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https://mathoverflow.net/questions/396256
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0
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Suppose $f, g: \mathbb R \to \mathbb R$ are continuous, non negative functions with $f \leq g$.
Fix some $T > 0$, and denote by $X^f$ the stochastic integral $\int\_{[0, T]} f(s) \ dW\_s$, where $W\_s$ is a standard Brownian motion. Similarly write $X^g$ for the corresponding integral of $g$.
Write $F^f$ for the cumulative distribution function of $|X^f|$, that is $F^f (x) := P(|X^f| < x)$. Similarly write $F^g$ for the corresponding cumulative distribution function of $|X^g|$.
>
> **Question:** Is it true that $F^f \geq F^g$?
>
>
>
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https://mathoverflow.net/users/173490
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A monotonicity formula for the stochastic integral with respect to Brownian motion
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$X^f$ and $X^g$ are Gaussian random variables with zero mean and variances $$\mathbb{E}(|X^f|^2) = \int\_0^T|f(t)|^2dt \leq \int\_0^T|g(t)|^2dt = \mathbb{E}(|X^g|^2).$$
Thus, we know the probability density functions of both random variables. Comparison is up to the reader.
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2
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https://mathoverflow.net/users/164350
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396262
| 163,643 |
https://mathoverflow.net/questions/396243
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6
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I am seeking for an Artin $k$-algebra (especially for group algebra) which is infinite-dimensional over some field $k$. It's known that any complex group algebra has trivial Jacobson radical. So I have the following question:
Is there a countable discrete infinite group $G$ over which the group algebra $\mathbb{C} G$ is semisimple?
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https://mathoverflow.net/users/134942
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Is there a countable discrete infinite group $G$ over which the group algebra $\mathbb{C} G$ is semisimple?
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The answer is negative as any semisimple Hopf algebra is finite-dimensional. More generally, the same conclusion is true for all Artinian Hopf algebras. See e.g. [Liu and Zhang - Artinian Hopf algebras are finite dimensional](https://www.ams.org/journals/proc/2007-135-06/S0002-9939-07-08711-4/home.html).
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8
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https://mathoverflow.net/users/14653
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396264
| 163,645 |
https://mathoverflow.net/questions/396250
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5
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For the function $f\_1(x)=\dfrac{ax+b}{a'x+b'},\quad a'\neq0$ , we have $$f\_1'(x)=\dfrac{\begin{vmatrix}{a} && {b} \\ {a'} && {b'}\end{vmatrix}}{(a'x+b')^2}$$
For $f\_2(x)=\dfrac{ax^2+bx+c}{a'x^2+b'x+c'},\quad a'\neq0$, we have
$$f\_2'(x)=\dfrac{{ \begin{vmatrix}{a} && {b} \\ {a'} && {b'}\end{vmatrix} }x^2+2{ \begin{vmatrix}{a} && {c} \\ {a'} && {c'}\end{vmatrix} }x+{ \begin{vmatrix}{b} && {c} \\ {b'} && {c'}\end{vmatrix} }}{(a'x^2+b'x+c')^2}$$
Can we generalize the formula containing determinants to find $f\_n'(x)$?
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https://mathoverflow.net/users/171386
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General formulas for derivative of $f_n(x)=\dfrac{ax^n+bx^{n-1}+cx^{n-2}+\cdots}{a'x^n+b'x^{n-1}+c'x^{n-2}+\cdots},\quad a'\neq0$
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You can easily extend this, but for $n\geq 3$ you will end up with more than one term per monomial:
For two functions $f$, $g$ rewrite the quotient rule using a determinant
$$\frac{d}{dx} \frac{f}{g} = \frac{\frac{df}{dx}g-f \frac{dg}{dx}}{g^2} = \frac{\begin{vmatrix} \frac{df}{dx} & f \\ \frac{dg}{dx} & g \end{vmatrix}}{g^2}$$
Now assume that $f(x) := a\_nx^n+\dots + a\_0$, $g(x) :=b\_n x^n+ \dots + b\_0$ are polyonomials. Then $\frac{df}{dx}$ and $\frac{dg}{dx}$ can be calculated explicitly and you can use the multilinearity of the determinant to split it by monomials:
$$\frac{d}{dx} \frac{f}{g} = \frac{\begin{vmatrix} \sum\_{k=0}^n a\_k k x^{k-1} & \sum\_{j=0}^n a\_j x^j \\ \sum\_{k=0}^n b\_k k x^{k-1} & \sum\_{j=0}^n b\_j x^j \end{vmatrix}}{g^2} = \frac{\sum\_{k=0}^n \sum\_{j=0}^n k\begin{vmatrix} a\_k & a\_j \\ b\_k & b\_j \end{vmatrix} x^{k+j-1} }{g^2}$$
Now in the last sum for $k=j$ the determinant vanishes and if $k\neq j$ the same determinant occurs again with flipped sign if their roles are reversed. So you can only count the cases $j<k$ and get
$$\frac{d}{dx} \frac{f}{g} = \frac{\sum\_{k=0}^n \sum\_{j=0}^{k-1} (k-j)\begin{vmatrix} a\_k & a\_j \\ b\_k & b\_j \end{vmatrix} x^{k+j-1} }{g^2} $$
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6
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https://mathoverflow.net/users/51695
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396267
| 163,647 |
https://mathoverflow.net/questions/396209
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10
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I want to calculate the number of solutions to the quadratic equation $$x\_1^2+\dots+x\_m^2=0$$ where $m$ is odd (a given number) and $x\_i\in\mathbb{Z}/p^n$ for a given prime number $p$ and a given positive integer $n$.
I guess one can consider the projective variety over the $p$-adic field $\mathbb{Q}\_p$ and count the point of some kind of projection but I didn't make it.
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https://mathoverflow.net/users/nan
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What's the number of solutions of the quadratic equation $x_1^2+\dots+x_m^2=0$ over finite ring $\mathbb{Z}/p^n$?
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$\newcommand\Z{\mathbf{Z}}$Here's a solution for odd $p$ ($p=2$ seems to have specific complications), granted the case $n=1$.
>
> Lemma 1: ($p$ odd) Let $u\_n$ be the number of solutions of $\sum\_{i=1}^mx\_i^2=0$ in $\Z/p^n\Z$ such that $(x\_1,\dots,x\_m)$ is not $(0,\dots,0)$ mod $p$. Then, for $n\ge 1$, $u\_n=p^{(n-1)(m-1)}u\_1$.
>
>
>
Proof: it is enough to prove that $u\_n=p^{m-1}u\_{n-1}$ whenever $n\ge 2$, namely showing that every such solution mod $p^{n-1}$ has exactly $p^{m-1}$ lifts that are solutions mod $p^n$.
Let $(x\_1,\dots,x\_m)$ be a solution mod $p^{n-1}$, that is nonzero mod $p$. Lift it mod $p^n$, so we get $a\_1,\dots,a\_n$ in $\Z/p^n\Z$ such that $M:=\sum\_{i=1}^ma\_i^2$ belongs to $p^{n-1}\Z/p^n\Z$. The set of possible lifts of this solution mod $p^{n-1}$ is given by the $(a\_1+p^{n-1}x\_1,\dots,a\_n+p^{n-1}x\_n)$, for $x\_i\in \Z/p\Z$ such that $\sum\_{i=1}^m(a\_i+p^{n-1}x\_i)^2=0$ (in $\Z/p^n\Z$). This rewrites $M+2p^{n-1}\sum\_{i=1}^m a\_i x\_i=0$. Write $-M/2=p^{n-1}M'$ (since $2$ is invertible mod $p^n$), with $M'\in\Z/p\Z$. Then this rewrites $\sum\_{i=1}^{m} x\_i a\_i=M'$. This affine hyperplane in $(\Z/p\Z)^m$ has $p^{m-1}$ elements. Thus the given solution mod $p^{n-1}$ has exactly $p^{m-1}$ lifts that are solutions.$\Box$
>
> Lemma 2: ($p$ odd). Let $v\_n$ be the number of solutions of $\sum\_{i=1}^mx\_i^2=0$ in $\Z/p^n\Z$. Then $v\_n=\sum\_{0\le 2k<n}p^ku\_{n-2k}+p^{m\lfloor n/2\rfloor}$.
>
>
>
Proof: first, for $k\ge n/2$, every $m$-tuple in $p^k\Z/p^n\Z$ is a solution. This yields the last term. Now consider another solution: for some unique $k<n/2$, it belongs to $(p^k\Z/p^n\Z)^m\smallsetminus (p^{k+1}\Z/p^n\Z)^m$. Write it $(p^kx\_1,\dots,p^kx\_m)$ with $(x\_1,\dots,x\_m)\in (\Z/p^{n-k}\Z)^m$ nonzero mod $p$. Then it is a solution if and only if $\sum\_{i=1}^mx\_i^2=0$ modulo $p^{n-2k}$. The number of such solutions mod $p^{n-2k}$, is thus, by Lemma 1, equal to $u\_{n-2k}$, and all their lifts mod $p^{n-k}$ are the solutions, so there are $p^ku\_{n-2k}$.$\Box$
Thus, writing $q=p^{m-1}$, $$v\_{2n}=u\_{2n}+pu\_{2n-2}+\dots+p^{n-1}u\_2+p^{nm}$$
$$=(q^{2n-1}+\dots +p^{n-2}q^3+p^{n-1}q)u\_1+p^{nm}=q\frac{q^{2n}-p^n}{q^2-p}u\_1+p^{nm},$$ $$v\_{2n-1}=(q^{2n-2}+\dots +p^{n-2}q^2+p^{n-1})u\_1+p^{m(n-1}=\frac{q^{2n}-p^n}{q^2-p}u\_1+p^{m(n-1)}.$$
---
For $p=2$ it's a bit different. Let $w\_n$ be the number of solutions (it also depends on $m$).
$n=1$ is obvious: the equation reduces to $\sum\_{i=1}^m x\_i=0$, which has $2^{m-1}$ solutions.
$n=2$: the number of solutions is $2^ma(m)$, where $a(m)=\sum\_{0\le 4i\le m}\binom{m}{4i}$ counts the number of subsets of $\{1,\dots,m\}$ of cardinal in $4\Z$. This is an OEIS sequence: [A038503](https://oeis.org/A038503).
In general, for $n\ge 2$ the set of solutions is invariant by addition by $2^{n-1}\Z/2^n\Z$, so we have $w\_n=2^mw'\_n$: here we can view $(x\_1,\dots,x\_m)\mapsto \sum\_{i=1}^mx\_i^2$ as a map $\Phi:(\Z/2^{n-1}\Z)^m\to\Z/2^n\Z$, and $w'\_n$ is the cardinal of $\Phi^{-1}(\{0\})$. So $w'\_2=a(m)$.
$n=3$: start from a solution in $(\Z/2\Z)^m$ of $\Phi=0\_{\Z/4\Z}$, and lift it to $(a\_1,\dots,a\_m)\in (\Z/4\Z)^m$. We have to count the solutions of the form $(a\_1+2x\_1,\dots,a\_m+2x\_m)$, satisfying $\Phi=0$ mod 8. Writing $\sum a\_i^2=4M$, this means the equality $\sum\_i a\_ix\_i+x\_i^2=M$ (this being mod 2). This yields a discussion. If some $a\_i$ is zero mod 2, the set of solutions is a coset of a subgroup of index 2, so there are $2^{m-1}$ solutions.
If all $a\_i$ are 1 mod 2, the value does not depend on the lift, and modulo 8, $\sum a\_i^2$ is then equal to $m$. Hence this yields no solution if $m\notin 8\Z$, and $2^m$ solutions (in $\Z/4\Z$) if $m\in 8\Z$. Thus $w'\_3=2^{m-1}w'\_2$ if $m\notin 8\Z$, and $w'\_3=2^{m-1}(w'\_2-1)+2^m$ if $m\in 8\Z$.
$n=4$: start from a solution in $(\Z/4\Z)^m$ of $\Phi=0\_{\Z/8\Z}$, and lift it to $(a\_1,\dots,a\_m)\in (\Z/8\Z)^m$. We have to count the solutions of the form $(a\_1+4x\_1,\dots,a\_m+4x\_m)$, satisfying $\Phi=0$ mod 16. Writing $\sum a\_i^2=8M$, this means the equality $\sum\_i a\_ix\_i=M$ (this being mod 2). This yields a discussion. If some $a\_i$ is 1 mod 2, the set of solutions is a coset of a subgroup of index 2, so there are $2^{m-1}$ solutions.
If all $a\_i$ are 0 mod 2, the value does not depend on the lift and we need another argument. We count the solutions $(2x\_1,\dots,2x\_m)$ with $x\_i\in\Z/4\Z$. The equation is $4\sum\_{i=1}^mx\_i^2=0$ mod 16, that is, $\sum\_{i=1}^mx\_i^2=0$ mod 4. So there are $w\_2$ such solutions.
We thus need to single out the remaining solutions in the $n=3$ case: those $0$ mod $2$ are the $(2x\_1,\dots,2x\_m)$, $x\_i\in\Z/2\Z$ with $4\sum\_{i=1}^mx\_i^2=0$ mod $8$, that is, $\sum x\_i=0$ mod $2$. There are $2^{m-1}$ such solutions, hence $w'\_3-2^{m-1}$ other solutions.
Hence $w'\_4=w\_2+2^{m-1}(w'\_3-2^{m-1})$.
$n\ge 4$: start from a solution in $(\Z/2^{n-2}\Z)^m$ of $\Phi=0\_{\Z/2^{n-1}\Z}$, and lift it to $(a\_1,\dots,a\_m)\in (\Z/2^{n-1}\Z)^m$. We have to count the solutions of the form $(a\_1+4x\_1,\dots,a\_m+4x\_m)$, satisfying $\Phi=0$ mod 16. Writing $\sum a\_i^2=8M$, this means the equality $\sum\_i a\_ix\_i=M$ (this being mod 2). This yields a discussion. If some $a\_i$ is 1 mod 2, the set of solutions is a coset of a subgroup of index 2, so there are $2^{m-1}$ solutions.
If all $a\_i$ are 0 mod 2, the value does not depend on the lift and we need another argument. We count the solutions $(2x\_1,\dots,2x\_m)$ with $x\_i\in\Z/4\Z$. The equation is $4\sum\_{i=1}^mx\_i^2=0$ mod 16, that is, $\sum\_{i=1}^mx\_i^2=0$ mod 4. So there are $w\_2$ such solutions.
We thus need to single out the remaining solutions in the $n=3$ case: those $0$ mod $2$ are the $(2x\_1,\dots,2x\_m)$, $x\_i\in\Z/2\Z$ with $4\sum\_{i=1}^mx\_i^2=0$ mod $8$, that is, $\sum x\_i=0$ mod $2$. There are $2^{m-1}$ such solutions, hence $w'\_3-2^{m-1}$ other solutions.
Hence $w'\_4=w\_2+2^{m-1}(w'\_3-2^{m-1})$.
---
Edit: let me do $p=2$ in general. For the induction to work I need to consider $u\_n$ as in odd $p$ case: those solutions that are not identically $0$ modulo $2$. Recall that $a(m)$ is the number of subsets of $[m]$ of cardinal in $4\Z$.
Then following the above argument, for $m\ge 1$,
* $u\_1=2^{m-1}-1$;
* $u\_2=2^m(a(m)-1)$;
* $u\_3=2^{2m-1}(a(m)-1)$ if $m\notin 8\Z$, and $u\_3=2^{2m-1}a(m)$ if $m\in 8\Z$;
* $u\_n=2^{m-1}u\_{n-1}$ for $n\ge 4$. Thus, for $n\ge 3$, $u\_n=2^{(n-1)(m-1)+1}(a(m)-1)$ if $m\notin 8\Z$ or $n=2$, and $u\_n=2^{(n-1)(m-1)+1}a(m)$ if $m\in 8\Z$.
Then, discussing the largest $k$ such that the solution is divisible by $2^k$, we get (if I computed correctly):
$$w\_{n}=\sum\_{0\le 2k<n}2^ku\_{n-2k}+2^{m\lfloor n/2\rfloor}.$$
So for $m\notin 8\Z$, $$w\_{2n}=2^m(a(m)-1)\frac{2^{(2m-2)n}-2^n}{2^{2m-2}-2}+2^{mn};$$
$$w\_{2n+1}=(a(m)-1)2^{2m-1}\frac{2^{2(m-1)n}-2^n}{2^{2(m-1)}-2}+3.2^{n(m-1)}-2^n.$$
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6
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https://mathoverflow.net/users/14094
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396268
| 163,648 |
https://mathoverflow.net/questions/396234
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2
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Let $M$ be the space of right continuous functions $\ell: \mathbb R\_+\to [0,1]$ that are non increasing s.t. $\ell(0)=0$. Define the map $\Gamma : M\to M$ by $\Gamma[\ell](t):=\mathbb P[\tau^{\ell}>t]$ for all $\ell \in M$ and $t\ge 0$, where $\tau^{\ell}:=\inf\{t\ge 0: X^{\ell}\_t\le 0\}$ and
$$X^{\ell}\_t:=1+t+\int\_0^t\frac{1}{1+\ell(s)}dW\_s,\quad \forall t\ge 0.$$
Here $(W\_t)\_{t\ge 0}$ denotes a Brownian motion. Let $M$ be endowed with the topology as follow: $\ell^n$ converges to $\ell$ in $M$ iff $\lim\_{n\to\infty}\ell^n(t)=\ell(t)$ for all points of continuity of $\ell$. Can we prove the continuity of $\Gamma$ with respect to this topology?
Remark : To prove $\{\inf\_{0\le s\le t}X^{\ell}\_s\le 0\}=\{\inf\_{0\le s\le t}X^{\ell}\_s< 0\}$, it suffices to use Lévy's characterization. More precisely, we can write $X^{\ell}\_t=1+t+B\_{\langle X^{\ell}\rangle\_t}\equiv 1+t+B\_{L(t)}$, where $B$ denotes a Brownian motion and $L(t):=\int\_0^t ds/(1+\ell(s))^2$. Therefore
$$\inf\_{0\le s\le t}X^{\ell}\_t = \inf\_{0\le u\le L(t)}\{1+L^{-1}(u)+B\_t\},$$
which implies the desired result as $\inf\_{0\le u\le L(t)}\{1+L^{-1}(u)+B\_t\}$ admits a density.
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https://mathoverflow.net/users/261243
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On the continuity of map $\Gamma$
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I believe we can. Let $\ell\_n \to \ell$ in your topology, and fix $t\_0 \in \mathbb R\_+$. We show that $\Gamma(\ell\_n)(t\_0) \to \Gamma(\ell)(t\_0)$. We work over the interval $[0, T]$ with $T > t\_0$.
**Step 1:** We first note that $\ell\_n$ converges to $\ell$ in measure.
Indeed, let $\varepsilon > 0$ be arbitrary. As $\ell$ is nonincreasing, $\ell$ contains only jump discontinuities and for any $n > 0$, there exist only finitely many jumps with magnitude greater than $\frac{\varepsilon}{2^j}$. Cover these jumps with finitely many closed intervals $C\_i^j$ of total length less than $\frac{\varepsilon}{2^j}$. Divide the complement into intervals $D\_i^j$ on which $\ell$ varies by no more than $\frac{\varepsilon}{2^j}$, and consider the partition $\mathcal P\_j ;= C\_i^j \cup D\_i^j$. Applying the pointwise convergence of $\ell\_n$ to $\ell$ near the endpoints of the $C\_i^n$ and the monotonicity of $\ell\_n$ and $\ell$ now allows us to conclude.
**Step 2:** We show that $X^{\ell\_n} \to X^{\ell}$ uniformly in probability.
Namely, that $$P(\sup\_{s \in [0, T]} | X\_t^{\ell\_n} - X\_t^{\ell}|
\geq \frac{1}{2^k}) \to 0.$$
To see this, note that we can write
$$X^{\ell}\_t:=1+t+\int\_0^t\frac{1}{1+\ell(s)} + g\_n(s) \ dW\_s \, \quad \forall t\ge 0.$$
with $g\_n := \frac{1}{1 + \ell\_n(s)} - \frac{1}{1 + \ell(s)} \to 0$ in $L^2$. Whereby the aforementioned convergence follows from the convergence in measure of $\ell\_n$ to $\ell$.
Note that $X\_s^{\ell\_n} - X\_s^{\ell}$ is a martingale, and that $X\_T^{\ell\_n} - X\_T^{\ell}$ is Gaussian with mean $0$, and variance $||g\_n||\_{L^2}$.
And so by Doob’s (sub)martingale inequality and symmetry of the Gaussian distribution, we have
$$P(\sup\_{s \in [0, T]} | X\_s^{\ell\_n} - X\_s^{\ell}|
\geq \frac{1}{2^k}) \leq 2^{k+1} E[(X\_T^{\ell\_n} - X\_T^{\ell})^+] \to 0,$$
as $n \to \infty$.
**Step 3:** Conclusion.
Let $\varepsilon > 0$ be arbitrary. We note that by continuity of $X\_n^{\ell}$ we can write the event $\{\tau^{\ell} > t\}$ as $$\bigcup\_{k \in \mathbb N} A\_k := \bigcup\_{k \in \mathbb N} \{X\_s \geq \frac{1}{k}, \ \forall s \in [0,t\_0]\}.$$
By continuity from below, we have that for some $k\_0 \in \mathbb N$ that $P(A\_{k\_0}) > P(\{\tau^{\ell} > t\}) - \frac{\varepsilon}{2}$.
Since $X^{\ell\_n} \to X^{\ell}$ uniformly in probability, for all large enough $n$, we have
$$P(\sup\_{s \in [0, T]} | X\_s^{\ell\_n} - X\_s^{\ell}|
\geq \frac{1}{2k\_0}) < \frac{\varepsilon}{2},$$
so that for all $n > N$ we have
$$\Gamma(\ell\_n)(t\_0) = P(\{\tau^{\ell\_n} > t\_0\}) > P(A\_{k\_0} \cup \{\sup\_{s \in [0, T]} | X\_s^{\ell\_n} - X\_s^{\ell}| > \frac{1}{2k\_0} \}^c) > P(\{\tau^{\ell} > t\_0\}) - \varepsilon$$
For the reverse inequality, we argue similarly - we note that, up to $P$-null sets, we can write the event $\{\tau^{\ell} \leq t\_0\}$ as
$$\bigcup\_{i \in \mathbb N} \{\inf\_{s \in [0, t\_0)}\ X\_s^{\ell} \leq -\frac{1}{i}\},$$
In order for the above representation to hold, we need to show that $\{\inf\_{0\le s\le t}X^{\ell}\_s\le 0\}=\{\inf\_{0\le s\le t}X^{\ell}\_s< 0\}$, up to a $P$-null set.
For this, it suffices to use Lévy's characterization. More precisely, we can write $X^{\ell}\_t=1+t+B\_{\langle X^{\ell}\rangle\_t}\equiv 1+t+B\_{L(t)}$, where $B$ denotes a Brownian motion and $L(t):=\int\_0^t ds/(1+\ell(s))^2$. Therefore
$$\inf\_{0\le s\le t}X^{\ell}\_t = \inf\_{0\le u\le L(t)}\{1+L^{-1}(u)+B\_t\},$$
which implies the desired result as $\inf\_{0\le u\le L(t)}\{1+L^{-1}(u)+B\_t\}$ admits a density.
Thus with a similar calculation, we obtain $$\Gamma(\ell\_n)(t) < P(\{\tau^{\ell} < t\_0\}) + \varepsilon$$
for all large enough $n$.
Since $\varepsilon$ was arbitrary, we conclude that $\Gamma(\ell\_n)(t\_0) \to \Gamma(\ell)(t\_0)$, as was to be shown.
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5
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https://mathoverflow.net/users/173490
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396269
| 163,649 |
https://mathoverflow.net/questions/396271
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6
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Let $\underline{\Omega}\_X^{\bullet}$ denote the Deligne -- Du Bois complex of a normal variety $X$. What kind sigularities satisfy $gr^k\underline{\Omega}\_X^{\bullet}[k] \simeq \Omega\_X^{[k]}$ where $\Omega\_X^{[k]} := j\_\*\Omega^k\_{X^{\operatorname{reg}}}$ and $j\colon X^{\operatorname{reg}} \hookrightarrow X$ is the inclusion of the regular locus of $X$? ADDED LATER: Is there even a map like $\Omega\_X^{[k]} \to gr^k\underline{\Omega}\_X^{\bullet}[k]$ to begin with?
Recall that the case $k=0$ is by definition the Du Bois singularities and in order to define it, one does not need to assume that $X$ is normal. Log canonical singularities (normal) are known to be Du Bois (<https://arxiv.org/abs/0902.0648>).
EDIT: As Donu pointed out, when $X$ is smooth $h^{-k}\omega\_X^{\bullet} = 0$ for $k\neq -\dim X$ but $gr^k\underline{\Omega}\_X^{\bullet}[k] = \Omega\_X^k$. So the following question does not make sense.
One could perhaps ask more generally when is $gr^k\underline{\Omega}\_X^{\bullet}[k] \simeq h^{-k}\omega\_X^{\bullet}$ where $\omega\_X^{\bullet}$ is the dualizing complex of $X$?
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https://mathoverflow.net/users/164620
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Is there a generalization of Du Bois singularities?
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Good question. Your first question has a positive answer when $X$ has finite quotient singularities, i.e. locally analytically $\mathbb{C}^n/G$, with $G\subset GL\_n(\mathbb{C})$ finite. This was essentially proved in Du Bois' original paper "Complexe de de Rham filtré...", see also Steenbrink "Mixed Hodge structure on the vanishing cohomology". I have a hunch (see added comment) this true when $G$ is reductive. There are some other classes as well, like toric singularities. I can say more later, or perhaps Sándor will.
**Added** A better reference for examples would be the book by Navarro Aznar et. al. "Hyperrésolutions cubiques et descente cohomologique".
Also I overlooked your second question: Think about the case when $X$ is smooth. It's almost never true. Perhaps you meant something else?
**2nd edit** I suppose I may as well make my old [notes](https://www.math.purdue.edu/~arapura/preprints/singform.pdf) public. See conjecture 2.8 of that for a more precise statement of what I'm expecting.
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8
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https://mathoverflow.net/users/4144
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396272
| 163,650 |
https://mathoverflow.net/questions/396182
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4
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Let $\mathbf{C}$ be a closed symmetric monoidal category (I probably need even less than this; the examples I have in mind are simply the category of modules over a commutative ring and the category of sets) and let $\omega$ be an (arbitrary) object in $\mathbf{C}$ which I will term the “dualizing object”. Define a contravariant functor $D\colon \mathbf{C} \to \mathbf{C}$ taking $X$ to $[X,\omega]$ (internal Hom) and $X\to Y$ to the composition map $[Y,\omega] \to [X,\omega]$: let us call $DX$ the “dual” of $X$.
Now let $T = D^2$ be the covariant functor taking an object to its “bidual”. Call $\eta\colon 1\_{\mathbf{C}}\to T$ the natural transformation $\eta\_X \colon X \to D^2X = [[X,\omega], \omega]$ obtained from the evaluation map $X \otimes [X,\omega] \to \omega$. And define a natural transformation $\mu\colon T^2 \to T$ by letting $\mu\_X \colon D^4 X \to D^2 X$ be $D(\eta\_{DX})$.
**Fact:** $(T,\eta,\mu)$ is a monad.
This is probably a well-known observation, and it is certainly not difficult (although it is tedious to check, or at least I found it tedious, having to go up to the sexiesdual(!) $D^6X$ of $X$).
* Does this monad have a name? (The “bidual monad” perhaps?) Is there a standard reference for it?
* What are some “natural” occurrences, if any, of algebras for this monad?
(This came up to me by asking myself whether the sequence of iterated biduals $T^n X$ of an object stabilizes: the fact that $T$ is a monad says that, somehow, even though $T^2$ and $T$ are not the same, there is still a form of idempotency to $T$ in the existence of $\mu$.)
|
https://mathoverflow.net/users/17064
|
The bidualizing monad
|
It has been explained by Maxime Ramzi in the comments that this monad simply arises from the adjunction $[-,\omega] \vdash [-,\omega]^{\mathrm{op}}$.
As for the name, it's called the double dualization monad. The classical reference is
>
> A. Kock, *On double dualization monads*, Math. Scand. 27 (1970), 151-165, [pdf](https://www.mscand.dk/article/download/10995/9016)
>
>
>
The double dualization monad classifies algebra structures on a given object, see Theorem 3.2 in Kock's paper. The special case for $\mathbf{Set}$ appeas as Proposition 3.14 in
>
> E. Manes, *Monads of sets*, Handbook of algebra. Vol. 3. North-Holland, 2003, 67-153, [link](https://www.sciencedirect.com/science/article/pii/S1570795403800591)
>
>
>
The double dualization monad also appears in Linton's "contravariant representation theorem", see for example Theorem 3.53 in Manes' article. The classical reference for this is
>
> F.E.J. Linton, *Applied functorial semantics I*, Annali di Matematica Pura ed Applicata 86 (1970), 1–14, [pdf](https://link.springer.com/content/pdf/10.1007/BF02415703.pdf)
>
>
>
The double dualization monad for $\mathbf{Vect}\_k$ and the object $k$ has been discussed at [MO/104777](https://mathoverflow.net/questions/104777/what-are-the-algebras-for-the-double-dualization-monad).
|
7
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https://mathoverflow.net/users/2841
|
396278
| 163,651 |
https://mathoverflow.net/questions/382866
|
2
|
Let $G$ be a compact group with finite-dimensional, real representations $\phi$ and $\psi$ on $V$ and $W$ respectively. (e.g. $V = \mathbb{R}^m$, $W = \mathbb{R}^n$.)
Is it true that, as is the case for finite groups, the dimension of the intertwiner space of the two representations is equal to the inner product of the characters? That is, do we still have
$$
\dim \text{Hom}\_G(V, W) = \int\_G \text{d}\lambda(g) \text{Tr}(\phi(g))\text{Tr}(\psi(g))
$$
where $\lambda$ is the Haar measure on $G$?
As an aside, do the proposed representations always exists? Do they exist if I insist they are unitary/orthogonal?
|
https://mathoverflow.net/users/102255
|
Dimension of intertwiner space: finite-dimensional representations of compact groups
|
Yes, this formula holds, and such representations exist.
If $\mathbb K$ is either $\mathbb R$ or $\mathbb C$ and $G$ is a compact topological group with $\mathbb K$-representations $V$ and $W$, then $Hom\_{\mathbb K}(V,W)$ is another representation of $G$ satisfying
$$ Hom\_G(V,W) = Hom\_{\mathbb K}(V,W)^G.$$
For any representation $U$ of $G$, we have the formula
$$ \dim U^G = \int\_G tr(g|\_U) d\lambda(g).$$
This follows from considering the integral operator
$$ e\_U = \int\_G g|\_U d\lambda(g),$$
which is an idempotent in $End(U)$ projecting onto the invariants $U^G$. Hence $tr(e\_U) = \dim U^G$.
If $V$ and $W$ have characters $\phi$ and $\psi$, then the trace of $g \in G$ on $Hom(V,W)$ is $\phi(g^{-1})\psi(g)$. For compact groups, the representation is conjugate to a unitary representation, and hence $\phi(g^{-1}) = \overline{\phi(g)}$. Assuming your representation is real, this yields your formula.
Finite-dimensional unitary representations over $\mathbb C$ exist by the Peter-Weyl theorem. Restricting scalars to $\mathbb R$ gives real finite-dimensional orthogonal representations.
|
1
|
https://mathoverflow.net/users/125523
|
396288
| 163,652 |
https://mathoverflow.net/questions/396119
|
6
|
*I'd like to close a gap left open in [an old question of mine](https://mathoverflow.net/questions/285651/undetermined-games-of-overdetermined-type); I've tweaked the terminology to be a bit nicer.*
For a (boldface) pointclass $\Gamma$ and a payoff set $G\subseteq\omega^\omega$, say that $G$ is $\Gamma$-**narrow** iff there is some equivalence relation $E\in\Gamma$ such that $E$ has $\le\omega\_1$-many classes and $G$ is $E$-invariant. Gabe Goldberg showed that, provably in $\mathsf{ZFC+\neg CH}$, there is a $(\Sigma^1\_1\wedge\Pi^1\_1)\vee(\Sigma^1\_1\wedge\Pi^1\_1)$-narrow set whose associated Gale-Stewart game is undetermined.
My question is whether this bound can be improved. For example, is there $\mathsf{ZFC+\neg CH}$-provably a $\Pi^1\_1\vee\Sigma^1\_1$-narrow undetermined game? What about $\Sigma^1\_1\wedge\Pi^1\_1$?
Briefly, the motivation for this is that when $\mathsf{CH}$ fails, an assumption of $\Gamma$-narrowness prevents an easy construction of an undetermined game via diagonalization against strategies since there are more strategies than there are basic facts needed to determine the whole payoff set. Producing $\mathsf{ZFC+\neg CH}$ examples of $\Gamma$-narrow undetermined games, especially for "tame" $\Gamma$s, seems to require meaningfully more effort than straight diagonalization. For this reason I'd also be interested in the situation of "$\le\omega\_1$" is replaced with "$<2^{\aleph\_0}$," although I am primarily interested in the precise version above.
|
https://mathoverflow.net/users/8133
|
Are there "very narrow" undetermined games?
|
Solovay's coded club game is $\Pi^1\_1\vee\Sigma^1\_1$-narrow. So $\mathsf{ZFC}$ proves the existence of a $\Pi^1\_1\vee\Sigma^1\_1$-narrow undetermined game. I don't know about $\Pi^1\_1\wedge\Sigma^1\_1.$
It's just optimizing the analysis in the other answer you linked to, [Undetermined games of "overdetermined" type](https://mathoverflow.net/questions/285651/undetermined-games-of-overdetermined-type/285922#285922). I'll use the variant where the sequences aren't required to be increasing; see Hachtman and Palumbo's notes "Notes on determinancy" [http://homepages.math.uic.edu/~shac/determinacy/determinacy2015.pdf](http://homepages.math.uic.edu/%7Eshac/determinacy/determinacy2015.pdf) Definition 14.6.
To streamline the analysis I will change the indexing to treat both players at once, so a point in Baire space $z$ corresponds to a single sequence $((z))\_0,((z))\_1,\dots$ instead of the usual two sequences $((x\_n)\_{n\in\omega},(y\_n)\_{n\in\omega}).$ The conversion is given by $((z))\_{2n}=(x)\_n$ and $((z))\_{2n+1}=(y)\_n.$ I'll regard the $((z))\_n$ as binary relations on $\omega\times\omega,$ so they can be called isomorphic if they are isomorphic as binary relations. $|.|$ means order-type.
Define partial equivalence relations $S,P$ on Baire space by:
* $z S z'$ iff: for some $n,$ both $((z))\_n$ and $((z'))\_n$ fail to be well-founded total orders, and $((z))\_m$ is isomorphic to $((z'))\_m$ for all $m<n.$
* $z P z'$ iff: $((z))\_n$ and $((z'))\_n$ are well-founded total orders for all $n,$ and $\sup\_n(|((z))\_n|+1)=\sup\_n(|((z'))\_n|+1).$
The domains of $S$ and $P$ partition Baire space. So $S\cup P$ is a total equivalence relation. The winner of the coded club game with play $z$ is a function of the equivalence class of $z$ (and the stationary co-stationary set $A\subseteq \omega\_1$ that defines the club game).
$S$ has $|\sum\omega\_1^n|=\aleph\_1$ equivalence classes. $P$ has $|\omega\_1\setminus (\omega+1)|=\aleph\_1$ equivalence classes.
$S$ is $\Sigma^1\_1.$ A witness consists of infinite decreasing sequences in $((z))\_n$ and $((z'))\_{n}$ and isomorphisms $((z))\_m\cong((z'))\_m$ for $m<n.$ It's ok for the witness not to use the minimal possible $n,$ because larger $n$ give a stronger equivalence.
$P$ is $\Pi^1\_1.$ A antiwitness consists of an infinite decreasing sequence in some $((z))\_n$ or $((z'))\_n,$ or a sequence of non-surjective embeddings $f\_n:((z))\_n\to ((z'))\_{m}$ for some fixed $m,$ or a sequence of non-surjective embeddings $f\_n:((z'))\_n\to ((z))\_{m}$ for some fixed $m.$
|
1
|
https://mathoverflow.net/users/164965
|
396290
| 163,653 |
https://mathoverflow.net/questions/396297
|
11
|
My question is about the Hamming Weight (or number of 1's in binary expansion) of $a\_n = \frac{2^{\varphi(3^n)}-1}{3^n}$ [A152007](https://oeis.org/A152007)
For example, $a\_3 = 9709 = (10110111101001)\_2 $ has nine 1's in binary expansion
I guess the answer is $3^{(n-1)}$ but I can't prove it
Is that correct?
|
https://mathoverflow.net/users/36456
|
Number of 1's in binary expansion of $a_n = \frac{2^{\varphi(3^n)}-1}{3^n}$
|
It is, indeed, correct. Notice first that $2-(-1)=3$ is divisible by $3$, so by lifting-the-exponent lemma the number
$$
A=\frac{2^{3^{n-1}}-(-1)^{3^{n-1}}}{3^n}=\frac{2^{3^{n-1}}+1}{3^n}
$$
is an integer. Notice also that for $n>0$ it has less than $3^{n-1}$ binary digits. Assume that it has $m$ binary digits. We have
$$
\frac{2^{\varphi(3^{n})}-1}{3^n}=\frac{2^{2\cdot 3^{n-1}}-1}{3^n}=A(2^{3^{n-1}}-1).
$$
Next, let $l=3^{n-1}-m$, $B=2^l-1$ and $C=2^m-A$. We claim that your number's binary expansion looks like a concatenation of $A-1$, $B$ and $C$, where we also add some zeros in expansion of $C$ until we get exactly $m$ digits. So, we should have
$$
a\_n=C+2^mB+2^{m+l}(A-1).
$$
This equality is true, because
$$
C+2^mB+2^{m+l}(A-1)=2^m-A+2^m(2^l-1)+2^{m+l}(A-1)=
$$
$$
=(A-1)(2^{m+l}-1)+2^{m+l}-1=A(2^{m+l}-1)=A(2^{3^{n-1}}-1).
$$
Now, if the sum of digits of $A-1$ is equal to $s$, then the sum of digits of $C$ is $m-s$ and the sum of digits of $B$ is, of course, $l$, so the sum of digits of $a\_n$ is
$$
s+m-s+l=m+l=3^{n-1},
$$
as needed.
For the particular case $n=3$, described in the post, $A=10011\_2=19=\frac{2^9+1}{27}$ and $C=2^5-A=13=01101\_2$ (notice the added zero)
|
22
|
https://mathoverflow.net/users/101078
|
396300
| 163,656 |
https://mathoverflow.net/questions/396125
|
2
|
Let $M$ be a 3-dimensional complex manifold, and $\Lambda$ a discrete lattice in $\mathbb C^2$. Suppose there is a holomorphic submersion $f:M\to\mathbb{C}^2/\Lambda$ with fibers given by 1-dimensional compact complex manifolds. And these fibers form the leaves of a 1-dimensional holomorphic foliation $\mathcal{W}$.
Then can we have a more specific description of $M$? Is $M$ a fiber bundle or even a trivial bundle over $\mathbb{C}^2/\Lambda$? Can we say something about the leaves of $\mathcal{W}$?
|
https://mathoverflow.net/users/167284
|
Complex fibration over complex torus
|
**Added.** Thanks to the comment of abx below I understood that there was a big gap in the reasoning, and $M$ doesn't need to be a fiber bundle.
**Example.** I'll construct an example $S$ of a complex surface that admits a submersion to an elliptic curve $E$ but that is not a fiber bundle over $E$. Then $M$ can be taken as $S\times E'$, where $E'$ is any elliptic curve (and $M$ admits a submersion to $E\times E'$).
So, let's start with the elliptic curve $E$ and any complex curve $C\_1$ (say of genus $>2$) that admits a fixed point free involution $\sigma\_1$. Now take any degree two cover $C\_2$ of $E$ with ramifications (so that the genus of $C\_2$ is $>1$). Let $\sigma\_2$ be the corresponding involution of $C\_2$. Finally consider the quotient of $C\_1\times C\_2$ by $\mathbb Z\_2 $ that is acting on the $C\_1$ factor by $\sigma\_1$ and on the $C\_2$ factor by $\sigma\_2$. The resulting suface $(C\_1\times C\_2)/\mathbb Z\_2$ admits a submersion to $E$.
**(corrected) Old answer.** The fact that such a manifold is a submersion implies that all fibers are smooth. As the above example shows, the submersion doesn't need to be a fiber bundle, but in case it *is* a fiber bundle something can be said.
Namely, one can say that the fiber bundle is isotrivial, i.e. all fibers are isomorphic curves. Indeed, we can associate to such a manifold a holomorphic map from $\mathbb C^2$ to the corresponding Teichmüller space, and since the latter space is a bounded domain, the map is constant.
At the same time we can't claim that the fibration is trivial, it can be isotrivial. Indeed, one can find curves $\Sigma$ that admit a non-trivial holomorphic action of $\mathbb Z^4$ on them. Then we can take a quotient of $\Sigma\times \mathbb C^2$ by such an action. It should not be hard to classify all such actions for small $2g-2=\chi(\Sigma)$, but I guess that for larger $g$ this might be difficult.
|
1
|
https://mathoverflow.net/users/943
|
396303
| 163,658 |
https://mathoverflow.net/questions/396310
|
5
|
Let $\ n\ $ be an arbitrary natural number ($\ 1\ 2\ \ldots).\ $ Then
* $\ n\ $ is coarse $\ \Leftarrow:\Rightarrow\ $ there exists a prime divisor $p$ of $\ n\ $ such that $\ p^3>n.$;
* $\ n\ $ is a p-cube $\ \Leftarrow:\Rightarrow\ $ the positive cubical root of $\ n\ $ is a prime number;
* $\ n\ $ is fine $\ \Leftarrow:\Rightarrow\ p^3<n\ $
for every prime divisor $\ p\ $ of $\ n$.
**Example:** Natural $\ 64\ $ and
$$ 4095=64^2-1\ = 3^2\cdot 5\cdot 7\cdot13 $$
are both fine. However,
$$ 4097=64^2+1=17\cdot241 $$
is coarse.
**QUESTION** Does there exist a fine natural number $\ n\ $ such that both $\ n^2-1\ $ and $\ n^2+1\ $ are fine too? (*My guess: perhaps NOT*).
Also, I don't expect that there is any p-cube $\ n\ $ such that both $\ n^2-1\ $ and $\ n^2+1\ $ are fine.
On the other hand, I believe that there are infinitely many coarse $\ n\ $ such that both $\ n^2-1\ $ and $\ n^2+1\ $ are fine (as rare as they may be).
|
https://mathoverflow.net/users/110389
|
Can all three numbers $\ n\ \ n^2-1\ \ n^2+1\ $ be fine (as opposed to coarse)?
|
$n = 2673$ has largest prime factor $11$ whose cube is $1331$.
$n^2 - 1 = 7144928$ has largest prime factor $191$ whose cube is $6967871$.
$n^2 + 1 = 7144930$ has largest prime factor $61$ whose cube is $226981$.
|
10
|
https://mathoverflow.net/users/46140
|
396318
| 163,661 |
https://mathoverflow.net/questions/396071
|
1
|
This question is an offshoot of [this closely related MO question](https://mathoverflow.net/questions/393738).
Here, we consider the Diophantine equation
$$m^2 - p^k = 2^r t,$$
where $r \geq 2$ and $\gcd(2,t)=1$.
Furthermore, we place the following restrictions:
$$p \equiv k \equiv 1 \pmod 4$$
$$2^r \neq t$$
$$2^r t > \frac{3373m^2}{3375} \approx {10}^{750}$$
$$\min(t,2^r) < m < \dfrac{1+\sqrt{1+2^{r+2}t}}{2} < \max(t,2^r) \text{ based on mathlove's answer }$$[here](https://math.stackexchange.com/a/4177978/28816).
Basically, we are trying to check whether there is a solution to the said Diophantine equation satisfying $p^k < m$, and all of the other constraints above. (Thus, we would like to determine whether there is a counterexample to our conjecture that $m < p^k$.)
From this [answer to a closely related MSE question](https://math.stackexchange.com/a/3555207/28816), we know that
$$m = 9, p = 5, k = 1, r = 2, t = 19$$
is a counterexample to $m < p^k$, except that it does not satisfy the third restriction above, which must necessarily be satisfied by an odd perfect number $p^k m^2$ with special prime $p$.
We tried to search for other counterexamples via [Sage Cell Server](https://sagecell.sagemath.org/), but it is currently unable to do this search in the range that we require.
Could somebody out there with more computing power, and more adept programming skills in e.g. Python, lend us a hand please?
|
https://mathoverflow.net/users/10365
|
On the Diophantine equation $m^2 - p^k = 2^r t$, where $r \geq 2$ and $\gcd(2,t)=1$
|
I think that
$$m=10^{375}+1,p=5,k=1,r=2,t=25\cdot 10^{748}+5\cdot 10^{374}-1\tag1$$
is a solution.
*Proof* :
When $m=10^{375}+1,p=5,k=1$, we have
$$m^2-p^k=(10^{375}+1)^2-5\equiv 1-5\equiv 4\pmod 8$$
from which $r=2$ and
$$t=\frac{(10^{375}+1)^2-5}{2^r}=25\cdot 10^{748}+5\cdot 10^{374}-1$$
follow.
Now, (1) satisfies
* $r \geq 2$
* $\gcd(2,t)=1$
* $2^r \neq t$
* $p \equiv k \equiv 1 \pmod 4$
* $p^k\lt m$
* $\dfrac{3373m^2}{3375} \approx {10}^{750}$
Now, (1) satisfies $2^r t > \dfrac{3373m^2}{3375}$ since
$$2^rt-\frac{3373m^2}{3375}=m^2-5-\frac{3373m^2}{3375}=\frac{2m^2-5\times 3375}{3375}\gt 0$$
(1) satisfies $\min(t,2^r) < m$ since
$$m-\min(t,2^r)=m-2^r=m-4\gt 0$$
(1) satisfies $m < \dfrac{1+\sqrt{1+2^{r+2}t}}{2}$ since
$$\begin{align}m < \dfrac{1+\sqrt{1+2^{r+2}t}}{2}&\iff 2m-1\lt \sqrt{1+2^{r+2}t}
\\\\&\iff 2m-1\lt \sqrt{1+4(m^2-5)}
\\\\&\iff (2m-1)^2\lt 1+4(m^2-5)
\\\\&\iff m\gt 5\end{align}$$
which does hold.
(1) satisfies $\dfrac{1+\sqrt{1+2^{r+2}t}}{2} < \max(t,2^r)$ since
$$\begin{align}\dfrac{1+\sqrt{1+2^{r+2}t}}{2} < \max(t,2^r)&\iff \dfrac{1+\sqrt{1+16t}}{2} < t
\\\\&\iff \sqrt{1+16t}\lt 2t-1
\\\\&\iff 1+16t\lt (2t-1)^2
\\\\&\iff t\gt 5\end{align}$$
which does hold.
So, it follows that $(1)$ is a solution.
|
1
|
https://mathoverflow.net/users/34490
|
396321
| 163,663 |
https://mathoverflow.net/questions/396326
|
16
|
The following real $2 \times 2$ matrix has determinant $1$:
$$\begin{pmatrix}
\sqrt{1+a^2} & a \\
a & \sqrt{1+a^2}
\end{pmatrix}$$
The natural generalisation of this to a real $2 \times 2$ block matrix would appear to be the following, where $A$ is an $n \times m$ matrix:
$$\begin{pmatrix}
\sqrt{I\_n+AA^T} & A \\
A^T & \sqrt{I\_m+A^TA}
\end{pmatrix}$$
Both $I\_n+AA^T$ and $I\_m+A^TA$ are positive-definite so the positive-definite square roots are well-defined and unique.
Numerically, the determinant of the above matrix appears to be $1$, for any $A$, but I am struggling to find a proof. Using the Schur complement, it would suffice to prove the following (which almost looks like a commutativity relation):
$$A\sqrt{I\_m + A^TA} = \sqrt{I\_n + AA^T}A$$
Clearly, $A(I\_m + A^TA) = (I\_n + AA^T)A$. But I'm not sure how to generalise this to the square root. How can we prove the above?
|
https://mathoverflow.net/users/119987
|
Proof that block matrix has determinant $1$
|
Write the SVD of $A$, say $A=PDQ^T$ with $D$ diagonal with non-negative entries and $P\in O(n),Q\in O(m)$. Then $\sqrt{I\_n + AA^T} = P\sqrt{1+D^2}P^T$
and $\sqrt{I\_m+ A^TA} = Q\sqrt{1+D^2}Q^T$. This gives
$$
\begin{pmatrix}
\sqrt{I\_n + AA^T} & A \\ A^T& \sqrt{I\_m+A^TA}
\end{pmatrix}
=
\begin{pmatrix}
P & 0 \\
0 & Q
\end{pmatrix}
\begin{pmatrix}
\sqrt{I\_n + D^2} & D \\ D & \sqrt{I\_m+D^2}
\end{pmatrix}
\begin{pmatrix}
P^T & 0 \\
0 & Q^T
\end{pmatrix}.
$$
Up to permutation, the matrix in the middle is diagonal by block with $n$ blocks given by 2x2 matrices of the same form as in the question.
|
25
|
https://mathoverflow.net/users/141760
|
396329
| 163,666 |
https://mathoverflow.net/questions/396314
|
4
|
Let $\mathcal{H}$ be a separable Hilbert space and let $TC( \mathcal{H})$, $HS(\mathcal{H})$ be the space of trace-class operators and Hilbert-Schmidt operators on $\mathcal{H}$. Recall that these space are Banach spaces and that $HS(\mathcal{H})$ is a even a Hilbert space. For a Banach space $X$ let $B(X)$ be the space of bounded operators on $X$.
For $a \in B(TC( \mathcal{H}))$ since $TC( \mathcal{H})$ is dense in $HS(\mathcal{H})$ there is at most bounded linear extension, but it may not be bounded.
For example one may consider $a(x) = Tr(x)y $ where $y \neq 0$ is a trace-class operator.
Now, consider the set
\begin{align\*}
\mathcal{A} = \left \{ a \in B(TC( \mathcal{H})) \mid \bar a \in B(HS(\mathcal{H})) \right \}
\end{align\*}
what can I say about this space?
It is a vector space and by the example it is not all of $ B(TC( \mathcal{H}))$.
Can I for example define an involution by taking
$\tilde a = \bar a^\*\mid\_{TC}$?
|
https://mathoverflow.net/users/143779
|
Which operators on the trace-class operators extend to operators on Hilbert-Schmidt operators?
|
No, this space is not closed under involution. Choose $A \in TC$ and $B \in HS\setminus TC$ and consider the rank one operator $T \mapsto \langle T, B\rangle A$ on $HS$. This restricts to a bounded operator on $TC$ but its adjoint doesn't.
|
6
|
https://mathoverflow.net/users/23141
|
396332
| 163,668 |
https://mathoverflow.net/questions/396344
|
2
|
Let $(\mathbb R^{1+2},\eta)$ be Minkowski with the metric $\eta= -dt^2+(dx^1)^2+(dx^2)^2$. Suppose $\Sigma$ is a smooth timelike hypersurface and denote by $h$ the second fundamental form on $\Sigma$. Assume that
$$ h(N,N) \geq 0 \quad \forall N\in L\_p\Sigma \quad \text{and}\quad p \in \Sigma,$$
where $L\_p\Sigma= \{N \in T\_p\Sigma\,:\, \eta(N,N)=0\}$. Assume also that there is **no** point $p$ on $\Sigma$ with the property that $h(N,N)>0$ for all nontrivial $N \in L\_p\Sigma$. Does it follow that $\Sigma$ is part of a hyperboloid?
|
https://mathoverflow.net/users/50438
|
Hyperboloids in Minkowski geometry
|
By hyperboloid, you seem to mean the one-sheeted surface $\{ |x|^2 - t^2 = 1\}$; this surface is umbilical with the induced metric proportional to the mean curvature, and so $h(N,N) = 0$ for all $N\in L\_p\Sigma$.
Before answering your question, a couple comments:
1. It is meaningless to specify $h(N,N) \geq 0$ with a particular sign: the choice of the sign of the mean curvature scalar depends on the choice of orientation of the surface.
2. If you have a time-like surface in $\mathbb{R}^{2,1}$, the set $L\_p\Sigma$ is the union of two lines. Locally (and in fact globally under mild assumptions) there exists two non-vanishing linearly independent vector fields $\ell, n$ generating $L\_p\Sigma$. Your requirements is essentially stating that $h(\ell,\ell)h(n,n) = 0$ and $h(\ell,\ell)+h(n,n)$ is signed.
Certainly it is not necessary for $\Sigma$ to be part of the hyperboloid for this to hold. As Leo Moos pointing out in the comment, just taking $\Sigma$ to be a time-like hyperplane guarantees that $h \equiv 0$ and hence your requirement is satisfied. More generally, if you take any time-like surface $\Sigma$ with the property that it is ruled by null lines, then this guarantees one of $h(\ell,\ell)$ or $h(n,n)$ vanishes identically. It then suffices to bend the surface so that the other family of null curves accelerate only in one way.
For example: you can let $\Sigma$ be the graph $\{x^1 = f(t, x^2)\}$ where $f(t,x^2) = g(t - x^2)$ with $|g'| < 1$ and $g'' \geq 0$.
|
4
|
https://mathoverflow.net/users/3948
|
396349
| 163,672 |
https://mathoverflow.net/questions/330771
|
7
|
Let $G$ be a compact Lie group. The classical Peter-Weyl theorem shows that $L^2(G)$ splits into an orthogonal direct sum of irreducible finite-dimensional unitary representations of $G$. This is a powerful statement as it allows to answer questions about functions on $G$ in terms of matrix coefficients of irreducible representations.
I was wondering if there exist a decomposition of the space $\mathfrak{X}(G)$ of vector fields on $G$ that has a similar spirit than the Peter-Weyl theorem. In particular, I was hoping that the gradient map $C^\infty(G) \to \mathfrak{X}(G)$ (with respect to the Riemannian metric induced by the Killing form of $G$) has a nice behavior with respect to the decompositions of the spaces on both sides. I'm a bit vague here because I don't know what I can hope for. In the best case, the gradient map is diagonalized (similar to how the Laplacian is diagonalized by the classical Peter-Weyl theorem).
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https://mathoverflow.net/users/17047
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Peter–Weyl theory for vector fields
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Let $\mathfrak g \to \mathfrak{X}(G)$ be the inclusion of right-invariant vector fields on $G$ into vector fields on $G$. Then we have an isomorphism $C^\infty(G) \otimes\_{\mathbb R} \mathfrak g \to \mathfrak{X}(G)$ of left $C^\infty(G)$-modules defined by $f \otimes\_{\mathbb R} x \mapsto fx$. It is $G \times G$-equivariant. Applying Peter-Weyl to $C^\infty(G)$ gives the desired decomposition. The gradient $C^\infty(G) \to \mathfrak{X}(G)$ is $G\times G$-equivariant if the metric is biinvariant.
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6
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https://mathoverflow.net/users/125523
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396360
| 163,677 |
https://mathoverflow.net/questions/396368
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1
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Does there exist a Borel (or even continuous) function $f:\mathcal{C}\to\mathcal{C}$, where $\mathcal{C}$ is the Cantor set (or Cantor space $2^\omega$) such that for every nonempty closed perfect set $P\subseteq\mathcal{C}$, $f|P$ maps surjectively onto $\mathcal{C}$?
Such functions (on $\mathbb{R}$) are called *perfectly everywhere surjective* here: <https://core.ac.uk/download/pdf/83599431.pdf>, but the maps constructed there rely on, in essence, a well-ordering of $\mathbb{R}$ and are likely far from any kind of measurability.
Hoping that the self-similarity of the Cantor set could be exploited here.
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https://mathoverflow.net/users/16107
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A Borel perfectly everywhere surjective function on the Cantor set
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As suggested by the comment above, the answer is no. The relevant fact is that every nonmeager subset of $\mathcal{C}$ with the Baire property (in particular, any Borel set) contains a nonempty closed perfect set.
Suppose that there was a Borel function $f$ with this property. Observe that for every $p\in\mathcal{C}$, $f^{-1}(\{p\})$ is a comeager Borel set: $f^{-1}(\{p\})$ is a Borel set which meets every nonmeager set, since nonmeager subsets of $\mathcal{C}$ must contain perfect sets, so its complement is meager.
But then, if we take $p\in\mathcal{C}$, $f^{-1}(\{p\})$ contains a perfect set which maps only onto $p$, a contradiction.
This argument works for any function which is Baire measurable, a similar argument can be given for Lebesgue measurable.
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3
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https://mathoverflow.net/users/16107
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396371
| 163,682 |
https://mathoverflow.net/questions/192314
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12
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I'm broadly interested in notions of "generic presentability" - when a given object exists in *every* forcing extension of the universe by some fixed forcing, at least up to the appropriate notion of equivalence. Sometimes this is boring - [per Solovay](https://mathoverflow.net/questions/155915/who-proved-sets-in-every-generic-are-already-in-the-ground-model?noredirect=1&lq=1), the only "generically presentable sets up to equality" are those already in $V$ - but other times it can be more interesting. In particular, the appropriate notion of "generically presentable countable structure" is nontrivial ([1](https://www.cambridge.org/core/journals/journal-of-symbolic-logic/article/computable-structures-in-generic-extensions/A13DB0FC4DB8F0BD7F9F2D9FFC18DFCE), [2](https://onlinelibrary.wiley.com/doi/abs/10.1002/malq.201400094)).
I'd like to ask about an intermediate notion, that of **generically presentable cardinals** (or if one prefers, generically presentable sets up to *equipollence* instead of equality):
>
> A generically presentable cardinal is a pair $(\nu,\mathbb{P})$ where $\nu$ is a $\mathbb{P}$-name and we have $$\Vdash\_{\mathbb{P}^2}\nu[G\_0]\equiv\nu[G\_1].$$
>
>
>
Over $\mathsf{ZFC}$, the generically presentable cardinalities are boring in the sense that for every generically presentable cardinal $(\nu,\mathbb{P})$ there is some $a\in V$ such that $\Vdash\_\mathbb{P}\nu[G]\equiv\check{a}$. However, this uses choice in a crucial way (by choice we WLOG have $\Vdash\_\mathbb{P}\nu[G]\in Ord$, and now we just observe that forcing adds no new ordinals). So this leaves the following question open:
>
> Is there a $V\models ZF$ containing a generically presentable cardinal $(\nu,\mathbb{P})$ such that $$\Vdash\_{\mathbb{P}^2}\forall a\in V(\nu[G\_0]\not\equiv a),$$ or at least such that $$\Vdash\_{\mathbb{P}}\forall a\in V(\nu[G\_0]\not\equiv a)?$$
>
>
>
Note that per Laver/Woodin, this makes sense: the ground model is appropriately definable in its forcing extensions. (And the two questions are indeed distinct, since two non-equipollent sets can become equipollent after further forcing. So an affirmative answer to the former implies an affirmative answer to the latter, but not obviously conversely.)
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https://mathoverflow.net/users/8133
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A new cardinality living in every forcing extension?
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Partial answer. There is a $V\models ZFA$ containing a generically presentable cardinal $(\nu,\mathbb{P})$ such that for all $a\in V$ we have $\Vdash\_{\mathbb{P}^2}\nu[G\_0]\not\equiv\check{a}.$ This should be equivalent to your first condition, assuming the first condition can even be stated i.e. the ground model is definable in the ZFA setting.
I am not sure about ZF models. (But I *don’t* see a reason why Pincus’ transfer theorems *don’t* apply, using the “surjectively boundable” condition for example. I think we only need to consider $a$ that don’t admit a surjection to $\omega\_1,$ because the below model should give $\Vdash\_{\mathbb{P}^2}\text{there is no surjection }\nu[G\_0]\to\omega\_1.$ And the forcing relation should be equivalent to some fairly tame sentence, surely?)
$
\DeclareMathOperator{dom}{dom}
\DeclareMathOperator{rng}{rng}
\DeclareMathOperator{supp}{supp}
$
Take $V$ to be the basic Fraenkel model, with set of atoms $A.$ Every set $x\in V$ has a minimal finite support $\supp(x)\subset A.$ So $x$ is fixed by permutations that fix each element of $\supp(x).$
Conditions $p\in \mathbb P$ are bijections such that $\dom p$ and $\rng p$ are disjoint finite subsets of $A.$ Graphically, imagine a finite set of vertex-disjoint directed edges using the vertex set $A.$ As usual $p\leq q$ if $p\supseteq q.$
Take $\nu$ to be the name for $\dom \bigcup G.$
I’ll make use of the dense subset $\mathbb Q\subset \mathbb P^2$ consisting of pairs $p=(p\_0,p\_1)$ such that $\dom p\_0 \cup \rng p\_0 = \dom p\_1 \cup \rng p\_1.$ Graphically, these conditions are a finite union of cycles (possibly 2-cycles) using the vertex set $A.$ In each cycle edges are alternately colored $0$ and $1,$ and every edge has a direction. (Note these aren’t “directed cycles”, just cycles with directed edges - the direction on different edges are unrelated.)
The interesting part is that these cycles have no orientation-reversing symmetry. Graphically, a reflection of a regular polygon would have to pass through the midpoint of an edge, which is ruled out by the directions, or pass through a vertex, which is ruled out by the coloring. Pick an isomorphism-invariant choice function for orientations of alternating-colored cycles with directed edges. This is kosher because we can define these using only pure sets, and in fact this is completely finitary. This gives a bijection $\nu[G\_0]\to \nu[G\_1]$ by identifying each set with edges of the appropriate color, and sending each edge in a cycle one step along the chosen orientation.
The rest of the argument is just the usual trick of making use of compatible conditions related by symmetries. Suppose for contradiction that there exists $z\in V$ and $p\in\mathbb Q$ and a $\mathbb Q$-name $\dot{f}$ such that $$p\Vdash\text{$\dot{f}$ is a bijection $\check{z} \to \nu[G\_0]$}$$
where $\Vdash=\Vdash\_{\mathbb Q}.$
By refining $p$ if necessary we can assume that the supports of $z$ and $\dot{f}$ are contained in $A\_0:=\dom p\cup \rng p.$ There exists $q\leq p$ and elements $x\_a\in z$ for $a\in A\_0$ such that $q\Vdash \bigwedge\_{a\in A\_0}\dot{f}(\check{x\_a})=\check{a}.$ Consider a permutation $\pi$ fixing $A\_0$ but moving every other atom used by $q,$ and every $x\_a,$ to unused atoms. Then $q$ and $\pi q$ are compatible, which shows that $\supp(\check{x\_a})\subseteq A\_0,$ so the choice of $q$ was irrelevant:
$$p\Vdash \bigwedge\_{a\in A\_0}\dot{f}(\check{x\_a})=\check{a}.$$
Pick an element $x\in z\setminus \{x\_a:a\in A\_0\}.$ Pick $r\leq p$ such that each element of $\supp(x)\setminus A\_0$ is in a distinct 2-cycle of $r.$ Let $A\_1$ be the set of elements of cycles in $r$ that use an element of $\supp(x)\setminus A\_0.$ Refine $r$ if necessary such that: for all $y=\pi x$ where $\pi$ fixes $A\_0$ and $\supp y\subseteq A\_1,$ there exists $a\_y\in A$ with $$r\Vdash \dot{f}(\check y)=\check{a\_y}$$
For each such $a=a\_y$ we have $a\not\in A\_0.$ Suppose for contradiction that $a\not\in A\_1.$ Pick $\pi$ fixing $A\_0\cup A\_1$ but sending all other atoms used by cycles in $r$ to unused atoms, and such that $a\neq \pi a.$ Then $\pi r$ and $r$ are compatible, which gives the contradiction $r\cup \pi r\Vdash \check a=\pi \check a.$ So $a\in A\_1.$
Each $a\_y$ lies in the set $\dom r\_0\cap A\_1$ of order $|A\_1|/2,$ but there are at least $\binom{|A\_1|}{|A\_1|/2}> |A\_1|/2$ choices of $y$ (with $\binom{0}{0}=1>0$) because sets with distinct supports are distinct.
I think this all generalizes to $\kappa$-closed conditions, using a generalization of the basic Fraenkel model to supports of order $<\kappa.$ However, the cycles can then be infinite, so the choice function for orientations now seems to genuinely need choice, for pairs of countable sets of reals.
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1
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https://mathoverflow.net/users/164965
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396375
| 163,683 |
https://mathoverflow.net/questions/396381
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1
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Suppose that $A = M\_n(\mathbb{C})$ be the algebra of $n\*n$ matrices over $\mathbb{C}$.
If com(A) = {$B \in M\_n(\mathbb{C}); AB = BA$}, then what is the $dim(com(A))?$
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https://mathoverflow.net/users/137242
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Dimension of commutant
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This is known for a general field by a theorem of Frobenius:
Let $F$ be a field and $V$ a finite dimensional $F$-vector sapce with a linear operator A. When $f\_i(X)$ denote the invariant factors of $A$ (such that $f\_i(X)$ divides $f\_{i+1}(X)$, then the dimension is equal to $\sum\limits\_{i=1}^{k}{(2k-2i+1)deg(f\_i(X))}$.
See Theorem 5.15 in the book "Algebra: An approach via module theory" by Adkins and Weintraub.
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6
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https://mathoverflow.net/users/61949
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396383
| 163,686 |
https://mathoverflow.net/questions/396378
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2
|
*This note is related to*
[Can all three numbers $\ n\ \ n^2-1\ \ n^2+1\ $ be fine (as opposed to coarse)?](https://mathoverflow.net/questions/396310/can-all-three-numbers-n-n2-1-n21-be-fine-as-opposed-to-coarse)
---
Let
$$ m\ n\ \in\ \mathbb N\_{\_{>1}}\ :=\ \{x\in\mathbb Z: x>1\} $$
be arbitrary. Let $\ P(n)\ $ be the largest prime divisor of $n$.
**Definition:** Molecularity of $n$ is
$$ M(n)\ :=\ \log\_{P(n)}(n) $$
Instantly,
**Theorem**
* $ M(n) \ge 1;$
* $ M(n)=1\quad\Leftrightarrow\quad p\ $ is a prime;
* $ M(n^k)\ =\ k\cdot M(n)\qquad $ (for every $\ k=1\ 2\ \ldots);$
* $ M(m\cdot n)\ \le\ M(m)+M(n);$
* $ M(m\cdot n)=M(m)+M(n)\quad\Leftrightarrow\quad
P(\gcd(m\ n))\ =\ P(m\cdot n). $
For instance:
$$ n>3\quad\Rightarrow\quad M(n^2-1)\ <\ M(n-1)+M(n+1) $$
**Question 1:** What is
$$ \sup\_{n>2}\ \min(M(n-1)\,\ M(n)\,\ M(n+1))\quad ?$$
**Question 2:** What is
$$ \inf\_{n>2}\ \frac1{M(n-1)}+\frac1{M(n)}+\frac1{M(n+1)}
\quad? $$
The ever-sharper bounds would be greatly appreciated.
I dare, this time with a greater probability, that the above sup is $\ \le 4,\ $ and that $4$ cannot be actually attained.
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https://mathoverflow.net/users/110389
|
Molecularity $\ M(n)$
|
For every $u>0$, there exists $n$ such that each of $P(n-1)$, $P(n)$, $P(n-1)$ is less than $n^u$. This was proved by Eggleton and Selfridge (Consecutive integers with no large prime factors, J. Austral. Math. Soc. Ser. A 22 (1976), 1–11). In fact their proof is constructive (see pp. 2-3 of their paper). It follows that the supremum in Question 1 is infinite, while the infimum in Question 2 is zero.
I should add that this phenomenon also holds for an arbitrary long string of consecutive integers. For example, there exists $n$ such that each of $P(n-50)$, $P(n-49)$, ..., $P(n+50)$ is less than $n^u$. This was proved by Balog and Wooley (On strings of consecutive integers with no large prime factors, J. Austral. Math. Soc. Ser. A 64 (1998), 266–276).
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5
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https://mathoverflow.net/users/11919
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396385
| 163,687 |
https://mathoverflow.net/questions/395939
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9
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Might there be a good historical reference on Egyptian number theory ($ \sim 2000$ B.C.)? The following online reference by a professor at the UCLA indicates that they were aware of the Pythagorean theorem [1]. This makes me wonder whether Egyptian scientists and engineers might have done fundamental work in number theory as well.
In particular, I'd like to know whether they were aware of Euclid's theorem of the infinitude of primes.
References:
-----------
1. Allen Klinger. Right Triangles - Pythagorean Theorem
. [http://web.cs.ucla.edu/~klinger/dorene/math1.htm](http://web.cs.ucla.edu/%7Eklinger/dorene/math1.htm)
2. Thomas Eric Peet. Mathematics in Ancient Egypt. 1931.
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https://mathoverflow.net/users/56328
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Egyptian number theory
|
According to the chapter on Egyptian mathematics and astronomy in the book by O. Neugebauer, *The Exact Sciences in Antiquity* (1951, 1957), Egyptian mathematics only involved basic arithmetic with positive integers and a restricted class of fractions. So the answer to your question appears to be "no".
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2
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https://mathoverflow.net/users/106467
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396388
| 163,690 |
https://mathoverflow.net/questions/396390
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4
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Let $\mathbf X := (X, \mathcal S, \mu, T)$ be an ergodic measure preserving system with finite measure such that for every increasing sequence $\{n\_k\}$ of natural numbers with positive lower density, we have that for all $f \in L^1(X)$,
$$\frac{1}{N} \sum\_{i=0}^{N-1} f(T^{n\_i} x) \to \int\_X f d\mu$$
as $n \to \infty$ for a.e. $x \in X$.
>
> **Question:** Is $\mathbf X$ necessarily weakly mixing?
>
>
>
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https://mathoverflow.net/users/173490
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A sufficient condition for an ergodic system to be weakly mixing
|
I think $\mathbf{X}$ has to be weakly mixing.
In order to prove it, note that one characterization of weak mixing is that the Koopman operator $U\_Tf:= f\circ T$ does not have any eigenvectors on the space $L^{2}\_{0}$ of mean zero square-integrable functions. Suppose then that there is an eigenvector $f$ such that $f\circ T = \lambda f$. Your assumption says that for any increasing sequence $(n\_k)$ of positive lower density we have
$$
\lim\_{N\to \infty} \left(\frac{1}{N} \sum\_{k=0}^{N-1} \lambda^{n\_k}\right) f(x) = 0
$$
for a.e. $x\in X$. It means that $\lim\_{N\to \infty} \frac{1}{N} \sum\_{k=0}^{N-1} \lambda^{n\_k} = 0$. We now have to find an appropriate sequence $(n\_k)$ for which this convergence does not hold.
Note that the sequence $(\lambda^n)$, where $\lambda = \exp(i\varphi)$, is an orbit of the rotation by an angle $\varphi$ on the unit circle. If $\lambda$ is a root of unity of order $m$ then we can take $n\_k:= km$ so that $\lambda^{n\_k}=1$ -- we have convergence to $1$. If $\lambda$ is not a root of unity then we have an irrational rotation, so the orbit is equidistributed on the unit circle. In particular it means that the set of natural numbers for which $\lambda^n$ lands in a small neighbourhood around $1$ has positive lower density. We can use this subset as our sequence $(n\_k)$ and then $\frac{1}{N} \sum\_{k=0}^{N-1} \lambda^{n\_k}$ cannot converge to $0$.
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3
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https://mathoverflow.net/users/24953
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396394
| 163,691 |
https://mathoverflow.net/questions/396384
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1
|
Let $B=\left\{ \left(x,y\right)\in\mathbb{R}^{2}:x^{2}+y^{2}<1\right\} $
be the unit ball in $\mathbb{R}^{2}.$
Can we construct a subharmonic
function $f:B\rightarrow\left[-\infty,0\right]$ such that
$$
0<\int\_{\widetilde{B}}\left(1-x^{2}-y^{2}\right)^{-2}dV<\infty,
$$
where $\widetilde{B}=\left\{ \left(x,y\right)\in B:-1<f\left(x,y\right)\right\} $? Here $dV$ is the standard Lebesgue measure on $\mathbb{R}^{2}$.
I thought that in order to answer this we need to control (understand) the growth
on sublevel sets of a subharmonic function.
The question was already asked on mathstackexchange and received no answers.
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https://mathoverflow.net/users/310212
|
A subharmonic function with a growth property
|
The answer is "yes". Let $E$ be some Jordan region in the unit disk on which
$$\int\_E(1-x^2-y^2)^{-2}dxdy<\infty,$$
and such that $E$ contains $[0,1)$, and the closure of $E$ is contained in the open unit disk, except the point $1$. Let $\phi$ be a conformal map of $E$ onto the right half-plane, such that $\phi(1)=\infty$.
Let $u=\Re \phi$. Then $u$ is positive and harmonic n $E$, zero on $\partial E\backslash\{1\}$. Extend $u$ to the whole unit disk by setting is equal to $0$
outside $E$. Then take $f=u-2.$
As you see from this example, the weight $(1-|z|^2)^{-2}$ is not very relevant. You can replace it by anything you like.
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1
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https://mathoverflow.net/users/25510
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396408
| 163,698 |
https://mathoverflow.net/questions/396338
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5
|
For a Banach space $X$ let $S\_X$ denote its unit sphere and let $\mathrm{Iso}\_0(X)$ denote the group of rotations of $X$, that is isometries fixing the origin. There is a natural continuous action $\mathrm{Iso}\_0(X)\curvearrowright S\_X$.
When $X=L^p([0,1])$ the group $\mathrm{Iso}\_0(X)$ is Polish, so we can ask how complex the orbit equivalence relation induced on $S\_X$ is in the descriptive set theoretic sense. By a result of Pełczyński and Rolewicz every orbit is dense, but I don´t know much apart from this.
Are orbits meagre? Does this equivalence relation admit classification by countable structures? What are good references for this kind of questions and for more general questions along those lines (for example when $X$ is the Gurarij space)?
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https://mathoverflow.net/users/49381
|
How complex is the orbit equivalence relation of $\mathrm{Iso}_0(X)\curvearrowright S_X$ for $X=L^p([0,1])$?
|
The orbit structure is extremely simple. If $p=2$, there is one orbit (the isometry group of a Hilbert space acts transitively), whereas for $p\neq 2$ there are exactly $2$ orbits: the (classes of) functions that do not vanish on a set of positive measure and its complement, the functions that do vanish on a set of positive measure.
This is certainly well-known, but I do not remember where this is written (I thought that it might in a paper by Ferenczi and Rosendal, but I could not locate it). Let me provide the proof in the interesting case $p\neq 2$.
Let me first prove that if two functions $f$ and $g$ in $S\_X$ do not vanish, then they are in the same orbit. The two probability spaces $([0,1], |f|^p d\lambda)$ and $([0,1], |g|^p d\lambda)$ (where $\lambda$ is the Lebesgue measure) are standard complete atomless probability spaces. They are therefore isomorphic. This implies that there is a bimeasurable bijection $\phi \colon [0,1] \to [0,1]$ which sends $|f|^p d\lambda$ to $|g|^p d\lambda$. The three maps
* $h \in L^p(d\lambda) \mapsto \frac{h}{f} \in L^p(|f|^p d\lambda)$,
* $h \in L^p(|f|^p d\lambda) \mapsto h \circ \phi^{-1} \in L^p(|g|^p d\lambda)$,
* $h \in L^p(|g|^p d\lambda) \mapsto gh \in L^p(d\lambda)$.
are all surjective isometries (here we use that $|f|>0$ and $|g|>0$ a.s.). Their composition $h \mapsto g \frac{h \circ \phi^{-1}}{f \circ \phi^{-1}}$ is therefore a linear isometry which maps $f$ to $g$.
If $f$ and $g$ both vanish (say on $A$ and $B$ respectively), we can find a linear isometry from $L^p(A,d\lambda)$ onto $L^p(B,d\lambda)$ and combine it with a linear isometry $L^p([0,1]\setminus A) \to L^p([0,1]\setminus B)$ given by the previous case to find a isometry of $L^p([0,1])$ that maps $f$ to $g$.
The converse (that a function that almost surely does not vanish cannot be in the orbit of a function that vanishes) follows from the Banach-Lamperti theorem, which characterizes the linear isometries of $L^p([0,1])$: all such isometries are of the form
$$ Uf(x) =\omega(x) f(T^{-1}(x))$$
for a measurable bijection $T \colon [0,1] \to [0,1]$ that preserves the class of the Lebesgue measure $\lambda$, and a function $\omega\colon[0,1] \to \mathbf{C}^\*$ satisfying $|\omega(x)|^p = \frac{d T\_\*\lambda}{d\lambda}(x)$.
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5
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https://mathoverflow.net/users/10265
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396412
| 163,700 |
https://mathoverflow.net/questions/396356
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2
|
We consider $2a$ - periodic smooth solutions for
\begin{eqnarray\*}
-\Delta u+V(x)\,u=0\qquad\hbox{in}\:[-a,a]
\end{eqnarray\*}
We assume that $V$ is smooth and even (i.e. $V(-x)=V(x)$). We also assume that (up to multiplication with a real number) there exists only one odd $2a$ - periodic solution.
Can one say anything about the number of even $2a$ - periodic solutions?
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https://mathoverflow.net/users/79956
|
Even and odd solutions for the Schrödinger equation
|
Because of the uniqueness of the initial value problem, there can be at most two solutions, i.e., if we have one odd $2a$-periodic solution, then there can be at most one more even $2a$-periodic solution. For example, for $V(x)=-(\pi /a)^2 $, we have the odd solution $\sin \pi x/a $ and the even solution $\cos \pi x/a $. On the other hand, for generic $V(x)$, the eigenvalues are simple, i.e., if we have an odd $2a$-periodic solution, there is no additional even $2a$-periodic solution. An example is the [Mathieu equation](https://dlmf.nist.gov/28.2#vi), rescaled to $2a$-periodicity, where $V(x)=(\pi^2/(2a)^2 ) (2q\cos (\pi x/a)-\lambda (q))$, with nonzero $q$ and associated eigenvalue $\lambda (q)$; cf. [Ince's Theorem](https://dlmf.nist.gov/28.5#i).
In summary, there are either zero or one even $2a$-periodic solutions.
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6
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https://mathoverflow.net/users/134299
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396418
| 163,702 |
https://mathoverflow.net/questions/396416
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2
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Cohomology ring and cup product can be defined on simplicial complex (ie a triangulation of a manifold). Can we define cohomology ring and cup product on a more general complex? In particulate, I am interested in defining cohomology ring and cup product on a complex and its dual complex. The dual of a simplicial complex, in general, is not a simplicial complex, which causes a problem.
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https://mathoverflow.net/users/17787
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Cohomology ring on non-simplicial complex
|
If $H^\ast(-,R)$ is cohomology with coefficients in a ring, the cup product may be defined purely via the functoriality of $H^\ast(-,-)$ and certain compatibilities of tensor products. There is a map
$$H^\ast(X,A) \otimes H^\ast(Y,B) \to H^\ast(X \times Y, A \otimes B)$$
coming from taking the tensor product of cochain complexes. Then, when $R$ is a ring, we have the multiplication $m: R \otimes R \to R$ and the diagonal map $\Delta: X \to X \times X$ defined by $\Delta(x) = (x,x)$. Combining these with the above gives a map
$$ H^\ast(X,R) \otimes H^\ast(X,R) \to H^\ast(X\times X, R \otimes R) \to H^\ast(X\times X, R) \to H^\ast (X,R)$$
where the middle map is $m\_\ast$ and the final map is $\Delta^\ast$.
For more details, see May, *A Concise Course in Algebraic Topology*, p. 139.
There is also a construction of cup products on singular chains. See Hatcher, *Algebraic Topology*, §3.2.
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1
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https://mathoverflow.net/users/125523
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396422
| 163,703 |
https://mathoverflow.net/questions/396414
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1
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I am reading a paper '**[Periodic Nonlinear Schrodinger Equation and Invariant Measures](https://math.mit.edu/classes/18.158/gibbs-1D.pdf)**' written by **J.Bourgain**. And I am wondering if I can have some help from this website.
My question is an inequality at **(3.18)** of the paper. The inequality is, with $\lambda>1$
$$\sum\_{M>M\_0} e^{CM-c\sigma\_M^2 M^{1+\frac{2}{p}}\lambda^2}< e^{-cM\_0^{1+\frac{2}{p}}\lambda^2}$$
where the index $M$ is a dyadic [the form of $2^k$] and an arbitrary $M\_0>0$ be given. It is mentioned in the paper that he let $\sigma\_M=M^{-\frac{1}{p}}+\left( \frac{M\_0}{M} \right)^{\frac{1}{2}}$ for $M>M\_0$. Here, I also assume that the positive constants $c$ and $C$ are changing.
I have been trying to figure out this inequality and, using $\sigma\_M$, I was only able to figure out the inequality below.
\begin{align\*}
\sum\_{M>M\_0} e^{CM-c\sigma\_M^2 M^{1+\frac{2}{p}}\lambda^2}
&= \sum\_{M>M\_0} e^{CM-c(M+M\_0M^{2/p}+2M\_0M^{1/2+1/p})\lambda^2} \\
&< \sum\_{M>M\_0} e^{-c(M\_0M^{2/p}+2M\_0^{1+1/p})\lambda^2}\\
&< \sum\_{M>M\_0} e^{-c(M\_0^{1+2/p}+2M\_0^{1+1/p})\lambda^2}
\end{align\*}
where the first inequality is due to the fact that $\lambda>1$ and $c$ and $C$ can be modified. And the second inequality is due to the fact that $M>M\_0$.
Actually, I don't think the second inequality is good to use for this estimate as it would be independent of $M$ so the sum would be infinity.
I hope to figure out it... I thank in advance for the answer or any hints.
|
https://mathoverflow.net/users/127918
|
An inequality between sum of exponential functions wrt dyadic index
|
$\newcommand\si\sigma\newcommand\la\lambda$It seems to be (tacitly) assumed in the paper that $p\ge2$ -- see the display between (3.11) and (3.12).
Also, $c$ and $C$ seem to be (tacitly) assumed in the paper to be positive real constants.
Note that
$$M^{-2/p}+M\_0/M\le\si\_M^2\le2M^{-2/p}+2M\_0/M.$$
So,
$$\begin{aligned}
&\sum\_{M>M\_0} \exp\{CM-c\si\_M^2 M^{1+2/p}\la^2\} \\
&\le\sum\_{M>M\_0} \exp\{CM-c\la^2(M+M\_0M^{2/p})\} \\
&<\exp\{-c\la^2 M\_0^{1+2/p}\}\sum\_{M>M\_0} \exp\{CM-c\la^2 M\} \\
&< \exp\{-c\la^2 M\_0^{1+2/p}\},
\end{aligned}$$
as desired, if $\la>0$ is large enough (as assumed at the end of the proof, right after (3.19)).
---
On the other hand, if $C\ge2c\la^2(1+M\_0)$ , then
$$\begin{aligned}
&\sum\_{M>M\_0} \exp\{CM-c\si\_M^2 M^{1+2/p}\la^2\} \\
&\ge\sum\_{M>1+M\_0} \exp\{CM-2c\la^2(M+M\_0M^{2/p})\} \\
&\ge\sum\_{M>1+M\_0} \exp\{CM-2c\la^2(1+M\_0)M)\}=\infty.
\end{aligned}$$
|
1
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https://mathoverflow.net/users/36721
|
396425
| 163,704 |
https://mathoverflow.net/questions/396413
|
12
|
I'm an analyst who needs to use Deligne's Theorem 8.4 in **1**, but I feel lost in the maze of definitions, and I don't trust my geometric intuition here.
>
> Theorem 8.4: Let $Q$ be a polynomial in $n$ variables and of degree $d$ over $\mathbb{F}\_q$, $Q\_d$ the homogeneous part of degree $d$ of $Q$, and $\psi:\mathbb{F}\_q\to\mathbb{C}^\*$ a non-trivial additive character over $\mathbb{F}\_q$. Suppose that:
>
>
> * $d$ is comprime to $p$;
> * The hypersurface $H\_0$ in $\mathbb{P}^{n-1}\_{\mathbb{F}\_q}$ defined by $Q\_d$ is smooth.
>
>
> Then
> $$\Big|\sum\_{x\_1,\ldots,x\_n\in\mathbb{F}\_q}\psi(Q(x\_1,\ldots,x\_n))\Big|\le (d-1)^nq^{n/2}$$
>
>
>
I'm only interested in the case $q = p$ a prime. I need clarification about the statement.
>
> How are smooth hypersurfaces defined?
>
>
>
I guess smoothness here is equivalent to $\nabla Q\_d(x) \neq 0$ for every $x\in \overline{\mathbb{F}}\_q\setminus\{0\}$ (I'm using $x\cdot\nabla Q\_d = dQ\_d$). Am I right?
Some books I've glanced (p. ej. Shafarevich or Hartshorne) define smoothness over closed fields, and in [these notes](https://www.jmilne.org/math/CourseNotes/LEC.pdf) (p. 91) it says that we should extend the field, I think. All sources define "smoothness" in two ways, one using [regular rings](https://en.wikipedia.org/wiki/Regular_scheme) and other using the [gradient](https://en.wikipedia.org/wiki/Smooth_scheme). These definitions, which should be equivalent in my case, seem to work over any field, so do I really need to extend to $\overline{\mathbb{F}\_q}$? It would be nice to find references about all this which I could understand.
I had the idea that to define smoothness we had to take $H\_0 = \{Q\_d = 0\}\subset \mathbb{P}^{n-1}(\overline{\mathbb{F}}\_q)$, define the ideal $I\_H = \langle P\rangle$ of polynomials vanishing there, where $P$ is a polynomial, and then to verify that $\nabla P$ doesn't vanish. It seems wrong. If I take $Q = X\_1^d$, $d\ge 2$, it seems that Deligne's theorem doesn't hold, even though $\{X\_1 = 0\}$ looks smooth.
>
> If $Q\_d\in\mathbb{Z}[X\_1,\ldots,X\_n]$ and $\nabla Q\_d(x)\neq 0$ for every $x\in\mathbb{C}^n\setminus\{0\}$, does Theorem 8.4 hold for all but finitely many primes?
>
>
>
From what I've gathered until now, arithmetic-geometers find it trivial (should I too?), probably they'd point to proposition A.9.1.6 in **2**, but again, I don't know if all these definitions of smoothness coincide. I'd be nice (for me at least) to have a proof without using schemes, but [it is probably painful](https://mathoverflow.net/a/59129/90189).
I stated my question over the complex numbers because it is closed, but can I weaken the assumptions? On the other hand, is the converse true?
I'd appreciate any reference or help.
References
----------
**1** Deligne, Pierre. La conjecture de Weil. I. (French) Inst. Hautes Études Sci. Publ. Math. No. 43 (1974), 273--307.
**2** Hindry, Marc; Silverman, Joseph H. Diophantine geometry. An introduction. Graduate Texts in Mathematics, 201. Springer-Verlag, New York, 2000.
|
https://mathoverflow.net/users/90189
|
Deligne's theorem on exponential sums
|
Yes, smoothness is equivalent to the gradient being nonzero for every $x \in \overline{\mathbb F}\_q^n \setminus \{0\}$.
I would define smoothness of the hypersurface defined by $Q\_d$ as the condition that for each such $x$, either the gradient or the value of $Q\_d$ at $x$ is nonzero, but as you note, by homogeneity and the fact that $d$ is prime to $p$, if the value is nonzero then the gradient is nonzero.
Checking over $\overline{\mathbb F}\_q$ and not $\mathbb F\_q$ is really necessary.
For example, fix a basis for the field extension $\mathbb F\_{q^n}$, giving a bijection between $\mathbb F\_{q^n}$ and $\mathbb F\_q$. The norm map that sends each element of $\mathbb F\_{q^n}$ to the determinant over $\mathbb F\_q$ of its multiplication action on $\mathbb F\_{q^n}$ can be expressed as a polynomial of degree $n$ in the $n$ coordinates. Set $d=n$ and $Q\_= Q\_d=$ this polynomial. Then because the norm is a surjective homomorphism of multiplicative groups, there are $\frac{q^n-1}{q-1}$ elements of norm $a$ for each $a \in \mathbb F\_q^\times $, so
$$\sum\_{x \in \mathbb F\_q^n } \psi(Q(x))= \sum\_{x\in \mathbb F\_{q^n}} \psi (\operatorname{Norm}(x))= 1 + \frac{q^n-1}{q-1} \sum\_{x \in \mathbb F\_q^\times} \psi(x) = 1 - \frac{q^n-1}{q-1} $$ which does not satisfy Deligne's bound for $n>2$ and $q$ large.
However, $Q\_d$ never vanishes at any $x \in \mathbb F\_q^n$, and thus neither does its gradient. So checking over the algebraic closure is really necessary!
You can avoid working over the algebraic closure only by using definitions that involve checking point-by-point. For example, smoothness is equivalent to the claim that the ideal generated by $Q\_d$, and $\frac{ \partial Q\_d}{\partial x\_i}$ for all $i$ contains $x\_1^N, \dots, x\_n^N$ for some $N$.
---
You are right that we do not want to pass to the vanishing set of $Q\_d$ and then to the ideal - we just want to look at $Q\_d$ itself and whether it satisfies the gradient condition. The vanishing *scheme* of $Q\_d$ is sensitive to taking powers in this sense, which is why this is referred to as a smoothness condition on the hypersurface rather than a smoothness condition on $Q\_d$. If you like, this is equivalent to smoothness of the vanishing set plus the fact that $Q\_d$ isn't divisible by the square of a nontrivial polynomial.
---
It is indeed true that if the gradient of $Q\_d$ is nowhere vanishing over $\mathbb C^n \setminus \{0\}$, then it is nowhere vanishing on $\overline{\mathbb F}\_p^n \setminus \{0\}$ for all but finitely many $p$. The same is true for any algebraically closed field of characteristic zero.
The proof actually doesn't require scheme there. Here are two approach:
(1) By the Nullstellensatz, this implies that the ideal generated $\frac{ \partial Q\_d}{\partial x\_1}, \dots, \frac{ \partial Q\_d}{ \partial x\_n}$ contains a power of the ideal $(x\_1,\dots, x\_n)$ and in particular contains $x\_1^N, \dots, x\_n^N$ for some $N$. So some $\mathbb C[x\_1,\dots,x\_n]$-linear combination of those derivatives is equal to $x\_1^N$, some linear combination is $x\_2^N$, etc.
We can choose a linear map $\mathbb C \to \mathbb Q$ whose composition with $\mathbb Q \to \mathbb C$ is the identity. (In fact we only need to do this for the finite-dimensional subset of $\mathbb C$ generated by the coefficients of the linear combination.) Plugging the coefficients of all the polynomials in $\mathbb C[x\_1,\dots,x\_n]$ into this map, we see that some $\mathbb Q[x\_1,\dots,x\_n]$-linear combination of those derivatives is equal to $x\_1^N$, and so on.
Clearing denominators, we see that there is a natural number $M$ such that some $\mathbb Z[x\_1,\dots,x\_n]$-linear combination of those derivatives is equal to $M x\_1^N$, $\mathbb Z[x\_1,\dots,x\_n]$-linear combination of those derivatives is equal to $M x\_2^N$, and so on.
It follows for all $p$ not dividing $M$ that, if $x\_1,\dots, x\_n \in \overline{\mathbb F}\_p$ are not all zero, then one of these derivatives is nonvanishing, as desired.
(2) There exists a universal polynomial, the discriminant, in the coefficients of a homogeneous polynomial $f$ of degree $d$, which vanishes if and only if the derivatives of $f$ all vanish at some nonzero point over an algebraically closed field of characteristic not dividing $d$. This can be obtained as the [Macauley resultant](https://en.wikipedia.org/wiki/Resultant#Macaulay%27s_resultant) of the partial derivatives of $f$.
If the nonvanishing condition holds over $\mathbb C$, then the discriminant is nonzero, so it is nonzero mod $p$ for all but finitely many $p$, so the nonvanishing condition holds for all but finitely many $p$.
|
17
|
https://mathoverflow.net/users/18060
|
396428
| 163,705 |
https://mathoverflow.net/questions/396411
|
1
|
Let $d,m \to \infty$ (integers) with $m/d \to \rho \in (0, \infty)$. Let $C$ be a $d \times d$ psd matrix with $trace(C)=\mathcal O(1)$, and let $w\_1,\ldots,w\_m$ be iid uniformly distributed on the unit-sphere in $\mathbb R^d$. Consider the quartic form
$$
F := \frac{1}{m}\sum\_{j,\ell=1}^m (w\_j^\top w\_\ell)(w\_j^\top C w\_\ell).
$$
>
> **Question.** What are good probabilistic lower and upper-bounds for $F$ only in terms of $\rho$ and the eigenvalues of $C$ ?
>
>
>
For example, the solution for the case where $C$ is **diagonal** will already be very helpful.
Isotropic example
-----------------
Thanks to this post <https://mathoverflow.net/a/334219/78539>, we know that if $C = (1/d) I\_d$, then $F = m^{-1}\|WW^\top\|\_F^2 = m^{-1}\sum\_{j}\lambda\_j(W W^\top)^2\overset{a.s}{\to} \langle \lambda^2\rangle\_{\text{MP}(1/\rho)}$ (if I haven't made some scaling errors), where $\text{MP}(\gamma)$ is the Marchenko-Pastur law with parameter $\gamma$.
|
https://mathoverflow.net/users/78539
|
Probabilistic lower and upper-bounds for a certain random quartic form involving gaussian random matrices
|
Assume iid $N(0,1)$ entries, assume $C$ diagonal, and focus first on the non-diagonal terms:
$G=\sum\_j \sum\_{l\ne j} w\_j^Tw\_l w\_j^TCw\_l
= \sum\_{j\ne l, ik} w\_{ji}w\_{li} c\_i w\_{jk} w\_{lk}$.
Write this quantity as
$$
\begin{split}
G=\sum\_{j\ne l, i\ne k} w\_{ji}w\_{li} c\_i w\_{jk} w\_{lk}
&+
\sum\_{j\ne l, i=k} (w\_{ij}^2-1)(w\_{lj}^2 -1)c\_i
\\
&+(w\_{ij}^2-1)c\_i +
(w\_{lj}^2 -1)c\_i
+c\_i
\end{split}
$$
This is a decomposition in uncorrelated polynomials (any two terms are uncorrelated), so that the second moment is
$$
E[(G-m(m-1)trace[C])^2]=\sum\_{j\ne l, i \ne k} c\_i^2
+
\sum\_{j\ne l, i}(E[(Z^2-1)^2]^2 + 2 E[(Z^2-1)^2])c\_i^2.
$$
$$= m(m-1)\|C\|\_F^2((d-1)+E[(Z^2-1)^2]^2 + 2E[(Z^2-1)^2]).$$
The dominant term is of order $m^2d \|C\|\_F^2$, while the mean is $m(m-1)trace[C]$. Hence $G/E[G]-1$ converges to 0 in probability (or in L2) provided that $E[G]^2 \gg Var[G]$, that is,
$$
m^2 trace[C] \gg \|C\|\_F m \sqrt{d}.
$$
For the diagonal terms, we have
$\sum\_j d w\_j^TCw\_j + \sum\_{j} (\|w\_j\|^2-d)w\_j^TCw\_j$.
The second term is negligible compared to the first one if you use $\chi^2$ concentration (e..g, Bernstein inequality) for $\|w\_j\|^2-d$, while the first term has mean $md trace[C]$ and variance $2md^2\|C\|\_F^2$.
Again, the mean dominates the standard deviation if and only if
$$
m d ~trace[C] \gg \sqrt m d \|C\|\_F.$$ This is equivalent to the condition on the non-diagonal terms if $m\asymp d$.
Edit: since $\|C\|\_F^2 \le trace[C]^2$ for $C$ psd, these conditions are always satisfied.
|
1
|
https://mathoverflow.net/users/141760
|
396431
| 163,708 |
https://mathoverflow.net/questions/396427
|
1
|
The Erdős–Rado notation $a \rightarrow (b)^c\_d$ is common in partition calculus / combinatorial set theory, as well as its negation $a \not\rightarrow (b)^c\_d$. In that field, is there a standard way to read them out loud?
|
https://mathoverflow.net/users/310472
|
Pronunciation: the Erdős–Rado partition notation
|
Community wiki because it is answered over at [MSE](https://math.stackexchange.com/questions/674961/how-to-pronounce-the-partition-relation).
[source](https://books.google.nl/books?id=tCATRdtV_cwC)
|
3
|
https://mathoverflow.net/users/11260
|
396432
| 163,709 |
https://mathoverflow.net/questions/393532
|
5
|
While studying the so-called [Mittag-Leffler Polynomials](https://mathworld.wolfram.com/Mittag-LefflerPolynomial.html), denoted $M\_n(x)$, I was looking into the sequence $\frac1{n!}M\_n(n)$ which takes the following form
$$a\_n:=\sum\_{k=1}^n\binom{n-1}{k-1}\binom{n}k2^k.$$
>
> **QUESTION 1.** Let $\nu\_2(m)$ denote the [$2$-adic valuation](https://en.wikipedia.org/wiki/P-adic_order) of $m\in\mathbb{N}$. Is this true?
> $$\nu\_2(a\_n)=\begin{cases} \,\,\,\,\,\,1 \qquad \,\,\,\,\,\text{if $n$ is odd} \\
> 3\nu\_2(n) \qquad \text{if $n$ is even}.
> \end{cases}
> $$
> **QUESTION 2.** Is it true that $a\_n$ is never divisible by $5$, for $n\geq1$?
>
>
>
**Postscript.** I've recently solved QUESTION 2, so only QUESTION 1 remains.
|
https://mathoverflow.net/users/66131
|
Power of $2$ dividing a specialized Mittag-Leffler polynomial
|
I will confine myself to Question 1 since you mentioned that you know how to do Question 2. Also the case when $n$ is odd is easy, and let us restrict to $n$ being divisible exactly by $2^r$ with $r\ge 1$, and we need to show that the exact power of $2$ dividing $a\_n$ is $3r$. Thus in what follows we may discard any terms that are divisible by a power of $2$ larger than $3r$. For example, we can restrict attention to the terms with $k\le 3r$ in the sum. The case $r=1$ can be easily checked by hand, and we assume below that $r\ge 2$, and $k \le 3r$.
Consider the $k$-th term in the sum defining $a\_n$:
$$
2^{k} \binom{n}{k} \binom{n-1}{k-1} = 2^k \frac{n}{k} \prod\_{j=1}^{k-1} \Big(1 -\frac{n}{j}\Big)^2.
$$
Let us now expand the product above, discarding any terms that are divisible by $2^{3r+1}$. Note that $n/j$ is a $2$-adic integer, since the power of $2$ dividing $j$ is at most $\lfloor \log\_2(k-1)\rfloor \le \lfloor \log\_2 (3r-1)\rfloor \le r$.
We first observe that when expanding the product out terms that ``have $n^4$" in them (so using $3$ factors of $n/j$ in the product) can be omitted, and therefore terms that have higher powers of $n$ may also be omitted. Indeed a term using $3$ factors of $n/j$ in the product is divisible by a power of $2$
$$
\ge k + r -v\_2(k) + 3r - 3\lfloor \log\_2(k-1)\rfloor.
$$
A small calculation shows that $k-v\_2(k)-3\lfloor \log\_2(k-1) \rfloor \ge -1$ for all $k\ge 1$, and therefore the above is $\ge 3r+r-1 \ge 3r+1$, as desired.
Now consider terms with $n^3$ in them (so using $2$ terms of the form $n/j$ from the product). We claim that there is exactly one such term that is relevant, and this happens only in the case $k=4$, and the two terms from the product are (both) $n/2$. Indeed the power of $2$ in term using two factors $n/j$ is
$$
\ge k + r-v\_2(k) + 2 r -2 \lfloor \log\_2 (k-1)\rfloor,
$$
and we can check that $k-v\_2(k) -2 \lfloor \log\_2(k-1)\rfloor \ge 1$ for $k\ge 1$, except in the case $k=4$. In the case $k=4$ we can quickly check that the only relevant term with two factors $n/j$ must be $(n/2)(n/2)$.
Putting all these observations together, we find that $a\_n$ equals (after omitting any terms divisible by $2^{3r+1}$)
\begin{align\*}
&\sum\_{k\le 3r} 2^k \frac{n}{k} \Big( 1 - 2 \sum\_{j=1}^{k-1} \frac{n}{j}\Big) + 2^4 \frac{n}{4} \frac n2 \frac n2
\\
&= n \sum\_{k\le 3r} \frac{2^k}{k} -n^2 \sum\_{k\le 3r} \frac{2^k}{k}
\sum\_{j=1}^{k-1}\Big( \frac 1j + \frac{1}{k-j}\Big) +n^3\\
&= n \sum\_{k\le 3r} \frac{2^k}{k} -n^2 \sum\_{k\le 3r} \sum\_{j=1}^{k-1} \frac{2^k}{j(k-j)} + n^3.
\end{align\*}
Now here it is convenient to extend the sums over $k$ to infinity, noting that the extra terms are divisible by $2^{3r+1}$. (Thus one checks that for $k>3r$ one has $r+k -v\_2(k) \ge 3r+1$, and that $2r + k -2 \lfloor \log\_2 (k-1)\rfloor \ge 3r+1$.)
Now comes the magical bit. In the $2$-adics one has
$$
\sum\_{k=1}^{\infty} \frac{2^k}{k} =0,
$$
and therefore also
$$
\sum\_{k=1}^{\infty} 2^k \sum\_{j=1}^{k-1} \frac{1}{j(k-j)} = \Big( \sum\_{k=1}^{\infty} \frac{2^k}{k} \Big)^2 =0. \tag{\*}
$$
We are then left with $a\_n$ being $n^3$ up to multiples of $2^{3r+1}$, which is what we want.
For the evaluation of $\sum\_{k=1}^{\infty} 2^k/k$, note that if $|x|\_2 <1$ then the series
$$
\sum\_{k=1}^{\infty} \frac{x^k}{k}
$$
converges in the $2$-adics, and this clearly "looks like" $-\log(1-x)$. In our case we'd get $-\log (1-2) = \log (-1) =0$ (upon "using" $\log (-1) + \log (-1) = \log 1 = 0$). This last bit can all be made precise; a lovely write up of what is involved can be found in [Keith Conrad's notes](https://kconrad.math.uconn.edu/blurbs/gradnumthy/infseriespadic.pdf) (see Example 8.10 there).
|
6
|
https://mathoverflow.net/users/38624
|
396433
| 163,710 |
https://mathoverflow.net/questions/396420
|
1
|
Let $F(x,y)$ be a real polynomial in two variables of degree $d$. How many roots can $F(x,e^x)$ have? In other words, is there a bound one can place on the number of intersection points of $F(x,y)=0$ and $y=e^x$?
|
https://mathoverflow.net/users/nan
|
Roots of $F(x,e^x)$
|
Such a function has at most $\frac{d(d + 3)}{2}$ roots. In fact, let us show by induction on $k$ the following general statement:
Let $p\_0, p\_1, \dots, p\_k$ be polynomials of degree at most $n\_0, n\_1, \dots, n\_k$ respectively, not all of which are 0. Then, the function $f(x) = \sum\_{j = 0}^{k} p\_j (x) e^{j x}$ has at most $n\_0 + n\_1 + \cdots + n\_k + k$ zeroes. In the case where $p\_i$ is 0 we take $n\_i = -1$.
For $k = 0$ the claim is obvious. Suppose we have shown the claim for $k - 1$. Say that $f$ has $n$ zeroes. Notice that $f'(x) = p\_0 '(x) + \sum\_{j = 1}^{k} \left( p\_j '(x) + j p\_j (x) \right) e^{j x}$ satisfies the conditions of the claim, where the free coefficient has degree which is 1 smaller, and the degrees of the rest of the coefficients are the same. However, by Rolle's theorem, $f'(x)$ has at least $n - 1$ zeroes. Taking the $n\_0 + 1$-th derivative, dividing by $e^{x}$ and using the induction hypothesis we prove the claim.
In our case, we have $\mathrm{deg} (p\_i) \leq d - i$ and so our function has at most $\frac{d(d + 3)}{2}$ roots.
|
4
|
https://mathoverflow.net/users/88679
|
396434
| 163,711 |
https://mathoverflow.net/questions/396364
|
5
|
I know that the group structure on Hom sets can be recovered from biproducts if they exit. Indeed, if $f, g : A \to B$ are two maps then there is a uniquely defined map $f \oplus g : A \oplus A \to B \oplus B$ and then there are diagonal and codiagonal maps giving
$$ A \to A \oplus A \to B \oplus B \to B $$
so we get a composition law on Hom sets.
However, if products/coproduct/biproducts don't exist in my category, then I don't see why the Ab-enrichment should be a "property" and not a "structure." Therefore, I am wondering if it possible for some wacky pre-additive category without biproducts to have multiple Ab-enriched structures?
|
https://mathoverflow.net/users/154157
|
Does an Ab-enriched category have a unique Ab-enrichment?
|
As Martin Brandenburg and Maxime Ramzi suggest, it is easy to construct examples on small categories.
For example, a one object Ab-enriched category is exactly a ring. The category corresponds to the monoid (multiplication) of the ring and the Ab-enrichment to the addition law. There are monoids which have multiple abelian group structures that make them a ring.
Even more simply, consider the category with two objects $x,y$ and $n$ parallel morphisms $x \to y$. We need two morphisms $0, \mathrm{id} \in \mathrm{Hom}(x,x)$ so we may set $\mathrm{End}(x,x) \cong \mathbb{Z}$ as a ring generated by $\mathrm{id}$. Then we set $\mathrm{Hom}(y,x) = 0$ to be the trivial group. The only nontrivial compositions are $\mathrm{Hom}(y, y) \times \mathrm{Hom}(x, y) \to \mathrm{Hom}(x,y)$ and $\mathrm{Hom}(x,y) \times \mathrm{Hom}(x,x) \to \mathrm{Hom}(x,y)$ which must be the unique $\mathrm{Z}$-module structure on any abelian group because the generator $\mathrm{id}$ acts as a unit under composition.
Therefore, an Ab-enrichment is exactly an abelian group structure on $\mathrm{Hom}(x,y)$ of which there are many.
|
11
|
https://mathoverflow.net/users/154157
|
396441
| 163,712 |
https://mathoverflow.net/questions/396426
|
3
|
Is there a Borel function $f:2^\omega\to\omega^\omega$ such that for every nonempty closed perfect set $P\subseteq 2^\omega$, $f|P$ is a dominating family of functions in $\omega^\omega$?
This is a refinement of the question I asked in: [A Borel perfectly everywhere surjective function on the Cantor set](https://mathoverflow.net/questions/396368/a-borel-perfectly-everywhere-surjective-function-on-the-cantor-set). I think the argument which yields a "no" answer there would also give that there is no function $2^\omega\to\omega^\omega$ with comeager, or possibly even nonmeager, image on every perfect set. But while every nonmeager family of functions in $\omega^\omega$ is unbounded, the converse (even for dominating families) is not true.
|
https://mathoverflow.net/users/16107
|
A Borel perfectly everywhere dominating family of functions
|
The answer is no. By Lusin's Theorem there exists $A \subseteq 2^\omega$ closed and of positive measure such that $f \restriction A$ is continuous. Since $A$ cannot be countable it must contain a perfect set $P$. Now $f[P] \subseteq \omega^\omega$ is compact, hence bounded in $\omega^\omega$.
|
6
|
https://mathoverflow.net/users/134910
|
396443
| 163,714 |
https://mathoverflow.net/questions/396436
|
9
|
Judging by the compact regular case, and more generally the spatial case, regular projectivity of locales, resp. regular injectivity of frames, must have something to do with $\neg p\lor\neg\neg p$ and $\neg(x\land y)\to(\neg x\lor\neg y)$. On the other hand, existence of locales without points shows that the terminal locale is not regular projective. I am convinced somebody should already have found out which locales are regular projective, but who?
The same question about projectivity with respect to *arbitrary* epimorphisms of locales is probably easier but less interesting. Still, I don't know anything about that either.
PS As Simon Henry pointed out, I should rather pick some pullback stable class of locale epimorphisms. I guess I don't know an answer about any of them (except maybe the proper ones), so please just choose an as large as possible class of nicely behaved epimorphisms of your choice - say, quotients, or of effective descent, or triquotients, etc.
|
https://mathoverflow.net/users/41291
|
What are projective locales / injective frames?
|
So the short answer is that there is no non-empty projective locales for essentially any reasonable class of epimorphisms you can think of (except maybe proper maps).
The problem is that there exists a family of non-trivial Boolean locales $B\_\kappa$ indexed by infinite cardinal numbers $\kappa$, such that the only locale $X$ that has map to all the $B\_\kappa$ is the empty locale
The map $B\_\kappa \to 1$ are open surjections, so in particular they are effective descent map, stable regular epimorphism and triquotient map, so pretty much any class of epimorphism you might think about will contains these.
But if $X$ was projective with respect to any class containing these cover, then the unique map $X \to 1$ would lift to maps $X \to B\_\kappa$ for all $\kappa$ which contradict our claim above.
Explicitely, $B\_\kappa$ can be taken to be any (non-trivial) Boolean locale that "collaps the cardinal $\kappa$ to $\omega$" I mean by that, if $p:B\_\kappa \to 1$ denotes the unique map, then $p^\*\kappa \simeq p^\* \omega$ as sheaves over $B\_\kappa$ . For example, $B\_\kappa$ can be taken to be the double negation sublocale of the locale of injective functions $\omega \to \kappa$. The fact that $B\_\kappa$ is non trivial then follows from the fact that it is dense in this locale of injective function. (See details in edit below)
And a locale $X$ having functions to all the $B\_\kappa$ would collapse all infinite cardinals at the same time (in the sense that all $p^\*\kappa$ for infinite cardinal $\kappa$ would be isomorphic), which is impossible as a locale can't collapse to $\omega$ cardinal much larger than itself. THe following are very loose bound that shows it, though experts on forcing surely known much better bounds:
If $X$ is a non-degenerate locales, then the total number of locale section of $p^\* \kappa$ is larger than $\kappa$ as every element of $\kappa$ gives a globale section, and it is smaller than the function space $\kappa^{\mathcal{O}(X) \times 2^{\mathcal{O}(X)}}$ as every locale section can be written as: you chose a cover of its domain of definition (a special subset of $\mathcal{O}(X)$) and then for each element of that subset, you chose an element of $\kappa$.
So for any locale $X$, picking a $\kappa$ to be larger than $\omega^{\mathcal{O}(X) \times 2^{\mathcal{O}(X)}}$ we get that $p^\* \omega$ can't have as much section as $p^\* \kappa$ and hence they can't be isomorphic, so $X$ can't collaps any cardinal bigger than this to $\omega$.
---
Here is some clarification on the construction of locales $B\_\kappa$. This is a fairly standard observation, but I'm strugling to find a reference, so given that it is fairly simple, I'll write the details.
We fixe $\kappa$ some infinite cardinal number.
We start $I\_\kappa$ the locale that classifies injection $i:\omega \to \kappa$, so that a map $X \to I\_\kappa$ is the same as the data of an injective map $p^\* \omega \to p^\* \kappa$.
It is easy to write a propositional geometric theory of such injections (it has base proposition $R\_{x,y}$ for $x \in \omega$ and $y \in \kappa$ which is interpreted as $i(x)=y$ and all the axioms that make this into an "injective" functional relation)
$I\_\kappa$ is non-trivial because it has plenty of points.
Now, consider the (open) sublocale $V\_y \subset I\_\kappa$ for $y \in \kappa$ that clasifies these injection that further satisfies $\exists x \in \omega, i(x)=y$.
$V\_y$ is dense: indeed, the basic open of $I\_\kappa$ are the finite intersection of $R\_{x,z}$ and for any finite intersection $\cap R\_{x\_i,z\_i}$ of these, if it is non-degenerate you can explicitely construct a point of it that is also in $V\_y$ (take a function that send $x\_i$ to $z\_i$ and some other value to $y$, if its impossible it means the intersection is empty).
Now the intersection of all $V\_y$ is hence a dense sublocale. (an intersection of a familly of dense sublocale is dense).
By definition this intersection classifies bijection from $\omega \to \kappa$. So this is exactly the $T\_\kappa$ I mentioned in the comment.
Alternatively, you can define $B\_\kappa$ to be the double negation sublocale of $I\_\kappa$, which is hence included in all the $V\_y$, so that $B\_\kappa \subset T\_\kappa$ also "collaps the cardinal $\kappa$ to $\omega$).
Both $B\_\kappa$ and $T\_\kappa$ are non-trivial because they are dense in $I\_\kappa$ which is non-trivial.
|
12
|
https://mathoverflow.net/users/22131
|
396444
| 163,715 |
https://mathoverflow.net/questions/396220
|
1
|
Let $F$ be an algebraically closed field of characteristic $p$ equipped with a nonarchimedean dense absolute value $|\cdot|:F \rightarrow \mathbb{R}\_{\ge 0}$ with respect to which $F$ is complete. Let $\mathcal{O}\_{F}$ denote the ring of integers of $F$.
>
> First define the product norm $|\cdot|\_{prod}$ on $F\otimes \_{\mathbb F\_p} F$ in the following way. If $c\in F\otimes \_{\mathbb F\_p} F$, then
>
>
> $$|c|\_{prod}:=\inf\left\{\max\_{1\le i\le n}\{|c\_{1,i}||c\_{2,i}| \}\ : \ c=\sum^{n}\_{i=1}c\_{1,i}\otimes c\_{2,i}, \text{where } c\_{1,i}, c\_{2,i}\in F\right\}$$
> On the other hand, define $|\cdot|'\_{prod}$ on the subring $\mathcal{O}\_{F}\otimes \_{\mathbb F\_p}
> \mathcal{O}\_{F}$ in the following way. If $d\in \mathcal{O}\_{F}\otimes \_{\mathbb F\_p}
> \mathcal{O}\_{F}$, then
>
>
> $$|d|'\_{prod}:=\inf\left\{\max\_{1\le i\le n}\{|d\_{1,i}||d\_{2,i}| \}\ : \ d=\sum^{n}\_{i=1}d\_{1,i}\otimes d\_{2,i}, \text{where } d\_{1,i}, d\_{2,i}\in \color{red}{\mathcal{O}\_{F}}\right\}$$
>
>
>
>
> In both definitions the infimum is taken over all the possible ways to write the element as a sum of pure tensors in the respective rings.
> Is it true that $|\cdot|\_{prod}$ and $|\cdot|'\_{prod}$ coincide when restricting them to elements of $\mathcal{O}\_{F}\otimes \_{\mathbb F\_p}
> \mathcal{O}\_{F}$?
>
>
>
My first guess is that the answer is true. It is clear that $|c|\_{prod}\le |c|'\_{prod}$ for $c\in \mathcal{O}\_{F}\otimes\_{\mathbb{F}\_{p}}\mathcal{O}\_{F}$, since there are more ways to write $c$ as a sum of pure tensors in $F$.
I am able to show that they do coincide on pure tensors in $\mathcal{O}\_{F}\otimes \_{\mathbb F\_p}
\mathcal{O}\_{F}$, more precisely I showed that $|x\otimes y|\_{prod}=|x\otimes y|'\_{prod}=|x|\cdot|y|$ for $x,y\in \mathcal{O}\_{F}$, but I am not able to prove that in the general case.
|
https://mathoverflow.net/users/105386
|
Product absolute value in rings of integers
|
The natural map $i \colon \mathcal O\_F \to F$ is an injective map of $\mathbb F\_p$-vector spaces, hence we can choose a splitting, i.e. an $\mathbb F\_p$-linear map $s \colon F \to \mathcal O\_F$ such that $s \circ i$ is the identity.
We have $|s(x)| \leq |x|$ since if $x \in \mathcal O\_F$ then $|s(x) | = |x|$ and if $x \notin \mathcal O\_F$ then $|s(x)| \leq 1 \leq |x|$.
By $\mathbb F\_p$-linearity, if $$c = \sum\_{i=1}^n c\_{1,i} \otimes c\_{2,i}$$ then $$ s \otimes s(c) = \sum\_{i=1}^n s(c\_{1,i}) \otimes s(c\_{2,i})$$ and if $c \in \mathcal O\_F \otimes \mathcal O\_F$ then by definition $$s \otimes s(c)=c.$$
Combining these, we have
$$\begin{aligned}|c|\_{prod}
&=\inf\left\{\max\_{1\le i\le n}\{|c\_{1,i}||c\_{2,i}| \}\ : \ c=\sum^{n}\_{i=1}c\_{1,i}\otimes c\_{2,i}, \text{where } c\_{1,i}, c\_{2,i}\in F\right\}
\\
&\ge \inf\left\{\max\_{1\le i\le n}\{|s(c\_{1,i})||s(c\_{2,i})| \}\ : \ c=\sum^{n}\_{i=1}c\_{1,i}\otimes c\_{2,i}, \text{where } c\_{1,i}, c\_{2,i}\in F\right\}
\\
&= \inf\left\{\max\_{1\le i\le n}\{|s(c\_{1,i})||s(c\_{2,i})| \}\ : \ s\otimes s (c)=\sum^{n}\_{i=1}s(c\_{1,i})\otimes s(c\_{2,i}), \text{where } c\_{1,i}, c\_{2,i}\in F\right\}
\\
&= \inf\left\{\max\_{1\le i\le n}\{|s(c\_{1,i})||s(c\_{2,i})| \}\ : \ c=\sum^{n}\_{i=1}s(c\_{1,i})\otimes s(c\_{2,i}), \text{where } c\_{1,i}, c\_{2,i}\in F\right\}
\\
&\ge \inf\left\{\max\_{1\le i\le n}\{|d\_{1,i}||d\_{2,i}| \}\ : \ c=\sum^{n}\_{i=1}d\_{1,i}\otimes d\_{2,i}, \text{where } d\_{1,i}, d\_{2,i}\in {\mathcal{O}\_{F}}\right\}
\\
&= |c|'\_{prod}\end{aligned}$$
which together with the trivial inequality proves the identity.
|
3
|
https://mathoverflow.net/users/18060
|
396446
| 163,716 |
https://mathoverflow.net/questions/396126
|
9
|
I would have a proof of the following fact; but it's a bit clunky, and am wondering if one can get a more elegant one (and/or improve the constants). I couldn't find this anywhere, and searching properties of the Skellam distribution didn't help much either.
>
> Let $X\sim\operatorname{Poi}(\lambda)$ and $Y\sim\operatorname{Poi}(\mu)$ be two independent random variables. Then
> $$
> \max(|\lambda-\mu|, \frac{1}{40}\min(\sqrt{\lambda+\mu}, \lambda+\mu)) \leq \mathbb E[|X-Y|] \leq\min(|\lambda-\mu|+\sqrt{\lambda+\mu}, \lambda+\mu)
> $$
>
>
>
*Here is the current proof I came up with:*
The upper bound follows from Cauchy–Schwarz, as $\mathbb E[(X-Y)^2] = (\lambda-\mu)^2 + \lambda+\mu$; and by the triangle inequality, as $\mathbb{E}[|X-Y|]\leq \mathbb{E}[X+Y]$. The lower bound $|\lambda-\mu|$ is a direct consequence of Jensen's inequality; we now proceed to establish the second term.
* Consider first the case $\lambda+\mu\geq 1$. We will use Paley–Zygmund, noting that for any $c\in(0,1)$ we have
$$
\mathbb{E}[|X-Y|] \geq c\sqrt{\mu+\lambda}\cdot\mathbb{P}\{|X-Y|\geq c\sqrt{\mu+\lambda} \}
$$
and
$$
\begin{align\*}
\mathbb{P}\{|X-Y|&\geq c\sqrt{\mu+\lambda}\} \\ &= \mathbb{P}\{(X-Y)^2\geq c^2(\mu+\lambda)\} \\
&\geq \mathbb{P}\{(X-Y)^2\geq c^2\mathbb{E}[(X-Y)^2]\}\\
&\geq (1-c^2)^2\frac{\mathbb{E}[(X-Y)^2]^2}{\mathbb{E}[(X-Y)^4]}\\
&= \frac{ (1-c^2)^2(\lambda+\mu+(\lambda-\mu)^2)^2}{\lambda+\mu+6(\lambda^2+\mu^2)+6(\lambda-\mu)^2(\lambda+\mu)+(\lambda-\mu)^2+(\lambda-\mu)^4}\\
&\overset{(\dagger)}{\geq} (1-c^2)^2 \frac{(\lambda+\mu)^2+(\lambda-\mu)^4}{(\lambda+\mu)+10(\lambda+\mu)^2+3(\lambda-\mu)^4}\\
&\overset{(\ddagger)}{\geq} \frac{(1-c^2)^2}{11}
\end{align\*}
$$
where $(\dagger)$ is using the AM-GM inequality and $(a-b)^2\leq (a+b)^2$; and $(\ddagger)$ relies on $\lambda+\mu \leq (\lambda+\mu)^2$, which holds since $\lambda+\mu\geq 1$.
Taking $c=1/\sqrt{5}$ concludes the proof of this case, as then $\frac{c(1-c^2)^2}{11} > \frac{1}{40}$.
* If $\lambda+\mu \leq 1$, we will use properties of the modified Bessel function of the first kind $I\_0$ (namely, that it is nondecreasing on $[0,\infty)$, with $e^{-x}I\_0(x) \geq 1-\frac{x}{2}$ for $x\in[0,1]$) to conclude, as
$$
\begin{align}
\mathbb{E}[|X-Y|] &\geq
1 - \mathbb{P}\{|X-Y|=0\} \\
&= 1- e^{-(\lambda+\mu)} I\_0(2\sqrt{\lambda\mu}) \\
&\geq 1- e^{-(\lambda+\mu)} I\_0(\lambda+\mu)
\geq \frac{\lambda+\mu}{2}
\end{align}$$
where the first inequality is the AM-GM inequality.
|
https://mathoverflow.net/users/37266
|
Bounds on the expectation of $|X-Y|$ for $X,Y$ Poisson
|
$\newcommand{\la}{\lambda}$Let us show that
\begin{equation\*}
E|Z|\ge J(1)\min[c,\sqrt c\,], \tag{1}
\end{equation\*}
where
\begin{equation\*}
Z:=X-Y,\quad c:=\la+\mu,
\end{equation\*}
\begin{equation\*}
J(x):=\frac2\pi\,\int\_0^\infty\frac{1-\exp\{-x\, (1-\cos t)\}}{t^2}\,dt.
\end{equation\*}
Mathematica's command NIntegrate[] produces
\begin{equation\*}
J(1)=0.67390\dots.
\end{equation\*}
So, the constant factor $J(1)$ is an about 27-times improvement of the corresponding coefficient $\frac1{40}$ in the OP. Moreover, the constant factor $J(1)$ is the best possible.
Inequality (1) is based on the Zolotarev identity
\begin{equation\*}
E|Z|=\frac2\pi\,\int\_0^\infty\frac{1-\Re Ee^{itZ}}{t^2}\,dt;
\end{equation\*}
see e.g. formula (3.26) in [this paper](https://link.springer.com/article/10.1007/s10959-016-0709-1) or formula (48) in the corresponding [arXiv preprint](https://arxiv.org/pdf/1603.07365). In our case,
\begin{equation\*}
\Re Ee^{itZ}=\exp\{-c(1-\cos t)\}\cos((\la-\mu)\sin t)\le\exp\{-c(1-\cos t)\};
\end{equation\*}
Moreover, the latter inequality will turn into the equality when $\la=\mu$.
So,
\begin{equation\*}
E|Z|\ge J(c), \tag{2}
\end{equation\*}
and the latter inequality will turn into the equality when $\la=\mu$.
Note that, for each real $z>0$,
\begin{equation\*}
\frac{1-e^{-zx}}x=\int\_0^z e^{-xu}\,du
\end{equation\*}
is decreasing in $x$. So, $J(x)/x$ is decreasing in real $x>0$. Moreover, as shown in [this answer](https://mathoverflow.net/a/396573/36721), $J(x)/\sqrt x$ is increasing in real $x>0$. So, $J(c)\ge cJ(1)$ if $c\in(0,1]$ and $J(c)\ge\sqrt c\,J(1)$ if $c\in[1,\infty)$.
Thus, (1) follows from (2). Moreover, (1) turns into the equality when $\la=\mu=1/2$ (so that $c=1$).
|
9
|
https://mathoverflow.net/users/36721
|
396447
| 163,717 |
https://mathoverflow.net/questions/396448
|
3
|
We say that an invariant measure $\mu$ on some symbolic space $\Sigma$ has local product structure if there is a measurable function $\psi: \Sigma \rightarrow(0, \infty)$ such that the restriction is of the form
$$
\mu\_{|[I]}=\psi(\mu^{+} \times \mu^{-}),$$
where $\psi$ is continuous and positive, and $\mu^{+}$ and $\mu^{-}$ are the projections of $\mu\_{|[I]}$ to the spaces of one-sided sequences indexed by positive and negative integers, respectively.
Let $T:\Sigma \to \Sigma$ be a full shift. Are all $T-$ invariant measures have the local product structure?if not, could you please give an example ?
Update: Let $f:M \to M$ be a $C^{1+\alpha}$ diffeomorphism. Is it true that all hyperbolic measures have local product structure?
|
https://mathoverflow.net/users/127839
|
Does full shift have the local product structure?
|
No. Take $\mu$ to be a measure supported on a Sturmian shift corresponding to some irrational rotation $R\_\alpha$. If $\mathcal F^+$ denotes the $\sigma$-algebra generated by the coordinates in $\mathbb Z\_{\ge 0}$ and $\mathcal F^-$ denotes the $\sigma$-algebra generated by coordinates in $\mathbb Z\_{<0}$, then conditional on $\mathcal F^+$, $\mu$ is supported on a single point (there is an almost everywhere one-one correspondence between negative and positive tails). This means $\mu$ is very far from a product of $\mu^-$ and $\mu^+$ (in which for each positive tail, all negative tails occur with conditional probability $\psi$).
|
4
|
https://mathoverflow.net/users/11054
|
396449
| 163,718 |
https://mathoverflow.net/questions/396174
|
8
|
$\newcommand{\SH}{\mathit{SH}}\newcommand{\Z}{\mathbb Z}$Let $G$ be a compact Lie group and $\lambda\in
H^4(BG;\Z)$. The data $(G, \lambda)$ determine a 3d topological field theory called Chern-Simons theory, except not
quite: there is an obstruction to defining it on general closed, oriented $3$-manifolds called the *anomaly*.
[Freed-Teleman](https://arxiv.org/abs/1212.1692) characterize anomalies of $n$-dimensional field theories as
$(n+1)$-dimensional invertible field theories, which have been classified. I think in this case the anomaly
field theory should be unitary, so is classified up to isomorphism by a torsion element of
$\mathrm{Hom}(\Omega\_4^{\mathrm{SO}}, \mathbb C^\times)$, by a theorem of
[Freed-Hopkins](https://arxiv.org/abs/1604.06527). If I choose $G$ and $\lambda$, is the isomorphism type of the
anomaly field theory, as a bordism invariant, known?
I'm actually interested in a slightly more general story, where $\lambda\in\SH^4(BG)$. (Here $\SH$ is a generalized
cohomology theory called *supercohomology*: $\pi\_0\SH = \Z$, $\pi\_1\SH = \Z/2$, and the $k$-invariant is nonzero.
When $G$ is simple and simply connected, using $\SH$ instead of $H$ amounts to choosing a half-integer rather than
an integer.) Then there is a 3d spin TFT called spin Chern-Simons theory, which is again anomalous. Now the anomaly
is a torsion element of $\mathrm{Hom}(\Omega\_4^{\mathrm{Spin}}, \mathbb C^\times)$.
**For spin Chern-Simons theories, is the isomorphism type of the anomaly known? If not, is there an explicit
conjectured description?**
I'm primarily interested in the spin case when $G$ is a torus, but any information (e.g. $G$ simple and simply
connected, only for the oriented case, etc.) is helpful.
|
https://mathoverflow.net/users/97265
|
Formula for the anomalies of spin Chern-Simons theories?
|
This is not a direct answer to your question, but I think it's relevant.
One way of thinking about the anomaly for ordinary (oriented) Chern-Simons theories is that it's the evaluation of the associated 3+1-dimension Crane-Yetter theory (what Freed would call the anomaly theory) on a generator of 4d oriented bordism, i.e. $CP^2$. Choosing the standard handle decomposition of $CP^2$, one calculates that
$$
Z\_{3+1}(CP^2) = D^{-1/2} \sum\_a \theta\_a d\_a^2 =: C
$$
(The sum is over simple objects $a$ of the modular category associated to the Chern-Simons theory, $D$ is the global dimension of that category, $\theta\_a$ is the ribbon twist of $a$, and $d\_a$ is the quantum dimension (loop value).) The bordism invariant in this case is $C^{\sigma(W)}$, where $W$ is a closed 4-manifold and $\sigma$ is its signature.
Imitating this approach for 2+1-dimensional Spin Chern-Simons TQFTs we can again evaluate the associated 3+1-dimensional Spin Crane-Yetter theory on a generator of 4-dimensional Spin bordism, such as the K3 surface. So
$$
C := Z\_{3+1}(K3) = \;\; ???
$$
The reason I wrote ??? above rather than an explicit formula is that the simplest handle decomposition of the K3 surface (which is in some sense the simplest generator of 4d Spin bordism) has 22 2-handles. Translating that complicated handle structure into a formula involving quantum dimensions, ribbon twists, $6j$-symbols, etc. would not be very enlightening.
One way around this difficulty would be to enlarge the class of manifolds (on which the TQFT is defined) from Spin manifolds to "characteristic pairs" -- pairs $W^k \supset V^{k-2}$ such that $V$ is a Poincare dual to the second Stiefel-Whitney class of $W$ and $W\setminus V$ is equipped with a spin structure which does not extend over $V$. (See [this paper](https://www.maths.ed.ac.uk/%7Ev1ranick/papers/kirbytaylorpin.pdf) by Kirby and Taylor for details.) The generators of the 4-dimensional characteristic pair bordism group are much simpler: $(CP^2,CP^1)$ and $(S^4,RP^2)$. The central charge associated to $(S^4,RP^2)$ can be normalized away. The central charge associated to $(CP^2,CP^1)$ looks very similar to the oriented case:
$$
C:= Z\_{3+1}(CP^2, CP^1) = D^{-1/2} \sum\_v \frac{\theta\_v d\_v^2}{|\mbox{End}(v)|}
$$
Here the sum is over only "vortex" simple objects (associated to the pair $(D^2,pt)$). $|\mbox{End}(v)|$ denotes the dimension of the endomorphism algebra of $v$ -- 2 for Majorana simple objects and 1 for ordinary simple objects.
Can we always enlarge the domain of definition of a Spin TQFT associated to a fermionic modular category to characteristic/vortex pairs? I think that a [recent paper](https://arxiv.org/abs/2105.15167) of Johnson-Freyd and Reutter implies the answer is "yes", and Corey Jones, David Aasen and I think we have a more constructive proof (which we currently are in the process of writing up).
More details on the above approach can be found in [these](https://mathinstitutes.org/videos/videos/16665) two [talks](http://scgp.stonybrook.edu/video_portal/video.php?id=3238), and [slides](https://canyon23.net/math/talks/SCGP%20talk%20201706%20compressed.pdf) are here.
|
8
|
https://mathoverflow.net/users/284
|
396453
| 163,720 |
https://mathoverflow.net/questions/396377
|
5
|
The following question concerns that without $ZF+DC$, can every function be "a little bit" continuous?
>
> **Question** Is it consistent with $ZF+DC$ that for any function $f:[0,1]\to [0,1]$ and any positive measure set $A$, there is a positive measure closed set $B\subseteq A$ so that $f^{-1}(B)$ is closed?
>
>
>
How about in Solovay's model?
|
https://mathoverflow.net/users/14340
|
Continuity of real functions
|
Any Borel $f:[0, 1] \to [0, 1]$ satisfying $f^{-1}[\{y\}]$ is dense in $[0, 1]$ for every $y \in [0, 1]$ is a counterexample. For example, $f(x) = \limsup\_n \frac{x\_1 + x\_2 + \dots + x\_n}{n}$ where $x = 0.x\_1 x\_2 \dots$ is the binary expansion of $x$.
|
7
|
https://mathoverflow.net/users/310787
|
396465
| 163,724 |
https://mathoverflow.net/questions/396291
|
2
|
So, I asked a similar question at math stackexchange and was directed here. Please let me know if this question is better suited elsewhere.
Let $U$ and $V$ be (infinite-dimensional) Banach spaces. Assume we have a sequence $(u\_n)$ in $U$ converging weakly to $u\_0$ and a nonlinear function $F:U \to V$, where we can assume for instance that $\|F(u)\|\_V \leq C \|u\|\_U$, it is not really important for the question.
Moving on to the question, in $L^p(0,1)$, weak continuity of a function $\psi: \mathbb{R} \to \mathbb{R}$ implies that that $\psi$ is affine (see 2.10 in [these lecture notes](https://www-m7.ma.tum.de/foswiki/pub/M7/Analysis/PDE2-10/July_25.pdf) )
I am wondering if there are similar results for other Banach (or Hilbert spaces), that weak continuity implies some sort of triviality? I suspect there is no general theorem, but I am particularly concerned with spaces such as $L^2([a,b],V)$ where $V$ is a reflexive, separable Banach space (perhaps even finite-dimensional), or $C^1([a,b],V)$, $W^{k,p}([a,b],V)$. In the first case my guess is that it does carry over and weakly continuous functions are affine, in the other cases, I am not so sure.
|
https://mathoverflow.net/users/308734
|
Weakly continuous function implies some sort of triviality
|
From the question and from the OP's [related question](https://math.stackexchange.com/q/4180682/793015) on Mathematics StackExchange, I infer that the OP is in general interested in the weak continuity of nonlinear mappings. So here are two general facts the seem to be relevant:
Let $X$, $Y$ be Banach spaces (over the same field). Assume that $X$ is infinite-dimensional and that $Y$ is non-zero. Then:
(a) There exists a norm-continuous nonlinear mapping $F: X \to Y$ that is not weakly continuous.
(b) There exists a non-linear mapping $G: X \to Y$ that is weakly continuous.
*Proof.* (a) Choose a non-zero vector $y \in Y$ and set $F(x) = \|x\| \cdot y$ for each $x \in X$. Obviously, $F$ is norm-continuous. But if $F$ were weakly continuous, then the preimage of the set $\{y\}$ under $F$ - which is the unit sphere in $X$ - would be weakly closed. However, the unit sphere an an infinite-dimensional Banach space is [never weakly closed](https://math.stackexchange.com/q/153889/793015).
(b) Let $x' \in X'$ be a non-zero bounded linear functional on $X$ and fix a non-zero vector $y \in Y$. We set $G(x) = \langle x', x \rangle^2 \cdot y$ for each $x \in X$. Then $G$ is not linear, but it is weakly continuous since it is the composition of the weakly continuous mappings
$$
X \overset{x'}{\longrightarrow} \mathbb{F} \; \overset{s \mapsto s^2}{\longrightarrow} \; \mathbb{F} \; \overset{t \mapsto t \cdot y}{\longrightarrow} \; Y
$$
(where $\mathbb{F} \in \{\mathbb{R}, \mathbb{C}\}$ denotes the underlying scalar field of $X$ and $Y$). $\square$
**Remarks.**
(1) The mapping $F$ from (a) is not even sequentially weakly continuous. This follows from the fact that, in an infinite-dimensional Banach space $X$, there always exists a sequence $(x\_n)$ in the unit sphere that converges weakly to $0$.
(2) It is maybe worthwhile to recall that, for a *linear* mapping $T$ between two Banach spaces $X$ and $Y$, the following are equivalent:
(i) $T$ is norm continuous (i.e., bounded).
(ii) $T$ is weakly continuous.
(iii) $T$ is sequentially weakly continuous.
(Assertion (i) implies (ii) due to the existence of the dual operator $T'$, (ii) obviously implies (iii), and (iii) implies (i) due to the closed graph theorem.)
*Disclaimer.* Actually, I think that this would be a better fit for Mathematics StackExchange, but since the OP was directed to MathOverflow from there, it thought would be a bit unfair to send them back without an answer.
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1
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https://mathoverflow.net/users/102946
|
396469
| 163,726 |
https://mathoverflow.net/questions/396424
|
2
|
Consider the drifted Brownian motion $X\_t=1+\lambda(t)+W\_t$, where $\lambda: \mathbb R\to [0,\infty)$ with $1\le \lambda'(t)\le 2$ and $(W\_t)\_{\ge 0}$ denotes a Brownian motion. Define the hitting time
$$\tau:=\inf\{t\ge 0: X\_t\le 0\}$$
and further the function $p(t):=\mathbb P[\tau>t]$ for all $t\ge 0$. Can we show that $t\mapsto p(t)$ is Lipschitz?
|
https://mathoverflow.net/users/261243
|
Lipschitz continuity of $\mathbb P[\tau>t]$ with respect to $t$
|
I claim this is not a complete answer. I wonder whether it can be improved to obtain the Lipschitz continuity.
Define
$$\left(\frac{d\mathbb Q}{d\mathbb P}\right)\_t := \exp\left(-\int\_0^t \lambda'(s)dW\_s-\frac{1}{2}\int\_0^t(\lambda'(s))^2ds\right).$$
Then $B\_t:=W\_t+\lambda(t)$ is a Brownian motion under $\mathbb Q$. Thus,
$$p(t) = \mathbb E^{\mathbb Q}\left[\left(\frac{d\mathbb P}{d\mathbb Q}\right)\_{t}{\bf 1}\_{\{\inf\_{0\le r\le t}(1+\lambda(r)+W\_{r})>0\}}\right] = \mathbb E^{\mathbb Q}\left[\left(\frac{d\mathbb P}{d\mathbb Q}\right)\_{t}{\bf 1}\_{\{\inf\_{0\le r\le t}(1+B\_{r})>0\}}\right].$$
For every $0\le t<t+\Delta t$, it holds
$$0\le p(t)-p(t+\Delta t) = \mathbb E^{\mathbb Q}\left[\left(\frac{d\mathbb P}{d\mathbb Q}\right)\_{t+\Delta t}\left({\bf 1}\_{\{\inf\_{0\le r\le t}(1+B\_{r})>0\}}-{\bf 1}\_{\{\inf\_{0\le r\le t+\Delta t}(1+B\_{r})>0\}}\right)\right].$$
Using Holder inequality for $a, b>1$ with $1/a+1/b=1$, one has
$$p(t)-p(t+\Delta t)\le \mathbb E^{\mathbb Q}\left[\left(\frac{d\mathbb P}{d\mathbb Q}\right)\_{t+\Delta t}^a\right]^{1/a} \left(\mathbb Q\left[\inf\_{0\le r\le t}B\_{r}>-1\right]-\mathbb Q\left[\inf\_{0\le r\le t+\Delta t}B\_{r}>-1\right]\right)^{1/b}.$$
Let $\Phi:\mathbb R\to (0,1)$ be the CDF defined by
$$\Phi(x):=\int\_{-\infty}^x\frac{1}{\sqrt{2\pi}}e^{-z^2/2}dz.$$
Then we finally obtain
$$p(t)-p(t+\Delta t)\le \mathbb E^{\mathbb Q}\left[\left(\frac{d\mathbb P}{d\mathbb Q}\right)\_{t+\Delta t}^a\right]^{1/a}2^{1/b} \left( \Phi\left(\frac{1}{\sqrt{t}}\right)-\Phi\left(\frac{1}{\sqrt{t+\Delta t}}\right)\right)^{1/b}.$$
It follows from some tedious computation, we have
$$ \mathbb E^{\mathbb Q}\left[\left(\frac{d\mathbb P}{d\mathbb Q}\right)\_{t+\Delta t}^a\right]^{1/a} \le \exp(2(a-1)(t+\Delta t))$$
and
$$2^{1/b} \left( \Phi\left(\frac{1}{\sqrt{t}}\right)-\Phi\left(\frac{1}{\sqrt{t+\Delta t}}\right)\right)^{1/b} \le \left(\frac{2}{\pi t^3}\right)^{1/2b} \exp\left(-\frac{1}{2bt}\right) \Delta t^{1/b}.$$
We conclude thus the (local) $1/b-$Holder continuity. Is there any chance to have the Lipschitz continuity?
|
1
|
https://mathoverflow.net/users/261243
|
396484
| 163,730 |
https://mathoverflow.net/questions/396483
|
4
|
Is there a usable bound for the minimal index of a proper subgroup in a finite simple group of Lie type in terms of its rank and the characteristic (or even cardinality) of its field of definition?
One way to get such a bound would be to give a lower bound for the dimension of a non-trivial representation (over $\mathbb Q$, or over $\mathbb C$ — I don't know if these are the same for these groups). Possibly such a bound is obvious from a good knowledge of Deligne–Lusztig theory but this is something i would like to avoid if possible.
My naïve guess would be $p^r$ (where $r$ is the rank), from the example of $\mathrm{PSL}\_{r+1}(\mathbb F\_p)$ where there is a subgroup of index roughly $p^r$ (the maximal parabolic with Levi component $\mathrm{GL}\_r$) and it seems hard (impossible?) to come up with subgroups of lower index. For $r=1, 2$ one can actually prove that $p^r$ is the right order using a lower bound on the dimension of a nontrivial representation (proven by elementary means), I don't know whether this generalises to higher ranks.
|
https://mathoverflow.net/users/32210
|
Subgroups and representations of finite groups of Lie type
|
Explicit values for the minimum degree of a primitive permutation representation of a simple group of Lie type can be found in Table 4 of this paper:
*Guest, Simon; Morris, Joy; Praeger, Cheryl E.; Spiga, Pablo*, [**On the maximum orders of elements of finite almost simple groups and primitive permutation groups.**](http://dx.doi.org/10.1090/S0002-9947-2015-06293-X), Trans. Am. Math. Soc. 367, No. 11, 7665-7694 (2015). [ZBL1330.20002](https://zbmath.org/?q=an:1330.20002).
The arXiv version is available [here](https://arxiv.org/pdf/1301.5166.pdf). As the authors explain, the entries in the table come from various sources, some of which were found to have the odd mistake. So far as I am aware no mistakes are known for the table as given.
One consequence of the values given in the table is that the degree of a primitive permutation representation of ${\rm PSL}\_{r+1}(q)$ is bounded below by $q^r$ except when $(q,r)=(9,1)$.
**Edit 18 March 2022**: Apparently the entry for $E\_7(q)$ in the paper above is wrong -- there is a $q^5-1$ in the formula which should be $q^5+1$. For the record the correct formula is given in this paper:
*Vasil’ev, A. V.*, [**Minimal permutation representations of finite simple exceptional groups of types (E\_6), (E\_7), and (E\_8)**](https://eudml.org/doc/187826), Algebra Logika 36, No. 5, 518-530 (1997); translation in Algebra Logic 36, No. 5, 302-310 (1997). [ZBL0941.20006](https://zbmath.org/?q=an:0941.20006).
|
5
|
https://mathoverflow.net/users/801
|
396489
| 163,731 |
https://mathoverflow.net/questions/396488
|
2
|
Let $\ell \geq 5$ be a prime. Show that there exists an imaginary quadratic field $K$ with odd fundamental discriminant:
(a) $\ell$ inert in $K$,
(b) $(\ell, h\_{K})=1$.
Remark. Without requiring the discriminant being odd, the existence due to Horie-Onishi.
|
https://mathoverflow.net/users/311108
|
Imaginary quadratic fields with $\ell$-indivisible class number
|
Here's an elementary argument. For $\ell < 41$, $K = \mathbb{Q}(\sqrt{-163})$ works. For $\ell = 41$, $K = \mathbb{Q}(\sqrt{-3})$ works. Assume then that $\ell \geq 43$.
Choose an integer $1 \leq n \leq \ell - 1$ so that $\left(\frac{-n}{\ell}\right) = -1$ and let $k \in \{ 0, 1, 2, 3 \}$ be the unique integer so that $n+k \ell \equiv 3 \pmod{4}$. Write $-(n+k \ell) = d r^{2}$ with $d$ squarefree. Note that $d \equiv 1 \pmod{4}$ so $d$ is an odd fundamental discriminant. Moreover, $\left(\frac{d}{\ell}\right) = \left(\frac{-n}{\ell}\right) = -1$ and so $\ell$ is inert in $\mathbb{Q}(\sqrt{d})$.
The convexity bound for $L(1,\chi\_{d})$ and Dirichlet's class number formula gives the bound $h(d) \leq \frac{2}{\pi} \sqrt{|d|} \log(|d|)$. Thus,
$$
h(d) \leq \frac{2}{\pi} \sqrt{|d|} \log(|d|) < \frac{2}{\pi} \sqrt{4 \ell} \log(4 \ell),
$$
and it is straightforward to see that this is $< \ell$ if $\ell \geq 43$. Hence $(\ell,h\_{\mathbb{Q}(\sqrt{d})}) = 1$.
Note: In a previous version of this answer, I mention a result of Wiles [here](https://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/jlms/jdv031), which states that if $\ell \geq 5$ is a prime and $S$ is a finite set of odd primes written as a disjoint union $S = S\_{-} \cup S\_{0} \cup S\_{+}$ with the properties that
$\bullet$ $q \in S\_{-} \implies q \not\equiv 1 \pmod{\ell} \text{ and } q \not\equiv 3 \pmod{4}$, and
$\bullet$ $q \in S\_{0} \implies q \not\equiv 1 \pmod{\ell}$, and
$\bullet$ $q \in S\_{+} \implies q \not\equiv -1 \pmod{\ell}$,
then there is a fundamental discriminant $d < 0$ so that the class number of $\mathbb{Q}(\sqrt{d})$ is coprime to $\ell$ and $\left(\frac{d}{q}\right) = -1$ for $q \in S\_{-}$, $\left(\frac{d}{q}\right) = 0$ for $q \in S\_{0}$, and $\left(\frac{d}{q}\right) = 1$ for $q \in S\_{+}$.
However, Wiles's result does not apply since the elements of $S\_{-}$, $S\_{0}$ and $S\_{+}$ must be odd.
|
2
|
https://mathoverflow.net/users/48142
|
396495
| 163,734 |
https://mathoverflow.net/questions/396474
|
2
|
The following fraction shows up when trying to show consistency of the OLS slope estimator in a simple linear regression on a log-log scale where the window of observation changes as the sample size $T$ increases and I want to find a constant $c > 0$ such that
$$
\frac{\sum\_{k = 1}^{N(T)} \vert x\_{k, T} - \bar{x}\_{N(T)} \vert}{\sum\_{k = 1}^{N(T)} (x\_{k, T} - \bar{x}\_{N(T)})^2} \leq c
$$
if $T$ is large enough, where $$
\begin{align\*}
x\_{k, T} &= \log(T^{\varepsilon} + k - 1), \ \varepsilon \in (0, 1), \\
\bar{x}\_{N(T)} &= \frac{1}{N(T)} \sum\_{k = 1}^{N(T)} x\_{k, T}, \\
N(T) &= (m - 1) T^{\varepsilon},\ m > 1
\end{align\*}
$$
So far, I have tried some form of figuring out for which $k$ it holds that $\vert x\_{k, T} - \bar{x}\_{N(T)} \vert \leq 1$ and then using that the absolute value is less or equal to its square which helps to bound some part of the fraction but I do not now how to deal with the remaining part.
Another idea of mine is to rewrite the fraction by multiplying both the numerator and denominator by $1 / N(T)$ such that the numerator is the mean absolute deviation of a (non-random) sample $(x\_{1, T}, \ldots, x\_{N(T), T})$ from its mean whereas the denominator is the mean squared deviation. I was hoping that there maybe is some knowledge about the asymptotic behavior of both numerator and denominator but so far I did not find anything. Also, things are a bit tricky because as $T$ increases the sample not only becomes larger but changes altogether.
Finally, another idea is to use the rewritten fraction from the previous approach and try to show that both numerator and denominator are asymptotically equivalent to integrals of the form $\int\_{T^{\varepsilon}}^{mT^{\varepsilon}}...dx$ which are hopefully somewhat easy to evaluate.
However, I am not quite sure if such an asymptotic equivalence is feasible.
|
https://mathoverflow.net/users/302666
|
Asymptotic bound of quotient of absolute and squared deviation from mean
|
$\newcommand{\ep}{\varepsilon}\newcommand{\num}{\operatorname{num}}\newcommand{\den}{\operatorname{den}}\newcommand{\R}{\mathbb R}$Let
$$x\_k:=x\_{k,T}=\ln(S+k-1),\quad S:=T^\ep,\quad N:=N(T).$$
Let us ensure that $N$ is an integer, assuming, more generally, just that
\begin{equation}
N\sim(m-1)S.
\end{equation}
We have to show that eventually (that is, for all large enough $S>0$)
\begin{equation}
\frac{\num}{\den}\le c<\infty,
\end{equation}
where
\begin{equation}
\num:=\sum\_1^N|x\_k-\bar x|,\quad \den:=\sum\_1^N(x\_k-\bar x)^2,\quad \bar x:=\frac1N\,\sum\_1^Nx\_k,
\end{equation}
and
$c$ may depend only on $m\in(1,\infty)$.
Note that $\sum\_1^N|x\_k-a|$ is convex in $a\in\R$. Hence,
\begin{align\*}
\num&\le\max\Big(\sum\_1^N(x\_k-x\_1),\sum\_1^N(x\_N-x\_k)\Big) \\
&\le\max\Big(\sum\_1^N\ln\Big(1+\frac{k-1}S\Big),\sum\_1^N\ln\Big(1+\frac{N-K}S\Big)\Big) \\
&\le\max\Big(\sum\_1^N\frac{k-1}S,\sum\_1^N \frac{N-K}S\Big) \\
&=\frac{N(N-1)}{2S}\le\frac{(m-1)^2}{2+o(1)}\,S.
\end{align\*}
On the other hand, by Jensen's inequality for the concave function $\ln$,
\begin{equation}
\bar x=\frac1N\,\sum\_1^N\ln(S+k-1)\le\ln\Big(\frac1N\,\sum\_1^N(S+k-1)\Big)
=\ln(S+(N-1)/2).
\end{equation}
So,
\begin{align\*}
\den&\ge\sum\_{3N/4\le k\le N}(x\_k-\bar x)^2 \\
&\ge\sum\_{3N/4\le k\le N}\ln^2\Big(1+\frac{k-1-(N-1)/2}{S+(N-1)/2}\Big) \\
&\ge\frac N{4+o(1)}\,\ln^2\Big(1+\frac{N/4}{S+(m-1)S/2}\Big) \\
\ &\sim\frac{(m-1)S}4\,\ln^2\Big(1+\frac{m-1}{2m+2}\Big).
\end{align\*}
Thus,
\begin{equation}
\frac{\num}{\den}\le(2+o(1))(m-1)\Big/\ln^2\Big(1+\frac{m-1}{2m+2}\Big),
\end{equation}
as desired.
|
1
|
https://mathoverflow.net/users/36721
|
396497
| 163,735 |
https://mathoverflow.net/questions/396487
|
1
|
Apologies if this question is a little vague.
I have seen written in a few places that the space of projective structures on a Riemann surface is an affine space modelled on the space of holomorphic quadratic differentials. Firstly, is this true for real curves and higher dimensional projective structures? Secondly, is there an analogous result for other Cartan-geometric structures (or G-structures) that the space of such structures is an affine space modelled on some appropriate vector bundle?
Any references or pointers in the right direction greatly appreciated.
|
https://mathoverflow.net/users/163024
|
Understanding the space of structures
|
The space of affine connections is an affine space, since any two affine connections, say with Christoffel symbols $\Gamma^i\_{jk}$ and $\bar\Gamma^i\_{jk}$ differ by the difference of their Christoffel symbols $\Gamma^i\_{jk}-\bar\Gamma^i\_{jk}=a^i\_{jk}$, so the difference determines a tensor $a^i\_{jk}dx^j dx^k\partial\_i$. If they are torsion free this tensor will be symmetric. So both the moduli space of affine connections and of affine torsion free connections are affine spaces, modelled on $T^\* \otimes T^\* \otimes T$ and $S^2T^\* \otimes T$ respectively.
Every affine connection has the same unparameterized geodesics as its associated torsion free connection with symmetrized Christoffel symbols $\bar\Gamma^i\_{jk}=(1/2)(\Gamma^i\_{jk}+\Gamma^i\_{kj})$ (proof: write out the geodesic equations). Every torsion free connection with Christoffel symbols $\Gamma^i\_{jk}$ has the same unparameterized geodesics as another torsion free connection just when the second has Christoffel symbols $\Gamma^i\_{jk}+a\_j\delta^i\_k+a\_k\delta^i\_j$ for any $1$-form $a\_idx^i$, in any local coordinates. So the space of normal projective connections is the quotient space of the space of affine connections, which we can assume to be torsion free, by the space of $1$-forms. If they are torsion free with the same geodesics, it will be of this form $a^i\_{jk}=a\_j\delta^i\_k+a\_k\delta^i\_j$. So the space of normal projective connections is therefore an affine space modelled on the quotient space, i.e. on the affine space of traceless symmetric tensors $a^i\_{jk}dx^j dx^k\partial\_i$, $a^i\_{jk}=a^i\_{kj}$, $a^i\_{ik}=0$, which we can denote $T \otimes\_0 S^2 T^\*$, or something like that. This discussion works the same for the real normal projective connections and for the holomorphic ones, so we see that the moduli space of holomorphic normal projective connections on any compact complex manifold is finite dimensional, modelled on the vector space $H^0(M,T \otimes\_0 S^2 T^\*)$.
On a complex manifold, a holomorphic affine connection exists just when the Atiyah class of the tangent bundle vanishes. Less obviously: a holomorphic projective connection exists just when a normal holomorphic projective connection exists, both just when the tracefree part of the Atiyah class of the tangent bundle vanishes.
|
3
|
https://mathoverflow.net/users/13268
|
396501
| 163,737 |
https://mathoverflow.net/questions/396435
|
3
|
**Problem summary:** I'm trying to get some intuition for what the moduli space of objects for a dg-category (as in [this paper by Brav and Dyckerhoff](https://arxiv.org/abs/1812.11913)) actually looks like/how to give an alternative identification of the moduli of objects in certain cases.
**Edited to add:** Adding as per David's suggestion below: [the original paper on the topic by Toën and Vaquié](http://www.numdam.org/item/ASENS_2007_4_40_3_387_0/), and [this paper on a coherent version of the construction by Lowrey that I was previously unaware of](https://arxiv.org/abs/1110.5117).
---
Let $\mathcal{C}$ be a compactly generated dg-category. Then the moduli of objects is defined such that on an affine $U$, $\mathcal{M}\_{\mathcal{C}}(U)$ is the space of exact functors from compact objects $\mathcal{C}^c$ in $\mathcal{C}$ to perfect complexes on $U$. When $\mathcal{C}$ is smooth, such functors are corepresented by *left proper* objects $c\in \mathcal{C}^c\otimes\text{Perf}(U)$ -- this means that inner hom out of $c$ is a continuous functor $\mathcal{C}^c\otimes\text{Perf}(U) \to \text{Perf}(U)$ with a continuous right adjoint.
I'd like to get a better grasp on what these left proper objects actually look like. For instance, if $\mathcal{C} = \text{QCoh}(T^\ast C)$ for $C$ a smooth projective curve, then $\mathcal{M}\_\mathcal{C}$ should be (I think) the moduli space of Higgs sheaves on $C$. If $\mathcal{C} = k[x]\text{-mod}$, then I think that being left proper should be related to some sort of ''finite support'' condition. But I'm not sure how to actually *prove* either of these.
To be concrete, let's suppose that $A$ is a regular Noetherian $k$-algebra for $k$ a field of characteristic zero, and consider th dg-category $A\text{-mod}$. If I have everything straight in my head (already a bold assumption), then I think $\text{Hom}\_A(M,-): A\text{-mod}\to k\text{-mod}$ is continuous with a right adjoint precisely if $M$ is a perfect complex. I also think that in this case the right adjoint is the functor $\text{Hom}\_k(M^\vee, -)$ where $M^\vee = \text{Hom}\_A(M,A)$. If this is all the case, then what additional constraint is imposed on $M$ by continuity of its right adjoint?
|
https://mathoverflow.net/users/143797
|
Intuition for points of the moduli of objects for a dg-category
|
[It would be good to mention the original paper of Toën-Vaquié https://arxiv.org/abs/math/0503269 where these moduli spaces of objects are defined, and maybe that of Lowrey for the ``coherent" version https://arxiv.org/abs/1110.5117 ]
Having a continuous right adjoint (in the compactly generated presentable / "large" dg category setting) is equivalent to preserving compact objects. So you're asking for perfect A-modules to be taken to bounded complexes of vector spaces, which geometrically means taking Hom from a sheaf with proper support (so cohomology is finite dimensional) -- or as you say, in the affine setting, this implies having finite support. So for example in the Higgs sheaves context you're only seeing Higgs sheaves which are proper over $C$ -- i.e. supported on spectral varieties that are finite over the curve, as opposed to "going off to infinity" in the cotangent bundle (like Higgs sheaves with poles do).
There are many references for sheaves with proper support, not sure what the best one is, but in any case <https://arxiv.org/abs/1312.7164> talks about representability results for these kinds of proper functors (taking proper functors to Vect exchanges Coh and Perf).
|
2
|
https://mathoverflow.net/users/582
|
396514
| 163,742 |
https://mathoverflow.net/questions/396513
|
3
|
Let $R$ be a ring and let $\mathcal{C}$ be the category of perfect $R$-complexes. Suppose that $$S=\bigoplus\_{i=1}^{\infty}R$$
Let us define $\mathcal{D}$ the smallest thick category generated by $S$.
Clearly $\mathcal{C}$ is a full subcategory of $\mathcal{D}$
The natural embedding $i:\mathcal{C}\rightarrow \mathcal{D} $ induces a morphism in algebraic $K$-theory given by $K(i):K(\mathcal{C})\rightarrow K(\mathcal{D}) $
My question is the following: What can be said about $K\_{n}(\mathcal{C})\rightarrow K\_{n}(\mathcal{D}) $ for each $n\in \mathbf{N}$, is it an injective/surjective homomorphism?
|
https://mathoverflow.net/users/165456
|
Algebraic K-theory of a category containing all perfect complexes
|
Following Achim's comment, it suffices to show that one can apply the usual Eilenberg swindle argument.
Maybe it's not super clear to the OP that D is closed under countable direct sums of any single object (which is all that one needs for the Eilenberg swindle).
The point is as follows : let $E$ be the subcategory of $D$ on those objects $X$ such that $\bigoplus\_{i\in\mathbb N} X \in D$. Then $E$ is a thick subcategory of $D$, and it contains $S$ (because $\mathbb N\times \mathbb N \sim \mathbb N$), thus it contains $D$. In particular $\bigoplus\_\mathbb N : D\to D$ is well-defined and allows the Eilenberg swindle argument to go through.
For the sake of self-containment, let me recall how this argument goes : $\bigoplus\_\mathbb N \simeq \mathrm{id}\_D \oplus \bigoplus\_\mathbb N$.
Therefore by additivity, $\mathrm{id}\_{K(D)} + K(F) \simeq K(F)$ (where $F = \bigoplus\_\mathbb N$), and so $\mathrm{id}\_{K(D)} \simeq 0$, in particular $K(D)=0$.
|
8
|
https://mathoverflow.net/users/102343
|
396515
| 163,743 |
https://mathoverflow.net/questions/396452
|
3
|
I should preface this question by saying that I strongly suspect the answer is negative, but I couldn't find the counterexample myself.
Say we are working on the unit disc $D \subset \mathbf{R}^n$, where we are given two uniformly elliptic operators with coefficients $A^{ij}$ and $a^{ij}$. (These may be as regular as one needed, for instance at least $C^1$. Moreover for definiteness one might be the identity matrix, say $a^{ij} = \delta^{ij}$.)
Let $U$ and $u$ be two solutions respectively of the divergence-form equations $D\_i(A^{ij} D\_j U) = 0$ and $D\_i(a^{ij} D\_j u) = 0$. (Again $U$ and $u$ can be taken as regular as one desires.) We consider their difference $v:= U - u$.
**Question.** Is there a function $\lambda: D \to [0,1]$ so that $v$ satisfies the 'intermediate' equation $D\_i((\lambda A^{ij} + (1 - \lambda) a^{ij}) D\_j v) = 0$?
A negative answer would follow for example from the failure of $v$ to satisfy some property of elliptic PDE, the maximum principle being the obvious candidate. As mentioned above, I couldn't find the example that demonstrates this failure. (I suspect there might even exist one where $A^{ij}$ and $a^{ij}$ are constant.)
|
https://mathoverflow.net/users/103792
|
Does the difference of solutions of two unrelated PDE solve an 'intermediate' equation?
|
$U = 2x^2-y^2$ and $u = x^2-2y^2$ solve constant-coefficient elliptic equations, but $U - u = x^2 + y^2$ has an interior minimum and thus cannot solve an elliptic equation.
|
4
|
https://mathoverflow.net/users/16659
|
396518
| 163,745 |
https://mathoverflow.net/questions/396274
|
1
|
Let $M$ be a type III$\_1$ factor. Suppose $\rho$ is a normal state on $M$, given any $c\in [0,2]$, can we find a normal state $\rho'$ on $M$ such that $\|\rho-\rho'\|=c$? Or can we find a sequence of normal states $\rho\_n$ such that $\|\rho-\rho\_n\|\to c$ as $n\to \infty$?
|
https://mathoverflow.net/users/153196
|
Normal states on a type III$_1$ factor
|
I don't think you can get $c = 2$, but you can do this for any $c \in [0,2)$. Given $\epsilon > 0$, find a projection $p \in M$ with $\rho(p) < \epsilon$ and then find $\rho'$ supported on $p$. Then $$(\rho - \rho')(1 - 2p) = 1 - 2\rho(p) - 1 + 2 = 2(1 - \rho(p)),$$ showing that $\|\rho - \rho'\|$ can be arbitrarily close to $2$. Then linearly interpolating between $\rho$ and $\rho'$ gives you normal states of any intermediate distance to $\rho$.
|
3
|
https://mathoverflow.net/users/23141
|
396523
| 163,747 |
https://mathoverflow.net/questions/396521
|
7
|
In [arxiv:0909.3140](https://arxiv.org/abs/0909.3140) the $G$-extensions of a fusion category $\mathcal{D}$ are classified via monoidal 2-functors $G \to \underline{\underline{BrPic}}(\mathcal{D})$. A crucial part of the classification is Theorem 7.7 that shows that equivalence classes of such functors are in 1-1 correspondence with equivalence classes of $G$-extensions of $\mathcal{D}$. However in the proof they never show that the tensor categories they obtain from such functors are actually rigid. I was also unable to find this statement anywhere else in the paper, it is not even mentioned. Is there some obvious proof of this that I am missing or are they in fact classifying not necessarily rigid $G$-extensions? (The latter would however contradict Definition 2.1, which defines a G-extension to be in particular a fusion category, hence rigid.)
|
https://mathoverflow.net/users/122206
|
Why are $G$-Extensions of fusion categories rigid, when constructed via monoidal 2-functors $G \to \underline{\underline{BrPic}}(\mathcal{D})$?
|
Great question! Yes, this is missing from the original paper and is certainly needed for the main result. It is true though. There's two different proofs currently in the literature, [Deshpande and Mukhopadhyay](https://arxiv.org/pdf/1909.10799.pdf) Corollary 2.11 and [Davydov and Nikshych](https://arxiv.org/pdf/2006.08022.pdf) proof of Theorem 8.5.
You can also prove it using an argument I learned from Theo Johnson-Freyd, namely given an object $X$ in $C\_g$ you can realize $C\_g$ as $A$-mod in $C\_e$ with $X$ corresponding to $A$ itself, then $C\_{g^{-1}}$ can be identified with mod-$A$ and the dual object will be $A$ again and the evaluation and coevaluation maps are given by multiplication and the unit of the algebra structure on $A$.
As to your last two sentences, rigidity is pretty essentially to the whole setup, otherwise the graded parts don't need to be invertible and then you leave the world of homotopy theory. For example, take the semisimple category with two objects 1 and X with tensor product defined by $X \otimes X = 0$ (aka the universal counterexample to all questions involving rigidity), this is $\mathbb{Z}/2\mathbb{Z}$ graded but the odd part is not invertible as a bimodule over the even part.
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7
|
https://mathoverflow.net/users/22
|
396529
| 163,749 |
https://mathoverflow.net/questions/396505
|
2
|
I apologize in advance if this question is not up to the level of research level questions on Math overflow. I am a complete outsider to invariant theory/representation theory and would like someone more knowledgeable than me to direct me to where I should read.
Let $Sym^k(\mathbb{R}^n)$ be the vector space of symmetric multilinear real valued maps in $k$ arguments. The orthogonal group, $O(n)$, acts on $\mathbb{R}^n$ and hence gives a natural action on $Sym^k(\mathbb{R}^n)$. Let $\mathcal{A}$ be the algebra of polynomials with domain $Sym^k(\mathbb{R}^n)$ and codomain $\mathbb{R}$. After a brief literature review, I believe Hilbert and others' work on classical invariant theory gives that the subalgebra of $\mathcal{A}^{O(n)}$ of polynomials that are constant on the orbits of action of $O(n)$ on $\mathcal{A}$ is finitely generated and "separates" the orbits.
**Question:** Is there an explicit finite list of generators of the algebra of $\mathcal{A}^{O(n)}$ ?
I seem to have an answer in case $k$ is even (assuming I didn't make any mistakes with identifications or other details):Set $k=2b$. A symmetric $\mathbb{R}$-multilinear map in $k$ arguments can (?) be naturally identified/thought of as a symmetric (?) endomorphism of $\otimes\_{i=1}^{b}\mathbb{R}^n$, finally take your polynomials invariants to be the cofficents of the characteristic equation of that endomorphism on $\otimes\_{i=1}^{b}\mathbb{R}^n$. My strategy is relying on the fact that any symmetric matrix with real entries can be diagonalized by an orthonormal eigenbasis
**Question:** Is the above argument correct for the case $k$ even, and what about the case $k$ is odd ? I admit I did not work out all details/identifications of the above argument yet so I imagine I could be wrong, but I will work them out asap.
Thank you,
|
https://mathoverflow.net/users/32135
|
$O(n)$ Polynomial invariant of symmetric tensors
|
Just in case this wasn't obvious from Abdemalek Abdesselam's answer, the following is a well-known fact. The algebra of invariant polynomials $\mathcal{A}^{O(n)}$ consists of linear combinations of all possible total contractions of any number of copies of the symmetric tensors from $Sym^k(\mathbb{R}^n)$. These contractions can be represented by $k$-valent graphs (draw a $k$-valent vertex for each copy of the symmetric tensor, with each attached leg representing an index, then connect all pairs of legs that correspond to an index contraction). Then, obviously, the algebra $Sym^k(\mathbb{R}^n)$ is generated by the connected graphs of this type (multiplication of invariants corresponds to the disjoint union of corresponding graphs). So you get an explicit lit of generators (the connected graphs), but this list is infinite. By the work of Hilbert, as you mention, there is a finite list of generators, but finding say a graphical representation of these generators is the difficult problem, with the answer (when known) being highly dependent on $n$ and $k$, as mentioned in the other answer.
The above well-known is a special case of the so-called *First Fundamental Theorem* of the invariant theory of $O(n)$. It can be challenging to find a clean statement of it of sufficient generality. We happen to have included such a statement in Appendix C of [this paper](https://arxiv.org/abs/1710.01937) with a self-contained proof. Actually, we stated it for the group $O(1,n-1)$, but the $O(n)$ case is completely analogous and even easier due to its compactness.
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4
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https://mathoverflow.net/users/2622
|
396530
| 163,750 |
https://mathoverflow.net/questions/396519
|
3
|
*This was [asked at MSE](https://math.stackexchange.com/questions/4182108/how-complicated-must-a-real-with-an-easy-total-computability-problem-be-to-com) without success. Granted, a bounty is still ongoing there, but it doesn't look like it will be answered.*
---
For (noncomputable) $A\subseteq\omega$ let $\tilde{A}=\{e: \varphi\_e^A\mbox{ is total and }\exists c(\varphi\_e^A\simeq\varphi\_c)\}$. A priori $\tilde{A}$ is $\Sigma^0\_3(A)$ ("$\varphi\_e^A$ is total and there is some $c$ such that on all inputs we eventually see agreement between $\varphi\_c$ and $\varphi\_e^A$"). However, this bound isn't sharp in general: [if $A$ is sufficiently Cohen generic then $\tilde{A}$ is $\Pi^0\_2(A)$](https://math.stackexchange.com/a/4179424/28111) *(of course we can't do better than this: $\tilde{A}$ is always $\Pi^0\_2(A)$-hard)*.
However, a fair amount of genericity (at a glance, $2$-genericity) is needed for that argument. This raises the question of how hard it must be to compute a real (nontrivially) satisfying "$\tilde{A}$ is not $\Sigma^0\_3(A)$-complete." Specifically:
>
> Does every noncomputable $\Delta^0\_2$ set $A$ satisfy "$\tilde{A}$ is $\Sigma^0\_3$-complete"?
>
>
>
Genericity-based arguments almost certainly won't be useful here, since there aren't even any $\Delta^0\_2$ *weak* $2$-generics.
|
https://mathoverflow.net/users/8133
|
Is there a $\Delta^0_2$ real with "easy total computability problem"?
|
It seems to me that if $G$ is 1-generic and recursive in $0'$ then $\tilde{G}$ is Boolean $\Sigma^0\_2$, which is sufficient to conclude that $\tilde{G}$ is not $\Sigma^0\_3(G)$-complete.
First, show that if (1) there is a condition $p$ in $G$ such that there is no $\varphi\_e$-splitting pair of conditions extending $p$ and (2) for every condition $q$ in $G$ and every $m$, there is an extension $r$ of $q$ which makes $\varphi\_e(m)$ converge, then $e$ is in $\tilde{G}$. We can exhibit a $\varphi\_c$ for $\varphi\_e(G)$ by observing that the value of $\varphi\_e(G)$ at $m$ can be recursively determined by searching for any extension of $p$ that gives a value at $m$. Second, show that if $e$ is in $\tilde{G}$ then (1) and (2) must hold. Obtaining (1) uses the 1-genericity of $G$ and obtaining (2) uses the totality of $\varphi\_e(G)$.
Since $G$ is $\Delta^0\_2$, Condition (1) is $\Sigma^0\_2$ and (2) is $\Pi^0\_2$, and so $\tilde{G}$ is Boolean $\Sigma^0\_2$.
|
5
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https://mathoverflow.net/users/31026
|
396540
| 163,752 |
https://mathoverflow.net/questions/396423
|
5
|
Let $v\in\mathcal{C}^1(\mathbb{R}^n,\mathbb{R}^n)$ $(n\geq 1)$ a velocity field such that every solution $(x\_t)\_{t\geq 0}$ of $(d/dt)x\_t=v(x\_t)$ is periodic. Denote, for a non-stationary point $x\in\mathbb{R}^n$ (meaning $v(x)\neq0$), by $T(x)$ the period of such a solution $(x\_t)\_{t\geq0}$ such that $x(0)=x$.
**Quetion:** Is $T$ continuous over the set of non-stationnary points?
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https://mathoverflow.net/users/159940
|
Continuity of the period for a periodic dynamical system
|
The answer is trivially negative already for $n=3$: Start with the flow following along the "long" lines of a thickened (say with disc-shaped cut) Möbius strip. You can imagine this like a thick “rope” bent to a loop, making half a rotation on its way, and the flow follows the fibres of the rope (with a fixed speed).
Obviously, the fibre in the "center" of the rope has half of the period than the nearby fibres.
To extend the flow to $\mathbb R^3$, decrease the speed smoothly near the "boundary" of the rope to $0$, and then extend the field by letting it $0$ outside of the rope.
However, this extension works only, because the question explicitly admits stationary points. If one wants to exclude stationary points, there is a topological obstacle in the extension of this particular flow to $\mathbb R^3$. After an embedding into $\mathbb R^4$ this obstacle trivially vanishes, as the Möbius strip can be "unentangled" in $\mathbb R^4$, so in $\mathbb R^4$ there is obiously a non-singular counterexample.
I conjecture that for $n=3$ there is no non-singular counterexample at all, but I do not know how to prove this.
|
4
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https://mathoverflow.net/users/165275
|
396548
| 163,757 |
https://mathoverflow.net/questions/396537
|
1
|
Let $k$ be a field of characteristic 0 and let $\varphi:\mathfrak{g}\rightarrow\mathfrak{f}$ and $\psi:\mathfrak{h}\rightarrow\mathfrak{f}$ be maps of Lie algebras. Is there a reference showing that the pullback (in the category of Lie $k$-algebras) of $\mathfrak{g}$ and $\mathfrak{h}$ along these maps is given by the following formula
>
> $\mathfrak{g}\times\_\mathfrak{f}\mathfrak{h}=\{(x,y)\in\mathfrak{g}\times\mathfrak{h}\mid \varphi(x)=\psi(y)\}$
>
>
>
EDIT: Just to be extra clear. I do know how to prove it, but I cannot find a place in the literature where this is proved or stated and I was wondering if you knew of one.
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https://mathoverflow.net/users/226648
|
Pullback of Lie algebras
|
A reference for pullbacks for modules over a ring: Proposition 5.11, p.222 in "An Introduction to Homological Algebra" by Rotman. For Lie $k$-algebras (or just $k$-algebras), you could refer to this and say the same construction/proof works.
|
1
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https://mathoverflow.net/users/38068
|
396559
| 163,762 |
https://mathoverflow.net/questions/396563
|
10
|
I am working on a article in poset theory. In that article, I am defining a subposet of a poset. The definition is following:
Let $P$ be a finite poset. A subposet $P'$ of $P$ is called **closed under covering** if for every $x,y \in P'$ with $x\lessdot y$ in $P'$, we have $x\lessdot y$ in $P$. Here, $x \lessdot y$ means $x$ is covered by $y$.
I want to know weather the above definition is already in the literature? If yes, then what these subposets are called? If not, then the name I have given is correct? or what should I call such subposets?
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https://mathoverflow.net/users/311595
|
A definition in poset theory
|
I recall seeing in various sources the terminology "cover preserving embedding" and "cover preserving subposet". Googling it now (<https://www.google.com/search?q=poset+%22cover+preserving%22>) brings some 4000 results, many of which are research articles (with some repetitions - I am not implying there are 4000 distinct articles on this topic).
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10
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https://mathoverflow.net/users/1306
|
396566
| 163,765 |
https://mathoverflow.net/questions/347835
|
10
|
* A space $X$ is said to be *sequential* if whenever $A \subset X$ is not closed then $A$ contains a sequence converging to a point outside of $A$.
* A space $X$ is said to have *countable tightness* if for every non-closed set $A \subset X$ and every point $x \in \overline{A} \setminus A$ there is a countable subset $C$ of $A$ such that $x \in \overline{C}$.
Clearly, every sequential space has countable tightness, and not every space of countable tightness is sequential. The Moore-Mrowka problem asks whether every compact Hausdorff space of countable tightness is sequential.
This problem is known to independent of ZFC. Balogh proved that every compact space of countable tightness is sequential under PFA and the one-point compactification of Ostaszewski's example under $\Diamond$ of a locally compact countably compact hereditarily separable perfectly normal non-compact space, is an example of a compact space of countable tightness which is not sequential.
*Balogh, Zoltán*, [**On compact Hausdorff spaces of countable tightness**](http://dx.doi.org/10.2307/2046929), Proc. Am. Math. Soc. 105, No. 3, 755-764 (1989). [ZBL0687.54006](https://zbmath.org/?q=an:0687.54006).
Even though it is not sequential, the one-point compactification of Ostaszewski space enjoys another natural convergence-type property known as pseudoradiality. A space is *pseudoradial* if whenever $A \subset X$ is not closed then $A$ contains a transfinite sequence (that is, a well-ordered net) converging outside of $A$.
$MA\_{\omega\_1}$ negates the existence of Ostaszewski space for two reasons: 1) because it implies that every countably compact perfectly normal space is compact (Weiss) or 2) because it implies that every locally compact hereditarily separable space is hereditarily Lindelof (Szentmiklossy). However, $MA\_{\omega\_1}$ is not strong enough to imply a positive solution to the Moore-Mrowka problem (see Balogh's paper).
>
> QUESTION: Assume $MA\_{\omega\_1}$. Is it true then that every compact pseudoradial space of countable tightness is sequential?
>
>
>
NOTE: 1) A byproduct of Szentmiklossy's result is that the Moore-Mrowka problem has a positive answer for hereditarily separable spaces under $MA\_{\omega\_1}$ (every hereditarily Lindelof space has points $G\_\delta$ and every compact space with points $G\_\delta$ is first-countable, and hence, sequential).
2) A Hausdorff (non-compact) pseudoradial non-sequential space of countable tightness was constructed in ZFC by Simon and Tironi.
*Simon, Petr; Tironi, Gino*, [**Two examples of pseudo-radial spaces**](https://eudml.org/doc/17445), Commentat. Math. Univ. Carol. 27, 155-161 (1986). [ZBL0596.54005](https://zbmath.org/?q=an:0596.54005).
|
https://mathoverflow.net/users/11647
|
A variant of the Moore-Mrowka problem
|
This question was answered in the negative by Alan Dow and Istvan Juhász in a recent preprint.
[On the cardinality of separable pseudoradial spaces.](https://webpages.uncc.edu/%7Eadow/mioduszewskiDec4.pdf)
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5
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https://mathoverflow.net/users/11647
|
396572
| 163,768 |
https://mathoverflow.net/questions/396511
|
3
|
In the comments section of [this question](https://mathoverflow.net/questions/301882/on-functors-preserving-monoid-objects#comment879728_301882) there was a question that I don't know if it has been asked on the site. It is well-known and easily proved that lax monoidal functors preserve monoids. So the question, which is about the converse, in my words is the following
>
> Given a functor between monoidal categories that preserves monoids and morphisms of monoids on these categories, does it follow that the functor is lax monoidal?
>
>
>
I am interested in this question because in my research I've encountered a functor between the underlying collections of operads that do preserve the operad structure as well as the morphisms of operads. Since operads are monoids on the category of collections (or $\mathbb{S}$-modules for the symmetric case) with the composite of collections as monoidal product, I was expecting this functor to be lax monoidal, but I haven't been able to prove it.
If the answer to my above question is positive it must mean that I have made some mistake and I would post a question related to that particular case. If not, I am fine with my functor preserving just operads and its morphisms. In fact, I believe that this implication is not going to be true in general, because the associativy axiom of monoidal categories relates 3 (in general different) objects while the axioms of monoids only involve a single object. But maybe there are some nice sufficient condition for this implication to hold.
A similar question that I might post separated is related to the fact that symmetric lax monoidal functors not only preserve monoids on the category but also operads on the category (here I am using the monoidal product of the underlying category, not the composite, which is eventually defined in terms of the monoidal product of the underlying category). So a natural question again is whether a functor that preserves operads and their morphisms is automatically symmetric lax monoidal.
|
https://mathoverflow.net/users/144957
|
Functors that preserve monoids
|
A simple class of counterexamples can be found among [thin](https://ncatlab.org/nlab/show/thin+category) [small](https://ncatlab.org/nlab/show/small+category) [strict monoidal categories](https://ncatlab.org/nlab/show/strict+monoidal+category), i.e. preordered monoids. These are just sets equipped with a preorder and a monoid structure such that the multiplication is increasing in each variable. For simplicity, let us only look at partial orders.
Let $P$ be a partially ordered monoid. A monoid in $P$ is an element $a$ of the underlying set satisfying $1 \leq a$ and $a^2 \leq a$ (and hence $a^2=a$). (So *here*, a monoid structure is merely a property.) If $1$ happens to be the largest element of $P$, it follows that $1$ is the only monoid in $P$. It follows that if $f : P \to Q$ is an increasing map with $f(1)=1$, it preserves monoids. The question is if we can conclude that $f$ is lax monoidal, which means that $f(a) \cdot f(b) \leq f(a \cdot b)$ holds for all $a,b \in P$. We cannot.
For example, let $P = Q = ([0,\infty],\geq,+,0)$ (this example appears in the definition of [Lawvere metric spaces](http://www.tac.mta.ca/tac/reprints/articles/1/tr1abs.html), which are just $P$-enriched categories). The monoidal unit is $0$ and it is the largest element with respect to $\geq$. Consider the map $f(a):=a^2$. The lax monoidal condition would mean $a^2 + b^2 \geq (a+b)^2$, which is not true.
*Edit.* Alternatively, if $P$ is any partially ordered group, $1$ is the only monoid in $P$ (since $a^2=a \implies a=1$). The map $f : P \to P$, $f(a):=a^2$ is increasing and satisfies $f(1)=1$, but it is lax monoidal iff $a^2 b^2 \leq (ab)^2$ or all $a,b$ iff $ab \leq ba$ for all $a,b$ iff $ ab=ba$ for all $a,b$ iff $P$ is commutative. But [there are](https://math.stackexchange.com/questions/353160) non-abelian partially ordered groups.
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6
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https://mathoverflow.net/users/2841
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396577
| 163,770 |
https://mathoverflow.net/questions/396590
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2
|
I am trying to find the asymptotic behavior (with respect to N) of the integral $$ \frac{2}{\sqrt{\pi}}\int\_0^\infty \varPhi^{N-2}(p)e^{-p^2}\ dp. $$ In Rényi and Sulanke's paper *Uber die konvexe Hulle von n zufaillig gewahlten Punkten* they use the relation $$ \varPhi(p) = 1 - \frac{e^{\frac{-p^2}{2}}}{\sqrt{2\pi}p\left(1 + \frac{\theta\_p}{p^2}\right)} $$ for $p>1$, where $0< \theta\_p < 1$. Can anyone help me understand where this second equation comes from?
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https://mathoverflow.net/users/311932
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Approximation of $\Phi (p)$
|
Equality
$$ \varPhi(p) = 1 - \frac{e^{\frac{-p^2}{2}}}{\sqrt{2\pi}p\left(1 + \frac{\theta\_p}{p^2}\right)} $$
for some $\theta\_p\in(0,1)$ can be rewritten as
$$r\_2(p):=\frac1{p+1/p}<r(p):=\frac{1-\varPhi(p)}{e^{-p^2/2}/\sqrt{2\pi}}<r\_1(p):=\frac1p,\tag{1}$$
which actually holds for all real $p>0$ and is a special case of a well-known result -- see e.g. formula (1.8) (with $m=1$) in [this paper](https://eudml.org/doc/122588).
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1
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https://mathoverflow.net/users/36721
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396597
| 163,775 |
https://mathoverflow.net/questions/396595
|
6
|
In my old high school notebook (20 years ago), the following inequality appears with its proof:
$$1+\cos x + \frac{1}{2}\cos 2x + \cdots + \frac{1}{n}\cos nx \geq 0$$
for any real $x$ and positive integer $n$.
I am not the one that created this inequality. So the my question is where references for this inequality can be found.
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https://mathoverflow.net/users/161614
|
Need a reference for a trigonometric inequality
|
According to the last sentence on page 16 of [this paper](https://www.dcce.ibilce.unesp.br/%7Edimitrov/papers/main.pdf), this inequality was proved by
W. H. Young, On certain series of Fourier, Proc. London Math. Soc. (2) 11
(1912), 357–366.
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9
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https://mathoverflow.net/users/36721
|
396599
| 163,776 |
https://mathoverflow.net/questions/394266
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4
|
Let $p:E\to B$ and $p':E'\to B$ be fibrations. It is [well known](https://en.wikipedia.org/wiki/Fiber-homotopy_equivalence) that if $f:E\to E'$ a fibrewise map that is also a homotopy equivalence, then it is a fibrewise homotopy equivalence.
What about the more general situation of fibrations $p:E\to B$ and $p':E'\to B'$ over different bases? A *fibre-preserving map* from $p$ to $p'$ is a pair of maps $f:E\to E'$ and $\overline{f}:B\to B'$ such that $\overline{f}\circ p=p'\circ f$, and a *fibre-preserving homotopy* between such maps is a pair of homotopies $H:E\times I\to E'$ and $\overline{H}:B\times I\to B'$ such that the following diagram commutes:
$\require{AMScd}$
\begin{CD}
E\times I @> H >> E'\\
@V p\times\operatorname{Id} V V @VVp'V\\
B\times I @> \overline{H}>> B'
\end{CD}
One can then easily define a *fibre-preserving homotopy equivalence* from $p$ to $p'$.
>
> Question: Is there an example of a fibre-preserving map of fibrations
> \begin{CD}
> E @> f >> E'\\
> @V p V V @VVp'V\\
> B @> \overline{f}>> B'
> \end{CD}
> such that $f$ and $\overline{f}$ are homotopy equivalences, but the pair $(f,\overline{f})$ is not a fibre-preserving homotopy equivalence, i.e. does not admit a fibre-preserving homotopy inverse?
>
>
>
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https://mathoverflow.net/users/8103
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fibre-preserving homotopy equivalence
|
The answer to the question as asked is no: a fibre-preserving map of fibrations in which the maps of total and base spaces are homotopy equivalences is neccessarily a fibre-preserving homotopy equivalence (also known as a *homotopy equivalence of fibrations*). A reference was supplied in the comments by Gustavo Granja, to Peter May's book *A Concise Course in Algebraic Topology*, where the statement appears as a Proposition on page 53. (The proof, although not given in detail, does appear to be a straightforward dualization of the corresponding result for cofibrations, proved on page 48.)
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1
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https://mathoverflow.net/users/8103
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396607
| 163,779 |
https://mathoverflow.net/questions/396547
|
2
|
Let $(M,\omega,H)$ be a Hamiltonian system and assume that $\gamma$ is a periodic orbit on a regular energy hypersurface. Then the regular orbit cylinder theorem (see for example *Abraham/Marsden: Foundations of Mechanics* Theorem 8.2.2 or the book by *Hofer/Zehnder* on Hamiltonian Dynamics Proposition 2 on page 110) states the following:
>
> If the flow of the Hamiltonian vector field $X\_H$ at $\gamma(0)$ has precisely two Floquet multipliers equal to $1$, then $\gamma$ is contained in a whole one-parameter family of periodic orbits, i.e. an orbit cylinder.
>
>
>
Now there is a delicate Analysis of different bifurcation scenarios and various stability properties (see *Abraham/Marsden* chapter 8). My question: **How "generic" is the condition of having precisely two Floquet multipliers equal to $1$?** Equivalently, how likely is it that a Poincaré section map has precisely one eigenvalue equal to $1$?
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https://mathoverflow.net/users/98139
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On the existence of regular orbit cylinders
|
The two multipliers correspond to the exponential growth rate of perturbations 1). Along the periodic orbit, and 2). Normal to energy hypersurface.
Unless there is additional symmetry in the system, two multipliers = 1 is the generic situation. A common example of symmetry is integral of motion of Hamiltonian system (of which the energy is an example). For each additional constant to motion, you should expect one more unity multiplier.
See more details in : Continuation of periodic orbits in conservative and
Hamiltonian systems, Physica D, 2003
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1
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https://mathoverflow.net/users/230017
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396618
| 163,781 |
https://mathoverflow.net/questions/396613
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2
|
Non-extendable 2D TQFTs correspond to finite dimensional Frobenius algebras [1].
How about 3D TQFTs? While the answer is clear for the extended ones (e.g. (3,2,1) TQFTs almost correspond to modular tensor categories [2]), I have not seen any discussion for (3,2) TQFTs.
More precisely, can one classify the functors
$$Cob\_{3,2}^{oriented} \to (Vect\_\mathbb{C})?$$
**Reference**
* [1] [Cohomology rings and 2D TQFTs](https://mathoverflow.net/questions/8772/cohomology-rings-and-2d-tqfts)
* [2] K. Walker's answer [here](https://mathoverflow.net/questions/386/do-all-3d-tqfts-come-from-reshetikhin-turaev/)
* [3] related - [Example of a non-extendable TQFT?](https://mathoverflow.net/questions/290880/example-of-a-non-extendable-tqft)
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https://mathoverflow.net/users/124549
|
Non-extendable 3D TQFTs
|
Check out Andras Juhasz' paper: <https://arxiv.org/pdf/1408.0668.pdf>
Specifically, Theorem 1.10:
>
> There is an equivalence between the symmetric monoidal category
> of (2+1)-dimensional TQFTs and the category of J-Algebras.
>
>
>
J-Algebras are somewhat cumbersome to define.
They are graded vector spaces equipped with a Frobenius-like structure, together with an action of the genus $g$ mapping class group on the $g$-th graded piece.
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7
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https://mathoverflow.net/users/5690
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396621
| 163,782 |
https://mathoverflow.net/questions/396622
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4
|
Let $X$ be a CAT(0) space, $x \in X$ and $\xi$ be a point in the boundary at infinity of $X$. Denote by $\rho(t)$ the geodesic ray (parameterized by arc length) starting from $x$ and asymptotic to $\xi$. Let $y$ be a point close to $x$ and let $\overline{xy}$ denote the geodesic segment connecting $x$ and $y$. I would like to estimate the angle (at $x$) between $\overline{xy}$ and $\rho$, in terms of the Busemann function
$$b\_{\xi}(y) = \lim\_{t \to \infty} d(y, \rho(t)) -t.
$$
Precisely, I would like to say that if $b\_{\xi}(y)>0$, then the angle is obtuse.
Can this be true? I’m interested in the case where $X$ is a Euclidean building, but I was hoping it is a known fact (either true or false) about CAT(0) spaces.
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https://mathoverflow.net/users/127739
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Angles and Busemann function in CAT(0)
|
The answer is "no" even in the hyperbolic plane $\mathbb{H}^2$.
Consider the horocycle about $\xi$ through $x$: this is the set of points such that $b\_{\xi}(z)=0$. Let $\gamma$ be the geodesic through $x$ tangent to the horocycle, and take $y$ to be any point on $\gamma$ (except $x$ itself). Then the angle between $\overline{xy}\subset\gamma$ and $\rho$ is $\pi/2$, but $b\_{\xi}(y)>0$ since $y$ is outside the horodisc enclosed by the horocycle.
However, similar geometric arguments show that a slightly weaker statement is true in $\mathbb{H}^2$, which may be covered by your assumption that $y$ is 'close enough' to $x$, as follows. For any $\theta<\pi/2$ there is an $\epsilon>0$ such that, if $d(x,y)<\epsilon$ and the angle between $\rho$ and $\overline{xy}=\theta$, then $b\_{\xi}(y)<0$.
|
5
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https://mathoverflow.net/users/1463
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396636
| 163,787 |
https://mathoverflow.net/questions/396633
|
0
|
Suppose $X, Y$ are $L^1$ random variables, and $X\_t$ and $Y\_t$ are real valued stochastic processes with $X\_t, Y\_t \in L^1$ for all $t$ such that the following convergences hold:
i) $X\_t \to X$, $Y\_t \to Y$ a.s and in $L^1$.
ii) $E(X\_t| Y\_t) \to E(X)$ a.s.
iii) $E(Y\_t|X\_t) \to E(Y)$ a.s.
Does it follow that $X$ and $Y$ are independent?
|
https://mathoverflow.net/users/173490
|
Independence of limits of asymptotically independent processes
|
The title mentions asymptotic independence, but I don't see any sort of asymptotic independence condition here. For random variables $X$ and $Y$, the conditions $E(X|Y)=E(X)$ a.s. and $E(Y|X)=E(Y)$ a.s. don't imply that $X$ and $Y$ are independent.
For your question, you can take $(X,Y)$ to be uniformly distributed on the eight values $(1,1), (1,-1), (-1,1), (-1,-1), (2,2), (2,-2), (-2,2), (-2,-2)$. Observing $X$ gives you information about the magnitude of $Y$, but not its sign (and vice versa). Then let $X\_t=X$ and $Y\_t=Y$ for all $t$.
|
0
|
https://mathoverflow.net/users/5784
|
396639
| 163,788 |
https://mathoverflow.net/questions/396600
|
8
|
Let $\alpha$ be an exterior product of a harmonic and a parallel form on a Riemannian manifold. Then $\alpha$ is known to be harmonic. I have heard that this is an old result due to R. Bott, but I could never find a reference. I would be very grateful for any pointers to the early literature.
|
https://mathoverflow.net/users/3377
|
reference to a theorem about a product of harmonic and parallel forms
|
One place where (a generalization of) the desired result is stated explicitly is in a 1973 paper by J. H. Sampson, [On a theorem of Chern](https://www.jstor.org/stable/1996588). Sampson gives a simplified proof of Chern's main result in his 1957 paper *On a generalization of Kähler geometry* (Algebraic geometry and topology. A symposium in honor of S. Lefschetz, pp. 103–121. Princeton University Press, Princeton, N. J., 1957).
There is also a very nice 1962 exposition (in French) of Chern's theorem by André Weil, [Un théorème fondamental de Chern en géométrie riemannienne](http://www.numdam.org/item/SB_1961-1962__7__273_0.pdf), in *Séminaire N. Bourbaki, 1962, exp. no 239, p. 273-284*.
The form in which Sampson and Weil state Chern's result is as follows:
*Theorem:* If a Riemannian manifold $(M^n,g)$ has holonomy $H\subseteq\mathrm{O}(n)$, then the $g$-Laplacian commutes with all of the linear operators on $\Omega^\*(M)$ constructed from the ring of $H$-equivariant linear maps $L:\Lambda^\*(\mathbb{R}^n)\to\Lambda^\*(\mathbb{R}^n)$.
The desired result is a special case of this theorem, since, if $\pi$ is a $g$-parallel $p$-form, then the holonomy $H$ of $g$ is contained in the stabilizer of $\pi$, and hence, by the above result, the operator $L(\alpha) = \alpha\wedge\pi$ commutes with the Laplacian of $g$. Of course, this implies the desired result since $\Delta(\alpha\wedge\pi) = \Delta\bigl(L(\alpha)\bigr) = L\bigl(\Delta(\alpha)\bigr) = 0$ when $\Delta(\alpha) = 0$, but is stronger.
*Comments:* Chern proved the above theorem in his 1957 paper for the ring of operators $L$ as above that are degree-preserving, but, in fact, this implies the more general result.
Both Chern and Sampson phrase the theorem in terms of $H$-structures without torsion, but, of course, this is the same as Riemannian manifolds with holonomy contained in $H$, a point explicitly made by Weil.
Sampson does not refer to Weil's article; perhaps he was unaware of it. Meanwhile, Sampson goes on to apply Chern's idea to other Laplacians and proves new results.
There are earlier results in special cases by André Lichnerowicz, e.g. *Généralisations de la géométrie kählérienne globale.* (French) Colloque de Géométrie Différentielle, Louvain, 1951, pp. 99–122. Georges Thone, Liège; Masson & Cie., Paris, 1951, but neither his articles nor those of Chern, Sampson, or Weil mention Bott.
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8
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https://mathoverflow.net/users/13972
|
396647
| 163,790 |
https://mathoverflow.net/questions/396644
|
1
|
Fix $n\geq 2$ and let $\mathbb{H}^{n}=\mathbb{R}\_{+}\times \mathbb{S}^{n-1}$ be the hyperbolic space be defined as a Riemannian manifold equipped with the Riemannian metric $$g=dr^{2}+\sinh^{2}rd\theta^{2},$$ where $dr^{2}$ is the standard metric on $\mathbb{R}\_{+}$ and $d\theta^{2}$ is the standard metric on the sphere $\mathbb{S}^{n-1}$. Denote with $\mu$ the Riemannian measure on $\mathbb{H}^{n}$ and with $\sigma$ the Riemannian measure of co-dimension $1$ on hypersurfaces on $\mathbb{H}^{n}$. It is a known fact that $\mathbb{H}^{n}$ admits the Isoperimetric inequality $$\sigma(\partial \Omega)\geq f(\mu(\Omega)),$$ for all precompact open sets $\Omega\subset \mathbb{H}^{n}$ with smooth boundary, where $f$ is defined by $$f(v)=\left\{\begin{array}{cl}v, &v\geq 1\\ v^{\frac{n-1}{n}}, &v\leq 1.\end{array}\right.$$
Let us fix some compact set $K\subset \mathbb{H}^{n}$ and consider $\mathbb{H}^{n}\setminus K$ as a manifold. Does the same Isoperimetric inequality now hold for precompact open sets $\Omega\subset \mathbb{H}^{n}\setminus K$ with smooth boundary and if yes, where can I find a reference for this statement? If this is not known for the hyperbolic space $\mathbb{H}^{n}$, is a similar statement known for other Riemannian manifolds, for example $\mathbb{R}^{n}$?
Thanks in advance for your help!
|
https://mathoverflow.net/users/163368
|
Isoperimetric inequality for exterior domains on $\mathbb{H}^{n}$
|
I might be missing something, but if you require $\Omega$ to be precompact in $\mathbb{H}^n \setminus K$, then $K$ makes no difference for the purpose of determining the measures and you can use the original inequality. The same if you only require pre-compactness in $\mathbb{H}^n$ but include the measure of $\partial \Omega \cap K$.
But if you do neither and only consider bounded $\Omega$ and their relative boundary in $\mathbb{H}^n \setminus K$, then generally there is no isoperimetric inequality: Pick any set $\Omega$ of positive measure and $K := \partial \Omega$, then $$\sigma(\partial \Omega \setminus K) = 0 < f(\mu(\Omega)).$$
|
1
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https://mathoverflow.net/users/51695
|
396649
| 163,792 |
https://mathoverflow.net/questions/396656
|
0
|
Given a sequence $u\_k\in W^{1,p}(B\_1)\cap C^{\alpha}(B\_1)$ such that $\|u\_k\|\_{C^{\alpha}(B\_1)}\le 1$ for all $k\in \mathbb N$. Suppose we have
$$
u\_k \rightharpoonup u\;\;\mbox{weakly in $W^{1,p}(B\_1)$}
$$
and we also have
$$
u\_k \rightarrow u\_0\;\;\mbox{in $C^{\alpha}(B\_1)$}.
$$
Can I imply from the above two informations that $u\_k\rightarrow u\_0$ strongly in $W^{1,p}(B\_1)$?
|
https://mathoverflow.net/users/68370
|
Does a weakly convergent sequence in $W^{1,p}(B_1)$ which also converges in $C^{0,\alpha}(B_1)$ converges strongly in $W^{1,p}(B_1)$?
|
I don't think so. Define a sequence of 'zig-zag' functions $(f\_n)$ on $[0,1]$ as follows. Let $f\_n$ be $\frac{2}{n}$-periodic with $f\_n(x) = x$ when $x \in [0,\frac{1}{n}]$ and $f\_n(x) = \frac{2}{n}- x$ when $x \in [\frac{1}{n},\frac{2}{n}]$. Every function is Lipschitz with $\lvert f\_n \rvert\_\infty \leq 1/n$ and $\lvert f\_n' \rvert\_\infty = 1$.
Moreover the sequence converges to zero weakly in $W^{1,p}$ for every $p \in (1,\infty)$ and strongly in $C^{0,\alpha}$ for every $\alpha \in (0,1)$, while $\lvert f\_n \rvert\_{1,p} \geq 1$ for every $n$.
|
2
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https://mathoverflow.net/users/103792
|
396661
| 163,797 |
https://mathoverflow.net/questions/396662
|
3
|
The title is meant to be punchy, but also a tongue-in-cheek acknowledgement of the prevalence of ‘reduce’-derived words in this area. (Unfortunately, I overlooked the fact that the question in the title is the opposite of the one in the body. Fortunately, [@LaurentMoret-Bailly's answer](https://mathoverflow.net/a/396667) is quite clear about this: ‘yes’ to the question in the title, ‘no’ to the question below.)
Let $k$ be a separably closed field, let $\overline k/k$ be an algebraic closure, and suppose that $G$ is a connected $k$-group scheme of finite type.
I know that $G\_\text{red}$ need not be geometrically reduced, but the standard examples of this have $G\_\overline k$ unipotent. Is that for convenience, or is it necessary?
To put it more precisely, suppose that $(G\_\overline k)\_\text{red}$ is a *reductive* $\overline k$-group scheme. Does it follow that $G\_\text{red}$ is geometrically reduced?
|
https://mathoverflow.net/users/2383
|
Can non-geometrically reduced reduced subschemes happen for reductive groups?
|
No, it does not follow that $G\_\mathrm{red}$ is geometrically reduced (thus the answer to the title question is yes).
Let $p=\mathrm{char}(k)>0$ and let $H=\mathbb{G}\_a \rtimes \mathbb{G}\_m$ be the group of affine transformations of $\mathbb{A}^1\_k$.
Let $(x,y)\in H$ act on $\mathbb{A}^1\_k$ by the "twisted" action $z\mapsto x^p+y^p z$. If $z\in k$ is not a $p$-th power, its stabilizer $G$ is easily seen to be reduced, but $G\_{\overline{k}}$ is isomorphic to the stabilizer of the origin which is $\alpha\_p \rtimes \mathbb{G}\_m$.
In particular, $(G\_{\overline{k}})\_\mathrm{red}$ is isomorphic to $\mathbb{G}\_m$.
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4
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https://mathoverflow.net/users/7666
|
396667
| 163,800 |
https://mathoverflow.net/questions/313825
|
19
|
Let $X$ be a smooth, Riemannian manifold. It is known that the geometry of $X$ can be recovered from its heat kernel $k\_{t}(x,y)$, using Varadhan's Lemma: $\displaystyle\lim\_{t \to 0} t \log k\_{t}(x,y) = -\frac{1}{4}d^{2}(x,y)$. Since the heat kernel $k\_{t}(x,y)$ can be expressed in terms of eigenfunctions and eigenvalues,
$$k\_{t}(x,y) = \sum\_{i=0}^{\infty}e^{-\lambda\_{i}t}\phi\_{i}(x)\phi\_{i}(y)$$
we can say that the knowledge of the eigenvalues and eigenfunctions of $X$ determines its geometry.
Now, the following result of Bates: <https://arxiv.org/pdf/1605.01643.pdf> tells us that there is a constant $d(X)$, depending on the dimension and geometry of $X$, such that the map $X \to \mathbb{R}^{d(X)}$ sending $x \in X$ to $\langle \phi\_{0}(x), \cdots, \phi\_{d(X)}(x)\rangle$ is injective.
My question therefore is: we see that a map to Euclidean space using finitely many eigenfunctions is enough to recover the topological type of $X$. Can we also recover its metric (up to some simple transformation)? What if we used infinitely many eigenfunctions? In either case we do not have access to the eigenvalues.
|
https://mathoverflow.net/users/13341
|
Do eigenfunctions determine the geometry of a manifold? If so, do finitely many suffice?
|
Here is a sketch of an idea of how to show that the set $\mathcal{E}(g)\subset C^\infty(M)$ of *all* the eigenfunctions of the metric $g$ on a compact manifold $M$ determines $g$ up to a constant multiple. (Note that I do not assume that the corresponding eigenvalues are given, which would be easier.) Clearly, this is the best one can hope for, since for any constant $c>0$, $\mathcal{E}(cg) = \mathcal{E}(g)$. In fact, I think it's very likely that knowing a sufficiently 'large' finite subset of $\mathcal{E}(g)$ should be sufficient, but that remains to be seen.
The basic idea is this: Let $J^k(M)\to M$ denote the vector bundle consisting of the $k$-jets of smooth functions on $M$. When $M$ has dimension $n$, the bundle $J^k$ has rank ${n+k}\choose n$ over $M$. Given $g$, it is easy to show that there is a closed quadratic cone bundle $Q(g)\subset J^3(M)$ of codimension $n$ such that the $3$-jets of all the local eigenfunctions of $g$ lie in $Q(g)$. In fact, the $3$-jets of local eigenfunctions fill out $Q(g)$.
Note that $Q(g)$ is *not* a linear subbundle of $J^3(M)$ precisely because we are not specifying the eigenvalues of the eigenfunctions. Of course, the $3$-jets of local eigenfunctions with eigenvalue $\lambda$, fill out a linear subbundle $Q(g,\lambda)\subset J^3$, but the union of the $Q(g,\lambda)$ as $\lambda$ varies is not a linear subbundle. However, it is easy to show that, for any $x\in M$, the subset $Q\_x(g)\subset J^3\_x$ is the zero locus of an ideal generated by $n$ polynomials homogeneous of degree $2$ on the vector space $J^3\_x$. ($Q\_x(g)$ is not a smooth variety in $J^3\_x$ but the singular locus is quite small.)
I claim that the subbundle $Q(g)\subset J^3$ determines $g$ up to a constant factor (assuming that $M$ is connected). Here is why: Let $Q^0(g)\subset Q(g)$ denote the subset consisting of those $3$-jets in $Q(g)$ whose $0$-jet vanishes. The projection $\pi^2\_3(Q^0(g)\_x)$ of $Q^0(g)\_x$ into $J^2\_x$ has codimension $n{+}1$ in $J^2\_x$, cut out by the linear equation that says that the $0$-jet vanishes and the $n$ quadratic equations that then turn out to all be multiples of a *single* linear equation on the space of $2$-jets whose $0$-jet vanishes, as is easy to verify in local coordinates.
Consequently, it follows that there exist second order, elliptic differential operator of the form
$$
L u = a^{ij}\,\frac{\partial^2u}{\partial x^i\,\partial x^j} + b^i\,\frac{\partial u}{\partial x^i} + c\,u
$$
(with $(a^{ij})$ positive definite) such that every local eigenfunction of $g$ satisfies an equation of the form $Lu = \phi(u)u$ where $\phi(u)$ is a smooth function (that could depend on $u$), and, moreover, $L$ is unique up to scalar multiplication and the addition of a $0$-th order term.
Finally, the requirement that there exist a function $f>0$ such that $fL$ can be expressed in the divergence form
$$
(fL)(u) = |h|^{-1/2}\frac{\partial}{\partial x^i}\left(|h|^{1/2}h^{ij}\,\frac{\partial u}{\partial x^j}\right)
$$
is easily seen to imply that $h = cg$ for some constant $c\not=0$. Thus, $g$ can be recovered, up to a constant multiple, from knowledge of the subbundle $Q(g)\subset J^3$, as claimed.
Finally, what one would expect is that, when $M$ is compact, if we now look at the $3$-jets of the elements of $\mathcal{E}(g)$, i.e., the global eigenfunctions of $g$, that a sufficiently large subset will determine sufficiently many points in each $J^3\_x$ that they will determine $Q(g)\_x$, which, after all, is known to be cut out by $n$ homogeneous quadratic polynomials for each $x$. (Of course, the number of elements of $\mathcal{E}(g)$ needed to do this at each point could be large, even for $n=2$, but it will be finite.) Assuming such a 'density' result, $\mathcal{E}(g)$ will determine $Q(g)\subset J^3$, which, as we have seen, will determine $g$ up to a constant multiple.
|
4
|
https://mathoverflow.net/users/13972
|
396678
| 163,802 |
https://mathoverflow.net/questions/228359
|
23
|
Let $\mathbb{F}$ be a field. The Tits building for $\text{SL}\_n(\mathbb{F})$, denoted $T\_n(\mathbb{F})$, is the simplicial complex whose $k$-simplices are flags
$$0 \subsetneq V\_0 \subsetneq \cdots \subsetneq V\_k \subsetneq \mathbb{F}^n.$$
The space $T\_n(\mathbb{F})$ is $(n-2)$-dimensional, and the Solomon-Tits theorem
says that in fact $T\_n(\mathbb{F})$ is homotopy equivalent to a wedge of $(n-2)$-dimensional spheres. The Steinberg representation of $\text{SL}\_n(\mathbb{F})$, denoted $\text{St}\_n(\mathbb{F})$, is $\widetilde{H}\_{n-2}(T\_n(\mathbb{F});\mathbb{C})$.
This is one of the most important representations of $\text{SL}\_n(\mathbb{F})$; for instance, if $\mathbb{F}$ is a finite field of characteristic $p$, then $\text{St}\_n(\mathbb{F})$ is the unique nontrivial irreducible representation of $\text{SL}\_n(\mathbb{F})$ whose dimension is a power of $p$.
The only proof I know that $\text{St}\_n(\mathbb{F})$ is an irreducible representation of $\text{SL}\_n(\mathbb{F})$ when $\mathbb{F}$ is a finite field uses character theory, and thus does not work for $\mathbb{F}$ infinite (in which case $\text{St}\_n(\mathbb{F})$ is an infinite-dimensional representation of the infinite group $\text{SL}\_n(\mathbb{F})$).
Question: For an infinite field $\mathbb{F}$, is $\text{St}\_n(\mathbb{F})$ an irreducible representation of $\text{SL}\_n(\mathbb{F})$? If not, is it at least indecomposable?
---
EDIT 2: In the previous edit, I said that I accepted an answer that did not answer the question as stated. However, this has now changed since Andrew Snowden and I have written a paper giving a complete answer to this question.
---
EDIT: I accepted an answer, but I am particularly interested in the field $\mathbb{Q}$, which is not covered by that answer. This case is interesting to me because it arises when studying the cohomology of $\text{SL}\_n(\mathbb{Z})$; indeed, in this case the Tits building forms the boundary of the Borel–Serre bordification of the associated symmetric space and the Steinberg representation (as I defined it above) provides the "dualizing module" for $\text{SL}\_n(\mathbb{Z})$. See Section 2 of my paper
* T. Church, B. Farb, A. Putman *A stability conjecture for the unstable cohomology of $\text{SL}\_n(\mathbb{Z})$, mapping class groups, and $\text{Aut}(F\_n)$*,
in "Algebraic Topology: Applications and New Directions", pp. 55–70,
Contemp. Math., **620**, Amer. Math. Soc., Providence, RI. doi:[10.1090/conm/620/12366](https://doi.org/10.1090/conm/620/12366), arXiv:[1208.3216](https://arxiv.org/abs/1208.3216)
for a discussion of this and references. It is also available on my webpage [here](http://www.nd.edu/%7Eandyp/papers).
|
https://mathoverflow.net/users/317
|
Is the Steinberg representation always irreducible?
|
Andrew Snowden and I managed to finally answer this question in our paper "The Steinberg representation is irreducible", available [here](https://arxiv.org/abs/2107.00794). As you might guess from the title, we prove that the Steinberg representation over an infinite field is always irreducible. In fact, we prove something much more general that applies to arbitrary reductive groups over infinite fields, and also allows arbitrary coefficients for the Steinberg module.
It's worth also mentioning another recent paper by Galatius--Kupers--Randal-Williams called "$E\_{\infty}$-cells and general linear groups of infinite fields", available [here](https://arxiv.org/abs/2005.05620). One of their results says that the Steinberg representation for $\text{GL}\_n$ (as discussed in this question) is indecomposable, i.e. is not the nontrivial direct sum of two subrepresentations. For infinite fields, the Steinberg representation is infinite-dimensional, so this is weaker than being irreducible. However, I think their proof is quite beautiful and worth reading even if it gives a weaker result.
---
EDIT: My attention has been drawn to two earlier papers:
N. Xi, Some infinite dimensional representations of reductive groups with Frobenius maps, Sci. China Math. 57 (2014), no.~6, 1109--1120.
R. Yang, Irreducibility of infinite dimensional Steinberg modules of reductive groups with Frobenius maps, J. Algebra 533 (2019), 17--24.
These focus on connected reductive group $\mathbf{G}$ over the algebraic closure $k=\overline{\mathbb{F}}\_q$ of a finite field $\mathbb{F}\_q$. For instance, we could have $\mathbf{G}(k) = \text{GL}(n,k)$ as in the question. Their main theorem says that the Steinberg representation of $\mathbf{G}$ is irreducible with coefficients in any field. Xi's paper handles the case when the coefficients have characteristic $0$ or $\text{char}(k)$, and Yang's paper handles other characteristics.
|
9
|
https://mathoverflow.net/users/317
|
396684
| 163,804 |
https://mathoverflow.net/questions/396280
|
1
|
In the paper Floer cohomology of lagrangian intersections and pseudo-holomorphic disks in page $1009$ the authors claim that a map $w:D^2\rightarrow S^2$ with $w|\_{\partial D^2}\subset L$, where $L$ is the equator, and such that $\mu(w)=2$, where $\mu$ is the maslov index of the map, cannot be multiply covered and hence it will be injective in the interior of $D^2$.
I have tried proving this , but I am getting nowhere.
Following the paper "Relative frames on $J$-holomorphic curves" the definition of multicovered that I found is that there exists a simple disk $v:(D,\partial D)\rightarrow (S^2,L) $ and a surjective map $p:(D,\partial D)\rightarrow (D,\partial D)$ continuous on $D$, holomorphic on the interior , satisfying $p^{-1}(\partial D)=\partial D$ and $u=v\circ p$ thus $[u]=m[v]$ in relative homology with $m\geq 1$.
And so using this and the fact that the minimal maslov number is $2$ will force $m=1$ and $[u]=[v]$.
However from here I am not sure out to prove that $u$ is bijective in the interior of $D^2$.
Any help or reference where I can look this up is appreciated, thanks in advance.
|
https://mathoverflow.net/users/nan
|
Maslov index equal to $2$ implies that the disk is not multiply covered
|
First, I assume you want $w$ to be holomorphic (or else it isn't true).
If the target is really $S^2$ with boundary on the equator $L$ then this should be easy to prove directly. Take a point $p$ not on $L$. We can define an intersection number between $w$ and $p$ using the intersection pairing between $H\_2(S^2,L)$ and $H\_0(S^2\setminus L)$. By positivity of intersections, any point $q$ in the preimage of $p$ under $w$ contributes positively to this intersection number (with multiplicity 1 if and only if $w$ is transverse to $p$ at $q$). Let's write $[N]$ and $[S]$ for the Northern and Southern hemispheres considered as (Maslov 2) classes (in fact generators) in $H\_2(S^2,L)$. The class of $w$ is $a[N]+b[S]$ for some $a$ and $b$ with $a+b=1$ because $\mu(w)=2$. The intersection number of $w$ with $p$ is $a$ if $p\in N$ and $b$ if $p\in S$. Since both of these must be nonnegative and sum to 1, we get $[w]=[N]$ or $[S]$. Moreover, the preimage of each point in $N$ (or $S$) is a single point and the derivative of $w$ is nonzero there to get intersection multiplicity 1 at $p$.
Moreover, the preimage of any point in $N$ (or $S$) has the same intersection number with the disc, so $w$ is really just a bijective parametrisation of a hemisphere.
In fact, this works for any embedded circle on the sphere (the argument didn't depend on areas). I'm sure there are other, simpler, ways to see it that don't explicitly invoke relative homology; this is just one that sprang to mind.
|
0
|
https://mathoverflow.net/users/10839
|
396697
| 163,808 |
https://mathoverflow.net/questions/396682
|
0
|
Consider a minimal smooth conic bundle $S$ of dimension two. Assume that there are two curves $C,F$ on $S$ such that $C^2 < 0$ and $F^2 = 0$. Let $D$ be a pseudoeffective divisor on $S$ such that $D\sim\_{\mathbb{Q}} aC+bF$ (numerical equivalence) where $a,b\in\mathbb{Q}$.
Is it true then that $mD$ must be an integral effective divisor for some $m\geq 1$?
|
https://mathoverflow.net/users/14514
|
Pseudoeffective divisors on surfaces
|
If I understand the question correctly, then the Picard number of $S$ is 2 and hence the existence of the curve $F$ implies that the cone of effective curves is closed, so every pseudoeffective divisor is actually effective.
**Added to answer Friedrich's question in the comments:** Let $C\subseteq S$ be a(n effective) curve with negative self-intersection. First of all, as the intersection number of different irreduicble components is non-negative, at least one of the irreducible components of $C$ must have negative self-intersection and hence we might as well assume that $C$ is irreducible.
However, then $C$ is non-negative on every irreducible curve other than itself. Intersecting with $C$ is a linear functional on the space of curves, and by the previous observation the hyperplane defined by $C\cdot (\ \ )=0$ separates $C$ from all other irreducible curves. This implies that $C$ must be on the boundary of the effective cone. (In particular, this also means that on a surface with Picard number $2$, there can be at most two irreducible curves with negative self-intersection. An example with two is a general degree 4 surface in $\mathbb P^3$ containing a conic. I leave it for the reader to figure out what the two curves with negative self-intersection are...)
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2
|
https://mathoverflow.net/users/10076
|
396701
| 163,810 |
https://mathoverflow.net/questions/396715
|
1
|
Let $k$ be a field; $X$ a projective scheme over $k$; $\mathcal A$ an ample $\mathcal O\_X$-module. Sometimes we want to apply induction on dimension by finding an effective Cartier divisor $D$ such that $\mathcal O\_X(D) \cong \mathcal A$. However, the problem is, whether $D$ always exists, i.e., a regular section of $\mathcal A$ exists, under some assumptions on $\mathcal A$, for example, if we assume that $\mathcal A$ is generated by global sections, is it true that then $D$ exists? If it is not true, then maybe we have to assume that $X$ is integral, then any section of $\mathcal A$ is regular. But now we run into another problem, whether $D$ can be chosen to be integral again? If $D$ is not integral, then we cannot apply the induction hypothesis, then maybe we also have to assume that $X$ is smooth in order to apply the Bertini theorem to make sure $D$ is again smooth and integral. But sometimes smoothness is too strong for $X$ to hold, if there is a more suitable way to deal with this kind of problem.
|
https://mathoverflow.net/users/129738
|
Effective Cartier divisors corresponding to a line bundle
|
If $\mathcal A$ is generated by global sections and $k$ is infinite then a regular section must exist. For each irreducible component of $X$, global sections vanishing on that component form a positive-codimension subspace of global sections. For non-reduced $X$, global sections vanishing on the induced reduced subscheme of any associated prime also find a positive-codimension subspace.
Over an infinite field, we may avoid arbitrarily many positive codimension subspaces with a single vector.
If $k$ is finite then this is not true. We could take $X$ to be the union of all $\mathbb F\_q$-rational lines in $\mathbb P^2$ and $\mathcal A = \mathcal O(1)$. Then using the ideal sheaf of $X$ we can check that all global sections of $\mathcal A$ are the obvious global sections of $\mathcal O(1)$ on $\mathbb P^2$, each of which vanishes on a line.
But probably, in the situation you're thinking of, it's fine to assume $k$ is infinite.
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2
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https://mathoverflow.net/users/18060
|
396719
| 163,812 |
https://mathoverflow.net/questions/396596
|
12
|
In searching for a counterexample in homological stability, I came across the following question:
>
> Is there a known example of a finitely presented group $G$, so that the group ring $\mathbb{Z}[G]$ has infinite Bass [stable rank](https://encyclopediaofmath.org/wiki/Stable_rank)?
>
>
>
|
https://mathoverflow.net/users/157284
|
Group ring with infinite stable rank
|
**Yes**, the integral group ring $\mathbb{Z}[F\_2]$ of the free group $F\_2$ on two generators has **infinite** stable rank.
This can be deduced from [1, Corollary 3.6]:
>
> There exists a cyclic $\mathbb{Z}[F\_2]$-module $M$ with the following property.
> For every $N \ge 1$, there exists an epimorphism $\theta\_N: (\mathbb{Z}[F\_2])^N \twoheadrightarrow M$ such that $(\mathbb{Z}[F\_2])^N$ cannot be generated by $N$ elements one of which is contained in $\ker \theta\_N$.
>
>
>
It should be easy to see that the cyclic $\mathbb{Z}[F\_2]$-module $M$ above has infinite stable rank. (Actually, explicit non-stable unimodular rows of length $N$ for every $N \ge 2$ can be extracted from the proof).
Then combine the previous result with [2, Lemma 11.4.6]
>
> If $n$ is in the stable range of a finitely generated right $R$-module $M$ then $n$ is in the stable range of any factor module $M/K$.
>
>
>
to conclude that the stable rank of $\mathbb{Z}[F\_2]$ is infinite.
---
[1] M. Evans, "Presentations of groups involving more generators than necessary", 1992.
[2] J. McConnell and J. Robson, "Noncommutative Noethering rings", 1987.
|
7
|
https://mathoverflow.net/users/84349
|
396729
| 163,814 |
https://mathoverflow.net/questions/396727
|
6
|
I am trying to understand the relation between Wigner's $3j$-symbols (or Clebsch-Gordan coefficients) and matrix coefficients of intertwiners. I am new to this topic and need some help to understand it properly.
So, let us start with the Lie algebra $\mathfrak{sl}(2,\mathbb{C})$, which is the complexification of $\mathfrak{su}(2)$. Now, as usual, all the irreducible representations of $\mathfrak{sl}(2,\mathbb{C})$ can be described by spins $j\in\mathbb{N}\_{0}/2$ and have dimension $2j+1$. Now it is a general fact that for a simple and complex Lie algebra, ever finite-dimensional irreducible representation can be labeled with a heighest weight $\Lambda$, which in the case of $\mathfrak{sl}(2,\mathbb{C})$ are given by $2j$. In general, the tensor product of two such heighest weight modules is fully reducible and hence we can write
$$V\_{\Lambda}\otimes V\_{\Lambda^{\prime}}=\bigoplus\_{i}C\_{\Lambda\Lambda^{\prime}}^{\Lambda\_{i}}V\_{\Lambda\_{i}}$$
with some coefficients, usually called multiplicities or Littlewood-Richardson coefficients. Now in order to relate this to intertwiners, we first of all know, according to the Lemma of Schur, that the space of intertwiners between two irreducible finite-dimensional representations of a complex Lie algebra $\mathfrak{g}$ is either $0$-dimensional (if they are not isomorphic) or $1$-dimensional (if they are isomorphic). As a consequence, we get that the Littlewood-Richardson coefficients are given by the dimension of the space of intertwiners from $V\_{\Lambda\_{i}}$ to $V\_{\Lambda}\otimes V\_{\Lambda^{\prime}}$, i.e.
$$C\_{\Lambda\Lambda^{\prime}}^{\Lambda\_{i}}=\mathrm{dim}\_{\mathbb{C}}(\mathrm{Int}(V\_{\Lambda\_{i}},V\_{\Lambda}\otimes V\_{\Lambda^{\prime}}))$$
So in the case of $\mathfrak{sl}(2,\mathbb{C})$ (or equivalently $\mathfrak{su}(2)$), the coefficients are hence given by
$$C\_{\Lambda\Lambda^{\prime}}^{\Lambda\_{i}}=\mathrm{dim}\_{\mathbb{C}}(\mathrm{Int}(V\_{2j\_{i}},V\_{2j}\otimes V\_{2j^{\prime}}))$$
So far so good. In a textbook (Fuch, Schweigert — *Symmetries, Lie algebras and representations*), they then say that using this correspondence, it is clear that the Clebsch-Gordan coefficients for fixed $J$ are the matrix coefficients of intertwiners in $\mathrm{Int}(V\_{2j\_{i}},V\_{2j}\otimes V\_{2j^{\prime}})$ for "a definite choice of basis". Can anyone explain this step to me in more detail? Furthermore, what does "a definite choice of basis" mean in this context? Also, what is then the correspondence with the $3j$-symbols?
Thank you very much!
EDIT: In order to clarify the conventions I use, let me add some more details about the Clebsch-Gordan coefficients: Let $J\_{\pm},J\_{0}$ be the three generators of $\mathfrak{sl}(2,\mathbb{C})$, i.e.
$$[J\_{0},J\_{\pm}]=\pm J\_{\pm}\hspace{1cm}\text{and}\hspace{1cm}[J\_{+},J\_{-}]=2J\_{0}$$
Then there is a basis $\{\mid j,m\rangle\}\_{-j\leq m\leq j}$ of $V\_{2j}$ satisfying
$$J\_{0}\mid j,m\rangle=m\mid j,m\rangle$$
$$J\_{\pm}\mid j,m\rangle=\sqrt{j(j+1)-m(m\pm 1)}\mid j,m\pm 1\rangle$$
In order to define the Clebsch-Gordan coefficients, one usually defines to different bases of the tensor product $V\_{2j}\otimes V\_{2j^{\prime}}$:
(1) The tensor product of the bases described above, i.e. $\{\mid j\_{1},j\_{2},m\_{1},m\_{2}\rangle\}\_{-j\_{1}\leq m\_{1}\leq j\_{1},-j\_{2}\leq m\_{2}\leq j\_{2}}$
where $\mid j\_{1},j\_{2},m\_{1},m\_{2}\rangle:=\mid j\_{1},m\_{1}\rangle\otimes\mid j\_{2},m\_{2}\rangle$.
(2) Define the total spin $\vec{J}:=\vec{J}\_{1}+\vec{J}\_{2}$. Then there is a basis $\{\mid J,M\rangle:=\mid j\_{1},j\_{2},J,M\rangle\}$ satisfying
$$\vec{J}^{2}\mid J,M\rangle=J(J+1)\mid J,M\rangle$$
Then the Clebsch-Gordan coefficients are the coefficients of the change of basis matrix, i.e. $\langle j\_{1},j\_{2},m\_{1},m\_{2}\mid J,M\rangle$
|
https://mathoverflow.net/users/259525
|
Relations between $3j$-symbols and intertwiners
|
There is a [standard basis](https://en.wikipedia.org/wiki/Angular_momentum_operator#Orbital_angular_momentum_in_spherical_coordinates) for finite-dimensional $SO(3)$ (and hence $\mathfrak{su}(2)$) representations, denoted by $\left| j, m\right\rangle$; the total spin $j\ge 0$ labels the representation, the magnetic quantum number $m=-j,\ldots,j$ labels the basis. If you look at the first few equations in the Wikipedia article on [$3j$-symbols](https://en.wikipedia.org/wiki/3-j_symbol), one can immediately realize that they give the following formula for the (unique up to a scalar multiple) intertwiners in this basis:
$$ I\_{j\_1,j\_2}^{j\_3} \colon \left| j\_3, m\_3\right\rangle \mapsto \sum\_{m\_1=-j\_1}^{j\_1} \sum\_{m\_2=-j\_2}^{j\_2} (-1)^{-j\_1+j\_2+m\_3} \sqrt{2j\_3+1} \begin{pmatrix} j\_1 & j\_2 & j\_3 \\ m\_1 & m\_2 & -m\_3 \end{pmatrix} \left|j\_1, m\_1\right\rangle \otimes \left|j\_2, m\_2\right\rangle.$$
|
6
|
https://mathoverflow.net/users/2622
|
396737
| 163,816 |
https://mathoverflow.net/questions/396706
|
1
|
Suppose we are given a sequence $\phi\_k$ of traces (i.e. functions defined on boundary $\partial B\_1$) such that
$$
\phi\_k \rightarrow 0 \;\mbox{in $L^{\infty}(\partial B\_1)$}
$$
(one can consider $C^{\alpha}$ convergence if required).
Now we consider affine subspaces Sobolev space $W^{1,p}(B\_1)$
$$
W^{1,p}\_{\phi\_k}(B\_1):= \Big \{ v\in W^{1,p}(B\_1): \text{Trace}(v) =\phi\_k \Big \}
$$
Let $v\_k\in W^{1,p}\_{\phi\_k}(B\_1)$ be such that $v\_k$ has least $W^{1,p}(B\_1)$ norm in the space $W^{1,p}\_{\phi\_k}(B\_1)$. That is $v\_k$ is the minimizer of the following convex functional
$$
J\_k(v):= \int\_{B\_1}(|\nabla v|^p +|v|^p)\,dx,\;\;v\in W^{1,p}\_{\phi\_k}(B\_1).
$$
We can see that $v\_k$ satisfies the following Euler-Lagrange equation of $J\_k$ :
$$
\begin{cases}
-\Delta\_p v\_k = |v\_k|^{p-2}v\_k\;\;\mbox{in $B\_1$}\\
v\_k =\phi\_k\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mbox{ on $\partial B\_1$}.
\end{cases}
$$
Question is since the traces $\phi\_k$ are tending to Zero in $L^{\infty}(\partial B\_1)$, is there any result which shows that the smallest elements in $W^{1,p}\_{\phi\_k}(B\_1)$ also tend to zero in **strong $W^{1,p}(B\_1)$** topology?
We have to show that
$$
v\_k \rightarrow 0 \;\;\mbox{in $W^{1,p}(B\_1)$}.
$$
Isn't it the case that the smallest possible Sobolev norm for an element in $W^{1,p}\_{\phi\_k}(B\_1)$ has to be controlled by some norm of $\phi\_k$.
The above question can also be seen in terms of $\Gamma$ convergence where we try to show that $J\_k \xrightarrow{\Gamma} 0$ in weak $W^{1,p}(B\_1)$ topology.
Moreover, we can also see it as eigenvalue problem for $p$-Laplacian.
Thank you in advance.
|
https://mathoverflow.net/users/68370
|
Characterization on smallest element in affine Sobolev subspace
|
It seems to depend on the strength assumed of the convergence of the $\phi\_k$. This is a partial answer, where in some parts it is assumed that $p > n - 1$. Under this hypothesis, the convergence does not hold if only $\lvert \phi\_k \rvert\_{L^\infty(\partial B\_1)} \to 0$. However if $\lvert \phi\_k \rvert\_{C^{0,\beta}(\partial B\_1)} \to 0$ for some $\beta \in (1 - \frac{1}{p},1)$ then also $\lvert v\_k \rvert\_{W^{1,p}(B\_1)} \to 0$, regardless of the relative size of $p$ and $n$.
The trace operator is instrumental in seeing this: recall that there is a constant $C > 0$ so that
$$
\lvert \mathrm{tr} \, u \rvert\_{W^{1-1/p,p}(\partial B\_1)} \leq C \lvert u \rvert\_{W^{1,p}(B\_1)}
$$
for all $u \in W^{1,p}(B\_1)$. Here and throughout $W^{1-1/p,p}(\partial B\_1)$ is a fractional Sobolev-Slobodeckij space. Recall also that there is an extension operator $W^{1,1-1/p}(\partial B\_1) \to W^{1,p}(B\_1)$ that is a right inverse to the trace.
In particular, if there were the strong convergence $\lvert v\_k \rvert\_{W^{1,p}(B\_1)} \to 0$ then also
$$
\lvert \phi\_k \rvert\_{W^{1-p,p}(\partial B\_1)} \to 0.
$$
This is strictly stronger than $L^\infty$-convergence because $W^{1-1/p,p}(\partial B\_1)$ embeds into the Holder space $C^{0,\alpha}(\partial B\_1)$, where $\alpha \in (0,1 - \frac{1}{p} -\frac{n-1}{p}] = (0,1-\frac{n}{p}]$. Therefore, if one chose a sequence of traces so that
$$
\lvert \phi\_k \rvert\_{C^{0,\alpha}(\partial B\_1)} \not\to 0
$$
for some exponent $\alpha$ in this range then the convergence $\lvert v\_k \rvert\_{W^{1,p}(B\_1)} \to 0$ would be impossible. (This is where the hypothesis $p > n - 1$ is used: when $p < n - 1$ then instead $W^{1-p,p}(\partial B\_1) \subset L^s(\partial B\_1)$ for all $s \in (0,\frac{(n-1)p}{n-1-p}]$.)
However, if $\beta > 1 - \frac{1}{p}$ and one assumes $$\lvert \phi\_k \rvert\_{C^{0,\beta}(\partial B\_1)} \to 0$$ then the convergence $\lvert v\_k \rvert\_{W^{1,p}(B\_1)} \to 0$ is guaranteed. This is because the 'reverse' inclusion $C^{0,\beta}(\partial B\_1) \subset W^{1-1/p,p}(\partial B\_1)$ holds. Therefore $\phi\_k \in W^{1-1/p,p}(\partial B\_1)$; let $u\_k \in W^{1,p}(B\_1)$ be its image under the extension operator. By the above there is $C > 0$ so that
$$
\lvert u\_k \rvert\_{W^{1,p}(B\_1)} \leq C \lvert \phi\_k \rvert\_{C^{0,\beta}(\partial B\_1)} \quad \text{for all $k$}.$$
By minimality, the same holds with $v\_k$ replacing $u\_k$, and therefore $\lvert v\_k \rvert\_{W^{1,p}(B\_1)} \to 0$ also. (Note that the assumption that $p > n-1$ is not needed here.)
|
2
|
https://mathoverflow.net/users/103792
|
396741
| 163,818 |
https://mathoverflow.net/questions/396754
|
2
|
Let $X$ be an open complex manifold, e.g., the complement of a simple normal crossing divisor $D$ in a (smooth) projective manifold $M$. Let $T^{1,0}X$ be the holomorphic tangent bundle of $X$. Let $K \subset X$ be a compact set with $U \subset X$ a sufficiently small open neighbourhood of $K$. Let $V$ be a holomorphic vector field on $K$, i.e., a holomorphic section $V\in H^0(K,T^{1,0}X)$. Does there exist an extension $\widetilde{V} \in H^0(U, T^{1,0}X)$ such that $\widetilde{V} = V$ on $K$?
Note that I'm not asking whether the sheaf of holomorphic vector fields is a fine sheaf, I'm only asking whether holomorphic vector fields defined on compact sets admit local extensions to open neighbourhoods. I'm also aware of the failure of the tubular neighbourhood theorem to exist in the holomorphic category (in complete generality), but the literature on such a theorem appears quite vast and formidable.
|
https://mathoverflow.net/users/nan
|
Local extension of holomorphic vector fields
|
By Theorem II.9.5 in Bredon's Sheaf Theory,
for any closed subset $K$ of a paracompact space $X$
and for any sheaf of abelian groups $F$ on $X$,
the canonical map $$\mathop{\rm colim} F(U) \to F(K)$$ is an isomorphism, where $U$ runs over all open neighborhoods of $K$.
In particular, every element of $F(K)$ is the restriction of some element of $F(U)$, where $U$ is an open neighborhood of $K$.
|
5
|
https://mathoverflow.net/users/402
|
396755
| 163,824 |
https://mathoverflow.net/questions/396751
|
1
|
The first chapter of this [paper](http://www.numdam.org/article/ASENS_1974_4_7_2_181_0.pdf) (Gersten’s conjecture and the homology of schemes) defines a "Poincaré duality theory with supports" is a *twisted cohomology* theory satisfying certain properties on page 184.
The second chapter of this paper mentioned some examples. The first one is the étale cohomology (page 185) which is well-known for its extra structure of Galois representation by Tate twist. But the second one is the algebraic de Rham cohomology (page 187). What is the twist structure they mean on it? They referred to Hartshorne's paper, but I have not seen any twist construction in it.
|
https://mathoverflow.net/users/119770
|
Twist cohomology theory of algebraic de Rham cohomology
|
Nothing in the axioms says that the Tate twist should be nontrivial: it could be that $H^k(X,n)=H^k (X,m) $ for all $n,m $. This will be the case in both algebraic de Rham and Betti cohomology.
If you want the Tate twist in the de Rham theory to be nontrivial, you must keep track of extra structure, for example the comparison isomorphism with Betti cohomology, which changes under Tate twist.
|
3
|
https://mathoverflow.net/users/1310
|
396761
| 163,825 |
https://mathoverflow.net/questions/396735
|
2
|
Let $X^n$ and $X$ be stochastic processes defined by
$$X^n\_t=1+\int\_0^tb\_n(s)ds+\int\_0^t\sigma\_n(s)dW\_s \quad\mbox{and}\quad X\_t=1+\int\_0^tb(s)ds+\int\_0^t\sigma(s)dW\_s,$$
where $b\_n, \sigma\_n, b, \sigma$ are uniformly bounded measurable functions s.t.
$$\lim\_{n\to\infty}\sup\_{0\le t\le T}|b\_n(t)-b(t)|=0=\lim\_{n\to\infty}\sup\_{0\le t\le T}|\sigma\_n(t)-\sigma(t)|,\quad \mbox{for all } T>0.$$
Consider the stopped processes $(X^n\_{\tau\_n\wedge t})\_{t\ge 0}$ and $(X\_{\tau\wedge t})\_{t\ge 0}$, where $\tau\_n:=\inf\{t\ge 0: X^n\_t\le 0\}$ and $\tau:=\inf\{t\ge 0: X\_t\le 0\}$. Can we prove $X^n\_{\tau\_n\wedge t}$ converges to $X\_{\tau\wedge t}$ in law for all (or almost every) $t\ge 0$?
|
https://mathoverflow.net/users/261243
|
Convergence in law of stopped stochastic processes
|
I believe convergence in law holds for all $t \geq 0$. The proof proceeds in three steps.
**Step 1:** Note that by the dominated convergence theorem for stochastic integrals (see, for example [Theorem 7](https://almostsuremath.com/2010/01/11/properties-of-the-stochastic-integral/) here), we have that $X\_n$ converges to $X$ in the ucp topology, that is, $\lim\_{n \to \infty} P(\sup\_{t \in [0, T]} |X\_t^n - X\_t| > \varepsilon) = 0$ for all $\varepsilon > 0$, $T \geq 0$.
**Step 2:** With similar reasoning as Step 3 [here](https://mathoverflow.net/questions/396234/on-the-continuity-of-map-gamma/396269#396269), we can then show that the above convergence implies that $\tau\_n$ converges to $\tau$ in probability. (Can provide details here if needed)
**Step 3:** We claim that the above two convergences combined are enough to imply, for every $t > 0$, convergence in law of $X\_{\tau\_n \wedge t}^n$ to $X\_{\tau \wedge t}$.
The remainder will be dedicated to the proof of Step 3.
To show this, we need to show that $E(f(X\_{\tau\_n \wedge t}^n)) \to E(f(X\_{\tau \wedge t}))$ for all bounded continuous $f$. We argue as follows:
Fix $t \geq 0$, and let $\varepsilon > 0$ be arbitrary. Choose $M > 0$ large enough so that $P(X\_{\tau \wedge t} > M) < \varepsilon ||f||\_{L^{\infty}}$.
By uniform continuity of $f$ on $[0, M+1]$, there exists some $0 < \delta < 1$ such that $|f(x) - f(y)| < \varepsilon$ whenever $x, y \in [0, M+1]$ are such that $|x - y| < \delta$.
By convergence in probability of $\tau\_n$ to $\tau$, we deduce that $\tau\_n \wedge t$ converges in $L^1$ to $\tau \wedge t$.
This implies $|X\_{\tau\_n \wedge t}^n - X\_{\tau \wedge t}^n|$ converges to $0$ in probability.
To see this, note that by the Markov inequality, we have that $\mathbb P(|X^n\_{\tau\_n\wedge t}-X^n\_{\tau\wedge t}|>\epsilon)\le \mathbb E[|X^n\_{\tau\_n\wedge t}-X^n\_{\tau\wedge t}|^2]/\epsilon^2$.
We estimate
$\mathbb E[|X^n\_{\tau\_n\wedge t}-X^n\_{\tau\wedge t}|]^2 = \mathbb E[(\int\_{\tau\_n \wedge t}^{\tau \wedge t} \sigma\_n (s) dW\_s]^2)] = \mathbb E[\int\_{\tau\_n \wedge t}^{\tau \wedge t} \sigma\_n (s)^2 ds] <C \mathbb E[|\tau\_n \wedge t - \tau \wedge t|],$
for some constant $C$ independent of $n$.
Here the last line follows by the uniform boundedness of $\sigma\_n^2$.
Hence we have
$$P(|X^n\_{\tau\_n\wedge t}-X^n\_{\tau\wedge t}|>\epsilon) < C\_0 \mathbb E[|\tau\_n \wedge t - \tau \wedge t|],$$
with the constant $C\_0$ independent of $n$, and so the LHS goes to $0$, as $n \to \infty$, as was to be shown.
Now, by ucp convergence of $X^n$ to $X$, we may find some $N\_0 > 0$ such that $P(\{|X\_{\tau \wedge t}^n - X\_{\tau \wedge t}| > \frac{\delta}{2}\}) < \frac{ \varepsilon}{||f||\_{L^{\infty}}}$.
To see the above, note that by step 1, we may take $N\_0$ to be such that $ P(\sup\_{s \in [0, t]} |X\_s^n - X\_s| > \frac{\delta}{2}) < \frac{ \varepsilon}{||f||\_{L^{\infty}}}$ for all $n > N\_0$. Since $\tau \wedge t \leq t$, and the aforementioned convergence is uniform on $[0, t]$, the statement follows.
By convergence of $|X\_{\tau\_n}^n - X\_{\tau}^n|$ to $0$ in probability, we may choose $N\_1$ such that $P(|X\_{\tau\_n \wedge t}^n - X\_{\tau \wedge t}^n| > \frac{\delta}{2}) < \frac{\varepsilon}{||f||\_{L^\infty}}.$
Thus whenever $n > \max(N\_0, N\_1)$, by the triangle inequality, we have with probability greater than $1 - \frac{3 \varepsilon}{||f||\_{L^{\infty}}}$ that $|X\_{\tau\_n \wedge t}^n - X\_{\tau \wedge t}| < \delta$, and so $|f(X\_{\tau\_n \wedge t}^n) - f(X\_{\tau \wedge t})| < \varepsilon$.
Finally we compute
$E(f(X\_{\tau\_n \wedge t}^n)) - E(f(X\_{\tau \wedge t}))$
$ \leq E(|f(X\_{\tau\_n \wedge t}^n) - f(X\_{\tau \wedge t})|$
$< (1 - 3 \varepsilon||f||\_{L^{\infty}})\varepsilon + 3 \frac{\varepsilon}{ ||f||\_{L^{\infty}}}||f||\_{L^\infty} $
$<\varepsilon + 3 \varepsilon$
$= 4\varepsilon.$
Since $\varepsilon> 0$ was arbitrary, we conclude.
|
2
|
https://mathoverflow.net/users/173490
|
396763
| 163,826 |
https://mathoverflow.net/questions/396757
|
3
|
Let $p$ be a prime number and $q=p-1$. I’m trying to prove that the nonzero coefficients $a\_{qk}$ ($k\ge1$) of the power series
$$ \sum\_{k\ge1} a\_{qk} z^{qk} := \left( \sum\_{k\ge0} \frac{z^{qk+1}}{(qk+1)!} \right)^q $$
satisfy the congruence
$$ a\_{qk} \cdot (qk)! \equiv -1 \mod p. $$
*I’ve managed to work out that:* There is a combinatorial formula for this power series. First of all, the series being raised to the $q$th power can be written
$$ f(z) = \frac1q \sum\_{1\le i\le q} \zeta^{-i} e^{\zeta^iz} = \sum\_{k\ge0} \frac{z^{qk+1}}{(qk+1)!} $$
where $\zeta$ is a primitive $q$th root of unity. And its $q$th power can be expressed in terms of the derivative
$$ f'(z) = \frac1q \sum\_{1\le i\le q} e^{\zeta^iz} = \sum\_{k\ge0} \frac{z^{qk}}{(qk)!} $$
by the formula
$$
f(z)^q = \frac1{q^{q}} \sum {q\choose i\_1,\dots,i\_{q}} {\big|C\_q\cdot(i\_1,\dots,i\_q)\big|} \zeta^{-\big(1\cdot i\_1+2\cdot i\_2+\cdots+q\cdot i\_{q}\big)} f'\Big(\big(i\_1\zeta^1+\cdots+i\_{q}\zeta^{q}\big)z\Big) $$
where the sum ranges over $C\_{q}$-orbits of weak compositions $i\_1+\cdots+i\_{q}=q$, $i\_k\ge0$, and where $\big|C\_q\cdot(i\_1,\dots,i\_q)\big|$ denotes the size of the $C\_q$-orbit of the weak composition.
It is straightforward to show that Fermat's little theorem extends to $\mathbf{F}\_p[\zeta]/(\zeta^q-1)$ in the sense that the Frobenius is the identity, and it follows that $q$th powers in this ring are fixed by exponentiation by any positive integer power.
This reduces proving the claim to verifying it for the coefficient of $z^q$, i.e. verifying the congruence
$$
-1 \equiv \sum {q\choose i\_1,\dots,i\_{q}} {\big|C\_q\cdot(i\_1,\dots,i\_q)\big|} \cdot \zeta^{-\big(1\cdot i\_1+2\cdot i\_2+\cdots+q\cdot i\_{q}\big)} \Big(i\_1\zeta^1+\cdots+i\_{q}\zeta^{q}\Big)^q \mod p.
$$
I’ve done this by computer for the primes up to $p=17$ but haven’t found a general argument.
|
https://mathoverflow.net/users/313592
|
On the arithmetic of powers of subseries of the exponential series
|
Your argument is in fact almost complete. It reduces the problem to checking the normalized coefficient $a\_q$ of $z^q$ is congruent to $-1$ mod $p$, but since $$ f = z + \frac{z^{q+1}}{(q+1)!} + \dots ,$$ we have $$f^q = z^q + \frac{ q z^{2q}}{ (q+1)!} + \dots$$ and so $$a\_q = q! \equiv -1 \mod p.$$
---
Here is an alternate "bijective" presentation of the same argument. We can interpret $a\_n$ combinatorially as the number of $q$-colorings of the numbers from $1$ to $n$ such that each color is used a number of times congruent to $1$ mod $q$.
For $n \geq p$, there is an action of $\mathbb Z/p$ on this set of colorings by rotating the colorings of the numbers $n+1-p, n+2-p, \dots, n$. The number of colorings is congruent mod $p$ to the number of fixed points of this action. A coloring is fixed if the last $p$ numbers have the same color.
Since $q=p-1$, removing the last $q$ numbers in such a coloring doesn't affect the congruent mod $q$ of the number of times each coloring is used, and this gives a bijection between fixed points and colorings of the first $n-q$ numbers satisfying the same congruence condition.
So $a\_n \equiv a\_{n-q} \mod p$ for $n>q$ and $a\_q$ is the number of $q$-colorings of $\{1,\dots, q\}$ with all colors occurring exactly once, which is $q! \equiv -1 \mod p$.
|
2
|
https://mathoverflow.net/users/18060
|
396778
| 163,831 |
https://mathoverflow.net/questions/396768
|
5
|
$\DeclareMathOperator\SO{SO}$Suppose we have a (continuous) linear action of $\SO(n,\mathbb R)$ on a vector space $\mathbb R^N$. Consider the ring of invariants $A\subset \mathbb R[x\_1,\ldots, x\_N]$, which is an $\mathbb R$-algebra. Is it true that the orbits of the $\SO(n,\mathbb R)$ action are in one-to-one correspondence with the $\mathbb R$-algebra homomorphisms $A\to\mathbb R$?
It is clear that for each orbit of the action we get such an evaluation homomorphism, by the value of an invariant polynomial on the orbit. But is the inverse true? Is there some nice book that considers that type of setting, over the real numbers?
|
https://mathoverflow.net/users/13441
|
Invariant theory over $\mathbb R$
|
The answer depends on what you mean by "one-to-one correspondence". Is it bijective or just injective? Robert Bryant's (standard) argument shows that $\mathbb R^N/\mathrm{SO}(n)\to \mathrm{AlgHom}\_{\mathbb R}(A,\mathbb R)$ is injective. In general, this map is very far from being surjective, though. In other words, not every Homomorphism $A\to\mathbb R$ is an evaluation map.
Already the standard action of $\mathrm{SO}(n)$ on $\mathbb R^n$ is a counterexample. In this case, $A=\mathbb R[q]$ where $q(v)=\|v\|^2$ is the norm square function. Thus a homomorphisms $f:A\to\mathbb R$ is given by the value $f(q)$ which can be an arbitrary real number. But the image of $\mathbb R^n/\mathrm{SO}(n)$ is obviously only $\mathbb R\_{\ge0}$.
So what is the "correct quotient"? The answer is "It depends". A topologist would say it is the orbit space. The embedding into $\mathrm{AlgHom}\_{\mathbb R}(A,\mathbb R)$ gives the orbit space the structure of a semialgebraic set on which you have notions of continuous or smooth function. A theorem of G. Schwarz states that all continuous/smooth invariants are pull-backs of continuous/smooth function on the orbit space.
An algebraist would prefer to call $\mathrm{AlgHom}\_{\mathbb R}(A,\mathbb R)$ to be the quotient since it classified all closed orbits defined over $\mathbb R$ regardless whether the contain a real point or not.
A very nice paper calculating the image of $\mathbb R^N/\mathrm{SO}(n)\to \mathrm{AlgHom}\_{\mathbb R}(A,\mathbb R)$ (in particular) in terms of inequalities is Procesi, Claudio; Schwarz, Gerald: Inequalities defining orbit spaces.
Invent. Math. 81 (1985), 539–554
|
9
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https://mathoverflow.net/users/89948
|
396780
| 163,832 |
https://mathoverflow.net/questions/396769
|
3
|
The last few days I am trying my best to understand a part of a proof from [these](https://www.win.tue.nl/%7Erhofstad/percolation_randomgraphs_rev.pdf) lecture notes on page 14:
[Picture of the relevant part](https://i.stack.imgur.com/0297Xl.png)
The setting is percolation on a regular tree with degree $r$, $C\_{BP}(x)$ denotes the cluster of $x$ and $h(\cdot)$ is the distance of the vertex from the origin. It is clear to me why $\theta\_n$ satisfies the recursion (1.57).
However, I was not able to show that (1.57) together with $p\_c=1/(r-1)$ implies that $\theta\_n=(C\_\rho+o(1))/n$ for some
constant $C\_\rho>0$.
After some looking around I have also found this problem as an exercise with the hint to consider $v\_n=1/\theta\_n$ and performing induction on $n$. Unfortunately I am really stuck and don't see what to do. Maybe someone can help me. Is there hope for an explicit formula for $\theta\_n$ or $v\_n$? Thank's a lot for your help and time.
|
https://mathoverflow.net/users/313721
|
Deriving an asymptotic statement from a recursion
|
$\newcommand{\ep}{\varepsilon}$Let
\begin{equation\*}
x\_n:=\theta\_n,\quad s:=r-1>1,\quad b:=\frac{s-1}{2s}>0,
\end{equation\*}
\begin{equation\*}
f(x):=1-(1-x/s)^s,
\end{equation\*}
so that
\begin{equation\*}
x\_n=f(x\_{n-1})
\end{equation\*}
for natural $n$, with $x\_0\in[0,1]$. Without loss of generality, $x\_0\in(0,1]$.
We have $f'(x)=(1-x/s)^{s-1}<1$ for $x\in(0,1]$ and hence $f(x)<x$ for $x\in(0,1]$, so that $x\_n$ is decreasing to some limit $x\_\infty=f(x\_\infty)\in[0,1]$. Therefore and because $f(x)<x$ for $x\in(0,1]$, we have $x\_\infty=0$, so that $x\_n\downarrow0$ (as $n\to\infty$).
Next, $f(x)=x-(b+o(1))x^2$ as $x\downarrow0$. So,
\begin{equation\*}
x\_n=x\_{n-1}-a\_n x\_{n-1}^2
\end{equation\*}
for some $a\_n\to b$ and all natural $n$. Letting now
\begin{equation\*}
c\_n:=nx\_n,
\end{equation\*}
we have
\begin{equation\*}
c\_n=\frac n{n-1}\,c\_{n-1}-\frac{b\_n}n\,c\_{n-1}^2 \tag{1}
\end{equation\*}
for some
\begin{equation\*}
b\_n\to b. \tag{2}
\end{equation\*}
Since $c\_n/n=x\_n\to0$, by (1) and (2), we get the crucial conclusion that
\begin{equation\*}
\frac{c\_n}{c\_{n-1}}\to1. \tag{2.5}
\end{equation\*}
Take now any $h\in(0,1)$. Informally, we are going to show that the sequence $(c\_{n-1})$ is mainly confined between the left "moving barrier" $(c^{-h}\_n)$ and the right "moving barrier" $(c^h\_n)$, where
\begin{equation\*}
c^{-h}\_n:=\frac n{n-1}\,\frac{1-h}{b\_n},\quad c^h\_n:=\frac n{n-1}\,\frac{1+h}{b\_n}.
\end{equation\*}
Moreover, by (2.5) and (2), the jumps of the sequence $(c\_{n-1})$ from between these two moving barriers
to the left or right of both of these moving barriers will be of negligible magnitudes.
Indeed, if, for some $n\ge3$, we have
\begin{equation\*}
c\_{n-1}\le c^{-h}\_n, \tag{3}
\end{equation\*}
then, by (1),
\begin{equation\*}
\frac{c\_n}{c\_{n-1}}\ge1+\frac h{n-1}>1. \tag{4}
\end{equation\*}
Therefore and because $\prod\_{j=2}^\infty(1+\frac h{j-1})=\infty$, we will have $c\_n\to\infty$ if (3) holds for all natural $k\ge n$, in place of $n$, that is, if $c\_{k-1}\le c^{-h}\_k$ for all natural $k\ge n$. However, in view of (2), $c^{-h}\_k\to\frac{1-h}b<\infty$ as $k\to\infty$. So, (3) cannot hold for all natural $k\ge n$, in place of $n$. So,
there will be some natural $m=m\_n\ge n$ such that
\begin{equation\*}
c\_{n-1}\le c^{-h}\_n,\dots,c\_{m-1}\le c^{-h}\_m,\ c\_m>c^{-h}\_{m+1}
\end{equation\*}
and
\begin{equation\*}
c\_{n-1}<\cdots<c\_{m-1}<c\_m.
\end{equation\*}
Informally, if $(c\_{n-1})$ ventures to the left of the (left) moving barrier $(c^{-h}\_n)$, it is returned, in a finite number of steps and in a monotonic manner, to the right of the moving barrier $(c^{-h}\_n)$.
On the other hand, similarly, if, for some $n\ge3$, we have
\begin{equation\*}
c\_{n-1}\ge c^h\_n, \tag{3a}
\end{equation\*}
then, by (1),
\begin{equation\*}
\frac{c\_n}{c\_{n-1}}\le1-\frac h{n-1}<1. \tag{4a}
\end{equation\*}
Therefore, in view of (2) and because $\prod\_{j=2}^\infty(1-\frac h{j-1})=0$, there will be some natural $k=k\_n\ge n$ such that
\begin{equation\*}
c\_{n-1}\ge c^h\_n,\dots,c\_{k-1}\ge c^h\_k,\ c\_k<c^h\_{k+1}
\end{equation\*}
and
\begin{equation\*}
c\_{n-1}>\cdots>c\_{k-1}>c\_k.
\end{equation\*}
Informally, if $(c\_{n-1})$ ventures to the right of the (right) moving barrier $(c^h\_n)$, it is returned, in a finite number of steps and in a monotonic manner, to the left of the moving barrier $(c^h\_n)$.
Recalling now (2.5) and (2), we conclude that, for any $h\in(0,1)$,
\begin{equation\*}
\frac{1-h}b\le\liminf\_n c\_n\le\limsup\_n c\_n\le\frac{1+h}b;
\end{equation\*}
that is, $c\_n\to1/b$; that is, $\theta\_n=x\_n\sim1/(nb)$, as desired.
|
3
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https://mathoverflow.net/users/36721
|
396788
| 163,834 |
https://mathoverflow.net/questions/396685
|
1
|
Let $N\_{n}=(1/n)\_{i=1,j=1}^{n}$ be the $n\times n$-matrix where all the entries are equal.
Suppose $n>0$. Let $\delta\_{n}$ be the least natural number such that $N\_{n}$ can be factored as $N\_{n}=A\_{1}\dots A\_{k}$ for some $k$ and $A\_{1},\dots,A\_{k}$ such that if $1\leq i\leq k$, then $A\_{i}$ is the convex combination of at most $\delta\_{n}$ many permutation matrices. Is $\delta\_{n}=2$ for all $n>1$? If not, then what are the constants $\delta\_{n}$? If $\delta\_{n}$ is difficult to calculate exactly, what are some upper or lower bounds on the constants $\delta\_{n}$?
Observe that $\delta\_{n}\leq n$ since $N\_{n}$ is the convex combination of $n$ permutation matrices.
Lemma: $\delta\_{a\_{1}\dots a\_{r}}\leq\max(\delta\_{a\_{1}},\dots,\delta\_{a\_{r}})$.
Proof: Suppose that $N\_{a\_{i}}=A\_{i,1}\dots A\_{i,k}$ where each $A\_{i,j}$ is the convex combination of at most $\delta\_{a\_{i}}$ many permutation matrices.
$$N\_{n}=N\_{a\_{1}}\otimes\dots\otimes N\_{a\_{r}}$$
$$=(N\_{a\_{1}}\otimes I\_{a\_{2}\dots a\_{r}})(I\_{a\_{1}}\otimes N\_{a\_{2}}\otimes I\_{a\_{3}\dots
a\_{r}})\dots(I\_{a\_{1}\dots a\_{r-1}}\otimes N\_{a\_{r}}).$$
However, for $1\leq i\leq r$, we have
$$I\_{a\_{1}\dots a\_{i-1}}\otimes N\_{a\_{i}}\otimes I\_{a\_{i+1}\dots a\_{r}}=
(I\_{a\_{1}\dots a\_{i-1}}\otimes A\_{i,1}\otimes I\_{a\_{i+1}\dots a\_{r}})\dots
(I\_{a\_{1}\dots a\_{i-1}}\otimes A\_{i,k}\otimes I\_{a\_{i+1}\dots a\_{r}}).$$
Q.E.D.
As a consequence, we conclude that $\delta\_{n}\leq p$ whenever $n>1$ and $p$ is the largest factor of $n$. However, I have shown that $\delta\_{3}=2$ experimentally, so the reverse inequality does not always hold.
This is a continuation of the series of questions including [this question](https://mathoverflow.net/questions/396525/how-quickly-can-irreducible-aperiodic-convex-combinations-of-permutation-matrice) and [this question](https://mathoverflow.net/questions/396352/convex-combinations-a-of-n-times-n-permutation-matrices-such-that-every-entr).
|
https://mathoverflow.net/users/22277
|
Factorizing the doubly stochastic matrix where all entries are equal such that the factors are all convex combinations of few permutation matrices
|
I claim that $\delta\_{n}=2$ for all $n$.
>
> Lemma: For each $n$, and each vector $[x\_{1},\dots,x\_{n}]^{T}$ with
> $x\_{1}+\dots+x\_{n}=0$, there are matrices $A\_{1},\dots,A\_{k}$ where
> each $A\_{i}$ is the convex combination of $2$ permutation matrices and
> where $A\_{1}\dots A\_{k}[x\_{1},\dots,x\_{n}]^{T}=\mathbf{0}$.
>
>
>
Proof: Let $r$ be the number of non-zero entries in the vector $[x\_{1},\dots,x\_{n}]^{T}$. Then we shall prove this result by induction on $r$. The result is clear when $r=0,r=1$, so assume $r>1$. Then there is a permutation matrix $A$ such that $A[x\_{1},\dots,x\_{n}]^{T}=[y\_{1},\dots,y\_{r-1},y\_{r},0,\dots,0]^{T}$ such that $y\_{r-1}>0,y\_{r}<0$.
Let $t=\frac{y\_{r-1}}{y\_{r-1}-y\_{r}}$.
Then $$(t\cdot I\_{n}+(1-t)\cdot\rho\_{(r-1,r)})[y\_{1},\dots,y\_{r-1},y\_{r},0,\dots,0]^{T}$$
$$=t[y\_{1},\dots,y\_{r-1},y\_{r},0,\dots,0]^{T}+(1-t)[y\_{1},\dots,y\_{r-2},y\_{r},y\_{r-1},0,\dots,0]^{T}=[y\_{1},\dots,y\_{r-2},y\_{r-1}+y\_{r},0,\dots,0]^{T}.$$
Therefore, there are matrices $A\_{1},\dots,A\_{k}$ where
$$A\_{1}\dots A\_{k}[y\_{1},\dots,y\_{r-2},y\_{r-1}+y\_{r},0,\dots,0]^{T}=\mathbf{0}$$ and $A\_{i}$ is the convex combination of 2 permutation matrices for $1\leq i\leq k$. Therefore, if we set $A\_{k+1}=t\cdot I\_{n}+(1-t)\cdot\rho\_{(r-1,r)}$ and
$A\_{k+2}=A$, then $A\_{1}\dots A\_{k+2}[x\_{1},\dots,x\_{n}]^{T}=\mathbf{0}$ and $A\_{i}$ is the convex combination of 2 permutation matrices whenever $1\leq i\leq k+2$. Q.E.D.
>
> Theorem: For all $n>1$, we have $N\_{n}=A\_{1}\dots A\_{k}$ for some $k$
> and matrices $A\_{1},\dots,A\_{k}$ such that $A\_{i}$ is the convex
> combination of $2$ permutation matrices for $1\leq i\leq k$.
>
>
>
Proof: Observe that if $A$ is a doubly stochastic matrix, then $A=N\_{n}$ if and only if $A(x\_{1},\dots,x\_{n})^{T}=\mathbf{0}$ whenever $x\_{1},\dots,x\_{n}$ are real numbers with $x\_{1}+\dots+x\_{n}=0$.
Let $V$ be the space of all column vectors $[x\_{1},\dots,x\_{n}]^{T}$ with
$x\_{1}+\dots+x\_{n}=0$.
Claim: For all $r$ with $0\leq r<n$, there are matrices $A\_{1}\dots A\_{k}$ where for $1\leq i\leq k$, the matrix $A\_{i}$ is the convex combination of two permutation matrices and where $\text{dim}(V\cap\text{null}(A))\geq r$ where $A=A\_{1}\dots A\_{k}$.
Proof: We prove this claim by induction on $r$. The case when $r=0$ is trivial. Suppose now that $A=A\_{1}\dots A\_{k}$ where $\text{dim}(V\cap\text{null}(A))\geq r$ and where $A\_{i}$ is the convex combination of two $n\times n$ permutation matrices for $1\leq i\leq r$. Now, if $\dim(V\cap\text{null}(A))=n-1$, then the induction step is complete. If $\dim(V\cap\text{null}(A))<n-1$, then let $[x\_{1},\dots,x\_{n}]^{T}\in V\setminus\text{null}(A)$, and let
$[y\_{1},\dots,y\_{n}]^{T}=A[x\_{1},\dots,x\_{n}]^{T}$. Then there are matrices $A\_{-r}\dots A\_{0}$ such that $A\_{-r}\dots A\_{0}[y\_{1},\dots,y\_{n}]^{T}$ and $A\_{i}$ is the convex combination of two permutation matrices for $-r\leq i\leq 0$. Therefore, if $B=A\_{-r}\dots A\_{0}A\_{1}\dots A\_{k}$, then $\dim(V\cap\text{null}(B))\geq r+1$.
Our claim has been proven.
The result follows from our claim in the case that $r=n-1$. Q.E.D.
We are in fact able to produce the following explicit factorization of $N\_{n}$
Let $$C\_{n,r}=\frac{1}{r+1}(I\_{n}+r\rho\_{(r,r+1)}).$$ Let $$D\_{n,r}=C\_{n,1}\dots C\_{n,r-1}.$$
>
> Theorem: Suppose $n>0$. Then $N\_{n}=D\_{n,n}\dots D\_{n,2}.$
>
>
>
Proof outline: We use induction. Using the induction hypothesis, we have
$$D\_{n,n-1}\dots D\_{n,2}=\begin{bmatrix}N\_{n-1} & \mathbf{0}\\ \mathbf{0}&I\_{1}
\end{bmatrix}.$$
Set $A=D\_{n,n-1}\dots D\_{n,2}$. Then $\dim(\ker(A)\cap V)=n-2$.
Now, $[-1,\dots,-1,n-1]^{T}\in V\setminus\ker(A)$, however, $$D\_{n,n-1}A[-1,\dots,-1,n-1]^{T}=D\_{n,n-1}[-1,\dots,-1,n-1]^{T}=\mathbf{0},$$ so
$\dim(\ker(D\_{n,n-1}A)\cap V)>n-2$, but this is only possible if
$V\subseteq\ker(D\_{n,n-1}A)$. This implies that $D\_{n,n-1}A=N\_{n}$. Q.E.D.
There are other decompositions as well.
>
> Theorem: Suppose $n>0$. Then $D\_{n,n}^{n-1}=N\_{n}$.
>
>
>
|
0
|
https://mathoverflow.net/users/22277
|
396795
| 163,835 |
https://mathoverflow.net/questions/396797
|
6
|
Let $k$ be an algebraically closed field. Let $H\_1, H\_2$ be two smooth hypersurfaces of the same degree $d$ in $P^n\_k$. Let $U\_1,U\_2$ be their complements respectively. Are $U\_1,U\_2$ isomorphic as algebraic varieties?
In $n=1,d=1$ case this is true, because the complement of any point is isomorphic to $A^1$.
But $n=2$ case I guess this might be false. I want to prove that if $U\_1,U\_2$ are isomorphic then they must be induced by an automorphism of $P^n$, but this seems hard. I read something about the 'complement problem' on [enter link description here](http://archive.numdam.org/item/SB_1994-1995__37__295_0/), but this seems to be a more complicated question, and it focuses on $A^n$ instead.
Maybe the $n=2,d=3$ case is easier? In this case, elliptic curves are isomorphic if and only if they have the same $j$-invariant. Can we read this from its complement?
Are there any solutions/counterexamples? Any comments are welcome!
|
https://mathoverflow.net/users/177957
|
Open complement of hypersurfaces
|
The answer is no. Perhaps the simplest case is $n=2$, $d=4$. There is a unique double covering $\pi \_i:S\_i\rightarrow \mathbb{P}^2$ branched along $H\_i$. If $U\_1$ and $U\_2$ are isomorphic, $S\_1$ and $S\_2$ are isomorphic; then $H\_1$ and $H\_2$ are isomorphic, because $H\_i$ is the branch locus of the morphism $\pi \_i$, which is given by the anticanonical system.
|
9
|
https://mathoverflow.net/users/40297
|
396802
| 163,839 |
https://mathoverflow.net/questions/396762
|
5
|
Is there a well-defined notion of connection on a measurable bundle of Hilbert spaces?
|
https://mathoverflow.net/users/78032
|
Connection on a Hilbert bundle
|
This is an answer to the refined question formulated in the comments:
the base space is the unitary dual of a Lie group $G$.
The definition can be carried out in the setting of stacks in groupoids (or simplicial sets) on the site of cartesian spaces ($\def\R{{\bf R}} \R^n$ with smooth maps, for all $n≥0$).
Specifically,
we define a stack $R\_G$ that gives the unitary dual of a Lie group $G$
and a stack $B\_∇$ that gives bundles of Hilbert spaces with connectiom.
Then a morphism $R\_G→B\_∇$ is precisely a bundle of Hilbert spaces
with connection over $R\_G$.
To define the stack $R\_G$ that gives the unitary dual of a Lie group,
assign to a cartesian space $T$ the following groupoid $R\_G(T)$.
Objects of $R\_G(T)$ are given by a Hilbert space $H$ together with a smooth $T$-indexed
family of irreducible unitary representations of $G$ on $H$.
This is simply a smooth map $\def\Hom{\mathop{\rm Hom}} f\colon T→\Hom(G,U(H))$ (landing in irreps), where $U(H)$ is equipped with the ultraweak topology and $\Hom(G,U(H))$ is the space of continuous group homomorphisms equipped with the compact-open topology. Smooth means that the adjoint map $T⨯G→U(H)$ composed with the inclusion $U(H)→B(H)$ is smooth as a map from a finite-dimensional smooth manifold to a topological vector space $B(H)$.
Morphisms of $R\_G(T)$ from $(H,f)$ to $(H',f')$ are given by a smooth map $h\colon T→U(H,H')$ that intertwines the action of $G$.
Here $U(H,H')$ is the topological space of unitary isomorphisms $H→H'$ equipped with the ultraweak topology.
Next, we define stacks $B$ and $B\_∇$ of bundles of Hilbert spaces,
(equipped with connection in the case of $B\_∇$) as follows.
Given a cartesian space $T$, we define a groupoid $B\_∇(T)$ as follows.
Objects of $B\_∇(T)$ are pairs $(H,∇)$, where $H$ is a Hilbert space
and $∇\colon T→Hom(T,I(H))$ is a smooth map,
where $I(H)$ denotes the space of unbounded skew-adjoint operators on $H$
(equivalently, one-parameter unitary groups on $H$)
and $\Hom(T,I(H))$ is the space of linear maps $T→I(H)$.
Morphisms of $B\_∇(T)$ from $(H,∇)$ to $(H',∇')$ are smooth maps $p\colon T→U(H,H')$
such that $∇'=p^\*\theta+\mathop{\rm Ad}\_{p^{-1}}∇$, where $\theta$ is the Maurer–Cartan form and Ad denotes the adjoint action.
The stack $B$ is defined analogously, but dropping $∇$ and the condition on $p$.
We have a canonical forgetful map $B\_∇→B$.
The obvious forgetful map $W\colon R\_G→B$ defines the canonical bundle $W$
over $R\_G$.
In particular, the fiber of $W$ of a point of $R\_G$ given
by an irreducible representation $ρ$ of $G$ is simply $ρ$ itself.
Now, a connection on $W$ is a lift of the map $W\colon R\_G→B$
through the forgetful map $B\_∇→B$.
|
3
|
https://mathoverflow.net/users/402
|
396805
| 163,842 |
https://mathoverflow.net/questions/396794
|
2
|
Let's say that a (right) module $M$ is *well complemented* if every non-zero submodule of $M$ has an indecomposable direct summand (by the way, is there a better or more standard name for this property?). For instance, every module of finite [uniform dimension](https://en.wikipedia.org/wiki/Uniform_module#Uniform_dimension_of_a_module) is well complemented.
>
> **Question.** Is the regular right module $R\_R$ of a von Neumann regular ring $R$ well complemented?
>
>
>
As a recall, a ring $R$ is *von Neumann regular* if, for every $x \in R$, there exists $y \in R$ such that $x = xyx$.
|
https://mathoverflow.net/users/16537
|
Every non-zero submodule of $R_R$ has an indecomposable direct summand: True when $R$ is von Neumann regular?
|
The answer is no. Take a compact totally disconnected space $X$ with no isolated points, like the Cantor set. Let $K$ be any field and let $R$ be the ring of locally constant functions $f\colon X\to K$ with pointwise operations. This is a commutative von Neumann regular ring. The idempotents of $R$ are precisely the characteristic functions $1\_K$ of clopen sets $K$. An orthogonal decomposition of $1\_K$ into idempotents corresponds to writing $K$ as a disjoint union of clopen sets. Since $X$ has no isolated points, if $K$ is a nonempty clopen set, there are $x\neq y\in K$. Then we can find a clopen subset $K'$ of $K$ with $x\in K'$ and $y\notin K'$. Thus $1\_K = 1\_{K'}+1\_{K\setminus K'}$ is a decomposition into orthogonal idempotents. If follows that $R$ has no primitive idempotents and hence no indecomposable summands (as an indecomposable summand is of the form $eR$ with $e$ primitive).
|
4
|
https://mathoverflow.net/users/15934
|
396806
| 163,843 |
https://mathoverflow.net/questions/396214
|
7
|
A nonempty subset $D$ of a group $G$ is called
$\bullet$ *decomposable* if $D\subseteq DD$, that is every element $x\in D$ is can be written as the product $x=yz$ of some elements $y,z\in D$;
$\bullet$ *product-one* if there exists $n\in\mathbb N$ and pairwise distinct elements $x\_1,\dots,x\_n\in D$ such that $x\_1\cdots x\_n=1$.
>
> **Problem 1.** Let $D$ be a finite decomposable subset of a group. Is $D$ product-one?
>
>
>
**Remark 1.** For commutative groups this problem was posed by [Gjergji Zaimi](https://mathoverflow.net/q/16857/61536) and solved affirmatively by [Lev, Nagy, and Pach](https://mathoverflow.net/a/380683/61536).
**Remarks 2.** For some non-commutative groups like generalized dihedral groups the answer to Problem is also affirmative, see my [partial answer](https://mathoverflow.net/a/396764/61536) below. This partial answer suggests the following
>
> **Problem 2.** Let $G$ be a group containing an Abelian subgroup of index 2. Is every finite decomposable set in $G$ product-one?
>
>
>
|
https://mathoverflow.net/users/61536
|
Product-one sets in non-commutative groups
|
GAP shows that the groups SmallGroup(27,3), SmallGroup(27,4), SmallGroup(36,11), SmallGroup(39,1) SmallGroup(48,3) do contain many 5-element decomposable sets, which are not product-one. So, the lower bound 5 for the smallest cardinality of a counterexample, obtained by @YCor in his answer, is the best possible.
Below I write down 5-element decomposable non-product-one sets found by GAP in the groups
SmallGroup(27,3): [ f1, f2, f1 \* f2, f1^2 \* f2, f1 \* f2^2 ]
SmallGroup(27,4): [ f1, f2, f1 \* f2 \* f3, f1^2 \* f2 \* f3, f2^2 \* f3^2 ]
SmallGroup(36,11): [ f1, f2 \* f3, f1^2 \* f3, f1 \* f2^2 \* f3, f1^2 \* f2^2 \* f4 ]
SmallGroup(39,1): [ f1, f2, f1 \* f2, f1^2 \* f2, f2^4 ]
SmallGroup(48,3): [ f1, f2, f1 \* f2, f2 \* f3, f1^2 \* f2 ]
These 5 groups are the only groups of order $\le 50$ that contain decomposable non-product-one sets.
|
3
|
https://mathoverflow.net/users/61536
|
396817
| 163,849 |
https://mathoverflow.net/questions/396746
|
4
|
Let $(R,\mathfrak m,k)$ be an Artinian Gorenstein local ring such that $$\mu(\mathfrak m)=2, \quad\mathfrak m^2\ne 0,\quad\text{and}\quad \mathfrak m^3=0.$$
Then, is it true that every non-maximal ideal of $R$ is principal?
*Thoughts*: We have $\mathfrak m^2 \subseteq (0:\_R \mathfrak m)$. Since $R$ is Artinian Gorenstein, so $\dim\_k (0:\_R \mathfrak m)=1$, hence $\dim\_k \mathfrak m^2=1$. Thus, $\mathfrak m^2, (0:\_R \mathfrak m)$ are principal ideals. For an arbitrary ideal $I\subsetneq \mathfrak m$, we have
$$\begin{array}{r@{}l}
\mu(I)
&=\dim\_k(I/\mathfrak mI)\\
&\le \dim\_k(\mathfrak m/\mathfrak m I)-1\\
&=\dim\_k(\mathfrak m/\mathfrak m^2)+\dim\_k(\mathfrak m^2/\mathfrak m I)-1\\
&\le \mu(\mathfrak m)+1-1\\
&=2,
\end{array}
$$
which is just one off from being principal.
[Also note that since $\mu(\mathfrak m) -\dim R=2$ and $R$ is Gorenstein, so a result of Serre implies $R$ is complete intersection]
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https://mathoverflow.net/users/158239
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Ideals in Artinian Gorenstein local ring $(R,\mathfrak m)$ with $\mu(\mathfrak m)=2, \mathfrak m^2\ne 0$ and $\mathfrak m^3=0$
|
We use the fact that in an Artinian Gorenstein ring, any ideal contains the socle. The assumption tells us that the socle of $A$ is $\mathfrak m^2$, which is principal.
Let $I\neq (0)$ be a non-maximal ideal. If $I=\mathfrak m^2$, we are done. Otherwise, $I$ strictly contains $\mathfrak m^2$. Thus $\mathfrak mI\neq 0$, but then $\mathfrak mI \supset\mathfrak m^2$. On the other hand as $I\subset \mathfrak m$, $\mathfrak mI = \mathfrak m^2$.
We have $I/\mathfrak mI = I/\mathfrak m^2$ is a non-zero proper subspace of $\mathfrak m/\mathfrak m^2$ which has $k$-dimension 2, so it has $k$-dimension $1$.
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4
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https://mathoverflow.net/users/2083
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396824
| 163,851 |
https://mathoverflow.net/questions/396820
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0
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Let $n=3^m$ for some positive integer $m$. Let $G\leq S\_n$ be a transitive permutation group on $n$ letters. Denote the largest normal subgroup of $G$ with odd order by $O\_{2'}(G)$. My question is the following:
Does there exist $G$ such that $G/O\_{2'}(G)\cong A\_4$ or $S\_4$ for suitable $m$?
|
https://mathoverflow.net/users/134942
|
Is there a permutation group satisfying the following property?
|
Let $\Omega=\{1,2,3,4,5,6,7,8,9\}$ and let $G\leq\mathrm{Sym}(\Omega)$ be the group generated by the following permutations:
* $(1, 2, 9)$
* $(4, 5)(7, 8)$
* $(1, 4, 7)(2, 5, 8)(3, 6, 9)$
* $(3, 6)(4, 7)(5, 8)$
This is an example with $n=9$, with $G/O\_{2'}\cong S\_4$. If you want $G/O\_{2'}\cong A\_4$, remove the last generator.
To get larger examples, we can simply use direct products: let $H$ be a group of odd order acting transitively on $\Omega'$, with $|\Omega'|=3^{m-2}$ (for example, an elementary abelian $3$-group acting regularly). We can view $G\times H$ as a permutation group on $\Omega\times\Omega'$ with coordinate wise action, and it has the required properties.
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5
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https://mathoverflow.net/users/22377
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396830
| 163,852 |
https://mathoverflow.net/questions/396615
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4
|
Let $C$ be a class of plane curves, regarded as subsets of $\mathbb{R}^2$ (parametrization won't matter), I'm thinking for example of splines or algebraic subsets. Let $D$ be a class of topological discs bounded by plane curves from a possibly different class, for example circles, ellipses, closed splines.
Now if I take a $c \in C$ and a $d \in D$ I can form the Minkowski sum $c + d$ (simply all sums of points from $c$ and $d$). Its boundary will again be a curve (possibly not connected). My question is:
>
> Are there "interesting" classes $C$ and $D$ that ensure that the boundary of $c + d$ will again lie in $C$? Is this kind of closure behavior ever considered?
>
>
>
By "interesting" I mean that the classes should be small enough, so that each curve can be described by finite data, but not so small that the statement becomes trivial. For instance taking both classes to be $C^1$ is too large and taking $D$ to consist just of points is too small.
The origin for this question is the following: the sum $c + d$ models tracing the curve $c$ with a pen whose shape is $d$. At some point I learned that Knuth's Computer Modern typeface was turned into OpenType by first producing high resolution pixel versions and then vectorizing these. I was surprised at first because to me Metafont was vector graphics as was OpenType. But the point is that Knuth drew Computer Modern tracing curves with pens (so the letters are described as $c + d$) while OpenType uses the outlines (the boundary of $c + d$). The fact that the transition is not purely formal suggests that splines do not have the above closure property. On the other hand, some vector graphics software can transform a traced curve into an outline (perhaps inexactly?). I'm not interested practical solutions but in whether there is a class of curves that works theoretically.
|
https://mathoverflow.net/users/5339
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Classes of curves closed under Minkowsky sum
|
There are lots of finite dimensional curve families that are closed under Minkowski sum with circular disks. Here's a way to construct examples:
First, choose a finite dimensional space $\mathcal{C}$ of smooth, functions that contains the constants. Then, for any $f\in\mathcal{C}$, consider the curve $X\_f$ defined by
$$
X\_f(\theta) = \int\_0^\theta \begin{pmatrix}\cos t\\\sin t\end{pmatrix}\,f(t)\,\mathrm{d}t.
$$
Let $C$ be the family of curves plust their translations in the plane.
(If one wants the family to be also invariant under rotation, one should require that $\mathcal{C}$ be invariant under translation.)
Then the Minkowski sum of $X\_f$ and the disk of radius $r$ will be bounded by the curves
$$
X\_{f\pm r}(\theta) = p+ \int\_0^\theta \begin{pmatrix}\cos t\\\sin t\end{pmatrix}\,(f(t)\pm r)\,\mathrm{d}t,
$$
which, by construction, also belong to the family $\mathcal{C}$.
Of course, these curves may be singular, and if one wants to avoid that one might want to discard the locus in $X\_f$ where $f$ vanishes or where $f\pm r$ vanishes. This can also be done by restricting the range of $r$ and restricting the allowable $f$ to an appropriate open set in $\mathcal{C}$.
This can be generalized considerably: Let $\mathcal{D}$ be a finite dimensional vector space of $2\pi$-periodic functions that contains the constants with the property that, for all $f\in\mathcal{D}$ we have
$$
\int\_0^{2\pi} \begin{pmatrix}\cos\theta\\\sin\theta\end{pmatrix}\,f(\theta)\,\mathrm{d}\theta=0.
$$
For example, $\mathcal{D}$ might be the span of $\{1,\cos2\theta,\sin2\theta,\ldots,\cos m\theta,\sin m\theta\}$ for some integer $m\ge 2$.
Now let $\mathcal{D}^+\subset\mathcal{D}$ denote the set of those elements of $\mathcal{D}$ that are positive. Then, for $f\in\mathcal{D}^+$, the curve
$$
Y\_f(\theta) = \int\_0^\theta \begin{pmatrix}\cos t\\\sin t\end{pmatrix}\,f(t)\,\mathrm{d}t
$$
will be $2\pi$-periodic and will bound a convex region $R\_f$ in the plane. Let $D$ be the (finite dimensional) set of such convex regions $R\_f$. Now assume, as before, that $\mathcal{C}$ is a finite dimensional subspace of smooth functions that contains $\mathcal{D}$. Then the Minkowski sum of a curve $X\_f$ for $f\in\mathcal{C}$ and a region $R\_g$ for $g\in\mathcal{D}^+$ will be bounded by the curves $X\_{f\pm g}$, which, since $\mathcal{D}\subset\mathcal{C}$, will still be a curve in $C$.
|
3
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https://mathoverflow.net/users/13972
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396850
| 163,856 |
https://mathoverflow.net/questions/392187
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3
|
I am reading "Natural and Gauge Natural Formalism for Classical Field Theory" by Lorenzo Fatibene and I am really confused by his definition of a connection in local coordinates.
Let's say we have a principal bundle $\mathcal{P}=(P, M, \pi ; G)$ and the isomorphism $T\_{e} L\_{p}: \mathfrak{g} \longrightarrow V\_{p}(\pi): T\_{A} \mapsto \lambda\_{A}(p)$. He fixes a point $p=[x, g]\_{(\alpha)}$. $\theta\_{(L)}^{A}=\bar{L}\_{a}^{A}(g) \mathrm{d} g^{a}$ is a local basis of left invariant 1-forms dual to $\lambda\_{A}=L\_{A}^{a}(g) \partial\_{a}$ where the two matrices are inverses of each other. Then the connection 1-form can be written as
\begin{equation}
\bar{\omega}\_{p}=\left[\theta\_{(L)}^{A}(p)+\mathrm{Ad}\_{B}^{A}\left(g^{-1}\right) \omega\_{\mu}^{B}(x) \mathrm{d} x^{\mu}\right] \otimes T\_{A}.
\end{equation}
Pulling back with a section, he gets
\begin{equation}\sigma^{\*} \bar{\omega}=\left[\bar{L}\_{a}^{A}(g) \partial\_{\mu} g^{a}(x)+\mathrm{Ad}\_{B}^{A}\left(g^{-1}\right) \omega\_{\mu}^{B}(x)\right] \mathrm{d} x^{\mu} \otimes T\_{A}.
\end{equation}
However, then he goes on saying "We remark that the local gauge $\sigma$ also induces a local trivialization of $\mathcal{P}$. In the induced local trivialization, the section $\sigma$ has the expression $\sigma: x^{\mu} \mapsto\left(x^{\mu}, e\right)$ and the vector potential is of the form"
\begin{equation}
\sigma^{\*} \bar{\omega}=\omega\_{\mu}^{A}(x) \mathrm{d} x^{\mu} \otimes T\_{A}
\end{equation}
He also states that the induces connection is of the form
\begin{equation}
\omega=\mathrm{d} x^{\mu} \otimes\left(\partial\_{\mu}-\omega\_{\mu}^{A}(x) \rho\_{A}\right)
\end{equation}
1. My first question is how one can derive the expression for $\bar{\omega}\_{p}$?
2. My second question is why every other book on the subject I know uses $\sigma^{\*} \bar{\omega}=\omega\_{\mu}^{A}(x) \mathrm{d} x^{\mu} \otimes T\_{A}$ as the definition of a form in local coordinates; even this seems not to be true for a general section.
3. I am also not sure how one can derive the form of the induced connection.
Cross posted: <https://math.stackexchange.com/questions/4128191/local-coordinates-of-one-form-on-a-principal-bundle>
|
https://mathoverflow.net/users/209074
|
Local coordinates of one form on a principal bundle
|
Since you cross-posted, I'll cross-answer my reply from MSE. I'M rather new to the MO forum so I could use some reputation...
* I assume by $T\_A$ you mean a set of basis vectors of $\mathfrak{g}$.
* I assume by $\omega$ (without bar) you mean the connection of an associated vector bundle, associated via some representation of $G$.
For (1.):
By definition a connection form is a differential form $\mathcal{A}\in\Omega^1(P,\mathfrak{g})$, which is both of type $\text{Ad}$, i.e. $(R\_g)^\*\mathcal{A}=\text{Ad}\circ\mathcal{A}$, and in a point $p\in P$ the map $\mathcal{A}\big|\_p:T\_pP\to\mathfrak{g}$ has a certain behavior on vertical tangent vectors. That is, by being of type $\text{Ad}$, a connection form is determined already on a whole fiber by only one single value in this fiber. On the other hand, the difference between two connection forms is also a $1$-form of type $\text{Ad}$ and vanishes on vertical tangent vectors (since both have the same prescribed values on these). Such $1$-forms with the latter property are called horizontal. Conversely, you can show that for any connection form $\mathcal{A}$ and any horizontal $1$-form $\omega$ of type $\text{Ad}$, also $\mathcal{A}+\omega$ is a connection form. In other words, the set of connection forms is an affine space with spatial structure of the space of horizontal $1$-forms of type $\text{Ad}$.
Now, on a trivial bundle $U\times G$ you have a canonical flat connection form, which essentially does only what it should on vertical tangent vectors, and nothing else. Up to some issues of identification, I think this should be more or less what you have labeled $\theta\_{(L)}$, or $\theta\_{(L)}^A$ in some basis of $\mathfrak{g}$. Moreover, if you have a local trivialization $\phi:\pi^{-1}(U)\to U\times G$, then you can pull back the distinguished connection form on $U\times G$ to $\pi^{-1}(U)$.
Hence, your local trivialization did distinguish a connection form on $\pi^{-1}(U)$. As we have seen above, any other connection form, restricted to $\pi^{-1}(U)$, is then our distinguished one, plus some horizontal $1$-form of type $\text{Ad}$ on $\pi^{-1}(U)$, call it $\eta$. The local trivialization of $\pi^{-1}(U)$, moreover, distinguishes a section $s$ in $\pi^{-1}(U)$ (the one you have mentioned), and a $1$-form of type $\text{Ad}$ is fully determined by its values on the range of this section (which contains one point in each fiber). Then, if you pull back this $1$-form through a section $s:U\to\pi^{-1}(U)$, you end up with a $1$-form $s^\*\eta$ on $U$ with values in $\mathfrak{g}$ and your original $\eta$ may be fully recovered from only the values of $s^\*\eta$, namely, more or less, by the second term of your formula for $\bar{\omega}\_p$ (where you additionally expanded in the basis $T\_A$).
Taken together, locally on some $U$ on which both the principal bundle admits a trivialization and the basis manifold admits a chart, your formula is just expanding an arbitrary connection form into the basis $\text{d}x^\mu$, the basis $T\_A$ and the part (first term) with the distinguished connection form on $\pi^{-1}(U)$.
For (2.):
I'm not sure how to answer this. To me it seems that these "every other books" concentrate on the non-trivial part of a connection form which I have discussed above, and just call this the connection form. This is quite sloppy as not always possible, but suitable for many applications. From your first sentence I conclude that you head up to physics applications. Note that, e.g. Minkowski space is contractible (just $\mathbb{R}^4$), and thus every vector bundle (and the associated frame bundles) over it is trivializable. Therefore, such a bundle admits a distinguished (global) connection form on the frame bundle and it suffices to regard any other connection form only by its deviation from the distinguished one. Also if you head to GR applications, a metric also distinguishes a connection form on the frame bundle and we have the same situation.
For (3.):
As I said, I assume you mean by $\omega$ the connection on an associated vector bundle (probably on the adjoint bundle, as you expand $\omega$ with in indices $A$. Is that right?). Also on a vector bundle you can show that the set of connections is an affine space, but now with the spatial structure of $1$-forms on the basis manifold, which take values in the endomorphism bundle of the given vector bundle. Moreover, you find a one-on-one association between connections on a vector bundle, and connection forms on its frame bundle. Given a local trivialisation of the former, you also have a local trivialization of the latter. Moreover, given a local trivialization of the vector bundle you can distinguish a certain connection/covariant derivative, which is essentially just the Cartan derivative $\text{d}$ (concretely acting in your formula by $\partial\_\mu$), pulled back through the local trivialization. This connection is, under the above one-on-one association, mapped to the trivial flat connection form on the respective part of the frame bundle, if you do everything correctly (I know, that's a stupid thing to say...). Moreover, this one-on-one association maps the deviation from the respective distinguished connection/connection forms to each other (where the latter is first pulled back through the right section, above called $s^\*\eta$ to reproduce your forms $\omega\_\mu$ or $\omega\_\mu^A$, resp.). That's why you find your $\omega\_\mu^A$ again in this connection, as deviation of your given connection form the distinguished trivial connection acting by $\partial\_\mu$.
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4
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https://mathoverflow.net/users/276879
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396853
| 163,858 |
https://mathoverflow.net/questions/396843
|
1
|
I am studying the notes of Sorger concerning the moduli problem of principal bundles over curve <https://inis.iaea.org/collection/NCLCollectionStore/_Public/38/005/38005695.pdf> and there is something I don't quite understand. He considers the moduli stack of principal $G$-bundles over a curve $G$, and it is denoted by $\mathcal{M}\_{G,C}$, being $\mathcal{M}\_{G,C}(S)$ the groupoid of principal $G$-bundles over $C\times S$ for each $\mathbb{C}$-scheme $S$. In Proposition 3.6.8, he considers "the universal principal $G$-bundle on $C\times\mathcal{M}\_{G,C}$" and I don't have a clue about what this is. I know that giving a principal bundle over a stack is to give for each $\mathbb{C}$-scheme $S$ a principal bundle over $S$, for each morphism of schemes $S\rightarrow S'$ an isomorphism between the principal bundle over $S$ and the pullback of the principal bundle over $S'$ by the map $S\rightarrow S'$, and in such way that the cocycle compatibility condition holds. I would really like to know what is the universal principal bundle in this case. Thank you for your time and wisdom in advance.
|
https://mathoverflow.net/users/140062
|
Universal principal bundle on stack
|
Expanding on abx’s comment, let’s suppose $\mathcal{M}\_{G,C}$ was representable as a scheme. Then for any $\mathbb{C}$-scheme $S$ there is a canonical isomorphism
$Hom\_\mathbb{C}(S, \mathcal{M}\_{G,C}) \cong \{\text{Groupoid of principal }G\text{-bundles over }C \times S\}$.
Now take $S = \mathcal{M}\_{G,C}$, and consider the element of the LHS given by the identity map on $\mathcal{M}\_{G,C}$. The corresponding element of the RHS is precisely the “the universal principal $G$-bundle on $C\times\mathcal{M}\_{G,C}$". It is called this because any other principal $G$-bundle on some $C \times T$ can be uniquely realized as a pullback of this one via some morphism $T\rightarrow \mathcal{M}\_{G,C}$.
Now in reality, $\mathcal{M}\_{G,C}$ is not a scheme but merely an algebraic stack. However, there is still a notion of morphisms $T\rightarrow \mathcal{M}\_{G,C}$ for $\mathbb{C}$-schemes $T$, and some stacky notion of bundle on it which pulls back to honest bundles on schemes. I can’t fill in the details but I hope this gives some intuition.
|
2
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https://mathoverflow.net/users/5513
|
396862
| 163,861 |
https://mathoverflow.net/questions/396856
|
9
|
Let $X$ be a smooth complex algebraic variety endowed with a $\mathbb{C}^\*$ action. We assume also to have an antiholomorphic involution $\sigma$ over $X$ such that it anticommutes with the action above i.e $$\sigma(t \cdot x)=\bar{t}\cdot \sigma(x) .$$
Let us also assume that the $\mathbb{C}^\*$-action respects the properties in order to get a well behaved Byalinicki-Birula decomposition.
For every $x\in X$ there exists $\lim\_{t \to 0} t \cdot x$ and the fixed point set is a projective variety. (I think $X$ is called semiprojective in this case).
In this case, we know that the Poincare polynomial of $X$ can be expressed as $$P(X,t)=\sum\_{i \in I}P(F\_i,t)t^{d\_i} $$ where $$X^{\mathbb{G}\_m}=\bigsqcup\_{i \in I} F\_i $$ and $F\_i$ are connected and $d\_i$ are some positive integers associated to the action.
The fixed point set $X^{\sigma}$ of the antiholomorphic involution is a smooth real manifold. The hypothesis tells us that $\sigma(X^{\mathbb{G}\_m})=X^{\mathbb{G}\_m}$. Is it still true somehow that $$P(X^{\sigma},t)=\sum\_{i \in I}P(F\_i^{\sigma},t)t^{h\_i} $$ or not ?
Is this known in the literature? I'm totally new to real algebraic geometry.
EDIT: The answer below indicates this is not true in general. However, I'd be interested in the following more specific situation. $X$ should be given the structure of an hyperkahler manifold with complex structures $I,J,K$ such that the complex algebraic variety we are looking at is the one induced by $I$.
The involution $\sigma$ should then be antiholomorphic with respect to structure $I,J$ and holomorphic with respect to $K$. The $\mathbb{C}^\*$ action should be algebraic with respect to the structure induced by $I$ while in general it is not clear what happens with respect to the other structures.
The setting to think of is that of Non Abelian Hodge theory: it is known that we have an hyperkahler manifold $M$ such that with respect to $I$ it is the moduli space of stable higgs bundle $(\mathcal{E},\phi)$ of fixed rank $n$ and degree $d$. There we have the action $$t(\mathcal{E},\phi)=(\mathcal{E},t\phi) .$$ With respect to the structure $J$ is the de Rham moduli space of connections and with respect to $K$ is the associated (twisted)character variety.
As suggested below, it is likely that hyperkahler structure should help because of parity of dimension of the cells involved somehow, but I wasn't able to prove this neither to find some references.
|
https://mathoverflow.net/users/146464
|
Bialynicki-Birula decomposition for real analytic varieties
|
No, consider the following $\mathbb{C}^\*$-action on $\mathbb{CP}^2$ : $$z.[z\_0:z\_1:z\_{2}] = [z\_0:z.z\_1:z^2.z\_2] ,$$ along with the antiholomorphic map $\sigma ([z\_0:z\_1:z\_2]) = [\bar{z\_{0}}:\bar{z\_{1}}:\bar{z\_2}]$. One map check that $\sigma$ anticommutes with the $\mathbb{C}^\*$-action.
This is the nicest situation that one could wish for in terms of the BB decomp, etc.
Then the fixed point set of the $\mathbb{C}^\*$-action is the three points $[1:0:0],[0:1:0],[0:0:1]$, the fixed point set of the involution is $\mathbb{RP}^2 \subset \mathbb{CP}^2$. So, the statement about the Poincare polynomials cannot hold for any choice of $h\_i$ since the total Betti numbers are different (3 and 1 respectively).
Answer to edited question:
Even when the fixed point set is orientable the statement is false. Consider the action on $\mathbb{CP}^n$ $z.[z\_{0}: z\_{1} : \ldots : z\_{n}] = [z\_{0}: z z\_{1} : \ldots : z^n z\_{n}] $ where $n>1$ is odd, and the antiholomorphic involution given by conjugating all of the co-ordinates (as above). Then the the fixed point set of the involution is $\mathbb{RP}^n$ which is an orientable manifold with total Betti number $2$, and the fixed point set of the torus action is $n+1$ points, with total Betti number $n+1$.
|
6
|
https://mathoverflow.net/users/99732
|
396864
| 163,862 |
https://mathoverflow.net/questions/396705
|
2
|
Consider the resolvent operator $ R(z) := (-\Delta - z)^{-1}$ of the Laplace operator on $L^2(\mathbb R^d)$, where $z\in \rho(-\Delta) = \mathbb C \setminus \mathopen [0, \infty)$.
For $p \geq 1$, let $\lVert \cdot \rVert\_p$ denote the Schatten $p$-norm on the space of compact operators and let $1\_{\Gamma\_n}(x)$ denote multiplication by the indicator function of some cube $\Gamma\_n := n + [0,1]^d, n \in \mathbb Z^d$. For a given $E>0$, I am interested whether
$$\sup\_{\substack{z = E+i\epsilon \\ -1 \leq \epsilon \leq 1}}\lVert 1\_{\Gamma\_n}(x) R(z) 1\_{\Gamma\_n}(x) \rVert\_p < \infty \tag{1}$$
for some suitable $p=p(d)$, presumably all $p>d/2$.
So far I could not find any reference proving this, the problem being the supremum in front: In the book *Trace Ideals and their Applications* by Barry Simon, one can find the bound $$\lVert f(x) g(-i\nabla) \rVert\_p \leq C \lVert f \rVert\_p \lVert g \rVert\_p.$$
Noting that $\lVert 1\_{\Gamma\_n}(x) R(z) 1\_{\Gamma\_n}(x) \rVert\_p \leq \lVert 1\_{\Gamma\_n}(x) R(z) \rVert\_p$ and applying the above inequality to $f(x) := 1\_{\Gamma\_n}(x)$ and $g\_z(x):= \frac{1}{\lvert x \rvert^2 - z}$ yields
$$\lVert 1\_{\Gamma\_n}(x) R(z) 1\_{\Gamma\_n}(x) \rVert\_p \leq C \lVert g\_z \rVert\_p,$$
which is finite if $p>d/2$. Unfortunately, this is not enough for $(1)$ since the expression blows up as $\operatorname{Im }z = \epsilon \to 0$.
Still, I think $(1)$ should be true. For example, if $d=3$, we know that the kernel of $R(z)$ is given by
$$R(x,y;z) = \frac{1}{4\pi \lvert x - y \rvert} e^{-\sqrt{-z} \lvert x -y \rvert}$$
so that we can explicitly compute the Hilbert-Schmidt norm
$$\lVert 1\_{\Gamma\_n}(x) R(z) 1\_{\Gamma\_n}(x) \rVert\_2^2 = \int\_{\Gamma\_n} \int\_{\Gamma\_n} \lvert R(x,y;z) \rvert^2 \, dx dy \leq C \int\_{\Gamma\_n} \int\_{\Gamma\_n - y} \frac{1}{\lvert x \rvert^2} \, dx dy < \infty$$
uniformly in $z$. However, this does not work in $d>3$, where a higher $p$-norm would be needed (I assume $p>d/2$ as suggested by the above). Any help is appreciated!
|
https://mathoverflow.net/users/271621
|
Finiteness of Schatten $p$-norm of truncated free resolvent
|
I found a positive answer to this question in this [paper](https://arxiv.org/abs/1404.2817) of Frank and Sabin: *Restriction theorems for orthonormal functions, Strichartz inequalities, and uniform Sobolev estimates*, Theorem 12.
|
0
|
https://mathoverflow.net/users/271621
|
396865
| 163,863 |
https://mathoverflow.net/questions/396668
|
15
|
It is well known that, for a small category $\mathbf A$, the category $\widehat{\mathbf A} = [\mathbf A^\circ, \mathbf{Set}]$ of presheaves on $\mathbf A$ together with the Yoneda embedding $\mathbf A \to \widehat{\mathbf A}$ exhibits $\widehat{\mathbf A}$ as the cocompletion of $\mathbf A$ under small colimits.
Where was this first observed? If an observation first appears independently of a proof, I would be interested in knowing where a proof first appears too.
Bunge's 1966 thesis *Categories of Set-Valued Functors* seems a likely candidate, as it is concerned with properties of presheaf categories. However, I have been unable to obtain a copy of the thesis, so I do not know whether it appears here, and it is not mentioned in Bunge's summary of thesis, *Regular categories*.
|
https://mathoverflow.net/users/152679
|
Original reference for categories of presheaves as free cocompletions of small categories
|
The earliest reference I can find to the universal property of the presheaf construction is Remark 2.29 of Ulmer's [Properties of Dense and Relative Adjoint Functors](https://core.ac.uk/download/pdf/82101024.pdf) (1968). However, the proof is only lightly sketched, and in the introduction Ulmer states:
>
> As an application of relative adjoints we will show in a subsequent paper that every category $\mathbf M'$ admits a free right complete category.
>
>
>
As far as I can tell, this paper never appeared. Note that Ulmer actually considers the universal property for arbitrary (possibly large) categories, by taking small presheaves rather than arbitrary presheaves.
There is an earlier reference for the universal property of the Ind-completion (i.e. cocompletion under filtered colimits) in Proposition 8.7.3 of SGA4 (dated 1963–1964, but published in 1972), which suggests the universal property of free cocompletion was also known as this time, though an explicit statement does not appear.
In the enriched context, the universal property first appears as Theorem 2.11 of Lindner's [Morita equivalences of enriched categories](http://www.numdam.org/item/?id=CTGDC_1974__15_4_377_0) (1974), where the Lindner attributes the unenriched result to Ulmer.
|
12
|
https://mathoverflow.net/users/152679
|
396866
| 163,864 |
https://mathoverflow.net/questions/396855
|
1
|
This question seems to be related to Theorem IX.7.28 in [J. Jacod and A. Shiryaev's *Limit theorems for stochastic processes* (2013)](https://www.google.co.kr/books/edition/Limit_Theorems_for_Stochastic_Processes/mSD4CAAAQBAJ?hl=ko&gbpv=1&printsec=frontcover), and it is very important to prove asymptotic properties of my statistical estimator.
The situation can be much like simplified as follows.
**Here is a question.**
Let our discretized process be
\begin{align}
X^n\_t = \sum\_{i=1}^{[nt]} \left(W\_{\frac{i}{n}} - W\_{\frac{i-2}{n}}\right),
\end{align}
where $W$ is a standard Wiener process.
My intuition says that it would converge stably in law to $2W$ (in the Skorokhod space $\mathbb{D}([0,1])$), i.e.,
\begin{align}
X^n \xrightarrow{\mathcal{L}-s} 2W.
\end{align}
But I'm not sure it is ok to apply Theorem IX.7.28 to show this, because the predictable part (denote it as $B^n\_t$) converges to Wiener process itself which has "infinite variation":
\begin{align}
B^n\_t &:= \sum\_{i=1}^{[nt]} \mathbb{E}\_{\frac{i-1}{n}} \left[ W\_{\frac{i}{n}} - W\_{\frac{i-2}{n}} \right] \\
&= \sum\_{i=1}^{[nt]} \left(W\_{\frac{i-1}{n}} - W\_{\frac{i-2}{n}}\right) \\
&= W\_{\frac{[(n-1)t]}{n}} \xrightarrow{\mathbb{P}} W\_t,
\end{align}
while Jacod and Shiryaev's theorem says that $B^n\_t$ should converge to a predictable "finite variation" process $B\_t$.
So how can I show this?
**Here is a theorem.**
Without considering the truncation function or jumps of $X$, the theorem says:
>
> **IX.7.28 Theorem.** For every càdlàg process $X$, we use the following notation:
> \begin{align\*}
> X\_t^n = X\_{[nt]/n},\qquad\qquad \Delta\_i^n X = X\_{i/n} - X\_{(i-1)/n} = \Delta X^n\_{i/n},
> \end{align\*}
> We also consider the discretized process of the form
> \begin{align\*}
> X\_t^n = \sum\_{i=1}^{[nt]} \chi\_i^n,
> \end{align\*}
> where each $\chi\_i^n$ is $\mathcal{F}\_{i/n}$-measurable.
> Assume that each $\chi\_i^n$ is square-integrable, and $X$ is a continuous and $\mathbb{E}[|X\_t|^2] < \infty$ for all $t$ with the canonical decomposition $X\_t = B\_t + M\_t$ where $B\_t$ is predictable finite variation process and $M\_t$ is square-integrable local martingale.
>
>
> Suppose also that for all $t>0$ and all uniformly integrable martingale $N$ which are orthogonal to $X$ we have
> \begin{align\*}
> \sup\_t \left| \sum\_{i=1}^{[nt]} \mathbb{E} \_{\frac{i-1}{n}}[\chi\_i^n] - B\_t \right| &\xrightarrow{\mathbb{P}} 0, \\
> \sum\_{i=1}^{[nt]} \mathbb{V}\_{\frac{i-1}{n}}[\chi\_i^n] &\xrightarrow{\mathbb{P}} \langle M, M \rangle\_t + \langle w \cdot W', w \cdot W' \rangle\_t \\
> \sum\_{i=1}^{[nt]} \mathbb{E}\_{\frac{i-1}{n}}[\chi\_i^n \Delta\_i^n M] &\xrightarrow{\mathbb{P}} \langle M, M \rangle\_t, \\
> \sum\_{i=1}^{[nt]} \mathbb{E}\_{\frac{i-1}{n}}[\chi\_i^n \Delta\_i^n N] &\xrightarrow{\mathbb{P}} 0.
> \end{align\*}
> Then there is a very good canonical Wiener extension of $(\Omega, \mathcal{F}, \{\mathcal{F}\_t\}, \mathbb{P})$ with Wiener process $W'$ and a continuous $X$-biased $\mathcal{F}$-progressive conditional martingale PII $X'$ on this extension such that
> \begin{align\*}
> X^n \xrightarrow{\mathcal{L}-s} X' = X + w \cdot W',
> \end{align\*}
> where $w \cdot W' = \int w\,dW'$ and $w$ is a predictable process.
>
>
>
If I have any misunderstanding please let me know.
Any help will be appreciated.
Thanks,
|
https://mathoverflow.net/users/159685
|
Convergence of discretized process when its predictable part converges to infinite variation process
|
Let $h\_i:=W\_{i/n}-W\_{(i-1)/n}$ and $n\_t:=\lfloor nt\rfloor$, so that $t-1/n\le n\_t/n\le t$. Then
$$X^n\_t=\sum\_{i=1}^{n\_t}(h\_{i-1}+h\_i)
=\sum\_{i=1}^{n\_t}h\_{i-1}+\sum\_{i=1}^{n\_t}h\_i=W\_{n\_t/n-1/n}-W\_{-1/n}+W\_{n\_t/n},$$
whence
$$|X^n\_t-2W\_t|\le|W\_{n\_t/n-1/n}-W\_t|+|W\_{-1/n}|+|W\_{n\_t/n}-W\_t|$$
and, for each real $u>0$,
$$P(\sup\_{0\le t\le 1}|X^n\_t-2W\_t|>3u)\le2nP(\max\_{0\le s\le2/n}|W\_s|>u)+P(|W\_{-1/n}|>u).$$
Next, $P(|W\_{-1/n}|>u)=P(|W\_1|>u\sqrt n)\to0$ (as $n\to\infty$) and
$$
\begin{aligned}
2nP(\max\_{0\le s\le2/n}|W\_s|>u)
&=2nP(\max\_{0\le t\le1}|W\_t|>u\sqrt{n/2}) \\
&\le2nP(\max\_{0\le t\le1}W\_t>u\sqrt{n/2}) \\
&=4nP(W\_1>u\sqrt{n/2}) \\
&\le2n\exp\{-u^2n/4)\to0.
\end{aligned}
$$
So, $X^n\to2W$ in probability in $C[0,1]$ and hence in probability in $D[0,1]$ and hence stably in $D[0,1]$, as desired.
|
1
|
https://mathoverflow.net/users/36721
|
396868
| 163,865 |
https://mathoverflow.net/questions/396827
|
3
|
Consider the symmetric group $S\_n$ under the uniform distribution. For integer $k > 1$, suppose we draw $k$ elements $s\_1, \dots, s\_k$ independently at random. What is the probability that there exists at least one rearrangement of the $s\_i$ that composes to the identity?
|
https://mathoverflow.net/users/173490
|
Probability that k randomly drawn permutations can be arranged to compose to the identity
|
Only a partial answer, which however is too long for a comment: Let $p\_{n,k}$ denote the given probability. Then we have
(1) $p\_{n,1} = p\_{n,2} = \frac{1}{n!}$, $p\_{n,3} = \frac{2\cdot n!-p(n)}{(n!)^2}$.
(2) $p\_{1,k}=1$, $p\_{2,k}=\frac{1}{2}$.
(3) $\frac{1}{n!}\leq p\_{n,k}\leq \frac{(k-1)!}{n!}$.
(4) For $k$ fixed, most likely $p\_{n,k}\sim \frac{(k-1)!}{n!}$ for large $n$.
(5) $p\_{3,4} = \frac{77}{216}$, $p\_{3,5} = \frac{139}{324}$, $p\_{3,6} = \frac{101}{216}$, $p\_{4,4} = \frac{35}{216}$, $p\_{4,5} = \frac{3257}{10368}$ and $p\_{5,4} = \frac{9533}{216000}$.
Clearly $p\_{n,1}=\frac{1}{n!}$ and for $k\geq 2$
\begin{equation\*}
p\_{n,k} = \frac{1}{(n!)^k} \sum\_{x\_1, \ldots, x\_{k-1}\in \Sigma\_n} f(x\_1, \ldots, x\_{k-1})
\end{equation\*}
where $f(x\_1, \ldots, x\_{k-1})$ denotes the number of elements $x\_k$ in the symmetric group $\Sigma\_n$ such that $x\_{\sigma(1)} x\_{\sigma(2)} \ldots x\_{\sigma(k)}=1$ for some rearrangement $\sigma\in \Sigma\_k$. This equation is equivalent to $x\_k^{-1} = x\_{\tau(1)} x\_{\tau(2)} \ldots x\_{\tau(k-1)}=1$ for some $\tau\in \Sigma\_{k-1}$, i.e. $x\_k$ should be the inverse of some rearrangement of the product $x\_1 x\_2\ldots x\_{k-1}$. Since the number of rearrangements is between $1$ and $(k-1)!$ we obtain (3) and (1) for $k=1$, $2$. If $n\leq 2$, $\Sigma\_n$ is abelian so there is exactly $1$ rearrangement, proving (2).
Moreover for generic elements $x\_1, \ldots, x\_{k-1}$ we have $f(x\_1, \ldots, x\_{k-1}) = (k-1)!$ indicating why (4) should hold. (I havent written down the details, hence the most likely above). Finally consider $p\_{n,3} = \frac{1}{(n!)^3} \sum\_{x\_1, x\_2\in \Sigma\_n} f(x\_1,x\_2)$. Here $f(x\_1,x\_2)$ equals $1$ if $x\_1x\_2=x\_2 x\_1$ and $2$ otherwise. Thus, with $X = \{ x,y\in \Sigma\_n | xy=yx\}$ we have $p\_{n,3} = \frac{1}{(n!)^3} \left(1\cdot |X| + 2\cdot ((n!)^2-|X|)\right)$. Note that
\begin{equation\*}
|X| = \sum\_{x\in \Sigma\_n} |C\_G(x)| = \sum\_{x\in \Sigma\_n} \frac{|\Sigma\_n|}{|x^{\Sigma\_n}|},
\end{equation\*}
where $x^{\Sigma\_n}$ denotes the conjugacy class of $x$. The last sum clearly equals $n!$ times the number of conjugacy classes in $\Sigma\_n$, i.e. $n! p(n)$. Plugging this in gives the desired formula for $p\_{n,3}$ proving the last part of (1). Finally, point (5) was obtained by direct computer calculation.
|
3
|
https://mathoverflow.net/users/65801
|
396880
| 163,869 |
https://mathoverflow.net/questions/396833
|
1
|
$$
x\_{n}=\sum^{n-1}\_{i=0} {a\_i x\_{n-1-i}}
$$
where
$$
\sum^{+\infty}\_{i=0} {a\_i}=1,1>a\_i>0,1>x\_i>0
$$
In fact, the specific problem (comes from probability theory) I want to solve is that:
$0<d<0.2$ is a constant.
$E\_n(p) \in C[0,1]$is a function of p, $E\_0(p)=p$, and $E\_n(p)$ can be defined by：
$$
E\_n(p)=a\_n(p)+\sum^{n-1}\_{i=0} {a\_{n-1-i}(p) E\_i(p)}
$$
$$
if p>1-5d, then\ a\_n(p)=0,
$$$$
otherwise,\ a\_n(p)=l\_n(p)
$$
where
\begin{eqnarray}
l\_n(p)=
\begin{cases}
p, & n=0 \cr
\prod^{n-1}\_{k=0}{(1-p-d k)} (p+d n), & 1\leq n\leq 4 \cr
\prod^{4}\_{k=0}{(1-p-d k)} (1-p-5d)^{n-5} (p+5d),& n \geq 5 \cr
\end{cases}
\end{eqnarray}
I am sure that there is a function $u \in C[0,1]$ s.t.
\begin{eqnarray}
\lim\limits\_{n\to+ \infty}|| E\_n(p)-u(p)||\_{max}=0
\end{eqnarray}
but I do not know how to prove the convergence.
It is bounded, but it seems that the sequence is not monotonous.I tried some elementary methods to simplify the question, but we still need to prove $E\_{n+1}-E\_{n}$->0.
I found that $E\_{n+1}-E\_{n}$ is not decreasing, and I have no good ideas now.
By the way, I guess there is no easy(elementary) expression, but I cannot prove the existence yet.
|
https://mathoverflow.net/users/151339
|
How to prove the convergence of this kind of sequence?
|
If $p>1-5d$, then $a\_n(p)=0$ for all $n\ge1$ and hence $E\_n(p)=0$ for all $n\ge1$.
So, without loss of generality $0\le p\le1-5d$. Then, letting $E\_{-1}:=1$, for all $n\ge0$ we get
\begin{equation\*}
E\_n=\sum\_{i=-1}^{n-1}l\_{n-1-i}E\_i,
\end{equation\*}
where $E\_n:=E\_n(p)$ and $l\_n:=l\_n(p)$.
Here is the key point: Using the fact that $(l\_5,l\_6,\dots)$ is a geometric progression, one can check that
\begin{equation\*}
E\_{n+1}=\sum\_{j=0}^5 b\_j E\_{n-j} \tag{1}
\end{equation\*}
for all $n\ge4$, where
\begin{equation\*}
b\_j:=l\_j+(p+5d-1)l\_{j-1}.
\end{equation\*}
Moreover, one can check that
\begin{equation}
\text{$b\_0,\dots,b\_5$ are $>0$ and $\sum\_{j=0}^5 b\_j=1$. }\tag{2}
\end{equation}
The [general solution](https://en.wikipedia.org/wiki/Recurrence_relation#Roots_of_the_characteristic_polynomial) of the [linear difference equation](https://en.wikipedia.org/wiki/Linear_difference_equation#Characteristic_equation_and_roots) (1) is given by
\begin{equation\*}
E\_n=\sum\_{k=0}^5 C\_k(n) z\_k^n, \tag{3}
\end{equation\*}
where the $z\_k$'s are the roots of the characteristic polynomial
\begin{equation\*}
P(z):=z^6-\sum\_{j=0}^5 b\_j z^{5-j}=z^6-\sum\_{j=0}^5 b\_{5-j}z^j
\end{equation\*}
and the $C\_k(n)$'s are polynomials in $n$ (whose degrees are nonzero if the corresponding roots $z\_k$ are multiple ones). Moreover, in view of (2), we have the following:
(i) $1$ is a root of $P(z)$; say, $z\_0=1$.
(ii) The multiplicity of $z\_0=1$ is $1$, since $P'(1)=6-\sum\_{j=0}^5 b\_j(5-j)\ge6-5\sum\_{j=0}^5 b\_j=1>0$. So, $C\_0:=C\_0(n)$ does not depend on $n$.
(iii) $|z\_k|<1$ for all $k=1,\dots,5$: Indeed, if $0=P(z)=z^6-\sum\_{j=0}^5 b\_{5-j}z^j$ for some complex $z$ with $|z|\ge1$, then, by the triangle inequality, $|z|^6\le\sum\_{j=0}^5 b\_{5-j}|z|^j\le\sum\_{j=0}^5 b\_{5-j}|z|^5=|z|^5$, so that $|z|\le1$ and hence $|z|=1$. Moreover, similarly the condition $|z|=1$ for a root $z$ of $P(z)$ implies the contradictory inequality $|z|^6<|z|^5$ unless the complex numbers $1$ and $z$ lie on the same ray emanating from $0$ -- that is, unless $z=1$.
It follows from (3) and (i)--(iii) that
\begin{equation}
E\_n\to C\_0 \tag{4}
\end{equation}
as $n\to\infty$. Of course, $E\_n$ and $C\_0$ may/will depend on $d,p$. The convergence in (4) holds for each pair $(d,p)$ satisfying the OP conditions $0<d<1/5$ and $0\le p\le1-5d$.
|
1
|
https://mathoverflow.net/users/36721
|
396896
| 163,873 |
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