\theta\_0}=\frac{2(p^{k+1}-1)}{n(p-1)}.\tag{1}$$
In contrast, for $j\in\{1,\dots,n-1\}$ we have
$$|e^{i\theta\_j}+e^{-i\theta\_j}|=2|\cos\theta\_j|=|\lambda\_j/\sqrt{p}|<p^{1/2}+p^{-1/2},$$
which eventually yields that $p^{-1/2}<|e^{i\theta\_j}|\leq 1$ by the line below (4.11). So in this case
$$\frac{\sin((k+1)\theta\_j)}{\sin\theta\_j}=
\frac{e^{-i(k+1)\theta\_j}-e^{i(k+1)\theta\_j}}{e^{-i\theta\_j}-e^{i\theta\_j}}
=o(p^{k/2}),\qquad k\to\infty,$$
whence also
$$\frac{2p^{k/2}}{n}\cdot\frac{\sin((k+1)\theta\_j)}{\sin\theta\_j}=o(p^k),\qquad k\to\infty.\tag{2}$$
Combining $(1)$ and $(2)$, we get
$$\frac{2p^{k/2}}{n}\sum\_{j=0}^{n-1}\frac{\sin((k+1)\theta\_j)}{\sin\theta\_j}=\frac{2(p^{k+1}-1)}{n(p-1)}+o(p^k),\qquad k\to\infty.$$
In the light of (4.16), this implies the display above (4.20).
**2.** Regarding your second question, let us start from
$$\sum\_{j=1}^{n-1}\frac{\sin((k+1)\theta\_j)}{\sin\theta\_j}=O(p^{k\epsilon}).$$
We can rewrite this as
$$\sum\_{j=1}^{n-1}\frac{e^{-i(k+1)\theta\_j}-e^{i(k+1)\theta\_j}}{e^{-i\theta\_j}-e^{i\theta\_j}}=O(p^{k\epsilon}).\tag{3}$$
Note that here $|e^{i\theta\_j}|\leq 1\leq|e^{-i\theta\_j}|$ by the line below (4.11). Let us record the smallest of these positive numbers as
$$r:=\min\left(|e^{i\theta\_1}|,\dots,|e^{i\theta\_{n-1}}|\right).$$
By $(3)$, the power series
$$f(z):=\sum\_{k=0}^\infty z^k\sum\_{j=1}^{n-1}\frac{e^{-i(k+1)\theta\_j}-e^{i(k+1)\theta\_j}}{e^{-i\theta\_j}-e^{i\theta\_j}}$$
converges in the open disk $D(0,p^{-\epsilon})$ for any $\epsilon>0$, hence also in the union of these disks, which is $D(0,1)$. In particular, $f(z)$ is a holomorphic function in $D(0,1)$. On the other hand, in the smaller open disk $D(0,r)$ we can evaluate $f(z)$ as
\begin{align\*}
f(z)&=\sum\_{j=1}^{n-1}\sum\_{k=0}^\infty\frac{z^k e^{-i(k+1)\theta\_j}-z^k e^{i(k+1)\theta\_j}}{e^{-i\theta\_j}-e^{i\theta\_j}}\\
&=\sum\_{j=1}^{n-1}\frac{e^{-i\theta\_j}(1-ze^{-i\theta\_j})^{-1}-e^{i\theta\_j}(1-ze^{i\theta\_j})^{-1}}{e^{-i\theta\_j}-e^{i\theta\_j}}\\
&=\sum\_{j=1}^{n-1}\frac{1}{(z-e^{i\theta\_j})(z-e^{-i\theta\_j})}.\end{align\*}
So the right-hand side extends holomorphically from $D(0,r)$ to $D(0,1)$, which means that its poles $e^{\pm i\theta\_j}$ cannot lie in $D(0,1)$. So each $e^{i\theta\_j}$ lies on the unit circle (rather than in the open unit disk), whence $\theta\_j\in\mathbb{R}$.
| 3 | https://mathoverflow.net/users/11919 | 432390 | 174,974 |
https://mathoverflow.net/questions/432387 | 2 | I've been playing around numerically with Haar random $\text{CUE}$ unitary matrices of size $N$ by $N$, with $N$ around $1000$. If I "truncate" the matrix by keeping the upper left $fN$ by $fN$ block for some fixed $f$ independent of $N$, setting all other entries to zero, the resulting matrix is no longer unitary. In fact, one can argue that the eigenvalues of the truncated matrix will with high probability now be strictly within, rather than on, the unit circle. The distribution of the eigenvalues is [well-known](https://arxiv.org/abs/0911.5645).
However, from my numerics, it appears that despite the truncation, $(2f-1)N$ singular values are still **exactly** $1$ for $f \geq 1/2$. When $f\leq 1/2$, all the singular values are less than one.
Now, the above seems true for generic unitaries, but there are fine-tuned exceptions. The simplest exception comes from taking the unitary $U$ to be the identity. Then there are exactly $fN>(2f-1)N$ unit singular values.
My question is, how can I see that the truncated $\text{CUE}$ matrices have, with high probability, $(2f-1)N$ singular values that are identically $1$?
| https://mathoverflow.net/users/153549 | The singular values of truncated Haar unitaries | Your numerical findings have a simple explanation: Let me denote the upper left block of the $N\times N$ unitary matrix by $R$ and the upper right block by $T$; the matrix $R$ has dimension $fN\times fN$, the upper right block has dimension $fN\times(1-f)N$; unitarity requires that
$$RR^\dagger+TT^\dagger=I,$$
with $\dagger$ indicating the conjugate transpose and $I$ the identity matrix. For $f\geq 1/2$ the matrix $TT^\dagger$ has rank $\leq (1-f)N$, so it has at least $fN-(1-f)N=(2f-1)N$ eigenvalues equal to zero.
Hence for $f\geq 1/2$ the matrix $RR^\dagger$ has at least $(2f-1)N$ eigenvalues equal to 1, or equivalently, $R$ has $\geq (2f-1)N$ singular values equal to unity. For generic matrices the $\geq$ will be an equality.
| 4 | https://mathoverflow.net/users/11260 | 432391 | 174,975 |
https://mathoverflow.net/questions/416294 | 11 | $\DeclareMathOperator\GL{GL}\DeclareMathOperator\SO{SO}$I know a little bit about complex representation theory of finite reductive groups as $\GL\_n(q),\SO\_n(q)$ etc via Deligne-Lusztig induction and so on. If I correctly understood, there's another geometric way to build the characters (at least in the $\GL\_n$ case) via the so-called character sheaves. I've just have read very vaguely something about this and it seems to me strictly related to the same circle of idea which brings up Springer correspondence etc: I'd be particularly interested in the link between the twos.
Are there any book or introductory references which treat this construction? I'd be particularly interested just in the $\GL\_n$ case and in a focus towards examples maybe. I tried to read the original articles by Lusztig but they are maybe a bit too general/ difficult for what I had in mind.
| https://mathoverflow.net/users/146464 | Reference for character sheaves over $\mathrm{GL}_n(q)$ | I am not exactly sure what you have read already, but how about this set of notes, from a course given by Victor Ostrik in Luminy in 2010 (notes by Geordie Williamson)?
* [Character sheaves, tensor categories and non-Abelian Fourier transform](http://people.mpim-bonn.mpg.de/geordie/Ostrik.pdf)
It says in $\S~2.1$ that, given a reductive algebraic group $G/\mathbb{F}\_q$, the goal is to construct a class $\widehat{G}$ of irreducible perverse sheaves on $G$ such that their characteristic functions give characters of $G(\mathbb{F}\_q)$, which as far as I understand is what you are asking about :-)
If you are interested in the relations with the Springer correspondence, you can consult:
* Anne-Marie Aubert, [Character sheaves and generalized Springer correspondence](https://projecteuclid.org/journals/nagoya-mathematical-journal/volume-170/issue-none/Character-sheaves-and-generalized-Springercorrespondence/nmj/1114631876.full)
| 3 | https://mathoverflow.net/users/481162 | 432392 | 174,976 |
https://mathoverflow.net/questions/432389 | 3 | **Notation**:
$L/K$, finite extension of global fields
$K^\times$, unit group of $K$
$L^\times$, units group of $L$
$\mathbb{A}\_L^\times$, ideles of $L$
$N\_{L/K}$, the norm map
The **knot group** of an extension of global fields $L/K$ is defined as the quotient group of 'local norms' by 'global norms':
$$\mathfrak{K}(L/K):=\frac{K^\times \cap N\_{L/K}(\mathbb{A}\_L^\times)}{N\_{L/K}(L^\times)}.$$
**Question**: In what way is this related to knots?
**Remarks**: 1. The terminology *knot group* appears to have been introduced by Arnold Scholz, in the papers *Totale Normenreste, die keine Normen sind, als Erzeuger nichtabelscher Körpererweiterungen*. Part I (1936), Part II (1940). (I have not looked into these papers yet.)
2. I don't think this is related to the usual analogy between knots and number fields (but I could be wrong). For example, the usual knot group of a knot $\mathcal{K}$ in $S^3$ is the fundamental group $\pi\_1(S^3\setminus \mathcal{K})$. The analog of this should be something like $\pi\_1(\text{Spec}(\mathcal{O}\_K)\setminus \mathfrak{p})$, where $\mathcal{O}\_K$ is the ring of integers of a global field $K$ and $\mathfrak{p}\subset \mathcal{O}\_K$ is a prime ideal.
| https://mathoverflow.net/users/92433 | Knot group of a field extension | Arnold Scholz was fond of a colorful language in mathematics. I don't think there's any connection to actual knots except perhaps for a faint reference to the Gordian knot, which is difficult to solve without a striking idea.
| 6 | https://mathoverflow.net/users/3503 | 432393 | 174,977 |
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