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https://mathoverflow.net/questions/441352 | 6 | Dale Peterson famously gave a series of lectures on the quantum cohomology of flag varieties $G/P$ at MIT in 1997. These lectures are often cited in subsequent papers by other authors on the subject (where many of the proofs appear for the first time in the literature), as Peterson himself never published any of this work to my understanding.
I have always assumed that these citations were to acknowledge Peterson with the result, and not specifically referring to a written resource. However, [this paper](https://alco.centre-mersenne.org/articles/10.5802/alco.242/) references a *specific lecture* from the series. This leads me to ask:
1. Does an actual resource from Peterson's lecture series exist, in typed/written/photocopied/etc. form?
2. If so, is that resource obtainable in an "ethical" way (meaning, not going against Peterson's intentions in terms of distribution)?
| https://mathoverflow.net/users/138296 | Peterson's quantum cohomology of G/P lectures | I put it as an answer to keep the page from jumping to the front from time to time:
There is this typed version, see the last two lines of the last page:
[http://math.soimeme.org/~arunram/Resources/QuantumCohomologyOfGPL16-18.html](http://math.soimeme.org/%7Earunram/Resources/QuantumCohomologyOfGPL16-18.html)
Following Sam Hopkins's comment, the link to the first lecture is:
[http://math.soimeme.org/~arunram/Resources/QuantumCohomologyOfGPL1-5.html](http://math.soimeme.org/%7Earunram/Resources/QuantumCohomologyOfGPL1-5.html)
and additionally, there are handwritten notes in a link provided by him.
| 2 | https://mathoverflow.net/users/3903 | 441466 | 178,179 |
https://mathoverflow.net/questions/441329 | 2 | Let a non trivial $\alpha\in \pi\_m(S^n)$ with a finite order $|\alpha|$. Write $G\_\alpha(S^m, S^n)$ for the path-component determined by $\alpha$ of the mapping space $G(S^m, S^n)$ of free maps. Next, take a natural number $k>1$ relatively prime to $|\alpha|$. Then, for a self-map $f\colon S^{n-1} \to S^{n-1}$ with degree $k$, consider the induced map $(\Sigma f)\_\* \colon G\_\alpha(S^m, S^n) \to G\_{k\alpha}(S^m, S^n)$.
Is it true that $(\Sigma f)\_\*$ is a homotopy equivalence?
| https://mathoverflow.net/users/35872 | Path component of the mapping spaces $G(S^m, S^n)$ | This is not true. The short version is that even though $\alpha$ has order relatively prime to $k$, it's still a whole path component of $G(S^m,S^n)$. The homotopy groups of $G\_\alpha(S^m,S^n)$ then still contain a lot of $k$-torsion that tends to be killed by this degree-$k$ map.
$\require{AMScd}$
---
Pick basepoints $p \in S^m$ and $q \in S^n$ (fixed by $\Sigma f$). Evaluation at $p$ determines a map of fibration sequences
$$
\begin{CD}
(\Omega^m S^n)\_\alpha @>>> G\_\alpha(S^m,S^n) @>>> S^n\\
@V\Omega^m \Sigma f VV @V (\Sigma f)\_\* VV @VV\Sigma f V\\
(\Omega^m S^n)\_{k \alpha} @>>> G\_{k\alpha}(S^m,S^n) @>>> S^n\\
\end{CD}
$$
where $(\Omega^m S^n)\_\alpha$ is the path component of the chosen element $\alpha$. In particular, the long exact sequence on homotopy groups implies that $(\Omega^m S^n)\_\alpha \to G\_\alpha(S^m,S^n)$ is an isomorphism on homotopy groups in degrees less than $(n-1)$. Therefore, it will suffice to find an example where this map on homotopy groups is not an isomorphism in this range.
For any basepoint-preserving map $g: S^n \to S^n$, we also have a commutative diagram
$$
\begin{CD}
(\Omega^m S^n)\_0 @>{\Omega^m g}>> (\Omega^m S^n)\_0\\
@V {t\_\alpha} V V @VV {t\_{g\_\* \alpha}} V\\
(\Omega^m S^n)\_\alpha @>>{\Omega^m g}> (\Omega^m S^n)\_{g\_\* \alpha}
\end{CD}
$$
where $t\_\beta: \Omega^m S^n \to \Omega^m S^n$ is the homotopy equivalence $x \mapsto \beta \ast x$ induced by loop multiplication. As a result, we can check if the map $\Omega^m (\Sigma f)$ determines an isomorphism $\pi\_\*(\Omega^m S^n)\_\alpha \to \pi\_\* (\Omega^m S^n)\_{k\alpha}$ by checking if it induces an isomorphism at the zero component. But at the zero component, we have a natural isomorphism
$$
\pi\_d (\Omega^m S^n)\_0 \cong \pi\_{d+m} S^n
$$
for any $d > 0$.
Therefore, we need to find:
* a nontrivial element $\alpha \in \pi\_m S^n$ of finite order,
* a $k > 1$ relatively prime to the order of $\alpha$, and
* a $d$ with $0 < d < n-1$ such that a degree $k$ map $S^n \to S^n$ does not induce an isomorphism on $\pi\_{d+m} S^n$.
As one way to make this easier on ourselves, if $d+m \leq 2n-2$, then $\pi\_{d+m} S^n$ is in the stable range (up to $\pi\_{2n-2}$), and a degree $k$ map $S^n \to S^n$ induces multiplication by $k$ on $\pi\_{d+m}$.
---
Let's take $n \geq 5$ and $m = n+1$. Choose $\alpha$ be the Hopf element $\eta \in \pi\_{n+1} S^n$. Since $n > 2$, this element has order $2$, so we can pick $k=3$ (hence $\eta = 3\eta$). The above isomorphisms say
$$
\pi\_2 G\_\eta(S^{n+1},S^n) \cong \pi\_2 (\Omega^{n+1} S^n)\_\eta \cong \pi\_{n+3} S^n.
$$
This is in the stable range because $n \geq 5$, and so a degree $3$ map on $S^n$ induces multiplication-by-$3$ on $\pi\_{n+3} S^n$. Finally, this group is isomorphic to $\mathbb{Z}/24$ by classical calculations of stable homotopy groups of spheres, so multiplication-by-3 is not an isomorphism on it.
| 2 | https://mathoverflow.net/users/360 | 441470 | 178,181 |
https://mathoverflow.net/questions/441465 | 3 | I have the following problem. It is well known that $H^\ast(D\_8,\mathbb{Z}/2)\cong \mathbb{F}\_2[x,y,w]/(xy=0)$ with $|x|=|y|=1$ and $|w|=2$ (see Adem,Milgram "Cohomology of finite groups"). So we get that.
$$
H^2(D\_8,\mathbb{Z}/2)=\mathbb{F}\_2\langle x^2, y^2,w\rangle.
$$
I wanted to ask if there is any method/reference for describing extensions corresponding to particular cohomology classes. I am especially interested in classes $x^2+y^2+w$ and $x^2+w$.
Also, a broader question is - are there any general, or more general, methods of constructing extensions given a class in the second cohomology of a group? Any reference or help would be appreciated.
EDIT: $D\_8$ here is the group of symmetries of a square.
| https://mathoverflow.net/users/123432 | Identifying group extension from cohomology class of $D_8$ | Here is a start for your specific questions. It is easy to see that under the inclusion $C\_4 < D\_8$, $x$ and $y$ both map to the 1 dimensional class and $w$ maps to the nonzero 2 dimensional class. So one concludes that a group $P$ of order 16 fitting into a central extension
$$ C\_2 \rightarrow P \rightarrow D\_8$$
corresponds to an element of $H^2(D\_8)$ that involves $w$ if and only if $P$ has a cyclic subgroup of order 8, as $C\_8$ is the group fitting into the nontrivial central extension
$$ C\_2 \rightarrow C\_8 \rightarrow C\_4.$$
Checking the wonderful website GroupNames ([https://people.maths.bris.ac.uk/~matyd/GroupNames/index.html](https://people.maths.bris.ac.uk/%7Ematyd/GroupNames/index.html)) one sees that there are only three groups of order 16 that both have $D\_8$ (called $D\_4$ there) as a quotient and $C\_8$ as a subgroup: $D\_{16}$, $Q\_{16}$, and $SD\_{16}$.
I think that $w$ corresponds to $D\_{16}$ since I think one gets the right cohomology ring. (See another fun website: <https://users.fmi.uni-jena.de/cohomology/16web/index.html>) Symmetry considerations make me then guess that $x^2+y^2+w$ corresponds to $Q\_{16}$ and $x^2+w$ corresponds to $SD\_{16}$.
| 6 | https://mathoverflow.net/users/102519 | 441481 | 178,187 |
https://mathoverflow.net/questions/441467 | 3 | $G$ is a group of odd order, $\sigma$ is an automorphism of $G$, and $\sigma^2=\mathrm{id}$.
I want to find an example to show that
$G\_s= \{ g \in G \mid \sigma \left( g \right)= g^{-1} \} $ might not be a subgroup of $G$.
$G$ has to be a non-Abelian group.
If $G\_s$ is a group, it should be an Abelian group.
I tried some examples (the orders of most examples are not very large), but I didn't find counterexamples.
| https://mathoverflow.net/users/151339 | Find an example where a subset of “inverse fixed points“ is not a subgroup | For an odd prime $p$, let $$G = \langle a,b \mid a^p=b^p=[[a,b],a]=[[a,b],b]=1 \rangle$$ be an extraspecial group of order $p^3$ and exponent $p$.
Then there is an automorphism $\sigma$ of $G$ with $\sigma^2=1$, $\sigma(a)=a^{-1}$, $\sigma(b)=b^{-1}$, and $\sigma([a,b])=[a,b]$.
Since $a,b \in G\_s$ but $G\_s \ne G$, it is clear that $G\_s$ cannot be a subgroup of $G$. In fact $|G\_s| = p^2$.
| 12 | https://mathoverflow.net/users/35840 | 441483 | 178,188 |
https://mathoverflow.net/questions/441460 | 0 | I am trying to proof the uniqueness of a maximum for a two-dimensional function (well behaved, twice differentiable, domain $R^2$, etc.), yet cannot compute the exact derivatives or the Hessian.
I have $f(x,y) = g(x,y) - bx - cy$ and know that $g\_{x}>0, \ g\_{y}>0, \ g\_{xx}<0$ and $g\_{yy}<0$, but do not know $g\_{xy}$. Also, $b>0$ and $c>0$.
Is that sufficient structure to say anything about the existence of single/multiple maxima? $g(x,y)$ is a function without close-form solutions, but if there was a particular property missing to claim uniqueness, I could try and proof that too.
Any help is much appreciated - if you have any pointers to relevant text-books or papers, please let me know. So far I was using Sundaram's book on Optimization Theory.
| https://mathoverflow.net/users/476037 | Identify maxima for 2-Dimensional Function without knowing cross-derivative | Of course, nothing definite can be said here. E.g., let $b=c=1$ and $g(x,y)=x+y-x^2-y^2+axy$ for some real $a$ and all real $x,y$, so that $f(x,y)=-x^2-y^2+axy$.
Then, if $|a|<2$, then $(0,0)$ is the only point of (local and global) maximum of $f$. If $|a|>2$, then $(0,0)$ is a saddle point of $f$ and there is no point of local or global maximum of $f$. If $|a|=2$, then there are infinitely many points of local (and global) maximum of $f$. So, here everything hinges on $g\_{xy}=a$.
| 2 | https://mathoverflow.net/users/36721 | 441489 | 178,190 |
https://mathoverflow.net/questions/441504 | 1 | Let $ X , S $ be proper varieties over $ \mathbb{C} $ and $ \pi : X \rightarrow S $ be a smooth, proper morphism with relative canonical line bundle $ K\_{X/S} $. If $ L $ is a $ \pi $-relatively ample line bundle on $ X $, is there a class of examples where it is true that $ H^q(X, K\_{X/S} \otimes L) = 0 $ for all $ q \ge 1 $?
| https://mathoverflow.net/users/152391 | Relative Kodaira Vanishing? | The natural analogue of Kodaira vanishing is $R^q\pi\_\*(K\_{X/S} \otimes \mathscr L) = 0$ for $q > 0$, which follows from Kodaira vanishing plus 'cohomology and base change' [Hartshorne, Thm. III.12.11(a)].
Thus the Leray spectral sequence for $\pi$ collapses on the $E\_2$ page, giving
$$H^q(X,K\_{X/S} \otimes \mathscr L) \cong H^q\big(S,\pi\_\*(K\_{X/S} \otimes \mathscr L)\big).$$
Then the question becomes a positivity question for $\pi\_\*(K\_{X/S}\otimes \mathscr L)$. The situation is underdefined to answer this: the $\pi$-ampleness hypothesis on $\mathscr L$ is preserved by tensoring with $\pi^\* \mathscr M$ for any line bundle $\mathscr M$ on $S$, and the projection formula gives
$$\pi\_\*(K\_{X/S}\otimes \mathscr L \otimes \pi^\*\mathscr M) \cong \pi\_\*(K\_{X/S} \otimes \mathscr L) \otimes \mathscr M.$$
In particular, taking $\mathscr M = \mathcal O\_S(d)$ for $d \gg 0$ gives the required vanishing by Serre vanishing.
Note that $\mathscr L\otimes \pi^\* \mathcal O\_S(d)$ is ample for $d \gg 0$, but this doesn't seem to help. At any rate, the result cannot possibly be true in general without some sort of 'absolute' restriction on $\mathscr L$.
| 4 | https://mathoverflow.net/users/82179 | 441507 | 178,196 |
https://mathoverflow.net/questions/441486 | 2 | Suppose $X$ is a smooth projective complex variety, connected of dimension $n$.
Let $a$ be an algebraic correspondence in $A^n(X\times X)$, the group of cycles modulo homological equivalence in $H^{2n}((X\times X)(\mathbf{C}),\mathbf{Q})$. This is a ring with the usual composition of correspondences.
$a$ acts on the cohomology of $X$ by the Künneth formula, as the graded linear map of degree zero $a^\*$.
>
> * Is $a^\*$ compatible with cup products?
> * Do we have, for another algebraic correspondence $b$, $(a\circ b)^\*=b^\*\circ a^\*$?
>
>
>
I expect the answer to the first question to be no, and the answer to the second question to be yes. It would be great to collect a few examples.
For instance, I’m thinking if $X$ is an abelian variety and $f$ is the graph of multiplication by $N>1$, then $H^2\times H^i\to H^{i+2}$ should give counterexamples to the first question for $i\le n-1$ by taking the characteristic polynomial $p(T)$ of $f^\*$ on $H^{i+2}$ and choosing $a=p(f)$.
| https://mathoverflow.net/users/nan | Cup products and correspondences | It is not true that $a^\*(\alpha \smile \beta) = a^\*\alpha \smile a^\*\beta$:
**Example.** Let $X = \mathbf P^2$, and let $a$ be the Künneth projector onto $H^2(\mathbf P^2) \subseteq H^\*(\mathbf P^2)$. Explicitly, $a$ can be represented by the algebraic cycle $\Delta\_{\mathbf P^2} - \mathbf P^2 \times x - x \times \mathbf P^2$ for any point $x \in \mathbf P^2$. If $\alpha \in H^2(\mathbf P^2)$ is the class of a line, then $a^\*(\alpha \smile \alpha) = a^\*(x) = 0$, but $a^\*\alpha \smile a^\*\alpha = \alpha \smile \alpha$ is the class of a point.
The answer to the second question is positive, and this is in fact purely a linear algebra statement. Indeed, the construction of $a \circ b$ in $A^n(X \times X)$ is in terms of pullbacks, pushforwards, and intersection products, all of which are preserved by the cycle class map $\operatorname{cl} \colon A^n(X \times X) \to H^{2n}(X \times X)$. This is explained for instance in §3 of Kleiman's chapter *The standard conjectures* in the 1991 Seattle Motives proceedings [Kleiman].
---
**References.**
[Kleiman] S. L. Kleiman, *The standard conjectures*. In: [Motives](http://dx.doi.org/10.1090/pspum/055.1). Proceedings of the summer research conference on motives, Seattle WA, July 20-August 2, 1991. American Mathematical Society. Proc. Symp. Pure Math. **55**.1, p. 3-20 (1994). [ZBL0820.14006](https://zbmath.org/?q=an:0820.14006).
| 1 | https://mathoverflow.net/users/82179 | 441510 | 178,197 |
https://mathoverflow.net/questions/441503 | 2 | Suppose that $X,Y$ are nonnegative random variables satisfying, for each $t>0$,
$$ P(X>t,Y>t)\le P(X>t)P(Y>t).$$
Is there a standard term for this type of negative dependence?
(Edit: I've seen
the condition $E[XY] \le E[X]E[Y]$ called *negative covariance*.
Edit 2: the nonnegativity condition really isn't essential. I care more about the negative dependence condition.)
| https://mathoverflow.net/users/12518 | Name of type of negative dependence | It is called negative quadrant dependence. See Chapter 5 of "An Introduction to Copulas" by R. B. Nelson.
| 3 | https://mathoverflow.net/users/25622 | 441511 | 178,198 |
https://mathoverflow.net/questions/441519 | 0 | Following this question: [Can we get that $ P(N^{2/3}(\lambda\_N-\lambda\_{N-1})\le c)\ge 1-\epsilon$?](https://mathoverflow.net/questions/436173/can-we-get-that-pn2-3-lambda-n-lambda-n-1-le-c-ge-1-epsilon).
We know that for $\lambda\_N\le \lambda\_{N\_1}\le \dots le\lambda\_1$ (eigenvalues of GOE matrix)
>
> $$
> \lim\_{N\to\infty}P(N^{2/3}(\lambda\_N-2)\le s\_1,\dotsc, N^{2/3}(\lambda\_{N-k+1}-2)\le s\_k)=F\_{\beta, k}(s\_1,\dotsc, s\_k).
> $$
>
>
>
Can we say that for any $\epsilon>0$, there exists a constant $c>0$ such that
$$
P(N^{2/3}(\lambda\_N-\lambda\_{N-1})\ge c)\ge 1-\epsilon?
$$
| https://mathoverflow.net/users/168083 | Does there exist a constant $c>0$ such that $$ P(N^{2/3}(\lambda_N-\lambda_{N-1})\ge c)\ge 1-\epsilon? $$ | The probability density function of the spacing $\delta\_N=\lambda\_N-\lambda\_{N-1}$ of the eigenvalues $\lambda\_N$ and $\lambda\_{N-1}$ at the edge of the spectrum decays linearly for $\delta\_N\ll N^{-2/3}$, with a slope that is independent of $N$. So no matter how small $\epsilon$, you can always find a $c>0$ such that $P(N^{2/3}\delta\_N\le c)\le \epsilon$, or equivalently, such that $P(N^{2/3}\delta\_N\ge c)\ge 1-\epsilon$. The coefficient $c$ will depend on $\epsilon$ but it will not depend on $N$.
| 1 | https://mathoverflow.net/users/11260 | 441526 | 178,200 |
https://mathoverflow.net/questions/441532 | 4 | In the paper "Convex Sets of Cardinals", Truss mentioned a result of Jech:
>
> If $M$ is a countable transitive model of ZFC, and $(P,<)∈M$ is a poset, then there exists a Cohen extension of $M$ such that $(P,<)$ is isomorphic to a set of cardinalities of that model.
>
>
>
The result is referenced to to "On ordering of cardinalities", I found several other mention of this result, also referencing "On incomparable cardinals" by Takahashi. (while this is not my main question, I failed to find both papers, so if someone knows where I can locate them I would love to know).
In his answer [here](https://math.stackexchange.com/questions/837831/partial-order-on-cardinalities-without-the-axiom-of-choice), Asaf said:
>
> We can show that given a model of ZFC, every partial order in that model, and in fact the entire model itself, can be embedded into the cardinals of a larger model
>
>
>
Strengthening the result from above.
Those result led me to think about the internal variation of the question, instead of looking at posets in a model and extending it to a model with enough cardinals, looking at the posets in the universe and asking if the universe has enough cardinals:
* Is it consistent with ZF that for every partially ordered set $(P,<)$ there exists a set of cardinals that is isomorphic to $(P,<)$?
+ The same question but with the schema statement about definable partially ordered classes as well
+ The same question but in NBG and about partially ordered classes
In other words, is it possible in ZF that the cardinals capture all possible orders?
This also leads to 2 variations of dual questions:
* Is there a definable class $C$ of partial orders such that if there exists $(P,<)∈C$ that does not embeds to the cardinals, then the axiom of choice holds?
* Does there exists a minimal definable class $C$ of partial orders such that if non of the orders in $C$ embeds into the cardinals, then the axiom of choice holds?
Both of those variations are unfortunately trivialized by looking $C=\{(\{a,b\}, ∅)\}$
| https://mathoverflow.net/users/113405 | Exactly how much (and how little) can partial ordered sets (classes) embed to the cardinalities | Yes. This was essentially proved by Honsel and Forti in the 1980s by analysing a model that generalises the Cohen model (essentially, the one Monro used to show it can be consistent for Dedekind finite sets to have large Lindenbaum numbers).
In my preprint Iterated Failures of Choice I provide a different construction for the same result.
The point is that we can essentially embed $V$ with the subset relation into the cardinals.
| 6 | https://mathoverflow.net/users/7206 | 441533 | 178,203 |
https://mathoverflow.net/questions/441536 | 7 | Are there any formulas for the irreducible off-diagonal elements $E^{\lambda}\_{ij}$ in the Gelfand-Tsetlin basis of the symmetric group algebra $\mathbb{C}[S\_n]$?
Here is the context for my question. There exists well-known formula for *minimal idempotents* (sometimes also called *primitive idempotents*) in the Gelfand-Tsetlin basis (in the sense of Okounkov-Vershik approach) of the symmetric group algebra $\mathbb{C}[S\_n]$ originally due to Murphy ["A new construction of Young's seminormal representation of the symmetric group", J. Algebra, 69 (1981), 287-297]:
$$
E\_T=\prod\_{k=1}^n \prod\_{c \neq c\_k(T)} \frac{J\_k-c}{c\_k(T)-c},
$$
where $T$ is a path in the Bratteli diagram of $\mathbb{C}[S\_n]$ (or equivalently, *standard Young tableaux* - SYT), $J\_k$ is the $k$-th Jucys-Murphy element of $\mathbb{C}[S\_n]$, $c\_k(T)$ is the content of the box $k$ in SYT $T$, and $c$ runs over all possible contents in all SYTs of size $n$.
If we denote by $\lambda$ the Young diagram of the SYT $T$ and use the index $i$ to refer to the SYT $T$ as a basis vector in the Gelfand-Tsetlin basis of the irreducible representation $\lambda$, then we can think about $E\_T$ as being an orthogonal projector $E^{\lambda}\_{ii}$ onto diagonal entry $i$ in the block $\lambda$ under Artin-Wedderburn isomorphism $\mathbb{C}[S\_n] \cong \bigoplus\_{\lambda} \text{End}(V^\lambda)$, where $V^\lambda$ is the irreducible representation labeled by $\lambda$. This naturally raises the question of the existence of a formula for the *off-diagonal* "elementary matrices" $E^{\lambda}\_{ij}$ as elements of $\mathbb{C}[S\_n]$.
| https://mathoverflow.net/users/146845 | Formula for the off-diagonal "elementary matrices" in the Gelfand-Tsetlin basis of the symmetric group algebra? | Yes, these exist and are known as Young's orthogonal matrix units.
To construct $E\_{T,T'}$ note that $E\_T \,\mathbb{C}[S\_n] \, E\_{T'}$ is one dimensional. Thus it suffices to pick any $x\in \mathbb{C}[S\_n]$ such that $E\_T \, x\, E\_{T'}$ is nonzero, for example $x \in S\_n$ the unique permutation that sends $T'$ to $T$. (One can also use the Baxterised elements to exchange boxes labelled $i,i+1$ in the tableaux successively to get from $T'$ to $T$.)
For the $q$-case (Hecke algebra) their construction is given in Prop 2.1 of [Ram and Wenzl, "Matrix units for centralizer algebras", J Algebra
145 (1992) 378-395](https://www.sciencedirect.com/science/article/pii/002186939290109Y).
| 9 | https://mathoverflow.net/users/45956 | 441543 | 178,205 |
https://mathoverflow.net/questions/441540 | 2 | (In following we are working in "classical" complex setting: i.e. all involved schemes are considered to be varieties over $k=\mathbb{C}$)
Let $X \subset \mathbb{P}^n$ be irreducible *surface* and $L $ some general $(n-3)$-plane disjoint from $X$.
We consider now the projection map $f = \pi\_L: X \to \mathbb{P}^2$ from $L$.
Let $Z \subset X$ be the locus of *critical points*, i.e. points $p \in X$ that are singular or such that the tangent map $df\_p: T\_pX \to T\_{f(p)}\mathbb{P}^2$ has a non trivial kernel
$\text{Ker}(df\_p) \neq 0 $. It's a fact that this is a proper closed subset of $X$.
Let $B = f(Z)$ be it's image in $\mathbb{P}^2$ which by properness of $f$ has to be closed too. $B$ is from naive point of
view given as union of a plane curve and a finite collection of some points.
There is an interesting remark in Harris' book *[Algebraic Geometry](http://www.springer.com/it/book/9780387977164)*, on p 290 in proof of Prop. 18.10
that one knows moreover that $B$ has *no* $0$-dimensional components. Unfortunately this fact is not needed for the rest
of the proof, so the author not gave a justification for this.
*Question:* How to check this last statement? Ot looks rather counterintuitive, since thinking of contractions of curves one might expect $B$ might have $0$-dimensional components.
Thoughts: It seems not to be true for all generically finite projective morphisms $f: X \to \mathbb{P}^2$ of surfaces,
since for example the blowup at center $0 \in \mathbb{P}^2$ gives $B=0$. So if the statement is true then it must be based on
special structure of projections from linear subspaces, but I not find an argument. I also not know how "deep" this result is, ie which "tools" are required to prove it. Even if the quoted book is for undergrades, Harris often quotes there some facts going far beyond the scope of the book, and so I'm not sure if this property of $B$ is a result of advanced research or can it be seen with rather "elmentary" tools.
| https://mathoverflow.net/users/108274 | Generically finite projection $\pi_L: X \to \mathbb{P}^2$ from plane $L$ and critical points | The morphism $f$ is not just generically finite but finite. I think this resolves all the difficulties.
To see this, since $f$ is proper it just suffices to check it is quasi-finite, and thus to check the fiber over a point can't contain a curve. But the fiber over a point is a $(n-2)$-plane containing $L$, and any curve in an $(n-2)$-plane intersects every $(n-3)$-plane, so if such a curve were contained in the fiber it would contradict the assumption that $X$ is disjoint from $L$.
Alternately, we can argue that $\mathbb P^n \setminus L \to \mathbb P^2$ is affine and thus $f$ is affine, and, since affine and proper, is finite.
| 1 | https://mathoverflow.net/users/18060 | 441546 | 178,207 |
https://mathoverflow.net/questions/441542 | 4 | A *symmetric function* is a function $f:\mathbb R^n\to \mathbb R$ such that $f(x\_1,\ldots,x\_n)=f(\sigma(x\_1,\ldots,x\_n))$ for every permutation $\sigma\in S\_n.$
The most commonly encountered symmetric functions are polynomial functions.
**Question 1:** Is it true that, given a symmetric function $f:\mathbb R^n\to \mathbb R,$ there exist symmetric polynomials $p\_1,\ldots,p\_m:\mathbb R^n\to \mathbb R$ and functions $h:\mathbb R^m\to \mathbb R$ and $g\_1,\ldots,g\_n:\mathbb R\to \mathbb R$ such that $f$ can be factorized as $$f(x\_1,\ldots,x\_n)=h(p\_1(g\_1(x\_1),\ldots, g\_1(x\_n)),p\_2(g\_2(x\_1),\ldots, g\_2(x\_n)),\ldots,p\_m(g\_m(x\_1),\ldots, g\_m(x\_n)))?$$
**Question 2:** If $f$ has some regularity, e.g., continuous, $C^\infty$ or analytic, can we find the above $g\_i$ and $h$ to be also regular?
To illustrate the idea behind the question I furnish an example.
If $f:\mathbb R^2\to \mathbb R$ with $$f(x,y)=(x^2+y^2)\ln(\frac{xy}{x+y})+\sin(x)+\sin(y),$$ we have $m=4$, $p\_1(x,y)=x^2+y^2,$ $p\_2(x,y)=xy$, $p\_3(x,y)=x+y$, $p\_4(x,y)=x+y$, $g\_1,g\_2,g\_3=\operatorname{id}\_{\mathbb R}$, $g\_4(x)=\sin(x)$ and $h(x,y,z,t)=x\ln(\frac{y}{t})+t$.
Notice that **Question 1** seems to be true thanks to an argument similar to that in the [answer](https://math.stackexchange.com/a/3800574) of Qi Zhu to [Can every symmetric function be written as some function of a sum?](https://math.stackexchange.com/questions/3800563/can-every-symmetric-function-be-written-as-some-function-of-a-sum).
**Question 3:** does this argument works here?
| https://mathoverflow.net/users/165036 | Can every symmetric function be factorized through symmetric polynomials? | $\newcommand\R{\mathbb R}$For any $(x\_1,\dots,x\_n)\in\R^n$,
$$P(x):=\prod \_{i=1}^{n}(x-x\_{i})=\sum \_{k=0}^{n}(-1)^{k}e\_k(x\_1,\dots,x\_n)x^{n-k},$$
where the $e\_k$'s are the [elementary symmetric polynomials](https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial) in $x\_1,\dots,x\_n$. Therefore and because "the roots of a polynomial are determined by its coefficients", there is a function $g\colon\R^m\to[\R^n]$ such that
$$[x\_1,\dots,x\_n]=g(e\_0(x\_1,\dots,x\_n),\dots,e\_n(x\_1,\dots,x\_n))$$
for all $(x\_1,\dots,x\_n)\in\R^n$, where $m:=n+1$,
$$[x\_1,\dots,x\_n]:=\{(x\_{\pi(1)},\dots,x\_{\pi(n)})\colon\pi\in S\_n\},$$
$$[\R^n]:=\{[x\_1,\dots,x\_n]\colon(x\_1,\dots,x\_n)\in\R^n\},$$
and $S\_n$ is the set of all permutations of the set $\{1,\dots,n\}$.
On the other hand, if a function $f\colon\R^n\to\R$ is symmetric, then there clearly exists a function $u\colon[\R^n]\to\R$ such that
$$f(x\_1,\dots,x\_n)=u([x\_1,\dots,x\_n])$$
for all $(x\_1,\dots,x\_n)\in\R^n$.
Thus,
$$f(x\_1,\dots,x\_n)=h(e\_0(x\_1,\dots,x\_n),\dots,e\_n(x\_1,\dots,x\_n)) \tag{1}\label{1}$$
for all $(x\_1,\dots,x\_n)\in\R^n$, where $h:=u\circ g$. This answers your Question 1. $\quad\Box$
---
As an illustration of the previous consideration, here is the particular case of identity \eqref{1} for the symmetric function $f\colon\R^2\to\R$ given by the formula $f(x,y):=x\sin y+y\sin x$:
$$x\sin y+y\sin x \\
=\Big(-\frac{p(x,y)}2+\sqrt{\frac{p(x,y)^2}4-q(x,y)}\Big) \\
\times\sin\Big(-\frac{p(x,y)}2-\sqrt{\frac{p(x,y)^2}4-q(x,y)}\Big) \\
+\Big(-\frac{p(x,y)}2-\sqrt{\frac{p(x,y)^2}4-q(x,y)}\Big) \\
\times\sin\Big(-\frac{p(x,y)}2+\sqrt{\frac{p(x,y)^2}4-q(x,y)}\Big) \\
$$
for all $(x,y)\in\R^2$, where $p(x,y):=x+y$ and $q(x,y):=xy$ are elementary symmetric polynomials in $x,y$.
---
It is now an easy exercise to show that $h$ is continuous if $f$ is continuous. Hints: (i) note that the polynomial $P$ is monic and use a bound on the roots of a monic polynomial in terms of the bounds on its coefficients; (ii) using compactness and the continuity of the value of a polynomial with respect to the coefficients and the argument of the polynomial, show that the map $g$ is continuous; (iii) the map $f\mapsto u$ is clearly continuous in any reasonable topology; (iv) so, $h=u\circ g$ is continuous.
| 2 | https://mathoverflow.net/users/36721 | 441548 | 178,208 |
https://mathoverflow.net/questions/441550 | 2 | Let $\Theta \subset \mathbb{R}^n, \mathcal{X} \subset \mathbb{R^m}$, and suppose that $C: \Theta \rightrightarrows \mathcal{X}$ is a correspondence defined by $f: \Theta \times \mathcal{X}\to \mathbb{R}^d$ as follows:
$$
C(\theta) = \{ x\in \mathcal{X} \mid f\_1(\theta, x) \geq 0, \dots, f\_d(\theta, x) \geq 0 \}
$$
Is it true that 1) if $f$ is continuous then $C$ also is, and 2) if $C$ is continuous then $f$ also is?
(Here continuity for correspondences is defined as upper and lower hemicontinuity.)
| https://mathoverflow.net/users/160399 | Can continuous correspondence be represented via continuous functions? | Neither implication holds.
Let $\Theta=\mathcal{X}=[-1,1]$.
First, define $f$ by $f(x,y)=xy$. Then $C$ is not lower hemicontinuous. Indeed, $$\big\{\theta\mid C(\theta)\cap (-1,0)\neq\emptyset\big\}=[-1,0]$$
is not open.
Next, let $f$ be any discontinuous function with nonnegative values. Then $C$ is constant with value $[-1,1]$ and, therefore, continuous.
| 2 | https://mathoverflow.net/users/35357 | 441557 | 178,212 |
https://mathoverflow.net/questions/441438 | 5 | Let $\epsilon\_1, \dots, \epsilon\_n$ be random signs, equiprobably in $\{-1, 1\}$, independently.
Let $S\_k = \sum\_{j=1}^k \epsilon\_j$. I am wondering what is known about the expectation
$$
\mathbb{E}\Big[\max\_{k \leq n} |S\_k| \Big].
$$
It can be seen as the maximum distance from the origin over $n$ steps of a simple random walk which moves left or right from the origin with equal probability.
A naive bound is via Lévy's maximal inequality, which implies that the quantity above is bounded above by $2 \sqrt{n}$. Can the constant $2$ be improved?
| https://mathoverflow.net/users/121486 | Maximum distance from origin of simple random walk | If you just care about the asymptotics, it is indeed just $(1+o(1))\sqrt{\pi n/2}$, where the $o(1)$ term decays like $\tilde{O}(1/n^{1/4})$; this can be done using the approach that I suggested of using the natural embedding to compare to Brownian motion at the obvious stopping times, the fact that the Brownian motion does not fluctuate by more than 1 between stopping times, and the fact that $\vert T\_n - n\vert=\tilde{O}(\sqrt{n})$ with very high probability by Bernstein's inequality applied to the (subexponential) increments in stopping times. I can make this more rigorous if needed.
If you really do care about the best constant $C^\*$ such that $\mathbb{E}[\max\_{k\leq n} S\_k]\leq C\sqrt{n}$ for *every* $n$, not just for large enough $n$, there's a couple of approaches; it's true that $C^\*$ is at most twice the optimal constant when you take the absolute values outside the max because $\max\_{k\leq n}\vert \sum\_{i=1}^k \varepsilon\_i\vert\leq \vert\max\_{k\leq n} \sum\_{i=1}^k \varepsilon\_i\vert+\vert\min\_{k\leq n} \sum\_{i=1}^k \varepsilon\_i\vert$, which can be treated exactly via the reflection principle. This should (at least morally) get you something like $2\sqrt{2/\pi}\approx 1.596$. Just for variety, here's a slightly different approach that gives a slightly better constant of $\pi/2\approx 1.571$.
Simply note that if $g\_1,\ldots,g\_n$ are i.i.d. standard Gaussians, then by Jensen's inequality,
\begin{align\*}
\mathbb{E}\left[\max\_{k\leq n} \left\vert \sum\_{i=1}^k \varepsilon\_i \right\vert\right] &= \sqrt{\frac{\pi}{2}} \mathbb{E}\left[\max\_{k\leq n} \left\vert \sum\_{i=1}^k \varepsilon\_i \mathbb{E}[\vert g\_i\vert] \right\vert\right]\\
&\leq \sqrt{\frac{\pi}{2}}\mathbb{E}\left[\max\_{k\leq n} \left\vert \sum\_{i=1}^k \varepsilon\_i\vert g\_i\vert \right\vert\right]\\
&=\sqrt{\frac{\pi}{2}}\mathbb{E}\left[\max\_{k\leq n} \left\vert \sum\_{i=1}^k g\_i \right\vert\right]\\
&\leq \sqrt{\frac{\pi}{2}}\mathbb{E}\left[\max\_{t\leq n} \left\vert B\_t\right\vert\right]\\
&=\pi/2\cdot \sqrt{n},
\end{align\*}
using the symmetry of standard Gaussians and upper bounding by a coupled Brownian motion and appealing to the MSE solution.
| 3 | https://mathoverflow.net/users/170770 | 441560 | 178,214 |
https://mathoverflow.net/questions/441552 | 1 | I wonder if we can replace the binomial terms in Bernstein polynomial with multinomial terms. To be more specific, given a vector $f=(f\_1, f\_2,\dotsc, f\_M)\in \mathbb{Z}^M$ such that $\forall m, f\_m\ge0, \sum\_mf\_m=N\in\mathbb{Z}$, define
$F\_{N,f}(\sigma)=\frac{N!}{\prod\_{m} f\_{m}!}\prod\_{m:\sigma\_m>0}(\sigma\_{m})^{f\_{m}}$, where $\sigma\in\Delta^{M-1}$.
I think when $M=2$, this is Bernstein polynomial. Is there a name for $F\_{N, f}$ for cases $M>2$ and what are its properties?
| https://mathoverflow.net/users/136078 | Is there a name for an extension of Bernstein basis polynomial based on multinomial distribution? | $\newcommand{\de}{\delta}\newcommand{\De}{\Delta}\newcommand\R{\mathbb R}\newcommand{\om}{\omega}\newcommand{\si}{\sigma}$Such polynomials are called [multivariate Bernstein polynomials](https://www.google.com/search?q=multivariate%20Bernstein%20polynomials&oq=multivariate%20Bernstein%20polynomials&aqs=chrome..69i57.448j0j1&sourceid=chrome&ie=UTF-8).
Many of the [properties of Bernstein polynomials](https://en.wikipedia.org/wiki/Bernstein_polynomial#Properties) should be extendible to these multivariate versions.
Perhaps the most interesting of the properties of Bernstein polynomials is that they provide an explicit bound on the uniform approximation error of continuous functions by polynomials.
Let us show that such a property holds in the multivariate case as well.
Let $g\colon\De^{M-1}\to\R$ be any continuous function with the modulus of continuity $\om$ given by the formula
\begin{equation\*}
\om(\de):=\sup\{|g(x)-g(y)|\colon\|x-y\|\_2\le\de\}
\end{equation\*}
for real $\de>0$, where $\|\cdot\|\_2$ is the Euclidean norm.
For any $\si=(\si\_1,\dots,\si\_M)\in\De^{M-1}$, let a random $M$-tuple $X=(X\_1,\dots,X\_M)$ have the multinomial distribution with parameters $N;\si\_1,\dots,\si\_M$. Then
\begin{equation\*}
F\_{N,f}(\si)=P(X\_1=f\_1,\dots,X\_M=f\_M)
\end{equation\*}
for $f=(f\_1,\dots,f\_M)\in\{0,1,\dots\}^M$ with $\sum\_{j=1}^M f\_j=N$. So,
\begin{equation\*}
p\_{N,f}(\si):=Eg\Big(\frac XN\Big)=\sum\_f g\Big(\frac fN\Big)F\_{N,f}(\si),
\end{equation\*}
and $p\_{N,f}$ is a polynomial; of course, here $\sum\_f$ denotes the summation over all $f=(f\_1,\dots,f\_M)\in\{0,1,\dots\}^M$ with $\sum\_{j=1}^M f\_j=N$.
Moreover, for any real $\de>0$ and all $\si\in\De^{M-1}$,
\begin{equation\*}
\begin{aligned}
|p\_{N,f}(\si)-g(\si)|&\le E\Big|g\Big(\frac XN\Big)-g(\si)\Big| \\
&=E\Big|g\Big(\frac XN\Big)-g(\si)\Big|\,1\Big(\Big\|\frac XN-\si\Big\|\_2\le\de\Big) \\
&+E\Big|g\Big(\frac XN\Big)-g(\si)\Big|\,1\Big(\Big\|\frac XN-\si\Big\|\_2>\de\Big) \\
&\le\om(\de)+2\|g\|\_\infty\,P\Big(\Big\|\frac XN-\si\Big\|\_2>\de\Big) \\
&\le\om(\de)+\frac{2\|g\|\_\infty}{\de^2}\,E\Big\|\frac XN-\si\Big\|\_2^2 \\
&\le\om(\de)+\frac{2\|g\|\_\infty}{\de^2}\,\sum\_{j=1}^M E\Big(\frac{X\_j}N-\si\_j\Big)^2 \\
&=\om(\de)+\frac{2\|g\|\_\infty}{\de^2\,N}\,\sum\_{j=1}^M \si\_j(1-\si\_j) \\
&\le\om(\de)+\frac{2\|g\|\_\infty}{\de^2\,N},
\end{aligned}
\tag{1}\label{1}
\end{equation\*}
where $\|g\|\_\infty:=\max\_{\si\in\De^{M-1}}|g(\si)|$.
The penultimate inequality and the last equality in \eqref{1} follow because the $X\_j$'s are negatively correlated random variables with respective binomial distributions with parameters $N,\si\_j$, and the last equality in \eqref{1} follows because $\si\in\De^{M-1}$.
So, $\limsup\_{N\to\infty}\|p\_{N,f}(\si)-g(\si)\|\_\infty\le\om(\de)$, for any real $\de>0$. Since $g$ is continuous, it is uniformly continuous on $\De^{M-1}$, so that $\om(\de)\to0$ as $\de\downarrow0$. We conclude that
\begin{equation}
\lim\_{N\to\infty}\|p\_{N,f}(\si)-g(\si)\|\_\infty=0,
\end{equation}
as claimed. $\quad\Box$
| 1 | https://mathoverflow.net/users/36721 | 441564 | 178,216 |
https://mathoverflow.net/questions/441562 | 0 | Consider two $n-$dimensional random vectors $u$ and $v$ uniformly distributed on the sphere. Define $X\_n :=u\cdot v$. Note that as $n\to \infty$, $\sqrt{n}X\_n \to N(0,1)$ as $n\to \infty$. Fix $\epsilon>0$ (very small).
Can we show that for every $\delta>0$, there exist constants $\alpha>0, \beta>0$ so that
$$
\lim\_{n\to \infty}P\left(\frac{X\_n^{-2}-1}{\epsilon^{-2}-1}<\alpha n^\beta\right)\ge 1-\delta.
$$
Or can we revise this upper bound for $\frac{X\_n^{-2}-1}{\epsilon^{-2}-1}$ depending on $n$.
Since we know that the order $X\_n=O\_p(n^{-1/2})$, then the order of $\frac{X\_n^{-2}-1}{\epsilon^{-2}-1}$ is about $O\_p(n)$. But I am stuck on how get the strict upper bound. (Maybe this question would be helpful: <https://math.stackexchange.com/questions/4593238/can-we-find-c1-so-that-px-le-frac-epsilonc-ge-1-delta?>)
---
Let $Y\sim N(0,1)$ (hence we can write $X\_n=n^{-1/2}Y$. Note that
$$\begin{align\*}
\lim\_{n\to \infty}P\left(\frac{X\_n^{-2}-1}{\epsilon^{-2}-1}<\alpha n^\beta\right)&=P\left(n\frac{Y^{-2}-1}{\epsilon^{-2}-1}<\alpha n^\beta\right)\\&=P\left(nY^{-2}<\alpha(\epsilon^{-2}-1)n^{\beta}+1\right)\\&=P\left(Y^2>\frac{n}{\alpha(\epsilon^{-2}-1)n^{\beta}+1}\right)\end{align\*}
$$
I am not sure if we can apply the concentration result of the Gaussian variable to find proper $\alpha, \beta>0$ so that this probability larger than $1-\delta$.
| https://mathoverflow.net/users/168083 | Can we show that for every $\delta>0$, there exist constants $\alpha>0, \beta>0$ so that the following inequality holds with high probability? | $\newcommand\al\alpha\newcommand\be\beta\newcommand\ep\epsilon\newcommand\de\delta$Let $\al=\beta=1$. Let $Z\_n:=\sqrt n\,X\_n$, so that $Z\_n\to Z\sim N(0,1)$ in distribution. Then for real $\ep>0$
$$
P\Big(\frac{X\_n^{-2}-1}{\ep^{-2}-1}<\al n^\be\Big)
=P\Big(|Z\_n|>\sqrt{\frac n{1+(\ep^{-2}-1)n}}\,\Big) \\
\to P\Big(|Z|>\sqrt{\frac1{\ep^{-2}-1}}\,\Big)
\ge 1-\de
$$
if $\ep>0$ is small enough, depending on $\de$. $\quad\Box$
---
To address a comment by the OP, alternatively we can take any real $\be>1$ and then
$$
P\Big(\frac{X\_n^{-2}-1}{\ep^{-2}-1}<\al n^\be\Big)
=P\Big(|Z\_n|>\sqrt{\frac n{1+(\ep^{-2}-1)n^\be}}\,\Big) \\
\to P(|Z|>0)=1\ge 1-\de.
$$
| 1 | https://mathoverflow.net/users/36721 | 441567 | 178,218 |
https://mathoverflow.net/questions/441559 | 2 | A rational normal curve $C \subset \mathbb{P}\_k^d$ (assume $k= \mathbb{C}$) can be defined usually up to projective equivalence in two equivalent ways:
* smooth irreducible nondegenerate curve $C \subset \mathbb{P}^d$ of minimal degree $\text{deg}(C)=d$
* projectively equivalent to the image of the Veronese map
$$ \nu\_d: \mathbb{P}^1 \to \mathbb{P}^d, \ \ \
[X\_0:X\_1] \mapsto [X\_0^d: X\_0^{d-1}X\_1:..., X\_1^d] $$
The second definition can be reprased as that there exist a $\mathbb{C}$-basis of the vector space of homogeneous polynomials $p\_0,p\_1,..., p\_d$ of degree $d$ in two variables $X\_0, X\_1$ such that $C$ is the image under
$$ P: \mathbb{P}^1 \to \mathbb{P}^d, \ \ \
[X\_0:X\_1] \mapsto [p\_0(X): p\_1(X):..., p\_d(X)] $$
In Harris' book *[Algebraic Geometry](http://www.springer.com/it/book/9780387977164)*, on page 14 there is an alternative presumably equivalent construction of a rational normal curve:
Start by choosing $d$ codimension two linear
spaces $\Lambda\_i \cong \mathbb{P}^{d-2} \subset \mathbb{P}^{d}$. The family $\{H\_i(\lambda)\}$ of hyperplanes in $\mathbb{P}^{d}$ containing $\Lambda\_i$ is
then parameterized by $\lambda \in \mathbb{P}^1$; choose such parameterizations, subject to the condition
that for each $\lambda$ the planes $H\_1(\lambda), ... , H\_d(\lambda)$ are independent, i.e., intersect in a
point $ p(\lambda)$. Claim: It is then the case that the locus of these points $p(\lambda)$ as $\lambda$ varies in $\mathbb{P}^1$ is a rational normal curve.
*Question:* Why this construction gives a rational normal curve $\bigcup\_{\lambda \in \mathbb{P}^{1}} H\_1(\lambda) \cap ... \cap H\_d(\lambda)$?
I found [here](https://math.stackexchange.com/questions/4170453/steiner-construction-of-rational-normal-curve) a rather promising approach discussing identical problem: in the linked discussion is showed how one can construct explicitly the point $p(\lambda)= [p\_0(\lambda): p\_1(\lambda):...: p\_d(\lambda)]$ in terms of certain homogenous polynomials $p\_i(\lambda)$ of degree $d$ in coordinates $(\lambda)=\lambda\_0, \lambda\_1$.
The idea is quite simple. Consider the $(d+1) \times (d+1)$ matrix $A(\lambda)$ where each $j$-th row of $A(\lambda)$ is encoded by the $j$-th family of hyperplanes as follows. The parametrized family $H\_j(\lambda)$ of hyperplanes which contain $\Lambda\_i \cong \mathbb{P}^{d-2}$ is given as vanishing loci of $\lambda\_0S\_j +\lambda\_1T\_j$, where
the two linear forms $S\_j, T\_j$ in variables $Y\_0,..., Y\_d$ determine $\Lambda\_j$. The $j$-th row of $A(\lambda)$ is encoded in
$$a\_0(\lambda)Y\_0+ a\_1(\lambda)Y\_1 +...+ a\_d(\lambda)Y\_d= \lambda\_0S\_j +\lambda\_1T\_j $$
The $(d+1)$-row is filled with zeroes. The adjugate matrix $P(\lambda)$ having $d$-minors of $A$ as entries satisfies $A \cdot P=P \cdot A= \text{det}(A) \cdot E\_{d+1} $ and $\text{det}(A) $ is zero. The $(d+1)$-th column of $P(\lambda)$ is the unique point $p(\lambda)=[p\_0(\lambda): p\_1(\lambda):...: p\_d(\lambda)]$ with homogeneous $p\_i(\lambda)$ of degree $d$.
One expects finally to obtain the rational normal curve as image of
$$[X\_0:X\_1] \mapsto [p\_0(X): p\_1(X):..., p\_d(X)] $$
but it isn't clear why the $p\_i(\lambda)$ are linearly *independent*. If they aren't, we obtain a degenerate curve in the image, that's not what we want.
| https://mathoverflow.net/users/108274 | Equivalent characterizations of rational normal curve | Let $\mathbb{P}^d = \mathbb{P}(V)$. A hyperplane in $\mathbb{P}(V)$ corresponds to an epimorphism $V \to k$, a pencil of hyperplanes to an epimorphism
$$
V \otimes \mathcal{O}\_{\mathbb{P}^1} \to \mathcal{O}\_{\mathbb{P}^1}(1),
$$
and a collection of $d$ pencils to a morphism
$$
f \colon V \otimes \mathcal{O}\_{\mathbb{P}^1} \to \mathcal{O}\_{\mathbb{P}^1}(1)^{\oplus d}.
$$
The condition imposed on the pencils means that this is an epimorphism, therefore its kernel $\mathrm{Ker}(f)$ is a line subbundle in $V \otimes \mathcal{O}\_{\mathbb{P}^1}$, and computing its determinant it is easy to see that
$$
\mathrm{Ker}(f) \cong \mathcal{O}\_{\mathbb{P}^1}(-d).
$$
The embedding $\mathrm{Ker}(f) \to V \otimes \mathcal{O}\_{\mathbb{P}^1}$ induces a degree $d$ map $\mathbb{P}^1 \to \mathbb{P}(V)$, and its image is precisely the curve in question. By this construction, it is a rational normal curve of degree $d$.
EDIT. To show that the curve is nondegenerate, consider the exact sequence
$$
0 \to
\mathcal{O}\_{\mathbb{P}^1}(-d) \to
V \otimes \mathcal{O}\_{\mathbb{P}^1} \stackrel{f}\to
\mathcal{O}\_{\mathbb{P}^1}(1)^{\oplus d} \to
0
$$
obtained above. Dualizing it, we obtain
$$
0 \to
\mathcal{O}\_{\mathbb{P}^1}(-1)^{\oplus d}
\stackrel{f^\vee}\to
V^\vee \otimes \mathcal{O}\_{\mathbb{P}^1} \to
\mathcal{O}\_{\mathbb{P}^1}(d) \to
0.
$$
Since the sheaf $\mathcal{O}\_{\mathbb{P}^1}(-1)$ has no cohomology, the induced morphism
$$
V^\vee =
H^0(\mathbb{P}^1,
V^\vee \otimes \mathcal{O}\_{\mathbb{P}^1}) \to
H^0(\mathbb{P}^1, \mathcal{O}\_{\mathbb{P}^1}(d))
$$
is an isomorphism. This means that the curve is nondegenerate.
| 4 | https://mathoverflow.net/users/4428 | 441593 | 178,229 |
https://mathoverflow.net/questions/441595 | 7 | I have difficulty even in finding a Russian version of the next paper:
"Aleksandrov, A. D., Almost everywhere existence of the second differential of a convex
function and some properties of convex surfaces connected with it (in Russian).
Uchenye Zapiski Leningrad. Gos. Univ., Math. Ser. 6 (1939), 3–35"
This is a quite famous paper which also appears in other questions in MathOverflow, e.g., [Aleksandrov's proof of the second order differentiability of convex functions](https://mathoverflow.net/questions/353950/aleksandrovs-proof-of-the-second-order-differentiability-of-convex-functions). Since Leningard has changed its name to St. Petersburg, I am not sure what's the current name of this journal.
Thank you very much!
| https://mathoverflow.net/users/113353 | Asking for an English version of Aleksandrov's famous 1939 paper in Convex Geometry | This paper is contained in the 1-st volume of Aleksandrov's Selected works. This has an English translation:
Alexandrov, A. D.
Reshetnyak, Yu. G. (ed.)
Selected works. Part 1: Selected scientific papers. Ed. by Yu. G. Reshetnyak and S. S. Kutateladze, transl. from the Russian by P. S. V. Naidu. (English) Zbl 0960.01035
Amsterdam: Gordon and Breach Publishers. x, 322 p.
I have not seen the translation but in the Russian
original, the paper is on pages 171-207.
Remark. Unfortunately, the English translation does not contain this paper. The table of contents of the English translation is available here:
[https://www.amazon.com/Alexandrov-Selected-Works-Part-Mathematics-ebook/dp/B07BFS986K](https://rads.stackoverflow.com/amzn/click/com/B07BFS986K)
It only has 322 pages, while the Russian original contains 803 pages.
The Russian original is available for a free download in Genesis library. When searching, be aware that his name has two different spellings in English: Alexandrov and Aleksandrov. The Russian spelling is Александр Данилович Александров.
| 13 | https://mathoverflow.net/users/25510 | 441597 | 178,230 |
https://mathoverflow.net/questions/441599 | 3 | $\newcommand\tetra[2]{{^{#1}{#2}}}$In a recent discussion on the Tetration Forum (see <https://math.eretrandre.org/tetrationforum/showthread.php?tid=1703&page=2>), it has been pointed out how my results on the constancy of the "congruence speed" of the integer tetration $\tetra b a$ (a peculiar property of hyper-$4$ described in [The congruence speed formula](https://arxiv.org/abs/2208.02622)), would imply that $\lim\_{b \rightarrow \infty} \tetra b a$ is always an element of the set $\mathbb{Q}\_p$.
At the time, I did not fully realize the power of this breakthrough, since my original goal was only to find a function $V(a) : \mathbb{Z}^+-\{M\} \rightarrow \mathbb{Z}^+$, where $M = \{a : a \not\equiv 0 \pmod {10} \land a \neq 1\}$, but after a little search, I have seen that this is a general open problem, explicitly solved only for a few cases (since 1953, for $a:=2$). Furthermore, I have not managed to find any proof that even $\lim\_{b \rightarrow \infty} \tetra b 6$ is a rational $p$-adic number (or rather an irrational one, disproving the aforementioned claim).
Now, my first concern is if I am wrong and $\lim\_{b \rightarrow \infty} \tetra b 6$ has already been proven to be an element of $\mathbb{Q}\_p$ (or not), while my general request here is to know more about the implications of Equation 16 on page 15 of [Ripà and Onnis - Number of stable digits of any integer tetration](https://arxiv.org/abs/2210.07956) on the above mentioned general open problem (and related topics).
| https://mathoverflow.net/users/481829 | $\lim_{b \rightarrow \infty} {^{b}a} \in \mathbb{Q}_p$ for any $a \in \mathbb{Z}^+$? | A sequence given by $x\_1=a$, $x\_{n+1}=a^{x\_n}$, where $a$ is a positive integer, is eventually constant modulo every positive integer $T$. This is widely known.
A short proof. We induct in $T$, so assume that $T>1$ and the claim is proven for smaller numbers. Denote $T=T\_1T\_2$, where $T\_1$ has prime divisors which divide $a$, and $T\_2$ is coprime with $a$. Modulo $T\_1$, obviously $x\_n$ become equal to 0. Modulo $T\_2$, the value of $x\_n$ is determined by the remainder of $x\_{n-1}$ modulo $\varphi(T\_2)<T\_2$, thus by induction hypothesis it is eventually constant.
I hope that this is what was asked.
| 9 | https://mathoverflow.net/users/4312 | 441602 | 178,232 |
https://mathoverflow.net/questions/441608 | 5 | McCoy's theorem (one of them) says that for any commutative ring $A$, $f\in A[x]$ is a zero-divisor iff it's annihilated by a scalar in $A$.
There's a widespread [proof by contradiction](https://math.stackexchange.com/a/83171/223002). There's also a constructive proof using the Dedekind-Mertens lemma about polynomial content ([Lombardi & Quitté p91, 2.3](https://arxiv.org/pdf/1605.04832.pdf)).
Is there a direct constructive proof that does not make use of Dedekind-Mertens, or even define content? If $fg=0$ then what is the scalar which annihilates $f$?
| https://mathoverflow.net/users/69037 | Constructive proof of univariate McCoy theorem without Dedekind-Mertens? | The question you're asking might not be the question you want to ask. "What is the scalar" should be "what is the nonzero scalar", because $0$ wouldn't be interesting, right? But we don't know how to find any nonzero elements of $A$ in the first place, and they might not even exist ($A$ can be trivial). So the most natural constructive version of McCoy's theorem is actually the contrapositive statement: "If a polynomial $f \in A\left[x\right]$ has the property that the map $A \to A\left[x\right]$ that sends each scalar $r$ to $rf$ is injective, then the map $A\left[x\right] \to A\left[x\right]$ that sends each polynomial $g$ to $gf$ is also injective". If you state the theorem in this form, then the proof by contradiction that you referenced can be straightforwardly translated into an induction proof (as usual in constructivism, don't speak of "the degree" of a polynomial, but just talk about polynomials of degree $\leq k$, because you don't always know which coefficients are zero).
Now, could there be a stronger constructive version of McCoy's theorem, which would give some approximation to an "explicit zero-dividing witness", or maybe a tuple of such? I'll think about it, but let me just say that this would not be my standard interpretation of the theorem.
**EDIT:** Yes, there could be a stronger constructive version. But it would look somewhat awkward. It would have the following form: "Assume that $f, g \in A\left[x\right]$ are two polynomials satisfying $fg = 0$. Then, there exists a finite list $\left(u\_1, u\_2, \ldots, u\_n\right)$ of elements of $A$ with the following two properties:
(1) Each coefficient of $g$ belongs to this list.
(2) For each $i \in \left\{1,2,\ldots,n\right\}$, if we have $u\_1 = u\_2 = \cdots = u\_{i-1} = 0$, then $u\_i f = 0$."
Such a finite list would non-constructively be a witness for the regularity (= non-zero-divisorness) of $f$, because property (1) would ensure that some $i$ satisfies $u\_i \neq 0$, and then (2) would ensure that if we choose the smallest such $i$, then $u\_i$ kills $f$.
For example, if $f = a + bx$ and $g = c + dx$, then $\left(bc, c, d\right)$ is such a list.
I am not sure if such a list exists in the general case, but I consider it plausible and wonder what its smallest size could be. If my hunch is right, the entries of this list are products of coefficients of $f$ and $g$, so there should be a combinatorics behind this all.
| 5 | https://mathoverflow.net/users/2530 | 441611 | 178,235 |
https://mathoverflow.net/questions/441612 | 3 | I am trying to understand something which is probably basic for experts so I am sorry if this is not suited for this forum.
Let $\mathcal{O}\_n$ denote the ring of germs at $0 \in \mathbb{R}^n$ of real-analytic functions. It is known that $\mathcal{O}\_n$ is a [Noetherian ring](https://en.wikipedia.org/wiki/Noetherian_ring) (see [1, Ch 3, Thm 3.8]). I understand from reading related bibliography that this should imply (quite easily?) the following result, but I cannot find a reference where it is stated exactly as below.
**Theorem?** Let $f \in \mathcal{O}\_n$ and $\varphi\_1, \dotsc, \varphi\_q \in \mathcal{O}\_n$. Then there exist $g\_1, \dotsc, g\_q \in \mathcal{O}\_n$ such that $f = g\_1 \varphi\_1 + \dotsb + g\_q \varphi\_q$ if and only if there exists a formal solution at the origin (i.e. formal series $\hat{g}\_1, \dotsc, \hat{g}\_q$ in $n$ variables such that $\hat{f} = \hat{g}\_1 \hat{\varphi}\_1 + \dotsb + \hat{g}\_q \hat{\varphi}\_q$ where $\hat{f}$ and $\hat\varphi\_i$ denote the Taylor series of $f$ and $\varphi\_i$ at 0).
I have read statements with $q > 1$ (such as [1, Ch 6, Th 1.1']) where the assumptions must hold in a whole vicinity of the origin (not just at the origin as above) and statements with $q = 1$ (see [2]) where the assumptions hold only at the origin (even in the quasianalytic setting).
My understanding is that the above theorem should follow rather easily from the Noetherianity of the ring of germs at 0, but I somehow failed to prove it by myself. **So, does the above result hold and is it stated somewhere?**
1. Malgrange, *Ideals of differentiable functions*.
2. Jan Nowak, *[On division of quasianalytic function germs](https://doi.org/10.1142/S0129167X13501115)*
| https://mathoverflow.net/users/50777 | Does Noetherianity imply division theorem? | This is a consequence of the noetherianity of $\mathcal{O}\_n$, in some way.
Let me write $A\_n$ for the ring of real formal formal series around the origin, i.e., $A\_n=\mathbb{R}[[t\_1,\dotsc,t\_n]]$. Let $I$ be the ideal of $\mathcal{O}\_n$ generated by $\varphi\_1,\dotsc,\varphi\_q$. Then your question is equivalent to whether
$$ \mathcal{O}\_n\cap IA\_n = I.$$
Let $\mathfrak{m}$ be the maximal ideal of $\mathcal{O}\_n$. Then $A\_n$ is the completion of $\mathcal{O}\_n$ at with respect to the $\mathfrak{m}$-adic topology. The fact that $\mathcal{O}\_n$ is noetherian now means that $A\_n$ is faithfully flat over $\mathcal{O}\_n$ (see [the Stacks project, Lemma 10.97.3](https://stacks.math.columbia.edu/tag/00MC), for instance). This means that the evident map $\mathcal{O}\_n/I\to (\mathcal{O}\_n/I)\otimes\_{\mathcal{O}\_n}A\_n\cong A\_n/IA\_n$ is injective, and it follows that $\mathcal{O}\_n\cap IA\_n = I$.
| 3 | https://mathoverflow.net/users/86006 | 441621 | 178,239 |
https://mathoverflow.net/questions/441634 | 7 | A classic result, of Murray and Von Neumann I believe, is that if $\mathcal M\subseteq B(H)$ is a factor then the $\*$-homomorphism $\pi : \mathcal M \odot \mathcal M' \rightarrow B(H)$ given by $\pi(x\_1\otimes x\_2) = x\_1x\_2$ on simple tensors is injective.
On the other hand, if $A \subseteq B(H)$ is a commutative C\*-algebra then this same multiplication map from $A\odot A \rightarrow B(H)$ is clearly not injective.
Question: For $A,B\in B(H)$ commuting C\*-algebras, when is the multiplication map $A\odot B\rightarrow B(H)$ injective?
Are there any conditions on one or both of the C\*-algebras that ensure this? E.g. $A$ simple? I've been looking but cannot find anything.
| https://mathoverflow.net/users/76593 | When is the multiplication map of the algebraic tensor product of C*-algebras injective? | For a commuting pair of C\*-algebras $A,B\subset B(H)$, the multiplication map is injective if and only if there are no nonzero elements $a\in A$ and $b\in B$ with $ab=0$. In particular, if $A$ is simple (and non-degenerate on $H$), then the multiplication map is injective.
This is because the pure states on C\*-algebras are "excised"; for any pure state $\phi$ on $A$, there is a net $e\_i$ in $A$ with $0\le e\_i\le 1$ and $\|e\_i\|=1$ that satisfies $\|e\_iae\_i - \phi(a)e\_i^2\|\to0$ for all $a\in A$. It follows that any pure states on $\phi$ on $A$ and $\psi$ on $B$ give rise to a state $\phi\times\psi$ on $C^\*(A,B)$ as long as $ab\neq0$ for any nonzero $a\in A\_+$ and $b\in B\_+$ (which immediately implies $\|ab\|=\|a\| \|b\|$ for every pair $(a,b)\in A\_+\times B\_+$ by functional calculus). See "Another proof of Proposition 3.4.7" in my book with Brown (p. 82).
Also, at the algebraic level, it is known that if $A$ is a central simple algebra (any simple C\*-algebra is central simple), then every ideal in the algebraic tensor product $A \otimes B$ (over the relevant field) is of the form $A \otimes I$ for some ideal $I$ in $B$. See e.g., Drozd & Kirichenko "Finite Dimensional Algebras" Theorem 4.3.2.
| 10 | https://mathoverflow.net/users/7591 | 441638 | 178,244 |
https://mathoverflow.net/questions/440731 | 20 | Let $V\_0, V\_1, \dots, V\_n$ denote a series of finite-dimensional vector spaces. We write $v\_i : = \dim V\_i$ for $i=0, 1, \dots, n$. I am thinking of these as real vector spaces, but I think the answer to my question ultimately doesn't depend on that.
Consider the affine space $A^N := \oplus\_{i=0}^{n-1} \mbox{Hom}(V\_{i+1},V\_i)$. In $A^N$, consider the variety $W$ of $n$-tuples of linear maps $(\phi\_1, \phi\_2, \dots, \phi\_n)$ such that $\phi\_i \circ \phi\_{i+1} =0$ for every $1 \le i \le n-1$. Here, every point in $W$ represents a chain complex on the given vector spaces.
**What is $\dim W$, as a function of $v\_0, v\_1, \dots, v\_n$?**
Here is what I know so far. For every $n$-tuple of non-negative integers $(r\_1, r\_2, \dots, r\_n)$, we can consider $W(r\_1, r\_2, \dots, r\_n)$ defined to be the subvariety where $\mbox{rank} \, \phi\_i = r\_i$ for every $1 \le i \le n$. It is not hard to see that this subvariety is nonempty if and only if $r\_i + r\_{i+1} \le v\_i$ for $i=0, 1, \dots$, so we will make this assumption going forward. (We make the convention that $r\_0 =r\_{n+1} = 0$.)
There are formulas for the dimensions of these subvarieties. In particular,
$$ \dim W(r\_1, r\_2, \dots, r\_n) = \sum\_{i=0}^n (v\_i-r\_{i+1})(r\_i+r\_{i+1}).$$
This can be found, for example, as Lemma 2.3 in the paper ["On the Variety of Complexes", by De Concini and Strickland, which appeared in Advances in Mathematics in 1981](https://www.sciencedirect.com/science/article/pii/S0001870881800047). They say that this is a well-known result, and they are including a proof for the sake of completeness. (Comment: they actual consider the subvarieties where $\mbox{rank} \, \phi\_i \le r\_i$, but these have the same dimension, so the above formula still holds.)
So one way to answer my question is that
$$ \dim W = \max \{ \dim W(r\_1, \dots, r\_n)\},$$
as $(r\_1, \dots, r\_n)$ ranges over all possible values. However, this is a little bit unsatisfying. It reduces the question to a quadratic optimization problem, over integer points in the convex polytope cut out by the linear inequalities:
1. $0 \le r\_i$ for every $i$
2. $r\_i + r\_{i+1} \le v\_i$
We observe that $\dim W(r\_1, r\_2, \dots, r\_n)$ is strictly increasing in each $r\_i$. However, this observation does not solve the optimization problem by itself. In fact, there may be many subvarieties that have equal, maximum dimension.
For example, if $(v\_0, v\_1, \dots, v\_6) = (10,10,10,10,10,10,10)$, then the ranks that maximize dimension are $(7,3,4,6,2,8)$, $(7,3,5,5,2,8)$, $(7,3,5,5,3,7)$, $(8,2,5,5,2,8)$, $(8,2,5,5,3,7)$, and $(8,2,6,4,3,7)$.
In special cases, I can solve this exactly. For example, it seems more tractable when the dimensions of the vector spaces are all equal $v\_0 = v\_1 = \dots = v\_n$, although even then the cases of $n$ even and $n$ odd seem to work out quite differently.
I am curious if there are some things that make $\dim W$ more amenable to computation than this, in the general setting. Is there either something that greatly simplifies the quadratic optimization problem, or another way entirely to compute $\dim W$?
De Concini and Stickland mention at the beginning of their paper that the varieties $W$ and $W(r\_1, \dots, r\_n)$ are called Buchsbaum–Eisenbud varieties of complexes.
| https://mathoverflow.net/users/4558 | What is the dimension of the variety of chain complexes? | There is a general special case where the problem has a nice answer. The situation is when the dimensions of the vector spaces are equal in pairs, or in other words $\dim V\_{2i}=\dim V\_{2i+1}$ for $0\le i\le \lfloor (n-1)/2 \rfloor$.
For convenience, we keep the auxiliary $r\_{0}=r\_{n+1}=0$, and in the case of even $n$, we also add $r\_{n+2}=v\_{n}$ and $v\_{n+1}=v\_n$. One can regroup the terms in the formula for $\dim W(r\_1, r\_2, \dots, r\_n)$ as follows:
$$\dim W(r\_1, r\_2, \dots, r\_n) = \sum\_{i=0}^n (v\_i-r\_{i+1})(r\_i+r\_{i+1})$$
$$=\sum\_{i=0}^{\lfloor n/2\rfloor} v\_{2i} v\_{2i+1}-
\frac12\sum\_{i=0}^{\lfloor n/2\rfloor}(r\_{2i}-r\_{2i+2})^2-\sum\_{i=0}^{\lfloor n/2 \rfloor}(v\_{2i+1}-r\_{2i}-r\_{2i+1})(v\_{2i}-r\_{2i+1}-r\_{2i+2})
+\frac{1}{2}r\_{2\lfloor\frac{n}{2}\rfloor+2}^2.$$
However, in our special case, the third summation is equal to $$\sum\_{i=0}^{\lfloor n/2 \rfloor}(v\_{2i+1}-r\_{2i}-r\_{2i+1})(v\_{2i}-r\_{2i+1}-r\_{2i+2})=\sum\_{i=0}^{\lfloor n/2 \rfloor}(v\_{2i}-r\_{2i}-r\_{2i+1})(v\_{2i+1}-r\_{2i+1}-r\_{2i+2}),$$
and we know that this must vanish at a maximum of $\dim W(r\_1, r\_2, \dots, r\_n)$, by the observation that our dimension is monotonic in each $r$, and thus, for each $i$, either $r\_{2i}+r\_{2i+1}\le v\_{2i}$ or $r\_{2i+1}+r\_{2i+2}\le v\_{2i+1}$ turns into an equality (otherwise you could increase $r\_{2i+1}$ by a small amount).
Hence, our problem reduces to minimizing the expression
$$\sum\_{i=0}^{\lfloor n/2\rfloor}(r\_{2i}-r\_{2i+2})^2 \tag{\*}$$
subject to our restrictions. This can be done by applying Cauchy-Schwarz, which tells us that the differences $r\_{2i}-r\_{2i+2}$ would have to be as close as possible to each other.
When $n$ is odd, the maximum can be achieved by taking $r\_0=r\_2=\cdots=r\_{n+1}=0$, and the remaining values will be $r\_{2i+1}=v\_{2i}=v\_{2i+1}$ for $0\le i\le \frac{n-1}{2}$.
When $n$ is even we have $r\_0=0$ and $r\_{n+2}=v\_n$. We can visualize our restriction as saying that the data points $(i, r\_{2i})$ must lie below the points $(i,\min(v\_{2i}, v\_{2i-1}))$ for $1\le i\le \frac{n}{2}$ together with endpoints $(0,0)$ and $(\frac{n}{2}+1,v\_n)$. Applying Cauchy-Schwarz in intervals shows that the (real) maximum is achieved when the points $(i,r\_{2i})$ are on the Newton polygon (convex hull) of the points $(i,\min(v\_{2i}, v\_{2i-1}))$, and the integral maximum is obtained by taking $r\_{2i+2}-r\_{2i}$ to be as close as possible to the slope of the corresponding edge of the Newton polygon.
There will be choices of taking either the ceiling or the floor of the slope, but we must make sure the final sum of all assigned values to the $r\_{2i+2}-r\_{2i}$'s comes out to $v\_n$. In the example of $(v\_0,v\_1,\dots,v\_6)=(10,10,\dots,10)$, we want $r\_{2i+2}-r\_{2i}$ to be as close as possible to the slope $10/4=2.5$ so we can choose either $2$ or $3$, since the final sum is $10$ the choices have to be some ordering of $2,2,3,3$, and each of the $6$ orderings corresponds to one of the maxima you listed.
| 5 | https://mathoverflow.net/users/2384 | 441643 | 178,247 |
https://mathoverflow.net/questions/436099 | 7 | Are there peer-reviewed journals that focus on "fun mathematics"?
By this I mean fun things that do involve nontrivial mathematics and which I think other mathematicians would enjoy reading in their leisure time but that are not interesting from the point of view of contemporary mathematical research.
For instance say that I have found a winning strategy for a folklore game that it is not as complicated as chess or Go but also not so trivial that you would figure it out in one afternoon. Or maybe say that I found some nice mathematical "magic trick" that was not known before. Where could I submit these things?
Of course there are journals that just accept more or less anything but this is not what I am looking for.
| https://mathoverflow.net/users/36563 | Are there journals for "fun mathematics"? | I agree this usually goes under the name "recreational mathematics." I'd suggest looking at The Mathematical Gazette as well, which has a range of articles including short notes and problems.
| 2 | https://mathoverflow.net/users/87779 | 441647 | 178,249 |
https://mathoverflow.net/questions/441636 | 3 | Given a quadratic program,
$$\begin{array}{ll} \text{minimize} & \displaystyle \frac12 x^TAx + b^Tx \\ \text{subject to} & Cx \le d \end{array}$$
Suppose $A \succ 0$, so the program strongly convex. The question is, is the solution $x^\*$ continuous with respect to the weights $A$ and $b$ ?
If we only have equality constraints, this is easy since we can simply write out the system of equations using the Lagrangian and the solution is determined by the system. However, with the inequality constraints, we need to deal with active / inactive constraints. It is not obvious to me how we can analyze the change in the solution when some inactive constraints become active or vice versa.
Some related reference:
1. [Perturbation Analysis of Optimization Problems, J. Frédéric Bonnans, Alexander Shapiro](https://link.springer.com/book/10.1007/978-1-4612-1394-9)
Update: I found the following reference characterizes the problem. Basically, we need regularity of the linear system.
[Continuity of the optimal value function in indefinite quadratic programming, Tam, N.N. Journal of Global Optimization 23, 43–61 (2002)](https://link.springer.com/content/pdf/10.1023/a:1014011906646.pdf)
| https://mathoverflow.net/users/500057 | Sensitivity of the solution of QP with respect to parameters | $\newcommand\R{\mathbb R}\newcommand\tz{\tilde z}\newcommand{\de}{\delta}$Yes, the minimizer $x\_{A,b}$ of $\frac12 x^TAx + b^Tx$ subject to $Cx\le d$ is
continuous with respect to $A$ and $b$ -- provided that the set $X:=\{x\in\R^n\colon Cx\le d\}$ is nonempty (otherwise, such a minimizer does not exist).
Indeed, let $M^+$ denote the set of all (symmetric) positive definite $n\times n$ real matrices. In what follows, $A,A\_k$ are in $M^+$, and $b,b\_k$ are in $\R^n=\R^{n\times1}$, where $k$ is any natural number.
Using the substitution $x=f\_{A,b}(z):=A^{-1/2}z-A^{-1}b$, we rewrite the problem as
\begin{equation\*}
\text{minimize } z^T z \quad \text{over } z\in Z\_{A,b},
\end{equation\*}
where
\begin{equation\*}
Z\_{A,b}:=\{z\in\R^n\colon Cf\_{A,b}(z)\le d\}=f\_{A,b}^{-1}(X).
\end{equation\*}
So, $Z\_{A,b}$ is a non-empty closed subset of $\R^n$ and hence there is a unique minimizer, say $z\_{A,b}$, of $z^T z$ over $z\in Z\_{A,b}$. So, $x\_{A,b}:=f\_{A,b}(z\_{A,b})=A^{-1/2}z\_{A,b}-A^{-1}b$ is the unique minimizer of $\frac12 x^TAx + b^Tx$ subject to $Cx\le d$.
So, it remains to show that $z\_{A,b}$ is continuous in $A,b$.
To do this, suppose that $A\_k\to A$ and $b\_k\to b$ (as $k\to\infty$). Then
\begin{equation\*}
z\_k:=f\_{A\_k,b\_k}^{-1}(f\_{A,b}(z\_{A,b}))\in Z\_{A\_k,b\_k} \tag{0a}\label{0a}
\end{equation\*}
and
\begin{equation\*}
z\_k\to z\_{A,b}, \tag{0b}\label{0b}
\end{equation\*}
so that
\begin{equation\*}
z\_k^T z\_k\to z\_{A,b}^T z\_{A,b}=m\_{A,b}:=\min\{z^T z\colon z\in Z\_{A,b}\}
\end{equation\*}
and hence
\begin{equation\*}
z\_{A\_k,b\_k}^T z\_{A\_k,b\_k}=m\_{A\_k,b\_k}\le m\_{A,b}+o(1); \tag{1}\label{1}
\end{equation\*}
in particular, the sequence $(z\_{A\_k,b\_k})$ is bounded.
Therefore and because $A\_k\to A$ and $b\_k\to b$, we see that for
$$\tz\_k:=f\_{A,b}^{-1}(f\_{A\_k,b\_k}(z\_{A\_k,b\_k}))$$
we have $\tz\_k-z\_{A\_k,b\_k}\to0$. Therefore and because $(z\_{A\_k,b\_k})$ is bounded, we have $\tz\_k^T \tz\_k-z\_{A\_k,b\_k}^T z\_{A\_k,b\_k}\to0$.
Also, $\tz\_k\in Z\_{A,b}$. So,
\begin{equation\*}
m\_{A,b}\le \tz\_k^T \tz\_k=z\_{A\_k,b\_k}^T z\_{A\_k,b\_k}+o(1)=m\_{A\_k,b\_k}+o(1). \tag{2}\label{2}
\end{equation\*}
By \eqref{1} and \eqref{2},
\begin{equation\*}
m\_{A\_k,b\_k}\to m\_{A,b}. \tag{3}\label{3}
\end{equation\*}
To obtain a contradiction, suppose now that $z\_{A\_k,b\_k}\not\to z\_{A,b}$. Then, passing to a subsequence, without loss of generality (wlog) assume that $|z\_{A\_k,b\_k}-z\_{A,b}|\ge2\de$ for some $\de>0$ and all $k$, where $|\cdot|$ is the Euclidean norm. So, by \eqref{0b}, wlog
\begin{equation\*}
|z\_{A\_k,b\_k}-z\_k|\ge\de \tag{4}\label{4}
\end{equation\*}
for all $k$. Because (i) the set $Z\_{A\_k,b\_k}$ is convex and (ii) the minimizer $z\_{A\_k,b\_k}$ of $z^T z$ over $z\in Z\_{A\_k,b\_k}$ is in $Z\_{A\_k,b\_k}$ and (iii) $z\_k\in Z\_{A\_k,b\_k}$ by \eqref{0a}, we see that $w\_k:=\frac12\,(z\_{A\_k,b\_k}+z\_k)\in Z\_{A\_k,b\_k}$. Using now (i) the parallelogram identity and (ii) the definitions of $z\_{A\_k,b\_k}$ and $m\_{A,b}$ and (iii) \eqref{0b} and \eqref{4}, we get
\begin{equation\*}
\begin{aligned}
4|w\_k|^2&=2|z\_{A\_k,b\_k}|^2+2|z\_k|^2-|z\_{A\_k,b\_k}-z\_k|^2 \\
&\le 2m\_{A\_k,b\_k}+2(m\_{A,b}+o(1))-\de^2 \\
&\le 4m\_{A\_k,b\_k}+o(1)-\de^2,
\end{aligned}
\end{equation\*}
so that for all large enough $k$ we have $w\_k^T w\_k=|w\_k|^2<m\_{A\_k,b\_k}$,
which contradicts the condition $w\_k\in Z\_{A\_k,b\_k}$, in view of the definition of $m\_{A\_k,b\_k}$.
Thus, $z\_{A\_k,b\_k}\to z\_{A,b}$, which proves that $z\_{A,b}$ is continuous in $A,b$. $\quad\Box$
| 3 | https://mathoverflow.net/users/36721 | 441648 | 178,250 |
https://mathoverflow.net/questions/441127 | 4 | Let $G$ be an abelian group and let $q:G \to \mathbb{Q/Z}$ be a quadratic form, i.e. $q(a)=q(-a)$ and $b(x,y)=q(x+y)-q(x)-q(y)$ is a bihomomorphism. On vector spaces, when people speak about the kernel of a quadratic form they mean the radical of $\frac{b}{2}$ which is obviously not possible here, at least not straight-forwardly. Taking the radical of $b$ itself is not sufficient since in general we don't have a quadratic form $\bar{q}:G/\mathrm{Rad}(b) \to \mathbb{Q/Z}$, s.t. $\pi^\*\bar{q}=q$ (for example let $G=\mathbb{Z}$ and $q(n)=\frac{n}{2} + \mathbb{Z}$). Hence my question:
Is there some definition/construction of the kernel of quadratic forms on abelian groups?
| https://mathoverflow.net/users/58211 | Is there a good notion of kernels of quadratic forms on abelian groups? | As requested, I post my comment (with mild changes) as answer:
In the theory of quadratic forms over a field of characteristic $2$, the radical of a quadratic form $q$ is sometimes defined as $R(q)=\{x\in \mathrm{rad}(b):q(x)=0\}$, where $b$ is the polar form of $q$. This should work in your situation as well: $R(q)$ is a subgroup of $G$ and $q$ factors via a $\mathbb{Q}/\mathbb{Z}$-valued quadratic map on $G/R(q)$.
The subgroup $R(q)$ should coincide with the kernel suggested in @YCor's comment.
| 2 | https://mathoverflow.net/users/86006 | 441655 | 178,251 |
https://mathoverflow.net/questions/441637 | 8 | $\DeclareMathOperator\vcd{vcd}\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Inn{Inn}\DeclareMathOperator\Out{Out}$Here I mean $\vcd(G)$ to be the virtual cohomological dimension of $G$. Some possible examples of what I mean by $H$ are $H=\Aut(G)$, $\Inn(G)$ or $\Out(G)$. I could extend this to other natural, more topological examples as well.
I've wondered about this and wondered if anyone actually speculated about this since the work in the 70s and 80s studying this problem where G is either a surface group (Harer, Penner, Thurston, etc.) or a free group (Culler, Vogtmann, Charney?, Bestvina, etc.). Of course we have the work of Borel, Serre and others for arithmetic groups, where G could be free abelian, or probably polycyclic-by-finite.
I would reframe and expand this question as follows:
Let $G$ have $\vcd(G)$ finite so we have a finite dimensional model $X$ of $K(G,1)$.
How can we construct a suitably good finite dimensional model $Y$ of $K(\Aut(G),1)$, for example?
As a side question, are there groups $G$ with $\vcd(G)$ finite but infinitely generated center $Z(G)$? What happens with $\Inn(G)=G/Z(G)$ then?
| https://mathoverflow.net/users/495429 | If G is a finitely generated group with vcd(G) finite, is vcd(H) finite for H, where H is an automorphism group of G? | Sorry, it is Rips' and not Mikhailova's construction: I should not comment when I am half-asleep.
Let me start with the Rips construction.
Let $Q$ be a finitely presented group. Rips in [1] constructed $C'(1/\lambda)$-[small cancellation](https://en.wikipedia.org/wiki/Small_cancellation_theory) groups $G$ (with arbibtarily large $\lambda$) and normal finitely generated subgroups $N< G$ such that $G/N\cong Q$. For $\lambda\ge 7$ the group $G$ will be hyperbolic.
A nice exposition of the Rips construction and its generalizations can be found in these two blog-posts: [here](https://berstein.wordpress.com/2011/02/27/the-rips-construction-i) and [here](https://berstein.wordpress.com/2011/03/02/the-rips-construction-ii). Actually, the Rips construction is quite flexible and one can make choices so that no defining relator of $G$ is a proper power; hence, the presentation complex of $G$ is aspherical. In particular, $G$ is torsion-free and is 2-dimensional. The subgroup $N$, therefore, is also 2-dimensional. However, the group $Q$ can be taken to have infinite virtual cohomological dimension.
We, thus, obtain a finitely generated 2-dimensional group $N$ such that $Out(N)$ contains $Q$ and thus has infinite vcd.
I am not sure how to find examples where $Aut$ has infinite vcd.
[1] *Rips, E.*, [**Subgroups of small cancellation groups**](https://doi.org/10.1112/blms/14.1.45), Bull. Lond. Math. Soc. 14, 45-47 (1982). [ZBL0481.20020](https://zbmath.org/?q=an:0481.20020).
**Edit.** A modification of the Rips construction in [2] yields examples where $Out(N)\cong Q$ even though it is not need for your question.
[2] *Bumagin, Inna; Wise, Daniel T.*, [**Every group is an outer automorphism group of a finitely generated group.**](https://doi.org/10.1016/j.jpaa.2004.12.033), J. Pure Appl. Algebra 200, No. 1-2, 137-147 (2005). [ZBL1082.20021](https://zbmath.org/?q=an:1082.20021).
| 4 | https://mathoverflow.net/users/39654 | 441675 | 178,256 |
https://mathoverflow.net/questions/441576 | 5 | Suppose that $f \colon [0, 1] \to \mathbb{R}$ is a $C^\infty$ function satisfying the constraints
$$
f(0) = f'(0) = f(1) = f'(1) = 0, \quad \mbox{and} \quad \int\_0^1 (f''(y))^2 \, dy \leq 1.
$$
Denote this class of functions $\mathcal{F}.$
I want to know what is the best approximation one can give of the $L^2$ norm of $f$ in terms of the evaluation of $f$ on a uniform grid. Basically, I want to know what we can say about
$$
\sup\_{f \in \mathcal{F}}
\Big\{\int\_0^1 f^2(y) \, dy - \frac{1}{n}\sum\_{i=1}^n f^2(i/n) \Big\}.
$$
My conjecture is that the error should go down as $n^{-4}$ since the quadratic interpolant should have this error, but I had difficulty checking this to be the case.
---
**Addendum:** I have added an additional periodicity constraint, primarily because as indicated by the argument from Iosif below, one can apply the Euler-Maclaurin formula to obtain $n^{-p}$ for any $p$ provided that $f \in C^\infty$ is periodic on $[0, 1]$. Hence, let's make the problem easier. We assume periodicity for the function and first order derivative. The naive application of Euler-Maclaurin gives an upper bound on the quantity above of $O(1/n^2)$ uniformly over the class. However, I cannot construct an $f$ that actually achieves this, subject to my constraints.
| https://mathoverflow.net/users/121486 | Optimal constant to compare $L^2$ norm of smooth function on $[0, 1]$ to a grid | By the [Euler–Maclaurin formula](https://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula#The_formula) (with $p=4$, $m=0$, and $g(x):=\frac1n\,f^2(\frac xn)$ in place of $f(x)$ there in the formula),
$$d\_n(f):=\int\_0^1 f^2(y) \, dy - \frac{1}{n}\sum\_{i=1}^n f^2(i/n) \\
=-\frac1{2n}\,f^2(1)-\frac1{6n^3}\,f(1)f'(1)+O(n^{-4})\le O(n^{-4})$$
for each $f\in\mathcal F$.
---
However, one cannot get a constant factor $O(n^{-4})$ good for all $f\in\mathcal F$ at once. Specifically, for $n\ge2$,
$$\sup\_{f\in\mathcal F}d\_n(f)\ge d\_n(f\_n)
=\frac{1}{\pi ^4 (2 n+1)^2}\sim\frac1{4\pi^4 n^2} \tag{1}\label{1}$$
as $n\to\infty$, where
$$f\_n(x):=\frac1{\pi^2}\,
\Big(\sin (\pi x)-\frac{\sin (\pi (2 n+1) x)}{(2 n+1)^2}\Big),$$
so that $\int\_0^1(f\_n'')^2=1$ and $f\_n(0)=0=f\_n(1)$.
(Note that $f\_n\notin\mathcal F$, since $f\_n'(0)\ne0$ and $f\_n'(1)\ne0$. However, $f\_n$ can be approximated however closely by functions in $\mathcal F$ with respect to the norm given by the formula
$$\|f\|^2=\max\_{[0,1]}(f^2)+\int\_0^1(f'')^2.$$
More generally, this approximation shows that $\sup\_{f\in\mathcal F}d\_n(f)$ will not change if the conditions $f'(0)=0=f'(1)$ are removed from the definition of $f\in\mathcal F$.)
---
On the other hand, even without the conditions $f(1)=f'(1)=0$ on $f\in\mathcal F$ (added later by the OP), one can see that
$$\sup\_{f\in\mathcal F}d\_n(f)\le \frac1{72 n^5}+\frac1{12 n^2}\sim\frac1{12 n^2} \tag{2}\label{2}$$
as $n\to\infty$, so that the lower bound on $\sup\_{f\in\mathcal F}d\_n(f)$ in \eqref{1} is sharp up to a universal positive real constant factor.
To get \eqref{2}, use the [Euler–Maclaurin formula](https://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula#The_formula) again, but this time with $p=2$, which together with the condition $\int\_0^1(f'')^2\le1$ yields
$$d\_n(f)\le-\frac1{2n}\,f^2(1)-\frac1{6n^3}\,f(1)f'(1)+\frac1{12 n^2} \tag{3}\label{3}$$
for all $f\in\mathcal F$. Using the condition $\int\_0^1(f'')^2\le1$ again, now together with the condition $f'(0)=0$, we get $|f'(1)|\le\int\_0^1|f''|\le1$, whence
$$-\frac1{2n}\,f^2(1)-\frac1{6n^3}\,f(1)f'(1)
\le-\frac1{2n}\,f^2(1)+\frac1{6n^3}\,|f(1)|\le\frac1{72 n^5},$$
so that \eqref{2} follows from \eqref{3}.
| 2 | https://mathoverflow.net/users/36721 | 441682 | 178,259 |
https://mathoverflow.net/questions/441663 | 1 | It comes from estimates for wave equations.
>
> For any $ u=u(t,x)\in C\_{0}^{\infty}(\mathbb{R}^{1+2}) $, which is a smooth compactly supported function, prove that
> $$
> \|u\|\_{L^{\infty}(\mathbb{R}^{1+2}\,\,)}\leq C\|\partial\_{x\_1}\square u\|\_{L^1(\mathbb{R}^{1+2}\,\,)},
> $$
> where $ \square $ is the wave operator $ \partial\_t^2-\Delta $ and $ C $ is independent of $ u $.
>
>
>
>
> Given an example that there is not constant $ C $ such that
>
> $$
> \|u\|\_{L^{\infty}(\mathbb{R}^{1+2}\,\,)}\leq C\|\partial\_t\square u\|\_{L^1(\mathbb{R}^{1+2}\,\,)}.
> $$
>
>
>
Here is my try. Note that $ u $ is compactly supported, can consider $ v(t,x)=u(t+T,x) $ with $ T\gg 1 $ if necessary, I find that I can assume that $ u(0,x)=0 $ and $ \partial\_tu(0,x)=0 $. Then by using the formula of the solution for wave equations, I have
$$
u(t,x)=\int\_{0}^t\int\_{y\_1^2+y\_2^2\leq 1}\frac{sF(t-s,x\_1-sy\_1,x\_2-sy\_2)}{2\pi\sqrt{1-y\_1^2-y\_2^2}}dy\_1dy\_2ds,
$$
where $ F=\square u $. However, I cannot go on. Can you give me some references or hints?
| https://mathoverflow.net/users/241460 | How to prove $ \|u\|_{L^{\infty}}\leq C\|\partial_1\square u\|_{L^1} $ for any $ u\in C_0^{\infty}(\mathbb{R}^{1+2}) $? | Let $E$ be a fondamental solution of $\partial\_{x\_1}\square$. Then you have for $u$ compactly supported
$$
u=u\ast \delta=u\ast (\partial\_{x\_1}\square E)= (\partial\_{x\_1}\square u)\ast E,
$$
so that
$
\Vert u\Vert\_{L^\infty}\le \Vert \partial\_{x\_1}\square u\Vert\_{L^1}
\Vert E\Vert\_{L^\infty}
$
and
it is now enough to prove that $E$ can be chosen bounded:
take
$$
E\_0=\frac{1}{2π}H(t-\vert x\vert)(t^2-\vert x\vert^2)^{-1/2}, \quad H=\mathbf 1\_{\mathbb R\_+},
$$
the fondamental solution of $\square$ in two-space dimensions. We take
$$
E=\int\_0^{x\_1} E\_0(t, y, x\_2) dy,
$$
and noting that
$
\int\_0^a\frac{dy}{\sqrt{a^2-y^2}}
$
is bounded, you get the sought result.
| 2 | https://mathoverflow.net/users/21907 | 441683 | 178,260 |
https://mathoverflow.net/questions/441661 | 2 | In the literature, there are a number of results called *Rosenlicht's lemma*, but I am talking about the following one:
Let $T$ be a torus over a, in my case, number field $k$. Denote by $\bar{k}[T]$ the ring of regular functions on $T$, and by $\widehat{T} := \mathrm{Hom}\_{\bar{k}}(T\_{\bar{k}},\mathbb{G}\_m)$ the group of characters of $T$, where $T\_{\bar{k}} := T \times\_k \bar{k}$. Rosenlicht's lemma states that one has $$\widehat{T} \cong \bar{k}[T]^\*/\bar{k}^\*.$$
Now let $C$ be a smooth projective curve of genus $\geq 1$ over $k$ and for some $n \geq 1$, identify the closed points $p\_0,...,p\_n$ to obtain a singular curve $C'$ with an $(n+1)$-fold singularity $Q$. Then one has a morphism from the smooth affine curve $X :=C' \setminus \{Q\}$ into the *generalized Jacobian variety* $S$ of $C$ w.r.t. the divisor $D = p\_0+...+p\_n$, sitting in the middle of the exact sequence
$$0 \rightarrow T \rightarrow S \rightarrow A \rightarrow 0,$$
where $T$ is a torus of dimension $n$ and $A$ is the usual Jacobian variety of $C$, assuming of course we have an embedding $C \hookrightarrow J$. The geometric points of $\widehat{T}$ is dual $\mathrm{Div}\_{\bar{D}}^0(\bar{C})$. This is the subgroup of the group of divisors of $\bar{C}$, supported on $\bar{D}$ and algebraically equivalent to zero, i.e., of degree zero. By Rosenlicht's lemma, one has $\bar{k}[T]^\*/\bar{k}^\* \cong \mathrm{Div}\_{\bar{D}}^0(\bar{C})$.
**Question 1.** Does one have $\widehat{S} \cong \bar{k}[S]^\*/\bar{k}^\*$?
I have a feeling this is true, but I cannot explain precisely why, only provide some intuition, sorry if it doesn't make sense. Since $S$ is the extension of $A$ by $T$, for a regular function on $S$ on has to consider it on the abelian variety part and the torus part. For the case of $A$, all regular functions are constant and so $\bar{k}[A]^\*/\bar{k}^\* = 0$ and so we can sort of ignore it. Thus we only need to consider the torus part which Rosenlicht's lemma tells us that $\bar{k}[S]^\*/\bar{k}^\* \cong \widehat{S}$.
Here I am not saying that $\widehat{S} \rightarrow \widehat{T}$ is an isomorphism, but I think this is an inclusion? Intuitively, a regular function on the whole of $S$ can always restrict to a regular function on $T$.
**Question 2.** One has an exact sequence of Galois modules
$$1 \rightarrow \bar{k}^\* \rightarrow \bar{k}[X]^\* \rightarrow \mathrm{Div}\_{\bar{D}}^0(\bar{C}) \rightarrow \mathrm{Pic}\_{\bar{D}}^0(\bar{C}),$$
where $X$ is the affine curve defined above. Could $\widehat{S}$ be the subgroup $$\mathrm{ker}(\mathrm{Div}\_{\bar{D}}^0(\bar{C}) \rightarrow \mathrm{Pic}\_{\bar{D}}^0(\bar{C}))?$$
| https://mathoverflow.net/users/172132 | Is there a Rosenlicht's lemma for semi-abelian varieties? | I am not sure which Lemma of Rosenlicht you mean. In "Toroidal algebraic groups. Proc. Amer. Math. Soc. 12 (1961), 984–988" he proves:
**Theorem 3:**
Let $\phi:\Gamma\to G$ be an everywhere defined rational map from
a connected algebraic group $\Gamma$ into a toroidal algebraic group $G$, with
$\phi(e)=0$. Then $\phi$ is a homomorphism.
Apply this to $G=\mathbf G\_m$. Since a morphism $\phi:\Gamma\to\mathbf G\_m$ is just a unit the theorem implies that $\chi:=\phi(e)^{-1}\phi$ is automatically a character. Thus
$$k[\Gamma]^\times=\operatorname{Hom}(\Gamma,\mathbb G\_m)\times k^\*$$ for any connected algebraic group, even nonlinear and noncommutative ones.
Does this settle **Question 1**?
| 3 | https://mathoverflow.net/users/89948 | 441685 | 178,261 |
https://mathoverflow.net/questions/441087 | 19 | **TL;DR.** Given a topological space $X$, is there a natural way to "induce" a topology on $\mathcal{P}(X)$ from the topology of $X$ in such a way that 1) all the basic operations of set theory (intersections, unions, direct images, etc.) become continuous and 2) the topologies on $X$ and $\mathcal{P}(X)$ are compatible in a certain sense?
---
Given a topological space $X$, there are a couple of ways to assign a topology to its powerset $\mathcal{P}(X)$ that do not depend specifically on $X$. For example:
* We can view $\mathcal{P}(X)$ as the hom $\operatorname{Hom}(X,2)$, put either the discrete, indiscrete, or Sierpiński topology on $2$, and consider the compact-open topology;
* We can take the Alexandroff topology with respect to $\subset$ or $\supset$;
* There are the so-called [hit-and-miss topologies](https://encyclopediaofmath.org/wiki/Hit-or-miss_topology), of which the Vietoris and Fell topologies are examples.
* (A non-example would be to define a topology on $\mathcal{P}(\mathbb{R})$ using the order of $\mathbb{R}$.)
Given an assignment of topologies on $\mathcal{P}(X)$ from topologies on $X$ as above, we could consider the following sets of niceness conditions:
1. The map $\iota\colon X\to\mathcal{P}(X)$ given by $x\mapsto\{x\}$ is continuous.
2. Binary union ${\cup}\colon\mathcal{P}(X)\times\mathcal{P}(X)\to\mathcal{P}(X)$ is continuous.
3. Binary intersection ${\cap}\colon\mathcal{P}(X)\times\mathcal{P}(X)\to\mathcal{P}(X)$ is continuous.
4. Difference ${\setminus}\colon\mathcal{P}(X)\times\mathcal{P}(X)\to\mathcal{P}(X)$ is continuous.
5. Arbitrary union ${\bigcup}\colon\mathcal{P}(\mathcal{P}(X))\to\mathcal{P}(X)$ is continuous.
6. Arbitrary intersection ${\bigcap}\colon\mathcal{P}(\mathcal{P}(X))\to\mathcal{P}(X)$ is continuous.
7. If $f\colon X\to Y$ is a continuous map of topological spaces, then so are its direct and inverse images
\begin{align\*}
f\_{\*} &{}\colon\mathcal{P}(X)\to\mathcal{P}(Y),\\
f^{-1} &{}\colon\mathcal{P}(Y)\to\mathcal{P}(X).
\end{align\*}
I'm also interested in the following two sets of additional conditions that are a bit more specific:
**More compatibility conditions between $X$ and $\mathcal{P}(X)$:**
8. For each $x\in X$, if $\{x\}$ is closed in $X$, then $\{\{x\}\}$ is closed in $\mathcal{P}(X)$.
9. If $S\subset X$ is closed in $X$, then $\{S\}$ is closed in $\mathcal{P}(X)$.
10. If $U\subset X$ is open in $X$, then $\{U\}$ is open in $\mathcal{P}(X)$.
**Another niceness requirement for the topology on $\mathcal{P}(X)$:**
11. Given any monoid structure $(\star,1\_X)$ on $X$ making it into a topological monoid, the map
$$\circledast\colon\mathcal{P}(X)\times\mathcal{P}(X)\to\mathcal{P}(X)$$
given by
$$U\circledast V:=\{uv\in X\ |\ u\in U,v\in V\}$$
is continuous.¹
12. (Implies 11 by Tobias's [second comment](https://mathoverflow.net/questions/441087/existence-of-a-really-nice-topology-on-the-powerset-of-a-topological-space#comment1137836_441087).) Given topological spaces $X$ and $Y$, the map
$$\mathcal{P}(X)\times\mathcal{P}(Y)\to\mathcal{P}(X\times Y)$$
given by $(U,V)\mapsto U\times V$ is continuous.
13. Given topological spaces $X$ and $Y$, the bijection
$$\mathcal{P}(X)\times\mathcal{P}(Y)\to\mathcal{P}(X\sqcup Y)$$
given by $(U,V)\mapsto U\cup V$ is a homeomorphism.
14. Given topological spaces $X$ and $Y$, the isomorphism of suplattices
$$\mathcal{P}(X)\otimes\mathcal{P}(Y)\to\mathcal{P}(X\times Y)$$
is a homeomorphism, where $\otimes$ denotes the tensor product of suplattices (see [Eric Wofsey's answer](https://math.stackexchange.com/a/3892072) to [Concrete description of the tensor product of suplattices?](https://math.stackexchange.com/questions/3891985/concrete-description-of-the-tensor-product-of-suplattices)).
**Question I.** Does there exist a powerset topology satisfying (at least but not necessarily only) conditions 1–7? What about (1–7+11), 1–8, 1–9, or 1–10?
**Question II.** Given a topological space $X$, what is the finest topology on $\mathcal{P}(X)$ satisfying conditions 1–6? What about (1–6 + 11–14), or these plus any of 8–10? Lastly, in case this topology turns out to be definable in a way that is independent of $X$ (like the Vietoris topology), do we also have 7?
---
**Footnotes.**
¹The motivation for this is that $\circledast$ is the zero-categorical analogue of ordinary Day convolution, and we may compute it via a completely analogous coend formula when viewing subsets $U$ of $X$ as functions $\chi\_U\colon X\to\{\mathrm{true},\mathrm{false}\}$, namely $\chi\_U\circledast\chi\_V=\int^{x,y\in X}\mathrm{Hom}\_{X}(-,xy)\times\chi\_U(x)\times\chi\_V(y)$.
| https://mathoverflow.net/users/130058 | Existence of a *really* nice topology on the powerset of a topological space | 1-7 together are pretty strong. You aren't going to have a procedure for doing this without most of the powerset topologies being indiscrete.
First note that if $X$ is a set and $\tau$ is at topology on $\mathcal{P}(X)$ satisfying 2-4, then the function $A \mathbin{\Delta} B = (A \cup B) \setminus (A \cap B)$ is continuous, which implies that for fixed $B\subseteq X$, the map $A \mapsto A \mathbin{\Delta} B$ is a homeomorphism. Therefore for any $A$ and $B$ in $\mathcal{P}(X)$, there is a homeomorphism of $(\mathcal{P}(X),\tau)$ taking $A$ to $B$ (namely the map $C \mapsto C \mathbin{\Delta} (A \mathbin{\Delta} B)$).
When I say a topology is *non-trivial*, I mean specifically that it is not indiscrete.
>
> **Lemma.** If $X$ is a set and $(\mathcal{P}(X),\tau)$ satisfies 2-4, then the closure $F$ of $\{\varnothing\}$ is a filter (i.e., $A \subseteq B \in F \Rightarrow A \in F$ and $A\in F \wedge B \in F \Rightarrow A \cup B \in F$). In particular, $\tau$ is non-trivial if and only if the $X \notin F$.
>
>
>
*Proof.* Suppose that $A \notin F$ and $B \supseteq A$. Since $\cap$ is continuous, we have that $G = \{C : C \cap A \in F\}$ is closed. Clearly $\varnothing \in G$, so $F \subseteq G$. Furthermore, $B \notin G$, so $B \notin F$.
For showing that $F$ is closed under unions, first note that it is sufficient to show it for disjoint $A,B \in F$. So assume that $A,B \in F$ and $A$ and $B$ are disjoint. Since $C \mapsto C \mathbin{\Delta} B$ is a homeomorphism, we have that $A \in F$ if and only if $A \mathbin{\Delta} B \in \overline{\{B\}}$ (since $\varnothing \mathbin{\Delta} B = B$). Therefore $A \mathbin{\Delta} B \in \overline{\{B\}} \subseteq F$, as required.
For the final statement. Since $\mathcal{P}(X)$ has a transitive homeomorphism group, $\tau$ is non-trivial if and only if $F$ is not all of $\mathcal{P}(X)$. Since $F$ is a filter, this happens if and only if $X \notin F$. $\square\_{\text{Lemma}}$
>
> **Proposition.** For any topological space $X$, if $(\mathcal{P}(X),\tau\_1)$ and $(\mathcal{P}(\mathcal{P}(X)),\tau\_2)$ satisfy 1-5 and 7 and $\tau\_1$ is non-trivial, then $X$ is discrete.
>
>
>
*Proof.* To hopefully make this proof a little bit easier to read, I'm going to denote the empty set as $\varnothing\_1$ when we're thinking about it as an element of $\mathcal{P}(X)$ and as $\varnothing\_2$ when we're thinking about it as an element of $\mathcal{P}(\mathcal{P}(X))$.
Let $F\_1$ be the $\tau\_1$-closure of $\{\varnothing\_1\}$, and let $F\_2$ be the $\tau\_2$-closure of $\{\varnothing\_2\}$. Since $\tau\_1$ is non-trivial, $X \notin F\_1$ by the lemma. Since $\bigcup: \mathcal{P}(\mathcal{P}(X)) \to \mathcal{P}(X)$ is continuous, we have that $\bigcup^{-1}(F\_1)$ is a closed set in $\mathcal{P}(\mathcal{P}(X))$. Since $\varnothing\_2 \in \bigcup^{-1}(F\_1)$, we have that $F\_2 \subseteq \bigcup^{-1}(F\_1)$, i.e., if $A \in F\_2$, then $\bigcup A \in F\_1$. Therefore, in particular, $\{X\} \notin F\_2$.
By 7, every homeomorphism $f$ of $\mathcal{P}(X)$ induces a homeomorphism $f\_\ast$ of $\mathcal{P}(\mathcal{P}(X))$. Each such homeomorphism $f\_\ast$ fixes $\varnothing\_2$ and takes singletons to singletons (specifically, for any $A \in \mathcal{P}(X)$, $f\_\ast(\{A\}) = \{f(A)\}$). By the discussion before the lemma, the homeomorphism group of $\mathcal{P}(X)$ is transitive. Therefore $\{A\} \notin F\_2$ for every $A \in \mathcal{P}(X)$ and so $F\_2 = \{\varnothing\_2\}$ (because $F\_2$ does not contain one particular singleton, $\{X\}$, so it does not contain any singletons, since it needs to be invariant under any homeomorphisms of $\mathcal{P}(\mathcal{P}(X))$ fixing $\varnothing\_2$).
Consider the function $(x,y) \mapsto \{x\} \cap \{y\}$ from $\mathcal{P}(X) \times \mathcal{P}(X) \to \mathcal{P}(\mathcal{P}(X))$. By 1 and 3, this is continuous. Since $\{\varnothing\_2\}$ is closed, this implies that the off diagonal $\{(x,y) \in \mathcal{P}(X) \times \mathcal{P}(X): x \neq y\}$ is closed and therefore the diagonal is open. It's relatively straightforward to show that if the diagonal of $Y^2$ is open, then $Y$ has the discrete topology. Therefore $\tau\_1$ is discrete. Repeating the argument again (since now we know that $\{\varnothing\_1\}$ is closed) gives that the topology on $X$ is discrete. $\square$
| 4 | https://mathoverflow.net/users/83901 | 441689 | 178,264 |
https://mathoverflow.net/questions/441610 | 10 | In Peter Johnstone's 1979 paper [On a topological topos](http://plms.oxfordjournals.org/content/s3-38/2/237.full.pdf), he proposed the topos of sheaves on the full subcategory of topological spaces spanned by the single object $\mathbb{N}\_\infty$, the one-point compactification of the natural numbers, as a convenient [big topos](https://ncatlab.org/nlab/show/big+and+little+toposes) whose objects are a sort of space. He showed that this topos (now sometimes known as [Johnstone's topological topos](https://ncatlab.org/nlab/show/Johnstone%27s+topological+topos)) has the following nice properties:
1. It is a local topos, i.e. its global sections functor (whose left adjoint constructs "discrete spaces") has a fully faithful right adjoint, which constructs "indiscrete spaces". (But it is not locally connected, hence not "cohesive".)
2. It contains the category of sequential topological spaces as a full subcategory closed under limits (and indeed reflective) --- although sequential spaces are *coreflective* in all topological spaces, so limits in the former don't coincide with limits in the latter.
3. The embedding of sequential spaces preserves colimits arising from open covers, closed covers, and quotients of sequentially closed equivalence relations. In particular, therefore, it preserves the construction of CW-complexes.
4. The internally-constructed [real numbers object](https://ncatlab.org/nlab/show/real+numbers+object) coincides with the usual (sequential) space of real numbers with its usual topology.
5. Geometric realization is the inverse image part of a geometric morphism from this topos to the topos of simplicial sets.
Much more recently, Barwick and Haine have introduced the topos of [pyknotic sets](https://ncatlab.org/nlab/show/pyknotic+set) with, it seems to me, a similar motivation. (The Scholze-Clausen category of [condensed sets](https://ncatlab.org/nlab/show/condensed+set) is related, but not a topos, so for purposes of this question I'm not as interested in it.) In some ways, pyknotic sets feel to me like an extension of Johnstone's topological topos to non-sequential notions of convergence. By definition, it consists of sheaves on a category of compact Hausdorff topological spaces, which includes $\mathbb{N}\_\infty$. And the paper [Pyknotic objects, I](https://arxiv.org/abs/1904.09966) proves that the topos of pyknotic sets has analogues of properties 1 and 2 above, and partly 3:
1. Pyknotic sets is a local topos.
2. It contains the category of compactly generated spaces as a full subcategory closed under limits, while compactly generated spaces are coreflective in all topological spaces.
3. This embedding preserves sequential colimits of compactly generated spaces whose colimit is $T\_1$.
Thus, I wonder whether the topos of pyknotic sets shares any of the other nice properties of Johnstone's topological topos. Namely:
* Does the embedding of compactly generated spaces preserve any more colimits, such as those arising from open or closed covers? In particular, does it preserve the construction of CW-complexes?
* Does the real numbers object in pyknotic sets coincide with the usual (compactly generated) space of real numbers with its usual topology?
* Is geometric realization the inverse image part of a geometric morphism from pyknotic sets to simplicial sets?
| https://mathoverflow.net/users/49 | Properties of pyknotic sets | Let me recall a little bit of the background. The question is about the relation between topological spaces and pyknotic sets, and properties of the topos of pyknotic sets. Recall that pyknotic sets are sheaves on the site of compact Hausdorff spaces, for the Grothendieck topology generated by finite families of jointly surjective maps. (So this includes, notably, finite covers by closed subsets; but because any open cover can be refined by a finite closed cover, open covers are also covers in this Grothendieck topology.) "Pyknotic" as opposed to "condensed" refers to one way of addressing the issue that the site is large; in this case, by fixing universes. The issue is inconsequential for the whole discussion, so let me not discuss it further.
For any topological space $X$, one can define a pyknotic set $\underline{X}$, whose $S$-valued points are the continuous maps from $S$ to $X$, for any compact Hausdorff space $S$.
On compactly generated topological spaces, this functor is fully faithful. By its definition, it also commutes with all limits. The first part of the question concerns the question to what extent it commutes with colimits.
The first type of colimits are geometric realizations. Assume you have a topological space $X$ that is written as the quotient of some topological space $Y$ by the equivalence relation $R=Y\times\_X Y\subset Y\times Y$ (with induced topology). Under which conditions is the natural map $\underline{Y}/\underline{R}\to \underline{Y/R}=\underline{X}$ an equivalence?
We do not want this to be the case in all situations, as for "bad" equivalence relations (with dense orbits), the topological quotient space is quite bad, and the pyknotic one is actually much better. But often, the topological construction is nice, so we want it to be preserved.
To check whether a map of pyknotic sets is an isomorphism, it suffices to check that it is an isomorphism on $S$-valued points where $S$ is an extremally disconnected compact Hausdorf space -- this is because any compact Hausdorff space can be covered by an extremally disconnected one. Moreover, evaluation at extremally disconnected sets commutes with all sifted colimits, and in particular with quotients by equivalence relations. So we have to see whether for all such $S$, the map
$$\mathrm{Cont}(S,Y)/\mathrm{Cont}(S,R)\to \mathrm{Cont}(S,X)$$
is bijective. Note that the map is necessarily injective -- if two continuous maps $f\_1,f\_2: S\to Y$ induce the same map to $X$, then they induce a continuous map to $R$. Thus, only surjectivity is at stake. Thus,
>
> The map $\underline{Y}/\underline{R}\to \underline{Y/R}=\underline{X}$ is an isomorphism if and only if for all extremally disconnected sets $S$, any continuous map $S\to X$ lifts to a continuous map $S\to Y$. Equivalently, for any compact Hausdorff space $S$ and any continuous map $S\to X$, there is a surjective map of compact Hausdorff spaces $S'\to S$ such that the composite map $S'\to S\to X$ lifts to a continuous map $S'\to Y$.
>
>
>
In particular, if $Y$ is a finite disjoint union of closed subsets of $X$ (corresponding to gluing $X$ from finitely many closed subsets), or if $Y$ is a disjoint union of open subsets of $X$ (corresponding to gluing $X$ from open subsets), then this holds. But there are much more general situations.
[**Edit:** Some details about why this is true for those two cases. If $Y$ is a finite disjoint union of closed subsets of $X$ (covering $X$), then after pullback along any continuous map $S\to X$, one gets a finite disjoint union of closed subsets of $S$ (covering $S$). This is itself a compact Hausdorff space, surjective over $S$, which hence admits a splitting (as $S$ is extremally disconnected). If $Y$ is a disjoint union of open subsets of $X$, then after pullback along any continuous map $S\to X$, one gets a disjoint union of open subsets of $S$, and as remarked above, this can be refined by a finite disjoint union of closed subsets of $S$ covering $S$, reducing us to the previous case.]
Another interesting type of colimit is a filtered colimit. Filtered colimits of topological spaces tend to be reasonable only when the transition maps are closed immersions, so let us assume that. Thus, consider a filtered index set $I$ and a diagram $X\_i$, $i\in I$, of compactly generated topological spaces, with transition maps closed immersions, and let $X=\mathrm{colim}\_i X\_i$ be their colimit (in compactly generated topological spaces). Under what conditions is the natural map
$$ \mathrm{colim}\_i \underline{X\_i}\to \underline{X} $$
an isomorphism?
Again, this can be checked on $S$-valued points for extremally disconnected $S$, and again injectivity is automatic, so the question is about surjectivity.
>
> For filtered colimits, the map $\mathrm{colim}\_i \underline{X\_i}\to \underline{X}$ is an isomorphism if and only if for any compact Hausdorff space $S$ with a continuous map $S\to X$, there is some $i$ such that $S\to X$ factors over $X\_i\subset X$.
>
>
>
(The factorization is necessarily continuous, as $X\_i\subset X$ is a closed subset.)
Unfortunately, this is not the case for all colimits, but it is often the case, as discussed in some detail in Proposition A.15 of Schwede's *[Global homotopy theory](https://www.math.uni-bonn.de/people/schwede/global.pdf)*. In particular, it is always the case for sequential colimits.
A counterexample is given by writing $[0,1]$ as the filtered colimit of all of its countable closed subsets -- this is a colimit in (compactly generated) topological spaces, but it is not a colimit in pyknotic sets. Note that it would also be a colimit in Johnstone's topos, but I find it more realistic that $[0,1]$ is actually "larger" than this filtered colimit.
Now there are two questions about the pyknotic topos. First, whether the internally constructed reals agree with the externally defined object $\underline{\mathbb R}$. As a rule of thumb, such agreements tend to always hold -- internal constructions in pyknotic sets automatically acquire the "correct"/"expected" pyknotic structure. (This is one advantage of the pyknotic/condensed formalism over classical topological spaces -- usually, whenever you do some mathematical construction, you have to specify the topology by hand; but pyknotic structures just come along for the ride.)
For the internal real numbers, let me consider Cauchy reals first. This is the quotient of the group of Cauchy sequences of rational numbers (with given convergence) by the group of null sequences of rational numbers (with given decay), both constructed internally in pyknotic abelian groups. The internally constructed group of Cauchy sequences of rational numbers takes any extremally disconnected $S$ to sequences of locally constant functions $f\_n: S\to \mathbb Q$ such that $|f\_n-f\_{n+1}|\leq 2^{-n}$; and in there one can consider the subspace of null sequences, where necessarily in addition $|f\_n|\leq 2^{1-n}$. It is an elementary exercise to see that their quotient is precisely the group of continuous maps $S\to \mathbb R$ for the usual topology on $\mathbb R$. In fact, one is just looking at the completion of the normed abelian group $\mathrm{Cont}(S,\mathbb Q)$, which is indeed the Banach space $\mathrm{Cont}(S,\mathbb R)$.
[**Edit:** Some details about this verification. First, given any such sequence $f\_n$, one can define a function $f\_\infty: S\to \mathbb R$ by taking $s\in S$ to the limit of the Cauchy sequence $f\_n(s)$. The function $f\_\infty$ is continuous: If $s\in S$ is any element and $f\_n$ is constant in the open neighborhood $U$ of $s$, then for all $s'\in U$, one has
$$|f\_\infty(s)-f\_\infty(s')|\leq |f\_\infty(s)-f\_n(s)|+|f\_n(s)-f\_n(s')|+|f\_\infty(s')-f\_n(s')|\leq 2^{1-n} + 0 + 2^{1-n}\leq 2^{2-n}.$$
If $f\_\infty=0$, then we must have $|f\_n(s)|\leq 2^{1-n}$ for all $n$ and hence the $f\_n$ lie in the subspace of null sequences. It remains to see that any continuous function $f: S\to \mathbb R$ can be written as a limit of locally constant functions $f\_n: S\to \mathbb Q$ with the required convergence. It suffices to show that for any $n$ one can find a locally constant function $f\_n: S\to \mathbb Q$ such that $|f-f\_n|\leq 2^{-n-1}$. To do so, consider for any $s\in S$ the open subset $U\_s\subset S$ of all $s'$ such that $|f(s)-f(s')|<2^{-n-1}$. These open sets cover, so there is a finite subcover. As $S$ is totally disconnected, this can be refined by a finite disjoint union of open and closed subsets. So there are finitely many open and closed subsets $U\_i\subset S$ with points $s\_i\in U\_i$ such that for all $s'\in U\_i$, one has $|f(s\_i)-f(s')|<2^{-n-1}$. Now define $f\_n$ to be equal to $f(s\_i)$ on $U\_i$.]
In the case of Dedekind reals, one would instead send any such $S$ to the Dedekind cuts over $S$, which essentially partition the locally constant functions $S\to \mathbb Q$ into a "left" and a "right" part. Again, it is an elementary exercise to see that all of them arise from some continuous map $S\to \mathbb R$.
[**Edit:** Some more details about this verification. Again, given such a Dedekind cut over $S$, one can define a function $f: S\to \mathbb R$ by taking any $s\in S$ to the value determined by the Dedekind cut in the fibre at $s$. To see that this is continuous, note that if $q<f(s)$ for some rational number $q$, then there must be some locally constant function $g: S\to \mathbb Q$ that is a lower bound for $f$, and such that $q<g(s)<f(s)$. As $g$ is locally constant, this implies that $q<g(s)=g(s')<f(s')$ for $s'$ in an open neighborhood $U$ of $s$. Similarly, upper bounds spread to neighborhoods. Conversely, note that any continuous function $f$ determines a unique Dedekind cut, as those locally constant functions to $\mathbb Q$ that lie below resp. above $f$.]
Finally, there was the question whether geometric realization defines a geometric morphism of topoi from pyknotic sets to simplicial sets. Taking away the jargon, this is just the question whether geometric realization commutes with finite limits. As it commutes with all colimits, and finite limits commute with filtered colimits, this reduces directly to the case of simplicial sets with finitely many nondegenerate simplices. But for such simplicial sets, the geometric realization is just a compact Hausdorff space, and the question of commutation with finite limits is independent of the context one works in (in fact, isomorphisms of compact Hausdorff spaces can be checked on underlying sets), and known to be true.
| 16 | https://mathoverflow.net/users/6074 | 441690 | 178,265 |
https://mathoverflow.net/questions/441681 | 4 | Consider the 1-dimensional Burger's equation on a finite interval $I=(0,1)$,
$$u\_t+uu\_x=u\_{xx}$$
with initial condition
$$u(x,0)=f(x)$$
and boundary conditions
$$u(0,t)=A(t) \qquad u(1,t)=B(t).$$
>
> **QUESTION.** Suppose that $A(t)$ and $B(t)$ are periodic functions, of integer period $m$ and $n$, respectively. Is it true that the solution to the above IBVP should be periodic in time? If so, what is the period in terms of $m$ and $n$?
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>
>
| https://mathoverflow.net/users/66131 | Periodicity and Burger's equation | Of course **not**. For instance if $A=B\equiv0$ (these are periodic), then the solution decays to $0$ as $t\to+\infty$.
Instead, and this is classical in dynamical system theory, if $A$ and $B$ are periodic as in your assumption, then you might be able to prove that there exists an initial data $f$ for which the solution is periodic with the same period $p:={\rm lcm}(m,n$). And it should be unique, because the solution mapping $S(t):u(0)\mapsto u(t)$ is $L^1$-contracting. The existence relies upon two ingredients: 1) denoting $M=\max\{\|A\|\_\infty,\|B\|\_\infty\}$, the set of measurable functions $a$ such that $|a(x)|\le M$ a.e. is sent into itself by $S(p)$, 2) $S(p)$ is compact in the $L^1$-topology.
| 5 | https://mathoverflow.net/users/8799 | 441692 | 178,266 |
https://mathoverflow.net/questions/441633 | 4 | I found a statement involving a homeomorphism $f:X\to X$ of a compact metric space $X$, with Lipshitz coefficient 1, i.e., a non-expansive map, and cannot think of an example where $f$ is not an isometry. Must it be?
| https://mathoverflow.net/users/172802 | Is every 1-Lipschitz homeomorphism $f:X\to X$ from a compact metric space to itself an isometry? | It follows from 1.6.15(1) in "A Course in Metric Geometry" by Burago, Burago, and Ivanov and 1.6.15(2) is a more general statement:
>
> Any distance-noncontracting map from a compact metric space to itself is an isometry.
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>
This statement is needed in the proof that Gromov--Hausdorff metric is a metric.
Also check the solution of 1.12 in my ["Pure metric geometry"](https://arxiv.org/abs/2007.09846) it is based on a different idea.
| 5 | https://mathoverflow.net/users/1441 | 441693 | 178,267 |
https://mathoverflow.net/questions/441707 | 13 | It's probably a really naive question, but I didn't find any references. Given a category $V$ and we know that it is equivalent to the category $\mathbf{Vect}(X)$ of smooth vector bundles over a smooth manifold $X$, under what assumptions could we determine $X$? Or is this generally hopeless?
| https://mathoverflow.net/users/104710 | Reconstructing base manifold from its category of smooth vector bundles | Yes, it is possible to recover the manifold through the following steps:
* [Smooth Serre-Swan theorem](https://ncatlab.org/nlab/show/smooth+Serre-Swan+theorem): $\mathbf{Vect}(X)$ is equivalent to the category of finitely generated projective modules over $C^\infty(X)$.
* If $R$ is any commutative ring, then $R$ is canonically isomorphic to the ring of natural endomorphisms of the identity functor on the category of finitely generated projective modules. Thus, we obtain $C^\infty(X)$ as the ring of natural endomorphisms of $\mathrm{id}\_{\mathbf{Vect}(X)}$.
* Finally, it is [well-known](https://math.stackexchange.com/a/1765040/745634) that a manifold can be recovered from its ring of smooth functions. More precisely, the functor $C^\infty$ from manifolds to the opposite category of commutative rings is fully faithful.
| 23 | https://mathoverflow.net/users/27013 | 441709 | 178,270 |
https://mathoverflow.net/questions/441713 | 2 | In the context of generic embeddings, we fix a set $Z$, an ideal $I$ on $Z$, and $G$ a generic ultrafilter on $\mathcal{P}(Z) / I$. In $V[G]$, we can define the generic ultrapower $Ult(V, G)$ and an elementary embedding from $V$ into $Ult(V, G)$. If $I$ is precipitous then $Ult(V, G) \cong M$ for some transitive class $M$ of $V[G]$, so we also have $j : V \rightarrow M$ an elementary embedding. Now if $I$ has the disjointing property then in particular, $M^\omega \subset M$, which is a nice closure property to have.
I am interested in the scenario where one iterates this ultrapower construction in a larger forcing extension. Specifically, we consider $Z = \omega\_1$ and $I = \mathrm{NS}\_{\omega\_1}$, the non-stationary ideal on $\omega\_1$. If $I$ is precipitous then in $V^{Col(\omega, 2^{\omega\_1})}$, $(V; \in, I)$ becomes generically iterable. Fix $G$ a $Col(\omega, 2^{\omega\_1})$-generic filter over $V$ and consider a generic iteration $\mathfrak{I}$ of $(V; \in, I)$ of length $\omega\_1^{V[G]}$ in $V[G]$. This yields an elementary embedding $j: V \rightarrow M$, where $M$ is the final iterate of $\mathfrak{I}$. Here $crit(j) = \omega\_1^V$ and $j(\omega\_1^V) = \omega\_1^{V[G]}$. In this case, what additional properties does $I$ need to have, in order to guarantee $M^\omega \subset M$ in $V[G]$?
| https://mathoverflow.net/users/29231 | Closure properties of elementary embeddings resulting from generic iterations | It is impossible that $M^{\omega}\subseteq M$ in $V[G]$. The map $j$ is continuous at $\omega\_2^V$, i.e. $\omega\_2^M=\sup\_{\alpha<\omega\_2^V} j(\alpha)$ (this can be seen by induction along the length of the iteration). As $\omega\_2^V$ is countable in $V[G]$, this means that $\omega\_2^M$ has countable cofinality in $V[G]$, but of course $\omega\_2^M$ is regular in $M$.
| 3 | https://mathoverflow.net/users/125703 | 441726 | 178,273 |
https://mathoverflow.net/questions/441714 | 8 | Does $x\_0=1/3$ lead to periodicity in the logistic map $x\_{k+1}=4x\_k(1-x\_k)$?
I believe it does not, but this is equivalent to proving that $(2\pi)^{-1}\arcsin(\sqrt{1/3})$ is irrational. I am wondering if there are any known results. After all, $1/3$ is the most rudimentary seed that (I suspect) leads to non-periodicity in the fully chaotic logistic map. Or maybe not?
| https://mathoverflow.net/users/140356 | Does $x_0=1/3$ lead to periodicity in the logistic map $x_{k+1}=4x_k(1-x_k)$? | The orbit of $1/3$ is infinite. You can show this via the [$3$-adic valuation $\nu\_3$](https://en.wikipedia.org/wiki/P-adic_valuation).
Let us show by induction that $\nu\_3(x\_n) = -2^{n}$:
We have $\nu\_3(x\_0) = \nu\_3(1/3) = -1 = -2^0$.
Now $\nu\_3(x\_{n+1}) = \nu\_3(4 x\_{n} (1 - x\_{n})) =
\nu\_3(4)+ \nu\_3(x\_n) + \nu\_3(1 - x\_n)$.
We have $\nu\_3(4) = 0$, and since $\nu\_3(x\_n)$ is negative by induction, we have
$\nu\_3(1 - x\_n) = \nu\_3(x\_n)$, so overall $\nu\_3(x\_{n+1}) = 2\nu\_3(x\_n) =-2^{n+1}$.
From this is is clear that the orbit must be infinite.
A similar argument shows that $p/q$ has infinite orbit for $p, q$ coprime, $q$ divisible by some odd prime.
| 20 | https://mathoverflow.net/users/500150 | 441730 | 178,276 |
https://mathoverflow.net/questions/441735 | 1 | We are considering the following problem:
Given an integer $n$ and a sequence of integers $r\_i,\ 1\le i\le n$, with $0\le r\_i\le n-1$ does there exists a symmetric matrix $A$ such that the diagonal elements of $A$ are $0$'s, the off-diagonal elements are only $-1$ and $1$ and for each $i$ there are exactly $r\_i$ $1$'s on the $i$-th row of $A$? How do we construct such a matrix? How do we construct all such matrices (up to some equivalence relation).
For example, if $n$ is even and half of $r\_i$ are equal to $n$, the other half to $n-1$. How do we construct such matrices? What about the case when $\{r\_i\}\_{i=1}^n$ is a constant sequence? When such a matrix exists?
The problem is related to the existence of certain codes with given distances between the codewords.
| https://mathoverflow.net/users/500153 | On the existence of symmetric matrices with prescribed number of 1's on each row | Replace each $-1$ with $0$. Then the matrix you look for is the adjacency matrix of a graph with a given degree sequence $r\_1, r\_2, \dots$. Reconstruction of such a graph (and its adjacency matrix) is known as the *graph realization* problem. E.g., see [Wikipedia](https://en.wikipedia.org/wiki/Graph_realization_problem) or [this webpage](https://piperfw.github.io/degreeSequences/) for additional details and solutions.
| 3 | https://mathoverflow.net/users/7076 | 441737 | 178,279 |
https://mathoverflow.net/questions/441739 | 3 | Recently, a problem in my research has appeared and now I need to construct some algebraic numbers with special properties (related to its degree and some other fields extensions).
Now, in order to help me, I would like to prove the following result:
**Proposition.** *Let $\alpha$ be a real algebraic number and let $n>4$ be a positive integer, then the polynomial $X^n-p$ is irreducible over $K:=\mathbb{Q}(\alpha)$, for all large enough prime number $p$.*
I tried to use some splitting fields, discriminantes properties, (un)ramified primes to prove it, but I was not able to do it.
Any suggestion is very welcomed.
Thanks in advance.
| https://mathoverflow.net/users/120084 | Irreducibility of polynomials over some number fields | **Lemma.** *Let $K$ be any number field, and $p$ a prime unramified in $K$. Then $X^n-p$ is irreducible over $K$.*
*Proof.* It suffices to show that the field $L = K(\sqrt[n\ \ ]{p})$ has degree $n$ over $K$. Let $\mathfrak q \subseteq \mathcal O\_L$ be a prime above $p$, and let $\mathfrak p = \mathcal O\_K \cap \mathfrak q$ be its image in $\operatorname{Spec} \mathcal O\_K$. Since $\mathcal O\_L$ contains $\mathbf Z[\sqrt[n\ \ ]{p}]$, we have $e\_{\mathfrak q/p} \geq n$. But $K$ is unramified above $p$, so $e\_{\mathfrak p/p} = 1$. We conclude that $e\_{\mathfrak q/\mathfrak p} = n$, so $[L:K] \geq n$. The reverse inequality is clear. $\square$
| 11 | https://mathoverflow.net/users/82179 | 441741 | 178,280 |
https://mathoverflow.net/questions/441740 | -2 | Recall the definition of [cardinal definable](https://mathoverflow.net/questions/412541/is-every-set-being-cardinal-definable-consistent-with-zf-negation-of-choice) sets, to re-iterate:
$Define: X \text { is cardinal definable} \iff \\\exists \text { cardinal } \kappa \, \exists \text { cardinals } \lambda\_1,.., \lambda\_n <^\rho \kappa \ \exists \phi : \\ X=\{ y \in V\_{\rho(\kappa)} \mid \phi^{V\_{\rho (\kappa)}} (y,\lambda\_1,..,\lambda\_n)\}$
Where: $\lambda\_i <^\rho \kappa \iff \rho(\lambda\_i) < \rho(\kappa)$, and $\rho$ is the rank function; and "*cardinal*" is defined after Scott's as an equivalence class under *bijection* of sets of the lowest possible rank.
That every set is cardinal definable is proved consistent with the failure of choice (see [here](https://mathoverflow.net/a/412591/95347)), on the other hand, it is also proved consistent with choice (see [here](https://mathoverflow.net/a/412536/95347)).
>
> Now, working in $\sf ZF$, if we say that every cardinal definable set admits a choice function, would that entail full $\sf AC$?
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> If we work in $\sf ZF-Reg.$, would cardinal definable choice imply $\sf AC$?
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| https://mathoverflow.net/users/95347 | Does cardinal definable choice imply AC? | Over ZF yes, it does.
Let $X$ be any family of nonempty sets, and let $κ$ be cardinal such that $ρ(κ)>ρ(X)$, then $V\_{ρ(κ)}\setminus\{∅\}$ is cardinal definable, hence has a choice function that induce a choice function on $X$.
In ZF-Regularity Scott trick doesn't work anymore, in fact Lévy has shown that it is consistent with ZF-Regularity that there is no assignment to each set a cardinality in any way (that is: it is consistent that there is no $C:V→V$ such that $C(x)=C(y)⇔|x|=|y|$ (even with parameters)).
In e.g. ZF-Regularity+AFA were you do have possible definition of cardinality for every set, similar idea to what I wrote above works: You start with with a set $X$, and define $X'=X\cap WF$ and $X^∘=X\setminus X'$, by the above $X'$ has a choice function, for $X^∘$ define $Y$ to be the class of accessible pointed graphs who are pictures for elements is in $Y$. Let $f:Y∩WF→X^∘$ be a function that sends $y$ to the unique element in $X^∘$ that $y$ is it's picture.
Note that $f''(Y∩WF)=X^∘$, and in fact that there exists a *set* $Z\subseteq Y∩WF$ such that $f''Z=X^∘$ (We don't need choice to get this $Z$, we can use Scott's trick, because $Y∩WF\subseteq WF$).
From the above $Z$ and $\bigcup Z$ are well-orderable, so let $x\in X^∘$, and $(G,E)∈Z$ be the minimal graph with the unique decoration $π$ such that $x\in\pi'' G$, and let $p$ be the minimal element (in $\bigcup Z$) such that such that $p E\pi^{-1}(x)$, and so $π(p)∈x$ and we have a choice function on $X=X'⊔X^∘$
| 4 | https://mathoverflow.net/users/113405 | 441750 | 178,282 |
https://mathoverflow.net/questions/441751 | 0 | Suppose we are given a scheme $S$ and two vector bundles $V$ and $W$ over $S$. Is it always true that $\mathbb{P}(V)\cong \mathbb{P}(W)$ implies that $V\cong W$ as $S$-schemes?
If the statement is false then what is the most general condition on $S$ for which it becomes true?
| https://mathoverflow.net/users/99988 | Are projective bundles corresponding to non-isomorphic vector bundles always non-isomorphic? | This is not true as stated, because for any line bundle $L$ on $S$ one has
$$
\mathbb{P}(V \otimes L) \cong \mathbb{P}(V),
$$
but this is the only issue. Indeed, if $X = \mathbb{P}(V) \stackrel{p}\to S$ and $S$ is connected, the relative Picard group $\mathrm{Pic}(X/S)$ is cyclic, and if $H$ is a lift to $\mathrm{Pic}(X)$ of its relatively ample generator, then
$$
p\_\*\mathcal{O}\_X(H) \cong V^\vee,
$$
and since $H$ is uniquely defined up to twist by $p^\*(\mathrm{Pic}(S))$, it follows that $V$ is uniquely defined up to twist by $\mathrm{Pic}(S)$.
| 7 | https://mathoverflow.net/users/4428 | 441755 | 178,283 |
https://mathoverflow.net/questions/441295 | 3 | Are there any results that allow us to characterize affine schemes via morphisms to/from other schemes?
Suppose we have a category $\mathcal{C}$ which is equivalent to $\mathbf{Sch}$, the category of schemes. We do not know anything about the equivalence. We want to find the objects $X \in \mathcal{C}$ such that these objects are equivalent exactly to the affine schemes in $\mathcal{C}$.
Is this possible? If not, can we modify the problem slightly (e.g. replace $\mathbf{Sch}$ with a reasonable subcategory, or assume the existence of some reasonable functors) to get an answer? Or is the impossibility fundamental?
| https://mathoverflow.net/users/104710 | Characterizing affine schemes via morphisms | As pointed out in the comments, [this answer](https://mathoverflow.net/a/349015/82179) implies much more: for a category $\mathscr C$ equivalent to $\mathbf{Sch}\_{/X}$ for any $X \in \mathbf{Sch}$, there is an explicit formula producing a functor $F\_{\mathscr C} \colon \mathscr C \to \mathbf{Sch}$ that only depends on the equivalence type of $\mathscr C$, and such that for $F\_{\mathbf{Sch}\_{/X}} \colon \mathbf{Sch}\_{/X} \to \mathbf{Sch}$ is naturally isomorphic to the forgetful functor (in an explicit way).
In particular, if $\mathscr C$ is equivalent to $\mathbf{Sch}$ (the case $X = \operatorname{Spec} \mathbf Z$), this gives an explicit equivalence $F \colon \mathscr C \stackrel\sim\to \mathbf{Sch}$ that only depends on $\mathscr C$ up to equivalence. So if you want to know whether an object $X \in \mathscr C$ 'is' affine (say under *any* equivalence $\mathscr C \stackrel\sim\to \mathbf{Sch}$), it suffices to check that $F(X) \in \mathbf{Sch}$ is affine.
Of course this is not very useful in practice, since the construction of $F$ is a bit involved, and then you still need a criterion to check whether $F(X)$ is affine.
If you restrict to quasi-compact $k$-schemes, then $X$ is affine if and only if the infinite product $X^{\mathbf N}$ exists in $\mathbf{Sch}\_{/k}$; see [this answer](https://mathoverflow.net/a/65923/82179). On the other hand, it is certainly too optimistic to hope that a scheme $X$ is affine if and only if $X^I$ exists for every set $I$. For instance, if we define $X = \coprod\_{p \text{ prime}} \operatorname{Spec} \mathbf F\_p$, then $X \to \operatorname{Spec} \mathbf Z$ is a monomorphism, so $X^I \cong X$ for any nonempty set $I$. But $X$ is not affine since it is not quasi-compact.
It would be nice to find a clean criterion in general, but this requires some genuine new ideas. I have some thoughts, but in my opinion, this is maybe not the most pressing question in algebraic geometry...
| 4 | https://mathoverflow.net/users/82179 | 441763 | 178,286 |
https://mathoverflow.net/questions/441715 | 1 | $\newcommand{\Ex}{\mathbb E}$ I'm reading an argument in the proof of *Proposition 3.8.* in the paper [Nonlinear self-stabilizing processes - I Existence, invariant probability, propagation of chaos](https://www.sciencedirect.com/science/article/pii/S0304414998000180).
---
Let $X\_0$ and $X\_0^{\prime}$ be two independent real-valued random variables, having the same distribution. We consider two independent $1$-dimensional Brownian motions $B$ and $B'$. Let $X$ and $X'$ be solutions of
$$
X\_t=X\_0+B\_t-\frac{1}{2} \int\_0^t b\left(s, X\_s\right) \mathrm{d} s,
$$
and
$$
X\_t^{\prime}=X\_0^{\prime}+B\_t^{\prime}-\frac{1}{2} \int\_0^t b\left(s, X\_s^{\prime}\right) \mathrm{d} s,
$$
where $b:\mathbb R\_{\ge 0} \times \mathbb R \to \mathbb R$ is regular enough. Let
$$
Y\_t := X\_t-X\_t^{\prime}
\quad \text{and} \quad
\mu\_n(t) := \Ex \left ( \left|Y\_t\right|^n \right), \quad n \geqslant 2 .
$$
Then $Y$ is a semi-martingale with decomposition
$$
Y\_t=Y\_0+B\_t-B\_t^{\prime}-\frac{1}{2} \int\_0^t\left(b\left(s, X\_s\right)-b\left(s, X\_s^{\prime}\right)\right) \mathrm{d} s .
$$
We apply the Itô's formula and take the expectation and the derivative. We obtain
$$
\mu\_{2 n}^{\prime}(t)=n\left\{2(2 n-1) \mu\_{2 n-2}(t) - \Ex \left[Y\_t^{2 n-1}\left(b\left(t, X\_t\right)-b\left(t, X\_t^{\prime}\right)\right)\right]\right\}.
\tag{1}\label{1}
$$
---
**My understanding** By Itô's lemma,
$$
\begin{align\*}
& (Y\_t)^{2n} - (Y\_0)^{2n} \\
= & 2n \int\_0^t (Y\_s)^{2n-1} \mathrm{d} Y\_s + \frac{2n(2n-1)}{2} \int\_0^t (Y\_s)^{2n-2} \mathrm{d} \langle Y \rangle \_s.
\end{align\*}
$$
We have
$$
\begin{align\*}
\mathrm{d} Y\_s &= \mathrm{d} B\_s - \mathrm{d} B'\_s- \frac{\left(b\left(s, X\_s\right)-b\left(s, X\_s^{\prime}\right)\right) }{2} \, \mathrm{d} s, \\
\mathrm{d} \langle Y \rangle\_s &= \mathrm{d} \langle B - B' \rangle \_s = 2 \, \mathrm{d} s.
\end{align\*}
$$
Hence
$$
\begin{align\*}
\Ex[(Y\_t)^{2n}] - \Ex [(Y\_0)^{2n}] &= -n \int\_0^t \Ex [ (Y\_s)^{2n-1} \left(b\left(s, X\_s\right)-b\left(s, X\_s^{\prime}\right)\right)] \, \mathrm{d} s \\
& \qquad + 2n(2n-1) \int\_0^t \Ex [ (Y\_s)^{2n-2} ] \, \mathrm{d} s.
\end{align\*}
$$
So
$$
\begin{align\*}
\mu\_{2n} (t) - \mu\_{2n}(0) &= -n \int\_0^t \Ex [ (Y\_s)^{2n-1} \left(b\left(s, X\_s\right)-b\left(s, X\_s^{\prime}\right)\right)] \, \mathrm{d} s \\
& \qquad + 2n(2n-1) \int\_0^t \mu\_{2n-2} (s) \,\mathrm{d} s.
\tag{2}\label{2}
\end{align\*}
$$
>
> Could you please explain how to go from $\ref{2}$ to $\ref{1}$?
>
>
>
| https://mathoverflow.net/users/99469 | How to obtain this differential relation about moments of a stochastic process? | I believe you meant lemma 3.8. Here at the final step of your calculation, you just apply [Lebesgue-differentiation theorem](https://en.wikipedia.org/wiki/Lebesgue_differentiation_theorem) since those quantities are continuous and the integral is Lebesgue. That will give you the $\mu'(0)$.
So to get the $\mu'(t\_{0})$, I would apply Ito's formula with starting points $t=t\_{0}$ as opposed to $t=0$. So the integrals will be $\int\_{t\_{0}}^{t\_{0}+t}$. Meaning apply Ito's lemma separately to $t\_{0}$ and to $t\_{0}+t$ and then take their difference.
| 2 | https://mathoverflow.net/users/99863 | 441764 | 178,287 |
https://mathoverflow.net/questions/441762 | 3 | Let $P$ be a positive, self-adjoint (unbounded) operator in a Hilbert space $H$ with $0\notin \sigma(P)$. Consider its spectral decomposition
$$P = \int\_{\sigma(P)} t dE(t).$$
Since $0 \notin \sigma(P)$, we can define (the functional calculus)
$$P^{-}= \int\_{\sigma(P)} t^{-1}dE(t).$$
This is an unbounded operator with domain
$$\left\{\xi \in H: \int\_{\sigma(P)} t^{-2}dE\_{\xi, \xi}(t) < \infty\right\}.$$
Is there a relation between this operator and the inverse operator $P(T)\to H: P(\xi)\mapsto \xi$? Are they equal? For example, it is not even clear to me that they have the same domains.
| https://mathoverflow.net/users/216007 | Unbounded positive self-adjoint without $0$ in its spectrum: can we construct its inverse using functional calculus? | $\newcommand\si\sigma\newcommand\D{\mathscr D}$Yes, $P^-=P^{-1}$.
Indeed, the resolvent set of $P$ is open and $0$ is in this set. So, $\si(P)$ is bounded away from $0$, and hence the domain $\D(P^-)$ of $P^-$ is the entire space $H$ -- because, by Theorem 13.24 (a) in [1],
$$\D(P^-)=\D\_g:=\Big\{x\in H\colon \int\_{\si(P)}g^2\,dE\_{x,x}<\infty\Big\},$$
where $g(t):=1/t$. So, by Theorem 13.24 (b) in [1] with $g$ as above and $f(t):=t$, and in view of Definition 12.17 in [1] of a resolution of the identity,
$$PP^-=\Psi(f)\Psi(g)=\Psi(fg)=\Psi(1)=I\_H,$$
where $I\_H$ is the identity operator on $H$ and $\Psi(h):=\int\_{\si(P)}h\,dE$. (Note the obvious typo in Theorem 13.24 (b) in [1], which has $\D\_{fg}\in\D\_g$ instead of $\D\_{fg}\subseteq\D\_g$.)
Thus, $P^-=P^{-1}$, as claimed.
[1] Functional Analysis: Rudin, Second Edition
| 4 | https://mathoverflow.net/users/36721 | 441773 | 178,292 |
https://mathoverflow.net/questions/441777 | 3 | Let $X$ be a space and $\mathcal U$ a cover of $X$. Instead of Čech cohomology, I would like to take the following construction:
let
$$I= \{ \text{finite nonempty intersections of elements of }\,\mathcal U\},$$
which is a poset, and now I can take the nerve $N(I)$ in the sense of category theory,
whose $n$-simplices are chains of length $n+1$ in $N(I)$.
Given a sheaf $\mathcal F$ on $X$, I get a local system of coefficients $F$ on $N(I)$ by taking $F(\sigma) = \mathcal F(\min \sigma)$.
I would like to relate $H^\*(N(I), F)$ to the sheaf cohomology $H^\*(X,\mathcal F)$. Do you know how to do this?
My hope is that this is sheaf cohomology if we assume enough acyclicity about $\mathcal F$ and $I$. This complex smells similar to the Čech complex, but I am not sure of the general relation. If $X$ was a simplicial complex and $\mathcal U$ the covering by star neighborhoods, then we recover Čech cohomology with respect to star neighborhoods in the barycentric subdivision.
In my situation $X$ is a quasi-compact separated scheme, $\mathcal U$ is a cover by affine opens, and $\mathcal F$ is a quasi-coherent sheaf. But, the same setup might work if we just assume $\mathcal F$ is acyclic on every element of $I$.
| https://mathoverflow.net/users/125523 | Čech-like cohomology with the “other nerve” | This construction as stated gives exactly the barycentric subdivision of the Čech nerve of $\mathcal{V}$, as any simplicial complex yields a face poset and the categorical nerve of that is the barycentric subdivision of the original complex. There are several alternative nerve constructions that can be used; see H. Abels and S. Holz, Higher generation by subgroups, J. Alg, 160, (1993), 311– 341, for a discussion on several of them. In your fairly classical case there is also the Vietoris nerve which was explored by Dowker in Homology Groups of Relations, Annals of Maths, 56, (1952), 84 – 95.
| 5 | https://mathoverflow.net/users/3502 | 441786 | 178,297 |
https://mathoverflow.net/questions/441789 | 3 | It is well know that the convergence in distributions does not necessarily imply convergence in expectation, but implies convergence in expectation of bounded continuous functions.
Let $\{X\_n\}$ be a sequence of random variables that converge in distribution to $X$. I would like to ask two examples as follows.
1. An example such that $\mathbb{E}[X\_n]$ does not convergence to $\mathbb{E}[X]$.
2. An example such that $\mathbb{E}[X\_n^k]$ does not convergence to $\mathbb{E}[X^k]$ for all $k = 1, 2,\ldots$. That is, the sequence does not converge in all the moments, not just a or a few fixed moments. Suppose that all their moments exist.
| https://mathoverflow.net/users/170508 | Examples of convergence in distribution not implying convergence in moments | Let $P(X\_{n} = n) = \frac{1}{n}$ and $P(X\_{n} = 0) = 1 - \frac{1}{n}$. Then $X\_{n}$ converges in distribution to $X=0$, but $\mathbb{E}(X\_{n}^{k}) = \frac{1}{n} n^{k} + 0 = n^{k-1} \not\xrightarrow{n\to\infty} 0$ for each $k \in \mathbb{N}$.
**UPDATE:**
If you prefer continuous random variables on $\mathbb{R}$, you can "smoothen out" the previous example:
Let $X \sim N(0,1)$ and $X\_{n}$ have the probability density
$$
\rho\_{X\_{n}}(x)
=
\frac{1}{n} \cdot \frac{1}{\sqrt{2\pi}}\, \exp(-(x-n)^2/2) +
\left(1-\frac{1}{n}\right) \cdot \frac{1}{\sqrt{2\pi}}\, \exp(-x^2/2).
$$
Then $X\_{n} \stackrel{\mathrm{d}}{\to} X$.
Now denote the moments of $X$ by $A\_{k} := \mathbb{E}(X^{k})$ and, for $m \geq 0$, let $A\_{k,m} := \mathbb{E}((X+m)^{k}) \geq A\_{k} + m^k$.
It follows for each $k\in\mathbb{N}$:
\begin{align\*}
\mathbb{E}(X\_{n}^{k})
&=
\int\_{\mathbb{R}} x^{k} \rho\_{X\_{n}}(x)\, \mathrm dx
\\
&=
\frac{1}{n} A\_{k,n} + \left(1-\frac{1}{n}\right) A\_{k}
\\
&\geq
\frac{1}{n} (A\_{k}+n^{k}) + \left(1-\frac{1}{n}\right) A\_{k}
\\
&=
n^{k-1} + A\_{k}
\\
&\not\xrightarrow{n\to\infty}
A\_{k}.
\end{align\*}
| 6 | https://mathoverflow.net/users/143828 | 441791 | 178,300 |
https://mathoverflow.net/questions/441711 | 4 | Assume that $\sigma\in S\_n$ has the cycle type $(p,.,p,1,..,1)$ where $p>2$ is a prime and the numbers of $1$ maybe $0$. If $\sigma\_1$ and $\sigma\_2$ are chosen uniformly in the conjugacy class of $\sigma$. Assume the cycle type of $\sigma\_1 \sigma\_2=(k\_1,k\_2,.,k\_l)$ . Is there a lower bound $B$ such that $$\mathrm{Pr}\big(\exists j\in \{1,2,..,l\}, \text{ s.t. } k\_j \bmod 2=0\big)\geq B \geq 1/T(n),$$ where $T$ is a polynomial.
| https://mathoverflow.net/users/482299 | A probability problem in the conjugacy classes of symmetric group | Let $kp$ be the size of the support of $\sigma$. Let $1,2,3,4$ be four points of the ground set. The probability that $\sigma\_1$ maps $1 \mapsto 2$ is $kp/n(n-1)$, because there is a $kp/n$ chance that $1$ is in the support and conditionally a $1/(n-1)$ chance that $1 \mapsto 2$. Conditional on that, the probability that $3$ is in the support is $(kp-2)/(n-2)$ and then a $1/(n-2)$ chance that $3 \mapsto 4$ (because $3$ can map to anything apart from $2$ and $3$). Therefore the probability that $\sigma\_1$ maps $1\mapsto2$ and $3 \mapsto4$ is $kp(kp-2)/n(n-1)(n-2)^2 \ge 1/n^4$. Likewise the probability that $\sigma\_2$ maps $2 \mapsto 3$ and $4 \mapsto 1$ is also at least $1/n^4$. The probability that both these happen is at least $1/n^8$ and in this case $\sigma\_1\sigma\_2$ has a $2$-cycle.
| 5 | https://mathoverflow.net/users/20598 | 441798 | 178,303 |
https://mathoverflow.net/questions/441794 | 6 | Let $\mu$ be a Borel measure on the real line $\mathbb{R}$ taking values in a separable Banach algebra $A$. Assume that $\mu$ is such that the total variation measure $|\mu|$ is finite. Let $f$ be a bounded measurable function on the real line $\mathbb{R}$ taking values in a Banach algebra $A$. Is there a general theory that makes sense of integrals of the form
$$
\int\_{\mathbb{R}} f(x)\mu(dx) \in A.
$$
I would be also interested in special cases: $A$ a sub-algebra of $B(H)$, where $H$ -- a Hilbert space or $A$ a commutative algebra.
| https://mathoverflow.net/users/47256 | Integration in Banach algebra | I believe OP's setting is an instance of the more general so called *bilinear integral*, defined for the following data:
* a measurable space $(S,\mathfrak{S})$;
* a function $f: S \to X$, where $X$ is a normed linear space;
* an *additive* measure $\mu: \mathfrak{S} \to Y$, where $Y$ is another normed linear space;
* a bilinear map $\beta: X \times Y \to Z$, where $Z$ is a Banach space;
Then, for $E \in \mathfrak{S}$, one can consider
$$
\int\_E \beta(f(s),\mu(ds))
$$
In OP's setting, the bilinear mapping is given by the multiplication in the Banach algebra.
The canonical reference is *Bartle - A General Bilinear Vector Integral (1955)*, but some further references, that de facto establish it as a theory, are:
* *Dunford - A Bilinear Vector Integral (1975)*
* *Freniche, Garcia-Vazquez - The Bartle Bilinear Integration and Carleman Operators (1998)*
* *Jefferies - Bilinear Integrals and Radon-Nikodym Derivatives*
* *Jefferies - Some Recent Applications of Bilinear Integration*
A generalization of Bartle's integral is obtained and substantially expanded upon in Dobrakov's series of 13 papers titled *"Integration in Banach Spaces"*. There he considers $f: S \to X$, where $X$ is a Banach space, and $\mu: \mathfrak{S} \to BL(X,Y)$, where $Y$ is another Banach space, but this reduces to OP's setting via the adjoint action of the Banach algebra.
**EDIT:** Apologies, forgot to mention that Panchapagesan has a kind of a nice survey/summary on Dobrakov's integral, titled *"On the Distinguishing Features of the Dobrakov Integral (1991)"*, which should help navigate Dobrakov's series of papers. Moreover, Panchapagesan apparently has a whole **book** *"The Bartle-Dunford-Schwartz Integral"* that is relatively recent (from 2000 or so), but I am not sure to what extent it addresses OP's setting, but it might be worth looking into.
| 12 | https://mathoverflow.net/users/1849 | 441803 | 178,304 |
https://mathoverflow.net/questions/77845 | 9 | This question is not directly related to, but was inspired by, [this question](https://mathoverflow.net/questions/77583/is-the-free-product-of-arbitrarily-many-copies-of-mathbbz-and-mathbb). We know that a finitely generated residually nilpotent group is residually of prime-power order. However, we may need to use different primes for different elements. Classes of groups for which residual nilpotence forces there to be a single prime that will do for all elements (i.e., for which the group in question must be residually $p$-finite, for some $p$) seem to be interesting, and include, for instance, free products of cyclic groups.
**Is there a (non-cyclic) one-relator group that is residually nilpotent, but is not residually a finite $p$-group, for any prime number $p$?**
Such a group must be torsion-free, with trivial centre.
| https://mathoverflow.net/users/1392 | Is there a residually nilpotent one-relator group that is not residually a finite p-group for any prime p? | The answer is yes, such a group exists: this is the main result of [1]. In fact, one can take the Baumslag-Solitar group $\operatorname{BS}(p^r, -p^r) = \langle a, b \mid ba^{p^r}b^{-1} = a^{-p^r} \rangle$ where $p$ is an odd prime and $r \geq 1$. These are residually nilpotent, but are not residually a finite $q$-group for any prime $q$.
In particular, the smallest example of this form is $\operatorname{BS}(3, -3) = \langle a, b \mid ba^3b^{-1}a^3 = 1 \rangle$.
(Note that OP's question was posed by McCarron [2] in 1996).
${}$
[1] *Moldavanskii, D. I.*, [**Residual nilpotence of groups with one defining relation**](https://doi.org/10.1134/S0001434620050090), translation from Mat. Zametki 107, No. 5, 752-759 (2020). [ZBL07215590](https://zbmath.org/?q=an:07215590).
[2] *McCarron, James*, [**Residually nilpotent one-relator groups with non-trivial centre**](https://www.ams.org/journals/proc/1996-124-01/S0002-9939-96-03148-6/S0002-9939-96-03148-6.pdf), Proc. Amer. Math. Soc. 124, No. 1 (1996).
| 6 | https://mathoverflow.net/users/120914 | 441807 | 178,306 |
https://mathoverflow.net/questions/441800 | 0 | $z\_i=f+a\_i+\epsilon\_i$ ，where $f\sim N(\bar{f},\sigma\_{f}^2)$ ; $a\_i\sim N(\bar{a\_{i}},\sigma\_{a}^2)$; $\epsilon\_i\sim N(0,\sigma\_{\epsilon}^2)$. We can see the signals $\{z\_i\}$ where $i\subseteq {1,2,……,M}$, and note that $f$ is a common term for all signals. The question is we want to know the **Posterior mean** of $a\_i$ for any $i$.
The outcome in that paper is as follows but I don't understand it. I want a more detailed process.
\begin{array}{l}
E\left[a\_{i} \mid\left\{z\_{i}\right\}\_{i=1, \ldots, M}\right] \\
=\bar{a}\_{i}+\frac{\sigma\_{a}^{2}\left(\sigma\_{a}^{2}+\sigma\_{\varepsilon}^{2}+(M-1) \sigma\_{f}^{2}\right)}{\left(\sigma\_{a}^{2}+\sigma\_{\varepsilon}^{2}\right)\left(\sigma\_{a}^{2}+\sigma\_{\varepsilon}^{2}+M \sigma\_{f}^{2}\right)}\left(z\_{i}-\bar{z}\_{i}\right)-\frac{\sigma\_{a}^{2} \sigma\_{f}^{2}}{\left(\sigma\_{a}^{2}+\sigma\_{\varepsilon}^{2}\right)\left(\sigma\_{a}^{2}+\sigma\_{\varepsilon}^{2}+M \sigma\_{f}^{2}\right)} \sum\_{j \neq i}\left(z\_{j}-\bar{z}\_{j}\right)
\end{array}
| https://mathoverflow.net/users/500213 | How does this Bayesian updating work $z_i=f+a_i+\epsilon_i$ | $\newcommand{\bR}{\mathbb{R}}
\newcommand{\one}{\mathbf{1}}
\newcommand{\diag}{\textrm{diag}}
\newcommand{\Id}{\textrm{Id}}$
I guess you assume independence of the random variables $f,a,\varepsilon$, which I am going to use.
**General strategy** for Gaussian conditioning in linear models:
Let $a\sim N(\bar{a},Q\_{a})$ on $\bR^{n}$ and $z = Aa+\eta$, where $A\in\bR^{m \times n}$ and $\eta \sim N(\bar{\eta},Q\_{\eta})$ on $\bR^{m}$ is independent of $a$.
Then consider the joint distribution of $a$ and $z$,
$$
\begin{pmatrix}
a\\ z
\end{pmatrix}
\sim
N\left(
\begin{pmatrix}
\bar{a}\\ A\bar{a}+\bar{\eta}
\end{pmatrix}
,
\begin{pmatrix}
Q\_{a} & Q\_{a}A^{\top} \\ A Q\_{a} & Q\_{\eta} + A Q\_{a} A^{\top}
\end{pmatrix}
\right)
$$
and apply the Gaussian conditioning formula (see e.g. [here, scroll down to Conditional distributions](https://en.wikipedia.org/wiki/Multivariate_normal_distribution)) to obtain the conditional mean
$$
(\ast)\qquad
\bar{a}^{z}
=
\bar{a} + (Q\_{a}A^{\top}) (Q\_{\eta} + A Q\_{a} A^{\top})^{-1} (z-A\bar{a}-\bar{\eta}).
$$
**Answer to your question:**
Now let us consider your particular case and denote $\one\_{m} = (1,\dots,1)^{\top} \in \bR^{m}$ and a diagonal matrix with the values $s\_{1},\dots,s\_{m}$ on the diagonal by $\diag(s\_{1},\dots,s\_{m})$ and by $\diag(s)$ if all $s\_{j}$ equal $s$.
The first nontrivial step is to identify $\eta$ correctly: $z = A a + \eta$, where $A = \mathrm{Id\_{M}}$ and
$$
\eta
=
\underbrace{(\one \ \Id\_{M})}\_{=: B} \begin{pmatrix}
f\\
\varepsilon\_{1}\\
\vdots\\
\varepsilon\_{M}
\end{pmatrix}
\sim
N\left(
\bar{f}\one , B \, \diag(\sigma\_{f}^2,\sigma\_{\varepsilon}^{2},\dots,\sigma\_{\varepsilon}^{2}) \, B^{\top}
\right).
$$
(Here we used the fact that $y = Bx$, $x\sim N(\bar{x},Q\_{x})$ implies $y\sim N(B\bar{x},BQ\_{x}B^{\top})$; I can provide details on this if required.)
Then, a simple (but tedious) computation shows that
\begin{align\*}
Q\_{\eta} + AQ\_{a}A^{\top}
&=
\diag(\sigma\_{a}^{2} + \sigma\_{\varepsilon}^{2}) + \sigma\_{f}^{2} \one \one^{\top},
\\
(Q\_{\eta} + AQ\_{a}A^{\top})^{-1}
&=
\frac{\diag(\sigma\_{a}^{2} + \sigma\_{\varepsilon}^{2} + M \sigma\_{f}^{2}) - \sigma\_{f}^{2} \one \one^{\top}}{(\sigma\_{a}^{2} + \sigma\_{\varepsilon}^{2})(\sigma\_{a}^{2} + \sigma\_{\varepsilon}^{2} + M \sigma\_{f}^{2})}
\end{align\*}
Note that finding the inverse might be hard, but showing that its the inverse, once you have it, is easy (but tedious): Simply multiply the two matrices above to obtain $\Id\_{M}$.
If you plug this into $(\ast)$ and write it out componentwise you obtain the desired result.
Hope this helps. If you have any questions, don't hesitate to ask in the comments.
**Remarks:**
* $(\ast)$ holds similarly if $\bR^{m}$ and $\bR^{n}$ are replaced by arbitrary separable Hilbert space (possibly infinite-dimensional). In this case replace "transpose" $A^{\top}$ by "adjoint" $A^{\ast}$.
* There is also a formula for the conditional covariance matrix/covariance operator, which, somewhat surprisingly, does not dependent of $z$ (a particular property of Gaussian conditioning):
$$
Q\_{a}^{z}
=
Q\_{a} - Q\_{a}A^{\top} (Q\_{\eta} + AQ\_{a}A^{\top})^{-1} A Q\_{a}.
$$
$a|z \sim N(\bar{a}^{z},Q\_{a}^{z})$ has again a normal distribution.
| 1 | https://mathoverflow.net/users/143828 | 441810 | 178,307 |
https://mathoverflow.net/questions/441591 | 6 | Recall that for $P$ a forcing order and $\dot{Q}$ a $P$-name for a forcing order, the termspace forcing $T(P,\dot{Q})$ consists of minimal-rank names for elements of $\dot{Q}$, ordered by $\dot{q}'\leq\dot{q}$ iff $1\_{P}\Vdash\dot{q}'\leq\_{\dot{Q}}\dot{q}$. In his chapter in the Handbook of Set Theory, James Cummings states the following result:
If $\kappa$ is inaccessible, $P$ is $\kappa$-cc. and $\dot{Q}$ is forced to be $\kappa$-cc., $T(P,\dot{Q})$ is $\kappa$-cc.
and attributes it to the paper "More saturated ideals" by Matthew Foreman. However, I was unable to find the above statement in the paper. Later on, Cummings proves it, but only for measurable (or at least Jonsson) $\kappa$.
Is there a direct proof (or a different source) for the exact result above?
---
### Edit:
Philipp Lücke in the comments gave an argument that the stated result is actually false. If $|P|<\kappa$ and $\kappa$ is weakly compact, it holds. So another interesting question would be:
>
> Does the above statement hold for (not necessarily weakly compact) $\kappa$ if $|P|<\kappa$?
>
>
>
| https://mathoverflow.net/users/138274 | Proof (or reference) about the cc-ness of termspace forcing | The answer is consistently negative, and it seems likely that it is actually always negative.
Let us look at the property: for every $|\mathbb P| < \kappa$, if $\Vdash\_{\mathbb{P}} \dot{\mathbb{Q}}$ is $\kappa$-c.c. then $T(\mathbb P, \dot{\mathbb Q})$ is $\kappa$-c.c.
It is known that for $\kappa$ which is weakly compact, this property holds.
Indeed, if $A = \langle \dot{q}\_i \mid i < \kappa\rangle$ is an antichain, then for every $i < j$ there is $p\in \mathbb{P}$ such that $p \Vdash \dot{q}\_i \perp \dot{q}\_j$. This gives us a coloring of pairs of ordinals below $\kappa$ with $|\mathbb P|$ many color. As $\kappa$ is assumed to be weakly compact, there is a homogeneous set $H$ with a fixed color $p$. So $p$ forces $\langle \dot{q}\_i \mid i \in H\rangle$ to be an antchain in $\mathbb{Q}$, contradicting the chain condition hypothesis.
Let us assume that this property holds. In particular, it means that the product of less than $\kappa$ many copies of a $\kappa$-c.c. forcing $\mathbb{Q}$ is $\kappa$-c.c.: Let $\mathbb{P}$ be the atomic forcing with $\theta$ many atoms, and let $\mathbb{Q}$ be a $\kappa$-c.c. forcing. Then, for every $\vec q \in \mathbb{Q}^\theta$ we can assign a name $\dot{u}$ which is forced to be $\vec{q}(\alpha)$ if and only if $\min G\_P$ is the $\alpha$-th atom. Clearly, this gives us a way to translate an antichain in $\mathbb{Q}^\theta$ into an antichain in $T(\mathbb{P}, \check{\mathbb{Q}})$.
In the paper, "Knaster and friends I: Closed colorings and precalibers", by Lambie-Hanson and Rinot, they define the combinatorial principle $U(\kappa,\mu,\theta,\chi)$. Let us focus on the case $U(\kappa, 2, \omega, 2)$. They proved that this case (and stronger ones) holds for successor cardinals, or if $\square(\kappa)$ holds [so in $L$, it fails exactly for weakly compact cardinals].
Moreover, they show (for example) that $U(\kappa,2,\omega,2)$ implies the existence of a $\kappa$-c.c. forcing which its $\omega$-th power is not $\kappa$-c.c., and they conjecture that $\neg U(\kappa,2,\omega,2)$ implies that $\kappa$ is weakly compact.
| 7 | https://mathoverflow.net/users/41953 | 441814 | 178,310 |
https://mathoverflow.net/questions/441820 | 9 | Is it known if there are any examples of a finitely generated group $G$ such that:
1. $G$ has a finite index subgroup $H$ which is free-by-cyclic
2. $G$ itself is **not** free-by-cyclic
3. $G$ is torsion-free
Since subgroups of free-by-cyclic groups are free-by-cyclic, one may strengthen (1) and ask that $H$ is normal in $G$. It is then fairly easy to construct groups that satisfy (1) and (2) by extending $H$ under any finite group. However, I couldn't come up yet with an example satisfying all three conditions. I've already know such a group must satisfy some properties:
* By a combination of Serre's and Stallings-Swan's theorems, such a group must have cohomological dimension 2.
* Since $H/[H,H]$ has a finite index image in $G/[G,G]$, $G$ must have infinite abelianization.
* In particular, $G$ admits homomorphisms onto $\mathbb{Z}$, all of whose kernels must have cohomological dimension exactly $2$. So $G$ must be a semidirect product $K \rtimes \mathbb{Z}$ for some group $K$ of cohomological dimension $2$.
| https://mathoverflow.net/users/117514 | Torsion-free virtually free-by-cyclic groups | The group $$G=\langle a, b, x, y\mid [a, b]^2=[x, y]^2\rangle$$
is a torsion-free group which is not free by cyclic. However, $G$ is free-by-$D\_{\infty}$ and so virtually free-by-cyclic (containing an index-two subgroup which is free-by-cyclic).
This example is from the paper Baumslag, Fine, Miller and Troeger, *Virtual properties of cyclically pinched one-relator groups.* Int. J. Alg. Comp. (2009).
* Firstly, $G$ is torsion-free as it is a free product with amalgamation of two torsion-free groups.
* Secondly, $G$ is not free-by-cyclic. Both $[a, b]$ and $[x, y]$ are contained in the commutator subgroup. These elements are non-equal but their squares are equal. Hence, the commutator subgroup does not have unique roots, and so it not free. Hence, any map to $\mathbb{Z}$ has non-free kernel.
* Finally, $G$ is free-by-$D\_{\infty}$ by Theorem 4 of the above-mentioned paper. The idea of the proof is to map $G$ onto $D\_{\infty}=\langle c, d\mid c^2=1, c^{-1}dc=d^{-1}\rangle$ by $\phi(a)=c=\phi(x)$ and $\phi(b)=d=\phi(y)$, and then prove that $\phi$ has free kernel. The index-two subgroup $\phi^{-1}(d)$ of $G$ is therefore free-by-cyclic.
| 12 | https://mathoverflow.net/users/6503 | 441826 | 178,315 |
https://mathoverflow.net/questions/441677 | 1 | Let $X$ be a smooth projective complex variety and $t\in H^{2p}(X,\mathbf{Q})(p)$ a rational cohomology class.
Assume that there exist
$$t\_1,\ldots,t\_N\in H^{2\*}(X,\mathbf{Q})(\*)$$
algebraic classes (i.e. in the image of the cycle map $cl : A^\*(X)\to H^{2\*}(X,\mathbf{Q})(\*)$) and an algebraic subgroup $$G\subset\text{Aut}(H^\*(X,\mathbf{Q}))\times\text{GL}(\mathbf{Q}(1))$$ (automorphisms as a graded $\mathbf{Q}$-algebra) fixing the $t\_1,\ldots, t\_N$ and largest with respect to this property, such that $G$ also fixes $t$.
>
> Is $t$ algebraic?
>
>
>
From papers of Milne, one can deduce this is true under the variational Hodge conjecture. However, I wonder if this is already known unconditionally.
| https://mathoverflow.net/users/nan | Cohomology classes fixed by algebraic automorphism subgroups | Essentially you are asking whether the category of homological motives generated by $X$ is tannakian, which is true if and only if homological equivalence coincides with numerical equivalence (Jannsen, Deligne). This coincidence is known for abelian varieties (Lieberman) but not much else, so I think the answer to your question is No in general.
| 1 | https://mathoverflow.net/users/500240 | 441831 | 178,316 |
https://mathoverflow.net/questions/441813 | 1 | Let $k$ be a number field and $\mathbb{G}\_m$ be the multiplicative group sheaf. For an algebraic group $G$, we define the character group $\widehat{G}:= \mathrm{Hom}\_{\bar{k}}(\bar{G},\mathbb{G}\_{m,\bar{k}})$, where $\bar{G}$ is the base change to an algebraic closure of $k$. This is a contravariant functor and I believe that it is right exact since $\mathbb{G}\_{m,\bar{k}}$ when viewed as a group is divisible.
Now suppose we have the following exact sequence of algebraic $k$-groups
$$0 \rightarrow T \rightarrow S \rightarrow A \rightarrow 0,$$
where $T$ is a torus and $A$ an abelian variety. Applying the character group functor we get
$$0 \rightarrow \widehat{A} \rightarrow \widehat{S} \rightarrow \widehat{T} \rightarrow 0$$
if nothing I did above is wrong. Since $A$ is an abelian variety, it has no non-constant morphism to the affine variety $\mathbb{G}\_{m,\bar{k}}$ and since such a character has to be a group homomorphism, we must have $\widehat{A} = 0$.
This will leave us with $\widehat{S} \cong \widehat{T}$ which is strange to me. Furthermore, doesn't the group $\mathrm{Ext}^1\_{\bar{k}}(\bar{A},\mathbb{G}\_{m,\bar{k}})$ contain all extensions of $\bar{A}$ by $\mathbb{G}\_{m,\bar{k}}$ which are classes of semi-abelian varieties? Why do we only have the trivial extension in this case?
I'm not sure what could've possibly went wrong. Could it be that taking the functor does not actually give rise to an exact sequence in the first place? Pretty sure I've seen something similar done for taking character groups, but maybe I didn't apply it correctly.
| https://mathoverflow.net/users/172132 | Character group functor of an exact sequence of algebraic groups | The functor isn't right exact. For example,
$\mathrm{Ext}^1(A,\mathbb{G}\_m)\simeq$
the set of primitive elements in $H^1(A,\mathcal{O}\_A^{\times})$.
| 4 | https://mathoverflow.net/users/500240 | 441843 | 178,321 |
https://mathoverflow.net/questions/441838 | 20 | As is probably clear from my previous questions, I am coming to "condensed mathematics" from the naive perspective of a category theorist, without much knowledge of the intended applications in algebraic geometry and functional analysis. I hope therefore that this question is not *too* naive, but I haven't found a place yet where it is addressed clearly for a "lay" audience.
According to my current understanding, there are four categories that achieve something similar:
* The category of $\kappa$-condensed sets, for some cardinal $\kappa$, which are sheaves on the site of compact Hausdorff spaces of cardinality $<\kappa$.
* The category of pyknotic sets, which are the $\kappa$-condensed sets when $\kappa$ is an inaccessible.
* In the opposite direction (making the site smaller), Johnstone's topological topos, which is the category of sheaves on the full subcategory of spaces containing the point and the one-point compactification of $\mathbb{N}$.
* The category of condensed sets, which is the colimit of the categories of $\kappa$-condensed sets over all $\kappa$, or equivalently the category of "small sheaves" on the large site of all compact Hausdorff spaces.
I want to understand the differences between these categories, and why and in what situations one might choose one over the others. Specifically:
1. Johnstone's topological topos seems closely related to the category of $\aleph\_1$-condensed sets. (The references I've seen restrict $\kappa$ to be a strong limit cardinal, but at least the definition seems to make sense for any cardinal.) It seems too much to hope for that they would be equivalent, but are they related by an adjunction at least? How "close" are they?
2. On that note, why is $\kappa$ usually restricted to be a strong limit cardinal?
3. To a pure category theorist, the first three categories have the obvious advantage over the fourth that they are toposes. In fact they are even *local* toposes: their forgetful functor to sets has a right adjoint as well as a left adjoint. Why might one nevertheless choose to work with the non-local non-topos of condensed sets instead of one of these three toposes?
4. Relatedly, for what applications is it *not* sufficient to work with $\kappa$-condensed sets for a fixed small $\kappa$, like the smallest uncountable strong limit $\beth\_\omega$? Or, for that matter, Johnstone's topological topos? In particular, are there desirable *abstract* properties that these categories lack? Or are there constructions that would give the "wrong" result when performed in these categories (and if so, in what sense)? Or is it that there are important examples of compactly generated spaces that aren't "$\kappa$-compactly generated" for small $\kappa$ and thus don't embed fully-faithfully in these categories?
| https://mathoverflow.net/users/49 | Condensed vs pyknotic vs consequential | Some comments:
Regarding 1): They are quite different. Johnstone actually uses a very general notion of "cover" in his sequential topos -- his site is a full subcategory of metrizable profinite sets (=$\aleph\_1$-small profinite sets=countable limits of finite sets=sequential Pro-category of finite sets), but not all of his covers are covers in the condensed/pyknotic sense. I'm not sure the more general covers he allows are of much relevance for his positive results, but they preclude his topos from having enough points. (All the other choices have enough points.) It also means that the generating object $\mathbb N\cup\{\infty\}$ is not actually a quasicompact object in his topos.
So a better comparison would be between the version of Johnstone's topos that restricts to the finitary covers. This admits a geometric morphism from $\aleph\_1$-condensed sets. Now $\aleph\_1$-condensed sets actually admit a description very similar to this version of Johnstone's topos, but replacing $\mathbb N \cup \{\infty\}$ with the Cantor set, which is the universal metrizable profinite set (i.e. surjects onto any other). But the Cantor set is much bigger than $\mathbb N\cup\{\infty\}$! This has some important consequences, for example $[0,1]$ is quasicompact in $\aleph\_1$-condensed sets, but very much fails to be so in Johnstone's topos. Such quasicompactness is used all over the place in our arguments. For example, an extremely important property is the following:
You can consider CW complexes as ($\aleph\_1$-)condensed sets. Now any topos has its inherent notion of cohomology, so you can take the resulting cohomology of CW complexes. Then, in the condensed world:
>
> Applied to CW complexes, the internal notion of cohomology agrees with singular cohomology.
>
>
>
I believe this would fail in Johnstone's topos (correct me if I'm wrong!). And I hope you agree that this is a very desirable property. It's the starting point for seeing that the internal notion of group cohomology of all sorts of topological/condensed groups agrees with the various (ad hoc!) notions of continuous group cohomology you can find in the literature.
On the other hand, almost everything we do in condensed sets could also be done already with $\aleph\_1$-condensed sets; in fact, I'm contemplating switching to that setting for some things. One nasty issue is that while condensed abelian groups have enough projectives, there are no nontrivial ones that are *internally* projective. But in $\aleph\_1$-condensed abelian groups, $\mathbb Z[\mathbb N\cup\{\infty\}]$ *is* internally projective!
Regarding 2): As you observe, the theory basically works for any $\kappa$. One thing one might like is that as you increase $\kappa$, the pullback functors are fully faithful. While I don't know whether that's always true, it's true at least when the cardinals are either regular or strong limits. And the reason for choosing strong limits is that in that case, one has enough compact projective objects (the extremally disconnected profinite sets), which are very useful (even if not ultimately necessary) for building the theory.
Regarding 3): The main reason is the desire to avoid artificial choices. Let me elaborate by switching to the next question:
Regarding 4): One thing we prove early on is a general Pontrjagin duality on locally compact abelian groups (even on the derived level). But this requires that there are as many discrete abelian groups as there are compact abelian groups. If you work with $\kappa$-condensed abelian groups, the Pontrjagin duality would force one to restrict not only to $\kappa$-small compact abelian groups, but also to $\kappa$-small discrete abelian groups.
Also, if you really want to say that compactly generated topological spaces embed into condensed sets, without implicitly actually talking about $\kappa$-compactly generated ones, again you need to go this colimit over all $\kappa$.
But in practice, mostly everything is $\aleph\_1$-compactly generated, and you can just work with $\aleph\_1$-condensed sets (or the much larger category of $\beth\_\omega$-condensed sets, where you have enough compact projectives).
But as I said above, you absolutely cannot work with Johnstone's topos, the space $\mathbb N\cup\{\infty\}$ is just too small.
| 19 | https://mathoverflow.net/users/6074 | 441847 | 178,322 |
https://mathoverflow.net/questions/441835 | 5 | This question is a question about nomenclature more than anything. I have shown all the math, but I don't know what to search for for similar results. In such a sense, it is more so a reference request, or a request for someone to point me in the right direction.
We start with the function:
$$
f\_1(z) = \sum\_{n=0}^\infty \frac{z^n}{1+z^n} \frac{1}{2^n}\\
$$
Which is holomorphic for $|z| <1$. And the second function is:
$$
f\_2(z) = \sum\_{n=0}^\infty \frac{z^n}{1+z^n} \frac{1}{2^n}\\
$$
Which is holomorphic for $|z| > 1$.
Both of these functions are expressed by the same summation, but on different domains. Using the usual concept of "Analytic Continuation" the functions $f\_1$ and $f\_2$ are not related. They are analytic on non-intersecting domains. So, to call one a continuation of the other is a fallacy to begin with.
To make matters worse; both functions $f\_1$ and $f\_2$ have a "wall" of singularites along $|z| =1$. Where for all $q^n =-1$ for all $n \in \mathbb{N}$ we have that $f\_1(q) = \infty = f\_2(q)$, which are simple poles. There is no domain $D \subset \mathbb{C}$ which intersects $|z| =1$ where either $f\_1$ or $f\_2$ are holomorphic--because there are a dense amount of singularities on this line. So any standard way of analytic continuation is hopeless.
BUT! And this is a big but. Let us take the contour $C = \{ \zeta \in \mathbb{C}\,|\, |\zeta| = 2\}$. And let's take $|z|<1$. Then:
$$
f\_1(z) = \frac{1}{2\pi i} \int\_C \frac{f\_2(\zeta)}{\zeta-z}\,d\zeta\\
$$
This is a pretty involved result, which basically amounts to. All of the residues of the poles along the "wall" of singularities sum to zero when added up. And all that's left is the function for $|z|<1$. This is one of those perfect moments where everything cancels out and $f\_2(z)$ acts as an analytic continuation of $f\_1$. Despite the fact "analytic continuation" is meaningless.
I like to write this as:
$$
f(z) = \frac{1}{2\pi i} \int\_{|z| = R} \frac{f(\zeta)}{\zeta - z}\,d\zeta\\
$$
For all $|z| < \min(R,1)$ for all $R \neq 1$.
From this, I'd like to ask my question. I'd like to think of $f(z) = f\_1(z) = f\_2(z)$, where $f(z) : \mathbb{C}/\mathcal{U} \to \mathbb{C}$; where $\mathcal{U} = \{z \in \mathbb{C}\,|\, |z| =1\}$. But I know this is kind of nonsense. But it's not entirely nonsense, because Cauchy's integral theorem still works... It's not really an analytic continuation--but it kinda (..?) is.
Any help is greatly appreciated! Any pointing in the right direction is also greatly appreciated!
| https://mathoverflow.net/users/133882 | Analytic continuation for disjoint domains | As I already hinted at in my comment, your formula is based on a miscalculation (the only alternative is that $f\_1$ can actually be continued past $|z|=1$, which even without closer analysis looks highly suspicious since the summands have poles at $n$th roots of $-1$, as you pointed out). Here's a direct quick test that confirms this:
Take $z=0$. We can integrate $\int\_C \frac{f\_2(\zeta)}{\zeta}\, d\zeta$ term by term, so let's look at (for $n\ge 1$)
\begin{align\*}
\int\_{|\zeta|=2 } \frac{\zeta^{n-1}}{\zeta^n+1}\, d\zeta & =2^ni\int\_0^{2\pi}\frac{e^{in t}}{2^ne^{int}+1}\, dt = \frac{2^ni}{n}\int\_0^{2\pi n} \frac{e^{ix}}{2^n e^{ix}+1}\, dx \\
& = i\int\_0^{2\pi} \frac{e^{ix}}{e^{ix}+2^{-n}}\, dx
=\int\_{|w|=1} \frac{dw}{w+2^{-n}} = 2\pi i .
\end{align\*}
I use the standard parametrization $z=Re^{it}$ of the circle here (we can also get this more quickly by right away substituting $u=\zeta^n$). Also, when $n=0$, the integral equals $\pi i$.
Hence
$$
\frac{1}{2\pi i}\int\_{|\zeta|=2} \frac{f\_2(\zeta)}{\zeta}\, d\zeta = \frac{1}{2}+\sum\_{n\ge 1} 2^{-n}= \frac{3}{2} \not= f\_1(0)= \frac{1}{2} .
$$
| 4 | https://mathoverflow.net/users/48839 | 441853 | 178,323 |
https://mathoverflow.net/questions/441844 | 4 | I heard that the Langlands functoriality conjecture implies the Ramanujan conjecture for $GL(2)$. especially for the Maass form.
There are various versions of the Langlands functoriality concerning to which groups are associated.
I am wondering which version of the Langlands functorial conjecture could prove the Ramanujan conjecture for $GL(n)$ completely?
| https://mathoverflow.net/users/29422 | Which Langlands functoriality conjecture implies the original Ramanujan conjecture? | Let $F$ be a number field, let $\mathbb{A}\_F$ be the ring of adeles of $F$, and let $\mathcal{A}(n)$ be the set of cuspidal automorphic representation of $\mathrm{GL}\_n(\mathbb{A}\_F)$ with unitary central character. For $\pi\in\mathcal{A}(n)$, I will express the generalized Ramanujan conjecture (GRC) for $\pi$ as the conjectural bound
$$|\lambda\_{\pi}(\mathfrak{a})|\ll\_{n,\epsilon}\mathrm{N}\mathfrak{a}^{\epsilon}$$
for all $\epsilon>0$, where $\lambda\_{\pi}(\mathfrak{a})$ is the Hecke eigenvalue at $\mathfrak{a}$. One way to approach GRC is to study the moments
$$\sum\_{\mathrm{N}\mathfrak{a}\leq x}|\lambda\_{\pi}(\mathfrak{a})|^{2k},$$
where $k\geq 1$ is a natural number. When $k=1$, this is bounded from above by
$$\sum\_{\mathrm{N}\mathfrak{a}\leq x}\lambda\_{\pi\times\widetilde{\pi}}(\mathfrak{a}),$$
where $\pi\times\pi'$ denotes the Rankin-Selberg convolution. The work of Jacquet, Piatetski-Shapiro, and Shalika establishes the basic properties of the Rankin-Selberg $L$-function $L(s,\pi\times\widetilde{\pi})$. These properties (particularly the absolute convergence of the Euler product that defines $L(s,\pi\times\widetilde{\pi})$ in the region $\mathrm{Re}(s)>1$) imply the first nontrivial bound: $|\lambda\_{\pi}(\mathfrak{a})|\ll\_{n,\epsilon}\mathrm{N}\mathfrak{a}^{1/2+\epsilon}$.
To study $k=4$, the pertinent Dirichlet series (which is conjecturally an $L$-function with an analytic continuation and functional equation) is $L(s,\pi\times\widetilde{\pi}\times\pi\times\widetilde{\pi})$. For $k=6$, we need $L(s,\pi\times\widetilde{\pi}\times\pi\times\widetilde{\pi}\times\pi\times\widetilde{\pi})$. Hopefully the pattern is clear. We expect each Dirichlet series in this sequence converges absolutely for $\mathrm{Re}(s)>1$. That would suffice to prove GRC (take $k$ to be sufficiently large in terms of $\epsilon$).
Such a region of absolute convergence follows immediately if all of these Dirichlet series are in fact products of $L$-functions of cuspidal automorphic representations (as above). Equivalently, each of these higher-order convolutions are isobaric sums of cuspidal automorphic representations. One way to express the Langlands functoriality conjecture for $\mathrm{GL}\_n$, at least as it pertains to the question, is that if $\pi\in\mathcal{A}(n)$ and $\pi'\in\mathcal{A}(n')$, then there exists an isobaric sum of cuspidal automorphic representations, say $\pi\boxtimes\pi'$, such that
$$L(s,\pi\times\pi')=L(s,\pi\boxtimes\pi').$$
| 9 | https://mathoverflow.net/users/111215 | 441856 | 178,324 |
https://mathoverflow.net/questions/441848 | 1 | I happen to encounter the following inequality which I need to prove:
$$\left(k+(1+k)\left(1-\frac{1}{e}\left(1+\frac{1}{k}\right)^k\right)\right)\log\left(1+\frac{1}{k}\right)>1,$$ for $k\in\mathbb{Z}^{+}$.
My current idea is to expand $$k+(1+k)\left(1-\frac{1}{e}\left(1+\frac{1}{k}\right)^k\right)=k+\frac{1}{2}+\frac{1}{24k}+O\left(\frac{1}{k^2}\right).$$
It seems like $$\left(1+\frac{1}{k}\right)^{k+\frac{1}{2}}>e$$ is always true? But not sure how can I proceed with the remaining $O\left(\frac{1}{k^2}\right)$ terms.
| https://mathoverflow.net/users/136078 | An inequality about e | We can actually prove this directly for $k \ge 7$ and it should be easy to check it for $k<7$
We note that for $x < 1$ we have $x/2-x^2/3+x^3/4-x^4/5... \ge x/2-x^2/3$ (series converges absolutely and grouping in pairs the remainder is positive).
For $y<1$ we also have $1-e^{-y} \ge y-y^2/2$ again by using the Taylor series and grouping
So with $x=1/k, k \ge 2$ we have, $$1-\frac{1}{e}\left(1+\frac{1}{k}\right)^k=1-e^{k\log(1+1/k)-1}=1-e^{-x/2+x^2/3-x^3/4+x^4/5 \ldots} \ge 1-e^{-x/2+x^2/3}$$
Also $$1-e^{-x/2+x^2/3} \ge x/2-x^2/3-(x/2-x^2/3)^2/2=x/2-11x^2/24+x^3/6-x^4/18$$
So $(1+k)\left(1-\frac{1}{e}\left(1+\frac{1}{k}\right)^k\right) \ge \frac{1}{2}+\frac{1}{24 k}-\frac{7}{24 k^2}+\frac{1}{9 k^3}-\frac{1}{18 k^4} > \frac{1}{2}$ for $k \ge 7$
Hence $\left(k+(1+k)\left(1-\frac{1}{e}\left(1+\frac{1}{k}\right)^k\right)\right)\log\left(1+\frac{1}{k}\right) > (k+\frac{1}{2})\log\left(1+\frac{1}{k}\right), k \ge 7$
But if $f(x)=(x+\frac{1}{2})\log\left(1+\frac{1}{x}\right), x>1$ we have $f'(x)=\log\left(1+\frac{1}{x}\right)-\frac{1}{2x}-\frac{1}{2(x+1)}$ and then $f''(x)=-\frac{1}{x(x+1)}+\frac{1}{2x^2}+\frac{1}{2(x+1)^2}> 0$ so $f'$ increasing hence $f'<0$ (it is $0$ at infinity), hence $f$ decreasing so $f(x) >\lim\_{y\to \infty}f(y)=1$ so indeed $(k+\frac{1}{2})\log\left(1+\frac{1}{k}\right)>1$ and finally $$\left(k+(1+k)\left(1-\frac{1}{e}\left(1+\frac{1}{k}\right)^k\right)\right)\log\left(1+\frac{1}{k}\right)>1, k \ge 7$$
| 2 | https://mathoverflow.net/users/133811 | 441858 | 178,325 |
https://mathoverflow.net/questions/441871 | 0 | Consider a $N\times N$ normalized matrix sample from GOE (the definition see [https://www.lpthe.jussieu.fr/~leticia/TEACHING/Master2019/GOE-cuentas.pdf](https://www.lpthe.jussieu.fr/%7Eleticia/TEACHING/Master2019/GOE-cuentas.pdf)). If we apply the following result of the edge of the spectrum,
>
> If we denote the $k$ largest eigenvalues by $\lambda\_N,\lambda\_{n-1},··· ,\lambda\_{N-k+1}, $ then for Gaussian ensembles the joint distribution function of rescaled eigenvalues has the limit:
> $$
> \lim\_{N\to\infty}P(N^{2/3}(\lambda\_N-2)\le s\_1,\dots, N^{2/3}(\lambda\_{N-k+1}-2)\le s\_k)=F\_{\beta, k}(s\_1,\dots, s\_k)
> $$
> where $F\_{\beta, k}(s\_1,\dots, s\_k)$ is the Tracy-Widom distribution.
>
>
>
then we will get the following results by continuous mapping theorem:
$$\lambda\_N-\lambda\_{N-k+1}=O\_P(N^{-2/3})$$
Now, if we ordering all eigenvalues by
$|\sigma\_N|\ge |\sigma\_{N-1}|\ge \dots \ge |\sigma\_1|$.
**I would like have the similar result that for every $\epsilon>0$, there exists constants $C>0,\alpha>0$ so that
$$
P\left(N^{\alpha}\left(\frac{|\sigma\_N|}{|\sigma\_{N-k+1}|}-1\right)\le C\right)\ge 1-\epsilon
$$**
I am not if we can take $\alpha=2/3$?
| https://mathoverflow.net/users/168083 | The probability upper bound on the ratio of the eigenvalues | The Tracy-Widom distribution says that the level spacing near the edge of the spectrum at $\pm 2$ is of order $N^{-2/3}$, hence we may define
$$|\sigma\_N|\equiv 2+N^{-2/3}\delta,\;\;|\sigma\_{N-k+1}|\equiv 2+N^{-2/3}\delta',$$
with $\delta,\delta'$ of order $N^0$. Now consider
$$\Delta=N^{2/3}\left(\frac{|\sigma\_N|}{|\sigma\_{N-k+1}|}-1\right)=\frac{\delta-\delta'}{2+N^{-2/3}\delta'}=\tfrac{1}{2}(\delta-\delta')+{\cal O}(N^{-2/3}).$$
Normalisation requires that $\lim\_{C\rightarrow\infty}P(\Delta\leq C)=1$, so for each $\epsilon>0$ we can find a constant $C>0$ so that
$$P(\Delta\leq C)\geq 1-\epsilon,$$
which is the desired inequality. The constant $C$ will depend on $\epsilon$, but it will be independent of $N$ for large $N$.
| 1 | https://mathoverflow.net/users/11260 | 441873 | 178,327 |
https://mathoverflow.net/questions/441861 | 0 | Suppose that for all $n$ natural numbers, $d\_{n}$ is a pseudometric on set $X
$. Define $d=\sum\_{n=1}^{\infty }a\_{n}\frac{d\_{n}}{1+d\_{n}}$, where $\left(
a\_{n}\right) $ is a sequence of positive numbers such that $\sum\_{n\in
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
}a\_{n}<\infty $.
It is claimed that the uniformity $\mathcal{U}\_{d}$ generated by
pseudometric $d$ is the same the uniformity $\mathcal{U}\_{P}$ generated by
pseudometrics $P=\left\{ d\_{n}:n\in
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
\right\} $, that is $\mathcal{U}\_{d}=\mathcal{U}\_{P}$.
At page 237 in Topology Book by W. W. Fairchild, C. I. Tulcea,
<https://archive.org/details/topology0000fair/page/236/mode/2up>
| https://mathoverflow.net/users/86099 | A question about uniformities generated by pseudometrics | One direction:
let $\varepsilon>0$ and taken $N$ such that $\sum\_{n>N}a\_n<\frac12\varepsilon$. Take $\delta>0$ such that $\delta\cdot\sum\_{n\le N}a\_n<\frac12\varepsilon$. Now if $d\_n(x,y)<\delta$ for all $n\le N$ then
$$
d(x,y)< \delta\cdot\sum\_{n\le N}a\_n+\sum\_{n>N}a\_n<\varepsilon
$$
(because $d\_n(x,y)/(1+d\_n(x,y))\le d\_n(x,y)$ always).
For the second direction: you have $a\_n\cdot d\_n(x,y)/(1+d\_n(x,y))\le d(x,y)$. Then from $d(x,y)\le\frac12a\_n$ you get $d\_n(x,y)/(1+d\_n(x,y))\le\frac12$ and then $d\_n(x,y)\le1$. And from this you get $\frac12d\_n(x,y)\le d\_n(x,y)/(1+d\_n(x,y))$. In summary: if $d(x,y)\le\frac12a\_n$ then $\frac{a\_n}2d\_n(x,y)\le d(x,y)$, so you have an explicit formula for $\delta$ in terms of $\varepsilon$ to show uniform continuity of the identity.
| 1 | https://mathoverflow.net/users/5903 | 441875 | 178,328 |
https://mathoverflow.net/questions/441877 | 2 | [Here](https://mathoverflow.net/questions/371010/how-to-calculate-inverse-of-sum-of-two-kronecker-products-with-specific-form-eff) and [here](https://www.sciencedirect.com/science/article/pii/S0024379587903144?via%3Dihub), specific ways to address the equation in $x$, for $N=2$, are given:
$$\sum\_{i=1}^N (A\_i\otimes B\_i)x=c$$
Is anything know about the case $N>2$?
I am looking in fact for an efficient solution to the above type of linear system. Such structure may arise from space-time algorithms applied to parabolic, non-linear problems.
---
In fact, the system I am interested in has the following structure:
$\sum\_{j,l,a,b}M\_{j,a}^iT\_{l,b}^k\nu^1\_{a,b}x^1\_{j,l}+\sum\_{j,l,a,b,o}S\_{j,a}^iD\_{l,b,o,j}^k\nu^2\_{a,b} x^2\_{j,l}=f^{i,k}$
As you can see, the $\nu$ factors prevent me from writing the system as in the link I posted. Also, the matrix $D$ contains the index $j$, which gives further problems. My idea was to decompose $\nu$ as a sum of Kronecker products (approximately, and carry out a similar procedure for $D$, I won't go into details), hoping to obtain a better structure.
By doing so, we see that we obtain the original system I decsribed above. So, to answer to @Nathaniel, $N$ should be large so that the approximation of e.g. $\nu$ in terms of a sum of Kronecker product, is good. I don't have a specific number in mind.
As for the size of $A\_i, B\_i$, they are square, of side $1e4$, $1e3$ respectively. I suppose then that $N$ will be much smaller than these numbers.
| https://mathoverflow.net/users/155442 | Linear system with sum of Kronecker products | The recent state of the art is described in section 7.2 of Simoncini, V. "Computational methods for linear matrix equations." *SIAM Rev.* **58**, 377 (2016), <https://doi.org/10.1137/130912839>. Your equation is equivalent to equation (2) in that reference. A lightly reformatted quote from there:
>
> Equation (2) is very difficult to analyze in its full generality, and
> necessary and sufficient conditions for the existence and uniqueness
> of the solution explicitly based on $\{A\_i\}$, $\{B\_i\}$ are hard to
> find, except for some very special cases.
>
>
>
| 1 | https://mathoverflow.net/users/1847 | 441893 | 178,332 |
https://mathoverflow.net/questions/441870 | 0 | (**Preliminaries:**) 1.) Let $S\subset\mathbb{R}^n$ and define $\mathcal{M}(S) = \{\text{$\mu$ a Borel measure}: \text{$0 < \mu(S) < \infty$ and $\mathrm{support}(\mu)\subset S$}\}$.
2.) Define the Fourier transform of a measure as $\hat{\mu}(x):= \int\_{\mathbb{R}^n}e^{-2\pi i\xi \cdot x}d\mu(\xi)$.
3.) Define the Fourier dimension of a set $S\subset\mathbb{R}^n$ as
$$\mathrm{dim}\_FS := \sup\{s\in \left[0,n\right]: \exists \mu\in \mathcal{M}(S):\forall x\in\mathbb{R}^n:\left|\hat{\mu}(x)\right|\leq \left|x\right|^{-s/2}\}$$
(**Remark:**) I was a bit hesitant to post this question here since there is a non-zero chance for it being completely trivial, but as I (honest to God) could not find any discussion on this in books such as Mattila's *Fourier Analysis and Hausdorff Dimension* or *Geometry of Sets and Measures in Euclidean Spaces* or in exercise sections of such relevant books, here I am.
(**Question:**) Given $-\infty<a<b<\infty$, what is the Fourier dimension of the interval $\left[a, b\right]\subset\mathbb{R}$ and what measure $\mu\in \mathcal{M}\left(\left[a, b\right]\right)$ gives it?
My naïve first thought was to just use the one dimensional Lebesgue measure restricted to $\left[a, b\right]$. However, then for $x\neq 0$ we get
$$|\hat{\mu}(x)|^2 = \left|\frac{i}{2\pi x}\left(e^{-2\pi i xb} - e^{-2\pi ixa}\right)\right|^2 = \frac{1 - \cos\left(2\pi x(b - a)\right)}{2\pi^2x^2}$$
whence $|\hat{\mu}(x)|^2 \leq \left|x\right|^{-s}\Longleftrightarrow 1 - \cos\left(2\pi x(b - a)\right)\leq 2\pi^2\left|x\right|^{2-s}, s\in \left[0,1\right]$
By taking $e.g. b = 4, a = -2, x = 0.05$ we see that if $\left[a, b\right]$ were to have a Fourier dimension equal to one, then $\mu$ will not give it as the LHS is equal to $\approx 1.309$ while the RHS is equal to $\approx 0.987$.
Any ideas how the measure $\mu$ should be constructed? Also, do you happen to know a good source which works through a bit more elementary examples (like this) of the Fourier dimension?
| https://mathoverflow.net/users/172696 | Calculating the Fourier dimension of a real interval $\left[a, b\right]$ | If $-\infty\le a<b\le\infty$, then $\dim\_F[a,b]=\infty$.
Indeed, without loss of generality $[a,b]=[-1,1]$. Take any natural $n$ and let
$$\nu\_n:=\frac1{c\_n}\,\mu\_{1/2}\*\cdots\*\mu\_{1/2^n},$$
where $\mu\_L$ is the uniform distribution on $[-L,L]$ and $c\_n:=\prod\_{k=1}^n 2^k$. Then for all real $t\ne0$
$$\hat\nu\_n(t):=\int\_{-\infty}^\infty e^{itx}\nu\_n(dx)
=\frac1{c\_n}\,\prod\_{k=1}^n\hat\mu\_{1/2^k}(t)
=\frac1{c\_n}\,\prod\_{k=1}^n\frac{\sin(t/2^k)}{t/2^k}$$
and hence
$$|\hat\nu\_n(t)|\le\frac1{c\_n}\,\prod\_{k=1}^n\frac1{|t|/2^k}=|t|^{-n},$$
for all natural $n$. Also, $\nu\_n$ is a nonzero measure with support contained in $[-1,1]$. So, $\dim\_F[-1,1]=\infty$. $\quad\Box$
---
This assumes the more natural definition
$$\mathrm{dim}\_FS := \sup\{s\in \left[0,\infty\right): \exists \mu\in \mathcal{M}(S)\ \forall x\in\mathbb{R}^d\setminus\{0\}\ \left|\hat{\mu}(x)\right|\leq \left|x\right|^{-s/2}\}$$
instead of the strange definition
$$\mathrm{dim}\_FS := \sup\{s\in \left[0,d\right]: \exists \mu\in \mathcal{M}(S):\forall x\in\mathbb{R}^d:\left|\hat{\mu}(x)\right|\leq \left|x\right|^{-s/2}\}$$
in the OP.
Using the definition in the OP (with $\mathbb{R}^d$ corrected to $\mathbb{R}^d\setminus\{0\}$), we would of course get $\dim\_F[-1,1]=1$.
I also used a slightly different definition of the Fourier transform, without the extra factor $-2\pi$ in the exponent. Of course, this rescaling does not affect $\dim\_F$.
| 2 | https://mathoverflow.net/users/36721 | 441899 | 178,335 |
https://mathoverflow.net/questions/441834 | 3 | Usually, Berkovich analytic spaces are derived from some Banach rings (or chains of Banach rings) over a completely normed field $k$ through Berkovich spectrum. But when the base field is the complex number $\Bbb{C}$, there is another way of getting analytic spaces, namely gluing open sets in $\Bbb{C}^n$ via biholomorphic transition maps. However, with the "gluing method" we get **more spaces**, for example a non-algebraic complex $2$-torus that admits no non-constant meromorphic functions and thus not the Berkovich spectrum of any $\Bbb{C}$-Banach rings.
What happens if we apply the "gluing method" to open sets of the $n$-dimensional Berkovich affine space $\Bbb{A}^n\_k$? *(For simplicity, only $k=\Bbb{C}\_p$, the $p$-adic complex numbers with $p$-adic metric, are considered.)* Do we also get **compact** "non-algebraic" analytic spaces, in the sense that the function field has a small transcendental degree over $k$, or it can not be realized as the underlying Berkovich space of any $k$-algebraic varieties?
**Edit:** As Wojowu comments, there do exist non-algebraic torus analogues for $\Bbb{C}\_p$-Berkovich spaces. So I'm now asking for a **simply connected** example. (For $\Bbb{C}$-analytic spaces we have non-algebraic $K3$ surfaces.)
| https://mathoverflow.net/users/166298 | "Non-algebraic" Berkovich spaces | Indeed, every complex manifold is locally isomorphic to an open in $\mathbf{C}^n$. More generally, we define complex analytic spaces as those locally isomorphic to a locally ringed space of the form $(Z, i^{-1}\mathcal{O}\_U/I)$ where $U$ is an open in $\mathbf{C}^n$, $I$ is the ideal in the sheaf $\mathcal{O}\_U$ of holomorphic functions defined by finitely many sections $f\_1, \ldots, f\_r$, where $Z\subseteq U$ is the vanishing locus of the $f\_i$, and $i\colon Z\to U$ is the closed embedding.
Following Berkovich, one can define analytic spaces over any Banach ring $A$ in basically the same way. We define $\mathbf{A}^n\_A$ as the set of multiplicative real-valued seminorms on the polynomial ring $A[T\_1, \ldots, T\_n]$ which are bounded on $A$ by the given Banach norm. We give it the coarsest topology making the maps $|\cdot| \mapsto |f|$ ($f\in A$) continuous. We endow it with a structure sheaf $\mathcal{O}$ defined using a certain completion of rational functions (see Berkovich's first book of Lemanissier-Poineau for details). We can then define $A$-analytic spaces as in the previous paragraph.
For $A=\mathbf{C}$ with the Euclidean norm, we recover complex analytic spaces, while for $A$ being a non-Archimedean field, we recover the category of "boundaryless" (or "partially proper") Berkovich spaces. For more general $A$, the study of such spaces is very difficult (see the work of Poineau and Berger for the case $A=\mathbf{Z}$).
However, even for $A=\mathbf{C}\_p$, it is not true that a smooth Berkovich space is locally isomorphic to an open in $\mathbf{A}^n$. For example, the Berkovich analytification of a smooth projective curve over $\mathbf{C}\_p$ of positive genus and good reduction will not have this property: it will have a unique point (of "type 2") which does not admit the open disc as a neighborhood.
Turning to your question. It is indeed easy to produce "non-algebraic" Berkovich spaces over $\mathbf{C}\_p$ by taking a smooth projective variety $X\_0$ over $k=\overline{\mathbf{F}}\_p$ with unobstructed deformations and $H^2(X\_0, \mathcal{O}\_{X\_0})\neq 0$ and taking a "random" formal deformation $\mathfrak{X}$ of $X\_0$ over a complete discrete valuation ring $W$ with residue field $k$, embedding $W$ into $\mathbf{C}\_p$, and taking $X$ to be the Berkovich space associated to the rigid-analytic generic fiber of the formal scheme $\mathfrak{X}$. For a generic choice of the deformation, we will have ${\rm Pic}(X)=0$, so in particular $X$ will not be isomorphic to the analytification of any algebraic variety over $\mathbf{C}\_p$. This applies to abelian varieties of dimension $>1$ and to K3 surfaces.
However, if $X\_0$ is not rational, the $X$ resulting from the above construction will never have the property you imposed that it is locally isomorphic to an open subset of the affine space, because the generic point of $X\_0$ will have a unique preimage in $X$ without such a neighborhood.
To achieve this, we need to be more clever. We pick $X\_0$ to be semistable (or log smooth over the log point $\mathbf{N}\to k$) and maximally degenerate. Think of the product of several Tate curves. In this case, the generic fiber of a "random" smoothing over $W$ will still be nonalgebraic, but will admit a Tate-Raynaud uniformization by $(\mathbf{C}\_p^\*)^g$, and hence will have the property you wanted.
Finally, you asked for a simply connected example. Here are some indications how this could go. Consider the Dwork family of K3 surfaces over $\mathbf{Z}\_p$ given by
$$ p(x^4 + y^4 + z^4 + t^4) = xyzt $$
where $p>5$. This is projective alright. Consider a random smooth formal scheme $\mathfrak{X}$ over $\mathbf{Z}\_p$ which is isomorphic to this one modulo $p^2$, and let $X$ be the associated Berkovich space over $\mathbf{C}\_p$, a non-algebraic rigid-analytic K3 surface. After a suitable blowup (see Section 1 in Harris, Shepherd-Barron, Taylor "A family of Calabi-Yau varieties and potential automorphy"), the model $\mathfrak{X}$ is Zariski-locally isomorphic to
$$ \mathfrak{U} = {\rm Spf}(\mathbf{Z}\_p\langle x, y, z\rangle/(p-xyz)) $$
and hence every point of $X$ admits an open neighborhood (maybe in the weaker sense of an affinoid neighborhood) isomorphic to an open in the generic fiber $U$ of $\mathfrak{U}$. But $U$ itself embeds as an open into $(\mathbf{A}^2)^{\rm an}$, and we are done.
**Postscriptum.** Since you seem to be interested in the homotopy type of the Berkovich space $X$, in the above example it should be homotopy equivalent to $S^2$, as the dual complex of the special fiber is a $2$-sphere. I am sure that the people working on "non-Archimedean SYZ" (following the ideas of Konstevich and Soibelman, Gross and Siebert) have figured it all out in details.
| 5 | https://mathoverflow.net/users/3847 | 441901 | 178,336 |
https://mathoverflow.net/questions/441894 | 3 | Let $G$ be a discrete group, and let $\mathcal{F}$ be a family of subgroups of $G$ (closed under conjugation and taking subgroups). Then we may define the geometric and cohomological dimensions of $G$ with respect to $\mathcal{F}$, denoted $\operatorname{gd}\_\mathcal{F}(G)$ and $\operatorname{cd}\_\mathcal{F}(G)$ respectively, with reference to classifying spaces for $G$-actions with isotropy in $\mathcal{F}$ and the homological algebra of modules over the $\mathcal{F}$-restricted orbit category. (If you've read this far, you probably already know the definitions.)
A well known result credited to Lueck and Meintrup says that $\operatorname{gd}\_\mathcal{F}(G)\le \max\{3,\operatorname{cd}\_\mathcal{F}(G)\}$. See Theorem 0.1(a) [here.](https://www.him.uni-bonn.de/lueck/data/lm.pdf)
I can understand the general framework of the proof given by Lueck and Meintrup, but there is a technical Lemma 2.7 about chain complexes which they do not prove (referring only to pp 279-280 of Lueck’s book [Transformation Groups and Algebraic K-Theory](https://www.him.uni-bonn.de/lueck/data/lueck_Transformation_Groups_and_Algebraic_K-theory.pdf), where he seems to be working in a more general framework to prove something about finite domination). Lemma 2.7 contains the statement "If $D\_\*$ is homotopic to a finite-dimensional free complex, then $C’\_\*$ can be chosen to be finite-dimensional" when it seems to me that what is needed is the stronger "If $D\_\*$ is homotopic to a $d$-dimensional free complex, then $C’\_\*$ can be chosen to be $d$-dimensional".
Does this stronger statement follow from the argument in Lueck’s book (which I therefore need to understand)? Is there any other source where this argument is written up in full?
| https://mathoverflow.net/users/8103 | Geometric vs cohomological dimension with families - on a proof of Lueck and Meintrup | **Note** This is based on misreading the question as being about projective resolutions, not free ones. I had deleted it but the OP found it helpful so I have undeleted.
I have managed to cobble this out of the Brown references I gave in now deleted comments. In this answer all chain complexes are assumed to be bounded below.
Suppose that $D\_\ast$ is a chain complex of projectives that is homotopy equivalent to a complex vanishing in dimensions $>n$. Then $D\_\ast$ is homotopy equivalent to a complex of projectives $C\_\ast$ vanishing in dimensions $>n$. Moreover, one can choose $C\_\ast$ so that $C\_q=D\_q$ for $q<n$ and $C\_n=D\_n/B\_n$ where $B\_n$ is the $n$-boundaries of $D\_\ast$. In particular, if $D\_\ast$ consists of finitely generated projectives then the same is true for $C$.
First of all we will use the standard fact that if you have a chain map between (bounded below) chain complexes which are degreewise projective, then it induces a homotopy equivalence if and only if it is an isomorphism on homology.
Define $C\_\ast$ as proposed above. Clearly the quotient map $D\_\ast\to C\_\ast$ which is $0$ for in degree $q>n$, the identity for $q<n$ and the projection $D\_n\to D\_n/B\_n$ gives a homology isomorphism since $D\_\ast$ is homotopy equivalent to a complex vanishing in dimension $>n$. It remains to show that $C\_n$ is projective. Consider the cochain complex $K^\ast=\mathrm{Hom}(D\_\ast,B\_n)$. Since $D\_\ast$ is homotopy equivalent to a chain complex vanishing in degree $n+1$, we have $H^{n+1}(K^\ast)=0$. Clearly $d\_{n+1}\colon D\_{n+1}\to B\_n$ is an $(n+1)$-cocycle in $K^{n+1}$, and hence there is $g\colon D\_n\to B\_n$ with $g\circ d\_{n+1}=d\_{n+1}$. Then $g$ is a retraction onto $B\_n$ and so $D\_n= B\_n\oplus \ker g$. Thus $\ker g\cong D\_n/B\_n=C\_n$ is projective, as required.
| 1 | https://mathoverflow.net/users/15934 | 441908 | 178,340 |
https://mathoverflow.net/questions/441829 | 3 | Let $V$ be a finite-dimensional vector space over $\mathbb{R}$ equipped with an inner product $\omega(-,-)$. One standard fact is that there is an induced inner product on $\wedge^k V$. For instance, this shows up when you're setting up Hodge theory.
All the constructions I've seen in books construct the inner product on $\wedge^k V$ by first choosing an orthonormal basis $\{e\_1,\ldots,e\_n\}$ for $(V,\omega)$, and then declaring that the basis $e\_{i\_1} \wedge \cdots \wedge e\_{i\_k}$ for $\wedge^k V$ where the $i\_j$ range over increasing sequences $1 \leq i\_1 < \cdots < i\_k \leq n$ is an orthonormal basis for $\wedge^k V$.
This strikes me as pretty unnatural; in particular, you then have to do a calculation to prove that $O(V,\omega)$ acts on $\wedge^k V$ by orthogonal transformations.
Does anyone know a good coordinate-free way to do this? I know a partial solution. Namely, you can view $\omega(-,-)$ as giving an isomorphism $\iota\colon V \rightarrow V^{\ast}$, and we then get an isomorphism
$$\wedge^k V \stackrel{\wedge^k \iota}{\longrightarrow} \wedge^k V^{\ast} \cong \left(\wedge^k V\right)^{\ast}.$$
This gives a nondegenerate bilinear form on $\wedge^k V$. However, while it is easy to see that this bilinear form is symmetric, it is not obvious that it is positive-definite (and the calculation you have to do for this is no easier than what I'm trying to avoid!).
Another feature I'd like from a construction is uniqueness: namely, the inner product on $\wedge^k V$ should (up to rescaling) be the unique inner product such that $O(V,\omega)$ acts by orthogonal transformations on $\wedge^k V$.
| https://mathoverflow.net/users/500239 | Coordinate free way to construct inner product on exterior powers | I'm going to answer my own question, summarizing the comments and adding a little more from my own reflections (marked community wiki, though it doesn't matter since this is not a registered account and I can't earn reputation from it).
The first observation (from Tom Goodwillie) is that it is actually very easy to see that the inner product I wrote down is positive definite. Indeed, it is almost immediate that if $e\_1,\ldots,e\_n$ is an orthonormal basis for $V$, then the $e\_{i\_1} \wedge \cdots \wedge e\_{i\_k}$ form an orthonormal basis for $\wedge^k V$.
The second observation builds on what Igor Khavkine pointed out. Namely, it is obvious from pure thought that the inner product I wrote down is either positive definite or negative definite. This is actually a general phenomena, as follows.
**Theorem**: Let $G$ be a compact Lie group and let $W$ be a finite-dimensional irreducible real representative of $G$. Then $W$ has a $G$-invariant inner product $\omega(-,-)$, and if $\omega'(-,-)$ is any $G$-invariant symmetric bilinear form on $W$ then there exists some $r \in \mathbb{R}$ such that $\omega'(-,-) = r \cdot \omega(-,-)$. In particular, if $\omega'(-,-)$ is nonzero it is either positive definite or negative definite.
The existence of $\omega(-,-)$ follows from the usual averaging argument (this is where we use the fact that $G$ is compact, which is not used in the rest of the proof). As for $\omega'(-,-)$, since $\omega(-,-)$ is nondegenerate there exists a linear map $f\colon V \rightarrow V$ such that $\omega'(x,y) = \omega(x,f(y))$ for all $x, y \in V$. For $g \in G$, we have
$$\omega'(x,y) = \omega'(gx,gy) = \omega(gx,f(gy)) \quad \text{and} \quad \omega'(x,y) = \omega(x,f(y)) = \omega(gx, g f(y))$$
for all $x,y \in V$. From this and the nondegeneracy of $\omega(-,-)$, we see that $f(gy) = g f(y)$ for all $y \in V$. Since $V$ is irreducible, Schur's Lemma implies that there exists some $r \in \mathbb{R}$ with $f(y) = r y$ for all $y \in V$. In particular,
$$\omega'(x,y) = \omega(x,f(y)) = \omega(x,r y) = r \cdot \omega(x,y),$$
as desired.
| 3 | https://mathoverflow.net/users/500239 | 441911 | 178,342 |
https://mathoverflow.net/questions/441752 | 2 | I am working with subsets of $[n]$ of the form $(A+B)\cap A$, where $A+B$ is a sumset. Namely, I am interested if there are nonempty sets $B$ such that whenever $A$ covers a positive proportion of $[n]$, the set $(A+B)\cap A$ also covers a positive proportion of $[n]$. In fact it would be nice if the proportion is the same, or tends to the same limit. (What I mean by "positive proportion" only really makes sense when we take $n$ to infinity; I suppose if $A$ and $B$ are regarded as (possibly infinite) subsets of ${\bf N}$, then this is their asymptotic density.)
For instance $B = \{1\}$ is not an example, since if $A$ is the set of even integers in $[n]$, then $A$ covers one half of $[n]$ but $(A+B)\cap A = \emptyset$. The specific example I'm considering at the moment is the set $B = \{1, 2, 6, 24, 120,\ldots,\}$ of factorials, so information particular to this example would also be appreciated, if more general information is not known. But this problem seems natural enough that I thought it might have a name, I just didn't know what to search online. As always, thank you all in advance for the help!
**Edit.** I thought I should illustrate a nontrivial example where $B$ is the set of factorials and the densities of $A$ and $(A+B)\cap A$ are the same. Let $p$ be a prime and let $a$ be a residue modulo $p$. Then the set $A$ of all integers congruent to $a$ modulo $p$ has asymptotic density $1/p$. Now since only finitely many members of $B$ are not divisible by $p$, if I'm not wrong the set $(A+B)\cap A$ should have the same asymptotic density as $A$. Is this true for all $A$, not just this example?
**Second edit.** Following Terry Tao's hint below, we recall the definition of *intersective*: $B$ is intersective if for all $A$ of positive upper density, we have $(A-A)\cap B \ne\emptyset$. Here is a proof that this is equivalent to my condition on $B$:
*Proof.* Let $B\subseteq {\bf N}$ and suppose that there is $A$ of positive upper density such that
$(A+B)\cap A$ has density strictly less than that of $A$. Then $A\setminus (A+B)$ has positive upper
density. Now let $x,y\in A\setminus(A+B)$ and set $z = x-y$, so $z\in A\setminus(A+B)-A\setminus(A+B)$.
Since $x\notin A+B$ and $y\in A$, from $x=y+z$ we conclude that $z\notin B$. Thus we find that
$$\bigl(A\setminus(A+B)-A\setminus(A+B)\bigr) \cap B = \emptyset;$$
in other words, $B$ is not intersective.
On the other hand, if $B$ is not intersective, then there exists $A$ of positive upper density such that
$(A-A)\cap B = \emptyset$. For such an $A$, we must have $(A+B)\cap A = \emptyset$, since if $a\in (A+B)\cap A$,
then $a = a'+b$ for $a'\in A$, $b\in B$, and then $a-a'=b$. ▮
So one part of my question is settled. But I am still interested in knowing whether the set of factorials is intersective or not (and would happily accept this as an answer to the question). Forgive me if this is trivial but I am not able to see it right away. The set $S$ given by the greedy algorithm (suggested by mathworker21 below) seems dense, but this is because for a long time it consists of elements congruent to one of $\{0,3,7,10,14,17,21\}$ modulo $25$, which is fine (and has density $7/25=0.28$) until we get to $5040$, which is congruent to $15 = 0-10$. The phenomenon is more obvious when we start the greedy algorithm with $0$ instead of $1$, in which case for a long time we get numbers congruent to either $0$ or $3$ modulo $7$ (this has density $2/7 = 0.2857\dots$), and this works because the factorials $<5040$ are all congruent to one of $\{1,2,3,6\}$ modulo 7, and $\{0,3\} - \{0,3\} = \{0,3,4\}$ in ${\bf Z}/(7)$. What happens after $5040$ is still somewhat mysterious to me, but the proportion dips below $0.28$ and I cannot convince myself that it won't go to $0$.
| https://mathoverflow.net/users/168142 | Sets with certain property concerning density of sumsets | The set of factorials $B$ is not intersective, hence there is a set $A$ of positive density for which $A$ and $A+B$ are disjoint.
Indeed, every natural number $n$ has a unique "factorial base" representation
$$ n = \sum\_{k=1}^\infty a\_k k!$$
with $0 \leq a\_k \leq k$ and all but finitely many of the $a\_k$ non-zero. If we let $E$ be the set of such numbers for which we do not have $a\_k=k$ for two consecutive values of $k$, then $E$ has positive density (this is a simple sieve theoretic calculation using the fact that $\sum\_{k=1}^\infty \frac{1}{k(k+1)}$ converges). Now we partition $E$ into four classes depending on the parity of the two quantities
$$ \sum\_{k \geq 1, \hbox{ even}} (k+1) a\_{k+1} + a\_k$$
and
$$ \sum\_{k \geq 1, \hbox{ odd}} (k+1) a\_{k+1} + a\_k.$$
Note that adding $k!$ to $n$ with $k$ even would flip the parity of the first sum (noting that $(k+1) a\_{k+1} + a\_k$ is strictly less than $k(k+1)$ by definition of $E$), and adding $k!$ to $n$ with $k$ odd would flip the parity of the second sum. Thus none of the four classes in $E$ contains a pair $n, n+k!$. By the pigeonhole principle, one of these classes has positive upper density (indeed it looks straightforward to prove that all four classes have positive density), and the claim follows.
| 4 | https://mathoverflow.net/users/766 | 441912 | 178,343 |
https://mathoverflow.net/questions/441907 | 0 | Let $G$ be a group and let $\mathbb{G}$ be the associated one object category. Is there an explicit presentation of representable functors from $\mathbb{G} \to $**Set**? If so how does the Yoneda lemma look like explicitly in this setting?
| https://mathoverflow.net/users/160055 | Yoneda lemma for one object categories | Let me write $\def\B{\mathbf{B}}\B G$ for what you call $\mathbb{G}$.
You can check that presheaves on $\B G$ are precisely the right $G$-sets: the unique point of $\B G$ is sent to some set $X$, and functoriality defines a group homomorphism $\def\op{\mathrm{op}}G^\op\to\operatorname{Aut}(X)$.
In particular, the (unique) representable functor corresponds to the $G$-set $G$ itself with the action given by right multiplication.
The Yoneda Lemma in this setting then says, for any $G$-set $X$, that elements of $X$ correspond naturally to $G$-equivariant maps $G\to X$: send $x\in X$ to the map $g\mapsto x.g$, and send a function $f:G\to X$ to $f(1\_G)$.
**Edit:** to be a bit more explicit about the naturality, the group $G$ acts on both sides of this correspondence (the action on $X$ is given by the fact that $X$ is a $G$-set, and the action on $G$-equivariant functions $G\to X$ is componentwise), and the naturality of the Yoneda lemma says that this correspondence respects these $G$-actions.
| 2 | https://mathoverflow.net/users/160838 | 441913 | 178,344 |
https://mathoverflow.net/questions/441321 | 6 | Where can I find the construction for a **skew Hadamard matrix** of order 756?
According to multiple papers (e.g. [Koukouvinos and Stylianou - On skew-Hadamard matrices](https://www.sciencedirect.com/science/article/pii/S0012365X07004013?via%3Dihub) and [Seberry - On skew Hadamard matrices](https://ro.uow.edu.au/cgi/viewcontent.cgi?article=2003&context=infopapers)) this matrix has been constructed. However, none of the papers I have found give a reference to the actual construction that is being used to obtain it. In particular, the paper by Seberry (published in 1978) is the oldest I could find which states that the construction is known.
Note that I also couldn't find a construction for a skew Hadamard matrix of order 292 (see [this question](https://mathoverflow.net/questions/439195/construction-of-skew-hadamard-matrix-of-order-292)), which also appeared as known for the first time in that same 1978 paper, so it is possible that they were created using the same construction.
| https://mathoverflow.net/users/498306 | How to construct a skew Hadamard matrix of order 756? | The matrix can be constructed using Theorem 7 from [Seberry - On skew Hadamard matrices](https://ro.uow.edu.au/cgi/viewcontent.cgi?referer=&httpsredir=1&article=2003&context=infopapers), by setting $m = 1$ and $n = 28$.
The amicable Orthogonal Design of type $((1, 27); (28))$ are constructed from amicable hadamard matrices of order 28 (construction taken from [(v, k, λ) Configurations and Hadamard matrices - Wallis](https://www.cambridge.org/core/journals/journal-of-the-australian-mathematical-society/article/v-k-configurations-and-hadamard-matrices/DB40CEDE6B57B928253CF7A47750FEF4), for the relation between amicable Hadamard matrices and amicable OD see Lemma 5.4 from [Orthogonal Designs - Seberry](https://link.springer.com/book/10.1007/978-3-319-59032-5)).
Lastly, the OD of type $(1, 1, 26)$ is constructed using the Goethals-Seidel Array and sequences taken from Table G.7 of [Orthogonal Designs - Seberry](https://link.springer.com/book/10.1007/978-3-319-59032-5)
| 4 | https://mathoverflow.net/users/498306 | 441915 | 178,345 |
https://mathoverflow.net/questions/441393 | 4 | In Zermelo-Fraenkel set theory without the Axiom of Choice (AC), it is consistent to say that there are models in which:
1. there are vector spaces without a basis;
2. the field of complex numbers $\mathbb{C}$ only has two automorphisms (identity and complex conjugation).
Before I even ask my main question, I want to ask two "pre-questions" (as I am not a logician):
* is my formulation "in Zermelo-Fraenkel set theory without the Axiom
of Choice (AC), it is consistent to say that there are models in
which ..." formally correct? (And if not, what is a formal correct
statement ?);
* for the second statement above, what is a precise reference in which I can find this statement?
Now for my main question: is it also true that it is consistent to say that there are models of ZF set theory without AC, in which every vector space over $\mathbb{C}$ *has* a base (or even stronger: in which *dimension* is well defined), and in which $\mathbb{C}$ also has precisely two field automorphisms?
| https://mathoverflow.net/users/12884 | Automorphisms of vector spaces and the complex numbers without choice | This is not a full answer, but it is too long to be a comment.
Let $B(F)$ for field $F$ be the statement "every vector space over $F$ has a basis" and let $AL19(F)$ be the statement "for every vector space $V$ over $F$, every generating subset of $V$ contains a basis", $AL20(F)$ means "for every vector space $V$ over $F$, every independent subset of $V$ is contained in a basis".
In 2012 Paul Howard and Eleftherios Tachtsis said in [1] that both:
* There exists a field $F$ such that "$B(F)\implies AC$"
* There exists a field $F$ such that "$B(F)\;\not\!\!\!\implies AC$"
Are open, in particular *if the answer of your question is positive, then it is an open problem*.
On the other hand, Paul Howard had proven in [2] that $(∃F\ s.t.\ AL19(F))⇒AC$, in particular, the strengthening of $B(ℂ)$ to $AL19(ℂ)$ does imply AC and hence imply that there are wild automorphisms for $ℂ$.
Similarly, both [1] and [2] claim that in [3,4] it was proven that $(∃F\ s.t.\ AL20(F))⇒MC$ (which over ZF implies $AC$), and hence the dual strengthenin of $B(ℂ)$ to $AL20(ℂ)$ also imply that there are wild automorphisms for $ℂ$ (although from quick glance over [3,4] I couldn't see this result, a proof of this result, together with the result of the previous paragraph can be found in [5]).
>
> [1] *Howard, Paul; Tachtsis, Eleftherios*, [**On vector spaces over specific fields without choice**](https://doi.org/10.1002/malq.201200049), Math. Log. Q. 59, No. 3, 128-146 (2013). [ZBL1278.03082](https://zbmath.org/?q=an:1278.03082).
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>
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>
> [2] *Howard, Paul*, [**Bases, spanning sets, and the axiom of choice**](https://doi.org/10.1002/malq.200610043), Math. Log. Q. 53, No. 3, 247-254 (2007). [ZBL1121.03064](https://zbmath.org/?q=an:1121.03064).
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>
> [3] *Armbrust, M. K.*, [**An algebraic equivalent of a multiple choice axiom**](https://doi.org/10.4064/fm-74-2-145-146), Fundam. Math. 74, 145-146 (1972). [ZBL0234.04011](https://zbmath.org/?q=an:0234.04011).
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> [4] *Bleicher, M. N.*, [**Some theorems on vector spaces and the axiom of choice**](https://doi.org/10.4064/fm-54-1-95-107), Fundam. Math. 54, 95-107 (1964). [ZBL0118.25503](https://zbmath.org/?q=an:0118.25503).
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> [5] Rubin, H., & Rubin, J. E. (1985). Algebraic Forms. In [*Equivalents of the axiom of choice, II*](https://www.elsevier.com/books/equivalents-of-the-axiom-of-choice-ii/rubin/978-0-444-87708-6) (p. 122). North-Holland.
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| 4 | https://mathoverflow.net/users/113405 | 441918 | 178,346 |
https://mathoverflow.net/questions/440661 | 2 | Let $f:[0, \infty)\to [0, \infty)$ be non-decreasing and satisfy for all $t>t\_{0}$, $$f(t)+C\int\_{t\_{0}}^{t}f^{\gamma}(s)ds\leq \frac{1}{t-t\_{0}}\int\_{t\_{0}}^{t}f(s)ds,$$ where $0<\gamma<1$ and $C>0$.
Of course, if $f$ is differentiable and the somewhat similar differential inequality $$f'+Cf^{\gamma}\leq 0$$ holds, we have that $f(t)\equiv 0$ for all $t\geq T$ for some finite $T>0$. which follows from comparing $f$ with the function $g(t)=a(T-t)^{1/(1-\gamma)}$ where $a=((1-\gamma)C)^{1/(1-\gamma)}$.
**My question**: Is it possible to show that that $f(t)\equiv 0$ for all $t\geq T$ for some finite $T>0$ which does not use differentiation of $f$, that is, only the inequality in the first display math holds?
The differential inequality is, for example, used in the paper **Bénilan, Philippe, and Michael G. Crandall, "The Continuous Dependence on $p$ of Solutions of $u\_{t}— \Delta \varphi(u) = 0$" Indiana University Mathematics Journal 30.2 (1981): 161-177** to show finite extinction time of a solution $u(x, t)$ of a non-linear parabolic PDE. There, $f$ is an integral over $\mathbb{R}^{n}$ of some power of $u$ and the differentiation is with respect to $t$.
Thanks in advance!
| https://mathoverflow.net/users/163368 | Property so that $f(t)\equiv 0$ for all $t\geq T$ for some finite $T>0$? | As $f$ is non-decreasing, the integral inequality implies that $f$ is $0$.
**Proof.** The integral inequality implies, in particular, that the same inequality is satisfied for $C=0$, so
$$
\int\_{t\_0}^t f(t) \, ds = (t-t\_0)f(t) \le \int\_{t\_0}^t f(s) \, ds
$$
for all $t \ge t\_0$. Now fix $t > t\_0$. Then it follows that the non-increasing function $g: [t\_0,t] \ni s \mapsto f(t) - f(s) \in [0,\infty)$ has integral $\le 0$ over the interval $[t\_0,t]$, and thus this integral is actually $0$. So $g(s) = 0$ for all $s \in (t\_0,t]$ and hence, $f(t) = f(s)$ for all $s \in (t\_0,t]$. Since $t > t\_0$ was arbitrary, it follows that $f$ is constant on $(t\_0,\infty)$.
Using the integral inequality from the question again, but now with $C > 0$, one can readily check that the constant value of $f$ on $(t\_0,\infty)$ is $0$. As $f$ is non-decreasing, we conclude that $f$ is constantly $0$ on $[0,\infty)$. $\square$
| 1 | https://mathoverflow.net/users/102946 | 441954 | 178,357 |
https://mathoverflow.net/questions/441956 | 1 | Let $f:[0, \infty)\to [0, \infty)$ be **non-increasing** and satisfy for all $t>t\_{0}$, $$f(t)+C\int\_{t\_{0}}^{t}f^{\gamma}(s)ds\leq \frac{1}{t-t\_{0}}\int\_{t\_{0}}^{t}f(s)ds,$$ where $0<\gamma<1$ and $C>0$.
**My question**: Is it true that $f(t)\equiv 0$ for all $t\geq T$ for some finite $T>0$?
**Remark**: If $f$ is differentiable and the somewhat similar differential inequality $$f'+Cf^{\gamma}\leq 0$$ holds, we have that $f(t)\equiv 0$ for all $t\geq T$ for some finite $T>0$. This follows from comparing $f$ with the function $g(t)=a(T-t)^{1/(1-\gamma)}$ where $a=((1-\gamma)C)^{1/(1-\gamma)}$.
See also [Property so that $f(t)\equiv 0$ for all $t\geq T$ for some finite $T>0$?](https://mathoverflow.net/questions/440661/property-so-that-ft-equiv-0-for-all-t-geq-t-for-some-finite-t0) for the case when $f$ is non-decreasing.
Thanks in advance!
| https://mathoverflow.net/users/163368 | Integral inequality implies $f(t)\equiv 0$ for all $t\geq T$ for some finite $T>0$? | Since $f$ is nonnegative and nonincreasing, we have $f(t)\to L$ as $t\to\infty$ for some $L\in[0,\infty)$. Then from your first display we get
$$L+C\int\_{t\_{0}}^\infty f^{\gamma}(s)\,ds\le L.$$
So, $f=0$ on $(t\_0,\infty)$, as desired.
| 3 | https://mathoverflow.net/users/36721 | 441959 | 178,358 |
https://mathoverflow.net/questions/441898 | 3 | Let $X$ be a compact metric space, and let $K\_X$ be the set of non-empty closed subsets of $X$, equipped with the $\sigma$-algebra
$$ \mathcal{B}(K\_X) \ := \ \sigma(\{C \in K\_X : C \cap U = \emptyset\} \, : \, \text{open } U \subset X ) . $$
(This is precisely the Borel $\sigma$-algebra of the Hausdorff metric.) Define the equivalence relation $\,\sim\,$ on $K\_X$ to be topological equivalence (i.e. $C\_1 \sim C\_2$ if and only if there exists a homeomorphism $\,h \colon C\_1 \to C\_2$), and let
$$ \pi \colon K\_X \to \tfrac{K\_X}{\sim} $$
be the natural projection.
>
> Does there exist a standard $\sigma$-algebra on $\frac{K\_X}{\sim}$ such that $\pi$ is measurable?
>
>
>
| https://mathoverflow.net/users/15570 | Can the set of compact metrisable topologies naturally be equipped with the structure of a standard Borel space? | In most cases the answer is no:
A much stronger result was proved in
J. Zielinski: The complexity of the homeomorphism relation between compact metric spaces, Adv. Math. 291 (2016), 635–645.
It is shown that the homeomorphism equivalence relation on compact subspaces of the Hilbert cube X=Q is Borel bireducible to the "universal orbit" equivalence relation and hence it can not be Borel reduced to the equality of points in a standard Borel space.
Also it is known that if $X$ is an uncountable compact metric space then the homeomorphism equivalence relation between its compact subspaces is complicated as well (as complicated as isomorphism relation of countable graphs). Hence the answer is no in this case too.
In the case $X$ is countably infinite compact metric space it is homeomorphic to a countable ordinal number and thus, there are only countably types of closed subspaces of $X$ up to homeomorphism. In this case the answer to your question is clearly positive.
| 4 | https://mathoverflow.net/users/128723 | 441962 | 178,359 |
https://mathoverflow.net/questions/441931 | 0 | Here is the definition of the frog model we are interested in:
"... consider the homogeneous tree $\mathbb{T}\_{d}$, that is, the rooted tree in which each
vertex has (is connected by edges to) $d + 1$ neighbours. One frog is put at each vertex and
all but the one of the root start inactive. Active frogs perform simple symmetric random
walks on $\mathbb{T}\_{d}$, for a geometric (parameter $1 - p$) number of steps, activating the inactive
frogs of the visited vertices. After its geometric number of steps, the active frog “dies”: it remains inactive forever. The process survives if infinitely many frogs are activated.
For any $p\in[0,1]$, we denote the law of the process by $\mathbb{P}\_{p}$. Naturally, if $p = 0$ then the frog of the root dies and $\mathbb{P}\_{p}(\text{survival}) = 0$, while on the other hand, if $p = 1$, frogs won't die and $\mathbb{P}\_{p}(\text{survival}) = 1$. Moreover, it is clear that $\mathbb{P}\_{p}(\text{survival})$ is non-decreasing in $p$, and we can define the critical parameter for the model on $\mathbb{T}\_{d}$ as
\begin{align\*}
p\_{c} = p\_{c}(d) := \inf\{p\in[0,1] : \mathbb{P}\_{p}(\text{survival}) > 0\}."
\end{align\*}
My questions are: what is the probability space associated to the probability measure $\mathbb{P}\_{p}$? How can one describe the event "survival" rigorously? How do we prove that $\mathbb{P}\_{p}(\text{survival})$ is non-decreasing in $p$?
I am talking about [this](https://arxiv.org/pdf/2206.13695.pdf) paper.
Any references are welcome.
| https://mathoverflow.net/users/476677 | What is the probability space corresponding to the probability measure $\mathbb{P}_{p}$ in the context of this paper? | (1) As you read the probability literature, you'll soon discover that people tend not to be very specific about the sample space that they are working with. The details of that space are not important; what matters is the collection of observables (events and random variables) and their joint distribution. Any probability space that supports such a distribution is then fine. So the $\mathcal{F}$ and $\mathbb{P}$ of the probability space $(\Omega, \mathcal{F}, \mathbb{P})$ get much more attention than the $\Omega$. Here for example, it's enough that you have countably many simple symmetric random walks (one for each site of the tree), and countably many Geometric($p$) random variables, all of these independent.
[Sometimes some "canonical" choice of space might suggest itself; e.g. if you were studying bond percolation on a graph with edge set $E$, then you might explicitly take $\Omega=\{0,1\}^E$; but even then it's basically a matter of taste.]
(2) In the particular case of the frog model you describe, starting from a single active frog, one can see that in any bounded time-interval, there will be only finitely many jumps. Consequently there's actually a straightforward way to set up the model as a continuous-time Markov chain on a countable state space. The event of "survival" is then just the event that this Markov chain avoids some given state for ever.
(3) You might like to consider more general frog models, e.g. where an infinite number of frogs might be alive at the same time. Now the state space becomes uncountable, and the times of jumps may become dense, and things become a bit more subtle. However, I don't think there are any particular difficulties about the frog model beyond what you already need to construct more well-known interacting particle systems such as exclusion processes, contact processes, voter models, etc.
For a fairly rigorous account of the construction of such models, the books of Liggett might suit you well, e.g. [Stochastic Interacting Systems](https://link.springer.com/book/10.1007/978-3-662-03990-8). There are a ton of other resources introducing interacting particle systems (and many people who work with them, e.g. from a biology or physics viewpoint, are comfortable with a more informal approach to the construction).
(4) Monotonicity properties for interacting particle systems, like the one you mention, can be proved in various ways. A frequent approach involves coupling; you couple one copy of the process with parameter $p$ and another copy with parameter $p'>p$ in such a way that the path of the $p'$-process always lies above that of the $p$-process in some suitable sense. Again, if you read up about exclusion processes, contact processes, etc, you will find lots of examples of this sort of argument!
| 3 | https://mathoverflow.net/users/5784 | 441977 | 178,364 |
https://mathoverflow.net/questions/441942 | 2 | I have a block matrix
$$M=\begin{bmatrix}
I\_0& I\_1& \cdots& I\_1\\
I\_2& I\_0& \ddots& \vdots\\
\vdots& \ddots& \ddots& I\_1\\
I\_2& \cdots& I\_2& I\_0\\
\end{bmatrix}\_{n \times n}$$
with
$$I\_0=\begin{bmatrix}
0& 1\\
1& 0\\
\end{bmatrix}, \qquad
I\_1=\begin{bmatrix}
0& 1\\
-1& 0\\
\end{bmatrix},\qquad
I\_2=\begin{bmatrix}
0& -1\\
1& 0\\
\end{bmatrix}.$$
I want to find all its eigenvalues $\{\lambda\_1,\lambda\_2,\ldots,\lambda\_{2n}\}$, where $\lambda\_1 < \lambda\_2 < \cdots < \lambda\_{2n}$.
Due to the chiral symmetry, we can find $\lambda\_i=-\lambda\_{2n+1-i}$ for all $i$.
| https://mathoverflow.net/users/495317 | Eigenvalues of a specific matrix | For the signed circulant matrix
$$U:=\left[\begin{matrix}
& 1 & & & \\
& & \ddots & & \\
& & & 1 &\\
-1 & & & &
\end{matrix}\right]
\mbox{ in }
M\_n(\mathbb{C}),$$
one has
$$M= 1 \otimes I\_0 + (U+U^2+ \cdots + U^{n-1}) \otimes I\_1
\mbox{ in }
M\_n(\mathbb{C})\otimes M\_2(\mathbb{C}).$$
For $\omega:=\exp\frac{i\pi}{n}$,
the unitary matrix $U$ has eigenvalues
$\{ \omega^k : k=1,3,5,\ldots,2n-1\}$
and eigenvectors
$v\_k:=[\begin{smallmatrix} 1 & \omega^k & \omega^{2k} & \cdots &\omega^{(n-1)k}\end{smallmatrix}]^{\mathrm{T}}/\sqrt{n}$ in $\ell\_2^n$.
Accordingly, the matrix $M$ is decomposed into the direct sum of
$$I\_0+ (\omega^k+\omega^{2k}+ \cdots + \omega^{(n-1)k})I\_1
=\left[\begin{matrix}
0 & \lambda\_k\\
\overline{\lambda\_k} & 0
\end{matrix}\right]
\mbox{ acting on }
\mathbb{C}v\_k \otimes \ell\_2^2,$$
where
$$\lambda\_k=\frac{2}{1-\omega^k}.$$
The eigenvalues of $M$ are
$\pm|\lambda\_k|$, $k=1,3,\ldots,2(n-1)$
with eigenvectors
$ v\_k \otimes
[\begin{smallmatrix} 1 & \pm \mathrm{sgn}(\overline{\lambda\_k}) \end{smallmatrix}]^{\mathrm{T}}/\sqrt{2}$ in $\ell\_2^n \otimes \ell\_2^n$.
Note that since $|\lambda\_k|=|\lambda\_{2n-k}|$, all eigenvalues
except for $1$ (corresponding to the case when $n$ is odd and $\lambda\_n=-1$) have multiplicity $2$.
| 4 | https://mathoverflow.net/users/7591 | 441988 | 178,369 |
https://mathoverflow.net/questions/441980 | 14 | It is not too hard to find examples of finite groups which have fewer automorphisms than one of their subgroups. For example $\mathcal D\_4 \times \mathbb Z/2\mathbb Z$ (where $\mathcal D\_4$ is the dihedral group of order $8$) has at most (in fact exactly) $64$ automorphisms but contains a subgroup isomorphic to $\left(\mathbb Z/2\mathbb Z\right)^3$, which has $\operatorname{GL}\_3(\mathbb F\_2)$ as automorphism group, with $168$ elements.
Every example I found was non-abelian, so I wondered if a finite abelian group can have less automorphism than one of its subgroup? [Hillar and Rhea - Automorphisms of finite Abelian groups](https://arxiv.org/abs/math/0605185) gives the number of automorphisms of an abelian $p$-group, but it's not clear to me if that function is non-decreasing with cardinality.
| https://mathoverflow.net/users/133679 | Finite abelian groups with fewer automorphisms than a subgroup | From the Hiller-Rhea formula
$$|\operatorname{Aut} H\_p| = \prod\_k (p^{d\_k} - p^{k-1}) \prod\_j (p^{e\_j})^{n-d\_j} \prod\_i (p^{e\_i-1})^{n-c\_i+1},$$
given an abelian $p$-group of type $p^{e\_1}\cdots p^{e\_n}$ where $1 \leq e\_1 \leq \cdots \leq e\_n$, you can try to increment one of the exponents $e\_l$ by 1, and see how things change. Here, $d\_k = \max \{l | e\_l = e\_k\}$ and $c\_k = \min \{ l | e\_l = e\_k \}$.
Case 1: $e\_{l-1} < e\_l$ and $e\_l + 1 < e\_{l+1}$. The only changes to the formula are $(p^{e\_l})^{n-d\_l}$ and $(p^{e\_l-1})^{n-c\_l+1}$, and yield a strict increase.
Case 2: $e\_l$ is isolated, but $e\_l + 1$ is equal to $m$ other values. Then, $d\_l$ jumps by $m$, and $c\_i$ decrease by $1$ for $m$ entries. We end up adding an exponent of $2n - 2l + 1 + \epsilon > 0$, where $\epsilon$ is a way to ignore the $p^{k-1}$ in the first product.
Case 3: $e\_l$ is equal to $m$ other values, but $e\_l+1$ is isolated. Then, $d\_i$ drops by $1$ for $m$ entries, and $c\_l$ jumps by $m$. We add an exponent of $2n-2l+1 - \epsilon > 0$.
Case 4: $e\_l$ is equal to $m$ other values, and $e\_l+1$ is equal to $r$ other values. Then, $d\_i$ drops by $1$ for $m$ entries, $d\_l$ jumps by $r$, $c\_i$ decrease by $1$ for $r$ entries, and $c\_l$ jumps by $m$. We add an exponent of $2n-2l+1 - \epsilon > 0$.
In conclusion, a subgroup of a finite abelian group always has strictly smaller automorphism group (**Edit**: with the exception of a group of order $2$ times an odd number, and the subgroup of index 2, where we have equality). The minimal case is when $e\_1 = \cdots = e\_n$. Then, incrementing $e\_n$ multiplies the order by $\frac{p^n(p-1)}{p^n-1}$, which approaches 1 when $p=2$ and $n$ is large.
| 19 | https://mathoverflow.net/users/121 | 441999 | 178,372 |
https://mathoverflow.net/questions/441994 | 3 | In thinking about my question [here](https://mathoverflow.net/questions/441982) for the Linial arrangement, the following limit arose:
$$ \lim\_{n\to\infty}\frac{(n-1)\sum\_{k=0}^n {n\choose k}(k+1)^{n-2}}
{\sum\_{k=0}^n {n\choose k}(k+1)^{n-1}}. $$
Is this limit finite? What is its value? It might be around $2.27\cdots$.
| https://mathoverflow.net/users/2807 | A limit involving the quotient of two sums | First of all, it seems like the value of the limit is more like $1.27...$, and not $2.27...$.
Using some heuristics outlined below it is possible to find the limit:
$$
a=1.278464542761...,
$$
where $a$ is the root of $(x-1)e^x=1$ (can be expressed in terms of Lambert W-function).
The approach is standard, based on saddle point approximation. First, one notes that the main contribution to both sums come from large values of $k\gtrapprox n/2$. When both $k$ and $n$ are large then one has the asymptotics <https://en.wikipedia.org/wiki/Binomial_coefficient#Bounds_and_asymptotic_formulas> :
$$
\binom{n}{k}=\sqrt{\frac{n}{2\pi k(n-k)}}\frac{n^n}{k^k(n-k)^{n-k}}.
$$
Second, replace the sums with Riemann integral. Using saddle point approximation one can find that the main contribution to the sums come from values of $k$ that give maximum to the function (here the contibution from square root factors are omitted, but they can be taken into account without altering the final result):
$$
e^{(n-\epsilon)\ln(x+1)-x\ln x-(n-x)\ln (n-x)}
$$
where $\epsilon$ is 1 ir 2. This gives equation for $k\_m$:
$$
\frac{x\_m}{n-x\_m}=e^{\frac{n-\epsilon}{x\_m+1}},
$$
which is asymptotically the same as
$$
\frac{1}{z-1}=e^{z}, \quad z=n/x\_m.
$$
Then the ratio of two sums is approximated around the saddle point as
$$
\frac{(n-1)g(x\_m)I}{(x\_m+1)g(x\_m)I}\to a
$$
where $g=\binom{n}{x\_m}(x\_m+1)^{n-2}$, and $I$ is the Gaussian integral around the saddle point, which is the same both for the numerator and denominator sums.
Of course, this analysis can be made rigorous using all the usual big O notation, etc...
| 12 | https://mathoverflow.net/users/82588 | 442008 | 178,374 |
https://mathoverflow.net/questions/441797 | 8 | I am a PhD student in the represention theory of finite groups. One of my friends and I solved all exercises in the book [I M Isaacs - Algebra A Graduate Course](https://bookstore.ams.org/view?ProductCode=GSM/100) except for the following exercise in Chapter 9(transfer theory):
**(Exercise 9.4)**
Let $G$ be a finite group and $P\in {\rm Syl}\_p(G)$ and $N={\rm N}\_G(P)$. Suppose $z\in {\rm Z}(N)\cap P$ and $z\notin P'$. Show that $z\notin G'$.
HINT: Use the Transfer evaluation lemma. Note that if $tz^nt^{-1}\in P$, then $tz^nt^{-1}=z^n$ by Lemma 9.12.
It is equivalent to the following proposition:
**Proposition**
Let $G$ be a finite group and $P\in {\rm Syl}\_p(G)$, then $ {\rm Z}({\rm N}\_G(P))\cap P\cap G'\le P'$.
**Lemma 9.12 (Burnside)**
Let $P\in {\rm Syl}\_p(G)$ and suppose that $x,y\in {\rm C}\_G(P)$ are conjugate in G. Then $x$ and $y$ are conjugate in ${\rm N}\_G(P)$.
In this problem, it is hard to show that if $tz^nt^{-1}\in P$, then $tz^nt^{-1}=z^n$.
I asked a professor in our university, he can't solve it as well.
Here is my idea about this question:
By contradiction, suppose that $z\in G'$, then $z\in {\rm Z}(N)\cap P\cap G'$. In order to use Lemma 9.12，We can choose two elements $tz^nt^{-1}$ and $z^n$，So it suffices to show that $tz^nt^{-1}\in{\rm C}\_G(P)$. But I don't know whether it is true or not. Maybe some results about fusion and focal subgroups are useful.
I checked this question for some classes of finite groups. I guess that it is not true for simple groups.
Could you give us some help? Thank you very much.
| https://mathoverflow.net/users/138348 | I M Isaacs Algebra Exercise 9.4 | I think the exercise is incorrect (and so is the hint).
There was a question before about this same exercise at math.stackexchange; the answer there gives a counterexample. Link: [Math.SE](https://math.stackexchange.com/questions/4577477/fusion-in-the-normaliser-of-a-sylow-subgroup/).
| 7 | https://mathoverflow.net/users/38068 | 442010 | 178,375 |
https://mathoverflow.net/questions/441990 | 1 | Let $N(h)$ be the number of solutions of the following linear diophantine equation:
\begin{equation}
x\_1 + 2x\_2 + 3x\_3 + \dots + (h-1)x\_{h-1} = 6h-6;
\end{equation}
where $h\geq 2$ and solution means a vector $(z\_1,\dots,z\_{h-1})$ of non-negative integers satisfying the equation.
Does there exist a formula for $N(h)$ or at least an explicit expression for the behavior of $N(h)$ for $h\mapsto +\infty$?
| https://mathoverflow.net/users/14514 | Solutions of a linear diophantine equation | As Max Alekseyev observed, $N(h)$ is the number of partitions of $6h-6$ into at most $h-1$ parts. A complicated asymptotic formula exists for this quantity, as a special case of a result of Szekeres (1953). See [this paper by Canfield](https://www.combinatorics.org/ojs/index.php/eljc/article/view/v4i2r6/pdf) for more detail.
| 3 | https://mathoverflow.net/users/11919 | 442012 | 178,376 |
https://mathoverflow.net/questions/441792 | 9 | There are many objects in mathematics that have the term "chiral" in their name, for instance, chiral algebra by Beilinson and Drinfeld, chiral de Rham complex, chiral Koszul duality etc. Some people told me that chiral algebras are $2$-dimensional analogue of associative algebras, which are considered to be $1$-dimensional. However, I don't understand its precise meaning since the definition of a vertex operator algebra is so complicated. Does the term chiral has something to do with this $2$-dimensionality?
For a vertex operator algebra $V$, Yongchang Zhu constructed an associative algebra $A(V)$ out of $V$, such that there is a bijection between the set of isomorphism classes of irreducible positive energy representations of $V$ and that of simple $A(V)$-modules. For an associative algebra $A$, Tomoyuki Arakawa calls $V$ to be the *chiralization* of $A$ if $A\simeq A(V)$ as associative algebras. What's the meaning of chiralization here?
There are some other explanations for the term chiral that I have ever heard. For example, in electromagnetism, chirality means the handedness of electromagnetic waves associated with their polarization. Some others also told me that in the $2$-dimensional setting, chiral means holomorphic.
I want to know the geometry/physics behind the term chiral. A philosophical answer is welcome, but a mathematical/physical answer is better.
| https://mathoverflow.net/users/466793 | What is the meaning of chiral in the context of vertex algebras? | A vertex operator algebra describes the algebra of local operators in the chiral part of a 2d CFT. Typically one sees a VOA depending on a complex coordinate $z$. To describe a full 2d CFT, you would need to also include an "anti-chiral" VOA depending on a conjugate coordinate $\bar{z}$. So by considering only a single vertex algebra depending on one complex variable, you are only considering a "chiral half" of the CFT.
In physics more generally, people will often refer to "chiral algebras" of local operators in other types of theories of different dimensions as well, so in that field the terminology is quite broad, but relates to the chirality of a theory in a traditional sense.
Mathematicians have since extended and generalized a number of things relating to vertex algebras. For example Beilinson-Drinfeld chiral algebras generalise vertex algebras, the Chiral de Rham complex and chiral differential operators are VOA versions of differential forms and differential operators etc... In the past, constructions such as these were the only examples of mathematically well-defined and well-studied chiral algebras (in the physics sense of algebras of local operators). Hence in the mathematical community, I suspect it became practice to name constructions relating to vertex algebras and BD chiral algebras "chiral" since this was the only real example of a chiral algebra of local operators they were looking at in those communities.
In Arakawa's work, "chiralization" essentially means going from an associative algebra type object to a VOA type object. More generally, I believe chiralization/adding the word "chiral" to something in these communities will describe working with something like an affine or loopy version of a previously known construction.
For example, taking the Weyl algebra of differential operators, one can consider its VOA counter-part which is the $\beta\gamma$ VOA, also known as "chiral differential operators". The modes of this VOA satisfy similar relations to the Weyl algebra and the Weyl algebra can be recovered from it in various constructions.
I guess the point to emphasise though is that the word "chiral" itself is often not-literal in the mathematics community, but is rather describing something like "VOA-ization" (which in BD language is chiralization).
| 2 | https://mathoverflow.net/users/88421 | 442020 | 178,378 |
https://mathoverflow.net/questions/442017 | 5 | Given a homeomorphism between complex manifolds, $f : X → Y$, is it then true that the rational Pontrjagin class $p\_1(X) \in H^4(X,\mathbb Q)$ equals the pull-back $f^\* p\_1(Y)$?
If $X$ and $Y$ are compact, then I understand that this is the famous Novikov result. I am, however, unsure if the result holds in the non-compact setting. I am aware of papers that claim the result for "smooth manifolds" -- but I have not been able to find out if "manifolds" are meant to be compact by the authors.
The spaces $X$ and $Y$ that I have in mind are Zariski-open subsets of complex-algebraic varieties, and therefore topologically harmless. Would that be of any help?
| https://mathoverflow.net/users/33531 | Topological invariance of rational Pontrjagin classes for non-compact spaces | I am not sure about the reference, but here is an argument: the map $BO\to BTOP$ induces an isomorphism on rational cohomology, as mentioned, e.g., on p.2 of [Dalian notes on rational Pontryagin classes](https://arxiv.org/pdf/1507.00153.pdf) by Weiss. The topological rational Pontryagin class $p\_i$ is an element in $H^{4i}(BTOP;\mathbb Q)$ that corresponds to the usual Pontryagin class under the isomorphism. For a topological manifold $M$ its tangent microbundle is a homotopy class of maps $\tau\_M: M\to BTOP$. By definition, the Pontryagin class of $M$ is $\tau^\*\_M p\_i$, the $\tau\_M$-image of $p\_i$ under the map $\tau\_M$ in rational cohomology. If $h: N\to M$ is a homeomorphism, then $\tau\_M\circ h$ and $\tau\_N$ are homotopic as maps from $N$ to $BTOP$. Since homotopic maps induce the same map on cohomology, $h^\*$ sends $\tau^\*\_M p\_i$ to $\tau^\*\_N p\_i$. Thus the real work is in showing that $BO\to BTOP$ is an isomorphism on rational cohomology.
| 8 | https://mathoverflow.net/users/1573 | 442025 | 178,380 |
https://mathoverflow.net/questions/316226 | 11 | Let $G$ be a finite $p$- group and let $\varphi$ be an automorphism of $\mathbb{F}\_pG$ as $\mathbb{F}\_p$-algebras and let $n = \sum\_{g\in G} g$ be the norm element. Does it follow that $\varphi(n)=n$?
Since the $\mathbb{F}\_p$-vector space spanned by $n$ is exactly the annihilator of the maximal ideal, it follows that this vector space is characteristic and thus $\varphi(n)=\lambda n$ for some $\lambda \in \mathbb{F}\_p^\*$ and thus we have a group homomorphism:
$Aut(\mathbb{F}\_pG)\rightarrow \mathbb{F}\_p^\*$. So the question is whether this map is always trivial.
| https://mathoverflow.net/users/3969 | Is the norm element characteristic in modular group rings? | I know this is an old question, but I think the answer is yes. The idea is to pass to the associated graded for the Jennings filtration, and look at the $p$-restricted Lie algebra $L(G)$ you get that way. The element $n$ corresponds to the socle element in the restricted universal enveloping algebra, which is one dimensional. Taking a basis of homogeneous elements, $x\_1,\dotsc,x\_m$, of $L(G)$, the norm element is $(x\_1\dotsm x\_m)^{p-1}$. An automorphism of $G$ acts as a linear transformation on $L(G)$, and multiplies $x\_1\dotsm x\_m$ by a constant that depends on that linear transformation (modulo commutators further down the filtration, and squares), and this constant is an element of $\mathbb{F}\_p^\*$. Taking $(p-1)$st powers, the action is trivial on $n$, because those commutators and squares drop off the end. This argument needs checking, but I think it's right.
Edit 20 March 2023: I found a clean way to present the proof, which can be found at [The socle of the group algebra of a finite $p$-group](https://arxiv.org/abs/2303.09940).
| 4 | https://mathoverflow.net/users/460592 | 442037 | 178,384 |
https://mathoverflow.net/questions/442049 | 2 | We call the natural number $n$ a partition number $\iff$
$$
\exists d | n: \gcd\left(d,\frac{n}{d}\right)=1 \text{ and } \Omega(d) = \Omega\left(\frac{n}{d}\right)\;,
$$
where $\Omega$ counts the prime divisors with multiplicities.
Then we have:
$n= p\_1^{a\_1} \cdots p\_r^{a\_r}$ is a partition number $\iff S = (a\_1,\cdots,a\_r)$ has a solution to the [partition problem](https://en.wikipedia.org/wiki/Partition_problem) which is known to be [NP-complete](https://en.wikipedia.org/wiki/NP-completeness).
I wanted to test the following heuristic idea:
1.) Space $\equiv$ Time. (for computations)
2.) Most strings are incompressible, because only finitely many space below.
$\rightarrow$ Most problems should not be solved faster (in time) than some boundary determined by the Kolmogorov complexity (=space).
To make this idea more precise I decided to test it for the following three problems:
Determine if a number $n$ is even. Determine if a number $n$ is prime. Determine if a number $n$ is a partition number.
For each of this problem I wrote a computer program to output a bit $=1$ if the number $n$ is even/prime/partition number, or $0$ otherwise.
Intuitively, if it was easy or if there was a pattern to these bits, then it should be exploited by an algorithm to be fast on this problem. To measure how easy these problems are, I suggest to use the Kolmogorov complexity or a zip program. We get in increasing order:
Result:
Even numbers are compressible to nearly 100%.
Prime numbers are compressible to about 59%.
Partition numbers are compressible to about 4%.
(Random bits are compressible to about 0%.)
Here is some Sagemath code which does the computation. I have zipped the files by hand afterwards and compared the file size.
```
def Omega(n):
return sum([valuation(n,p) for p in prime_divisors(n)])
def omega(n):
return len(prime_divisors(n))
def checkIfPartitionNumber(x):
return any([Omega(d)==Omega(x/d) for d in divisors(x) if gcd(d,x/d)==1])
def isPrime(x):
return is_prime(x)
def getWordUpTo(n,func=checkIfPartitionNumber):
s = ''
for x in range(1,n+1):
if func(x):
s+="1"
else:
s+="0"
return s
def toFile(n,func=isPrime,fname="is_prime.bin"):
from bitarray import bitarray
ba = bitarray(getWordUpTo(n,func))
print(ba)
#rint(dir(ba))
with open(fname,"wb") as f:
ba.tofile(f)
toFile(n=100000,func=lambda x: x%2, fname="is_even.bin")
toFile(n=100000,func=is_prime, fname="is_prime.bin")
toFile(n=100000,func=checkIfPartitionNumber, fname="is_partition_number.bin")
toFile(n=100000,func=lambda x: randint(0,1), fname="random_bits.bin")
```
I am interested if this idea is being pursued in the literature and if so a reference would be nice to share. Or maybe this idea is not pursued because you can think of some problem which violates the hypothesis of this heuristic? (That would also be nice to share.) Or maybe you would like me to test a problem you can think of if it fits the hypothesis of this question?
**Update**:
What does "Space $\equiv$ Time" mean?
What I try to do is to measure the "search space" of an algorithm considered and if the resulting string can be compressed by gzip or zip. For instance, in each time step $t$ we have for a Turing machine a configuration $c\_t$, which we could encode as a string $s\_t$ and write down all those strings $s\_1,\cdots,s\_T$ to a different tape. Later these "string of configurations" s:=$s\_1 \cdot s\_2 \cdots \cdot s\_T$ (concatenated) , will be compressed with gzip or zip to give a compressed string $s\_c$. Intuitively, if $s$ it is strongly compressible $|s\_c|<<|s|$, then (?) it means that the algorithm / the Turing machine could be improved, because it considers repetetive configurations. If the string $s$ appears random because $|s\_c| \equiv |s|$, then intuitively there can not be much done to improve the runtime of the algorithm / Turing machine on this input.
Since this is just a heuristic it is difficult to get more formal or concrete. I hope this clarifies the idea. (It is slightly different from the implementation described above, but [here you can find some SAGEMATH implementation of this idea](https://sagecell.sagemath.org/?z=eJyVGGtv2zbwe4H-By5BENJRHNt9YAuqAtuaGCmGtqi7DKhhGJJFK0wkShVlx-mv3x0pSpSsNG0-tNK933dyxNckLDYlv8yKFZ9lyZbPNqHi5WyT0hk7f_6MwN8V8QnAaBHImNOES0AxZnDrrCDviZDE8Cl6ZbnwT6yJQknz24WmvEXK98z3DVQYqEAoKrhiw0is17zgcsUpQt6DIkce_hW83BSSfCk23CAqwGWQKE7I82fPn0XgVszltAjyG0Wl_8LKuAZP5ovGcq3ZuDX25MnY1VXba_Cii9fyhkGecxlRQN_akOSFkCW9ZpbIQKdTrbsNRB0XbvSuXQ3IoX2gSZblykeXWdvALTJfd6yaDoMoWm55UfId3c5HC_ZD_HjREcpR6EWvUB5BKDiI9Hib7zBPspJO2VDdZPeUNS4a322cpqyVs-nUSRcvgpKbnP2bf8noB_-ljUYcdxIn3cR9aCcmjod8V6I2pwZYW3EcW8UfUx4HgD9v4bE-59sg2QSlyCSVXs603hz1QoJTvozEVqisQNkYCCMt65WGLbPHVPOsbvjq7mr9KShKgdo-bNKQF3TXERLIBzo31kbQQeZpdxYZwyI0rJa-Y9h68SqikYckvj82NhqNoZCqLPbMPDgY3mZCUsTtjNgdir2SJY850kN7xgLKdNJYH9ypJfoWJKJ8WJZclY3cCLMmq7SBQZL4PpkQnT94euEkzelqj0QOxxEwAO3IcB2RF87bGzLZF6HnQC2jABNem8f7G5FwQgtySsaMDAYg-I1PpCMhsmVaOJVdWWH5XFsAdmJh_VPKGBM5RpKT2iDX58hkx8RUqE9YLfslsB_rHWs6qPwvKyLdOtJbb-TKr-RYKQpicXxsnpPE6SiR5llRkhJoK30bucQ3oBkZSBh2OnD3g9EZeiHqQhu0gbXzG1lJxY4IVRtlNEJ4KrIGmSS2o1s8YWjzFVbQQ7QsQcuSpLLnUHt-4lc1nzh7K0TCMHQMBzokCgfj9riw1nnKUtqol9klVFUn4N5aBin3D4RJFx-C8gMboHWRpdCBZVAUwYMNvX2vgh1AiCyI7ie2vWnCwHqvXyNRIKiC3YvyhmQQJKpt8g7uwwNGAkXWbr6CYZmt0Y-1LqfGSaPBtWA8Gnm2rphLWtUghmMmvnP1KzHZK8BDBc6Xtgbxv2HOi_VylUFtwChiLT6Red9Fjg546kH9dJirOvPI_aYELT1x7ksHENt4Q-k9aSIwLcMH6FWUgXHWLxab8nSJdgNSZMO_EHX1kbq5qxwbfhW5rjTL4ZE0i7h_fH_skVWW5gVXCpaHX9NffVq-u7j8588vF-90whHh5Bxfh_eFwClSVMXRGFtZoCCRy5BDs6CFENshBAmB2Zr20gZrcH2P1NqMIFyqnLK6ZOuRUAfzlDTJ73ahyZdjlufodauxXZdQdF836R1PEl4snf0ktsoZa_isRJziAm82UFYSOhniao0gbwpiKSMiB3p5KHbemvSGa4ZyyNkZLJhTIg3sM8BoqsdO2lrVUq_qFKXN2kMHFz7eISu9YGaad4W8f2cpNI--TBT9zJxDs1kgjb_tdezofczpeguqThRUs4x1DEi1QhV7agv3RaRWY93GEM2dENnlAor0lq0WcdpYYKL22D2vl8xnppn7lrzonL09sfWI6J78kV1FK9ZGqBU6qfPV5bDx7GLw82ilk9tR4gTSXgd7sY10vg_tAvJhMMOfmblJkIZRQHbnZHc08UgzevmWSzN5m6LpF2HntMvtDO6n2PuP2pYwi1pKjfs5uY1nkOYIl9PIG7NaLgKzdAnDWjnynhYG9qhvm6DAm8s10gB_1TRK6emYDQbVjc7w_maDcTcRy0Rkm62AJn0kBHUza2rcLnzlnNfQCXUz6d5ZHFZXTloV2rY54Mz1m76tOudo4vvuzYqy0rOzSU-fpIx05WrZJ_7Y-f6L6hNrWxF9A5Hb6lCBx28n42ac_OZPBgOaY5QoPOFDX4vuEbFHBk3VIBye9-9Qe07nzFXRd1FWMkM9uOz1M-8mwP02mr_2Jr97L_94vVg4H72txO19Z0kdfK_5NLL_Qj8k-lCY19124NlqaZ6sIQfAjpWnOeyZVdF4HdoFOaw7ujUdtCVPl27dC3uCwYb25MWxq82qf7rBt-p7QM1F5XRZ6rMJPa5heQ7fn6W7k1H2nTvVX7U_NXpvgtYp0r5JoWMHgzvTsPrEa7rNPpmsJ1lMGzkno-GIeZWyszG8oZQG79TQDjd9HzfpsGuo-wuK1ut-L1XhsN1gUTQREqrZID0dQr8sWfu3F5hS-leT3h9WXjF3iGHC4pi1ppg8JwY8l6fjxTANytWNkDEcbm9H9Ry7CZQtg6WlMDPsfzvRoDs=&lang=sage&interacts=eJyLjgUAARUAuQ==).)
| https://mathoverflow.net/users/165920 | Using Kolmogorov complexity to measure difficulty of problems? | It is worth mentioning that *variants of* Kolmogorov complexity have been investigated in the context of the study of pseudorandomness.
See [On One-Way Functions from NP-Complete Problems by Yanyi Liu and Rafael Pass](https://drops.dagstuhl.de/opus/volltexte/2022/16598/) ([arXiv link](https://arxiv.org/abs/2009.11514)).
It is probably worth mentioning as well that Quanta magazine has [written a popular science article on the paper](https://www.quantamagazine.org/researchers-identify-master-problem-underlying-all-cryptography-20220406/).
| 5 | https://mathoverflow.net/users/101207 | 442054 | 178,388 |
https://mathoverflow.net/questions/442058 | -4 | I have been working on a problem concerning the "line of sight" from a fixed integer co-ordinate — let's say $(0,0)$ — to a variable co-ordinate $(a,b)$. Having a line of sight means that there are no other integer co-ordinates on a straight line connecting the two points.
This is a a well documented problem and there isn't very much that's worth exploring on it. However, when investigating the long run behaviour, I noticed that if were look at the probability that a uniformly chosen point has a line of sight as the co-ordinate approaches infinity (I define this more formally below), then the proof that this probability approaches $0$ relies on the fact that the probability that these the $x$ and $y$ co-ordinates are co-prime also approaches $0$. This is a fact that is clearly true (at least intuitively), however, I have struggled to construct (or find) a proof of this fact.
More formally, my question asks the following:
>
> Prove that for $a,b \sim \operatorname{Uniform}\{n, \space n+1, \dotsc \space 2n\}$, the following is true:
>
>
> $$ \mathbb{P}(\{ \text{$a$ and $b$ are co-prime} \}) \rightarrow0$$
> as $n \rightarrow \infty$.
>
>
>
As stated above, this is clearly true. What I am after is either a proof or a reference to a proof of the above claim so that I can justify this result rigorously.
I would be grateful for any contributions here.
---
Note: as discussed in the comments, I am aware of the fact that for any two randomly chosen positive integers, the probability that they are coprime will be $6 / \pi^2$ [(see proof here)](https://math.stackexchange.com/questions/64498/probability-that-two-random-numbers-are-coprime-is-frac6-pi2), however, the application of this to my question is not immediately clear me to beyond providing me with the intuition for my conjecture.
| https://mathoverflow.net/users/497178 | Proving that $P($$\{\text{$a$ and $b$ are co-prime}$ }$)=0$ for $a,b$ following the Uniform distribution over $[n, 2n]$ as $n \rightarrow \infty$ | I seem to get the probability is still the usual $1/\zeta(2)$ by the usual inclusion/exclusion argument, but possibly there's a mistake in the following calculation:
$$
\begin{aligned}
\frac{1}{n^2}\sum\_{a,b\in[n,2n]} \;\;\sum\_{d\mid(a,b)} \mu(d)
&= \frac{1}{n^2}
\sum\_{1\le d\le 2n} \mu(d)\sum\_{\substack{A,B\\ A\in[n/d,2n/d]
\\ B\in[n/d,2n/d] \\}} 1 \quad\text{setting $a=Ad$ and $b=Bd$,}\\
&= \frac{1}{n^2}\mu(d) \sum\_{1\le d\le 2n} \left(\frac{n}{d} + O(1)\right)^2 \\
&= \frac{1}{n^2} \sum\_{1\le d\le 2n} \mu(d)\frac{n^2}{d^2} + O\left(\frac{n}{d}\right) \\
&= \sum\_{1\le d\le 2n} \mu(d)\frac{1}{d^2}
+ O\left( \frac{1}{n} \sum\_{1\le d\le 2n}\frac{1}{d}\right) \\
&= \frac{1}{\zeta(2)} + O\left(\frac{\log n}{n}\right).
\end{aligned}
$$
| 6 | https://mathoverflow.net/users/11926 | 442059 | 178,389 |
https://mathoverflow.net/questions/442063 | 2 | Let $(E, d)$ be a metric space. For $\varepsilon>0$, we define two notions of $\varepsilon$-covering number as follows, i.e.,
* $N\_\varepsilon^o (E)$ is the smallest number of *open* balls whose radii are $\varepsilon$ that cover $E$.
* $N\_\varepsilon^c (E)$ is the smallest number of *closed* balls whose radii are $\varepsilon$ that cover $E$.
Then $N\_\varepsilon^o (E)$ is not necessarily equal to $N\_\varepsilon^c (E)$, for example, take $E = \{0, 1\}$ and $\varepsilon=1$. However, if $D$ is a dense subset of $E$, then $N\_\varepsilon^c (E) = N\_\varepsilon^c (D)$.
Let $(E, d)$ and $(E', d')$ be metric spaces. The spaces $E$ and $E'$ are said to be isometric (denoted by $E \cong E'$) if there is a bijective isometry between them.
I would like to ask if any of below statements is true, i.e.,
>
> 1. If $N\_\varepsilon^o (E) = N\_\varepsilon^o (E')$ for all $\varepsilon>0$, then $E \cong E'$.
> 2. If $N\_\varepsilon^o (E) = N\_\varepsilon^o (E')$ and $N\_\varepsilon^c (E) = N\_\varepsilon^c (E')$ for all $\varepsilon>0$, then $E \cong E'$.
>
>
>
Thank you so much for your elaboration!
| https://mathoverflow.net/users/99469 | Are two metric spaces isometric if they have the same $\varepsilon$-covering numbers for all $\varepsilon>0$? | No. For $0 < \delta \leq 2$ let $E\_\delta$ be the metric space
consisting of three points $A,B,C$ with
$d(A,B) = d(A,C) = 1$ and $d(B,C) = \delta$.
I claim that the $E\_\delta$ for $1 \leq \delta \leq 2$ all have
the same covering numbers for all radii $\varepsilon$.
Indeed in every such $E\_\delta$
all open $\varepsilon$-balls with $\varepsilon\leq 1$
and all closed $\varepsilon$-balls with $\varepsilon < 1$
contain only their centers, so it takes $3$ such balls to cover the space;
but $E\_\delta$ is covered by
a single open $\varepsilon$-ball centered at $A$
once $\varepsilon > 1$,
and also by a single closed $\varepsilon$-ball centered at $A$
once $\varepsilon \geq 1$. Clearly no two $E\_\delta$ with distinct $\delta$ are isometric.
| 6 | https://mathoverflow.net/users/14830 | 442064 | 178,391 |
https://mathoverflow.net/questions/441724 | 15 | The ordinary generating function $T\_n=T\_n(x)$ for the $n$-ary trees satisfies the functional equation
$$
T\_n=1+xT\_n^n.
$$
This is usually defined for $n\ge 0$, but the functional equation can be extended to negative $n$. Writing
$$
T\_{-n}=1+xT\_{-n}^{-n}
$$
and dividing through by $T\_{-n}$, we obtain that
$$
T\_{-n}^{-1}=1-x(T\_{-n}^{-1})^{n+1},
$$
i.e.
$$
T\_{-n}(x)=\frac{1}{T\_{n+1}(-x)}.
$$
What would be a natural way to interpret this combinatorially? I.e. what are "$n$-ary trees" for negative $n$, why do we get the extra $1$ degree, etc.
| https://mathoverflow.net/users/113161 | A combinatorial interpretation for $n$-ary trees for negative $n$ | Here's an explanation of the combinatorial meaning of $T\_{-n}(x)$.
The combinatorial interpretation $T\_n(x)$ is that it counts $n$-ary trees. More precisely, it counts ordered trees in which every vertex has 0 or $n$ children, and each internal vertex (with $n$ children) is weighted $x$ and each leaf is weighted 1. Let's mark each edge from a vertex to its $i$th child with $i$, and then delete all the leaves (together with their incident marked edges). The original tree can easily be reconstructed from this reduced tree. What we now have is an ordered tree in which the edges from each vertex to its children are marked with some subset of $[n]=\{1,2,\dots,n\}$ in increasing order from left to right.
If we remove all the marks we obtain an underlying ordered tree. Given an ordered tree, how many ways are there to mark it to obtain a tree counted by $T\_n(x)$? For each vertex with $k$ children, we can assign marks to the edges to its children in $\binom{n}{k}$ ways. So for an ordered tree with $m$ vertices, if the numbers of children of the vertices are $k\_1,k\_2,\dots, k\_m$ then the number of ways of marking this tree is $\binom{n}{k\_1}\binom{n}{k\_2}\cdots \binom{n}{k\_m}$. So the coefficient of $x^m$ in $T\_n(x)$ is the sum of these products of binomial coefficients over all ordered trees on $m$ vertices. If $m>0$ and we replace $n$ by $-n$ this product of binomial coefficients becomes
$$\binom{-n}{k\_1}\binom{-n}{k\_2}\cdots \binom{-n}{k\_m}=(-1)^{m-1}\binom{n+k\_1-1}{k\_1}\binom{n+k\_2-1}{k\_2}\cdots \binom{n+k\_m-1}{k\_m},$$ since $k\_1+\cdots +k\_m = m-1$. But $\binom{n+k-1}{k}$ is the number of ways of marking the edges from a vertex to its $k$ children with elements of $[n]$ so that the marks are weakly increasing from left to right, but with repeated marks allowed. Thus $1-T\_{-n}(-x)$ counts ordered trees (with at least one vertex) in which the edges joining each vertex to its children are marked with elements of $[n]$ so that the marks are weakly increasing from left to right. I'll call these trees *$n$-colored trees*. (Such trees have appeared in various places in the literature; the only reference I can recall offhand is to a paper of mine and S. Seo, [*A refinement of Cayley's formula for trees*](https://doi.org/10.37236/1884).)
Thus if we set $U\_n(x) = 1-T\_{-n}(-x)$ then $U\_n(x)$ is the generating function for nonempty $n$-colored trees. It is also not hard to prove this algebraically. In the defining equation $T\_{-n}(x) =1-xT\_{-n}(-x)^{-n}$, we replace $T\_{-n}(x)$ with $1-U\_n(x)$ and we obtain
$$U\_n(x) =\frac{x}{\bigl(1-U\_n(x)\bigr)^n}$$
from which the combinatorial interpretation of $U\_n(x)$ is clear.
Alex's identity
$$
T\_{-n}(x)=\frac{1}{T\_{n+1}(-x)}
$$
may be rewritten as $$T\_{n+1}(x)=\frac{1}{1-U\_n(x)}.$$
I'm sure that there is a reasonably straightforward bijective proof of this identity, but I didn't work it out. (This post is long enough!)
| 12 | https://mathoverflow.net/users/10744 | 442068 | 178,394 |
https://mathoverflow.net/questions/442070 | 2 | Let $(X, d)$ be a compact metric space.
* We say that $\{x\_1, \cdots, x\_n\} \subseteq X$ is an $\varepsilon$-**covering** of $X$ if for any $x \in X$, there exists $i \in \{1, \ldots, n\}$ such that $d(x, x\_i) \leq \varepsilon$. Let
$$
\operatorname{Cov} (X, \varepsilon) := \min \{n: \exists \varepsilon \text {-covering of } X \text { with size } n\}
$$
be the $\varepsilon$-covering number of $X$.
* We say that $\{x\_1, \cdots, x\_n\} \subseteq X$ is an $\varepsilon$-**packing** of $X$ if $d(x\_i, x\_j)>\varepsilon$ for all distinct $i, j$. Let
$$
\operatorname{Pack} (X, \varepsilon) := \max \{n: \exists \varepsilon \text {-packing of } A \text { with size } n\}
$$
be the packing number of $A$.
Let $(X, d)$ and $(X', d')$ be metric spaces. The spaces $X$ and $X'$ are said to be isometric (denoted by $X \cong X'$) if there is a bijective isometry between them.
@Noam gave below example in his [answer](https://mathoverflow.net/questions/442063/are-two-metric-spaces-isometric-if-they-have-the-same-varepsilon-covering-num):
>
> For $\delta \in (0, 2]$, let $E\_\delta$ be the metric space consisting of three points $A,B,C$ with $d(A,B) = d(A,C) = 1$ and $d(B,C) = \delta$. Then for all $\delta, \delta' \in [1, 2]$,
> $$
> \operatorname{Cov} (E\_\delta, \varepsilon) = \operatorname{Cov} (E\_{\delta'}, \varepsilon) =
> \begin{cases}
> 3 & \text{if} \quad \varepsilon < 1, \\
> 1 & \text{if} \quad \varepsilon \ge 1.
> \end{cases}
> $$
>
>
>
This example shows that $\operatorname{Cov} (X, \varepsilon) = \operatorname{Cov} (X', \varepsilon)$ for all $\varepsilon>0$ does not necessarily imply $X \cong X'$. Back to @Noam example, it's clear that
$$
\operatorname{Pack} (E\_1, 1) = 0 \neq 2= \operatorname{Pack} (E\_{2}, 1).
$$
I would like to ask if below statement is true, i.e.,
>
> If $\operatorname{Cov} (X, \varepsilon) = \operatorname{Cov} (X', \varepsilon)$ and $\operatorname{Pack} (X, \varepsilon) = \operatorname{Pack} (X', \varepsilon)$ for all $\varepsilon>0$, then $X \cong X'$.
>
>
>
Thank you so much for your elaboration!
| https://mathoverflow.net/users/99469 | Are two metric spaces isometric if they have the same $\varepsilon$-covering and $\varepsilon$-packing numbers for all $\varepsilon>0$? | Certainly no. Consider metric spaces on $n$ points and all distance 1 and 2. There are $2^{n^2/2+o(n^2)}$ such spaces. But only polynomially many different covering and packing functions.
| 10 | https://mathoverflow.net/users/4312 | 442072 | 178,395 |
https://mathoverflow.net/questions/442035 | 6 | Let $ \mathbb{P}$ be any directed, well-founded poset of cofinality $ \aleph\_{n+1}$, where $n$ is a natural.
Can we write it as an increasing union $ \mathbb{P} = \bigcup\_{\alpha < \omega\_{n+1} } \mathbb{P}\_\alpha $
where each $\mathbb{P}\_\alpha $ is a directed poset of cofinality $\aleph\_n$ ?
| https://mathoverflow.net/users/495743 | Poset as union of posets of lower cofinality | Yes, and well-foundedness is irrelevant (this is working in ZFC). Let $\mathbb{P}$ be the poset, with ordering $\leq$. Let $f:[\mathbb{P}]^2\to\mathbb{P}$ be a function such that for all $x,y\in\mathbb{P}$, we have $x\leq f(\{x,y\})$ and $y\leq f(\{x,y\})$.
Let $c:\aleph\_{n+1}\to\mathbb{P}$ be cofinal in $\mathbb{P}$, and injective. For each limit ordinal $\alpha<\aleph\_{n+1}$, let $\bar{\mathbb{Q}}\_\alpha$ be the restriction of $\mathbb{P}$ to the $f$-closure of $c``\alpha$, and let $\mathbb{Q}\_\alpha$
be the $\leq$-downward closure of $\bar{\mathbb{Q}}\_\alpha$.
Certainly $\mathbb{P}$ is the monotone increasing union of the $\mathbb{Q}\_\alpha$'s, and each $\mathbb{Q}\_\alpha$ is a directed partial order of cofinality $\leq\aleph\_n$. Also the sequence of $\mathbb{Q}\_\alpha$'s is continuous (at limits of limits). And $\bar{\mathbb{Q}}\_\alpha$ is a cofinal subset of $\mathbb{Q}\_\alpha$ of cardinality $\leq\aleph\_n$. No $\mathbb{Q}\_\alpha$ is cofinal, so there is a cofinal set $C\subseteq\aleph\_{n+1}$ such that for all $\alpha\in C$, we have (i) $\bar{\mathbb{Q}}\_\alpha\not\subseteq\mathbb{Q}\_\beta$
for all $\beta<\alpha$, and (ii) $\mathrm{cof}(\alpha)=\aleph\_n$.
But note that for each $\alpha\in C$,
$\mathbb{Q}\_\alpha$ has cofinality $\aleph\_n$. (If $X\subseteq\mathbb{Q}\_\alpha$
has cardinality $<\aleph\_n$ and is cofinal in $\mathbb{Q}\_\alpha$,
then since $\bar{\mathbb{Q}}\_\alpha$
is cofinal in $\mathbb{Q}\_\alpha$,
note that $X\subseteq$ the $\leq$-downward closure of $\bar{\mathbb{Q}}\_\alpha$, but $\bar{\mathbb{Q}}\_\alpha=\bigcup\_{\beta<\alpha}\bar{\mathbb{Q}}\_\beta$,
so there is $\beta<\alpha$
such that $X\subseteq$ the $\leq$-downward closure of $\bar{\mathbb{Q}}\_\beta$,
i.e. $X\subseteq\mathbb{Q}\_\beta$, but then $\mathbb{Q}\_\beta=\mathbb{Q}\_\alpha$, contradiction.)
| 6 | https://mathoverflow.net/users/160347 | 442080 | 178,398 |
https://mathoverflow.net/questions/442081 | 0 | For a one-dimensional $f(x)$, the dilation operator $f(ax)$ can be expressed as $\exp(g(D))f(x)$, where $g$ is a closed-form function. This is easily checked by e.g. formal Taylor series expansion.
However, it is not clear what to do in the multidimensional case- even if one restricts $U$ to be orthogonal with determinent 1, all attempts to expand $f(xU)$ result in a Taylor series in $xU$, which cannot be simplified further without explicitly expanding it in terms of $x$, a very complicated process. I'm not a big fan of Occam's razor, but I believe there should be a multidimensional $g(D)$ that works even in the multidimensional case. What should I do here?
| https://mathoverflow.net/users/113020 | Differential form of the multidimensional "orthogonal dilation" operator | In 3 dimensions, rotations, i.e., transformations corresponding to orthogonal $U$ with determinant 1, are generated by the (orbital) angular momentum operator $\vec{L} $ with components $L\_i =-i \epsilon\_{ijk} x\_j \,\partial / \partial x\_k $. By Euler's rotation theorem, any given such transformation can be effected by rotating around a specific axis $\vec{e} $ by an angle $\alpha $. Then, the desired rotation operator is
$$
\exp \left(-i \alpha \ \vec{e} \cdot \vec{L} \right) \ .
$$
In other than 3 dimensions, there isn't, of course, such an intuitive description in terms of a vector axis and an angle, but the modification is purely on the level of the rotation theorem -- once this is adapted, one will still then generate the rotations using the antisymmetric tensor operator $x\_j \,\partial / \partial x\_k - x\_k \,\partial / \partial x\_j $.
| 1 | https://mathoverflow.net/users/134299 | 442084 | 178,399 |
https://mathoverflow.net/questions/441609 | 2 | In Jacob Rasmussen's paper [Khovanov homology and the slice genus](https://arxiv.org/pdf/math/0402131.pdf), he states as Corollary 3.6 that $s(\mathfrak s\_o)=s(\mathfrak s\_{\bar o})=s\_{min}(K)$, where $s$ is the $s$-grading and $\mathfrak s\_o,\mathfrak s\_{\bar o}$ are Lee's canonical generators for her homology theory $Kh'$. I don't really understand why this follows from the previous lemma. And in fact, I feel that at least one of $s(\mathfrak s\_o),s(\mathfrak s\_{\bar o})$ should be $s\_{max}$. After all, if any state $S$ can be written as $a\mathfrak s\_o+b\mathfrak s\_{\bar o}$, then we should have $s(S)\le\text{max}(s(\mathfrak s\_o),s(\mathfrak s\_{\bar o}))$. I contemplated that this could've been a typo in the corollary statement, but I still don't see why $s(\mathfrak s\_o)=s(\mathfrak s\_{\bar o})$, and I also don't see how this relates to the lemma.
Can someone explain the proof of this corollary, or at least explain why my "contradiction" to it is incorrect? Thanks.
| https://mathoverflow.net/users/146012 | Corollary in Rasmussen's paper about $s$-grading of Lee's canonical generators | Let $ C ( D) $ denote the Lee chain complex and $ Kh ' ( K ) $ its homology, $ q $ denote the grading on $ C ( D) $ (associated to the filtration) and $ s $ the induced grading on $ Kh ' ( K ) $.
Recall that $ \mathfrak{s}\_\overline{o} $ is obtained from $ \mathfrak{s}\_o $ by interchanging $ r = v\_+ + v\_- $ and $ g = v\_- - v\_+ $ on all circles of the oriented smoothing of $ D $. As $ q ( r ) = q ( g ) $ it follows that $ q ( \mathfrak{s}\_o ) = q ( \mathfrak{s}\_\overline{o} ) $ and that $ s ( \mathfrak{s}\_o ) = s ( \mathfrak{s}\_\overline{o} ) ~\text{mod}~ 4 $. By Lemma 3.5 $ Kh' ( K ) $ is supported in s-degrees $ 1 ~\text{mod}~ 4 $ and $ 3 ~\text{mod}~ 4 $, so that $ s ( \mathfrak{s}\_o ) = s ( \mathfrak{s}\_\overline{o} ) $ exactly.
The filtration on $ C ( D ) $ is upward: $ F^{i+1} C ( D ) \subset F^{i} C ( D ) $. As $ s ( \mathfrak{s}\_o ) = s ( \mathfrak{s}\_\overline{o} ) $ we must have $ s ( \mathfrak{s}\_o \pm \mathfrak{s}\_\overline{o} ) \geq s ( \mathfrak{s}\_o ) , s ( \mathfrak{s}\_\overline{o} ) $, so that $ s ( \mathfrak{s}\_o ) = s ( \mathfrak{s}\_\overline{o} ) = s\_{\text{min}} ( K ) $.
| 3 | https://mathoverflow.net/users/61064 | 442085 | 178,400 |
https://mathoverflow.net/questions/442078 | 1 | Let
$$\ell(n)=\left\lfloor\log\_2 n\right\rfloor$$
Let
$$T(n,k)=\left\lfloor\frac{n}{2^k}\right\rfloor\operatorname{mod}2$$
Here $T(n,k)$ is the $(k+1)$-th bit from the right side in the binary expansion of $n$.
Let $a(n)$ be the sequence of positive integers such that we start from $A:=0$ and then for $k=0..\ell(n)$ we iterate:
1. If $T(n,k)=1$, then $A:=\left\lfloor\frac{A}{2}\right\rfloor$;
otherwise $A:=A+1$;
2. $A:=A+1$.
Then $a(n)$ is the resulting value of $A$.
For example for $n=18=10010\_2$ we have:
1. $A:=0$;
2. $T(n,0)=0$, $A:=A+1=1$, $A:=A+1=2$;
3. $T(n,1)=1$, $A:=\left\lfloor\frac{A}{2}\right\rfloor=1$, $A:=A+1=2$;
4. $T(n,2)=0$, $A:=A+1=3$, $A:=A+1=4$;
5. $T(n,3)=0$, $A:=A+1=5$, $A:=A+1=6$;
6. $T(n,4)=1$, $A:=\left\lfloor\frac{A}{2}\right\rfloor=3$, $A:=A+1=4$.
Then $a(18)=4$.
Let
$$R(n,k)=\sum\limits\_{j=2^{n-1}}^{2^n-1}[a(j)=k]$$
I conjecture that
1. $R(n,k)=0$ if $n<1$ or $k>n$;
2. $R(n,k)=1$ if $k=1$ or $k=n$;
3. $R(n,k)=R(n-1,k-1)+R(n-1,2(k-1))+R(n-1,2k-1)$ otherwise.
To verify this conjecture one may use this PARI prog:
```
a(n) = my(A=0); for(i=0, logint(n, 2), if(bittest(n, i), A\=2, A++); A++); A
R1(n) = my(v); v=vector(n, i, sum(k=2^(n-1), 2^n-1, a(k)==i))
R(n, k) = if(k==1, 1, if(k<=n, R(n-1, k-1) + R(n-1, 2*(k-1)) + R(n-1, 2*k-1)))
R2(n) = my(v); v=vector(n, i, R(n,i))
test(n) = R1(n)==R2(n)
```
Is there a way to prove it? Is there a suitable closed form for $R(n,k)$?
| https://mathoverflow.net/users/231922 | Coefficients of number of the same terms which are arising from iterations based on binary expansion of $n$ | In other words, if $(b\_\ell b\_{\ell-1}\dots b\_0)\_2$ is the binary representation of $n$, then
$$a(n) = g(g(\dots g(g(0,b\_0),b\_1)\dots ),b\_{\ell-1}), b\_\ell),$$
where
$$g(A,b) = \begin{cases} A+2, &\text{if } b=0;\\
\left\lfloor \frac{A+2}2\right\rfloor, &\text{if } b=1.
\end{cases}$$
Consider a number triangle obtained from $A=0$ by iteratively applying $g(\cdot,0)$ and $g(\cdot,1)$:
$$
\begin{gathered}
0 \\
1 \ \ \ \ \ \ \ \ 2 \\
1 \ \ \ \ 3 \ \ \ \ 2 \ \ \ \ 4 \\
1\ 3\ 2\ 5\ 2\ 4\ 3\ 6 \\
\dots
\end{gathered}
$$
Let $f(n,k)$ be the multiplicity of $k$ at the level $n\in\{0,1,2\dots\}$ in this triangle.
It is easy to see that each number $k\geq 1$ in this triangle may result only from the following numbers in the previous row: $2k-2$, $2k-1$, or $k-2$, implying that $f$ satisfies the recurrence formula:
$$f(n,k) = \begin{cases}
\delta\_{k,0}, & \text{if }n=0; \\
f(n-1,2k-2) + f(n-1,2k-1) + f(n-1,k-2), & \text{if }n>0.
\end{cases}$$
The quantity $R(n,k)$ accounts for numbers $k$ in the $n$th row, but only for those that resulted from $g(\cdot,1)$, that is
$$R(n,k) = f(n-1,2k-2) + f(n-1,2k-1).$$
Expanding this formula using the recurrence for $f$, we get
\begin{split}
R(n,k) &= f(n-2,4k-5) + f(n-2,4k-6) + f(n-2,2k-4) \\
&\quad + f(n-2,4k-3) + f(n-2,4k-4) + f(n-2,2k-3) \\
&= R(n-1,2k-2) + R(n-1,2k-1) + R(n-1,k-1).
\end{split}
QED
| 2 | https://mathoverflow.net/users/7076 | 442095 | 178,404 |
https://mathoverflow.net/questions/441910 | 4 | Let $\Omega$ be a Polish locally compact space and $(\Omega, \mathscr{F}, \mathbb{P})$ be a probability space. Consider a measurable map
\begin{align\*}
\theta\colon T\times \Omega &\to \Omega\\
(t,\omega)&\mapsto \theta\_{t}\omega
\end{align\*}
such that $\theta\_{t+s}=\theta\_{t}\circ \theta\_{s}$ for
every $s, t\in T$ and $\theta\_{t}$ preserves $\mathbb{P}$ for every $t\in T$, that is, $\theta\_{t}\mathbb{P}=\mathbb{P}$ for every $t\in T$. In other words,
$(\theta\_{t})$ is a measure preserving dynamical system on $(\Omega, \mathscr{F}, \mathbb{P})$.
**Definition 1 (Random Dynamical Systems and Invariant Measures)**:
Let $(\Omega, \mathscr{F}, \mathbb{P})$ and $(\theta\_{t})$ as above and consider a compact metric space $X$ (endowed with its Borel $\sigma$-algebra $\mathscr{B}$).
A measurable map
\begin{align\*}
\varphi\colon T\times \Omega\times X &\to X\\
(t,\omega,x)& \mapsto \varphi(t,\omega)x
\end{align\*}
is called a *random dynamical system* on $X$ over the measure preserving dynamical system $(\theta\_{t})$ if
a) $x\mapsto \varphi(t,\omega)x$ is continuous for $\mathbb{P}$-almost every $\omega$ and every $t\in T$.
b) $\varphi(t+s, \omega)=\varphi(t,\theta\_{s}\omega)\circ \varphi(s,\omega)$ for all $t,s\in T$, and $\varphi(0,\omega)=id$, for $\mathbb{P}$-almost every $\omega$.
A random dynamical system $\varphi$ induces a dynamical system over $\Omega\times X$ given by
$$
\Theta(t)(\omega, x)=(\theta(t)\omega, \varphi(t,\omega)x),\quad t\in {T}
$$
called the *skew product* associated with $\,\varphi$.
A probability measure $\mu$ on $\Omega\times X$ is said to be a $\varphi$-*invariant measure* for the random dynamical system $\varphi$, if it satisfies
1. $\Theta(t)\mu=\mu$ for all $t\in T$;
2. $\Pi\mu=\mathbb{P}$, where $\Pi$ is the natural projection on the first factor defined by $\Pi(\omega,x)=\omega$.
Let $\mu$ be a $\varphi$-invariant measure. Since $\Pi\mu=\mathbb{P}$, we can consider the disintegration $d\mu=\mu\_{\omega}\, d\mathbb{P}(\omega)$ of $\mu$ with respect to $\mathbb{P}$. A necessary and sufficient condition for $\mu$ to be a $\varphi$-invariant measure is that for every $t\in T$
\begin{equation}\label{inva}
\varphi(t,\omega)\mu\_{\omega}=\mu\_{\theta\_{t}\omega}
\end{equation}
for $\mathbb{P}$-almost every $\omega$.
**Definiton 2:** We say that $\mu$ is a **random Dirac measure** if
there is measurable map $\Phi\colon \Omega\to X$ such that
$$
\mu\_{\omega}=\delta\_{\Phi(\omega)}
$$
for $\mathbb{P}$-almost every $\omega$.
**MY QUESTION:** I am looking for examples of Random Dynamical Systems (References of Papers) in which there is a finite($\geq$ 2) number of ergodic $\varphi$-invariant measures,
which are, additionally, random Dirac measures.
I am most interested in the continuous time cases, but examples (References of Papers)in the discrete time setting is also welcome.
| https://mathoverflow.net/users/98969 | Finite number of ergodic random Dirac measures | Let $\theta$ be a discrete time dynamical system as in your question. And for each $\omega$, let $A(\cdot)$ be a measurable map from $\Omega$ into $\text{GL}(d,\mathbb R)$ such that $\log \|A(\cdot)\|$ and $\log\|A(\omega)^{-1}\|$ are integrable.
Let me also assume $\mathbb P$ is ergodic. Then the multiplicative etgodic theorem guarantees that there exist $\lambda\_1>\ldots>\lambda\_k$; multiplicities $d\_1,\ldots,d\_k$ summing to $d$ and subspaces $V\_1(\omega),\ldots,V\_k(\omega)$ of dimensions $d\_1,\ldots,d\_k$ respectively whose direct sum is $\mathbb R^d$ such that:
* $A(\omega)V\_i(\omega)=V\_i(\sigma\omega)$;
* if $x\in V\_i(\omega)\setminus\{0\}$, then $(1/n)\log\|A^{(n)}(\omega)v\|\to\lambda\_i$.
The $\lambda\_i$ are called Lyapunov exponents and the $V\_i(\omega)$ are called Oseledets spaces. A common case is when the multiplicities are all 1. This is called simple Lyapunov spectrum.
Given such a system, one can build a natural map from $\Omega\times P\mathbb R^{d-1}$ to itself by
$\bar \Theta(\omega,v)=(\theta\omega, A(\omega)x)$, where matrices act on the projective space in the obvious way.
In the case where the original bicycle has simple Lyapunov spectrum, the delta measures sitting on the directions of the Oseledets spaces are exactly random Dirac measures as you are asking for.
There are also continuous time versions of all of this. These systems arise very naturally in the context of differentiable maps of manifolds or differentiable flows on manifolds.
| 1 | https://mathoverflow.net/users/11054 | 442096 | 178,405 |
https://mathoverflow.net/questions/431840 | 0 | We have a inhomogeneous continous $K$-State Markov chain $X(t)$ with transition intensity matrix $Q(t)$. Therefore its entries are:
$$q\_{ij}(t)= \lim\_{\delta \to 0} \frac{1}{\delta} \mathbb{P}(X(t+\delta) = j | X(t) = i).$$
The transistion probabilities are in a matrix $P(s,t)$ with entries
$$P\_{ij}(s,t) = \mathbb{P}(X(t) = j | X(s) = i).$$
If we use the Chapman-Kolmogorov equation we can get the Kolmogorov forward
equation as
$$\frac{d}{dt}P(s,t) = P(s,t)Q(t)$$
This can also be written as
\begin{align}
P(s,t) &= I+\int\_s^tP(s,u-)Q(u)du \\
&= I+\int\_s^tP(s,u-)dA(u),
\end{align}
where $I$ is the identity matrix and $A(u)$ the cumulative transition intensities.
Now every literature states that one can find a solution to this Integral as the product-limit
$$P(s,t) = \prod\_{u \in (s,t]}\bigl(I+dA(u)\bigr).$$
I don´t see why this is true and how someone can derive this product-limit of the integral representation from the Kolmogorov forward equation.
| https://mathoverflow.net/users/476504 | Inhomogeneous Markov chains and the product-integral as a solution to the Kolmogorov forward equation | I see it as follows: $$\frac{\partial}{\partial t}P(s,t)=P(s,t)Q(t) \rightarrow P(x,y+\Delta y)=P(x,y)+P(x,y)Q(y)\Delta y+o(\Delta y)\ ,$$ so we also have $$P(x,y+\Delta y)=P(x,y)[I+Q(y)\Delta y]+o(\Delta y) \rightarrow P(y,y+\Delta y)=I+Q(y)\Delta y+o(\Delta y)\ .$$Now, as $P(a,c)=P(a,b)P(b,c)$ whenever $a\le b \le c$, if we divide the interval between $s$ and $t$ in $N$ equal intervals with amplitude $\Delta t$ we can write $$P(s,t)=P(s,s+\Delta t)P(s+\Delta t,s+2\Delta t)\ldots P(t-\Delta t,t)=\\ [I+Q(s)\Delta t +o(\Delta t)][I+Q(s+\Delta t)\Delta t +o(\Delta t)]\ldots [I+Q(t-\Delta t)\Delta t +o(\Delta t)]= \\ = [I+Q(s)\Delta t][I+Q(s+\Delta t)\Delta t]\ldots [I+Q(t-\Delta t)\Delta t)]+N\; o(\Delta t) .$$Since $\Delta t=(s-t)/N$ the quantity $N\; o(\Delta t)=o(s-t)$ tends to zero as $\Delta t$ tends to zero, and so we can write: $$ P(s,t)=\lim\_{N\to\infty}\prod\_{k=0}^{N-1}[I+Q(s+k\Delta t)\Delta t]$$where $\Delta t=(s-t)/N$ as usual, which is, I think, equivalent to $$P(s,t) = \prod\_{u \in (s,t]}\bigl(I+dA(u)\bigr).$$
It is not difficult to prove, starting from the productorial formula, that, if $Q(x)Q(y)=Q(y)Q(x)$ for every $x$ and $y$ in $(s,t)$ then we also have $$P(s,t)=\exp\left[\int\_s^t Q(u)du\right]\ .$$
| 0 | https://mathoverflow.net/users/470349 | 442098 | 178,406 |
https://mathoverflow.net/questions/442104 | 8 | I'm trying to come up with a good explanation for my students of why the [finite lattice representation problem](https://en.wikipedia.org/wiki/Finite_lattice_representation_problem) is difficult. I've already shown that the "greedy approach" to representing the lattice drawn in [this post](https://mathoverflow.net/questions/85724/given-a-lattice-l-with-n-elements-are-there-finite-groups-h-g-such-that-l-c) as the congruence lattice on a finite algebra - namely, realize the lattice as a sublattice of a lattice of partitions on some large enough finite set $X$, and then look at the algebra on $X$ given by all functions respecting each partition in that sublattice - breaks down.
I'd now like to explain why this *doesn't* lead to an easy *negative* solution. Towards that end:
>
> Is there a not-too-hard-to-verify example of a bounded lattice $L$ such that
>
>
> * $L$ is isomorphic to the congruence lattice of some finite algebra, but
> * the smallest $n$ such that $L$ embeds into the bounded lattice of partitions of $\{1,...,n\}$ is strictly less than the smallest size of an algebra with congruence lattice $\cong L$?
>
>
>
Essentially, a concrete example of this would give a convincing argument that the failure of the greedy strategy doesn't actually tell us much.
| https://mathoverflow.net/users/8133 | Example of trickiness of finite lattice representation problem? | $M\_4$, the modular lattice of height two with four atoms is an example.
$M\_4$ arises as the subgroup lattice of the symmetric group on $3$ letters, hence it arises as the congruence lattice of a regular $S\_3$-set. This is an algebra with $6$ elements. There is no representation of $M\_4$ as the congruence lattice of a finite algebra $\mathbf{A}$ where $\mathbf{A}$ has fewer than $6$ elements. This is proved in
[A minimal congruence lattice representation for $M\_{p+1}$](https://www.cambridge.org/core/journals/journal-of-the-australian-mathematical-society/article/abs/minimal-congruence-lattice-representation-for-mathbbmp1/F1031B09770B7060128554C0A2E86EB5)
Roger Bunn, David Grow, Matt Insall, Philip Thiem
J. Aust. Math. Soc. 108 (2020), 332-340
Their main result is that if $p$ is an odd prime, then a minimal congruence lattice representation of $M\_{p+1}$ has size $2p$, so a minimal congruence lattice representation of $M\_4 = M\_{3+1}$ has size $2\cdot 3 = 6$.
A few years ago I posted [a shorter proof](https://arxiv.org/abs/2008.04784) of the Bunn-Grow-Insall-Thiem theorem to the arxiv.
On the other hand, it is possible to embed $M\_4$ into the lattice of partitions of a $5$-element set, and $5<6$. The following four partitions of $\{1,2,3,4,5\}$ generate such an $M\_4$:
12/34/5
13/25/4
14/35/2
15/24/3
| 8 | https://mathoverflow.net/users/75735 | 442105 | 178,408 |
https://mathoverflow.net/questions/442099 | 5 | Let $\mathscr{C}$, $\mathscr{D}$, and $\mathscr{E}$ be (infinity) categories, and assume we're given a Cartesian diagram: $\require{AMScd}$
\begin{CD}
\mathscr{C} \times\_{\mathscr{E}} \mathscr{D} @>\operatorname{pr}\_1>> \mathscr{D}\\
@V \operatorname{pr}\_2 V V @VV \varphi V\\
\mathscr{C} @>>\psi> \mathscr{E}
\end{CD}
Now let $\mathbf{C}$ be some other cocomplete infinity category and let $F: \mathscr{D} \to \mathbf{C}$ be a functor.
Since $\mathbf{C}$ is cocomplete, we can consider the left Kan extension $\varphi\_!(F)$ of $F$ along $\varphi$. By definition, $\varphi\_!$ is the left adjoint of the restriction functor $$\varphi^\*: \operatorname{Fun}(\mathscr{E},\mathbf{C}) \longrightarrow \operatorname{Fun}(\mathscr{D},\mathbf{C}),$$ where $\operatorname{Fun}(-,-)$ denotes the category of functors between two categories. Similarly, we can consider the left Kan extension $(\operatorname{pr}\_2)\_!(\operatorname{pr}\_1)^\*(F)$ of the pullback of $F$ to the fiber product along $\operatorname{pr}\_2$.
By the universal property of left Kan extensions, there is a natural transformation $$\beta\_F:(\operatorname{pr}\_2)\_!(\operatorname{pr}\_1)^\*(F) \longrightarrow \psi^\* \varphi\_!(F)$$
and I would like to know if there are reasonable conditions we can put on the categories and functors involved so that $\beta\_F$ is an isomorphism. Probably this doesn't always hold, but my hope is that there are some nice situations in which it does.
| https://mathoverflow.net/users/101861 | Base change isomorphism for left Kan extensions | I believe one set of conditions is for either $\varphi$ to be proper or $\psi$ to be smooth. The dual of this (using right Kan extensions rather than left Kan extensions) is proven by Cisinski in "Higher Categories and Homotopical Algebra", Theorem 6.4.13.
For completeness, a map between simplicial sets $p : X \to Y$ is said to be proper when pullback along any pullback of $p$ preserves final maps (right cofinal in Lurie's terminology). More precisely, $p$ is proper if for any pair of cartesian squares in the following form, if $f$ is final then $g$ is final:
$\require{AMScd}$
\begin{CD}
X'' @>g>> X' @>>> X\\
@V V V @V V V @V V p V \\
Y'' @>>f> Y' @>>> Y
\end{CD}
A map is smooth just when its opposite is proper.
| 6 | https://mathoverflow.net/users/76636 | 442113 | 178,411 |
https://mathoverflow.net/questions/442107 | 5 | The unknot and the Hopf link are (as far as I know) the only links whose complements have abelian fundamental groups. Are there more examples whose complement have amenable fundamental group?
| https://mathoverflow.net/users/89741 | Amenable link groups | No. Suppose that $K$ is a non-trivial knot. Then its knot genus is at least one. Thus $X = S^3 - K$ contains a $\pi\_1$-essential surface (with boundary) of genus at least one. Thus $\pi\_1(X)$ contains a free group of rank at least two.
The story for links is similar.
---
A more subtle proof uses the geometrisation theorem. Every link is either a split link, a torus link, a satellite link, or a hyperbolic link. Torus link complements (other than of the unknot and Hopf link) fibre over a base orbifold with negative Euler characteristic - this gives the desired free group. Hyperbolic knot complements have free groups in their fundamental groups for geometric reasons. Split (satellite) links have an essential sphere (torus) in their complement - so cut and induct.
| 7 | https://mathoverflow.net/users/1650 | 442120 | 178,414 |
https://mathoverflow.net/questions/442114 | 7 | I apologise in advance for what must be a naive question. Let $\mathcal O\_K$ be the ring of integers of the algebraic number field $K.$ Let $p$ be a rational prime, and factorize $$(p)=\mathfrak p\_1^{e\_1}\cdots\mathfrak p\_r^{e\_r}$$ where the $\mathfrak p\_i$ are primes in $\mathcal O\_K.$ Let $k\_i=\mathcal O\_K/\mathfrak p\_i$ be the residue fields for $1\le i\le r.$
I've seen a statement without proof that $\mathcal O\_K / (p)$ is ring isomorphic to the sum of truncated polynomial rings $\bigoplus\_1^r k\_i[t]/(t^{e\_i}).$ It looks like a standard result, but I can't find a proof, could anyone point out a source? I see a simple proof where $\mathcal O\_K$ is $p$-monogenic (i.e. $|\mathcal O\_K:\mathbb Z[\alpha]|$ is prime to $p$ for some $\alpha\in\mathcal O\_K,$ the usual condition for Dedekind's criterion to be applied) though I would still welcome a source to check my reasoning, but I understand the result holds without that assumption.
| https://mathoverflow.net/users/313687 | Quotients of number fields by certain prime powers | I remember working this out 25 years ago. The main idea is to view both rings as quotient rings of completions: $\mathcal O\_K/\mathfrak p^e \cong \widehat{\mathcal O\_{\mathfrak p}}/\widehat{\mathfrak p}^e$ and $k[t]/(t^e) \cong k[[t]]/(t^e)$. Then the structure of the ring of integers of local fields, in characteristic $0$ and characteristic $p$, will help us solve the problem.
For a nonzero prime ideal $\mathfrak p$ in $\mathcal O\_K$ lying over $p$ and $m \geq 1$, the quotient ring $\mathcal O\_K/\mathfrak p^m$ is unchanged up to isomorphism if we replace $\mathcal O\_K$ by its localization at $\mathfrak p$ or by its completion at $\mathfrak p$ (and the modulus also changes to the ideal it generates in the localization or completion). Also, the ramification index $e = e(\mathfrak p|p)$ and residue field degree $f = f(\mathfrak p|p)$ are unchanged by localizing or completing at $\mathfrak p$.
Let $A = \widehat{\mathcal O\_{\mathfrak p}}$ be the completion of $\mathcal O\_K$ at $\mathfrak p$, so in $A$ we can write $p = \pi^e u$ for some uniformizer $\pi$ and $u \in A^\times$. Then
$\mathcal O\_K/\mathfrak p^e \cong A/(\pi^e) = A/(p)$. (In this step it's important that the exponent $e$ is the ramification index of $\mathfrak p$ over $p$.) The ring $A$ is the ring of integers of the completion $K\_\mathfrak p$. Even though $\mathcal O\_K$ need not be monogenic, completions *are* monogenic: the ring of integers of every finite extension of $\mathbf Q\_p$ has a power basis over $\mathbf Z\_p$, so $A = \mathbf Z\_p[\alpha]$ for some $\alpha \in A$. That there is a power basis over $\mathbf Z\_p$ is the key fact you were missing.
Let $\alpha$ have minimal polynomial $F(x)$ in $\mathbf Z\_p[x]$, so $F(x)$ is irreducible over $\mathbf Z\_p$ and
$A \cong \mathbf Z\_p[x]/(F(x))$ as rings. Viewing both sides as $\mathbf Z\_p$-modules and computing their $\mathbf Z\_p$-ranks shows $[K\_\mathfrak p:\mathbf Q\_p] = \deg F$, so
$$
\deg F = e(\mathfrak p|p)f(\mathfrak p|p) = ef.
$$
Using $F(x)$,
$$
\mathcal O\_K/\mathfrak p^e \cong A/(p) = \mathbf Z\_p[\alpha]/(p) \cong \mathbf Z\_p[x]/(p,F(x)) \cong \mathbf F\_p[x]/(\overline{F}(x)).
$$
The mod $p$ reduction $\overline{F}(x)$ in $\mathbf F\_p[x]$ has a factorization into monic irreducibles. All monic irreducible factors of $\overline{F}(x)$ are the same, because if that were not the case then we could write $F(x) \equiv G(x)H(x) \bmod p$ for nonconstant monic $G(x)$ and $H(x)$ where $\gcd(G \bmod p,H \bmod p) = 1$ in $\mathbf F\_p[x]$, and then by Hensel's lemma (the one about lifting relatively prime factorizations, not just about lifting a simple root) we'd get a factorization of $F(x)$ in $\mathbf Z\_p[x]$ into nonconstant monic factors, which contradicts the irreducibility of $F(x)$ over $\mathbf Z\_p$. So in $\mathbf F\_p[x]$ we must have $\overline{F}(x) = Q(x)^d$ for some monic irreducible $Q(x)$ in $\mathbf F\_p[x]$ and $d \geq 1$. That means
$$
\mathcal O\_K/\mathfrak p^e \cong A/(p) \cong \mathbf F\_p[x]/(Q(x)^d)
$$
as rings.
We will now show
$$
d = e = e(\mathfrak p|p), \ \ \deg Q = f = f(\mathfrak p|p).
$$
Let $k = \mathcal O\_K/\mathfrak p$, which is the residue field of $\mathcal O\_K$ at $\mathfrak p$, so $\dim\_{\mathbf F\_p}(k) = f$ by definition. Residue fields are unchanged up to isomorphism by completion, so $k \cong A/(\pi)$. The local ring $A/(p) = A/(\pi^e)$ has maximal ideal $(\pi)/(\pi^e)$ and residue field $A/(\pi)$, while the local ring $\mathbf F\_p[x]/(Q(x)^d)$ has maximal ideal $(Q(x))/(Q(x)^d)$ and residue field $\mathbf F\_p[x]/(Q(x))$. Isomorphic local rings have isomorphic residue fields, so $k \cong \mathbf F\_p[x]/(Q(x))$. Computing $\mathbf F\_p$-dimensions of both sides,
$$
f = \deg Q.
$$
Returning to $F$, which is monic and reduces mod $p$ to $Q(x)^d$, we can now say
$$
\deg F = \deg \overline{F} = d\deg Q = d f
$$
and we already saw $\deg F = ef$, so
$$
d = e.
$$
Thus
$$
\mathcal O\_K/\mathfrak p^e \cong A/(p) \cong \mathbf F\_p[x]/(Q(x)^e), \ \ \deg Q = f.
$$
The last step is to show $\mathbf F\_p[x]/(Q(x)^e) \cong k[t]/(t^e)$ as rings, so
$$
\mathcal O\_K/\mathfrak p^e \cong A/(p) \cong \mathbf F\_p[x]/(Q^e) \cong k[t]/(t^e).
$$
In fact we'll show
$$
\mathbf F\_p[x]/(Q(x)^m) \cong k[t]/(t^m)
$$
for all $m \geq 1$. We need $m = e$ only to identify these rings with $\mathcal O\_K/\mathfrak p^e$, but the rings are isomorphic to each other for all $m \geq 1$. (Note $\mathcal O\_K/\mathfrak p^m$ has characteristic $p$ if and only if $\mathfrak p^m \mid p\mathcal O\_K$, forcing $m \leq e$, but the rings $\mathbf F\_p[x]/(Q(x)^m)$ and $k[t]/(t^m)$ have characteristic $p$ for all $m \geq 1$, so there's nothing unusual about them winding up as isomorphic to each other for all $m$.) I will describe two methods, the first one being more concrete.
Method 1: The elements of $k[t]/(t^m)$ are uniquely expressible as
$$
c\_0 + c\_1t + \cdots + c\_{m-1}t^{m-1} \bmod t^m
$$
with $c\_j \in k$. Inside $\mathbf F\_p[x]/(Q^m)$, the elements are uniquely expressible as
$$
a\_0(x) + a\_1(x)Q(x) + \cdots + a\_{m-1}(x)Q(x)^{m-1}
$$
with $a\_j(x) = 0$ or $\deg(a\_j(x)) < \deg Q$. But this way of writing the elements of $\mathbf F\_p[x]/(Q^m)$ is **terrible** in order to set up a ring isomorphism with $k[t]/(t^m)$ since those base $Q$ digits $a\_j(x)$ are not multiplicatively closed (in contrast to the $c\_j$ in $k$). What we need to do is find a copy of the field $k$ of order $p^f$ inside $\mathbf F\_p[x]/(Q^m)$.
The polynomial $t^{p^f} - t$ splits completely over the field $\mathbf F\_p[x]/(Q)$, so for each $m \geq 1$ the $Q$-adic Hensel's lemma tells us we can lift each of those roots mod $Q$ uniquely to a root of $t^{p^f}-t$ in $\mathbf F\_p[x]/(Q^m)$. Set
$$
k\_m := \{b(x) \bmod Q^m : b(x)^{p^f} \equiv b(x) \bmod Q^m\},
$$
which is the roots of $t^{p^f}-t$ in $\mathbf F\_p[x]/(Q^m)$. This set is closed under addition and multiplication and each nonzero element is invertible: if $b(x) \not\equiv 0 \bmod Q^m$ then $b(x) \not\equiv 0 \bmod Q$ (the only lift of $0 \bmod Q$ as a root is $0 \bmod Q^m$), so $\gcd(b(x),Q^m) = 1$. So $k\_m$ is a field of order $p^f$ inside $\mathbf F\_p[x]/(Q^m)$ and in fact it's the only such field inside $\mathbf F\_p[x]/(Q^m)$ thanks to the unique lifting of roots of $t^{p^f}-t$ from modulus $Q$ to modulus $Q^m$.
Since $k\_m$ inside $\mathbf F\_p[x]/(Q^m)$ is a set of representatives for $\mathbf F\_p[x]/(Q)$, we can write every element of $\mathbf F\_p[x]/(Q^m)$ uniquely as
$$
b\_0(x) + b\_1(x)Q(x) + \cdots + b\_{m-1}(x)Q(x)^{m-1} \mod Q(x)^m
$$
where $b\_j(x) \bmod Q(x)^m \in k\_m$. This way of writing
the elements of $\mathbf F\_p[x]/(Q^m)$ looks just like the usual way of writing the elements of $k[t]/(t^m)$ and it shows $k\_m[Q \bmod Q^m]$ fills up $\mathbf F\_p[x]/(Q^m)$
Since $k$ and $k\_m$ are fields of equal size $p^f$ there is a field isomorphism $\varphi\_m \colon k \to k\_m$ (in fact there are $m$ isomorphisms, but that doesn't matter). Extend $\varphi\_m$ to a ring homomorphism $k[t] \to \mathbf F\_p[x]/(Q^m)$ by mapping $t$ to $Q \bmod Q^m$:
$$
\sum\_{i} c\_it^i \mapsto \sum\_i \varphi\_m(c\_i)Q^i \bmod Q^m.
$$
This is surjective since $\mathbf F\_p[x]/(Q^m)$ is generated as a ring by $k\_m$ and $Q \bmod Q^m$. Since $t^m \mapsto Q^m \bmod Q^m = 0$, we get an induced surjective ring homomorphism $k[t]/(t^m) \to \mathbf F\_p[x]/(Q^m)$. Both have order $p^{fm}$, so this is a ring isomorphism.
Method 2: Since $Q$ is irreducible in $\mathbf F\_p[x]$, the ring $\mathbf F\_p[x]/(Q^m)$ is unchanged up to isomorphism if we replace $\mathbf F\_p[x]$ with its $Q$-adic completion $\mathbf F\_p[x]\_Q$:
$$
\mathbf F\_p[x]/(Q^m) \cong \mathbf F\_p[x]\_Q/(Q^m)
$$
as rings.
The residue field of $\mathbf F\_p[x]\_Q$ is isomorphic to $\mathbf F\_p[x]/(Q)$, which is isomorphic to $k$. The completion $\mathbf F\_p[x]\_Q$ is the ring of integers of the $Q$-adic field completion $\mathbf F\_p(x)\_Q$, and the structure of local fields of positive characteristic says they are all isomorphic to the formal Laurent series field over the residue field. Thus $\mathbf F\_p(x)\_Q \cong k((t))$ as valued fields, so their rings of integers are isomorphic: $\mathbf F\_p[x]\_Q \cong k[[t]]$. The isomorphism identifies powers of the maximal ideal on both sides,
and the maximal ideal of $\mathbf F\_p[x]\_Q$ is $(Q)$ since $Q$ is a uniformizer in the $Q$-adic completion. Thus
$$
\mathbf F\_p[x]\_Q/(Q^m) \cong k[[t]]/(t^m)
$$
for each $m \geq 1$. Both sides simplify to quotient rings of $\mathbf F\_p[x]$ and $k[t]$, just as $\mathbf Z\_p/(p^m) \cong \mathbf Z/(p^m)$:
$$
\mathbf F\_p[x]/(Q^m) \cong k[t]/(t^m).
$$
Here is an illustration of Method 1.
**Example**. Let $Q(x) = x^2 - 2$ in $\mathbf F\_5[x]$. Then
$\mathbf F\_5[x]/(Q^3) \cong \mathbf F\_{25}[t]/(t^3)$. To write down an explicit ring isomorphism we need a copy of $\mathbf F\_{25}$ in $\mathbf F\_5[x]/(Q^3)$. We can view $\mathbf F\_{25}$ as $\mathbf F\_5(\alpha)$ where $\alpha^2 = 2$. One root of $t^2 - 2$ in $\mathbf F\_5[x]/(x^2-2)$ is $x \bmod x^2-2$, and the unique lifting of this to a root of $t^2 - 2$ in $\mathbf F\_5[x]/(Q^3)$ is
$$
r = x + xQ + 4xQ^2 = x + x(x^2-2) + 4x(x^2-2)^2 \bmod Q^3
$$
by an explicit search.
Therefore the copy of $\mathbf F\_{25}$ inside $\mathbf F\_5[x]/(Q^3)$ is $\mathbf F = \mathbf F\_5 + \mathbf F\_5r$ and
$$
\mathbf F\_5[x]/(Q^3) = \mathbf F + \mathbf F Q + \mathbf F Q^2 \bmod Q^3.
$$
The right side naturally looks like $\mathbf F\_{25}[t]/(t^3)$ and we get an isomorphism $\mathbf F\_{25}[t]/(t^3) \to \mathbf F\_5[x]/(Q^3)$ once we decide on an isomorphism between $\mathbf F\_{25}$ and $\mathbf F$.
| 15 | https://mathoverflow.net/users/3272 | 442125 | 178,415 |
https://mathoverflow.net/questions/442124 | 15 | **INTRODUCTION**. I am teaching a course in Harmonic Analysis. In class, very often I find myself stressing out the fundamental property that the functions
$$
e\_n(x)=\exp(2\pi i n x), \quad \text{where }\quad x\in \mathbb R / \mathbb Z$$
are simultaneous eigenfunctions of all translations. By this I mean that, defining
$$\label{1}\tag{1}
\tau\_y f(x):=f(x-y), \quad \text{for }x, y\in\mathbb R/ \mathbb Z$$
it holds that
$$\label{2}\tag{2}
\tau\_y e\_n=\lambda\_{y,n}e\_n, \quad \text{for all }y\in\mathbb R/\mathbb Z.$$
---
**QUESTION**. Let us now replace $\mathbb R/\mathbb Z$ with the sphere $\mathbb S^2$. By analogy with (1), define
$$
\tau\_A f(\omega):=f(A^{-1}\omega), \quad \text{for }A\in \mathrm{SO}(3),\ \omega\in\mathbb S^2.$$
Is there a non-constant $f\in L^2(\mathbb S^2)$ such that
$$\label{3}\tag{3}
\tau\_Af=\lambda\_A f, \quad \text{for all }A\in \mathrm{SO}(3)?$$
---
**REMARKS**. From a more general standpoint, the simultaneous diagonalization property \eqref{2} of $e\_n$ is a common feature of all locally compact abelian groups. Indeed, if $G$ denotes such a group, then we can define
$$
\tau\_g f(h):=f(g^{-1}h), \quad g,h\in G,\ f\in L^2(G), $$
yielding a unitary representation of $G$. And since $G$ is abelian, all irreducible subrepresentations must be one-dimensional, or in other words, they must be eigenspaces of all $\tau\_g$ simultaneously. This is precisely the diagonalization phenomenon observed above in the case $G=\mathbb R / \mathbb Z$.
The present question occured to me while I was trying to find a concrete, pedagogical example of the failure of this simultaneous diagonalization in the non-abelian case.
---
**ANSWER**. I can actually provide an answer to the question, which unfortunately I find unsatisfactory for reasons that I will explain below. The answer is the following.
We claim that no non-constant function $f$ satisfying \eqref{3} exists. To prove this, introduce the projectors
$$ P\_{n} f(\omega):=\int\_{\mathbb S^2} f(\nu)L\_n(\omega\cdot \nu)\ d\sigma(\nu),$$
where $L\_n$ is the Legendre polynomial of degree $n$, with an appropriate normalization (the exact value of which is irrelevant here). For each $n\in \mathbb N\_{\ge 0}$, $P\_n f$ is a spherical harmonic of degree $n$. It is clear that $P\_n$ commutes with $\tau\_A$ for all $A\in\mathrm{SO}(3)$, so by \eqref{3}
$$
\tau\_A\left( P\_n f\right) = P\_n \tau\_Af= \lambda\_AP\_n f. $$
So, if a non-constant $f$ satisfying the property \eqref{3} existed, we could find a spherical harmonic $Y\_n=P\_n f$ of degree $n>0$ that satisfies the same property. But this cannot be, for in that case $Y\_n$ would span an irreducible subrepresentation of the space $\mathbb{Y}\_n$ of spherical harmonics of degree $n$ and we know that $\mathbb{Y}\_n$ is irreducible.
The only possibility is that $f$ is a constant, for in that case $P\_n f=0$ except for $n=0$. $\Box$
---
**CONCLUDING REMARK**. The answer provided above is unsatisfactory for my teaching purposes, since it is too technical and requires the introduction of several concepts, such as spherical harmonics and irreducible subrepresentations. Ideally, I would like an answer that is geometrical in nature and which directly exposes the role of the non-Abelianity of $SO(3)$.
| https://mathoverflow.net/users/13042 | Is there a non-constant function on the sphere that diagonalizes all rotations simultaneously? | For $\mathrm{C}^1$-functions, the argument can be reduced to the circle case: Write the sphere as a union of circles meeting at the poles. On each circle the considered function is equivariant under all rotations, therefore its restriction to the circle must be of the form $t\mapsto C\exp(i\lambda t)$ for some real numbers $C,\lambda$. Since the circles meet at the poles and the function is continuously differentiable, the $C$'s and $\lambda$'s agree for all the circles. To see that $\lambda$ must be zero, apply the argument again with a different choice of poles, for example a pair of antipodal points in the original equator.
| 27 | https://mathoverflow.net/users/58125 | 442129 | 178,416 |
https://mathoverflow.net/questions/442071 | 1 | [Crossposted](https://math.stackexchange.com/q/4651807/339790) on Mathematics SE
---
I've seen in many optimisation papers the statement that general non-convex optimisation problem is NP-hard. If we assume that non-convex optimisation is in NP class, it can be shown, as I remember, that some NP-complete problems can be relaxed to corresponding continuous non-convex optimisation problems, which makes general non-convex optimisation problem NP-complete. But we assumed that it is in NP class, which is not obvious for me — indeed, it would mean that solution of multi-extremum problem can be verified in polynomial time. But it seems to be too strong and even practically important statement, which makes me sad that I do not know this fact (honestly, some important problems would be simpler that we think if it holds). In short — can one show that solution of non-convex optimisation problem can be verified in polynomial time, or it is in general more complicated problem than any in NP?
| https://mathoverflow.net/users/479332 | Is non-convex optimisation really in NP class? | As noted in a [comment by Emil Jeřábek](https://mathoverflow.net/questions/442071/is-non-convex-optimisation-really-in-np-class#comment1140811_442071), $\mathsf{NP}$ is a class of decision problems, so on the face of it, an optimization problem cannot be in $\mathsf{NP}$ for the rather trivial reason that it is the wrong type of thing. Nevertheless, people often do talk about $\mathsf{NP}$-complete optimization problems; what they mean is that the *decision version* of the optimization problem is $\mathsf{NP}$-complete:
>
> $(\*)$ Given an instance of the problem and a number $k$, does there exist a solution with cost at most $k$?
>
>
>
For most problems, the $(\*)$ is *polynomially equivalent* to the problem of computing the optimal cost. Namely, if you can solve $(\*)$ in polynomial time, then you can compute the optimal cost by performing a binary search on the value of $k$. Furthermore, for most problems, $(\*)$ is almost trivially in $\mathsf{NP}$; to confirm a "yes" answer, someone can provide a solution with cost $\le k$, and you can compute the cost and confirm that the cost is indeed $\le k$. Convexity or non-convexity does not have much to do with it.
There are, however, some objections that you can raise. The recipe in the preceding paragraph only tells you how to compute the optimal *cost* (if you know how to solve the decision version), and does not tell you how to compute an optimal *solution*. This objection can be met in the case of problems that are *self-reducible*. I won't define self-reducibility formally, but for example, maybe your variables are all 0-1 variables, and if you set one of the variables equal to 0 or 1, what you get is another (slightly smaller) instance of the same type of problem. In that case, once you have the optimal cost $k$, you can compute a solution by choosing some variable, setting it to 0 (say), and asking whether the resulting smaller problem admits a solution with cost $\le k$. If so, then permanently set it to 0; otherwise, permanently set it to 1. Either way, you can then proceed to the next variable, and set the values of all the variables one by one. Again, convexity or non-convexity has little to do with it.
You could also object that being "polynomially equivalent" in the above sense is too coarse an equivalence relation. I referred to $(\*)$ as **the** decision version, but what about this?
>
> $(\dagger)$ Given an instance of the problem and a number $k$, is the optimum cost **exactly equal to** $k$?
>
>
>
Maybe in your opinion, it is $(\dagger)$ and not $(\*)$ that should be called **the** decision version of the problem. (This seems to be what you're getting at with your question about the optimum solution being verifiable in polynomial time.) While $(\dagger)$ and $(\*)$ are polynomially equivalent in the sense that if you can solve either one in polynomial time then you can solve the other in polynomial time, it does not immediately follow that $(\dagger)$ is in $\mathsf{NP}$ even if $(\*)$ is in $\mathsf{NP}$. And in fact, for many optimization problems, $(\dagger)$ is in [$\mathsf{DP}$](https://complexityzoo.net/Complexity_Zoo:D#dp) but not necessarily in $\mathsf{NP}$. So in this sense, you can question whether it is really appropriate to say that an optimization problem is in $\mathsf{NP}$ just because $(\*)$ is in $\mathsf{NP}$. However, once again, note that this subtlety really has nothing to do with convexity versus non-convexity.
| 4 | https://mathoverflow.net/users/3106 | 442139 | 178,420 |
https://mathoverflow.net/questions/442138 | -7 | Is the group of symmetries of the rectangle-not-square related to the Klein bottle mathematically?
The reason I am asking is because I want to put a Klein bottle coffee cup in a joke about V\_4 and wonder if this would be a good idea.
| https://mathoverflow.net/users/498715 | Is the Klein group related to the Klein bottle? | No, according to wikipedia ([bottle](https://en.wikipedia.org/wiki/Klein_bottle), [group](https://en.wikipedia.org/wiki/Klein_four-group)) they're both named after Felix Klein, but appeared in different papers on completely different topics. The bottle comes from his [notes](https://www.kleinbottle.com/The%20First%20Klein%20Bottle.html) from 1882 and the group in [Lectures on the icosahedron and the solution of equations of the fifth degree](https://archive.org/details/cu31924059413439/mode/2up).
Klein group is also a bit of a pun, since Klein means small in German.
| 3 | https://mathoverflow.net/users/22 | 442140 | 178,421 |
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