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https://mathoverflow.net/questions/397443
|
6
|
Consider $n$ i.i.d spherically distributed random vectors $z\_1 ,\cdots , z\_n \sim \text{Unif}(\mathbb{S}^{d-1})$. What is the best lower bound on $n$ for which whp there exists a constant $c>0$ such that the following bound holds for all $v\in \mathbb{R}^d\backslash \{0\}$:
\begin{equation}
cn\leq \left\vert\left\{i:\langle z\_i,v \rangle>0 \right\} \right\vert
\end{equation}
|
https://mathoverflow.net/users/59151
|
Almost evenly distributed spherical random vectors
|
$\newcommand\PP{\mathbb P}$
Surely $n\_\min \lesssim d$, because it works for $c = 1/4$ and $n=160d$.
We use that the number of "distinct" $v$ with respect to the classifiers $\textrm{sgn}\langle \cdot, z\_i \rangle$ is
$$ \sum\_{i=0}^{d-1} \binom{n-1}{i} \le \left( \frac{ne}{d} \right)^d $$
The proof can be found in
[Bürgisser and Cucker (2013), Lemma 13.7]. (Anyone has a better reference?)
Let $X = \lvert \left\{ i ~:~ \langle z\_i, e\_1 \rangle \right\} \rvert$ for a basis vector $e\_1$.
Then, by the union bound
$$ \PP \left[ ~\exists v ~~ \lvert \left\{ i ~:~ \langle z\_i, v \rangle > 0 \right\}\rvert < cn \right]
\\
\le \left( \frac{ne}{d} \right)^d \PP \left[ X < cn \right]
\\
\le \exp(d\log(n) - \frac{n}{16} + d - d \log{d})
\\
= \exp(\log(160)d - 9d) \xrightarrow{d \to \infty} 0$$
where we used the Chernoff bound on $\PP[X < cn]$, in the form
$$ \PP\left[X \le (1 - \delta) \mathbb E[X] \right] \le \exp(-\frac12 \delta^2 \mathbb E[X]). $$
As noted in the comment by Sandeep Silwal, $n\_{\text{min}} \ge d$ due to the strict inequality sign in the question. So the answer to the original question is $n\_{\text{min}} = \Theta(d)$.
|
6
|
https://mathoverflow.net/users/122628
|
397464
| 164,076 |
https://mathoverflow.net/questions/396191
|
4
|
I am able to show that any $k$-dimensional subspace of $\mathbf{R}^{Ck\log(k)}$ must contain a unit vector $x$ such that $\|x\|\_{\infty} \ge c\sqrt{1/\log(k)}$ for a small enough constant $c$.
But is there a $k$-dimensional subspace of $\mathbf{R}^{Ck\log(k)}$ such that *every* nonzero vector $x$ in the subspace satisfies $\|x\|\_{\infty} \ge (1/\text{poly}(\log(k)))\|x\|\_2$? And is there any random rotation that can be applied to a given $k$-dimensional subspace such that the subspace obtained after rotating has this property with some constant probability?
|
https://mathoverflow.net/users/307001
|
Subspaces with all vectors having large $\|x\|_{\infty}/\|x\|_2$ value
|
Since this hasn't been answered, I think the answer is no.
Indeed, suppose that there exists a $k$-dimensional subspace $V$ of $\mathbb{R}^n$ such that for all $\mathbf{x}\in V$,
\begin{equation}
\frac{\|\mathbf{x}\|\_2}{D}\leq \|\mathbf{x}\|\_{\infty}\leq \|\mathbf{x}\|\_2,
\end{equation}
for some value of $D\geq 1$.
Equivalently, there exists a linear transformation $T:\mathbb{R}^k\to \mathbb{R}^n$ that maps onto $V$ and preserves the $2$-norm (namely, let the columns be an orthonormal basis of $V$), so that this is equivalent to
\begin{equation}
\frac{\|\mathbf{x}\|\_2}{D}\leq \|T\mathbf{x}\|\_{\infty}\leq \|\mathbf{x}\|\_2 \quad\forall \mathbf{x}\in \mathbb{R}^k.
\end{equation}
By considering the rows of $T$, this means there exists $n$ vectors $\mathbf{t}\_1,\ldots,\mathbf{t}\_n\in \mathbb{R}^k$ with at most unit length such that for any vector $\mathbf{x}\in S^{k-1}$,
\begin{equation}
\max\_{i\in [n]} \vert \langle \mathbf{t}\_i,\mathbf{x}\rangle\vert\geq \frac{1}{D}.
\end{equation}
Consider a random vector $\mathbf{X}$ drawn uniformly from $S^{k-1}$. The distribution of $\vert \langle \mathbf{t}\_i/\|\mathbf{t}\_i\|\_2,\mathbf{X}\rangle\vert$ does not depend on $\mathbf{t}\_i$, and by concentration of measure on the sphere, the probability that $\vert \langle \mathbf{t}\_i,\mathbf{X}\rangle\vert$ exceeds $O\left(\sqrt{\frac{\log n}{k}}\right)$ is less than $1/n$ (as $\|\mathbf{t}\_i\|\_2\leq 1$). If $D\leq c\sqrt{\frac{k}{\log n}}$ for some sufficiently small constant $c$, we obtain a contradiction as there exists a vector violating the desired inequality. Therefore, $k\leq CD^2\log n$ for some large enough constant $C$. For your regime of $D=\text{polylog}(n)$, this unfortunately implies a polylogarithmic upper bound on the dimension of $k$.
|
4
|
https://mathoverflow.net/users/170770
|
397467
| 164,078 |
https://mathoverflow.net/questions/397418
|
5
|
$\DeclareMathOperator\AE{AE}\DeclareMathOperator\Lip{Lip}$Let $\AE(X)$ denote the Arens-Eells space on a Banach space $X$. Consider the map:
$$
\begin{aligned}
\delta: X & \rightarrow \AE(X)
\\
x&\mapsto \delta\_x
\end{aligned}
$$
Is the map $\delta$ ever Gâteaux (or Fréchet) differentiable?
---
Recall that $\AE(X)$ is the/a [pre-dual](https://arxiv.org/abs/1611.01812) of the Banach space $\Lip\_0(X)$ whose elements are Lipschitz functions sending $0\in X$ to $0\in \mathbb{R}$ (with norm sending any $f\in \Lip\_0(X)$ to its (unique) [Lipschitz constant](https://en.wikipedia.org/wiki/Lipschitz_continuity) $\Lip(f)$), $\delta\_x$ denotes the evaluation map defined on Lipschitz functions $f\in \Lip\_0(X)$ by
$$
\delta\_x(f):= f(x),
$$
and $\AE(X)$ is normed using the [dual-norm](https://en.wikipedia.org/wiki/Dual_norm) construction; i.e.:
$$
\|F-G\|:=\inf\_{f \in \Lip\_0(X),\, \Lip(f)\leq 1} F(f)-G(f).
$$
|
https://mathoverflow.net/users/318661
|
Differentiability of the map $x\mapsto \delta_x$ in the Arens-Eells/Lipschitz-free space
|
This fails for $X = \mathbb{R}$, and hence for every nonzero Banach space, since they all contain copies of $\mathbb{R}$. If the map $t \mapsto \delta\_t$ were differentiable in either sense then for every bounded linear functional $F$ on $AE(\mathbb{R})$ the map $t\mapsto F(\delta\_t)$ would be differentiable. Recalling that the dual of $AE(\mathbb{R})$ is ${\rm Lip}\_0(\mathbb{R})$, differentiability at $t$ would imply that every Lipschitz function on $\mathbb{R}$ is differentiable at $t$, which is obviously false.
|
7
|
https://mathoverflow.net/users/23141
|
397476
| 164,082 |
https://mathoverflow.net/questions/383874
|
5
|
In several sources (for instance on page 58 of the first ed. of Crandall & Pomerance book on prime numbers or at the end of [this paper](https://www.tandfonline.com/doi/abs/10.1080/00029890.2007.11920459) by J. H. Jaroma), I have seen a result that goes like this:
>
> Let $p$ be an odd prime congruent to $-1$ modulo $4$. Then $2p+1$ is a
> prime iff $(2p+1) \mid (2^{2p}-1)$.
>
>
>
Do you know if the hypothesis $p \equiv -1 \pmod 4$ may be removed? If the result is still valid for primes congruent to $1$ modulo $4$, I wonder why it is that it is not mentioned in the sources I referred to above...
|
https://mathoverflow.net/users/99957
|
On a result of Euler on pseudoprimes
|
Yes, the result holds for every odd prime number $p$... I certainly find it somewhat "strange" that it is only stated for primes congruent to $-1$ modulo $4$ in several places:
**Proposition.** Let us suppose that $p$ is an odd prime number and that $2p+1$ divides $2^{2p}-1$. Then, $2p+1$ is a prime number.
*Proof.* (I learnt it from J. I. Restrepo) For the sake of contradiction, let us suppose that $2p+1$ is not a prime number and that $q$ is a prime number dividing $2p+1$. From the hypothesis and Fermat's little theorem we have that
\begin{eqnarray}
2^{2p} \equiv 1 \pmod{q}\\
2^{q-1} \equiv 1 \pmod{q}.
\end{eqnarray}
It follows from these congruences that $\mathrm{ord}\_{q}(2)=:\mathfrak{o}$ is a common positive divisor of $2p$ and $q-1$; since there are only four positive divisors of $2p$ and $q-1 \leq (p-\frac{1}{2})<p$, we get that $\mathfrak{o}=1$ or $\mathfrak{o}=2$. Given that $q$ is a prime number, $\mathfrak{o}$ can't be equal to $1$. Hence, $\mathfrak{o}=2$ and $q=3$.
From what we have established in the above paragraph, we obtain that $2p+1=3^{\ell}$ for some $\ell \in \mathbb{Z}^{+} \setminus \{1\}$. This implies that $9 \mid (2^{2p}-1)$ which is an absurdity because $9 \mid (2^{n}-1)$ iff $6 \mid n$. **Q.E.D.**
|
6
|
https://mathoverflow.net/users/1593
|
397479
| 164,083 |
https://mathoverflow.net/questions/397477
|
4
|
Say that a structure $\mathcal{M}$ is **amorphic** iff for every finite $\overline{a}\in\mathcal{M}$ and bi-infinite $X\subseteq\mathcal{M}$ there is some automorphism $\alpha\in Aut(\mathcal{M})$ fixing $\overline{a}$ pointwise but not respecting $X$ (that is, either $\alpha[X]\not=X$ or $\alpha[\mathcal{M}\setminus X]\not=\mathcal{M}\setminus X$). An amorphic structure can be "given an amorphous copy" in a symmetric extension of the universe; precisely, there is a symmetric extension $V\subset N\subset V[G]$ and a structure $\mathcal{A}\in N$ such that $\mathcal{A}$'s underlying set is amorphous in $N$ but $V[G]\models\mathcal{A}\cong\mathcal{M}$.
Now say that a countable complete theory $T$ is $\kappa$-amorphic iff there is a $\mathcal{M}\models T$ with $\vert\mathcal{M}\vert=\kappa$, and amorphic iff $T$ is $\kappa$-amorphic for some $\kappa$. For example, the empty theory is trivially $\kappa$-amorphic for every $\kappa$. I'm curious about what implications exist between the various amorphicities. To get started, my main question is:
>
> Does amorphic imply $\omega$-amorphic?
>
>
>
Note that $\omega$-amorphicity is a $\Sigma^1\_3(L\_\omega)$ property, while on the face of things amorphicity is $\Sigma^0\_2(V)$, so this would be a great improvement in terms of the complexity of amorphicity. I vaguely recall a negative result due to Shelah here, but I can't find it (and "due to Shelah" doesn't really narrow the search space much).
|
https://mathoverflow.net/users/8133
|
Sizes of "nearly amorphous" models
|
Amorphicity implies strong minimality and $\omega$-categoricity, which together imply $\kappa$-amorphicity for any $\kappa$.
Assume that $T$ is amorphic. To see that $T$ is $\omega$-categorical, we proceed by induction. It is clear that the type space $S\_1(T)$ must be finite, otherwise we could form a bi-infinite set $\bigvee$-definable over a finite set of parameters, which would spoil amorphicity of any given model of $T$. Now suppose that we know that $S\_n(T)$ is finite. Assume for the sake of contradiction that $S\_{n+1}(T)$ is infinite. Then there must be some type $p(\bar{x}) \in S\_n(T)$ with infinitely many extensions to an $(n+1)$-type. Since $S\_n(T)$ is finite, $p(\bar{x})$ must be realized in any model of $T$. Fix a model $\mathcal{M}$ and let $\bar{a}$ realize $p(\bar{x})$. If there is some formula $\varphi(x,\bar{a})$ such that $\varphi(\mathcal{M},\bar{a})$ is bi-infinite, then we have a failure of amorphicity, so assume that for every formula $\varphi(x,\bar{y})$, $\varphi(\mathcal{M},\bar{a})$ is either finite or co-finite. Since there are infinitely many extensions of $p(\bar{x})$ to an $(n+1)$-type, we must have that $S\_1(\bar{a})$ is infinite and scattered (i.e., has ordinal Cantor-Bendixson rank or equivalently contains no perfect subset), but this implies that there are infinitely many isolated points in $S\_1(\bar{a})$, so we can again form a bi-infinite set $\bigvee$-definable over a finite set of parameters, which spoils amorphicity. Therefore it must be the case that $S\_{n+1}(T)$ is finite, and thus by induction we have that $S\_n(T)$ is finite for all $n < \omega$. So by the Engeler–Ryll-Nardzewski–Svenonius theorem, $T$ is $\omega$-categorical.
It is clear that if $\mathcal{M}$ is amorphic, then it can have no bi-infinite definable subsets, since any such subset defined by a formula $\varphi(x,\bar{a})$ would have $\bar{a}$ witnessing the failure of amorphicity. So we have that $\mathcal{M}$ is minimal. Finally, $\omega$-categorical theories eliminate the quantifier $\exists^\infty$ (i.e., for any formula $\varphi(\bar{x},\bar{y})$, there is a formula $\psi(\bar{y})$ such that for any $\bar{a}$, $\varphi(\bar{x},\bar{a})$ has infinitely many solutions if and only if $\psi(\bar{a})$ holds) and minimal sets are always strongly minimal in such theories, so $T$ is strongly minimal.
To see that strong minimality and $\omega$-categoricity imply $\kappa$-amorphicity for any $\kappa$, note that any such theory is totally categorical and therefore has every model saturated. Saturated models are homogenous, so the only automorphism invariant sets over a finite tuple are those actually definable over it.
|
6
|
https://mathoverflow.net/users/83901
|
397481
| 164,084 |
https://mathoverflow.net/questions/397455
|
3
|
Suppose $H$ is a Hilbert space with orthonormal basis $\{e\_i\}\_{i\in \mathbb N}$. To every operator $T$, we associate a infinite matrix $[T\_{ij}]$, where $T\_{ij}=\left<Te\_j,e\_i\right>$. We know that for any trace class operator $T$, the trace norm is $||T||\_1=\operatorname{Tr}(|T|) $.
Q). Suppose $T$ is a trace class operator and $S$ is such that its matrix entries are either equal to the matrix entries of $T$ or they vanish (possibly at infinite number of points). Can I say that $||S||\_1\leq ||T||\_1$? If not, is there any finite upper bound to such $S$ obtained from $T$?
|
https://mathoverflow.net/users/145729
|
Trace norm of operators obtained by restricting the matrix of a trace class operator
|
Here's an algorithm for testing an ad-hoc conjecture $C$ about Hilbert space operators. :-)
0. Set up the runtime environment correctly by loading the information "Most conjectures are false" into short term memory.
1. Test $C$ against the zero and the identity operator.
2. Test $C$ against finite-dimensional diagonal matrices.
3. Test $C$ against multiplication operators on $\ell^2$ and $L^2$.
4. Test $C$ against the following $2 \times 2$-matrices:
$$
\begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix},
\quad
\begin{pmatrix}
1 & 1 \\
0 & 1
\end{pmatrix},
\quad
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix},
\quad
\begin{pmatrix}
0 & 1 \\
-1 & 0
\end{pmatrix},
\quad
\begin{pmatrix}
1 & 1 \\
1 & 1
\end{pmatrix}.
$$
5. Test $C$ against simple modifications (appropriate to the setting of $C$) of the matrices from Step 4.
6. Write a computer programm to test $C$ against randomly generated $2 \times 2$-matrices; make sure to restrict the matrices that your random generator creates to the set of matrices that occur in $C$.
7. Repeat Step 6 with $3 \times 3$-matrices.
8. If you have not found a counterexample yet, there might be a reason to believe that $C$ holds.
Of course this should not be taken completely seriously - but often it works.
In my experience, for many ad-hoc conjectures the algorithm stops at Step 5 or earlier. The question from the OP adds another data point to this pattern:
The matrix
$$
T =
\begin{pmatrix}
1 & 1 \\
1 & 1
\end{pmatrix}
$$
has trace norm $2$, but the matrix
$$
S =
\begin{pmatrix}
1 & 1 \\
1 & 0
\end{pmatrix}
$$
has trace norm
$$
\frac{1 + \sqrt{5}}{2} + \frac{\lvert 1 - \sqrt{5} \rvert}{2} = \sqrt{5} > 2.
$$
|
10
|
https://mathoverflow.net/users/102946
|
397482
| 164,085 |
https://mathoverflow.net/questions/397478
|
1
|
Let $\mathbb{P}$ be a Borel probability measure on $\mathbb{R}^n$ with $\int\_{x \in \mathbb{R}^n} \|x\|^p\mathbb{P}(dx)<\infty$. For which $p\in [1,\infty)$ (other than $p=2$) is the following optimality set non-empty:
$$
X(\mathbb{P}):=\left\{
x\in \mathbb{R}^n:\, \int\_{u \in \mathbb{R}^n}\|u-x\|^p\mathbb{P}(du)=\inf\_{x'\in \mathbb{R}^n}
\int\_{u \in \mathbb{R}^n}\|u-x'\|^p\mathbb{P}(du)
\right\}.
$$
More importantly, for which such $p$ does there exist a continuous selector $S:\mathcal{P}\_p(\mathbb{R}^n)\rightarrow \mathbb{R}^n$ with:
$$
S(\mathbb{P})\in X(\mathbb{P})
?
$$
Note, here, we equip $\mathcal{P}\_p(\mathbb{R}^n)$ with the $L^p$-Wasserstein distance.
This question is related to [this post](https://mathoverflow.net/questions/397420/holder-continuous-barycenter-maps).
|
https://mathoverflow.net/users/36886
|
$L^p$-barycenters via continuous selectors
|
There exists a minimizer for all $p\in [1,\infty[$. To see that, let $R > 0$ be big enough so that
$$
\int\_{B\_R(0)}\|u\|^pd\mathbb{P}(u) > \frac{1}{2}\int\_{\mathbb{R}^n}\|u\|^pd\mathbb{P}(u).
$$
Then if $x > B\_{3R}(0)$, we see that
$$
\int\_{\mathbb{R}^n}\|u-x\|^pd\mathbb{P}(u) > \int\_{\mathbb{R}^n}\|u\|^pd\mathbb{P}(u)
$$
and hence is
$$
\inf\_{x\in\mathbb{R}^n}\int\_{\mathbb{R}^n}\|u-x\|^pd\mathbb{P}(u) = \inf\_{x\in B\_{3R}(0)}\int\_{\mathbb{R}^n}\|u-x\|^pd\mathbb{P}(u).
$$
Now, just note that $x\mapsto \int\_{\mathbb{R}^n}\|u-x\|^pd\mathbb{P}(u)$ is continuous so it attains a minimum on the compact set $B\_{3R}(0)$. Moreover, we can say that the minimizer is unique since the function is convex (that follows from the convexity of $x\mapsto \|x\|^p$).
|
2
|
https://mathoverflow.net/users/313861
|
397484
| 164,086 |
https://mathoverflow.net/questions/397472
|
4
|
Some time ago I was trying to find a closed form formula for the number of tuples $(a\_k)\_{k=1}^{n+s}$ of non-negative integers satisfying following conditions:
1. $\sum\_{k=1}^{n+s} a\_k = n$,
2. $\forall m \in \mathbb{N}\_0 \quad m < n \implies \sum\_{k=1}^{m+s} a\_k > m$,
where $n \in \mathbb{N}\_0 = \mathbb{N} \cup \{0\}$ and $s \in \mathbb{N}$. I think (a document I've written in Polish wasn't peer reviewed, so it'd just be my claim) I proved that the number of the mentioned tuples is given by
$$ \left( \binom{n +s}{n} \right) \frac{s}{n+s},$$
where symbol $\left( \binom{a}{b} \right)$ is a so-called multiset coefficient defined by
$$ \forall a,b \in \mathbb{N}\_0 \quad \left( \binom{a}{b} \right) =
\begin{cases}
\binom{a+b-1}{b}, &\text{ for }a, b \in \mathbb{N}, \\
0, &\text{ for } a = 0, b \in \mathbb{N}, \\
1, &\text{ for } a \in \mathbb{N}\_0, b = 0.
\end{cases} $$
If we were to substitute $s = 1$ in the formula for the tuples, we would get the formula for the $n^\text{th}$ Catalan number.
Also, using the mentioned formula I claimed that the following holds:
$$ \forall s \in \mathbb{N} \quad \sum\_{n=0}^\infty \left( 2^{-(2n+s)} \left( \binom{n +s}{n} \right) \frac{s}{n+s} \right) = 1. $$
I would like to ask whether either of these results are known (if they are correct) and if so, where could I find some paper regarding it. The proof (, which, I hope, is correct) of the first claim which I've written in the mentioned document is rather cumbersome, so I would be interested if this result could be proven in some relatively easy way.
|
https://mathoverflow.net/users/170491
|
Generalization of Catalan numbers
|
Let $a\_k$ be such a sequence and define $\lambda\_k := n - \sum\_{i=1}^{k}a\_i$ for $k=1,\ldots,n+s-1$. Then $\lambda = (\lambda\_1,\ldots,\lambda\_{n+s-1})$ is a partition with $\lambda\_1 \leq n$ and $\lambda\_{s+m} \leq n- 1- m$ for all $0 \leq m < n$. Its transpose partition $\lambda^t = (\lambda^t\_1,\ldots,\lambda^t\_{n})$ satisfies $\lambda^t\_i \leq n+s-2-i$ for all $i=1,\ldots,n$. In this way we obtain a bijection between your sequences $a\_k$ and partitions contained in (i.e., with Young diagram is contained in) the "staircase shape" partition $(n+s-2,n+s-3,\ldots,s)$.
There are various formulas for the number of partitions contained in a given partition shape: for instance, a determinantal formula due to MacMahon. However, in your particular case the staircase shape has a straight product formula for this number, which is special. This is probably vast overkill, but the number of partitions contained in the staircase shape is the same as the number of plane partitions of this shape, with entries at most one. In general, the number of plane partitions of staircase shape with entries at most $m$ has a product formula, due to Proctor; see Corollary 4.1 of his paper "Odd symplectic groups" (<https://doi.org/10.1007/BF01404455>) or exercerise 7.101 of Stanley's Enumerative Combinatorics, Vol. 2.
|
4
|
https://mathoverflow.net/users/25028
|
397488
| 164,088 |
https://mathoverflow.net/questions/397456
|
6
|
Consider a bump function supported in the ball of radius $1$, that is $\psi:\mathbb R^n\to\mathbb R$ such that
1. $\ \psi(x)>0$ for $|x|<1$
2. $\ \psi(x)=0$ for $|x|\geq 1$
3. $\ \psi\in C^\infty$.
Is it possible to find such a function $\psi$ that satisfies also one of the following conditions? For all $i,j=1,\dots,n$
4. $\ \displaystyle \frac{\partial\_{x\_i}\psi(x)}{\psi(x)}\to 0\ $, $\ \displaystyle\frac{\partial\_{x\_i}\partial\_{x\_j}\psi(x)}{\psi(x)}\to0\ $ as $|x|\to1\ $
5. $\ \displaystyle\lim\frac{\partial\_{x\_i}\psi(x)}{\psi(x)}\in\mathbb R\ $, $\ \displaystyle\lim\frac{\partial\_{x\_i}\partial\_{x\_j}\psi(x)}{\psi(x)}\in\mathbb R\ $ as $|x|\to1\ $.
Intuitively this would mean that both $\psi$ and its derivatives vanish approaching the boundary of the ball, but the derivatives vanish faster than $\psi$ (or at least not slower).
A typical example of function satisfying conditions 1.-3. is given by
$$ \psi\_0(x) = \begin{cases} e^{-\frac{1}{1-|x|^2}} &\textrm{ if }|x|<1 \\[2pt] 0 &\textrm{ if }|x|\geq1\end{cases} $$
but 4.,5. are clearly not satisfied by $\psi\_0$ since
$$ \frac{\nabla\psi\_0(x)}{\psi\_0(x)} \,=\, \frac{2x}{(1-|x|^2)^2}\,\xrightarrow[|x|\to1]{}\,\infty \,.$$
|
https://mathoverflow.net/users/58793
|
Can I find a bump function $\psi$ such that $\nabla\log\psi$ vanishes too?
|
Elaborating the comment by Wojowu: If we take a look at $n=1$ and $\psi\in C^\infty\_{\text c}(\mathbb R)$ is a function satisfying conditions 1., 2. and 3. of your question, then for every $x\in]-1,1[$, we have, by smoothness of $\ln\psi$ on $]-1,1[$ and the fact that $\frac{\psi'}{\psi}$ is continuous on every $[-K,K]$ for $K\in]0,1[$,
$$\ln\psi(x)=\ln\psi(0)+\int\_0^x\frac{\psi'(s)}{\psi(s)}\,\mathrm ds.$$
Now, if we had $$\lim\_{s\to 1}\frac{\psi'(s)}{\psi(s)}=r$$ for any real number $r\in\mathbb R$, then the function $$s\mapsto\frac{\psi'(s)}{\psi(s)}$$ would be well-defined and continuous on $[0,1]$, and therefore we would have $$\lim\_{x\to1}\ln\psi(x)=\ln\psi(0)+\int\_0^1\frac{\psi'(s)}{\psi(s)}\,\mathrm ds\in\mathbb R,$$ which is impossible since $\lim\_{x\to1}\psi(x)=0$.
---
Similarly, for a natural number $n>1$, any $\phi\in C\_{\text c}^\infty(\mathbb R^n)$ with your properties 1., 2. and 3. defines a function
$$\psi:\mathbb R\to\mathbb R, t\mapsto\phi(t,0,0,\dots, 0)$$ and we can proceed our argumentation as above.
|
6
|
https://mathoverflow.net/users/129831
|
397503
| 164,091 |
https://mathoverflow.net/questions/397498
|
11
|
I start by saying that I am not an expert in this field and I apologize if the question is too elementary.
Let $K$ be a knot in $S^3$. I denote by $\pi\_1(K)$ the knot group, which is the fundamental group of its exterior:
$$ \pi\_1(K) = \pi\_1(S^3 \smallsetminus K) .$$
The *minimal number of generators* of a knot $K$ is the minimal number of generators of $\pi\_1(K)$.
I am searching for knots with an arbitrarily high minimal number of generators. In particular:
1. I found in <https://arxiv.org/abs/1007.3175> , Lemma 4.17, a reference: *Goodrick, R. (1968). Non-simplicially collapsible triangulations of In*. In this article, the author proves that the connected sum of $n$ copies of a two-bridge knot is a $m$-bridge knot with $m$>$n$. A sharper result should hold from *Schultens, Jennifer (2003). Additivity of bridge numbers of knots*. I cannot understand, though, how this should prove the statement. In particular, by [Knot theory question: bridge number vs. min generators of fundamental group of complement](https://mathoverflow.net/questions/103049/knot-theory-question-bridge-number-vs-min-generators-of-fundamental-group-of-c) , this should not imply the result. Maybe I am missing something in the article, I did not go through the details. However, the article is quite dated and there is, hopefully, a simple way to prove this fact nowadays. So the first question is: is there a simple way to prove that there are knots with an arbitrarily high minimal number of generators?
2. The techniques used seem to rely on the connected sum. What if we search for *prime* knots with an arbitrarily high minimal number of generators?
3. What if we search for *hyperbolic* knots with an arbitrarily high minimal number of generators?
Thank you in advance for the attention.
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https://mathoverflow.net/users/128408
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Knot groups with big number of generators
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If $\pi\_1(S^3\setminus K)$ has a presentation with $n$ generators then its representation variety $\mathrm{Hom}(\pi\_1(S^3\setminus K),SL\_2(\mathbb{C}))$ is a subvariety of $(SL\_2(\mathbb{C}))^n$, which has complex dimension $3n$, so any component will have complex dimension at most $3n$. So if you want the minimal number of generators to be arbitrarily large, you just have to find knots with high-dimensional representation varieties. (You can replace $SL\_2(\mathbb{C})$ with other groups if you prefer.)
Cooper and Long ("Remarks on the A-polynomial of a knot", section 8) show how to construct hyperbolic knots where there are components of arbitrarily large dimension. Start with some $n$-fold connected sum, and realize it as the closure of a braid $\beta$ such that the union of the closure $\hat\beta$ with its braid axis is hyperbolic. Lift $\hat\beta$ to the $p$-fold cyclic branched cover of $S^3$, branched over the braid axis -- this will again be $S^3$, since the axis is unknotted -- to get a new knot, which is hyperbolic for all large enough $p$. (This knot will be the closure of the braid $\beta^p$.) Then the representation variety of this new knot turns out to have a component of dimension at least $n$ as well.
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12
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https://mathoverflow.net/users/428
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397508
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https://mathoverflow.net/questions/397510
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4
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Following M. Ruzhansky and V. Turunen's book [Pseudo-Differential Operators and Symmetries](http://www.math.nagoya-u.ac.jp/%7Erichard/teaching/s2017/Ruzhansky_Turunen.pdf), in Chapter 3, Definition 3.1.25 (page 304), the space of periodic distributions is defined as follows (paraphrasing):
>
> **Definition 3.1.25 (Periodic Distributions)** The space of periodic distributions is defined as dual space $\mathcal{D}'(\mathbb{T}^n) = \mathcal{L}(C^{\infty}(\mathbb{T}^n),\mathbb{C})$ (i.e. of continuous linear operators $u:C^{\infty}(\mathbb{T}^n) \to \mathbb{C}$), where we write $u(\varphi) = \langle u , \varphi\rangle$ for any $\varphi\in C^{\infty}(\mathbb{T}^n)$. Furthermore, for any $\psi\in C^\infty(\mathbb{T}^n)$ or $L^p(\mathbb{T}^n)$, define the associated distribution to $\psi$ in the canonical sense, i.e. $u\_{\psi} \in \mathcal{D}'(\mathbb{T}^n)$ where for any $\varphi \in C^\infty(\mathbb{T^n})$
> \begin{align}u\_{\psi}(\varphi) = \langle u\_{\psi}, \varphi\rangle=\int\_{\mathbb{T}^n}\psi(x) \varphi(x) dx.\end{align}
>
>
>
Given the above definition, as with standard distributions, we can define a Fourier transform of periodic distributions with respect to Fourier transforms of functions.
>
> **Definition 3.1.8 (Periodic Fourier Transform)** Let $\mathcal{F}\_{\mathbb{T}^n} : C^{\infty}(\mathbb{T^n}) \to \mathcal{S}(\mathbb{Z}^n) $ be defined as
> $$ \begin{align} (\mathcal{F}\_{\mathbb{T}^n}f)(\xi):=\int\_{\mathbb{T}^n} f(x) e^{-2\pi i x \cdot \xi} dx \end{align}$$
> for $f \in C^{\infty}({\mathbb{T}^n})$, and its inverse be defined as
> $$\begin{align} (\mathcal{F}\_{\mathbb{T}^n}^{-1}h)(x) = \sum\_{\xi \in \mathbb{Z}^n} h(\xi) e^{2\pi i x \cdot \xi}\end{align}$$
> for $h\in\mathcal{S}({\mathbb{Z}^n})$.
>
>
>
>
> **Definition 3.1.27 (Fourier Transform of Periodic Distributions)** For any $u\in\mathcal{D}'(\mathbb{T}^n)$, define the Fourier transform $\mathcal{F}\_{\mathbb{T}^n}:\mathcal{D}'(\mathbb{T}^n) \to \mathcal{S}'(\mathbb{Z}^n)$ as
> $$\begin{align}\langle \mathcal{F}\_{\mathbb{T}^n} u, \varphi \rangle = \langle u, \mathcal{F}\_{\mathbb{T}^n}^{-1} \varphi(-\cdot) \rangle\end{align} \tag{1}$$
> for any $\varphi \in C^{\infty}(\mathbb{T}^n)$
>
>
>
**QUESTION**
------------
My question is, is the definition of the Fourier transform on periodic distributions, as given in Definition 3.1.27, **correct**? I feel that the definition should not involve the inverse Fourier transform, but instead be defined as
$$\begin{align} \langle \mathcal{F}\_{\mathbb{T}^n} u, \varphi \rangle = \langle u, \mathcal{F}\_{\mathbb{T}^n} \varphi \rangle. \tag{2}
\end{align}$$
**Reason**
----------
The reason I feel that this should be the case is because, if we consider the periodic Dirac distribution at zero, defined as $\delta\_0 \in \mathcal{D}'(\mathbb{T}^n)$ where
$$
\begin{align}\langle \delta\_0, \varphi\rangle = \varphi(0)\end{align}
$$
for any $\varphi \in C^{\infty}(\mathbb{T}^n)$, then its Fourier transform using $(1)$ gives
$$
\begin{align}
\underbrace{\langle \mathcal{F}\_{\mathbb{T}^n} \delta\_0, \varphi \rangle}\_{\in \mathbb{C}} &= \langle \delta\_0, \mathcal{F}\_{\mathbb{T}^n}^{-1} \varphi (-\cdot) \rangle \\
&= \mathcal{F}\_{\mathbb{T}^n}^{-1} \varphi (0) \\
&= \sum\_{\xi \in \mathbb{Z}^n} \varphi(\xi),
\end{align}
$$
where the right hand side is infinite since $\varphi$ is periodic. So this doesn't make sense. If we instead use $(2)$, then we obtain the following:
$$
\begin{align}
\underbrace{\langle \mathcal{F}\_{\mathbb{T}^n} \delta\_0, \varphi \rangle}\_{\in \mathbb{C}} &= \langle \delta\_0, \mathcal{F}\_{\mathbb{T}^n} \varphi \rangle \\
&= (\mathcal{F}\_{\mathbb{T}^n} \varphi) (0)\\
&= \int\_{\mathbb{T}^n} \varphi(x) dx \\
&= \langle 1, \varphi \rangle
\end{align}
$$
i.e. $\mathcal{F}\_{\mathbb{T}^n} \delta\_0 = 1$ in the sense of periodic distributions, as we would expect.
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https://mathoverflow.net/users/160454
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Fourier transform of periodic distributions
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Actually, the definition you gave in the post differs from the one in the book. The test function $\varphi$ should lie in $\mathcal S(\mathbb Z^n)$, not in $C^\infty(\mathbb T^n)$, since the $\mathcal F\_{\mathbb T^n}$ maps the second of these spaces into the first, so for $\varphi\in C^\infty(\mathbb T^n)$ the expression $\mathcal F\_{\mathbb T^n}^{-1}\varphi$ is undefined, as you correctly noticed. So, the correct formula would be
$$
\langle\mathcal F\_{\mathbb T^n}u,\varphi \rangle=\langle u,\iota\,\circ\,\mathcal F\_{\mathbb T^n}^{-1}\varphi \rangle,
$$
where $\iota(f)(x)=f(-x)$ and $\varphi \in \mathcal S(\mathbb Z^n)$.
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https://mathoverflow.net/users/101078
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397515
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https://mathoverflow.net/questions/397502
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1
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In ${\sf ZFC}$ it can be easily proved that we cannot have infinitely descending sequences of cardinalities, that is, the following statement does **not** hold:
>
> (DescSeq) There is a set $A$ a map $\alpha: \omega \to {\cal P}(A)$ such that for all $n\in \omega$ we have $\alpha(n+1) \subseteq \alpha(n)$, and there is **no injective** function from $\alpha(n)$ into $\alpha(n+1)$.
>
>
>
It seems to be unknown whether $\neg$(DescSeq) implies the Axiom of Choice.
**Question.** In ${\sf ZF}$, are there any implications between $\neg$(DescSeq), the [Boolean Prime Ideal Theorem](https://en.wikipedia.org/wiki/Boolean_prime_ideal_theorem), and the [Partition Principle](https://karagila.org/2014/on-the-partition-principle/)?
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https://mathoverflow.net/users/8628
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Strictly descending sequences of sets, the Partition Principle, and the Boolean Prime Ideal Theorem
|
You are asking about three choice principles, two of which have practically no research around them. Mainly due to the lack of tools we have for dealing with them.
The most you can find is the following paper,
>
> *Howard, Paul; Tachtsis, Eleftherios*, [**No decreasing sequence of cardinals**](http://dx.doi.org/10.1007/s00153-015-0472-5), Arch. Math. Logic 55, No. 3-4, 415-429 (2016). [ZBL1339.03038](https://zbmath.org/?q=an:1339.03038).
>
>
>
Where the authors show that $\sf AC\_{WO}$ does not imply that there are no decreasing sequences of cardinals (which they take to mean actual sets, rather than the cardinals of these sets), at least in $\sf ZF$ (in $\sf ZFA$ we can say ever so slightly more).
This is really not a whole lot. As for $\sf BPI$, we know that it holds in Cohen's model, where there is a decreasing sequence of sets, simply because there are infinite Dedekind-finite sets. But since we don't have any models of $\sf ZF+\lnot AC$ where we know that the cardinals are well-founded, we cannot prove it will not imply $\sf BPI$, mainly because it may very well imply $\sf AC$ for all we know.
So, you either need to come up with a completely new proof that somehow $\sf BPI$ follows from the well-foundedness of the cardinals, or prove that it implies choice outright, or better yet: develop new tools for positive results about cardinal structure in $\sf ZF$ that will let you deal with this sort of choice principle.
---
**Appendix A.**
One can ask about decreasing chains of sets/cardinals versus well-foundedness of sets/cardinals.
In $\sf ZF+DC$ the four questions are equivalent, and given a sequence of cardinals we can choose (using $\sf DC$) the sets to match. So it all falls back to be the same.
If there is a Dedekind-finite set, then there is a decreasing sequence of cardinals, but there is also one where we cannot match a decreasing sequence of sets. However, there are *other* sequences of cardinals where we do have a decreasing sequence of sets to match.
So the real question about this is really "*Suppose that $\sf ZF+\lnot DC+DF=F$ holds, are the four principles still equivalent?*"
Again, due to the lack of tools, this is frustratingly difficult to answer. Quite possibly, the assumption of $\sf DC$ can be weakened to $\sf AC\_\omega$, which closes the gap a bit further, but still not enough to conclude the answer is positive.
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https://mathoverflow.net/users/7206
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397517
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https://mathoverflow.net/questions/397511
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5
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Let $X$ be a geodesically complete Riemannian manifold (we may assume that $X$ is simply connected and negatively curved, although I don't think it matters). Given a closed, convex subset $K \subset X$, there is a result by Rolf Walter that the $\epsilon$-neighbourhood of $K$ (denoted $K\_{\epsilon}$) has $C^{1,1}$-regular boundary, i.e. the topological boundary of $K\_{\epsilon}$ is a $C^{1,1}$-submanifold of $X$. I was wondering if the regularity can be improved if we allow for more deformations of the convex set. Specifically, I would like to know the following:
${\bf Question:}$ Given a closed, convex subset $K \subset X$ and $\epsilon > 0$, does there exist a convex closed $K'$ such that $K \subset K' \subset K\_{\epsilon}$, and $\partial K'$ is a $C^2$-submanifold of $X$?}
Considering that Walter's result goes back to the 70s (Rolf Walter, "Some analytical properties of geodesically convex sets"), this seems like a question whose answer should be known, but I'm struggling to find anything. All methods I could come up with to deform $K$ so that its boundary becomes more smooth while keeping the set convex rely on the second fundamental form, which requires that we make the boundary $C^2$ first.
I should empathize that I do not want to assume that $K$ is compact. (I'm fine assuming that there is a cocompact action by isometries on $K$ though.)
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https://mathoverflow.net/users/319208
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Improving regularity of the boundary of a convex set in Riemannian manifolds
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The answer is "yes" we used a similar argument in our ["An optimal lower curvature..."](https://arxiv.org/abs/1303.5884).
Let me sketch the proof.
I will assume that curvature is negative.
The argument will use the existence of cocompact isometric action.
Note that the function $f=\mathrm{dist}^2\_K$ is strongly convex in $K\_{2\cdot\varepsilon}\backslash K\_\varepsilon$.
Let us smooth $f$, applying the argument of Green and Wu [Theorem 2(a) in "$C^∞$ convex functions..."].
Here we have to use the cocompact action; otherwise there might be problems.
Take $K'$ to be set bounded by a level set of smoothed $f$ that lies in $K\_{2\cdot\varepsilon}\backslash K\_\varepsilon$.
This way we get $K'$ with $C^\infty$-smooth boundary.
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2
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https://mathoverflow.net/users/1441
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397519
| 164,098 |
https://mathoverflow.net/questions/397428
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4
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I am trying understand if there is a relation between two formulations of the spontaneous symmetry breaking.
The first is provide by Derdzinski in his book "Geometry of the standard model of elementary particles" in which we have a vector bundle $(E, M, \mathbb{C}^2)$ with a usual inner product in each fiber, and the Higgs field is identified with a global section $\phi: M \rightarrow E$ with constant norm $|\phi| = v$. Then, we consider the subbundle (a line bundle) generated by $\phi^\perp$ as a eletromagnetic bundle.
The second approach is the natural approach in terms of the principal bundles. Given a principal bundle $(P, M, G= SU(2) \times U(1))$, we consider the eletromagnetic as a subbundle which is given by a global section $h: M \rightarrow P/U(1)$ as guaranteed by the Kobayashi & Namizu page 57-58 in their book "Foundations of differential geometry vol. 1". $P/U(1)$ is a bundle with typical fiber $G/U(1) \simeq SU(2)$.
My problem is that the line bundle $\phi^\perp$ (it's frame bundle) don't seem to have nothing related to the principal bundle of the second case. The better relation that I get see is: as $\phi$ belongs to orbit $Gv$ for all $x\in M$, which is diffeomorphic to $G/ U(1) \simeq SU(2)$, then we can use the global section $\phi$ to obtain a global section of the bundle $P/U(1)$. But, I don't get relate the $\phi^\perp$ subbundle.
>
> I believe there is a deep structure behind these things. Both have
> further results that seem to be correct. Can anyone help me?
>
>
>
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https://mathoverflow.net/users/166778
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Relationship between two bundles approaches of spontaneous symmetry breaking
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Before continuing, let me make some algebraic observations.
1. We can view $U(1)$ as a subgroup of $U(2)$ via the injective homomorphism $\iota : U(1) \to U(2)$ given by
$$
\forall z \in U(1), \quad \iota(z) := \begin{pmatrix} 1&0\\0&z \end{pmatrix};
$$
moreover, for all orthonormal $\{v,w\} \subset \mathbb{C}^2$, so that $[v\vert w] \in U(2)$, we have
$$
\forall z \in U(1), \quad [v\vert w]\cdot \iota(z) = [v \vert zw],
$$
so that the map $(g \mapsto (g\_{11},g\_{21})^T) : U(2) \to S^3 \subset \mathbb{C}^2$ descends to a left $U(2)$-equivariant diffeomorphism $F : U(2)/U(1) \to S^3$.
2. We can view $U(1) \times SU(2)$ as a double cover of $U(2)$ via the surjective homomorphism $\mu : U(1) \times SU(2) \to U(2)$ given by
$$
\forall (z,g) \in U(1) \times SU(2), \quad \mu(z,g) := zg,
$$
where $\ker\mu = \langle (-1,-I\_2) \rangle \cong \mathbb{Z}\_2$. As we’ll soon see, you’ll actually have to work with $U(2) \cong (U(1) \times SU(2))/\mathbb{Z}\_2$ instead of $U(1) \times SU(2)$, but note that the induced Lie algebra homomorphism $\mu\_\ast : \mathfrak{u}(1) \oplus \mathfrak{su}(2) \to \mathfrak{u}(2)$ is an isomorphism.1
Let $M$ be a smooth manifold and let $E \to M$ be a Hermitian vector bundle of rank $2$, which is precisely your data $(E,M,\mathbb{C}^2)$. Let $P \xrightarrow{\pi} M$ be the orthonormal frame bundle of the Hermitian vector bundle $E$, which is a principal $U(2)$-bundle; a further reduction of the structure group to $U(1) \times SU(2)$ would be nontrivial extra structure that we don’t actually need. Modulo this subtlety, you’re correct, one can use the canonical isomorphisms
$$
P \times\_{U(2)} \mathbb{C}^2 \to E, \quad P \times\_{U(2)} (U(2)/U(1)) \to P/U(1)
$$
and the resulting identifications of global sections with equivariant maps to construct an explicit bijection between the following sets (which may or may not be empty):
* the set of global sections $\phi$ of $E$ with $\lvert \phi \rvert = 1$;
* the set of all global sections of the associated fibre bundle $P/U(1) \to M$.
Given a global section $\phi$ of $E$ with $\lvert \phi \rvert = 1$, I claim that the reduction of the structure group of $P$ induced by the corresponding global section of $P/U(1)$ recovers the circle bundle of the line bundle $\phi^\perp$.
Now, suppose that $\phi$ is a global section of $E$ with $\lvert \phi \rvert = 1$; let $\tilde{\phi} : P \to \mathbb{C}^2$ be the corresponding $U(2)$-equivariant map, so that global section $\psi$ of $P/U(1)$ induced by $\phi$ corresponds to the $U(2)$-equivariant map $\tilde{\psi} : P \to U(2)/U(1)$ given by
$$
\forall p \in P,\quad \tilde{\psi}(p) := F^{-1} \circ \tilde{\phi}(p) = \begin{pmatrix} \tilde{\phi}(p)\_1 & -\overline{\tilde{\phi}(p)\_2} \\ \tilde{\phi}(p)\_2 & \overline{\tilde{\phi}(p)\_1} \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 \\ 0 & U(1) \end{pmatrix}.
$$
The section $\psi$ of $P/U(1) \to M$ yields a reduction of the structure group of $P$ to $U(1)$, i.e., the principal $U(1)$-bundle
$$
\tilde{P} := \{ p \in P \,\mid\, p \cdot U(1) = \psi(\pi(p))\}
$$
satisfying $P \cong \tilde{P} \times\_{U(1)} U(2)$ as principal $U(2)$-bundles on $M$. By unpacking definitions and canonical isomorphisms and applying observation 1, one can check that
$$
\forall x \in M, \quad \tilde{P}\_x := \{(\phi\_x,w) \, \mid \, w \in \phi\_x^\perp, \, \lvert w \rvert = 1\}.
$$
But now, this is clearly isomorphic as a principal $U(1)$-bundle over $M$ to the circle bundle
$$
Q := \{ w \in \phi^\perp \mid \lvert w \rvert = 1\}
$$
of $\phi^\perp$ via the isomorphism $f : Q \to \tilde{P}$ given by
$$
\forall x \in M, \, \forall w \in Q\_x, \quad f(w) := (\phi\_x,w).
$$
---
1 Compare the Standard Model, where the famed group $U(1) \times SU(2) \times SU(3)$ is actually a $6$-fold cover of the true structure group $(U(1) \times SU(2) \times SU(3))/\mathbb{Z}\_6$. Since the covering homomorphism of Lie groups induces an explicit isomorphism of Lie algebras, one can formally work with the group $U(1) \times SU(2) \times SU(3)$ at the price of introducing fractional charges.
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https://mathoverflow.net/users/6999
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397520
| 164,099 |
https://mathoverflow.net/questions/397505
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5
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I'm reading the proof of monotonicity formula from *A Course in Minimal Surfaces* by Colding-Minicozzi. The theorem says
>
> Suppose $\Sigma^k \subset \mathbb{R}^n$ is a minimal submanifold and $x\_0\in\mathbb{R}^n$; then for all $0<s<t$,
> $$
> \frac{\mathrm{Vol}(B\_t(x\_0)\cap\Sigma)}{t^k} - \frac{\mathrm{Vol}(B\_s(x\_0)\cap\Sigma)}{s^k} \\
> = \int\_{(B\_t(x\_0)\setminus B\_s(x\_0))\cap\Sigma} \frac{|(x-x\_0)^N|^2}{|x-x\_0|^{k+2}},
> $$
> where $B\_t(x\_0)$ is the ball in $\mathbb{R}^n$ with center $x\_0$ and radius $t$, and $(\cdot)^N$ the projection onto the normal space of $\Sigma$.
>
>
>
In the proof, they define $f(x)=|x-x\_0|$ and then argue that
$$
\mathrm{Vol}(\{f\le s\}\cap\Sigma) = \int\_{\{f\le s\}\cap\Sigma} |\nabla\_\Sigma f|^{-1} |\nabla\_\Sigma f| d\mathcal{H}^k = \int\_0^s \int\_{f=\tau} |\nabla\_\Sigma f|^{-1} d\mathcal{H}^{k-1} d\tau
$$
My question is about the first equality. Here the authors implicitly assumed that
$$
|\nabla\_\Sigma f|>0 \quad \mathcal{H}^k\mathrm{-a.e.} \text{ on } \Sigma,\label{1}\tag{$\*$}
$$
but didn't give a proof.
>
> How do I prove \eqref{1}?
>
>
>
My idea: Since $\Delta\_\Sigma f^2 = 2k>0$, $f^2$ is a subharmonic function on $\Sigma$. It suffices to show that the set of critical points of a smooth subharmonic function has $\mathcal{H}^k$ measure $0$. But I can't find a reference for the latter.
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https://mathoverflow.net/users/319148
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A question on the monotonicity formula for minimal submanifolds
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A stronger property is true: the critical set of a strictly subharmonic function $f: \Sigma \to \mathbf{R}$ is locally contained inside a codimension one submanifold. Explicitly: for every critical point $x \in \Sigma$ there is $r > 0$ and a $(k-1)$-dimensional submanifold $\Gamma \subset \Sigma \cap D\_r(x)$ so that $\{ \nabla f = 0 \} \cap D\_r(x) \subset \Gamma$. (Here $D\_r(x)$ is the ball of radius $r$ in the intrinsic metric on the surface.)
Let $x \in \Sigma$ be such a critical point. In local normal coordinates one can identify $\nabla^2 f(x) = (\partial^2\_{ij} f(x))$. This matrix has positive trace because $\Delta f(x) > 0$. Therefore at least one partial derivative $\nabla \partial\_i f(x) \neq 0$, and by the implicit function theorem the zero set $\{ \partial\_i f = 0 \}$ is a $C^1$ embedded surface in a small enough neighbourhood of $x$. We conclude by setting $\Gamma = \{ \partial\_i f = 0 \} \cap D\_r(x)$ and observing that $\{ \nabla f = 0 \} \cap D\_r(x) \subset \Gamma$.
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4
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https://mathoverflow.net/users/103792
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397521
| 164,100 |
https://mathoverflow.net/questions/397501
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11
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If $\mathbf{G}$ is an algebraic group defined over $\mathbb{Q}$, a subgroup of $\mathbf{G}(\mathbb{Q})$ is *arithmetic* if it is commensurable to $\mathbf{G}(\mathbb{Q}) \cap \operatorname{GL}\_n(\mathbb{Z})$ where some representation $\mathbf{G} < \operatorname{GL}\_n$ has been chosen (and the definition is made so that the choice does not matter).
Fur the purpose of this question let us call a subgroup $\Gamma$ of $\mathbf{G}(\mathbb{Q})$ *strictly arithmetic* if there exists a group $\mathbb{Z}$-scheme $\mathbf{G}\_\mathbb{Z}$ with generic fiber $\mathbf{G}$ such that $\Gamma = \mathbf{G}\_\mathbb{Z}(\mathbb{Z})$.
I was recently asked the natural question whether strictly arithmetic is the same as arithmetic. I suspect that the answer is "no". More specifically arithmetic groups can be arbitrarily small (for instance have arbitrarily large covolume in $\mathbf{G}(\mathbb{R})$) while I suspect that this is not true of strictly arithmetic groups. But I don't know enough about group schemes to underpin that intuition. So I'm asking here:
**Original question: Are there (resp. what are) examples of arithmetic groups that are not strictly arithmetic?**
The original question was answered in the comments by David Loeffler using a different obstruction so let me (following YCor's suggestion in the comments) specifically ask:
**Additional question: Do there exist "arbitrarily small" strictly arithmetic subgroups, for instance in the sense that the covolume or injectivity radius in $\mathbf{G}(\mathbb{R})$ is arbitrarily large?**
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https://mathoverflow.net/users/5339
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Arithmetic groups and integral points of integral structures
|
**First question** (do non-strictly-arithmetic subgroups exist?):
Any "strictly arithmetic" subgroup in your sense will, in particular, be a congruence subgroup, i.e. the intersection of $G(\mathbb{Q})$ with an open compact subgroup in $G(\mathbb{A}\_f)$. Since non-congruence subgroups exist in $SL\_2 / \mathbb{Q}$, and in lots of other groups too, these are examples of arithmetic subgroups which are not strictly arithmetic.
**Second question** (can strictly arithmetic subgroups be small?):
Start with your favourite $GL\_n$-embedding $\iota$, defining some strictly arithmetic $\Gamma$. Take some $g \in G(\mathbb{Q})$ which isn't in $\Gamma$, and consider the embedding into $GL\_{2n}$ sending $h$ to the block-diagonal matrix $\begin{pmatrix} \iota(h) \\& \iota(g^{-1} h g))\end{pmatrix}$. This defines a new $\mathbb{Z}$-model of $G$ whose integral points are $\Gamma \cap g \Gamma g^{-1}$. If $G = SL\_2$, and probably for just about any $G$ which isn't abelian, the index of $\Gamma \cap g \Gamma g^{-1}$ in $\Gamma$ can be made arbitrarily large by a suitable choice of $\Gamma$.
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11
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https://mathoverflow.net/users/2481
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397522
| 164,101 |
https://mathoverflow.net/questions/397496
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1
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On page 87 of the book *Hyperbolic Conservation Laws in Continuum Physics* by C. M. Dafermos, there is a theorem which I summarise as follows
>
> **Theorem.** (Theorem 4.5.2 in the book.) Let $U$ be a weak solution to the conservation law $\partial\_t U + \text{div }G(U) =0,$ with initial data $U\_0$ and $U\in\mathcal O\subseteq \mathbb R^n, x\in \mathbb R^m, G$ an appropriate matrix-valued function. Suppose that $U\_0 - \bar U \in L^2$ for some constant $\bar U,$ and $\bar U$ satisfy the entropy condition for entropy $\eta.$ If we normalise $\eta$ so that $\eta(\bar U ) =0, D\eta(\bar U) =0,$ then $$S(t) = \int\_{\mathbb R^m} \eta (U(x,t)) dx$$ is a decreasing function of $t.$
>
>
>
I am worried about the statement of this theorem because the only assumption made on $G$ is that it is smooth in the image of $U.$ (This is stated at the beginning of the chapter.) No other technical assumptions are given. This seems insufficient to control the growth of $Q,$ and there are problems that integrals of $Q$ diverging.
If we look into the proof, there is a sentence: "fix $s$ sufficiently large so that $s\eta \geq |Q|$ for all output of the function $U.$" **I do not know why such $s$ must exist.** We know that $DQ\_\alpha = D\eta \cdot DG\_\alpha,$ by the definition of entropy flux. So if $G\_\alpha$ grow very fast, we **cannot** expect $Q$ to be dominated by $\eta.$ Apparently some bound on $G$ need to be assumed; but nearly nothing is assumed in the book. (Of course we can normalise so that $Q(\bar U) =0$. But if we do so, there is no guarantee that $U$ will always stay close to $0$ at infinity.)
**So, are we implicitly assuming some conditions on $G$, which are not stated in the book?**
|
https://mathoverflow.net/users/121404
|
Assumptions on the flux of a conservation law required to obtain an entropy inequality
|
I just quickly read the proof you mentioned, and I think what is meant is following:
1. Note that in Section 4.3 it is noted that any weak solution $U$ may be renormalized to be a continuous (in weak\* topology) mapping from $[0,T)\to L^\infty$.
2. For the argument, one proves a statement to be true on $[0,T)$ by proving it true on $[0,\tau]$ for all $\tau\in [0,T)$. The interval $[0,\tau]$ is compact.
3. This suggests that $\{Q(U(t,x)): (t,x)\in [0,\tau]\times \mathbb{R}^m\}$ should be essentially bounded. (We only need the comparison to hold a.e. since the desired result is integrated.)
4. You should be more worried then about where $\eta$ is small, and not where $Q$ is large; this is resolved by the convexity of $\eta$.
|
0
|
https://mathoverflow.net/users/3948
|
397524
| 164,102 |
https://mathoverflow.net/questions/397534
|
0
|
How to simulate a process $S\_t=\sum\_{0\leq s\leq t}\Delta\_s,$ where $\Delta\_s$ is a Poisson point process with values in $(0,\infty)$ and with characteristic measure $\Pi(dx)=\frac{\alpha}{\Gamma(1-\alpha)}x^{-1-\alpha}dx, \alpha=0.5,1,1.5.$ This means for every Borel set $B\subset (0,\infty),$ the counting process $N\_\cdot^B=Card\{s\in [0,\cdot]:\Delta\_s\in B\}$ is a Poisson process with intensity $\Pi(B).$
I am a fresh student on simulation. It would be nice if you can give me a program (Matlab or Mathematica) of a simple case. Thank you very much in advance.
|
https://mathoverflow.net/users/172842
|
How to simulate Poisson point process
|
I assume we know how to simulate a Poisson point process with constant intensity in an interval (e.g. by considering partial sums of i.i.d. exponential variables.)
That allows you to simulate a standard Poisson point process in a rectangle $[a,b] \times [0,d]$ by simulating an intensity $d$ Poisson process in $[a,b]$ and assigning the resulting points $\{x\_i\}\_{i=1}^T$ uniform heights $\{y\_i\}\_{i=1}^T$ in $[0,d]$ to obtain $\{(x\_i,y\_i)\}\_{i=1}^T$.
To simulate (with a small error $<\epsilon$ in total variation) a Poisson point process with slowly decreasing integrable intensity $f(x)dx$ in $[0,\infty)$, first pick an increasing sequence $\{N\_k\}\_{k \ge 0}$ with $N\_0=0$, e.g. $N\_k=2^k-1$. Then pick $m$ large enough so that $\int\_{N\_m}^\infty f(x) \, dx <\epsilon$, so we can neglect points beyond $N\_m$. Then for each $k=0,1,\ldots m-1$ independently simulate a standard Poisson process in a rectangle $[N\_k,N\_{k+1}] \times [0,f(N\_k)]$. For each $k$ this yields points $(x\_i(k),y\_i(k))$
for $1 \le i \le T\_k$. Reject those points for which $y\_i(k)>f(x\_i(k))$.
Note that with $f$ decreasing polynomially and $N\_k$ increasing geometrically, the rejection probability is bounded away from 1. The $x$ coordinates of the surviving points will yield a sample from a Poisson process with intensity $f(x)\, dx $ in $[0,N\_m]$.
|
1
|
https://mathoverflow.net/users/7691
|
397541
| 164,107 |
https://mathoverflow.net/questions/397554
|
6
|
$ \def \CZF {\mathbf {CZF}}
\def \IZF {\mathbf {IZF}}
\def \A {\mathcal A}
\def \then {\mathrel \rightarrow}
\def \r {\mathrel \Vdash}
\DeclareMathOperator \V V $
In "Realizability for Constructive Zermelo-Fraenkel Set Theory", Michael Rathjen shows that a notion of realizability due to Charles McCarty works well for $ \CZF $ (Constructive Zermelo-Fraenkel set theory), a more restricted theory than $ \IZF $ (Intuitionistic Zermelo-Fraenkel set theory), which is the theory of concern in McCarty's work.
The notion is defined in terms of the realizability (class) structure $ \V ( \A ) $ over an applicative structure $ \A $. For $ e \in | \A | $ and a sentence $ \phi $ in the language of set theory extended by adding constants denoting members of $ \V ( \A ) $, realizability of $ \phi $ by $ e $ is defined recursively on the subformula tree of $ \phi $, and the notation "$ e \r \phi $" is used, read as "$ e $ realizes $ \phi $". The definition for the cases where $ \phi $ is not atomic is similar to the definitions for corresponding cases in the other well-known notions of realizability, e.g. that of Kleene defined for the theories of arithmetic. It is the definition of realizability for atomic sentences which I can't quite understand:
\begin{align\*}
e \r a \in b \iff& \exists c \big( \langle ( e ) \_ 0 , c \rangle \in b \land ( e ) \_ 1 \r a = c \big) \\
e \r a = b \iff& \forall f , d \Big( \big( \langle f , d \rangle \in a \then ( e ) \_ 0 f \r d \in b \big) \land {} \\ &\qquad\qquad \big( \langle f , d \rangle \in b \then ( e ) \_ 1 f \r d \in a \big) \Big)
\end{align\*}
Here $ a , b \in \V ( \A ) $, juxtaposition is used to denote application in the structure $ \A $, $ \langle . , . \rangle $ is a fixed pairing function, and $ ( . ) \_ 0 $ and $ ( . ) \_ 1 $ are the corresponding projections.
Struggling to understand how this definition works, I ended up asking myself the following questions, which I couldn't figure out, and thus I decided to ask here.
1. Isn't this definition circular? Something of the form $ e \r a = b $ appears in the definition of $ e \r a \in b $, and vice versa.
2. Is this definition sensitive to the minimal language of set theory? More specifically, if one extends the language by adding function symbols to the language, say a unary symbol denoting the union of members of a set, would the notion cease to work? Would one need to break the case of atomic sentences into cases where the form of the terms appearing in the sentence is taken into account? Or would it be similar to Kleene's realizability where the atomic sentences are treated regardless of the addition and multiplication symbols appearing in the terms?
3. In case where adding function symbols does not affect the way the definition works, does constructivity of the intended function really matter? This question comes for example from the fact that $ \IZF $ contains power sets, which may not be considered constructive (as they are rejected in $ \CZF $). To make this more specific and go even beyond $ \IZF $, would the notion of realizability work if we add a binary function symbol $ \chi $ to the language, with the intended meaning of the characteristic (class) function of membership, and add the following axioms to the language (one can either add $ \varnothing $ and $ \{ . \} $ to the language, or modify the following sentences in the obvious way so that they don't contain these symbols)?
* $ \forall x , y ( x \in y \then \chi ( x , y ) = \{ \varnothing \} ) $
* $ \forall x , y ( \neg x \in y \then \chi ( x , y ) = \varnothing ) $
---
*Rathjen, Michael*, [**Realizability for constructive Zermelo-Fraenkel set theory**](https://www1.maths.leeds.ac.uk/%7Erathjen/KLRend.pdf), Stoltenberg-Hansen, Viggo (ed.) et al., Logic colloquium ’03. Proceedings of the annual European summer meeting of the Association for Symbolic Logic (ASL), Helsinki, Finland, August 14–20, 2003. Wellesley, MA: A K Peters; Urbana, IL: Association for Symbolic Logic (ASL) (ISBN 1-56881-293-0/hbk; 1-56881-294-9/pbk). Lecture Notes in Logic 24, 282-314 (2006). [ZBL1102.03053](https://zbmath.org/?q=an:1102.03053).
*McCarty, Charles*, [**Realizability and recursive set theory**](http://dx.doi.org/10.1016/0168-0072(86)90050-3), Ann. Pure Appl. Logic 32, 153-183 (1986). [ZBL0631.03035](https://zbmath.org/?q=an:0631.03035).
|
https://mathoverflow.net/users/76416
|
Realizability for constructive Zermelo-Fraenkel set theory
|
For your first question, the definition of $e\Vdash x\in y$ and $e\Vdash x=y$ seems circular, but $\mathsf{CZF}$ provides a way to avoid the circularity, called *inductive definition*.
>
> **Definition.** An *inductive definition* is a class $\Phi\subseteq \mathcal{P}(V)\times V$. For each inductive definition $\Phi$, define
> $\Gamma\_\Phi(X)=\{a \mid \exists Y\subset X : (Y,a)\in \Phi\}$.
>
>
> (Note that $X$ may be a class, but $Y$ must be a set.)
>
>
>
Sometimes we write $X/\_\Phi a$ or $X\vdash\_\Phi a$ instead of $(X, a)\in \Phi$, to emphasize the analog between deduction and inductive definition. In that view, we may think $\Gamma\_\Phi(X)$ the least class containing $X$ that is closed under $\Phi$-deduction.
It is known that $\mathsf{CZF}$ admits arbitrary inductive definition (unlike $\mathsf{KP}$ only admit $\Sigma$-ones):
>
> **Theorem.** (Class Inductive Definition Theorem) Let $\Phi$ be an inductive definition. Then there is a unique $\Gamma\_\Phi$-least fixed point $I(\Phi)$, i.e., $I(\Phi)$ is least among classes such that $\Gamma\_\Phi(I(\Phi))\subseteq I(\Phi)$.
>
>
>
But justifying the circular definition with an inductive definition is sometimes a bit tricky. In our case, it seems that we need a 'mutual' inductive definition to justify it.
Fortunately, it turns out that this is not the case: we may remove every formula of the form $f\Vdash x\in y$ in the definition of $e\Vdash a=b$ by replacing $f\Vdash x\in y$ to its definition, so we have a recursive definition of $e\Vdash a=b$. Then we can define $e\Vdash a\in b$ from $f\Vdash x=y$.
---
For your second question, I do not get the point of your question clearly, but if my understanding ― whether adding a new predicate or function into the language affects the definability of $\Vdash$ ― is correct, then I think the answer is yes in general. The reason is that Class Inductive Definition Theorem works for arbitrary $\Phi$, although adding functional symbols would require more justifications. For example, consider the case of adding the binary unpaired order function $\{,\}$. For each $a,b\in V^{\mathcal{A}}$ for a pca $\mathcal{A}$, define
$$\mathsf{up}(a,b) = \{\langle\underline{0}, a\rangle,\langle\underline{1}, b\rangle\}.$$
Now interpret $\{,\}$ by making use of $\mathsf{up}$, that is, for each $a,b\in V^\mathcal{A}$, take $\{a,b\}^{V(\mathcal{A})} := \mathsf{up}(a,b)$. This means we will replace every occurrence of $\{a,b\}$ in the realizability relation to $\mathsf{up}(a,b)$.
A more dramatic example would be given by [Chen and Rathjen](https://www1.maths.leeds.ac.uk/%7Erathjen/Lifschitz.pdf): they defined $\Vdash$ over the extension of $\mathsf{CZF}$, which contains natural numbers and relevant operations as primitive notions. (Also it looks like a mere combination of the realizability à la McCarty and that of Kleene.)
---
For the last question, I think the answer is yes if the ambient theory ($\mathsf{CZF}$ in our case) proves the functionality of our desired function. The following theorem seems relevant (Proposition 9.6.2 of [Aczel and Rathjen](https://www1.maths.leeds.ac.uk/%7Erathjen/book.pdf).)
>
> **Theorem.** Let $T$ be a theory which comprises $\mathsf{BCST}$ (e.g. $\mathsf{CZF}$). Suppose $T \vdash ∀x ∃!yΦ(x, y)$.
> Let $T\_Φ$ be obtained by adjoining a function symbol $F\_Φ$ to the language, extending
> the schemata to the enriched language, and adding the axiom $∀x Φ(x, F\_Φ(x ))$.
> Then $T\_Φ$ is conservative over $T$.
>
>
>
Focusing on your example, in fact, we can define $\chi$ over $\mathsf{CZF}$: take $\chi(x,y)=\{0\mid x\in y\}$. By using this definition, we have a conservative extension of $\mathsf{CZF}$, and we may simply reduce $\chi$ in the extended language to a corresponding definition over $\mathsf{CZF}$.
|
8
|
https://mathoverflow.net/users/48041
|
397559
| 164,115 |
https://mathoverflow.net/questions/397523
|
3
|
On page 92 of the book Hyperbolic Conservation Laws in Continuum Physics by C. M. Dafermos, there is a theorem 4.6.1 which says
>
> Under some assumptions, suppose a sequence of solutions $U\_{\mu\_k}$ to a conservation law with viscosity term **converges boundedly almost everywhere** on $\mathbb R^m \times [0,T)$ to some function $U,$ and $\mu\_k \to 0.$ Then $U$ is a weak solution to the conservation law without viscosity term. Moreover, $U$ is entropy admissible.
>
>
>
**Question:** What does "converges boundedly almost everywhere" mean?
Note: $U\_\mu$ should be smooth according to theory of heat equations, but no other assumptions are made about $U\_\mu.$
|
https://mathoverflow.net/users/121404
|
What does it mean by "converges boundedly"?
|
This should mean that $U\_{\mu\_k} \to U$ almost everywhere on $\mathbb{R}^m \times [0,T)$, and moreover the sequence of functions $U\_{\mu\_k}$ is uniformly bounded:
$$\sup\_k \sup\_{(x,t) \in \mathbb{R}^m \times [0,T)} |U\_{\mu\_k}(x,t)| < \infty.$$
I suppose that the functions $U\_{\mu\_k}$ take their values in $\mathbb{R}$ or $\mathbb{C}$ or some other obvious normed space.
This is standard terminology; see for instance [Section 10.5](https://books.google.com/books?id=1qd6DwAAQBAJ&lpg=PA458&ots=39gz2Zw3JG&dq=converge%20%22boundedly%22&pg=PA458#v=onepage&q=converge%20%22boundedly%22&f=false) of:
>
> *Ghorpade, Sudhir R.; Limaye, Balmohan V.*, [**A course in calculus and real analysis**](http://dx.doi.org/10.1007/978-3-030-01400-1), Undergraduate Texts in Mathematics. Cham: Springer (ISBN 978-3-030-01399-8/hbk; 978-3-030-01400-1/ebook). ix, 538 p. (2018). [ZBL1403.26001](https://zbmath.org/?q=an:1403.26001).
>
>
>
In context, it is probably fine to replace the sup over $(x,t)$ with the essential supremum.
|
2
|
https://mathoverflow.net/users/4832
|
397561
| 164,116 |
https://mathoverflow.net/questions/397560
|
5
|
I have the polynomial $f(x) = x^2-x+1$ and I am wondering if there is a positive prime value $p$ such that $f(p),f^2(p),f^3(p)\dots$ are all prime.
I have ran some computer simulations and I feel like the answer should be "no" ( because looking at the map $x^2-x+1 \bmod p$ I get that the expected number of prime divisors of the first $M$ values should be larger than $M$). I feel that my analysis is not very good however.
Does anyone know an approach which could be more fruitful?
Note: $f^2(x) = f(f(x))$
Edit: I would be happy with any sort of reference regarding a polynomial staying inside the primes under iteration, if there isn't anything particularly useful which can be said about this sort of thing I understand that as well).
|
https://mathoverflow.net/users/24478
|
Checking if polynomial can be iterated and only take prime values
|
This is not an answer to your question, but will point you toward work on the number theoretic properties of such sequences. Iteration of $x^2-x+1$ starting at $a=2$ is called the [Sylvester sequence](https://en.wikipedia.org/wiki/Sylvester%27s_sequence). A theorem about primes that divide the terms in such sequences was proved by Rafe Jones ([The density of prime divisors in the arithmetic dynamics of quadratic polynomials. *J. Lond. Math. Soc.* (2) **78** (2008), no. 2, 523–544. MR2439638](https://arxiv.org/abs/math/0612415)) One of his results is that if $k\in\mathbb Z$ with $k\notin\{0,2\}$, then the set of primes dividing the integers in the orbit of any $a\in\mathbb Z$ under iteration of $x^2+kx-1$ has density $0$.
In a different direction, one can prove that the numbers in your sequence, for any starting prime $p$, grow quadratically exponentially, and in fact
$$ \lim\_{n\to\infty} \frac{1}{2^n} \log f^n(p) \quad\text{converges
to a positive real number.} $$ Thus your sequence is growing very rapidly, which means that we have very few tools at our disposal to prove statements of the sort you ask.
|
5
|
https://mathoverflow.net/users/11926
|
397566
| 164,118 |
https://mathoverflow.net/questions/397572
|
2
|
For real $s>0$, let
$$S(s):=\sum\_{n=-\infty}^\infty e^{-n^2/(2s^2)}
=\vartheta \_3\left(0,e^{-1/(2 s^2)}\right),$$
where $\vartheta$ is the elliptic theta function.
Plotting suggests that the identity
\begin{equation}
S(s)=s\sqrt{2\pi}
\end{equation}
is true at least for $s\ge3/2$. Is it indeed?
This conjecture, with a plot, appeared as a part of [this answer](https://mathoverflow.net/a/397437/36721), but seems to warrant separate posting. Mathematica cannot prove or disprove this identity.
|
https://mathoverflow.net/users/36721
|
An identity for the elliptic theta function
|
To give an answer, adding to my comments, your formula doesn’t hold true, although the error is exponentially small as $s\to \infty$, as can be seen by Poisson summing, which transforms your sum to
$$\sqrt{2\pi}s\left(1+\mathcal{O}(e^{-2\pi^2 s^2})\right).$$
|
11
|
https://mathoverflow.net/users/152473
|
397574
| 164,119 |
https://mathoverflow.net/questions/397532
|
4
|
Let $X,Y$ be two Banach spaces.
A bounded operator $A$ is Fredholm if $\ker A$ and $\mathrm{coker} A$ are finite dimensional. Denote by $Fred(X,Y) \subset \mathcal{L}(X,Y)$ the space of Fredholm operators endowed with the subspace topology (hence the operator norm).
It is well known that $Fred(X,Y)$ is an open subset of $\mathcal{L}(X,Y)$.
>
> Q. What is the closure of $Fred(X,Y)$ in $\mathcal{L}(X,Y)$? ($X,Y$ infinite dimensional)
>
>
>
I would be happy to have an answer even for the case $X=Y$ separable Hilbert space.
|
https://mathoverflow.net/users/99042
|
Closure of the space of Fredholm operators
|
$\DeclareMathOperator\Ker{Ker}$
>
> Let $\mathcal{H}$ be a separable Hilbert space. Then a bounded operator $T$ is **not** in the norm closure of Fredholm operators iff either $T$ or $T^\*$ has a finite-dimensional kernel and has a closed image of infinite codimension.
>
>
>
(Note that in particular, the only operators that are in the closure of Fredholm operators but not in the closure of invertible operators, are Fredholm operators of nonzero index.)
*Proof:* Let $T$ not be in the norm closure of Fredholm operators. So it is not in the norm closure of invertible operators. By the [Bouldin result](https://www.ams.org/journals/proc/1990-108-03/S0002-9939-1990-1000147-9/S0002-9939-1990-1000147-9.pdf) mentioned previously, $T$ has a closed image and $\dim(\Ker(T))$ and $\dim(\Ker(T^\*))$ differ. If both are finite, $T$ is Fredhlom. So one is finite and the other is infinite.
Conversely, suppose that $T$ satisfies the condition. Up to adjoint, we can suppose that $T$ has closed range, has finite-dimensional kernel and infinite-dimensional cokernel. So there exists a block-decomposition under which we can write $T$ as
$$T=\begin{pmatrix} A\_0 & 0 \\ 0 & 0\end{pmatrix},$$
where the right-hand column is finite-dimensional, $A\_0$ is invertible, and both rows are infinite-dimensional. Let $T'$ be norm-close to $T$: write it as
$$T'=\begin{pmatrix} A & b \\ C & d\end{pmatrix},$$
so $A$ being close to $A\_0$, it is invertible. Suppose by contradiction that $T'$ is Fredholm. So there exists a bounded operator $S$ such that $TS-I$ has finite rank. Write $S=\begin{pmatrix} E & F \\ g & h\end{pmatrix}$ in the transpose block decomposition. (Small letters emphasize the fact that they have either finitely many rows or columns.) Then
$$T'S=\begin{pmatrix} AE+bg & AF+bh \\ CE+dg & CF+dh\end{pmatrix},$$
this time in the symmetric block decomposition corresponding to the row decomposition of $T$ (hence with both blocks infinite-dimensional. Hence $AF+bh$ has finite rank. Since $h$ has finite rank, we deduce that $AF$ has finite rank. Since $A$ is invertible, we deduce that $F$ has finite rank. But then $CF+dh$ has finite rank, which contradicts that $CF+dh-I$ has finite rank.
|
3
|
https://mathoverflow.net/users/14094
|
397578
| 164,120 |
https://mathoverflow.net/questions/397547
|
3
|
Any real symmetric matrix $A$ can be written as $A=SDS^T$ for some diagonal matrix $D$ and invertible matrix $S$. Let's fix $D$ to be the (diagonal) inertia matrix of $A$, which has an entry $1, -1, 0$ for each positive, negative, and zero eigenvalue of $A$.
My question is, what is the space of invertible matrices $S$ such that $A=SDS^T$?
|
https://mathoverflow.net/users/150898
|
The invertible matrices $S$ that satisfy $A=SDS^T$
|
Let us first consider the case $D=I$. An invertible matrix $S$ can be (uniquely) factorized as $S=RQ$, where $R$ is upper triangular with positive diagonal entries and $Q$ is orthogonal, $QQ^\top=I$ (this is QR decomposition). Then $SS^\top = RQQ^\top R^\top = RR^\top$. On the other hand, by Cholesky decomposition, every positive definite symmetric matrix $A$ can be uniquely decomposed as $A=RR^\top$, where $R$ is as above. So in this case the space of such $S$ is $R\cdot O(n)$. This reflects the fact that the space of positive definite matrices is isomorphic (in a suitable sense) to $GL(n,\mathbb{R})/O(n)$ (through the mapping $X\mapsto XX^\top$).
*What follows lacks some (many) details and should definitely not be taken for granted.*
Now for a general non-singular $D$ we would like to use the same consideration, and for that we need $QDQ^\top=D$. This means that $Q$ is an element of the corresponding [indefinite orthogonal group](https://en.wikipedia.org/wiki/Indefinite_orthogonal_group) $O(D)$.
The analogue for QR decomposition in this case is called HR decomposition (sometimes also hyperbolic QR). Denote $\mathcal{X}$ the set of all matrices $X$ such that the leading principal minors of $X^\top D X$ have the same signs as the corresponding minors of $D$. Then every matrix $X\in\mathcal{X}$ has a decomposition $X=HR$, where $HDH^\top=D$ and $R$ is upper triangular with positive entries (similarly, one can take $R$ to be lower triangular).
If the leading principal minors of $A$ are non-zero, it posesses [LDL decomposition](https://en.wikipedia.org/wiki/Cholesky_decomposition#LDL_decomposition_2) $A=LCL^\top$, where $L$ is lower triangular with $1$ on the diagonal and $C$ is diagonal. This can be rewritten as $A=L'DL'^\top$, where $D$ is as above (up to a different order of the diagonal entries) and $L'$ is lower triangular with positive entries on the diagonal (namely, the square roots of the absolute values of the diagonal entries of $C$).
Again, if $D$ is ordered as the inertia matrix obtained from LDL decomposition, $A = SDS^\top = LDL^\top$, hence $D = L^{-1}SDS^\top L^{-\top}$ and $L^{-1}S \in O(D)$. If $D$ is ordered differently, the answer differs by a multiplication with a permutation matrix.
The relation with the HR decomposition is based on two observations: 1) the entries of $C$ can be chosen as $C = \operatorname{diag}(\Delta\_1,\frac{\Delta\_2}{\Delta\_1},\ldots,\frac{\Delta\_n}{\Delta\_{n-1}})$, where $\Delta\_i$ are the leading principal minors of $A$ (this is Jacobi method for the diagonalization of quadratic forms); 2) the upper-left corner submatrix of $LML^\top$ for any $M$ and lower-triangular $L$ is the product of the corresponding corner submatrices, and so in case the diagonal entries of $L$ are positive, the leading principal minors of $LML^\top$ and $M$ have the same signs.
I am not sure what to do in case $D$ is singular. Apparently, one can devise some sort of combination of HR and thin QR decompositions, but I do not see this in the literature.
|
3
|
https://mathoverflow.net/users/5018
|
397583
| 164,122 |
https://mathoverflow.net/questions/397592
|
6
|
It is a standard fact in the representation theory of finite groups that for $G,H$ finite groups, all of the irreducible representations of $G \times H$ are the external tensor product of irreps of $G$ and $H$. Today I was talking to a friend about profinite groups and it got me thinking: "Is (some version of) this result still true?" The fact that so many results from the finite case carry over makes me think that this **could be true**, but I have no idea how to go about proving it. The standard proof for finite groups uses a counting argument to show that they are all of this form, so certainly some higher-level techniques will be required.
Since we're considering profinite groups, we will definitely want to restrict ourselves to continuous representations on topological vector spaces. If the statement is not true in this generality, are there adjectives we can add that make it true? What if our representations are unitary, or the profinite groups are (topologically) finitely-generated? Any results, no matter the number of hypotheses, would be of interest to me.
|
https://mathoverflow.net/users/175051
|
Irreducible representations of product of profinite groups
|
This is not even true for finite groups, in this generality, and not even in characteristic $0$. Consider, for example, the group $Q\_8 \times C\_3$, where $Q\_8$ is the quaternion group and $C\_3$ is cyclic of order $3$, and consider $\mathbb{Q}$-representations of this direct product. The standard representation $\rho$ of $Q\_8$ is not realisable over $\mathbb{Q}$, only $\rho\oplus \rho$ is. $C\_3$ has an irreducible $\mathbb{Q}$-representation $\chi$, given by the sum of the two non-trivial irreducible complex characters of $C\_3$. Now, $\rho\otimes \chi$ *is* realisable over $\mathbb{Q}$ and defines a simple $\mathbb{Q}[G]$-module, but it is not of the form $V\otimes W$ for any $\mathbb{Q}[Q\_8]$-module $V$ and $\mathbb{Q}[C\_3]$-module $W$.
If you wanted to restrict to finite dimensional representations over $\mathbb{C}$, then the statement will be true also for profinite groups, because any continuous complex finite dimensional representation of a profinite group will factor through a finite quotient.
|
7
|
https://mathoverflow.net/users/35416
|
397593
| 164,126 |
https://mathoverflow.net/questions/396589
|
1
|
Let $X$ be a closed manifold. $g:X\rightarrow \mathbb{R}$ be a smooth function ,$\alpha$ a section of a line bundle with discrete zeros and $c>0$ a constant, then Kazdan-Warner's work says that the following equation has an unique solution for$f$:
\begin{align\*}
2\Delta f+\frac{e^g\lvert\alpha\lvert^2}{4}e^{5f}=c
\end{align\*}
I am interested in the asymptotic behaviour of the solution when we scale $\alpha$ by $\lambda\alpha$ for some constant $\lambda$ and take $\lambda\rightarrow\infty$ and $\lambda\rightarrow 0$. Any idea, answer or reference is most welcome. Here $\Delta=d^\*d$.
|
https://mathoverflow.net/users/131004
|
Asymptotic behaviour of solution of Kazdan-Warner equations
|
It's actually quite easy and I completely missed it. If $f$ is the solution of the original equation:
\begin{align\*}
2\Delta f+\frac{e^g\lvert\alpha\lvert^2}{4}e^{5f}=c
\end{align\*}
and say $f\_\lambda$ is the solution of the perturbed equation:
\begin{align\*}
2\Delta f+\frac{e^g\lvert\lambda\alpha\lvert^2}{4}e^{5f}=c
\end{align\*}
Then notice just defining $f\_\lambda=f-\frac{2}{5}\text{ln}|\lambda|$ solves the second equation.
|
0
|
https://mathoverflow.net/users/131004
|
397601
| 164,129 |
https://mathoverflow.net/questions/397608
|
9
|
I've been studying the paper [An estimate of the remainder in a combinatorial central limit theorem](https://doi.org/10.1007/BF00533704) by Bolthausen, which proves the Berry Essen theorem using Stein's method:
Let $\gamma$ be the absolute third moment of a random variable $X$, and let $X\_{i}$ be iid with the same law as $X$. Let $S\_{n}=\sum\_{i}^{n}X\_{i}$, and suppose $E(X)=0$, $E(X^{2})=1$.
The goal is to find some universal constant $C$ such that $|P(S\_{n} \leq x) - P(Y\leq x)| \leq C\frac{\gamma}{\sqrt{n}}$.
Let $\delta(n,\gamma) = \sup\_{x}|P(S\_{n} \leq x) - P(Y\leq x)|$. We would like to bound $\sup\_{n}\frac{\sqrt{n}}{\gamma}\delta(n, \gamma)$.
In the proof the following bound is derived:
$\delta(n,\gamma) \leq c\frac{\gamma}{\sqrt{n}}+\frac{1}{2}\delta(n-1,\gamma)$ where $c$ is a universal constant. Noting that $\delta(1,\gamma) \leq 1$, the author claims that the result is implied. However when I try to use induction to get the result the constant $C$ increases without bound $n$ grows. If anyone has studied this paper before, I would love to hear from you.
|
https://mathoverflow.net/users/116781
|
Induction arising in proof of Berry Esseen theorem
|
Let $a\_n = \frac{\sqrt{n}}{\gamma} \delta \left( n, y \right)$. The bound you have stated implies that
$$a\_n \leq c + \frac{2}{3} a\_{n - 1}$$
where I replaced $\frac{\sqrt{n}}{\sqrt{n - 1}}$ with $\frac{4}{3}$ which is certainly true for $n > 2$. Therefore,
$$a\_n \leq c + \frac{2}{3} a\_{n - 1} \leq c \left( 1 + \frac{2}{3} \right) + \left( \frac{2}{3} \right)^2 a\_{n - 2} \leq \cdots \leq c \frac{1}{1 - \frac{2}{3}} + a\_2$$
which is bounded as required.
|
14
|
https://mathoverflow.net/users/88679
|
397610
| 164,132 |
https://mathoverflow.net/questions/397620
|
2
|
$$I\_n(t)=\int\_0^t\frac{1}{\left(x^5+1\right)^n}dx.$$
What is the relation between $I\_{n+1}(t)$ and $I\_n(t)$?
Can it be done with integration by parts?
|
https://mathoverflow.net/users/319917
|
Given the integral. What's the relation between $I_{n+1}(t)$ and $I_n(t)$?
|
We have
$$
I\_{n+1}(t)=\left(1-\frac{1}{5n}\right)I\_n(t)+\frac{t}{5n(t^5+1)^n},
$$
which is also compatible with Carlo Beenakker's comment above.
Indeed, integrating by parts we get
$$
I\_n(t)=\int\_0^t \frac{dx}{(x^5+1)^n}=\frac{t}{(t^5+1)^n}-\int\_0^txd\left(\frac{1}{(x^5+1)^n}\right)=
$$
$$
=\frac{t}{(t^5+1)^n}+\int\_0^tx\frac{5nx^4}{(x^5+1)^{n+1}}dx=\frac{t}{(t^5+1)^n}+\int\_0^t\frac{5n((x^5+1)-1)}{(x^5+1)^{n+1}}dx=
$$
$$
=\frac{t}{(t^5+1)^n}+5nI\_n(t)-5nI\_{n+1}(t),
$$
so
$$
5nI\_{n+1}(t)=\frac{t}{(t^5+1)^n}+(5n-1)I\_n(t)
$$
and we get the desired relation.
|
7
|
https://mathoverflow.net/users/101078
|
397626
| 164,136 |
https://mathoverflow.net/questions/397622
|
6
|
In [this thesis](http://theses.gla.ac.uk/182/) by Martin Hamilton on
Finiteness Conditions in Group Cohomology there is on page 11 a reference to following result:
**Theorem 1.2.14.** If $G$ is a *torsion-free* group and $H$ is a
subgroup of *finite index*, then
$$ \operatorname{cd} H = \operatorname{cd} G $$
where $\operatorname{cd} G $ is the cohomological dimension of $G$
defined as the **projective dimension**
of $\mathbb{Z}$ considered as $\mathbb{Z}G$-module with
trivial $G$ action, i.e. $g.1=1$ for every $g \in G$.
That is $\operatorname{cd} G =
\operatorname{proj.dim}\_{\mathbb{Z}G} \mathbb{Z}$ and
the latter is defined as the minimal length of all projecive
resolutions
$$ 0 \to P\_n \to P\_{n-1} \to ... \to P\_1 \to P\_0 \to \mathbb{Z} \to 0 $$
of projective $\mathbb{Z}G$-modules $P\_j$.
In the thesis the author gave as reference
Jean-Pierre Serre's publication "Cohomologie des groupes discrets", can be found in this Bourbaki collection band:
<https://www.springer.com/gp/book/9783540057208>
Unfortunatelly, this result cannot be found in this publication. So my concern
is where I can find a complete proof of the quoted Theorem above.
|
https://mathoverflow.net/users/108274
|
Cohomological dimension of torsion-free groups and its subgroups
|
This is Theorem 3.1, p. 190, in Brown, "Cohomology of groups". He also attributes it to Serre.
As a remark, this is the reason that the virtual cohomological dimension (vcd) is well-defined.
|
13
|
https://mathoverflow.net/users/5339
|
397627
| 164,137 |
https://mathoverflow.net/questions/396848
|
6
|
Do there exist pairs of $n$-dimensional closed Einstein manifolds $(M\_1,g\_1)$ and $(M\_2,g\_2)$, $n\ge 3$, such that the connected sum $M\_1\#M\_2$ carries an Einstein metric which is conformal to $g\_1$ and $g\_2$ on the summands?
|
https://mathoverflow.net/users/312063
|
Einstein metrics on connected sums
|
There are no nontrivial examples with $n\ge3$ beyond what I mentioned in my comment above, namely, either a conformal connected sum of a compact space form $(M\_1,g\_1)$ with the standard round $n$-sphere with the connected sum $M\_1\# M\_2$ being homothetic to $(M\_1,g\_1)$ or the case where both $M\_1$ and $M\_2$ are diffeomorphic to a (possibly exotic) $n$-sphere with homothetic metrics $g\_1$ and $g\_2$ and the resulting connected sum being again homothetic to $(M\_1,g\_1)$. (Note that [there are many inequivalent Einstein metrics on manifolds homeomorphic to odd-dimensional spheres](https://arxiv.org/pdf/math/0309408.pdf).)
To begin, consider the following question: Given an Einstein $n$-manifold $(M^n,g)$ (that is connected but not necessarily complete), with Einstein constant $\lambda$ (i.e., $\mathrm{Ric}(g)= (n{-}1)\lambda g$), what are the possibilities for Einstein metrics on $M$ that are conformally equivalent to $g$?
By [the well-known formula](https://en.wikipedia.org/wiki/List_of_formulas_in_Riemannian_geometry#Ricci_tensor) for the Ricci tensor of $\tilde g = u^{-2}\,g$, where $u>0$ is a function on $M$,
$$
\mathrm{Ric}(\tilde g)
= \mathrm{Ric}(g) + (n{-}2)\,u^{-1}\,\nabla(\mathrm{d}u)
- \bigl(u^{-1}\,\Delta u + (n{-}1)\, u^{-2}\,|\mathrm{d}u|^2\bigr)\, g\,,
$$
it follows that $\tilde g$ is also Einstein if and only if $u$ is a non-vanishing function on $M$ that satisfies the equation $\nabla(\mathrm{d}u) = v\,g$ for some function $v$. Differentiating this equation, one finds that $v+\lambda u$ must be locally constant. Thus, by the connectedness of $M$, there exists a constant $c$ such that $\nabla(\mathrm{d}u) = (c{-}\lambda u)\,g$. Consequently, $\Delta u = n(\lambda u{-}c)$, which will be useful below.
Moreover, one sees that the Einstein constant of $\tilde g$ is $\tilde\lambda = 2c\,u - \lambda\,u^2 - |\mathrm{d}u|^2$.
(Also, one sees that the vector field $\nabla u$ must annihilate the Weyl tensor of $g$, i.e., $W(\nabla u, X,Y,Z) = 0$ for all vector fields $X,Y,Z$ on $M$. However, that will play no role in this argument.)
For example, if $(M^n,g)=\bigl(\mathbb{R}^n,|\mathrm{d}x|^2\bigr)$ is the standard flat metric, then $\tilde g = u^{-2}\,g$ is Einstein (on the open set where $u$ is nonzero) if and only if $u = p\,|x|^2 + 2\,q{\cdot}x + r$
for some constants $p,r\in\mathbb{R}$ and vector $q\in\mathbb{R}^n$, not all zero.
This shows how to essentially 'linearize' the problem of describing the conformal multiples of an Einstein metric that are, themselves, Einstein:
On the bundle $E = \mathbb{R}\oplus\mathbb{R}\oplus T^\*\!M$ over $M$, consider the connection $D$ that, for any vector field $X$ and any section $(c,u,\alpha)$ of $E$, satisfies
$$
D\_X\begin{pmatrix}c, & u, & \alpha\end{pmatrix}
= \begin{pmatrix}\mathrm{d}c(X),& \mathrm{d}u(X)-\alpha(X),&
\nabla\_X\alpha - (c-\lambda\,u) X^\flat\end{pmatrix}
$$
where $X^\flat$ is the $1$-form that satisfies $X^\flat(Y) = g(X,Y)$
for all vector fields $Y$ on $M$.
Then we can rephrase the above equation on $u$ (and $c$) by saying that $u^{-2}\,g$ is Einstein if and only if there is a $c$ such that $(c,u,\mathrm{d}u)$ is a $D$-parallel section of $E$.
Now, it turns out that, *except for the metrics of constant sectional curvature on the $n$-sphere*, for a compact, connected Einstein manifold $(M,g)$, the space of global $D$-parallel sections of $E$ consists only of the sections of the form $(\lambda u,\,u,\,0)$ where $u$ is constant. (Such sections are obviously $D$-parallel.)
To see this, first note that if $(M^n,g)$ is compact, connected and Einstein with a non-positive Einstein constant, then the only global $D$-parallel sections of $E$ are those with $u$ constant. This is because the defining equation implies $\Delta u = n(\lambda u {-} c)$. When $\lambda=0$, this clearly implies $c=0$ and that $u$ be constant. When $\lambda<0$, this implies that $\lambda u {-} c$ is an eigenvalue of $\Delta$ with negative eigenvalue $n\lambda$ and hence must be $0$.
Meanwhile, if one supposes that $\lambda>0$, the equation $\nabla(\mathrm{d}u) = (c{-}\lambda u)\,g$ implies that, if $p$ is a critical point of $u$, then the Hessian of $u$ at $p$ is $\bigl(c{-}\lambda u(p)\bigr)$ times the metric at $p$. If $c{-}\lambda u(p)=0$, then the $D$-parallel section $(c,u,\mathrm{d}u)$ agrees with the $D$-parallel section $(\lambda\,u(p), u(p),0)$ at $p$ and hence must equal it, i.e., $u$ must be constant. On the other hand, if $c{-}\lambda u(p)\not=0$, then $p$ is a non-degenerate critical point of $u$ that is either a maximum or a minimum. Thus, on a connected, compact Einstein manifold $(M^n,g)$
the only critical points of a nonconstant $u$ such that $(c,u,\mathrm{d}u)$ is $D$-parallel for some $c$ are either strict maxima or minima. By the Mountain Pass Lemma, it follows that $u$ can only have one maximum and one minimum, which implies that $M$ is homeomorphic to a $n$-sphere.
Suppose that such a non-constant $u$ exists on $(M^n,g)$. By adding a constant to $u$, we can suppose that $u$ is strictly positive on $M$ and by scaling, we can assume that $\lambda=1$. Now by using the structure equations, one can prove that, on a punctured neighborhood of a maximum $p\in M$, $g$ can be written in the 'sine-cone' form $g = \mathrm{d}r^2 + (\sin r)^2\, h$, where $r$ is the distance from $p$, $h$ is an Einstein metric on $S^{n-1}$ with Einstein constant $1$, and $u = a + b \cos r$ for some constants $a$ and $b$. However, unless $h$ is conformally flat, the above sine-cone metric cannot be smooth at $r=0$. Thus, $h$ is conformally flat, which implies that $g$ is also conformally flat. Hence, the only possibility for $(M,g)$ is the round $n$-sphere, as was to be shown.
Now, for any connected Einstein $(M^n,g)$, the sheaf of $D$-parallel sections of $E$ is well-behaved: There is an integer $k$ satisfying $1\le k\le n{+}2$ such that, for any 1-connected open set $U\subset M$, the dimension of the vector space $\mathcal{P}(U)\subset\Gamma(U,E)$ consisting of $D$-parallel sections of $E$ over $U$ is $k$. This follows from the well-known result of DeTurck and Kazdan that $g$, being Einstein, is real-analytic in harmonic coordinates: Since the linear connection $D$ is perforce real-analytic, any germ of a $D$-parallel section defined on an open neighborhood of $p\in M$ can be real-analytically continued as a $D$-parallel section along any real-analytic curve that starts at $p$.
With all this understood, let's look at what might reasonably described as a 'conformal connected sum' of compact Einstein manifolds in dimension $n\ge 3$. I'll take it to be this: Three compact Einstein $n$-manifolds $(M\_i,g\_i)$ for $i=0, 1, 2$, with respective Einstein constants $\lambda\_i$, together with smoothly embbeded compact submanifolds $N\_i\subset M\_i$ where $N\_0$ is diffeomorphic to $[0,1]\times S^{n-1}$, and $N\_i$ is diffeomorphic to the $n$-ball $B^n$ for $i=1,2$, and, finally, where $M\_0{\setminus}N\_0$ is conformally diffeomorphic (as Riemannian manfolds) to the disjoint union of $M\_1{\setminus}N\_1$ and $M\_2{\setminus}N\_2$. (Without lost of generality, one can take this diffeomorphism to be the identity map, so assume this.)
Given these hypotheses, there exist smooth functions $u\_i: M\_i{\setminus}N\_i\to (0,\infty)$ for $i=1,2$, such that $g\_0 = {u\_i}^{-2} g\_i$ on $M\_i{\setminus}N\_i\subset M\_0{\setminus}N\_0$. Thus, there exist constants $c\_i$ such that $(c\_i,u\_i,\mathrm{d}u\_i)$ is a $D\_i$-parallel section of $E\_i$ on $M\_i{\setminus}N\_i$. Since $u\_i$ is defined on the complement of the smoothly embedded ball $N\_i\subset M\_i$ in $M\_i$, the 1-connectedness of $N\_i$ implies that this $D\_i$-parallel section extends uniquely as a $D\_i$-parallel section of $E\_i$ on all of $M\_i$. In particular, $u\_i$ extends to all of $M\_i$ satisfying $\nabla(\mathrm{d}u\_i)=(c\_i{-}\lambda\_i\,u\_i)g\_i$ for $i=1,2$.
Consequently, for $i=1$ or $2$, $u\_i$ must be constant unless $(M\_i,g\_i)$ is the round $n$-sphere.
If both $u\_1$ and $u\_2$ are constant, then by scaling $g\_1$ and $g\_2$ by constants, we can assume that $g\_0 = g\_i$ on $M\_i{\setminus}N\_i\subset M\_0{\setminus}N\_0$ for $i=1$ and $2$, and hence $\lambda\_0=\lambda\_1=\lambda\_2$. Moreover, because $N\_i$ is simply-connected and $g\_0$ and $g\_i$ are real-analytic, it is not hard to show that the isometric inclusion $M\_i{\setminus}N\_i\subset M\_0{\setminus}N\_0$ extends to an isometry $\iota\_i:(M\_i,g\_i)\to (M\_0,g\_0)$, and the compactness of $M\_i$ and $M\_0$ then implies that $\iota\_i$ is a diffeomorphism, mapping $N\_i$ diffeomorphically onto $M\_0{\setminus}M\_i = N\_0\cup (M\_{3-i}{\setminus}N\_{3-i})$. Thus, $M\_0$, $M\_1$, and $M\_2$ are all isometric and homeomorphic to an $n$-sphere. While $(M\_0,g\_0)$ is indeed a 'conformal connected sum' of $(M\_1,g\_1)$ and $(M\_2,g\_2)$ in the above sense, these ought to be regarded as trivial cases. As remarked at the beginning, though, there are many, many such examples that are not conformally flat.
If, say, $u\_1$ is constant and $u\_2$ is not, we can still reduce to the case that $g\_0=g\_1$ on $M\_1{\setminus}N\_1\subset M\_0{\setminus}N\_0$, and hence $\lambda\_0=\lambda\_1$. Again, because $N\_1$ is simply-connected and $g\_0$ and $g\_1$ are real-analytic, it follows that the isometric inclusion $M\_1{\setminus}N\_1\subset M\_0{\setminus}N\_0$ extends to an isometry $\iota\_1:(M\_1,g\_1)\to (M\_0,g\_0)$, and the compactness of $M\_1$ and $M\_0$ then implies that $\iota\_1$ is a diffeomorphism, mapping $N\_1$ diffeomorphically onto $M\_0{\setminus}M\_1 = N\_0\cup (M\_2{\setminus}N\_2)$. Meanwhile, $(M\_2,g\_2)$ must be isometric to the round $n$-sphere, which is conformally flat, implying that $(M\_0,g\_0)$ is conformally flat. Of course, examples such as this do exist with all of the $(M\_i,g\_i)$ being conformally flat, and $u\_2$ non-constant, in which one attaches a conformally flat bubble to a compact space form, but $(M\_0,g\_0)$ is isometric (up to a constant multiple) to $(M\_1,g\_1)$.
Finally, if neither $u\_1$ nor $u\_2$ is constant, then, $(M\_1,g\_1)$ and $(M\_2,g\_2)$ are both round $n$-spheres, so $M\_0$ is as well.
|
9
|
https://mathoverflow.net/users/13972
|
397631
| 164,139 |
https://mathoverflow.net/questions/397630
|
3
|
Given a convolution integral
$$
g(y) =\int\_a^b\varphi(y-x)f(x)dx=\int\_{-\infty}^{+\infty}\varphi(y-x)f(x)\mathbb{I}\_{[a,b]}(x)dx
$$
where
* $\varphi(x)= \frac{1}{\sqrt{2\pi}}\exp{\left(-\frac{x^2}{2}\right)}$ a gaussian function
* $f:x\in[a,b] \to \Bbb R$ a known function
I'm seeking a fast algorithm that allows to accurately approximate the function $g(y)$ over an interval $y\in [c,d]$.
I think this problem may be known in image processing or signal processing. Perhaps there exists a clever method using
* the property that $\varphi$ is a Gaussian function and/or
* $g(y)$ is a convolution of two function $\phi$ and $f\cdot \mathbb{I}\_{[a,b]}$.
**My attempt:**
Denote $\Delta y= \frac{d-c}{N\_y} $ , $\Delta x= \frac{b-a}{N\_x} $, $(y\_i,x\_i) = (c+i\cdot \Delta y, a+ i\cdot \Delta x)$
Approximate the integal g(y) by $(g(y\_i))\_{i=1,..,N}$, with
$$g(y\_i)=\Delta x \cdot\sum\_{j=1}^{N\_x} \varphi(y\_i -x\_j) f(x\_j) \text{for } i=1,...,N\_y$$
The complexity of this method is $\mathcal{O}(N\_x N\_y)$, which is slow.
I tried to set $\Delta x = \Delta y$ for arranging the terms $\varphi(y\_i-x\_j)\_{i,j}$ but it's difficult because there are cases where $(d-c) \gg (b-a)$ (or $(d-c) \ll (b-a)$) so if I fixed $N\_y$ for example, $N\_x$ becomes too many or too little.
|
https://mathoverflow.net/users/62193
|
Fast computation of convolution integral of a gaussian function
|
Convolution with a Gaussian kernel of an $n$-point function has $n^2$ complexity, while Fourier transformation (FFT), multiplication, and inverse Fourier transformation is only of complexity $n\log n$. Here is a [Python code](https://scipy-lectures.org/intro/scipy/auto_examples/solutions/plot_image_blur.html) for the two-dimensional case.
|
5
|
https://mathoverflow.net/users/11260
|
397632
| 164,140 |
https://mathoverflow.net/questions/397618
|
2
|
Let $X\sim\text{Hypergeometric}(n,k,m)$ and $Y\sim\text{Hypergeometric}(\binom{n}{2},\binom{k}{2},M)$, where $n>k>m$ are natural numbers and $M = \binom{m}{2}$. Consider $Z = \binom{X}{2}$. I want to show that $Y$ is stochastically dominated by $Z$.
Note that $X$ and $Y$ can be written as $X =\sum\_{i=1}^mX\_i$ and $Y=\sum\_{i=1}^MY\_i$ where $X\_i\sim\text{Bern}(k/n)$ and $Y\_i\sim\text{Bern}(\binom{k}{2}/\binom{n}{2})$, and the $X\_i$'s and $Y\_i$'s are not independent. The above stochastic dominance inequality to be shown translates to
$$
\sum\_{i=1}^MY\_i\leq \binom{X}{2} = \sum\_{i<j}X\_iX\_j.
$$
where the inequality is in the stochastic dominance sense. I cannot find a natural coupling for the left and right hand side summation terms.
|
https://mathoverflow.net/users/131426
|
Hypergeometric random variables domination
|
In general, $Y$ is not stochastically dominated by $Z$.
Indeed, suppose that $n>k>m\to\infty$ and $p:=\dfrac n{n+k}\to p\_\*\in[1/2,1)$. Then $EX=mp$ and $Var\,X\le mpq$, where $q:=1-p$. So, $\sqrt{Var\,X}\le\sqrt{mpq}=o(EX)$. So, $X$ is concentrated near $EX=mp\to\infty$ and hence $Z/m^2=X(X-1)/(2m^2)$ is concentrated near $$\dfrac{p^2}2\to\dfrac{p\_\*^2}2.$$
Similarly, letting $q\_\*:=1-p\_\*$, we see that $Y/m^2$ is concentrated near
$$\frac M{m^2}\, \dfrac{n^2/2}{n^2/2+k^2/2}\sim\dfrac{p^2}{2(p^2+q^2)}\to\dfrac{p\_\*^2}{2(p\_\*^2+q\_\*^2)}>\dfrac{p\_\*^2}2.$$
So, here $Y$ is not stochastically dominated by $Z$.
---
The above consideration was made assuming the parametrization of the hypergeometric distribution such that the population size is $n+k$. Assuming now the parametrization such that the population size is $n$, the answer is still no in general.
Indeed, suppose that $\infty\leftarrow m^2=o(\min(k,n-k))$ and $p:=k/n$ stays away from $0$ and $1$. Then the hypergeometric probabilities for $X$ are uniformly asymptotically equivalent to the corresponding probabilities of the binomial distribution with parameters $m,p$ -- cf. e.g. this [standard consideration](https://onlinelibrary.wiley.com/doi/pdf/10.1002/9781119536963.app4). Therefore and by the normal approximation to the binomial distribution,
$$X\approx N(mp,mpq)=N(EX,mpq),\tag{1}$$
again with $q:=1-p$. So, by the [delta method](https://en.wikipedia.org/wiki/Delta_method#Univariate_delta_method),
$$Z=X(X-1)/2\approx N(EZ,(mp)^2mpq).$$
On the other hand, similarly to (1),
$$Y\approx N(EY,\tfrac12\,m^2\,p^2(1-p^2)).$$
Also, as noted in a comment by the OP, $EZ=EY=:\mu$.
Thus, $Z$ and $Y$ have the same asymptotic mean $\mu$, but the asymptotic variance of $Z$ is much greater than that of $Y$, so that $Y$ is much more concentrated near $\mu$ than $Z$ is. Therefore, neither $Y$ is stochastically dominated by $Z$ nor vice versa.
|
3
|
https://mathoverflow.net/users/36721
|
397641
| 164,142 |
https://mathoverflow.net/questions/397639
|
5
|
Does there exist a continuous time martingale $X\_t$ not a.s. constant in $t$ that is almost surely everywhere differentiable?
|
https://mathoverflow.net/users/173490
|
Does there exist an almost surely differentiable martingale?
|
The answer is no.
Indeed, if a martingale is a.s. everywhere differentiable, then its [quadratic variation](https://en.wikipedia.org/wiki/Quadratic_variation#Definition) is a.s $0$. So, by the [Burkholder--Davis--Gundy inequality](https://en.wikipedia.org/wiki/Quadratic_variation#Martingales), the martingale is a.s. constant.
---
**Details:** Suppose that $X:=(X\_t)\_{t\in[0,1]}$ is an almost surely (a.s.) everywhere differentiable martingale. Replacing $X\_t$ by $X\_t-X\_0$, without loss of generality let us assume that $X\_0=0$. Take any real $a>0$ and consider the bounded martingale $X^a:=(X\_t^a)\_{t\ge0}$, where $X\_t^a:=X\_{\min(t,T\_a)}$ and $T\_a:=\inf\{t\in[0,1]\colon|X\_t|=a\}$, with $\inf\emptyset:=\infty$.
For any real-valued function $f$ on $[0,1]$, define its quadratic variation by the formula
$$[f]:=\limsup\sum\_{j=1}^n(f(t\_j)-f(t\_{j-1}))^2,$$
where the $\limsup$ is taken over all "partitions" $0=t\_0<\cdots<t\_n=1$ of $[0,1]$ as $\max\_{1\le j\le n}(t\_j-t\_{j-1})\to0$.
By the [Burkholder--Davis--Gundy inequality](https://en.wikipedia.org/wiki/Quadratic_variation#Martingales),
\begin{equation}
c\,E([X^a]^{1/2})\le EM^a\le C\,E([X^a]^{1/2}), \tag{BDG}
\end{equation}
where $c$ and $C$ are universal positive real constants and $M^a:=\max\_{t\in[0,1]}|X\_t^a|$.
By the first one of inequalities (BDG), $[X^a]<\infty$ a.s. So, using (i) Corollary 23 in [this paper](https://www.ams.org/journals/tran/2011-363-08/S0002-9947-2011-05209-8/S0002-9947-2011-05209-8.pdf), (ii) the remark on line 3 of page 4228 of the same paper that $\mu\_f=0$ iff $f\in V\_2^0$, (iii) the obvious identity $\mu\_f([0,1])=[f]$, and (iv) the definition of $V\_2$ as the set of all functions $f$ with $[f]<\infty$, we conclude (as in [this answer](https://mathoverflow.net/a/398042/36721)) that $[X^a]=0$ a.s. Therefore, by the second one of inequalities (BDG), $M^a=0$ a.s. for each $a$ and hence $X=0$ a.s., as desired.
|
13
|
https://mathoverflow.net/users/36721
|
397642
| 164,143 |
https://mathoverflow.net/questions/118092
|
32
|
Let $n \in \mathbb{N}$. Is it true that for any $a, b, c \in \mathbb{N}$ satisfying
$1 < a, b, c \leq n-2$ the symmetric group ${\rm S}\_n$ has elements of order $a$ and $b$
whose product has order $c$?
The assertion is true at least for $n \leq 10$, see
[here](https://stefan-kohl.github.io/problems/permutation_product_examples.txt).
**Update on Sep 2, 2015:** On Aug 10, 2015
[Joachim König](https://mathsci.kaist.ac.kr/%7Ejkoenig/) has posted
a [preprint](https://arxiv.org/abs/1508.02170) to the arXiv which gives
a positive answer to the question. Assuming that this preprint is correct,
this completely answers the question -- and thus also Problem 18.49 in the
[Kourovka Notebook](https://arxiv.org/abs/1401.0300).
**Update on Jun 18, 2014:** The assertion is true at least for $n \leq 50$, see
[here](https://stefan-kohl.github.io/problems/permutation_product_examples_50.txt) (4MB text file).
The list of examples in GAP-readable format can be found
[here](https://stefan-kohl.github.io/problems/permutation_product_examples_50.g).
**Added on Dec 11, 2013:** This question will appear as
Problem 18.49 in:
Kourovka Notebook: *Unsolved Problems in Group Theory*. Editors V. D.
Mazurov, E. I. Khukhro. 18th Edition, Novosibirsk 2014.
**Added on Nov 24, 2013:** Is there really not enough known about, say, the
class multiplication coefficients of ${\rm S}\_n$ to answer this question?
**Text of the question as of Feb 12, 2013:**
This question is a follow-up on [Order of elements](https://mathoverflow.net/questions/118035/order-of-elements) .
Derek Holt's answer to that question is nice, but it seems that the degree of the permutations it gives is a lot larger than necessary.
So, given natural numbers $m, n, k > 1$, what is the smallest $d$ such that the symmetric group of degree $d$ has elements of order $m$ and $n$ whose product has order $k$? -
Clearly if the largest of the numbers $m$, $n$, $k$ is prime, then $d$ must be at least
$\max(m,n,k)$, and there are some cases where $d$ actually must be larger. However a quick computation suggests that $d = \max(m,n,k) + 2$ might work always. - But does this or a similar bound hold?
EDIT: Smallest-degree examples for all $m, n, k \leq 8, m \leq n$ can be found
[here](https://stefan-kohl.github.io/problems/mnk_examples.txt).
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https://mathoverflow.net/users/28104
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Order of products of elements in symmetric groups
|
The main theorem in a paper of G. A. Miller [1] is the following:
>
> THEOREM. If $l, m, n$ are any three integers greater than unity, of which we
> call the greatest $k$, it is always possible to find three substitutions $(L, M, N)$ of $k + 2$ or some smaller number of elements and of orders $l, m, n$ respectively such that $LM=N$.
>
>
>
Which gives a positive answer to your question.
Papers [3], [4], [5] by Brenner and Lyndon give a different proof of Miller's result, and consider the problem of finding for $l,m,n > 1$ the smallest $d$ such that $S\_d$ contains permutations $x,y$ with $(|x|, |y|, |xy|) = (l,m,n)$. Other related results are e.g. in [2] and [6].
Papers [8], [9] also give a proof of the theorem by Miller. Also in [7], although it seems without assuming permutations of degree $\leq \max(l,m,n)+2$ in some cases.
The main theorem of [10] constructs for $l,m,n > 1$ elements $A,B \in PSL(2,q)$ (for suitable $q$) such that $|A| = l$, $|B| = m$, and $|AB| = n$. The construction pointed out by Derek Holt [here](https://mathoverflow.net/questions/118035/order-of-elements) is similar but a bit shorter. Yet another construction with matrices is in [11].
References:
[1] [MR1505829](https://mathscinet.ams.org/mathscinet-getitem?mr=1505829) G. A. Miller, On the Product of Two Substitutions. Amer. J. Math. 22 (1900), no. 2, 185–190. [JSTOR](https://www.jstor.org/stable/2369754)
[2] [MR1505882](https://mathscinet.ams.org/mathscinet-getitem?mr=1505882) G. A. Miller, Groups Defined by the Orders of Two Generators and the Order of their Product. Amer. J. Math. 24 (1902), no. 1, 96–100. [JSTOR](https://www.jstor.org/stable/2370009)
[3] [MR0767585](https://mathscinet.ams.org/mathscinet-getitem?mr=767585) J. L. Brenner, R. C. Lyndon, A theorem of G. A. Miller on the order of the product of two permutations. I. Jñānābha 14 (1984), 1–16. [link](http://docs.vijnanaparishadofindia.org/jnanabha/jnanabha_volume_14_1984/1.pdf)
[4] [MR0809272](https://mathscinet.ams.org/mathscinet-getitem?mr=809272) J. L. Brenner, R. C. Lyndon, A theorem of G. A. Miller on the order of the product of two permutations. II. The minimal degree in case a=2. Indian J. Math. 26 (1984), no. 1-3, 105–133 (1985).
[5] [MR0748120](https://mathscinet.ams.org/mathscinet-getitem?mr=748120) J. L. Brenner, R. C. Lyndon, A theorem of G. A. Miller on the order of the product of two permutations. III. The minimal degree in case a>2. Pure Appl. Math. Sci. 20 (1984), no. 1-2, 37–51.
[6] [MR0743150](https://mathscinet.ams.org/mathscinet-getitem?mr=743150) J. L. Brenner, R. C. Lyndon, The orbits of the product of two permutations. European J. Combin. 4 (1983), no. 4, 279–293. [DOI](https://doi.org/10.1016/S0195-6698(83)80024-9)
[7] [MR0053937](https://mathscinet.ams.org/mathscinet-getitem?mr=53937) R. H. Fox, On Fenchel's conjecture about F-groups. Mat. Tidsskr. B 1952 (1952), 61–65. [JSTOR](https://www.jstor.org/stable/24530069)
[8] [MR3508006](https://mathscinet.ams.org/mathscinet-getitem?mr=3508006) J. König, A note on the product of two permutations of prescribed orders. European J. Combin. 57 (2016), 50–56. [DOI](https://doi.org/10.1016/j.ejc.2016.03.006)
[9] [MR3694453](https://mathscinet.ams.org/mathscinet-getitem?mr=3694453) J. Pan, On a conjecture about orders of products of elements in the symmetric group. J. Pure Appl. Algebra 222 (2018), no. 2, 291–296. [DOI](https://doi.org/10.1016/j.jpaa.2017.04.001)
[10] [MR0283093](https://mathscinet.ams.org/mathscinet-getitem?mr=283093) R. D. Feuer, Torsion-free subgroups of triangle groups. Proc. Amer. Math. Soc. 30 (1971), 235–240. [DOI](https://doi.org/10.1090/S0002-9939-1971-0283093-6)
[11] [MR0207852](https://mathscinet.ams.org/mathscinet-getitem?mr=207852) J. Mennicke, Eine Bemerkung über Fuchssche Gruppen. Invent. Math. 2 (1967), 301–305. [DOI](https://doi.org/10.1007/BF01425406)
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https://mathoverflow.net/users/10146
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397646
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https://mathoverflow.net/questions/397308
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4
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For positive integers $n$ and $d$ satisfying $d = n-1$, let the $d$-dimensional regular simplex of side-length $\sqrt{2}$ be $X = \{(x\_1, x\_2, \cdots, x\_n) \in \mathbb{R}^n: x\_1+x\_2+\cdots + x\_n = 1, x\_i \ge 0\}$. How many translates of the set $\frac12 X = \{\frac12 x: x \in X\}$ are necessary to cover $X$? Rotation is not allowed!
The volume bound yields a lower bound of $2^d = 2^{n-1}$. The best upper bound I can get is the following construction: It's obvious that $n$ translates of $(1-\frac{1}{n})X$ can cover $X$ (this follows from the fact that among any $n$ nonnegative real numbers with sum $1$, at least one of them must be $\ge \frac{1}{n}$). Iterating this, we get that $n^k$ translates of $(1-\frac{1}{n})^k X$ can cover $X$. Choosing $k = cn$ yields an upper bound of $n^{cn} \approx d^{cd}$ for some constant $c$.
The vast gulf between the lower bound and the upper bound invites the question: Is the true answer exponential in $n$? That is, does there exist $c$ such that $c^n$ translates always suffice to cover $X$?
Idea #1 for improving the lower bound: If we can find points $P\_1, P\_2, \cdots, P\_r \in X$ with the property that $\| P\_i - P\_j \|\_{L\_1} > 1$ for $i \not= j$, then it's easy to show that no two points $P\_i$ can be part of the same translate of $\frac12 X$, implying that the answer is at least $r$. How might we construct such points $P\_i$? I tried a random maximal packing using translates of $\frac14 X - \frac14 X$, but this appears not to work. Perhaps a suitably chosen lattice? Or take a sparse subset of the $n!$ points $(p\_1, p\_2, \cdots, p\_n)$ satisfying $\{p\_1, p\_2, \cdots, p\_n\} = \{\frac12, \frac14, \frac18, \cdots, \frac{1}{2^{n-1}}, \frac{1}{2^{n-1}}\}$?
Idea #2 for improving the lower bound: If we can construct a weight function $\rho: X \rightarrow \mathbb{R}\_{\ge 0}$, then a lower bound would be $$\frac{\int\_X \rho}{\sup\_v \int\_{\frac12 X + v} \rho}.$$ (Inspired by the proof that one cannot cover a unit disk with a collection of strips the sum of whose widths is $< 2$.) But I don't think the simple weight function $\rho(x\_1,x\_2,\cdots,x\_n) = x\_1^2 + x\_2^2 + \cdots + x\_n^2$ will work, as it concentrates all the weight away from the center of $X$.
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https://mathoverflow.net/users/130843
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How many regular d-dimensional simplices of side length 1/2 are required to cover a regular d-dimensional simplex of side length 1?
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The following argument can probably be optimized, but it's the easiest I see at the moment. I think that you can cover $X$ by $8^n$ translates of $X/2$.
Fix $d$. Define by $\vec{v}$ the vector $(1, \ldots, 1)$, and by $\vec{v}^\perp$ the orthogonal subspace to $\vec{v}$ in $\mathbb{R}^d$.
Clearly the "filled simplex" $X' = \{(x\_1, \ldots, x\_n) \ : \ \sum x\_i \leq 1, x\_i \geq 0\}$ contains the cube $[0, 1/n]^n$. Therefore, $X'/2$ contains the cube
$C = [0, \frac{1}{2n}]^n$, which implies that the projection of $C$ to $\vec{v}^\perp$ is contained in the projection of $X'/2$ to $\vec{v}^{\perp}$, which is a translate of $X/2$.
Finally, note that $X$ can be covered by translates of $C$ as follows: for each vector $\vec{m} = (m\_1, \ldots, m\_n)$ of nonnegative integers, take an associated translate $C + \frac{\vec{m}}{2n}$. These translates cover the positive octant in $\mathbb{R}^d$ and so clearly cover $X$. A translate $C + \frac{\vec{m}}{2n}$ has nonempty intersection with $X$ iff $n \leq \sum m\_i \leq 2n$. The number of such translates needed to cover $X$ is thus bounded from above by the number of $\vec{m}$ summing to less than or equal to $2n$, which is ${3n \choose n}$ and so less than $2^{3n} = 8^n$.
But then after projecting to $\vec{v}^\perp$, the projections of the fewer than $8^n$ translates of $C$ cover a translate of $X$, and each projection of a translate of $C$ is contained in a translate of $X/2$ as argued above. So, $X$ is covered by $8^n$ translates of $X/2$.
UPDATE: I claim that $X$ can in fact be covered by $\binom{2d}{d}$ translates of $X/2$. To see this, I want to use an affine transformation to consider $Y = \{(x\_1, \ldots, x\_d) \ : \ x\_i \geq 0, \sum x\_i \leq 2d\}$ instead; clearly the answers for $X$ and $Y$ are the same.
Now, we simply note that any $\vec{y} \in Y$ can be written as $\vec{v} + \vec{w}$, where $\vec{v} \in (\mathbb{N}\_0)^d$ is the coordinatewise floor of $\vec{y}$. If the sum of $\vec{v}$ is less than or equal to $d$, we leave it alone. If it's greater than $d$, then we redefine $\vec{v}$ to be any coordinatewise smaller vector with entries in $\mathbb{N}\_0$ and sum exactly $d$ (and define $\vec{w} = \vec{y} - \vec{v}$).
In either case, $\vec{w}$ has nonnegative entries and sum at most $d$; in the former case this is because all entries are in $[0,1)$, and in the latter case since its sum is exactly the sum of $\vec{y}$ minus $d$. Therefore, $\vec{w} \in Y/2$.
The set of possible $\vec{v}$ is exactly the set $S = \{(n\_1, \ldots, n\_d) \ : \ n\_i \in \mathbb{N}\_0, \sum n\_i \leq d\}$. $S$ is in bijective correspondence with the set of $d$-element subsets of $\{1, \ldots, 2d\}$; for any such
subset $T$ with elements $t\_1 < t\_2 < \ldots t\_d$, define
$\phi(T) = (n\_1, \ldots, n\_d)$ by $n\_1 = t\_1 - 1$ and $n\_i = t\_i - t\_{i-1} - 1$ for $i > 1$. Therefore, $|S| = \binom{2d}{d}$.
We've shown that $Y \subset S + (Y/2)$, so $Y$ is covered by $\binom{2d}{d}$ translates of $Y/2$.
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https://mathoverflow.net/users/116357
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397648
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https://mathoverflow.net/questions/397636
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I was searching for a response on the internet but I was not able to find out an explicit answer.
It is known that if $\mathbb{P}^n \subset \mathbb{P}^N$ is embedded linearly then the normal bundle $N\_{\mathbb{P}^n/\mathbb{P}^N}\cong \mathcal{O}\_{\mathbb{P}^n}(1)^{\oplus (N-n)}$.
This can be proved for example via Koszul complex.
My question now is the following: if we embed $\mathbb{P}^n$ into $\mathbb{P}^N$ with higher degree, for example with the Veronese embedding
$$v\_d:\mathbb{P}^n \rightarrow \mathbb{P}^N:=\mathbb{P}(H^0(\mathcal{O}\_{\mathbb{P}^n}(d)))$$
what is it the normal bundle $N\_{v\_d(\mathbb{P}^n)/ \mathbb{P}^N}$? It is possible that could be $\mathcal{O}\_{v\_d(\mathbb{P}^n)}(d)^{\oplus(N-n)}$?
I was trying some Koszul approach like in the linear case but for $d>1$ I'm not able anymore to control the free resolution of the Veronese varieties.
Thanks in advance.
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https://mathoverflow.net/users/146431
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Normal bundle to Veronese varieties $v_d(\mathbb{P}^n)$ into $\mathbb{P}(H^0(\mathcal{O}_{\mathbb{P}^n}(d)))$
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The normal bundle $N$ of the Veronese embedding $\mathbb{P}(V) \to \mathbb{P}(S^dV)$ can be described by the exact sequence
$$
0 \to V \otimes \mathcal{O}(1) \to S^dV \otimes \mathcal{O}(d) \to N \to 0,
$$
where the first arrow is the unique nonzero $\mathrm{GL}(V)$-equivariant morphism.
Alternatively, one can describe the normal bundle as an iterated extension of symmetric powers of the tangent bundle $T$ of $\mathbb{P}(V)$ --- there is a filtration on $N$ with associated graded of the form
$$
\mathrm{gr}\_\bullet(N) = \bigoplus\_{i=2}^d S^i T.
$$
Finally, let me give a couple of explicit examples. If $\dim(V) = 2$ one has
$$
N\_{\mathbb{P}(V)/\mathbb{P}(S^dV)} \cong S^{d-2}V \otimes \mathcal{O}(d+2),
$$
and if $d = 2$ one has
$$
N\_{\mathbb{P}(V)/\mathbb{P}(S^dV)} \cong S^2T.
$$
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6
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https://mathoverflow.net/users/4428
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397649
| 164,146 |
https://mathoverflow.net/questions/397651
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1
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Let $Y$ be degree 5 index two prime Fano threefold. Let $\mathcal{E}$ and $\mathcal{Q}$ be the tautological sub and quotient bundle on $Y$. It is not hard to show that there is a short exact sequence:
$$0\rightarrow\mathcal{E}\xrightarrow{p}\mathcal{Q}^{\vee}\rightarrow I\_L\rightarrow 0$$, where $L$ is a line on $Y$. This is because one can show the map $p$ is injective by slope stability of $\mathcal{E}$ and $\mathcal{Q}^{\vee}$ and slope of them are $-\frac{1}{2},-\frac{1}{3}.$
But I was wondering whether the following exact sequence also exist or not?
$$0\rightarrow I\_L\rightarrow\mathcal{Q}\xrightarrow{\pi}\mathcal{E}^{\vee}\rightarrow 0$$
The problem to show the map $\pi$ is surjective is that the cokernel of $\pi$ could be rank zero sheaf, for example $\mathcal{O}\_L$, then there is no contradiction by arguing with stability and slope.
The similar thing happened in Gushel-Mukai threefold, there is a short exact sequence $$0\rightarrow\mathcal{E}\xrightarrow{p}\mathcal{Q}^{\vee}\rightarrow I\_C\rightarrow 0$$ with $C$ being a conic by the same stability argument and note that the zero locus of section of $\mathcal{Q}$ is either two points or a conic(in both special and ordinary GM case, with the argument slightly different)
But I was wondering whether the following exact sequence exist?
$$0\rightarrow I\_C\rightarrow\mathcal{Q}\xrightarrow{\pi}\mathcal{E}^{\vee}\rightarrow 0.$$
I thought it exists, but then I located a mistake in the argument, say the cokernel of $\pi$ could be $\mathcal{O}\_L$ and the image of $pi$ can be a torsion free semistable sheaf $E\in M(2,1,5)$.
Maybe there is a very simple reason that such short exact sequence does not exist, which I miss it?
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https://mathoverflow.net/users/41650
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A short exact sequence on del Pezzo threefold and Gushel-Mukai
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These sequences do not exist, because the kernel of an epimorphism of locally free sheaves is itself locally free, while the ideal sheaf of a curve on a threefold is not locally free.
Instead, there are distinguished triangles of the same form, but whose first terms are $I\_L^\vee$ and $I\_C^\vee$, the **derived** duals of the ideals. Note that these are complexes with two cohomology sheaves, which means that instead of a short exact sequence one has a four term exact sequence in each case.
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https://mathoverflow.net/users/4428
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397655
| 164,147 |
https://mathoverflow.net/questions/397576
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1
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Let $R$ be (assumed to be commutative, Noetherian) a regular local ring. Let $A$ be a direct limit of $R$-smooth algebras, such that the transition maps are $R$-étale.
Let $U= Spec(B)$ be an affine open subscheme of $Spec(A)$.
Further, assume that A and B are Noetherian (since it might happen that A is not necessarily Noetherian as noted at [Are essentially smooth schemes noetherian?](https://mathoverflow.net/questions/204745/are-essentially-smooth-schemes-noetherian)).
Is it true that $B$ can be written as a direct limit of $R$-smooth algebras with transition maps $R$-étale.
By [Popescu's desingularization theorem](https://en.wikipedia.org/wiki/Popescu%27s_theorem), it follows that $B$ is a direct limit of $R$-smooth algebras. But I suppose $R$-étale transition maps may not be guaranteed.
Also, can we put further restrictions on the base ring $R$, so that such a statement as above would be true?
Any comments are much appreciated!
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https://mathoverflow.net/users/157738
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Open affine subscheme of a direct limit of smooth algebras
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Turning the comments into an answer (CW). Write $A=\varinjlim\_{i\in I} A\_i$ and let $X=\operatorname{Spec}(A)$, $X\_i=\operatorname{Spec}(A\_i)$ and $U=\operatorname{Spec}(B)\subseteq X$. Every point $x\in U$ has an open neighborhood of the form $\operatorname{Spec}(A[f^{-1}])\subseteq U$ for some $f\in A$. Since $U$ is quasi-compact, we have $U=\bigcup\_{j=1}^n \operatorname{Spec}(A[f^{-1}\_j])$ for some $f\_1, \ldots, f\_n\in A$. In particular, we have an exact sequence
$$
0\to B \to \prod\_j A[f\_j^{-1}] \to \prod\_{j,k} A[(f\_j f\_{k})^{-1}].
$$
Changing the index set $I$, we may assume that it has a smallest element $0$ and that $f\_1, \ldots, f\_n\in A\_0$. Let $U\_i\subseteq X\_i$ denote the union of the opens $\operatorname{Spec}(A\_i[f\_j^{-1}])$ for $j=1, \ldots, n$. Writing $B\_i=\mathcal{O}(U\_i)$, we then have short exact sequences
$$
0\to B\_i \to \prod\_j A\_i[f\_j^{-1}] \to \prod\_{j,k} A\_i[(f\_j f\_{k})^{-1}].
$$
Since for $f\in A\_0$, we have $A[f^{-1}]= \varinjlim\_i A\_i[f^{-1}]$, and because colimit is exact, taking the colimit of the above exact sequences and comparing with the previous one we obtain
$$
B \simeq \varinjlim B\_i.
$$
Now each $U\_i$ is smooth and the maps $U\_i\to U\_{i'}$ are etale for $i\geq i'$. So we can conclude if we show that the $U\_i$ are affine for $i\gg 0$. But this follows from [SP Tag 01Z6](https://stacks.math.columbia.edu/tag/01Z6).
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https://mathoverflow.net/users/3847
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397665
| 164,153 |
https://mathoverflow.net/questions/397629
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34
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What I mean to ask is this:
given an irreducible **cubic** polynomial $P(X)\in \mathbb{Z}[X]$ is there always a **quadratic** $Q(X)\in \mathbb{Z}[X]$ such that $P(Q)$ is reducible (as a polynomial, and then necessarily the product of 2 irreducible cubic polynomials)?
I did quite some testing and always found a $Q$ that does the job. For example:
$P=aX^3+b,\quad Q=-abX^2,\quad P(Q)=-b(a^2bX^3-1)(a^2bX^3+1)$
$P=aX^3-x+1,\quad Q=-aX^2+X,\quad P(Q)=-(a^2X^3-2aX^2+X-1)(a^2X^3-aX^2+1)$
and a particular hard one to find:
$P=2X^3+X^2-X+4,\quad Q=-8X^2+5X+1,\quad P(Q)=(16X^3-18X^2+X+3)(64X^3-48X^2-11x-2)$
Could there be a formula for $Q$ that works for all cases?
It feels to me that this may have a really basic Galois theoretic proof or explanation, but I can't figure it out.
**Update.** Maybe a general formula for $Q$ is close. For $P=aX^3+cX+d$ taking $Q=-adX^2+cX$ works.
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https://mathoverflow.net/users/2480
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Does any cubic polynomial become reducible through composition with some quadratic?
|
You should refer to Lemma 10 (page-233) in [this](https://eudml.org/doc/204826) paper by Schinzel where he proves that for any polynomial $F(x)$ of degree $d$ we have a polynomial $G(x)$ of degree $d-1$ such that their composition is reducible.
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https://mathoverflow.net/users/160943
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397669
| 164,155 |
https://mathoverflow.net/questions/397660
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2
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Let $X$ be a projective variety with two morphisms $f:X\rightarrow Y$ and $g:X\rightarrow Z$ with irreducible fibers of positive dimension. Assume that $Pic(X) = f^{\*}Pic(Y)\oplus g^{\*}Pic(Z)$. Then if $D$ is a divisor on $X$ we can write $D = f^{\*}D\_Y + g^{\*}D\_Z$, where $D\_Y,D\_Z$ are divisors on $Y$ and $Z$ respectively.
If $D$ is effective are then $D\_Y$ and $D\_Z$ effective as well?
This holds for instance when $X = \mathbb{P}^n \times \mathbb{P}^m$ is a product and $f,g$ are the projections onto the factors.
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https://mathoverflow.net/users/14514
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A question on effective divisors
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This is false. The original post, included below, had some mistakes.
The example of @Pop is better, and in fact that example is where I started. It is straightforward to modify that example into an example satisfying the constraints. If @Pop wants to add an answer, then I am happy to delete this answer.
Let $Y$ be the projective plane. Fix a point $p$ in $Y$, and denote by $Z'$ the $1$-dimensional projective space parameterizing lines $L$ in $Y$ that contain $p$. Let $Z$ be the product of $Z'$ with a projective space $W$ of dimension $\geq 1$. Let $X'$ be the parameter of ordered pairs $(q,L)$ of a points of $Y$ and a line $L$ in $Y$ that contains both $p$ and $q$. Let $X$ be the product of $X'$ with $W$. Then, as a subvariety of the product $Y\times Z$ considered as a projective space bundle (of relative dimension $2$) over $Z$, the variety $X$ is a projective space subbundle over $Z$ (of relative dimension $1$). Thus, the pullback homomorphism from $\text{Pic}(Y)\oplus \text{Pic}(Z)$ to $\text{Pic}(X)$ is an isomorphism.
Now let $D'$ be the divisor in $X'$ parameterizing pairs $(q,L)$ such that $q$ equals $p$, and let $D$ be $D'\times W$. The normal bundle of $D'$ in $X'$ is anti-ample. Thus, the normal bundle of $D$ in $X$ is not nef. For the reason mentioned above, the divisors $D\_Y$ and $D\_Z$ are not both effective.
**Original post**.
Here are the details. Fix a vector space $V$ of dimension $4$ together with a linear subspace $U$ of dimension $1$. Denote by $\text{Flag}(1,3;V)$ the partial flag variety parameterizing ordered pairs $(A,B)$ of a $1$-dimensional linear subspace $A$ contained in a $3$-dimensional linear subspace $B$ contained in $V$. Denote by $X$ the closed subvariety of $X$ parameterizing ordered pairs such that $U$ is contained in $B$.
Let $Y$ be $\mathbb{P}V$, the parameter space of $1$-dimensional linear subspaces $A$ of $V$. Let $Z$ be the linear $2$-plane in $\text{Grass}(3,V)$ that parameterizes $3$-dimensional subspaces $B$ of $V$ that contain $U$, i.e., the dual projective space of $V/U$. Denote by $f$ and $g$ the forgetful morphisms from $X$ that remember only $A$, respectively $B$.
The fiber of $f$ over every point of $Y$ other than $[U]$ is a projective space $\mathbb{P}^1$. The fiber of $f$ over $[U]$ is the full projective space $Z = \mathbb{P}(V/U)^\vee \cong \mathbb{P}^2$. In all cases, the fiber is irreducible of positive dimension. The fiber of $g$ over every point of $Z$ is a P^1 $\mathbb{P}^2$.
In fact, the embedding of $X$ in $Y\times Z$ realizes $X$ as a projective space subbundle (of relative dimension 1 $2$)over $Z$ inside the (constant) projective space bundle $Y\times Z$ over $Z$ (of relative dimension $2$). From this and the formula for the Picard group of a projective space bundle, it is straightforward to see that the Picard group of $X$ is the isomorphic image under pullback of $\text{Pic}(Y)\oplus \text{Pic}(Z)$.
Now consider the divisor $D$ in $X$ parameterizing pairs $(A,B)$ such that $A$ equals $U$. The divisor $D\_Y$ is effective, linearly equivalent to a hyperplane class in the projective space $Y$. However, the divisor $D\_Z$ is not effective. In fact, it is linearly equivalent to the negative of the hyperplane class in the projective space $Z$.
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https://mathoverflow.net/users/13265
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397676
| 164,158 |
https://mathoverflow.net/questions/397674
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1
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Let $M \subset \mathbb{R}^d$ be some $C^2$ submanifold and $f:M \rightarrow \mathbb{R}$ be some $C^2$ function.
Since $f$ is $C^2$, there is $U$ a neighborhood of $M$ and $F:U \rightarrow \mathbb{R}$ a local $C^2$ extension of $f$ such that $F\_{M}= f$.
The gradient of $f$ at $x$ can then be defined as $\nabla f(x) = P\_x \nabla F(x)$, where $P\_x$ is the projection onto the tangent space of $M$ at $x$. This definition doesn't depend on the local extension $F$.
**Question:** Can $\nabla f : M \rightarrow \mathbb{R}^d$ be extended to a $C^1$ function on a neighborhood of $M$? And if so, is there a simple way to prove it?
Intuitively, since $f$ is $C^2$, it seems to me that the answer should be positive.
If $M$ is $C^3$, then I managed to prove this result using the fact that, $\pi\_{M}$, the projection onto $M$ is $C^2$. In this case $F = f \circ \pi\_{M}$ is $C^2$ (on a neighborhood of $M$) and its gradient is indeed equal to $\nabla f$ (and is $C^1$)
However, i don't know how to approach the case where $M$ is only $C^2$. Any help would be greatly appreciated.
|
https://mathoverflow.net/users/294260
|
Does the gradient of a twice differentiable function on a submanifold can be extended to a differentiable vector field?
|
Suppose your submanifold is a graph of a function $\mathbb{R}^{k}\to\mathbb{R}^{d-k}$.
Then you can extend your vector field by moving it in the directions on $ \mathbb{R}^{d-k}$; the obtained vector field is as smooth as the original.
In general case you can cover $M$ by such graphs, extend the vector field separately for each graph and mix the results using a partition of unity.
|
4
|
https://mathoverflow.net/users/1441
|
397677
| 164,159 |
https://mathoverflow.net/questions/397672
|
1
|
Does there exist infinite primes $p$ such that either $(p^a-1)/2$ or $(p^a+1)/2$ is a prime power for some integer $a\geq 2$?
|
https://mathoverflow.net/users/134942
|
Find primes satisfying specific properties
|
For $(p^2+1)/2$ solutions are $7,41,239,63018038201,19175002942688032928599$.
For $(p^2-1)/2$ solution is $3$.
The Pell equation $x^2 - 2 y^2 = \pm 1$ has infinitely many integer solutions, not sure the primality constraints leaves only the above.
|
0
|
https://mathoverflow.net/users/12481
|
397680
| 164,161 |
https://mathoverflow.net/questions/397682
|
7
|
I have a smooth, compact complex surface $X$, and I need an explicit formula for the Euler characteristic
$$\chi(X, \, S^n \Omega^1\_X),$$
where $S^n$ denotes the symmetric product, in terms of $c\_1(X), c\_2(X)$.
I know how to do the computation, by using the splitting principle in order to calculate the Chern classes $c\_i(S^n \Omega^1\_X)$ and then the Hirzebruch-Riemann-Roch formula, and I did it for small values of $n$. However, extracting a formula for general $n$ seems quite tedious.
>
> **Question.** Is there a reference where I can find the value of $\chi(X, \, S^n \Omega^1\_X)$? Related to this: is there a reference
> for the Chern classes $c\_1(S^n \Omega^1\_X)$, $c\_2(S^n \Omega^1\_X)$?
>
>
>
Notice that I need an *exact* formula, not an asymptotic one (which I already know).
|
https://mathoverflow.net/users/7460
|
Exact formula for $\chi(X, \, S^n \Omega^1_X)$
|
As you say, formulae for $c\_1(\Omega\_X^1)$ and $c\_2(\Omega\_X^1)$ can be obtained from the splitting principle. The following is a more general version of the calculation in [this answer](https://mathoverflow.net/a/351571/21564).
>
> **Lemma:** Let $V \to X$ be a rank two complex vector bundle. Then $c\_1(S^nV) = \frac{1}{2}n(n+1)c\_1(V)$ and $c\_2(S^nV) = \frac{1}{24}(n-1)n(n+1)(3n+2)c\_1(V)^2 + \frac{1}{6}n(n+1)(n+2)c\_2(V)$.
>
>
>
**Proof:** By the splitting principle, there is a map $p : Y \to X$ such that $p^\*$ is injective on integral cohomology and $p^\*V \cong L\_1\oplus L\_2$, so $p^\*(S^nV) \cong S^n(p^\*V) \cong S^n(L\_1\oplus L\_2)$. In general, we have $S^n(E\_1\oplus E\_2) \cong \bigoplus\_{i+j=n} S^i(E\_1)\otimes S^j(E\_2)$, but for a line bundle $L$ we have $S^k(L) = L^k$ (every tensor on a one-dimensional vector space is symmetric). Therefore
$$S^n(L\_1\oplus L\_2) \cong \bigoplus\_{i+j=n} S^i(L\_1)\otimes S^j(L\_2) \cong \bigoplus\_{i+j=n}L\_1^i\otimes L\_2^j \cong \bigoplus\_{i=0}^nL\_1^i\otimes L\_2^{n-i},$$
so we have
\begin{align\*} c\_1(S^n(L\_1\oplus L\_2)) & = c\_1\left(\bigoplus\_{i=0}^nL\_1^i\otimes L\_2^{n-i}\right)\\
&= \sum\_{i=0}^nc\_1(L\_1^i\otimes L\_2^{n-i})\\
&= \sum\_{i=0}^nic\_1(L\_1) + (n-i)c\_1(L\_2)\\
&= \left(\sum\_{i=0}^ni\right)c\_1(L\_1) + \left(\sum\_{i=0}^nn-i\right)c\_1(L\_2)\\
&= \tfrac{1}{2}n(n+1)(c\_1(L\_1) + c\_1(L\_2))\\
&= \tfrac{1}{2}n(n+1)c\_1(L\_1\oplus L\_2).\end{align\*}
That is, $p^\*(c\_1(S^nV)) = p^\*\left(\frac{1}{2}n(n+1)c\_1(V)\right)$ so the claim for $c\_1$ follows by the injectivity of $p^\*$.
For $c\_2$ we have
\begin{align\*}
&\, c\_2(S^n(L\_1\oplus L\_2))\\
&=\, c\_2\left(\bigoplus\_{i=0}^nL\_1^i\otimes L\_2^{n-i}\right)\\
&=\, \sum\_{0 \leq i < j \leq n}c\_1(L\_1^i\otimes L\_2^{n-i})c\_1(L\_1^j\otimes L\_2^{n-j})\\
&=\, \sum\_{0 \leq i < j \leq n}(ic\_1(L\_1) + (n-i)c\_1(L\_2))(jc\_1(L\_1) + (n-j)c\_1(L\_2))\\
&=\, \sum\_{0 \leq i < j \leq n}ijc\_1(L\_1)^2 + [i(n - j) + (n-i)j]c\_1(L\_1)c\_1(L\_2) + (n-i)(n-j)c\_1(L\_2)^2.\end{align\*}
By writing $\displaystyle\sum\_{0 \leq i < j \leq n}$ as $\displaystyle\sum\_{j=1}^n\sum\_{i=0}^{j-1}$ and using formulae for the sum of the first $k$ integers, squares, and cubes, one can show that
$$\sum\_{0 \leq i < j \leq n}ij = \sum\_{0\leq i < j \leq n}(n-i)(n-j) = \tfrac{1}{24}(n-1)n(n+1)(3n+2)$$
and
$$\sum\_{0 \leq i < j \leq n}i(n-j)+(n-i)j = \tfrac{1}{12}n(n+1)(3n^2+n+2),$$
so
\begin{align\*}
&\, c\_2(S^n(L\_1\oplus L\_2))\\
&=\, \tfrac{1}{24}(n-1)n(n+1)(3n+2)(c\_1(L\_1)^2 + c\_1(L\_2)^2)\\
&\, \ \quad + \tfrac{1}{12}n(n+1)(3n^2+n+2)c\_1(L\_1)c\_1(L\_2)\\
&=\, \tfrac{1}{24}(n-1)n(n+1)(3n+2)(c\_1(L\_1)^2 + 2c\_1(L\_1)c\_1(L\_2) + c\_1(L\_2)^2)\\
&\, \ \quad + \left[\tfrac{1}{12}n(n+1)(3n^2+n+2) - 2\tfrac{1}{24}(n-1)n(n+1)(3n+2)\right]c\_1(L\_1)c\_1(L\_2)\\
&=\, \tfrac{1}{24}(n-1)n(n+1)(3n+2)(c\_1(L\_1) + c\_1(L\_2))^2\\
&\, \ \quad + \tfrac{1}{12}n(n+1)[3n^2+n+2-(n-1)(3n+2)]c\_1(L\_1)c\_1(L\_2)\\
&=\, \tfrac{1}{24}(n-1)n(n+1)(3n+2)c\_1(L\_1\oplus L\_2)^2 + \tfrac{1}{6}n(n+1)(n+2)c\_2(L\_1\oplus L\_2).\\
\end{align\*}
As before, the claim for $c\_2$ follows from the injectivity of $p^\*$.$\quad\square$
Let $x\_n$, $y\_n$, and $z\_n$ denote the expressions of $n$ in the formulae above so that $c\_1(S^nV) = x\_nc\_1(V)$ and $c\_2(S^nV) = y\_nc\_1(V)^2 + z\_nc\_2(V)$. Then we have
\begin{align\*}
\operatorname{ch}(S^n\Omega\_X^1) &= \operatorname{rank}(S^n\Omega\_X^1) + c\_1(S^n\Omega\_X^1) + \tfrac{1}{2}(c\_1(S^n\Omega\_X^1)^2 - 2c\_2(S^n\Omega\_X^1))\\
&= \tbinom{n+2-1}{n} + x\_nc\_1(\Omega\_X^1) + \tfrac{1}{2}(x\_n^2c\_1(\Omega\_X^1)^2 - 2y\_nc\_1(\Omega\_X^1)^2 -2z\_nc\_2(\Omega\_X^1))\\
&= \tbinom{n+1}{n} - x\_nc\_1(X) + \tfrac{1}{2}(x\_n^2c\_1(X)^2 - 2y\_nc\_1(X)^2 -2z\_nc\_2(X))\\
&= n + 1 - x\_nc\_1(X) + \tfrac{1}{2}((x\_n^2 - 2y\_n)c\_1(X)^2 - 2z\_nc\_2(X))
\end{align\*}
so
\begin{align\*}
&\, \chi(X, S^n\Omega\_X^1)\\
&=\, \int\_X\operatorname{ch}(S^n\Omega\_X^1)\operatorname{Td}(X)\\
&=\, \int\_X\left(n + 1 - x\_nc\_1(X) + \tfrac{1}{2}((x\_n^2-2y\_n)c\_1(X)^2 - 2z\_nc\_2(X))\right)\cdot\\
&\qquad\qquad\qquad\left(1 + \tfrac{1}{2}c\_1(X) + \tfrac{1}{12}(c\_1(X)^2+c\_2(X))\right)\\
&=\, \int\_X\tfrac{1}{12}(n+1)(c\_1(X)^2 + c\_2(X)) -\tfrac{1}{2}x\_nc\_1(X)^2 + \tfrac{1}{2}((x\_n^2-2y\_n)c\_1(X)^2 - 2z\_nc\_2(X))\\
&=\, \left(\tfrac{1}{12}(n+1) - \tfrac{1}{2}x\_n + \tfrac{1}{2}(x\_n^2-2y\_n)\right)\int\_Xc\_1(X)^2 + \left(\tfrac{1}{12}(n+1) - z\_n\right)\int\_Xc\_2(X)\\
&=\, \tfrac{1}{12}(n+1)(2n^2-2n+1)\int\_Xc\_1(X)^2 - \tfrac{1}{12}(n+1)(2n^2+4n-1)\int\_Xc\_2(X).\end{align\*}
This can further be expressed in terms of the Euler characteristic $\chi(X)$ and signature $\sigma(X)$ using the fact that $\int\_Xc\_1(X)^2 = 2\chi(X) + 3\sigma(X)$ and $\int\_Xc\_2(X) = \chi(X)$:
$$\chi(X, S^n\Omega\_X^1) = \tfrac{1}{12}(n+1)(2n^2-8n+3)\chi(X) + \tfrac{1}{4}(n+1)(2n^2-2n+1)\sigma(X).$$
|
16
|
https://mathoverflow.net/users/21564
|
397705
| 164,168 |
https://mathoverflow.net/questions/397557
|
2
|
Mendelson, in *Introduction to Mathematical Logic*, 4th ed, 1997, had a more elegant approach to comprehension than predecessors, in my opinion.
With $x\in\mathbf{V}$ short for $\exists y(x\in y)$, and $\alpha$ any formula in the language of set theory (possibly without =), use the axiom schemas:
SE: $\exists y(y=\{x|\alpha\})$ and CA: $\forall x(x\in\{x|\alpha\}\leftrightarrow x\in \mathbf{V}\wedge \alpha)$
A class is a set just if it is a member of $\mathbf{V}$. Mendelson goes on and develops NBG set theory on the basis of these, and further assumptions.
*Question:* Have others explored other set theories, with an approach as this?
|
https://mathoverflow.net/users/37385
|
Have others explored Mendelson's approach to comprehension?
|
Yes! There are other set theories explored along this way generally speaking.
(1) Quine's Mathematical Logic $\sf ML$ adopted a similar approach on top of his $\sf NF$, and easily one can get a similar treatment on top of $\sf NFU$. See: Quine, Willard Van Orman (1951), Mathematical logic (Revised ed.), Cambridge, Mass.: Harvard University Press, ISBN 0-674-55451-5, MR 0045661
(2) Randall Holmes in his theory about symmetric extensions also used a similar approach. <https://randall-holmes.github.io/Drafts/symmetryrevisited.pdf>
I also used this approach in a closely related article on: <https://arxiv.org/abs/2012.08299>
(3) [Vopenka](https://plato.stanford.edu/entries/settheory-alternative/#VopeAlteSetTheo)'s alternative set theory
(4) Holmes [Pocket](https://plato.stanford.edu/entries/settheory-alternative/#PockSetTheo) set theory
|
4
|
https://mathoverflow.net/users/95347
|
397712
| 164,171 |
https://mathoverflow.net/questions/397719
|
6
|
Let $\mathbb H= \{z \in \mathbb C\,:\, \textrm{Re}\,(z)\geq 0\}$ and for $j=1,2$ suppose that $F^{(j)}:\mathbb H\to \mathbb H$ is defined via
$$ F^{(j)}(z) = \sum\_{k=1}^{\infty} \frac{a\_k^{(j)}}{z+\lambda^{(j)}\_k},$$
where $|a\_{k}^{(j)}|\leq k^{-2}$ and $0<\lambda^{(j)}\_1<\lambda^{(j)}\_2<...$ with $\lim\_{k\to \infty} \lambda^{(j)}\_k=\infty$. Suppose that
$$ F^{(1)}(n)=F^{(2)}(n)\quad \text{for all $n \in \mathbb N$}.$$
Does it follow that $F^{(1)}(z)=F^{(2)}(z)$ for all $z \in \mathbb H$?
|
https://mathoverflow.net/users/50438
|
Does it follow that $F^{(1)}(z)=F^{(2)}(z)$ for all $z \in \mathbb H$?
|
The answer is positive. Indeed, $f=F^{(1)}-F^{(2)}$ is a bounded analytic function in the right half-plane (this follows from your conditions $|a\_k|\leq k^{-2}$ and $\lambda\_k\to\infty$). But a bounded analytic function cannot
be zero at positive integers, unless it is identically equal to zero.
This follows from the [Blaschke condition](https://en.wikipedia.org/wiki/Blaschke_product) on zeros of bounded analytic functions.
Your assumption that $F$ maps the right half-plane into itself is redundant, and the assumption that $\lambda\_k>0$ can be very much relaxed; all you need is that your functions are bounded in the right half-plane.
|
10
|
https://mathoverflow.net/users/25510
|
397728
| 164,175 |
https://mathoverflow.net/questions/397726
|
10
|
Let $f\_n: [0, 1] \to \mathbb R$ be a sequence of positive functions in $L^\infty$ (hence a fortiori in $L^1$) that are equibounded in $L^\infty$ norm - that is $\sup\_{n \in \mathbb N} \|f\_n\|\_{L\_\infty} \leq M$ for some $M > 0$.
Is it true that there exists some absolute positive constant $c < 1$ such that
$$\inf\_{n\_k} \sup\_{i, j > N} \|f\_{n\_i} - f\_{n\_j}\|\_{L^1} \leq cM$$
for all such sequences $f\_n$?
Where the first infimum is taken over all increasing sequences $n\_k$ of naturals.
|
https://mathoverflow.net/users/173490
|
On equibounded sequences in $L^\infty$
|
**Edit:** I improved the constant to $c = \frac{2}{3}$. (Later edit: But the optimal constant turns out to be $c = \frac{1}{2}$, see [Yuval Peres' answer](https://mathoverflow.net/a/397845/102946).)
**Answer:** Yes, we have
$$
\inf\_{(n\_k)} \sup\_{i,j \in \mathbb{N}} \|f\_{n\_i} - f\_{n\_j}\|\_{L^1} \le \frac{2}{3} M, \label{1}\tag{$\ast$}
$$
for each sequence $(f\_n)$ in $(L^1)\_+$ whose sup norm is bounded by $M$. So we can choose $c = \frac{2}{3}$.
To see this, let $[0,\mathbf{1}] \subseteq L^\infty$ denote the positive unit ball in $L^\infty$.
**Lemma.** Three functions $f\_1, f\_2, f\_3 \in [0,\mathbf{1}]$ cannot have mutual $L^1$-distances that are all strictly larger than $\frac{2}{3}$.
*Proof.*
Set $g\_1 = |f\_1 - f\_2|$, $g\_2 = |f\_1 - f\_3|$ and $g\_3 = |f\_2 - f\_3|$.
For any three numbers $r\_1,r\_2,r\_3 \in [0,1]$, the sum of their three mutual distances in $\mathbb{R}$ is at most $2$.
Hence, $\int g\_1 + \int g\_2 + \int g\_3 \le 2$, which shows that it can't happen that all three functions $g\_k$ have norm strictly larger than $\frac{2}{3}$. $\square$
*Proof of the claim.*
We may, and will, assume that $M=1$. Assume for a contradiction that we can find a sequence $(f\_n)$ in $[0,\mathbf{1}]$ such that the infimum in the question is strictly larger than $\frac{2}{3}$.
Then there exists $n\_0$ such that $\|f\_{n\_0} - f\_n\|\_{L^1} > \frac{2}{3}$ for infinitely many $n$ (otherwise we could recursively construct a subsequence $(f\_{n\_k})$ such that the supremum in \eqref{1} is no more than $\frac{2}{3}$); let's denote the set of these $n$ by $J$.
For any two $j,k \in J$, it follows from the lemma that $\|f\_j - f\_k\| \le \frac{2}{3}$. Thus, you can take the elements of $J$ to be the indices of your wanted subsequence $(f\_{n\_k})$. Contradiction, since we assumed no such subsequence exists. $\square$
**Remark.** It's easy to see that the constant $\frac{2}{3}$ is optimal for the lemma (divide $[0,1]$ into three distjoint intervals $I\_k$ of measure $\frac{1}{3}$ and define $f\_k = \mathbf{1} - \mathbf{1}\_{I\_k}$), but I don't know whether it is optimal for the answer to the question.
|
10
|
https://mathoverflow.net/users/102946
|
397752
| 164,182 |
https://mathoverflow.net/questions/397750
|
5
|
Let $\mathcal{C}$ be a rigid, monoidal category. Can I talk about $\mathcal{C}$ as having a unique, well-defined, dualizing functor (i.e. one that maps objects and morphisms onto their respective duals)?
What is clear to me is that dual objects are unique up to unique isomorphism. However, all that seems to tell me is that dualizing functors are also unique up to unique isomorphism. In particular, the question of whether or not any dualizing functors exist never seems to come up. Is it not possible to imagine a rigid category which doesn't admit a functorial mapping of its objects onto their duals?
|
https://mathoverflow.net/users/137577
|
Does rigidity imply a unique dualizing functor?
|
For rigid symmetric monoidal categories, there is in fact always a duality functor, unique up to isomorphism (without symmetry you would have to specify what you mean by "dual").
Here is a possible proof of existence:
Consider the following category $\tilde C$, its objects are quadruples $(x,y,\eta,\epsilon)$ where $x,y$ are objects of $C$ and $\eta: 1\_C\to x\otimes y$ is a morphism in $C$, and $\epsilon : y\otimes x\to 1\_C$ is another morphism, together exhibiting $y$ as a dual of $x$.
A morphism $(x\_0,y\_0,\eta\_0,\epsilon\_0)\to (x\_1,y\_1,\eta\_1,\epsilon\_1)$ is a pair of morphisms, $f :x\_0\to x\_1, g : y\_1\to y\_0$ such that $g$ "is dual to $f$" in the following sense: the composite $$y\_1\cong y\_1\otimes 1\_C \to y\_1\otimes x\_0\otimes y\_0\to y\_1\otimes x\_1\otimes y\_0 \to 1\_C\otimes y\_0\cong y\_0$$
is equal to $g$, where I've used the unitors of $C$, the map $\eta\_0$, then $f$, and $\epsilon\_1$ .
Composition is the obvious one, where you have to check that the "dual" of $f\circ f'$ is the dual of $f'\circ$ the dual of $f$. This check is essentially going to come up in any proof, and is the essential part of the proof - you should try to do that check.
I then claim the following two things: the forgetful functor $\tilde C\to C, (x,y,\eta,\epsilon)\mapsto x$ and $(f,g)\mapsto f$ is an equivalence of categories.
The fact that it is essentially surjective comes from the fact that $ C$ is rigid, and fully faithfulness comes from the fact that for any $f$ there is a unique $g$ which works (namely, the composite that I described !).
Then you get the following zigzag $C\overset\sim\leftarrow \tilde C\to C^{op}$ where the second arrow is projection to $y$ (and $g$)
Now, equivalences have quasi-inverses (if you have the axiom of choice, or a canonical choice of a dual and duality data for any $x$ - which you don't need to be functorial *a priori*), so that gives you a well defined functor $D: C\to C^{op}$ which is obviously an equivalence.
Unicity (up to natural isomorphism) is not hard to prove, you should try that !
It gets a little bit more complicated for higher categories. A functor $D$ is always definable, in a similar way as above (although the definition of $\tilde C$ needs to be changed a bit, because you need to specify a $2$-morphism witnessing one of the triangle identities), but saying what it does on objects and morphisms is no longer enough to fully determine it as a functor - the obstruction is exactly the same obstruction as the unicity of a functor which is the identity on objects and on $1$-morphisms.
|
9
|
https://mathoverflow.net/users/102343
|
397753
| 164,183 |
https://mathoverflow.net/questions/397761
|
5
|
>
> Is it true that any subalgebra of singular matrices have a common null-vector?
>
>
>
In other words, is it true that, for any subalgebra $\cal S$
of the algebra of linear operators in a finite-dimensional vector space over a field,
$$
\bigcap\_{A\in\cal S}\ker A=\{0\}\quad\hbox{implies that}
\quad\ker A=\{0\}\hbox{ for some $A\in\cal S\;$? }
$$
(I am interested in finite prime fields mostly.)
Note that for subspaces (instead of subalgebras) [nothing similar is true](https://mathoverflow.net/q/47362/24165). But for algebras, Burnside's theorem gives me some hope...
|
https://mathoverflow.net/users/24165
|
Subalgebras of singular matrices
|
It's false. Take the subalgebra of $M\_3(K)$ generated by the matrices $\begin{bmatrix} 0 & 0&0\\ 1 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}$ and $\begin{bmatrix} 0 & 0 & 0\\ 0 &1&0\\ 1& 0&1\end{bmatrix}$. These two elements form a two element right zero semigroup and so the algebra they generate is just their span which is $2$-dimensional and obviously singular since all elements have first row zero. These two matrices have no common zero except zero.
Conceptually, let $S$ be the two-element right zero semigroup $\{a,b\}$. Then $KS$ acts faithfully on the left of $KS^1$ where $S^1$ is obtained by adjoining an identity. But these two right zeroes have no common vector they annihilate. For $a(c\_11+c\_2a+c\_3b) = c\_1a+c\_2a+c\_3b$ and $b(c\_11+c\_2a+c\_3b) =c\_1b+c\_2a+c\_3b$ and for these both to be zero you need $c\_3=0=c\_2$ and you need $c\_1=-c\_3$ and $c\_1=-c\_2$ and so $c\_1=c\_2=c\_3=0$. The representation is by singular matrices since $1$ is not in $KS\cdot KS^1$.
**Update** I guess you can go one dimension lower and consider the algebra of matrices of the form $\begin{bmatrix}a&b\\0&0\end{bmatrix}$.
|
5
|
https://mathoverflow.net/users/15934
|
397766
| 164,188 |
https://mathoverflow.net/questions/397770
|
7
|
We all know the series expansion
$$\log 2=\sum\_{n=1}^{\infty}\frac{(-1)^{n-1}}n. \tag1$$
I also am able to use the method of [Wilf-Zeilberger](https://en.wikipedia.org/wiki/Wilf%E2%80%93Zeilberger_pair) to the effect that
$$\log 2=3\sum\_{n=1}^{\infty}\frac{(-1)^{n-1}}{n\binom{2n}n2^n}. \tag2$$
>
> **QUESTION.** Can you provide yet another proof of the formula in (2)?
>
>
>
**Remark.** My motivation for this question goes beyond this particular series, hoping it paves a way forward in my study.
**Postscript.** After those generous replies (see below), it appears that the idea rests on
$$\log\left(1+\frac1x\right)=2\sinh^{-1}\left(\frac1{2\sqrt{x+x^2}}\right)$$
so that we may put $x=1$ to obtain (1) and (2).
To reveal the background: (2) is found from (1) by a "series acceleration" method which does not even stop there. In fact, stare at these two
\begin{align\*}\log 2&=3\sum\_{n=1}^{\infty}\frac{14n-3}{\binom{2n}2\binom{4n}{2n}2^{2n+1}}, \tag3 \\
\log 2&=3\sum\_{n=1}^{\infty}
\frac{(171n^2 - 111n + 14)(-1)^{n-1}}{\binom{3n}3\binom{6n}{3n}2^{3n+1}}
\tag4 \end{align\*}
One may now ask: can you furnish an alternative proof for the formulae (3) or (4)?
|
https://mathoverflow.net/users/66131
|
In search of an alternative proof of a series expansion for $\log 2$
|
Since you wish to develop techniques, you might want to consider the more general form
$$S\_k=\sum\_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^k\binom{2n}n2^n}.$$
The arcsine representation
$$\arcsin^2z=\frac12\sum\_{n=1}^\infty\frac{(2z)^{2n}}{n^2{2n \choose n}}$$
directly gives
$$S\_2=\tfrac{1}{2}\ln^2 2,$$
(substitute $z=2^{-3/2}i$), upon differentiation one finds
$$S\_1=\tfrac{1}{3}\ln 2,$$
$$S\_0=\tfrac{1}{9}+\tfrac{4}{27}\ln 2,$$
and upon integration,
$$S\_3=\tfrac{1}{4}\zeta (3)-\tfrac{1}{6}\ln^3 2 ,$$
$$S\_4=4\operatorname{Li}\_4\left(\tfrac12\right)-\tfrac72\zeta(4)+\tfrac{13}4\ln2\zeta(3)-\ln^22\zeta(2)+\tfrac5{24}\ln^42.$$
This method apparently fails to give a closed form expression for $k>4$, see this [MSE posting](https://math.stackexchange.com/q/3307589/87355).
|
13
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https://mathoverflow.net/users/11260
|
397774
| 164,191 |
https://mathoverflow.net/questions/397741
|
2
|
Let $f\in W^{1,2}\_{\text{loc}}(\mathbb R^2)$. Here, $W^{1,2}\_{\text{loc}}(\mathbb R^2)$ denotes the usual Sobolev space. More explicitly, $f:\mathbb R^2\to\mathbb R$ is a function such that, for every relatively compact open set $U\subset\mathbb R^2$,
* $f\vert\_U\in L^2(U)$ ;
* there exist $g\_1,g\_2\in L^2(U)$ such that $$\int\_U f\partial\_1\phi=-\int\_U g\_1 \phi,\text{ and }\int\_U f\partial\_2\phi=-\int\_U g\_2 \phi$$
for all test functions $\phi\in C\_{\text{c}}^\infty(U)$.
---
**My question.** Is the function $F:\mathbb R^2\to\mathbb R$, defined by
$$F(x,y)=\int\_0^y f(x,t)\,\mathrm dt$$
continuous?
More precisely stated, does there exist a function $\tilde F\in C(\mathbb R^2)$ such that $F=\tilde F$ Lebesgue-almost everywhere? (Note that the function $F$ is not well-defined at every point since $f$ is only defined as an equivalence class modulo "being equal almost everywhere".)
---
Note that (cf. Brezis *Functional Analysis, Sobolev Spaces and Partial Differential Equations*, Lemma 8.2) the function $F$ is continuous in the variable $y$ and, weakly, $\partial\_2 F=f$. However, I don't even see whether $F$ needs to be continuous in $x$.
*Remark.* If we had for instance $f\in W^{1,3}\_{\text{loc}}(\mathbb R^2)$, then it would be clear that $F$ is continuous, since, by Morrey's inequality (see Evans *Partial Differential Equations*, chapter 5.6.2, Theorem 4), the space $W^{1,p}(\mathbb R^n)$ can be embedded into $C^0(\mathbb R^n)$ whenever $p>n$. But my case is $p=n$, so this Theorem doesn't apply.
|
https://mathoverflow.net/users/129831
|
Is the parameter-dependent integral of a Sobolev function continuous?
|
I believe this holds more generally—here is the attempt I propose. Consider a function $f \in W^{1,p}\_{\mathrm{loc}}(\mathbf{R}^2)$ for some $p > 1$. Since the second variable is fixed in the problem, we can take $y = 1$ and define $F(x) = \int\_0^1 f(x,t) \mathrm{d} t$, outside of some negligible subset in $\mathbf{R}$.
The claim is that this function inherits from $f$ the property that
$$
F \in W^{1,p}\_{\mathrm{loc}}(\mathbf{R}),$$
from which the desired conclusion follows.
Let $I \subset \mathbf{R}$ be a finite interval. Then $$\int\_I \lvert F \rvert^p = \int\_I \Big \lvert \int\_0^1 f(x,t) \mathrm{d}t \Big\rvert^p \mathrm{d} x
\leq \int\_I \int\_0^1 \lvert f(x,t) \rvert^p \mathrm{d} t \mathrm{d} x,$$
so that $F \in L\_{\mathrm{loc}}^p(\mathbf{R})$.
In the same vein, given $h \in \mathbf{R}$ let $\tau\_h F: x \mapsto F(x-h)$. Then
\begin{eqnarray\*}
\int\_I \lvert \tau\_h F - F \rvert^p
&=& \int\_I \Big \lvert \int\_0^1 f(x+h,t) - f(x,t) \mathrm{d} t \Big \rvert^p \mathrm{d} x \\
&\leq& \int\_I \int\_0^1 \lvert f(x+h,t) - f(x,t) \rvert^p \mathrm{d} t \mathrm{d} x.
\end{eqnarray\*}
In other words
$$
\lvert \tau\_h F - F \rvert\_{L^p(I)} \leq \lvert \tau\_{he\_1}f - f \rvert\_{L^p(I \times [0,1])},$$
where $\tau\_{he\_1}$ is the translate of $f$ in the direction of the standard basis vector $e\_1 \in \mathbf{R}^2$.
The characterisation of Sobolev functions in terms of difference quotients means that there is $C > 0$ so that $$ \lvert \tau\_{he\_1}f - f \rvert\_{L^p(I \times [0,1])} \leq C h$$ for small enough $h$, which in turn implies that $F \in W^{1,p}\_{\mathrm{loc}}(\mathbf{R})$.
|
1
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https://mathoverflow.net/users/103792
|
397779
| 164,194 |
https://mathoverflow.net/questions/397777
|
4
|
For which integers $n>1$ is there a set of positive integers $S\subseteq \mathbb{N}$ with $n$ elements, and for every $s\in S$ the set $S\setminus\{s\}$ can be partitioned into two subsets with equal sum?
|
https://mathoverflow.net/users/8628
|
Finite subsets of $S\subseteq \mathbb{N}$ such that $S\setminus\{s\}$ can be partitioned with equal sum
|
As conjectured by bof, the answer is all odd $n \geq 7$.
*Proof.* Let $S$ be a set of positive integers such that $S \setminus \{s\}$ can be partitioned into two sets of equal sum for all $s \in S$. By parity considerations, note that all elements of $S$ are either all odd or all even. If all elements of $S$ are even, then the set obtained from $S$ by dividing every element by $2$ is also a valid set. Repeating this argument, we may assume that all elements of $S$ are odd. This implies that $n$ must be odd. Clearly, $n \notin \{1,3\}$ and bof has shown $n=5$ is also impossible (see comment below). To complete the proof we now show that every odd $n \geq 7$ is possible.
As noted by bof, $S=\{1,3,5,7,9,11,13\}$ shows that $n=7$ is possible, and it is easy to check $S=\{1,3,5,7,9,11,13,15,17\}$ is also valid. We will show by strong induction on $k$ that $S=\{1,3, \dots, 4k+1\}$ is a valid set. If $s \in \{1, 3, \dots, 4k-7\}$, then $\{1, 3, \dots, 4k-7\} \setminus \{s\}$, can be partitioned into two equal sum sets by induction. Since $\{4k-5, 4k-3, 4k-1, 4k+1\}$ can also be partitioned into two equal sum sets, we are done. If $s \in \{4k-5, 4k-3\}$, then by induction $\{1, 3, \dots, 4k-3\} \setminus \{s\}$ can be partitioned into two equal sum sets $S\_1$ and $S\_2$. Suppose that $1 \in S\_1$. Then $(S\_1 \setminus \{1\}) \cup \{4k+1\}$ and $S\_2 \cup \{1, 4k-1\}$ are equal sum subsets of $S \setminus \{s\}$. Suppose $s=4k-1$. By induction, there is a partition of $\{1, 3, \dots, 4k-5\}$ into two equal sum sets $S\_1$ and $S\_2$. Suppose $4k-5 \in S\_1$, Then there must be some $\ell \in S\_1$ such that $\ell-2 \in S\_2$. We obtain the required partition of $S \setminus \{s\}$ by swapping $\ell$ and $\ell-2$, adding $4k+1$ to $S\_1$, and adding $4k-3$ to $S\_2$. Finally, suppose $s=4k+1$. By induction, there is a partition of $\{1, 3, \dots, 4k-7\} \cup \{4k-3\}$ into two equal sum sets $S\_1$ and $S\_2$. Suppose $4k-7 \in S\_1$. Then there must be some $\ell \in S\_1$ such that $\ell-2 \in S\_2$. The required partition of $S \setminus \{4k+1\}$ is obtained by swapping $\ell$ and $\ell-2$, adding $4k-1$ to $S\_1$, and adding $4k-5$ to $S\_2$.
|
9
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https://mathoverflow.net/users/2233
|
397794
| 164,198 |
https://mathoverflow.net/questions/397800
|
5
|
Let $f(x),g(x),p(x)$ be non-constant polynomials with rational coefficients.
Is it true that for all $f$ exist $g,p$ such that $p(x)^2 \mid f(g(x))$?
Partial results:
$f(g(x))$ is divisible by square iff the discriminant of $f(g(x))$ is zero.
For variables $z\_i$, write $g\_0(x)=\sum\_{i=0}^n z\_i x^i$.
Then the discriminant of $f(g\_0(x))$ is polynomial $D$ in variables $z\_i$.
Solution $D=0$ will find $p(x)$, but finding points on variety is hard.
If we allow $g,p$ to be with coefficients algebraic integers, the solution
is easy: in $D$ fix all but one variables say $z\_0$ to be integers and work in the
number field with defining polynomial $D(z\_0)$.
As corollary to positive solution, we have $f(g(x))$ reducible.
|
https://mathoverflow.net/users/12481
|
When $p(x)^2 \mid f(g(x))$?
|
Yes, it is true.
Let $f\_0$ be an irreducible divisor of $f$. It suffices to find $g$ such that $f\_0^2$ divides $f\_0(g(x))$ (which, in turn, divides, $f(g(x))$).
Try to choose $g(x)=x+h(x)f\_0(x)$. Then $f\_0(g(x))=f\_0(x+h(x)f\_0(x))\equiv f\_0(x)+f\_0'(x)h(x)f\_0(x) \pmod {f\_0^2(x)}$, and we need $1+f\_0'(x)h(x)$ to be divisible by $f\_0$. Since $f\_0'$ and $f\_0$ are coprime (as $f\_0$ is irreducible and $0\leqslant \deg f\_0'<\deg f\_0$), such $h$ exists.
Actually by the same reasoning once we found $g\_k$ for which $f\_0^k$ divides $f\_0(g\_k)$, where $k\geqslant 1$, we may find $h$ such that $f\_0(g\_k+f\_0^kh)$ is divisible by $f\_0^{k+1}$.
|
15
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https://mathoverflow.net/users/4312
|
397803
| 164,200 |
https://mathoverflow.net/questions/397760
|
2
|
In Fulton's intersection theory, example 1.7.1, he mentioned an example that contradicts to the splitting of cycles with respect to irreducible components. Consider the subscheme $X$ in $\mathbb{A}^3$ defined by $(zx,zy)$, and consider the Cartier divisor $E$ defined by $z-x$. Then should the cycle of $E$ should be the y-axis with an additional zero point? Or. it is just the y-axis itself.
|
https://mathoverflow.net/users/130556
|
Cycle of non-equidimensional scheme
|
Probably this is more than what you were looking for. I hope I haven't made silly blunders.
$X$ has two components (the x-y plane) $X\_1=V(z)$ and (the z-axis) $X\_2=V(x, y)$ with geometric multiplicities 1 each.
As in the Lemma 1.7.2, the RHS is $1[E\cap X\_1]+1[E\cap X\_2]=1[(0, 0, 0)]+1[y-axis]$.
Now the LHS: the fundamental cycle $[E]=2[(0, 0, 0)]+1[y-axis]$.
(To see this: $\mathcal{O}(E)=\dfrac{k[x,y,z]}{(xz, yz, x-z)}\simeq \dfrac{k[x, y]}{(x^2, xy)}$.
So $E$ has two irreducible components corresponding to the ideals $(x, z)$ and $(x, y, z)$ in $k[x, y, z]$ (or the ideals $(x)$ and $(x, y)$ in $k[x, y]$ via the above isomorphism.
The geometric multiplicity of $E$ at (the origin) $V((x, y))$ is given by the length $l(\mathcal{O}\_{E, V((x, y)})$ as $\mathcal{O}\_{E, V((x, y)}$-module. $\mathcal{O}\_{E, V((x, y)}\simeq \Bigg(\dfrac{k[x, y]}{(x^2, xy)}\Bigg)\_{(x, y)}.$ This has a filtration $\Bigg(\dfrac{k[x, y]}{(x^2, xy)}\Bigg)\_{(x, y)}\supset (x, y) \supset \{0\}$ such that the consecutive quotients are simple $\Bigg(\dfrac{k[x, y]}{(x^2, xy)}\Bigg)\_{(x, y)}$-modules.
So the length $l(\mathcal{O}\_{E, V((x, y)})=2$.)
The other multiplicity can be calculated similarly.
All these equalities are in the graded group $Z\_\*(X)$.
Thus, the 'LHS' and 'RHS' in Lemma 1.7.2 above don't agree in $Z\_{\*}(X)$.
|
1
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https://mathoverflow.net/users/157738
|
397807
| 164,202 |
https://mathoverflow.net/questions/397809
|
7
|
Let $f\in C^2(\Bbb R^m), f\geq 0$, Hessian matrix of $f$ is upper bounded by some constant $C$. Do we have $|\nabla f|\leq \alpha \sqrt{f}$ for some $\alpha$, even if the Hessian matrix is degenerate?
|
https://mathoverflow.net/users/321329
|
A property of $C^2$ functions
|
$\newcommand\R{\mathbb R}$Let $\R:=R$. Suppose that $|f''(x)(h,h)|\le C|h|^2$ for all $x$ and $h$ in $\R^m$ -- this is how we interpret the condition "Hessian matrix of $f$ is upper bounded by some constant $C$". Of course, here $f''(x)$ is the bilinear form that is the second derivative of $f$ at $x$, so that $f''(x)(h,k)=h^\top H(x)k$ for all $x,h,k$ in $\R^m$, where $H(x)$ is the Hessian matrix of $f$ at $x$.
Consider first the case $m=1$. Take any $x\in\R$ and any real $h>0$. Then
$$0\le f(x+h)\le f(x)+f'(x)h+Ch^2/2,$$
whence
$$f'(x)\ge-\frac{f(x)}h-\frac C2\,h=-K\sqrt{f(x)}$$
if $h=\sqrt{2f(x)/C}$,
where $K:=\sqrt{2C}$. Similarly (or by the left-right symmetry), $f'(x)\le K\sqrt{f(x)}$, and hence
$$|f'(x)|\le K\sqrt{f(x)} \label{1}\tag{1}$$
for all $x\in\R$.
Now take any natural $m$. Considering the restrictions of $f$ to all straight lines in $\R^m$, we see that \eqref{1} holds for all $x\in\R^m$, where now $|f'(x)|$ denotes the norm of the linear form $f'(x)$ that is the derivative of $f$ at $x$, so that $f'(x)(h)=h^\top\,\nabla f(x)$ for all $h\in\R^n$ and hence
$|f'(x)|=|\nabla f(x)|$.
Thus, the desired conclusion holds, in general.
|
9
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https://mathoverflow.net/users/36721
|
397813
| 164,204 |
https://mathoverflow.net/questions/397816
|
5
|
Consider the PDE
$$\partial\_t f(x,t) = \langle q(x), \nabla \rangle f(t,x) + p(x),$$
with Schwartz initial data $f(0,x) = f\_0(x) \in \mathscr S(\mathbb R^n).$
I am wondering then if $q$ and all its derivatives are polynomially bounded and $p$ is Schwartz, too:
Does there exist a solution to this equation that decays faster than any polynomial in space $x$ at any fixed time $t>0$?
This sounds plausible to me, but I am not sure how one argues for such an equation. I assume it must be a classical question.
As there was apparently some confusion about the meaning of this question, let me ask it again:
Fix a time $t>0$, then as a function of $x$, does the solution decay faster than any polynomial? This seems to be true in your case for example, as it is just a translation of a Schwartz function.
|
https://mathoverflow.net/users/150564
|
Linear transport equation with unbounded coefficients
|
No. If e.g. $n=1$, $p=0$, and $Bq(x)=1$ for all $x$, then $f(t,x)=f\_0(t+x)$, which does not decay along the lines $\{(t,x)\colon t+x=c\}$ for real $c$.
---
The OP has changed the question, now looking for decay only in $x$, faster than any polynomial, for each $t>0$. Then the above answer is no longer valid.
However, then the answer is still no, in general; here, we just need to change the space variable. E.g., let $n=1$ and $Bq(x)=x^2+1$ for all $x$. Then
\begin{equation}
f(t,x)=f\_0(\tan(t+\tan^{-1}x)),
\end{equation}
which is not even defined at any point $(t,x)$ such that $t+\tan^{-1}x=\pi/2$. For each $t\notin\pi\mathbb Z$, the solution $f$ will explode to $\pm\infty$ at all the points of the form $x=(-1)^k\cot t$ for $k\in\mathbb Z$.
|
5
|
https://mathoverflow.net/users/36721
|
397820
| 164,206 |
https://mathoverflow.net/questions/397747
|
6
|
The field of Puiseux series over an algebraically closed field of characteristic zero is also an algebraically closed field, and furthermore it has a valuation so that our Puiseux series can be tropicalized.
Is the tropicalization of the solutions of a monic polynomial equation over them the same as the solution of the tropicalization of the polynomial equation?
I.e. informally, can we get the degree of the largest pole of the series that solves the equation only from the information of the largest poles in the coefficients of the polynomial?
|
https://mathoverflow.net/users/174368
|
Does solving polynomial equations commute with tropicalization? (particularly for the field of Puiseux series)
|
Yes.
This is normally expressed in terms of the [Newton polygon](https://en.wikipedia.org/wiki/Newton_polygon) of the polynomial. Specifically, given an arbitrary field $K$ with a valuation, and a polynomial $f(x) = a\_n x^n + a\_{n-1} x^{n-1} + \dots + a\_1 x + a\_0$, the Newton polygon of $f$ is the lower side of the convex hull of the set of points $(i, v(a\_i))$.
If $K$ is algebraically closed, then the valuations of the $n$ roots of $f$ are exactly minus the slopes of the Newton polygon over the $n$ intervals $[0,1], [1,2], \dots, [n-1,n]$.
The most straightforward way I know to check this is to first, completely factor $f$ into linear factors, i.e. $a\_n \prod\_{i=1}^n (x- \alpha\_i)$, with the $\alpha\_i$ in order of increasing valuation, write $a\_i$ in terms of the $\alpha\_j$, and then check that $$v(a\_{n-k}) \geq v(a\_0) + \sum\_{i=1}^k v(\alpha\_i)$$ with equality if $v(\alpha\_i) < v(\alpha\_j)$, so that the Newton polygon is contained in the polygon with slopes $- v(\alpha\_1),\dots, -v(\alpha\_n)$, and contains the corners of that polygon, hence is equal to that polygon. This inequality and equality can be checked using the definition of valuation.
Similarly, one can check that the tropicalization of a polynomial equation is a piecewise linear function with corners at minus the slopes of the Newton polygon (in fact, it is dual to the Newton polygon in a certain sense, so the edges of the tropicalization correspond to vertices of the Newton polygon and vice versa), and so the solutions of the tropicalization are also minus the slopes of the Newton polygon.
|
6
|
https://mathoverflow.net/users/18060
|
397821
| 164,207 |
https://mathoverflow.net/questions/397826
|
3
|
I've been self studying differential geometry for a little while now (4-6 months). I am learning from Lee's *Introduction to Smooth Manifolds*, and I just don't quite get the point of the subject. Why do we study the constructions that we study, such as differential forms, submanifolds, vector bundles, etc. ? What is the goal of differential geometry, i.e., what is the motivation for studying it?
Edit: As Will Sawin suggested, I will describe things I find motivating. I find beautiful structures, unsolved problems, and a general goal of what we are trying to do in the subject to be motivating. For instance, the motivation in topology (point-set) is to find topological invariants of a given space. This leads us to connectedness, compactness, the fundamental group, etc. In this example, we have a general rule for what we are trying to accomplish. What is this such rule in differential geometry?
|
https://mathoverflow.net/users/167759
|
What's the point of differential geometry?
|
In Physics, most theories can be formulated in a differential geometric framework:
1. General relativity: Space-time is modeled as a 4d-pseudo-Riemannian manifold. Frederic Schuller has an excellent set of lectures on this: <https://www.youtube.com/watch?v=7G4SqIboeig>
2. Electromagnetism: Has an elegant formulation in terms of differential forms. The generalized Stokes theorem for differentiable manifolds is critical here.
3. Quantum mechanics and Quantum field theory also have differential geometric formulations. I believe Frederic Schuller speaks more on this in his public Quantum Mechanics lecture series.
4. Hamiltonian (classical) mechanics models the state-space (momentum + position) as a manifold.
The key benefit of formulating physical theories in terms of an abstract (coordinate-free) manifold is that this ensures that if two scientists in different coordinate frames develop the "same" theory for some physical phenomenon then the two theories will agree with one another (i.e. are equivalent up to a change in coordinates).
In Statistics, the general theory of statistical efficiency theory is built upon (infinite-dimensional) differential geometry (See Bickel, Klaassen, Ritov, Wellner, (1998). Efficient and Adaptive Estimation for Semi Parametric Models. Chicago Journal of Theoretical Computer Science). Specifically, the theory models the statistical model (space of probability distributions) as a Hilbert manifold where the tangent spaces are now Hilbert spaces. The theory then considers the pathwise derivative of smooth functions/parameters of the Hilbert manifold where the pathwise derivative can be viewed as a linear mapping on the tangent spaces. In this case, the statistical model is truly an abstract manifold with charts being the densities of probability distributions that are dominated by some measure $\mu$. Since probability distributions are not dominated by a single measure, there is no global coordinate chart for a fully nonparametric statistical model. The tools of abstract manifold theory allow one to rigorously formulate the notion of a smooth statistical model and develop statistical theory without restricting ourselves to working with densities (which are arbitrary as they depend on the choice of dominating measure). Loosely speaking, this ensures that if two people derive the same statistical theory in the density space but with different dominating measures, we can be confident that their theories will agree. (This is also the motivation in physics). Here is one document I could find that goes over some of the differential geometric formulation of efficiency theory: [http://www.stat.columbia.edu/~bodhi/Talks/SPThNotes.pdf](http://www.stat.columbia.edu/%7Ebodhi/Talks/SPThNotes.pdf). See section 5, especially.
There is also a more parametric-focused subfield of statistics called information geometry that leverages euclidean/finite-dimensional differential geometry. Also, in recent years, manifold learning has become of great interest. For example, in brain-imaging, the brain is modeled as either a 2d or 3d manifold. People have generalized a number of well-known machine learning algorithms to this setting. Differential geometry plays a key role here.
Computer science and optimization.
Functional gradient descent and optimization/gradient descent on manifolds are current research areas that leverage geometry. The gradient flow on a manifold is a common object here. See this presentation for instance: <https://www.math.cmu.edu/users/slepcev/Chi_squared_SVGD_presentation.pdf>. Also, convex geometry/optimization borrows a lot of ideas from differential geometry (e.g. tangent cones).
Personally, my biggest motivation for studying differential geometry was to understand smooth maps on smooth (possibly infinite-dimensional) surfaces. The notions of "paths" on manifolds and derivatives of functions along paths are critical for this. Also, viewing the (path-wise) derivative of a function as a linear mapping on the tangent spaces (and push-forwards and pull-backs) is a powerful abstraction.
|
16
|
https://mathoverflow.net/users/317937
|
397828
| 164,209 |
https://mathoverflow.net/questions/397824
|
2
|
Let $M,N$ be smooth closed manifolds acted by a finite group $G$. Let $f\colon M\to N$ be a $C^1$-smooth $G$-equivariant map.
**Is it true that for any $\varepsilon>0$ there exists a $C^\infty$-smooth $G$-equivariant map
$$f\_\varepsilon\colon M\to N$$ such that $\|f-f\_{\varepsilon}\|\_{C^1}<\varepsilon$, where the $C^1$-norm is taken with respect to some Riemannian metrics on $M$ and $N$?**
A reference would be most helpful.
|
https://mathoverflow.net/users/16183
|
Approximation of $C^1$-smooth equivariant maps by infinitely smooth ones
|
One option is to use the harmonic map flow developed by Eels and Sampson [1]. In a certain sense this is a (non-linear) analog of the heat equation for maps $M \to N$.
Endow the manifolds $M$ and $N$ with two smooth Riemannian metrics $g$ and $h$, which additionally we may assume invariant under the action of $G$. Then you can evolve $u\_0 := f$ along a family of mappings $(u(t,\cdot) \mid 0 \leq t < \epsilon)$ which solve the so-called harmonic map flow for some short time.
Specifically
\begin{equation}
\begin{cases}
\partial\_t u &=& \tau\_g(u) \\
u(0,\cdot) &=& f
\end{cases}
\end{equation}
where $\tau\_g(u) = \mathrm{tr} \nabla \mathrm{d} u$ is known as the tension field.
This is the gradient flow for the energy functional \begin{equation}E(u) = \int\_M \lvert \mathrm{d} u \rvert^2 \, \mathrm{d vol}\_g . \end{equation} The (short-time) existence, uniqueness and smoothness of the solution is guaranteed by virtue of the PDE being strictly parabolic.
In particular, for all $t \in (0,\epsilon)$, $u(t,\cdot): M \to N$ is smooth, and as $t \to 0$,
\begin{equation}
\lvert u(t,\cdot) - f \rvert\_{C^1} \to 0.
\end{equation}
The equivariance of $u$ under $G$—given that of the initial datum $f$—follows from the uniqueness of the solution. This is because the PDE is also invariant under the action of the group.
For an explicit confirmation, let $\gamma \in G$ be arbitrary and define $u\_1: x \mapsto \gamma u\_0(\gamma^{-1} x)$. Then on the one hand the family $(\gamma u(t,\gamma^{-1} \cdot) \mid 0 \leq t < \epsilon)$ is the harmonic map flow from $u\_1$. On the other hand $u\_1 = u\_0$ by assumption, and therefore
\begin{equation}
\gamma u(t,\gamma^{-1} \cdot) = u(t,\cdot) \text{ for all $0 \leq t < \epsilon$.}
\end{equation}
[1] J. Eels and J.H. Sampson. *Harmonic mappings of Riemannian manifolds.* Amer. J. Math. 86 (1964) 109-160.
|
4
|
https://mathoverflow.net/users/103792
|
397829
| 164,210 |
https://mathoverflow.net/questions/397696
|
9
|
$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}$Consider the lattices in $\SL(2,\mathbb R)(\mathbb Z^2)$ up to rotation. The space of such lattices can be identified with the modular surface $\mathcal M:=\SO(2)\backslash \SL(2,\mathbb R)/\SL(2,\mathbb Z)$. We can then define a family of functions $f\_t(A):\mathcal M\rightarrow \mathbb R^+$ by
$$f\_t(A)=\sum\_{ v\in A(\mathbb Z^2)} e^{-t \|v\|\_2},$$ for $t>0$, where $\|v\|\_2$ denotes the Euclidean norm.
I conjecture that for all $t>0$, $f\_t(A)$ has a unique minimum at the lattice which can be tiled by equilateral triangles.
I have computational evidence which suggests that this is true; I have plotted approximations of $f\_t(A)$, $$f\_{t,N}(A):=\sum\_{v\in A(\mathbb Z\_N^2)} e^{-t\|v\|\_2},$$ where $\mathbb Z\_N:= \{m\in\mathbb Z:-N\leq m\leq N\}$, and the conjecture seems to hold. Note that for $t$ small and $N$ small, $f\_{t,N}(A)$ is not necessarily minimized by the equilaterally tiled lattice. However, for a fixed $t>0$, it appears that there exists an $N\_t>0$ such that for all $N>N\_T$, $f\_{t,N}(A)$ is minimized by the equilaterally tiled lattice.
|
https://mathoverflow.net/users/126628
|
Which unimodular lattices $L\subset \mathbb R^2$ minimize $f_t(L):=\sum_{ v\in L} e^{-t \|v\|_2}$? (for parameters $t>0$)
|
The answer is that $f\_t(A)$ is uniquely minimized at the hexagonal lattice (up to rotation).
The comment by Marco Golla led me to the following paper by Laurent Bétermin which proves the result in a more general setting: <https://arxiv.org/abs/1502.03839>
|
5
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https://mathoverflow.net/users/126628
|
397832
| 164,212 |
https://mathoverflow.net/questions/397830
|
3
|
Given the equation [here](https://mathoverflow.net/questions/397816/linear-transport-equation-with-unbounded-coefficients), I would like to ask the following relaxed question:
Consider the PDE
$$\partial\_t f(x,t) = \langle q(x), \nabla \rangle f(t,x) + p(x),$$
with Schwartz initial data $f(0,x) = f\_0(x) \in \mathscr S(\mathbb R^n).$
I am wondering then if $q$ is Lipschitz and $p$ is Schwartz as $f\_0$, does there exist a solution $(t,x)\mapsto f(t,x)$ for this equation that decays faster than any polynomial in the space variable $x$ for any fixed time $t>0$?
|
https://mathoverflow.net/users/150564
|
Linear transport equation with Lipschitz conditions
|
The anwer is **yes**.
**Preliminary observations:**
Let us first consider the case $p=0$.
Since $q$ is globally Lipschitz on $\mathbb{R}^n$, say with Lipschitz constant $L$, all solutions of the ODE $\dot x = q(x)$ on $\mathbb{R}^n$ exist globally and, if $(\varphi\_t)\_{t \in \mathbb{R}}$ denotes the induced flow in $\mathbb{R}^n$, each function $\varphi\_t: \mathbb{R}^n \to \mathbb{R}^n$ is Lipschitz continuous with constant $e^{|t|L}$ (this follows from Grönwall's lemma).
Since we assumed $p=0$, the solution of your PDE is given by
$$
f(t,x) = f\_0(\varphi\_t(x)) \tag{\*}\label{1}
$$
for each $t$ and $x$.
Hence, if $f\_0$ decays faster than every polynomial, so does $f(t, \cdot)$ for each fixed time $t$.
Let us now phrase this in a more functional analytic language: For each integer $k \ge 0$, consinder the weight function $w\_k: \mathbb{R}^n \to [1,\infty)$ given by $w\_k(x) =1 + \lvert x \rvert^k$, and consider the Banach space $E\_k$ of all continuous functions $f: \mathbb{R}^n \to \mathbb{R}$ which satisfies $w\_k f \in C\_0(\mathbb{R}^n; \mathbb{R})$; we endow the space $E\_k$ with the weighted supremum norm $\|\cdot\|\_{w\_k}$ given by
$$
\|f\|\_{w\_k} = \|w\_k f\|\_\infty.
$$
It follows from $\eqref{1}$ that the *Koopman group* $(T\_k(t))\_{t \in \mathbb{R}}$ given by
$$
T\_k(t)f = f \circ \varphi\_t
$$
for each $f \in E\_k$ is a well-defined one-parameter group of bounded linear operators on $E\_k$. Moreover, the orbits of the group are continuous with respect to the weak topology on $E\_k$ (use that (i) $E\_k$ is isomorphic to $C\_0(\mathbb{R}^n; \mathbb{R})$ via multiplication with $w\_k$, that (ii) for each $x \in \mathbb{R}^n$, the mapping $t \mapsto \varphi\_t(x)$ is continuous, and (iii) the Riesz representation theorem for measures on $C\_0(\mathbb{R}^n; \mathbb{R})$).
Hence, by a standard result from $C\_0$-semigroup theory, the the group $(T\_k(t))\_{t \in \mathbb{R}}$ is even continuous with respect to the strong operator topology, i.e., it is a $C\_0$-group on $E\_k$.
It is trivial, but important to observe, that the groups $(T\_k(t))\_{t \in \mathbb{R}}$ act consistently on the spaces $E\_k$.
Using these preliminary observations, we can show:
**Theorem.** Let $p$ be continuous and decay faster than every polynomial.
(i) For each $f\_0 \in C\_0(\mathbb{R}^n; \mathbb{R})$ there exists a unique mild solution $f: \mathbb{R} \to C\_0(\mathbb{R}^n; \mathbb{R})$ of the Cauchy problem
$$
\dot f = \langle q, \nabla \rangle f(t) + p, \qquad f(0) = f\_0.
$$
(ii) If $f\_0$ decays faster than every polynomial and $f: \mathbb{R} \to C\_0(\mathbb{R}^n; \mathbb{R})$ denotes the mild solution from (i), then for each fixed time $t \in \mathbb{R}$, the function $f(t)$ also decays faster (in space) than every polynomial.
*Proof.*
Fix $k \ge 0$. Since $(T\_k(t))$ is a $C\_0$-group on $E\_k$ and since $p \in E\_k$, it follows that, if $f\_0 \in E\_k$, the PDE has a unique mild solution $f: [0,\infty) \to E\_k$ given by
$$
f(t) = T\_k(t)f\_0 + \int\_0^t T\_k(s) p \, ds,
$$
where the integral is a Bochner integral in $E\_k$. Applying this to $k=0$, we obtain (i).
Moreover, these solutions are consistent on the $E\_k$-spaces (i.e., if $f\_0 \in E\_{k+1} \subseteq E\_k$, then the mild solutions in $E\_k$ and $E\_{k+1}$ are the same). If $f\_0$ decays faster than every polynomial, then we have $f\_0 \in E\_k$ for each $k$. Hence, for each time $t$ and each $k$, we have $f(t) \in E\_k$. Thus, for each time $t$, the function $f(t)$ decays faster in space than any polynomial. $\square$
|
4
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https://mathoverflow.net/users/102946
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397843
| 164,217 |
https://mathoverflow.net/questions/397846
|
0
|
I encounter an integer programming problem like this:
Suppose a student needs to take exams in n courses {math, physics, literature, etc}. To pass the exam in course i, the student needs to spend an amount of effort e\_i on course i. The student can graduate if she/he passes 60% of the n courses (courses have different weights). The objective is to allocate her/his efforts to different courses such that the student can graduate with the minimal amount of efforts spent on courses.
I think this problem is similar to bin covering problem when there is only one bin. The formulation is simple. Use x\_i\in{0,1} to denote whether the student allocates effort to course i. Let w\_i denote the weight of course i in calculating the final score.
Min \sum x\_i e\_i
s.t. \sum x\_i w\_i >= 60% \* n (or some other predetermined threshold)
My question is, is there a simple heuristic solution for this problem?
|
https://mathoverflow.net/users/321632
|
Integer programming for bin covering problem
|
This is equivalent to a special case of the [0-1 knapsack problem](https://en.wikipedia.org/wiki/Knapsack_problem), and the [greedy heuristic](https://en.wikipedia.org/wiki/Knapsack_problem#Greedy_approximation_algorithm) suggested by @TonyHuynh is well known but not necessarily optimal for the general case, which you can solve exactly with [dynamic programming](https://en.wikipedia.org/wiki/Knapsack_problem#0-1_knapsack_problem).
In your special case, where each $x\_i$ has the same constraint coefficient, greedy is optimal.
|
1
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https://mathoverflow.net/users/141766
|
397848
| 164,220 |
https://mathoverflow.net/questions/397802
|
0
|
(*I asked this question a couple of days back on Stackexchange but with no success, it seems elementary, but I am struggling to go about attempting it.*)
Let $X$ be a smooth geometrically integral variety over a number field $k$. We denote by $\bar{k}[X]^\*$ the group of invertible functions on $\bar{X}$, and let $$G = \varinjlim\_{n \in \mathbb{Z}\_{>0}}I\_n$$
where $I\_n=\bar{k}[X]^\*$ for all $n$ and for $m$ such that $m|n$, we have $$I\_m \rightarrow I\_n:\bar{k}[X]^\* \rightarrow \bar{k}[X]^\*:x \mapsto x^{n/m}.$$
**Question 1.** How do we show that this direct limit is $\bar{k}[X]^\* \otimes\_\mathbb{Z} \mathbb{Q}$? Most examples of computing direct limits are related to multiplicative maps between $\mathbb{Z}$, so I'm struggling to find any similarities in methods.
Furthermore, we have
$$H^i(k,\bar{k}[X]^\*\otimes \mathbb{Q}) \cong H^i(k,\bar{k}[X]^\*) \otimes \mathbb{Q},$$ where $H^i(k,-)$ denotes the Galois cohomology functor.
**Question 2.** How do we compute $H^0(k,\bar{k}[X]^\*)$, i.e., the invertible functions fixed by Galois action?
**Question 3.** I believe that $H^i(k,\bar{k}[X]^\*)$ is torsion for $i \geq 1$, is there any direct way to show this?
|
https://mathoverflow.net/users/172132
|
The direct limit of invertible functions on a variety
|
As for the first question:
It is an elementary exercise that $\mathbb{Q}$ is the colimit of the diagram of copies of $\mathbb{Z}$ indexed by the positive integers with the divisibility relation. The transition maps are given by multiplying with the corresponding fraction, and a number $z$ in the $n$th copy represents $z/n$.
Since the tensor product is cocontinuous in both variables, it follows that for every abelian group $A$ the tensor product $A \otimes \mathbb{Q}$ is the colimit of the diagram of copies of $A \otimes \mathbb{Z} \cong A$ with the evident transition maps.
|
1
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https://mathoverflow.net/users/2841
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397857
| 164,224 |
https://mathoverflow.net/questions/397859
|
2
|
Given an union-closed family of sets $\mathcal{F}$, with $n = \vert\mathcal{F}\vert$ and thus $n \choose 2$ unordered couples of distinct sets $\{A, B\}$, $A,B \in \mathcal{F}$, I would like to compute a good lower bound for the number of couples such that $A \subset B$ or $B \subset A$ as a function of $n$, i.e.:
$$\text{Number of couples such that } A \subset B \text{ or } B \subset A \ge f(n)$$
Intuitively for a couple $\{A, B\}$ with $A \not\subset B$ and $B \not\subset A$, there are two couples $\{A, C\}$ and $\{B, C\}$, $C = A \cup B$, with $A \subset C$ and $B \subset C$, however two couples $\{A, B\_1\}$ and $\{A, B\_2\}$ may share the same $\{A, C = A \cup B\_1 = A \cup B\_2\}$, so this does not seem to help.
Any hint?
|
https://mathoverflow.net/users/136218
|
Lower bound for sets couples in an union-closed family such that $A \subset B$ or $B \subset A$
|
In terms of $n$ alone, and lacking any extra constraints, I think $n-1$ is the best lower bound you can get.
It is a lower bound, because if you take $A = \bigcup {\cal F}$, then for all $B \in {\cal F} \setminus \{A\}$ you have $B \subset A$, and this gives you $n-1$ pairs.
The bound is reached with the union-closed family
$$
{\cal F} = A \;\cup\; \{B\_i \;:\; i=1,\ldots,n-1\},
$$
where $A = \{1,2,\ldots,n-1\}$, and $B\_i = A\setminus\{i\}$.
|
3
|
https://mathoverflow.net/users/171662
|
397860
| 164,225 |
https://mathoverflow.net/questions/397811
|
4
|
Is it true that, for any subalgebra $\cal S$
of the algebra of linear operators in a finite-dimensional vector space $V$ over a field,
$$
\bigcap\_{A\in\cal S}\ker A=\{0\}\hbox{ and }
\bigcup\_{A\in\cal S}A(V)=V
\quad\hbox{implies that}
\quad\hbox{some $A\in\cal S$ is non-singular? }
$$
(A more naive version of this question [was answered by Benjamin Steinberg](https://mathoverflow.net/a/397766/24165).)
|
https://mathoverflow.net/users/24165
|
Subalgebras of singular matrices (less naive version)
|
Let $S$ b be the set of 3 x 3 matrices whose lower left 2 x 2 block equals zero. Then $S$ is an algebra satisfying the conditions, but containing no invertible matrix.
|
4
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https://mathoverflow.net/users/nan
|
397861
| 164,226 |
https://mathoverflow.net/questions/397862
|
4
|
Let $X$ be a smooth scheme of finite type over $\mathbb{Z}$ (or let's say a finitely generated $\mathbb{Z}$ algebra). To each prime $p \in \mathbb{Z}$ we can consider the $\mathbb{F}\_p$ variety $$X\_{\mathbb{F}\_p}=X \times\_{\mathbb{Z}} \mathbb{F}\_p$$ and the $\overline{\mathbb{F}\_p}$ variety $$X\_{\overline{\mathbb{F}\_p}}=X \times\_{\mathbb{Z}} \overline{\mathbb{F}\_p}=X\_{\mathbb{F}\_p}\times\_{\mathbb{F}\_p} \overline{\mathbb{F}\_p}$$ (I'm omitting spec everywhere to lighten the notations a little bit).
In this way, the etale cohomology $H^{\*}(X\_{\overline{\mathbb{F}\_p}},\overline{\mathbb{Q}}\_{\ell})$ gets endowed with a Frobenius morphism and we can consider the associated increasing weight filtration $W^{i}\_m$.
If one now considers $X\_{\mathbb{C}}=X \times\_{\mathbb{Z}}\mathbb{C}$ this is a smooth complex algebraic variety. We now from comparison theorems that $$H^{\*}\_{etale}(X\_{\mathbb{C}},\overline{\mathbb{Q}}\_{\ell})=H^{\*}\_{simp}(X\_{\mathbb{C}}(\mathbb{C}),\overline{\mathbb{Q}}\_{\ell}) $$ where on the right side we have the usual simplicial cohomology.
We moreover know that for $p >>>1$ we will have $$H^{\*}\_{etale}(X\_{\mathbb{C}},\overline{\mathbb{Q}}\_{\ell}) \cong H^{\*}(X\_{\overline{\mathbb{F}\_p}},\overline{\mathbb{Q}}\_{\ell}) .$$ In this way, we can endow the cohomology $H^{\*}\_{simp}(X\_{\mathbb{C}}(\mathbb{C}),\overline{\mathbb{Q}}\_{\ell}) $ and so the cohomology with complex coefficients $H^{\*}\_{simp}(X\_{\mathbb{C}}(\mathbb{C}),\mathbb{C}) $ with a weight filtration which I'm denoting $W^{\*}\_{m,p}$.
On the other side, Deligne's theory of mixed Hodge structures, endowes $H^{\*}\_{simp}(X\_{\mathbb{C}}(\mathbb{C}),\mathbb{C})$ with another weight filtration $W^{\*}\_{m}$. Do these two filtrations coincide in general? Does the filtration $W\_{m,p}$ depends on the prime chosen?
I know that there are some invariants which relate the two filtrations. One can define the E-polynomial for example $$E\_p(X\_{\mathbb{C}},q)=\sum\_{m,k}(-1)^k dim \frac{W^{k}\_{m,p}}{W^{k}\_{m-1,p}}q^{m} $$ and analogously $$E(X\_{\mathbb{C}},q)=\sum\_{k}(-1)^k\sum\_{i+j=r} h^{i,j;k}q^r $$ where $$h^{i,j;k}=dim Gr\_W^{i+j} Gr\_F^{i} H^{k}(X\_{\mathbb{C}},\mathbb{C})$$ are the Hodge numbers.
One can show that $E\_p(X\_{\mathbb{C}},q)=E(X\_{\mathbb{C}},q)$ using the additivity with respect to locally closed decompisition of both polynomials and the statement for projective varieties which is true as everything is pure.
|
https://mathoverflow.net/users/146464
|
Comparison of weight filtration on cohomology of complex manifold
|
Yes, the $\ell$-adic weight filtration is compatible with the weight filtration in mixed Hodge theory under the comparison isomorphism. These facts go back to Deligne, and are described in his announcement *Poids dans la cohomologie des variétiés algébriques* ICM 1974. Finding a detailed proof is bit harder though...
**Added remarks**
You can take a look at Huber's *Mixed motives and their realizations in derived categories*. Although she proves something more general, which involves quite a bit of overhead. Unfortunately, I don't know of a simple complete account in the literature for what you are asking about. So why don't I simple do it here. Since you ask in the comments about the case when $X$ is projective, let me just focus on that.
Supppose that $X$ is a complex projective variety. Choose finitely generated field of definition $K$.
By resolution of singularities, one can construct a smooth projective simplicial scheme $\pi\_\bullet:X\_\bullet\to X$ which satisfies cohomological descent. See Brian Conrad's notes on cohomological descent for a detailed construction. Then one has a spectral sequence
$$E\_1^{pq} = H^q(X\_{p})\Rightarrow H^{p+q}(X)$$
for either "Betti" or $\ell$-adic cohomology. In the first case, the filtration on the abuttment is the weight filtration for the MHS, essentially by construction (cf Deligne Hodge III). In the second case, this is a spectral sequence of $Gal(\bar K/K)$-modules. The term $E\_1^{pq}$ is pure of weight $q$ by the Weil conjectures, so the same holds for $E\_\infty^{pq} = Gr\_W^pH^{p+q}(X,\mathbb{Q}\_\ell)$. That's it.
|
7
|
https://mathoverflow.net/users/4144
|
397863
| 164,227 |
https://mathoverflow.net/questions/397810
|
2
|
Let $X,Y$ be complete metric spaces, and let $\Sigma:X\times Y\rightarrow Y$ be a continous mapping which satisfies the following property: there exists a $C<1$, such that for all $x\in X$ and $y\_{1},y\_{2}\in Y$ one has
$d(\Sigma(x,y\_{1}),\Sigma(x,y\_{2}))\leq Cd(y\_{1},y\_{2})$. The fixed point theorem for complete metric spaces then implies that for all $x\in X$, there exists a unique $y(x)\in Y$ such that $\Sigma(x,y(x))=y(x)$. It is possible to prove that the correspondence $x\mapsto y(x)$ is a continuous map of metric spaces $X\rightarrow Y$.
So far I stated what I know about general metric spaces. Now furtherly assume that in particular, $X,Y$ are Fréchet spaces, and that $\Sigma$ is a smooth map of Fréchet manifolds.
My question is this: can we conclude under these additional assumptions that the correspondence $x\mapsto y(x)$ is actually a smooth map between Fréchet manifolds? I haven't found any reference for this, nor could I find a way to extend the continuous-metric proof to the smooth setting.
I will be extremely thankful for guidance or reference.
|
https://mathoverflow.net/users/144247
|
Smooth dependence in the fixed point theorem between complete Fréchet manifolds
|
Deane Yang's comment shows that the premise of my question naively fails even in finite dimension. However, thanks to his insight, I think that I managed to figure it out in the Fréchet setting as well. For reference's sake, here is how I think my question falls into the setting of Theorem 3.3.1 in Richard Hamilton's article on the Nash-Moser inverse function theorem (1982):
Suppose that $X,Y$ are tame Fréchet spaces (which they are in my original setting, since they are both section spaces of vector bundles over a compact Riemannian manifold with boundary, equipped with the $C^{\infty}$-topology. More details are in Hamilton's article) and $\Sigma$ is a tame map (in my original setting, a nonlinear differential operator between these spaces). Consider the map $A=\Sigma-I\_{Y}$, where $I\_{Y}(x,y)=y$. Then the correspondence $x\mapsto y(x)$ in the above becomes the condition $A(x,y(x))=0$. Here is where Deane's comment about the assumption on the differential of $\Sigma$ enters into play: we need to assume that the partial directional derivative (also known as partial Gateaux derivative) $D\_{x}A(x,y)=D\_{x}\Sigma(x,y):X\rightarrow X$ is a surjective linear map whenever $A(x,y)=0$. Under this assumption, and a certain "good guess" for an inverse which Hamilton explains about in his article, if $A(x\_{0},y\_{0})=0$ for some $(x\_{0},y\_{0})$, we can find a neighbourhood of $(x\_{0},y\_{0})$ for which every $x$ has a $y$ such that $A(x,y)=0$, and the solution map $x\mapsto y(x)$ restricted to this neighbourhood is a smooth tame map.
Of course the situation in the case where $X,Y$ are Banach spaces (which includes the finite dimensional case) is much simpler as we only need to use the standard implicit function theorem. Thank you Deane!
|
2
|
https://mathoverflow.net/users/144247
|
397866
| 164,228 |
https://mathoverflow.net/questions/397876
|
25
|
[Google N-Gram shows](https://books.google.com/ngrams/graph?content=temperate%20distribution%2Ctempered%20distribution&year_start=1800&year_end=2019&corpus=26&smoothing=3&direct_url=t1%3B%2Ctemperate%20distribution%3B%2Cc0%3B.t1%3B%2Ctempered%20distribution%3B%2Cc0) that both "tempered distribution" and "temperate distribution" are used in English, but the first version significantly prevails, and usage of the second term declines.
Schwartz himself seems to have used this term for the first time in his paper of 1951, *Analyse et synthèse harmoniques dans les espaces de distributions*,
Canad. J. Math. **3** (1951), 503–512, doi:[10.4153/CJM-1951-051-5](https://doi.org/10.4153/CJM-1951-051-5), and the French term was "distributions tempérées".
Google translates "tempérées" as "temperate".
I am not a native English speaker, but I understand "temperate" as a synonym of "moderate", which makes sense to me as a name of those distributions. For example, we say "temperate climate", not "tempered climate".
While "tempered" seems to be related to metallurgy, namely to a process making steel hard. What does tempering of steel have to do with distributions?
Why does the name "tempered" win in English?
|
https://mathoverflow.net/users/25510
|
Why are distributions "tempered"?
|
>
> Can someone explain, why in English the name "tempered" wins?
>
>
>
Presumably because that’s how the inventor himself translated it (French past participle to English past participle), on e.g. p. 188 of
*Schwartz, Laurent*, Mathematics for the physical sciences, Collection enseign. des sciences. ADIWES Internation Series in Mathematics. Paris: Hermann & Cie.; Reading, Mass. etc.: Addison-Wesley Publishing Company. 357 p. (1966). [ZBL0151.34001](https://zbmath.org/?q=an:0151.34001):
>
> A distribution $\mathrm T$ (that is to say a continuous linear form on $\mathscr D$) is termed
> a *tempered distribution* if it may be extended to a continuous linear form on $\mathscr S$.
>
>
>
The usage was apparently already well-established by 1956 when G. Mackey [reviewed](//mathscinet.ams.org/mathscinet-getitem?mr=84713) a French paper of F. Bruhat and wrote (first occurrence of the term in MathSciNet):
>
> The representations concerned are assumed to yield “tempered representations” when restricted to the Abelian normal subgroup being studied. Here tempered means being not too badly unbounded in a precise sense suggested by Schwartz’s definition of tempered distribution.
>
>
>
|
13
|
https://mathoverflow.net/users/19276
|
397883
| 164,233 |
https://mathoverflow.net/questions/397865
|
5
|
Let $T$ be a transitive permutation group in $S\_n$, embedded in $GL\_n(F)$ as permutation matrices. Let $D$ be the group of diagonal matrices in $GL\_n(F)$. Let $G$ be the group generated by $T$ and $D$. That is, $G$ is a subgroup of the monomial group in $GL\_n(F)$.
Question: is $G$ irreducible as a matrix group?
Probably this is a very basic question, but I cannot figure it out or find a reference...
Thank you for your help!
|
https://mathoverflow.net/users/8012
|
on the group generated by transitive permutation groups and diagonal groups
|
This is just a summary of the answers in the comments.
If $|F|=2$, then the vector $(1,1,\ldots,1)$ spans a subspace invariant under $G$, so the group is reducible (assuming that $n>1$).
Otherwise, if $|F|>2$ then, under the action of $D$, the natural module $V$ is the sum $V\_1 \oplus V\_2 \cdots \oplus V\_n$ of $1$-dimensional irreducible submodules which are mutually nonisomorphic. (The mutual nonisomorphism follows from the fact the actions of $D$ on these submodules have different kernels.) It follows that the only submodules under the action of $D$ are sums of subsets of $\{V\_1,\ldots,V\_n\}$.
But, the action of $T$ is transitive, so the $V\_i$ are permuted transitively by $T$, and none of these subspaces except for $V$ itself is invariant under $T$. So the action of $G$ is irreducible.
|
6
|
https://mathoverflow.net/users/35840
|
397885
| 164,234 |
https://mathoverflow.net/questions/397793
|
2
|
Let $G$ be a finite solvable group and $F(G)$ its Fitting subgroup. If $F(G)$ is a $p$-subgroup, is $G$ always a split extension over $F(G)$?
|
https://mathoverflow.net/users/134942
|
Is $G$ always a split extension over Fitting subgroup under certain hypothesis?
|
This has been answered in the comments, but here is a summary.
There are counterexamples. One such (suggested by Geoff Robinson) is $\texttt{SmallGroup}(48,28)$, with $F(G) \cong Q\_8$. A similar example is $\texttt{SmallGroup}(48,30)$, but here $F(G)$ is elementary abelian of order $8$.
You asked for examples with $|F(G)|$ odd. One such is $\texttt{SmallGroup}(648,531)$, with $F(G$ extraspecial of order $27$. A similar example is $\texttt{SmallGroup}(648,531)$, but here $F(G)$ is elementary abelian of order $27$.
In both of those last two examples, $G/F(G) \cong {\rm SL}(2,3)$, and a corresponding example with $p=5$ would not be solvable. But fortunately you have not requested an example with $p>3$.
|
5
|
https://mathoverflow.net/users/35840
|
397887
| 164,235 |
https://mathoverflow.net/questions/397886
|
6
|
Cross-post from [math.sx](https://math.stackexchange.com/questions/4198077/convergence-criterion-in-the-domain-of-an-unbounded-operator).
My question is somewhat close to [this](https://math.stackexchange.com/questions/1653867/is-a-self-adjoint-operator-continuous-on-its-domain) one, but the counterexamples given there do not apply here.
**Setup.** Given a Hilbert space $\mathcal H$, a closed operator $A$ and a convergent sequence $(x\_n)\_{n\in\mathbb N}\subset \mathcal D(A)$, I want to ensure that $\lim\_{n\to\infty}x\_n \in \mathcal D(A)$.
It is *not* necessary that $\lim\_{n\to\infty} Ax\_n = Ax$.
My intuition says that something like a uniform upper bound on $\|Ax\_n\|$ should be a sufficient condition. Let me illustrate that in the following setting for *normal* operators.
**Lemma.** Let $A$ be a normal operator on $\mathcal H$, $x\_n$ a convergent sequence in $\mathcal D(A)$ and assume $\liminf\_{n\to\infty}\|Ax\_n\|<\infty$. Then $\lim\_{n\to\infty} x\_n \in \mathcal D(A)$.
*Proof.* We use the spectral theorem and [Fatou's lemma](https://en.wikipedia.org/wiki/Fatou%27s_lemma#Extensions_and_variations_of_Fatou%27s_lemma) for weakly convergent measures. Let $E\_A$ be the projection valued measure associated to $A$. Then
$$ \int\_{\mathbb C} |\lambda|^2 d\langle x,E\_A(\lambda)x\rangle \le \liminf\_{n\to\infty} \int\_{\mathbb C} |\lambda|^2 d\langle x\_n,E\_A(\lambda)x\_n\rangle = \liminf\_{n\to\infty}\|Ax\_n\|^2<\infty. $$
Hence, $x\in\mathcal D(A)$. $\square$
**Question.** What extensions of above lemma are known? Does anyone have a counterexample? References or own proofs both are warmly welcome.
|
https://mathoverflow.net/users/166168
|
Convergence criterion in the domain of an unbounded operator
|
If you have a uniform upper bound on $\|Ax\_n\|$ then you can extract a weakly convergent subsequence $Ax\_{n\_k}$. Denote $\lim\_n Ax\_{n\_k} = y\_{\infty}$ to be the weak limit. Since $A$ is closed, it is also weakly closed. Since $A$ is a weakly closed operator, we have weakly $y\_{\infty} = Ax\_{\infty}$ and $x\_{\infty} \in D(A)$ as desired. (Noting that $\lim\_n x\_{n\_k} = \lim\_n x\_{n} =: x\_{\infty}$ strongly and weakly).
Actually, this shows that every subsequence of $ Ax\_{n}$ contains a further subsequence that converges weakly to the same $y\_{\infty}$. Thus, we can actually conclude that $A x\_n$ converges to $Ax\_{\infty}$ weakly. If we assume the stronger condition that $\|Ax\_n\| \rightarrow \|Ax\_{\infty}\| < \infty$, which implies $\|Ax\_n\|$ is uniformly bounded, then we have $Ax\_n \rightarrow A{x\_\infty}$ strongly as well.
Edit: I incorrectly stated/implied that uniform boundedness of $\|Ax\_n\|$ gave a (strongly) convergent subsequence of $Ax\_n$. This actually only gives a weakly convergent subsequence, which suffices for the proof.
|
6
|
https://mathoverflow.net/users/317937
|
397889
| 164,236 |
https://mathoverflow.net/questions/397267
|
3
|
In this [paper](https://link.springer.com/content/pdf/10.1007/BF00416848.pdf) Podles introduced a $2$-parameter family of $q$-deformed spheres $S\_{q,c}$ that are now called the "Podles spheres". The case of $c=0$ is very special and is known as the "standard Podles sphere". All algebras can be realised as subalgebras of the quantum group $SU\_q(2)$ and for this case $c=0$ we have a well known description of $S\_{q,0}$ as the space of invariants of $U({\frak h})$ the Hopf-subalgebra of $U\_q({\frak sl}\_2)$ generated by the elements $K$ and $K^{-1}$. What happens for the other cases? Is there a general family of algebras $U\_c({\frak h})$ such that $S\_{q,c}$ is the space of invariants of $U\_c({\frak h})$?
|
https://mathoverflow.net/users/153228
|
Nonstandard Podles spheres as $U_c(\frak{h})$ invariants
|
Yes, there is a one parameter family of coideal subalgebra of $U\_q(\mathfrak{sl}\_2)$ that give the Podleś sphere algebras as their coinvariants. The generators of those coideals are given in [1]. More conceptually, there is a duality between the coideals of "function algebra" and those of "universal enveloping algebra" of quantum groups [2].
[1] *Noumi, Masatoshi; Mimachi, Katsuhisa*, [**Askey-Wilson polynomials as spherical functions on (\mathrm{SU}\_q(2))**](http://dx.doi.org/10.1007/BFb0101182), Quantum groups, Proc. Workshops, Euler Int. Math. Inst. Leningrad/USSR 1990, Lect. Notes Math. 1510, 98-103 (1992). [ZBL0743.33011](https://zbmath.org/?q=an:0743.33011).
[2] *Dijkhuizen, Mathijs S.; Koornwinder, Tom H.*, [**Quantum homogeneous spaces, duality and quantum 2-spheres**](http://dx.doi.org/10.1007/BF01278478), Geom. Dedicata 52, No. 3, 291-315 (1994). [ZBL0818.17017](https://zbmath.org/?q=an:0818.17017).
|
3
|
https://mathoverflow.net/users/9942
|
397893
| 164,237 |
https://mathoverflow.net/questions/397896
|
9
|
*Throughout assume $\mathsf{ZFC}$ + "There is a proper class of inaccessible cardinals." I'm also happy to strengthen the large cardinal hypothesis if that would help.*
Say that a model $M\models\mathsf{ZFC}$ is **powerful** iff every end extension satisfying $\mathsf{ZFC}$ is a top extension. The transitive powerful models are exactly the $V\_\alpha$s where $\alpha$ is a worldly cardinal: that every $V\_{\mathsf{worldly}}$ is powerful is trivial, and the large cardinal hypothesis above gives the other half of the result.
The ill-founded situation is on the other hand completely unclear to me:
>
> Is there an ill-founded powerful model of $\mathsf{ZFC}$?
>
>
>
I don't really see where to start with this. Certainly there are no *countable* powerful models of $\mathsf{ZFC}$, ill-founded or otherwise, since we can (genuinely) force over such models; however, I don't see any useful tools for analyzing uncountable ill-founded models. The existence of rigid ill-founded models *(e.g. the substructure of any ill-founded model of $\mathsf{ZFC+V=L}$ consisting of the parameter-freely-definable elements)* does make the existence of powerful models feel not *totally* implausible, but that said I don't see how any of the techniques for building examples of the former type are useful for attempting to build an example of the latter.
|
https://mathoverflow.net/users/8133
|
Are there ill-founded "maximally wide" models of $\mathsf{ZFC}$?
|
Back in the mid 1980s I remember convincing myself (alas, in unpublished work) that *there is an ill-founded model $M$ of ZFC that has no end extension to another model of ZFC.* Such a model $M$ by default is powerful and technically answers the question.
My unpublished work above used techniques employed in the following related results from the same time period:
**1.** There are ill-founded models $M$ *of any given consistent extension of* ZFC that have no top extension to another model of ZFC; see Theorem 1.5 of [this paper of mine](https://www.ams.org/journals/tran/1984-283-02/S0002-9947-1984-0737894-1/S0002-9947-1984-0737894-1.pdf). Note that top extensions are referred to as rank extensions in the paper.
**2.** There are ill-founded models $M$ *of any given consistent extension of* ZFC such that there are no $N \supsetneq M$ such that $N$ is a *forcing* end extension of $M$. This was noted in [this other paper of mine](https://www.researchgate.net/publication/38382376_Conservative_Extensions_of_Models_of_Set_Theory_and_Generalizations#fullTextFileContent) (see the remark after Theorem 1.5, which itself relates to Theorem 1.4).
In (1) and (2) above, the cardinality of $M$ can be arranged to be $\aleph\_1$. This is optimum since every countable model of ZF (including ill-founded ones) has lots of top extensions and forcing extensions
|
7
|
https://mathoverflow.net/users/9269
|
397902
| 164,240 |
https://mathoverflow.net/questions/397909
|
3
|
I hope not to be too simplistic.
I read about this monotonicity formula [A question on the monotonicity formula for minimal submanifolds](https://mathoverflow.net/questions/397505/a-question-on-the-monotonicity-formula-for-minimal-submanifolds)
I noticed that the monotonicity formula is often used in regularity theory for surfaces, declined in different versions, like for varifold (Allard 1972) or for currents or finite perimeter sets.
My question is what is the main utility of the monotonicity formula? I noticed the results showing that the density exists at every point and it is greater or equal than 1. But how does this help as well? Thanks!
|
https://mathoverflow.net/users/170982
|
Main utility of the monotonicity formula for generalized surfaces
|
A basic answer is that "the monotonicity formula places constraints on the shape of a minimal surface" e.g., you cannot have a lot of area concentrated in a ball if then later there is a (relatively) small amount of area. This, along with the convex hull property, already tells you a lot about the possible shape of a minimal surface.
However, I think you are looking for a concrete application. Here is one sample application that is somewhat different from what I have described above:
Suppose that $\Sigma\_i,\Sigma \subset B\_2(0) \subset \mathbb{R}^3$ are smooth embedded minimal surfaces and for any $f \in C^0\_c(B\_2(0))$ it holds that
$$
\tag{\*} \int\_{\Sigma\_i} f|\_{\Sigma\_i} \to \int\_\Sigma f|\_\Sigma
$$
as $i\to\infty$ (this is a very weak notion of convergence of surfaces). Then, if $x\_i \in \Sigma\_i\cap B\_1(0)$ has $x\_i\to x$ then $x\in\Sigma$.
Proof: Assume $x\not \in \Sigma$. Then, there is $B\_{3\epsilon}(x)$ disjoint from $\Sigma$. Choose a bump function $f=1$ on $B\_{2\epsilon}(x)$ and $f$ vanishing outside of $B\_{3\epsilon}(x)$. Now, by (\*) we have that
$$
\int\_{\Sigma\_i} f|\_{\Sigma\_i}\to 0.
$$
Thus, for $i$ large, $B\_{\epsilon}(x\_i)\subset B\_{2\epsilon}(x)$, so we have arranged that
$$
|\Sigma\_i\cap B\_\epsilon(x\_i)|\leq \int\_{\Sigma\_i} f|\_{\Sigma\_i} \to 0.
$$
On the other hand **the monotonicity formula** implies that
$$
\frac{|\Sigma\_i\cap B\_\epsilon(x\_i)|}{\pi \epsilon^2} \geq \lim\_{r\to0}\frac{|\Sigma\_i\cap B\_r(x\_i)|}{\pi r^2} = 1
$$
since $\Sigma\_i$ is smooth (and thus nearly flat on small scales). This is a contradiction.
---
Note that if the $\Sigma\_i$ are not minimal (i.e., if they don't satisfy the monotonicity formula), it is easy to find a counterexample to the above result by taking a flat disk $\Sigma$ and forming $\Sigma\_i$ by gluing on a lot of "tentacles" which all have small area (and thus small contribution to the integral as in (\*)). Thus, this is a nontrivial result.
(Note that I have not attempted to prove the most general version of this result, if you want you might see Simon's GMT book or other sources on minimal surfaces.)
There are many related applications of the monotonicity formula. A simple yet powerful one consists of White's proof of the Allard regularity theorem. See Section 1.1 here <https://annals.math.princeton.edu/wp-content/uploads/annals-v161-n3-p07.pdf> or Theorem 7.8 here [http://web.stanford.edu/~ochodosh/MinSurfNotes.pdf](http://web.stanford.edu/%7Eochodosh/MinSurfNotes.pdf)).
|
5
|
https://mathoverflow.net/users/1540
|
397911
| 164,241 |
https://mathoverflow.net/questions/397919
|
0
|
$\DeclareMathOperator\FSym{FSym}$Let $\FSym(\mathbb{N})$ denote the finitary symmetric group on the set of natural numbers. How many Sylow $p$-subgroups does $\FSym(\mathbb{N})$ have for any prime $p$? Countably or uncountably many?
|
https://mathoverflow.net/users/98061
|
Sylow $p$-subgroups of FSym($\mathbb N$)
|
Uncountably (continuum) many. A $p$-Sylow subgroup (or at least some of them) determines a nested partition (into $p$-element subsets, into $p^2$-element subsets, etc), and hence determines a partition into $p$-element subsets. There are continuum many such partitions and all are conjugate under the permutation group.
|
2
|
https://mathoverflow.net/users/14094
|
397920
| 164,245 |
https://mathoverflow.net/questions/397643
|
11
|
A famous theorem of Birman and Series says that if $S$ is a compact hyperbolic surface, then the set of points that lie on simple geodesics is nowhere dense and has Hausdorff dimension one; in particular, it has measure zero. This is proved in
Birman, Joan S.; Series, Caroline,
Geodesics with bounded intersection number on surfaces are sparsely distributed.
Topology 24 (1985), no. 2, 217–225.
**Question**: Is this true for compact oriented surfaces equipped with metrics of variable negative curvature?
|
https://mathoverflow.net/users/317
|
Birman-Series for variable negative curvature
|
Here is another argument which reduces the general result to the constant curvature case:
For any negatively curved metric $g$ on $S$, the set of geodesics of the universal cover $\tilde{S}$ is canonically identified with $\partial\_\infty \tilde{S}^{(2)}$, the set of pairs of distinct points in the Gromov boundary of $\tilde{S}$, in which $\overline{\Lambda}$ (the subset of geodesics that are embedded in $S$) is a closed subset. Note that the Gromov boundary and the set $\Lambda$ are quasi-isometric invariants, so they don't really depend on the choice of the metric $g$. What choosing the metric $g$ does is it puts a $\mathcal C^1$ structure on $\partial\_\infty \tilde{S}^{(2)}$.
Let $g\_0$ be another metric of constant curvature of minus one. Then Birman and Series' theorem tells you that $\overline{\Lambda} \subset \partial\_\infty \tilde{S}^{(2)}$ has Hausdorff dimension $0$ with respect to the $\mathcal C^1$ structure defined by $g\_0$. Now, one can prove that the two $\mathcal C^1$ structures on $\partial\_\infty \tilde{S}^{(2)}$ induced respectively by $g$ and $g\_0$ are Hölder regular with respect to each other, which implies that $\overline{\Lambda}$ also has Hausdorff dimension zero for the $\mathcal C^1$ structure defined by $g$.
|
7
|
https://mathoverflow.net/users/173096
|
397927
| 164,246 |
https://mathoverflow.net/questions/377971
|
8
|
Consider an "ambiguous" function class
$F^\star\subseteq\{0,1,\star\}^X$ (i.e., $F$ consists of Boolean functions acting on a set $X$ with some missing values, indicated by $\star$).
We say that
$F^\star$ *shatters* a set $S\subseteq X$
if $F^\star(S)\supseteq\{0,1\}^S$.
Define $VC(F^\star)$ as the maximal size of any shattered set (possibly, $\infty$).
We say that $\bar f\in\{0,1\}^X$ is a *disambiguation*
of $f^\star\in F^\star$ if
the two functions agree on $x\in X$
whenever $f^\star(x)\neq\star$.
We say that $\bar F\subseteq\{0,1\}^X$ is a disambiguation of
$F^\star$ if each $\bar f\in \bar F$
is a disambiguation of
*some* $f^\star\in F^\star$
and *every* $f^\star\in F^\star$
has a disambiguated representative $\bar f\in \bar F$.
Conjecture: There is a universal constant $c$ such that for any ambiguous
$F^\star$ there is a disambiguation $\bar F$ such that
$$
VC(\bar F)
\le c
VC(F^\star)
.$$
Note:
This open problem was posed here:
<https://arxiv.org/abs/1810.02180> .
It is known that $c$ must be $>1$, and Lemma 6.2 therein provides an analog of Sauer's lemma for $F^\star$.
|
https://mathoverflow.net/users/12518
|
VC-dimension of disambiguated classes
|
This has been disproved in Theorem 11 of [Alon, Hanneke, Holzman, and Moran](https://arxiv.org/abs/2107.08444). The proof is short and elegant (building on recent deep results of others').
|
5
|
https://mathoverflow.net/users/955
|
397928
| 164,247 |
https://mathoverflow.net/questions/397918
|
5
|
Consider the equation
$$
\begin{equation}
\frac{\partial^2f}{\partial x\partial y}=f
\end{equation}
$$
on a Jordan domain (i.e. the interior of a simple, closed curve on the plane). The equation is hyperbolic, but we cannot formulate a Cauchy problem in the usual sense since the domain is finite, so the question is
>
> What kind of initial value problem can we formulate on such a domain ? Which initial data along the boundary do when need to establish existence ?
>
>
>
|
https://mathoverflow.net/users/171439
|
Linear hyperbolic PDE on compact two dimensional domain
|
Generally, you want there to be a *non-characteristic transversal*, i.e., a (let's say, smooth) curve $C$ in your domain $D$ such that each segment of each line $x=x\_0$ in $D$ is connected and meets $C$ exactly once transversely and each segment of each line $y=y\_0$ in $D$ is connected and meets $C$ exactly once transversely. Then you get existence and uniqueness for the non-characteristic initial value problem in which you specify $f$ and its normal derivative along $C$.
For example, $D$ could be a rectangle $a\le x\le b$ and $c\le y\le d$ and $C$ could be the graph of a surjective smooth map $g:[a,b]\to[c,d]$ with $g'>0$ on $[a,b]$. Solving this particular case (as Riemann did) gives the solution for any non-characteristic transversal $C$ in a domain $D$.
|
4
|
https://mathoverflow.net/users/13972
|
397932
| 164,250 |
https://mathoverflow.net/questions/397942
|
11
|
Let $\mathsf{A}$ be an abelian category and $\mathsf{B}$ be a full abelian subcategory. More often than not, instead of being interested in the derived category $\mathsf{D}(\mathsf{B})$, we are interested in the full subcategory $\mathsf{D}\_{\mathsf{B}}(\mathsf{A})$ of $\mathsf{D}(\mathsf{A})$ composed of the complexes whose cohomology is in $\mathsf{B}$. (The cases of quasi-coherent / coherent sheaves and of holonomic $\mathcal{D}$-modules come to mind.)
This seems counter-intuitive for me since we usually have more tools for dealing with objects in the category $\mathsf{B}$ which could be applied to $\mathsf{D}(\mathsf{B})$. While both categories are usually equivalent, I feel that people usually think that $\mathsf{D}\_{\mathsf{B}}(\mathsf{A})$ is the *right* object. Why is it so?
|
https://mathoverflow.net/users/131975
|
Why is $\mathsf{D}_{qc}(X)$ the right notion, instead of $\mathsf{D}(\mathsf{QCoh}(X))$?
|
The triangulated category $\mathsf{D}\_{\mathsf{B}}(\mathsf{A})$ can be promoted to a stable $\infty$-category. One of the many interests of working with stable $\infty$-categories is that we have a reasonable [theory of descent](https://mathoverflow.net/q/385397/1017) for them: we can define sheaves of stable $\infty$-categories (it is possible to formulate such concepts using the language of dg categories). However, the assignment
$$X\mapsto \mathsf{D}(\mathsf{QCoh}(X))$$
is unfortunately not a sheaf of stable $\infty$-categories on the big site of all schemes (for the Zariski or fppf topology, say) and its sheafification leads to the assignment
$$X\mapsto \mathsf{D}\_{qc}(X)\, .$$
We have $\mathsf{D}(\mathsf{QCoh}(X))\cong \mathsf{D}\_{qc}(X)$ locally (in fact for $X$ any separated and quasi-compact scheme), so that $\mathsf{D}\_{qc}(X)$ gets many of the nice features of $\mathsf{D}(\mathsf{QCoh}(X))$ by local-to-global principles.
|
18
|
https://mathoverflow.net/users/1017
|
397946
| 164,255 |
https://mathoverflow.net/questions/397938
|
3
|
$\newcommand\la{\langle}\newcommand\ra{\rangle}$Let $G=\mathbb{Z}/p^{i\_1}\times\cdots\times\mathbb{Z}/p^{i\_r}$ with $i\_1\leq\ldots\leq i\_r$ be a finite abelian $p$-group. Then there can be many choices of generators $\{x\_1,\ldots,x\_r\}$ such that the order of $x\_j$ is $p^{i\_j}$ and $G=\la x\_1\ra\times\cdots\times \la x\_r\ra$.
Let $H$ be a subgroup of $G$. Then $H$ is of the same form with less or equal number of factors.
Does there exist a choice of generators $\{x\_1,\ldots,x\_r\}$ of $G$ as above such that $H$ is a product of subgroups of $\la x\_j\ra$?
If it is not true, is there an easy counterexample?
|
https://mathoverflow.net/users/304053
|
Structures of subgroups of a finite abelian p-group
|
Let $G$ = ${\mathbb Z}/2 \oplus {\mathbb Z}/8$, and let $H$ be the cyclic subgroup of order $4$ generated by the element $h = (\bar{1},\bar{2})$.
There is no element $g \in G$ with $2g = h$, and so $H$ cannot be a subgroup of a cyclic direct summand of $G$ of order $8$. And clearly it cannout be a subgroup of a summand of order $2$, so the answer to the question is no, and this is a counterexample.
|
7
|
https://mathoverflow.net/users/35840
|
397953
| 164,256 |
https://mathoverflow.net/questions/397950
|
1
|
Let $\Omega\subset \mathbb R^d$ be a smooth, bounded domain. Let $(e\_n)\_{n\geq 0}\subset L^2(\Omega)$ be the Hilbert basis generated by the Dirichlet-Laplacian eigenfunctions, i-e $-\Delta e\_n=\lambda\_n e\_n$ with zero boundary conditions.
We know that $\lambda\_n\to+\infty$.
For $s\geq 0$ let me denote the "spectral" $H^s$ norm
$$
\|f\|^2\_{H^s}=\sum\limits\_{n\geq 0} \lambda\_n^{s}|f\_n|^2
\hspace{1cm}\mbox{for }f=\sum\limits\_{n\geq 0} f\_ne\_n,
$$
and finally for $N\geq 1$ let me denote the projector onto $E\_N=span(e\_0,\dots,e\_{N-1})$ as
$$
P\_N f=\sum\limits\_{n\leq N-1} f\_n e\_n.
$$
---
**Fact 1**: for any fixed $f\in H^s$ there holds $(1-P\_N)f\to 0$ in $H^s$, which can be stated as "$1-P\_N\to 0$ pointwise on $H^s$". This is easy to check, since
$$
\|(1-P\_N)f\|\_{H^s}^2=\sum\limits\_{n\geq N}|\lambda\_n|^{s}|f\_n|^2\to 0
$$
as the remainder of a convergent series.
---
**Fact 2**: for any $r<s$ there holds
$$
\|(1-P\_N)f\|\_{H^r}\leq \lambda\_{N}^{(r-s)/2}\|f\|\_{H^s},
$$
which can be stated as "$(1-P\_N)\to 0$ in the $H^r$ norm uniformly on any $H^s$ ball."
Indeed, one can write immediately
$$
\|(1-P\_N)f\|\_{H^r}^2
=\sum\limits\_{n\geq N}|\lambda\_n|^{r-s}|\lambda\_n|^s|f\_n|^2
\leq |\lambda\_N|^{r-s}\sum\limits\_{n\geq N}|\lambda\_n|^s|f\_n|^2
\leq |\lambda\_N|^{r-s} \|f\|\_{H^s}^2
$$
---
>
> **Question:** can we extend the uniform convergence on arbitrary $H^r$-compact sets? More explicitly, is it true that
> $$
> \sup\limits\_{f\in K}\|(1-P\_N)f\|\_{H^r}\to 0
> \qquad\mbox{as }N\to\infty
> $$
> for any $H^r$-compact set $K$? Fact 2 ecactly guarantees that this holds at least if $K$ is an $H^s$ ball, which is indeed $H^r$-compact classical Sobolev embedding since $r<s$. What about more generic compact sets?
>
>
>
I suspect this is well-known but for some reason I could not find anything on the subject (me not being a specialis in spectral analysis certainly does not help).
|
https://mathoverflow.net/users/33741
|
uniform convergence of $H^r$ projectors on compact sets?
|
If $(T\_n)$ is a sequence of uniformly bounded, linear operators in a Banach space $X$ nd $T\_nx→0$ for every $x∈X$, then the convergence is uniform on a compact set $K$. Just fix $ϵ>0$ and cover $K$ with a finite number of balls $B(x\_i,ϵ)$ and use $∥T\_nx∥≤∥T\_n(x−x\_i)∥+∥T\_n x\_i∥$.
|
2
|
https://mathoverflow.net/users/150653
|
397955
| 164,257 |
https://mathoverflow.net/questions/397968
|
3
|
Let $X$ be a smooth projective variety over $\mathbf{F}\_p$, call $\overline{X}$ the base change to $\overline{\mathbf{F}}\_p$, and denote by $F$ the base change to $\overline{X}$ of the absolute Frobenius of $X$ over $\mathbf{F}\_p$.
Call $A$ the Chow ring of cycles up to homological equivalence (defined using, say, $\ell$-adic cohomology, i.e. the image of the cycle map $Z^\*(\overline{X}\times \overline{X})\otimes\_{\mathbf{Z}}\mathbf{Q}\to H^{2\*}(\overline{X}\times \overline{X},\mathbf{Q}\_{\ell})(\*)$).
Call $R$ the $\mathbf{Q}$-subalgebra of $A$ generated by the classes of the graphs of the endomorphisms $\{F^n, n\ge 1\}$ of $\overline{X}$.
>
> Is anything at all known about $R$?
>
>
>
Examples of questions I'd be interested in are
* is $R$ known to not be $A$?
* is $R$ a domain, or even a field?
* is $R$ normal?
* does $R$ contain the Lefschetz class?
* is $R$ expected to contain the inverse of the Lefschetz class, perhaps assuming some standard conjectures.
|
https://mathoverflow.net/users/nan
|
Subrings of Chow rings
|
Plenty!
$R$ is generated, as a ring, by $F$. So its structure as a ring is going to be $\mathbb Q(\alpha)/f(\alpha)$, where $f$ is the minimal polynomial of $F$. Because you are using homological equivalence, $f$ is just the least common multiple of the minimal polynomials of the action of $F$ on the various cohomology groups of $X$.
By Deligne's RH, these polynomials are coprime, so it is in fact the product of the minimal polynomials on the different cohomology groups.
Thus $R$ is never a domain unless $X$ is a point.
Since the graphs of powers of $F$ have dimension $n$, $R$ is not equal to $A$ unless every homologically nontrivial cycle on $X \times X$ occurs in degree $n$, i.e. unless $X$ is a point.
Since normal rings are by definition domain, it's not normal.
Because the Lefschetz class lives in dimension $n-1$ and its inverse lies in dimension $n+1$, neither one is contained in this ring since neither has dimension $n$.
|
10
|
https://mathoverflow.net/users/18060
|
397971
| 164,260 |
https://mathoverflow.net/questions/397973
|
1
|
I want to solve the optimization problem
$$
\text{minimize }g(x) \quad \text{subject to} \quad \Vert x\Vert\_{\infty}/\Vert x\Vert\_{2} \le s
$$
for $x\in\mathbb{R}^d$ and $s\in(0,\infty)$.
The function $g:\mathbb{R}^d\to\mathbb{R}$ is (strongly) convex and Lipschitz smooth.
I know, that I could probably try to find saddle points of the corresponding Lagrangian but I would like to know, if there is a faster or more elegant way.
Do you know of a similar problem, that has been considered before?
|
https://mathoverflow.net/users/75500
|
Was a quotient of two norms considered as a constraint to a convex optimization problem before?
|
As @Mark L. Stone commented, that constraint isn't convex (and therefore not a convex optimization problem). You could instead consider the different constraint:
$$\|x\|\_{\infty} \leq sM$$
$$\|x\|\_{2} \leq M$$
which is convex. Note that the elements $x$ satisfying $\|x\|\_{\infty} \leq s \|x\|\_2$ and $\|x\|\_2 \leq M$ satisfy this constraint. So the solution $\tilde{x}^\*$ of this new optimization problem will do at least as well (in terms of the criteria $g$) as the solution $x^\*$ of your nonconvex optimization problem as long as $\|x^\*\|\_{2} \leq M$ (of course, you might be overfitting).
From a practical perspective, I can't imagine there is any loss in using this constraint. Also, this type of constraint has been used in practice (e.g. Lasso/Ridge/ElasticNet regression with a lower/upper bound on the coefficients, bounded Lipschitz regression).
As a note, the nonconvex constraint is only interesting/well-defined if $s \in [1/\sqrt{d},1)$. This is because we always have $\|x\|\_{\infty} \leq \|x\|\_2$ and $\|x\|\_2 \leq \sqrt{d}\|x\|\_{\infty} \leq \sqrt{d}s\|x\|\_2$. Also, it is not clear to me apriori that the original nonconvex constraint offers any kind of meaningful regularization.
Edit: Actually if you set $s=1/\sqrt{d}$ then the nonconvex constraint is equivalent to $\|x\|\_2 = \|x\|\_{\infty}$ which implies $x$ has only one nonzero element. So in that sense, it might give sparse solutions for small $s$. This is interesting as $\|\cdot\|\_2$ regularization alone only shrinks and does not give sparse solutions.
|
5
|
https://mathoverflow.net/users/317937
|
397975
| 164,262 |
https://mathoverflow.net/questions/397880
|
11
|
I'm interested in how do people that work in birational geometry view their field — specifically, what are the kinds of geometric questions (as opposed to commutative-algebraic questions) that interest them?
Often I hear that the main objective of birational geometry is to classify algebraic varieties up to birational isomorphism. This is an interesting mathematical question, but do the people who study it view it as a geometric question or as a commutative-algebraic one?
As a concrete example, we can take the birational classification of surfaces: it gives you very nice answers to questions like "when can we find a function with certain properties between two surfaces? what are the properties of these functions?" However, I don't see what kinds of geometric questions (i.e. questions about shapes) this theory gives you answers for, given that surfaces birational to each other can nevertheless look so radically different from each other.
Do people who study birational geometry view their field as geometry or as algebra? What are the kinds of geometric questions that interest them?
I originally asked this question [on math.se](https://math.stackexchange.com/questions/4202459/soft-question-motivation-for-birational-geometry), but was advised to move it here.
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https://mathoverflow.net/users/321982
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Motivation for birational geometry
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>
> what are some interesting properties of varieties that are preserved under birational transforms?
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I will answer the question for smooth projective varieties (certainly a geometrically nice class of varieties) specifically.
(1) For each $k$, the dimension of the space of global holomorphic/algebraic differential $k$-forms is a birational invariant. This is one of the most important invariants that is very meaningful both algebraically and geometrically, and it itself plays a significant role in the classification of birational types of surfaces.
(2) The fundamental group $\pi\_1(X)$ is a birational invariant. This is certainly one of the most fundamental topological invariants of the variety.
(3) For $Y$ any variety not containing rational curves, the set of maps $X \to Y$ (or the moduli space parameterizing maps) is a birational invariant.
Of course, maybe a more general answer is that finding interesting properties of varieties preserved under birational transforms is (by definition) one of the main areas of study of birational geometry! This is often done with the goal of, say, proving a particular variety is not rational, but you can also view these as more new invariants which you now know you understand after determining the birational type of a given variety.
Another more general answer is "often, almost everything". In particular, for many birational equivalence classes (almost all in the case of surfaces), we can identify inside the birational equivalence class a single variety $Y$ such that every smooth projective variety $X$ in the equivalence class must in fact map birationally to $Y$. We can, for many purposes, think of the variety $X$ as just $Y$ with a bit of extra elaboration added. Finding the most general statement of this type is a primary goal of the minimal model program.
---
To answer your original question, the birational classification of surfaces forms a starting point to answering the majority of questions about the existence of a surface with a particular property. If called upon to construct a surface with some unusual features, an experienced algebraic geometer will, if an obvious strategy for constructing one doesn't present itself, automatically start running through "Can a rational surface have this property? What about ruled? K3? General type? Elliptic? ...", considering blow-ups of these as appropriate.
Such a coarse classification is needed for getting the lay of the land, ruling out wide swathes of terrain to direct your attention to more fertile fields for growing surfaces of a particular type. If you had to start listing "What about the plane blown up at one point? What about the plane blown up at two points? What about the plane blown up at three points? ..." and so on, you'd never get anywhere, except for those particular problems where cleverly blowing up the plane is the linchpin to a brilliant solution.
I am not a birational geometer, but I gained a great appreciation for the birational classification of surfaces by playing around with different problems, on MO and elsewhere. I can't imagine organizing information on surfaces in a useful way without it.
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https://mathoverflow.net/users/18060
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397977
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https://mathoverflow.net/questions/397710
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In light of [Knot groups with big number of generators](https://mathoverflow.net/questions/397498/knot-groups-with-big-number-of-generators), I was wondering...
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> **Question 1** What is the minimal number of generators of the fundamental group of a [satellite knot](https://en.wikipedia.org/wiki/Satellite_knot)?
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Another more specific question
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> **Question 2** If we take a twisted Whitehead double of a nontrivial knot, what is the minimal number of generators of the resulting knot group?
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https://mathoverflow.net/users/114032
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Minimal number of generators of satellite knot groups
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Knot groups of satellite knots can have rank 2, and only the trivial knot can have rank 1, so 2 is the minimal possible number of generators.
The knot group of any tunnel number one knot has a presentation with 2 generators and 1 relator. [Morimoto and Sakuma](https://doi.org/10.1007/BF01446565) classified all satellite knots which have tunnel number one. In their notation (see section 1.7 of their article), I believe that the tunnel number one knot $K(8,3;p,q)$ will be some Whitehead double of the $(p,q)$-torus knot. Thus there exist Whitehead doubles of torus knots whose knot groups have rank 2.
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https://mathoverflow.net/users/126206
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397978
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https://mathoverflow.net/questions/397980
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Let $X$ be a curve (proper smooth variety of dimension $1$) over $\mathbf C$; $\mathcal L$ an invertible $\mathcal O\_X$-module; $r = \dim\_{\mathbf C}(\mathrm H^0(X; \mathcal L)) - 1$.
If $\mathcal L$ is very ample, then $\mathcal L(-P\_1 - \cdots - P\_{r - 1})$ is generated by global sections for $r - 1$ general points $P\_1, \dots, P\_{r - 1}$. In fact, let $i: X \to \mathbf P^r$ be the closed immersion corresponding to $\mathcal L$, then $\mathcal L(-P\_1 - \cdots - P\_{r - 1})$ is globally generated if and only if the $(r - 2)$-dimensional linear subspace of $\mathbf P^r$ passing through $P\_1, \dots, P\_{r - 1}$ does not pass through an $r$-th point of $X$, and this holds for general $r - 1$ points by the trisecant lemma.
I wonder whether $\mathcal L(-P\_1 - \cdots - P\_{r - 1})$ is generated by global sections for $r - 1$ general points $P\_1, \dots, P\_{r - 1}$ if $\mathcal L$ is only assumed to be globally generated.
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https://mathoverflow.net/users/129738
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Line bundle $\mathcal L(-P_1 - \cdots - P_{r - 1})$ on a curve being globally generated for $r - 1$ general points
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No, this is not true. Take for instance a smooth plane curve $C$, and a double covering $\pi :X\rightarrow C$ branched along $k$ points, with $k> \frac{1}{2}\deg(C) $. Put $\mathscr{L}:=\pi ^\*\mathscr{O}\_C(1)$. Then $H^0(X,\mathscr{L})=\pi ^\*H^0(C,\mathscr{O}\_C(1))$, so the map $X\rightarrow \mathbb{P}^2$ defined by $\mathscr{L}$ factors through $\pi $. For any point $P$ of $X$, the linear system $\lvert \mathscr{L}(-P)\rvert$ has a fixed point, namely the point $Q\neq P$ such that $\pi (Q)=\pi (P)$.
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https://mathoverflow.net/users/40297
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397983
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https://mathoverflow.net/questions/397933
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2
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Let $\alpha$ be an irrational number, and $R\_\alpha$ be the rotation by $\alpha$, that is $R\_\alpha(x)=x+\alpha\bmod 1$.
S. Ferenczi in his survey [Systems of finite rank. Colloq. Math. 73 (1997), no. 1, 35--65. MR1436950] states (Thm. 5): *Every irrational rotation is of rank at most two by intervals (without spaces).* For a proof he refers to a more general result (Thm. 7) stating that *An ergodic exchange of $s$ intervals is of rank at most $s$ by intervals, without spacers.*
As far as I understand, it means that I can find a nested family of intervals $J\_1\supset J\_2\supset \ldots$ such that $J\_n=F\_n\cup G\_n$, where $F\_n=[a\_n,b\_n)$ and $G\_n= [b\_n,c\_n)$ are left-closed, right-open intervals on the unit circle (interpreted here as $[0,1)$), and for each $n=1,2,\ldots$ there are positive integers $h\_n^F$ and $h^G\_n$ such that
$$
[0,1)=\bigcup\_{j=0}^{h\_n^F-1}R\_{\alpha}^j(F\_n) \cup \bigcup\_{j=0}^{h\_n^G-1}R\_{\alpha}^j(G\_n)
$$
is a disjoint union, and hence
$$
\{R\_{\alpha}^j(F\_n):j=0,1,\ldots,h\_n^F-1\} \cup \{R\_{\alpha}^j(G\_n):j=0,1,\ldots,h\_n^G-1\}
$$
is a partition refining the previous one
$$
\{R\_{\alpha}^j(F\_{n-1}):j=0,1,\ldots,h\_{n-1}^F-1\} \cup \{R\_{\alpha}^j(G\_{n-1}):j=0,1,\ldots,h\_{n-1}^G-1\}.
$$
So here are my questions:
1. Is the above interpretation correct? (Hope it is, but due to a certain vaguesness in defining ``rank by intervals'' I am not 100% sure.)
2. What can be said about $h\_n^F$ and $h^G\_n$ and their relation to each other as $n\to\infty$?
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https://mathoverflow.net/users/24676
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Irrational rotations are rank 2 by intervals without spacers
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As was pointed out, the answer is related to the continued fraction expansion. If $\alpha=\frac{1}{c\_1+\frac1{c\_2}...}$, we fix rational approximations $\frac{p\_k}{q\_k}=\frac{1}{c\_1+\frac1{c\_2+...\frac1{c\_k}}}$. $q\_k+q\_{k-1}$ iterated preimages ( or images) of a point decompose the unit circle in the wished intervals that form two towers of heights $q\_k$ and $q\_{k-1}$. For example, $T^i(0)$, $i=0..(q\_k+q\_{k-1}-1)$, decompose unit circle into interval
* $[T^i(0),T^{i+q\_{k-1}}(0))$, $i=0..q\_k-1$,
* $[T^{i+q\_k}(0),T^{i}(0))$, $i=0..q\_{k-1}-1$.
The orientation switches with the parity of $k$. But it is likely not important for you. So, the heights are $q\_k$ and $q\_{k-1}$, possibly the best sequence related with $\alpha$ you can play with. Let me recall, that $q\_{k+1}=c\_{k+1}q\_k+q\_{k-1}$. The ratio can be as close to 1 as possible, along the subsequence, or unbouded in other case, or constant e.g. for Fibonnacci.
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https://mathoverflow.net/users/101832
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397992
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https://mathoverflow.net/questions/397914
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The polytope algebra was defined by P. McMullen in "The polytope algebra" Adv. Math. 78 (1989) as follows.
Let us denote by $\Pi'\_\mathbb{R}$ the quotient of the free abelian group generated by the symbols $[P]$, where $P\subset \mathbb{R}^n$ is an arbitrary convex compact polytope, by the subgroup generated by elements
\begin{eqnarray\*}
(1)\, [P\cup Q]+[P\cap Q]-[P]-[Q] \mbox{ where } P,Q,P\cup Q \mbox{ are convex compact polytopes};\\
(2)\, [P+x]-[P] \mbox{ where } x\in \mathbb{R}^n.
\end{eqnarray\*}
Similarly let us define the analogous group $\Pi\_{\mathbb{Q}}'$ generated by convex compact polytopes with *rational* vertices and the same relations (1)-(2) (in the relation (2) one assumes $x\in\mathbb{Q}^n$).
**We have the obvious group homomorphism $\Pi\_{\mathbb{Q}}'\to \Pi'\_\mathbb{R}$. Is it injective?** A reference would be helpful.
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https://mathoverflow.net/users/16183
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The polytope algebras generated by polytopes with rational vs arbitrary vertices
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Looking more carefully at the above McMullen's paper, I realized that the question has the positive answer due to Theorem 3 in the paper.
McMullen constructs homomorphisms $\Pi'\_{\mathbb{F}}\to \mathbb{F}$ (where $\mathbb{F}= \mathbb{R},\mathbb{Q}$) which all together separate points, i.e. if all of them vanish on some element, then the element vanishes. Let $u$ be a non-zero linear functional. For a polytope $P$ denote
$$P\_u:=\{x\in P|\, u(x)=\max\_{y\in P}u(y).\}$$
Let us define the homomorphism $f\_u\colon \Pi'\_{\mathbb{F}}\to \mathbb{F}$ on generators by $f\_u([P]):=vol(P\_u)$ where $vol$ is $(n-1)$-dimensional volume. McMullen shows that $f\_u$ is well defined.
This construction can be generalized recursively. Let $u\_1,\dots,u\_k$ be linearly independent linear functionals. Define $P\_{u\_1,\dots,u\_k}:=(P\_{u\_{1},\dots,u\_{k-1}})\_{u\_k}$. Define the homomorphism
$f\_{u\_1,\dots,u\_k}\colon \Pi'\_{\mathbb{F}}\to \mathbb{F}$ by
$$f\_{u\_1,\dots,u\_k}([P]):=vol(P\_{u\_{1},\dots,u\_{k}}),$$
where $vol$ is $(n-k)$-dimensional volume.
**Theorem 3.**(McMullen) All the homomorphisms $f\_{u\_1,\dots u\_k}$ separate points in
$ \Pi'\_{\mathbb{F}}$, i.e. if on some element $w\in \Pi'\_{\mathbb{F}} $ all such homomorphisms vanish then $w=0$.
Let us show how Theorem 3 answers (immediately) the question in the post. Let $w\in\Pi'\_{\mathbb{Q}} $ belongs to the kernel of the homomorphism in the question.
That implies that for all $k$-tuples of linearly $\mathbb{R}$-independent real linear functionals $u\_1,\dots,u\_k$ one has $f\_{u\_1,\dots,u\_k}(w)=0$. Now if $u\_1,\dots,u\_k$ are rational linear functionals which are linearly independent over $\mathbb{R}$ then they are linearly independent over $\mathbb{Q}$. For all of such $k$-tuples all $f\_{u\_1,\dots ,u\_k}$ separate points in $\Pi'\_{\mathbb{Q}}$. Hence $w=0$.
QED.
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https://mathoverflow.net/users/16183
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397996
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https://mathoverflow.net/questions/380003
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The random graph is the Fraisse limit of the class of finite graphs, the random directed graph is the Fraisse limit of the class of directed graphs, a directed graph is just a set with a binary relation.
It's easy to see that the random directed graph interprets the random graph, in fact the second is a reduct of the first. I am curious to know about the other direction. I don't see an easy way to do it, or an easy way to rule it out. More generally, I would be curious have pointers to research on when countable homogeneous structures interpret other countable homogeneous structures. I don't know if anyone has thought about this, I'm not very familiar with that terrain.
Motivation: I am doing some work with a weak notion of interpretability. I show that the random graph weakly interprets the random directed graph, and I use this as a fairly important lemma. So I'm curious to know if anything is known/obvious to experts about actual interpretations. If there is an actual interpretation then it would be a bit silly to spend a page constructing a weak interpretation.
I am actually thinking about the random $k$-ary hypergraph and the random $k$-ary relation.
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https://mathoverflow.net/users/152899
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Does the random graph interpret the random directed graph?
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No, the random graph cannot interpret the random binary relation.
I’ll just answer the title question with the goal of illustrating a technique; I haven't considered $k$-ary structures. The approach is to use the property of least supports to set up a counting argument. Some equivalences are discussed at [Least supports and weak elimination of imaginaries](http://atoms.mimuw.edu.pl/?p=1281). Combined with $\omega$-categoricity we get a useful classification of all definable equivalence relations. I’ll use the following result with $M$ being the random graph.
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> **Fact.** [1, Lemma 2.7(ii)] If $X\subseteq M^n$ is invariant, there is a unique smallest set $D\subset M$ such that $X$ is $D$-invariant.
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“$D$-invariant” means $X$ is preserved setwise by all automorphisms that fix $D$ elementwise. “Invariant” means $D\_0$-invariant for some finite $D\_0.$ Since $M$ has an $\omega$-categorical theory, invariant is the same as definable, and $D$-invariant is the same as $D$-definable. The proof in [1] uses the free amalgamation property. I’m not sure whether this special case is obvious or not.
Fix:
* a finite set $M\_0\subset M$
* an $M\_0$-definable set $X\subseteq M^n$
* an $M\_0$-definable equivalence relation ${\sim}\subseteq X\times X,$ and
* an $M\_0$-definable relation $R\subseteq X\times X$ that respects ${\sim}$ i.e. it’s the preimage of a relation $R/{\sim}\subseteq (X/{\sim})\times(X/{\sim}).$
We aim to show that $(X/{\sim}, R/{\sim})$ is not isomorphic to the random binary relation.
For each complete $n$-type $p$ over $M\_0$ let $X\_p=\{x\in X:M\Vdash p(x)\}.$ Let $P=\{p:X\_p\neq\emptyset\}.$ Pick representatives $x\_p\in X\_p$ for each $p\in P.$ Let $x\_p/{\sim}$ denote the ${\sim}$-equivalence class of $x\_p.$ By the Lemma, for each $p$ there is a unique smallest set $D$ such that $x\_p/{\sim}$ is $D$-definable. We must have $D\subset \operatorname{rng}(x\_p).$ Let $I\_p\subset \{1,\dots,n\}$ be a minimal set of indices such that $\operatorname{rng}(x|I\_p)=D$ - this step is necessary because $x\_p$ might have repeated components. Let $\hat{p}$ denote the type of $x|I\_p$ over $M\_0,$ implicitly reindexing if necessary, or allowing types to use free variables $x\_i,i\in I\_p$ instead of the usual contiguous $x\_1,\dots,x\_{|I\_p|}.$ Note that $I\_p$ and $\hat p$ do not depend on the choice of $x\_p\in X\_p.$
By uniqueness, whenever $x,y\in X\_p$ have different supports, i.e. $\operatorname{rng}(x|I\_p)\neq \operatorname{rng}(y|I\_p),$ then $x\not\sim y.$ So $x/{\sim}$ is coded by $x|I\_p$ modulo perhaps a permutation group acting on the indices of $I\_p.$
For each $p,q\in P$ define $R\_{p,q}$ on vectors $\hat{x}\in X^{I\_p}$ and $\hat{y}\in X^{I\_q}$ by
$$R\_{p,q}(\hat x,\hat y)\iff (\exists x\in X\_p)(\exists y\in X\_q)(x\supseteq \hat x \wedge y \supseteq \hat y \wedge R(x,y))$$
Then for $x\in X\_p$ and $y\in X\_q$ we have $R\_{p,q}(x|I\_p,y|I\_q)\iff R(x,y)$ because the choice of extension doesn’t affect the equivalence classes: by definition of $I\_p$ any automorphism fixing $x|I\_p$ will fix $x/{\sim},$ and similarly for $y|I\_q.$ The relation $R\_{p,q}$ is $M\_0$-definable.
Let $d=\max\_{p\in P}|I\_p|.$ If $d=0$ then $X/{\sim}$ is finite, which is absurd. The $d=1$ case could be handled by symmetry, but the following counting argument happens to cover this case too. So assume $d\geq 1.$
Pick $N$ large enough that $$4^{N^d}>|P|(N^d+2^{Nd})^d.$$
Pick $p$ with $|I\_p|=d.$ Use the extension property to pick $dN$ distinct vertices $x\_{i,j}$ where $i\in I\_p$ and $1\leq j\leq N,$ satisfying $M\Vdash \hat p((x\_{i,{j\_i}})\_{i\in I\_p})$ for each $j: I\_p\to \{1,\dots,N\}.$ In graph theory terms we’re blowing up each vertex $(x\_p)\_i$ into $N$ vertices. It doesn’t matter whether there are edges between $x\_{i,j}$ and $x\_{i,k}$ for the same $i.$
For each $q\in P$ I estimate that there are at most $(N^d+2^{Nd})^d$ types over $M\_0$ realized by tuples of the form $((x\_{i,j})\_{i\in I\_p,1\leq j\leq N},y)$ with $y\in X^{I\_q}$ and $M\Vdash \hat{q}(y).$ This is because the type of the tuple is determined by the binary relations, and the only freedom is how the $1$-type of each $y\_k$ over $M\_0$ extends to a $1$-type over $M\_0\cup \{x\_{i,j}:i\in I\_p, 1\leq j\leq N\}.$ There are at most $N^d+2^{Nd}$ such extensions: either $y\_k=x\_{i,j}$ for some $(i,j),$ or else there are the $2^{Nd}$ choices of whether or not each $\{x\_{i,j},y\_k\}$ is an edge.
Each $y\in X$ determines values $z\_y(j)\in \{1,2,3,4\}$ for each $j: I\_p\to \{1,\dots,N\},$ according to whether $R\_{p,q}(x\_{i,j\_i},y)$ and $R\_{q,p}(y,x\_{i,j\_i})$ are false/false, false/true, true/false or true/true. By the estimate in the previous paragraph, there are at most $|P|(N^d+2^{Nd})^d$ distinct functions of the form $z\_y.$ So there is some $z\in 4^{N^{I\_p}}$ not equal to $z\_y$ for any $y.$ In terms of the original interpretation, we can pick $\xi^j\in X\_p$ with $\xi^j\_i=x\_{i,j\_i}.$ The equivalence classes $\xi^j/{\sim}$ are distinct because they have different supports. The extension property for the random binary relation requires that there exists $y\in X$ such that $R(\xi^j, y)$ and $R(y,\xi^j)$ are false/false etc according to $z(j).$ But then $z=z\_y,$ contradicting the choice of $z.$
[1] *Macpherson, Dugald; Tent, Katrin*, [**Simplicity of some automorphism groups.**](http://dx.doi.org/10.1016/j.jalgebra.2011.05.021), J. Algebra 342, No. 1, 40-52 (2011). [ZBL1244.20002](https://zbmath.org/?q=an:1244.20002).
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https://mathoverflow.net/users/164965
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398026
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https://mathoverflow.net/questions/398029
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This [question](https://mathoverflow.net/q/398024/11260) on a theorem in information theory called *Mrs. Gerber's lemma* piqued my curiosity. Who is this individual, and why the "mrs." ? A quick Google search was not informative, although it did produce a Mr. Gerber's lemma ([arXiv:1802.05861](https://arxiv.org/abs/1802.05861)) -- can someone enlighten me?
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https://mathoverflow.net/users/11260
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Who is Mrs. Gerber?
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Check out the original reference "A theorem on the entropy of certain binary sequences and applications - I" by Wyner and Ziv: <https://doi.org/10.1109/TIT.1973.1055107>. Footnote 2 on page one explains
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> This result is known as “Mrs. Gerber’s Lemma” in honor of a certain lady whose presence was keenly felt by the authors at the time this research was done.
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I'm not sure you're going to get more of an explanation than that.
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https://mathoverflow.net/users/25028
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398030
| 164,278 |
https://mathoverflow.net/questions/397787
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Let $g,\hat{g}$ be two Riemannian metrics with volume forms $dv\_g$, $dv\_{\hat{g}}$, respectively. A standard estimate in the subject is the following: $$\text{tr}\_g(\hat{g}) \leq \text{tr}\_{\hat{g}} (g)^{n-1}\frac{dv\_{\hat{g}}}{dv\_g},$$ where $n$ is the dimension.
In particular, if $g$ and $\hat{g}$ are related by some Monge-Ampere equation, one can often control $\text{tr}\_{g}(\hat{g})$ by $\text{tr}\_{\hat{g}}(g)$.
**Question:** Has anyone seen alternatives to this inequality? That is, suppose one has control of the ratio of volume forms, and one of the traces, is there another way of getting an estimate on the other trace?
To make the question more precise, fix a point $p$ such that $g(p) = \delta(p)$ (Euclidean metric) and $\hat{g}(p) = \lambda\_i \delta\_{ij}(p)$. Then the displayed inequality reads: $$\text{tr}\_g(\hat{g}) = \sum\_i \lambda\_i \leq \left( \sum\_i \lambda\_i^{-1} \right)^{n-1} \prod\_i \lambda\_i,$$ which appears to be some formulation of the arithmetic-geometric mean inequality.
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https://mathoverflow.net/users/nan
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Alternative to well-known trace estimate in Riemannian geometry?
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There appear to be no alternatives, following the answer given by River Li over on MSE to [the more pedestrian formulation of this question](https://math.stackexchange.com/questions/4203635/powers-of-am-gm-hm-triples/4203956#4203956).
For posterity, let me give the details here: Let $g,\hat{g}$ be two Riemannian metrics. The estimate is local, so fix a point $p$ and choose coordinates such that $g$ is Euclidean at $p$ and $\hat{g}$ is diagonal with eigenvalues $\lambda\_i$. The original estimate $$\text{tr}\_g(\hat{g}) \leq \text{tr}\_{\hat{g}}(g)^{n-1} \frac{dv\_{\hat{g}}}{dv\_g}$$ is then written $$\sum\_i \lambda\_i \leq \left( \sum\_i \lambda\_i^{-1} \right)^{n-1} \prod\_i \lambda\_i.$$
Let $\text{AM} := \frac{1}{n} \sum\_i \lambda\_i$ denote the arithmetic mean, $\text{GM} := \left( \prod\_i \lambda\_i \right)^{\frac{1}{n}}$ denote the geometric mean, and $\text{HM} := n \left( \sum\_i \lambda\_i^{-1} \right)^{-1}$ denote the harmonic mean. Expresed in terms of these means, the aforementioned inequality is $$n \text{AM} \leq \frac{1}{n^{n-1}} \text{HM}^{1-n} \text{GM}^n,$$
or equivalently, $$\text{AM} \cdot \text{HM}^{n-1} \leq n^{-n} \text{GM}^n.$$
The original question, namely, whether there is an alternative to this inequality can therefore be more precisely formulated as, given constants $a,b,c, \lambda \in \mathbb{R}^+$ such that $$\text{AM}^a \text{HM}^b \leq \lambda \text{GM}^c,$$ are the only admissible candidates those which appeared in the original inequality?
In the answer to this [MSE question](https://math.stackexchange.com/questions/4203635/powers-of-am-gm-hm-triples/4203956#4203956), River Li gives us the following: Let $n \geq 2$ be an integer. Then $$\text{AM}^a \cdot \text{HM}^b \leq \lambda \text{GM}^{a+b}$$ if and only if $a/b \geq n-1$.
Note that the formulation in the MSE question has the inequality reversed, but is equivalent (take $\hat{g}$ to be Euclidean, and take $g$ to be diagonal).
If one spells out the condition $\frac{a}{b} \geq n-1$, one recovers that the only admissible powers are those in the original inequality.
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https://mathoverflow.net/users/nan
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398036
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https://mathoverflow.net/questions/397892
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1
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Are there any known upper bounds on the number of maximal independent sets in a hypergraph? I'm aware that simple graphs have an upper bound of $O(3^{n/3})$. How about on the number of independent sets?
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https://mathoverflow.net/users/322046
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Number of maximal independent sets in a hypergraph
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The family of maximal independent sets of a hypergraph has the property that no member of the family is contained in another. Such a family of sets is called an *antichain* or a *clutter* or a *Sperner family*. By [Sperner's theorem](https://en.wikipedia.org/wiki/Sperner%27s_theorem), a Sperner family of subsets of an $n$-element set has at most $\binom n{\lfloor n/2\rfloor}$ members. In particular, a hypergraph $H=(V,E)$ of order $|V|=n$ has at most $\binom n{\lfloor n/2\rfloor}$ maximal independent sets. This bound is attained when $E$ is the set of all $\left(\lfloor n/2\rfloor+1\right)$-element subsets of $V$.
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https://mathoverflow.net/users/43266
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398039
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https://mathoverflow.net/questions/398037
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14
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Consider the following sequence defined as a sum
$$a\_n=\sum\_{k=0}^{n-1}\frac{3^{3n-3k-1}\,(7k+8)\,(3k+1)!}{2^{2n-2k}\,k!\,(2k+3)!}.$$
>
> **QUESTION.** For $n\geq1$, is the sequence of rational numbers $a\_n$ always integral?
>
>
>
|
https://mathoverflow.net/users/66131
|
Integrality of a sequence formed by sums
|
Let $A(x) = \sum\_{n=1}^\infty a\_n x^n$ and let
$$S(x) = \sum\_{k=0}^\infty (7k+8)\frac{(3k+1)!}{k!\,(2k+3)!} x^k.$$
Then the formula for $a\_n$ gives
$A(x) = R(x)S(x)$,
where
$$R(x) = \frac{1}{3}\biggl(\frac{1}{1-\frac{27}{4} x} -1\biggr).$$
A standard argument, for example by Lagrange inversion, gives
$$S\left(\frac{y}{(1+y)^3}\right)=\frac{4+y}{3(1+y)^2}.$$
A straightforward computation gives
$$R\left(\frac{y}{(1+y)^3}\right) = \frac{9y}{(4+y)(1-2y)^2}.$$
Thus
$$A\left(\frac{y}{(1+y)^3}\right)=\frac{3y}{(1-2y)^2(1+y)^2}.$$
Since the power series expansion of $y/(1+y)^3$ starts with $y$ and has integer coefficients, its compositional inverse has integer coefficients, so $A(x)$ does also.
|
31
|
https://mathoverflow.net/users/10744
|
398040
| 164,282 |
https://mathoverflow.net/questions/397707
|
8
|
For a real-valued function $f$ on $[0,1]$, define its quadratic variation by the formula
$$[f]:=\limsup\sum\_{j=1}^n(f(t\_j)-f(t\_{j-1}))^2,$$
where the $\limsup$ is taken over all "partitions" $0=t\_0<\cdots<t\_n=1$ of $[0,1]$ as $\max\_{1\le j\le n}(t\_j-t\_{j-1})\to0$.
If $f$ is continuously differentiable or, more generally, uniformly Hölder continuous, then, of course, $[f]=0$. On the other hand, if $f(x)=x^2\cos(1/x^4)$ for $x\ne0$ and $f(0)=0$, then $f$ is differentiable, but $[f]=\infty$.
Suppose now that $f$ is differentiable and $[f]<\infty$. Does it then necessarily follow that
$$\sum\_{j=1}^n\Big(f\Big(\frac jn\Big)-f\Big(\frac{j-1}n\Big)\Big)^2\to0$$
as $n\to\infty$ (or maybe even that $[f]=0$)?
|
https://mathoverflow.net/users/36721
|
A dichotomy for the quadratic variation of differentiable functions?
|
The paper linked formulates quadratic variation in a measure-theoretic framework. The references therein may also be of interest. As a disclaimer, I did not read this paper very closely, nor is this a research area I am familiar with. I imagine being able to access measure-theoretic tools might offer some interesting approaches to solving this problem.
In the paper, the authors assign positive finite (Hausdorff) measures to functions of bounded quadratic variation. Their Theorem 20 proves that for every positive finite measure on $[0,1]$, there is a function of finite quadratic variation that generates this measure (and I believe has quadratic variation related to the variation norm of the measure. So a nonzero finite measure would correspond with a nonzero but finite quadratic variation function).
The paper: <https://www.ams.org/journals/tran/2011-363-08/S0002-9947-2011-05209-8/S0002-9947-2011-05209-8.pdf>
Title: HAUSDORFF MEASURES AND FUNCTIONS OF BOUNDED QUADRATIC VARIATION
Authors: D. APATSIDIS, S. A. ARGYROS, AND V. KANELLOPOULOS
Abstract:
To each function $f$ of bounded quadratic variation we associate
a Hausdorff measure $\mu\_f$ . We show that the map $f \mapsto \mu\_f$ is locally Lipschitz
and onto the positive cone of $\mathcal{M}[0,1]$. We use the measures $\{\mu\_f : f \in V\_2\}$ to
determine the structure of the subspaces of $V\_2^0$ which either contain $c\_0$ or the 2
square stopping time space $S^2$.
Chapter 3 is of most interest I think.
Edit
----
So I think I totally misunderstood Theorem 22 of the paper (I have removed my incorrect interpretation). The theorem states that the set $X:= \{x \in [0,1]: f \text{ is differentiable at } x\}$ has measure 0 under the quadratic variation measure $\mu\_f$ (assuming $f$ has finite quadratic variation to begin with). If $f$ is differentiable on $[0,1]$ then we have $X = [0,1]$ and therefore $\mu\_f([0,1]) = 0$. This in particular implies the quadratic variation of $f$ is zero (since $\mu\_f([0,1])$ equals the quadratic variation). I believe this is Corollary 23 of the paper, which actually proves the stronger claim that if a function of finite quadratic variation is continuous and has only countably many points of non-differentiability then the function has quadratic variation zero. So it looks like the answer is indeed affirmative.
|
5
|
https://mathoverflow.net/users/317937
|
398042
| 164,283 |
https://mathoverflow.net/questions/394333
|
0
|
Suppose $Y$ is a random variable in $\mathbb{R}^d$, and we want to find the covering number
\begin{equation\*}
\mathcal{F} = \big\{ F\_{Y|W} (y | W) : y \in \mathbb{R}^d \big\}
\end{equation\*}
where $W$ is another random variable in $\mathbb{R}^k$ and $F\_{Y|W} (y | W)$ is the conditional distribution function. Denote $P\_{Y, W}$ is the law of $(Y, W)$, what we want to find is the covering number $N(\epsilon, \mathcal{F}, L\_2(P\_{Y, W}))$.
At the first glimpse, this question is quite simple, since if we treat $W$ as a determined number, then
\begin{equation\*}
\mathcal{F} = \overline{\operatorname{conv}}(\mathcal{G}),
\end{equation\*}
where $\mathcal{G} := \big\{ 1(Y \leq y) : y \in \mathbb{R}^d \big\}$ with $N(\epsilon, \mathcal{G}, L\_2(P\_{Y})) \lesssim \epsilon^{- 2d}$ (This is because $F\_{Y|W} (y | W) : \mathbb{R}^d \rightarrow [0, 1]$ is a distribution function). Then by the result for convex hulls, we have
\begin{equation\*}
\log N \big( \epsilon, \mathcal{F}, L\_2 (P\_Y)\big) \lesssim \epsilon^{-2 d/ (d + 1)}.
\end{equation\*}
However, I think the randomness of $W$ complex this problem: what we want is $N(\epsilon, \mathcal{F}, L\_2(P\_{Y, W}))$ rather than $N ( \epsilon, \mathcal{F}, L\_2 (P\_Y))$. And I try to find some articles about this, but it is proved to be futile.
So can anyone help me with this question. Thanks in advance!
|
https://mathoverflow.net/users/153595
|
Covering number of the conditional distribution function
|
You need a different approach. Each function in your function space can be written as
$$F\_{Y|W}(y|W) = \int 1(s \leq y) P(Y = ds|W)$$
for some $y$. Thus,
$$\|F\_{Y|W}(y\_2|W) - F\_{Y|W}(y\_1|W)\|\_{L^1} = E\_{P\_W}|F\_{Y|W}(y\_2|W) - F\_{Y|W}(y\_1|W)| \leq E\_{P\_W}\int 1(y\_1 \leq s \leq y\_2) P(Y = ds|W) \lessapprox \|1(Y \leq y\_2) - 1(Y \leq y\_1)\|\_{L^1}.$$
A similar argument probably works for $L^2$.
In particular, your function space is obtained by a Lipschitz transformation applied to the function space
$$\mathcal{F}\_{ind} = \{s \mapsto 1(s\leq y): y \in \mathbb{R}\}.$$
So your covering number is essentially that of $\mathcal{F}\_{ind} $.
|
1
|
https://mathoverflow.net/users/317937
|
398044
| 164,284 |
https://mathoverflow.net/questions/398035
|
4
|
By Fermat's Last Theorem, there are no solutions to the Diophantine equation $a^n + b^n = c^n$ for $a,b,c > 0$ and $n>2$. Beal's conjecture allows the exponents to be different (but also $>2$ ). Is the lack of solutions because there is not enough wiggle room? (Squares are too abundant, but what about cubes and so on?)
Let's call a positive integer *N-power-min* if the smallest exponent in its prime factorization is $N$.
For $N=3$, I've downloaded a list of [ABC Triples](https://www.math.leidenuniv.nl/%7Edesmit/abc/abctriples_below_1018.gz) and the first solutions were:
$271^{3} + 2^{3} 3^{5} 73^{3} = 919^{3}$
and
$3^{4} 29^{3} 89^{3} + 7^{3} 11^{3} 167^{3} = 2^{7} 5^{4} 353^{3}$
Nothing with $N>3$ was found.
Does the ABC conjecture (or a related one) imply that for any $N > 2$ , we have a limited or no solutions $a+b=c$ with a,b,c *N-power-min* ?
|
https://mathoverflow.net/users/323465
|
Numbers with large prime exponents and the ABC conjecture
|
If $a,b,c$ are $N$-power min then $\operatorname{rad}(abc) \leq (abc)^{1/N} \leq c^{ 3/N}$
and the $abc$ conjecture implies that $$c< K\_\epsilon \operatorname{rad}(abc)^{1+\epsilon} \leq K\_\epsilon c^{ (3/N)(1+ \epsilon)} $$ so $$ c< K\_\epsilon^{ \frac{1}{ 1 - (3/N)(1+\epsilon)}}$$
But if $c$ is $N$-power-min then $c \geq 2^N$, so for $N$ sufficiently large depending on $\epsilon, K\_\epsilon$, this inequality cannot be satisfied.
For $N$ sufficiently large depending on $\epsilon, K\_\epsilon$, these inequalities cannot be satisfied.
So certainly the abc conjecture implies for *some* $N>2$, we have no solutions.
But, as stated, it doesn't imply (by any easy argument) that for *every* $N>2$ we have no solutions, nor does it even imply for any particular $N$ that there is no solution, since the exact value of $K\_\epsilon$ is not given in the statement (as was explained by Mark Sapir in the comments).
---
There probably are infinitely many solutions for $N=3$, albeit very sparse ones. You just need to search for an elliptic curve of the form $a^4 x^3 + b^4 y^3 + c^4 z^3$ of positive rank, then find various $x,y,z$ solutions.
|
6
|
https://mathoverflow.net/users/18060
|
398047
| 164,285 |
https://mathoverflow.net/questions/398004
|
1
|
The game $G(N,M)$ is played:
$N$ ($N\geq 2$) is the number of players, labeled $1$~$N$. In the beginning they have a pot with some chips in it. Players move alternatively in the order from $1$ to $N$. In their move, a player announces an integer $C$ and toss a fair coin: if head, $C$ more chips are added to the pot; if tail, $C$ chips are removed from the pot, where $1\leq C\leq $ the current number of chips in the pot.
The game ends on two conditions:
1. The pot is empty after a player's move, in which case that player loses the game, and everyone else wins.
2. There're $M$ or more chips in the pot, in which case everyone wins.
Communication is not allowed, and **we assume a player's choice of integer $C$ is a function only of the current number of chips in the pot**. Formally, in game $G(N,M)$ a player's strategy is a function $f: \{1,2,...,M-1\} \longmapsto \{1,2,...,M-1\}$, with the restriction $f(x)\leq x, \forall x$.
**Question:** Is there always an equilibrium for $G(N,M)$? If so, what can we say about the equilibria? Is it feasible to search for an equilibrium of, say $G(3,1000)$?
---
**Edit**: Notice that the strategy of always betting all but one chips can't be an equilibrium for many games. For example in $G(3,5)$, if the other 2 players stick to that strategy and there are 4 chips, you're better off betting 1 rather than 3.
|
https://mathoverflow.net/users/75935
|
Is there an equilibrium for this non-zero-sum game?
|
Edited on 24-July-2021 to reflect the requirement that the equilibrium is in pure stationary strategies.
The game you present is a stochastic game: the number of chips in the pot and the identity of the player whose turn it is to move serve as a state variable. Since the number of chips in the pot is bounded (between 0 and M), there are finitely many states and actions to each player. In fact, the game is a stochastic game with perfect information: the players move alternately, so there are no simultaneous moves.
Such games have (a) an equilibrium that do not involve randomization, that is, the choice of the number of chips is deterministic yet it depends on past play, see Thuijsman and Raghavan, Perfect Information Stochastic Games and Related Classes, International Journal of Game Theory, 1997, 26, 403-408.
They also have (b) a symmetric stationary equilibrium that involves randomization, that is, the choice of the number of chips is random and depends only on the current state, see Fink, Equilibrium in a Stochastic
n-Person Game, Journal of Science of the Hiroshima University, Series A (mathematics), 1964, 28(1), 89-93.
You, however, are interested in stationary equilibria that involve no randomization. The theory does not guarantee that such equilibria exist.
|
1
|
https://mathoverflow.net/users/64609
|
398051
| 164,288 |
https://mathoverflow.net/questions/397640
|
5
|
Let $X$ be a nice scheme (additional assumptions could be added), and let $Et(X)$ be its (Artin-Mazur) etale homotopy type. I am looking for a/the scheme $Y$ over $X$ whose etale homotopy type $Et(Y)$ will be the topological universal cover of $Et(X)$. By definition $Et(X)$ is the geometric realization of a simplicial set, and it was pointed out to me that if $R \rightarrow S$ is the universal cover of a simplicial set $S$, then the geometric realization $|R|$ is the topological universal cover of $|S|$.
What is the meaning of "Universal cover of a simplicial set"; Is there a reference for that, and also for the second assertion?. How could we apply this to find $Y$?
**Edit:** If we want to avoid the simplicial method. Suppose that there is a scheme $Y$ over $X$ such that
1. The first etale homotopy group $\pi\_1^{et}(Y)$ is trivial.
2. For all $n \geq 2$ we have $\pi\_n^{et}(Y) \simeq \pi\_n^{et}(X)$.
Are these conditions sufficient to state that $Et(Y)$ is the topological universal cover of $Et(X)$?
|
https://mathoverflow.net/users/144181
|
Construction of the universal covering space of the etale homotopy type $Et(X)$
|
Such an "étale universal cover" exists at least if $X$ is Noetherian and geometrically unibranch (and for all qcqs $X$ if one considers profinite étale homotopy types).
**Background.** I will regard the étale homotopy type of a scheme as an object in the $\infty$-category $\mathrm{Pro}(\mathcal S)$ of pro-spaces. In the $\infty$-category $\mathcal S$, the universal cover of a pointed space $(X,x)$ can be characterized as the initial object in the $\infty$-category of pointed $0$-truncated morphisms to $(X,x)$ (i.e., morphisms with discrete fibers). Similarly, one can define $0$-truncated morphisms in $\mathrm{Pro}(\mathcal S)$, and every pointed object admits a universal cover. References for the étale homotopy type and for $n$-truncated/$n$-connected morphisms in $\mathrm{Pro}(\mathcal S)$ are Sections E.2 and E.4.2 in [Spectral Algebraic Geometry](https://www.math.ias.edu/%7Elurie/papers/SAG-rootfile.pdf).
Let $\mathcal S\_{<\infty}$ be the $\infty$-category of truncated spaces and let $\mathrm{Et}\colon \mathrm{Sch} \to \mathrm{Pro}(\mathcal S\_{<\infty})$ be the [protruncated étale homotopy type](https://arxiv.org/pdf/1812.11637.pdf), which is the homotopy-coherent incarnation of the construction of Artin and Mazur. Under the equivalence
$$\mathrm{Pro}(\mathcal S\_{<\infty}) \simeq \mathrm{Fun}^\mathrm{acc,lex}(\mathcal S\_{<\infty},\mathcal S)^\mathrm{op},$$
$\mathrm{Et}(X)$ is the functor $\mathcal S\_{<\infty}\to\mathcal S$ sending a truncated space $K$ to the global sections $\Gamma(X\_\mathrm{et},K)$ of the constant étale sheaf on $X$ with value $K$.
**Construction.** Let $X$ be a scheme with a geometric point $x\colon \operatorname{Spec} k\to X$, with $k$ separably closed. Consider the category $\mathrm{FEt}(X,x)$ of pointed finite étale covers of $X$, that is, factorizations of $x$ through a finite étale morphism $X'\to X$. This category has finite limits and in particular is cofiltered (it is also essentially small). Thus, the limit $\tilde X$ of the forgetful functor
$$
\mathrm{FEt}(X,x) \to \mathrm{Sch}\_{/X}
$$
exists (note that $\tilde X$ depends on $x$, so the notation is abusive). This is a natural candidate for the "universal cover" of $(X,x)$.
Now, the statement that $\mathrm{Et}(\tilde X)\to\mathrm{Et}(X)$ is the universal cover of the pointed pro-space $(\mathrm{Et}(X),x)$ is equivalent to the following three conditions: 1) the map $\mathrm{Et}(\tilde X)\to\mathrm{Et}(X)$ is $0$-truncated, 2) $\mathrm{Et}(\tilde X)$ is connected, and 3) $\pi\_1^\mathrm{et}(\tilde X,x)$ is trivial.
**Results.**
1) always holds if $X$ is quasi-compact and quasi-separated.
First, I claim that the functor $\mathrm{Et}$ preserves the limit of the diagram defining $\tilde X$. Under the above equivalence of $\infty$-categories, this is the statement that for $K$ a truncated space, $\Gamma((-)\_\mathrm{et},K)$ transforms this limit into a colimit, which is a standard property of étale cohomology with respect to inverse limits of qcqs schemes. [Here it is important that $K$ is truncated, otherwise this may not be true.]
Then, if $p\colon X'\to X$ is finite étale, I claim that the morphism $\mathrm{Et}(p)\colon \mathrm{Et}(X')\to \mathrm{Et}(X)$ is $0$-truncated. In fact, it is the pullback of a morphism of groupoids $\pi\colon\Xi'\to\Xi$ with finite discrete fibers. To see this, note that the morphism of étale $\infty$-topoi induced by $p$ is itself the pullback of such a morphism $\pi$. The point is then that for any space $K$, $\Gamma(X'\_\mathrm{et}, K)$ is $\Gamma(X\_\mathrm{et},p\_\*K)$ and $p\_\*K$ is a locally constant sheaf on $X$ in a strong sense: it is the pullback of the sheaf $\pi\_\*K$ on $\Xi$ (the fact that the fibers of $\pi$ are finite spaces is used here, to commute the pushforward with the pullback). One can thus apply Proposition 2.15 in [Higher Galois theory](http://www.mathematik.ur.de/hoyois/papers/highergalois.pdf) to compute $\Gamma(X\_\mathrm{et},p\_\*K)$ in terms of the étale homotopy type of $X$. Unpacking this formula gives $\mathrm{Et}(X')=\mathrm{Et}(X)\times\_\Xi\Xi'$. [Here, $K$ need not be truncated, so $p$ induces a $0$-truncated morphism on actual étale homotopy types, and $X$ need not even be qcqs.]
**ETA:** If $X$ is locally noetherian, another proof of this claim is Lemma 2.1 in Schmidt-Stix, [Anabelian geometry with etale homotopy types](https://arxiv.org/pdf/1504.01068.pdf).
Finally, since $0$-truncated morphisms are stable under limits, $\mathrm{Et}(\tilde X)\to\mathrm{Et}(X)$ is also $0$-truncated.
2) also holds if $X$ is qcqs. In this case $\tilde X$ is connected, as any clopen subset of $\tilde X$ lifts to a clopen subset of some finite étale $X'\to X$.
3) holds if we assume moreover that the pro-group $\pi\_1^\mathrm{et}(X,x)$ is profinite, e.g., $X$ is Noetherian and geometrically unibranch. Then $\pi\_1^\mathrm{et}(\tilde X,x)$ is also profinite by 1), so 3) is equivalent to the statement that every finite étale cover of $\tilde X$ is trivial, which holds by construction.
Note that the additional assumption for 3) is not needed if we pass to the profinite completions. That is, if $X$ is qcqs, then $\mathrm{Et}(\tilde X)^\wedge \to \mathrm{Et}(X)^\wedge$ is the universal cover of the profinite étale homotopy type $(\mathrm{Et}(X)^\wedge,x)$.
**Example.** Let $X=\mathbb G\_m$ over a field $k$ of characteristic zero, pointed at $1$ by an algebraic closure $\bar k$ of $k$. Then $\tilde X=\operatorname{Spec}\bar k[t^{\pm 1}][t^{1/n}, n\geq 2]$.
|
6
|
https://mathoverflow.net/users/20233
|
398056
| 164,291 |
https://mathoverflow.net/questions/398067
|
2
|
Knowing that $\omega\Subset\Omega\subset\mathbb{R}^2$ (compactly included) are two open and bounded sets with $C^2$ boundary, is it true that for any function $\phi\_0:\overline{\omega}\to\mathbb{R},\ \phi\_0\in C^1(\overline{\omega})$ ($\overline{\omega}$ is the closure of $\omega$) we can find an extension $\phi:\Omega\to\mathbb{R}$ with $\phi\in C^1\_{c}(\Omega)$ (compactly supported in $\Omega$)?
**Motivation**
If this type of result is true than we can obtain simple formulas for perimeter of implicitly defined curves in $\mathbb{R}^2$, putting $\phi\_0$ the unit outer normal vector to a regular curve (which is defined in a neighborhood of the boundary of $\omega$). See here: [Perimeter continuity of $BV$ sets on any sequence from $W^{1,1}$](https://mathoverflow.net/questions/398001/perimeter-continuity-of-bv-sets-on-any-sequence-from-w1-1)
**What did I do?**
I proved by standard methods (using cut-off functions and convolution) that we can obtain a mollifying sequence $\phi\_n,\ n\in\mathbb{N}^\*$ compactly supported in $\Omega$ that tends to $\phi\_0$ in $L^1(\Omega)$, but I cannot prove that we can indeed have an extension.
|
https://mathoverflow.net/users/61629
|
$C^1$ extension with compact support
|
$\newcommand\de\delta\newcommand\Om\Omega\newcommand\om\omega\newcommand\R{\mathbb R}$The answer is yes. Indeed, by [Whitney's theorem](http://www.ams.org/tran/1934-036-01/S0002-9947-1934-1501735-3/S0002-9947-1934-1501735-3.pdf), there is a function $f\in C^1(\mathbb R^2)$ whose restriction to $\overline\omega$ is $\phi\_0$. Now take any open set $\Omega\_0$ such that $\omega\Subset\Omega\_0\Subset\Omega$ and then any
function $g\in C^1(\mathbb R^2)$ such that $g=1$ on $\overline\omega$ and $g=0$ on $\mathbb R^2\setminus\Omega\_0$. It remains to let $\phi$ be the restriction of $gf$ to $\Om$. (The conditions that $\om$ and $\Om$ have $C^2$ boundaries and that $\Om$ be bounded are not needed. The condition that $\om$ be bounded follows from $\om\Subset\Om$.)
---
**Detail:** The set $\Om\_0$ and the function $g$ can be constructed as follows. Since $\om\Subset\Om$, there is some real $\de>0$ such that the closure $\overline{\om\_{2\de}}$ of the $(2\de)$-neighborhood $\om\_{2\de}$ of $\om$ is contained in $\Om$. Let then $\Om\_0:=\om\_{2\de}$ and $g:=1\_{\om\_\de}\*\psi\_\de$, where $\psi\_\de$ is any nonnegative function in $C^1(\R^2)$ supported on the ball of radius $\de$ in $\R^2$ centered at $0$ and such that $\int\_{\R^2}\psi\_\de=1$.
|
3
|
https://mathoverflow.net/users/36721
|
398079
| 164,296 |
https://mathoverflow.net/questions/398080
|
1
|
I am trying to see them as subfield $\mathbb{Q}(\zeta\_n).$ I feel it is a tiring job by using SageMath. Moreover, I am ending up with the abelian cubic field with the class number $1.$
I appreciate any alternative methods.
|
https://mathoverflow.net/users/131448
|
How do I find abelian cubic extension over $\mathbb{Q}$ with class number more than 1?
|
Günter Lettl, A lower bound for the class number of certain cubic fields, Math. Comp. 46, #174 (April 1986) 659-666, has abstract,
Let $K$ be a cyclic number field with generating polynomial $$x^3-{a-3\over2}x^2-{a+3\over2}x-1$$ and conductor $m$. We will derive a lower bound for the class number of these fields and list all such fields with conductor $m=(a^2+27)/4$ or $m=(1+27b^2)/4$ and small class number.
Lawrence C. Washington, Class numbers of the simplest cubic fields, Math. Comp. 48, #177 (Jan. 1987) 371-384, has abstract,
Using the "simplest cubic fields" of D. Shanks, we give a modified proof and an extension of a result of Uchida, showing how to obtain cyclic cubic fields with class number divisible by $n$, for any $n$.
See also Jaclyn Lang, [Properties](https://www.math.univ-paris13.fr/%7Elang/mathesis.pdf) of a family of cyclic cubic fields.
Ennola and Turunen, On cyclic cubic [fields](https://www.ams.org/journals/mcom/1985-45-172/S0025-5718-1985-0804947-3/S0025-5718-1985-0804947-3.pdf), tabulate class numbers and units in cyclic cubic fields with conductor less than $16000$.
<http://www.lmfdb.org/NumberField/?start=0°ree=3&galois_group=C3&count=20> lists over 4,000 cyclic cubic number fields, with class number.
|
4
|
https://mathoverflow.net/users/3684
|
398082
| 164,297 |
https://mathoverflow.net/questions/398059
|
2
|
We know that if an operator has $L^2$-kernel, then it is Hilbert-Schmidt.
Is there a similar simple criterion to detect compact operators?
In particular, I'd like to know the following: Let $f$ be a Schwartz function on ${\mathbb R}^2$ with $\mathrm{supp}(f)\subset{\mathbb R}\times J$ for some compact Interval $J$.
Let
$$
k(x,y)=f(e^x,x-y)
$$
Is the operator $T:L^2({\mathbb R})\to L^2({\mathbb R})$, $T(\phi)(x)=\int\_{\mathbb R}k(x,y)\phi(y)\,dy$ a compact operator?
|
https://mathoverflow.net/users/nan
|
Compactness of integral operators
|
The answer is no, in general. Assume for example that $J=[0,1]$ so that $|f| \leq C \chi\_{\bf R \times [0,1]}$. Then $|T\phi(x)| \leq C \int\_{x-1}^x |\phi(y)|\, dy \leq C\int\_0^1 |\phi(x+y)|\, dy$ and Minkowsky inequaility for integrals gives $\|T\phi\_2\| \leq C\|\phi\|\_2$. So boundedness follows without any smoothness assumption. Assume now that $f=\chi\_{[0,1]\times[0,1]}$. Then $T\phi(x)=\int\_{x-1}^x \phi (y)\, dy$ for $x \leq e$ and this is not a compact operator (consider $\chi\_{[-n-1,-n]}$). To get a smooth counterexample it suffices to consider a smooth f with compact support which dominates $\chi\_{[0,1]\times[0,1]}$. If $S$ is the corresponding integral operator, then $S\phi \geq T\phi$ for positive $\phi$ and $S$ is not compact.
|
3
|
https://mathoverflow.net/users/150653
|
398107
| 164,305 |
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