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https://mathoverflow.net/questions/449864 | 2 | Let $\pi$ be an automorphic representation of $\textrm{GL}\_n$. Associated to $\pi$, we can define the standard $L$-function $L(s, \pi)$.
My question is: what is the difference between $L(s, \pi)$ and the $L$-function attached to a **cuspidal** automorphic representation of $\textrm{GL}\_n$ ? For example, are there any properties of these two $L$-functions which are different?
| https://mathoverflow.net/users/167708 | Question on automorphic $L$-functions | Let us restrict to automorphic representations of $\mathrm{GL}\_n$ over $\mathbb{Q}$ with arbitrary $n$ and unitary central character.
If $\pi$ is an irreducible cuspidal representation, then $L(s,\pi)$ is entire unless $\pi\cong|\det|^{it}$ in which case $L(s,\pi)=\zeta(s+it)$ has a simple pole at $s=1-it$. Let us call these $L$-functions cuspidal.
If $\pi=\pi\_1\boxplus\dotsb\boxplus\pi\_r$ is an isobaric sum of irreducible cuspidal representations, then $L(s,\pi)$ factors uniquely into cuspidal $L$-functions as $L(s,\pi\_1)\dotsb L(s,\pi\_r)$. In particular, $L(s,\pi)$ can have several poles. See Liu-Ye: Weighted Selberg orthogonality and uniqueness of factorization of automorphic $L$-functions, Forum Math. 17 (2005), 493-512.
If $\pi$ is a general automorphic representation, then $L(s,\pi)$ differs from a product of cuspidal $L$-functions by finitely Euler factors. See Jacquet: Principal $L$-functions of the linear group, In: Automorphic forms, representations and $L$-functions, Part 2, 63-86, Proc. Sympos. Pure Math., XXXIII, Amer. Math. Soc., Providence, R.I., 1979.
| 2 | https://mathoverflow.net/users/11919 | 449938 | 181,034 |
https://mathoverflow.net/questions/449858 | 17 | What are the torsion units of the ring $R\_n:=\mathbb{Z}[x]/(1+x+x^2+\cdots+x^{n-1})$? Since $x^n = 1$ in $R\_n$ it is clear that all elements of the form $\pm x^i$ are torsion units. Is this all of them?
Of course, if $n$ is prime then $R\_n$ is the ring of integers of the $n^{th}$ cyclotomic field and the result is true in this case.
For small composite $n$ is it not hard to verify the statement numerically. The Chinese Remainder Theorem gives an injective ring map
$$
\Psi: R\_n\rightarrow \prod\_{\substack{k|n \\ k\neq 1}}\mathbb{Z}[\zeta\_k]
$$
where $\mathbb{Z}[\zeta\_k]$ is the ring of integers of the $k^{th}$ cyclotomic field. For small $n$ one can enumerate all of the torsion units on the right-hand side, choose $\mathbb{Z}$-bases for both sides, and via a simple matrix calculation deduce which torsion units on the right-hand side are in the image of $\Psi$. I've carried out this calculation for all composite $n\leq 30$ and verified that the only torsion units in $R\_n$ are indeed the elements $\pm x^i$. I've been unable to turn this into a proof for all $n$ as the inverse of the matrix of $\Psi$ has no obvious general form.
The corresponding statement is indeed true for the similar ring $S\_n:=\mathbb{Z}[x]/(x^n-1)$, which is the group ring of the cyclic group of order $n$. The result in this case goes back to at least Higman [1], and is not difficult to show. Indeed, let $\phi: S\_n\rightarrow M\_n(\mathbb{C})$ denote the left regular representation of $S\_n$, and suppose $u = a\_0 + a\_1x + \cdots + a\_{n-1}x^{n-1}$ with $a\_i\in\mathbb{Z}$ satisfies $u^m = 1$. Multiply $u$ by an appropriate power of $x$ so that $a\_0 \neq 0$. We have $\mathrm{tr}(\phi(x^i)) = 0$ for all $i \neq 0$, and hence $\mathrm{tr}(\phi(u)) = a\_0n$. On the other hand, since $\phi(u)$ is an $n\times n$ matrix of finite order we must have $|\mathrm{tr}(\phi(u))|\leq n$ and therefore $a\_0 = \pm 1$. This implies that we have equality $|\mathrm{tr}(\phi(u))| = n$ and therefore $\phi(u)$ is a scalar matrix, from which we see that $\phi(u) = a\_0I\_n$. Since $\phi$ is injective we must have $u = a\_0$, which completes the proof.
The difficulty in applying this argument to the ring $R\_n$ is that the left regular representation results in a much weaker inequality. Specifically, if $\rho:R\_n\rightarrow M\_{n-1}(\mathbb{C})$ denotes the left regular representation of $R\_n$ we now have $\mathrm{tr}(\rho(x^i)) = -1$ if $i\neq 0$. So if $u = a\_0 + a\_1x + \cdots + a\_{n-2}x^{n-2}$ with $a\_i\in\mathbb{Z}$ satisfies $u^m = 1$, we can only say that
$$
|a\_0(n-1) - \sum\_{i = 1}^{n-2}a\_i|\leq n - 1
$$
from which I can deduce very little about the individual $a\_i$.
[1] *Higman, G.*, [**The units of group-rings.**](https://doi.org/10.1112/plms/s2-46.1.231), Proc. London math. Soc. (2) 46, 231-248 (1940). [ZBL66.0104.04](https://zbmath.org/?q=an:66.0104.04).
| https://mathoverflow.net/users/507811 | Torsion units of the ring $\mathbb{Z}[x]/(1+x+x^2+\cdots+x^{n-1})$ | I think, they are all of them. Let me be more concrete and accurate than in the initial answer. But this also makes the answer sometimes boring. Shortcuts are welcome.
Let $f$ be a torsion unit, that is, $f$ is represented by a polynomial with integer coefficients (again denoted by $f$) of degree at most $n-2$ such that $f^N-1$ is divisible by $1+x+\ldots+x^{n-1}$ for certain positive integer $N$.
Denote by $\Theta(n)$ the set of primitive roots of unity of degree $n$, and by $T(n)=\sqcup\_{1<d,d|n} \Theta(d)$ the set of all roots of unity of degree $n$ other than 1. Thus, we are given that $f(\xi)$ is a root of unity for all $\xi\in T(n)$.
**Lemma 1.** If $w\in \Theta(a)$, $g(z)\in \mathbb{Z}[z]$ and $g(w)\in \Theta(b)$, then either $b$ divides $a$ or $a$ is odd and $b$ divides $2a$.
**Proof.** Denote $c={\rm LCM}(a,b)$ and choose $\eta\in \Theta(c)$. Then we may find integers $k,\ell$ such that $w=\eta^k$, $g(w)=\eta^\ell$. So, the polynomial $P(x):=g(x^k)-x^\ell$ has a root $x=\eta$. Thus any element of $\Theta(c)$ is a root of $P$, since cyclotomics are irreducible. For any integer $t$ coprime to $c$ we have $\eta^t\in \Theta(c)$, so $g(w^t)-(g(w))^t=P(\eta^t)=0$. Try to choose $t$ such that $w=w^{t}$ but $g(w)^t\ne g(w)$, this would yield a contradiction. Our requirements for $t$ are, in other words: $a$ divides $t-1$; but $b$ does not divide $t-1$. Assume that $b$ does not divide $a$, and, if $a$ is odd, then $b$ does not divide $2a$. This allows to choose a prime $p$ such that $s:=\nu\_p(b)$ (notation means that $p^s$ is the maximal power of $p$ which divides $b$) satisfies $s>\nu\_p(a)$ and either $p$ is odd, or $p=2$ and $s\geqslant 2$. Using Chinese remainders theorem, choose $t$ be congruent to $1+p^{s-1}$ modulo $p^s$ and $t$ be congruent to 1 modulo $c/p^{s}$. Then $t$ is coprime with $c$, $a$ divides $t-1$ but $b$ does not divide $t-1$. This yields a necessary contradiction.
Let now $a>1$ be a divisor of $n$, choose $w\in \Theta(a)$. Let $g(w)\in \Theta(b)$.
If $a$ is odd and $b$ is an even divisor of $2a$, then $-g(w)\in \Theta(b/2)$. Thus in all cases of Lemma 1 we have $f(w)=\pm w^m$ for certain $m$. Then this equation $f(x)=\pm x^m$ holds for all $x\in \Theta(a)$, since cyclotomics are still irreducible.
So, for each divisor $a>1$ we have a sign $\varepsilon(a)=\pm 1$ and an exponent $m(a)$ such that $f(x)\equiv \varepsilon(a) x^{m(a)} \pmod{\Phi\_a(x)}$. Now we need to prove that these signs and exponents for different $a$ are consistent. The idea (not very deep) is to prove this by induction on $n$. Let $p\_1<p\_2<\ldots<p\_r$ be all prime divisors of $n$. By induction, we may suppose that for certain signs $\delta\_i$ and exponents $m\_i$ we have $\varepsilon(a)x^{m(a)}=\delta\_i x^{m\_i}$ for all divisors $a>1$ of $n/p\_i$ and all $x\in \Theta(a)$ (in other words, that $f(x)\equiv \delta\_ix^{m\_i} \pmod{1+x+\ldots+x^{n/p\_i-1}}$).
We proceed with proving the consistency of signs $\delta\_i$ and exponents $m\_i$. Let's start with signs. Assume that $\delta\_i=-\delta\_j$ for some $i<j$. This yields that $x^{m\_i}+x^{m\_j}$ is divisible by $1+x+\ldots+x^{n/(p\_ip\_j)-1}$. Substituting $x=1$, we get that $2$ is divisible $n/(p\_ip\_j)$, that is, $n=p\_ip\_j$ or $n=2p\_ip\_j$. In the latter case, if $1<i<j$, then replacing the pair $i<j$ to $1<i$ or $1<j$ in the above argument, we get a contradiction. So, $n=p\_ip\_j$ or $n=4p\_j$. If $p\_1=2$ and $n=2p\_2$, then the signs may be chosen consistently because $-1=x$ on $T(2)$. The cases $n=pq$ with odd primes $p\ne q$ and $n=4p$ with odd prime $p$ are to be considered separately, this is done at the end of this answer.
Now for the exponents, assuming that the signs are consistent (say, $\delta\_i=1$ for all $i$). We know that $x^{m\_i}-x^{m\_j}$ is divisible by $\tau:=1+x+\ldots+x^{n/(p\_ip\_j)-1}$. That is, $m\_i$ and $m\_j$ are congruent modulo $n/(p\_ip\_j)$ (powers of $x$ are periodic modulo the polynomial $\tau$ with period $n/(p\_ip\_j)$). For every $i=1,\ldots,r$, denote $\alpha\_i:=\nu\_{p\_i}(n)$. Then there exists an integer $T$ congruent to $m\_j$, $j\ne i$, modulo $p\_i^{\alpha\_i}$, and congruent to $m\_i$ modulo $p\_i^{\alpha\_i-1}$. For such $T$ we have $f(x)\equiv x^{T} \pmod {1+x+\ldots+x^{n/p\_i-1}}$ for all $i=1,\ldots,r$.
So, the signs and exponents are consistent for all proper divisors of $n$. Without loss of generality, $f(\xi)=1$ for all $\xi\in \sqcup\_{1<d<n,d|n} \Theta(d)$. It remains to prove that if also $f(\xi)=\varepsilon \xi^a$ for $a\in \Theta(n)$, this is also consistent (not necessarily by trivial $\varepsilon=1$, $a=0$: it may be $n=p^\alpha$, $\varepsilon=1$, $\nu\_p(a)=\alpha-1$, or $n=4$, $f(x)=-x=x^3$, in the latter case the formula $-x$ works on the whole $T(4)$, but $x^3$ does not.)
If $n$ is even, we may replace $\varepsilon=-1$ to $x^{n/2}$ modulo $\Phi\_n(x)$, so, if $n$ is even, we may suppose that $\varepsilon=1$.
If $n$ is odd and $\varepsilon=-1$, then the polynomial $f^n \pmod {1+x+\ldots+x^{n-1}}$ (which is of course again a torsion with integer coefficients) takes the values $-1$ on $\Theta(n)$ and 1 on $T(n)\setminus \Theta(n)$.
If $\varepsilon=1$ and $n$ is not a prime power, then the only consistent variant is $a$ divisible by $n$. Assuming that this is not the case, we may a prime $p$ for which $\nu\_p(a)<\nu\_p(n)$ and taking integers $u,v$ such that $au+nv=n/p$ (they exists since $n/p$ is divisible by a GCD of $n$ and $a$) replace $f$ to $f^ux^{nv}$, this allows to assume that $a=n/p$. If $n=p^\alpha$ is a prime power, then $a$ divisible by $p^{\alpha-1}$ is consistent, and, analogously, we may suppose that $a=p^{\alpha-2}$ for odd $p$ or for $p=2$, $\alpha\geqslant 3$ (in the case of $n=4$ everything is consistent, see above).
So, we should exclude three cases for $f(x)$ restricted to $\Theta(n)$ subject to $f(x)=1$ on $T(n)\setminus \Theta(n)$:
1. $f(x)=-1$ for odd $n$;
2. $f(x)=x^{n/p}$, when $p$ is a prime divisor of $n$ and $n$ is not a prime power;
3. $f(x)=x^{p^{\alpha-2}}$ for $n=p^\alpha$, $n>4$.
We use the interpolational formula
$$
1-f(x)=\frac1n\left(\sum\_{\xi\in \Theta(n)}(1-f(\xi))\xi\cdot \frac{x^n-1}{x-\xi}-
\sum\_{\xi\in \Theta(n)}(1-f(\xi))\xi(
1+x+\ldots+x^{n-1})\right).
$$
And we use that the sum of elements of the set $\Theta(n)$ is $\mu(n)$, and that the $k$-th powers of $\Theta(n)$ uniformly cover the set $\Theta(n/{\rm GCD}(n,k))$.
Look at $1-f(0)$.
In the case 1), we get $$1-f(0)=\frac{2}n(\varphi(n)-\mu(n))$$
which is not an integer (since $n$ is not prime).
In the case 2), we get
$$1-f(0)=\frac{1}n\left(\varphi(n)+\frac{\varphi(n)}{p-1}-\mu(n)+\sum \xi^{n/p+1}\right).$$
Since ${\rm GCD}(n/p+1,n)$ is either 1 or $p$, the sum $\sum \xi^{n/p+1}$ equals either to $\mu(n)$, or to $(p-1)\mu(n/p)$. In the first case
$$
\varphi(n)+\frac{\varphi(n)}{p-1}-\mu(n)+\sum \xi^{n/p+1}=
\frac{p}{p-1}\varphi(n)=n\prod\_{q\, \text{is prime},q\ne p,q|n}(1-1/q)
$$
which clearly is not divisible by $n$ (recall that $n$ is not a prime power, so the product over $q$ is strictly smaller than 1).
In the second case, since $\mu(n)+\mu(n/p)=0$, we need $$n\prod\_{q\, \text{is prime},q\ne p,q|n}(1-1/q)+p\mu(n/p)$$
to be divisible by $n$. This is the case only if $n=2p$ (otherwise it is too small). But $p$ divides $n/p+1=3$, so $p=3$, $n=6$. In this case the coefficient of $x^4$ in $1-f(x)$ is not an integer.
In the case 3), the number $p^{\alpha-2}+1$ is coprime to $n=p^\alpha$, thus $\sum \xi(1-f(\xi))=0$. We have $\sum \xi^j=0$ for $j$ not divisible by $p^{\alpha-1}$, but if $\nu\_p(j)=\alpha-1$ then $\sum \xi^j=-p^{\alpha-1}$. Thus, between the coefficients of $1-f(x)$ we may find $\pm\frac1n\cdot p^{\alpha-1}=\pm 1/p$, a contradiction.
Finally, the cases of not consistent signs with $n=4p$ and $n=pq$. Well, they are similar to the above analysis.
If $n=pq$ with distinct odd primes $p$, $q$, and $f(x)=x^a$ on $\Theta(p)$, $f(x)=-x^b$ on $\Theta(q)$, $f(x)=x^c$ on $\Theta(pq)$, then replacing $f$ to $f^n$ we may suppose that $f$ equals 1 on $\Theta(p)$ and $\Theta(pq)$ and $f=-1$ on $\Theta(q)$. Thus
$$
1-f(x)=\frac2{pq}\left(\sum\_{\xi\in \Theta(q)} \xi\cdot \frac{x^n-1}{x-\xi}+(1+x+\ldots+x^{n-1})\right)
$$
and $1-f(0)=2/p$, a contradiction.
Finally, if $n=4p$, $f(x)=x^a$ on $T(2p)=\Theta(2)\sqcup \Theta(p)$, $f(x)=-x^b$ on $T(4)=\Theta(4)\sqcup \Theta(2)$, $f(x)=x^c$ on $\theta(4p)$ (for $\Theta(4p)$ the sign may be chosen as we wish by the already used "$-1=x^{2p}$ on $\Theta(4p)$" trick). As before, we may suppose that $a=0$ and that $c$ is divisible by $p$ (otherwise replace $f$ to $(f/x^a)^p$). Also, $b$ is odd since $1=f(-1)=(-1)^{b+1}$. Next point is that $c$ must be even. Assume that, on the contrary, $c$ is odd multiple of $p$. Choose $w\in \Theta(4p)$, then $-w\in \Theta(2p)$ and $\rho:=\frac{f(w)-f(-w)}{w-(-w)}$ must be an algebraic integer. We have $f(-w)=1$ and $f(w)=\pm i$, thus $w\rho=(-1\pm i)/2$ is an algebraic integer, but it is not. So, $c$ is divisible by $2p$. Then the polynomial $g(x):=x^{c}-f(x)$ equals 0 on $T(2p)\sqcup \Theta(4p)$, and takes values $\pm 1\pm i$ at two elements of $\Theta(4)$. Therefore (reducing as usually $g$ modulo $1+x+\ldots+x^{n-1}$) we get
$$
g(x)=\frac1{4p}\left(\sum\_{\xi\in \Theta(4)} g(\xi)\cdot \xi\cdot \frac{x^{4p}-1}{x-\xi}-\sum\_{\xi\in \Theta(4)}\xi g(\xi)(1+x+\ldots+x^{4p-1})\right).
$$
This does not have integer coefficients simply because the coefficients are too small.
| 13 | https://mathoverflow.net/users/4312 | 449940 | 181,035 |
https://mathoverflow.net/questions/449618 | 3 | **Definition 1**: A Hadamard matrix is an $n\times n$ matrix $H$ whose entries are either $1$ or $-1$ and whose rows are mutually orthogonal.
**Definition 2**: A matrix $A$ is half-skew-centrosymmetric if there exist two square matrices $B$ and $C$ of order $n$ such that
\begin{equation}
A = \begin{bmatrix}
B & R C R \\
C & -R B R
\end{bmatrix}.
\end{equation}
where $R$ is the reverse identity matrix.
One day I found that these two definitions can be considered simultaneously. For example, \begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix} is the simplest half-skew-centrosymmetric Hadamard matrix.
Here comes my questions:
>
> First, can you give some references about half-skew-centrosymmetric Hadamard matrices? Are there other names for these matrices?
>
>
>
>
> Second one is about construction methods. It is easy to figure out how to construct a half-skew-centrosymmetric Hadamard matrix of order $2^k\cdot n$ based on a Hadamard matrix of order $n$ by using a variant of Sylvester's construction.
> Can you propose more methods to construct half-skew-centrosymmetric Hadamard matrices?
>
>
>
>
> Third, can you prove the following conjecture or give a counter-example?
>
>
> **Conjecture**: A half-skew-centrosymmetric Hadamard matrix exists for $n=2$ or $n$ is a multiple of $4$.
>
>
>
| https://mathoverflow.net/users/369335 | On the half-skew-centrosymmetric Hadamard matrices | Let $H\_n$ be an $n×n$ Hadamard matrix and $R\_n$ the $n×n$ reverse identity matrix.
The matrix $X= \begin{pmatrix}
H\_n & R\_nH\_n \\
H\_n & -R\_nH\_n
\end{pmatrix}$ has entries of length $1$ and $$XX^\* = 2nI\_{2n} + ((nI\_n - R\_nH\_nH\_n^\*R\_n) \otimes R\_2)$$ which is simply $2nI\_{2n}$ so it is a Hadamard matrix. Permute the last $n$ columns with $R\_n$ and you have it in the form you give. This gives half-skew-centrosymmetric Hadamard matrices of twice the size of a Hadamard matrix. (It also works for complex Hadamard matrices or even Hadamard matrices over \*-rings)
Another construction (edit): suppose we apply the [Paley construction](https://en.m.wikipedia.org/wiki/Paley_construction) II to a finite field with $q=4k+1$ elements giving an $2n×2n$ Hadamard matrix $H= \begin{pmatrix}
H\_1 & H\_2 \\
H\_3 & H\_4
\end{pmatrix} $ after permuting the odd columns and rows to the first $n$ rows/columns, the even as last $n$ AND put the first even row/column at the last position. Define $X := X\_{i,j} = (i-j)^{0.5(q-1)}$ (in the finite field this is $\pm 1$ with zero diagonal and symmetric because $-1$ is a square). Note that $X\_{i,j+k}=X\_{j+k,i}=X\_{n-i+k,n-j}$. Then for Paley construction II, using $j$ as all-ones vector $$H\_2^T = H\_3 = \begin{pmatrix}
j & X-I \\
-1 & j^T
\end{pmatrix}$$ $$H\_1 = \begin{pmatrix}
1 & j^T \\
j & X+I
\end{pmatrix}$$ $$H\_4 = \begin{pmatrix}
-X-I & -j \\
-j^T & -1
\end{pmatrix}$$The $H\_i$ now have the properties $R\_nH\_2R\_n = H\_3$ and $-R\_nH\_1R\_n = H\_4$ due to our note above. Thus $H$ is a half-skew-centrosymmetric Hadamard matrix!
| 2 | https://mathoverflow.net/users/490128 | 449943 | 181,036 |
https://mathoverflow.net/questions/449898 | 2 | Let $e\_d$ be the $d$-th standard-basis vector in the Hilbert space $H=l\_2(\mathbb{N})$.
Let $h(n) = J\_2(n)$ be the second Jordan totient function, defined by:
$$J\_2(n) = n^2 \prod\_{p|n}(1-1/p^2)$$
Define:
$$\phi(n) = \frac{1}{n} \sum\_{d|n}\sqrt{h(d)} e\_d.$$
Then we have:
$$ \left < \phi(a),\phi(b) \right > = \frac{\gcd(a,b)^2}{ab}=:k(a,b)$$
The vectors $\phi(a\_i)$ are linearly independent for each finite set $a\_1,\cdots,a\_n$ of natural numbers, [since (page 1,2 in the notes)](https://drive.google.com/file/d/1vPz4llzvv9dup3RcJxvCCgrE-RLXwu4h/view)
$$\det(G\_n) = \prod\_{i=1}^n \frac{h(a\_i)}{a\_i^2} $$
is not zero, where $G\_n$ denotes the Gram matrix.
We want to look at $a\_1,\cdots,a\_n = 1,\cdots,n$ and get:
$$d(n):=\det(G\_n) = \prod\_{i=1}^n \frac{h(i)}{i^2} = \prod\_{k=1}^n \prod\_{p|k} (1-1/p^2) = \prod\_{p \le n} (1-1/p^2)^{\operatorname{floor}(n/p)}$$
Supposing now, that there exist only finitely many primes $p\_1,\cdots,p\_r$ we get for $d(n)$:
$$d(n) = \prod\_{i=1}^r (1-1/p\_i^2)^{\operatorname{floor}(n/p\_i)}$$
Consider now the number $N = p\_1 \cdots p\_r$ then we have:
$$\operatorname{floor}(N/p\_i) = \operatorname{floor}((N+1)/p\_i)$$
hence also:
$$d(N) = d(N+1)$$
But this [should be empirically impossible](https://sagecell.sagemath.org/?z=eJwtjkEKhTAMRPeCd8gy0aqty483EQWhrQj-tES9v62aWc6bzFjnwVpk-pUFpBN3XsIQJVgc0TSmi3NPVeX3EAS5iwQ-CETYMrT93YFcG5qoLF5ll7MrC68Ojeq1TsD3PkX4RFZPZa1brZ72BAzvDLoBhYQm7g==&lang=sage&interacts=eJyLjgUAARUAuQ==), if the function $d(n)$ can be shown to be monotonically decreasing. Notice also that the volume $\operatorname{vol}(n) = \sqrt{d(n)}$ is the volume spaned by the vectors $\phi(k), k=1,\cdots,n$. Hence maybe a geometric inequality could be applied in this setting?
**Question:
Is it possible to show that $d$ is monotonically decreasing in $n$ and thereby giving a geometric proof of the infinitude of primes?**
**Edit**: With the help of @Mark the proof is now complete! I still do not understand the downvote yet. :-(
| https://mathoverflow.net/users/165920 | A geometric proof that there are infinitely many primes? | You have claimed twice that
$$
d(n)= \prod\_{k = 1}^n \prod\_{p\mid k}(1-p^{-2})
$$
If this is true, your desired claim immediately holds, namely $d(n+1) <d(n)$.
This is because the quotient
$$
\frac{d(n+1)}{d(n)} = \frac{\prod\_{k = 1}^{n+1}\prod\_{p\mid k}(1-p^{-2})}{\prod\_{k = 1}^n\prod\_{p\mid k}(1-p^{-2})} = \prod\_{p\mid (n+1)}(1-p^{-2}).
$$
Note that for any non-trivial divisor $p\mid (n+1)$, we have that $p>1$, and therefore $0 < p^{-2} < 1$, and $0<1-p^{-2} < 1$.
It therefore follows that the product of these numbers in $(0,1)$ is in $(0,1)$, i.e.
$$\frac{d(n+1)}{d(n)} < 1\implies d(n+1) < d(n),$$
as desired.
| 4 | https://mathoverflow.net/users/101207 | 449944 | 181,037 |
https://mathoverflow.net/questions/449954 | 3 | Assume we have a Borel probability measure $\mu$ in $\mathbb{R}^n$ and a sequence of $\mu$ distributed I.I.D. random variables $x\_n$. Is there a limit formula for $supp(\mu)$, something like closure of limit points of $x\_n$ or similar, which allows kind of asymptotic recovering of $supp(\mu)$ from observed data.
| https://mathoverflow.net/users/34984 | Recovering measure support from the sequence of I.I.D random variables | By f.i. <https://encyclopediaofmath.org/wiki/Support_of_a_measure> $supp(\mu)$ is the smallest closed set $C \subset \mathbb{R}^n$ such that $\mu(C^c) = 0$, $C^c$ the complement of $C$. Let $y := (y\_n)\_{n \in \mathbb{N}}$ be any sample of $(x\_n)$ (independent realizations of $\mu$) and $A := \{y\_n \colon n \in \mathbb{N}\}$, $A\_N := \{y\_n \colon n \leq N\}$ for $N \in \mathbb{N}$. Then by Borel-Cantelli with probability $1$ $A \subset C$, $A\_N \uparrow A$ and $A$ is dense in $C$, in particular $\bar A = C$. Thus each sequence of iid random variables recovers the support (a.s.).
| 4 | https://mathoverflow.net/users/100904 | 449958 | 181,041 |
https://mathoverflow.net/questions/449948 | 0 | I am aware that in a finite dimensional vector space, any two norms are equivalent.
However, I cannot really figure out how "universal" the equivalence constants are.
To be specific, let us think of the space $L^2\Bigl([0,1],\mathbb{R} \Bigr)$ of periodic real-valued functions on $[0,1]$. Denote its inner product by $\langle , \rangle\_{L\_2}$.
Let us choose "any triplet" of "smooth" functions $f\_1, f\_2, f\_3 : [0,1] \to \mathbb{R}$ which are orthonormal with resept to $\langle , \rangle\_{L\_2}$ and denote their linear span as
\begin{equation}
V\_{f\_1,f\_2,f\_3}:= \langle f\_1, f\_2, f\_3 \rangle \subset L^2\Bigl([0,1],\mathbb{R} \Bigr).
\end{equation}
Then, $V\_{f\_1,f\_2,f\_3}$ is clearly just $3$-dimensional and therefore all norms must be equivalent on it.
That is, even the Sobolev $H^1$ norm, denoted as $\lVert \cdot \rVert\_{H\_1}$ must be equivalent to the $L^2$ norm $\lVert \cdot \rVert\_{L^2}$, which means that there are constants $c, C>0$ satisfying
\begin{equation}
c \lVert f \rVert\_{H\_1}\leq \lVert f \rVert\_{L^2} \leq C \lVert f \rVert\_{H\_1}
\end{equation}
for all $f \in V\_{f\_1,f\_2,f\_3}$.
Now, my question is that, are these constants $c$ and $C$ "universal"?
That is, if I choose another triplet of $L^2$-orthonormal smooth functions $g\_1, g\_2, g\_3$ entirely different from $f\_1, f\_2, f\_3$ above and consider $V\_{g\_1,g\_2,g\_3}$, do we still have
\begin{equation}
c \lVert g \rVert\_{H\_1}\leq \lVert g \rVert\_{L^2} \leq C \lVert g \rVert\_{H\_1}
\end{equation}
for all $g \in V\_{g\_1,g\_2,g\_3}$ and the same constants $c, C$?
OK I will clarify my question: I wonder if there exists a pair $(c,C)$ validating the above equivalence relation for "any" choice of $L^2$-orthonormal triplet of smooth functions. In this sense, the pair $(c,C)$ is "universal".
| https://mathoverflow.net/users/56524 | Norm equivalence in finite dimensions - is the equivalence "universal" if the dimension is fixed? | I am turning my comment into an answer.
If I've understood your question correctly, the constants $(c,C)$ cannot be "universal" in your sense for rather trivial set-theoretical reasons.
Suppose that the norm inequality is valid for every 3-dimensional vector subspace for the same constant choice of constants $(c,C)$, where 3 really is an arbitrary fixed positive integer. Then, since every element of the infinite-dimensional vector space is in particular an element of a 3-dimensional vector subspace, the norm inequality with $(c,C)$ is valid on the whole infinite-dimensional vector space. Hence both norms must be equivalent as norms on the whole infinite-dimensional vector space.
In other words, two inequivalent norms can't become equivalent with the same constants on all $n$-dimensional vector subspaces.
| 4 | https://mathoverflow.net/users/1849 | 449960 | 181,042 |
https://mathoverflow.net/questions/449922 | 2 | If we coin a theory in $\mathcal L\_{\omega\_1, \omega}$ that begins with constructing pure true well founded finite sets, then the set of all true well founded hereditarily finite sets, then builds up stages of $L$ on top of it using the infinitary machinery (depicted below), then restrict iteration to be secured only below the first weakly inaccessible cardinal. Then would there be finitary first order sentences that this theory cannot decide upon?
Language $\mathcal L(=,\in)\_{\omega\_1, \omega}$.
$\textbf{Extensionality: } \forall z (z \in x \leftrightarrow z \in y) \to x=y$
$\textbf{Empty: } \exists x \forall y (y \not \in x)$
$\textbf{Define: } x=\varnothing \iff \forall y (y \not \in x)$
$\textbf{Finite construction: } \bigwedge\_{n \in \omega} \forall v\_0..\forall v\_n \exists x: x=\{v\_0,..,v\_n \} $
Define the finite stages of $L$ as:
$L\_0 = \varnothing$
$L\_{n+1} = \{ x \mid x=\varnothing \lor \bigvee\_{n \in \omega} (\exists v\_0,..,v\_n \in L\_\alpha: x=\{v\_0,..,v\_n\} ) \} $
Now the first infinite stage is:
$L\_\omega= \{x \mid \bigvee\_{n \in \omega} x \in L\_n\}$
$\textbf{Finite Foundation: } \\ \forall x \in L\_\omega: \neg [ \bigwedge\_{n \in \omega} (\exists v\_0..\exists v\_n: \bigwedge\_{i \in n} (v\_{i+1} \in v\_i) \land v\_0 \in x)]$
Up till now this $L\_\omega$ is the set of all true well founded hereditarily finite sets, its one (up to isomorphism) in all models of this theory. This entails that this theory is *arithmetically complete*.
Now we proceed to build the rest of stages of $L$:
$\textbf{Bounded Separation: } \forall A \exists x: x= \{y \in A \mid \phi^A \} $; where $\phi^A$ is formula bounded by $A$, in which "$x$" doesn't occur, and $y$ is its sole free variable.
$\textbf {Constructible Power: } \forall A \exists B:\\ B=\{x \mid \bigvee x=\{y \in A \mid \Phi^A \}\} $
Where $ \Phi^A$ is all formulas bounded by $A$ having $y$ as their sole free variable.
Now we come to define successor stages:
$\textbf {Define: } L\_{\alpha+1} = \{x \mid \bigvee x= \{y \in L\_\alpha \mid \Phi^{L\_\alpha}(y) \} \} $
Where $ \Phi^{L\_\alpha}(y)$ is all formulas bounded by $L\_\alpha$ having $y$ as their sole free variable.
$\textbf{Replacement: }$ if $f$ is a function from ordinals to ordinals, then: $\forall \lambda \exists x: x=\{y \in L\_{f({\alpha})} \mid \alpha \in \lambda \}$
Now, we come to define limit stages of $L$ as:
$\textbf{Define: } (\not \exists \kappa: \lambda=\kappa+1) \to L\_\lambda= \{y \in L\_\kappa \mid \kappa \in \lambda \}$
$\textbf{Cardinals: } \forall \kappa \exists \lambda: \kappa < \lambda$
Where $<$ is for cardinal strict smaller than.
$\textbf{Size: } \not \exists \kappa: \operatorname {inaccessible}(\kappa)$
Where inaccessible means a regular limit of regular ordinals.
$\textbf {Restriction: } \forall x \exists \alpha: x \in L\_\alpha$
My point is that we've achieved $V=L$ here so all models of this theory are constructible, there are no models of this theory that prove existence of an inaccessible set. So it answers to all large cardinal properties in the negative. We cannot have inner models since $L$ is the minimal inner model. Forcing fails here. Also we have this theory arithmetically complete so we cannot have statements like Godels or Rossers or the alike, also being arithmetically complete it settles all question about consistency of theories since those are arithmetical statements.
>
> So, does this theory decide on every sentence in $\mathcal L\_{\omega, \omega}$? That is, is it finitary first order complete?
>
>
>
| https://mathoverflow.net/users/95347 | Is this theory finitary first order complete? | Let $\mathbb{K}$ be the class of well-founded models of $\mathsf{ZFC+V=L}$ + "There is no inaccessible cardinal." This is a subclass of the model class of your theory, but under mild hypotheses its associated theory $$Th(\mathbb{K}):=\bigcap\_{\mathfrak{M}\in\mathbb{K}} Th(\mathfrak{M})$$ is not complete. For example, suppose $\kappa$ is inaccessible. Then by by downward Lowenheim-Skolem + condensation, there is a countable $\alpha$ such that $L\_\alpha\models\mathsf{ZFC}$. But since $L\_\alpha\in L\_\kappa$, this means that $L\_\kappa\models$ "$\mathsf{ZFC}$ has a countable transitive model," whereupon by elementarity $L\_\alpha$ satisfies the same. This, plus well-foundedness of the ordinals, gives us lots of non-elementarily-equivalent elements of $\mathbb{K}$: for each $n\in\omega$, let $\alpha\_n$ be the unique ordinal whose corresponding level of $L$ satisfies $\mathsf{ZFC}$ "There are exactly $n$ levels of $L$ satisfying $\mathsf{ZFC}$." (And of course we can do lots more along the same lines.)
* Of course, it takes some work to see that "There are exactly $n$ levels of $L$ satisfying $\mathsf{ZFC}$" can be expressed in the language of set theory. This is however a standard result/technique in set theory which you should definitely spend time mastering if you're interested in this topic. If memory serves, Kunen's book (specifically, the section "Defining definability") gives a good summary of the topic.
Note, though, that - again under mild hypotheses - your theory has more models than just those in $\mathbb{K}$. This is because $\mathcal{L}\_{\infty,\omega}$ cannot capture well-foundedness; in particular, if there are arbitrarily large levels of $L$ satisfying $\mathsf{ZFC}$, then there is a non-well-founded model of $\mathsf{ZFC}$ satisfying every well-foundedness principle $\mathcal{L}\_{\infty,\omega}$ can express.
| 4 | https://mathoverflow.net/users/8133 | 449963 | 181,043 |
https://mathoverflow.net/questions/449961 | 1 | Are there set theories that extend some complete infinitary language $\mathcal L\_{\kappa, \lambda}$, prove all axioms of $\sf ZFC$, and are finitary $\textbf{FOL}$ complete? That is, every sentence in $\mathcal L(=,\in)\_{\omega,\omega}$ is decidable.
| https://mathoverflow.net/users/95347 | Are there strong set theories written in infinitary language, that are finitary FOL complete? | To avoid triviality (e.g. "The true $\mathcal{L}\_{\kappa,\lambda}$-theory of $V$") let's look specifically for theories which $(1)$ consist of adding to $\mathsf{ZFC}$ a single infinitary sentence and $(2)$ have that single sentence being "reasonably simple." As long as we stick to $\mathcal{L}\_{\omega\_1,\omega\_1}$ this last condition has a natural interpretation, namely that we want a single **computable** sentence (once our syntax gets too big it's unclear what this should mean).
Even with this restriction, the answer is **yes**. In $\mathcal{L}\_{\omega\_1,\omega\_1}$ we can express true well-foundedness, so the class of well-founded models of "$\mathsf{ZFC}$ + '$\mathsf{ZFC}$ has no well-founded model'" is $\mathcal{L}\_{\omega\_1,\omega\_1}$-elementary. But this class has a single element (up to isomorphism), so a fortiori its theory is $\mathsf{FOL}$-complete.
* Similarly, $\mathsf{ZFC}$ + "The ordertype of the levels of $L$ satisfying $\mathsf{ZFC}$ is $\omega^2+17$" + $\mathcal{L}\_{\omega\_1,\omega\_1}$-wellfoundedness is categorical so a fortiori $\mathsf{FOL}$-complete. You have to go quite a ways up the $L$-hierarchy before you get a level which is not $\mathcal{L}\_{\omega\_1,\omega\_1}$-"identifiable."
---
What if we try to work in the much weaker (and better behaved) logic $\mathcal{L}\_{\omega\_1,\omega}$? Well, here things are trickier. To start off, here's an easy affirmative answer to a weak version of the question. Every countable structure is pinned down up to isomorphism by a single $\mathcal{L}\_{\omega\_1,\omega}$-sentence (this is **[Scott's isomorphism theorem](https://people.math.wisc.edu/%7Eslempp/conf/dag17/Knight.pdf)**), so letting $\sigma$ be the sentence so characterizing the Cohen-Shepherdson minimal wellfounded model of $\mathsf{ZFC}$ we get that $\mathsf{ZFC+\sigma}$ is categorical. However, this $\sigma$ is in no way computable. In fact, no model of $\mathsf{ZFC}$ can have a computable Scott sentence, so if we demand that our single sentence be computable then the "categoricity trick" won't work. Fortunately, you only asked for $\mathsf{FOL}$-completeness; if memory serves there's a trick that gives an affirmative answer here too, but I'll have to recall the details.
| 3 | https://mathoverflow.net/users/8133 | 449965 | 181,044 |
https://mathoverflow.net/questions/449934 | 2 | Let $p:E\to B$ be a locally trivial fibration of connected, non-compact smooth manifolds. Let $U\subset E$ be a connected open subset and $p|\_U:U\to p(U)$ has connected diffeomorphic fibers.
Can we conclude that $p|\_U$ is again a locally trivial fibration?
This question is related to my other question on fibering a certain hyperplane arrangement complement, I posted before. I find this general version interesting in its own right.
| https://mathoverflow.net/users/126243 | Restriction of a fibration to an open subset with connected diffeomorphic fibers | Here is a counterexample inspired by algebraic geometry:
**Example.** Let $E \to B$ be the first projection $\mathbf C^2 \to \mathbf C$ (or $\mathbf R^4 \to \mathbf R^2$, if you like), and let $U \subseteq \mathbf C^2$ be the complement of the divisor $\{(x,y) \in \mathbf C^2\ |\ xy = 1\}$ and the origin $\{(0,0)\}$. This is Zariski connected (as it is dense in an irreducible space) and hence also connected in the classical topology. Each fibre of $U \to \mathbf C$ is isomorphic to $\mathbf C^\times$: above nonzero points $x \in \mathbf C$ we removed the point $(x,1/x)$, and above $0$ we removed the point $(0,0)$.
But the restriction $f \colon U \to \mathbf C$ of $p$ to $U$ is not a fibre bundle. Indeed, for any open ball $V \subseteq \mathbf C$ around $0$, the fibre $f^{-1}(V)$ satisfies $H^3(f^{-1}(V),\mathbf Z) \neq 0$, so $f^{-1}(V)$ is never homeomorphic to $V \times \mathbf C^\times$.
| 4 | https://mathoverflow.net/users/82179 | 449966 | 181,045 |
https://mathoverflow.net/questions/449949 | 4 | I’ve been self-studying axiomatic systems for classical logic for a while. The standard Hilbert/Mendelssohn/Lukasiewicz axiomatizations were a bit tough for me to get used to without using the Deduction Theorem, but now I’m confident with those systems.
I recently learned about Meredith’s axiom for classical logic:
$((((A \to B) \to (\neg C \to \neg D)) \to C) \to E) \to (E \to A) \to D \to A$.
Could anyone give me tips for constructing proofs in this system? It seems obvious that one just needs to find the correct substitution of an instance of the axiom into the antecedent of another instance of the axiom, but I can’t seem to get any further than that strategy. Any help would be appreciated.
| https://mathoverflow.net/users/498245 | How to use Meredith’s axiom for classical logic? | See <https://us.metamath.org/mpeuni/meredith.html> and the links there for the proofs you want.
| 4 | https://mathoverflow.net/users/3684 | 449971 | 181,047 |
https://mathoverflow.net/questions/449663 | 5 | A (very?) naïve question, but I didn't get an answer on math.se: so here goes ….
In his original [ETCS](https://doi.org/10.1073/pnas.52.6.1506) paper, Lawvere states a categorial version of choice for sets in this form: "If the domain of $f$ has elements, then there exists $g$ such that $fgf = f$". So let's say, generalizing
(1) A category has Choice-1 iff for any $f\colon X \to Y$ where $X \not\cong 0$ [with $0$ initial], there exists a $g \colon Y \to X$ such that $f \circ g \circ f = f$.
Another more familiar version of choice for sets is that any surjection has a right inverse (or left inverse, depending on your preferred handedness convention!). Generalizing, let's say
(2) A category has Choice-2 iff for any epic $f\colon X \to Y$ there exists a $g \colon Y \to X$ such that $f \circ g = 1\_Y$.
But (forgive a senior moment!) I'm embarrassingly unclear about the conditions — the weakest/most natural/nicest conditions? — under which a category which has Choice-2 has Lawvere's Choice-1. Is there a standard story about this which I've missed?
| https://mathoverflow.net/users/14111 | Versions of Choice in categories | If we take Choice-1 in Lawvere's original form with "has (global) elements" in place of "is not initial", then Choice-2 follows from Choice-1 in any [Boolean category](https://ncatlab.org/nlab/show/Boolean+category). Factor $f:X\to Y$ as an epi $h : X \twoheadrightarrow Z$ followed by a mono $m:Z\hookrightarrow Y$. Then $h$ is a split epi by Choice-2, while $m$ is complemented by Booleanness, and therefore a split mono since $Z$ has a global element. So we have $s:Z\to X$ with $h s = 1\_Z$, and $r:Y\to Z$ with $r m = 1\_Z$, and defining $g = s r$ we have $f g f = (m h) (s r) (m h) = m (h s) (r m) h = m h = f$.
In a topos, Booleanness also follows from Choice-1 by the [Diaconescu-Goodman-Myhill theorem](https://ncatlab.org/nlab/show/Diaconescu-Goodman-Myhill+theorem). So Choice-2 also follows from Lawvere's original Choice-1 in any topos.
| 4 | https://mathoverflow.net/users/49 | 449974 | 181,049 |
https://mathoverflow.net/questions/449978 | 10 | The standard algorithm to compute the factorial function $N!$ via repeated multiplications has complexity $\mathcal{O}(N)$, in the model in which each operation costs 1, no matter how many digits the operands have. Intuitively, it would seem that this cost could not be improved.
However, how can we be sure? What if there is some clever equivalent algorithm with complexity $\mathcal{O}(\log(N))$, or $\mathcal{O}(N^{0.998})$? If so, how could we know it is possible, or find an example? If not, why not?
| https://mathoverflow.net/users/161776 | Can $N!$ be computed in less than $\mathcal{O}(N)$ operations? | The thing that you are asking is the required number of integer operations for computing $N!$ starting only from $1$ and $N$, which is usually referred to as *the straight-line complexity* of $N!$ denoted by $\tau(N!)$. (or [BSS model](https://en.wikipedia.org/wiki/Blum%E2%80%93Shub%E2%80%93Smale_machine), though I am not sure they are indeed equivalent; they are at least highly related.)
And your main question can be rephrased in terms of the standard big-O notation as follows:
$$
\tau(N!) = \Omega(N), \text{ or } \tau(N!) =O(N^{1-\epsilon}) \text{ for some constant }\epsilon?
$$
The answer is the latter. Indeed, there is a known algorithm showing $\tau(N!)=O(\sqrt{N}\log^2 N)$ and if we just want to find a *multiple* of $N!$ then there is a subexponential algorithm $\log N$, though I never checked about the algorithms.
The most relevant references I know are as follows:
1. "On the Ultimate Complexity of Factorials" by Qi Cheng, which shows the subexponential algorithm for computing a multiple of factorial,
2. "ON THE INTRACTABILITY OF HILBERT’S NULLSTELLENSATZ AND AN ALGEBRAIC VERSION OF “NP != P?”" by Michael Shub and Steve Smale, which emphasizes this problem.
In fact, according to the second reference, the complexity of factorial is indeed an important problem known to be connected to the *algebraic* version of the P=NP problem. Namely, if any multiplication of $N!$ cannot be computed in time $O({\rm poly}\log(N))$, then the algebraic version of $P\neq NP$ is true.
| 11 | https://mathoverflow.net/users/171820 | 449983 | 181,050 |
https://mathoverflow.net/questions/449964 | 6 | Let $\varphi(n):=(-1)^{n+1}(n+1)2^{2n}$.
I am able to prove the following identity (${\color{red}{\mathbf{LHS}}}$=infinite series, ${\color{blue}{\mathbf{RHS}}}$=finite sum)
\begin{align\*}
{\color{red}{\frac{1}{\varphi(n)}\sum\_{k\geq1}\frac{(-1)^k}{\binom{3n+k+2}{2n+2}}}}
&={\color{blue}{\sum\_{j=1}^n\frac{(-1)^j(28j^2+24j+4)}{\binom{3j}j j (3j+2)(3j+1)\,2^{2j}}+3-4\ln 2}}. \tag1
\end{align\*}
Replacing $n=0$ in equation (1) leads to $\displaystyle\sum\_{j\geq1}\frac{(-1)^j}{\binom{j+2}2}=4\ln 2-3$.
Letting $n\rightarrow\infty$ in (1) unearths the less trivial result
$$\sum\_{j=1}^{\infty}\frac{(-1)^j(28j^2+24j+4)}{\binom{3j}j j (3j+2)(3j+1)\,2^{2j}}=4\ln 2-3.$$
>
> **QUESTION.** How can one exercise an analytic continuation of the ${\color{blue}{\mathbf{RHS}}}$ in (1), ${\color{purple}{\text{in the variable $n$}}}$, so that we would be able to compute at $n=-\frac23$ and derive (from it) a value for the infinite series on the ${\color{red}{\mathbf{LHS}}}$, i.e.
> $$\frac1{\varphi(-\frac23)}\sum\_{k\geq1}(-1)^k\binom{k}{\frac23}^{-1}\,?$$
>
>
>
| https://mathoverflow.net/users/66131 | A need for analytic continuation of a finite sum function | As for the sum $$\sum\_{k\geq1}(-1)^k\binom{k}{\frac23}^{-1}$$
one can evaluate it by means of the Beta function integral, like in this recent [computation](https://mathoverflow.net/questions/449776/closed-form-of-an-infinite-series/449810#449810).
$$\sum\_{k\geq1}(-1)^k\binom{k}{\frac23}^{-1}=\frac23\sum\_{k\geq1}(-1)^k \frac{\Gamma\big(k+\frac13\Big)\Gamma\big(\frac23\big)}{\Gamma(k+1)}=\frac23\sum\_{k\geq1}(-1)^kB\Big(k+\frac13,\frac23\Big)=$$
$$=\frac23\sum\_{k\geq1} \int\_0^1(-1)^k {x^{k-\frac23}}{(1-x)^{-\frac13}}dx=-\frac23 \int\_0^1\frac{1}{1+x} \Big(\frac{x }{1-x}\Big)^{1/3}dx=$$
The latter integral is easily rationalised by changing variable; we get $$=-\int\_0^\infty\frac{2t^3}{(t^3+1)(2t^3+1)}dt=\int\_0^\infty\frac{2dt}{2t^3+1 } -\int\_0^\infty\frac{2dt}{ t^3+1}=$$$$=(\sqrt[3]4-2)\int\_0^\infty\frac{dt}{ t^3+1}$$ whence $$\sum\_{k\geq1}(-1)^k\binom{k}{\frac23}^{-1}=
\frac{2\pi}{3\sqrt3} (\sqrt[3]4-2)$$
which evaluates to $-0.4989..$.
| 5 | https://mathoverflow.net/users/6101 | 449986 | 181,052 |
https://mathoverflow.net/questions/449846 | 2 | Consider the following random walk $(y\_t)\_{t \in \mathbb Z\_+}$:
$$y\_t = y\_{t-1} + u\_t,\quad (u\_t)\_{t \in \mathbb Z\_+} \overset{iid}{\sim} N(0,1), \quad (t \in \mathbb Z\_+)$$
where $y\_0, u\_1, u\_2,...$ are independent.
We know that the process is not stationary and non-ergodic. On the other hand, if $|a|< 1$, we have:
$$y\_t = a y\_{t-1} + u\_t,\quad (u\_t)\_{t \in \mathbb Z\_+} \overset{iid}{\sim} N(0,1), \quad (t \in \mathbb Z\_+)$$
is stationary and ergodic. Consider then the following sequence of stochastic processes: let $(a\_n)\_{n \in \mathbb N}$ be a sequence of real numbers such that, $|a\_n|< 1$ for all $n \in \mathbb N$ and $a\_n \to 1$, as $n \to \infty$. So, for every $n \in \mathbb N$, define the process $(y\_t^n)\_{t \in \mathbb Z\_+}$ as:
$$y^n\_t = a\_n y^n\_{t-1} + u^n\_t,\quad (u^n\_t)\_{t \in \mathbb Z\_+} \overset{iid}{\sim} N(0,1), \quad (t \in \mathbb Z\_+)$$
So, I would like to know which metric we can use to determine whether the sequence of stochastic processes $(y^n\_t)\_{t \in \mathbb{Z}\_+}$ converges to the random walk $(y\_{t})\_{t \in \mathbb{Z}\_+}$ defined above.
| https://mathoverflow.net/users/479236 | A question about convergence of stochastic processes converging to a random walk | Assume, naturally, that for each $n$ we have $y\_0^n\to y\_0$ (as $n\to\infty$) in distribution and $y\_0^n$ is independent of $(u^n\_t)$.
Then for each $T=0,1,\dots$ we have $Y^n\_T\to Y\_T$ in distribution, where $Y^n\_T:=(y^n\_0,\dots,y^n\_T)$ and $Y^n\_T:=(y\_0,\dots,y\_T)$. This follows because (say) for all $t=0,1,\dots$
$$y^n\_t=a\_n^t y^n\_0+\sum\_{k=1}^t a\_n^{t-1-k}u\_k$$
(which latter can be proved by induction on $t$), so that for all $T=0,1,\dots$
$$Y^n\_T=(y^n\_0,\dots,y^n\_T)\to\Big(y\_0,y\_0+u\_1,\dots,y\_0+\sum\_{k=1}^T u\_k\Big)
=(y\_0,\dots,y\_T)=Y\_T$$
in distribution.
The latter convergence is equivalent to the convergence of $d(P\_{Y^n},P\_Y)$ to $0$, where
$$d(P\_{Y^n},P\_Y):=\sum\_{T=0}^\infty c\_T\,
\frac{d\_{LP}(P\_{Y^n\_T},P\_{Y\_T})}{1+d\_{LP}(P\_{Y^n\_T},P\_{Y\_T})},$$
where (i) $(c\_T)\_{T=0}^\infty$ is any summable sequence of positive numbers and
(ii) $P\_{Y^n\_T}$ and $P\_{Y\_T}$ are, respectively, the distributions of $Y^n\_T$ and $Y\_T$ in $\mathbb R^{T+1}$ and $d\_{LP}$ is the [Lévy–Prokhorov distance](https://en.wikipedia.org/wiki/L%C3%A9vy%E2%80%93Prokhorov_metric#Properties) between such distributions.
Any metrics on the sets of all probability distributions over $\mathbb R^T$ can be used here in place of $d\_{LP}$. If $y\_0$ and the $y\_0^n$'s are Gaussian, an especially convenient metric seems to be the Wasserstein $W\_2$ metric, for which there is a rather simple [explicit expression](https://projecteuclid.org/journals/michigan-mathematical-journal/volume-31/issue-2/A-class-of-Wasserstein-metrics-for-probability-distributions/10.1307/mmj/1029003026.full) in the Gaussian case.
| 1 | https://mathoverflow.net/users/36721 | 449995 | 181,055 |
https://mathoverflow.net/questions/449999 | 0 | In the Milnor and Moore paper, "[On the structure of Hopf algebras](https://doi.org/10.2307/1970615)" proposition 1.7 said the following:
>
> 1. $A$ a connected $K$-algebra.
> 2. $N$ a left $A$ module that is connected as a $K$-graded module i.e. there is an isomorphism $\eta\_N:K\to N\_0$ which results in an
> augmentation $\varepsilon\_N:N\to K$.
> 3. $C=K\otimes\_A N$, here $K$ is considered as $A$-rightmodule by the action $K\otimes A \xrightarrow{\cong} A \xrightarrow{\varepsilon\_A} K$.
> 4. $\Delta :N \to N\otimes C$ a morphism of left $A$ modules.
> 5. $\pi:N\to C$ the canonical epimorphism. Let $f:C\to N$ such that $\pi f = \operatorname{id}\_C$. In this paper they used $C$ the object letter to indicate
> the identity morphism.
> 6. $(\varepsilon\_N\otimes C)\circ \Delta =\pi$.
> 7. $(N\otimes \varepsilon\_C)\circ \Delta = N$.
> 8. $A\otimes C \xrightarrow{i\otimes C} N\otimes C $ is a monomorphism.
>
>
> If $\tilde{f}$ is the composition $$A\otimes C
> \xrightarrow{A\otimes f} A \otimes N \xrightarrow{\varphi\_N} N$$ then $\tilde f$ is an isomorphism.
>
>
>
The map $i$ is defined as the composition
$$A\xrightarrow{\cong} A\otimes K \xrightarrow{A\otimes \eta\_N} A\otimes N \xrightarrow{\varphi\_N} N$$
and $\varphi\_N$ is the left action of $A$ on $N$.
**My first question:** in the process to show that $\tilde{f}$ is a
monomorphism, they showed $\Delta \tilde{f}$ is a monomorphism.
They define a filtration on $$F\_p(A\otimes C) = \sum\_{q\leq p} A\otimes C\_q \text{ and } F\_p(N\otimes C) = \sum\_{q\leq p} N\otimes C\_q$$
Then they said: "let $E^0 (A\otimes C)$ be the associated bigraded module."
Is this the grading associated with the filtration i.e. $\operatorname{gr}\_p(A\otimes C) = F\_p/F\_{p-1}$? Or it is just the grading on $A$ and $C$ as $K$-modules? I guess it is the second since they said $E^0\_{p,q} (A\otimes C) = A\_p \otimes C\_q$.
After that they said we identify $E^0(A\otimes C)$ with $A\otimes C$. Then why did they introduce the $E^0$?
**Second question:** How come the two following morphisms are the same:
$$A\otimes C \xrightarrow{A\otimes f} A\otimes N \xrightarrow{\varphi\_N} N\to\_{\Delta} N\otimes C$$
and $i\otimes C$ which is
$$A\otimes C \xrightarrow{\cong\otimes C } A\otimes K\otimes C \xrightarrow{A\otimes \eta\_N\otimes C } A\otimes N \otimes C \xrightarrow{\varphi\_N} N\otimes C. $$
| https://mathoverflow.net/users/77914 | Milnor and Moore paper "On the structure of Hopf algebra" proposition 1.7 , filtration and grading | Although $E^0(A\otimes C)$ can be identified with $A\otimes C$ as objects, this is not compatible with morphisms. If $g\colon A\otimes C\to A\otimes C$ is a map of graded groups, then the map $g\colon A\_i\otimes C\_j\to(A\otimes C)\_{i+j}$ can be expressed as a sum of morphisms $g\_k\colon A\_i\otimes C\_j\to A\_{i+k}\otimes C\_{i-k}$. If $g$ preserves the filtration then $g\_k=0$ for all $k<0$, and $E^0(g)=g\_0$, whereas $g=\sum\_{k\geq 0}g\_k$.
In the case of interest, we have $g=\Delta\circ\widetilde{f}$. For $a\in A\_i$ and $c\in C\_j$ we have $g(a\otimes c)=\Delta(a\,f(c))$, but $\Delta$ is a map of $A$-modules by condition 4, so $g(a)=a\,\Delta(f(c))$. Here $\Delta(f(c))$ can be written as $\sum\_{k=0}^jx\_k$ with $x\_k\in N\_k\otimes C\_{j-k}$ so $g\_k(a\otimes c)=ax\_k$ and $E^0(g)(a\otimes c)=g\_0(a\otimes c)=ax\_0$. On the other hand, $\epsilon$ is zero on $N\_k$ for $k>0$ and $N\_0=K$ so $x\_0$ is essentially the same as $(\epsilon\otimes 1)(\Delta(f(c)))$, which is $\pi(f(c))=c$ by condition 6. Here we have implicitly identified $C$ with $K\otimes C\leq A\otimes C$; if we unwrap that we get $x\_0=1\otimes c$ and $ax\_0=a\otimes c$ in $A\otimes C$. This means that $E^0(\Delta\circ\widetilde{f})$ is the identity.
| 2 | https://mathoverflow.net/users/10366 | 450008 | 181,060 |
https://mathoverflow.net/questions/450010 | 0 | Let $A$ and $B$ be two compact convex sets (which may be assumed to be polytopes) in $\mathbb R^n$ such that $A\cap B\ne\emptyset$. Is it then always true that either $A\cap\text{ext}B\ne\emptyset$ or $B\cap\text{ext}A\ne\emptyset$, where $\text{ext}$ denotes the set of all extreme points of a set?
| https://mathoverflow.net/users/36721 | On the extreme points of two convex sets | No: consider the line segments $\{0\}\times[-1,1]$ and $[-1,1]\times\{0\}$ in $\mathbb{R}^2$.
| 3 | https://mathoverflow.net/users/2363 | 450011 | 181,062 |
https://mathoverflow.net/questions/449894 | 3 | Let $A\_{ijl}(t,x) : [0,\infty) \times \mathbb{R}^n \to [\mathbb{R}^n]^3$ be a smooth tensor field. That is, $i,j,l \in \{1,2,3, \cdots, n\}$
Further assume that $A\_{ijl}(t,x)=A\_{jil}(t,x)$ for all $(t,x) \in [0,\infty) \times \mathbb{R}^n$.
Also, let $0=\lambda\_1< \lambda\_2 < \cdots < \lambda\_n$ be some fixed numbers and finally suppose that $A\_{ijl}(t,x)$ satisfies the following ODE:
\begin{equation}
\partial\_t \sum\_{l=1}^n A\_{ijl}(t,x)= -\sum\_{l=1}^n \lambda^2\_l A\_{ijl}(t,x) \text{ with } \sum\_{l=1}^n A\_{ijl}(0,x)=\delta\_{ij}
\end{equation}
Then, at least when $t=0$, we have
\begin{equation}
\partial\_t \sum\_{l=1}^n A\_{ssl}(t,x)\mid\_{t=0}=-\lambda^2\_s
\end{equation}
for all $s=1,2, \cdots, n$ while
\begin{equation}
\partial\_t \sum\_{l=1}^n A\_{ijl}(t,x)\mid\_{t=0}=0
\end{equation}
for $i \neq j$.
From such information, is it possible to conclude, together with the smoothness assumption, that
\begin{equation}
\sum\_{l=1}^n A\_{ijl}(t,x)=e^{-\lambda\_i^2 t} \text{ if } i=j \text{ and } 0 \text{ otherwise}
\end{equation}
for all $t \geq 0$?
It seems quite plausible for me, but I cannot really justify my guess for extrapolation to $t>0$. Could anyone please help me?
Edit : I add one more condition that $A\_{ijl}(0,x)=\delta\_{il}\delta\_{jl}$ so that the above explanation makes sense.
| https://mathoverflow.net/users/56524 | An ODE for tensor - possibility of the equation together with the initial condition at $t=0$ deciding the solution for all $t>0$ | These equations are not sufficient to determine $\sum\_l A\_{ijl}$ for $n>1$.
Here is a counterexample of a solution of the problem in the OP which contradicts the conjectured solution:
set $n=2$, $\lambda\_1=0$, $\lambda\_2=1$; all elements of $A\_{ijl}$ are identically zero, except
$$A\_{111}=1-t-t^2/2,\;\;A\_{112}=t,$$
$$A\_{221}=t,\;\;A\_{222}=-1+2e^{-t}.$$
---
More generally, you can take any $F\_{ij}(t)=\sum\_l A\_{ijl}(t)$, and define
$$A\_{ijl}=-\left(\sum\_{k=2}^n\lambda\_k^{-2}\right)^{-1}F'\_{ij},\;\;l\geq 2,$$
$$A\_{ij1}=F\_{ij}+(n-1)\left(\sum\_{k=2}^n\lambda\_k^{-2}\right)^{-1}F'\_{ij}.$$
---
| 1 | https://mathoverflow.net/users/11260 | 450020 | 181,065 |
https://mathoverflow.net/questions/449993 | 5 | Question:
Do all connected Lie groups have dense torsion-free subgroups?
Context :
Let $ R\_\alpha \in SO\_2(\mathbb{R}) $ be a rotation by $ \alpha/2\pi $. If $ \alpha $ is irrational, then $ R\_\alpha $ generates a dense torsion free subgroup of $ SO\_2(\mathbb{R}) $.
Let $ R\_{\alpha,z} \in SO\_3(\mathbb{R}) $ be a rotation by $ \alpha/2 \pi $ about the $ z $ axis. Let $ R\_{\beta,x} \in SO\_3(\mathbb{R}) $ be a rotation by $ \beta/2\pi $ about the $ x $ axis. If $ \alpha, \beta $ are algebraically independent then the two rotations $ R\_{\alpha,z}$ and $ R\_{\beta,x} $ generate a torsion-free dense subgroup of $ SO\_3(\mathbb{R}) $.
It is my guess more generally that: Let $ R\_{\alpha\_i} \in SO\_n(\mathbb{R}) $ for $ i=1, \dots, n-1 $ be rotations where each $ R\_{\alpha\_i} $ is a block diagonal matrix consisting of all $ 1 $s on the diagonal except a $ 2 \times 2 $ rotation matrix by an angle of $ \alpha\_i/2\pi $ between the standard basis vectors $ e\_i, e\_{i+1} $. If $ \alpha\_1, \dots, \alpha\_{n-1} $ are algebraically independent then $ \langle R\_{\alpha\_1}, \dots,R\_{\alpha\_{n-1}} \rangle $ is a dense torsion-free subgroup of $ SO\_n(\mathbb{R}) $.
It is worth noting that every simply connected solvable Lie group is contractible and thus torsion free. So all simply connected solvable Lie groups trivially have a dense torsion free subgroup. $ \widetilde{SL}\_2(\mathbb{R}) $ is a contractible, and thus torsion-free, Lie group. So $ \widetilde{SL}\_2(\mathbb{R}) $ also trivially has a dense torsion-free subgroup.
---
cross-posted from MSE <https://math.stackexchange.com/questions/4721843/does-every-connected-lie-group-have-a-dense-torsion-free-subgroup>
| https://mathoverflow.net/users/387190 | Does every connected Lie group have a dense torsion-free subgroup? | Let $G$ be a connected Lie group. We know that $G$ has a real analytic structure for which the law is analytic, and we fix it. Also fix a left Haar measure and a smooth positive function of integral 1, thus defining a fully-supported probability $\mu$ on $G$.
Let $F\_n$ be the free group on $n$ given generators, and let $R\_n(G)$ be the set of words $w\in F\_n$ that vanish on $G$ (i.e., $w\in F\_n$ such that $w(g\_1,\dots,g\_n)=1$ for all $g\_1,\dots,g\_n\in G$). Then $R\_n(G)$ is a fully characteristic subgroup of $F\_n$ (i.e., stable under all endomorphisms), and in particular is a normal subgroup (this is true for an arbitrary group $G$).
For $w\in F\_n$, write $M\_w=\{(g\_1,\dots,g\_n)\in G^n :w(g\_1,\dots,g\_n)=1\}$. Then $M\_w$ is closed in $G^n$, and since the law is real-analytic and $G$ is connected, for $w\notin R\_n(G)$, $M\_w$ has $\mu$-measure zero. Hence the complement $J\_{n,G}$ of $\bigcup\_{w\notin R\_n(G)}M\_w$ in $G^n$ is (G$\_\delta$-)dense in $G^n$. For $(g\_1,\dots,g\_n)\in J\_G$, the group generated by $\{g\_1,\dots,g\_n\}$ is isomorphic to $F\_n/R\_n(G)$.
[Note: the latter fact is false in the group $\mathrm{O}(2)$, which is not connected, with the word $w(g)=g^2$.]
Next let us check that $F\_n/R\_n(G)$ is torsion-free. Suppose that $w\in F\_n$ and $w^d\in R\_n(G)$, $d\ge 1$. The image of the word map $w:G^n\to G$ is connected and contained in the set $T\_d(G)=\{g\in G:g^d=1\}$ of $d$-torsion elements in $G$, and contains $1$ (image of $(1,1,\dots,1)$). But in any Lie group, $1$ is isolated in $T\_d(G)$. Hence we deduce that $w\in R\_n(G)$. This proves torsion-freeness.
It follows that a $\mu$-generic sequence $(g\_n)$ generates a torsion-free subgroup. But also, clearly, a $\mu$-generic sequence is dense. Hence a $\mu$-generic sequence generates a dense torsion-free subgroup.
[I think that for $d$ large enough (maybe $d=\dim(G)+1$ always works), a $\mu$-generic $d$-tuple generates a dense subgroup. But this would require more work; I don't immediately see a reference working.]
| 12 | https://mathoverflow.net/users/14094 | 450022 | 181,067 |
https://mathoverflow.net/questions/449991 | 3 | Let $E/\Bbb{Q}$ be an elliptic curve. Let $D$ be a square free negative integer.
It is conjectured that 50％ of twist of elliptic curve $E\_D$ has rank $0$ and $50％$ has rank $1$.
But is some particular case known or conjectured?
I'm particularly interested in the case $E:y^2=x^3+17x$.
Are there infinitely many twists $E\_D$( $D≡5\bmod8$) which has $\operatorname{rank}(E\_D/K)\ge 1$ ?
Reference is also appreciated. Thank you for your help.
P.S.
Thanks to Nulhomologous, by parity conjecture, what I should ask is 'Are there infinitely many twists $E\_D(D≡5\bmod8$) which has $\operatorname{rank}(E\_D/K)=2$ ?'
| https://mathoverflow.net/users/144623 | Infinitely many elliptic curve with twist rank more than $1$ in specific case | Let me turn my comment into an answer. There are indeed infinitely many such twists with a non-torsion point. By Nagell–Lutz, it suffices to produce infinitely many different squarefree integers $D \equiv 5 \pmod 8$ for which there exist $x,y \in \mathbf Q$ such that $y^2 = x^3+17D^2x$ and $x$ and $y$ are not both integers. Writing $X = Dx$ and $Y = D^2y$, we get the equivalent equation $DY^2 = X^3+17X$ (but beware that the Nagell–Lutz criterion does not apply in these coordinates).
One way to produce values of $D$ is as follows: there are infinitely many primes $p > 5$ such that $-17$ is a square modulo $p$. For such a prime $p$, pick $m \in \mathbf N$ such that $v\_p(m^2+17) = 1$ and $m \equiv 5 \pmod 8$. Pick $n \in \mathbf N$ with $4n \equiv 1 \pmod p$ and $n \equiv 1 \pmod 8$. Set $X = \tfrac{m}{4n}$, and write $X^3+17X$ as $DY^2$ with $D \in \mathbf Z$ squarefree and $Y \in \mathbf Q$. Finally, set $x = \tfrac{X}{D}$ and $y = \tfrac{Y}{D^2}$.
We claim that $(x,y)$ is a non-torsion point on $E\_D$, that $p \mid D$, and that $D \equiv 5 \pmod 8$. Since we can carry out this process for infinitely many primes $p$, we conclude that the numbers $D$ obtained this way also form an infinite set. It is clear that $x$ is not an integer, so $(x,y)$ cannot be a torsion point.
To see that $p \mid D$, note that $v\_p(DY^2) = v\_p(X^3+17X) = v\_p(X^2+17) = 1$, so $v\_p(D) = 1$ by definition of $D$. To see that $D \equiv 5 \pmod 8$, note that
$$DY^2 = X^3+17X = X(X^2+17) = \tfrac{m}{4n}\cdot \tfrac{m^2+272n^2}{16n^2}.$$
Since $m$ and $n$ are odd, we see that $v\_2(Y^2) = -6$ and $v\_2(D) = 0$. Factoring out all even denominators, the above equation reads
$$D \cdot (2^3Y)^2 = \tfrac{m}{n} \cdot \tfrac{m^2+272n^2}{n^2}.$$
Since $2^3Y$ is a unit in $\mathbf Z\_{(2)}$, its square is 1 modulo 8. Since $m$ and $n$ are odd, we get $\tfrac{m^2+272n^2}{n^2} \equiv 1 \pmod 8$, and we have $\tfrac{m}{n} \equiv 5 \pmod 8$ by the choice of $m$ and $n$. Putting everything together gives $D \equiv 5 \pmod 8$. $\square$
| 2 | https://mathoverflow.net/users/82179 | 450029 | 181,069 |
https://mathoverflow.net/questions/450035 | 2 | Let $K$ be a number field, $\overline{K}$ an algebraic closure, and $X$ be a positive dimensional finite type $K$-scheme.
Could there exist a proper subfield $L\subset\overline{K}$ such that the natural inclusion
$$X(L)\hookrightarrow X(\overline{K})$$
is surjective?
If so, what sorts of conditions can we put on $X$ to ensure that this can't happen? I'm also curious about the answer for more general fields $K$.
| https://mathoverflow.net/users/88840 | Varieties whose residue fields do not generate the algebraic closure of the ground field | No this never happens: Without loss of generality, $X$ is affine. The case $X=\mathbb{A}^n\_K$ is obvious, the general case then follows from Noether normalization and going up.
| 5 | https://mathoverflow.net/users/50351 | 450039 | 181,071 |
https://mathoverflow.net/questions/450044 | 0 | Let $M$ and $N$ be Riemannian manifolds such that $\pi:M\to N$ is a surjective Riemannian submersion, i.e. for each $x\in M$,
$$\langle \pi\_{\*x}(v),\pi\_{\*x}(w) \rangle\_{\pi(x)} = \langle p(v), p(w) \rangle\_x$$
where $p: T\_xM = \text{Ker}(\pi\_{\*x}) \oplus \text{Ker}(\pi\_{\*x})^\perp \to \text{Ker}(\pi\_{\*x})^\perp$ is the orthogonal projection. For instance $M = G$ is a Lie group with a right-invariant Riemannian metric, $N= G/H$ is a homogeneous space with the metric induced by $G$, and $\pi: G\to G/H$ is the projection.
By the implicit function theorem we know that for $x\in M$, there is a neighbourhood $U$ of $\pi(x)$ and a smooth function $f:U \to M$ such that $f(\pi(x)) = x$ and $\pi\circ f = id\_U$. My question is, can $U$ and $f$ be chosen such that $f$ is an isometric embedding? If yes, how so?
**Some ideas:** For $x\in M$, $$\pi\_{\*x}|\_{\text{Ker}(\pi\_{\*x})^\perp}: \text{Ker}(\pi\_{\*x})^\perp \to T\_{\pi(x)}N$$
is an isometry. So asking for $f$ to be an isometric embedding is equivalent to asking that for all $x'\in U$, the map
$$Df: T\_{\pi(x')}N \to T\_{x'}M = \text{Ker}(\pi\_{\*x'}) \oplus \text{Ker}(\pi\_{\*x'})^\perp$$
vanishes on the $\text{Ker}(\pi\_{\*x'})$ coordinate. I can see that the implicit function theorem allows us choose $f$ so that $Df|\_x$ satisfies this, but I don't know whether it is possible to do so locally around $x$.
**Intuition:**
For $\pi: \mathbb{R^2}\to \mathbb{R}$, projection onto the first coordinate, there is a map $f:\mathbb{R}\to \mathbb{R^2}$ such that $\pi \circ f = id\_\mathbb R$. For example we can take $f(x) = (x, g(x))$ for any function $g:\mathbb R \to \mathbb R$. However $f(x) = (x,a)$ for constant $a\in \mathbb R$ have the special property that they are isometric embeddings. So I'm interested to know if we can always do this locally in the general setting above, and if so how.
Any help is appreciated!
| https://mathoverflow.net/users/506774 | Local isometric embedding right inverse to a Riemannian submersion | You can do so if and only if the *horizontal distribution* $x \mapsto \ker(\pi\_\*)^\perp$ is integrable. This is equivalent to the vanishing of O'Neill's $A$-tensor for the Riemannian submersion:
$$
\langle \nabla\_X Y, U \rangle = 0,
$$
for all vector fields $X,Y,U$ with $X,Y$ horizontal (i.e. tangent to $\ker( \pi\_\*)^\perp$, $U$ vertical (i.e. tangent to $\ker(\pi\_\*)$). Here $\nabla$ is the Levi-Civita connection of $M$.
Since this tensor is skew-symmetric in $X,Y$, this is always satisfied for instance when $\dim N = 1$.
See e.g. Besse, "Einstein manifolds", Chapter 9.
| 6 | https://mathoverflow.net/users/14708 | 450046 | 181,073 |
https://mathoverflow.net/questions/431453 | 1 | Let $ G $ be a compact topological group which is quasisimple in the sense that
$$
[G,G]=G
$$
and
$$
G/Z(G)
$$
is simple as an abstract group. Must $ G $ be a Lie group?
This is a follow-up question to
<https://math.stackexchange.com/questions/4537401/compact-simple-group-which-is-not-a-lie-group>
By Peter-Weyl theorem there must be a nontrivial continuous homomorphism from $ G/Z(G) $ to some $ U\_n $. Since $ G/Z(G) $ is abstractly simple then this is an embedding and so we realize $ G/Z(G) $ as a compact and thus closed subspace of some $ U\_n $. Thus $ G/Z(G) $ is a closed subgroup of $ U\_n $ so it is a Lie group. So the question is equivalent to: let the compact topological group $ G $ be a perfect central extension of a compact simple Lie group. Must $ G $ be a Lie group?
| https://mathoverflow.net/users/387190 | Is every compact quasisimple group a Lie group? | The group $G/Z(G)$ being simple, is a Lie group by Peter-Weyl. Hence it is either finite or connected. If it is finite, $G$ has center of finite index, hence has a finite derived subgroup. Since $G$ is perfect, this means that $G$ is finite.
Now suppose that $G/Z(G)$ is connected. Let $H$ be any Lie quotient of $G$ that is a cover of $G/Z(G)$. Then the image of $Z(G)$ in $H$ is central, and since the quotient has trivial center, we deduce that this image is precisely $Z(H)$. In particular, we see that $H=Z(H)H^0$. Since $G$ is perfect, so is $H$, and we deduce that $H=H^0$. Thus $H$ is a connected cover of $G/Z(G)$. Since $G/Z(G)$ has a finite fundamental group, say of order $d$, we deduce that $|Z(H)|\le d$.
This concludes: indeed if by contradiction $|Z(G)|>d$, then there exists a Lie quotient $H$ of $G$ in which the image of $Z(G)$ has cardinal $>d$ (by Peter-Weyl) and the above yields a contradiction.
| 3 | https://mathoverflow.net/users/14094 | 450050 | 181,074 |
https://mathoverflow.net/questions/449975 | 5 | I do not understand the proof of Variant 4.2.3.16 of *Higher Topos Theory* by Jacob Lurie, and I need help.
---
Variant 4.2.3.16 asserts the following:
>
> ($\diamond$) Let $K$ be a finite simplicial set. There is a cofinal map $N(A)\to K$, where $A$ is a finite poset.
>
>
>
The proof proceeds as follows:
(**1**). Consider the following property for a simplicial set $K$:
($\ast$) Every nondegenerate simplex $\Delta^n\to K$ is a monomorphism.
If K satisfies ($\ast$), then $K$ satisfies the conclusion of ($\diamond$).
(**2**). We show that, for each finite simplicial set $K$, there is a cofinal map $\widetilde{K}\to K$, where $\widetilde{K}$ is finite and satisfies ($\ast$).
---
I am fine with step (1). But I don't understand the proof of (2).
Let me explain Lurie's proof of (2). He argues that we can prove (2) by induction on the number of nondegenerate simplices of $K$. According to him, this is because if $K$ can be written as $K=K\_0\amalg \_{\partial\Delta^n} \Delta^n$, where $K\_0$ satisfies ($\ast$), then $K$ also satisfies ($\ast$). To prove this, he choose a cofinal map $\widetilde{K\_0}\to K\_0$, where $\widetilde{K}\_0$ is a finite simplicial set satisfying ($\ast$), and claims that the map $$\widetilde{K}=(\widetilde{K}\_0\times \Delta^n)\amalg \_{\partial \Delta^n} \Delta^n\to K$$
witnesses property ($\ast$).
The problem is, **I do not understand what the map $\partial \Delta^n\to \widetilde{K}\_0$ used in the definition of $\widetilde{K}$ is.** Can someone explain what this map is? (Or does anyone know how to prove (2) or ($\diamond$) in different ways?)
Thanks in advance.
| https://mathoverflow.net/users/144250 | Cofinal maps from posets (HTT, 4.2.3.16) | [Rephrasing my comment as an answer]
While I cannot speak for the actual proof, two points are worth noting. First, a similar construction appears in Kerodon (<https://kerodon.net/tag/02QA>) but there (1) $\widetilde{K} \to K$ is required to be a trivial fibration but (2) it doesn't restrict correctly to when $K$ is finite.
The second point is that Kerodon also includes a stronger proof of this same fact (<https://kerodon.net/tag/02MU>). The construction is quite different but should suffice as a replacement.
| 2 | https://mathoverflow.net/users/76636 | 450057 | 181,077 |
https://mathoverflow.net/questions/450061 | 2 | Assume that C is a stable infinity category; $SH\_{fin}$ is the homotopy category of finite spectra. Is there a canonical bi-functor (action? module structure?) $SH\_{fin}\times hC \to hC$?
Is there any "canonical" formulation of this statement; what about the properties of this bi-functor? It possibly follows from Proposition 4.8.12.8 of Lurie's Higher Algebra, but I don't really understand what the latter says and do not want to read the whole book.
| https://mathoverflow.net/users/2191 | Does the homotopy category of finite spectra act on stable homotopy categories? | Yes: Since $\mathcal{C}$ is stable, $\operatorname{Fun}(\mathcal{C},\mathcal{C})$ is stable, too. In particular, it has finite colimits, so $\operatorname{Ind}\operatorname{Fun}(\mathcal{C},\mathcal{C})$ has all colimits. So we get a unique colimit-preserving functor $\mathcal{S} \to \operatorname{Ind}\operatorname{Fun}(\mathcal{C},\mathcal{C})$ taking the one-point space to the identity functor, and this extends further to a unique exact functor $\operatorname{Sp}\to \operatorname{Ind}\operatorname{Fun}(\mathcal{C},\mathcal{C})$. Passage to compact objects gives a functor $\operatorname{Sp}^\omega \to \operatorname{Fun}(\mathcal{C},\mathcal{C})$, or equivalently $\operatorname{Sp}^\omega \times \mathcal{C}\to \mathcal{C}$.
| 7 | https://mathoverflow.net/users/39747 | 450068 | 181,079 |
https://mathoverflow.net/questions/449932 | 2 | Let $A$ be a finite-dimensional $\*$-algebra over $\mathbb R$. We say that an element $x \in A$ is *positive definite* if $x$ admits an inverse and if $x = y y^\*$ for some $y \in A$. Does every such $x$ admit a $z \in A$ such that $x = z^2$? If someone has references for me to read, I would appreciate that.
I think the Wedderburn-Malcev theorem can help here, as well as Hermite interpolation. But non-commutativity seems like an obstacle.
| https://mathoverflow.net/users/75761 | Do positive-definite elements in finite-dimensional $*$-algebras over $\mathbb R$ always admit square roots? | I'll assume that a $\*$-algebra just means an $\mathbb{R}$-algebra with a linear operation satisfying $(xy)^\*=y^\*x^\*$ and $x^{\*\*}=x$ (as at <https://ncatlab.org/nlab/show/star-algebra>). If so, you can just take $A=\mathbb{R}\times\mathbb{R}$ with $(x\_0,x\_1)^\*=(x\_1,x\_0)$. Then take $x=(-1,-1)$ and $y=(1,-1)$ so $x=yy^\*$ and $x$ is invertible but $x$ has no square root. If you want any more axioms then you should specify them.
| 6 | https://mathoverflow.net/users/10366 | 450072 | 181,082 |
https://mathoverflow.net/questions/450075 | 0 | What are the rational solutions to the equation
$$
y^3 = x^4 + x,
$$
in particular, are there any (finite) solutions other than $(x,y)=(0,0)$ and $(-1,0)$?
Context: This is the simplest-looking example of a Picard curve, that is, curve $y^3=P(x)$, where $P(x)$ is a polynomial of degree 4. The rank of its Jacobian is 0, the simplest case possible. Hashimoto and Morrison <https://arxiv.org/abs/2002.03291> treated more difficult case of rank $1$. For rank $0$, Jackson Morrow [$y^3 = x^4 + x + 2$, and existence of rational points on rank 0 Picard curves](https://mathoverflow.net/questions/407677) presented some Magma code that can prove that there are no rational points (if this is indeed the case). However, the curve $y^3 = x^4 + x$ has some rational points, so deeper analysis seems to be required. Hence the question.
| https://mathoverflow.net/users/89064 | $y^3=x^4+x$, and computing all rational points on rank $0$ Picard curves | Let $x = \frac{a}{d}$, $y = \frac{b}{d}$, $\gcd(a, d, b) = 1$.
$$b^3d = a^4 + ad^3$$
Suppose $p \mid a$ and $p \mid d$ for prime $p$. Then $b$ is not divisible by $p$.
$$\nu\_p(d) = \nu\_p(a^4 + ad^3) = \nu\_p(a) + \nu\_p(a^3 + d^3) = 4\nu\_p(a)$$
The last equality is true because $\nu\_p(a) < \nu\_p(d)$. Hence, we can make the substitution $a = a\_1c$, $d = d\_1c^4$; $(a\_1, d\_1, c)$ are pairwise coprime.
$$b^3d\_1 = a\_1^4 + a\_1d\_1^3c^9$$
$d\_1 \mid a\_1$. But this is only possible if $d\_1 = \pm 1$.
$(\pm b)^3 = a\_1(a\_1^3 \pm c^9)$
Whence $a\_1^3 \pm c^9$ must be a cube, which contradicts Fermat's theorem.
| 5 | https://mathoverflow.net/users/507773 | 450081 | 181,085 |
https://mathoverflow.net/questions/449920 | 2 | If someone has gone through the Griffiths' paper ``On the periods of certain rational integrals: I,'' could you help me to understand Lemma 8.10?
I don't get why $\eta\in Z^{q,k+1}(l-1)$; although $\eta$ is surely a closed form of type $(q,0)+\cdots+(q-k-1,k+1)$ with a pole of order $l-1$, I think we have to say more to conclude $\eta\in Z^{q,k+1}(l-1)$.
In the paper's notation, $B^{q,k}(l)$ denotes the space of differential forms $\varphi$ on $\mathbb{P}^n$ which are of type $(q,0)+\cdots+(q-k,k)$, are $C^\infty$ on $\mathbb{P}^n-V$, and which have the local property that $f^l\varphi$ and $f^{l-1}df\wedge\varphi$ are $C^\infty$ if $f=0$ is a local holomorphic defining equation for $V$.
$Z^{q,k}(l)$ is the space of closed forms in $B^{q,k}(l)$.
In Lemma 8.7, we write $\varphi\in B^{q,k}(l)$ as $\varphi=d\psi+\eta$ locally, where $\psi,\eta$ have poles of order $l-1$ along $V$, $\psi$ is of type $(q-1,0)+\cdots+(q-1-k,k)$, and $\eta$ is of type $(q,0)+\cdots+(q-k-1,k+1)$.
In Lemma 8.9, we glue them by using a partition of unity to obtain $\varphi=d\psi+\eta$ globally.
Lemma 8.10 contains a claim that if $\varphi\in Z^{q,k}(l)$ and $l>1$ then $\eta\in Z^{q,k+1}(l-1)$, and this is what I am in trouble with.
Indeed, taking a look at the construction of $\eta$ in Lemma 8.7 and 8.9, we can say that $\eta$ is of type $(q,0)+\cdots+(q-k-1,k+1)$ and has a pole of order $l-1$ ,i.e. $f^{l-1}\eta$ is $C^\infty$. In addition, since $\varphi=d\psi+\eta$ is a closed form, so is $\eta$. However, how can we show that $f^{l-2}df\wedge\eta$ is $C^\infty$? It seems a stronger condition than just having a pole of order $l-1$.
In the same reason, I don't get why the statement $\psi-\psi'\in B^{q-1,k}(l-1)$ in the proof of Lemma 8.10 holds. Excuse me if I'm asking a dumb question, for I'm not from the background of mathematics. Any comments are helpful.
| https://mathoverflow.net/users/507853 | A specific question on the Griffiths' paper: the reduction of the pole order | In fact, it just follows from that $\eta$ has a pole of order $l-1$ and is $\textit{closed}$: We have the identity $$f^{l-2}df\wedge\eta=\frac{1}{l-1}d(f^{l-1})\wedge\eta=\frac{1}{l-1}d(f^{l-1}\wedge\eta),$$
which is $C^{\infty}$ since $f^{l-1}\wedge\eta$ is $C^{\infty}$.
| 2 | https://mathoverflow.net/users/74322 | 450085 | 181,088 |
https://mathoverflow.net/questions/450048 | 2 | First of all: I apologise in advance for if my question will be arid, wrong written or even nonsensical.
I was at a talking with a professor last week, and the question of "Entanglement and Algebraic Geometry" came out.
What emerged really fascinated me, so I'm here to ask more clarifications, explanations or even references.
Here is what I "understood" and wrote down on a piece of paper (that is, heard and hence reporting here):
"Say we have two systems made of states. It's easy to understand what entanglement is from a geometric point of view. Those systems can the thought as objects in $\mathbb{P}^2(\mathbb{C}^3)$ (projective space). The states they are made of can be pure or not pure. We can see the pure states are basically quadrics in $\mathbb{P}^2(\mathbb{C}^3)$, while the non pure states are what are called the entangled ones".
So I'm asking you: can you explain me better the above statement, assuming it's not meaningless? Or provide me some reference, paper, books...
Thank you so much!
| https://mathoverflow.net/users/88816 | Entanglement, quadrics and $\mathbb{P}^2(\mathbb{C}^3)$ | $\DeclareMathOperator\SL{SL}\DeclareMathOperator\PSL{PSL}$Let me just make a couple of comments on the question to clean up the confusion about the correct dimension of the projective space — it should be $\mathbb{P}^3(\mathbb{C})$. The state space of a qubit is the Hilbert space $\mathbb{C}^2$, and the $n$-qubit is obtained as tensor product of $n$ copies of $\mathbb{C}^2$. There are different choices for symmetry group, one possible is SLOCC (stochastic local operations with classical communication) which for the $n$-qubit would be the group $\SL\_2(\mathbb{C})^{\times n}$ acting on $(\mathbb{C}^2)^{\otimes n}$. For the 2 qubits, this would be $\SL\_2(\mathbb{C})\times\SL\_2(\mathbb{C})$ acting on $\mathbb{C}^2\otimes\mathbb{C}^2$. The action can be identified with the group acting on $2\times 2$-matrices by left and right multiplication. The orbits of the group on the projective space associated to $\mathbb{C}^2\otimes\mathbb{C}^2$ (that is the $\mathbb{P}^3$ that should appear in the question) correspond to the different states the system can be in.
Now algebraic geometry tells us that there are two orbits for the action of $\SL\_2(\mathbb{C})\times\SL\_2(\mathbb{C})$ acting on $\mathbb{P}^3$. One open orbit, isomorphic to $\PSL\_2(\mathbb{C})$, and one closed orbit isomorphic to $\mathbb{P}^1\times\mathbb{P}^1$ — that is the quadric mentioned in Tom Ducat's [comment](https://mathoverflow.net/questions/450048/entanglement-quadrics-and-mathbbp2-mathbbc3#comment1163095_450048). The closed orbit can be obtained as the orbit of an elementary tensor, so it consists of pure states. The open orbit consists of the entangled states, it is the orbit of the so-called Bell state.
There is further literature on this, people have been working on using geometric invariant theory etc to understand polynomial invariants which separate different entangled states in $n$-qubits. My knowledge there is a bit dated, but it seemed that 4 qubits is the state of the art.
| 6 | https://mathoverflow.net/users/50846 | 450090 | 181,091 |
https://mathoverflow.net/questions/450095 | 4 | Let $X\_k$ be i.i.d. Cauchy random variables with parameters $0,1$. For each $N$ define the process $Y\_N$ by $$Y\_N(t)=\frac{1}N\sum\_{k=1}^{\lfloor tN\rfloor}X\_k+\text{piecewise linear interpolation}.$$
Note that for each grid point, the sum of Cauchy random variables is another Cauchy random variable. I am interested in the convergence of this process. Is the limit continuous? Is it Hölder? We cannot apply Kolmogorov continuity criterion due to lack of moments.
| https://mathoverflow.net/users/479223 | Hölder continuity of process from Donsker like theorem with Cauchy random variables | Indeed this falls a bit outside of the standard theory. As mentioned in this MSE answer to [Does a random walk with infinite mean ever converge to anything?](https://math.stackexchange.com/questions/1577043/does-a-random-walk-with-infinite-mean-ever-converge-to-anything), in particular for Cauchy too in the second answer, one has to use the stable FCLT to get that the limit of this random walk is the [Cauchy process](https://en.wikipedia.org/wiki/Cauchy_process). For example, as mentioned in the comments from these notes ["Stochastic-process limits"](http://www.columbia.edu/%7Eww2040/jumps.html) by Whitt, we can use theorem 4.5.3.
| 5 | https://mathoverflow.net/users/99863 | 450097 | 181,092 |
https://mathoverflow.net/questions/450025 | 17 | Let $X$ and $Y$ be sets. It is undecidable in ZFC whether $2^{|X|} = 2^{|Y|}$ implies $|X| = |Y|$ (in Cohen's original model for ZFC + $\neg$CH, one has $2^{\aleph\_0} = 2^{\aleph\_1}$). What if we restrict our attention to the *finite* parts of $X$ and $Y$?
**Question.** Do $X$ and $Y$ have the same cardinality if the families of *finite* subsets of both sets do?
As noted by Nik Weaver [in a comment](https://mathoverflow.net/questions/450025/do-x-and-y-have-the-same-cardinality-if-their-families-of-finite-subsets-do#comment1162981_450025), the answer is yes when we assume in the background to be working with the axioms of ZFC. But, what about ZF?
I feel this must be either basic or very well known (to those who know it very well).
| https://mathoverflow.net/users/16537 | Do $X$ and $Y$ have the same cardinality if their families of finite subsets do? | It can be proved in $\mathsf{ZF}$ that ``for all cardianls $\mathfrak{a},\mathfrak{b}$, if $\mathrm{fin}(\mathfrak{a})=\mathrm{fin}(\mathfrak{b})$, then $\mathfrak{a}=\mathfrak{b}$'' implies $\mathsf{AC}$. Here $\mathrm{fin}(\mathfrak{a})$ denotes the cardinality of the set of all finite subsets of a set which is of cardinality $\mathfrak{a}$.
Assume that, for all cardianls $\mathfrak{a},\mathfrak{b}$, if $\mathrm{fin}(\mathfrak{a})=\mathrm{fin}(\mathfrak{b})$, then $\mathfrak{a}=\mathfrak{b}$. We prove $\mathsf{AC}$ by showing that, for an arbitrary limit ordinal $\lambda$, $\mathrm{V}\_\lambda$ is well-orderable. Let $\mathfrak{a}$ be the cardinality of $\mathrm{V}\_\lambda$ and let $\kappa$ be the Hartogs number of $\mathrm{V}\_\lambda$, i.e., the first ordinal that cannot be mapped injectively into $\mathrm{V}\_\lambda$. Note that every finite subset of $\mathrm{V}\_\lambda$ is an element of $\mathrm{V}\_\lambda$, so $\mathrm{fin}(\mathfrak{a})=\mathfrak{a}$.
It suffices to prove that $\mathrm{fin}(\mathfrak{a}+\kappa)=\mathrm{fin}(\mathfrak{a}\cdot\kappa)$, since then by our assumption we would have $\mathfrak{a}+\kappa=\mathfrak{a}\cdot\kappa$, and thus $\mathfrak{a}$ is well-ordered by a lemma of Tarski. Clearly, $\mathrm{fin}(\mathfrak{a}+\kappa)=\mathrm{fin}(\mathfrak{a})\cdot\mathrm{fin}(\kappa)=\mathfrak{a}\cdot\kappa$. It remains to show that $\mathrm{fin}(\mathfrak{a}\cdot\kappa)=\mathfrak{a}\cdot\kappa$. It is proved as follows.
Let $u$ be an arbitrary finite subset of $\mathrm{V}\_\lambda\times\kappa$.
Then $\mathrm{ran}(u)=\{\alpha\mid\exists x(\langle x,\alpha\rangle\in u)\}$ is a finite subset of $\kappa$, and thus there is a unique isomorphism $h\_u$ of $\mathrm{ran}(u)$ onto some natural number. Note that $u$ is uniquely determined by $\langle\mathrm{ran}(u),\{\langle x,h\_u(\alpha)\rangle\mid\langle x,\alpha\rangle\in u\}\rangle$, so
$$
\mathrm{fin}(\mathfrak{a}\cdot\kappa)\leqslant\mathrm{fin}(\kappa)\cdot\mathfrak{a}=\mathfrak{a}\cdot\kappa,
$$
which completes the proof.$\quad\square$
| 16 | https://mathoverflow.net/users/101817 | 450104 | 181,095 |
https://mathoverflow.net/questions/449946 | 1 | Let $G$ be a $p$-adic reductive group and $Z\subseteq G$ the center. Let $\pi$ be an irreducible admissible representation of $G$. By Schur's lemma, it is easy to show that there is a group homomorphism $\omega\_\pi:Z\to \mathbb{C}^\times$, namely the central character. But how does one to prove that this group homomorphism is indeed continuous? My understanding is that a character on $Z$ is a continuous homomorphism, not just a homomorphism.
| https://mathoverflow.net/users/32746 | Continuity of central character | Let $K\subset G$ be a compact open subgroup such that $V^K\ne0$. Now $Z$ acts on $V^K$ by the central character $\omega\_\pi$, and the action is trivial on $Z\cap K$. Thus $\omega\_\pi$ is trivial on the compact open subgroup $Z\cap K$, hence is continuous.
| 1 | https://mathoverflow.net/users/123673 | 450105 | 181,096 |
https://mathoverflow.net/questions/449600 | 1 | Let $G$ be a $k$-partite $k$-uniform hypergraph with at least $dn^k$ many edges. I want a lower bound on the number of $K\_{2, 2,\, \ldots\,,2}$ in $G$, preferably something like $\gamma n^{2k}$ for some constant $\gamma$ independent of $n$.
My ideas: I can get down to a minimum degree subhypergraph from the edge-density by cleaning up vertices with low degrees. Starting from the edge-density I can also get down to a hypergraph where all vertices have a partite minimum co-degree either $0$ or $\beta n^{k-1}$. But I have no idea of how to do both of that together. I have also been thinking of some kind of induction approach but have not been able to prove it with that.
| https://mathoverflow.net/users/485879 | Counting $K_{2, 2, \,\ldots\,,2}$ in a $k$-partite $k$-uniform hypergraph | You can use the doubling method (a.k.a. Cauchy-Schwarz)
For simplicity, suppose we are working in the graph case. So, let $G$ be a graph with bipartition $X\cup Y$ with $|X|=|Y|=n$ and suppose that $e(G)=\Omega(n^2)$. Let $f(x,y)$ be the indicator function of the edges of $G$ and let
$$A=\sum\_{x\in X, y\in Y}f(x,y) $$
Using the Cauchy-Schwarz inequality, we have
$$A^2=\left(\sum\_{x\in X}\sum\_{y\in Y}f(x,y)\right)^2\le n\sum\_{x\in X}\left(\sum\_{y\in Y}f(x,y)\right)^2=n\sum\_{x\in X}\sum\_{y,y'\in Y}f(x,y)f(x,y'),$$
where the RHS is $n$ times the the number of "cherries" (copies of $K\_{1,2}$) in $G$. This operation is sometimes called "doubling" since we are duplicating the variable $y$. Now we double the variable $x$ by squaring again.
$$A^4\le n^2\left(\sum\_{y,y'\in Y}\sum\_{x\in X}f(x,y)f(x,y')\right)^2\le n^4\sum\_{x,x'\in X, y,y'\in Y}f(x,y)f(x,y')f(x',y)f(x',y'),$$
and thus the RHS is $n^4$ times the number of $K\_{2,2}$. Then the number of $K\_{2,2}$ is of order $n^{-4}A^4=\Omega(n^4)$. The same proof works for $k$-uniform hypergraphs but doing "doublings" $k$ times.
| 3 | https://mathoverflow.net/users/507998 | 450115 | 181,097 |
https://mathoverflow.net/questions/319554 | 8 | Among the good asymptotic bounds in coding theory in the MRRW bound. It is obtained by using the linear programming problem of Delsarte's and providing a solution. The LP problem is
>
> Suppose $C \subset \mathbb{F}\_2^n $ is a code such $d(C)\ge d$. Let
> $\beta(x) = 1+ \sum\_{k=1}^{n} y\_k K\_k (x)$ be a polynomial such that
> $y\_k \ge 0$ but $\beta(j) \le 0$ for $j=d, d+1,\dots ,n$. Then, we have that $|C| \le \beta(0)$.
>
>
>
Here $K\_k(x)$ are the Kravchuk polynomials. In the proof of the MRRW bound, upto scaling, they basically come up with the following polynomial $\beta$ for a general $n$.
$$\beta(x) =\frac{1}{a-x} \left[ K\_t(a) K\_{t+1}(x) - K\_{t+1}(a)K\_{t}(x) \right]^{2}$$
After using the Christoffel-Darboux formula the values of $t$ and $a$ are adjusted to make it optimal.
There is no justification for why such a polynomial was chosen other than that it works. Is there anything more that can be said over why this polynomial was chosen?
| https://mathoverflow.net/users/94546 | How did they come up with the MRRW bound? | I don't know what they were thinking in the past, but there are more modern points of view these days.
1. We can view the Delsarte LP as the symmetrization of a very large LP (or as the symmetrization of the [Lovász theta function](https://en.wikipedia.org/wiki/Lov%C3%A1sz_number) SDP on the hypercube graph with edges between strings of distance $\le d$). The Kravchuk polynomials arise naturally in this way since they are the symmetrization of the Boolean Fourier characters. Symmetrization makes the program exponentially smaller, but it does not change the value.
2. The candidate solution to the Delsarte LP has the form $(a-x) F(x)^2$ after applying Christoffel-Darboux to the candidate $\beta(x)$ in the question. The term $(a-x)$ enforces the sign constraints of the LP, and $\widehat{F}(x)$ is chosen to be an approximate eigenfunction to the matrix $a\cdot\text{Id} - A$. Here $A$ is the adjacency matrix of the standard Boolean hypercube, then symmetry-reduced into a matrix on $\{0,1, \dots, n\}$.
The approximate eigenfunction condition guarantees that the MacWilliams identities are satisfied.
See "An Elementary Proof of the First LP Bound on the Rate of Binary Codes" by Nati Linial and Elyassaf Loyfer for more on this perspective (for disclosure, they are my collaborators)
| 2 | https://mathoverflow.net/users/48204 | 450137 | 181,105 |
https://mathoverflow.net/questions/450133 | 7 | Let $X$ be a connected qcqs scheme. We say that $X$ is a (étale) $K(\pi,1)$ if for every locally constant constructible abelian sheaf $\mathscr{F}$ on $X$ and every geometric point $\overline{x}$ the natural morphisms
$$\mathrm{H}^i(\pi\_1(X,\overline{x}),\mathscr{F}\_x) \to \mathrm{H}^i\_\text{ét}(X,\mathscr{F})$$
are isomorphisms for all $i\geq 0$. (See section 4 on [P. Achinger's Wild Ramification and $K(\pi,1)$ spaces](https://arxiv.org/pdf/1701.03197.pdf) for a quick review on this.)
There's a natural *de Rham* analog of this story. Let $X$ be a smooth algebraic variety over a characteristic zero field $k$. The category $\textsf{DE}(X/k)$ of vector bundles on $X$ with integrable connection is tannakian. In particular, a fiber functor $\omega$ induces an equivalence of categories $\omega:\textsf{DE}(X/k)\to \textsf{Rep}\_\text{fd}(\Gamma)$ for some linear algebraic group $\Gamma$. (The index fd means finite-dimensional representations.) Then, we say that $X$ is a (de Rham) $K(\pi,1)$ if for every vector bundle with integrable connection $\mathscr{E}$ and every fiber functor $\omega$ we have an isomorphism
$$\mathrm{H}^i(\Gamma,\omega(\mathscr{E}))\simeq\mathrm{H}^i\_\text{dR}(X,\mathscr{E}),$$
where the cohomology on the left is algebraic group cohomology (as in chapter 15 of Milne's book Algebraic Groups), for all $i\geq 0$.
I can prove that (as in the étale context) the isomorphism above always holds for $i=0$ but I know nothing more about this story. What are some classes of algebraic varieties that are de Rham $K(\pi,1)$'s? Is a variety an étale $K(\pi,1)$ if and only if it's a de Rham $K(\pi,1)$? (Or perhaps there's only one implication?)
**In general, what's known about de Rham $K(\pi,1)$'s?**
| https://mathoverflow.net/users/131975 | Is anything known about de Rham $K(\pi,1)$'s? | I had a derelict project (joint with Javier Fresan) trying to study this notion. We didn't prove much and we got stuck with the more interesting questions.
The isomorphism you wrote always holds for $i=1$. An easy way to see this is to compare both sides to extensions of $\mathcal{O}$ by $\mathcal{E}$.
If $X$ is proper (or if we restrict to regular connections), the comparison question becomes group-theoretic. Namely, we have a finitely presented group $\Gamma = \pi\_1^{\rm top}(X)$, its profinite completion $\widehat{\Gamma} = \pi\_1^{\rm et}(X)$, and its pro-algebraic completion $\Gamma\_{\rm alg} = \pi\_1^{\rm dR}(X)$. Cohomology comparison results (Deligne LNM 163 for de Rham, SGA4 for etale) give you isomorphisms
$$ H^i\_{\rm dR}(X, \mathcal{E}) \simeq H^i(X, \mathbf{V}) $$
and
$$ H^i\_{\rm et}(X, \mathcal{L}) \simeq H^i(X, \mathcal{L}) $$
where $\mathcal{E}$ is a regular connection corresponding to the local system $\mathbf{V}$, and where $\mathcal{L}$ is an etale local system of (pro)finite abelian groups (and also the corresponding Betti local system). The comparison maps from group cohomology to sheaf cohomology and the above maps, together with similar comparison maps on the level of group cohomology, give a $2\times 3$ commutative diagram that I don't remember how to draw on MO. You might be able to deduce something from this diagram. Recall that $\Gamma$ is "good" if its group cohomology (with coefficients in a finite module) agree with the group cohomology of $\widehat{\Gamma}$. By analogy, let us call $\Gamma$ "cute" if the same holds for every $\mathbf{C}$-representation, with $\widehat{\Gamma}$ replaced with $\Gamma\_{\rm alg}$. With this you may be able to conclude something like: if $\Gamma$ is cute and good, then $X$ is an etale $K(\pi,1)$ if it is a de Rham $K(\pi, 1)$.
So, for example, I think abelian varieties and curves of genus $>0$ are $K(\pi,1)$ in all three senses.
If $X$ is not proper, and you allow irregular connections, then the de Rham fundamental group is nasty while the etale one is reasonable, and we expect no connection.
For $\mathbf{A}^n$, one can use the fibration trick (Deligne-Laumon) as in characteristic $p$, and I believe we can conclude that $\mathbf{A}^n$ is a de Rham $K(\pi, 1)$.
I have no idea what happens for arbitrary smooth affines. At least, the trick of reducing everything to $\mathbf{A}^n$ using finite etale maps completely fails.
| 9 | https://mathoverflow.net/users/3847 | 450138 | 181,106 |
https://mathoverflow.net/questions/450111 | 2 | Suppose that $f$ is a holomorphic function on a domain $D$ in $\mathbb{C}^n$, $\partial D$ is smooth, and $f$ is $C^1$ on $\partial D$. Then, the Bochner-Martinelli formula states that
$f(z) = \int\_{\partial D} f(\zeta) \omega(\zeta, z)$,
where $\omega(\zeta, z)$ is the Bochner-Martinelli kernel. I am wondering if a $C^1$ function satisfying this formula is necessarily holomorphic.
This question arose when I was trying to prove that if a sequence of $C^1$ functions $f\_n:D\rightarrow\mathbb{C}$ satisfies $f\_n\rightarrow f$ uniformly and $\partial\bar{f}\_n\rightarrow0$ uniformly, then $f$ is holomorphic. Is this statement correct?
| https://mathoverflow.net/users/167284 | Inverse of Bochner–Martinelli formula | The result you need is a now classical result of Aronov and Kytmanov [see for example [1] chapter 4 §15.1, p. 161, theorem 15.1]: if $D$ is a bounded domain in $\Bbb C^n$ with piecewise smooth boundary and $f\in C^1(\bar D)$ then
$$
f\in \mathscr{O}(D) \iff f(z) = \int\_{\partial D} f(\zeta) \omega(\zeta, z)
$$
Therefore if you have a sequence of $C^1$ functions $\{f\_n\}\_{n\in\Bbb N}$ which are representable as Bochner-Martinelli integrals converging uniformly to a function $f$ representable in the same way, both each single function in the sequence and their limit function are holomorphic in $D$.
**Note**
Aizenberg and Kytmanov proved that the same result is true also for $f\in C^0(\bar{D})$, provided the boundary of $D$ satisfies a higher regularity requirement (namely $\partial D\in C^2$, see [1] chapter 4 §15.2, p. 162, theorem 15.4 for the details).
**Reference**
[1] Alexander M. Kytmanov (1995) [1992], *The Bochner-Martinelli integral and its applications*, Birkhäuser Verlag, pp. xii+305, doi:10.1007/978-3-0348-9094-6, ISBN 978-3-7643-5240-0, [MR1409816](https://mathscinet.ams.org/mathscinet-getitem?mr=1409816), [Zbl 0834.32001](https://zbmath.org/0834.32001).
| 3 | https://mathoverflow.net/users/113756 | 450154 | 181,111 |
https://mathoverflow.net/questions/450162 | 5 | I note that Mathematica could yield the identity
$$\int\_0^1\frac{\log(1+x^2(x-1)/2)}{x^2(x-1)}dx=\frac{\pi(\pi-4)-12\log^22+24\log2}{16}.\tag{1}\label{1}$$
But I don't know how Mathematica got this.
**Question.** How to prove \eqref{1} manually?
Your comments are welcome!
| https://mathoverflow.net/users/124654 | For a manual evaluation of a definite integral | With Mathematica, one can actually find an antiderivative $F=G+H$ of the function $f$, where
$$f(x):=\frac{\ln(1+x^2(x-1) /2)}{x^2(x-1)},$$
$$G(x):=-\text{Li}\_2\left(\frac{1-x}{2}\right)-\frac{1}{2}
\text{Li}\_2\left(-(x-1)^2\right)+\text{Li}\_2(-x)+\text{Li}\_2\left(\left(\frac{1}{2}-\frac{i}{2}\right) x\right)+\text{Li}\_2\left(\left(\frac{1}{2}+\frac{i}{2}\right) x\right),$$
\begin{gather}
H(x):=\frac{\left(x \ln \left(\frac{x-1}{x}\right)+1\right) \ln \left(\frac{1}{2} (x-1)
x^2+1\right)}{x}+\ln (x+1) \\
+\ln (x) \left(\ln
\left(1-\left(\frac{1}{2}+\frac{i}{2}\right) x\right)+\ln
\left(1-\left(\frac{1}{2}-\frac{i}{2}\right) x\right)+\ln (x+1)\right) \\
-\ln (x-1)
\left(\ln (i x+(1-i))+\ln \left(\frac{1}{2} ((1+i)-i x)
(x+1)\right)\right) \\
-\frac{1}{2} \ln ((x-2) x+2)+\tan ^{-1}(1-x).
\end{gather}
This can be verified by differentiation, since $\text{Li}'\_2(x)=-\ln(1-x)/x$.
The desired result now follows because
$$G(0)=\frac{\ln ^2(2)}{2}-\frac{\pi ^2}{24},\quad G(1)=\frac{1}{48} \left(\pi ^2-12 \ln ^2(2)\right),$$
$$\lim\_{x\to0}H(x)=(\pi - \ln4)/4,\quad \lim\_{x\to1}H(x)=\ln2.$$
---
The indefinite integral of $f$ can be found "more manually" as follows. Integrate $f$ by parts, to get an integrand of the form $R(x)+R\_0(x)\ln x+R\_1(x)\ln(1-x)$ instead of $f(x)=\dfrac{\ln(1+x^2(x-1) /2)}{\cdots}$, where $R(x),R\_0(x),R\_1(x)$ are certain rational expressions. Use then partial fraction decomposition (involving certain complex constants) to reduce the integrand to ones of the form $\frac{\ln u}{u+a}$, and note that
$$\int \frac{\ln u}{u+a}\,du=
\text{Li}\_2\left(-\frac{u}{a}\right)+\ln(u) \ln\left(\frac{a+u}{a}\right)+C.$$
| 9 | https://mathoverflow.net/users/36721 | 450164 | 181,115 |
https://mathoverflow.net/questions/449709 | 20 | I recently discovered a proof of the following. Let $p$ be a prime that's $1 \bmod {3}$. Suppose that $p$ is not represented by the principal quadratic form $(1,9,81)$ of discriminant $-243$ (The first three $p$ that are represented by this form are $61$, $67$, and $73$). Then neither $3p$ nor $3p^2$ is a sum of two rational cubes.
The proof is elementary (at the undergraduate level) in the following sense. In the statement of the theorem I impose two further conditions on $p$, each of which like the non-representability by $(1,9,81)$ is easily checked for any given $p$. The first of these supplementary conditions in fact holds for all $p$, while the second (that $3$ is not a cube in $\mathbb{Z}/p\mathbb{Z}$) is equivalent to $p$ not being representable by $(1,9,81)$. But the assertions of this last sentence are cubic reciprocity results authored by Eisenstein and Gauss.
**Question 1**
In view of this last paragraph, my result could have been proved even as early as the time of Gauss. But I think it may be new - I've seen no reference in Sylvester. Does anyone have a historical reference?
**Question 2**
When $p$ is $61$, $67$ or $73$, then $3p$ is indeed the sum of two rational cubes. So it's natural to ask whether the converse to my theorem holds. When $p$ is represented by $(1,9,81)$ is it true that $3p$ and $3p^2$ are sums of two rational cubes? This has something of the flavor of Sylvester's question about representing primes congruent to $4$, $7$ or $8 \bmod{9}$ (and their squares) as sums of two rational cubes. Before the work of Elkies, Dasgupta-Voight and Kriz which (I hope) have settled the question there was $3$-descent work by Selmer that "reduced" things to the parity conjecture. What does the approach of Selmer reveal here? And has anyone looked at the elliptic curves explicitly, and tried to use the Heegner point approach to produce solutions to the equations?
| https://mathoverflow.net/users/6214 | Writing $3p$ when $p \equiv 1 \pmod{3}$ as a sum of two rational cubes. Is this result new? And what about its converse? | For question 1.
The condition $p$ is represented by $(1,9,81)$ or equivalently by $(1,1,61)$ is equivalent to the condition that $p\equiv 1\mod 3$ and $3 \mod p$ is not a cube (this is exercise 9.10 in Cox's book 'Primes of the forms...').
And Satge's paper 'Groupes de Selmer et corps cubiques' already proved under this condition, neither $3p$ nor $3p^2$ is a sum of two rational cubes.
For question 2.
I checked using Sagemath that for $p=61,67,73,103,151,193,271$ the rank of the related elliptic curve of $3p$ is 2 while for $p=307$ the rank is 0. So the converse is not true. Similar thing also occurs for $3p^2$. However for $p\equiv 2,5\mod 9$, it is proved that $3p$ and $3p^2$ are both sums of two cubes by constructing the Heegner point in the paper <https://link.springer.com/content/pdf/10.1007/s00208-022-02370-3.pdf>
| 6 | https://mathoverflow.net/users/144225 | 450175 | 181,117 |
https://mathoverflow.net/questions/449839 | 7 | $\DeclareMathOperator\dom{dom}$Sorry to bother the community again with these type of questions about power series, I am ready to delete the question if it is not suitable.
**Definition:** I say a function $h$ is $(m,n)$-representable by power series iff $0\in \dom(h)$, $h(0)=0$, and $h$ is a restriction (to a smaller domain) of some power series (centered at $0$) that takes as input $m$ real variables and outputs a vector in $\mathbb{R}^n$. In other words, $0\in \dom(h)\subseteq B\_r(0)$ (where $r$ is the radius of convergence of the power series), and for every $x\in \dom(h)$ we have that $h(x)$ equals the evaluation of the power series at the point $x$.
**Question:** Let $f$ be $(a,b)$-representable by power series. Let $g$ be $(b,c)$-representable by power series. It is given that $\dom(g\circ f)$ (i.e. $\{x\in \dom(f)\mid f(x)\in \dom(g)\}$) is path connected. It is also given that $\dom(f),\dom(g)$ are path connected. Must $g\circ f$ be $(a,c)$ representable by power series?
I edited my question to include the hypothesis of path connectivity of domains of $f,g$
| https://mathoverflow.net/users/32135 | Composition of power series is power series? | Maybe I misunderstood the question, is this a counterexample?
Let $c>1$, put $f(x) = \sin x$ and $g(x) = \tfrac{x}{c^2+x^2}$. Then $f$ is representable by a power series on $\mathbb{R}$, $g$ is representable by a power series on $(-c,c)$ and $f(\mathbb{R}) \subset (-c,c)$.
However, the meromorphic function $g \circ f$ has a pole at $i \sinh^{-1}(c)$, so the power series of $g \circ f$ is only convergent on $(- \sinh^{-1}(c), \sinh^{-1}(c))$, not on all of $\mathbb{R}$.
| 9 | https://mathoverflow.net/users/297 | 450185 | 181,120 |
https://mathoverflow.net/questions/450093 | 2 | One of the problems that has come up during my research concerns $K\_4$-simple groups (simple groups with $4$ prime divisors). The only (potentially) infinite family of groups satisfying this condition is $PSL(2,q)$, and I was interested in exactly what values of $q$ satisfied this condition. After writing a quick Python program, I got this output (where each ordered pair denotes $(q, |PSL(2,q)|)$:
$$[(11, 660), (13, 1092), (16, 4080), (19, 3420), (23, 6072), (25, 7800), (27, 9828), (31, 14880), (32, 32736), (37, 25308), (47, 51888), (49, 58800), (53, 74412), (73, 194472), (81, 265680), (97, 456288), (107, 612468), (127, 1024128), (128, 2097024), (163, 2165292), (193, 3594432), (243, 7174332), (257, 8487168), (383, 28090752), (487, 57750408), (577, 96049728), (863, 321367392), (1153, 766403712), (2187, 5230175508), (2593, 8717209632), (2917, 12410213148), (4373, 41812719372), (8192, 549755805696), (8747, 334616519988), (131072, 2251799813554176), ...$$
There seems to be a huge gap in $q$ after $8747$, going immediately to $2^{17} = 131072$ and then skipping another couple hundred thounsand or so. The condition on $q$ is that it is a prime power satisfying $q(q^2-1) = 2^{\alpha\_1}3^{\alpha\_2}p^{\alpha\_3}r^{\alpha\_4}$ for primes $r > p > 3$. Yet, I found it quite surprising that these gaps exist. Does anyone know why? Are there arbitrarily large gaps in values of $q$?
This is more number-theoretic than group-theoretic, but I am definitely interested in that aspect as well. Thanks in advance!
| https://mathoverflow.net/users/507796 | Sparsity of q in groups PSL(2,q) that are K_4-simple | Let me expand my earlier comment to a partial answer of "why sparsity?". It is impossible for $|{\rm PSL}(2,p^{m})|$ to have four or fewer different prime factors when
$p$ is an odd prime which is neither Fermat nor Mersenne and $m > 1$ is an integer. In fact, if $m >1$ is not a power of $2$, we will see that
whenever $p >3$ is prime, then $|{\rm PSL}(2,p^{m})|$ has at least $5$ different prime divisors.
(These two facts "explain" why the non-squarefree $q$ appearing in your Python output are all either powers of $2$ or powers of Fermat or Mersenne primes, and why we only have $q$ of the form $p^{2n+1}$ ( for positive $n$) appearing for $p = 2$ or $p =3).$
If $m$ is itself even, then $p^{2m}-1$ is divisible by $p^{4}-1$. It is easy to check that $\frac{p-1}{2}, \frac{p+1}{2}$ and $\frac{p^{2}+1}{2}$ are pairwise coprime. Also $\frac{p^{2}+1}{2}$ is odd, while one of $\frac{p \pm 1}{2}$ is even, and the other is odd. Hence if the odd prime $p$ is neither a Fermat prime nor a Mersenne prime, then $p^{4}-1$ is even, and has at least three different odd prime factors, so has $4$ or more different prime factors (none of which is $p$). Thus
$|{\rm PSL}(2,p^{2n})|$ has $5$ or more different prime factors whenever $p$ is an odd prime which is neither Fermat nor Mersenne, and $n$ is any positive integer.
If $m$ is not a power of $2$, then $m$ is divisible by some odd prime $r$. We note below that that $p^{2r}-1$ has four or more prime factors whenever $r$ is an odd prime and $p$ is a prime greater than $3$.
For of the four integers $\frac{p-1}{2}, \frac{p+1}{2}, \frac{p^{r}-1}{p-1}$ and $\frac{p^{r}+1}{p+1},$ exactly one is even. Since $p >3$, the product $\frac{p+1}{2}\frac{p-1}{2}$ has at least two different prime factors, since $p-1$ and $p+1$ can't both be powers of $2$ for any prime $p$ greater than $3$.
Note also that Zsygmondy's Lemma tells us that there are primes $s$ and $t$ such that $p+s\mathbb{Z}$ has multiplicative order $r$ in the units of $(\mathbb{Z}/s\mathbb{Z})^{\times}$ and $p+t\mathbb{Z}$ has multiplicative order $2r$ in the units of $(\mathbb{Z}/t\mathbb{Z})^{\times}$. Then $s$ and $t$ are different (and both odd) and neither of them divides $p^{2}-1.$ Hence $p^{2r}-1$
has at least $4$ different prime divisors, and $|{\rm PSL}(2,p^{m})|$ has at least five prime divisors.
| 5 | https://mathoverflow.net/users/14450 | 450187 | 181,121 |
https://mathoverflow.net/questions/450182 | 6 | I put forward a hypothesis in number theory, it is as follows.$ \sigma\_1(n)=\sigma\_1(m)=p$, where $\sigma\_1$ is the divisor sum function, $n,m\in \mathbb N$, and $p$ is prime. I recently noticed and assumed that the pattern $ \sigma\_1(16)=\sigma\_1(25)=31$ is the only one. And what do you think about this?
| https://mathoverflow.net/users/508058 | Are there infinite numbers of the form $\sigma_1(n)=\sigma_1(m)=p$, or is there only one? | As mentioned in the comments, it suffices to assume that $m,n$ are prime powers. Then we are looking at an equation of the form
$$\displaystyle u^a + \cdots + u + 1 = v^b + \cdots + v + 1.$$
The expressions on either side are cyclotomic polynomials. If either one is reducible (over $\mathbb{Z}$), then it is exceedingly unlikely that they will evaluate to a prime. Thus we may assume, without losing very much, that both of the cyclotomic polynomials are irreducible. This then requires both $a,b$ to be of the form $p-1$ for some prime $p$.
Note that $16 = 2^4 = 2^{5-1}$ and $25 = 5^2 = 5^{3-1}$ are both of this form.
We are then looking for integer points on a family of plane curves of the form given by
$$\displaystyle x^{p-1} + \cdots + x = y^{q-1} + \cdots + y, x,y \in \mathbb{P}$$
where $\mathbb{P}$ denotes the set of prime numbers.
Results of this type, with one variable exponent, are known in the literature; see for example the papers of Bennett and Dahmen ([Klein forms and the generalized superelliptic equation](https://%20%20%20https://annals.math.princeton.edu/2013/177-1/p04)) and Bennett and Siksek ([Rational points on Erdős–Selfridge superelliptic curves](https://www.cambridge.org/core/journals/compositio-mathematica/article/rational-points-on-erdosselfridge-superelliptic-curves/1C53006D003585C6F4746BE4E0162836)). However, these do not rule out the possible existence of infinitely many solutions with two variable exponents.
It would be interesting to see if the family of hyperelliptic curves corresponding to $q = 3$ in the above family can be treated using the methods in the two papers above. That is, are there only finitely many integers $x,y,p$ (with $p$ variable) such that
$$\displaystyle y^2 + y = x^{p-1} + \cdots + x?$$
| 5 | https://mathoverflow.net/users/10898 | 450188 | 181,122 |
https://mathoverflow.net/questions/450189 | 1 | This is soft question. I decided to ask my question on MathOverflow rather than on academia StackExchange because I believe that the community here is more equipped to answer the question.
The [Graduate Journal of Mathematics](https://gradmath.org/) mentions in its mission-statement that it takes inspiration from a similar journal that was discontinued in 2000: Le Journal des Elèves de l’Ecole Normale Supérieure de Lyon. I have searched about the later journal "Le Journal des Elèves de l’Ecole Normale Supérieure de Lyon" in google but found no active webpage for it.
* Can someone tell me about the journal "Le Journal des Elèves de l’Ecole Normale Supérieure de Lyon" ?
* Why it was discontinued ?
There are there other journals which were also discontinued for publication. For example, [LMS Journal of Computation and Mathematics](https://www.lms.ac.uk/publications/jcm) was discontinued in 2015.
>
> So what causes some mathematical journals to discontinue?
>
>
>
My main motivation for the above questions is the following:
>
> Suppose someone publish an article in a journal and later the journal get closed. How would people evaluate the work from a journal which is closed ?
>
>
>
| https://mathoverflow.net/users/493164 | What causes some mathematical journals to discontinue? | At <https://futurelms.wordpress.com/2016/01/14/save-the-lms-journal-of-computation-and-mathematics/> you will find discussion of the decision to close the LMS Journal of Computation and Mathematics. My knowledge is based solely on a brief scan of the documents linked from that page. It seems that some people felt that the papers in the journal were not of high quality and were seldom read. The decision to close was narrowly confirmed by a vote at a special general meeting of the LMS.
| 5 | https://mathoverflow.net/users/10366 | 450194 | 181,125 |
https://mathoverflow.net/questions/450200 | 0 | Let $F$ be the set of all integers $n>1$ such that in the [Fibonacci sequence](https://en.wikipedia.org/wiki/Fibonacci_sequence) modulo $n$, the value $0$ occurs infinitely often. What is the value of $\lim\sup\_{n\to\infty}\frac{|F\cap\{0,\ldots,n\}|}{n+1}$?
| https://mathoverflow.net/users/8628 | Density of "Fibonacci friends" | The value $0$ always occurs infinitely often regardless of the value $n>1$. To see this, consider the mapping $L\_n:\mathbb{Z}\_n^2\rightarrow\mathbb{Z}\_n^2$ defined by $L\_n(x,y)=(y,x+y)$. Then the mapping $L\_n$ is invertible. Then if we set $F\_{0,n}=[0]\_n,F\_{1,n}=[1]\_n$ and $F\_{i+2,n}=F\_{i+1,n}+F\_{i,n}$, then $F\_{i,n}$ is the $i$-th Fibonacci number modulo $n$. On the other hand,
$L\_n(F\_{i,n},F\_{i+1,n})=(F\_{i+1,n},F\_{i+2,n})$, so $L\_n^m(F\_{i,n},F\_{i+1,n})=(F\_{i+m,n},F\_{i+m+1,n})$ for all $m\geq 0$. Since $L\_n$ is invertible, there must be some $m>0$ where $([0]\_n,[1]\_n)=L^m\_n([0]\_n,[1]\_n)=L^m\_n(F\_{0,n},F\_{1,n})=(F\_{m,n},F\_{m+1,n})$, so $F\_{m,n}=[0]\_n$.
This is in contrast to the Foobonacci sequence where the function $M\_n$ is not injective. Define the Foobonacci sequence $FO\_{1,n}=1,FO\_{2,n}=1,FO\_{i+2,n}=FO\_{i+1,n}\*\_nFO\_{i,n}$ where $\*\_n$ is the self-distributive operation in the $n$-th Laver table. For example, in Hexadecimal ($104\_{\text{hex}}=260\_{\text{dec}}$), we have (according to my artificial intelligence calculations but without proof) $(FO\_{1,104},\dots,FO\_{7,104})=(1,1,2,3, F00...008, \mathbf{F00...00B}, 100...00)$, so we call this sequence the Foobonacci sequence.
The mapping
$M\_n:A\_n^2\rightarrow A\_n^2$ defined by $M\_n(x,y)=(x\*\_ny,x)$ (where $A\_n$ denotes the $n$-th Laver table) is not invertible, and for all $n>1$, there is no $t>2$ where $FO\_{n,t}=1$, but for all $n>1$, we eventually have
$FO\_{n,2t+1}=2^n$ for sufficiently large $t$.
| 10 | https://mathoverflow.net/users/22277 | 450201 | 181,128 |
https://mathoverflow.net/questions/450030 | 5 | Denote by $D(r)$ the disc at the origin of radius $r>1$. Denote by $P\_m$ the set of polynomials of degree $m$. Since $P\_m$ is finite dimensional, there is a constant $C(m,r)$ such that
$$
\|p\|\_{L^{\infty}(D(r))}
\leq
C(m,r)
\|p\|\_{L^1(-1,1)}
$$
for all $p \in P\_m$.
I'd like some estimates for $C(m,r)$, especially the growth in $m$?
| https://mathoverflow.net/users/73890 | The constant of the reverse Hölder inequality for polynomials | I presume that $D(r)$ refers to a disk in the complex plane, in which case the maximum value of a polynomial (like any analytic function) is achieved on the boundary of $D(r)$. If $\|p\|\_{L^\infty(D(r))} \le C \|p\|\_{L^1(-1,1)}$, it means that the ball $\|p\|\_{L^1(-1,1)} \le 1/C$ fits entirely into the ball $\|p\|\_{L^\infty(D(r))}\le 1$ and by convexity it is enough to check that condition for the extreme points of the $L^1(-1,1)$-ball. So bounding $C$ from above is the same as bounding $\|p\|\_{L^1(-1,1)}$ from below with $\|p\|\_{L^\infty(D(r))} = 1$.
By a result of
>
> *Garkavi, A. L.*, The unit sphere of the space of polynomials with integral metric, [Mat. Zametki 1, 299-304 (1967)](http://mi.mathnet.ru/mzm9417). [ZBL0156.36603](https://zbmath.org/?q=an:0156.36603), [MR209815](https://mathscinet.ams.org/mathscinet/article?mr=209815).
>
>
>
the extreme points of the $L^1(-1,1)$-ball for polynomials of degree $m$ are those that have exactly $m$ roots in the interval $[-1,1]$ counting multiplicity. We can represent such a polynomial with $\|p\|\_{L^\infty(D(r))} = 1$ as
$$
p(x) = \frac{1}{N} (x-a\_1) \cdots (x-a\_m) ,
$$
where the normalization factor $N=(z-a\_1)\cdots(z-a\_m)$ for some complex $|z|=1$ and all the roots $a\_i \in [-1,1]$, while
$$
\|p\|\_{L^1(-1,1)} = \frac{2}{N} \int\_{-1}^1 \prod\_{i=1}^m |x-a\_i| \frac{dx}{2} \ge \frac{1}{C} .
$$
So to bound $1/C$ from below, it is enough to bound each factor from below.
For $N$, it is easy to see that $1/N \ge 1/(r+1)^m$, realized by $p(x) = (x+1)^n$ for instance.
For the integral, using the convexity of $\exp(x)$, we get
$$\begin{aligned}
\int\_{-1}^1 \prod\_{i=1}^m |x-a\_i| \frac{dx}{2}
&\ge \exp\left(\int\_{-1}^1 \sum\_{i=1}^m \ln|x-a\_i| \frac{dx}{2}\right) \\
&= \exp\left(\frac{1}{2} \sum\_{i=1}^m \left[(1-a\_i)\ln|(1-a\_i)/e|+(1+a\_i)\ln|(1+a\_i)/e|\right]\right) \\
&\ge \exp(-m) ,
\end{aligned}$$
where the last inequality is saturated when all $a\_i=0$.
Putting together the pieces, we find that
$$
C \le 2 e^m (r+1)^m .
$$
This ~~improves the bound from~~ is an alternative bound to [the other answer](https://mathoverflow.net/a/450139). It gets rid of the the $(m+1)$ factor, but is is generally worse because the base exponent $e(r+1) > (2r+1)$.
| 4 | https://mathoverflow.net/users/2622 | 450211 | 181,131 |
https://mathoverflow.net/questions/450228 | 1 | Let $f, g \in \mathbb{Z}[x]$ be coprime polynomials.
I am interested in an upper bound for
$$
N(B) = \# \{ x \in [-B, B] \cap \mathbb{Z}: f(x)\mid g(x) \}.
$$
I assume there must be something known about this quantity... If someone could provide me a reference it would be appreciated. Thank you
ps I assume $\deg f > 0$.
| https://mathoverflow.net/users/84272 | Number of integers $x \leq B$ such that $f(x)\mid g(x)$ for coprime polynomials $f,g$ | The assumption that $f(X)$ and $g(X)$ are relatively prime means that there is
a positive integer $R\_{f,g}=\operatorname{Resultant}(f,g)$ and polynomials $a(X)$ and $b(X)$ in $\mathbb Z[X]$ so that
$$ a(X)f(X) + b(X)g(X) = R\_{f,g}. $$
Hence if $x\in\mathbb Z$ satisfies $f(x)\mid g(x)$, then $f(x)\mid R$. Since $\bigl|f(x)\bigr|\to\infty$ as $|x|\to\infty$, this shows that your quantity $N(B)$ is bounded as $B\to\infty$. A relatively trivial bound is simply
$$ N(B) \le \#\Bigl\{x\in\mathbb Z : \bigl| f(x) \bigr| \le R\_{f,g} \Bigr\}. $$
One can bound $R\_{f,g}$ in terms of the coefficients of $f$ and $g$ and their degrees, if you want a more explicit bound.
| 6 | https://mathoverflow.net/users/11926 | 450229 | 181,136 |
https://mathoverflow.net/questions/450219 | 2 | Let $f\geq 0$ be a Lipschitz function and let $(L\_t)\_{t\geq 0}$ be an $\alpha$-stable Lévy process ($0<\alpha<2$, possibly multivariate). Consider the process given by $$dX\_t=-\nabla f(X\_t)dt+\sigma dL\_t$$ where $\sigma>0$ is a constant. This SDE is reminiscient of Langevin Dynamics, where we usually let the process be driven by Brownian Motion instead of a Lévy process.
I am interested in seeing which results from the "Brownian setting" extend to the setting with $\alpha$-stable Lévy processes as described above.
Most importantly, my main question is: **Does $X\_t$ admit a stationary distribution?**
When the process is driven by Brownian Motion, [it is known that there exists a stationary distribution](https://math.stackexchange.com/questions/4200043/langevin-equation-and-convergence-to-stationary-solutions-free-energy-sde-fpe), for which we can also find a closed form expression for the density.
Example: Ornstein-Uhlenbeck Process
===================================
If we consider the case $f(x)=\frac12 |x|^2$ we get the Ornstein-Uhlenbeck process driven by the Lévy process $L$. There it is known, see e.g. [Topics in Infinitely Divisible Distributions and Lévy Processes](https://link.springer.com/book/10.1007/978-3-030-22700-5) Theorem 2.17, that the process admits a stationary distribution.
| https://mathoverflow.net/users/498406 | Stationary Distribution of Langevin Dynamics driven by Lévy Process | There are several generalizations of the Brownian case results to general L'evy processes. For instance, under a classical log-concave assumption on $U$ we have a convergence to equilibrium in Wassertein distance to a unique stationary distribution. More precisely, consider the following SDE
$$
dX^x\_t=-\nabla U(X^x\_t) dt +dN\_t, \quad X^x\_0=x \in \mathbb R^n
$$
where $N$ is any L'evy process such that $\mathbb{E} \left( \sup\_{t\in [0,T]} | N\_t |^p \right) <+\infty$ for some $p>1$. For stable processes one therefore asks that $\alpha>1$. Denote $P\_t$ the semigroup of $X\_t$.
**Theorem:** *Assume that there exists $a>0$ such that $\nabla^2 U \ge a$ (uniformly
in the sense of quadratic forms). Then, there exists a unique
probability measure $\mu$ in the Wasserstein space
$\mathcal{P}\_p(\mathbb R^n)$ such that for every $t \ge 0$, $\mu P\_t =
\mu$. Moreover, for every $t \ge 0$, and $\nu \in
\mathcal{P}\_p(\mathbb R^n)$ one has,
$$
W\_p ( \nu P\_t, \mu ) \le e^{-at} W\_p ( \nu , \mu ).
$$
Therefore $X\_t$ converges exponentially fast to the invariant distribution in the Wasserstein distance $W\_p$.*
This can be proved using similar arguments as in Theorem 4.9 in the paper [Transport inequalities for Markov kernels and their applications](https://arxiv.org/pdf/2004.02050.pdf) which treated the case $p=2$.
Here are the main steps.
Let $J\_t=\frac{\partial X\_t^x}{\partial x}$ be the first variation process. Let $f$ be a $C^1$ and bounded Lipschitz function.
Since $P\_tf(x)=\mathbb{E}( f(X\_t^x))$, by the chain rule we have
$$
\nabla P\_t f (x)=\mathbb{E}\left( J\_t^\* \nabla f(X\_t^x)\right).
$$
Therefore, by Holder inequality,
$$
| \nabla P\_t f (x) | \le \mathbb{E}\left( | J\_t^\*|^p\right)^{1/p}
\mathbb{E}\left( | \nabla f(X\_t^x) |^q\right)^{1/q}.
$$
where $q$ is the conjugate exponent of $p$.
Observe that
$$
dJ\_t=-\nabla^2 U(X^x\_t) J\_t dt, \quad J\_0=\mathbf{Id}\_{\mathbb R^n}.
$$
From the assumption $\nabla^2 U \ge a$ this yields
$$
| J\_t^\*| \le e^{-at }.
$$
One concludes $\mathbb{E}\left( | J\_t^\*|^p\right) \le e^{-pat }$ and therefore
$$
| \nabla P\_t f (x) | \le e^{-at} P\_t (| \nabla f |^q)(x)^{1/q}.
$$
By Kuwada duality, this yields that for every $\nu\_0,\nu\_1 \in \mathcal{P}\_p(\mathbb
R^n)$,
$$
W\_p ( \nu\_0 P\_t, \nu\_1 P\_t ) \le e^{-at} W\_p ( \nu\_0 , \nu\_1 ).
$$
A fixed point argument in the Wasserstein space allows then to conclude.
| 3 | https://mathoverflow.net/users/48356 | 450232 | 181,139 |
https://mathoverflow.net/questions/450222 | 6 | [A previous question on the categorical nature of ultraproducts](https://mathoverflow.net/questions/11261/is-the-ultraproduct-concept-fundamentally-category-theoretic) had great answers, mostly categorically characterizing ultraproducts in the category of $L$-structures and *homomorphisms* for a fixed signature $L$. This is understandable, given one of the most important use of ultraproducts is to give a syntax-free characterization of elementary equivalence.
Here I would like to ask sort of its opposite: can one characterize ultraproducts in the category of $L$-structures and *elementary embeddings*? (This is related to the question of whether or not ultraproducts are a genuinely model-theoretic notion.)
This sounds like what experts of AECs study, but I was not able to find relevant literature.
*Addendum*: Answers focusing on ultra\_powers\_ are also welcome. (In fact, that was my original motivation.)
| https://mathoverflow.net/users/310424 | Ultraproducts in the category of structures and elementary embeddings | Since you asked about ultraproducts, and not ultrapowers, let me argue that the answer must be negative. The reason is that the category of $L$-structures under elementary embeddings is partitioned into disconnected components by the theories of those structures, since elementary embeddings must preserve the theory of the structure and so there can be no morphisms in this category that connect structures with different theories.
But I may form the ultraproduct $\prod M\_i/U$ of structures $M\_i$ with all different theories, in such a way that the ultraproduct itself had a different theory than each of them. So there will be no morphisms at all that relate any of the $M\_i$'s to each other or to the ultraproduct. There will be no diagrams in which the $M\_i$'s participate along with the ultraproduct, except diagrams that are disconnected into components in the same way as the category is by the theories of the models. For example, there can be no universal property in this category relating the $M\_i$'s to the ultraproduct or to models connected by morphisms to the ultrapower or each other, since there are no morphisms at all relating these structures to any other structures in common—they simply live on different islands. In these circumstances, it seems impossible to find a category-theoretic characterization of the ultraproduct in the category of elementary embeddings.
| 8 | https://mathoverflow.net/users/1946 | 450237 | 181,141 |
https://mathoverflow.net/questions/450225 | 1 | Last October, I learned from [Benjamin Steinberg's answer](https://mathoverflow.net/questions/432657/structure-theorem-for-a-class-of-idempotent-monoids-where-xy-x-or-xy-y/432691#432691) to another question of mine that a semigroup $S$ is called *breakable* if $xy \in \{x, y\}$ for all $x, y \in S$. Let's now say that $S$ is an *almost breakable* semigroup if $xy \in \{x, y\}$ or $yx \in \{x, y\}$ for all $x, y \in S$ (by the way, is there a more standard name for these objects?). Every breakable semigroup is almost breakable (that's obvious), while the converse need not be true. Also, an almost breakable semigroup is idempotent (whence the set of its units is either empty or trivial).
>
> **Question.** Assume $S$ is an almost breakable semigroup and call a pair $(x,y)$ of elements of $S$ *irregular* if $xy \notin \{x, y\}$. Do there exist an integer $n \ge 1$ and $n$ elements $x\_1, \ldots, x\_n \in S$ such that the pair $(x\_i, x\_{i+1})$ is irregular for each $i \in [\![1, n ]\!]$, where $x\_{n+1} := x\_1$? If so, we will say that the $n$-tuple $(x\_1, \ldots, x\_n)$ is an $n$-cycle (in $S$) and $S$ contains an $n$-cycle.
>
>
>
Of course, an almost breakable semigroup contains neither 1- nor 2-cycles. Moreover, experiments in Maple have shown that the answer is no for every almost brekable semigroup of order $\le 6$.
| https://mathoverflow.net/users/16537 | Cycles in almost breakable semigroups | Here is a self-contained version of the argument that there are no cycles, avoiding using the structure of bands. Suppose that $S$ is almost breakable.
Claim 1. If $SxS=SyS$, then both $xy,yx\in \{x,y\}$ and $xyx=x$, $yxy=y$.
Pf. Without loss of generality, assume that $yx=y$ (the other cases follow from renaming or working with the opposite semigroup, which is also almost breakable). Trivially $yxy=y$. Write $x=uyz$. Then $xz=x$ and $xyx=xyxz=xyz=uyzyz=uyz=x$. Thus $x=xyx=xy$.
Claim 2. If $x,y\in S$, then $SxS\subseteq SyS$ or vice versa.
Pf. Either $xy\in \{x,y\}$ or $yx\in \{x,y\}$.
Suppose now that $(x\_1,\ldots, x\_n)$ is a cycle. Without loss of generality, we may assume that $Sx\_1S$ is maximal among the $Sx\_iS$. Assume inductively that $Sx\_1\supsetneq Sx\_2\supsetneq \cdots\supsetneq Sx\_k$ with $1\leq k<n$. Then $x\_{k+1}x\_k\in \{x\_k,x\_{k+1}\}$. If $x\_{k+1}x\_k=x\_k$, then $x\_kx\_{k+1}x\_k=x\_k$ and $x\_kx\_{k-1}x\_k=x\_k$ (the latter since $x\_{k-1}\in Sx\_k$). Thus $Sx\_kx\_{k+1}S=Sx\_kS=Sx\_{k-1}x\_kS$. So by Claim 1, $x\_kx\_{k+1}\cdot x\_{k-1}x\_k\in \{x\_kx\_{k+1},x\_{k-1}x\_k\}$. But then $$x\_kx\_{k+1}\cdot x\_{k-1}x\_k=x\_k(x\_kx\_{k+1}\cdot x\_{k-1}x\_k)x\_k\in x\_k\{x\_kx\_{k+1},x\_{k-1}x\_k\}x\_k=\{x\_kx\_{k+1}x\_k,x\_kx\_{k-1}x\_k\}=\{x\_k\}$$ by another application of Claim 1 since $Sx\_kS=Sx\_kx\_{k+1}S=Sx\_{k-1}x\_kS$. Thus $x\_k\in \{x\_{k-1}x\_k,x\_kx\_{k+1}\}$, a contradiction to the definition of a cycle. Thus $x\_{k+1}x\_k=x\_{k+1}$ and so $Sx\_k\supseteq Sx\_{k+1}$. These left ideals cannot be equal by Claim 1, and so $Sx\_k\supsetneq Sx\_{k+1}$.
Thus we have $Sx\_1\supsetneq Sx\_2\supsetneq\cdots\supsetneq Sx\_n$. But then $x\_nx\_1 = x\_nx\_{n-1}x\_1=\cdots=x\_nx\_{n-1}\cdots x\_2x\_1=x\_nx\_{n-1}\cdots x\_2=\cdots =x\_nx\_{n-1}=x\_n$, a contradiction to $(x\_n,x\_1)$ being irregular.
**Original answer.** It seems there are no cycles. First note that if $S$ is almost breakable, then the principal ideals in $S$ form a chain, for if $x,y\in S$, then either $xy\in \{x,y\}$ or $yx\in \{x,y\}$ and so either $SxS\subseteq SyS$ or conversely. Two elements that generate the same principal ideal are called $\mathscr J$-equivalent.
I will use a little bit of structural semigroup theory of bands. Maybe this can be avoided. Each $\mathscr J$-class of a band is a rectangular band (satisfies the identity $xyx=x$). If a rectangular band is not a left or right zero semigroup, then it contains elements $x,y$ with $xy\notin \{x,y\}$ and $yx\notin \{x,y\}$. Thus each $\mathscr J$-class of an almost breakable semigroup is a left zero semigroup or a right zero semigroup.
Let $(x\_1,\ldots, x\_n)$ be a cycle and assume without loss of generality that $x\_1$ generates the largest principal ideal amongst these elements. Then since $x\_2x\_1\in \{x\_1,x\_2\}$, we must have $x\_2x\_1=x\_2$ by choice of $x\_1$ (for if $x\_2x\_1=x\_1$, then these elements are $\mathscr J$-equivalent and by the above remarks form a right zero semigroup, but then $x\_1x\_2=x\_2\in \{x\_1,x\_2\}$). Assume inductively that $x\_ix\_{i-1}=x\_i$ for $1\leq i\leq k<n$. Notice that $x\_{k-1}x\_k\neq x\_k$, but $x\_kx\_{k-1}x\_k = x\_k$ and so $x\_{k-1}x\_k$ and $x\_k$ are in the same $\mathscr J$-class and this $\mathscr J$-class is a left zero semigroup. Suppose that $x\_{k+1}x\_k=x\_k$. Then $x\_kx\_{k+1}\neq x\_k$ and $x\_kx\_{k+1}x\_k=x\_k$. It follows that $x\_k$ is $\mathscr J$-equivalent to $x\_kx\_{k+1}$ and they generate a right zero semigroup. But we already saw that the $\mathscr J$-class of $x\_k$ is a left zero semigroup. So $x\_{k+1}x\_k=x\_{k+1}$. Thus we have that $x\_{k+1}x\_k=x\_{k+1}$ for all $1\leq k\leq n-1$.
It follows that $x\_nx\_{n-1}\cdots (x\_2x\_1)=x\_n\cdots (x\_3x\_2)=\cdots=x\_nx\_{n-1}=x\_n$. Therefore, $x\_nx\_1=x\_n$, contradicting that $(x\_n,x\_1)$ is irregular.
| 1 | https://mathoverflow.net/users/15934 | 450244 | 181,142 |
https://mathoverflow.net/questions/450243 | 2 | Let G be an undirected, simple graph containing distinct vertices x and y. Let P,Q,R be three distinct paths in G from x to y. We can assume the graph G is only those paths (any vertex in G is in one of P,Q, and R and same with any edge in G).
Assume there is no vertex (other than x and y) such that P,Q, and R contain that vertex. Does this guarantee us a cycle which contains both x and y? (The cycle does not have to be simply two of P,Q, and R but may be created by two x-y paths which are made from the union of P,Q, and R).
My feeling is it does, but I am having trouble proving it rigorously.
Mostly I have tried to solve it by starting with the three paths then adjusting the paths so that P and R never touch. Start at x and work towards y, when two paths, say P and Q, intersect at a vertex v we can swap the subpath from v to y of P with the v-y subpath of the Q. But this is tough because it may affect a vertex on P that was closer to x than v.
Another thought was more combinatorial: If we have two x-y paths which intersect each other at a vertex which is not x or y then they do not create a cycle which contains both x and y. So if we start with three paths that have some intersections among them, do these intersections make more paths which would also need to intersect each other for there to be no cycle containing x and y? Seems like the number of paths made by intersections would grow larger than the number of intersections necessary to stop any two x-y paths from forming a cycle, but I am not sure how to concretely show that to be true.
Thanks in advance!
| https://mathoverflow.net/users/114995 | If you have three paths from vertex x to vertex y, when are you guaranteed a cycle which contains both x and y? | Yes, by Menger theorem. There exists either a vertex $z\notin \{x, y\}$ after removing which there is no path from $x$ to $y$ remained, or two disjoint paths from $x$ to $y$.
| 1 | https://mathoverflow.net/users/4312 | 450250 | 181,144 |
https://mathoverflow.net/questions/450241 | 13 | This came up in the comments to [an answer of Joel's](https://mathoverflow.net/questions/450222/ultraproducts-in-the-category-of-structures-and-elementary-embeddings). Suppose $\mathcal{M}\_i$ ($i\in I$) are elementarily equivalent structures in the same fixed signature and $\mathcal{U}$ is an ultrafilter on $I$. Must some $\mathcal{M}\_i$ elementarily embed into $\prod\_{i\in I}\mathcal{M}\_i/\mathcal{U}$?
I'm happy to add further restrictions to make this question easier, such as setting $I=\omega$ or requiring the language of the structures to be finite.
| https://mathoverflow.net/users/8133 | Can ultraproducts avoid all "factor structures"? | Here is another example, which is inspired by James's answer, but I find this one a little simpler.
Let $T$ be the theory of an equivalence relation $\sim$ with infinitely many classes, all infinite, plus countably many constants $c\_0,c\_1,c\_2,\ldots$, taken from different equivalence classes. This is a complete theory, which can be seen by elimination of quantifiers.
Let $M\_n$ be a model with all equivalence classes countable, except the equivalence class of $c\_n$, which we take of size $\mathfrak{c}^+$. Let $M=\prod\_n M\_n/\mu$ be the ultraproduct of the $M\_n$ by a nonprincipal ultrafilter $\mu$ on $\omega$.
The size of the equivalence class of $c\_n$ in $M$ is determined by functions from $\omega$ to the corresponding $[c\_n]^{M\_i}$, which are all countable except for $M\_n$. Since no one factor is relevant for the nature of $M$, it follows that $[c\_n]^M$ has size continuum. Thus, $M\_n$ does not embed in $M$.
So none of the factors embed into the ultrapower, as desired.
**Optimal size.** For the reasons mentioned in [Emil's comment below](https://mathoverflow.net/questions/450241/can-ultraproducts-avoid-all-factor-structures#comment1163705_450257), if CH holds this size $\mathfrak{c}^+$ is the best possible. I am unsure whether there could be a smaller counterexample if CH fails.
**Analyzing the ultrapower further.** I find it interesting to notice a few things about the ultrapower structure — in fact we can describe it exactly. It will have a bunch of other equivalence classes not containing any of the constants $c\_n$. Any function $f$ with $f(n)$ not almost always chosen from the class of the same constant $c\_k$ will represent a new equivalence class in the ultrapower. These will all have size continuum also (and there are continuum many of them), except for the one equivalence class arising from functions $f$ with almost always $f(n)\in[c\_n]^{M\_n}$, choosing from the big class at these coordinates. This one equivalence class in the ultrapower will have size $(\mathfrak{c}^+)^\omega=\mathfrak{c}^+$.
So the ultraproduct $M$ consists of continuum many equivalence classes of size continuum, with countably many of them occupied by constants $c\_n$, plus one more equivalence class of size $\mathfrak{c}^+$.
**Generalizing to arbitrary ultrafilters.** The construction generalizes to any uniform ultrafilter $\mu$ on any set $I$. We have an equivalence relation with infinitely many classes, all infinite, plus constants $c\_i$ for each $i\in I$, and inequivalent. Let $M\_i$ be a model with all equivalence classes countable, except for $[c\_i]^{M\_i}$, which should have size larger than $\omega^I$. The ultrapower $M=\prod\_i M\_i/\mu$ will have $[c\_i]^M$ of size at most $\omega^I$, and so no $M\_i$ will embed into $M$.
| 19 | https://mathoverflow.net/users/1946 | 450257 | 181,146 |
https://mathoverflow.net/questions/450163 | 3 | Kolmogorov tightness criterion says that if $X\_N$ is a sequence of continuous process with $X\_N(0)=0$ and $E[[X\_N(t)-X\_N(s)|^p]\leq C\_p |t-s|^{1+\beta}$ then for all $\gamma\in (0,\beta/p)$ we have that the laws of $X\_N$ are tight on the Holder space $C^\gamma$.
There are some easy examples where the assumption is not necessary. First of all, if $X\_N(t)=Z \psi(t)$ for all $N$ where $Z$ is a random variable with no moments and $\psi$ is as smooth as you want, then $P(\|X\_N\|\_\gamma>K)\to 0$ uniformly in $N$ as $K\to\infty$ and we have tightness on any Holder space.
I am wondering how we get around the lack of moments to prove tightness. For example, suppose that for all pairs $(s,t)$ we have $\lim\_{K\to\infty} P(\{|X\_N(t)-X\_N(s)|>K|t-s|^\gamma\})=0$ uniformly in $N$, then can we conclude tightness on $C^{\gamma'}$ for any $\gamma'<\gamma$?
Generally how you prove KTC is through Garsia-Rodemich-Rumsey inequality but it is hard to handle without moments.
| https://mathoverflow.net/users/479223 | Version of Kolmogorov tightness criterion without moments | Unfortunately, this does not hold, in the sense that you cannot conclude tightness even for any particular $\gamma’ < \gamma$. This very simple modification of my example [here](https://mathoverflow.net/questions/449705/garsia-rodemich-rumsey-without-markov) is a counterexample (I’ve just replaced the $\gamma$ in the exponent with $\gamma’$):
For any fixed positive integer $n \geq 22$, let $X^n$ be defined as follows - uniformly at random pick a point $p$ in, $[\frac{1}{4}, \frac{3}{4}]$, and define
$$
X^n :=
\begin{cases}
0 & \text{on } [0, p], \\
n(t - p)^{\gamma’} & \text{on } (p, p + \frac{1}{n^{2/\gamma’}}], \\
\frac{1}{n} & \text{on } (p + \frac{1}{n^{2/\gamma’}}, 1]. \\
\end{cases}
$$
Then
$$\lim\_{K\to\infty} P(\{|X\_n(t)-X\_n(s)|>K|t-s|^\gamma\})=0$$
indeed uniformly in $n, s, t$ but $\| X^n \|\_{\gamma’} = n$ almost surely, which tends to $\infty$ and so tightness in Holder space is not possible.
I can do the calculation in full if needed, but morally the reason why this counterexample works is this:
The requested limit above can be shown to hold even if $X\_n$ were the process that was discontinuous at $p$ with a jump of size $\frac{1}{n}$. Of course we have assumed $X\_n$ continuous, but it follows that you could have $X\_n$ continuous but make the bump essentially as bad as you want in $\gamma’$ Holder norm and still have that limit hold.
I second the suggestion of Thomas Kojar to try to look at the particular process and see if we can come up with some criterion that fits your purpose.
| 3 | https://mathoverflow.net/users/173490 | 450260 | 181,148 |
https://mathoverflow.net/questions/450254 | 1 | Let $n$ be a multiple of $4$, is there any $n \times n$ negacyclic Hadamard matrix? If yes - how to construct it? If no - why?
Here an $n \times n$ nega-cyclic matrix is a square matrix of the form:
\begin{align}
\begin{bmatrix}
x\_1 & x\_2 & \cdots & x\_{n-1} & x\_n \\
-x\_n & x\_1 & \cdots & x\_{n-2} & x\_{n-1} \\
\vdots & \vdots& \ddots & \vdots & \vdots \\
-x\_3 & -x\_4 & \cdots & x\_1 & x\_2 \\
-x\_2 & -x\_3 & \cdots & -x\_n & x\_1 \\ \end{bmatrix}.
\end{align}
An $n \times n$ Hadamard matrix is a matrix whose entries are either $1$ or $-1$ and whose rows are mutually orthogonal.
| https://mathoverflow.net/users/369335 | One question about nega-cyclic Hadamard matrices | Such matrices do not exist as from the parity consideration already first two rows cannot be orthogonal.
| 2 | https://mathoverflow.net/users/7076 | 450274 | 181,152 |
https://mathoverflow.net/questions/450273 | 7 | It is known that the only elementary abelian $2$-groups (finite and nonfinite) in $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ are in fact finite and cyclic – that is to say, they are of order $2$.
(Here, $\overline{\mathbb{Q}}$ is the algebraic closure of $\mathbb{Q}$ inside $\mathbb{C}$.)
Question: Does one need the Axiom of Choice to acquire this property ?
| https://mathoverflow.net/users/12884 | Involutions in the absolute Galois group (and the Axiom of Choice) | No, you shouldn't need any choice for this, and it should still be true if you replace $\overline{\mathbb{Q}}$ with any other algebraic closure of $\mathbb{Q}$. Let $K$ be a field (which in our application will be $\overline{\mathbb{Q}}$) and let $G$ be a finite group of automorphisms of $K$. Then $K/K^G$ is always a degree $\#(G)$ extension. (See, e.g., [Milne - Fields and Galois theory](https://www.jmilne.org/math/CourseNotes/FT.pdf) Corollary 2.14 and Theorem 3.4.)
The [Artin–Schreier theorem](https://kconrad.math.uconn.edu/blurbs/galoistheory/artinschreier.pdf) says that, if $K/F$ is a finite degree extension and $K$ is algebraically closed, then $[K:F] \leq 2$.
Combining the two statements, if $G$ is a finite subgroup of $\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$, then $\#(G) \leq 2$.
There is no use of choice in the proof of either of the big tools that I'm using.
| 11 | https://mathoverflow.net/users/297 | 450279 | 181,154 |
https://mathoverflow.net/questions/450280 | 4 | Let $0 < \varepsilon < 1$. A natural number $x$ is called an $\varepsilon$-square if $x = ab$, $a, b \in \mathbb{N}$ and $(1 - \varepsilon)b \le a \le b$. Denote by $f(N)$ the number of $\varepsilon$-squares on the interval $[1, N]$. Is it true that $\lim\_{N\to\infty} \frac{f(N)}{N} = 0$?
There is at least $cN$, for some constant $c > 0$, pairs of natural numbers $(a, b)$ such that $(1 - \varepsilon)b \le a \le b$ and $ab \le N$. But the product of such pairs can define the same $\varepsilon$-squares. For example, we can take $x = \prod\_{i = 1}^{2m}p\_i$, where $p\_i$ are prime numbers close enough to each other. Then there are $\binom{2m}{m}$ representations $x = ab$.
| https://mathoverflow.net/users/507773 | Density of a set of natural numbers which are the product of close numbers | If $x\leq N$ is an $\varepsilon$-square, then $x=ab$ and $a\leq b\leq \sqrt{x/(1-\varepsilon)}=:K$. So, $x$ should appear in $K\times K$ multiplication table. It is known, however, that most numbers below $K^2$ are not in this multiplication table, see [Erdős–Tenenbaum–Ford constant](https://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Tenenbaum%E2%80%93Ford_constant). If $M(K)$ is the number of distinct elements in multiplication table $K\times K$, then for $K\to +\infty$ we have
$$
M(K)=\frac{K^2}{(\ln K)^{\delta+o(1)}},
$$
where $\delta\approx 0.086$. In particular, for any $\varepsilon>0$ there are $o(K^2)=o(N)$ numbers below $N$, which are $\varepsilon$-squares.
| 6 | https://mathoverflow.net/users/101078 | 450283 | 181,156 |
https://mathoverflow.net/questions/450233 | 0 | I came across this partial sum which I cannot find reasonable bounds on; I feel this must be known in the literature, but I do not know where to look. Here is the problem:
Let $s\in (0,1)$ and consider the binomial sum
$$
\boldsymbol{(\*)}\qquad \sum\_{k=1}^N\, \binom{N}{k}\, \frac1{k^s}
$$
are there any known good upper and lower bounds on $\boldsymbol{(\*)}$?
| https://mathoverflow.net/users/491352 | Two-Sided Bounds on Binomial Sum | $\newcommand{\Si}{\Sigma}$We have to lower- and upper-bound
\begin{equation\*}
\Si\_N:=\sum\_{k=1}^N\,\binom Nk\, \frac1{k^s}.
\end{equation\*}
Note that
\begin{equation\*}
\Si\_N=2^N EX^{-s}\,1(X\ge1), \tag{10}\label{10}
\end{equation\*}
where $X$ is a random variable with the binomial distribution with parameters $N,1/2$.
By Jensen's inequality applied to the convex function $[1,\infty)\ni x\mapsto x^{-s}$,
\begin{equation\*}
EX^{-s}\,1(X\ge1)=P(X\ge1)\,E(X^{-s}|X\ge1) \\
\ge P(X\ge1)\,(E(X|X\ge1))^{-s}. \tag{20}\label{20}
\end{equation\*}
Also,
\begin{equation\*}
E(X|X\ge1)=\dfrac{EX\,1(X\ge1)}{P(X\ge1)}=\dfrac{EX}{P(X\ge1)}=\dfrac{N/2}{P(X\ge1)} \tag{30}\label{30}
\end{equation\*}
and $P(X\ge1)=1-2^{-N}$. So, by \eqref{10}, \eqref{20}, and \eqref{30}, we get a lower bound on $\Si\_N$:
\begin{equation\*}
\Si\_N\ge L\_N:=M\_N\,(1-2^{-N})^{1+s}, \tag{33}\label{33}
\end{equation\*}
where
\begin{equation\*}
M\_N:=\frac{2^N}{(N/2)^s}. \tag{35}\label{35}
\end{equation\*}
To get an upper bound on $\Si\_N$, write
\begin{equation\*}
EX^{-s}\,1(X\ge1)=E\_1+E\_2, \tag{40}\label{40}
\end{equation\*}
where
\begin{equation\*}
E\_1:=EX^{-s}\,1(X\ge\tfrac N2\,(1-h)),\quad \\ E\_2:=EX^{-s}\,1(1\le X<\tfrac N2\,(1-h)),
\end{equation\*}
and $h\in(0,1-2/N]$. Next,
\begin{equation\*}
E\_1\le(\tfrac N2\,(1-h))^{-s}. \tag{50}\label{50}
\end{equation\*}
By Hoeffding's inequality,
\begin{equation\*}
E\_2\le P(1\le X<\tfrac N2\,(1-h)) \\
\le \exp(-\tfrac2N\,(\tfrac{Nh}2)^2)=e^{-h^2 N/2}=N^{-c^2/2} \tag{60}\label{60}
\end{equation\*}
if $h=c\sqrt{\frac{\ln N}N}\in(0,1-2/N]$. So, in view of \eqref{10}, \eqref{40}, \eqref{50}, \eqref{35}, and \eqref{60}, we get an upper bound on $\Si\_N$:
\begin{equation\*}
\Si\_N\le U\_N:=M\_N\,\Big[\Big(1-c\sqrt{\frac{\ln N}N}\,\Big)^{-s}
+2^{-s}N^{s-c^2/2}\Big].
\tag{70}\label{70}
\end{equation\*}
Fixing now any real $c>\sqrt{2s}$, letting $N\to\infty$, and looking at \eqref{33} and \eqref{70}, we see that $L\_N\sim M\_N\sim U\_N$ and hence
\begin{equation\*}
\Si\_N\sim L\_N\sim U\_N\sim M\_N=\frac{2^N}{(N/2)^s}.
\end{equation\*}
Thus, our lower and upper bounds $L\_N$ and $U\_N$ on $\Si\_N$ are asymptotically exact for large $N$.
| 2 | https://mathoverflow.net/users/36721 | 450284 | 181,157 |
https://mathoverflow.net/questions/450272 | 2 | I have this integral that comes from my research with some Fourier Transforms of spectrum functions:
$$ G(\tau) = \int\_{0}^{\infty} e^{-\Lambda x} x^n e^{i \tau ( c\_1 - c\_2 e^{-c\_3 x} ) } dx $$
where $c\_1, c\_2, c\_3, \Lambda, n > 0$.
The only way for me now is to use a series expansion of the term $e^{i \tau ( c\_1 - c\_2 e^{-c\_3 x} ) }$ and take the first few terms. However, I was wondering if there is a smarter way to do it.
====== EDIT =========
Here, $i = \sqrt{-1}$.
| https://mathoverflow.net/users/489481 | Is this integral solvable analytically? | there is a closed form solution for
$$I\_n = \int\_{0}^{\infty} e^{-\Lambda x} x^n e^{i \tau ( c\_1 - c\_2 e^{-c\_3 x} ) } dx$$
for integer $n$, for example, for $n=0$:
$$I\_0=\frac{1}{\lambda (c\_3+\lambda)}e^{i c\_1 \tau}$$
$$\qquad\times \left[(c\_3+\lambda) \, \_1F\_2\left(\frac{\lambda}{2 c\_3};\tfrac{1}{2},\tfrac{\lambda}{2 c\_3}+1;-\tfrac{1}{4} c\_2^2 \tau^2\right)-i c\_2 \lambda \tau \, \_1F\_2\left(\tfrac{\lambda}{2 c\_3}+\tfrac{1}{2};\tfrac{3}{2},\tfrac{\lambda}{2 c\_3}+\tfrac{3}{2};-\tfrac{1}{4} c\_2^2 \tau^2\right)\right].$$
for larger integer $n$ the expressions are similar, in terms of hypergeometric functions, but much longer.
| 3 | https://mathoverflow.net/users/11260 | 450286 | 181,159 |
https://mathoverflow.net/questions/450218 | 1 | Consider the spatially homogenous Boltzmann equation $$\partial\_t f\_t = Q^+(f\_t,f\_t) - f\_t.$$ A semi-explicit representation formula for solutions of this Boltzmann equation can be written as (see for instance [Villani's monograph](https://cedricvillani.org/sites/dev/files/old_images/2012/07/B01.Handbook.pdf))
$$
f\_t = \mathrm{e}^{-t}\sum\_{n=1}^\infty \left(1-\mathrm{e}^{-t}\right)^{n-1} Q^+\_n(f\_0) \label{1}\tag{1}
$$ where the $n$-fold nonlinear operator $Q^+\_n$ is defined recursively by
$$Q^+\_1(f\_0) = f\_0,\quad Q^+\_n(f\_0) = \frac{1}{n-1}\,\sum\_{k=1}^{n-1} Q^+\left(Q^+\_k(f\_0),Q^+\_{n-k}(f\_0)\right).$$
The author mentioned that one can easily check that \eqref{1} indeed solves the Boltzmann equation by plugging it into the integral formulation of the Boltzmann equation, which reads as $$f\_t(v) = f\_0(v)\,\mathrm{e}^{-t} + \int\_0^t \mathrm{e}^{-(t-s)}\,Q^+(f\_s,f\_s)\,\mathrm{d}s.$$ However, I have no clue as to how one can verify that the Wild's sum representation \eqref{1} indeed solves the Boltzmann equation. Thus any help in filling in the missing details will be greatly appreciated.
| https://mathoverflow.net/users/163454 | Wild's sum for Boltzmann's equation | Completely ignoring all convergence issues, this really is just following your nose.
Plugging in the representation you give, you want to check
$$ \sum\_{n = 1}^\infty (1 - e^{-t})^{n-1} Q\_n^+(f\_0) \overset{?}{=} f\_0 + \int\_0^t e^{s} Q^+(f\_s,f\_s) ~ds $$
Using the representation you give again to replace $f\_s$, you rewrite this as checking (using that $Q^+$ is bilinear)
$$ \sum\_{n = 1}^\infty (1-e^{-t})^{n-1}Q\_n^+(f\_0) \overset{?}{=} \\Q\_1^+(f\_0) + \sum\_{n = 1}^\infty \sum\_{m = 1}^\infty \int\_0^t e^{-s} (1 - e^{-s})^{n-1} (1-e^{-s})^{m-1} Q^+(Q\_n^+(f\_0), Q\_m^+(f\_0)) ~ds $$
The integral can now be explicitly evaluated, as $\frac{d}{ds} (1-e^{-s})^k = k (1 - e^{-s})^{k-1} e^{-s}$, so we reduce to
$$ \sum\_{n = 1}^\infty (1-e^{-t})^{n-1}Q\_n^+(f\_0) \overset{?}{=} Q\_1^+(f\_0) + \sum\_{n,m = 1}^\infty \frac{1}{n+m-1} (1 - e^{-t})^{n+m-1} Q^+(Q\_n^+(f\_0), Q\_m^+(f\_0)) $$
Rewrite the sum in terms of $n$ and $k = n+m$, you get
$$ \sum\_{n = 1}^\infty (1-e^{-t})^{n-1}Q\_n^+(f\_0) \overset{?}{=} Q\_1^+(f\_0) + \sum\_{k = 2}^\infty (1 - e^{-t})^{k-1}\underbrace{\frac{1}{k-1} \sum\_{n = 1}^{k-1} Q^+(Q\_n^+(f\_0), Q\_{k-n}m^+(f\_0))}\_{= Q\_k^+(f\_0)} $$
and so equality follows.
Convergence can be checked for small $t$: if the bilinear form satisfies the bound $|Q^+(f,g)| \leq C|f||g| $, then by induction you can show that $|Q\_n^+(f\_0)| \leq C^{n-1} |f\_0|^n$. For $|t|\ll 1$ you have that $|(1-e^{-t})| \ll 1$ and hence the representation converges as it is dominated by a geometric series. And so for $t$ sufficiently small all of the interchanges of integrals and sums, and reindexing of summation, are valid in the formal computation above.
| 2 | https://mathoverflow.net/users/3948 | 450289 | 181,161 |
https://mathoverflow.net/questions/450086 | 8 | Suppose $M\_1^\#$ exists and is $\omega\_1$-iterable.
Is it consistent that we can go to a generic extension $V[G]$ where $M\_1^\#$ is no longer $\omega\_1$-iterable?
Or "worse" $M\_1^\#$ is no longer 2-iterable in Neeman's sense?
I suspect the answer is yes and that it will be relatively obvious. So more generally, I'm looking for places where problems like these might be addressed.
| https://mathoverflow.net/users/9324 | Destroying the iterability of $M_1^\#$ | Andreas Lietz has already pointed out that $\omega\_1$-iterability can fail in a generic extension. One can also consistently get $(\omega+1)$-iterability to fail:
Suppose there is a transitive model of ZFC + "$M\_1^\#$ exists and is $(0,\omega\_1)$-iterable", and let $M$ be such with minimal ordinal height. It is easy to see we may assume that $M=L\_\alpha[X]$ for some set $X\in M$ of ordinals, where $\alpha=\mathrm{OR}^M$, and we can take $X\subseteq\lambda$ for some strong limit cardinal $\lambda$ of $M$, and $X$ coding $V\_\lambda^M$. Let $N=(M\_1^{\#})^M$.
Let $G$ be $(M,\mathrm{Coll}(\omega,\lambda))$-generic.
I claim that in $M[G]$, $N$ is not $(\omega+1)$-iterable with respect to trees based on $N|\delta^N$ (the Woodin of $N$).
For suppose otherwise. We have $M[G]=L\_\alpha[x]$ for a certain real $x$,
and we can take $x$ to code $X$ in a simple manner. So since $\mathcal{P}(\delta^N)\cap N$ is countable in $M[G]$, working in $M[G]=L\_\alpha[x]$, we can form a Neeman genericity iteration on $N$ (see Theorem 3.3 of "AN INNER MODELS PROOF OF THE KECHRIS-MARTIN THEOREM"), producing a tree $\mathcal{T}$ on $N$ of length $\omega$, such that if $b$ is any $\mathcal{T}$-cofinal wellfounded branch, then there is a $\mathrm{Coll}(\omega,\delta^{M^{\mathcal{T}}\_b})$-generic $g\in L\_\alpha[x]$
with $x\in M^{\mathcal{T}}\_b[g]$.
But letting $\kappa$ be the critical point of the active extender of $M^{\mathcal{T}}\_b$, then note that $\kappa<\alpha$ and $L\_\kappa[x]\models$ ZFC and $X\in L\_\kappa[x]$. Therefore $L\_\kappa[X]\models$ ZFC. But since $\kappa<\alpha$,
we have $L\_\kappa[X]\subseteq M$,
so $V\_\lambda^{L\_\kappa[X]}=V\_\lambda^{M}$,
so $L\_\kappa[X]\models$ ZFC + "$M\_1^{\#}$ exists and is $(0,\omega\_1)$-iterable",
contradicting the minimality of $\alpha$.
However, one should note that if $M\models$ ZFC + "$M\_1^{\#}$ exists and is $(0,\omega\_1)$-iterable" and there is a larger model $W$ of ZFC with $M\subseteq W$
and $\mathrm{OR}^M=\mathrm{OR}^W$
and $(M\_1^{\#})^M$ is fully iterable in $W$
(so $(M\_1^{\#})^M=(M\_1^{\#})^W$)
then for any set generic extension $M[G]$ with $\mathbb{R}^{M[G]}\subseteq W$, we will have that $(M\_1^{\#})^M$ is $(0,\omega+1)$-iterable in $M[G]$. This is just by $\Sigma^1\_2$ absoluteness between $M[G]$ and $W$: given a length $\omega$ tree $\mathcal{T}\in M[G]$, there is a $\mathcal{T}$-cofinal wellfounded branch $b\in W$, but this is just a $\Sigma^1\_2$ assertion about $\mathcal{T}$ which is true in $W$, hence true in $M[G]$. Of course, the "sharp" at the top of $M^{\mathcal{T}}\_b$ need not be iterable.
| 5 | https://mathoverflow.net/users/160347 | 450299 | 181,165 |
https://mathoverflow.net/questions/450294 | 5 | Let $s\_{\lambda}(x\_1,\dots,x\_k)$ be the [Schur polynomial](https://en.wikipedia.org/wiki/Schur_polynomial) associated to the partition $\lambda=(\lambda\_1\geq\lambda\_2\geq\cdots\geq\lambda\_k>0)$.
Among the many things involved with these polynomial, I was exploring the number of (distinct) monomials appears in them. I was rather amused by what I noticed.
>
> **QUESTION.** If $\lambda^{(k)}$ denotes the staircase partition $(k,k-1,\dots,1)$, is it true that the number of monomials, denoted $a\_n=\#s\_{\lambda^{(k)}}(x\_1,\dots,x\_k)$, equals the [number of forests on n labeled nodes](https://oeis.org/A001858)?
>
>
>
| https://mathoverflow.net/users/66131 | Enumerating monomials in Schur polynomials | A monomial $x\_1^{a\_1}\cdots x\_{n}^{a\_n}$ will appear in the expansion of the Schur polynomial $s\_{\lambda}(x)$ if and only if $(a\_1,a\_2,...,a\_n)\le (\lambda\_1,\lambda\_2\dots, \lambda\_n)$ in the dominance (majorization) order. This is equivalent to saying that $(a\_1,a\_2,...,a\_n)$ is a lattice point in the convex hull of all the points obtained by permuting the coordinates $(\lambda\_1,\lambda\_2\dots, \lambda\_n)$.
The lattice points in the convex hull of all the permutations of the vector $(n,n-1,...,1)$ are in bijection with forests on n labeled nodes, as pointed out in the OEIS link. (In fact, if you follow the Stanley reference there you will see a more general statement about score vectors and forests on graphs.)
| 5 | https://mathoverflow.net/users/2384 | 450312 | 181,169 |
https://mathoverflow.net/questions/450308 | 0 | [Nancy Cartwright](https://en.wikipedia.org/wiki/Nancy_Cartwright_(philosopher)) introduced an interesting distinction with regard to modeling of physical phenomena. According to Cartwright, a mathematical theory is not applied directly to such phenomena. Rather, one first builds a basic mathematical model of the phenomenon in question, in a step Cartwright refers
to as *phenomenological model-building* (I think she uses "phenomenological" in the sense of "empirical" but I may be wrong). Only then does one exploit a full-fledged mathematical theory, applied not to the original phenomenon but rather to the basic mathematical model. Since I am not familiar with the relevant literature, I am wondering if this distinction has been explored further in recent work.
The relevant papers are the following:
Cartwright, Nancy; Shomar, Towfic; Su'arez, Mauricio. The tool box
of science. Tools for the building of models with a superconductivity
example. Poznan Studies in the Philosophy of the Sciences and the
Humanities 44 (1995), 137--149.
Su'arez, Mauricio; Cartwright, Nancy. Theories: Tools versus models.
Studies in History and Philosophy of Science Part B: Studies in
History and Philosophy of Modern Physics, Vol. 39, Issue 1 (2008),
62--81. <https://doi.org/10.1016/j.shpsb.2007.05.004>.
| https://mathoverflow.net/users/28128 | Nancy Cartwright's dichotomy | Cartwright's case study, model building for the theory of superconductivity, has been explored further in the Ph.D.thesis [The Role of Concrete Models in the Revolution in Superconductivity](https://uwspace.uwaterloo.ca/bitstream/handle/10012/9818/Chattoraj__Ananya.pdf;sequence=1) (A. Chattoraj, 2015).
More generally, one of Cartwright's former students, Roman Frigg, has developed this approach in a monograph, [Models and theories: A philosophical inquiry](https://library.oapen.org/bitstream/handle/20.500.12657/57620/9781000609530.pdf?sequence=1) (2022).
| 3 | https://mathoverflow.net/users/11260 | 450313 | 181,170 |
https://mathoverflow.net/questions/450300 | 2 | This is probably a simple question, maybe more suited for MSE. In the coarea formula, you have
$$\int\_{{\mathbb{R}}^n} g (x) |\nabla f(x)|\, dx= \int\_\mathbb{R} \left(\int\_{\{f=t\}} g d \mathcal{H}^{n-1} \right)dt ,$$
with $f$ Lipshitz and $g$ Borel (positive maybe?). Anyway the question is: how do you define the integral $\int\_{\{f=t\}} g d \mathcal{H}^{n-1} $? When you do the theory of Sobolev spaces it's often remarked that traces of $L^p$ functions make no sense, and here you're integrating generic measurable functions basically. I'm sure I (used to) know this but I am not seeing it right now.
| https://mathoverflow.net/users/109382 | Definition of integral over level sets in coarea formula | Check out section 3.4 in
*Evans, Lawrence Craig; Gariepy, Ronald F.*, Measure theory and fine properties of functions, Textbooks in Mathematics. Boca Raton, FL: CRC Press (ISBN 978-1-4822-4238-6/hbk). 309 p. (2015). [ZBL1310.28001](https://zbmath.org/?q=an:1310.28001).
I won't reproduce all the details, but a key statement is Lemma 3.5, which states
>
> If $f:\mathbb{R}^n\to\mathbb{R}^m$ is Lipschitz, $n \geq m$, and $A\subseteq \mathbb{R}^n$ is $\mathcal{L}^n$ measurable, then $A\cap f^{-1}\{y\}$ is $\mathcal{H}^{n-m}$ measurable for $\mathcal{L}^m$ almost every $y$.
>
>
>
Basically, while for a fixed single level set the trace makes no sense, for almost every level set the trace makes sense. And so you are fine if you are integrating things (or as long as you are working in a context where having $\int\_{\{f =t\}} g d\mathcal{H}^{n-1}$ defined for almost every $t$ is good enough).
| 6 | https://mathoverflow.net/users/3948 | 450314 | 181,171 |
https://mathoverflow.net/questions/449311 | 1 | Let $M$ be a closed complex Kähler manifold, $dim\_{\mathbb C} M = n\geq 2$, with a Kähler form $\omega$. Assume $U\subset M$ is a Stein domain with a smooth boundary and $f: U\to [0;1]$ is a smooth exhausting function for $U$ such that $f^{-1} (1) = \partial U$ and $L:=f^{-1} (0)$ is a smooth totally real closed submanifold of $U$, $dim\_{\mathbb R} L = n$. Does there exist a (smooth) strictly $\omega$-plurisubharmonic function $\phi: M\to {\mathbb R}$ such that $C:=\min\_{\partial U} \phi - \max\_L \phi$ is arbitrarily large?
Note that taking an arbitrary smooth extension of $f$ to $M$ and multiplying it by a sufficiently small positive constant, one can find $\phi$ with a positive, albeit small, $C$.
| https://mathoverflow.net/users/102829 | Restrictions of strictly $\omega$-plurisubharmonic functions to a Stein domain in a closed Kahler manifold | An answer courtesy of Vincent Guedj (for the terminology see the book "Degenerate Complex Monge-Ampère Equations" by V.Guedj and A.Zeriahi):
No, $C$ cannot be made arbitrarily large - there is an upper bound on it depending only on $L$.
The proof goes as follows.
Since $L$ is totally real, it is not locally pluripolar (this is an old result of
Sadullaev from the 1970s, also see [here](https://arxiv.org/pdf/1207.3312.pdf) for a more precise result). Consequently, $L$ is not $PSH (X,\omega)$-polar (indeed, if there had existed a locally defined $\omega$-psh function equal to $-\infty$ locally on $L$, then adding to it a local Kähler potential of $\omega$ one would have got a locally defined psh function equal to $-\infty$ locally on $L$, which is impossible since $L$ is not locally pluripolar).
Now assume that all smooth $\omega$-psh functions $\phi$ from the question are normalized so that $\max\_L \phi = 0$. Then, since $L$ is not $PSH (X,\omega)$-polar, Corollary 9.18 from the book quoted above implies that there exists $K>0$, depending on $L$, such that $\max\_M \phi \leq K$ for all such $\phi$. (Strictly speaking, Corollary 9.18 yields an upper bound on the $L^1$-norm of $\phi$ but using Proposition 8.5 in the book one translates it to an upper bound on $\max\_M \phi$). Clearly, for all such $\phi$ the constant $K$ bounds $C$ from above:
$$C=\min\_{\partial U} \phi − \max\_L \phi = \min\_{\partial U} \phi \leq \max\_M \phi\leq K.$$
| 0 | https://mathoverflow.net/users/102829 | 450319 | 181,172 |
https://mathoverflow.net/questions/450253 | 4 | Assume that we have heavy-tailed distribution $F(x)$ such that
\begin{align}
F(x)=\mathbb{P}[X\geq x]=x^{-0.5}.
\end{align}
Then, we produce $N$ independent samples $X\_1,X\_2,\ldots,X\_N$ from this distribution. Assuming that $n(N)\leq N$ is a function of $N$, we pick the $n(N)$ largest amounts among $X\_1,X\_2,\ldots,X\_N$ and produce the summation as
\begin{align}
S\_{n(N)} = \sum\_{\ell=1}^{n(N)} X\_{i\_\ell},
\end{align}
Here, $i\_1, \ldots, i\_{n(N)}$ represent the indices corresponding to the $n(N)$ largest values among the $N$ variables. The question is whether there exists a specific choice of $n(N)$ such that as $N$ approaches infinity, the ratio $n(N)/N$ tends to zero, and we have
\begin{align}
\lim\_{N\rightarrow\infty}\frac{S\_{n(N)}}{S\_{N}}= 1.
\end{align}
In other words, does a very small subset of variables represent the whole of them? How can I think about these kinds of problems to solve them?
| https://mathoverflow.net/users/68835 | Does a subset with small cardinality represent the whole set? | The probability that all samples are less than $N^{19/10}$ is $(1-N^{-19/20})^{N}$ that tends to 0. The expected number of samples greater than $N^{1/2}$ is $N^{3/4}$, thus, the probability that we have more than $N^{4/5}$ such samples is by Chebyshev inequality at most $N^{-1/20}$,also tends to 0. Therefore, with probability tending to 1 the sum of all but $N^{4/5}$ largest samples does not exceed $N^{3/2}$ while the sum of $N^{4/5}$ largest samples (and even one single largest sample) is at least $N^{19/10}$. It yields that you may take $n(N)=N^{4/5}$. Of course this is not optimal.
| 4 | https://mathoverflow.net/users/4312 | 450332 | 181,176 |
https://mathoverflow.net/questions/450317 | 4 | I am developing a sort of standard representation for profinite quandles. This involves profinite groups a lot, actually. In one part of my construction the filtered diagram used to construct a profinite group becomes important.
Suppose we have two profinite groups $G\_1$ and $G\_2$, with proper, dense subgroups $\Gamma\_1<G\_1$ and $\Gamma\_2<G\_2$ with a topological isomorphism $\Gamma\_1\cong\Gamma\_2$. Does this imply that $G\_1\cong G\_2$? Note that this is certainly not true for topological spaces, but I hope the unique form of compactness offered by profinite things helps the argument some. If this is not true, is there a known counterexample?
| https://mathoverflow.net/users/508126 | Profinite groups with isomorphic proper, dense subgroups are isomorphic | Let $G$ be a compact group and $H$ a dense subgroup. I claim that $H$, as topological group, determines $G$. For simplicity, let me assume that $G$ is metrizable.
Note that a sequence $(h\_n)$ in $H$ converges in $G$ if and if $h\_n^{-1}h\_m\to 1$ when $n,m\to\infty$, and two such sequences $(h\_n)$, $(h'\_n)$ have the same limit if and only if $(h\_n^{-1}h'\_n)$ converges to 1. Thus one can define $G\_H$ as set of Cauchy sequences modulo this relation, which is an equivalence relation, endow it with the group law (noting that being Cauchy passes to products and that the product factors through the equivalence relation). In $G\_H$, a sequence $(h\_n^k)\_k$ of Cauchy sequences converges to $1$ if and only if for every neighborhood $V$ of $1$ there exists $k\_0$ such that for every $k\ge k\_0$ there exists $n\_k$ such that $h\_n^k\in V$ for all $n\ge n\_k$. Thus there exists at most one metrizable group topology on $G\_H$ for which these are the converging sequences to $1$.
From this it follows that any two compact [metrizable] groups with isomorphic dense subgroups (isomorphic as topological groups) are isomorphic as topological groups.
The general case (no metrizability) can be done by a suitable use of filters. Also compactness is not fully used; local compactness is enough, and also it should be enough to suppose some kind of completability, but I haven't seriously checked and I'm not sure what uniformity is best to use.
| 4 | https://mathoverflow.net/users/14094 | 450335 | 181,178 |
https://mathoverflow.net/questions/450357 | 1 | Let $P$ and $Q$ be two distributions over a sample space $\Omega$ which I would like to show are close under some choice of distance function. So far I have managed to show that there exists a subset $S\subseteq \Omega$ such that:
* $P(S)$ is large, say, at least $(1-\varepsilon)$
* the conditional distributions of $P$ and $Q$ over $S$, denoted $P\_S(x)=P(x)/P(S)$ and $Q\_S(x)=Q(x)/Q(S)$ are pointwise close, i.e. $P\_S(x)\in [1\pm\varepsilon]\;Q\_S(x)$ for any $x\in S$.
My question is: does this correspond to a distributional closeness between $P$ and $Q$ under any well-known divergence? I realize that the above notion of closeness is asymmetric as the first point above is with respect to one of the distributions.
| https://mathoverflow.net/users/508178 | What is this distributional closeness? | $\newcommand\Om\Omega$No. E.g., for natural $n$, suppose that $\Om=[n]:=\{1,\dots,n\}$, $S=[n-1]$, $P(x)=\frac1n$ for $x\in\Om$, $Q(x)=\frac1{n^2}$ for $x\in S$, and $Q(n)=1-\frac{n-1}{n^2}$.
Then your conditions hold for $\varepsilon=\frac1n$, $P$ is uniform over $\Om$, but (for large $n$) almost all $Q$-mass is at the one point, $n$.
---
A probability distribution is a measure. So, you should write $P(\{x\})$ instead of $P(x)$, assuming that the singleton sets $\{x\}$ are in the underlying $\sigma$-algebra -- which, looking at the context of your post, appears to be the largest $\sigma$-algebra over a discrete set $\Om$. However, in the answer above I used your notations, such as $P(x)$. Also, $P\_S(x)$ is a number, not a distribution, conditional or not.
| 1 | https://mathoverflow.net/users/36721 | 450358 | 181,182 |
https://mathoverflow.net/questions/450094 | 3 | I am currently going through Shimura's paper on half-integer weight modular forms. I would like to understand given a -expansion of half-integral weight modular forms of arbitrary level and character, how to compute the effect of the Hecke operator and the Atkin-Lehner operator/Fricke involution in SAGE/Magma. I am a beginner in SAGE/Magma and
The only documentation I found is related to computing basis of weight k/2 and character chi.
If anyone has any sources on or explanations as to how to compute how q-expansions are transformed under the action of Hecke operators and Atkin-Lehner/Fricke involution, it would be greatly appreciated!
| https://mathoverflow.net/users/86441 | Computations of half-integer forms in SAGE/Magma | This can be done using PARI/GP, which can deal with spaces of modular forms of half-integral weight. Given a modular form $f$ of weight $k$ (possibly half-integral), the command *mfslashexpansion* can compute the $q$-expansion of $f |\_k g$ for any $g \in \mathrm{GL}\_2^+(\mathbf{Q})$. This relies on a floating-point method but works well in practice. One of the ideas is to multiply your half-integral weight modular form by the weight $1/2$ modular form $\theta$, whose behaviour under the Atkin-Lehner involution is known. This reduces to the case of integral weight, which is still difficult, but one other idea is to write the form as a linear combination of pairwise products of Eisenstein series. The behaviour of Eisenstein series under the Atkin-Lehner involution is known, which gives the result.
You can look at PARI/GP's [users manual](https://pari.math.u-bordeaux.fr/doc.html) (see the section *Modular forms*). If you want to know the mathematics behind the algorithms, you can read [this article](https://arxiv.org/abs/1810.00547) by Belabas and Cohen.
| 7 | https://mathoverflow.net/users/6506 | 450364 | 181,183 |
https://mathoverflow.net/questions/450349 | 1 | *This question (and a second part) have been [asked at MSE](https://math.stackexchange.com/questions/4718733/an-infinitely-conditioned-state-in-a-c-algebra) and gone through two bounties without an answer. I have been beating my head at it for a while without success*.
Let $\mathcal{A}$ be a unital $\mathrm{C}^\*$-algebra. Let $f\in \mathcal{A}$ self-adjoint with spectrum $\sigma(f)$. The bidual $\mathcal{A}^{\*\*}$ contains the spectral projections of $f$, denoted $\mathbf{1}\_{E}(f)$, for Borel $E\subseteq \sigma(f)$. Let $\lambda\in\sigma(f)$, $\varepsilon>0$ and define
$$p\_\varepsilon:=\mathbf{1}\_{(\lambda-\varepsilon,\lambda+\varepsilon)}(f).$$
Suppose that $\varphi$ is a state on $\mathcal{A}$, with an extension to a state on $\mathcal{A}^{\*\*}$ denoted $\omega\_\varphi$. Suppose further, and this is to be a key assumption, that for all $\varepsilon>0$, $$\omega\_\varphi(p\_\varepsilon)>0\qquad(\star).$$
If the following is defined, it defines a state. But is it well-defined?
**Question: Let $g\in\mathcal{A}$. Under the assumption $(\star)$, does the following limit exist:**
$$\lim\_{\varepsilon\to 0^+}\frac{\omega\_\varphi(p\_\varepsilon\,g\,p\_\varepsilon)}{\omega\_\varphi(p\_\varepsilon)}.$$
| https://mathoverflow.net/users/35482 | Conditioning a $\mathrm{C}^*$-algebra state with infinite precision | No, this limit doesn't exist in general. Here's a commutative counterexample.
Let $A = L^\infty[0,1]$, $f(x) = x$, $\phi =$ integration against Lebesgue measure. Let $g$ be the indicator function of the set $\bigcup [10^{-(2n + 1)}, 10^{-2n}] = [.1, 1] \cup [.001, .01] \cup [.00001, .001] \cup \cdots$.
Taking $\lambda = 0$, we get that $p\_\epsilon$ is the indicator function of $[0,\epsilon]$. Then $\frac{\phi(p\_\epsilon g)}{\phi(p\_\epsilon)}$ takes the same value on $\epsilon =$ any even negative power of 10, and the same value on $\epsilon =$ any odd negative power of 10, and those two values aren't the same.
| 3 | https://mathoverflow.net/users/23141 | 450365 | 181,184 |
https://mathoverflow.net/questions/450321 | 1 | Let $X=\text{Sp}(A)$ be an affinoid $K$ space, where $K$ is a $p$-adic field. If $f\_0, f\_1,..., f\_s \in A$ generate the unit ideal then we can define the rational subdomain $U= X(f\_0, f\_1..., f\_s) = \{ x \in X: \vert f\_i(x) \vert \leq \vert f\_0(x) \vert \text{ for } i=1...s \}$ of $X$ with coordinate ring $\mathcal{O}(U)= A \langle x\_1,...,x\_s \rangle/(f\_1-f\_0x\_1, ..., f\_s-f\_0x\_s)$.
If $\rho \in \sqrt{\vert K^\times \vert}$ and $\rho >1$ then, by definition, there exists a natural number $n$ such that $\rho^n= \vert t \vert$ for some $t \in K^\times$. We can define another rational subdomain $U(\rho)= X(\rho f\_0, f\_1..., f\_s) = \{ x \in X: \vert f\_i(x) \vert \leq \rho \vert f\_0(x) \vert \text{ for } i=1...s \}$ of $X$. I think its coordinate ring is $\mathcal{O}(U(\rho))= A \langle t^{-1}x\_1,..., t^{-1}x\_s \rangle/(f^n\_1-f^n\_0 x\_1, ..., f^n\_s-f^n\_0 x\_s)$.
As $U \subset \subset\_{X} U(\rho)$ we have, in particular, a sheaf restriction map
$ r\_{U(\rho)U}: \mathcal{O}(U(\rho)) \rightarrow \mathcal{O}(U)$. I think this is just given by $t^{-1} x\_i \mapsto x\_i$ for all $i=1...s$. *I have been trying to work out whether this restriction map is necessarily injective*. So far, my attempts have been direct, i.e. trying to prove that the kernel is zero, but I haven't got anywhere. Irritatingly, I also cannot find any counterexample to the statement I am trying to prove (I might be being stupid).
Does anybody know whether what I am trying to prove is true? If so, could you please point me towards a proof? I would also be very interested to know whether the more general statement $U \subset \subset\_{X} V$ (wide open neighbourhood) with $U$ an affinoid subdomains of $X$ implies that the structure sheaf restriction map $\mathcal{O}(V) \rightarrow \mathcal{O}(U)$ is injective. My original question is a special case of this one. I had a go at the special case first because things seemed more explicit!
Thank you so much in advance for any help!
| https://mathoverflow.net/users/498675 | Injectivity of sheaf restriction maps for wide open neighbourhoods of rational subdomains | If your space has several connected components, then a rational domain may well isolate one of them and you could get a non-injective map.
For an explicit example, you can choose $A = \mathbb{Q}\_p \langle pT \rangle/(T(pT-1))$. Its spectrum has exactly two points: $0$ and $1/p$. To find a example of a non-injective map, you can consider the Weierstrass domains $\{|T| \le 1\}$ (which contains only $0$) and $\{|T| \le 1/p\}$ (which is the whole space).
Also, the map you write is wrong. When $n=1$, you want to send $x\_i$ to $x\_i$ in order to preserve the relations in the quotient.
| 0 | https://mathoverflow.net/users/4069 | 450373 | 181,187 |
https://mathoverflow.net/questions/450331 | 3 | If I have a braided tensor category that's unitary and modular, then how does the unitarity and modularity constrain the fusion multiplicities?
I know that if $a,b,c \in ob({C})$ satisfy the fusion rule $a \otimes b = \oplus\_c N^c\_{ab} c $ then $N^c\_{ab} \in \mathbb{Z}\_+$ for all possible non-trivial fusion channels. Does being unitarity and modular guarantee that $N^c\_{ab} = 0,1$ for all fusion channels?
I require it specifically for the category Rep(D(G)), which I know to be unitary modular. If it's in general not true, but is for this specific category, I'd still be happy!
Edit: <https://arxiv.org/pdf/2306.05560.pdf> computes the fusion rules of Rep(D(G)) to have $N^c\_{ab} =0,1$ for all dihedral and dicyclic groups, which lends some credence to this hypothesis
| https://mathoverflow.net/users/146495 | Does unitarity and modularity constrain fusion multiplicities to be 0,1? | This is false for $D(G)$, when $G$ is sufficiently complicated. For a finite group $G$, the representation category of $D(G)$ has irreducible objects parametrized by pairs $(g, V)$ where $g$ is a conjugacy class representative in $G$ and $V$ is an irreducible representation of the centralizer of $g$. The monoidal structure for pairs of the form $(1,V)$ coincides with the tensor product of representations of $G$, so we can find a counterexample just by finding a finite group $G$ and a pair of irreducible representations whose tensor product has multiplicity greater than 1.
For example, when $G = A\_4$, we have $N\_{V,V}^V = 2$ when $V$ is the unique irreducible representation of dimension 3, i.e., $V \otimes V$ contains $V \oplus V$.
You can check this in GAP:
```
t := CharacterTable(AlternatingGroup(4));
Irr(t)[4]; # yields [3,-1,0,0]
ScalarProduct(t,Irr(t)[4],[9,1,0,0]); # yields 2
```
| 7 | https://mathoverflow.net/users/121 | 450379 | 181,190 |
https://mathoverflow.net/questions/450239 | 2 | Let $k$ be a field of characteristic $0$. Let $R$ be a Noetherian local normal domain containing $k$. Also assume that $R$ is the homomorphic image of a Gorenstein ring of finite dimension, hence $R$ admits a Dualizing complex.
Let $Y$ be a Gorenstein normal scheme over Spec$(k)$ of finite Krull-dimension.
If there exists a proper birational map $f: Y \to \text{Spec}(R)$ such that $R^i f\_\* \mathcal O\_Y=0, \forall i>0$, then is it true that $R$ is Cohen-Macaulay?
If this is not true in general, what if I also assume $Y$ is regular?
As $R$ is normal, $R$ is of course Cohen-Macaulay when $\dim R \leq 2$. So we only have to think about $\dim R \geq 3$.
| https://mathoverflow.net/users/386496 | Proper birational morphism from a Gorenstein normal scheme to a normal local domain, with trivial higher direct images, implies Cohen-Macaulay? | I started writing this last night, but didn't finish. In addition to [Jason Starr's answer](https://mathoverflow.net/a/450348/33088), you can use my new vanishing theorems to remove the assumption that $R$ is essentially of finite type over a field. Note that the special case when $R$ is essentially of finite type over a field of characteristic zero also follows from Kempf's criterion for rational singularities [[Kempf 1973](https://doi.org/10.1007/BFb0070319), p. 50].
---
In short, if $Y$ is assumed to be regular, then $R$ will be Cohen–Macaulay. In fact, we have the following stronger result.
**Theorem.** *Let $R$ be a Noetherian ring and let $f\colon Y \to \operatorname{Spec}(R)$ be a proper birational morphism such that $Y$ is regular and $R^if\_\*\mathcal{O}\_Y = 0$ for all $i > 0$. Suppose $R$ contains $\mathbf{Q}$. Then, every localization of $R$ is pseudo-rational, and in particular $R$ is Cohen–Macaulay.*
*Proof.* Since the conclusion can be checked locally, it suffices to consider the case when $(R,\mathfrak{m})$ is local. Set $\hat{Y} := Y \times\_{\operatorname{Spec}(R)} \operatorname{Spec}(\hat{R})$, and consider the Cartesian diagram
$$\require{AMScd}\begin{CD}
\hat{Y} @>>> Y\\
@V\hat{f}VV @VVfV\\
\operatorname{Spec}(\hat{R}) @>>> \operatorname{Spec}(R)
\end{CD}$$
By base change, $\hat{f}$ is proper, and by flat base change, $\hat{f}$ is birational [[EGAI$\_\text{new}$](https://mathscinet.ams.org/mathscinet/article?mr=3075000), Proposition 3.9.9] and $R^i\hat{f}\_\*\mathcal{O}\_{\hat{Y}} = 0$ for all $i > 0$. Note that $\hat{R}$ is normal by [[Lipman 1969](http://www.numdam.org/item/?id=PMIHES_1969__36__195_0), Remark 16.2] and that the pseudo-rationality of $\hat{R}$ would imply that $R$ is pseudo-rational [[Murayama 2022](https://doi.org/10.1112/S0010437X21007715), Proposition 4.20]. Since $\hat{Y}$ is regular by [[EGAIV$\_2$](http://www.numdam.org/item/PMIHES_1965__24__5_0/), Lemme 7.9.3.1], we may replace $R$ by $\hat{R}$ and $Y$ by $\hat{Y}$ to assume that $R$ is complete local.
We now prove the special case when $(R,\mathfrak{m})$ is complete local. First, by my version of Grauert-Riemenschneider vanishing [[Murayama](https://arxiv.org/abs/2101.10397), Theorem B(i)], we know that $R$ has rational singularities (see, e.g., [[Kollár 2013](https://doi.org/10.1017/CBO9781139547895), Definition 2.76] for the definition). Since $R$ having rational singularities is equivalent to saying that $R$ is pseudo-rational in our setting [[Murayama](https://arxiv.org/abs/2101.10397), Remark 7.4], we are done.
We can also show that $R$ is Cohen–Macaulay directly. Since Cohen–Macaulayness can be checked locally, it suffices to consider the case when $(R,\mathfrak{m})$ is local. Since $R$ is normal, the map $R \overset{\sim}{\to} f\_\*\mathcal{O}\_Y$ is an isomorphism [[EGAIII$\_1$](http://www.numdam.org/item/?id=PMIHES_1961__11__5_0), Corollaire 4.3.12]. Combined with the assumption that $R^if\_\*\mathcal{O}\_Y = 0$ for all $i > 0$, we have the quasi-isomorphism $R \overset{\sim}{\to} \mathbf{R}f\_\*\mathcal{O}\_Y$.
Applying $\mathbf{R}\Gamma\_{\mathfrak{m}}(-)$, we obtain
$$\mathbf{R}\Gamma\_{\mathfrak{m}}(R) \overset{\sim}{\longrightarrow} \mathbf{R}\Gamma\_{\mathfrak{m}}(\mathbf{R}f\_\*\mathcal{O}\_Y) \cong \mathbf{R}\Gamma\_{f^{-1}(\{\mathfrak{m}\})}(\mathcal{O}\_Y).$$
Now taking $i$-th cohomology modules, we obtain the isomorphisms
$$H^i\_{\mathfrak{m}}(R) \overset{\sim}{\longrightarrow} H^i\_{\mathfrak{m}}(\mathbf{R}f\_\*\mathcal{O}\_Y) \cong H^i\_{f^{-1}(\{\mathfrak{m}\})}(\mathcal{O}\_Y).$$
Since the local cohomology modules on the right vanish for all $i < \dim(R)$ by my version of Grauert–Riemenschneider vanishing (in its dual formulation) [[Murayama](https://arxiv.org/abs/2101.10397), Theorem B\*(i)], we see that $R$ is Cohen–Macaulay. $\blacksquare$
| 3 | https://mathoverflow.net/users/33088 | 450383 | 181,191 |
https://mathoverflow.net/questions/450337 | 0 | Let $C$ be a 1-dimensional complex manifold whose universal covering is provided by the half-plane $\mathcal{H}=\{z \in \mathbb{C} \mid \operatorname{Im}z>0\}$. The symmetric product $C^{(n)} = C^n / S\_n$ is a complex manifold. My question is: is the universal covering of $C^{(n)}$ provided by $\mathcal{H}^{(n)}$? Or is it some different quotient of $\mathcal{H}^n$?
| https://mathoverflow.net/users/505150 | Universal covering of symmetric product | In fact, the universal cover of $C^{(n)}$ will not be $\mathcal H^n$ once $n \gg 0$. Indeed, if $C$ is a compact Riemann surface of genus $g \geq 2$ (so the universal cover is $\mathcal H$) with a base point $x$, then the Riemann–Roch theorem implies that the map
\begin{align\*}
C^{(n)} &\to \operatorname{Jac}\_C = \operatorname{Pic}^0\_C\\
(x\_1,\ldots,x\_n) &\mapsto [x\_1 + \ldots + x\_n - nx]
\end{align\*}
is a $\mathbf P^{n-g}$-bundle if $n > 2g-2$ (see e.g. [CS86, Ch. VII, Rmk. 5.6(c)]). So the universal cover of $C^{(n)}$ is a $\mathbf P^{n-g}$-bundle over $\mathbf C^g$, which is very far from $\mathcal H^n$ (e.g. it is not Kobayashi hyperbolic since $C^{(n)}$ is not Brody hyperbolic).
---
**References.**
[CS86] G. Cornell, J. H. Silverman (eds.), [*Arithmetic geometry*](https://doi.org/10.1007/978-1-4613-8655-1). Springer-Verlag, 1986. [ZBL0596.00007](https://zbmath.org/?q=an:0596.00007).
| 6 | https://mathoverflow.net/users/82179 | 450393 | 181,194 |
https://mathoverflow.net/questions/450395 | 1 | Let $p:I\to Cat\_{\infty}$ be a diagram of infinity categories, where $I$ is a small Kan complex. Let $C:=\lim p$ be the limit of $p$. For any two objects $x,y\in C$ and $i\in I$, let $x\_i,y\_i\in C\_i=p(i)$ be the corresponding objects in $C\_i$. Then we have a new diagram $q\_{x,y}: I\to \mathcal{S}$ given by $q\_{x,y}(i)=Map\_{C\_i}(x\_i,y\_i)$, where $\mathcal{S}$ is the infinity category of spaces.
Question: Is it always true that $Map\_C(x,y)\simeq\lim q\_{x,y}$?
| https://mathoverflow.net/users/153842 | Limits of infinity categories and mapping spaces | Yes. To see this, let us make the preliminary observation that it suffices to prove that this holds for products and pullbacks since we can decompose a general limit into these two special cases.
Let us begin with products. If $I$ is discrete then we want to show that $\prod\_i \hom\_{\mathcal{C\_i}}(x\_i,y\_i) \cong \hom\_{\prod\_i \mathcal{C\_i}}((x\_i)\_{i \in I},(y\_i)\_{i \in I})$. To see this, note that we can realize the limit as the honest product of quasicategories in simplicial sets. By considering the model of $\hom\_{\mathcal{D}}(d\_0,d\_1)$ given by $\mathbf{1} \times\_{\mathcal{D}} \mathcal{D}^{\Delta^1} \times\_{\mathcal{D}} \mathbf{1}$ the conclusion follows immediately as the two $\infty$-groupoids admit isomorphic models as simplicial sets.
Next, we want to show that the same is true for pullbacks. Again, we can rectify our diagram, this time into cospan of $\infty$-categories $\mathcal{C}\_0 \to \mathcal{C}\_{01} \leftarrow \mathcal{C}\_1$. We can replace $\mathcal{C}\_0$ and $\mathcal{C}\_0 \to \mathcal{C}\_{01}$ such that the latter becomes a fibration by changing $\mathcal{C}\_0$ up to equivalence. In this case, the desired homotopy limit is realized by the actual pullback and exactly the same argument as for products applies (exponentials and limits commute with limits in simplicial sets).
| 3 | https://mathoverflow.net/users/76636 | 450398 | 181,195 |
https://mathoverflow.net/questions/450368 | 8 | A special case of a theorem of Brian Scott (from [*On the existence of totally inhomogeneous spaces*](https://www.ams.org/journals/proc/1975-051-02/S0002-9939-1975-0375262-5/S0002-9939-1975-0375262-5.pdf)) is that there is a size-continuum set $S\subset\mathbb{R}$ such that if $x,y\in S$ are distinct then $S\setminus\{x\}\not\cong S\setminus\{y\}$. The proof relies heavily on the axiom of choice, and at a glance any such $S$ must be quite wild. However, I don't immediately see how to show that such an $S$ must fail any of the standard regularity properties (e.g. measurability).
>
> **Question**: Is the existence of such an $S$ consistent with $\mathsf{ZF+DC+AD}$?
>
>
>
"Obviously" the answer is negative ...
| https://mathoverflow.net/users/8133 | Can totally inhomogeneous sets of reals coexist with determinacy? | In *Rigid Borel sets and better quasiorder theory* (Logic and combinatorics, Proc. AMS-IMS-SIAM Conf., Arcata/Calif. 1985, Contemp. Math. 65, 199-222 (1987), [zbMath review here](https://zbmath.org/0646.03045)) Fons van Engelen, Arnold Miller, and John Steel showed that the only rigid Borel sets are the singletons; the proof yields the same for all sets of reals if one assumes $\mathsf{AD}$.
So the answer to the present question is negative.
| 8 | https://mathoverflow.net/users/5903 | 450402 | 181,198 |
https://mathoverflow.net/questions/450412 | 0 | Let $p \in \mathbb{Z}$ be prime and $K / \mathbb{Q}$ be a finite Galois extension. The Galois group $G$ of $K$ acts on the primes of $\mathcal{O}\_K$ over $p$. Do we know any statistical information about the distribution of isomorphism classes of these actions as $p$ ranges over all unramified primes? By this I mean that we define two actions of $G$ on sets $X$, $Y$ to be isomorphic if there exists a $G$-equivariant bijection from $X$ to $Y$, and ask how often a given isomorphism class occurs (for your favorite definition of “often”). Have questions in this vein been studied before? If we can allow the base field to be a number field different from $\mathbb{Q}$ that’s even better.
| https://mathoverflow.net/users/91041 | Statistics of action of Galois group of number field on primes over unramified rational primes | This is an elaboration of Chris Wuthrich's comment. Let $p$ be unramified (i.e. $p$ does not divide the discriminant of the Galois extension $K / \mathbb{Q}$), and let $\mathfrak{P}$ be a prime in $\mathcal{O}\_K$ over $p$ with decomposition group $D\_\mathfrak{P}\leq G$. The primes in $\mathcal{O}\_K$ over $p$ correspond bijectively to the left cosets in $G/D\_\mathfrak{P}$, and the action of $G$ on them corresponds to the left action of $G$ on $G/D\_\mathfrak{P}$. Now $D\_\mathfrak{P}$ is the cyclic group generated by $\mathrm{Frob}\_\mathfrak{P}$, and changing $\mathfrak{P}$ results in conjugating $\mathrm{Frob}\_\mathfrak{P}$. Hence the question boils down to how often the conjugacy class of $\mathrm{Frob}\_\mathfrak{P}$ equals a given conjugacy class $C\subset G$. By the Chebotarev density theorem, the density of such primes $p$ equals $|C|/|G|$.
| 4 | https://mathoverflow.net/users/11919 | 450414 | 181,200 |
https://mathoverflow.net/questions/450417 | 9 | Let $F\_n$ be the free group on letters $\{x\_1,\ldots,x\_n\}$ and let $X\_n$ be the (reduced) outer space of rank $n$. Points of $X\_n$ thus correspond to pairs $(G,\mu)$, where $G$ is a finite connected metric graph of total edge-length $1$ with no valence $1$ or $2$ vertices and no separating edges and $\mu\colon F\_n \rightarrow \pi\_1(X\_n)$ is the conjugacy class of an isomorphism.
The space $X\_n$ was introduced in the paper
M. Culler and K. Vogtmann, [Moduli of graphs and automorphisms of free groups](https://doi.org/10.1007%2FBF01388734), Invent. Math. 84 (1986), no. 1, 91–119.
The main theorem of this paper is that $X\_n$ is contractible. They prove this in the following way. Let $W\_0$ be the following set of elements of $F\_n$:
* $x\_i$ with $1 \leq i \leq n$, and
* $x\_i x\_j$ and $x\_i x\_j^{-1}$ with $1 \leq i < j \leq n$.
For each $w \in W\_0$, and each point $(G,\mu)$ of $X\_n$, let $|(G,\mu)|\_w$ be the length of the shortest unbased loop in $G$ representing the conjugacy class $w$. Define
$$|(G,\mu)| = \sum\_{w \in W\_0} |(G,\mu)|\_w.$$
They use this as a sort of "Morse function" on $X\_n$, and study its level sets. One of their key results is as follows. A **rose** in $X\_n$ is a point $(R,\mu)$, where $R$ is a metric graph with $1$ vertex and $n$ edges, each of length $1/n$. Let $(R\_0,\mu\_0)$ be the rose where $\mu\_0$ identifies the loops in $R\_0$ with the basis elements $x\_1,\ldots,x\_n$.
**Proposition** (Prop 6.2.5 in the above paper): The rose $(R\_0,\mu\_0)$ is the unique minimum of the above Morse function among roses, i.e., if $(R,\mu)$ is another rose then $|(R\_0,\mu\_0)| \leq |(R,\mu)|$ with equality if and only if $(R,\mu) = (R\_0,\mu\_0)$.
I remark that Prop 6.2.5 has the opposite inequality to the one I wrote, but I think this is a typo and the proof seems to give the above. The proof of this proposition is pretty simple and convincing.
However, it seems to be contradicted in the following paper:
J. Smillie and K. Vogtmann, [Length functions and outer space](http://doi.org/10.1307/mmj/1029004602), Michigan Math J. 39 (1992) 485–493.
In fact, Theorem 1 of this paper seems to say that not only is the above proposition false, but it remains false if $W\_0$ is replaced by any other finite set of elements of $F\_n$. The authors of this paper do not say anything about this contradicting anything in the previous paper (even though they share an author, so presumably if I am right they were aware of this).
**Question**: Am I misinterpreting any of these papers? If not, what exactly is going on here?
| https://mathoverflow.net/users/508240 | Morse theory on outer space via the lengths of finitely many conjugacy classes | You don't misunderstand, it's a subtle point that I'm sure I'll get wrong here too. You might find the proof of a slightly more general statement in Krstić and Vogtmann's "Equivariant Outer Space and automorphisms of free-by-finite groups" illuminating: basically the idea is that for any rose (I believe in fact for any marked metric graph, but let's not get ahead of ourselves), there exists a finite set of conjugacy classes whose lengths pin down that particular rose completely.
The problem, or the result of the Smilie–Vogtmann paper, is that the lengths of *those same conjugacy classes* will not work for *every* marked graph at once. The reason there's no contradiction is that the *minimizing* marked graph is unique, while the others need not be.
| 13 | https://mathoverflow.net/users/135175 | 450421 | 181,202 |
https://mathoverflow.net/questions/450297 | 4 | My question comes from a computation in the paper [Central limit theorem for Maxwellian molecules and truncation of Wild expansion](https://citeseerx.ist.psu.edu/document?repid=rep1&type=pdf&doi=827b55efbcc4a2eaf1c12526590ab0d067f0f650). Specially, consider the following Boltzmann equation
$$\frac{\partial f}{\partial t} = Q^+(f,f) - f \label{1}\tag{1}$$ with $f(t=0) = F$. Introduce the notation $f \circ f = Q^+(f,f)$ which is commonly referred to as the Wild convolution, where $Q^+(\cdot,\cdot)$ is some bilinear collision operator. If we define a map $f \mapsto \Phi(f)$ by
$$\Phi(f)(t) = \mathrm{e}^{-t}F + \int\_0^t \mathrm{e}^{-(t-s)}f\circ f(s)\mathrm{d}s \label{2}\tag{2},$$ then $f(t)$ solves the Boltzmann equation \eqref{1} exactly when $\Phi(f) = f$. Thus, to construct a solution to \eqref{1} with $f(t=0) = F$, the aforementioned paper suggests that we put $f\_{(0)} = 0$, and then define $f\_{(j+1)} = \Phi\left(f\_{(j)}\right)$ for all $j\geq 1$. It is easy to check that $f\_{(1)}(t) = \mathrm{e}^{-t}F$ and $$f\_{(2)}(t) = \mathrm{e}^{-t}F + \int\_0^t \mathrm{e}^{-(t-s)}f\_{(1)}\circ f\_{(1)}(s)\mathrm{d}s = \mathrm{e}^{-t}F + \mathrm{e}^{-t}(1-\mathrm{e}^{-t})F\circ F.$$ However, the authors also computed $f\_{(3)}$ and claimed that $$f\_{(3)}(t) = \mathrm{e}^{-t}F + \mathrm{e}^{-t}(1-\mathrm{e}^{-t})F\circ F + \mathrm{e}^{-t}(1-\mathrm{e}^{-t})^2\left(\frac{1}{2}F\circ (F\circ F) + \frac{1}{2}(F\circ F)\circ F\right),$$ which confuses me a lot since I do not understand why in the expression for $f\_{(3)}$ there is no term involving $(F\circ F)\circ (F\circ F)$. Any help is greatly appreciated!
---
Remark 1: Given Professor Terrence Tao's comment, it is **unlikely** that the authors made a typo in the computation of $f\_{(3)}$, which in my personal opinion should also include a term $$\frac{1}{3}\mathrm{e}^{-t}(1-\mathrm{e}^{-t})^3 (F\circ F)\circ (F\circ F) \label{3}\tag{3} $$ I reach out to this claim based on a statement in the same paper (near formula (1.15) in the version provided in this post) saying that ``McKean's expression for $f\_{(n)} - f\_{(n-1)}$ is (1.15)'', but the right side of (1.15) only involves the $n$-fold Wild convolution $Q^+\_n(F)$. This means that if the formula for $f\_{(3)}$ indeed includes the term \eqref{3} which I think the authors have somehow missed, then $f\_{(3)}$ (with $n=3$) will involve $(F\circ F)\circ (F\circ F)$, which is a member of $Q^+\_4(F)$. At this point I am super confused... By the way, I also need to mention that McKean's paper are usually written in the 1950s or 1960s and his papers are usually pretty short (or say "condensed") and hard to digest in my personal opinion...
---
Remark 2: Professor Terrence Tao's answer is very helpful, but my concern is from a different perspective. It is a classical result (see for instance [Villani's monograph](https://cedricvillani.org/sites/dev/files/old_images/2012/07/B01.Handbook.pdf))) that a semi-explicit representation formula for solutions of the Boltzmann equation \eqref{1} can be written as
$$f\_t = \mathrm{e}^{-t}\sum\_{n=1}^\infty \left(1-\mathrm{e}^{-t}\right)^{n-1} Q^+\_n(f\_0) \label{4}\tag{4} $$ where the $n$-fold nonlinear operator $Q^+\_n$ is defined recursively by
$$Q^+\_1(f\_0) = f\_0,\quad Q^+\_n(f\_0) = \frac{1}{n-1}\,\sum\_{k=1}^{n-1} Q^+\left(Q^+\_k(f\_0),Q^+\_{n-k}(f\_0)\right).$$ Thus, using Professor Terrence Tao's notation, we have that $$f\_t = \lim\_{n \to \infty} f\_{[n]} = \lim\_{n \to \infty} \mathrm{e}^{-t}\sum\_{k=1}^n \left(1-\mathrm{e}^{-t}\right)^{k-1} Q^+\_k(f\_0) .$$ From a motivation to construct the solution formula \eqref{4} instead of checking directly that \eqref{4} indeed solves \eqref{1}, we would proceed as in the Wild's iterative scheme $f\_{(j+1)} = \Phi(f\_{(j)})$ described before. What is funny is that we have $f\_{(0)} = f\_{[0]} = 0$, $f\_{(1)} = f\_{[1]}$, $f\_{(2)} = f\_{[2]}$, while $f\_{(3)} > f\_{[3]}$. In some heuristic sense, it is not clear at all as to how to arrive at the "ansatz" (or say "guess") \eqref{4} at the first place if we just want to set/take $$f\_t := \lim\_{n \to \infty} f\_{[n]}.$$ By the way, I tried email Professor Eric Carlen some year ago asking about a (different) question I have regarding one of his papers and I never got replied (which occurs very often in general I guess), so I would not try to bother him again.
| https://mathoverflow.net/users/163454 | Iterated Duhamel's formula for solutions of Boltzmann equation | I took a closer look at the manuscript. If one lets $f\_{[n]}$ denote the quantity implicitly defined by (1.15), then it appears to me that this is indeed slightly different from $f\_{(n)}$ in that some terms in the expansion of $f\_{(n)}$ are missing in $f\_{[n]}$, leading to the inequalities
$$ f\_{[n]} \leq f\_{(n)} \leq f.$$
On the other hand, this doesn't seem to impact the main results of the paper, which then control the size of $f - f\_{[n]}$, and hence $f - f\_{(n)}$.
It is probably worth following up with one of the authors of the paper for confirmation, though.
| 5 | https://mathoverflow.net/users/766 | 450422 | 181,203 |
https://mathoverflow.net/questions/450441 | 2 | Let $M$ be a von Neumann algebra, $P(M)$ be its projection lattice, and $\mathcal{F}$ a proper filter on $P(M)$. Does there exist a state $\varphi$ (not necessarily normal) s.t. $\varphi(p) = 1$ for all $p \in \mathcal{F}$? Does this hold even more generally, say for the projection lattices of $AW^\*$-algebras?
| https://mathoverflow.net/users/504602 | Defining states on von Neumann algebras from filters on the projection lattices | Filters are directed downward. Given a filter $F$, for every $p\in F$ let $\phi\_p$ be a state that takes the value $1$ on $p$, then find a cluster point of the net $(\phi\_p)\_{p\in F}$. This will be a state that takes the value $1$ on everything in $F$. I think this works fine for $AW{}^\*$-algebras.
| 4 | https://mathoverflow.net/users/23141 | 450444 | 181,206 |
https://mathoverflow.net/questions/450428 | 4 | Trying to find and answer to this question, I have encountered two more-studied problems.
The first is to find when a Banach space admits an **equivalent** uniformly convex norm. [The answer](https://mathoverflow.net/a/30458/58082) is that for example separable spaces always do, but nonseparable spaces might not. This does not satisfy me because I only want the existence of a *continuous* strictly convex norm. I am happy for it to generate a strictly weaker topology.
I have also found [this](https://www.um.es/beca/papers/UniformlyConvexFunctionOnBanachSpaces.pdf) paper that says that is we are interested in uniform and not strict convexity, then even admitting a single uniformly convex function is enough to make the space equivalent to a uniformly convex one. However this is about uniform convexity which is stronger than strict convexity.
The second problem is about compact convex subsets of a Banach space admitting a continuous strictly convex function. Hervé proved this is the case iff the set is metrisable. The proof is Theorem I.4.3 of "Erik M.Alfsen Compact Convex Sets and Boundary Integrals". This does not sound useful to me either, because all the sets I'm interested in are both noncompact and metrisable.
Does anyone know whether every Banach space admits a continuous strictly convex norm? Or do I need to put on my learning goggles and give the paper a detailed read?
| https://mathoverflow.net/users/58082 | Does every Banach space admit a continuous (not necessarily equivalent) strictly convex norm? | If $\|\cdot\|\_1$ is a continuous strictly convex norm on $(X,\|\cdot\|\_0)$, then $\|x\|\_2=\|x\|\_0 + \|x\|\_1$ defines an strictly convex norm on $X$ equivalent to $\|\cdot\|\_0$. Therefore, a space like $\ell\_\infty/c\_0$ that does not admit an equivalent strictly convex norm, it does not admit a continuous strictly convex norm.
J. Bourgain proved that $\ell\_\infty/c\_0$ does not admit an equivalent strictly convex norm [here.](https://www.ams.org/journals/proc/1980-078-02/S0002-9939-1980-0550499-2/S0002-9939-1980-0550499-2.pdf)
| 13 | https://mathoverflow.net/users/39421 | 450445 | 181,207 |
https://mathoverflow.net/questions/434602 | 9 | I'm looking for a reference that covers things like the lemma below - it doesn't have to be the exact statement I'm going to give, anything in the general ballpark would probably be useful. Or if you know a very short proof of the lemma - that would be interesting too.
So, I have a map $\pi: E \to U$ in a category $C$ (let's say $C$ has finite limits) that I want to think as a kind of "small maps classifier", that is I'm going to say that a map is "$\pi$-small" if it is a pullback of $\pi: E \to U$ (or maybe if it is locally a pullback of $E\to U$ for some notion of locality).
I'm interested in figuring out if small maps are closed under composition. In general, this has no reason to be true of course, but if the polynomial endofunctor $P\_\pi$ induced by $\pi$ is a cartesian monad (which is often the case) then this should imply that $\pi$-small maps are closed under-composition. More precisely - I think we have the following:
**Lemma :** Assume the map $\pi$ is exponentiable, and let $P\_\pi$ the polynomial endofunctor induced by $\pi$ (that is $P\_\pi(X) = \pi\_\*(X \times E)$ seen as an endofunctor $C \to C$). Then the following are equivalent:
1. $\pi$-small maps (that is pullback of $\pi$) are closed under composition.
2. There exists a cartesian natural transformation $P\_\pi \circ P\_\pi \to P\_\pi$.
I'm mostly interested in (2) => (1) and I'm fine with assuming that $P\_\pi \circ P\_\pi \to P\_\pi$ is part of a monad structure. But as I said, anything in that spirit would be of interest. I feel there are many area of category theory where this could have been studied (Models of type theory, algebraic set theory, theory of polynomial functor, etc...) but I couldn't find it in the literature... The proof isn't awfully hard, but unless I'm missing something it require lots of stuff (like properties of polynomial functors and the formula for their composition, with some tricky computation on top of this)
| https://mathoverflow.net/users/22131 | Reference request: a lemma on universes and polynomial monads | This is mentioned in Remark 13 in
>
> Steve Awodey: *Natural models of homotopy type theory*, January 2017, [arXiv:1406.3219](https://arxiv.org/abs/1406.3219)
>
>
>
and then followed up in detail in:
>
> Steve Awodey, Clive Newstead: *Polynomial pseudomonads and dependent type theory*, February 2018, [arXiv:1802.00997](https://arxiv.org/abs/1802.00997)
>
>
>
as well as in:
>
> Clive Newstead: *Algebraic models of dependent type theory*, March 2021, [arXiv:2103.06155](https://arxiv.org/abs/2103.06155)
>
>
>
| 9 | https://mathoverflow.net/users/123877 | 450446 | 181,208 |
https://mathoverflow.net/questions/450410 | 4 | Let $C^{j\_3 m\_3}\_{j\_1 m\_1 j\_2 m\_2}$ be the standard Clebsch–Gordan coefficients of $\operatorname{SU}(2)$. They obey the orthogonality relation
$$ \sum\_{j\_3} \sum\_{m\_3} \left(C^{j\_3 m\_3}\_{j\_1 m\_1 j\_2 (m\_3 - m\_1)} \right)^2 = 1.$$
My question is about what can be said if I remove the sum over $j\_3$. Does there exist a bound for
$$ \max\_{j\_3} \sum\_{m\_3} \left(C^{j\_3 m\_3}\_{j\_1 m\_1 j\_2 (m\_3 - m\_1)} \right)^2 $$
in terms of $m\_1$, $j\_1$ and $j\_2$? Of course it is $\leq 1$, but I am interested in whether this expression decays in $j\_1$ and $j\_2$, or some combination thereof. Positive or negative statements, or a reference, would be very useful.
| https://mathoverflow.net/users/104213 | Single sum of squares of Clebsch–Gordan coefficients | Consider the sum of [Clebsch–Gordan coefficients](https://en.wikipedia.org/wiki/Clebsch%E2%80%93Gordan_coefficients),$^\ast$
$$J= \sum\_{m\_3=-j\_3}^{j\_3} \left(C^{j\_3,m\_3}\_{j\_1, m\_1; j\_2, (m\_3 - m\_1)} \right)^2$$
with $2j\_1,2j\_2,2j\_3\in\mathbb{N}$ and $2m\_1\in\mathbb{Z}$. For an nonvanishing sum we also need $j\_3\in\{|j\_1-j\_2|,|j\_1-j\_2|+1,\ldots j\_1+j\_2-1, j\_1+j\_2\}$ and $m\_1\in\{-j\_1,-j\_1+1,\ldots,j\_1-1, j\_1\}$.
With these restrictions the sum is independent of $m\_1$, so I may set $m\_1=j\_1$. The sum also increases monotonically with increasing $j\_3$, reaching its maximal value $J\_{\rm max}$ for $j\_3=j\_1+j\_2$, hence
$$J\_{\rm max}=\sum\_{m=-j\_2}^{j\_2} \left(C^{j\_1+j\_2,j\_1+m}\_{j\_1, j\_1; j\_2, m} \right)^2=\sum\_{m=-j\_2}^{j\_2}\frac{\Gamma (2 j\_2+1) \Gamma (2 j\_1+j\_2+m+1)}{(1+2j\_2)\Gamma (2 j\_1+2 j\_2+1) \Gamma (j\_2+m+1)}$$
$$\qquad=\frac{1+2j\_1+2j\_2}{(1+2j\_1)(1+2j\_2)}.$$
---
$^\ast$ Mathematica normalizes these coefficients such that $\sum\_{j\_3}J=1+2j\_2$. Here I use the normalization of the OP, where $\sum\_{j\_3}J=1$.
| 7 | https://mathoverflow.net/users/11260 | 450447 | 181,209 |
https://mathoverflow.net/questions/450360 | 3 | Based on a method that apparently seems to be widely used in computational chemistry (cf <https://en.wikipedia.org/wiki/Anisotropic_Network_Model>)
Trying to build a very simple model with 3 atoms linked together in an equilateral triangle, Suppose the 2d coordinates of the 3 atoms are as follows:
$(x\_0, y\_0) = (0, 0)$
$(x\_1, y\_1) = (1/2, \frac{\sqrt{3}}{2})$
$(x\_2, y\_2) = (1, 0)$
Computing the Hessian matrix as described in the wikipedia article above I get the following matrix entries for the Hessian matrix H.
| | | | | | |
| --- | --- | --- | --- | --- | --- |
| 0 | 0 | 1/4 | $\sqrt{3}/4$ | 1 | 0 |
| 0 | 0 | $\sqrt{3}/4$ | 3/4 | 0 | 0 |
| 1/4 | $\sqrt{3}/4$ | 0 | 0 | 1/4 | -$\sqrt{3}/4$ |
| $\sqrt{3}/4$ | 3/4 | 0 | 0 | -$\sqrt{3}/4$ | 3/4 |
| 1 | 0 | 1/4 | -$\sqrt{3}/4$ | 0 | 0 |
| 0 | 0 | -$\sqrt{3}/4$ | 3/4 | 0 | 0 |
So I am expecting the matrix to have 3 eigenvalues equal to zero, corresponding to the 3 rigid body motions in 2 dimensions (2 translations one rotation).
However this is not the case, the 3 lowest eigenvalues are 1/2 and a multiplicity 2 eigenvalue around -0.65...
So what is wrong? Did I misinterpret the formula in this wikipedia article?
Isn't this the case that 3 eigenvalues should exactly be equal to 0? Or does this only occur asymptotically with large molecules as the number of atoms gets bigger?
In advance many thanks for any help!
| https://mathoverflow.net/users/22279 | Elastic network model Hessian rigid body motion 0 eigenvalues | Your construction of the stiffness matrix (Hessian or Kirchhoff matrix) is not correct. The $2\times 2$ diagonal blocks should not be zero,$^\ast$ these are minus the sums of the other blocks in the same row, $H\_{ii}=-\sum\_{j\neq i}H\_{ij}$, where $H\_{ij}=\begin{pmatrix}
(x\_j-x\_i)^2&(x\_j-x\_i)(y\_j-y\_i)\\
(y\_j-y\_i)(x\_j-x\_i)&(y\_j-y\_i)^2\end{pmatrix}$ denotes the $2\times 2$ block in position $i,j$.
Hence the correct stiffness matrix $H$ in your case $(x\_0, y\_0) = (0, 0),\;\;(x\_1, y\_1) = (\tfrac{1}{2}, \tfrac{1}{2}\sqrt{3}),\;\;(x\_2, y\_2) = (1, 0)$ is
$$H=\left(
\begin{array}{cccccc}
-\tfrac{5}{4} & -\tfrac{\sqrt{3}}{4} & \tfrac{1}{4} & \tfrac{\sqrt{3}}{4} & 1 & 0 \\
-\tfrac{\sqrt{3}}{4} & -\tfrac{3}{4} & \tfrac{\sqrt{3}}{4} & \tfrac{3}{4} & 0 & 0 \\
\tfrac{1}{4} & \tfrac{\sqrt{3}}{4} & -\tfrac{1}{2} & 0 & \tfrac{1}{4} & -\tfrac{\sqrt{3}}{4} \\
\tfrac{\sqrt{3}}{4} & \tfrac{3}{4} & 0 & -\tfrac{3}{2} & -\tfrac{\sqrt{3}}{4} & \tfrac{3}{4} \\
1 & 0 & \tfrac{1}{4} & -\tfrac{\sqrt{3}}{4} & -\tfrac{5}{4} & \tfrac{\sqrt{3}}{4} \\
0 & 0 & -\tfrac{\sqrt{3}}{4} & \tfrac{3}{4} & \tfrac{\sqrt{3}}{4} & -\tfrac{3}{4} \\
\end{array}
\right)$$
Its eigenvalues are $(-3,-3/2,-3/2,0,0,0)$; indeed, three eigenvalues are equal to zero.
---
$^\ast$ See, for example, equation (3) of [Anisotropic network model: systematic evaluation and a new web interface](https://academic.oup.com/bioinformatics/article/22/21/2619/251345)
| 3 | https://mathoverflow.net/users/11260 | 450453 | 181,212 |
https://mathoverflow.net/questions/450220 | 9 | Let $x>0$ and consider the integral
$$I(x):=\int\_0^\infty \frac{e^{i r}}{r^{\frac{1}{2}}} \int\_0^\infty \frac{e^{-s}}{s^{\frac{1}{2}}} \frac{r}{sx+\sqrt{sxr}+r} \, ds \, dr.$$
I am trying to determine the asymptotic behavior of $I(x)$ as $x\rightarrow+\infty$.
Note that $\lim\_{x\rightarrow+\infty}I(x)=0$.
Here is why:
---
Notice that
$(r,s)\mapsto{e^{i r}}{r^{-\frac{1}{2}}}{e^{-s}}{s^{-\frac{1}{2}}}\frac{r^{2}}{s^2 x^2+sxr+r^2}$
is dominated by ${r^{-\frac{1}{2}}}{e^{-s}}{s^{-\frac{1}{2}}}$ which is an $L^{1}([0,1]\times [0,+\infty[)$ function.
On $[1,+\infty[\times [0,+\infty[$ we need an integration by parts w.r.t. $r$ to justify the passing of the limit as $x\rightarrow+\infty$. Doing so, we get
\begin{align}
& \int\_1^\infty\frac{e^{i r}}{r^{\frac{1}{2}}} \int\_0^\infty \frac{e^{-s}}{s^{\frac{1}{2}}} \frac{r}{sx+\sqrt{sxr}+r} \, ds \, dr \\[6pt]
= {} & {i e^{i }}\int\_{0}^{\infty}\frac{e^{-s}}{s^{\frac{1}{2}}}\frac{1}{sx+\sqrt{sx}+1} \, ds \\[6pt]
& {}+\frac{1}{2}\int\_1^\infty \frac{e^{i r}}{r^{\frac{3}{2}}} \int\_0^\infty \frac{e^{-s}}{s^{\frac{1}{2}}} \frac{r}{sx+\sqrt{sxr}+r} \, ds \, dr \\[6pt]
& {}-\int\_1^\infty \frac{e^{i r}}{r^{\frac{3}{2}}} \int\_0^\infty \frac{e^{-s}}{s^{\frac{1}{2}}} \left(\frac{r}{sx+\sqrt{sxr}+r} -\frac{\frac{1}{2}\sqrt{sxr}+r}{(sx+\sqrt{sxr}+r)^2} \right)\, ds \, dr \tag{$\*\*$}
\end{align}
Simple remarks on the formula $(\*\*)$:
(1) The limit (taken at computing the boundary terms) and derivative under the $s$-integral are both justified by
the fact that $e^{-s}/\sqrt{s}\in L^1 ([0,+\infty[)$.
(2) A factor $1/r$ has been pulled outside the $s$-integral.
More importantly,
(3) The integrand in this formula, as a function of $(r,s)$, is absolutely dominated by
$r^{-3/2}s^{-1/2}e^{-s}$ which is an $L^1( \left[1,+\infty\right[\times [0,+\infty[ )$ function.
---
What is the precise the asymptotic behavior of $I(x)$ as $x\to+\infty\text{ ?}$
I am probably wrong, but I tried rescaling/changing order of integration/Taylor-expanding the rational factor in any variable. Nothing seems to work.
| https://mathoverflow.net/users/116555 | Asymptotic behavior of a certain oscillatory integral | We can evaluate $I(x)$ explicitly, and then asymptotically.
Indeed, using the substitution $s=ru/x$, we get
\begin{equation\*}
I(x)=\frac1{\sqrt x}\lim\_{R\to\infty}J\_R(x), \tag{1}\label{1}
\end{equation\*}
where
\begin{equation\*}
\begin{aligned}
J\_R(x)&:=\int\_0^R dr\,e^{ir}\int\_0^\infty \frac{du}{\sqrt u\,(u+\sqrt u+1)}e^{-ru/x} \\
&=\int\_0^\infty \frac{du}{\sqrt u\,(u+\sqrt u+1)}\int\_0^R dr\,e^{(i-u/x)r} \\
&=\int\_0^\infty \frac{du}{\sqrt u\,(u+\sqrt u+1)}\frac{1-e^{(i-u/x)R}}{u/x-i}.
\end{aligned}
\end{equation\*}
Next, (i) for any real $u,x>0$ we have $\dfrac{1-e^{(i-u/x)R}}{u/x-i}\to\dfrac1{u/x-i}$ as $R\to\infty$, (ii) for any real $u,x,R>0$ we have $\Big|\dfrac{1-e^{(i-u/x)R}}{u/x-i}\Big|\le\dfrac2{|u/x-i|}\le2$, and (iii) for any real $x>0$ we have $\displaystyle{\int\_0^\infty \frac{du}{\sqrt u\,(u+\sqrt u+1)}\,2<\infty}$.
So, by dominated convergence,
\begin{equation\*}
\lim\_{R\to\infty}J\_R(x)=J(x), \tag{2}\label{2}
\end{equation\*}
where
\begin{equation\*}
\begin{aligned}
&J(x):=\int\_0^\infty \frac{du}{\sqrt u\,(u+\sqrt u+1)}\frac1{u/x-i} \\
&=\frac{x \left(-18 \ln x+\pi \left(8 i \sqrt{3} x+(9-9 i) \sqrt{2} \sqrt{x}-\frac{18
\sqrt[4]{-1}}{\sqrt{x}}+4 \sqrt{3}+9 i\right)\right)}{18 (-1+x (x-i))}
\end{aligned}
\tag{3}\label{3}
\end{equation\*}
(note that the integrand in \eqref{3} is rational in $\sqrt u\,$, so that one can use partial fraction decomposition to get \eqref{3}).
So, as $x\to\infty$,
\begin{equation\*}
J(x)\to c:=\frac{4 i \pi }{3 \sqrt{3}},
\end{equation\*}
so that, by \eqref{1},
\begin{equation\*}
I(x)\sim \frac c{\sqrt x}.
\end{equation\*}
| 8 | https://mathoverflow.net/users/36721 | 450458 | 181,213 |
https://mathoverflow.net/questions/450186 | 4 | Let $ O(n) $ be the manifold of orthornormal matrix, i.e.
$$
O(n)=\{A\in\mathbb{R}^{n\times n}:A^TA=I\}.
$$
Then $ O(n) $ is a submanifold of $ \mathbb{R}^{n\times n} $. On $ O(n) $, there is a Riemannian metric on $ O(n) $ induced by Euclidean metric of $ \mathbb{R}^{n\times n} $. I want to consider the geodesics on it. I know that if $ n=2 $, then $ O(n) $ can be seen as $ \mathbb{S}^1 $. However the case is much more hard if $ n\geq 3 $. Can you give some hints or references?
| https://mathoverflow.net/users/241460 | Geodesics on orthogonal matrix | A direct computation shows that for each $T\in O(n)$, the map $L\_T : \mathbb{R}^{n\times n} \to \mathbb{R}^{n\times n}$ given by
$$
A \mapsto TA,
$$
is an isometry. This isometry preserves the submanifold $O(n)$, thus it's also an isometry of $O(n)$ when the latter is endowed with the induced metric. The same applies for the map $R\_T : \mathbb{R}^{n\times n} \to \mathbb{R}^{n\times n}$,
$$
A \mapsto AT.
$$
It follows that the induced metric on $O(n)$ is *bi-invariant.* For such metrics on Lie groups it is well-known that geodesics through the identity are given by orbits of one-parameter subgroups: see for instance problem 3 in Ch.3 of Do Carmo's Riemannian geometry (2nd Ed.), which includes some hints.
| 2 | https://mathoverflow.net/users/14708 | 450463 | 181,214 |
https://mathoverflow.net/questions/450462 | 1 | This question is about the content of [this paper](https://arxiv.org/abs/math/0301343) by J. Bourgain, N. Katz, T. Tao.
---
In the final step (page 18) of the proof of Szemerédi-Trotter type theorem, we have already known
$$|A''+A''|\lesssim N^{\frac{1}{2}+C\epsilon}$$
Similarly, by the Balog-Szemerédi-Gowers Theorem, there exists $A^{\star\star}\subseteq A'\setminus\{0\}$ with $|A^{\star\star}|\gtrsim N^{\frac{1}{2}-C\epsilon}$ such that
$$|A^{\star\star}\cdot A^{\star\star}|\lesssim N^{\frac{1}{2}+C\epsilon}$$
My question is why we have $A^{\star\star}=A''$. Do I understand it wrong?
Actually, I have tried to let the above $A^{\star\star}$ be a subset of $A’'\setminus\{0\}$ (by using the Balog-Szemerédi-Gowers Theorem) but it seems that we don’t have (or I cannot prove)
$$\left|\left\{(t,x\_0,x\_1)\in B\times A\times A’’:(1-t)x\_0+tx\_1\in A,t\neq0,1\right\}\right|\gtrsim N^{\frac{3}{2}-C\epsilon}$$
(The similar inequality (16) for $A’'$)
Could anyone help me understand the proof? Thanks for any help!
| https://mathoverflow.net/users/508148 | Szemerédi–Trotter type theorem in finite field | I think you're correct that (18) should read:
$$|A'' + A''| \lesssim N^{1/2+C\epsilon}.$$
The display before (16) tells you that for each $x\_1 \in A' $, (and hence $x\_1 \in A'' \subseteq A'$) one has:
$$|\{(t,x\_0) \in B \times A : (1-t)x\_0 + t x\_1 \in A; t \neq 0, 1 \}| \gtrsim N^{1-C \epsilon}$$
Since $|A''|\gtrsim N^{1/2}$, after adjusting implicit constants, summing over elements in $A'$, gives us the estimate
$$|\{(t,x\_0,x\_1) \in B \times A \times A'' : (1-t)x\_0 + t x\_1 \in A; t \neq 0, 1 \}| \gtrsim N^{3/2-C \epsilon}$$
On the other hand, using (14) (e.g. $|A| \lesssim N^{1/2}$), the pigeonhole principle gives us (after adjusting implicit constants again) that for some $x\_0$ (in fact the typical $x\_0 \in A$) one has:
$$|\{(t,x\_1) \in B \times A'' : (1-t)x\_0 + t x\_1 \in A; t \neq 0, 1 \}| \gtrsim N^{1-C \epsilon}.$$
This is the claim in the display following (18). One can then return to the argument as written.
Also let me point out that these arguments have been simplified and improved since the breakthrough paper of Bourgain-Katz-Tao. For instance, one can replace the repeated applications to BSG with a geometric argument. See, for instance, the paper of Stevens and de Zeeuw: <https://arxiv.org/abs/1609.06284>.
| 1 | https://mathoverflow.net/users/630 | 450465 | 181,215 |
https://mathoverflow.net/questions/450466 | 3 | I'm trying to derive a very basic result stated in several books on random matrix theory (e.g. Terry Tao's book and Potters & Bouchaud's book).
Given a symmetric matrix $A \in \mathbb{R}^{N \times N}$, with eigenvalues $\lambda\_1, \dots, \lambda\_n$, define a complex-valued function $g : \mathbb{C} \mapsto \mathbb{C}$ as follows:
$$
g(z) = \frac{1}{N} \text{Tr} [ (zI - A)^{-1} ] = \frac{1}{N} \sum\_{n=1}^N \frac{1}{z - \lambda\_n}
$$
This is called the Stieltjes transform of $A$. For sufficiently large $z$, it is claimed that the Taylor expansion can be written as (see eq. 2.22 in Potters & Bouchaud):
$$
g(z) = \sum\_{k=0}^\infty \frac{1}{z^{k + 1}} \frac{1}{N} \text{Tr}[A^k]
$$
Both books emphasize that this works *for sufficiently large* $z$ and I suspect this is the key piece of information I'm missing. Because If I try to expand around $z = 0$, I get:
$$
g(z) = \sum\_{k=0}^\infty \frac{z^k}{k!} \cdot g^{(k)}(0) = \sum\_{k=0}^\infty \frac{z^k}{k!} \sum\_{n=1}^N \frac{(-1)^k \cdot k!}{N \cdot (-\lambda\_n)^k} = \sum\_{k=0}^\infty \frac{z^k}{N} \sum\_{n=1}^N \frac{1}{\lambda\_n^k} = \sum\_{k=0}^\infty z^k \frac{1}{N} \text{Tr}[A^{-k}]
$$
which is more-or-less the reciprocal of what I was hoping for. This is a total stab in the dark, but is there some justification for expanding instead like...
$$
g(z) = \sum\_{k=0}^\infty \frac{k!}{z^k \cdot g^{(k)}(0)}
$$
| https://mathoverflow.net/users/59128 | Taylor expansion of Stieltjes Transform | You certainly can't expand around $z=0$ if "$z$ has to be sufficiently large". Expand around $1/z =0$:
$$
\frac{1}{N} \sum\_{n=1}^{N}\frac{1}{z-\lambda\_{n} } =
\frac{1}{N} \frac{1}{z} \sum\_{n=1}^{N}\frac{1}{1-\lambda\_{n}/z }
=\frac{1}{N} \frac{1}{z} \sum\_{n=1}^{N} \sum\_{k=0}^{\infty } \frac{\lambda\_{n}^{k} }{z^k } = \sum\_{k=0}^{\infty } \frac{1}{z^{k+1} } \frac{1}{N} \mbox{Tr } [A^k ]
$$
(it seems your quoted result has an error in the summation index - it should start at $k=0$, not $k=1$).
| 8 | https://mathoverflow.net/users/134299 | 450468 | 181,216 |
https://mathoverflow.net/questions/450429 | 5 | I am interested in the interplay between the Playfair and Proclus Axioms in linear spaces.
By a *[linear space](https://mathworld.wolfram.com/LinearSpace.html)* I understand a pair $(X,\mathcal L)$ consisting of a set $X$ and a family $\mathcal L$ of subsets of $X$ such that the following axioms are satisfied:
(L1) for any distinct points $x,y\in X$ there exists a unique set $L\in\mathcal L$ containing the points $x,y$;
(L2) every set $L\in\mathcal L$ contains at least two distinct points.
Elements of the family $\mathcal L$ are called *lines*. By the axiom (L1), for every distinct points $p,q\in X$ there exists a unique line containing those points. This unique line will be denoted by $L(a,b)$. Also we put $L(a,a)=\{a\}$.
Given a linear space $(X,\mathcal L)$, consider the following (well-known) axioms:
**[Playfair](https://en.wikipedia.org/wiki/Playfair%27s_axiom):** For every line $L\in\mathcal L$, points $a,b\in L$, $c\in X\setminus L$ and $x\in L(a,c)\setminus L$ there exists a unique point $y\in L(c,b)$ such that $L(x,y)\cap L=\emptyset$;
**[Proclus](https://mathworld.wolfram.com/ProclusAxiom.html):** For every lines $L\in \mathcal L$, points $a,b\in L$, $c\in X\setminus L$, $x\in L(a,c)\setminus\{c\}$, $y\in L(c,b)\setminus\{c\}$ and $z\in L(x,y)$, if $L(a,b)\cap L(x,y)=\emptyset$, then $L(c,z)\cap L(a,b)\ne\emptyset$.
>
> **Question.** Does **Playfair** imply **Proclus** in linear spaces?
>
>
>
| https://mathoverflow.net/users/61536 | Does Playfair imply Proclus? | I think the following construction gives a counterexample. It stems from the observation that the Playfair axiom is quite weak in the case where all lines only have three points (it produces some pairs of parallel lines, but doesn't force any new intersections between lines).
The simplest geometry in which lines have three points is a vector space ${\bf F}\_3^n$ over the field of three elements, where the lines $L$ take the form $\{a,b,c\}$ where $a,b,c \in {\bf F}\_3^n$ are distinct with $a+b+c=0$. This geometry turns out to obey both the Playfair and Proclus axioms, so we don't have a counterexample yet. But one can modify this geometry in a flexible fashion by a skew-product construction. Let $f: {\bf F}\_3^n \to {\bf F}\_3^m$ be an arbitrary even function with $f(0)=0$. We can then define a geometry on ${\bf F}\_3^n \times {\bf F}\_3^m$ by defining the lines to be the set of triples $\{ (x,x'), (y,y'), (z,z')\}$ of three distinct points $(x,x'), (y,y'), (z,z')$ in ${\bf F}\_3^n \times {\bf F}\_3^m$ with
$$ x+y+z=0$$
and
$$ x'+y'+z' = f(y-x)$$
(note that $f(y-x)=f(x-y)=f(z-y)=f(y-z)=f(x-z)=f(z-x)$ since $f$ is even and $x,y,z$ are in arithmetic progression). One easily checks that this is a linear system. It also obeys the Playfair axiom. To see this, one just has to check that if $(a,a'), (b,b'), (c,c')$ are not collinear, $(x,x')$ is the third point on $L((a,a'), (c,c'))$, and $(y,y')$ is the third point on $L((b,b'), (c,c'))$, then $L((a,a'),(b,b'))$ and $L((x,x'),(y,y'))$ do not intersect. Suppose for contradiction that they intersected in $(z,z')$. Then we have the equations
$$ a + b + z = a + c + x = b + c + y = x + y + z = 0$$
$$ a' + b' + z' = f(b-a)$$
$$ a' + c' + x' = f(c-a)$$
$$ b' + c' + y' = f(c-b)$$
$$ x' + y' + z' = f(y-x).$$
The first equation (noting that one can divide by $2$ in ${\bf F}\_3^n$) implies that $z=c$, $b=x$, $a=y$, so that the right-hand sides of the next four equations all agree; this then implies that $z'=c', b'=x', a'=y'$, and then $(a,a'), (b,b'), (c,c')$ are collinear, a contradiction.
On the other hand, the Proclus axiom does not need to be obeyed. Take three non-collinear points $(a,a'), (b,b'), (c,c')$, let $(x,x')$ be the third point in $L((a,a'), (c,c'))$, let $(y,y')$ be the third point in $L((b,b'), (c,c'))$, let $(z,z')$ be the third point in $L((a,a'), (b,b'))$, and let $(w,w')$ be the third point in $L((x,x'), (y,y'))$. Playfair gives that $(w,w'), (z,z'), (c,c')$ are distinct; Proclus would force these three points to be collinear. This gives the equations
$$ a+c+x=b+c+y=a+b+z=x+y+w=w+z+c=0$$
$$ a'+c'+x' = f(c-a)$$
$$ b'+c'+y' = f(c-b)$$
$$ a'+b'+z' = f(b-a)$$
$$ x'+y'+w' = f(y-z)$$
$$ w'+z'+c' = f(z-w).$$
The last four equations imply
$$ f(c-a)+f(c-b) - f(b-a) - f(y-z) + f(z-w) = 3c' = 0.$$
On the other hand, $y-z = a-c$ and $z-w = a+b+c$ so we simplify to
$$ f(c-b) - f(b-a) + f(a+b+c) = 0.$$
But one can easily construct an $f$ for which this equation fails for some non-collinear $a,b,c$, so the Proclus axiom can fail.
| 6 | https://mathoverflow.net/users/766 | 450472 | 181,217 |
https://mathoverflow.net/questions/450457 | -2 | If $u : \mathbb{T}^3 \to \mathbb{R}$ is a smooth function on the $3$-dimensional torus $\mathbb{T}^3$, I wonder it is possible to reverse the Sobolev interpolation inequality in the sense that
\begin{equation}
\lVert u \rVert\_2 \lVert \Delta u \rVert\_2 \leq C\lVert \nabla u \rVert\_2^2
\end{equation}
for some constant $C>0$ independent of $u$. Here $\lVert \cdot \rVert\_2$ is the usual $L^2$ norm with respect to the Lebesgue measure.
Could anyone please help me?
| https://mathoverflow.net/users/56524 | Inverse of Sobolev interpolation inequality : $\lVert u \rVert_2 \lVert \Delta u \rVert_2 \leq C\lVert \nabla u \rVert_2^2$? | This inequality is incorrect for an essential reason. Assume it is true for smooth functions, then using an approximation by convolution we would conclude that it is true for Sobolev spaces and hence it would imply that functions in $W^{1,2}$ belong to $W^{2,2}$ so $W^{1,2}=W^{2,2}$. Then by an inductive argument we would conclude that $W^{1,2}=W^{2,\infty}$ and hence every $W^{1,2}$ is $C^\infty$ smooth by the Sobolev embedding.
| 4 | https://mathoverflow.net/users/121665 | 450473 | 181,218 |
https://mathoverflow.net/questions/450471 | 4 | We say a field $F$ has the property $\*$ if the equation $x^2 + y^2=-1$ has no solution in $F$. For an example if $F$ is a subfield of real numbers then $F$ satisfies $\*$. On the other hand if $ F $ is a finite field then by Chevalley-Warning theorem the equation $ x^2 + y^2 =-1 $ always has a solution.
I want to know which fields that satisfy property $ \* $. Does this type of field belong to a special class of field?
| https://mathoverflow.net/users/215016 | Fields in which $ -1 $ can't be written as sum of two square elements | In the notation of Lam's *Quadratic forms over fields*, the *Stufe* (a German word) or *level* (its English translation) $s(F)$ of a field is the minimal $n$ such that $-1$ is the sum of $n$ squares.
A theorem of Pfister says that $s(F)$ is always a power of $2$ (or $\infty$).
So you are taking of fields $F$ such that $s(F) > 2$ (or $s(F) \geq 4$, if you are willing to use Pfister's theorem).
| 9 | https://mathoverflow.net/users/105957 | 450476 | 181,219 |
https://mathoverflow.net/questions/450470 | 0 | Given a prime $P$, an integer $A$ $(0\leq A<P$), and a set of legal positions (encoded as a binary mask $\text{mask}$), is there an efficient algorithm to find a number $B$ that has the same modulus as $A$ and (when $B$ is represented in its binary form) has ones only in legal positions. In other words, $B$ satisfies the following two constraints:
1. $A = B\mod P$
2. `B & (~mask) == 0`
A related problem that might be simpler is **restricting the illegal positions (~mask)** to be sparse. In this situation, a brute force search seems can finish in $O(2^k)$ steps where $k = \text{number of ones in (~mask)}$ by trying $A +P t$ for random $t$.
**Some thoughts**:
The problem seems can be reduced to **solving a linear equation on a finite field with bound constraints** if we consider the assignment of bits for $B$ in each continuous regions of ones in the $\text{mask}$ as a set of independent variables $v\_i$ with bound constraints $0\leq v\_i < 2^{s\_i}$ where $s\_i$ is the length of continuous ones.
For example, suppose the $\text{mask}$ is `1110011100111`; it is equivalently to solve the following linear equation $\mod P$:
$$A = v\_1 + 2^5\times v\_2 + 2^{10}\times v\_3\mod P$$
with constraints:
$$ 0\leq v\_1, v\_2, v\_3 < 2^3$$
But I don't know if this problem can be solved efficiently.
(Edit:
The original problem seems too hard without any restrictions. I am not sure if the following extra background makes this problem easier.
We can assume that all legal positions are below $O(\log(P))$ instead of $O(P)$. For example, we may have $P\sim 2^{64}$, and the mask is about $128$ bits long.)
| https://mathoverflow.net/users/508296 | Algorithm to find a number B with same modulus as A with prime P and specific binary positions set to zero | If you interpret the bit mask as encoding a finite set $S = \{2^{b\_i}\}$ of powers of $2$, you are precisely asking whether there exists a subset of $S$ which sums to $A$ modulo $p$. This is known as the modular subset-sum problem, for algorithms, see for example
<https://arxiv.org/pdf/2008.10577.pdf>
Since there are primes such that all elements of $\mathbb{Z}/p$ can be written as powers of $2$, this problem also does not seem to be easier than modular subset-sum. If I read the introduction to the linked paper correctly, there are conjectural lower bounds on the runtime that look like $p^{1-\varepsilon}$, so it doesn't seem to get better than polynomial in $p$.
(EDIT: The modular subset-sum problem is phrased as an existence statement, whereas you want to find an explicit subset. At least for the dynamic programming algorithm sketched in the beginning of the linked paper, which gets you $O(dp)$ time where $d$ is the digit sum of your mask, this should not be a problem, since instead you can tag every element of the intermediate sum set $S\_i$ by one choice of subset (or $B$) that gets you there. I haven't checked whether the other algorithms are similarly constructive)
| 2 | https://mathoverflow.net/users/39747 | 450480 | 181,221 |
https://mathoverflow.net/questions/450482 | 0 | For a script I have to evaluate the associated Legendre polynomial of second kind $Q^0\_{n}(z)$.
Until now, I was using an implementation that is based on the definition in equation 8.702 in the Book "[Table of Integral, Series and Products](http://fisica.ciens.ucv.ve/%7Esvincenz/TISPISGIMR.pdf)" by Gradshteyn.
When I insert $n=-3/2$, Matlab returns NaN, and for the term that is related to the hypergeometric function -Inf is returned. Also the book says that the expression for $Q^0\_{n}(z)$ looses its meaning for $n=-3/2$. Are there other possibilities to obtain $Q^0\_{-3/2}(z)$ especially for real valued $z>1$. From the expressions in Gradshteyn I expect real valued solutions however the Mathematica function LegendreQ returns complex values. Where does this come from?
Many thanks for any help
| https://mathoverflow.net/users/508182 | How are the Legendre Polynomials of second kind for negative degrees defined? | It helps to rewrite the expression from Gradshteyn,
$$Q\_\nu^0(z)=\frac{ \Gamma \left(\frac{1}{2}\right) \Gamma (\nu+1)\, \_2F\_1\left(\frac{\nu}{2}+1,\frac{\nu}{2}+\frac{1}{2};\nu+\frac{3}{2};\frac{1}{z^2}\right)}{2^{\nu+1}z^{\nu+1} \Gamma \left(\nu+\frac{3}{2}\right)},$$
in terms of the [regularized hypergeometric function](https://mathworld.wolfram.com/RegularizedHypergeometricFunction.html),
$$Q\_\nu^0(z)=\sqrt{\pi } (2z)^{-\nu-1} \Gamma (\nu+1) \, \_2\tilde{F}\_1\left(\tfrac{\nu}{2}+1,\tfrac{\nu}{2}+\tfrac{1}{2};\nu+\tfrac{3}{2};\frac{1}{z^2}\right).$$
This representation remains well-defined for $\nu=-3/2$. You can then transform back to the ordinary hypergeometric function, by means of the identity
$$\, \_2\tilde{F}\_1(a,b;0;x)=a b x \, \_2F\_1(a+1,b+1;2;x),$$
arriving at
$$Q\_{-3/2}^0(z)=\frac{\pi \, \_2F\_1\left(\frac{5}{4},\frac{3}{4};2;\frac{1}{z^2}\right)}{4 \sqrt{2} z^{3/2}}.$$
| 2 | https://mathoverflow.net/users/11260 | 450486 | 181,222 |
https://mathoverflow.net/questions/440658 | 3 | It is well known that some dispersive non--linear equations admit traveling wave solutions
$$
u(t,x)=u\_0(x-ct)\in L^2\_x\,, \qquad (t,x)\in \mathbb{R}\times \mathbb{R}\,\text{ or }\, \mathbb{R}\times\mathbb{T}\,,
$$
where $u\_0$ is the profile and $c$ is a real constant.
Sometimes these traveling waves can be obtained as ground states (minimizers) of the energy, mass... functionals of the equation, leading to say that there exists an $L^2$--threshold below (or above) which the existence of traveling waves cannot occur.
**My question :** Do there exist nonlinear dispersive PDEs, which have traveling waves with large *and* small $L^2$ norms ?
| https://mathoverflow.net/users/498602 | Does there exist always an $L^2$ threshold below (or above) which a traveling waves of a nonlinear dispersive PDE cannot exist? | I recently came across a [paper on arXiv](https://arxiv.org/abs/2307.01592) that addresses the question. I am sharing it here if it may be of interest to someone else.
The author seems to consider a nonlocal nonlinear schrödinger equation, referred to as *the Calogero-Sutherland DNLS equation*. And she finds periodic (in the space variable) traveling waves with small and large $L^2$-norms.
See Remark 1.3 (1) of [page 5](https://arxiv.org/pdf/2307.01592.pdf).
| 1 | https://mathoverflow.net/users/498602 | 450490 | 181,223 |
https://mathoverflow.net/questions/449278 | 7 | I am trying to work through a supposedly simple counterexample given in papers by [Love](https://academic.oup.com/jlms/article-abstract/s1-26/1/1/966397?redirectedFrom=fulltext) and [Gehring](https://www.jstor.org/stable/1990790) regarding a $p$-power generalization of bounded variation and absolute continuity.
Let $p > 1$. For a function $f: [0,1] \to \mathbb{R}$, the $p$-variation is defined by
$$
\text{Var}\_p f = \sup \left(\sum\limits\_{j=1}^n |f(u\_j)-f(u\_{j-1})|^p \right)^{\frac{1}{p}},
$$ where the supremum is taken over all partitions $0 = x\_0 < x\_1 < \dots < x\_n = 1$ of $[0,1]$. Likewise, we can define the $p$-variation on a smaller subinterval $[a,b] \subseteq [0,1]$ which we denote $\underset{[a,b]}{\text{Var}\_p} f$. The space of all functions with finite $p$-variation is denoted $BV\_p([0,1])$ or just $BV\_p$ since we are fixing the domain.
Likewise, we can define a generalized version of absolute continuity. We will say $f \in AC\_p$ if for every $\varepsilon > 0$, there exists $\delta > 0$ such that $\left( \sum\limits\_{j=1}^n |f(b\_j) - f(a\_j)|^p \right)^{\frac{1}{p}} < \varepsilon$ whenever $\{(a\_j,b\_j)\}$ is a collection of non-overlapping subintervals of $[0,1]$ such that $\left( \sum\limits\_{j=1}^n (b\_j - a\_j)^p \right) < \delta$. Much like in the classic $p = 1$ case, one can show $AC\_p \subseteq BV\_p$. However, a continuous function of bounded $p$-variation need not be $p$-absolutely continuous, like the Cantor staircase function demonstrates in the $p = 1$ case.
Both of the papers I mentioned cite
$$
u(x) = \sum\limits\_{k = 0}^\infty \frac{1}{2^{k/p}} \cos(2^k\pi x)
$$ as a counterexample which is continuous, has finite $p$-variation, and is not $p$-absolutely continuous (which is what I'm struggling to show). I have tried to work directly with the definition in addition to some alternative characterizations I found, but I haven't been successful so far. For example, Love's paper shows that if $f \in AC\_p$, then $\underset{[x,x+h]}{\text{Var}\_p} f = o(h^{\frac{1}{p}})$ for almost all $x$. Likewise, Gehring's paper mentions that if $f$ has finite $p$-variation and finite derivative except on possibly a countable set, then $f$ is $p$-absolutely continuous. It's worth pointing out that naively differentiating the terms defining $u$ yields a diverging series for all $x$.
I still think working directly with the negation of the $AC\_p$ condition is the way to go. A key observation I made is that for $N$ large, the dyadic partition $x\_j = j2^{-N}, j = 0, \dots, 2^N$ can satisfy the $\delta$ requirement (which obviously can't happen if $p = 1$). More specifically, for any $\delta > 0$, we require $N$ to be large enough such that $2^{-N/q} < \delta$, where $q$ is the conjugate of $p$ satsifying $\frac{1}{p} + \frac{1}{q} = 1$. I then would need to show
$$
\left( \sum\limits\_{j=1}^{2^N} |u(x\_j) - u(x\_{j-1})|^p \right)^{\frac{1}{p}} \geq 1
$$ or some other constant independent of $\delta$ (but not necessarily $p$). Using a dyadic partition allows for some nice reindexing of the series defining $u$, but I haven't figured out a good use yet. Part of the difficulty is that I need to directly work with cosines and bound the differences from below, rather than above.
| https://mathoverflow.net/users/118997 | A counterexample showing $BV_p \neq AC_p$ | So first of all, what is claimed in Love's paper is slightly different, It says that for a sufficient large choice of the parameter $c>0$ the function
$$ g(x): = \sum\_{n=0}^\infty c^{-n/p}\cos(c^n \pi x) $$ is of vounded $p$-variation but not $AC\_p$. One way to see it is that to notice that the Weierstrass type function
$$ W\_{a,b}(x) = \sum\_{n=1}^\infty b^{-na}\cos(b^n \pi x), $$
where $0<a<1, b \in \mathbb{N}$ is Holder continuous of order $\alpha: = -\log\_b(a)$ but not Higher. I think the best source for this is Hardy's paper "[Weierstrass's Non-Differentiable Function](https://www.jstor.org/stable/1989005)" Theorem 1.32. Although you can also read it off the Littlewood Paley characterization of Holder functions. Hence if $a=1/p, b=c$, this shows that for $c$ sufficiently large your function $g$ cannot satisfy
$$ |g(x)-g(x+h)| \leq Var^p\_{[x,x+h]} g = o(h^{1/p}) $$
suffices to take $ p\ln(p)<\ln(c)$.
| 1 | https://mathoverflow.net/users/153260 | 450494 | 181,226 |
https://mathoverflow.net/questions/450492 | 7 | Let $A, B, C$ be finite Abelian groups fitting in a short exact sequence
$$
1 \rightarrow A\overset{\iota}{\rightarrow} B\overset{\pi}{\rightarrow} C\rightarrow 1
$$
This determines a class $[\epsilon]\in H^2(C,A)$ measuring the failure of the sequence to split:
$$
s(c\_1)+s(c\_2)=s(c\_1+c\_2)+\iota(\epsilon (c\_1,c\_2))
$$
where $s:C\rightarrow A$ is any section of $\pi$ (different sections gives different representatives of $[\epsilon]$). By Pontryagin duality we have a dual sequence
$$
1 \rightarrow C^\vee\overset{\pi ^\vee}{\rightarrow} B^\vee\overset{\iota ^\vee}{\rightarrow} A^\vee\rightarrow 1
$$
and thus a dual class $\epsilon^\vee \in H^2(A^\vee,C^\vee)$.
I think this should be naturally defined explicitly in terms of $\epsilon$, but I don't see how. Part of my problem is that the only "explicit" equation in which $\epsilon$ enters is through the section $s$, which however does not gives something natural under duality. How can I write down an explicit formula for
$$
\epsilon^\vee(\alpha\_1,\alpha\_2) c \in \mathbb{R}/\mathbb{Z} \ , \ \ \ \ \ \alpha\_1,\alpha\_2 \in A^\vee \ , \ \ c\in C
$$
?
| https://mathoverflow.net/users/495347 | Pontryagin dual of a group-cohomology class | Ok, I think I worked this out based on my last comment, writing down the usual double complex for $\operatorname{Ext}(A^\vee,C^\vee)$ using a projective resolution of $A^\vee$ and an injective resolution of $C^\vee$ at the same time, and doing the diagram chase. Here's how the resulting map works:
Given $\varepsilon: C\times C\to A$, interpret this as pairing $\overline{\varepsilon}: C\times C\times A^\vee \to \mathbb{R}/\mathbb{Z}$, $(c\_1,c\_2,\varphi) = \varphi(s(c\_1,c\_2))$.
Using injectivity of $\mathbb{R}/\mathbb{Z}$ and that $\varepsilon$ is a symmetric cocycle, one sees that there exists $\beta: C\times A^\vee\to \mathbb{R}/\mathbb{Z}$ with
$$
\overline{\varepsilon}(c\_1,c\_2,\varphi) = \beta(c\_1,\varphi) + \beta(c\_2,\varphi) - \beta(c\_1+c\_2,\varphi).
$$
Now we form $\overline{\varepsilon^\vee}: C\times A^\vee\times A^\vee\to \mathbb{R}/\mathbb{Z}$ via
$$
\overline{\varepsilon^\vee}(c,\varphi\_1,\varphi\_2) = \beta(c,\varphi\_1) + \beta(c,\varphi\_2) - \beta(c,\varphi\_1+\varphi\_2).
$$
Then we let $\varepsilon^\vee: A^\vee \times A^\vee \to C^\vee$ be the map $(\varphi\_1,\varphi\_2) \mapsto \overline{\varepsilon^\vee}(-, \varphi\_1,\varphi\_2)$, a diagram chase using that $\overline{\varepsilon}$ was additive in the $A^\vee$ argument shows that this is additive in the remaining argument, i.e. really is an element of $C^\vee$.
A further diagram chase shows that the choice of $\beta$ influences $\varepsilon^\vee$, but not its cohomology class.
EDIT: After reverse engineering everything through homological algebra, let me provide a completely self-contained proof.
For an abelian extension $0\to A\to B\to C \to 0$ with section $s$ and cocycle $\varepsilon(c\_1,c\_2) = s(c\_1)+s(c\_2)-s(c\_1+c\_2)$, there exists $\beta: C\times A^\vee \to \mathbb{R}/\mathbb{Z}$ with $\varphi(\varepsilon(c\_1,c\_2)) = \beta(c\_1,\varphi) + \beta(c\_2,\varphi) - \beta(c\_1+c\_2,\varphi)$, using that every abelian extension by $\mathbb{R}/\mathbb{Z}$ splits (i.e. that it is an injective object of the category of abelian groups).
Explicitly, $B$ is identified with $A\times C$ with group structure $(a\_1,c\_1) + (a\_2,c\_2) = (a\_1+a\_2-\varepsilon(c\_1,c\_2), c\_1+c\_2)$.
Now for every $\varphi\in A^\vee$, note that
$$
(a,c) \mapsto \varphi(a) - \beta(c,\varphi)
$$
is a homomorphism $B\to \mathbb{R}/\mathbb{Z}$, which we denote $s^\vee(\varphi)$. This determines a section $s^\vee: A^\vee\to B^\vee$. We have
$$
(s^\vee(\varphi\_1) + s^\vee(\varphi\_2) - s^\vee(\varphi\_1+\varphi\_2))(a,c) = -\beta(c,\varphi\_1) - \beta(c,\varphi\_2) + \beta(c,\varphi\_1+\varphi\_2).
$$
This is an element of $B^\vee$ which is zero in $A^\vee$, so it comes from $C^\vee$, in particular is additive in $c$ (which we could also check directly). So it determines a cocycle of $A^\vee$ with values in $C^\vee$, and this is $-\varepsilon^\vee$ from above (apologies for the sign error above, I'm sure this comes from the double complex story.)
| 7 | https://mathoverflow.net/users/39747 | 450505 | 181,230 |
https://mathoverflow.net/questions/450498 | 2 | (Reposted from [MSE](https://math.stackexchange.com/questions/4720776/weighted-sobolev-spaces-and-decay) after no responses)
Introduce the following weighted Sobolev space norm on $\mathbb{R}^n$ (common in the study of hyperbolic PDE):
$$
\|u\|\_{H\_{k,\delta}}^2 = \sum\_{0 \leq i \leq k} \int\_{\mathbb{R}^n} \langle x \rangle^{2(\delta + i)}|\nabla^i u|^2 \, dx.
$$
This is found (for instance; I'm sure they're found in other places) in Christodoulou & Klainerman's stability of Minkowski spacetime, and in Choquet-Bruhat's book on general relativity and the Einstein equations. Here, $\langle x \rangle = (1 + |x|^2)^{1/2}$.
They behave nicely and have their own set of embedding theorems (see Choquet-Bruhat, *General Relativity and the Einstein Equations*, Appendix I, Theorem 3.4). The idea of their construction seems to be that $|u|$ has a certain (integrated) decay rate (related to $\delta$), and each derivative of $u$ behaves one power of decay better.
Let us focus on the case $\delta = 0$, so $u \in H\_{k, 0}$. This means, in particular, that $\langle x \rangle^\delta u \in L^2$. (However, this is not the same as the usual Sobolev space $H^k$, since recall each derivative of $u$ gains a power of decay in our case. This does not happen with classical Sobolev spaces.). Now, from learning about power functions, integration, and the $p$-test in calculus class, one might expect this (heuristically) to imply that $|u| \lesssim \langle x \rangle^{-n/2}$, as this is the "borderline" case of integrability. Of course, this is in general false. But the weighted Sobolev embeddings tell us that if $k > n/2$, then in fact
$$
\sup\_{x \in \mathbb{R}^n} \langle x \rangle^{n/2} |u| \lesssim \|{u}\|\_{H\_{k,\delta}}.
$$
So we recover the decay one might heuristically have expected!
**My question:** It seems to me like the derivative estimates (i.e. the fact that $\langle x \rangle^{i}\nabla^i u \in L^2$) did not come into the estimate here: in fact, if $u \in H\_{100, 0}$ versus $u \in H\_{100000,0}$, one cannot improve the decay of $u$ just by having "more derivatives". (For instance let $u(x) = \langle x \rangle^{-q}$.) It seemed like once you had $u \in H\_{k,0}$ with $k > n/2$, that was all you need, and any more derivatives are "useless." Does this seem right?
And on a related note, the decay could be guessed at from the zeroth-order behavior, i.e. just from $u \in L^2$ one expects $|u| \lesssim \langle x \rangle^{-n/2}$. The higher derivatives seemed like they were just there to rule out highly irregular behavior, like narrow/tall spikes that might spoil this. So, my **second question:** is the same embedding true if one just assumes $u \in L^2$ and $u \in H^k$ with $k > n/2$, in the sense of classical Sobolev spaces? That is, can the Sobolev embeddings for classical Sobolev spaces (without weights) be improved to give *decay*?
| https://mathoverflow.net/users/147016 | Weighted Sobolev Spaces and Decay | Question 1 (that higher derivatives are not used) is **yes**.
Question 2 (getting decay without weights) is **no**.
Without weights, let $u$ be a compactly supported smooth function. Let $f\_k(x) = u(x - k v) + u(x + kv)$ where $v$ is a unit vector. The family $f\_k$ is uniformly bounded in any classical $H^s$ space. But the family $f\_k$ is NOT uniformly decaying.
---
To better understand the behavior of these weighted Sobolev spaces, you want to first split it into a compact part ($|x| < 1$) and the remainder. In the compact part, you just use the standard Sobolev theory, since decay does not care about the compact region.
Outside, it is more convenient to think of these spaces as defined with respect to the differential operators $|x|\partial$ instead of with respect to $\partial$. The advantage of $|x|\partial$ is that they are scale invariant: if you consider mappings of $\mathbb{R}^n$ to itself given by $x\mapsto \lambda x$, then the operator $|x|\partial$ pushes forward to itself. This means that the $H\_{s,\delta}$ spaces are *scaling homogeneous*. (For standard Sobolev spaces, the $\mathring{H}^k$ portion has a different scaling from the $L^2$ portion.) It is this scaling homogeneity that enables us to get decay from Sobolev embedding.
(For comparison, the standard $H^s$ spaces are translation invariant, which the $H\_{s,\delta}$ spaces are not. Which you want to leverage depends on which problem you are solving.)
| 2 | https://mathoverflow.net/users/3948 | 450506 | 181,231 |
https://mathoverflow.net/questions/325055 | 2 | If $X$ is a self-distributive algebra, then define $x^{[n]}$ for all $n\geq 1$ by letting $x^{[1]}=x$ and $x^{[n+1]}=x\*x^{[n]}$. The motivation for this question comes from the following fact about self-distributivity on one generator.
>
> **Theorem:** Suppose that $X$ is a self-distributive algebra generated by one element. Then for all $x,y\in X$, there are $m,n$
> with $x^{[m]}=y^{[n]}$.
>
>
>
Define the Fibonacci terms by letting $t\_{1}(x,y)=y,t\_{2}(x,y)=x,t\_{n+2}(x,y)=t\_{n+1}(x,y)\*t\_{n}(x,y)$.
An algebra $(X,\*,1)$ is said to be a reduced permutative algebra if $x\*(y\*z)=(x\*y)\*(x\*z),x\*1=1,1\*x=x$ for all $x,y,z\in X$ and where for all $x,y\in X$, there is some $n$ where $t\_{n}(x,y)=1$. We say that a reduced permutative algebra $(X,\*,1)$ is critically simple if whenever $\simeq$ is a congruence, either $\simeq$ is the trivial congruence or $x\simeq 1$ for some $x\neq 1$.
Suppose that $(X,\*,1)$ is a finite reduced permutative algebra and $x,y\in X$ and suppose that $p\geq 0$. Then does there a reduced permutative algebra $(Y,\*,1)$, natural numbers $m,n$ and a homomorphism $\phi:Y\rightarrow X$ and $x',y'\in Y$ where $\phi(y')=y,\phi(x')=x$ and $x'^{[m]}=y'^{[n]}$ and where
$x'^{[m+p]}\neq 1$? What if $(X,\*,1)$ is critically simple?
| https://mathoverflow.net/users/22277 | Attraction in Laver tables | I have a method of generating counterexamples even if $(X,\*,1)$ is critically simple. In this post, all algebras will be assumed to be finite reduced permutative
self-distributive algebras. To do this, we provide an example of an algebra algebra $X$ generated by $x,y$ where there does not exist a algebra $Y$ generated by $x\_1,y\_1$ and a surjective homomorphism $\phi:Y\rightarrow X$ with $\phi(x\_1)=x,\phi(y\_1)=y$ and where $Y$ has more critical points than $X$.
The examples of the algebras $X$ are quite rare, and I currently only know of one technique for producing such an algebra. Let $X\_1$ be the algebra with only one element which we shall denote by $x\_1$ and also by $y\_1$. We then by induction build a sequence of critically simple algebra $(X\_1,\dots,X\_n)$ where $|\text{crit}[X\_j]|=j$ for $1\leq j\leq n$ and where $X\_j$ is generated by 2 elements $x\_j,y\_j$ where $y\_j\*y\_j=1$ and where there is a surjective homomorphism $\phi\_j:X\_j\rightarrow X\_{j-1}$ with $\phi(x\_j)=x\_{j-1},\phi(y\_j)=y\_{j-1}$, and we quite the induction process when we cannot extend the sequence $(X\_1,\dots,X\_n)$ any further in the sense that there does not exist an algebra $X\_{n+1}$ with $\text{crit}[X\_{n+1}]=n+1$ and a surjective homomorphism $\phi\_{n+1}:X\_{n+1}\rightarrow X\_n$ along with generators $x\_{n+1},y\_{n+1}$ such that $\phi\_{n+1}(x\_{n+1})=x\_n,\phi\_{n+1}(y\_{n+1})=y\_n$ and $y\_{n+1}\*y\_{n+1}=1$.
After we recursively build our sequence of critically simple algebras $(X\_1,\dots,X\_n)$, we extend the sequence some more to obtain more critically simple algebras $(X\_n,\dots,X\_m)$ and homomorphisms
$\phi\_j:X\_j\rightarrow X\_{j-1}$ where $|\text{crit}[X\_j]|=j$ and $\phi\_j(x\_j)=x\_{j-1},\phi\_j(y\_j)=y\_j$ and where $X\_j$ is generated by $x\_j,y\_j$ until we cannot extend $X\_m$ any more to an algebra $X\_{m+1}$ generated by $x\_{m+1},y\_{m+1}$ with $|\text{crit}[X\_{m+1}]|=m+1$ and $\phi\_{m+1}(x\_{m+1})=x\_m,\phi\_{m+1}(y\_{m+1})=y\_m$.
In this case, $Y\_m$ with the generators $x\_m,y\_m$ is our counterexample.
To go from $X\_j$ to $X\_{j+1}$, we use my (possibly inefficient) backtracking algorithm that takes a critically simple algebra $X$ and generating set $(x\_a)\_{a\in A}$ and returns all possible (up-to generator preserving isomorphism) critically simple algebras $Y$ with generating sets $(y\_a)\_{a\in A}$ and homomorphisms $\phi:Y\rightarrow X$ with
$\phi(y\_a)=x\_a$ for $a\in A$ and where $|\text{crit}[Y]|=|\text{crit}[X]|+1$.
For critically simple algebras $X$ with multiple generators $(x\_a)\_{a\in A}$, I have no algorithm or theory other than my exhaustive search that finds all critically simple algebras $Y$ generated by $(y\_a)\_{a\in A}$ along with a homomorphism
$\phi:Y\rightarrow X$ with $\phi(y\_a)=x\_a$ for $a\in A$, and I have no way of knowing what critically simple algebras $Y$ might look like, so I have no way of obtaining a different kind of counterexample.
| 2 | https://mathoverflow.net/users/22277 | 450507 | 181,232 |
https://mathoverflow.net/questions/450353 | 0 | Given a rational polyhedral fan $\Sigma$ in $\mathbb{R}^d$ (say with full-dimensional support), its barycentric subdivision $\mathrm{bar}(\Sigma)$ is obtained by performing star-subdivision at the barycenters
$$
\rho\_\sigma = \sum\_{\tau \in \sigma(1)} \rho\_\tau
$$
of its cones $\sigma$ (where $\rho\_\tau$ are the primitive generators of the rays $\tau$ of $\sigma$, and the star-subdivisions are performed in order of decreasing dimension on the cones $\sigma \in \Sigma$).
**Question:** Is it true that for any fan $\Sigma'$ with support $|\Sigma'|=|\Sigma|$ there exists $n \geq 0$ such that the $n$-th iterated barycentric subdivision $\mathrm{bar}^{\circ n}(\Sigma)$ is a refinement/subdivision of $\Sigma'$?
For my purposes it would also suffice to restrict to the case where $\Sigma$ consists of the the positive orthant $\sigma\_d = \mathbb{R}\_{\geq 0}^d$ and its faces.
**Example:** For $d=2$ the positive orthant is subdivided at the rays spanned by
$$
(1,1); (2,1), (1,2); (3,1), (3,2), (2,3), (1,3); \ldots
$$
in the successive barycentric subdivisions. The generators $(x,y)$ of these rays exactly run through the primitive integer vectors in the interior of $\sigma\_2$, and the ratios $x/y$ form the entries of the [Stern-Brocot tree](https://en.wikipedia.org/wiki/Stern%E2%80%93Brocot_tree), which enumerates all rational numbers in $(0,1)$. In particular, it is the case that any fan $\Sigma'$ with support $\sigma\_2$ will eventually be refined by some iterated barycentric subdivision of $\sigma\_2$ (since the slopes of the rays of $\Sigma'$ appear at some point in the Stern-Brocot tree).
The behaviour on the rays generalizes: for $\sigma\_d$, the ray generators of the iterated barycentric subdivisions run through the primitive vectors in $\sigma\_d$, so at least the $1$-skeleton of $\Sigma'$ will eventually be captured by this process.
| https://mathoverflow.net/users/69630 | Iterated barycentric subdivision cofinal in system of subdivisions? | It turns out that the answer is "No" for dimension $d \geq 3$.
Indeed, for the counter-example in dimension $3$ take $\Sigma$ given by $\sigma\_3$ and its faces, and consider the fan $\Sigma'$ obtained by subdividing $\Sigma$ along the hyperplane $H=-x\_1 + 2 x\_2 + x\_3 = 0$. The fact that $H$ has both positive and negative coefficients implies that it meets the interior of $\sigma\_3$.
Now the first barycentric subdivision of $\Sigma$ will contain the cone
$$
\overline \sigma = \left\{x\_1' \cdot \pmatrix{1\\0\\0} + x\_2'\cdot \pmatrix{1\\1\\1} + x\_3' \cdot \pmatrix{1\\1\\0} : x\_1', x\_2', x\_3' \geq 0\right\}
$$
Pulling back the equation $H$ to the coordinates $x\_1', x\_2', x\_3'$ on $\overline \sigma$ we get
$$H' = -(x\_1' + x\_2' + x\_3') +2(x\_2'+x\_3')+x\_2' = -x\_1' + 2 x\_2' + x\_3',$$
which has the *same* coefficients $(-1,2,1)$ as $H$. So we observe that $H$ meet the interior of the maximal cone $\overline \sigma$ of the first barycentric subdivision of $\Sigma$, and thus $\mathrm{bar}(\Sigma)$ does not refine $\Sigma'$. But in fact since the first barycentric subdivision contains a cone on which the equation of $H$ has not changed, in fact any iterated barycentric subdivision also contains such a cone, so none of these will refine $\Sigma'$.
Since $\sigma\_3$ appears as a face of $\sigma\_d$ for $d > 3$, it's easy to see that this also gives a counterexample in arbitrary dimension $d$.
| 0 | https://mathoverflow.net/users/69630 | 450511 | 181,233 |
https://mathoverflow.net/questions/450521 | 9 | So, I ask whether from the ZFC axioms one can prove X that *every uncountable set has strictly more than continuum many subsets*, or whether X is independent of the ZFC axioms. Note that (within ZFC) the continuum hypothesis implies X and hence "not X" is not provable in ZFC if ZFC is consistent.
The above question arose from [this MO-question](https://mathoverflow.net/q/450262) to which an answer is given by Problem 10 on page 99 in Richard M. Dudley's book *Real Analysis and Probability*, Wadsworth 1989. However, in view of Problem 9 on page 387 *loc.cit.* it seems that some additional hypothesis (e.g. CH or the weaker X above) is missing in Dudley's problem setting. If X is provable within ZFC, then without any additional assumptions we have the result that *every Borel map from any separable metric space to a metric space has separable range*. If X is not provable, then Dudley's hints for a possible proof seem to require some ad hoc assumption for the proof to succeed.
| https://mathoverflow.net/users/12643 | Within ZFC, is $2^{\aleph_0}<2^{\aleph_1}$ provable/independent? | The assertion that $2^{\aleph\_0}=2^{\aleph\_1}$ is known as [Luzin's hypothesis](https://encyclopediaofmath.org/wiki/Luzin_hypothesis), and was presented by Luzin as an alternative to Cantor's continuum hypothesis.
This is now known to be independent of ZFC by the method of forcing (assuming ZFC is consistent).
Namely, under the GCH, which is true in the constructible universe, we have $2^{\aleph\_0}=\aleph\_1<2^{\aleph\_1}$ and therefore the negation of Luzin's hypothesis.
But in Cohen's model of $\neg\text{CH}$, obtained by forcing over $L$ to add $\aleph\_2$ many Cohen reals, Luzin's hypothesis is true. The basic reason is that one can count in the ground model the number of "nice" names for subsets of $\aleph\_1$ in the forcing extension, and there are only $\aleph\_2$ many such names in the ground model. So $2^{\aleph\_1}=\aleph\_2$ in the Cohen model.
One can produce in just the same way models of ZFC in which all the $2^{\aleph\_n}$'s are equal for finite $n$, or indeed, where $2^\kappa=2^{\aleph\_0}$ for all cardinals $\kappa<2^{\aleph\_0}$. Indeed, this situation is a consequence of [Martin's axiom](https://en.wikipedia.org/wiki/Martin%27s_axiom), which can hold even when $2^{\aleph\_0}$ is quite large, as large as desired.
Models where Luzin's hypothesis is true provide counterexamples to the powerset size axiom, the principle asserting
$$\kappa<\lambda\implies2^\kappa<2^\lambda$$
This is a principle that [many people find very natural](https://mathoverflow.net/a/6594/1946), but it is independent of ZFC.
But Luzin's hypothesis is not equivalent to the failure that principle, since by [Easton's theorem](https://en.wikipedia.org/wiki/Easton%27s_theorem), we can by forcing control the continuum function $\kappa\mapsto 2^\kappa$ on the infinite regular cardinals in a very flexible manner.
For example, we could have the GCH holding up to $\aleph\_{17}$, but then $2^{\aleph\_n}=\aleph\_{\omega\_1+5}$ for all natural numbers $n>17$.
| 22 | https://mathoverflow.net/users/1946 | 450522 | 181,235 |
https://mathoverflow.net/questions/450518 | 2 | Let $T \in \mathbb{R}$ be large and $\rho$ be a non-trivial zero of the Riemann zeta function. Assume that $|\rho|=|\rho\_T| \approx T$ and let $\varepsilon\_T \approx \frac{\log \log T}{\log T}$. Is it true that
$$|\zeta(1-\rho\_T + \varepsilon\_T)|=o(1)$$
uniformly for $T \geq T\_0$?
| https://mathoverflow.net/users/507786 | On the upper bound for $|\zeta(s)|$ near the zeta zeros | If $\rho\_T$ is a zero of $\zeta(s)$, so is $1-\rho\_T$ by the functional equation, and $|1-\rho\_T|\approx T$ (depending on what you mean by $\approx$), so the $1-\ldots$ is superfluous.
A Taylor expansion of $\zeta(s)$ at $\rho\_T$ gives
$$
\zeta(\rho\_T+\epsilon)=\zeta^\prime(\rho\_T)\cdot \epsilon+O(\epsilon)^2.
$$
Even assuming the Riemann Hypothesis, all that is known is that
$$
\zeta^\prime(\rho\_T)=O(T^{\epsilon^\prime})
$$
for any $\epsilon^\prime$ (via Lindelof and the Cauchy Integral Formula for derivatives), so there's not much hope for a uniform bound.
---
Edit: I'm expanding on DavidFarmer's comment below. Selberg proved that for $E\subset \mathbb R$, we have that the limit as $T\to\infty$ of
$$
\frac{1}{T}\,\mu(T\le t\le 2T\,|\,\log|\zeta(1/2+i t)|/(1/2\log\log t)^{1/2}\in E)
$$
where $\mu$ is Lebesgue measure, is equal to the integral over $E$ of a Gaussian Random Variable with mean $0$ and standard deviation $1$:
$$
\frac{1}{\sqrt{2\pi}}\int\_E \exp(-x^2/2)\, dx.
$$
which is the basis for David's comment that $|\zeta(1/2+it)|$ is equal $1$ "on average". Since the count of zeros up to height $T$ is $\sim T/(2\pi)\log(T)$, at height $T$ the average spacing between zeros is $2\pi/\log(T)$. So the Hardy function $Z(t)$ is like $\sin(t \log(T)/2)$ near height $T$, and it's derivative has a multiplier of $\log(T)$.
| 9 | https://mathoverflow.net/users/6756 | 450526 | 181,237 |
https://mathoverflow.net/questions/450531 | 4 | Assume spaces are regular.
A space is [$\sigma$-compact](https://topology.pi-base.org/properties/P000017) if and only if the second player in the [Menger game](https://en.wikipedia.org/wiki/Selection_principle#The_Menger_game) has a winning Markov strategy (relying on only the most recent move of the opponent and the round number) [[source]](https://dml.cz/handle/10338.dmlcz/146790).
The proof there nearly translates directly to this conjecture: a space is [hemicompact](https://topology.pi-base.org/properties/P000111) if and only if the second player in the $K$-Rothberger game has a winning Markov strategy, where the $K$-Rothberger game is played with $k$-covers (each compact set is contained in some open set) and P2 may choose only one set per turn rather than finitely-many.
But there's a gotcha: while in the proof for the Menger game, we had $x\not\in\bigcap\_{\mathcal U}\bigcup\sigma(\mathcal U,n)\Rightarrow x\not\in\bigcup\sigma(\mathcal U,n)$ for some open cover $\mathcal U$, I don't believe it's true that we'd obtain a $k$-cover $\mathcal V$ for which $K\not\subseteq\sigma(\mathcal V,n)$ just because $K\not\subseteq\bigcap\_{\mathcal V}\sigma(\mathcal V,n)$.
So if that cannot be fixed, what counterexample exists?
| https://mathoverflow.net/users/73785 | Does there exist a non-hemicompact regular space for which the 2nd player in the $K$-Rothberger game has a winning Markov strategy? | I believe no such example exists. In fact, finite selections do the trick, so I'll detail the case for both finite and single selections (though it seems the argument for single selections below requires $T\_1$).
>
> **Claim.** For a regular space $X$, if the second player has a Markov winning strategy in the $k$-Menger game on $X$, then $X$ is hemicompact. (This doesn't require the $T\_1$ assumption.)
>
>
>
*proof.* According to Theorem 30 of [the arxiv version of this paper](https://arxiv.org/pdf/2102.00296.pdf), the $k$-Menger game on $X$ is equivalent to the Menger game on $\mathbb K(X)$, the space of compact subsets of $X$ with the Vietoris topology. To detail a bit more here, note that Lemma 19 shows that any $k$-cover of $\mathbb K(X)$ can be refined by a $k$-cover on $X$ by sets of the form $[V]$ where $V$ is open in $X$ and $[V]$ is the set of all compact subsets of $V$. In particular, any open cover of $\mathbb K(X)$ can be closed under finite unions to produce a $k$-cover of $\mathbb K(X)$ and then refined by a $k$-cover of $X$ in the indicated way. This will allow us to transfer the Markov winning strategy for the second player in the $k$-Menger game on $X$ to a Markov winning strategy in the Menger game on $\mathbb K(X)$.
According to [Michael's paper on spaces of subsets](https://www.ams.org/journals/tran/1951-071-01/S0002-9947-1951-0042109-4/S0002-9947-1951-0042109-4.pdf), $X$ is regular if and only if $\mathbb K(X)$ is regular (no requirements on $T\_1$ here).
Suppose that the second player has a Markov winning strategy in the $k$-Menger game on a regular space $X$. Then the second player has a Markov winning strategy in the Menger game on $\mathbb K(X)$ in the following way. Given an open cover $\mathscr U$ of $\mathbb K(X)$, we can form the collection of finite unions, which is a $k$-cover. For any selection of finite sets of the form $[V]$ in the specified refinement, we can collect a finite collection of members of $\mathscr U$ whose union covers the union of the finite collection of sets of the form $[V]$. Then, when the second player produces a $k$-cover of $X$, the corresponding finite selections from open covers of $\mathbb K(X)$ form an open cover. So the second player has a winning Markov strategy in the Menger game on $\mathbb K(X)$.
By the equivalence with $\sigma$-compactness mentioned in the question, since $\mathbb K(X)$ is regular, $\mathbb K(X)$ is $\sigma$-compact. By a straightforward argument detailed in Theorem 40 [this paper](https://arxiv.org/pdf/1812.06379.pdf), $X$ is hemicompact. The argument relies on 2.5.2 of [Michael's paper](https://www.ams.org/journals/tran/1951-071-01/S0002-9947-1951-0042109-4/S0002-9947-1951-0042109-4.pdf) which doesn't rely on any separation axiom assumptions. $\square$
>
> The case for single selections (this portion requires $T\_1$)
>
>
>
On the other hand, Theorem 39 of [this paper](https://arxiv.org/pdf/1812.06379.pdf) gives a pretty general equivalence with single selections and predetermined strategies for the first player in a point-open-like game. Then using the duality results of [Clontz' Dual Selection Games](https://www.sciencedirect.com/science/article/abs/pii/S0166864120300031) should establish the equivalence with Markov winning strategies for the second player in the $k$-Rothberger game and being hemicompact.
| 4 | https://mathoverflow.net/users/57800 | 450536 | 181,243 |
https://mathoverflow.net/questions/450539 | 4 | Let $G$ be a finite nonabelian simple group.
We call $G$ a $K\_3$-group if $|G|=p^aq^br^c$ where $p,q,r$ are distinct primes and $a,b,c$ are positive integers.
**My question is: Is there a CFSG-free proof of the classification of simple $K\_3$-groups?**
Any explanation, references, suggestion and examples are appreciated.
| https://mathoverflow.net/users/44312 | CFSG-free proof for classifying simple $K_3$-group | By the Feit-Thompson theorem, one of the primes has to be two. Then by John Thompson's N-group classification (1970-1973), the three primes have to come from a very short list of triples. The various cases were then worked out by Geoff Mason, David Wales, and Jeff Leon in the seventies. This should give you enough information to get the references from MR.
| 6 | https://mathoverflow.net/users/460592 | 450542 | 181,245 |
https://mathoverflow.net/questions/450544 | 1 | Let $$f(z)=(1+z)^{3/4}-\left(\frac{3}{8}+\frac{\sqrt{3}}{4}\right)^{1/4}-\frac{\left(3 z+\sqrt{6} \sqrt{-1+z^2}\right)^{3/4}}{\left(2 \left(2+\sqrt{3}\right)\right)^{3/4}}.$$ Is there a simple proof that this function is positive for $z\ge 1$?
| https://mathoverflow.net/users/504719 | An inequality for a real function | Make the substitution $u=(1+z)^{3/4}$, so that $u\ge2^{3/4}$, $z=u^{4/3}-1$, and the inequality in question becomes
$$F(u):=f(u^{4/3}-1)>0. \tag{1}\label{1}$$
For $z=u^{4/3}-1>1$
$$F''(u)=\frac{5 u^2 \sqrt{z-1}+\sqrt{6} \left(2 z^2+3 z-1\right)}{\left(2
\left(2+\sqrt{3}\right)\right)^{3/4} (z-1)^{3/2} (z+1) \left(\sqrt{6} \sqrt{z^2-1}+3
z\right)^{5/4}}, $$
which is manifestly $>0$. So, $F(u)$ is convex in $u\ge2^{3/4}$. Also, for $u\_\*:=\frac{212}{100}$ we have $F'(u\_\*)=0.00227\ldots>0$. So,
$$\min\_{u\ge2^{3/4}}F(u)
\ge\min\_{u\ge2^{3/4}}[F(u\_\*)+F'(u\_\*)(u-u\_\*)] \\
=F(u\_\*)+F'(u\_\*)(2^{3/4}-u\_\*)=0.0579\ldots>0.$$
Thus, \eqref{1} is proved. $\quad\Box$
| 1 | https://mathoverflow.net/users/36721 | 450580 | 181,255 |
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