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https://mathoverflow.net/questions/450591 | 1 | In the book Fourier Analysis and Self-adjointess of Reed and Simon in the proof of the
Feynman-Kac formular the author states that for any $V\in L^\infty (\mathbb{R}^3)$ there is a sequence $(V\_n)\_n$ of continuous and compact supported functions, which pointwise almost everywhere approximate $V$.
I do know some results concerning $C\_c(X)$ being dense in $L^p(X)$-spaces with respect to the $p$-norms. This holds only if $p<\infty$ (Rudin Real and Complexanalysis). Of course we are talking about a different topology. But I cannot find a good explanation why the property I stated holds. Actually to make it more precise, but probably an easy modification, the author wants the sequence to be uniformly bounded by $\|V\|\_\infty$.
| https://mathoverflow.net/users/508379 | Any $L^\infty (\mathbb{R}^3)$ can be approximated pointwise almost everywhere by continuous function with compact support | Let $V\in L^\infty$ and let $V\_\varepsilon=V\*\varphi\_\varepsilon$ be a standard approximation by convolution. Since $V\in L^1$ on any bounded set, it follows that $V\_\varepsilon\to V$ in $L^1$ on bounded sets. Since a convergence in $L^1$ implies a.e. convergence on a subsequence, for every bounded set you find a subsequence that converges a.e. Then using a standard diagonal argument you find a subsequence that converges a.e. in $\mathbb{R}^3$.
| 7 | https://mathoverflow.net/users/121665 | 450593 | 181,260 |
https://mathoverflow.net/questions/450592 | 2 | Let $$\chi\_S(x,y)=\begin{cases}1&\text{ if }0< x<y< 1\\0&\text{ else }\end{cases}$$
be the indicator function of the simplex $S=\{(x,y)\in (0,1)^2:x<y\}$. I am interested in an explicit partition of unity. That is, I am looking for continuous positive functions $f\_n:\mathbb R^2\to \mathbb R$ so that $\sum\_{n=1}^\infty f\_n(x,y)=\chi\_S(x,y).$
Does anyone have a reference to such an example? I suspect this should be relatively standard but I have been unable to find anything.
| https://mathoverflow.net/users/479223 | Partition of unity of simplex | It is easy to construct an explicit continuous (or even smooth) partition of unity over $\mathbb R^2$. Map this partition of unity to the continuous (or smooth) partition of the indicator $\chi\_Q$ of the open square $Q:=(-1,1)^2$ via the bi-smooth map $\mathbb R^2\ni(x,y)\mapsto\frac2\pi\,(\arctan x,\arctan y)\in Q$. Finally, use an explicit homeomorphism from (open convex set) $Q$ onto (open convex set) $S$ to get an explicit continuous partition of $\chi\_S$.
| 3 | https://mathoverflow.net/users/36721 | 450595 | 181,261 |
https://mathoverflow.net/questions/450585 | 4 | For natural $n$, let
\begin{equation}
p\_n:=2^{1-n}\sum\_{v=1}^l \binom l{(v+l)/2}1(v\equiv l)
\sum\_{u=1-v}^{v-1}\binom k{(u+k)/2}1(u\equiv k), \tag{1}\label{1}
\end{equation}
where $k:=\lfloor(n+1)/2\rfloor$, $l:=\lfloor n/2\rfloor$, and $a\equiv b$ means that $a-b$ is even.
>
> Can this expression for $p\_n$ be simplified? In particular, is it true that $p\_{4m+3}=1/2$ for $m=0,1,\dots$? Is it true that $p\_n\le1/2$ for all $n$?
>
>
>
A correct and complete answer to any one of these questions will be considered a correct and complete answer to this entire post.
---
The expression for $p\_n$ in \eqref{1} was obtained in this [previous answer](https://mathoverflow.net/a/450578/36721).
| https://mathoverflow.net/users/36721 | On a double sum involving binomial coefficients | The formula for $p\_n$ was derived to be a probability relating two independent transformed binomials. In particular,
$$p\_n = \mathbb{P}(|U\_n|<|V\_n|)$$
where
$$ U\_n = a\_1 + \dots a\_k \hspace{20pt} V\_n = b\_1 + \dots + b\_l $$
and $k = \lfloor (n+1)/2 \rfloor, l = \lfloor n/2 \rfloor$.
When $n=2m$ then $k = l = m$ and so $U\_{2m}$ and $V\_{2m}$ are i.i.d. In this case we can exploit symmetry; $\mathbb{P}(|U\_n| < |V\_n|) = \mathbb{P}(|V\_n| < |U\_n|)$, and so we can write:
$$ p\_{2m} = \frac{1}{2} \left ( 1 - \mathbb{P} ( |U\_{2m}| = |V\_{2m}| ) \right ).$$
In general, for discrete i.i.d. random variables $X, Y$ taking values in $\mathbb{N}$, we have:
$$\mathbb{P}(X=Y) = \sum\_{n = 0}^\infty \mathbb{P}(X = n)^2.$$
We can continue by splitting into two further cases depending on whether $m$ is even or odd, i.e. whether $n$ is $0$ or $2$ mod 4. The case $n = 4m + 2$ is easier; recall that $U\_{4m+2} \overset{d}{=} 2B\_{2m+1} - (2m+1)$, where $B\_{2m+1} \sim \text{Binomial}(2m+1,1/2)$. So
\begin{align}
\mathbb{P}(|U\_{4m+2}| = |V\_{4m+2}| ) & = \sum\_{p=0}^m \mathbb{P}(B\_{2m+1} = (m-p) \text{ or } (m+p+1) )^2 \\
&= 2^{-4m} \sum\_{p=0}^m {2m+1 \choose p}^2\\
&= 2^{-4m-1} {4m + 2 \choose 2m+1}.
\end{align}
The case $n = 4m$ is similar, but the probability is slightly different since one has to take into account the event $\{|U\_{4m}| = 0\}$. Then we have:
\begin{align}
\mathbb{P}(|U\_{4m}| = |V\_{4m}|) &= 2^{-4m} \left ( {2m \choose m}^2 + 4 \sum\_{p=0}^{m-1} {2m \choose p}^2 \right ) \\
&=2^{-4m} \left (2 {4m \choose 2m} - {2m \choose m}^2 \right).
\end{align}
Now we have the case where $n$ is odd. $U\_n$ and $V\_n$ are not i.i.d, but instead $U\_n \overset{d}{=} V\_n + a$, where $a$ is an independent Rademacher. Let $V\_n'$ be an i.i.d. copy of $V\_n$. Then using the law of total probability we can write:
$$p\_n = \frac{1}{2} \left ( \mathbb{P}(|V\_n+1| < |V\_n'|) + \mathbb{P}(|V\_n-1| < |V\_n'|) \right ).$$
When $n = 4m+3$, $V\_n$ is odd and in particular greater than 1 in absolute value. Thus, $|V\_n+a| \overset{d}{=} |V\_n| + a$ as adding a Rademacher will not change the sign of $V\_n$. Then we have
\begin{align}
p\_{4m+3} &= \frac{1}{2} \left ( \mathbb{P}(|V\_n| + 1 < |V\_n'|) + \mathbb{P}(|V\_n| < |V\_n'| + 1) \right ) \\
&= \frac{1}{2} \mathbb{P}(|V\_n| \neq |V\_n'| + 1)\\
&= \frac{1}{2}
\end{align}
since $|V\_n|$ is always odd.
When $n = 4m+1$, $V\_n$ is even and so once again the limiting factor is the event where $V\_n = 0$.
If we condition on the events where $V\_n$ and $V\_n'$ are zero and nonzero, respectively, we obtain the following expression:
\begin{align}
p\_{4m+1} &= \mathbb{P}(V\_n' \neq 0) \left (\mathbb{P}(V\_n = 0) + \frac{1}{2} \mathbb{P}(V\_n \neq 0) \right ) \\
&= \frac{1}{2}(1-\mathbb{P}(V\_n = 0)^2) \\
&= \frac{1}{2} \left ( 1 - 2^{-4m} {2m \choose m}^2 \right ).
\end{align}
Overall we have the following:
\begin{equation}
p\_{4m+k} =
\begin{cases}
\frac{1}{2} \left ( 1 - 2^{-4m} \left (2 {4m \choose 2m} - {2m \choose m}^2 \right) \right) & k = 0 \\
\frac{1}{2} \left ( 1 - 2^{-4m} {2m \choose m}^2 \right ) & k = 1\\
\frac{1}{2} \left ( 1 - 2^{-4m-1} {4m + 2 \choose 2m+1} \right) & k = 2\\
\frac{1}{2} & k = 3
\end{cases}
\end{equation}
In particular, $p\_n \leq 1/2$ for all $n$.
| 5 | https://mathoverflow.net/users/507348 | 450598 | 181,262 |
https://mathoverflow.net/questions/450600 | 9 | How to prove that there can't exist a countable set $\{A\_1,A\_2,\dots\}\subset \mathcal{L}(\mathbb{R})$ (where $\mathcal{L}(\mathbb{R})$ denotes the family of all Lebesgue measurable sets) such that $\sigma(\{A\_1,A\_2,\dots\})=\mathcal{L}(\mathbb{R})$?
(I know that the cardinality of $\mathcal{L}$ is equal to the cardinality of all subsets of $\mathbb{R}$ and hence larger then the cardinality of the borel sets but I'm not familiar with any deeper hierarchy results. Is there a more or less elementary proof?)
| https://mathoverflow.net/users/508383 | How to prove that the Lebesgue $\sigma$-Algebra is not countably generated? | Let me mount the kind of cardinality argument to which you allude.
You had asked for a proof that the $\sigma$-algebra of Lebesgue measurable sets is not countably generated. But in fact, a much stronger claim is true:
**Theorem.** The $\sigma$-algebra of all Lebesgue measurable sets is not generated by any family of size continuum.
**Proof.** First, I claim that the $\sigma$-algebra generated by any family of size at most continuum itself has size at most continuum. (In particular, every countably generated $\sigma$-algebras have size at most continuum.)
The reason is that we can perform the generation process concretely in a transfinite process of length $\omega\_1$. We begin with the generating family, which has size at most continuum. At each stage, we form the next collection by taking all possible countable unions, countable intersections, and complements of what we already have; at limit stages of the process, we gather together all the sets we have formed before that stage. If we began with at most continuum many sets, then since there are only at most continuum many countable subsets of a set of size continuum, it follows that the next stage will also have size at most continuum. And at countable limit stages, we have a countable union of previous families, all of size at most continuum, and so the collection at that stage also has size at most continuum.
By stage $\omega\_1$, we have generated the $\sigma$-algebra, since all countable collections from what we have at that stage have already appeared altogether at a countable stage. So we have $\omega\_1$ many stages, each with at most size continuum many sets, making the generated family of size at most continuum, which establishes the initial claim.
Second, since every subset of a Lebesgue measure zero set is measurable, and there are sets, such as the Cantor set, which have measure zero and size continuum, it follows that there are $2^{2^{\aleph\_0}}$ many Lebesgue measurable sets. This is strictly larger than the continuum, and so by the above it cannot be generated by any family of size at most continuum. $\Box$
| 15 | https://mathoverflow.net/users/1946 | 450601 | 181,263 |
https://mathoverflow.net/questions/450553 | 7 | * Let $a(n)$ be [A204262](https://oeis.org/A204262) i.e. [permanent](https://en.wikipedia.org/wiki/Permanent_(mathematics)) of the matrix $n\times n$ with elements $\min(i,j)$.
* Let
$$
f\_{n,\ell}(x)=g\_{n,\ell}(x)+f\_{n,\ell-1}(\ell)-g\_{n,\ell}(\ell), \\
g\_{n,\ell}(x)=\int (n-\ell)^2 f\_{n-1,\ell}(x)\,dx, \\
f\_{n,0}(x)=n!x^n
$$
* Let
$$
R(n,q)=\sum\limits\_{j=0}^{q+1}\binom{q+1}{j}(j+1)R(n-1,j), \\
R(0,q)=1
$$
I conjecture that
$$R(n,0)=f\_{n+1,n+1}(0)=a(n+1).$$
Here is the PARI/GP prog to check it numerically:
```
a(n)=matpermanent(matrix(n, n, i, j, min(i, j)))
f_upto(n)=my(v1); v1=vector(n+1, i, i--; i!*x^i); for(i=1, n, for(j=i, n, my(A=intformal((j-i)^2*v1[j])); v1[j+1] = A + subst(v1[j+1] - A, x, i))); v1
R_upto(n)=my(v1, v2, v3, v4); v1=vector(n+1, i, 1); v2=v1; v3=vector(n+1, i, 0); v3[1]=1; v4=vector(n, i, vector(i+1, j, binomial(i, j-1)*j)); for(i=1, n, for(q=0, n-i, v2[q+1]=sum(j=0, q+1, v4[q+1][j+1]*v1[j+1])); v1=v2; v3[i+1]=v1[1];); v3
test1(n)=f_upto(n)==concat(1, R_upto(n-1))
test2(n)=f_upto(n)==vector(n+1, i, a(i-1))
```
Is there a way to prove it?
| https://mathoverflow.net/users/231922 | Remarkable recursions for the A204262 | I can show the first identity $R(n,0) = f\_{n+1,n+1}(0)$, as a consequence of the more general identity
$$ R(n,q) = \frac{1}{(q+1)!} f\_{n+q+1,n}(n+1)\tag{1}\label{1}$$
for $n,q \geq 0$. Indeed, note that $g\_{n+1,n+1} \equiv 0$ and thus $f\_{n+1,n+1}(0) = f\_{n+1,n}(n+1)$, so $R(n,0) = f\_{n+1,n+1}(0)$ is basically a special case of \eqref{1}.
One can prove \eqref{1} by induction on $n$. For $n=0$ we have
$$ R(0,q) = 1 = \frac{1}{(q+1)!} f\_{q+1,0}(1)$$
by hypothesis. For $n>0$, we see from Taylor expansion (and noting from induction that $f\_{n+q+1,n}$ is a polynomial of degree $q+1$) that
$$ \frac{1}{(q+1)!} f\_{n+q+1,n}(n+1) = \sum\_{j=0}^{q+1} \frac{1}{(q+1)! j!} f\_{n+q+1,n}^{(j)}(n).$$
But from the recursive definition of $f\_{n,\ell}$ and the fundamental theorem of calculus one has
$$ f\_{n,\ell}'(x) = (n-\ell)^2 f\_{n-1,\ell}(x)$$
and hence on iterating
$$ f\_{n+q+1,n}^{(j)}(x) = \left(\frac{(q+1)!}{(q+1-j)!}\right)^2 f\_{n+q+1-j,n}(x)$$
so that
$$ \frac{1}{(q+1)!} f\_{n+q+1,n}(n+1) = \sum\_{j=0}^{q+1} \frac{(q+1)!}{(q+1-j)! (q+1-j)! j!} f\_{n+q+1-j,n}(n).$$
But from the recursive definition of $f\_{n,\ell}$ and the induction hypothesis we have
$$ f\_{n+q+1-j,n}(n) = f\_{n+q+1-j,n-1}(n) = (q+2-j)! R(n-1,q+1-j)$$
and hence
$$ \frac{1}{(q+1)!} f\_{n+q+1,n}(n+1) = \sum\_{j=0}^{q+1} \frac{(q+1)!}{(q+1-j)! j!} (q+2-j) R(n-1,q+1-j).$$
Replacing $j$ by $q+1-j$ we obtain
$$ \frac{1}{(q+1)!} f\_{n+q+1,n}(n+1) = \sum\_{j=0}^{q+1} \binom{q+1}{j} (j+1) R(n-1,j)$$
recovering \eqref{1} by the recursive definition of $R$.
I discovered \eqref{1} by working backwards from the desired identity $R(n,0) = f\_{n+1,n+1}(0)$ to try to write $R(n,1) = \frac{1}{2} (R(n+1,0) - R(n,0))$ and then $R(n,2) = \frac{1}{3} (R(n+1,1) - R(n,0) - 4 R(n,1))$ in terms of the $f\_{n,\ell}$ in as simple a fashion as possible (using the ODE form $f\_{n,\ell}' = (n-\ell)^2 f\_{n-1,\ell}$, $f\_{n,\ell}(\ell) = f\_{n,\ell-1}(\ell)$ of the recursion for $f$ to eliminate the role of the indefinite integral $g$), at which point the general pattern became clear. Presumably a similar technique, using the multilinearity and symmetry properties of the permanent to express $R(n,1)$, then $R(n,2)$, etc., as a suitable permanent starting from the hypothesis $R(n,0) = a(n+1)$, will then give a similar identity relating $R(n,q)$ with the permanent of a suitable matrix which can then be established by induction to complete the proof, but I will leave this for someone else to try. EDIT: actually it may he easier to try to discover how $f\_{n,\ell}(x)$ can be expressed neatly as a permanent, as it has a simpler recursion.
SECOND EDIT: OK, I have found the permanental formula for $f\_{n,\ell}$:
$$ f\_{n,\ell}(x) = \mathrm{per}( A\_{n,\ell} + x B\_{n,\ell} ) \tag{2} \label{2}$$
where $A\_{n,\ell}$ is the $n \times n$ matrix with entries $\min(i,j) 1\_{\min(i,j) \leq \ell}$, and $B\_{n,\ell}$ is the $n \times n$ matrix with entries $1\_{\min(i,j) > \ell}$. (Guessing this formula for $\ell=n$ and $\ell = n-1$ was not too difficult, but it took a while to realize that $f\_{n,\ell}$ could be placed in this form for $\ell = n-2$. Again, at this point the pattern became clear.)
Indeed, one has
$$ f\_{n,0}(x) = x^n \mathrm{per}( B\_{n,0} ) = n! x^n $$
(since $B\_{n,0}$ is the all $1$'s matrix), and from the identity
$$ A\_{n,\ell} + \ell B\_{n,\ell} = A\_{n,\ell-1} + \ell B\_{n,\ell-1}$$
we have $f\_{n,\ell}(\ell) = f\_{n,\ell-1}(\ell)$. Finally, from multilinearity, cofactor expansion and the fact that the $n-1 \times n-1$ minor formed by deleting the $i^{th}$ row and $j^{th}$ column from $A\_{n,\ell} + x B\_{n,\ell}$ is equal to $A\_{n-1,\ell} + x B\_{n-1,\ell}$ when $\min(i,j) > \ell$, we have
$$ f'\_{n,\ell}(x) = (n-\ell)^2 f\_{n-1,\ell}(x)$$
(the factor $(n-\ell)^2$ coming from the number of 1's appearing in $B\_{n,\ell}$) and this is all we need to verify \eqref{2} recursively. Replacing $n,\ell$ by $n+1$ we obtain $f\_{n+1,n+1}(0) =a(n+1)$ as conjectured.
Combining \eqref{1}, \eqref{2} we also have the nice identity
$$ R(n,q) = \frac{1}{(q+1)!} \mathrm{per}( \min( i, j, n+1 ) )\_{1 \leq i,j \leq n+q+1}.$$
| 10 | https://mathoverflow.net/users/766 | 450607 | 181,264 |
https://mathoverflow.net/questions/132618 | 7 | Ogg characterized the finitely many N such that $X\_0(N)\_{\mathbb{Q}}$ is hyperelliptic, and Poonen proved in "Gonality of modular curves in characteristic p" that for large enough N, $X\_0(N)\_{\mathbb{F}\_p}$ is not hyperelliptic.
**Question**: Are there any N such that $X\_0(N)\_{\mathbb{Q}}$ is not hyperelliptic but for some p not dividing N $X\_0(N)\_{\mathbb{F}\_p}$ is hyperelliptic?
I'm also interested in this question for other modular curves of the form $X\_H$, where H is a congruence subgroup.
| https://mathoverflow.net/users/2 | Hyperelliptic modular curves in characteristic p | I already gave the currently accepted answer to this question around 10 years ago. And the answer can be summarised as "Yes, here is an example". However, after 10 years I wanted to come back to this question and answer it again, but now in the opposite direction. Namely, as it turns out the example that I found above is the only example. To be more precise:
**Theorem.** Let $N$ be a positive integer, $p$ a prime not dividing $N$, and $X\_H$ a modular curve with of level N corresponding to a congruence subgroup $\Gamma\_1(N) \subseteq H \subseteq \Gamma\_0(N)$. Suppose that $X\_{H,\mathbb F\_p}$ is hyperelliptic but $X\_{H, \mathbb Q}$ is not, then $p=2$, $N=37$ and $H$ is the congruence subgroup generated by $\Gamma\_1(N)$ and $d\_4$ where $d\_4$ is a matrix representing the diamond operator $\langle 4 \rangle$.
This is Theorem 1.3 of <https://arxiv.org/abs/2307.04864>. In that theorem, the hyperellipticity is considered over $\overline {\mathbb F}\_p$ and $\mathbb C$. However, since all the curves $X\_H$ under consideration have a rational point, they are hyperelliptic over $\mathbb F\_p$ if and only if they are $\overline {\mathbb F}\_p$ and similar over $\mathbb Q$ and $\mathbb C$.
The proof is consisting of two steps, the first one is using the Castelnuovo-Severi inequality to establish an apriori upper bound on $N$, see Proposition 3.6 of loc. cit. The second step is checking for hyperellipticity by looking at the space of quadrics that vanish on the image of the canonical map, as in my other answer.
| 5 | https://mathoverflow.net/users/23501 | 450617 | 181,270 |
https://mathoverflow.net/questions/450627 | 5 | This might be well-known to experts. I was just teaching a course where we went through some parts of Quillen's theorem computing the higher algebraic K-theory of finite fields. Denote by $\mathbb F\_q$ the field with $q$ elements.
>
> **Theorem (Quillen)** Let $q=p^k$ for some prime $p$ and $k \in \mathbb N\_{\geq 1}$. $$K\_{2i-1}(\mathbb F\_q) = \mathbb Z/(q^i-1)\mathbb Z \quad \mbox{and} \quad K\_{2i}(\mathbb F\_q)= 0.$$
>
>
>
Canonically, $K\_1(\mathbb F\_q) = \mathbb F\_q^\times$, so that the action of ${\rm Aut}(\mathbb F\_q)$ on $K\_1(\mathbb F\_q)$ is clear. Since higher Milnor K-theory of finite fields vanishes, this does not help to understand the action on the higher K-groups.
>
> **Question:** How does ${\rm Aut}(\mathbb F\_q)$ act on $K\_{2i-1}(\mathbb F\_q)$?
>
>
>
Quillen's proof relies on the choice of an embedding $\bar {\mathbb F}\_q^\times \to S^1$, which makes the isomorphism somewhat non-canonical, as far as I understand. Anyway, my guess would be that the Frobenius $F(x) = x^p$ acts as $a \mapsto p^i a$ on $\mathbb Z/(q^i-1)\mathbb Z$, but I do not know how to prove this.
| https://mathoverflow.net/users/8176 | Galois action on algebraic K-theory of finite fields | $K(\overline{\mathbb{F}\_p})$ is, after completion at any prime $\ell \neq p$, equivalent to $ku^\wedge\_\ell$. It's true that the equivalence relies on a non-canonical embedding, but the conclusion that the homotopy groups of the $\ell$-completion are a polynomial ring, is independent of the concrete isomorphism. Since the homotopy groups of $K(\overline{\mathbb{F}\_p})^\wedge\_\ell$ are $\ell$-divisible and concentrated in odd degrees, we get
$$
\pi\_{2n} K(\overline{\mathbb{F}\_p})^\wedge\_\ell = \operatorname{Hom}(\mathbb{Q}\_\ell/\mathbb{Z}\_\ell, \pi\_{2n-1}K(\overline{\mathbb{F}\_p})).
$$
We know the Frobenius acts by multiplication by $p$ on $\pi\_1K(\overline{\mathbb{F}\_p})\cong \overline{\mathbb{F}\_p}^\times$, so it acts by multiplication by $p$ on $\pi\_2 K^\wedge\_\ell$, hence by multiplication by $p^n$ on $\pi\_{2n} K^\wedge\_\ell$, hence by multiplication by $p^n$ on $\pi\_{2n-1} K(\overline{\mathbb{F}\_p})$ as you expected (by going over all $\ell\neq p$). The finite field case follows from this by comparison along the canonical map between them.
| 9 | https://mathoverflow.net/users/39747 | 450628 | 181,271 |
https://mathoverflow.net/questions/450573 | 2 | There is a well know theorem by Coven and Hedlund, in [Sequences with minimal block growth](https://link.springer.com/article/10.1007/BF01762232), stating that the complexity function of an aperiodic sequence\configuration $\omega\in \mathcal{A}^{\mathbb{Z}}$ is at least linear, where the complexity function $c\_\omega(\cdot):\mathbb{N}\to \mathbb{N}$ of $\omega$, is the number of distinct words of given length occurring in $\omega$. I saw that it is referred to sometimes as the Morse-Hedlund complexity gap as well.
Since we can define configurations in a Bernoulli shift $\mathcal{A}^G$ for general groups $G$, I was wondering whether there are known results for complexity gaps of this form, for more general groups. Something like if $G$ contains an element of infinite order, then there exists a strictly monotonic function $g:[0,\infty)\to [1,\infty)$ satisfying $g(r)\overset{r\to \infty}{\to}\infty$, such that any configuration $\omega\in \mathcal{A}^G$ whose stabilizer is trivial must have complexity satisfying $c\_\omega(r)=O\big( g(r) \big)$.
This seems like a natural question and I was wondering whether there is literature relating to that in certain non-Abelian cases?
| https://mathoverflow.net/users/143153 | Morse-Hedlund\Coven-Hedlund theorem for non-Abelian groups | Your motivation is about aperiodicity implying high complexity, but you ask whether aperiodicity implies an upper bound on complexity, which is a little strange. Probably there is a typo, and $O(g(r))$ should be $\Omega(g(r))$?
In any case, I'll give a simple generalization of the Morse-Hedlund theorem to all groups.
Definition. Let $G$ be any group, and let $D\_1, D\_2, D\_3, ...$ be a family of finite subsets of $G$, such that $|D\_i| \geq 1$. We say $(D\_i)\_i$ satisfies property (C) (for "connected") if the following holds: For $i \geq 1$, define a graph $\mathcal{G}\_i$ with nodes $G$, and an edge between $g$ and $h$ if and only if $gD\_i \cup hD\_i \subset kD\_{i+1}$ for some $k \in G$. Then for all $i$, the graph $\mathcal{G}\_i$ is connected.
For $G = \mathbb{Z}$, we can take $D\_i = [1,i]$. For general groups we can take balls of increasing radius $r$ with respect to any finite symmetric generating set.
Let $X \subset A^G$ be any subshift (closed set invariant under $G$-translations). Then we can measure complexity of $X$ by counting the finite sets $P\_i = \{x|\_{D\_i} \;|\; x \in X\}$.
>
> Theorem. If $X$ is an infinite subshift, $(D\_i)\_i$ have property (C), and $P\_i$ are the associated pattern sets, then $|P\_n| \geq n+1$ for all $n$.
>
>
>
Proof. Clearly $|P\_1| \geq 2$. Otherwise, since $|D\_1| \geq 1$, in particular all configurations in $X$ have the same symbol at the origin, thus $|X| \leq 1$ and $X$ is not infinite.
We now show $|P\_{i+1}| > |P\_i|$, from which $|P\_n| \geq n+1$ immediately follows by induction.
Suppose the contrary, that $|P\_{i+1}| \leq |P\_i|$ for some $i$. Since $D\_i \subset D\_{i+1}$, we actually have $|P\_{i+1}| = |P\_i|$, and we see that by shift-invariance that for any $x \in X$, the restriction $x|\_{gD\_i}$ uniquely determines $x|\_{hD\_{i+1}}$ whenever $gD\_i \subset hD\_{i+1}$.
I claim that then $x|\_{D\_i}$ uniquely determines $x$ for all $x \in X$. This is more or less immediate from (C): Write $N \subset \mathcal{G}\_i$ for all $g$ such that $x|\_{D\_i}$ determines $x|\_{gD\_i}$ uniquely. Note that ``$x|\_{aD\_i}$ determines $x|\_{bD\_i}$ uniquely'' is a transitive relation.
By definition $N$ contains $1\_G$. It is also a connected component of $\mathcal{G}$: if we have the edge $(g, h)$ then $gD\_i \cup hD\_i \subset kD\_{i+1}$ for some $k \in G$, thus $x|\_{gD\_i}$ determines $x|\_{kD\_{i+1}}$, in particular $x|\_{gD\_i}$ determines $x|\_{hD\_i}$. Square.
In the case of $\mathbb{Z}$, with the choice $D\_i = [1, i]$ this precisely recovers the Morse-Hedlund theorem. In the case of $\mathbb{Z}^2$ if we use rectangles for $D\_i$ that grow alternately in the two dimensions, then this gives only a linear lower bound on complexity of squares, so it falls far behind what is known about Nivat's conjecture.
| 3 | https://mathoverflow.net/users/123634 | 450636 | 181,276 |
https://mathoverflow.net/questions/450623 | 3 | Consider for $i=1,\ldots, N\ge2$
$$X^i\_t=x\_i+W^i\_t,\quad \forall t\ge 0,$$
where $x\_1,\ldots, x\_N\in (0,\infty)$ and $W^1,\ldots, W^N$ are independent Brownian motions. Denote by $\tau\_i$ the first hitting time of $X^i$ at zero, i.e.
$$\tau\_i:=\inf\big\{t\ge 0: X^i\_t\le 0 \big\}.$$
How to prove (rigorously) $\mathbb P[\exists i\neq j \mbox{ such that } \tau\_i=\tau\_j]=0$?
| https://mathoverflow.net/users/493556 | Can independent Brownian motions hit zero at the same time? | Your question is asking whether two Brownian motions can both first hit zero simultaneously. In fact we can say something stronger; for $N$ independent Brownian motions, the set of times where each Brownian motion hits 0 are pairwise disjoint.
When $N=2$, this is equivalent to asking whether a standard two-dimensional Brownian motion starting at $(x\_1,x\_2)$ ever hits $(0,0)$; it is known that, almost-surely, it will not.
For $N>2$, by the union bound, the probability is less than the sum over $i \neq j$ of the probability that a BM starting at $(x\_i,x\_j)$ ever hits $(0,0)$, so this probability is also 0.
| 9 | https://mathoverflow.net/users/507348 | 450641 | 181,277 |
https://mathoverflow.net/questions/27716 | 40 | Fan Chung and Ron Graham's book *Erdos on Graphs: His Legacy of Unsolved Problems* (A. K. Peters, 1998) collects together all of Erdos's open problems in graph theory that they could find into a single volume, complete with bounties where applicable. Of course Erdos posed many other open problems in combinatorics and number theory that do not appear in this book. I once heard a rumor that some people were working on a project to publish a similar but more comprehensive book or series of books, covering *all* of Erdos's open problems, but I don't know if the rumor is true. Does such a compilation exist? If not, is there *anything* else like this besides Chung and Graham's book?
| https://mathoverflow.net/users/3106 | Does there exist a comprehensive compilation of Erdos's open problems? | Recently, Thomas Bloom created a website dedicated exactly to this:
<https://www.erdosproblems.com/>
It currently lists 214 problems, both open and closed. They are all tagged and some problems carry additional information. The list still evolves and if a problem is missing you might want to contact Thomas directly.
| 7 | https://mathoverflow.net/users/31469 | 450643 | 181,278 |
https://mathoverflow.net/questions/450606 | 2 | Sounds like a trivial question, but could not find any answer other than the fact that there are many ways to define it. My problem is this: I look at different elevation maps,
1. some with sharp mountains,
2. some rather flat, or
3. with steep but consistent slopes with little gradient variations.
For simplicity, let us assume that we are dealing with mathematical functions instead, infinitely differentiable everywhere. In order words, I am not interested here in metrics related to chaos and entropy: that would be the subject of a different question, but you are welcome to discuss it as well. In one dimension, a smoothness metric frequently cited is
$$ S(f, [a, b]) = \int\_a^b |f''(x)|^2 dx. $$
How do you generalize this in a way that makes sense for my problem? All gradients or Jacobians are discrete in my case, but ignore this for simplicity. I tested the following:
$$ S(f, D) = \int\_D \big\|\nabla\|\nabla(f(z))\|\big\| dz.$$
where $\|\cdot\|$ is $L\_2$ norm. Sounds ugly and I did not use $\|\cdot\|^2$, yet I get decent results. My guess is that metrics based on the Jacobian may make more sense. What would you suggest, and why?
| https://mathoverflow.net/users/140356 | What are the best definitions for smoothness of a 2D curve (real-valued function)? | $S(f,[a,b])=\int\_a^b f''(x)^2\,dx$ is, not a measure of smoothness, but rather a measure of nonlinearity (or, better, of non-affinity or, one might say, of "sharp-turn'ness") of a function $f$. Indeed, $S(f,[a,b])$ takes its smallest value $0$ if and only if $f$ is affine on $[a,b]$. (Of course, here I use the term "measure" not in a measure-theoretic sense.)
$S(f,[a,b])$ is used as the penalty term in [cubic spline regression](https://en.wikipedia.org/wiki/Smoothing_spline#Cubic_spline_definition). As even this name, "cubic spline regression", suggests, the resulting estimate of an unknown estimated function is a cubic spline -- which is of course not infinitely smooth.
The cubic spline regression is naturally and faithfully generalized to the [multidimensional case](https://en.wikipedia.org/wiki/Smoothing_spline#Multidimensional_splines), that is, for functions $f$ defined on subsets $D$ of $\mathbb R^d$ for $d\ge2$, by using a penalty term of the form
$$\int\_D \|f''(x)\|^2\,dx,$$
where (for our purposes here) $f''(x)$ is the Hessian matrix of $f$ at a point $x$ and $\|\cdot\|$ is a matrix norm. In the just linked section of the Wikipedia article, they use the Frobenius matrix norm, which is one of simplest to compute, but one can also use any other matrix norm instead.
Thus, the natural suggestion is to use $\int\_D \|f''(x)\|^2\,dx$ as a generalization of $S(f,[a,b])=\int\_a^b f''(x)^2\,dx$.
Concerning discrete settings, use naturally the discrete counterparts of the second partial derivatives, and the corresponding summation instead of the integration.
| 2 | https://mathoverflow.net/users/36721 | 450645 | 181,279 |
https://mathoverflow.net/questions/450584 | 3 | A *grating* is a notion in algebraic topology from the 1940 introduced by Alexander. Cartan extended it as follows.
A grating (carapace in french) is defined by a topological space $X$, a module (or a differential ring) $A$ and for each $x \in X$, a surjective morphism $\varphi\_x$ from $A$ to some quotient $A\_x$ such that
* for all $\alpha \in A$ and $x \in X$ such that $\varphi\_x(\alpha) = 0$, we have $\varphi\_y(\alpha) = 0$ for all $y$ close to $x$.
* If $\varphi\_x(\alpha) = 0$ for all $x\in X$, then $\alpha = 0$.
In his seminar [Carapaces](http://www.numdam.org/item/SHC_1950-1951__3__A18_0/) from the 1950, Cartan showed that gratings correspond to (pre-)sheaves on $X$ and this is the reason why, I guess, the notion of grating is not used anymore.
The sheaf $\mathcal F$ associated to a grating $(A, \varphi\_x)$ is built by taking the $A\_x$ as the stalks. This, I understand. The grating associated to a sheaf is built from the module of sections. Cartan does not give any other details in the [seminar](http://www.numdam.org/item/SHC_1950-1951__3__A18_0/). I don't understand what $A$ should be, hence my question:
>
> What is the module $A$ associated to the sheaf $\mathcal F$?
>
>
>
The module $A\_x$ must be the stalk of the sheaf. So $A$ can't be the module of global sections, since it does not always surject on the stalks. My guess would be something like the inverse limit of all the ${\mathcal F}(U)$, where $U$ is an arbitrary open subset of $X$. But I have never seen this limit used before in that context.
| https://mathoverflow.net/users/6129 | Sheaves and gratings | I think you may have misunderstood what Cartan was doing. In modern language he defined functors $\Gamma$ from sheaves to gratings and $\mathcal{F}$ from gratings to sheaves but he did not claim that these are mutually inverse equivalences of categories (even allowing for the apparent anachronism). He did write down a natural transformation from the identity functor on gratings to the functor $\Gamma \circ F$ explained at the end of section 4 of the seminar you link to.
He called objects $A$ where the map $A\to \Gamma\circ F(A)$ is an isomorphism *complete gratings* and argues that $\Gamma$ of any sheaf is a complete grating. I guess we should understand this as meaning that $\Gamma$ is right adjoint to $F$ but I haven't worked through the details to check this is true.
To directly answer your specific question, I think the module $A$ underlying $\Gamma(\mathcal{F})$ is in fact the space of global sections $\mathcal{F}(X)$ and your argument that it cannot be so fails since we don't have $A\_x=\mathcal{F}\_x$ in general. Rather, I think if you work through Cartan's argument, $A\_x$ can be defined to be the image of $\mathcal{F}(X)$ in $\mathcal{F}\_x$.
| 2 | https://mathoverflow.net/users/345 | 450646 | 181,280 |
https://mathoverflow.net/questions/450605 | 0 | Given a centrally symmetric convex body $K$ in the plane (with smooth boundary), it is easy to see that there exists a norm function $g:\mathbb{R}^2\to \mathbb{R}\_{\geq 0}$ for which $K$ is the unit ball.
For the polar body $K^{\circ}$ we also have a norm function $h:\mathbb{R}^2\to \mathbb{R}\_{\geq 0}$ for which $K^{\circ}$ is the unit ball. I have the following question:
Is $x\times \nabla g(x)=y\times \nabla h(y)$ whenever $(x,y)$ is s.t. $g(x)=h(y)$? Where $\nabla$ denotes the gradient of the function (it makes sense outside the origin in this case) and $x\times y:=x\_1y\_2-x\_2y\_1$.
This is certainly true for the K the unit ball in the $\ell^2$-norm.
Oddly enough, if we change $\times$ by the usual inner product this assertion follows from Euler's theorem on homogeneous functions.
Thanks in advance.
| https://mathoverflow.net/users/334733 | Norm functions induced by convex bodies | I think you need to assume that $K$ has a smooth boundary and is strictly convex to ensure that $g$ and $h$ are differentiable outside $0$.
Anyway, I do not think that the result is true. Assume that $K$ is the unit ball associated to the $\ell^p$-norm with $p>1$. Then $K^\*$ is the unit ball associated to the $\ell^q$-norm with $q>1$ such that $1/p+1/q=1$.
The functions $g$ and $h$ are the $\ell^p$-norm and the $\ell^q$-norm, so everything can be computed explicitly.
For every $x \in \mathbb{R}^2$, set
$$x^{p-1}:=(|x\_1|^{p-1}\mathrm{sign(x\_1)},|x\_2|^{p-1}\mathrm{sign(x\_2)}).$$
Then $\nabla (g^p)(x) = (p|x\_1|^{p-1}\mathrm{sign(x\_1)},p|x\_2|^{p-1}\mathrm{sign(x\_2)}) = px^{p-1}$. By the chain rule, if $x \ne 0$,
$$\nabla g(x) = \frac{1}{p}(g^p(x))^{1/p-1}\nabla (g^p)(x) = \frac{x^{p-1}}{g(x)^{p-1}}$$
$$x \times \nabla g(x) = \frac{x\_1x\_2^{p-1}-x\_2x\_1^{p-1}}{g(x)^{p-1}}.$$
My impression is that $g(x)=h(y)$ does not imply $x \times \nabla g(x) = y \times \nabla h(y)$. A counterexample may be found by taking $x=(1,0)$ and $y=2^{-1/q}(1,1)$. I did not push the computations forward.
| 2 | https://mathoverflow.net/users/169474 | 450650 | 181,281 |
https://mathoverflow.net/questions/293563 | 3 | Can one use lattice basis reduction algorithms, such as LLL over (low-rank) module lattices over rings of number fields of degree greater than 1? Is there any work on lattice reductions over Euclidean rings (for example, using BKZ, lattice enumeraiton, etc)?
| https://mathoverflow.net/users/24541 | Lattice basis reduction over rings of number fields | Apparently, this was studied in the following work: <https://eprint.iacr.org/2019/1035>
| 0 | https://mathoverflow.net/users/24541 | 450653 | 181,282 |
https://mathoverflow.net/questions/450657 | 6 | If I'm not mistaken, the question I put on the title used to be on this site, but I'm not being able to find it at all. I'm therefore reposting it so that someone can either give me the old link or help me answer this question anyway.
If $\tau$ is a topology on the real numbers, then saying that $(\mathbb{R}, \tau, +)$ is a compact Hausdorff topological group is equivalent to saying that $Hom\_{\tau}(\mathbb{R}, S^1)$ is discrete, by Pontryagin Duality. I initially thought that we could just look at subgroups of $Hom(\mathbb{R}, S^1)$ and that their duals would give us what we want, but this probably doesn't work since the subspace topology of $Hom(\mathbb{R}, S^1)$ doesn't coincide with the topology of uniform convergence on compact sets, in the general case. So I'm lost.
I am convinced that one of the answers in the old link used this Pontryagin Duality approach though.
| https://mathoverflow.net/users/494553 | Topologies that turn the real numbers into a compact Hausdorff topological group | A compact abelian group $A$ is the same as the Pontryagin dual of a discrete abelian group $B$. The group $A$ is divisible ($nA=A$ for all $n\ge 1$) if and if $B$ is torsion-free, and $A$ is torsion-free iff $B$ is divisible. So $A$ is divisible torsion-free iff $B$ is divisible torsion-free, i.e., $B\simeq\mathbf{Q}^{(I)}$ for some set $I$. In this case, denoting by $P$ the Pontryagin dual of $\mathbf{Q}$, we have $A\simeq P^I$.
The cardinal of $P$ is continuum, so the cardinal of $P^I$ is continuum if $I$ is nonempty and countable (in particular, for $I$ singleton, i.e. for $A=P$). In this case, it is abstractly isomorphic to the group of real numbers. Of course, it is better *not* to identify $P$ with real numbers since the identification is very non-canonical (and not explicitable).
| 6 | https://mathoverflow.net/users/14094 | 450659 | 181,284 |
https://mathoverflow.net/questions/450570 | 4 | I believe that a catenoid supports a parametrization $\sigma : U \subset \mathbb{R}^2 \rightarrow \mathbb{R}^3$ that forms a conjugate system (i.e., $\sigma\_{uv} \in\mathrm{span}(\sigma\_u, \sigma\_v)$) with the additional property that the coordinate curves are geodesics. I tried the following computational approach in building such a prameterization. I chose the following usual parameterization of a catenoid:
$$ \sigma = \left(\begin{array}{c} \cosh(u) \cos(v),\\
\cosh(u) \sin(v),\\
u\end{array}\right) $$
I assumed that there is a coordinate transformation $u = \phi(x,y)$ and $ v = \psi(x,y)$ such that they would give me the result I need.
Upon computing the following conditions that ensure the geodesic-conjugate properties
* conjugate property $\det(\sigma\_x,\sigma\_y,\sigma\_{xy}) = 0$
* geodesic x-coordinates $\det(N,\sigma\_x,\sigma\_{xx}) = 0$
* geodesic y-coordinates $\det(N,\sigma\_y,\sigma\_{yy}) = 0$
where $N$ is the Gauss map, I arrived to the following system of PDEs:
* $\ln \left(\psi\_{x}\right)-\ln \left(\phi\_{x}\right) =
\ln \left(\psi\_{y}\right)-\ln \left(\phi\_{y}\right)
$
* $\frac{{\partial}}{{\partial}y}\ln \left(\psi\_{y}\right)-\frac{{\partial}}{{\partial}y}\ln \left(\phi\_{y}\right)
=
\frac{\left(\phi\_{y}^{2}+\psi\_{y}^{2}\right) \tanh \left(\phi \right)}{\phi\_{y}}$
* $\frac{{\partial}}{{\partial}x}\ln \left(\psi\_{x}\right)-\frac{{\partial}}{{\partial}x}\ln \left(\phi\_{x}\right)=\frac{\left(\phi\_{x}^{2}+\psi\_{x}^{2}\right) \tanh\left(\phi \right)}{\phi\_{x}}$
Can someone clarify how I can continue from here? Unfortunately I don't know much about PDEs but I still believe that there should be an answer to this.
| https://mathoverflow.net/users/109420 | Building a geodesic conjugate parameterization on catenoid | I thought about your problem and realized that there is no coordinate parametrization of the catenoid with the properties that you want.
Here is my argument: First, note that, in the given $uv$-parametrization, the first and second fundamental forms of the immersion are
$$
I= \cosh(u)^2 (\mathrm{d}u^2 + \mathrm{d}v^2)
\quad\text{and}\quad
I\!I = \mathrm{d}u^2- \mathrm{d}v^2.
$$
We want to find two (possibly local) foliations by geodesics,
say $\mathcal{F}\_i$ for $i=1,2$, such that the tangent vectors to the leaves of one foliation are conjugate to the tangent vectors to the leaves of the other foliation. What this means is that,
if $X$ is a vector field tangent to the leaves of $\mathcal{F}\_1$ and $Y$ is a vecor field tangent to the leaves of $\mathcal{F}\_2$, then $I\!I(X,Y)=0$.
Of course, we can assume, without loss of generality, that $X$ and $Y$ are unit vector fields, i.e., $I(X,X)=I(Y,Y)=1$.
Now, given any unit tangent vector field to the surface, say, $Z$, the condition that the flow lines of $Z$ be geodesics is that $Z^\flat$, the dual $1$-form, be closed.
Thus, we require that $X$ and $Y$ be unit vector fields so that $I(X,X)=I(Y,Y)=1$ and that $X^\flat$ and $Y^\flat$ be closed, and we also require that $I\!I(X,Y) = 0$, since this is the conjugate property. If
$$
X^\flat = \cosh u\,(\cos\alpha\,\mathrm{d}u +\sin\alpha\,\mathrm{d}v),
$$
for some angle function $\alpha$, then
the equations $I(Y,Y)=1$ and $I\!I(X,Y) = 0$ force
$$
\pm Y^\flat = \cosh u\,(\sin\alpha\,\mathrm{d}u +\cos\alpha\,\mathrm{d}v).
$$
Note that $\cos(2\alpha)\not=0$, otherwise, $X\pm Y = 0$, and the two foliations will not be transverse, so they can't be the level sets of the desired coordinate functions.
So we need that the following $1$-forms be closed:
$$
\xi = \cosh u\,(\cos\alpha\,\mathrm{d}u +\sin\alpha\,\mathrm{d}v) \quad\text{and}\quad
\eta = \cosh u\,(\sin\alpha\,\mathrm{d}u +\cos\alpha\,\mathrm{d}v).
$$
If we take the exterior derivatives of these 1-forms and set them equal to zero, we find that $\alpha$ must satisfy the equation
$$
\mathrm{d}\alpha = \frac{\tanh u}{\cos 2\alpha}\,
\bigl(\mathrm{d}v - \sin 2\alpha\, \mathrm{d}u\bigr).
$$
But now applying the exterior derivative to both sides of this relation, we see that
$$
0 = \frac{1-2\cosh^2u}{\cosh^2u\,\cos 2\alpha}\,\mathrm{d}u\wedge\mathrm{d}v.
$$
Since $1-2\cosh^2u$ is never zero, we have a contradiction, so $\alpha$ cannot exist.
**Remark:** For comparison, it might be helpful to look at the case of the helicoid, parametrized by $(u\cos v, u\sin v, v)$. Here we have
$$
I= \mathrm{d}u^2 + (u^2{+}1)\mathrm{d}v^2
\quad\text{and}\quad
I\!I = \frac{2 du\,dv}{\sqrt{u^2{+}1}}.
$$
now, reasoning as above, we seek an angle $\alpha$ such that the $1$-forms
$$
\xi = \cos\alpha\,\mathrm{d}u +\sin\alpha\,\sqrt{u^2{+}1}\,\mathrm{d}v
\quad\text{and}\quad
\eta = \cos\alpha\,\mathrm{d}u -\sin\alpha\,\sqrt{u^2{+}1}\,\mathrm{d}v
$$
are closed. The equation $\mathrm{d}(\xi+\eta)=2\,\mathrm{d}(\cos\alpha\,\mathrm{d}u)=0$ implies that $\alpha$ must
be a function of $u$ and the equation $\mathrm{d}(\xi-\eta)=2\,\mathrm{d}(\sin\alpha\,\sqrt{u^2{+}1}\,\mathrm{d}v)=0$ then implies
that $\sin\alpha\,\sqrt{u^2{+}1} = c$ for some constant $c\not=0$.
Thus,
$$
\xi = \sqrt{\frac{u^2{+}1{-}c^2}{u^2{+}1}}\,\mathrm{d}u +c\,\mathrm{d}v
\quad\text{and}\quad
\eta = \sqrt{\frac{u^2{+}1{-}c^2}{u^2{+}1}}\,\mathrm{d}u -c\,\mathrm{d}v,
$$
defined on the range where $u^2{+}1{-}c^2>0$ (which is everywhere if $|c|<1$). The dual vector fields are
$$
X\_\pm = \sqrt{\frac{u^2{+}1{-}c^2}{u^2{+}1}}\,
\frac{\partial}{\partial u} \pm \frac{c}{u^2{+}1}\,\frac{\partial}{\partial v},
$$
so we choose $x$ and $y$ so that $\mathrm{d}x$ and $\mathrm{d}y$
each annihilate one of $X\_\pm$, say
$$
\mathrm{d}x = \mathrm{d}v
+\frac{c\,\mathrm{d}u}{\sqrt{(u^2{+}1{-}c^2)(u^2{+}1)}}
\quad\text{and}\quad
\mathrm{d}y = \mathrm{d}v
-\frac{c\,\mathrm{d}u}{\sqrt{(u^2{+}1{-}c^2)(u^2{+}1)}},
$$
which gives a 1-parameter family of solutions to the problem.
| 8 | https://mathoverflow.net/users/13972 | 450668 | 181,288 |
https://mathoverflow.net/questions/323192 | 1 | A finite reduced Laver-like algebra is a finite algebra $(X,\*,1)$ that satisfies the identities $1\*x=x,x\*1=1,x\*(y\*z)=(x\*y)\*(x\*z)$ and where there is a natural number $n$ and a function $\mathrm{crit}:X\rightarrow n+1$ where
1. $\mathrm{crit}(x)=n$ if and only if $x=1$,
2. $\mathrm{crit}(x\*y)=\mathrm{crit}(y)$ whenever $\mathrm{crit}(y)<\mathrm{crit}(x)$, and
3. $\mathrm{crit}(x\*y)>\mathrm{crit}(y)$ whenever
$\mathrm{crit}(x)\leq\mathrm{crit}(y)<n$.
We say that a finite reduced Laver-like algebra $(X,\*,1)$ is critically simple if whenever $\simeq$ is a congruence on $(X,\*,1)$ where there are $x,y\in X$ with $x\simeq y,x\neq y$, there is some $z\in X\setminus\{1\}$ with $z\simeq 1$.
We say that $(X,\*,1)$ is critically subsimple if $x,y\in X,x\*x=1,y\*y=1,\mathrm{crit}(x)=\mathrm{crit}(y)$ implies that $x=y$ for each $x,y\in X$. Every critically simple finite reduced Laver-like algebra is critically subsimple.
Suppose that $(X,\*,1)$ is a critically subsimple finite reduced Laver-like algebra generated by $(x\_{a})\_{a\in A}$ where $A$ is finite. Then does there exist a critically simple finite reduced Laver-like algebra $(Y,\*,1)$ generated by $(y\_{a})\_{a\in A}$ along with a homomorphism $\phi:Y\rightarrow X$ such that $\phi(y\_{a})=x\_{a}$ for each $a\in A$ and an element $c\in Y$ such that if $y\_{1},y\_{2}\in Y$, then $\phi(y\_{1})=\phi(y\_{2})$ if and only if $c\*y\_{1}=c\*y\_{2}$?
| https://mathoverflow.net/users/22277 | Is every critically subsimple Laver-like algebra a quotient of a critically simple Laver-like algebra on the same number of generators? | No. A critically subsimple Laver-like algebra is not necessarily a quotient of a critically simple Laver-like algebra with the same number of generators and more critical points. Our strategy for producing counterexamples is to exhibit finite reduced Laver-like algebras $X$ generated by $(x\_a)\_{a\in A}$ such that there does not exist a finite reduced Laver-like algebra $Y$ generated by $(y\_a)\_{a\in A}$ such that $Y$ has more critical points than $X$ and such that there is a homomorphism $\phi:Y\rightarrow X$ with $\phi(y\_a)=x\_a$ for $a\in A$.
Given a finite reduced Laver-like algebra $X$ and a generating set $(x\_a)\_{a\in A}$, I have obtained an algorithm for finding the class $\text{Cov}(X,(x\_a)\_{a\in A})$ of all finite reduced Laver-like algebras $Y$ together with a generating set $(y\_a)\_{a\in A}$ such that $Y$ has $1$ more critical point than $X$ and where there is a homomorphism $\phi:Y\rightarrow X$ with $\phi(y\_a)=x\_a$ for $a\in A$.
Producing finite reduced Laver-like algebras where $\text{Cov}(X,(x\_a)\_{a\in A})=\emptyset$ is not trivial since examples of algebras $(X,(x\_a)\_{a\in A})$ with $\text{Cov}(X,(x\_a)\_{a\in A})=\emptyset$ are difficult to obtain, and I have not been able to obtain any algebras $(X,(x\_a)\_{a\in A})$ with $\text{Cov}(X,(x\_a)\_{a\in A})=\emptyset$ without explicitly looking for such an algebra. I only know of one way of constructing such an algebra.
The way I constructed an algebra was to first perform a brute force search for a small reduced Laver-like algebra $(X,\*)$ generated by two elements $x,y$ with $y\*y=1$ and where if $\phi:Y\rightarrow X$ is a homomorphism between two Laver-like algebras and $Y$ is generated by $a,b$ and $\phi(a)=x,\phi(b)=y$, then $Y,X$ both have the same number of critical points. After, we have obtained the minimal Laver-like algebra $(X,\*)$, we produce a sequence $(X\_1,\dots,X\_n)$ where $X\_1=(X,(x,y))$, and $X\_{j+1}\in\text{Cov}(X\_j)$ for $1\leq j<n$, and where $\text{Cov}(X\_j)=\emptyset$. If say that two finite Laver-like algebras $X,Y$ are critically equivalent if there are congruences both denoted by $\simeq$ on $X,Y$ where $X/\simeq$ is isomorphic to $Y/\simeq$ and where $X,Y,X/\simeq,Y/\simeq$ all have the same number of critical points. We find some $Z$ that is critically equivalent to $X\_n$ but which is not critically subsimple and then $Z$ is our desired counterexample.
| 1 | https://mathoverflow.net/users/22277 | 450670 | 181,289 |
https://mathoverflow.net/questions/448764 | 3 | Let $G$ be a split reductive group over a nonarchimedean local field $F$ (I'm particularly interested in the case of $\operatorname{GSp}\_{2n}$).
---
>
> Given a parahoric subgroup $K \subset G(F)$, and a parabolic subgroup
> $P$, is there a "nice" group-theoretic description of the orbits of
> $K$ on the flag variety $\mathcal{F}(F) = P(F) \backslash G(F)$?
>
>
>
---
I have the following conjecture: if $T$ is a split maximal torus in $G$ and $\widetilde{W} = N\_G(T)(F) / T(\mathcal{O}\_F)$ is the extended affine Weyl group, then $\widetilde{W}$ is generated by a set $S$ of simple reflections, and there is a bijection between standard parahorics and proper subsets of $S$. I think that if $K\_J$ is a standard parahoric corresponding to some $J \subset S$, then
$$ P \backslash G / K = W\_{M\_P} \backslash W\_G / \pi(W\_J) $$
where $W\_J$ is the subgroup of $\widetilde{W}$ generated by $J$, and $\pi(W\_J)$ is its image in the usual Weyl group. This conjecture is true if $K$ is contained in $G(\mathcal{O}\_F)$. Does anyone know if it is true in general?
| https://mathoverflow.net/users/2481 | Orbit of a parahoric subgroup on a flag variety | Let $G$ a reductive group over a nonarchimedean local field $F$.
Let $P\_0$ be a minimal parabolic subgroup of $G$ and $A$ a maximal split torus contained in $P\_0$.
The normalizer $N\_G(A)(F)$ acts on the apartment of $A$ via the extended affine Weyl group,
which contains the affine Weyl group $W\_\mathrm{aff}$ with finite index.
Let $I$ be an Iwahori subgroup of $G(F)$ compatible with $P\_0$.
By the Iwasawa decomposition, the double cosets in $P\_0(F)\backslash G(F)/I$
are represented by the elements of the Weyl group $W := N\_G(A)(F)/C\_G(A)(F)$.
Let $\pi$ be the natural projection $W\_\mathrm{aff}\rightarrow W$.
It's not hard to show that if $w\in W$ and $s\in W\_\mathrm{aff}$
is a simple reflection (relative to $I$), then
\begin{equation}
\tag{1}\label{parabolic\_mult}
P\_0(F) w\pi(s) I\subseteq P\_0(F) w I s I \subseteq P\_0(F) w\pi(s) I \sqcup P\_0(F) wI,
\end{equation}
and if furthermore $s\in W$, then
\begin{equation}
\tag{2}\label{iwahori\_mult}
P\_0(F) sw I\subseteq P\_0(F) s P\_0(F) w I \subseteq P\_0(F) sw I\sqcup P\_0(F) wI.
\end{equation}
Now let $P$ be a parabolic subgroup of $G$ containing $P\_0$,
and let $K$ be a parahoric subgroup of $G(F)$ containing $I$.
Then
\begin{equation}
\tag{3}\label{bruhat}
P(F) = \bigsqcup\_{u\in W\_{M\_P}}P\_0(F)uP\_0(F),
\end{equation}
where $W\_{M\_P}$ is the Weyl group of the standard Levi component of $P$, and
\begin{equation}
\tag{4}\label{iwahori}
K = \bigsqcup\_{v\in W\_K}IvI,
\end{equation}
where $W\_K$ is the Coxeter subgroup of $W\_\mathrm{aff}$ consisting of those elements with representatives in $K$.
Given $w\in W$, consider the double coset $P(F)wK$. From (\ref{bruhat}) and (\ref{iwahori}),
$$P(F)wK = \bigcup\_{u\in W\_{M\_P},v\in W\_K}P\_0(F)uP\_0(F)wIvI.$$
Applying (\ref{parabolic\_mult}) one simple generator of $W\_K$ at a time,
we get that this union is equal to
$$\bigcup\_{u\in W\_{M\_P},v\in W\_K}P\_0(F)uP\_0(F)w\pi(v)I.$$
Applying (\ref{iwahori\_mult}) one simple generator of $W\_{M\_P}$ at a time,
we find that this is equal to
$$\bigcup\_{u\in W\_{M\_P},v\in W\_K}P\_0(F)uw\pi(v)I.$$
It follows from the Iwasawa decomposition that $w,w'\in W$ represent the same double coset in
$P(F)\backslash G(F)/K$ if and only if $w,w'$ lie in the same double coset in $W\_{M\_P}\backslash W/\pi(W\_K)$.
| 3 | https://mathoverflow.net/users/507750 | 450683 | 181,293 |
https://mathoverflow.net/questions/450680 | 4 | Let $(X,\succeq)$ be a poset. I have the following two questions:
1. Is it true that there exists a finite complemented distributive lattice (a Boolean lattice) $(S, \succeq^\*)$ such that $X\subseteq S$ and $\succeq^\*$ agrees with $\succeq$ over $X$? If not, are there conditions on $(X,\succeq)$ that guarantee this?
2. If the answer to question 1 is true, then is it true that $(S, \succeq^\*)$ is "unique" conditional on the cardinality of $S$? That is, if $(S\_1, \succeq^\*\_{1})$ and $(S\_2, \succeq^\*\_{2})$ satisfy the conditions described in question 1 above and $\lvert S\_1\rvert=\lvert S\_2\rvert$, then $(S\_1, \succeq^\*\_{1})$ is isomorphic to $(S\_2, \succeq^\*\_{2})$? If not, are there any uniqueness results along these lines?
I'm not a mathematician, but this question has come up in my research. Sorry if this is obvious, and thank you for your help!
| https://mathoverflow.net/users/98626 | Is every finite poset a subset of a finite complemented distributive lattice? | As Sam mentioned in the comments, the answer to question 1 is yes. One can map every condition $p$ in the partial order to the lower cone, the set $S\_p=\{q\in P\mid q\leq p\}$ of conditions below $p$. It is clear that $q\leq p\iff S\_q\subset S\_p$, and so this maps $P$ into the powerset algebra, which is a Boolean algebra. This idea works for any poset, whether finite or infinite.
Meanwhile, the answer to question 2 is no for infinite cardinalities. The particular Boolean algebra that we had used was atomic, but we may find a larger Boolean algebra of the same infinite cardinality by embedding that Boolean algebra inside an atomless algebra. Simply view every atom as sitting atop a copy of the countable atomless algebra, and then generate the corresponding Boolean algebra. This will have the same infinite cardinality, while still accepting an embedded copy of $P$, but this larger Boolean algebra, being atomless, is not isomorphic to the earlier one, which is.
For finite cardinalities, however, of course the answer to question 2 is yes, since every finite Boolean algebra is isomorphic to a finite powerset algebra, and so if two of them have the same size (must be $2^n$), then they are isomorphic.
| 6 | https://mathoverflow.net/users/1946 | 450696 | 181,297 |
https://mathoverflow.net/questions/450554 | 1 | Suppose that $u$ is $C^1$ in $[0,\pi/2]$ and $u(0)=u’(\pi/2)=0$. I want to derive the following Poincaré inequality
$$
\int\_0^{\pi/2} u(x)^2\,dx \leq \int\_0^{\pi/2} u’(x)^2\,dx.
$$
Since $u(0)=0$, we have that $u(x) = \int\_0^x u’(s)\,ds$. Hence
\begin{align}
u^2(x) & \leq \biggl( \int\_0^x u’(s)\,dx\biggr)^2 \\
& \leq x\int\_0^x u’(s)^2\,ds
\end{align}
by Cauchy-Schwarz inequality. It follows that
\begin{align}
\int\_0^{\pi/2} u^2(x)\,dx
& \leq \int\_0^{\pi/2} \int\_0^x xu’(s)^2\,ds\,dx \\
& = \int\_0^{\pi/2} \int\_s^{\pi/2} xu’(s)^2\,dx\,ds \\
& = \int\_0^{\pi/2} \biggl(\frac{\pi^2}{8} - \frac{s^2}{2}\biggr) u’(s)^2\,ds \\
& \leq \frac{\pi^2}{8} \int\_0^{\pi/2} u’(s)^2\,ds.
\end{align}
As you can see, the constant $1$ (better than $\pi^2/8$) can not be obtained by using only the condition $u(0)=0$.
My question is how to utilize the other boundary condition $u’(\pi/2)=0$ to derive this better constant. I have seen this conclusion in one paper and the author said that the constant can be obtained by considering the eigenvalue problem
\begin{cases}
-u’’ = \lambda u \\
u(0)=u’(\pi/2)=0
\end{cases}
and calculating that $1$ is the minimal eigenvalue of the above equation. What’s the principle behind this?
| https://mathoverflow.net/users/480661 | Poincaré inequality in dimension one with mixed boundary condition | We want to show that
$$\int\_0^{\pi/2} u(x)^2\,dx \le \int\_0^{\pi/2} u'(x)^2\,dx \tag{1}\label{1}$$
for all $u\in C^1[0,\pi/2]$ such that $u(0)=0$ and $u'(\pi/2)=0$.
Let us show that **\eqref{1} holds even without any condition on $u'(\pi/2)$**.
We have the [Poincaré inequality](https://en.wikipedia.org/wiki/Poincar%C3%A9_inequality#The_Poincar%C3%A9_constant)
$$\int\_0^\pi w(x)^2\,dx \le \int\_0^\pi w'(x)^2\,dx \tag{2}\label{2}$$
for all $w\in W\_{0}^{{1,2}}(0,\pi)$.
Take any $u\in C^1[0,\pi/2]$ such that $u(0)=0$. For $x\in[0,\pi]$, let
$$w(x):=u(x)\,1(x\le\pi/2)+u(\pi-x)\,1(x>\pi/2). \tag{3}\label{3}$$
Then $w\in W\_{0}^{{1,2}}(0,\pi)$. So, by \eqref{3} and \eqref{2},
$$\int\_0^{\pi/2} u(x)^2\,dx
=\frac12\int\_0^\pi w(x)^2\,dx \le \frac12\int\_0^\pi w'(x)^2\,dx=\int\_0^{\pi/2} u'(x)^2\,dx,$$
which proves \eqref{1}. $\quad\Box$
---
*Remark 1:* That the condition $u'(\pi/2)=0$ can be dropped should be clear from the very beginning, because any $u\in C^1[0,\pi/2]$ such that $u(0)=0$ can be however closely approximated in $ W^{{1,2}}(0,\pi/2)$ by a function $v\in C^1[0,\pi/2]$ such that $v(0)=0$ and $v'(\pi/2)=0$.
*Remark 2:* In view of what has been said here, your statement "As you can see, the constant $1$ (better than $\pi^2/8$) can not be obtained by using only the condition $u(0)=0$" cannot be true. The "reason" why this statement of yours does not follow from your reasoning is that the Cauchy--Schwarz inequality in your reasoning and the inequality resulting from dropping $-\frac{s^2}2$ in the integrand cannot both turn into the corresponding equalities, for any nonzero $u\in C^1[0,\pi/2]$. Indeed, the equality in the Cauchy--Schwarz inequality in your reasoning holds only if $u'$ is almost everywhere equal to a constant $c$, and $c=0$ together with the condition $u(0)=0$ would imply $u=0$.
| 2 | https://mathoverflow.net/users/36721 | 450698 | 181,298 |
https://mathoverflow.net/questions/450676 | 0 | I am confused on [the Lemma 5.7 (Lipschitz maximal inequality) here](https://web.math.princeton.edu/%7Ervan/APC550.pdf). Let me first restate the definition and the lemma:
**Def**
$\{X\_t\}\_{t\in T}$ is called Lipschitz for metric $d$ on $T$ if there exists a random variable $C$ such that
$$|X\_t-X\_s|\leq Cd(t,s),\text{ for all }t,s\in T.$$
**Lemma**
Suppose $\{X\_t\}\_{t\in T}$ is a Lipschitz process and is $\sigma^2$-subgaussian for every $T$. Then $$E[\sup\_{t\in T}X\_t]\leq \inf\_{\epsilon>0}\{\epsilon E[C]+\sqrt{2\sigma^2\log N(T,d,\epsilon)}$$
To prove the lemma, we split $X\_t$ into two parts, i.e. $$\sup\_{t\in T}X\_t\leq \sup\_{t\in T}\{X\_t-X\_{\pi(t)}\}+\sup\_{t\in T}X\_{\pi(t)}$$ then take expectation we have $$E[\sup\_{t\in T}X\_t]\leq \epsilon E[C]+\sqrt{2\sigma^2\log|N|}$$
I am a bit confused on the applicability of this lemma on random process that its $E[\sup X\_t]$ is negative.
In the proof, we use the Lipschitz bound to control $\sup\_{t\in T}\{X\_t-X\_{\pi(t)}\}\leq Cd(t,\pi(t))$. Clearly, this bound $Cd(t,\pi(t))$ is non-negative, thus the the resulting bound $ \inf\_{\epsilon>0}\{\epsilon E[C]+\sqrt{2\sigma^2\log N(T,d,\epsilon)}$ in the lemma is also non-negative.
But, what if the expectation of sup of random process is negative? Then the bound given by the lemma is very loose.
For example, I guess this process $\sum\_{i=1}^n(-100+w\_i)\sin(\theta\_i)$ where $w\_i$ is iid standard gaussian would have negative expectation of supreme.
Am I misunderstood something on this lemma?
| https://mathoverflow.net/users/494410 | Lipschitz maximal inequality for random process | $\newcommand\ep\epsilon\newcommand\si\sigma\newcommand\td{\tilde d}$You wrote: "I am a bit confused on the applicability of this lemma on random process that its $E[\sup X\_t]$ is negative."
One way to use this result, whether $E\sup\_{t\in T} X\_t$ is negative or not, is as follows. We have
>
> $$E\sup\_{t\in T}X\_t\le \inf\_{\ep>0}\big\{\ep EC+\sqrt{2\si^2\ln N(T,d,\ep)}\big\} \tag{10}\label{10} $$
> given that
> $$|X\_t-X\_s|\le Cd(t,s)\text{ for all }(t,s)\in T\times T. \tag{20}\label{20}$$
>
>
>
Here $N(T,d,\ep)$ is the smallest number of balls of radius $\ep$ in metric $d$ needed to cover the set $T$, and $C>0$ is a random variable.
By appropriately truncating the $X\_t$'s and $d$ (or by going through the proof of this highlighted result), here we may allow $d$ to be a generalized metric -- such that $d(t,s)$ may take the value $\infty$ for some $(t,s)\in T\times T$.
Let a set $S$ be a copy of the set $T$ **disjoint** from $T$, so that we have a bijection $f$ from $S$ onto $T$. For all $u\in U:=T\cup S$, let
$$Y\_u:=\begin{cases}
X\_u&\text{ if }u\in T, \\
-X\_{f(u)}&\text{ if }u\in S.
\end{cases}
$$
Then
$$\sup\_{u\in U}Y\_u=\max\Big(\sup\_{u\in T}X\_u,\sup\_{u\in S}(-X\_{f(u)})\Big) \\
=\max\Big(\sup\_{t\in T}X\_t,\sup\_{t\in T}(-X\_t)\Big)
=\sup\_{t\in T}|X\_t|. \tag{30}\label{30}
$$
Extend the generalized metric $d$ on $T\times T$ to the generalized metric $\td$ on $U\times U$ by the formula
$$\td(u,v):=\begin{cases}
d(u,v)&\text{ if }(u,v)\in T\times T, \\
d(f(u),f(v))&\text{ if }(u,v)\in S\times S, \\
\infty&\text{ otherwise }.
\end{cases}
$$
Then $|Y\_u-Y\_v|\le C\td(u,v)\text{ for all }(u,v)\in U\times U$; cf. \eqref{20}. So, by the highlighted result,
$$E\sup\_{u\in U}Y\_u\le \inf\_{\ep>0}\big\{\ep EC+\sqrt{2\si^2\ln N(U,\td,\ep)}\big\}. \tag{40}\label{40}$$
But $N(U,\td,\ep)\le2N(T,d,\ep)$ for real $\ep>0$.
So, by \eqref{30} and \eqref{40},
$$E\sup\_{t\in T}|X\_t|\le \inf\_{\ep>0}\big\{\ep EC+\sqrt{2\si^2\ln (2N(T,d,\ep))}\big\}. $$
| 1 | https://mathoverflow.net/users/36721 | 450703 | 181,300 |
https://mathoverflow.net/questions/450679 | 3 | For an $n \times n$ matrix $M$, the $\infty\to 1$ and [cut](https://en.wikipedia.org/wiki/Matrix_norm#Cut_norms) norms are given by
$$\|M\|\_{\infty \to 1} := \max\limits\_{x, y \in \{\pm 1\}^n} \sum\limits\_{i, j} m\_{i, j} x\_i y\_j, \qquad \|M\|\_{\square} := \max\limits\_{A, B} \left|\sum\limits\_{i \in A, j \in B} m\_{ij}\right|.$$
By expanding to a matrix so that the row and column sums are $0,$ both norms become essentially the same, i.e.,
$$\|M\|\_{\infty \to 1} = 4 \|M\|\_{\square}.$$
The idea is that due to the zero sums, the $1$'s in $x, y$ will represent $A, B$ respectively. This observation and usage of integer programming allows us to approximate the cut norm within a constant factor in polynomial time.
For finite sets $X, Y,$ we say $X < Y$ if $\max X < \min Y.$ I am interested in the following measurement: $$f(M) = \max\limits\_{A < B < C \\ |A|=|B|=|C|} \left(\sum\limits\_{i \in A, j \in C} m\_{ij} - \sum\limits\_{i \in B, j \in C} m\_{ij} \right).$$ Motivation: It's a measure of how much $M$ fails to increase as we move towards the diagonal from above, the useful property being that $f(M) \ge 0$ and $f(M) = 0$ if and only if $M$ increases towards the diagonal. The function $f$ also behaves well with the cut norm.
I can replace $C$ with $y \in \{-1,1\}^n$ and do the same row sum trick. It may be possible to replace $A$ and $B$ with $x \in \{-1, 0, 1\}^n$ and then use the column sum trick. However, there is a major problem: we are only considering $x$ with the same number of $1$s as $-1$s and where all the $-1$'s come after the $1$s.
Each such $x$ corresponds bijectively to a size $2m$ subset of $\{1, 2, \dots, n\}$ by making the first $m$ elements $1,$ the last $m$ elements $-1,$ and the rest $0.$ Thus, ignoring the constraint from $|C|,$ there are $\binom{n}{0} + \binom{n}{2} + \dots = 2^{n-1}$ ways to choose $A, B.$ To choose all of $A, B, C,$ there are of course $\binom{n}{0}+\binom{n}{3}+\dots = \frac{2^n + 2 \cos(n\pi/3)}{3}$ ways.
The cut norm was nicely reduced due to the natural correspondence between the $2^n$ subsets of $\{1, 2, \dots, n\}$ and the elements of $\{-1,1\}^n.$ Since we have neither $2^n$ nor $3^n$ choices for $A, B, C,$ plus the choices are not as clean, I am unable to turn computing $f$ into an integer programming problem. Is it possible in the case $M \ge 0$? It may be possible for all $M$ but I only care about non-negative matrices anyways.
| https://mathoverflow.net/users/127521 | How to turn $\{-1, 0, 1\}$-valued optimization problem into integer program? | Given $n \times n$ matrix $M$, you want to find $A,B,C \subset \{1,\dots,n\}$ to maximize
$$\sum\limits\_{i \in A, j \in C} m\_{ij} - \sum\limits\_{i \in B, j \in C} m\_{ij}$$
subject to $A < B < C$ and $|A|=|B|=|C|$.
Introduce binary decision variables $a\_i, b\_i, c\_i \in \{0,1\}$ to indicate whether $i\in A, i\in B, i\in C$, respectively. Introduce decision variable $d$ to represent the common cardinality. The problem is to maximize
$$\sum\_{i=1}^n \sum\_{j=1}^n m\_{ij} (a\_i c\_j - b\_i c\_j)$$
subject to linear constraints
\begin{align}
a\_i + b\_j &\le 1 &&\text{for $i \ge j$} \\
b\_i + c\_j &\le 1 &&\text{for $i \ge j$} \\
\sum\_i a\_i &= d \\
\sum\_i b\_i &= d \\
\sum\_i c\_i &= d \\
\end{align}
This is an integer quadratic programming (IQP) problem, but you can linearize by introducing binary variables $u\_{ij}$ and $v\_{ij}$ to replace the products $a\_i c\_j$ and $b\_i c\_j$, respectively. The resulting integer linear programming (ILP) problem is to maximize
$$\sum\_{i=1}^n \sum\_{j=1}^n m\_{ij} (u\_{ij} - v\_{ij})$$
subject to linear constraints
\begin{align}
a\_i + b\_j &\le 1 &&\text{for $i \ge j$} \\
b\_i + c\_j &\le 1 &&\text{for $i \ge j$} \\
\sum\_i a\_i &= d \\
\sum\_i b\_i &= d \\
\sum\_i c\_i &= d \\
a\_i &\ge u\_{ij} &&\text{for all $i$ and $j$} \tag1\label1 \\
c\_j &\ge u\_{ij} &&\text{for all $i$ and $j$} \tag2\label2 \\
a\_i + c\_j - 1 &\le u\_{ij} &&\text{for all $i$ and $j$}\\
b\_i &\ge v\_{ij} &&\text{for all $i$ and $j$}\\
c\_j &\ge v\_{ij} &&\text{for all $i$ and $j$}\\
b\_i + c\_j - 1 &\le v\_{ij} &&\text{for all $i$ and $j$} \tag3\label3
\end{align}
For nonnegative $m\_{ij}$, you can omit constraints \eqref{1}, \eqref{2}, and \eqref{3}.
| 2 | https://mathoverflow.net/users/141766 | 450718 | 181,305 |
https://mathoverflow.net/questions/450523 | 8 | My question is motivated by [this one](https://mathoverflow.net/q/316985), but within real matrices instead of complex ones.
>
> ${\bf Sym}\_n(\mathbb R)$ is a vector space of dimension $N=\frac{n(n+1)}2$. Equipped with the scalar product $\langle A,B\rangle={\rm Tr}(AB)$, this is a Euclidean space. Does there exist a basis $(R\_1,\ldots,R\_N)$, made of orthogonal matrices only ?
>
>
>
In this situation, each element $U\_j$ satisfies $(U\_j)^2=I\_n$ and therefore is of the form $2\pi\_j-I\_n$, where $\pi\_j$ is an orthogonal projector, over some subspace $E\_j$ of $\mathbb R^n$.
Two observations. On the one hand, the answer is **No** when $n=3$. The basis may not contain $\pm I\_3$, because $\langle I\_3,R\_j\rangle=\pm1\ne0$ for $R\_j\ne\pm I\_3$. Therefore one may choose each $R\_j$ of the form $2x\_jx\_j^T-I\_3$ where $x\_j$ is a unit vector. The condition that $\langle R\_j,R\_k\rangle=0$ for $j\ne k$ writes $|\langle x\_j,x\_k\rangle|=\frac12$. Therefore $\|x\_j-x\_k\|=1$ or $\sqrt3$. The twelve unit vectors $\pm x\_j$ are such that the pairwise distances are either $1$, $\sqrt3$ or $2$ (for opposite vectors). In short, they should form a regular icosahedron (each vertex has five neighbours) inscribed in the unit sphere, whose edges have unit length ! This is false, since the edge of this icosahedron equals
$$\sqrt{\frac{10-2\sqrt5}5}.$$
On the other hand, if we denote $d\_j=\dim E\_j={\rm Tr}\pi\_j$, the orthogonality condition is
$$4{\rm Tr}(\pi\_j\pi\_k)=2d\_j+2d\_k-n,\qquad j\ne k.$$
We deduce that
$$\langle\pi\_i-\pi\_j,\pi\_k-\pi\_\ell\rangle=0$$
whenever $i,j,k,\ell$ are pairwise distinct. Therefore ${\bf Sym}\_n(\mathbb R)$ contains a subset made of $\frac{N(N-1)}2$ elements (the $S\_{ij}=\pi\_i-\pi\_j$), in which each element is orthogonal to $\frac{(N-2)(N-3)}2$ other ones. Can such a rich configuration exist ?
**Edit**. Here is an interesting remark. The case $n=3$, studied above, suggested that $I\_n$ is unlikely to belong to the basis (though it does for $n=2$). Thus let us raise the side question:
>
> does there exist such a base $\cal B$, containing $R\_1=I\_n$ ?
>
>
>
If so the other elements $R\_i=2\pi\_i-I\_n$, being orthogonal to $I\_n$, satisfy $\dim E\_i=\frac n2$. In particular $n$ is even. Now, the orthogonality of $R\_i$ and $R\_j$ gives
$$4{\rm Tr}(\pi\_i\pi\_j)=2\frac n2+2\frac n2-n=n.$$
Let now $k\in[1,N-1]$ be an integer, and $J\subset[2,N]$ have cardinal $k$. Let us form
$$H\_J=\sum\_{j\in J}\pi\_j,$$
which is positive semi-definite. Then
$${\rm Tr}H\_J=k\frac n2,\qquad{\rm Tr}(H\_J)^2=k\frac n2+k(k-1)\frac n4=k\frac n4 (k+1).$$
Let $r\_J$ be the rank of $H\_J$. By Cauchy-Schwarz (written for the eigenvalues of $H\_J$), we have
$$({\rm Tr}H\_J)^2\le r\_J{\rm Tr}(H\_J)^2,$$
which gives
$$r\_J\ge n\frac k{k+1}.$$
If $k=n$, then this implies $r\_J\ge\frac{n^2}{n+1}>n-1$ and thus $r\_J\ge n$. In other words:
>
> Suppose $R\_1=I\_n$. For every $J\subset[2,N]$ of cardinal $n$, the symmetric matrix $H\_J=\sum\_{j\in J}\pi\_j$ is positive definite.
>
>
>
Of course this sum of $n$ terms is relatively small, compared to the $N-1=\frac{(n-1)(n+2)}2$ involved projectors $\pi\_j$.
**Redit**. The argument above actually extends to the general case (when $\cal B$ does not necessarily contain $I\_n$). Let $J\subset[1,N]$ be of cardinal $m$, and denote
$$H\_J=\sum\_J\pi\_j.$$
If $D\_J={\rm Tr}H\_J=\sum\_Jd\_j$, then ${\rm Tr}(H\_J)^2=m(D\_J-\frac{n(m-1)}4)$. Cauchy-Schwarz yields
$$D\_J^2\le r\_Jm\left(D\_J-\frac{n(m-1)}4\right).$$
The polynomial $X^2-r\_Jm\left(X-\frac{n(m-1)}4\right)$ thus has real roots. Writing that its discriminant is non-negative, we obtain
$$r\ge n\frac{m-1}m.$$
For $m=n+1$, this gives $r>n-1$, that is $r=n$. Therefore
>
> The sum of $n+1$ distinct $\pi\_j$'s is positive semi-definite.
>
>
>
| https://mathoverflow.net/users/8799 | Orthogonal basis of ${\bf Sym}_n(\mathbb R)$, made of orthogonal matrices | Here is an example with $n=4$. (ADDED BELOW: An example for any power of $2$)
I identify $\mathbb R^4$ with the quaternions, and describe $10$ subspaces such that any two of the resulting orthogonal projections satisfy the required condition $4Tr(\pi\pi')=2d+2d'-4$. One of the ten is the zero-dimensional subspace, and the other nine are $2$-dimensional. I need $Tr(\pi\pi')=1$ for any two of these nine.
Recall that in the group of unit quaternions there is a subgroup of order $24$. It contains (and normalizes) the order $8$ group $\lbrace \pm 1,\pm i,\pm j, \pm k\rbrace$. The other sixteen elements are the quaternions of the form
$$
\frac{\pm 1\pm i \pm j\pm k}{2}.$$
Dividing by the subgroup $\lbrace \pm 1\rbrace$, we get a group $G$ of order $12$ with a normal subgroup $V$ of order $4$. The elements of $G$ correspond to $12$ one-dimensional vector subspaces. The four lines belonging to $V$ are perpendicular to each other, and within each of other two cosets of $V$ in $G$ the four lines are again perpendicular to each other. This gives us eighteen two-dimensional subspaces, each spanned by two perpendicular lines. These come in nine orthogonal pairs. Choose one from each pair.
When $V$ and $V'$ are two of the nine and $\pi$ and $\pi'$ are the corresponding projections, then:
Case 1. $V$ and $V'$ might both come from the same coset of $V$ in $G$. A typical example is: $V$ spanned by $\lbrace 1,i\rbrace$ and $V'$ spanned by $\lbrace 1,j\rbrace$. The trace here is $1$ (eigenvalues $1,0,0,0$).
Case 2. $V$ and $V'$ come from different cosets. Here a typical example is $V$ spanned by $\lbrace 1,i\rbrace$ and $V'$ spanned by $\lbrace 1+i,j+k\rbrace$. The trace is again $1$ (eigenvalues $\frac{1}{2},\frac{1}{2},0,0$).
ADDED
Now that I think it over, a more direct way to describe these nine symmetric-and-orthogonal matrices is to say that they are the maps $x\mapsto axb$, where $\lbrace a,b\rbrace\subset \lbrace i,j,k\rbrace$.
GENERALIZATION
The answer is yes when $n=2^k$ for any $k\ge 0$. In the basis that I will specify, each matrix is a monomial matrix, that is, the product of a permutation matrix (for a very particular kind of permutation) and a diagonal matrix whose eigenvalues are $\pm 1$.
Let $V$ be a $k$-dimensional vector space over $\mathbb F\_2$. I will make a set of "$V$ by $V$ matrices", i.e. functions $A:V\times V\to \lbrace 1,-1,0\rbrace\subset \mathbb R$. Each of these will be an orthogonal matrix, the inner product of two of them will always be zero, and he number of elements of the set will be $\frac{n(n+1)}{2}$.
If $H\subset V$ is either $V$ or a codimension one vector subspace, and if $c\in H$, then define the matrix $A\_{H,c}$ by $A\_{H,c}(v,v+c)=1$ if $v\in H$ and $A\_{H,c}(v,v+c)=-1$ if $v\notin H$, and $A\_{H,c}(v\_1,v\_2)=0$ if $v\_2\neq v\_1+c$. Note that this is symmetric because $c\in H$.
The inner product of $A\_{H,c}$ and $A\_{H',c'}$ is $0$ if $c\neq c'$, because in that case there is no $(v\_1,v\_2)$ such that both $A\_{H,c}(v\_1,v\_2)$ and $A\_{H',c'}(v\_1,v\_2)$ are non-zero.
The inner product of $A\_{H,c}$ and $A\_{H',c}$ is $0$ if $H\neq H'$ because exactly one half of the vectors of $V$ are in the symmetric difference of $H$ and $H'$, insuring that one half of the non-zero terms $A\_{H,c}(v,v+c)A\_{H',c}(v,v+c)$ in the inner product are $-1$ while the other half are $+1$.
The total number of pairs $(H,c)$ with $H=V$ is $n$, and the total number of pairs $(H,c)$ with $H\neq V$ is $\frac{(n-1)n}{2}$. This adds to $\frac{n^2+n}{2}$.
| 8 | https://mathoverflow.net/users/6666 | 450721 | 181,306 |
https://mathoverflow.net/questions/450615 | 4 | Is there a smooth embedding of $S^2$ into some Euclidean space that is equivariant with respect to a linear representation of $PSL(2,\mathbb C)$?
A counterexample to a more general question can be found [here.](https://mathoverflow.net/questions/149867/is-there-an-analogue-of-mostow-palais-equivariant-embedding-theorem-for-noncompa)
| https://mathoverflow.net/users/1573 | An analogue of Mostow-Palais equivariant embedding theorem for the group of conformal automorphisms of the 2-sphere | The answer is 'no'. In fact, a stronger statement is true: If $V$ is a finite dimensional vector space and $G\subset\mathrm{GL}(V)$ is a (connected) non-compact simple Lie group, then the only bounded orbits of $G$ are fixed points. *A fortiori*, the only compact $G$-orbits are fixed points.
To see this, it's enough to prove it for $G = \mathrm{SL}(2,\mathbb{R})$, since every connected non-compact simple Lie group contains a copy of $\mathrm{SL}(2,\mathbb{R})$ (or its quotient $\mathrm{SO}(2,1)$).
Moreover, it is enough to prove the claim when $V$ is an irreducible representation of $\mathrm{SL}(2,\mathbb{R})$, so assume that. Thus, $V = S^d(W)$ where $W\simeq\mathbb{R}^2$ is the standard $2$-dimensional representation of $\mathrm{SL}(2,\mathbb{R})$, with basis elements $x$ and $y$. The diagonal subgroup of $\mathrm{SL}(2,\mathbb{R})$, a copy of $\mathbb{R}^\times$, acts on $x$ and $y$ as $r \cdot x = rx$ and $r\cdot y = r^{-1}y$ for $r\in\mathbb{R}^\times$. An element of $S^d(W)$ is a polynomial in $x$ and $y$, homogeneous of degree $d$, and we have $r\cdot (x^k y^{d-k}) = r^{2k-d} x^k y^{d-k}$. Thus, the only element of $S^d(W)$ that could have a bounded $G$-orbit would be $x^{d/2}y^{d/2}$ (and that exists only when $d$ is even). However, unless $d=0$, the element $(xy)^{d/2}$ (when it exists) has its $G$-orbit containing polynomials that are not multiples of $(xy)^{d/2}$, and hence their $G$-orbits are not bounded, so the $G$-orbit of $(xy)^{d/2}$ is not bounded either.
| 4 | https://mathoverflow.net/users/13972 | 450738 | 181,311 |
https://mathoverflow.net/questions/450699 | 2 | How to prove that the following expression is positive for $u\ge 0$ and $q\in(0,\pi/3)$. $$\frac{2 \sqrt{2} ((2+u) (1+\cos(q)))^{3/4}}{3^{3/4}}-\frac{2^{3/4} (1+\sec(q))}{\sqrt{3} \sec(q)^{3/4}}-\left(1+u+\sqrt{-1+(1+u)^2} \sqrt{1-\frac{4 \sin(q)^2}{3}}\right)^{3/4}$$
| https://mathoverflow.net/users/504719 | An inequality involving 2 variables | Letting $c:=\cos q$, we have $c\in[1/2,1]$. Writing $\sec q=1/c$ and $\sin^2q=1-c^2$, and multiplying the big expression by $3^{3/4}c^{1/4}$, we see that the inequality in question can be rewritten as
\begin{equation\*}
f(c,u)\overset{\text{(?)}}>0, \tag{10}\label{10}
\end{equation\*}
where $(c,u)\in[1/2,1]\times[0,\infty)$,
\begin{equation\*}
f(c,u):=\big(1-(r\_1(c,u)+r\_2(c,u))\big)\,2\sqrt2\, (c (1 + c)^3 (2 + u)^3)^{1/4},
\end{equation\*}
\begin{equation\*}
r\_1(c,u):=\frac{\sqrt[4]{3}}{2^{3/4}}\, \sqrt[4]{\frac{c+1}{c (u+2)^3}},\quad
r\_2(c,u):=\frac1{2 \sqrt{2}}\,
\left(\frac{\sqrt{3} \sqrt{\left(4 c^2-1\right) u (u+2)}+3 u+3}{c u+2
c+u+2}\right)^{3/4}.
\end{equation\*}
So, it is enough to show that
\begin{equation\*}
r\_1(c,u)+r\_2(c,u)\overset{\text{(?)}}<1. \tag{20}\label{20}
\end{equation\*}
Note that $r\_1(c,u)$ is decreasing in $c\in[1/2,1]$ and in $[0,\infty)$. So,
\begin{equation\*}
r\_1(c,u)\le r\_1(1/2,73/10)<1-\frac{3^{3/4}}{2 \sqrt{2}}\quad
\text{for }(c,u)\in[1/2,1]\times[73/10,\infty).
\end{equation\*}
Next, for all $(c,u)\in[1/2,1]\times[0,\infty)$
\begin{equation\*}
\begin{aligned}
&\big((2\sqrt2\,r\_2(c,u))^{4/3}-3\big) \\
&\times (c+1) (u+2) \left(\sqrt{3} \sqrt{\left(4 c^2-1\right) u (u+2)}+3 c u+6 c+3\right) \\
&=\left(3-3 c^2\right) u^2+6 \left(2 c^2+3 c+1\right) u+9 (2 c+1)^2,
\end{aligned}
\end{equation\*}
which is manifestly $>0$, whence $r\_2(c,u)<\dfrac{3^{3/4}}{2 \sqrt{2}}$.
So, we have \eqref{20} and hence \eqref{10} for $(c,u)\in[1/2,1]\times[73/10,\infty)$.
It remains to prove \eqref{10} for $(c,u)\in[1/2,1]\times[0,73/10]$.
To do this, note that
\begin{equation}
f(c,u)=f\_1(c,u)-f\_2(c,u),
\end{equation}
where
\begin{equation}
f\_1(c,u):=2 \sqrt{2} \sqrt[4]{c (c+1)^3 (u+2)^3}-2^{3/4} \sqrt[4]{3} (c+1),
\end{equation}
\begin{equation}
f\_2(c,u):=\sqrt[4]{c \left(\sqrt{3} \sqrt{\left(4 c^2-1\right) u (u+2)}+3 u+3\right)^3}.
\end{equation}
Clearly, $f\_2(c,u)$ is increasing in $c\in[1/2,1]$ and in $[0,\infty)$.
Also clearly, $f\_1(c,u)$ is increasing in $u\in[0,\infty)$.
Also,
\begin{equation}
g\_1(c,u):=\partial\_c f\_1(c,u)=\frac1{\sqrt{2}}
\Big(\frac{(4 c+1) (u+2)^{3/4}}{\sqrt[4]{c^3 (c+1)}}-2 \sqrt[4]{6}\Big)
\ge g\_1(c,0)
\end{equation}
and
\begin{equation}
\partial\_c g\_1(c,0)=-\frac{3}{2\ 2^{3/4} c^{7/4} (c+1)^{5/4}}<0.
\end{equation}
So, $g\_1(c,u)\ge g\_1(c,0)\ge g\_1(1,0)=2.78\ldots>0$. So, $f\_1(c,u)$ is increasing in $c\in[1/2,1]$.
So, $f\_1(c,u)$ is increasing in $c\in[1/2,1]$ and in $[0,\infty)$.
So, "partitioning" the rectangle $[0,73/10]$ into $20\times200=4000$ congruent rectangles, with vertices $(c\_i,u\_j):=(\frac{1}{2}+\frac{1}{2}\frac{i}{20},\frac{73}{10}\frac{j}{200})$,
for all $(c,u)\in[1/2,1]\times[0,73/10]$ we get
\begin{equation}
\begin{aligned}
f(c,u)
\ge\min\big\{f\_1 & \big(c\_{i-1},u\_{j-1}\big)
-f\_2\big(c\_i,u\_j\big) \colon \\
&i=1,\dots,20,\,j=1,\dots,200\big\}=0.0247\ldots>0.
\end{aligned}
\end{equation}
This completes the proof of \eqref{10}. $\quad\Box$
| 1 | https://mathoverflow.net/users/36721 | 450740 | 181,312 |
https://mathoverflow.net/questions/450756 | 4 | Let $G$ be a group and $\pi$ be a finite-dimensional (not necessarily unitary) representation of $G$ on a complex Hilbert space $H$. We shall say that $\pi$ is completely reducible if there exists a decomposition of $H$ into *orthogonal* irreducible sub-representations of $\pi$.
>
> **Question 1.** Suppose $(\pi\_1, H\_1)$, $(\pi\_2, H\_2)$ are completely reducible representations of $G$. Then is the tensor product $(\pi\_1 \otimes \pi\_2, H\_1 \otimes H\_2)$ also completely reducible?
>
>
>
Note, this question is different from the below question because in that question it does not require the direct sum decomposition to be orthogonal since the group is simply acting on a vector space.
[Semisimple representations of discrete groups](https://mathoverflow.net/questions/257583/semisimple-representations-of-discrete-groups)
Though, I had a read of the wonderful answer by nfdc23 to the above question which explains that the crux of proof uses the fact that all finite-dimensional linear representations of a reductive smooth affine group over a complex vector space are completely reducible (where this notion is weaker than the one that I have defined in this question). This leads me to my second question.
>
> **Question 2.** Are there any "large" family of groups such that all their finite-dimensional (*insert here any "mild" adjectives*) representations on a complex Hilbert space are completely reducible in the sense defined in this question?
>
>
>
| https://mathoverflow.net/users/165204 | When are finite-dimensional representations on Hilbert spaces completely reducible? | I don't know about Q1 off the top of my head, but I think that for Q2 you are unlikely to get a good answer. Of course I may have a different view from you as to what "mild" adjectives are reasonable to impose.
The reason I say this is that in your definition of completely reducible you are requiring the summands of the decomposition of $\pi$ to be orthogonal, yet you are allowing $\pi$ to be non-unitary. Therefore I can always take a reducible unitary representation $\sigma: G \to {\mathcal U}(H)$ and then conjugate it with a non-unitary invertible operator on $H$ which will almost surely mess up the the orthogonality inside the decomposition of $\sigma$.
For a concrete example: let $G=\{\pm 1\}$, fix $\varepsilon>0$, and let $\pi$ be the representation of $G$ on ${\bf C}^2$ which sends $-1$ to $\begin{pmatrix} 1 & \varepsilon \\ 0 & -1 \end{pmatrix}$. Since the eigenvectors of this matrix are not orthogonal to each other, $\pi$ does not decompose into orthogonal summands.
Another example showing that "nice" groups can fail to have the property you seek: fix $\varepsilon>0$ and let $\pi: {\bf Z} \to {\rm GL}\_2({\bf C})$ be the representation which sends $1$ to $\begin{pmatrix} 1 & \varepsilon \\ 0 & 1 \end{pmatrix}$. This representation is reducible but not decomposable; however, I am not sure if you are ruling out such examples with one of your unspecified "mild" adjectives.
By the way, if there are no examples other than the trivial representation, which satisfy the conditions in Q2 then your Q1 will have a positive answer. This is why I think Q1 is actually not such a good question unless you pin down what your intended definition of "completely reducible" should be.
| 7 | https://mathoverflow.net/users/763 | 450761 | 181,314 |
https://mathoverflow.net/questions/450759 | 0 | In a Banach space X, given a norm one bounded linear functional $f$ and $c\in \mathbb{C}\backslash \{0\}$, define $H = \big\{ x\in X \,\vert\, f(x) = c\big\}$ and $\inf H$ = $\inf\_{h\in H} \|h\|$.
1. Is there a necessary condition for $\inf H = \vert\, c\,\vert$?
| https://mathoverflow.net/users/151332 | Infimum of norms of elements in a hyperplane | I am assuming your linear functional means "complex linear", then $\inf H = |c|$ will always be true if $\lVert f \rVert = 1$.
On the one hand, since $\lVert f \rVert = 1$, we can choose a sequence $\{x\_n\}$ of norm $1$ elements in $X$ such that $|f(x\_n)| > 0$ and $|f(x\_n)| \rightarrow \lVert f \rVert = 1$. Replacing $x\_n$ with $\dfrac{|f(x\_n)|}{f(x\_n)} \cdot x\_n$, we can assume that $f(x\_n) \in (0, + \infty)$ and $f(x\_n) \rightarrow 1$. Apparently $\dfrac{c}{f(x\_n)} \cdot x\_n \in H$ and $\left\lVert \dfrac{c}{f(x\_n)} \cdot x\_n \right\rVert = \dfrac{|c|}{|f(x\_n)|} \rightarrow |c|$, so $\inf H \le |c|$.
On the other hand, if $\inf H < |c|$, then there exists a sequence $\{ y\_n \} \subseteq H$, $f(y\_n) = c$ and $\lVert y\_n \rVert \rightarrow \inf H$, which leads to a contradiction that $|c|=|f(y\_n)| \le \lVert f \rVert \lVert y\_n \rVert \rightarrow \inf H < |c|$.
About your second question, is there really any difference between real and complex hyperplanes in a complex Banach space? I am confused about your motivation of asking this question.
This qusetion is more suitable for MSE, though.
| 1 | https://mathoverflow.net/users/166298 | 450765 | 181,317 |
https://mathoverflow.net/questions/450766 | 2 | Let $\mathbb{T}^3=(\mathbb{R}/\mathbb{Z})^3$ be the three-dimensional torus with sides identified. That is, I am considering the unit box $[0,1]^3$ with periodic boundary conditions.
In this case, I am (hopefully) trying to find an explicit formula for the Green function for the Laplacian $-\Delta$.
That is, what would a function $G$ on $\mathbb{T}^3$ satisfying $-\Delta G(x)=\delta^3(x)$ look like explicitly?
I am aware that on whole $\mathbb{R}^3$, $G(x)= \frac{1}{4\pi \lvert x \rvert}$.
According to standard PDE references, I think I need to find the corrector function $\phi^y(x)$ defined by
\begin{equation}
-\Delta\_x \phi^y(x)=0 \text{ for } x \in (0,1)^3 \text{ and } \phi^y(x)=\frac{1}{4\pi \lvert x-y \rvert} \text{ for } x \in \partial \mathbb{T}^3
\end{equation}
together with $y \in \mathbb{T}^3$ and $y \neq x$.
However, I have great difficulty solving the above Poisson equation, and moreover, I am not sure if the resulting $\phi^y(x)$ will even be a periodic function.
Could anyone please help me?
Edit: I forgot the fact that the periodic boundary conditions allow zero modes and the Laplacian does not have an inverse under this condition. So, I am restricting the domain of the Laplacian to $L^2$ space over $\mathbb{T}^3$ with no zero modes. Then will it make a difference?
| https://mathoverflow.net/users/56524 | Any formula or estimates the Green function for the Laplacian in $3D$ periodic box? | Well it depends on what you mean by "explicit". Let $(\varphi\_k)\_k \subset L^2(\mathbb{S}^1)$ be the eigenfunctions of the Laplacian on $\mathbb{S}^1$, these have an explicit form that comes by solving the relative ODE, and let $(\lambda\_k)\_k$ be the relative eigenvalues. Then $G\_{\mathbb{S}^1}(p,q)=\sum\_{k\ge 1} \frac{\varphi\_k(p)\varphi\_k(q)}{\lambda\_k} $ and I highly doubt this has a form more explicit that this, meaning some form that does not involve an infinite sum or other similar operations. Since $\mathbb{T^3}=\mathbb{S}^1\times \mathbb{S}^1\times \mathbb{S}^1 $ you can write the Green function of $\mathbb{T}^3$ as a (triple) sum involving only the eigenfunctions $(\varphi\_k)\_k$ and eigenvalues $(\lambda\_k)\_k$, I don't this there is a form more explicit that this sum.
| 5 | https://mathoverflow.net/users/508380 | 450768 | 181,318 |
https://mathoverflow.net/questions/450706 | 18 | $\DeclareMathOperator\Sha{Sha}$I calculated the Tate–Shafarevich group $\Sha(E/K)[2]$ of the elliptic curve $E:y^2=x^3+17x$ over $K=\Bbb{Q}(\sqrt{-37})$.
I calculated that by hand and I reached the conclusion that $\Sha(E/K)[2]$ has order $1$ or $4$.
Magma code
```
K<b>:=QuadraticField(-37);
A:=EllipticCurve([K!0,0,0,17,0]);
Sel2:=TwoSelmerGroup(A); Sel2;
K<b>:=QuadraticField(-37);
A:=EllipticCurve([K!0,0,0,17,0]);
Tor:=TorsionSubgroup(A); Tor;
E := EllipticCurve([0,0,0,17,0]); // Define your elliptic curve E
Q := QuadraticTwist(E, -37); // Compute the quadratic twist of E by -37
rank := Rank(Q); // Calculate the rank of the quadratic twist rank;
```
outputs
```
Abelian Group isomorphic to Z/2 + Z/2 + Z/2 Defined on 3 generators in
supergroup:
Sel2.1 = $.1 + $.5
Sel2.2 = $.7
Sel2.3 = $.8 Relations:
2*Sel2.1 = 0
2*Sel2.2 = 0
2*Sel2.3 = 0 Abelian Group isomorphic to Z/2 Defined on 1 generator Relations:
2*Tor.1 = 0 \
2
```
Thus Magma reads $\Sha(E/K)[2]\cong ((\Bbb{Z}/2\Bbb{Z})^3/((\Bbb{Z} /2\Bbb{Z})\times (\Bbb{Z}/2\Bbb{Z})^2 )\cong 0$
($E\_{tor}$ is isomorphic to $\Bbb{Z}/2\Bbb{Z}$ and last '2' means rank of twist $E\_{-37}$ over $\Bbb{Q}$ is $2$. $rank(E/\Bbb{Q})=0$, thus $rank(E/\Bbb{Q}(\sqrt{-37})=rank(E/\Bbb{Q})+rank(E\_{-37}/\Bbb{Q})=0+2=2$.
On the other hand, Sage Math's code thanks to Chris Wuthrich's [comment](https://mathoverflow.net/questions/450213/does-17x4y2-1-have-solution-in-bbbq-2-sqrt-5#comment1163693_450213) on [Does $17x^4+y^2=-1$ have solution in $\Bbb{Q}\_2(\sqrt{-5})$?](https://mathoverflow.net/questions/450213/does-17x4y2-1-have-solution-in-bbbq-2-sqrt-5):
```
sage: E = EllipticCurve([17,0])
sage: K.<t> = QuadraticField(-37)
sage: EK = E.base_extend(K)
sage: EK.simon_two_descent(verbose=4)
```
outputs $\Sha(E/K)[2]$ is nontrivial.
Why does this discrepancy occur and which is the correct Tate–Shafarevich group?
| https://mathoverflow.net/users/144623 | Discrepancy in Magma's calculation and Sage's of elliptic curve? | Well spotted. This is a problem. After quite a bit of fiddling I found that the error is in Denis Simon's script used by Sage.
In fact, when executed with higher values of the parameters so that the script finds the rational points, it also prints the correct information.
The command
```
K.<t> = QuadraticField(-37)
E = EllipticCurve([K(17),K(0)])
E.simon_two_descent(lim1=10,lim3=50,limtriv=10, verbose=3)
```
prints correctly
```
[E(K):phi'(E'(K))] >= 2
#S^(phi')(E'/K) = 8
#III(E'/K)[phi'] <= 4
[E'(K):phi(E(K))] = 8
#S^(phi)(E/K) = 8
#III(E/K)[phi] = 1
#III(E/K)[2] <= 4
#E(K)[2] = 2
#E(K)/2E(K) >= 8
2 <= rank <= 4
```
When executed with lower parameter it produces the incorrect information
```
2 <= #III(E/K)[2] <= 8
```
I believe the script actually performs the correct calculations, but the part that draws these conclusions about the Tate-Shafarevich group has a bug in it. With the (correct) information that has been calculated before that line, the correct conclusion would be that $Ш(E/K)[2]$ has between $1$ and $2^3$ elements.
Magma is consistent. The following
```
K<b> := QuadraticField(-37);
A:=EllipticCurve([K!0,0,0,17,0]);
"Torsion::", TorsionSubgroup(A);
"Selmer Group::", TwoSelmerGroup(A);
"Rank::", Rank(A);
"Generators::", Generators(A);
T := A![0,0];
phi := TwoIsogeny(T);
"Phi-Selmer group ::", SelmerGroup(phi);
phihat := DualIsogeny(phi);
"Phi hat =Selmer group ::", SelmerGroup(phihat);
```
returns the information about the descent by isogenies and for the full $2$-Selmer group, rather than only the ones by isogenies.
Let $\varphi \colon E\to E'$ is the isogeny with kernel $\{O,(0,0)\}$. The $\phi$-Selmer group $\operatorname{Sel}^{\phi}(E/K)$ has dimension $3$ and so does the $\hat\phi$-Selmer group $\operatorname{Sel}^{\hat{\phi}}(E'/K)$. The $2$-Selmer group $\operatorname{Sel}^{2}(E/K)$ is also of dimension $3$, which means that the image of $\operatorname{Sel}^{\hat{\phi}}(E'/K) \to Ш(E'/K)/\phi(Ш(E/K))$ has dimension $2$. One finds that $Ш(E'/K)[\hat\varphi]$ is of order $2$, but the Tate-Shafarevich group of $E/K$ has a trivial $2$-part.
This is constistant with the Birch and Swinnerton-Dyer conjecture, I believe.
| 10 | https://mathoverflow.net/users/5015 | 450776 | 181,321 |
https://mathoverflow.net/questions/450619 | 5 | Let $P\_{n}(x)$ the $n-th$ [Legendre polynomial](https://en.wikipedia.org/wiki/Legendre_polynomials). It is well-knonw that $$\int\_{-1}^1 P\_n(x) P\_m(x) P\_h(x) \, dx=2\left(\begin{array}{ccc}
n & m & h\\
0 & 0 & 0
\end{array}\right)^{2}\tag{1}$$ where $\left(\begin{array}{ccc}
n & m & h\\
0 & 0 & 0
\end{array}\right)$ is the [Wigner 3-j symbol](https://en.wikipedia.org/wiki/3-j_symbol). I know that there exist some possible choices for the generalization of the Wigner 3-j symbol to complex numbers but my question is: is it possible to extend othe Wigner 3-j symbol to complex numbers such that type (1) equation holds? In other words, is it possible to find a generalization of the 3-j symbols such that $$\int\_{-1}^1 P\_a(x)P\_b(x)P\_c(x) \, dx = f(a,b,c) \left(\begin{array}{ccc}
a & b & c\\
0 & 0 & 0
\end{array}\right)^{2}$$
for a suitable function $f$ depending on $a,b,c\in\mathbb{C}$ and where $P\_{a}(x)$ is the $C^{2}(-1,1)$ solution of the Legendre differential equation with parameter $a\in\mathbb{C}$, that is,
$$P\_{a}(x)=\,\_{2}F\_{1}\left(a+1,-a;1;\frac{1-x}{2}\right)?$$
| https://mathoverflow.net/users/68301 | Generalized Wigner 3-j symbol and Legendre functions | **Summary:** this is a topic of active research [1,2,3,4]; the answer to the general case is not known; answers exist for $a\in\mathbb{C}$, $b,c,\in\mathbb{N}$, and for $a,b,c\in\mathbb{C}$ with $b=c$.
---
**Case 1:** Consider first the case that one of the three parameters $a,b,c$ is complex, while the other two are still integers. For the Legendre function I use the representation
$$P\_\mu(z)=\, \_2F\_1\left(-\mu,\mu+1;1;\frac{1-z}{2}\right),\;\;|1-z|<2,$$
and for the 3j symbol squared
$$\left(\begin{array}{ccc}
a & b & c\\
0 & 0 & 0
\end{array}\right)^{2}=\frac{\binom{a+b-c}{\frac{1}{2} (a+b-c)} \binom{a-b+c}{\frac{1}{2} (a-b+c)} \binom{-a+b+c}{\frac{1}{2} (-a+b+c)}}{(a+b+c+1) \binom{a+b+c}{\frac{1}{2} (a+b+c)}}.$$
Then
$$\int\_{-1}^1 P\_{a}(x)P\_{b}(x)P\_c(x)\,dx=2\cos ^2\left(\tfrac{1}{2} \pi (a+b+c)\right)\left(\begin{array}{ccc}
a & b & c\\
0 & 0 & 0
\end{array}\right)^{2},$$
$$\text{for}\;\;a\in\mathbb{C},b\in\mathbb{N},c\in\mathbb{N}.$$
(I checked this with Mathematica, which can evaluate the integral in closed form.)
So in this case of one complex variable the function $f(a,b,c)$ in the OP equals $2\cos ^2\left(\tfrac{1}{2} \pi (a+b+c)\right)$. Note that if all three variables are integers, the $\cos^2$ factor can be set to unity, since the 3j symbol equals zero for $a+b+c$ odd.
**Case 2:** If all three parameters are complex, a closed form expression in terms of generalized 3j symbols (or equivalently, generalized Clebsch-Gordan coefficients) exists for the integral
$$T\_{\mu,\nu}=\int\_{-1}^1 P\_{\mu}(x)P\_{\nu}(x)P\_\nu(-x)\,dx,\;\;\mu,\nu\in\mathbb{C}.$$
Note the sign change of one argument of the Legendre function. For integer $\nu=n$ one would simply have $P\_n(-x)=(-1)^nP\_n(x)$, but for noninteger $\nu$ the sign change is essential.
The result, derived by Yajun Zhou [1] (see also [2,3,4]) is given below, it is lengthy. A simple case is $\mu=\nu$, when we have
$$\int\_{-1}^1 P\_{\nu}(x)P\_{\nu}(x)P\_\nu(-x)\,dx=\frac{1}{3} \bigl(2+3 \cos \pi \nu+\cos 2 \pi \nu\bigr)\left(\begin{array}{ccc}
\nu & \nu & \nu\\
0 & 0 & 0
\end{array}\right)^{2},\;\;\nu\in\mathbb{C}.$$
For integer $\nu$ the integral vanishes for $\nu$ odd, and for $\nu$ even the usual formula is recovered.
Here is the general result:
\begin{align}
T\_{\mu,\nu}={}&\frac{2}{\pi^2}\frac{\sin(\mu\pi)\sin(\nu\pi)}{\mu(\mu+1)}\left[ \, \frac{1}{\nu}\,{\_4F\_3}\left(\left.\begin{array}{c}
1,\frac{1-\mu}{2},\frac{\mu+2 }{2},-\nu \\[4pt]
\frac{2-\mu }{2},\frac{\mu+3 }{2},1-\nu \\
\end{array}\right| 1\right) -\frac{1}{\nu+1} \,{ \_4F\_3}\left(\left.\begin{array}{c}
1,\frac{1-\mu}{2},\frac{\mu+2 }{2},\nu+1\ \\[4pt]
\frac{2-\mu }{2},\frac{\mu+3 }{2},\nu+2 \\
\end{array}\right| 1\right) \right].
\end{align}
The combination of two hypergeometric functions ensures the symmetry $T\_{\mu,\nu}=T\_{-\mu-1,\nu}=T\_{\mu,-\nu-1}$, required by $P\_{\mu}(z)=P\_{-\mu-1}(z)$.
---
1. Yajun Zhou, [Legendre
Functions, Spherical Rotations, and Multiple Elliptic Integrals](https://arxiv.org/abs/1301.1735)
2. Marco Cantarini, [A
note on Clebsch-Gordan integral, Fourier-Legendre expansions and
closed form for hypergeometric series](https://arxiv.org/abs/2107.11969)
3. John Campbell, [New Clebsch-Gordan-type integrals involving threefold products of complete elliptic integrals](https://arxiv.org/abs/2302.05819)
4. Yajun Zhou, [On Some Integrals Over the Product of Three Legendre Functions](https://arxiv.org/abs/1304.1606)
| 4 | https://mathoverflow.net/users/11260 | 450782 | 181,322 |
https://mathoverflow.net/questions/450640 | 4 | Let $\Omega$ be a measurable space equipped with a $\sigma$-ideal $\mathcal{N}$ (though of as the "null sets"). Define the compact Hausdorff space
$$ \tilde{\Omega} := \mathrm{Spec}(L^\infty(\Omega)), $$
where $\mathrm{Spec}$ denotes the Gelfand spectrum. In [this question](https://mathoverflow.net/questions/99091/spectrum-of-l-inftyx-mu), Terry Tao mentions the following (universal?) property of $\tilde{\Omega}$: "any [modulo-null-sets-equivalence class of?] Baire-measurable map[s] from $\Omega$ to a compact Hausdorff space $K$ can be uniquely identified with a continuous map from $\tilde{\Omega}$ to K." I haven't been able to find a precise statement or proof in the given references (or elsewhere, where would I have to look precisely?). If $K$ is a compact subset of the real numbers, then this is a consequence of Gelfand duality. How would one proceed in the general case?
**EDIT:** Let's ask a (maybe) simpler question. Assume that $\Omega$ is itself a compact Hausdorff space equipped with a Radon probability measure. If necessary, suppose also that $\Omega$ and $K$ are metrizable. Then, via Gelfand duality, we obtain a continuous map
$f: \tilde{\Omega} \to \Omega $. $\tilde{\Omega}$ also inherits the probability measure from $\Omega$ and $f$ is measure preserving. Now the question becomes: Is $f$ a mod $0$ isomorphism of probability spaces? (If this were the case, then a version of the cited segment would be true.)
| https://mathoverflow.net/users/103549 | Representing measurable map to compact space as a continuous map | I think I have found a precise formulation and proof of the cited statement.
Let's suppose for simplicity that $(\Omega, \Sigma, \mathbb{P})$ is a probability space. Define the set of $K$-valued random variables up to almost sure equality (with $K$ some compact space),
$$ \mathrm{RV}(\Omega, K) := \{X: \Omega \to K \mid X \text{ Baire measurable} \}/(=\_{a.e.}).$$
With $\tilde{\Omega}$ as in the question, the desired statement is:
**Theorem.** Let $K$ be a compact metrizable space. Then the ("Gelfand-duality-induced") map
$$ \Psi\_K: \mathrm{RV}(\Omega, K) \to C(\tilde{\Omega}, K), \;\;\; X \mapsto (\tilde{\Omega}\ni \omega \mapsto \Gamma^{-1}((C\_b(K) \ni f \mapsto \omega(f(X))), $$
is a bijection, where
$$ \Gamma^{-1}: \mathrm{Spec}(C\_b(K)) \to K $$
is the inverse Gelfand transform.
**Proof.** As noted already in the question, the claim follows from Gelfand duality when $K$ is a compact subset of $\mathbb{C}$. Moreover, for any compact space $L$, we have that
$$ \mathrm{RV}(\Omega, L^\mathbb{N}) \to \mathrm{RV}(\Omega, L)^\mathbb{N}, \;\;\; X \mapsto (\pi\_n(X))\_{n\in \mathbb{N}}, $$
is bijective. (For injectivity, we use that we consider a *countable* product $L^\mathbb{N}$, while for surjectivity, we use the Stone-Weierstrass theorem and compactness of $L$.) This yields the claim for the case of $K = L^\mathbb{N}$ with $L\subseteq \mathbb{C}$ (cf. the approach suggested by Christian Remling).
Finally, for the general case note that since $K$ is compact and metrizable, it admits an embedding,
$$ \iota: K \to [0,1]^\mathbb{N}, $$
into the Tychonov cube (Tychonov's embedding theorem), as well as a surjection (epimorphism in the category $\mathsf{CompHaus}$ of compact Hausdorff spaces),
$$ \pi: \{0,1\}^\mathbb{N}\to K,$$
from Cantor space (Hausdorff-Alexandrov theorem). This gives us the following commutative diagram (in $\mathsf{Set}$):
$\require{AMScd}$
\begin{CD}
\mathrm{RV}(\Omega, \{0,1\}^\mathbb{N}) @>>> \mathrm{RV}(\Omega, K) @>>> \mathrm{RV}(\Omega, [0,1]^\mathbb{N}) \\
@V{\Psi\_{\{0,1\}^\mathbb{N}}}VV @V{\Psi\_K}VV @V{\Psi\_{[0,1]^\mathbb{N}}}VV \\
C(\tilde{\Omega}, \{0,1\}^\mathbb{N}) @>{C(\tilde{\Omega}, \pi)}>> C(\tilde{\Omega}, K) @>{C(\tilde{\Omega},\,\iota)}>> C(\tilde{\Omega},\, [0,1]^\mathbb{N})
\end{CD}
Since $\tilde{\Omega}$ is extremely disconnected, it is projective in $\mathsf{CompHaus}$ (Gleason's theorem), so $C(\tilde{\Omega}, \pi)$ is surjective. With this, the square on the left gives us surjectivity of $\Psi\_K$, while the square on the right yields injectivity.
| 0 | https://mathoverflow.net/users/103549 | 450783 | 181,323 |
https://mathoverflow.net/questions/450742 | 4 | Let $p$ be an odd prime. What's the condition on $q$ for
$$
p^{1+2r}\cdot\operatorname{Sp}(2r,p)\leqslant \operatorname{GU}(p^r,q)\;?
$$ I did some computation and seemed that $q\equiv -1$(mod $p$) does give the embedding. I feel that there is some work already done about it, am I right? Or it is an obvious question and I'm being silly. I did check Kleidman and Liebeck's book on maximal subgroups [1], but it just hasn't provided much help. Thank you.
**Reference**
[1] Peter Kleidman, Martin Liebeck, *The subgroup structure of the finite classical groups* London Mathematical Society Lecture Note Series, 129. Cambridge: Cambridge University Press, pp. x+303 (1990), ISBN:0-521-35949-X, [MR1057341](https://mathscinet.ams.org/mathscinet-getitem?mr=1057341), [Zbl 0697.20004](https://zbmath.org/0697.20004).
| https://mathoverflow.net/users/488802 | Condition on $q$ for inclusion $p^{1+2r}\cdot\operatorname{Sp}(2r,p)\leqslant \operatorname{GU}(p^r,q)$ | Too long for a comment. Your conjecture is correct: $q\equiv-1\pmod p$ is a necessary and sufficient condition. This follows directly from the two lemmas below.
It's not a silly question, but it can be answered with some standard tools of representation theory, and a couple of facts about $Sp\_{2r}(p)$: It is perfect and has trivial Schur multiplier. (I'll ignore the case $r=1$, $p=3$, for which $Sp\_{2r}(p)\cong SL\_2(3)$; subgroups of $GU\_3(q)$ are well-known, thanks to H. H. Mitchell.) Let $q$ be a prime power which is relatively prime to the odd prime $p$. Let $r$ be a positive integer.
Consider $p^{1+2r}$, which refers, I assume, to the extraspecial $p$-group of exponent $p$ and cardinality $p^{1+2r}$. Let $q$ be a power of the prime $s$, $s\ne p$.
Lemma 1. $P:=p^{1+2r}$ embeds in $L:=GU\_{p^r}(q)$ (via a monomorphism $\rho$, say) if and only if $q\equiv-1\pmod p$.
Necessity: By the well-known characteristic $0$ representation theory of $P$, and the well-known connection between characteristic $0$ and characteristic $s$ representation theories of $P$ (since $s\ne p$), any faithful $p^r$-dimensional representation $\rho$ of $P$ in characteristic $s$ is absolutely irreducible. Then $\rho(Z(P))$ consists of scalar matrices. The scalar subgroup of $GU\_m(q)$ for any $q$ is cyclic of order $q+1$, so $p=|Z(P)|$ divides $q+1$.
Sufficiency: Suppose that $p$ divides $q+1$. Let $P\_0$ be an elementary abelian subgroup of $P$ of maximal order. Then $|P\_0|=p^{r+1}$ and $Z(P)\le P\_0$. Write $P\_0=Z(P)\times P\_1$ and let $\sigma:P\_0\to GU\_1(q)$ be a representation with kernel $P\_1$. Such a representation exists since $p$ divides $q+1$. Induce $\sigma$ to $P$ to obtain an embedding $\rho:P\to GU(p^r,q)$.
Lemma 2. Let $V$ be a $p^r$-dimensional vector space over $GF(q^2)$ equipped with a nondegenerate hermitian form. (There exists a unique such $V$, up to isometry.) Then any embedding $\rho:P\to GU(V)$ extends to an embedding $\sigma:P:Sp\_{2r}(p)\to GU(V)$.
To prove Lemma 2, first observe that there is an extension to $\sigma\_0:P:Sp\_{2r}(p)\to GL(V)\cong GL\_{p^r}(q^2)$.
To see this, observe that for every $g\in Sp\_{2r}(p)$, the representations $\rho$ and $\rho^g$ are equivalent over $GL\_{p^r}(q^2)$. Here $\rho^g(x)=\rho(g^{-1}xg)$. Since $g$ centralizes $Z(P)$, $\rho^g$ and $\rho$ have the same character, proving the equivalence; and both are absolutely irreducible. By definition of equivalence, there is an $X(g)\in GL(V)$ such that $\rho^g(x)=X(g^{-1})\rho(x)X(g)$ for all $x\in P$. By the absolute irreducibility and Schur's Lemma, $X(g)$ is uniquely determined by this condition, up to multiplication by a scalar. Then (making arbitrary choices for each $X(g)$) $X$ is a projective representation of $Sp\_{2r}(p)$. Now the theory of projective representations, plus the facts that $Sp\_{2r}(p)$ has trivial Schur multiplier, implies that the $X(g)$'s, $g\in Sp\_{2r}(p)$, may be chosen so that $X$ is a genuine representation. Then define $$\sigma\_0(xg)=\rho(x)X(g)$$
for every $x\in P$ and $g\in Sp\_{2r}(p)$. Since $\rho$ and $X$ are representations, and $\rho(g^{-1}xg)=X(g)^{-1}\rho(x)X(g)$, $\sigma\_0$ is a representation of the semidirect product.
Finally, again by Schur's Lemma, $\rho(P)$ preserves a unique hermitian form $H$ on $V$, up to multiplication by a scalar in $GF(q)$. This implies that for each $g\in Sp\_{2r}(p)$, $X(g)$ carries $H$ to $\lambda(g)H$ for some $\lambda(g)\in GF(q)^\times$. Since $\sigma\_0$ is a representation, so is $\lambda$. But $Sp\_{2r}(p)$ is perfect, so $\lambda(g)=1$ for all $g\in Sp\_{2r}(p)$. That is, $\sigma\_0$ preserves $H$, i.e., $\sigma\_0(P:Sp\_{2r}(p))\le GU\_{p^r}(q)$. And $\sigma\_0$ obviously extends $\rho$, as required.
| 7 | https://mathoverflow.net/users/99221 | 450784 | 181,324 |
https://mathoverflow.net/questions/450781 | 4 | In an infinite-dimensional Banah space $(X, \|\cdot\|)$ with a countable Schauder basis $\{x\_n\}$, define:
$$
F\_r: \operatorname{Span}(\{x\_n\}) \rightarrow \operatorname{Span}(\{x\_n\}), \hspace{0.3cm} \sum\_{i\leq N} \lambda\_i x\_i \mapsto \sum\_{i\leq N} \operatorname{Re}(\lambda\_i) x\_i
$$
and:
$$
F\_i: \operatorname{Span}(\{x\_n\}) \rightarrow \operatorname{Span}(\{x\_n\}), \hspace{0.3cm} \sum\_{i\leq N} \lambda\_i x\_i \mapsto \sum\_{i\leq N} \operatorname{Im}(\lambda\_i) x\_i
$$
1. Can $F\_r$ be always continuously extended to the entire $X$?
Next, define another norm $\|\cdot\|\_1$ in $\operatorname{Span}(\{x\_n\})$ as $\|x\|\_1 = \|F\_r(x)\| + \|F\_i(x)\|$ and let $X\_1$ be the completion of $\operatorname{Span}(\{x\_n\})$ with respect to $\|\cdot\|\_1$. Notice that in the case when $F\_r$ could be continuously extended, $F\_i$ could be too and we could find $M>0$ such that $\|x\|\_1 \leq M\|x\|$ for each $x\in X$. Then:
2. In the case when $F\_r$ (or $F\_i$) could be continuously extended, can $\|\cdot\|$ an $\|\cdot\|\_1$ be equivalent?
**Update**: Thanks for losifs' answers, if we define $\|\cdot\|\_r = \|F\_r(x)\|$, then in the completion of the complex span of $\{e\_n\}$ with respect to $\|\cdot\|\_r$, say $(\ell')^2$, we have $\ell^2\subseteq (\ell')^2$ but $\|\cdot\|\_r$ is not equivalent to $\|\cdot\|$.
Similar to how the idea of proving the real version of the Hahn-Banach Theorem is extended to the complex version one, when we deal with problems in $X^{\ast}$, we could always start from a real bounded linear functional. However, I am not sure if, when dealing with problems related to Banach space geometry, it can be assumed what holds in a real Banach space will also hold in a complex one.
| https://mathoverflow.net/users/151332 | The real and the imaginary part of a vector | $\newcommand{\R}{\mathbb R}\newcommand{\ep}{\varepsilon}\newcommand{\C}{\mathbb C}$No, in general the map $F\_r$
\begin{equation\*}
\sum\_1^n w\_j b\_j\mapsto \sum\_1^n \Re(w\_j) b\_j \tag{10}\label{10}
\end{equation\*}
from the span of the $b\_j$'s into itself cannot be continuously extended to the entire $X$, where $(b\_j)$ is a Schauder basis of $X$, $n$ is any natural number, and the $w\_j$'s are any complex numbers.
Indeed, let $X$ be the complex space $\ell^2$ and let $(e\_k)$ be the standard basis of $X=\ell^2$. For any natural $n$, any $c\in\C^n=\C^{n\times1}$ with norm $|c|=1$, and any $\ep\in(0,1)$, let
\begin{equation\*}
Z:=Z^{n,c,\ep}:=(\ep-1)I\_n+cc^\*=:(z\_{jk}=z\_{jk}^{n,c,\ep}\colon(j,k)\in[n]\times[n]),
\end{equation\*}
where $I\_n$ is the $n\times n$ identity matrix, $c^\*$ is the complex conjugate of the $n\times1$ row matrix $c^\top$, and $[n]:=\{1,\dots,n\}$. The eigenvectors of $Z$ are the nonzero multiples of $c$ (with eigenvalue $\ep$) and the nonzero vectors in $\C^n$ orthogonal to $c$ (with eigenvalue $\ep-1$). So, the matrix $Z$ is nonsingular.
For $q=0,1,\dots$, let $K^q:=\{2^q,\dots,2^{q+1}-1\}$ and then let
\begin{equation\*}
b\_j:=\sum\_{k\in K^q}z\_{jk}^q e\_k \tag{20}\label{20}
\end{equation\*}
for $j\in K^q$, where $z\_{jk}^q:=z\_{jk}^{2^q,c^q,\ep^q}$, $\ep^q\in(0,1)$,
$c^q:=x^q+iy^q$, and $x^q$ and $y^q$ are any vectors in $\R^{K^q}$ such that $|x^q|=|y^q|=1/\sqrt2$ and $(y^q)^\top x^q=0$, so that $|c^q|=1$.
Since the matrix $Z^q:=Z^{2^q,c^q,\ep^q}$ is nonsingular for each $q$, we see that $(b\_j)$ is a Schauder basis of $X=\ell^2$.
Next, writing $c^q=(c^q\_k\colon k\in K\_q)$, by \eqref{20} we have
\begin{equation\*}
\sum\_{k\in K\_q}c^q\_k b\_k=\sum\_{k\in K\_q}(Z^qc^q)\_k e\_k=\ep^q\sum\_{k\in K\_q}c^q\_k e\_k
=\ep^q c^q,
\end{equation\*}
so that for all $q$
\begin{equation\*}
\Big|\sum\_{k\in K\_q}c^q\_k b\_k\Big|=\ep^q. \tag{30}\label{30}
\end{equation\*}
On the other hand, for all $q$ we have
\begin{equation\*}
|\Im(Z^q x^q)|=|x^q|^2|y^q|=A:=\frac1{2\sqrt2}>0
\end{equation\*}
and
\begin{equation\*}
\sum\_{k\in K\_q}\Re(c^q\_k) b\_k
= \sum\_{k\in K\_q}x^q\_k b\_k
=\sum\_{k\in K\_q}(Z^qx^q)\_k e\_k=Z^qx^q,
\end{equation\*}
and hence
\begin{equation\*}
\Big|\sum\_{k\in K\_q}\Re(c^q\_k) b\_k\Big|
\ge|\Im(Z^q x^q)|=A>0. \tag{40}\label{40}
\end{equation\*}
Letting now $\ep^q\downarrow0$ as $q\to\infty$, we see from \eqref{30} and \eqref{40} that $\big|\sum\_{k\in K\_q}c^q\_k b\_k\big|\to0$ while $\big|\sum\_{k\in K\_q}\Re(c^q\_k) b\_k\big|
\ge|\Im(Z^q x^q)|=A>0$. So, the map $F\_r$ cannot be continuously extended to the entire $X$. $\quad\Box$
| 4 | https://mathoverflow.net/users/36721 | 450789 | 181,325 |
https://mathoverflow.net/questions/450791 | 12 | Essential undecidability of PA says every complete extension of PA includes a non r.e. set of new "axioms," all undecidable in PA. In that sense lots of sentences of PA are undecidable in PA. And it is trivial to design a measure on, say sentences of PA with Godel number below $n$, where the set of decidable sentences has measure 1 (for all $n$ above the Godel number of the first decidable sentence, which would be 0=0 for the usual Godel numberings). Just take the counting measure while restricting the count to decidable sentences. Similar cheap tricks will give other uninformative results.
But is there a useful, nontrivial way to define the density of decidable sentences in PA?
The answer seems to be the state of the art. It has nothing special to do with PA of course. It holds for any incomplete theory, just using any theorem in place of $0=0$ and any refutable sentence in place of $0=1$. Chaitin's "heuristic principle" uses complexity to suggest that in fact almost all true sentences are independent in any recursive extension of PA, as cited in discussion of [How many of the true sentences are provable?](https://mathoverflow.net/questions/4454/how-many-of-the-true-sentences-are-provable/7902#7902) But people do not seem persuaded by this.
| https://mathoverflow.net/users/38783 | Is there a useful measure of density of decidable sentences in PA? | Asymptotic density seems a very natural measure. The density of a set of sentences is the limit as $n\to\infty$ of the proportion of those sentences of length at most $n$ amongst all sentences of length at most $n$.
(One should use a formalism that has only finitely many sentences of a given length—for example, one can use variable symbols $x$, $x'$, $x''$, and so on, where the prime counts as another symbol, instead of indices; or else just stratify the sentences as a union of finite sets, such as by Gödel codes.)
Having density 1 means that as one considers longer and longer sentences, the proportion of them in the set goes to 100%. Having density 0 means that as one considers longer and longer sentences, the proportion of them in the set goes to 0.
We similarly get a notion of lower density and upper density by using $\liminf$ and $\limsup$.
Using this measure, it turns out that the set of theorems, the set of refutable sentences, and the set of independent sentences (over some fixed theory) all have a lower density bounded away from zero and an upper density bounded away from 1.
For example, every sentence of the form $\varphi\vee( 1=1)$ is a theorem, and this is a positive density set. Every sentence of the form $\varphi\wedge (1=0)$ is refutable, and this is a positive density set. And every sentence of the form $(\varphi\vee (1=1))\wedge\psi$ is independent, where $\psi$ is some fixed independent sentence, and this also is a positive density set.
| 12 | https://mathoverflow.net/users/1946 | 450792 | 181,326 |
https://mathoverflow.net/questions/450804 | 6 | Disclaimer: I'm a relative beginner in this area. I'm trying to prove that if one has a commutative ring $R$ and a prime number $p$, then there is an exact sequence of the form
$$\DeclareMathOperator\THH{THH}
0 \rightarrow \pi\_n(\THH(R)\_{h\mathbb{T}})/p\pi\_n(\THH(R)\_{h\mathbb{T}}) \rightarrow \pi\_n(\THH(R; \mathbb{F}\_p)\_{h\mathbb{T}}) \rightarrow \pi\_{n-1}(\THH(R)\_{h\mathbb{T}})[p] \rightarrow 0.
$$
My basic intuition is that arguing formally should go quite far. It would suffice to show that
$$
\THH(R)\_{h\mathbb{T}} \xrightarrow{\times p} \THH(R)\_{h\mathbb{T}} \rightarrow \THH(R; \mathbb{F}\_p)\_{h\mathbb{T}}
$$
is a fibration, for one could then take the long exact sequence in homotopy groups and then do some manipulations to quickly obtain the result. I'm not really sure how to show that one gets a fibration out of this however, or if one should take an entirely different approach. It does resemble the standard exact sequence one gets in the Bockstein spectral sequence, except $\THH(R)\_{h\mathbb{T}}$ is far from needing to be torsion-free and that wouldn't help give a fibration anyway. Thanks in advance.
| https://mathoverflow.net/users/508592 | An exact sequence involving THH | Let's extract a clear question (about spectra in general) from your question, and then answer it. Let $E$ be any spectrum.
There is the degree $p$ map $p:S\to S$ from the sphere spectrum to itself. Smashing with $E$, this gives a map $E\to E$; we also call this map $p$. It multiplies elements of $\pi\_n(E)$ by $p$. Denote its spectrum cofiber by $E/pE$. The cofibration sequence $E\to E\to E/pE$ is also a fibration sequence, because we are talking about spectra. From the exact sequence
$$
\dots \to \pi\_n(E)\to \pi\_n(E)\to \pi\_n(E/pE)\to \pi\_{n-1}(E)\to \pi\_{n-1}(E)\to \dots
$$
you get what I think you want: an exact sequence
$$
0\to coker(p)\to \pi\_n(E/pE)\to ker(p)\to 0
$$
where the group on the left is
$$
coker (p:\pi\_n(E)\to \pi\_n(E))=\pi\_n\otimes \mathbb F\_p
$$
(it could also be called $\pi\_n(E)/p$) and the group on the right is
$$
ker (p:\pi\_{n-1}(E)\to \pi\_{n-1}(E))=Tor(\pi\_{n-1},\mathbb F\_p)
$$
(I think you are also calling it $\pi\_{n-1}(E)[p]$.)
$\pi\_n(E/pE)$ is called the $n$th mod $p$ homotopy group of $E$.
Curiously, it can have elements of order $4$ if $p=2$.
| 8 | https://mathoverflow.net/users/6666 | 450809 | 181,333 |
https://mathoverflow.net/questions/450530 | 3 | Consider the numbers $$a\_n=\frac{1}{n+1}\sum\_{k=0}^{n}\frac{2^{k-1}\binom{n+1}{k}B\_k}{2^{s+k-1}-1}, \ n\geq0,$$ where $s\neq1;0;-1;-2;-3;...$ is a fixed real number, and the $B\_k$ are the Bernoulli numbers ($B\_1=-1/2$).
I am trying to show that the power series $\sum\_{n=0}^{+\infty} a\_n x^n$ converges. Numerical calculations indicate that its radius of convergence is $1$, but I have been unable to prove it analytically.
| https://mathoverflow.net/users/109569 | Convergence of a power series | Let $s > 1$. Using Faulhaber's formula and some well known properties of the Bernoulli numbers, we have $$a\_n = \frac{1}{n+1} \sum\_{k = 0}^n 2^{-s} \left(\sum\_{j = 0}^{+\infty} 2^{-(s+k-1) j}\right) \binom{n+1}{k} B\_k = \frac{2^{-s}}{n+1} \sum\_{j = 0}^{+\infty} 2^{-(s-1) j} \sum\_{k = 0}^n (2^j)^{-k} \binom{n+1}{k} B\_k = \frac{2^{-s}}{n+1} \sum\_{j = 0}^{+\infty} 2^{-(s-1) j} 2^{-(n+1)j} (B\_{n+1}(2^j)-B\_{n+1}) = 2^{-s} \sum\_{j = 0}^{+\infty} 2^{-(n+s)j} \sum\_{k = 0}^{2^j - 1} k^n.$$ Since $2^{-(n+s)j} k^n$ is non-negative, Fubini's theorem leads to $$a\_n = 2^{-s} \sum\_{k = 0}^{+\infty} k^n \sum\_{j = \lceil \log\_2(k+1)\rceil}^{+\infty} 2^{-(n+s)j} = 2^{-s} \sum\_{k = 0}^{+\infty} \frac{k^n}{2^{(s+n) \lceil \log\_2(k+1)\rceil}} \frac{1}{1-2^{-(s+n)}} = 2^{-s} \sum\_{k = 0}^{+\infty} b\_{k,s}(n).$$ We have the obvious inequality (using $\lceil x \rceil \geq x$) $b\_{k,s}(n) \leq \frac{1}{(k+1)^s (1-2^{-s})} (\*)$. Now, define $i \in \mathbb{N}$ such that $2^i \leq k \leq 2^{i+1}-1$, one has $$b\_{k,s}(n) = \frac{1}{2^{s(i+1)}(1-2^{-(s+n)})} \left(\frac{k}{2^{i+1}}\right)^n \underset{n \to +\infty}{\to} 0.$$ Using $(\*)$, we can apply the DCT which gives $a\_n \underset{n \to +\infty}{\to} 0$. This shows that the radius of convergence of your series is greater than $1$. On the other hand, we have (since $b\_{k,s}(n) \geq 0$) $$\sum\_{k = 0}^{+\infty} b\_{k,s}(n) \geq \sum\_{k = 0}^{+\infty} b\_{2^k-1,s}(n) = \sum\_{k = 0}^{+\infty} \frac{(2^k-1)^n}{1-2^{-(s+n)}} 2^{-(s+n)k} \geq \sum\_{k = 0}^{+\infty} \left(1-2^{-k}\right)^n 2^{-\lceil s \rceil k} = B\_n(\lceil s \rceil)$$. This $B\_n(j)$ is well known as a supertelescoping series (it is analogous to the asymptotic probability of a tie for first place). Indeed, one has for $j \geq 2$ $$(n+1) B\_n(j) \geq \sum\_{k=0}^{+\infty} 2^{-(j-1)(k+1)} \int\_{1-2^{-k}}^{1-2^{-(k+1)}} (n+1)x^{n} dx = \sum\_{k=0}^{+\infty} 2^{-(j-1)(k+1)} ((1-2^{-(k+1)})^{n+1} - (1-2^{-k})^{n+1}) = \sum\_{k=0}^{+\infty} (1-2^{-(j-1)})2^{-(j-1) (k+1)}(1-2^{-(k+1)})^{n+1} = (1-2^{-(j-1)}) B\_{n+1}(j-1).$$ By induction, we get $$B\_n(j) \geq B\_{n+j-1}(1) \prod\_{i = 1}^{j-1}\frac{1-2^{-i}}{n+i}.$$ Since $B\_{n}(1) \geq \sum\_{k=0}^{+\infty} ((1-2^{-(k+1)})^{n+1} - (1-2^{-k})^{n+1}) = 1$, we obtain $a\_n \geq 2^{-s} \prod\_{i = 1}^{\lceil s \rceil -1}\frac{1-2^{-i}}{n+i} \geq \frac{C}{n^{\lceil s \rceil}}$ for some $C > 0$ and thus the radius of convergence of your series is exactly $1$.
| 2 | https://mathoverflow.net/users/508605 | 450816 | 181,334 |
https://mathoverflow.net/questions/450817 | -2 | I want to coin a notion of "strong provability", to be defined as:
$S$ is strongly provable in $T$ if and only if there is a Gödel code of its proof in $T$ that is strictly smaller than any Gödel code of a proof of its negation in $T$
Formally:
$ S \text { is strongly provable in } T \iff \\ \exists x: \operatorname {Proof}\_T(x, \ulcorner S \urcorner ) \land \forall y (\operatorname {Proof}\_T (y, \operatorname {neg}(\ulcorner S \urcorner)) \implies x < y ) $
Now let $T \vdash S$ be the usual metatheoretic statement of $S$ being syntactically proved from $T$
Then is it provable that:
$(T\vdash S) \iff S \text { is strongly provable in } T$
The forward direction is clear, but the opposite is what irks me? Can we have a sentence whose smallest code of its proof (not the metatheoretic) is a non-standard natural, and every proof of its negations is strictly lager than it? I mean in this case it'll be indeterminate syntactically speaking, and the statements would be false, yet is this possible, can we have a theory that spells that and be consistent?
| https://mathoverflow.net/users/95347 | Can we have consistent theories stating opposing provability statements that are non-standardly coded? | This idea in play here is due to Rosser and is the main idea behind the [Gödel-Rosser theorem](https://en.wikipedia.org/wiki/Rosser%27s_trick).
Specifically, Rosser proposes to consider the sentence $\rho$ asserting that for every proof of $\rho$ in PA, there is a smaller proof of $\neg\rho$. In your terminology, $\rho$ asserts its own non-strong-provability. Such a sentence can be constructed by the fixed-point lemma. The conclusion is that if PA is consistent, then $\rho$ is independent of PA. It cannot be provable, since then it would have a proof of some specific length, and PA would prove that some smaller number would be a proof of $\neg\rho$, but by consistency none of those numbers can actually code a proof. And it cannot be refutable, since then $\neg\rho$ would have a proof of some specific length, and so PA would have to prove that one of the smaller numbers is a proof of $\rho$, which again can't happen by consistency.
The forward direction of your biconditional, which you say is "clear", is false if $T$ is inconsistent, since $T$ will prove every $S$, but it will not strongly prove every $S$. Meanwhile, the converse direction is true, since if a statement is strongly provable, it is provable.
Meanwhile, regarding the issues in your final paragraph, it may be interesting to consider the case of the Rosser sentence. In some models of PA, there is a proof of $\rho$ that is smaller than any proof of $\neg\rho$; and in other models of PA, there is a proof of $\neg\rho$ that is smaller than any proof of $\rho$. In both cases, those proofs are nonstandard, since $\rho$ is actually independent of PA.
| 7 | https://mathoverflow.net/users/1946 | 450818 | 181,335 |
https://mathoverflow.net/questions/450810 | 2 | Let $F$ be a p-adic field. A character on $F^\times$ is defined as a *continuous* group homomorphism $F^\times\longrightarrow\mathbb{C}^\times$. But is there any way to construct a non-continuous group homomorphism $F^\times\longrightarrow\mathbb{C}^\times$? Is there even any such homomorphism?
| https://mathoverflow.net/users/32746 | Non-continuous group homomorphism from p-adic field to C* | Since $\mathbf C^\times$ is a divisible group, Zorn’s lemma tells us for every abelian group $G$ and subgroup $H$ that each group homomorphism $H\to \mathbf C^\times$ extends (somehow) to a group homomorphism $G\to\mathbf C^\times$.
Use $G = F^\times$ and $H = \langle u\rangle$, where $u$ is in $\mathcal O\_F^\times$ and not a root of unity, so $H$ is isomorphic to $\mathbf Z$. Start with the homomorphism $H\to \mathbf C^\times$ where $u^k \to 2^k$ and extend to a homomorphism with domain $F^\times$ by Zorn’s lemma. This homomorphism is not continuous since $\mathcal O\_F^\times$ is compact but the homomorphism on this subgroup of $F^\times$ is discontinuous because its image in $\mathbf C^\times$ is unbounded and thus not compact.
| 7 | https://mathoverflow.net/users/3272 | 450822 | 181,336 |
https://mathoverflow.net/questions/450787 | 13 | From [Wikipedia](https://en.wikipedia.org/wiki/Second-order_arithmetic#Projective_determinacy) (I couldn't find the original source):
>
> $\text{ZFC} + \{\text{there are $n$ Woodin cardinals: $n$ is a natural number}\}$ is conservative over $\text{Z}\_2$ with projective determinacy.
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(where projective determinacy is interpreted as a *schema* in the language of second order arithmetic).
How much determinacy do we actually need if we want to get up to ZFC level? Or more precisely, **for which pointclasses $\mathbf{S}$ in the [projective hierarchy](https://en.wikipedia.org/wiki/Projective_hierarchy) is $\mathbf{\text{ZFC} + S\text{-determinacy}}$ conservative over $\mathbf{\text{Z}\_2 + S\text{-determinacy}}$?** (This is equivalent to saying that if a statement in the language of second order arithmetic is also a theorem of ZFC, it is a theorem of $\text{Z}\_2 + S\text{-determinacy}$.)
$S$ is at least borel (since ZFC proves [borel determinacy](https://en.wikipedia.org/wiki/Borel_determinacy_theorem)).
| https://mathoverflow.net/users/65915 | How much determinacy do you need for second order arithmetic to be as strong as ZFC? | Because ZFC proves soundness of $\text{Z}\_2$, no consistent finite extension of $\text{Z}\_2$ proves all second order arithmetic statements that are provable in ZFC (for example, the statement "the conjunction of the new axioms implies their own consistency with $\text{Z}\_2$"). Moreover, for each fixed $n$, $Σ^1\_n$ soundness can be expressed by a single sentence in $\text{Z}\_2$, so infinite extensions with $Σ^1\_n$ sentences (fixed $n$) would not work either. I suspect that $\text{Z}\_2 + \mathbf{Σ^1\_n}$ determinacy (fixed $n$) does not even prove the countable choice schema. It is only by using statements of unbounded quantifier alternation depth that we get the quoted conservation result.
With regard to consistency strength, $\text{Z}\_2 + Σ^1\_1-\text{Det}$ is at a least a strong as ZFC. Going further, over $\text{Z}\_2$, $\mathbf{Σ^1\_1}-\text{Det}$ (note the boldface) is equivalent to every real having a sharp.
I suspect that $\text{Z}\_2 ⊢ Σ^1\_1-\text{Det} ⇔ 0^\#$. Now, $\text{Z}\_2+ 0^\#$ proves $Σ^1\_1-\text{Det}$ and $\operatorname{Con}(\text{ZFC} + ∀α < ω\_1 \, α-\text{Erdős cardinal})$. However, the current proof of $Σ^1\_1-\text{Det} ⇒ 0^\#$ does not work in $\text{Z}\_2$. That proof yields a real $x$ such that every $x$-admissible ordinal is an $L$-cardinal, which makes $ω\_1$ inaccessible in $L$. However, getting $0^\#$ from such an $x$ requires going beyond $\text{Z}\_2$ (and even third order arithmetic), and over $\text{Z}\_2$ such an $x$ is merely equiconsistent with ZFC (see [Harrington's principle over higher order arithmetic](https://arxiv.org/abs/1503.04000) by Cheng and Schindler). For $\mathbf{Σ^1\_1}-\text{Det}$, we get ZFC in $L[r]$ for every real $r$, so we get the sharps.
| 11 | https://mathoverflow.net/users/113213 | 450839 | 181,342 |
https://mathoverflow.net/questions/450838 | 2 | The question is in the title.
Equation $\sum\_{i=1}^n x\_i^3 = 0$ has no non-trivial integer solutions for $n=3$. For $n=4$, there are known descriptions of all integer/rational solutions, see
Choudhry, Ajai. "On equal sums of cubes." The Rocky Mountain journal of mathematics (1998): 1251-1257.
My question is about the next case, $n=5$. Because the equation is homogeneous, integer and rational solutions are related in an obvious way. Some families of solutions are known, see for example [this link](https://dxdy.ru/topic134512.html) and [this one](https://artofproblemsolving.com/community/c3046h1872888_five_dice), but the question is to describe ALL solutions.
| https://mathoverflow.net/users/89064 | Describe all integer/rational solutions to $x^3+y^3+z^3+t^3+s^3=0$ | This is not possible in any meaningful way.
In fact the variety you describe defines a smooth cubic threefold $X$ in $\mathbb{P}^4$. By a famous theorem of Clemens and Griffiths these are not even rational varieties over $\mathbb{C}$. This means that these are no way to parametrise the $\mathbb{C}$-points, let alone the $\mathbb{Q}$-points.
Here are some partial attempts which fail.
Firstly this variety $X$ is unirational over $\mathbb{Q}$ which means that it admits a dominant rational map $\mathbb{P}^3 \dashrightarrow X$ over $\mathbb{Q}$. This gives a Zariski dense set of rational points on $X$. But it is impossible to write all rational points this way, even allowing finitely many maps. This follows from the fact that $X(\mathbb{Q})$ is not thin (see Theorem 3.1 of [1]).
Next one expects that $X$ satisfies weak approximation. This means that $X(\mathbb{Q})$ should be dense in $X(\mathbb{R})$ and $X(\mathbb{Q}\_p)$ for all primes $p$. So one can ``describe'' rational points by stipulating that you have rational points close to some given collection of real and $p$-adic points. But this is a partial solution, and in any case no one is able to prove that weak approximation holds with current tools.
This is nothing special about your variety $X$. All these properties are expected to hold for any rationally connected variety with a rational point which is not a rational variety. (In general one needs to replace weak approximation by weak weak approximation.)
[1] Demeio, Julian Lawrence. Elliptic fibrations and the Hilbert property. Int. Math. Res. Not. IMRN 2021, no. 13, 10260--10277.
| 19 | https://mathoverflow.net/users/5101 | 450843 | 181,344 |
https://mathoverflow.net/questions/450850 | 1 | O. Baues and F. Gruewald in their paper "AUTOMORPHISM GROUPS OF POLYCYCLIC-BY-FINITE GROUPS AND ARITHMETIC GROUPS" have stated that a polycyclic-by-finite group has a unique maximal finite normal subgroup (see page 216, Section 1.2). I was wondering if somone could give a proof for that? Does this result still hold for finitely presented amenable groups, in particular elementary amenable ones?
**My try**: Let $G$ be a virtually polycyclic group. Since a virtually polycyclic group is finitely presented, it is countbale. So assume that $N\_1 ,N\_2 ,\ldots $ are finite normal subgroups of $G$. For each inetegr $n\geq 1$, $N\_1 N\_2 \cdots N\_n$ is a finite subgroup. On other hand, a given virtually polycyclic group has only finitely many finite subgroups up to isomorphism. Hence the ranks of finite subgroups are bounded. Thus there are only at most $n\_G$, where $n\_G$ is an integer depending only on $G$, possible different finite subgroups $N\_i$ in $G$.
| https://mathoverflow.net/users/114476 | Does a finitely presented amenable group contain a unique maximal finite normal subgroup? | In a group $G$, the polyfinite radical $W(G)$ is the union of all finite normal subgroups. So, $G$ has a maximal finite normal subgroup iff $W(G)$ is finite.
You're asking whether $W(G)$ is finite for every finitely presented elementary amenable group. The answer is negative. There are indeed finitely presented solvable groups in which the center has infinite torsion (the torsion part of the center is clearly contained in the polyfinite radical, which is actually the torsion part of the FC-center).
I'm not sure what are the simplest known examples, but here are ones: for a prime $p$, the group of matrices
$$\begin{pmatrix}1 & a\_{12} & a\_{13} & a\_{14}\\ 0 & t^{n\_1}(t+1)^{n\_2} & a\_{23} & a\_{24} \\ 0 & 0 & t^{n\_3}(t+1)^{n\_4} & a\_{34}\\ 0 & 0 & 0 & 1\end{pmatrix}: a\_{ij}\in\mathbf{F}\_p[t,(t+t^2)^{-1}],n\_k\in\mathbf{Z}.$$
| 4 | https://mathoverflow.net/users/14094 | 450852 | 181,348 |
https://mathoverflow.net/questions/450842 | 3 | The following comes from Definition 2 in Pavel Pudlak, "A new proof of the congruence lattice representation theorem," *Algebra Universalis* **6** (1976), 269-275.
Let $X$ be a set. Let $F$ be a family of functions from $X$ to itself containing the identity map and closed under composition.
Define a binary relation on the family of one- or two-element subsets of $X$ as follows. Let $a,b,c,d\in X$. We will say that $\{a,b\}$ *dominates* $\{c,d\}$ if there are $n\in\mathbb N\_0$, $u\_0,\dots,u\_n\in X$, and $f\_1,\dots,f\_n\in F$ such that $u\_0=c$, $u\_n=d$, and $\{f\_i(a),f\_i(b)\}=\{u\_{i-1},u\_i\}$ for $i=1,\dots,n$.
Why is domination transitive?
| https://mathoverflow.net/users/51389 | Why Is Pudlak's relation on the family of one- or two-element subsets of a set transitive? | I don’t see what the problem is supposed to be; you just compose the functions witnessing the two domination relations in the obvious way:
Fix $c=u\_0,\dots,u\_n=d$ and $f\_1,\dots,f\_n\in F$ such that $\{f\_i(a),f\_i(b)\}=\{u\_{i-1},u\_i\}$ for each $i=1,\dots,n$.
Fix $e=v\_0,\dots,v\_m=f$ and $g\_1,\dots,g\_m\in F$ such that $\{g\_j(c),g\_j(d)\}=\{v\_{j-1},v\_j\}$ for each $j=1,\dots,m$.
Then the fact that $\{a,b\}$ dominates $\{e,f\}$ is witnessed by the sequence
$$e=v\_0=w\_{1,0},w\_{1,1},\dots,w\_{1,n}=v\_2=w\_{2,0},w\_{2,1},\dots,w\_{m,n}=v\_m=f,$$
where
$$w\_{j,i}=\begin{cases}g\_j(u\_i)&\text{if }g\_j(c)=v\_{j-1}\text{ and }g\_j(d)=v\_j,\\
g\_j(u\_{n-i})&\text{otherwise.}\end{cases}$$
We have $\{w\_{j,i-1},w\_{j,i}\}=\{h\_{j,i}(a),h\_{j,i}(b)\}$, where
$$h\_{j,i}=\begin{cases}g\_j\circ f\_i&\text{if }g\_j(c)=v\_{j-1}\text{ and }g\_j(d)=v\_j,\\
g\_j\circ f\_{n+1-i}&\text{otherwise}\end{cases}$$
is in $F$ as it is closed under composition.
| 3 | https://mathoverflow.net/users/12705 | 450863 | 181,350 |
https://mathoverflow.net/questions/450796 | 1 | Let $X\_{t}=\sum\_{i=1}^n(1+s\cdot w)\sin(t\_i)$ where $t\in T=[-\pi/2,\pi/2]^n/\{\vec 0\}$, $w\sim\mathbb{N}(0,1)$, $s$ is a scalar denoting the strength of Gaussian noise. How to find the condition on $s$ such that $X\_t$ is strictly positive with high probability? i.e. when $n\rightarrow\infty$,
$$P(\inf X\_t>0)\rightarrow 1.$$
| https://mathoverflow.net/users/494410 | the infimum of a random process | We have $X\_t=Y\sum\_{i=1}^n\sin(t\_i)$, where $Y:=1+sw$,
$t=(t-1,\dots,t\_n)\in T^n:=[-\pi/2,\pi/2]^n/\{\vec 0\}$, $w\sim N(0,1)$, and $s$ is a real number.
So,
$$\inf\_{t\in T^n}X\_t=\min\_{t\in T^n}X\_t=-n|Y|;$$
this minimum is attained at $t=(-\frac\pi2,\dots,-\frac\pi2)$ if $Y\ge0$ and at $t=(\frac\pi2,\dots,\frac\pi2)$ if $Y<0$.
If $s=0$, then $Y=1$. If $s\ne0$, then $Y$ is normal random variable. So, in any case, $P(Y\ne0)=1$.
So, for any real $s$,
$$P(\inf\_{t\in T^n}X\_t>0)=0\not\to1.$$
| 0 | https://mathoverflow.net/users/36721 | 450867 | 181,351 |
https://mathoverflow.net/questions/450860 | 1 | Let $S$ be a finite semigroup and $K$ a field of characteristic 0 (we can assume the complex numbers for simplicity).
>
> Question: Is there a characterization when the center of the semigroup algebra $KS$ is one-dimensional and a basis is given by the idendity of $KS$ (meaning $KS$ has an identity)?
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If it helps, we can assume that $S$ is a band, meaning that $x^2=x$ for all $x \in S$.
Having trivial center seems to be rare but there are still many examples.
Here the number of semigroups with trivial center with $n \geq 2$ elements for small $n$:
0,5,9,251,1478.
Here the number of bands with trivial center with $n \geq 2$ elements for small $n$:
0,3,4,49,142.
| https://mathoverflow.net/users/61949 | Semigroup algebras with one dimensional center | Let me answer what goes on with bands. I'll assume the band is a monoid for simplicity. I think everything works so long as the band is unital but this can get messier to describe. Details of what I'm saying can be found in [Margolis, S., & Steinberg, B. (2012). Quivers of monoids with basic algebras. Compositio Mathematica, 148(5), 1516-1560.](https://www.cambridge.org/core/journals/compositio-mathematica/article/quivers-of-monoids-with-basic-algebras/39F94B0FF1357609C65687DBA4D307D4) and some of our later papers.
The algebra of a band is split basic over every field so it is isomorphic to the algebra of a bound quiver.
A left regular band is a semigroup satisfying the identities $x^2=x, xyx=xy$. These identities say that repetition is useless. For monoids, the first identity is a consequence of the second. Left regular bands are precisely the bands $S$ for which $aS=bS$ implies $a=b$ (so left divisibility is an order). Therefore, a left regular band is ordered by $a\leq b\iff aS\subseteq bS\iff ba=a$.
If $M$ is a left regular band, then the principal left ideals of $M$ are closed under intersection and hence form a lattice $\Lambda(M)$. The vertices of the quiver of $M$ (if you work with left modules) is $\Lambda(M)$. If $Ma,Mb$ are two principal left ideals, then there are no arrows from $Ma$ to $Mb$ unless $Ma\subsetneq Mb$. If $Ma\subsetneq Mb$, then the number of edges from $Ma$ to $Mb$ is one less than the number of connected components of the restriction of the Hasse diagram of $M$ to those elements $m\in M$ with $Ma\subseteq Mm$ and $m<b$. So as you can see this is already a bit messy to say when you get a connected quiver.
A right regular band is defined dually: $x^2=x$ and $xyx=yx$ for all $x,y$. The quiver of a right regular band is computed dually. The principal right ideals form the vertices, there are no arrows $aM$ to $bM$ unless $bM\supsetneq aM$, in which case you count the number of connected components minus one of the Hasse diagram for the order $a\leq b$ if $ab=a$ of the set of $m$ with $m<b$ and $aM\subseteq mM$.
A regular band is a band satisfying $xyxzx=xyzx$. In other words $a\mapsto xax$ is a retraction from $M$ to $xMx$. It turns out that if $M$ is a band and $K$ is a field, then the map $M\to KM\to KM/J(KM)^2$ (where $J(KM)$ is radical) factors through a regular band $M'$ (the universal regular band image of $M$). Hence the quiver of $KM$ and $KM'$ are the same, and so one is reduced to the case of regular bands.
Every regular band $M$ embeds in a direct product $M\_l\times M\_r$ where $M\_l$ is a left regular band that has lattice of principal left ideals isomorphic to the lattice of principal two-sided ideals of $M$ and $M\_r$ is a right regular band that has lattice of principal right ideals isomorphic to the lattice of principal two-sided ideals of $M$. The quiver of $M$ is obtained by superimposing the quivers of $M\_l$ and $M\_r$ on each other (identifying their vertices via the bijections above). So the vertices of the quiver can be identified with the two-sided principal ideals of $M$. Arrows going up in the order are computed via the left regular band image $M\_l$ and arrows going down in the order are computed via the right regular band image $M\_r$. In particular, the quiver need not be acyclic.
How you would every turn this into a simple characterization of connectedness of the quiver is beyond me.
Here is a good example. Let $S$ be the semigroup $\{1,2\}\times \{1,2\}$ with multiplication $(i,j)(k,l)=(i,l)$. Let $M=S\cup I$ where $I$ is an identity. Then the quiver of $M$ has two vertices $v,w$ and edges $x\colon v\to w$ and $y\colon w\to v$. There is a single relation $yx=0$ (where I compose paths from right to left because I do left modules).
**Added.** In a paper with Margolis and Saliola, we showed the path algebra of an acyclic quiver is always the monoid algebra of a left regular band.
| 1 | https://mathoverflow.net/users/15934 | 450871 | 181,354 |
https://mathoverflow.net/questions/450788 | 3 | Let $f(x)\in\mathbb{Z}[x]$ be a polynomial of degree $d$ and naive height (maximum of the absolute values of the coefficients) at most $H$. Is there anything known about the number of prime factors of the discriminant of $f$? Particularly upper bounds?
[This](https://mathoverflow.net/questions/11880/what-primes-divide-the-discriminant-of-a-polynomial) question is similar, but more restrictive, in that it asks about palindromic polynomials and asks for the prime factors explicitly.
| https://mathoverflow.net/users/477216 | Number of prime factors of a polynomial discriminant | The discriminant is $O(H^{2(d-1)})$ so its number of prime factors is bounded by $$ (1 + o(1)) \frac{ \log (H^{2(d-1)})}{\log \log (H^{2(d-1)})} = (2(d-1) + o(1)) \frac{ \log H}{\log \log H}$$ by the asymptotics for primorials (a consequence of the prime number theorem).
This is not too far from optimal (in the worst case) since the for any set of primes whose product is less than $H$, we can choose a polynomial of height $\leq H$ whose discriminant is divisible for all those primes by the Chinese remainder theorem, and such a set of primes can have size $(1+o(1))\frac{\log H}{\log \log H}$ by the same primorial asymptotics, so the upper bound is sharp (in the worst case) to within a factor of $2(d-1)$.
However, based on your comments, you may be interested in the average case also. For each $p<H$, the fraction of polynomials for which $p$ divides the discriminant is approximately $\frac{1}{p}$, so the average number of prime factors of the discriminant less than $H$ is approximately $\sum\_{p<H} \frac{1}{p} \approx \log \log H$, and the maximum number of prime factors of size $\geq H$ is $2(d-1)$ by the $O(H^{2(d-1)})$ bound for the discriminant, so the average number of prime factors is $\log \log H + O\_d(1)$.
| 3 | https://mathoverflow.net/users/18060 | 450874 | 181,355 |
https://mathoverflow.net/questions/450858 | 8 |
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> **Question:**
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> what is the geometric interpretation of the subject of machine learning and/or deep learning?
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Being "forced" to have a closer look at the subject, I have the impression that it all boils down to approximately reconstruct the characteristic function that decides about set-membership of the points of the feature space.
To me the essential "innovation" seems to be that a function must calculated on basis of the signum of its values for a certain sample of coordinates, that resemble the "training" set.
Say we are given finite sets $C\subset\mathcal{C}$ and $D\subset\bar{\mathcal{C}}$ my impression is that the goal is to find a function, whose zero-set partitions the feature space so that the signum of the function values complies with characteristic function of $\mathcal{C}$ with high probability; is that impression right?
\*\*Addendum:\*\*
my final "working assumption" about what Machine Learning, resp. Deep Learning is actually about, is to play a sophisticated version of [battleship](https://en.wikipedia.org/wiki/Battleship\_(game)) in disguise.
The analogues of the ship geometries are more complicated shapes also in higher dimensions and spaces with non-euclidean distance metrics.
The whole subject of research is to devise methods to sink the opponent's ships faster.
| https://mathoverflow.net/users/31310 | Geometric formulation of the subject of machine learning | There are several geometric aspects of machine learning.
1. You can think of the goal of ML as function approximation. Your question mentions reconstructing a characteristic function, but people often look at approximating other types of functions as well ("regression" as opposed to "classification"). The geometry here comes from defining what approximation means, i.e., the "loss function."
2. Many models (i.e., members of a parameterized family of functions) themselves have a natural kind of geometry: that is, you can look at distances between models based on their parameters, rather than the functions they compute. A Riemannian metric on parameters is useful in defining gradients for the learning process.
3. In deep learning, one often looks at representations of data in a geometric way. A classic example is word embeddings, which associate words with points in a high-dimensional space, so that geometric operations (say, addition and subtraction) correspond to meaningful linguistic operations (say, analogical reasoning).
Taking a geometric point of view on ML is incredibly powerful! That said, it's also just one part of the subject, and thinking of ML only in terms of pure geometry is like thinking of probability only in terms of measure theory: not technically wrong, but not always the most helpful viewpoint.
| 10 | https://mathoverflow.net/users/1227 | 450877 | 181,357 |
https://mathoverflow.net/questions/449817 | 1 | Let $A,B,C$ be Banach spaces and $m\,:\,A\times B\to C$ be a bilinear map such that $\|m(a,b)\|\leq \textrm{const}\,\|a\|\|b\|$. We denote by $\mathcal{S}(\mathbb{R}^d)$ be the standard space of Schwartz test functions and we denote by $\mathcal{S}(\mathbb{R}^d,B)$ the space of Schwartz test functions valued in $B$.
Let $F$ be a linear and continuous functional $\mathcal{S}(\mathbb{R}^d)\to A$. For $g\in \mathcal{S}(\mathbb{R}^d,B)$ of the form $g=\sum\_{i=1}^N f\_i b\_i$, where $f\_i\in\mathcal{S}(\mathbb{R}^d)$, $b\_i\in B$ we define
$$
\langle F,g\rangle\_m := \sum\_{i=1}^N m(\langle F,f\_i\rangle,b\_i)\in C.
$$
Does the above bracket extend uniquely to all $g\in \mathcal{S}(\mathbb{R}^d,B)$?
| https://mathoverflow.net/users/47256 | Banach space valued distributions and test functions | Yes.
The tensor product $\mathcal S(\mathbb R^d)\otimes B$ can be identified with a subspace of $\mathcal S(\mathbb R^d,B)$ by denoting by $g\otimes b$ the function $x\mapsto g(x)b$ for all $g\in\mathcal S(\mathbb R^d)$ and $b\in B$ and this is the space you're first defining $g\mapsto\langle F,g\rangle\_m$ on.
Then, since $\mathcal S(\mathbb R^d,B)$ is the completion of $\mathcal S(\mathbb R^d)\otimes B$ for the projective tensor product topology, your question reduces to showing that the bilinear mapping
$$\begin{array}{ccc}\mathcal S(\mathbb R^d)\times B&\longrightarrow&C\\ (g,b)&\longmapsto&m(\langle F,g\rangle,b)\end{array}$$
is continuous. But this is clear since both $F$ and $m$ are continuous.
| 2 | https://mathoverflow.net/users/508539 | 450881 | 181,359 |
https://mathoverflow.net/questions/450873 | 1 | Let $X\_{t}=\sum\_{i=1}^n(1+s\cdot w\_i)t\_i\sin(t\_i)$ where $t\in T=[-\pi/2,\pi/2]^n/\{\vec 0\}$, $w\_i$ are iid standard gaussian variables, $s$ is a scalar denoting the strength of Gaussian noise.
How to find the condition on $s$ such that $X\_t$ is strictly positive with high probability? i.e. when $n\rightarrow\infty$,
$$P(\inf X\_t>0)\rightarrow 1.$$
**What I tried:**
I tried to prove $P(X\_t>0)\rightarrow 1$ for every $t\in T$. This can be computed exactly by using the symmetric property of gaussian and gaussian tail bound. However this gives a unsatisfactory bound on $s$. Thus I am seeking for 'epsilon-net' type idea as follows.
I've already found covering consitituted by uniformly sampled points in the region $T$ as the centers of the balls with $\epsilon$ radius. Also, on the random centers, I prove that $P(X\_t>0)\rightarrow 1$ where $t\in N$ and $s<\sqrt{n}$. This nice bound in $s$ gives me a hope that it also apply to the whole region $T$ instead of only sampled points. Thus I came up with this epsilon net argument to deal with the points that are not on the net
\begin{equation}
\begin{aligned}
P(\inf\_{t\in T} X\_t>0)&=1-P(\inf\_{t\in T} X\_t<0)\\&=1-P(\inf\_{t\in T} (X\_t-X\_{\pi(t)}+X\_{\pi(t)})<0)\\&\geq 1-P(\inf\_{t\in T} (X\_t-X\_{\pi(t)})<0)+P(\inf X\_{\pi(t)}<0)\\
\end{aligned}
\end{equation}
The second probability term $P(\inf X\_{\pi(t)}<0)$ can be lower bounded by union bound.
However, I was stuck on bounding the first probability term.
Normally what we did is $P(\sup\_{t\in T} X\_t-X\_{\pi(t)}>0)\leq P(\sup\_{t\in T}|X\_t-X\_{\pi(t)}|>0)$ and assuming the process is Lipschitz, i.e. $|X\_t-X\_{\pi(t)}|\leq C|t-\pi(t)|$ where $C$ is a random variable, then the probability can be further bounded by $P(C\epsilon>0)$.
Following the same procedure here, assuming we have $|X\_t-X\_{\pi(t)}|<C|t-\pi(t)|$. Then we have the lower bound of the first term $$P(\inf\_{t\in T} (X\_t-X\_{\pi(t)})<0)\leq P(\inf\_{t\in T} C|t-\pi(t)|>0).$$
Clearly, this bound is $1$. Then we would have no chance to prove the first probability goes to $0$ when $n\rightarrow \infty$. I think the issue is that we lost too much when doing $X\_t-X\_{\pi(t)}<C|t-\pi(t)|$.
Following [this book](https://web.math.princeton.edu/%7Ervan/APC550.pdf), I also tried to look into chaining (dealing with probabilistic Lipschitz bound) and slicing (dealing with $E(X\_t)\neq 0$ case). But I think the issue can not be resolved by applying these more technical methods.
This problem looks quite the same as what epsilon net method, chaining trying to prove. But it seems that they do not apply and I was not able to see what exactly the issue is. Is it because I am actually asking a totally different question than the common one, i.e. $P(\sup X\_t>x)<e^{-x^2}$ where $x>0$?
| https://mathoverflow.net/users/494410 | Lower bounding the infimum of a random process | This will not hold under any condition on $s$.
Indeed, by the continuity of $X\_t$ in $t$,
$$\inf\_{t\in T} X\_t=\inf\_{t\in[-\pi/2,\pi/2]^n} X\_t
\le X\_{\vec 0}=0.$$
So, for any real $s$,
$$P(\inf X\_t>0)=0\not\to1.$$
| 0 | https://mathoverflow.net/users/36721 | 450883 | 181,360 |
https://mathoverflow.net/questions/450880 | 0 | Assume that $u~ \colon \mathbb{R}\_+ \to [-M,M]$ is a bounded continuously differentiable function such that $u(0) = 0$ and $$u(t) \leq \int\_0^t \lambda(s)~u(s)~\mathrm{d}s + C \label{1}\tag{1}$$ where $\lambda ~ \colon \mathbb{R}\_+ \to [-\infty,-\alpha]$ is a strictly negative function with $\lambda(t) \leq -\alpha < 0$ for all $t \geq 0$, and $C > 0$ is a fixed constant. Moreover, suppose that $-u$ satisfies the same inequality \eqref{1} as $u$, i.e., $$-u(t) \leq -\int\_0^t \lambda(s)~u(s)~\mathrm{d}s + C \label{2}\tag{2}.$$ My question is, is it possible to deduce from \eqref{1} and \eqref{2} that $$\lvert u(t)\rvert \leq C~\mathrm{e}^{-\alpha t}~?$$ If not, is it possible to deduce (at least) that $u(t) \to 0$ as $t \to \infty~?$
| https://mathoverflow.net/users/163454 | On the validity of a certain Grönwall-type inequality | No. A counterexample is $C=3$, $\lambda(t)=-1$, and $u(t)=\sin t$ for all $t$.
| 2 | https://mathoverflow.net/users/36721 | 450884 | 181,361 |
https://mathoverflow.net/questions/450888 | 3 | Let $K$ be a field. We consider the polynomial $f(x) = x^4 + 1$. It is known that $f(x)$ is irreducible over $\mathbb{Q}$ but reducible over any finite field. Thus $ f(x)$ is reducible over any field $ K $ having positive characteristic.
I am interested in determining for which fields $f(x)$ remains irreducible, based on the case $\text{char}(K) = 0$.
| https://mathoverflow.net/users/215016 | irreducibility of the polynomial $ x^4 +1 $ | Among $-1$, $2$, and $-2$, there are $0$, $1$ or $3$ squares. This determines the irreducible factorization of $X^4+1$.
1. If none is a square, it is irreducible;
2. If only $-1=i^2$ is a square, the irreducible factorization is $(X^2-i)(X^2+i)$;
3. If only $-2=u^2$ is a square, the irreducible factorization is $(X^2+uX-1)(X^2-uX-1)$;
4. If only $2=v^2$ is a square, the irreducible factorization is $(X^2+vX+1)(X^2-vX+1)$;
5. If all are squares and char is $\neq 2$, the polynomial is split, with simple roots $i^k(1+i)/v$, $k=0,1,2,3$;
6. If the char is 2 (so all are squares), the polynomial is $(X-1)^4$.
---
Added: here's a proof. The char. 2 case (Case 6) being trivial, I assume $2\neq 0$. First, if there is a root $\theta$ in the field, then the polynomial splits with distinct roots $\theta^k$, $k=\pm 1,\pm 3$ (with $\theta^{-3}=-\theta$). In particular, if the polynomial is not irreducible, it is product of two degree 2 polynomials.
If $\theta$ is a root (possibly in a larger field) and the polynomial is not irreducible, the polynomial therefore splits with one of the following factors: $A=(X-\theta)(X-\theta^{-1})$, $B=(X-\theta)(X-\theta^3)$, $C=(X-\theta)(X+\theta)$. We observe that $A=X^2+vX+1$ with $v^2=2$, $B=X^2+uX-1$, $C=X^2-i$ with $v^2=2$, $u^2=-2$, $i^2=-1$. Hence if $2$ resp $1$ resp $-1$ is a square then $A$ resp $B$ resp $C$ yields a factorization in the ground field. If all are squares, $(1+i)/u$ is a root and the polynomial is totally split.
Conversely if $-2$ has no square root then $A$ is not a polynomial in the ground field and hence the polynomial is not totally split, etc, and if none of the three is a square, none of the possible degree two factors occurs and hence the polynomial is irreducible. Thus all of 1, 2, 3, 4, 5 are proved.
NB in char. $3$, we have $2=-1$ but this is no contradiction: just Cases 2 and 4 are excluded (and 1 is excluded as well). But actually in char $p>0$, for given $p$, at most one of 2, 3, 4 is possible and 1 is excluded, so char. $3$ is not really special.
| 12 | https://mathoverflow.net/users/14094 | 450900 | 181,366 |
https://mathoverflow.net/questions/450890 | 16 | I asked this on Math Stack Exchange, but it didn't get a single answer. So, I am now asking it here. Let our signature be that of a single binary operation $+$. I define the constant identity to be $x+y=z+w$. The commutative identity is, of course, the well-known identity $x+y=y+x$. I wonder, is there an identity strictly between those two, meaning, is there a single identity $E$ such that the constant identity implies $E$, but not conversely, and $E$ implies the commutative identity, but not conversely?
| https://mathoverflow.net/users/43439 | Is there an identity between the commutative identity and the constant identity? | Yes:
$$(x + x) + y = y + x$$
The constant identity implies this because both sides are $+$es. This does not imply the constant identity because it is true about any set with an operation that is commutative, associative, and idempotent (meaning $x + x = x$ for all $x$), the smallest nontrivial example is $\{x,y\}$ where $x + x = x + y = y + x = x$ and $y + y = y$.
This implies commutativity. $(x + x) + (x + x) = (x + x) + x = x + x$, so $x + x$ is idempotent. $(x+x)+y=((x+x)+(x+x))+y=y+(x+x)$, so $x + x$ commutes with everything. $(x + x) + (y + y) = (y + y) + x = x + y$, and $(x + x) + (y + y) = (y + y) + (x + x)$ because $x+x$ commutes with everything, so $x + y = y + x$, so $+$ is commutative.
This is not implied by commutativity, because (for example) addition of natural numbers is commutative but does not satisfy this identity.
| 25 | https://mathoverflow.net/users/508667 | 450905 | 181,369 |
https://mathoverflow.net/questions/450799 | 6 | I need to prove the following statement.
Let $ n, g, m, a ,t$ be integers. Prove that the following statement is true for all $ n \geq g(1+2m)+1 $, $ g\geq 2t $, $ m\geq t $, $ 0\leq a <t $, and $ t>0 $:
$$\sum\_{l=0}^{m}(-1)^l\frac{\binom{m}{l}}{\binom{n/g-l}{m+1}}\left[\frac{\binom{n-2t}{gl-a}}{\binom{n}{gl}}-\frac{\binom{n-2t}{gl-2t+a}}{\binom{n}{gl}}\right]=0.$$
It works for multiple different values of the parameters. I tried induction or some combinatorial identities but could not come with a rigorous proof. Any idea might be helpful.
| https://mathoverflow.net/users/508578 | A summation involving fraction of binomial coefficients | First let's notice that
\begin{split}
\frac{\binom{m}l}{\binom{n/g-l}{m+1}} &= \binom{n/g}l\binom{n/g-l-m-1}{m-l} \frac{m!(m+1)!\Gamma(n/g-2m)}{\Gamma(n/g+1)} \\
&=(-1)^{m-l}\binom{n/g}l\binom{2m-n/g}{m-l} \frac{m!(m+1)!\Gamma(n/g-2m)}{\Gamma(n/g+1)}
\end{split}
and similarly
$$\frac{\binom{n-2t}{gl-a}}{\binom{n}{gl}} = \binom{gl}{a}\binom{n-gl}{2t-a} \frac{a!(2t-a)!(n-2t)!}{n!}.$$
Hence, the identity in question is equivalent to
$$\sum\_{l=0}^m \binom{n/g}l\binom{2m-n/g}{m-l} \binom{gl}{a}\binom{n-gl}{2t-a}$$
being invariant under replacement of $a$ with $2t-a$.
Multiplying the last expression by $x^a$ and summing over $a=0..2t$, we get the coefficient of $y^{2t}$ in
$$\sum\_{l=0}^m \binom{n/g}l\binom{2m-n/g}{m-l} (1+xy)^{gl} (1+y)^{n-gl},$$
and doing same after replacement of $a$ with $2t-a$, we get the coefficient of $y^{2t}$ in
$$\sum\_{l=0}^m \binom{n/g}l\binom{2m-n/g}{m-l} (1+y)^{gl} (1+xy)^{n-gl}.$$
Since $a$ and $t\leq m$ are arbitrary, we need to prove that the last two expressions as series in $x,y$ are equal modulo $y^{2m+1}$. Equivalently, we need to show that $F(x,y)\equiv G(x,y)\pmod{y^{2m+1}}$, where:
\begin{split}
F(x,y) &:= [z^m]\ (1+z(1+xy)^g)^{n/g} (1+z(1+y)^g)^{2m-n/g} (1+y)^{n-mg}\\
&= [z^m]\ \left(\frac{(1+y)^g+z(1+xy)^g}{1+z}\right)^{n/g} (1+z)^{2m},
\end{split}
and
\begin{split}
G(x,y) &:= [z^m]\ (1+z(1+y)^g)^{n/g} (1+z(1+xy)^g)^{2m-n/g} (1+xy)^{n-mg} \\
&=[z^m]\ \left(\frac{(1+xy)^g+z(1+y)^g}{1+z}\right)^{n/g} (1+z)^{2m}.
\end{split}
---
Using [Lagrange–Bürmann formula](https://en.wikipedia.org/wiki/Lagrange_inversion_theorem#Lagrange%E2%80%93B%C3%BCrmann_formula), we derive the expession
$$F(x,y) = [w^m]\ \left( \frac{A+B}2 + \frac{A-B}2\sqrt{1-4w} \right)^{n/g}\frac1{\sqrt{1-4w}},$$
where $A:=(1+y)^g$ and $B:=(1+xy)^g$, and the expression for $G(x,y)$ is obtained by exchanging $A$ and $B$.
It remains to notice that $A-B$ is a multiple of $y$ and in the expansion
$$F(x,y) = (-4)^m\left( \frac{A+B}2 \right)^{n/g} \sum\_{j=0}^{2m} \binom{n/g}j \left( \frac{A-B}{A+B} \right)^j \binom{j/2-1/2}{m} + O(y^{2m+1})$$
the terms with odd $j$ are zero, while for even $j$ the corresponding terms in $F$ and $G$ coincide.
QED
| 3 | https://mathoverflow.net/users/7076 | 450906 | 181,370 |
https://mathoverflow.net/questions/450879 | 1 | Let $\texttt{R}$ be an $\texttt{E}$-infinity ring and let $\texttt{M,N}$ be $\texttt{E}$-infinity modules. Under what conditions do we have
$$ \texttt{[M, N] ≅ [M,R] ⊗ N}$$
Under ordinary circumstances (with modules), we can show that a canonical map is an isomorphism using rank or dimension. What is the easiest approach here?
One application is to the (higher) trace $\mathbb{Tr}$:
$$ \texttt{R → [M, M] → [M,R] ⊗ M → R}.$$
| https://mathoverflow.net/users/30211 | [M,N]≅ [M,R] ⊗ N for E-infinity modules | See page 70 of EKMM. The map $$F\_R(M,N) \wedge\_R F\_R(M',N') \rightarrow F\_R(M \wedge\_R M', N \wedge\_R N')
$$ is an isomorphism if $M,M'$ are strongly dualizable or if $M$ is strongly dualizable and $N=R$. Being strongly dualizable is equivalent to being a wedge summand of a finite cell $R$ module.
| 3 | https://mathoverflow.net/users/134512 | 450908 | 181,372 |
https://mathoverflow.net/questions/450897 | 3 | For each $n \in \mathbb{N}$, the Hermite function $\psi\_n : \mathbb{R} \to \mathbb{R}$ is a Schwartz function defined by
\begin{equation}
\psi\_n(x):=(-1)^n(2^n n!\sqrt{\pi})^{-1/2} e^{x^2/2} \frac{d^n}{dx^n}e^{-x^2}
\end{equation}
according to the Wikipedia article : <https://en.wikipedia.org/wiki/Hermite_polynomials#Differential-operator_representation>
Moreover, it is written there that each Hermite function satisfies the $L^\infty$ bound
\begin{equation}
\lvert \psi\_n(x) \rvert \leq \pi^{-1/4} \text{ for all } x \in \mathbb{R}, n \in \mathbb{N}
\end{equation}
which is known as the Cramer's inequality.
Now, I am trying to find a bound of
\begin{equation}
\sup\_{x \in \mathbb{R}} \lvert x^m \psi\_n(x) \rvert
\end{equation}
for any given $n,m \in \mathbb{N}$.
More specifically, I am curious about the asymptotic behavior of this supremum as $n \to \infty$ while $m$ is fixed. Would it remain bounded or diverge?
I searched for references myself but cannot get an idea about resolving this issue. Could anyone please help me?
| https://mathoverflow.net/users/56524 | $L^\infty$ bound of $x^m \psi_n(x)$ where $\psi_n$ is a Hermite function and $m,n \in \mathbb{N}$ - extension from Cramer's inequality | $\newcommand\ep\varepsilon$According to the 6th display in [this section](https://en.wikipedia.org/wiki/Hermite_polynomials#Asymptotic_expansion), for any fixed $\ep>0$ we have
$$\psi\_n(x)=(2/\pi)^{1/4}(\pi n)^{-1/4}(\sin t)^{-1/2}s\_n(x)$$
if $n\to\infty$, $x=\sqrt{2n+1}\,\cos t$, and $t\in[\ep,\pi-\ep]$, and
$$s\_n(x):=\sin\Big(\frac{3\pi}4+\frac{2n+1}4\,(\sin2t-2t)\Big)+O(1/n).$$
Fix now any $t\in(0,\pi/2)$ such that $\sin2t-2t$ is irrational, and let $x\_n:=\sqrt{2n+1}\,\cos t$. Then $\limsup\_n s\_n(x\_n)=1$ and hence
$$\limsup\_n x\_n^m\, s\_n(x\_n)=\infty$$
for each integer $m\ge1$.
So, for each integer $m\ge1$,
\begin{equation}
\sup\_{x\in\mathbb R}|x^m\,\psi\_n(x)|\text{ is unbounded in $n$}.
\end{equation}
---
Using [Szegő, formula (8.22.14)](https://people.math.osu.edu/nevai.1/SZEGO/szego=szego1975=ops=OCR.pdf) (say with $t=0$), we see that, moreover, for each real $m>1/12$,
\begin{equation}
\lim\_n\sup\_{x\in\mathbb R}|x|^m\,|\psi\_n(x)|=\infty.
\end{equation}
| 3 | https://mathoverflow.net/users/36721 | 450910 | 181,373 |
https://mathoverflow.net/questions/450928 | 6 | Circumstances: I'm studying Grothendieck's Galois Theory and recently encountered a proposition that discussed the stability of coproducts under pullback. And I found the page [pullback-stable colimit](https://ncatlab.org/nlab/show/pullback-stable+colimit) in nLab.
On the mentioned nLab page, pullback-stability wasn't defined for individual colimits but for colimits with the same shape. However, I think it can be also defined for individual colimits like following.
>
> For a functor $G: D \to C$, we say that $\mathrm{colim}\_D\ G$ is stable under the pullback
> if for all pullback diagrams of $(\mathrm{colim}\_{D}\ G) \times\_Z Y,$ the canonical morphism
>
>
> $$\mathrm{colim}\_{d∈D}\ (G(d) \times\_Z Y)
> \to
> (\mathrm{colim}\_{d∈D}\ G(d))\times\_Z Y$$
>
>
> is an isomorphism.
>
>
>
So why isn't pullback-stability defined for individual colimits? Or are there any papers available that define pullback-stability for individual colimits?
| https://mathoverflow.net/users/508703 | Why isn't pullback-stability defined for individual colimits but for colimits with the same shape? | **Pullback-stability *is* sometimes considered for individual colimits, or at least, smaller classes than “all colimits of shape $D$”.** However, it’s most often used in settings where it holds for large classes of colimits, so authors usually define it just at the level of generality they need. In particular, the nlab article you link seems to be based in large part on Lurie’s treatment in *Higher Topos Theory*, §6.1.1.1; there, Lurie’s motivation is presenting the Giraud conditions for $\infty$-toposes, so he defines pullback-stability of colimits in the form and generality he wants for that.
One place where it’s used for smaller classes of colimits is in the study of [adhesive categories](https://ncatlab.org/nlab/show/adhesive+category), as studied in e.g. [Garner, Lack, *On the axioms for adhesive and quasi-adhesive categories*, 2011](https://arxiv.org/abs/1108.2934), particularly in the condition “pushouts along monomorphisms exist, and are stable under pullback”. But, as discussed in that paper, adhesive categories satisfy rather more: such pushouts are [van Kampen](https://ncatlab.org/nlab/show/van+Kampen+colimit). And this I think is the other reason why pullback-stability isn’t often considered for individual colimits: **in the most-studied settings that have only some colimits pullback-stable, one wants not just stability, but the stronger condition of van Kampen-ness.** And the [nlab page for *van Kampen colimits*](https://ncatlab.org/nlab/show/van+Kampen+colimit#universality_and_descent) *does* define it for individual colimits, and notes that one half of that definition can be seen as a definition of pullback-stability:
>
> The condition (1) ⇒ (2) is precisely the statement that the colimit of $G$ is universal, i.e. preserved by pullback.
>
>
>
Many sources, e.g. the above paper of Garner and Lack, talk about pullback-stability of colimits without explicitly defining it. This I think is because it’s essentially always considered in settings where enough pullbacks exist that they can be viewed as pullback functors between slices $f^\* : \mathcal{C}/X \to \mathcal{C}/Y$; so **preservation of colimits under pullback is viewed just as a special case of preservation by a functor** (this is explicit in e.g. Lurie’s Def. 6.1.1.2) — which is defined for individual colimits in many standard references. And this again comes out equivalent to the “universality” implication in the definition of van Kampen, and to your proposed definition in the question.
| 12 | https://mathoverflow.net/users/2273 | 450938 | 181,383 |
https://mathoverflow.net/questions/450948 | 10 | Context: I've just started reading Tate's thesis. In it, we start with a local field $k.$ The aim of the section is to describe the structure of the character groups of $k^+$ (the additive group) and $k^\*$(the multiplicative group). But for some reason when looking at the character group for $k^+$, we are looking only for the characters $\chi: k^{+} \to S^1$, where $S^1$ is the circle group but in $k^\*$, we are looking at quasi characters $\chi^\prime:k^\* \to \mathbb{C}^\*$. Why are we doing this? @anon [answered](https://math.stackexchange.com/a/868840) a related question, [Characters of a Group: two definitions](https://math.stackexchange.com/questions/868835/characters-of-a-group-two-definitions), on Math StackExchange regarding this but it really doesn't help much.
| https://mathoverflow.net/users/508721 | Why are we defining character groups differently for additive and multiplicative group in Tate's thesis? | I think the basic reason for the apparently different definitions boils down to the different topologies of $k^{\*}$ versus $k^{+}$.
For simplicity of discussion, let's consider the case that $k= \mathbb{Q}\_p$. If $\psi: \mathbb{Q}\_p^{+}$ is a continuous additive character, then it must have image in the unit circle. A simple way to see this is to use that $\mathbb{Z}\_p$ is compact, so its image under the continuous map $\psi$ is a compact subgroup of $\mathbb{C}^{\*}$. It therefore lies in $S^1$. One can apply the same argument to the compact subgroups $p^{-1} \mathbb{Z}\_p$, $p^{-2} \mathbb{Z}\_p$, and so on.
On the other hand, one can easily define continuous quasi-characters of $\mathbb{Q}\_p^\*$ that do not have image in the unit circle. For instance, take $\chi(p) = p$ and $\chi(x) = 1$ for $x \in \mathbb{Z}\_p^{\*}$. It is still true that $\mathbb{Z}\_p^\*$ is a compact subgroup of $\mathbb{Q}\_p^\*$, but now we have $\mathbb{Q}\_p^\* = \bigcup\_{j \in \mathbb{Z}} p^j \mathbb{Z}\_p^{\*}$, and the character is free to send $p$ anywhere in $\mathbb{C}^\*$.
| 19 | https://mathoverflow.net/users/2627 | 450953 | 181,391 |
https://mathoverflow.net/questions/450966 | 4 | Fix natural numbers $d,N$ and a polynomial $\Delta \in \mathbb{C}[x\_1,\ldots,x\_d]$. Let $S\_{d,N}$ be the set of field extensions $K/ \mathbb{C}(x\_1,\ldots,x\_d)$ such that
1. The degree $[K: \mathbb{C}(x\_1,\ldots,x\_d)]$ is bounded by $N$.
2. The discriminant of $K/ \mathbb{C}(x\_1,\ldots,x\_d)$ is $\Delta$.
3. $K$ is generated by an element whose minimal polynomial has coefficients in $\mathbb{C}[x\_1,\ldots,x\_d]$ that have degrees at most $N$.
Question: Is $S\_{d,N}$ finite?
| https://mathoverflow.net/users/4690 | Finiteness of number of extensions with bounded degree and discriminant | Yes, and I don't think you need 3. Let $D\subset \mathbb{C}^d$ be the locus where $\Delta $ vanishes. You are looking at étale covers of a fixed degree $n\ (\leq N)$ of $\mathbb{C}^d\smallsetminus D$, up to birational isomorphism. Such covers are classified by an action of $\pi \_1(\mathbb{C}^d\smallsetminus D)$ on a set with $n$ elements. Since $\pi \_1(\mathbb{C}^d\smallsetminus D)$ is finitely generated there are only finitely many possibilities.
| 8 | https://mathoverflow.net/users/40297 | 450972 | 181,395 |
https://mathoverflow.net/questions/450974 | 13 | Let $A$ be a $N \times N$ symmetric positive semi-definite matrix with $N \geq 2$. Let $D$ be a diagonal matrix of dimension $N$. I would like to measure how much $A$ "is far" from $D$, i.e. I am trying to find a way to quantify how $A$ differs from a diagonal matrix. More generally, the aim is to come up with a measure that captures the overall "level of orthogonality" of $A$, such that if I have another positive semi-definite matrix $B$, with the same dimension, I can argue that $A$ is more or less close to being orthogonal than $B$, which means I am interested in a relative statement. To be more explicit: I would like to quantify how $A$ is close to being an orthogonal matrix, and how $B$ is close to being an orthogonal matrix, not how $A$ and $B$ are orthogonal to each other.
Any suggestion (even partial) is welcome
| https://mathoverflow.net/users/508747 | Measuring the "distance" of a matrix from a diagonal matrix | **Q:** *Is there a measure that captures the overall "level of orthogonality" of two matrices $A$ and $B$.*
You can collect the $N^2$ elements of $A$ and $B$ into a pair of vectors $a$, $b$, and then take the inner product $(a,b)=\sum\_{ij}\bar{A}\_{ij}B\_{ij}={\rm tr}\,A^\ast B$. This would generalize the usual measure of orthogonality of pairs of vectors to pairs of matrices.
I understand this to be the motivation for your question of the "distance of a matrix $A$ from a diagonal matrix $D$"; for that distance you could just use the Frobenius norm, ${\rm tr}\,(A-D)(A-D)^\ast$.
| 15 | https://mathoverflow.net/users/11260 | 450976 | 181,396 |
https://mathoverflow.net/questions/445334 | 5 | **Background on heights**
Consider $P = [a: b] \in \mathbb{P}^1(\mathbb{Q})$, where $a,b$ are coprime integers. We define the naive (multiplicative) height as
$$H(P) = \max \{|a|, |b|\}$$
We can change the coordinates of $\mathbb{P}^1$, which would induce a different height function. For example, consider the automorphism $\varphi: \mathbb{P}^1 \to \mathbb{P}^1$ that sends $[x: y] \to [x: x+y]$. Then pullback height $H' = H \circ \varphi$ is then
$$H'(P) = \max\{|a|, |a+b|\}$$
These two heights are related; we can show that
$$\log H'(P) = \log H(P) + O\_{\varphi}(1)$$
**Questions**
In point counting questions, it is pretty reasonable to count points in the set
$$N(H'; B) := \# \{P \in \mathbb{P}^1(\mathbb{Q}): H'(P) \leq B \}$$
for some parameter $B$. The $O(1)$ relation above tells us that $N(H'; B)$ and $N(H, B)$ should be of same order of magnitude as $B \to \infty$, but does not say anything about how the leading coefficient changes. This seems to mean the naive height is not intrinsic in some sense.
* Is there a sense where the naive height $H$ is "canonical"/"intrinsic", where it is invariant under automorphisms of $\mathbb{P}^1$? The main thing I am shooting for here is the preservation of leading constant, not just the order of magnitude.
* If there's no such sense, does it mean we should always work with log height instead of multiplicative height?
+ As an example, Schanuel's theorem says that as $B \to \infty$,
$$N(H; B) := \#\{P \in \mathbb{P}^1(\mathbb{Q}): H(P) \leq B\} \sim \frac{2}{\zeta(2)} B^2$$
In that case, is there any arithmetic interpretation for the leading coefficient $\frac{2}{\zeta(2)}$, or is that the wrong framing and I should just look at $\log N(H; e^B)$ instead?
Thanks.
| https://mathoverflow.net/users/479911 | How should multiplicative height on projective space interact with automorphisms? | The naive height is not at all intrinsic. It is just a convenient choice to work with for notational simplicity. If one is doing things properly one should be counting rational points of bounded height with respect to every choice of adelic metric on the line bundle $\mathcal{O}(1)$. For different choices of adelic metric one obtains a different leading constant in general; this is expected and closely related to Peyre's notation of equidistribution of rational points.
Logarithmic heights are only used on varieties with very sparse sets of rational points, for example elliptic curves or abelian varieties. For Fano varieties the natural choice of height are the usual exponential heights. But again one should really do this with respect to every choice of adelic metric.
I would recommend reading some of the survey papers of Peyre on Manin's conjecture for Fano varieties, they are very informative.
| 1 | https://mathoverflow.net/users/5101 | 450993 | 181,399 |
https://mathoverflow.net/questions/450973 | 6 | This is probably a $y=f(x)$ question, but I searched several times on the MathOverflow without success so I decided to explicitly ask for the help of other members: please feel free to ask me to remove the question if inappropriate.
Nearly a month or so, in a MathOverflow answer (I guess so, but at this point I should consider the possibility I am wrong) to a question on (computational) number theory I read about a C library for arbitrary precision arithmetic which was successfully used in several research papers, including one where the Authors proved the existence of a soliton solution to a PDE by proving the strict positivity of a term by estimating its size as being not less than a $21$ bit mantissa multiplied by $10^{-5000}$ (I am fairly sure of this fact but, as already said, I should consider the possibility I am wrong). I went to the library web page and saved the link in my bookmarks: then a crash occurred (yes, also experienced electronics engineers lose data...) and up until now I was not able to find the former Q&A.
**Question**: can someone provide me a reference to this library?
| https://mathoverflow.net/users/113756 | On a fast high precision numerical analysis C library | Since you speak about mathematical proofs, probably you don't want an arbitrary-precision library, but a verified computation library based on interval arithmetic.
Maybe [Arb](https://arblib.org/)? Or [boost-interval](https://www.boost.org/doc/libs/1_82_0/libs/numeric/interval/doc/interval.htm)?
And maybe the post you remembered is about Warwick Tucker [proving](https://mathoverflow.net/a/123683/1898) the existence of the Lorenz attractor.
| 13 | https://mathoverflow.net/users/1898 | 450995 | 181,400 |
https://mathoverflow.net/questions/450992 | 0 | I am a bit puzzled about some probable implication in a paper.
Let $\varphi:\mathbb{P}^1\_K\to \mathbb{P}^1\_K$ be a rational map, where $K$ is a number field and let $\alpha\in K$ be such that $\varphi(\alpha)\neq\infty$. Then is it true that $\alpha$ is an exceptional point (backward orbit of $\alpha$ under $\varphi$ is finite) implies $\alpha$ is a preperiodic point (forward orbit of $\alpha$ under $\varphi$ is finite)? If this is not true providing an example will be appreciated.
| https://mathoverflow.net/users/481562 | A question about the backward orbit and forward orbit of a rational map in 1-dimensional projective space | Let $\alpha\_0=\alpha$, and for each $n\ge1$, choose a point $$\alpha\_n\in\phi^{-1}(\alpha\_{n-1})\subset \mathbb P^1(\overline K).$$ (The map $\phi$ is surjective on $\mathbb P^1(\overline K)$.) The points $\alpha\_n$ are in the backward orbit of $\alpha$, which you have assumed is finite. Hence I can find some $m>n\ge0$ such that $\alpha\_m=\alpha\_n$. Then
$$
\alpha = \phi^m(\alpha\_m) = \phi^m(\alpha\_n) = \phi^{m-n}\circ\phi^n(\alpha\_n)
= \phi^{m-n}(\alpha).
$$
Hence $\alpha$ is periodic.
| 3 | https://mathoverflow.net/users/11926 | 450998 | 181,401 |
https://mathoverflow.net/questions/451007 | 8 | Let $[\omega]^\omega$ denote the collection of infinite subsets of $\omega$, and let $c:[\omega]^\omega\to\{0,1\}$ be a function. We say that $a\in [\omega]^\omega$ is **monochromatic** with respect to $c$ if the restriction $c|\_{{\cal P}(a)\cap [\omega]^\omega}$ of $c$ to the infinite subsets of $a$ is constant. Moreover, $c$ is said to be **Ramsey** if there is $a\in [\omega]^\omega$ such that $a$ is monochromatic with respect to $c$.
Using the [Axiom of Choice](https://en.wikipedia.org/wiki/Axiom_of_choice) it is possible to construct [a non-Ramsey function $c:[\omega]^\omega\to\{0,1\}$](https://dominiczypen.wordpress.com/2023/07/18/non-ramsey-functions-property-b-and-the-axiom-of-choice/).
**Question.** Does the existence of a non-Ramsey function $c:[\omega]^\omega\to\{0,1\}$ imply some (weak) form of (AC) that cannot be proved in ${\sf (ZF)}$?
| https://mathoverflow.net/users/8628 | Non-Ramsey functions $c:[\omega]^\omega\to\{0,1\}$ and the Axiom of Choice | The keyword for searching is "infinite exponent partition relation."
The following article, for example, has many interesting results concerning the infinite exponent partition relation $\omega\to(\omega)^\omega$, which asserts that every coloring function has your Ramsey property, and its connection with weak forms of choice.
*Kleinberg, E. M.; Seiferas, J. I.*, [**Infinite exponent partition relations and well-ordered choice**](https://doi.org/10.2307/2272066), J. Symb. Log. 38, 299-308 (1973). [ZBL0274.04004](https://zbmath.org/?q=an:0274.04004).
In particular, they prove that under well-ordered choice, which is a weakening of the axiom of choice to well-ordered index sets (e.g. countable choice is an instance of this), then case $\omega\to(\omega)^\omega$ is the only possible case.
Meanwhile, it is mentioned there that Adrian Mathias, in his dissertation, proved that the infinite exponent partition relation $\omega\to(\omega)^\omega$ is consistent with countable choice and indeed with dependent choice. Consequently, if ZF is consistent, those choice principles are insufficient to produce a non-Ramsey function.
| 8 | https://mathoverflow.net/users/1946 | 451019 | 181,408 |
https://mathoverflow.net/questions/450997 | 1 | Let $G$ be a discrete nonamenable countable group acting on a standard probability space $(X,\mu)$ through measure-preserving transformations.
Is the action of $G$ always amenable?
(Amenable action, in the sense of Zimmer, which is equivalent to the action being orbit-equivalent to some action of the integers on $(X,\mu)$)
I do not see a proof, and am unfortunately unable to find a counterexample either. The closest thing I get, though perhaps not quite related, is that [Bowen and Nevo](https://doi.org/10.48550/arXiv.1103.2519) proved the existence of pointwise ergodic families for amenable equivalence relations (and hence amenable actions), and [Tao](https://doi.org/10.48550/arXiv.1505.04725) has an example of an action of $\mathbb{F}\_2$ where the family of increasing subsets of words of the same length does not form a pointwise ergodic family.
| https://mathoverflow.net/users/173694 | Nonamenable p.m.p. action on a standard probability space | A measure preserving action of a non-amenable group on a finite measure space is always **non-amenable**, if this is what you meant.
It is pretty obvious if one uses the most natural definition of an amenable action in terms of asymptotically equivariant families of probability measures. This definition (due to Renault) is equivalent to the orginal definition of Zimmer (given in terms of a fixed point property), but is much more constructive and easier to deal with (this is the reason why some of the first papers on amenability of actions are much longer than necessary). The definition you use is a combination of either of these definitions with a number of quite deep results culminating in the Connes - Feldman - Weiss theorem.
Indeed, according to Renault definition, an action of a countable group $G$ on a measure space $(X,m)$ is amenable if there exists a family of measurable maps $x\mapsto\lambda\_n^x$ from $X$ to the space of probability measures on $G$ that is **asymptotically equivariant** in the sense that
$\|g\lambda\_n^x - \lambda\_n^{gx}\|\to 0$ almost surely for any $g\in G$. If the measure $m$ is invariant, then one can easily verify that the averages $\lambda\_n=\int\lambda\_n^x\,dm(x)$ are asymptotically invariant (i.e., $\|g\lambda\_n-\lambda\_n\|\to 0$ for any $g\in G$), and therefore the group $G$ is amenable.
By using Renault's definition one also immediately sees that **any** (no matter measure preserving, or not) action of an amenable group is amenable (just take an asymptotically invariant sequence $\lambda\_n$ on $G$ and put $\lambda\_n^x\equiv\lambda\_n$). Thus, for probability measure preserving actions amenability of the acting group is equivalent to amenability of the action. However, for actions without an invariant measure the situation is different, and there exist amenable actions of non-amenable groups.
PS Ergodic theorems *per se* are not that much related to amenability of actions. What is more important is isoperimetric properties of the action, which, indeed, can be used to characterize amenability (in a way similar to the isoperimetric characterization of amenable groups).
| 2 | https://mathoverflow.net/users/8588 | 451021 | 181,409 |
https://mathoverflow.net/questions/451018 | 1 | Let $\Theta(x,t)$ be the Jacobi-Theta function:
\begin{equation}
\Theta(x,t):=1+\sum\_{n=1}^\infty e^{-\pi n^2 t} \cos(2\pi n x)
\end{equation}
Usually, the heat equation with the periodic boundary conditions is solved by the "spatial" convolution with $\Theta(x,t)$. However, I wonder what would happen if we perform convolution with respect to $t$ as well.
More specifically, let $f(x,t) : [0,1] \times [0,\infty) \to \mathbb{R}$ be a everywhere-defined function such that
1. $f(\cdot,t) : [0,1] \to \mathbb{R}$ is continuous and satisfies $f(0,t)=f(1,t)$ for each fixed $t \in [0,\infty)$.
2. $\int\_0^T \lvert f(x,t) \rvert dt <\infty$ for any $0<T<\infty$ and $x \in [0,1]$.
Then, define a "spacetime" convolution between $f$ and $\Theta$ as
\begin{equation}
(\Theta \* f)(x,t):=\int\_{t/2}^t \int\_0^1 \Theta(x-y,t-\tau)f(y,\tau)dy d\tau
\end{equation}
for $x \in [0,1]$ and $t \in (0,\infty)$.
Now, my question is that: is this $(\Theta \* f)$ still jointly smooth on $ [0,1] \times (0, \infty)$ with respect to $(x,t)$? Note that I excluded $0$ from the domain for $t$, to avoid possible complications.
I am aware that $\Theta(x,t) \to \sum\_{n \in \mathbb{Z}} \delta(x-n)$ in the sense of distribution as $t \to 0^+$. So, I am still concerned about the time integral in the above convolution.
Could anyone please clarify for me?
| https://mathoverflow.net/users/56524 | Convolution with the Jacobi Theta-function on "both the space and time variables" - still jointly smooth? | $\newcommand\Th\Theta\newcommand{\Z}{\mathbb Z}$The answer here is no.
E.g., for $(x,t)\in[0,1]\times[0,\infty)$ let
\begin{equation}
f(x,t):=2x\,1(x<1/2)+(2-2x)\,1(x\ge1/2).
\end{equation}
Then
\begin{equation}
(\Th\*f)(x,t)=\frac t4-\frac2{\pi^3}\sum\_{n=1}^\infty a\_n\cos2n\pi x
=\frac t4-\frac{c\_t}{\pi^3}\,g\_t(x),
\end{equation}
where
\begin{equation}
a\_n:=a\_{t,n}:=(1+(-1)^{n-1})\frac{1-e^{-\pi n^2t/2}}{n^4},
\end{equation}
\begin{equation}
g\_t(x):=\sum\_{n\in\Z}p\_n e^{2\pi ixn},
\end{equation}
$p\_n:=a\_n/c\_t$ for $n\ne0$, $p\_0:=0$,
$c\_t:=2\sum\_{n=1}^\infty a\_n$, so that $p\_n\ge0$ for all $n$ and $\sum\_{n\in\Z} p\_n=1$.
So, $g\_t$ is the characteristic function (c.f.) of a random variable $X\_t$ such that $P(X\_t=2\pi n)=p\_n$ for $n\in\Z$. Note that $EX^4=\infty$. So, the fourth derivative of $g\_t$ at $0$ does not exist (see e.g. [Theorem 2.3.1](https://rads.stackoverflow.com/amzn/click/com/0852641702)). So, $(\Th\*f)(x,t)$ is not smooth in $x$ at $x=0$ and hence is not jointly smooth in $(x,t)\in[0,1]\times(0,\infty)$. $\quad\Box$
| 2 | https://mathoverflow.net/users/36721 | 451024 | 181,411 |
https://mathoverflow.net/questions/450488 | 2 | Moufang identities
$$x(y⋅xz)=(xy⋅x)z,$$
$$(zx⋅y)x=z(x⋅yx),$$
$$xy⋅zx=x(yz⋅x)$$
are identities deeply related with alternativity (since setting $z=1$ one recovers left and right alternativity), while a Moufang plane is a strictly geometrical concept that encodes the fact that the group of automorphisms that fixes every point of a line acts transitively on the points of the plane, not on the line.
I know it is a well established fact that Moufang identities are deeply related to the geometric properties of the Moufang plane, but I could not find any direct proof. Can someone explain the main lines of the proof or at least give me a reference that directly addresses the question?
| https://mathoverflow.net/users/83165 | Moufang identities and Moufang plane | This is discussed and proved in detail in *Hall, Marshall jun.*, The theory of groups, New York: The Macmillan Company. xiii, 434 p. (1959). [ZBL0084.02202](https://zbmath.org/?q=an:0084.02202), specifically in chapter 20 "Group Theory and Projective Planes", and in there in section 20.5 "Moufang and Desarguesian Planes".
See also <https://www.ams.org/notices/200710/tx071001294p.pdf> for a quick overview of the idea, esp. the definition of "ternary rings" and how to use them to coordinatize projective planes.
| 4 | https://mathoverflow.net/users/8338 | 451025 | 181,412 |
https://mathoverflow.net/questions/450990 | 7 | Let $f:\mathbb{C} \to \mathbb{D}$ be a functor of 2-categories and let $\operatorname{Fun}(\mathbb{C},\operatorname{Cat})^{\operatorname{lax}}$ denote the 2-category of functors and lax natural transformations.
Is it true in general that restriction along $f$,
$f^\*:\operatorname{Fun}(\mathbb{D},\operatorname{Cat})^{\operatorname{lax}} \to \operatorname{Fun}(\mathbb{C},\operatorname{Cat})^{\operatorname{lax}}$
preserves all lax (weighted) limits?
| https://mathoverflow.net/users/117760 | Preservation of lax limits in categories of functors and lax natural transformations | I believe this functor preserves all *oplax* limits, and more generally all "$l$-rigged limits" in the sense of [Enhanced 2-categories and limits for lax morphisms](https://arxiv.org/abs/1104.2111) by Steve Lack and myself. (I don't think it's reasonable to ask it to preserve *lax* limits, since these 2-categories don't in general even *have* lax limits.)
To prove this, using the technology of that paper, observe that $\mathrm{Fun}({C},\mathrm{Cat})^{\rm lax}$ is $T\_{{C}}\rm Alg\_l$ for a 2-monad $T\_{{C}}$ on the 2-category $\mathrm{Cat}^{\mathrm{ob}{C}}$ of families of categories indexed by the objects of ${C}$. This 2-category underlies an $\mathcal{F}$-category $T\_C \mathbb{A}\mathrm{lg}\_l$ for a corresponding $\mathcal{F}$-monad $T\_{C}$ on $\mathrm{Cat}^{\mathrm{ob}{C}}$, which includes the data of the strict natural transformations as well and their embedding into the lax ones.
The main result of the above-cited paper is that the forgetful $\mathcal{F}$-functor $U\_C:T\_C \mathbb{A}\mathrm{lg}\_l\to\mathrm{Cat}^{\mathrm{ob}{C}}$ creates $l$-rigged limits, so that $T\_C \mathbb{A}\mathrm{lg}\_l$ has these limits and $U\_C$ preserves them. These are a class of $\mathcal{F}$-enriched weighted limits which include "oplax limits whose generating projections are strict and detect strictness". (Steve had already proven in [Limits for Lax Morphisms](https://link.springer.com/article/10.1007/s10485-005-2958-5) that oplax limits of this sort lift in the 2-categorical case.)
Now $f:C\to D$ induces a commutative square of $\mathcal{F}$-functors involving $U\_C$ and $U\_D$ and the restriction functors $f^\*: \mathrm{Cat}^{\mathrm{ob}{D}} \to \mathrm{Cat}^{\mathrm{ob}{C}}$ and $f^\* :T\_D \mathbb{A}\mathrm{lg}\_l \to T\_C \mathbb{A}\mathrm{lg}\_l$. The former $f^\*$ preserves all limits since they are pointwise, hence the composite $T\_D \mathbb{A}\mathrm{lg}\_l \to \mathrm{Cat}^{\mathrm{ob}{D}} \to \mathrm{Cat}^{\mathrm{ob}{C}}$ preserves $l$-rigged limits. Therefore, so does the composite $T\_D \mathbb{A}\mathrm{lg}\_l \to T\_C \mathbb{A}\mathrm{lg}\_l\to \mathrm{Cat}^{\mathrm{ob}{C}}$. This means that the comparison maps for $l$-rigged limits induced by $f^\* :T\_D \mathbb{A}\mathrm{lg}\_l \to T\_C \mathbb{A}\mathrm{lg}\_l$ are inverted by $U\_D$. But the forgetful $\mathcal{F}$-functor $U\_D$ is conservative, so these comparison maps are already isomorphisms in $T\_D \mathbb{A}\mathrm{lg}\_l$, i.e. $f^\*$ preserves $l$-rigged limits.
| 6 | https://mathoverflow.net/users/49 | 451034 | 181,416 |
https://mathoverflow.net/questions/451049 | 1 | Let $B$ be a standard Brownian motion. Its characteristic exponent (or Fourier transform) is easily calculated to be
$$ \mathbb E [e^{ixB\_t}] = e^{-\frac 12 x^2 t}. $$
Now I want to apply a Girsanov transformation with density $\exp(\int\_0^t A\_s d B\_s - \int\_0^t A\_s^2 ds)$ for some predictable process $A\_s$ satisfying Novikov's condition. It is well-known that $B\_t-\int\_0^t A\_sds$ is Brownian motion under the new measure. However, I am more interested in the question what the characteristic exponent of the *old* Brownian motion under the new measure is. Is there an explicit way to calculate this? Same question, but differently phrased: Is there an elegant way to calculate
$$ \mathbb E [ e^{\int\_0^t A\_s dB\_s} e^{ixB\_t}]? $$
| https://mathoverflow.net/users/166168 | Characteristic exponent after Girsanov transformation | By Doob-Dynkin we have that $A\_t=f(t,\{B\_s\}\_{0\leq s\leq t})$.
If $\mu\_0$ is the law of Brownian motion then under the measure $\mu=\exp\left(\int\_0^T A\_sdB\_s-1/2\int\_0^T A\_s^2 ds\right)\mu\_0$ we have that the law of $B$ is the law of the solution to the (possibly path dependent) SDE $dX\_t=f(t, \{X\_s\}\_{0\leq s\leq t}))dt+dB\_t$. So computing the characteristic function is the same as finding the law of $X$.
| 4 | https://mathoverflow.net/users/479223 | 451056 | 181,419 |
https://mathoverflow.net/questions/451039 | 2 | This posting is related to the [answer](https://mathoverflow.net/a/450985/95347) to this question.
Lets extend the language of $\sf PA$ with a monadic symbol "$\vdash$", add to the formula formation rules, the rule:
* if $(\phi)$ is a sentence in the language of $\sf PA$, then $(\vdash \phi)$ is a formula.
Now add all of the usual axioms of $\sf PA$ with induction restricted to the language of $\sf PA$, i.e. doesn't use the symbol "$\vdash$".
To be noted is that $\vdash$ can occur in the logical axioms, so we for example have $\neg (\neg(\vdash A)) \iff (\vdash A)$, for any sentence $A$ in the language of $\sf PA$
Add the following axioms:
**Axioms:** if $A$ is an axiom [logical or extra-logical] of $\sf PA$, then: $$ \vdash A$$
**Modus Ponens:** Let $A;B$ be sentences in the language of $\sf PA$, then: $$ (\vdash A) \land (\vdash (A \to B)) \to (\vdash B)$$
**Truth:** if $A$ is a sentence in the language of $\sf PA$, then: $$ (\vdash A) \to A $$
/
Call this theory $\sf PA+ \vdash + T$, while $\sf PA +\vdash$ denotes the theory without the last axiom.
Here, this theory won't have any of its models satisfying "$ (\vdash A) $ for every sentence $A$". The addition of the last axiom would bar that.
What I've noticed is that if we add the sentence $\neg \rho$ (i.e. the negation of the Rosser sentence) as an axiom, then $\sf PA+ \vdash + T + \neg \rho$ would actually prove: $$(\neg \rho) \not\to (\vdash \rho)$$
And so, all models of $\sf PA+ \vdash + T+ \neg \rho$ would falsify the implication schema present in the prior [posting](https://mathoverflow.net/q/450889/95347) for the instance of $\rho$. While with $\sf PA+ \vdash + \neg \rho $ it only says that $\neg \rho$ is provable and $(\vdash \rho)$ is not proved, so it fails to prove the instance of that implication for $\rho$, rather than manage to reject it. However, still $\sf PA+ \vdash + \neg \rho$ answered the question, because if $(\vdash \rho)$ is not provable, then adding $\neg (\vdash \rho)$ as an axiom is consistent, and in that theory the implication would be falsified, so it is one step short of this theory.
Seeing that, my question is:
>
> Does $\sf PA+ \vdash + T$ have higher consistency strength than $\sf PA+ \vdash$?
>
>
>
| https://mathoverflow.net/users/95347 | Do these two provability theories over PA differ in consistency strength? | No, both of these theories are equiconsistent with PA.
The reason is that given any model of PA, we may expand it to a model in the language with $\vdash$ by interpreting $\vdash\psi$ as true in the model exactly when PA proves $\psi$. That is, we interpret $\vdash\psi$ inside the model as being the same as provability in the metatheory. This fulfills all your axioms, including the truth axiom, since every theorem of PA is true in this model. And so every model of PA expands to a model of PA + $\vdash$ + T.
The argument shows that your theories are both conservative over PA for assertions in the language of arithmetic, which is a very strong way of being equiconsistent.
Because your theory does not have induction in the language with your symbol $\vdash$, what it does is simply copy the metatheoretic proofs over into the model. Every model of PA+$\vdash$ is simply a model of PA equipped with metatheoretic provability of some (possibly inconsistent) extension of PA. Every model of PA + $\vdash$ + T is a model $M$ of PA together with metatheoretic provability of some extension of PA that is true in $M$.
| 6 | https://mathoverflow.net/users/1946 | 451057 | 181,420 |
https://mathoverflow.net/questions/451055 | 0 | This problem comes from this commutative algebra problem
>
> Let $R$ be a commutative ring with identity, $I$ is a finite generated
> ideal of $R$ such that $I^2=I$, then exists $e\in R$ such that $I=Re$.
>
>
>
This problem can be proved using the Nakayama lemma, but the condition for finite generated is necessary? If I isn't finitely generated, does the conclusion still hold? follow
>
> Let $R$ be a commutative ring with identity, $I$ is an
> ideal of $R$ such that $I^2=I$, then exists $e\in R$ such that $I=Re$.
>
>
>
How to prove this? Or there is a counter-example to disprove.
| https://mathoverflow.net/users/476484 | Can every idempotent ideal be generated by an idempotent? | I have a counterexample from analysis. Let $R= C[0,1]$, the ring of real-valued continuous functions on the unit interval, and let $$I = \{ f \in C[0,1] : f(0) = 0 \}.$$
Then $I$ cannot be generated by any idempotent $p$, for if it were $p$ would have to satisfy $p(0) =0$ and $p(x) = 1 \ (x \neq 0)$, contradicting continuity. However you can check that $I^2 = I$: if $f\in I$ and $f(x) \geq 0 \ (x \in [0,1])$, then $f = (\sqrt{f})^2$ shows that $f \in I^2$; for general $f \in I$, decompose the function into its positive and negative parts.
| 4 | https://mathoverflow.net/users/507821 | 451066 | 181,423 |
https://mathoverflow.net/questions/451069 | 1 | Let $G(t, x) := \frac{1}{\sqrt{4 \pi t}} \exp\left( -\frac{x^2}{4 t }\right)$ for all $(t, x) \in (0, T) \times \mathbb{R}$ be the fundamental solution to the heat equation $\partial\_tu = \partial\_{xx}u$.
Evans PDE book (not verbatim but its clear from the context) states that
$$
\lim\_{\delta \downarrow 0} \int\_{\mathbb{R}} G(\delta,y) f(t-\delta, x-y)~\mathrm{d}y = f(x, y),
$$
if $f \in C\_c((0, T) \times \mathbb{R})$. I am wondering if this holds true a.e. if $f \in L^2((0, T); L^2(\mathbb{R})) \cap L^\infty((0, T), L^\infty(\mathbb{R}))$. My first impulse was to approximate $f$ by some $g\_\varepsilon \in C\_c^\infty((0, T) \times \mathbb{R})$ and to use triangle inequality a couple of times. Then, the only term that I am not sure about yet is
$$
\int\_{\mathbb{R}} G(\delta, y) \big(f(t-\delta, x-y) - g\_\varepsilon(t-\delta, x-y) \big)~\mathrm{d}y.
$$
So of course, for fixed $\delta$, this gets small as $\varepsilon \downarrow 0$. But does it get small uniformly in $\delta$? Remember, that afterwards I want to let $\delta \downarrow 0$?
I would be grateful for any input.
| https://mathoverflow.net/users/500621 | Does convolution with heat kernel converge to pointwise evaluation? | What you want cannot be true. Elements of $L^2((0,T); L^2(\mathbb{R}))\cap L^\infty((0,T),L^\infty(\mathbb{R}))$ can be discontinuous everywhere. Without doing any "time integration" you cannot get the "pointwise in time" limit.
For an example: let $k$ be a fixed, non-trivial, $C^\infty\_0(\mathbb{R})$ function. And define
$$ f(t,x) = \begin{cases}
0 & t\not\in \mathbb{Q} \\
k(x) & t \in \mathbb{Q}\end{cases}$$
This function is in the function space you specified, being a.e. equal to the 0 function. However, if $x$ is such that $k(x) \neq 0$, then for any $t$ the limit
$$ \lim\_{\delta \searrow 0} \int G(\delta,y) f(t-\delta,x-y) ~\mathrm{d}y $$
does not converge, as it has both $0$ and $k(x)$ as accumulation points.
---
Incidentally, this also shows that it is **not** true that if $g\_\epsilon$ is a smooth function approximating $f$ in the $L^2((0,T); L^2(\mathbb{R}))\cap L^\infty((0,T),L^\infty(\mathbb{R}))$ sense, then for fixed $\delta$ you can always make
$$ \int G(\delta,y) [ f(t-\delta, x-y) - g\_{\epsilon}(t-\delta,x-y) ] ~\mathrm{d}y $$
small just by taking $\epsilon$ sufficiently small. (Here you can take $g \equiv 0$.)
| 3 | https://mathoverflow.net/users/3948 | 451074 | 181,427 |
https://mathoverflow.net/questions/451047 | 1 | Consider a family of stochastic processes $dX^h\_t=(g(X^h\_t)+h(s))\,dt+dW\_t$ and a functional $I\_f:h(s) \rightarrow E[f(X\_t^h)] $. I would like to compute the kernel of the derivative of this functional with respect to $h$ using the Girsanov formula or Malliavin calculus. For example, Gisranov's formula gives
\begin{align}
E[f(X^h\_t)] & = E \left[ f(X\_t) \exp \left( \int\_0^t h(s) \, dW-\frac{1}{2}\int\_0^t h(s)^2 \, ds \right) \right] \\[8pt]
& \sim E\left[f(X\_t) \left(\int\_0^t h(s)\, dW\_s+1\right)\right].
\end{align}
Malliavin calculus also gives the same weight: $\int\_0^t h(s) \, dW\_s$. So the question is, how can I proceed from that? If the response function function $R(s,t)$ defined by $$E\left[f(X\_t) \int h(s) \, dW\_s\right] \sim \int h(s)R(s,t)\, ds + o(h^2),$$ then I would expect that formally $R(s,t) = E\left[f(X\_t)\dot W\_s \right]$. Is there a way to make this statement precise or express $E\left[ f(X\_t) \dot W\_s \right]$ in a more common way?
| https://mathoverflow.net/users/508809 | Linear response for SDE | For the original question of computing the kernel of the derivative of this functional with respect to h, I am more leaning on just regular Frechet derivative (if h is some deterministic function) and using the Girsanov formula to study the difference $I\_{f}(h+\epsilon v)-I\_{f}(h)$. For just the exponentials we have
$$\frac{1}{\epsilon}\exp \left( \int\_0^t h(s) \, dW-\frac{1}{2}\int\_0^t h(s)^2 \, ds \right)\cdot\left[\exp\left( \epsilon\left(\int\_0^t v(s) \, dW-\int\_0^t h(s)v(s) \, ds\right)-\frac{\epsilon^{2}}{2}\int\_0^t v(s)^2 \, ds\right) -1\right]. $$
From here only the first order survives and so in the limit we get
$$\exp \left( \int\_0^t h(s) \, dW-\frac{1}{2}\int\_0^t h(s)^2 \, ds \right)\cdot\left[\int\_0^t v(s) \, dW-\int\_0^t h(s)v(s) \, ds\right]. $$
i.e.
$(D\_{h}I\_{f}(h),v)=E\left[f(X\_{t})\exp \left( \int\_0^t h(s) \, dW-\frac{1}{2}\int\_0^t h(s)^2 \, ds \right)\cdot\left[\int\_0^t v(s) \, dW-\int\_0^t h(s)v(s) \, ds\right]\right].$
From here it is a bit hard to proceed without information on $f$.
Does this work for you or do you want something more?
In terms of getting an expression
$$(D\_{h}I\_{f}(0),v)=\int\_{0}^{t}R(r,t)v(r) dr$$
for some $R(u,t)$ kernel, we get inspiration from the work ["Mathematical foundation of nonequilibrium fluctuation-dissipation theorems for inhomogeneous diffusion processes with unbounded coefficients"](https://arxiv.org/pdf/1708.09744.pdf).
In particular, we apply Ito's formula to $f$
$$f(X\_{t})=f(x\_0)+\int\_{0}^{t}f'(X\_{r})g(X\_{r})dr+\int\_{0}^{t}f'(X\_{r})dW\_{r}+\frac{1}{2}\int\_{0}^{t}f''(X\_{r})dr.$$
By [Ito isometry](https://math.stackexchange.com/questions/3169779/the-expectation-of-the-product-of-two-stochastic-integrals)
$$(D\_{h}I\_{f}(0),v)=E\left[f(X\_{t})\int\_0^t v(s) \, dW\right]=\int\_{0}^{t}E\left[f'(X\_{r})\right]v(r) dr.$$
| 1 | https://mathoverflow.net/users/99863 | 451081 | 181,428 |
https://mathoverflow.net/questions/451075 | 0 | Let us consider the sequence space $c\_0$ with the equivalent norm
$$\Vert x \Vert^2 = \max\_{i\ge1} \vert x^i \vert^2 + \sum\_{i=2}^{\infty} 2^{-i+1} \vert x^i \vert^2 $$
for $x=(x^1,x^2,\ldots)\in c\_0$.
Let us take two sequences $(x\_{2n})$ and $(x\_{2n+1})$ in $c\_0$ such that $x\_{2n} \xrightarrow{w} x$ and $x\_{2n+1} \xrightarrow{w} y$ and $I\_n:=2\Vert x\_{2n} \Vert^2 + 2 \Vert x\_{2n+1} \Vert^2 - \Vert x\_{2n} + x\_{2n+1} \Vert^2 =0$ for all $n$. Does it then follow that $x=y$?
This question can be restated simply as follows: Take any $x$ and $y$ in $c\_0$ such that $I(x,y):=2\Vert x \Vert^2 + 2 \Vert y \Vert^2 - \Vert x+y \Vert^2 =0$. Does it then follow that $x=y$?
Or even more simply: Take any $x$ and $y$ in $c\_0$ such that $\Vert x+y \Vert=\|x\|+\|y\|$. Does it then follow that $x$ and $y$ are linearly dependent?
| https://mathoverflow.net/users/494605 | Renorming on a separable Banach space | The answer is no. E.g., let $x\_{2n}=x:=(\frac12,\frac12,\frac13,\frac14,\ldots)$ and $x\_{2n+1}=y:=(0,\frac12,\frac13,\frac14,\ldots)$ (for all $n$). Then (i) $x\_{2n}=x\to x$ and $x\_{2n+1}=y\to y$ in any sense and (ii) $\|x\_{2n}+x\_{2n+1}\|=2\|x\_{2n}\|=2\|x\_{2n+1}\|$ and hence $I\_n=0$ for all $n$. However, $x\ne y$. $\quad\Box$
| 0 | https://mathoverflow.net/users/36721 | 451089 | 181,429 |
https://mathoverflow.net/questions/451092 | 7 | Let $FA\_\kappa (\mathbb{P})$ be the claim that for every family $\mathscr{D}$ of dense sets in the poset $\mathbb{P}$ with $\vert \mathscr{D} \vert = \kappa $ there is a filter $G$ such that for all $D \in \mathscr{D} $, we have $G \cap D \neq \emptyset$. So basically what I am asking for is a forcing axiom for a single poset, rather than for a family of posets as one usually does.
For which couples $(\kappa, \mathbb{P})$ is the assumption $FA\_\kappa (\mathbb{P})$ consistent?
| https://mathoverflow.net/users/495743 | Forcing axiom for a single poset | There are several immediate things to say.
Some instances are outright provable in ZFC:
* $\newcommand\FA{\text{FA}}\FA(\omega,\newcommand{\P}{\mathbb{P}}\P)$ is a theorem for every poset $\P$.
* $\FA(\kappa,\P)$ holds whenever $\P$ is ${<}\kappa$-closed, and indeed, merely ${\prec}\kappa$-strategically closed is sufficient.
* $\FA(\kappa,\P)$ fails whenever $\P$ is splitting and $\kappa\geq 2^{|\P|}$.
Other instances are consistent, because they follow from certain forcing axioms. Indeed, most of the standard forcing axioms are simply equivalent to the FA principle individually for a wide class of forcing notions:
* Under Martin's axiom, $\FA(\kappa,\P)$ holds for every c.c.c. $\P$ and $\kappa<2^{\aleph\_0}$. And indeed, MA is equivalent to the collection of these assertions.
* Under MM, $\FA(\omega\_1,\P)$ holds for every stationary-preserving poset $\P$, and indeed, MM is equivalent to the collection these assertions.
* Under PFA, $\FA(\omega\_1,\P)$ holds for every proper forcing $\P$, and indeed, PFA is equivalent to the collection of these assertions.
The axiom $\text{MA}(\text{Cohen})$ asserts $\FA(\kappa,\text{Cohen})$ for Cohen forcing for all $\kappa$ below the continuum, and this axiom appears now and then in applications. For example, I had used this axiom in my paper, [Infinite Wordle and the Mastermind numbers](https://arxiv.org/abs/2203.06804).
The *Martin number* $\mathfrak{m}(\P)$ is the smallest $\kappa$ such that $\FA(\kappa,\P)$ fails, and this is amongst a very useful cardinal characteristic, particularly for Cohen forcing or other common forcing notions. As Will mentions in the comments, these Martin numbers interact usefully with other common cardinal characteristics.
For someone such as yourself who seems interested in looking at $\FA(\kappa,\P)$ for specific and well-known $\P$, I would recommend investigating those relations as a way to explore the broader context of your question.
| 9 | https://mathoverflow.net/users/1946 | 451093 | 181,432 |
https://mathoverflow.net/questions/450951 | 5 | For Grassmannians, the Schubert cells can be indexed by certain Young Tableaux, whose partition determines the dimensions of intersections of the chosen subspace with the standard complete flag. For example, up to change in convention, in $Gr(2,4)$, the subspace $V = \langle e\_2,e\_4\rangle$ may belong to $X\_{(1,0)}$, since, if $F\_\bullet$ is the standard complete flag, then $dim(V \cap F\_j) = (0,1,1,2)$, while for a generic 2-dimensional subspace, $V\_{gen}$, $dim(V\_{gen} \cap F\_j) = (0,0, 1,2)$. One scheme for associating the cells to partitions is to record the difference (between $V$ and $V\_{gen}$) between the position of the first appearance of 1 in the "intersection dimension vector" as the first element of the partition, and the difference between the position of the first appearance of 2 as the second entry, hence $(1,0)$. (This was explained to me only one time in person, so please let me know if I've got some aspect of this process wrong.) So if I am given the algebraic data (partition), I can determine the geometric data (intersection dimension vector) and vice versa.
For complete flags, the Schubert cells are naturally indexed by $S\_n$, which admits a bijection to pairs of same-shape standard Young tableaux via Robinson-Schensted correspondence. Is there a geometric understanding of these SYT's in a way analogous to the above? i.e. For $Fl(m)$, we have a $2\times m-1$ ($m-1$ if we ignore the final column, which will always just be $\begin{pmatrix} m \\ m \end{pmatrix}$) matrix of intersection dimensions, $dim(E\_i\cap F\_j)$ for $i,j \in \{1,2,\dots,m-1\}$. If I know the intersection dimension matrix of a certain flag, can I determine the pair of SYT of its Schubert cell or vice versa?
| https://mathoverflow.net/users/131046 | geometric meaning to pairs of SYT indexing for the basis of cohomology ring of full flag variety | Since, surprisingly, there are still no answers or even comments, let me note that the answer to the last question is well known to be "yes": the Schubert cell containing a flag $(E\_1,\dots,E\_m)$ is indeed determined by the values $\dim(E\_i\cap F\_j)$.
Specifically, if $\sigma$ is the corresponding permutation, then $\dim(E\_i\cap F\_j)$ is the number of values no greater than $j$ among $\sigma(1),\dots,\sigma(i)$. It is evident that this data determines the permutation uniquely, so the boring method here is to determine the permutation and then apply the insertion algorithm. The question whether there is a substantially more direct relation between the dimensions and the pair of tableaux is curious but I wouldn't count on it.
| 4 | https://mathoverflow.net/users/19864 | 451096 | 181,433 |
https://mathoverflow.net/questions/450980 | 3 | I am having trouble in understanding Theorem 8.2 of [Adams's book](https://press.uchicago.edu/ucp/books/book/chicago/S/bo21302708.html) and the application afterwards of constructing the spectral sequence. I think I should prove somehow that the spectral sequence in this case will be
$$\varprojlim\_{a}{}^{p}E^{q}(X\_{a})\implies E^{p+q}(X)$$
where $X\_{a}$ are the skeletons of $X$, $\varprojlim\_{a}{}^{p}$ is the $p-$th right derived limit as defined in the same chapter of Adams.
As far as I understood, we start with a spectral sequence with the first page being $E\_{1}:=E^{p+q}(X\_{p},X\_{p-1})$. So, I have many questions.
1. How do we prove that the second page is given by $\varprojlim\_{a}{}^{p}E^{q}(X\_{a})$?
2. What is the notion of "convergence" here? Is it the condition one he mentions which says $E\_{\infty}^{p,q}\to\varprojlim\_{r}E\_{r}^{p,q}$ is isomorphism? And how are the maps $E\_{r+1}^{p,q}\to E\_{r}^{p,q}$ defined and why being monomorphism implies that the limit exists?
3. What are exactly the filtration quotients of $E^{p+q}(X)$ in condition 3? And how is the exact sequence constructed (the exact sequence in condition 3)?
4. How do we even use theorem 8.2 in Adam's book to provide the exact sequence? This means, how do we verify condition (ii) of this theorem?
| https://mathoverflow.net/users/160383 | Spectral sequence in Adam's book, Theorem 8.2 | This is not a complete answer, but it is too long for a comment. I will assume you are interested in the spectral sequences in Section 8 of Adams, not the Atiyah-Hirzebruch spectral sequences of Section 7. There are two different spectral sequences there, depending on whether you start with a CW complex filtered by its skeleta (or any linearly ordered system of subcomplexes) as just before items 1-3:
$$
0 = X\_{-1} \subset X\_0 \subset X\_1 \subset \dots \subset X,
$$
or a CW complex "which is the union of a directed set of subcomplexes $X\_{\alpha}$" (end of the section). For the first of these the higher derived functors vanish. For the second, they need not.
Answers to some of your questions:
1. Not an answer, but a note that when the higher derived functors $\lim^n$ are zero for $n \geq 2$, then a spectral sequence of the form at the end of the section will degenerate to a short exact sequence as in Proposition 8.1. So try to generalize the proof of 8.1 to the situation of an arbitrary directed system.
2. The convergence condition is items 1-3, which he proves (Theorem 8.2) is equivalent to just item 2, and item 2 is typically verified using a Mittag-Leffler condition (exercise just before Prop 8.1).
3. The spectral sequence should converge to $E^{p+q}(X)$, and this is filtered by the groups $E^{p+q}(X\_p)$. That is, we have maps
$$
E^{p+q}(X) \to E^{p+q}(X\_p)
$$
and I think we want to define $F^{p,q}$ by the exact sequence
$$
0 \to F^{p,q} \to E^{p+q}(X) \to E^{p+q}(X\_p).
$$
Then $E^{p,q}\_{\infty}$ will be the kernel of $F^{p,q} \to F^{p-1,q+1}$.
4. Not sure what you mean by "the exact sequence." Adams' Theorem 8.2 might be the same as [Boardman's](https://www.uio.no/studier/emner/matnat/math/MAT9580/v21/dokumenter/conm239-3597.pdf) Theorem 7.4. Take a look.
| 6 | https://mathoverflow.net/users/4194 | 451100 | 181,435 |
https://mathoverflow.net/questions/451094 | 8 | Skolem wrote an interesting paper in 1960 about the use of a three-valued logic to create a set theory with true comprehension that solves Russell's paradox. That paper is [A set theory based on a certain 3-valued logic](https://doi.org/10.7146/math.scand.a-10600).
He basically uses what most today would call the strong Kleene three-valued logic. The tables are:
\begin{gather\*}
\begin{array}{c|c|c|c}
a \wedge b& 0 & ½ & 1 \\
\hline
0 & 0 & 0 & 0 \\
\hline
½ & 0 & ½ & ½ \\
\hline
1 & 0 & ½ & 1 \\
\end{array}
\\
\begin{array}{c|c|c|c}
a \vee b& 0 & ½ & 1 \\
\hline
0 & 0 & ½ & 1 \\
\hline
½ & ½ & ½ & 1 \\
\hline
1 & 1 & 1 & 1 \\
\end{array}
\\
\begin{array}{c|c}
a & \lnot a \\
\hline
0 & 1 \\
\hline
½ & ½ \\
\hline
1 & 0 \\
\end{array}
\end{gather\*}
So we have that $a \wedge b = \min(a,b)$, $a \vee b = \max(a,b)$, and $\lnot a = 1-a$. He then goes on to show how you can solve Russell's paradox with this system and retain true comprehension, as the Russell set $R$ simply has $R \in R = ½$. He also allows non-bounded quantifiers with $\forall$ as an infinite $\min$ and $\exists$ as an infinite $\max$.
There seems to be a real snag right away with these kinds of three-valued systems in defining singletons (and thus ordered pairs, etc.), and the subtlety is all about if equality is taken as primitive and how one can use it. Skolem, for instance, doesn't allow equality in comprehension, nor provides for any kind of primitive singleton operator, so we can't define $\{x\} := \{y \mid y = x\}$.
In two-valued logic, we can do this via extensionality instead: here we will use the symbol $x \approx y$ to mean $\forall z: (z \in x) \leftrightarrow (z \in y)$. There is a question here, however, regarding what "$\leftrightarrow$" means. It turns out there is *only* one function in this three-valued logic extending the usual bidirectional on Boolean values:
$$
\begin{array}{c|c|c|c}
a \leftrightarrow b& 0 & ½ & 1 \\
\hline
0 & 1 & ½ & 0 \\
\hline
½ & ½ & ½ & ½ \\
\hline
1 & 0 & ½ & 1 \\
\end{array}
$$
This is, again, the unique bidirectional in the system Skolem is using. If we try to build the "weak singleton" using this, which would be $\{x\}\_W = \{y \mid y \approx x\}$, we can get some strange results. For starters, if $x$ itself has any elements in it with membership value $½$, then $x \in \{x\}\_W$ will evaluate to $½$ — it won't even be in its own weak singleton. If $x$'s membership function is purely Boolean and doesn't have any such elements, then for any $y$ which has element $z$ such that $z \in y$ takes value $½$, then we know that the above biconditional will evaluate to $½$ no matter what, and when quantified over all such $z$ we will either $0$ or $½$. Thus, for any $x$, it will always contain "extra" elements with membership value $½$, and it may even contain itself with membership value $½$.
If one tries adding *any* other bidirectional to the system, it will lose the property that all unary functions have fixed points and we will be able to rebuild Russell's paradox. With the Łukaszewicz bidirectional, for instance, we can rebuild it as $\{x \mid (x \in x) \leftrightarrow \lnot(x \in x)\}$.
Thus we are not guaranteed that singletons exist in this set theory — or if they do, you can't build them in comprehension — and thus it would seem to be very difficult to form other sets depending on them such as ordered pairs: $\{\{a\}, \{a,b\}\}$, and thus functions, etc.
One could, of course, try to "solve" this problem by simply permitting equality in comprehension to begin with. However, this can also causes problems, unless one is very careful: there's *another* version of Russell's paradox: $\{x \mid x \neq x \cup \{x\}\}$. This could perhaps be solved by not allowing union in comprehension, or perhaps prohibiting "nested comprehensions" in some form — not sure how that would work.
Summarizing all of this: it would seem like not being able to form a singleton would make for a very strange set theory, one which I cannot figure out how you could possibly do math in. On the other hand, Skolem doesn't seem to think it's a problem, and I know there are many other people who have studied non-classical logics in set theory. **Is there some way that these things can actually be useful, even though you can't build what we would normally think of as singletons?** Or, is there some standard way to somehow let you build singletons, and thus more complex sets from them, without making everything inconsistent?
| https://mathoverflow.net/users/24611 | Using three-valued logic for set theory with singletons and equality | One way to adopt three-valued truth in set theory is with paraconsistent set theory, which typically uses Kleene three-valued logic, but with the understanding that truth value 1/2 is regarded as "both true and false," and so in particular, both 1/2 and 1 are designated as instances of truth (and this logic is then called the logic of paradox). The method generalizes to paraconsistent model theory, not just set theory.
My favored way to think about it is that we are to take any two models $M\_0$, $M\_1$ in the same language, with the same domain. We define their paraconsistent mixture $M$ as follows: $[\![\varphi(\vec a)]\!]$ for any atomic formula $\varphi$ and individuals $\vec a$ in the domain has value $1$, if both $M\_i\models\varphi(\vec a)$, and $0$ if both $M\_i\models\neg\varphi(\vec a)$, and otherwise $[\![\varphi(\vec a)]\!]=1/2$. Thus, we mix together the atomic truths of the two models. One can just as easily use any collection of models.
Then, we extend the truth values by the truth-functional definition of Kleene logic. The result is a truth-functional account of 3-valued truth in the mixture model $M$.
Some accounts of paraconsistent set theory or paraconsistent model theory focus on the two-model case, but present the situation of every atomic relation being specified by its positive instances and its negative instances, setting up a duality between truth ad falsity. In a relational language, this amounts the same to the mixture as I specified it, with two models, but I regard the mixture idea as a more general approach to paraconsistent model theory, since it works with more than two models and with other logics.
A non-truth-functional approach simply defines that $[\![\varphi(\vec a)]\!]$ has value $1$, if $\varphi(\vec a)$ is true in all the models, value $0$ if it is false in all the models, and value 1/2 otherwise. This violates truth functionality, since if $\varphi$ is true in some and false in others, then $[\![\varphi\vee\neg\varphi]\!]$ has value 1 in this semantics, but value 1/2 in the truth-functional semantics.
There is an interesting connection with forcing, in the sense that for any algebra $\newcommand\A{\mathbb{A}}\A$ of truth values, including the 3-valued truth case, one can form the class $V^{\A}$ of all $\A$-names, as in forcing, and then define $\A$-valued truth $[\![\varphi(\vec\tau)]\!]$, just as in the Boolean-valued approach to forcing, except that $\A$ may not be a Boolean algebra. Things work out quite well in general. I had worked out a theory of this, but just as I was getting ready to complete my paper, I noticed the following great paper of Löwe and Tarafder, which did it even better:
* *Löwe, Benedikt; Tarafder, Sourav*, [**Generalized algebra-valued models of set theory**](https://doi.org/10.1017/S175502031400046X), Rev. Symb. Log. 8, No. 1, 192-205 (2015). [ZBL1375.03066](https://zbmath.org/?q=an:1375.03066).
This paper is really great.
| 9 | https://mathoverflow.net/users/1946 | 451103 | 181,436 |
https://mathoverflow.net/questions/451085 | 4 | Suppose I work over $\mathbb{C}$. Is it known for which locally convex topological vector spaces $V$, we have $\text{Ext}^i(V, \mathbb{C})=\{0\}$ for all $i>0$, working with the type of Ext groups that Wengenroth uses. For instance, is it true when $V$ is Frechet and nuclear?
| https://mathoverflow.net/users/3396 | Ext groups of locally convex topological vector spaces | The Hahn-Banach theorem implies that $\mathbb C$ is an injective object in the category of complex locally convex topological vector spaces. Therefore, Ext$^i(V,\mathbb C)=0$ for every locally convex space $V$.
| 8 | https://mathoverflow.net/users/21051 | 451122 | 181,442 |
https://mathoverflow.net/questions/451123 | 0 | This question seem a bit elementary, but I find it more subtle than its looks. So, I post the question here.
Let $f(x,t) : [0,2\pi] \times [0,1] \to \mathbb{C}$ be a function such that $f(0,t)=f(2\pi,t)$ for all $t \in [0,1]$. Also, further suppose that $f(x,t)$ is $C^\infty$ in $x$ for each $t \in [0,1]$ and jointly continuous in $(x,t) \in [0,2\pi] \times [0,1]$.
Then, by Fourier expansion, I see that
\begin{equation}
f(x,t)=\sum\_{n \in \mathbb{Z}} a\_n(t) e^{in x}
\end{equation}
where each $a\_n(t)= \int\_0^{2\pi} f(x,t) e^{-inx} dx$ is continuous on $[0,1]$.
Now, by $C^\infty$ assumption with respect to $x$, we observe that $a\_n(t)=O\_t(1/\lvert n\rvert^k)$ for any $k \in \mathbb{N}$. Here I used the notation $O\_t$ to emphasize that this estimate depends on $t \in [0,1]$.
I wonder if $x$-derivatives of $f(x,t)$ are continuous in $t$ by any chance, since the Fourier coefficients $a\_n(t)$ are all continuous on the compact interval $[0,1]$.
It seems that
\begin{equation}
\lvert a\_n(t) \rvert \leq \frac{C\_{n,k}}{ \lvert n \rvert^k}
\end{equation}
for all $t \in [0,1]$, $n \in \mathbb{Z}$ and $k \in \mathbb{N}$ for some positive constants $C\_{n,k}$, but I do not see how to control $C\_{n,k}$'s.
Could anyone please help me?
| https://mathoverflow.net/users/56524 | If $f(x,t)=\sum_{n \in \mathbb{Z}} a_n(t) e^{in x}$ is $C^\infty$ in $x$ and all $a_n(t)$ continuous, $x$ derivatives of $f$ are continuous in $t$? | I am going to intepret the question as: given an $f$ satisfying the conditions in the question, must $\partial\_x f$ be continuous as a function on $[0,2\pi]\times [0,1]$?
@Echo's suggestion seems to give a concise, explicit counterexample. However, I thought it might be worth spelling out a line of thought that can be adapted to other contexts.
---
The first thing to note is that Fourier analysis is really a red herring. Indeed, while we know that functions with a certain degree of smoothness have Fourier coefficients decaying at a certain rate, in general one loses information and hence cannot use the decay of Fourier coefficients to get back the same level of smoothness. (For instance $o(1/n)$ decay is satisfied by Fourier coefficients of all $C^1$ functions on the circle, but you cannot use $o(1/n)$ decay to deduce that the original function is in $C^1$.) Also, periodicity is not really important, since you are interested in continuity and smoothness which are local not global properties.
So I will produce an example of a function $F: [0,1]\times [0,1]\to {\mathbb R}$ which is (jointly) continuous, is $C^\infty$ in the first variable, but such that $\partial\_1 F$ is not jointly continuous on $[0,1]\times [0,1]$. Once you have such an $F$, one can extend it by reflection to get a function $F:[-1,1]\times [0,1] \to {\mathbb R}$ with all the same properties and also satisfying $F(-1,y)=F(1,y)$ for all $y$. It should then be clear how to rescale the domain to get a function that satisfies your original requirements.
The key is that once one starts trying to disprove the claim/question that you describe, it is natural to ask what a counterexample should look like. We want $\partial\_1 F$ to be discontinuous as a function on $[0,1]\times [0,1]$ even though $\partial\_1 F(\cdot, y)$ should be very nice for each fixed $y$. This suggests looking at one of the standard examples which shows separate continuity does not imply joint continuity. Let's pick the following function off the shelf:
$$
g(x,y) = \begin{cases} \dfrac{xy}{x^2+y^2} & \text{ if $(x,y) \neq (0,0)$ } \\ 0 & \text{ if $x=y=0$} \end{cases}
$$
This is not continuous at $(0,0)$, since $g(x,x)= \frac{1}{2}$ for all $x>0$.
Of course $g(\cdot ,0)$ is the zero function on $[0,1]$, while for any $y\in (0,1]$ the function $g(\cdot, y)$ is the quotient of a polynomial by a non-vanishing polynomial hence is smooth.
We now take this as our Ansatz for $\partial\_1 F$. That is, let us define
$$
F(x,y) = \int\_0^x g(u,y) \, du.
$$
Integration improves continuity/smoothness properties, so for each $y\in [0,1]$ we still have $F(\cdot y) \in C^\infty[0,1]$. Therefore all I need to do is to show the following result.
CLAIM. $F:[0,1]\times [0,1] \to {\mathbb R}$ is continuous.
Proof. Observe that $g$ is continuous at every point of $[0,1]\times [0,1]$ except the origin, and so the same behaviour is true for $F$, again because integrating will only improve things. It remains to show that $F$ is continuous at the origin.
Note that $F(0,\cdot)=0$. So $F(x,y) - F(0,0) = F(x,y) - F(0,y)$.
But since $0\leq g \leq 1$ on our given domain, this is bounded in modulus by $|x|$, and so as $(x,y)\to (0,0)$ we see that $|F(x,y)-F(0,0)| \leq |x| \to 0$, as required.
$\qquad\Box$
---
As a final remark: I defined $F$ in this way to show that it is not necessary to have an "explicit formula" to do calculations and prove properties of $F$. However, of course we know by calculus how to "antidifferentiate" $g(\cdot, y)$, so we could take $F(x,y) = \frac{1}{2}y\log (x^2+y^2)- y\log y$.
| 5 | https://mathoverflow.net/users/763 | 451136 | 181,446 |
https://mathoverflow.net/questions/451139 | 5 | I was looking for a complex extension of the surreals and then I found $\operatorname{No}[i]$, what are its properties and how do I express $x+yi$ in the $a | b$ notation?
| https://mathoverflow.net/users/508896 | What are the properties of $\operatorname{No}[i]$? | The surreal complex field $\text{No}[i]$, known as the [surcomplex field](https://en.wikipedia.org/wiki/Surreal_number#Surcomplex_numbers), is a proper-class-sized set-saturated algebraically closed field. It is universal for all fields of characteristic 0. Indeed, under global choice, every class field structure in characteristic 0 embeds as a subfield of $\text{No}[i]$. It is the algebraic closure of the transcendental field extension of $\mathbb{Q}$ by a proper class of algebraically independent transcendentals.
| 5 | https://mathoverflow.net/users/1946 | 451143 | 181,449 |
https://mathoverflow.net/questions/451149 | 5 | The axiom $W\_\kappa$, for $\kappa$ a cardinal, is the statement that for all sets $X$, either $|X|\leq\kappa$ (that is, there is an injection $X\to\kappa$) or $\kappa\leq|X|$. Is there literature on the dual notion $W^\*\_\kappa$ that for all $X$, $|X|\leq^\*\kappa$ (that is, there is a surjection $\kappa\to X$, or $X$ is empty) or $\kappa\leq^\*|X|$?
| https://mathoverflow.net/users/478588 | References for the axiom of surjective comparability | Not too much that I'm aware of. It comes up in the case of $\omega$, since that would imply that even if Dedekind finite sets exist, they can at least be mapped onto $\omega$ (equivalently, "the power set of any infinite set is Dedekind infinite").
In the general case, this isn't as helpful, and there's not too much we can say (except, perhaps, there are no $\kappa$-amorphous sets).
Do note, of course, that if $\kappa$ is an $\aleph$, then $|X|\leq^\*\kappa$ if and only if $|X|\leq\kappa$, so the real difference is in the other direction, which we can restate as "any set that cannot be well ordered can be mapped onto $\kappa$".
| 5 | https://mathoverflow.net/users/7206 | 451150 | 181,453 |
https://mathoverflow.net/questions/451110 | 1 | I have asked a similar question on MSE but I did not receive any replies, so I am reposting here in case it is more appropriate (though I have slightly generalized the question).
As an example consider the Laplacian operator. The inverse of the Laplacian is given by
$$(-\Delta)^{-1} u(x) = C \int\_{\mathbb{R}^n} u(x-y) \frac{1}{|y|^{n-2}} dy$$
where $n$ is the dimension of $\mathbb{R}^n$.
I would like to learn more about such operators, as I have often seen their formulas stated but not explained. For example, we can motivate fractional powers of the Laplacian by use of the Fourier transform. I am looking for a similar explanation but for the inverse.
Some of the questions I have been thinking about are:
1. How are these explicit inverses derived? In the case of the Laplacian, is there a reason its inverse is identical to its fundamental solution?
2. Under what conditions and for which function spaces are the inverses valid?
3. In the case of a differential operator, I assume that we gain some degree of regularity by applying its inverse, but how is this proven? What is the image of this operator?
I have struggled to find this information in any textbooks. Can anyone suggest some textbooks or references in this direction? Thanks.
| https://mathoverflow.net/users/498931 | Reference request: inverse of differential operators | There are many possible answers to this question, depending on the type of the differential operator, the domain of the functions, and whether you want to impose any additional conditions such as specifying boundary data.
For example, if you assume that the differential operator has constant coefficients, then the [Malgrange-Ehrenpreis theorem](https://en.wikipedia.org/wiki/Malgrange%E2%80%93Ehrenpreis_theorem) shows that such a Green's function exists on $\mathbb{R}^n$.
In general, given a linear differential operator $P$ on a domain $D \subset \mathbb{R}^n$, the existence of an inverse operator is essentially equivalent to showing that there exists a constant $C$ such that for any compactly supported functions $f$ and $u$ on $D$, there is an a priori bound of the form
$$ \|u\|\_1 \le C\|\_f\|\_2, $$
where $\|\cdot\|\_1$ and $\|\cdot\|\_2$ are appropriately chosen norms, depending on what $P$ is.
If $P$ is elliptic on a domain $D$ and suitable assumptions are made on both $P$ and $D$, then an inverse operator exists (but is not unique) and is a pseudodifferential operator (a generalization of differential operators defined using the Fourier transform).
If $P$ is hyperbolic on a domain $D$ and suitable assumptions are made on both $P$ and $D$, then an inverse operator exists and is a Fourier integral operator (a generalization of pseudodifferential operators, also defined using the Fourier transform).
Analogous but different stories hold for parabolic, hypoelliptic, and many other types of differential operators. The problem is that there is no systematic approach that unifies all of these results. Each type is handled using an ad hoc approach tailored to the assumptions.
Back in the 60's and 70's, there was a big effort by many, including Hormander, Nirenberg, Treves, to identify which differential operators have inverse operators. However, the effort never completely succeeded, and everyone started focusing more on nonlinear PDEs.
| 3 | https://mathoverflow.net/users/613 | 451157 | 181,455 |
https://mathoverflow.net/questions/451014 | 5 | Let $F:\mathbf{\Delta}\to\mathcal{S}\_{\leqslant n-1}$ be a cosimplicial object in the $\infty$-category of $(n-1)$-truncated spaces. Is it always a right Kan extension of its restriction along $\mathbf{\Delta}^{\leqslant n}\hookrightarrow\mathbf{\Delta}$? (Then one might call $F$ to be $n$-coskeletal.)
I want to use this to show that *$\mathbf{\Delta}^{\leqslant n}\_{/[m]}:=\mathbf{\Delta}^{\leqslant n}\times\_{\mathbf{\Delta}}\mathbf{\Delta}\_{/[m]}$ is $n$-connected ($\pi\_{<n}\simeq 0$), for every $[m]\in\mathbf{\Delta}$* (which I view as a finite version of Quillen’s theorem A). Other ways to confirm is also fine (without solving the above question).
One might replace $\mathbf{\Delta}$ with $\mathbf{\Delta}\_s$ everywhere, e.g. by (proof of) [HTT, Lemma 6.5.3.8]; in the case $m\leqslant n$, it is easy to see that it has a final object hence is in fact contractible.
| https://mathoverflow.net/users/42571 | Connectedness of truncated version of cosimplicial indexing category | I don't believe the first claim is true, but I can give a somewhat formal argument for the connectivity of $\Delta\_{\le n} \times\_{\Delta} \Delta\_{/m}$.
Let us write $u$ for the inclusion $\Delta\_{\le n} \to \Delta$. The crucial result follows the following special case of Appendix A of <https://arxiv.org/abs/2207.09256>:
**Theorem 1.** Given a diagram $F : \Delta \to \mathcal{S}\_{\le n-1}$ the canonical map $\lim\_{\Delta} F \to \lim\_{\Delta\_{\le n}} u^\*F$ is invertible.
(This may already be the result you were after, but we can carry this through to actually obtain the connectivity result you requested.)
Fix an arbitrary $F : \Delta \to \mathcal{S}\_{\le n-1}$. By Theorem 1, we know that the following is an equivalence:
\begin{align\*}
&\lim F
\\
&\cong \hom\_{\mathbf{Fun}(\Delta,\mathcal{S})}(\mathbf{1}, F)
\\
&\to \hom\_{\mathbf{Fun}(\Delta\_{\le n},\mathcal{S})}(u^\*\mathbf{1}, u^\*F)
\\
&\cong \hom\_{\mathbf{Fun}(\Delta\_{\le n},\mathcal{S})}(\mathbf{1},
u^\*F)
\\
&\cong \lim u^\*F
\end{align\*}
Transposing, we conclude that $\hom\_{\mathbf{Fun}(\Delta,\mathcal{S})}(\mathbf{1}, F) \to \hom\_{\mathbf{Fun}(\Delta,\mathcal{S})}(u\_!u^\*\mathbf{1}, F)$ is an equivalence and, consequently, that $u\_!u^\* \mathbf{1} \to \mathbf{1}$ is orthogonal to any $(n-1)$-truncated object in $\mathbf{Fun}(\Delta, \mathcal{S})$. Accordingly, $u\_!u^\*\mathbf{1} = u\_!\mathbf{1}$ is $n$-connected. We therefore conclude that $(u\_!\mathbf{1})(m) : \mathcal{S}$ is $n$-connected for any $m : \Delta$. Let us unfold this using the formula for left Kan extensions:
$$
(u\_!\mathbf{1})(m) =
\mathrm{colim}\_{
\Delta\_{\le n}
\times\_{\Delta}
\Delta\_{/m}
} \mathbf{1}
= L(\Delta\_{\le n} \times\_{\Delta} \Delta\_{/m})
$$
The conclusion then follows.
Incidentally, this is just slight alteration of the usual proof of one direction of Quillen's theorem A. The other direction (that suitably connective fibers implies $n$-cofinality) also [holds](https://mathoverflow.net/questions/236578/difficulties-with-descent-data-as-homotopy-limit-of-image-of-%c4%8cech-nerve/236600#236600).
| 2 | https://mathoverflow.net/users/76636 | 451158 | 181,456 |
https://mathoverflow.net/questions/451153 | 3 | $\quad$Let $\mathcal{Sch}/\mathbb C$ denote the category of schemes over $\mathbb C$. For an arbitrary $X\in\mathcal{Ob}(\mathcal{Sch}/\mathbb C)$, Deligne in his [Article](http://publications.ias.edu/sites/default/files/Number19.pdf) defined a polarizable Hodge complex by replacing $X$ by a smooth simplicial scheme $\tilde{X}\_\bullet$, and then compactifying by $\tilde{X}\_\bullet\rightarrow \bar{\tilde{X}}\_\bullet$, such that $\tilde{D}=\bar{\tilde{X}}\_\bullet\setminus\tilde{X}\_\bullet$ is a simplicial normal crossing divisor. Deligne’s construction then furnished a cosimplicial polarizable Hodge complex, and by taking the corresponding simple complex, one obtains a polarizable Hodge complex $\underline{\mathrm{R}\Gamma}(X,A)\in \mathcal{Ob}(\mathrm{D}\_{\mathcal H^p}^b)$. Along this way, one can construct a variant with compact supports $\underline{\mathrm{R}\Gamma}\_c(X,A)\in \mathcal{Ob}(\mathrm{D}\_{\mathcal H^p}^b)$ for $X\in \mathcal{Ob}(\mathcal{Sch}\_\ast/\mathbb C)$ where $\mathcal{Sch}\_\ast/\mathbb C$ is category of schemes over $\mathbb C$ and proper morphism. For a closed subscheme $Y\subset X$ there is a distinguished triangle $$…\rightarrow\underline{\mathrm{R}\Gamma}\_c(X\setminus Y,A)\rightarrow \underline{\mathrm{R}\Gamma}\_c(X,A)\rightarrow \underline{\mathrm{R}\Gamma}\_c(Y,A)\rightarrow …$$$\quad$So now we take a further consider, is there a homological counterpart with compact supports and a Borel-Moore version with all standard functoriality properties leads to define the absolute Hodge cohomology of a scheme?
---
$\quad$For any question about the notations, please check [here(1)](https://i.stack.imgur.com/S3Y6K.jpg) [here(2)](https://i.stack.imgur.com/GHjDH.jpg) [here(3)](https://i.stack.imgur.com/MKSpa.jpg)
| https://mathoverflow.net/users/508433 | Geometric Interpretation of absolute Hodge cohomology | In fairness, some of the credit for the general construction of absolute cohomology should go to Beilinson (see his *Notes on Absolute cohomology*).
I think you'll find the homology version, that you are asking about, discussed in Janssen's article, *Deligne homology, Hodge D-conjecture, and motives*.
| 4 | https://mathoverflow.net/users/4144 | 451159 | 181,457 |
https://mathoverflow.net/questions/451161 | 4 | Let $j: K\to K′$ be a categorical equivalence of simplicial sets. By [HTT, Remark 2.1.4.11], we have a Quillen equivalence (with the covariant model structures)
$$
j\_!:\mathsf{sSet}\_{/K}\rightleftarrows\mathsf{sSet}\_{/K'}:j^\*.
$$
On the other hand, we have the functor $\mathbf{\Delta}\to\mathsf{sSet}, [n]\mapsto\Delta^n$, thus a functor $\mathbf{\Delta}\_{/K}\to\mathbf{\Delta}\_{/K'}$ on categories of simplices.
I want to know: is this functor also a categorical equivalence (i.e., equivalence of ordinary categories)?
The initial motivation for asking this is: we have $\operatorname{colim}((\mathbf{\Delta}\_{/K})^{\rm op}\xrightarrow{\*}\mathcal{S})\simeq\operatorname{colim}\_{[n]\in\mathbf{\Delta}^{\rm op}}K\_n$, viewing each $K\_n$ as a discrete/$0$-truncated object in $\mathcal{S}$ (I think, by Bousfield-Kan). I want to see if this can also be identified with $\operatorname{colim}(K\xrightarrow{\*}\mathcal{S})$ in a canonical way (in the $\infty$-category of spaces $\mathcal{S}$) so that $\operatorname{colim}\_{[n]\in\mathbf{\Delta}^{\rm op}}K\_n$ in $\mathcal{S}$ depends on $K$ only up to categorical equivalence, which I can confirm when $K$ is an $\infty$-category.
(\* means constant functor with value \*.)
| https://mathoverflow.net/users/42571 | Categorical equivalences vs. categories of simplices | No it is not. But there is a replacement for this. There is another model of $\infty$-categories: marked simplicial sets. If $X$ is a simplicial set with a set of marked $1$-simplices $S$, the fibrant replacement of $(X,S)$ is a model of the localization by $S$ of the $\infty$-category corresponding to $X$ (as a mere simplicial set). There is always a canonical map $\tau\_K:\Delta\_{/K}\to K$ sending an $n$-simplex over $K$ to its value at $n$. We can mark those $1$-simplices of $\Delta\_{/K}$ that are sent to the identites in $K$. Then the map $\tau\_K:\Delta\_{/K}\to K$ is an equivalence in the model structure of marked simplicial sets (where the marked simplices in $K$ are the identities). By a 2 out of 3 property, a morphism of simplicial sets $K\to K'$ is a categorical equivalence if and only if the induced functor $\Delta\_{/K}\to\Delta\_{/K'}$ is a weak equivalence of marked simplicial sets. The main properties of the comparison map $\tau\_K:\Delta\_{/K}\to K$ needed to prove what I claim above are proved in [Kerodon](https://kerodon.net/tag/01NC) or in my book [Higher categories and homological algebra](https://cisinski.app.uni-regensburg.de/CatLR.pdf) (Prop. 7.3.15) for instance. The model structure on marked simplicial sets is discussed in HTT, of course.
| 13 | https://mathoverflow.net/users/1017 | 451164 | 181,459 |
https://mathoverflow.net/questions/451079 | 3 | **Background:** If $\mathcal{H}$ is a Hilbert space and $U:\mathcal{H}\rightarrow\mathcal{H}$ is a unitary operator, then for each $f \in \mathcal{H}$ the sequence $(\langle U^nf,f\rangle)\_{n = 1}^\infty$ is positive definite, so it can be viewed as the Fourier coefficients of some measure $\mu\_f$ on the torus $\mathbb{T}$. If $f,g \in \mathcal{H}$ are such that $\mu\_f\perp\mu\_g$, i.e., $\mu\_f$ and $\mu\_g$ are mutually singular, then $\langle f,g\rangle = 0$. The most prominent example of this is when $\mu\_f$ is a discrete measure and $\mu\_g$ is a continuous measure, as this corresponds to the compact-weak mixing decomposition of $\mathcal{H}$.
My goal is to find a proof or reference of the generalization of this fact to the setting of second countable locally compact type I groups $G$. In particular, we recall that for such groups $G$, the dual space $\widehat{G}$ of unitary equivalence classes of irreducible representations of $G$ has the structure of a standard measure space with respect to the Mackey-Borel structure. Now if $\pi$ is a representation of $G$ on some separable Hilbert space $\mathcal{H}$, then there exists a measure $\mu$ on $\widehat{G}$ so that we can write $\pi = \int^{\oplus}\_{\widehat{G}}\pi\_pd\mu(p)$. This is essentially Theorem 7.40 in 'A course in abstract harmonic analysis' by G. Folland, or Theorem 8.6.6 in '$C^\*$-Algebras' by J. Dixmier. We say that a set $A \subseteq \widehat{G}$ *carries* $\pi$ if $\mu(\widehat{G}\setminus A) = 0$ (Definition 8.6.9 from Dixmier).
In conclusion, I would like a proof or a reference for the following conjecture.
**Conjecture:** Let $\pi$ be a representation of $G$ on $\mathcal{H}$, and let $A$ and $B$ be closed subspaces of $\mathcal{H}$ that are invariant under the action of $\pi$. If $\pi|\_A$ and $\pi|\_B$ are carried by disjoint sets, then $A\perp B$.
P.S. This feels like it should be an easy consequence of the uniqueness portion of Theorem 7.40 of Folland's book, but I have been having difficulty making the details work.
| https://mathoverflow.net/users/508836 | Spectral disjointness of unitary representations of Type I groups and orthogonality | A friend provided an answer to my question in a private communication. I have added some details and will present it below in my own words.
**Additional notation:** Below I use $\mu\_{\pi}$ to refer to the measure $\mu$ from my initial post with respect to a particular representation $\pi$. I also use $[\mu]$ to denote the equivalence class of measures that are mutually absolutely continuous with $\mu$.
Let us assume for the sake of contradiction that $A\not\perp B$ and let $P:A\rightarrow B$ denote the unique orthogonal projection. Since $A$ and $B$ are $\pi$-invariant, we see that for any $g \in G$, the map $P\_g := \pi(g)P\pi(g)^{-1}:A\rightarrow B$ is also an orthogonal projection, so $P$ and $\pi$ commute. Letting $A = \text{ker}(P)\oplus N\_1$ and $B = \text{Im}(P)\oplus N\_2$, we see that $P:N\_1\rightarrow N\_2$ is a bijection between two $\pi$-invariant subspaces. Recalling that $P$ and $\pi$ commute, one can easily verify that $P':N\_1\rightarrow N\_2$ given by $P'f = \frac{||f||}{||Pf||}Pf$ is a unitary operator that commutes with $\pi$, so $\pi\_1 := \pi|\_{N\_1}$ and $\pi\_2 := \pi|\_{N\_2}$ are unitarily equivalent representations of $G$, so $[\mu\_{\pi\_1}] = [\mu\_{\pi\_2}]$. Letting $\pi\_k := \pi|\_{\text{ker}(P)}$ and $\pi\_I := \pi|\_{\text{Im}(P)}$, we see that $[\mu\_{f\_1}] = [\mu\_{\pi\_k}]+[\mu\_{\pi\_1}]$ and $[\mu\_{f\_2}] = [\mu\_{\pi\_I}]+[\mu\_{\pi\_2}]$, which yields the desired contradiction.
| 0 | https://mathoverflow.net/users/508836 | 451172 | 181,460 |
https://mathoverflow.net/questions/451004 | 5 | Let $a>0$ and consider the operator
$$Tf(t,x)= \int\_{\mathbb{R}^{n}}e^{ i x\cdot \xi} e^{i t \lvert\xi\rvert^{a}} \widehat{f}(\xi) \, d\xi.$$
When $a=2$, the function $Tf$ solves the Cauchy problem associated with the homogeneous Schrödinger equation:
$$i \, \partial\_t u+i \, \Delta u=0,\qquad u(0)=f.$$
A classical problem is to study the pointwise convergence of the solution $u(t,x)$ to $f(x)$ for almost every $x$ as $t\to 0$. To this end, it is useful to study the estimate
$$\lVert Sf\rVert\_q \leq C \lVert f\rVert\_{H^s}, \tag1\label{1}$$
where
$$Sf(x)=\sup\_{0<t<1}\lvert Tf(t,x)\rvert$$
and $H^s$ is the usual Sobolev space of functions $f\in \mathcal{S}^{\prime}$ such that $\lVert(1+\lvert\xi\rvert^s)\widehat{f}\rVert\_2<\infty$.
My question:
I have seen this step in many papers on this problem and I can not make it rigorous:
To prove \eqref{1}, it suffices to prove that
$$\lVert Rf\rVert\_q\leq C \lVert f\rVert\_{H^s}, \tag2\label{2}$$
where
$$Rf(x)= \int\_{\mathbb{R}^n}e^{ i x\cdot \xi} e^{i t(x) \lvert \xi\rvert^a} \widehat{f}(\xi) \, d\xi.$$
and $t:\mathbb{R}^n\to (0,1)$ is a measurable function. Is the pointwise relation
$\lvert Sf(x)\rvert\leq C \lvert Rf(x)\rvert$ true and why ?
Remark: I have just found an [answer](https://mathoverflow.net/a/136758) to a very closely related question [To give an estimate for the maximal function associated to the Schrödinger group by using a measurable selector function](https://mathoverflow.net/questions/135143/to-give-an-estimate-for-the-maximal-function-associated-to-the-schr%c3%b6dinger-group).
It is very interesting. It treats the estimate \eqref{1} with the $L^p(\mathbb{R}^{n})$ norm on the left side replaced by a an $L^p(B)$ where $B$ is a ball. I will try to adapt the solution to the global estimate.
| https://mathoverflow.net/users/116555 | Maximal operator estimates for the Schrödinger equation | $\newcommand\dotcup{\mathbin{\dot\cup}}$I shared your confusion until a year ago, when Markus Haase explained to me that \eqref{2} can actually be rewritten as assertion (ii) in the following theorem, which I find much clearer.
**Theorem.** Let E be a Banach space, let $(\Omega,\mu)$ be a measure space, let $p \in [1,\infty)$ and let $I$ be a non-empty index set. For each $i \in I$ let $T\_i: E \to L^p := L^p(\Omega,\mu)$ be a bounded linear operator. The following are equivalent:
(i) The operator family $(T\_i)\_{i \in I}$ satisfies a maximal inequality, i.e., there is a constant $C \ge 0$ with the following property: for every $f \in E$ there exists $0 \le h \in L^p$ of norm $\lVert h\rVert\_{L^p} \le C \lVert f\rVert\_{E}$ such that $\lvert T\_if\rvert \le h$ for all $i \in I$ (where the inequality is to be understood pointwise almost everywhere).
(ii) There exists a constant $C$ with the following property:
for every finite measurable partition $\Omega = A\_1 \dotcup \dotsb \dotcup A\_n$, all indices $i\_1, \dotsc, i\_n \in I$, and every $f \in E$ one has $\Bigl\lVert \sum\_{k=1}^n 1\_{A\_k} T\_{i\_k} f \Bigr\rVert\_{L^p} \le C \lVert f\rVert\_{E}$.
*Proof.*
"(i) $\Rightarrow$ (ii)"
This implication is straighforward to check.
"(ii) $\Rightarrow$ (i)"
Let $\mathcal{F}$ denote the set of all non-empty finite subsets of $I$, which is directed with respect to set inclusion.
Fix $f \in E$.
For each $F \in \mathcal{F}$ the vector $0 \le \bigvee\_{i \in F} \lvert T\_if\rvert \in L^p$ (where $\lvert\cdot\rvert$ denotes the pointwise almost everywhere modulus and $\bigvee$ denotes the pointwise almost everywhere supremum) is norm bounded by $C\lVert f\rVert\_{E}$. To see this, enumerate the elements of $F$ as $i\_1, \dotsc, i\_n$ and choose a measurable partition of $\Omega$ into sets $A\_1, \dotsc, A\_n$ such that
$$
\bigvee\_{i \in F} \lvert T\_if\rvert
=
\sum\_{k=1}^n 1\_{A\_k} \lvert T\_{i\_k} f\rvert
=
\Bigl\lVert \sum\_{k=1}^n 1\_{A\_k} T\_{i\_k} f \Bigr\rVert
.
$$
Since the net $\bigl(\bigvee\_{i \in F} |T\_if|\bigr)\_{F \in \mathcal{F}}$ in $L^p$ is increasing and norm bounded by $C\lVert f\rVert\_{E}$, it is norm convergent to a vector $0 \le h \in L^p$ of norm $\le C\lVert f\rVert\_{E}$ (here was used that $p \in [1,\infty)$).
Clearly, $h$ satisfies the property claimed in (i). $\quad \square$
**Remarks.**
(a) Assertion (i) is a convenient way to write down a maximal inequality if the index set $I$ is not assumed to be countable and if one does not make any regularity assumptions regarding the dependence of $T\_i$ on $i$. Writing down the maximal operator before knowing whether a maximal inequality holds leads to measurability problems in this case. (But once the maximal inequality is established, one can define the maximal operator by using the supremum within the ordered space $L^p$ rather than the almost everywhere supremum.)
In cases where the maximal operator can be reasonably defined as a pointwise supremum assertion (i) is equivalent to the $L^p$-boundedness of the maximal operator.
(b) In the situation of the question, assertion (ii) of the theorem is precisely the estimate \eqref{2} for simple functions $t: \mathbb{R}^n \to (0,1)$. So this shows, in particular, that it suffices to consider simple functions in \eqref{2}.
This does not only resolve all measurability issues regarding $t$; rather, in the situation of the theorem such a measurability is not even defined since the index set $I$ is not assumed to carry a $\sigma$-algebra.
In the same vein, note that assertion (ii) makes sense in this general setting, while for an object such as $T\_{i(x)}f(x)$ for a non-simple function $i: \Omega \to I$ it is not even clear how to define it if the $T\_i$ do not have any particular structure.
(c) The measure space $(\Omega, \mu)$ is not assumed to be $\sigma$-finite. In fact, the proof shows that the theorem does not really rely on the structure of $L^p$-spaces, but stays true on Banach lattices where every increasing norm bounded net is norm convergent (these are the so-called *KB-spaces*, of which $L^p$ for $p \in [1,\infty)$ are examples).
To make sense of (ii) one needs to replace the multiplication with indicator functions with band projections in this abstract setting.
(d) The proof shows that $C$ can be chosen as the same number in (i) and (ii).
(e) Fun fact: The existence of $C$ is actually redundant in (i).
Indeed, if one only assumes that for each $f \in E$ there exists $0 \le h \in L^p$ such that $\lvert T\_i f\rvert \le h$ for all $i \in I$, then there automatically exists a number $C \ge 0$ such that $h$ can always be chosen to satisfy $\lVert h\rVert\_{L^p} \le C \lVert f\rVert\_{E}$.
This can be shown by means of the closed graph theorem, see for instance (warning: shameless self-promotion ahead) Theorem 4.1 in the preprint [Order boundedness and order continuity properties of positive operator semigroups](https://arxiv.org/abs/2212.00076v1) with Michael Kaplin.
| 3 | https://mathoverflow.net/users/102946 | 451174 | 181,461 |
https://mathoverflow.net/questions/450895 | 2 | Suppose $X\_1,\ldots,X\_n\sim \text{i.i.d.} \operatorname N(\mu,\sigma^2).$ Consider these functions of the foregoing random variables:
* Any weighted average of $X\_1,\ldots,X\_n,$ with constant (i.e. non-random) weights not depending on $\mu$ or $\sigma^2.$
* $\operatorname{median}\{X\_1,\ldots,X\_n\}.$ (If $n$ is even, take this to be the mean of the two middle observations.)
* $(\max\{X\_1,\ldots,X\_n\} + \min\{X\_1,\ldots,X\_n\})/2$
* $\operatorname{mean} \{X\_i : i\in I\},$ where $I\subset\{1,\ldots,n\}$ is randomly chosen from among all $2^n-1$ non-empty subsets, in a way that is independent of the values of $X\_1,\ldots,X\_n.$
* $\operatorname{mean} \{X\_i : i\in I\},$ where $I\subset\{1,\ldots,n\}$ is randomly chosen from among all $2^n$ subsets, in a way that does depend on $X\_1,\ldots,X\_n$ but for which this mean nonetheless has the properties stated below. (The third item in this list is a special case of this.)
* $\operatorname{median}\{X\_i : i\in I \}$ with $I$ as in either of the last two situations above.
All of these functions of $X\_1,\ldots,X\_n$ have the following properties:
* The value of each can be found without knowing the values of $\mu$ or $\sigma^2,$ i.e. each of these is a "statistic."
* Each has expected value $\mu$ regardless of the values of $\mu$ and $\sigma^2,$ i.e. each is an unbiased estimator of $\mu.$
**My question is** whether there is a way to classify all such functions. Can we describe some set of functions without mentioning anything about $\{\operatorname N(\mu,\sigma^2) : \mu\in\mathbb R, \sigma^2\in[0,+\infty)\}$ or any other family of probability distributions and show that those and only those have the properties stated in the last two bullet points above?
| https://mathoverflow.net/users/6316 | Classification of all unbiased estimators of the mean of a normal population | $\newcommand{\R}{\mathbb R}\newcommand{\si}{\sigma} $At least in your setting, statistics (also called estimators) are random variables of the form $T(X)$, where $T$ is a Borel-measurable function from $\R^n$ to $\R$ and $X:=(X\_1,\ldots,X\_n)$.
An estimator $T(X)$ of $\mu$ is called unbiased (for $\mu$) if
\begin{equation}
E\_{\mu,\si^2}T(X):=\int\_{\R^n}dx\,T(x)f\_{\mu,\si^2}(x)=\mu
\end{equation}
for all $(\mu,\si^2)\in\R\times(0,\infty)$, where $f\_{\mu,\si^2}$ is the (joint) density of $X$, so that
\begin{equation}
f\_{\mu,\si^2}(x)=(2\pi\si^2)^{-n/2}\exp-\sum\_1^n\frac{(x\_i-\mu)^2}{2\si^2}
\end{equation}
for $x:=(x\_1,\ldots,x\_n)\in\R^n$.
E.g.,
\begin{equation}
\bar X:=\frac1n\,\sum\_1^n X\_i
\end{equation}
is unbiased (for $\mu$).
So, $T(X)$ is unbiased for $\mu$ iff $E\_{\mu,\si^2}(T(X)-\bar X)=0$ for all $(\mu,\si^2)\in\R\times(0,\infty)$ -- that is, iff the function $\R^n\ni x=(x\_1,\ldots,x\_n)\mapsto\bar x:=T(x)-\frac1n\,\sum\_1^n x\_i\in\R$ is orthogonal in $L^2(\R)$ to all the normal density functions $f\_{\mu,\si^2}$, for all $(\mu,\si^2)\in\R\times(0,\infty)$.
This rather trivial characterization of unbiased estimators of $\mu$ for $N(\mu,\si^2)$ is apparently the least non-tautological characterization possible.
Indeed, the most comprehensive study of unbiased estimators is apparently the two-volume [book by Voinov and Nikulin](https://books.google.com/books?id=xgH1CAAAQBAJ&printsec=copyright#v=onepage&q&f=false). There is no non-tautological characterization of all unbiased estimators of any function of the parameter of **any** parametric family of distributions there.
In particular, in the long table of unbiased estimators of functions of the parameter $(\mu,\si^2)$ of the family of normal distributions, the only unbiased estimator of $\mu$ listed there is $\bar X$.
Indeed, all the other unbiased estimator of $\mu$ for the normal family do not really matter. Indeed, by the completeness, $\bar X$ is almost surely (a.s.) the only unbiased estimator of $\mu$ that is a function of a minimal sufficient statistic. So, by the [Blackwell--Rao](https://en.wikipedia.org/wiki/Rao%E2%80%93Blackwell_theorem)--[Lehmann–Scheffé](https://en.wikipedia.org/wiki/Rao%E2%80%93Blackwell_theorem#Completeness_and_Lehmann%E2%80%93Scheff%C3%A9_minimum_variance) theory, $\bar X$ is the (a.s. unique) uniformly minimum variance unbiased estimator of $\mu$; actually, $\bar X$ is the best unbiased estimator of $\mu$ with respect to any convex [loss function](https://en.wikipedia.org/wiki/Loss_function).
| 2 | https://mathoverflow.net/users/36721 | 451188 | 181,463 |
https://mathoverflow.net/questions/451186 | 6 | For context, I heard that the Atiyah-Segal formalism of quantum field theory is equivalent to the traditional Wightman approach (at least within the scope of certain theories). However, I could not find a precise proof of this, only physical arguments.
| https://mathoverflow.net/users/164131 | Is Segal's notion of conformal field theory a quantum field theory in the sense of Wightman axioms? | My understanding is that Segal invented his formalism (which was then adapted by Atiyah) by thinking about the same thing Wightman was thinking about: formalising the theory of local operators. In hindsight, the theory Segal was groping towards was essentially the theory of [factorization algebra](https://www.cambridge.org/core/books/factorization-algebras-in-quantum-field-theory/9597AFE8E8767F8F38A73C74B8F2501B) of Costello and Gwilliam. Segal was unable to come up with the Weiss locality axiom in factorization algebra: his approach with cobordisms was the best locality he could think of. In hindsight, Segal was extremely prescient, as functorial/cobordism field theory has proved more powerful than you would be able to tell from 2D examples. For example, functorial field theory can organically talk about extended operators, whereas factorization algebra, as currently defined, has trouble axiomatizing them.
For comparison, factorization algebras as we currently know them evolved out of [Beilinson–Drinfeld](https://www.ams.org/publications/authors/books/postpub/coll-51)'s work on conformal field theory chiral algebras, which globalize the vertex operator algebras first defined mathematically by [Frenkel–Lepowsky–Meurman](https://shop.elsevier.com/books/vertex-operator-algebras-and-the-monster/frenkel/978-0-12-267065-7) and [Borcherds](https://www.pnas.org/doi/10.1073/pnas.83.10.3068). These are, in turn, essentially Wightman QFTs, implemented for 2D Euclidean-signature conformal field theory.
At the technical level, Segal formalism and Wightman formalism are different. They have different locality conditions, different analytic conditions, and different algebraic conditions. But "formalism" is a vague term. One can, and should, consider ways to adapt formalisms to meet examples. Formalisms are as much about worldviews and mindsets as they are about definitions.
As far as I am aware, in spacetime dimension $>2$, there is no real comparison between these formalisms, except perhaps in the topological case where the comparison becomes much more formal. One issue is a lack of understanding of even what either formalism should be in the nontopological case. Bare categories are not a good setting in which to describe the operator products of nontopological extended operators, because categories are inherently "topological" in the sense that their composition laws are associative and do not depend on the geometry of the pasting/composition diagrams. The Wightman axioms as they exist today only describe the composition of local operators, not extended operators, and these are insufficient to completely describe the quantum field theory. On the cobordism side, it is a technically hard challenge to marry geometry with unitality. Some of these challenges were resolved by [Grady–Pavlov](https://arxiv.org/abs/2111.01095), but I find their geometric cobordism category not to accommodate the examples I would consider vital. (Not that every formalism must accommodate every example!) Most notably, Grady–Pavlov do still have lots of adjunctibility/dualizability built into their theory, and this is hard to justify on a moral level.
In the special case of 2D conformal field theory, where all the formalisms are under the best analytic and algebraic control (well, 1D is even better: the formalisms are all under complete control and there is nothing formal left to do), one can hope to establish a collection of adjectives to add to both formalisms so that all the examples you want enjoy those adjectives, and so that those adjectives are enough to make the two formalisms equivalent. I believe that [Tenner](https://tener.cc/research.html) has done the most meaningful work in this direction.
I mentioned parenthetically that the 1D case is fully controlled. This is a very good example to think through carefully. Stated more strongly, it is a very good example for *you*, gentle reader, whoever you are, to think through carefully. What is a 1D Segal QFT? (What is a 1D modular functor?) What is a 1D Wightman QFT? How do these compare to the textbook pictures with names like "Schrödinger" and "Heisenberg"? What role does functional analysis play? (What role does the Stone–von Neumann theorem play?) What role does Wick rotation play? And so on. Will you learn something new about quantum mechanics from doing this? Probably not, depending on how much you already "know". But still you, and I really mean *you*, should think through this carefully: your conclusions might not match someone else's (for example, me), and there is a value to having diverse conclusions.
| 9 | https://mathoverflow.net/users/78 | 451189 | 181,464 |
https://mathoverflow.net/questions/451086 | 5 | $\DeclareMathOperator\SO{SO}$Recently I came across this old [MSE post](https://math.stackexchange.com/questions/4267430/measuring-distance-in-son) or [this paper (w.o. proof)](https://epubs.siam.org/doi/10.1137/S0895479801383877) discussing the geodesic distance on $\SO(n)$ when it is equipped with the left-invariant Riemannian metric given by $g\_{I\_n}(X,Y):=-\operatorname{tr}(XY)$ for $X,Y\in T\_{I\_n}\SO(n)\leftrightarrow \mathfrak{so}(n)$. The post states that, the *length/geodesic metric* induced by $g$ is
$$
d(A,B)^2:= \sum\_{i=1}^d \theta\_i^2
$$
where $\theta\_i=\ln(\lambda\_{i}(A^{\top}B))$, for any $A,B\in \SO(n)$, where $\{\lambda\_{i}(A^{\top}B)\}\_{i=1}^d$ are the eigenvalues of $A^{\top}B$. Where can I find a proof, or a reference, to this fact?
| https://mathoverflow.net/users/491352 | Geodesic distance on $\mathrm{SO}(n)$ | The formula holds with a minus sign on the RHS and only for small enough distances. Here's a short proof: by invariance you may set $A = I\_n$. Then for $B$ close enough to $I\_n$, the unit-speed minimising geodesic from $I\_n$ to $B$ is a one-parameter subgroup $\gamma(s) = \exp(s \Xi)$, for some $\Xi \in \mathfrak{so}(n)$ with ${\rm tr}\, \Xi^2 = -1$. We thus have $B = \exp(d \Xi)$, where $d = d(I\_n, B)$, and the eigenvalues satisfy $\theta\_i = \ln \lambda\_i = d \xi\_i$, where $\xi\_i$ are the eigenvalues of $\Xi$. Your formula follows since $\sum \xi\_i^2 = -1$. (Notice the minus sign on the RHS.)
| 4 | https://mathoverflow.net/users/14708 | 451196 | 181,466 |
https://mathoverflow.net/questions/451191 | 4 | Question:
Are there any well known actions of braid groups on trees? For example is there some action of a braid group $ B\_n $ on a $ p $ regular tree for some $ p $ such that the action is transitive on vertices and also transitive on directed edges?
Context:
We often understand a group by its actions. For a finite group or compact group, the theory of its unitary representations is very well behaved. Even for a general reductive algebraic group its representation theory is well behaved.
For finite groups we can also learn things by thinking of the finite group acting as automorphisms of a graph.
I'm not used to working with infinite groups that aren't algebraic groups over some field. Braid groups are an example of infinite groups that aren't algebraic groups over any field. Although there is a theory of unitary representations of Braid groups (e.g. certain values of the Burau representation and Lawrence–Krammer representation) the image of the Braid group is not a closed subgroup of the unitary group and so it is a bit beyond my intuition.
Since the Braid group is an infinite "discrete" group in my intuition it might be related to the automorphisms of an infinite tree.
Cross-posted from MSE <https://math.stackexchange.com/questions/4729118/action-of-braid-groups-on-regular-trees>
| https://mathoverflow.net/users/387190 | Action of braid groups on regular trees | In the article *A group theoretic criterion for property FA*, Culler and Vogtmann notice that
>
> for $n \geq 5$, the braid group $B\_n$ has property $A \mathbb{R}$,
>
>
>
meaning that every *non-trivial* (i.e. without a global fixed point) action of $B\_n$ on a tree has an invariant bi-infinite geodesic. As a consequence, the only regular tree on which $B\_n$ can act transitively is the line. The proof given in the paper, available [online](https://www.ams.org/journals/proc/1996-124-03/S0002-9939-96-03217-0/S0002-9939-96-03217-0.pdf), is short and elementary.
**For n=3:** About $B\_3= \langle \sigma\_1, \sigma\_2 \mid \sigma\_1 \sigma\_2 \sigma\_1 = \sigma\_2 \sigma\_1 \sigma\_2 \rangle$, one can notice that the relation is equivalent to $(\sigma\_1 \sigma\_2)^3= (\sigma\_1 \sigma\_2 \sigma\_1)^2$. Thus, setting $a:=\sigma\_1 \sigma\_2$ and $b:= \sigma\_1 \sigma\_2 \sigma\_1$, one finds the alternative presentation $\langle a,b \mid a^3=b^2 \rangle$ of $B\_3$. This implies that $B\_3$ is a non-trivial amalgamated sum $\mathbb{Z} \underset{\mathbb{Z}}{\ast} \mathbb{Z}$. As such, it admits an edge-transitive action on a tree.
**For n=4:** As mentioned by Ishan Banerjee in the comments, there is a surjective morphism $B\_4 \twoheadrightarrow B\_3$, so the previous action of $B\_3$ on a tree yields a similar action of $B\_4$ (with an infinite kernel though). The morphism consists in taking the presentation
$$\langle \sigma\_1, \sigma\_2, \sigma\_3 \mid [\sigma\_1, \sigma\_3]=1,\ \sigma\_1 \sigma\_2 \sigma\_1 = \sigma\_2 \sigma\_1 \sigma\_2,\ \sigma\_2 \sigma\_3 \sigma\_2 = \sigma\_3 \sigma\_2 \sigma\_3 \rangle$$
of $B\_4$ and setting $\sigma\_1= \sigma\_3$.
| 12 | https://mathoverflow.net/users/122026 | 451204 | 181,467 |
https://mathoverflow.net/questions/451201 | 0 | Let $R=k[x\_1,...,x\_n]$ be a polynomial ring over a field. Let $f$ be a homogeneous polynomial and $I=(f\_1,...,f\_m)$ a homogeneous ideal.
With Macaulay2, one can compute the Groebner basis of $I$ when $k=\mathbb{Q}$ or when $k=\mathbb{F}\_p$. This helps one decide whether $f$ is in $I$ or not. This is of course just a rephrase of the ideal membership problem.
**Question 1:** Suppose we know with M2 that $f\in I$ when $k$ is the field of rationals or the finite field of prime cardinality. Is it true that $f\in I$ when $k$ is a different field?
*Naive partial answer:* Sometimes. Say we run M2 for $k=\mathbb{Q}$ and obtain $f\in I$. Then we can then compute the presentation:
$$
f=g\_1f\_1+\cdots+g\_mf\_m.
$$
Suppose all the coefficients of $g\_1,...,g\_m$ are integers. Then we can conclude that $f\in I$ no matter what $k$ is because the integers transfer nicely to other fields. On the other hand, suppose all the coefficients of $g\_1,...,g\_m$ are (integer) multiples of $1/2$. Then we can conclude that $f\in I$ for any $k$ as long as $char(k)\neq 2$.
Here is a variation of Ques 1:
**Question 2:** Suppose we know with M2 that $f\notin I$ when $k$ is the field of rationals or the finite field of prime cardinality. Is it true that $f\notin I$ when $k$ is a different field?
For this one, I do not even have a naive answer. In fact, I'm not even sure how to approach it. My naive way would be assuming that
$$
f=g\_1f\_1+\cdots + g\_mf\_m.
$$
Then obtain polynomial equations and try to find a contradiction. This should work well if $f,f\_1,...,f\_m$ are of the same degree, and it gets harder when there is a big difference between the degrees.
**Question:** Are there other approaches for these problems?
The examples I'm working on: Let $X,Y$ be square generic matrices of the same size. COnsider the ideal $I=I\_1(XY-YX)$, i.e., $I$ is generated by all entries of $XY-YX$. Let $f$ be a $n$-minor of the matrix concatenation $(X|Y)$, for some $n$. Is it true that $f\in I$?
Of course, with brute force, one may determine this when $n=2$. When $n$ is larger, M2 can show you some answers. But the questions come up: how about when $k$ is a field where you can't run on M2?
| https://mathoverflow.net/users/103381 | Ideal membership and change of fields | The ideal membership question is completely algorithmic, and indeed you can solve is using the reduced Gröbner basis of your ideal (for any ordering of your choice).
The reduced Gröbner basis of $(f\_1,\ldots,f\_m)$ is computed via the Buchberger algorithm, and it is clear from the way the algorithm works, that everything happens over the subfield of $k$ generated by the coefficients of $f\_1,\ldots,f\_m$.
I am assuming that the coefficients of $f\_1,\ldots,f\_m, f$ are integers, since otherwise your question does not make much sense to begin with. In that case, what said above ensures that the reduced Gröbner basis is defined over the prime subfield of $k$, so it is indeed enough to figure it out over $\mathbb{Q}$ and all $\mathbb{F}\_p$ to conclude the result for any field.
| 2 | https://mathoverflow.net/users/1306 | 451205 | 181,468 |
https://mathoverflow.net/questions/451202 | 4 | Consider the number of integer partitions of $n$, usually denoted by $p(n)$ and generated by
$$\sum\_{n\geq0}p(n)x^n=\prod\_{k\geq1}\frac1{1-x^k}.$$
I have encountered an interesting enumeration.
Take all partitions of $n+2$ which do not use $1$ as a part. Then, count all different integers appearing in every such partition, call this number $u(n)$.
For example, say $n=4$ and hence $\{6,51,42,411,33,321,3111,222,2211,21111,111111\}$ are the partitions of $n+2=6$. Of these, $\{6,42,33,222\}$ avoid the integer $1$. The different integers here are $6,4,2,3,2$ for a total of $u(4)=5$.
>
> **QUESTION.** Is it true that $u(n)=p(n)$ for all $n\geq1$?
>
>
>
| https://mathoverflow.net/users/66131 | Number of partitions of $n$ and number of different integers in 1-avoiding partitions | Note that $u(n)=|\Theta(n+2)|$ where $\Theta(k)$ is the set of partitions of $k$ without 1's with one part being labelled (i.e., multisets of integers greater than 1 summing up to $k$ with one labelled element, like $\{{\bf 6}\},\{{\bf 4},2\},\{{\bf 2},4\},\{{\bf 3},3\}, \{{\bf 2},2,2\}$ for $k=6$).
Bijection from partitions of $n$ to $\Theta(n+2)$.
Take a partition $\lambda$ of $n$, let it have $s$ parts equal to 1. Remove them and add a labelled part $s+2$.
| 7 | https://mathoverflow.net/users/4312 | 451212 | 181,474 |
https://mathoverflow.net/questions/451220 | 1 | Let $f(t) : [0,\infty) \to \mathbb{R}$ be an $L^1\_\text{loc}$ function.
Then, I wonder if the following series
\begin{equation}
\sum\_{n=1}^\infty e^{-n^2 T} \int\_0^T e^{n^2 t} \lvert f(t)\rvert
\, dt
\end{equation}
converges, where $T \in (0, \infty)$ is fixed.
This seems related to Dawson integral. Could anyone please help me?
Addition : I see that there can certainly exist $T \in (0,\infty)$ where the above series diverges. However, since $f$ is locally $L^1\_\text{loc}$ on whole $[0,\infty)$, I hope that the above series converges for a.e. $T$. Would this be possible at least?
| https://mathoverflow.net/users/56524 | Does $\sum_{n=1}^\infty e^{-n^2 T} \int_0^T e^{n^2 t} \lvert f(t)\rvert \, dt$ converge for $L^1_\text{loc}$ $f : [0,\infty) \to \mathbb{R}$? | The updated version is indeed correct. I will show that the sum is finite for almost all $T\in [1, 2]$ but it should be applicable for all $T\in (0, \infty)$.
The reason why the result can fail for a particular $T$ is that our function $f$ can resemble too closely $\delta$-function at the point $T$ so that we will be summing ones and get infinity. But in that case for (say) all smaller $T$ we will just have sum of zeroes and the sum will be finite. So, we have to somehow find balance between different $T$'s, and the natural way to do it is to introduce some averaging.
First, let us turn your sum into a more human-friendly form. By Fubini we have (I use $f(t)$ in place of $|f(t)|$ since we can always assume that $f$ is positive)
$$\sum\_{n = 1}^\infty e^{-n^2T}\int\_0^T e^{n^2t}f(t)dt = \int\_0^T f(t)\sum\_{n = 1}^\infty e^{n^2(t-T)}dt.$$
The sum that we have here is a (half of) the theta-function, and asymptotics of it is known: it is proportional to $\frac{1}{\sqrt{T-t}}$ (at least in the regime $0 < T-t < 1$ that we will consider). In particular, since it is not in $L^\infty$, we can find $f$ such that for a given $T$ the integral would be infinite. But now we will use many $T$'s at once by integrating this over $T\in [1, 2]$. I will also assume that $f(t) = 0, t < 1$ for simplicity, for $T > 1$ this does not affect convergence issues at all. We have
$$\int\_1^2 \int\_1^T \frac{f(t)}{\sqrt{T-t}}dtdT = \int\_1^2f(t)\int\_t^2\frac{1}{\sqrt{T-t}}dTdt = \int\_1^2f(t) 2\sqrt{2-t}dt \le \\2\int\_1^2 f(t)dt < \infty,$$
since $f(t)$ is in $L^1\_{loc}$. In particular, our sum is finite for almost all $T\in [1, 2]$.
Note that it is crucial that we have $n^2$ and not $n$, otherwise we would've had to integrate singularity of order $\frac{1}{T-t}$ which would give us unbounded $\log(T-t)$.
| 5 | https://mathoverflow.net/users/104330 | 451228 | 181,479 |
https://mathoverflow.net/questions/451219 | 1 | I'm looking for a reference and a proof of the following version (or eventually a more general version) of the Central Limit Theorem for complex random variables.
**Theorem.** *Let $Z\_1, Z\_2, \dots, Z\_n$ be independent identically distributed complex Goodman (see Note 1) random variables with mean $\mu$ and complex variance $\sigma^2$ (see Note 2). Then as $n \to \infty$
$$(\star)\quad\frac{\sum\_{i=1}^n Z\_i - \mu n}{(\sigma^2 n)^{1/2}}$$
converges in law to a standard complex normal random variable.*
**Note 1.** A complex random variable $Z = X + iY$ ($X, Y \in \mathbb{R}$) is *Goodman* if the covariance of $X$ and $Y$ is zero.
**Note 2.** The complex variance of a complex random variable $Z$ is defined as $\mathbf{E}\left[(Z - \mathbf{E}[Z])\overline{(Z - \mathbf{E}[Z])}\right]$.
The only references I was able to find are:
1. Theorem 7.3, Jonathan Shuster, Properties of the Gaussian Gaussian Distribution (this is an apparently unpublished B.Sc. thesis).
2. Exercise 2.2.14, Terrence Tao, Topics in Random Matrix Theory (this is an exercise without solution).
I would like a published reference, possibly a book, containing a proof.
In general, I'm a bit surprise of not finding books devoted to complex random variables. Perhaps this is due to the fact that many theorems on complex random variables follow easily from theorems on real random variables.
Furthermore, I don't understand why the condition on begin Goodman is needed ((1) provides a counter-example but I don't find it enlightening). Since ($\star$) involves only additive properties on the complex random variables, why one cannot deduce the Central Limit Theorem for complex random variables just be identifying $Z$ with the real vector $(X, Y)$ and using the Central Limit Theorem for vectors?
Thank you for the help.
| https://mathoverflow.net/users/503895 | Reference request and clarification for Central Limit Theorem for complex random variables | On the "Furthermore":
The central limit theorem for vectors involves the variance-covariance matrix. Let $Z = X+iY$ be a complex random variable (with mean $0$ for simplicity). If $X,Y$ have variance-covariance matrix:
$$\begin{pmatrix} \mathbf E[X^2] & \mathbf E[XY] \\ \mathbf E[XY] & \mathbf E [Y^2] \end{pmatrix} = \begin{pmatrix} a & b \\ b & c \end{pmatrix} $$
then the complex variance is $$\mathbb E [ (X+iY) (X- i Y) ]= \mathbb E[ X^2+Y^2] =a+c$$ so knowing the complex variance does not determine the variance-covariance matrix and thus doesn't determine the limiting distribution.
Your definition of Goodman is equivalent to $b=0$. Even under $b=0$ knowing $a+c$ does not determine the full matrix. The reference by Shuster gives a different definition of Goodman, which is equivalent to $b=0$ and $a=c$. It is clear from the matrix perspective why such an assumption is necessary (since every symmetric positive semidefinite matrix can in fact appear). (The reference by Tao doesn't use the notion of Goodman random variables, but has a similar assumption.)
| 3 | https://mathoverflow.net/users/18060 | 451229 | 181,480 |
https://mathoverflow.net/questions/451242 | 7 | A try to capture the informal notion of "this sentence is not provable in less than $n$-many steps" where $n$ is a concrete natural.
Use the diagonal lemma to coin the following sentences, per each concrete natural $n$:
$\zeta\_n \iff \ulcorner \zeta\_n \urcorner \text { is not provable in }< S\_n(0)\text{-many steps}$
Where $S\_n(0)$ is the $n^{th}$ successor of zero.
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If that can exist then neither $\zeta\_n$ nor $\neg \zeta\_n$ would be provable in the underlying theory if that theory is consistent. That is, it demonstrates incompleteness without the need of the assumption of $\omega$-consistency. That is, similar to the Rosser sentence in that respect.
| https://mathoverflow.net/users/95347 | Can this provide an example of incompleteness under the assumption of mere consistency? | No, quite the opposite in fact. Namely, if we have PA as the background theory, or some other minimal theory of arithmetic, and this theory is consistent, then all the sentences $\zeta\_n$ are true!
Let me assume that each $\zeta\_n$ asserts of itself that it has no proof in PA (or some other fixed sufficient background theory) in less than $n$ steps. Let us assume this counting is done in a systematic definite manner, perhaps simply by using Gödel codes, so that there are only finitely many such proofs for any given $n$.
The sentence $\zeta\_n$ cannot be provable in fewer than $n$ steps, since if it were, then we could prove this fact, that there was such a proof. But this would be proving the opposite of the assertion made by $\zeta\_n$ itself, and so we will have proved a contradiction, contrary to consistency.
Thus, no $\zeta\_n$ is provable by a short proof. And so every sentence $\zeta\_n$ is true.
But let me draw out a further consequence of this interesting situation. Precisely because there is in fact no short proof of $\zeta\_n$, this is something that admits of concrete proof in PA. We can simply enumerate all the proofs using fewer than $n$ steps, and prove of each of them that it is not a proof of $\zeta\_n$ (since in fact it is not). Thus, every $\zeta\_n$ is provable in PA, but only by a very long proof and not by any short proof.
Thus we have a speed-up phenomenon for proofs. Assuming consistency, no sentence $\zeta\_n$ has a short proof in PA, but they all have long proofs in PA. But they have short proofs in PA+Con(PA), because we just explained the argument above, and this can be formalized in PA.
I have recently noticed that one can use these observations to provide an alternative simple proof of the second incompleteness theorem (and so this restores somewhat your broader aim by another means). Namely, if PA proved Con(PA), then that proof would have some definite length. Consider $\zeta\_n$ where $n$ is much larger than this, and where $n$ has a comparatively short description. Above we showed how to provide a short proof from PA+Con(PA) that $\zeta\_n$ is true. By combining that argument with the given proof of Con(PA) in PA, we would obtain a short proof of $\zeta\_n$ in PA itself, which we have argued is impossible. So PA does not prove Con(PA).
I wrote on this in May 2023 in an essay on my substack, [Infinite Liars](https://www.infinitelymore.xyz/i/122078297/the-second-incompleteness-theorem). See also my [Twitter thread](https://twitter.com/JDHamkins/status/1654099992263503872) in which I describe the proof of the second incompletness theorem this way.
Meanwhile, the consideration of the sentences $\zeta\_n$ and the basic speed-up phenomenon seems to be due to:
* *Pudlák, Pavel*, On the length of proofs of finitistic consistency statements in first order theories, Logic colloq. ’84, Proc. Colloq., Manchester/U.K. 1984, Stud. Logic Found. Math. 120, 165-196 (1986). [ZBL0619.03037](https://zbmath.org/?q=an:0619.03037). [Citeseer pdf](https://citeseerx.ist.psu.edu/document?repid=rep1&type=pdf&doi=54952ff2ef51c3bea49b4dfaffd656a372f048db)
| 11 | https://mathoverflow.net/users/1946 | 451243 | 181,485 |
https://mathoverflow.net/questions/451252 | 18 | Let $G = \langle A \mid R \rangle$ be a finitely presented group, given by a finite presentation. If $G$ is abelian, then we can verify this fact: simply verify the fact that $[a, b] = 1$ for all generators $a, b \in A$ (this can be done by enumerating all words equal to $1$ in the group). That is, the property "being abelian" is a semi-decidable property (however, the property of "being non-abelian" is not semi-decidable, by the [Adian–Rabin theorem](https://en.wikipedia.org/wiki/Adian%E2%80%93Rabin_theorem), cf. also [this translation](https://arxiv.org/abs/2208.08560)). I recently bumped into the following question(s), but couldn't figure out any obvious way to answer either:
Is the property of being solvable semi-decidable? Is the property of **not** being solvable semi-decidable?
That is, is it possible to verify that a group is solvable? Is it possible to verify that a group is *not* solvable? The same questions above can also be asked for solvability of bounded derived length; for example, can one verify whether $G$ is a metabelian group (i.e. derived length $\leq 2$)? For reference, nilpotency is semi-decidable, i.e. there is a procedure for verifying nilpotency, see the article ["Verifying nilpotence" by Charles Sims](https://doi.org/10.1016/S0747-7171(87)80002-0).
| https://mathoverflow.net/users/120914 | Is solvability semi-decidable? | The Adian–Rabin theorem, although it is not usually stated like this, says that Markov properties are not co-semi-decidable, thus « not being metabelian » or « not being solvable » are not semi-decidable.
However, whether « being solvable » or « being solvable of derived length $\le n$ » is semi-decidable is an open problem.
Giving a precise classification of these properties in terms of the arithmetical hierarchy is stated as an open problem in Miller’s survey article [Decision Problems for Groups — Survey and Reflections](https://doi.org/10.1007/978-1-4613-9730-4_1).
It is natural to expect that « being solvable of derived length $\le n$ » should be $\Pi\_2$ complete —i.e. equivalent to an infinite conjunction of semi-decidable properties, but not semi-decidable. Also, « being solvable » should be $\Sigma\_3$ complete.
I think that these questions are both important and very basic and they definitely deserve some attention.
| 17 | https://mathoverflow.net/users/153080 | 451265 | 181,491 |
https://mathoverflow.net/questions/451260 | 2 | Let $\omega$ denote the first infinite ordinal (also known as the set of natural numbers). We call a set $E\subseteq {\cal P}(\omega)$ a *deck* if for all $n\in \omega$, the set $E$ contains exactly one member of cardinality $n$. (More formally, $E$ is a deck if for all $n\in \omega$ we have $|\{e\in E: |e| = n\}| = 1$.)
If $H=(V,E)$ is a [hypergraph](https://en.wikipedia.org/wiki/Hypergraph) and $k\neq\emptyset$ is a cardinal, we say that a map $c:V\to\kappa$ is a *coloring* if for any $e\in E$ with $|e|>1$ we have that the restriction $c|\_e:e\to\kappa$ is not constant. The smallest cardinal $\kappa> 0$ such that there is a coloring $c:V\to \kappa$ is said to be the *chromatic number* $\chi(H)$ of $H$.
**Question.** Given $k\in\omega\setminus\{0\}$, is there a deck $E\subseteq{\mathcal P}(\omega)$ such that for $H=(\omega, E)$ we have $\chi(H) = k$?
| https://mathoverflow.net/users/8628 | Possible chromatic numbers of a hypergraph on $\omega$ with a deck of edges | The answer is plainly negative for $k=1$ and positive for $k=2$. The answer is negative for $k\gt2$ because $(\omega,E)$ is $2$-colorable whenever the edge-set $E$ is a "deck," i.e., contains just one edge of size $n$ for each integer $n\ge2$. To see this consider a random $2$-coloring of $\omega$; the probability that it fails to be a proper coloring of $(\omega,E)$ is strictly less than $\sum\_{n=1}^\infty2^{-n}=1$. Alternatively, color the vertices recursively, First spoil the edge of size $2$ by coloring one vertex red and one vertex blue. Next spoil the edge of size $3$ by coloring at most one more vertex in each color. When we come to coloring the edge of size $n$, we will already have colored at most $n-1$ vertices in each color, so we can continue the process.
| 3 | https://mathoverflow.net/users/43266 | 451276 | 181,494 |
https://mathoverflow.net/questions/451287 | 3 | Let $\mathcal g\_1$ denote the usual Godel sentence defined as: $$ \mathcal g\_1 \iff \neg\exists x:\operatorname {Proof}\_T(x, \ulcorner \mathcal g\_1 \urcorner)$$
Lets suppose that $\sf T$ is consistent (metatheoretically), effectively generated, extends $\sf PA $, and complete.
Accordingly, $T \vdash \neg \mathcal g\_1$, that is, there is a natural $x$ that codes a proof of $\mathcal g\_1$. That is, formally we do have: $$T \vdash \exists x: \operatorname {Proof}\_T(x, \ulcorner \mathcal g\_1 \urcorner)$$
Let's call the smallest such Godel code by $\mathcal g^{\* 1}$
This is necessarily a non-standard natural.
Now we take another sentence $\mathcal g\_2$ that is defined as:
$\mathcal g\_2 \iff \forall x \, (\operatorname {Proof}\_T(x,\ulcorner \mathcal g\_2 \urcorner) \to x > \mathcal g^{\*1})$
This is what I call *remote provability*, so the above is a capture of the informal notion "every proof of this sentence is a remote proof", in other words "if this sentence is provable then it is remotely provable", where a remote proof means one that is coded by a non-standard natural. Actually the above sentence even adds more the claim that $\mathcal g\_2$ can only be remotely provable by exceeding $\mathcal g\_1$ provability.
Since $T$ is complete, then it will prove $\neg \mathcal g\_2$, and lets call the smallest Gödel code of a proof of $\mathcal g\_2$ as $\mathcal g^{\*2}$.
Now we do have $\mathcal g^{\*2} < \mathcal g^{\*1}$.
Let $\mathcal g\_3$ states it can only be remotely provable more than $\mathcal g\_2$ is, that is:
$\mathcal g\_3 \iff \forall x \, (\operatorname {Proof}\_T(x,\ulcorner \mathcal g\_3 \urcorner) \to x > \mathcal g^{\*2})$
And so on.. If we continue this for every concrete natural $n$, then we'll end up having $$ \mathcal g^{\*1} > \mathcal g^{\*2} > \mathcal g^{\*3} > ...$$
But, this cannot happen, since its provable in $\sf PA$ that there is no descending sequence of naturals. I say that because I think $(\exists n: x=\mathcal g^{\*n})$ is expressible in the language of $\sf PA$.
So, this must end up at some natural call it $\lambda$, i.e. we have: $\mathcal g^{\*\lambda+1} \not < \mathcal g^{\*\lambda}$.
But, by then we'll have: $T\not \vdash \neg \mathcal g\_{\lambda+1}$, proving the first incompleteness theorem.
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| https://mathoverflow.net/users/95347 | Can we use remote provability to prove the first incompleteness theorem sans $\omega$-consistency? | It doesn't contradict PA to have such a descending sequence. For example, you could instead simply keep substracting one from $g^{\*1}$, and this would be provably descending, but there is no contradiction in that — it doesn't mean somehow that at some stage, substracting one gives a larger number.
The point is that such descending sequences could be part of a nonstandard finite sequence, making your $\lambda$ nonstandard. But you have been considering only the standard part of the sequence, and you lack a uniform argument that handles the sequence in its entirety. For example, in the steps where you argue it is descending, it is important that the sentences themselves are standard and have standard proofs.
This issue seems to break your argument.
Meanwhile, let me add that since you've asked several questions about avoiding $\omega$-consistency in the incompleteness theorem, the Gödel-Rosser version of the incompleteness completely avoids Gödel's now-extraneous $\omega$-consistency assumption. Also, there are proofs of the incompleteness theorem using the undecidability of the halting problem that avoid $\omega$-consistency.
Let me add that there is some affinity of your idea with the proof I give of the universal algorithm result. See my paper [The modal logic of arithmetic potentialism and the universal algorithm](https://arxiv.org/abs/1801.04599). In the main argument, one searches for the smallest proof of a certain statement. And having found it, one searches for a proof of a related next statement, using a strictly smaller fragment of PA — the new proof is necessarily larger. And then so on. In this way, a (possibly nonstandard) finite sequence is generated, and it is consistent with PA that the sequence is any sequence at all. What I find related to your idea here is that the proofs are necessarily getting longer, while the fragment of PA that is used is getting smaller. This is what ensures that the process stops growing at some (possibly nonstandard) stage.
| 9 | https://mathoverflow.net/users/1946 | 451293 | 181,499 |
https://mathoverflow.net/questions/451297 | 1 | Suppose $X$ is a singular quasi-projective curve over the complex numbers, and $X'$ is a good nonsingular compactification of a resolution of singularities $Y\to X$. Let $D$ be the complement of $Y$ in $X'$.
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The answer should be "yes" using a combination of spectral sequences. I tried to first compute the hypercovering-to-cohomology spectral sequence for $Y\to X$, and then the weight spectral sequence for $Y\subset X'$ (and this is really just a Gysin long exact sequence), but I couldn't quite work them out.
| https://mathoverflow.net/users/nan | Cohomology of singular curves | You don't need all this machinery in this case (unless your goal was to understand the machinery). You have to exact sequences
$$0\to W\_0\to H^1(X) \to H^1(Y)\to 0$$
$$0 \to H^1(X')\to H^1(Y)\to \oplus\_{p\in D} \mathbb{Q}(-1)\to H^2(Y)\to 0$$
The last sequence (which is Gysin) can be shortened (non canonically) to
$$0 \to H^1(X')\to H^1(Y)\to \mathbb{Q}(-1)^{|D|-1}\to 0$$
The first sequence is easier to understand by drawing a picture and using homology. Then $W\_0$ is dual to the span of loops which get destroyed when the singular points of $X$ are pulled apart in the normalization $Y$.
| 3 | https://mathoverflow.net/users/4144 | 451300 | 181,500 |
https://mathoverflow.net/questions/451302 | 2 | Let $\mathcal{R}$ be a $1$-category. Assume that one has a pushout of representable $1$-presheaves $\mathrm{y} A \cup\_{\mathrm{y} B} \mathrm{y} C$ in $\mathsf{PSh}(\mathcal{R})$. Under which conditions does then the inclusion $\mathsf{PSh}(\mathcal{R}) \to \mathcal{P}(\mathcal{R})$ into $\infty$-preheaves preserve this pushout?
The reason why I am asking this, is the claim in Notation 8.3 of [Barwick‘s and Schommer-Pries‘ unicity paper](https://arxiv.org/pdf/1112.0040.pdf#page19) which says that a specific class (called $S\_{0}$) of specific morphisms in $\mathcal{P}(\mathcal{R})$ induced by some pushouts in $\mathcal{R}$ is the essential image under the above functor of a similarly defined class in $\mathsf{PSh}(\mathcal{R})$. For that to make sense, I think that the above must hold for the pushouts appearing there. Note also that in the situation, we know that the pushout $A \cup\_{B} C$ exists in $\mathcal{R}$.
| https://mathoverflow.net/users/145805 | Inclusion of $1$-presheaves into $\infty$-presheaves preserves pushouts? | The inclusion from discrete presheaves to spaces doesn't preserve all pushouts, even when $\mathcal R$ is a point.
But it does preserves some pushouts. The usual condition is to check that at least one of $yB \to yA$ or $yB \to yC$ is a levelwise monomorphism of sets. Then because the Kan-Quillen model structure is left proper with monomorphisms for cofibrations, it follows that the pushout is a levelwise pushout of spaces, and hence a pushout of presheaves of spaces.
Note that for $yB \to yA$ (say) to be a levelwise monomorphism of sets is equally for $B \to A$ to be a monomorphism in $\mathcal R$.
I haven't checked carefully, but I suspect that this criterion suffices to imply that all the pushouts considered by Barwick and Schommer-Pries are preserved by the inclusion from discrete presheaves to presheaves of spaces.
| 6 | https://mathoverflow.net/users/2362 | 451307 | 181,503 |
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