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https://mathoverflow.net/questions/444373 | 4 | Apologies in advance if this question is obvious/not research level.
>
> Let $\preceq$ be the consistency strength relationship on theories. Working over $ZF$ or $ZFC$, is there some large cardinal notion $\psi(\alpha)$ with an ordinal parameter such that there exists some ordinal $\kappa$ such that $ZF(C)+\psi(\alpha)$ is consistent relative to some stronger theory for all $\alpha<\kappa$ and $$ZF(C)+\psi(\alpha)\prec ZF(C)+\psi(\alpha+1)$$ for all $\alpha<\bigcup\kappa$, but $ZF(C)+\psi(\kappa)$ is inconsistent? What about a $\psi'(\alpha)$ such that $ZF(C)+\psi'(\alpha)$ is consistent relative to some stronger theory and $$ZF(C)+\psi'(\alpha)\prec ZF(C)+\psi'(\alpha+1)$$ for all set-sized $\alpha$, but $ZF(C)+\psi'(O\_n)$ is inconsistent?
>
>
>
The motivation is having a large cardinal notion that gets 'as close to inconsistency' as possible without becoming inconsistent, in a precise sense.
| https://mathoverflow.net/users/92164 | Cofinal inconsistency | The answer is yes.
One can make easy artificial examples. For example, let $\psi(\gamma)$ assert that there are (only) finitely many inaccessible cardinals, but at least $\gamma$ many.
From the assumption of infinitely many inaccessible cardinals, we can prove the consistency of ZFC+$\psi(n)$ for any particular finite $n$, and furthermore the consistency strength of $\psi(n)$ is steadily increasing as $n$ increases, since in fact $\psi(n+1)$ implies $\text{Con}(\text{ZFC}+\psi(n))$. But $\psi(\omega)$ is clearly inconsistent, so this is an instance of your requested phenomenon with $\kappa=\omega$.
One can make many other similar such examples using larger $\kappa$, simply by taking $\psi(\gamma)$ to assert that there are fewer than $\kappa$ many LC of a certain type, but at least $\gamma$ many.
| 5 | https://mathoverflow.net/users/1946 | 444376 | 179,174 |
https://mathoverflow.net/questions/444371 | 9 | Let $\sigma\_0(n)$ be the divisor counting function:
$$\sigma\_0(n) = \sum\_{d \vert n} 1.$$
I ran some numerical experiments that showed when $p$ is prime, the function $\sigma\_0(n)$ is equidistributed mod $p$. That is, for any residue class $a \mod p$,
$$\lim\_{X \to \infty} \dfrac{ \vert \{ n<X: \sigma\_0(n) \equiv a \mod p \} \vert }{X} = \dfrac{1}{p}.$$
Is this fact correct? If so, could anyone sketch a proof / provide a reference for a proof?
| https://mathoverflow.net/users/394740 | Is the divisor counting function equidistributed mod $p$? | $\newcommand{\Y}{\mathfrak{X}\_p(X)}$I haven't checked all the details on the application of Selberg–Delange below, so it's possible something I am saying is nonsense, but even after accounting for the correction noted by Noam Elkies in the [comments](https://mathoverflow.net/questions/444371/is-the-divisor-counting-function-equidistributed-mod-p#comment1147153_444371), it appears to be the case that $\sigma\_0(n)$ is *not*, in general, equidistributed among the non-zero congruence classes mod $p$ unless $2$ is a primitive root modulo $p$ (see [A001122](https://oeis.org/A001122) on OEIS). In fact, we have that
$$\lim\_{X \to \infty}\biggl(\frac{1}{\Y}\sum\_{n\leqslant X} 1\_{\sigma\_0(n) \equiv a \mod p}\biggr) = \frac{1}{p-1}\biggl(1+\frac{1}{\delta\_p}\sum\_{\substack{\chi \neq \chi\_0\\\chi(2) = 1}} \overline{\chi}(a)G\_{\chi}(1)\biggr),$$
where the sum runs over all nontrivial Dirichlet character mod $p$ with $\chi(2) = 1$,
$$\Y = \sum\_{n\leqslant X} 1\_{p \nmid \sigma\_0(n)}, \qquad \delta\_p = \lim\_{p \to \infty} \frac{\Y}{X},$$
are the counting function and density of the set of $n$ with $p \nmid \sigma\_0(n)$ and $G\_\chi$ is defined below. Note if $n$ is squarefree, then $\sigma\_0(n) = 2^{\omega(n)}$, where $\omega(n)$ is the prime-divisor counting function, and hence $p \nmid \sigma\_0(n)$ for $p > 2$. Since the squarefrees have positive density, this implies that $\Y \gg X$ and $\delta\_p > 0$.
When $2$ is a primitive root modulo $p$, the only character with $\chi(2) = 1$ is the trivial one, so this would imply equidistribution in that case. But, for example, if you take $p = 7$ or $p = 17$, and consult the table of values of Dirichlet characters mod $p$, then you'll find that the only characters which have $\chi(2) = 1$ are the trivial character and the quadratic character; from this it follows that $\sigma\_0(n)$ is biased towards being a quadratic residue over being a nonresidue. The number of characters that will occur in this sum for a given prime $p$ is $1$ less than [A001917](https://oeis.org/A001917) on OEIS.
Roughly speaking, this is because $\sigma\_0(p) = 2$, and hence the characters which have $\chi(2) = 1$ give a main term contribution. Here's a sketch:
Guided by Weyl's criterion for $(\mathbb{Z}/p\mathbb{Z})^\times$, suppose $(a,p)=1$, and note that orthogonality of Dirichlet characters reads
$$1\_{n \equiv a \mod p} = \frac{1}{p-1}\sum\_{\chi} \overline{\chi}(a) \chi(n),$$ where the sum is over all Dirichlet characters mod $p$. Thus,
$$\frac{1}{\Y}\sum\_{n\leqslant X} 1\_{\sigma\_0(n) \equiv a \mod p} = \frac{1}{p-1} + \frac{1}{p-1}\sum\_{\chi\neq \chi\_0} \overline{\chi}(a)\left( \frac{1}{\Y} \sum\_{n\leqslant X} \chi(\sigma\_0(n)) \right), \tag{$\star$}\label{star}$$
where we have separated the contribution of the trivial character. It thus suffices to study
$$M\_\chi(X) = \sum\_{n\leqslant X} \chi(\sigma\_0(n)), $$
which is a mean-value of multiplicative function. The characters which satisfy $M\_\chi(X) = o(X)$ do not contribute a main term to \eqref{star} since $\Y \gg X$, while the characters with $M\_\chi(X) \asymp X$ do.
Standard multiplicative number theory techniques apply here. To put this into effect, define the Dirichlet series
$$F\_\chi(s) = \sum\_{n\geqslant 1} \frac{\chi(\sigma\_0(n))}{n^s},$$
so that an application of Perron's formula gives that
$$\sum\_{n\leqslant X} \chi(\sigma\_0(n)) = \frac{1}{2\pi i} \int\_{2-i\infty}^{2+i\infty} F\_\chi(s) X^s \frac{ds}{s}.$$
Investigating the Euler product of $F\_\chi$, we find
$$F\_\chi(s) = \prod\_p \biggl(\sum\_{k=0}^\infty \frac{\chi(\sigma\_0(p^k))}{p^{ks}}\biggr) = \prod\_p \biggl(\sum\_{k=0}^\infty \frac{\chi(k+1)}{p^{ks}}\biggr) = \zeta(s)^{\chi(2)} G\_\chi(s),$$
where $G\_\chi(s)$ is convergent in $\Re(s) > 1/2$. An application of the Selberg-Delange method (see Chapter 5 of [Tenenbaum's "Introduction to analytic and probabilistic number theory"](https://mathscinet.ams.org/mathscinet-getitem?mr=3363366), in particular Theorem 5.2 with $z = \chi(2)$, $N = 1$ and $F = F\_\chi$) should then give that
$$ \sum\_{n\leqslant X} \chi(\sigma\_0(n)) = \frac{X (\log X)^{\chi(2) - 1}}{\Gamma(\chi(2))} \biggl(G\_\chi(1) + O\Big(\frac{1}{\log X}\Big)\biggr),$$
from which it is clear that if $\chi(2) \neq 1$, then the expression on the right is $o(X)$ while, if $\chi(2) = 1$, then we get something of size $\asymp X$. Putting this back into \eqref{star}, and concentrating on the main terms proves the claim.
| 17 | https://mathoverflow.net/users/37327 | 444381 | 179,177 |
https://mathoverflow.net/questions/444379 | 0 | I was looking at the fresnel integral $S(x)=\int^x\_0\sin(s^2)ds$. From reading I learned that this integral approaches $\frac{1}{2} \sqrt{\frac{\pi}{2}}$ as $x \rightarrow \infty$. Through messing around on desmos, I found an excellent estimate: $S\_2(x)=\frac{\sin(x^2)+2x^2\cos(x^2)}{-4x^3}+\frac{1}{2}\sqrt{\frac{\pi}{2}}$, which is not a good approximation near $x=0$, but quickly converges to $S(x)$ as $x\rightarrow \infty$. For my purposes, combining this with the taylor approximation for $S$ centered at $x=0$ yeilds sufficiently accurate results. While it is clear through solving the limit that $S\_2$ will arrive at the same equilibrium value as $S$ at infinity, I neither know how to prove, or how to describe the sense in which it converges to the integral curve's shape over that range. I would appreciate if someone could shed some light on this.
Thanks
| https://mathoverflow.net/users/502440 | How to prove approximation for fresnel integral converges | If you want to build an asymptotics, write
$$S(x)=\int^x\_0\sin(s^2)\,ds=\frac{1}{2}\sqrt{\frac{\pi }{2}}-\int\_x^\infty\sin(s^2)\,ds.$$ Let $s=\sqrt x$
$$\int\_x^\infty\sin(s^2)ds=\frac 12\int\_{x^2}^\infty\frac{\sin (t)}{\sqrt{t}}\,dt.$$ Integrate by parts a few times and using the bounds, you will obtain
\begin{gather\*}
\int\_{x^2}^\infty\frac{\sin (t)}{\sqrt{t}}\,dt=\frac{\cos \left(x^2\right)}{x}+\frac{\sin \left(x^2\right)}{2 x^3}-\frac{3 \cos \left(x^2\right)}{4
x^5}+\dotsb \\
S(x)=\frac{1}{2}\sqrt{\frac{\pi }{2}}-\frac{\cos \left(x^2\right)}{2x}-\frac{\sin \left(x^2\right)}{4 x^3}+\frac{3 \cos \left(x^2\right)}{8
x^5}+\dotsb.
\end{gather\*}
Use $x=12.34$ : the above gives $0.622816621$ to be compared to the value of
$0.622816642$.
**Edit**
Making the problem more general, we can write
\begin{multline\*}
\color{blue}{S(x)=\frac{1}{2}\sqrt{\frac{\pi }{2}}-\frac{\cos \left(x^2\right)}{2x}\Bigl[1+\sum\_{n=1}^\infty (-1)^n \frac {(4n)!}{(2n)!\, (16x^4)^n}\Bigr]-{}} \\
\color{blue}{\frac{\sin \left(x^2\right)}{4x^3}\Bigl[1+\frac 12\sum\_{n=1}^\infty (-1)^n \frac{(4 n+2)!}{ (2 n+1)!\, (16x^4)^n}\Bigr]}
\end{multline\*}
which is extremely good even for rather small values of $x$.
| 3 | https://mathoverflow.net/users/42185 | 444384 | 179,178 |
https://mathoverflow.net/questions/444099 | 4 | Let $A$ be a unital $C^{\ast}$-algebra and $\{ f\_i: A \rightarrow A\_i \}\_i$ a finite collection of morphisms of unital $C^{\ast}$-algebras, such that the associated map $A \rightarrow \prod\_i A\_i$ is injective. Let $g: A \rightarrow B$ be a morphism of unital $C^{\ast}$-algebras. Then, there is an induced collection of maps $\{ B \rightarrow A\_i \ast\_A B \}\_i$, where $A\_i \ast\_A B$ is the amalgamated free product (pushout). This collection induces a map $f: B \rightarrow \prod\_i A\_i \ast\_A B$. Is $f$ injective?
$\textbf{Motivation:}$ I am trying to use the extensive Grothendieck topology in the category of unital $C^{\ast}$-algebras.
| https://mathoverflow.net/users/130868 | Property of pushouts in the category of unital $C^{\ast}$-algebras | Unfortunately, the assertion is, in general, not true even if we just have a single (injective) unital $\*$-homomorphism $f\_1\colon A\to A\_1$ and another (non-injective) unital $\*$-homomorphism $g\colon A\to B$. The reason is that amalgamated free products with non-injective maps can be trivial: $B\*\_A A\_1=0$ in certain cases.
For an example, take any simple unital $C^\*$-algebra $A\_1$, for example $A\_1=M\_2(\mathbb{C})$ the $2\times 2$-matrices, and $A$ any unital $C^\*$-subalgebra of $B$ with $\dim(A)>1$ and which has a character $\chi\colon A\to \mathbb{C}$, for example the diagonal matrices $D\cong \mathbb{C}^2$ in $M\_2(\mathbb{C})$.
Now take $B=\mathbb{C}$ with the character $g=\chi\colon A\to B$ above, and take $f\_1$ to be the embedding $A\subseteq A\_1$.
In this situation, the amalgamated free product
$$A\_1\*\_A B$$
is canonically isomorphic to the quotient $A\_1/J$, where $J$ is the ideal of $A\_1$ generated by the differences $\chi(a)1\_{A\_1}-a$ with $a\in A$. Since $\dim(A)>1$, this ideal is not zero, so that $J=A\_1$ because $A\_1$ is simple and therefore
$$A\_1\*\_A B\cong A\_1/J =0.$$
In particular, for the specific example mentioned above:
$$M\_2(\mathbb{C})\*\_{\mathbb{C}^2}\mathbb{C}=0.$$
| 4 | https://mathoverflow.net/users/75215 | 444385 | 179,179 |
https://mathoverflow.net/questions/443919 | 1 | $\DeclareMathOperator\Hom{Hom}$For which unital $C^{\ast}$-algebras $A$ does it hold that for all compact Hausdorff $S$ we have the bijection:
\begin{align\*}
\Hom(A, C(S)) \cong \Hom(S, \Hom (A, \mathbb{C}))?
\end{align\*}
This holds for all commutative unital $C^{\ast}$, but it seems to hold for all $C^{\ast}$ algebras with $\Hom(A, \mathbb{C}) = \varnothing$.
Does this hold in general? Are there some easy counter-example? Are there some characterizations of the algebras for which this property holds?
| https://mathoverflow.net/users/130868 | Adjunction via Gelfand duality | $\DeclareMathOperator\Hom{Hom}$
Yes, this is true, and the proof is elementary: let us write $\Omega(A):=\Hom(A,\mathbb{C})$ for the space of characters of $A$, viewed as a subspace of the unit ball of the dual $A^\*$, and endowed with the weak\*-topology (i.e., the topology of pointwise convergence). This is a compact Hausdorff space if $A$ is a unital $C^\*$-algebra (in general it is locally compact Hausdorff, if $A$ is not unital).
We claim that there exists a natural bijection
$$\Hom(A,C(S))\cong \Hom(S,\Omega(A)),$$
where the left hom-set is in the category of unital $C^\*$-algebras with unital $\*$-homomorphisms, and the right hom-set is in the category of compact Hausdorff spaces.
Given $f\in \Hom(A,C(S))$, we define $\tilde f\in \Hom(S,\Omega(A))$ by $\tilde f(s)(a):=f(a)(s)$ for all $s\in S$, $a\in A$. It is clear that
$$\Hom(A,C(S))\ni f\mapsto \tilde f\in \Hom(S,\Omega(A))$$
is a well-defined injective map. To see that it is surjective, take $g\in \Hom(S,\Omega(A))$ and define $f(a)(s):=g(s)(a)$. Then $\tilde f=g$, so that the map above is also a surjection, therefore a bijection, as desired.
| 1 | https://mathoverflow.net/users/75215 | 444388 | 179,180 |
https://mathoverflow.net/questions/444362 | 1 | If I have a smooth surface $M$ (2D embedded in 3D), under what conditions I can assure that there exists a finite collection of charts $\{U\_i, \phi\_i \}\_{i}$, with $\phi\_i : U\_i \to M$, such that its directional derivatives have always the same magnitud and that they are orthogonal set of coordinates?, i.e.
$$
|\partial\_x \phi\_i| = |\partial\_y \phi\_i| \\
\partial\_x \phi\_i \cdot \partial\_y \phi\_i = 0
$$
Trivially flat surfaces have this property, and I suspect that is the general case, but I'm not really sure.
| https://mathoverflow.net/users/130126 | Requirement of parametrization of surfaces | You can always do this, but it's not as simple as using some kind of ODE (such a flow of vector fields) to construct such charts.
First, assume that your surface is connected and simply-connected. Then it's either compact, in which case it's a $2$-sphere, or it's diffeomorphic to $\mathbb{R}^2$. In the compact, case, the uniformization theorem implies that it's conformally equivalent to the standard $2$-sphere, i.e., there is a smooth diffeomorphism $f:S^2\to M$ such that the $f$-pullback of the induced metric on $M$ is a multiple of the standard metric on $S^2$. Now composing $f$ with the inverses of the usual stereographic projections (which are conformal) of the sphere to the plane from the north and south poles gives you two coordinate patches on $M$ that are conformal. In the non-compact case, the uniformization theorem says that there is a single smooth diffeomorphism $\phi:U\to M$ that is conformal, i.e., the $\phi$-pullback of the induced metric on $M$ is a multiple of the standard metric on $U\subset\mathbb{R}^2$. Here, by the Riemann mapping theorem, we can either take $U$ to be $\mathbb{R}^2$ or to be the unit disk.
To cover the general case, you only need to know that $M$ has a finite open cover by simply-connected open sets. This is a standard fact about 2-dimensional surfaces.
| 3 | https://mathoverflow.net/users/13972 | 444395 | 179,182 |
https://mathoverflow.net/questions/444394 | 0 | Let $F$ be a field, $G$ be a finite group and $\alpha \in Z^2(G, F^\*)$ . The twisted group algebra $F^{\alpha}G$ is a $F$-algebra with $F$ vector basis, $\{\bar g : g \in G \},$ and multiplication defined by $\bar x \bar y = \alpha(x,y) \overline{xy}$ for $x,y \in G$ and extended distributively.
Let $F^{\alpha} G$ and $F^{\beta}G$ be twisted group algebras with bases $\{\bar g : g \in G\}$ and $\{\tilde{g} : g \in G\}$ respectively. We say that $F^{\alpha}G$ and $F^{\beta}G$ are said to be **equivalent** if there exist an $F$-algebra isomorphism $\psi : F^{\alpha}G \to F^{\beta} G$ and a map $t : G \to F^\*$ such that $\psi(\bar g) = t(g) \tilde{g}.$
Are there are examples twisted group algebras which are isomorphic as $F$ algebras but not equivalent?
There is a equivalent criterion which says that $F^{\alpha} G$ and $F^{\beta}G$ are equivalent iff $\alpha$ and $\beta$ are in same cohomology class. So we need to find examples where $H^2(G,F^\*)$ is non trivial. For example cyclic groups, certain abelian groups such as $\mathbb{Z}\_n\times \mathbb{Z}\_m$ with $(m,n)=1$ would not work when $F$ is algebraically closed field. I am also not aware of any result which gives the wedderburn decomposition of $F^{\alpha} G$ when $G$ is abelian . The case when $G$ is abelian, there might be potential counter examples.
| https://mathoverflow.net/users/502460 | Examples of isomorphic non-equivalent twisted group algebras | Let $G={\mathbb Z}/3 \times {\mathbb Z}/3$ and $\alpha$ be the $2$-cocycle corresponding to the extraspecial group of exponent $3$ and order $27$. Then the twisted group algebras defined using $\alpha$ and $-\alpha$ are isomorphic, by swapping the two copies of ${\mathbb Z}/3$.
| 2 | https://mathoverflow.net/users/460592 | 444396 | 179,183 |
https://mathoverflow.net/questions/444169 | 3 | In [this math.stackexchange question](https://math.stackexchange.com/q/4668840/111012) Adam Rubinson asked (I paraphrase):
>
> Given a natural number $r$, what is the least number $n$ such that every strictly increasing sequence of $n$ real numbers has a subsequence $x\_1,x\_2,\dots,x\_r$ of length $r$ whose sequence of differences $x\_2-x\_1,x\_3-x\_2,\dots,x\_r-x\_{r-1}$ is (nonstrictly) monotonic? (E.g., if $r=4$ then $n=7$.)
>
>
>
I can't recall having seen this before, but it seems like a reasonably natural question, so it must be discussed somewhere in the literature. Where?
**Edit.** I'm not sure the question should be closed as a duplicate, seeing as I posted it as a reference request, and I still don't know where or whether this result appears as a traditional publication.
I would have expected this to be a "classical" problem. It's hard to believe that it's not among the thousands of problems Paul Erdős wrote about.
| https://mathoverflow.net/users/43266 | A combinatorial problem about sequences of numbers | As pointed out in a comment by [Vladimir Dotsenko](https://mathoverflow.net/users/1306/vladimir-dotsenko), the problem I asked about was posed and solved in the classical paper by Erdős and Szekeres, A combinatorial problem in geometry, *Compositio Math.* 2 (1935), 463–470 ([pdf](http://www.numdam.org/item/CM_1935__2__463_0.pdf)); see the discussion beginning with "We solve now a similar problem" on p. 468.
In another comment [Darij Grinberg](https://mathoverflow.net/users/2530/darij-grinberg) pointed out that a generalization to possibly non-monotonic sequences was the subject of a 2012 [math overflow question](https://mathoverflow.net/questions/90128/erd%C5%91s-szekeres-for-first-differences) by [Seva](https://mathoverflow.net/users/9924/seva) which was answered by [Sergey Norin](https://mathoverflow.net/users/8733/sergey-norin).
| 1 | https://mathoverflow.net/users/43266 | 444399 | 179,185 |
https://mathoverflow.net/questions/444407 | 9 | Have there been any attempts to extend the "F\_un" analogy to the representation theory of finite groups?
If I might be allowed some speculation:
If combinatorics can be regarded as analagous to linear algebra over the "field with one element", then are characters of finite groups over the "field with one element" simply permutation characters? Perhaps the analogous notion of irreducibility in this case would be transitivity?
In the complex character theory of finite groups, two irreducible characters are identical if and only if representations that afford them are equivalent. Perhaps by developing a representation theory of finite groups over the field with one element, we would find some analogous way of comparing "F\_un characters" to check if two *permutation* representations of a given group are equivalent.
| https://mathoverflow.net/users/502468 | Representations of finite groups over the "field with one element" | The table of marks, as defined by Burnside, has rows indexed by the transitive permutations $X$ and columns indexed by the conjugacy classes of subgroups $H\leqslant G$. The entry corresponding to $X$ and $H$ is $|X^H|$, the number of fixed points.
The Burnside ring $A(G)$ is the Grothendieck ring of permutation representations, with disjoint union as the addition and direct product as the multiplication. The ring homomorphisms from $A(G)$ to $\mathbb Z$ are exactly the fixed point functions in the table of marks, and they separate permutation representations. This is very analogous to character theory.
The ring $A(G)$ was extensively studied by Dress, who computed the index of the image of $A(G)$ in the sum of copies of $\mathbb Z$, one for each conjugacy class of subgroups, under the sum of these homomorphisms. There are many applications of $A(G)$ in representation theory, especially induction theorems and related topics.
Dress also discovered that there are no non-trivial idempotents in $A(G)$ if and only if $G$ is soluble. There was some speculation that one might be able to approach the odd order theorem this way, but nobody really got very far with it.
Burnside rings play a role in a lot of applications of group theory. In stable homotopy theory, for example, the stable cohomotopy of the classifying space $BG$ is compared with the completion of $A(G)$. This is the Segal conjecture, which was proved back in the eighties. This is analogous with the Atiyah completion theorem, which compares the unitary $K$-theory of $BG$ with the completion of the character ring.
| 13 | https://mathoverflow.net/users/460592 | 444415 | 179,189 |
https://mathoverflow.net/questions/444416 | 4 | $\DeclareMathOperator\cl{cl}$Let $X$ be a topological space and let $Y$ be a dense subspace of $X$. Suppose
that $R\left( X\right) $ denotes all regular closed subsets of $X$.
Question 1: $R\left( Y\right) \longrightarrow R\left( X\right) $, $A\rightarrow \cl\_{X}A$ is bijective.
Question 2: If $A$ is regular closed in $Y$, then $\cl\_{X}A$ is the unique
regular closed subset of $X$ with $A=Y\cap \cl\_{X}A$.
Can you answer these questions?
| https://mathoverflow.net/users/86099 | A question about regular closed sets | The answer to both of these questions is **Yes.** And this result can generalize to point-free topology and I consider this result to be more natural in the context of point-free topology.
The following observations were produced earlier by Mehmet Onat, so let me paraphrase those arguments.
Observation: If $U$ is an open subset of $X$ and $Y$ is dense in $X$, then
$\text{Cl}\_X(U\cap Y)=\text{Cl}\_X(U)$.
Proof: Clearly $\text{Cl}\_X(U\cap Y)\subseteq\text{Cl}\_X(U)$. For the converse direction, suppose that $a\in\text{Cl}\_X(U)$. Then whenever $O$ is an open subset of $X$ that contains $a$, we have $O\cap U\neq\emptyset$. However, since $Y$ is dense in $X$, we also know that $O\cap U\cap Y\neq\emptyset$. Therefore, $a\in\text{Cl}\_X(U\cap Y)$, so we may conclude the converse $\text{Cl}\_X(U)\subseteq \text{Cl}\_X(U\cap Y).$ $\square$
Claim: Suppose that $Y$ is dense in $X$. If $C\in R(Y)$, then $\text{Cl}\_X(C)\in R(X)$.
Proof: If $C\in R(Y)$, then there is some open subset $U\subseteq X$ where
$$C=\text{Cl}\_Y(U\cap Y)=\text{Cl}\_X(U\cap Y)\cap Y=\text{Cl}\_X(U)\cap Y.$$
Therefore, $$\text{Cl}\_X(C)=\text{Cl}\_X(\text{Cl}\_X(U)\cap Y)\supseteq
\text{Cl}\_X(U\cap Y)=\text{Cl}\_X(U)=\text{Cl}\_X(\text{Cl}\_X(U))$$
$$\supseteq\text{Cl}\_X(\text{Cl}\_X(U)\cap Y)=\text{Cl}\_X(C),$$
so $\text{Cl}\_X(C)=\text{Cl}\_X(U)$ which is regular closed. $\square$
Claim: If $C\in R(X)$, then $C\cap Y\in R(Y)$
Proof: Since $C\in R(X)$, we have $C=\text{Cl}\_X(U)$ for some open $U\subseteq X$. Therefore, $$R(Y)\ni\text{Cl}\_Y(U\cap Y)=\text{Cl}\_X(U\cap Y)\cap Y=\text{Cl}\_X(U)\cap Y=C\cap Y.$$ $\square.$
If $X$ is a topological space and $Y\subseteq X$ is dense, then let
$i\_{Y,X}:R(Y)\rightarrow R(X)$ be the mapping defined by
$i\_{Y,X}(C)=\text{Cl}\_X(C)$ and define a mapping $j\_{X,Y}:R(X)\rightarrow R(Y)$ by letting $j\_{X,Y}(C)=C\cap Y$.
Claim: The mapping $i\_{Y,X}$ is injective. More generally, if $C,D$ are distinct closed subsets of $Y$, then $\text{Cl}\_X(C)\neq\text{Cl}\_X(D)$.
Proof: We can assume that
$x\_0\in C\setminus D$. Then since $Y\setminus D$ is open in $Y$, there is an open set $U\subseteq X$ where $U\cap Y=Y\setminus D$. Therefore, since
$U\cap D=\emptyset$, we have $U\cap\text{Cl}\_X(D)=\emptyset$ as well, so
$x\_0\in\text{Cl}\_X(C)$, but $x\_0\not\in\text{Cl}\_X(D)$. Therefore, the mapping $R(Y)\rightarrow R(X),C\mapsto\text{Cl}\_X(C)$ is injective. Mehmet Onat observed that injectivity also follows from the fact that $C=\text{Cl}\_Y(C)=Y\cap\text{Cl}\_X(C)$. $\square$
Claim: If $C\in R(X)$, then $i\_{Y,X}(j\_{X,Y}(C))=C$. Therefore, the mappings $i\_{Y,X},j\_{X,Y}$ are inverses.
Proof: Clearly, $\text{Cl}\_X(C\cap Y)\subseteq C$. For the converse direction, suppose that $x\_0\in C$. Suppose now that $U$ is an open subset of $X$ with $x\_0\in U$. Then since $C=\overline{C^\circ}$, we know that
$U\cap C^\circ\neq\emptyset$. Since $U\cap C^\circ\neq\emptyset$ and $U\cap C^\circ$ is open, we know that $U\cap C^\circ\cap Y\neq\emptyset$ since $Y$ is dense in $X$. Therefore, since $U$ is an arbitrary neighborhood of $x\_0$, we have $x\_0\in\overline{C^\circ\cap Y}\subseteq\overline{C\cap Y}$. Therefore,
$C=\text{Cl}\_X(C\cap Y)$. Q.E.D.
**Forcing and point-free topology**
Let $X$ be a regular space, and let $D$ be the intersection of all open dense subsets of $X$. If $X$ has no isolated points and is $T\_1$, then $D$ is empty, but $D$ has virtual points, and $R(X)$ is isomorphic to the lattice of closed subsets of the space $D$. The virtual points of the space $D$ live inside forcing extensions $V[G].$ For example, the Boolean valued model $V^{R[X]}$ always adds points to the space $D$.
If $L$ is a frame, then let $B\_L=\{x^{\*\*}\mid x\in L\}=\{x^\*\mid x\in L\}$ where $^\*$ is the pseudocomplement operation. Then $B\_L$ is a complete Boolean algebra which is the point-free analogue to the lattice of all regular open (and also the lattice of regular closed) subsets of $X$. We say that a sublocale $S$ of a frame $L$ is dense if $0\in S$. The frame $B\_L$ is also a sublocale of $L$, and $B\_L$ is the smallest dense sublocale of $L.$ A frame is fit if and only if each sublocale is the intersection of open sublocales, so in a fit frame $L$, the Boolean algebra $B\_L$ is the intersection of all open sublocales of $L$. If $S$ is a sublocale of a frame $L$, then the Heyting operation on $S$ is the same as the Heyting operation on $L$. In particular, if $S$ is a dense sublocale of $L$, then the pseudocomplement operation on $S$ coincides with the pseudocomplement operation on $S$. Therefore, if $S$ is a dense sublocale of a frame $L$, then $B\_S=B\_L$.
| 4 | https://mathoverflow.net/users/22277 | 444419 | 179,190 |
https://mathoverflow.net/questions/444392 | 0 | Assume $A\in \mathbb{R}^{n\times n}$ with each entry being i.i.d. bounded r.v. in $[a,b]$, is $\Vert A\Vert\_2$ is sub-Gaussian?
Intuitively, since $\{A\_{ij}\}\_{i,j=1,...,n}$ is bounded, then
$$\Vert A \Vert\_2 = \sup\_{\Vert v \Vert = 1} \vert v^TA^TAv\vert = \sup\_{\Vert v \Vert = 1}\vert\sum\_{i,j}v\_iv\_j(\sum\_k A\_{ki}A\_{kj})\vert\leq \max(a^2,b^2)$$
Then $\Vert A\Vert\_2$ is bounded so that it is sub-Gaussian. I doubt whether there exists any problem in my guess above because sub-Gaussian is much weaker than "bounded" condition.
| https://mathoverflow.net/users/500967 | Spectral norm of matrices of bounded random variables | Your chain of equality is slightly off, the idea is there however. I think that the following one is correct :
\begin{align\*}
\| A \|\_2^2 &= \sup\_{\| v \|\_2=1} v^TA^TAv\\
&= \sup\_{\|v \|\_2 = 1} \left| \sum\_{i,j} v\_i v\_j \left( \sum\_{k} A\_{ki} A\_{kj} \right) \right|\\
&\leq \sup\_{\|v \|\_2 = 1} \sum\_{i,j} |v\_i|\cdot |v\_j|\cdot \left| \sum\_{k} A\_{ki} A\_{kj} \right|\\
&\leq n \cdot \max(a^2,b^2)\sup\_{\|v \|\_2 = 1} \|v\|\_1^2\\
&=n^2\cdot \max(a^2, b^2)
\end{align\*}
Note that all inequalities can be achieved ($A\_{i,j}=\max(a,b)$ for all $i,j$ and $v\_i=\frac{1}{\sqrt{n}}$ for all $i$) and therefore your bound cannot really be correct.
Now this yields that $0\leq \| A \|\_2 \leq n\cdot\max(|a|,|b|)$, and therefore $\|A \|\_2$ is a sub-Gaussian.
| 0 | https://mathoverflow.net/users/492816 | 444420 | 179,191 |
https://mathoverflow.net/questions/444134 | 5 | Suppose that a finite group $G$ admits a Frobenius group of
automorphisms $F H$ with kernel $F$ and complement $H$ such that $F$
acts without nontrivial fixed points (that is, such that $C\_G(F)=1$).
It is proved by Belyaev and Hartley in [Centralizers of finite nilpotent subgroups in locally finite groups](https://doi.org/10.1007/BF02367023) that $G$ is a solvable group.
So, there are some papers studied on the Fitting height of $G$, exponent of $G$, rank of $G$.
For example, a paper [Fitting height of a finite group with a Frobenius group of automorphisms](https://doi.org/10.1016/j.jalgebra.2012.05.011) by E.I. Khukhro.
I am curious about the following question.
**Question: If $G$ is an abelian group which is $FH$-indecomposable and $F$ acts fixed-point-freely on $G$, I am wondering whether it is true that $G$ is homocyclic $p$-group.**
Obviously, $G$ is a $p$-group for some prime $p$.
If the action of $FH$ on $G$ is coprime, then a result of M. Harris shows that $G$ is homocyclic without assuming that $FH$ being a Frobenius group.
So, we may assume that $p\mid |H|$.
If $G$ is an elementary abelian group, then $G$ has a basis which is permuted by $H$ by Theorem 15.16 in "[Character theory of finite groups](https://doi.org/10.1090/chel/359)" by I.M. Isaacs.
Here is an example of a very special case: $G$ is an abelian $2$-group which is $FH$-indecomposable and $FH\cong S\_3$.
Then $$G=C\_G(H)\times C\_G(H)^x$$ where $F=\langle x\rangle$.
Since $F$ acts fixed-point-freely on $G$, $C\_G(H)$ is a cyclic subgroup.
So, $G$ is a homocyclic 2-group. Also, power map defines isomorphism of $FH$-chief factors of $G$.
Any explanation, references, suggestion and examples are appreciated.
| https://mathoverflow.net/users/44312 | Finite abelian group admits a Frobenius group of automorphism | Let $V$ be an $\mathbb{F}\_p[X]$-module such that $p\mid |G|$.
Then $V$ is not necessarily completely reducible.
A classical example is: $X\cong C\_p$ and $V\cong C\_p \times C\_p$ and $XV$ is extraspecial $p$-group of order $p^3$.
Richard Lyons's [idea](https://mathoverflow.net/questions/444134/finite-abelian-group-admits-a-frobenius-group-of-automorphism#comment1147095_444134) is: when a group $X$ acting completely reducibly on every $X$-invariant elementary abelian subquotient of $A$, $A$ must decompose into a direct product of $X$-invariant homocyclic groups.
Here is a proof based on the idea of Richard Lyons.
**Proposition** Let $A$ be a finite abelian $p$-group on which a group $X$ acts.
Suppose that that for every $X$-invariant subquotient $B/C$ of $A$ $($i.e., both $B$ and $C$ are $X$-invariant)
such that $B/C$ is elementary abelian, $X$ acts completely reducibly on $B/C$.
Then $A$ is a direct product of $X$-invariant homocyclic subgroups.
**Proof** Let $A$ be a counterexample of minimal possible order.
Then $A$ is $X$-indecomposable.
Let $\overline{A}=A/\Phi(A)$, and observe that $X$ acts completely reducibly on $\overline{A}$.
Then
$$\overline{A}=\overline{B}\times \overline{C}$$
where $\overline{B}$ and $\overline{C}$ is completely reducible such that $\Phi(A)=B\cap C$, $\exp(B)=\exp (A)$.
By the minimality of $A$, $B$ and $C$ are both direct product of $X$-invariant homocyclic subgroups,
and, without loss of generality, we may assume that $B$ is homocyclic.
Also, $\Phi(B)=\Phi(A)$.
Let $F$ be a maximal homocyclic subgroup of $A$ containing $B$.
Then $\Phi(B)\leq \Phi(F)\leq \Phi(A)$.
However, since $\Phi(B)=\Phi(A)$, $\Phi(F)=\Phi(B)$, and hence $B=F$ (as $F$ is homocyclic).
By Krull-Remak-Schmidt's theorem, $A=B\times V$ where $V$ is elementary abelian (as $\Phi(A)=\Phi(B)$),
i.e. $A=B\Omega\_1(A)$, where $\Omega\_1(A)$ is a subgroup of $A$ generating by every element of order 2 in $A$.
By the assumption of this proposition, $\Omega\_1(A)$ is completely reducible.
Therefore, $\Omega\_1(A)=\Omega\_1(B)\times D$, where $\Omega\_1(B)$ and $D$ are $X$-invariant.
Observe that $B\cap D=\Omega\_1(B)\cap D=1$, and hence $A=B\times D$ is $X$-decomposible, a contradiction.
I guess Dave Benson's [idea](https://mathoverflow.net/questions/444134/finite-abelian-group-admits-a-frobenius-group-of-automorphism#comment1146669_444134) is trying to show that: under the assumption of this question, every $\mathbb{F}\_p[FH]$-module $M$ such that $C\_M(F)=1$ is completely reducible.
The proof of Theorem 15.16 in "Character theory of finite groups" (Isaacs) use the same idea suggested by Dave Benson.
| 1 | https://mathoverflow.net/users/44312 | 444433 | 179,195 |
https://mathoverflow.net/questions/444421 | 3 | There are two proofs of
$$\sum\_{n=1}^\infty \frac{1}{n^s}=\prod\_{p \text{ prime}}\frac{1}{1-p^{-s}},\quad \Re (s)\gt 1$$
which I'm aware of. I'll call the first one the **Sieve proof** and the second one will be the **Factorization proof**. Both of them use infinitude of primes (at least I think so).
1. Sieve proof
By sieving, we see that for every prime $q$
$$\left(\prod\_{p\,\text{prime}\le q}\left(1-\frac{1}{p^s} \right) \right) \zeta (s)=\sum\_{n;2,3,\ldots ,q\,\nmid\, n}\frac{1}{n^s},\quad \Re (s)\gt 1$$
where the sum is over all $n\in\mathbb{N}$ that are not divisible by the primes from $2$ to $q$. Choosing $r\gt 1$ such that $|s|\gt r$ gives
$$\begin{align}\left|\left(\prod\_{p \text{ prime}\le q}\left(1-\frac{1}{p^s}\right)\right)\zeta (s)-1\right|&=\left|\sum\_{n;n\ne 1 \& 2,3,\ldots q,\,\nmid\, n}\frac{1}{n^s}\right|\\
&\le \sum\_{n;n\ne 1 \& 2,3,\ldots q,\\,\nmid\, n}\frac{1}{n^r}\\
&\le \sum\_{n=q}^\infty \frac{1}{n^r},\quad \Re (s)\gt 1.\end{align}$$
**Because of infinitude of primes**, we can let $q\to\infty$, and by Cauchy's criterion for series $\lim\_{q\to\infty}\sum\_{n=q}^\infty \frac{1}{n^r}=0$, so
$$\left(\prod\_{p\,\text{prime}}\left(1-\frac{1}{p^s}\right)\right)\zeta (s)=1,\quad \Re (s)\gt 1.$$
2. Factorization proof (this proof is taken from *The Theory of the Riemann Zeta-function* by Titchmarsh)
We have
$$\prod\_{p\le P}\left(1+\frac{1}{p^s}+\frac{1}{p^{2s}}+\cdots\right)=1+\frac{1}{n\_1^s}+\frac{1}{n\_2^s}+\cdots$$
where $n\_1,n\_2,\ldots$ are those integers none of whose prime factors exceed $P$. It follows that
$$\begin{align}\left|\zeta (s)-\prod\_{p\le P}\left(1-\frac{1}{p^s}\right)^{-1}\right|&=\left|\zeta (s)-1-\frac{1}{n\_1^s}-\frac{1}{n\_2^s}-\cdots\right|\\
&\le \frac{1}{(P+1)^{\Re (s)}}+\frac{1}{(P+2)^{\Re (s)}}+\cdots\end{align}$$
This tends to $0$ as $P\to\infty$ (again **using infinitude of primes**) if $\Re (s)\gt 1$ and the product formula follows.
In [their comment](https://math.stackexchange.com/questions/4674095/proof-of-eulers-product-for-zeta-s-without-using-infinitude-of-primes#comment9880111_4674095) to [this question](https://math.stackexchange.com/questions/4674095/proof-of-eulers-product-for-zeta-s-without-using-infinitude-of-primes) the user J.G. commented that the proofs actually don't use infinitude of primes (which he didn't explain – so I'm very confused and I'm writing this question) and his comments contradict a textbook proof by Amann and Escher (see a [related question here](https://math.stackexchange.com/questions/4165001/zeta-function-product-in-amann-and-escher)) and a [Wikipedia proof](https://en.wikipedia.org/wiki/Proof_of_the_Euler_product_formula_for_the_Riemann_zeta_function#Another_proof) – notice "up to some **prime number limit** $q$".
This is not a duplicate of [this question](https://math.stackexchange.com/questions/3746532/proof-of-infinitude-of-primes-by-eulers-product-formula-is-circular) because the OP there fails to distinguish between the fundamental theorem of arithmetic and infinitude of primes.
No one answered on MSE so I'm posting here.
| https://mathoverflow.net/users/502484 | Do all rigorous proofs of Euler's product for $\zeta (s)$ use infinitude of primes? | Neither proof uses the infinitude of primes. Here is why ($p$ will always denote a prime number).
1. Sieve proof.
Proceed as you indicated, but don't assume that $q$ is prime. We still have that
$$\left|\left(\prod\_{p\le q}\left(1-\frac{1}{p^s}\right)\right)\zeta (s)-1\right|=\left|\sum\_{\substack{n>1\\\forall p:\ p\mid n\ \Rightarrow\ p>q}}\frac{1}{n^s}\right|
\le \sum\_{n>q}\frac{1}{n^{\Re(s)}},\qquad \Re (s)\gt 1.$$
The right-hand side tends to zero as $q\to\infty$, done.
2. Factorization proof
Proceed as you indicated, and recall that
$$\left|\zeta (s)-\prod\_{p\le P}\left(1-\frac{1}{p^s}\right)^{-1}\right|
=\left|\sum\_{\exists p:\ p\mid n\ \&\ p>P}\frac{1}{n^s}\right|
\le \sum\_{n>P}\frac{1}{n^{\Re(s)}},\qquad \Re (s)\gt 1.$$
The right-hand side tends to zero as $P\to\infty$, done.
| 14 | https://mathoverflow.net/users/11919 | 444435 | 179,196 |
https://mathoverflow.net/questions/444447 | 2 | I'm trying to find formulas for the finite difference approximation "Five-points-stencil" of the first derivative for non-constant grid spacing. It's needed for the outermost left and right points. I tried to search for it but can't find it.
| https://mathoverflow.net/users/500658 | Finite difference approximation | Given pairwise distinct real numbers $x\_0,\dots,x\_4$, one can approximate $f'(x\_0)$ by a linear combination $a\_0f(x\_0)+\cdots+a\_4f(x\_4)$ so that
$$g\_j'(x\_0)=a\_0g\_j(x\_0)+\cdots+a\_4g\_j(x\_4)$$
for $g\_j(x):=x^j$ and $j\in\{0,\dots,4\}$.
Solving the resulting system of equations for $a\_0,\dots,a\_4$, we get
$$a\_0=-(a\_1+\cdots+a\_4)$$
and
$$a\_i=\frac{\prod\limits\_{j\in\{0,\dots,4\}\setminus\{0,i\}}(x\_0-x\_j)}
{\prod\limits\_{j\in\{0,\dots,4\}\setminus\{i\}}(x\_i-x\_j)}$$
for $i\in\{1,\dots,4\}$.
(Here it does not matter whether $x\_0$ is an outermost point or not.)
| 2 | https://mathoverflow.net/users/36721 | 444457 | 179,200 |
https://mathoverflow.net/questions/444448 | 2 | I am new to this area and I am a bit confused by the literature. Is there a strong maximum principle for CR functions over domains in a CR manifold, please? If so, could someone please state it (together with its hypotheses etc.) and provide a reference to a proof (or maybe just prove it if it is not too long)? In the mean time, I will keep on digging in the literature, but in any case, it may be of interest to others too.
Edit 1: it looks like the answer is no, according to ex. 3.2.5 in [CR functions (LibreText Mathematics), by Jiří Lebl](https://math.libretexts.org/Bookshelves/Analysis/Tasty_Bits_of_Several_Complex_Variables_(Lebl)/03%3A_CR_Functions/3.02%3A_CR_Functions). Actually, this makes sense, because for example, in the simplest case where our CR manifold is a real hypersurface in $\mathbb{C}^{n+1}$ given by $z\_{n+1} - \bar{z}\_{n+1} = 0$, then being a CR function is, if I understand the definition well, just being holomorphic in the $z\_i$, for $i = 1, \ldots, n$, but it says nothing about how the function depends on $x\_{n+1}$ ($x\_{n+1}$ is the real part of $z\_{n+1}$).
But in my case, the function is also real analytic. So allow me to ask if there is a strong maximum principle for *real analytic* CR functions on a domain inside a CR manifold. I'd also be happy to assume that (if it helps), at each point $p \in M$, the $T\_p^{(1,0)}$ bundle of $M$ has complex rank $n$, where the dimension of $M$ is $2n+1$ (so that, in some sense, $M$ is like a real hypersurface inside some complex $n+1$ dimensional manifold).
| https://mathoverflow.net/users/81645 | Is there a maximum principle for CR functions over domains inside CR manifolds? | It's not really about real-analytic or smooth. It is really about the CR structure of the manifold. In your case, the manifold is given by $\operatorname{Im} z\_{n+1} = 0$, so any CR function is just a function holomorphic in the first $n$ variables, as you noted. Hence, any CR function on your manifold that achieves a maximum will be constant along that "leaf", that is, if $f$ is your CR function on your manifold and achieves a maximum at the origin, then $z' \mapsto f(z',0)$ (where $z' \in {\mathbb{C}}^n$) is constant, but that's the best you can get whether real-analytic or not.
Try to find a counterexample to a strong maximum principle by considering the sphere, which is a wonderful CR manifold. Your counterexample will be a polynomial, so definitely real-analytic. It will be the restriction of an entire function.
On the other hand for a manifold such as the Lewy surface, $\operatorname{Im} z\_3 = |z\_1|^2-|z\_2|^2$, you can prove a maximum principle. For this, look at the chapter in the book that talks about extension and attaching discs. Then note that the attached discs will force the extended holomorphic function to have a maximum at the origin and the standard maximum principle for holomorphic functions applies. To see this, consider an analytic disc attached to the manifold, suppose your function is already extended as a holomorphic function, and note that the one-dimensional maximum principle tells you that the function inside the disc is bounded by its values on the boundary, which are the values of your CR function on the CR manifold.
| 3 | https://mathoverflow.net/users/2783 | 444464 | 179,203 |
https://mathoverflow.net/questions/444432 | 4 | I am looking for a proof of the following statement without using full power of [Chevalley's theorem](https://stacks.math.columbia.edu/tag/00FE) on constructible sets. We say a domain $A$ is $0$*-open* if $\{(0)\}$ is open in $\operatorname{Spec}(A)$. Equivalently, there is an element $x\in A$ such that $A\_x$ is a field. And equivalently, either the only prime is $(0)$, or there is a non-zero element $x\in A$ contained in all non-zero primes $\mathfrak p\subset A$.
>
> For domains $A$ and $B$, let $A\hookrightarrow B$ be injective and of
> finite type. Then $A$ is $0$-open if $B$ is $0$-open.
>
>
>
The above statement could be proved using Chevalley's theorem as below:
If $B$ is $0$*-open*, then let $B\_y$ be a field for some $y\in B$ and the composition $A\hookrightarrow B\hookrightarrow B\_y$ would be injective and finite type. By [this](https://stacks.math.columbia.edu/tag/00FG) elementary lemma, there is a nonzero element $x\in A$ such that $A\_x\hookrightarrow B\_y$ is injective and of finite presentation. Chevalley's theorem would give us that $A\_x$ is $0$-open.
If $A\_x$ is 0-open, then there is a non-zero element $x'\in A$ contained in all non-zero primes $\mathfrak p$ not containing $x$. The element $xx'\neq 0$ is then contained in all non-zero primes of $A$. Thus $A$ is $0$-open, as desired.
I am cross posting the [same question](https://math.stackexchange.com/questions/4672282/localization-at-a-point-is-a-field) from MSE to here.
| https://mathoverflow.net/users/132430 | Finite type injective ring map between domains preserves the open point $(0)$ | Here's a more down to earth argument that uses the weak Nullstellensatz¹ instead of Chevalley's theorem:
**Lemma.** *Let $\phi \colon A \hookrightarrow B$ be an injective ring homomorphism of finite type between integral domains, and assume there exists a nonzero element $y \in B$ such that $B\_y$ is a field. Then there exists a nonzero element $x \in A$ such that $A\_x$ is a field.*
*Proof.* If $A \hookrightarrow B$ is of finite type, then so is $A \hookrightarrow B\_y$. Replacing $B$ by $B\_y$, we may assume that $B$ is a field. Then $A \hookrightarrow B$ is a finite type ring homomorphism, hence so is $\operatorname{Frac} A \hookrightarrow B$. The weak Nullstellensatz then says that $\operatorname{Frac} A \hookrightarrow B$ is a finite extension, so there exists a nonzero element $x \in A$ such that the ring homomorphism $A\_x \hookrightarrow B\_x = B$ is finite.
Now if $\mathfrak p \subseteq A\_x$ is a nonzero prime ideal, then 'going up' [Tag [00GU](https://stacks.math.columbia.edu/tag/00GU)] for the inclusion $(0) \subseteq \mathfrak p$ shows that there exists a nonzero prime ideal $\mathfrak q \subseteq B$ with $\mathfrak q \cap A\_x = \mathfrak p$. This is impossible since $B$ is a field, so we conclude that $A\_x$ is a field. $\square$
**Remark.** You could also try to use generic flatness instead of generic finiteness (this is possibly more natural for this question). But the lemma that a finite type flat map is open usually depends on Chevalley's theorem, or at least gets very close to proving Chevalley. By contrast, the proof above only uses 'going up' without discussing closedness or openness of morphisms of schemes.
---
¹ The weak Nullstellensatz says that if $K \to L$ is a finite type ring homomorphism between fields, then it is actually finite.
| 2 | https://mathoverflow.net/users/82179 | 444471 | 179,208 |
https://mathoverflow.net/questions/444321 | 4 | It is known that
\begin{equation\*}
\tan x=\sum\_{k=1}^{\infty}\frac{2^{2k}\bigl(2^{2k}-1\bigr)}{(2k)!}|B\_{2k}|x^{2k-1}, \quad |x|<\frac{\pi}{2}
\end{equation\*}
and
\begin{equation\*}
\ln\tan x=\ln x+\sum\_{k=1}^{\infty}\frac{2^{2k}\bigl(2^{2k-1}-1\bigr)}{k(2k)!}|B\_{2k}|x^{2k}, \quad 0<x<\frac{\pi}{2},
\end{equation\*}
where $B\_{2k}$ denotes the classical Bernoulli numbers. From the second series expansion, we acquire
\begin{equation\*}
\ln\frac{\tan x}{x}=\sum\_{k=1}^{\infty}\frac{2^{2k}\bigl(2^{2k-1}-1\bigr)}{k(2k)!}|B\_{2k}|x^{2k}, \quad |x|<\frac{\pi}{2}.
\end{equation\*}
My question is:
What and where is the series expansion of the function $\ln\bigl(\frac{\tan x}{x}-1\bigr)$ or $\ln(\tan x-x)$ around $x=0$?
References
1. I. S. Gradshteyn and I. M. Ryzhik, *Table of Integrals, Series, and Products*, Translated from the Russian, Translation edited and with a preface by Daniel Zwillinger and Victor Moll, Eighth edition, Revised from the seventh edition, Elsevier/Academic Press, Amsterdam, 2015; available online at <https://doi.org/10.1016/B978-0-12-384933-5.00013-8>.
| https://mathoverflow.net/users/147732 | What or where is the series expansion of the function $\ln\bigl(\frac{\tan x}{x}-1\bigr)$ or $\ln(\tan x-x)$ around $x=0$? | Let
\begin{equation\*}
f(x)=\begin{cases}
\ln\dfrac{\tan x-x}{x^3}, & 0<|x|<\dfrac{\pi}{2};\\
-\ln3, & x=0.
\end{cases}
\end{equation\*}
Then the even function $f(x)$ has the Maclaurin power series expansion
\begin{equation}\label{ln-tan-x-cubic-ser-expans}
\begin{aligned}
f(x)&=-\ln3-\sum\_{k=1}^{\infty}\frac{3^{2k}D\_{2k}}{(2k)!}x^{2k}\\
&=-\ln3+\frac{2 x^2}{5}+\frac{43 x^4}{525}+\frac{524 x^6}{23625}+\frac{40897 x^8}{6063750} +\frac{19393844 x^{10}}{8868234375}+\dotsm
\end{aligned}
\end{equation}
for $|x|<\frac{\pi}{2}$, where
\begin{align\*}
D\_{2k}&=\begin{vmatrix}
0 & Q\_0 & 0 & \dotsm&0& 0& 0\\
Q\_1 & 0 & Q\_0 & \dotsm& 0&0& 0\\
0 & Q\_1 & 0 & \dotsm& 0& 0&0\\
\dotsm & \dotsm & \dotsm & \ddots&\dotsm& \dotsm& \dotsm\\
Q\_{k-1} & 0 & \binom{2k-3}{1}Q\_{k-2} & \dotsm& 0 & Q\_0 & 0\\
0 & Q\_{k-1} & 0 & \dotsm& \binom{2k-2}{2k-4}Q\_1 & 0 & Q\_0\\
Q\_{k} & 0 & \binom{2k-1}{1}Q\_{k-1} & \dotsm& 0 & \binom{2k-1}{2k-3}Q\_1 & 0
\end{vmatrix}\\
&=\begin{vmatrix}
e\_{i,j}
\end{vmatrix}\_{2k\times2k}, \quad k\ge1,\\
e\_{i,j}&=\begin{cases}
0, & (i,j)=(2\ell-1,1),\quad 1\le\ell\le k;\\
Q\_\ell, & (i,j)=(2\ell,1),\quad 1\le\ell\le k;\\
0, & 1\le i\le j-2\le2k-2;\\
0, & i-j=2\ell, \quad 0\le\ell\le k-1;\\
\binom{i-1}{j-2}Q\_{\ell}, & i-j=2\ell-1, \quad 0\le\ell\le k-1,\quad j\ge2,
\end{cases}\\
Q\_m&=\frac{2^{2m+2}\bigl(2^{2m+4}-1\bigr)}{(m+1)(m+2)(2m+1)(2m+3)}|B\_{2m+4}|, \quad m\ge0,
\end{align\*}
and $B\_{2m+4}$ denotes the Bernoulli numbers generated by
\begin{equation\*}
\frac{z}{\operatorname{e}^z-1}=\sum\_{k=0}^\infty B\_k\frac{z^k}{k!}=1-\frac{z}2+\sum\_{k=1}^\infty B\_{2k}\frac{z^{2k}}{(2k)!}, \quad |z|<2\pi.
\end{equation\*}
As corollaries of the above Maclaurin power series expansion, we deduce the series expansions
\begin{equation\*}
\ln(\tan x-x)=-\ln3+3\ln x-\sum\_{k=1}^{\infty}\frac{3^{2k}D\_{2k}}{(2k)!}x^{2k}, \quad 0<x<\frac{\pi}{2}
\end{equation\*}
and
\begin{equation\*}
\ln\biggl(\frac{\tan x}{x}-1\biggr)
=-\ln3+2\ln|x|-\sum\_{k=1}^{\infty}\frac{3^{2k}D\_{2k}}{(2k)!}x^{2k}, \quad 0<|x|<\frac{\pi}{2}.
\end{equation\*}
Since the Bernoulli numbers can be expressed by closed-form formulas, so the sequence $D\_{2k}$ is surely a closed-form formula.
About closed-form formulas for the Bernoulli numbers and about some conclusions on the tangent function, I would like to recommend the following papers.
References
1. Xue-Yan Chen, Lan Wu, Dongkyu Lim, and Feng Qi, *Two identities and closed-form formulas for the Bernoulli numbers in terms of central factorial numbers of the second kind*, Demonstratio Mathematica **55** (2022), no. 1, 822--830; available online at <https://doi.org/10.1515/dema-2022-0166>.
2. Feng Qi and Jacques Gelinas, *Revisiting Bouvier's paper on tangent numbers*, Advances and Applications in Mathematical Sciences **16** (2017), no. 8, 275--281.
3. Feng Qi and Bai-Ni Guo, *An explicit formula for derivative polynomials of the tangent function*, Acta Universitatis Sapientiae Mathematica **9** (2017), no. 2, 348--359; available online at <https://doi.org/10.1515/ausm-2017-0026>.
4. Feng Qi, *Derivatives of tangent function and tangent numbers*, Applied Mathematics and Computation **268** (2015), 844--858; available online at <https://doi.org/10.1016/j.amc.2015.06.123>.
5. Jiao-Lian Zhao, Qiu-Ming Luo, Bai-Ni Guo, and Feng Qi, *Remarks on inequalities for the tangent function*, Hacettepe Journal of Mathematics and Statistics **41** (2012), no. 4, 499--506.
6. Chao-Ping Chen and Feng Qi, *A double inequality for remainder of power series of tangent function*, Tamkang Journal of Mathematics **34** (2003), no. 4, 351--355; available online at <https://doi.org/10.5556/j.tkjm.34.2003.236>.
7. <https://math.stackexchange.com/a/4248328>
8. <https://math.stackexchange.com/a/4248341>
9. <https://math.stackexchange.com/a/4248488>
10. <https://math.stackexchange.com/a/4254493>
11. <https://math.stackexchange.com/a/4254500>
12. <https://math.stackexchange.com/a/4256893>
13. <https://math.stackexchange.com/a/4256913>
14. <https://math.stackexchange.com/a/4256915>
15. <https://math.stackexchange.com/a/4656534>
| -1 | https://mathoverflow.net/users/147732 | 444485 | 179,210 |
https://mathoverflow.net/questions/444484 | 3 | I know that maximal ideals of $C[0, 1]$ corresponds to singleton. Also, using Zorn's lemma one can construct a prime ideal in $C[0, 1]$ which is not maximal.
>
> Is there any $\textbf{closed}$ prime ideal of $C[0, 1]$ which is not maximal?
>
>
>
Any references or ideas?
| https://mathoverflow.net/users/129638 | Closed prime ideal in $C[0, 1]$ | No. Note that an ideal in a commutative ring with identity is prime if and only if the quotient ring is an integral domain.
Now consider $C[0,1]$. It is known that the closed ideals in this Banach algebra are all of the form $J\_F = \{ f\in C[0,1] \colon {f\vert}\_F =0 \}$ for some closed subset $F\subseteq [0,1]$, and $C[0,1]/J\_F \cong C(F)$. If $F$ contains two distinct points then $C(F)$ is not an integral domain (just take a "spike" which peaks at $x$ and is supported on a small neighbourhood, and then do likewise for $y$).
The same should be true with $[0,1]$ replaced by any compact Hausdorff space (in the last part of the proof, Urysohn's lemma will produce functions with the desired properties).
---
Update: following requests in the comments, here is a characterization of the closed ideals in $C(X)$, where $X$ is any compact Hausdorff space. Since I am doing this off the top of my head and writing for analysts, I'm going to use excluded middle freely; I'm sure the arguments could be reformulated to reduce or remove this.
Note that I do not need any ${\rm C}^\ast$-algebra theory or functional calculus, and several ingredients below will work for any regular Banach function algebra.$\newcommand{\hull}{\rm hull}$
Given an ideal $J\subseteq C(X)$ let $\ker(J)=\{ x\in X \colon f(x)=0 \,\forall\,f\in J\}$. Given a subset $S\subseteq X$ let $\hull(S)=\{ f\in C(X) \colon f(x)=0 \,\forall\,x\in S\}$. Note that $\ker(J)$ is always closed in $X$ and $\hull(S)$ is always norm-closed in $C(X)$
One can check that $\ker$ and $\hull$ form a Galois connection (adjunction between appropriate posets) but we only need the prosaic fact that $\ker(\hull(J))\supseteq J$. In fact, since $\ker$ of any subset is norm-closed, $\overline{J}\subseteq \ker(\hull(J))$. What I will now show is that this last inclusion is an equality.
So let $a\in \ker(\hull(J))$ and let $\varepsilon>0$. Let $U=\{ x\in X \colon |a(x)| <\varepsilon\}$; this is an open neighbourhood of $\hull(J)$.
Suppose we can produce $b\in J$ such that $b\geq 0$ and $b(x) \geq 1$ for all $x\in X\setminus U$.
Consider $a\cdot (\varepsilon +b)^{-1}b$, which belongs to $J$. Note that if $x\in U$ then
$$
\left\lvert a(x) - \dfrac{a(x)b(x)}{\varepsilon+b(x)} \right\rvert \leq |a(x)| < \varepsilon
$$
and if $x\in X\setminus U$ then
$$
\left\lvert a(x) - \dfrac{a(x)b(x)}{\varepsilon+b(x)} \right\rvert
= \dfrac{\varepsilon \lvert a(x)\rvert}{\varepsilon+b(x)} \leq |a(x)| \dfrac{\varepsilon}{\varepsilon+1}
$$
Putting these together gives
$$
\lVert a - a\cdot(\varepsilon+b)^{-1}b \rVert\_\infty \leq \max\left(\varepsilon, \dfrac{\varepsilon}{\varepsilon+1}\lVert a\rVert\_\infty\right).
$$
Since $\varepsilon>0$ is arbitrary, we can conclude that $a\in \overline{J}$.
It remains to produce $b\in J$ with the desired properties, which we do using a routine compactness argument. For each $x\in X\setminus U$, since $x\notin \hull(J)$ we can pick $f\_x\in J$ such that $f\_x(x)=2$, and then let $U\_x$ be an open neighbourhood of $x$ such that $|f\_x(y)|^2 > 1$ for all $y\in X\setminus U\_x$. By compactness we can extract a finite subcover, relabelled as $U\_1,\dots, U\_m$ with corresponding $f\_1,\dots, f\_n \in J$ that satisfy $|f\_i(y)|^2 > 1$ for all $y\in U\_i$. Now put $b= f\_1 \overline{f\_1} + \dots + f\_n\overline{f\_n} \in J$. Clearly $b\geq 0$ and $b(y) \geq 1$ for all $y\in X\setminus U$, as required.
**Remark:** pursuing these ideas further one can show that the Jacobson topology = hull-kernel topology on $X$ (analogue of Zariski topology) coincides with the Gelfand topology. This is definitely not true for general commutative semisimple Banach algebras.
| 6 | https://mathoverflow.net/users/763 | 444488 | 179,212 |
https://mathoverflow.net/questions/444472 | 1 | Classical case:
Let $\{a,b\}$ be linearly independent set over $\mathbb Q$ and $\{e^{at},e^{bt}\}$ be linearly independent set over $\mathbb Q[[t]]$. Suppose $P(x,y)$ is a polynomial over $\mathbb Q$. Then it seems $P(e^{a},e^{b})=0$ implies $P=0$, because
if $P(x,y)=Ax^2+Bxy+Cy^2 \in \mathbb Q[x,y]$. Then,
\begin{align}& P(e^{a},e^{b})=0 \\
\Rightarrow & Ae^{2a}+Be^{a+b}+Ce^{2b}=0 \\ \Rightarrow &A=B=C=0,~\text{because $a,b$ are linearly independent} \end{align}
The same is true for any arbitrary polynomials $P(x,y) \in \mathbb Q[x,y]$, because $P(e^{a},e^{b})$ can be written as linear combinations of exponentials of the form $e^{c}$, where $c=n\_1a+n\_2b$, $n\_1,n\_2 \in \mathbb N$. But since $\{a,b\}$ is linearly independent, $P=0$.
>
> Does the same hold for $p$-adic exponential as well ?
>
>
>
$p$-adic case:
Let $\exp\_p$ denotes the $p$-adic exponential function satisfying $\exp\_p(x+y)=\exp\_p(x) \exp\_p(y)$ for $|x,y|\_p<p^{-\frac{1}{p-1}}$.
Let $\{a,b\}$ be linearly independent set over $\mathbb Q$ and $\{\exp\_p(at), \exp\_p(bt)\}$ be linearly independent set over $\mathbb Q[[t]]$. Suppose $P(x,y)$ is a polynomial over $\mathbb Q$. Then it seems $P(\exp\_p(a), \exp\_p(b))=0$ implies $P=0$, because
if $P(x,y)=Ax^2+Bxy+Cy^2 \in \mathbb Q[x,y]$. Then,
\begin{align}& P(\exp\_p(a), \exp\_p(b))=0 \\
\Rightarrow & A\exp\_p(2a)+B \exp\_p(a+b)+C\exp\_p(2b)=0 \end{align}
I think, as $a,b$ are linearly independent, the $p$-adic exponential $\exp\_p(2a),~\exp\_p(2b)$ and $\exp\_p(2(a+b))$ will not cancel, so we must have $A=B=C=0$. In other word, $P=0$. The same will be true for arbitrary polynomial.
So the answer seems to be positive, if there is no flaw in my argument.
I appreciate your comments.
| https://mathoverflow.net/users/493164 | Does $P(\exp_p(a),\exp_p(b))=0$ imply $P=0$, where $\exp_p(\cdot)$ is $p$-adic exponential? | As written, both the classical and $p$-adic statements are false. Let $a=\log(9)$ and $b=\log(17)$ (classical or $2$-adic log). Then they are linearly independent over $\mathbb{Q} $ since otherwise we would find integers $n, m$ with $9^n=17^m$. But $P(\exp(a), \exp(b))=0$ for the nontrivial polynomial $P(x, y) = 17x-9y$.
(Note that the power series $\exp(at), \exp(bt)$ are still linearly independent as asked in the question.)
| 3 | https://mathoverflow.net/users/39747 | 444495 | 179,216 |
https://mathoverflow.net/questions/444497 | 2 | Let $G$ be a locally compact group, $K$ a compact subgroup in $G$, and $\mu\_K$ the normalized Haar measure on $K$:
$$
\mu\_K(K)=1.
$$
Let us denote by $\widetilde{\mu\_K}$ the measure on $G$ defined as the functional on ${\mathcal C}(G)$ by the formula
$$
\widetilde{\mu\_K}(u)=\int\_K u(t) \ \mu\_K(dt), \qquad u\in{\mathcal C}(G)
$$
Suppose now that $K$ and $L$ are two normal compact subgroups in $G$, and $H$ is the closed normal subgroup in $G$ generated by $K$ and $L$:
$$
H=\operatorname{Gr}(K\cup L)
$$
I have two questions:
>
> 1. Is it true that $H$ is always compact?
>
>
>
and
>
> 2. Suppose $H$ is compact, does the following equality hold:
> $$
> \widetilde{\mu\_K}\*\widetilde{\mu\_L}=\widetilde{\mu\_H}
> $$
> ?
>
>
>
| https://mathoverflow.net/users/18943 | Haar measures of compact subgroups | The answer to both questions seems yes to me.
First, only assume that $K$ is a normal subgroup of $G$. Then, $(\text{Ad } g)\_{g \in L}$ defines a continuous action of $L$ by automorphisms of $K$ and we can define the semidirect product group $K \rtimes L$, which is just the set $K \times L$ with product
$$(x,g) \cdot (y,h) = (x (gyg^{-1}),gh) \; .$$
Then $K \times L$ is a compact group. Because the Haar measure $\mu\_K$ of $K$ is invariant under every continuous automorphism of $K$, we get that $\mu\_K \times \mu\_L$ is the Haar measure of $K \times L$.
The product map $\theta : K \times L \to G : \theta(x,g) = xg$ is a continuous group homomorphism. The image $\theta(K \times L)$ is compact and equals the group considered in the question. The Haar measure on this group is given by $\theta\_\*(\mu\_K \times \mu\_L)$, which is equal to $\widetilde{\mu\_K} \* \widetilde{\mu\_L}$.
Note that $\theta(K \times L) = K \cdot L$ equals the subgroup of $G$ generated by $K$ and $L$.
Finally, if both $K$ and $L$ are normal in $G$, then for every $g \in G$, we have that
$$g \cdot (K \cdot L) \cdot g^{-1} = (gKg^{-1}) \cdot (gLg^{-1}) = K \cdot L$$
so that $K \cdot L$ is normal in $G$. This means that $K \cdot L$ equals the group $H$ in the question.
| 1 | https://mathoverflow.net/users/159170 | 444504 | 179,220 |
https://mathoverflow.net/questions/444500 | 5 | We assume that we have a $\alpha$-Hölder continuous function $f$ on an interval $[0,1]$ with $f(0)=0$.
I am wondering if there exists an explicit construction of a sequence $f\_{n} \in C\_c^{\infty}(\mathbb R)$ such that
$$\lVert f-f\_n\rVert\_{C^{\alpha}([0,1])} \le \frac{1}{n}$$
and $\lvert f\_n(x)\rvert \le \lvert f(x)\rvert$ on $[0,1]$. The usual convolution idea does not respect the last condition.
| https://mathoverflow.net/users/496243 | Approximation of Hölder continuous functions "from below" | It is not possible: your condition $|f\_n|<|f|$ implies
that the zero set of $f\_n$ is contained in the zero set of $f$. So $f(x)=|x|^{\alpha}$ cannot be approximated by a smooth function $f\_n$, since $f\_n(x)=(c+o(1))x$, and $\sup|f(x)-f\_n(x)|\geq (1+o(x))|x|^\alpha$ for some small $|x|$.
| 7 | https://mathoverflow.net/users/25510 | 444506 | 179,222 |
https://mathoverflow.net/questions/444481 | 1 | Is the integral
$$
t^2\left(\iint\_{\mathbb{R}^2}|\xi|e^{\frac{-a\xi^4t}{\xi^2+b}}\,d\xi\_1\,d\xi\_2\right)$$ bounded when $t\rightarrow\infty$? Here
* $\xi=(\xi\_1,\xi\_2)\in\mathbb{R}^2$,
* $|\xi|=\sqrt{\xi\_1^2+\xi\_2^2}$,
* $a$ and $b$ are fixed positive numbers.
| https://mathoverflow.net/users/502529 | Whether the integral $t^2(\iint_{\mathbb{R}^2}|\xi|e^{\frac{-a\xi^4t}{\xi^2+b}} \,d\xi_1 \,d\xi_2)$ is bounded? | Assuming that by $\xi^2$ and $\xi^4$ you meant, respectively, $|\xi|^2$ and $|\xi|^4$, switching to polar coordinates, and making the substitution $r=ut^{-1/4}$, we have
$$
I(t):=t^2\left(\iint\_{\mathbb{R}^2}|\xi|e^{\frac{-a\xi^4t}{\xi^2+b}}\,d\xi\_1\,d\xi\_2\right)
=2\pi t^2\int\_0^\infty dr\,r^2\exp\Big(-\frac{tar^4}{r^2+b}\Big)
=2\pi t^{5/4}J(t),$$
where
$$J(t):=\int\_0^\infty du\,u^2\exp\Big(-\frac{au^4}{t^{-1/2}u^2+b}\Big).$$
By dominated convergence, as $t\to\infty$,
$$J(t)\to J(\infty):=\int\_0^\infty du\,u^2\exp\Big(-\frac{au^4}{b}\Big)\in(0,\infty).$$
So,
$$
I(t)\sim2\pi J(\infty) t^{5/4}\to\infty$$
as $t\to\infty$. In particular, $I(t)$ is unbounded.
| 1 | https://mathoverflow.net/users/36721 | 444511 | 179,224 |
https://mathoverflow.net/questions/444515 | 4 | A locally compact quantum group (in the sense of Vaes-Kustermans) consists of the data $(M, \Delta, \varphi, \psi)$ with $M$ a von Neumann algebra, $\Delta: M \to M \overline{\otimes} M$ a normal unital $\*$-homomorphism satisfying coassociativity and $\varphi, \psi: M\_+ \to [0, \infty]$ normal, semifinite, faithful weights satisfying left and right invariance.
One often writes $M= L^\infty(\mathbb{G})$ and refers to $\mathbb{G}$ as the locally compact quantum group.
The motivating example for the latter notation comes from the situation where $G$ is a locally compact group, $M= L^\infty(G)=L^\infty(G, \lambda)$, $\Delta: L^\infty(G)\to L^\infty(G)\overline{\otimes} L^\infty(G) \cong L^\infty(G\times G)$ given by $\Delta(f)(s,t) = f(st)$ and $$\varphi: L^\infty(G)\to \mathbb{C}: f \mapsto \int\_G f d\lambda, \quad \psi: L^\infty(G)\to \mathbb{C}: f \mapsto \int\_G fd\rho$$
where $\lambda$ is left Haar measure and $\rho$ is right Haar measure.
However, I see some problems. For example, for a general locally compact group $G$, it need not even be true that $L^\infty(G, \lambda)$ is a von Neumann algebra (acting on $L^2(G, \lambda)$) or it does not even need to be a $W^\*$-algebra (if it is a $W^\*$-algebra, then it is automatically true that integration is a normal weight). Of course, if $G$ is $\sigma$-compact, then everything works out nicely. So concretely, my question is:
If $G$ is a non $\sigma$-compact locally compact group, how exactly can we view it as a locally compact quantum group? Do we have to re-define $L^\infty(G, \lambda)$ as is usually done in harmonic analysis?
| https://mathoverflow.net/users/216007 | Every locally compact group gives rise to a locally compact quantum group | In the general, not necessarily $\sigma$-compact setting, one has to interpret $L^\infty(G,\lambda)$ as the von Neumann algebra of locally measurable functions that are bounded outside a locally null set. This is very well explained in Section 2.3 of Folland's "A course in abstract harmonic analysis".
Also note that as explained in Proposition 2.4 of that same book, every locally compact group $G$ admits a $\sigma$-compact subgroup $G\_0 \subset G$ that is open. So measure theoretically, we are always in the quite tame situation of a (possibly uncountable) disjoint union of copies of $G\_0$.
| 5 | https://mathoverflow.net/users/159170 | 444520 | 179,227 |
https://mathoverflow.net/questions/444525 | 4 | According to [this comment](https://mathoverflow.net/questions/444490/ask-for-a-proof-of-an-inequality-involving-the-bernoulli-numbers#comment1147661_444490) and [this comment](https://mathoverflow.net/questions/444490/ask-for-a-proof-of-an-inequality-involving-the-bernoulli-numbers#comment1147671_444490), a positive answer to [this recent question](https://mathoverflow.net/q/444490/36721) (about Bernoulli numbers) would be sufficient to prove the following:
>
> $r:=f/g$ is increasing on $(0,\pi/2)$ from $5/6$ to $1$, where
> \begin{equation}
> f(x):=\ln\frac{\tan x}x,\quad g(x):=\ln\frac{\tan x-x}{x^3/3}.
> \end{equation}
>
>
>
However, the mentioned sufficient condition seems much, much harder to prove than the actually desired highlighted claim above. In the answer below, the highlighted claim will be proved, using a so-called l'Hospital-type rule for monotonicity.
| https://mathoverflow.net/users/36721 | On the monotonicity of the ratio of two logarithmic expressions | The values $5/6$ and $1$ for $r(0+):=\lim\_{x\downarrow0}r(x)$ and $r(\frac\pi2-):=\lim\_{x\uparrow\pi/2}r(x)$ are easy to get by the l'Hospital rule for limits.
It remains to show that $r=f/g$ is increasing on $(0,\pi/2)$.
Note that $f(0+):=\lim\_{x\downarrow0}f(x)=0$ and $g(0+)=0$. Next, for real $t>0$ we have $0<\arctan t<t$ and
\begin{equation}
G\_1(t):=g'(\arctan t)\frac{(t-\arctan t) \arctan t}{3 + t^2}
=\arctan t-\frac{3t}{3+t^2}.
\end{equation}
In what follows, by default, $t\in(0,\infty)$.
Next, $G\_1(0)=0$ and $G'\_1(t)=4 t^4/((1 + t^2) (3 + t^2)^2)>0$. So, $G\_1>0$ (on $(0,\infty)$) and hence $g'>0$ (on $(0,\pi/2)$). So, by the l'Hospital-type rule for monotonicity given by [Proposition 4.1](https://www.emis.de/journals/JIPAM/images/157_05_JIPAM/157_05.pdf), it is enough to show that the "derivative ratio"
\begin{equation}
r\_1(t):=\frac{f'(\arctan t)}{g'(\arctan t)}=\frac{f\_1(t)}{g\_1(t)}
\end{equation}
is increasing (in $t\in(0,\infty)$), where
\begin{equation}
f\_1(t):=\frac{(t - \arctan t) (-t + (1 + t^2) \arctan t)}{1 + t^2},
\end{equation}
\begin{equation}
g\_1(t):=\frac{t (-3 t + (3 + t^2) \arctan t)}{1 + t^2}.
\end{equation}
Note that
\begin{equation}
G\_2(t):=g\_1'(t)\frac{(1 + t^2)^2}{3 + t^4}=\frac{t (-3 + t^2)}{3 + t^4} +\arctan t,
\end{equation}
so that $G\_2(0)=0$ and $G'\_2(t)=(8 t^4 (3 + t^2))/((1 + t^2) (3 + t^4)^2)>0$. So, $G\_2>0$ and hence $g\_1'>0$.
Also, $f\_1(0+)=0$ and $g\_1(0+)=0$.
So, by the same l'Hospital-type rule for monotonicity, it is enough to show that the second "derivative ratio"
\begin{equation}
r\_2(t):=\frac{f\_1'(t)}{g\_1'(t)}=\frac{f\_2(t)}{g\_2(t)}
\end{equation}
is increasing, where
\begin{equation}
f\_2(t):=\frac{t^2 (t + (-1 + t^2)\arctan t)}{3 + t^4},
\end{equation}
\begin{equation}
g\_2(t):=\frac{t (-3 + t^2) + (3 + t^4) \arctan t}{3 + t^4}.
\end{equation}
Next, $g'\_2(t)=(8 t^4 (3 + t^2))/((1 + t^2) (3 + t^4)^2)>0$ and $f\_2(0+)=0=g\_2(0+)$.
So, using once again the same l'Hospital-type rule for monotonicity, we see that it is enough to show that the third "derivative ratio"
\begin{equation}
r\_3(t):=\frac{f\_2'(t)}{g\_2'(t)}
=\frac{3 t + 6 t^3 - t^5 + (-3 + 3 t^2 + 7 t^4 + t^6) \arctan t}{4 t^3 (3 + t^2)}
\end{equation}
is increasing. Let
\begin{equation}
r\_{31}(t):=r'\_3(t)\frac{4 t^4 (3 + t^2)^2}{(3 + t^4) (9 + 2 t^2 + t^4)}
=\frac{t (-9 + t^2)}{9 + 2 t^2 + t^4} + \arctan t.
\end{equation}
Then $r\_{31}(0)=0$ and $r'\_{31}(t)=(32 t^4 (3 + t^2))/((1 + t^2) (9 + 2 t^2 + t^4)^2)>0$, so that $r\_{31}>0$ and hence $r\_3$ is indeed increasing. $\quad\Box$
| 5 | https://mathoverflow.net/users/36721 | 444526 | 179,229 |
https://mathoverflow.net/questions/444524 | 2 | Side note: so far neither Bard nor ChatGPT has managed to do this correctly, even when I show the errors.
I have 4N players ( N = 4 or N = 5 suffices) and want to set up three rounds of play. In each round, there will be N games played (four players per game). I want to set up the groupings so that no two players appear together more than once, regardless of whether they are teammates or not.
Is there a reasonably simple non-exhaustive search algorithm that will set up these rounds for me?
The first round, of course is easy:
(16 players, each row is the players in one game)
1,2,3,4
5,6,7,8
9,10,11,12
13,14,15,16
To clarify: in a given round, every player is in one of the foursomes. All games (foursomes) are played simultaneously in a given round.
So in the second round, player 1 must not be in the same game as any of 2,3,4 ;and so on. I can't tell whether the [Hungarian](https://en.wikipedia.org/wiki/Hungarian_algorithm) or [Round Robin](https://en.wikipedia.org/wiki/Round-robin_item_allocation) algos can't do this or just that the "AI" software can't do it correctly.
**Another thought**
Based on an external suggestion; not sure this holds in all cases.
First divide the players into two equal sets. It's relatively easy to create unique pairs for the rounds in each of these "brackets." Then any combination of pairs from BracketA with pairs from BracketB will yield the groups of four that I'm looking for. Is this valid?
| https://mathoverflow.net/users/50013 | Optimal algorithm for a "round robin" doubles tournament? | For $N=5$ and more generally $N$ relatively prime to $6$, it is easy to make $N$ rounds.
Name the players $(i,j)$ for $0 \leq i \leq 3$ and $0 \leq j \leq N-1$. In the $a$-th round, take the elements of the $b$-th quadruple to be $\{ (0,b \bmod N), (1,a+b \bmod N), (2, 2a+b \bmod N), (3, 3a+b \bmod N) \}$.
Suppose, to the contrary that $(i\_1, j\_1)$ and $(i\_2, j\_2)$ play together in both rounds $a$ and $a'$. Since $(i,j\_1)$ and $(i,j\_2)$ never play in the same round, we must have $i\_1 \neq i\_2$. Then the equation that $i\_1$ and $i\_2$ play together in round $a$ says that $j\_1 - a i\_1 \equiv j\_2 - a i\_2 \equiv N$ and, likewise, $j\_1 - a' i\_1 \equiv j\_2 - a' i\_2 \bmod N$. So $(a-a') (i\_1 - i\_2) \equiv 0 \bmod N$. But $i\_1 - i\_2 \in \{ \pm 1, \pm 2, \pm 3 \}$ and $\text{GCD}(N,6)=1$, so this implies that $a=a'$. $\square$
---
A more pictorial description. I'll use bridge instead of tennis, because the 4-sides of a table make for good nomenclature. Arrange $N$ bridge tables in a cricle. At each of $N$ tables, let one player sit at north, one at east, one at south and one at north.
After the first set of games, the north players stay put, the east players move on one table, the south players move on two tables and the west players move on three table, each staying in their same position (north, east, west or south). Repeat this $N$ times.
| 3 | https://mathoverflow.net/users/297 | 444528 | 179,230 |
https://mathoverflow.net/questions/444531 | 11 | Mertens' first theorem states that
$$
\sum\_{p \leq n} \frac{\log p}{p} = \log n + O(1).
$$
I read in [this paper](https://tenenb.perso.math.cnrs.fr/PPP/PropStatFriables.pdf) that the following variant is "classical":
$$
\sum\_{p \leq n} \frac{\log p}{p - 1} = \log n - \gamma + o(1).
$$
Could anyone provide a reference (or a proof) to the "classical" variant?
| https://mathoverflow.net/users/502584 | Mertens-like theorem | This lies beyond Mertens, in the sense that this variant actually implies the Prime Number Theorem, as will be explained below, while Mertens' theorem is weaker than the PNT.
I sketch below a complex analytic proof of the variant, due to Landau.
---
Let $\Lambda$ be the von Mangoldt function, defined as $\Lambda(p^k)=\log p$ if $p$ is a prime and $k \ge 1$, and $\Lambda(n)=0$ otherwise. Note that your sum
$$S\_1(n):=\sum\_{p \le n} \frac{\log p}{p-1}$$
is very close to
$$S\_2(n):= \sum\_{m \le n} \frac{\Lambda(m)}{m}.$$
Indeed,
$$S\_2(n)-S\_1(n) = \sum\_{p \le n} \log p \sum\_{1 \le k \le \log\_p n} p^{-k}- \sum\_{p \le n} \log p\sum\_{k \ge 1} p^{-k} = \sum\_{p \le n} \log p \sum\_{k >\log \_p n} p^{-k}.$$
The inner $k$-sum is always $\ll \min\{1/n, 1/p^2\}$, so that this difference is
$$\ll \sum\_{p \le n} \log p \min\{1/n, 1/p^2\}\ll \sum\_{m \le \sqrt{n}} \frac{\log m}{n} + \sum\_{\sqrt{n}<m \le n} \frac{\log m}{m^2} \ll \frac{\log n}{\sqrt{n}}=o(1).$$
We see it suffices to estimate $S\_2(n)$.
---
The reason that I introduced $\Lambda$ is that the PNT is usually proved with $\Lambda$. The Dirichlert series of $\Lambda$ is $-\zeta'(s)/\zeta(s)$, so Perron's formula gives
$$S\_2(n) = \frac{1}{2\pi i}\int\_{(1)}-\frac{\zeta'}{\zeta}(s+1) n^s \frac{ds}{s}+O((\log n)/n).$$
The main term $\log n - \gamma$ arises as the residue of the double pole of the integrand at $s=0$. Indeed, near $s=0$ we have the Laurent expansion
$$\begin{align}-\frac{\zeta'}{\zeta}(s+1)n^s/s&= s^{-2} (1-\gamma s + O(s^2)) (1+s \log n + O(s^2 \log^2 n)) \\
&= s^{-2}(1+(\log n-\gamma)s+O(s^2)).
\end{align}$$
Here we made use of
$$(\star)\, -\frac{\zeta'}{\zeta}(s) = \frac{1}{s-1} -\gamma + O(s-1)$$
near $s=1$. To justify the above heuristic one needs, as in the classical proof of the PNT, to 1) apply a truncated version of Perron's formula and 2) shift the contour to the left using a zero-free region. This is why I believe Hadamard and de la Vallée Poussin had possibly known this variant.
Landau, in his book "Handbuch der Lehre von der Verteilung der Primzahlen" (1911) used Perron's formula to work out an explicit formula for $\sum\_{m \le n} \Lambda(m)/m^s$ for general $s$. For $s=1$ it corresponds to $S\_2(n)$. A reference with proof is Lemma 4.1 [in this preprint](https://arxiv.org/abs/2211.08973) (specialize to $s=1$). In particular, letting
$$\Delta(n):=S\_2(n)-( \log n- \gamma)$$
then, assuming $n$ is a positive integer,
$$\Delta(n)=\frac{\Lambda(n)}{2n}- \sum\_{\rho} \frac{n^{\rho-1}}{\rho-1} \ll \exp(-c(\log n)^{3/5}(\log \log n)^{-1/5})$$
for some $c>0$.
Here the sum is over the zeros of $\zeta$, and the inequality follows from the Vinogradov--Korobov zero-free region. Under RH, $\Delta(n)\ll n^{-1/2}\log^2 n$.
---
The Laurent expansion $(\star)$ is not mysterious; it's equivalent to the Laurent expansion
$$(\star \star)\, \zeta(s) = \frac{1}{s-1} + \gamma +O(s-1)$$
at the same point, which in turn is equivalent to the Harmonic sum $\sum\_{m \le n} 1/n$ being equal to $\log n + \gamma + O(1/n)$. See Corollary 1.16 in Montgomery and Vaughan's book, "Multiplicative Number Theory I" for a rigorous derivation of $(\star \star)$.
---
Here is a more accessible proof. Given an arithmetic function $f=\alpha\*\beta$ we have
$$\sum\_{n \le x} f(n)=\sum\_{n \le x} \alpha(n) \sum\_{m\le x/n} \beta(m).$$
Applying this with $\log = \mathbf{1}\*\Lambda$ where $\log$ is the natural logarithm and $\mathbf{1}$ is the constant function taking the value $1$, we find
$$\sum\_{n \le x} \log n = \sum\_{n \le x} \Lambda(n)\lfloor x/n \rfloor.$$
This identity goes back to Chebyshev. Dividing both sides by $x$ we see
$$\sum\_{n \le x} \frac{\Lambda(n)}{n} =x^{-1}\sum\_{n \le x} \log n +x^{-1} \sum\_{n \le x} \Lambda(n) \{x/n\}$$
where $\{t\} \in [0,1)$ is the fractional part of $t$. A very weak version of Stirling's approximation tells us
$$x^{-1}\sum\_{n \le x} \log n=\log x - 1 + O\left( \frac{\log x}{x}\right).$$
To establish $S\_2(x) =\log x- \gamma+o(1)$ it remains to show
$$x^{-1} \sum\_{n \le x} \Lambda(n) \{x/n\} = 1-\gamma + o(1).$$
By the Prime Number Theorem, $\Lambda =1$ 'on average', so we would expect the sum to be
$$=x^{-1} \sum\_{n \le x} \{x/n\}+o(1)$$
which indeed tends to $1-\gamma$ (see Exercise 2.1.1.1, page 39 in Montgomery--Vaughan, which is attributed to de la Vallée Poussin; this is a restatement of the asymptotic $\sum\_{n \le x} d(n) = x\log x + (2\gamma-1)x+o(x)$ which was known to Dirichlet). To make this formal we just need to show that the difference
$$\sum\_{n \le x} (\Lambda(n)-1) \{x/n\}$$
is $o(x)$ which can be done via the Prime Number Theorem, see Exercise 8.1.1.1, page 248, in Montgomery--Vaughan. In the same exercise they use Axer's theorem to show that $\sum\_{n\le x}\Lambda(n) \sim x$ (PNT) is in fact equivalent to $S\_2(n) =\log n- \gamma+o(1)$.
| 16 | https://mathoverflow.net/users/31469 | 444534 | 179,231 |
https://mathoverflow.net/questions/444533 | 1 | I have a question about the following statement from the article
* Alves, M., Rivera, J.M., Sepúlveda, M., Villagrán, O.V. and Garay, M.Z., *The asymptotic behavior of the linear transmission problem in viscoelasticity*, Math. Nachr. **287** (2014) pp 483-497, doi:[10.1002/mana.201200319](https://doi.org/10.1002/mana.201200319), ([author pdf](http://www.im.ufrj.br/rivera/Art_Pub/ViscoLocal.pdf))
They state on page 8 that:
Let $\alpha = \dfrac{\rho\_1\lambda^2}{\kappa\_1 + i\kappa\_2\lambda}$ where $\rho\_1, \kappa\_1, \kappa\_2 > 0$. then
$$
\operatorname{Coth}(i\alpha L) \to 1, \ \ \text{ when } \ \ n \to \infty
$$
where $L > 0$.
Well, firstly, I think they are changing $\alpha$ to $\alpha\_{n}$ and leaving $n$ hidden. And doing some mathematical manipulations, I'm getting to
Let $x\_n = \Re(\alpha\_n L)$ and $y\_n = \Im(\alpha\_n L)$. Using some properties, I can get
$$
\operatorname{Coth}(i\alpha\_n L) = \dfrac{\cos(x\_n)\sin(x\_n)}{\cosh^2 (y\_n) - \cos^2(x\_n)} - \dfrac{\tanh(y\_n)} {\frac{\cos^2(x\_n)}{\cosh^2 (y\_n)} - 1}
$$
From the equality above can I conclude what I want? Why?
| https://mathoverflow.net/users/481556 | $\operatorname{Coth}(\alpha_n a) \to i$ when $n \to \infty $ | $\newcommand\la\lambda\newcommand\al\alpha\newcommand\ka\kappa\newcommand\th\theta$First of all, you copied the expression for $\al$ incorrectly. On p. 489 of the paper linked by you (the fourth display from the bottom of the page), we find
$$\al=r(\la)e^{i\th/2},$$
where $\mathbb R\ni\la=\la\_n\to\infty$,
$$r(\la):=\frac{\rho\_1^{1/2}\la}{\sqrt[4]{\ka\_1^2+\ka\_2^2\la^2}}\to\infty,
\quad e^{i\th/2}\to\frac{1-i}{\sqrt2},
$$
and $\rho\_1,\ka\_1,\ka\_2$ are fixed positive real numbers.
So,
$$|e^{-2i\al L}|=|\exp(-2iLr(\la)e^{i\th/2})|
=\exp(\Re(-2iLr(\la)e^{i\th/2})) \\
=\exp(-(\sqrt2+o(1))\,Lr(\la))\to0.$$
Thus,
$$\coth(i\al L)=\frac{1+e^{-2i\al L}}{1-e^{-2i\al L}}\to\frac{1+0}{1-0}=1,$$
as desired.
| 2 | https://mathoverflow.net/users/36721 | 444538 | 179,232 |
https://mathoverflow.net/questions/444480 | 1 | I'm interested in the computations of the Goresky-Hingston product (defined <https://arxiv.org/abs/0707.3486>)
on the cohomology of the relative free loop space on the circle (or better yet, their extension to absolute cohomology). They give a full computation for all $S^n, n\geq 3$, and give some description of all manifolds whose geodesics are closed. Yet somehow I cannot read off the computation for $S^1$ (as someone who is well outside this area). Any pointers would be appreciated!
| https://mathoverflow.net/users/166758 | Goresky-Hingston product on cohomology of the relative free loop space on $S^1$ | Here is a sort of provocative, quick, partial answer. Take it as a pointer. Let $R$ be a commutative ring and consider $R[x]$ the polynomial ring on one variable $x$. For any $f \in R[x]$ we may consider the following discrete version of the derivative:
$$D\_q(f)= \frac{f(qx)-f(x)}{qx-x}$$
where $q$ here is thought of as a formal variable, so that $D\_q(f)$ is an element in the polynomial ring $R[q,x]$ in two variables.
In fact, we have an identification $R[q,x] \cong R[x] \otimes R[x]$ and under this identification $D\_q$ gives rise to the coproduct
$$\Delta \colon R[x] \to R[x] \otimes R[x]$$
determined by the formula
$$\Delta(x^m)= \sum\_{i=0}^{m-1} x^i \otimes x^{m-1-i}.$$
This is a coproduct of degree $-1$ sometimes called the "quantum derivative". It satisfies a Leibniz rule with respect to the multiplication of polynomials, namely, it defines an "infinitesimal bialgebra" structure on $R[x]$. I think after making certain identifications and extensions - you may want to add $x^{-1}$ as well - this construction should determine the Goresky-Hingston coproduct on $H\_\*(LS^1;R)$, the homology of the free loop space of $S^1$ and and consequently the product on $H^\*(LS^1;R)$. Note there is no need to work modulo constant loops in this particular case.
| 2 | https://mathoverflow.net/users/5450 | 444544 | 179,235 |
https://mathoverflow.net/questions/444393 | 17 | Thinking on the theory NBG (of von Neumann–Bernays–Gödel) I arrived at the conclusion that it is contradictory using an argument resembling Russell's Paradox. I am sure that I made a mistake in my arguments (because NBG cannot be contradictory, as all mathematicians believe), but I can not find the exact place where the error happened.
The [theory NBG](https://en.wikipedia.org/wiki/Von_Neumann%E2%80%93Bernays%E2%80%93G%C3%B6del_set_theory) is finitely axiomatizable theory whose undefined notions are "class" and "element". A class is called a *set* if it is an element of some other class. The axioms of NBG imply the existence of the universal class $\mathbf V$ containing all sets as elements. Axioms of NBG allow to make the following six basic operations over classes $X$, $Y$:
* **The difference:** $X\setminus Y=\{z:z\in X\wedge z\notin Y\}$;
* **The Cartesian product:** $X\times Y=\{\langle x,y\rangle:x\in X\wedge y\in Y\}$;
* **The transposition:** $X^{-1}=\{\langle y,x\rangle:\langle x,y\rangle\in X\}$;
* **The cyclic permutation:** $X^\circlearrowright=\{\langle\langle z,x\rangle,y\rangle:\langle\langle x,y\rangle,z\rangle\in X\}$;
* **The domain:** $\DeclareMathOperator\dom{dom}\dom[X]=\{x:\exists y\;\langle x,y\rangle\in X\}$.
* **The membership:** $X\_\in=\{\langle x,y\rangle\in X:x\in y\}$.
A class $X$ is called *constructible* if it can be constructed from the universal class $\mathbf V$ applying finitely many basic operations over classes. For example, the empty set is constructible because $\emptyset=\mathbf V\setminus\mathbf V$. [Gödel's Class Existence Theorem](https://en.wikipedia.org/wiki/Von_Neumann%E2%80%93Bernays%E2%80%93G%C3%B6del_set_theory) implies that a class is constructible if it can be described by a formula with a unique parameter $\mathbf V$ and all quantifiers running over elements of $\mathbf V$. It is clear that there are only countably many constructible classes.
All possible compositions of basic operations can be effectively enumerated by the set $T=\bigcup\_{n\in\omega}7^{2^{<n}}$ of 7-labeled full binary trees of finite height.
The idea of this enumeration is as follows. First, enumerate the basic operations over classes $X$, $Y$:
\begin{align\*}
& G\_0(X,Y):=X, \\
& G\_1(X,Y):=X\setminus Y, \\
& G\_2(X,Y)=X\times Y, \\
& G\_3(X,Y):=X^{-1}, \\
& G\_4(X,Y)=X^\circlearrowright \\
& G\_5(X,Y):=\dom[X], \\
& G\_6(X,Y)=X\_\in.
\end{align\*}
Let $2^{<\omega}=\bigcup\_{n\in\omega}2^n$ be the full binary tree and for every $n\in\omega$, let $2^{<n}=\bigcup\_{k\in n}2^k$ be the full binary tree of height $n$.
For every $k\in\{0,1\}$, consider the function $\vec k:2^{<\omega}\to 2^{<\omega}$, $\vec k:f\mapsto \{\langle 0,k\rangle:\langle i+1,v\rangle:\langle i,v\rangle\in f\}$ and observe that $\vec k[2^n]\subseteq 2^{n+1}$.
For every $n\in\omega$, function $\lambda:2^{<n}\to 7=\{0,1,2,\dotsc,6\}$, and class $X$, consider the class $G\_\lambda(X)$ defined by the recursive formula $G\_\lambda(X)=X$ if $n=0$ and $G\_\lambda(X)=G\_{\lambda(0)}(G\_{\lambda\circ \vec 0}(X),G\_{\lambda\circ \vec 1}(X))$ if $n>0$. So, $G\_\lambda$ represents a composition of the basic operations taken in the order suggested by the labels at the vertices of the binary tree $2^{<n}$.
A class $X$ is constructible if and only if $X=G\_\lambda(\mathbf V)$ for some $n\in\omega^\star$ and $\lambda\in 7^{2^{<n}}$. Here $\omega^\star$ is the set of "standard" numbers in the model of NBG. So, the constructibility of classes is an external notion to the model of NBG. Here we assume that NBG is not contradictory and fix some model of NBG. This model contains an element $\omega$ whose elements are natural numbers in the model. Among those natural numbers, there are standard natural numbers, which are successors of the empty set (from the viewpoint of the universe in which the model of NBG lives).
Let $T=\bigcup\_{n\in\omega}7^{2^{<n}}$ be the set of all 7-labeled binary trees of finite height. This set is a countable constructible subset of the model.
**Claim.** There exists a constructible class $C\subseteq T\times\mathbf V$ such that for every standard number $n\in\omega^\star$ and every 7-labeled tree $\lambda\in 7^{2^{<n}}$, the class $C\_\lambda=\{x\in \mathbf V:\langle\lambda,x\rangle\in C\}$ coincides with the class $G\_\lambda(\mathbf V)$.
Assume for a moment that the Claim is proved. The constructibility of the class $C$ implies the constructibility of the class $\Lambda=\{\lambda\in T:\langle\lambda,\lambda\rangle\notin C\}$.
Then there exists a standard number $n\in\omega^\star\subseteq\omega$ and a 7-labeled tree $\lambda\in 7^{2^{<n}}$ such that $\Lambda=C\_\lambda:=\{x\in\mathbf V:\langle\lambda,x\rangle\in C\}$. Now we have a paradox of Russell's type:
* if $\lambda\in\Lambda$, then $\lambda\in C\_\lambda$ and hence $\langle\lambda,\lambda\rangle\in C$ and $\lambda\notin\Lambda$;
* if $\lambda\notin\Lambda$, then $\langle\lambda,\lambda\rangle\in C$ and hence $\lambda\in C\_\lambda=\Lambda$.
In both cases, we obtain a contradiction.
*Proof of Claim.* Let $\mathbf{On}$ be the class of ordinals and $(V\_\alpha)\_{\alpha\in\mathbf{On}}$ be the von Neumann hierarchy, which can be identified with the class $\bigcup\_{\alpha\in\mathbf{On}}\{\alpha\}\times V\_\alpha$. It can be shown that both these classes are constructible (because they can be defined by formulas in which quantifiers run over elements of the universal class $\mathbf V$). It can be shown that for every standard number $n\in\omega^\star\subseteq\omega$, every 7-labeled tree $\lambda\in 7^{2^{<n}}$ and every ordinal $\alpha$ there exists an ordinal $\beta$ such that $V\_\alpha\cap G\_\lambda(\mathbf V)=V\_\alpha\cap G\_\lambda(V\_\gamma)$ for all ordinals $\gamma\ge\beta$.
Applying the Theorem of Recursion, for every ordinals $\alpha,\beta$ one can construct a class $C\_{\alpha,\beta}\subseteq T\times\mathbf V$ such that for every standard number $n\in\omega^\star\subseteq\omega$ and 7-labeled tree $\lambda\in 7^{2^{<n}}$ we have $\{x\in \mathbf V:\langle \lambda,x\rangle\in C\_{\alpha,\beta}\}=V\_\alpha\cap C\_\lambda(V\_\beta)\}$. Moreover, the recursive definition of the indexed sequence $(C\_{\alpha,\beta})\_{\alpha,\beta\in\mathbf{On}}$ shows that it is constructible as a subclass of $\mathbf{On}\times\mathbf{On}\times T\times\mathbf V$ (because it is defined by a formula whose quantifiers run over elements of the universal set). For every ordinal $\alpha$ consider the class $$C\_\alpha=\{\langle\lambda,x\rangle\in T\times\mathbf V:\exists \beta\in\mathbf{On}\;\forall \gamma\in \mathbf{On}\; (\beta\le \gamma\to \langle \lambda,x\rangle \in C\_{\alpha,\gamma})\}.$$ The constructibility of the family $(C\_{\alpha,\beta})\_{\alpha,\beta\in \mathbf{On}}$ implies the constructibility of the set indexed family $(C\_\alpha)\_{\alpha\in\mathbf {On}}$ (identified with the subclass $\bigcup\_{\alpha\in\mathbf{On}}\{\alpha\}\times C\_\alpha$ of the class $\mathbf{On}\times (T\times\mathbf V)$. The constructibility of the indexed family $(C\_\alpha)\_{\alpha\in\mathbf{On}}$ implies the constructibility of the class
$$C=\{\langle \lambda,x\rangle\in T\times\mathbf V:\exists \alpha\in\mathbf{On}\;\langle\lambda,x\rangle\in C\_\alpha\},$$ which has the property, required in the Claim. $\square$
So
>
> **Question.** In which place does this argument proving the inconsistency of NBG contain a gap?
>
>
>
| https://mathoverflow.net/users/61536 | A contradiction in the Set Theory of von Neumann–Bernays–Gödel? | The comments to the question, especially those by Emil Jeřábek and Joel Hamkins make it clear that the proposed inconsistency proof breaks down because the recursive construction carried out in the proposed proof of inconsistency cannot be implemented in NBG.
As pointed out first by Mostowski in his 1950 paper [Some impredicative definitions in set theory](https://eudml.org/doc/213208), when it comes to recursive constructions, the problem with NBG (as opposed to the much stronger system KM of Kelley and Morse) is that, the scheme of induction over the natural numbers is not provable in NBG. I would like to outline Mostowski's reasoning here since it sheds light on why the recursion proposed by Taras Banakh need not succeed in an arbitrary model of NBG.
The main idea is that within NBG we can describe, via a second order definition (i.e., by quantifying over classes of NBG) a subset $I$ of the ambient natural numbers of the NBG model such that NBG can prove that $I$ is closed under predecessors, contains $0$, and is closed under successors, but NBG cannot prove that $I$ coincides with the set $\omega$ of natural numbers (in the sense of NBG).
The basic idea is devilishly simple: let $I$ consist of numbers $k$ such that there is a class $S$ (in the ambient model of NBG) such that $S$ is a $k$-satisfaction class, i.e., $S$ satisfies Tarski's compositional clauses for a satisfaction predicate for the structure $(\mathbf{V},\in)$ (in the sense of the ambient NBG model) for *formulae of depth at least* $k$; here the depth of a formula $\varphi$ is the length of the longest path in the "parsing tree" of $\varphi$ (also known as the "formation tree" or "syntactic tree" of $\varphi$).
With $I$ defined as above, we can specify a unary formula $\sigma(x)$ in the language of NBG such that, provably in NBG, $\sigma(x)$ satisfies Tarski's compositional clauses for all set-theoretical formula whose depth (as defined above) lies in $I$. *In particular, $\sigma(x)$ correctly decides the truth of all set-theoretical sentences in the ambient ZF model of NBG since for each standard natural number $k$, NBG can prove that $k \in I$.*
$\sigma(x)$ expresses: $x$ is a set-theoretical formula (with parameters) $\varphi(m\_1,...,m\_s)$, $\mathrm{depth}(\varphi) \in I$, and there is a class $S$ (in the sense of NBG) such that $\varphi(m\_1,...,m\_s) \in S$ and $S$ is a $k$-satisfaction predicate for $k=\mathrm{depth}(\varphi)$. Note that, provably in NBG, any two $k$-satisfaction classes agree on the truth-evaluation of formulae of depth at most $k$.
>
> Mostowski's construction unveiled a rather intriguing
> phenomena: there can be a theory $T^+$ [in this case NBG] which possesses a truth-predicate for
> a theory $T$ [in this case ZF], and yet the formal consistency of $T$ is unprovable in
> $T^+$ (due to the lack of sufficient formal induction in $U$ [because Con(ZF) is not provable in GB, since GB is set-theoretically conservative over ZF, and of course ZF cannot prove Con(ZF)].
>
>
>
With the above definition of $\sigma$ at our disposal, we can write a formula $\theta$ in the language of NBG that asserts that every class is definable in the sense of $\sigma$ in the following sense:
**Theorem.** *If $(\mathcal{M},\frak{X})$ is a model of NBG, where $\mathcal{M}=(M,E)$ is a model of ZF, and $\frak{X}$ is a subset of the powerset of $M$ that specifies the classes of the model, and $M$ is $\omega$-standard, then $(\mathcal{M},\frak{X}) \models \theta$ iff $(\mathcal{M},\frak{X})$ is spartan, i.e., $\frak{X}$ consists of subsets of $M$ that are parametrically definable in $\mathcal{M}$.*
**N.B.** In the above, the $\omega$-standardness hypothesis is only used in the left-to-right direction, in other words, *$\theta$ holds in every spartan model of* NBG.
Besides the Mostowski paper referenced above, readers interested in the history of this subject might want to also examine John Myhill's 1952 paper [The hypothesis that all classes are nameable.](https://www.pnas.org/doi/epdf/10.1073/pnas.38.11.979)
| 21 | https://mathoverflow.net/users/9269 | 444549 | 179,238 |
https://mathoverflow.net/questions/444551 | 8 | This question is motivated by my preceding [MO-question](https://mathoverflow.net/q/444393/61536) on (in)consistency of [NBG theory of classes](https://en.wikipedia.org/wiki/Von_Neumann%E2%80%93Bernays%E2%80%93G%C3%B6del_set_theory).
Let $\varphi(x,Y,C)$ be a formula of NBG with free parameters $x,Y,C$ and all quantifiers running over sets. Godel's Class Existence Theorem ensures that for every classes $Y,C$ the class $\{x:\varphi(x,Y,C)\}$ does exist. On the other hand, the existence of the class $\{x:\exists Y\;\varphi(x,Y,C)\}$ cannot be proved in NBG, see [the answer](https://mathoverflow.net/a/444549/61536) of @AliEnayat to [this MO-question](https://mathoverflow.net/q/444393/61536).
>
> **Question.** Can on prove that in NBG for every finite set $F$ and every class $C$ the set $\{x\in F:\exists Y\;\varphi(x,Y,C)\}$ exists?
>
>
>
| https://mathoverflow.net/users/61536 | The existence of definable subsets of finite sets in NBG | The answer is in the negative. Let $\mathcal{M}$ be an $\omega$-nonstandard model of ZF, and $\mathfrak{X}$ be the collection of parametrically definable subsets of $\mathcal{M}$. Let $I$ be the cut defined in my answer to [the other MO question](https://mathoverflow.net/questions/444393/a-contradiction-in-the-set-theory-of-von-neumann-bernays-g%C3%B6del/444549?noredirect=1#comment1147821_444549). Note that by Tarski's undefinability of truth theorem, in the model $(\mathcal{M},\mathfrak{X})$ of NBG, the cut $I$ exactly corresponds to the collection of *standard* natural numbers. Also note that the definition of $I$ is $\Sigma^1\_1$.
Let $c$ be a nonstandard natural number of $\mathcal{M}$, and let $F$ be the finite set (in the sense of $\mathcal{M}$) of the predecessors of $c$. Then the collection of members of $F$ that belong to $I$ has a $\Sigma^1\_1$-definition, but it does not exist in the model (since it is a subset of natural numbers of the model that is bounded above but has no maximum).
In the above, if one wishes to do away with the nonstandard parameter $c$ in the construction of the counterexample, one can choose $\mathcal{M}$ to satisfy $\lnot \mathrm{Con(ZF)}$, and use "the shortest proof of the inconsistency of $\mathrm{Con(ZF)}$" instead of $c$.
| 14 | https://mathoverflow.net/users/9269 | 444553 | 179,239 |
https://mathoverflow.net/questions/444507 | 9 | Let $\alpha\in H^2(B\operatorname{PSU}(N) ; \mathbb{Z}\_N)$ be the obstruction class for lifting a $\operatorname{PSU}(N)$-bundle to an $\mathrm{SU}(N)$-bundle. Note that $\operatorname{PSU}(N)\cong \operatorname{SU}(N)/\mathbb{Z}\_N$ is the quotient group of $\operatorname{SU}(N)$ by its center, and $B\operatorname{PSU}(N)$ is the classifying space.
The short exact sequence
$$
0 \rightarrow \mathbb{Z}\_q \rightarrow \mathbb{Z}\_{q \cdot N} \rightarrow \mathbb{Z}\_N \rightarrow 0
$$
invokes the Bockstein map $\textrm{Bock}:H^2(B \operatorname{PSU}(N) ; \mathbb{Z}\_N) \rightarrow H^3\left(B \operatorname{PSU}(N) ; \mathbb{Z}\_q\right)$.
The image $\textrm{Bock}(\alpha)\in H^3\left(B \operatorname{PSU}(N) ; \mathbb{Z}\_q\right)$ of the obstruction class under the Bockstein map is a Postnokov class. My question is under which condition (on $q$ and $N$) the Postnikov class $\textrm{Bock}(\alpha)$ is non-trivial. Is there a classification or a study to tell whether the Postnikov class in $H^3\left(B \operatorname{PSU}(N) ; \mathbb{Z}\_q\right)$ is non-trivial?
**Background on physics**. The non-triviality of the Postnikov class will tell us the 2-group structure of 1-form symmetry and flavor symmetry in three dimensions, and this will tell equivalence classes of line defects in three dimensions.
| https://mathoverflow.net/users/17644 | Non-triviality of a Postnikov class in $H^3\left(B \operatorname{PSU}(N) ; \mathbb{Z}_q\right)$ | Consider the map of short exact sequences
$$\begin{array}{ccccccccc} 0 & \rightarrow & \mathbb{Z} & \xrightarrow{\cdot N} & \mathbb{Z} & \xrightarrow{mod N} & \mathbb{Z}\_N & \rightarrow & 0\\
\downarrow & & \downarrow & & \downarrow & & \downarrow{=} & & \downarrow\\
0 & \rightarrow & \mathbb{Z}\_q & \xrightarrow{1 \mapsto N} & \mathbb{Z}\_{Nq} & \xrightarrow{mod N} & \mathbb{Z}\_N & \rightarrow & 0\end{array}$$
and the stretch of long exact sequence in cohomology from the top row:
$$H^2(BPSU(N); \mathbb{Z}) \rightarrow H^2(BPSU(N); \mathbb{Z}\_N) \xrightarrow{\beta} H^3(BPSU(N); \mathbb{Z})$$
Now, from the fibration $B\mathbb{Z}\_N \to BSU(N) \to BPSU(N)$ we have that $BPSU(N)$ is simply connected and $H\_2(BPSU(N);\mathbb{Z}) \cong \pi\_2(BPSU(N)) \cong \mathbb{Z}\_N$. Then from universal coefficients we get $H^2(BPSU(N); \mathbb{Z})= 0$ and $H^2(BPSU(N); \mathbb{Z}\_N) \cong \mathbb{Z}\_N$. Furthermore, we have that $\pi\_3(BPSU(N)) = \pi\_2(SU(N)) = 0$ surjects onto $H\_3(BPSU(N);\mathbb{Z})$
by Hurewicz, giving us $H^3(BPSU(N);\mathbb{Z}) \cong \mathbb{Z}\_N$ again by universal coefficients.
Therefore the integral Bockstein above is an injection of $\mathbb{Z}\_N$ into itself and hence an isomorphism, sending 1 to an invertible element of $\mathbb{Z}\_N$. Now we use the commutativity of
$$\begin{array}{ccc} H^2(BPSU(N);\mathbb{Z}\_N) & \xrightarrow{\beta} & H^3(BPSU(N);\mathbb{Z}) \\ \downarrow{=} & & \downarrow \\ H^2(BPSU(N); \mathbb{Z}\_N) &\xrightarrow{\beta} & H^3(BPSU(N);\mathbb{Z}\_q) \end{array}$$
to see that $1 \in H^2(BPSU(N);\mathbb{Z}\_N)$ is in the kernel of the lower Bockstein (the one you are interested in) if and only if $1$ (or equivalently any invertible element in $\mathbb{Z}\_N$) gets sent to 0 under mod $q$ reduction, i.e. if and only if $1$ is $q$ times another integral class, i.e. if and only if $q$ is invertible modulo $N$.
That is, the Bockstein is non-trivial if and only if $q$ and $N$ have a prime factor in common (or $q = 0$).
| 5 | https://mathoverflow.net/users/104342 | 444559 | 179,241 |
https://mathoverflow.net/questions/444561 | 4 | A group is called minimal non-abelian if it is non-abelian and all proper subgroups are abelian.
Does this notion also exist with Lie groups or algebras? As an example, consider the Lie algebra defined by the generators $\{e\_1,e\_2\}, [e\_1,e\_2]=e\_2$. Do you have a paper at hand?
| https://mathoverflow.net/users/11504 | Minimal non-abelian groups -> Lie groups/algebras | These Lie algebras were called *semiabelian* by some authors. Some old papers are [On the structure of simple-semiabelian
Lie-algebras](https://projecteuclid.org/journals/pacific-journal-of-mathematics/volume-111/issue-2/On-the-structure-of-simple-semiabelian-Lie-algebras/pjm/1102710571.pdf) by Farnsteiner and [On simple semiabelian $p$-adic lie algebras](https://www.tandfonline.com/doi/pdf/10.1080/00927879208824390) by myself.
| 6 | https://mathoverflow.net/users/18739 | 444562 | 179,242 |
https://mathoverflow.net/questions/442177 | 9 | Let $G$ be a directed graph.
Call a vertex $v$ in $G$ *central* if there exists $\Theta(n^2)$ distinct pairs of vertices $(u,w)$ such that $v$ lies on some path from $u$ to $w$. We do not care whether these paths are shortest or not, nor whether other paths avoiding $v$ exist.
Question: Do strongly connected digraphs always admit a central vertex?
| https://mathoverflow.net/users/16485 | Does a strong digraph always admit a vertex that lies on some path between $\Theta(n^2)$ pairs of vertices? | In [this manuscript](https://arxiv.org/abs/2304.03567), Bessy, Thomassé, and Viennot proved the following stronger property:
Let $D$ be a strongly connected directed graph, then there is a vertex $v$ in $D$ such that there exists an in-tree towards $v$ and an out-tree from $v$ that are vertex-disjoint (apart from sharing $v$ itself) and contain at least $n/6$ vertices each.
As a result, $v$ lies on a path between at least $n^2/36$ pairs of vertices.
| 5 | https://mathoverflow.net/users/16485 | 444571 | 179,246 |
https://mathoverflow.net/questions/444575 | 5 | Let $R$ be a local ring with field of fractions $K$, maximal ideal $\mathfrak{m}$ and residue field $\kappa = R/\mathfrak{m}$. Let $R^\mathrm{sh}$ be a strict henselization of $R$, and let $L$ be the field of fractions of $R^\mathrm{sh}$.
1. Is $L$ a maximal unramified extension of $K$ with respect to $\mathfrak{m}$? By this I mean the union (inside a separable closure of $K$) of all extensions that are unramified at $\mathfrak{m}$.
2. Is $R^\mathrm{sh}$ the integral closure of $R$ in $L$?
| https://mathoverflow.net/users/37368 | Alternative description of strict henselization | (Assuming $R$ is not complete)
For 2, the answer is no, as suggested in a now-deleted comment of LSpice.
For $k$ of characteristic not $2$, with $R$ the localization of $k[t]$ at $0$, the element $\frac{1}{ \sqrt{1+t}+1 }$ is in the strict henselization but not integral. This is the root of the polynomial $$(1-x)^2 - x^2(1+t) = 1 - 2x - t x^2,$$ which reduces mod $t$ to a polynomial with distinct roots and thus has a root in the henselization but which is clearly not divisible by any monic polynomial.
For 1, I think the answer depends on what you mean by "that are unramified at $\mathfrak m$". Recall that an extension of fields may have multiple places lying over $\mathfrak m$, with some ramified and others unramified. You have to choose a lift of $\mathfrak m$ to every extension contained in the separable closure in a consistent way, i.e. choose an extension of the $\mathfrak m$-adic valuation to the separable closure, and then consider the union of the unramified extensions.
The point being that every étale extension of (one-dimensional) local rings comes with a choice of a lift of a valuation (i.e. the one associated to the maximal ideal of the extended ring) and is unramified for this choice, and conversely if you have a field extension with an unramified choice of lift that gives an étale ring extension.
On the other hand, if $R$ is complete (i.e. already non-strictly henselian) then every unramified extension is associated to a separable extension of the residue field and thus has a unique lift of the maximal ideal and is integral, so the answer to both questions is yes.
| 9 | https://mathoverflow.net/users/18060 | 444576 | 179,248 |
https://mathoverflow.net/questions/415167 | 3 | I am reading Stopple's *A Primer of Analytic Number Theory*. On page 234, the Mellin Transform $\mathcal{M} f$ of a function $f$ is defined as
$$\mathcal{M} f (s) = \int\_1^{\infty} f(x) x^{-s - 1} dx$$
His goal is to use the injectivity of the Mellin Transform (if $\mathcal{M} f = \mathcal{M} g$ then $f = g$) to prove Von Mangoldt's Formula (also on page 234). However, one of the functions he considers has infinitely many jump discontinuities. Therefore, as far as I can see, it looks like he could have replaced it by any other function agreeing with it up to a zero measure set.
So, my question is about the nature of the injectivity of the Mellin Transform: in which sense are $f$ and $g$ equal if $\mathcal{M} f = \mathcal{M} g$?
| https://mathoverflow.net/users/78173 | Injectivity of the Mellin Transform and discontinuities | The answer is implicit in what you wrote. One can define an equivalence relation $f\sim g$ if $f$ and $g$ are equal except on a set of measure 0, then define Mellin transform on equivalence classes instead of functions. For context, the book was intended for undergraduates who have seen no more than a typical calculus sequence, certainly not measure theory. There are much better books for advanced students. (I like Montgomery & Vaughan "Multiplicative Number Theory").
By the way @Brunalt is correct that typically the Mellin transform is defined as an integral from $0$ to $\infty$, not $1$ to $\infty$. In the book it is only applied to functions which are $0$ for $0\le x\le 1$, so including that in the integral only obfuscates. For the typical reader (who is not going on to a PhD in Mathematics) this alternate definition suffices.
| 3 | https://mathoverflow.net/users/6756 | 444591 | 179,251 |
https://mathoverflow.net/questions/444592 | 8 | Let $f\_n(x)=1+x+x^{\sqrt{2}}+x^{\sqrt{3}}+x^{\sqrt{4}}+\cdots+x^{\sqrt{n}}$ be a sequence of functions on the interval $[0, 1]$. Is there a good closed form approximation for such a function ( whatever "good" means), analogous to the well-known expression for the sum of geometric progression.
Such an approximation is needed to get an asymptotics on the positive root of equation $f\_n(x)=\ln(n)$ for large $n$.
| https://mathoverflow.net/users/34984 | Approximation of pseudogeometric progression | For natural $n$ and $x\in(0,1)$, one has
$$L\_n(x)\le f\_n(x)\le U\_n(x), \tag{10}\label{10}$$
where
$$U\_n(x):=1+I\_n(x), \tag{20}\label{20}$$
$$I\_n(x):=\int\_0^n x^{\sqrt t}\,dt
=2\frac{1+x^{\sqrt{n}} \left(\sqrt{n} \ln x-1\right)}{\ln^2 x},$$
$$L\_n(x):=I\_{n+1}(x). \tag{30}\label{30}$$
Next, fixing any real $c>0$ and letting $n\to\infty$, for $x=1-c/\sqrt{\ln n}$ we will have
$$\sqrt n\ln x
\sim-c\sqrt{n/\ln n}\to-\infty, $$
$$x^{\sqrt{n}}=\exp(\sqrt n\ln x)=\exp(-(c+o(1))\sqrt{n/\ln n}),$$
and hence
$$I\_n(1-c/\sqrt{\ln n})\sim\frac2{c^2}\,\ln n.$$
Similarly (or because $1-c/\sqrt{\ln(n+1)}=1-(c+o(1))/\sqrt{\ln n}\;$), we have
$$I\_{n+1}(1-c/\sqrt{\ln n})\sim\frac2{c^2}\,\ln n.$$
So, by \eqref{20} and \eqref{30},
$$U\_n(1-c/\sqrt{\ln n})\sim\frac2{c^2}\,\ln n\sim L\_n(1-c/\sqrt{\ln n}).$$
Hence, in view of \eqref{10},
$$f\_n(1-c/\sqrt{\ln n})\sim\frac2{c^2}\,\ln n.$$
So, for the only zero $x=x\_n$ of the strictly increasing function $(0,1)\ni u\mapsto f\_n(u)-\ln n$ we have
$$x\_n=1-\sqrt{\frac{2+o(1)}{\ln n}}$$
as $n\to\infty$.
| 10 | https://mathoverflow.net/users/36721 | 444596 | 179,254 |
https://mathoverflow.net/questions/444598 | 8 | We assume that we have a $\alpha$-Hölder continuous function $f$ on an interval $[0,1].$
I am wondering if there exists an explicit construction of a sequence $f\_{n} \in C\_c^{\infty}(\mathbb R)$ such that
$$\lVert f-f\_n\rVert\_{C^{\beta}([0,1])} \le \frac{1}{n}$$
for fixed $\beta<\alpha$ and $\lvert f\_n(x)\rvert \le \lvert f(x)\rvert$ on $[0,1]$. The usual convolution idea does not respect the last condition. In an earlier post, I mistakingly took $\beta=\alpha.$
| https://mathoverflow.net/users/496243 | Smooth approximation of Hölder functions "from below" | $\newcommand\R{\mathbb R}\newcommand{\al}{\alpha}\newcommand{\de}{\delta}\newcommand{\J}{\mathcal J}
\newcommand{\be}{\beta}\newcommand{\ep}{\varepsilon}$Yes, such a construction exists.
Indeed, take any real $\ep>0$. For a real $C\ge0$ and $\al\in(0,1]$, let us say that a function $g$ is $(C,\al)$-Hölder on a set $S\subseteq[0,1]$ if $|g(y)-g(x)|\le C|y-x|^\al$ for all $x,y$ in $S$.
Without loss of generality (wlog), the function $f$ is $(1,\al)$-Hölder on $[0,1]$.
Let $Z:=\{z\in I:=[0,1]\colon f(z)=0\}$ and $N:=I\setminus Z$. Wlog, $\{0,1\}\in Z$ (otherwise, extend $f$ appropriately to an interval $[A,B]\supset I$ so that $f(A)=0=f(B)$ and then shrink the interval $[A,B]$ to $I$). So, $N=\bigcup\_{J\in\J}J$ for some (countable) set $\J$ of pairwise disjoint nonempty open subintervals of $I$.
It is enough to construct a smooth function $f\_\ep$ such that for each $J\in\J$
\begin{equation\*}
\|f-f\_\ep\|\_{C^\be(J)}\le100\ep \tag{10}\label{10}
\end{equation\*}
and
\begin{equation\*}
|f\_\ep|\le|f|\text{ on } J. \tag{20}\label{20}
\end{equation\*}
To begin such a construction, take any real $\de>0$. Let $\J\_\de$ denote the set of all intervals $J\in\J$ of length $>\de$. Of course, the set $\J\_\de$ is finite. Let
\begin{equation\*}
f\_\ep(x):=0\text{ for }x\in I\setminus\bigcup\_{J\in\J\_\de}J.
\end{equation\*}
If $\de$ is small (which will be henceforth assumed), then on $I\setminus\bigcup\_{J\in\J\_\de}J$ the function $f-f\_\ep=f$ is small and $(1,\al)$-Hölder. So, $f-f\_\ep=f$ is $(\de\_1,\be)$-Hölder on $I\setminus\bigcup\_{J\in\J\_\de}J$ for a small $\de\_1$. This follows because $|y-x|^\al<<|y-x|^\be$ if $\be\in(0,\al)$ (as given) and $|y-x|<<1$, whereas $|y-x|^\al\asymp|y-x|^\be\asymp1$ if $|y-x|\asymp1$. (We write $E<<F$ if $E=o(F)$, $E\ll F$ if $E=O(F)$, and $E\asymp F$ if $E\ll F\ll E$.)
We shall build the function $f\_\ep$ separately on each interval $J=(a,b)\in\J\_\de$, in such a manner that
\begin{equation\*}
\text{$f\_\ep=0$ near the endpoints of $J$. } \tag{30}\label{30}
\end{equation\*}
Then the smoothness of $f\_\ep$ on each interval $J\in\J$ will be enough for the smoothness of $f\_\ep$ on the entire interval $I$.
Take indeed any interval $J\in\J\_\de$.
Note that (i) $f(a)=0=f(b)$ and (ii) either $f>0$ on $(a,b)$ or $f<0$ on $(a,b)$. Wlog, $f>0$ (on $(a,b)$). Moreover, $f$ is continuous. So, $M:=f(x\_\*)\ge f(x)$ for some $x\_\*\in J$ and all $x\in J$.
For any $c\in(0,M]$, the points
\begin{equation\*}
x\_+(c):=\min\{x\in[x\_\*,b)\colon f(x)\le c\},\quad x\_-(c):=\max\{x\in(a,x\_\*]\colon f(x)\le c\}
\end{equation\*}
are well defined. Moreover,
\begin{equation\*}
f\ge c\text{ on }J\_c:=[x\_-(c),x\_+(c)],\quad f(x\_\pm(c))=c. \tag{40}\label{40}
\end{equation\*}
Let
\begin{equation\*}
g:=(f-c)1\_{J\_c}. \tag{45}\label{45}
\end{equation\*}
Then $g$ is $(1,\al)$-Hölder on $\R$ and hence so is
\begin{equation\*}
g\_\eta:=g\*K\_\eta, \tag{47}\label{47}
\end{equation\*}
where, for each $\eta\in(0,\de)$, we let $K\_\eta$ be any smooth nonnegative function supported on the interval $[-\eta,\eta]$ such that $\int K\_\eta=1$.
Moreover, for $x\in J\_c=[x\_-(c),x\_+(c)]$,
\begin{align}
g\_\eta(x)&=\int\_{-\eta}^\eta du\,K\_\eta(u)(f(x-u)-c)1(x-u\in J\_c) \notag \\
&\le\eta^\al+\int\_{-\eta}^\eta du\,K\_\eta(u)(f(x)-c)1(x-u\in J\_c) \notag \\
&\le\eta^\al+(f(x)-c)\le f(x) \notag
\end{align}
assuming that $\eta^\al\le c$, which will be indeed assumed henceforth.
Further, for $x\in[x\_+(c),x\_+(c)+\eta]$,
\begin{align}
g\_\eta(x)&=\int\_{-\eta}^\eta du\,K\_\eta(u)(f(x-u)-c)1(x-u\in J\_c) \notag \\
&=\int\_{-\eta}^\eta du\,K\_\eta(u)(f(x-u)-f(x\_+(c))1(x-u\in J\_c) \notag \\
&\le\int\_{-\eta}^\eta du\,K\_\eta(u)(x\_+(c)-(x-u))^\al 1(x-u\in [x\_-(c),x\_+(c)]) \notag \\
&\le\int\_{-\eta}^\eta du\,K\_\eta(u)\eta^\al 1(x-u\in [x\_-(c),x\_+(c)])
\le\eta^\al. \notag
\end{align}
On the other hand, again for $x\in[x\_+(c),x\_+(c)+\eta]$, we have $f(x)\ge f(x\_+(c))-(x-x\_+(c))^\al\ge c-\eta^\al\ge\eta^\al\ge g\_\eta(x)$ assuming that
\begin{equation\*}
2\eta^\al\le c, \tag{50}\label{50}
\end{equation\*}
which will be indeed assumed henceforth.
Similarly, $g\_\eta(x)\le\eta^\al$ for $x\in[x\_-(c)-\eta,x\_-(c)]$ given \eqref{50}.
Also, $g\_\eta(x)=0\le f(x)$ for $x\in J\setminus J\_c$.
We conclude that
\begin{equation}
g\_\eta\le f\text{ on }J.
\end{equation}
So, letting
\begin{equation\*}
f\_\ep:=g\_\eta\text{ on }J=(a,b), \tag{60}\label{60}
\end{equation\*}
we see that $f\_\ep$ is smooth on $J=(a,b)$, and conditions \eqref{20} and \eqref{30} hold.
Moreover, $g\_\eta$ is $(1,\al)$-Hölder on $\R$ and hence $f\_\ep$ is $(1,\al)$-Hölder on $J$.
So, it remains to check that $f\_\ep$ is uniformly close to $f$ on $J$, also uniformly in $J\in\J\_\de$.
By \eqref{45}, on $J\_c$ the function $g$ is uniformly close to $f$ if $c$ is small enough, which will be henceforth assumed.
Also, on $J\setminus J\_c$ we have $0\le f\le\max((b-x\_+(c))^\al,(x\_-(c)-a)^\al)$, which is small (since $c$ was assumed to be small), and hence $f$ is close to $g$ (which is $0$ on $J\setminus J\_c$). So, $g$ is uniformly close to $f$ on $J$, uniformly in $J\in\J\_\de$.
Finally, because $\eta$ is small, and in view of \eqref{60} and \eqref{47}, we conclude that indeed $f\_\ep$ is uniformly close to $f$ on $J$, uniformly in $J\in\J\_\de$. $\quad\Box$
| 4 | https://mathoverflow.net/users/36721 | 444611 | 179,257 |
https://mathoverflow.net/questions/444610 | 2 | I have a matrix $A$ $(n\times n)$ with eigenvalues $\lambda\_i$, then I add another matrix to it as: $A+xx^\top$ where $x$ $(n\times 1)$ is a column vector.
and also $A=yy^\top$ with $y$ a $(n-1)$ rank matrix, so A is symmetric.
Is there any way to calculate the new eigenvalues and eigenvectors of $(xx^\top+yy^\top)$ using the information from the vector $x$ and the eigenvalues and eigenvectors of $yy^\top$？
| https://mathoverflow.net/users/502666 | Eigenvalues of a rank-one update of a symmetric matrix | Given eigenvectors and eigenvalues of the symmetric matrix $A$, you can transform to a basis where $A$ is a diagonal matrix $D$; the vector $x$ in that basis transforms to $\tilde{x}$. Then the problem is that of computing the eigenvalues of a rank-one update $D+\tilde{x}\tilde{x}^\top$ of a diagonal matrix. An efficient algorithm is given in a [paper](https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.454.9868&rep=rep1&type=pdf) by Golub, page 325. Once you have the eigenvalues the eigenvectors follow directly from the [Bunch–Nielsen–Sorensen formula.](https://en.wikipedia.org/wiki/Bunch%E2%80%93Nielsen%E2%80%93Sorensen_formula)
| 5 | https://mathoverflow.net/users/11260 | 444618 | 179,258 |
https://mathoverflow.net/questions/444608 | 16 | I am trying to follow the beautiful notes by Peter Scholze on condensed mathematics (<https://www.math.uni-bonn.de/people/scholze/Condensed.pdf>)
I am noting that the hard time that I am getting is a lack of basis in Grothendieck topologies, sites, and so on, things that are not from the theory of condensed math itself but prerequisites.
So I would like to know if is there a standard book, or nice notes used for the introduction on such matters, with more concrete examples of the objects defined.
Thank you very much for any suggestions.
| https://mathoverflow.net/users/43027 | Reference request for condensed math | Dagur Ásgeirsson has written [a text](https://dagur.sites.ku.dk/files/2022/01/condensed-foundations.pdf) to fill this gap:
>
> We discuss in some detail the prerequisites for each of the first four chapters of Scholze's "Lectures on Condensed Mathematics". Some proofs are
> given in more detail or slightly altered, but all the main ideas are
> the same. In particular, I claim no originality of the results. It is
> my hope, however, that I have succeded in creating a more accessible
> presentation of the foundations of condensed mathematics, along with
> the prerequisites, than has previously been available.
>
>
>
| 20 | https://mathoverflow.net/users/11260 | 444624 | 179,261 |
https://mathoverflow.net/questions/444626 | 2 | Let $G(n,l)$ denote the set of connected graphs with $n$ vertices and $l$ edges and let $G\_0(n,l)$ denote the elements of $G(n,l)$ without leaves. It is easily seen that $G\_0(n,n-1)$ is empty, since $G(n,n-1)$ is the set of trees, and $G\_0(n,n)$ consists of all circular grpahs of length $n$.
**Question:** How can we characterize all elements of $G\_0(n,n+1)$ (maybe also $G\_0(n,n+2)$) as belonging to some easily described classes of graphs and what would those classes look like? (in analogy to circle graphs for $G\_0(n,n)$)
Sorry for the soft question, I guess someone must have already characterized such graphs but I can't find any sources, so I would also be thankful for references to previous work in this direction.
Any help is much apprechiated.
| https://mathoverflow.net/users/409412 | Characterization of graphs without leaves | The elements of $G\_0(n,n+k)$ for $k\geq 1$ can be characterized as follows. Consider any partition $e\_1+e\_2+\cdots+e\_v = 2k$ of $2k$ into $v\leq n$ parts and any multigraph $S$ with $v$ vertices of degrees $2+e\_1, 2+e\_2, \ldots,2+e\_v$ and $v+k$ edges. Then any graph $H\in G\_0(n,n+k)$ is obtained by distributing $n-v$ vertices of degree two on the edges of $S$, for a suitable choice of partition and multigraph $S$. The multigraph $S$ is sometimes called the skeleton of $H$. In the case $k=1$, therefore three different multigraphs $S$ occur (pictorially: $\infty$, $\theta$ and $0\!\!-\!\!0$)
| 2 | https://mathoverflow.net/users/47484 | 444628 | 179,262 |
https://mathoverflow.net/questions/444552 | 2 | I would be very grateful for any references I might be led to, from a categorical point of view for the functors:
* $\textsf{Spec}\_{\mathscr{Z}\textrm{arisky}}(-)$, related to $\mathcal{O}(-)$, which leads to $\mathcal{O}(\textsf{Spec}\_{\mathscr{Z}\textrm{arisky}}(-))$,
* $\textsf{Spec}\_{\mathscr{G}\textrm{elfand}}(-)$, related to $\mathcal{C}(-,\mathbb{C})$, which leads to $\mathcal{C}(\textsf{Spec}\_{\mathscr{G}\textrm{elfand}}(-),\mathbb{C})$
These references would deal with "good" equivalences of categories, at the levels of algebraic geometry and operator algebras, which would be raised in a purely unifying categorical construction.
Many thanks in advance for any knowledge of books and articles written with this in mind.
---
In particular, dear @DmitriPavlov, thank you very much for your articles, your high categorical point of view and your answer, 01/13/2022, 11:51 p.m., to the internal question which prompted my question from personal research related to the relationship between Gelfand spectrum and Zariski spectrum, to which your answer to question : <https://mathoverflow.net/a/413776/502369> begins to reply.
You wrote this answer very brilliantly in terms of category theory and localization for this question of the relationship between the Gelfand spectrum and the Zariski spectrum, and the corresponding locales, topologies and sheaves, which therefore coincide in a more general setting .
I am also very sensitive to your arguments, Dear @DmitriPavlov, which appear in your article on Gelfand duality, stating the equivalences of categories with Hyperstonean spaces, Von Neumann algebras and three other categories, as well as in your text on category theory.
My first general question about possible references, therefore, continues, Dear @DmitriPavlov, with the following two questions, if you allow me to address you directly, but also to anyone who would be kind enough to answer them:
First :
>
> I have not yet found the statement of your answer by
> localization and construction of the sheaf, in your own publications
> and available texts, or elsewhere.
>
>
> I found links with localization and construction of the sheaf, in
> particular in the Remark 12.4.19 by Henning Krause: *Homological
> Theory of Representations*, Cambridge University Press, and the
> Ziegler spectrum and the Zarisky spectrum by Henning Krause: *The
> spectrum of a module category*, Memoirs of the AMS, volume 149 ,
> Number 707, which both send to Melvin Hochster: *Prime ideal structure
> in commutative rings*, and to Mike Perst: *Remarks on elementary
> duality*, 1993. But the link with C\*-algebras, spatial locales and the
> categorification of the subject of your answer in very few words which
> are very enlightening, leads me, if you allow me, to ask you for
> references for this construction, which would possibly be your own
> publications themselves, where I would not have found yet this
> statement, or in some other papers and references?
>
>
> Moreover, would this luminous construction come from bibliographical
> references that could even be cited?
>
>
> It would indeed be very valuable for me to go deeper into the study of
> this question which arises for me in categorical terms. Thank you all
> in advance for your enlightning.
>
>
>
Secondly :
>
> It seems that it was asked to you earlier on the same MathOverFlow
> page: <https://mathoverflow.net/a/413776/502369>, but I'll rephrase it
> for all, because you cite, Dear @DmitriPavlov, many application cases,
> including Von Neumann algebras and Hyperstonean spaces in a
> categorification of Measure Theory, to which you devote an equally
> brilliant article.
>
>
> Thus: What could be the minimum assumptions about the category for
> this construction of localization and structural sheaf to remain
> rigorously perfect? But maybe this question is contained in the
> references that could be indicated, about the previous question.
>
>
>
Thank you in advance for your attention to my question and your interest in answering it.
Best regards
| https://mathoverflow.net/users/502369 | Unifying categorical equivalences and dualities for the functors : Gelfand spectrum and Zariski spectrum, Structural sheaf and Continuous sections |
>
> I have not yet found the statement of your answer by localization and construction of the sheaf, in your own publications and available texts, or elsewhere.
>
>
>
To the best of my knowledge, there is nothing published on this topic yet.
I once wrote some very rough unfinished notes on how to treat the algebraic, differential, and holomorphic cases simultaneously: <http://dmitripavlov.org/notes/cart.pdf>, which do discuss the construction of the spectrum and the structure sheaf, but do not go much further.
I am not sure how general this approach is, for example, I have not tried to use it with continuous and/or measurable maps, or connect it to C*-algebras or von Neumann algebras, although such a connection might exist. (I expect the construction of the structure sheaf to go through just fine for commutative C*-algebras and von Neumann algebras.)
>
> What could be the minimum assumptions about the category for this construction of localization and structural sheaf to remain rigorously perfect?
>
>
>
That's a good question, and I am pretty sure the answer has not been written up (yet?).
In principle, it should not be too difficult to look at the construction of the Zariski spectrum and sheaf and abstract away the relevant properties that make the construction work. I would not be surprised if Malcev/protomodular/homological/semiabelian categories make an appearance at this point (see the book of Borceux and Bourn), since these are designed as abstractions of the usual categories of groups/modules/rings from algebra. But the precise conditions have not been pinned down yet, I think.
| 2 | https://mathoverflow.net/users/402 | 444646 | 179,270 |
https://mathoverflow.net/questions/444630 | 1 | Let $G$ be a locally compact abelian group and $f:G \to \mathbb{C}$ a function. Its Fourier transform (when it exists) is defined to be
$$\widehat{f}(\chi) = \int\_G f(g) \bar{\chi}(g) \mathrm{d} g,$$
Here $\chi \in \widehat{G} := \mathrm{Hom}(G, S^1)$ and $\mathrm{d} g$ is a choice of Haar measure on $G$.
Why does $\bar{\chi}$ appear and not $\chi$? Is this for historical reasons or is there a mathematical reason?
In practice one considers all characters $\chi$ at once (for example when performing Poisson summation), so it doesn't really make a difference including the complex conjugate in the definition.
| https://mathoverflow.net/users/5101 | Why complex conjugate in definition of the Fourier transform? | This is an $L^2$ product: consider for instance the most classical Fourier expansion of $\mathbb Z$ periodic functions (or distributions) of one real variable.
You have for $f$ locally square integrable
$$
f(x)=\sum\_{k\in \mathbb Z}\langle f, e\_k\rangle\_{L^2} e\_k(x),
$$
that is
$
f(x)=\sum\_{k\in \mathbb Z}\left(\int\_0^1 f(y){e^{-2iπ ky}} dy\right) e^{2iπ kx},
$
simply since
$
\langle f, e\_k\rangle\_{L^2}=\int\_0^1 f(y)\overline{e\_k(y)}dy.
$
| 2 | https://mathoverflow.net/users/21907 | 444652 | 179,272 |
https://mathoverflow.net/questions/431456 | 6 | Edit: I'm specializing this question to the compact case I'll ask about the noncompact case as a new question.
Let $ G $ be a compact connected semisimple Lie group.
Do there always exist two finite order elements of $ G $ which generate a dense subgroup?
Example:
$$
\frac{i}{\sqrt{2}}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}
$$
and
$$
\begin{bmatrix} \overline{\zeta\_{16}} & 0 \\ 0 & \zeta\_{16} \end{bmatrix}
$$
generate a dense subgroup of $ SU\_2 $.
This question is partially inspired by
[Generating finite simple groups with $2$ elements](https://mathoverflow.net/questions/59213/generating-finite-simple-groups-with-2-elements)
which shows that every finite simple group is 2-generated. Indeed even every finite quasisimple group is 2-generated [Is every finite quasi-simple group generated by 2 elements?](https://mathoverflow.net/questions/254164/is-every-finite-quasi-simple-group-generated-by-2-elements) (however this fails for infinite simple groups:
$ \mathrm{PSL}\_2(\mathbb{Q}) $ is not 2-generated indeed not even finitely generated).
This is a cross-post of <https://math.stackexchange.com/questions/4537024/dense-subgroups-generated-by-two-finite-order-elements>
an answer posted there points out that every compact semisimple Lie group can be topologically generated by 2 infinite order elements.
| https://mathoverflow.net/users/387190 | Semisimple compact Lie group topologically generated by two finite order elements | You might want to look at MR0034766 (Kuranishi 1949, [link](https://projecteuclid.org/journals/kodai-mathematical-journal/volume-1/issue-5-6/Two-elements-generations-on-semi-simple-Lie-groups/10.2996/kmj/1138833534.full)). It deals with the connected semisimple case, which might be what you're after.
There's also a 1999 Proc AMS paper by M. Field, ([freely accessible at AMS site](https://www.ams.org/journals/proc/1999-127-11/S0002-9939-99-04959-X/) — MR number for subscribers: MR1618662), which shows that the set of pairs generating a dense open subset of $G$ is non-empty Zariski open in $G \times G$ when $G$ is compact connected semisimple. This should lead to a proof that the elements can be taken to have finite order in this case, as the set of pairs of elements of finite order is dense in $G\times G$. Of course, in the non-compact case some extra condition on $G$ is required.
| 5 | https://mathoverflow.net/users/460592 | 444664 | 179,277 |
https://mathoverflow.net/questions/444422 | 7 | A classical theorem of Kronecker says that the sequence $(\{\alpha\_1 n\}, \{\alpha\_2 n\},\dots,\{\alpha\_d n\})$ ($n \in \mathbb{N}$) is uniformly distributed in $[0,1)^d$ provided that $1,\alpha\_1,\alpha\_2,\dots,\alpha\_d$ are linearly independent over $\mathbb{Q}$. More generally, for a polynomial $p(x)$ (or a $d$-tuple of polynomials $p\_1(x),p\_2(x),\dots,p\_d(x)$) a theorem of Weyl completely describes the distribution of $\{p(n)\}$ ($n \in \mathbb{N}$) (respectively, $(\{p\_1(n)\}, \{p\_2(n) n\},\dots,\{p\_d(n)\})$). More generally still, the distribution of bounded *generalised polynomials* - i.e., expressions built up using polynomials, and the operations of addition, multiplication and fractional part - is completely understood thanks to work of Bergelson and Leibman. Generalised polynomials include, for instance, expressions such as $\{\alpha n \{\beta n \} \}$.
I'm interested in the behaviour of expressions that resemble generalised polynomials, but also include division. (In the application that I have in mind it is sufficient to consider relatively simple expressions of this type - such as $\{ n g(n)/h(n) \}$ where $g,h$ are generalised polynomials bounded by $1$ and $h(n) \neq 0$ for all $n \in \mathbb{N}$.) It seems to me that nothing is known about expressions like that, although I'd be delighted to be proven wrong. With this in mind, I would like to ask about the first non-trivial instance:
*Is the sequence $\{n/\{ \sqrt{2}n\} \}$ uniformly distributed in $[0,1)$? Is the set of its values at least dense in $[0,1)$? What about sequences $\{\alpha n/\{ \beta n\} \}$, where $\alpha$ is non-zero and $\beta$ is irrational?*
| https://mathoverflow.net/users/14988 | Is the sequence $\{n/\{ \sqrt{2}n\} \}$ uniformly distributed in $[0,1)$? | I'll show that $\{ n / \{ \alpha n \} \}$ is equidistributed under a certain Diophantine condition on $\alpha$ which holds generically (and in particular for $\sqrt 2$), but the proof goes through verbatim for $\{ \beta n / \{ \alpha n \} \}$.
A very reasonable guess is that $n$ and $\{ n \alpha \}$ are "independent", in the sense that for any rational function $p(x, y)$, if $\{ p(x, y) \}$ is equidistributed (as $N \to \infty$) for $0 \leq x \leq N$ and $0 \leq y \leq 1$, then the sequence $\{ p(n, \{ n \alpha \} ) \}$is equidisitributed for any irrational $\alpha$. I haven't thought about how much this proof can be adapted to this more general question.
The Diophantine condition on $\alpha$ is that
$$
\liminf\_{p, q \to \infty} q^{\frac{7}{3}} \left\lvert \alpha - \frac{p}{q} \right\rvert = \infty
$$
For a generic number $\alpha$ this condition holds with $\frac{7}{3}$ replaced by $2 + \epsilon$.
---
The general idea is this: take $q$ such that $q \alpha$ is very close to an integer (say, the denominator of a convergent in the continued fraction). Then, $\frac{n + k q}{\{ (n + k q) \alpha) \}} = \frac{n + k q}{\{ n \alpha \} + k \varepsilon}$ and so we can understand the behavior of this sequence better than the original sequence, and show that along almost all arithmetic progressions of difference $q$ the sequence is equidistributed.
Expanding the denominator using the formula for a geometric progression, it turns out that the sequence is very close to a quadratic polynomial in $k$, and these are easily shown to be equidistributed using Weyl differencing. Now for the details:
---
First some definitions and preliminaries.
Let's look at the sequence of $\{ n / \{ \alpha n \} \}$ for $1 \leq n \leq N$. Take $p, q$ such that $q \alpha - p = \varepsilon$ is small (we will choose them precisely later). For now, we will require that $q = o(N)$. Let $\ell$ be the largest integer such that $q \ell \leq N$. Since $q = o(N)$ we will show equidistribution of $\{ n / \{ \alpha n \} \}$ for $1 \leq n \leq q \ell$ and this will be sufficient.
Since $\{ \alpha n \}$ is equidistributed, for the rest of the proof we will work only with the subset of $n$ such that $c \leq \{ \alpha n \} \leq 1 - c$ for some arbitrarily small positive constant $c$. This will make things easier because the function $\frac{1}{x}$ behaves nicer when $x$ is bounded away from $0$.
Now, take some $1 \leq n \leq q$ and look at $\frac{n + k q}{\{ \alpha n + k q \alpha \}} = \frac{n + k q}{\{ \alpha n \} + k \varepsilon}$ for $0 \leq k < \ell$ (we have assumed here that $\lvert \ell \varepsilon \rvert < c$). Notice that
$$
\frac{n + k q}{\{ \alpha n \} + k q} = \frac{1}{\{ \alpha n \}} \cdot \frac{n + k q}{1 + k \frac{\varepsilon}{\{ \alpha n \}}} =
$$
$$
\frac{1}{\{ \alpha n \}} \cdot (n + k q) \cdot \left( 1 - k \frac{\varepsilon}{\{ \alpha n \}} + k^2 \frac{\varepsilon^2}{\{ \alpha n \}^2} + \mathcal{O} \left( \ell^3 \varepsilon^3 \right) \right) =
$$
$$
\frac{n}{\{ \alpha n \}} + \left( \frac{q}{\{ \alpha n \}} - \frac{\varepsilon}{\{ \alpha n \}^2} \right) \cdot k - \frac{q \varepsilon}{\{ \alpha n \}^2} \cdot k^2 + \mathcal{O} (\ell^3 q \varepsilon^2)
$$
Assume that $\ell^3 q \varepsilon^2 = o(1)$. Then, it is sufficient to show equidistribution (for $0 \leq k < \ell$) of the quadratic sequence
$$
\frac{n}{\{ \alpha n \}} + \left( \frac{q}{\{ \alpha n \}} - \frac{\varepsilon}{\{ \alpha n \}^2} \right) \cdot k - \frac{q \varepsilon}{\{ \alpha n \}^2} \cdot k^2
$$
The technique of Weyl differencing says that for a sequence $a\_n$, if $a\_{n + h} - a\_n$ is equidistributed for every $h$ then $a\_n$ is equidistributed. Because our sequence is a quadratic polynomial, and so what we need to show is that for almost all $n < q$ the sequence
$$
\frac{q \varepsilon}{\{ \alpha n \}^2} \cdot k
$$
is equidistributed for $0 \leq k < \ell$.
---
It is easy to see that a necessary condition for this sequence to be equidistributed for almost all $n$ is $q \varepsilon \ell \to \infty$. The following simple fact is useful:
Suppose that $q \varepsilon \ell \to \infty$.
1. If $q \varepsilon = o(1)$ then the sequence is equidistributed for all $n$.
2. If $q \varepsilon = \Theta(1)$ then the sequence is equidistributed for almost all $n$.
Proof:
1. We have an arithmetic progression with tiny step size which wraps around the interval a number of times which goes to infinity, and so this is clearly equidistributed.
2. The exceptional set of $\beta$ which are $\Theta(1)$ such that $\beta k$ is not equdistributed is a finite set of very small intervals around rational numbers of very small denominator, and from the equidistribution of $\{ \alpha n \}$ we see that a negligible proportion of $n$'s fall into this exceptional set.
Either way, we see that when $q \varepsilon \ell \to \infty$ and $q \varepsilon = \mathcal{O}(1)$ the sequence is equidistributed, so we will try to choose $q, \varepsilon$ like this.
---
As an example, suppose for a moment that the continued fraction of $\alpha$ has bounded coefficients, as happens in the case of $\sqrt 2$. In this case, we have convergents of any size we want, and so we can take $q$ such that $\ell$ grows arbitrarily slowly to infinity and $q \varepsilon = \Theta (1)$, which proves equidistribution in this case by the above lemma.
---
Unfortunately, sometimes there are large coefficients in the continued fraction and so we have to work a bit more. Here is the general method.
Take $t$ growing to infinity. By Dirichlet's theorem, there are some $q\_0, p\_0 \leq N t$ such that $\lvert \varepsilon\_0 \rvert = \lvert q\_0 \alpha - p\_0 \rvert \leq \frac{1}{t N}$. We will take $q = r q\_0, \ \ell = \frac{N}{r q\_0}$ for some $r$ which we will choose now.
Firstly, we will want $\varepsilon = r \varepsilon\_0$ (that is, for the closest integer to $r q\_0 \alpha$ to be $r p\_0$) and for this we need $r \lvert \varepsilon\_0 \rvert < \frac{1}{2}$. In the proof we have required $\lvert \ell \varepsilon \rvert < c$ and $\ell^3 q \varepsilon^2 = o(1)$ which is equivalent to
$$
\frac{N \lvert \varepsilon\_0 \rvert}{q\_0} < c \iff N \left\lvert \alpha - \frac{p\_0}{q\_0} \right\rvert < c
$$
and
$$
\frac{N^3 \varepsilon\_0^2}{q\_0^2} = o(1) \iff N^3 \left\lvert \alpha - \frac{p\_0}{q\_0} \right\rvert^2 = o(1) \iff N^{\frac{3}{2}} \left\lvert \alpha - \frac{p\_0}{q\_0} \right\rvert = o(1)
$$
The first condition obviously follows from the second. We do have one additional condition on $r$, which is $\ell \to \infty$ (which actually is a condition on $r$).
It is easy to see that $\lvert q\_0 \varepsilon\_0 \rvert \leq 1$. Let $r$ be the smallest positive integer such that $\lvert q \varepsilon \rvert = r^2 \lvert q\_0 \varepsilon\_0 \rvert \to \infty$. From the fact mentioned above, for such $r$ the sequence is equidistributed (assuming $\ell \to \infty$).
The only thing that is left now is to check that in this case $\ell \to \infty$. If $r = 1$ then obviously $\ell \to \infty$. Otherwise, $r \approx \frac{1}{\sqrt{\varepsilon\_0 q\_0}}$
$$
\ell = \frac{N}{r q\_0} \approx \frac{N \sqrt{\varepsilon\_0}}{\sqrt{q\_0}} = N \left\lvert \alpha - \frac{p\_0}{q\_0} \right\rvert^{\frac{1}{2}}
$$
This tends to infinity iff $\ell^2 \approx N^2 \left\lvert \alpha - \frac{p\_0}{q\_0} \right\rvert \to \infty$.
Thus, all that is left is to show that for every $N$ there exist $p\_0, q\_0$ such that $N^{\frac{3}{2}} \left\lvert \alpha - \frac{p\_0}{q\_0} \right\rvert \to 0$ but $N^2 \left\lvert \alpha - \frac{p\_0}{q\_0} \right\rvert \to \infty$ which follows in an elementary manner from our Diophantine condition on $\alpha$ at the beginning and choosing $t \approx N^{-\frac{1}{7}}$.
---
Some comments:
1. I think the Diophantine condition can be relaxed all the way to requiring $\alpha$ not to be a Liouville number by taking higher order terms in the geometric series expansion, and applying iterated Weyl differencing.
2. The Diophantine condition occurs because when $q$ is very small compared to $N$, the sequences become long and very complicated, which makes them more difficult to control. Perhaps one could divide the interval into a bunch of more short arithmetic progressions. The fact that the numbers appearing are bigger might be compensated by the fact that the denominator changes very slowly, but I'm not sure of this.
| 3 | https://mathoverflow.net/users/88679 | 444668 | 179,278 |
https://mathoverflow.net/questions/444665 | 8 | One way to stratify the large cardinal hierarchy between I3 and I1 is by using second-order elementary embeddings. We may view $j\colon V\_{\lambda+1}\to V\_{\lambda+1}$ as a second-order embedding $j\colon (V\_\lambda,V\_{\lambda+1})\to (V\_\lambda,V\_{\lambda+1})$, and it is known that $\mathrm{I2}(\lambda)$ is equivalent to there is an $\Sigma^1\_1$-elementary embedding $j\colon (V\_\lambda,V\_{\lambda+1})\to (V\_\lambda,V\_{\lambda+1})$, or equivalently, $\Sigma\_1$-elementary embedding over $V\_{\lambda+1}$ to itself.
(A sentence is $\Sigma^1\_1$ if it is of the form $\exists X\subseteq V\_\lambda \phi^{V\_\lambda}(x)$.)
Laver proved that $\Sigma^1\_{2n-1}$-elementary embedding is also $\Sigma^1\_{2n}$-elementary for $n\ge 1$, and if there is a $\Sigma^1\_{2n+1}$-elementary embedding $j\colon V\_\lambda\to V\_\lambda$, then we can find $\lambda'<\lambda$ and a $\Sigma^1\_{2n-1}$-elementary $k\colon V\_{\lambda'}\to V\_{\lambda'}$.
We may turn our view to third-order embeddings $j\colon V\_{\lambda+2}\to V\_{\lambda+2}$, or alternatively, $\Sigma\_n$-elementary embeddings over $V\_{\lambda+2}$. Of course, full elementary embedding $j\colon V\_{\lambda+2}\to V\_{\lambda+2}$ is incompatible with $\mathsf{ZFC}$, but it does not automatically rule out all $\Sigma\_n$-elementary embeddings over $V\_{\lambda+2}$ to itself. On the other hand, coding Woodin's proof into $V\_{\lambda+2}$ seems to exclude $\Sigma\_n$-elementary embeddings over $V\_{\lambda+2}$ for some $n$ (maybe $n\ge 2$?)
>
> **Question.** Is a $\Sigma\_1$-elementary embedding $j\colon V\_{\lambda+2}\to V\_{\lambda+2}$ compatible with $\mathsf{ZFC}$?
>
>
>
| https://mathoverflow.net/users/48041 | The consistency of $\Sigma_1$-elementary embeddings $j\colon V_{\lambda+2}\to V_{\lambda+2}$ over $\mathsf{ZFC}$ | The answer is no, it is not consistent. If $j : V\_{\lambda+2}\to V\_{\lambda+2}$ is $\Sigma\_1$-elementary, then $\mathcal U = \{A\subseteq P(V\_\lambda) : j[V\_\lambda]\in j(A)\}$ is a normal fine ultrafilter concentrating on $\sigma$ such that $\text{ot}(\sigma\cap \lambda) = \lambda.$ Then taking an ultrapower of the universe, we obtain an elementary embedding $j\_\mathcal U : V \to M$ with $j\_{\mathcal U}[V\_\lambda] = [\text{id}]\_{\mathcal U}\in M$ and $j\_{\mathcal U}(\lambda) = \lambda$, contrary to Kunen's inconsistency theorem.
For more information (and in the ZF context), see section 4.3 (Reinhardt ultrafilters) of my joint paper ["Periodicity in the cumulative hierarchy"](https://ems.press/content/serial-article-files/26941) with Farmer Schlutzenberg. The work in this section is for the most part due to Farmer.
| 9 | https://mathoverflow.net/users/102684 | 444676 | 179,284 |
https://mathoverflow.net/questions/444653 | 21 | There is an easy proof of the PNT, just in a few lines, in the book by Julian Havil, "Gamma", pages 201-202. Specifically, Von Mangoldt's formula, which is very easy to derive:
$$
\psi(x) = x - \ln(2\pi) - \frac{1}{2}\ln(1 - x^{-2}) - \sum\_{\zeta(\rho) = 0} \frac{x^{\rho}}{\rho},
$$
implies the PNT, since from this equation it follows that $\psi(x)/x \to 1$.
I am a Physicist and my Mathematician colleagues regard this proof as non-elementary. I've never seen easier proof of the PNT than this one. Any other proofs of the PNT without complex analysis are very lengthy, tedious and difficult to understand (and actually non-elementary, therefore). But why this a very simple proof is regarded as non-elementary? Unfortunately, I could not get a clear answer. Could you please clarify?
EDIT: Von Mangoldt's formula itself is not sufficient to immediately claim that $\psi(x)/x \to 1$. However, we can consider [Newman's proof](https://www.tandfonline.com/doi/pdf/10.1080/00029890.1997.11990704) of the PNT. This proof is much easier and shorter than any existing non-elementary proofs of the PNT. Although this proof involves very basics of the complex analysis, it is, nevertheless, regarded as a non-elementary proof. Therefore, the difference between elementary and non-elementary proofs of the PNT is the presence of the complex analysis. I hope, I am not mistaken.
| https://mathoverflow.net/users/502718 | What is the difference between elementary and non-elementary proofs of the Prime Number Theorem? | To complement Will Sawin's answer, in the specific context of the prime number theorem, there are historically well-established notions of "elementary" and "non-elementary" proofs, stemming from Hardy's 1921 lecture:
>
> No elementary proof of the prime number theorem is known, and one may ask whether it is reasonable to expect one. Now we know that the theorem is roughly equivalent to a theorem about an analytic function, the theorem that Riemann’s zeta function has no roots on a certain line. A proof of such a theorem, not fundamentally dependent on the theory of functions, seems to me extraordinarily unlikely. It is rash to assert that a mathematical theorem cannot be proved in a particular way; but one thing seems quite clear. We have certain views about the logic of the theory; we think that some theorems, as we say ‘lie deep’ and others nearer to the surface. If anyone produces an elementary proof of the prime number theorem, he will show that these views are wrong, that the subject does not hang together in the way we have supposed, and that it is time for the books to be cast aside and for the theory to be rewritten.
>
>
>
The history behind is that while up-to-multiplicative constant bounds for the prime counting function were obtained, by Chebyshev, by an ingenious but *elementary* counting arguments, all the proofs of PNT known in 1921 were a lot more involved and went via Riemann's zeta function and it analytic properties.
An elementary proof was found in 1948 by Selberg and Erdős:
>
> In this paper will be given a new proof of the prime-number theorem, which is elementary in the sense that it uses practically no analysis, except the simplest properties of the logarithm.
>
>
>
This became a big deal, in a large part, because it answered the above-quoted challenge of Hardy. The shortest proof nowadays is [Newman's proof,](https://www.tandfonline.com/doi/pdf/10.1080/00029890.1997.11990704?casa_token=nqD5CeXgqLwAAAAA:-4CZupo4hWsPL-lLYDcWd_OstRk04V2JZnUK2dpQ6cdbhUZynZ_TqZ4e6hnfCc_C_313YPuwoLB-yUc) although using complex analysis, it is much more "elementary" than the proofs known to Hardy; it is an interesting question how would Hardy judge it.
| 29 | https://mathoverflow.net/users/56624 | 444677 | 179,285 |
https://mathoverflow.net/questions/444694 | 7 | For *weakly cohesive toposes*, there exists a notion of *contractability*, and toposes with a subobject classifier $\Omega$ that is contractible are of special interest (see [here](https://ncatlab.org/nlab/show/sufficiently+cohesive+topos#cohesively_connected_truth)).
It occured to me that one could view probability theory as a theory of (convex) interpolation of truth, which replaces boolean truth $\Omega = \{0,1\}$ by 'continuous truth' $\Omega = [0,1]$, so I wonder, is there an example of a topos with a subobject classifier given (in a suitable sense) by the real unit interval?
| https://mathoverflow.net/users/3824 | Topos with $\Omega = [0,1]$? | Here's an example which, I'm afraid, is not very interesting (and may not match your notion of “suitable sense”): let $X$ be the topological space $\mathopen]0,1\mathclose[$ (the open interval) with the (not at all separated) topology given by the $U\_x := \mathopen]0,x\mathclose[$ for $x \in [0,1]$ (with $U\_0 = \varnothing$, obviously). Since $U\_x \cap U\_y = U\_{\min(x,y)}$ and
$$
\bigcup\_{i\in I} U\_{x\_i} = U\_{\sup\{x\_i : i\in I\}}
$$
this is indeed a topology. Of course, $\mathcal{O}(X) := \{U\_x : x\in[0,1]\}$ can be identified with $[0,1]$ as an ordered set (indeed, frame). The topos of sheaves on $(X, \mathcal{O}(X))$ then has this as subobject classifier.
| 6 | https://mathoverflow.net/users/17064 | 444696 | 179,291 |
https://mathoverflow.net/questions/444625 | 1 | Let $(F,|.|)$ be a complete algebraically closed field. Let $x$ be the point of type 5 corresponding to the unit open disc of the adic affine line over $F$. Can we obtain a concrete description of the complete residue field of $x$ (as for point of type 2 or 3)? Does the local ring of $x$ coincide with the Robba ring?
| https://mathoverflow.net/users/117853 | Complete residue field of a point of type 5 | A point of type 5 corresponds to a valuation of rank 2, so I am not sure if the meaning of completion is completely clear here. I think that the usual thing is to complete with respect to the associated rank 1 valuation. Then, you would get back the complete residue field at the associated type 2 point (the Gauss point in your case).
Your point of type 5 is closed and is in the closure of the Gauss point. As a result, any neighborhood of the type 5 point will also contain the Gauss point, so you will get something strictly smaller than the Robba ring. It is contained in the bounded Robba ring, but still strictly smaller I guess. For a rather concrete description, you can have a look at the notes by Florent Martin here: <https://florentguymartin.github.io/pdf/intro_adic.pdf>.
| 5 | https://mathoverflow.net/users/4069 | 444698 | 179,292 |
https://mathoverflow.net/questions/444615 | 3 | For a Lie group $G$, we can define a principal $G$ bundle as a submersion of manifolds $\pi:P \to X$ equipped with a free right $G$-action on $P$ that is transitive on the fibres over $X$.
What goes wrong with an analogous definition for 2-groups? For now, we can think of 2-groups as weak monoidal groupoid with monoidal inverses. The autoequivalences of a category $\operatorname{Aut}(C)$ form a 2-group. One can define a right action of a 2-group on a category $C$ as a monoidal functor from the 2-group to $\operatorname{Aut}(C)$. I will also consider only essentially finite two groups.
Now given a 2-group $G$, let $\pi:\mathfrak{P} \to \mathfrak{X}$ be a representable submersion of stacks over the category of manifolds (called $\mathrm{Man}$) equipped with the étale site. Suppose we define a "principal $G$ bundle" by an action of G on the category $\mathfrak{P}$ (on the right) such that the functor
\begin{gather\*}
\mathfrak{P} \times G \to \mathfrak{P} \times\_{\mathfrak{X}} \mathfrak{P} \\
(p,\gamma) \mapsto (p\cdot\gamma,p)
\end{gather\*}
is an equivalence of categories over $\mathrm{Man}$. Note that the action is over $\mathrm{Man}$, i.e. $p$ and $p\cdot \gamma$ are over the same object.
I could not find such a definition in the literature. Does something go wrong? If such a definition is available, please share a reference.
| https://mathoverflow.net/users/18080 | Categorifying the definition of a principal $G$ bundle | The definition you are looking for is precisely Def. 6.1.5 in:
*Nikolaus, Thomas; Waldorf, Konrad*, [**Four equivalent versions of nonabelian gerbes**](https://doi.org/10.2140/pjm.2013.264.355), Pac. J. Math. 264, No. 2, 355-420 (2013). [ZBL1286.55006](https://zbmath.org/?q=an:1286.55006).
This definition is based on the work of Toby Bartels ([Higher gauge theory I: 2-bundles](https://arxiv.org/abs/math/0410328)) and Christoph Wockel ([Principal 2-bundles and their gauge 2-groups](https://arxiv.org/abs/0803.3692)). The main advantage of the definition in the paper of Thomas Nikolaus and me is that it is as simple as possible: the action functor is a smooth functor (as opposed to a bibundle, or anafunctor), and the action is strict. Yet, principal 2-bundles in this sense are classified by Giraud's non-abelian cohomology, and are equivalent to bundle gerbes.
| 7 | https://mathoverflow.net/users/3473 | 444700 | 179,293 |
https://mathoverflow.net/questions/444662 | 10 | Igusa defined a genus 2 Siegel modular form $\chi\_{10}$, which vanishes on the Humbert surface $G\_{1}$ (the image of a "degenerate" Hilbert modular surface, the product of modular curves, inside the Siegel modular threefold, see for instance page 218 of van der Geer's Hilbert Modular Surfaces).
This should be related to Borcherds products, and in particular I have seen it mentioned (for instance in page 47 of [this paper](https://www.mathematik.tu-darmstadt.de/media/algebra/homepages/bruinier/publikationen/BY1h.pdf) of Bruinier and Yang) that $\chi\_{10}$ is up to a power of 2 the square of a Borcherds product. Is there a place where I can see this explicitly worked out? I would like to see, for example, the product expansion as well as which weakly holomorphic form this Borcherds product is lifted from, etc.
| https://mathoverflow.net/users/85392 | Igusa's $\chi_{10}$ and Borcherds products | The product expansion and the weakly holomorphic (Jacobi) form it lifts from appear as Example 2.4 in Gritsenko, Nikulin - *Automorphic Forms and Lorentzian Kac--Moody Algebras II*, Internat. J. Math. 9(2) (1998), 201--275.
[(arXiv)](https://arxiv.org/abs/alg-geom/9611028).
The square root of $\chi\_{10}$ is denoted there by $\Delta\_5(Z)$.
| 4 | https://mathoverflow.net/users/502756 | 444704 | 179,295 |
https://mathoverflow.net/questions/444701 | 11 | I came up with this question when trying to give a more detailed answer to a question by Tim Campion in [a comment](https://mathoverflow.net/questions/424356/examples-of-statements-that-are-valid-in-every-spatial-topos?rq=1#comment1121603_435325) to Ingo Blechschmidt's answer to [Examples of statements that are valid in every spatial topos](https://mathoverflow.net/q/424356/41291).
There is an obvious internalized form of Zorn's lemma for toposes. Basically, you can use the Mitchell-Bénabou language to spell out the following:
given a poset $(P,\leqslant)$, consider the object of pairs $(C,u)$ where $C$ is a chain in $P$ and $u$ is an upper bound of $C$ in $P$. We say that the *internal Zorn's lemma* **IZ** holds if whenever the projection $(C,u)\mapsto C$ from this object to all chains of $P$ is epi, the object of maximal elements of $P$ is "as inhabited as $P$ itself", that is, has the same support as $P$.
If you do not care for a more rigorous formulation, skip everything until the questions.
Here is this more rigorous formulation. Given a poset $P$ in a topos $\mathscr S$, we can form the objects $\max(P)\rightarrowtail P$ of maximal elements of $P$ and $\operatorname{chains}(P)\rightarrowtail\Omega^P$ of chains of $P$. We can also form the object of upper-bounded-chains of $P$, call it, say, $\operatorname{ubc}(P)$: it is uniquely determined by saying that $\hom(X,\operatorname{ubc}(P))$ must be in one-to-one correspondence with pairs $(C,u)$, where $C\rightarrowtail X\times P$ is a subobject of $X\times P$ and $u:X\to P$ is a morphism, such that $C$ is a chain and $u$ is an upper bound of $C$, if one considers $u$ as an element of $X^\*(P)$ and $C$ as a subobject of $X^\*(P)$, in the slice topos $\mathscr S/X$.
Clearly there is a canonical projection $\pi:\operatorname{ubc}(P)\to\operatorname{chains}(P)$, given by sending $(C,u)$ to $C$.
We can then formulate the internal version **IZ** of the Zorn lemma as follows:
>
> If $\pi:\operatorname{ubc}(P)\to\operatorname{chains}(P)$ is epi, then $\max(P)$ has the same support as $P$; that is, the image of $\max(P)\to1$ is the same as the image of $P\to1$ (where $1$ is the terminal object of $\mathscr S$).
>
>
>
Minimal question: do all toposes satisfy this?
I suspect that, arguing internally, one might deduce from **IZ** internal choice **IC** which in turn implies booleannes, but somehow I don't see how to actually do it.
Extended question (again in case the answer to the minimal question is negative): there might be more sophisticated internalizations of the Zorn's lemma. For example, one can consider, for an object $P$, the object $\operatorname{Orders}(P)$ of partial orders on $P$ and then internalize the statement "partial orders with all chains upper-bounded are included in partial orders having a maximal element". Is there a form which would be weaker, in the sense that it holds for some topos which does not satisfy **IZ**?
Here is a version about which I would like to ask specifically. This actually corresponds on the "classical" side to the variant of the Zorn's lemma with requiring upper bounds for *nonempty* chains only.
Call an internal poset $(P,\leqslant)$ *internally inductive* if for any object $X$ and any $C\rightarrowtail X\times P$ which is a chain of $X^\*(P)$ in $\mathscr S/X$, the object $U\_C\rightarrowtail X^\*(P)$ of upper bounds of $C$ has support no less than $C$. That is, the image of the composite $U\_C\rightarrowtail X\times P\to X$ contains the image of $C\rightarrowtail X\times P\to X$, where $X\times P\to X$ is the projection.
We then say that **IIZ** holds in $\mathscr S$ if for every internally inductive poset $P$, the object $\max(P)$ has the same support as $P$, as above.
And the specific instance of my Extended question is,
>
> Which toposes satisfy **IIZ**? How does it compare to **IZ**?
>
>
>
**Important correction**
As Gro-Tsen points out in a comment, this does not make much sense unless I restrict to Grothendieck toposes with Axiom of Choice holding in my set theory. Slightly more generally, one may consider toposes bounded over a topos with **AC**. Maybe still more generality is possible, but let us stick to this for definiteness.
| https://mathoverflow.net/users/41291 | Do all toposes satisfy the internal Zorn's lemma? | Assuming the law of excluded middle internally your formulation of Zorn's lemma is equivalent to the axiom of choice by the usual argument. Now there are Boolean Grothendieck topos in which the axiom of choice fails. Hence they don't satisfies Zorn lemma either.
I think the simplest example of this is to take $G$ to be a non-discrete but pro-discrete localic/topological group (typically $G = \mathbb{Z}\_p$ the group of p-addic integer) and consider $\mathcal{T}$ the topos of sets endowed with a continuous (hence smooth) action of $G$ (that is the stabilizer of each point is an open subgroup)
Assuming LEM in the base, $\mathcal{T}$ is clearly boolean (sub-object are $G$-stable subsets and set theoretic complement preserve subobject).
But $\mathcal{T}$ doesn't satisfies either internal nor external choice. For exemple for $G = \mathbb{Z}\_p$ the infinite product of all the $\mathbb{Z}/p^k\mathbb{Z}$ with their canonical action is the initial object. (more generally, the product of the $G/U$ for $U$ in a fundamental system of open subgroup is empty unless $G$ is discrete).
**Edit:** However it should be noted that, assuming Zorn lemma in the base topos, an appropriate version of Zorn lemma is known to holds in every **Localic** Grothendieck topos (but the topos mentioned above is not localic of course). This is proved as Proposition D4.5.14 in Sketches of an elephant.
More precisely, one take the Zorn lemma to be the statement: "every inductie poset has a maximal element" but where "inductive" is defined as the internal statement "every chain as an upper bound". More precisely prop D4.5.14 shows that if P is an (internally) inductive poset in a localic topos, then it has an (internally) maximal element given by a global section. In particular, it proves the internal statement "there exists a maximal element" is valid.
One also have the stronger version of the internal Zorn Lemma where the quantification over "all inductive poset" is interpreted internally using the Stack semantics, because every slice of a localic topos is localic, and we can apply D4.5.14 in each slice.
The paper by Bell linked in Peter Lumsdain comment, also prove a similar result but for an (at least in appearance) weaker version of Zorn lemma where "inductive" is taken as "every chain has a least upper bound" instead of an upper-bound. Bell explicitly says that he doesn't know how to prove the stronger version of Zorn lemma, but looking at the proof of D4.5.14 it seems to me that the whole point is that both version of Zorn lemma are constructively equivalent: given any poset the poset of chains ordered by inclusion is inductive in Bell's stronger sense, so assuming Bell's version of ZL one get a maximal chain in every poset and an upper bound for a maximal chain has to be a maximal element.
| 9 | https://mathoverflow.net/users/22131 | 444719 | 179,301 |
https://mathoverflow.net/questions/444723 | 10 | [This is a cross-post](https://math.stackexchange.com/questions/4602412/derivative-without-extrema-is-monotone) from Math.SE.
The question was asked there 3 months ago but didn't receive much attention aside from one comment asking for clarification. I feel like it might be non-trivial and so I decided to repost the question here.
Let $f\colon(a,b)\to\mathbb R$ be differentiable on $(a,b)$ such that $f'$ has no local extrema on $(a,b)$. Is it true that $f'$ is necessarily monotone on $(a,b)$?
If $f'$ is assumed to be continuous on $(a,b)$ then this is definitely true. Indeed, by continuity of $f'$ each segment $[x\_1,x\_2]\subset(a,b)$ contains points of global maximum and minimum of $f'$ on this segment which must be attained at the endpoints of the segment (otherwise $f'$ will have a local extremum), so $\min\{f'(x\_1),f'(x\_2)\}\leq f'(x)\leq\max\{f'(x\_1),f'(x\_2)\}$ for any points $x\_1<x<x\_2$ in $(a,b)$, and the latter property can be shown to be equivalent to monotonicity.
However, derivatives can be discontinuous, and I'm not sure how to proceed in that case. I couldn't find similar problems on both Math.SE and MO or elsewhere on the internet, but the question seems so natural that it probably had been considered and researched by some analysts of 20th century.
| https://mathoverflow.net/users/494118 | Derivative without extrema is monotone | The following is copied from A. Bruckner, *Differentiation of Real Functions* (2d ed, 1994), at the end of chapter 6 (“The Zahorski Classes”): I didn't check the details carefully, but it seems to answer your question. (Remarks in brackets are mine. All typos are also likely to be mine.)
>
> A continuous function defined on $[a,b]$ must assume a maximum and a minimum on $[a,b]$. The corresponding statement for derivatives is false. (Consider, for example, a derivative $g$ whose graph is similar to the graph of $g\_0 = \sin(1/x)$, $g\_0(0)=0$ except that $g$ never quite achieves the values $1$ or $-1$.) More surprising is the fact that a derivative need not achieve any *local extrema*. To see this, consider the function $f$ of Example 1.1 of Chapter 1 [reproduced below]. This function has the Darboux property [i.e., the intermediate value property], but is not in Baire class one. If we redefine $f$ so as to equal $0$ at *every* point of $C$, then the resulting function $g\_1$ is in $\mathcal{B}\_1$ [Baire class $1$], but no longer has the Darboux property. Now let $C\_2$ be a nowhere dense perfect subset of $[0,1]$ such that each point of $C\_1$ [I think $C\_1 := C$ here] (other than $0$ and $1$) is a bilateral point of accumulation of $C\_2$, while $0$ is a point of accumulation [of] $C\_2$ from the right and $1$ is a point of accumulation from the left. Let $g\_2$ be defined relative to $C\_2$ as $g\_1$ was defined relative to $C\_1$. Continue the process, arriving at a sequence of nowhere dense perfect sets $\{C\_k\}$ such that $0\in C\_k$, $1\in C\_k$ and every other point of $C\_k$ is a bilateral point of accumulation of $C\_{k+1}$ and such that $\bigcup C\_k$ is dense in $[0,1]$. To each set $C\_k$ corresponds a function $g\_k$. Let $g(x) = \sum g\_k(x)/2^k$.
>
>
> Since $g$ is the uniform sum of functions in $\mathcal{B}\_1$, $g$ is also in $\mathcal{B}\_1$. Even though none of the functions $g\_k$ is in $\mathcal{D}$ [functions with the Darboux property], their sum is. To verify this, we can apply any of the several parts of Theorem 1.1 of Chapter 2 [characterizing Darboux functions among those in Baire class $1$]. Furthermore, it is easy to verify that $g$ attains no local extrema.
>
>
> By Theorem 7.1 of Chapter 2 [a theorem of Maximoff], there exists a homeomorphism $h$ of $[0,1]$ onto itself such that the function $g\circ h$ is approximately continuous. Since this function is bounded, it is also a derivative. Since $g$ attains no local extrema, neither does $g\circ h$. We summarize as a theorem:
>
>
> **Theorem 3.1.** *There exists a bounded approximately continuous derivative which achieves no local maximum and no local minimum.*
>
>
>
Example 1.1 of chapter 1 is this:
>
> Let $C$ be the Cantor set in $[0,1]$. If $(a,b)$ is an interval contiguous to $C$, we define $f(x) = (2(x-a)/(b-a))-1$ for $x\in[a,b]$. Otherwise let $f(x)=0$. Then $f$ is a Darboux function which is discontinuous at every point of $C$.
>
>
>
| 13 | https://mathoverflow.net/users/17064 | 444724 | 179,304 |
https://mathoverflow.net/questions/444727 | 5 | Let $k$ be a characteristic zero field and consider the category $(\mathsf{Sch}/k)\_\text{fppf}$ of schemes over $k$ with the fppf topology. I know that $\underline{\operatorname{Hom}}(\mathbb{G}\_a,\mathbb{G}\_m)=\widehat{\mathbb{G}}\_a$. (Here we see all the groups as abelian sheaves on $(\mathsf{Sch}/k)\_\text{fppf}$.)
**Is it true that the Ext sheaves $\underline{\operatorname{Ext}}^i(\mathbb{G}\_a,\mathbb{G}\_m)$ vanish for $i>0$?**
I know that [Br] shows that the Ext *groups* $\operatorname{Ext}^i(\mathbb{G}\_a,\mathbb{G}\_m)$ vanish for $i>0$. But I don't know how to deduce what I want from this. (This is very close to the problem in [Vanishing of $\text{Ext}^2$ sheaf from abelian variety to multiplicative group](https://mathoverflow.net/questions/247161/vanishing-of-textext2-sheaf-from-abelian-variety-to-multiplicative-group).)
Reference:
* [Br] L. Breen - [Extensions of Abelian Sheaves and Eilenberg–Maclane Algebras](https://doi.org/10.1007/BF01389887)
| https://mathoverflow.net/users/131975 | Do we have $\underline{\operatorname{Ext}}^i_\text{fppf}(\mathbb{G}_a,\mathbb{G}_m)=0$ for $i>0$? | This is false, see Remark 2.2.16 of [Rosengarten - Tate Duality In Positive Dimension Over Function Fields](https://arxiv.org/pdf/1805.00522.pdf#page34) in which a nontrivial local extension of $\mathbb{G}\_a$ by $\mathbb{G}\_m$ is constructed. However, the same paper shows that the first Ext-sheaf vanishes in positive characteristic.
| 6 | https://mathoverflow.net/users/101861 | 444742 | 179,308 |
https://mathoverflow.net/questions/444300 | 4 | Let $A$ be a supersingular elliptic curve over $\mathbb{Z}/p\mathbb{Z}$ and $\mathcal{O}$ an order in an imaginary quadratic field contained in the quaternion algebra $\operatorname{End}(A)$, then by Deuring's lifting lemma, $A$ lifts to an elliptic curve $E$ over $\bar{\mathbb{Q}}$ with complex multiplication by an order $\mathcal{O}'$ containing $\mathcal{O}$. For a reference, see "supersingular j-invariants as singular moduli mod-p" by M. Kaneko. My question is about whether this sort of assertion about lifting can be made quantitative in the following sense.
Let $\mathcal{S}$ be the set of all imaginary quadratic fields $K=\mathbb{Q}(\sqrt{-D})$, and $\mathcal{S}\_A$ the subset of $\mathcal{S}$ consisting of all imaginary quadratic fields $K$ such that there exists an elliptic curve $E$ over $\bar{\mathbb{Q}}$ with complex multiplication by $\mathcal{O}\_K$ such that $E$ is a lift of $A$. Given a real number $x>0$, set $\mathcal{S}(x)$ to consist of $K=\mathbb{Q}(\sqrt{-D})$ such that $D$ is positive, squarefree and $D\leq x$; set $\mathcal{S}\_A(x):=\mathcal{S}\_A\cap \mathcal{S}(x)$. The sets $\mathcal{S}(x)$ and $\mathcal{S}\_A(x)$ are finite, set $\#\mathcal{S}(x)$ (resp. $\#\mathcal{S}\_A(x)$) to denote the cardinality of $\mathcal{S}(x)$ (resp. $\mathcal{S}\_A(x)$). Then, what can be said about the relative upper density $\limsup\_{x\rightarrow \infty} \frac{\#\mathcal{S}\_A(x)}{\#\mathcal{S}(x)}$? Is it positive?
There are only finitely many elliptic curves $A$ over $\mathbb{Z}/p\mathbb{Z}$, perhaps these densities are the same for all such $A$. This would be some sort of statement about the equi-distibution of CM j invariants with respect to mod-$p$ reduction. Working explicitly with quarterion algebras (as is done in loc. cit.), it seems to be difficult to derive such a statement. However, perhaps there are tools from arithmetic geometry to prove such a statement? In fact, I'm looking for asymptotic lower bounds for $\#\mathcal{S}\_A(x)$ as $x\rightarrow \infty$. I'm hoping to do better than simply show that the set $\mathcal{S}\_A$ is infinite.
The reason why I'm interested in this sort of statement has to do with the average behavior of the classical Iwasawa $\lambda$-invariant of an imaginary quadratic field. A recent preprint by M. Stokes <https://arxiv.org/abs/2302.09594> establishes the connection between the $\lambda$-invariant of the cyclotomic $\mathbb{Z}\_p$-extension of an imaginary quadratic field $K$ in which $p$ splits, and the number of points on the mod-$p$ reduction of any CM elliptic curve $E$ with complex multiplication by $\mathcal{O}\_K$.
EDIT: Let $\mathcal{S}$ be the set of all imaginary quadratic fields $K=\mathbb{Q}(\sqrt{-D})$ such that $p$ is inert in $K$. The set $\mathcal{S}(x)$ consists of $K=\mathbb{Q}(\sqrt{-D})$ in $\mathcal{S}$ such that $D$ is positive squarefree and $D\leq x$. This seems to be setting in which this question is nontrivial, although it doesn't quite square with the context in which Stokes' result is proven.
| https://mathoverflow.net/users/470091 | Quantitative lifting for mod-p elliptic curves to characteristic zero CM elliptic curves | Let $E$ be a supersingular elliptic curve in characteristic $p$. Then
$A = \mathrm{End}(E)$ is a maximal order in the quaternion algebra
$D/\mathbf{Q}$ ramified exactly at $p$ and $\infty$. If $\alpha \in A \smallsetminus \mathbf{Z}$, then Deuring's theorem guarantees that the pair $(E,\alpha)$ can be lifted to a characteristic zero, and the lift will necessarily be a CM elliptic curve. If $K$ is an imaginary quadratic field then $\mathcal{O}\_K = \mathbf{Z}[\alpha]$ for some $\alpha$, so the question as to whether $E$ can be lifted to an elliptic curve with CM by the full ring of integers $\mathcal{O}\_K$ is equivalent to asking that there is an inclusion
$$\mathcal{O}\_K \rightarrow A.$$
A necessary condition is that $K \hookrightarrow D$. This is equivalent to the condition that $K$ splits $D$, which is equivalent to the condition that $p$ does not split in $K$. So from now on restrict to such $p$.
Consider for convenience of exposition the case when $K$ has even discriminant so one can choose $\alpha = \sqrt{-\Delta\_K}$. Then $\alpha \in A$ if and only if there exists a trace zero element in $A$ with norm $\Delta\_K$. The trace zero elements form a rank three module (over $\mathbf{Z}$), and so the condition that $\Delta\_K$ be a norm is equivalent to asking that $\Delta\_K$ is represented by the corresponding ternary quadratic form $P\_A(x,y,z)$. Understanding what numbers are represented by ternary quadratic forms is a tricky problem in general. In this setting, the ternary quadratic forms $P\_A$ depend on $A$. However, these forms are all locally the same for all finite places. So the easy thing to compute is how many ways an integer is represented by $P\_A(x,y,z)$ for some $A$. But this is just the number of CM elliptic curves with endomorphisms by
$\mathcal{O}\_K$ which is the class number $h\_K$. This is why things are much simpler when there is only one supersingular point which occurs for only finitely many $p$. When $p=2$, for example, then $D$ is the Hamilton quaternions and $P\_A(x,y,z) = x^2 + y^2 + z^2$, and the number of ways of representing a prime in this form is directly related to class numbers, as was understood by Gauss. For larger $p$, there will be a similar formula for the number of ways of representing numbers in terms of *some* $P\_A$, but understanding the individual factors is more complicated.
In more general settings of ternary definite quadratic forms,
The best one can hope for is that an integer $\Delta\_K$ is represented by such a form as long as it is locally representable and $D$ is sufficiently large. (Small $D$ are always going to be difficult.) Even this is too much to ask for general ternary quadratic forms (a good introductory reference is here [https://personal.math.ubc.ca/~cass/siegel/hanke-ternary.pdf](https://personal.math.ubc.ca/%7Ecass/siegel/hanke-ternary.pdf)).
Fortunately it turns out for this particular case that we are in the best situation, and the conclusion is indeed that, fixing $p$, lifts exist for any $K$ where $p$ does not split as long as $\Delta\_K \gg\_p 1$. But even more is true. We know there are $h\_K \gg D^{1/2 - \varepsilon}$ CM elliptic curves by the ring of integers of $K$. It turns out that there reductions are distributed uniformly among the supersingular points as $\Delta\_K \rightarrow \infty$. This follows from Theorem 3 of Michel's paper (<https://annals.math.princeton.edu/wp-content/uploads/annals-v160-n1-p05.pdf>)
**Some Variations** There are a number of other variations that could be asked. The question (somewhat artificially) considered supersingular elliptic curves over $\mathbf{F}\_p$. One could insist on finding CM lifts which are themselves defined over $\mathbf{Z}\_p$. To be precise, since $p$ is inert in $K$, these lifts do not actually have CM over the base field, but they
are lifts over $\mathbf{Z}\_p$ which have CM over some extension. For convenience let us also restrict to $K/\mathbf{Q}$ of prime discriminant. Then one can ask:
**Problem:** For a fixed $p$, let $K/\mathbf{Q}$ range over imaginary quadratic fields of prime discriminant. What is the distribution of which supersingular elliptic curves over $\mathbf{F}\_p$ lift to a (potentially) CM elliptic curve over $\mathbf{Z}\_p$?
To be more specific, we know that when $K$ has prime discriminant then $h\_K$ is odd. Let $H\_K$ be the class field. The decomposition group at $p$ of $\mathrm{Gal}(H\_K/\mathbf{Q})$
surjects onto $\mathrm{Gal}(K/\mathbf{Q})$ because $p$ is inert in this field; since $h\_K$ is odd, this implies that the decomposition group in $\mathrm{Gal}(H\_K/\mathbf{Q})$ is (any of the) reflections. Since the $j$-invariant of any CM elliptic curve is fixed by exactly one reflection, it follows that there will be a *unique* elliptic curve over $\mathbf{Z}\_p$ which has (potential) CM by $\mathcal{O}\_K$. The mod-$p$ reduction of this curve then gives a unique supersingular elliptic curve over $\mathbf{F}\_p$ determined by $\Delta\_K$. How are these reductions distributed as $\Delta\_K$ ranges over all admissible primes?
To make this even more explicit, let $p=11$, so there are two supersingular elliptic curves with $j=0$ and $j = 1728$.
Let $p \equiv 3 \bmod 4$ be a prime such that $11$ is inert in $\mathbf{Q}(\sqrt{-p})$, so $p \not\equiv 2,6,7,8,10 \bmod 11$. There is a class polynomial $P(x)$ of degree $h\_K$, and this will have a unique root defined over $\mathbf{Z}\_{11}$. For what primes $p$ is this root $0 \bmod 11$ and for what primes is it $1728 \bmod 11$? (This question came up naturally from discussions with David Speyer in the comments). For example, with $\Delta\_K = -3, -23, -31, -47$, the reductions are $j = 0,0,0,1728$. The first case is obvious, the modular equations in the next two cases are
$$t^3 + 3491750t^2 - 5151296875t + 12771880859375,$$
$$t^3 + 39491307t^2 - 58682638134t + 1566028350940383,$$
and the last case is computed in the comments). The default guess in this case is presumably that they each occur roughly half the time.
The question of understanding supersingular isogenies is something that seems to be intensively studied for cryptography,
so these questions may well be known.
| 2 | https://mathoverflow.net/users/491858 | 444748 | 179,310 |
https://mathoverflow.net/questions/444709 | 5 | I have a concrete question about continuous functions on $X = [0,1]^\omega$ (with the product topology).
The map $f:X\to [0, 1]$ given by $(x\_i)\mapsto \prod x\_i$ is well-defined and Borel but not continuous.
Suppose instead that a *continuous* function $g: X\to [0, 1]$ satsifies $g \ge f$ on $X$. What can be said about $g$?
In particular, I would like to know if the following statements are true (it is clear that 1 implies 2):
1. For $S$ a subset of $\omega$, let $f\_S(x) = \prod\_{s\in S} x\_s$. Then $g\ge c\cdot f\_S$ everywhere for some finite set $S$ (EDIT: and some $c > 0$).
2. If $x$ satisfies $x\_i > 0$ for all $x$, then $g(x) > 0$.
| https://mathoverflow.net/users/502762 | Continuous functions on $[0,1]^\omega$ and a product lower bound | The second question also has the negative answer:
Take any sequence $z=(z\_n)\_{n\in\omega}\in[0,1]^\omega$ with $f(z)=0$. On the Hilbert cube $[0,1]^\omega$, consider the metric $d(x,y)=\max\_{n\in\omega}\frac{|x\_n-y\_n|}{2^n}$.
Let $c=(c\_n)\_{n\in\omega}$ be the constant sequence with $c\_n=\frac12$ for all $n$. It is clear that $f(c)=0$. For every $n\in\mathbb Z$, let $$B\_n=\{x\in [0,1]^\omega:d(x,c)< 2^{-n}\}=\{x\in[0,1]:\forall k\le n;(|x\_k-\tfrac12|<2^{k-n}\}$$ be the open ball of radius $2^{-n}$ centered at the point $c$. Observe that $B\_n=[0,1]^\omega$ for $n\le 0$.
For every $n\in\omega$ consider the continuous function $$\lambda\_n:[0,1]^\omega\to[0,1],\quad \lambda\_n:x\mapsto \inf\{d(x,y):y\in B\_{n+1}\cup([0,1]^\omega\setminus B\_{n-1})\},$$ and observe that for every $x\in [0,1]^\omega\setminus\{c\}$ the set $\{n\in\omega:\lambda\_n(x)>0\}$ is not empty and contains at most three numbers. This fact can be used to show that the function $\lambda=\sum\_{n\in\omega}\lambda\_n$ is continuous and $\lambda(x)>0$ for all $x\in [0,1]^\omega\setminus\{c\}$.
For every $n\in\omega$ consider the continuous function $f\_n:[0,1]^\omega\to[0,1]$, $f\_n:(x\_k)\_{k\in\omega}\mapsto\prod\_{k\in n}x\_k$. It is clear that $f(x)=\lim\_{n\to\infty}f\_n(x)=\inf\_{n\in\omega}f\_n(x)$ for every $x\in[0,1]^\omega$.
It is easy to see that $\sup\_{x\in B\_n}f\_n(x)\le \prod\_{k\le n}(\frac12+2^{k-n})\to 0$. Then the function $g:[0,1]^\omega\to[0,1]$ defined by $$g(x)=\begin{cases}
\sum\_{n\in\omega}\frac{\lambda\_n(x)f\_n(x)}{\lambda(x)}&\mbox{if $x\ne c$};\\
0&\mbox{if $x=c$};
\end{cases}
$$
is continuous, and $g\ge f$ because for every $x\in[0,1]^\omega\setminus\{c\}$ the value $g(x)$ belongs to the convex hull of the set $\{f\_n(x):n\in\omega\}\subseteq [f(x),1]$.
| 6 | https://mathoverflow.net/users/61536 | 444758 | 179,312 |
https://mathoverflow.net/questions/444756 | 4 | Specifically, I'm wondering, if X and Y are Hausdorff, and Y is compactly generated, does it follow that C(X,Y), with the compact-open topology, is compactly generated?
Edit: answered as written, but curious about other conditions that do imply the compact-open topology is compactly generated. E.g. if we strengthen the assumption to Y being locally compact?
| https://mathoverflow.net/users/83073 | Under what conditions is the compact-open topology compactly generated? | Not necessarily: consider the compactly generated space $Y=\mathbb R^\infty=\varinjlim \mathbb R^n$, which is the direct limit of Euclidean spaces. Then for the countable discrete space $X=\omega$ the function space $C(X,Y)$ is homeomorphic to $(\mathbb R^\infty)^\omega$ and hence is not sequential and so is not compactly generated. To see that the space $(\mathbb R^\infty)^\omega$ is not sequential, one should apply the known fact that the product $\mathbb R^\infty\times\mathbb R^\omega$ is not sequential.
| 9 | https://mathoverflow.net/users/61536 | 444759 | 179,313 |
https://mathoverflow.net/questions/444757 | 0 | Let $H$ be a Hilbert space and let $a,b\in B(H)$ be such that
$${\rm Tr}(l(a))={\rm Tr}(r(a))<\infty , {\rm Tr}(l(b))={\rm Tr}(r(b))<\infty,$$
$${\rm Tr}(l(a)+l(b)) ={\rm Tr}(l(\alpha\_1 a +\alpha\_2 b ))$$
for any $0\ne \alpha\_1,\alpha\_2\in \mathbb{R}$, where ${\rm Tr}$ denotes the standard trace on $B(H)$ and $l(\cdot),r(\cdot)$ denote the left and the right-support respectively.
Can we say that $l(a)l(b)=r(a)r(b)=0$?
If we replace $B(H)$ with $\ell\_\infty$, then the above equality is clearly true. However, I am not sure about the case for $B(H)$.
| https://mathoverflow.net/users/91769 | Does ${\rm Tr}(l(a)+l(b)) ={\rm Tr}(l(\alpha_1 a +\alpha_2 b ))$ imply that $l(a)l(b)=r(a)r(b)=0$? | This already fails for $2 \times 2$ matrices. Take the rank one matrices $a = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ and $b = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$. For all $\alpha \neq 0$ and $\beta \neq 0$, the matrix $\alpha a + \beta b$ is invertible, so that $\text{Tr}(l(\alpha a + \beta b)) = 2$. Also $\text{Tr}(l(a) + l(b)) = 2$. But the ranges of $a$ and $b$ are not orthogonal.
| 3 | https://mathoverflow.net/users/159170 | 444766 | 179,314 |
https://mathoverflow.net/questions/444768 | 1 | Let $G = (V, E)$ be a finite, simple, undirected graph with $V \cap E = \emptyset$. The *total graph* $T(G)$ is defined on the vertex set $V \cup E$ and its edge set is given by $$E(T(G)) = E \cup \big\{\{e, f\}: e, f \in E\text{ and } |e\cap f| = 1\big\}\cup \big\{\{v, e\}: v\in V, e\in E, v\in e\big\}.$$
**Question.** If there is a Hamiltonian path in $G$, is there a Hamiltonian path in $T(G)$?
| https://mathoverflow.net/users/8628 | Hamiltonian path in total graph | The edges incident to $v \in V$ form a clique in $T(G)$, so the Hamiltonian path $v\_1, v\_2, v\_3, \ldots, v\_n$ in $G$ can be lifted into a Hamiltonian path in $T(G)$ as follows:
1. Insert the edges to get $v\_1, \{v\_1, v\_2\}, v\_2, \{v\_2, v\_3\}, v\_3, \ldots, v\_n$.
2. In arbitrary order, insert every edge which is not already included in the path immediately after one of its vertices.
| 3 | https://mathoverflow.net/users/46140 | 444769 | 179,315 |
https://mathoverflow.net/questions/444750 | 1 | $X$ is in the form of exponential family i.e.
$$\mathbb{P\_\theta}x = h(x)e^{\langle \theta,T(x)\rangle-\phi(\theta)}$$
where $\theta\in \mathbb{R}^d$. If $\nabla\phi(\theta)$ is L-Lipschitz i.e.
$$\vert\nabla\phi(\theta\_1) - \nabla\phi(\theta\_2)\vert \leq L\vert\theta\_1-\theta\_2\vert,$$
how can we prove $Z = \langle v, T(X)\rangle$ for fixed $v$ satisfying $\Vert v\Vert\_2=1$ is sub-Gaussian?
I thought $Z$ should be like the linear combination of $X$, but I failed to reach what the Lipschitz condition can control here.
| https://mathoverflow.net/users/500967 | Is the main part of certain exponential family sub-Gaussian? | $\newcommand\th\theta\newcommand\la\lambda\newcommand\R{\mathbb R}$Note that for $t\in\R^d$ we have
$$M\_\th(t):=E\_\th e^{t\cdot T(X)}
=\int\_{\R^d}dx\,h(x)e^{(t+\th)\cdot T(x)-\phi(\th)}
=e^{\phi(t+\th)-\phi(\th)},$$
where $\cdot$ denotes the dot product. So,
$$E\_\th T(X)=\nabla M\_\th(0)=\nabla\phi(\th)$$
and, for any real $\la$,
$$E\_\th e^{\la(Z-E\_\th Z)}
=E\_\th\exp\{\la v\cdot T(X)-\la v\cdot E\_\th T(X))\} \\
=\exp\{\phi(\la v+\th)-\phi(\th)-\la v\cdot \nabla\phi(\th)\}.
\tag{1}\label{1}$$
By the mean-value theorem, for some $a\in(0,1)$ depending on $\la,v,\th$, we have
$$\phi(\la v+\th)-\phi(\th)=\la v\cdot\nabla\phi(a\la v+\th).$$
Also, by the Lipschitz condition on $\nabla\phi$,
$$|\nabla\phi(a\la v+\th)-\nabla\phi(\th)|\le La\la|v|\le L\la.$$
Thus,
$$\phi(\la v+\th)-\phi(\th)-\la v\cdot \nabla\phi(\th)
=\la v\cdot(\nabla\phi(a\la v+\th)-\nabla\phi(\th))
\le L\la^2$$
and hence, by \eqref{1},
$$E\_\th e^{\la(Z-E\_\th Z)}
\le e^{L\la^2},$$
so that $Z$ is (uniformly) [sub-Gaussian](https://en.wikipedia.org/wiki/Sub-Gaussian_distribution#More_equivalent_definitions) for all $\th$.
| 1 | https://mathoverflow.net/users/36721 | 444776 | 179,317 |
https://mathoverflow.net/questions/444781 | 6 | #### Background/Motivation
A *theory* T over a signature(language) Σ is a set of formulae over Σ. These formulae are called the *non-logical axioms* of T.
To talk about what is provable in T we can agree on Hilbert calculus and first-order logic.
A theory $ T' $ over $ Σ' $ is said to be an *extension* of a theory $ T $ over $ Σ $ if
* every sort/relation symbol/function symbol in $ Σ $ is also a sort/relation symbol/function symbol in $ Σ' $
* Every non-logical axiom of $ T $ is a non-logical axiom of $ T' $.
An extension $ T' $ of $ T $ is said to be *conservative* (over $ T $) if every formula over $ T $ that is provable in $ T' $ is also provable in $ T $
$ T' := T\_1 \cup T\_2 $ is a theory with signature $Σ\_1 \cup Σ\_2 $ and $Σ\_1 \cap Σ\_2 = Σ$ and is exactly the disjoint union of the nonlogical axioms of $T\_1$ and $T\_2$. (By this I mean that the instances of the axioms are united, and axiom schemas are not automatically applied to each other's signatures but rather kept separate and treated as an infinite set of axioms. Every non-logical axiom in $T'$ is a nonlogical axiom in $T\_1$ or $T\_2$)
---
#### Question
If $T\_1$ and $T\_2$ are extensions of a theory $T$, then so is $T\_1 \cup T\_2$ . If $T\_1$ and $T\_2$ are conservative, and therefore add no new theorems to $ T $ , is the same true for $T'$ ?
>
> Is $ T' := T\_1 \cup T\_2 $ a conservative extension of $ T $, if $T\_1$ and $T\_2$ are conservative extensions of $T$ ?
>
>
>
And if not, what would a simple counterexample be?
---
#### Attempts
So far I have
* considered the weaker question (due to conservative over T ⇒ consistent over T):
>
> Is $ T' $ a consistent extension of $ T $, if $T\_1$ and $T\_2$ are conservative extensions of $T$ ?
>
>
>
* tried to find or construct counterexamples specifically for this purpose but failed
* tried to prove the statement by induction over a hilbert-style proof tree. The main problem were the logical axioms at the leaves, which being axiom schemes, allow for a "mixing" of the signatures, making it difficult to "split" the prooftree into separate branches for $T\_1$ and $T\_2$. I still find this attempt promising but haven't made much progress yet. I also took into account the conservativity of a language extension.
* found no sources on this, especially since taking the union of theories as I have seems uncommon, and because it seems most material on (conservative) extensions takes a model theoretic rather than proof-theoretic approach.
| https://mathoverflow.net/users/502824 | Is the union of two conservative extensions of a theory conservative? | Yes, this is true (and somewhat nontrivial). That is, if $T$ is a theory in a language $\Sigma$, and $T\_1$ and $T\_2$ are conservative extensions of $T$ in languages $\Sigma\_1$ and $\Sigma\_2$ (respectively) such that $\Sigma\_1\cap\Sigma\_2=\Sigma$, then $T\_1\cup T\_2$ is a conservative extension of $T$. This is a form of [Robinson’s joint consistency theorem](https://en.wikipedia.org/wiki/Robinson%27s_joint_consistency_theorem).
To derive it from a more common formulation of the joint consistency theorem that requires $T$ to be complete, let $\phi$ be any $\Sigma$-sentence such that $T\nvdash\phi$; we will show $T\_1\cup T\_2\nvdash\phi$. Since $T\nvdash\phi$, there exists a complete consistent $\Sigma$-theory $T'\supseteq T\cup\{\neg\phi\}$. Since $T\_1$ and $T\_2$ are conservative over $T$, the theories $T\_1\cup T'$ and $T\_2\cup T'$ are consistent. But then $T\_1\cup T\_2\cup T'$ is also consistent by the joint consistency theorem, hence $T\_1\cup T\_2\nvdash\phi$.
The requirement $\Sigma\_1\cap\Sigma\_2=\Sigma$ is essential, otherwise there are very simple counterexamples: e.g., let $T$ be the empty theory in the empty language, $T\_1$ the theory $\{\exists x\,R(x)\}$ in language $\{R(x)\}$, and $T\_2=\{\neg\exists x\,R(x)\}$ in the same language.
| 11 | https://mathoverflow.net/users/12705 | 444784 | 179,321 |
https://mathoverflow.net/questions/444778 | 11 | Let $k$ be a field and let $R$ be a commutative (standard) graded $k$-algebra, that is, $R=\bigoplus\_{n=0}^\infty R\_n$ with $R\_0=k$ (and $R=k[R\_1]$). The Hilbert function $h\_R:\mathbb{N}\rightarrow \mathbb{N}$ is given by $h\_R(n)=\dim\_k R\_n$. When $n\gg 0$, it is known that $h\_R(n)$ agrees with a polnyomial $p\_R(n)$ in $n$ with rational coefficients.
In many theorems in combinatorics, polynomials which count objects have combinatorially sensible meaning when evaluated at negative numbers. For example, the chromatic polynomial $\chi\_G(x)$ of a graph, which counts the number of proper colorings of $G$ with $x$ colors, has the delightful property that $\chi\_G(-1)$ is the number of acyclic orientations of $G$.
One other example, intimately related to the Hilbert function, is that of Ehrhart polynomials. The Ehrhart polynomial $E\_P(n)$ of a convex integral polytope $P$ inside $\mathbb{R}^m$ is the number of lattice points of $\mathbb{Z}^m$ inside the $n$th dilate $nP$. It is also known that $E\_P(-n)$ is the number of interior lattice points (up to sign) of $nP$. For some (all?) polytopes, the Ehrhart polynomial agrees with the Hilbert polynomial of an associated affine semigroup ring.
In general, or in the specific context of affine semigroup rings $R$, is there some sensible (algebraic or combinatorial) meaning to $p\_R(-n)$, or even $p\_R(-1)$?
| https://mathoverflow.net/users/159030 | Hilbert polynomials of graded algebras evaluated at negative numbers | This answer is essentially the same as that of Phil Tosteson, written
before I saw that post. I also mention a non-Cohen-Macaulay example at
the end.
If $R$ is Cohen-Macaulay (but not necessarily generated in degree
one), then $R$ has associated with it a *canonical module*
$\Omega(R)$ which can be graded so its Hilbert function agrees with
$(-1)^d p\_R(-n)$ for $n$ sufficiently large, where $d$ is the Krull
dimension of $R$ or $\Omega(R)$. If $R$ has $n$ generators then it can
be regarded as a module over the polynomial ring
$A=k[x\_1,\dots,x\_n]$. One can then define $\Omega(R) =
\mathrm{Ext}^{n-d}\_A(R,A)$. If $R$ is not Cohen-Macaulay, then there are
"correction terms" to the formula $p\_R(-n) = (-1)^d\mathrm{HQ}(\Omega(R))$,
where $\mathrm{HQ}$ denotes Hilbert quasipolynomial. Namely,
$$ p\_R(-n) = (-1)^d \sum\_{i=0}^d
(-1)^i\mathrm{HQ}\left( \mathrm{Ext}^{n-d+i}\_A(R,A)\right). $$
(If $R$ is Cohen-Macaulay, then only the term indexed by $i=0$ doesn't
vanish.) I don't know where this result is stated in precisely this
form, but it is equivalent to Theorem 6.4 of my book *Combinatorics
and Commutative Algebra*, second ed. Theorem 8.2 gives an example,
stated in terms of Hilbert series rather than Hilbert quasipolynomials.
| 11 | https://mathoverflow.net/users/2807 | 444790 | 179,325 |
https://mathoverflow.net/questions/444658 | 10 | Consider the homotopy category $\mathrm{hoSp}$ of spectra. It has a full subcategory $\mathrm{hoSp}\_{\geq 0}$ of connective spectra, equivalently of infinite loop spaces, equivalently $E\_\infty$-group spaces.
The inclusion $\mathrm{hoSp}\_{\geq 0} \hookrightarrow \mathrm{hoSp}$ has a right adjoint, sending a spectrum $T$ to its connective cover $T\langle0\rangle$. In other words, for any connective spectrum $A$, we have
$$ [A, T] = [A, T\langle 0\rangle].$$
I'm wondering about intermediate categories $\mathrm{hoSp}\_{\geq 0} \hookrightarrow \mathcal{C} \hookrightarrow \mathrm{hoSp}$ for which the inclusion $\mathrm{hoSp}\_{\geq 0} \hookrightarrow \mathcal{C}$ also has a left adjoint. Explicitly: for which spectra $T$ is there a connective spectrum $T'$ such that
$$[T,A] = [T', A]$$
for all connective $A$? This $T'$ would be some sort of universal "quotient" of $T$ that quotients it down to being connective.
Dually, I can consider the full subcategory $\mathrm{hoSp}\_{\leq 0} \hookrightarrow \mathrm{hoSp}$ of coconnective spectra. This inclusion has a left adjoint given by the truncation $T \mapsto \tau\_{\leq 0}T$. On which intermediate category does this inclusion also have a right adjoint?
I am also interested in variations of these questions, e.g. working with the $(\infty,1)$-categories rather than their homotopy categories, or looking at the inclusion of spectra with homotopy in some bounded range $[0,n]$ (perhaps $n=0$, say) among spectra with homotopy bounded on only one side, or looking at spectra with some finiteness built in.
---
**Codicil:** The accepted answer below resolves the $\infty$-categorical variation of this question. Since the homotopy-categorical version is different, I have posted it as [new question](https://mathoverflow.net/questions/444845/).
| https://mathoverflow.net/users/78 | Which spectra have a universal connective quotient? | This answer is about the $\infty$-categorical variant. This is a fancy way to say: on *spaces* of maps, the natural map
$$
Map(T',A) \to Map(T,A)
$$
is an equivalence for any connective $A$. Note that for any bounded-below spectrum $B$, we can use $Map(-,B) \simeq \Omega^n Map(-,\Sigma^n B)$ to conclude that the natural map above is an equivalence for $B$ as well, and hence by taking homotopy classes of maps we get that
$$
[T',B] \to [T,B]
$$
is an equivalence for any bounded-below $B$.
---
I first claim: a map $T \to T'$ induces an equivalence $[T', B] \to [T,B]$ for all bounded-below $B$ if and only if it is an integral homology isomorphism.
To see that this is necessary:
Consider $B$ of the form $K(Q,n)$ for $Q$ an injective abelian group -- these are bounded-below. Then the universal coefficient theorem implies that the natural map
$$
Hom(H\_n(T'), Q) \to Hom(H\_n(T), Q)
$$
is isomorphic to the natural map
$$
H^n(T', Q) \cong [T', K(Q,n)] \to H^n(T,Q) \cong [T, K(Q,n)],
$$
and hence is an isomorphism. If this is true for all injective abelian groups $Q$, for a generic abelian group $M$ we can apply it to an injective resolution of $M$ and find that the natural map
$$
Hom(H\_n(T'), M) \to Hom(H\_n(T), M)
$$
is an isomorphism.
To see that this is sufficient:
Suppose we have a map $T \to T'$ which is an integral homology isomorphism, and consider the class of spectra $B$ such that $[T',B] \to [T,B]$ is an isomorphism.
* This class contains Eilenberg-Mac Lane spectra, by the universal coefficient theorem.
* This class is closed under fibers. If $B' \to B \to B''$ is a fiber sequence of spectra, then we get long exact sequences upon applying $[T,-]$ or $[T',-]$, and if the maps $[T',B] \to [T,B]$ and $[T',B''] \to [T',B'']$ are isomorphisms then so is $[T',B'] \to [T,B']$ by the five-lemma.
* This class is closed under spectra that have finitely many nonzero homotopy groups, by inductively applying this to the fiber sequences $K(\pi\_n B, n) \to P\_n B \to P\_{n-1} B$ in their Postnikov tower.
* This class is closed under products.
* This class is closed under homotopy limits of towers. If we have a tower $\dots \to B\_2 \to B\_1 \to B\_0$ of objects in this class, then the homotopy limit is part of a fiber sequence
$$
holim (B\_i) \to \prod B\_i \xrightarrow{1-\text{shift}} \prod B\_i
$$
and so the homotopy limit is in the class.
* This class contains all bounded-below spectra, because any bounded-below $B$ is the homotopy limit of its Postnikov tower $\dots \to P\_2 B \to P\_1 B \to P\_0 B$, and the individual $P\_n B$ all have only finitely many nonzero homotopy groups.
---
Now I claim: the collection of $T$ for which such a connective $T'$ exists consists of the spectra $T$ such that the integral homology $H\_\*(T)$ is concentrated in nonnegative degrees. This is necessary: because $T \to T'$ is an integral homology isomorphism, the isomorphism $H\_n(T) \to H\_n(T')$ forces $H\_n(T) \cong H\_n(T') \cong 0$ for $n < 0$.
Suppose that $H\_\* T = 0$ in negative degrees. An equivalent formulation is that the smash product $H\Bbb Z \wedge T$ is connective. If this is true, then we can form the $H\Bbb Z$-Adams tower. Briefly, construct a fiber sequence $J \to \Bbb S \to H\Bbb Z$ (hence $J$ is 0-connected) and form the tower
$$
\Bbb S/J \wedge T \leftarrow \Bbb S/J^2 \wedge T \leftarrow \Bbb S/J^3 \wedge T \leftarrow \dots
$$
The homotopy limit $T'$ is called the $H\Bbb Z$-nilpotent completion of $T$, and it has a natural map $T \to T'$.
Something special happens in this case, however, that I'll just briefly describe. The associated graded, in degree $p$, is $\Bbb S/J \wedge J^{p} \wedge T \simeq (H\Bbb Z \wedge T) \wedge J^p$. This is $p$-connected because $J^p$ is $p$-connected and $H\Bbb Z \wedge T$ is connective. As a result, by induction this nilpotent completion is connective. Moreover, smashing with $H\Bbb Z$ can be moved inside the homotopy limit because of this connectivity growth, and the map $H\Bbb Z \wedge T \to H\Bbb Z \wedge T'$ turns out to be a weak equivalence (true for Adams towers whenever smashing can be moved inside the limit); so $T \to T'$ is a homology isomorphism.
---
As a note, for this to happen we need for the fiber of $T \to T'$ to have trivial homology; this is impossible unless it is contractible or it is unbounded below. This is certainly interesting but carries it outside the usual library of basic examples.
| 9 | https://mathoverflow.net/users/360 | 444797 | 179,328 |
https://mathoverflow.net/questions/444754 | 2 | Let $m$ be a positive integer satisfying $\dfrac{m(m+1)}{4}\in \mathbb{Z}$. Show that there exists a positive integer $t$ and $t$ positive integers $m\_1,m\_2,\cdots,m\_t$ such that $$\begin{cases} \sum\limits\_{k=1}^{t}m\_k=m\\ \sum\limits\_{k=1}^{t}\dfrac{m\_k(m\_k+1)}{2}=\dfrac{m(m+1)}{4} \end{cases}.$$
P.S. This question comes from Problem F in The 2022 ICPC Asia Shenyang Regional Contest (<https://codeforces.com/gym/104160/problem/F>). One of the key steps in this problem is to find a set of solutions to the above indeterminate equations. And I noticed that $\dfrac{m(m+1)}{4}$ cannot be arbitrarily replaced by other positive integers, which may make the proposition no longer true.
| https://mathoverflow.net/users/502801 | The existence of solutions of a system of indeterminate equations | In the comments OP proposed a greedy algorithm to represent a given positive integer $A$ as the sum of triangular numbers whose indices sum to $m$, and applied it to $A = \frac{m(m+1)}4$. I will prove that it indeed always produces a solution, and thus the given system is soluble for any positive integer $m$ with $\frac{m(m+1)}4\in\mathbb Z$.
---
Essentially, the algorithm goes from a pair $(m,A)$ with $m\leq A$ to the pair $(m-t,A-\tfrac{t(t+1)}2)$ for the largest integer $t$ such that $m-t \leq A-\tfrac{t(t+1)}2$, or equivalently
$$(\star)\qquad \frac{t(t-1)}2\leq A-m < \frac{t(t+1)}2.$$ The value for $t$ can be given explicitly as
$t = \left\lfloor\tfrac{1 + \sqrt{8(A-m)+1}}2\right\rfloor.$
Notice that $t\geq 1$, and furthermore $t\geq 2$ as soon as $m<A$.
The algorithm converges and produces a solution when/if it reaches the pair $(0,0)$. Let's start with the following claim.
**Claim.** The greedy algorithm converges for any given pair of integers $(m,A)$ satisfying $0\leq m\leq A \leq \frac{(m-1)^2}4$.
**Proof.** Proof is done by induction on $m$. In the base cases $m< 43$ the claim is verified computationally. Let's prove the claim for $m\geq 43$, assuming that it's proved for all smaller values of $m$.
The algorithm from the given pair $(m,A)$ goes to the pair $(m',A')$ where $m':=m-t$ and $A':=A-\frac{t(t+1)}2$ with $t\geq 1$ defined by the formula above. Clearly, $m'<m$. Hence, for the induction assumption to work it remains to verify that
$0\leq m'\leq A'\leq \frac{(m'-1)^2}4$.
It can be seen that inequality $0\leq m'$ (ie. $t\leq m$) follows from the inequality $A \leq \frac{(m-1)^2}4$ coupled with $(\star)$.
The inequality $m'\leq A'$ follows from the definition of $t$.
Finally, the inequality $A' \leq \frac{(m'-1)^2}4$, given that $(\star)$ implies $A'<m$, would follow from $m \leq \frac{(m-t-1)^2}4$, for which we need $t \leq m-1-2\sqrt{m}$. Since $A\leq \frac{(m-1)^2}4$, from the explicit formula for $t$, we have
$$t\leq \frac{1 + \sqrt{2(m^2-6m+1)+1}}2<\frac{1+\sqrt{2}(m-3)}2\leq m-1-2\sqrt{m},$$
where the last inequality holds since $m\geq 43$. QED
---
While Claim is not directly applicable to $A = \frac{m(m+1)}4$, we show that for $m\geq 50$ after one iteration the pair $(m',A'):=(m-t,A-\frac{t(t+1)}2)$ does satisfy the condition of Claim. Similarly to the above, it's enough to show that $t \leq m-1-2\sqrt{m}$. Here the explicit formula for $t$ implies
$$t\leq \frac{1 + \sqrt{2(m^2-3m)+1}}2<\frac{1+\sqrt{2}(m-1.5)}2\leq m-1-2\sqrt{m},$$
where the last inequality holds for $m\geq 50$. Hence, our Claim implies that the algorithm converges on the input $(m,\frac{m(m+1)}4)$ for all $m\geq 50$. For $m<50$, the algorithm convergence on the input $(m,\frac{m(m+1)}4)$ is verified computationally.
| 2 | https://mathoverflow.net/users/7076 | 444800 | 179,329 |
https://mathoverflow.net/questions/444796 | 1 | (all morphism here means birational)
(the ground field is "small", but I don't think it should matter)
Here is the picture. I have a morphism $f:\mathbb{P}^{1}\rightarrow\mathbb{P}^{1}$. I want to lift this up to a morphism $\tilde{f}:C\rightarrow E$ where $C$ is a curve and $E$ is an elliptic curve with projection $\pi\_{C}:C\rightarrow\mathbb{P}^{1}$ and $\pi\_{E}:E\rightarrow\mathbb{P}^{1}$, so that we have a commutative diagram $f\circ\pi\_{C}=\pi\_{E}\circ\tilde{f}$. Now, most of these are not fixed. A few things are fixed: the degree of $\pi\_{C}$ is fixed at some numbers $d$ independent of $f$, the degree of $\pi\_{E}$ is fixed at 2, but other than that $C,E,\pi\_{C},\pi{E},\tilde{f}$ we are allowed to freely choose dependent on $f$.
So my actual question is: is there any $d$ in which this lifting can be done "often", in the sense that it is non-special, ie. failure to lift is not a generic property? Or is this just always false?
I previously tried this for $d=2$, $C$ being (another) elliptic curve, and $\pi\_{C}$ be the x-coordinate, but this gives a strong restriction on $f$, that is ramification points of $f$ must lift to $2n$-torsion points on $C$, and thus knowing $2$ points gives us at most $4n^{2}$ possible elliptic curves.
| https://mathoverflow.net/users/502838 | Are morphism from $\mathbb{P}^{1}$ to itself often liftable to a morphism from a curve to an elliptic curve with bounded degree? | $\newcommand{\CC}{\mathbb C}$If there is a commutative diagram as requested, then there is also one with $C$ and $\pi\_C$ replaced by $C'$ and $\pi\_{C'}$, where $\pi\_{C'}$ has degree $2$ by letting $C'$ be a connected component of the fiber product of $f$ and $\pi\_E$.
Or in terms of function fields, with $x$ a transcendental over $\CC$ and $t=f(x)$: The fields $\CC(x)$ and $\CC(E)$ are intermediate fields of $\CC(t)$ and $\CC(C)$. Now define $C'$ by the new intermediate field $\CC(C')=\CC(x, E)$. Note that $[\CC(C'):\CC(x)]=2$, so $\pi\_{C'}$ has degree $2$.
So one is reduced to the case $d=2$. And there is always a cover as requested: Start with any $E$ and $\pi\_E$, let $L$ be an algebraic extension of $\CC(t)$ which contains $\CC(x)$ and $\CC(E)$, and define $C$ by $\CC(C)=\CC(x,E)\subseteq L$.
But for generic functions $f$, like Morse functions, the genus of $C$ goes to infinity which $\operatorname{deg} f$. So there is probably little use of the lift $\tilde f$.
| 3 | https://mathoverflow.net/users/18739 | 444801 | 179,330 |
https://mathoverflow.net/questions/444804 | 2 | Let $G$ be a simple graph on $n$ vertices.
Prove that, for every $k\in\{2,\dots,n\}$, there exist $k$ distinct vertices in $G$ whose degrees differ by at most $k-2$.
| https://mathoverflow.net/users/501463 | There exist $k$ vertices with degree differences at most $k - 2$ | This result was originally proved by [Erdős-Chen-Rousseau-Schelp (1993)](https://www.sciencedirect.com/science/article/pii/S0195669883710231) using the [Erdős-Gallai theorem](https://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Gallai_theorem). A more direct proof (which seems to be identical to the one in 1001's response) was given by [Caro-Lauri-Zarb (2018)](https://arxiv.org/abs/1806.08303). The latter paper appeared in Bull. Inst. Combin. Appl. 85 (2019), 79–91.
| 6 | https://mathoverflow.net/users/11919 | 444810 | 179,334 |
https://mathoverflow.net/questions/444811 | 0 | Recently I have made some interesting observations on the limit $$\lim\_{k\rightarrow \infty}{\sum\_{n=1}^{k}{\dfrac{(-1)^{n-1}\biggl( \cos \left(\beta\ln(n)\right)\biggr)}{n^{\alpha}}}}. $$ When this limit exist denoteits convergence point by $\zeta$. If we define $f\_k$ to be the partial sums of this series:
$$f\_k(\alpha,\beta):= \sum\_{n=1}^{k}{\dfrac{(-1)^{n-1}\biggl( \cos \left(\beta\ln(n)\right)\biggr)}{n^{\alpha}}} $$ I want to prove that $$f\_{2k}(\alpha,\beta)\le \zeta \le f\_{2k+1}(\alpha,\beta)$$ for all values k. One approach is to prove them separately. First show $\zeta-f\_{2k}(\alpha,\beta)\ge 0$ and then $f\_{2k+1}(\alpha,\beta) - \zeta \le 0$.
From this we can easily obtain small bounds on the actual value of the zeta function and hopefully show that the there is only one value of $\alpha$ such that both the limits $$\lim\_{k\rightarrow \infty}{\sum\_{n=1}^{k}{\dfrac{(-1)^{n-1}\biggl( \cos \left(\beta\ln(n)\right)\biggr)}{n^{\alpha}}}}$$ and $$\lim\_{k\rightarrow \infty}{\sum\_{n=1}^{k}{\dfrac{(-1)^{n-1}\biggl( \sin\left(\beta\ln(n)\right)\biggr)}{n^{\alpha}}}}$$ are both simultaneously zero. We should be able to bound zeta by the first few terms.
| https://mathoverflow.net/users/353746 | proving inequality in Riemann zeta function | The inequality you want to prove is false. For example,
$$f\_5(1/2,1)=0.6096\dots,$$
while
$$\lim\_{k\to\infty}f\_k(1/2,1)=0.6398\dots$$
| 2 | https://mathoverflow.net/users/11919 | 444812 | 179,335 |
https://mathoverflow.net/questions/442718 | 5 | Let $\mathcal{U}\subset \mathbb{R}\times \mathbb{C}$ a neighborhood of $(0,0)$, and $f:\mathcal{U}\to \mathbb{C}$ differentiable in the first variable and holomorphic in the second variable, with $f(0,0)=0$. I want to locally express the zeros of $f$ as one or more curves $z=z(x)$ with $z(0)=0$. The hypothesis I want to eliminate is that $\partial\_z f(0,0)\neq 0 $. Instead, suppose that for some $k,h\in \mathbb{N}$ with $k\geq 2$, as $(x,z) \to (0,0)$ we have
$$ f(x,z)= z^k+ x^h+ h.o.t.,$$
where $h.o.t.$ stands for higher order terms in $x$ *or* $z$, i.e. they are $o(z^k)$ *or* $o(x^h)$ as $(x,z)\to (0,0)$.
I wish to prove that there is at least one *continuous* (not necessarily differentiable) curve $z(x)$ on $\mathcal{U}$ such that $f(x,z(x))=0$ in a neighborhood of $x=0$. Stronger statement: there are $k$ curves $z\_j(x)$ such that in a neighborhood of $(0,0)$ we have $f(x,z)=0\iff z=z\_j(x)$ for some $j=0,\dots, k-1$.
Could someone point at a specific result from the literature which implies the above (or explain why it doesn't hold)?
**Example**: At least if $f$ is a polynomial in $z$ then the result must hold. If $n\geq k$ is the degree then the polynomial has $n$ roots but at $(0,0)$ the root only has multiplicity $k$. Because the roots of polynomials depend continuously on the coefficients, which in turn are continuous with respect to $x$, this gives rise to exactly $k$ continuous roots $z\_1(x),\dots, z\_k(x)$ with $z\_j(0)=0$.
**Attempts at proof**.
>
> The issue is to generalize the result to functions which aren't just
> polynomials in $z$. My efforts at trying to transform $f$ so
> that the standard IFT can be applied haven't been successful:
>
>
> * Define $g(x,z):=f(x,z)^{1/k}\_j$, then $g$ is no longer differentiable in $(0,0)$ and hence does not satisfy the
> assumptions of the IFT. Also, the limit of $f$ to $(0,0)$
> does not exist.
> * One can apply the IFT to $\partial\_z^{k-1}f$ but this is useless. We could even assume by induction that the statement holds for $k'<k$,
> but the curves where $\partial\_z^{k'}f$ vanishes are different for
> each $k'=1,\dots, k-1$.
> * Dividing by $(z)^{k-1}$ doesn't work as the limit of $f$ in $(0,0)$ no longer exists.
>
>
> The problem is also equivalent to finding a (continuous, differentiable for $x\neq 0$) solution $z(x)$
> to the I.V.P. $$z'(x)\partial\_z f(x,z(x))=-\partial\_x f
> (x,z(x)),\qquad z(0)=0.$$ In standard IFT, we can divide by
> $\partial\_z f$ and apply the standard local existence and uniqueness
> result. Here we cannot because $\partial\_z f(0,0)=0$. I do not
> know any existence theorem that would apply in this case. We could
> perturbate the initial condition by a $w\in \mathbb{C}$,
> obtaining a curve $z\_w(x)$ such that for $w$ small enough, $$ |f(x,z\_w(x))|=|f(0,w)|\leq 2|C\_1| |w|^k.
> $$ Here the map $w\mapsto z\_w(x)$ is continuous on
> $\mathbb{C}\setminus \{{0\}}$, but this is not enough to conclude that
> the limit as $w\to 0$ exists.
>
>
>
[This question](https://mathoverflow.net/questions/85490/implicit-function-theorem-at-a-singular-point) provides a positive answer to the case $k=2$ but the proof does not extend to higher $k$. I suspect the result might be a special case of [this paper](https://arxiv.org/pdf/1311.0088.pdf) but there is too much algebra for me to understand even the statements. Maybe someone could confirm whether they can be applied or not?
| https://mathoverflow.net/users/125758 | Implicit function theorem with singularities of any order | [Rouché's theorem](https://en.wikipedia.org/wiki/Rouch%C3%A9%27s_theorem) provides a simple solution.
Let $f\_1(x,z)=z^k+x^h$ and $f\_2(x,z)=h.o.t.$. For each $x$, the functions $f\_1$ and $f\_2$ are holomorphic in $z$. We know $f\_2(x,z)=o(z^k)+o(x^h)$, so there exist $z\_0,x\_0>0$ such that for $|z|\leq z\_0$ and $|x|\leq x\_0$ we have
$$ |f\_2(x,z)|\leq \frac{1}{4}(|z|^k+|x|^h).$$
(Note that this is weaker than the type of inequality discussed in the comments).
Fix now any $|x|\leq x\_0$ small enough so that
$$r:= (2|x|^h)^{1/k}\leq z\_0. $$
For $|z|=r$ we then have
$$|f\_2(x,z)|\leq \frac{1}{4}\left( |z|^k+|x|^h\right)= \frac{3}{8}r^k< \frac{1}{2}r^k= |z|^k-|x|^h\leq |f\_1(x,z)|. $$
Therefore by Rouché's theorem, we deduce that $f\_1(x,\cdot)$ and $f(x,\cdot)=f\_1(x,\cdot)+f\_2(x,\cdot)$ have the same number of zeroes in $B\_r(0)$. Since $f\_1(x,\cdot)$ has $k$ zeroes, thus $f(x,\cdot)$ has $k$ zeroes $z\_1(x),\dots,z\_k(x)$ in $B\_r(0)$. In particular, we have
$$\lim\_{x\to 0}|z\_j(x)|\leq \lim\_{x\to 0}(2|x|^h)^{1/k}=0,\qquad j=1,\dots, k. $$
This proves that the $z\_j$ are at least continuous at the origin. In order to extend this to a neighborhood it should suffice to apply Rouche's theorem to balls centered on the individual roots of $f\_1$, but I am content with this. Besides, remarkably, the above proof does not require any regularity in $x$.
| 1 | https://mathoverflow.net/users/125758 | 444814 | 179,336 |
https://mathoverflow.net/questions/444822 | 8 | Are there any uses of generating functions within logic, in particular to count how many models exists for a given theory $T$, say in FOL?
The concrete problem I'm hoping to apply this to involves counting the number of states witnessed by a given program. In this application some ad-hoc combinatorial arguments can be made for any instance, but it feels like there should exist a general construction scaffolded by the logic underlying our PL.
My far-reaching hope is that such functions not only exist, but are compositional in a reasonable sense. One way this could manifest: perhaps we build one such generating function for each axiom in our theory, and the generating function for the entire theory is just the product over each axiom.
Or perhaps they compose in a different manner, but the core idea remains that each connective in our logic induces an operation on these generating functions.
Regardless of my hope bearing fruit, I'd love to hear about any resources involving generating functions in logic
| https://mathoverflow.net/users/165086 | Use of generating functions in logic | Not sure if the lambda calculus or 2-SAT counts for you as "logics" but here is a couple of recent papers presenting bijections between these formal structures of mathematical logic and "more classical" combinatorial objects. The bijections transfer certain properties of the former objects to the properties and patterns in graphs and maps, use the generating functions to enumerate them and study their asymptotic distribution.
[Asymptotic Distribution of Parameters in Trivalent Maps and Linear Lambda Terms](https://arxiv.org/abs/2106.08291)
by Olivier Bodini, Alexandros Singh, Noam Zeilberger
[Exact enumeration of satisfiable 2-SAT formulae](https://arxiv.org/abs/2108.08067)
by Sergey Dovgal, Élie de Panafieu, Vlady Ravelomanana.
| 5 | https://mathoverflow.net/users/31830 | 444830 | 179,340 |
https://mathoverflow.net/questions/444829 | 5 | Given a diagram of sets $D\colon\mathcal{C}\to\mathsf{Set}$, we have a bijection ([Proof](https://math.stackexchange.com/a/3838971))
$$\operatorname{colim}(D) \cong \pi\_0 (\textstyle\int\_\mathcal{C}D).$$
1. Is there any known application or significance of the homotopy groups of $\int\_\mathcal{C}D$ (meaning the homotopy groups of the geometric realisation of $\int\_\mathcal{C}D$ at a point $x\in\pi\_0(\textstyle\int\_\mathcal{C}D)$)?
2. What about its "fundamental monoid" $\pi\_0(\mathrm{Hom}\_{\mathrm{N}\_{\bullet}(\int\_\mathcal{C}D)}(x,x))$ at a point $x\in\pi\_0(\textstyle\int\_\mathcal{C}D)$? (Is the fundamental group of the geometric realisation of $\textstyle\int\_\mathcal{C}D$ at $x$ just the groupification of this?)
3. Can these be described in terms of other well-known notions in ($\infty$-)category theory?
| https://mathoverflow.net/users/130058 | Homotopy groups of categories of elements as higher colimits | To answer these questions, the best is to note that $|\int\_C D|$, the geometric realization of this total category, is equivalently the colimit of $D$, *viewed as a functor with values in the $\infty$-category of spaces*, namely, along the inclusion $Set\to Spaces$.
So
1. Insofar, as you believe that this colimit has significance, so do its homotopy groups. For example, if $C= \Delta^{op}$, $D$ is a simplicial set and $|\int\_{\Delta^{op}} D|$ is the homotopy type of this simplicial set; if $C= BG$ for some group $G$, $D$ is a set with a $G$-action, and this homotopy type is $D\_{hG}$, the homotopy orbits of $D$ under this action, etc.
2. The "fundamental monoid" is maybe not so relevant, in a sense because it is easily computed. First, note that it depends on more than a class in $\pi\_0$ ($\pi\_0$ inverts all morphisms), but really on an object in $\int\_C D$. Second, given such an object $x$, it lives in a fiber over some $c\in C$, and consequently, $\hom\_{\int\_CD}(x,x) = \coprod\_{f\in \hom\_C(c,c)} \hom\_{D(c)}(f\_!x,x)$ where $f\_! : D(c)\to D(c)$ is $D(f)$. The monoid structure is "fiberwise". But in a sense it can be easily read off of $D$, whereas understanding $|\int\_C D|$ is typically more complicated.
In particular, note that the group-completion of this monoid is *not* (in general) $\pi\_1(|\int\_CD|, x)$, because the latter also involves "visiting" other objects than $x$. For example if $D$ is the constant functor with value a point, then $\int\_C D=C$, and you'd be claiming that for an arbitrary $C$, $\pi\_1(|C|,c) = $ the goup-completion of $\hom\_C(c,c)$. But any group can be realized as $\pi\_1(|P|)$ for a *poset* $P$ where, in particular, $\hom\_P(p,p) = $ a point for any $p\in P$.
3. I think the very first part of my question answers this one.
| 13 | https://mathoverflow.net/users/102343 | 444836 | 179,341 |
https://mathoverflow.net/questions/444837 | 0 | If $G = (V,E)$ is a finite, connected, simple, undirected graph, is there a [Hamiltonian path](https://en.wikipedia.org/wiki/Hamiltonian_path) in the [line graph](https://en.wikipedia.org/wiki/Line_graph) $L(G)$ of $G$?
| https://mathoverflow.net/users/8628 | Hamiltonian path in the line graph of a connected graph | No. Consider the graph $G = (V, E)$ with
$$V = \{0,1,2,3,4,5,6\}\quad \text{and} \quad E = \{\{0,1\}, \{1, 2\}, \{0,3\}, \{3, 4\}, \{0,5\}, \{5, 6\}\}.$$
Note that its line-graph has three vertices of degree 1. Therefore $L(G)$ has no Hamiltonian path.
| 5 | https://mathoverflow.net/users/502833 | 444838 | 179,342 |
https://mathoverflow.net/questions/376467 | 90 | Take a plane curve $\gamma$ and a disk of fixed radius whose center moves along $\gamma$. Suppose that $\gamma$ always cuts the disk in two simply connected regions of equal area. Is it true that $\gamma$ must be a straight line?
| https://mathoverflow.net/users/167834 | Does this property characterize straight lines in the plane? | It seems $\gamma:\mathbb{R}\to\mathbb{R}^2$ (I assume $\gamma$ is injective and continuous) is indeed a line. My argument is very similar to the [one](https://mathoverflow.net/a/377074/172802) by Ilkka Törmä (I thought I could write a shorter one, but it grew longer than expected). As in their answer, I will take the radius from the question to be $1$ and $D(p)$ will denote the disk of radius $1$ centered at $p$.
First suppose that $\gamma$ is bounded. Consider the point $p\_M=(x\_M,y\_M)\in\overline{\gamma(\mathbb{R})}$, with $x\_M$ being the maximal $x$ coordinate of all points of $\overline{\gamma(\mathbb{R})}$, and $y\_M$ the maximal $y\in\mathbb{R}$ such that $(x\_M,y)\in\overline{\gamma(\mathbb{R})}$. Then $D(p\_M)\setminus\overline{\gamma(\mathbb{R})}$ has a connected component of area $>\frac{\pi}{2}$, and the same is true of $D(p)\setminus\overline{\gamma(\mathbb{R})}$ for points $p$ in $\gamma$ close to $p\_M$, so $\gamma$ cannot be as in the question.
So we can assume that $\gamma$ is not bounded. This implies that for any $p\in\mathbb{R}^2$, $\gamma^{-1}(D(p))$ is a union of bounded intervals, and if $p\in\gamma(\mathbb{R})$, as explained in the other [answer](https://mathoverflow.net/a/377074/172802), $\gamma^{-1}(D(p))$ is just one bounded open interval, because if not $\gamma$ would divide $D(p)$ into more than $2$ connected components. Using this, it is not hard to prove by contradiction that $\gamma^{-1}(\overline{D(p)})$ is bounded for all $p\in\gamma(\mathbb{R})$.
**Lemma 1:** For any real numbers $a<b$, $\gamma([a,b])$ is contained in the convex closure of $X:=\gamma([a,b])\cap\left(\overline{D(\gamma(a))}\cup\overline{D(\gamma(b))}\right)$.
*Proof:* Note first that, for any point $p\in\gamma([a,b])\setminus X$, the interval $\gamma^{-1}(D(p))$ is strictly contained in $[a,b]$, so $D(p)$ only intersects $\gamma(\mathbb{R})$ in $\gamma([a,b])$. Also note that, as $X$ is compact, the convex closure of $X$ is the intersection of all closed half planes containing it.
So if the lemma was not true, there would be some closed half plane $H$ containing $X$ but not containing $\gamma([a,b])$. Composing with an isometry if necessary, we can suppose $H=\{(x,y)\in\mathbb{R}^2;x\leq0\}$. Now consider the point $p\_M=(x\_M,y\_M)\in\gamma([a,b])$ given by $x\_M=\max\{x\in\mathbb{R};(x,y)\in\gamma([a,b])\}$ and $y\_M=\max\{y\in\mathbb{R};(x\_M,y)\in\gamma(a,b)\}$. Then $D(p\_M)\setminus\gamma([a,b])$ has a connected component of area $>\frac{\pi}{2}$, but $p\_M\not\in H$, so $p\_M\not\in X$, so $D(p\_M)$ only intersects $\gamma(\mathbb{R})$ in some interval strictly contained in $\gamma([a,b])$. So $D(p\_M)\setminus\gamma(\mathbb{R})$ has a connected component of area $>\frac{\pi}{2}$, a contradiction. $\square$
**Lemma 2:** $\gamma(\mathbb{R})$ is contained between two parallel lines $l,l'$.
*Proof:* For each $v\in\mathbb{S}^1$ let $R\_v$ be the region between the two lines that have direction $v$ and are at distance $2$ of $\gamma(0)$; more precisely, $p\in R\_v$ iff $d(\gamma(0),p)^2-\left(\langle v,p-\gamma(0)\rangle\right)^2\leq4$. Consider the directions $v\_n=\frac{\gamma(n)-\gamma(-n)}{|\gamma(n)-\gamma(-n)|}$. Note that for all naturals $n$, by the previous lemma $\gamma(0)$ is in the convex hull of $\overline{D(\gamma(-n))}\cup\overline{D(\gamma(n))}$, so $\overline{D(\gamma(-n))}\cup\overline{D(\gamma(n))}$ is contained in $R\_{v\_n}$, so by the previous lemma again, $\gamma([-n,n])$ is contained in $R\_{v\_n}$. Then taking a limit point $v$ of the $v\_n$, $\gamma(\mathbb{R})$ has to be contained in $R\_v$: this is because for any $x\in\mathbb{R}$, $\gamma(x)$ is in $\bigcap\_{n\geq|x|}R\_{v\_n}$, that is, $d(\gamma(0),\gamma(x))^2-\left(\langle v\_n,\gamma(x)-\gamma(0)\rangle\right)^2\leq4$ for all $n\geq|x|$, so $d(\gamma(0),\gamma(x))^2-\left(\langle v,\gamma(x)-\gamma(0)\rangle\right)^2\leq4$, that is, $\gamma(x)$ is in $R\_v$. $\square$
So, maybe after a rotation, we can suppose that $\gamma(\mathbb{R})$ is contained between two horizontal lines. Now, let $[a,b]$ be the minimal closed interval such that $\gamma$ is contained in the horizontal strip $\{a\leq y\leq b\}$
**Lemma 3:** There is some $p\in \gamma(\mathbb{R})$ with $y$ coordinate $a$.
If this lemma is true and $\gamma$ satisfies the question, then $\gamma(\mathbb{R})$ has to contain all the line $\{y=a\}$: if not, there would be some $p=(x,a)\in\gamma(\mathbb{R})$ such that the interval $[x-1,x+1]\times\{a\}$ is not contained in $\gamma(\mathbb{R})$, so $D(p)\setminus\gamma(\mathbb{R})$ would have a connected component of area $>\frac{\pi}{2}$, a contradiction.
*Proof:* We will prove it by contradiction, suppose the lemma is not true. By minimality of the interval $[a,b]$, there is a sequence of points $\gamma(t\_n)=(x\_n,y\_n)$ of $\gamma(\mathbb{R})$ such that $y\_n\to a$. The falsehood of lemma 3 implies that $|t\_n|\to\infty$, and then, the facts that $|t\_n|\to\infty$ and that $\gamma^{-1}(D(\gamma(x)))$ is bounded for all $x$ implies that $|x\_n|\to\infty$ (if not, consider an accumulation point of $x\_n$). Also note that $\gamma^{-1}\left(\overline{D(\gamma(0))}\right)$ is bounded, so $\gamma(\mathbb{R})\cap\overline{D(\gamma(0))}$ is contained in some rectangle $R=[c,d]\times[a+\varepsilon,b]$ for some $\varepsilon>0$. Now we will prove that $\gamma([0,\infty))\subseteq\{y\geq a+\delta\}$ for some $\delta>0$ by dividing in two cases; the same proof works for $\gamma((-\infty,0])$, contradicting the minimality of $[a,b]$.
* Case $1$: There are no sequences $x\_n,y\_n,t\_n$ as in the paragraph above such that $t\_n\to\infty$ (that is, we don't always have $t\_n\to-\infty$). Then it is obvious that $\exists\delta>0$ such that $\gamma([0,\infty))\subseteq\{y\geq a+\delta\}$.
* Case $2$: There are sequences $t\_n,x\_n,y\_n$ as above with $t\_n\to\infty$. Then by lemma $1$, $\gamma([0,t\_n])$ is contained for all $n$ in the convex closure of $\gamma([0,t\_n])\cap\left(\overline{D(\gamma(0))}\cup\overline{D(\gamma(t\_n))}\right)\subseteq R\cup\overline{D(\gamma(t\_n))}$. Note that $R$ is a fixed rectangle with $y$-coordinates in $[a+\varepsilon,b]$, and $\overline{D(\gamma(t\_n))}$ is a sequence of disks of radius $1$
centered at points $(x\_n,y\_n)$, where $y\_n\in[a,b]$ $\forall n$ and $|x\_n|\to\infty$. So any fixed point $p=(x,y)$ with $y<a+\varepsilon$ cannot be in the convex hull of $R\cup\overline{D(\gamma(t\_n))}$ for big enough $n$: if that was the case (see $(\*)$ below), then there would be a segment $s$ from some point $r\in R$ to some point $q\in \overline{D(\gamma(t\_n))}$ passing through $p$, so the slope of the segments $qp$ and $pr$ would be the same. However, the slopes of segments from $p$ to points of $R$ are bounded below, while the slopes from $p$ to points of $\overline{D(\gamma(t\_n))}$ converge to $0$ when $n\to\infty$, so for big enough $n$, the segment $s$ cannot exist.
So any point $p=(x,y)\in\gamma([0,\infty))$ has to satisfy $y\geq a+\varepsilon$, as we wanted. $\square$
$(\*)$ I have used that if $X,Y$ are convex subsets of $\mathbb{R}^2$, then any point $p$ in the convex closure of $X\cup Y$ is in a segment between a point of $X$ and a point of $Y$: indeed, as $p$ is in the convex hull of $X,Y$, there are points $x\_i\in X,y\_j\in Y$ and coefficients $a\_i,b\_j\geq0$ with $\sum a\_i+\sum b\_j=1$ such that $p=\sum\_i a\_ix\_i+\sum\_j b\_jy\_j$. Suppose $\sum a\_i,\sum b\_j>0$, if not $p$ is in $X$ or in $Y$. So $p$ is in fact an affine combination of the point $\sum\_i \frac{a\_i}{\sum\_ka\_k}x\_i$ of $X$ and the point $\sum\_j\frac{b\_j}{\sum\_kb\_k}y\_j$ of $Y$: $p=\left(\sum\_ia\_i\right)\left(\sum\_i \frac{a\_i}{\sum\_ka\_k}x\_i\right)+\left(\sum\_jb\_j\right)\left(\sum\_j\frac{b\_j}{\sum\_kb\_k}y\_j\right)$.
| 2 | https://mathoverflow.net/users/172802 | 444859 | 179,351 |
https://mathoverflow.net/questions/444865 | 1 | The general $4$-deg and some $8$-deg (such as the [Schein octic](https://mathoverflow.net/q/145145/12905)) when a linear transformation is done so their $x^{n-1}$ term vanishes can have a neat solution as,
$$x = \sqrt{z\_1}+\sqrt{z\_2}+\sqrt{z\_3}$$
$$x = \sqrt{z\_1}+\sqrt{z\_2}+\dots+\sqrt{z\_7}$$
where the $z\_i$ are the roots of their cubic and septic resolvents, respectively. Of course, the sign of the square root is chosen appropriately. At first, I assumed this is only doable when the resolvent $2^m-1$ is a Mersenne prime.
However, inspecting $12$-deg, it seems it is also doable if its Galois group is the **smallest sporadic group**, the Mathieu group $M\_{11}$, of order 7920. (Or, in Magma notation, $12T272$.) For example, the $12$-deg,
$$x^{12} - 21x^{10} - 20x^9 + 210x^8 + 12x^7 - 670x^6 + 108x^5 + 1305x^4 - 940x^3 - 189x^2 - 120x + 4 = 0$$
its $11$-deg resolvent (a $11T6$),
$$z^{11} - 126 z^{10} + 4815 z^9 - 36180 z^8 - 680625 z^7 - 1853982 z^6 + 3094497 z^5 + 280910160 z^4 + 1168901280 z^3 - 16329427200 z^2 + 30682457856 z -119664^2 = 0$$
and the two are related by,
$$x\_k = \frac{\sqrt{z\_1}+\sqrt{z\_2}+\dots+\sqrt{z\_{11}}}6$$
analogous to the ones for $4$-deg and $8$-deg.
**Question:** So, given a $12$-deg whose Galois group is the Mathieu group $M\_{11}(12)$. If needed, do a linear transformation such that the $x^{11}$ term vanishes, hence the sum of the roots equals zero. Then is it true there is an ordering of its roots $x\_k$ such that,
$$z=(x\_1+x\_2+x\_3+x\_4+x\_5+x\_6)^2 = (x\_7+x\_8+x\_9+x\_{10}+x\_{11}+x\_{12})^2$$
and $z$ is a root of an $11$-deg equation?
| https://mathoverflow.net/users/12905 | On reducing degree-$12$ equations with Mathieu group $M_{11}$ to its degree-$11$ resolvent? | The answer is yes: $M\_{11}$ in its action of degree $12$ has a subgroup of index $11$ (some people call it $M\_{10}$) such that there is a $6$-element subset $A$ of the $12$ points which $M\_{11}$ acts on such that $A^g=A$ or $A\cap A^g=\emptyset$ for all $g\in M\_{10}$. From that the claim follows immediately.
| 5 | https://mathoverflow.net/users/18739 | 444866 | 179,352 |
https://mathoverflow.net/questions/444401 | 17 | Does a irrational number $x > 1$ exist such that $\{x^n \} \le \frac{1}{2}$ for
all positive integers $n$ ?
$x=1+ \sqrt 2$ holds for $n$ odd, but not in even
| https://mathoverflow.net/users/174530 | Fractional part power | The OP asks for an instance of what Dubickas [1] has called a ${\cal Z}$-number: A real number $x>1$ for which there exists a real $\xi\neq 0$ such that $\{\xi x^n\}<1/2$ for every integer $n$.
An example [2] of an irrational ${\cal Z}$-number is $x=\tfrac{1}{2}(7+\sqrt{41})$, when $\{\xi x^n\}$ with $\xi=\tfrac{1}{4}(2+3x)$, converges to $1/4$ for $n\rightarrow\infty$.
The OP asks for an irrational ${\cal Z}$-number with $\xi=1$. It is known [3] that there exists no algebraic non-integer $x>1$ such that $\{x^n\}$ converges to 0. A stronger result (theorem 4 of [2]) indicates that a non-integer ${\cal Z}$-number with $\xi=1$ cannot be a [Pisot number](https://en.wikipedia.org/wiki/Pisot%E2%80%93Vijayaraghavan_number). I am not aware of any results for non-algebraic numbers.
1. [Even and odd
integral parts of powers of a real number](https://doi.org/10.1017/S0017089506003090), A. Dubickas (2006).
2. [On the
limit points of the fractional parts of powers of Pisot numbers](https://www.emis.de/journals/AM/06-2/am1306.pdf),
A. Dubickas (2006).
3. [On a
question of G. Kuba](https://link.springer.com/article/10.1007/s000130050442), F. Luca (2000).
| 11 | https://mathoverflow.net/users/11260 | 444868 | 179,353 |
https://mathoverflow.net/questions/444798 | 1 | My research has brought me to the following linear parabolic second order PDE:
$$ \frac{\partial^2}{\partial x^2}\Psi(t,x)=c(t,x)\frac{\partial}{\partial t}\Psi(t,x) $$
for $c(t,x)=-\frac{t}{x}$ and $x\in (0,\infty)$ and time $t\in (0, 1).$
Using the ansatz $\Psi(t,x)=X(x)Y(t)$ this can be simplified to ODE's and solved.
However, an important solution class is realized in the following way:
Let $(M,g)$ be the Minkowski plane with null coordinates. Take the isometry $f:M \to \zeta $ with $f(x,y)=(e^x,e^y).$ Take the natural Cauchy foliation of $\zeta$ given by $\log x \log y = s.$ Essentially when you furnish $\zeta$ with $g$ and the induced measure by means of the volume form, a natural solution to the above equation becomes clear:
$$ \Psi(t,x)=\exp \frac{x}{\log t} $$
Just like the free Schrodinger equation can be seen as a heat equation with an imaginary constant, there is an analogous PDE here too:
$$ \frac{\partial^2}{\partial x^2}\Psi(t,x)=\frac{t}{ix} \frac{\partial}{\partial t}\Psi(t,x) $$
We get a similar looking solution:
$$ \Psi(t,x)=\exp \frac{ix}{\log t} $$
>
> Are there any papers in the literature that deal with these specific PDE's where I can read more about them?
>
>
>
If there are no papers in the literature, as I suspect, what is the interpretation of these equations and my solution, and how might they be physically relevant?
| https://mathoverflow.net/users/411249 | Physical relevancy of two curious PDE's | Sorry to bring bad news. Usually, physicists are interested in the Cauchy problem: given initial data, determine the evolution of the state.
Alas, it is well known that PDE's whose order in the time variable (here 2) exceeds the order in the space variable (here 0) are not globally well-posed. Well, you can invoque Cauchy-Kovalevska's theorem in the analytic class, but it is useless for practical applications.
| 2 | https://mathoverflow.net/users/8799 | 444872 | 179,355 |
https://mathoverflow.net/questions/444699 | 8 | I know Jensen developed a forcing notions in $L$ that adds a unique, minimal and $\Delta\_3^1$ $L$-generic real. In his paper [Definable sets of minimal degree](https://zbmath.org/?q=an:0245.02055) he says that Solovay had already shown the consistency relative to $\mathsf{ZF}$ of $``V$ is the constructible closure of a real which is the unique solution of a $\Pi\_2^1$ predicate (hence the real is $\Delta\_3^1$)$"$. I searched a bit, but I couldn't find an exhaustive presentation of Solovay's result.
My questions are:
* Is there a paper/thesis where the abovementioned result due to Solovay is explained?
* Are there other known forcing notions adding an $L$-generic real which is the unique solution of a $\Pi\_2^1$ predicate?
EDIT: Solovay's result is fully discussed [here](https://doi.org/10.1016/S0049-237X(08)71932-3), even though it would be nice to find a more modern account. The second question still stands.
| https://mathoverflow.net/users/141146 | Forcing a unique $\Delta_3^1$ generic real | Since the first question has already been answered in the EDIT at the end of the question, I will focus on the second question.
The short answer to second question is: Yes, a good source is Sy Friedman's paper *The $\Pi^1\_2$-conjecture*, Journal of the American Mathematical Society, Vol. 3, No. 4 (Oct., 1990), pp. 771-791.
In the above paper, Friedman *negatively* settled a conjecture of Robert Solovay, a conjecture that stated: Assuming the existence of $0^\sharp$ , there are no $\Pi^1\_2$ singletons $s$ such that $s \in \mathrm{L}[0^\sharp]$ and yet $0^\sharp \notin \mathrm{L} [s]$. Friedman's construction is labyrinthine; a key ingredient of it is Jensen's technique of [Coding the Universe](https://www.cambridge.org/core/books/coding-the-universe/39F95AD061AE2D5A3DFB5F7B175DA7A2), and the important related work of René David in his paper *A very absolute $\Pi^1\_2$ real singleton*, Ann. Math. Logic 23 (1982), no. 2-3, 101–120 (1983).
| 8 | https://mathoverflow.net/users/9269 | 444876 | 179,357 |
https://mathoverflow.net/questions/432429 | 1 | Let $X$ be a projective normal variety over $\mathbb C$, I have several questions about semi-stable sheaves:
**Question 1.** Suppose that $E$ is a pure sheaf such that $HN\_\*(E)$ is the Harder-Narasimhan filtration of $E$. Let $H$ be an ample divisor and $D \in |aH|$ be a general element for $a\gg 1$. Then is $HN\_{\*}(E)|\_D$ the HN-filtration for $E|\_D$?
(I guess the statement is false because we even cannot guarantee that $HN\_{\*}(E)|\_D$ is a filtration (i.e. the restriction of inclusion may not be inclusion). But the statement might be true if $D$ is replaced by a general complete intersection curve.)
**Question 2.** If $E, F$ are (slope) stable sheaves, then is $E \otimes F$ still stable?
**Question 3.** Suppose that $0=E^0 \subset E^1\subset \cdots \subset E^k=E$ is a filtration of $E$ such that $G^i=E^i/E^{i-1}$ is semi-stable with slopes $\mu(G^i)$ strictly decreasing. Then is above filtration the NH-filtration?
(The answer may be no. But I was wondering if there is a non-constructive way to formulate NH-filtration? If there is a such why, please indicate how it goes.)
Thank you very much!
| https://mathoverflow.net/users/29730 | Some question about (semi-)stable sheaves | Concerning your **Questions 1** & **2**, since you adress to slope semistable sheaves and Metha-Ramanathan type results, I think every sheaf you mean is in fact torsion free (i.e. pure of dimension $\dim X$), right? Under this assumption, **Question 1** is true if $D$ is replaced by a general complete intersection curve and **Question 2** is also true.
Your **Question 3** is just the definition of the Harder-Narasimhan filtration (thus is true), c.f. [HL10, Definition 1.3.2, p.16].
Let me explain how to prove your **Question 1** & **2**:
* For **Question 1** (the version for general complete intersection curve), the point is that a general complete intersection curve $C$ is disjoint from the non-locally free loci (which is of codimension $\geqslant 2$ by torsion-freeness) of the factor sheaves of the HN-filtration, then the inculsions and factor sheaves are preserved under the restriction to $C$. Indeed, let
$$E\_0\subset E\_1\subset\cdots \subset E\_r=E$$
be the HN-filtration of $E$. Then for each $i$, since $E\_i/E\_{i-1}$ is locally free near $C$, the restriction to $C$ of the exact sequence
$$0\to E\_{i-1}\to E\_i\to (E\_i/E\_{i-1})\to 0,$$
remains exact. The remaining thing is to use Metha-Ramanthan (c.f. [HL10, Theorem 7.2.1, p.197]) to show the slope semistability of each $(E\_i/E\_{i-1})|\_C$. This result should also hold over algebraically closed fields of positive characteristic, since Metha-Ramanthan holds for arbitrary characteristic.
* As for **Question 2**, again let $C$ be a general complete intersection curve, then $E|\_C$ and $F|\_C$ are both locally free and slope stable (by Metha-Ramanthan, c.f. [HL10, Theorem 7.2.8, p.202]). And so is $(E\otimes F)|\_C\simeq E\_C\otimes F|\_C$, as a consequence of Narasimhan-Seshadri theorem (Ulenbeck-Yau in the curve case); there is also a purely algebraic proof for this fact.
Reference(s):
[HL10] Daniel Huybrechts & Manfred Lehn: The Geometry of Moduli Spaces of Sheaves (2nd ed.), Cambridge: CUP, 2010,
| 1 | https://mathoverflow.net/users/502863 | 444884 | 179,360 |
https://mathoverflow.net/questions/444846 | 4 | Lemma 3.2.1 in [Baker, González-Jiménez, González, Poonen, "Finiteness theorems for modular curves of genus at least 2", Amer. J. Math. 127 (2005), 1325–1387.](https://math.mit.edu/%7Epoonen/papers/finiteness.pdf)
[enter image description here](https://i.stack.imgur.com/ecXLP.png)
I don't understand why "all supersingular elliptic curves over $\bar{F\_p}$ are isogenous".
Could anyone help me?
This problem arises from [Supersingular elliptic curves and their "functorial" structure over F\_p^2](https://mathoverflow.net/questions/18982/supersingular-elliptic-curves-and-their-functorial-structure-over-f-p2).
Moreover, I found some similar question: [Isogenies between elliptic curves and their endomorphism rings](https://math.stackexchange.com/questions/2364125/isogenies-between-elliptic-curves-and-their-endomorphism-rings) and [Isogenies between supersingular elliptic curves](https://mathoverflow.net/questions/86444/isogenies-between-supersingular-elliptic-curves).
| https://mathoverflow.net/users/502709 | Why all supersingular elliptic curves over $\bar{\mathbb{F}_p}$ are isogenous? | This can be proved in several ways (this can be found in the literature as Lemma 42.1.11 in Voight's book on Quaternion algebras, or Proposition 5.2 in <https://arxiv.org/pdf/2005.01537v1.pdf>). Typically the proof relies on Tate's isogeny theorem (1966) which states, among other things, that if two elliptic curves over a finite field have the same number of rational points, then they are isogenous.
$$
\newcommand{\End}{\mathrm{End}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\F}{\mathbb{F}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\C}{\mathbb{C}}
$$
---
**Proposition.**
Let $p$ be a prime.
Let $E,E'$ be two supersingular elliptic curves over $\F\_{p^2}$. Then $E,E'$ are isogenous over $\F\_{p^{24}}$.
(Note that the $j$-invariant of supersingular elliptic curves over $\overline{\F\_p}$ belongs to $\F\_{p^2}$, we may assume (up to $\overline{\F\_p}$-isomorphism) that $E,E'$ are defined over $\F\_{p^2}$).
---
**Proof.**
Write $\alpha, \beta \in \C$ for the eigenvalues of the $p^2$-Frobenius map on the $\ell$-adic Tate module of $E$ for some $\ell \neq p$, so that $|E(\F\_{p^{2r}})| = p^{2r} + 1 - (\alpha^r + \beta^r)$ for every $r \geq 1$.
Since $E$ is supersingular, its trace $t\_1 := \alpha + \beta$ is $\equiv 0 \pmod p$. By Hasse bound we have $|t\_1| \leq 2 \sqrt{p^2} = 2p$, so we have $t\_1 \in \{ -2p, -p, 0, p, 2p \}$.
In each case, $\alpha, \beta$ can be computed explicitly as roots of a quadratic polynomial (characteristic polynomial of Frobenius):
* If $t\_1 = -2p$, then $X^2 + 2pX + p^2 = (X+p)^2
= 0$ yields $\alpha = \beta = -p$.
* If $t\_1 = -p$, then $X^2 + pX + p^2 = 0$ yields $\alpha,\beta =
\frac{-p + \sqrt{p^2 - 4p^2}}{2} = p \cdot \dfrac{-1 \pm i
\sqrt{3}}{2}$
* If $t\_1 = 0$, then $X^2 + p^2 = 0$ yields $\alpha = - \beta =
ip$.
* If $t\_1 = p$, then $X^2 - pX + p^2 = 0$ yields $\alpha,\beta =
p \cdot \dfrac{1 \pm i \sqrt{3}}{2}$
* If $t\_1 = 2p$, then $(X - p)^2 = 0$ yields $\alpha = \beta =
p$.
In all cases, we observe that $\alpha^{12} = \beta^{12} = p^{12}$ (i.e. $\alpha,\beta$ are equal to $p$ times some $12$-th root of unity [in fact either of order $1,2,3,4$ or $6$]). This proves that
$$
|E(\F\_{p^{24}})| =
p^{24} + 1 - (\alpha^{12} + \beta^{12})
=
p^{24} + 1 - 2 \cdot p^{12}
=
|E'(\F\_{p^{24}})|
$$
so that $E,E'$ are isogenous over $\F\_{p^{24}}$ by Tate's theorem. This concludes the proof. $\blacksquare$
| 2 | https://mathoverflow.net/users/84923 | 444889 | 179,362 |
https://mathoverflow.net/questions/444886 | 4 | Let $f:\mathbb{R}^n\rightarrow\mathbb{R}$ be a convex function. Denote $X^\*$ as the set of minimizers of $f$ and assume $X^\*$ is unbounded. Is it possible that $\|g\_x\|$ is unbounded when $d(x,X^\*)$ is bounded where $g\_x\in\partial f(x)$. Here $\partial f(x)$ denotes the set of subgradient vectors of $f$ at $x$ and $d(x,X^\*)$ denotes the distance between point $x$ and set $X^\*$.
---
I know $X^\*$ is a closed convex set and I am thinking whether it is true that $\|g\_x\|$ is bounded given $d(x,X^\*)=1$. It must be true if $X^\*$ is bounded, but not sure when it is unbounded. Also, does it help if we further assume $f$ is differentiable, so that $\partial f(x)=\{\nabla f(x)\}$?
---
**Edit**: I realized that what I really want to ask is whether it is possible that $\|g\_x\|$ is unbounded but $d(x,X^\*)$ is bounded. It is not equivalent to the boundedness of the ratio of $\|g\_x\|$ and $d(x,X^\*)$.
| https://mathoverflow.net/users/490600 | An upper bound of gradient norm for convex functions near minimizer | The function $q$ might be unbounded even in 1-neighborhood of $X^\*$; here is an example of such function $f$ on the $(x,y)$-plane.
Let $\phi(t)=|t|-t$.
Choose a sequence $x\_n\to \infty$.
For each $n$ consider function $f\_n(x,y)=\phi[-2\cdot x\_n\cdot(x-x\_n)+ (y-x\_n^2)]$.
Let $f=\max\_n\{f\_n\}$.
| 3 | https://mathoverflow.net/users/1441 | 444895 | 179,364 |
https://mathoverflow.net/questions/444893 | 4 | [Rademacher’s formula](https://en.m.wikipedia.org/wiki/Partition_function_(number_theory)) for the partition function allows fast computation using high precision arithmetic, but requiring a lot of memory. [Here](https://fredrikj.net/blog/2014/03/new-partition-function-record) is an example computation of $p(10^{20})$ by Fredrik Johansson.
**Question:** Is there a $p$-adic or modular analogue that allows fast computation modulo an arbitrary small prime, with low memory (related only to the size of the prime)?
| https://mathoverflow.net/users/22930 | Fast computation of the partition function modulo a prime | To the best of my knowledge, there is no known method to compute $p(n)$ modulo a small prime using less than $n^{1/2+o(1)}$ time or memory, except for those cases where a Ramanujan-like congruence applies.
| 5 | https://mathoverflow.net/users/4854 | 444897 | 179,365 |
https://mathoverflow.net/questions/444870 | 2 | Suppose an $n$-dimensional process $(X\_t)\_{0 \leq t \leq 1}$ satisfies an SDE of the form:
$$dX\_t = u\_t(X\_t) \,dt + dB\_t, ~~X\_0 = 0$$
where $(B\_t)\_{t\geq 0}$ is a Brownian motion with $B\_1 \sim N(0,K)$, and $K$ is positive definite.
Does anyone have a reference for simple conditions on $u\_t$ that will ensure that $\operatorname{law}(X\_t)$ has full support on $\mathbb{R}^n$?
For example, for $t>0$, if there exists a finite constant $C\_t$ such that $|u\_s(x)| \leq C\_t(|x|+1)$ for all $0 \leq s\leq t$, does this ensure $\operatorname{law}(X\_t)$ has full support?
Of course, some regularity on $u\_t$ is needed; consider the standard Brownian bridge, which has $X\_1 = 0$ a.s., but $\operatorname{law}(X\_t)$ has full support for each $0< t<1$.
| https://mathoverflow.net/users/99418 | When does a solution to SDE have full support? | I am writing down a very similar answer to [this one](https://mathoverflow.net/a/443534/129074), that I wrote for the similar question [*SDE with non-degenerate diffusion visits every point*](https://mathoverflow.net/q/438425/129074). I am not sure how closely it answers your question, since it is not a reference for the exact result you quote, but it does follow as a corollary.
The support of the whole trajectory $X$ in $\mathcal C([0,\infty),\mathbb R^n)$ is described by the so-called Stroock-Varadhan support theorem. In [*On the support of diffusion processes with applications to the strong maximum principle*](https://projecteuclid.org/proceedings/berkeley-symposium-on-mathematical-statistics-and-probability/Proceedings-of-the-Sixth-Berkeley-Symposium-on-Mathematical-Statistics-and/Chapter/On-the-support-of-diffusion-processes-with-applications-to-the/bsmsp/1200514345?tab=ChapterArticleLink), their Theorem 3.1 ensures that the (unique) solution to the martingale problem associated with $u\_t+\frac12\Delta\_K$ has full support provided $u:[0,\infty)\times\mathbb R^d\to\mathbb R^d$ is bounded measurable.
A strong solution to the equation $\mathrm dX\_t=u\_t(X\_t)\mathrm dt + \mathrm dB\_t$ will always be a solution to the martingale problem, so this answers your question in the case where (1) you have conditions guaranteeing existence of a strong solution and (2) $u$ is bounded measurable (you can also define the solution as a solution to the martingale problem, in which case (2) is sufficient for existence and uniqueness).
A natural case that doesn't quite fit the above is when $u$ is unbounded but you have a unique strong and weak solution up to some explosion time using some other argument $\tau$. We can still rely on this theorem in more general situations, and to illustrate my point I will consider the classical case where $u$ is continuous in $(t,x)$, locally Lipschitz in $x$ (for instance $u$ is $\mathcal C^1$ in $(t,x)$). Then one can define the solution $X^R>0$ to the equation
$$ \mathrm dX^R\_t = \big(0\vee(2R-|X^R\_t|)\wedge1\big)u\_t(X^R\_t)\mathrm dt + \mathrm dB\_t, $$
which is defined for all times and coincides with $X$ until one of the two exists the ball of radius $R$. This means that
$$\mathbb P(X\_t\in U)\geq\mathbb P(X\_t\in U\text{ and }\forall s\leq t,|X\_s|<R)=\mathbb P(X^R\_t\in U\text{ and }\forall s\leq t,|X^R\_s|<R)$$
for $R$ large enough. By the theorem above, we know that this last probability must be positive, and so is the first one.
A similar argument shows that your condition $|u\_s(x)|\leq C\_t(1+|x|)$ is enough, provided $u$ is measurable, and your notion of solution coincides with the unique martingale problem solution.
I should add that these results rely on applying a [Girsanov argument](https://en.wikipedia.org/wiki/Girsanov_theorem), making a change of probability with density (if I am not mistaken with my use of $K$ and $K^{-1}$)
$$\exp\left(\int\_0^tu\_s(X\_s)K^{-1}\mathrm dB\_s-\frac12\int\_0^tu\_s\mathrm ds\right).$$
| 3 | https://mathoverflow.net/users/129074 | 444902 | 179,368 |
https://mathoverflow.net/questions/444885 | 1 | Suppose there are $n$ IID random variables denoted as $X=(X\_1,\dots, X\_n)$, they follow Laplace distribution with parameter $\lambda$, denoted as $Lap(\lambda)$. That is,
$$f(x)=\frac{1}{2\lambda}\exp (-\frac{|x|}{\lambda})$$
$$
F(x)= \begin{cases}\frac{1}{2} \exp \left(\frac{x}{\lambda}\right) & \text { if } x<0 \\ 1-\frac{1}{2} \exp \left(-\frac{x}{\lambda}\right) & \text { if } x \geq 0\end{cases}
$$
Let $M=\max(X)-\min(X)$, how can we compute the expectation of $M$, that is $E(M)$, or could we provide an upper bound of $E(M)$?
I have tried to compute the cdf and pdf of $M\_1=\max(X), M\_2=\min(X)$. Then we have
$$F\_{M\_1}(x)=(F(x))^n, F\_{M\_2}=1-(1-F(x))^n$$
However, the expectation computation seems difficult, and I did not figure it out.
Besides, I found this problem is related to *Extreme Value Theory*, but I am not a statistic student and I don't know much about that.
| https://mathoverflow.net/users/327644 | Expected (maximum minus minimum) of Laplacian random variables | $\newcommand\la\lambda$By rescaling, it is enough to consider the case $\la=1$ (multiplying the resulting expectation by $\la$ at the end). By symmetry, $Y:=\max X$ and $-\min X$ are identically distributed. So, for $M\_n:=M$ we have
\begin{equation\*}
EM\_n=2EY. \tag{10}\label{10}
\end{equation\*}
Next, $Y=\max(0,Y)-\max(0,-Y)$ and hence
\begin{equation\*}
EY=I\_1-I\_2, \tag{20}\label{20}
\end{equation\*}
where
\begin{equation\*}
\begin{aligned}
I\_1&:=E\max(0,Y)=\int\_0^\infty dx\,P(Y>x)=\int\_0^\infty dx\,(1-(1-F(x)^n),\\
I\_2&:=E\max(0,-Y)=\int\_0^\infty dx\,P(-Y>x)=\int\_0^\infty dx\,F(-x)^n.
\end{aligned}
\end{equation\*}
Here I use a standard trick:
\begin{equation\*}
\int\_0^\infty dx\,P(Y>x)=\int\_0^\infty dx\,E\,1(Y>x) \\
=E\int\_0^\infty dx\,1(Y>x)
=E\int\_0^{\max(0,Y)} dx=E\max(0,Y).
\end{equation\*}
Further,
\begin{equation\*}
I\_2=\int\_0^\infty dx\,(e^{-x}/2)^n=\frac{2^{-n}}n \tag{30}\label{30}
\end{equation\*}
and
\begin{equation\*}
\begin{aligned}
I\_1&=\int\_0^\infty dx\,(1-(1-e^{-x}/2)^n) \\
&=-\int\_0^\infty dx\,\sum\_{j=1}^n\binom nj(-e^{-x}/2)^j \\
&=-\sum\_{j=1}^n\binom nj\frac{(-1/2)^j}j
=\frac{n}{2} \, \_3F\_2\left(1,1,1-n;2,2;\frac{1}{2}\right),
\end{aligned}
\end{equation\*}
where $\_3F\_2$ is the hypergeometric function.
Thus, for $\la=1$,
\begin{equation\*}
EM\_n=n \, \_3F\_2\Big(1,1,1-n;2,2;\frac{1}{2}\Big)-\frac1{n2^{n-1}}
\end{equation\*}
and
\begin{equation\*}
EM\_n=\la\Big(n \, \_3F\_2\Big(1,1,1-n;2,2;\frac{1}{2}\Big)-\frac1{n2^{n-1}}\Big)
\end{equation\*}
for any real $\la>0$.
---
It follows from the [previous answer](https://mathoverflow.net/a/378667/36721) (thanks to Matt F. for reminding us of it), formulas \eqref{10}, \eqref{20}, \eqref{30}, and the "rescaling" remark that
\begin{equation\*}
l\_n:=\gamma+\ln(n/2)-\frac1{n{2^n}}<\frac{EM\_n}{2\la}<u\_n:=1+\ln(n/2)-\frac1{n{2^n}},
\end{equation\*}
where $\gamma=0.577\ldots$ is the Euler gamma.
It is clear now that
\begin{equation\*}
\frac{EM\_n}{2\la}\sim l\_n\sim u\_n\sim\ln(n/2)\sim\ln n,
\end{equation\*}
so that the upper bound $u\_n$ and the lower bound $l\_n$ on $\frac{EM\_n}{2\la}$ are each asymptotically exact.
| 2 | https://mathoverflow.net/users/36721 | 444903 | 179,369 |
https://mathoverflow.net/questions/444899 | 3 | Let $U\subset\mathbb{R}^{d}$ be an open set. The singular support of a distribution $u\in\mathcal{D}^{\prime}(U)$ is defined to be the compliment of the set of points, which have a neighbourhood in which $u$ can be identified with a smooth function. In a paper [Fourier integral operators. I](https://doi.org/10.1007/BF02392052) of Hörmander, it is claimed that
$$\DeclareMathOperator\singsupp{sing supp}\singsupp(u)=\bigcap\_{\varphi\in C^{\infty}\_{c}(U),\varphi u\in C^{\infty}(U)}\{x\in U\mid \varphi(x)=0\}.$$
However, I don't see why it is the case. Does anyone know how to see that this is the case?
My attempt:
The direction $\Leftarrow$ is quite clear I think: Let $x\notin\singsupp(u)$. Then there is an open neighbourhood $N$ of $x$ on which $u\vert\_{N}\in C^{\infty}$. Let $\chi\in C\_{c}^{\infty}(U)$ be supported in $N$ such that $\chi(x)\neq 0$. Then $\chi u\vert\_{N}\in C^{\infty}(U)$ and hence $x\notin\bigcap\_{\varphi\in C^{\infty}\_{c}(U),\varphi u\in C^{\infty}(U)}\{x\in U\mid \varphi(x)=0\}$.
What is left is the other direction. I essentially have to show that if there exists a test function $\varphi\in C^{\infty}\_{c}(U)$ such that $\varphi u\in C^{\infty}(U)$ and $\varphi(x)\neq 0$, then $x\notin\singsupp(u)$.
| https://mathoverflow.net/users/199422 | Singular support: equivalent definition | The basic idea is the one put forward by Pierre PC's comment above. More precisely, let $u,\varphi,x$ be as in the last paragraph of the OP. There is no loss of generality in assuming that $\varphi(x)=\lambda>0$. Then by continuity of $\varphi$ at $x$ there is an open neighborhood $V\ni x$, $V\subset U$ such that $\varphi(x')>\frac{\lambda}{2}$ for all $x'\in V$, so that by the chain rule $\psi=\frac{1}{\varphi|\_V}\in C^\infty(V)$. One then clearly has that $u|\_V=\psi(\varphi u)|\_V\in C^\infty(V)$ (*exercise:* check this!), thus proving the direction $\Rightarrow$ of the claim.
By the way, this (together with the reasoning used in the OP to get the $\Leftarrow$ direction of the claim) is essentially the same argument used to prove a similar characterization of the support of a distribution: $$\mathrm{supp}\,u = \bigcap\_{\varphi\in C^\infty\_c(U),\phi u\equiv 0}\{x\in U\ |\ \phi(x)=0\}\ .$$
| 5 | https://mathoverflow.net/users/11211 | 444909 | 179,370 |
https://mathoverflow.net/questions/444882 | 10 | Using the definitions from Peter May's [A Concise Course in Algebraic Topology](https://www.math.uchicago.edu/%7Emay/CONCISE/ConciseRevised.pdf), a topological space $X$ is weak Hausdorff if for every compact Hausdorff space $K$ and continuous function $f:K\to X$, $f(K)$ is closed in $X$. A subset $A$ of $X$ is said to be compactly closed if for any compact Hausdorff space $K$ and continuous map $f:K\to X$, $f^{-1}(A)$ is closed in $K$. Finally, $X$ is said to be a $k$-space if every compactly closed subset is closed.
In the lemma at the top of page 40, May gives a characterization of weak Hausdorff spaces which can be paraphrased as such:
>
> A $k$-space $X$ is weak Hausdorff if and only if the diagonal $\Delta\_X$ is compactly closed in $X\times X$ under the product topology.
>
>
>
I was searching for a counterexample to this lemma when the $k$-space assumption is omitted. I came up with a proof of this lemma, and the forward direction didn't require the $k$-space assumption. So in other words, I am asking
>
> What is an example of a topological space $X$ such that $\Delta\_X$ is compactly closed in $X\times X$, but $X$ is not weak Hausdorff?
>
>
>
By the lemma, such an example must not be a $k$-space.
| https://mathoverflow.net/users/502850 | Space with compactly closed diagonal but which is not weak Hausdorff | The space $X$ constructed in Theorem 1.5 of this [preprint](https://arxiv.org/abs/2211.12579) has the required properties. This space contains a non-closed compact metrizable subspace $K$, so is not weakly Hausdorff.
On the other hand, the space $X$ is $k\_2$-metrizable, which means that $X$ is the image of a metrizable space $M$ under a continuous bijective proper map $f:M\to X$ such that for every continuous map $g:H\to X$ from a compact Hausdorff space $H$, the map $f^{-1}\circ g:H\to M$ is continuous. This property implies that the diagonal $\Delta\_X$ is compactly closed in $X\times X$.
| 11 | https://mathoverflow.net/users/61536 | 444913 | 179,372 |
https://mathoverflow.net/questions/444906 | 5 | In the paper, *A polytope related to empirical distributions, plane trees, parking functions, and the associahedron*, [Pitman and Stanley](https://arxiv.org/pdf/math/9908029.pdf) studied the $n$-dimensional polytope
$$\Pi\_n(\mathbf x)=\{y\in\Bbb{R}^n: y\_i\geq0, y\_1+\cdots+y\_i\leq x\_1+\cdots+x\_i, \text{for all $1\leq i\leq n$}\}.$$
They noted the results
\begin{align\*}
\text{Vol}(\Pi\_n(\mathbf x))
&=\frac1{n!}\sum\_{\mathbf{k}\in \mathbf{K}\_n}\binom{n}{k\_1,\dots,k\_n}\,\,x\_1^{k\_1}\cdots x\_n^{k\_n} \\
&=\det\left(\frac{\chi(j-i+1\geq0)}{(j-i+1)!}\left(\sum\_{h=1}^ix\_h\right)^{j-i+1}\right)\_{i,j=1}^n
\end{align\*}
where $\chi$ is the *indicator function* and
$$\mathbf{K}\_n=\{\mathbf{k}\in\mathbb{Z}^n: k\_i\geq0, k\_1+\cdots+ k\_n=n, k\_1+\cdots k\_i\leq i, \text{for all $1\leq i\leq n$}\}.$$
The set $\mathbf{K}\_n$ has cardinality $\vert\mathbf{K}\_n\vert=\frac1{n+1}\binom{2n}n$. Consider instead the determinant
$$\det\left(\frac{\chi(j-i+1\geq0)}{(j-i+1)!}\cdot x\_i^{j-i+1}\right)\_{i,j=1}^n.$$
A *composition* of $n$ is a finite sequence of positive integers summing to $n$. To help us calculate the latter determinant, we introduce the set of compositions of length-$n$
$$\mathcal{B}\_n=\{y\in\mathbb{Z}^n: y\_i\geq0, \text{$y$ is a composition of $n$ with $0$ suffixes padded if necessary}\}.$$
For example, $\mathcal{B}\_2=\{20, 11\}$ and $\mathcal{B}\_3=\{300, {\color{red}{210}}, 120, 111\}$. Note $\vert\mathbf{B}\_n\vert=2^{n-1}$.
**QUESTION.** Is this true?
$$\det\left(\frac{\chi(j-i+1\geq0)}{(j-i+1)!}\cdot x\_i^{j-i+1}\right)\_{i,j=1}^n
=\frac1{n!}
\sum\_{\mathbf{k}\in \mathcal{B}\_n}(-1)^{\#(\mathbf{k})}\binom{n}{k\_1,\dots,k\_n} \,\,x\_1^{k\_1}\cdots x\_n^{k\_n},$$
where $\#(\mathbf{k})$ stands for the number of zeroes in $\mathbf{k}$.
**REMARK.** One may find it convenient to work with the alternative formulation
$$\det\left(\frac{\chi(j-i+1\geq0)}{(j-i+1)!}\cdot x\_i^{j-i+1}\right)\_{i,j=1}^n
=\prod\_{k=1}^n\frac{k!}{(1+k)!}\cdot \det\left(\binom{1+j}{i}\cdot x\_i^{j-i+1}\right)\_{i,j=1}^n.$$
${\color{blue}{POSTSCRIPT}}$. Max Alekseyev's comment requires some adjustment in the set $\mathcal{B}\_n$. So, here is (I hope) the correct construction: $\mathbf{y}=(y\_1,\dots,y\_n)\in\mathcal{B}\_n$ iff $y\_1>0$; $y\_i\in\mathbb{Z}\_{\geq0}$ for all $i$; when $\mathbf{y}$ is read (cyclically) $y\_1\rightarrow y\_2\rightarrow\cdots\rightarrow y\_n\rightarrow y\_1$, each $y\_i\neq0$ is followed by $y\_i-1$ zeroes. Clearly, $y\_i\leq n$ for all $i$.
For example, $\mathcal{B}\_2=\{20, 11\}$ and $\mathcal{B}\_3=\{300, {\color{red}{201}}, 120, 111\}$ and $\mathbf{B}\_4=\{4000,3001,2020,2011,1300,1201,1120,1111\}$. Observe that $\vert \mathcal{B}\_n\vert=2^{n-1}$, still.
| https://mathoverflow.net/users/66131 | Catalan sequences vs composition sequences | Given the corrected definition of $\mathcal{B}\_n$ in the Postscriptum, the equality does hold and can be proved by induction on $n$.
For $n=1$, the equality is trivial. For $n>1$, let's expand the determinant in the left-hand size of the equality over the first column to get
$$\det\left(\frac{\chi(j-i+1\geq0)}{(j-i+1)!}\cdot x\_i^{j-i+1}\right)\_{i,j=1}^n
= x\_1 \det\left(\frac{\chi(j-i+1\geq0)}{(j-i+1)!}\cdot x\_i^{j-i+1}\right)\_{i,j=2}^n - \int\_0^{x\_1} dx\_2 \det\left(\frac{\chi(j-i+1\geq0)}{(j-i+1)!}\cdot x\_i^{j-i+1}\right)\_{i,j=2}^n.$$
By the induction, the right-hand side in the expansion equals
$$\frac1{n!}
\sum\_{\mathbf{k}\in \mathcal{B}\_{n-1}}(-1)^{\#(\mathbf{k})}\binom{n}{1,k\_1,\dots,k\_{n-1}} \,\,x\_1 x\_2^{k\_1}\cdots x\_n^{k\_{n-1}} - \frac1{n!}
\sum\_{\mathbf{k}\in \mathcal{B}\_{n-1}}(-1)^{\#(\mathbf{k})}\binom{n}{k\_1+1,k\_2,\dots,k\_{n-1}} \,\,x\_1^{k\_1+1}x\_3^{k\_2}\cdots x\_n^{k\_{n-1}}$$
$$=\frac1{n!}
\sum\_{\mathbf{t}\in \mathcal{B}\_n}(-1)^{\#(\mathbf{t})}\binom{n}{t\_1,\dots,t\_n} \,\,x\_1^{t\_1} x\_2^{t\_2}\cdots x\_n^{t\_n},$$
where the two sums in the left-hand side correspond to the following two cases in the right-hand size:
* $t\_1=1$ and $t\_{i+1} = k\_i$ for $i=1,2,\dots,n-1$; and
* $t\_1=k\_1+1>1$, $t\_2=0$, and $t\_{i+1} = k\_i$ for $i=2,\dots,n-1$.
QED
| 7 | https://mathoverflow.net/users/7076 | 444917 | 179,374 |
https://mathoverflow.net/questions/444588 | 5 | Let $\lambda$ denote the Liouville function, and let $L(x):=\sum\_{n \leq x} \lambda(n)$ be the Liouville sum. Define $c$ to be the supremum of the real parts of the zeros of the Riemann zeta function. It is a straightforward exercise to show that for any $\varepsilon>0$, one has $L(x) =\Omega\_{\pm }(x^{c-\varepsilon})$ as $x \rightarrow \infty$. Is this also true when $x$ runs through some special set, specifically the set of primes?
| https://mathoverflow.net/users/501735 | Asymptotics of the Liouville sum at the primes | This should be true. By a Corollary II of a [result of Pintz](http://matwbn.icm.edu.pl/ksiazki/aa/aa42/aa4214.pdf) (with not too much work, one can get this to work for the Liouville function in place of Mobius), we have that
$$\sum\_{Y/(100\log Y)\le n\le Y} |L(n)|\gg Y^{1 + c - \varepsilon},$$
in your notation. We also have that for $n$ in this range, $L(n) \ll Y^{c + \varepsilon}$, so we get that with
$$\mathcal S = \{Y/(100\log Y)\le n\le Y : |L(n)|\ge Y^{c - 10\varepsilon}\},$$ we have that
$\#\mathcal S\gg Y^{1-2\varepsilon}.$ At this point, we wish to show there is a prime within $Y^{c - 10\varepsilon}/2$ of an element of $\mathcal S$. This should follow comfortably from results on primes in almost all short intervals as $c\ge\frac{1}{2}$.
| 4 | https://mathoverflow.net/users/40983 | 444921 | 179,376 |
https://mathoverflow.net/questions/444925 | 14 | Let $f(x\_1, \cdots, x\_n) \in \mathbb{R}[x\_1, \cdots, x\_n]$ be a polynomial. Define property $\mathbf{P}$ to be the property that there exists a compact set $K \subset \mathbb{R}^n$ and a positive number $\kappa$ such that for all $\mathbf{x} \in \mathbb{R}^n \setminus K$ we have $f(\mathbf{x}) > \kappa$. Clearly, $f$ having property $\mathbf{P}$ implies $2d = \deg f$ is even. For a positive number $R$ define
$$\displaystyle m\_f(R) = \inf\_{R \leq \lVert \mathbf{x} \rVert\_2 \leq 2R} f(\mathbf{x}).$$
Is the following statement true?
Suppose $f \in \mathbb{R}[x\_1, \cdots, x\_n]$ has property $\mathbf{P}$ and $\lim\_{R \rightarrow \infty} m\_f(R) = \infty$. Does there exist $1 \leq r \leq d = (\deg f)/2$ and a positive number $\delta$ such that
$$\displaystyle f(\mathbf{x}) - \delta \left(\sum\_{j=1}^n x\_j^{2r} \right)$$
has property $\mathbf{P}$?
Edit: As remarked in the comments by MattF, the supposition that $\lim\_{R \rightarrow \infty} m\_f(R) = \infty$ supersedes property $\mathbf{P}$.
| https://mathoverflow.net/users/10898 | Real polynomials that go to infinity in all directions: how fast do they grow? | The assumption "$\lim\_{R\to\infty}m\_f(R)\to\infty$" already includes $\bf P$ as remarked in comments, and it is the same as $\lim\_{\|x\|\to\infty} f(x)=+\infty$, that is: **$f$ is a coercive polynomial**. You are asking whether a coercive polynomial in $n$ variables needs to go to infinity at least quadratically: the answer is no: *the order of coercivity* of a coercive polynomial may be an arbitrarily small positive number.
Check e.g. this article, [How fast do coercive polynomials grow?](http://num.math.uni-goettingen.de/preprints/files/2017-1.pdf) (section 4 for examples). In particular (proposition 19), even in two variables, for any $k\in\mathbb N\_+$,
$$f(x,y)=x^2+(y-x^{2k})^2$$
is coercive and has order of coercivity $1/k$, that is $f(z)/{\|z\|^\theta}\to\infty$ as $\|z\|\to+\infty$, for $\theta<1/k$, but for no $\theta>1/k$ (just because along the curve $y=x^{2k}$ one has $f(x,y)=x^2=y^{1/k}$, so that $f(z)/{\|z\|^{1/k}}=O(1)$)
| 27 | https://mathoverflow.net/users/6101 | 444938 | 179,381 |
https://mathoverflow.net/questions/444943 | 7 | Let $Sets$ be the category of finite sets and all maps. I have come accross several example of functors $F : Sets^{op} \to Sets$ which satisfy the condition below:
-- There exists an integer $k$ such that, whenever $X$ is a finite set, with subsets $U\_1, U\_2, U\_3...$ whose union is $X$ *and with* $|U\_i| \ge k$, then we have the usual glueing condition: given elements $a\_i \in F(U\_i)$ with $a\_i|\_{U\_i\cap U\_j} = a\_j|\_{U\_i\cap U\_j}$, there exists a unique $a \in F(X)$ with $a |\_{U\_i} = a\_i$.
(I hope the meaning of the "restriction" notation is obvious.)
So far I have taken this, informally, to mean that $F$ "is determined by what it does to sets of size $\le k$" (whenever $|X| > k$, we can recover $F(X)$ from all the sets $F(U)$ where $U$ runs through the subsets of $X$ of size $k$). However, it would be instructive to have a more conceptual understanding of what's going on.
Obviously the condition is almost the sheaf condition, but of course, the "subsets of size $\ge k$" do not form a Grothendieck topology...
I know of one way to rephrase this: given a contravariant functor $G$ from $C\_k$, the category of sets of size $\le k$, to $Sets$, you can extend it to a functor $R(G)$ on $Sets^{op}$ in such a way that $R$ is a right adjoint to the "truncation" functor, which takes $F$ to its restriction $F\_{\le k}$ to $C\_k^{op}$. And my condition is that $F$ is isomorphic to $R(F\_{\le k})$ (I think this is called the $k$-th "coskeleton" of $F$). However, I don't really know what to make of this.
>
> Does anyone know of a good conceptual framework for this situation ? Perhaps in terms of sheaves and Grothendieck topologies?
>
>
>
Edit : I like the answer by T; Goodwillie below.
>
> If there is such a framework, then what can I learn about $F$, concretely?
>
>
>
Edit : let me more specific. For a given $F$, I'm trying to look at the minimal $k$ such that the above holds, and I'm hoping that $k$ is a good measure of the "complexity" of $F$, with a low value of $k$ meaning that $F$ is "easier" to understand. For example you get $k=1$ for $F(X) = Hom(X, S)$ for a fixed $S$; I think the converse holds, with $S=F(\{x\})$, which is defined up to a canonical bijection.
So I would like to
(1) Substantiate the hope that $k$ is a good quantity to look at. Any interpretation à la $n$-lab is good for this, and by now I'm convinced that it's reasonable. Extra comments are welcome of course.
(2) compute $k$ concretely in some examples. Which is why any observations about the consequences of the above condition may help put some bounds on $k$.
| https://mathoverflow.net/users/37021 | not quite the sheaf condition | Consider the Grothendieck topology in which a cover of $X$ is a way of expressing $X$ as a union of sets $U$ with $|U|\le k$. It seems to me that you are talking about sheaves for this topology.
This looks wrong (because I have "$\le$" where you had "$\ge$"), but I think it's right.
EDIT: This was nonsense. I should have said:
For each $k\ge 0$ there is a topology on the category of (all) sets, in which a cover of $X$ by subsets $U$ means that every subset of $X$ having at most $k$ elements belongs to some $U$. Your presheaves are the sheaves for this topology.
| 8 | https://mathoverflow.net/users/6666 | 444948 | 179,383 |
https://mathoverflow.net/questions/444963 | 1 | Consider the matrix
$$A\_2:= \begin{pmatrix} a & b\_1 \\ b\_2 & a\end{pmatrix}.$$
Let $\sigma\_2 = \begin{pmatrix} 0 & i \\ -i & 0 \end{pmatrix}$, then
$$\sigma\_2 A\_2 \sigma\_2 = \begin{pmatrix} a & -b\_2 \\ -b\_1 & a\end{pmatrix}.$$
I wonder if I have the matrix
$$A\_3:= \begin{pmatrix} a & b\_1 & 0 \\ b\_2 & a & c\_1 \\ 0 & c\_2 & a \end{pmatrix}$$
if there is an analogous matrix
$\sigma\_3$ that works for all possible choices of coefficients in $A\_3$ such that
$$\sigma\_3 A\_3 \sigma\_3^{-1} = \begin{pmatrix} a & -b\_2 & 0 \\ -b\_1 & a & -c\_2 \\ 0 & -c\_1 & a \end{pmatrix}.$$
That there exists one such matrix for each set of coefficients is clear, since the eigenvalues of the two $3x3$ matrices are the same. I am looking for one that works for all choices.
| https://mathoverflow.net/users/496243 | Matrix transformation that always works? | The answer is no. Indeed, let $T:=\sigma\_3=(t\_{ij}\colon i,j\in\{1,2,3\})$, $A:=A\_3$, and
$$B:=\begin{pmatrix} a & -b\_2 & 0 \\ -b\_1 & a & -c\_2 \\ 0 & -c\_1 & a \end{pmatrix}.$$
Then the equality in question would imply that
$$TA=BT \tag{1}
\label{1}$$
for all $a,b\_1,b\_2,c\_1,c\_2$. Solving system \eqref{1} of linear equations for the $t\_{ij}$'s, we get
$$b\_1 t\_{11}+c\_2 t\_{13}+b\_2 t\_{22}=0,\quad
c\_1 t\_{12}+b\_2 t\_{23}=0$$
for all $b\_1,b\_2,c\_1,c\_2$.
It follows that
$$t\_{11}=t\_{13}=t\_{22}=0,\quad
t\_{12}=t\_{23}=0,$$
so that $t\_{11}=t\_{13}=t\_{12}=0$ and hence $\det T=0$, which precludes the desired identity $TAT^{-1}=B$. $\quad\Box$
---
In fact, one can see that, if \eqref{1} holds for all $a,b\_1,b\_2,c\_1,c\_2$, then necessarily $T=0$.
| 2 | https://mathoverflow.net/users/36721 | 444967 | 179,387 |
https://mathoverflow.net/questions/444970 | 0 | Let $\{ g\_n \} $ be a frame in a separable Hilbert space $H$. Then the frame operator $S:H\to H$ defined as
\begin{equation} S f := \sum\_{n=1}^\infty (f,g\_n)g\_n \end{equation}
is a Hilbert space isomorphism i.e., continuous and bijective. Then one can define the canonica dual frame of $g\_n$ as the frame $S^{-1}g\_n=:f\_n$. It is quite straightforward to prove that this is in fact a frame. Where I am having trouble is the following claim from the artile "[Revisiting Landauʼs density theorems for Paley–Wiener spaces](https://reader.elsevier.com/reader/sd/pii/S1631073X12001392?token=B133329B64EF5D9474F0A08D0637E56463A5D50C211624546E3F3CCE96078C75BF90A4FFE8AD8C78552964F1BA66CE30&originRegion=eu-west-1&originCreation=20230417084448)" in page 5 the authors claim that furthermore $|\langle f\_n,g\_n \rangle | \leq 1 $. The reference that they give does not give a proof of this claim. A proof is given in Lemma 6.14 of "[What is variable bandwidth](https://onlinelibrary.wiley.com/doi/pdf/10.1002/cpa.21694)", but to me this (two lines) proof is not transparent.
I still think that the claim holds, but I was not able to prove it.
| https://mathoverflow.net/users/153260 | A property of the canonical dual frame in a Hilbert space | I assume your inner product is conjugate linear in the second factor, based on your assertion that the operator $S$ is a Hilbert space iso. Then
$$ \langle f\_n, g\_n\rangle = \langle f\_n, S f\_n \rangle = \langle f\_n, \sum\_k \langle f\_n, g\_k\rangle g\_k\rangle = \sum\_k |\langle f\_n, g\_k\rangle|^2 $$
is real and non-negative.
Subtract $|\langle f\_n,g\_n\rangle|^2$ from both sides, you find
$$ |\langle f\_n,g\_n\rangle| - |\langle f\_n,g\_n\rangle|^2\geq 0$$
which can only hold if $|\langle f\_n,g\_n\rangle| \leq 1$.
| 1 | https://mathoverflow.net/users/3948 | 444976 | 179,390 |
https://mathoverflow.net/questions/444950 | 3 | (The following definitions are meant to be standard and are reproduced for completeness of the question.) A **frame** is a partially ordered set in which every finite subset has a greatest lower bound (“meet”, denoted $x\_1\wedge\cdots\wedge x\_n$, or $\top$ for the empty meet) and every subset has a least upper bound (“join”, denoted $\bigvee\{x\_i : i\in I\}$), and such that the latter distributes over the former, which amounts to demanding $z \wedge \bigvee \{x\_i : i \in I\} = \bigvee\{(z\wedge x\_i) : i \in I\}$ (in fact, arbitrary meets exist in a frame, but they generally fail to distribute over joins). (An important source of examples of frames are the lattice of open sets on a topological space, in which case finite meets and arbitrary joins are simply finite intersections and arbitrary unions.) The **Heyting operation** on a frame $L$, denoted $\Rightarrow$, is defined as $(x\Rightarrow z) := \bigvee\{y : x\wedge y \leq z\}$ (this sup is, in fact, a max). A **nucleus** on a frame $L$ is a map $j\colon L\to L$ such that ① $j(x\wedge y) = j(x)\wedge j(y)$ (from which it follows that $j$ is order-preserving), ② $x \leq j(x)$, and ③ $j(j(x)) = j(x)$ (these three conditions are understood universally quantified over free variables in $L$). The set of nuclei on $L$, with the pointwise order (viꝫ., $j\_1\leq j\_2$ when $j\_1(x)\leq j\_2(x)$ for all $x\in L$) is itself a frame, denoted $N(L)$; its top element is $x \mapsto \top\_L$ and bottom $x \mapsto x$. Meet of nuclei can be defined pointwise: $(j\_1\wedge j\_2)(x) = j\_1(x) \wedge j\_2(x)$ (in fact, the same holds for arbitrary meets). Joins of nuclei, however, are more difficult to compute in general, although it is true that the fixset $L\_j := \operatorname{im} j = \{x\in L : j(x)=x\}$ of $j := \bigvee\{j\_i : i\in I\}$ is $\bigcap\_{i\in I} L\_{j\_i}$ (so $j(x)$ can be expressed as the smallest element $\geq x$ of $L$ which is fixed by every $j\_i$). See [Escardó, “Joins in the Frame of Nuclei” (2003)](https://doi.org/10.1023/A:1023555514029) for details and references about nuclei in general and how to compute their join.
My question concerns the computation of the Heyting operation in the frame $N(L)$ of nuclei over a frame $L$. So $j\_1\Rightarrow j\_2$ is defined as the largest nucleus $k$ such that $j\_1(x) \wedge k(x) \leq j\_2(x)$ for all $x$ (in particular, we have $(j\_1\Rightarrow j\_2)(x) \leq (j\_1(x) \Rightarrow j\_2(x))$). But this is not an equality in general.
To provide at least one interesting example of this, for an arbitrary element $a\in L$ we two naturally occurring nuclei, $j\_a\colon x\mapsto a\vee x$ (the “complementary closed sublocale” nucleus associated to $a$) and $j^a\colon x\mapsto (a\Rightarrow x)$. They turn out to be complementary in $N(L)$ (see, e.g., Fourman & Scott, “Sheaves in Logic”, p. 302–401 in Fourman, Mulvey & Scott, *Applications of Sheaves (Durham 1977)*, Springer LNM **753** (1979), §2.18), in the sense that $j\_a \vee j^a = \top\_{N(L)}$ and $j\_a \wedge j^a = \bot\_{N(L)}$; in particular, we have $j^a = (j\_a \Rightarrow \bot\_{N(L)})$, which is unsurprising, but also $j\_a = (j^a \Rightarrow \bot\_{N(L)})$, which is more surprising as the inequality $a\vee x \leq ((a\Rightarrow x) \Rightarrow x)$ is generally not an equality (the RHS here is not a nucleus in $x$, although it *is* a nucleus in $a$ which caused me some level of confusion).
**Question:** How can we compute the value $(j\_1\Rightarrow j\_2)(x)$ of the Heyting operation for $j\_1,j\_2 \in N(L)$ two nuclei (and $x\in L$)? Or how can we, at least, say something beyond the inequality $(j\_1\Rightarrow j\_2)(x) \leq (j\_1(x) \Rightarrow j\_2(x))$ remarked above? Pretty much anything that can be said about $(j\_1\Rightarrow j\_2)(x)$ interests me, including the special cases $j\_2 = \bot\_{N(L)}$, and/or when $L$ is the lattice of open sets on a topological space $X$ and $j\_1,j\_2$ are defined by reflecting to open subsets of subspaces $E\_1,E\_2 \subseteq X$.
| https://mathoverflow.net/users/17064 | Computing the Heyting operation on the frame of nuclei | First, every nucleus is the join of those of the form $j^x\land j\_y$.
Namely,
$$
j=\bigvee\_xj^x\land j\_{jx}
$$
Hm, this is too confusing. Let me change notation and write: $u\_a$ instead of $j^a$; $v\_a$ instead of $j\_a$. Thus,
$$
j=\bigvee\_xu\_x\land v\_{jx}.
$$
It then follows
$$
j\_1\Rightarrow j\_2=\bigwedge\_x[(u\_x\land v\_{j\_1x})\Rightarrow j\_2].
$$
Next, since $u\_x$ and $v\_x$ are complements of each other, $u\_x\Rightarrow j=v\_x\lor j$ and $v\_x\Rightarrow j=u\_x\lor j$. Thus $[u\_x\Rightarrow j]a=j(a\lor x)$ and $[v\_x\Rightarrow j]a=x\Rightarrow ja$ for every $a$, $x$ and every nucleus $j$.
To show that $j(x\lor-)$ is a nucleus,
$$
j(x\lor j(x\lor a))=jj(x\lor a)=j(x\lor a)
$$
since $x\leqslant jx\leqslant j(x\lor a)$.
And, to show that $x\Rightarrow j-$ is a nucleus,
$$
x\Rightarrow j(x\Rightarrow ja)=x\Rightarrow(x\Rightarrow ja)=x\Rightarrow ja
$$
since $L\_j$ is an exponential ideal, i. e. $b\in L\_j$ implies $x\Rightarrow b\in L\_j$ for any $x$.
This gives
$$
[j\_1\Rightarrow j\_2]a=\bigwedge\_x j\_1x\Rightarrow j\_2(a\lor x),
$$
which can be also replaced by
$$
\tag{\*}
[j\_1\Rightarrow j\_2]a=\bigwedge\_{x\geqslant a}j\_1x\Rightarrow j\_2x,
$$
as well as
$$
[j\_1\Rightarrow j\_2]a=\bigwedge\_x(a\Rightarrow x)\Rightarrow(j\_1x\Rightarrow j\_2x).
$$
One can also use that every nucleus is a meet of nuclei of the form $w\_x$, where $w\_x(a)=(a\Rightarrow x)\Rightarrow x$. Namely,
$$
j=\bigwedge\_{x\in L\_j}w\_x.
$$
Using this, we can further replace $(\*)$ with
$$
[j\_1\Rightarrow j\_2]a=\bigwedge\_{x\geqslant a,y\in L\_{j\_2}}j\_1x\Rightarrow((a\Rightarrow y)\Rightarrow y)$$
or as well with
$$
[j\_1\Rightarrow j\_2]a=\bigwedge\_{x\geqslant a,y\in L}j\_1x\Rightarrow((a\Rightarrow j\_2y)\Rightarrow j\_2y).$$
The version $(\*)$ is the most concise but sometimes other versions might be more useful.
For references - all this must be in the book by Picado and Pultr, for example.
| 1 | https://mathoverflow.net/users/41291 | 444977 | 179,391 |
https://mathoverflow.net/questions/444964 | 2 | Do there exist integers $(x,y,z)\neq (0,0,0)$ such that
$$
(a) \quad 2x^5+3y^5=6z^5
$$
$$
(b) \quad x^5+3y^5=7z^5
$$
Here is a short motivation. Equation $ax^d + by^d=cz^d$ is trivial for $d=1$, solved by Lagrange for $d=2$, investigated in depth by Selmer <https://doi.org/10.1007/BF02395746> for $d=3$. For $d=4$, at least the case $a=b=1$ is well-studied, see e.g. Section 6.6 of Cohen's famous book "Tools and Diophantine Equations, Vol. 1". However, I cannot find any references about the case $d=5$. The listed equations are the smallest ones with $d=5$ for which I cannot find neither small solution $(x,y,z)\neq (0,0,0)$ nor local obstructions modulo small $p$ (not sure for until which $p$ I should check), hence the question. If you can point me any references for the case $d\geq 5$ (or $d=4$ with general $a,b,c$), this would also be interesting.
| https://mathoverflow.net/users/89064 | Existence of rational points on generalized Fermat quintics | Both curves have no rational points.
Curves of this shape map to genus 2 curves of the form $y^2 = a x^5 + b$
(one can make $a = 1$ if one likes), by quotienting out by the group
of automorphisms generated by $(x:y:z) \mapsto (\zeta x:\zeta^{-1} y:z)$
where $\zeta$ is a primitive fifth root of unity (one can pick any two
coordinates for the scaling to act on). For curves of this type, Magma can
fairly easily compute the Mordell-Weil group (group of rational points
on the Jacobian variety). In both cases, for the first curve I tried
(using the quotient as above) the Mordell-Weil group has rank 1.
One can then use Chabauty's method (also implemented in Magma) to determine
the set of rational points on the genus 2 curve. In each case, there
are three points, but they do not lift to rational points on the given
curve.
(EDIT:) Using the quotients via the action on $x$ and $z$ leads to
curves whose Jacobians have finite Mordell-Weil groups, leading
to a simpler computation.
The three points one finds on the quotient curves correspond
to $x = 0$, $y = 0$ and $z = 0$. So if these are the only points
there, then one can easily check that there are no rational
points on the original curves.
| 11 | https://mathoverflow.net/users/21146 | 444982 | 179,393 |
https://mathoverflow.net/questions/444951 | 3 | I have a question regarding a conjecture due to H. L. Montgomery on the number of primes in short intervals. The conjecture apparently arises from probabilistic reasoning upon assuming the Riemann Hypothesis and some statistical randomness across the ordinates $\gamma$ of the nontrivial zeros of the Riemann zeta function. Let $\psi(x)=\sum\_{n\le x}\Lambda(n)$ denote the (second) Chebyshev function. Then (considering only $\varepsilon>0$), he makes the following
**Conjecture. (Montgomery)**
$$\psi(x+h)-\psi(x)=h+O\_{\varepsilon}(h^{\frac{1}{2}}x^{\varepsilon})$$
for $2\le h\le x$.
**My question: Which heuristic arguments support this conjecture (excluding numerical verification)?**
The conjecture appears in several places:
(A) H. L. Montgomery, "Problems concerning prime numbers", Proceedings of symposia in pure mathematics, Vol. XXVIII, pp.307-310, Mathematical developments arising from Hilbert Problems (1976), AMS. Providence, Rhode island. [See p.309]
(B) D. A. Goldston, "On a result of Littlewood concerning prime numbers", Acta Arithmetica, Vol.40 (1982), pp. 263-271. [See p.269]
(C) H. L. Montgomery, R. C. Vaughan, Multiplicative Number Theory I. Classical Theory, Cambridge University Press (2007) [Conjecture 13.4, p.422]
**Following the argument of Montgomery and Vaughan (p.422),**
is something I manage up to a certain point, but I'm not sure how to "get there". Specifically, using the explicit formula for $\psi(x)$ on the form
$$\psi(x)=x-\sum\_{\rho}\frac{x^{\rho}}{\rho}-\frac{\zeta'}{\zeta}(0)-\frac{1}{2}\log(1-x^{-2})+\frac{1}{2}\Lambda(x) \hspace{5mm}(x>1),$$
we can write
$$\psi(x+h)-\psi(x)=h-\sum\_{|\gamma|\le T}C(\rho)+\lim\_{U\to \infty}\sum\_{T<|\gamma|\le U}C(\rho)+ O\big(\hspace{-0.2mm}\log\hspace{0.4mm}\max(x,h)\big)$$
for $x,h\ge 2$, say, where
$$C(\rho)=\frac{(x+h)^{\rho}-x^{\rho}}{\rho} \ll \min\Big(hx^{\beta-1}, \frac{x^{\beta}}{|\gamma|}\Big).$$
Now assume that the Riemann hypothesis is true, and write
\begin{align\*}C(\rho)=&\;\int\_{x}^{x+h}t^{\rho-1}dt=\int\_{x}^{x+h}x^{\rho-1}dt+\int\_{x}^{x+h}t^{\rho-1}-x^{\rho-1}dt\\
=&\; hx^{\rho-1}+\int\_{0}^{h}(x+t)^{\rho-1}-x^{\rho-1}dt\\
=&\; hx^{\rho-1}+\int\_{0}^{h}(\rho-1)\int\_{0}^{t}(x+z)^{\rho-2}dzdt\\[1mm]
=&\; hx^{-\frac{1}{2}+\gamma i}+O\big(h^2x^{-\frac{3}{2}}|\gamma|\big). \hspace{30mm} (\dagger)
\end{align\*}
Ignoring the error here for the moment and taking $T=x/h$, then
$$\sum\_{|\gamma|\le x/h}hx^{-\frac{1}{2}+\gamma i}=\frac{h}{\sqrt{x}}\sum\_{|\gamma|\le x/h}\text{e}^{\gamma \log(x)i}.$$
Now, if we were to replace the $\gamma \log x$-s with independent and identically distributed uniform random variables $(Y\_n)\_{n=1}^{\infty}$ on the interval $[0,2\pi)$ , then, as this [this post](https://mathoverflow.net/questions/89478/magnitude-of-the-sum-of-complex-i-u-d-random-variables-in-the-unit-circle)
shows, we could conjecture that the above sum behaves like
$$\frac{h}{\sqrt{x}}\sum\_{n\ll N(x/H)}\text{e}^{Y\_n i}\ll\_{\varepsilon} \frac{h}{\sqrt{x}}N\big(\frac{x}{h}\big)^{1/2+\varepsilon}\ll\_{\varepsilon} \frac{h}{\sqrt{x}}\Big(\frac{x}{h}\log\big(\frac{x}{h}\big)\Big)^{\frac{1}{2}+\varepsilon}=h^{\frac{1}{2}-\varepsilon}x^{\varepsilon}\big(\hspace{-0.2mm}\log \frac{x}{h}\big)^{\frac{1}{2}+\varepsilon},$$
as in Montgomery's conjecture.
This is the point where I get stuck. For indeed, there are two unresolved sizes here. The first is the contribution of the large zeros:
$$\sum\_{|\gamma|>x/h}C(\rho), \hspace{15mm} (\ddagger)$$
and the second is the contribution of the error term in $(\dagger)$. The contribution of the error in $(\dagger)$ may not be too hard to resolve, but I am quite unable to show that the sum $(\ddagger)$ is of order $O\_{\varepsilon}(h^{\frac{1}{2}}x^{\varepsilon})$. For example, using the explicit formula for $\psi(x)$ on the (commonly stated) form
$$\psi(x)=x-\sum\_{\rho}\frac{x^{\rho}}{\rho}+O\Big(\frac{x\log^2(xT)}{T}+\log x\Big) \hspace{4mm} (x,T\ge 2),$$
I obtain
$$\sum\_{|\gamma|>x/h}C(\rho)\ll \frac{(x+h)\log^2((x+h)\frac{x}{h})}{x/h}\ll h\log^2 x$$
if $2\le h\le x$, which is not $O\_{\varepsilon}(h^{\frac{1}{x}}x^{\varepsilon})$ unless $h\ll x^{2\varepsilon}(\log x)^{-4}$ (and this does not permit $2\le h\le x$ if, say, $0<\varepsilon<1/2$). The problem may be that the explicit formula with error term used here is unconditional, and that a better formula assuming RH should be employed.
Montgomery and Vaughan say about this (on p.422), that "The contribution of zeros with $|\gamma|>x/h$ can be attenuated by employing a smoother weight, but no amount of smoothing will eliminate the smaller zeros." By smoothing, they here likely mean that the explicit formula for $\sum\_{n\le x}\Lambda(n)=\sum\_{n=1}^{\infty}\Lambda(n)1\_{(x,x+h]}(n)$ should be replaced by an explicit formula for
$$\sum\_{n=1}^{\infty}\Lambda(n)w(n),$$
where $w(n)=w(n;x,h)$ is a `weight function'. This weight function should be such that it gives a useful explicit formula (meaning that the contribution of the $\rho$-s in the right hand side decays rapidly as a function of $|\gamma|$), but also approximates the indicator function $1\_{(x,x+h]}(n)$ to such an extent that $\sum\_{n}\Lambda(n)w(n)$ is close to $\psi(x+h)-\psi(x)$.
I have been playing around with such explicit formulas lately, including the formulas
\begin{align\*}
\frac{1}{k!}\sum\_{n\le x}\Lambda(n)(x-n)^{k}=&\; \frac{x^{k+1}}{(k+1)!}-\frac{x^{k}}{k!}\frac{\zeta'}{\zeta}(0)-\sum\_{\rho}\frac{x^{\rho+k}}{\rho(\rho+1)\cdots (\rho+k)}+ \sum\_{0\le j\le (k-1)/2}\frac{x^{k-2j-1}}{(2j+1)!(k-2j-1)!}\frac{\zeta'}{\zeta}(-2j-1)\\[1.5mm]
+&\; (-1)^{k}\sum\_{j>k/2}x^{k-2j}\frac{(2j-k-1)!}{(2j)!}+\sum\_{0<j\le k/2} \frac{x^{k-2j}}{(2j)!(k-2j)!}\Big(\frac{1}{2}\frac{\zeta''}{\zeta'}(-2j)-\log x+\sum\_{\substack{r=-2j\\ r\ne 0}}^{k-2j}r^{-1}\Big) \hspace{5mm} (x\ge 1, k\in \mathbb{N}^{+}), \\[2mm]
\frac{1}{\Gamma(\xi+1)}\sum\_{n<x}\Lambda(n)(x-n)^{\xi}=&\; \frac{x^{\xi+1}}{\Gamma(\xi+2)}-\sum\_{\rho}\frac{x^{\rho+\xi}\Gamma(\rho)}{\Gamma(\rho+\xi+1)}-\frac{x^{\xi}}{\Gamma(\xi+1)}\frac{\zeta'}{\zeta}(0)+\sum\_{j=0}^{\infty} \frac{x^{\xi-2j-1}}{\Gamma(2j+2)\Gamma(\xi-2j+1)}\cdot \frac{\zeta'}{\zeta}(-2j-1)\\
-&\;\sum\_{j=1}^{\infty} \frac{x^{\xi-2j}}{\Gamma(2j+1)\Gamma(\xi-2j+1)}\Big(-C\_{\text{Eul}}+\sum\_{k=1}^{2j}\frac{1}{k}+\frac{1}{2}\frac{\zeta''}{\zeta'}(-2j)+\log(x)-\psi^{(0)}(\xi-2j+1)\Big) \hspace{5mm} (x\ge 1, \text{Re }\xi>0, \xi \not \in \mathbb{Z}), \\[2mm]
\sum\_{n\le x}\Lambda(n)\log(x/n)=&\;x-\sum\_{\rho}\frac{x^{\rho}}{\rho}-(\log 2\pi)\log x-(\frac{\zeta'}{\zeta})'(0)-\frac{1}{4}\sum\_{k=1}^{\infty}\frac{x^{-2k}}{k^2} \hspace{5mm} (x>1),\\[2mm]
\sum\_{n=1}^{\infty}\Lambda(n)\text{e}^{-n/z}=&\; z-\sum\_{\rho}\Gamma(\rho)z^{\rho}-\text{e}^{-1/z}\log 2\pi -(-1+\cosh 1/z)\log z+\sum\_{k=1}^{\infty}(-1)^{k}\frac{\zeta'}{\zeta}(k+1)\frac{z^{-k}}{k!}\\
-&\;\sum\_{k=0}^{\infty}\frac{\Gamma'}{\Gamma}(2k+2)\frac{z^{-2k-1}}{(2k+1)!} \hspace{5mm} (\text{Re }z>0).
\end{align\*}
However, I am not able to get the desired result. Indeed, I am not able to combine the weight functions provided in the formulas above, to get a weight function approximating $1\_{(x,x+h]}(n)$ to a reasonable extent while giving a reasonable explicit formula. This may be because:
1. I have not found any literature explaining what would constitute a good smoothing of $1\_{(x,x+h]}(n)$ in this case (i.e. how does the smoothness and cut-off play a role)
2. Taking a weighted sum of one of the explicit formulas above, one could likely approximate $1\_{(x,x+h]}(n)$ by expressions of the form $\sum\_{n=1}^{N}a\_nw(n,x\_n,h\_n)$. Here I am not at all sure which function spaces I would try to do an approximation in (e.g. which norm).
These investigations are related to my Master's thesis on primes in short intervals, and I
would highly appreciate if anyone could comment on how Montgomery's conjecture can be backed up. (And possibly also what the goal/strategy of the smoothing process should be, if you would be kind enough to explain).
Sincerely, R.
| https://mathoverflow.net/users/153908 | What heuristic arguments support Montgomery's conjecture for primes in short intervals? | The motivation for many conjectures about the distribution of primes $p\in[x,x+h]$ in intervals of length $o(\sqrt{x}\log x)$ comes from studying the mean square of $|\psi(x+h)-\psi(x)-h|$. For example, for $\theta\in[0,1]$, Selberg considered the integral
$$J(x,\theta):= \frac{1}{x}\int\_{x}^{2x}|\psi(t+\theta t)-\psi(t)-\theta t|^2 dt.$$
Selberg proved that RH suffices to conclude that if $1/x\leq\theta\leq x$, then
$$J(x,\theta) \ll x\theta (\log (2/\theta))^2,$$
which means that if $x$ is large and $1/x\leq \theta\leq x$, then for all $t\in[x,2x]$ outside of a subset of Lebesgue measure $o(x)$, we have
$$\psi(t+\theta t)-\psi(t)=\theta t +O(\sqrt{\theta x}\log x).$$
One might now conjecture that the subset of Lebesgue measure $o(x)$ is in fact empty. However, this contradicts a result of Maier, who used the Maier matrix method to prove that if $A>1$, then
$$\limsup\_{x\to\infty}\frac{\psi(x+(\log x)^A)-\psi(x)}{(\log x)^{A}}>1$$
and
$$\liminf\_{x\to\infty}\frac{\psi(x+(\log x)^A)-\psi(x)}{(\log x)^{A}}<1.$$
So what is the correct minimal length in which we obtain an asymptotic prime number theorem? Is it $\exp((\log x)^{1/3})$? $\exp((\log x)^{1/2})$? $\exp((\log x)^{3/4})$? There is no "scientific" way to determining what the right threshold is (at least that I am aware of), so at this point we conjecture that whatever the correct threshold should be (that is, the least permissible value of $\theta x$), it should still be less than $x^{\epsilon}$ for any fixed $\epsilon>0$. This is an alternate formulation of Montgomery's conjecture in the original post.
| 4 | https://mathoverflow.net/users/111215 | 444993 | 179,395 |
https://mathoverflow.net/questions/444996 | 18 | I recently learnt that the proof of the classical theorem "$\mathsf{AD}$ $\implies$ $\aleph\_1$ is measurable" uses computability theory tools (or at least one of its proofs does so). I'm interested in more examples of this. More precisely:
1. What are some other well-known results in set theory that use non-trivial tools from computability theory to prove?
2. What are some well-known results in computability theory that use non-trivial tools in set theory to prove?
I'm more interested in classical results, but modern examples of such are also welcome. If possible, I would also like a reference (textbook/paper) for each example.
| https://mathoverflow.net/users/146831 | Theorems in set theory that use computability theory tools, and vice versa | Here are several examples.
* There is a natural affinity between forcing and many constructions in the Turing degrees. Specifically, many constructions of degrees by meeting requirements in succession can be seen as meeting dense sets in a suitable forcing notion, with the result that the computability construction amounts to a genericity argument. For example, to prove there are incomparable degrees, one should build the oracles by initial segment (so, Cohen forcing), and extend them generically in such a manner that neither is computable relative to the other as oracle. There are dozens of examples like this, and I recommend the forcing manner of understanding such constructions in computability. This is similar to the forcing genericity way of looking at the Fraïssé limit of a class of finite structures.
* Another example of this type: there must be strongly independent antichains in the degrees (no element is computable from the sum of all the others), since if you add infinitely many Cohen reals, they are like that, and the assertion that it exists is $\Sigma^1\_1$, hence absolute. From such an antichain, you get universality as an easy consequence — every countable order embeds into the degrees.
* The previous analogy is very strong in certain cases, and was historically the way the results were proved. For example, Sacks construction of a minimal degree in the Turing degrees translates directly to the proof of the minimality feature of Sacks forcing.
* The concept of Turing degrees is directly analogous to many other degree notions in set theory. For example, many arguments translate to the degrees of constructibility, where $x\sim y$ iff $L[x]=L[y]$. For example, the role of iterated Sacks forcing has been used (especially in work of Marcia Groszek and others) to produce models of set theory realizing a specified structure for the degrees of constructibility. This method also appears in my paper with Groszek on the [Implicitly constructible universe](https://arxiv.org/abs/1702.07947). And similarly with the arithmetic and hyperarithmetic degrees.
* The notion of computability is generalized to hypercomputation and E-recursion, in a way that is deeply connected with descriptive set theory.
* The notion of computability is also generalized by [Infinite time Turing machines](https://www.jstor.org/stable/2586556), which is also deeply connected with descriptive set theory. That notion in turn is generalized by ordinal computation, which provides an alternative presentation of the L-hierarchy in computability-theoretic terms.
* In my paper, [Forcing as a computational process](https://arxiv.org/abs/2007.00418), my co-authors and I investigate the computable model theory of forcing, looking into how one can compute a forcing extension from an oracle for a given model.
* There is a fun argument of mine for the nonlinearity of Turing degrees using set theory: if the degrees are linear, then since every initial segment is countable, the cofinality would have to be $\omega\_1$, but since there are continuum many degrees, this implies CH. But CH fails in a forcing extension. The assertion that there are incomparable degrees is $\Sigma^1\_1$, hence absolute. So there must be incomparable degrees in the ground model.
* Another argument, which I was just reminded of by Jason Chen (but I think it appears elsewhere here on MO): the Turing degree relation $x\leq\_T y$ is arithmetically definable and hence Borel, and since it has countable sections, it must have measure 0. The inverse relation also therefore has measure 0. So there is a measure one set of pairs $(x,y)$ not related by relative computability. So the degrees are nonlinear.
| 21 | https://mathoverflow.net/users/1946 | 444997 | 179,397 |
https://mathoverflow.net/questions/444687 | 5 | I am sorry that the following question is elementary. I have not received an answer from my post at [Math Stack Exchange](https://math.stackexchange.com/questions/4676078/is-the-interior-of-the-tensor-product-of-two-convex-cones-equal-to-the-tensor-pr).
In the following question, all cones are convex and contain the origin. Let $C \subset \mathbb{R}^{m}$ be a cone consisting of $m$-dimensional column vectors, and $D \subset \mathbb{R}^{n}$ another cone of $n$-dimensional column vectors.
Given any subsets $A \subset \mathbb{R}^{m}$ and $B\subset \mathbb{R}^{n}$, define $A \otimes B \subset \mathbb{R}^{m \times n}$ as the set of $m \times n$ matrices of the form $x\_1 {y\_1}^T + \dotsb + x\_q {y\_q}^T$, where $x\_1, \dotsc, x\_q \in A$ and $y\_1, \dotsc, y\_q \in B$, $q\geq 1$ is an arbitrary positive integer, and $^T$ denotes the transpose.
Is it true that
$$(C \otimes D)^\circ = C^\circ \otimes D^\circ,$$
where $^\circ$ denotes the interior relative to the respective euclidean topologies?
---
Here's an example where the said equality holds. Take $C = {\mathbb{R}\_{\ge 0}}^m$ to be the cone of all $m$-dimensional column vectors whose every entry is a nonnegative real number. Similarly, take $D = {\mathbb{R}\_{\ge 0}}^n$. Then $C \otimes D = {\mathbb{R}\_{\ge 0}}^{m \times n}$ is the cone of $m \times n$ matrices whose every entry is nonnegative. Indeed, it suffices to show that $\mathbf{E}\_{ij}$, the matrix whose $(i, j)$th entry is $1$ and has $0$s everywhere else, lies in $C \otimes D$, because $C \otimes D$ is closed under taking nonnegative linear combinations. But $\mathbf{E}\_{ij} = \vec{e\_i}\vec{e\_j}^T$, where $\vec{e\_i} \in C$ and $\vec{e\_j} \in D$. Therefore $(C \otimes D)^\circ = {\mathbb{R}\_{> 0}}^{m \times n}$ is the set of all $m \times n$ matrices whose every entry is strictly positive.
Noting that $C^\circ = {\mathbb{R}\_{> 0}}^m$ consists of all $m$-dimensional vectors whose entries are all strictly positive real numbers, and similarly $D^\circ = {\mathbb{R}\_{> 0}}^n$, we get $C^\circ \otimes D^\circ = {\mathbb{R}\_{> 0}}^{m \times n}$ also.
| https://mathoverflow.net/users/136356 | Is the interior of the tensor product of two convex cones equal to the tensor product of their respective interiors? | Yes, $(C \otimes D)^\circ = C^\circ \otimes D^\circ$ is correct.
Let us start by proving the inclusion $C^\circ \otimes D^\circ \subseteq (C \otimes D)^\circ$. To this end, it is enough to show that $C^\circ \otimes D^\circ$ is open. For a given point $z = x\_1 {y\_1}^T + \dotsb + x\_q {y\_q}^T \in C^\circ \otimes D^\circ$, we can use convexity and openness to replace $x\_1$ by a convex combination of $n$ linearly independent points in $C$, and similarly for $y\_1$. In this way, we can assume without loss of generality that the $x\_i$'s span $\mathbb{R}^m$, and similarly the $y\_i$'s span $\mathbb{R}^n$. Therefore by nudging each one of these points a little (without leaving $C^\circ$ or $D^\circ$), we can move in any direction in $C^\circ \otimes D^\circ$, and it follows that the latter set is open.
Furthermore, the inclusion $C^\circ \otimes D^\circ \subseteq (C \otimes D)^\circ$ is dense, because $C^\circ \otimes D^\circ$ is dense even in $C \otimes D$ by the obvious termwise approximation argument.
To now arrive at the claimed equality, recall that [every convex open set is the interior of its closure](https://math.stackexchange.com/questions/360187/is-convex-open-set-in-mathbbrn-is-regular). This applies in particular to both $C^\circ \otimes D^\circ$ and $(C \otimes D)^\circ$. This gives
$$
C^\circ \otimes D^\circ = \overline{C^\circ \otimes D^\circ}^\circ = \overline{(C \otimes D)^\circ}^\circ = (C \otimes D)^\circ,
$$
where the second step is the density from the previous paragraph.
| 3 | https://mathoverflow.net/users/27013 | 445012 | 179,401 |
https://mathoverflow.net/questions/444989 | 2 | I'm working on a problem about Artin groups, and to simplify this problem I want to take a quotient that allow us to go to an easier Artin group, but I'm not sure if the quotient is well defined. This is the statement of my problem:
Let $A\_\Gamma$ be an Artin group, where $\Gamma$ is a complete graph, $f:A\_\Gamma\to\mathbb{Z}$ a group homomorphism and $a\in V(\Gamma)$ such that $f(a)=1$. The artin group is generated by the vertices of the graph, and there are two types of generators: the $v\in V(\Gamma)$ such that $f(v)=0$ and the $v\in V(\Gamma)$ such that $f(v)\neq0$. My objective is to define a normal subgroup $N\trianglelefteq A\_\Gamma$ such that $A\_\Gamma/N\cong A\_{\Gamma'}$ where $\Gamma'\subset\Gamma$ is the induced subgraph by the $v\in V(\Gamma)$ such that $f(v)=0$ and $a$, i.e. we want a new Artin group given by colapsing all vertices with non-zero image to $a$. To do so, I wanted to define $N$ as the normal subgroup generated by the set:
$$\lbrace v^{-1}a^{f(v)} \mid v\in V(\Gamma),f(v)\neq0\rbrace$$
(Remark: I $N\leq\ker(f)$ if possible, that is why I need to define those ''weird'' generators instead of $v^{-1}a$)
It is obvious that in the quotient $A\_\Gamma/N$ all the $v\in V(\Gamma)$ with $f(v)\neq 0$ are identified with some power of $a$, which allow us to get rid of those generators. However, I can't see if this quotient is indeed the Artin group I want or sth different.
Any hint will be thanked.
Edit: Inspired by @MoisheKohan I was able to see that I can reduce the problem to the right-angled case by first taking a quotient with the normal subgroup generated by the commutators of the generators. In this situation, since $\Gamma$ is a complete graph $A\_\Gamma\cong\mathbb{Z}^n$, where $n$ is the number of generators of $\mathbb{Z}$, so the problem should be easier.
| https://mathoverflow.net/users/482329 | Quotient of an Artin group is an Artin group | (Edit: Missed that the graph should be complete, now it is.)
If I'm understanding all this correctly, the answer is "no". Take three vertices, $a$, $b$, and $c$. Take an edge labeled 4 from $a$ to $b$, an edge labeled 4 from $a$ to $c$, and an edge labeled 2 from $b$ to $c$. Take $f$ to be $f(a)=f(b)=1$, $f(c)=0$. So now you're modding out that $a$ and $b$ get identified. In this quotient, $a$ and $c$ suddenly commute, which is not what the induced subgraph $\Gamma'$ wants (it's $a$ and $c$ with an edge labeled 4). If you want to phrase it in terms of quotient maps instead of normal subgroups, the problem is that the quotient map isn't well defined: you have that $b$ and $c$ commute, but their images are $a$ and $c$, which don't commute.
| 3 | https://mathoverflow.net/users/164670 | 445018 | 179,403 |
https://mathoverflow.net/questions/444949 | 1 | **Question**
Given a random field $X(t)$ where the parameter space $T\subset\mathbb{R}\_N$. Is there result regarding the concentration of the random field? For example
$\mathbb{P}\{\|X(t)-\mathbb{E}\{X(t)\}\|\geq t\}\leq \text{small}$ where $\|\cdot\|$ is certain norm, maybe $L\_p$ norm.
**Take the following random function as an example**
$f(x):=f(x\_1,x\_2,x\_3)=a\_1\sin(x\_1-x\_2)+a\_2\sin(x\_2-x\_3)$, where $a\_1=1+w\_1,a\_2=1+w\_2)$ and $w\_1,w\_2$ are independent standard Gaussian scalar. The coefficients $a\_1,a\_2$ are independent, and $x\_1,x\_2\in\mathbb{R}$.
The expectation of the random polynomial is $\mathbb{E}f(x)=\sin(x\_1-x\_2)+\sin(x\_2-x\_3)$, which is not random.
Then my question is the concentration of random function to its expectation $$\mathbb{P}\{\|a\_1\sin(x\_1-x\_2)+a\_2\sin(x\_2-x\_3)-\big(\sin(x\_1-x\_2)+\sin(x\_2-x\_3)\big)\|\_{L\_p}\geq t\}\leq \text{small}$$
**What I tried**
I read the book *random fields and geometry - Robert J. Adler, Jonathan E. Taylor*, and found it focus on (at least in the chapter 2) the exeedence probability
$$\mathbb{P}\{\sup\_{t\in T}f(t)\geq u\}$$
I did not find concentration properties of random field in this book and other materials on random fields. I am not sure whether it is because I did not recognize them or there actually have no result regarding this type.
Anyone could help? Thanks!
| https://mathoverflow.net/users/147940 | concentration of random field to its expectation function | $\newcommand\R{\mathbb R}$For real $x\_1,x\_2,x\_3$, let
\begin{equation}
g(x\_1,x\_2,x\_3):=\sin(x\_1-x\_2),\quad h(x\_1,x\_2,x\_3):=\sin(x\_2-x\_3).
\end{equation}
The request is to bound
\begin{equation}
P(Y\ge t)
\end{equation}
from above, where
\begin{equation}
Y:=\|w\_1 g+w\_2 h\|\_{L^p}
\end{equation}
and $w\_1,w\_2$ are independent standard normal random variables.
If $\|\cdot\|\_{L^p}$ is understood here as the standard norm on $L^p(\R^3)$, then $Y=\infty$ with probability $1$. This follows because, say, (i) $h(x\_1,x\_2,x\_3)$ is nonzero, continuous, and periodic in $x\_3$; (ii) the function $|\cdot|^p$ is continuous and has nonzero variation on any interval of nonzero length; and (iii) $P(w\_2\ne0)=1$.
So, for the problem to make sense, assume that the norm $\|\cdot\|\_{L^p}$ is the standard norm on $L^p(T)$, where $T$ is a measurable subset of $\R^3$ such that $G:=\|g\|\_{L^p(T)}\in(0,\infty)$ and $H:=\|h\|\_{L^p(T)}\in(0,\infty)$.
Then, by Minkowski's inequality, for $p\in[1,\infty]$,
\begin{equation}
\begin{aligned}
P(Y\ge t)&\le P(G|w\_1|+H|w\_2|\ge t) \\
&\le P(G|w\_1|\ge\tfrac G{G+H}\,t)+P(H|w\_2|\ge\tfrac H{G+H}\,t) \\
&=2P(|w\_1|\ge\tfrac t{G+H}) \\
&\le2\exp(-\tfrac{t^2}{2(G+H)^2})
\end{aligned}
\end{equation}
for real $t\ge0$.
---
It was assumed above that both $G$ and $H$ are $>0$.
If exactly one of the norms $G$ and $H$ is $0$ (and still $G<\infty$ and $H<\infty$), then, similarly, $P(Y\ge t)\le\exp(-\tfrac{t^2}{2(G+H)^2})$ for real $t\ge0$. If $G=H=0$, then, clearly, $P(Y\ge t)=0$ for real $t>0$. So, when the condition that $G>0$ and $H>0$ does not hold, we have better upper bounds than $2\exp(-\tfrac{t^2}{2(G+H)^2})$.
---
Suppose now that
\begin{equation}
Y:=\|w\_1 g\_1+\cdots+w\_n g\_n\|\_{L^p},
\end{equation}
where $w\_1,\dots,w\_n$ are independent standard normal random variables and the $g\_j$'s are deterministic functions in $L^p$ with $G:=\max\_j\|g\_j\|\_{L^p}\in(0,\infty)$. Then for $p\in[2,\infty)$, by [Theorem 3.3](https://projecteuclid.org/journals/annals-of-probability/volume-22/issue-4/Optimum-Bounds-for-the-Distributions-of-Martingales-in-Banach-Spaces/10.1214/aop/1176988477.full) (with $\Gamma=1$; see also [typo correction](https://projecteuclid.org/journals/annals-of-probability/volume-27/issue-4/Correction--Optimum-bounds-for-the-distributions-of-martingales-in/10.1214/aop/1022874833.full)) and [Proposition 2.1](https://projecteuclid.org/journals/annals-of-probability/volume-22/issue-4/Optimum-Bounds-for-the-Distributions-of-Martingales-in-Banach-Spaces/10.1214/aop/1176988477.full), we have the following Bernstein-type bound:
\begin{equation}
P(Y\ge t)\le2\exp\Big(-\frac{t^2}{B^2+t+B\sqrt{B^2+2t}}\Big)
\end{equation}
for real $t\ge0$, where $B:=\sqrt{(p-1)G^2 n}$. In particular, if $t=O(\sqrt n)$ or, more generally, $t=o(n)$ as $n\to\infty$, then
\begin{equation}
P(Y\ge t)\le2\exp\Big(-\frac{t^2}{(2+o(1))B^2}\Big)
=2\exp\Big(-\frac{t^2}{(2+o(1))(p-1)G^2 n}\Big),
\end{equation}
so that we have a normal-type upper bound on $P(Y\ge t)$ with an asymptotically correct constant factor $2+o(1)$ in the denominator of exponent (consider the special case when the $g\_j$ is the constant $1$ on (say) the cube $[0,1]^d$, so that one may take here $p=2$).
| 1 | https://mathoverflow.net/users/36721 | 445031 | 179,407 |
https://mathoverflow.net/questions/445028 | 2 | Is it posible to find the article "Pósa Lajos, A prímszámok egy tulajdonságáról (Hungarian), Mat. Lapok 11 (1960), 124-129" on the internet?
| https://mathoverflow.net/users/169583 | Pósa's inequality for a product of consecutive primes | Yes, here is the archive of [Matematikai Lapok](http://real-j.mtak.hu/view/journal/Matematikai_Lapok.html).
| 2 | https://mathoverflow.net/users/11919 | 445041 | 179,409 |
https://mathoverflow.net/questions/445044 | 2 | Given an integer $n\geq2$, consider the following integral
$$I\_n:=\int\_0^1nx^{n-1}\sqrt{\left\vert \frac{\log(1-x)}{\log n}\right\vert} \, dx.$$
**QUESTION.** Is this true? It appears to be so.
$$\lim\_{n\rightarrow\infty}I\_n=1.$$
| https://mathoverflow.net/users/66131 | Asymptotics of an integral requested | Change variables to $y=n(1-x)$, which gives
$$
I\_n=\int\limits\_{0}^{n}\left(
1-\frac{y}{n}\right)^{n-1}
\sqrt{1-\frac{\log y}{\log n}}\ {\rm d}y\ ,
$$
then dominated convergence should allow passing to the limit inside the integral
so that
$$
\lim\_{n\rightarrow \infty} I\_n=\int\limits\_{0}^{\infty} e^{-y}\ {\rm d}y=1\ .
$$
| 9 | https://mathoverflow.net/users/7410 | 445048 | 179,412 |
https://mathoverflow.net/questions/445043 | 6 | Let $u\in L^1(\mathbb{S}^{d-1})$. I want to show that
\begin{align\*}
\lim\_{|\xi|\to \infty}
\int\_{\mathbb{S}^{d-1}}(1-\cos(\xi\cdot w))u(w)d \sigma\_{d-1}(w)
= \int\_{\mathbb{S}^{d-1}}u(w)d \sigma\_{d-1}(w).
\end{align\*}
Basically, the question can be reduced into showing that
\begin{align\*}
\lim\_{|\xi|\to \infty}
\int\_{\mathbb{S}^{d-1}}\cos(\xi\cdot w)u(w)d \sigma\_{d-1}(w)
= 0.
\end{align\*}
This looks like a Riemann-Lebesgue lemma. But I don't know how to tackle it.
I intuitively guessed this from the classical Riemann-Lebesgue Lemma which infers that
\begin{align\*}
\lim\_{|\xi|\to \infty}
\int\_{B\_1(0)}(1-\cos(|\xi| z\cdot x))u(x)d x
= \int\_{B\_1(0)}u(x)dx\quad \text{for fixed $z\in \Bbb R^d$}.
\end{align\*}
More generally, if $f$ is $T^d$-periodic, then $f\_\lambda(x)= f(\lambda x)$ weakly converge in $L^p$ to its mean value as $\lambda\to\infty$ that is
$$f\_\lambda \rightharpoonup \bar f,\quad \quad \bar f=\frac{1}{T^d}\int\_{[0,T]^d}f(x) dx.$$
Is there any good reference for this type of limit?
Any help is welcome
| https://mathoverflow.net/users/112207 | Computing a limit on the unit sphere: Riemann Lebesgue? | The key fact here is the (surprising, initially, but well known) (power) decay of $\widehat{\sigma}(\xi)$. If $u\in C^{\infty}(S)$, we can extend to a function $u\_0\in C^{\infty}\_0(\mathbb R^d)$, and then
$$
\widehat{u\,d\sigma}=\widehat{u\_0\,d\sigma}=\widehat{u\_0}\*\widehat{\sigma}
$$
still decays. See [here](https://en.wikipedia.org/wiki/Convolution_theorem#Convolution_theorem_for_tempered_distributions) for the general version of the convolution theorem needed here.
We can then extend this to arbitrary $u\in L^1$ by the argument from the traditional Riemann-Lebesgue lemma: given $\epsilon>0$, pick a $v\in C^{\infty}(S)$ with $\|u-v\|\_1<\epsilon$. Since $|\widehat{u\, d\sigma}(\xi) -\widehat{v\, d\sigma}(\xi)|<\epsilon$ and $\widehat{v\, d\sigma}\to 0$, we also have $|\widehat{u\, d\sigma}(\xi)|<2\epsilon$ for all large $\xi$.
| 9 | https://mathoverflow.net/users/48839 | 445055 | 179,414 |
https://mathoverflow.net/questions/445058 | 3 | Consider an ideal $I = \langle f\_1,\dotsc,f\_n\rangle$ in the ring $k[x\_1,\dotsc,x\_m]$.
Define the $i$-th elimination ideal to be $I\_i = I \cap k[x\_{i+1},\dotsc,x\_m]$.
For any two polynomials $f$ and $g$ in $I$, the resultant $\operatorname{Res}(f,g,x\_1)$ belongs to the first elimination ideal $I\_1$.
1. Is there a way to represent $I\_1$ as $\langle r\_1,\dotsc,r\_l\rangle$, where the basis $r\_j$ are obtained as resultants of polynomials in $I$ with respect to $x\_1$?
2. Is there a way to represent each elimination ideal in this way? (I imagine by iterating the possible representation of point 1?)
| https://mathoverflow.net/users/48526 | Resultants and elimination theory | A bit long for a comment, so posted as an answer, although this is really a comment.
For 1) I would rather try the following. Assuming the $f$'s have the same degree in $x\_1$,
introduce formal variables $s\_1,\dotsc,s\_n$ and $t\_1,\dotsc,t\_n$, and take the binary resultant $\operatorname{Res}(s\_1f\_1+\dotsb+s\_nf\_n,t\_1f\_1+\dotsb+t\_nf\_n,x\_1)$. This is a monstrous polynomial in the $s,t$ variables. I think the coefficients of this polynomial should be in $I\_1$ and may give a basis. If the $f$'s don't have the same degree, you may have to make two copies of them, multiply one copy by powers of $x\_1$ and the other copy by powers of $1-x\_1$, so they all have the same degree, then do the above with $2n$ instead of $n$.
| 3 | https://mathoverflow.net/users/7410 | 445061 | 179,416 |
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