parent_url
stringlengths 37
41
| parent_score
stringlengths 1
3
| parent_body
stringlengths 19
30.2k
| parent_user
stringlengths 32
37
| parent_title
stringlengths 15
248
| body
stringlengths 8
29.9k
| score
stringlengths 1
3
| user
stringlengths 32
37
| answer_id
stringlengths 2
6
| __index_level_0__
int64 1
182k
|
|---|---|---|---|---|---|---|---|---|---|
https://mathoverflow.net/questions/398083
|
3
|
I would like to estimate from above the following sum
$$
\sum\_{1 \leq x\_1 < X} .. . \sum\_{1 \leq x\_n < X} \frac{\prod\_{1 \leq i \leq n } \phi(x\_i)}{\mathrm{lcm}(x\_1, .., x\_n)^a}.
$$
$\phi$ is the Euler totient function and $a$ is a positive integer less than $2n$. A trivial estimate would be $\ll X^{2n - a}$. Is there a way to get a better bound? Thank you!
|
https://mathoverflow.net/users/84272
|
Estimating $\sum_{x_i < X} \prod_i \phi(x_i)/ \mathrm{lcm}(x_i)^a$
|
One can improve on $X^{2n-a}$ as long as $(a,n)\neq(1,1)$ (for $a=n=1$, the sum grows like $X/\zeta(2)$ so there's no room for improvement). Let us introduce $$f\_n(m) := \# \{ (x\_1,\ldots,x\_n) : \mathrm{lcm}(x\_1,\ldots,x\_n) = m\},$$ which satisfies $f\_n(m) \le \tau(m)^n \ll\_{n,\varepsilon} m^{\varepsilon}$ where $\tau$ is the usual divisor function. First suppose that $a >n$. We have $$\frac{\prod\_{i=1}^{n} \phi(x\_i)}{\mathrm{lcm}(x\_1,\ldots,x\_n)^a} \le \frac{\prod\_{i=1}^{n} \phi(x\_i)}{\max\_{1\le i \le n} x\_i^n} \frac{1}{\mathrm{lcm}(x\_1,\ldots,x\_n)} \le \frac{1}{\mathrm{lcm}(x\_1,\ldots,x\_n)}$$
which gives the upper bound $$< \sum\_{1 \le m < X^{na}} \frac{f\_n(m)}{m} = X^{o(1)}.$$
This is optimal since we have the lower bound $\ge 1$. We may assume $n \ge a$ from now on. Your sum is $$< X^n \sum\_{1 \le m < X^{na}} \frac{f\_n(m)}{m^a} \ll\_{a,n} X^{n+o(1)}.$$
If $n\neq a$ this already beats $X^{2n-a}$.
We now sketch how one can do better than $X^{n+o(1)}$. In section 3 of R. R. Hall's ``The distribution of squarefree numbers'' (Reine Angew. Math. 394 (1989), 107–117), the author introduces `total decomposition sets', which help him study a sum related to yours with $a=2$ and $n \ge 2$ (see his Lemma 3). Modifying the proof of Lemma 3 slightly, we obtain the bound
$$\ll\_{a,n} \begin{cases} X^{n-\frac{n(a-1)}{n-1}+o(1)} & \text{if }n > a,\\ X^{1+o(1)} & \text{if }n=a,\end{cases}$$
which beats $2n-a$ as long as $(a,n) \neq (1,1)$.
The dependence on $a,n$ can be made explicit but is quite horrible. To see that $n=a$ and $a=1$ are optimal consider the contribution of $x\_1=x\_2=\ldots=x\_n$.
|
1
|
https://mathoverflow.net/users/31469
|
398110
| 164,306 |
https://mathoverflow.net/questions/398109
|
2
|
Let $f\in L^p(\mathbb{R})$ and define $f\_\theta(x)=f(x-\theta)$. I would like to compute (or at least lower bound) the following:
$$
\inf\_{\theta\ne\theta'}\frac{\Vert f\_\theta - f\_{\theta'}\Vert\_p}{|\theta-\theta'|}.
$$
In particular, I want to understand how this depends on $f$, and would like a bound that depends explicitly on $f$. This is also where the properties of $f$ come in: The weaker the assumptions the better, but e.g. if there a nice bound that depends (say) on the deriviatives of $f$, then we can assume the needed regularity.
My suspicion is that there is an easy counterexample to show this can be rather poorly behaved even for smooth functions, but I have not been creative enough so far.
|
https://mathoverflow.net/users/323995
|
Lower bounds on translates of a function
|
For any real $p\ge1$ and any real $t\ne0$,
$$\frac{\|f\_t-f\_0\|\_p}{|t-0|}
\le\frac{\|f\_t\|\_p+\|f\_0\|\_p}{|t|}
=\frac{2\|f\|\_p}{|t|}\to0$$
as $|t|\to\infty$.
So, the least lower bound in question is always $0$.
|
3
|
https://mathoverflow.net/users/36721
|
398111
| 164,307 |
https://mathoverflow.net/questions/395996
|
8
|
Consider the following statement:
***If a vector space has a basis then its dual vector space also has a basis.***
It is not an axiom of ZF. It clearly follows from the Axiom of Choice. But it is also strictly weaker than the Axiom of Choice.
I have convinced myself that this statement holds in a variety of models of set theory with atoms (e.g. it holds in the first Fraenkel-Mostowski model, in the second Fraenkel-Mostowski model, in the ordered Fraenkel-Mostowski model, etc.).
I am wondering:
***What is the strength of the above statement?***
I am also interested in any reasonable variant of the question --- e.g. what if we restrict to a particular field $\mathbf{K}$? In case $\mathbf{K} = 2$, the statement can be rephrased as:
*For any set $A$ the vector space $\mathcal{P}(A)$ has a basis.*
Can the general statement be reduced to this special one over two-element field $2$?
---
Let us work out some examples in ZFA. For simplicity, I shall assume that our vector spaces are over the field of real numbers.
1. Let $A$ be the set of all atoms in the first Fraenkel-Mostowski model. Then the vector space $\mathbb{R}^A$ consists of all symmetric functions $f \colon A \rightarrow \mathbb{R}$. Let us assume that the support of $f$ is $A\_0$. Then $f$ has to be constant on $A \setminus A\_0$ and arbitrary on $A\_0$. Therefore, the set of vectors:
$$\{1, a\_1^\*, a\_2^\*, \cdots \} \approx A \sqcup 1$$
where:
* 1 is the constant function, i.e. $1(a)=1$
* $a^\*$ is the characteristic function, i.e. $a^\*(b) = [a = b]$
is a basis for $\mathbb{R}^A$
2. Let $Z\_\*$ be the set of all atoms in the second Fraenkel-Mostowski model (thought of as non-zero integer numbers with symmetry given as multiplication by $-1$). A symmetric function $f \colon Z\_\* \rightarrow \mathbb{R}$ must satisfy $f(-n) = f(n)$ for all but finitely many $n$. Let us consider *classical* (i.e.~in the real world) vector space $\mathbb{R}^{\mathcal{N}\_\*}$, where $\mathcal{N}\_\*$ is the set of *positive* natural numbers. Let us consider standard vectors $e\_1, e\_2, \cdots$ and extend them (in any way) to a basis of $\mathbb{R}^{\mathcal{N}\_\*}$. Let us denote the set of these additional vectors by $\Lambda$, i.e. $\{e\_1, e\_2, \cdots\} \cup \Lambda$ is a basis for $\mathbb{R}^{\mathcal{N}\_\*}$. For every vector $\lambda \in \Lambda$ define a symmetric function $\overline{\lambda} \colon Z\_\* \rightarrow \mathbb{R}$ as follows:
$$\overline{\lambda}(z) = \lambda(|z|)$$
I claim that the set of vectors:
$$\{\overline{\lambda} \colon \lambda \in \Lambda\} \cup \{z^\* \colon z \in Z\_\*\} \approx \Lambda \sqcup Z\_\*$$
forms a basis for $\mathbb{R}^{Z\_\*}$
Indeed, it is clear that this set generates $\mathbb{R}^{Z\_\*}$. To see that the vectors are linearly independent consider the following two cases:
* $\overline{\lambda} = \sum\_{i=1}^k a\_i\overline{\lambda\_i} + \sum\_{i=1}^l b\_i z^\*\_i$ where all vectors are distinct. In particular, this implies that:
$$\overline{\lambda}(n) = \sum\_{i=1}^k a\_i\overline{\lambda\_i}(n) + \sum\_{i=1}^l b\_i z^\*\_i(n)$$
for every positive natural number $n$. But this reduces to: $$\lambda(n) = \sum\_{i=1}^k a\_i \lambda\_i(n) + \sum\_{i=1}^l b\_i e\_{z\_i}(n)$$
where $e\_{z\_i} \equiv 0$ if $z\_i < 0$. But this leads to the contradiction with the assumption that $\{e\_1, e\_2, \cdots\} \cup \Lambda$ are linearly independent.
* $z^\* = \sum\_{i=1}^k a\_i\overline{\lambda\_i} + \sum\_{i=1}^l b\_i z^\*\_i$ where all vectors are distinct and $z < 0$ (the case $z> 0$ is symmetric). In particular, this implies that:
$$z^\*(-n) = \sum\_{i=1}^k a\_i\overline{\lambda\_i}(-n) + \sum\_{i=1}^l b\_i z^\*\_i(-n)$$ for every positive natural number $n$. This reduces to: $$e\_{-z}(n) = \sum\_{i=1}^k a\_i \lambda\_i(n) + \sum\_{i=1}^l b\_i e\_{-z\_i}(n)$$
This leads to the contradiction with the assumption that $\{e\_1, e\_2, \cdots\} \cup \Lambda$ are linearly independent.
3. Let $Q$ be the set of all atoms in the ordered Fraenkel-Mostowski model. Consider a symmetric function $f \colon Q \rightarrow \mathbb{R}$. Then there is a finite decomposition $Q = I\_1 \sqcup I\_2 \sqcup \cdots I\_n$ on intervals $I\_k$, such that $f$ is constant on each $I\_k$. Therefore, the set of vectors:
$$\{1, p\_1^\*, p\_2^\*, \cdots, p\_1^{<}, p\_2^{<}, \cdots \} \approx Q \sqcup Q \sqcup 1$$
where:
* 1 is the constant function, i.e. $1(a)=1$
* $p^\*$ is the characteristic function, i.e. $p^\*(q) = [p = q]$
* $p^{<}$ is the open down set of $p$, i.e. $p^{<}(q) = [p > q]$
generates $\mathbb{R}^Q$. Moreover, these vectors are linearly independent: if $S$ is any non-empty finite set of the above vectors, then there exists vector $m \in S$ and an atom $p$ with $m(p) = 1$ such that for every $s \in S \setminus \{m\}$ we have that $s(p) = 0$, so $m$ cannot be a linear combination of $S \setminus \{m\}$.
|
https://mathoverflow.net/users/13480
|
If a vector space has a basis then its dual vector space has a basis
|
Here are two statements you might find interesting:
1. With base field $k\subseteq\mathbb C$ your axiom implies that any free $\mathbb Z$-action has a choice of representatives.
2. For each prime $p,$ there is an $\omega$-categorical theory such that the associated permutation model contains a set $X$ with no basis for $\mathbb F\_p^X$ - this addresses a suggestion in the comments.
I’ll use ZFA throughout, blithely ignoring the question of whether your axiom implies AC over ZF.
---
A free $\mathbb Z$-action on a set $X$ is (generated by) a bijection $T:X\to X$ such that $T^ix=x$ implies $i=0.$ Let $k\subseteq \mathbb C.$ Let $X$ be a set such that $k^X$ has a basis $\mathcal B.$ We will show that every free $\mathbb Z$-action on $X$ has a choice of orbit representatives.
Why might this be interesting? It seems like a non-trivial consequence because the conclusion does not directly refer to linear algebra. The conclusion for all $X$ implies the axiom $\omega\zeta\mathrm{AC}$ introduced in [1], Howard-Rubin Form 119, which is equivalent to: every free $\mathbb Z$-action with a countable number of orbits has a set of representatives. It might not seem like a huge amount of strength, but then the evidence you’ve gathered suggests your axiom is quite weak (not pejoratively!), at least over ZFA.
Restricting to a single orbit, we are given a free transitive $\mathbb Z$-action generated by $T:A\to A$ and we want to pick an element of $A,$ uniformly in $(A,T).$
Let $\delta\_b\in (k^X)^\*$ for $b\in\mathcal B$ be the dual vectors, so every $v\in k^X$ is a unique $k$-linear sum $\sum\_{b\in\mathcal B} \delta\_b(v) b$ with finitely many non-zero coefficients. Let $[A]$ be the indicator function of $A.$ For each $a\in A$ define $f\_a\in k^X$ by $f\_a(T^na)=n$ and $f\_a(x)=0$ for $x\not\in A.$ Define
$$c(a)=\sum\_{b\in \mathcal B} |\delta\_b(f\_a)|.$$
Then $c(T^na)\to\infty$ as $|n|\to\infty$ for fixed $a,$ because
$$|\delta\_b(f\_{T^na})| = |\delta\_b(f\_a) - n\delta\_b([A])|\to\infty$$
for any $b$ such that $\delta\_b([A])\neq 0.$
The set $C=\operatorname{argmin} c\subset A$ is therefore finite, and we can take the least element in the ordering generated by $a<Ta.$ $\square$
---
Take the $\omega$-categorical theory to be the Fraïssé limit of the class of bilinear forms $B:X\times \Phi\to \mathbb F\_p,$ where $X$ and $\Phi$ are finite-dimensional vectors spaces over $\mathbb F\_p.$ These are structures in the language with unary symbols $X,\Phi,\mathbb F\_p,$ field operations, two sets of vector space operations, and a binary operation for $B.$ The finite support permutation model $N$ of this structure will therefore contain a bilinear form which I’ll also call $B:X\times\Phi\to\mathbb F\_p.$
I’ll need a “unique minimal support” property. Unique support arguments often require a lot of care. To make life as easy as possible I’ve tried to prove only the exact statement that the argument needs. Define an “acl support” to be a pair $(X\_0,\Phi\_0)$ where $X\_0$ is a finite subspace of $X$ and $\Phi\_0$ is a finite subspace of $\Phi.$ (“acl” stands for algebraically closed, the same as definably closed here). A set $x\in N$ is “supported by” $(X\_0,\Phi\_0)$ if every automorphism that fixes $(X\_0,\Phi\_0)$ elementwise also fixes $x.$
**Lemma.** Every $f\in \mathbb F\_p^X$ in $N$ is supported by
a unique minimal acl support $\operatorname{supp}(f).$
*Proof:* $f$ is supported by $(X\_0,\Phi\_0)$ if and only if: $f(x)=f(x’)$ whenever $x$ and $x’$ have the same $1$-type over $(X\_0,\Phi\_0).$ These $1$-types (extending the partial type of elements of $X$) have an explicit description: there are exactly $|X\_0|+|\Phi\_0|.$ corresponding to the points of $X\_0$ and the $1$-types defined by the formula $$(\bigwedge\_{x\_0\in X\_0}x\neq x\_0)\wedge (\bigwedge\_{\phi\in\Phi\_0}B(x,\phi)=\hat x(\phi))$$ for each $\hat x\in \Phi\_0^\*.$
Suppose $f$ is supported by both $(X\_0,\Phi\_0)$ and $(X\_1,\Phi\_1).$ We will show that $f$ is supported by the intersection of these supports, which clearly implies the Lemma. Consider $x,x’\in X$ with the same $1$-type over $(X\_0\cap X\_1,\Phi\_0\cap \Phi\_1)$; we want to show $f(x)=f(x’).$ The case $x\in X\_0\cap X\_1$ is immediate. If $x\in X\_0\setminus X\_1$ then we can replace $x$ by $x’’$ where $x’’$ has the same $1$-type over $(X\_1,\Phi\_1)$ but $x’’\not\in X\_0.$ Since $f$ is supported by $(X\_1,\Phi\_1),$ we know $f(x)=f(x’’).$ So we can reduce to the case $x\not\in X\_0\cup X\_1,$ and by a similar argument we can reduce to the case $x’\not\in X\_0\cup X\_1.$
Now pick $x’’\in X\setminus (X\_0\cup X\_1)$ such that $B(x’’,\phi)=B(x,\phi)$ for all $\phi\in\Phi\_0,$ and $B(x’’,\phi)=B(x’,\phi)$ for all $\phi\in\Phi\_1.$ Then $f(x)=f(x’’)$ because $f$ is supported by $(X\_0,\Phi\_0).$ And $f(x’’)=f(x’)$ because $f$ is supported by $(X\_1,\Phi\_1).$ Therefore $f(x)=f(x’’)$ as required. $\square$
**Corollary.** If a finite set $\mathcal S\subset\mathbb F\_p^X$ is supported by $(X\_0,\Phi\_0),$ then every element of $\mathcal S$ is supported by $(X\_0,\Phi\_0).$
*Proof.* Let $Z=\bigcup\_{f\in\mathcal S} \operatorname{supp}(f).$ Then $Z$ is a finite subset of $X\cup \Phi$ supported by $(X\_0,\Phi\_0).$ Suppose for contradiction that there exists $z\in Z\setminus (X\_0\cup \Phi\_0).$ There are infinitely many $z’$ with the same $1$-type as $z$ over $(X\_0,\Phi\_0).$ So we can find an automorphism fixing $(X\_0,\Phi\_0)$ elementwise but sending $z$ to some $z’\not\in Z,$ a contradiction. $\square$
Suppose for contradiction that $N$ has a basis $\mathcal B$ of $\mathbb F\_p^X.$ Work in the full universe of ZFCA. $\mathcal B$ is supported by some $(X\_0,\Phi\_0),$ which can be taken to be vector subspaces but are not assumed minimal or unique in any sense.
Pick two elements of $\Phi$ that map to linearly independent vectors in $\Phi/\Phi\_0.$ Their span $\Phi\_1$ has dimension $2.$ Let $L$ be the set of $p+1$ one-dimensional subspaces of $\Phi\_1.$ Each $\ell\in L$ determines a function $f\_{\ell}\in \mathbb F\_p^X$:
* $f\_{\ell}(x)=0$ if $B(x,\phi)=0$ for all $\phi\in \ell$
* $f\_{\ell}(x)=1$ otherwise
For any $x\in X$ the kernel $\{\phi\in\Phi\_1: B(x,\phi)=0\}$ either has dimension $1,$ in which case $f\_\ell(x)=0$ for exactly one $\ell,$ or has dimension $2,$ in which case $f\_\ell(x)=0$ for all $\ell.$ Thus
$$\sum f\_\ell = 0.$$
Let $F\_0\subset \mathbb F\_p^X$ denote the space of functions that are supported by $(X\_0,\Phi\_0).$ Let $F\_\ell$ denote the space of functions that are supported by $(X\_0,\operatorname{span}(\Phi\_0\cup \ell)).$ Then $F\_\ell\cap F\_{\ell’}=F\_0$ for distinct $\ell,\ell’$ - consider the unique minimal acl support of functions in $F\_\ell\cap F\_{\ell’}.$
Let $\delta\_b(f\_\ell)$ denote the coefficient of the basis function $b\in\mathcal B$ when expanding $f\_\ell$ in the basis $\mathcal B.$ $f\_\ell$ is not in the span of $F\_0,$ because it is possible to find $x,x’$ satisfying $f\_\ell(x)=0$ and $f\_\ell(x’)=1$ with the same $1$-type over $(X\_0,\Phi\_0),$ specifically $x,x’\not\in X\_0$ and $B(x,\phi)=B(x’,\phi)=0$ for all $\phi\in\Phi\_0.$ By the Corollary applied to the set $\{b:\delta\_b(f\_\ell)\neq 0\},$ every $b$ with $\delta\_b(f\_\ell)\neq 0$ lies in $F\_\ell.$ Pick any $\ell$ and $b\in F\_\ell\setminus F\_0$ with $\delta\_b(F\_\ell)\neq 0.$ For $\ell’\neq\ell,$ we have $b\not\in F\_{\ell’}$ and hence $\delta\_b(f\_{\ell’})=0.$ We can finally state the contradiction to the assumption of a basis of $\mathbb F\_p^X$:
$$0\neq \delta\_b(f\_\ell)=\delta\_b(\sum\_{\ell’} f\_{\ell’}) = 0.$$
---
[1] *van Douwen, Eric K.*, [**Horrors of topology without AC: A nonnormal orderable space**](http://dx.doi.org/10.2307/2045582), Proc. Am. Math. Soc. 95, 101-105 (1985). [ZBL0574.03039](https://zbmath.org/?q=an:0574.03039).
|
1
|
https://mathoverflow.net/users/164965
|
398116
| 164,310 |
https://mathoverflow.net/questions/398106
|
2
|
Let $Y$ be a compact Riemann surface and $B$ a finite subset of $Y$. It is a standard fact that isomorphism classes of holomorphic ramified covers $f:X\rightarrow Y$ of degree $d$ with branch points in $B$ are in a correspondence with homomorphisms $\rho:\pi\_1(Y-B)\rightarrow S\_d$ with transitive image modulo conjugation by elements of the permutation group $S\_d$. Writing a formula for $f$ from the knowledge of $B\subset Y$ and $\rho$ is often hard, e.g. the task of recovering a Belyi map from its dessin where $|B|=3$. I am interested in the case of $X=Y=\Bbb{CP}^1$, and some points from $B$ moving in the Riemann sphere. Here is an example:
* Consider rational functions $f:\Bbb{CP}^1\rightarrow\Bbb{CP}^1$ of degree $3$ with four simple critical points that have $1,\omega,\bar{\omega}$ among their critical values $\left(\omega={\rm{e}}^{\frac{2\pi{\rm{i}}}{3}}\right)$, thus $B=\{1,\omega,\bar{\omega},\beta\}$ with $\beta$ varying in a punctured sphere. To fix an element in the isomorphism class, we can pre-compose $f$ with a suitable Möbius transformation so that $1$, $\bar{\omega}$ and $\omega$ are the critical points lying above $1$, $\omega$ and $\bar{\omega}$ respectively: $f(1)=1, f(\bar{\omega})=\omega, f(\omega)=\bar{\omega}$. A normal form for such functions is
$$
\left\{f\_\alpha(z):=\frac{\alpha z^3+3z^2+2\alpha}{2z^3+3\alpha z+1}\right\}\_\alpha.
$$
A simple computation shows that the fourth critical point is $\alpha^2$, and hence $\beta=\beta\_\alpha=:f\_\alpha(\alpha^2)=\frac{\alpha^4+2\alpha}{2\alpha^3+1}$. Here is my question:
**Why $\beta$ is not a degree one function of $\alpha?$ Shouldn't the knowledge of the branch locus and the monodromy determine $f\_\alpha(z)$ in the normalized form above? I presume the monodromy does not change because there are only finitely many possibilities for it and this is a continuous family.**
To monodromy of $f\_\alpha$ is a homomorphism
$$
\rho\_\alpha:\pi\_1\left(\Bbb{CP}^1-\{1,\omega,\bar{\omega},\beta\_\alpha\}\right)\rightarrow S\_3
$$
where small loops around $1,\omega,\bar{\omega},\beta\_\alpha$ generate the fundamental group, and are mapped to transpositions in $S\_3$ whose product is identity and are not all distinct. So I guess my question is how can such a discrete object vary with $\alpha$; and if it doesn't, why the assignment $\alpha\mapsto\beta(\alpha)$ is not injective. The degree of this assignment is four, and there are also four conjugacy classes of homomorphisms
$\rho:\langle\sigma\_1,\sigma\_2,\sigma\_3,\sigma\_4\mid\sigma\_1\sigma\_2\sigma\_3\sigma\_4=\mathbf{1}\rangle\rightarrow S\_3$ with ${\rm{Im}}(\rho)$ being a transitive subgroup of $S\_3$ generated by transpositions $\rho(\sigma\_i)$:
$$
\sigma\_1\mapsto (1\,2),\sigma\_2\mapsto (1,2),\sigma\_3\mapsto (1,3), \sigma\_4\mapsto (1,3);\\
\sigma\_1\mapsto (1\,2),\sigma\_2\mapsto (1,3),\sigma\_3\mapsto (1,2), \sigma\_4\mapsto (2,3);\\
\sigma\_1\mapsto (1\,2),\sigma\_2\mapsto (1,3),\sigma\_3\mapsto (1,3), \sigma\_4\mapsto (1,2);\\
\sigma\_1\mapsto (1\,2),\sigma\_2\mapsto (1,3),\sigma\_3\mapsto (2,3), \sigma\_4\mapsto (1,3).
$$
Is it accidental that the degree of $\alpha\mapsto\beta(\alpha)$ is the same as the number of possibilities for the monodromy representations compatible with our ramification structure?
|
https://mathoverflow.net/users/128556
|
Recovering a family of rational functions from branch points
|
The reason for this phenomenon is that you are afflicted with a serious mathematical condition, that being:
Your monodromy has monodromy.
---
To be less cryptic, the key thing is that the fundamental group is not actually generated by small loops around these four points, except if by "small loop around $x$" you mean "a path from the base point to $x$, then a small loop around $x$, then following the same path back" - these do generate the fundamental group.
When you move the point $\beta$, you will have to continuously adjust these paths - the reason you need to adjust the path to $\beta$ is clear, but also the other paths may need to be adjusted if $\beta$ would otherwise pass through them.
If you bring $\beta$ around a loop and back to its starting value, these paths have no reason to be the same paths they were before, so while the conjugacy classes of the loops around the four points will be the same in your chosen map $\rho$ from $\pi\_1$ to $S\_3$, the actual values of the loops around the four points under $\rho$ could be different.
So, in the general case, what you get is a map from the moduli space of possible covers to the curve parameterizing values of $\beta$ (or a higher-dimensional version, if more points vary) which is locally constant but not (usually) constant, and where the fiber consists of all monodromy representations which have fixed local conjugacy classes.
So it is not an accident at all that the degree is the same as the number of possibilities - it just reflects the fact that this covering map is connected, which is a common occurrence but not universal - there are known invariants which sometimes separate different components.
|
5
|
https://mathoverflow.net/users/18060
|
398125
| 164,316 |
https://mathoverflow.net/questions/398132
|
3
|
**Problem set up:**
Let $\mathbf X := (X, \mathcal A, \mu)$ be a standard probability space.
We say that a measure preserving transformation $T$ on $\mathbf X$ is *$\varepsilon$-almost weakly mixing* if for every $\delta > \varepsilon$, and every pair of non-null measurable sets $A, B \in \mathcal A$, there exists an $N > 0$ such that for all $n > N$, we have $$\frac{1}{n} \sum\_{k = 1}^{n} |\mu(T^{-k}A \cap B) - \mu(A)\mu(B)| < \delta \mu(A)\mu(B).$$
We say a measure preserving transformation $G$ on $X$ *admits an ergodic component* if there exists some non-null measurable subset $E$ of $X$ such that $G(E) \subset E$, and the “restricted system” ($\mathbf E, G\_{|E})$, with $\mathbf E := (E, \mathcal A\_{|E}, \mu\_{|E})$ is ergodic. Here $\mathcal A\_{|E}$ is the restricted sigma algebra, and $\mu\_{|E}$ is defined by $\mu\_{|E}(A) := \mu(A \cap E)/\mu(E)$.
>
> **Question:** Does there exist some $\varepsilon > 0$ such that any $\varepsilon$-almost weakly mixing transformation $T$ on $\mathbf X$ admits an ergodic component?
>
>
>
**Remark:** This is a potential sharpening of an [earlier result](https://mathoverflow.net/questions/392667/does-an-almost-mixing-transformation-admit-a-non-null-ergodic-component).
|
https://mathoverflow.net/users/173490
|
Does an “almost weakly mixing” transformation admit a non-null ergodic component?
|
Your question is equivalent to asking whether $T$ being $\epsilon$-almost weak mixing implies that the invariant $\sigma$-algebra, $\mathcal I$ contains atoms. The answer is yes.
I will prove the contrapositive. I claim if $\mathcal I$ contains no atoms, then $T$ is not $\epsilon$-almost weak mixing for any $\epsilon>0$. To see this, notice that there are invariant sets of arbitrarily small measure. Let $A$ be an invariant set and let $B=A$. Then for each $k$, $|\mu(T^{-k}A\cap B)-\mu(A)\mu(B)|=\mu(A)-\mu(A)^2=\mu(A)(1-\mu(A))=\frac {1-\mu(A)}{\mu(A)}\mu(A)$.
In particular, $T$ is not $\epsilon$-almost weak mixing.
|
5
|
https://mathoverflow.net/users/11054
|
398133
| 164,318 |
https://mathoverflow.net/questions/398138
|
2
|
In Qing Liu's Algebraic Geometry and Arithmetic Curve, we have the following proposition(6.3.13):
Let $S$ be a locally Noetherian scheme. Let $X$, $Y$ be smooth schemes over $S$. Then any immersion $f:X\rightarrow Y$ of $S$-schemes is a regular immersion, and we have a canonical exact sequence on $X$:
$0\rightarrow \mathcal{C}\_{X/Y}=f^\*(\mathscr{I}\_X/\mathscr{I}\_X^2)\rightarrow f^\*\Omega\_{Y/X}\rightarrow \Omega\_{X/S}\rightarrow 0$
**For this question**: Suppose $Y\rightarrow X$ is a birational morphism of regular schemes. Suppose we have an embedding $i:Y\hookrightarrow W$ and $W$ is smooth over $X$. Then one can show that the sheaf of ideal $\mathscr{I}\_Y$ of $\mathcal{O}\_W$ defining $Y$ is generated by regular sequence. If we define $\mathcal{C}\_{Y/W}=i^\*(\mathscr{I}\_Y/\mathscr{I}\_Y^2)$, then $\mathcal{C}\_{Y/W}$ is a locally free sheaf of rank n on Y, where n is the relative dimension of $W$ over $X$
**I am wondering whether the following sequence is left exact given the fact that $Y$ is not smooth over $X$.
$0\rightarrow \mathcal{C}\_{Y/W}\rightarrow i^\*\Omega\_{W/X}\rightarrow \Omega\_{Y/X}\rightarrow 0$**
If this is exact, then we may define the Jacobian ideal $\mathcal{J}\_{Y/X}$ as the ideal sheaf $\bigwedge^n\mathcal{C}\_{Y/W}\otimes (\bigwedge^n i^\*\Omega\_{W/X})^{-1}$ of $\mathcal{O}\_Y$. One can show that this definition is independent of $W$ but if the above sequence is not exact, then I don't know whether this is still an ideal sheaf of $\mathcal{O}\_Y$. Thank you.
|
https://mathoverflow.net/users/nan
|
Construction of Jacobian Ideal
|
I think this is true: since $Y\rightarrow X$ is birational, the homomorphism $\mathcal{C}\_{Y/W}\rightarrow i^\*\Omega\_{W/X}$ is bijective on a dense open subset of $Y$. Thus its kernel is torsion, hence zero since $\mathcal{C}\_{Y/W}$ is locally free.
|
3
|
https://mathoverflow.net/users/40297
|
398140
| 164,320 |
https://mathoverflow.net/questions/398033
|
2
|
Let FSym$(\mathbb{N})$ denote the finitary symmetric group on the natural numbers. Are all Sylow p-subgroups of FSym$(\mathbb{N})$ isomorphic (maybe to the iterated wreath product $C\_p\wr C\_p\wr\dots$ of cyclic $p$-group $C\_p$)?
It is well-known that the Sylow $p$-subgroups of $S\_{p^k}$, for the positive integer $r$, are all isomorphic to $C\_p\wr C\_p\wr\dots\wr C\_p$ (with $k$ copies of $C\_p$) [see for example Proposition 19.10 of the book “A Course in Group Theory” by J.H. Humphreys].
|
https://mathoverflow.net/users/98061
|
Isomorphic Sylow p-subgroups of FSym$(\mathbb{N})$
|
$\DeclareMathOperator\FSym{FSym}\DeclareMathOperator\Sym{Sym}\newcommand\N{\mathbf{N}}$The answer is no, as already mentioned in the comments: a transitive Sylow subgroup, or more generally any subgroup without finite orbit, has a trivial center (clear, since for a nontrivial element, its centralizer preserves its support), while any Sylow with a finite orbit of cardinal $\ge p$ has a nontrivial center (and they indeed exist in $\FSym(X)$ for any infinite $X$).
The main question seems to classify $p$-Sylow subgroups. Let us provide a full reduction to the transitive case.
**Proposition 1**. *Let $P$ be a $p$-Sylow, $(X\_i)$ its orbit decomposition, $P\_i$ the image of $P$ in $\FSym(X\_i)$. Then $P=\bigoplus\_i P\_i$, and all finite $X\_i$ have a $p$-power cardinal.*
Indeed, $P\subset\bigoplus\_i P\_i$ (as elements have finite support), and by maximality, $P=\bigoplus\_i P\_i$. The second assertion is clear from the finite case.$\Box$
For the converse:
**Proposition 2**. \*Let $X$ be a set with a partition $(X\_i)$ such that each finite $X\_i$ has a $p$-power cardinal. Let $P\_i$ be a transitive $p$-Sylow in $\FSym(X\_i)$. Then $P=\bigoplus P\_i$ is a $p$-Sylow in $\FSym(X\_i)$ iff for each $n<\infty$, the number $n\_i$ of $i$ such that $|X\_i|=p^n$ is $<p$.
The condition is clearly necessary, using the finite case. Conversely, suppose it holds. Choose $f\notin \bigoplus P\_i$ such that $\langle P,f\rangle$ is a $p$-subgroup. Let $J$ be the set of $i$ such that $f(X\_i)\neq X\_i$, and $X'=\bigcup\_{i\in J}X\_i$. Note that $J$ is not empty, and that $X'$ is $\langle P,f\rangle$-invariant.
a) Suppose that $X'$ is finite. The assumptions $n\_i<p$ implies that $\prod\_{i\in J}P\_i$ is a $p$-Sylow in the finite group $\Sym(X')$, so we get a contradiction.
b) So $X'$ is infinite; since $J$ is finite (as the support of $f$ is finite), there is $i\in J$ with $X\_i$ infinite. Let $W$ be the (finite) support of $f$ and $W\_i=W\cap X\_i$. By a classical lemma of B.H. Neumann, there exists $s\in P\_i$ such that $s(W\_i)\cap W\_i=\emptyset$. Then for some $k\ge 1$, some cycle of $f$ visits a point in $X'-X\_i$, then $k$ points in $X\_i$, and then goes back to $X'-X\_i$. By direct verification, the commutator $f'=[s,f]=sfs^{-1}f^{-1}$ contains a $(2k+1)$-cycle whose support meets $X\_i$ in a subset of cardinal $2k$. Apply the same with $f'$: find $s'\in P\_i$ so that $f''=[s',f']$ contains a $(4k+1)$-cycle. Since $f'$ and $f''$ are both in a $p$-Sylow, we deduce that both $2k+1$ and $4k+1$ are $p$-powers. Since $1<(4k+1)/(2k+1)<2\le p$, we deduce a contradiction. $\Box$
This entirely reduces to understanding transitive Sylow subgroups.
And indeed it's known that all transitive $p$-Sylow are conjugate when $X$ is infinite countable. This is due to Ivanyuta *(Sylow p-subgroups of the countable symmetric group. (Russian) Ukrain. Mat. Ž 15 1963 240–249)*, see also *[here](https://doi.org/10.1016/j.jalgebra.2015.12.019)*.
(To complete the picture, one should prove that for $X$ uncountable there's no transitive $p$-Sylow.) So a conjugacy class of $p$-Sylow is determined by some cardinal $n\_\omega$ (the number of infinite orbits), and for each $i$, the number $0\le n\_i<p$ of orbits of cardinal $p^i$.
Actually it is claimed in both papers that $p$-Sylow subgroups of $\FSym(X)$ are isomorphic iff they're conjugate: this is slightly false: they are isomorphic iff they have the same $n\_i$ for each $0<i\le\omega$ (but $n\_0$ can vary).
|
3
|
https://mathoverflow.net/users/14094
|
398147
| 164,323 |
https://mathoverflow.net/questions/398143
|
4
|
There is this (folklore?) fact: for a commutative ring $R$, the category of $R$-modules is equivalent to the category of internal abelian groups in the slice category $\operatorname{Commutative rings}/R$.
I would like to replace here $\operatorname{Commutative rings}$ with some ($\infty$-?)category $\bf X$ of symmetric monoidal (should I say E$\_\infty$?) stable $\infty$-categories. Then presumably "internal abelian groups" is not relevant, as stable $\infty$-categories have their own intrinsic sort of "abelian group" structure. On the other hand, also presumably, just taking slice over $R$ would not give any notion of module readily: for whatever kind of morphism $f:R'\to R$ of monoidal $\infty$-categories, I don't see any sensible way to define some action of $R$ on the fibre of $f$ (or is there any?).
So my question is what kind of structures in ${\bf X}/R$ would provide good notion of $R$-module in the stable $\infty$-setting? Or maybe there are different notions of module and some of them must be captured in entirely different ways?
|
https://mathoverflow.net/users/41291
|
Is there essentially unique notion of module over monoidal stable $\infty$-categories?
|
The ∞-categorical analog of the fact you mention can be found in *Higher Algebra*, corollary 7.3.4.14:
>
> Let $\operatorname{CAlg}$ be the category of $E\_\infty$-rings and $A\in \operatorname{CAlg}$. Then there is a canonical equivalence
>
>
> $$\operatorname{Sp}(\operatorname{CAlg}\_{/A})\simeq \operatorname{Mod}\_A$$
>
>
> between the stabilization of the ∞- category of $E\_∞$-rings with a map to $A$ and the ∞-category of $A$-modules.
>
>
>
Note that when $A$ is a discrete commutative ring, $\operatorname{Mod}\_A$ is the usual derived category of $A$ (so not the 1-category of $A$-modules).
In fact a similar result holds for algebras over any coherent operad (ibid. Theorem 7.3.4.13). So one would be tempted to take $\operatorname{Sp}(\mathcal{C}\_{/A})$ to be the definition of "$A$-modules" for $A\in\mathcal{C}$. It is not clear at all that this is a well behaved notion though: in the case where $\mathcal{C}$ is the ∞-category of animated rings (i.e. the ∞-category obtained by inverting the weak equivalences in the category of simplicial rings), this **does not** recover the natural notion of module: see for example the discussion in Section 25.3.3 of Lurie's *Spectral Algebraic Geometry*.
|
11
|
https://mathoverflow.net/users/43054
|
398148
| 164,324 |
https://mathoverflow.net/questions/398008
|
5
|
This question is a repost of the following: <https://math.stackexchange.com/questions/4195805/sections-of-a-polar-action-are-totally-geodesic>. I've decided it to post it here because it didn't seem to get much traction there.
Suppose $G\curvearrowright M$ is an isometric action of a Lie group on a complete Riemannian manifold $M$, and assume it is polar. This means that the action is proper and there exists a closed (hence complete) embedded submanifold $\Sigma\subseteq M$ (called a section) which meets all orbits orthogonally. It is well known that $\Sigma$ is totally geodesic, but I have not found a convincing proof of that fact.
There is a special case where the last statement is easy to prove: the second fundamental form vanishes at regular (and exceptional) points. Indeed, given any $p\in \Sigma$ such that $G\cdot p$ has maximal dimension, $v\in T\_{p}(\Sigma)=\nu\_{p}(G\cdot p)$ and $\xi\in T\_{p}(G\cdot p)$, we can find an element $X$ in the Lie algebra $\mathfrak{g}$ of $G$ such that
$$\xi=X^{\*}(p), \quad X^{\*}(q)=\dfrac{d}{dt}\bigg|\_{t=0}\operatorname{Exp}(tX)\cdot q.$$
The vector field $X^{\*}$ is Killing, so that its covariant derivative is antisymmetric. Let $\mathbb{II}$ be the second fundamental form of $\Sigma$. Then $\mathbb{II}(v,v)$ is tangent to $G\cdot p$ and $\langle \mathbb{II}(v,v),\xi \rangle=-\langle v,\nabla\_{v}X^{\*} \rangle=0$, so $\mathbb{II}(v,v)=0$. Polarizing, we get $\mathbb{II}=0$.
The usual argument for proving that sections are totally geodesic at all points revolves around the fact that regular points are dense in $\Sigma$. My problem is that all proofs that I found seem to lack crucial details for it. Here are some examples:
1. In "Lie Groups and Geometric Aspects of Isometric Actions", by Alexandrino and Bettiol, it is stated in Exercise 4.9 that the density follows from Kleiner's Lemma (cf Lemma 3.70), but I can't get the connection between the lemma and this fact.
2. In "Critical Point Theory and Submanifold Geometry", by Palais and Terng, the authors state that density follows from the theory of Riemannian submersions, without giving further details.
3. In "Polar Manifolds and Actions", by Grove and Ziller, the authors state that singular points are isolated along any geodesic, because of the Slice Theorem. This is because if $\gamma$ is any geodesic of $\Sigma$, and $t\_{0}$ lies in the closure of $\{ t\in \mathbb{R}\colon \gamma(t)\in M\_{R} \}$, then $\gamma(t\_{0}-\varepsilon)$ and $\gamma(t\_{0}+\varepsilon)$ have the same isotropy for sufficiently small $\varepsilon>0$ (again, because of the Slice Theorem), but I don't understand why this is the case.
4. Another method of proof is proposed in the book "Submanifolds and Holonomy", by Berndt, Console and Olmos. I have an (almost full) solution, which needs to prove the following crucial fact: if $p\in \Sigma$ and there is an open subset $\Omega\subseteq \Sigma$ such that all orbits of the points in $\Omega$ have the same (nonprincipal) type, then $T\_{p}(\Sigma)$ is pointwise fixed by the slice representation. If somebody could give a proof of this last fact, I would also accept it as an answer.
Could somebody elaborate on any of the methods of proof proposed above (preferably the first or the third), or give a reference to a detailed proof of the fact that regular points are dense?
Thank you in advance!
|
https://mathoverflow.net/users/175146
|
Sections of a polar action are totally geodesic
|
I will try to answer your question posed in (4). From now on, $G$ is a Lie group acting properly and isometrically on a connected Riemannian manifold $M$. First, we need a little preparation.
**Fact.** Let $Q \subseteq M$ be an orbit of $G$. There exists $\varepsilon > 0$ such that the exponential map is defined on $N^\varepsilon Q$ and sends it diffeomorphically onto a neighborhood of $Q$, which we denote $U^\varepsilon Q$ (here $NQ$ is the normal bundle of $Q$ and $N^\varepsilon Q$ denotes the subset of vectors in it of length $< \varepsilon$). Given any such $\varepsilon$ and $p \in Q$, we write $S^\varepsilon\_p$ for $\exp(N^\varepsilon\_p Q)$. It then follows that $S^\varepsilon\_p$ is a slice at $p$.
This is a special case of the tubular neighborhood theorem, when our submanifold is homogeneous, meaning that for any $p, q \in Q$ there exists an isometry $\varphi \in I(M)$ with $\varphi(Q) = Q$ that sends $p$ to $q$. Generally, unless $Q$ is compact, we need to allow our tubular neighborhood to have varying thickness, but if $Q$ is homogeneous, we can make the neighborhood uniform. This is the content of exercise 2.11.2 in the book "Submanifolds and Holonomy" that you mentioned in your question (see also definition 6.2 in [Michor's notes](https://www.researchgate.net/publication/2628583_Isometric_Actions_Of_Lie_Groups_And_Invariants)). I can sketch a proof here if needed.
Let's now agree that whenever we say the word "slice", we mean a slice of the form $S^\varepsilon\_p$ as above. Here is a simple but vital
**Corollary.** Let $p \in M$ be any, and let $S^\varepsilon\_p$ be a slice at $p$. Then $G\_q \subseteq G\_p$ for any $q \in S^\varepsilon\_p$.
*Proof.* Take any $\varphi \in G\_q$. Since isometries commute with the exponential map, $\varphi(S^\varepsilon\_p) = S^\varepsilon\_{\varphi(p)}$. These two slices have a point in common, namely $\varphi(q) = q$, so they coincide, i.e. $\varphi$ preserves $S^\varepsilon\_p$. But $G \cdot p \cap S^\varepsilon\_p = \{p\}$, so $\varphi$ preserves $p$. $\square$
Now we assume that $M$ is complete and the action of $G$ is polar. Let $\Sigma$ be a section. Let me remark here that some authors only require a section to be an immersed complete submanifold, not necessarily closed or embedded (for example, C. Gorodski does so [in this paper](https://link.springer.com/article/10.1023/B:GEOM.0000013806.62086.72)). As far as I understand, such a pathology can occur, for example, for certain polar actions on symmetric spaces of compact type. In any case, everything works even for such a general definition because any immersed submanifold is locally (in its own topology!) an embedded one. So let us assume there is an open subset $\Omega \subset \Sigma$ consisting of nonregular points of the same type. WLOG, we may assume $\Omega$ is an embedded submanifold of $M$. Pick $p \in \Omega$ and let $S^\varepsilon\_p$ be a slice at $p$. Shrinking $\Omega$ if needed, we may assume $\Omega \subseteq U^\varepsilon Q$, where $Q = G \cdot p$. *A posteriori*, $\Omega$ simply lies in $S^\varepsilon\_p$ because it intersects $Q$ orthogonally and is totally geodesic, but we don't know that at this point, so we need to somehow "project" $\Omega$ to $S^\varepsilon\_p$. Consider the map $\pi \colon G \times S^\varepsilon\_p \twoheadrightarrow U^\varepsilon Q, (g, q) \mapsto gq$. This is a smooth surjective submersion and in fact it is just quotienting by the free proper right action
$$
G\_p \curvearrowright G \times S^\varepsilon\_p, \quad h \cdot (g,q) = (gh,h^{-1}q).
$$
In other words, $\pi$ is a principal $G\_p$-bundle (in fact, it is one of the chief properties of slices that $\pi$ realizes $U^\varepsilon Q$ as $G \times\_{G\_p} S^\varepsilon\_p = (G \times S^\varepsilon\_p)/G\_p$, which is the fiber bundle over $G/G\_p \cong Q$ with a fiber $S^\varepsilon\_p$ associated to the principal $G\_p$-bundle $G \twoheadrightarrow G/G\_p \cong Q$). Our principal bundle $\pi$ can be restricted to the submanifold $\Omega$ of $U^\varepsilon Q$:
$$
\widehat{\pi} \colon \; \pi^{-1}(\Omega) \twoheadrightarrow \Omega.
$$
Clearly, $\widehat{\pi}$ is a principal $G\_p$-bundle in its own right. In particular, it is a smooth surjective submersion, which we'll use in a second. Now it's time to use the corollary above. I claim that if $(g,q), (g',q') \in \pi^{-1}(\Omega)$ lie over the same point in $\Omega$, then $q = q'$. Indeed, since $q \in S^\varepsilon\_p$, we have $G\_q \subseteq G\_p$ by the corollary above. On the other hand, $q$ has the same orbit type as $gq = \pi(g,q) \in \Omega$, which, by our assumption, coincides with the orbit type of $p$. By definition, it means that $G\_q$ and $G\_p$ are conjugate, so we deduce that $G\_q = G\_p$. But $(g',q')$ and $(g,q)$ differ by some element $h$ from $G\_p = G\_q$: $(g',q') = (gh, h^{-1}q) = (gh, q)$, so the assertion follows. Now, consider the composite map
$$
\pi^{-1}(\Omega) \hookrightarrow G \times S^\varepsilon\_p \twoheadrightarrow S^\varepsilon\_p.
$$
We've just proved that it is constant on the fibers of $\widehat{\pi}$, hence, by the characteristic property of smooth surjective submersions, it factors through $\widehat{\pi}$ to a smooth map $f \colon \Omega \to S^\varepsilon\_p$. One readily checks that $f$ is an injective immersion. This map $f$ is our desired "projection" of $\Omega$ to $S^\varepsilon\_p$. Write $\Omega\_1 \subseteq S^\varepsilon\_p$ for the image of $f$. As we have already shown in the assertion above, the action of $G\_p$ on $S^\varepsilon\_p$ fixes $\Omega\_1$ pointwise, hence the slice representation $G\_p \curvearrowright N\_pQ$ fixes $T\_p\Omega\_1$ pointwise as well. Actually, since
$$
\dim \Omega\_1 = \dim \Sigma = \mathrm{cohom}(G \curvearrowright M),
$$
this suffices to derive a contradiction, for the subspace of invariants of the slice representation at a nonregular point cannot be that large, but let me formally finish the proof by showing that $T\_p\Omega\_1 = T\_p\Sigma$. Denote $\mathrm{Lie}(G\_p) = \mathfrak{k} \subseteq \mathfrak{g} = \mathrm{Lie}(G)$, and let $\sigma \colon \mathfrak{g} \twoheadrightarrow T\_pQ$ stand for the differential of the orbit map $G \twoheadrightarrow Q, \; g \mapsto gp,$ at $e$ (it sends $X \in \mathfrak{g}$ to the value of the corresponding Killing vector field at $p$). Clearly, $\ker \sigma = \mathfrak{k}$. We make two simple observations. First,
$$
T\_{(e,p)} \pi^{-1}(\Omega) = \mathfrak{k} \oplus T\_p \Omega\_1 \subseteq \mathfrak{g} \oplus N\_pQ = T\_{(e,p)} (G \times S^\varepsilon\_p).
$$
Second, the differential of $\pi$ at $(e,p)$ is given simply by
$$
\sigma \oplus \mathrm{Id}\_{N\_pQ} \colon \; \mathfrak{g} \oplus N\_pQ \to T\_pQ \oplus N\_pQ = T\_pM.
$$
Therefore, this differential sends $T\_{(e,p)} \pi^{-1}(\Omega)$ onto $T\_p\Omega\_1$. But at the same time,
$$
d\pi(T\_{(e,p)} \pi^{-1}(\Omega)) = d\widehat{\pi}(T\_{(e,p)} \pi^{-1}(\Omega)) = T\_p \Omega = T\_p \Sigma,
$$
which completes the proof. $\square$
**Remark.** What you're saying in (3) is true for geodesics in $\Sigma$, i.e. geodesics normal to the orbits, but it is false for general geodesics. Consider the action of $\mathrm{SO}(2)$ on $\mathbb{R}^3$ by rotations around the $z$-axis. It is a polar action, and its singular orbits are nothing but points on the $z$-axis. So the geodesic running along the $z$-axis consists entirely of singular points.
|
2
|
https://mathoverflow.net/users/132126
|
398154
| 164,325 |
https://mathoverflow.net/questions/398129
|
1
|
Denote by $\mathcal L$ the set of continuously differentiable real valued functions on $[0, 1]$ with Lipschitz continuous derivative. Does there exist a Borel measurable function $ f: [0, 1] \times \mathbb R \to \mathbb [0, \infty) $ such that
$$\inf\_{g \in \mathcal L} \int\_{0}^{1} f(t, g(t)) \ dt < \inf\_{h \in C^2([0, 1])} \ \int\_{0}^{1} f(t, h(t)) \ dt?$$
*Note: Here the integrals are allowed to take the value $+\infty$.*
|
https://mathoverflow.net/users/173490
|
Lavrentiev phenomenon between $C^1$ + Lipschitz derivative and $C^2$
|
Edited.
Let $g\_0^\prime(x)=|x-1/2|,\; g\_0(x)=\int\_0^xg\_1(t)dt,\; 0\leq x\leq1$.
Then $g\_0$ has continuus derivative, namely $g\_0^\prime$, which is Lipschitz,
but the second derivative is discontinuous. Let $L=\{(x,y):y=g\_0(x)\}$
be the graph of $g\_0$.
Now define $f(x,y)=0$ when $0\leq x\leq 1, y=g\_0(x)$ and
$$f(x,y)=(\mathrm{dist}(x,y),L)^{-3}+1$$
otherwise.
Evidently $f$ is measurable, and
$\int\_0^1 f(x,g\_0(x))dx=0$, while $\int\_0^1f(x,g(x))dx>c$ for every $C^2$ function $g$, where $c$ is an absolute constant.
Of course, the last fact requires an accurate proof, but on the other hand, it seems evident. Moreover, one can replace $3$ in the exponent
by some larger constant, to make it more evident. The idea is that when the graph of $g$ is too close to $L$, the integral is large.
|
1
|
https://mathoverflow.net/users/25510
|
398163
| 164,329 |
https://mathoverflow.net/questions/398156
|
3
|
Automorphisms of connected, reductive groups are well understood: the outer automorphism group is an essentially combinatorial object associated to the root datum. I am trying to understand automorphisms of possibly *disconnected* reductive groups, by which I just mean groups whose identity components are reductive.
As in the connected case, one fixes a pinning and uses it to reduce the study of the full automorphism group to a simpler object. In this case, one winds up studying the subgroup of the disconnected group that fixes the pinning, whose identity component is central in the identity component of the original group.
All that is by way of motivation of the following question: suppose that $G$ is a smooth, affine algebraic group whose identity component is a torus. Do we have any concrete description of the automorphisms of $G$, or at least of its ‘identity component’? I put ‘identity component’ in quotes because I'm not too comfortable with discussing group schemes not of finite type; so let me give a work-around entirely in terms of varieties. The identity component of $G$ acts on $G$ by inner automorphisms. If $\Gamma$ is a smooth, connected, affine algebraic group that acts on $G$, then does the action factor through a map $\Gamma \to G^\circ/\mathrm C\_{G^\circ}(G)$?
|
https://mathoverflow.net/users/2383
|
Automorphisms of étale-by-torus groups
|
For any group $\Gamma$ that acts on $G$, $\Gamma$ acts on $G^{\circ}$ and on $G/G^{\circ}$. Since $G/G^{\circ}$ is finite and $G^{\circ}$ is a torus, both their automorphism groups are discrete groups, thus because $\Gamma$ is connected, the action of $\Gamma$ on them is trivial.
Automorphisms of $G$ fixing $G^{\circ}$ and $G/G^{\circ}$ are classified by cocycles in $C^1 ( G/G^{\circ}, G^{\circ})$. Let's do this explicitly so we can make sure this works schematically.
For $f$ such an automorphism of $G$, the function that sends $g \in G$ to $f(g) g^{-1}$ is valued in $G^\circ$ and invariant under right multiplication by $G^{\circ}$, thus defines a function $G /G^{\circ} \to G^{\circ}$, which geometrically gives a map from $\Gamma$ to a power of the torus $G^{\circ}$, whose image is contained in the closed subscheme satisfying the cocycle condition.
The closed subscheme of coboundaries is isomorphic to $G^{\circ}/C\_{G^{\circ}}(G)$, so it suffices to show $\Gamma$ is contained in this subscheme. Since $\Gamma$ is connected, it suffices to show this cohomology group is finite.
Because this cohomology group is a subquotient of a torus, to show it is finite, it suffices to show it is $|G/G^{\circ}|$-torsion. To do this we can use the two maps $ G^\circ \to G^{\circ} \otimes \mathbb Z[G/G^{\circ}]\to G^\circ$, whose composition is multiplication by $|G/G^{\circ}|$, to verify that multiplication by $|G/G^{\circ}|$ factors through a map to a projective $G/G^{\circ}$-module and thus acts as zero on cohomology in nonzero degree, as desired.
|
3
|
https://mathoverflow.net/users/18060
|
398166
| 164,330 |
https://mathoverflow.net/questions/398001
|
3
|
In the book of *S.Osher & R.Fedkiw - Level set methods and dynamic implicit surfaces*, at page 15, is stated without proof, a formula like that:
$$\text{Per}\_{\Omega}(\omega)=\lim\_{\varepsilon\to 0} \int\limits\_{\Omega}\delta\_{\varepsilon}(\phi(x))\ |\nabla\phi(x)|\ dx,$$
where
* $\Omega\subset\mathbb{R}^2$ is an open and bounded set,
* $\omega=\{x\in\Omega\ |\ \phi(x)>0\}$,
* $\phi:\Omega\to\mathbb{R}$ is a smooth level set function.
* $\text{Per}\_{\Omega}(\omega)$ is the perimeter of $\omega$ that lies inside $\Omega$ (the *relative* perimeter).
* $\delta\_{\varepsilon}:\mathbb{R}\to\mathbb{R}$, is a smooth approximation of the $\delta$-Dirac (generalized) function, precisely
$$
\delta\_{\varepsilon}(x)=\dfrac{\varepsilon}{\pi (\varepsilon^2+x^2)}\;.
$$
Now, denoting by $\bigvee\_{\Omega} f$ the total variation of $f$ in $\Omega$, here is **my question:**
>
> **Is it true that for any $\chi\_{\omega}\in BV(\Omega)$ ($\chi\_\omega$ being the characteristic function of $\omega$ - so by this condition we require $\omega$ to be a set with finite perimeter), whenever $\chi\_n\to \chi\_{\omega}$ in the $L^1(\Omega)$ norm and $0\leq \chi\_n\leq 1, \chi\_n\in W^{1,1}(\Omega)$ for each $n$, we have that:**
> $$\bigvee\_{\Omega} \chi\_{\omega}=\lim\limits\_{n\to\infty} \bigvee\_{\Omega}\chi\_n(x)\;\;\boldsymbol{?}$$
>
>
>
---
**Notes**
* It is known that:
$$\bigvee\_{\Omega} f = \sup\left\{\,\int\limits\_{\Omega} f(x)\text{div}(\varphi(x))\ dx\ |\ \varphi\in C^{\infty}\_{c}(\Omega;\mathbb{R}^2)\ \text{and} \ \Vert\varphi\Vert\_{L^{\infty}(\Omega;\mathbb{R}^2)}\leq 1\right \}$$
* It is also known that for $\chi\in W^{1,1}(\Omega)\subset BV(\Omega)$ we have that:
$$\bigvee\_{\Omega} \chi=\int\limits\_{\Omega} |\nabla\chi(x)|\ dx.$$
* We have that $\text{Per}\_{\Omega}(\omega)=\bigvee\_{\Omega}\chi\_{\omega}$ for any $\omega\subseteq\Omega$.
* Also, one part of the inequality is well-know (lower semicontinuity of the total variation), so:
$$\bigvee\_{\Omega} \chi\_{\omega}\leq\liminf\limits\_{n\to\infty} \bigvee\_{\Omega}\chi\_n$$
* In the cited book, at page 10, Theoreme 1.3 there's an approximation result of the $BV$ functions with $C^{\infty}\_c$ functions. Maybe it is useful.
* I tried to prove it. It is correct for many examples that I take, but I can't figure out why: intuitively I understand it but technically I'm blocked.
**Motivation**
I'm interested in this type of formulas because if it is indeed true we will have that:
$$\text{Per}\_{\Omega}(\omega)=\lim\limits\_{\varepsilon\to 0}\int\_{\Omega} H'\_{\varepsilon}(\phi(x))|\nabla\phi(x)|\ dx,
$$
for any function $H\_{\varepsilon}:\mathbb{R}\to\mathbb{R},\ H\_{\varepsilon}\in W^{1,1}(\mathbb{R})$ that approximates in $L^1(\Omega)$ the *Heaviside function*
$$
H(x)=\begin{cases} 1,\ x>0 \\ 0,\ x<0\end{cases}.
$$ In the above example,
$$
H\_{\varepsilon}(x)=\dfrac{1}{2}+\dfrac{1}{\pi}\cdot\tan^{-1}(x/\varepsilon),$$
for each $\varepsilon>0$ and $\delta\_{\varepsilon}(x)=H'\_{\varepsilon}(x)$.
|
https://mathoverflow.net/users/61629
|
Perimeter continuity of $BV$ sets on any sequence from $W^{1,1}$
|
This seems too strong, regardless of which set $\omega \subset \Omega$ one works with. We suppose that $\omega$ has bounded perimeter, so that $\chi\_\omega \in BV(\Omega)$.
Let $(f\_n \mid n \in \mathbf{N})$ be a sequence of functions in $W^{1,1}(\Omega)$ with
\begin{equation}
\lvert f\_n \rvert\_{L^1} \to 0
\text{ and }
V(f\_n) \geq 3 \, \mathrm{Per}(\omega).
\end{equation}
Let moreover $g\_n \in W^{1,1}(\Omega)$ be a sequence of functions
so that
\begin{equation}
\lvert g\_n - \chi\_\omega \rvert\_{BV} \to 0;
\end{equation}
for example $g\_n$ can be obtained by a mollification argument. Then the two sequences combined have
\begin{equation}
\lvert f\_n + g\_n - \chi\_\omega \rvert\_{L^1} \to 0
\text{ but }
V(f\_n + g\_n) \geq V(f\_n) - V(g\_n) \geq 3/2 \mathrm{Per}(\omega).
\end{equation}
We construct Lipschitz functions $f\_n$ on $\Omega$ that are independent of $y$. Specifically let, given $A > 0$,
\begin{equation}
f\_n(x,y) =\begin{cases}
Ax & \text{ on $[0,\frac{1}{2n}]$} \\
A/n - Ax & \text{ on $[\frac{1}{2n},\frac{1}{n}]$}
\end{cases}
\end{equation}
and extend this to be $\frac{1}{n}$-periodic in $x$.
Then $\lvert f\_n \rvert\_{L^\infty} \leq \frac{A}{2n} \to 0$ but the total variation is constant equal to $\int\_\Omega \lvert D f\_n \rvert = A \lvert \Omega \rvert$. To conclude it only remains to take $A > 3 \mathrm{Per}(\omega)/\lvert \Omega \rvert$.
|
1
|
https://mathoverflow.net/users/103792
|
398171
| 164,333 |
https://mathoverflow.net/questions/395216
|
4
|
Let's say we have a principal bundle $(P,B,\pi;G)$ and associated bundle $E=P \times\_{(G,\rho)}V$and $Ad(P)=P\times\_{(G,Ad)} \mathfrak{g}$ the adjoint bundle. The Yang-Mills-Higgs action (without potential) is
\begin{equation}
\int\_M(- \frac{1}{2}\langle F^A, F^A \rangle\_{\operatorname{Ad}(P)} +\langle d\_A \phi, d\_A \phi \rangle\_E - m^2 \langle \phi, \phi \rangle\_E )d\nu\_g
\end{equation}
with $\phi \in \Gamma(E)$. If we vary the equation, i.e. add $A+t \omega$ and $\phi+t \alpha$ and then take $d/dt$ the equations of motion are
\begin{equation}
\delta\_A F^A=j \quad \delta\_A d\_A \phi=0
\end{equation}
with the codifferential $\delta\_A$ and the implicit defined current
\begin{equation}
\langle j, \alpha \rangle = -\langle d\_A \phi,\rho\_\*(\alpha) \phi \rangle
\end{equation}
In Differential Geometry And Mathematical Physics 2 by Gerd Rudolph and Matthias Schmidt it says that if we take the associated bundle, i.e. $E$ to be the adjoint bundle, the first equation of motion becomes
\begin{equation}
\delta\_A F^A=[d\_A \phi,\phi]
\end{equation}
I have two equations:
1. How can I derive $j=[d\_A \phi,\phi]$ form the general equation for the current. And how do I express it in local coordinates to get something similar to the currents in physics.
2. Is there some Binachi-identity for the curvature form with the codifferential, i.e. $\delta\_A \delta\_A F^A=0$. This somehow should be the case, since the current should be conserved, i.e. $\delta\_A j=0$
|
https://mathoverflow.net/users/209074
|
The Yang-Mills Higgs Lagrangian
|
1. The first question has been already answered in a comment by NicAG, but let us repeat the 3-line long argument here for completeness:
>
> For the adjoint bundle , ρ∗=ad(⋅)(⋅)=[⋅,⋅]. Hence $⟨j,α⟩=−⟨d\_Aϕ,[α,ϕ]⟩$. Because of ad-invariance, the commutator can be 'moved': $⟨j,α⟩=⟨[d\_Aϕ,ϕ],α⟩$. The rest its then just non degeneracy.
>
>
>
**Edit:** changed the sign in the latter formula because so [does](https://en.wikipedia.org/wiki/Killing_form#Properties) 'moving' the commutator.
**Remark.** Possible confusion (see the comments) may come from another common formula for the current borrowed here from a different context:
$$j\_μ=i(ϕ^†D\_μϕ−(D\_μϕ^†)ϕ).$$
In fact the latter formula is not applicable here because in the extremely abstract setting considered in the book, there is simply no complex structure, hence no complex conjugation. But if we come down to Earth and recall that G is a *compact* Lie group (this is assumed in the 3rd line of Section 7.2 `Yang–Mills–Higgs Systems' in the book in question), hence, ``more or less'', subset of $U(n)$, hence any element $\phi$ of the Lie algebra satisfies $ϕ^†=-ϕ$, then we come to the same expression (up to constant factor). Then, plugging in the expressions
$$
D\_μϕ=(d\_Aϕ)\_\mu=\frac{\partial\phi}{\partial x^\mu}+A\_\mu\phi-\phi A\_\mu,
$$
one gets the expression for the current in local coordinates.
2. The identities $δ\_Aj=0$ and $δ\_Aδ\_AF\_A=0$ indeed hold with a one-line proof: for each element $\alpha$ of the Lie algebra we have
$$
\langle δ\_Aδ\_AF\_A,\alpha\rangle=\langle F\_A,d\_Ad\_A\alpha\rangle=\langle F\_A,[\alpha,F\_A]\rangle=\langle \alpha,[F\_A,F\_A]\rangle=0.
$$
However this is not called ``the Bianchi identity'', the name being already reserved for the identity $d\_AF\_A=0$ (Eq. 7.2.6 in the book in question).
**Remark.** One should not be confused (see the comments) by another common formula $d\_Ad\_A\phi=F\wedge \phi$, which is valid for a 'vector' $\phi$ but *not* for a 'matrix' $\phi$ (more formally, for the fundamental representation but not for the adjoint representation).
**Remark.** In fact the material of the book is not properly exposed in the question. For instance, in the right-hand side of the equation of motion $δ\_Ad\_Aϕ=0$ there should be $mϕ$ or $-mϕ$ instead. Notice that here the Higgs potential $V(\phi)=m^2|\phi|^2$ does *not* vanish. But in the absence of higher degree terms in the potential, the action should not be called Yang-Mills-Higgs action: it is rather action for a Klein-Gordon field coupled with a Yang-Mills field.
**Remark.** Maybe the deep source of all those confusions is an extremely abstract style of most literature on the subject. Let me try to advertise [a paper](https://arxiv.org/abs/1709.04788) with an elementary introduction of gauge theory. See Sections 2.4-2.5 there. It deals with a lattice formulation, thus to get a deep understanding, one even does not need to know what a derivative is.
|
4
|
https://mathoverflow.net/users/108862
|
398175
| 164,335 |
https://mathoverflow.net/questions/398180
|
4
|
$\newcommand\P{\mathcal P}$A "partition" $P$ (of the interval $[0,1]$) is a finite sequence $(t\_0,\dots,t\_n)$ such that $0=t\_0<\cdots<t\_n=1$; then the mesh of $P$ is $\|P\|:=\max\_{1\le j\le n}(t\_j-t\_{j-1})$.
Fix any sequence $\P:=(P\_k)$ of "partitions" $P\_k=(t\_{k,0},\dots,t\_{k,n\_k})$ such that $\|P\_k\|\to0$ (as $k\to\infty$).
For a real-valued function $f$ on $[0,1]$, define its quadratic variation (with respect to $\P$) by the formula
$$[f]\_\P:=\limsup\_{k\to\infty}\sum\_{j=1}^{n\_k}(f(t\_{k,j})-f(t\_{k,j-1}))^2.$$
Suppose now that $f$ is differentiable and $[f]\_\P<\infty$. Does it then necessarily follow that
$[f]\_\P=0$?
This question is a modification of the (now answered, positively) previous question [A dichotomy for the quadratic variation of differentiable functions?](https://mathoverflow.net/q/397707/36721). The difference is that now the sequence $\mathcal P$ is fixed.
|
https://mathoverflow.net/users/36721
|
A narrower dichotomy for the quadratic variation of differentiable functions?
|
Using your previous example of $f(x) = x^2 \cos(x^{-4})$. Note that on any interval $[\epsilon,1]$ the function is continuously differentiable, and hence has quadratic variation 0.
Construct $P\_k$ so that the following points are contained in the partition:
1. 0
2. $\frac{1}{\sqrt[4]{\pi}} \ell^{-1/4}$ for natural $\ell$ between $k$ and $2k$.
3. choose a partition of $[(k\pi)^{-1/4}, 1]$ such that the quadratic variation of $f$ on that interval is less than 1/10, and such that the points are no further than $1/k^{1/4}$ apart.
With this the sum relative to $P\_k$ of
$$ \sum (f(t\_{k,n}) - f(t\_{k,n-1}))^2 $$
evaluates to approximately $\ln(2)$.
|
5
|
https://mathoverflow.net/users/3948
|
398183
| 164,336 |
https://mathoverflow.net/questions/398128
|
12
|
Atiyah duality is the equivalence $M/\partial M \simeq (M^{-T(M)})^\vee$, i.e. the Spanier-Whitehead dual of the space $M/\partial M$ is the Thom complex of the stable normal bundle of $M$. The theorem is proven by taking an appropriate embedding $i$ of $M$ into a sphere $S^d$, identifying the complement with the Thom complex of the normal bundle of the embedding, and constructing a duality map $M\_+ \wedge M^{N(i)}\rightarrow S^d$ using the embedding.
In the case $M$ is a compact, framed manifold of dimension $n$, it is possible to describe a duality map $M\_+ \wedge M\_+ \rightarrow S^n$ without reference to an embedding by appealing to the fact that if we collapse everything outside a small ball around $p \in M$, we can use the framing to continuously identify it with $S^n$. This is the adjoint of a duality map $M\_+ \wedge M\_+ \rightarrow S^n$.
$\bf Question:$ Is it possible to create such a duality map for a framed manifold with nonempty boundary, namely one that does not invoke an embedding?
|
https://mathoverflow.net/users/134512
|
Atiyah duality without reference to an embedding
|
Here is another short construction which is much simpler and just takes a few lines.
1. Let $M$ be a closed $n$-manifold. Consider the diagonal $M \to M \times M$. It is an embedding. Take its Pontryagin-Thom construction to get a map
$$
M\_+ \wedge M\_+ \to M^\tau
$$
(we have identified a tubular neighborhood of the diagonal with the total space of the tangent bundle).
2. If $M$ is (stably) framed, then $M^\tau \simeq M\_+ \wedge S^n$ (stably). Then we have the map $M\_+ \wedge S^n \to S^n$ induced by smashing $M\_+ \to \text{pt}\_+$ with $S^n$.
3. The composition
$$
M\_+ \wedge M\_+ \to M^\tau \simeq M\_+ \wedge S^n \to S^n
$$
is what you want: it's a duality map.
This can be seen on the level of homology (but it is enough to check a map is a duality map on the level of homology).
4. If $M$ is compact with non-empty boundary $\partial M$, then there is a map of pairs
$$
(M\times M, M\times \partial M) \to (M\times M,M\times M - U)
$$ where $U$ is a tubular neighborhood of the diagonal.
To get this map, one might choose a collar neighborhood $C$ of $\partial M$ and thereafter identify $M$ with $M-C$. Then
$M-C $ and $\partial M$ are disjoint.
The map of pairs determines a map of quotients
$$
M\_+ \wedge M/\partial M \to M^\tau \, ,
$$
and one may then proceed as in the empty boundary case, assuming that $M$ is framed, to obtain a map
$$
M\_+ \wedge M/\partial M \to S^n
$$
which will be an $S$-duality.
|
9
|
https://mathoverflow.net/users/8032
|
398188
| 164,337 |
https://mathoverflow.net/questions/398153
|
4
|
The set theory with atoms ([ZFA](https://ncatlab.org/nlab/show/ZFA)), is a modified version of set theory, and is characterized by the fact that it admits objects other than sets, atoms. Atoms are objects which do not have any elements.
I have thusfar found two applications of ZFA-Set Theory. Firstly, ZFA can be used to [prove](https://www.sciencedirect.com/science/article/pii/S0049237X08717567) the independence of the Axiom of Choice.
Secondly, one can show a [correspondence](https://mathoverflow.net/questions/119667/models-of-zfa-corresponding-exactly-with-a-particular-class-of-groups) between certain ZFA-models and a certain class of groups.
Both of these applications of ZFA make use of so-called [permutation models](https://en.m.wikipedia.org/wiki/Permutation_model#CITEREFFraenkel1922).
I'd like to know what other applications ZFA might have. I would particularly be interested in applications that do not make use of permutation models.
|
https://mathoverflow.net/users/324481
|
Applications of ZFA-Set Theory
|
Although you prefer models other than permutation models, let me point out an appearance of permutation models, particularly the basic Fraenkel model, at the border between computer science and logic. The topic concerns operators that bind variables, like $\forall$ and $\exists$ in logic, the $\lambda$ in lambda-calculus, and $\int$ in calculus. The actual variables used with such an operator don't matter, i.e., bound variables can be renamed, as long as there are no clashes. That suggests thinking of variables as atoms in a ZFA universe, thinking of the renaming process as permuting the atoms, and requiring each formula to be invariant under enough permutations (all permutations fixing the formula's free variables). The theory built up from these ideas is called "nominal sets".
|
12
|
https://mathoverflow.net/users/6794
|
398193
| 164,340 |
https://mathoverflow.net/questions/398187
|
11
|
It is well-known that when passing to $\infty$-categories the notion of commutativity gets replaced by an infinite array of notions of commutativity: $\mathbb{E}\_{1}$, $\mathbb{E}\_{2}$, ..., $\mathbb{E}\_{\infty}$. This is already apparent when passing from sets to categories and $2$-categories:
* For sets, we have monoids ($\mathbb{E}\_{1}$) and commutative monoids ($\mathbb{E}\_2=\cdots=\mathbb{E}\_\infty$);
* For categories, we have monoidal ($\mathbb{E}\_{1}$), braided ($\mathbb{E}\_{2}$), and symmetric monoidal categories ($\mathbb{E}\_{3}=\mathbb{E}\_{4}=\cdots=\mathbb{E}\_{\infty}$);
* For $2$-categories, we have monoidal ($\mathbb{E}\_{1}$), braided ($\mathbb{E}\_{2}$), sylleptic ($\mathbb{E}\_{3}$), and symmetric monoidal categories ($\mathbb{E}\_{4}=\mathbb{E}\_{5}=\cdots=\mathbb{E}\_{\infty}$).
A similar phenomenon happens to bilinearity:
* A morphism $f\colon A\times B\to C$ of commutative monoids is **bilinear** if, for each $a,a'\in A$ and each $b,b'\in B$, we have
\begin{gather\*}
f(a,b+b') = f(a,b)+f(a,b'),\\
f(a+a',b) = f(a,b)+f(a',b),\\
f(1\_A,b) = 1\_C,\\
f(a,1\_B) = 1\_C.
\end{gather\*}
* For categories, these relations are replaced by morphisms: we say that a **strong bilinear structure** on a functor $F\colon\mathcal{C}\times\mathcal{D}\to\mathcal{E}$ of symmetric monoidal categories is a collection of isomorphisms
\begin{align\*}
F^{\mathsf{bil}}\_{A,B\otimes\_{\mathcal{D}}B'} &\colon F(A,B\otimes\_{\mathcal{D}}B') \longrightarrow F(A,B)\otimes\_{\mathcal{E}}F(A,B'),\\
F^{\mathsf{bil}}\_{A\otimes\_{\mathcal{C}}A',B} &\colon
F(A\otimes\_{\mathcal{C}}A',B) \longrightarrow F(A,B)\otimes\_{\mathcal{E}}F(A',B),\\
F^{\mathsf{bil}}\_{\mathbf{1}\_{\mathcal{C}},B} &\colon F(\mathbf{1}\_{\mathcal{C}},B) \longrightarrow \mathbf{1}\_{\mathcal{E}},\\
F^{\mathsf{bil}}\_{A,\mathbf{1}\_{\mathcal{D}}} &\colon F(A,\mathbf{1}\_{\mathcal{D}}) \longrightarrow \mathbf{1}\_{\mathcal{E}}
\end{align\*}
satisfying coherence conditions.
**Questions:**
1. Is there a similar array of notions of bilinearity in higher algebra?
2. In particular, can we speak of "$\mathbb{B}\_{k}$-morphisms of spectra"?
3. Tensor products can be characterised/defined as universal bilinear maps; do we also have intermediate tensor products corresponding to "$\mathbb{B}\_{k}$"-bilinearity?
|
https://mathoverflow.net/users/130058
|
Intermediate notions of bilinearity in higher algebra
|
Let me clarify a bit what I meant in [my comment](https://mathoverflow.net/questions/398187/intermediate-notions-of-bilinearity-in-higher-algebra#comment1020095_398187) on how the notion of bilinearity will depends on "how commutative" are $A$, $B$ and $C$, and this is one way to define a hierarchy of notion of bilinear maps in higher algebras.
The idea is that up to equivalence of $\infty$-categories, (and let's say in a cartesian monoidal $(\infty,1)$-category for simplicity), an $\mathbb{E}\_{n+k}$-algebra is the same as an $\mathbb{E}\_n$-algebra in the (cartesian monoidal) category of $\mathbb{E}\_{k}$-algebra.
So if $A$ is an $\mathbb{E}\_k$-algebra, $B$ is an $\mathbb{E}\_n$-algebra and $C$ is an $\mathbb{E}\_{n+k}$-algebras then I can define a $(n,k)$-bilinear map $A \times B \to C$ to be a morphism of $\mathbb{E}\_k$-algebra from $A$ to the $\mathbb{E}\_k$-algebra $\operatorname{Map}\_{\mathbb{E}\_n}(B,C)$.
Where by $\operatorname{Map}\_{\mathbb{E}\_n}(B,C)$ I mean the space of morphisms of $\mathbb{E}\_n$-algebra morphism from $B$ to $C$, which has an $\mathbb{E}\_k$-algebra structure induced by the fact that $C$ is an $\mathbb{E}\_k$-algebra when seen as an object of the category of $\mathbb{E}\_n$-algebra.
But this is very different from the notion you mention in [your coment](https://mathoverflow.net/questions/398187/intermediate-notions-of-bilinearity-in-higher-algebra#comment1020102_398187) when instead one goes to "lax/colax" notion of bilinearity.
|
9
|
https://mathoverflow.net/users/22131
|
398194
| 164,341 |
https://mathoverflow.net/questions/397624
|
1
|
Consider any 3D body with an axis of rotational symmetry (e.g. cone, cylinder...) and packing the 3d space efficiently with infinitely many copies of this body. Is the following claim valid?
**Claim:** The densest packing with any such body is necessarily such that all units are aligned along or opposite to the same direction
Proving such a claim will greatly limit the possibilities that need to be considered to find the densest packing. The 2D analog of the claim above would involve bodies with a reflection symmetry.
Note: This question was recorded at <https://nandacumar.blogspot.com/2019/02/on-packing-with-axi-symmetric-bodies.html>
|
https://mathoverflow.net/users/142600
|
On packing axisymmetric bodies in 3D
|
Your claim is false for axially-symmetric ellipsoids: when restricted to all have their axis of symmetry in the same direction, they cannot pack more densely than spheres (in terms of volume fraction). However, they can in fact pack more densely than spheres, as discussed in [Donev et al](https://dx.doi.org/10.1103/PhysRevLett.92.255506) and references therein.
|
6
|
https://mathoverflow.net/users/20186
|
398206
| 164,345 |
https://mathoverflow.net/questions/395592
|
9
|
Fix $(R,m)$ a complete DVR of mixed characteristic $(0,p)$ with perfect residue field, and consider finite flat commutative group schemes $G = Spec(A)$ over $R$. One can associate a differential invariant to $G$, an integer $d \geq 0$, in two ways:
1. the "absolute different", such that the dual $A^\*$ of $A$ under the $R$-trace pairing is equal to $m^{-d}A$, or
2. the "cotangent length", equal to the $R$-length of the cotangent space $e^\*\Omega\_{G/R}$ (where $e$ is the unit section).
My question is: why is the *cotangent length* $d$ additive in short exact sequences of $G$?
A direct argument for the cotangent length would be great, though I'm open to an argument that compares the two definitions and proves additivity for the absolute different. For what it's worth, the comparison is clear in some cases by the "differential characterization of the different", and the additivity for the absolute different is clear in some cases by the "transitivity of the different", but I don't know where either of these classical facts is stated in the full generality needed here.
(Context: In Raynaud's famous "... type $(p,\ldots,p)$" paper, the theorem in Section 4.1 computes the action of inertia on the determinant of the generic fiber of $G$ in terms of the *absolute different*. The argument given proceeds in two steps. First, in the case where $G$ is simple, the result is explicitly verified using the classification worked out earlier in the paper. I have no issues with this argument, and "differential characterization of the different" even applies to compare the two definitions of $d$ in this case. Second, there is a reduction to the simple case by dévissage. But the required additivity of $d$ is never explicitly addressed, and moreover my needs would prefer the cotangent definition. Hence the question.)
|
https://mathoverflow.net/users/367
|
Cotangent spaces of finite flat group schemes in short exact sequences
|
By Proposition 5.1(i) in [Mazur and Roberts, Local Euler characteristics, *Invent. Math.* **9** (1970), 201-234](https://doi.org/10.1007/BF01404325), $G$ fits in an exact sequence $0 \to G \to A \to B \to 0$, where $A$ and $B$ are smooth commutative affine group schemes over $R$ of the same relative dimension, say $n$. Then the cotangent space $e^\* \Omega\_{G/R}$ is isomorphic to the cokernel of the induced homomorphism $\psi \colon e^\* \Omega\_{B/R} \to e^\* \Omega\_{A/R}$ of free $R$-modules of rank $n$. Its length $d(G)$ equals the valuation of $\det \psi$.
Given a short exact sequence $0 \to G' \to G \to G'' \to 0$, one can choose *compatible* resolutions of $G'$, $G$, $G''$ by using the same pushout constructions one would use for modules, to get a $3 \times 3$ commutative diagram with short exact rows and columns. The homomorphisms $\psi'$, $\psi$, $\psi''$ define a morphism of short exact sequences of cotangent modules, so $(\det \psi) = (\det \psi')(\det \psi'')$ as ideals of $R$. Hence $d(G)=d(G')+d(G'')$.
|
4
|
https://mathoverflow.net/users/2757
|
398211
| 164,347 |
https://mathoverflow.net/questions/398088
|
2
|
When we talk about the theory of variation of Hodge structures, we always assume that the central fiber is a Kähler manifold $X$, then consider a family of deformations $\pi:\mathcal X\to B$ and the period map $\mathcal P:B\to Grass(b^{p,k},H^k(X,\mathbb C))$, $b^{p,k}=dim F^pH^k(X,\mathbb C)$.
What if we replace the Kähler manifold by a $\partial\bar\partial$-manifold and consider a holomorphic family of $\partial\bar\partial$-manifolds $\pi:\mathcal X\to B$?
Recall a $\partial\bar\partial$-manifold is a compact complex manifold which satisfies for any $\partial$, $\bar\partial$-closed $d$ exact $(p,q)$ form $\alpha$, there this a $(p-1,q-1)$ form $\beta$ such that $\alpha=\partial\bar\partial \beta$.
Is the theory of variation of Hodge structures remains the same? And is there also a period map $\mathcal P:B\to Grass(b^{p,k},H^k(X,\mathbb C))$?
If so, is the period map holomorphic and the Griffiths transversality still holds?
In my opinion, first we should define what $F^pH^k(X,\mathbb C)$ means for a non-Kähler manifold, since for a Kähler manifold, we have the Kähler identity $\Delta=2\Delta\_{\bar\partial}=2\Delta\_{\partial}$ from which we deduce the Hodge decomposition: $H^k(X,\mathbb C)=\oplus\_{p+q=k}H^{p,q}(X)$, so we can define
$F^pH^k(X,\mathbb C)=H^{p,k-p}(X)\oplus H^{p+1,k-p-1}(X)\oplus...\oplus H^{k,0}(X)$ which is obviously a subspace of $H^k(X,\mathbb C)$,
but for a non-Kähler manifold, there is no Kähler identities. But from [AT13](https://link.springer.com/article/10.1007/s00222-012-0406-3), we know that the Bott-Chern cohomology $H^{p,q}\_{BC}(X,\mathbb C):=\frac{ker\partial\cap ker\bar\partial}{im\partial\bar\partial}$ is isomorphic to Dolbeault cohomology $H\_{\bar\partial}^{p,q}(X)$ and $H\_{\partial}^{p,q}(X)$, so if we define $F^pH^k(X,\mathbb C)=H^{p,k-p}\_{BC}(X,\mathbb C)\oplus H^{p+1,k-p-1}\_{BC}(X,\mathbb C)\oplus...\oplus H^{k,0}\_{BC}(X,\mathbb C)$, is it reasonable to treat it as a subspace of $H^k(X,\mathbb C)$ and define a period map as in the Kähler case?
Has anyone thought it before? or any reference about this issue?
|
https://mathoverflow.net/users/99826
|
Period map for $\partial\bar\partial$-manifolds
|
Let me start with a disclaimer that I think the following facts are true, but I'm doing this over coffee and I haven't checked the details carefully. First, I'll redefine $F^pH^k(X,\mathbb{C})$ to be the space of de Rham classes represented by the sum of $(p', p'-k)$ forms with $p'\ge p$, or equivalently as
$$F^pH^k(X,\mathbb{C})= im(H^k(\Omega\_X^{\ge p})\xrightarrow{\iota} H^k(X,\Omega\_X^\bullet))$$
Then the $\partial\bar\partial$-lemma is sufficient to guarantee the filtration is *strict* in the sense that the maps above are injective (or equivalently that the Hodge to de Rham spectral sequence degenerates); **edit** see remark 5.21 of Deligne, Griffiths, Morgan, Sullivan, *Real homotopy theory of Kähler manifolds*. Then, if I understand the notation of the paper you linked, this should probably give decomposition in terms of BC cohomology as you wrote. The other thing I want to remark is that if $\mathcal{X}\to B$ is a smooth proper family such that the fibres satisfy $\partial\bar\partial$-lemma, then usual arguments should imply that
$$F^p R^kf\_\*\Omega\_{\mathcal{X}/B}^\bullet= im(R^kf\_\*\Omega\_{\mathcal{X}/B}^{\ge p}\xrightarrow{\iota}R^kf\_\*\Omega\_{\mathcal{X}/B}^\bullet) $$
satisfies Griffiths transversality etc. So in this sense, things work. However, you would be missing the polarization, which need for most of the deeper results about the Griffiths period map.
|
1
|
https://mathoverflow.net/users/4144
|
398228
| 164,349 |
https://mathoverflow.net/questions/398210
|
1
|
I'm encountering a lot of problems when dealing with the root of unity sheaf $\mu\_\infty := \mathrm{colim}\_n\mu\_n$.
Let $X$ be a smooth geometrically integral variety over a number field $k$. Although we have the canonical inclusion $\mu\_\infty \subset \mathbb{G}\_m$, the cohomology groups with coefficients in the latter sheaf are better understood. For example, in Galois cohomology for our given $k$ we know that
$$H^1(k,\mathbb{G}\_m) = 0,\,\,H^2(k,\mathbb{G}\_m) = \mathrm{Br}(k), \,\, H^3(k, \mathbb{G}\_m) = 0.$$
For etale cohomology on $X$, we have
$$H^0(X,\mathbb{G}\_m) = k[X]^\*, \,\, H^1(X,\mathbb{G}\_m) = \mathrm{Pic}(X), \,\, H^2(X,\mathbb{G}\_m) = \mathrm{Br}(X).$$
Now consider the spectral sequence
$$H^p(k,H^q(\bar{X},\mu\_\infty)) \implies H^{p+q}(X,\mu\_\infty).$$
One obtains a long exact sequence of low degree terms:
$$0 \rightarrow H^1(k,H^0(\bar{X},\mu\_\infty)) \rightarrow H^1(X,\mu\_\infty) \rightarrow H^0(k,H^1(\bar{X},\mu\_\infty)) \rightarrow H^2(k,H^0(\bar{X},\mu\_\infty))$$
$$\rightarrow \mathrm{Ker}[H^2(X,\mu\_\infty) \rightarrow H^0(k,H^2(\bar{X},\mu\_\infty))] \rightarrow H^1(k,H^1(\bar{X},\mu\_\infty)) \rightarrow H^3(k,H^0(\bar{X},\mu\_\infty)).$$
**Question 2.** Does $H^2(\bar{X},\mu\_\infty)$ have trivial Galois action? If so, is it true for all $H^i(\bar{X}, \mu\_\infty)$?
This question came about because one of the papers I'm reading defined some term to be the kernel of $H^2(X,\mu\_\infty) \rightarrow H^2(\bar{X},\mu\_\infty)$, so I have a feeling it came from this spectral sequence.
There are many other questions I can think of but I would like to end off with this:
**Question 3.** Is there some connection between $H^1(X,\mu\_\infty)$ and $\mathrm{Pic}(X)$?
**EDIT.** I've removed Question 1 due to some confusion.
|
https://mathoverflow.net/users/172132
|
Cohomology with coefficients in $\mu_\infty$
|
**Question 2.** Does $H^2(\bar{X},\mu\_\infty)$ have trivial Galois action? If so, is it true for all $H^i(\bar{X}, \mu\_\infty)$?
The answers are "often not" and "almost never".
The exact sequence $1 \to \mu\_{\infty} \to \mathbb G\_m \to \mathbb G\_m \otimes \mathbb Q \to 1$ you mention gives a map on cohomology $\operatorname{Pic}(\overline{X} ) \to \operatorname{Pic} (\overline{X} )\otimes \mathbb Q \to H^2 (\overline{X} , \mu\_{\infty} )$.
Now $\operatorname{Pic} (\overline{X} )$ will map to a finitely-generated abelian group, the Neron-Severi group, with big divisible kernel, the Picard variety. When we tensor with $\mathbb Q$, we lose the torsion part of the finitely generated abelian group, and the divisible kernel will be canceled by the map from $\operatorname{Pic} (\overline{X} )$, but the rank part is preserved.
Let's say this finitely generated abelian group is $\mathbb Z^r$ times something torsion. We'll get $\mathbb Q^r$ appearing as a quotient of $\operatorname{Pic} (\overline{X} )\otimes \mathbb Q$ which modulo the original $\mathbb Z^r$ gives a $\mathbb Q^r / \mathbb Z^r$ inside $H^2 (\overline{X} , \mu\_{\infty} )$.
The Galois action on this is trivial if and only if the Galois action on the rank part of the original Neron-Severi group is trivial. That will be true for some varieties and not others. The simplest example where it fails is a quadric hypersurface $x^2-y^2 + z^2 - d w^2$ where $d \in k$ is not a perfect square. Then Picard will be $\mathbb Z^2$ with Galois group elements swapping the two copies, and $H^2( \overline{X}, \mu\_{\infty})$ will be $(\mathbb Q/\mathbb Z)^2$ with a similar swap.
For general $H^i (\overline{X}, \mu\_{\infty})$, note that $\mathbb Q\_\ell / \mathbb Z\_\ell (1)$ is a summand of $\mu\_{\infty}$ so $H^i ( \overline{X} , \mathbb Q\_\ell / \mathbb Z\_\ell (1 ))$ is a summand of $H^i (\overline{X}, \mu\_{\infty})$.
We have a short exact sequence $0\to \mathbb Z\_\ell(1) \to \mathbb Q\_\ell(1) \to \mathbb Q\_\ell/\mathbb Z\_\ell (1)\to 0$ giving a long exact sequence
$$ H^i ( \overline{X} , \mathbb Z\_\ell (1) ) \to H^i(\overline{X}, \mathbb Q\_\ell(1)) \to H^i ( \overline{X} , \mathbb Q\_\ell / \mathbb Z\_\ell (1 ))$$
Here $H^i ( \overline{X} , \mathbb Z\_\ell (1) )$ will look like $\mathbb Z\_\ell^n$ plus some torsion, which we can see will force $H^i(\overline{X}, \mathbb Q\_\ell(1))$ to look like $\mathbb Q\_\ell^n$ and $H^i ( \overline{X} , \mathbb Q\_\ell / \mathbb Z\_\ell (1 ))$ to look like $(\mathbb Q\_\ell/\mathbb Z\_\ell)^n$ plus some torsion coming form $H^{i+1} ( \overline{X} , \mathbb Z\_\ell (1) )$.
Thus, if the Galois action on $H^i(X, \mathbb Q\_\ell(1))$ is nontrivial, the Galois action on $H^i ( \overline{X} , \mathbb Q\_\ell / \mathbb Z\_\ell (1 ))$ is nontrivial.
By the Weil conjectures, all the eigenvaluse of a Frobenius element (at a prime $q$ at which $X$ has good reduction) in the Galois group of $k$ on $H^i (\overline{X}, \mathbb Q\_\ell)$ have size $q^{i/2}$, so all the eigenvalues on $H^i (\overline{X}, \mathbb Q\_\ell(1))$ have size $q^{i/2-1}$. So the eigenvalues are never equal to $1$ unless $i=2$.
In other words, in every other degree, the Galois action is always nontrivial unless the cohomology group vanishes.
(The Weil conjectures are, unsurprisngly, more advanced than what you're learning right now, but keeping what they tell you in mind can be helpful for intuition.)
|
5
|
https://mathoverflow.net/users/18060
|
398232
| 164,351 |
https://mathoverflow.net/questions/398220
|
1
|
Let $R$ be an $I$-adically separated and complete valuation ring, with $I$ finitely generated.
It is used a few times in Bosch, [Lectures on Formal and Rigid Geometry](https://www.math.purdue.edu/%7Etongliu/seminar/rigid/Bosch.pdf) e.g. first lines of pg. 164, Cor. 5 and Cor. 6 (their condition (V) is what I stated above) that
>
> If an $A$ module has no $I$ torsion then it is flat over $R$.
>
>
>
I don't see why this is true. Any suggestions / references would be appreciated.
---
What I thought: We know $A$ is flat over domain $R$ iff it is $R$-torsion free.
If the statement were true: $I$-torsion free $\Rightarrow$$R$-torsion free.
This would hold if $I$ is a maximal ideal but otherwise I don't see why.
|
https://mathoverflow.net/users/97321
|
Flatness criterion for $I$-adic ring: $I$-torsion free
|
In Section 7.3 it is assumed that $I$ is the ideal of definition or $R$. It follows that the $I$-adic topology is separated, so (because $R$ is a valuation ring) every nonzero ideal of $R$ contains some power of $I$, so everything that is $R$-torsion is $I$-torsion.
|
2
|
https://mathoverflow.net/users/18060
|
398240
| 164,354 |
https://mathoverflow.net/questions/398243
|
5
|
Given a group $G$, find subsets $A,B$ such that $G=A\sqcup B$ and $A$ and $B$ are closed under multiplication: $x,y\in A$ (corr. $x,y\in B$) implies $xy\in A$ (corr. $xy\in B$).
For example, if $G$ is finite then all splitting are trivial: if $1\in A$ then $A=G$ and $B=\emptyset$.
If $G=\mathbb{Z}^d$ then any splitting is determined by a halfspace $\Pi^+$ in $\mathbb R^d\supset\mathbb Z^d$: $A=\Pi^+\cap\mathbb Z^d$. (More precisely, by a flag of halfspaces $\Pi^+\supset \Pi\_1^+\supset\dots$, because we can distribute the points of the hyperplane $\partial\Pi^+$ between $A$ and $B$.)
What about the free group $G=F\_d$? Are there splitting which are not induced by the projection $F\_d\to\mathbb Z^d$?
|
https://mathoverflow.net/users/90980
|
Splitting a group into two subsets closed under multiplication
|
This is a way to rediscover a quite well-studied class of groups:
**Proposition** *Let $G$ be a group. Equivalent statements: (a) $G$ admits a partition $G=A\sqcup B$ with $1\in A$, $B$ nonempty, and both $A,B$ subsemigroups. (b) $G$ admits a nontrivial order-preserving action on some totally ordered set [which can be chosen to be the real line, or $\mathbf{Q}$, if $G$ is countable] (c) $G$ has a nontrivial left-orderable quotient.*
The equivalence between (b) and (c) is classical. If $G$ has a non-trivial orderable quotient $p:G\to Q$, there is (by definition) a submonoid $K$ of $Q$ such that $K\cap K^{-1}=\{1\_Q\}$ and $K\cup K^{-1}=Q$; pulling back to $G$ yields the desired decomposition.
Conversely, suppose that $G=A\cup B$ as above (with $1\in A$). Define $H=A\cap A^{-1}$: this is a subgroup (call it "core" of the decomposition).
Observe that $HB=BH=B$: indeed if $h\in H$ and $b\in B$, suppose by contradiction $a:=hb\in A$: so $b=ah^{-1}\in AA\subset A$, contradiction, same contradiction if $bh\in A$.
On $G/H$ define a total order by $gH\le g'H$ if $g^{-1}g'\in A$. Note that this doesn't depend on the choices of $g,g'$ modulo $H$ on the right. This relation is $G$-invariant, and is easily seen to be a total order. It is not reduced to a singleton (since $H=A$ would force $B$ empty, since the complement of a subgroup can't be closed under multiplication, unless empty).
*Edit:* this argument shows something more precise, not only describing the class of groups, but describing these decompositions:
**Proposition 2**: *Let $G$ be a group. (1) If $G$ acts on a totally ordered set $X$ in a order-preserving and $x\in X$, define $A=\{g\in G:gx\le x$ and $B=\{g\in G:gx>x\}$. Then $G=A\sqcup B$ is a decomposition with the given axioms (and $1\in A$). (2) Conversely, any such decomposition of $G$ occurs in this way (for some such action and $x$, where we can assume in addition the action to be transitive).*
Note that $B$ is empty iff $G$ fixes $x$.
---
Note that this shows that if $G$ has such a decomposition, it has one for which the core is a normal subgroup (while in general the core need not be normal).
---
Also, since non-abelian free groups are themselves left-orderable, they admit such decomposition that are not of the prescribed form (namely have such a decomposition with trivial core).
|
10
|
https://mathoverflow.net/users/14094
|
398250
| 164,357 |
https://mathoverflow.net/questions/398241
|
5
|
I am reading the book *Applications of Diophantine Approximation to Integral Points and Transcendence* by Zannier and Corvaja and, after their proof of the Chevalley-Weil theorem, in Example 3.8 they suggest that it is possible to prove the weak Mordell-Weil theorem using Chevalley-Weil. Honestly, I can't see that, but I am very curious. Does anyone have an idea about it?
Thank you in advance!
|
https://mathoverflow.net/users/167909
|
Weak Mordell-Weil for EC using Chevalley-Weil theorem
|
The multiplication-by-$m$ map $[m]:E\to E$ is unramified, so there exists a finite set of primes $S$, depending only on $E$ and $m$, so that for every $P\in E(K)$, the field generated by the coordinates of the points in $[m]^{-1}(P)$ is unramified outside of $S$. (This is where we use Chevelley-Weil.) The degree of that extension is also bounded as a function of $m$. It follows from standard results in algebraic number theory that there are only finitely many such fields. Hence $[m]^{-1}\bigl(E(K)\bigr)$ is contained in $E(L)$, where the extension $L/K$ is a **finite** extension.
I'll let you read elsewhere the fact that it suffices to prove the weak Mordell-Weil theorem under the assumption that $E[m]\subset E(K)$. Under that assumption, and with notation as in the first paragraph with $L=K\bigl([m]^{-1}\bigl(E(K)\bigr)\bigr)$, there is a well-defined injective homomorphism
$$
E(K)/mE(K) \longrightarrow \operatorname{Hom}\bigl(\operatorname{Gal}(L/K),E[m]\bigr)
$$
defined by
$$ P \longmapsto \bigl(\sigma\mapsto \sigma(Q)-Q\bigr)
\quad\text{for any choice of $Q\in E(L)$ satisfying $[m](Q)=P$.}
$$
Since $L/K$ is a finite extension from Chevalley-Weil, and since $E[m]$ is a finite group, this gives the finiteness of $E(K)/mE(K)$.
|
12
|
https://mathoverflow.net/users/11926
|
398252
| 164,358 |
https://mathoverflow.net/questions/398167
|
5
|
Let $\Omega\subset\mathbb{R}^2$ a open and bounded set with smooth boundary and $\phi:\Omega\to\mathbb{R}$ a smooth function such that:
$\bullet$ $\phi^{-1}(0)\neq\emptyset$;
$\bullet$ $\nabla\phi(x)\neq 0$ on a neighborhood $W\subset\Omega$ of the curve $\phi^{-1}(0)$.
WLOG we can assume that $W=\{x\in\mathbb{R}^2\ |\ |\phi(x)|<\varepsilon\_0\}\subset\Omega$.
How can we prove that:
$$\lim\limits\_{\varepsilon\to 0^+} \mathcal{H}^1\big (\{x\in\Omega\ |\ \phi(x)=\varepsilon\}\big )=\mathcal{H}^1\big (\{x\in\Omega\ |\ \phi(x)=0\}\big )\;\; ?$$
Here $\mathcal{H}^1$ denotes the Hausdorff 1-dimensional measure.
In the article of *L.Modica -The gradient theory of phase transitions and the minimal interface criterion* that can be found here: [https://www.math.cmu.edu/~tblass/CNA-PIRE/Modica1987.pdf](https://www.math.cmu.edu/%7Etblass/CNA-PIRE/Modica1987.pdf) this property is proved only for the **signed distance function** (see Lemma 3, at page 8), but from more examples that I take it seems to be valid for many other level functions.
|
https://mathoverflow.net/users/61629
|
Continuity of Hausdorff measure on level sets
|
As Leo Moos suggested in the comments, in any dimension $d$, this is a simple consequence of the implicit function theorem.
The implicit function theorem implies that every point $x\in\phi^{-1}(0)$ has a neighborhood $\Omega\_x$ that is a diffeomorphic image $\varphi\_x(Q\_{\delta\_x})$ of a box $Q\_{\delta\_x}=\{|y\_i|\leq\delta\_x,\;1\leq i \leq d\}$, such that $\phi(\varphi(y))=|\nabla \phi(x)|\cdot y\_1$, and such that $\varphi\_x(0)=x$ and $D\varphi\_x(0)$ is a rotation. (Details: assume wlog that $x=0$, and, by rotating, that $\nabla \phi (0)/|\nabla \phi (0)|=e\_1,$ the first basis vector. Consider the function $g:\mathbb{R}^d\to \mathbb{R}^d$ defined by $g(x)=(\phi(x)/|\nabla \phi (0)|,x\_2,\dots,x\_d)$. Then $Dg(0)$ is the identity, and $\varphi\_x$ is the inverse $g^{-1}$ provided by the inverse function theorem.) Given $\varepsilon>0$, by choosing smaller $\delta\_x$ if necessary, we can ensure that the restriction of $\varphi\_x$ to the leaves $\{y\_1=h\}$ distorts the $(d-1)$-area by no more than $1+\varepsilon$, i. e., in $Q\_{\delta\_x}$,
$$1-\varepsilon<\left(\det\_{2\leq i,j\leq d}\left(\partial\_{y\_i}\varphi\cdot\partial\_{y\_j}\varphi\right)\right)^\frac12\leq 1+\varepsilon.$$
By compactness, we can choose a finite cover $\Omega\_i=\Omega\_{x\_i}$, $i=1,\dots,n$ of $\phi^{-1}(0)$. Then, for $\epsilon\_0>0$ small enough, we have $W\_{\epsilon\_0}:=\{x:|\phi(x)|\leq\epsilon\_0\}\subset \cup\_{i=1}^n\Omega\_i$. Put $\Omega\_0=\Omega\setminus W\_{\epsilon\_0}$, then $\Omega\_0,\Omega\_1,\dots,\Omega\_n$ is a finite cover of $\Omega$, and we can pick a partition of unity $f\_0,\dots,f\_n$ subordinate to that cover. We have for $|h|<\epsilon\_0$,
$$
\mathcal{H}^{d-1}(\phi^{-1}(h))=\sum\_{i=1}^n\int f\_i d\mathcal{H}^{d-1}(\phi^{-1}(h)),
$$
hence, by changing the variable,
$$
(1-\varepsilon)\sum\_{i=1}^n\int\_{y\_1=h} f\_i\circ\varphi\_i\leq \mathcal{H}^{d-1}(\phi^{-1}(h))\leq (1+\varepsilon)\sum\_{i=1}^n\int\_{y\_1=h} f\_i\circ\varphi\_i.
$$
As $h\to 0$, the terms in the right-hand side tend to
$$
(1+\varepsilon)\int\_{y\_1=0}f\_i\circ\varphi\_i\leq (1+\varepsilon)^2\int f\_i d\mathcal{H}^{d-1}(\phi^{-1}(0)),
$$
and similarly for the LHS. This means that
$$
(1-\varepsilon)^2\mathcal{H}^{d-1}(\phi^{-1}(0))\leq\liminf\_{h\to 0} \mathcal{H}^{d-1}(\phi^{-1}(h)),
$$
$$
\limsup\_{h\to 0} \mathcal{H}^{d-1}(\phi^{-1}(h))\leq (1+\varepsilon)^2\mathcal{H}^{d-1}(\phi^{-1}(0)),
$$
and since $\varepsilon>0$ is arbitrary, we are done.
|
6
|
https://mathoverflow.net/users/56624
|
398254
| 164,359 |
https://mathoverflow.net/questions/398238
|
1
|
It is known that the first order error term in the Shannon entropy formula for a binomial distribution is $1/n$ (for example, see the Wikipedia page [Binomial distribution](https://en.wikipedia.org/wiki/Binomial_distribution)), where in the limit $n \to \infty$ the entropy of a binomial approaches the Gaussian one. There are quite technical papers which have calculated the entropy of a binomial distribution exactly; however, I would like to see this error correction $1/n$ to the Gaussian in a simpler way. So, I do the following.
[Here](http://scipp.ucsc.edu/%7Ehaber/ph116C/NormalApprox.pdf), in Equation (4), the binomial distribution in the limit $n \to \infty$ has been written as $$f(x) = \frac{1}{\sqrt{2\pi npq}} e^{- (x - np)^2/2npq} \bigg[1+\mathcal{O}\Big(\frac{1}{\sqrt{n}}\Big)\bigg].$$.
In order the limit $n \to \infty$ to be well-defined and $f(x)$ doesn't vanish, we standardize $f(x)$ by defining $\mu \equiv np$, $\sigma \equiv \sqrt{npq}$, and $g(x) \equiv \sigma f(\sigma x+\mu)$, then we have: $$g(x) = \frac{1}{\sqrt{2\pi}} e^{- x^2/2} \bigg[1+\mathcal{O}\Big(\frac{1}{\sqrt{n}}\Big)\bigg],$$ where the term outside the bracket is a Gaussian with mean $0$ and standard deviation $1$.
Now, we can calculate the differential Shannon entropy of $g(x)$ as $S = -\int dx \, g(x) \ln g(x)$, which reads $$S = \frac{1}{2} \ln 2 \pi e + \mathcal{O}\Big(\frac{1}{\sqrt{n}}\Big).$$
The first term is correctly the entropy of the standard Gaussian but the error is of the order $1 / \sqrt{n}$. If we assume the result quoted in Wikipedia is correct, what is wrong with my approach?
**Edit 1 (Comment on answer 1)**
By the method of moment-generating functions, one can show that the standardized binomial distribution, $B(n, p)$, approaches the moment-generating function of the standard Gaussian, $\mathcal{N}(0,1)$, when $n \to \infty$. Thus, one expects the entropy of the former approaches $(1/2) \ln 2 \pi e$, as $n \to \infty$, I'm looking for the error terms of this expansion.
|
https://mathoverflow.net/users/nan
|
Obtaining the error term of binomial distribution's entropy from the differential entropy of a Gaussian distribution
|
The confusion about $1/\sqrt n$ versus $1/n$ corrections is addressed below. First, for reference, let me quote the relevant results, all following from this [source](https://core.ac.uk/download/pdf/82776425.pdf).
The Gaussian entropy is $S\_0=\tfrac{1}{2}\ln(2\pi \sigma^2)+\tfrac{1}{2}$, the leading correction $\delta S$ is of order $1/n$ and depends on the distribution.
• For the binomial distribution, with variance $\sigma^2=npq$ one has
$\delta S=\frac{1}{3n}-\frac{1}{12\sigma^2}.$ This vanishes if $p=q=1/2$, the leading order correction is then of order $1/n^2$.
• For the Poisson distribution, with variance $\sigma^2=n\lambda$ one has
$\delta S=-\frac{1}{12\sigma^2}.$
• For the negative binomial distribution, with variance $\sigma^2=nqp^{-2}$ one has
$\delta S =\frac{1}{6n}-\frac{1}{12\sigma^2}.$
More generally, for a discrete distribution with variance $\sigma^2=n\kappa\_2$ and third cumulant $n\kappa\_3$ one has
$$\delta S=-\frac{\kappa\_3^2}{12\kappa\_2^3 n}.$$
If the third cumulant vanishes the correction becomes of higher order in $1/n$.
---
Let me try to make contact with the calculation in the OP, to see where the confusion arises. I rescale $z=(x-np)/\sqrt{npq}$ to write the expansion
$$f(x) = \frac{1}{\sqrt{2\pi npq}} e^{- (x - np)^2/2npq} \bigg[1+\mathcal{O}\Big(\frac{1}{\sqrt{n}}\Big)\bigg]$$
of the binomial distribution around the Gaussian in the form
$$\tilde{f}(z)=\frac{1}{\sqrt{2\pi npq}} e^{-z^2/2}\bigg[1+h(z)n^{-1/2}\bigg].$$
In the large-$n$ limit, the sum over the discrete variable $x$ may be replaced by $\sqrt{npq}\int\_{-\infty}^\infty dz$. We thus find the entropy
$$S=-\sqrt{npq}\int\_{-\infty}^\infty \tilde{f}(z)\ln \tilde{f}(z)\,dz.$$
Now comes the key point: the function $h(z)$ is odd in $z$ (see eq. 4.3 in the cited reference). The order $1/\sqrt n$ correction then vanishes since it is an integral over an odd function, and the next term of order $1/n$ is the first nonzero correction.
|
0
|
https://mathoverflow.net/users/11260
|
398260
| 164,362 |
https://mathoverflow.net/questions/398261
|
1
|
Let $X$ be a multivariate normal $\mathcal{N}(\mu, \Sigma^2)$ and let $X$ be anisotropic, that is I am considering $\Sigma$ to be a diagonal matrix but the elements on the diagonal might be different.
I am interested in finding the distribution of $X/\|X\|\_2$.
As a start let $X$ be isotropic. Then $\|X\|\_2^2$ will be Gamma distributed. But then $\|X\|\_2$ will perhaps have to follow what I found to be a Nakagami distribution (<https://en.wikipedia.org/wiki/Nakagami_distribution>). So I need to find the ratio of a normal and this Nakagami distribution. However for the anisotropic case $\|X\|\_2^2$ will not be Gamma but a mixture of Gamma distributions (<https://stats.stackexchange.com/questions/72479/generic-sum-of-gamma-random-variables>) and this seems more complicated to be honest.
On the other hand it feels intuitively somewhat similar to a truncated normal distribution, but with the truncation happening over a unit ball and this also seems complicated.
Any ideas/ references/ hints/ thoughts
Thanks
|
https://mathoverflow.net/users/325572
|
Norm contrained Gaussian distribution
|
In two dimensions, with
$$\mu=\begin{pmatrix} m \\ n \end{pmatrix},\ \
\Sigma=\begin{pmatrix} v & 0 \\ 0 & w \end{pmatrix},$$
an integration over all possible radii gives the distribution of $X/\|X\|\_2$ as
$$f(\cos t,\sin t)=\frac{
1+\sqrt{\pi}u
\exp(u^2)
(1+\text{erf}(u))
}{
\exp(a)c\pi \sqrt{v w}
}
$$
where
$$a=\frac{m^2}{2v}+\frac{n^2}{2w},\ \
b=\frac{m\cos t}{v}+\frac{n\sin t}{w},\ \
c = \frac{2\cos^2t}{v}+\frac{2\sin^2 t}{w},\ \
u = \frac{b}{\sqrt{c}}$$
and the expression is indeed complicated enough to deter more exact computations.
|
3
|
https://mathoverflow.net/users/nan
|
398272
| 164,365 |
https://mathoverflow.net/questions/398251
|
20
|
In Zhu's [Coherent sheaves on the stack of Langlands parameters](https://arxiv.org/abs/2008.02998) theorem 4.7.1 relates the cohomology of the moduli stack of shtukas to global sections of a certain sheaf on the stack of global Langlands parameters:
>
> Let $H\_{I,V}^i=H^iC\_c\left(\mathrm{Sht}\_{\Delta(\bar\eta),K},\mathrm{Sat}(V)\right)$. By [Xu20, Xu1, Xu2], the natural Galois action and the partial Frobenii action together induce a canonical $W\_{F,S}^I$-action on $H\_{I,V}^i$. The following statement can be regarded as a generalization of the main construction of [LZ].
>
>
> **Theorem 4.7.1**. *Assume that $k=\mathbb Q\_\ell$ and regard $\mathrm{Loc}\_{^cG,F,S}$ as an algebraic stack over $\mathbb Q\_\ell$. Then for each $i$, there is a quasi-coherent sheaf $\mathfrak A\_K^i$ on $^{cl}\mathrm{Loc}\_{^cG,F,S}$, equipped with an action of $H\_K$, such that for every finite dimensional representation $V$ of $(^cG)^I$, there is a natural $(H\_K\times W\_{F,S}^I)$-equivariant isomorphism*
> $$
> H\_{I,V}^i\cong\Gamma(^{cl}\mathrm{Loc}\_{^cG,F,S},(w\_{F,S}V)\otimes\mathfrak A\_K^i)\tag{4.26}
> $$
> *where $w\_{F,S}V$ is the vector bundle on $\mathrm{Loc}\_{^cG,F,S}$ equipped with an action by $W\_{F,S}^I$ as in Remark 2.2.7.*
>
>
>
This is a generalization of a construction in section 6 of Lafforgue-Zhu's [Décomposition au-dessus des paramètres de Langlands elliptiques](https://arxiv.org/abs/1811.07976) (also recapitulated in remark 8.5 of V. Lafforgue's [ICM survey](https://arxiv.org/abs/1803.03791)). The idea is that a regular function of the Langlands parameter determines an endomorphism of $H\_{\lbrace 0\rbrace,\mathrm{Reg}}$, a subspace of the cohomology of the moduli of shtukas with one leg, with relative position of the modification bounded by the regular representation of $\widehat{G}$ (see section 7 of [4](https://arxiv.org/abs/1803.03791)).
Let $V$ be a representation of $\widehat{G}$, let $x\in V$, $\xi\in V^{\*}$. Any function $f$ of $\widehat{G}$ can be expressed as a matrix coefficient $\langle\xi,g.x\rangle$, and for $\gamma\in\mathrm{Gal}(\overline{F}/F)$ the functions $F\_{f,\gamma}:\sigma\mapsto f(\sigma(\gamma))$ are supposed to topologically generate all such functions on the stack of Langlands parameters, as $f$ and $\gamma$ vary.
Letting $\underline{V}$ be the underlying vector space of the representation $V$, the tensor product $H\_{\lbrace 0\rbrace,\mathrm{Reg}}\otimes \underline{V}$ will have an action of $\mathrm{Gal}(\overline{F}/F)$, obtained from an isomorphism with $H\_{\lbrace 0,1\rbrace,\mathrm{Reg}\boxtimes V}$ (see remark 8.5 of [4](https://arxiv.org/abs/1803.03791)).
Since the data of $F\_{f,\gamma}$ is the same as the data of $x$, $\xi$, and $\gamma$, we may now construct the endomorphism of $H\_{\lbrace 0\rbrace,\mathrm{Reg}}$ as follows:
$$H\_{\lbrace 0\rbrace,\mathrm{Reg}}\xrightarrow{\mathrm{Id}\otimes x} H\_{\lbrace 0\rbrace,\mathrm{Reg}}\otimes \underline{V}\xrightarrow{\gamma} H\_{\lbrace 0\rbrace,\mathrm{Reg}}\otimes \underline{V}\xrightarrow{\mathrm{Id}\otimes\xi} H\_{\lbrace 0\rbrace,\mathrm{Reg}}$$
This allows us to construct a sheaf of $\mathcal{O}$-modules on the stack of Langlands parameters, whose global sections is $H\_{\lbrace 0\rbrace,\mathrm{Reg}}^{i}$.
In several other places in [1](https://arxiv.org/abs/2008.02998), Zhu also hints at the hope of finding an analogue of theorem 4.7.1 for the cohomology of Shimura varieties, following ongoing work in progress of Zhu and Emerton on an analogous stack of Langlands parameters for number fields.
How might such an analogue proceed? In the setting of Shimura varieties, as opposed to shtukas, it appears we do not have the notion of legs, or of bounding relative positions of modifications with representations of $\widehat{G}$ (I do not know whether there are any existing analogies). How might such an endomorphism be constructed?
|
https://mathoverflow.net/users/85392
|
Cohomology of Shimura varieties and coherent sheaves on the stack of Langlands parameters
|
The anticipated analogue is as follows:
There is a map $f: \mathcal X \to \prod\_{v \in S} \mathcal X\_v,$
where $\mathcal X$ is the stack of $p$-adic representations of $G\_{E,S}$ into ${}^LG$ (the $L$-group over $E$ of some group $G$ that is part of a Shimura datum, with reflex field $E$) unramified outside $S$, and each $\mathcal X\_v$
is the corresponding local version at the place $v$. (We choose $S$ to include all primes $v$ over $p$, and over $\infty$. And probably any primes where $G$ is ramified.)
Choosing a level structure at each place $v$ of $S$ will give rise to a coherent sheaf $\mathfrak A\_v$ on $\mathcal X\_v$ as in Xinwen's paper. (Strictly speaking, his paper only treats the case where $v \not\mid p$
or $\infty$, but there should be sheaves for the other $v$ too. The case of $v \mid \infty$ is conceptually a bit mysterious, but there are some ideas. The case of $v \mid p$ is related to the conjectural $p$-adic Langlands program; in this case $\mathcal X\_v$ will be an appropriate EG-type stack.)
We can then form $\mathfrak A := \boxtimes\_{v \in S} \mathfrak A\_v$,
the exterior tensor product, a coherent sheaf on $\prod\_{v \in S} \mathcal X\_v$.
We can consider $f^! \mathfrak A$ on $\mathcal X$.
The Hodge cocharacter $\mu$ from the Shimura datum gives rise to a representation $V\_{\mu}$ of ${}^LG$, and composing this with the universal
representation over $\mathcal X$, we obtain a locally free sheaf $\mathcal V$
of Galois representations over $\mathcal X$.
Now we have the conjectural formula:
$$R\Gamma\_c(\text{ Shimura variety at given level}, \mathbb Z\_p)[\text{shift by dimension of Shimura variety}]
$$
$$ = R\Gamma(\mathcal X, f^!\mathfrak A \otimes \mathcal V).$$
The analogous formula in the function field case, for cohomology of stacks of shtuka, is stated in Xinwen's paper, and this is the Shimura varieties version.
The isomorphism should be compatible with Galois and Hecke actions on the two sides.
The main difference with the shtuka case is that we just have the single $V\_{\mu}$ that we can consider, so there is less structure in the Shimura variety case than in that case. (We can get a bit more structure by allowing non-trivial coefficient systems in the cohomology, or even going to $p$-adically completed cohomology; this will be reflected in the precise choice of sheaf $\mathfrak A\_v$ at the places $v|p$.)
This will all be discussed in (hopefully) forthcoming papers of mine and Xinwen's, and of mine, Toby Gee, and Xinwen.
|
13
|
https://mathoverflow.net/users/169863
|
398274
| 164,367 |
https://mathoverflow.net/questions/398282
|
5
|
I am struggling to understand what an invariant section with respect to a linearization of a line sheaf is. In Geometric Invariant Theory, given a $k$-scheme $X$ (being $k$ an algebraically closed field of characteristic zero) where a reductive algebraic group $G$ acts on by $\sigma$, and a line bundle $L\rightarrow X$ over $X$, a linearization on $L$ is just a lift of the $G$-action on $X$. This is equivalent to give an isomorphism of sheaves of $\mathcal{O}\_{X\times G}$-modules
$$\sigma^{\ast}\mathcal{L}=p\_{1}^{\ast}\mathcal{L}$$
satisfying the 1-cocycle condition, where $\mathcal{L}$ is the invertible sheaf associated to $L$ and $p\_{1}:X\times G\rightarrow X$ is the projection on the first factor. Mumford also says that from this definition we obtain a dual action of $G$ on the group of global sections of the line bundle
$$H^{0}(X,\mathcal{L})\rightarrow H^{0}(G,\mathcal{O}\_{G})\otimes H^{0}(X,\mathcal{L})$$
So my questions are:
1. What is a dual action of the group?
2. Can we talk about invariant sections of the line sheaf? How?
3. We have a correspondence between sections of an invertible sheaf, and sections of the associated line bundle, that is, map of schemes $X\rightarrow L$ which are sections of the projection map $L\rightarrow X$. In this sense, what is an invariant section of the line bundle?
Thank you very much for your comments in advance. It is really hard for a new geometer to understand fully what is inside of Mumford's head.
|
https://mathoverflow.net/users/140062
|
Invariant section of a linearized sheaf
|
**Question:** "What is a dual action of the group?"
**Answer:** For simplicity, if $X:=Spec(A)$ and $G:=Spec(R)$ is a linear algebraic group over a field $k$ acting on $X$ via $\sigma^\*: G\times\_k X \rightarrow X$ you get a "dual action"
$$\sigma: A \rightarrow R\otimes\_k A.$$
If $L\in Pic(A)$ is an invertible module with $\mathcal{L}:=\mathbb{V}(L^\*):=Spec(Sym\_A^\*(L^\*)):=Spec(B)$, a $G$-linearization of $L$ gives an action
$$\gamma^\*: G\times \mathcal{L}\rightarrow \mathcal{L}$$
and a "dual action"
$$\gamma: B \rightarrow R\otimes\_k B.$$
There are canonical maps $A\rightarrow B$ and $R\otimes\_k A \rightarrow R\otimes\_k B$ and all diagrams are to commute.
The map $\sigma$ is a local version of your map on global sections
$$H^{0}(X,\mathcal{L})\rightarrow H^{0}(G,\mathcal{O}\_{G})\otimes H^{0}(X,\mathcal{L}).$$
**Question:** "Can we talk about invariant sections of the line sheaf? How? We have a correspondence between sections of an invertible sheaf, and sections of the associated line bundle, that is, map of schemes X→L which are sections of the projection map L→X. In this sense, what is an invariant section of the line bundle?"
**Answer:** You formulate the notions "invariant function" and "invariant section" using the dual actions $\sigma, \gamma$ defined above.
**Example:** If $X \subseteq \mathbb{P}^n\_k$ is a quasi projective scheme with an action $\sigma^\*:G\times\_k X \rightarrow X$, and where you can give $X$ an open affine cover $U\_i:=Spec(A\_i)$ of $G$-invariant subschemes, you get induced dual actions
$$ \sigma\_i:A\_i \rightarrow R\otimes\_k A\_i.$$
You use these dual actions $\sigma\_i$ to study the action $\sigma$ and the "quotient" $X/G$.
When $G$ is "finite" such an affine open cover always exist. Moreover the quotient $X/G$ "exist" and you define it locally as $Spec(A\_i^G)$.
**Example:** if $C:=\mathbb{P}^1\_k$ is the projective line, there is an action of the symmetric group $S\_n$ on $C^{\times\_k n}$, and you may prove that if $n\neq char(k)$ there is an isomorphism
$$Sym^n(C):=(C^{\times\_k n})/S\_n \cong \mathbb{P}^n\_k.$$
You embed the product $C^{\times\_k n}$ into a projective space and use an open affine $S\_n$-invariant cover to construct the quotient $Sym^n(C)$ and to prove it is isomorphic to projective $n$-space. So you cannot construct the symmetric product $Sym^n(C)$ as $Spec(A^G)$ for some group $G$ acting on a ring $A$, since projective space is not affine.
**Example:** An affine example and a "set theoretic paradox". If $G:=S\_n$ is the symmetric group on $n$ letters acting on the polynomial ring $A:=k[x\_1,..,x\_n]$ it follows the invariant ring $A^G \cong k[s\_1,..,s\_n] \subseteq A$ where $s\_i$ are the elementary symmetric polynomials. The polynomials $s\_i$ are algebraically independent over $k$ hence $A^G$ is a polynomial ring and there is an isomorphism
$$ \pi: \mathbb{A}^n\_k \rightarrow \mathbb{A}^n\_k/G \cong \mathbb{A}^n\_k.$$
The map of schemes $\pi$ gives at the level of topological spaces a surjective and non-injective endomorphism
$$\pi: \mathbb{A}^n\_k \rightarrow \mathbb{A}^n\_k$$
of affine $n$-space. Since the ring extension $A^G \subseteq A$ is an integral extension, it follows $\pi$ is surjective. Given any prime ideal $\mathfrak{p} \subseteq A$, it follows the orbit of $\mathfrak{p}$ under the action of $G$ has $n!$ elements in general. Hence the map $\pi$ is not injective. To some people (not all) this is a "paradoxical situation".
**Note:** When working with schemes over fields that are not algebraically closed you will need this construction. Some people are "sloppy" when considering such actions and do not use the dual action.
|
6
|
https://mathoverflow.net/users/nan
|
398295
| 164,371 |
https://mathoverflow.net/questions/398012
|
0
|
>
> Donsker's invariance principle:
> Let $X\_1,X\_2,...$ be i.i.d. real-valued random variables with mean 0 and variance 1. We define $S\_0=0$ and $S\_n= X\_1+ ... + X\_n$ for $n \geq 1$. To get a process in continuous time, we interpolate linearly and define for all $t \geq 0$
> $$
> S\_t = S\_{[t]}+ (t-[t])(S\_{[t]+1}- S\_{[t]}).
> $$
> Then we define for all $t \in [0,1]$
> $$
> S^\*\_n(t)= \frac{S\_{nt}}{\sqrt{n}}.
> $$
> Let $C[0,1]$ be the space of real-valued continuous function defined on $[0,1]$ and endow space with the supremumnorm. Then $(S^\*\_n(t))\_{0 \leq t \leq 1}$ can be seen as a random variable taking values in $C[0,1]$. Now let $\mu\_n$ be its law on that space of continuous functions and let $\mu$ be the law of Brownian motion on $C[0,1]$. Then the following holds:
>
>
>
>
> **Theorem (Donsker):** The probability measure $\mu\_n$ converges weakly to $\mu$, i.e. for every $F: C([0,1]) \rightarrow \mathbb{R}$ bounded and continuous,
> $$
> \int F d\mu\_n \rightarrow \int F d\mu
> $$
> as $n \rightarrow \infty$.
>
>
>
But for a two-dimensional case, the 'coupling version' is as following.
$\textbf{'coupling version'}$: Fix a square $S$ of size $s$. Fix $x\in nS$. Let $X$ be a random walk starting from $x$ until it exits the square $nS$ and let $B$ be a Brownian motion until it exits $nS$. For $\forall \epsilon>0$, then there exists $N>0$ such that $n\ge N$, one can couple $X$ and $B$ so that
$$d(X,B)\le \epsilon n.$$
My questions:
(1) Can we extended the classical Donsker's to the two-dimensional case?
(2) Is there any reference for the proof of the 'coupling version'?
|
https://mathoverflow.net/users/168083
|
How to prove the coupling version of the Donsker's Invariance Principle?
|
Yes, this can be done in any dimension but the two dimensional case is especially simple. Rotating a simple RW in two dimensions by 45 degrees yields a random walk where the $x$ and $y$ coordinates are independent. Couple each coordinate to a BM using Skorokhod embedding to obtain the desired coupling.
|
0
|
https://mathoverflow.net/users/7691
|
398314
| 164,376 |
https://mathoverflow.net/questions/398306
|
-1
|
I would say no. Because a constant curvature manifold is symmetric. Any 2 dimensional subspace should be like a hyperboloid or a sphere. These objects do not have torsion.
|
https://mathoverflow.net/users/105352
|
Do exist constant curvature manifolds (hyperbolic or elliptic) with torsion?
|
You should consider the following example:
Let $M^3=\mathrm{SU}(2)\simeq S^3$ endowed with its biïnvariant metric $g$ (unique up to a constant multiple, let's fix this by requiring that the $g$ has constant sectional curvature equal to $1$). Now consider the unique connection $\nabla$ for which the left-invariant vector fields on $SU(2)$ are parallel. This connection is compatible with the biïnvariant metric and it has torsion, since $T(X,Y) = -[X,Y]\not=0$ when $X$ and $Y$ are linearly independent left-invariant vector fields. This metric has the same geodesics as the Levi-Civita connection, and every $2$-plane has sectional-curvature $1$.
The symmetry group of $(M^3,g,\nabla)$ is $\mathrm{SO}(4)$, the maximum possible. So this is an example of a constant curvature manifold with torsion.
As É. Cartan and J. A. Schouten showed in 1926 (*On Riemannian geometries admitting an absolute parallelism*), this example generalizes to biïnvariant metrics on compact simple Lie groups endowed with the unique connection for which the left-invariant vector fields are parallel, though they no long actually have constant sectional curvature.
However, they also show in that paper that there is a connection $\nabla$ on the $7$-sphere $S^7$ of unit octonions that is flat, compatible with the natural metric of constant curvature on $S^7$, and invariant under the group $\mathrm{Spin}(7)\subset\mathrm{SO}(8)$. Because $\mathrm{Spin}(7)$ acts transitively on $3$-planes in $\mathbb{O}\simeq\mathbb{R}^8$, it follows that this geometry $(S^7, g,\nabla)$ also has constant curvature.
|
11
|
https://mathoverflow.net/users/13972
|
398323
| 164,380 |
https://mathoverflow.net/questions/398316
|
6
|
This question was [posted](https://math.stackexchange.com/posts/4199651/edit) on Math Stack Exchange, but did not attract an answer. Here is the question:
Informal Description
====================
Let me start with an example. Let $X$ be the set $\{a, b, c, d, e\}$ and $E$ be the set $\{a, b, c\}$. Let $f$ be a function with domain $X$.
Then the mapping that sends $E$ to $\{f(a), f(b), f(c) \}$ is called the direct image of $E$ under $f$, denoted by $f\_\*(E)$.
Now, let $W$ be the set $\{\{a, b\}, \{c\}\}$. What do you call the mapping that sends $W$ to $\{\{f(a), f(b)\}, \{f(c)\}\}$ ?
Informally, this mapping replaces every element of $X$ *"inside"* $W$ with its image under $f$.
Slightly More Formal Description
================================
Let us assume the [ZFA-axiomatisation](https://ncatlab.org/nlab/show/ZFA#:%7E:text=ZFA%20is%20a%20variant%20of,made%20up%20of%20other%20elements.) of Set Theory, a variant of ZF that allows non-set objects called atoms. Let $A$ be a set of atoms.
Define $P^0(A)$ = A.
And for all $n \in \mathbb{N} \colon$ define $P^n(A)= \mathscr{P}(P^{n-1}(A)) \cup A $ ,
where $\mathscr{P}(-)$ denotes the powerset of a set.
And define $P^{\infty}(A)$ as $\bigcup\_{n \in \mathbb{N}} P^{n}(A)$
Let $X$ be a set of atoms. Let $f$ be a function with domain $X$. We could then inductively define a function $f\_{\*\*}$ such that $dom(f\_{\*\*}) = P^{\infty}(X) \backslash X$ , and $f\_{\*\*}$ has the property that for all $Z \in dom(f\_{\*\*}) \colon f\_{\*\*}(Z) = \{f\_{\*\*}(T) | T \in Z \backslash X \} \cup \{f(x) | x \in Z \cap X \}$.
Examples
--------
* Let $X = \{a, b, c, d, e\}$ be a set of atoms. Then $\{a, \{b\}\} \in P^{\infty}(X) \backslash X$ and $f\_{\*\*}(\{a, \{b\}\}) = \{f(a), \{f(b)\}\}$
* Let $X = \{a, b, c, d, e\}$ be a set of atoms. Then $\{\{ \{c \} \}\} \in P^{\infty}(X) \backslash X$ and $f\_{\*\*}(\{\{\{c \} \}\}) = \{\{\{f(c) \} \}\}$
So $f\_{\*\*}$ can, in a sense, be thought of as a generalisation of the direct image. My quesiton is, what do you call this concept? And can you point me towards any resources regarding it? I would prefer if the resource makes use of the ZFA-axiomatisation, but I would also accept any other resource.
Sidenote
--------
I am aware that this concept comes up in [Permutation Models](https://en.m.wikipedia.org/wiki/Permutation_model#CITEREFFraenkel1922) for the case when $f$ is a bijective endofunction.
|
https://mathoverflow.net/users/324481
|
What do you call the generalisation of the direct image?
|
This idea is commonly used in set theory with atoms. I'm not sure whether it has a standard name, but I would be inclined to call it the *natural extension* of $f$ to sets.
There is no need to stop the iteration at $\omega$ as you do, for one can continue the cumulative hierarchy through the ordinals. If $A$ is any class of atoms, you can define the well-founded cumulative hierarchy over $A$ by transfinite recursion, just like the ordinary cumulative hierarchy, like this:
$\newcommand{\WF}{{\rm WF}}$
$$\WF\_0(A)=A,$$
$$\WF\_{\alpha+1}(A)=\mathcal P(\WF\_\alpha(A)),$$
$$\WF\_\lambda(A)=\bigcup\_{\alpha<\lambda}\WF\_\alpha(A)$$
Specifically, we start with just the atoms, and then iteratively add all subsets of what we've got so far.
It is straightforward to see that every function $\sigma:A\to V$ defined on the atoms extends naturally to a function $\bar\sigma$ defined on the entire hierarchy over those atoms by the following recursion:
$$\bar\sigma(x)=\begin{cases}
\sigma(x)&\text{if $x\in A$,}\\
\bar\sigma[x]&\text{otherwise}
\end{cases}$$
You can view this as a $\in$-recursion or alternatively as a recursion on ordinals, defining $\bar\sigma$ on each $\WF\_\alpha(A)$. This definition agrees with yours (on finite levels), since we apply the original function to the atoms, and otherwise apply the pointwise image recursively.
Another way to think about the natural extension is that every set is having the atoms in its hereditary membership replaced with their values as specified by the original function. So this is a kind of hereditary application of the original function.
You asked for references, and so let me mention that we use exactly this idea in our paper:
* *Daghighi, Ali Sadegh; Golshani, Mohammad; Hamkins, Joel David; Jeřábek, Emil*, [**The foundation axiom and elementary self-embeddings of the universe**](http://www.arxiv.org/abs/1311.0814), Geschke, Stefan (ed.) et al., Infinity, computability and metamathematics. Festschrift celebrating the 60th birthdays of Peter Koepke and Philip Welch. London: College Publications (ISBN 978-1-84890-130-8/hbk). 89-112 (2014). [ZBL1358.03070](https://zbmath.org/?q=an:1358.03070).
In the proof of theorem 2, for example, we have a function $\sigma$ defined on the atoms (for us it is Quine atoms, but ZFA urelements would work the same), and then extend this to all of $\WF(A)$ by recursion.
In the case that the original map was a permutation of the atoms, then the natural extension of it is an automorphism of the cumulative hiearchy over those atoms.
Another application of the idea arises in the Jech-Sochor embedding theorem, which can be found in Jech's book Set Theory.
|
4
|
https://mathoverflow.net/users/1946
|
398329
| 164,385 |
https://mathoverflow.net/questions/398317
|
3
|
>
> **QUESTION.** Let $x>0$ be a real number or an indeterminate. Is this true?
> $$\sum\_{n=0}^{\infty}\frac{\binom{2n+1}{n+1}}{2^{2n+1}\,(n+x+1)}=\frac{2^{2x}}{x\,\binom{2x}x}-\frac1x.$$
>
>
>
**POSTSCRIPT.** I like to record this presentable form by Alexander Burstein:
$$\sum\_{n=0}^{\infty}\frac{\binom{2n}n}{2^{2n}(n+x)}=\frac{2^{2x}}{x\binom{2x}x}.$$
|
https://mathoverflow.net/users/66131
|
Proving a binomial sum identity
|
\begin{align}\sum\_{n=0}^{\infty}\frac{\binom{2n+1}{n+1}}{2^{2n+1}\,(n+x+1)}&= \int\_0^1\sum\_{n=0}^{\infty}\frac{\binom{2n+2}{n+1}y^{n+x}}{2^{2n+2}}\,{\rm d}y\\&=\int\_0^1 y^{x-1}\big((1-y)^{-1/2}-1\big){\rm d}y\\&=B\left(x,\frac12\right)-\frac1x\end{align}
and the rest follows from the properties of [beta function](https://en.wikipedia.org/wiki/Beta_function).
|
9
|
https://mathoverflow.net/users/7076
|
398338
| 164,391 |
https://mathoverflow.net/questions/398267
|
3
|
Let $\omega\_{\text{WP}}$ denote the Weil-Petersson metric associated to a family of Calabi-Yau manifolds. That is, let $f : X \to Y$ be a surjective holomorphic map with connected fibres such that, over the regular locus, the fibres $X\_y$ of $f^{\circ} : X^{\circ} \to Y^{\circ}$ are smooth compact Kähler manifolds with $c\_1 =0$ in the $H^2\_{\text{DR}}(X\_y, \mathbb{R})$.
Let $D$ be the associated classifying space, which we know is a symmetric space of non-compact type. Let $\omega\_H$ denote the invariant metric on $D$, which is referred to as the *Hodge metric*. [Lu-Sun](https://arxiv.org/pdf/math/0510021.pdf) showed that the Ricci curvature of $\omega\_{\text{WP}}$ is pinched according to: $$- \omega\_H \leq \text{Ric}(\omega\_{\text{WP}} ) \leq \omega\_H,$$ and $2\omega\_{\text{WP}} \leq \omega\_H$.
*Question:* Is $\text{Ric}(\omega\_{\text{WP}})$ honestly bounded from below, i.e., does there exist a constant $C>0$ such that $$\text{Ric}(\omega\_{\text{WP}}) \geq - C \omega\_{\text{WP}}?$$
I'd be interested in any results concerning the Ricci curvature of the Weil-Petersson metric, even for non-Calabi-Yau families.
|
https://mathoverflow.net/users/105103
|
Ricci curvature of the Weil-Petersson metric?
|
A positive answer to this question is given in part (2) of Theorem 3.1. of C.-L. Wang's paper [Curvature properties of the Calabi-Yau moduli](https://www.math.uni-bielefeld.de/documenta/vol-08/18.pdf).
The specific reference is:
Wang, C.-L., *Curvature properties of the Calabi-Yau moduli*, Documenta Math., **8** (2003) 577-590.
|
0
|
https://mathoverflow.net/users/105103
|
398340
| 164,393 |
https://mathoverflow.net/questions/398268
|
18
|
The recent [article on Quanta](https://www.quantamagazine.org/how-many-numbers-exist-infinity-proof-moves-math-closer-to-an-answer-20210715/)
(by Natalie Wolchover)
concerning $\aleph\_1$ vs. $\aleph\_2$ suggests that there is
excitement within that community:
>
> Juliette Kennedy: "It’s one of the most intellectually exciting, absolutely dramatic things that has ever happened in the history of mathematics."
>
>
>
Another instance is the [Fargues/Scholze advances](https://arxiv.org/abs/2102.13459)
on "Geometrization of the local Langlands correspondence,"
which has the Langlands world excited:
>
> Eva Viehmann: "It’s really changed everything. These last five or eight years, they have really changed the whole field."
>
>
>
This makes me wonder if there is something like a [heat map](https://en.wikipedia.org/wiki/Heat_map) for all of mathematics,
which would show the areas with a lot of excitement.
It seems difficult to capture this via arXiv postings,
but that is an obvious starting point.
Has anyone pursued this?
|
https://mathoverflow.net/users/6094
|
Heat map of current mathematics
|
<https://paperscape.org/> is a 'heat map' of the arxiv if you color the graph by age. Unfortunately, its ability to detect links between mathematics papers is a bit lacking compared to physics papers for some reason, but it still gives a very interesting view of the subject.
|
23
|
https://mathoverflow.net/users/152678
|
398347
| 164,395 |
https://mathoverflow.net/questions/398343
|
1
|
Let $W$ be a standard Brownian motion on a probability space $(X, \mathcal F, \mathbb P)$ let and $\mathcal F\_t$ its natural filtration.
For $\varepsilon > 0, T \in [0, \infty)$ let $A\_{\varepsilon, t}$ be the event $\{\sup\_{s \in [0, T]} |W\_s|\ < \varepsilon\}$.
Let $X\_t$ be the solution to the SDE
$$dX\_t = \sigma(t, X\_t) dW\_t$$
where $\sigma$ is nice enough for a solution to exist.
For each $\varepsilon, T$ define the probability measure $\mathbb Q\_{\epsilon, T}$ by
$$\mathbb Q\_{\varepsilon, t} (E) = \frac{\mathbb P(E \cap A\_{\varepsilon, T})}{\mathbb P(A\_{\varepsilon, T)}}.$$
>
> **Question:** Is $X$ a local $\mathcal F\_t$ martingale under $\mathbb Q\_{\varepsilon, T}$?
>
>
>
**Remark:** We note that $\mathbb Q\_{\varepsilon, T}$ is *not* equivalent to $\mathbb P$, so that we cannot apply Girsanov’s theorem.
|
https://mathoverflow.net/users/173490
|
Is the integral against a Brownian motion conditioned to stay bounded a local martingale?
|
In case $\sigma = 1$ the claim is that the conditioned process itself is a martingale, however, as it paths are the paths of a brownian motion, it will have quadratic variation t, and therefore, it is an ordinary brownian motion.
|
2
|
https://mathoverflow.net/users/143907
|
398348
| 164,396 |
https://mathoverflow.net/questions/398288
|
7
|
Assume $X$ is a Tychonoff space. Then $A(X)$ is the free topological abelian group over $X$. I know that $A(X)$ is Hausdorff and the canonical embedding from $X$ to $A(X)$ is a topological embedding.
Now consider the subgroup $N:=2A(X)=\left\{g+g\ \colon\ g\in A(X)\right\}$.
The quotient $B(X):=A(X)/N$ is the free topological Boolean group. I want to show that $B(X)$ is Hausdorff which is equivalent to say that $N$ is closed in $A(X)$. Unfortunately, I do not know how show the closedness of $N$ since I can only describe the topology on $A(X)$ by the universal property and I do not know how that helps me here... Or is this maybe not true for a general Tychonoff space $X$ and one needs additional properties on $X$ like hereditarily disconnected?
I have found a reference on Free Boolean Groups
Genze, L.V. Free Boolean topological groups. Vestn. Tomsk. Gos. Univ. 2006, 290, 11–13
which maybe could answer my question but I am unable to find this text on math.sci.net (or anywhere else).
Thank you!
|
https://mathoverflow.net/users/326011
|
Why are free Boolean topological groups Hausdorff?
|
You can use the universality property with the following Boolean group as codomain: $B$ is the measure algebra over the unit interval (the quotient of the $\sigma$-algebra of Lebesgue measurable sets by the ideal of sets of measure zero), with symmetric difference as operation this is a Boolean group and $d(A,B)=\lambda(A\Delta B)$ defines a metric that will turn it into a topological group.
This group contains a copy of the unit interval, namely the set $\{[0,t]:0\le t\le 1\}$.
Since $X$ is Tychonoff you now have enough continuous functions to separate all non-trivial words from the empty word.
This survey by Sipacheva gives more information: [*Free Boolean Topological Groups*, Axioms 4 (2015) 492-517](https://doi.org/10.3390/axioms4040492)
|
8
|
https://mathoverflow.net/users/5903
|
398353
| 164,398 |
https://mathoverflow.net/questions/398375
|
2
|
Let $f: X \to Y$ be a locally trivial fibration between locally compact spaces with fiber $F$. It is well known that for a constant sheaf $A\_X$ on $X$, the higher direct images $R^n f\_\* A\_X$ are locally constant, with stalk $H^n(F, A)$. This can be seen from the description of said higher direct images as the sheaf associated to the presheaf
$$ U \mapsto H^n(f^{-1}(U), A\_X|\_{f^{-1}(U)}). $$
According to Iversen, *Cohomology of Sheaves*, Theorem VII.1.4, the stalks of the higher direct images with compact support have a similar description (for an arbitrary continuous map $f: X \to Y$ and an arbitrary sheaf $\mathcal{F}$)
$$ (R^n f\_! \mathcal{F})\_y = H^n\_c(f^{-1}(y), \mathcal{F}). $$
My question is then: if $f$ is a locally trivial fibration, does the same result hold? That is, are the sheaves $R^n f\_! A\_X$ local systems on Y? The same method of proof will not do, since the analogous candidate for a presheaf does not have well-defined restriction maps. And if you try to modify the proof that Iversen gives for the result on the stalks, things seem to break down from the begining, so I don't know how I could prove this.
---
**UPDATE:** Since we want to prove a local result, we may assume that we have a trivial fibration $B \times F \to B$. Using proper base change on the diagram
$$ \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int\_0^1}\right.}
\begin{array}{ccc}
B \times F & \xrightarrow{\pi\_1} & B \\
\da{\pi\_2} & & \da{a\_B} \\
F & \xrightarrow{a\_F} & \{\*\}
\end{array}$$
gives an isomorphism
$$ a\_B^\* \circ R^n a\_{F,!} (A\_F) \cong R^n \pi\_{1,!} \circ \pi\_2^\* (A\_F), $$
which by adjointness gives a map
$$ H^n\_c(F, A) = R^n a\_{F,!} (A\_F) \to a\_{B,\*} \circ R^n \pi\_{1,!} \circ \pi\_2^\* (A\_F) = \Gamma(B, R^n \pi\_{1,!} (A\_{B\times F})). $$
This map should be an isomorphism if the result were true (it is equivalent to the result) but I don't see how to prove that it actually is an isomorphism.
|
https://mathoverflow.net/users/173545
|
Higher direct image with compact support of a constant sheaf
|
With the information added in the last update, the result is already proved once you realize that the unit of the adjunction
$$ id \to a\_{B,\*} a\_B^\* $$
is actually an isomorphism on the category of abelian groups if the space $B$ is connected.
|
1
|
https://mathoverflow.net/users/173545
|
398379
| 164,405 |
https://mathoverflow.net/questions/398380
|
0
|
For $s\in (0,1)$, is there are an explicit expression for
$$(-\Delta)^s \left(\frac{1}{\left(1+|x|^2\right)^s}\right)?$$
Edit: My goal is to show that for the function $u(x)=\frac{1}{\left(1+|x|^2\right)^s}$ defined on the ball $B(0,R)$ where $R>1$ we have $(-\Delta)^s u(x)\geq c(n,s) u^2(x)$ where $c(n,s)>0$ is constant dependent on $n$ and $s.$ Any references/hints on how to do this will be much appreciated.
|
https://mathoverflow.net/users/68232
|
Explicit expression for the fractional Laplacian of $1/(1+|x|^2)^s$
|
Unless I am making a typo, the result is:
$$ 2^{2s} \frac{\Gamma(\tfrac n2+s) \Gamma(2s)}{\Gamma(\tfrac n2)^2} {\_2F\_1}(\tfrac n2+s, 2s, \tfrac n2, -|x|^2), $$
where $n$ is the dimension and ${\_2F\_1}$ is the Gauss's hypergeometric function. See Table 1 on page 168 in my survey [1], or Corollary 2 in the original paper [2].
References:
[1] Mateusz Kwaśnicki, *Fractional Laplace Operator and its Properties*, in: A. Kochubei, Y. Luchko, *Handbook of Fractional Calculus with Applications. Volume 1: Basic Theory*, De Gruyter Reference, De Gruyter, Berlin, 2019, <https://doi.org/10.1515/9783110571622-007>
[2] Bartłomiej Dyda, Alexey Kuznetsov, Mateusz Kwaśnicki, *Fractional Laplace operator and Meijer G-function*, Constructive Approx. 45(3) (2017): 427–448, <https://doi.org/10.1007/s00365-016-9336-4>
---
*Edit:* The hypergeometric function is continuous on $(-\infty, 1)$, and — as described, for example, in [this DLMF entry](https://dlmf.nist.gov/15.12#i) — we have
$$ {\_2F\_1}(a, b, c, -|x|^2) = \frac{p}{|x|^{2a}} \, {\_2F\_1}(a', b', c', |x|^{-2}) + \frac{q}{|x|^{2b}} \, {\_2F\_1}(a'', b'', c'', |x|^{-2}) $$
for some (explicit) constants $p, q, a', b', c', a'', b'', c''$. In particular,
$$ {\_2F\_1}(a, b, c, -|x|^2) \sim \frac{p}{|x|^{2a}} + \frac{q}{|x|^{2b}} $$
as $|x| \to \infty$. Therefore, as long as $n > 4s$, we have
$$ (-\Delta)^s [(1 + |x|^2)^{-2s}] \sim C\_{n,s} |x|^{-4s} $$
as $|x| \to \infty$, for some (explicit) constant $C\_{n,s}$. Obviously, $(-\Delta)^s [(1 + |x|^2)^{-2s}]$ is continuous, so we immediatiely get
$$ |(-\Delta)^s [(1 + |x|^2)^{-2s}]| \leqslant C\_{n,s}' (1 + |x|^2)^{-2s} . $$
In order to get a similar lower bound, one needs to check that $C\_{n,s} > 0$ (which seems to be the case when $n > 4s$, as long as I did not make any mistake), and that $(-\Delta)^s [(1 + |x|^2)^{-2s}]$ is never zero. I am no expert here, but [another DLMF entry](https://dlmf.nist.gov/15.13) seems to suggest that the total number of zeroes of the hypergeometric function is $$\lfloor s \rfloor + \tfrac12 + \tfrac12 \operatorname{sign}(\Gamma(-s) \Gamma(2s) \Gamma(\tfrac n2+s) \Gamma(\tfrac n2-2s)) = 0 ,$$ as long as, again, $n > 4s$.
(Double check for typos.)
|
4
|
https://mathoverflow.net/users/108637
|
398382
| 164,406 |
https://mathoverflow.net/questions/398278
|
10
|
It is well-known that the free symmetric monoidal category on one object is the [category $\mathbb{F}$ of finite sets and bijections](https://ncatlab.org/nlab/show/FinSet). This is supposed to be the categorification of the monoid of natural numbers, and its algebraic $K$-theory is given by the sphere spectrum $\mathbb{S}$, the free symmetric monoidal $\infty$-groupoid *with inverses*.
Is there a natural description of the free symmetric monoidal $\infty$-groupoid (resp. $\infty$-category, $(\infty,\infty)$-category) on one object?
|
https://mathoverflow.net/users/130058
|
What is the free symmetric monoidal $\infty$-category on one object?
|
Yes, it is the same as $\mathbb{F}$.
As John Baez points out, it is the same as the free symmetric monoidal $\infty$-groupoid on one object. (This can also be seen by playing around with the adjoints between groupoids and categories).
Symmetric monoidal $\infty$-groupoids are the same as $E\_\infty$-spaces (careful: not assumed to be "grouplike"). So we want the free $E\_\infty$-space on a point. As a space this is homotopy equivalent to $\sqcup\_{n\geq0} B \Sigma\_n$. Note all the hom spaces are actually 1-types, so it can be modeled as a 1-groupoid, namely $\mathbb{F}$. It might seem surprising at first that the free symmetric monoidal $\infty$-category is actually an ordinary category. However on reflection we see that it follows from the fact that the spaces in the $E\_\infty$-operad are actually 1-types.
Equivalently, as John says, it is the space of all finite subsets of $\mathbb{R}^\infty$. If you pick an identification of $\mathbb{R}^\infty$ with the "interior" of an infinite cube, then you get a natural $E\_\infty$ structure on this model of the space, too.
You can also see that it is $\mathbb{F}$ in the way that Denis suggests in his comment to the OP.
|
17
|
https://mathoverflow.net/users/184
|
398384
| 164,407 |
https://mathoverflow.net/questions/398265
|
5
|
Picard $n$-groupoids are expected to model stable homotopy $n$-types. So far this has been proved for $n=1$ in
>
> Niles Johnson, Angélica M. Osorno, *Modeling stable one-types*. Theory Appl. Categ. 26 (2012), No. 20, 520–537; [arXiv:1201.2686](https://arxiv.org/abs/1201.2686).
>
>
>
and for $n=2$ in
>
> Nick Gurski, Niles Johnson, Angélica M. Osorno, *The $2$-dimensional stable homotopy hypothesis*. Journal of Pure and Applied Algebra, Volume 223, Issue 10, 2019, Pages 4348-4383. [arXiv:1712.07218](https://arxiv.org/abs/1712.07218).
>
>
>
In particular, for $n=1$, Johnson and Osorno have also shown that the free Picard groupoid on one object $\mathbb{S}$ models the $1$-truncation of the sphere spectrum. In detail, this is the Picard groupoid where
* The objects of $\mathbb{S}$ are the integers;
* For $m,n\in\mathrm{Obj}(\mathbb{S})$, we have
$$
\mathrm{Hom}\_{\mathbb{S}}(m,n)
\overset{\mathrm{def}}{=}
\begin{cases}
\emptyset &\text{if $m\neq n$,}\\
\mathbb{Z}\_{2} &\text{if $m=n$;}
\end{cases}
$$
* The monoidal structure $\oplus$ on $\mathbb{S}$ is given by addition of integers;
* The symmetry of $\mathbb{S}$ at $(A,B)$ is the morphism $\beta^{\mathbb{S},\oplus}\_{m,n}\colon m\oplus n\to n\oplus m$ defined by
$$
\beta^{\mathbb{S},\oplus}\_{m,n}
\overset{\mathrm{def}}{=}
\begin{cases}
0 &\text{if $mn$ is even,}\\
\eta\_{m+n} &\text{if $mn$ is odd,}
\end{cases}
$$
where $\eta\_{m+n}$ is the unique non-zero element of $\mathrm{Hom}\_{\mathbb{S}}(n,n)$.
In section 3 of the same paper, they also show that there is a symmetric monoidal functor
$$\xi\colon\mathbb{F}\longrightarrow\mathbb{S}$$
from the symmetric monoidal category of finite sets and bijections $\mathbb{F}$, the categorification of the monoid of natural numbers $\mathbb{N}$, defined on objects by the inclusion $\mathbb{N}\hookrightarrow\mathbb{Z}$ and on morphisms by the sign map $\mathrm{sgn}\colon\Sigma\_{n}\to\mathbb{Z}\_{2}$.
**Questions.** Here are some questions around these topics:
* Free Picard $n$-groupoids on one-object are supposed to model the $n$-truncations of the sphere spectrum. Is there some sense in which free symmetric monoidal $n$-categories on one object model the $n$-truncations of the "directed sphere spectrum"?
* The zeroth and first truncations of the sphere spectrum are given respectively by the abelian group of integers $\mathbb{Z}$ and by the Picard groupoid $\mathbb{S}$ above. What are the second and third truncations of the sphere spectrum, or, equivalently, how can we explicit describe the free Picard $2$- and $3$-groupoids on one object?
* The free symmetric monoidal $n$-categories on one object for $n=0,1,2$ are given by the commutative monoid $\mathbb{N}$ of natural numbers, the symmetric monoidal category $\mathbb{F}$, and the discrete monoidal bicategory on $\mathbb{F}$ (see [arXiv:1210.1174](https://arxiv.org/abs/1210.1174), Corollary 1.11). What is an explicit description of the free symmetric monoidal tricategory on one object?
|
https://mathoverflow.net/users/130058
|
Categorical models for truncations of the sphere spectrum
|
I don't understand what you mean about the "directed sphere" so will focus on the other questions.
The free Picard $n$-category on one object has a description as a bordism $n$-category. Specifically it has:
* Objects are stably framed 0-manifolds;
* 1-Morphisms are stably framed 1-dimensional bordisms;
* 2-morphisms are stably framed 2-dimensional bordism between bordisms;
etc
until level $n$ where we take equivalence classes of stably framed $n$-dimensional bordisms between bordisms between ... and the equivalence relation is up to one more layer of stable bordism.
Using the Pontyragin-Thom construction, one can see that this is just the same as the fundamental $n$-groupoid of $\Omega^\infty S^\infty$.
The free (non-Picard) symmetric monoidal $n$-category is always just $\mathbb{F}$ when $n\geq 1$. This is explained in my answer to you other recent question. <https://mathoverflow.net/a/398384/184>
|
7
|
https://mathoverflow.net/users/184
|
398385
| 164,408 |
https://mathoverflow.net/questions/397900
|
1
|
Let $\log(1),\log(2),\log(3),\log(4)...\log(n)$ be approximated by fractions generated by the truncated sums:
$k=0$
$c=1$
$$\text{log1}=\sum\_{n=0}^k \frac{0}{(1 n+1)^c}=0$$
$$\text{log2}=\sum \_{n=0}^k \left(\frac{1}{(2 n+1)^c}-\frac{1}{(2 n+2)^c}\right)=\frac{1}{2}$$
$$\text{log3}=\sum \_{n=0}^k \left(\frac{1}{(3 n+1)^c}+\frac{1}{(3 n+2)^c}-\frac{2}{(3 n+3)^c}\right)=\frac{5}{6}$$
$$\text{log4}=\sum \_{n=0}^k \left(\frac{1}{(4 n+1)^c}+\frac{1}{(4 n+2)^c}+\frac{1}{(4 n+3)^c}-\frac{3}{(4 n+4)^c}\right)=\frac{13}{12}$$
Letting $k \rightarrow \infty$ makes the sums above converge to the precise values of $\log(n)$ by their Dirichlet generating function:
$$\log(n) = \lim\limits\_{s \rightarrow 1} \zeta(s)\left(1 - \frac{1}{n^{(s - 1)}}\right)$$
The alternating series which is valid in the critical strip can be written as, and is set to zero:
$$\frac{1}{\left(e^{\text{log1}}\right)^s}-\frac{1}{\left(e^{\text{log2}}\right)^s}+\frac{1}{\left(e^{\text{log3}}\right)^s}-\frac{1}{\left(e^{\text{log4}}\right)^s}+...(-1)^{(n+1)}\frac{1}{\left(e^{\log(n)}\right)^s}=0$$
The truncated alternating series:
$$\sum\_{n=1}^{n=4}(-1)^{(n+1)}\frac{1}{\left(e^{\log(n)}\right)^s}=0$$
is solved for $s$ in Mathematica 8.0.1 with the program:
```
(*start*)
Clear[log1, log2, log3, log4, n, k, s];
c = 1;
k = 0;
log1 = Sum[0/(1*n + 1)^c, {n, 0, k}];
log2 = Sum[1/(2*n + 1)^c - 1/(2*n + 2)^c, {n, 0, k}];
log3 = Sum[1/(3*n + 1)^c + 1/(3*n + 2)^c - 2/(3*n + 3)^c, {n, 0, k}];
log4 = Sum[
1/(4*n + 1)^c + 1/(4*n + 2)^c + 1/(4*n + 3)^c - 3/(4*n + 4)^c, {n,
0, k}];
$MaxRootDegree = 1000;
Last[Solve[
1/(E^(log1))^s - 1/(E^(log2))^s + 1/(E^(log3))^s -
1/(E^(log4))^s == 0, s]];
FullSimplify[%]
(*end*)
```
which gives the output:
$$\left\{s\to -4 i \pi +\log \left(\text{Root}\left[\text{$\#$1}^{10}-3 \text{$\#$1}^9+3 \text{$\#$1}^8+23 \text{$\#$1}^7+40 \text{$\#$1}^6 \\-2 \text{$\#$1}^5+42 \text{$\#$1}^4+12 \text{$\#$1}^2+1\&,10\right]\right)\right\} \label{1}\tag{$\*$}$$
which has the form:
$$\left\{s\to -\text{integer } i \pi +\log \left(\text{polynomial root}\right)\right\}$$
Now the following number also has a similar form:
$$7 \pi -\text{Log}\left[\frac{7}{2} e^{-7 \pi /2}+\frac{5}{2} e^{-5 \pi /2}+\frac{3}{2} e^{-3 \pi /2}+e^{5 \pi /2}+2 \pi \right] = \\ 14.1347251415...$$
$$14.1347251417...$$
with the value of the actual first Riemann zeta zero `Im[ZetaZero[1]]` right below it for comparison.
This can't be all coincidence.
**Question:**
>
> How does Mathematica arrive at the output in \eqref{1}$?
>
>
>
In the edit at 19.7.2021 in <https://mathematica.stackexchange.com/q/63541/328>
I made a guess at the solution for a simpler case which is not the alternating series, but one that also has $1/(e^{\text{fraction}})^s$ as its terms.
|
https://mathoverflow.net/users/25104
|
Are the Riemann zeta zeros of the form $-\text{integer } i \pi +\log \left(\text{polynomial root}\right)$?
|
The actual question asked here is a *Mathematica* question, so this is not really the right site for it, but here goes. Suppressing the extraneous notation in $c$ and $k$, the expression in $s$ is
$$
-e^{-13 s/12}+e^{-5 s/6}-e^{-s/2}+1
$$
which is a polynomial of degree 13 in $x=e^{-s/12}$:
$$
- x^{13} + x^{10} - x^6+1=-(x-1)(x^2+x+1)(x^{10}+x^3+1)
$$
The Mathematica code 'Solve' tries to find roots of this polynomial in $x$, and the $\log$ expressions will then solve for $s$. (I suspect the OP is showing output from a different choice of $c$ and/or $k$, because already I'm seeing something different.). The first two factors are solvable by radicals of course. The OP has asked for the 'last' solution. To avoid introducing new variables which may conflict with user defined variables, it is merely expressed as Root[1#^10+1#^3+1&,10]. Here 1# is just the variable name (Mathematica can do this in more than one variable - the others would be 2#, 3#, etc.). The & at the end is just a way of saying'Think of this as a function of 1#' - a 'pure function' in the language of the *Mathematica* documentation . The 10 indicated the last root.
---
Addendum: the -integer $\pi i$ in the title is misleading of course. the questions is not about the zeros of $\zeta(s)$, but the imaginary parts of the zeros, which are real numbers. The $\exp(\text{integer}\pi i)$ is just $\pm 1$, and Mathematica is just trying to give the most general form of the solution when one uses Log to solve for $s$.
|
6
|
https://mathoverflow.net/users/6756
|
398389
| 164,409 |
https://mathoverflow.net/questions/398388
|
1
|
The classification of finite simple groups has been called one of the great intellectual achievements of humanity, but I don't even know one single application of it. Even worse, I know a lot of applications of simple *modules* over some ring/algebra $A$, but I can barely know an application of them for finite simple groups. When studying modules, one has, for example,
1. If $S$ and $T$ are distinct simple modules, then $\operatorname{Hom}(S,T) = 0$, and one can enhance this using Jordan-Holder to prove that, if $M$ and $N$ are modules whose Jordan-Holder decomposition don't have common factors, then $\operatorname{Hom}(M,N)=0$. We may use this, for example, to try to compute some cohomology, also;
2. The simple modules form a basis of the $K\_0$ group, and therefore if we're interested in, for example, the multiplicative structure of $K\_0$ it's enough to compute the (tensor) product of simple modules;
3. If the algebra $A$ is basic (i.e. every simple representation is $1$-dimensional), which happens for path algebras, then simple modules have a group structure with respect to the tensor product (so they are an analogue for the Picard group).
For finite simple groups, the only application I know is for the (non)-solubility of polynomials, and it's a quite particular example which uses only $S\_n$ and $A\_n$. So I have two questions:
1. What are some (concrete) applications of (finite simple groups + Jordan-Holder) for general finite groups?
2. What are some (concrete) applications of the classifications of finite simple groups?
|
https://mathoverflow.net/users/146933
|
Why are finite simple groups useful?
|
There's an entire book on this subject, "Applying the Classification of Finite Simple Groups: A User’s Guide" by Stephen D. Smith, published through the AMS, though you can find a draft version [here](http://homepages.math.uic.edu/%7Esmiths/book.pdf).
The applications are not as simple as they are for modules, but many questions can be settled by invoking the classification of finite simple groups. For example, you can invoke the classification to in turn classifying 2-transitive groups. The only known proof of the Schreier conjecture relies on the classification. The book has many more applications.
|
9
|
https://mathoverflow.net/users/3711
|
398393
| 164,411 |
https://mathoverflow.net/questions/398263
|
5
|
Definition
==========
Call a [partition](https://en.wikipedia.org/wiki/Partition_(number_theory)) $\lambda$ of an even integer $2n$ "black-white balanced" if the following equivalent conditions are satisfied:
* In the usual (Ferrers-)[Young diagram](https://en.wikipedia.org/wiki/Young_tableau#Diagrams) of $\lambda$, if you color alternating squares black and white in a checkerboard pattern, there are an equal number of black and white squares.
* The Young diagram of $\lambda$ can be covered by dominoes.
* The partition $\lambda$ has empty 2-core.
* If you sort the parts of $\lambda$ (as is usual), then the number of odd parts in even places is equal to the number of odd parts in odd places. (A part $\lambda\_i$ is odd if the number $\lambda\_i$ itself is odd; the place is odd or even based on the parity of the index $i$.)
* If you sort the parts of $\lambda$, then $\sum\_i (-1)^i(1-(-1)^{\lambda\_i}) = 0$.
The number of black-white balanced partitions of $2n$ is [OEIS sequence A000712](http://oeis.org/A000712), which begins $1$, $2$, $5$, $10$, $20$, $36$, $65$, $110$. The name of this sequence is "Number of partitions of n into parts of 2 kinds". Alternatively, it is the convolution of the partition numbers with themselves. The description above appears on the linked page as "Also equals number of partitions of 2n in which the odd parts appear as many times in even as in odd positions."
Question
========
In other words, there appears to be a bijection between black-white balanced partitions $\lambda$ of $2n$ and ordered pairs of partitions ($\lambda\_1$, $\lambda\_2$) of a *combined* $n$ (that is, if $\lambda\_1$ is a partition of $n\_1$ and $\lambda\_2$ is a partition of $n\_2$, then $n=n\_1+n\_2$).
Can anyone describe this bijection, or provide a reference for it?
Remarks
-------
I'm fairly sure I'm just a Google search away from finding a reference, but I just haven't succeeded yet. I found [this question on Mathematics Stack Exchange](https://math.stackexchange.com/questions/345529/bijection-between-number-of-partitions-of-2n-satisfying-certain-conditions-with) and [this identical question on MathOverflow](https://mathoverflow.net/questions/125709/bijection-between-number-of-partitions-of-2n-satisfying-certain-conditions-with) asking about a bijection between *strict* (ie, distinct parts) black-white balanced partitions of $2n$ and unrestricted partitions of $n$ (not pairs thereof). The paper ["Balanced partitions" by Sam Vandervelde](http://myslu.stlawu.edu/%7Esvanderv/balance.pdf) also proves this result.
|
https://mathoverflow.net/users/1079
|
Bijection from "black-white balanced" partitions to pairs of partitions
|
[Richard Stanley](https://mathoverflow.net/users/2807/richard-stanley) and [Sam Hopkins](https://mathoverflow.net/users/25028/sam-hopkins) have answered this question in the comments. (Thanks!)
Professor Stanley mentions that this is a special case of exercise 7.59(e) in *Enumerative Combinatorics*, volume 2:
>
> 7.59(e) Let $\mu$ be a $p$-core. Let $Y\_{p,\mu}$ be the set of all partitions whose $p$-core is $\mu$. Define $\lambda \le \nu$ in $Y\_{p,\mu}$ if $\lambda$ can be obtained from $\nu$ by removing border strips of size $p$. Show that $Y\_{p,\mu} \cong Y^k$, where $Y$ denotes Young's lattice. Deduce that if $f\_{\mu}(n)$ is the number of partitions of $n$ with $p$-core $\mu$, then $$ \sum\_{n\ge 0} f\_{\mu}(n)x^n = x^{\lvert \mu\rvert} \prod\_{i\ge 1} (1-x^{pi})^{-p}. $$
>
>
>
(Presumably the $k$ in $Y^k$ is meant to be a $p$ as in $Y^p$.) The proof sketched by the exercise is very nice and uses an infinite binary sequence called the [*code* of a partition $\lambda$](https://www.google.com/books/edition/Catalan_Numbers/IRIHBwAAQBAJ?hl=en&gbpv=1&dq=%22infinite+binary+sequence%22+%22integer+partition%22&pg=PA192&printsec=frontcover). The bijection from this question can be obtained by considering every other bit from this binary sequence.
Professor Hopkins mentions that my desired bijection is known under the name "$2$-quotient". Specifically, every partition $\lambda$ has a $2$-core $\alpha$, which is a single partition, and a $2$-quotient $(\beta\_1, \beta\_2)$, which is an ordered pair of partitions. Two key facts are the following:
* $\lvert \lambda\rvert = \lvert \alpha \rvert + 2\lvert \beta\_0\rvert + 2\lvert \beta\_1\rvert. $
* The triple $(\alpha; \beta\_0, \beta\_1)$ uniquely determines $\lambda$.
In the case of this question, the $2$-core $\alpha$ is empty (by assumption), so the $2$-quotient $(\beta\_0, \beta\_1)$ is precisely the desired bijection. One way of describing the $2$-quotient is by coloring the Young diagram of $\lambda$ in a checkerboard pattern. Then, $\beta\_0$ (say) is obtained by considering the subdiagram of $\lambda$ restricted to those rows ending in a black square and those columns ending in a white square. Oppositely, $\beta\_1$ is obtained by restricting to those rows ending in a white square and those columns ending in a black square.
One reference for this is James and Kerber, "The Representation Theory of the Symmetric Group," page 83, Theorem 2.7.30.
|
3
|
https://mathoverflow.net/users/1079
|
398399
| 164,417 |
https://mathoverflow.net/questions/398386
|
3
|
Let $A$ be a pre-$C^\*$-algebra, i.e. $A$ satisfies all axioms for a $C^\*$-algebra except completeness. In other words, $A$ is an involutive algebra with a $C^\*$-norm.
We say that $x \in A$ is positive (notation: $x \ge 0$) if there exists $a\in A$ with $x = a^\*a$ and on self-adjoint elements we define the usual relation $x \le y \iff y-x \ge 0$.
Given a positive element $c \ge 0$ in $A$, and $a \in A$, does the inequality
$$a^\*ca \le \|c\| a^\*a$$
hold? For $C^\*$-algebras, this result is well-known. A quick proof is given by faithfully representing $A \subseteq B(H)$ and using $c \le \|c\|1$. Does the same result continue to hold for pre $C^\*$-algebras?
---
Context question: In the book "Hilbert $C^\*$-modules" by Lance on p4, it is claimed that all results proved so far continue to hold for inner product modules defined over a pre-$C^\*$-algebra. The above inequality is used in the proof of proposition 1.1 and I wanted to check that it still works for pre $C^\*$-algebras.
---
Bonus question: Is the sum of two positive elements again positive?
|
https://mathoverflow.net/users/216007
|
The inequality $a^*ca \le \|c\| a^*a$ in a pre-$C^*$-algebra
|
Given your definitions, the first one is an easy no. If $a = 1$ then it says $c \leq \|c\|$, which fails when $A$ is the polynomials in $C[0,1]$: let $c$ be the polynomial $x$, then $\|c\|= 1$ and $1-x$ is not of the form $p\bar{p}$ for any polynomial $p$.
For the second question, take $a = x^2 + y^2$ and $b = z^2$ in the polynomials on $[0,1]^3$. Then $a = (x + iy)(x- iy)$ and $b = z\cdot z$ are both positive, but their sum is not. (Suppose you had a polynomial $p$ with $p\bar{p} = x^2 + y^2 + z^2$. Then $p$ has to be linear in $x$, $y$, and $z$ and the lack of $xy$ and $xz$ cross terms forces the arguments of the coefficients of $y$ and $z$ to be the same or differ by $\pi$, which in either case creates a nonzero $yz$ cross term.)
But I think the usual definition of positivity in a pre C${}^\*$- algebra is that it is positive in the completion, which would make both answers trivially yes.
|
9
|
https://mathoverflow.net/users/23141
|
398410
| 164,426 |
https://mathoverflow.net/questions/398360
|
6
|
I'm interested in finding solutions a fourth order version of the standard wave equation in $d$ dimensional Minkowski spacetime $\mathcal{M}^d$. Defining $\Box := \partial\_0^2 - \sum\_{i = 1}^{d-1} \partial\_{i}^2$, I want to find solutions to $$\Box^2\Phi(x) = 0,$$ which are not also solutions to $\Box \Phi(x) = 0$, and which transform as scalars under Lorentz transformations. I'm mostly interested in solutions for $d=4$, but I don't think that's likely to be relevant at the level of analysis I'm currently working at so I want to stick to general $d$ if possible.
I know that in solving $\Box \Phi(x) = 0$ with boundary conditions that require some kind of fast enough fall off at infinity, then the solution space can be taken to be Lorentz invariant $L^2$ integrable functions. A basis of solutions for functions with these boundary conditions can be given by plane waves $\Phi\_k(x) := e^{i k\cdot x}$ where $k^2 = 0$. My understanding is that as $\Box \Phi(x) = 0$ is seperable, Sturm-Liouville theory tells us that for some choices of boundary conditions we can find a function space of solutions to the equation spanned by some basis of solutions.
(Disclaimer: I don't have a very in-depth understanding of analysis. I can see that $\Phi\_k \not\in L^2(\mathcal{M}^d)$, so I understand that at some level calling this a `basis' for the space is not correct. I'm happy to call it this given that two such functions are orthogonal in the sense that $\int\_{\mathcal{M}^d}dx \;\Phi\_k(x)\Phi\_{k'}(x) \propto \delta^d(k-k')$, and that any (I think?) function in this space can be decomposed in terms of an integral over all possible $\Phi\_k$ by the Fourier transform. I appreciate that to answer my question it may be necessary to go to some more detailed level of analysis where it doesn't make sense to think of $\Phi\_k$ as a basis.)
I note that, given any $f$ such that $\Box f(x) = 0,$ we can construct a solution to $\Box^2 \Phi(x) = 0$ by taking $\Phi(x) = a \cdot x f(x)$, where $a$ is a vector. For some choices of $a$ this new solution may also satisfy $\Box\left( a\cdot x f(x)\right) = 0,$ but I think it's always possible to choose $a$ such that $\Box\left( a\cdot x f(x)\right) \neq 0.$
Then my question is, given a function space of solutions corresponding to a choice of boundary conditions to $\Box \Phi(x) = 0$, is there a canonical way to extend this to a larger function space of soluitons to $\Box^2 \Phi(x) = 0$, which includes functions which satisfy $\Box^2 \Phi(x) = 0$ and $\Box \Phi(x) \neq 0$, which can be expanded in terms of a basis of functions which are orthogonal under some inner product? If not a canonical way, is there some set of ways of doing this?
So then in the case of $L^2$ functions, is it possible to do something like extend to functions which increase like $x$ as $x\rightarrow \infty$? Then I could imagine that this new space could be spanned by $\Phi\_k$, and $\chi\_{k,a}(x) := a \cdot x e^{i k \cdot x}$, with some inner product under which $\Phi\_k, \Phi\_{k'}$, $\chi\_{k,a}$ and $\chi\_{k',a'}$ are orthogonal to each other? I think what I've written here is probably a little naive, but I'm hoping there may be some kind of result similar to this which makes sense.
After some searching online I've found that there exists a fourth order Sturm-Liouville theory, but I wasn't able to find anything accesible for me to read, and it appears to me that $\Box^2 \Phi(x) = 0$ is anyway not seperable, at least in Cartersian coordinates. Any references to a simple introduction or review article of fourth-order Sturm-Lioville theory would be appreciated, as well as any local coordinate transformation which makes $\Box^2\Phi(x) = 0$ seperable, if you think this would be relevant to my problem.
I appreciate that you could read this question and say, 'why would you expect there to be a canonical way to extend solution spaces for $\Box \Phi(x) = 0$ to $\Box^2 \Phi(x) = 0$ with boarder boundary conditions?'. So here is some evidence that I have from solving the equation in 2D.
In 2D in lightcone coordinates $u = x^0 + x^1, v = x^0-x^1$ the wave operator factorises so that $\Box = \partial\_u\partial\_v$. This makes it simple to write down d'Alembert's general solution to $\Box\Phi(x) = 0$ for all possible boudary conditions, $\Phi(x) = f^+(x^0 + x^1) + f^-(x^0-x^1).$
It's also simple to solve $\Box^2\Phi(x) = 0$ in this case. Writing a seperable ansatz $\Phi(u,v) = U(u)V(v)$ the equation becomes $U'' V''=0$, which is solved non-trivially by either $U' = 0$, $V' = 0$, $U''=0$ or $V''=0$. Then the general solution is given by $\Phi(u,v) = f^+\_1(u) + v f^+\_2(u) + f^-\_1(v) + u f^-\_2(v).$ Introducing null vectors $k\_+ = k(1,1)$ and $k\_- = q(1,-1)$, then we see that $u \propto k\_+ \cdot x$, and $v \propto k\_- \cdot x.$ Then the general solution can be written as a sum over solutions of the form $$\Phi(x) = f\_1(k\cdot x) + a \cdot x f\_2(k \cdot x),$$ where $k^2 = 0$ and $a$ is any vector. If $a \propto k$ then the second term solves $\Box\Phi(x) = 0$, otherwise it produces new solutions that satisfy $\Box^2\Phi(x) = 0$ and $\Box\Phi(x) \neq 0$. (I took a linear combination of the lightcone coordinate solutions to produce the new solution). I believe this to be the general solution to $\Box^2\Phi(x) = 0$ in 2D for all possible boundary conditions, and I've intentionally written it in a form that extends in a natural way to general dimension $d$. So this is some kind of motivation for why I think my question may have a solution; it appears to me that what I'm asking in general dimensions happens in 2D based on the form of the general solution I gave here.
|
https://mathoverflow.net/users/171026
|
Space of solutions to a fourth order wave equation
|
You talk about the non-separability of the $\Box^2 \phi = 0$ equation, which I don't understand. Each plane wave $e^{ik\cdot x} = \prod\_{j=0}^{d-1} e^{i k\_j x^j}$ is already in separated form with the components of the null vector $k=(k\_j)$ playing the role of the separation constants.
But rather than get into technicalities about what does or does not constitute a separable equation or a basis of separated solutions, why not just solve the equation via the Fourier transform?
Recall first the ordinary wave equation. If $\phi(x)$ is smooth and grows no faster than polynomially in any direction, its Fourier transform $\hat{\phi}(k)$ exists as a distribution (a subclass of [tempered distributions](https://en.wikipedia.org/wiki/Distribution_(mathematics)#Tempered_distribution) in this case). The equation $\Box \phi(x) = 0$ translates to $k^2 \hat{\phi}(k) = 0$, where of course $k^2 = k\_0^2 - \sum\_{j=1}^{d-1} k\_j^2$. No non-vanishing continuous, let alone smooth function of $k$ can satisfy that condition. Hence, $\hat{\phi}(k)$ must be a distribution. A natural candidate is $\hat{\phi}(k) = f(k) \delta(k^2)$, with $f(k)$ an at least continuous function, because of the simple identity $u \delta(u) = 0$. The function $f(k)$ is of course not unique, any functions $f\_1,f\_2$ such that the difference $f\_1(k)-f\_2(k)$ vanishes smoothly on the locus of $k^2$ define the same $\hat{\phi}(k)$. The level of regularity (including behavior at infinity) of $f(k)$ translates in a certain way to the level of regularity of $\phi(x)$ via the Fourier transform. The above parametrization of $\hat{\phi}(k)$ is actually exhaustive when these regularity classes are appropriately fixed. To see how the data in $f(k)$ translates to initial data for $\phi(x)$, try the $d=1$ example.
Now, on to the squared wave equation. Under the Fourier transform, $\Box^2 \phi(x) = 0$ translates to $(k^2) \hat{\phi}(k) = 0$. Taking derivatives of the $\delta$-function identity, we can get $u \delta'(u) + \delta(u) = 0$ or more importantly $u^2 \delta'(u)=0$. Hence, a natural candidate for a solution is
\begin{align\*}
\hat{\phi}(k) &= f(k) \delta(k^2) + g(k) \delta'(k^2) , \\
&= F(k) \delta(k^2)
+ ia\cdot\partial\_k [G(k) \delta(k^2)]
\end{align\*}
for functions $f(k), g(k)$ of appropriate regularity and $G(k) = g(k)/(2ia\cdot k)$, $F(k) = f(k) - (a\cdot\partial\_k) G(k)$. To make it easier to go between $g(k)$ and $G(k)$, the vector $a$ should not be null, so that $a\cdot k$ only vanishes at $k=0$. The corresponding formula in real space is
$$
\phi(x) = \phi\_1(x) + (a\cdot x) \phi\_2(x) ,
$$
where $\phi\_1(x), \phi\_2(x)$ are independent solutions of the wave equation, $\Box \phi\_{1,2}(x)=0$. This is I think the parametrization of solutions that you were looking for. As before, it is a matter of figuring out the right regularity classes for $F(k), G(k)$ and the corresponding initial data for $\phi(x)$ to make sure that the above parametrization is exhaustive.
|
3
|
https://mathoverflow.net/users/2622
|
398411
| 164,427 |
https://mathoverflow.net/questions/398436
|
2
|
The intersection of (countably many) 'spheres' in a Hilbert space can be non-empty. If we make this situation moving real analytically, the mid points and the radii, can it happen that the intersection becomes empty while it is not empty on an open set of the parameter space?
Precisely:
Let $H$ be a (separable) complex Hilbert space and $ z\_i:(0,2)\to H$ and $r\_i:(0,2)\to (\delta,1)$ be **real analytic maps** for $i=1,2,...$, where $\delta>0$. Furthermore $z\_i(t)\not=z\_j(t)$ for all $t$ and $i\not =j$. The set $M\_t=\{w\in H\ \vert\
\langle w,z\_i(t) \rangle =r\_i(t)\ \forall i\}$ **is assumed to be** not empty for $t\in (0,1)$. Is it possible that $M\_1$ *is* empty?
Remark:
I am aware that the sets I call 'spheres' in fact are not spheres. I couldn't think of a better word.
Thanks for any hint!
|
https://mathoverflow.net/users/109905
|
Intersection of 'spheres' in Hilbert space with respect to real analytically moving mid points
|
**This is my *new* answer for the edited question**
Here is a counterexample. Let $H = \mathbb{C}^2$ and let $z\_1(t) = (1,t)$ and $z\_i(t) = (2,2)$ for all $i > 1$ and for all $t$. Also let $r\_i(t) = 1$ for all $i$ and $t$.
Then $M\_t$ is the set of points $(z,w) \in \mathbb{C}^2$ such that $z + tw = 1$ and $2z + 2w = 1$.
As long as $t \neq 1$, there is a unique point $(z,w)$ which satisfies this system of linear equations (i.e. there is one point in $M\_t$, so it is not empty).
However, when $t = 1$ the system becomes
$$
z + w = 1 = 2 (z + w),
$$
which clearly has no solutions, so $M\_1$ is empty.
For what it's worth, the 'spheres' you talk of are simply hyperplanes in Hilbert space (affine spaces of codimension 1).
---
**This is my *old* answer to this question, before it was edited**
If I'm interpreting your question right, then yes it is possible for $M\_1$ to be empty (and in fact it is possible for $M\_t$ to be empty for all $t \in (0,1)$ as well).
For example, consider $H = \mathbb{C}$, $z\_i(t) = i$ and $r\_i(t) = c$ for all $i$ and $t$, where $c$ is some number in $(\delta, 1)$.
Then $M\_t$ is the set of points $w \in \mathbb{C}$ such that $w i = c$ for all $i = 1, 2, \dotsc$, and of course there are no such $w$.
|
1
|
https://mathoverflow.net/users/175976
|
398439
| 164,438 |
https://mathoverflow.net/questions/386213
|
13
|
>
> It should be the case that, in some appropriate sense
> $$\pi (x)\sim \operatorname{Ri}(x)-\sum\_{\rho}\operatorname{Ri}(x^{\rho}) \tag\*{(4)}$$
> with $\operatorname{Ri}$ denoting the *Riemann function* defined:
> $$\operatorname{Ri}(x)=\sum\_{m=1}^\infty \frac{\mu (m)}{m}\operatorname{li}\left(x^{\frac{1}{m}}\right). \tag\*{(5)}$$
> This relation $(4)$ has been called "exact" [in Ribenboim's *The New Book of Prime Number Records*], yet we could not locate a proof in the literature; such a proof should be nontrivial, as the conditionally convergent series involved are problematic. In any case relation $(4)$ is quite accurate, and furthermore the Riemann function $\operatorname{Ri}$ can be calculated efficiently (...) The sum in $(4)$ over **critical zeros** is not absolutely convergent, and furthermore the phases of the summands interact in a frightfully complicated way.
>
>
>
—from *Journal of Computational and Applied Mathematics by Borwein et al.*
>
> Of profound importance, Bernhard Riemann proved that the prime-counting function is exactly
> $$\pi (x)=\operatorname{R}(x)-\sum\_{\rho}\operatorname{R}(x^{\rho})$$
> where
> $$\operatorname{R}(x)=\sum\_{n=1}^\infty \frac{\mu (n)}{n}\operatorname{li}\left(x^{\frac{1}{n}}\right),$$
> (...) $\rho$ indexes **every zero** of the Riemann zeta function, and $\operatorname{li}\left(x^{\frac{\rho}{n}}\right)$ is not evaluated with a branch cut but instead considered as $\operatorname{Ei}\left(\frac{\rho}{n}\ln x\right)$. Equivalently, if the trivial zeros are collected and the sum is taken **only over the non-trivial zeros** $\rho$ of the Riemann zeta function, then $\pi (x)$ may be written
> $$\pi (x)=\operatorname{R}(x)-\sum\_{\rho}\operatorname{R}(x^{\rho})-\frac{1}{\ln x}+\frac{1}{\pi}\arctan\frac{\pi}{\ln x}.$$
>
>
>
—from *Wikipedia's Prime counting function article* (before 30/7/2021)
**Questions:**
1. According to Borwein and Ribenboim, the index in $(4)$ should run only over **non-trivial zeros**. According to Wikipedia, the index in $(4)$ should run over **all zeros**. Wikipedia states that if the sum runs only over non-trivial zeros, then we add the $\ln$ and $\arctan$ terms, which is even more confusing. So what's true?
2. I'm pretty sure that Riemann did not prove the formula $(4)$. He only **proposed** a "weaker" form of it, namely
$$\pi (x)=\sum\_{m=1}^\infty \frac{\mu (m)}{m}J\left(x^{\frac{1}{m}}\right)$$
where
$$J(x)=\operatorname{li}(x)-\sum\_{\rho}\operatorname{li}\left(x^{\rho}\right)+\int\_x^\infty \frac{dt}{t(t^2-1)\ln t}-\ln 2$$
and where $\rho$ runs over all non-trivial zeros. The formula was proven by Mangoldt, not Riemann. The Wikipedia article is wrong in that historical fact, isn't it?
3. Even though the proof of $(4)$ is nowhere to be found in the literature, Raymond Manzoni provided a partial proof [here](https://math.stackexchange.com/questions/269997/two-representations-of-the-prime-counting-function/282848#282848). I call it partial because it is unknown whether the series converges at all: How could that be settled down?
*Note:* When I refer to the formula $(4)$ in this question, I assume $=$ instead of $\sim$.
**Riesel and Gohl**
Riesel and Gohl show in *Some Calculations Related to Riemann's Prime Number Formula* that
$$\sum\_{n=1}^N \frac{\mu (n)}{n}\left(\int\_{x^{\frac{1}{n}}}^\infty \frac{\mathrm dt}{t(t^2-1)\ln t}-\ln 2\right)=\frac{1}{2\ln x}\sum\_{n=1}^N \mu (n)+\frac{1}{\pi}\arctan\frac{\pi}{\ln x}+\epsilon (x,N)$$
where
$$\epsilon (x,N)=-\sum\_{n=N+1}^\infty \frac{\mu (n)}{n}\left(\frac{1}{2}\ln\ln x+C\right)+\frac{1}{2}\sum\_{n=N+1}^\infty \frac{\mu (n)\ln n}{n}+\sum\_{n=N+1}^\infty \frac{\mu (n)}{n}\sum\_{k=1}^\infty \frac{1}{\pi k}\arctan\frac{\ln x}{n\pi k}$$
where
$$C=\sum\_{k=1}^\infty \frac{1}{\pi k}\arctan\frac{1}{\pi k}+\int\_1^\infty \frac{\mathrm du}{u(e^{2u}-1)}-\ln 2+\frac{1}{2}.$$
Now, $\epsilon\to 0$ as $N\to\infty$.
If $\sum\_{n=1}^\infty \mu (n)=-2$, then
$$\sum\_{n=1}^\infty \frac{\mu (n)}{n}\int\_{x^{\frac{1}{n}}}^\infty \frac{\mathrm dt}{t(t^2-1)\ln t}=\frac{1}{\pi}\arctan\frac{\pi}{\ln x}-\frac{1}{\ln x}.$$
But this uses the zeta-regularized result $\sum\_{n=1}^\infty \mu (n)=-2$. It is one way of assigning a finite value to a divergent sum, obtained by just plugging $s=0$ in
$$\frac{1}{\zeta (s)}=\sum\_{n=1}^\infty \frac{\mu (n)}{n^s}.$$
In general, this assigning of values is arbitrary. There are lots of ways to do that. I don't understand how can this arbitrary choice produce an "empirically-correct" result about the distribution of prime numbers. In fact, the paper of Riesel and Gohl implies the mysterious formula
$$\pi (x)= \operatorname{R}(x)-\sum\_{\rho}\operatorname{R}(x^{\rho})-\frac{1}{\ln x}+\frac{1}{\pi}\arctan\frac{\pi}{\ln x}.$$
For the empirical results, see [here](http://www.primefan.ru/stuff/primes/table.html#theory).
**Questions [continued]:**
4. What makes zeta-regularization (or other regularizations producing the same value) the only regularization that seems to produce the "empirically-correct" result? Or is it the case that the result is ultimately wrong?
*Note*: $\operatorname{li}x$ is to be interpreted as the Cauchy principal value of $\operatorname{Ei}\ln x$.
This question has been on [MSE](https://math.stackexchange.com/questions/3615529/is-pi-x-operatornamerx-sum-rho-operatornamerx-rho-correct-a) for some time, but it still doesn't have any answers, so I decided to post it here.
**Edit:**
The Wikipedia formula (before 30/7/2021)
$$\pi (x)= \operatorname{R}(x)-\sum\_{\rho}\operatorname{R}(x^{\rho})-\frac{1}{\ln x}+\frac{1}{\pi}\arctan\frac{\pi}{\ln x}$$
turned out to be a misinterpretation of Riesel and Gohl.
|
https://mathoverflow.net/users/175751
|
Is $\pi (x)=\operatorname{R}(x)-\sum_{\rho}\operatorname{R}(x^{\rho})$ correct at all?
|
*Preliminaries.* We will denote the slight modifications of the prime counting function and the prime power counting function with $\pi^\*$ and $J^\*$ respectively, which assume a halfway step at discontinuities. Furthermore we will denote the non-trivial zeros of the zeta function with $\rho$ and all zeros i.e. the trivial and non-trivial zeros with $\varrho$ in questions two to four.
Q1. It depends on what the meaning of $\sim$ in the statement $\pi^\* (x)\sim \operatorname{Ri}(x)-\sum\_{\rho}\operatorname{Ri}(x^{\rho})$ is. Putting convergence questions aside, which we will discuss in Q3, we indeed have an equality if the sum runs over all zeros of $\zeta$. If we omit the trivial zeros, the statement might still be correct, if $\sim$ denotes asymptotic equivalence since $\sum\_n\operatorname{Ri}(x^{-2n})=o(1).$ This is because
$$\left\lvert\lim\_{x\to\infty}\sum\_n\sum\_m \frac{\mu(m)}{m}\operatorname{li}(x^{-2n/m})\right\rvert \leqslant \lim\_{x\to\infty}\sum\_n\sum\_m \frac{1}{m}\left\lvert\operatorname{li}(x^{-2n/m})\right\rvert\overset{\text{(Tonelli)}}{=}0.$$
Q2. "Proposed" might not be the best fitting word. Riemann showed and proved essential ideas of his, while leaving out certain steps, which caused a certain lack of rigor. Hence, Riemann's proof was not complete. Quoting Edwards, p.38, 1.19 QUESTIONS UNRESOLVED BY RIEMANN:
>
> Riemann evidently believed that he had given a proof of the product formula for $\zeta(s)$, but, at least from the reading of the paper given above, one cannot consider his proof to be complete, and, in particular, one must question Riemann's estimate of the number of roots $\rho$ in the range $\{0 < \operatorname{Im} \rho < T\}$ on which this proof is based. It was not until 1893 that Hadamard proved the product formula, and not until 1905 that von Mangoldt proved the estimate of the number of roots in $\{0 < \operatorname{Im} \rho < T\}$.
>
>
>
Q3. We want to prove that $\pi^\*(x)=\operatorname{R}(x)-\sum\_{\varrho} \operatorname{R}(x^{\varrho})$ follows from $\pi^\*(x)=\sum\_{m=1}^{\infty}\frac{\mu(m)}{m}J^\*(x^{1/m})$. Define $\pi\_X^\*\colon \mathbb R\_{>1}\to\mathbb N$ as $$x\mapsto\sum\_{m\leqslant X}\frac{\mu(m)}{m}J^\*(x^{1/m}),$$
where $X>1$. If $X>\log\_2(x)$, then $\pi\_X^\*(x)=\pi^\*(x)$, since $J^\*(x)=0$ for $x<2$. After applying Möbius inversion on $J^\*$ we get that for all $X>\log\_2(x)$
$$\pi^\*(x)=\sum\_{m\leqslant X}\frac{\mu(m)}{m}\operatorname{li}(x^{1/m})-\sum\_{\varrho}\sum\_{m\leqslant X}\frac{\mu(m)}{m}\operatorname{li}(x^{\varrho/m})-\log 2\sum\_{m\leqslant X}\frac{\mu(m)}{m}.$$
We were able to change the order of summation involving the conditional convergent series since the series over $m$ is finite. Now the idea is not to look at the case $X\to\infty$ separately. There exists an $N\in\mathbb N$ for every $X>1$ so that $\pi^\*\_X(x)=\pi^\*\_N(x)$. The sequence $(\pi\_N^\*)\_{N\in\mathbb N}$ converges since it is Cauchy.
Q4. Riesel's and Göhl's objective was to find an approximation and not an exact form of the last two terms of $\pi\_N(x)$, $\sum\_{n=1}^N \frac{\mu (n)}{n}\left(\int\_{x^{\frac{1}{n}}}^\infty \frac{\mathrm dt}{t(t^2-1)\ln t}-\ln 2\right)$. The integral is equivalent to $\sum\_m\operatorname{li}(x^{-2n/m})$. They eventually showed that it is equal to $\frac{1}{2\ln x}M(N)+\frac{1}{\pi}\arctan\frac{\pi}{\ln x}+\epsilon (x,N),$ where $M$ denotes the Mertens function and $\epsilon$ is an error term, which tends to zero as $N\to\infty$. In order to make the calculation more efficient they replace the Mertens function with $-2$ and argue on page 979 as follows:
>
> It is thus advantageous to choose such a value of $N$ that the sums $g\_k$ become comparatively small. It also is advantageous to have $\sum\_1^N\mu(n) = -2$ at the same time, since (32) then has the order of magnitude only $= O((\log x)^{-3})$ instead of $O((\log x)^{-1})$.
>
>
>
Here no zeta regularization was used. The triple sums in the error term are dependent on $g\_k$. There exist infinitely many $N$ such that $M(N)=-2$. This immediately follows from the bounds $\liminf~M(n)/\sqrt n < -1.009$ and $\limsup~M(n)/\sqrt n > 1.06$, which were used to disprove the Mertens conjecture.
|
11
|
https://mathoverflow.net/users/175826
|
398456
| 164,444 |
https://mathoverflow.net/questions/398461
|
0
|
Let $M$ be a monoid, and let $x\in M$. One says that $x$ is *periodic* if
$$x^{i+j}=x^j$$
for some integers $i\geq 1$ and $j\geq 0$.
An easy division algorithm argument shows that if $m$ is the smallest value of $i$ where this happens (for some $j$), and similarly $n$ is the smallest value of $j$ where this happens (for some $i$), then $x^{m+n}=x^n$. (So those minimal values work together.)
Moreover, given such an $m$ and $n$, the displayed equality holds if and only if $m|i$ and $n\leq j$.
Question 1: Is there a standard reference for these basic facts in the monoid setting?
Question 2: Is there a standard name for $m$ and $n$?
Question 3: Are there special names for the periodic property when $n=0$ or when $n=1$? (I've seen them called "torsion units" and "potents" [generalizing "idempotents"] in the ring-theoretic setting.)
|
https://mathoverflow.net/users/3199
|
More vocabulary for periodic elements in monoids
|
Question 1: See Clifford and Preston, volume 1.
Question 2: $m$ is the period, $n$ is called the index of the element. See this [Wikipedia text](https://en.wikipedia.org/wiki/Monogenic_semigroup).
Question 3: If $n=1$, the element is called a group element of finite order. If $n=0$, it is called a unit of finite order. Or you can just call it an element of index 1 (resp. 0).
|
2
|
https://mathoverflow.net/users/157261
|
398462
| 164,446 |
https://mathoverflow.net/questions/398363
|
10
|
I want to prove that weak descent of a $1$-category implies the classical Giraud axioms.
More precisely, let $\mathsf{C}$ be a cocomplete, finitely complete $1$-category. We say that $\mathsf{C}$ satisfies weak descent if the following conditions are satisfied:
* $(\mathbf{D1}a)$-(Universal coproducts): Given a collection of objects $\{ Y\_i \}\_{i \in I}$, let $Y = \coprod\_i Y\_i$. Let $f: X \to Y$ be a morphism, and let $X\_i = Y\_i \times\_Y X$. Then the induced map $\coprod\_i X\_i \to X$ is an isomorphism,
* $(\mathbf{D1}b)$-(Universal pushouts): Given a span $Y\_0 \leftarrow Y\_1 \to Y\_2$, let $Y = Y\_0 \coprod\_{Y\_1} Y\_2$. Let $f: X \to Y$ be a morphism and let $X\_i = Y\_i \times\_{Y} X$. Then the induced map $X\_0 \coprod\_{X\_1} X\_2 \to X$ is an isomorphism.
* $(\mathbf{D2}a)$-(Effective coproducts): Given a collection of maps $\{ f\_i: X\_i \to Y\_i \}$, let $X = \coprod\_i X\_i$, and $Y = \coprod\_i Y\_i$, and let $f: X \to Y$ be the coproduct $\coprod\_i f\_i$. Then the natural maps $X\_i \to Y\_i \times\_Y X$ are isomorphisms for each $i$.
* $(\mathbf{D2}b)$-(Weak effective pushouts): Given a map of spans:
$\require{AMScd}$
\begin{CD}
X\_0 @<<< X\_1 @>>> X\_2\\
@Vf\_0VV @Vf\_1VV @Vf\_2VV\\
Y\_0 @<<< Y\_1 @>>> Y\_2
\end{CD}
such that each square is a pullback square, let $X = X\_0 \coprod\_{X\_1} X\_2$ and $Y = Y\_0 \coprod\_{Y\_1} Y\_2$, and let $f:X \to Y$ denote the induced map of pushouts. Then the natural maps $X\_i \to Y\_i \times\_Y X$ are regular epimorphisms.
Condition $(\mathbf{D2}b)$ is the real difference between $1$-topoi and $\infty$-topoi, and I am trying to better understand this comparison. Now recall the classical Giraud Axioms:
* $(\mathbf{G1})$ Coproducts are disjoint, namely $A \times\_{A \coprod B} B \cong \varnothing$,
* $(\mathbf{G2})$ For any morphism $f: X \to Y$, the base change functor $f^\*: \mathsf{C}\_{/Y} \to \mathsf{C}\_{/X}$ preserves colimits,
* $(\mathbf{G3})$ Equivalence relations are effective.
Rezk [sketches](https://faculty.math.illinois.edu/%7Erezk/leeds-lectures-2019.pdf) how to prove that $(\mathbf{D1}) = (\mathbf{D1}a) \wedge (\mathbf{D1}b)$ is equivalent to $(\mathbf{G2})$, and that $(\mathbf{D2}a) \implies (\mathbf{G1})$.
My suspicion is that $(\mathbf{D2}b) \implies (\mathbf{G3})$ or is possibly equivalent to it, but I can't quite see how to prove it. I also suspect that the way one can prove it by is proving that $(\mathbf{D2}b)$ is equivalent to the coequalizer defining equivalence relations being effective. Namely if $R \xrightarrow{(s,t)} X \times X$ is an equivalence relation, then $R \rightrightarrows X \to X/R$ is a coequalizer iff $X/R \cong R \coprod\_{R \coprod R} X$, and my idea is to show that having condition $(\mathbf{D2}b)$ hold, but this time up to isomorphism rather than regular epi but only for pushouts of equivalence relations as above, and this would be equivalent to $(\mathbf{D2}b)$ and from this somehow it would be easier to see that it implies $(\mathbf{G3})$, but I've had no luck with this either.
Any ideas or insight would be appreciated.
|
https://mathoverflow.net/users/124010
|
Weak descent and effective equivalence relations
|
I find it a bit surprising, but I think you are correct. The proof I have is maybe a little too long for MO, so I'm only sketching it, but I'll be happy to provide more details if needed.
First one observe a form of "weak descent" for coequalizer diagram:
**Lemma:** Assume that the category $C$ satistifes all four condition of the op, then for any pair of coequalizer diagram $Y\_0 \rightrightarrows Y\_1 \to \text{coeq }Y\_i$ and $X\_0 \rightrightarrows X\_1 \to \text{coeq }X\_i$ and a *cartesian* natural transformation $X\_0 \to Y\_0$ , $X\_1 \to Y\_1$; the natural map $X\_1 \to Y\_1 \times\_{\text{coeq } Y\_i} (\text{coeq } X\_i)$ is a regular epimorphism.
**Proof:** We use that a coequalizer $Y\_0 \rightrightarrows Y\_1$ can be written as a pushout $Y\_1 \coprod\_{Y\_0 \coprod Y\_1} Y\_1$. Using that coproduct are disjoint and universal, one obtains that if $X\_i \to Y\_i$ is cartesian the map of spans from $X\_1 \leftarrow X\_0 \coprod X\_1 \rightarrow X\_1$ to $Y\_1 \leftarrow Y\_0 \coprod Y\_1 \rightarrow Y\_1$ is also cartesian, indeed, $(Y\_0 \coprod Y\_1) \times\_{Y\_1} X\_1 = (Y\_0 \times\_{Y\_1} X\_1) \coprod (Y\_1 \times\_{Y\_1} X\_1) = X\_0 \coprod X\_1$.
and one can apply condition $(D2b)$ to this pushout.
With a bit more work, one can actually prove this also for $X\_0 \to..$ and not just for coequalizer but for all colimits.
---
Now, we apply this to coequalizer of equivalence relation in a way inspired from the usual proof that descent for all colimits (in $\infty$-topos) implies that equivalence realtion are effective.
All the claim below are proved in the same way: they are clear for sets and we prove them for a general category with finite limits by interpreting everything in terms of "generalized" elements (or if you prefer by doing everything in terms of presheaves).
Consider now the case of an equivalence relation $R \rightrightarrows X$. Let $R^{(2)}$ be the subobject of $X^3$ corresponding to $\{ x\_1,x\_2,x\_3 \in X^3 | x\_1 R x\_2 \text{ and } x\_2 R x\_3 \}$
$R^{(2)}$ has two maps to $R$ that sends $(x\_1,x\_2,x\_3)$ to $(x\_1,x\_2)$ and $(x\_1,x\_3)$ and this makes $R^{(2)}$ into an equivalence relation on $R$. The coequalizer $R/R^{(2)}$ is $X$ because there is a split coequalizer diagram $R^{(2)} \rightrightarrows R \rightarrow X$ (with $X \to R$ and $R \to R^{(2)}$ defined in the obvious way).
Finally, we have a cartesian transformation of equalizer: $(R^{(2)} \rightrightarrows R) \to ( R \rightrightarrows X)$ where $R^{(2)} \to R$ is $(x\_1,x\_2,x\_3) \mapsto (x\_2,x\_3)$ and $R \to X$ is $(x\_1,x\_2) \mapsto x\_2$.
It then follows from the version of $(D2b)$ for coequalizer proved above that $R \to X \times\_{E} X$ is a regular epimorphism (where $E =X/R$ is the coeqalizer).
But given that both $R$ and $X \times\_E X$ are subobject of $X \times X$, this map is both a regular epimorphism and a monomorphism, hence it is an isomorphisms.
|
5
|
https://mathoverflow.net/users/22131
|
398463
| 164,447 |
https://mathoverflow.net/questions/397925
|
8
|
I recently stumbled across a quote of Fang-Hua Lin that I have trouble understanding [1, page 42].
>
> It is a well-known fact that a weakly converging sequence of stationary integral currents may have a limit which is not a stationary current.
>
>
>
**Question.** How should I interpret this quote? What does Lin mean by a 'stationary current', and which sequence demonstrates this 'well-known fact'?
My initial guess would be that an integral current $T$ is 'stationary' if the varifold $\lvert T \rvert$ obtained by forgetting orientations is stationary. If I am not mistaken, this should mean that $\partial T = 0$? However my impression is that a flat limit $T$ would be 'stationary' in this sense of the word.
[1] F.-H. Lin. Mapping problems, fundamental groups and defect measures. Acta Math. Sin. 15 (1999), 25-52.
|
https://mathoverflow.net/users/103792
|
How to interpret this quote of Lin?
|
I believe the following sequence demonstrates the failure of flat limits to be stationary. This would be consistent with the natural interpretation of the quote, meaning: a current $T$ is called *stationary* if the varifold $\lvert T \rvert$ is.
(A quick side remark before the construction: on second thought whether $\partial T = 0$ or not seems seems unrelated to the stationarity of $\lvert T \rvert$. For example, a triple junction has boundary as a current, but is stationary as a varifold. A slightly simpler variant of the example below would see the current $S$ replaced with a triple junction.)
That being said, the currents in the constructed sequence $(T\_n \mid n \in \mathbf{N})$ are one-dimensional cycles in the unit disc $D \subset \mathbf{R}^2$: $\partial T\_n = 0$ for all $n$. They converge weakly as currents to another cycle, say $T\_n \to T$ as $n \to \infty$. Most important: $\lvert T\_n \rvert$ is stationary for all $n$, but $\lvert T \rvert$ is not.
To construct the sequence, let $\{ v\_1,\dots,v\_6 \} \subset \partial D$ be unit vectors with
\begin{equation}
v\_1 + \cdots + v\_6 = 0,
\end{equation}
but which do not match up into antipodal pairs. For example
\begin{equation}
-v\_1 \not \in \{ v\_1,\dots,v\_6 \}.
\end{equation}
* Let $S \in I\_1(D)$ be the current supported in the union of the segments $\{ t v\_i \mid 0 \leq t \leq 1 \}$, oriented so that
$\partial S = 0.$
The associated varifold $\lvert S \rvert$ is stationary by construction.
* Let $L$ be the current supported in the segment $\{ tv\_1 \mid -1 \leq t \leq 1 \}$, which we orient in the *opposite* direction. In other words
\begin{equation}
\{ t v\_1 \mid 0 < t \leq 1 \} \cap \mathrm{spt} \, (S + L) = \emptyset.
\end{equation}
This too has $\partial L = 0$ and $\lvert L \rvert$ stationary.
The orientations are chosen so as to ensure that the current $T := S + L$ is not stationary; this is because $-v\_1 + \cdots + v\_6 = -2v\_1 \neq 0$.
Next we consider a sequence of positive angles $\theta\_n \to 0$. We use these angles to rotate $L$, forming a sequence of currents
\begin{equation}
R\_{\theta\_n \#} L \to L
\text{ as $n \to \infty$.}
\end{equation}
As long as $\theta\_n$ is small enough that $\mathrm{spt} \, R\_{\theta\_n \#} L \cap \mathrm{spt} \, S = \{ 0 \}$, the cycles $T\_n := S + R\_{\theta\_n \#} L$ are stationary. However
\begin{equation}
T\_n = S + R\_{\theta\_n \#}L \to S + L = T \text{ as $n \to \infty$}
\end{equation}
in the current topology, which was pointed out above is not stationary.
|
2
|
https://mathoverflow.net/users/103792
|
398464
| 164,448 |
https://mathoverflow.net/questions/398449
|
1
|
I am trying to study the asymptotic behavior of $k^{th}$ order statistic of $n$ i.i.d chi-square distribution. Let $X\_1, \cdots , X\_n$ be i.i.d $\chi^2\_1$ random variables and $X\_{(k:n)}$ be the $k^{th}$ order statistic of these random variables. I do know that,
$$
\frac{X\_{(n:n)}}{\log n} \overset{p}{\to} 2\; \text{as $n \to \infty$}.
$$
But now I am interested in studying the behavior of $X\_{(k:n)}$ where $\frac{k}{n}\to 1$ as $n \to \infty$. My conjecture is that,
$$
\frac{X\_{(k:n)}}{\log n} \overset{p}{\to} 2 \; \text{as $n \to \infty$}.
$$
My idea is to show that $P( \frac{X\_{(k:n)}}{\log n} <a ) \to 0$ if $a<2$ and $P( \frac{X\_{(k:n)}}{\log n} <a ) \to 1$ if $a>2$. Any help will be appreciated.
|
https://mathoverflow.net/users/151115
|
$k^{\text{th}}$ maxima of $n$ i.i.d chi-square random variables
|
$\newcommand{\ep}{\varepsilon}
\newcommand{\pp}{\overset p\to}$Let $Y\_k:=X\_{(k:n)}$, where $n-1\ge k\sim n$. The correct asymptotics for $Y\_k$ is as follows:
\begin{equation\*}
Y\_k/l\_{n,k}\pp2,\tag{1}
\end{equation\*}
where
\begin{equation\*}
l\_{n,k}:=\ln\frac n{n-k},
\end{equation\*}
so that $l\_{n,k}\to\infty$.
In particlular, if $n-k=O(n^\ep)$ for each $\ep>0$, then (1) does imply $Y\_k/\ln n\pp2$.
To prove (1), note first that for all real $x>0$
\begin{equation\*}
P(Y\_k\le x)=P(N\_x\le n-k),\tag{1.5}
\end{equation\*}
where
\begin{equation\*}
N\_x:=\sum\_{i=1}^n 1(X\_i>x),
\end{equation\*}
so that the random variable $N\_x$ is binomial with parameters $n$ and
\begin{equation\*}
p:=P(X\_1>x)=2(1-\Phi(\sqrt x)),
\end{equation\*}
where $\Phi$ is the standard normal cdf, so that
\begin{equation\*}
p=e^{-x/(2+o(1))}\to0 \tag{2}
\end{equation\*}
as $x\to\infty$.
Fix any real $a>0$. Choosing then $x=a\,l\_{n,k}$ and using (2), we see that
$np<<n-k$ if $a>2$ and
$np>>n-k$ if $a<2$;
we write $A<<B$ and $B>>A$ if $A/B\to0$.
So, if $a>2$, then, by Markov's inequality,
\begin{equation\*}
P(N\_x\le n-k)=1-P(N\_x>n-k)\ge1-\frac{np}{n-k}\to1.
\end{equation\*}
If now $a<2$, then, by Chebyshev's inequality,
\begin{equation\*}
P(N\_x\le n-k)=P(N\_x-np\le-(np-(n-k))
\le\frac{np(1-p)}{(np-(n-k))^2}\sim\frac{np}{(np)^2}\to0.
\end{equation\*}
So, for $x=a\,l\_{n,k}$, in view of (1.5), $P(Y\_k\le x)$ converges to $1$ or $0$ depending on whether $a>2$ or $a<2$. Thus, (1) is proved.
|
1
|
https://mathoverflow.net/users/36721
|
398466
| 164,449 |
https://mathoverflow.net/questions/398457
|
0
|
Suppose that I have 2 CW-complexes $A\subset B $ such chat thé inclusions is a retract.
Let $H\_{\ast}$ be the ordinary homology with integral coefficients.
Let
$$\mu: B\times B \rightarrow B $$
be continiuous map such that:
* the restriction map $\mu:A\times A\rightarrow A$ is well defined.
* There exists an element $b\in B$ such that $\mu(b,b')=\mu(b',b)=b'$ for any $b'\in B$
* there exists an element $a\in A$ such that $[a]=[b] $ in $H\_{0}(B)$.
**Question:** Clearly the map $\mu$ induces a map in homology $\bullet:H\_{\ast}(A)\otimes H\_{\ast}(A)\rightarrow H\_{\ast}(A)$. Does it follow that $$[a]\bullet[a\_{n}]=[a\_{n}]\bullet[a]= [a\_{n}]$$
for any $[a\_{n}]\in H\_{n}(A)$
|
https://mathoverflow.net/users/141114
|
Retractions, homology and multiplication
|
The role of $C$ in the problem is irrelevant.
Let $r : C \to B$ be a retraction, and set
$$m := r\circ \mu : B \times B \to B .
$$ Set $\ast := b$ and think of it as the basepoint of $B$. Then the restriction of $m$ to the wedge $B\vee\_\ast B$ is the fold map $B \vee\_\ast B\to B$ (so $B$ is an $H$-space).
Since $[a] = [b]$ in $H\_0(B)$, the restriction of $m$ to
$$
B\vee\_a B := a \times B \cup B\times a
$$
is homotopic to the the fold map on $B \vee\_a B$. Choose
a retraction $s: B\to A$. Then the composition
$$
A \vee\_a A \to B \vee\_a B \overset{m}\to B \overset{s}\to A
$$
is homotopic to the fold map. But this last composite coincides with the restriction of the map $\mu: A\times A \to A$ to $A\vee\_a A$.
The result you seek follows immediately from this by taking homology.
|
2
|
https://mathoverflow.net/users/8032
|
398470
| 164,451 |
https://mathoverflow.net/questions/398020
|
7
|
I'm looking at chapter 4 of Waterhouse's *"Abelian varieties over finite fields"*; and Theorems 4.1 and 4.2 seem to use the following fact:
>
> Suppose that $E/\mathbb{F}\_q$ is an elliptic curve over a finite field with $q=p^n$ elements and let $\pi\_E$ denote the $q$th-power Frobenius map acting on $E$. Suppose that $\pi\_E$ does not act like multiplication-by-$N$ for any integer $N$ so that $\pi\_E = [\alpha]\_E$ where $\alpha$ is a root of the polynomial
> $$x^2 - \mathrm{tr}(\pi\_E)x + q$$
> Then $E$ is ordinary if and only if $p$ is splits in $\mathbb{Q}(\alpha)$.
>
>
>
Waterhouse uses some very general theory (namely a theorem of Honda/Tate and the general theory of abelian varieties) and I find his reasoning/terminology vague. Does anybody know of a more down to earth proof for this result?
Lang has a proof utilitizing the fact that $End(E)\to End(T\_p(E))$ is injective; but i'm trying to find proofs that don't use local methods
**EDIT:** Update on attempt: Here is a proof attempt; perhaps someone can help me finish it off.
We prove $E$ is supersingular if and only if $p$ is non-split in $\mathbb{Q}(\alpha)$.
First suppose that $p$ is non-split. If $p$ is inert, then since $\alpha$ has norm $q = p^n$, it follows that $p$ divides $(\alpha)$ and so $E[p]\subseteq E[\pi\_E] = \{O \}$; thus $E$ is supersingular. If $p$ ramifies then $p$ divides the discriminant of $\mathbb{Q}(\alpha)$ and so $p$ divides $\mathrm{tr}(\pi\_E)^2 - 4q$, which implies that $p$ divides $\mathrm{tr}(\pi\_E)$ and so $E$ is supersingular;
Conversely, suppose that $E$ is supersingular. Suppose that $p$ splits as $(p) = \mathfrak{p}\overline{\mathfrak{p}}$. Then $(\alpha) = \mathfrak{p}^a\overline{\mathfrak{p}}^b$ where $a+b=n$. However, since $E$ is supersingular $\alpha^N \in \mathbb{Z}$ for some positive integer $N$. Which implies that $(\alpha)^N = \mathfrak{p}^{Na}\overline{\mathfrak{p}}^{Nb}$ and so $Na=Nb$ since $(\alpha^N) = \overline{(\alpha^N)}$. Therefore $a=b$ and so $(\alpha) = \mathfrak{p}^a \overline{\mathfrak{p}}^a = (p^a)$ where $a=n/2$. This implies that $\alpha = \zeta p^{a}$ for some unit $\zeta$. This is where I am stuck.
Any help would be appreciated.
|
https://mathoverflow.net/users/103423
|
Supersingular curves over $\mathbb{F}_q$ and the splitting of $p$
|
Here is a solution that avoids explicit use of $\mathbf{Q}\_p$ and in particular does not require knowing that $(\operatorname{End} E) \otimes \mathbf{Q}\_p$ is a division ring. The key is to use *inseparable degree of endomorphisms*.
Identify $\alpha$ with $\pi\_E$. Let $a = \operatorname{tr} \alpha \in \mathbf{Z}$.
Let $\mathcal{O}$ be the ring of integers of $\mathbf{Q}(\alpha)$.
Let $\mathcal{O}' := (\operatorname{End} E) \cap \mathbf{Q}(\alpha)$.
For $\beta \in \mathcal{O}'$, the inseparable degree $\deg\_{\text{i}} \beta$ is multiplicative in $\beta$. The $\beta \in \mathcal{O}'$ with $\deg\_{\text{i}} \beta \ge p^m$ are the $\beta$ that factor through the $p^m$-power Frobenius morphism $F\_m \colon E \to E^{(p^m)}$; they form an additive subgroup. Thus $\beta \mapsto \log\_p \deg\_{\text{i}} \beta$ defines a valuation $v \colon \mathcal{O}' \to \mathbf{Z} \cup \{\infty\}$. Extend $v$ to the fraction field $\mathbf{Q}(\alpha)$.
**Case 1: $E$ is ordinary.** Then $p \nmid a$, so $x^2-ax+q$ mod $p$ has two distinct factors, so $p$ splits in $\mathbf{Q}(\alpha)$.
**Case 2: $E$ is supersingular.** Then $p \mid a$, so the equality $\alpha^2 - a \alpha + q = 0$ yields $\alpha^2 \in p \mathcal{O}$.
First consider the case $\mathcal{O}'=\mathcal{O}$. If $\beta \in \mathcal{O}$ satisfies $v(\beta) \ge 2n$, then $\beta$ factors through $F\_{2n} = \alpha^2 \in p \mathcal{O}$, so $\beta \in p \mathcal{O}$.
This implication for all $\beta$ shows that $v$ is the unique place above $p$.
In general, write $(\mathcal{O}:\mathcal{O}')=p^e d$ with $p \nmid d$.
If $\beta \in \mathcal{O}$ satisfies $v(\beta) \ge (e+1)2n$,
then the endomorphism $(p^e d) \beta$ is divisible (in $\mathcal{O}'$ or in $\mathcal{O}$)
by $\alpha^{2(e+1)}$ and hence by $p^{e+1}$, but $p \nmid d$,
so $\beta \in p\mathcal{O}$.
Again, this shows that $v$ is the unique place above $p$.
|
5
|
https://mathoverflow.net/users/2757
|
398482
| 164,459 |
https://mathoverflow.net/questions/398505
|
2
|
Let us recall this fact. Let $G$ be a semisimple algebraic group over $\mathbb C$ and let $V,V'$ be two irreducible $G$-representations. We denote by $X,X'$ the unique closed $G$-orbits contained in $\mathbb P V, \mathbb P V'$ respectively. We know that if
$$
\mathbb P V \supset X \cong X' \subset \mathbb P V'
$$
as projective $G$-varieties, then $\mathbb PV \cong \mathbb PV'$ as projective spaces. In particular, $\dim V=\dim V'$.
I want to understand the inverse direction: if I have two irreducible $G$-representations $W,W'$ of the same dimension, should I conclude that the closed $G$-orbits $Y \subset \mathbb P W, Y' \subset \mathbb P W'$ are isomorphic as projective $G$-varieties?
|
https://mathoverflow.net/users/147236
|
Do representations of same dimension implies isomorphic closed orbits?
|
No, this is not true. For instance the symplectic group
$$
G = \mathrm{Sp}(V),
$$
where $V$ is a symplectic vector space of dimension 6
has two irreducible representations of dimension 14:
$$
V\_1 = \wedge^2V^\vee / \langle \omega \rangle,
\qquad\text{and}\qquad
V\_2 = \wedge^3V^\vee / (V^\vee \wedge \omega),
$$
where $\omega$ is the symplectic form. These representations are not isomorphic, as well as the corresponding closed orbits, which are the isotropic Grassmannians
$$
X\_1 = \mathrm{IGr}(2,V)
\qquad\text{and}\qquad
X\_2 = \mathrm{IGr}(3,V).
$$
By the way, the inverse is also not true unless the line bundles associated with the embeddings of $X \to \mathbb{P}(V)$ and $X' \to \mathbb{P}(V')$ are identified by the isomorphism $X \cong X'$.
|
4
|
https://mathoverflow.net/users/4428
|
398506
| 164,464 |
https://mathoverflow.net/questions/398481
|
3
|
Let $\mathfrak{A}$ be a C\*-algebra, and let $\phi \in \mathfrak{A}^\*$ be a self-adjoint bounded linear functional on $\mathfrak{A}$. Then there exists a unique pair $\phi^+, \phi^-$ of positive bounded linear functionals on $\mathfrak{A}$ such that $\phi = \phi^+ - \phi^-$ and $\| \phi \| = \left\| \phi^+ \right\| - \left\| \phi^- \right\|$.
My question is: If $\phi$ is tracial (i.e. $\phi(xy)=\phi(yx)$ for all $x, y \in \mathfrak{A}$), then are $\phi^+, \phi^-$ necessarily tracial?
|
https://mathoverflow.net/users/62469
|
Jordan decomposition of tracial functionals on a C*-algebra
|
Assuming $\mathfrak{A}$ is unital, every algebra element is a linear combination of unitaries, so a linear functional being tracial is equivalent to $\phi = \phi \circ \operatorname{Ad} u$ for every unitary $u$. In this case, for every unitary $u$ we have
$$
\phi = \phi \circ \operatorname{Ad} u = \phi^+ \circ \operatorname{Ad} u - \phi^- \circ \operatorname{Ad} u.
$$
By the uniqueness of the decomposition you have $\phi^\pm = \phi^\pm \circ \operatorname{Ad} u$ and so $\phi^{\pm}$ are tracial.
|
4
|
https://mathoverflow.net/users/2085
|
398507
| 164,465 |
https://mathoverflow.net/questions/398519
|
0
|
For any set $X$ we let $[X]^2 = \big\{\{x, y\}: x\neq y \in X\big\}$. Consider the following statement:
>
> (S) : If $G =(V,E)$ is a simple, undirected graph such that $E \neq [X]^2$, then there is $e^\* \in [X]^2 \setminus E$ such that $G \not \cong (V, E\cup\{e^\*\})$.
>
>
>
For finite graphs, (S) is true since adding any edge changes the degree sequence. Does (S) hold for infinite graphs as well?
|
https://mathoverflow.net/users/8628
|
Edge-adding conjecture for graphs
|
No - for a countexample, take the disjoint union of all possible finite graphs, each repeated countably many times.
There is a similar counterexample even if we ask that the graph be connected - just add all possible edges between vertices from different 'copies'.
|
5
|
https://mathoverflow.net/users/385
|
398521
| 164,468 |
https://mathoverflow.net/questions/398526
|
2
|
In a [posting](https://math.stackexchange.com/questions/4206809/is-countability-absolute-over-supertransitive-models-of-set-theory?noredirect=1#comment8734837_4206809) to mathstackexchange I've alluded to the concept of supertransitive model. Now $M$ is a supertransitive model of a set $Q$ of first order sentences, denoted by $M \models^{sptr} Q$, is defined as $M$ being a set obeying $\sf ZFC$ rules and that is supertransitive (i.e.; transitive and has all subsets of all of its elements being among it elements too) and that satisfy $Q$.
Now every $V\_\kappa$ where $\kappa$ is a wordly cardinal, would serve as a supertransitive model of $\sf ZFC$, and apparantly there is no upper limit to that. However, is there a lower bound on the size of a supertransitive model of a theory? Especially of $\sf ZFC$ itself, i.e. is there a lower bound on the cardinality of a supertranstive model $M$ of $\sf ZFC$?
|
https://mathoverflow.net/users/95347
|
Is there a lower bound on the size of a supertransitive model of ZFC?
|
If $M$ is supertransitive and satisfies $\sf ZFC$, then $\omega\in M$, and more importantly, $V\_\omega\in M$.
Now by recursion, if $\alpha$ is an ordinal in $M$, then $V\_\alpha\in M$ as well.
Therefore $M$ must agree on the $V\_\alpha$ hierarchy, and therefore it must have the form $V\_\kappa$ for a worldly cardinal $\kappa$, or $M=V$.
So the smallest supertransitive model is the least worldly cardinal.
|
6
|
https://mathoverflow.net/users/7206
|
398527
| 164,470 |
https://mathoverflow.net/questions/398533
|
2
|
Let $R$ be a PID with field of fraction $K$.
Let $L$ be a lattice with non-degenerate quadratic form $q:L\times L \to R$.
Let
$$
L^\* = \{x \in L\otimes K \text{ s.t. } q(x,l) \in R \text{ for all } l \in L \}.
$$
By integrality of $q$, we have $L \subseteq L^\*$.
I heard the following
**Claim.** The unique decomposition of the quotient $L^\*/L$ given by the structure theorem of modules over principal ring is exactly the one given by the Smith normal form of $q$.
I think it should pretty much follow from the definitions, but I'm a bit confused about how the argument goes. I checked OMeara's and Cassel's books but without success. I'd very much welcome a reference.
Thanks!
|
https://mathoverflow.net/users/148575
|
Reference request: Given a non-degenerate integral quadratic lattice $L,q$ over a PID, the quotient $L^*/L$ is given by SNF of $q$
|
The following more general statement is easier to prove:
Let $L\_1, L\_2$ be lattices with a nondegenerate bilinear form $b: L\_1 \times L\_2 \to R$. Let $$L\_1^\* = \{ x \in L\_2 \otimes K \textrm{ s.t. } b(l,x) \in R \textrm{ for all } l \in L\_1 \}$$
Claim: The unique decomposition of the quotient $L\_1^\*/L\_2$ given by the structure theorem of modules over a principal ring is exactly the one given by the Smith normal form of $b$.
Proof: Putting a matrix in Smith normal form involves acting by invertible matrices on the left and the right, which corresponds to invertible changes of coordinates in $L\_1$ and $L\_2$, which do not affect the quotient $L\_1^\* /L\_2$. So we may assume that $b$ is already in Smith normal form - in particular, is a diagonal matrix with diagonal entries $a\_1,\dots, a\_n$. Then if $e\_1,\dots, e\_n$ is a basis for $L\_2$, $a\_1^{-1} e\_1,\dots, a\_n^{-1} e\_n$ is a basis for $L\_1^\*$, and so $$L\_1^\*/L\_2 = \prod\_i a\_i^{-1} e\_iR /e\_iR =\prod\_i R/a\_i,$$ as desired.
|
3
|
https://mathoverflow.net/users/18060
|
398536
| 164,474 |
https://mathoverflow.net/questions/398525
|
5
|
Let $f:X\rightarrow Y$ be a flat morphism of smooth projective varieties, and $\mathcal{L}$ an effective and ample line bundle on $Y$. For a divisor $A\in H^0(Y,\mathcal{L})$ set $X\_A := f^{-1}(A)$.
Let $D\subset X$ be a divisor such that $D\_{|X\_A}$ (the restriction of $D$ to $X\_A$) is big for $A\in H^0(Y,\mathcal{L})$ general. Under these conditions, might $D$ be not pseudo-effective?
|
https://mathoverflow.net/users/nan
|
Divisors whose restriction is big
|
Consider a product $X = \mathbb{P}^n\times\mathbb{P}^1$, with projections $g:X\rightarrow\mathbb{P}^n$ and $f:X\rightarrow\mathbb{P}^1$.
Set $H\_1:= g^{\*}\mathcal{O}\_{\mathbb{P}^n}(1)$ and $H\_2:= f^{\*}\mathcal{O}\_{\mathbb{P}^1}(1)$. The effective cone of $X$ is closed and generated by $H\_1,H\_2$.
Now, take a divisor $D = aH\_1+bH\_2$ with $a > 0$ and $b < 0$. Then $D\_{|f^{-1}(p)} = \mathcal{O}\_{\mathbb{P}^n}(a)$, which is ample since $a > 0$, for all $p\in\mathbb{P}^1$. However, since $b < 0$ the divisor $D$ is not pseudo-effective.
|
4
|
https://mathoverflow.net/users/14514
|
398537
| 164,475 |
https://mathoverflow.net/questions/398414
|
24
|
For a structure $\mathcal{X}=(X;...)$, say that a cardinal $\kappa$ is **$\mathcal{X}$-detectable** iff there is some sentence $\varphi$ in the language of $\mathcal{X}$ together with a fresh unary predicate symbol $U$ such that for all $A\subseteq X$, the expansion of $\mathcal{X}$ gotten by interpreting $U$ as $A\subseteq X$ satisfies $\varphi$ iff $\vert A\vert\ge\kappa$.
For example, $\omega\_1$ is $(\omega\_1;<)$-detectable since the uncountable subsets of $\omega\_1$ are exactly the unbounded ones. By contrast, [Alex Kruckman observed](https://math.stackexchange.com/a/4203688/28111) that by a result of Robinson no uncountable cardinal is $\mathcal{R}=(\mathbb{R};+,\times)$-detectable.
I'm interested in the expansion $\mathcal{R}\_\mathbb{N}:=(\mathbb{R};+,\times,\mathbb{N})$ of $\mathcal{R}$ gotten by adding a predicate naming the natural numbers (equivalently, adding all projective functions and relations). Since we can talk about one real enumerating a list of other reals, $\omega\_1$ is $\mathcal{R}\_\mathbb{N}$-detectable ("there is no real enumerating all elements of $U$"). More pathologically, if $\mathfrak{c}=2^\omega$ is regular and there is a projective well-ordering of the continuum of length $\mathfrak{c}$ then $\mathfrak{c}$ is $\mathcal{R}\_\mathbb{N}$-detectable. So for example it is consistent with $\mathsf{ZFC}$ that $\omega\_2$ is $\mathcal{R}\_\mathbb{N}$-detectable.
I'm curious whether this type of situation is the only way to get $\mathcal{R}\_\mathbb{N}$-detectability past $\omega\_1$. There are multiple ways to make this precise, of course. At present the following two seem most natural to me:
* Is it consistent with $\mathsf{ZFC}$ that there are at least two distinct regular cardinals $>\omega\_1$ which are $\mathcal{R}\_\mathbb{N}$-detectable?
* Is it consistent with $\mathsf{ZFC}$ that there is a singular cardinal which is $\mathcal{R}\_\mathbb{N}$-detectable?
Note that an affirmative answer to either question requires a large continuum, namely $\ge\omega\_3$ and $\ge\omega\_{\omega+1}$ respectively. Although my primary interest is in *first-order* definability, I'd also be interested in answers for other logics which aren't too powerful (e.g. $\mathcal{L}\_{\omega\_1,\omega}$).
|
https://mathoverflow.net/users/8133
|
Detecting uncountable cardinals in $(\mathbb{R};+,\times,\mathbb{N})$
|
For the first question (distinct regular cardinals $>\aleph\_1$): Force ZFC + MA + $2^{\aleph\_0}=\aleph\_3$ over $L$ in the usual way (see Jech, Theorem 16.13; note the forcing is ccc and it forces MA + $2^{\aleph\_0}=\aleph\_3$, which is all we need here). Then in $L[G]$, $\aleph\_2$ and $\aleph\_3$ are both $\mathcal{R}\_{\mathbb{N}}$-detectable.
$\aleph\_2$: By MA$\_{\aleph\_1}$, every $\omega\_1$-sequence of reals is coded via almost disjoint forcing with respect to the canonical almost disjoint sequence $\left<A\_\alpha\right>\_{\alpha<\omega\_1}$ in $L$. This a.d. sequence is lightface projective (in the standard codes for countable ordinals), so the relation "$y$ is a real enumerated in the $\omega\_1$-sequence of reals $\vec{z}\_x$ coded by $x$" is lightface definable (over $\mathcal{R}\_{\mathbb{N}}$). So just let the statement $\varphi$ be "$A$ is uncountable and there is no real $x$ such that every element of $A$ is enumerated in $\vec{z}\_x$" (the "uncountable" part is dealt with as in the original post). Then $\varphi$ is true exactly when $A\subseteq\mathbb{R}$ has cardinality $\geq\aleph\_2$.
$\aleph\_3$: (It doesn't seem obvious to me that there is a lightface projective wellorder of $\mathbb{R}$ in $L[G]$, so we seem to need another argument than that in the original post.) Let $A\subseteq\mathbb{R}$ with $A\in L[G]\models$"$A$ has cardinality $\leq\aleph\_2$".
Then we can definably talk about ordered pairs of reals and $A^2$ over $(\mathcal{R}\_{\mathbb{N}},A)$, and we can talk about subsets of $A^2$ coded by reals $x$, again via disjoint forcing, but this time with respect to the set $(A^2)'$,
where the prime ' means that we convert the family $A^2$ into a disjoint family $(A^2)'$ in the usual manner. I.e., although we had a wellordered family $\left<A\_\alpha\right>\_{\alpha<\omega\_1}$ in the previous case, this is not relevant. The almost disjoint forcing for coding a subset of $A^2$ is ccc (in fact $\sigma$-centered), and there is an $\aleph\_2$-sized family of dense sets which ensures that the generic real codes a given set $\subseteq A^2$, so by MA$\_{\aleph\_2}$ we will have a real coding any given $X\subseteq A^2$). Note that there is a wellorder of $A$ in ordertype $\leq\omega\_2$, and this is a subset of $A^2$, so we have a code for it, and moreover, every proper segment of this wellorder has cardinality $\leq\aleph\_1$. Since "$\geq\aleph\_2$" is already known to be detectable, hence so is "$\leq\aleph\_1$", so we can detect whether there is such a wellorder of a given $A$. I.e. let $\psi$ be the statement (in the augmented language with symbol $\dot{A}$) saying "there is a real $x$ which codes a subset $X\subseteq\dot{A}^2$ with respect to the family $(\dot{A}^2)'$,
$X$ is a wellorder of $\dot{A}$, every proper segment of $X$ has cardinality $\leq\aleph\_1$". Note that given any $A\subseteq\mathbb{R}$ in $L[G]$, we have $(\mathcal{R}\_{\mathbb{N}},A)\models\psi$ iff $A$ has cardinality $\leq\aleph\_2$ in $L[G]$. Therefore $\aleph\_3$ is also $\mathcal{R}\_{\mathbb{N}}$-delectable.
---
Edit: For the second question: Proceed as above but forcing MA + $2^{\aleph\_0}=\aleph\_{\omega+1}$. Then all cardinals $\kappa\leq\aleph\_{\omega+1}$ are $\mathcal{R}\_{\mathbb{N}}$-detectable in $L[G]$. For $\aleph\_n$ where $n<\omega$ this is basically as above. However, the complexity of the formulas used for the $\aleph\_n$'s seems to increase with $n$, when done just as above, so this doesn't seem to immediately yield $\aleph\_{\omega}$. Instead we can use a slight variant. We first observe that "$\leq\aleph\_\omega$" is detectable: Note that $A\subset\mathbb{R}$ has cardinality $\leq\aleph\_{\omega}$ iff there is a wellorder of $A$ in order type $\leq\omega\_{\omega}$, and any such wellorder will be coded by a real (via a.d. forcing as before). We can assert that the wellorder $<^\*$ has ordertype $\leq\omega\_{\omega}$ by saying that there is a sequence $\left<x\_n\right>\_{n<\omega}\subseteq A$ which is cofinal in $<^\*$ and such that $x\_0$ has only countably many predecessors and for each $n<\omega$ and each $y\in A$ with $x\_n<^\*y<^\*x\_{n+1}$, the set of predecessors of $y$ and the set of predecessors of $x\_n$ have the same cardinality, as witnessed by a bijection coded by some real.
It follows that "$\geq\aleph\_{\omega+1}$" is detectable. To get "$\geq\aleph\_\omega$", note that $A$ has card $\geq\aleph\_\omega$ iff $A$ has card $\geq\aleph\_{\omega+1}$ or there is a wellorder of $A$ exactly in ordertype $\omega\_\omega$, and the latter condition can be expressed as above, together with the extra requirement that there is no real coding a bijection between the predecessors of $x\_n$ and those of $x\_{n+1}$, for each $n$.
|
13
|
https://mathoverflow.net/users/160347
|
398547
| 164,476 |
https://mathoverflow.net/questions/398371
|
6
|
**Notation.**
1. $\mathsf{Topoi}$ is the bicategory of topoi, geometric morphisms and natural transformations between left adjoints.
2. $\mathsf{Topoi}\_{\text{ess}}$ is the bicategory of topoi, essential geometric morphisms and natural transformations between left adjoints.
3. $\mathsf{Presh}$ is the full subcategory of $\mathsf{Topoi}$ spanned by presheaf topoi.
4. $\mathsf{Presh}\_{\text{ess}}$ Is the full subcategory of $\mathsf{Topoi}\_{\text{ess}}$ spanned by presheaf topoi.
**Questions.**
1. Is there a reference for the bicategorical properties of $\mathsf{Topoi}\_{\text{ess}}$, $\mathsf{Presh}$, $\mathsf{Presh}\_{\text{ess}}$?
2. Which (pseudo)(co)limits are preserved by the inclusion $\mathsf{Presh}\_{\text{ess}} \subset \mathsf{Topoi}$? (This is my motivating question.)
|
https://mathoverflow.net/users/104432
|
Stability properties of essential geometric morphisms
|
This is only a partial answer. With '(co)limit' I will always mean pseudo(co)limit.
1. If $\mathcal{C}$ and $\mathcal{D}$ are Cauchy-complete, then the category of essential geometric morphisms $\mathbf{PSh}(\mathcal{C}) \to \mathbf{PSh}(\mathcal{D})$ (and geometric transformations between them) is equivalent to the opposite of the category of functors $\mathcal{C} \to \mathcal{D}$ (and natural transformations between them). This is in “Sketches of an Elephant”, Part A, Example 4.1.4 and Lemma 4.1.5. So in this sense $\mathsf{Presh}\_\mathrm{ess}$ is a full subcategory of $\mathsf{Cat}^\mathrm{co}$, the bicategory of small categories, functors, and natural transformations (with the natural transformations in the opposite direction).
2. By using the above, I think it follows that $$\mathrm{colim}\_i\, \mathbf{PSh}(\mathcal{C}\_i) ~\simeq~ \mathbf{PSh}(\mathrm{colim}\_i \,\mathcal{C}\_i)$$ in $\mathsf{Presh}\_\mathrm{ess}$, as long as $\mathrm{colim}\_i\, \mathcal{C}\_i$ is still Cauchy-complete. In particular, coproducts are computed as $\bigsqcup\_i \mathbf{PSh}(\mathcal{C}\_i) \simeq \mathbf{PSh}(\bigsqcup\_i \mathcal{C}\_i)$, and this agrees with the coproduct in $\mathsf{Topoi}$. So the inclusion $\mathsf{Presh}\_\mathrm{ess} \subset \mathsf{Topoi}$ preserves coproducts, in particular the initial object.
Similarly, I think that $$\mathrm{lim} \, \mathbf{PSh}(\mathcal{C}\_i) \simeq \mathbf{PSh}(\mathrm{lim}\_i \mathcal{C}\_i)$$ in $\mathsf{Presh}\_\mathrm{ess}$, as long as $\lim\_i \mathcal{C}\_i$ is still Cauchy-complete. In particular, the terminal object is $\mathbf{PSh}(1) \simeq \mathbf{Sets}$, just like in $\mathsf{Topoi}$. Also, the product of $\mathbf{PSh}(\mathcal{C})$ and $\mathbf{PSh}(\mathcal{D})$ is $\mathbf{PSh}(\mathcal{C}\times\mathcal{D})$. This is also the product in $\mathsf{Topoi}$ (see Pitts, ["On product and change of base for toposes"](http://www.numdam.org/item/CTGDC_1985__26_1_43_0/)). I don’t know whether $\mathsf{Presh}\_\mathrm{ess} \subset \mathsf{Topoi}$ preserves pullbacks. Examples seem to suggest that pullbacks are preserved, I would be very interested in a proof (here by 'pullback' I mean pseudo-pullback).
**Update:** here is an example showing that arbitrary products are not preserved. Consider the categories $(\mathcal{C}\_n)\_{n \in \mathbb{N}}$ with $\mathcal{C\_n}$ given by the discrete category on two objects, for each $n \in \mathbb{N}$. Then in $\mathsf{Presh}\_\mathrm{ess}$ the product $\prod\_{n \in \mathbb{N}} \mathbf{PSh}(\mathcal{C\_n})$ is given by $\mathbf{PSh}(\mathcal{D})$, where $\mathcal{D} \simeq \prod\_{n \in \mathbb{N}} \mathcal{C}\_n$ is the discrete category with $2^{|\mathbb{N}|}$ objects. However, in $\mathsf{Topoi}$ the product is $\prod\_{n \in \mathbb{N}} \mathbf{PSh}(\mathcal{C\_n}) \simeq \mathbf{Sh}(X)$, where $X$ is the Cantor space (the product in the category of toposes/locales agrees with the product in the category of topological spaces, because it is a countable product of completely metrizable spaces, see Isbell, "Atomless parts of spaces").
**Update 2:** I believe the inclusion $\mathsf{Presh}\_\mathrm{ess} \subset \mathsf{Cat}^\mathrm{co}$ has a left adjoint given by $\mathcal{C} \mapsto \mathbf{PSh}(\mathcal{C})$. This can be used to show that in $\mathsf{Presh}\_\mathrm{ess}$ the colimit of $\mathbf{PSh}(\mathcal{C}\_i)$'s, with each $\mathcal{C}\_i$ Cauchy-complete, is given by $\mathbf{PSh}(\mathcal{D})$ where $\mathcal{D} = \mathrm{colim}\, \mathcal{C}\_i$ is the colimit in $\mathsf{Cat}^\mathrm{co}$ (the category $\mathcal{D}$ does not have to be Cauchy-complete in order for this to work).
To show that $\mathbf{PSh}(\mathcal{D})$ is also the colimit in $\mathsf{Topoi}$ (so colimits are preserved), we can use that colimits in $\mathsf{Topoi}$ are computed as the limit in $\mathsf{Cat}$ of the corresponding diagram of inverse image functors (see [here](https://mathoverflow.net/q/59862/)). Further, in $\mathsf{Cat}$ we have that $$\mathrm{lim}\_i\, \mathbf{PSh}(\mathcal{C}\_i) \simeq \mathrm{lim}\_i \, \mathrm{Fun}(\mathcal{C}\_i^\mathrm{op}, \mathbf{Sets}) \simeq \mathrm{Fun}(\mathrm{colim}\_i\, \mathcal{C}\_i^\mathrm{op}, \mathbf{Sets}) \\ \simeq \mathrm{Fun}(\mathcal{D}^\mathrm{op}, \mathbf{Sets}) \simeq \mathbf{PSh}(\mathcal{D}).$$
Here we use that $\mathbf{Fun}(-,-)$ sends colimits to limits in the first argument.
The proof above is based on the answer by Yonatan Harpaz [here](https://mathoverflow.net/a/322625).
|
5
|
https://mathoverflow.net/users/37368
|
398558
| 164,480 |
https://mathoverflow.net/questions/398176
|
3
|
What is the generating function of $f\_{m,n}$?
$ f\_{m,n} = \begin{cases} 0 , & \text{if $m<0 $ or $ n<0$ }; \\
f\_{n,m} , & \text{ if $n<m$}; \\
1, & \text{ if $0=m$ and $ n\in\{0,1\} $}; \\
f\_{0 ,n-1}+ f\_{1,n-1}, & \text{ if $0=m$ and $ n>1 $}; \\ f\_{m-1 ,n}+ f\_{m ,n-1}+ f\_{m-1,n-1}, & \text{ if $0<m\in \{n,n-1\} $}; \\
f\_{m-1 ,n}+ f\_{ m ,n-1}+ f\_{m-1,n-1} + f\_{m+1,n-1},& \text{ if $0<m<n-1$}.
\end{cases}
$
|
https://mathoverflow.net/users/168671
|
A generating function related to the Delannoy numbers
|
Consider the following generating function:
$$F(x,y) := \sum\_{n=0}^\infty \sum\_{m=0}^n f\_{m,n} x^n y^m$$
and its diagonal
$$D(z) := \sum\_{n=0}^\infty f\_{n,n} z^n.$$
Then the recurrence implies the following functional equation:
$$2(1-x-y-xy)F(x,y) - 2x\frac{F(x,y)-F(x,0)}y = 1 + (1-xy-2y)D(xy).$$
I'm not yet sure if it can be solved explicitly.
---
It's also worth to mention a couple of boundary properties for $F(x,y)$:
$$[y^1]\ F(x,y) = \left.\frac{F(x,y)-F(x,0)}y\right|\_{y=0} = \frac{(1-x)F(x,0)-1}x$$
and
$$\sum\_{n=1}^\infty f\_{n-1,n} z^n = \frac{(1-z)D(z)-1}2.$$
---
**ADDED.** Introducing another generating function:
$$G(x,y) := F(x,\frac{y}{x}) = \sum\_{n=0}^\infty \sum\_{m=0}^n f\_{m,n} x^{n-m} y^m$$
we get a self-contained functional equation:
$$2(y-xy-y^2-xy^2-x) G(x,xy) = y + (y-2y^2-xy^2) G(0,xy) – 2x G(x,0).$$
|
2
|
https://mathoverflow.net/users/7076
|
400574
| 164,485 |
https://mathoverflow.net/questions/398572
|
4
|
In the prominent examples of weak double categories (or bicategories), the weak composition is typically defined by a universal property (specifically, it's usually the tensor product of some kind of bimodules), and as such it's not a uniquely designated arrow.
However, the currently accepted definitions also require *specific assignments* as the weak composition, and impose the coherence axioms on them.
Is this question treated in the literature?
I suggest to simply replace the 'specific assignments', i.e. the functor which is performing the weak composition by a *functorial profunctor*.
What would then happen to the coherence axioms?
---
Let me spell out a possible start:
Let $Gr(C)$ denote the category of internal directed graphs of a category $C$ (i.e. diagrams of two parallel arrows), and let $P:Gr(C)\to Gr(C)$ be the '*path monad*' which assigns the graph with the same vertices and all paths of the original graph as edges to an arbitrary graph.
An internal category is just an algebra for this monad $P$:
That is, it's a graph $G:E\overset{s,t}\to V$ equipped with a 'composition' $c:P(G)\to G$ that behaves appropriately with the monad structure.
Specifically, if $C=Cat$, we arrive to strict double categories, and the composition is ultimately (determined by) a *functor* $c:E\_{P(G)}\to E$.
So, instead of a functor, what would happen if we considered the weak composition operation as a (functorial) profunctor $c':E\_{P(G)}\not\to E$?
[A profunctor $u:A\not\to B$ is a functor $u:A^{op}\times B\to Set$, it's functorial if $u(a,-)$ is representable for each object $a\in A$. Note however, that there are no single representing objects designated in $B$ for objects of $A$.]
Finally, a naive question:
Isn't it possible that if the conditions of being a $P$-algebra could be properly translated in this setting, then we can get rid of the coherence axioms for the weak composition?
|
https://mathoverflow.net/users/44339
|
Proper weakness condition of double categories
|
There's a general theory of this sort of "replacing algebraic structure by universal properties", called [generalized multicategories](https://ncatlab.org/nlab/show/generalized+multicategory). The simplest case is for an ordinary monoidal category, in which case the corresponding "virtual" structure is a [multicategory](https://ncatlab.org/nlab/show/multicategory), which has "multi-maps" in terms of which the tensor product can be characterized by a universal property. The analogous thing for a double category is known as a [virtual double category](https://ncatlab.org/nlab/show/virtual+double+category). The general notion of generalized multicategory is quite similar to your sketched idea in terms of a "pro-algebra" for a monad.
|
6
|
https://mathoverflow.net/users/49
|
400585
| 164,488 |
https://mathoverflow.net/questions/398275
|
1
|
Let $R\_{\theta}$ be the rotation by an angle $\theta$.
Is it then true that for multi-indices $\alpha$ of fixed order $j$ and any smooth function $f$ we have
$$\sum\_{\vert \alpha \vert=j}(R\_{\theta}z)^{\alpha} \partial^{\alpha}f(x) = \sum\_{\vert \alpha \vert=j}(z)^{\alpha} (R\_{-\theta}\partial)^{\alpha}f(x)$$
It is true for $j=1$, which just follows since $R\_{-\theta}$ is the adjoint map of $R\_{\theta}.$
|
https://mathoverflow.net/users/325814
|
Generalized adjoint operation valid?
|
If I am understanding your notation correctly (which admittedly, I may not be), I believe this is already false for $j = 2$. Let me write:
$$ R\_{\theta} = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \qquad f = f(x,y) \qquad z = \begin{bmatrix} u \\ v \end{bmatrix} $$
Then for $j = 2$, the LHS is:
\begin{align\*}
\text{LHS} &=
(u\cos\theta - v\sin\theta)^2 \frac{\partial^2 f}{\partial x^2} \\
&\hspace{1cm} + (u\cos\theta - v\sin\theta) (u\sin\theta + v\cos\theta) \frac{\partial^2 f}{\partial x \partial y} \\
&\hspace{1cm} + (u\sin\theta + v\cos\theta)^2 \frac{\partial^2 f}{\partial y^2}
\end{align\*}
and the RHS is:
\begin{align\*}
\text{RHS} &=
u^2 \left(
\cos^2\theta \frac{\partial^2 f}{\partial x^2}
+ 2\sin\theta\cos\theta \frac{\partial^2 f}{\partial x \partial y}
+ \sin^2\theta \frac{\partial^2 f}{\partial y^2}
\right) \\
&\hspace{1cm} + uv \left(
-\sin\theta \cos\theta \frac{\partial^2 f}{\partial x^2}
+ (\cos^2\theta - \sin^2\theta) \frac{\partial^2 f}{\partial x \partial y}
+\sin\theta \cos\theta \frac{\partial^2 f}{\partial y^2}
\right) \\
&\hspace{1cm} + v^2 \left(
\sin^2\theta \frac{\partial^2 f}{\partial x^2}
- 2\sin\theta \cos\theta \frac{\partial^2 f}{\partial x \partial y}
+ \cos^2\theta \frac{\partial^2 f}{\partial y^2}
\right)
\end{align\*}
If you fully expand out the products, you will find that these two quantities differ by:
$$ \text{LHS} - \text{RHS} = \sin\theta\cos\theta \left(
(v^2-u^2) \frac{\partial^2 f}{\partial x \partial y}
- u v \left(\frac{\partial^2 f}{\partial x^2} - \frac{\partial^2 f}{\partial y^2}\right)
\right) $$
This is certainly not identically zero, so $\text{LHS} \ne \text{RHS}$ in general.
|
0
|
https://mathoverflow.net/users/61479
|
400589
| 164,490 |
https://mathoverflow.net/questions/400586
|
4
|
Let $\Sigma$ be a Riemann surface and let $n,d$ be two relatively prime integers. We can consider different moduli spaces related to those. On one hand we have:
-$M\_{Dol}$ the moduli space of stable Higgs field of rank $n$ and degree $d$
-$M\_B$ moduli space of (twisted) representations of fundamental group of $\Sigma$ with central monodromy $e^{\frac{2 \pi d}{n}}$
-$M\_{dR}$ moduli space of stable meromorphic connections with one pole with residue $-\frac{2 \pi i d}{n}$.
In the case $n=1$ there is an explicit description of the first twos:
$$M\_{Dol}=Jac(\Sigma) \times \mathbb{C}^g $$ and $$M\_B=(\mathbb{C}^\*)^g .$$
I struggled to find an explicit description for de Rham moduli space however.
I know this is biholomorphic to $M\_B$ but it's not the same as an algebraic variety.
Is there a well known algebrogeometric description of it? Is there a concrete description of the Riemann Hilbert map too in this case?
EDIT: The general question was motivated by the following way more specific one. What happens at the moduli space if we consider a map (not necessarily holomorphic) $f:\Sigma \to \Sigma$ and we look at the map on the moduli space $$(\mathcal{E},\nabla) \to (f^\*\mathcal{E},f^\*\nabla) .$$
I was trying to get a glimpse of the geometrical properties of this kind of morphism(for example I think it should be holomorphic but I can't see why) via the identifications in the answer(which are very well written anyway).
|
https://mathoverflow.net/users/146464
|
Explicit example de Rham moduli space of connections
|
Yes, there is a description of the rank 1 de Rham moduli space, and you can find it for example in Goldman's notes: Higgs Bundles and Geometric Structures on Surfaces.
The deRham moduli space is given by an affine holomorphic bundle over the Jacobian with underlying holomorphic bundle the cotangent bundle, where the Atiyah class is determined by the natural symplectic form (see Tyurins book on Theta functions,..).
You can proof all this by applying classical Hodge theory of harmonic 1-forms on Riemann surfaces. As opposed to the higher rank case, the gauge group has different connected components classified by $H^1(\Sigma,\mathbb Z).$ Flat line bundle are given by closed complex 1-forms. By factoring out the identity component of gauge transformations you obtain the first deRham cohomology group $H^1(\Sigma,\mathbb C)$, but taking the quotient by the full gauge group you obtain $H^1(\Sigma,\mathbb C)/H^1(\Sigma,\mathbb Z).$ The Riemann Hilbert map is given by integration and exponentiation.
|
6
|
https://mathoverflow.net/users/4572
|
400590
| 164,491 |
https://mathoverflow.net/questions/398058
|
0
|
[The collecting numbers](https://cses.fi/problemset/task/2216) problem in the CSES problem set has a [greedy solution](https://www.youtube.com/watch?v=lhhHCZYoJvs) where we compare the position of a number x with the position of x-1. If pos(x) < pos(x-1) then we increment rounds because this means that x can't be in the same round as x-1. But what if x can go in another round that doesn't have x-1.
For ex:
suppose there are rounds like this:
```
… x-2
… x-1, x
```
So till here, I have chosen all numbers in ascending order. Meaning x-2 was put in one round, then pos(x-1) < pos(x-2) so it was put in new round. Then pos(x) > pos(x-1) so it was put in same round as x-1. Now, for x+1, suppose pos(x+1) < pos(x). According to this algorithm, x+1 would go in a new round. But what if pos(x+1) > pos(x-2) ?? It could have gone in the round that has x-2. But instead, we created a new round.
**How can we be sure this doesn’t happen?**
Because all we know is pos(x-1) < pos(x-2) and pos(x-1) < pos(x) but we don’t know the relation between pos(x) and pos(x-2). It could be that pos(x) > pos(x-2) and now when we choose x+1, it may be that pos(x+1) > pos(x-2) so it could go in the round with x-2 which would mean we don't have to create a new round for x+1
My approach was to count the **minimum number of increasing subsequences in the array**. And the minimum of that would be the answer (i.e. the number of rounds required). This is passing the sample test case and 2/3 test cases. But it doesn't work on the 3rd case.
Code:
```
int solve()
{
int n;
cin >> n;
vector<int> a;
for(int i=0;i<n;++i)
{
int x;
cin >> x;
x = -x;
auto ind = upper_bound(all(a), x) - a.begin();
if(ind < a.size())
a[ind] = x;
else
a.emplace_back(x);
}
return a.size();
}
```
My logic here is, instead of finding the minimum number of increasing subsequences, I find the minimum number of **decreasing** subsequences. To do this, I convert each number to the negative of itself. This way, an increasing subsequence like [4, 5, 8, 12] will become a decreasing subsequence like this: [-4, -5, -8, -12]
|
https://mathoverflow.net/users/323644
|
How does the greedy algorithm for CSES problem collecting numbers work?
|
I was actually confused about the problem itself. The problem stated that we had to collect numbers in an increasing order in each round. I assumed that they didn't have to be consecutive. For ex: in one round, I could have [2, 4]
But this is not the case. In the above example, I will have to pick 3 only after 2 in that round, otherwise I will have to start a new round. And no other number can now be added to that round, even if their position is after 2 and they are increasing.
|
0
|
https://mathoverflow.net/users/323644
|
400601
| 164,495 |
https://mathoverflow.net/questions/400604
|
11
|
Let $M$ be a closed connected smooth manifold. We define the degree of symmetry of $M$ by $N(M):=\sup\_\limits{g}\mathrm{dim}\,\mathrm{Isom}(M,g)$, where $g$ is a smooth Riemannian metric on $M$ and $\mathrm{Isom}$ is the isometry group of the Riemannian manifold $(M,g)$.
The torus $T^n$ does not admit a Riemannian metric with positive scalar curvature and has $N(T^n)\neq 0$.
Is there a closed connected manifold $M$ with $N(M)=0$ such that $M$ admits a metric with positive scalar curvature? That is, does admitting a metric with positive scalar curvature imply its degree of symmetry is nonzero?
|
https://mathoverflow.net/users/90512
|
Scalar curvature and the degree of symmetry
|
It seems that there are examples. By a theorem of Gromov and Lawson every simply connected manifold of dimension $n \geq 5$ which is not spin admits a metric of positive scalar curvature.
There are many examples of simply connected, non-spin, closed $6$-manifolds which cannot admit a smooth circle action, constructed by Puppe. Theorem 7 of <https://arxiv.org/pdf/math/0606714.pdf>.
Then, since the isomotetry group of a closed manifold is a compact Lie group, if $N(M)>0$ then taking a maximal torus gives a non-trivial circle action, which contradicts the above. So every metric has isometry group of dimension $0$.
Edit: A specific example would be a quartic $3$-fold $X \subset \mathbb{CP}^4$. It admits a metric with positive Ricci curvature (since it is Fano), or alternatively since it is not spin we can apply Gromov-Lawson. It does not admit any smooth circle action due to a Theorem of Dessai and Wiemeler <https://arxiv.org/pdf/1108.5327.pdf>.
|
18
|
https://mathoverflow.net/users/99732
|
400605
| 164,496 |
https://mathoverflow.net/questions/398541
|
13
|
Thirty or so years ago, someone showed me a clever proof of the Fundamental Theorem of Arithmetic that did not make use of the lemma "If $p\mid ab$ then $p\mid a$ or $p\mid b$". I'm unable to reconstruct the argument; all I remember is that it used induction and that it didn't generalize to other number rings. Can anyone provide such a proof, or provide other offbeat elementary proofs of unique factorization of natural numbers into primes?
|
https://mathoverflow.net/users/3621
|
Nonstandard proofs of the fundamental theorem of arithmetic
|
To summarize the comments, this is also known as Zermelo's proof. A version can be found on [wikipedia](https://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic#Uniqueness_without_Euclid%27s_lemma). I will give the proof here to avoid link rot.
The proof is by contradiction. If FTA did not hold, then use the well ordering principle to select the smallest number $s$ which can be factored in two distinct ways into products of primes, $s = p\_1p\_2 \dots p\_m = q\_1q\_2\dots q\_n$. No $p$ can be equal to a $q$, for otherwise $s/p = s/q$ would give a smaller example, violating minimality.
Assume without loss of generality that $p\_1 < q\_1$. Let $P = p\_2 \dots p\_m$ and $Q = q\_2 \dots q\_n$. Then $s = p\_1P = q\_1Q$, so that
$$t=(q\_1-p\_1)Q = p\_1(P-Q) < s$$
Since $t$ is less than $s$, then by minimality of $s$ there is only one way to factor it into a product of primes, namely the prime factorization of $q\_1-p\_1$ times the (known) prime factorization of $Q$. Since $p\_1$ is a prime factor of $t$, $p\_1$ must either be one of the prime factors of $q\_1 - p\_1$ or be one of the prime factors $q\_i$ of $Q$. We already said that $p\_1$ is not equal to any of the $q\_i$, so $p\_1$ is one of the prime factors of $q\_1-p\_1$. In particular, $p\_1$ divides $q\_1$. But $q\_1$ is a prime not equal to $p\_1$, a contradiction.
|
21
|
https://mathoverflow.net/users/1106
|
400607
| 164,497 |
https://mathoverflow.net/questions/400606
|
7
|
Given a compact Hausdorff space $X$ and a continuous mapping $\varphi: X \to X$. We denote by $C(X)$ the space of continuous functions $f: X \to \mathbb{C}$. A probability measure $\mu$ on the Borel-$\sigma$-algebra of $X$ is said to be ergodic for $\varphi$ if it is $\varphi$-invariant, i.e.,
$$\int\_X f \, d\mu = \int\_X f \circ \varphi \, d\mu$$
for all $f \in C(X)$, and if all Borel sets $A \subseteq X$ with $\phi(A) \subseteq A$ satisfy $\mu(A) \in \{0,1\}$.
It would be of interest to know how large the set of all ergodic measures $\mu$ that have a support $\mathrm{supp}(\mu)$ that contains at least 2 elements, let's denote it by $M^\mathrm{erg}\_{\varphi, \geq 2}(X)$, can become. More precisely, is $M^\mathrm{erg}\_{\varphi, \geq 2}(X)$ always countable? And if not in general, are there conditions that assure that $M^\mathrm{erg}\_{\varphi, \geq 2}(X)$ is countable, e.g., $\varphi$ is a homeomorphism, $X$ is metrizable or both.
|
https://mathoverflow.net/users/169719
|
Count of non-trivial ergodic measures of a topological dynamical system
|
Suppose $X$ is the unit circle and $\varphi$ is the doubling map (multiplicatively, $X = \{ z\in \mathbb{C} : |z| = 1\}$ and $\varphi(z) = z^2$, or additively, $X = \mathbb{R}/\mathbb{Z}$ and $\varphi(x) = 2x \mod \mathbb{Z}$). Then the set of ergodic measures is uncountable and in fact is path-connected and dense in the set of all invariant measures (with the weak\* topology). See [the answers at this other question](https://mathoverflow.net/questions/83981/connectedness-of-space-of-ergodic-measures) for more details and references. That's written in terms of the full shift (and more generally, shifts of finite type) but the simplex of invariant measures is the same for the doubling map and the full two-shift.
Note that invertibility (or lack thereof) plays no role here; a non-invertible system has the same simplex of invariant measures as its natural extension, which is invertible, so the story is the same if you consider homeomorphisms.
I'll mention that all those examples are "high complexity" systems in the sense of having positive topological entropy, exponential growth of periodic orbits, etc. One gets rather different behavior for various classes of "low complexity" systems such as interval exchange transformations, translation surfaces, and subshifts of linear growth; see for example the following paper, which gives some results and references from the symbolic point of view.
*Cyr, Van; Kra, Bryna*, [**Counting generic measures for a subshift of linear growth**](http://dx.doi.org/10.4171/JEMS/838), J. Eur. Math. Soc. (JEMS) 21, No. 2, 355-380 (2019). [ZBL1437.37020](https://zbmath.org/?q=an:1437.37020).
|
8
|
https://mathoverflow.net/users/5701
|
400609
| 164,499 |
https://mathoverflow.net/questions/400613
|
4
|
I was interested in the question of figuring out the rational homotopy type of mapping spaces (regular or rational) between two algebraic varieties over $\mathbb{C}$. I encountered the following paper ["Contractibility of the space of rational maps"](http://people.math.harvard.edu/%7Egaitsgde/GL/Contractibility.pdf). The paper for the most part is beyond me, but there seems to be a theorem that if $Y$ is a connected affine scheme covered by opens of $\mathbb{A}^n$ for some $n$. The space $Map\_{rat}(X,Y)$ is contractible. One question that I had and I wasn't able to find an explicit description in the paper. What is the topology on $Map\_{rat}(X,Y)$? Is it the same topology induced by the compact open topology on the space of continuous maps?
On paragraph above section 0.5.7., it mentions that this might be true under milder condition that requires $Y$ to be only smooth and birational.
I was wondering whether there are further results in this direction. I was particularly interested in the space of rational maps from $\mathbb{A}^m$ to something like $Sym^{\infty}(\mathbb{P}^r)$, is there any way to figure out its homotopy type maybe by considering a specific CW complex structure on $Sym^{\infty}(\mathbb{P}^r)$ and then seeing how each part is glued to each other after taking the $Map\_{rat}(\mathbb{A}^m,-)$.
|
https://mathoverflow.net/users/127776
|
Rational homotopy type of rational mapping spaces
|
In Gaitsgory's setup, $Rat(X,Y)$ is not really a topological space (its an algebraic prestack). If you want to think about it topologically, then it is probably best to think of its associated homotopy type.
By definition, this homotopy type obtained as a homotopy colimit over the category $fset^{op}$ of finite sets and surjections of the following functor to $Top$: $$I \mapsto \{ f \in X^I, ~ g:(X-f(I)) \to Y \text{ regular}\},$$ where a surjection $h : I \to J$ acts by $(f,g) \mapsto (f \circ h, g)$.
A slightly tricky thing to extract is how the space assigned to $I$ is topologized, since it is inherently infinite dimensional. The idea is that it is naturally a union of finite dimensional topological spaces, which can be identified as schemes, and then the whole space is topologized as a colimit of the topological spaces underlying those schemes. More concretely, consider the space of regular maps $X - f(I) \to Y$ that have poles of order at most $n$ and take the union as $n \to \infty$.
Note that this whole setup is not at all equivalent to the setting that KhashF mentioned. In his sense, the space of degree $d$ rational maps $f: \mathbb P^1 \to \mathbb P^n$ is the same as the space of degree $d$ algebraic maps $f: \mathbb P^1 \to \mathbb P^n$, which is far from contractible. (In fact Segal proved that as $d \to \infty$ the homology of this space converges to the space of continuous maps $\mathbb P^1 \to \mathbb P^n$, which has nontrivial homology).
|
3
|
https://mathoverflow.net/users/131945
|
400616
| 164,500 |
https://mathoverflow.net/questions/400621
|
0
|
Given 2 finite sets $S$ and $M$, with $\operatorname{card}(S) \geq \operatorname{card}(M)$, and an item $z \not\in M$. There is an unknown function $f: S \to M \cup \{z\}$, which is known to be one-to-one for all $s \in S$ for which $f(s) \in M$ (i.e. for which $f(s) \neq z$). The goal is to find $f$. To this end, I can query an oracle by sending it a question $Q \subseteq S$, and getting back from it answer $A = f(Q) \subseteq M \cup \{z\}$. Obviously, I could use the trivial strategy and sequentially ask the questions $Q = \{s\}$ over all $s \in S$, but querying the oracle is very costly. Is there a questioning strategy that is less costly than the trivial one? Or can one prove that there is no strategy less costly than the trivial one?
|
https://mathoverflow.net/users/332154
|
Mapping problem reminiscent of Mastermind
|
Since $S$ is finite, we can label its elements $s\_1, s\_2, \ldots, s\_n$. Then take as query sets $S\_1 = \{ s\_i \mid i \,\&\, 1 = 1 \}$ where $\&$ represents bitwise conjunction; $S\_2 = \{ s\_i \mid i \,\&\, 2 = 2 \}$, $S\_k = \{ s\_i \mid i \,\&\, 2^{k-1} = 2^{k-1} \}$. This gives $\lceil \lg \operatorname{card}(S) \rceil$ queries. Then the element of $S$ which maps to $m \in M$ can be found by taking the bitwise disjunction of the powers of 2 from the queries whose answers contained $m$.
|
1
|
https://mathoverflow.net/users/46140
|
400629
| 164,504 |
https://mathoverflow.net/questions/398257
|
10
|
Let $G=F\_2$ be the free group of rank $2$. Is there a constant $c>0$ such that the word length $|[u,v]|$ of every commutator $[u,v]=uvu^{-1}v^{-1}$ where $u,v\in G$, $|u|,|v|>0$ is at least $c(|u|+|v|)$ unless $[u,v]=[u\_1,v\_1]$ for some $u\_1,v\_1$ with $|u\_1|+|v\_1|<|u|+|v|$?
|
https://mathoverflow.net/users/157261
|
Length of commutators in the free group
|
Victor Guba sent me a proof. The proof is based on the old result by Wicks, Wicks, N. J. Commutators in free products. J. London Math. Soc. 37 (1962), 433–444., which describes all words which are commutators in a free group (Lemma 5 in the paper). By that result, a word is a commutator iff it is a conjugate of a reduced word of the form $ABCA^{-1}B^{-1}C^{-1}$. This immediately implies that one can take $c=1/4$ (possibly bigger).
|
7
|
https://mathoverflow.net/users/157261
|
400630
| 164,505 |
https://mathoverflow.net/questions/400634
|
15
|
Green and Tao's version of Freiman's theorem over finite fields ([doi:10.1017/S0963548309009821](https://www.doi.org/10.1017/S0963548309009821)) is as follows:
If $A$ is a set in $\mathbb{F}\_2^n$ for which $|A+A| \leqslant K|A|$, then $A$ is contained in a subspace of size $2^{2K+O(\sqrt{K}\log(K))}|A|$.
Does anybody know of a version of this bound with an **explicit** constant for the error term, or just an explicit exponent with more or less the same order of magnitude? For example, replacing the $O(\cdot)$ term by another multiple of $K$ might be sufficient for my purposes. (I would like to apply the bound to some specific sets $A$ and values of $n$. Roughly speaking, my $n$ range from about $50$ to $100$, and my $K$ range from about $50$ to $1000$, in case that helps.)
There is an earlier result of Ruzsa and Green, in which the upper bound is given explicitly as $K^22^{2K^2-2}$, but this is far too large for what I need it for.
I suppose one might be able to infer an explicit bound by reading Green and Tao's proof with enough care, but of course it would be easier not to have to do this.
|
https://mathoverflow.net/users/332227
|
Explicit constant in Green/Tao's version of Freiman's Theorem?
|
Even-Zohar (On sums of generating sets in $\mathbb{Z}\_2^n$, Combin. Probab. Comput. 21 (2012), no. 6, 916–941, available at <https://arxiv.org/abs/1108.4902>) has proved a completely explicit and sharp version of Frieman's theorem in $\mathbb{F}\_2^n$ - see Theorem 2 of that paper.
As a corollary, one gets that if $\lvert A+A\rvert\leq K\lvert A\rvert$ then $A$ is contained a subspace of size at most
$$ (1+o(1))\frac{4^K}{2K}\lvert A\rvert.$$
(Precise bounds with no hidden constants are in the Theorem 2, but are quite cumbersome to write out here - but they are trivial to calculate for any specific values of $K$.)
|
16
|
https://mathoverflow.net/users/385
|
400640
| 164,509 |
https://mathoverflow.net/questions/400622
|
4
|
Suppose I have a topological space $X$ with a collection of closed subsets $X\_\tau$ for $\tau \in P$ where I think of $P$ as a poset with $\tau \leq \lambda \iff X\_\tau \subset X\_\lambda$. Is there some nice (algorithmic?) way to get information about the (compactly supported) cohomology of $U\_{\tau} = X\_\tau - \bigcup\_{\lambda<\tau}X\_\lambda$ in terms of the cohomology of the $X\_\tau$?
For instance, if we I just one closed subset $Z \subset X$, then there is a long exact sequence:
$$\dots \to H^n\_c(U) \to H^n\_c(X) \to H^n\_c(Z) \to \dots$$
Is there something similar in the general case? I can certainly think of some inductive process for building up the relevant cohomology starting from the lowest strata by a combination of the excision sequence and Mayer-Vietoris but it would be nice if there was succinct description.
|
https://mathoverflow.net/users/58001
|
Can I reconstruct the cohomology from a collection of open sets?
|
I wrote a paper on precisely this question:
[A spectral sequence for stratified spaces and configuration spaces of points.](https://msp.org/gt/2017/21-4/p16.xhtml)
Geom. Topol. 21 (2017), no. 4, 2527–2555.
|
7
|
https://mathoverflow.net/users/1310
|
400645
| 164,511 |
https://mathoverflow.net/questions/400644
|
2
|
**Motivation.** I was trying to prove that whenever $G$ is a simple, undirected graph and $\kappa< \chi(G)$ is a cardinal, then there is an induced subgraph with chromatic number exactly $\kappa$. This is easy to do when $\chi(G)$ is finite. For $\chi(G)$ infinite, my strategy is to consider the collection of induced subgraphs colorable with $\kappa$ colors and pick a maximal element (with respect to $\subseteq$), which must have chromatic number $\kappa$. For finding a maximal element, I tried the usual approach with [Zorn's Lemma](https://en.wikipedia.org/wiki/Zorn%27s_lemma) - without success. Which leads me to the question below.
**Formulation of question.** For any set $X$, let ${X \choose 2} = \big\{\{x,y\}:x\neq y\in X\big\}$. Let $G=(V,E)$ be a simple, undirected graph with infinite chromatic number. For $S\subseteq V$ we let $G[S]:= (S, E \cap {S \choose 2}).$
Let $\kappa$ be a cardinal with $0 < \kappa < \chi(G)$. Suppose ${\cal W}$ is a collection of subsets of $V$ such that for all $W, W'\in {\cal W}$ we have $W\subseteq W'$ or $W'\subseteq W$, and for every $W\in{\cal W}$ there is a $\kappa$-coloring of $G[W]$.
Is there a $\kappa$-coloring of $G[\bigcup {\cal W}]$?
|
https://mathoverflow.net/users/8628
|
Is the union of a chain of $\kappa$-colorable subgraphs $\kappa$-colorable?
|
For a counterexample, let $G$ be the complete graph on the ordinal $\omega\_1$, let the $W$'s be the countable ordinals, and let $\kappa=\aleph\_0$.
|
9
|
https://mathoverflow.net/users/75735
|
400647
| 164,512 |
https://mathoverflow.net/questions/400648
|
2
|
Over projective space, it is well-known that given a graded $S^\bullet$-module $M\_\bullet$, where $S^\bullet = k[x\_0, \dots, x\_N]$, there is a unique minimal free resolution
$$
\cdots \to \bigoplus\_q S^\bullet(-q) \otimes B\_{p, q} \to \cdots \to \bigoplus\_q S^\bullet(-q) \otimes B\_{1, q} \to \bigoplus\_q S^\bullet(-q) \otimes B\_{0, q} \to M \to 0
$$
Over an arbitrary variety $X$ over $k$ (assumed to be projective and smooth if necessary), if I define a locally free resolution of a coherent $\mathcal O\_X$-module $\mathcal F$
$$
\mathcal E^\bullet \to \mathcal F \to 0
$$
to be **minimal** if for each local ring $\mathcal O\_{X, x}$, the complex $\mathcal E^\bullet \otimes\_{\mathcal O\_{X, x}} \kappa(x)$ has zero differentiations (Matsumura, (18.E)).
I wonder whether a minimal locally free resolution of $\mathcal F$ exists, and whether it is unique if exists.
|
https://mathoverflow.net/users/129738
|
Minimal free resolution over arbitrary varieties
|
This is not a good definition, because the complex $\mathcal{E}^\bullet \otimes\_{\mathcal{O}\_{X,x}} \kappa(x)$ computes $\mathrm{Tor}\_i^{\mathcal{O}\_{X,x}}(\mathcal{F}, \kappa(x))$, so if this complex has zero differential and at least two terms, it follows that
$$
\mathrm{Tor}\_1^{\mathcal{O}\_{X,x}}(\mathcal{F}, \kappa(x)) \ne 0
$$
for any point $x \in X$. But any coherent sheaf is locally free at general point, which contradicts the above inequality.
|
4
|
https://mathoverflow.net/users/4428
|
400650
| 164,513 |
https://mathoverflow.net/questions/398476
|
9
|
Let $M$ be an open orientable three-manifold such that $\pi\_1 (M)$ is isomorphic to the fundamental group of a closed orientable surface $S\ncong \mathbb{S}^2$. Furthermore, suppose that $\tilde{M} \cong \mathbb{R}^3$. Is it true that $M \cong S \times \mathbb{R}$?
Without making any assumptions about $\tilde{M}$, there are counterexamples to the question, ([see for instance...](https://math.stackexchange.com/questions/734107/a-3-manifold-with-fundamental-group-isomorphic-to-a-surface-group) ), but in those cases the universal cover is not very nice.
[As remarked by several commentators, you need $S$ to be orientable. I added this hypothesis and fixed some of the notation. - Sam]
|
https://mathoverflow.net/users/148805
|
Noncompact three-manifold with fundamental group isomorphic to a surface group
|
I'll restate the question for the convenience of the reader.$\newcommand{\RR}{\mathbb{R}}$
>
> Suppose that $M$ is a non-compact, connected, oriented three-manifold without boundary, with universal cover homeomorphic to $\RR^3$. Suppose that $F$ is a compact, connected, oriented surface with genus at least one. Suppose that the fundamental groups of $M$ and $F$ are isomorphic. Is $M$ homeomorphic to $F \times \RR$?
>
>
>
The answer is "no". We can build a counterexample using the paper [Some examples of exotic non-compact three-manifolds](https://academic.oup.com/qjmath/article-abstract/40/4/481/1531319) by Scott and Tucker. Here I closely follow their wording and notation, starting at the bottom of page 488.
Consider their example manifold $M\_5$. This is a three-manifold with boundary a closed surface $F$. The universal cover of $M\_5$ is homeomorphic to $\RR^2 \times \RR\_{\geq 0}$, and the inclusion of $F$ into $M\_5$ is a homotopy equivalence. However, $M\_5$ is not "almost compact" (that is, it is not tame), and $M\_5$ is not homeomorphic to $F \times \RR\_{\geq 0}$.
Now double $M\_5$ across its boundary to obtain $M = D(M\_5)$. This does not change the fundamental group. Note that $M$ has two ends, separated by (the image of) $F$. By Scott-Tucker, neither end is tame. Thus $M$ is not homeomorphic to $F \times \RR$. However, the universal cover of the double is (in this case) the double of the universal cover; thus it is $\RR^2 \times \RR$, as desired.
|
8
|
https://mathoverflow.net/users/1650
|
400657
| 164,515 |
https://mathoverflow.net/questions/398528
|
4
|
Let $M$ and $N$ be von Neumann algebras, and $\mathcal{H}$ a cyclic $M-N$ correspondence with unit cyclic vector $\xi$. For which $\eta\in \mathcal{H}$ is the bimodule map extending $\xi\mapsto \eta$ a closable operator on $\mathcal{H}$? Is it closable for every $\eta$?
|
https://mathoverflow.net/users/6269
|
Closability of a natural bimodule map between cyclic correspondences of von Neumann algebras
|
Here is a down-to-Earth counter-example. Set $N=\mathbb C$ and $M=\mathcal B(K)$ with $H=K\otimes K$ and $M$ acting on the first tensor factor (should perhaps be $K\otimes\overline K$ but this will not affect the argument). Let $K$ have orthonormal basis $(e\_n)$ and take for example $\xi = \sum\_n n^{-1} e\_n\otimes e\_n$.
Then $\xi$ is also separating, and so the module map sending $\xi$ to $\eta$ will exist for any $\eta$. (In general, this need not be the case I think, so already in complete generality the bimodule map sending $\xi$ to $\eta$ might not be well-defined.) I now make some choices: let
$$ \eta = \Big( \sum\_n n^{-3/4} e\_n \Big) \otimes e\_1. $$
Consider the matrix unit $e\_{m,n}$ which sends $e\_n$ to $e\_m$, so our operator is
$$ T:e\_{m,n}\cdot \xi = n^{-1} e\_m \otimes e\_n \mapsto
e\_{m,n}\cdot\eta = n^{-3/4} e\_m \otimes e\_1. $$
Hence $T:e\_m \otimes e\_n \mapsto n^{1/4} e\_m \otimes e\_1$. Consider the sequence $(e\_1\otimes\alpha\_k) = (k^{-1/4} e\_1\otimes e\_k) \rightarrow 0$ in $H$, while
$$ T(\alpha\_k) = k^{-1/4} k^{1/4} e\_1\otimes e\_1 = e\_1\otimes e\_1, $$
for all $k$. Thus $T$ is not closable.
|
1
|
https://mathoverflow.net/users/406
|
400659
| 164,516 |
https://mathoverflow.net/questions/400667
|
2
|
Let $S$ be a smooth projective surface of Picard rank $\rho(S)$ over a field $K$, and $\overline{S}$ its algebraic closure.
Take a point $p\in\overline{S}$ and denote by $\overline{X}$ be blow-up of $\overline{S}$ at $p$. Is it possible to chose the point $p$ in such a way that $\overline{X}$ is the algebraic closure of a surface $X$ defined over $K$ and having Picard number equal to $\rho(S)$?
|
https://mathoverflow.net/users/nan
|
Blow-ups of surfaces over a field
|
The answer is yes for some $S$ and no for other $S$. I think the answer is no for most $S$ for a reasonable definition of "most".
For a positive example, take $S$ to be $\mathbb P^2$ blown up at a point, of Picard rank $2$. Then $\overline{X}$ will be $\mathbb P^2$ blown up at two points, which is the base change to the algebraic closure of $\mathbb P^2$ blown up at a degree two point, which has Picard rank $2$. (Here I am making the mild assumption that $K$ has a separable quadratic extension.)
For a negative example, take $S$ to be a simple abelian surface with no extra endomorphisms, so its Picard rank is $1$. Then $X$ will always have a nontrivial map defined over $K$ to its Albanese variety (or its minimal model), which means its Picard rank must be at least $2$ (an ample divisor on the target of the map plus a curve must be contracted).
A similar argument should work for any surface of nonnegative Kodaira dimension and Picard rank $1$.
|
2
|
https://mathoverflow.net/users/18060
|
400671
| 164,518 |
https://mathoverflow.net/questions/400678
|
1
|
If $B\in\mathbb{R}^{n\times e}$ is the incidence matrix corresponding to a graph with $n$ vertices and $e$ edges, we know that $BB^T\in\mathbb{R}^{n\times n}$ is the graph Laplacian matrix.
I am curious that if there is any special meaning of the matrix $B^TB\in\mathbb{R}^{e\times e}$.
|
https://mathoverflow.net/users/178204
|
Incidence matrix in a graph, meaning of $B^TB$
|
Well, "special" is maybe an overstatement, because the meaning is really boring: If we consider edges of an undirected graph as two-element subsets of its vertex set, then
$$(B^T B)\_{e,e'} = \sum\_v B\_{v,e} B\_{v,e'} = \#\{v | v \text{ is incident to $e$ and $e'$}\}=|e\cap e'|$$
which is $2\cdot 1\_{n\times n}+$ the adjacency matrix of the line graph of $G$.
And and a slightly more complicated, but similarly boring formula can be derived directly from the definition if you consider directed graphs instead.
|
4
|
https://mathoverflow.net/users/3041
|
400679
| 164,521 |
https://mathoverflow.net/questions/400649
|
8
|
Let $G = (V,E)$ be a simple, undirected graph with $\chi(G)$ infinite. Given a cardinal $\kappa$ with $0 < \kappa < \chi(G)$, is there an induced subgraph $S$ of $G$ with $\chi(S) = \kappa$?
**What I tried:** Let ${\cal S}$ be the collection of all subgraphs of $G$ colorable with $\kappa$ colors. I hoped to find a maximal element $M$ in ${\cal S}$ using [Zorn's Lemma](https://en.wikipedia.org/wiki/Zorn%27s_lemma), establishing $\chi(M) = \kappa$. But this approach [does not work](https://mathoverflow.net/questions/400644/is-the-union-of-a-chain-of-kappa-colorable-subgraphs-kappa-colorable).
|
https://mathoverflow.net/users/8628
|
Induced subgraphs of any given smaller chromatic number
|
Not necessarily. Komjáth showed that it is consistent that there is a graph of chromatic number $\aleph\_2$ which does not have a subgraph (not just induced) of chromatic number $\aleph\_1$. See P. Komjáth, Consistency results on infinite graphs, Israel J. Math., 61 (1988), pp. 285-294.
|
15
|
https://mathoverflow.net/users/332589
|
400684
| 164,522 |
https://mathoverflow.net/questions/398430
|
1
|
According to [Riemann surfaces, dynamics and
geometry](http://people.math.harvard.edu/%7Ectm/home/text/class/harvard/275/09/html/base/rs/rs.pdf) by C. McMullen (Course notes), the definition for a quadratic differential $\phi$ on a Riemann surface $X$ is given by
$$
\|\phi\|\_p = \left(\int\_X \rho^{2-2p} |\phi|^p\right)^{1/p}, \|\phi\|\_\infty = \sup\_X \frac{|\phi|}{\rho^2},
$$
where $\rho |dz|$ is the hyperbolic metric.
For a Beltrami coefficient $\mu,$
$$
\|\mu\|\_p = \left(\int\_X \rho^{2} |\mu|^p\right)^{1/p}, \|\mu\|\_\infty =\sup\_X |\mu|.
$$
I am confused about the mysterious exponents like $2-2p$ and I am not sure how the definition of $L^p$ can pass "smoothly" to $L^\infty.$ I also do not know why we are dividing $\rho^2$ in the first expression but not in the second one.
**Why must we put that certain power of $\rho$ into the expression?**
|
https://mathoverflow.net/users/121404
|
Motivation for the definition of $L^p$ norm for quadratic and Beltrami differentials
|
The powers of $\rho$ are necessary to make the integrals well-defined. In probably excessive detail:
1. **Because $X$ is a Riemann surface, the integrands here need to be $(1,1)$-forms.** This is just the familiar fact (change of variables theorem) that integrals over (oriented) manifolds are only well-defined on objects $\omega$ which are given in local coordinates by functions $\omega(x)$ satisfying the transformation rule
$$
\omega(x) = \det(D\psi\_x) \omega(y)
$$
whenever $y = \psi(x)$ a smooth change of coordinates. If $M = X$ is a Riemann surface, the coordinate changes $w = \psi(z)$ are holomorphic, so that $\det(D\psi) = |\psi'|^2$ (Cauchy-Riemann), and thus the above rule can be rewritten as
$$
\phi(z) = |\psi'(z)|^2 \phi(w) = \psi'(z) \overline{\psi'(z)} \ \phi(w).
$$
This condition can be summarized in the language of Riemann surface $(n, m)$-tensors (see Lyubich's [book](http://www.math.stonybrook.edu/%7Emlyubich/book.pdf), p. 99) by saying that $\phi$ can be integrated over $X$ iff it is a $(1,1)$-form, i.e. an object with local form
$$
\phi = \phi(z) \ dz \ d\bar{z} = \phi(z) \ |dz|^2.
$$
2. **Quadratic differentials are $(2, 0)$-forms. Beltrami forms are $(-1, 1)$-forms.** Take $\phi$ a quadratic differential. Then $|\phi|^p$ is a $(p, p)$-form. In local coordinates:
$$
\phi = \phi(z) \ dz^2 \implies
|\phi|^p = |\phi(z) dz^2|^p = |\phi(z)|^p |dz|^{2p} = |\phi(z)|^p dz^p d\bar{z}^p.
$$
In particular, $|\phi|$ is a $(1, 1)$-form, so can be integrated over $X$. This shows that the $L^1$-norm on quadratic differentials is canonically defined, even when $X$ is not hyperbolic (and thus has no preferred conformal metric), in the way you'd expect:
$$
||\phi||\_1 = \int\_X |\phi|.
$$
When $p \neq 1$, however, $|\phi|^p$ is *not* a $(1,1)$-form, so its integral over $X$ is not defined, and thus the naive definition of the $L^p$-norm for quadratic differentials will not work! Fortunately, this issue can be remedied in the presence of a conformal metric $\rho = \rho(z) |dz|$, because we can adjust the tensor type of our candidate integrand using powers of $\rho$. In particular, multiplying by $\rho^{2-2p}$ gives
$$
\rho^{2-2p} |\phi|^p = \rho(z)^{2-2p} |dz|^{2-2p} |\phi(z)|^p |dz|^{2p} = \rho(z)^{2-2p} |\phi(z)|^p |dz|^{2},
$$
which *is* a $(1,1)$-form, and so can be integrated over $X$. Therefore we can define an $L^p$-norm on quadratic differentials by the expression you gave above:
$$
||\phi||\_p = \left( \int\_X \rho^{2-2p} |\phi|^p \right)^{1/p}.
$$
Moreover, though this definition appears to depend on an arbitrary choice of $\rho$, when $X$ is a *hyperbolic* Riemann surface there is a canonical choice, namely the unique conformal metric of constant curvature -1. So, in the hyperbolic case with $\rho$ taken to be this distinguished metric, it makes sense to refer to the above definition as *the* $L^p$-norm on quadratic differentials. The Beltrami case works similarly.
The construction here is exactly analogous to the problem of defining $L^p$-norms for spaces of *functions* on a smooth manifold $M$. The $L^\infty$-norm of a function is, of course, defined as usual by
$$
||f||\_\infty = \displaystyle\text{ess sup}\_M |f|.
$$
But without further structure, the $L^p$-norms for $p < \infty$ are not well-defined, precisely because functions cannot be integrated over $M$. After fixing a Riemannian metric on $M$, however, the $L^p$-norm of a function can be defined in the obvious way by integrating against the associated volume form:
$$
||f||\_p = \left( \int\_M |f|^p \ \text{dVol} \right)^{1/p}
$$
Taking this approach in our context leads to the same result as above. That is, we could have alternatively begun by modifying $\phi$ by $\rho$ to obtain a *function* on $X$, then applied the Riemannian definition given above to our special case. In detail, the local computation
$$
\dfrac{|\phi|}{\rho^2} = \dfrac{|\phi(z)||dz|^2}{\rho(z)^2 |dz|^2} = \dfrac{|\phi(z)|}{\rho(z)^2},
$$
shows that $\dfrac{|\phi|}{\rho^2}$ is a function on $X$. Then we can define
$$
||\phi||\_\infty = \displaystyle\text{ess sup}\_X \dfrac{|\phi|}{\rho^2}$$
and, using the volume form $\rho^2 = \rho(z)^2 |dz|^2$ associated to the conformal metric $\rho$, define
$$||\phi||\_p = \left( \int\_X \left( \dfrac{|\phi|}{\rho^2} \right)^p \ \rho^2 \right)^{1/p} = \left( \int\_X \rho^{2-2p} |\phi|^p\right)^{1/p}
$$
for $p < \infty$. Voila!
|
3
|
https://mathoverflow.net/users/174177
|
400692
| 164,525 |
https://mathoverflow.net/questions/400680
|
12
|
Suppose that $\{I\_m\}$ is a sequence of pairwise disjoint intervals in $\mathbb{Z}$. The well known Rubio de Francia's inequality says that for any function $f\in L^p(\mathbb{T})$, $2\le p<\infty$, we have
\begin{equation}
\Big\| \Big( \sum\_{m} |(\hat{f}\chi\_{I\_m})^{\vee}|^2 \Big)^{1/2} \Big\|\_{L^p}\lesssim \|f\|\_{L^p}.
\end{equation}
The constant which hides under the sign ''$\lesssim$'' depends only on $p$.
Using duality, it is not difficult to see that this inequality is equivalent to the following:
\begin{equation}
\Big\| \sum\_j f\_j \Big\|\_p\lesssim \Big\|\Big(\sum\_j |f\_j|^2\Big)^{1/2}\Big\|\_p, \qquad 1<p\le 2,
\label{RdFclass}
\end{equation}
where the functions $f\_j$ are such that $\mathrm{supp} \hat{f}\_j\subset I\_j$ and $\{I\_j\}$ are pairwise disjoint intervals in $\mathbb{Z}$. The inequality in such form also holds for $p=1$ as it was shown by Bourgain and for $p<1$ (this is the result of Kislyakov and Parilov).
My question is the following: can the second inequality (for $1<p\le 2$) hold for *arbitrary* pairwise disjoint sets $I\_j$ instead of the intervals? The first one can't for obvious reasons but I couldn't find a simple counterexample for the second one (probably there should be a simple counterexample and I just don't see something).
|
https://mathoverflow.net/users/69086
|
A generalization of Rubio de Francia's inequality
|
The answer is "no" for any $p<2$ (obviously the inequality holds for $p=2$), but the construction I have is rather indirect (analogous to how the Hardy-Littlewood majorant conjecture is disproved, see e.g., [this paper](https://arxiv.org/abs/1203.2378)). A shame, because restriction theory (and other related areas of harmonic analysis) would be a *lot* easier if such a powerful inequality was true!
Suppose your claim was true for some $p<2$, that is to say that
$$ \| \sum\_j f\_j \|\_{L^p({\bf T})} \lesssim \| (\sum\_j |f\_j|^2)^{1/2} \|\_{L^p({\bf T})}$$
whenever $f\_j$ are one-dimensional trigonometric polynomials with disjoint Fourier supports. This implies a higher dimensional analogue
$$ \| \sum\_j f\_j \|\_{L^p({\bf T}^d)} \lesssim \| (\sum\_j |f\_j|^2)^{1/2} \|\_{L^p({\bf T}^d)} \quad (1)$$
whenever $f\_j$ are $d$-dimensional trigonometric polynomials with disjoint Fourier supports, with constant independent of $d$. This follows from applying the equidistribution observation
$$ \int\_{{\bf T}^d} F(x\_1,\dots,x\_d)\ dx\_1 \dots dx\_d = \lim\_{N \to \infty} \int\_{{\bf T}} F( x, Nx, N^2 x, \dots, N^{d-1} x)\ dx$$
for any continuous $F$ (easily checked first for Fourier phases, then the general case follows from Stone-Weierstrass and a limiting argument) to express both sides of (1) as limiting values of one-dimensional counterparts, and applying the one-dimensional hypothesis.
From (1) and the [tensor power trick](https://terrytao.wordpress.com/2008/08/25/tricks-wiki-article-the-tensor-product-trick/) we can now eliminate the constant and conclude that
$$ \| \sum\_j f\_j \|\_{L^p({\bf T})} \leq \| (\sum\_j |f\_j|^2)^{1/2} \|\_{L^p({\bf T})}$$
whenever $f\_j$ are trigonometric polynomials with disjoint Fourier supports. In particular
$$ \| 1 + f \|\_{L^p({\bf T})} \leq \| (1+|f|^2)^{1/2} \|\_{L^p({\bf T})}$$
whenever $f$ is a trigonometric polynomial of mean zero. By a limiting argument this inequality must then hold for all bounded measurable $f$ of mean zero, thus
$$ \int\_{\bf T} |1+f|^p - pf - 1 \leq \int\_{\bf T} |1+f^2|^{p/2} - 1$$
whenever $f: {\bf T} \to {\bf R}$ is real-valued of mean zero. I've subtracted terms on both sides to make the expressions quadratic or higher order in $f$. Applying this inequality to $f = c (1\_{[0,\varepsilon]} - \varepsilon)$ for any real $c$ and small $\varepsilon$, dividing by $\varepsilon$, and then taking the limit $\varepsilon \to 0$, we conclude that
$$ |1+c|^p - pc - 1 \leq |1+c^2|^{p/2} - 1$$
for any real $c$. Setting $c = -x$ for a large $x$, we see from Taylor expansion and the hypothesis $p<2$ that
$$ |1+c|^p - pc - 1 = x^p + px + o(x)$$
and
$$ |1+c^2|^{p/2} - 1 = x^p + o(x)$$
and we obtain a contradiction for $p$ large enough.
|
13
|
https://mathoverflow.net/users/766
|
400694
| 164,526 |
https://mathoverflow.net/questions/400668
|
4
|
The following problem seems a very hard one, is it known? It has a resemblance to the lonely runner conjecture. I am guessing.
In the plane let $v\_i$ be $n$ unit vectors no two of them are colinear. Take $P\_0$ any point in the plane and construct a successive set of points $P\_i$ such that for every $i$, $1\le i\le n$ the segments $P\_{i-1}P\_{i}=v\_j$ for some $j$. Every vector $v\_j$ is used once. Can we choose the vectors so that the obtained path from $P\_0\to P\_n$ is either a simple cycle or has no crossings (planar)?
|
https://mathoverflow.net/users/121643
|
Computational (conjecture) choices for a path
|
Let $S=\Sigma v\_i$. If $S=0$, sort the vectors according to their angle along the unit circle. Then the corresponding closed path traces the boundary of a convex polygon.
In fact, the vectors $v\_i$ can be of arbitrary length.
If $S\neq 0$, then add an auxiliary vector $v\_{n+1}=-S$ and proceed as in the first case. Finally remove the segment given by the vector $v\_{n+1}$.
|
11
|
https://mathoverflow.net/users/24076
|
400697
| 164,528 |
https://mathoverflow.net/questions/400663
|
4
|
I'm looking for a finite-dimensional Hopf algebra (over any field) that is unimodular, has unimodular dual, but is not involutory. Is there such a thing?
Here's what I know:
* By Radford's formula, the antipode $S$ must have order 4.
* Suzuki [has constructed](https://www.jstor.org/stable/43685968) unimodular Hopf algebras over any field that are not involutory (and where $S^2$ is not even inner). My calculations so far suggest that these are not counimodular, but I still need to double-check.
* It is well-known that semisimplicity implies unimodularity. Therefore finding a cosemisimple unimodular Hopf algebra that is not involutory would be even stronger (or dually, a semisimple counimodular one).
* A [result](https://arxiv.org/abs/q-alg/9712033) of Etingof and Gelaki is that "semisimple cosemisimple" implies involutory, so this combination would be too strong.
|
https://mathoverflow.net/users/27013
|
Hopf algebra that is unimodular and counimodular but not involutory
|
It turns out that an explicit example of such a Hopf algebra already appears at the end of Radford's seminal paper [The Order of the Antipode of a Finite Dimensional Hopf Algebra is Finite](https://www.jstor.org/stable/2373888) as Example 2.
The example is a bit too complicated to reproduce here, so let me just note that it is 8-dimensional, can be defined and has the desired properties over any field, and happens to be isomorphic to its opposite.
|
1
|
https://mathoverflow.net/users/27013
|
400705
| 164,532 |
https://mathoverflow.net/questions/400615
|
4
|
Let $A$ be an (abstract) affine plane. We call $A$ a translation plane if the group of translations acts transitively on the set of points (Axiom 4a in Artin's book "Geometric algebra").
Desargues' "little" theorem characterizes translation planes by the following geometric assertion:
Let $l\_1, l\_2, l\_3$ be distinct parallel lines with distinct points $A\_i,B\_i$ on $l\_i$ such that the line $A\_1A\_2$ is parallel to $B\_1B\_2$ and $A\_2A\_3$ is parallel to $B\_2B\_3$. Then $A\_1A\_3$ is parallel to $B\_1B\_3$ (see theorems 2.16, 2.17 in Artin's book).
If the lines $l\_1, l\_2, l\_3$ are required to meet in a point, then we call the statement above just Desargues' theorem. Artin shows that Desargues' theorem is characterized by Axiom 4b: Whenever $P, Q, R$ are distinct points on a line, then there exists a dilatation sending $P$ to $Q$ and fixing $R$.
Unfortunately, Artin does not indicate whether axioms 4a and 4b are independent.
In fact, in [Stevenson, Projective Planes, theorems 5.2.5 and 5.3.1] it is shown (via a lengthy detour to projective planes) that Desargues' theorem indeed implies Desargues' little theorem! Given the elementary geometric meanings, I'm wondering if there is a purely "affine" proof of this implication. I am aware that there is a nice "affine" proof of Hessenberg's theorem asserting that Pappus planes satisfying Desargues' theorem.
|
https://mathoverflow.net/users/332108
|
How does the affine Desargues theorem imply the little Desargues theorem?
|
I just found a marvelous proof in the style I was looking for in Bennett's book "Affine and projective geometry" (see the corollary on page 60).
|
3
|
https://mathoverflow.net/users/332108
|
400719
| 164,534 |
https://mathoverflow.net/questions/400732
|
10
|
Vopenka's principle implies the existence of weakly compact cardinals (a proper class of them, I believe). My question is whether Vopenka's principle is consistent with the assertion that the universe itself is weakly compact. Alternatively, can a Vopenka cardinal be weakly compact?
There are several versions of this question, depending on how classes are treated in the formulation of the two statements. I'm hopeful that the answer is not too different for the different formulations.
I suspect that VP does not *imply* that ORD is weakly compact. After all, VP [implies](https://doi.org/10.1142/S0219061321500240) that ORD is Woodin, and I've read that the first Woodin cardinal is not weakly compact. So if consistent, the conjunction VP + ORD is weakly compact might have higher consistency strength than either of its conjuncts.
|
https://mathoverflow.net/users/2362
|
Is Vopenka's Principle + "ORD has the tree property" consistent?
|
Yes, a Vopenka cardinal can be weakly compact, at least assuming the consistency of a huge cardinal (though this is certainly a bit of an overkill). A huge cardinal is a weakly compact (in fact, measurable) Vopenka cardinal.
EDIT: Actually almost huge cardinals suffice to get measurable Vopenka cardinals. Theorem 24.18 of Kanamori's "Higher Infinite" says that if $\kappa$ is almost huge, then there is a normal ultrafilter $U$ on $\kappa$ such that there are $U$-many $\alpha < \kappa$ such that $\alpha$ is a Vopenka cardinal. The argument actually shows that there are $U$-many $\alpha$ such that $\alpha$ is Vopenka AND measurable. To see why, note that part (b) of that theorem follows from the fact that if $j: V \to M$ witnesses the almost-hugeness of $\kappa$, then $M \models$ "$\kappa$ is Vopenka". But $M$ also models "$\kappa$ is measurable" (in fact $U$ itself is in $M$), because
1. $\kappa$ is measurable in $V$
2. $j(\kappa)$ is inaccessible in $V$; so in particular all $\kappa$-complete ultrafilters on $\kappa$ are in $V\_{j(\kappa)}$; and
3. $V\_{j(\kappa)} \subset M$ because $M$ is closed under $<j(\kappa)$ sequences.
EDIT 2 (regarding Tim's comment): actually $\kappa$ itself is a Vopenka cardinal, because ``$\kappa$ is a Vopenka cardinal" is absolute between any transitive models that have the same powerset of $\kappa$ (and $V$ and $M$ in the above argument have the same powerset of $\kappa$, and $M$ believes that $\kappa$ is a Vopenka cardinal).
|
11
|
https://mathoverflow.net/users/26319
|
400739
| 164,536 |
https://mathoverflow.net/questions/400743
|
7
|
As Hilbert spaces, $L^2(\mathbb{R}^2)$ and $L^2(\mathbb{R})$ are isomorphic. Of course the isomoprhism is vastly not unique. I wonder if there are any particularly nice explicit isomorphisms. E.g. I wonder if there is an integral transform
$$
f(x,y) \mapsto (K f)(z)=\int dx\, dy K(x,y,z) f(x,y)
$$
with a nice explicit kernel $K(x,y,z)$ which maps $L^2(\mathbb{R}^2)$ isometrically onto $L^2(\mathbb{R})$? Any example would be appreciated.
|
https://mathoverflow.net/users/38654
|
Explicit isomorphism between $L^2(\mathbb{R}^2)$ and $L^2(\mathbb{R})$?
|
The following result was obtained by an "explicit" construction in [1]. It is related to the comment of Terry Tao. A modification of the argument allows one to replace the cube by the whole space.
>
> **Theorem.** If $k\geq n$ and $1\leq p\leq \infty$, then there is an isometric isomorphism $\Phi: L^p([0,1]^k)\to L^p([0,1]^n)$ such that $\Phi(u)$ is continuous on $(0,1)^n$ for each $u\in L^p([0,1]^k)$ that is continuous on $(0,1)^k$.
>
>
>
I do not know if the result is true for $k<n$.
During the editorial corrections one of the results in the paper (the Homeomorphic Measures Theorem) has been stated incorrectly; the erratum is available at <https://sites.google.com/view/piotr-hajasz/research/publications?authuser=0>
**[1] P. Hajłasz, P. Strzelecki**, How to measure volume with a thread. *Amer. Math. Monthly* 112 (2005), no. 2, 176–179.
|
4
|
https://mathoverflow.net/users/121665
|
400748
| 164,539 |
https://mathoverflow.net/questions/400726
|
4
|
For a $k$-Hopf algebra $H$ and element $h \in H$ is called **grouplike** is $\Delta(h) = h \otimes h$ and $\epsilon(h)=1\_k$ ($\epsilon$ is the counit). The identity $1\_H$ is clearly grouplike, but in general non-trivial grouplike elements may fail to exist. See this question for Is there a name for a Hopf algebra whose only grouplike element if the identity?
|
https://mathoverflow.net/users/326091
|
Name for a Hopf algebra whose only grouplike element is the identity?
|
There is a one-to-one correspondence between the grouplike elements and the simple, $1$-dim subcoalgebras. So if the only grouplike element is $1\_H$, then there is a unique $1$-dim simple subcoalgebra (which is $k\cdot 1\_H$). In that case, the HA is - by definition- called: *connected HA*.
Also, notice that a connected HA is the same thing as an *irreducible HA*.
(Although if we confine ourselves at the level of coalgebras, the connected ones are a subset of the irreducibles. Actually, at the level of coalgebras, connected=pointed+irreducible, but i think that this is another story).
**Edit:** Motivated by the OP's comment, regarding the use of the term *connected* under the presence of an algebra grading, i think it would be useful to mention the following: the term connected is used in the encyclopedia link provided (<https://encyclopediaofmath.org/wiki/Hopf_algebra>) in the sense of a *connected graded algebra*. The way i am using the term *connected HA* in this post is different and is in the sense of a *connected coalgebra*. These are two different notions of connectedness with non-trivial interactions between them. I think that in contemporary literature the second use tends to be more "standard". For more details, i believe it would be interesting to take a look at: [arXiv:1601.06687v1 [math.RA]](https://arxiv.org/abs/1601.06687v1)
**Edit 2:** In my understanding, the question essentially provides another way to view the notion of a *connected coalgebra* as dual to the notion of a *[connected algebra](https://en.wikipedia.org/wiki/Connected_ring)* (in the ungraded setting):
Since a connected algebra (ring) -in the ungraded setting- is one which has no non-trivial idempotents (other than $0$ and $1$) in the same way a connected coalgebra is one which has no non-trivial grouplikes (other than $1$).
Here, we are essentialy viewing the notion of an idempotent ($g\cdot g=g$) element of an algebra as dual to the notion of a grouplike ($\Delta(g)=g\otimes g$) element of a coalgebra.
|
3
|
https://mathoverflow.net/users/85967
|
400761
| 164,545 |
https://mathoverflow.net/questions/400763
|
10
|
**Observation:** Every $\aleph\_1$-directed colimit $\varinjlim\_i X\_i$ of finite sets is finite.
**Proof:**
Because the $X\_i$'s are finite, the Mittag-Leffler condition holds, so by passing to the diagram of essential images, we may assume that the transition maps are injective. Therefore the cardinalities of the $X\_i$ must be bounded (here is where we use $\aleph\_1$-directednes), and by passing to a cofinal sequence we may assume that the cardinalities are constant. By the pigeonhole principle, all the transition maps are bijections. So the colimit is given by evaluation at any of its terms, and is finite.
**Question:** Is every $\aleph\_1$-directed colimit of finitely-generated abelian groups finitely-generated? How about not-necessarily-abelian groups?
More generally, let $\mathcal C$ be a locally finitely-presentable category. Is every $\aleph\_1$-directed colimit of finitely-presentable objects finitely-presentable?
|
https://mathoverflow.net/users/2362
|
Which abelian groups are $\aleph_1$-filtered colimits of finitely-generated abelian groups?
|
Yes. This can be rephrased as: let $G$ be an abelian group [resp. group] with a chain of f.g. subgroups $G\_\alpha$ for $\alpha<\omega\_1$. Is $G$ f.g.?
The answer is yes for abelian groups:
The answer is clearly yes if $G\_\alpha=G$ for large $\alpha$. Otherwise, up to extract we can suppose that all inclusions are proper. Then $\bigcup\_n G\_n$ is an infinitely generated group. But then it is contained in $G\_\omega$, which is therefore not f.g.
The answer is no for groups. Indeed, every countable group embeds into a f.g. group, so it is easy to define by induction such a directed limit of non-surjective injections, embedding the colimit into a f.g. group at each countable ordinal.
---
By the way, there exist familiar groups that are directed limit over a $\omega\_1$-complete nets (nets in which countable subsets are bounded) of f.g. groups. An example is the group of permutations of any infinite set (Galvin 1995).
|
8
|
https://mathoverflow.net/users/14094
|
400768
| 164,547 |
https://mathoverflow.net/questions/398025
|
3
|
Let $(X\_t)\_{0\le t\le 1}$ be a continuous Markov martingale (with respect to its natural filtration) s.t. $X\_0=0$ and $X\_1\in\{-1,1\}$. Can we prove the existence of some measurable function $\sigma: [0,1]\times \mathbb R\to\mathbb R\_+$ s.t.
$$X\_t=\int\_0^t\sigma(s,X\_s)dW\_s,\quad \forall 0\le t\le 1?$$
Here $(W\_t)\_{0\le t\le 1}$ denotes some Brownian motion.
|
https://mathoverflow.net/users/261243
|
Question on the martingale representation theorem
|
No. There are at least two reasons for that.
First, the underlying probability space $(\Omega, \mathcal F)$ and the natural filtration can be too small for the Brownian motion to exist. Indeed: consider the case when $X\_t$ is constant for $t \geqslant \tfrac12$, and $\Omega$ is just the space of paths of $X\_t$. If $W\_t$ and $\sigma$ existed, then $\sigma\{W\_t : t < \tfrac12\}$ would contain $\sigma\{X\_t : t < \tfrac12\} = \mathcal F \supseteq \sigma\{W\_t : t < 1\}$, and so $\sigma\{W\_t : t < \tfrac12\} = \sigma\{W\_t : t < 1\}$, a contradiction.
Another, more important, reason is that $X\_t$ can be too rough to be an Itô integral with respect to a Brownian motion. To see this, consider an arbitrary martingale $X\_t$, and modify its time using the Cantor's (devil's staircase) function $\phi(t)$: the process $X\_{\phi(t)}$ is again a continuous martingale, but it is piecewise constant on a set of full Lebesgue measure, so the corresponding function $\sigma$ would have to be zero almost everywhere, a contradiction.
---
That said, I believe the following is likely to be true: if $X\_t$ is a continuous martingale *with respect to some Brownian filtration*, and additionally $X\_t$ is a Markov process, then it is an Itô integral with respect to the underlying Brownian motion, with the integrand of the form $\sigma(s, X\_s)$.
|
5
|
https://mathoverflow.net/users/108637
|
400788
| 164,552 |
https://mathoverflow.net/questions/400779
|
8
|
In [page 6, RH Equivalence 5.3](https://web.archive.org/web/20120731034246/http://aimath.org/pl/rhequivalences). An equivalence of the Riemann Hypothesis says that
$$\sum\_{\rho} \frac{1}{|\rho|^2} =\sum\_{\rho} \frac{1}{\rho (1{-}\rho)}= 2 + \gamma - \log 4\pi$$
where $\rho$ is over nontrivial zeros of the Riemann zeta function.
It's not hard to see that RH implies
$$\sum\_{\rho} \frac{1}{|\rho|^2}= 2 + \gamma - \log 4\pi.$$
But conversly I don't see how the equality above implies RH and there is no reference in [page 6, RH Equivalence 5.3](https://web.archive.org/web/20120731034246/http://aimath.org/pl/rhequivalences).
|
https://mathoverflow.net/users/159935
|
A question on an equivalence of RH
|
Note that if $1/2< \sigma <1, t \in \mathbb R$ one has $\frac{2\sigma-1}{\sigma^2+t^2} < \frac{2\sigma-1}{(1-\sigma)^2+t^2}$.
By a little manipulation, one gets:
$\frac{2\sigma}{\sigma^2+t^2} + \frac{2(1-\sigma)}{(1-\sigma)^2+t^2} < \frac{1}{\sigma^2+t^2} + \frac{1}{(1-\sigma)^2+t^2} $
But if RH is false and there is $\rho=\sigma+it, 1/2<\sigma<1$, the above gives that
$2\Re{\frac{1}{\rho}}+2\Re{\frac{1}{1-\rho}}=2\Re{\frac{1}{\rho (1-\rho)}} < \frac{1}{|\rho|^2}+\frac{1}{|1-\rho|^2}$, so if we group together the four terms $\frac{1}{\rho (1-\rho)}, {\frac{1}{\bar \rho (1-\bar \rho)}}$ corresponding to the four roots $\rho, 1-\rho, \bar \rho, 1-\bar \rho$ we get that their sum is srictly less than the sum of the corresponding reciprocal of the respective four roots square modulus, so RH false implies $\sum\_{\rho} \frac{1}{|\rho|^2} >\sum\_{\rho} \frac{1}{\rho (1{-}\rho)}$ and the equivalence is established
Edit later - per comments - note that if $\Re \rho =1/2$ then $\bar \rho=1-\rho$ so roots group naturally in pairs only and $\frac{1}{\rho (1-\rho)}=\frac{1}{|\rho|^2}$ so the corresponding (two) terms on both sides of the equality $\sum\_{\rho} \frac{1}{|\rho|^2} =\sum\_{\rho} \frac{1}{\rho (1{-}\rho)}$ are equal
When there is a root with $\Re \rho\_0 >1/2, t>0$ the roots group into four as noted $\rho\_0, \bar \rho\_0, 1-\rho\_0, 1-\bar \rho\_0$ and now the corresponding terms in $\sum\_{\rho} \frac{1}{|\rho|^2}$ are $2(\frac{1}{|\rho\_0|^2}+\frac{1}{|1-\rho\_0|^2})$, while the terms in the sum $\sum\_{\rho} \frac{1}{\rho (1{-}\rho)}$ are also four and since they are conjugate in pairs add to $2(2\Re{\frac{1}{\rho}}+2\Re{\frac{1}{1-\rho}})=4\Re{\frac{1}{\rho (1-\rho)}}$ and the inequality above applies
|
12
|
https://mathoverflow.net/users/133811
|
400789
| 164,553 |
https://mathoverflow.net/questions/400795
|
7
|
Selmer's curve is the equation $3x^3 +4y^3 +5z^3=0$. This equation is famous for having non-trivial solutions in every completion of $\mathbb{Q}$ but only having the trivial solution in the rationals. This curve has been discussed on Mathoverflow before such as [here](https://mathoverflow.net/questions/47442/diophantine-equation-with-no-integer-solutions-but-with-solutions-modulo-every) and [here](https://mathoverflow.net/questions/2779/proof-of-no-rational-point-on-selmers-curve-3x34y35z3-0). A nice proof that the curve always have solutions for all $p$-adics is in this writeup by [Kevin Buzzard](https://www.ma.imperial.ac.uk/%7Ebuzzard/maths/teaching/10Aut/M4P32/exsht3.pdf). I have two questions related to this curve.
**Question I:** How much worse can we get for cubics if the number of variable is increased. That is:
For every $n \geq 3$ is there a list of non-zero integers $a\_1, a\_2 \cdots a\_n$ such that the equation $$a\_1x\_1^3 +a\_2x\_2^3 \cdots a\_n x\_n^3 =0$$ has solutions in every completion of $\mathbb{Q}$ but no non-trivial integer solutions?
**Question II:** can we make a family of such equations which is nested? That is is there a sequence of non-zero integers $a\_1, a\_2, a\_3 \cdots $ such that for any $n \geq 3$ the equation $$a\_1x\_1^3 +a\_2x\_2^3 \cdots a\_n x\_n^3 =0$$ has solutions in every completion of $\mathbb{Q}$ but no non-trivial integer solutions? And if so, can we take $a\_1=3$, $a\_2=4$ and $a\_3=5$ (that is using Selmer's curve as the start of our family).
|
https://mathoverflow.net/users/127690
|
A family of Diophantine equations with no integer solutions but solutions modulo every integer
|
The answer is no. Heath-Brown has shown every cubic form over the integers in at least 14 variables represents zero nontrivially. The Wikipedia article on [Hasse principle](https://en.wikipedia.org/wiki/Hasse_principle#Cubic_forms) contains references as well as related results for forms in higher odd degrees.
|
18
|
https://mathoverflow.net/users/30186
|
400797
| 164,557 |
https://mathoverflow.net/questions/400801
|
1
|
Suppose that ${\bf x} \in\mathbb C^n$ is a complex random vector, we know the mean vector and covariance matrix of $\bf x$ are defined as follows:
$${\bf m}\_{\bf x} = \mathbb{E} ({\bf x}) \\
{\bf C}\_{\bf x x} = \mathbb{E} (({\bf x}-{\bf m}\_{\bf x})({\bf x}-{\bf m}\_{\bf x})^H)$$
How is 4th order cumulant of a ${\bf \text{complex random vector}}$ such as ${\bf x} \in\mathbb C^n$ defined?
|
https://mathoverflow.net/users/164342
|
How is 4th order cumulant of a complex random vector defined?
|
$\newcommand{\C}{\mathbb C}\newcommand{\ip}[2]{\langle #1,#2\rangle}$First of all, cumulants are defined, rather than derived.
Now, let $X:=\mathbf x$. Suppose $E\|X\|^m<\infty$ for some natural $m$. Let $Y:=X-EX$. Then for the respective characteristic functions $f$ an $g$ of $X$ and $Y$ we have
\begin{equation}
f(s)=Ee^{i\ip Xs}=e^{i\ip{EX}s}Ee^{i\ip Ys}=e^{i\ip {EX}s}g(s),
\end{equation}
where $s\in\C^n$ and $\ip st:=\sum\_{p=1}^ns\_p\overline{t\_p}$ for $s=(s\_1,\dots,s\_n)$ and $t=(t\_1,\dots,t\_n)$ in $\C^n$.
Then the cumulant of order $m$ of $X$ can be defined as the [$m$-linear form/tensor](https://en.wikipedia.org/wiki/Tensor#As_multilinear_maps) on $\C^m$ that is $i^{-m}$ times the $m$th derivative $\ell^{(m)}(0)$ at $0\in\C^m$ of the function $\ell:=\ln f$. So, for any $s\in\C^n$,
\begin{equation}
\ell(s)=\ln f(s)=i\ip{EX}s+\ln g(s)
\end{equation}
and, for any $a,b,c,d$ in $\C^n$,
\begin{equation}
g'(s)(a)=i\,E\ip Ya e^{i\ip Ys},\quad g'(0)(a)=0,
\end{equation}
\begin{equation}
g''(s)(a,b)=i^2\,E\ip Ya\ip Yb e^{i\ip Ys},
\end{equation}
\begin{equation}
g'''(s)(a,b,c)=i^3\,E\ip Ya\ip Yb\ip Yc e^{i\ip Ys},
\end{equation}
\begin{equation}
g''''(0)(a,b,c,d)=i^4\,E\ip Ya\ip Yb\ip Yc\ip Yd.
\end{equation}
So, for any $s\in\C^n$ and any $a,b,c,d$ in $\C^n$,
\begin{equation}
\ell'(s)(a)=\ip{EX}a+\frac1{g(s)}\,g'(s)(a),
\end{equation}
\begin{equation}
\ell''(s)(a,b)=-\frac1{g(s)^2}\,g'(s)(a)g'(s)(b)+\frac1{g(s)}\,g''(s)(a,b),
\end{equation}
\begin{multline\*}
\ell'''(s)(a,b,c)=\frac2{g(s)^3}\,g'(s)(a)g'(s)(b)g'(s)(c) \\
-\frac1{g(s)^2}\,g''(s)(a,c)g'(s)(b)-\frac1{g(s)^2}\,g''(s)(b,c)g'(s)(a)
-\frac1{g(s)^2}\,g''(s)(a,b)g'(s)(c) \\
+\frac1{g(s)}\,g'''(s)(a,b,c),
\end{multline\*}
\begin{multline\*}
\ell''''(0)(a,b,c,d)= \\
-g''(0)(a,c)g'(s)(b,d)-g''(0)(a,d)g'(s)(b,c)-g''(0)(a,b)g'(s)(c,d) \\
+g''''(0)(a,b,c,d).
\end{multline\*}
So, the cumulant of order $4$ of $X$ is the quadrilinear form/tensor on $\C^4$ that is $\ell''''(0)$, and this quadrilinear form is defined by the formula
\begin{multline\*}
\ell''''(0)(a\_1,\dots,a\_4)
=E\prod\_{j\in[4]}\ip Y{a\_j} \\
-\frac12\,\sum\_{J\in\binom{[4]}2}E\prod\_{j\in J}\ip Y{a\_j}\;E\prod\_{k\in J^{\mathsf c}}\ip Y{a\_k},
\end{multline\*}
where $a\_1,\dots,a\_4$ are in $\C^n$, $[4]:=\{1,2,3,4\}$, $J^{\mathsf c}:=[4]\setminus J$, and $\binom{[4]}2$ is the set of all subsets of cardinality $2$ of the set $[4]$.
So, the $n^4$ components of the tensor that is the cumulant of order $4$ of $X$ can be obtained by substituting for $a\_1,\dots,a\_4$ the standard basis vectors $e\_1,\dots,e\_n$ of $\C^n$. For instance, letting $(a\_1,\dots,a\_4)=(e\_p,e\_q,e\_q,e\_p)$ for distinct $p$ and $q$ in the set $[n]:=\{1,\dots,n\}$, we see that the $(p,q,q,p)$-component of the cumulant tensor is
\begin{equation\*}
EY\_p^2Y\_q^2-EY\_p^2\,EY\_q^2-2(EY\_pY\_q)^2.
\end{equation\*}
|
0
|
https://mathoverflow.net/users/36721
|
400806
| 164,562 |
https://mathoverflow.net/questions/400810
|
3
|
Let $K$ be a field which is a (transcendental) extension of $\mathbb{C}$. Let $L\_1, L\_2$ and $M\_1, M\_2$ be two field extensions of $K$ (not necessarily algebraic) such that $$L\_1 \otimes\_K L\_2 \cong M\_1 \otimes\_K M\_2$$
My question is: Is one of $M\_1$ or $M\_2$ a field extension (not necessarily finite) of $L\_1$?
|
https://mathoverflow.net/users/38832
|
On tensor product of field extensions
|
Not necessarily.
Let $p\_1,p\_2,p\_3,p\_4$ be primes, and let $A\_1,A\_2,A\_3,A\_4$ be extensions of $K$ of degrees $p\_1,p\_2,p\_3,p\_4$ respectively.
Let $$L\_1 = A\_1 \otimes\_K A\_2$$ $$L\_2 = A\_3 \otimes\_K A\_4$$ $$M\_1 = A\_1 \otimes\_K A\_3$$ $$M\_2= A\_2 \otimes\_K A\_4.$$
Then neither $M\_1$ nor $M\_2$ can be an extension of $L\_1$ or $L\_2$ because the degree of $M\_i$ is never a multiple of the degree of $L\_j$.
|
6
|
https://mathoverflow.net/users/18060
|
400811
| 164,563 |
https://mathoverflow.net/questions/400793
|
10
|
Let $C\_n=\frac1{n+1}\binom{2n}n$ be the all-familiar Catalan numbers. Then, the following identity has received enough attention in the literature (for example, [Lagrange Inversion: When and How](https://doi.org/10.1007/s10440-006-9077-7)):
\begin{equation}
\label1
\sum\_{k=0}^n\binom{2n-2k}{n-k}C\_k=\binom{2n+1}n \qquad \iff \qquad
\sum\_{i+j=n}\binom{2i}iC\_j=\binom{2n+1}n. \tag1
\end{equation}
I like to ask
>
> **QUESTION.** Is there a $q$-analogue of \eqref{1}? Possibly, a combinatorial proof of \eqref{1} would shed some light into this.
>
>
>
|
https://mathoverflow.net/users/66131
|
In search of a $q$-analogue of a Catalan identity
|
This identity is known as Jonah's formula (special case with $n\rightarrow 2n$ and $r\rightarrow n$, see "Catalan Numbers with Applications" by Thomas Koshy, pg. 325-326 for a combinatorial proof)
$$\sum\_{k=0}^r\binom{n-2k}{r-k}C\_k=\binom{n+1}r$$
and a $q$-analogue was obtained by Andrews in "$q$-Catalan identities" in the book "The legacy of Alladi Ramakrishnan in the Mathematical Sciences". It's Theorem 3, pg. 186.
$$\frac{(1+q^{n-r+1})}{(1+q^{r+1})}\left[ {\begin{array}{c}n+1\\r\end{array} } \right]\_{q^2}=-(-q\;;q)\_{n+1}\sum\_{k=0}^r\left[ {\begin{array}{c}n-2k\\r-k\end{array} } \right]\_{q^2}\frac{\textrm{C}\_{k+1}(-1;q)}{(-q\;;q)\_{n-2k}}q^{-k-1}$$
where $\textrm{C}\_n(\lambda,q)$ is a $q$-analogue of the Catalan numbers considered also by Andrews [here](https://www.sciencedirect.com/science/article/pii/0097316587900331).
$$\textrm{C}\_n(\lambda,q)=\frac{q^{2n}(-\lambda/q; q^{2})\_{n}}{(q^2;q^2)\_{n}}$$
In the paper, he says that the general strategy is to go from a binomial coefficient identity to a generalized hypergeometric identity, and then we can look for a $q$-analogue of the latter. In this case, he used the Pfaff-Saalschütz summation formula and then he searched for a $q$-analogue of this one with the help of Bailey's and Gasper and Rahman's books. I can't help much more, I'm not familiar with these kind of hypergeometric identities.
If $n\rightarrow 2n$ and $r\rightarrow n$, the limit $q\rightarrow 1$ recovers the identity (1).
|
9
|
https://mathoverflow.net/users/302667
|
400812
| 164,564 |
https://mathoverflow.net/questions/398094
|
3
|
Let $f\_i \in L^1 ([0, 1])$ be a sequence of functions equibounded in $L^1$ norm - that is, there exists some $M > 0$ such that $\|f\_i\|\_{L^1} < M$.
Define the functional $F: L^1([0, 1]) \to \mathbb R$ by
$$F(h) = \limsup\_{i \to \infty} \|f\_i - h\|\_{L^1}.$$
>
> **Question:** Does this functional admit a minimiser? Is the minimiser unique whenever it exists?
>
>
>
**Remarks:**
What I have tried so far is to attempt to apply the direct method of the calculus of variations.
Since the $f\_i$ are equibounded in $L^1$, it can be shown that $F$ is coercive, thus any minimising sequence is bounded in $L^1$ norm. In particular we have a weakly-\* converging subsequence, say $h\_n \overset{\*}{\to} h$.
The result would follow if we had weak-\* sequential lower semi continuity of $F$ at the minimiser - that is, that
$$\liminf\_{n \to \infty} F(h\_n) \geq F(h).$$
I could neither disprove this with a counterexample, nor prove it in generality.
**Edit:** As pointed out in the comments, weak-$\*$ convergence to an $L^1$ function isn’t guaranteed, only convergence to a measure.
|
https://mathoverflow.net/users/173490
|
Minimiser of a certain functional
|
As it has been already noted in the comments, the minimizer doesn't need to be unique. However, it always exists. It is not *terribly* hard to show but it is not trivial either, so I wonder why the question attracted so few votes.
The proof consists of two independent parts. The first one is that the limit of every minimizing sequence that converges almost everywhere (or just in measure) is a minimizer and the second one is that there exists a minimizing sequence converging almost everywhere. I will use the fact that we deal with a finite measure space though it should, probably, be irrelevant. WLOG, we may assume that $\|f\_j\|\_1\le 1$ for all $j$.
**Part 1**
Assume that $h\_n$ is a minimizing sequence and $h$ is its pointwise limit (or just limit in measure). Since we can assume WLOG that $\|h\_n\|\_1\le 3$ (to be a competitor, you need to perform not much worse than $0$, at the very least), we have $\|h\|\_1\le 3$ as well (by Fatou). Then, by the definition of the convergence in measure, we can write $h\_n=u\_n+v\_n$ where $u\_n$ converge to $h$ uniformly and $v\_n$ are supported on $E\_n$ with $m(E\_n)\to 0$ as $n\to\infty$. Also $\|v\_n\|\_1\le 7$, say.
Now, since $v\_n\in L^1$, we can find $\delta\_n>0$ such that for every set $E$ with $m(E)<\delta\_n$, we have $\int\_E|v\_n|<\frac 1n$, say. By induction, we can now choose a subsequence $n\_k$ such that
$$
\sum\_{q=k+1}^\infty m(E\_{n\_q})<\delta\_{n\_k}\,.
$$
Then $\|v\_{n\_k}\chi\_{\cup\_{q>k}E\_{n\_q}}\|\_1\to 0$ and, adding these parts to $u\_{n\_k}$, we see that we can (passing to a subsequence) assume that our minimizing sequence can be represented as $h\_n=U\_n+V\_n$ where $U\_n\to h$ in $L^1$ and $V\_n$ have disjoint supports $G\_n$.
Since correcting a minimizing sequence by a sequence tending to $0$ in $L^1$ results in a minimizing sequence, we can just as well assume that $h\_n=h+V\_n$.
If $\|V\_n\|\_1\to 0$ (even along a subsequence), we are done. Assume now that $\|V\_n\|\_1\ge \tau>0$. Then for every function $g$ of $L^1$-norm less than $4$, we have
$$
\|g\|\_1\le \max(\|g-V\_n\|\_1,\dots,\|g-V\_{n+N-1}\|\_1)
$$
for all $n$ as soon as $N>8/\tau$. Indeed, since the supports $G\_n$ of $V\_n$ are disjoint, there is $q\in\{0,\dots,N-1)$ such that $\int\_{G\_{n+q}}|g|\le \frac 1N\|g\|\_1< \frac\tau 2$, in which case subtracting $V\_{n+q}$ can only drive the norm up.
Applying this to the functions $g\_j=f\_j-h$, we conclude that
$$
\limsup\_{j\to\infty}\|f\_j-h\|\_1\le \limsup\_{j\to\infty}\max\_{0\le q\le N-1}\|f\_j-h\_{n+q}\|\_1= \max\_{0\le q\le N-1}\limsup\_{j\to\infty}\|f\_j-h\_{n+q}\|\_1
$$
for every $n$, i.e. $h$ is a minimizer in this case as well.
**Part 2**
Let $I$ be the infimum of our functional. For every $\varepsilon>0$ consider the set $X\_\varepsilon$ of all $L^1$-functions $h$ for which the value of the functional is at most $I+\varepsilon$. Clearly, it is a convex, non-empty, closed (in $L^1$) set and $X\_{\varepsilon'}\supset X\_{\varepsilon''}$ when $\varepsilon'\ge\varepsilon''$.
Fix some strictly convex non-negative function $\Phi(t)\le |t|$. To simplify the technicalities, I'll choose it by the conditions $\Phi(0)=\Phi'(0)=1$, $\Phi''(t)=\frac 2{\pi(1+|t|^2)}$ but pretty much any other choice will work as well.
Let
$$
J\_\varepsilon=\inf\_{h\in X\_\varepsilon}\int\Phi(h)\,.
$$
Clearly, $J\_\varepsilon$ is a bounded non-increasing function on $(0,1)$, say. Let $J$ be its limit at $0+$. Passing to an appropriate decreasing sequence $\varepsilon\_n\to 0+$ and re-enumerating $X\_n=X\_{\varepsilon\_n}, J\_n=J\_{\varepsilon\_n}$, we can assume that $J\_n\ge J-2^{-5n}$
We will choose a representative $h\_n$ of $X\_n$ for which $\int\Phi(h\_n)$ is not more than $2^{-5n}$ above $J$ and, thereby, not more than $2\cdot 2^{-5n}$ above its infimum over $X\_n$. Then for $m>n$, we have $h\_{n,m}=\frac{h\_n+h\_m}2\in X\_n$ (convexity of $X\_n$ plus inclusion $X\_n\supset X\_m$) and
$$
\int\Phi(h\_{n,m})\le \int\frac 12(\Phi(h\_n)+\Phi(h\_m))-\int\frac{(h\_n-h\_m)^2}{4\pi (1+|h\_n|^{2}+|h\_m|^{2})}
$$
(second order Taylor with the remainder in the Lagrange form), whence
$$
\int\frac{(h\_n-h\_m)^2}{4\pi(1+|h\_n|^2+|h\_m|^2)}\le 2\cdot 2^{-5n}
$$
regardless of $m>n$ (otherwise we would go below $J+2^{-5n}-2\cdot 2^{-5n}=J-2^{-5n}$, which is below the infimum over $X\_n$).
It remains to note that if $|h\_n|\le 2^n$ and $|h\_n-h\_m|>2^{-n}$, then the integrand is at least $\rm{const}\, 2^{-4n}$, so we get
$$
m(\{|h\_n|\le 2^n, |h\_n-h\_m|>2^{-n}\})\le \rm{Const}\,2^{-n}
$$
for all $m>n$ from where it follows at once that $h\_n(x)$ is a Cauchy sequence for almost all $x$ (recall that $\|h\_n\|\_1\le 3$, so the first condition excludes only a set of measure $3\cdot 2^{-n}$, then use Borel-Cantelli).
That's it. Feel free to ask questions if anything is unclear :-)
|
3
|
https://mathoverflow.net/users/1131
|
400816
| 164,565 |
https://mathoverflow.net/questions/400814
|
3
|
Let $R$ be a reduced Noetherian ring. Assume $R$ is quasi-excellent and Cohen-Macaulay.
>
> Is $R$ the quotient of a Gorenstein ring?
>
>
>
If the answer is yes, then $R$ has a dualizing complex. The question can, therefore, be rephrased into:
"is there an example of a reduced quasi-excellent Cohen-Macaulay ring that does not admit a dualizing complex?"
|
https://mathoverflow.net/users/nan
|
Quotients of Gorenstein rings
|
No, $R$ is not necessarily a quotient of a Gorenstein ring, because of the following:
**Example** [[Nishimura 2012](https://doi.org/10.1215/21562261-1503754), Example 6.1]**.** *There exists a two-dimensional Cohen–Macaulay factorial excellent local domain with a Gorenstein module, which has no dualizing (i.e., canonical) module.*
Nishimura attributes the construction to [[Weston 1988](https://doi.org/10.1016/0021-8693(88)90021-X)] and [[Ogoma 1982](https://mathscinet.ams.org/mathscinet-getitem?mr=643928), Section 4].
|
7
|
https://mathoverflow.net/users/33088
|
400822
| 164,567 |
https://mathoverflow.net/questions/400819
|
8
|
Can you prove or disprove the following claim:
>
> **Claim:**
> $$\frac{\sqrt{3} \pi}{24}=\displaystyle\sum\_{n=0}^{\infty}\frac{1}{(6n+1)(6n+5)}$$
>
>
>
The SageMath cell that demonstrates this claim can be found [here](https://sagecell.sagemath.org/?z=eJwrTbI1MgACa65i2-LSXI08WwOd0iQdQ30NDTOtPG1DTS0wbaqpqWnNVVCUmVeiUaxlqIfgFBaVaBhragVk6huZaFoDAD15FQQ=&lang=gp&interacts=eJyLjgUAARUAuQ==).
|
https://mathoverflow.net/users/88804
|
An infinite series that converges to $\frac{\sqrt{3}\pi}{24}$
|
Here is an elementary proof. We rewrite the series as
$$\frac{1}{4}\int\_0^1\frac{1-x^4}{1-x^6}\,dx=\frac{1}{8}\int\_0^1\frac{dx}{1-x+x^2}+\frac{1}{8}\int\_0^1\frac{dx}{1+x+x^2}.$$
It is straightforward to show that
\begin{align\*}
\int\_0^1\frac{dx}{1-x+x^2}&=\frac{2\pi}{3\sqrt{3}},\\
\int\_0^1\frac{dx}{1+x+x^2}&=\frac{\pi}{3\sqrt{3}},
\end{align\*}
so we are done.
|
27
|
https://mathoverflow.net/users/11919
|
400824
| 164,569 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.