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https://mathoverflow.net/questions/419906 | 1 | Suppose that $f$ is a continuous function on $[0,1]$. For $0<a<1$, if
$$ \varlimsup\_{\delta \rightarrow 0} \frac{\sup\_{0<\lvert y\rvert\leq \delta}\lvert f(x+y)-f(x)\rvert}{\delta^{a}} = \infty, $$
then, given any $\epsilon>0$, is it true that
$$ \varliminf\_{\delta \rightarrow 0} \frac{\sup\_{0<\lvert y\rvert\leq \delta}\lvert f(x+y)-f(x)\rvert}{\delta^{a+\epsilon}} = \infty? $$
| https://mathoverflow.net/users/152618 | A problem of the limit of $\frac{\sup_{0<\lvert y\rvert\leq \delta}\lvert f(x+y)-f(x)\rvert}{\delta^{a}}$ | This example is not continuous, but one can replace the jumps with linear pieces of fast growing slopes. Fix $0<\epsilon<a$. Define a sequence $(r\_j)\_{j\in\mathbb N}$ tending to zero inductively as follows:
$r\_1=1$ and $0<r\_{j+1}<r\_j$ so small that
$$
\frac{r\_{j+1}^{a-\epsilon}}{r\_j^{a+\epsilon}}\le 1.
$$
Then set
$$
f(t)=\begin{cases}0&t=0,\\ r\_{j+1}^{a-\epsilon}& r\_{j+1}<t\le r\_j,\\
1&t>1.
\end{cases}
$$
We have
$$
\limsup\frac{f(t)}{t^a}=\lim\_j\frac{f^+(r\_{j+1})}{r\_{j+1}^a}=\lim\_j\frac{r\_{j+1}^{a-\epsilon}}{r\_{j+1}^a}=\infty
$$
while on the other hand
$$
\liminf \frac{f(t)}{t^{a+\epsilon}}=\lim\_j\frac{f(r\_j)}{r\_j^{a+\epsilon}}=\frac{r\_{j+1}^{a-\epsilon}}{r\_{j}^{a+\epsilon}}\le 1.
$$
| 1 | https://mathoverflow.net/users/473423 | 419918 | 170,899 |
https://mathoverflow.net/questions/419929 | 3 | Let us define the following two stopping time $\tau\_B=\inf\{t\geq 0: X\_t\in B\}, \tau'\_B=\inf\{t> 0: X\_t\in B\}$, where $\tau\_B$ is entrance time and $\tau'\_B$ is hitting time. It is clear $\tau\_B=\tau'\_B, \mathbb{P}-a.s.$. By the book [Levy process](http://www.maphysto.dk/oldpages/events/LevyBranch2000/) by Bertoin P22, when $B$ is closed $\tau\_B=\tau'\_B$ may fail on $\partial B$. These point is called irregular point of B. When our starting point $x$ is irregular point of $B$, using Bulmenthal 0-1 law, $\mathbb{P}(\tau'\_B>0)=1$.
I don't understand how to use Bulmenthal 0-1 law? Is there someone to explain a little bit?
| https://mathoverflow.net/users/147009 | Blumenthal 0-1 law | $\newcommand\F{\mathcal F}\newcommand\N{\mathbb N}$First, some preliminary remarks:
1. Your link to Bertoin's book is not very good. Here is a link with a [better reference to the book](https://www.ams.org/bull/1998-35-04/S0273-0979-98-00761-7/S0273-0979-98-00761-7.pdf).
2. It is Blumenthal, not Bulmenthal.
3. It is not true that "when $B$ is closed $\tau\_B=\tau'\_B$ is failed on $\partial B$". This kind of failure may occur only at some (not all) points on $\partial B$, which are then called irregular points for $B$.
4. In Bertoin's book, $(X\_t)\_{t\ge0}$ is a right-continuous Lévy process, and $\F\_t$ is the completion of the sigma-field generated by $(X\_s\colon0\le s\le t)$. According to the book, let $T\_B:=\tau\_B$ and $T'\_B:=\tau'\_B$.
Now, to your question, "how to use Bulmenthal 0-1 law":
Let $B$ be a closed set. By part (iii) of Corollary 8 on p. 22 in Bertoin's book, $T'\_B$ is a stopping time and hence $[T'\_B=0]=[T'\_B\le0]\in\F\_0$, where $[\cdots]$ means the event $\cdots$. So, by the definition of $\F\_t$ cited above in Remark 4, we see that $P\_x(T'\_B=0)$ is $0$ or $1$ for any $x$. Suppose now that a point $x\in\partial B$ is irregular for $B$. Then $P\_x(T\_B=T'\_B)\ne1$, while $P\_x(T\_B=0)=1$. So, $P\_x(T'\_B=0)\ne1$. Thus, $P\_x(T'\_B=0)=0$.
In the above proof we did not explicitly use the Blumenthal 0-1 law, which states that the filtration $(\F\_t)$ is right-continuous at $0$, that is, $\F\_{0+}=\F\_0$. However, in the proof in Bertoin's book of the fact that $T'\_B$ is a stopping time he used (not quite explicitly) Proposition 4 on p. 18 in that book, which states that the filtration $(\F\_t)$ is right-continuous at all real $t\ge0$, and the proof of Proposition 4 on p. 18 in that book is based on Kolmogorov's 0-1 law. So, this reference in Bertoin's book to the Blumenthal 0-1 law is indeed imprecise and possibly confusing.
| 2 | https://mathoverflow.net/users/36721 | 419939 | 170,905 |
https://mathoverflow.net/questions/419924 | 4 | Let $A$ and $B$ be self-adjoint operators on some Hilbert space and $B$ is postive. Suppose we have $-B\leq A\leq B$.Is it true then that $\|A\|\_p\leq\|B\|\_p$ where $\|.\|\_p$ is the Schatten-$p$ norm defined as $\|A\|\_p:=(Tr(|A|^p)^{1/p}.$
| https://mathoverflow.net/users/136860 | A trace inequality between self-adjoint operators | Yes, this follows from the fact that $\|B\|\_p^p \geq \sum |\langle Be\_i, e\_i\rangle|^p$ for any orthonormal basis $(e\_i)$ (see [here](https://math.stackexchange.com/questions/2269975/is-this-inequality-on-schatten-p-norm-and-diagonal-elements-true)). If $(e\_i)$ diagonalizes $A$ then we have $\|A\|\_p^p = \sum |\langle A e\_i, e\_i\rangle|^p$, and also $|\langle Ae\_i,e\_i\rangle| \leq |\langle Be\_i, e\_i\rangle|$ for all $i$ because $-B \leq A \leq B$, and putting all that together yields $\|A\|\_p^p \leq \|B\|\_p^p$.
| 8 | https://mathoverflow.net/users/23141 | 419945 | 170,909 |
https://mathoverflow.net/questions/419903 | 0 | The diffusion equation with constant diffusion $D$ can be represented as:
\begin{equation}
\frac{\partial \phi(r, t)}{\partial t}=D \Delta \phi(r, t)
\end{equation}
where
* $\Delta$ is the Laplace operator and
* $\phi(r,t)$ represents a concentration at a point $r\in\mathbb{R}^n$ at time $t$.
When the diffusion is on a network, the Laplacian operator can be discretized and take the form of a matrix representation. The diffusion equation then takes the form:
\begin{equation}
\frac{d \phi\_{i}(t)}{d t}=D \sum\_{j} A\_{i j}\left(\phi\_{j}(t)-\phi\_{i}(t)\right)
\end{equation}
where now
* $\phi\_i(t)$ represents a concentration on the vertex $i$ at time $t$ and
* $A\_{ij}=1$ if there exists an edge between $i$ and $j$.
Consider now the case where the diffusion is not constant but is now a function depending on space and time: $D\to D(r,t)$. The diffusion equation simply is:
\begin{equation}
\frac{\partial \phi(r, t)}{\partial t}=\nabla \left[D(r,t) \nabla\phi(r, t)\right]
\end{equation}
**What happens to the network case now?** Writing out the discrete version of the Laplacian gives me:
\begin{equation}
\frac{d \phi\_{i}(t)}{d t}=\sum\_{j} A\_{i j}D\_{i}(t)\left(\phi\_{j}(t)-\phi\_{i}(t)\right)+"(\nabla D)(\nabla\phi)"
\end{equation}
But I have no idea how to discretize $\nabla$ and it feels wrong anyway. Intuitively I would expect something like:
\begin{equation}
\frac{d \phi\_{i}(t)}{d t}=-\phi\_{i}(t)+f\left(\sum\_{j} A\_{i j}\left(\phi\_{j}(t)(t)\right)\right)
\end{equation}
Where $f$ is some function related to $D$ so that we recover the non-linear behaviour of the continuous case.
What am I missing? [These notes](http://www.leonidzhukov.net/hse/2015/networks/lectures/lecture11.pdf) follow the approach I took, but are limited to constant diffusion. I was not able to find any lecture notes that cover non-linear diffusion on networks.
| https://mathoverflow.net/users/142153 | Non-linear diffusion on networks | $\newcommand{\R}{\mathbb R}$You do not need to "discretize $\nabla$". Also, you wrote the diffusion equation incorrectly. The correct version is this:
\begin{equation}
\frac{\partial f(r,t)}{\partial t}=\nabla\cdot[B(r,t)\,\nabla f(r,t)],
\end{equation}
where $f:=\phi$, $B:=D$, and $\cdot$ denotes the dot product. In the coordinate form, this equations is
\begin{equation}
\frac{\partial f(r,t)}{\partial t}=\sum\_{j=1}^n [B(r,t)\,(D\_j^2 f)(r,t)+(D\_j B)(r,t)\,
(D\_j f)(r,t)],
\end{equation}
where $D\_j$ is the operator of the partial differentiation with respect to the $j$th coordinate of $r\in\R^n$.
Now discretization becomes straightforward, by replacing the partial derivatives by the corresponding differences:
\begin{equation}
\frac{df\_i(r,t)}{dt}=
\sum\_j A\_{i,j}[B\_i(t)\,(f\_j(t)-f\_i(t))+
(B\_j(t)-B\_i(t))(f\_j(t)-f\_i(t))]
\end{equation}
or, simply,
\begin{equation}
\frac{df\_i(r,t)}{dt}=
\sum\_j A\_{i,j}B\_j(t)(f\_j(t)-f\_i(t)).
\end{equation}
More generally, we can write
\begin{equation}
\frac{df\_i(r,t)}{dt}=
\sum\_j A\_{i,j}B\_{i,j}(t)(f\_j(t)-f\_i(t)),
\end{equation}
where the $B\_{i,j}$'s are nonnegative functions. This will describe a general continuous-time random walk on the network. One may want to recall at this point that the diffusion equation describes an approximation of jump processes (which are continuous-time random walks) by processes continuous in space.
| 2 | https://mathoverflow.net/users/36721 | 419946 | 170,910 |
https://mathoverflow.net/questions/419888 | 8 | Let $\mathcal{A}$ be an abelian category, and let $X$ an object of $\mathcal{A}$. Recall that a *pseudoelement* of $X$ is an equivalence class of arrows $X\_1 \to X$, where $x\_1 \colon X\_1 \to X$ and $x\_2 \colon X\_2 \to X$ are equivalent if there is an object $P$ and *epimorphisms* $p\_1 \colon P \to X\_1$ and $p\_2 \colon P \to X\_2$ such that $x\_1 \circ p\_1= x\_2 \circ p\_2$. We write $x \in^\ast X$ to denote that $x$ is a pseudoelement of $X$. If $x \in^\ast X$ and $f \colon X \to Y$, then $f(x) \in^\ast Y$ is obtained by composition.
Pseudoelements are pretty nice, for example a morphism is equal to $0$ if and only if it sends all pseudoelements to $0$, and we can characterize monomorphisms (resp. epimorphisms) in terms of pseudoelements, as one expects. Similarly we can check if a sequence is exact by testing exactness on pseudoelements. Of course there are limitations, for example we can not test if two morphisms are equal on pseudoelements (in contrast with the Freyd–Mitchell embedding theorem).
A natural problem is what happens with pullbacks. Let $f \colon X \to Z$ and $g \colon Y \to Z$ be morphisms, and let $x \in^\ast X$ and $y \in^\ast Y$ such that $f(x) = g(y)$. It's easy to construct $s \in^\ast X \times\_Z Y$ such that $\pi\_1(s) = x$ and $\pi\_2(s) = y$, by universal property of the pullback. Now, we would like $s$ to be unique, and indeed Francis Borceux in his "Handbook of Categorical Algebra: Volume 2", Proposition 1.9.5, claims so. The proof is only sketched, and essentially Borceux says "All the relevant epimorphisms can, by successive pullbacks, be replaced by epimorphisms with the same domain, from which the claim follows". It's not completely clear how to write down the details, see for example [this](https://math.stackexchange.com/questions/3609374/replacing-a-family-of-epimorphisms-by-epimorphisms-with-the-same-domain-by-formi) stackexchange question, with a partial answer.
I've tried to prove the theorem by myself, essentially following Borceux's idea of constructing several pullbacks, but I am unable to finish the proof. To simplify the notation, let's consider the case of the product, so $Z=0$ in the above discussion. Let $a \colon A \to X \times Y \in^\ast X \times Y$ and $b \colon B \to X \times Y \in^\ast X \times Y$ be such that $\pi\_1(a)=\pi\_1(b)$ and $\pi\_2(a)=\pi\_2(b)$. So there are epimorphisms $a\_1 \colon Z\_1 \to A$, $b\_1 \colon Z\_1 \to B$, $a\_2 \colon Z\_2 \to A$ and $b\_2 \colon Z\_2 \to B$ such that $\pi\_1 \circ a \circ a\_1 = \pi\_1 \circ b \circ b\_1$ and $\pi\_2 \circ a \circ a\_2 = \pi\_2 \circ b \circ b\_2$. To construct the required epimorphisms we consider $P\_a = Z\_1 \times\_A Z\_2$ and $P\_b = Z\_1 \times\_B Z\_2$. We finally set $P = P\_a \times\_{Z\_1} P\_b$, where $P\_a \to Z\_1$ and $P\_b \to Z\_1$ are the first projections. We write $f\_a : P \to A := a\_1 \circ \pi\_1 \circ \pi\_1$ (two different $\pi\_1$) and $f\_b : P \to B := b\_1 \circ \pi\_1 \circ \pi\_2$. Then $f\_a$ and $f\_b$ are epimorphisms. We need to prove that $a \circ f\_a= b \circ f\_b$ and it is enough to prove that $\pi\_1 \circ a \circ f\_a= \pi\_1 \circ b \circ f\_b$ and $\pi\_2 \circ a \circ f\_a= \pi\_2 \circ b \circ f\_b$. The first one follows immediately by $\pi\_1 \circ a \circ a\_1 = \pi\_1 \circ b \circ b\_1$ and the commuting square of the pullback. I don't see how to prove the second equality, I tried essentially all the combinations of the two equalities we have and all the pullback squares, but it didn't work.
Of course one can try to construct different epimorphisms, but all the constructions I tried (even those more symmetric then the one above) end up with exactly the same equality to prove.
Does someone know a complete proof of this result? Thanks!
PS: I realized I am not able to prove it trying to formalize the result in Lean, in a lemma for the [Liquid Tensor Experiment](https://github.com/leanprover-community/lean-liquid/), you can see the actual Lean code [here](https://github.com/leanprover-community/lean-liquid/blob/ed8f03da6f8cdde28f936aeab8c0c21c457e45a7/src/for_mathlib/derived/les.lean#L84).
| https://mathoverflow.net/users/7845 | Pullback and pseudoelements | **The claim stated in the question is false, and the statement of Lemma 1.9.5 in Borceux is unclear, but seems wrong.** To be clear, the claim in this question is that for any Abelian category $\newcommand{\A}{\mathcal{A}}\A$, and maps $f:X \to Z$, $g : Y \to Z$, and pseudo-elements $[x]$, $[y]$ such that $f[x] = g[y]$, there is a $unique$ pseudo-element $[p]$ of $X \times\_Z Y$ with $\pi\_1[p] = [x]$, $\pi\_2[p]=[y]$. More concisely, the claim is that the “pseudo-elements” functor $\A \to \mathrm{Set}$ preserves pullbacks. (Borceux Lemma 1.9.5 claims that $[p]$ is “pseudo-unique”; from the “proof” sketch, I agree it seems like he intends this to mean “unique, as a pseudo-element”, but conceivably he had something else in mind.)
Work for concreteness in $\newcommand{\Ab}{\mathrm{Ab}}\Ab$. Then maps $x\_1 \colon X\_1 \to X$, $x\_2 \colon X\_2 \to X$ are equivalent as pseudo-elements if and only if they have the same image. (The “only if” direction is clear; for the “if”, note that in this case the projections $X\_1 \times\_X X\_2 \to X\_i$ are epi.) So **pseudo-elements of $X$ correspond to subobjects/subgroups of $X$**. Borceux notes this fact in the closing discussion of §1.9.
But now it’s easy to see this doesn’t preserve pullbacks. For instance, it doesn’t preserve the product $\newcommand{\Q}{\mathbb{Q}} \Q \times \Q$: the subgroups $\{ (x,x) | x \in \Q \}$ and $\{ (x,2x) | x \in \Q \}$ have the same images under each projection, but are not the same. In terms of pseudo-elements, the maps $s\_1, s\_2 : \Q \to \Q \times \Q$ given by $s\_i(x)=(x,ix)$ are not equal as pseudo-elements of $\Q \times \Q$, but their images under each projection are equal as pseudo-elements of $\Q$.
| 8 | https://mathoverflow.net/users/2273 | 419951 | 170,911 |
https://mathoverflow.net/questions/419823 | 4 | Let $a\_1,\ldots,a\_n \in \mathbb{C}$ be complex numbers that are the zeros of a real polynomial (meaning that the non-real ones come in complex conjugate pairs). Suppose that these numbers are such that
$$
s\_\lambda(a\_1,\ldots,a\_n) \ge 0
$$
for every partition $\lambda = (\lambda\_1 \ge \ldots \ge \lambda\_n)$, where $s\_\lambda$ denotes the [Schur polynomial](https://en.wikipedia.org/wiki/Schur_polynomial) associated to $\lambda$. **Do these inequalities imply that $a\_i \in \mathbb{R}\_+$ for all $i$?**
Here's what I know so far:
* The converse is obvious: if $a\_i \in \mathbb{R}\_+$ for all $i$, then the sum over Young tableaux formula immediately shows $s\_\lambda(a\_1,\ldots,a\_n) \ge 0$.
* It's enough to show that the $a\_i$ are real. Nonnegativity then follows by Descartes' rule of signs upon noting that the elementary symmetric functions $s\_{(1,\ldots,1)}$ are, up to alternating signs, the coefficients of the polynomial $\prod\_i (x-a\_i)$.
* The statement is true and easy to prove for $n = 2$: it's enough to postulate nonnegativity merely for $\lambda = (n,0,\ldots)$, i.e. on the complete homogeneous symmetric polynomials. On these,
$$
s\_{(n,0,\ldots)}(r e^{i\theta},r e^{-i\theta}) = \frac{(r e^{i\theta}))^{n+1} - (r e^{-i\theta})^{n+1}}{re^{i\theta}-re^{-i\theta}} = r^n \frac{\sin((n+1)\theta)}{\sin(\theta)},
$$
which for every nonzero $\theta$ is obviously negative for some $n$, proving the contrapositive statement.
This suggests that some Fourier analysis may be useful for the general case, or perhaps the [Harish-Chandra-Itzykson-Zuber formula](https://terrytao.wordpress.com/2013/02/08/the-harish-chandra-itzykson-zuber-integral-formula/), but I don't know how to go about it.
| https://mathoverflow.net/users/27013 | Nonnegativity locus of Schur polynomials | The answer to the main question is affirmative. The crucial result is due to M. Aissen, I. J. Schoenberg, and A. Whitney, [*J. Analyse Math.* **2** (1952), 93—103](https://link.springer.com/article/10.1007/BF02786970). For further details see the solution to Exercise 7.91(e) in *Enumerative Combinatorics*, vol. 2. (For some interesting information on A. Whitney, see [Biographical information on Anne Marie Whitney](https://mathoverflow.net/questions/320906/biographical-information-on-anne-marie-whitney))
| 9 | https://mathoverflow.net/users/2807 | 419953 | 170,912 |
https://mathoverflow.net/questions/419948 | 3 | Let $\Omega$ be an open set of $\Bbb R^d$: consider the following function spaces
* $H\_0^1(\Omega)$, i.e. the closure of $C\_c^\infty(\Omega)$ in $H^1(\Omega)$
* $H\_\*(\Omega)$, i.e. the closure of $C\_c^\infty(\Omega)$ in $H^1(\Bbb R^d)$.
* $H\_{\Omega}(\Omega)=\{u\in H^1(\Bbb R^d):\ u= 0 \text{ a.e on } \Omega^c\}$.
**Question**: does the above spaces coincide? If not when are they equal?
| https://mathoverflow.net/users/112207 | Possible way to define $H_0^1(\Omega)$ Sobolev spaces | The first two are equivalent, as the $H^1(\Omega)$ norm and $H^1(\mathbb{R}^d)$ norm coincide for $C^\infty\_c(\Omega)$ functions.
The third is in general different:
If you let $d = 1$ and $\Omega = \mathbb{R}\setminus \{0\}$,
you see that $H\_{\Omega}(\Omega) = H^1(\mathbb{R}) \supsetneq H^1\_0(\Omega)$. You can create similar examples in higher dimensions (by omitting a hyperplane instead of a point). On the other hand, if $\Omega$ is a Sobolev extension domain (see e.g. Leoni's *First Course in Sobolev Spaces*) then $u\in H^1\_0(\Omega) \iff$ extending $u$ by zero to the exterior gives an $H^1(\mathbb{R}^d)$ function. And in that case, the third is equivalent to the first two.
| 10 | https://mathoverflow.net/users/3948 | 419956 | 170,913 |
https://mathoverflow.net/questions/419847 | 6 | Fix constant $L,C>0$ and $k\geq 1$ and let $f\in W^{1,k}(\mathbb{R}^d,\mathbb{R}^n)$ with $\|f\|\_{W^{1,k}}\leq C$.
Is there a known estimate on the distance
$$
\|f - \operatorname{Lip}\_L(\mathbb{R}^d,\mathbb{R}^n)\|\_{L^1(\mathbb{R}^d,\mathbb{R}^n)},
$$
*depending on the constants $L,C,$ and $k$*, where $\operatorname{Lip}\_L(\mathbb{R}^d,\mathbb{R}^n)$ is the set of Lipschitz functions from $\mathbb{R}^d$ to $\mathbb{R}^n$ with Lipschitz constant *at-most* $L$?
| https://mathoverflow.net/users/36886 | Best approximation of L1 function by Lipschitz function | Let $(\rho\_\epsilon)\_{\epsilon>0}$ be a standard family of mollifiers, with $\rho\_\epsilon$ supported in the ball $B\_\epsilon(0)$. Since $\|\rho\_\epsilon\|\_{L^1}=1$ and $\|\rho\_\epsilon\|\_{L^\infty}=c\_d\epsilon^{-d}$, by interpolation we get $\|\rho\_\epsilon\|\_{L^{k'}}=c\_d^{1/k}\epsilon^{-d/k}$ (for the dual exponent $k'=\frac{k}{k-1}$). Hence,
$$\|\nabla(\rho\_\epsilon\*f)\|\_{L^\infty}=\|\rho\_\epsilon\*\nabla f\|\_{L^\infty}\le c\_d^{1/k}\epsilon^{-d/k}C$$
for your function $f$ (by Holder). In order to get an $L$-Lipschitz function you can take
$$\epsilon:=c\_d^{1/d}(C/L)^{k/d}=c\_d'(C/L)^{k/d}$$
for another constant $c\_d'$ depending only on $d$.
Then you can bound the distance of $\rho\_\epsilon\*f$ from $f$ as follows:
$$\|\rho\_\epsilon\*f-f\|\_{L^{k}}\le\epsilon\|\nabla f\|\_{L^{k}}\le c\_d'(C/L)^{k/d}.$$
Note: the question of measuring the distance in $L^1$ seems ill-posed, since $W^{1,k}$ does not embed into $L^1$ (unless $k=1$).
| 2 | https://mathoverflow.net/users/36952 | 419964 | 170,915 |
https://mathoverflow.net/questions/419718 | 6 | I came across these notes from a talk by Hoschter which talks about superheight of an ideal and it mentions Krull's ideal height theorem on P2-P3 in terms of superheight. Here are the notes: [http://www.math.lsa.umich.edu/~hochster/swb2.pdf](http://www.math.lsa.umich.edu/%7Ehochster/swb2.pdf). Does the Peskine–Szpiro intersection theorem imply Krull's ideal height theorem? I couldn't find any reference to this and wonder if anyone can either explain or point out some references about how to prove Krull's ideal height theorem from the Peskine–Szpiro intersection theorem.
| https://mathoverflow.net/users/144294 | Does the Peskine–Szpiro intersection theorem imply Krull's ideal height theorem? | * **Superheight Theorem:** Let $M$ be a non-zero finitely generated $R$-module over a Noetherian ring $R$. Then superheight(ann $M$)$\le$ projdim $M$.
-**The Superheight Theorem implies Krull's Height Theorem:** See page 6 of [Class Notes for Math 918: Homological Conjectures, Instructor
Tom Marley](https://digitalcommons.unl.edu/cgi/viewcontent.cgi?article=1005&context=mathclass) .
* **The Superheight Theorem implies the Intersection Theorem:** See page 6 of [Class Notes for Math 918: Homological Conjectures, Instructor
Tom Marley](https://digitalcommons.unl.edu/cgi/viewcontent.cgi?article=1005&context=mathclass) .
* **The Superheight Theorem follows from the New Intersection Theorem:** See the proof of Theorem 9.4.4 of [*Bruns, Winfried; Herzog, Jürgen*, Cohen-Macaulay rings. [ZBL0909.13005](https://zbmath.org/?q=an:0909.13005).].
* **The Superheight Theorem follows from the Intersection Theorem, at least in the case of equicharacteristic zero:** I do not know a reference for this, however I provide a proof for the equicharacteristic zero case:
Let $M$ be as in the statement of the Superheight Theorem. To prove the statement, as in the proof of Theorem 9.4.4 of [*Bruns, Winfried; Herzog, Jürgen*, Cohen-Macaulay rings. [ZBL0909.13005](https://zbmath.org/?q=an:0909.13005).], one can assume that $R\rightarrow S$ is a local homomorphism of local rings and $S$ is an $R$-algebra such that $(ann M)S$ is primary to the maximal ideal of $S$. Then passing to the completion, we can assume that $R$ and $S$ are both complete local rings (faithfully flat extension preserves the height and annihilator of a flat base change of $M$ is the extension of the annihilator of $M$ because $M$ is finite). Since $R$ and $S$ are both complete, so they admit coefficient fields and since $R$ (and $S$) has equicharacteristic zero so the coefficient field, $C\_R$, of $R$ maps into the coefficient field, $C\_S$ of $S$ by the map $R\rightarrow S$. Thus we can factor $R\rightarrow S$ through $R\widehat{\otimes}\_{C\_R}C\_S\rightarrow S$. Since $C\_R$ is the coefficient field of $R$, so $R\widehat{\otimes}\_{C\_R}C\_S$ is Noetherian (and complete local). Since $R\widehat{\otimes}\_{C\_R}C\_S$ and $S$ both have the same coefficient field $C\_S$, so $R\widehat{\otimes}\_{C\_R}C\_S\rightarrow S$ is a finite ring homomorphism (i.e. $S$ is module-finite over $R\widehat{\otimes}\_{C\_R}C\_S$). Thus, without loss of generality, we can assume that the ring homomorphsim $R\rightarrow S$ is a finite ring homomorphism ($M\otimes\_R(R\widehat{\otimes}\_{C\_R}C\_S)$ has the same projective dimension as of $M$ by flatness of the local homomorphism $R\rightarrow R\widehat{\otimes}\_{C\_R}C\_S$). Then since $(0:\_RM)S$ is primary to the maximal ideal of $S$, so $M\otimes\_RS$ has finite length, and thus the Peskine-Szpiro Intersection Theorem implies that $\text{height}\big((0:\_RM)S\big)=\dim(S)\le \text{projdim}(M)$, as was to be proved, because $S$ is module-finite.
For a reference for the details of the facts used about the complete tensor product and finiteness of $S$ over $R\widehat{\otimes}\_{C\_R}C\_S$, the factorization, the reason for the characteristic restriction of the statement (or some arguments around this restriction), or flatness of the complete tensor product $R\widehat{\otimes}\_{C\_R}C\_S$ over $R$ please see some facts/results of [arXiv:1911.11290](https://arxiv.org/abs/1911.11290), or/and Remark 5.2+Lemma 5.1 of [arXiv:1609.00095](https://arxiv.org/pdf/1609.00095.pdf).
| 1 | https://mathoverflow.net/users/127857 | 419971 | 170,918 |
https://mathoverflow.net/questions/419931 | 2 | I am currently reading the paper titled "Birational Geometry of Moduli spaces of Configurations of Points on the Line" by M.Bolognesi and A.Massarenti. I have following doubts in section 2.22.
Let $\mathcal{L}\_{2g}$ be the linear system of degree $2g+1$ hypersurfaces in $\mathbb{P}^{2g}$ passing through $2g+2$ general points say $p\_1,\cdots,p\_{2g+2}$ with multiplicity $2g-1$. Let $\mu\_g:\mathbb{P}^{2g}\dashrightarrow\Sigma\_{2g}\subset\mathbb{P}^N$ be the rational map induced by $\mathcal{L}\_{2g}$. Let $H\_{I}=H\_{i\_1,\cdots,i\_{g+1}}$ be the $g$-plane generated by $p\_{i\_1},\cdots,p\_{i\_{g+1}}\in \mathbb{P}^{2g}$ and let $J\subset I$ such that $|J|=g$. Then I have following questions.
1. Why the general element $D$ of $\mathcal{L}\_{2g}|\_{H\_I}$ must contain $H\_J$ with multiplicity $g(2g-1)-(g-1)(2g+1)=1$?
2. Why $\mu\_{g}|\_{H\_I}$ is the standard Cremona transformation on $\mathbb{P}^{g}$?
Any suggestions or reference related to questions is highly appreciated. Thanks in advance.
| https://mathoverflow.net/users/211682 | Question regarding linear system of projective space | The formula for the multiplicity of a linear system along a linear subspace is classical. You can find it for instance in Lemma 2.1 here <https://arxiv.org/pdf/1210.5175.pdf>
This will tell you that a general $D$ contains $H\_J$ with multiplicity one. Now, $D\_{|H\_I}$ has degree $2g+1$ and multiplicity $2g-1$ at $g+1$ general points. But you have a fixed component given by the $H\_J$, there are $g+1$ of them, and through each one of the $p\_i$ there are $g-1$ of these. So their union forms a hypersurface in $H\_I$ of degree $g+1$ having multiplicity $g$ at the $p\_i$. Once you remove this hypersurface from $D\_{|H\_I}$ you are left with a hypersurface of degree $2g+1 - (g+1) = g$ having multiplicity $2g-1-g = g-1$.
Hence, your linear system restricted to $H\_I\cong\mathbb{P}^g$ is (once you remove the part of codimension $1$ of the base locus) the linear system of hypersurfaces of degree $g$ having multiplicity $g-1$ at $g+1$ general points. This is exactly the linear system inducing the standard Cremona of $\mathbb{P}^g$. For instace, for $g = 2$ you get the conics through three points, for $g = 3$ the cubics of $\mathbb{P}^3$ with four double points, ecc...
| 2 | https://mathoverflow.net/users/14514 | 419973 | 170,919 |
https://mathoverflow.net/questions/419967 | 6 | It is discussed in this [question](https://mathoverflow.net/questions/316209/deforming-metrics-from-non-negative-to-positive-ricci-curvature) whether a simply-connected closed Riemannian manifold with non-negative Ricci curvature admits positive Ricci curvature, and the answer appears to be "no, there are counter-examples and known obstructions".
My question is this: are there any known examples of closed, simply-connected manifolds of non-negative **sectional** curvature that do not admit positive Ricci curvature? Are there known obstructions?
| https://mathoverflow.net/users/170859 | Is it known whether a closed simply-connected manifold of non-negative curvature admits positive Ricci? | No there are no such examples known. Most known examples of manifolds of nonnegative sectional curvature come from biquotients or cohomogeneity one manifolds. If these are simply connected they are known to admit metrics of positive Ricci curvature by a [result of Schwachhoefer and Tuschmann](https://link.springer.com/article/10.1007/s00208-004-0538-x). Also by a result of Aubin mentined in the thread you referenced a manifold of nonnegative Ricci curvature with Ricci positive somewhere admit positive Ricci everywhere. In the simply connected case the only known obstruction to positive Ricci is the scalar curvature one (also mentioned in that thread). If a closed spin manifold has nonvanishing $\hat A$ genus then it does not admit a metric of positive scalar curvature and it particular also doesn't admit a metric of positive Ricci curvature. But simply connected manifolds of nonnegetaive sectional curvature have scalar curvature positive somewhere (else they'd be flat and not simply connected). Their scalar curvature is also nonnegative. This implies that the $\hat A$ is zero in the spin case.
| 8 | https://mathoverflow.net/users/18050 | 419974 | 170,920 |
https://mathoverflow.net/questions/418942 | 6 | Consider the category $\operatorname{Solid}\_{\mathbf{Z}}$ of *solid* abelian groups in the sense of Clausen-Scholze. This category is a full subcategory of *condensed* abelian groups, $\operatorname{Cond}\_{\mathbf{Z}}$. These are, modulo set theoretical technicalities, abelian sheaves on the site of profinite sets, with finite families of jointly surjective maps as coverings. There is a left adjoint to the inclusion, and thus one can define a tensor product by tensoring in $\operatorname{Cond}\_{\mathbf{Z}}$ and then solidifying.
**Question**. Let $V$ be a pro-discrete topological abelian group. Note that $\mathbf{Z}((T))$ is solid, as it is the limit along open inclusions of pro-discrete spaces, and any map from a compact space must factor through some pro-discrete subspace. What is the solid tensor product of $V$ with $\mathbf{Z}((T))$? I presume it is $V\{T\}$, the module of two way infinite Laurent series over $V$, whose Laurent tail coefficients tend to $0$ in $V$? To prove this it would seem I need to prove that the solid tensor product commutes with certain cofiltered limits, which I have not been able to do.
| https://mathoverflow.net/users/nan | Solid tensor product of pro-discrete space with Laurent series | This is not true in general, the most important observation being that it fails already when $V$ is discrete. In that case $V\otimes^{\blacksquare} \mathbb Z((T))$ is just the usual algebraic tensor product. This agrees with $V((T))$ only if $V$ is finitely generated.
In a different direction, for those pro-discrete abelian groups $V$ that are limits of finitely generated abelian groups, a variant of the claim is true; namely, one gets $V\otimes^\blacksquare \mathbb Z((T)) = V((T))$ (but the Laurent tail is finite). Indeed, in that case $V$ is a compact object of the category of solid abelian groups, so has a finite resolution by product of copies of $\mathbb Z$, reducing on to that case; and then it follows from $\prod\_I \mathbb Z\otimes^\blacksquare \prod\_J \mathbb Z=\prod\_{I\times J} \mathbb Z$.
As Z. M observes in the comment, a related true statement is that $V\otimes\_{\mathbb Z\_\blacksquare} \mathbb Z[T]\_\blacksquare$ is given by $V\langle T\rangle$ (those series $\sum\_{i\geq 0} v\_i T^i$ for which $v\_i\to 0$ as $i\to \infty$). The tensor product here is not a solid tensor product, but a base change from solid $\mathbb Z$-modules to solid $\mathbb Z[T]$-modules (where being $\mathbb Z[T]$-solid is stronger than being solid over $\mathbb Z$).
| 4 | https://mathoverflow.net/users/6074 | 419975 | 170,921 |
https://mathoverflow.net/questions/419954 | 4 | For integers $A,B\geq 1$ we define the difference $\sigma(A)\sigma(B)-\sigma(AB)$, denoting it as $[A,B]$, where $\sigma(n)=\sum\_{1\leq d\mid n}d$ denotes the sum of divisors function. It is possible to get in closed-form the parity of the arithmetical function $[A,B]$, I mean $[A,B]\text{ mod }2$. I was studying propositions about this when I wondered about the veracity of some conjectures.
>
> **Question.** I would like to know if it you can to find a counterexample or well a (partial) proof for some of the following conjecture (please see bellow and comments). **Many thanks.**
>
>
>
**Conjecture 2.** Let $x\geq 1$ be an integer that satisfies $[x,x]=x$, then $x$ is a prime number.
Please if these conjectures are in the literature answer the question as a reference request, or add a comment.
I relegate two conjectures to an Appendix, y deleted the old Conjecture 1 since this is false (my problem was an implementation with three integer values in my program, instead two, the
I knew the counterexample, I wrote it in my notebook, but I relate this exception with an open question; any case Conjecture 1 has a bunch of counterxamples).
**Appendix:** Adding two conjectures from the genuine version of the post. When I can I'm going to revise the old Conjecture 1.
**Conjecture 3.** Let $x,y\geq 1$ be positive integers such that $y=\sigma(x)$, $\gcd(x,y)>1$, $xy$ isn't a perfect square, $\sigma(y)\equiv0\text{ mod }2$, the integer $x$ is a triangular number and $[x,y]\equiv 1\text{ mod }2$. Then $x$ is an even perfect number.
**Conjecture 4.** If there exists an integer $N>1$ such that $[N,\sigma(N)]=3\cdot[N,N]$ holds, then $N$ is an odd perfect number.
| https://mathoverflow.net/users/142929 | Around the equation $\sigma\left(\square\right)=\text{prime}$: counterexamples or a proof for some of these conjectures | Please restrict to one question per post (standard policy). Here is a proof of Conjecture 2.
For any $s\in\mathbb{C}$, the function $\sigma\_s(n):=\sum\_{d\mid n} d^s$ satisfies the Hecke multiplicativity relation
$$\sigma\_s(m)\sigma\_s(n)=\sum\_{d\mid(m,n)}d^s\sigma\_s\left(\frac{mn}{d^2}\right).$$
This is straightforward to prove by restricting $m$ and $n$ to be powers of the same prime, since $\sigma\_s$ is a multiplicative function. On a deeper level, $\sigma\_s(n)$ is the $n$-th Hecke eigenvalue of a certain Eisenstein series that is a Hecke eigenform.
At any rate, setting $s=1$ and $m=n=x$, we get
$$\sigma(x)^2=\sum\_{d\mid x}d\sigma\left(\frac{x^2}{d^2}\right).$$
Assume that $x\geq 2$. Then the right-hand side is at least $\sigma(x^2)+x$, because the divisors $d=1$ and $d=x$ are present. If the left-hand side equals this, then there are no further divisors, hence $x$ is a prime number.
**P.S.** I suggest that you open a new question for Conjecture 1, another one for Conjecture 3, and a third one for Conjecture 4. This question should be closed as I proved Conjecture 2.
| 5 | https://mathoverflow.net/users/11919 | 419982 | 170,923 |
https://mathoverflow.net/questions/419908 | 4 | On Voevodsky's paper 'Cancellation theorem', Lemma 4.8, he stated in the proof that the map
$$\begin{array}{ccc}\mathbb{G}\_m\times\mathbb{G}\_m&\longrightarrow&\mathbb{G}\_m\times\mathbb{G}\_m\\(x,y)&\longmapsto&(y,x^{-1})\end{array}$$
is $\mathbb{A}^1$-homotopic to the identity, i.e., one could find an explicit homotopy (finite correspondence)
$$h:\mathbb{G}\_m\times\mathbb{G}\_m\times\mathbb{A}^1\longrightarrow\mathbb{G}\_m\times\mathbb{G}\_m$$
such that $h(x,y,0)=(x,y), h(x,y,1)=(y,x^{-1})$.
But I don't see how to find such a map.
Alternatively, he wants to show that the map
$$\begin{array}{ccc}\mathbb{G}\_m\times\mathbb{G}\_m&\longrightarrow&\mathbb{G}\_m\times\mathbb{G}\_m\\(x,y)&\longmapsto&(y,x)\end{array}$$
is $\mathbb{A}^1$-homotopy to
$$\begin{array}{ccc}\mathbb{G}\_m\times\mathbb{G}\_m&\longrightarrow&\mathbb{G}\_m\times\mathbb{G}\_m\\(x,y)&\longmapsto&(x, y^{-1})\end{array}$$.
| https://mathoverflow.net/users/149491 | On the swapping map of $\mathbb{G}_m$ | Recall that $S\_t^1$ is the *reduced* motive $\tilde M(\mathbf G\_m)$ of $\mathbf G\_m$, obtained most explicitly as the kernel of the projector $M(\mathbf G\_m) \to M(\mathbf G\_m)$ given by $\operatorname{id}-1\_\*$ where $1 \colon \mathbf G\_m \to \mathbf G\_m$ is the constant map $1$. In particular, it is a direct summand of $M(\mathbf G\_m)$ such that $H^\*(\tilde M(\mathbf G\_m)) = H^1(\mathbf G\_m)$ for any Weil cohomology theory $H$.
For a scheme $X$ and morphisms $f\_1, \ldots, f\_n \colon X \to \mathbf G\_m$, write $\tilde f\_1, \ldots, \tilde f\_n \colon X \to \tilde M(\mathbf G\_m)$ for the compositions with the projection $M(\mathbf G\_m) \to \tilde M(\mathbf G\_m)$, and for simplicity write $[f\_1,\ldots,f\_n]$ for the morphism $\tilde f\_1 \otimes \ldots \otimes \tilde f\_n \colon X \to \tilde M(\mathbf G\_m)^{\otimes n}$. The claim in Lemma 4.8 is now:
**Lemma.** *Let $X$ and $f, g \colon X \to \mathbf G\_m$ be as above. Then the association $(f,g) \mapsto [f,g]$ is a skew-symmetric bilinear pairing in $f$ and $g$.*
In particular, this implies that $[f,g] = -[g,f] = [g,f^{-1}]$, as claimed. As cited in the proof of Lemma 4.8, this is really a relative version of (the easy part of) Proposition 3.4.3 of [[Suslin–Voevodsky](https://www.math.ias.edu/vladimir/sites/math.ias.edu.vladimir/files/susvoenew.pdf)].
*Proof of Lemma.* For simplicity, we will assume the base scheme $S$ is the spectrum of a field $k$, and $X$ is an integral smooth affine $k$-variety. (The ultimate application is to $X = \mathbf G\_m \times \mathbf G\_m$, so this hypothesis is harmless. These hypotheses are only used so that I know what the correct definitions are.)
If $f\_1,f\_2,g \colon X \to \mathbf G\_m$ are morphisms, define the correspondence $Z \subseteq (X \times \mathbf A^1) \times \mathbf G\_m$ by
$$y^2 - \big(t(f\_1(x)+f\_2(x))+(1-t)(1+f\_1(x)f\_2(x))\big)y + f\_1(x)f\_2(x) = 0$$
for $(x,t,y) \in X \times \mathbf A^1 \times \mathbf G\_m$. The projection to $X \times \mathbf G\_m$ is an isomorphism: the equation is linear in $t$, so we can eliminate $t$. Thus, $Z$ is integral. The projection $Z \to X \times \mathbf A^1$ is finite flat of rank $2$, so $Z$ is a finite correspondence $\phi \colon X \times \mathbf A^1 \to \mathbf G\_m$. Its fibre at $0$ is $[1] + [f\_1f\_2]$ and its fibre at $1$ is $[f\_1]+[f\_2]$, so tensoring with $g$ gives
$$[f\_1,g] + [f\_2,g] = [1,g] + [f\_1f\_2,g] \in \operatorname{Hom}\big(M(X),\tilde M(\mathbf G\_m)\otimes \tilde M(\mathbf G\_m)\big).$$
But $[1,g] = 0$ since $1 \colon X \to \mathbf G\_m$ induces the zero map when composed with the projector $\operatorname{id}-1\_\*$ on $\mathbf G\_m$. Thus, we see that $[f\_1f\_2,g] = [f\_1,g]+[f\_2,g]$, and likewise $[f,g\_1g\_2] = [f,g\_1]+[f,g\_2]$. This shows that the map $(f,g) \mapsto [f,g]$ is bilinear.
To see that it is alternating, consider the correspondence $Z \subseteq (X \times \mathbf A^1) \times (\mathbf G\_m \times \mathbf G\_m)$ given by
$$\left\{\begin{array}{l}
y\_1^2 - \big(t(f(x)+g(x))+(1-t)(1+f(x)g(x))\big)y\_1 + f(x)g(x) = 0, \\
y\_1 = y\_2,\end{array}\right.$$
for $(x,t,y\_1,y\_2) \in X \times \mathbf A^1 \times \mathbf G\_m \times \mathbf G\_m$. This is again an integral subscheme that is finite flat of rank $2$ over $X \times \mathbf A^1$, and it gives a correspondence $X \times \mathbf A^1 \to \mathbf G\_m \times \mathbf G\_m$ whose restrictions to $t=0$ and $t=1$ witness the relation
$$[fg,fg] = [f,f] + [g,g] \in \operatorname{Hom}\big(M(X),\tilde M(\mathbf G\_m) \otimes \tilde M(\mathbf G\_m)\big).$$
By bilinearity, we can rewrite the left hand side as $[f,f]+[f,g]+[g,f]+[g,g]$, so we conclude that $[f,g] + [g,f] = 0$, i.e. the pairing is skew-symmetric. $\square$
**Remark.** To actually produce a homotopy from $[f,g]$ to $[g,f^{-1}]$, we really have multiple components:
* A homotopy from $[f,f]+[f,g]+[g,f]+[g,g]$ to $[fg,fg]$ (even this requires multiple steps a priori, but see below for a simplification);
* A homotopy from $[fg,fg]$ to $[f,f]+[g,g]$;
* A homotopy from $[g,f] + [g,f^{-1}]$ to $0$.
Then some things cancel, some things don't, and we crucially use multiple times that constant maps $1 \colon X \to \mathbf G\_m$ induce the zero map $M(X) \to \tilde M(\mathbf G\_m)$.
In fact the first and second item on the list above can be simplified by using some cancellation, as the closed subschemes witnessing these homotopies are actually almost the same. The closed subscheme
$$\left\{\begin{array}{l}
y\_1^2 - \big(t(f(x)+g(x))+(1-t)(1+f(x)g(x))\big)y\_1 + f(x)g(x) = 0,\\
y\_2^2 - \big(t(f(x)+g(x))+(1-t)(1+f(x)g(x))\big)y\_2 + f(x)g(x) = 0,
\end{array}\right.$$
of $X \times \mathbf A^1 \times \mathbf G\_m \times \mathbf G\_m$ has two components: one where $y\_1 = y\_2$ that we used for $[fg,fg]$, and a second component $W$ witnessing $[f,g] + [g,f] = [1,fg]+[fg,1]$, which again is zero on the reduced motive $\tilde M(\mathbf G\_m) \otimes \tilde M(\mathbf G\_m)$. Using $W$ avoids redundancies, leaving only the need for something witnessing $[g,f] + [g,f^{-1}] = 0$. We ultimately end up with two subschemes of $X \times \mathbf A^1 \times \mathbf G\_m \times \mathbf G\_m$ that are both finite flat of degree $2$ over $X \times \mathbf A^1$.
If you want to consider the entire motive $M(\mathbf G\_m)$, then the same argument provides a homotopy from
$$(f \otimes g) + (g \otimes f) + (g \otimes 1) + (g \otimes 1)$$
to
$$(1 \otimes fg) + (fg \otimes 1) + (g \otimes f) + (g \otimes f^{-1}),$$
witnessed by the two degree $2$ correspondences explained above. The projector $(\operatorname{id}-1\_\*) \otimes (\operatorname{id}-1\_\*)$ on $M(\mathbf G\_m) \otimes M(\mathbf G\_m)$ sends $f \otimes g$ to
$$[f,g] := (f \otimes g) - (f \otimes 1) - (1 \otimes g) + (1 \otimes 1),$$
and one immediately sees that terms of the form $[f,1]$ or $[1,g]$ vanish. This gives a homotopy from $[f,g]+[g,f]$ to $[g,f]+[g,f^{-1}]$, and the result follows by cancelling $[g,f]$ on both sides.
| 10 | https://mathoverflow.net/users/82179 | 419986 | 170,924 |
https://mathoverflow.net/questions/419981 | 7 | In field theory, the following fact is used in the construction of splitting fields: *Given a field $F$ and an irreducible polynomial $f \in F[x]$, the quotient $F[\alpha]/(f(\alpha))$ is a field extension of $F$ which contains a root of $f$ (namely the congruence class of $\alpha$).*
Let $n$ be a positive integer and let $\lambda\_{1} + \dotsb + \lambda\_{k} = n$ be a partition of $n$. Does there exist a field $F$ and an irreducible polynomial $f \in F[x]$ of degree $\deg f = n+1$ such that, if we define $K := F[\alpha]/(f(\alpha))$, the factorization of $f$ in $K[x]$ is of the form $f = (x-\alpha) \cdot f\_{1} \dotsb f\_{k}$ where each $f\_{i} \in K[x]$ is irreducible and $\deg f\_{i} = \lambda\_{i}$?
| https://mathoverflow.net/users/15505 | Factorization of an irreducible polynomial in the field extension it defines | Let us show that for the partition $2+1+1=4$, there is no such $f$.
If $f$ were inseparable, then over $K$ it would factor as a constant times $(x-\alpha)^5$. If $f$ were separable, its Galois group $G$ would be a transitive subgroup of $S\_5$ containing a transposition, so $G=S\_5$, but then the stabilizer of a point would act transitively on the other four points, so $f$ would factor over $K$ into $x-\alpha$ and an irreducible polynomial of degree $4$.
| 15 | https://mathoverflow.net/users/2757 | 419988 | 170,925 |
https://mathoverflow.net/questions/419995 | 11 | In the little galaxy of Category Theory, *Friedrich Ulmer* is known for being one of the authors of **Lokal Präsentierbare Kategorien**, a book that laid the foundations for the theory of locally presentable categories.
Unlike his coauthor, Pierre Gabriel, we do not have much information about him, or at least I can't find much.
* From the Genealogy Project we know that he did his PhD in Zürich in 1964 under *Dold* and *van der Waerden*.
* According to Scopus, we only have 7 documents by Ulmer, even though other documents can be found with a bit of struggle on the internet. With a bit of help from google and semanticscholar we find a couple of more entrencies. zbMath lists a total of 12 publications by Ulmer.
I can't find a complete bibliography of his, nor I can tell whether he spent his life in Academia or outside of it.
Does anybody have better luck then me? Or happens to have more information? What I was looking for is something like [this one](https://www.sciencedirect.com/science/article/pii/S0022404996001582) or [this one](https://www.ams.org/notices/199810/mem-eilenberg.pdf). Of course, there might be nothing more to say.
| https://mathoverflow.net/users/104432 | Mathematical life of Friedrich Ulmer | I am not sure for how long and where you looked, but it takes less than 30 minutes to figure much more than you mention in your post. Apparently, he continued his academic career as Fritz Ulmer : if you look at the affiliation of the last indexed MathReviews paper
* *Localizations of endomorphism rings and fixpoints*, Journal of Algebra **43** Issue 2 (1976) Pages 529–551, <https://doi.org/10.1016/0021-8693(76)90125-3>,
you see that he was in Wuppertal, and then a quick Google search shows that "Fritz Ulmer = Friedrich Ulmer":
1. Look at <https://idw-online.de/de/news52962> and examine the part "Kontakt",
and
2. Look at <https://www.researchgate.net/profile/Fritz-Ulmer> .
The website <http://www.wahlprognosen-info.de/> seems to contain some of his interviews on matters of statistics etc., see <http://www.wahlprognosen-info.de/index2.htm?/archiv/BB7.htm> .
| 13 | https://mathoverflow.net/users/1306 | 419998 | 170,928 |
https://mathoverflow.net/questions/419960 | 7 | Does there exist a smooth, projective, complex algebraic variety $X$, with two cohomology classes $\alpha,\beta \in H^{\*}(X,\mathbb{Z})$ neither $\alpha$ nor $\beta$ is torsion but the product $\alpha \cup \beta$ is non-trivial and torsion?
| https://mathoverflow.net/users/99732 | Algebraic varieties with certain topological properties | Let $Y$ be a smooth complex projective surface whose cohomology group $H^2(Y;\mathbb{Z})$ has a nonzero torsion element $\gamma$ and whose torsion-free quotient has rank at least $2$, e.g., this holds for an Enriques surface. Let $D$ and $E$ be nontorsion elements whose cup product is zero. By the Hodge Index Theorem, such elements exist.
Let $X$ be $Y\times Y$. Let $\alpha$ be $\text{pr}\_1^\*\gamma +\text{pr}\_2^\*D$. Let $\beta$ be $\text{pr}\_2^\*E$.
| 8 | https://mathoverflow.net/users/13265 | 420004 | 170,931 |
https://mathoverflow.net/questions/420008 | 6 | For a function $f(x,y)$ on $\mathbb{R}^2,$ defined possibly outside the origin, write
$$\int\_\epsilon ' f \,dx\,dy : = \int\_{\mathbb{R}^2\setminus D\_\epsilon}f \, dx\,dy,$$
(the integral on the complement to the $\epsilon$-disk) and
$$\int' f \,dx \,dy : = \lim\_{\epsilon\to 0} \int'\_\epsilon f \,dx\,dy,$$ when defined.
We view $\mathbb{R}^2$ as the complex line with coordinate $z = x+iy$. Then I claim that the assignment $$\phi:f(x,y) \mapsto \int\_{\mathbb{R}^2}' f\cdot z^{-1} \,dx\,dy$$ makes sense as a functional on compactly supported, smooth functions (indeed, if $f$ is rotationally symmetric then $\phi(f) = 0$ even before taking the limit, and if $f(0) = 0$ then $f\cdot z^{-1}$ is bounded, hence integrable, at $0$; now in the space of smooth compactly supported functions, any function can be written as a rotationally symmetric function plus a function that vanishes at the origin). We can thus formally define a *distribution* $z^{-1}\in C^{-\infty}(\mathbb{R}^2)$ as $$z^{-1}: = \frac{\phi}{dx\,dy}.$$
Now write $\bar{\partial} = \partial\_x + i \partial\_y$ .Like any vector field, this acts on distributions, and so we have a distribution $\bar{\partial} z^{-1}.$ Since $z^{-1}$ is holomorphic where it is smooth, we must have $\bar{\partial} z^{-1}$ be a distribution supported at the origin. In fact, it is known to mathematical physicists that the result is a delta function at the origin: $$\bar{\partial} z^{-1} = 2\pi \delta\_0,$$ and this fact is useful in conformal field theory.
I would like to see a proof of this result (the physics sources I have seen do not prove this). In fact, I know how to give one using a direct calculation with polar coordinates: but this seems *ad hoc* and is not very satisfying to me.
I suspect there might be "nicer" proofs using one or more of the following three techniques, and I am hoping that more analytically literate MO users can provide them.
1. If one can show that $\bar{\partial} z^{-1}$ is determined by its values on holomorphic (near the origin) functions, the Cauchy residue formula would imply that $\bar{\partial} z^{-1} = 2\pi \delta\_0.$
2. I suspect there should be a proof using the stationary phase approximation.
3. This is probably overkill, but it would be nice if there were a proof using pseudodifferential operators.
| https://mathoverflow.net/users/7108 | Elegant proofs of $\bar{\partial}z^{-1} = 2\pi \delta_0$ | The identification of $\partial\_{\bar{z}}z^{-1}$ with a delta function follows directly from the [Cauchy–Pompeiu formula](https://en.wikipedia.org/wiki/Cauchy%27s_integral_formula#Smooth_functions)
$$f(\zeta) = \frac{1}{2\pi i}\int\_{\partial D} \frac{f(z) \,dz}{z-\zeta} - \frac{1}{\pi}\iint\_D \frac{\partial f(z)}{\partial \bar{z}} \frac{dx\wedge dy}{z-\zeta},\qquad\qquad(\ast)$$
for $f$ a complex-valued $C^1$ function on the closure of the disk $D$ in $\mathbb{C}$. (Note that $f$ need not be holomorphic, as in the usual Cauchy formula.) If the support of $f$ is within $D$ the boundary integral $\int\_{\partial D}$ does not contribute, while the area integral $\iint\_D$ states that$^\ast$
$$\frac{\partial }{\partial \bar{z}} \frac{1}{z-\zeta}=\pi\delta^{(2)}(z-\zeta).$$
*Historical note:* [Wikipedia](https://en.wikipedia.org/wiki/Cauchy%27s_integral_formula#cite_note-2) gives a [1905](http://archive.numdam.org/item/AFST_1905_2_7_3_265_0/) paper as the source of the formula ($\ast$), but I could not locate it there. I did find it in a [1913](https://doi.org/10.1007/BF03015607) paper by Pompeiu (equation 3):
The first term $h(z)$ is the boundary integral. The capital $S$ denotes the area integral, the function $\varphi$ is the derivative $\partial f/\partial \bar{\zeta}$.
---
$^\ast$ The OP has a factor $2\pi$ instead of $\pi$, because of a different definition of the [Wirtinger derivative](https://en.wikipedia.org/wiki/Wirtinger_derivatives). Here I follow the definition $\partial/\partial\bar{z}=\tfrac{1}{2}(\partial/\partial x+i\partial/\partial y)$, while in the OP there is no coefficient $\tfrac{1}{2}$.
| 9 | https://mathoverflow.net/users/11260 | 420015 | 170,936 |
https://mathoverflow.net/questions/419969 | 10 | My question is on whether or not there exists some monotone strictly decreasing sequence of positive numbers $c\_1>c\_2>\ldots$ such that given any $f$ which is a uniformly bounded holomorphic function in the right half of the complex plane with
$$ |f(k)|\leq c\_k \quad \forall\, k\in \mathbb N,$$
there holds $ f(z)=0$ on the right half plane.
| https://mathoverflow.net/users/50438 | On a variant of Carlson’s theorem | Condition
$$\lim\_{n\to\infty}\frac{\log|c\_n|}{n}=-\infty$$
is sufficient for $f=0$.
Since $f(z)=e^{-cz}$ and $c\_n=e^{-cn}$ satisfy all
conditions, we see that this is best possible in certain sense.
This follows for example from a (much more general) theorem of N. Levinson, Gap and density theorems, AMS, 1940, page 121.
Levinson's theorem allows some growth of $F$, and much more general class of
sequences instead of integers.
Remark. In fact Levinson generalizes a theorem of Vladimir Bernstein 1932
(Theorem 32 in Levinson's book), which also implies the result that I stated.
| 8 | https://mathoverflow.net/users/25510 | 420018 | 170,937 |
https://mathoverflow.net/questions/419947 | 1 | I have asked a related question on math.SE [here](https://math.stackexchange.com/questions/4417531/finite-free-resolution-of-a-finitely-generated-mathbb-zx-1-ldots-x-n-modul), but the notation is a bit different.
As the title says, I am interested in constructing a finite free resolution of a $\mathbb Z[x\_1,\dotsc,x\_n]$-module using a related finite free resolution of a $\mathbb Q[x\_1,\dotsc,x\_n]$-module. Let $R=\mathbb Z[x\_1,\dotsc,x\_n]$ and $R' = \mathbb Q[x\_1,\dotsc,x\_n]$. Let $M'$ be a submodule of $R'^k$. Since $R'$ is Noetherian, $M'$ is finitely generated. Consider a finite free resolution of $M'$:
$$
0 \longrightarrow R'^{k\_l} \overset{A\_l}\longrightarrow R'^{k\_{l-1}} \overset{A\_{l-1}}\longrightarrow \cdots \overset{A\_1}\longrightarrow R'^{k\_0}\overset{A\_0}\longrightarrow R'^k~.
$$
Here, $l \le n$ by Hilbert's syzygy theorem, $M' = \operatorname{im} A\_0$, and each matrix $A\_i$ can be chosen such that its elements are polynomials with *integer* coefficients. Moreover, at every step in constructing this free resolution, I choose the minimal generating set of least cardinality.
Let $M$ be the submodule of $R^k$ generated by the columns of $A\_0$. Since $R$ is also Noetherian, $M$ is also finitely generated and has a finite free resolution of length at most $n+1$ (this is proved in [Gamanda, Lombardi, Neuwirth, and Yengui - The syzygy theorem for Bézout rings](https://arxiv.org/abs/1905.08117)).
**Goal:** I want to construct a free resolution of $M$ using the above free resolution of $M'$.
Consider the complex
$$
0 \longrightarrow R^{k\_l} \overset{A\_l}\longrightarrow R^{k\_{l-1}} \overset{A\_{l-1}}\longrightarrow \cdots \overset{A\_1}\longrightarrow R^{k\_0}\overset{A\_0}\longrightarrow R^k~.
$$
**Question 1:** Is this complex exact? In other words, is it a free resolution of $M$?
For example, say $n=2$, and $M'$ is the ideal $(x\_1,x\_2)$ in $R'$. Then, the matrices $A\_0 = \begin{pmatrix} x\_1 & x\_2 \end{pmatrix}$, and $A\_1 = \begin{pmatrix} x\_2 \\ -x\_1 \end{pmatrix}$ give a free resolution of $M'$. In fact, they also give a free resolution of $M$, which is the ideal $(x\_1,x\_2)$ in $R$. I have tried several other examples and it always worked in the same way. (Most of the examples I worked out come from my research in physics. And for several reasons associated with my research, I believe that the answer to the above question is yes.)
**Attempt 1:** To prove that the above complex is exact, I thought I would use the following result of [Buchsbaum and Eisenbud - What makes a complex exact?](https://doi.org/10.1016/0021-8693(73)90044-6) for commutative Noetherian rings. It says the above complex is exact if and only if, for $i=0,\dotsc,l$,
1. $r\_i + r\_{i+1} = k\_i$, where $r\_i = \operatorname{rk} A\_i$ and $A\_{l+1} = 0$, and
2. depth of the ideal $I(A\_i)$ generated by the $r\_i \times r\_i$ minors of $A\_i$ is at least $i+1$ (not $i$, because of the way I indexed the complex).
The first condition is easy because a minor of $A\_i$ is nonzero over $R'$ if and only if it is nonzero over $R$. However, I am stuck at the second condition. Since the first complex over $R'$ is exact, for each $i$, there is an $R'$-regular sequence $(f\_1,\dotsc,f\_{i+1})$ of length $i+1$ in $I(A\_i;R')$ (here, $I(A\_i;R')$ is the ideal generated by $r\_i\times r\_i$ minors of $A\_i$ in $R'$). Without loss of generality, we can assume that each $f\_a$, for $a=1,\dotsc,i+1$, is a polynomial with integer coefficients. Then, it is clear that $(f\_1,\dotsc,f\_{i+1})$ is an $R$-regular sequence, but
**Question 2:** is $(f\_1,\ldots,f\_{i+1})$ an $R$-regular sequence in $I(A\_i;R)$?
While each $f\_a$ is an $R'$-linear combination of $r\_i\times r\_i$ minors of $A\_i$, it may not be an $R$-linear combination. In fact, there may not be any $R$-linear combinations of these minors generating the above regular sequence. This is where I am stuck.
It is clear that there is a least positive integer $m\_a$ such that $m\_a f\_a$ is an $R$-linear combination of the $r\_i \times r\_i$ minors of $A\_i$. If $\gcd(m\_a,m\_b) = 1$ for all $a\ne b$, then $(m\_1 f\_1,\dotsc,m\_{i+1} f\_{i+1})$ is an $R$-regular sequence in $I(A\_i;R)$. However, I am not sure how to show that $\gcd(m\_a,m\_b) = 1$ in general.
I would like to know if the answer to question 1 is known, or if there is a different approach to settle it. I am also interested in any counterexamples (I tried constructing some counterexamples but failed so far). In particular, a counterexample where $\operatorname{im} A\_1 = \ker A\_0$ over $R'$ but $\operatorname{im} A\_1 \subsetneq \ker A\_0$ over $R$ is enough. Note that the columns of $A\_1$ and $A\_0$ should be minimal generating sets with least cardinality of "$\ker A\_0$ over $R'$" and $M'$ respectively.
**Update 1:**
The answer to question 1 is yes when $n\le 1$. This is because, for $n\le 1$, $R'$ is a PID, so any submodule $M'$ of a free module $R'^k$ is also free. Choosing the columns of $A\_0$ to be a basis of $M'$, we have $\ker A\_0 = 0$ over $R'$. Therefore, $\ker A\_0 = 0$ over $R$ as well.
**Attempt 2:** Let $\mu\_S(N)$ denote the infimum of cardinalities of generating sets of $N$, an $S$-module, where $S$ is commutative Noetherian ring. Then, if $\mu\_R(\ker A\_0) = \mu\_{R'}(\ker A\_0)$ for any $A\_0$ defined as above, then by induction, the answer to question 1 is yes. Conversely, a counterexample can be obtained by finding an $A\_0$, defined as above, such that $\mu\_R(\ker A\_0) > \mu\_{R'}(\ker A\_0)$. I have been unsuccessful in constructing such a counterexample so far.
Note that $\operatorname{im} A\_i$ is always a torsion-free module over both $R$ and $R'$ because it is a submodule of a free module. I am not sure if this is helpful.
**Update 2:**
After looking at Aurora's [answer](https://mathoverflow.net/a/420023), which finds a counterexample to question 1, I am modifying the question to the following:
**Question 1':** Given an $A\_0$ associated with a minimal generating set of $M'$ with least cardinality, is it always the case that $\mu\_{R'}(\ker A\_0) = \mu\_R(\ker A\_0)$?
If yes, then there is always a choice of $A\_1$ such that $\operatorname{im} A\_1 = \ker A\_0$ over both $R'$ and $R$ (this is what I mentioned in Attempt 2 above). We can then proceed by induction to show that there is a choice of $A\_i$ for $i>0$ such that the complex is exact over both $R'$ and $R$.
| https://mathoverflow.net/users/149337 | Constructing a free resolution of a $\mathbb Z[x_1,\dotsc,x_n]$-module using a related free resolution of a $\mathbb Q[x_1,\dotsc,x_n]$-module | Let $\mathfrak{a}:=(X\_1-2X\_2,X\_1-2X\_3,X\_1)$ as ideal of $R$. Then the Koszul complex of the mentioned generating set of $\mathfrak{a}$ is not acyclic, because $X\_1-2X\_2,X\_1-2X\_3,X\_1$ is not a regular sequence. However, $X\_1-2X\_2,X\_1-2X\_3,X\_1$ forms a regular sequence in $R'$, thus the Koszul complex $K\_\bullet(\mathbf{a};R)\otimes\_RR'=K\_\bullet(\mathbf{a};R')$ is acyclic. To see the regular sequence property in $R'$ one can compute by Macaulay; or by hand $(X\_1-2X\_2,X\_1-2X\_3,X\_1)R'=(X\_1,X\_2,X\_3)R'$! To see the non-regular sequence property in $R$:
We have $X\_1(X\_2-X\_3)=(X\_1-2X\_2)(-X\_3)+(X\_1-2X\_3)(X\_2)$, thus $X\_2-X\_3\in (X\_1-2X\_2,X\_1-2X\_3):X\_1$, while $X\_2-X\_3\notin (X\_1-2X\_2,X\_1-2X\_3)$. Note that $X\_2-X\_3$ will be in the $2$-generated ideal after inverting $2$.
**For your new question**, set $$M:=\mathbb{Z}[X\_1,X\_2,X\_3]/(X\_1-2X\_2,X\_1-2X\_3,X\_1,X\_2^2,X\_3^2).$$ Then $M$ has projective dimension $4$ over $$R=\mathbb{Z}[X\_1,X\_2,X\_3];$$ because after localizing at $(2,X\_1,X\_2,X\_3)$ the module $M$ has depth $0$ and then apply the Auslander-Buchsbaum Formula as well as [Formula of Pdim and Localization](https://math.stackexchange.com/questions/630564/projective-dimension-and-localization). However, $M\otimes\_RR'=\mathbb{Q}$ has projective dimension 3 over $R'$. Thus it is impossible to obtain a minimal resolution over $R'$ for $R'\otimes\_RM$ (minimal in the sense that the length agrees with the projective dimension), by tensoring a free resolution of $M$ over $R$. It is also impossible to obtain a resolution of $M$ over $R$ from, simply, lifting to $R$ a minimal free resolution of $M\otimes\_RR'$ over $R'$.
| 3 | https://mathoverflow.net/users/127857 | 420023 | 170,938 |
https://mathoverflow.net/questions/419936 | 2 | This is more of a reference request in case anyone can direct me to the right literature. I asked originally on MathStack, but I was suggested to better post it here.
If you have an elliptic curve $E/\mathbb Q$, and you consider the $\mathbb Z\_p$ extension, $\mathbb Q\_{\infty}$, then we know that the rank over $\mathbb Q\_{\infty}$ is finite, which means that there must be a point in the tower that the rank stops growing. I wonder, are there any results that find exactly when this happens? Or can we at least find a number field in the tower above which the rank no longer grows, even if it's not the smallest one?
On a similar flavor, which perhaps may partly answer the above, on his paper "On L functions of elliptic curves and cyclotomic towers", Rohrlich proved the following: If $E/\mathbb Q$ a CM elliptic curve and $P$ a finite set of good primes and $L$ is the compositum of $\mathbb Q(\zeta\_{p^n})$ for all $n\geq 1$ and $p\in P$ then $E(L)$ is finitely generated by findng a number field $M'$ such that $E(L)=E(M')$. In particular, in the case that $P$ consists of a single prime, i.e $L=\mathbb Q(\zeta\_{p^\infty})$, he describes how to explicitly construct $M'$. These results should extend to non CM elliptic curves. So my question is have these results been extended and have these constructions been made precise?
| https://mathoverflow.net/users/124772 | Mordell-Weil rank growth in Iwasawa tower | Thanks to the work of Kato, and his construction of Euler systems for modular forms, this can be extended to all elliptic curves over $\mathbb{Q}$. See [*$p$-adic Hodge theory and values of zeta functions of modular forms*](http://www.numdam.org/item/AST_2004__295__117_0/), Astérisque 295 (2004), Theorem 14.4: for every abelian variety $A/\mathbb{Q}$ which is a quotient of $J\_1(N)$ for some $N \geq 1$, and for every $m \geq 1$, the abelian group $\bigcup\_{n \geq 1} A(\mathbb{Q}(\zeta\_{m^n}))$ is finitely generated.
Rohrlich's analytical result is valid for any newform of weight 2, CM or not, and you can proceed as Rohrlich to find the number field $M'$.
| 2 | https://mathoverflow.net/users/6506 | 420034 | 170,941 |
https://mathoverflow.net/questions/419962 | 7 | Let $f : X\rightarrow Y$ be a proper flat morphism (of schemes) with connected fibers over a smooth projective curve $Y$ over $\mathbb{C}$. Let $X\_{y\_0}$ denote a smooth fiber over $y\_0\in Y$.
If $f$ is smooth, then the topological Euler characteristic is multiplicative: $\chi(X) = \chi(Y)\cdot\chi(X\_{y\_0})$. In general, there is a [formula](http://www.math.lsa.umich.edu/%7Eidolga/sbore72.pdf) for the topological Euler characteristic which includes some correction terms for the singular fibers:
$$\chi(X) = \chi(Y)\chi(X\_{y\_0}) + \sum\_{y\in Y}(\chi(X\_y) - \chi(X\_{y\_0})$$
Does there exist an analogous formula for the holomorphic/algebraic Euler characteristic? I'm in particular interested in the case where $X\rightarrow Y$ is a family of curves over $Y$ (so $\dim X = 2$). To be precise, I'm looking for a formula for $\chi(X,\mathcal{O}\_X)$ in terms of geometric invariants of $Y$ and some fibral data. By fibral data I mean geometric properties which are local on the base $Y$ (e.g. geometric invariants of fibers, monodromy around singular fibers acting on the cohomology of a smooth fiber...etc).
As a nonexample, Riemann-Roch relates this Euler characteristic to the intersection number of $X\_{y\_0}$ with $K - X\_{y\_0}$ (which can be understood locally on $Y$) plus $\chi(X,\mathcal{O}\_X(X\_{y\_0}))$ (which does not appear to be local on $Y$).
**Added in edit:** To state a more precise question -- Is the holomorphic Euler characteristic $\chi(X,\mathcal{O}\_X)$ determined by $Y$ and data which is local on $Y$? Stated another way, do there exist maps $f : X\rightarrow Y$ and $f' : X'\rightarrow Y$ (satisfying the above conditions) such that
1. there is a covering in the etale topology $\{U\_i\rightarrow Y\}\_{i\in I}$ and isomorphisms $\phi\_i : X\_{U\_i}\rightarrow X'\_{U\_i}$, and
2. $\chi(X,\mathcal{O}\_X)\ne \chi(X',\mathcal{O}\_{X'})$.
Here I'm happy to assume that $X,X'$ are smooth over $\mathbb{C}$ (but I still want to allow singular fibers of $f,f'$). If it changes the answer, I'm also interested in the variation where $\{U\_i\}$ is instead an open covering in the analytic topology. I am also interested in results where we assume that $X,X'$ are moreover general type. If the euler characteristic is not determined by $Y$ and data local on $Y$, then I'm also interested in any contraints such data might impose on $\chi(X,\mathcal{O}\_X)$.
| https://mathoverflow.net/users/15242 | Relating the holomorphic Euler characteristic of a family of algebraic varieties to properties of the base and fibers | I'm not entirely sure what would constitute an answer. But here a few simple observations. Let me focus on what you seem be interested in, namely a projective family of connected curves over smooth projective curve $f:X\to Y$. Let me also assume $X$ smooth. Let $h$ be the genus of $Y$, and $g$ the genus of the general fibre. Then from the Leray spectral sequence, Riemann-Roch and Grothendieck duality
$$\chi(\mathcal{O}\_X)= (1-g)(1-h) - \deg R^1f\_\*\mathcal{O}\_X=\chi(\mathcal{O}\_Y)\chi(\mathcal{O}\_{X\_{y\_0}})+\deg f\_\*\omega\_{X/Y}$$
A theorem of Fujita then gives an inequality
$$\chi(\mathcal{O}\_X)\ge \chi(\mathcal{O}\_Y)\chi(\mathcal{O}\_{X\_{y\_0}})$$
which is something.
To say more, one would need to compute $\deg f\_\*\omega\_{X/Y}$. But I think this depends on more than local information at the bad fibres, which is what I think you are asking. The reason I say this, is that this degree can be positive and variable for Kodaira surfaces, which have no bad fibres at all. [**Added Comment** I guess there are various things called Kodaira surfaces. The ones I have in mind are of general type, and hence algebraic. See page 220 of *Compact Complex Surfaces* by Barth, Hulek, Peters and Van de Ven for further details.]
| 5 | https://mathoverflow.net/users/4144 | 420042 | 170,943 |
https://mathoverflow.net/questions/420046 | 4 | Let $f \in \mathbb{F}\_q[x\_1, \dots, x\_k]$ be a polynomial with $\deg f = n$, and let $\chi$ be a multiplicative character over $\mathbb{F}\_q$.
Is there any known bound, possibly with conditions about $f$ and $\chi$, for
$$\left|\sum\_{c\_1, \dots, c\_k \in \mathbb{F}\_q} \chi(f(c\_1, \dots, c\_k)) \right| ?$$
| https://mathoverflow.net/users/269936 | Bound for sum of multiplicative character calculated over multivariate polynomial | Yes, there are several known bounds. The following statement, due to Katz, has quite strict conditions on $f$, but gives a very strong result. It is perhaps the simplest statement that gives such a strong bound.
Suppose that the equation $f=0$ defines a nonsingular hypersurface in $\mathbb A^k$, and the degree $n$ part of $f$ defines a nonsingular hypersurface in $\mathbb P^{k-1}$. Suppose also that either $n$ is prime to $q$ or $\chi^n$ is trivial.
Then $$\left|\sum\_{c\_1, \dots, c\_k \in \mathbb{F}\_q} \chi(f(c\_1, \dots, c\_k)) \right| \leq (n-1)^k q^{k/2} $$
This is the main theorem of [Estimates for nonsingular multiplicative character sums](https://doi.org/10.1155/S1073792802106088) by Nick Katz.
| 7 | https://mathoverflow.net/users/18060 | 420052 | 170,946 |
https://mathoverflow.net/questions/420050 | 5 | Consider the set
$$\\\{ (A,B) \in \mathbb{P}^{n\times n-1} \times \mathbb{P}^{n\times n -1} : \text{im}(A) \subseteq \text{im}(B)\}.$$
That is, this is the set of pairs of square matrices $(A,B)$ so that the image of $A$ is contained in the image of $B$. Is this Zariski closed? I would be happy if this were at least true over an algebraically closed field.
I tried to write down explicit equations for this, but my first attempt would involved using determinants of submatrices of $A$, and would fail if $A$ was singular. I had thought that this would follow from the projection theorem for for projective varieties, but am unsure.
| https://mathoverflow.net/users/165301 | Is containment of images of linear maps Zariski closed? | I am making my comment an answer. The specified set is not Zariski closed. If it were, then its intersection with every Zariski closed subset $C$ would be relatively closed in $C$. But now let $C$ be the curve, a copy of the affine line, where the first component $A$ is held fixed as the identity $n\times n$ matrix, and the second component is a varying diagonal matrix whose first $n-1$ diagonal entries all equal $1$, yet whose last entry $t$ varies in a copy of the affine line. The intersection of $C$ with the specified set is a non-closed, dense Zariski open in $C$, namely the open subset where $t$ is invertible.
| 12 | https://mathoverflow.net/users/13265 | 420054 | 170,947 |
https://mathoverflow.net/questions/420057 | 5 | I am having some trouble trying to understand the proof of Theorem 7.2.5 in Bhatt and Scholze's paper [The pro-étale topology for schemes](https://arxiv.org/abs/1309.1198). Specifically, I don't quite understand why it was necessary to prove that $F: C \to \mathit{Sets}$ preserves connectedness of objects, and how viewing morphisms $f: X \to Y$ in $C$ as monomorphisms $\Gamma\_f: X \to X \times Y$ can help us prove that $F: C \to \pi\_1(C, F)\text-\mathit{Sets}$ is fully faithful (I could only show that $F: C \to \mathit{Sets}$ is fully faithful).
Any help is very much appreciated!
| https://mathoverflow.net/users/143390 | Infinite Galois equivalence | When $X$ is connected, we see that any graph $\Gamma\_f$ for $f \colon X \to Y$ is connected as well, so we get a bijection
\begin{align\*}
\mathscr C(X,Y) &\to \left\{\Gamma \subseteq X \times Y \text{ connected}\ \bigg|\ \Gamma \underset{\pi\_1}{\overset{\sim}\to} X \right\} \\
f &\mapsto \Gamma\_f.
\end{align\*}
The same holds in $G\text{-}\mathbf{Set}$ for any Noohi group $G$. Writing $G = \pi\_1(\mathscr C,F)$ for simplicity, we see that $\mathscr C(X,Y) \to \operatorname{Hom}\_G(FX,FY)$ is a bijection for $X$ connected because $F \colon \mathscr C \to G\text{-}\mathbf{Set}$ preserves connected components (and products, etc).
Now I suppose the general result follows from axiom (2) or (3) of infinite Galois categories (even though it's not completely spelled out what (3) means). But at the very least we know that any object $X$ is a coproduct $\coprod\_{i \in I} X\_i$ of connected objects $X\_i$. Since $F$ preserves colimits, we get
$$\begin{array}{ccccc}
\mathscr C(X,Y) & \cong & \mathscr C\Big(\coprod\limits\_{i \in I} X\_i,Y\Big) & \cong & \prod\limits\_{i \in I} \mathscr C\big(X\_i,Y\big)\\ & & & & \downarrow\wr\!\! \\ \operatorname{Hom}\_G(FX,FY) & \cong & \operatorname{Hom}\_G\bigg(\coprod\limits\_{i \in I} FX\_i,FY\bigg) & \cong & \prod\limits\_{i \in I} \operatorname{Hom}\_G\big(FX\_i,FY\big)
\end{array}$$
by the connected case presented above. $\square$
This type of argument might be considered 'standard' when dealing with Galois categories (see for instance [Tag [0BN0](https://stacks.math.columbia.edu/tag/0BN0) (7)]), which would explain why no details were given.
| 6 | https://mathoverflow.net/users/82179 | 420060 | 170,949 |
https://mathoverflow.net/questions/420000 | 2 | Let $d>1$ be not a square. Then the continued fraction expansion of $\sqrt d$ is $[a\_0; \overline{a\_1,\dots,a\_\ell}]$, where $a\_0=\lfloor \sqrt d\rfloor$ and $a\_\ell=2a\_0$.
Thus, $\ell=\ell(d)$.
About 30 years ago I heard a talk where $\ell(d)$ was somehow related to the class number of $\mathbb Q(d)$ -- in particular, to [the Gauss class number problem](https://en.wikipedia.org/wiki/Class_number_problem) which conjectures that there are infinitely many natural $d$ for which class number is 1.
Unfortunately, I don't remember how to restate this conjecture via $\ell(d)$. Any reference would be helpful.
| https://mathoverflow.net/users/8131 | Period of continued fraction expansion and class number | Claude Levesque, [On semi-reduced quadratic forms, continued fractions and class number](https://www.kurims.kyoto-u.ac.jp/%7Ekyodo/kokyuroku/contents/pdf/0998-7.pdf), quotes a theorem of Lu, H., On the class number of real quadratic fields, Scientia Sinica II (special number, 1979), 118-130, as follows:
Let $m>1$ be a squarefree integer. Then the class number of ${\bf Q}(\sqrt m)$ is one if and only if $$\theta+\sum\_{i=1}^{\ell}k\_i=\lambda\_1(m)+\lambda\_2(m)$$ where $\omega=(1+\sqrt m)/2$ if $m\equiv1\bmod4$, otherwise $\omega=\sqrt m$, the continued fraction for $\omega$ is $[k\_0,\overline{k\_1,\dots,k\_{\ell}}]$, $\theta$ is zero, one, or two depending on $m\bmod4$ and the parity of $\ell$ and $k\_{\ell/2}$, and $\lambda\_1(m)$ (respectively, $\lambda\_2(m)$) is the number of solutions in nonnegative integers of $x^2+4yz=\Delta$ (respectively, $x^2+4y^2=\Delta$), where $\Delta$ is $m$ if $m\equiv1\bmod4$, otherwise $4m$.
For the detailed definition of $\theta$, see the paper of Levesque. The paper goes on to prove related results.
Another paper that may be relevant is Louboutin, Mollin, & Williams, Class Numbers of Real Quadratic Fields, Continued Fractions, Reduced Ideals, Prime-Producing Quadratic Polynomials and Quadratic Residue Covers, Canadian Journal of Mathematics 44 (1992) 824-842. DOI:10.4153/CJM-1992-049-0.
| 2 | https://mathoverflow.net/users/3684 | 420061 | 170,950 |
https://mathoverflow.net/questions/368496 | 4 | Let $H$ be an infinite dimensional seperable Hilbert space. Is there an Irreducible involutive sub algebra $D$ of $B(H)$ with the following properties?:
1)For every open set $U\subset H$ and every Frechet differential map
$f:U \to H$ with $Df(x)\in D,\; \forall x\in U$, the mapping $f$ is automatically $C^{\infty}$
2)For every open set $U\subset H$, a uniform limit $f$ of a sequence of Frechet differential maps $f\_n:U \to H$ with $Df\_n(x)\in D$ is Frechet differentiable which satisfies $Df(x)\in D,\;\forall x\in U$.
The motivation comes from the concepet of holomorphic functions when we put (the real analogy) $ H=\mathbb{R}^2$ and $D=\left \{\begin{pmatrix} a&b\\ -b&a\end{pmatrix} \mid a,b\in \mathbb{R} \right \}$
| https://mathoverflow.net/users/36688 | A kind of holomorphicity of maps on Hilbert space | 1. In [this book](https://www.mat.univie.ac.at/%7Emichor/apbookh-ams.pdf), Thm 7.19 (1) $\Longleftrightarrow$ (9) answers this for $D$ the complex linear bounded maps, but you have to assume that $f$ is Gateaux differentiable with the derivative locally Lipschitz.
2. From Thm 7.17 (7) you can deduce an answer, but you have to adapt the definitions and strengthen the convergence at least at one point.
There is also the book:
* Dineen, Seán
Complex analysis in locally convex spaces. (English) Zbl 0484.46044
North-Holland Mathematics Studies, 57. Notas de Matematica (83). Amsterdam - New York - Oxford: North-Holland Publishing Company. XIII, 492 p.
| 1 | https://mathoverflow.net/users/26935 | 420066 | 170,953 |
https://mathoverflow.net/questions/420089 | 2 | Whether a complete non-compact non-flat Riemannian $n$-manifold $M$ with non-negative sectional curvature has Euclidean volume growth?
That is, whether there is a constant $C>0$ such that $\mathrm{Vol}(B\_x(r))\geq Cr^n$ for all $r>0$ and $x\in M$? Here $\mathrm{Vol}(B\_x(r))$ is the volume of the $r$-ball in $M$.
Since manifolds with non-negative Ricci curvature and Euclidean volume growth are studied a lot, I am curious about the non-negative sectional curvature case.
| https://mathoverflow.net/users/90512 | Non-negatively curved manifolds and the volume of balls | It is certainly not true that every complete nonflat open manifold of nonnegative curvature has Euclidean volume growth. Counterexamples are trivial to construct. Say, a capped cylinder. More generally any nonnegatively curved manifold $M^n$ with nontrivial soul has slower than Euclidean volume growth. Because its asymptotic cone at infinity has dimension strictly smaller than $n$ if the soul is not a point while manifolds with Euclidean volume growth have asymptotic cones of dimension $n$.
This also implies that any nonnegatively curved manifold with Euclidean volume growth is diffeomorphic to $\mathbb R^n$.
| 6 | https://mathoverflow.net/users/18050 | 420091 | 170,957 |
https://mathoverflow.net/questions/420088 | 14 | $\DeclareMathOperator\Aut{Aut}\newcommand\card[1]{\lvert#1\rvert}$So, after going over the classification of finite abelian groups in a class I was teaching this winter, I got curious about whether it could be used to obtain a 'nice' value for the [groupoid cardinality](https://ncatlab.org/nlab/show/groupoid+cardinality) of the class of abelian groups of fixed size, that is if one could compute :
$$ \sum\_{\card G=m} \frac{1}{\card{\Aut(G)}} $$
where the sum is over the isomorphism class of abelian groups with $m$ elements. That is just pure curiosity, I had absolutely no real motivation for this. One relatively quickly see that it is enough to compute it for $m=p^n$, so I started doing it for $p$, $p^2$ and $p^3$. Unsurprisingly, the results look messy at first, but then a quite surprising number of simplifications occured and I arrived at the formula :
$$ \sum\_{\card G=p^n} \frac{1}{\card{\Aut(G)}} = \prod\_{k=1}^n \frac{p^{k-1}}{p^k-1}. $$
I checked it by hand up to $n=5$.
In any case the simplicity of the formula suggests that there is a better way to compute this than going over all isomorphism class of abelian groups and computing $\Aut(G)$ and then taking the sum.
So basically I'm wondering if this formula is known, and if not if someone has an idea of how to prove it — ideally with an argument that doesn't involve summing over partitions of $n$.
Note : there are known formulæ for $\card{\Aut(G)}$ see for example the very end of [Hillar and Rhea - Automorphisms of finite Abelian groups](https://arxiv.org/abs/math/0605185).
| https://mathoverflow.net/users/22131 | Groupoid cardinality of the class of abelian p-groups | This is Corollary 3.8 in Cohen, H.; Lenstra, H. W., Jr. Heuristics on class groups. Number theory (New York, 1982), 26–36, Lecture Notes in Math., 1052, Springer, Berlin, 1984 (MR0750661).
In this paper you will find many formulas for averages of various functions over abelian groups weighted by inverse size of the automorphism group.
| 19 | https://mathoverflow.net/users/40821 | 420096 | 170,959 |
https://mathoverflow.net/questions/420093 | 2 | My question is:
>
> Is every Polish space image of a closed and continuous mapping with domain $\Bbb{N}^\Bbb{N}$?
>
>
>
Where a Polish space is a separable and completely metrizable space and where $\Bbb{N}^\Bbb{N}$ (aka Baire space) is the space of infinite sequences of natural numbers endowed with the product of the discrete topology over $\Bbb{N}$.
I know that every Polish space is image of a continuous bijection with domain a closed subset of $\Bbb{N}^\Bbb{N}$ (hence, in particular, it is a continuous image of $\Bbb{N}^\Bbb{N}$). But what happens if we require the mapping to be closed?
Moreover, in case the answer to the above question is "no", is there a counterexample? Can we characterize the Polish spaces that are images of a closed and continuous mapping with domain $\Bbb{N}^\Bbb{N}$?
Hints? Ideas?
Thanks!
| https://mathoverflow.net/users/141146 | Images of a closed and continuous mapping with domain $\Bbb{N}^\Bbb{N}$ | This question is answered positively in [On closed images of the space of irrationals](https://www.semanticscholar.org/paper/On-closed-images-of-the-space-of-irrationals-Engelking/726a02a1075858b78aa322f1dfb504e858d39cca), by Engelking.
| 3 | https://mathoverflow.net/users/172802 | 420099 | 170,960 |
https://mathoverflow.net/questions/420036 | 4 | Let $\pi(x;q,a)$ count the number of primes $\leq x$ congruent to $a$ mod $q$. The Brun-Titchmarsh Theorem states that for all $q< x$, $(a,q)=1$, we have
$$
\tag{1}
\pi(x;q,a) \leq \frac{2x}{\varphi(q)\log(x/q)}.
$$
Let $f(n) = 1$ if $n$ is prime and $0$ otherwise. Then we can rephrase $(1)$ as (almost) saying that
$$
\tag{2}
\sum\_{\substack{n\leq x\\ n\equiv a (q)}} f(n) \leq \frac{2}{\varphi(q)} \sum\_{\substack{n\leq x\\ (n,q)=1}} f(n).
$$
I'm curious if inequalities like these have been studied for other arithmetic functions $f$. There is a vast literature proving asymptotic formulas, i.e. things like
$$
\tag{3}
\sum\_{\substack{n\leq x\\ n\equiv a (q)}} f(n) \sim \frac{1}{\varphi(q)} \sum\_{\substack{n\leq x\\ (n,q)=1}} f(n),
$$
for various specific arithmetic functions (e.g. the divisor function or the indicator function of squarefree numbers or smooth numbers). However, I have not found much literature on inequalities like $(2)$. Thus my question:
**Are there any results that prove inequalities like $(2)$ for arithmetic functions other than the prime-indicator function?**
Any references and comments are most appreciated. Some thoughts and remarks about this general kind of problem:
* The fact that $(1)$ has $x/q$ instead of $x$ inside the $\log$ is not a mere technicality; it represents a deep barrier to improving the inequality. By analogy, inequalities like $(2)$ might only be provable in a slightly weaker form.
* Typically, one is more concerned with the range of validity and amount of uniformity of formulas of the type $(3)$, rather than the quality of the error terms (though these are certainly important as well). Given that the full conjectured range of validity and uniformity for formulas like $(3)$ are seldom known, inequalities like $(2)$ seem like an interesting avenue of research, as such inequalities (by analogy with $(1)$) may hold in wider ranges.
* My original motivation for this was thinking about smooth numbers in arithmetic progressions. In that case, the constant $2$ arises from thinking about what might happen if Vinogradov's conjecture on the least quadratic non-residue modulo a prime was false (basically, if $y$ is small enough, every $y$-smooth number is a quadratic residue, and these would (conjecturally) equidistribute in the $\varphi(p)/2$ available residue classes).
* Equidistribution results like $(3)$ often require complex/harmonic-analytic tools, such as the distribution of zeros of $L$-functions or estimates for Kloosterman sums (e.g. in the case of the divisor function). However, the Brun-Titchmarsh theorem (in its original form) uses only elementary sieve theory. If inequalities like $(2)$ can be proved without the use of such "heavy machinery," this would be another reason why they are interesting to study.
| https://mathoverflow.net/users/307675 | Generalizations of the Brun-Titchmarsh theorem | Yes, there are several results answering the question you ask about, in great generality.
1. Linnik, in his monograph on the dispersion method, proved an analogue of (2) for the $k$-th divisor function. Concretely,
$$\sum\_{\substack{n \le x \\ n \equiv a \bmod q}} d\_k(n) \ll \frac{x}{q} \left(\frac{\phi^{r-1}(q)}{q^{r-1}}\log x \right)$$
uniformly for $q \le x^{1-\varepsilon}$.
2. This was generalized by Peter Shiu (a student of Halberstam) in his Crelle paper ''A Brun-Titchmarsh theorem for multiplicative functions'' (J. Reine Angew. Math. 313 (1980), 161–170) to a wide class of multiplicative functions. Here the bound takes the form
$$\sum\_{\substack{n \le x \\ n \equiv a \bmod q}} f(n) \ll \frac{x}{q} \frac{1}{\log x}\exp\left( \sum\_{\substack{p \le x \\ p \nmid q}} f(p) \right)$$
uniformly for $q\le x^{1-\varepsilon}$, with an implied constant depending only on $f$ and $\varepsilon$. In fact, Shiu's result also allows you to add the additional restriction $n \in [x-y,x]$. Here $\frac{1}{\log x}\exp\left( \sum\_{\substack{p \le x \\ p \nmid q}} f(p) \right)$ arises as the mean value of $\alpha$ over integers coprime to $q$ (up to a multiplicative constant bounded away from $0$ and $\infty$), by e.g. applying Wirsing's theorem to $f(n) \cdot 1\_{(n,q)=1}$.
3. A further generalization was given by M. Nair and G. Tenenbaum in ''Short sums of certain arithmetic functions'' (Acta Math. 180 (1998), no. 1, 119–144), building on earlier work of Nair. In particular, they show that 'multiplicative' can be replaced by (a weak notion of) 'submultiplicaitve', dependence of the constant on $f$ can be made more specific, and the growth condition imposed by Shiu (which I did not mention) can be considerably weakened. Moreover, one can obtain similar bounds for (sub)multiplicative functions in *several variables* evaluated on *polynomial* arguments. This generalization leads to new results. A. Hildebrand, in his MathSciNet review calls this generalization 'far-reaching' and 'definitive'.
4. As far as I know, no author has tried to optimize the constant in the estimate (2) in the multiplicative setting.
| 3 | https://mathoverflow.net/users/31469 | 420100 | 170,961 |
https://mathoverflow.net/questions/420103 | 9 | I had been pointed to Ramanujan's 1912 article *Note on a set of simultaneous equations* in this [answer](https://mathoverflow.net/a/357941/31310) to my former question about the [Solvability of a system of polynomial equations](https://mathoverflow.net/questions/357138/solvability-of-a-system-of-polynomial-equations).
While the contents of the paper seemed a bit enigmatic at first reading, I'm now convinced that Ramanujan not only left it to the reader to recover the ideas that had led to the solution but also the motivation to investigate on that special system of polynomial equations.
>
> **Questions:**
>
>
> * is anything known about Ramanujan's motivation to work on that problem that somehow doesn't seem to fit his primary research interest; the situation seems analogous to Euler and his characteristic of polyhedra?
> * did Ramanujan himself ever refer to the paper in his later work?
> * is or was the paper of importance for mathematical research and what are remarkable examples?
> * have there been serious attempts to extend the solution method and/or to reduce more general types of systems of polynomial equations to the special case that Ramanujan has solved?
>
>
>
| https://mathoverflow.net/users/31310 | Impact of Ramanujan's Note on a set of simultaneous equations | *This refers to the third and fourth question in the OP.*
Ramanujan's 1912 paper addresses a problem similar to that considered by Sylvester in 1851 [1]. The method to solve the set of algebraic equations
$$\sum\_{k=1}^{n}x\_kz\_k^j=a\_j,\;\;0\leq j\leq 2n-1$$
in the unknown $x\_k$'s and $z\_k$'s is different in the two papers, but the final algorithm is similar.
This set of equations shows up in the socalled moment problem, to find a probability measure given its moments. That connection is explored in [The Sylvester-Ramanujan System of Equations and The Complex Power Moment Problem](https://doi.org/10.1023/B:RAMA.0000027196.19661.b7) by Yuri Lyubich.
[1] J.J. Sylvester, [On a remarkable discovery in the theory of canonical forms and of hyperdeterminants,](https://rcin.org.pl/dlibra/publication/139207/edition/114075/content) Phil. Magazine **2**, 391–410 (1851).
| 11 | https://mathoverflow.net/users/11260 | 420105 | 170,963 |
https://mathoverflow.net/questions/420107 | 7 | Let $A$ be a finitely generated $\mathbb{Z}$-algebra and let
$f: \operatorname{Spec} A \rightarrow \operatorname{Spec} \mathbb{Z}$ be the canonical map.
On pg. 53, Thm. 8.2 of <https://www.math.uni-bonn.de/people/scholze/Condensed.pdf>
one defines a functor $$f\_!: D(A\_\blacksquare) \rightarrow D(\mathbb{Z}\_\blacksquare)$$ where $A\_\blacksquare$ and $\mathbb{Z}\_\blacksquare$ denotes certain categories of solid modules.
On the same page, Scholze remarks that $f\_!$ does not preserve discrete objects in general?
**Is there an easy way to see that $f\_!$ can not possibly preserve discrete objects for general morphisms $f?$**
| https://mathoverflow.net/users/472750 | What is the reason for $f_!$ not preserving discrete objects? | You can see this by a direct calculation, for example in the most basic case $A=\mathbb{Z}[T]$. When you apply $f\_!$ to $A=\mathbb{Z}[T]$ itself, you get the object represented by the two-term complex
$$\mathbb{Z}[T]\to \mathbb{Z}((T^{-1})).$$
with $\mathbb{Z}[T]$ in degree $0$. (This is the "compactly supported cohomology of the structure sheaf" on $\mathbb{A}^1$; the Laurent series in $T^{-1}$ represents "functions defined in a neighborhood of $\infty$").
The map in this complex is injective, and the cokernel is $T^{-1}\mathbb{Z}[[T]]$, which is not discrete. In fact, $f\_!A$ is *predual* to an infinite discrete object, namely the usual algebraic Grothendieck dualizing complex of f. This is a general feature of the situation, and follows from the adjunction between $f\_!$ and $f^!$. Actually, $f\_!$ has a complementary property to preservation of discreteness: it preserves pseudocoherent objects.
| 6 | https://mathoverflow.net/users/480363 | 420139 | 170,977 |
https://mathoverflow.net/questions/420140 | 1 | Let $K$ be a number field and $\mathcal{O}\_K$ be its ring of integers. For any prime ideal $\mathfrak{p}$ in $\mathcal{O}\_K$ is it true that every residue class in $\mathcal{O}\_K/\mathfrak{p}$ contains an integer? I can prove that it is true if $\mathfrak{p}$ is unramified and has inertial degree 1, but not for general prime ideals. I kindly request your answers to this problem.
| https://mathoverflow.net/users/480366 | Integers in residue classes $\mathcal{O}_K/\mathfrak{p}$ | Let $p$ be the prime number that satisfies $p\mathbb{Z} = \mathbb{Z} \cap \mathfrak{p}$.
Then your claim is equivalent to the inclusion $\mathbb{F}\_p\subset \mathcal{O}\_K/\mathfrak{p}$ being an equality, in other words the prime $\mathfrak{p}$ having inertial degree $1$. So it is not true in general, for example it is not true for the prime ideal $3\mathcal{O}\_K$ if $K=\mathbb{Q}(i)$.
| 3 | https://mathoverflow.net/users/110362 | 420141 | 170,978 |
https://mathoverflow.net/questions/420084 | 5 | Let $X$ be an infinite discrete topological space. Is $$C\_b(X)=\{ f \colon X \to \mathbb{R} \text{ bounded }\}$$ a Jacobson ring ?
| https://mathoverflow.net/users/113200 | Ring of continuous functions is a Jacobson ring | First, let us consider the question of when the ring $C(X)$ of *all* continuous real-valued functions on a topological space $X$ (not necessarily discrete) is Jacobson, keeping in mind that $C\_b(X) = C(\beta X)$ where $\beta X$ is the Stone-Čech compactification.
In fact, every prime ideal of $C(X)$ is contained in a *unique* maximal ideal: see Gillman & Jerison, *Rings of Continuous Functions* (1960) theorem 7.15 on page 107. So $C(X)$ is a Jacobson ring iff every prime ideal is maximal, meaning that $X$ is a P-space, a rather strong condition which, [see the reference in this other answer](https://mathoverflow.net/questions/266808/when-c-x-is-zero-dimensional/266820#266820), is equivalent to the condition “every $f\in C(X)$ which vanishes at $p$ vanishes in some neighborhood of $p$“.
(Note for example that $C([0,1]) = C\_b([0,1])$ is *not* a Jacobson ring: the ring of functions which vanish in some neighborhood of $0$, while not prime itself, is contained in a prime ideal, which itself is contained in a a unique maximal ideal, namely that of functions which vanish at $0$.)
Now if $X$ is infinite discrete, it is indeed a P-space, so $C(X)$ is a Jacobson ring. However, $\beta X$ is *not* a P-space, as every compact P-space is finite (Gillman & Jerison, problem 4K.1 on page 63, and remarks following theorem 14.29 on page 212). So $C\_b(X) = C(\beta X)$ is *not* a Jacobson ring.
| 6 | https://mathoverflow.net/users/17064 | 420149 | 170,982 |
https://mathoverflow.net/questions/420156 | 1 | **Question:**
is there an established name for the set $\Big\lbrace\ f {\Large\ \boldsymbol{|}}\ f\in C^\infty\quad {\Large\boldsymbol{\land}}\quad \exists\,{k\in\mathbb{N}^+}:\frac{d^{i+k}}{dx^{i+k}}f(x)=\frac{d^{i}}{dx^{i}}f(x)\ \forall i\in\mathbb{N}^+ \Big\rbrace{\Large\text{?}}$
E.g. $f(x)=e^x\implies k=1;\ f(x)=\cos(x)\,\lor\,f(x)=\sin(x)\implies k=4$.
**Addendum:**
As there doesn't seem to be an established name already, I'm tempted to call that class of functions "**deri**vative-peri**odic**" or *deriodic* for short by means of inventing yet another [portmanteau](https://en.wikipedia.org/wiki/Portmanteau).
These *deriodic* functions can be viewed as generalizing the rationals to functions if the derivatives are interpreted as generalized digits.
These *deriodic* functions resemble good local approximations to smooth functions: $n$ *deriodic*s suffice to approximate a function's value and the first $2n-1$ of its derivatives for a given $x\_0$.
| https://mathoverflow.net/users/31310 | Functions with periodic sequence of derivative-values | Assuming that $\mathbb N^+$ is defined as $\{1,2,\dots\}$, the set of functions in questions is just the set of all solutions $f$ of all the simple linear ODE's $f^{(k+1)}=f'$ with $k\in\mathbb N^+$, that is, the set of all functions $f$ such that
$$f(x)=a+\sum\_{j=0}^{k-1} c\_{k,j}\exp\{e^{2\pi ij/k}x\}$$
for some $k\in\mathbb N^+$, some complex $a$ and $c\_{k,j}$'s, and all $x$.
(Whether this set of functions has a name I do not know.)
| 5 | https://mathoverflow.net/users/36721 | 420159 | 170,985 |
https://mathoverflow.net/questions/419734 | 4 | Let $X$ be an isolated, Gorenstein singularity of dimension at least $2$ and $\pi: \widetilde{X} \to X$ be a resolution of singularities. Let $E$ be the exceptional divisor and $E\_1,...,E\_r$ be the irreducible components of $E$. Can there exist integers $a\_1,...,a\_r$ not all zero such that $\mathcal{O}\_{\widetilde{X}}(\sum\_i a\_iE\_i) \cong \mathcal{O}\_{\widetilde{X}}$?
| https://mathoverflow.net/users/32151 | Do there exist linear relations between exceptional divisors | I am just posting my comment as an answer. For a quasi-compact, separated, irreducible, smooth $2$-dimensional algebraic space $\widetilde{Y}$ over an algebraically closed field $k$, for a connected, effective Cartier divisor $D$ that is a proper scheme and with irreducible components $(D\_i)\_{i=1,\dots,r}$, there exists a contraction $$f:\widetilde{Y}\to Y,$$ that maps $D$ to a $k$-point $p$ and that is an isomorphism from $\widetilde{Y}\setminus D$ to $Y\setminus\{p\}$ if and only if the following $r\times r$ intersection matrix is negative definite, $$(n\_{i,j})\_{1\leq i,j\leq r}, \ \ n\_{i,j} = (D\_i\cdot D\_j)\_{\widetilde{Y}}.$$
If $k=\mathbb{C}$ this is originally due to Hans Grauert.
MR0137127
Grauert, Hans
Über Modifikationen und exzeptionelle analytische Mengen.
Math. Ann. 146 (1962), 331–368.
In arbitrary characteristic, this is due to Michael Artin.
MR0146182
Artin, Michael
Some numerical criteria for contractability of curves on algebraic surfaces.
Amer. J. Math. 84 (1962), 485–496.
For $\widetilde{X}$ a quasi-projective, smooth, irreducible $k$-scheme of dimension $n\geq
2$, for a connected, effective Cartier divisor $E$ that is a proper scheme and with irreducible components $(E\_i)\_{i=1,\dots,r}$, for a surface $\widetilde{Y}$ in $X$ that is a complete intersection surface of $n-2$ sufficiently general and sufficiently ample divisors, the surface $\widetilde{Y}$ is a quasi-projective, smooth, irreducible $k$-scheme, the divisor $D=\widetilde{Y}\cap E$ in $\widetilde{Y}$ is a connected, effective Cartier divisor, and every divisor $D\_i = \widetilde{Y}\cap E\_i$ in $\widetilde{Y}\_i$ is an irreducible divisor.
If there exists a contraction of $\widetilde{X}$ to an algebraic space that contracts $E$ to a point $p$ and that is an isomorphism outside of $E$, resp. outside of this point, then the restriction of the contraction to $\widetilde{Y}$ is such a contraction. Then the intersection matrix is negative definite. In particular, since the rows (or equivalently, the columns) are linearly independent, it follows that the Cartier divisor classes $[D\_1],\dots,[D\_r]$ in the Néron-Severi group of $\widetilde{Y}$ are linearly independent. Since these are the pullbacks of the Cartier divisor classes $[E\_1],\dots,[E\_r]$ in the Néron-Severi group of $\widetilde{X}$, these classes are also $\mathbb{Z}$-linearly independent.
Finally, even when $\widetilde{X}$ is only quasi-compact, separated, and finitely presented, by Chow's Lemma, there exists a projective morphism,
$$ \rho:\widehat{X}\to \widetilde{X},$$ such that $\widehat{X}$ is quasi-projective. A general complete intersection surface in $\widehat{X}$ need not be smooth. However, we have resolution of singularities for $2$-dimensional schemes over an algebraically closed field, thus we can find a morphism $\widehat{Y}\to \widehat{X}$ playing the same role as the smooth surface above, where $\widehat{Y}$ is a connected, quasi-projective, smooth surface. If there is a nontrivial linear relation among the Cartier divisor classes $[E\_i]$ in $\widetilde{X}$, then this pulls back to a nontrivial linear relation among the pullback Cartier divisor classes on $\widehat{Y}$. By the argument above, the irreducible components of the exceptional locus on $\widehat{Y}$ are $\mathbb{Z}$-linearly independent. So, as above, the Cartier divisor classes $[E\_i]$ in the Néron-Severi group of $\widetilde{X}$ are also $\mathbb{Z}$-linearly independent.
| 2 | https://mathoverflow.net/users/13265 | 420161 | 170,986 |
https://mathoverflow.net/questions/418565 | 7 | This question was also asked [here](https://math.stackexchange.com/questions/4403054/tangent-bundle-of-a-tensor-product-bundle) on math-stackexchange.
Let $E\to M$ and $F\to M$ be vector bundles. The structure of their tangents $TE$ and $TF$ is well known. In particular, connectors map $K\_E: TE \to E\times\_M E$ and $K\_F: TF \to F\times\_M F$ induce isomorphisms $TE \simeq E \times\_M (E\oplus TM)$ and $TF \simeq F \times\_M (F\oplus TM)$ (I am using here a fibered product notation, rather than the equivalent pullbacks).
Consider now the vector bundle $E\otimes F\to M$. Its tangent can be characterized in the same way as the tangent bundle of every vector bundle. My question is whether there exists a canonical isomorphism of $T(E\otimes F)$ involving the tangent bundles $TE$ and $TF$. I would suspect a positive answer, which in particular looks like a "Leibniz rule". Also, what is the connector $K\_{E\otimes F}$ induced by $K\_E$ and $K\_F$? After all, covariant derivative of tensor products satisfy a Leibniz rule.
Here is yet an alternative formulation. How does the choice of horizontal bundles for $TE$ and $TF$ determine a horizontal bundle for $T(E\otimes F)$?
The same question can be asked regarding the vector bundle $\operatorname{Hom}(E,F)$.
In fact, the question seems reducible to the case of $M$ being a point, so that $E$ and $F$ are just vector spaces. Then, $TE \simeq E\times E$ and $TF \simeq F\times F$. If $E$ has dimension $m$ and $F$ has dimension $k$, then the tensor product $TE\otimes TF$ has dimension $4mk$, which is twice the dimension of $T(E\otimes F)$. How to proceed from here?
| https://mathoverflow.net/users/98733 | Tangent bundle of a tensor product bundle | Over a point:
$$
T(E\otimes F) = (E\otimes F)\oplus (E\otimes F) = E\otimes (F\oplus F)
$$
which is naturally isomorphic to $(E\oplus E)\otimes F$ using the canonical flip $E\otimes F = F\otimes E$. Likewise
$$
T\operatorname{Hom}(E,F) = \operatorname{Hom}(E,F)\oplus \operatorname{Hom}(E,F)=\operatorname{Hom}(E,TF)
$$
The Leibniz rule mixes the two representations: Consider first curve
$\sum\_i e\_i(t)\otimes f\_j(t)$; its velocity at $t=0$ is then
$$\Big(\sum\_i e\_i(0)\otimes f\_j(0), \sum\_i e\_i'(0)\otimes f\_j(0) + \sum\_i e\_i(0)\otimes f\_j'(0)\big).$$
Counting entries you have $2mk$.
It is more clear to consider a curve in terms of bases
$$
\sum\_{i,j} c\_{ij}(t)\; e\_i\otimes f\_j =
\sum\_j\Big(\sum\_{i} c\_{ij}(t)\; e\_i\Big)\otimes f\_j
= \sum\_{i} e\_i\otimes \Big(\sum\_jc\_{ij}(t)\;f\_j \Big),
$$
then its derivate via (footpoint, speed vector) is
$\Big(\sum\_{i,j} c\_{ij}(0)\; e\_i\otimes f\_j, \sum\_{i,j} c\_{ij}'(0)\; e\_i\otimes f\_j\Big)$. You see that you can move the function part from left to right which explains the isomorphism above.
For vector bundles it is similar: the $TM$-part should be there only once.
Added:
======
Now let $p\_E:E\to M$ and $p\_F:F\to M$ be vector bundles. Then
$$E\otimes F = \operatorname{Hom}(E^\*, F) = \operatorname{Hom}(F^\*,E)$$
where the last natural isomorphism is via transpose using $E^{\*\*}=E$.
Then
$$
T\operatorname{Hom}(E^\*,F) = \operatorname{Hom}(E^\*,TF) \xrightarrow{\operatorname{Hom}(E^\*,\pi\_F)} \operatorname{Hom}(E^\*,F),
$$
where the middle ${\operatorname{Hom}}$ abuses notation and uses unsaid conventions.
Note the second vector bundle structure
$$
\operatorname{Hom}(E^\*,TF)\xrightarrow{\operatorname{Hom}(E^\*,T(p\_F))} \operatorname{Hom}(E^\*,TM),
$$
see 8.12 ff of [this book](https://www.mat.univie.ac.at/%7Emichor/dgbook.pdf) or 6.11 in [that book](https://www.mat.univie.ac.at/%7Emichor/kmsbookh.pdf).
Your next question is essentially, how to write the induced connector $K\_{E\otimes F}: T(E\otimes F) \to (E\otimes F)\times\_M (E\otimes F)$ whose kernel would identify the pullback of $TM$ to $E\otimes F$ with the horizontal bundle. See 19.12 ff of [this book](https://www.mat.univie.ac.at/%7Emichor/dgbook.pdf) for background. Here we need a name for the canonical isomorphism $\rho:E\otimes TF = F\otimes TE$ (abuse of notation here). Then $K\_{E\otimes F} = Id\_E \otimes K\_F + \rho \circ Id\_F\otimes K\_E \circ \rho$.
Note that the horizontal bundle is not natural.
A remark to the formulation at end of your question: $TE$ is NOT a vector bundle over $M$, it has two vector bundle structures
$$
TM \xleftarrow{Tp} TE \xrightarrow{\pi\_E} E,
$$
and the chart changes over $M$ are quadratic (like for the Christoffel symbols).
So $TE\otimes TF$ does make sense only with a lot of abuse of notation and unsaid conventions.
| 2 | https://mathoverflow.net/users/26935 | 420166 | 170,989 |
https://mathoverflow.net/questions/420162 | 16 | 1. Is it true that if $A\_1 \vee A\_2 \vee .. \vee A\_n = B\_1 \vee B\_2 \vee .. \vee B\_m$, where $A\_i, B\_j$ are homotopy types of complexes not decomposable into a bouquet, then the multisets $A\_i$ and $B\_j$ coincide? That is, is it true that a commutative monoid of homotopy types decomposable into a bouquet of a finite number of indecomposable ones is freely generated by indecomposable ones.
2. Is it true that any finite complex decomposes into a finite number of indecomposable ones (and thus all finite complexes are included in the monoid above)? Countable complexes are not necessarily included in it, for example, any countable bouquet. But, for example, all spaces $K(G, n)$ are included, it seems.
| https://mathoverflow.net/users/148161 | Is the decomposition of the homotopy type of a complex into a bouquet unique? | In Hilton&Roitberg paper "On principal $S^3$-bundles over spheres" it's proven that if you have a prime order $p \neq 2,3$ class $\alpha$ in $\pi\_k(S^n)$ that is a suspension, then for a prime $q \neq \pm 1 \, mod \, p$ mapping cones $C(\alpha)$ and $C(q \cdot \alpha)$ satisfy $C(\alpha) \vee S^n \cong C(q \cdot \alpha) \vee S^n$.
Now it's easy to see that those cones are indecomposable, and give counterexample to 1.
For 2, I think Grushko theorem gives that all but finite number of summands are simply-connected, and then you have only finitely many generators in homology so overall any such sum has to be finite. I'll think about that, but my reasoning seems correct to me now.
| 17 | https://mathoverflow.net/users/81055 | 420171 | 170,992 |
https://mathoverflow.net/questions/420153 | 2 | I am interested in long-range percolation models with heavy-tailed degree distributions such as [DHH13](http://www.numdam.org/article/AIHPB_2013__49_3_817_0.pdf), [GLM21](https://www.cambridge.org/core/services/aop-cambridge-core/content/view/7B85863F4E9E3FB24BA77F538D1A871A/S0001867821000136a.pdf/div-class-title-percolation-phase-transition-in-weight-dependent-random-connection-models-div.pdf), [Y6](https://www.jstor.org/stable/pdf/27595763.pdf?casa_token=TPFzIgpV88EAAAAA:VSP9Vb3_17bZfYZOjo9STD5CK3Oc6Me2KcjAqajaMWzX_FrZVBQR3P0BjcIPH8R02FGjIFBIHGCJmv1O-ITwOckoiCZtFc_Yr3gH9CiY0jZ6cw7C8Z3T). The simplest example is scale-free percolation in which vertices are elements $\mathbb{Z}^d$, each vertex $x\in\mathbb{Z}^d$ carries and independent random heavy-tailed weight $W\_x>0$ and two vertices $x,y$ are joined by an edge with probability $1-\exp(W\_xW\_y/|x-y|^\alpha), \alpha>d$, independently (given the weights). If $W\_x$ has a first moment but no second moment, then the model is "robust" under Bernoulli percolation, in the sense that almost surely $p\_c^{\text{bond}}=p\_c^{\text{site}}=0$ for the random graph produced.
Some recent results are concerned with the question of recurrence and transience of simple random walk on the infinite clusters in these percolation models. Specifically, it was shown in [HH17](https://projecteuclid.org/journals/annals-of-applied-probability/volume-27/issue-4/Structures-in-supercritical-scale-free-percolation/10.1214/16-AAP1270.pdf), [GHMM19](https://arxiv.org/pdf/1911.04350.pdf) that robust infinite clusters are transient almost surely. The proofs are constructive but quite model specific and are based on a strategy from [B2](https://link.springer.com/content/pdf/10.1007/s002200200617.pdf) designed for long-range percolation without weights; they do not use robustness directly in an essential way.
My feeling is that the robustness property (in particular the "site version" $p\_c^{\text{site}}=0$) is so strong (for example it implies super-exponential growth) that it should imply transience for a much larger class of infinite random graphs than these rather specific models, for instance unimodular random measures concentrated on robust infinite graphs.
In the case of random trees, $p\_c^{\text{bond}}=0$ clearly implies transience but I have so far not found any results in that direction concerning non-tree like graphs.
My questions are:
1. Are there any results in the literature that shed light on the structure of random graphs with finite mean degree under the assumption $p\_c=0$ or maybe other "extreme" assumptions on growth or isoperimetry?
2. Is there an example of a (not necessarily random) graph with $p\_c^{\text{site}}=0$ that is recurrent?
| https://mathoverflow.net/users/479063 | Which infinite random graphs with percolation threshold $p_c=0$ are transient? | I will first construct a simple deterministic example of a recurrent graph with $p\_c^\mathrm{site}=0$ and then show how it can be modified to be a unimodular random rooted graph (see e.g. Nicolas Curien's lecture notes for the definition <https://drive.google.com/file/d/16qEMGJU2g01g4YWkYtVKTSqetk-SpRmy/view>)
Suppose we take a half-infinite line indexed by $\mathbb{N}=\{1,2,3,\ldots\}$ and, for each $n$, replace the vertex labelled $n$ with a complete graph of size $f(n)$ for some function $f:\mathbb{N}\to\mathbb{N}$ to be chosen later, where every vertex in the $n$th complete graph is connected to every vertex in the adjacent complete graphs at $n$ and $n-1$. In this example, site percolation occurs if there is at least one open vertex in the complete graph at $n$ for every sufficiently large $n$. From this and Borel-Cantelli, we see that $p\_c^\mathrm{site}<1$ when $f(n)=\Omega(\log n)$ and $p\_c^\mathrm{site}=0$ when $f(n)=\omega(\log n)$. If we take, say, $f(n)=\lceil(\log (n+2))\rceil^2$ then the resulting graph therefore has $p\_c^\mathrm{site}=0$ but is recurrent by the Nash-Williams criterion since there are $O((\log n)^4)$ edges between the $n$th and $(n+1)$th complete graphs and $\sum\_{n=2} (\log n)^{-4}=\infty$. (Note in particular that we have a lot of room in this construction!)
We can make something similar work as a unimodular random rooted graph by using the canopy trees instead of the half-infinite line. This is a pretty standard trick, you can see lots of instances of it in this paper of Omer Angel and myself <https://arxiv.org/abs/1710.03003>
Indeed, if we take the canopy tree and replace each vertex at height $n$ with a complete graph of size $\lceil (\log (n+2))\rceil^2$ then the resulting graph is recurrent but has $p\_c^\mathrm{site}=0$ for similar reasons to the line example. Since the probability that the root of the canopy tree is at height $n$ is $2^{-n}$, we can make this graph unimodular by biasing the height of the root vertex of the canopy tree by $\lceil (\log (n+2))\rceil^2$ then choosing uniformly one of the vertices of the complete graph at that vertex as the root of the new graph (this biasing makes sense since $\lceil (\log (X+2))\rceil^2$ is integrable when $X$ is a geometric random variable). Note that this graph also has rather light tailed degrees: The probability that the root has degree at least $n$ is roughly of order $2^{-\Theta(e^{\sqrt{n})}$.
Basically these examples work because replacing edges by parallel edges (or blowing up vertices into complete graphs) has a much stronger effect on percolation than it does on the random walk.
| 1 | https://mathoverflow.net/users/41827 | 420173 | 170,994 |
https://mathoverflow.net/questions/420158 | 47 | I would like to ask a question inspired by the title of a book by Sir [Roger Penrose](https://en.wikipedia.org/wiki/Roger_Penrose) ([1]). The germ of this is to ask about the role, if any, of the fashion in research of pure and applied mathematics.
I'm going to focus the post (and modulate my genuine idea) about an aspect that I think can be discussed here from an historical and mathematical point of view, according to the following:
>
> **Question.** I would like to know what are examples of remarkable achievements (in your research subject or another that you know) that arose against the general view/work of the mathematical community since the year 1900 up to the year 1975. *Refer* the literature if you need it. **Many thanks.**
>
>
>
**An example** is the mention that the author of [2] (as I interpret it) about [Lennart Carleson](https://en.wikipedia.org/wiki/Lennart_Carleson) and a conjecture due to [Lusin](https://en.wikipedia.org/wiki/Lusin%27s_theorem) in the second paragraph of page 671 (the article is in Spanish).
Your answer can refer to (for the research of pure or applied mathematics, and mathematical physics) unexpected proofs of old unsolved problems, surprising examples or counterexamples, approaches or mathematical methods that defied the contemporary (ordinary, mainstream) approaches, incredible modulizations solving difficult problems,... all these in the context of the question that is: **the proponents/teams of these solutions and ideas swam against the work of the contemporary mathematics that they knew at the time.**
\*You can refer to the literature for the statements of the theorems, examples, methods,... if you need it. Also from my side it is welcome if you want to add some of your own historical remarks about the mathematical context concerning the answer that you provide us: that's historical remarks (if there is some philosophical issue also) emphasizing why the novelty work of the mathematician that you evoke was swimming against the tide of the contemporary ideas of those years.
References:
-----------
[1] Roger Penrose, *Fashion, Faith, and Fantasy in the New Physics of the Universe*, Princeton University Press (2016).
[2] Javier Duoandikoetxea, *200 años de convergencia de las series de Fourier*, La Gaceta de la Real Sociedad Matematica Española, Vol. 10, Nº 3, (2007), pages 651-677.
| https://mathoverflow.net/users/142929 | Swimming against the tide in the past century: remarkable achievements that arose in contrast to the general view of mathematicians | After mathematicians had been been taught for decades that a consistent theory of the calculus based on infinitesimals was impossible, Abraham Robinson was certainly swimming against the tide when he proved otherwise.
Robinson, A. (1961): *Non-standard analysis*, Indagationes Mathematicae 23, pp. 432-440.
Robinson, A. (1966): *Non-standard Analysis*, North-Holland Publishing Company, Amsterdam.
| 48 | https://mathoverflow.net/users/18939 | 420175 | 170,996 |
https://mathoverflow.net/questions/420083 | 1 | I want to prove that "Given a bundle $E$, for any line bundle $L$ the projectivizations of $E$ and $E$ tensor $L$ are isomorphic i.e $P(E)≅P(E⊗L)$".
The statement can also be seen on the Wikipedia page of the projective bundle. The reference they give is for Hartshorne, Algebraic Geometry. But unfortunately, I am not aware of that.
By bundle, you can as well assume it to be a smooth vector bundle and their projectivization to be a fibre bundle (in the topological sense) with fibre 1-dim subspaces of the vector space fibres.
What I tried:-
At the fibre level, we can define it $[v]→[u⊗v]$ for an arbitrary non-zero element $u$ in the fibre of $L$ and $v$ in the fibre of $E$. This is clearly a well-defined map. Now, how can I get a continuous map at the total space level, at least locally we can do it as we have an explicit description of the projective bundles locally, will this suffice?
Any suggestions will be of great help.
Thanks and regards in advance
| https://mathoverflow.net/users/126899 | Projective bundle is stable under twisting by a line bundle | $\DeclareMathOperator{\Hom}{Hom}$There's a nice explicit map $P\Hom(L,E)\to P(E)$ that takes a homomorphism to its image. Then use that $E\otimes L \cong \Hom(L^\vee,E)$ where $L^\vee$ is the dual.
But if you work through what this does, it's simply the map $P(E\otimes L)\to P(E)$ that takes $[u\otimes v] \mapsto [u]$ for nonzero $u$ and $v$.
| 2 | https://mathoverflow.net/users/58888 | 420182 | 171,000 |
https://mathoverflow.net/questions/420183 | 8 | $\newcommand\Legendre{\genfrac(){}{}}$Let $p\equiv 1\pmod 4$ be a prime number, and $x\_{i}\ge 0$ be such that $$x\_{1}+x\_{2}+\dotsb+x\_{p}=1.$$
Show that
$$\sum\_{1\le i<j\le p}\Legendre{i-j}{p}x\_{i}x\_{j}\le\dfrac{p-1}{2p+6}$$
Here $\Legendre\cdot\cdot$ is the Legendre symbol.
This problem was encountered by a colleague of mine when he was writing a paper, and we couldn't prove this inequality. So I ask it. We found the constant in the right-hand side seems to be the best one because when $p = 5$ it can be reached:
$$p=5,x\_{1}=x\_{2}=0.5,x\_{3}=x\_{4}=\dotsb=0.$$
| https://mathoverflow.net/users/38620 | Prove an inequality related to sums of Legendre symbols | $\newcommand\Legendre{\genfrac(){}{}}$We have $$\sum\_{1\le i<j\le p}\Legendre{i-j}{p}x\_{i}x\_{j} \leq \frac{k-1}{2k }$$ where $k$ is the size of the largest clique in the Paley graph, and this is sharp.
Indeed, if the number of $i$ such that $x\_i>0$ is at most $k$ then
$$\sum\_{1\le i<j\le p}\Legendre{i-j}{p}x\_{i}x\_{j} \leq \sum\_{1\le i<j\le p}x\_{i}x\_{j} = \frac{1}{2} \sum\_{i \neq j} x\_i x\_j =\frac{1 - \sum\_i x\_i^2}{2} \leq \frac{ 1- k^{-1}}{2} = \frac{k-1}{2k }$$
by Cauchy-Schwarz.
Otherwise, there exist $i\_1,i\_2 $ with $\Legendre{i\_1-i\_2}{p} = -1$ and $x\_{i\_1},x\_{i\_2}>0$. Without loss of generality, we may assume $$\sum\_{j \neq i\_1} \Legendre{ i\_1 - j}{p} x\_j \geq \sum\_{j \neq i\_2} \Legendre{ i\_2 - j}{p} x\_j.$$
Let $y\_j = x\_j$ for $j\not\in\{i\_1, i\_2\}$, let $y\_j = x\_{i\_1} + x\_{i\_2} $ for $j=i\_1$, and let $y\_j=0$ for $j =i\_2$. Then $\sum\_j y\_j = \sum\_j x\_j=1$ and the number of nonzero $y\_j$ is at most the number of nonzero $x\_j$. Now, using the $i\neq j$ sum which is twice as large as the $i<j$ sum but much easier to work with, we have
$$\sum\_{1\leq i,j \leq p, i \neq j }\Legendre{i-j}{p}y\_{i}y\_{j} $$ $$= \sum\_{1\leq i,j \leq p, i \neq j }\Legendre{i-j}{p}x\_{i}x\_{j} + 2\sum\_{1 \leq j \leq p, j \neq i\_1}\Legendre{i\_1-j}{p}(y\_{i\_1}-x\_{i\_1}) x\_{j} + 2\sum\_{1 \leq j \leq p, j \neq i\_2}\Legendre{i\_2-j}{p}(y\_{i\_2}-x\_{i\_1}) x\_{j} + 2 \Legendre{ i\_1 i\_2}{p} (y\_{i\_1} - x\_{i\_1} ) (y\_{i\_2} -x\_{i\_2}) $$
$$= \sum\_{1\leq i,j \leq p, i \neq j }\Legendre{i-j}{p}x\_{i}x\_{j} + 2\sum\_{1 \leq j \leq p, j \neq i\_1}\Legendre{i\_1-j}{p} x\_{i\_2} x\_{j} - 2\sum\_{1 \leq j \leq p, j \neq i\_2}\Legendre{i\_2-j}{p}x\_{i\_2} x\_{j} - 2 \Legendre{ i\_1 i\_2}{p} x\_{i\_2}^2 $$
$$ > \sum\_{1\leq i,j \leq p, i \neq j }\Legendre{i-j}{p}x\_{i}x\_{j} .$$
Thus, we have increased your sum while reducing the number of nonzero $x\_i$'s. We may keep doing this until the number of nonzero $x\_i$'s is at most $k$, proving the claimed upper bound.
Sharpness follows from taking $x\_i=1/k$ for $i$ in a clique of size $k$ and $x\_i=0$ for all other $i$.
By the recent [breakthrough upper bound on the clique number of the Paley graph by Hanson and Petridis](https://arxiv.org/abs/1905.09134), we have $k \leq \lceil \sqrt{p/2} \rceil$. Plugging this in, we obtain your claimed bound for $p=5$ and do better for all larger $p$. (To do better, it suffices to have $k < \frac{p+3}{4}$.)
| 18 | https://mathoverflow.net/users/18060 | 420188 | 171,002 |
https://mathoverflow.net/questions/416761 | 3 | Assume all algebras are finite dimensional quiver algebras over a field (no restriction of generaltiy if the field is algebraically closed).
>
> Let A be a local Frobenius algebra.
> Is A isomorphic to its opposite algebra?
>
>
>
For non-Frobenius algebras this is false, see [Do you know which is the minimal local ring that is not isomorphic to its opposite?](https://mathoverflow.net/questions/376454/do-you-know-which-is-the-minimal-local-ring-that-is-not-isomorphic-to-its-opposi/376461?noredirect=1#comment1069416_376461) (where the current question remained open, see the answer and comment).
Is there an easy example of a (not necessarily local) Frobenius algebra that is not isomorphic to its opposite algebra?
| https://mathoverflow.net/users/61949 | Local Frobenius algebras and their opposite algebras | No. Consider, for example, the quantum complete intersection $A = k\langle X,Y\rangle/(X^2, Y^3, XY-qYX)$, $q\in k\setminus\{0\}$. This is a Frobenius local algebra (see Section 3 of *arXiv:0709.3029*), and the ideal $(X^2, Y^3, XY-qYX)$ in $k\langle X,Y\rangle$ is admissible.
Any isomorphism $f:A\to A^{\rm op}$ must satisfy $f(X) = aX+r$ and $f(Y) = bY + s$, where $a,b\in k\setminus\{0\}$ and $r,s\in\mathop{\rm rad}^2(A) = (XY, Y^2)$. Now
$$f(XY) = f(Y)f(X) = (bY + s)(aX +r) = abq^{-1}XY + t,\quad t\in\mathop{\rm rad}\nolimits^3(A)$$
but also
$$f(XY) = f(qYX) = qf(X)f(Y) = q(aX + r)(bY + s) = qabXY + u,\quad u\in\mathop{\rm rad}\nolimits^3(A)$$
Since $XY\notin\mathop{\rm rad}^3(A^{\rm op})$, this means that
$abq^{-1}XY = qab XY$ and thus $q=q^{-1}$ (because $a,b\ne0$).
So unless $q = \pm1$, the algebras $A$ and $A^{\rm op}$ are not isomorphic.
| 4 | https://mathoverflow.net/users/136180 | 420197 | 171,006 |
https://mathoverflow.net/questions/419157 | 0 | In this paper [N. Ghoussoub](https://www.degruyter.com/document/doi/10.1515/crll.1991.417.27/html),
the author claims the following version of Marino–Prodi perturbation, that is :
Let $H$ a Hilbert space.
Let $f\in C^2(H, \mathbb{R}),$ $K$ is a compact subset of $K\_c$ (that is, the set of critical points with critical value $c$) on which $d^2f$ is a Fredholm operator, for any $\epsilon\_1>0$ and $\epsilon\_2>0$ small enough, there exist a function $g\in C^2(H, \mathbb{R})$ such that :
* $\|f-g\|\_{C^2}<\epsilon\_1.$
* $f(x)=g(x)$ if $x$ outside $K^{2\epsilon\_2}$, that is, $2\epsilon\_2$-neighborhood of $K$.
* All the critical points of $g$ in $K^{\epsilon\_2}$ are non-degenerate and finite in number.
* Moreover, If $f$ is $G$-invariant functional for a a topological group $G$ on $H$, then $g$ can be chosen to be $G$-invariant.
In particular, if $G=Z\_2$, $f$ is even, hence $g$ is even.
The proof in [Solimini](https://link.springer.com/content/pdf/10.1007/BF01171757.pdf) for the first three conclusions is given here:
* First construct a smooth cut-off function $\phi(x)$, $\phi(x)$ equals $1$ on $K^{\epsilon\_2}$,
equals $0$ outside $K^{2\epsilon\_2}$.
* Set $g(x)=f(x)+\phi(x)y.x$, then by Sard-Smale Theorem, choosing a small $y$ such that $-y$ is a regular value of $df$, hence we complete the proof of the first three conclusions.
However, my question is how to construct a $G$-invariant $g$ when $f$ is $G$-invariant? In particularly, when $G=Z\_2$.
| https://mathoverflow.net/users/166368 | A question about Marino–Prodi perturbation | The proof for $G=Z\_2$ is given in [jde2016](http://dx.doi.org/10.1016/j.jde.2016.09.018) Appendix B Proposition B3.
| 0 | https://mathoverflow.net/users/166368 | 420199 | 171,007 |
https://mathoverflow.net/questions/420048 | 4 | Consider the following [bornologies](https://en.wikipedia.org/wiki/Bornology) $\mathbb{D},\mathbb{E}$ on the set $\mathcal{N}$ of all functions from $\mathbb{N}$ to $\mathbb{N}$:
* $\mathbb{D}=\{A: \exists f\in\mathcal{N}\forall g\in A\exists m\in\mathbb{N}\forall n>m(f(n)>g(n))\}$. (**Dominatable** sets)
* $\mathbb{E}=\{A: \exists f\in\mathcal{N}\forall g\in A\forall m\in\mathbb{N}\exists n>m(f(n)>g(n))\}$. (**Escapable** sets)
My question is when, set-theoretically speaking, these yield equivalent (= "bornomorphic") bounded structures on $\mathcal{N}$:
>
> Is it consistent that $\mathfrak{b}=\mathfrak{d}$ but there is *not* a bijection $i:\mathcal{N}\rightarrow\mathcal{N}$ such that the $i$-image of each set in $\mathbb{D}$ is in $\mathbb{E}$ and the $i$-preimage of each set in $\mathbb{E}$ is in $\mathbb{D}$?
>
>
>
Here $\mathfrak{b}$ and $\mathfrak{d}$ are the cardinal characteristics corresponding to escaping(/**b**ounding) and domination: the minimal cardinalities of sets of functions not dominated/escaped by any individual function, respectively. Clearly in order for $\mathbb{D}$ and $\mathbb{E}$ to be equivalent as bornologies we need $\mathfrak{b}=\mathfrak{d}$; I'm curious whether any additional information is captured by considering the bornological structure present.
| https://mathoverflow.net/users/8133 | Comparing bornologies for domination/escaping | Note that $\mathfrak{b}=\mathfrak{d}$ is equivalent to the existence of a $<^\ast$-increasing sequence $(f\_\alpha)\_{\alpha<\mathfrak{d}}$ which is cofinal in $(\mathcal{N},{<^\ast})$, where $f <^\ast g \iff \exists n\,\forall m \geq n\,{f(m) < g(m)}$.
For $\alpha<\mathfrak{d}$, let
$$\begin{aligned}
D\_\alpha &= \{ g \in \mathcal{N} \mid g <^\ast f\_\alpha \}, &
E\_\alpha &= \{ g \in \mathcal{N} \mid f\_\alpha \not<^\ast g \}.
\end{aligned}$$
Note that $D$ is dominatable iff $D \subseteq D\_\alpha$ for some $\alpha < \mathfrak{d}$, and that $E$ is escapable iff $E \subseteq E\_\alpha$ for some $\alpha < \mathfrak{d}$.
With some thinning of the cofinal sequence $(f\_\alpha)\_{\alpha<\mathfrak{d}}$ if necessary, we can make sure that for every $\alpha < \mathfrak{d}$ the sets
$$\begin{aligned}
&D\_\alpha \setminus {\textstyle\bigcup\_{\beta<\alpha} D\_\beta}, &
&E\_\alpha \setminus {\textstyle\bigcup\_{\beta<\alpha} E\_\beta},
\end{aligned}$$
each have size $\mathfrak{c}$. Working level-by-level, we can construct a bijection $\mathcal{N} \leftrightarrow \mathcal{N}$ which restricts to a bijection $D\_\alpha \leftrightarrow E\_\alpha$ for every $\alpha<\mathfrak{d}$. Such a bijection shows that the two bornologies are equivalent.
| 6 | https://mathoverflow.net/users/2000 | 420204 | 171,009 |
https://mathoverflow.net/questions/420198 | 3 | Given a matrix group $G$ by its generators i.e. $G =\langle A\_1,A\_2,...,A\_k \rangle \leq GL\_n(q)$, where each $A\_i$'s are matrix in $GL\_n(q)$
>
> **Q.** Does there exist a polynomial time (polynomial in input size) algorithm to find a minimal normal subgroup of $G$ if it exists otherwise it returns $G$ is simple?
>
>
>
There is an algorithm when group is given by permutation representation. Kindly share any reference if there is such algorithm for matrix group representation.
Thank you!
| https://mathoverflow.net/users/475097 | Algorithm to find a minimal normal subgroup of given group $G$ by matrix group representation | In the paper
```
Holt, D., Leedham-Green, C. R., & O'Brien EA (2020). Constructing composition factors for a linear group in polynomial time. JOURNAL OF ALGEBRA, 561, 215-236. 10.1016/j.jalgebra.2020.02.018
```
the authors consider the question of finding a composition series of a subgroup of ${\rm GL}(n,q)$ - I think that finding a minimal normal subgroup is of the same difficulty.
There is lots of earlier literature on this topic: Babai and various co-authors have made many theoretical contributions. The paper by Holt, Leedham-Green and O'Brien is geared more towards finding practical algorithms, and the $\mathtt{CompositionTree}$ function in Magma is very effective and improving all the time.
To have any chance of an affirmative answer to the question, you need to assume the availability of certain oracles. If your group is a cyclic subgroup of ${\rm GL}(n,q)$ of order dividing $q^n-1$, then you need to be able to factorize $q^n-1$ to proceed. There are also simple examples of subgroups of ${\rm GL}(2n,q)$ in which you need a discrete log oracle to decide whether the given group is $C\_p$ or $C\_p \times C\_p$ for a prime $p$ dividing $q^n-1$.
The current situation is that, subject to the availability of certain oracles, there is a polynomial-time Las Vegas algorithm to find a composition series of the group provided that it has no composition factors isomorphic to $^2B\_2(2^{2k+1})$, $^2F\_4(2^{2k+1})$, $^3D\_4(2^k)$, or $^2G\_2(3^{2k+1})$ for any $k$.
The problem with these families of exceptional groups is that there is currently no known polynomial-time constructive recognition algorithm, and it is to be hoped that this situation will be remedied in the future. But the case $^2G\_2(3^{2k+1})$ (Ree groups) looks particularly challenging.
Incidentally, the need for the oracles rarely if ever results in bottlenecks in practical calculations, mainly because of the amount of effort that has gone into finding effective implementations of the discrete log and integer factorization problems.
| 1 | https://mathoverflow.net/users/35840 | 420206 | 171,010 |
https://mathoverflow.net/questions/420216 | 2 | A tree $G$ on $n$ vertices $V=\{v\_1,...,v\_n\}$ is a connected undirected graph which is acyclic. For each tree $G$ one can split the set of vertices $V$ into two disjoint subsets $U,W \subset V$ such that $V = U \cup W$ and each edge in $G$ is between a vertex from $U$ and a vertex from $W$. In order to get uniqueness of $U$ and $W$ let's require $v\_1$ to be in $U$.
My question is if there is a known formula for the number of trees on $n$ vertices which satisfy $\#U = n\_1$ and $\#W = n\_2$ for given $n\_1,n\_2$ with $n\_1+n\_2=n$.
Any help is much appreciated.
| https://mathoverflow.net/users/409412 | Is there a formula for the number of trees with this extra condition? | If the trees are labelled, then each tree satisfying the condition corresponds to exactly one spanning tree of the bipartite graph $K\_{n\_1,n\_2}$. Therefore the answer is the number of spanning trees of the bipartite graph $K\_{n\_1,n\_2}$, which is $n\_1^{n\_2-1}n\_2^{n\_1-1}$ according to [this](https://math.stackexchange.com/questions/3781460/spanning-trees-of-the-complete-bipartite-graph).
| 5 | https://mathoverflow.net/users/480431 | 420218 | 171,012 |
https://mathoverflow.net/questions/420219 | 2 | Let $(X,Y)$ be a pair of random variables on a measure space $\mathcal T \subseteq \text{"subsets of }\mathbb R^2\text{"}$, with joint probability distribution $P$.
>
> We don't assume $X$ and $Y$ are independent!
>
>
>
Let $P\_X$ (resp. $P\_Y$) be the marginal distribution of $X$ (resp. $Y$), defined by $P\_X(A) := \int\_{p\_1(A)} \,dP$, where $\mathcal p\_k(A) := \{(z\_1,z\_2) \in \mathcal T \mid z\_k \in A\}$ defines the projection operator unto the $k$ coordinate of $\mathcal T$. Let $\Pi(X,Y)$ be the set of all couplings of $X$ and $Y$, i.e the set of all probability distributions on $\mathcal T$, with same margins as $P$. Finally, let $P\_X \otimes P\_Y \in \Pi(X,Y)$ be the independence coupling of $X$ and $Y$ defined by
$$
(P\_X \otimes P\_Y)(U) := P\_X(p\_1(U))\cdot P\_Y(p\_2(U)).
$$
Let $k$ and $n$ be positive integers, presumably, with $n \gg k$.
>
> **Question.** *Given $n$ independent copies $(X\_1,Y\_1),\ldots,(X\_n,Y\_n)$ of $(X,Y)$ (i.e an iid sample of size $n$ from the joint distribution $P$), what is a principled way to obtain an iid sample from $P\_X \otimes P\_Y$ of size $k$ ?*
>
>
>
| https://mathoverflow.net/users/78539 | Given iid samples from the joint distribution $P$ of pair of r.v.'s $(X,Y)$, how to get iid samples from independence coupling $P_X \otimes P_Y$? | It is unclear to me what "a principled way" could mean.
However, given $n$ iid pairs $(X\_1,Y\_1),\dots,(X\_n,Y\_n)$, it is easy to get $k:=\lfloor n/2\rfloor$ iid pairs $(X\_1,Z\_1),\dots,(X\_k,Z\_k)$ such that, for each $j\in\{1,\dots,k\}$, (i) the random variables $X\_j$ and $Z\_j$ are independent and (ii) $Z\_j$ equals $Y\_j$ in distribution:
Just let $Z\_j:=Y\_{k+j}$ for all $j\in\{1,\dots,k\}$.
---
If the joint probability distribution $P\_{X,Y}$ of the pair $(X,Y)$ is known, then it may be possible to get an iid sample of size $n$ from the distribution $P\_X\otimes P\_Y$ using an iid sample of (the same) size $n$ from the distribution $P\_{X,Y}$. This could be done by applying a transformation $T\colon\mathbb R^2\to\mathbb R^2$ to the pair $(X,Y)$ to get a pair $(U,V):=T(X,Y)$ with distribution $P\_X\otimes P\_Y$.
The transformation $T$ can apparently be obtained by discrete approximation, as follows. For each natural $n$, let $X\_n$ and $Y\_n$ be discrete random variables (r.v.'s), say each taking only finitely many values, such that $X\_n\to X$ and $Y\_n\to X$ in probability (as $n\to\infty$). Then $P\_{X\_n,Y\_n}\to P\_{X,Y}$, $P\_{X\_n}\to P\_X$, $P\_{Y\_n}\to P\_Y$, and $P\_{X\_n}\otimes P\_{Y\_n}\to P\_X\otimes P\_Y$ weakly.
Then for each natural $n$ there is a transformation $T\_n\colon\mathbb R^2\to\mathbb R^2$ such that the pair $(U\_n,V\_n):=T\_n(X\_n,Y\_n)$ has the distribution $P\_{X\_n}\otimes P\_{Y\_n}$. This follows because any discrete set can be transformed bijectively to a set on the real line.
If now the set $\{T\_n\colon n\in\mathbb N\}$ is compact in an appropriate sense, then, passing to a subsequence, from the pairs $(U\_n,V\_n)=T\_n(X\_n,Y\_n)$ with distributions $P\_{X\_n}\otimes P\_{Y\_n}$ one will get a pair $(U,V):=T(X,Y)$ with distribution $P\_X\otimes P\_Y$.
| 2 | https://mathoverflow.net/users/36721 | 420222 | 171,013 |
https://mathoverflow.net/questions/420177 | 0 | There is an ordered set $M$ with $N$ numbers in it (and $M\_n$ is the $n$-th number in $M$). Let $L\_k$ be the sum of the first $k$ numbers in $M$.
Consider equation $\sum\_{k=0}^{N-1} 3^k 2^{-L\_{k+1}} \equiv 1 \bmod 3^N$
For example, if $N=1 \Rightarrow 3^0 2^{-L\_1} \equiv 1 \bmod 3^1$ or $2^{-L\_1}\equiv 1 \bmod 3$, which is only possible if $-L\_1\equiv 0 \bmod 2 \Rightarrow M\_1 \equiv 0 \bmod 2$ (which is to say, "if $M$ contains $1$ number and solves the equation above, the number that $M$ contains must be even")
Is there a general way we can use to find all "modular restrictions" for all numbers in $M$ for any given size ($N$) of $M$? It gets really hard brute forcing even for $N=4$.
| https://mathoverflow.net/users/479568 | Is there a general way to solve this modular equation? | There seems to be no simple formula for the solutions to the given congruence. Still, they can be computed iteratively as follows.
First, we notice that $2^m = (3-1)^m \equiv (-1)^m \sum\_{i=0}^{N-1} \binom{m}{i} (-3)^i \pmod{3^N}$. Then the left-hand side of the congruence in question can be rewritten as
$$\sum\_{k=0}^{N-1} 3^k (-1)^{L\_{k+1}} \sum\_{i=0}^{N-1} \binom{- L\_{k+1}}{i} (-3)^i \equiv \sum\_{m=0}^{N-1} 3^m \sum\_{k=0}^m (-1)^{L\_{k+1}+m-k} \binom{- L\_{k+1}}{m-k} \pmod{3^N}$$
Next, we fix the parities of $L\_1,\dots,L\_N$, where $L\_1\equiv 0\pmod{2}$ and the others can be arbitrary. Then, we compute iteratively $L\_2,\dots,L\_{N-1}$ in parametric form by considering the above congruence modulo $3^l$ for $l=2,\dots,N$, dividing it by $3^{l-1}$, and solving the resulting linear congruence with respect to $L\_{l-1}$.
Finally, we obtain $M\_k = L\_k - L\_{k-1}$.
Hence, this approach constructs a complete solution as the union of $2^{N-1}$ parametric ones.
---
**Example for $N=4$.** Let's fix the parities as $(L\_1,L\_2,L\_3,L\_4)\equiv (0,1,0,1)\pmod{2}$. Then for $l=2$, have a congruence:
$$3^0\cdot (-1)^{L\_1+0-0} \binom{-L\_1}{0-0} + 3^1\cdot \big( (-1)^{L\_1+1-0} \binom{-L\_1}{1-0} + (-1)^{L\_2+1-1} \binom{- L\_2}{1-1}\big)\equiv 1\pmod{3^2},$$
that is
$$L\_1 - 1\equiv 0\pmod{3},$$
and thus $L\_1 = 4+6t$, where $t$ is an integer parameter.
Using the above result, we can simplify the congruence for $l=3$ to
$$\frac{L\_1 - 1}3 + (-1)^{L\_1+2-0} \binom{-L\_1}{2-0} + (-1)^{L\_2+2-1} \binom{- L\_2}{2-1} + (-1)^{L\_3+2-2} \binom{- L\_3}{2-2} \equiv 0\pmod{3}$$
that is
$$2t - L\_2 \equiv 0\pmod{3},$$
implying $L\_2 = 3 + 2t + 6s$, where $s$ is an integer parameter.
Next the congruence for $l=4$ simplifies to
$$t^2 + t + s + 1 + L\_3 \equiv 0\pmod{3},$$
implying $L\_3 = 2(t^2 + t + s + 1) + 6z$, where $z$ is an integer parameter.
Therefore, for the chosen parities, the solutions are described parametrically by
$$\begin{cases}
L\_1 = 4+6t, \\
L\_2 = 3 + 2t + 6s,\\
L\_3 \equiv 2(t^2 + t + s + 1) + 6z,\\
L\_4 \equiv 1+2w,
\end{cases}$$
and correspondingly
$$\begin{cases}
M\_1 = 4+6t, \\
M\_2 = 6s - 4t - 1,\\
M\_3 \equiv 6z + 2t^2 - 4s - 1,\\
M\_4 \equiv 2w - 2t^2 - 2t - 2s - 6z - 1,
\end{cases}$$
where $t,s,z,w$ are arbitrary integers.
The other $2^3-1 = 7$ parametric solutions (for the other parity choices) are constructed similarly.
| 0 | https://mathoverflow.net/users/7076 | 420232 | 171,016 |
https://mathoverflow.net/questions/420238 | -2 | I found a problem in my textbook and I have tried solving it, but I had no succes. The problem is:
Let $A$ and $B$ be $n \times n$ matrices with complex number entries. Given that $AB−BA$ is invertible and $(A−B)^2=I\_n$, where $I\_n$ is the identity matrix, prove that $tr(A)=tr(B)$ and that $n$ is even. How should I approach this problem? I have tried to actually compute the inverse of $AB−BA$ from $(A−B)^2$ but I had no succes. I can also say that the eignevalues of $A-B$ are $+1$ or $-1$, but how does that help?
| https://mathoverflow.net/users/480453 | Proving 2 matrices have the same trace | Let's call $C=A-B$. Then you have $C^2=I\_n$ and $BC-CB$ is invertible. You want to show that $C$ has an equal dimension of $1$ and $(-1)$ eigenspaces, which in turn implies both the equality $tr(A)=tr(B)$ and $n$ being even.
Pick an eigen basis of the space to assume that
$$
C=\left(
\begin{array}{c}I\_m & 0\\ 0&-I\_k\end{array}
\right),
~~
B=\left(
\begin{array}{c}B\_1 & B\_2\\ B\_3&B\_4\end{array}
\right)
$$
with the appropriate sizes of $B\_i$.
Then
$$
BC-CB=\left(
\begin{array}{c}0 & \*\\ \* &0\end{array}
\right).
$$
If $m>k$, then the first $m$ rows are linearly dependent and if $m<k$ then the last $k$ rows are.
| 3 | https://mathoverflow.net/users/38468 | 420242 | 171,019 |
https://mathoverflow.net/questions/420233 | 0 | $f\colon U\subset \mathbb{R}^{n}\longrightarrow\mathbb{R}^{m}$ is differentiable at $x\_{0}$ if there exist a linear transformation $T\colon \mathbb{R}^{n}\longrightarrow\mathbb{R}^{m}$, such that:
\begin{equation}
f(x\_{0}+h)=f(x\_{0})+T\cdot h +r(h) \quad\text{donde}\quad \lim\_{h\to 0}\frac{r(h)}{\Vert h\Vert}=0
\end{equation}
under this definition, I am trying to solve the following exercise.
Let $GL(\mathbb{R}^{n})=\{T\in \mathcal{L}(\mathbb{R}^{n},\mathbb{R}^{m}): T\quad \text{is invertible}\}$ and $Inv\colon GL(\mathbb{R}^{n})\longrightarrow GL(\mathbb{R}^{n})$ defined by $Inv(T)=T^{-1}$. Show that $Inv$ is differentiable and find $Inv'(T\_0)$.
Starting from the definition
\begin{align}
Inv(T\_0+H)-Inv(T\_0)&=(T\_0+H)^{-1}-T\_0^{-1}\\
&=(T\_0+H)^{-1}(I-(T\_0+H)T\_0^{-1})\\
&=(T\_{0}+H)^{-1}(I-T\_0T\_{0}^{-1}+HT\_{0}^{-1})\\
&=-(T\_0+H)^{-1}HT\_{0}^{-1}
\end{align}
However, I cannot identify who the candidate for $Inv'(T\_0)\cdot H$ and $r(H)$.
I also know that another way to calculate $Inv'(T\_0)\cdot H$ is
\begin{equation}
Inv'(T\_0)\cdot H=\lim\_{t\to 0}\frac{Inv(T\_{0}+tH)-Inv(T\_0)}{t}=\frac{d}{dt}(Inv(T\_{0}+tH))\big{|}\_{t=0}
\end{equation}
But I can't find it.
Any guidance is greatly appreciated
| https://mathoverflow.net/users/480450 | Application of the Frechet derivative | If $F:M^{n\times n}\to M^{n\times n}$ is differentiable, then for $A\in M^{n\times n}$, $DF(A)$ is a linear map from $M^{n\times n}$ to $M^{n\times n}$ and we denote by $DF(A)H$ value of this linear map at $H\in M^{n\times n}$.
If $F,G:M^{n\times n}\to M^{n\times n}$ are differentiable maps, then the product of the maps $F\cdot G:M^{n\times n}\to M^{n\times n}$ satisfies the product rule which is easy to prove:
$$
D(F\cdot G)(A)H=(DF(A)H)\cdot G(A) + F(A)\cdot (DG(A)H).
$$
Let $F:GL(\mathbb{R}^n)\to GL(\mathbb{R}^n)$, $F(A)=A^{-1}$. The formula for $A^{-1}$ shows that the entries of $F$ are rational functions and hence $F\in C^\infty$ (entries are $\pm$ minor determinant over matrix determinant and determinants are polynomials in coefficients).
Let $G(A)=A$. Since $G$ is the identity map, $DG(A)=Id$ i.e., $DG(A)H=H$. We have
$(F\cdot G)(A)=A^{-1}\cdot A=I$, so $F\cdot G$ is a constant map and hence $D(F\cdot G)=0$. Now. the profuct rule gives
$$
0=D(F\cdot G)(A)H=(DF(A)H)G(A)+F(A)(DG(A)H),
$$
$$
0=(DF(A)H)A+A^{-1}H
$$
and we get the formula for the derivative of the inverse map
$$
DF(A)H=-A^{-1}HA^{-1}.
$$
| 2 | https://mathoverflow.net/users/121665 | 420246 | 171,020 |
https://mathoverflow.net/questions/420247 | 6 |
>
> **Question:** For a smooth, bounded domain $\Omega\subset \mathbb R^d$, does there exist a function $u\in L^1(\Omega)$ such that
> $u\not\in L^\Phi(\Omega)$ for *any* Orlicz space $\Phi$?
>
>
>
---
For the definition of Orlicz spaces see [this wikipedia page](https://en.wikipedia.org/wiki/Orlicz_space#Formal_definition). In a nutshell, a Young function is a convex function $\Phi:\mathbb R^+\to \mathbb R^+$ which is superlinear at infinity and sublinear at the origin,
$$
\frac{\Phi(u)}{u}\xrightarrow[u\to 0]{ }0
\quad\mbox{and}\quad
\frac{\Phi(u)}{u}\xrightarrow[u\to +\infty]{ }+\infty
$$
and (with a slight abuse)
$$
u\in L^\Phi
\quad \mbox{iff}\quad
\Phi(|u|)\in L^1.
$$
The choice $\Phi(u)=u^p$ leads to the usual Lebesgue spaces $L^p(\Omega)$, so in some sense the Orlicz spaces allow to measure integrability on an arbitrary scale (instead of just pure powers). In particular, choosing $\Phi$ carefully allows somehow to measure fine regularity scales between $L^1$ and any other $L^p$ space. The archetypical and popular example is $L^1\log L^1$, corresponding to the choice $\Phi(u)=u[\log u]^+$
---
**Just a few thoughts** (before my question is abruptly downvoted and deemed inappropriate for MO): Of course for a given regularity scale, i-e for any fixed $\Phi$, it is easy to cook-up a function $u=u\_\Phi$ that belongs to $L^1$ but not to $L^\Phi$ (just as one can easily tweak the negative exponents $\alpha=\alpha(d,p)<0$ so that $u(x)=|x|^\alpha$ is in $L^1$ but not in $L^p$ for fixed $p>1$ in dimension $d$). The question is therefore: can a (necessarily pathological) function $u$ be $L^1$ but really no better, i-e $u$ does not belong to any better $L^\Phi$ space? (I would perhaps call that such a function "essentially $L^1$".) Again, it is easy to construct $u\in L^1$ that does not belong to $L^p$ for *any* $p>1$, but typically this function might belong to $L^1\log L^1$. And if it doesn't then it might belong to $L^1\log\log L^1$, and so on... The game is: can one "exhaust all the regulatity scales" except for the minimal $L^1$ requirement? This seems highly unplausible, as the set of all possible regularity scales is highly uncountable, but who knows, some hidden monotonicity might save the day (going "down the scales"?)
Following the reverse line of thoughts, another related question is: if $u\in L^1$, can one always find some $\Phi=\Phi\_u$ such that $u\in L^\Phi$ is in fact better than merely integrable? My intuition is that this should be blatantly false, but I realized that this is vaguely similar in spirit to [Lusin's theorem](https://en.wikipedia.org/wiki/Lusin%27s_theorem), stating that essentially a measurable function is continuous. So perhaps this second counterintuitive statement of mine holds in some sense?
---
**Some context:** it turns out that for one of my research problems I am given a family $(u\_\xi)\_{\xi\in G}\in \mathcal P(\Omega)$ of Borel probability measures over $\Omega$. For $\xi$ in a smaller (but uncountable) set $F\subset G$ I know that $u\_\xi=u\_\xi(x) \mathcal L(dx)$ is in fact absolutely continuous w.r.t. the Lebesgue measure $\mathcal L|\_\Omega$, and I would like to conclude (by a very intricate argument that I shall not discuss here) that this is the case also for $\xi\in G$. What I need for my argument to work is to find first some $\Phi$ such that $u\_\xi\in L^\Phi$ somehow uniformly in $\xi\in F$, and then I would be able to control what happens for $\xi \in G$ using the same scale $\Phi$. Somehow, what I am trying to do is turning a mere pointwise absolute continuity into a globally quantified absolute continuity (the quantification being measured on a putative $L^\Phi$ scale). I am fully aware that at first glance this seems hopeless, but thinking again about Lusin's theorem gave me a vague hope. In any case, I would equally love to be proven wrong or to get some positive insight (a reference, better still).
| https://mathoverflow.net/users/33741 | An $L^1$ function but (really) no better? | There is a much more general result of Vallée-Poussin from which a negative answer to your question follows.
Let $(X,\mu)$ be a measure space. We say that a family of function $\mathcal{F}\subset L^1(X)$ is *equi-integrable* if for every $\varepsilon>0$ there is $\delta>0$ such that
$$
\sup\_{f\in\mathcal{F}} \int\_E |f|\, d\mu<\varepsilon
\quad
\text{whenever } \mu(E)<\delta.
$$
>
> **Theorem (de la Vallee Poussin).** Let $(X,\mu)$ be a measure space with $\mu(X)<\infty$ and let $\mathcal{F}\subset L^1(X)$. Then $\mathcal{F}$ is equi-integrable if and only if there is a Young function $\Phi$, $\lim\_{t\to\infty}\Phi(t)/t=\infty$ such that
> $$
> \sup\_{f\in\mathcal{F}}\int\_X\Phi(|f|)\, d\mu\leq 1.
> $$
>
>
>
Clearly a family consisting of a single function is equi-integrable (by absolute continuity of the integral) so for each $f\in L^1$ there is $\Phi$ with $\Phi(f)\in L^1$.
Also we can adjust the function $\Phi$ around zero arbitrarily without changing integrability of the function $\Phi(f)$ so you can have the condition $\Phi(t)/t\to 0$ as $t\to 0^+$.
You can find the proof of the de la Vallee Poussin theorem in
**C. Dellacherie, P.-A. Meyer,** *Probabilities and potential.* C. Potential theory for discrete and continuous semigroups. Translated from the French by J. Norris. North-Holland Mathematics Studies, 151. North-Holland Publishing Co., Amsterdam, 1988.
**M.M. Rao, Z.D, Ren,** *Theory of Orlicz spaces.* Monographs and Textbooks in Pure and Applied Mathematics, 146. Marcel Dekker, Inc., New York, 1991.
| 12 | https://mathoverflow.net/users/121665 | 420248 | 171,021 |
https://mathoverflow.net/questions/420241 | 1 | Let $f:\mathbb R^d \to \mathbb R$ be a "sufficiently smooth" function. For simplicity, we may consider $f$ to be an affine function, i.e $f(x) \equiv b-x^\top w$, for some $(w,b) \in \mathbb R^{d}$. Let $\Phi:\mathbb R \to (0,1)$ be the standard Gaussian CDF defind by
$$
\Phi(t):= \frac{1}{\sqrt{2\pi}}\int\_{-\infty}^t e^{-s^2/2}\,dx,
$$
and let $\theta:\mathbb R \to \mathbb R$ be a the Heaviside unit-step funciton defined by
$$
\theta (t) = \begin{cases}
0,&\mbox{ if }t < 0,\\
1/2,&\mbox{ if }t = 0,\\
1,&\mbox{ if }t>0.
\end{cases}
$$
Define a scalar $s \in [0,1]$ by
$$
s:= \mathbb E\_X[\theta(f(X))] = \mathbb P(f(X)>0).
$$
Now, let $X$ be a random variable on $\mathbb R^d$ with "sufficiently smooth" probability density function $\rho$, and let $X\_1,\ldots,X\_n$ be $n$ iid copies of $X$. Given a bandwidth parameter $h=h\_n > 0$, define the random variable $\hat{s}\_{n,h} \in (0,1)$ by
$$
\hat{s}\_{n,h} := \frac{1}{n}\sum\_{i=1}^n\Phi\left(\frac{f(X\_i)}{h}\right).
$$
Finally, define the random variable $\Delta \in [0, 2)$ by
$$
\Delta\_{n,h} := |\hat{s}\_{n,h} - s|.
$$
Intuitively, one would expect "$\Delta\_{n,h} \to 0$" in the limit when $n \to \infty$ such that $h \to 0$ and $nh^d \to \infty$. My goal is to quantify this convergence by upper-bounding $\Delta\_{n,h}$, by a deterministic quantity $\varepsilon\_{n,h}$ which goes to zero.
>
> **Question 1.** *Is it possible to give good quantitative upper-bounds for $\Delta\_{n,h}$, in terms of $n$ and $h$?, which are valid with high-probability over the $X\_i$'s ?*
>
>
>
>
> **Question 2.** *Is it even true that $\Delta\_{n,h} \to 0$ ?*
>
>
>
| https://mathoverflow.net/users/78539 | Bound error in approximating $E_x [H(f(x))]$ with random $(1/n) \sum_{i=1}^n \Phi(f(x_i)/h)$ where $H$ is Heaviside function and $\Phi$ is normal CDF | $\newcommand{\si}{\sigma}$Let
\begin{equation\*}
Y:=f(X),\quad Y\_i:=f(X\_i),
\end{equation\*}
so that $Y,Y\_1,Y\_2,\dots$ are iid real-valued random variables (r.v.'s).
Assume that (i) the r.v. $Y$ has a density $p\_Y$ continuous at $0$ and (ii) $E|Y|^k<\infty$ for some real $k>0$. Except possibly for entering indirectly into these two conditions -- (i) and (ii), the dimension $d$ or any further specifics concerning $X$ and $f$ will play no role in what follows.
Indeed, then
\begin{equation\*}
s=P(Y>0),
\end{equation\*}
\begin{equation\*}
\hat s\_{n,h}=\frac1n\sum\_{i=1}^n\Phi(Y\_i/h),
\end{equation\*}
\begin{equation\*}
E\hat s\_{n,h}=E\Phi(Y/h)=:\mu\_h.
\end{equation\*}
Let $n\to\infty$ and $h\downarrow0$. Then $\Phi(Y/h)\to1(Y>0)$ in probability and hence, by dominated convergence,
\begin{equation\*}
\begin{aligned}
\mu\_h=E\Phi(Y/h)&\to p:=P(Y>0),\\
\si^2\_h:=Var\,\Phi(Y/h)&\to pq,\\
a\_h:=E|\Phi(Y/h)-\mu\_h|^3&\to pq(p^2+q^2),
\end{aligned}
\end{equation\*}
where $q:=1-p$; without loss of generality, $0<p<1$.
By the Berry--Esseen inequality, for all real $z\ge0$,
\begin{equation\*}
P\Big(\Big|\frac{\hat s\_{n,h}-\mu\_h}{\si\_h/\sqrt n}\Big|>z\Big)
\le2(1-\Phi(z))+\frac{a\_h/\si\_h^{3/2}}{2\sqrt n},
\end{equation\*}
whence
\begin{equation\*}
|\hat s\_{n,h}-\mu\_h|=O\_P(1/\sqrt n). \tag{1}\label{1}
\end{equation\*}
Let real $c>0$ vary so that $c\to\infty$ but $ch\to0$. Then
\begin{equation\*}
\begin{aligned}
|\mu\_h-s|&\le E|\Phi(Y/h)-1(Y>0)| \\
&\le P(|Y|<ch)+(1-\Phi(c))P(|Y|\ge ch) \\
&\lesssim 2p\_Y(0)ch+e^{-c^2/(2+o(1))} \frac{E|Y|^k}{(ch)^k}.
\end{aligned}
\end{equation\*}
Choosing now $c$ more specifically, so that $c\sim\sqrt{(2k+1)\ln\frac1h}$, we get
\begin{equation\*}
|\mu\_h-s|=O\big(h\sqrt{\ln\tfrac1h}\big).
\tag{1'}\label{1'}
\end{equation\*}
Thus, in view of \eqref{1} and \eqref{1'},
\begin{equation\*}
|\hat s\_{n,h}-s|=O\_P\Big(\frac1{\sqrt n}+h\sqrt{\ln\frac1h}\Big),
\tag{2}
\end{equation\*}
as $n\to\infty$ and $h\downarrow0$.
| 3 | https://mathoverflow.net/users/36721 | 420251 | 171,022 |
https://mathoverflow.net/questions/420254 | 1 | I am looking for an algorithm with polynomial complexity where, given a strongly connected edge-weighted digraph I can find the minimal subgraph which connects some root vertex v to a known set of other vertices.
As an example, given a strongly connected edge-weighted digraph with vertices labeled a-z, I want to find the minimal subgraph rooted at node j that includes nodes b, f, g, and p.
In my case I am going to be using this to determine an optimal pipeline for doing image manipulation (I already have a strongly connected edge-weighted digraph for this application).
| https://mathoverflow.net/users/480464 | Minimum edge-weighted directed subgraph in polynomial time | This is an instance of the Directed Steiner Network Problem and as such it's solvable in time $|V(G)|^{O(|T|)}$ as proved by [Feldman & Ruhl (2006)](https://doi.org/10.1137/S0097539704441241), where $G$ is the given graph and $T\subset V(G)$ is the given subset of vertices (terminals).
| 1 | https://mathoverflow.net/users/7076 | 420259 | 171,025 |
https://mathoverflow.net/questions/420245 | 6 | Q1 : If $X \to Y \to Z$ are maps of schemes, is there a relation such as
$$\omega\_{X/Z} \overset{?}{=} \omega\_{Y/Z}|\_X \overset{L}{\otimes} \omega\_{X/Y}$$
between their dualizing complexes? Or maybe some kind of distinguished triangle?
The reason I think so is that I'm told $\omega\_{X/Z}$ is the determinant $\det \mathbb L\_{X/Z}$ in the sense of determinants of complexes (if $X \to Z$ is l.c.i.). The cotangent complex always sits in a distinguished triangle
$$\mathbb L\_{Y/Z}|\_X \to \mathbb L\_{X/Z} \to \mathbb L\_{X/Y},$$
which leads me to my next question:
Q2 : Does the determinant of complexes send distinguished triangles to tensor products?
Deligne emphasizes this for split exact sequences of vector spaces in [La Determinante de Cohomologie](https://publications.ias.edu/sites/default/files/Number58.pdf): $\det (V \oplus W) = \det V \otimes \det W$ even in the sense of complexes. It's hard for me to imagine that wouldn't extend to triangles in the derived category. The stacks project 0FJW promises to add more details later, so maybe this is obvious.
This came up in a class I'm in. Let $F : X \to X^{(p)}$ be the frobenius. I think surjectivity of the map $F\_\* \omega\_X \to \omega\_{X^{(p)}}$ is one of the conditions to be "F-rational." I'd like to understand the cokernel or cone of this morphism in general to see the obstruction to F-rationality and I naively expected something like $\omega\_{X/X^{(p)}} = Hom(F\_\* \mathcal O\_X, \mathcal O\_{X^{(p)}})$.
| https://mathoverflow.net/users/86614 | Distinguished triangle of dualizing complexes and/or determinants? | A very good reference for these topics is
Lipman, Joseph: Notes on derived functors and Grothendieck duality. *Foundations of Grothendieck duality for diagrams of schemes*, 1–259, Lecture Notes in Math., **1960**, Springer, Berlin, 2009.
For your first question, the issue is the pseudo functoriality of $(-)^!$ together with the characterization of this functor in terms of its value in the structure sheaf.
In more detail, with $f : X \to Y$, $g : Y \to Z$, $h : X \to Z$ and $h = g \circ f$. Assume that all maps are *finite type separated map of noetherian schemes*. In this case, we have that $h^! \cong f^! \circ g^!$ (loc. cit. Th (4.8.1)).
Second If moreover $f$ is perfect, i.e. $\mathcal{O}\_X$ is relatively perfect over $Y$ then
$$
f^! \mathcal{F} \cong f^\* \mathcal{F} \otimes^L f^! \mathcal{O}\_Y
$$
(loc. cit. Th (4.9.4)).
By introducing the notation $\omega\_f = f^! \mathcal{O}\_Y$, (and similarly for $g$ and $h$) you get your desired result under the hypothesis mentioned. But beware: in full generality $\omega\_f$ is a complex not concentrated in a single degree unless the morphisms are Cohen-Macaulay.
Indeed, as a consequence of the previous discussion, we have the following chain of isomorphisms
$$
\omega\_h \cong h^! \mathcal{O}\_Z \cong f^! g^! \mathcal{O}\_Z \cong f^! \omega\_g \cong
$$
$$
\cong f^\* \omega\_g \otimes^L f^! \mathcal{O}\_Y \cong
f^\* \omega\_g \otimes^L \omega\_f
$$
As for the formula $\omega\_f \cong \det \mathbb L\_{f}$, it looks plausible to me under complete intersection hypothesis. I don't know of a published proof. And I don't think it holds under more general hypothesis because without the complete intersection condition, $\mathbb L\_{f}$ is not perfect.
Here I interpret $\det$ as something like $L\Lambda^n$, the derived exterior power, where $n$ denotes the relative dimension.
Finally, if $f$ is finite if follows from sheafified duality (loc. cit. Cor. (4.3.6)) that
$$
f^! \mathcal{F} \cong \mathbf{R}\mathcal{H}om(f\_\*\mathcal{O}\_X, \mathcal{F})^{\tilde{}}
$$
If you substitute by Frobenius you get you last formula, if I understand well.
| 4 | https://mathoverflow.net/users/6348 | 420273 | 171,028 |
https://mathoverflow.net/questions/420269 | 1 | I am having some problems to understand the meaning of the following theorem due to Roitmann. I found this theorem in Voisin's book: Hodge Theory and Complex Algebraic Geometry, Volume II, page 289.
Theorem 10.14.
The Albanese map
\begin{equation\*}
alb\_X:CH\_0(X)\_{hom}\rightarrow Alb(X)
\end{equation\*}
Induces an isomorphism on torsion points.
What does it mean that the isomorphism is on torsion points? what is an intuitive way to think about torsion points?
| https://mathoverflow.net/users/150116 | Meaning of torsion points in a Roitman's theorem | It occurs to me that your question shouldn't be taken literally, and is simply asking about the meaning of the theorem. To appreciate it, one can ask what $CH\_0(X)$ looks like. As a first attempt, map it to something more concrete like the Albanese (which is just a complex torus over $\mathbb{C}$). It is easy to see that the map is surjective, and an isomorphism for curves. In higher dimensions, Mumford showed that the kernel can be huge. Given this, Roitman's theorem that the Albanese map restricted to the torsion subgroups yields an isomorphism is pretty surprising. For example if $H\_1(X,\mathbb{Z})=0$, the theorem would imply that the torsion subgroup of $CH\_0(X)$ is trivial.
| 4 | https://mathoverflow.net/users/4144 | 420279 | 171,031 |
https://mathoverflow.net/questions/420280 | 3 | For sets $x, y$ we write $x\leq y$, if there is an injection $\iota: x \to y$, and we write $x \leq^\* y$ if either $x = \emptyset$ or there is a surjection $s: y \to x$. In ${\sf (ZF)}$ we have that $x \leq y$ implies $y \leq^\* y$.
Consider the following statements:
>
> Partition principle (PP): For all sets $x, y$ we have that $x \leq^\* y$ implies $x\leq y$.
>
>
>
>
> Dual Cantor-Bernstein (CB)\*: For all sets $x,y$, if $x\leq^\* y$ and $y \leq^\* x$, then there is a bijection $\varphi: x\to y$.
>
>
>
Via the "normal" [Cantor-Bernstein theorem](https://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem), which a theorem of ${\sf (ZF)}$, we can show that (PP) imples (CB)\* in ${\sf (ZF)}$.
It seems to be [open](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&cad=rja&uact=8&ved=2ahUKEwjR2u3Dk5H3AhWU8LsIHWS9C7IQFnoECAMQAQ&url=https%3A%2F%2Fprojecteuclid.org%2Fjournals%2Fnotre-dame-journal-of-formal-logic%2Fvolume-31%2Fissue-3%2FThe-dual-Cantor-Bernstein-theorem-and-the-partition-principle%2F10.1305%2Fndjfl%2F1093635502.full&usg=AOvVaw3lP42qtQ0HrR-MxttLfcuW) whether (CB)\* implies the [Axiom of Dependent Choice (DC)](https://en.wikipedia.org/wiki/Axiom_of_dependent_choice). Since (PP) is stronger than (CB)\*, this begs the question:
Does (PP) imply (DC) in ${\sf (ZF)}$?
| https://mathoverflow.net/users/8628 | Does the partition principle imply (DC)? | Yes. This is a combination of facts.
1. $\sf PP$ implies that if a set $X$ can be mapped onto an ordinal $\alpha$, then $\alpha$ injects into $X$. In other words, it implies that $\aleph^\*(X)=\aleph(X)$ for any set $X$.
2. $\sf AC\_{WO}$, that is the axiom of choice from families of sets indexed by an ordinal, is equivalent to the statement "For every $X$, $\aleph^\*(X)=\aleph(X)$".
3. $\sf AC\_{WO}$ implies $\sf DC$. This is due to the fact that if $T$ is a tree of height $\omega$ without cofinal branches, we can consider the various well-orderable subtrees of $T$, define a rank function on those and use that to define a rank function on $T$. Then, using $\sf AC\_{WO}$ we can show that it is impossible that $T$ itself is not truly well-founded, which would mean that it has a maximal element.
| 8 | https://mathoverflow.net/users/7206 | 420282 | 171,032 |
https://mathoverflow.net/questions/420281 | 2 | Let $R$ be a commutative ring, $M$ be an $R$-module, and $N$ be a submodule of $M$. Assume that both $M$ and $N$ are flat, so we can identify $N\otimes\_RN$, $M\otimes\_RN$, and $N \otimes\_RN$ as submodules of $M\otimes\_RM$.
Is it true that $N\otimes\_R N = (M\otimes\_RN)\cap (N\otimes\_R M)$? If not in general, under which conditions is this true?
| https://mathoverflow.net/users/nan | Intersection and tensor product of flat modules | This is not true in general: take $M=R$, $N=Rx$ for some $x\in R$. Then $N\otimes \_RN=Rx^2$ while $M\otimes \_RN=N\otimes \_RM=Rx$.
It is true if $M/N$ is flat: this follows from Proposition 7 of §2.6 in Bourbaki's *Commutative algebra*, ch. I.
| 7 | https://mathoverflow.net/users/40297 | 420285 | 171,034 |
https://mathoverflow.net/questions/420283 | 5 | Let $\mathfrak g$ be a semisimple Lie algebra over $\mathbb C$, $\rho : \mathfrak g \to \operatorname{End}(V)$ a finite-dimensional irreducible representation and $x \in \mathfrak g$ regular with centralizer $Z$.
Does $\rho(Z)$ consist of polynomials in $\rho(x)$?
Examples where I know this is true:
* ~~$x$ is semisimple: then $Z$ is a Cartan subalgebra, $\rho(x)$ acts by distinct scalars on the weight spaces $V\_\lambda$ and the claim follows from Lagrange interpolation. (This argument doesn't work: the scalars need not be distinct.)~~
* $\mathfrak g = \mathfrak{sl}\_n$ and $\rho = $ standard rep., because $x$ regular implies its minimal polynomial has degree $n$ so the (trace $0$) polynomials in $x$ give all of $Z$.
| https://mathoverflow.net/users/50929 | Does the centralizer of a regular element in a semisimple Lie algebra act by polynomials? | It is false for the same reason the first example is wrong. Take $\rho$ for example the adjoint representation. As soon as $\mathfrak g$ has rank at least $2$ you can find a regular semisimple $x$ such that some two roots (whose difference is not a root) of $Z$ coincide on $x$, while this is not true for generic $y \in Z$.
The standard representation for $\mathfrak{sl}\_n$ is special in that the weights of a Cartan subalgebra differ by roots (rather than linear combinations of such).
| 5 | https://mathoverflow.net/users/50929 | 420292 | 171,036 |
https://mathoverflow.net/questions/420208 | 4 | Recall that a basis $(x\_{n})\_{n}$ for a Banach space $X$ is called boundedly complete if for every scalar sequence $(a\_{n})\_{n}$ with $\sup\_{n}\|\sum\_{i=1}^{n}a\_{i}x\_{i}\|<\infty$, the series $\sum\_{n=1}^{\infty}a\_{n}x\_{n}$ converges. It is well-known that $c\_{0}$ has no boundedly complete basis. My question is to give a quantitative version of the fact. More precisely, for a bounded sequence $(x\_{n})\_{n}$, we set $$\textrm{ca}((x\_{n})\_{n})=\inf\_{n}\sup\_{k,l\geq n}\|x\_{k}-x\_{l}\|.$$ Then $(x\_{n})\_{n}$ is norm-Cauchy if and only if $\textrm{ca}((x\_{n})\_{n})=0$.
Let $(x\_{n})\_{n}$ be a basis for a Banach space $X$. We set
$$\textrm{bc}((x\_{n})\_{n})=\sup\Big\{\textrm{ca}((\sum\_{i=1}^{n}a\_{i}x\_{i})\_{n})\colon
(\sum\_{i=1}^{n}a\_{i}x\_{i})\_{n}\subseteq B\_{X}\Big\},$$ where $B\_{X}$ is the closed unit ball of $X$. Clearly, $(x\_{n})\_{n}$ is boundedly complete if and only if $\textrm{bc}((x\_{n})\_{n})=0$.
I have proved that the $\textrm{bc}$-values of the unit vector basis and the summing basis in $c\_{0}$ are both equal to $1$. Therefore, I have the following question.
Question. $\textrm{bc}((x\_{n})\_{n})\geq 1$ for every basis $(x\_{n})\_{n}$ in $c\_{0}$ ?
Thank you!
| https://mathoverflow.net/users/41619 | $c_{0}$ has no boundedly complete basis | Proposition.
Let $(x\_n)$ be a basis for a Banach space $(X,\|\cdot\|\_X)$ that is not boundedly complete. Then $bc(x\_n) \ge 1$.
Proof: Let $(Z,\|\cdot\|\_Z)$ be the Banach space of all sequences $A=(a\_n)$ of scalars for which
$$
\|A\|\_Z = \sup\_N \|\sum\_{n=1}^N a\_n x\_n\|\_X < \infty.
$$
For $N =1,2,...$, define the operator $R\_N$ on $Z$ by
letting the first $N-1$ coordinates of $R\_N(A)$ be zero and the other coordinates to agree with the coordinates of $A$ (so $R\_1(A)= A$). Let $A$ be any element of $Z$ s.t.
$$
!A! := \limsup\_N \|R\_N(A)\|\_Z >0.
$$
That such an $A$ exists is equivalent to the statement that $(x\_n)$ is not boundedly complete.
Now if $\|A\|\_Z$ were less than or equal to $ !A!$, then we would be done. That need not be true, but to complete the proof it is enough to show that
$$
\liminf\_N \|R\_N(A)\|\_Z -!R\_N(A)! \le 0,
$$
which is obvious from the definition of $!\cdot!$ and the observation that for every $N$, $!R\_N(A)! = !A!$.
Added 4/15/22.
Look at it this way. Since $!A! > 0$, divide by $!A!$ to see that WLOG $!A! = 1$. Now maybe
$\|R\_1(A)\|\_Z$ is much larger than one, but there is $N\_0$ so that for all $N>N\_0$,
$\|R\_N(A)\|\_Z$ is less than $1+\epsilon$. By replacing $A$ with $R\_{N\_0}(A)$, we can assume WLOG that $\|R\_N(A)\|\_Z$ is less than $1+\epsilon$ for all $N$ ("WLOG" because $!A!=!R\_N(A)!$ for all $N$). This means that ALL partial sums $\sum\_{n=k}^j a\_n x\_n$ in $X$ with any starting point have norm at most $1+\epsilon$, but arbitrarily far out you have partial sums $\sum\_{n=k}^j a\_n x\_n$ that have norm in $X$ at least $1-\epsilon$.
| 2 | https://mathoverflow.net/users/2554 | 420300 | 171,038 |
https://mathoverflow.net/questions/420272 | 7 | The minimal model program aims to find a minimal representative in the birational class of a given variety with reasonable singularities. Assuming this has been done, it seems natural to ask what these minimal models look like.
Is it feasible to ask for a classification of minimal (complex) varieties, say for threefolds? If so, has this been done? Or are there known wild examples which show that seeking a classification is too ambitious?
| https://mathoverflow.net/users/126543 | Is there a classification of minimal algebraic threefolds? | It depends what you mean by classification.
The key results for surfaces IMO are: 1) Any surface $S$ of general type has a canonical model given by $S\_{can}:={\rm Proj} R(K\_S)$ and a unique minimal model given by the minimal resolution of $S\_{can}$. 2) The canonical volume ${\rm vol}(S)=K\_{S\_{can}}^2$ is an integer and 3) $5K\_{S\_{can}}$ is very ample.
In particular for fixed $v$, canonical modules of surfaces of general type of volume ${\rm vol}(S)=v$ can be parametrized by a variety of finite type.
On the other hand, for any fixed invariants ${\rm vol}(S)$, $h^1(\mathcal O \_S)$ etc it can be impossibly hard to determine the moduli space of canonical models of such surfaces of general type.
All of this actually generalizes to all dimensions (with a few caveats / changes). 1) holds by Birkar-Cascini-Hacon-McKernan: $R(K\_X)=\oplus H^0(mK\_X)$ is finitely generated and we take $X\_{can}={\rm Proj}(R(K\_X))$; there are also minimal models which have terminal singularities. One canonical model can have multiple minimal models (but only finitely many and they are connected by flops). 2-3) By a result of of Hacon-McKernan, Takayama, Tsuji, for any fixed dimension, the canonical volumes ${\rm vol}(X)=K\_{X\_{can}}^{\rm dim(X)}$ belong to a discrete set and for fixed dimension $d$ and volume $v$ there is an integer $m$ such that $mK\_{X\_{can}}$ is very ample.
Explicit results are few and far apart (the work of J.Chen and M.Chen comes to mind).
| 15 | https://mathoverflow.net/users/19369 | 420303 | 171,040 |
https://mathoverflow.net/questions/420195 | 3 | I have a linear system
\begin{align\*}
\left[\begin{array}{cccc}
1 & 2 & 1 & -1 \\
3 & 2 & 4 & 4 \\
4 & 4 & 3 & 4 \\
2 & 0 & 1 & 5 \\
\end{array}\right]
\left[\begin{array}{c}
w \\ x \\ y \\z
\end{array}\right] =
\left[\begin{array}{c}
5 \\ 16 \\ 22 \\ 15
\end{array}\right],
\end{align\*}
whose matrix $\bf{A}$ is not diagonally dominant. What are the iterative methods that can be used to find the solution? I have tried Jacobi method, Gauss-Seidel method and SOR method but nothing works (the output diverges). The answer is ${\bf x}=[16,-6,-2,-3]^T$.
| https://mathoverflow.net/users/99453 | Iterative methods for linear system with non-diagonally dominant matrix | You can see your equation as $f(u)=Au-b$, with $f(u)=0$. You can use some numerical method to solve it. One way is to use the minimization problem
$$\min\_u g(u),\qquad g(u)=\frac{1}{2}\|f(u)\|^2.$$ This gives you the equation $$0=\nabla g(u)=A^T(Au-b)\quad \text{ or } \quad A^TAu=A^Tb$$ which gives you the possible minimizers.
To your matrix $A$ and vector $b$, you can use the [conjugate gradient method](https://en.wikipedia.org/wiki/Conjugate_gradient_method) to solve $$A^TAu=A^Tb,$$ and it [gives you the answer in four steps](https://www.mycompiler.io/view/2OmHyUE6x5m).
If you prefer to play with more steps, you can try [gradient descent method](https://en.wikipedia.org/wiki/Gradient_descent).
You can find more searching for "[\(Au=b\) gradient](https://www.searchonmath.com/result/?q=%5C%28Au%3Db%5C%29%20%20gradient%20) " on SearchOnMath.
**Note:** If you use the gradient descent method to minimize $h(u)=u^TAu-2u^Tb$ directly, you are going in the direction $$\frac{du}{dt}=- \nabla h(u)=-2(Au-b),$$ which is [unstable](https://en.wikipedia.org/wiki/Stability_theory#Non-linear_autonomous_systems), since $A$ has one negative eigenvalue and three positive ones.
| 1 | https://mathoverflow.net/users/478686 | 420305 | 171,041 |
https://mathoverflow.net/questions/420121 | -1 | In the paper [1] below, among other things, Carlitz introduced weighted Stirling numbers of the second kind $R(n,k,r)$. He also proved that the numbers $R(n,k,r)$ can be generated by
\begin{equation\*}%\label{S(n,k,x)-dfn}
\frac{(\textrm{e}^z-1)^k}{k!}\textrm{e}^{\lambda z}=\sum\_{n=k}^\infty R(n,k,\lambda)\frac{z^n}{n!}
\end{equation\*}
and can be explicitly expressed by
\begin{equation\*}%\label{S(n,k,x)-satisfy-eq}
R(n,k,r)=\frac1{k!}\sum\_{j=0}^k(-1)^{k-j}\binom{k}{j}(r+j)^n
\end{equation\*}
for $r\in\mathbb{R}$ and $n\ge k\ge0$.
Specially, when $\lambda=0$, the quantity $R(n,k,0)$ becomes the Stirling numbers of the second kind $S(n,k)$.
I guess that the identities
\begin{equation}\label{QGWSID1}\tag{SID1}
\sum\_{j=1}^k(-1)^j\binom{k}{j} \frac{R\bigl(2m+j-1,j,-\frac{j}2\bigr)}{\binom{2m+j-1}{j}}=0, \quad k,m\in\mathbb{N}
\end{equation}
and
\begin{equation}\label{QGWSID2}\tag{SID2}
\sum\_{j=1}^k(-1)^j\binom{k}{j} \frac{R\bigl(2m+j,j,-\frac{j}2\bigr)}{\binom{2m+j}{j}}=0, \quad k>m\ge1
\end{equation}
should be valid.
Stronger but simpler than \eqref{QGWSID1}, the identity
\begin{equation}\label{QGWSID3}\tag{SID3}
R\biggl(2m+j-1,j,-\frac{j}2\biggr)
=\frac{(-1)^j}{j!}\sum\_{\ell=0}^j(-1)^{\ell}\binom{j}{\ell}\biggl(\ell-\frac{j}{2}\biggr)^{2m+j-1}
=0
\end{equation}
for $k,m\in\mathbb{N}$ should be true.
These guesses \eqref{QGWSID1}, \eqref{QGWSID2}, and \eqref{QGWSID3} are related to series expansions at $x=0$ of the functions
$$
\biggl(\frac{\sin x}{x}\biggr)^r
\quad\text{and}\quad
\biggl(\frac{\sinh x}{x}\biggr)^r
$$
for real number $r\in\mathbb{R}$. For details, please read the arXiv preprint [2] below.
Could you please confirm or deny these identities \eqref{QGWSID1}, \eqref{QGWSID2}, and \eqref{QGWSID3} involving the weighted Stirling numbers of the second kind $R(n,k,r)$?
References
1. L. Carlitz, *Weighted Stirling numbers of the first and second kind, I*, Fibonacci Quart. **18** (1980), no. 2, 147--162.
2. Feng Qi, *Series expansions for any real powers of (hyperbolic) sine functions in terms of weighted Stirling numbers of the second kind*, arXiv (2022), available online at <https://arxiv.org/abs/2204.05612> or <https://doi.org/10.48550/arXiv.2204.05612>.
| https://mathoverflow.net/users/147732 | Could you please confirm or deny two identities involving weighted Stirling numbers of the second kind? | $R(n, k, -\tfrac k2)$ is just the central factorial number $T(n, k)$. (Given the definition of the central factorial numbers, it may be more natural to use them in your context than $R$).
Consider [A136630](https://oeis.org/A136630). We have $\operatorname{A136630}(n, k) = 2^{n-k} T(n,k)$ (this may be stated explicitly by Comtet, Riordan, or Charalambides; if not, it will follow from the egf) and the combinatorial interpretation that $\operatorname{A136630}(n, k)$ counts set partitions of $n$ elements into $k$ odd-sized blocks. This immediately gives (SID3), since an even-sized set cannot be partitioned into an odd number of odd-sized blocks, nor an odd-sized set partitioned into an even number of odd-sized blocks.
(SID2) is equivalent to the following theorem: if $1 \le m < k$ then $$\sum\_{j=1}^k (-1)^{k-j} \binom{2m+k}{2m+j} \operatorname{A136630}(2m+j, j) = 0$$
This theorem has a combinatorial proof: consider set partitions of $2m+k$ elements into $k$ odd-sized blocks where blocks of size $3$ or greater are coloured red and singleton blocks can be coloured red or blue. Then the sum counts such set partitions weighted by $(-1)^{\textrm{number of blue partitions}}$. ($j$ is the number of red partitions). Observe that partitions containing at least one singleton can be paired with the partition which differs only in the colour assigned to the singleton with the smallest element, so that the sum counts the number of partitions of $2m+k$ elements into $k$ odd-sized blocks of at least $3$ elements each. But if $k > m$ there are no such partitions, proving the theorem.
For the equivalence to (SID2) we rearrange $$\begin{eqnarray\*}
\textrm{LHS} &=& \sum\_{j=1}^k (-1)^j \binom{k}{j} \frac{R\bigl(2m+j,j,-\frac{j}2\bigr)}{\binom{2m+j}{j}} \\
&=& \sum\_{j=1}^k (-1)^j \frac{k!(2m)!}{(k-j)! (2m+j)!} T(2m+j,j) \\
&=& \frac{(-1)^k}{2^{2m} \binom{2m+k}{k}} \sum\_{j=1}^k (-1)^{k-j} \binom{2m+k}{2m+j} \operatorname{A136630}(2m+j, j) \\
\end{eqnarray\*}$$
| 2 | https://mathoverflow.net/users/46140 | 420309 | 171,044 |
https://mathoverflow.net/questions/420276 | 3 | I am considering the following wave equation (for $\phi=\phi(x,t)$)
$$
\phi\_{tt} - \Delta \phi = a |\nabla \phi|^2, \quad (x,t) \in \mathbb{R}^3 \times \mathbb{R}
$$
where $\nabla$ is just spatial gradient, i.e., $\nabla \phi= (\partial\_{x\_1} \phi, \partial\_{x\_2} \phi, \partial\_{x\_3} \phi)$ and $a \in \mathbb{R}$. And assume that initial data $\phi(x,0)$ and $\partial\_t \phi(x,0)$ are regular enough and compactly supported.
This PDE in some sense is the complementary case of the PDE in John's blow-up theorem, where we have
$$
\phi\_{tt} - \Delta \phi = (\partial\_t\phi)^2.
$$
Now, I have the following questions: is this PDE well-posed? Or the solution will blow-up? Is there any paper on this PDE in 3+1 dimensions?
| https://mathoverflow.net/users/468783 | On a nonlinear wave equation | By replacing $\phi$ by $-\phi$ you can set $\alpha$ to be positive (or negative, if you wish). By replacing $\phi$ by $\lambda \phi$ you can rescale away $\alpha$. So you can set $\alpha$ to be either $+1$ or $-1$ as you wish.
So for now let's consider the situation you are solving
$$ \partial^2\_{tt}\phi - \Delta \phi = |\nabla \phi|^2 $$
First observation: suppose $\phi$ is a $C^3$ solution with compactly supported initial data, then $\phi$ is uniformly bounded from below on $\mathbb{R}^3\times [0,\infty)$. This is due to the fact that on $\mathbb{R}^3$ the fundamental solution to the wave equation is signed (a fact also used by John in his proof), so that the solution to the above equation is pointwise $\geq$ the solution to the linear equation with the same initial data. The latter is globally uniformly bounded, and hence $\phi$ is uniformly bounded below.
Now $v = e^\phi - 1$; you have that $v$ is also $C^3$, and there exists $\beta$ such that $v + 1 > \beta$ uniformly on $\mathbb{R}^3\times [0\infty)$, using the uniform bound on $\phi$.
Now we compute
$$ (\partial^2\_{tt} - \Delta) v = e^\phi (\partial^2\_{tt} - \Delta) \phi + e^\phi (\partial^2\_t \phi - |\nabla\phi|^2) = \frac{1}{1+v}( \partial\_t v)^2 $$
The lower bound above ensures that the denominator on the RHS is never zero.
Suppose now that $\phi$ is globally bounded *above* also: then $(1+v)$ is bounded above globally by some $M > 0$. Then the condition for Theorem 2 in John's 1981 paper is satisfied and the result follows.
So the remaining question is whether it is possible for $\phi$ to blow-up only at infinity and not at any finite time. I think there may be a way to rule this out, but that would require checking some more technical details so I'll have to report back later.
| 5 | https://mathoverflow.net/users/3948 | 420320 | 171,046 |
https://mathoverflow.net/questions/420220 | 1 | Consider pseudo Gaussian densities for $0<s<t$ and $x,y\in\mathbb R$
$$f(s,x,t,y):=\frac{1}{\sqrt{2\pi A(s,t,y)}}\exp\left(-\frac{(y-x)^2}{2A(s,t,y)}\right)\quad\mbox{and} \quad g(s,x,t,y):=\frac{1}{\sqrt{2\pi B(s,t,y)}}\exp\left(-\frac{(y-x)^2}{2B(s,t,y)}\right),$$
where $A(s,t,y):=\int\_s^t k(u,y)/(1+a(u))du$ and $B(s,t,y):=\int\_s^tk(u,y)/(1+b(u))du$. I'm interested in the dependence of the difference $f-g$ on the parameters $a,b$. Assume $a, b: \mathbb R\_+ \to [0,1]$ are continuous, $k: \mathbb R\_+\times\mathbb R \to [1,2]$ is $1-$Lipschitz. Does there exist $C>0$ depending only on $T>0$ s.t.
\begin{eqnarray}
\left|\int\_0^\infty f(0,0,t,y)dy - \int\_0^\infty g(0,0,t,y)dy\right | &\le& Ct^{1/2}\|a-b \|\_t,\quad \forall 0<t\le T\quad (\ast) \\
\int\_0^t \left|\int\_0^\infty \partial\_sf(s,0,t,y)dy - \int\_0^\infty \partial\_s g(s,0,t,y)dy\right |ds &\le& C(t-s)^{1/2}\|a-b \|\_t,\quad \forall 0<s<t\le T\quad (\star),
\end{eqnarray}
where $\|a-b \|\_t:=\max\_{0\le u\le t}|a(u)-b(u)|$.
PS : $(\star)$ can be shown if
\begin{eqnarray}
\left|\int\_0^\infty \partial\_sf(s,0,t,y)dy - \int\_0^\infty \partial\_s g(s,0,t,y)dy\right | \le C(t-s)^{-1/2}\|a-b \|\_t.
\end{eqnarray}
Iosif Pinelis has shown in [How does the integral of pseudo Gaussian kernel on $(0,\infty)$ depend on its variance?](https://mathoverflow.net/questions/419673/how-does-the-integral-of-pseudo-gaussian-kernel-on-0-infty-depend-on-its-va) that
$$\left|\int\_0^\infty p(x)dx \int\_0^\infty f(s,x,t,y)dy - \int\_0^\infty p(x)dx\int\_0^\infty g(s,x,t,y)dy\right |\le C(t-s)^{1/2}\|a-b \|\_t,$$
where $p$ is some probability density on $(0,\infty)$.
| https://mathoverflow.net/users/nan | Questions on the integral of pseudo Gaussian kernel and its derivative on $(0,\infty)$ | $\newcommand{\De}{\Delta}\newcommand{\vpi}{\varphi}$Let
\begin{equation\*}
A:=A(y):=A(0,t,y),\quad B:=B(y):=B(0,t,y),
\end{equation\*}
so that
\begin{equation\*}
f(0,0,t,y)=\frac1{\sqrt{2\pi}}\,\vpi\_{A(y)}(y), \quad g(0,0,t,y)=\frac1{\sqrt{2\pi}}\,\vpi\_{B(y)}(y),
\end{equation\*}
where
\begin{equation\*}
\vpi\_a(u):=\frac1{\sqrt a}\,\exp\Big(-\frac{u^2}{2a}\Big).
\end{equation\*}
Note that
\begin{equation\*}
\frac{\partial}{\partial a}\vpi\_a(u)
=\frac12\Big(\frac{u^2}{a^{5/2}}-\frac1{a^{3/2}}\Big)\exp\Big(-\frac{u^2}{2a}\Big).
\end{equation\*}
So,
\begin{equation\*}
2\sqrt{2\pi}\,\int\_0^\infty dy\,[f(0,0,t,y)-g(0,0,t,y)] \tag{1}\label{1}
=\int\_0^1 dv\,I(v),
\end{equation\*}
where
\begin{equation\*}
I(v):=I\_1(v)-I\_2(v), \tag{2}\label{2}
\end{equation\*}
\begin{equation\*}
I\_1(v):=\int\_0^\infty dy\,H(y)
\frac{y^2}{c\_v(y)^{5/2}}\exp\Big(-\frac{y^2}{2c\_v(y)}\Big),
\end{equation\*}
\begin{equation\*}
I\_2(v):=\int\_0^\infty dy\,H(y)
\frac1{c\_v(y)^{3/2}}\exp\Big(-\frac{y^2}{2c\_v(y)}\Big),
\end{equation\*}
\begin{equation\*}
H:=B-A,
\end{equation\*}
\begin{equation\*}
c\_v:=A+v(B-A).
\end{equation\*}
Next is the crucial step:
\begin{equation\*}
I\_1(v)=I\_{11}(v)+I\_{12}(v), \tag{3}\label{3}
\end{equation\*}
where
\begin{equation\*}
I\_{11}(v):=\int\_0^\infty dy\,\exp\Big(-\frac{y^2}{2c\_v(y)}\Big)
\Big(\frac{y}{c(y)}-\frac{y^2c'(y)}{c(y)^2}\Big)
\frac{y}{c(y)^{3/2}}H(y),
\end{equation\*}
\begin{equation\*}
I\_{12}(v):=\int\_0^\infty dy\,\exp\Big(-\frac{y^2}{2c\_v(y)}\Big)
\frac{y^3c'(y)}{c(y)^{7/2}} H(y),
\end{equation\*}
\begin{equation\*}
c(y):=c\_v(y),
\end{equation\*}
and $c'$ is the derivative of the Lipschitz function $c=c\_v$; this derivative exists almost everywhere (a.e.), since $k$ is $1$-Lipschitz. Moreover,
\begin{equation\*}
|c'|\le t
\end{equation\*}
a.e.
Note also that for $y\ge0$ and $t>0$
\begin{equation\*}
t/2\le c(y)\le2t,\quad |H(y)|\ll t\,\De a,\quad |H'(y)|\ll t\,\De a,
\end{equation\*}
where
\begin{equation\*}
\De a:=\|a-b\|\_t
\end{equation\*}
and $E\ll F$ means that $|E|\le CF$ for some universal real constant $C$.
So,
\begin{equation\*}
|I\_{12}(v)|\ll\int\_0^\infty dy\,\exp\Big(-\frac{y^2}{4t}\Big)
\frac{y^3\,t}{t^{7/2}}\, t\,\De a \asymp\sqrt t\,\De a. \tag{4}\label{4}
\end{equation\*}
Integrating by parts, we have
\begin{equation\*}
\begin{aligned}
I\_{11}(v)&=-\int\_0^\infty dy\,\Big[\frac d{dy}\exp\Big(-\frac{y^2}{2c\_v(y)}\Big)\Big]
\frac{y}{c(y)^{3/2}}H(y) \\
&=\int\_0^\infty dy\,\exp\Big(-\frac{y^2}{2c\_v(y)}\Big)
\frac d{dy}\Big[\frac{y}{c(y)^{3/2}}H(y)\Big] \\
&=I\_2(v)+I\_{111}(v)-\frac32\,I\_{112}(v),
\end{aligned}
\tag{5}\label{5}
\end{equation\*}
where
\begin{equation\*}
\begin{aligned}
I\_{111}(v)&:=\int\_0^\infty dy\,\exp\Big(-\frac{y^2}{2c\_v(y)}\Big)
\frac{y}{c(y)^{3/2}}H'(y),
\end{aligned}
\end{equation\*}
\begin{equation\*}
\begin{aligned}
I\_{112}(v)&:=\int\_0^\infty dy\,\exp\Big(-\frac{y^2}{2c\_v(y)}\Big)
\frac{yc'(y)}{c(y)^{5/2}}H(y).
\end{aligned}
\end{equation\*}
Similarly to \eqref{4}, we get
\begin{equation}
|I\_{111}(v)|+|I\_{112}(v)|\ll \sqrt t\,\De a. \tag{6}\label{6}
\end{equation}
Collecting \eqref{1}, \eqref{2}, \eqref{3}, \eqref{4}, \eqref{5}, and \eqref{6}, we conclude that
\begin{equation\*}
\Big|\int\_0^\infty dy\,[f(0,0,t,y)-g(0,0,t,y)]\Big|\ll \sqrt t\,\De a,
\end{equation\*}
which proves the first inequality of the two ones in question.
The other inequality can apparently be proved similarly. Since this answer is already rather long and complicated, I will leave the other inequality as an exercise (or to be posted elsewhere).
| 1 | https://mathoverflow.net/users/36721 | 420321 | 171,047 |
https://mathoverflow.net/questions/420319 | 15 | The Hermitian matrices form a real vector space where we have a Lebesgue measure. In the set of Hermitian matrices with Lebesgue measure, how does it follow that the set of Hermitian matrices with repeated eigenvalue is of measure zero?
This result feels extremely natural but I do not see an immediate argument for it.
| https://mathoverflow.net/users/457901 | Why is the set of Hermitian matrices with repeated eigenvalue of measure zero? | Call $S$ the set of matrices with repeated eigenvalues and fix a hermitian matrix $A\not\in S$. In the vector space of hermitian matrices, any line through $A$ intersects $S$ in at most finitely many points. From this it easily follows that $S$ is negligible (using polar coordinates centered at $A$).
To check the claim, note that a line through $A$ consists of matrices of the form $M(t)=(1-t)A+tB$, for some $B\in S$. Hence, the characteristic polynomial of $M(t)$ has the form $\lambda^n+\sum\_{k=0}^{n-1}p\_k(t)\lambda^k$ for some polynomials $p\_k(t)$.
There is a polynomial expression of the coefficients $a\_k$ of a polynomial $\lambda^n+\sum\_{k=0}^{n-1}a\_k\lambda^k$ which vanishes precisely when it has repeated roots: indeed, calling $\alpha\_i$ the roots, you can consider the expression $\prod\_{i<j}(\alpha\_i-\alpha\_j)^2$. Since it's symmetric, it is in fact equal to some $P(a\_0,\dots,a\_{n-1})$. For our matrices, this means that $M(t)\in S$ precisely when $P(p\_0(t),\dots,p\_{n-1}(t))=0$. The left-hand side is a polynomial, which is not trivial since it does not vanish at $t=0$. Hence, $M(t)\in S$ only for finitely many $t$.
| 10 | https://mathoverflow.net/users/36952 | 420322 | 171,048 |
https://mathoverflow.net/questions/420326 | 3 | It is known that the property of being a worldly cardinal is not absolute (a cardinal $\kappa$ is worldly iff $V\_{\kappa} \vDash \textsf{ZFC}$). See [here](http://jdh.hamkins.org/worldly-cardinals-are-not-always-downwards-absolute/) and [here](https://mathoverflow.net/questions/130019/forcing-mildly-over-a-worldly-cardinal) for more.
This said, it is the case that the existence of two worldly cardinals has strictly greater consistency strength than the existence of one worldly cardinal, and any proof I can see of this will require that if $\kappa\_1 < \kappa\_2$ are worldly cardinals, then $V\_{\kappa\_2}$ must believe $\kappa\_1$ is worldly. (Easiest proof: show that $V\_{\kappa\_2}$ believes that $V\_{\kappa\_1} \vDash \textsf{ZFC}$.)
I wasn’t sure how to show that this was the case — I was thinking that maybe because $V\_{\kappa\_2}$ is transitive it must agree with $V$ on whether $V\_{\kappa\_1}$ models $\textsf{ZFC}$, but I’m not totally clear on/sure of this.
| https://mathoverflow.net/users/120461 | Why does the second smallest worldly cardinal believe the smallest worldly cardinal is worldly? | The truth of a first-order sentence $\varphi$ in a structure $\mathfrak{M}$ is absolute between $V$ (= reality) and sufficiently large transitive sets containing $\mathfrak{M}$. In particular, already $V\_{\kappa+2}$ correctly computes the full first-order theory of $V\_\kappa$ for each infinite ordinal $\kappa$, and even that is overkill. The key point is that truth of a first-order sentence in a structure is witnessed by an appropriate family of Skolem functions, and the relevant properties are all low-complexity.
Since $\kappa\_2>\kappa\_1+2$, the result follows.
(This is treated in more detail in Barwise's book *Admissible sets and structures*, if memory serves, but there really are no hidden surprises.)
---
EDIT (addressing Francois' comment above): It's also true that worldliness relativizes downwards to (but not necessarily upwards from!) $L$, but this isn't actually related to the above. The issue is "worldliness-in-$L$" is a property of $V\_\kappa^L$ (which for reasonably closed $\kappa$ is just $L\_\kappa$), not of the genuine $V\_\kappa$ ($=V\_\kappa^V$ if you like) considered inside $L$. So the key fact is Godel's theorem that if $V\_\kappa\models\mathsf{ZFC}$ then $L\_\kappa\models\mathsf{ZFC}$. But this actually involves a nontrivial analysis of $L$. And the general failure of *upwards* absoluteness ruins any hope of an application the other way.
(To see that worldliness need not be absolute upwards from $L$, consider what happens if $\kappa$ is worldly in $L$ but countable in $V$ - which can happen, e.g. by forcing over $L$ with $Col(\omega,\kappa)$.)
| 7 | https://mathoverflow.net/users/8133 | 420331 | 171,053 |
https://mathoverflow.net/questions/420317 | 2 | I'm trying to show that for an ordinary manifold $X$ and a supermanifold $S$, supermanifold morphisms $\varphi:S\to TX$ are one-to-one to the pairs $(f,F) $ where $f:C^\infty(X)\to C^\infty(S)$ is a super $\mathbb{R}$-algebra homomorphism and $F:C^\infty(X)\to C^\infty(S)$ is an even derivation with respect to $f$, i.e. $F$ is a parity-preserving $\mathbb{R}$-linear map and $F(ab)=F(a)f(b)+f(a)F(b)$ for any $a,b\in C^\infty(X)$.
Intuitively, $f$ is the pullback of smooth maps from the base manifold $X$ to $S$ and $F$ is a global section to the pullback vector bundle $f^\*TX$ on $S$. Hence if we let $p:TX\to X$ be the usual projection, then $f= \varphi^\*\circ p^\* $. The remaining data of $\varphi$ should give $F$, but I cannot see how.
If $S$ is also an ordinary manifold, then $F$ can be defined point-wisely. We can express $\varphi$ point-wisely by $\varphi(p)=(f(p),V\_{f(p)})$ for each $p\in S$, where $f(p)\in X$ and $V\_{f(p)}\in T\_{f(p)}X$. For any $a\in C^\infty(X)$, $F(a)\in C^\infty(S)$ is then the function that sends $p\in S$ to $V\_{f(p)}a\in \mathbb{R}$.
However, if $S$ is a supermanifold, then the point-wise definition no longer works. I tried to formulate the point-wise definition in a non-point-wise way but failed.
I guess that $F$ should be a composition $C^\infty(X)\xrightarrow{D}C^\infty(TX)\xrightarrow{\varphi^\*}C^\infty(S)$ for some natural derivation $D:C^\infty(X)\to C^\infty(TX)$ with respect to $p^\*$, but I cannot see any derivation that arises "naturally".
Thanks in advance for any help.
| https://mathoverflow.net/users/167862 | One-to-one correspondence between super morphisms $\varphi:S\to TX$ and pairs $(f:C^\infty(X)\to C^\infty(S) ,F\in Der_f(C^\infty(X),C^\infty(S))$ | The morphism $$\def\T{{\rm T}} φ:S→\T X$$ can be identified with the homomorphism of algebras $$\def\Ci{{\rm C}^∞} \Ci(\T X)→\Ci(S).$$
The algebra $\Ci(\T X)$ can be identified with the $\Ci$-symmetric algebra $$\def\CiSym{\mathop{\rm\Ci Sym}\nolimits} \CiSym\_{\Ci(X)}(\Gamma(\T^\*X))$$ of the $\Ci(X)$-module $\Gamma(\T^\* X)$ of smooth sections of the cotangent bundle of $X$.
Here $\Ci$-symmetric algebras are defined using exactly the same universal property as symmetric algebras, but working in the category of $\Ci$-rings instead of the category of commutative real algebras.
By the universal property of $\Ci$-symmetric algebras, the homomorphism
$$\CiSym(\Gamma(\T^\*X))→\Ci(S)$$ can be identified
with a homomorphism of $\Ci$-algebras (equivalently, commutative real algebras) $$f:\Ci(X)→\Ci(S)$$
together with a morphism of $\Ci(X)$-modules $$Ψ:\Gamma(\T^\*X)→\Ci(S).$$
The latter morphism can be identified with a derivation $$F:\Ci(X)→\Ci(S)$$
with respect to $f$: we set $\def\d{{\rm d}} F(g)=Ψ(\d g)$, where $\d g∈Γ(\T^\* X)$ is the differential of $g$.
This approach works equally well for ordinary manifolds, supermanifolds, ${\bf Z}$-graded manifolds, derived manifolds, etc.
| 2 | https://mathoverflow.net/users/402 | 420335 | 171,055 |
https://mathoverflow.net/questions/420337 | 4 | Let $E$ be a Landweber exact ring spectrum. That is, we have a map of homotopy ring spectra $MU\rightarrow E$ and an isomorphism of homology theories $E\_\*X\simeq MU\_\*X\otimes\_{MU\_\*}E\_\*$. Is the homotopy type of $E$ determined by the graded ring $E\_\*$?
My best guess is that the answer is "no" which would ideally be proved by finding a (graded) ring $R$ and two (graded), Landweber exact, formal group laws (graded ring maps) $e,f:MU\_\*\rightarrow R$, such that the spectra representing the two homology theories are not homotopy equivalent.
| https://mathoverflow.net/users/163893 | Are Landweber exact spectra determined by their coefficient ring? | Put $R\_\*=\mathbb{Z}\_{(p)}[x,y,y^{-1}]$ with $|x|=|y|=2$. Define $f,g\colon BP\_\*\to R\_\*$ by
\begin{align\*}
f(v\_i) &= \begin{cases}
y^{p-1} & \text{ if } i = 1 \\
0 & \text{ if } i > 1
\end{cases} \\
g(v\_i) &= \begin{cases}
x^{p-1} & \text{ if } i = 1 \\
y^{p^2-1} & \text{ if } i = 2 \\
0 & \text{ if } i > 2
\end{cases} \\
\end{align\*}
This gives Landweber exact ring spectra $A\_f$ and $A\_g$ of heights one and two respectively, so for a finite spectrum $X$ of type $2$ we have $A\_f\wedge X = 0 \neq A\_g\wedge X$, so $A\_f$ and $A\_g$ are not equivalent as spectra.
| 9 | https://mathoverflow.net/users/10366 | 420342 | 171,058 |
https://mathoverflow.net/questions/420328 | 9 | In differential geometry, Lie's theorems allow us to integrate any Lie algebra representation to a Lie group representation. The algebraic version of this is more complicated (and I'm not terribly familiar with it), but over a field of characteristic zero we can still get some results from Tannaka duality. This all fails in characteristic $p$, unfortunately. One of the algebraic motivations I've seen for the theory of formal groups is that they rectify the failure of Lie algebras to encode higher-dimensional differential information in positive characteristic, with this representation theoretic-issue given as an example of that failure. But do formal groups actually fix this particular problem? Do we have an adjunction between algebraic groups and formal groups, and can we use it to integrate (some) representations from formal groups to algebraic groups?
| https://mathoverflow.net/users/158123 | Can we use formal groups to recover Lie-theoretic representation theory in characteristic p? | The short answer is that the characteristic $p$ picture genuinely has more depth. In particular it's not correct to think of formal groups as "better" than Lie groups. In fact, they can be put on equal footing with each other.
I think it's helpful to put all the objects you're interested in in the same category. In this case it is the category of *group objects in formal schemes*, which I will call formal group objects. There is some notational confusion here. By rights, the category of formal group schemes should denote this category (which includes all algebraic groups as well). However, often when people say "formal group scheme" or "formal group" they mean *infinitesimal groups*, which are formal group schemes with a single point (and other times they mean something else more restrictive, such as formally smooth infinitesimal group schemes, which I don't want to consider separately). Here I'll hopefully avoid the notational confusion and use *group objects* for group objects in formal schemes and *infinitesimal groups* for group objects with one point.
Now in characteristic zero, every algebraic group $G$ has a unique infinitesimal subgroup $\hat{G}\subset G$ (as a group object) subject to the condition of having the same tangent space, i.e., the natural map $T\_e\hat{G}\to T\_e G$ being an isomorphism. In characteristic $p$, this is no longer the case. If $G$ is an algebraic group in characteristic $p$ then there is still a maximal sub-group object $\hat{G}\subset G$ corresponding to the "full" formal group of $G$, but there is another smaller object $$\hat{G}^{(1)}\subset \hat{G},$$ the first Frobenius neighborhood, which also has the same tangent space (there are also objects $\hat{G}^{(2)}, \dots,$ in addition to possible other intermediate subgroups). The group $\hat{G}^{(1)}$ has the nice property that its representations are equivalent to representations of the Lie algebra $\mathfrak{g}$ as a symmetric monoidal category, and this uniquely characterizes $\hat{G}^{(1)}.$ In fact if $\mathfrak{g}$ is an arbitrary Lie algebra in characteristic $p$, it might not integrate to a smooth algebraic group $G$, but it will always integrate to a group object $G^{(1)}$ that looks like a first Frobenius neighborhood.
Now even if you look at the "full" formal neighborhood $\hat{G}\subset G,$ you will not have anything resembling integration of representations. Indeed, if $G\to \text{End}(V)$ is a representation of a smooth algebraic group scheme, you do get induced representations of $\hat{G}$ and $\hat{G}^{(1)}$ (equivalently, $\mathfrak{g}$). But you can neither integrate representations of $\hat{G}^{(1)}$ to representations of $\hat{G}$ nor representations of $\hat{G}$ to representations of $G$ in any reasonable sense, even in one-dimensional cases.
| 12 | https://mathoverflow.net/users/7108 | 420346 | 171,061 |
https://mathoverflow.net/questions/420315 | 3 | Let $\mathcal{D}=\mathcal{D}\_X$ be the sheaf of rings of differential operators on a smooth algebraic curve $X$.
Since $\dim X=1$, the D-modules of the form $\mathcal{D}/\mathcal{D}L$ are necessarily holonomic. Conversely, I've verified that a holonomic D-module over $\mathbb{A}^1$ or $\mathbb{G}\_m$ is always cyclic; i.e., of the form $\mathcal{D}/I$ for some (not necessarily principal) left-ideal $I$.
(As W. Sawin nicely remarked, this is obvious whenever the holonomic D-module $M$ is simple. But even then, it's not clear to me that the left-ideal $I$ in $M=\mathcal{D}/I$ is principal.)
**I wonder if it's true that every simple holonomic D-module over $X$, assumed to be a curve, is of the form $\mathcal{D}/\mathcal{D}L$.** If this is false in general, is it true for $X=\mathbb{A}^1$? For $X=\mathbb{G}\_m$? For elliptic curves?
| https://mathoverflow.net/users/131975 | When are simple holonomic D-modules of the form $\mathcal{D}/\mathcal{D}L$? | If $X$ is a smooth proper curve of positive genus, then there exist simple $D$-modules which have no global sections at all. This is because every degree $0$ line bundle admits a flat connection (since the obstruction to a connection is the Atiyah class which vanishes). The line bundle with connection gives a $D$-module, and as soon as the module is trivial, has no global sections.
Since it has no global sections, it admits no nontrivial map from $\mathcal D$ and hence can't be a quotient of $\mathcal D$.
If $X$ is an affine curve, a $\mathcal D$-module is an honest module and not a sheaf of modules. So it has plenty of global sections. For a holonomic $D$-module, the space of proper submodules is finite-dimensional and each one has infinite codimension, so one can choose an element that is outside all of them, showing that the module has the form $\mathcal D/\mathcal D I$.
However, $I$ need not be principal, unless $X$ has genus $0$ and the module is simple. This is because the quotient by the ideal generated by a differential operator of degree $d$ has generic rank $d$.
If we choose a $D$-module of generic rank $0$, supported at a single point, say, then the generic rank is $0$, so if the ideal is principal it must be generated by some function on $X$, necessarily vanishing only at that point.
It follows that every module of the form $\mathcal D/ DL$ supported at a single point is simple, which is why the simple assumption is necessary, and that this can only happen if the point generates a finite subgroup of the Picard group, which is why the genus $0$ assumption is necessary.
So your desired property can only be true for curves of genus $0$.
| 6 | https://mathoverflow.net/users/18060 | 420353 | 171,065 |
https://mathoverflow.net/questions/420213 | 4 | Consider the general semilinear elliptic second-order PDE
$$
u\_t-\mathcal L u=f\left(t,x,u,\nabla u\right)
$$
where $\mathcal L$ is an elliptic linear operator (like minus the Laplace operator), $t \in [0,T],$ $x$ is in a bounded smooth domain $\Omega,$ and the boundary conditions are
$$
u(T,x)=c, \quad \forall x\in\Omega
$$
where $c$ is a constant. $f$ can be nonlinear in $u$ and $\nabla u$ with at most quadratic growth in both of them, and has no singularity in both of them and $x$. For example,
$$
f\left(t,x,u,\nabla u\right) = x^\alpha u^2 + ∇u \cdot ∇u
$$
is a possible form. Mind that this is a Cauchy problem with final data, a situation which fits with the "backward" heat operator $\partial\_t+\Delta$.
Suppose we know from the original problem that the solution to this PDE is bounded by two smooth $\mathcal C^2(\Omega)$ functions: $$\underline u \leq u \leq \overline u, \quad \forall (t,x)\in[0,T]\times \Omega.$$ My question is: is there a result (with possible conditions on $f$) to provide an existence result for a (weak) solution to this PDE ? any reference would be greatly appreciated.
| https://mathoverflow.net/users/480000 | Conditions for the existence of a solution to a semilinear second-order PDE with a-priori bounds | After a lot of reading, I came across the (enlightening) paper:
**On principally linear elliptic differential equations of the second order, Nagumo 1954**.
Basically, the (classical) results there shows that for any bounded domain $\Omega$ for your space variables, if you have lower and upper solutions for a quasi-linear / semi-linear PDE, and that your nonlinear operator in the PDE has at most quadratic growth in the gradient, then you can have (in this order) bounds on the solution and on the gradient in your bounded domain. I recommend reading it for simple scalar PDEs.
That being said, I did not need to find upper and lower solutions since I already had nice smooth à-priori bounds for my PDE. But for some simple PDEs, finding upper and lower solutions can be easy by trying constants, or simple form functions.
I wanted to leave this here in case someone with à-priori knowledge on the solution of their quasi-linear or semi-linear PDE and less than quadratic growth in their nonlinearity wanted some simple applicable result to prove existence, without having to go through all the weak-solution theory, which I think could take a few months to master.
| 2 | https://mathoverflow.net/users/480000 | 420368 | 171,068 |
https://mathoverflow.net/questions/420369 | 1 | I hope this is appropriate for the site. I am reading the paper "The analytic rank of a tensor" [S. Lovett, *Discrete Analysis* (2019), #7, 10 pp.] and am a bit confused in one of the applications sections. In the paper, a *tensor* is defined to be a multilinear map from $V^d\to {\bf F}$, where ${\bf F}$ is a field and $V$ is an $n$-dimensional vector space over ${\bf F}$.
However, later on, it is mentioned that to solve the cap-set problem, we define a tensor $T$ which captures the problem structure, then bound the largest *independent set* in $T$. This is a subset $A\subseteq [n]$ such that for all $i\_1,\ldots,i\_d\in A$, the coefficient $T\_{i\_1,\ldots,i\_d}$
of the tensor is nonzero if and only if $i\_1 = \cdots = i\_d$.
A *cap set* is a subset $A\subseteq {\bf F}\_3^n$ such that $x+y+z\ne 0$ whenever $x,y,z\in A$ are pairwise distinct.
From what I've seen, given $A\subseteq {\bf F}\_3^n$,
the "tensor" for the cap-set problem is
$$T(x,y,z) = \cases{1, & if $x=y=z$ and $x\in A$;\cr 0, & otherwise.}$$
If $A$ is a cap set, then this definition makes $A$ an independent set in $T$. However, I don't think that $T$ actually is a tensor, since if $x=y=z\in A$, then
$$f(x+x, y, z) = 0 \ne 2 = 1 + 1 = f(x,y,z) + f(x,y,z).$$
My conclusion is that I'm considering the wrong choice of $T$, but I can't seem to figure out the tensor alluded to by the author.
| https://mathoverflow.net/users/168142 | The cap set tensor in Lovett (2019) | The cap-set tensor is a tensor $V^3 \to \mathbb F\_3$ where $V$ is $\mathbb F\_3^{ \mathbb F\_3^n}$, not simply $\mathbb F\_3^n$. It is defined by
$$T (e\_a,e\_b,e\_c) = \begin{cases} 1 & \textrm{if }a+b+c = 0 \\ 0 & \textrm{if } a+b+c \neq 0 \end{cases}$$
and extended to arbitrary vectors by linearity.
The independent sets in this case are subsets of the index set $\mathbb F\_3^n$, not subsets of the vector space $\mathbb F\_3^{ \mathbb F\_3^n}$.
| 5 | https://mathoverflow.net/users/18060 | 420370 | 171,069 |
https://mathoverflow.net/questions/420361 | 12 | *This is a crosspost from [this MSE question](https://math.stackexchange.com/q/4395420/39599) from a year ago.*
---
>
> Consider the smooth four-manifold $M = (S^1\times S^3)\#(S^1\times S^3)\#(S^2\times S^2)$. Does $M$ admit a symplectic form?
>
>
>
If $\omega$ is a symplectic form, then the real cohomology class $[\omega]$ satisfies $[\omega]^2 = [\omega^2] \neq 0$. Note that $H^\*(M; \mathbb{R})$ has such classes.
Recall that a symplectic manifold admits an almost complex structure. In general, a closed four-manifold $N$ admits an almost complex structure if and only if there is $c \in H^2(N; \mathbb{Z})$ such that $c \equiv w\_2(N) \bmod 2$ and $c^2 = 2\chi(N) + 3\sigma(N)$. As $M$ is spin (i.e. $w\_2(M) = 0$), and $\chi(M) = \sigma(M) = 0$, the class $c = 0$ satisfies the two conditions so $M$ admits an almost complex structure.
In addition to the above, the vanishing of $w\_2(M)$, $\chi(M)$, and $\sigma(M)$ implies that $M$ is parallelisable by the Dold-Whitney Theorem. Moreover, it follows from our knowledge of complex surfaces that $M$ does not admit a complex structure, i.e. $M$ admits almost complex structures, but none of them are integrable.
One special aspect of symplectic manifolds in dimension four is that they have a non-trivial Seiberg-Witten invariant by a theorem of Taubes. When $b^+ \geq 2$, a non-trivial Seiberg-Witten invariant implies that no metric of positive scalar curvature exists, but in this case $b^+(M) = 1$ and $M$ does admit a metric of positive scalar curvature. I'm not sure if a deeper understanding of Seiberg-Witten theory could be used to provide a negative answer to the above question. In particular, I don't know if $M$ has a non-trivial Seiberg-Witten invariant.
| https://mathoverflow.net/users/21564 | Does $(S^1\times S^3)\#(S^1\times S^3)\#(S^2\times S^2)$ admit a symplectic form? | No, $M$ is not symplectic. Consider a double cover $\tilde{M}$ of $M$ along one of the $S^1$ components. Then it is not hard to prove that $\tilde{M}$ is diffeomorphic with $(S^1\times S^3)\#2(S^1\times S^3)\# 2 (S^2\times S^2)$. Now if $M$ were symplectic then you could pull back the symplectic structure on $\tilde{M}$. But notice that $\tilde M$ has a 3-sphere separating it into two copies with $b\_2^+ = 1$, so $SW(\tilde M) =0$, so this in contradiction with the fact that $\tilde M$ is symplectic.
| 14 | https://mathoverflow.net/users/33064 | 420371 | 171,070 |
https://mathoverflow.net/questions/420087 | 4 | I am trying to prove that the functor
\begin{align\*}
\mathrm{Top} &\longrightarrow \mathrm{Cond}(\mathrm{Set}) \\
X &\longmapsto \underline{X}
\end{align\*}
admits a left adjoint and it is the functor $T \mapsto T(\*)$ where $T(\*)$ has the quotient topology of the map $\bigsqcup\_{\underline{S} \rightarrow T} S \rightarrow T(\*)$ where the disjoint union runs over all profinite sets $S$ with a map to $T$.
To prove this I am trying to construct the following bijection. We have the map
\begin{align\*}
\phi: \mathrm{Hom}(T, \underline{X}) &\longrightarrow \mathrm{Hom}(T(\*), X) \\
f &\longmapsto f\_\* \ .
\end{align\*}
On the other hand, if $g \in \mathrm{Hom}(T(\*), X)$, then we can consider the map $\underline{g}: \underline{T(\*)} \rightarrow \underline{X}$. I am trying to construct a map $i: T \rightarrow \underline{T(\*)}$ to have an inverse for $\phi$:
\begin{align\*}
\psi: \mathrm{Hom}(T(\*), X) &\longrightarrow \mathrm{Hom}(T, \underline{X}) \\
g &\longmapsto \underline{g} \circ i \ .
\end{align\*}
Is there a way to construct $i$? How to prove this adjunction?
Edit: I have managed to construct $i$, but I still could not prove that $\phi$ and $\psi$ are mutually inverses.
I have constructed $i$ as the following: For each $S$ profinite we have the equivalence $\mathrm{Hom}(\underline{S},T) = T(S)$ thanks to the Yoneda Lemma. We can define
\begin{align\*}
i\_S: \mathrm{Hom}(\underline{S},T) &\longrightarrow \mathrm{Hom}(S, T(\*)) \\
\eta &\longmapsto \eta\_\* \ .
\end{align\*}
It is easy to prove that for $g \in \mathrm{Hom}(T(\*), X)$ we have $(\underline{g} \circ i)\_\* = g$, but have not managed to prove that for $f \in \mathrm{Hom}(T, \underline{X})$ we have $f = \underline{f\_\*} \circ i$. So it only remains to prove that for $\eta \in \mathrm{Hom}(\underline{S},T)$ we have
\begin{align\*}
f\_S (\eta) = f\_\* \circ \eta\_\*
\end{align\*}
I suspect this is due to the Yoneda Lemma but I could not prove it.
| https://mathoverflow.net/users/130868 | Adjunction between topological spaces and condensed sets | As Wojowu noted in the comments, one should really look at $T\_1$ topological spaces. Consider the functors
\begin{align\*}
G\!: \mathbf{Top}\_{T\_1} &\leftrightarrows \mathbf{Cond}\_\kappa:\!F\\
X &\mapsto \big(\underline X \colon S \mapsto \operatorname{Cont}(S,X)\big)\\
T(\*) &\leftarrow\!\shortmid T,
\end{align\*}
where $T(\*)$ is topologised with the quotient topology via the surjection
$$\pi \colon \coprod\_{(S,f \in T(S))} S \to T(\*)\tag{1}\label{1}$$
given on the component $f \in T(S)$ by $f \colon S \to T(\*)$. By the Yoneda lemma, this really means that $f \colon h\_S \to T$ is a morphism from the representable sheaf $h\_S$ to $T$, and the map $S \to T(\*)$ is the set-theoretic map $f\_{\{\*\}} \colon h\_S(\*) \to T(\*)$. But there is the consistent abuse of notation to denote $h\_S = \underline S$ as $S$. To see that (\ref{1}) is surjective, use $S = \{\*\}$.
As Dylan Wilson noted in the comments, the unit $\eta \colon 1 \to GF$ is given by the natural transformation
\begin{align\*}
(\eta\_T)\_S \colon T(S) &\to \operatorname{Cont}(S,T(\*)) \\
f &\mapsto \pi \circ \iota\_f,
\end{align\*}
where $\iota\_f \colon S \to \displaystyle\coprod\_{(S',f')} S'$ is the insertion of the coordinate corresponding to $(S,f)$. This gives the maps
\begin{align\*}
\phi\!: \operatorname{Hom}\_{\mathbf{Cond}}(T,\underline X) &\leftrightarrows \operatorname{Cont}(T(\*),X) :\!\psi \\
f &\mapsto f\_{\{\*\}} \\
g \circ \eta\_T &\leftarrow\!\shortmid g.
\end{align\*}
It remains to check that these are inverses. If $g \colon T(\*) \to X$ is continuous, then $\phi(\psi(g)) \colon T(\*) \to X$ is given by $\psi(g)\_{\{\*\}}$, i.e. the map taking $t \in T(\*)$ to $g \circ \pi \circ \iota\_f \colon \{\*\} \to X$. But $\pi \circ g \colon \{\*\} \to T(\*)$ is just (the constant map with value) the point $t$ (this is how we checked surjectivity of $\pi$ earlier!), so $\psi(g)\_{\{\*\}}$ takes $t$ to $g(t)$, i.e. $\phi(\psi(g)) = g$.
Conversely, given a natural transformation $f \colon T \to \underline X$, we get another natural transformation $\psi(\phi(f)) \colon T \to \underline X$. Write $g \colon T(\*) \to X$ for $\phi(f) = f\_{\{\*\}}$. If $S$ is extremally disconnected and $h \in T(S)$, then $\psi(g)\_S$ takes $h$ to the composition
$$S \overset{\iota\_h}\to \coprod\_{(S',h')} S' \overset\pi\to T(\*) \overset g\to X.$$
By definition, the composition $\pi \circ \iota\_h \colon S \to T(\*)$ is the map $h\_{\{\*\}} \colon h\_S(\*) \to T(\*)$. Thus $\psi(g)\_S(h)$ is the composition
$$h\_S(\*) \overset{h\_{\{\*\}}}\longrightarrow T(\*) \overset{f\_{\{\*\}}}\longrightarrow X.$$
This is the same thing as $(fh)\_{\{\*\}} \colon h\_S(\*) \to \underline X(\*)$, which is the continuous map $S \to X$ given by $f\_S(h) \in \underline X(S)$. We conclude that $\psi(\phi(f))\_S(h) = f\_S(h)$, and since $S$ and $h$ are arbitrary that $\psi(\phi(f)) = f$. $\square$
**Remark.** Morally what's going on here is the following: since $T(\*)$ has the quotient topology, a map $g \colon T(\*) \to X$ is continuous if and only if each of the compositions $gf \colon S \to X$ are continuous with $S$ extremally disconnected and $f \colon S \to T$ an $S$-point of $T$. Thus, a continuous map $T(\*) \to X$ is the same as continuous maps $S \to X$ for every $S \to T$, which is roughly what a natural transformation $T \to \underline X$ is.
| 2 | https://mathoverflow.net/users/82179 | 420373 | 171,072 |
https://mathoverflow.net/questions/420378 | 2 | Let $ X $ and $ Y $ be integral curves over some perfect field $ k $ and suppose that $ X $ is smooth. Moreover, let $ \pi\_1 : Y \to X $ and $ \pi\_2 : Y \to X $ be finite morphisms such that the function field $ k( Y ) $ of $ Y $ is the composite field $ \pi\_1^\*( k( X ) ) \cdot \pi\_2^\*( k( X ) ) $ where $ \pi\_i^\* : k( X ) \to k( Y ) $ denotes the pullback of $ \pi\_i $ for all $ i = 1, 2 $.
Finally, let $ q $ be a closed point on $ Y $ and let $ Q $ be a place in $ k( Y ) $ which dominates $ q $ (not sure if this is the right terminology; I mean that $ Q $ contains the maximal ideal in the local ring of $ q $).
Is it possible that $ Q $ is unramified in $ k( Y ) / \pi\_i^\*( k( X ) ) $ for all $ i = 1, 2 $ but $ q $ is a singular point on $ Y $?
Moreover, if this can happen:
Can this also happen if $ Y $ is even a projective curve which is contained in $ X \times\_k X $ and $ \pi\_1 $ and $ \pi\_2 $ are the restrictions of the canonical projections $ X \times X \to X $?
I don't believe that it matters. But just to be sure.
Thank you
| https://mathoverflow.net/users/132492 | Ramification and singular points | Sure. Take $X = \mathbb P^1$, $Y$ a nodal curve in $\mathbb P^1 \times \mathbb P^1$ given by some equation like $x^2 - y^2 + x y^2$. The point $(0,0)$ is singular but splits into two places in the function field.
Each of these two places is unramified since it is a degree $2$ extension in of both $k(x)$ and $k(y)$, so every pair of places lying over the same place of $k(x)$ or $k(y)$ is a pair of unramified places.
| 4 | https://mathoverflow.net/users/18060 | 420379 | 171,073 |
https://mathoverflow.net/questions/420382 | 2 | Let $s>1$ be a real number. We look at the zeta probability function / Zipf probability function defined as:
$$P(X = n) = \frac{1}{n^s \zeta(s)}$$
Suppose $f: \mathbb{N} \rightarrow \mathbb{R}$ is a function such that the following limit exists:
$$\lim\_{s \rightarrow 1} E(f(X\_s)) = \lim\_{s \rightarrow 1} \lim\_{N\rightarrow \infty} \sum\_{k=1}^N f(k) P(X\_s =k ) =$$
$$ = \lim\_{s \rightarrow 1} \lim\_{N\rightarrow \infty} \sum\_{k=1}^N f(k) \frac{1}{k^s \zeta(s)} = \lim\_{s \rightarrow 1} \frac{1}{\zeta(s)}\lim\_{N\rightarrow \infty} \sum\_{k=1}^N \frac{f(k)}{k^s} = \lim\_{s \rightarrow 1} \frac{D\_f(s)}{\zeta(s)}$$
where $D\_f(s) = \sum\_{k=1}^\infty \frac{f(k)}{k^s}$ is a Dirichlet series which converges for $s>1$ and the limit $= \lim\_{s \rightarrow 1} \frac{D\_f(s)}{\zeta(s)}$ is defined and exists.
**Question:** Is $\lim\_{s \rightarrow 1} E(f(X\_s)) = \lim\_{N \rightarrow \infty} \frac{1}{N} \sum\_{k=1}^N f(k)$?
If this can be proven, then the Prime number theorem would follow by plugging in $f(X\_s) = \lambda(X\_s)$ where $\lambda(n) = (-1)^{\Omega(n)}$ is the Liouville function, since it is known since Landau's doctoral thesis in 1899 that the right hand side
$$\lim\_{N \rightarrow \infty} \frac{1}{N} \sum\_{k=1}^N \lambda(k) = 0$$
is equivalent to the Prime number theorem.
But if the above question can be positively answered, then, since by Wikipedia we know that the left hand side is equal to:
$$\lim\_{s \rightarrow 1 }\frac{D\_{\lambda}(s)}{\zeta(s)} = \lim\_{s \rightarrow 1} \frac{\zeta(2s)}{\zeta(s)^2} = \frac{\zeta(2)}{\infty} = 0$$
so the right hand side would also be equal to $0$ and the prime number theorem would follow.
Intuitively for $s \rightarrow 1$ the probability
$$\lim\_{s \rightarrow 1} P(X\_s \equiv 0 \mod(n)) = \lim\_{s \rightarrow 1} \frac{1}{n^s} = \frac{1}{n}$$
equals the "probability that a random uniform natural number $X$ is divisible by $n$", which is $1/n$. So that is why I would expect the question to be positively answered, since it is a question about two expected values:
$$\lim\_{s \rightarrow 1} E(f(X\_s)) =^? \lim\_{N \rightarrow \infty} \frac{1}{N} \sum\_{k=1}^N f(k) = \lim\_{N \rightarrow \infty} E(f(X\_N))$$
where $X\_N$ is a random number $1 \le X\_N \le N$ drawn uniformly with probability $1/N$.
| https://mathoverflow.net/users/165920 | Is $\lim_{s \rightarrow 1} E(f(X_s)) = \lim_{N \rightarrow \infty} \frac{1}{N} \sum_{k=1}^N f(k)$? | The answer is no. E.g., if
$$f(k)=\sum\_{j=0}^\infty 1(2^{2j}<k\le2^{2j+1})$$
for natural $k$, then
\begin{equation}
Ef(X\_s)\to\frac13 \tag{1}\label{1}
\end{equation}
as $s\downarrow0$, whereas
$$\frac1n\,\sum\_{k=1}^n f(k)$$
will be forever oscillating between $\frac13$ and $\frac23$ as $n\to\infty$.
Indeed, let $s\downarrow0$. Then $\zeta(s)\sim\frac1{s-1}$ and
\begin{equation}
\begin{aligned}
\zeta(s)Ef(X\_s)&=\sum\_{k=1}^\infty\frac{f(k)}{k^s} \\
&=\sum\_{k=1}^\infty\frac1{k^s}\sum\_{j=0}^\infty 1(2^{2j}<k\le2^{2j+1}) \\
&=\sum\_{j=0}^\infty\sum\_{k=1}^\infty\frac1{k^s} 1(2^{2j}<k\le2^{2j+1}) \\
&=\sum\_{j=0}^\infty\Big(\int\_{2^{2j}}^{2^{2j+1}}\frac{dk}{k^s}
+O\Big(\frac1{2^{2sj}}\Big) \Big) \\
&=\sum\_{j=0}^\infty\frac{1-2^{1-s}}{s-1}\,2^{(1-s)2j}+O(1) \\
&\sim\frac1{3(s-1)},
\end{aligned}
\end{equation}
so that \eqref{1} is proved.
On the other hand, letting
\begin{equation}
s(n):=\sum\_{k=1}^n f(k)
\end{equation}
and letting a natural $m$ go to $\infty$, we have
\begin{equation}
\begin{aligned}
s(2^{2m+2})&=s(2^{2m+1}) \\
&=\sum\_{k=1}^{2^{2m+1}}\sum\_{j=0}^\infty 1(2^{2j}<k\le2^{2j+1}) \\
&=\sum\_{k=1}^{2^{2m+1}}\sum\_{j=0}^m 1(2^{2j}<k\le2^{2j+1}) \\
&=\sum\_{j=0}^m\sum\_{k=1}^{2^{2m+1}} 1(2^{2j}<k\le2^{2j+1}) \\
&=\sum\_{j=0}^m 2^{2j}\sim\frac43\,2^{2m},
\end{aligned}
\end{equation}
so that
\begin{equation}
\frac{s(2^{2m+2})}{2^{2m+2}}\to\frac13,
\end{equation}
whereas
\begin{equation}
\frac{s(2^{2m+1})}{2^{2m+1}}\to\frac23.
\end{equation}
So,
\begin{equation}
\frac1n\,\sum\_{k=1}^n f(k)=\frac{s\_n}n
\end{equation}
oscillates (at least) between $\frac13$ and $\frac23$ as $n\to\infty$, as was claimed.
| 3 | https://mathoverflow.net/users/36721 | 420388 | 171,075 |
https://mathoverflow.net/questions/420297 | 15 | Although I am by no means an applied mathematician, I like to occasionally explain applications of the math I teach to real world problems. Right now I am teaching some students about [longest increasing subsequences](https://en.wikipedia.org/wiki/Longest_increasing_subsequence) of permutations and their connection to the [Robinson-Schensted algorithm](https://en.wikipedia.org/wiki/Robinson%E2%80%93Schensted_correspondence).
Suppose you have some discrete time series data $x\_1,x\_2,\ldots$. This determines a permutation $\sigma = \sigma\_1, \sigma\_2,\ldots$ just by relative order of the data (if the data is continuous, or at least drawn from a wide range, it is very likely there will be no "ties" so we have an honest permutation). I remember once hearing (but don't remember where...) that checking the length of the longest increasing subsequence of $\sigma$ could be a way to detect if the data was observed in a random order or not.
But a cursory Googling (which gives a lot of info about longest increasing subsequences, connections to RS, "Ulam's problem" of computing the expected length, dynamic programming for finding a l.i.s., etc.) does not yield much about this application...
**Question**: Is comparing the length of the longest increasing subsequence of $\sigma$ to that of a random permutation a reasonable statistical test for randomness? Is there some literature about this?
Of course, for a more sophisticated test we could look at the shape of $\sigma$ under RS and compare it to the known Vershik–Kerov/Logan–Shepp limit shape.
| https://mathoverflow.net/users/25028 | Longest increasing subsequence as measure of randomness | The Longest Increasing Subsequence has been used as a test statistic for non-parametric tests by García and González-López in
Independence tests for continuous random variables based on the longest increasing subsequence, Journal of Multivariate Analysis, 2014 ([link](https://www.sciencedirect.com/science/article/pii/S0047259X14000335))
and also in [this (unpublished?) preprint](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.714.2983&rep=rep1&type=pdf). I don't know whether they were the first to consider this test statistic.
| 6 | https://mathoverflow.net/users/127599 | 420389 | 171,076 |
https://mathoverflow.net/questions/420304 | 2 | This is a noncommutative version of these three previous questions:
[differential operator power coefficients](https://mathoverflow.net/questions/80828/differential-operator-power-coefficients)
[Сlosed formula for $(g\partial)^n$](https://mathoverflow.net/questions/337330/%D0%A1losed-formula-for-g-partialn)
[A Leibniz-like formula for $(f(x) \frac{d}{dx})^n f(x)$?](https://mathoverflow.net/questions/415617/a-leibniz-like-formula-for-fx-fracddxn-fx)
Let $(R,\partial)$ be a noncommutative differential ring (unitality does not play a role), i.e.
we have $\forall a,b \in R$:
$$
\partial (ab) = \partial(a) b + a \partial (b).
$$
Fix a generic element ("regular function") $f \in R$. I am interested in an explicit description of the iterations
$$
(R\_f \partial)^n f,\ n \in \mathbb{N},
$$
where $R\_f$ denotes the multiplication operator by $f$ *from the right*. For example:
$$
(R\_f \partial) f = \partial(f) f
$$
and
$$
(R\_f \partial)^2 f = (R\_f \partial) (\partial(f) f) = \partial^2(f) f^2 + \partial(f)^2 f
$$
My question is thus:
>
> Is there a known explicit description for the words made out of the letters $\partial^n(f),\dotsc,\partial(f),f$ involved in the expansion of $(R\_f \partial)^n f$ and the coefficients in front of these words? Has this been investigated anywhere?
>
>
>
Ideally, I am looking for a description similar to Comtet's theorem in the commutative case cited by [Gjergji Zaimi](https://mathoverflow.net/a/80873) in [the first link mentioned above](https://mathoverflow.net/questions/80828/differential-operator-power-coefficients).
**Aside:** My setting is actually slightly more complicated than this. I only have a derivation "with a twist":
$$
\partial (ab) = \partial(a) \varphi(b) + a \partial(b),
$$
where $\varphi$ is an (injective, non-unital) ring endomorphism with $[\varphi,\partial]$ not being very illuminating, which only seems to complicate the combinatorics even further.
| https://mathoverflow.net/users/1849 | The combinatorics of $(f \partial)^n$ in the noncommutative setting? | In fact, Comtet's formula works almost directly in noncommutative case under an appropriate ordering of the products. Here is just a bit deeper look under the hood.
In umbral form $(R\_f \partial)^n f$ can be written as polynomial
$$f\_n(x\_1,\dots,x\_n) := x\_1(x\_1+x\_2)(x\_1+x\_2+x\_3)\cdots(x\_1+\dots+x\_n),$$
where in the expansion of $f\_n(x\_1,\dots,x\_n)$ each monomial $x\_1^{a\_1}x\_2^{a\_2}\cdots x\_n^{a\_n}$ corresponds to
$(\partial^{a\_1}f)(\partial^{a\_2}f)\cdots(\partial^{a\_n}f)f$.
The coefficient of $x\_1^{a\_1}x\_2^{a\_2}\cdots x\_n^{a\_n}$ in $f\_n(x\_1,\dots,x\_n)$ is nonzero only if the exponents satisfy the inequalities
$$a\_n + a\_{n-1} + \dots + a\_{n-k+1} \leq k,\qquad k\in\{1,2,\dots,n-1\},$$
with the total degree being $n$:
$$a\_n + a\_{n-1} + \dots + a\_1 = n,$$
in which case this coefficient is given by the formula:
\begin{split}
&\binom{1}{a\_n}\binom{2-a\_n}{a\_{n-1}}\binom{3-a\_n-a\_{n-1}}{a\_{n-2}}\cdots \binom{n-a\_n-a\_{n-1}-\dots-a\_2}{a\_1} \\
&=\frac{(2-a\_n)(3-a\_n-a\_{n-1})\cdots (n-a\_n-a\_{n-1}-\dots-a\_2)}{a\_1!a\_2!\cdots a\_n!}.
\end{split}
---
PS. It's worth to notice connection of $f\_n(x\_1,\dots,x\_n)$ to other combinatorial objects, such as $q$-factorial (for $x\_i=q^{i-1}$) and the generating function for Stirling numbers of first kind (for $x\_2=\dots=x\_n=1$).
| 1 | https://mathoverflow.net/users/7076 | 420390 | 171,077 |
https://mathoverflow.net/questions/420399 | 0 | I have the following setting:
Let $0 \leq r < 1$ and let $\{z\_i\}\_{i=1}^k$ be $k$ complex numbers such that $|z\_i| \leq r$ for all $i$.
Moreover, $r + \sum\_{i=1}^k 2Re(z\_i) \geq 0$
I am interested to know if the following is true:
$(1-r)\prod\_{i=1}^k |1-z\_i|^2 \leq 1$.
The above stems from trying to bound the determinant of $I-B$ where $B$ is nonnegative. One can think of $z\_i$ being the complex eigenvalues of $B$ and the reason for $|1-z\_i|^2$ is due to complex eigenvalues coming in conjugate pairs for real matrices. Also, $r$ is the spectral radius of the nonnegative $B$, and $r + \sum\_{i=1}^k 2Re(z\_i) \geq 0$ is due to the trace of a nonnegative matrix being nonnegative.
All ideas I have so far yield an upper bound greater than $1$, and I thought the expression was false. However, using Mathematica I could not find a counterexample to this, which makes me think it could be true.
The main ideas I used: triangle inequality, and arithmetic-geometric ineq., but could not get the 1 upper bound or a counterexample.
| https://mathoverflow.net/users/421678 | Proving maximum value of a determinant of $I - B$, where $B$ is nonnegative matrix | This is not true in general. E.g., take any $r\in(0,1)$ and let
$$z\_i:=x+iy,\quad x:=-\frac r{2k},\quad
y:=r\sqrt{1-\frac{1}{{4k^2}}}$$
for all $i$.
Then all the conditions on $r$ and $z\_i$'s hold, whereas
$$(1-r)\prod\_{i=1}^k|1-z\_i|^2=(1-r)(1+r^2+r/k)^k\to\infty \not\le 1$$
as $k\to\infty$.
Alternatively, let
$$z\_i:=ir$$
for all $i$. Then all the conditions on $r$ and $z\_i$'s hold, whereas
$$(1-r)\prod\_{i=1}^k|1-z\_i|^2=(1-r)(1+r^2)^k\to\infty \not\le 1$$
as $k\to\infty$.
| 2 | https://mathoverflow.net/users/36721 | 420400 | 171,080 |
https://mathoverflow.net/questions/420359 | 1 | For $N$ be a fixed natural number, define $w=e^{\frac{2\pi i}{N}}$ and $z=e^{\frac{\pi i}{N}}$, so that $z^2=w$. Let $D$ be the diagonal matrix $D=\operatorname{diag}(1,z,z^2,\ldots,z^{N-1})$ and $F$ be the Discrete Fourier matrix of order $N$, that is $F=\frac{1}{\sqrt{N}}(w^{kl})\_{0\leq k,l\leq N-1}$.
>
> Q. Does there exist any invertible matrix $A$ such that the following identity
> $$A(DF)A^{-1}=\alpha\_A DF$$
> holds for some complex number $\alpha\_A\neq1$.
>
>
>
| https://mathoverflow.net/users/84390 | A particular commutator of the discrete Fourier matrix | If such a pair $(A,\alpha\ne1)$ existed, then the spectrum of $DF$ would be invariant under the operation $\lambda\mapsto\alpha\lambda$, in particular $\alpha$ would be a root of unity.
The answer is therefore **negative** as soon as $N=2$, for then
$$D=\begin{pmatrix} 1 & 0 \\ 0 & i \end{pmatrix},\qquad
F=\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix},\qquad DF=\begin{pmatrix} 1 & 1 \\ i & -i \end{pmatrix}.$$
This matrix has distinct eigenvalues
$$\lambda\_\pm=\frac{1-i}2\left(1\pm i\sqrt{\frac72}\right).$$
However the ratio
$$\alpha:=\frac{\lambda\_+}{\lambda\_-}=\frac{\sqrt2+i\sqrt7}{\sqrt2-i\sqrt7}=\frac{-5+2i\sqrt{14}}9$$
is not a root of unity.
**Edit**. This is a general analysis. If $M:=DF$ is such as in the question, then $M$ is conjugated to $\alpha M$. Since $M$ is unitary (as the product of unitary matrices) its spectrum is made of non-zero numbers (actually of unit moduli). In particular $\alpha$ is a root of unity. The eigenvalues can be split into classes (taking in account their mutilpicities), which shows that the order $r\ge2$ of $\alpha$ divides $N$. All this shows that the characteristic polynomial $\chi\_M(X)$ actually depends only upon $X^r$.
Conversely, if $\chi\_M(X)=Q(X^r)$, then the spectrum is invariant under $\lambda\to \alpha\lambda$ with $\alpha^r=1$, hence $M$ is conjugated to $\alpha M$ (use the fact that $M$ is diagonalisable).
Thus a necessary and sufficient condition is that $\chi\_M$ writes as $Q(X^r)$ for some $r\ge2$ dividing $N$.
In particular a necessary condition is that
$${\rm Tr}M\quad\left(=\sum\_{k=0}^{N-1}z^{2k^2+k}\right),$$
the coefficient of $-X^{N-1}$, vanishes. This turns out to be **false** for every $N\le7$ and probably most of the time.
Therefore, I claim that the answer is **negative** for $N\le7$, and probably most of the time.
Calculation trick : because $M^{-1}=\bar F\bar D$ is conjugated to $\bar M$, we have
$$\bar\chi\_M(X)=\det(-M)^{-1}X^N\chi\_M(X^{-1}).$$
Thus we need only to check whether the first half of the polynomial $\chi\_M$ writes as a polynomial in $X^r$.
**Follow-up**. François Brunault's [answer](https://mathoverflow.net/q/420496) to a Q of mine sets up that the trace of $M$ never vanishes. Thus the answer to OP's question is **always negative**.
| 3 | https://mathoverflow.net/users/8799 | 420408 | 171,082 |
https://mathoverflow.net/questions/416173 | 5 | I am reading a paper [Cook and Forzani - Likelihood-Based Sufficient Dimension Reduction](http://dx.doi.org/10.1198/jasa.2009.0106) where the author uses the following result from matrix analysis but does not explain why it is true nor provide any reference.
More specifically, let $B \in \mathbb{R}^{p\times d}$ be a semi-orthogonal matrix, i.e $B^\top B = I\_d$, and $d < p$. Let $\Sigma$ and $\Delta$ denote two symmetric positive definite matrices such that $\Sigma - \Delta$ is also positive definite. What they claim is that
$$ \log \det \left\lvert B^\top \Sigma^{-1} B \right\rvert \leq \log \det \left\lvert B^\top \Delta^{-1} B \right\rvert.$$
Could someone point me in the direction of explaining why it is true?
I am thinking of using the Poincaré separation theorem, which provides bounds on the eigenvalues of the matrices on both the left and right-hand sides; however, I did not make any progress with it.
---
The claim is in the proof of Proposition 3, at the top of page 33. Here $\Sigma = \operatorname{Var}(X)$, and $\Delta = E (\operatorname{Var}(X\vert y))$, so $\Sigma-\Delta = \operatorname{Var}(E(X \vert y))$ is also positive definite. The $B$ in my question plays the role of $B\_0$ in the paper.
| https://mathoverflow.net/users/477109 | Log determinant of quadratic form | Since $\Sigma \succ \Delta$, by operator monotonicity we have $\Sigma^{-1} \prec \Delta^{-1}$ and thus $B^{\top}\Sigma^{-1}B \prec B^{\top}\Delta^{-1}B$. Since log on positive real is increasing, [the trace monotonicity](https://en.wikipedia.org/wiki/Trace_inequality#Convexity_and_monotonicity_of_the_trace_function) ([also Theorem 2.10 here](http://www.ueltschi.org/AZschool/notes/EricCarlen.pdf)) gives $\text{trace}\left(\log\left(B^{\top}\Sigma^{-1}B\right)\right) < \text{trace}\left(\log\left(B^{\top}\Delta^{-1}B\right)\right)$ where the log inside the last inequality is principal logarithm. Replacing $\text{trace}\left(\log\left(\cdot\right)\right)$ with $\log\det(\cdot)$, we get the desired inequality.
| 2 | https://mathoverflow.net/users/18526 | 420410 | 171,083 |
https://mathoverflow.net/questions/419858 | 4 | It is well known that a holomorphic vector field $z'=f(z), z\in \mathbb{C}$ does not have any limit cycle.See the last paragraph of this post
[Orbits space of real-analytic planar foliations](https://mathoverflow.net/questions/382871/orbits-space-of-real-analytic-planar-foliations)
One can imagine several reason for this fact. For example: The flow map is a holomorphic map. So if the limit cycle is of period $T$ then $\phi\_T$ admits an infinite number of fixed point hence it is identically equal to the identity map.
Now we ask the following question
>
> Is there a holomorphic vector field $z'=f(z)$ on an open region of $\mathbb{C}$ which admit an attractor homoclinic loop?
>
>
>
| https://mathoverflow.net/users/36688 | Can a holomorphic vector field have an attractor homoclinic loop? | The answer is 'no' for much the same reason that the OP indicates: the existence of a homoclinic or heteroclinic connection implies that neighboring trajectories are periodic.
First, one needs to have poles in $f$ if one wishes to have heteroclinic connections between (real) saddle singularities. Indeed if $f$ is holomorphic, then the stationnary points of the underlying real vector field are either centers, foci or multiple singularities locally conformally conjugate to $z'=\frac{z^{k+1}}{1+\mu z^k}$. None of these singularities can be part of a candidate limit (poly)cycle.
Next, consider the time-coordinate $$T(z)=\int\_{z\_\*}^z \frac{dz}{f(z)}.$$ Real-time trajectories of the vector field correspond to level sets $\mathrm{Im}(T)=\textrm{cst}$. The presence of a (poly)cycle $\Gamma$ connecting poles of $f$ implies that $\int\_\Gamma \frac{dz}{f(z)}=:\tau\in \mathbb{R}$ is the period of $\Gamma$.
Let $\gamma$ be a neighboring trajectory connecting a transverse section $\Sigma$ to itself "after making one turn". Without loss of generality one may choose $\Sigma$ as a small piece of $\mathrm{Re}(T)=\textrm{cst}$. Then there exists a small curve $\sigma\subset\Sigma$ such that the concatenation $\sigma\gamma$ is a closed loop. Cauchy's formula implies that $\int\_{\sigma\gamma} \frac{dz}{f(z)}=\tau$. But only $\sigma$ contributes to the imaginary part of the integral, therefore $\sigma$ is just one point and $\gamma$ is closed.
| 2 | https://mathoverflow.net/users/24309 | 420414 | 171,084 |
https://mathoverflow.net/questions/420423 | 2 | Suppose that $A\subseteq \mathbb{N}$ and suppose that you have an estimate of the form
$$
\sum\_{\substack{a\le x \\ a\in A}}f(a) \sim g(x).
$$
With this information is it possible to get an asymptotic estimate for $\sum\_{\substack{ab\le x \\ a,b\in A}}f(a)f(b)$? I would appreciate any reference in this kind of problem.
For example if $A$ is the set of prime numbers and $f(a)=1$, then we know that
$$
\sum\_{p\le x} 1 =\pi(x)\sim \frac{x}{\log{x}}.
$$
Now,
$$
\sum\_{pq\le x}1=\pi\_2(x)=\sum\_{j=1}^{\pi(x^{1/2})}\left[\pi\left(\frac{x}{p\_j}\right)-j+1\right]
$$
so that (after some calculations)
$$
\sum\_{pq\le x} 1 =\pi\_2(x)\sim \frac{x\log{\log{x}}}{\log{x}}
$$
So I was wondering if it's possible to generalize without relying too much on properties about $A$. The case I'm currently interested is when $A$ is the set of fundamental discriminants. In this case, we know that
$$
\sum\_{\substack{|D|\le x \\ D\in A}} 1\sim \frac{x}{\zeta(2)} .
$$
With this information is it possible to get an estimate for $\sum\_{\substack{|D\_1D\_2|\le x \\ D\_1,D\_2 \in A}}1$?
| https://mathoverflow.net/users/480636 | Asymptotic estimate for $\sum_{\substack{ab\le x \\ a,b\in A}}f(a)f(b)$ | I address your first (general) question, but I am sure it is applicable to your second (specialized) question. (Please restrict to one question per post to avoid confusion and frustration.)
You can try to apply Dirichlet's hyperbola method. First, you can forget about $A$, since you can always re-define $f$ as the restriction of $f$ to $A$. Then,
$$\sum\_{ab\leq x}f(a)f(b)=\sum\_{a\leq\sqrt{x}}f(a)\sum\_{b\leq x/a}f(b)+\sum\_{b\leq\sqrt{x}}f(b)\sum\_{a\leq x/b}f(b)-\sum\_{a,b\leq\sqrt{x}}f(a)f(b).$$
That is, if $s$ is the summatory function of $f$, then
$$\sum\_{ab\leq x}f(a)f(b)=2\sum\_{a\leq\sqrt{x}}f(a)s(x/a)-s(\sqrt{x})^2.$$
If $s$ is sufficiently regular (e.g. as in the case of the prime counting function), then the sum on the right-hand side can be evaluated rather directly in terms of $s$: write the sum as
$$\int\_{1-}^{\sqrt{x}} s(x/t)\,ds(t)=s(\sqrt{x})^2-\int\_{1-}^{\sqrt{x}}s(t)\,ds(x/t),$$
and approximate $s$ by a smooth function.
| 4 | https://mathoverflow.net/users/11919 | 420431 | 171,086 |
https://mathoverflow.net/questions/420415 | 4 | Let $ \Omega\_1 $ and $ \Omega\_2 $ be domains (open and connected) in $ \mathbb{R}^2 $. $ \psi:\Omega\_1\to\mathbb{R} $ and $ \phi:\Omega\_1\to\mathbb{R} $ are $ C^1 $ functions with two variables. Moreover, we assume that map $ (x,y)\to (\phi(x,y),\psi(x,y)) $ is homeomorphism from $ \Omega\_1 $ to $ \Omega\_2 $, i.e. the map $ (x,y)\to (\phi(x,y),\psi(x,y)) $ is continuous from $ \Omega\_1 $ to $ \Omega\_2 $ and has continuous inverse map. I want to ask that if I can obtain that the Jacobi determinant of the map, denoted as $ \frac{\partial(\phi,\psi)}{\partial(x,y)} $ is either non-positive or non-negative in $ \Omega\_1 $, i.e. either $ \frac{\partial(\phi,\psi)}{\partial(x,y)}\geq 0 $ for all $ (x,y)\in\Omega\_1 $, or $ \frac{\partial(\phi,\psi)}{\partial(x,y)}\leq 0 $ for all $ (x,y)\in\Omega\_1 $. I have tried by considering the image sets of a curve in $ \Omega\_1 $ and by using the connectedness of $ \Omega\_1 $ but failed. Can you give me some references or hints?
| https://mathoverflow.net/users/241460 | Does the homeomorphism have a non-negative or non-positive determinant? | Let $f=(\phi,\psi):\Omega\_1\to\Omega\_2$.
For every point $p\in\Omega\_1$ consider the curve $\gamma:t\mapsto p+\varepsilon e^{it}$, for $\varepsilon$ so small that the curve is contained in $\Omega\_1$. Let $n(p)$ be the winding number of $f\circ \gamma$ around $f(p)$. As $f$ is a homeomorphism, $n(p)$ is well defined (and an integer) and depends continuously on $p$, so as $\Omega\_1$ is connected, it is constant. So your statement is a direct consequence of the fact that $n(p)=1$ in points $p$ where $\frac{\partial(\phi,\psi)}{\partial(x,y)}(p)>0$ and $n(q)=-1$ in points $q$ where $\frac{\partial(\phi,\psi)}{\partial(x,y)}(q)<0$.
This can be generalized to higher dimensions using the local degree of $f$ instead of winding numbers.
| 7 | https://mathoverflow.net/users/172802 | 420432 | 171,087 |
https://mathoverflow.net/questions/420434 | 3 | Let $T: {\bf R}^n \rightarrow {\bf R}^n$ be an homeomorphism and $x$ a point in ${\bf R}^n$.
The positive orbit of $x$ is the set $\{T^n(x) \mid n \in {\bf N}\}$ and its
$\omega$-limit set is the set of accumulation points of its orbit.
$$\omega(x) = \{y \mid \exists \, n\_k \rightarrow +\infty \hbox{ such that } T^{n\_k}(x) \rightarrow y\}$$
>
> If the $\omega$-limit set of $x$ is bounded, does it imply that the positive orbit of $x$ is also bounded?
>
>
>
For flows, it is easy by a connectedness argument, but I am not sure that it holds
for homeomorphisms.
| https://mathoverflow.net/users/6129 | Boundedness of orbits and limit sets | We can modify the connectedness argument to show it also holds for homeomorphisms: suppose $\omega(x)$ is contained in an open ball $B(0,R)$, with $R>0$, and let $N$ be so big that $T(B(0,R))\subseteq B(0,N)$.
Then if the positive orbit of $x$ is not bounded, it must contain infinite points inside $B(0,N)\setminus B(0,R)$, which contradicts the fact that $\omega(x)\subseteq B(0,R)$.
| 6 | https://mathoverflow.net/users/172802 | 420436 | 171,088 |
https://mathoverflow.net/questions/420422 | 1 | Consider the semilinear wave equation in $[0,t\_0] \times \mathbb R^d$ :
$$\square u = \pm |u|^{p-1}u$$
With subcritical/critical power $1 < p \leq \frac{d+2}{d-2}$.
It is easy to show by energy method that any two $C^2$ classical solutions $u(t,x)$ & $v(t,x)$ with same initial conditions on $\{0\} \times B(x\_0, t\_0)$ will coincide on the backwards light cone $K(x\_0,t\_0) = \{(t,x) : 0 \leq t \leq t\_0, 0 \leq |x-x\_0| \leq t\_0-t\}$ (c.f. T. Tao, Nonlinear dispersive equations, Proposition 3.3).
I've seen this property being used for solutions in energy space $C^0([0,T], \dot{H}^1 \times L^2)$. I guess one should be able to generalize the above proposition using some approximation argument but I could not find a reference with a proof.
My main problem is that the proof using energy method uses the fact that a $C^2$ classical solution is locally bounded. This does not hold for energy space solutions.
My second problem is that if $u(t,x)$ is an energy space solution, then we can approximate $u(0,x)$, $\partial\_t u(0,x)$ and $\square u(t,x)$ by smooth functions $u\_{0,n}(x)$, $u\_{1,n}(x)$, $g\_n(t,x)$, and get an approximate classical solution $u\_n(t,x)$ to the inhomogeneous equation
\begin{align\*}
\square u\_n(t,x) &= g\_n(t,x), \quad (t,x) \in [0,t\_0] \times \mathbb R^d \\
u\_n(0,x) &= u\_{0,n}(x), \quad x \in \mathbb R^d \\
\partial\_t u\_n(0,x) &= u\_{1,n}(x), \quad x \in \mathbb R^d
\end{align\*}
but I don't know whether we can approximate $u(t,x)$ by classical solutions $u\_n(t,x)$ solving
\begin{align\*}
\square u\_n(t,x) &= \pm |u\_n|^{p-1}u\_n, \quad (t,x) \in [0,t\_0] \times \mathbb R^d \\
u\_n(0,x) &= u\_{0,n}(x), \quad x \in \mathbb R^d \\
\partial\_t u\_n(0,x) &= u\_{1,n}(x), \quad x \in \mathbb R^d
\end{align\*}
Any help or reference is welcomed.
| https://mathoverflow.net/users/130116 | Finite propagation speed for non-smooth solutions to nonlinear wave equation | You can approach this dually using that for classical solutions the finite propagation speed holds. This argument is similar in spirit to [this answer of mine for low regularity uniqueness for the linear wave equation](https://mathoverflow.net/a/361009/3948).
1
-
First write $w = u-v$. Then you see that $w$ solves a linear wave equation of the form
$$ \Box w = G(u,v) w $$
where the potential $G(u,v) \lesssim |u|^{p-1} + |v|^{p-1}$. **Pretend now that $G$ is a smooth function of $(t,x)$.**
2
-
Suppose for convenience that $x\_0 = 0$. You want to prove that $w(t,\cdot) |\_{B(0,t\_0 - t)} = 0$ for $t\in (0,t\_0)$. So it is enough to show that $\int w(t,x) f(x) ~dx = 0$ for every $f\in C^\infty\_c(B(0,t\_0-t))$. We will try to do so using a duality argument.
Fix $T\in (0,t\_0)$, and take $f\in C^\infty\_c(B(0,t\_0-T))$. Solve the wave equation
$$ \Box \varpi = G \varpi $$
with initial data $\varpi(0,T) = 0$ and $\partial\_t\varpi(0,T) = f$. **Under the assumption that $G$ is smooth**, we have that $\varpi$ is a smooth function with compact support for all time. In fact, for $t\in (0,T)$ we have that $\mathrm{supp}~ \varpi \subset B(0,t\_0 - t)$ using the finite speed of propagation for classical solutions.
3
-
That $w$ is an energy solution implies that the following identity holds for any test function $\varpi$:
$$ \int\_0^T \int\_{\mathbb{R}^d} \Box w \varpi - w \Box \varpi = \int\_{\mathbb{R}^d} w \partial\_t\varpi - \partial\_t w \varpi \Big|\_{t = 0}^T $$
Using the support property for $\varpi$ as derived above, you have that the right hand integral vanishes at $t = 0$ since $w,\partial\_t w$ vanishes on $B(0,t\_0)$. The integral at $t = T$ has only one term, and that is $\int w f$.
The left hand side however vanishes: since the two functions solve the same (linear) equation you have $LHS = \int Gw~v - w ~Gv = 0$.
4
-
The above works assuming that $G$ is smooth. In general, $G$ is not. Replace $G$ by $G\_\epsilon$ through mollification. And replace $\varpi$ correspondingly by $\varpi\_\epsilon$. Then it suffices to show that
$$ \int\_0^T \int\_{\mathbb{R}^d} (G - G\_\epsilon) w \varpi\_\epsilon \to 0 $$
Let me do the critical case for convenience; the argument should be similar for the subcritical cases with some adjustment of the exponents.
Noting that $p-1 = \frac{4}{d-2}$, by Strichartz inequality, $G$ belongs to space-time norm $L^1([0,T]; L^q\_x)$ for any $q\in [\frac{d}{2},d]$. And the mollification will converge in that norm. Since $w$ is energy class you have that it is uniformly bounded in $L^\infty\_t L^{2d/(d-2)}\_x$.
The Strichartz estimates together with Gronwell's inequality can be used to show that if $G\_\epsilon \in L^1\_t L^d\_x$, then
$$ \|\varpi\_\epsilon(t)\|\_{\dot{H}^1} \lesssim e^{C(T-t) \|G\_\epsilon\|\_{L^1\_t L^d\_x}} \|f\|\_{L^2} $$
This shows that $\|\varpi\_\epsilon(t)\|\_{L^\infty\_t L^{2d/(d-2)}}$ is uniformly bounded on the time interval $[0,T]$. And hence the fact that $G-G\_\epsilon$ converges to zero in $L^1\_t L^{d/2}$ gives that $|\int w(T,x) f(x) ~dx | = 0$ after taking the limit.
| 4 | https://mathoverflow.net/users/3948 | 420446 | 171,091 |
https://mathoverflow.net/questions/420441 | 6 | It is well-known that $e/\sqrt{2}$ is irrational.
Indeed, if it was rational, i.e. $p/q$ then $e^2/2 =p^2/q^2.$ Thus, $q^2e^2=2p^2,$ which would imply that $e$ is a root of $q^2x^2=2p^2.$
Now my question is: Does there exist a finite number of complex numbers $a\_1,...,a\_N$ all different from zero such that for every $n \in \mathbb N$
$$\sum\_{i=1}^N a\_i^n \in \mathbb Q \frac{e}{\sqrt{2}}?$$
| https://mathoverflow.net/users/457901 | Finite set of numbers whose powers sum up to irrational number | Such numbers do not exist. Indeed, let $p\_n$ (resp. $e\_n$) be the $n$-th power sum (resp. elementary symmetric polynomial) of the $a\_i$'s. Since $e\_n=0$ for $n>N$, [Newton's identities](https://en.wikipedia.org/wiki/Newton%27s_identities#Expressing_elementary_symmetric_polynomials_in_terms_of_power_sums) show that
$$\sum\_{m\_1 + 2m\_2 + \cdots + nm\_n = n \atop m\_1 \ge 0, \ldots, m\_n \ge 0} \prod\_{i=1}^n \frac{(-p\_i)^{m\_i}}{m\_i ! i^{m\_i}}=0,\qquad n>N.$$
Here $p\_i^{m\_i}$ is a rational multiple of $(e/\sqrt{2})^{m\_i}$. As $e/\sqrt{2}$ is transcendental, the contribution of the $n$-tuples $(m\_1,m\_2,\dotsc,m\_n)$ with fixed $m\_1+m\_2+\dotsb+m\_n$ is zero. There is only one $n$-tuple with $m\_1+m\_2+\dotsb+m\_n=n$, namely $(n,0,\dotsc,0)$, which shows that $p\_1=0$. Using this information, we can restrict the sum to $m\_1=0$. By choosing $n>N$ to be even, a similar reasoning yields that $p\_2=0$. By choosing $n>N$ to be divisible by $3$, we can infer that $p\_3=0$, and so on. So all the $p\_i$'s are zero, which then yields (again by Newton's identities) that all the $a\_i$'s are zero.
| 8 | https://mathoverflow.net/users/11919 | 420448 | 171,093 |
https://mathoverflow.net/questions/420449 | 0 | For simplicity, let us set $V\_n:=\{-1+1/2^n, \dotsc, 0 ,\dotsc, 1-1/2^n, 1\}$ with the periodic boundary conditions for each $n \in \mathbb{N}$ and think of the following vector space over this lattice:
\begin{equation}
M(V\_n):=\{f:V\_n \to \mathbb{C} \mid f \text{ is the restriction of some Schwartz function on }
\mathbb{R} \text{ to } V\_n \}.\end{equation}
Then, clearly, $M(V\_n)$ is a well-defined finite dimensional vector space over $\mathbb{C}$. Moreover, we can think of the direct limit of the collection $\{ M(V\_n)\}\_n$, which may be identified as the restriction of Schwartz space to the domain $(-1,1]$.
So, here are my questions. They are divided into two parts.
1. $M(V\_n)$ is set-theoretically identical to the "set of all mappings from $V\_n$ to $\mathbb{C}$". However, the "set of all mappings from $V\_n$ to $\mathbb{C}$" cannot form a direct limit of the above kind since they can approximate singular functions such as the Heaviside function. What makes the crucial difference?
2. A finite symmetric difference operator $\partial\_n$ on $M(V\_n)$ defined by
\begin{equation}
\partial\_n f(x):=2^{n+1}[f(x+1/2^n)-f(x-1/2^n)]
\end{equation}
is a linear operator. As $n \to \infty$, in what sense does this $\partial\_n$ converge to the ordinary continuum differential operator?
These two questions seem quite subtle to me…. Could anyone please clarify?
| https://mathoverflow.net/users/56524 | In what sense does the finite difference operator converge to the continuum differential operator in the hydrodynamic limit? | "In what sense does this $\partial\_n$ converge to the ordinary continuum differential operator?"
It converges in the sense that there exist injective linear maps $\iota\_n:M(V\_n)\to C^\infty\_\text{periodic}([-1,1])$ that take the $2^{n+1}$ eigenspaces of $\partial\_n$ to the first $2^{n+1}$ eigenspaces of $d/dx$, such that the $j$-th eigenvector of $\partial\_n$ converges pointwise on $[-1,1]\cap \mathbb Z[\tfrac12]$ to the $j$-th eigenvector of $d/dx$, as $n\to \infty$.
Here, the sentence "first $2^{n+1}$ eigenspaces of $d/dx$" requires some clarification:
The eigenspaces of $d/dx$ are ordered by increasing absolute value of the eigenvalue, with the (arbitrary) convention that $+a$ comes before $-a$ whenever $a>0$.
| 3 | https://mathoverflow.net/users/5690 | 420451 | 171,094 |
https://mathoverflow.net/questions/420450 | 1 | If $H=(V,E)$ is a [hypergraph](https://en.wikipedia.org/wiki/Hypergraph), and $\kappa \neq \emptyset$ is a cardinal, a map $c: V\to \kappa$ is said to be a *coloring* if the restriction $c|\_e: e\to \kappa$ is non-constant whenever $e\in E$ and $e$ has at least $2$ elements. The least non-empty cardinal such that there is a coloring onto that cardinal is said to be the *chromatic number* of $H$.
Let $[\omega]^\omega$ denote the collection of infinite subsets of $\omega$. For $n\in \omega\setminus\{0,1\}$, let $${\cal E}\_n =\{E\subseteq [\omega]^\omega: \chi(\omega, E) = n\}.$$
**Question.** For what values of $n$ does ${\cal E}\_n$ have maximal elements with respect to $\subseteq$?
| https://mathoverflow.net/users/8628 | Edge sets on $\omega$ maximal with respect to chromatic number | $\mathcal E\_n$ has maximal elements for every $n\ge1$.
Let $V\_1,\dots,V\_n$ be pairwise disjoint infinite sets such that $V\_1\cup\cdots\cup V\_n=\omega$ and let $E=\{e\in[\omega]^\omega:e\not\subseteq V\_i\text{ for all }i\in[n]\}$. Plainly $\chi(\omega,E)\le n$.
Let $c$ be any proper $n$-coloring of $(\omega,E)$. For each $i\in[n]$ there is some color $x\_i$ which occurs infinitely often in $V\_i$, and that color can not occur in $\omega\setminus V\_i$. Thus the only proper $n$-colorings of $(\omega,E)$ are the obvious ones, where the color classes are the sets $V\_i$. This shows that $\chi(\omega,E)=n$ and that $\chi(\omega,E')\gt n$ if $E\subsetneqq E'\subseteq[\omega]^\omega$.
| 3 | https://mathoverflow.net/users/43266 | 420456 | 171,098 |
https://mathoverflow.net/questions/420457 | 1 | I am interested in finding references that develop high probability suboptimality bounds for stochastic gradient descent (SGD) for **general** convex functions in the case where we return the average of the last $\alpha T$ iterates (suffix-averaging) where $\alpha\in(0,1)$ and $T$ denotes the total number of iterations. I have found [this](https://arxiv.org/pdf/1909.00843.pdf) source on suffix-averaging in the case of strongly-convex functions, but that's about it.
| https://mathoverflow.net/users/174439 | High probability bounds of SGD for general convex functions with suffix averaging | I didn't look very carefully. I found [this](https://www.cs.ubc.ca/%7Enickhar/papers/GradientDescent/GradientDescent.pdf) paper by Harvey et al (2018) demonstrating the result for general convex, 1-Lipschitz $f(\cdot)$.
| 1 | https://mathoverflow.net/users/174439 | 420458 | 171,099 |
https://mathoverflow.net/questions/420475 | 5 | I think the following statement is not true in the general situations, but consider it:
$R$ is a ring, $\mathfrak{p}$ is a prime ideal, then the unit group of $\dfrac{R}{\mathfrak{p}^nR}$ is isomorphic to $(\dfrac{R}{\mathfrak{p}R})^{\*}\times\dfrac{R}{\mathfrak{p}^{n-1}R}$.
This statement holds if $R=\mathbb{Z}$, and $\mathfrak{p}=p\mathbb{Z}$ for some odd prime number. What can we say if we consider $R=\mathcal{O}\_K$, where $\mathcal{O}\_K$ is the ring of integers of a number field? Can we say anything similar, if we consider the localization of $\mathcal{O}\_K$ at some prime ideal? I mean are there some sufficient conditions under which the above statement holds?
Is there any relation with the roots of unity? I mean can we do something similar to the following:
Let $K$ be a number field, and assume that the order of torsion elements in the multiplicative group $K^\*$ divides $n$. Then for any prime ideal $\mathfrak{p}$ with $\gcd(\mathfrak{p}, n)=1$ we have:
$$(\dfrac{\mathcal{O}\_K}{\mathfrak{p}^n})^\* \cong (\dfrac{\mathcal{O}\_K}{\mathfrak{p}})^{\*}\times\dfrac{\mathcal{O}\_K}{\mathfrak{p}^{n-1}}.$$
| https://mathoverflow.net/users/166540 | About the structure of unit groups appearing in number theory | In the case of the ring of integers of a number field, this problem is
studied and solved in complete detail in Section 4.2 of my book "Advanced topics in Computational number theory", Springer GTM 193. In particular,
your statement holds if $e<p-1$, where $p$ is the prime number below
the prime ideal and $e$ the ramification index. Thus, it holds if the prime
ideal is unramified and $p\ge3$.
| 7 | https://mathoverflow.net/users/81776 | 420477 | 171,105 |
https://mathoverflow.net/questions/420473 | 15 | In my answer to [this question](https://mathoverflow.net/q/420359), there appears the following sub-question about a sum of roots of unity. Denoting $z=\exp\frac{i\pi}N$ (so that $z^N=-1$), can the quantity
$$\sum\_{k=0}^{N-1}z^{2k^2+k}$$
vanish ?
The answer is clearly **No** for $N\le7$. I suspect that it never vanishes, which should answer definitively the above matricial question. Any idea of a general argument ?
| https://mathoverflow.net/users/8799 | Vanishing of a sum of roots of unity | For general $N$, we can reason by induction on the $2$-adic valuation of $N$. If $N$ is odd, GH from MO's answer shows that $S\_N :=\sum\_{k=0}^{N-1} \zeta\_{2N}^{2k^2+k} \neq 0$, where $\zeta\_{2N} = z = e^{\pi i/N}$ is a primitive $2N$-th root of unity. The same argument shows that for any odd $N$ and any $b \neq 1$, the following variant of $S\_N$ is nonzero:
\begin{equation\*}
S\_{N,b} :=\sum\_{k=0}^{N-1} \zeta\_{2N}^{2^b k^2+k} \neq 0.
\end{equation\*}
For convenience of the reader, the proof goes as follows. In the ring $\mathbb{Z}[z]/(2)$, we have
\begin{equation\*}
S\_{N,b}^{2^{b+3}} = \sum\_{k=0}^{N-1} z^{2^{b+3} (2^b k^2+k)} = \sum\_{k=0}^{N-1} \zeta\_N^{2^{2b+2} k^2+2^{b+2} k} = \sum\_{k=0}^{N-1} \zeta\_N^{(2^{b+1} k + 1)^2 - 1} = \zeta\_N^{-1} \sum\_{\ell=0}^{N-1} \zeta\_N^{\ell^2},
\end{equation\*}
and one conclude as in GH from MO's answer.
Now write $N=2^a M$ with $M$ odd, and assume $a \geq 1$. The Galois group of $\mathbb{Q}(\zeta\_{2N})/\mathbb{Q}(\zeta\_N)$ is of order 2, generated by the automorphism $\sigma : \zeta\_{2N} \mapsto \zeta\_{2N}^{1+N} = - \zeta\_{2N}$. Moreover, we have
\begin{equation\*}
\sigma(S\_{N,b}) = \sum\_{k=0}^{N-1} (-\zeta\_{2N})^{2^b k^2+k} = \sum\_{k=0}^{N-1} (-1)^k \zeta\_{2N}^{2^b k^2+k},
\end{equation\*}
so that
\begin{equation\*}
\frac12 (S\_{N,b}+\sigma(S\_{N,b})) = \sum\_{\substack{k=0 \\ k \textrm{ even}}}^{N-1} \zeta\_{2N}^{2^b k^2+k} = \sum\_{k=0}^{N/2-1} \zeta\_{2N}^{2^{b+2} k^2+2k} = \sum\_{k=0}^{N/2-1} \zeta\_N^{2^{b+1} k^2+k} = S\_{N/2, b+1}.
\end{equation\*}
By induction, we have $S\_{N/2,b+1} \neq 0$, which implies $S\_{N,b} \neq 0$.
| 15 | https://mathoverflow.net/users/6506 | 420496 | 171,113 |
https://mathoverflow.net/questions/418610 | 3 | Hartshorne defines (p. 257 of III.9) an associated point of a scheme $X$ as a point such that $\mathfrak{m}\_x$ is an associated prime of the local ring which he says is equivalent to the maximal ideal consisting of only zero divisors.
I am wondering if these are actually equivalent in the non-noetherian setting? I am aware that associated points are badly behaved and weakly associated point is the better notion in this generality but I am wondering if these two definitions (maximal ideal consists of zero divisors and maximal ideal is associated) are actually still equivalent.
| https://mathoverflow.net/users/154157 | Is Hartshorne's definition of associated points correct in the non-noetherian setting? | They aren't equivalent, unless Hartshorne's usage of "associated prime" is different from e.g. the definition in the Stacks project [[Definition 00LA](https://stacks.math.columbia.edu/tag/00LA)].
Indeed, [[Example 05AI](https://stacks.math.columbia.edu/tag/05AI)] gives the non-Noetherian local ring $A = k[x\_1,x\_2,\ldots]/(\{x\_i^2\}\_{i \geq 1})$ (for a field $k$) with no associated primes, but whose maximal ideal is locally nilpotent (i.e. every element in the maximal ideal is nilpotent).
Furthermore, the proposition afterwards (Hartshorne *Algebraic Geometry* Proposition III.9.7) is false without Noetherian hypotheses, again if "associated prime" has the definition as linked above, in the Stacks project. The proposition says that any scheme $X$ over an integral, regular, dimension $1$ scheme $Y$ is flat if and only if all associated points map to the generic point of $Y$.
Indeed, set $X = \mathrm{Spec} ~A$ where the ring $A$ is as above with $k$ of positive characteristic, and set $Y = \mathrm{Spec} ~\mathbb{Z}$. Then $X \rightarrow Y$ is not flat, but $X$ has no associated points so the hypotheses of the proposition are vacuously satisfied.
However, Hartshorne's claims here are ok if we interpret his "associated prime" as the notion called "weakly associated prime" in the Stacks project [[Section 0546](https://stacks.math.columbia.edu/tag/0546)]. These notions are equivalent for Noetherian rings.
| 4 | https://mathoverflow.net/users/217216 | 420507 | 171,115 |
https://mathoverflow.net/questions/420511 | 0 | Let $f(x)$ be an integer-valued polynomial (when $x\in \mathbb{Z}$, then $f(x)\in \mathbb{Z}$), and $a,b$ be positive integers, and $p$ be a prime number with $(a,p)=1$.
Show that
$$\sum\_{x=0}^{p-1}\left(\dfrac{f(ax+b)}{p}\right)=\sum\_{x=0}^{p-1}\left(\dfrac{f(x)}{p}\right)$$
If $f(x)=x$, it is well known
$$\sum\_{x=0}^{p-1}\left(\dfrac{ax+b}{p}\right)=\sum\_{x=0}^{p-1}\left(\dfrac{x}{p}\right)=0$$
| https://mathoverflow.net/users/38620 | Sums of Legendre symbols with integer-valued polynomials | The statment is false. For a counterexample, take $f(x)=\binom{x}{3}$, $p=3$, $a=2$, $b=0$.
The statement is true when $f$ has degree less than $p$, or when $f$ has integral coefficients.
| 8 | https://mathoverflow.net/users/11919 | 420512 | 171,117 |
https://mathoverflow.net/questions/420471 | 3 | EDITED:
A pair of finite simplical complexes are equivalent if and only if they are related by a finite sequence of the Pachner moves.
Is there a similar thing on finite cell complexes? That is, are there “related” notions of equivalence and “similar” theorems “reducing” such equivalence to finite sequences “combinatorial” moves? I am interested in any such examples, and I am not bothered if different examples require some side hypotheses (for instance, restricting to regular cell structures).
| https://mathoverflow.net/users/103418 | “Combinatorial” moves between cell complexes | 0. If a pair of finite simplicial complexes are PL manifolds, which are additionally PL homeomorphic, then there is a finite sequence of [bisteller flips](https://en.wikipedia.org/wiki/Pachner_moves) taking one to the other. (These are sometimes also called Pachner moves.)
1. [Kirby calculus](https://en.wikipedia.org/wiki/Kirby_calculus) on handle structures of four-manifolds.
2. Collapses and expansions (of CW complexes) generate the relation of [simple homotopy equivalence](https://en.wikipedia.org/wiki/Simple-homotopy_equivalence). (See Ryan's comments above.)
| 4 | https://mathoverflow.net/users/1650 | 420515 | 171,118 |
https://mathoverflow.net/questions/415627 | 3 | I looked in all textbooks on vector lattices (Riesz spaces) as well as ordered vector spaces, but couldn't find any mentions of neither inductive nor projective limit for these structures. Googling also didn't help. Since there is some ambiguity in terminology let me state the definition.
Let $\Gamma$ be a directed set and let $\left(E\_{\gamma}\right)\_{\gamma\in\Gamma}$ be a family of partially ordered vector spaces, such that for every $\alpha\le\beta$ there is a designated positive linear operator $T\_{\alpha\beta}:E\_{\alpha}\to E\_{\beta}$, such that $T\_{\gamma\gamma}=Id\_{E\_{\gamma}}$ and $T\_{\beta\gamma}T\_{\alpha\beta}=T\_{\alpha\gamma}$, for every $\alpha,\beta,\gamma$. The inductive limit then is an ordered space $E$ and a collection of positive linear operators $T\_{\gamma}:E\_{\gamma}\to E$ such that $T\_{\beta}T\_{\alpha\beta}=T\_{\alpha}$, and whenever $F,\left(S\_{\gamma}\right)\_{\gamma\in\Gamma}$ has this property, there is a unique $S:E\to F$ such that $S\_{\gamma}=ST\_{\gamma}$.
If we drop "ordered" and "positive", we will get the definition of the inductive limit for vector spaces. In fact, the inductive limit of ordered vector spaces can be constructed as the inductive limit of the underlying vector spaces endowed with the weakest linear order the makes all $T\_{\gamma}$'s positive. It is also not hard to show that if all $E\_{\gamma}$'s are vector lattices, and $T\_{\alpha\beta}$'s are lattice homomorphisms, then $E$ is itself a vector lattice.
>
> I need some permanence properties of this construction, which I suspect are known, so I am asking for a reference where it is studied systematically.
>
>
>
| https://mathoverflow.net/users/53155 | Reference on inductive (direct) limit of ordered vector spaces and vector lattices | There is a paper by Wolfgang Filter form 1980s.
W. Filter, Inductive limits of Riesz spaces, Proceedings of the International Conference held in Dubrovnik, June 23-27, 1987 (Bogoljub Stankovic, Endre Pap, Stevan Pilipovic, and Vasilij S. Vladimirov, eds.), Plenum Press, New York, 1988, pp. 383-392.
Unfortunately I cannot supply a link.
| 3 | https://mathoverflow.net/users/480716 | 420517 | 171,120 |
https://mathoverflow.net/questions/420530 | 1 | We know that in characteristic $0$, all algebraic series are differentiably finite.
Is this true in positive characteristic? I look at the [proof](https://www.e-periodica.ch/cntmng?pid=ens-001:1964:10::61), indeed we need to the
characteristic to be $0$ for the proof to work.
If it is not true in positive characteristic, is there a counter-example?
Thank you in advance!
| https://mathoverflow.net/users/122378 | Are algebraic power series in positive characteristics D-finite? | Over a field of characteristic $p$, the $p$-th derivative of any power series is $0$, and so every power series over a field of finite characteristic is $D$-finite.
| 6 | https://mathoverflow.net/users/88679 | 420540 | 171,127 |
https://mathoverflow.net/questions/419320 | 2 | Let $M$ be a large positive integer, $d$ an odd positive integer and $f: \mathbb{Z}\_{>0} \times \mathbb{Z}\_{>0} \to \mathbb{R}$. For a non-principal character $\chi\_d = \chi$ with modulus $d$, I am interested in the following type of sums.
$$S(\chi, f) = \sum\_{\substack{m = M \\ (m, d) = 1}}^{2M}f(m, d) \chi(m). $$
For $f \equiv 1$, we know that $\lvert S(\chi, f)\rvert \ll \sqrt{d}\log(d)$ by the Pólya–Vinogradov inequality. Now I am interested in bounding $\lvert S(\chi, f)\rvert$ when $$f(m, d) = \frac{\phi(m)}{m} \prod\_{\substack{p \textrm{ prime} \\ (p, md) = 1}} \left(1 - \frac{\rho\_{m}(p)}{p^2} \right)$$
where $\rho\_m(p) = 1 + \genfrac(){}{}m p$ for primes $p$ that do not divide $m$. $\chi\_d(m) = \genfrac(){}{}m d$, the Jacobi symbol.
I expect a similar result as the Pólya–Vinogradov inequality. Computational experiments for some $M$, $d$ values did agree with my expectation. Have these types of sums been studied earlier? Any help would be highly appreciated.
| https://mathoverflow.net/users/167999 | Pólya–Vinogradov like inequality for a character sum with Euler factors | Based on the ideas posted by @OfirGorodetsky and @tomos as comments to this post, I managed to compile this solution.
$$\begin{align\*}
S &= \sum\_{m = M}^{2M} \frac{\phi(m)}{m} \prod\_{p \nmid md}\left(1 - \frac{1}{p^2} - \frac{\chi\_p(m)}{p^2} \right) \; \chi\_d(m) \\
&= \sum\_{m = M}^{2M} \frac{\phi(m)}{m} \left( \sum\_{r \nmid md} \prod\_{p \nmid mdr} \left(1 - \frac{1}{p^2} \right) \mu(r) \frac{\chi\_r(m)}{r^2} \right) \chi\_d(m) \\
&= \sum\_{m = M}^{2M} \prod\_{p | m} \frac{1}{1 + \frac{1}{p}} \sum\_{r \nmid d} \prod\_{p \nmid dr} \left(1 - \frac{1}{p^2} \right) \mu(r) \frac{\chi\_{rd}(m)}{r^2} \\
&= \sum\_{r \nmid d} \left( \mu(r) \frac{1}{r^2} \prod\_{p \nmid dr} \left(1 - \frac{1}{p^2} \right) \right) \sum\_{m = M}^{2M} \left( \prod\_{p | m} \frac{p}{1 + p} \right)\chi\_{dr}(m) \end{align\*} $$
Now setting $f(m) = \prod\_{p | m} \frac{p}{1 + p}$ and $g = f\*1$, by the Mobius inversion formula we have that $g(m) = \sum\_{ab = m} f(a) \mu(b)$. It is easy to see by computation that $|g(m)| < \frac{1}{m}$. Hence we have,
$$\begin{align\*}
\left| \sum\_{m = M}^{2M} \left( \prod\_{p | m} \frac{p}{1 + p} \right)\chi\_{dr}(m)\right| &= \left| \sum\_{m = M}^{2M} \sum\_{ab = m} g(a)\chi\_{dr}(a) \chi\_{dr}(b) \right| \\ &= \left| \sum\_{a = 1}^{2M} g(a)\chi\_{dr}(a) \sum\_{b = M/a}^{2M/a} \chi\_{dr}(b) \right| \\ &\leq \sum\_{a = 1}^{2M} |g(a)| \left| \sum\_{b = M/a}^{2M/a} \chi\_{dr}(b) \right| \\ &\ll \sum\_{a = 1}^{2M} \frac{1}{a} \sqrt{dr} \log{(dr)} \ll \log{(M)} \sqrt{dr} \log{(dr)}
\end{align\*}$$
So now,
$$\begin{align\*}
|S| \ll \sum\_{r \nmid d} \frac{1}{r^2} \log{(M)} \sqrt{dr} \log{(dr)} \ll \log{(M)} \sqrt{d} \log{(d)}
\end{align\*}$$
The Euler factor involving the $\rho\_m$ function was too complicated to get a useful function applying Mobius inversion. So somehow had to expand out and dealt as shown.
| 0 | https://mathoverflow.net/users/167999 | 420546 | 171,129 |
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