parent_url
stringlengths 37
41
| parent_score
stringlengths 1
3
| parent_body
stringlengths 19
30.2k
| parent_user
stringlengths 32
37
| parent_title
stringlengths 15
248
| body
stringlengths 8
29.9k
| score
stringlengths 1
3
| user
stringlengths 32
37
| answer_id
stringlengths 2
6
| __index_level_0__
int64 1
182k
|
|---|---|---|---|---|---|---|---|---|---|
https://mathoverflow.net/questions/433819 | 6 | I am trying to generalize the Gödel sentence as follows.
Define a pair of sentence $A$ and $B$ such that:
\begin{gather\*}
A := \lnot \operatorname{Prov}(\hat B) \\
B := \operatorname{Prov}(\hat A)
\end{gather\*}
where $\hat A$ and $\hat B$ are the Gödel numbers of $A$ and $B$ respectively.
From the definition above I derive the following equation $\hat A = \text{the Gödel number of $\lnot \operatorname{Prov}(\widehat{\operatorname{Prov}(\hat A)}) $}$. My question is for a minimal system that the incompleteness theorems apply, whether this equation always has a fixed point?
| https://mathoverflow.net/users/480476 | Generalize the Gödel sentence requires a fixed point theorem | Yes, in fact every such finite self-referential system has a fixed-point solution, and this can be proved using the same methods usually used to prove the unary fixed-point lemma. Such systems were explored at great length by Raymond Smullyan in several of his books. (But your desired example requires only the ordinary fixed-point lemma, which will produce a sentence $A$ that it is provably equivalent to $\neg\text{Prov}(\text{Prov}(A))$.)
You can also find an account in my expository article, [A review of the Gödel fixed-point lemma with generalizations and applications](http://jdh.hamkins.org/a-review-of-the-goedel-fixed-point-lemma-with-generalizations-and-applications/), which gives this fixed-point result as well as others — the Gödel fixed-point lemma, the finite-system fixed-point lemma, the Gödel-Carnap fixed-point lemma, the Kleene recursion theorem, with applications to Turing's theory of computable numbers and so on. In particular, Lemma 3 in that paper is double-fixed point lemma:
**Lemma.** (Double-fixed point). $\newcommand{\gcode}[1]{\ulcorner\!#1\!\urcorner}$ Suppose that $A(x,y)$ and $B(x,y)$ are two formulas in the language of arithmetic, then there are sentences $\phi$ and $\psi$ such that PA proves the equivalences
$$\phi\iff A(\gcode{\phi},\gcode{\psi})$$
and
$$\psi\iff B(\gcode{\phi},\gcode{\psi}).$$
**Proof.** Let $\text{Sub}$ be the binary substitution operator, the
primitive recursive function such that
$\text{Sub}(\gcode{\eta(x,y)},n,m)=\gcode{\eta(\kern1pt\underline{n},\kern1pt\underline{m}\kern1pt)}$. Let
$\theta\_1(x,y)=A(\text{Sub}(x,x,y),\text{Sub}(y,x,y))$ and
$\theta\_2(x,y)=B(\text{Sub}(x,x,y),\text{Sub}(y,x,y))$. Let
$n=\gcode{\theta\_1(x,y)}$ and $m=\gcode{\theta\_2(x,y)}$. Finally, let
$\phi=\theta\_1(\kern1pt\underline{n},\underline{m}\kern1pt)$ and $\psi=\theta\_2(\kern1pt\underline{n},\underline{m}\kern1pt)$.
Observe that
\begin{eqnarray\*}
\phi &\iff& \theta\_1(\kern1pt\underline{n},\underline{m}\kern1pt)\\
&\iff& A(\text{Sub}(\kern1pt\underline{n},\underline{n},\underline{m}\kern1pt),\text{Sub}(\kern1pt\underline{m},\underline{n},\underline{m}\kern1pt))\\
&\iff& A(\gcode{\theta\_1(\kern1pt\underline{n},\underline{m}\kern1pt)},\gcode{\theta\_2(\kern1pt\underline{n},\underline{m}\kern1pt)})\\
&\iff& A(\gcode{\phi},\gcode{\psi}).\\
\end{eqnarray\*}
Also observe
\begin{eqnarray\*}
\psi &\iff& \theta\_2(\kern1pt\underline{n},\underline{m}\kern1pt)\\
&\iff& B(\text{Sub}(\kern1pt\underline{n},\underline{n},\underline{m}\kern1pt),\text{Sub}(\kern1pt\underline{m},\underline{n},\underline{m}\kern1pt))\\
&\iff& B(\gcode{\theta\_1(\kern1pt\underline{n},\underline{m}\kern1pt)},\gcode{\theta\_2(\kern1pt\underline{n},\underline{m}\kern1pt)})\\
&\iff& B(\gcode{\phi},\gcode{\psi}),
\end{eqnarray\*}
as desired.
$\Box$
Note that we can arrange that $\phi$ and $\psi$ are distinct simply by ensuring that $\theta\_1(n,m)$ and $\theta\_2(n,m)$ are not syntactically the same sentence,
such as by replacing $\theta\_1(x,y)$ with its conjunction, if necessary, while ensuring that $\theta\_2(x,y)$ does not have such a
form.
The lemma easily generalizes to any size system and indeed, to infinite systems of fixed points.
| 14 | https://mathoverflow.net/users/1946 | 433822 | 175,463 |
https://mathoverflow.net/questions/433812 | 2 | Given a topological space $(X,\tau)$, recall that a cover $\mathcal{U}$ of $X$ is locally finite if for every point $x\in \mathcal{U}$ has a neighborhood $U$ that intersects finitely many elements of $\mathcal{U}$.
Instead, we call a cover $\mathcal{U}$ **finitely intersecting** if every member of $\mathcal{U}$ intersect finitely many elements of $\mathcal{U}$.
Recall that $X$ is paracompact if every open cover $\mathcal{U}$ has a locally finite open refinement. Just for the purpose of this question, let us call $X$ **strongly paracompact** if every open cover $\mathcal{U}$ has a finitely intersecting open refinement.
My question is: does this notion coincide with paracompactness?
If the answer is no, I would be very courious to know more about this property. For example:
* Has strong paracompactness been studied before, and what is the right name for it?
* Is it true that all metrizable second countable spaces are strongly paracompact?
| https://mathoverflow.net/users/121875 | A stronger version of paracompactness | Lemma: Let $(U\_{\alpha})\_{\alpha\in A}$ be a finitely intersecting open cover of a space $X$. Then there is some partition $P$ of $A$ where $(\bigcup\_{\alpha\in R}U\_\alpha)\_{R\in P}$ is a partition of $X$ into clopen sets and where each $R\in P$ is finite or countable.
Proof: Let $E$ be the smallest equivalence relation on $A$ where if $U\_\alpha\cap U\_\beta\neq\emptyset$, then $(\alpha,\beta)\in E$. Let $P$ be the partition associated with $E$. Then since $(U\_\alpha)\_{\alpha\in A}$ is finitely intersecting, each $R\in P$ is finite or countable. Clearly, $(\bigcup\_{\alpha\in R}U\_\alpha)\_{R\in P}$ covers $X$. Furthermore, if $R,S\in P,R\neq S$, then
$$(\bigcup\_{\alpha\in R}U\_\alpha)\cap(\bigcap\_{\beta\in S}U\_\beta)
=\bigcup\_{\alpha\in R,\beta\in S}(U\_\alpha\cap U\_\beta)=\emptyset.$$
Therefore, $(\bigcup\_{\alpha\in R}U\_\alpha)\_{R\in P}$ is a partition into open (and hence clopen) sets. $\square$
Proposition: Every connected strongly paracompact space is Lindelof.
Proof: Let $X$ be a connected strongly paracompact space. Let $\mathcal{U}$ be an open cover of $X$. Let $\mathcal{V}$ be a finitely intersecting open refinement of $\mathcal{U}$. By the above lemma, $\mathcal{V}$ must be countable or finite, so $\mathcal{U}$ has a countable subcover. $\square$
| 4 | https://mathoverflow.net/users/22277 | 433823 | 175,464 |
https://mathoverflow.net/questions/433815 | 6 | Suppose $X$ is a smooth closed oriented 4-manifold, and $\Sigma\_1,\Sigma\_2$ are smoothly embedded compact oriented surfaces in $X$. Suppose they intersect transversally at two points with different sings, so that their algebraic intersection is zero: $[\Sigma\_1]\cdot [\Sigma\_2]=0$. Then can we take other representatives $\Sigma\_1', \Sigma\_2'$ with $[\Sigma\_i']=[\Sigma\_i]\in H\_2(X;\Bbb Z)$ so that $\Sigma\_1'\cap \Sigma\_2'$ is empty?
Here is a special case. Let $X=\Bbb CP^2\sharp \Bbb CP^2$ with $H\_2(X;\Bbb Z)=H\_2(\Bbb CP^2;\Bbb Z)\oplus H\_2(\Bbb CP^2;\Bbb Z)=\Bbb Z\oplus \Bbb Z$. Consider the classes $(1,1)$ and $(1,-1)$. Their algebraic intersection is zero. Can we take disjoint surfaces representing these classes?
| https://mathoverflow.net/users/164671 | Two surfaces in a 4-manifold whose algebraic intersection number is zero | Yes, this can be done by tubing one surface along the other.
Suppose that you have two intersection points $p\_+, p\_- \in \Sigma\_1 \cap \Sigma\_2$ of opposite signs. Suppose also that $\Sigma\_1$ and $\Sigma\_2$ are connected (which we can assume, without loss of generality).
Choose a path $\gamma \subset \Sigma\_1$ connecting $p\_+$ to $p\_-$ and avoiding all other intersection points with $\Sigma\_2$. The restriction to $\gamma$ of the normal bundle to $\Sigma\_1$ is trivial, and we can choose a cylinder $C = [-1,1] \times D^2$ embedded in $X$ such that $C \cap \Sigma\_2$ is two disc neighbourhoods $D\_\pm$ of $p\_\pm$, which comprise the top and bottom of the cylinder, $\{\pm1\}\times D^2 \subset C$. Call $S$ the sides of the cylinder, i.e. $[-1,1]\times S^1$.
Now replace $\Sigma\_2$ with $\Sigma\_2' = \Sigma\_2 \setminus (D\_+ \cup D\_-) \cup S$ (and secretly smooth corners along the way). This removes two points of intersection from $\Sigma\_2 \cap \Sigma\_1$, and keeps the same homology classes. ($C$ itself gives a 3-chain whose boundary is $\Sigma'\_2 - \Sigma\_2$.
Note that this operation increases the genus. If you wanted to do this while keeping the genus, you can't always do it. This is related to the minimal genus problem in 4-manifolds.
You can do it if, for instance, you find enough Whitney discs (allowing you to perform the so-called [Whitney trick](https://en.wikipedia.org/wiki/Whitney_embedding_theorem)).
| 10 | https://mathoverflow.net/users/13119 | 433824 | 175,465 |
https://mathoverflow.net/questions/433733 | 4 | I have been recently studying different methods to construct Hopf algebras.
In Theorem IX.2.3 of "*Quantum groups*" by Kassel the **bicrossed product** of a pair of matched bialgebras (or Hopf algebras) $X$ and $A$ is defined as the vector space $X \otimes A$ with unit $1 \otimes 1$ and the following structure:
$$ (x \otimes a) (y \otimes b) = \sum\_{(a) (y)} x \cdot \alpha(a', y') \otimes \beta(a^{''},y^{''}) \cdot b $$
$$ \Delta(x \otimes a) = \sum\_{(a) (y)} (x' \otimes a') \otimes (x'' \otimes a'') $$
$$ \epsilon(x \otimes a) = \epsilon(x) \epsilon(a) $$
where $\alpha: A \otimes X \to X$ and $\beta: A \otimes X \to A$ are the corresponding actions. The bicrossed product is denoted by $X \bowtie A$.
In Theorem 6.2.2 of "*Foundations of quantum group theory*" by Majid the **bicrossproduct** is defined in an alternative way. Here, $X \bowtie A$ is $X \rtimes A$ as an algebra and $X \ltimes A$ as a coalgebra, where these are defined in Proposition 1.6.6. The product of $X \rtimes A$ and therefore the product of $X \bowtie A$ is defined as
$$ (x \otimes a) (y \otimes b) = \sum\_{(a)} x \cdot \alpha(a',y) \otimes a'' b $$
The coproduct is defined dually and depending on $\beta$.
I thought these two constructions were the same but were just given slightly different names. However, if $X = k[G]$ and $A = k[H]$ are group algebras with non-trivial $\beta$ then the first construction yields the group algebra $k[G \bowtie H]$, but the second does not as it is not cocommutative in general.
Both books refer back to "*Matched pairs of groups and bismash products of Hopf algebras*" by Takeuchi. Here, **bismash products** are defined similarly to the bicrossedproduct by Majid (pp. 845-847).
**Question:** Are the *bismash product by Takeuchi* and the *bicrosspuct by Majid* the same construction, but different to the *bicrossed product by Kassel*?
| https://mathoverflow.net/users/493533 | Bicrossed and bismash product of Hopf algebras | The two construction are similar but different. In the former, we deform the algebra structure with the two actions but preserve the coalgebra structure. For instance, if $G$ and $H$ are groups we have that $\mathbb{C}[G \bowtie H] \cong \mathbb{C}[G] \bowtie \mathbb{C}[H]$. In the second construction, we deform both, the algebra and the coalgebra structure and the previous isomorphism is no longer true. I suspect that Kassel defines $X \bowtie A$ as $X^\* \bowtie A$ using Majid's and Takeuchi's products.
| 2 | https://mathoverflow.net/users/493533 | 433825 | 175,466 |
https://mathoverflow.net/questions/433695 | 6 | In trying to understand the higher algebraic geometry of the stable homotopy category, one thing I've come across repeatedly is the claim that one should only consider the Balmer spectrum of a tt-category whose objects are all compact. One argument for this is that a certain fundamental result (Theorem 2.14 of [this paper](https://arxiv.org/pdf/1912.08963.pdf)) depends on dualizability; but dualizability is not equivalent to compactness, or even stronger/weaker than compactness. Moreover, the result in question only becomes problematic because there will be non-radical ideals in the non-rigid case, and it's essentially a tt-version of the Nullstellensatz. I don't have any issue with non-radical ideals, myself, and in fact they're quite important for deformation theory, so I don't find this argument very convincing.
The reason this has come up is that I've been comparing and contrasting some localizations and completions. One famous result, for example, is that the category of *perfect* complexes over a quasicompact quasiseparated scheme $X$ has Balmer spectrum $X$. However, over a Noetherian ring $R$ (I'll take $R=\mathbb{Z}$ for concreteness), the same is true for the full derived category if we replace thick subcategories by localizing subcategories. The difference is that the irreducible thick subcategories of $D(\mathbb{Z})$ (which are to be thought of as the "residue class fields" at each prime tt-ideal) are the categories of (derived) $p$-*complete* complexes, whereas the irreducible localizing subcategories of $D^{\text{perf}}(\mathbb{Z})$ are the categories of $p$-*local* complexes. (Of course, p-completion doesn't live within the category of perfect complexes.)
This distinction arises, in particular, when we try to compare the tt-geometry of $D(\mathbb{Z})$ and the stable homotopy category. I recently [asked](https://mathoverflow.net/questions/431711/what-is-the-balmer-spectrum-of-the-p-complete-stable-homotopy-category) about the Balmer spectrum of $\operatorname{Sp}\_p^{\wedge}$, since I haven't been able to find any information about this but wanted to compare it to $D(\mathbb{Z})$ as is done (slightly less formally) in Barthel and Beaudry's chapter of the Handbook. I was told, once again, that it's problematic to apply the Balmer spectrum construction here; but, as discussed above, I don't see why that is.
Hence my question: does the Balmer spectrum really fail to describe the AG of stable symmetric monoidal infinity-categories containing non-compact objects? In particular, are there any specific examples of important results failing in this context?
EDIT: As Brian Shin pointed out to me, the most obvious difference is that one needs to replace thick subcategories with localizing subcategories.
| https://mathoverflow.net/users/158123 | How does the Balmer spectrum fail to describe the algebraic geometry of categories of non-compact objects? | After some discussion with Brian and reading the papers referenced by Balmer's survey, I realized what's going on here. The problem is not that the theory fails for "big" categories. Rather, it's that there are two possible theories: taking the thick primes of $C^{\omega}$, and taking the smashing primes of $C$. While we can exhibit the former as a retract of the latter (intersect a smashing prime with the compact objects/take the smashing subcategory generated by a thick prime), they are not in general equivalent. In fact, for the stable homotopy category, the question of whether they're the same is actually equivalent to the Telescope Conjecture. (Here the smashing primes are K(n)-localization, and the thick primes are finite localization.)
| 0 | https://mathoverflow.net/users/158123 | 433848 | 175,472 |
https://mathoverflow.net/questions/433847 | 5 | Let $ M $ be a compact connected manifold. The degree of symmetry of $ M $, denoted $ N(M) $, is the maximum of the dimensions of the isometry groups of all possible Riemannian structures on $ M $. See for example
<https://www.ams.org/journals/tran/1969-146-00/S0002-9947-1969-0250340-1/S0002-9947-1969-0250340-1.pdf>
Two metrics are considered to be equivalent if they are isometric up to a constant multiple. (see comment from Robert Bryant)
I'm interested in manifolds $ M $ for which there is a unique up to equivalence metric with isometry group of dimension $ N(M) $. My guess is that there is always such a unique metric for manifolds of the form $ G/H $ for $ G $ a compact connected simple Lie group and $ H $ a closed subgroup. Moreover I would imagine that this unique up to equivalence metric is just the pushforward of the unique up to scaling biinvariant metric on the compact connected simple Lie group $ G $.
I believe all spheres $ S^n, n \geq 2 $ have this property. And the unique maximum symmetry metric is the round metric.
What about the manifold $ M=\mathbb{C}P^n $ of real dimension $ 2n $? Is it the case that
$$
N(\mathbb{C}P^n) =n(n+2)
$$
And moreover is it true that every metric on $ \mathbb{C}P^n $ whose isometry group has maximum dimension must be equivalent to the Fubini-Study metric?
Edit: Ok so the guess about spaces $ G/H $ was a little ambitious and as Robert Bryant points out it is wrong. My new guess is that a unique up to equivalence metric exists for any irreducible compact symmetric space $ M $. (Edit: I originally left out "irreducible" which Robert Bryant pointed out in the comments makes this obviously false).
Ok I made this guess into a new a new question
[Unique maximum symmetry metric on irreducible compact symmetric space](https://mathoverflow.net/questions/433914/unique-maximum-symmetry-metric-on-irreducible-compact-symmetric-space)
| https://mathoverflow.net/users/387190 | Maximum symmetry metric on $ \mathbb{C}P^n $ | There's an easy counterexample to your guess: Let $M^6 = \mathrm{SU}(3)/\mathbb{T}^2$, where $\mathbb{T}^2\subset\mathrm{SU}(3)$ is the maximal torus (for example, the diagonal subgroup). In that case, there is a 3-parameter family of non-isometric metrics on $M^6$ that are invariant under $\mathrm{SU}(3)$, so they are not unique up to a constant (or even non-constant) scalar factor.
I imagine that $M^6$ does not carry a metric whose isometry group has dimension greater than $8=\dim\mathrm{SU}(3)$, but I don't have a proof handy.
On the other hand, it is true that any Riemannian metric on $\mathbb{CP}^n$ whose isometry group has dimension at least $n(n{+}2)$ must be isometric to a constant scalar multiple of the Fubini-Study metric. Here is one argument:
Suppose that a connected, compact group $G$ acts effectively and smoothly on $\mathbb{CP}^n$. Then, by averaging, there exists a $G$-invariant metric $g$. Moreover, since $H^2\_{dR}(\mathbb{CP}^n,\mathbb{R})\simeq\mathbb{R}$, it follows from the Hodge Theorem that there is a $g$-harmonic $2$-form $\omega$ that represents a generator of $H^2\_{dR}(\mathbb{CP}^n,\mathbb{R})$, and it is unique up to constant multiples. Since $G$ is connected, it follows that it must leave $\omega$ fixed. Moreover, because of the structure of the cohomology ring of $\mathbb{CP}^n$, the top-degree form $\omega^n$ must represent a generator of $H^{2n}\_{dR}(\mathbb{CP}^n,\mathbb{R})$. In particular, $\omega^n$ does not vanish identically.
Thus, there is a point $p\in\mathbb{CP}^n$ such that $\omega\_p\in \Lambda^2(T^\*\_pM)$ is a 2-form of full rank. Consider the stabilizer $G\_p\subset G$ of $p$. Since $G$ acts by isometries and $\mathbb{CP}^n$ is connected, $G\_p$ injects into $\mathrm{O}(T\_pM)$ by identifying $g\in G\_p$ with $g'(p):T\_pM\to T\_pM$. Moreover, $G\_p$ leaves $\omega\_p$ fixed. Thus, $G\_p$ must lie inside a subgroup of $\mathrm{O}(T\_pM)$ that fixes a complex structure $J:T\_pM\to T\_pM$ and hence must have dimension at most $\dim \mathrm{U}(n) = n^2$. Now, we have
$$
\dim G = \dim G\_p + \dim G/G\_p = \dim G\_p + \dim G{\cdot}p \le n^2 + 2n = n(n{+}2).
$$
If equality holds, then $\dim G\_p = n^2$ and $\dim G{\cdot}p = 2n = \dim \mathbb{CP}^n$. Thus, the orbit $G{\cdot}p$ is both open and closed in $\mathbb{CP}^n$, so $G$ acts transitively on $\mathbb{CP}^n$. It follows that $\omega$ is everywhere of full rank and, after scaling $\omega$ so that it has comass 1, we have that $\omega(u,v) = g(Ju,v)$ for a unique almost-complex structure $J$ on $\mathbb{CP}^n$ that is preservd by $G\_p$, which has the same dimension as the connected group $\mathrm{U}(g\_p,J\_p)\simeq \mathrm{U}(n)$. Thus, $G\_p = \mathrm{U}(g\_p,J\_p)$. Since $G\_p$ contains $-I\in\mathrm{U}(g\_p,J\_p)$, it follows that there is an element of $G$ that fixes $p$ and reverses all $g$-geodesics through $p$. Since $G$ acts transitively on $\mathbb{CP}^n$, it follows that $(\mathbb{CP}^n,g)$ is a Riemannian symmetric space. Using the classification, it follows that $G\simeq \mathrm{SU}(n{+}1)/Z$ (where $Z\simeq\mathbb{Z}\_{n+1}$ is the center of $\mathrm{SU}(n{+}1)$) and that the metric $g$ is, up to isometry, a constant multiple of the standard Fubini-Study metric.
| 13 | https://mathoverflow.net/users/13972 | 433850 | 175,473 |
https://mathoverflow.net/questions/433593 | 2 | Consider a union-closed family $\mathcal{F}$ of $n=\vert \mathcal{F} \vert$ sets, $n$ odd, and its family $\mathit{J}(\mathcal{F})$ of $m = \vert\mathit{J}(\mathcal{F})\vert$ basis sets (or $\cup$-irreducible sets, also called generators), and define $\mathcal{F\_a} = \{A \in \mathcal{F}: a \in A \}$, $\mathit{J\_a}(\mathcal{F}) = \{A \in \mathit{J}(\mathcal{F}) : a \in A \}$.
It is easy to prove by contradiction that if $\exists a$ such that $\vert\mathit{J\_a}(\mathcal{F})\vert \ge m-\Bigl\lfloor\log\_2{\frac{n+1}{2}}\Bigr\rfloor$, then $\vert \mathcal{F\_a} \vert \ge \frac{n}{2}$.
I have tried the two examples in ["The journey of the union-closed sets conjecture"](https://arxiv.org/abs/1309.3297) by Bruhn and Schaudt on section 6.3 "Limits of averaging" and both seem to satisfy the above requirement.
I understand it might not be easy, but any idea on how to construct an example of a union-closed family with average set size less than half of the universe and such that $\forall a$ $\vert\mathit{J\_a}(\mathcal{F})\vert \lt m-\Bigl\lfloor\log\_2{\frac{n+1}{2}}\Bigr\rfloor$?
| https://mathoverflow.net/users/136218 | How to find an example of a union-closed family with two given properties | Not a good example, but it might hint for a better one.
Take two integers $N=2k+1,M$. Let the basis sets be $\{x,x+1,...,x+k\},1\leq x\leq N$, taken modulo $N$ (identify $N\equiv 0$) and $\{1,2,...,N,a\_1,a\_2,...,a\_M\}$.
The union-closed family $\mathcal{F}$ contains all set of the form $\{x,x+1,...,x+l\},1\leq x\leq N,k\leq l\leq N-1$ and $\{1,2,...,N,a\_1,a\_2,...,a\_M\}$. So it's easy to see that $m=N+1,n=\frac{N(N-1)}{2}+2,J\_i(\mathcal{F})=\frac{N+1}{2},J\_{a\_j}(\mathcal{F})=1$. The average set size is $f(N)+\frac{M}{\frac{N(N-1)}{2}+2}$ ($f(N)$ is quite complicative to write, but it's independent of $M$).
Now we choosing $N,M$ such that $\frac{N-1}{2}$ is odd, $log\_2(\frac{N(N-1)}{2}+2)<\frac{N+1}{2}, f(N)+\frac{M}{\frac{N(N-1)}{2}+2}<\frac{N+M}{2}$ then $\mathcal{F}$ is the union-close family that we want.
| 2 | https://mathoverflow.net/users/432274 | 433861 | 175,478 |
https://mathoverflow.net/questions/433854 | 2 | I have recently run into a number of divergent oscillating integrals in various contexts. Thus, I have been led to desire general methods for assigning values to divergent oscillating integrals. All of the integrals I am interested in have the following form
$$ \int\_0^\infty f(x) \sin(x) dx \text{ or } \int\_0^\infty f(x) \cos(x) dx $$
Where $f(x)$ is usually an eventually monotonically increasing function.
##### Background
In the case where $f(x)$ grows like a polynomial I believe there are various approaches that all provide the same value. For instance, consider
$$
\begin{split}
\int\_0^\infty \sin(x) dx &= \frac{1}{2}\int\_0^\infty \left(\sin(x) + \sin(x)\right) dx \\
&= \frac{1}{2}\int\_0^{\pi} \sin(x)dx + \frac{1}{2}\int\_0^\infty \left(\sin(x) + \sin(x + \pi)\right) dx = 1.
\end{split}$$
Alternatively, in an analgous manner to applying a smooth cutoff function to a divergent series, we can also apply a smooth cutoff function to get
$$ \lim\_{\varepsilon\to 0} \int\_0^\infty \sin(x) (1-\varepsilon)^{1+x} dx = 1$$
An analogue to Cesaro summation for integrals also provides the same values for this integral.
We can apply these method to higher powers of $x$ as well. Doing this generates the following values
$$ \int\_0^\infty x^n \sin(x) = \cos\left( \frac{n \pi}{2}\right) \Gamma(n+1), \int\_0^\infty x^n \cos(x) = -\sin\left( \frac{n \pi}{2}\right) \Gamma(n+1)$$
Thus, if $f(x)$ has a power series presentation, we can write
$$\int\_0^\infty f(x) \sin(x) = \int\_0^\infty \sum\_{n=0}^\infty \frac{f^{n}(0)}{n!} x^n \sin(x) = \sum\_{n=0}^\infty (-1)^n f^{2n}(0)$$
This formula is a good start, but the series typically doesn't converge.
**Is there a general way to assign a value to divergent oscillating integrals, especially those where $f(x)$ grows at an exponential rate or faster?** Since it seems unlikely that there might be a way to assign a value to a general function-- are there intereting/broad categories of functions which can be assigned values?
| https://mathoverflow.net/users/146528 | Assigning values to divergent oscillating integrals | There is indeed such a method which is elementary and allows one to rigorously give a numerical value to such integrals, in particular, the value $1$ to $\int\_0^\infty \cos x\, dx$. This uses two facts:
1. Every continuous function has a primitive. In your case this is $\cos x$.
2. About 60 years ago, a notion of the limit of a distribution at a point (in your case at $0$ and $\infty$) was developed in an elementary context (i.e., at thelevel of a good calculus or real single variable
course).
Combining these concepts in the usual manner as in defining improper Riemann integrals, one obtains a notion of definite integrals of distributions which contains the classical ones and also ones of the type mentioned in your query.
The details can be found in many places—one of the most easily accessible ones (also in the mathematical sense) is the lecture notes on the theory of distributions by J. Sebastião e Silva
which are available (in english) at the site
jss100.campus.ciencias.ulisboa.pt.
Under the section
publicações
you will find
Vol. 3 Theory of Distributions,
which gives an introduction to distributions at sophomore level (without using functional analysis).
The relevant definitions and concepts for your query can be found in
Chapter 4–Limits and Integrals of Distributions.
| 1 | https://mathoverflow.net/users/494111 | 433868 | 175,481 |
https://mathoverflow.net/questions/433865 | 10 | I am not an expert in Differential Topology, so let me apologize if this question admits a straightforward answer. I checked some standard references, but I could not find one.
Let $M$ be a smooth $n$-manifold with boundary $N:= \partial M$, and assume that there exists a *continuous* retraction of $M$ onto $N$, namely, a continuous map $r \colon M \to N$ that is the identity on $N$.
>
> **Question.** Does there exist a *smooth* retraction $s \colon M \to N$? And what about a smooth retraction homotopic to $r$?
>
>
>
I am aware that, by Whitney Approximation Theorem, $r \colon M \to N$ is homotopic to a smooth map, but I do not see how to conclude from this that it is homotopic to a smooth retraction.
**Motivation.** This question arose when I was trying to mimic the standard proof of Brouwer Fixed Point Theorem (for *continuous* maps $f \colon \mathbb{D}^n \to \mathbb{D}^n$) by using de Rham cohomology, instead of singular homology. In the homology setting, the starting point is the non-existence of a continuous retraction $r \colon \mathbb{D}^n \to S^{n-1}$, which is obtained by contradiction looking at the functorial group homomorphisms induced by $r$ and by the inclusion $i \colon S^{n-1} \to \mathbb{D}^n$. But de Rham cohomology is only functorial with respect to *smooth* maps...
| https://mathoverflow.net/users/7460 | Is every retraction homotopic to a smooth retraction? | Using a collar of the boundary we resort to the case when $r\in C^0(M, \partial M)$ is given by the projection $\partial M\times I \to \partial M$ over the collar .
Since $r$ is smooth on an open neighbourhood of $\partial M$, for any $\epsilon>0$ we can find a smooth $g\in C^\infty(M,\partial M)$ such that $g=r$ in a neighbourhood of $\partial M$ and $|g-r|<\epsilon$.
For a reference see Prop. 3.11 of Milnor's lectures on Differential Topology 1958 (notes taken by James Munkres).
| 10 | https://mathoverflow.net/users/158806 | 433875 | 175,483 |
https://mathoverflow.net/questions/433876 | 1 | Let
* $X$ be a metric space,
* $\mathcal M(X)$ the space of all finite **signed** Borel measures on $X$, and
* $\mathcal C\_b(X)$ be the space of real-valued bounded continuous functions on $X$.
Then $\mathcal C\_b(X)$ is a real Banach space with supremum norm $\|\cdot\|\_\infty$. We endow $\mathcal M(X)$ with the total variation norm $[\cdot]$. Then $(\mathcal M(X), [\cdot])$ [is a Banach space](https://math.stackexchange.com/questions/4440407/the-space-of-finite-signed-measures-with-the-total-variation-norm-is-a-banach-sp). Let $\mathcal M(X)^\* := (\mathcal M(X))^\*$ be the continuous dual. Let $\mu\_n,\mu \in \mathcal M(X)$.
* We define the first type of weak convergence by
$$
\mu\_n \overset{1}{\rightharpoonup} \mu \overset{\text{def}}{\iff} \int\_X f \mathrm d \mu\_n \to \int\_X f \mathrm d \mu \quad \forall f \in \mathcal C\_b(X),
$$
Let $\sigma(\mathcal M(X), \mathcal C\_b(X))$ be the topology induced by $\overset{1}{\rightharpoonup}$.
* We define the second type of weak convergence by
$$
\mu\_n \overset{2}{\rightharpoonup} \mu \overset{\text{def}}{\iff} \varphi(\mu\_n) \to \varphi (\mu) \quad \forall \varphi \in \mathcal M(X)^\*,
$$
Let $\sigma(\mathcal M(X), \mathcal M(X)^\*)$ be the topology induced by $\overset{2}{\rightharpoonup}$.
Of course, we have $\mu\_n \overset{2}{\rightharpoonup} \mu \implies [\mu] \le \liminf\_n [\mu\_n]$. Also, we can prove [that](https://math.stackexchange.com/questions/4568256/does-mu-n-overset1-rightharpoonup-mu-necessarily-imply-mu-n-overset) $\mu\_n \overset{1}{\rightharpoonup} \mu \implies [\mu] \le \liminf\_n [\mu\_n]$.
>
> Are there some conditions (locally compact, separable, Polish,...) on $X$ that ensure [$\mu\_n \overset{1}{\rightharpoonup} \mu \implies \mu\_n \overset{2}{\rightharpoonup} \mu$] or [$\mu\_n \overset{2}{\rightharpoonup} \mu \implies \mu\_n \overset{1}{\rightharpoonup} \mu$]?
>
>
>
Thank you so much for your elaboration!
---
I posted [this](https://math.stackexchange.com/questions/4568256/does-mu-n-overset1-rightharpoonup-mu-necessarily-imply-mu-n-overset) question on MSE, but it seems to receive no answer. So I post it here.
| https://mathoverflow.net/users/99469 | Are there some conditions on a metric space $X$ such that these two types of weak converge of finite signed Borel measures on $X$ are related? | An example. $X = [0,1]$ with the usual metric. $\mathcal C[0,1] = \mathcal C\_b[0,1] = \mathcal C\_0[0,1]$. $\mathcal C[0,1]^\* = \mathcal M[0,1]$.
Let $\mu\_n$ be the unit point-mass at $1/n$ and $\mu$ the unit point-mass at $0$. Show $\mu\_n \overset{1}{\rightharpoonup} \mu $ is true but $\mu\_n \overset{2}{\rightharpoonup} \mu$ is false.
Whatever "some conditions" to insure $\big[\mu\_n \overset{1}{\rightharpoonup} \mu \implies \mu\_n \overset{2}{\rightharpoonup} \mu\big]$
are, they are not satisfied by $[0,1]$.
| 4 | https://mathoverflow.net/users/454 | 433880 | 175,485 |
https://mathoverflow.net/questions/433874 | 2 | Let $G$ be a compact (connected) semisimple Lie group. Let $G\_\mathbb{C}$ be the complexification of $G$.
Is $G$ a maximal compact subgroup of $G\_\mathbb{C}$?
| https://mathoverflow.net/users/172459 | Question about maximal compact subgroups of Lie groups | $\DeclareMathOperator\Lie{Lie}\newcommand\g{\mathfrak g}\newcommand\C{{\mathbb C}}$$\g = \Lie(G)$ is maximal among subalgebras of $\g\_\C = \Lie(G\_\C)$ on which the Killing form is negative definite, so $G$ is a maximal connected, compact subgroup of $G\_\C$. According to [@YCor](https://mathoverflow.net/questions/433874/question-about-maximal-compact-subgroups-of-lie-groups#comment1117360_433874), maximal compact subgroups of connected groups are connected, so $G$ is also a maximal compact subgroup.
| 5 | https://mathoverflow.net/users/2383 | 433885 | 175,487 |
https://mathoverflow.net/questions/433839 | 8 | EDIT: in this question, I was proposing a conjecture, Prop. 1. Fedor Pakhomov showed a counter-example. [In this new question](https://mathoverflow.net/questions/433954/computational-complexity-and-commuting-functions-improved-conjecture) I propose a slightly weaker conjecture that holds even for that example and seems to be still hard to prove. There is also an [older version](https://math.stackexchange.com/questions/4557690/computational-complexity-and-commuting-functions).
We have two functions:
$$
f: \{0,1\}^\* \to \{0,1\}^\*
$$
$$
g: \{0,1\}^\* \to \{0,1\}^\*
$$
that commute:
$$
f[g(x)] = g[f(x)]
$$
These two functions can be calculated in polynomial time (in the length of the input). Moreover, the outputs have the same length of the inputs: $|f(x)| = |x|$ and $|g(x)| = |x|$ .
A trivial example of functions that commute can be easily constructed by splitting the strings into two parts and defining:
$$
f(x,y) = ( h(x), y )
$$
and
$$
g(x,y) = ( x, l(y) )
$$
where the functions $h(x)$ and $l(y)$ can be calculated in polynomial time (in their inputs).
I was able to construct slightly more complex examples, but not much more complex. In all the examples, the evolution obtained by repeatedly applying $f$ seems to be independent of the evolution obtained by repeatedly applying $g$. More rigorously, in my examples, the following proposition holds:
**Proposition 1**
>
> There are two functions $n'$ and $m'$ depending on $n$ and $m$,
> at most polynomial, such that there is an algorithm that, for any
> integers $n$ and $m$, calculates the
> function $f^n[g^m(x)]$, operating in polynomial
> time (in the length of its input), and taking the following
> inputs: the binary representations of the numbers $n$ and $m$,
> $f^{n'}(x)$, and $g^{m'}(x)$.
>
>
>
**Important note** The expression $f^n$ means $f$ applied $n$ times. For example $f^2(x)$ means $f[f(x)]$, $f^3(x)$ means $f\{f[f(x)]\}$.
I remark that this happens *even if $n$ and $m$ increase exponentially in $|x|$.*
In the trivial example above, setting $n'=n$ and $m'=m$, we see that $f^n(x)= ( h^n(x), y )$ and $g^m(x) = (x, l^m(y) )$, from which it is easy to calculate $f^n[g^m(x,y)] = ( h^n(x), l^m(y) ) $.
The question is: is Prop. 1 a general theorem? Alternatively, is there a counter-example to Prop. 1?
Thanks to comments already received, I know that Prop. 1 holds for sure in the following cases:
1. if $f=h^a$ and $g=h^b$ (with $a$ and $b$ two natural numbers);
2. if $f^n$ can be calculated in polynomial time in the size (number of bits) of $n$.
Maybe the question is too difficult to be answered; thus any help is welcome.
| https://mathoverflow.net/users/138060 | Computational complexity and commuting functions | There is a counterexample to Proposition 1 iff $\mathsf{P}\ne\mathsf{PSPACE}$. The idea is to make a pair $f,g$ such that on certain inputs iterations of them individually are trivial, but their combination performs computation of a deciding algorithm for some $\mathsf{PSPACE}$-complete problem.
If $\mathsf{P}=\mathsf{PSPACE}$, then since $f^n(g^m(x))$ could be computed in polynomial space from $x$, we would be able to compute $f^n(g^m(x))$ in polynomial time.
Further assume that $\mathsf{P}\ne \mathsf{PSPACE}$. Let $L\subseteq \{0,1\}^{\star}$ be some $\mathsf{PSPACE}$-compete problem. Let $T$ be a Turing machine and $P(x)$ be a polynomial such that $T$ checks if $\alpha\in L$ using at most $P(|\alpha|)$ cells of the tape. For an appropriate polynomial $Q(x)$ that is strictly monotone as a function $\mathbb{N}\to\mathbb{N}$, we could naturally code all possible states of $T$ for computations on inputs $\alpha\in \{0,1\}^n$ as strings of the length $Q(n)$. Let $h$ be a polynomial time function preserving the lengths of strings such that whenever it maps codes of states of $T$ as above to the codes of states of $T$ after one step of computation (and we don't care what happen with the strings that are not codes as long as we preserve their lengths). For $\alpha,\beta,\gamma\in\{0,1\}^{Q(n)}$ let $\alpha'=\min(\alpha+1,2^{Q(n)}-1)$ and $\beta'=\min(\beta+1,2^{Q(n)}-1)$, where we treat strings of the length $Q(n)$ as codes for numbers $<2^{Q(n)}$, we put
$$f(\alpha\beta\gamma)=\alpha'\beta h^{\min(\alpha',\beta)-\min(\alpha,\beta)}(\gamma)\text{ and }g(\alpha\beta\gamma)=\alpha\beta'h^{\min(\alpha,\beta')-\min(\alpha,\beta)}(\gamma).$$
Clearly, $$f(g(\alpha\beta\gamma))=g(f(\alpha\beta\gamma)=\alpha'\beta'h^{\min(\alpha',\beta')-\min(\alpha,\beta)}(\gamma).$$
We don't care about the behavior's of $f,g$ on inputs of other forms (as long as we have commutation and length preservation). Since $\min(\alpha',\beta)-\min(\alpha,\beta)$ and $\min(\alpha,\beta')-\min(\alpha,\beta)$ are always either $0$ or $1$, we could make $f$ and $g$ polynomial time computable.
Assume for a contradiction that there is a polynomial time algorithm prescribed by Proposition 1. Let $u$ be the polynomial time function mapping $\alpha\in\{0,1\}^n$ to $u(\alpha)\in\{0,1\}^{Q(n)}$ that codes the initial state of $T$ for the computation on the input $\alpha$. Now for appropriate polynomial time $n'(x)$ and $m'(x)$ we should be able to compute in polynomial time $f^{2^{Q(|\alpha|)}}(g^{2^{Q(|\alpha|)}}(00u(\alpha)))$ from $2^{Q(|\alpha|)}$, $f^{n'(2^{Q(|\alpha|)})}(00u(\alpha))$ and $g^{m'(2^{Q(|\alpha|)})}(00u(\alpha))$; here I am abusing the notation and by $00u(\alpha)$ I mean the string of the length $3Q(n)$ coding triple consisting of $0,0$, and $u(\alpha)$. Clearly, $$f^{2^{Q(|\alpha|)}}(g^{2^{Q(|\alpha|)}}(00u(\alpha)))=(2^{Q(|\alpha|)-1})(2^{Q(|\alpha|)-1})h^{2^{Q(|\alpha|)}-1}(u(\alpha)),$$ $$f^{n'(2^{Q(|\alpha|)})}(00u(\alpha))= (\min(2^{Q(|\alpha|)-1},n'(2^{Q(|\alpha|)}))(0)u(\alpha)\text{, and}$$
$$g^{m'(2^{Q(|\alpha|)})}(00u(\alpha))= (0)(\min(2^{Q(|\alpha|)-1},m'(2^{Q(|\alpha|)}))u(\alpha).$$
Hence we could compute $h^{2^{Q(|\alpha|)}-1}$ in polynomial time from $\alpha$. But since $T$ on the input $\alpha$ terminates after at most $2^{Q(|\alpha|)}-1$ steps (simply due to the number of possible distinct states), $h^{2^{Q(|\alpha|)}-1}$ will always be the code of the terminal state of the computation of $T$ on the input $\alpha$. Hence we would be able to decide the problem $L$ in polynomial time. Contradiction.
| 5 | https://mathoverflow.net/users/36385 | 433887 | 175,488 |
https://mathoverflow.net/questions/433835 | 6 | Let $X$ be a metrizable topological space, $A\subseteq X\times X$ a nonempty closed subset which is reflexive, symmetric and transitive, $d:A\to \mathbb{R}\_+$ a continuous function that satisfies the metric axioms. Does there exist a metric on $X$ which extends $d$ and defines the topology on $X$?
If it makes any difference I can suppose $X$ locally compact second countable.
Edit: As explained in the comments, one needs to add the condition that $d$ defines the topology on $A$. If one restricts to $X$ compact then all such subtleties disappear, and continuity of an extension of $d$ suffices to ensure that it defines the topology.
Edit2: To clarify further the question.
There are two variants of the questions. For both we suppose $A$ is a closed equivalence relation, $d:A\to \mathbb{R}\_+$ continuous and satisfies the metric hypothesis.
Q1: Does there exists an extension $d:X\times X\to \mathbb{R}\_+$ which is continuous and a metric.
Q2: Does there exists an extension $d:X\times X\to \mathbb{R}\_+$ which is a metric and defines the topology on $X$.
Remark that if $X$ is compact Q1 and Q2 are the same.
Q2 is false without further hypothesis on $d$. The further hypothesis is as follows. Since $X$ is metrizable, it can be equipped with the notion of Cauchy sequence (depends on the choice of a metric). The additional hypothesis is that there exists a metric on $X$ such that if $x\_n$ is a sequence with $(x\_n,x\_m)\in A$ for all $n,m$, then $x\_n$ is Cauchy for the metric on $X$ if and only if it is Cauchy for the metric $d:A\to \mathbb{R}\_+$.
I am interested in both Q1 and Q2.
| https://mathoverflow.net/users/105656 | Extending a partially defined metric on a metrizable space | Here is a counterexample to Q2, with your stated extra condition.
Let $X$ consist of the half-open unit interval $(0,1]$ on the $x$-axis in the plane, together with the full unit interval $[0,1]$ at height $1$. That is,
$$X=B\sqcup T$$
where $B$ is the bottom line segment, having points of the form $(x,0)$ for $0<x\leq 1$, and $T$ is the top line segment, having points of the form $(x,1)$, where $0\leq x\leq 1$. We put the ordinary Euclidean metric on $X$ as embedded in the plane.
Let $A$ be the set of pairs arising from the equivalence relation relating each point on $B$ to the corresponding point on $T$ vertically above. So $A$ has the trivial pairs and the vertical pairs. This is a closed set in $X\times X$, since a convergent sequence of vertical pairs converges to a vertical pair. Note that the point $t=(0,1)$ at upper left is not part of any nontrivial vertical line in $X$, and appears in $A$ only reflexively.
Define $d$ on $A$ so that the vertical distance on the right side is $1$, but the vertical distances decay continuously to $0$ as one approaches the left side (and trivial distances are $0$). This is a continuous function on $A$ in the space $X\times X$, and it obeys the metric conditions on $A$ (note that there are no nontrivial triangles occuring in $A$).
This set $A$ trivially fulfills your Cauchy sequence condition, since if $x\_n$ is a sequence of points and $(x\_n,x\_m)\in A$ for all $n,m$, then every two points of the sequence must either be identical or vertically related, and so the sequence has at most two points on it. So it is Cauchy with respect to $d$ if and only if it is Cauchy in $X$.
But there can be no metric on $X$ extending $d$ and generating the topology of $X$, since the upper-left point $t$ is not in the closure of $B$ in $X$, but in any metrization of $X$ the point $t$ would have other points on $T$ that were close to points vertically below in $B$, according to $d$, so every neighborhood of $t$ would have to contain points in $B$. Contradiction.
| 3 | https://mathoverflow.net/users/1946 | 433888 | 175,489 |
https://mathoverflow.net/questions/433881 | 4 | There is a somewhat forgotten sieve-theoretic approach to the Goldbach conjecture, due to Buchstab et al, see e.g. pp.247-248 of R.D. James.
On p.247, James defines some function $F$ such that for any fixed $a \in \mathbb{N}$ and even $x \geq 6$:
1. $F(x ; 2, a, 1) = F(x; 2)$ with $a=1$ is the number of positive integers $n \leq x$ such that $n \equiv a\pmod{2}$. Thus $F(x; 2) = x/2$.
2. $F(x; 2, √x, a)=F(x; 2, √x)$ with $a=1$ is the number of odd positive integers $n<x$ (without double counting $n$ and $x-n$), such that each of $n$ and $x-n$ is either a prime or equal to 1. Thus if it could be shown that $F(x; 2, √x) \geq 2$, it would follow that there exists at least one representation $x= n+(x-n)$ whereby each of $n$ and $x-n$ is either a prime or equal to 1. Thus if $x-1$ is composite, it would suffice to show that $F(x; 2, √x) = F(x; 2, √x) \geq 1$.
On the bottom of p.248, James states that
$$
F(x; 2, √x) = F(x; 2) - 2\sum\_{r=1}^{k} F(x; 2p\_r, p\_{r-1}) = x/2 - 2\sum\_{r=1}^{k} F(x; 2p\_{r}, p\_{r-1})
,$$
where $p\_i$ denotes the $i$-th odd prime $\leq √x$. T. Kubalalika, in his preprint [2], lets $6 \leq x \equiv 2\pmod{4}$ where $x-1$ is composite. Now suppose that $x$ is a counterexample to Goldbach, so that $F(x; 2, √x)=0$. Putting this into the above equality gives
$$
x/2 = 2\sum\_{r=1}^{k} F(x; 2p\_r, p\_{r-1}),
$$
contradicting the fact that $x/2$ is odd. One therefore deduces that if $x\equiv 2\pmod{4}$ and $x-1$ is composite, then $x$ is a sum of two primes.
My question is, given the strength of Buchstab et al's sieve (as evidenced by how easily it leads to a proof of the above result), are there any modern improvements to it, such that it could possibly lead to even more powerful results ? A quick Google search seems to suggest that the sieve became forgotten as soon as the Hardy-Littlewood circle method lead to Vinogradov's 3-primes theorem.
**References**
[1] R. D. James, "[Recent progress in the Goldbach problem](https://www.ams.org/journals/bull/1949-55-03/S0002-9904-1949-09180-2/S0002-9904-1949-09180-2.pdf)" Bulletin of the American Mathematical Society 55, 246-260 (1949), [MR0028893](https://mathscinet.ams.org/mathscinet-getitem?mr=MR0028893), [Zbl 0034.02301](https://zbmath.org/?q=an%3A0034.02301).
[2] T. Kubalalika, "[On the binary Goldbach conjecture for certain even integers](https://figshare.com/articles/preprint/On_the_binary_Goldbach_conjecture/21342042)", figSHARE preprint.
| https://mathoverflow.net/users/493772 | On Buchstab et al's "forgotten" sieve and the Goldbach conjecture for certain integers | The result (aka Buchstab's identity) you mentioned is not forgotten. In modern sieve theory texts such as Halberstam & Richert's *Sieve Methods* and Friedlander & Iwaniec's *Opera de Cribro*, the identity is written as
$$
S(\mathcal A,z)=S(\mathcal A,w)-\sum\_{w\le p<z}S(\mathcal A\_p,p)
$$
where $S(\mathcal A,z)$ counts the integers in $\mathcal A$ that are free of prime divisors $<z$.
One of its generalization (Kuhn's weighted sieve) is used to prove Chen's theorem. When $N$ is a positive even integer, $\mathcal A=\{N-p:p\le N\}$, we have
$$
\begin{aligned}
r\_{1,2}(N)&=\#\{p\le N:n-p\text{ prime or product of two primes}\} \\
&>S(\mathcal A,N^{1/10})-
\frac12\sum\_{N^{1/10}\le p<N^{1/3}}S(\mathcal A\_p,N^{1/10})-\frac\Omega2+O(N^{9/10})
\end{aligned}\tag1
$$
in which
$$
\Omega=\#\{p\le N:N-p=p\_1p\_2p\_3,N^{1/10}\le p\_1<N^{1/3}\le p\_2<(N/p\_1)^{1/2}\}.
$$
By evaluating the right hand side using Jurkat-Richert's theorem and Selberg's sieve, Chen found out that for large $N$ there is
$$
r\_{1,2}(N)>{0.67N\over\log^2N}\prod\_{2<p|N}{p-1\over p-2}\prod\_{p>2}\left(1-{1\over(p-1)^2}\right).
$$
| 13 | https://mathoverflow.net/users/449628 | 433890 | 175,490 |
https://mathoverflow.net/questions/433561 | 1 | Conilpotent coenveloping coalgebra UC(T) of a conilpotent Lie coalgebra T is defined by an universal property, similar to usual enveloping algebra: it's a coassocative, conilpotent coalgebra UC(T) such that category of conilpotent reps of UC(T) is equivalent to the category of conilpotent reps of T. It can be constructed explicitly as $$\operatorname{colim} \,(U(T^\*)/I^k)^\*$$
Here $U(T^\*)$ is a usual enveloping algebra of a dual complete pro-finite dimensional pronilpotent Lie algebra $T^\*$. $I$ is its augmentation ideal.
---
There's a canonical filtration on any coaugmented conilpotent coalgebra C, with terms being kernels of maps $\bar{\Delta}^k : C \to (C/1)^{\otimes k}$. Let's call that filtration $J\_k UC$
---
There's an increasing filtration $V'\_iT$ by Lie subcoalgebras on every conilpotent Lie coalgebra, corresponding to a lower central filtration by ideals on the dual (complete, ..., etc) Lie algebra. Inclusions of filtration terms give us a **possibly** different filtration $V\_i UC(T)$; its terms are images of homomorphisms $UC(V'\_i T) \to UC(T)$, obtained by functoriality of $UC$. (Sidenote: those homomorphisms are usually injective, at least when T is flat over base ring they are.)
If I'm not mistaken, it's always true that $J\_i \subseteq V\_i$.
*Q: Is it true that $J\_i = V\_i$?*
| https://mathoverflow.net/users/81055 | Two (or less) filtrations on coenveloping coalgebra | Yes, both of those filtrations should be equal, at least when the ground field has characteristic $0$. I don't know if there is a good standard reference in the literature, but I did have to grapple with the dual version of this problem in a paper I wrote a couple of years back (see the appendix of arXiv:1909.05734, especially sections A.3 and A.4.1).
The result we ended up proving there concerned pro-nilpotent Lie algebras, i.e. Lie algebra objects in $\mathrm{pro}{-}\mathbf{Vec}$ (the pro-category of finite-dimensional vector spaces) whose descending central series is separated. Since $\mathrm{pro}{-}\mathbf{Vec}$ is dual to the category of all vector spaces, we know that the category of pro-nilpotent Lie algebras is dual to the category of conilpotent Lie coalgebras. Every pro-nilpotent Lie algebra $L$ has an associated completed universal enveloping algebra $U(L)$, which is a cocommutative Hopf algebra object in $\mathrm{pro}{-}\mathbf{Vec}$ which is pro-nilpotent (meaning that the filtration of $U(L)$ by powers of its augmentation ideal is separated).
From this dual perspective, the main result we want is then the following.
**Theorem:** For any pro-nilpotent Lie algebra $L$ (in characteristic $0$), there is a bijection between increasing filtrations on $L$ (by subobjects in $\mathrm{pro}{-}\mathbf{Vec}$, compatible with the Lie bracket, and indexed by non-positive integers) and increasing filtrations on $U(L)$ (by subobjects in $\mathrm{pro}{-}\mathbf{Vec}$, compatible with the Hopf algebra operations, and indexed by non-positive integers). This bijection sends a filtration on $L$ to the induced filtration on $U(L)$, and sends a filtration on $U(L)$ to the restriction of this filtration along the inclusion $L\hookrightarrow U(L)$. In particular, the descending central series on $L$ induces the filtration on $U(L)$ by powers of its augmentation ideal (since these are the smallest possible filtrations in either case).
This result is a consequence of a suitably completed and enriched version of the Milnor--Moore Theorem, which we proved as Theorem A.3.6. This says that if $\mathcal C$ is a $\mathbb Q$-linear complete tensor category, then there is an equivalence of categories between the category of pro-nilpotent Lie algebra objects in $\mathcal C$ and the category of pro-nilpotent cocommutative Hopf algebra objects in $\mathcal C$. This equivalence is given in one direction by $L\mapsto U(L)$, and in the other by $H\mapsto H^{\mathrm{prim}}$, where $H^{\mathrm{prim}}$ denotes the Lie algebra of primitive elements.
This can be applied to the category $\mathcal C$ of increasing non-positively filtered objects in $\mathrm{pro}{-}\mathbf{Vec}$, which tells us that there is an equivalence of categories between non-positively filtered pro-nilpotent Lie algebras and non-positively filtered pro-nilpotent cocommutative Hopf algebras, given by $L\mapsto U(L)$ with its induced filtration, and $H\mapsto H^{\mathrm{prim}}$ with its induced filtration. This in particular gives the claimed bijection between filtrations on $L$ and on $U(L)$.
---
**Some remarks**
* This feels like overkill to prove this particular result, and perhaps there is an easier way. But I don't think I found one when I thought about this.
* This idea of proving theorems for Lie algebra objects enriched in certain categories comes from Fresse's *Homotopy Theory of Operads* book. Though the results there don't seem to be directly applicable, since Fresse discusses these enriched results in categories with well-behaved colimits, which $\mathrm{pro}{-}\mathbf{Vec}$ does not have.
| 1 | https://mathoverflow.net/users/126183 | 433892 | 175,491 |
https://mathoverflow.net/questions/433889 | 1 | Setup
-----
To clarify, let constants $0 < a < b < \infty$, and $p \in \mathbb{N}$ be fixed. Further let $B \subset \mathbb{R}^{p}$ be a fixed compact support. We then define the space of bounded (probability) densities on $B$, as follows:
$$
\mathcal{F}\_{B}^{[a, b]} := \left\{f \colon B \to [a, b] \mid \int\_{B}{f} d\mu = 1, \text{$f$ measurable} \right\}
$$
Where $\mu$ is taken to be a normalizing (prob) measure. We can then define the $L\_{1}$- metric, and $L\_{2}$-metric on $\mathcal{F}\_{B}^{[a, b]}$, as follows:
\begin{aligned}
&\|f-g\|\_1=\int|f-g| d \mu \\
&\|f-g\|\_2^2=\int(f-g)^2 d \mu
\end{aligned}
Question
--------
The [lecture notes](http://www.stat.yale.edu/%7Eyw562/teaching/598/lec17.pdf) claim *without proof* (on top of page 4) that
\begin{equation}
\|f-g\|\_1^2 \gtrsim\|f-g\|\_2^2,
\end{equation}
where $\gtrsim$ means greater than or equal to, ignoring universal constants (which may depend on $a, b, B$).
This claim seems rather remarkable. Could anyone please prove this, if it is true? Or provide a suitable counterexample if it is false?
If it is true, what are the constants in the inequality (in terms of $a, b, B$)?
Additional Comments
-------------------
If this inequality is true, then I believe it would imply a [reverse Pinsker inequality](https://arxiv.org/pdf/1503.07118.pdf) on $\mathcal{F}\_{B}^{[a, b]}$ (see equation (10)), which is non-trivial since $B$ here is assumed to be compact, not just finite.
**Aside:** Although this has been [asked elsewhere](https://math.stackexchange.com/questions/4494978/l-1-distance-is-bigger-up-to-constants-than-l-2-distance-for-bounded-pdfs), the original specification there was slightly different and the counterexample given there does not apply (since they did not assume a *common* support set $B$). Moreover, the original math.SE question is now quite old and inactive. Since this is a research related question with no well citable proof, I believe it is fair to post here on math.overflow to settle the issue.
| https://mathoverflow.net/users/486152 | Lower bound $L_{1}$-metric with $L_{2}$-metric for bounded pdfs, on common support | This is false. If (let's say) $f-g=1$ on a set of measure $\epsilon$ and $|f-g|\simeq \epsilon$ otherwise (and note that we can still normalize them both), then $\|f-g\|\_1^2\simeq \epsilon^2$, $\|f-g\|\_2^2\simeq\epsilon$.
For a concrete example, you can take $\mu$ as Lebesgue measure on $B=[0,1]$ and then
$$
f(x) = \begin{cases} 2 & x<\epsilon \\ \frac{1-2\epsilon}{1-\epsilon} & x>\epsilon \end{cases} , \quad g(x)=1 .
$$
| 3 | https://mathoverflow.net/users/48839 | 433899 | 175,493 |
https://mathoverflow.net/questions/433898 | 7 | The sequence of polynomials
$$P\_n=\sum\_{k=0}^{\lfloor(2n-1)/3\rfloor}
\frac{(2n-2k-1)!(2n-2k-2)!}{k!(n-k)!(n-k-1)!(2n-3k-1)!}x^k$$
satisfies apparently the identities
$$0=\sum\_{j=0}^nP\_{n-j}(P\_j-(-x)^j)$$
for all $n\geq 2$. (The previous condition
$n\geq 1$ was incorrect, as pointed out in the answer of Ira Gessel.) (This has probably a WZ proof since it involves hypergeometric stuff, see the answers below.)
It is easy to see that $P\_1,P\_2,\ldots$ evaluates to the sequence $1,1,2,5,14,\ldots$ of Catalan numbers at $x=0$. Leading coefficients are also closely related to Catalan numbers and central binomial coefficients.
All roots of the polynomials $P\_n$ are apparently in the real interval
$(-\infty,-16/27)$ and the
largest root of $P\_n$ converges rather quickly (for $n\rightarrow \infty$) to the rational number $-16/27$.
*Has this sequence of polynomials appeared elsewhere? Other interesting properties?*
| https://mathoverflow.net/users/4556 | A sequence of polynomials related to Catalan numbers | I find a slightly different initial condition for the recurrence:
$$0=\sum\_{j=0}^nP\_{n-j}(P\_j-(-x)^j)$$
for $n\ne 1$; for $n=1$ the sum is $-1$. It's easy to derive a formula for the generating function
$\sum\_{n=0}^\infty P\_n(x) z^n$ from this recurrence. We find that, as noted by Tewodros,
$$\sum\_{n=0}^\infty P\_n(x) z^n = \frac{1-\sqrt{4z(1+xz)^2}}{2(1+xz)}.$$
In terms of the Catalan number generating function $c(u) = (1-\sqrt{1-4u})/(2u)$, this is
$z(1+xz) c\bigl(z(1+xz)\bigr)$.
From this we can easily derive the OP's formula for $P\_n(x)$ in the form
$$P\_n(x) = \sum\_i C\_i \binom{2i+1}{n-i-1}x^{n-i-1},$$
where $C\_i$ is the Catalan number $\frac{1}{i+1}\binom{2i}{i}$.
| 9 | https://mathoverflow.net/users/10744 | 433913 | 175,499 |
https://mathoverflow.net/questions/433911 | 1 | I apologize if this question is not suited for MathOverflow. This has been crossposted in MathStackExchange [here](https://math.stackexchange.com/questions/4568587/vector-subbundles-of-a-given-one-in-mathbbcp1) and it is related to some open questions on that site that remain unsolved.
I would like to understand and if possible classify all vector subbundles of a fixed given one $E$ on the complex projective line, in the holomorphic category. The Grothendieck decomposition theorem says that over $\mathbb{CP}^1$
$$ E \simeq \bigoplus\_{i=1}^{r} \mathcal{O}(k\_i) $$
where $r$ is the rank of $E$.
This is partially answered in [this question](https://math.stackexchange.com/questions/1619656/possible-subbundles-on-mathbbcp1) but unfortunately the general case is never covered there.
What are the general subtleties that arise (as the answer there suggests) in general? What kind of (probably cohomological) techniques could be used to tackle this problem?
To start somewhere, this is trivial for line bundles: $L' \subset L$ as holomorphic lines bundles implies that either $L' = 0$ or $L' = L$.
If this is trivial or there is some reference already addressing this, feel free to just point to that instead.
| https://mathoverflow.net/users/494147 | Vector subbundles of a given one in $\mathbb{CP}^1$ | $\sum\_{i=1}^r \mathcal O(a\_i)$ is a sub-bundle of $\sum\_{i=1}^s \mathcal O( b\_i)$ if and only if, for all $c$, (1) $\# \{ i \mid a\_i \geq c \} \leq \# \{i \mid b\_i\geq c\}$ and (2) if equality holds for one $c$ it holds for all greater $c$.
Proof of "only if": By twisting, we may assume $c=0$. Then $\sum\_{i, a\_i\geq 0} \mathcal O(a\_i)$ is the maximal globally generated sub-bundle of $\sum\_{i=1}^r \mathcal O(a\_i)$, thus a sub-bundle of the maximal globally generated sub-bundle $\sum\_{i, b\_i \geq 0} \mathcal O(b\_i)$ of $\sum\_{i=1}^s \mathcal O( b\_i)$. Hence we have the rank inequality (1) and, if the ranks are equal, the bundles must be isomorphic, giving (2).
Proof of "if": If $V, W$ are two vector bundles, if $V$ is a summand of $W$ then $V \oplus U$ is a summand of $W \oplus U$. Furthermore, our hypothesis holds for $V$ and $W$ (if and) only if holds for $V \oplus U$ and $W \oplus U$. Using this, we may reduce to the case when our two vector bundles have no common direct summand, i.e. when $a\_i \neq b\_j$ for all $j$.
We may assume the $a\_i$ and $b\_i$ are in nonincreasing order. In this case, $a\_j \leq b\_{j+1}$ for all $j$, since if $a\_j > b\_{j+1}$ then $\#\{ i \mid a\_i \geq a\_j \} \geq j \geq \# \{ i \mid b\_{i} \geq a\_j \}$ so by the (1) we have $\#\{ i \mid a\_i \geq a\_j \} =\# \{ i \mid b\_{i} \geq a\_j \}$ and then by (2) we have $a\_i = b\_i $ for all $i \leq j$, contradicting our assumption.
A map $\sum\_{i=1}^r \mathcal O(a\_i) \to \sum\_{i=1}^s \mathcal O(b\_i)$ is a matrix $M$ whose entry $M\_{ij}$ is a map from $\mathcal O(a\_j)$ to $\mathcal O(b\_i)$ and thus i section of $\mathcal O(b\_i-a\_j)$. The matrix has full rank at a point as long as some $r \times r $ minor is nonzero, so the map $V \to W$ is an inclusion of sub-bundles as long as the $r\times r$ minors have no common factors.
We choose a map $M$ such that $M\_{ij}=0$ unless $i=j$ or $i={j+1}$. Then the $r\times r$ minor from the first $r$ rows is $\prod\_{j=1}^r M\_{jj}$ and the $r \times r$ minor from the second $r$ rows is $\prod\_{j=1}^r M\_{(j+1)j}$. Since $a\_j \leq b\_{j+1} \leq b\_j$ for all $j$, both of these are products of sections of a positive power of $\mathcal O(1)$. We can then take all the $M\_{jj}$ to vanish only at $\infty$, and all the $M\_{(j+1)j}$ to vanish only at $0$, ensuring these two minors have no common roots, giving an inclusion of sub-bundles, as desired.
| 1 | https://mathoverflow.net/users/18060 | 433915 | 175,500 |
https://mathoverflow.net/questions/433866 | 1 | I have a two-dimensional vector space ${\mathbb C}^2$ with basis $e\_m, f\_1$ and action of ${\mathbb C}^\*$ by $t \cdot e\_m = t^m e\_m$ and $t \cdot f\_1 = f\_1$ and I have the projective line ${\mathbb P}^1$ which is the set of lines through the origin in this ${\mathbb C}^2$. I have the tautological line bundle $\Lambda$ on this ${\mathbb P}^1$ whose fiber over the line $L$ in ${\mathbb C}^2$ is the vector space $L$. I also have the trivial line bundle ${\mathcal O}$ on ${\mathbb P}^1$. Finally, I have the skyscraper sheaf ${\mathbb C}\_{e\_m}$ on ${\mathbb P}^1$ supported on the point of ${\mathbb P}^1$ which corresponds to the line spanned by $e\_m$ in ${\mathbb C}^2$.
Note that all of the sheaves $\Lambda$, $\mathcal O$ and ${\mathbb C}\_{e\_m}$ are ${\mathbb C}^\*$-equivariant.
Question: how to write a short exact sequence of ${\mathbb C}^\*$-equivariant sheaves on ${\mathbb P}^1$ which involves $\Lambda$, ${\mathcal O}$ and ${\mathbb C}\_{e\_m}$ possibly with twists of grading?
Namely, non-equivariantly we have a short exact sequence $0 \to \Lambda \to {\mathcal O} \to {\mathbb C}\_{e\_m} \to 0$, is this an exact sequence of equivariant sheaves, or do we need to add twists of grading somewhere to make this exact sequence equivariant?
| https://mathoverflow.net/users/198061 | Short exact sequence of equivariant line bundles on $\mathbb P^1$ | This is equivariant. The map $\Lambda \to \mathcal O$ explicitly on sections sends a section $(e\_m, f\_1) $ valued in $ L \subseteq \mathbb C^2$ to the section $f\_1$ of the trivial line bundle. This map is an isomorphism everywhere but the point corresponding to the line spanned by $e\_m$, and vanishes to order $1$ there. So the cokernel is the map to $\mathbb C\_{e\_m}$ given by evaluating at that point.
Both the projection and evaluation maps are manifestly $\mathbb G\_m$-equivariant.
| 3 | https://mathoverflow.net/users/18060 | 433919 | 175,502 |
https://mathoverflow.net/questions/433860 | 6 | $\DeclareMathOperator\Ho{Ho}\DeclareMathOperator\Hom{Hom}$There are (at least) seven kinds of morphism spaces for a simplicial set $X$:
1. The [left-pinched morphism space](https://kerodon.net/tag/01KW) $\Hom^L\_X(x,y)$,
2. The [right-pinched morphism space](https://kerodon.net/tag/01KW) $\Hom^R\_X(x,y)$,
3. The (non-pinched) [morphism space](https://kerodon.net/tag/01J4) $\Hom\_X(x,y)$,
4. The simplicial set $\Hom\_{\mathfrak{C}[X]}(x,y)$ where $\mathfrak{C}[X]$ is the rigidification of $X$,
5. The simplicial set $\Hom\_{\mathfrak{C}^{nec}[X]}(x,y)$ where $\mathfrak{C}^{nec}[X]$ is the [Dugger-Spivak rigidification](https://arxiv.org/abs/0910.0814) of $X$,
6. The simplicial set $\Hom\_{\mathfrak{C}^{hoc}[X]}(x,y)$ where $\mathfrak{C}^{hoc}[X]$ is the simplicial set defined in page 17 of [Dugger-Spivak's Rigidification of quasi-categories](https://arxiv.org/abs/0910.0814),
7. The simplicial set $\Hom^E\_X(x,y)$ defined in page 15 of [Dugger-Spivak's Mapping spaces in Quasi-categories](https://arxiv.org/abs/0911.0469).
Do these mapping spaces all agree on the "underived level"? I.e. do we have
$$
\pi\_0\Hom^L\_X(x,y)\cong\pi\_0\Hom^R\_X(x,y)\cong\pi\_0\Hom\_X(x,y)
$$
$$\cong\pi\_0\Hom\_{\mathfrak{C}[X]}(x,y)\cong\pi\_0\Hom\_{\mathfrak{C}^{nec}[X]}(x,y)\cong\pi\_0\Hom\_{\mathfrak{C}^{hoc}[X]}(x,y)
$$
$$
\cong\pi\_0\Hom^E\_X(x,y)
$$
for an *arbitrary* simplicial set $X$?
In particular, is this true in the special case where $x=y$? (~~Such a restriction helps already for the left/right pinched morphism spaces: by [Tag 01KZ](https://kerodon.net/tag/01KZ), we have $\Hom^L\_X(x,x)\cong\Hom^R\_X(x,x)^\mathrm{op}$, and thus $\pi\_0\Hom^L\_X(x,x)\cong\pi\_0\Hom^R\_X(x,x)$ since $\pi\_0(X^\mathrm{op})\cong\pi\_0(X)$.~~ **Edit: this is wrong, see R. van Dobben de Bruyn's answer below**)
| https://mathoverflow.net/users/130058 | Do the various notions of morphism spaces of simplicial sets agree on the underived level? | There is no hope of comparisons between $\pi\_0\operatorname{Hom}^L\_X(x,y)$ and $\pi\_0\operatorname{Hom}^R\_X(x,y)$ in general, even when $x=y$. (Note that the statement you cited says $\operatorname{Hom}^L\_{X^{\text{op}}}(x,x) \cong \operatorname{Hom}\_X^R(x,x)^{\text{op}}$, so it computes Homs in a different simplicial set.)
**Example.** Let $X$ be the coequaliser of $f,g \colon \Delta^0 \amalg \Delta^1 \rightrightarrows \Delta^2$, where $f$ maps $\Delta\_0$ to $0$ and $\Delta^1$ to $\operatorname{id}\_0$, and $g$ maps $\Delta\_0$ to $2$ and $\Delta^1$ to the arrow $0 \to 1$. The nondegenerate simplices in $X$ are:
* A single vertex $x$ (the image of $0$, $1$, and $2$ in $\Delta^2$).
* Two arrows $\alpha$ and $\beta$ (the images of $0 \to 2$ and $1 \to 2$; the third arrow $0 \to 1$ has become identified with a degenerate $1$-simplex).
* One $2$-simplex $h$ (the image of $0 \to 1 \to 2$), by definition a right homotopy $\alpha \stackrel\sim\to \beta$.
Thus we see that $\alpha$ and $\beta$ are right homotopic (hence homotopic in $\operatorname{Hom}\_X(x,x)$ as well), but not left homotopic as there are no other nondegenerate $2$-simplices. $\square$
In particular, any of the definitions that are self-dual (e.g. $\operatorname{Hom}\_{X^{\text{op}}}(x,y) = \operatorname{Hom}\_X(y,x)^{\text{op}}$) cannot agree with $\pi\_0\operatorname{Hom}^L\_X(x,y)$ or $\pi\_0\operatorname{Hom}^R\_X(x,y)$ in general. For instance, if $\pi\_0\operatorname{Hom}^L\_X(x,y) = \pi\_0\operatorname{Hom}\_X(x,y)$ for all $X$, then
\begin{align\*}
\pi\_0\operatorname{Hom}\_X^R(x,y) &= \pi\_0\operatorname{Hom}\_{X^{\text{op}}}^L(y,x)^{\text{op}} = \pi\_0 \operatorname{Hom}\_{X^{\text{op}}}(y,x)^{\text{op}} \\
&= \pi\_0 \operatorname{Hom}\_X(x,y) = \pi\_0 \operatorname{Hom}\_X^L(x,y),
\end{align\*}
which we saw is false (again feel free to assume $x=y$).
I believe all the others are self-dual (but I'm not familiar with all of them), so at least they don't agree with $\pi\_0\operatorname{Hom}\_X^L(x,y)$ or its dual. I'm not sure if any subset of the others agree in general, but you can probably make counterexamples like the one above. (The convenient coincidence is that finite simplicial sets are far removed from the Kan or inner horn conditions: from some dimension on, you can only produce degenerate simplices, which do a bad job filling horns.)
| 7 | https://mathoverflow.net/users/82179 | 433929 | 175,506 |
https://mathoverflow.net/questions/433923 | 2 | Find the function-constant pairs $\langle f(x),c\rangle$ that satisfy the differential equation below:
$$f'(x)=f(x+c),$$
where $c \in \mathbb{C}$ and $f(x) \in \mathbb{C}$.
I found two families of functions (i.e. exponential functions and sine/cosine functions) that satisfy this differential equation. For example, $f(x)=c\_1 \sin(x+c\_2)$ and $f(x)=c\_1 e^x$.
Can you come up with any more function families or prove no more functions exists?
| https://mathoverflow.net/users/369335 | One question about a specific first-order differential equation | Following Euler, let us look at solutions of the form $f(z)=e^{sz}$, where $s$ is a complex number. We obtain a transcendental equation
$$s=e^{cs},$$
which for every complex $c\neq 0$ has infinitely many complex solutions $s\_n$. Now any linear combination
$$\sum\_{n}a\_ne^{s\_nz}$$
with complex $a\_n$, and any limit of such linear combinations will satisfy your equation. This gives all solutions.
The method actually applies to a wide class of linear equations of "convolution type". See
Malgrange's approximation theorem (Theorem 16.4.1 in Hormander, Analysis of linear partial differential operators) and also these entries on MO: [Solve $f(x)=\int\_{x-1}^{x+1} f(t) \mathrm{d}t$](https://mathoverflow.net/questions/156312/),
[On equation $f(z+1)-f(z)=f'(z)$](https://mathoverflow.net/questions/114875/)
| 5 | https://mathoverflow.net/users/25510 | 433932 | 175,508 |
https://mathoverflow.net/questions/433928 | 2 | Let $G$ be a compact Lie group and Let $G\_\mathbb{C}$ be its complexification. Let $T$ be a maximal torus of $G$ and let $X$ be the quotient $G/T$.
Consider $H$ to be a Lie subgroup of $G$ and denote by $H\_\mathbb{C}$ its complexification. Let $x \in X$, denote by $H\_x$ and ${(H\_{\mathbb{C}})}\_x$ the stabilizers of $x$ in $H$ and in $H\_\mathbb{C}$ respectively.
Question : is $H/H\_x$ isomorphic to $H\_\mathbb{C}/{(H\_{\mathbb{C}})}\_x$?
| https://mathoverflow.net/users/172459 | Generalization of $G/T \simeq G_\mathbb{C}/B$ | No.
Take $G = SU(2)$, $G\_{\mathbb C} = SL\_2(\mathbb C)$, $G/B$ the complex projective line alias the sphere, $H$ the diagonal $U(1)$, $x$ any point other than the two fixed points of $H$, so that the orbit $H/H\_x$ is a circular slice of the sphere, and $H\_{\mathbb C}$ the diagonal $GL\_1(\mathbb C)$, so that the orbit $H\_{\mathbb C}/ (H\_{\mathbb C})\_x$ is the whole sphere minus two points.
The circle is not isomorphic to the whole sphere minus two points.
| 4 | https://mathoverflow.net/users/18060 | 433933 | 175,509 |
https://mathoverflow.net/questions/433667 | 7 | The famous [De Bruijn–Erdős theorem](https://en.wikipedia.org/wiki/De_Bruijn%E2%80%93Erd%C5%91s_theorem_(graph_theory)) and its [hypergraphs generalization](https://mathoverflow.net/questions/432595/de-bruijn-erd%C5%91s-theorem-for-hypergraphs) states the following.
**Theorem.** *Let $V$ be a set, and $E\subset2^V$ be a family of its subsets. Assume that every $e \in E$ is **finite** and that for some $k \in \mathbb{N}$, every **finite** sub-hypergraph of $H=(V,E)$ can be properly colored with $k$ colors. Then $H$ can also be properly colored with $k$ colors.*
It looks like that the standard proof based on [Tychonoff's theorem](https://en.wikipedia.org/wiki/Tychonoff%27s_theorem) doesn't work when the edges are allowed to be infinite. So, I wonder if the following analogue of De Bruijn–Erdős theorem holds.
>
> Let $V$ be a set, and $E\subset2^V$ be a family of its subsets. Assume that every $e \in E$ is **countable**. Moreover, assume that for some $k \in \mathbb{N}$ and for every **countable** $W \subset V$, the hypergraph $H(W)= (W,E(W))$ can be properly colored with $k$ colors, where $E(W) = \{e \in E: e \subset W\}$. Then $H=(V,E)$ can also be properly colored with $k$ colors.
>
>
>
Erdős and Hajnal proved in [[EH]](https://link.springer.com/article/10.1007/BF02066676) the following related result on families of bounded intersections.
**Theorem.** *Let $V$ be a set, and $E\subset2^V$ be a family of its subsets. Assume that every $e \in E$ is **countably infinite**. Moreover, assume that there exists $m \in \mathbb{N}$ such that $|e\_1\cap e\_2|<m$ for all $e\_1,e\_2 \in E$. Then $H=(V,E)$ can be properly colored with $2$ colors.*
Can this result be strengthened as follows to deal with all families of finite intersections?
>
> Let $V$ be a set, and $E\subset2^V$ be a family of its subsets. Assume that every $e \in E$ is **countably infinite**, and that $e\_1\cap e\_2$ is finite for all $e\_1,e\_2 \in E$. Then $H=(V,E)$ can be properly colored with $2$ colors.
>
>
>
[EH] P. Erdős & A. Hajnal, *On a property of families of sets*, Acta Math. Academiae Scientiarum Hungarica, 12, 87–123 (1964).
| https://mathoverflow.net/users/482790 | Two questions on infinite hypergraphs | The answer to the first question is also no, by a minor modification of the proof of the Elekes-Hoffmann result cited in the answer to the second question. In fact, we get the following:
**Theorem** There is a set $V$ of cardinality $2^{\aleph\_0}$ and a collection $E$ of countably infinite subsets of $V$ such that, for every countable $W \subset V$, $H(W) = (W, E(W))$ is 2-colorable, but $H=(V,E)$ cannot be properly colored with countably many colors.
**Proof** Let $V$ be the collection of all functions of the form $f:\beta \rightarrow \omega$, where $\beta$ is a countable ordinal. For each such $f$, let $A\_f$ be the set of all $n < \omega$ such that the preimage $f^{-1}(\{n\})$ is infinite. For each $n \in A\_f$, let $e\_{f,n} := \{f\} \cup \{f \restriction \alpha \mid \alpha \in f^{-1}(\{n\})\}$. Then let $E\_f := \{e\_{f,n} \mid n \in A\_f\}$. Note that each element of $E\_f$ is countably infinite and has a maximal element, namely $f$ itself. Finally, let $E := \bigcup\_{f \in V} E\_f$.
Let us first show that $H(W)$ is 2-colorable for every countable $W \subset V$. The point is that, for each such $W$, $E(W)$ is countable, since, by construction, $E(W) \subseteq \bigcup\_{f \in W} E\_f$, and each $E\_f$ is countable. Therefore, one can enumerate both $W$ and $E(W)$ in order-type $\omega$ and use these enumerations to inductively define a coloring $c:W \rightarrow \{0,1\}$, diagonalizing against the elements of $E(W)$ to make sure none of them are monochromatic for $c$ (here we use the fact that every element of $E(W)$ is infinite).
We finally show that $H$ cannot be properly colored with countably many colors. Fix a function $c:V \rightarrow \omega$. We will show that it is not a proper coloring of $H$. Let us define a function $g:\omega\_1 \rightarrow \omega$ by recursively specifying $g \restriction \beta$ for $\beta \leq \omega\_1$. If $\beta \leq \omega\_1$ is a limit ordinal and we have specified $g \restriction \alpha$ for all $\alpha < \beta$, then $g \restriction \beta = \bigcup\_{\alpha < \beta} g \restriction \alpha$. For the successor step, suppose that we have constructed $g \restriction \beta$. To define $g \restriction (\beta + 1)$, we must define $g(\beta)$; do so by setting $g(\beta) = c(g \restriction \beta)$.
Let $A$ be the set of all $n < \omega$ such that $g^{-1}(\{n\})$ is unbounded in $\omega\_1$, and let $\beta < \omega\_1$ be large enough so that:
* for all $\alpha < \omega\_1$, if $g(\alpha) \notin A$, then $\alpha < \beta$; and
* for all $n \in A$, $g^{-1}(\{n\}) \cap \beta$ is infinite.
Let $n = g(\beta)$, and let $f = g \restriction \beta$. By construction, we have $n = c(g \restriction \beta) = c(f)$. By the first bullet point above, we know that $n \in A$. By the second bullet point, we know that $f^{-1}(\{n\})$ is infinite, so $n \in A\_f$ and $e\_{f,n} = \{f\} \cup \{f \restriction \alpha \mid \alpha \in f^{-1}(\{n\})$ is in $E$. But for every $\alpha \in f^{-1}(\{n\})$, we have $c(f \restriction \alpha) = c(g \restriction \alpha) = g(\alpha) = n = c(f)$, so $e\_{f,n}$ is monochromatic for $c$, showing that $c$ is not a proper coloring of $H$.
| 5 | https://mathoverflow.net/users/26002 | 433957 | 175,516 |
https://mathoverflow.net/questions/433951 | 4 | I'm reading a proof of below theorem from [this paper](https://arxiv.org/pdf/2205.13207.pdf).
>
> **Theorem A.3.** Let $\Omega$ be a locally compact normal Hausdorff space. Let $\left\{\mu\_n\right\} \cup\{\mu\} \subset \mathcal{M}(\Omega)$ and assume that $\underset{n \rightarrow \infty}{\operatorname{v-lim}} \mu\_n=\mu$. Then for any open set $\Theta \subset \Omega$,
> $$
> |\mu|(\Theta) \leq \liminf \_{n \rightarrow \infty}\left|\mu\_n\right|(\Theta) .
> $$
> In particular, $\|\mu\| \leq \liminf \_{n \rightarrow \infty}\left\|\mu\_n\right\|$.
>
>
>
**Proof.** Let $\Theta \subset \Omega$ be open and $\varepsilon>0$. Since $\mu$ is inner regular and $\Omega$ is normal and locally compact, as a consequence of Urysohn's lemma [1, Lemma 2.46], there exists $f \in C\_c(\Omega)$ such that $|f| \leq 1, \operatorname{supp}(f) \subset \Theta$ and
$$
\int f \mathrm{~d} \color{red}{\mu} \geq \color{red}{|\mu|}(\Theta)-\varepsilon.
$$
Then by vague convergence of $\left\{\mu\_n\right\}$,
$$
|\mu|(\Theta)-\varepsilon \leq \int f \mathrm{~d} \mu=\lim \_{n \rightarrow \infty} \int f \mathrm{~d} \mu\_n \leq \liminf \_{n \rightarrow \infty} \int|f| \mathrm{d}\left|\mu\_n\right| \leq \liminf \_{n \rightarrow \infty}\left|\mu\_n\right|(\Theta)
$$
Now the result follows by letting $\varepsilon \downarrow 0$.
---
**My understading:** Below are the authors' related definitions. A finite signed measure $\mu$ is Radon if its variation $|\mu|$ is inner regular. In above proof, $\mu$ is Radon and thus $|\mu|$ is inner regular. So there is $f \in C\_c(\Omega)$ such that $f \le 1\_\Theta$ and
$$
\int f \mathrm{~d} \color{red}{|\mu|} \geq \color{red}{|\mu|}(\Theta)-\varepsilon.
$$
So I'm confused due to the appearance of $\color{red}{\mu}$ instead of $\color{red}{|\mu|}$ in the integral. Could you elaborate on my confusion?
---
**Authors' definitions:** Let
* $\Omega$ be a metric space and $\mathscr{B}(\Omega)$ its Borel $\sigma$-algebra.
* $C\_b(\Omega)$ the subspace of all real-valued bounded continuous functions on $\Omega$.
* $C\_0(\Omega)$ the subspace of all $f \in C\_b(\Omega)$ such that for any $\varepsilon>0$, there exists a compact set $K\_{\varepsilon}$ with $|f|<\varepsilon$ on $K\_{\varepsilon}^c$, and
* $C\_c(\Omega)$ the subspace of all $f \in C\_b(\Omega)$ such that $f$ has compact support.
For a signed measure $\mu$ on $(\Omega, \mathscr{B}(\Omega)$ ), we denote by $|\mu|$ its associated variation measure. A finite signed measure $\mu$ on $(\Omega, \mathscr{B}(\Omega))$ is called a finite signed Radon measure if **$|\mu|$ is inner regular**, i.e., for each $A \in \mathscr{B}(\Omega)$,
$$
|\mu|(A)=\sup \{|\mu|(K): K \in \mathscr{B}(\Omega), K \text { compact, } K \subset A\} .
$$
We denote the set of all finite signed Radon measures on $(\Omega, \mathscr{B}(\Omega))$ by $\mathcal{M}(\Omega)$ and the subset of all finite positive Radon measures by $\mathcal{M}^{+}(\Omega)$. We say that a sequence $\left\{\mu\_n\right\} \subset \mathcal{M}(\Omega)$ converges to $\mu \in \mathcal{M}(\Omega)$
* (a) weakly if $\int\_\Omega f \mathrm d \mu\_n \to \int\_\Omega f \mathrm d \mu$ for all $f \in C\_b(\Omega)$ and we write $\underset{n \rightarrow \infty}{\operatorname{w-lim}} \, \mu\_n= \mu$;
* (b) vaguely if $\int\_\Omega f \mathrm d \mu\_n \to \int\_\Omega f \mathrm d \mu$ for all $f \in C\_c(\Omega)$, and we write $\underset{n \rightarrow \infty}{\operatorname{v-lim}} \, \mu\_n=\mu$.
| https://mathoverflow.net/users/477203 | Vague convergence: confusion about the regularity of a signed Radon measure and that of its variation | $\newcommand{\Om}{\Omega}\newcommand{\Th}{\Theta}\newcommand{\B}{\mathscr B}\newcommand{\M}{\mathcal M}\newcommand\ep\varepsilon\newcommand{\de}{\delta}\newcommand{\R}{\mathbb R}$Take any $\mu\in\M(\Om)$, any open subset $\Th$ of $\Om$, and any real $\ep>0$. Let $\de:=\ep/4$.
By the [Hahn decomposition theorem](https://en.wikipedia.org/wiki/Hahn_decomposition_theorem), there is a partition of $\Om$ into Borel sets $D^\pm$ such that $D^+$ is a positive set for $\mu$ and $D^-$ is a negative set for $\mu$.
Let
\begin{equation\*}
A^\pm:=\Th\cap D^\pm. \tag{1}\label{1}
\end{equation\*}
Since $|\mu|$ is inner regular, there exist compact sets
\begin{equation\*}
K^\pm\subseteq A^\pm\text{ such that }|\mu|(A^\pm\setminus K^\pm)<\de. \tag{2}\label{2}
\end{equation\*}
Since $\Om$ is normal, there exist open subsets $U^\pm$ of $\Th$ such that
\begin{equation\*}
U^\pm\supseteq K^\pm\text{ and }U^+\cap U^-=\emptyset. \tag{3}\label{3}
\end{equation\*}
Since the sets $K^\pm$ are compact and $\Om$ is locally compact, without loss of generality the closures of the sets $U^\pm$ are compact.
By [Urysohn'slemma](https://en.wikipedia.org/wiki/Urysohn%27s_lemma), there exist continuous functions $f^\pm\colon\Om\to\R$ such that
\begin{equation\*}
0\le f^\pm\le1,\quad f^\pm=1\text{ on }K^\pm,\quad f^\pm=0\text{ on }\Om\setminus U^\pm. \tag{4}\label{4}
\end{equation\*}
Let
\begin{equation\*}
f:=f^+-f^-.
\end{equation\*}
Then $f^+f^-=0$, whence $|f|\le1$. Also, $f=0$ on $\Om\setminus(U^+\cup U^-)$. So, recalling that the closures of the sets $U^\pm$ are compact, we see that $f\in C\_c(\Om)$. Also, since $U^\pm$ are subsets of $\Th$, we have $|f|\le1\_\Th$.
It remains to show that
\begin{equation\*}
\int\_\Om f\,d\mu\ge|\mu|(\Th)-\ep. \tag{$\*$}\label{\*}
\end{equation\*}
To do this, note that, by \eqref{3}, \eqref{2}, and \eqref{1},
\begin{equation}
\begin{aligned}
|\mu|(U^-\setminus K^-)&\le|\mu|(\Th\setminus U^+\setminus K^-) \\
&=|\mu|(\Th)-|\mu|(U^+)-|\mu|(K^-) \\
&\le|\mu|(\Th)-|\mu|(K^+)-|\mu|(K^-) \\
&<|\mu|(\Th)-|\mu|(A^+)-|\mu|(A^-)+2\de=2\de.
\end{aligned}
\tag{5}\label{5}
\end{equation}
So, by \eqref{4}, \eqref{3}, \eqref{2}, \eqref{5}, and \eqref{1},
\begin{equation\*}
\begin{aligned}
\int\_\Om f\,d\mu&=\int\_{U^+} f^+\,d\mu-\int\_{U^-} f^-\,d\mu \\
&\ge\int\_{K^+} f^+\,d\mu-\int\_{K^-} f^-\,d\mu -\int\_{U^-\setminus K^-} f^-\,d\mu \\
&=\mu(K^+)-\mu(K^-) -\int\_{U^-\setminus K^-} f^-\,d\mu \\
&\ge\mu(K^+)-\mu(K^-) -|\mu|(U^-\setminus K^-) \\
&>\mu(A^+)-\de-\mu(A^-)-\de -2\de \\
&=|\mu|(\Th)-4\de=|\mu|(\Th)-\ep,
\end{aligned}
\end{equation\*}
so that \eqref{\*} is proved. $\quad\Box$
| 4 | https://mathoverflow.net/users/36721 | 433987 | 175,523 |
https://mathoverflow.net/questions/431870 | 3 | Let $\Omega$ be a bounded smooth domain,
$Lu = D\_i \left( a^{ij} (x) D\_ju \right)$, and two constants
$\lambda, \Lambda > 0$. Suppose the coefficient $a$ is measurable,
symmetric, and satisfies
$$
a^{ij} \xi\_i \xi\_j \ge \lambda \vert{\xi} \vert^2 \quad \text{ and} \quad
\sum\_{i,j}^{} \vert{a^{ij}(x)}\vert \le \Lambda^2,
$$
for all $x \in \Omega, \xi \in \mathbb{R}^n$. **However, $a$ can be discontinuous and not belong to any $VMO$ or $BMO$ spaces.**
Let $u \in W^{1,2}(\Omega)$ be a weak solution of $Lu = g$ for
$g \in L^{q/2},~ q > n$. Denote by
$\mathtt{data} = (\lambda,\Lambda,n,q)$. Then Theorem 8.15 in
Gilbarg-Trudinger, *Elliptic Partial Differential Equations of Second
Order*, says that
$$
\Vert u \Vert\_{L^{\infty}(\Omega)} \le C (\mathtt{data}) \left( \Vert u \Vert\_{L^2(\Omega)} + \Vert g \Vert\_{L^{q/2}(\Omega)} \right),
$$
and Theorem 8.24 in the same book says that for any $\Omega' \subset
\Omega$,
$$
\Vert u \Vert\_{C^{\alpha} \left( \Omega' \right)} \le K (\mathtt{data}, \text{dist}(\Omega',\Omega)) \left( \Vert u
\Vert\_{L^2(\Omega)} + \Vert g \Vert\_{L^{q/2}(\Omega)}\right).
$$
On page 214, after Theorem 8.37, the authors claim that solutions $w
\in W^{1,2}\_0(\Omega)$
of the eigenvalue problem
$$
Lw + \sigma w = 0,
$$
belong to $L^{\infty} (\Omega) \cap C^{\alpha}(\Omega)$, thanks to the
above theorems 8.15 and 8.24.
**My question:** If $2 \le n \le 4$, then by
Sobolev embedding, we have $w \in W^{1,2}\_0 \hookrightarrow L^{q/2}$
for some $q > n$. Therefore, we can apply theorems 8.15 and 8.24. What
about the case $n \ge 5$?
Any insights or references are appreciated!
Thank you.
| https://mathoverflow.net/users/82772 | Regularity of eigenfunctions of a self-adjoint differential operator in Gilbarg-Trudinger | I recommend [Luigi Orsina's Lecture Notes](https://www1.mat.uniroma1.it/people/orsina/AS1213/AS1213.pdf). They are beautifully written, and page 24 you will read Stampacchia's approach, which is (in my view) more elegant than Moser iterations and gives you the result you need in a jiffy.
| 2 | https://mathoverflow.net/users/40120 | 433995 | 175,526 |
https://mathoverflow.net/questions/433982 | 5 | $\newcommand{\U}{\mathcal{U}}$
$\newcommand{\F}{\mathcal{F}}$
$\newcommand{\D}{\mathcal{D}}$
$\newcommand{\C}{\mathcal{C}}$
For any infinite $X \subseteq \omega$, we define:
$$
\D\_X := \{Y \in [\omega]^\omega : Y \subseteq X \vee Y \cap X = \emptyset\}
$$
It's easy to see that $\D\_X$ is a dense open subset of $([\omega]^\omega,\subseteq)$ for all infinite $X$. Now let $\cal{C} \subseteq \lbrack\omega\rbrack^\omega$ be a collection of infinite sets. Consider the following statement:
>
> The statement $\mathsf{U}(\C)$ asserts that: If $\F$ is a filter on $\omega$ and $\F \cap \D\_X \neq \emptyset$ for all $X \in \cal{C}$, then $\F$ is an ultrafilter.
>
>
>
It's easy to verify that that $\mathsf{U}([\omega]^\omega)$ is true. Let $\mathfrak{u}'$ be the least cardinal such that there exists some $\cal{C} \subseteq \lbrack\omega\rbrack^\omega$ which $|\C| = \mathfrak{u}'$ and $\mathsf{U}(\C)$ holds. Is it consistent that $\mathfrak{u}' < \frak{c}$?
| https://mathoverflow.net/users/146831 | Minimum number of dense sets to make a filter an ultrafilter | No; $\mathfrak u'=\mathfrak c$.
To prove it, consider any $\mathcal C\subseteq[\omega]^\omega$ with cardinality $<\mathfrak c$. Working modulo finoite subsets of $\omega$ , and closing under (finitary) Boolean operations, we may assume that $\mathcal C$ is a Boolean subalgebra of $\mathcal P(\omega)/\text{fin}$, and of course a proper subalgebra because of the cardinality assumption. There are now two ways to finish the proof.
(1) Since the inclusion map $\mathcal C\to P(\omega)/\text{fin}$ is not surjective, its Stone dual (from $\beta\omega-\omega$ to the space of ultrafilters in $\mathcal C$) is not injective. So fix two distinct ultrafilters $U$ and $V$ on $\omega$ that have the same intersection with $\mathcal C$. Let $F=U\cap V$. For any $X\in\mathcal C$, either $X$ is in both $U$ and $V$ and therefore in $F$, or $\omega-X$ is in both $U$ and $V$ and therefore in $F$. Thus, $F$ meets $\mathcal D\_X$. But, since $U$ and $V$ are distinct, $F$ is not an ultrafilter.
(2) Fix some $A\subseteq\omega$ not in $\mathcal C$. (I'm still tacitly working mod finite.) Let
$$
I=\{X\in\mathcal C:A\cap X\in\mathcal C\},
$$
and note that $I$ is a proper ideal in $\mathcal C$. Let $F$ be an ultrafilter of $\mathcal C$ disjoint from $I$, and let $F'$ be the filter in $\mathcal P(\omega)/\text{fin}$ generated by $F$ (i.e., the upward closure of $F$ in $\mathcal P(\omega)/\text{fin}$). The fact that $F$ is ultra in $\mathcal C$ implies that $F'$ meets $\mathcal D\_X$ for each $X\in\mathcal C$. On the other hand, the fact that $F$ extends $I$ implies that every set in $F$ meets both $A$ and $\omega-A$; therefore the same holds for every set in $F'$, and thus $F'$ is not an ultrafilter.
| 7 | https://mathoverflow.net/users/6794 | 434012 | 175,531 |
https://mathoverflow.net/questions/434010 | 3 | Can we solve the follwing functional equation
$$f(xy)=g(x)h(y)+g(y)h(x)$$
on semigroups for unknown complex valued functions $f,g,h$ ?
| https://mathoverflow.net/users/494236 | A pexiderization of the sine addition law on semigroups | I corrected a typo on the right-hand-side of the equation in the OP, I'm unsure whether the left-hand-side has a typo, but the generalized sine addition law on semigroups is known in the form
$$g(xy)=g(x)h(y)+h(x)g(y),$$
so in terms of two unknown functions $g$ and $h$, generalizing
$$\sin(x+y)=\sin x\cos y+\cos x\sin y,$$
see [Ebanks - The sine addition and subtraction formulas on semigroups](https://doi.org/10.1007/s10474-021-01149-3) (2021).
| 5 | https://mathoverflow.net/users/11260 | 434015 | 175,533 |
https://mathoverflow.net/questions/433838 | 6 | Consider the $G(n,p)$ random graph model where $n$ is a ``large'' positive integer and $p\in (0,1)$. We may equip every realized random graph $G$ with its shortest path distance, making it into a (random) metric space $(G,d\_G)$. Since $G$ is finite then $(G,d\_G)$ is [doubling](https://en.wikipedia.org/wiki/Doubling_space).
Are there known estimates for the expected doubling constant of such a random graph?
| https://mathoverflow.net/users/36886 | Expected doubling constant of a random Erdős–Rényi graph | Let me assume $p > (1 + \varepsilon)(2 \ln n/ n)^{1/2}$, this in particular includes the case of constant $p \in (0,1)$.
If $p > (1 + \varepsilon)(2 \ln n/ n)^{1/2}$ then w.h.p. the binomial random graph $G(n,p)$ has diameter at most $2$. Also, for $p = \omega( \ln n / n)$ then w.h.p. every vertex has degree $(1 \pm o(1)) p n$. Thus the balls centered at any vertex have size $1$, $(1 \pm o(1)) p n$, and $n$; if the radius of the ball is $0, 1$ or at least $2$ respectively.
It seems that the only cases to consider are: how many singletons are needed to cover a radius-$1$ ball; and how many radius-$1$ balls are needed to cover the whole vertex set. In the first case any radius-$1$ ball trivially needs $(1 \pm o(1))pn$ singletons to cover it.
In the second case we are equivalently asking for the size of a minimum *dominating set* in $G(n,p)$, i.e. the minimum size of a set $S \subseteq V(G)$ where every vertex not in $S$ has a neighbour in $S$. This is known (see, e.g. [*Wieland, Ben; Godbole, Anant P.*, [**On the domination number of a random graph**](http://www.emis.de/journals/EJC/Volume_8/Abstracts/v8i1r37.html), Electron. J. Comb. 8, No. 1, Research paper R37, 13 p. (2001). [ZBL0989.05108](https://zbmath.org/?q=an:0989.05108).]) to be, w.h.p., concentrated around $(1 + o(1))\log\_q(n)$, where $q = 1/(1-p)$.
Then the doubling constant will depend on which of $(1 + o(1))\log\_q(n)$ or $(1 + o(1))pn$ is maximum. For constant $p \in (0,1)$ the former term is logarithmic and the second one is linear, so the doubling constant is $(1 + o(1))pn$.
| 2 | https://mathoverflow.net/users/45545 | 434020 | 175,534 |
https://mathoverflow.net/questions/72201 | 1 | Is there an explicit formula for the number of fourth powers mod n?
Finch & Sebah [1] give theorems, partially folklore, for squares and cubes mod n, but I don't know of a similar formula for higher powers.
[1] S. R. Finch and Pascal Sebah, [Squares and Cubes Modulo n](http://arXiv.org/abs/math.NT/0604465)
| https://mathoverflow.net/users/6043 | Number of biquadrates mod n | As stated in the comments ([1](https://mathoverflow.net/questions/72201/number-of-biquadrates-mod-n#comment182932_72201) [2](https://mathoverflow.net/questions/72201/number-of-biquadrates-mod-n#comment182983_72201) [3](https://mathoverflow.net/questions/72201/number-of-biquadrates-mod-n#comment183408_72201)), the counting function is multiplicative, so only the prime power case needs to be addressed. Last year, I derived a somewhat concise formula for $\lvert R\_k (p^m)\rvert$, where $R\_k (p^m)$ is the set of $k$-th power residues (not necessarily coprime to the modulus) modulo $p^m$. It was recently accepted for publication: [Counting general power residues](https://doi.org/10.7546/nntdm.2022.28.4.730-743). Here is the result, quoted from the paper:
>
> Let $\epsilon$ be the parity function. So for integers $t$, $
> \epsilon(t) = \begin{cases}
> 0 &\text{ if } 2\mid t\\
> 1 &\text{ if } 2\nmid t. \end{cases} $
>
>
> Let $p$ be a prime, and $k\ge 2$ and $m\ge 1$ be integers. Let $r$ be
> the remainder of $m$ upon division by $k$. Let \begin{align\*}
> \alpha &= \dfrac{p-1}{(k,p-1)},\\
> \beta &= (\nu\_p (k)+1)(1-\epsilon(k))(1-\epsilon(p))+\nu\_p(k)\epsilon (p),\\
> \gamma &=
> \begin{cases}
> k &\text{ if } k \mid m\\
> r &\text{ if } k\nmid m.
> \end{cases}. \end{align\*} Then \begin{align\*} \lvert R\_k (p^m)\rvert &= \alpha \cdot \left(\dfrac{p^k}{p^{\beta +1}}\cdot
> \dfrac{p^m-p^{\gamma}}{p^k-1}+
> \left\lceil\dfrac{p^{\gamma}}{p^{\beta+1}}\right\rceil\right)+1\\ &=
> \alpha \cdot \left\lceil\dfrac{1}{p^{\beta +1}}\cdot
> \dfrac{p^{m+k}-p^{\gamma}}{p^k-1}\right\rceil+1. \end{align\*}
>
>
> (Note that the $\dfrac{p^k}{p^{\beta +1}}\cdot
> \dfrac{p^m-p^{\gamma}}{p^k-1}$ term is necessarily an integer, so it
> can be absorbed into the ceiling term
> $\left\lceil\dfrac{p^{\gamma}}{p^{\beta+1}}\right\rceil$ as shown.)
>
>
>
Obviously the proof is what matters, but computational testing matched the results of the formulas in all cases tried.
**References:** My initial inspiration was a paper [Counting squares in $\mathbb Z\_n$](https://doi.org/10.1080/0025570X.1996.11996456) of Walter Stangl, who addressed the quadratic case. Also, my discussion with Arturo Magidin in [Is there a known formula for the number of $k^{\text{th}}$ power residues modulo $2^n$?](https://math.stackexchange.com/questions/3796985/is-there-a-known-formula-for-the-number-of-k-textth-power-residues-modulo) was helpful in the derivation. After submission to the first journal that I tried, a reviewer noted that similar formulas were mentioned by Maxim Korolev in [On the average number of power residues modulo a composite number](https://ui.adsabs.harvard.edu/abs/2010IzMat..74.1225K/abstract) ([DOI](https://doi.org/10.1070/IM2010v074n06ABEH002522)), who himself referred to a paper [The Number of kth Power Residues Modulo m](http://njsfdxzrb.paperonce.org/oa/darticle.aspx?type=view&id=200101001) of Ji Chungang.
**Note:** Given the content of the above references, the originality of my minor contribution rests only on unifying the various cases by noticing their similarities, summing a series that the others left unclosed, and expanded elementary exposition. I am not surprised that I was able to publish it only after trying a couple of other more well-known journals first, but it was a little surprising that the arXiv rejected posting it.
| 3 | https://mathoverflow.net/users/31084 | 434039 | 175,535 |
https://mathoverflow.net/questions/434035 | 9 | The following special cases are obvious:
1. A Grothendieck topos is concretizable ($F \mapsto \times\_i F(i)$)
2. A well-pointed topos is concretizable ($X \mapsto \rm{Hom}(1, X)$)
I looked at some more different examples and they are all obviously concretizable (but I'm just starting to learn topos theory). Is there an elementary topos from which there is no faitfull functor to the category of sets?
| https://mathoverflow.net/users/148161 | Is any elementary topos a concretizable category? | As Ivan Di Liberti suggests in the comments, according to Lemma 1.2 ([Freyd's paper, *Concretness*](https://doi.org/10.1016/0022-4049(73)90031-5), JPAA 1973) every regular well-powered category with equalizers is concretizable. In particular, every locally small elementary topos is concretizable.
| 12 | https://mathoverflow.net/users/148161 | 434043 | 175,538 |
https://mathoverflow.net/questions/434026 | 3 | For a given weighted graph $G = (V, E)$, there is a simple algorithm for finding the minimum weight circuit by running [Dijkstra's algorithm](https://en.wikipedia.org/wiki/Dijkstra%27s_algorithm#Applications) $|E|$ times.
Also for a matroid $M = (E, I)$ one can use the greedy Rado-Edmonds algorithm to find a basis of minimum weight.
I'm interested whether an algorithm for all weighted binary matroids, not just weighted graphic matroids, exists? Specifically an algorithm polynomial on the size of the ground set, using the independence oracle model?
| https://mathoverflow.net/users/494271 | Algorithm for finding a minimum weight circuit in a weighted binary matroid | The problem is NP-hard (even in the unweighted case) via a well-known connection to coding theory. Namely, if $A$ is the parity check matrix of a binary linear code $C$, then the distance of $C$ is the size of a shortest circuit in the binary matroid represented by $A$. In [The intractability of computing the minimum distance of a code](https://ieeexplore.ieee.org/document/641542), Vardy proved that computing the distance of a binary linear code is NP-hard.
However, for proper minor-closed classes of binary matroids, Geelen, Gerards, and Whittle conjecture that the problem is solvable in polynomial-time. See the paper [Computing girth and cogirth in perturbed graphic matroids](https://arxiv.org/abs/1504.07647) or [Finding Shortest Circuits in Binary Matroids](http://matroidunion.org/?p=1106) on the Matroid Union Blog for more information.
| 2 | https://mathoverflow.net/users/2233 | 434045 | 175,539 |
https://mathoverflow.net/questions/434054 | 8 | Given the famous Littlewood-Richardson rule, in terms of Schur polynomials:
$$s\_\mu s\_\nu=\sum\_\lambda c^{\lambda}\_{\mu\nu} s\_\lambda,$$
is there a classification of the cases where the LR coefficients are equal to 1?
| https://mathoverflow.net/users/166314 | When the Littlewood-Richardson rule gives only irreducibles? | The answer is **Yes**, but this requires some elaboration.
1. [Knutson-Tao-Woodward](https://arxiv.org/abs/math/0107011) prove Fulton's conjecture in $\S$6.1. In principle, you can follow the approach by [De Loera-McAllister](https://arxiv.org/abs/math/0501446) or [Mulmuley-Narayanan-Sohoni](https://www.emis.de/journals/JACO/Volume36_1/rg1w9655642r84t8.fulltext.pdf), to convert the result into a statement about certain polytopes being single points.
2. [Bürgisser-Ikenmeyer](https://arxiv.org/abs/1204.2484) make the answer much more explicit, see Prop. 3.13. Ikenmeyer further generalized this in to $c^\lambda\_{\mu\nu}>t$ in [his thesis](https://www3.math.tu-berlin.de/algebra/work/ikenmeyer_thesis.pdf), see Theorem 11.3.2.
| 14 | https://mathoverflow.net/users/4040 | 434055 | 175,541 |
https://mathoverflow.net/questions/434041 | 4 | Let $K$ be a number field i.e. a finite extension of $\mathbb{Q}$, $\overline{K}$ a fixed separable closure of $K$, and $G\_K:=\mathrm{Gal}(\overline{K}/K)$ the absolute Galois group of $K$. Let $S$ be a finite set of non-Archimedean places of $K$, or equivalently, a finite set of non-zero prime ideals in the ring of integers $\mathcal{O}\_K$ of $K$. We define
$$\overline{K}\_S:=\text{the maximal algebraic extension of $K$ in $\overline{K}$ unramified outside $S$}$$
and
$$G\_{K,S}:=\mathrm{Gal}(\overline{K}\_S/K).$$
At the beginning of Barry Mazur's part of *Modular Forms and Fermat’s Last Theorem*, he states:
>
> What is the "structure" of $G\_{K,S}$ - whatever that means? It is not even known whether or not $G\_{K,S}$ is finitely generated as a topological group (**although this has been conjectured to be the case by Shafarevich about thirty years ago**).
>
>
>
Does anyone have a source for this conjecture? Since this book was published in 1997, I guess the conjecture that Mazur is referring to would've been formulated in the mid-60s.
The only conjectures by Shafarevich that I could find are (1) the Tate-Shafarevich conjecture on the Tate-Shafarevich group, and (2) concerning the absolute Galois group of $\mathbb{Q}^{\text{ab}}$. I don't think either is equivalent to the above, but if so, what's the intuition behind that?
| https://mathoverflow.net/users/492396 | Shafarevich's conjecture on Galois groups over fields ramified at finitely many places | In the footnote on Page 11 of [Fernando Q. Gouvea: Deformations of Galois Representations],
>
> The reference is
>
>
> [I.R. Shafarevich, *Algebraic number fields*, Proceedings of the
> international Congress of Mathematicians, Stockholm, 1962 (Djursholm),
> Inst. Mittag-Leffler, 1963, Translated version reprinted in I.R.
> Shafarevich, *Collected Mathematical Papers* (Springer- Verlag, 1989),
> pp. 283-294, pp. 163-176.],
>
>
> but we should note that there Shafarevich simply asks whether it is
> the case that $G\_{K,S}$ is finitely generated for any number field
> (and whether the number of generators can be bounded in terms of the
> number of elements of $S$). His main reason for posting the question
> is that the analogous statement is true for function field over
> $\mathbb{C}$.
>
>
>
| 4 | https://mathoverflow.net/users/492970 | 434057 | 175,542 |
https://mathoverflow.net/questions/434061 | 5 | As the title says. Which finite projective planes admit a symmetric incidence matrix?
I am not an expert in the field at all, but I consulted with one. He claimed that $PG(2, \mathbb F\_q)$ can always have a symmetric matrix, but he was unsure about the non-Desarguesian planes.
I am mostly asking for reference (both for the Desarguesian and non-Desaruesian case), but any insight or (quick) proof is welcome.
| https://mathoverflow.net/users/161063 | Which finite projective planes can have a symmetric incidence matrix? | The key word here is "polarity". A polarity of a projective plane with point set $P$ and line set $L$ is a map $\pi$ from $P \cup L$ to itself mapping points to lines and lines to points, such that $\pi^2 = \operatorname{id}$ and such that a point $p$ is incident with a line $L$ if and only if the line $p^\pi$ is incident with the point $L^\pi$. The existence of a polarity $\pi$ is equivalent with the existence of a symmetric incidence matrix.
Not all projective planes admit polarities. One such non-example is given by the Hall planes (see, e.g., <https://en.wikipedia.org/wiki/Hall_plane>), because these projective planes are not self-dual.
For a reference, I would try looking in Hughes and Piper's 1973 book on Projective Planes, but I haven't checked this myself.
| 10 | https://mathoverflow.net/users/12858 | 434062 | 175,543 |
https://mathoverflow.net/questions/433974 | 5 | It is [well-known](http://www.scholarpedia.org/article/Attractor) in dynamical systems that the concept of "attractor" differs in the literature.
*My question is whether attractors need to be defined as subsets of $\omega$-limit sets of some point on the phase space, or can be **any** compact set in the phase space?* (What important results don't hold for the more general definition, which important example are (not) covered by the more restricted definition etc.?)
On one hand, in the link above, as well as in the well-known book "Handbook of Dynamical Systems Vol. 1A" by Hasselblatt & Katok, which are well-known authors and researchers, an attractor there can be any compact set.
On the other hand, in the book "Dynamical Systems and Chaos" by H. Broer, F. Takens, which are equally well-known names in the field, attractors are only defined as special types of $\omega$-limit sets.
Unfortunately, I know no reference that unifies these definitions and clarifies their relationship.
| https://mathoverflow.net/users/14101 | On the correct definition of attractors | This is just an extended comment, as I'm not even trying to recall all the uses of the term attractor in the literature, pointing out when equivalence holds or when there are some subtle differences depending, for instance, on the topology of the phase space or on properties of the map. Even the more basic definition of $\omega$-limit set of a certain subset $Y$ of the phase space is not defined completely uniformly in the literature when $Y$ is not just a finite set of points.
I just want to point out what in my view is a very clean, purely topological definition, from which it follows that attractors coincide with $\omega$-limits of inward sets. This is the approach I like best, since it naturally comes from the concept of stable, attracting set. Attractors are closed attracting sets which are strongly invariant, that is they coincide with their image.
**Definition:** Let $X$ be a compact metric space and $f$ a continuous map. An attractor is a nonempty, closed set $Y\subset X$ such that $f(Y)=Y$ and, for every $\epsilon>0$, there is $\delta$ such that all points $x$ such that $d(x,Y)<\delta$ will stay within distance $\epsilon$ (that is $d(f^n(x),Y)<\epsilon$ for every $n$) and verify $d(f^n(x),Y)\to 0$ when $n\to\infty$.
Assuming this definition you can easily prove the following:
**Fact:** Attractors are the $\omega$-limits of (nonempty) inward sets, that is of sets $Z\subset X$ such that $f(\overline{Z})\subseteq Z^\circ$.
See for instance Propositions 2.64-2.65 in Kurka, P. (2003). Topological and symbolic dynamics. Société mathématique de France.
You can also see Akin, Ethan. The general topology of dynamical systems. American Mathematical Soc., 1993, p. 43, Theorem 3, which contains some equivalent definitions of preattractor and attractor in the more general case when $f$ is a closed relation on the phase space. They are consistent with the definition above and the author points out the important link with the concept of chain-recurrence (see in particular point c)-1.).
| 3 | https://mathoverflow.net/users/167834 | 434066 | 175,545 |
https://mathoverflow.net/questions/434038 | 26 | There are several ways to generalize the notion of "algebraic closure" from fields to arbitrary commutative rings. A good overview is [On algebraic closures](https://www.jstor.org/stable/43737179) by R. Raphael. I am more interested in the notion suggested in [Stacks/0DCK](https://stacks.math.columbia.edu/tag/0DCK). In particular, I would like to know if there is a corresponding "algebraic closure" which is unique up to isomorphism - as in the case of fields.
A commutative ring $R$ is *absolutely integrally closed* if every monic polynomial over $R$ has a root in $R$. Equivalently, every monic polynomial over $R$ is a product of linear factors over $R$. (Notice that this differs from E. Enoch's notion of [totally integrally closed rings](https://www.ams.org/journals/proc/1968-019-03/S0002-9939-1968-0224600-9/S0002-9939-1968-0224600-9.pdf).)
It is easy to prove ([Stacks/0DCR](https://stacks.math.columbia.edu/tag/0DCR)) that every commutative ring $R$ has an integral extension $R \hookrightarrow R'$ such that $R'$ is absolutely integrally closed (and $R'$ is a free $R$-module, among other things). But clearly such an extension is not unique: Take any ideal $I \subseteq R'$ with $I \cap R = 0$, then $R \hookrightarrow R'/I$ has the same properties. So we have to make this extension tighter...
Recall that an extension $R \hookrightarrow R'$ is *tight* (aka *essential*) if for all ideals $I \subseteq R'$ we have $I \cap R = 0 \implies I =0$. If $R \hookrightarrow R'$ is any extension, there is always an ideal $J \subseteq R'$ which is maximal with the property $J \cap R = 0$ (Zorn's Lemma), and then $R \hookrightarrow R'/J$ is tight.
So for every commutative ring $R$ there is a tight integral ring extension $R \hookrightarrow R^{\text{aic}}$ such that $R^{\text{aic}}$ is absolutely integrally closed.
**Question.** Does this determine $R^{\text{aic}}$ up to isomorphism? If not, what properties do we need to add to make it unique?
The end result should be called the absolute integral closure of $R$, for short a.i.c. (Unfortunately, the name is already taken for [something else](https://arxiv.org/pdf/1409.0441.pdf) when $R$ is a domain.)
*Edit*. Actually the answer by Matthé proves that the a.i.c. in the case of a domain is what is already understood under this name, namely the integral closure inside the algebraic closure of the field of fractions. In this special case uniqueness holds.
*Edit.* Some simple facts:
(a) The a.i.c of a field is uniquely determined. It is the algebraic closure. (This is because every tight extension of a field is also a field.)
(b) If $R\_1,\dotsc,R\_n$ are commutative rings whose a.i.c. is uniquely determined, is straight forward to check that $R\_1 \times \cdots \times R\_n$ has a uniquely determined a.i.c. as well, with $(R\_1 \times \cdots \times R\_n)^{\text{aic}} = R\_1^{\text{aic}} \times \cdots \times R\_n^{\text{aic}}$.
(c) If $S \subseteq R$ is a multiplicative subset consisting of regular elements and $R^{\text{aic}}$ is some a.i.c. of $R$, then $S^{-1} R^{\text{aic}}$ is an a.i.c. of $S^{-1} R$. I'm not sure yet about uniqueness here, though.
| https://mathoverflow.net/users/2841 | Uniqueness of the "algebraic closure" of a commutative ring | Assume $R$ is reduced and $\mathrm{min}(R)$, the set of minimal prime ideals, is finite. Then $R$ has a *universal* aic, one in which every aic of $R$ can be embedded. We show this below.
Rings of this type have quite remarkable properties, and they include all Noetherian reduced rings. Domains are just the special case where $\mathrm{min}(R)$ is a one-point space. For if $\mathrm{min}(R)$ $=$ $\{\mathfrak{p}\}$, then $\mathfrak{p}$ $=$ $\bigcap\mathrm{min}(R)$ $=$ $\bigcap\mathrm{spec}(R)$ $=$ $\mathrm{nil}(R)$ $=$ $0$.
Let $R$ $\subseteq$ $S$ be a given tight integral extension, with $S$ an absolutely integrally closed ring. Then we have $\mathrm{nil}(S)\cap R$ $=$ $\mathrm{nil}(R)$ $=$ $0$, so by tightness $S$ must also be reduced.
For $\mathfrak{p}$ $\in$ $\mathrm{min}(R)$, we can find a $\mathfrak{P}$ $\in$ $\mathrm{spec}(S)$ with $\mathfrak{P}\cap R$ $=$ $\mathfrak{p}$, because the extension $R$ $\subseteq$ $S$ is integral. Take a minimal prime ideal $\tilde{\mathfrak{p}}$ $\subseteq$ $\mathfrak{P}$ of $S$. Then $\tilde{\mathfrak{p}}\cap R$ $=$ $\mathfrak{p}$. If $I$ $=$ $\bigcap\_{\mathfrak{p}\in\mathrm{min}(R)}\tilde{\mathfrak{p}}$, then $I\cap R$ $=$ $\bigcap\mathrm{min}(R)$ $=$ $\mathrm{nil}(R)$ $=$ $0$, hence $I$ $=$ $0$. So if $\mathfrak{Q}$ $\in$ $\mathrm{min}(S)$, since $\mathrm{min}(R)$ is finite, the product $\prod\_{\mathfrak{p}\in\mathrm{min}(R)}\tilde{\mathfrak{p}}$ exists and is contained in $I$ $=$ $0$ $\subseteq$ $\mathfrak{Q}$. Thus $\mathfrak{Q}$ must be one of the $\tilde{\mathfrak{p}}$. Therefore, $S$ is also "semiglobal", that is to say, $\mathrm{min}(S)$ $=$ $\{\tilde{\mathfrak{p}}\mid\mathfrak{p}\in\mathrm{min}(R)\}$ is finite.
Let $K$ and $L$ be the total rings of fractions of $R$ and $S$, respectively. So $K$ $=$ $\mathrm{reg}(R)^{-1}R$, where $\mathrm{reg}(R)$ is the set of regular elements of $R$. The prime ideals of $K$ are of the form $\mathfrak{p}K$, where $\mathfrak{p}$ $\in$ $\mathrm{spec}(R)$ with $\mathfrak{p}\cap\mathrm{reg}(R)$ $=$ $\varnothing$, i.e. $\mathfrak{p}$ consists of zero divisors of $R$. These $\mathfrak{p}$ are precisely the minimal prime ideals of $R$. For if $\mathfrak{p}$ $\in$ $\mathrm{min}(R)$ and $r$ $\in$ $\mathfrak{p}$, then $r$ $\in$ $\mathfrak{p}R\_\mathfrak{p}$. Being a localisation of a reduced ring, $R\_\mathfrak{p}$ is again reduced. So $0$ $=$ $\mathrm{nil}(R\_\mathfrak{p})$ $=$ $\bigcap\mathrm{spec}(R\_\mathfrak{p})$ $=$ $\mathfrak{p}R\_\mathfrak{p}$, for $\mathfrak{p}R\_\mathfrak{p}$ is the only prime ideal of $R\_\mathfrak{p}$. (Note that $R\_\mathfrak{p}$ therefore must be a field.) So $r$ $=$ $0$ in $R\_\mathfrak{p}$, and hence there is an $r'$ $\in$ $R-\mathfrak{p}$ for which $rr'$ $=$ $0$ in $R$. So $r$ is a zero divisor of $R$. Conversely, if $rr'$ $=$ $0$ and $r'$ $\ne$ $0$, then there is a $\mathfrak{p}$ $\in$ $\mathrm{min}(R)$ with $r'$ $\notin$ $\mathfrak{p}$, because $\bigcap\mathrm{min}(R)$ $=$ $0$. Therefore, $r$ $\in$ $\mathfrak{p}$. So if a prime $\mathfrak{q}$ of $R$ contains only zero divisors, it is contained in $\bigcup\mathrm{min}(R)$. But this a finite union, and so by the [prime avoidance lemma](https://stacks.math.columbia.edu/tag/00DS) the ideal $\mathfrak{q}$ is a minimal prime of $R$.
So $\mathrm{spec}(K)$ $=$ $\{ \mathfrak{p}K\mid\mathfrak{p}\in\mathrm{min}(R)\}$. And $R$ $\subseteq$ $K$, for if $r$ $\in$ $R$ becomes zero in $K$, there is an $r'$ $\in$ $\mathrm{reg}(R)$ with $rr'$ $=$ $0$, hence $r$ $=$ $0$.
Note that $\mathfrak{p}K\cap R$ $=$ $\mathfrak{p}$ when $\mathfrak{p}$ is a minimal prime. Indeed, if $r=u/v$ in $K$ with $u$ $\in$ $\mathfrak{p}$ and $v$ $\in$ $R$ regular, then there is a $w$ $\in$ $\mathrm{reg}(R)$ with $w(vr-u)$ $=$ $0$ in $R$, so $vr$ $=$ $u$ $\in$ $\mathfrak{p}$. If $v$ $\in$ $\mathfrak{p}$, then by the above $v$ is a zero divisor of $R$, contradiction. So $r$ $\in$ $\mathfrak{p}$. As a result, $R/\mathfrak{p}$ $\subseteq$ $K/\mathfrak{p}K$.
It follows that $K$ is a zero-dimensional reduced ring, that is, a von Neumann regular ring. For if $\mathfrak{p}$ and $\mathfrak{q}$ are minimals of $R$ with $\mathfrak{p}K$ $\subseteq$ $\mathfrak{q}K$, then $\mathfrak{p}$ $\subseteq$ $\mathfrak{q}K\cap R$ $=$ $\mathfrak{q}$, hence $\mathfrak{p}$ $=$ $\mathfrak{q}$. So if $\mathfrak{p}$ $\in$ $\mathrm{min}(R)$, we have $\mathfrak{p}K$ $\in$ $\mathrm{max}(K)$, and so $K/\mathfrak{p}K$ is a field. It is in fact the quotient field of $R/\mathfrak{p}$. The ring $L$ is also VNR, and it contains $S$ as a subring.
$R\subseteq S\subseteq L=\mathrm{reg}(S)^{-1}S$, and $\mathrm{reg}(R)$ $\subseteq$ $\mathrm{reg}(S)$. For if $rs$ $=$ $0$ with $r$ $\in$ $R$ and $s$ $\in$ $S-\{0\}$, then $r$ is in a minimal prime $\tilde{\mathfrak{p}}$ of $S$. But then $r$ $\in$ $\mathfrak{p}$ $\in$ $\mathrm{min}(R)$, so $r$ is a zero divisor in $R$. Thus $K$ $=$ $\mathrm{reg}(R)^{-1}R$ $\subseteq$ $\mathrm{reg}(R)^{-1}S$ (since localizations are flat) $\subseteq$ $\mathrm{reg}(S)^{-1}S$ $=$ $L$. Hence $K/\mathfrak{p}K$ is a subfield of $L/\tilde{\mathfrak{p}}L$. And the extension $K/\mathfrak{p}K$ $\subseteq$ $L/\tilde{\mathfrak{p}}L$ is algebraic, because $S$ is integral over $R$.
The natural map $K\to\prod\_{\mathfrak{p}\in\mathrm{min}(R)}K/\mathfrak{p}K$ is injective, for the kernel is the intersection of all prime ideals of $K$, and $K$ is reduced. As $\mathrm{dim}(K)$ $=$ $0$, by the CRT this is actually an isomorphism, and $K$ is a finite product of fields. It is easy to see that in fact $K/\mathfrak{p}K$ $\cong$ $K\_{\mathfrak{p}K}$ $\cong$ $R\_\mathfrak{p}$.
Let $\overline{K}$ $=$ $\prod\_{\mathfrak{p}\in\mathrm{min}(R)}C\_\mathfrak{p}$, where $C\_\mathfrak{p}$ is the algebraic closure of the field $K/\mathfrak{p}K$ (and of $L/\tilde{\mathfrak{p}}L$). This $\overline{K}$ may be regarded as the "algebraic closure" of $R$ (or, equally, of $K$, $S$ or $L$).
I claim that the integral closure $T$ of $R$ in $\overline{K}$ is tight over $R$. For let $I$ be a nonzero ideal of $T$, and take $0$ $\neq$ $t$ $\in$ $I$. For $\mathfrak{p}$ $\in$ $\mathrm{min}(R)$, denote the $\mathfrak{p}$-th coordinate of $t$ by $t\_\mathfrak{p}$, and pick a $\mathfrak{p}$ with $t\_\mathfrak{p}$ $\in$ $C\_\mathfrak{p}$ nonzero. Take a monic $f$ $\in$ $R[X]$ with $f(t)$ $=$ $0$ in $T$. If the constant term $f(0)$ is in $\mathfrak{p}$, it becomes zero in $K/\mathfrak{p}K$, hence in $C\_\mathfrak{p}$. As $t\_\mathfrak{p}$ is a root of the image of $f$ in $C\_\mathfrak{p}[X]$ and $C\_\mathfrak{p}$ is a field, $t\_\mathfrak{p}$ is also a root of (the image of) $g$ $=$ $(f-f(0))/X$ $\in$ $R[X]$. We then replace $f$ by $g$. If the new $f(0)$ is in $\mathfrak{p}$ again, repeat the process until $f(0)$ $\notin$ $\mathfrak{p}$.
By the finiteness of the minimal spectrum of $R$, the product of the minimal primes $\mathfrak{q}$ $\ne$ $\mathfrak{p}$ cannot be contained in $\mathfrak{p}$, so there exists a $c$ $\in$ $R$ that is in all minimals of $R$ except $\mathfrak{p}$. Put $h$ $:=$ $cf$. Then $h(t\_\mathfrak{p})$ $=$ $0$, and for $\mathfrak{p}$ $\ne$ $\mathfrak{q}$ $\in$ $\mathrm{min}(R)$, we have $c$ $=$ $0$ in $K/\mathfrak{q}K$, so $h$ $=$ $0$ in $C\_\mathfrak{q}[X]$. Hence $h(t)$ $=$ $0$ in $T$. But $h(0)$ $=$ $cf(0)$ $\notin$ $\mathfrak{p}$, and therefore $h(0)$ is a nonzero element of $tT$ $\subseteq$ $I$ that is in $R$. This settles tightness.
Since $T$ is integral over $R$ and clearly absolutely integrally closed, it is an absolute integral closure of $R$. The image of $S$ in $\overline{K}$ under the composition map $S\subseteq L\rightarrowtail\prod\_{\mathfrak{p}\in\mathrm{min}(R)}L/\tilde{\mathfrak{p}}L$ $\subseteq$ $\overline{L}$ $=$ $\overline{K}$ is integral over $R$, so it is contained in $T$.
An $e\in\overline{K}$ is idempotent iff for all $\mathfrak{p}$ $\in$ $\mathrm{min}(R)$ its $\mathfrak{p}$-th coordinate is either $0$ or $1$. Clearly, the idempotents are integral over $R$, hence they are in $T$, and there are $2^{|\mathrm{min}(R)|}$ of them. Denote the one with $e\_\mathfrak{p}$ $=$ $1$ and $0$ elsewhere by $e\_{(\mathfrak{p})}$. Then every idempotent is the sum of the elements of a subset of $\{e\_{(\mathfrak{p})}\mid\mathfrak{p}\in\mathrm{min}(R)\}$. And $e\_{(\mathfrak{p})}e\_{(\mathfrak{q})}$ $=$ $0$ when $\mathfrak{p}$ $\ne$ $\mathfrak{q}$. So the $e\_{(\mathfrak{p})}$ form what is known as a fundamental system of orthogonal idempotents.
Then $S=T$ iff the $e\_{(\mathfrak{p})}$ are all in $S$. For if they are, and $t$ $\in$ $T$, fix a $\mathfrak{p}$, and let $f$ $\in$ $R[X]$ be monic with $f(t)$ $=$ $0$. Then $f(t\_\mathfrak{p})$ $=$ $0$ in $C\_\mathfrak{p}$. Write $f$ $=$ $\prod\_{1\leq i\leq n}(X-s\_i)$ in $S[X]$, so that we have $f$ $=$ $\prod\_{1\leq i\leq n}(X-(s\_i)\_{\mathfrak{p}})$ in $C\_\mathfrak{p}[X]$. But then $t\_\mathfrak{p}$ $=$ $(s\_i)\_{\mathfrak{p}}$ for some $i$, because $C\_\mathfrak{p}$ is a field. Put $s$ $=$ $s\_i$. Since $e\_{(\mathfrak{p})}$ is in $S$, so is $e\_{(\mathfrak{p})}s$. Its $\mathfrak{p}$-th component is $t\_\mathfrak{p}$, and it has zero for the other $\mathfrak{q}$. So $e\_{(\mathfrak{p})}s$ is actually equal to $e\_{(\mathfrak{p})}t$. The sum of the $e\_{(\mathfrak{p})}$, taken over the $\mathfrak{p}$ $\in$ $\mathrm{min}(R)$, is equal to $1$. So $t$, which is therefore the sum of the $e\_{(\mathfrak{p})}t$, must be in $S$.
Note that, by integrality, lying-over holds for $S/R$. So when $R$ is a domain, there's a prime $\mathfrak{q}$ of $S$ with $\mathfrak{q}\cap R$ $=$ $0$. As $S$ is tight over $R$, we have $\mathfrak{q}$ $=$ $0$, so $S$ is a domain. Hence it contains $2^{|\mathrm{min}(R)|}$ idempotents, namely just the trivial ones, $0$ and $1$.
To sum up:
**1) $\boldsymbol{T}$ is the universal aic of $\boldsymbol{R}$. That is to say, it contains all other aic's.**
**2) An aic of $\boldsymbol{R}$ is isomorphic to $\boldsymbol{T}$ iff it contains $\boldsymbol{2^{|\mathrm{min}(R)|}}$ idempotents.**
**3) Domains have uniquely determined aic's.**
*Edit.* To this sum we can now add:
**4) If $\boldsymbol{k}$ is a field, then the ring $\boldsymbol{R=k[X,Y]/(XY)}$ has non-isomorphic aic's.**
Here, $\mathrm{min}(R)$ $=$ $\{\mathfrak{p},\mathfrak{q}\}$, with $\mathfrak{p}$ $=$ $(X)$ and $\mathfrak{q}$ $=$ $(Y)$, if we denote the images of $X$ and $Y$ in $R$ by the same symbols. And $K/\mathfrak{p}K$ $=$ $k(Y)$, with algebraic closure $C\_\mathfrak{p}$. The image of $R$ in $C\_\mathfrak{p}$ is the polynomial ring $k[Y]$. Let $T\_\mathfrak{p}$ be the integral closure of $k[Y]$ in $C\_\mathfrak{p}$, and $T\_\mathfrak{q}$ the integral closure of $k[X]$ in $C\_\mathfrak{q}$ $\cong\_k$ $C\_\mathfrak{p}$ (under $X\mapsto Y$). We then have $T$ $=$ $T\_\mathfrak{p}\times T\_\mathfrak{q}$ in $C\_\mathfrak{p}\times C\_\mathfrak{q}$ $=$ $\overline{K}$.
If $\overline{k}$ is the algebraic closure of $k$, there are ring homomorphisms $T\_\mathfrak{p}
\xrightarrow{\varphi}\overline{k}
\xleftarrow{\psi}T\_\mathfrak{q}$ extending the maps $\varphi:k[Y]\to \overline{k}\gets k[X]:\psi$ defined by $Y\mapsto$$\,0\,$↤$\,X$. For, the algebraic closure $C\_\mathfrak{p}$ results from the field $k(Y)$ by a transfinite series of field extensions, indexed by some ordinal number $\alpha$ (which we can take to be $\omega$ $=$ $\aleph\_0$ or the cardinal number of $k$, whichever is the largest of the two, for that must be the cardinality of the field $C\_\mathfrak{p}$). When $\gamma$ $\le$ $\alpha$ is a limit ordinal, the $\gamma$-th step simply consists of taking the union of the fields constructed in the earlier steps, plus taking the union of the maps $\varphi$ constructed. And when $\gamma$ $=$ $\beta+1$ is a successor ordinal, this step is the adjunction of a new algebraic element to the current field $F\_\beta$. By the induction hypothesis, we then have $\varphi:I\_\beta\to\overline{k}$, where $I\_\beta$ is the set of integers of $F\_\beta$ over $k[Y]$, and $F\_\gamma$ $=$ $F\_\beta[Z]/(g)$ for some monic irreducible $g$ $=$ $g(Z)$ $\in$ $F\_\beta[Z]$. If $u$ $\in$ $I\_\gamma-I\_\beta$ is a new integer and $f$ $\in$ $F\_\beta[Z]$ is the minimal polynomial of $u$ over $F\_\beta$, its coefficients are in $I\_\beta$. (They are elementary symmetric functions in the conjugates of $u$ over $F\_\beta$, and each of the latter is integral over $I\_\beta$.) Then $u$ can be identified with $Z\text{ mod }f$ $\in$ $I\_\beta[Z]/(f)$. Take a $v$ in $\overline{k}$ that is a zero of the image of $f$ in $\overline{k}[Z]$ under $\varphi$. Note that the homomorphism $\varphi:I\_\beta[Z]/(f)\to\overline{k}$, for which $Z\text{ mod }f\mapsto v$ and $\varphi\upharpoonright I\_\beta:I\_\beta\to\overline{k}$ is the map already built, is well defined. It is now clear that $\varphi$ extends to $\varphi:I\_\gamma\to\overline{k}$. Ultimately, for $\gamma$ $=$ $\alpha$, we obtain the desired $\varphi:T\_\mathfrak{p}$ $=$ $I\_\alpha\to\overline{k}$. And we can let $\psi:T\_\mathfrak{q}\to\overline{k}$ be the composition $T\_\mathfrak{q}\overset{\sim}{\underset{\text{nat}}{\to}}T\_\mathfrak{p}\underset{\varphi}{\to}\overline{k}$.
Now let $T\_0=\{(u,v)\in T=T\_\mathfrak{p}\times T\_\mathfrak{q}\mid\varphi(u)=\psi(v)\}$ be the pullback. It contains the image of $R$ in $T$. Indeed, every $r$ $\in$ $R$ is of the form $\lambda+Xf(X)+Yg(Y)$ with $\lambda$ $\in$ $k$ and $f$ and $g$ are univariate polynomials. This maps to $(\lambda+Yg(Y),\lambda+Xf(X))$ in $R/\mathfrak{p}\times R/\mathfrak{q}$ $=$ $k[Y]\times k[X]$ $\subseteq$ $T$, and thence to $(\lambda,\lambda)$ in $\overline{k}\times\overline{k}$ under $\varphi\times\psi$, since $\varphi(Y)$ $=$ $0$ $=$ $\psi(X)$. And $T\_0$ is integral and tight over $R$. For let $t\_0$ $=$ $(u,v)$ $\in$ $T\_0-R$. Say $u$ $\ne$ $0$. Then $Yt\_0$ $=$ $(Yu,0)$ is in $T\_0-\{0\}$, and we have $f(u)$ $=$ $0$ for some monic $f$ $=$ $f(Z)$ $\in$ $k[Y][Z]$ with $f(0)$ $\in$ $k[Y]$ non-zero, for example for the minimal polynomial $f$ of $u$ over $k(Y)$. If $f^\bullet$ $\in$ $R[X]$ denotes the same polynomial $f$ $\in$ $k[Y][Z]$ $\subseteq$ $R[Z]$, then $g$ $:=$ $Y^{\mathrm{deg}(f)}f^\bullet(Y^{-1}Z)$ $\in$ $R[Z]$ has $g(Yu)$ $=$ $0$ in $T\_\mathfrak{p}$ and $g(0)$ = $Y^{\mathrm{deg}(f)}f(0)$ $\ne$ $0$. But $Y$ $=$ $0$ in $T\_\mathfrak{q}$, so $g(0)$ vanishes in $T\_\mathfrak{q}$, hence $g(Yt\_0)$ $=$ $0$ in $T\_0$. This shows that $0$ $\ne$ $g(0)$ is in $Yt\_0T\_0\cap R$ $\subseteq$ $t\_0T\_0\cap R$.
Finally, $T\_0$ is an absolutely integrally closed ring. For let $f$ $\in$ $T\_0[Z]$ be monic, of degree $n$, say. Then $f$ $=$ $\prod\_{1\leq i\leq n}(Z-(u\_i,v\_i))$ in $T[Z]$ for suitable $u\_i$ $\in$ $T\_\mathfrak{p}$ and $v\_i$ $\in$ $T\_\mathfrak{q}$, as $T$ is absolutely integrally closed. But then $f$ $=$ $\prod\_{1\leq i\leq n}(Z-(u\_i,v\_{\pi(i)}))$ in $T[Z]$ $=$ $(T\_\mathfrak{p}\times T\_\mathfrak{q})[Z]$ for every permutation $\pi$ of the indices $1,\cdots,n$. For $i$ $<$ $n$, let $(a\_i,b\_i)$ $\in$ $T\_0$ be the coefficient of $Z^i$ in $f$. So $(a\_0,b\_0)$ $=$ $(-1)^n\prod\_{1\leq i\leq n}(u\_i,v\_i)$, and so on, up to $(a\_{n-1},b\_{n-1})$ $=$ $-\sum\_{1\leq i\leq n}(u\_i,v\_i)$. Then we have $(-1)^n\prod\_{1\leq i\leq n}\varphi(u\_i)$ $=$ $\varphi(a\_0)$ $=$ $\psi(b\_0)$ $=$ $(-1)^n\prod\_{1\leq i\leq n}\psi(v\_i)$, and so forth, up to $-\sum\_{1\leq i\leq n}\varphi(u\_i)$ $=$ $\varphi(a\_{n-1})$ $=$ $\psi(b\_{n-1})$ $=$ $-\sum\_{1\leq i\leq n}\psi(v\_i)$. It follows that in the ring $\overline{k}[Z]$ we have the equality $\prod\_{1\leq i\leq n}(Z-\varphi(u\_i))$ $=$ $\prod\_{1\leq i\leq n}(Z-\psi(v\_i))$, and hence $\varphi(u\_i)$ $=$ $\psi(v\_{\pi(i)})$ for all $i$, for some permutation $\pi$.
So $T\_0$ is an aic of $R$. But since $e\_{(\mathfrak{p})}$ $=$ $(1,0)$ is not in $T\_0$, the rings $T\_0$ and $T$ cannot be isomorphic. One is connected, while the other is not. Quod demonstrandum erat, et nunc demonstratum *est*.
**Edit** A paper based on the answer is now on [arXiv](https://arxiv.org/abs/2212.06738).
| 11 | https://mathoverflow.net/users/31923 | 434072 | 175,546 |
https://mathoverflow.net/questions/412068 | 14 | Let's call an $E\_\infty$-algebra $A$ in spaces free if there is a space $A\_0$ and an equivalence of $E\_\infty$-algebras:
$
\coprod\_{n \ge 0} (A\_0)^n\_{h\Sigma\_n} \simeq A.
$
Consider a diagram of $E\_\infty$-algebras in spaces
$$
A \longrightarrow B \longleftarrow C
$$
and assume that each of them is free in the above sense.
However, the maps $A \to B \leftarrow C$ are not required to be free, i.e. they don't have to send $A\_0$ to $B\_0$ or $C\_0$ to $B\_0$.
>
> In this setting, is the homotopy pullback $A \times\_B C$ always free?
>
>
>
---
I've convinced myself in a very roundabout way that this should be true, but would be very grateful for a proper proof or even better a reference.
I think a hands-on approach using the formula for the free $E\_\infty$-algebra might work, but the combinatorics will get quite tricky.
As an example, consider the case where $A, B, C$ are all free on a point and the maps $A \to B \leftarrow C$ both send the generator to twice the generator.
Then I think the resulting pullback $A \times\_B C$ is free on a space with infinitely many components.
Let me give some more details on what happens in this example. I'll think of all the involved 1-types as groupoids, so $A, B, C$ are all represented by the groupoid of finite sets and bijections $\mathrm{FB}$, which is symmetric monoidal under disjoint union.
In my example the two maps correspond to the functor $F: \mathrm{FB} \to \mathrm{FB}$ that sends $a \mapsto a \times \{0,1\}$.
Now the homotopy pullback can be computed in terms of groupoids. The resulting groupoid $\mathcal{G}$ has as objects triples $(a, b, \varphi)$ where $a, b \in \mathrm{FB}$ are finite sets and $\varphi: a \times \{0,1\} \cong b \times \{0,1\}$ is a bijection. Morphisms are tuples of bijections compatible with $\varphi$.
Define the set of connected components of such an object $(a, b, \varphi)$ as the pushout $\pi\_0(a,b,\varphi) = a \amalg\_{a \times \{0,1\}} b$ where the right-hand map is $\varphi$ composed with projection to $b$.
This defines a symmetric monoidal functor $\pi\_0: \mathcal{G} \to \mathrm{FB}$.
Let's say an object is connected if $\pi\_0(a,b,\varphi)$ is a point.
Then I claim that $\mathcal{G}$ is freely generated under disjoint union by connected objects.
This is probably easiest to see by thinking of the objects as directed graphs with red and blue edges, where the vertices are $a \times \{0,1\}$, the blue edges are $(a, 0) \to (a, 1)$, the red edges are $\varphi^{-1}(b,0) \to \varphi^{-1}(b,1)$. Only graphs where every vertex is incident to exactly one blue and one red edge are allowed.
Now this category is a free symmetric monoidal groupoid on the connected graphs, of which there are countably infinitely many.
| https://mathoverflow.net/users/91925 | Are free $E_\infty$-algebras in spaces closed under pullbacks? (as a full subcategory of all $E_\infty$-algebras in spaces) | ### Yes,
the full subcategory
$\mathrm{Mon}\_{\mathbb{E}\_\infty}(\mathcal{S})^{\rm free} \subset \mathrm{Mon}\_{\mathbb{E}\_\infty}(\mathcal{S})$
on those $\mathbb{E}\_\infty$-monoids that are equivalent to a free $\mathbb{E}\_\infty$-monoid, is **closed under all finite limits and retracts.**
This is Corollary 2.1.28 and 2.1.29 of my joint paper with Shaul Barkan. "[The equifibered approach to $\infty$-properads](https://arxiv.org/pdf/2211.02576.pdf)".
While I really like the proof we give, it has the disadvantage of being somewhat implicit:
it doesn't tell you at all what the pullback is free on.
So below I'll give a different proof that isn't in the paper, but might be in a future one.
---
### "Equifibered" maps
This proof uses some of the "equifibered" machinery developed in the aforementioned paper, so I'll recall the needed facts here.
(All of these have elementary proofs, see section 2.1 of the paper.)
We say that a morphism $f\colon M \to N$ in $\mathrm{Mon}\_{\mathbb{E}\_\infty}(\mathcal{S})$ is **equifibered** if the natural map
$$
M \times M \longrightarrow (N \times N) \times\_{N} M,
\qquad
(m\_1,m\_2) \mapsto ( (f(m\_1), f(m\_2)), m\_1+m\_2)
$$
is an equivalence.
(Here the pullback is taken over the addition map and $f$.)
Now one can show several facts about equifibered maps.
I'll write $\mathbb{F}(X) := \coprod\_{n \ge 0} X^n\_{h\Sigma\_n}$ for the free $\mathbb{E}\_\infty$-monoid on a space $X$.
1. For any map of spaces $g:X \to Y$ the induced free map $\mathbb{F}(g)\colon \mathbb{F}(X) \to \mathbb{F}(Y)$ is equifibered.
2. If $f$ is equifibered and $N$ is free, then $M$ is free and $f$ is a free map.
3. Equifibered maps are closed under limits as objects of the arrow category $\mathrm{Ar}(\mathrm{Mon}\_{\mathbb{E}\_\infty}(\mathcal{S}))$.
It follows that $M$ is free if and only if it admits an equifibered map to $\mathbb{F}(\*)$.
---
### The proof:
The idea is to reduce the proof to the following "universal" example:
$$
\mathbb{F}(\mathbb{F}(\*)) \xrightarrow{\ +\ }
\mathbb{F}(\*) \xleftarrow{\ +\ }
\mathbb{F}(\mathbb{F}(\*))
$$
**Claim:**
It suffices to show that the pullback of the "universal" cospan is free.
**Proof idea:**
An arbitrary cospan $f\colon \mathbb{F}(X) \to \mathbb{F}(Z) \leftarrow \mathbb{F}(Y) :\! g$ maps to the universal cospan by using the free map $\mathbb{F}(t): \mathbb{F}(Z) \to \mathbb{F}(\*)$ coming from $t: Z \to \*$
and then factoring the resulting
$$
\mathbb{F}(X) \xrightarrow{f} \mathbb{F}(Z) \xrightarrow{\mathbb{F}(t)} \mathbb{F}(\*)
\qquad \text{ as } \qquad
\mathbb{F}(X) \xrightarrow{\mathbb{F}((\mathbb{F}(t) \circ f)\_{|X})} \mathbb{F}(\mathbb{F}(\*)) \xrightarrow{\*} \mathbb{F}(\*).
$$
So we get a map of cospans and hence a map of pullbacks:
$$
\mathbb{F}(X) \times\_{\mathbb{F}(Z)} \mathbb{F}(Y)
\longrightarrow
\mathbb{F}(\mathbb{F}(\*)) \times\_{\mathbb{F}(\*)} \mathbb{F}(\mathbb{F}(\*))
$$
This map is equifibered because all of the maps between the cospans were free (and hence equifibered by 1) and equifibered maps are closed under pullbacks by 3.
(Note that this works even though the maps within the cospans aren't free.)
Now it follows from 2. that the general pullback is free if the universal one is.
**Proof of the universal case**
To check that the universal pullback is free we can rewrite it as a pullback of symmetric monoidal $1$-groupoids.
It is equivalent to:
$$
\mathrm{Ar}(\mathrm{Fin})^\simeq \xrightarrow{\ s\ }
\mathrm{Fin}^\simeq \xleftarrow{\ s\ }
\mathrm{Ar}(\mathrm{Fin})^\simeq.
$$
Here $\mathrm{Ar}(\mathrm{Fin})^\simeq$ is the maximal subgroupoid of the arrow category of finite sets and $s$ is the functor that sends an arrow to its source.
This pullback is equivalent to the following groupoid of diagrams:
$$
\mathrm{Fun}( \bullet \leftarrow \bullet \rightarrow \bullet, \mathrm{Fin})^\simeq
$$
Now one can check directly that this groupoid is freely generated by those diagrams of finite sets $A \leftarrow C \to B$ where $A \amalg\_C B$ is a single point.
---
### What is the pullback free on?
To answer this question we can first answer it for the universal pullback.
There I claimed that it is free on diagrams of finite sets $A \leftarrow C \to B$ where $A \amalg\_C B$ is a point.
These can be thought of as *connected* bi-partite graphs with vertices $A \amalg B$ and edges $C$.
The more general case will be fibered over this.
Now consider a cospan
$$
\mathbb{F}(X) \xrightarrow{\ f\ } \mathbb{F}(\*) \xleftarrow{\ g\ } \mathbb{F}(Y)
$$
This is not quite the most general case, but almost.
We can decompose $X = \coprod\_{n \ge 0} (X\_n)\_{h\Sigma\_n}$ where $X\_n$ is the fiber of $X \subset \mathbb{F}(X) \xrightarrow{f} \mathbb{F}(\*)$ at some point in $B\Sigma\_n \subset \mathbb{F}(\*)$ and similarly for $Y\_n$.
We can think of $X$ and $Y$ as symmetric sequences and for a finite set $D$ of size $n$ I'll write $X(D) := X\_n$.
Then the pullback is free on
$$
\operatorname\*{colim}\limits\_{A \xleftarrow{u} C \xrightarrow{w} B} \prod\_{a \in A} X(u^{-1}(a)) \times \prod\_{b \in B} Y(w^{-1}(b))
$$
where the colimit runs over connected bipartite graphs.
With a bit of imaginative thinking this can be identified with the *connected composition product for bisymmetric sequences* that Bruno Vallette introduces in his work on properads.
| 7 | https://mathoverflow.net/users/91925 | 434074 | 175,547 |
https://mathoverflow.net/questions/434047 | 3 | Let $(x\_n)\_{n=1}^N$ be a sequence taking values in $[1,2]$ with the property that
$x\_1<x\_2<...<x\_N$ and $$\frac1N \gtrsim \vert x\_j-x\_{j-1} \vert \gtrsim \frac1N.$$
We then define a function
$$f(x) = \sum\_{j=1}^{N} \frac{\alpha\_j}{x-x\_j},$$
where $\alpha\_j$ are positive numbers satisfying $1/N^2 \lesssim \alpha\_i \lesssim 1/N^2.$
I would like to show that any solution to $f(x)=-i$, where $i$ is the imaginary unit, satisfies $\Im(x) \gtrsim 1/N^2.$
The tildes in the inequalities mean we have these estimates up to a constant independent of $j$ and $N$!
| https://mathoverflow.net/users/457901 | Bounds on zeros of rational function | Let us drop the assumption $x\_j\in[1,2]$, it is not needed.
Proving the result by contradiction, denote our function by $f\_N$, suppose that $f\_N(z\_N)=-i$, and $\mathrm{Im}\ z\_N= 1/(N^2R\_N)$ where $R\_N\to\infty$. Since nothing depends on a shift in horizontal direction, one may assume without loss of generality that
$z\_N=i/(N^2R\_N)$, so our equation is
$$\sum\_{j=1}^N\frac{\alpha\_{j,N}N^2R\_N}{i-x\_{j,N}N^2R\_N}=-i.$$
Now consider what happens when $N\to\infty$.
Let us order $x\_j$ by increasing modulus.
Your condition
about $|x\_j-x\_{j+1}|>c/N$ implies that $|x\_{j,N}|\geq c\_1(j-1)/N$ for all $j$ except $j=1$.
Then the sum of all terms for $j\geq 2$ tends to $0$
as $N\to\infty$.
Indeed,
$$\sum\_{j=2}^N\left|\frac{\alpha\_{j,N}N^2R\_N}{i-x\_{j,N}N^2R\_N}\right|=
\sum\_{j=2}^N\frac{\alpha\_{j,N}N^2R\_N}{\sqrt{1+x\_{j,N}^2N^4R\_n^2}}$$
$$\leq
(c\_3/N)\sum\_{j=2}^N1/(j-1)\to 0.$$
So we conclude that the term with $j=1$ must tend to $-i$. But this is impossible, since
$$\frac{\alpha\_{1,N}R\_N N^2}{i-R\_N N^2x\_{1,N}}$$
$$=\frac{-iN^2\alpha\_{1,N}R\_N-\alpha\_{1,N}R\_N^2N^4x\_{1,N}}{1+R\_N^2N^4x\_{1,N}^2},$$
so by comparing the imaginary parts and using that $\alpha\_{1,N}N^2$ is bounded from above and below,
we obtain
$$R\_N\sim R\_N^2x\_{1,N}^2N^4\to\infty,$$
so $x\_{1,N}N^2\to 0$, while comparing imaginary parts gives $N^2x\_{1,N}\to\infty$. This contradiction proves the result.
| 4 | https://mathoverflow.net/users/25510 | 434078 | 175,548 |
https://mathoverflow.net/questions/434076 | 12 | A *pseudo-Kähler manifold* is a complex manifold $(X, I)$ endowed with a non-degenerate closed $(1, 1)$-form $\omega$. In that case, the symmetric tensor $g(\cdot, \cdot) = \omega(\cdot, I \cdot)$ is a [pseudo-Riemannian metric](https://en.wikipedia.org/wiki/Pseudo-Riemannian_manifold).
>
> **Question.** What are examples of compact complex manifolds which are pseudo-Kähler but not Kähler?
>
>
>
| https://mathoverflow.net/users/392184 | Non-Kähler pseudo-Kähler manifolds | I think that the easiest example of compact pseudo-Kähler manifold which does not admit any Kähler metric is the [Kodaira-Thurston manifold](https://www.jstor.org/stable/2041749?origin=crossref#metadata_info_tab_contents). See for instance the introduction of
*Yamada, Takumi*, [**Ricci flatness of certain compact pseudo-Kähler solvmanifolds**](https://doi.org/10.1016/j.geomphys.2011.06.006), J. Geom. Phys. 62, No. 5, 1338-1345 (2012). [ZBL1239.53100](https://zbmath.org/?q=an:1239.53100).
| 13 | https://mathoverflow.net/users/7460 | 434079 | 175,549 |
https://mathoverflow.net/questions/434003 | 2 | Let
* $\Omega$ be a metric space,
* $C\_b(\Omega)$ the space of all real-valued bounded continuous functions on $\Omega$, and
* $\mathcal{M}(\Omega)$ the space of all finite signed Borel measures on $\Omega$.
For $\mu \in \mathcal{M}(\Omega)$, we denote by $|\mu|$ its associated variation measure. We say that a sequence $\left\{\mu\_n\right\} \subset \mathcal{M}(\Omega)$ converges to $\mu \in \mathcal{M}(\Omega)$ weakly if $\int\_\Omega f \mathrm d \mu\_n \to \int\_\Omega f \mathrm d \mu$ for all $f \in C\_b(\Omega)$ and we write $\underset{n \rightarrow \infty}{\operatorname{w-lim}} \, \mu\_n= \mu$;
It is proved in [this answer](https://mathoverflow.net/a/433987/477203) that
>
> **Theorem** Let $\mu\_n,\mu\in \mathcal{M}(\Omega)$ such that that $\underset{n \rightarrow \infty}{\operatorname{w-lim}} \, \mu\_n=\mu$. Then for any open subset $\Theta$ of $\Omega$,
> $$
> |\mu|(\Theta) \leq \liminf \_{n \rightarrow \infty}\left|\mu\_n\right|(\Theta) .
> $$
>
>
>
The essential part of the proof is that given $\varepsilon>0$ there is $f \in C\_b(\Omega)$ such that
$$
|f| \le 1\_\Theta
\quad \text{and} \quad
\int\_\Omega f \mathrm d\mu\ge|\mu|(\Theta)-\varepsilon.
$$
Then by weak convergence of $(\mu\_n)$ we have
$$
|\mu|(\Theta)-\varepsilon \leq \int f \mathrm{~d} \mu=\lim \_{n \rightarrow \infty} \int f \mathrm{~d} \mu\_n \leq \liminf \_{n \rightarrow \infty} \int|f| \mathrm{d}\left|\mu\_n\right| \leq \liminf \_{n \rightarrow \infty}\left|\mu\_n\right|(\Theta).
$$
The result then follows by taking the limit $\varepsilon \to 0^+$.
---
**My understanding:** To have above inequalities, we use the fact that $\mu,\mu\_n$ are **real-valued**.
***Question:*** Can the above theorem be extended to the setting of complex Borel measures?
---
**Update:** Below is my failed attempt. It would be great if it can be fixed into a valid proof. I could not prove that
$$
\liminf \_{n \rightarrow \infty} \big (\left|\mu^1\_n\right|(\Theta) + \left|\mu^2\_n\right|(\Theta) \big ) \le \liminf\_n \left|\mu\_n\right|(\Theta) .
$$
**My attempt:** Let $\mu\_n, \mu$ are complex Borel measures on $\Omega$ such that $\underset{n \rightarrow \infty}{\operatorname{w-lim}} \, \mu\_n= \mu$. Assume $\mu\_n, \mu$ are decomposed into $\mu\_n =\mu\_n^1 + i \mu\_n^2$ and $\mu =\mu^1 + i \mu^2$ where $i$ is the imaginary unit and $\mu\_n^1, \mu\_n^2, \mu^1, \mu^2$ are finite signed Borel measures. We have
$$
\begin{align\*}
\underset{n \rightarrow \infty}{\operatorname{w-lim}} \, \mu\_n = \mu &\iff \int\_\Omega f \mathrm d \mu\_n^1 + i \int\_\Omega f \mathrm d \mu\_n^2 \to \int\_\Omega f \mathrm d \mu^1 + i \int\_\Omega f \mathrm d \mu^2 \quad \forall f \in C\_b(\Omega) \\
&\iff \int\_\Omega f \mathrm d \mu\_n^1 \to \int\_\Omega f \mathrm d \mu^1 \quad \text{and} \quad \int\_\Omega f \mathrm d \mu\_n^2 \to \int\_\Omega f \mathrm d \mu^2 \quad \forall f \in C\_b(\Omega) \\
&\iff \underset{n \rightarrow \infty}{\operatorname{w-lim}} \, \mu^1\_n = \mu^1 \quad \text{and} \quad \underset{n \rightarrow \infty}{\operatorname{w-lim}} \, \mu^2\_n = \mu^2 \\
&\implies |\mu^1|(\Theta) \leq \liminf \_{n \rightarrow \infty}\left|\mu^1\_n\right|(\Theta) \quad \text{and} \quad |\mu^2|(\Theta) \leq \liminf \_{n \rightarrow \infty}\left|\mu^2\_n\right|(\Theta)
\end{align\*}
$$
for all open subsets $\Theta$ of $\Omega$. From [this question](https://mathoverflow.net/questions/434000/complex-borel-measures-relation-between-the-total-variation-norm-of-a-measure-a), we have
$$
|\mu|(\Theta) \le |\mu^1|(\Theta) + |\mu^2|(\Theta).
$$
As such,
$$
|\mu|(\Theta) \le \liminf \_{n \rightarrow \infty}\left|\mu^1\_n\right|(\Theta) + \liminf \_{n \rightarrow \infty}\left|\mu^2\_n\right|(\Theta) \le \liminf \_{n \rightarrow \infty} \big (\left|\mu^1\_n\right|(\Theta) + \left|\mu^2\_n\right|(\Theta) \big ).
$$
| https://mathoverflow.net/users/477203 | Complex Borel measures: does $\mu_n \to \mu$ weakly imply $|\mu|(\Theta) \le \liminf_n |\mu_n|(\Theta)$ for every open subset $\Theta$? | $\newcommand{\Om}{\Omega}\newcommand{\Th}{\Theta}\newcommand{\B}{\mathscr B}\newcommand{\M}{\mathcal M}\newcommand\ep\varepsilon\newcommand{\de}{\delta}\newcommand{\R}{\mathbb R}$By the [polar decomposition of complex measures](https://en.wikipedia.org/wiki/Complex_measure#Variation_of_a_complex_measure_and_polar_decomposition), there is a Borel function $g\colon\Om\to[0,2\pi)$ such that
\begin{equation\*}
|\mu|(\Th)=\int\_\Om 1\_\Th e^{ig}\,d\mu. \tag{1}\label{1}
\end{equation\*}
Take any real $\ep>0$. Next, take any natural
\begin{equation\*}
m>\frac{2\pi |\mu|(\Th)}{\ep/2} \tag{2}\label{2}
\end{equation\*}
and any
\begin{equation\*}
\de\in\Big(0,\frac\ep{2(2m+1)}\Big). \tag{3}\label{3}
\end{equation\*}
For $j\in[m]:=\{1,\dots,m\}$, let
\begin{equation\*}
I\_j:=[\tfrac{2\pi(j-1)}m,\tfrac{2\pi j}m),\ A\_j:=\Th\cap g^{-1}(I\_j), \tag{4}\label{4}
\end{equation\*}
so that the $A\_j$'s are Borel sets forming a partition of $\Th$.
Since $\Om$ is a metric space and $|\mu|$ is a Borel measure, $|\mu|$ is regular. So, for each $j\in[m]$ there exist a closed set $F\_j$ and an open set $G\_j$ such that
\begin{equation\*}
F\_j\subseteq A\_j\subseteq G\_j\text{ and }|\mu|(G\_j\setminus F\_j)<\de, \tag{5}\label{5}
\end{equation\*}
so that the $F\_j$'s are (pairwise) disjoint and for
\begin{equation\*}
F:=\bigcup\_{j\in[m]}F\_j\text{ and }G:=\Th\setminus F \tag{6}\label{6}
\end{equation\*}
we have
\begin{equation\*}
|\mu|(G)=\sum\_{j\in[m]}|\mu|(A\_j\setminus F\_j)<m\de. \tag{8}\label{8}
\end{equation\*}
All metric spaces are normal. So, by [Urysohn's lemma](https://en.wikipedia.org/wiki/Urysohn%27s_lemma), for each $j\in[m]$ there exists a continuous function $h\_j\colon\Om\to\R$ such that
\begin{equation\*}
h\_j=1\text{ on }F\_j,\ h\_j=0\text{ on }G\_j^c:=\Om\setminus G\_j,\ 0\le h\_j\le1. \tag{9}\label{9}
\end{equation\*}
Let
\begin{equation\*}
h:=\sum\_{j\in[m]} \frac{2\pi j}m\,h\_j. \tag{10}\label{10}
\end{equation\*}
Then, by \eqref{6}, \eqref{10}, \eqref{9}, and \eqref{4}, on $F$ we have $0\le h-g\le\frac{2\pi}m$, and hence
\begin{equation\*}
|e^{ih}-e^{ig}|\le\frac{2\pi}m\quad \text{on}\ F. \tag{11}\label{11}
\end{equation\*}
Again by the regularity of $|\mu|$ and Urysohn's lemma, there exist a closed set $F\_0$ and a continuous function $h\_0\colon\Om\to\R$ such that
\begin{equation\*}
F\_0\subseteq\Th,\ |\mu|(\Th\setminus F\_0)<\de, \tag{12}\label{12}
\end{equation\*}
\begin{equation\*}
h\_0=1\text{ on }F\_0,\ h\_0=0\text{ on }\Th^c,\ 0\le h\_0\le1. \tag{13}\label{13}
\end{equation\*}
So, by \eqref{1}, \eqref{13}, \eqref{6}, \eqref{11}, \eqref{12}, \eqref{8}, \eqref{2}, \eqref{3},
\begin{equation\*}
\begin{aligned}
&\Big||\mu|(\Th)-\int\_\Om h\_0 e^{ih}\,d\mu\Big| \\
=&\Big|\int\_\Th e^{ig}\,d\mu-\int\_\Th h\_0 e^{ih}\,d\mu\Big| \\
\le&\int\_\Th |1-h\_0|\,d|\mu|+\int\_\Th|e^{ig}-e^{ih}|\,d|\mu| \\
=&\int\_\Th |1-h\_0|\,d|\mu|+\int\_F|e^{ig}-e^{ih}|\,d|\mu| +\int\_G|e^{ig}-e^{ih}|\,d|\mu| \\
\le&|\mu|(\Th-F\_0)+ \frac{2\pi}m\,|\mu|(F)+2|\mu|(G) \\
\le&\de+ \frac{2\pi}m\,|\mu|(\Th)+2m\de<\ep.
\end{aligned}
\end{equation\*}
So,
\begin{equation\*}
\begin{aligned}
|\mu|(\Th)&=\Re|\mu|(\Th) \\
&\le\ep+\Re\int\_\Om h\_0 e^{ih}\,d\mu \\
&=\ep+\lim\_n\Re\int\_\Om h\_0 e^{ih}\,d\mu\_n \\
&=\ep+\liminf\_n\Re\int\_\Om h\_0 e^{ih}\,d\mu\_n \\
&=\ep+\liminf\_n\Re\int\_\Th h\_0 e^{ih}\,d\mu\_n \\
&\le\ep+\liminf\_n|\mu\_n|(\Th).
\end{aligned}
\end{equation\*}
Letting $\ep\downarrow0$, we conclude that
\begin{equation\*}
|\mu|(\Th)\le\liminf\_n|\mu\_n|(\Th),
\end{equation\*}
as desired.
| 4 | https://mathoverflow.net/users/36721 | 434082 | 175,550 |
https://mathoverflow.net/questions/371012 | 4 | If $K$ is a field and $n \geq 1$ is such that $n \in K^{\times}$, then $H^1\_{et}(\mathrm{Spec}(K),\mu\_n)=K^{\times} / (K^{\times})^n$. This is easy to prove, see for instance Tamme, Etale Cohomology, Corollary 4.4.3. I assume that this is connected, perhaps even implies the main theorem from Kummer theory stating that, if $K$ has a primitive $n$th root of unity, then there is a correspondence between abelian extensions of exponent $n$ and subgroups of $K^{\times}/(K^{\times})^n$. I assume this is because $H^1\_{et}(\mathrm{Spec}(K),\mu\_n) = H^1\_{et}(\mathrm{Spec}(K),\mathbb{Z}/n\mathbb{Z})$ is the Galois group of the maximal abelian extension of exponent $n$ of $K$? If so, can you give me a reference for this result? Sorry if this question is too basic, I didn't touch etale cohomology for years.
| https://mathoverflow.net/users/2841 | Etale cohomology and Kummer theory | The question was answered by R. van Dobben de Bruyn in the comments.
>
> "Serre's Galois cohomology states this relation briefly in a remark
> (p. 73 in §II.1.2 of the second English edition), and refers to
> Bourbaki's Algèbre (Chap. V) for a statement of Kummer theory (but
> Bourbaki doesn't use Galois cohomology). In short: yes, this does
> imply Kummer theory with very little work."
>
>
>
| 0 | https://mathoverflow.net/users/2841 | 434084 | 175,551 |
https://mathoverflow.net/questions/433954 | 2 | History of the question. I was proposing a conjecture [here](https://mathoverflow.net/questions/433839/computational-complexity-and-commuting-functions/433887#433887), called Prop. 1. Fedor Pakhomov showed a counter-example. Here I am proposing a slightly weaker version of the conjecture, Prop. 2, that holds for that counter-example and is still apparently hard to prove.
An old version of the problem is also [here](https://math.stackexchange.com/questions/4557690/computational-complexity-and-commuting-functions).
Here is the question. We have two functions,
$
f: \{0,1\}^\* \to \{0,1\}^\*
$
and
$
g: \{0,1\}^\* \to \{0,1\}^\*
$,
that commute:
$$
f[g(x)] = g[f(x)]
$$
These two functions can be calculated in polynomial time (in the length of the input). Moreover, the outputs have the same length of the inputs: $|f(x)| = |x|$ and $|g(x)| = |x|$ .
A trivial example of functions that commute can be easily constructed by splitting the strings into two parts and defining:
$$
f(x,y) = ( h(x), y )
$$
and
$$
g(x,y) = ( x, l(y) )
$$
where the functions $h(x)$ and $l(y)$ can be calculated in polynomial time (in their inputs).
I was able to construct slightly more complex examples, but not much more complex.
An ingenious example is shown by Fedor Pakhomov as an answer to [my previous question](https://mathoverflow.net/questions/433839/computational-complexity-and-commuting-functions/433887#433887). However, in all the examples, the evolution obtained by repeatedly applying $f$ seems to be either independent of the evolution obtained by repeatedly applying $g$,
or exactly the same. More rigorously, in the examples I have seen, the following proposition holds:
**Proposition 2**
>
> For every polynomially-computable commuting functions $f$ and $g$
> preserving the string length, there is an algorithm that gets
> as input the binary
> representation of two integers, $n$ and $m$, and a string $x$;
> it calculates the
> function $f^n[g^m(x)]$, operating in polynomial
> time (in the length of its input); it
> accesses at most once an oracle that calculates
> $f^{n'}(y)$ and $g^{m'}(z)$ (both),
> with some desired $n'$, $m'$, $y$, and $z$.
>
>
>
**Important note** The expression $f^n$ means $f$ applied $n$ times. For example $f^2(x)$ means $f[f(x)]$, $f^3(x)$ means $f\{f[f(x)]\}$.
I remark that this happens *even if $n$ and $m$ increase exponentially in $|x|$.*
In the trivial example above, setting $n'=n$, $m'=m$, and $y=z=x$, we see that $f^n(x)= ( h^n(x), y )$ and $g^m(x) = (x, l^m(y) )$, from which it is easy to calculate $f^n[g^m(x,y)] = ( h^n(x), l^m(y) ) $.
The example shown by Fedor Pakhomov in the answer to [my previous question](https://mathoverflow.net/questions/433839/computational-complexity-and-commuting-functions/433887#433887) also satisfies Prop. 2. In that case, it is enough
to call an oracle that calculates $f^{n'}(y)$ ($g$ is not needed).
The question is: is Prop. 2 a general theorem? Alternatively, is there a counter-example to Prop. 2?
| https://mathoverflow.net/users/138060 | Computational complexity and commuting functions, examples and conjectures | This is not a proper answer. I will give a construction of a pair of functions $f,g$ assuming the access to some cryptographic function $e$ that probably should form a counterexample for Proposition 2 under some reasonable computability-theoretic assumptions which I will not properly specify. Unfortunately my knowledge of complexity theory is a bit too shaky and hence it is a bit too hard for me to properly isolate the complexity-theoretic and cryptographic assumptions.
Assume that there is a polynomial time-computable length-preservable bijection $e\colon \{0,1\}^\star \to \{0,1\}^\star$ such that its inverse, even when restricted to the sequences consisting just of zeroes, is not polynomial time computable. This essentially is a cryptographic bijective hash function. Given the current state of the field, clearly there are no proofs for such a thing to exist, I don't know if there are reasonable candidates for such a function or if it is known that such kind of function couldn't exist (although, the latter would be very surprising).
Now I will modify an example from my answer to the previous question [Computational complexity and commuting functions](https://mathoverflow.net/questions/433839/computational-complexity-and-commuting-functions/433887#433887)
I will construct functions $f\_1,g\_1$ that work in non-trivial way on strings of the form $\alpha\beta\gamma\delta\epsilon$, where $\alpha,\beta,\gamma\in \{0,1\}^Q(n)$ and $\delta,\epsilon\in\{0,1\}^n$. Naturally I will treat $\delta,\epsilon$ as elements of the additive group $\mathbb{Z}/\mathbb{Z}2^n$. Given an input $\alpha\beta\gamma\delta\epsilon$, let
1. $\alpha''=\alpha$ if $e(\delta)=0^n$ and $\alpha''=0$, otherwise;
2. $\beta''=\beta$ if $e(\epsilon)=0^n$ and $\beta''=0$, otherwise;
3. $\delta'=\delta$ if $e(\delta)=0^n$ and $\delta+1$, otherwise;
4. $\epsilon'=\epsilon$ if $e(\epsilon)=0^n$ and $\epsilon+1$, otherwise;
5. $\alpha'''=\min(\alpha''+1,2^{Q(n)}-1)$ if $e(\delta)=0^n$ and $\alpha'''=0$, otherwise;
6. $\beta'''=\min(\beta''+1,2^{Q(n)}-1)$ if $e(\epsilon)=0^n$ and $\beta'''=0$, otherwise;
7. $f\_1(\alpha\beta\gamma\delta\epsilon)=\alpha'''\beta''h^{\min(\alpha''',\beta'')-\min(\alpha'',\beta'')}(\gamma)\delta'\epsilon$;
8. $g\_1(\alpha\beta\gamma\delta\epsilon)=\alpha''\beta'''h^{\min(\alpha'',\beta''')-\min(\alpha'',\beta'')}(\gamma)\delta\epsilon'$.
The idea is essentially the same as in the previous example but now the precondition for two counter behaviour as in the previous example is that the unbounded search have already found a hard to find witness. Note that unless $e(\delta)=0^n$ (and $e^{-1}(0^n)$ should be hard to find), $f\_1^{n'}(\alpha\beta\gamma\delta\epsilon)$ will be polynomial-time computable from $e^{-1}(0^n)$. So the idea of why Proposition 2 should fail is that for polynomial-time computable $n',m',z,y$, the only kind of extra information that could be in general extracted from $f\_1^{n'}(y)$ and $g\_1^{m'}(z)$ are $e$-preimages of some strings consisting just of zeroes. This $e$-preimages shouldn't be of much help to compute instances of the $\mathsf{PSPACE}$-complete problem in vast majority of cases.
This of course, is just a rough idea of why I expect examples of this kind to be counterexamples for your Proposition 2. The problem with formulating proper assumptions here is that we want to exclude several possible non-trivial interactions between the function $e$ and the hardness of $\mathsf{PSPACE}$-complete problem of choice. I could formulate something, but it will look quite ugly. Hope that nevertheless this answer transmits the intuition of why I think the conjecture from the question to be false.
| 1 | https://mathoverflow.net/users/36385 | 434101 | 175,559 |
https://mathoverflow.net/questions/434096 | 2 | We know that a coherent sheaf on a scheme is determined by its restriction on certain open coverings (satisfying compatibility condition). Now I wonder how about a closed covering. To do so I started with simple cases on a smooth complex projective varieity $X$ of dimension $n$.
Taking $X$ itself to be the covering is trivial, so I want to start with coverings of smalll dimensions. Firstly I think about the closed points (covering of dimension zero), but they are not a covering; also coherent sheaf with same stalks can be different.
Then it comes to dimension one coverings: let $X$ a smooth projective variety which can be covered by lines. Given two coherent sheaves $\mathcal{F}$ and $\mathcal{G}$ on $X$ such that
$$\iota^\*\_L\mathcal{F}\cong\iota^\*\_L\mathcal{G}$$
for any line $\iota\_L:L\subset X$. What could one say about the relation between $\mathcal{F}$ and $\mathcal{G}$?
I think we should have $\mathcal{F}\cong\mathcal{G}$ for $X=\mathbb{P}^n$ by the structure of $\textbf{Coh}(\mathbb{P}^n)$ (edited: no we do not, here I should write $\mathbb{P}^1$ because the vector bundles on $\mathbb{P}^n$ are not necessarily decomposable).
What about the general case, for example a ruled surface?
I think such questions should be considered before. If so, any reference is welcome!
| https://mathoverflow.net/users/nan | Could certain closed covering determine a coherent sheaf? | It is not true even for $\mathbb{P}^n$. For instance, the tangent bundle $T\_{\mathbb{P}^n}$ restricts to each line as
$$
T\_{\mathbb{P}^n}\vert\_L \cong \mathcal{O}\_L(2) \oplus \mathcal{O}\_L(1)^{\oplus (n-1)},
$$
and on the other hand
$$
(\mathcal{O}\_{\mathbb{P}^n}(2) \oplus \mathcal{O}\_{\mathbb{P}^n}(1)^{\oplus (n-1)})\vert\_L \cong \mathcal{O}\_L(2) \oplus \mathcal{O}\_L(1)^{\oplus (n-1)},
$$
however $T\_{\mathbb{P}^n} \not\cong (\mathcal{O}\_{\mathbb{P}^n}(2) \oplus \mathcal{O}\_{\mathbb{P}^n}(1)^{\oplus (n-1)})$.
| 2 | https://mathoverflow.net/users/4428 | 434102 | 175,560 |
https://mathoverflow.net/questions/225235 | 0 | It is well known that for an Stochastic differential equation (on the real line) of the form:
$dX\_t = \mu(X\_t)dt + \sigma(X\_t)dW$
where $W$ is the standard Wiener process, the transition probability densities of the process can be related to a PDE of the form
$\frac{\partial u}{\partial t} = \frac{\partial^2}{\partial x^2}(a^2(x)u)-\frac{\partial}{\partial x}(v(x)u)$. Using standard PDE theory it can be proved that weak solutions of the PDE exist in suitably weighted Lesbesgue spaces for essentially bounded coefficients $a^2(x)$ and $v(x)$ with suitable growth conditions (and ellipticity). However almost all books/papers on SDEs I see that $\mu$ and $\sigma$ are taken to be at least continuous. Are there any counterexamples where solutions for PDEs exist but not for the corresponding SDE? Or even if solutions for the SDE exist is there a counterexample such that the sample paths are not continuous any more?
| https://mathoverflow.net/users/83631 | Weak solutions of linear parabolic PDEs and corresponding SDEs | One of the latest conditions on $\mu,\sigma$ are from ["A Numerical Method for SDEs with Discontinuous Drift"](https://arxiv.org/pdf/1503.08005.pdf):
>
> This result states that the SDE (1) ($dX\_t = \mu(X\_t)dt + \sigma(X\_t)dW$) admits a unique strong solution $X$ if the drift coefficient $\mu$ has finitely many discontinuity points and is piecewise Lipschitz continuous and the diffusion coefficient $\sigma$ is globally Lipschitz continuous and non-degenerate at the discontinuity points of $\mu$.
>
>
>
Also, as mentioned in the comments, one can even do with just bounded, measurable $\mu$ (cf ["A Numerical Method for SDEs with Discontinuous Drift"](https://arxiv.org/pdf/1503.08005.pdf)):
>
> In the case where the diffusion coefficient $\sigma$ is bounded, Lipschitz, and (partly) uniformly elliptic, and the drift coefficient $\mu$ is only bounded and measurable, the pioneering work by Zvonkin [15] and Veretennikov [13, 14] yields existence and uniqueness of the solution.
>
>
>
It seems to be the latest one based on overview here ["Existence, uniqueness and approximation of solutions of SDEs with superlinear coefficients in the presence of discontinuities of the drift coefficient'](https://arxiv.org/abs/2204.02343)
see here [What work has been done on SDE with diffusion coefficients of bounded variation in $\mathbb R^d$?](https://mathoverflow.net/questions/432599/what-work-has-been-done-on-sde-with-diffusion-coefficients-of-bounded-variation) too for more references.
A nice result is also in Revuz-Yor book "Continuous Martingales and Brownian Motion" in (1.14) Corollary:
>
> If $\sigma$ is a bounded function on the line such that $|\sigma(x,t)|\geq \epsilon>0$ and $\mu$ a bounded function on $R\_{+}\times R$ there is existence and uniquenss in law for
> the SDE.
>
>
>
| 0 | https://mathoverflow.net/users/99863 | 434120 | 175,569 |
https://mathoverflow.net/questions/434000 | 0 | Let $X$ be a metric space and $\mathcal B$ its Borel $\sigma$-algebra. For $B \in \mathcal B$ we denote by $\Pi(B)$ the collection of all finite measurable partitions of $B$, i.e.,
$$
\Pi(B)=\left\{\left(B\_{1}, \ldots, B\_{n}\right) \,\middle\vert\, n \in \mathbb{N^\*}, B\_{i} \in \mathcal B, B\_{i} \cap B\_{j}=\varnothing \text { for } 1 \leq i \neq j \leq n, \bigcup\_{i=1}^{n} B\_{i}=B\right\} .
$$
Let $\mu$ be a complex Borel measure on $X$. The variation $|\mu|$ of $\mu$ is defined by
$$
|\mu|(B) := \sup \left\{\sum\_{i=1}^{n}\left|\mu\left(B\_{i}\right)\right| \,\middle\vert\, \left(B\_{1}, \ldots, B\_{n}\right) \in \Pi(B)\right\} \quad \forall B \in \mathcal B.
$$
Let $[\mu] :=|\mu|(X)$ be the total variation norm of $\mu$. Let $\mu\_1, \mu\_2$ be the real and imaginary parts of $\mu$ respectively, i.e., $\mu = \mu\_1 + i\mu\_2$. Then $\mu\_1, \mu\_2$ are finite signed Borel measures on $X$. So for each $B \in \mathcal B$ we have
$$
\begin{align}
|\mu|(B) &= \sup \left\{\sum\_{i=1}^{n} \sqrt{|\mu\_1(B\_{i})|^2 + |\mu\_2(B\_{i})|^2} \,\middle\vert\, \left(B\_{1}, \ldots, B\_{n}\right) \in \Pi(B)\right\} \\
&\le \sup \left\{\sum\_{i=1}^{n} (|\mu\_1(B\_{i})| + |\mu\_2(B\_{i})|) \,\middle\vert\, \left(B\_{1}, \ldots, B\_{n}\right) \in \Pi(B)\right\} \\
&\le \sup \left\{\sum\_{i=1}^{n} |\mu\_1(B\_i)| \,\middle\vert\, \left(B\_{1}, \ldots, B\_{n}\right) \in \Pi(B)\right\} \\
&\quad + \sup \left\{\sum\_{i=1}^{n} |\mu\_2(B\_i)| \,\middle\vert\, \left(B\_{1}, \ldots, B\_{n}\right) \in \Pi(B)\right\} \\
&= |\mu\_1| (B) + \mu\_2(B).
\end{align}
$$
As such, $|\mu| \le |\mu\_1| + |\mu\_2|$. In particular, $[\mu] \le [\mu\_1] + [\mu\_2]$.
>
> I would like to ask if either $|\mu| \ge |\mu\_1| + |\mu\_2|$ or $[\mu] \ge [\mu\_1] + [\mu\_2]$ is true.
>
>
>
Thank you so much for your elaboration!
---
**Update:** Let's define a new variation on the space of complex Borel measures on $X$.
* For a finite signed Borel measure $\mu$, its new variation is $|\mu|' := |\mu|$.
* For a complex Borel measure $\mu = \mu\_1 + i\mu\_2$ with $\mu\_1, \mu\_2$ being its real and imaginary parts, its new variation is $|\mu|' := |\mu\_1| + |\mu\_2|$.
Then for any complex Borel measure $\mu$, we have
$$
\frac{1}{2} |\mu|' \le |\mu| \le |\mu|'.
$$
We define $[\mu]':= |\mu|' (X)$. Then $[\cdot]'$ is a norm on the space of complex Borel measures such that
$$
\frac{1}{2} [\cdot]' \le [\cdot] \le [\cdot]'.
$$
It follows that
* $|\mu|' = |\mu\_1|' + |\mu\_2|'$ and thus $[\mu]' = [\mu\_1]' + [\mu\_2]'$ for every complex Borel measure $\mu$ whose real and imaginary parts are $\mu\_1$ and $\mu\_2$ respectively.
* $[\cdot]$ and $[\cdot]'$ are [equivalent norms](https://en.wikipedia.org/wiki/Norm_(mathematics)#Equivalent_norms) on the space of complex Borel measures.
* $[\cdot]$ and $[\cdot]'$ coincide on the subspace of finite signed Borel measures.
| https://mathoverflow.net/users/477203 | Complex Borel measures: relation between the total variation norm of a measure and those of its real and imaginary parts | I represent below @NikWeaver's idea of improving $\frac{1}{2} [\cdot]' \le [\cdot]$ to get $\frac{1}{\sqrt 2} [\cdot]' \le [\cdot]$.
---
For complex number $z = x + iy$ with $x,y \in \mathbb R$, we have
$$
|z| \ge \frac{|x| +|y|}{\sqrt{2}} .
$$
Fix a Borel subset $B$ of $X$. Then
$$
\begin{align}
|\mu|(B) &= \sup \left\{\sum\_{i=1}^{n} \sqrt{|\mu\_1(B\_{i})|^2 + |\mu\_2(B\_{i})|^2} \,\middle\vert\, \left(B\_{1}, \ldots, B\_{n}\right) \in \Pi(B)\right\} \\
&\ge \frac{1}{\sqrt 2} \sup \left\{\sum\_{i=1}^{n} |\mu\_1(B\_{i})| + |\mu\_2(B\_{i})| \,\middle\vert\, \left(B\_{1}, \ldots, B\_{n}\right) \in \Pi(B)\right\}.
\end{align}
$$
It remains to prove that
$$
\begin{aligned}
&\sup \left\{ \sum\_{i=1}^{n} |\mu\_1(B\_{i})| + |\mu\_2(B\_{i})| \,\middle\vert\, (B\_{1}, \ldots, B\_{n}) \in \Pi(B)\right\} \\
= & \sup \left\{ \sum\_{i=1}^{n} |\mu\_1(B\_{i})| \,\middle\vert\, (B\_{1}, \ldots, B\_{n}) \in \Pi(B) \right\} + \sup \left\{\sum\_{i=1}^{n} |\mu\_2(B\_{i})| \,\middle\vert\, (B\_{1}, \ldots, B\_{n}) \in \Pi(B) \right\}.
\end{aligned}
$$
The direction $\le$ is obvious. Let's prove the reverse $\ge$. For each $n$, let
* $(B\_{1,1}, \ldots, B\_{1, \varphi\_n}) \in \Pi(B)$ such that $\sum\_{i=1}^{\varphi\_n} |\mu\_1(B\_{1, i})| \nearrow |\mu\_1| (B)$.
* $(B\_{2,1}, \ldots, B\_{2, \psi\_n}) \in \Pi(B)$ such that $\sum\_{i=1}^{\psi\_n} |\mu\_2(B\_{2, i})| \nearrow |\mu\_2| (B)$.
* $(B\_{3, 1}, \ldots, B\_{3, \lambda\_n}) :=\{B\_{1, i} \cap B\_{2, j} \mid i = 1, \ldots, \varphi\_n \text{ and } j = 1, \ldots, \psi\_n\} \in \Pi(B)$.
Then $(B\_{3, 1}, \ldots, B\_{3, \lambda\_n}) \in B$ is finer than both $(B\_{1,1}, \ldots, B\_{1, \varphi\_n})$ and $(B\_{2,1}, \ldots, B\_{2, \psi\_n})$. By triangle inequality, we have
$$
\sum\_{i=1}^{\varphi\_n} |\mu\_1(B\_{1, i})| \le \sum\_{i=1}^{\lambda\_n} |\mu\_1(B\_{3, i})|
\quad \text{and} \quad
\sum\_{i=1}^{\psi\_n} |\mu\_2(B\_{2, i})| \le \sum\_{i=1}^{\lambda\_n} |\mu\_2(B\_{3, i})|.
$$
It follows that
$$
\sum\_{i=1}^{\lambda\_n} |\mu\_1(B\_{3,i})| + |\mu\_2(B\_{3,i})| \ge \sum\_{i=1}^{\varphi\_n} |\mu\_1(B\_{1, i})| + \sum\_{i=1}^{\psi\_n} |\mu\_2(B\_{2, i})| .
$$
The claim then follows.
| 0 | https://mathoverflow.net/users/477203 | 434123 | 175,571 |
https://mathoverflow.net/questions/428422 | 7 | Let $G$ be a countable amenable group with a (left) Følner sequence $(F\_n)$. Let $\Gamma$ be a subset of $G$ that has density $1$ with respect to $(F\_n)$, in the sense that
$$
\lim\_{n \to \infty} \frac{\lvert\Gamma \cap F\_n\rvert}{\lvert F\_n\rvert} \ = \ 1.
$$
Now define
$$
\Gamma' := \{ (s,t) \in G^2 : st^{-1} \in \Gamma \}.
$$
Is it true that $\Gamma'$ has density $1$ in $G^2$ with respect to $(F\_n \times F\_n)$?
A starting point is the calculation
$$
\lvert\Gamma' \cap F\_n^2\rvert = \sum\_{\gamma \in \Gamma} \#\{(s,t) \in F\_n^2 : st^{-1} = \gamma \} \ = \ \sum\_{\gamma \in \Gamma} \lvert F\_n \cap \gamma^{-1} F\_n\rvert.
$$
In the special case $G = \mathbb{Z}$, $F\_n = [-n,n]$, the result is true, but my argument uses the geometry of $\mathbb{Z}$ quite strongly. The argument is: in this case, $\lvert F\_n \cap \gamma^{-1}F\_n\rvert = \max(2n+1 - \lvert\gamma\rvert, 0)$, so if $n$ is chosen large enough so that $\lvert\Gamma \cap F\_n\rvert > (1-\epsilon)|F\_n|$, then
$$
\sum\_{\gamma \in \Gamma} \lvert F\_n \cap \gamma^{-1} F\_n\rvert \ \geq \ \sum\_{\epsilon \cdot n < \lvert\gamma\rvert \leq 2n} (2n+1-\lvert\gamma\rvert) \ \geq \ (2n+1)^2 \cdot (1 - O(\epsilon)) \ = \ \lvert F\_n^2\rvert(1-O(\epsilon))
$$
as desired.
Is this result true for all countable amenable groups?
| https://mathoverflow.net/users/166445 | Density of “diagonal sets” in amenable groups | The answer to your question as stated is "no", but a variant of it is true (see the proposition below).
**Proof that the answer is "no":** Let $(F\_n)$ be the Følner sequence in $\mathbb{Z}$ given by $F\_n = [2^n, 2^n + n]$, and let $\Gamma = \bigcup\_{n \in \mathbb{N}}{F\_n}$ so that $\Gamma$ has full density along $(F\_n)$. Then $$|\Gamma' \cap (F\_n \times F\_n)| = \sum\_{t \in F\_n}{|\Gamma \cap (F\_n - t)|}.$$
But for $t \in F\_n$, $F\_n - t \subset F\_n - F\_n = [-n, n]$, so
$$\frac{|\Gamma' \cap (F\_n \times F\_n)|}{|F\_n|^2} \le \frac{|\Gamma \cap [-n,n]|}{n},$$
and it's easy to check that
$$\lim\_{n \to \infty}{\frac{|\Gamma \cap [-n,n]|}{n}} = 0.$$//
The key to this counterexample is that $F\_n - F\_n$ behaves very differently from $F\_n$ (in particular, they are disjoint sets). For the special example you considered (with $F\_n = [-n,n]$), the difference set $F\_n - F\_n = [-2n, 2n]$ has substantial overlap with $F\_n$, and that's what you were able to utilize.
For general Følner sequences, the basic idea behind your argument can still be adapted to prove the following weaker result:
**Proposition:** Let $G$ be a countable amenable group. If $\Gamma \subseteq G$ has full density along a Følner sequence $(F\_n)$, then there exists a Følner sequence $(\Phi\_n)$ in $G^2$ such that $\Gamma'$ has full density along $(\Phi\_n)$.
**Proof sketch:** For each $n \in \mathbb{N}$, choose $N\_n$ so that $F\_{N\_n}$ is almost-invariant with respect to shifts coming from $F\_n$, say
$$\frac{|F\_{N\_n} \cap (F\_{N\_n}t^{-1})|}{|F\_{N\_n}|} > 1 - \frac{1}{n}$$
for $t \in F\_n$.
By assumption, $|\Gamma \cap F\_{N\_n}| > (1 - o(1)) |F\_{N\_n}|$, so
$$|\Gamma' \cap (F\_{N\_n} \times F\_n)| = \sum\_{t \in F\_n}{|\Gamma \cap F\_{N\_n}t^{-1}|} \ge |F\_n| \left( 1 - o(1) - \frac{1}{n} \right) |F\_{N\_n}| = (1 - o(1)) |F\_{N\_n} \times F\_n|.$$//
The Proposition can also be proved another (in my opinion, easier) way using the fact that a set has upper Banach density equal to 1 (i.e., full density along some Følner sequence) if and only if the set is thick (contains a translate of every finite set).
**Proof sketch for thick sets:** Assume $\Gamma$ is thick in $G$. We want to show that $\Gamma'$ is thick in $G^2$. Let $F \subseteq G^2$ be a finite set. Then let $\tilde{F} = \{st^{-1} : (s, t) \in F\}$. This is a finite set, so there exists $x \in G$ such that $x\tilde{F} \subset \Gamma$, since $\Gamma$ is a thick subset of $G$. A simple calculation shows $(x,1)F \subset \Gamma'$.
| 2 | https://mathoverflow.net/users/171304 | 434124 | 175,572 |
https://mathoverflow.net/questions/434127 | 3 | Let $M = G/K$ be a compact homogeneous space, i.e. $G$ a connected compact Lie group, and $K$ a closed subgroup inside $G$ that contains no nontrivial normal subgroup of $G$.
1. If $r(G) > r(K)$ (here, $r$ means the rank of maximal torus), is it true that $G/K$ admits free $S^1$-action? If so, why? If not, what is a counterexample?
2. What if we further assume $K$ is connected?
(Note: In “Homogeneous Spaces with Non-Vanishing Euler Characteristics” by HC Wang (1949), it was proven that $\chi(G/K) \ne 0 \Leftrightarrow r(G) = r(K)$.)
3. If 1. is unknown, then, restricting to case when $G$ semisimple, is 1. known?
Any input would be really helpful. Thanks a lot!
| https://mathoverflow.net/users/166199 | Free $S^1$-action on compact homogeneous spaces | Here is a counterexample where $K$ is connected.
Let $W = SU\_3/SO\_3$ be the Wu manifold. This 5-dimensional manifold is simply-connected and has $H\_2(W;\Bbb Z) = \Bbb Z/2$, and in particular, $\pi\_2(W) = \Bbb Z/2$ by the Hurewicz theorem. Note that $r(SU\_3) = 2$ and $r(SO\_3) = 1$.
Now I claim that there is no free circle action on $W$. For if $M$ is any manifold with free circle action and quotient $X$, the projection defines a fiber-bundle $$S^1 \to M \to X.$$
* If $M$ is simply-connected, the long exact sequence of homotopy groups implies $X$ is, too.
* If $M$ is further a simply-connected 5-manifold, so that $X$ is now a simply-connected 4-manifold, observe that by Poincare duality $H\_2(X;\Bbb Z)$ is free abelian. The long exact sequence of homotopy groups gives an injection of $H\_2(W;\Bbb Z) \cong \pi\_2(W)$ into $\pi\_2(X) \cong H\_2(X;\Bbb Z)$. Thus if $M$ is a simply-connected 5-manifold which supports a free circle action, $\pi\_2(M)$ must be free abelian.
Because the Wu manifold has $\pi\_2(W) = \Bbb Z/2$, it does not support a free circle action. This answers your question in the negative, and unfortunately I do not see a way to salvage it by adding more conditions.
| 2 | https://mathoverflow.net/users/40804 | 434162 | 175,580 |
https://mathoverflow.net/questions/434132 | 2 | Let $K\_0$ and $K\_1$ be two knots in $S^3$. We say $K\_0$ and $K\_1$ are *concordant* if there exists a smoothly embedded annulus $A \subset S^3 \times [0,1] $ such that $\partial A = -(K\_0) \cup K\_1$.
Given two **non-trivial** concordant knots $K\_0$ and $K\_1$, assume that one of them is hyperbolic, say $K\_0$. Is it possible to show that $K\_1$ must be hyperbolic?
We may ask the similar question by changing the "hyperbolic" notion with "amphichiral". In other words, does the knot concordance preserve the hyperbolicity or amphichirality of the knot?
| https://mathoverflow.net/users/475366 | Knot concordance, hyperbolicity and amphichirality | Neither of these properties are preserved by concordance. As was pointed out in the comments, any hyperbolic (for the first question) or chiral (for the second question) knot which is concordant to the unknot will be a counter-example. For a specific example, the knot 6\_1 is slice (concordant to the unknot), but is hyperbolic and is not amphichiral.
In fact, in "Homology cobordisms, link concordances, and hyperbolic 3-manifolds", Myers showed that *every* knot is concordant to a hyperbolic knot.
| 6 | https://mathoverflow.net/users/173304 | 434167 | 175,582 |
https://mathoverflow.net/questions/433907 | 26 | *Update.* It's now on the [arXiv](https://arxiv.org/abs/2211.07508).
---
Some time ago I found my "own" proof of the [fundamental theorem of Galois theory](https://en.wikipedia.org/wiki/Fundamental_theorem_of_Galois_theory). You can find a pdf with the proof (link removed, see arXiv). It is quite short, self-contained, and uses a neat combinatorial argument:
>
> A field cannot be written as a union of finitely many proper subfields
>
>
>
Most users of mathoverflow can simply skip most of it and only read the combinatorial Lemma 3.3. which leads to Prop. 4.2, as well as Lemma 5.5 which leads to to Prop. 6.3. The rest is easy.
I wonder if this proof is new or not. For sure I have never seen it before, and I checked a bit the literature and couldn't find it so far. But also I am not really an expert on the history of algebra at all, and my days as an "active mathematician" are over anyway. Hopefully somebody else has a better overview?
| https://mathoverflow.net/users/2841 | A simple proof of the fundamental theorem of Galois theory | At a first glance your approach reminds me of [Meinolf Geck's](https://www.jstor.org/stable/10.4169/amer.math.monthly.121.07.637#metadata_info_tab_contents) American Mathematical Monthly article, see also the [arxiv version](https://arxiv.org/abs/1306.3853) of his article.
| 9 | https://mathoverflow.net/users/18739 | 434176 | 175,586 |
https://mathoverflow.net/questions/434161 | 3 | Considering a quotient singularity $\mathbb{C}^n/G,$ its crepant resolution $Y$ (i.e. having $c\_1(Y)=0$) has rational cohomology supported in even degrees only. This holds for many other resolutions of singularities, in particular for symplectic resolutions of conic symplectic singularities.
I am wondering whether there is a general statement saying that resolution $Y$, satisfying $c\_1(Y)=0$,
of a conic singular affine variety $X$ with isolated singularitity has $H^{odd}(Y,\mathbb{Q})=0$? Conic means that there is a $\mathbb{C}^\*$-action on $X$ that contracts it to a point.
Potentially ask that $X$ is a complete intersection.
| https://mathoverflow.net/users/114985 | Resolution of conical singularities have even-only cohomology? | Thanks to @Yosemite Stan, we have a counterexample, and actually many of them:
Pick a projective variety $Z$ with some odd-cohomology such that
$$c\_1(\omega\_Z)=-m c\_1(H), \text{ for some } m>0,$$
where $H$ is the bundle by which $Z$ embeds to $\mathbb{P}^n,$
and $\omega\_Z$ is the canonical bundle.
In other words, we have $c\_1(Z)=-c\_1(\omega\_Z)=m c\_1(i^\*\mathcal{O}(1)),$ where $i:Z \hookrightarrow \mathbb{P}^n$ is the embedding given by $H.$
In particular, a degree $d$-hypersurface in $\mathbb{P}^n$ having some odd cohomology works, with $m = n+1-d$ (due to adjunction formula).
Denoting by $\widetilde{Z}$ its image via the Veronese map
$\mathbb{P}^n\rightarrow\mathbb{P^{\binom{n+m}{m}-1}}$, define $X\subset \mathbb{C}^{\binom{n+m}{m}}$ to be the cone over $\widetilde{Z}$.
There is a natural resolution $Y$ of $X$ given by the $Y=\overline{\pi^{-1}(X\setminus 0)},$ where $\pi$
is the blow-up at ${0}\subset \mathbb{C}^{\binom{n+m}{m}}.$
As the blow-up at $0$ is $\mathcal{O}(-1)\rightarrow \mathbb{P^{\binom{n+m}{m}-1}},$ we have
$$Y=\mathcal{O}(-1)|\_\widetilde{Z} \cong \mathcal{O}(-m)|\_Z,$$ hence it has
$$c\_1(Y)=c\_1(i^\*\mathcal{O}(-m))+c\_1(Z)=-mc\_1(i^\*\mathcal{O}(1)+c\_1(Z)=0,$$ and cohomology equal to $Z,$ hence not supported in even degrees only.
In particular, choosing the quartic 3-fold $Z\_4\subset \mathbb{P}^4$, we have $b\_3(Z\_4)=60$ and $m=(4+1)-4=1,$
so $X=V(z\_0^4+\dots+z\_4^4)$ is an isolated singularity and (trivially) a complete intersection as well, and its resolution given by the blow up of $\mathbb{C}^5$ at the origin yields a counterexample.
| 2 | https://mathoverflow.net/users/114985 | 434179 | 175,588 |
https://mathoverflow.net/questions/370484 | 0 | I have been tyring to understand the first condition given in the link <https://en.wikipedia.org/wiki/Regularity_structure> for quite some time now, at least a year. I have posted a similar question in <https://math.stackexchange.com/questions/3524946/how-to-think-of-a-set-that-has-no-accumulation-point> but did not get any reply. Please help me if possible.
As I understand the regularity structure deals with stochastic partial differential equations and this means we deal with stochastic variable or functions which are continous but not differentiable, this leads to discrete type of sets and hence we get the condition "no accumulation point" is this correct?
If so, the question is why it has to be bounded from below only? I was expecting it to be bounded from below as well as above or bounded.
Few words why this condition is necessary in regularity structure would be highly appreciated?
**Edit:** The comment from Nate Eldredge made it clear why it is not bounded from above, but I am still keeping this question just in case someone write in very consice terms the conection bteween the index set A and the regularity structure.
| https://mathoverflow.net/users/71105 | When and why do we require the condition that :"a subset bounded from below and has no accumulation points?" | From the references mentioned in the comments , (section 4.4,[*"Renormalisation of parabolic stochastic PDEs"*](https://arxiv.org/pdf/1803.03044.pdf)) (RP) and (section 6.1,*"Introduction to regularity structures"*) (IRS), we first start from the pde fixed point problem
$$\Phi = G \ast (\xi − \Phi^3 ) + G\Phi\_0 ,$$
where $\*$ denotes space-time convolution and where we denote by $G \Phi\_0$
the harmonic extension of $\Phi\_0$ and we split $G = K + \hat K$ into singular kernel $K$ and smooth $\hat{K}$. Then we consider the analogous *abstract* fixed point problem ((4.22) in (RP) and (6.3) in (IRS))
$$\Phi = \mathcal{K} \bigl(\Xi - \Phi^3\bigr) + \hat{\mathcal{K}} \bigl(\Xi - \Phi^3\bigr) + G \Phi\_0 $$
where you see that by repeated insertion of $\Phi$, you get combinations of *abstract symbols* eg. $\Xi, \mathcal{I}\Xi, \mathcal{I}(\Xi) \mathcal{I}( \mathcal{I}(\Xi)^3)$ and $\mathcal{I}$ is the "abstract integration" map against that kernel.
So the question becomes: how to keep track and organize all those new abstract expressions. As you see in that section, he gives specific examples of regularity structures based on generating sets of the above symbols.
What about the index set?
A big issue with SPDEs is the presence of the white noise $\xi$ because it is a distribution and so when it shows up in an SPDE, it can make them ill-posed. This is what creates the negative regularity degree. So as in the example, we incorporate that by assigning negative numbers to $\Xi$, (the abstract symbol for noise) eg. in the previous example $deg(\Xi)=-\frac{5}{2}-\epsilon$ for some $\epsilon>0$.
On the other hand, what saves us is integrating against the kernel $\mathcal{K}$ because as you might know from the heat equation, integrating against a heat kernel immediately smoothens functions. A similar phenomenon happens here via the Schauder estimates improving the regularity of the solution pushing it to the positive side. So we assign positive increase: $|\mathcal{I} \tau| = |\tau| + 2$
So there is this battle between lowering/increasing regularity. The index set
contains all the degree of the above symbols starting from the most negative degree of $\Xi$ eg. above $\{-\frac{5}{2}-\epsilon,-\frac{5}{2}-\epsilon+2, -\frac{5}{2}-\epsilon+4,...\} $
So the requirement "bounded below" ensures that eventually with enough convolutions, we will get actual functions out of this process rather than distributions (using reconstruction operator and removal of appropriate constants).
| 1 | https://mathoverflow.net/users/99863 | 434181 | 175,589 |
https://mathoverflow.net/questions/434148 | 4 | $\DeclareMathOperator\MCG{MCG}$Let $\Sigma$ be a compact oriented surface, with empty or connected boundary. Let $\mathcal{O}$ the space of orbits of nontrivial simple closed curves on $\Sigma$ under $\MCG(\Sigma)$-action. (so, $\mathcal{O}$ has finitely many elements: the sets of nonseparating curves and the sets of separating curves of each possible genus)
If $f\in \MCG(\Sigma)$ is a noncentral element, is it true that for each $o\in \mathcal{O}$, there is a curve $c\in o$ such that the geometric intersection number $i(f(c),c)$ is non zero ?
Also, in the case where $\Sigma$ has one boundary component, I would like to ask the same question but for arcs (for the geometric intersection number, consider homotopy of arcs fixing the boundary and count only intersection points in the interior)
| https://mathoverflow.net/users/62201 | Action of noncentral mapping classes on curves or arcs on a surface | Yes, this is true - there are many ways to prove it, but I'll hit it with a hammer. Let $C(\Sigma)$ denote the [curve complex](https://en.wikipedia.org/wiki/Curve_complex) of $\Sigma$. Suppose not, then $f$ would map every curve $[c]$ in $\mathcal{O}\subset C(\Sigma)$ to $[f(c)]$ which has distance $\leq 1$ from $[c]$ since $i(c,f(c))=0$. Since the neighborhood of radius 1 of $\mathcal{O}$ is equal to $C(\Sigma)$, we see that $f$ is [quasi-isometric](https://en.wikipedia.org/wiki/Quasi-isometry) to the identity acting on $C(\Sigma)$. Schleimer and Rafi [showed that $Aut(C(\Sigma))\cong QI(C(\Sigma))$](https://arxiv.org/abs/0710.3794), hence $f$ acts trivially on $C(\Sigma)$. Then by [Ivanov's theorem](http://www.math.stonybrook.edu/%7Emlyubich/Archive/Geometry/Hyperbolic%20Geometry/Ivanov.pdf) $f$ is the identity (or central if $\chi(\Sigma)\geq -2$).
| 5 | https://mathoverflow.net/users/1345 | 434187 | 175,592 |
https://mathoverflow.net/questions/434185 | 1 | Let the function $f\colon [a,b] \to\mathbb{C}$ be Lipschitz and let $|f(a)| \geq c$ and $|f(b)| = c$. Is there a Lipschitz function $g$ such that $|g| \geq c,$ $g(a)=f(a),$ $ g(b)=f(b)$ and Lipschitz constant of $f-g$ is less than epsilon for any positive epsilon?
There should be some simple counterexample.
| https://mathoverflow.net/users/490101 | Construction of the Lipschitz function with a given Lipschitz constant and given two values | $\newcommand\ep\varepsilon$Yes, it is easy to construct a counterexample here.
Indeed, if $g\_\ep$ is such a function for each given real $\ep>0$ (so that $|g\_\ep| \geq c$, $g\_\ep(a)=f(a)$, $g\_\ep(b)=f(b)$, and the Lipschitz constant of $f-g\_\ep$ is less than $\ep$), then $g\_\ep\to f$ pointwise (as $\ep\downarrow0$). So, it would follow that $|f|\ge c$ everywhere.
However, it is very easy to construct a Lipschitz function $f\colon [a,b] \to\mathbb{C}$ such that $|f(a)|\geq c$ and $|f(b)| = c$, but $|f(x)|<c$ for some $x\in(a,b)$.
| 2 | https://mathoverflow.net/users/36721 | 434189 | 175,594 |
https://mathoverflow.net/questions/434144 | 4 | Let $\mathcal{M}$ be a
>
> locally finitely presentable model category, cofibrantly generated by
> two sets $\mathcal{I}$ and $\mathcal{J}$ of cofibrations and trivial
> cofibrations with presentable domain and codomain.
>
>
>
I know that weak equivalences and fibrations are stable by filtered colimits.
>
> 1. What can be said about cofibrations and trivial cofibrations?
> 2. Is there a class of good examples in which this is known to be
> true?
> 3. Are there additional axioms that can be imposed that ensure this?
>
>
>
| https://mathoverflow.net/users/139854 | When are filtered colimits of (trivial) cofibrations still (trivial) cofibrations? | If both cofibrations and weak equivalences are stable under filtered colimits, then so are trivial cofibrations. This happens for instance if $\mathcal{M}$ is a presheaf category on an elegant Reedy category (such as $\Delta$) with cofibrations the monomorphisms, whatever the weak equivalences are (see Cor. 3.4.41 & (the proof of) Prop. 8.2.9 [here](https://smf.emath.fr/publications/les-prefaisceaux-comme-modeles-des-types-dhomotopie)). This is also true for cochain complexes in a Grothendieck abelian category (with cofibrations as monomorphisms and quasi-isomorphisms as weak equivalences). Similarly, if you consider simplicial sheaves on a site. More generally, if you can define cofibrations and weak equivalences using functors which commute with filtered colimits (e.g. suitable kernels to detect monomorphisms in a topos, cohomology sheaves or sheaves of homotopy groups to detect weak equivalences), you have a good chance of survival. This kind of properties is easily seen to be preserved under left Bousfield localizations.
| 9 | https://mathoverflow.net/users/1017 | 434194 | 175,597 |
https://mathoverflow.net/questions/434183 | 3 | The following conjecture about analytic functions arose as a way to show the asymptotic growth for certain PDE solutions. As I am unfamiliar with any results of this type, I thought I'd ask here.
In some sense, this is an analytic continuation result, as it says that if measure of points close to $0$ on which the function is very small is big enough, then $f(0)$ must be small.
**Conjecture:**
Let $f\_n : (-1,1) \to \mathbb{R}$ analytic functions such that
$$|f\_n^{(m)}(0)| \leq C^n m!$$
for some $0 < C < \infty$. Suppose also that $f\_n(0) = 1$ for all $n$. Does there exist $\delta >0$ depending on $C$, such that
$$\liminf\_n | \{x \in [0,\delta] : |f\_n(x)| \leq e^{-n} \}| = 0,$$
where by $|A|$ we mean the Lebesgue measure of $A$?
| https://mathoverflow.net/users/146531 | Quantitative analytic continuation estimate for a function small on a set of positive measure | Unfortunately, no, as requested:
Take any sequence $\delta\_j\in(0,1)$ decaying to $0$, choose small $\mu\_j>0$ such that $\prod\_j \delta\_j^{\mu\_j}=e^{-1}$ and put $f\_n(z)=e^n\prod\_j B\_{\delta\_j}(z)^{[\mu\_j n]}$ where $B\_\delta(z)=\frac{\delta-z}{1-\delta z}$ is the usual Blaschke factor. Then $|f\_n(0)|\ge 1$ and the set $\{|f\_n\|\le e^{-n}\}$ contains an interval of fixed length around each $\delta\_j$ (the interval where $|B\_{\delta\_j}|\le e^{-3\mu\_j^{-1}}$, say) for sufficiently large $n$.
| 11 | https://mathoverflow.net/users/1131 | 434203 | 175,598 |
https://mathoverflow.net/questions/433978 | 1 | Cross post with mse
For example, let's say I have the following equations.
\begin{gather\*}
a^{x-1}+b^{x-1}=337 \\
a^{x}+b^{x}=1267 \\
a^{x+1}+b^{x+1}=4825 \\
a^{x+2}+b^{x+2}=18751.
\end{gather\*}
What we can notice is that we can rewrite the left-hand sides of all the equations as
$a^{x+k}(1+(b/a)^{x+k})$
for our values of $k$: $-1$, $0$, $1$, and $2$.
Let's label those formulas as $f(x)=a^x(1+(b/a)^x)$.
If we divide $f(x)^2$ by $f(x+1)f(x-1)$ we will get a new formula that is dependent on $x$ and $b/a$:
$$(1+(b/a)^x)^2/((1+(b/a)^{x-1})(1+(b/a)^{x+1}).$$
We can label $b/a$ as $d$ and $(b/a)^x$ as $f$ (probably should find better variable names).
And then we will have the equation
$$(1+f)^2/((1+f/d)(1+fd))=1267^2/(337
\* 4825).$$
Similarly. Another equation can be created for $f(x+1)^2/(f(x+2)f(x))$:
$$(1+f d)^2/((1+f \* d^2)(1+f))=4825^2/(18571
\* 1267).$$
While my first set of 4 (or only 3) equations was unsolvable by Wolfram|Alpha the second set is solvable and 2 pairs of values for $f$ and $d$ are returned.
One of them is
$f=4.21\ldots$ and $d=1.3333 $.
$\log(f)/\log(d)= 5$ which is the true value of $x$.
$d=a/b$ is $4/3$ with numerical error.
Finding $a$ and $b$ is pretty trivial from here and they are 4 and 3 respectively.
My question is can this method be generalized to more than 3 variables (meaning $x$ and more exponentials)?
On paper, it seems that this could be done with a slightly different technique to generate multiple equations based on the principle of removing the dependency on the least significant exponent. In practice, I wasn't quite able to do so yet.
Does anyone know of any research on this topic or if this question was already solved, and in general is there any technique to analyze the sum of exponentials based on multiple values?
To be more clear.
My idea is to take multiple points of a function of the form for example
$a^x+b^x+c^x$ (preferabliy if the number of exponentials is unkown)
Rewrite it as $a^x((b/a)^x+(c/a)^x+1)$
And find the points of the same x value for the function $(b/a)^x+(c/a)^x+1$ And Continue until only one exponential is left
Does anything like that exist or is even possible?
Like this method <https://en.m.wikipedia.org/wiki/Newton_polynomial>
But for exponentials
| https://mathoverflow.net/users/494220 | Method to solve system of exponential sums of the form $a^x+b^x=c$ given more equations than variables | We can use linear algebra to make this easier.
Consider the space of sequences of the form $A a^n + B b^n$. We can use several dual bases for this space. One is the obvious basis of maps that send $A a^n + B b^n$ to its coefficients, $A$ and $B$. If neither of $a$ and $b$ are zero and the two are distinct, then another is the basis of maps that send $A a^n + B b^n$ to its values at $n = 0$ and $n = 1$.
Now consider the self-map "increase $n$ by $1$", i.e. taking the sequences $A a^n + B b^n$ to the sequence $A a^{n + 1} + B b^{n + 1}$. In the basis of coefficients, this can be seen as a diagonal map with eigenvalues $a$ and $b$. In other words, it has the corresponding matrix
$A = \left[\begin{matrix}
a & 0 \\
0 & b
\end{matrix}\right]$
In the basis of values at $n = 0, 1$, it clearly moves the value for $n = 1$ into the value for $n = 0$, while doing something to give the new value for $n = 1$, so it has a matrix of the form
$B = \left[\begin{matrix}
0 & 1 \\
z & y
\end{matrix}\right]$. Note that $B$ is conjugate to $A$ - in fact, it's the companion matrix for the characteristic polynomial of $A$.
Now we can approach your specific values. It's not hard to see that we have the equations
$B \left[\begin{matrix} 337 \\ 1267 \end{matrix}\right] = \left[\begin{matrix} 1267 \\ 4825\end{matrix}\right]$ and $B \left[\begin{matrix} 1267 \\ 4825 \end{matrix}\right] = \left[\begin{matrix} 4825 \\ 18751\end{matrix}\right]$
Putting them together, we have that $B \left[\begin{matrix} 337 & 1267 \\ 1267 & 4825 \end{matrix}\right] = \left[\begin{matrix} 1267 & 4825 \\ 4825 & 18751\end{matrix}\right]$
We can then find $B$:
$B = \left[\begin{matrix} 1267 & 4825 \\ 4825 & 18751\end{matrix}\right]\left[\begin{matrix} 337 & 1267 \\ 1267 & 4825 \end{matrix}\right]^{-1} = \frac{1}{576} \left[\begin{matrix} 0 & 576 \\ -13247 & 5717\end{matrix}\right]$
But then we have the characteristic polynomial of $A$ - which the eigenvalues of $A$, the diagonal elements $a$ and $b$, must satisfy. So $a$ and $b$ are the roots of $c^2 - \frac{5717}{576} c + \frac{13247}{576} = 0$. Then $x$ is whichever value satisfies the original equations.
---
This process generalizes quite readily. If you have $m$ exponentials and $2m$ consecutive equations of the form $\sum\_j a\_j^i = C\_i$ (or even more generally, $\sum\_j A\_j a\_j^i = C\_i$), you can find $a\_j$ by writing a square matrix of the form
$\left[\begin{matrix}
C\_1 & C\_2 & \dots & C\_m \\
C\_2 & C\_3 & \dots & C\_{m + 1} \\
\vdots & \vdots & \ddots & \vdots \\
C\_m & C\_{m + 1} & \dots & C\_{2m - 1}
\end{matrix}\right]$
and a vector of the form $\left[\begin{matrix}
C\_{m + 1} & C\_{m + 2} & \dots & C\_{2m}
\end{matrix}\right]$, and multiplying the vector by the inverse of the matrix. This works as long as you actually need $m$ such exponentials - if you find that the first matrix isn't invertible, then decrease $m$ by $1$ and try again.
---
This is one of those funny times where it turns out to be easier to answer a more general question than to answer the specific question, because the more general question has a nicer structure than the specific question.
| 2 | https://mathoverflow.net/users/44191 | 434208 | 175,601 |
https://mathoverflow.net/questions/434204 | 6 | Let $n\ge2$ be a given positive integer, and $z\_{1},z\_{2},\cdots,z\_{n}\in \mathbb{C}$,such
$$|z\_{1}|^2+|z\_{2}|^2+\cdots+|z\_{n}|^2\ge n.$$
Prove or disprove
$$f\_{n}=\sum\_{j=1}^{n}\left|\sum\_{I\subseteq \{1,2,3,\cdots,n\},|I|=j}\prod\_{k\in I}z\_{k}\right|^2\ge 1$$
In the particular case when $n=2$, it can be proved that
\begin{align\*}f\_{2}&=|z\_{1}+z\_{2}|^2+|z\_{1}z\_{2}|^2=(|z\_{1}|^2+|z\_{2}|^2)+|z\_{1}z\_{2}|^2+2\Re(z\_{1}\overline{z\_{2}})\\
&\ge |z\_{1}|^2+|z\_{2}|^2+|z\_{1}z\_{2}|^2-2|z\_{1}z\_{2}|\\
&=(|z\_{1}|^2+|z\_{2}|^2-1)+(|z\_{1}z\_{2}|-1)^2\\
&\ge(|z\_{1}|^2+|z\_{2}|^2-1)\\
&\ge 1
\end{align\*}
| https://mathoverflow.net/users/38620 | a problem in complex-variable inequality | For a polynomial $Q(x)=\sum\_i q\_ix^i$, define $N(Q)=\sum\_i|q\_i|^2$. We need to show that $N(R)\geq 2$, where $R(x)=\prod\_i (x-z\_i)$.
For a polynomial $Q(x)=\sum\_{i=0}^k q\_ix^i$, define $Q^\*(x)=\sum\_{i=0}^k\overline{q\_i}x^{k-i}=x^k\overline{Q(\bar x^{-1})}$. Here is the lemma which I definitely saw somewhere.
**Lemma 1.** $N(FG)=N(FG^\*)$.
*Proof.* This can be shown directly, by expanding the brackets. However, a more neat way is to apply the discrete Fourier transform, assigning to a polynomial $Q$ of degree $<n$ the collection of values $Q(\zeta\_i)$, where $\zeta\_1,\dots,\zeta\_n$ are the $n$th degree roots of unity. The Plancherel identity claims
$$
N(Q)=\frac1n\sum\_{i=1}^n|Q(\zeta\_i)|^2.
$$
On the other hamd, we have
$$
|F(\zeta\_i)G^\*(\zeta\_i)|=\left|F(\zeta\_i)\overline{G(\bar\zeta\_i^{-1})}\right|
=|F(\zeta\_i)G(\zeta\_i)|,
$$
hence $N(FG)=N(FG^\*)$. $\Box$
Back to the problem, let $|z\_1|\leq|z\_2|\leq\dots\leq|z\_k|\leq 1\leq |z\_{k+1}|\leq\dots\leq |z\_n|$. Then, instead of $R$, we may consider the polynomial
$$
S(x)=\prod\_{i=1}^k(1-\bar z\_i x)\prod\_{i=k+1}^n(x-z\_i),
$$
as Lemma 1 claims $N(R)=N(S)$. In the polynomial $S(x)$, we are interested only in the coefficients of $x^n$ and $x^0$, whose absolute values are $|\prod\_{i=1}^kz\_i|$ and $|\prod\_{i=k+1}^nz\_i|$: the next Lemma shows they suffice.
**Lemma 2.** Assume that $a\_1,\dots,a\_k\leq 1$ and $a\_{k+1},\dots,a\_n\geq 1$ are nonnegative real numbers such that $\sum\_ia\_i^2=n$. Then $(a\_1\dots a\_k)^2+(a\_{k+1}\dots a\_n)^2\geq 2$.
*Proof.* Induction on the number of indices $i$ such that $a\_i\neq 1$. If all the $a\_i$ equal $1$, the claim is trivial. Also it is trivial if $a\_i=1$ for all $i\leq k$.
Assume that $a\_1<1$; then, without loss of generality, $a\_n>1$. Replace $a\_1$ and $a\_n$ with $b\_1\leq b\_n$ such that $b\_1^2+b\_n^2=a\_1^2+a\_n^2$, and one of the $b\_i$ equals $1$ (then $a\_1<b\_1\leq b\_n<a\_n$). Denote $B\_1=\prod\_{i=2}^k a\_i$ and $B\_n=\prod\_{i=k+1}^{n-1} a\_i$; the inductive hypothesis yields
$$
b\_1^2B\_1^2+b\_n^2B\_n^2\geq 2.
$$
On the other hand, we have
$$
\left(a\_1B\_1\right)^2+\left(a\_nB\_n\right)^2
=(b\_1B\_1)^2+(b\_nB\_n)^2+(a\_1^2-b\_1^2)B\_1^2+(a\_n^2-b\_n^2)B\_n^2\\
=(b\_1B\_1)^2+(b\_nB\_n)^2+(a\_1^2-b\_1^2)(B\_1^2-B\_n^2)
\geq (b\_1B\_1)^2+(b\_nB\_n)^2\geq 2,
$$
as desired. $\Box$
| 12 | https://mathoverflow.net/users/17581 | 434219 | 175,604 |
https://mathoverflow.net/questions/434200 | -1 | I posted this question on math stackexchange weeks ago, and it have not receive an answer yet after a bounty offer...
---
I've been recently playing around with the linear recurrence sequences. Consider the following recurrence equation:
$$ a\_n = c\_1a\_{n-1} + \cdots + c\_ka\_{n-k}, \quad \forall n > k $$
is equivalent to
$$\begin{pmatrix}a\_n \\ \vdots \\ a\_{n-k+1}\end{pmatrix}
= \begin{pmatrix} c\_1& c\_2& \dots& c\_{k-1}& c\_k \\ 1& 0& \dots& 0& 0\\ 0& 1& \dots& 0& 0\\ \vdots& \vdots& \ddots& 0& 0\\ 0& 0& \dots& 1& 0\end{pmatrix}\begin{pmatrix} a\_{n-1} \\ \vdots \\ a\_{n-k} \end{pmatrix} = \dots = \begin{pmatrix} c\_1& c\_2& \dots& c\_{k-1}& c\_k \\ 1& 0& \dots& 0& 0\\ 0& 1& \dots& 0& 0\\ \vdots& \vdots& \ddots& 0& 0\\ 0& 0& \dots& 1& 0\end{pmatrix}^{n-k} \begin{pmatrix} a\_k \\ \vdots \\ a\_1 \end{pmatrix}, \quad \forall n > k $$
denote the R.H.S. matrix as $M$ (without the exponent).
And I want to show such $M$, regardless of the choice of $c\_1, \dots , c\_k$ (with $c\_k \neq 0$), has only one Jordan block corresponding to each eigenvalue in its Jordan canonical form, by an alternative way.
Let $U^-1MU = \Lambda + N$ be its Jordan form with $\Lambda N = N \Lambda$. Suppose $M$ consists of $m$ different eigenvalues $\lambda\_1,\dots, \lambda\_m$ with algebraic multiplicity $\alpha\_1,\dots,\alpha\_m$; i.e. the characteristic polynomial is
$$ c\_M(x) = (x-\lambda\_1)^{\alpha\_1}\dots(x-\lambda\_m)^{\alpha\_m} $$
and the minimal polynomial reads
$$ (x-\lambda\_1)^{u\_1}\dots(x-\lambda\_m)^{u\_m}$$
then
$$\begin{pmatrix}a\_n \\ \vdots \\ a\_{n-k+1}\end{pmatrix}
=
U^{-1}\begin{pmatrix}
\lambda\_1^n& {n \choose 1}\lambda\_1^{n-1}& \dots& {n \choose \*} \lambda\_1^{n-\*}& 0& \dots& 0& \\
0& \lambda\_1^n& \dots& {n \choose {\*-1}}\lambda\_1^{n-\*+1}& 0& \dots& 0\\
\vdots& \vdots& \ddots& \vdots& \vdots& \vdots& \vdots\\
0& 0& \dots& \lambda\_1^n& 0& \dots& 0\\
0& 0& \dots& 0&\lambda\_1^n& \dots& \vdots\\
0& 0& \dots& 0& 0& \ddots& \vdots\\
0& 0& \dots& 0& 0& \dots& \lambda\_m^n\\
\end{pmatrix}
U \begin{pmatrix} a\_k \\ \vdots \\ a\_1 \end{pmatrix}, \; \forall n > k
$$
the product of three matrice R.H.S. each has all entries as a function of $n$ in the shape of $P\_1(n)\lambda\_1^n + ... + P\_m(n)\lambda\_m^n$, with each $\deg P\_i \leq u\_i-1 $(note that $u\_i \leq \alpha\_i$). Note that this equation holds as long as the first recurrence equation holds.
One may prove the following result using the fact that $M$ has only one Jordan block for each eigenvalue to derive the following fact:
>
> When $c\_1, \dots, c\_k$ and $a\_1. \dots, a\_k$ are chosen, there must uniquely exist $Q\_1, \dots, Q\_m$ with $\deg Q\_i \leq \alpha\_i-1$ such that the equation $a\_n = Q\_1(n)\lambda\_1^n + \dots + Q\_m(n)\lambda\_m^n$ always hold, and for all $\deg Q\_i \leq \alpha\_i-1$, the sequence given by $a\_n = Q\_1(n)\lambda\_1^n + \dots + Q\_m(n)\lambda\_m^n$ should also satisfy $a\_n = c\_1a\_{n-1} + \dots + c\_ka\_{n-k}$.
>
>
>
But I want to take the statement above for granted(one can verify the result by checking if the $k$ by $k$ matrix representation of the relation between coefficient of $P\_i$s and $a\_i$s is singular. And also there are other ways I know to derive this result) and prove that $M$ must only have one Jordan block for each eigenvalue(i.e. $u\_i = a\_i$).
Proof
=====
For fixed $c\_1, \dots, c\_k$, if $u\_i < \alpha\_i$ for some $i$, pick $a\_n = n^{\alpha\_i}\lambda\_i^n$. By the statement above, the recurrence relation holds, and hence (by the Jordan Form analysis) there exists $Q\_1, \dots, Q\_m$ such that $a\_n = Q\_1(n)\lambda\_1^n + \dots + Q\_m(n)\lambda\_m^n\, \; \forall n > k$ with $\deg Q\_i < u\_i$, but such expression is unique, and it does not contain the term $n^{\alpha\_i}\lambda\_i^n$, leading to a contradiction; thus there must only be one Jordan block corresponding to each $\lambda$.
Is this proof valid? I've seen other shorter proofs of it, and I would like to see if there are common concepts between this proof and the other proofs.
Short proof:
<https://math.stackexchange.com/questions/348498/jordan-basis-of-a-when-a-is-the-companion-matrix?rq=1>
Alternate proofs:
<https://math.stackexchange.com/questions/10216/the-characteristic-and-minimal-polynomial-of-a-companion-matrix>
| https://mathoverflow.net/users/198287 | Companion matrices must have geometric multiplicity one, linear recurrence sequence view | If $M$ had more than one Jordan block corresponding to some eigenvalue, then its minimal polynomial's degree would be smaller than $k$. This yields that *all* sequences satisfying your recurrence relation in fact satisfy a fixed smaller order linear recurrence relation (given by that minimal polynomial). But that is absurd according to the dimension argument.
| 1 | https://mathoverflow.net/users/17581 | 434220 | 175,605 |
https://mathoverflow.net/questions/433176 | 13 | I recently came across Kac algebra. They are roughly Hopf algebras and $C^\*$-algebras with compatible structures. It follows from Artin–Wedderburn theorem that every semisimple complex Hopf algebra can be seen as a $C^\*$-algebra structure, but I am unsure whether this structure will be compatible with the comultiplication. I guess that this is not the case, as I think that there are dimensions where Kac algebras have been classified, but semisimple Hopf algebras have not.
**Question:** Is every semisimple/finite-dimensional complex Hopf algebra a Kac algebra?
Something similar has been previously asked [here](https://mathoverflow.net/questions/324111/is-there-a-non-kac-complex-finite-dimensional-semisimple-hopf-algebra), but the question is unanswered. If this is unknown in general, it would also be helpful to know if this holds under some stronger assumptions (eg. for a given dimension).
EDIT: Lemma IV.8.2 in *Quantum Groups* by Kassel could be useful:
A Hopf algebra $H$ has a Hopf $\*$-algebra structure if and only if there exists an antilinear automorphism $\gamma$ of $H$ such that
$(i)$ the map $\gamma$ is a morphism of real algebras and an antimorphism of real coalgebras, and
$(ii)$ we have $\gamma^2 = (S \gamma)^2 = \text{id}\_H$.
| https://mathoverflow.net/users/493533 | Hopf algebras vs. Kac algebras | As pointed out in the comments all semisimple Hopf algebras of dimension up to $23$ are Kac algebras. So it seems like dimension $24$ is the smallest one where this is open.
Corollary 9.7 of "Weakly group-theoretical and solvable fusion categories" by Etingof, Nikshych and Ostrik says that a semisimple Hopf algebra of dimension $p q^2$ is either a Kac algebra or a twisted group algebra (by a twist corresponding to the subgroup $(Z/qZ)^2$) or the dual of a twisted group algebra.
| 1 | https://mathoverflow.net/users/493533 | 434225 | 175,606 |
https://mathoverflow.net/questions/434210 | 8 | For a natural number not a perfect square, is there always at least a prime number for which it is a primitive root?
[Artin's conjecture on primitive roots](https://en.wikipedia.org/wiki/Artin%27s_conjecture_on_primitive_roots) is that there are infinitely many such primes. Do we know for sure that there is at least one?
| https://mathoverflow.net/users/494432 | For a non-square, is there a prime number for which it is a primitive root? | We do not know that. It is typical for questions about the infinitude of primes satisfying some property that even proving the existence of at least one of those primes in sufficient generality is unproved. And sometimes it turns out that proving there is at least one such prime in sufficient generality (the type of thing what you're asking) would imply there are infinitely many such primes thanks to some trick using the sufficient generality.
For the case of Dirichlet's theorem, see Theorem A.1 [here](https://kconrad.math.uconn.edu/blurbs/ugradnumthy/prime-patterns-1.pdf). For the case of Bunyakosvky's conjecture and Schinzel's Hypothesis H see [here](https://mathoverflow.net/questions/226794/on-a-possible-equivalence-of-bunyakovsky-conjecture).
For the case of the Artin primitive root conjecture in its qualitative form (infinitely many primes, nothing about their density), here are two such arguments. They are based on correspondence with Paul Pollack.
**Theorem** *Assume the negative of each perfect square is a primitive root mod $p$ for at least one prime $p > 3$. Then the negative of each perfect square is a primitive root mod $p$ for infinitely many primes $p > 3$.*
It is reasonable to require $p > 3$ in the theorem, since if $-A^2$ is the negative of a perfect square and $3 \nmid A$ then $-A^2 \equiv 2 \bmod 3$ and thus $-A^2$ is a primitive root mod $3$. So the theorem would not be saying anything interesting if we allow $p = 3$.
*Proof*. Let $a = -A^2$ for $A$ in $\mathbf Z^+$ and let $S = S\_a$ be the set of primes $p > 3$ such that $a \bmod p$ is a primitive root, so $S$ contains at least one prime. To show $S$ is infinite, we'll argue by contradiction.
Assume $S$ is finite and let $L$ be the product of all odd primes $q$ that divide some $p-1$ where $p \in S$. Then $L$ is odd and $L \geq 1$. (The number $L$ is an empty product, and thus is $1$ by the usual convention for empty products, if each $p \in S$ has $p-1$ equal to a power of $2$ and thus has no odd prime factors.)
Set $b = a^L = (-A^2)^L = -(A^L)^2$, so $b$ is the negative of a perfect square. By the premise of the theorem, there is an odd prime $p > 3$ such that $b \bmod p$ is a primitive root. Since $a^L \bmod p$ generates $(\mathbf Z/p\mathbf Z)^\times$, so does $a \bmod p$, so $p \in S$. Since both $a \bmod p$ and $a^L \bmod p$ have order $p-1$, we have $(L,p-1) = 1$. If $p-1$ is not a power of $2$ then $p-1$ has an odd prime factor $q$, so $q \mid L$ by the definition of $L$, but $(L,p-1) = 1$ and $q \mid (p-1) \Longrightarrow (L,q) = 1$, which contradicts the condition $q \mid L$. Therefore $p-1 = 2^k$ for some $k$, making $p = 2^k + 1$.
Since $p > 3$ we have $k \geq 2$, so $p \equiv 1 \bmod 4$. By oddness of $L$, we have the Legendre symbol calculation
$$
-1 = \left(\frac{b}{p}\right) = \left(\frac{a^L}{p}\right) = \left(\frac{a}{p}\right)^L = \left(\frac{a}{p}\right) = \left(\frac{-A^2}{p}\right) = \left(\frac{-1}{p}\right)\left(\frac{A}{p}\right)^2.
$$
Thus $p \nmid A$ (otherwise the Legendre symbol $(A|p)$ would be $0$), so $(A|p)^2 = 1$. Also $(-1|p) = 1$ since $p = 1 \bmod 4$, so $-1 = 1\cdot 1 = 1$, a contradiction. That proves $S$ is infinite. $\Box$
Primes of the form $2^k+1$, which came up in the proof above, are Fermat primes and it is expected there are only finitely many of them (with the last one probably being $2^{16}+1$). That leads to the following second theorem with slightly different assumptions.
**Theorem**. *Assume for each $a$ in $\mathbf Z$ that's not $0, 1, -1$ or a perfect square, $a$ is primitive root mod $p$ for at least one odd prime $p$ that's not a Fermat prime. Then for each $a$ in $\mathbf Z$ that's not $0, 1, -1$ or a perfect square, $a$ is primitive root mod $p$ for infinitely many odd primes $p$ that are not Fermat primes.*
The hypothesis is reasonable in light of the Artin primitive root conjecture and the expectation that there are only finitely many Fermat primes.
*Proof*. Let $a$ be an integer that's not $0, 1, -1$ or a perfect square and let $S =S\_a$ be the set of odd primes $p$ that are not Fermat primes and $a \bmod p$ is a primitive root, so $S$ contains at least one prime that is not a Fermat prime. To show $S$ is infinite, we'll argue by contradiction.
Assume $S$ is finite and let $L$ be the product of all odd primes $q$ that divide some $p-1$ where $p \in S$, so $L$ is odd and $L \geq 1$. (In fact $L \geq 3$ since for each $p$ in $S$ there is an odd prime factor of $p-1$, as $p-1$ is not a Fermat prime.)
The integer $b = a^L$ fits the hypothesis of Artin's conjecture:
(1) $|b| = |a|^L \geq |a| \geq 2 > 1$, so $b$ is not $0, 1$, or $-1$.
(2) if $b$ were a square then $a$ would be too since $L$ is odd, but $a$ is not a square. Thus $b$ is not a square.
Therefore by the premise of the theorem, there is an odd prime $p$ that's not a Fermat prime and $b \bmod p$ is a primitive root. Since $a^L \bmod p$ generates $(\mathbf Z/p\mathbf Z)^\times$, so does $a \bmod p$, so $p$ is in $S$. Since both $a \bmod p$ and $a^L \bmod p$ have order $p-1$, we have $(L,p-1) = 1$. Assuming $p-1$ has an odd prime factor $q$, we have $q \mid L$ by the definition of $L$, but $(L,p-1) = 1$ and $q \mid (p-1) \Longrightarrow (L,q) = 1$, which contradicts the condition $q \mid L$. Therefore $p-1$ has no odd prime factor, so $p-1 = 2^k$ for some $k$, making $p = 2^k + 1$, which is a Fermat prime, and that contradicts the definition of $S$ (it contains no Fermat primes). That proves $S$ is infinite. $\Box$
It would be nice if the role of Fermat primes could be removed from the second theorem. Anyone who figures out a way to do that should edit this answer or (if you can't edit) leave a comment below.
| 24 | https://mathoverflow.net/users/3272 | 434228 | 175,607 |
https://mathoverflow.net/questions/434173 | 2 | **Question.** Let $f: \mathbf{R} \to \mathbf{R}$ be an analytic function. Is there a harmonic function $u$ on the circular cylinder $D \times \mathbf{R} \subset \mathbf{R}^3$ so that $u = f$ along the axis $\{ (0,0) \} \times \mathbf{R}$?
* The problem is obviously ill-posed in the sense of Hadamard because it is very underdetermined. Although it doesn't make sense as a 'Dirichlet problem', I think a (positive or negative) answer is possible.
* Probabilistic arguments seem tricky because the axis is too small for hitting times of Brownian motion to be defined.
| https://mathoverflow.net/users/103792 | 'Dirichlet problem' along axis for harmonic functions | Assuming the Taylor series of $f$ has an infinite radius of convergence, the sum
$$
\sum\_{k=0}^\infty \left(x^2+y^2\right)^kf^{(2k)}(z)\cdot \frac{(-1)^k}{4^k k!^2}
$$
converges absolutely and locally uniformly on $\mathbb{R}^3$ to a function $u(x,y,z)$ which is harmonic and satisfies $u(0,0,z)=f(z)$
| 3 | https://mathoverflow.net/users/49822 | 434229 | 175,608 |
https://mathoverflow.net/questions/434021 | 0 | Let $G$ be a compact Lie group with Lie algebra $\mathfrak{g}.$ Denote by $\mathfrak{g}^\*$ the dual space of $\mathfrak{g}$. Let $r$ be an element of $\mathfrak{g}^\*$ such that $G\_r$ the stabilizer of $r$ under the coadjoint action is a maximal torus of $G$. Denote by $\mathcal{O}\_r$ the coadjoint orbit of $G$ which pass through $r$.
$\mathcal{O}\_r$ is endowed with a 2-form which is a symplectic form
$$\omega\_\alpha(\hat{X},\hat{Y})= -\alpha([X,Y]), \alpha \in \mathfrak{g}^\*, \quad X,Y, \in \mathfrak{g}. $$
Often it is more convenient to choose
an inner product $\langle . , . \rangle$ on $\mathfrak{g}$ to identify $\mathfrak{g}^\*$ with $\mathfrak{g}$. Once such an inner product has been chosen, we can write the 2-form as
$$\omega\_\lambda(\hat{X},\hat{Y}) = −\langle \lambda, [X,Y] \rangle, \lambda, X, Y \in \mathfrak{g}.$$
If $G$ is semisimple, then we choose the killing form denoted $k$ to define $\omega$:
$$\omega\_\lambda(\hat{X},\hat{Y}) = −k(\lambda, [X,Y]).$$
>
> $\textbf{Question}$: In the case where $G$ is a compact connected Lie group (not necessarily semisimple), why does the 2-form
> $$\omega\_\lambda(\hat{X},\hat{Y}) = −k(\lambda, [X,Y]), \lambda, X, Y \in \mathfrak{g}.$$
> defined using the killing form defines a symplectic form on thecoadjoint orbit $ \mathcal{O\_r}$ of $G$ ?
>
>
>
Ps: This question appeared under bounty for 100 points here <https://math.stackexchange.com/questions/4563148/symplectic-form-of-a-coadjoint-orbit-of-a-compact-connected-lie-group> but didn't receive any answer so far.
| https://mathoverflow.net/users/172459 | Question about coadjoint orbits of compact connected Lie groups | $\DeclareMathOperator{\Tr}{Tr}
\DeclareMathOperator{\ad}{ad}
\DeclareMathOperator{\Lie}{Lie}
\newcommand{\g}{{\mathfrak g}}
\newcommand{\z}{{\mathfrak z}}
\newcommand{\s}{{\mathfrak s}}
\newcommand{\O}{{\mathcal O}}
\newcommand{\wh}{\widehat}
\newcommand{\wt}{\widetilde}$
Let $G$ be a connected compact Lie group, and let $\g$ denote its Lie algebra.
Write $\z$ for the center of $\g$ and set $\s=[\g,\g]$, which is a semisimple Lie algebra.
Then
$$ \g=\z\oplus\s.$$
Write $Z=Z(G)^0$ (the identity component of the center of $G$), $\ \wt S=[G,G]$, $S=G/Z(G)$.
Then $Z$ is a torus, whereas $\wt S$ and $S$ are semisimple Lie groups.
We can identify
$$ \z=\Lie Z,\quad \Lie S=\s=\Lie\wt S.$$
The group $G$ acts on $\g$ by adjoint representation.
Moreover, $G$ acts on $\g$ via the canonical surjective homomorphism
$\pi\colon G\to S$.
We write
$$ g\cdot X=s\cdot X,\quad\text{where} \ g\in G,\ s=\pi(g)\in S,\ X\in\g.$$
We write an element $X\in \g$ as
$$ X=X\_\z+X\_\s\ \quad \text{with}\ X\_\z\in\z,\ X\_\s\in \s\,.$$
Then
$$g\cdot X=X\_\z+g\cdot X\_\s=X\_\z+s\cdot X\_\s\,.$$
Let $\g^\*$ denote the dual space for $\g$.
Then
$$\g^\*=\z^\*\oplus \s^\*.$$
For $r\in \g^\*$ we may write
$$r=r\_\z+r\_\s\quad\text{with}\ r\_\z\in\z^\*, \ r\_\s\in \s^\*.$$
Then for $g\in G$ we have
$$g\cdot r=r\_\z+g\cdot r\_\s=r\_\z+s\cdot r\_\s\,.$$
Let $X\in\g$, $\ X=X\_\z+X\_\s\,$. Then $[X\_\z\,,X\_\s]=0$.
It follows that
$$\exp -tX=(\exp -tX\_\z)\cdot( \exp -tX\_\s) \quad\text{with}\
\exp -tX\_\z\in Z,\ \exp -tX\_\s\in\wt S,$$
whence
$$(\exp -tX)\cdot \alpha=\alpha\_\z+(\exp-t X\_\s)\cdot \alpha\_\s\quad
\text{for}\ \alpha=\alpha\_\z+\alpha\_\s\in \g^\*.$$
Write
$$ \wh X=\frac d{dt}\Big|\_{t=0}(\exp -tX)\cdot \alpha.$$
Then $\wh X=\wh{X\_\s}$, where
$$ \wh {X\_\s}=\frac d{dt}\Big|\_{t=0}(\exp -tX\_\s)\cdot \alpha\_\s.$$
For $r=r\_\z+r\_\z \in\g^\*$ write $\O\_r=G\cdot r$, $\ \O\_{r\_\s}=S\cdot r\_\s\,$. Then
$$ \O\_r=r\_\z+S\cdot r\_\s=r\_\z+\O\_{r\_\s}\,.$$
Thus for $\alpha=g\cdot r\in \O\_r$ we have
\begin{equation}\label{e:\*}
T\_\alpha(\O\_r)\cong T\_{\alpha\_\s}(\O\_{r\_\s}).\tag{$\*$}
\end{equation}
Consider the adjoint representation
$$\ad\colon \g\to\mathfrak{gl}(\g),\quad (\ad X)\cdot Y=[X,Y].$$
For $X=X\_\z+X\_\s\in\g$ we have $\ad X=\ad X\_\s\,$.
Consider the Killing form
$$ k\colon \g\times\g\to{\mathbb R}, \quad
(X,Y)\mapsto \Tr\!\big ((\ad X)\cdot (\ad Y)\big).$$
Then
$$ k(X,Y)=\Tr\!\big ((\ad X\_\s)\cdot (\ad Y\_\s)\big)=k(X\_\s\,,Y\_\s).$$
For $\lambda \in\g$, we define a skew-symmetric form on $T\_\alpha(\O\_r)$ by
$$\omega\_\lambda(\wh X,\wh Y)=-k(\lambda,[X,Y]).$$
Since $[X,Y]=[X\_\s\,,Y\_\s]$, for $\lambda=\lambda\_\z+\lambda\_\s$ we have
$$\omega\_\lambda(\wh X,\wh Y)=-k(\lambda,[X,Y])
=-k(\lambda\_\s,[X\_\s\,,Y\_\s])=\omega\_{\lambda\_\s}(\wh{X\_\s}\,, \wh{ Y\_\s\, }\,).$$
We can identify the tangent spaces
$T\_\alpha(\O\_r)$ and $T\_{\alpha\_\s}(\O\_{r\_\s})$ by \eqref{e:\*}.
Then the formula above is probably what you need.
| 2 | https://mathoverflow.net/users/4149 | 434234 | 175,610 |
https://mathoverflow.net/questions/433564 | 1 | Pair of sequences $\ v\_n\ $ and $\ U\_n\ $ of integers start as in the following table:
[\begin{array}{rrrrrrrrrr}
n= & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & \ldots \\
v\_n= & 0 & 2 & 5 & 10 & 17 & 37 & 50 & 82 & \ldots \\
U\_n= & 0 & 2 & 3 & 6 & 8 & 12 & 14 & 18 & \ldots
\end{array}]
These two sequences are defined as follows:
* $\ v\_0=U\_0=0;$
* $\ v\_n\in\mathbb N\ $ is the smallest natural number such that **none** of the consecutive $\,\ U\_{n-1}\!+\!1\,\ $ integers $\ v\_n\ \ldots\ v\_n\!+\!U\_{n-1}\ $ is powerful;
* $\ U\_n\ $ is the smallest natural number such $\ v\_n+U\_n\ $ **is** powerful.
Thus, we are looking at the ever longer maximal sequences of consecutive non-powerful sequences. One would like to know the behavior of these sequences:
**Question**: what are reasonable (as exact as possible, and easily computable) lower and upper bounds for terms $\ v\_n\ $ and $\ U\_n,\ $ and their asymptotic behavior?
Knowing roughly the number of powerful initegers $\ POW(x)\ $ that do not exceed $\ x\ $ (for every positive $\ x\in\mathbb R),\ $ we may deduce the average behavior of these sequences of non-powerful integers; the still harder challenge would be deducing the more delicate but consistent deviations from the regular statistical behavior.
| https://mathoverflow.net/users/110389 | Consecutive non-powerful integers | Infinitely often (and with positive density, as proved by Shiu) there are no powerful numbers between $n^2+1$ and $(n+1)^2-1$. Hence the maximal gap below $x$ is infinitely often of size $\sim2\sqrt x$. (This is more than can be deduced from the Erdős & Szekeres result, or the Bateman & Grosswald result, since $\zeta(3/2)/\zeta(3)>2.$)
$U\_n$ is largest if $v\_n$ is a square and the above interval is empty. So
$$
U\_n \le (\sqrt{v\_n}+1)^2 - v\_n = 2\sqrt{v\_n}+1
$$
with equality holding in that case. Deducing a lower bound is probably tantamount to finding gaps in [A336175](https://oeis.org/A336175), but I would be surprised if it was not $\sim2\sqrt{v\_n}$ as well.
| 4 | https://mathoverflow.net/users/6043 | 434236 | 175,611 |
https://mathoverflow.net/questions/434231 | -1 | I have a polynomial, and I want to get the conditions for the number of positive roots
What are the different methods out there to determine these conditions?
this is the polynomial:
f(g)=A1*g^5 + A2*g^4 + A3*g^3 + A4*g^2 + A5\*g + A6
and A1 to A6 are constants
I will appreciate any help
Warm regards
| https://mathoverflow.net/users/493074 | finding positive roots for a polynomial | Abel-Ruffini Theorem postulates the existence of polynomials of every degree above $4$ that are not solvable by radicals (see also Galois' results on the same topic at the beginning of the XIX century, about the necessary and sufficient condition to let a polynomial equation be solvable by radicals).
A good (simple to be read) online source is "Galois Theory: Polynomials of Degree 5 and Up" by Tokuei Higashino.
P.S. Without solving $f(g)=0$, you can use Descartes' Rule of Signs, stating that *the possible number of the positive roots of a polynomial is equal to the number of sign changes in the coefficients of the terms **or** less than the sign changes by a multiple of $2$*.
| -1 | https://mathoverflow.net/users/481829 | 434238 | 175,612 |
https://mathoverflow.net/questions/434242 | 1 | $\DeclareMathOperator\inc{inc}$Let $|X|=n$ and $\inc(X,\leq)=\{\{x,y\} : \neg (x\leq y)\wedge \neg (y\leq x)\}$, where $(X,\leq)$ is poset (possibly unconnected). Define the function:
$$\pi(n,m):=|\{(X,
\leq):\inc(X,\leq)=m\}/\cong|$$
, where $\cong$ is the relation of isomorphism of partial orders. It's obvious that if $m>\binom{n}{2}$, then $\pi(n,m)=0$. Is it true that for any $m\in [0,\binom{n}{2}]$ $\pi(n,m)>0$? I know that if $\sum\_{i=1}^k a\_i=n$ then
$$\pi\left(n,\sum\_{i=1}^{k-1}\left(a\_i\sum\_{j=i+1}^{k}a\_j\right)\right)>0$$
and if $1<m\leq n$ then $\pi(n,\binom{m}{2})\geq n-m+1$. I'm also interested in upper and lower bounds for $\pi(n,m)$. For example, if $n-3m+1=k>0$ then $\pi(n,m)\geq k$ . Maybe this MO question is related: [Poset of antichains of given cardinality](https://mathoverflow.net/questions/371744/poset-of-antichains-of-given-cardinality) .
| https://mathoverflow.net/users/175589 | The quantity of poset with a given number of pairs of incomparable elements | Yes. This is true. We shall prove this result by induction on $n$. Suppose that $n>0$. and $0\leq m\leq\binom{n}{2}$. If $m\leq\binom{n-1}{2}$, then there is some poset $X$ with $|X|=n-1$ and where $\text{inc}(X)=m$. Now attach a new element $1$ where $x\leq 1$ for $x\in X$ to obtain a poset $X\cup\{1\}$. Then $|X\cup\{1\}|=n$ and $\text{inc}(X)=m.$ Therefore, $\pi(n,m)>0$.
Let $r\leq\binom{n-1}{2}$. Suppose now that $|X|=n-1$ and $\text{inc}(X)=r$. Then let $c$ be a new element that is incomparable with all the elements in $X$. Then $|X\cup\{c\}|=n$ and $\text{inc}(X\cup\{c\})=r+n-1$. In particular, if we have $\binom{n}{2}\leq m\geq n-1$, then we can set $r=m-(n-1)$, so that $|X\cup\{c\}|=n$ and $\text{inc}(X\cup\{c\})=m$. We have covered the cases when $\binom{n}{2}\geq m\geq n-1$ and when $m\leq\binom{n-1}{2}$, and one can easily verify that this covers all possible cases $m$ with $0\leq m\leq\binom{n}{2}$.
**Alternate proof**
The result also follows from the one-point extension property of partial orders.
Theorem: Suppose that $(X,P)$ is a partially ordered set with $X$ finite. Let $(X,Q)$ be a partially ordered set with $P\subseteq Q,P\neq Q$. Then there exists a pair $(x',y')\in Q\setminus P$ where $P\cup\{(x',y')\}$ is a partial ordering.
Construction of $(x',y')$: Let $(x,y)\in Q\setminus P$. Suppose now that $(x',y')$ is a pair with $x'\leq\_P x,y\leq\_P y',x'\not\leq\_P y'$ and where
$$x''\leq\_P x',y'\leq\_P y''\Rightarrow(x''\leq\_P y''\,\text{or}\,x''=x',y''=y').$$
I gave a proof that $P\cup\{(x',y')\}$ is actually a partial order in [my other answer to another question here](https://mathoverflow.net/a/418100/22277).
Since every partial order can be extended to a linear order, for each $n$ and set $X$ with $|X|=n$, we have a sequence of partial orders $P\_0\subseteq\dots\subseteq P\_m$ where $m=\binom{n}{2}$ where $P\_0=\{(x,x)|x\in X\}$, and where $P\_m$ is a linear order, and where $|P\_{i+1}|=|P\_i|+1$ so that $|P\_i|=n+i$. In this case, $\text{inc}(X,P\_i)=m-i$.
| 1 | https://mathoverflow.net/users/22277 | 434248 | 175,614 |
https://mathoverflow.net/questions/433149 | 3 | Let $(X,d)$ be a connected geodesic metric space. When does there there exists a covering map $\pi:H\rightarrow X$ which is a *local-isometry* where $H$ is either a Hilbert space or a Euclidean space?
* **Strengthened:** If $(X,d)$ is simply connected and is the length space induced by a Riemannian geometry and if we only consider finite-dimensional $H$ then by [the classical Cartan-Hadamard Theorem](https://en.wikipedia.org/wiki/Cartan%E2%80%93Hadamard_theorem) we know that $(X,d)$ must be non-positive curvature.
* **Relaxed:** If instead we weaken the problem, so that H is only required to be Hadamard then $(X,d)$ must also be Hadamard (by the metric version of the Cartan-Hadamard Theorem).
However, what is know of $(X,d)$ if we specifically want $H$ to be Hilbert or Euclidean?
| https://mathoverflow.net/users/491352 | Spaces satisfying a strong Cartan-Hadamard theorem | Note that Hilbert spaces (of all dimensions finite or infinite) are the only geodesic spaces with extendable geodesics which are flat in the sense of Alexandrov.
Therefore $X$ has to have extendable geodesics + it has to be locally flat (in the sense of Alexandrov).
By Cartan--Hadamard theorem, these two conditions are also sufficient.
| 2 | https://mathoverflow.net/users/1441 | 434258 | 175,618 |
https://mathoverflow.net/questions/434217 | 7 | In the very first chapter of *Elements of* $\infty$*-category theory*, E. Riehl and D. Verity define their notion of an $\infty$-cosmos, which should axiomatise a category in which $\infty$-categories live. (So, for example, the category of quasi-categories is an example of an $\infty$-cosmos.) An $\infty$-cosmos is a category enriched over quasi-categories and equipped with a collection of maps called *isofibrations*, which should satisfy some properties.
Surely, in the $\infty$-cosmos of quasi-categories, the isofibrations correspond the usual notion of an isofibration of quasi-categories. The same holds in the $\infty$-cosmos of 1-categories.
Now, since I'm just learning quasi-categories for the first time (as is expected from a reader of this book, apparently), I have no intuition whatsoever for isofibrations. **Why is this class of functors so important as to be in the very definition of an $\infty$-category (i.e., an object of an $\infty$-cosmos)?** In particular, how should I think about isofibrations?
| https://mathoverflow.net/users/131975 | Intuition for isofibrations in $\infty$-categories | As has been mentioned, there's no homotopically meaningful content to the notion of an isofibration, since every map of $\infty$-categories is equivalent to an isofibration. So the point is really all in the definition of an $\infty$-cosmos: isofibrations are the kinds of maps between $\infty$-categories that you can take a strict pullbacks along, or take the strict limit of a countable tower of, and so on.
It is right at the heart of homotopical category theory that we often want a strict construction to model a more complicated, homotopy coherent construction, such as a homotopy pullback, basically because this saves us from carrying around lots of coherence isomorphisms in our arguments, replacing them with equalities. This can't be done for pullbacks along arbitrary $\infty$-functors, but for isofibrations, it works, which allows an $\infty$-cosmos to behave something like the category of fibrant objects in a simplicial model category. If you don't have any familiarity with model categorical arguments, then you'll get some as you read further in the book and see how the strict limit properties of isofibrations enable many of the arguments Riehl and Verity make. But in particular, don't worry too much about isofibrations, which are just a technical convenience–keep reading to get to the good stuff!
| 6 | https://mathoverflow.net/users/43000 | 434259 | 175,619 |
https://mathoverflow.net/questions/434252 | 4 | Given a set $X$, by a *tree in $X$* I mean a set $T$ of finite sequences of elements of $X$ which is closed under initial segments. It is *pruned* of every element has a proper extension, and *finitely branching* if every element has at most finitely many immediate successors.
If $X$ is countable, then trees in $X$ are subsets of the countable set $X^{<\infty}$ of all finite sequences in $X$, and thus can be viewed as elements of the space $2^{X^{<\infty}}$, which is homeomorphic to the Cantor space $2^\mathbb{N}$ and thus Polish. Moreover, in this topology, the set of trees is a Borel subset of $2^{X^{<\infty}}$. Alternatively, if we consider only pruned trees, then these can be identified with their sets of infinite branches, which are exactly the closed subsets of the Polish space $X^\mathbb{N}$ (where $X$ is discrete), and thus inherits a standard Borel structure from the Effros Borel space of closed subsets of $X^{\mathbb{N}}$. The finitely branching pruned trees then correspond to the compact subsets of $X^{\mathbb{N}}$, and thus can be viewed as a compact Polish space under the Vietoris or Hausdorff topology on the space of compact subsets of $X^{\mathbb{N}}$.
**My question:** If we replace $X$ in the above by $\mathbb{R}$ (or any other uncountable Polish space), is there a natural standard Borel structure on the set of finitely branching pruned trees in $\mathbb{R}$?
My gut instinct is no, and probably one can use something like the fact that there is no nice standard Borel structure on the space of all countable subsets of $\mathbb{R}$, but I don't see how at the moment.
| https://mathoverflow.net/users/16107 | Is there a standard Borel space of finitely branching real trees? | A natural way to represent a finitely branching tree over $\mathbb{R}$ is to separate the structure of the tree from the content (ie its labels from $\mathbb{R}$).
We can describe the structure of the tree by a function $d : \mathbb{N} \to \mathbb{N}$, where $d(n)$ is the number of children the $n$-th vertex in the tree has. We number the children by giving $0$ to the root, and always counting through an entire layer of the tree first, before moving on the next layer.
In addition, we have the labelling function $\ell : \mathbb{N} \to \mathbb{R}$ telling us which real number the $n$-th vertex of the tree carries. There is a condition to impose here, namely that if $n,n+1,\ldots,n + k$ are all siblings, then $\ell(n) < \ell(n+1) < \ldots < \ell(n+k)$.
This views the space of finitely branching tree as a $\Pi^0\_2$-subspace of the Polish space $\mathbb{N}^\mathbb{N} \times \mathbb{R}^\mathbb{N}$. As such, it is Polish itself. The space of pruned tree is in turn a $\Pi^0\_1$-subspace, thus again Polish.
If we were to try the same approach for countably-branching trees, the problem would be that defining a canonic order for the siblings becomes far more complex.
| 5 | https://mathoverflow.net/users/15002 | 434262 | 175,620 |
https://mathoverflow.net/questions/434250 | 0 | Suppose you have a set of objects *X* and a scoring function *f* (in which order does not matter; *f(x,y) = f(y,x)*) which works in the following way.
* Passing a viable pair of these objects to the function will return a real number between, say, 0 and 100.
* Pairing an object to itself will return 101. Think of this as not pairing at all.
* Pairing an unviable pair of these objects will return 203.
In other words, good pair score < bad pair score < no pair score < unviable pair score.
Evaluating *f* on any pair of objects *x,y* takes an equal amount of time.
How do you pair up objects in *X* to minimise the total sum of the scorings, and thereby create an optimal pairing of the objects in the set, in the most efficient way possible?
OR
Assuming we have a function *g* that takes *X* and returns an acceptable threshold, *g(X)*, what is the most efficient way to find a pairing on *X* with a score that is less than *g(X)* (assuming such a pairing is possible).
That is to say, the pairing doesn't need to be absolutely perfect, we're just looking for something that is "pretty good".
| https://mathoverflow.net/users/494467 | Pairing optimisation w.r.t. a given function, or at least close to optimised | If you negate the function, this is a [maximum weight matching](https://en.wikipedia.org/wiki/Maximum_weight_matching) problem in an undirected graph where each object is a node, and there are polynomial algorithms for both exact and approximate solutions.
| 0 | https://mathoverflow.net/users/141766 | 434263 | 175,621 |
https://mathoverflow.net/questions/418619 | 25 | *[Cross-posted from MSE.](https://math.stackexchange.com/questions/4231158/is-there-an-infinite-topological-space-with-only-countably-many-continuous-maps?noredirect=1)*
Is there an infinite countable topological space $X$ with only countably many continuous functions to itself?
It cannot be a metrizable space. Another large class of examples that I know of are [Alexandrov topologies](https://en.wikipedia.org/wiki/Alexandrov_topology), however each Alexandrov topology corresponds to a preorder, and the continuous maps between two Alexandrov topologies correspond to the morphisms between the preorders. An infinite countable preorder has always $2^{\aleph\_0}$ endomorphisms, hence I cannot find a counterexample there either. It also cannot be a filter (+ the empty set), because any function which restricts to the identity on a set in the filter is continuous (thanks to Eric Wofsey for this last fact).
Using the [$\pi$-Base](https://topology.pi-base.org/), an online database of topological spaces inspired by the book *Counterexamples in topology* and expanding it, I obtained [this list of possible spaces](https://topology.pi-base.org/spaces?q=Countable%20%2B%20%7EMetrizable%20%2B%20%7EFinite). I proved for every one of these spaces that there were too many continuous maps, except for the Relatively prime integer topology (also known as the Golomb space) and the Prime integer topology. The first one [was proved to have too many continuous maps](https://dml.cz/handle/10338.dmlcz/147548), and the second one is very similar to the first one, so I don't place much hope on it. We need to look somewhere else.
On MSE, Mirko indicated the existence of the following paper:
ADVANCES IN MATHEMATICS 29 (1978), 89-130
Constructions and Applications of Rigid Spaces, I
V. Kannan, M. Rajagopalan
<https://www.sciencedirect.com/science/article/pii/0001870878900063>
In it, it is proven (Theorem 2.5.6) that, for any cardinal $\kappa$, if $(2^\kappa)^+ < 2^{2^\kappa}$, then there is a Hausdorff topological space of cardinality $\kappa$ which is strongly rigid, i.e. such that any continuous endofunction is either constant or the identity, which is a lot stronger than what we are trying to prove.
| https://mathoverflow.net/users/469928 | Is there an infinite topological space with only countably many continuous functions to itself? | A partial answer: the only place where Kannan and Rajagopalan use the inequality $(2^\kappa)^+<2^{2^\kappa}$ is in the application of the Theorem on page 121. That theorem is a consequence of Corollary 10.15 in Comfort and Negrepontis' *The Theory of Ultrafilters*. However the particular case that they use can be proven without an appeal to that book.
They show that for their set $F$ one can find a partition $\mathcal{A}$ of $\kappa$ into $\kappa$ many sets of cardinality $\kappa$ such that $\bar A\cap F\neq\emptyset$ for all $A\in\mathcal{A}$. Using that the space $\{0,1\}^{2^\kappa}$ has a dense subset of cardinality $\kappa$ one can map $\kappa$ onto a dense subset of that cube such that $\mathcal{A}$ is the set of point-inverses of that map, call it $f$. Then $\beta f$ not only maps $\beta\kappa$ onto that cube it also maps $F$ onto it. Take a closed subset $K$ of $F$ such that $f$ is surjective on $K$ and irreducible.
For every $\alpha<2^\kappa$ let $I\_\alpha=\{\beta\in\kappa:\pi\_\alpha(f(\beta))=1\}$ and $J\_\alpha=\kappa\setminus I\_\alpha$.
Then $\bigl\{(I\_\alpha,J\_\alpha):\alpha<2^\kappa\bigr\}$ is an independent family; even independent modulo the filter $\mathcal{F}=\{X\subseteq\kappa:K\subseteq\bar X\}$.
The proofs of Theorems 2.2 and 2.7 in [K. Kunen, *Ultrafilters and independent sets*, Trans. Amer. Math. Soc. 172 (1972), 299–306.](https://doi.org/10.1090/S0002-9947-1972-0314619-7) go through with $\mathcal{F}$ as its starting point, so that $K$ contains a set of $2^\kappa$ many Rudin-Keisler incomparable ultrafilters.
Now specialize this to $\kappa=\omega\_0$ and you have a ZFC-construction of the space in Kannan and Rajagopalan's Theorem 2.5.6.
| 7 | https://mathoverflow.net/users/5903 | 434266 | 175,622 |
https://mathoverflow.net/questions/424584 | 11 | *(Below I conflate quantifiers and quantifier **symbols** in a couple places for readability; I can change that if that actually makes things less readable.)*
For the purposes of this question, an **$n$-ary quantifier** is a (class) function $\mathscr{Q}$ assigning to each (nonempty) set $X$ a family $\mathscr{Q}X$ of subsets of $X^n$ which is stable under permutations of $X$. Given a logic $\mathcal{J}$ and a quantifier $\mathscr{Q}$, let $\mathcal{J}^\mathscr{Q}$ be the extension of $\mathcal{J}$ gotten by "adding $\mathscr{Q}$;" the full definition is a bit tedious, but (for a structure $\mathfrak{M}$ with underlying set $M$) the key clause is $$\mathfrak{M}\models\mathscr{Q}x.\varphi(x)\quad\iff\quad\{a\in\mathfrak{M}:\mathfrak{M}\models\varphi(a)\}\in\mathscr{Q}(M).$$
We can then say that a quantifier $\mathscr{Q}$ is $\mathcal{J}$-definable iff there is a finite set $\Phi$ of $\mathcal{J}$-sentences augmented with a new quantifier symbol $\mathsf{Q}$ such that $\mathscr{Q}$ is the unique quantifier, when used to interpret $\mathsf{Q}$, which makes each sentence in $\Phi$ a tautology. For example, letting $\mathcal{L}\_0$ be the quantifier-free fragment of first-order logic, each of the standard unary quantifiers $\forall$, $\exists$, and $\exists!$ is $\mathcal{L}\_0$-definable:
* $\Phi\_\forall=\{\mathsf{Q}x.\top, (\mathsf{Q}x.U(x))\rightarrow U(a)\}$
* $\Phi\_\exists=\{\neg\mathsf{Q}x.\perp, U(a)\rightarrow(\mathsf{Q}x.U(x))\}$
* $\Phi\_{\exists!}=\{\mathsf{Q}x.U(x)\wedge U(a)\wedge U(b)\rightarrow a=b,\neg\mathsf{Q}x.\perp, \mathsf{Q}x.x=a\}$
(See [here](https://math.stackexchange.com/a/4470134/28111) for more discussion of this.) More interestingly (pathologically?), the "infinitely-many" quantifier $\exists^\infty$ is definable over full first-order logic; since it provides a possible solution to this question, I'll put the proof of this at the end of this answer.
I'm curious whether there is a quantifier which is not definable over $\mathcal{L}\_0$ but is definable over $\mathcal{L}\_0^{\mathscr{Q}\_1,...,\mathscr{Q}\_n}$ for some definable-over-$\mathcal{L}\_0$ quantifiers $\mathscr{Q}\_1,...,\mathscr{Q}\_n$. More snappily:
>
> **Main question**: Over $\mathcal{L}\_0$, if a quantifier is definable relative to definable quantifiers, is it definable?
>
>
>
A natural follow-up question is to understand what properties of a general logic $\mathcal{J}$ lead to an answer one way or the other, but $\mathcal{L}\_0$ seems like an already-interesting starting point. Note that although this question isn't directly about quantification over sets, there is a fundamentally higher-order aspect to this idea of quantifier definability, hence the "higher-order-logics" tag.
EDIT: perhaps the following sub-question may be more easily attacked:
>
> **Secondary question**: Can $\exists^\infty$ be defined over $\mathcal{L}\_0$ using a set of formulas each of which contains exactly one instance of "$\mathsf{Q}$"?
>
>
>
My hope is that this secondary question has a relatively easy negative answer, which would constitute helpful progress towards the expected answer ("yes, and $\exists^\infty$ does the job") to the main question.
---
A natural candidate
-------------------
I didn't notice this at first, but there is a natural candidate for a positive answer: the quantifier $\exists^\infty$. This is definable over $\mathsf{FOL}$ (and so is "two-step-definable" over $\mathcal{L}\_0$) as follows.
The sentences $$\mathsf{Q}x.U(x)\rightarrow\mathsf{Q}x.[U(x)\vee V(x)],$$ $$\neg\mathsf{Q}x.\perp,$$ and $$[\neg\mathsf{Q}x.U(x)]\rightarrow\neg\mathsf{Q}x.(U(x)\vee x=a)$$ are "tautologized" by exactly those $\mathscr{Q}$ which are "sub-quantifiers" of $\exists^\infty$; the first sentence corresponds to monotonicity, while the second and third rule out the sufficiency of a finite sets of satisfying instances.
Now let $\nu$ be a first-order sentence which only has infinite models not using the unary relation symbol $U$; the sentence $$\nu^U\rightarrow\mathsf{Q} x. U(x)$$ (where $\nu^U$ is the relativization of $\nu$ to $U$ as usual) gets us the rest of the way there.
This results in a finite defining set for $\exists^\infty$ ... over $\mathsf{FOL}$. However, there is no obvious way to bring this down to $\mathcal{L}\_0$, specifically because of the last step.
| https://mathoverflow.net/users/8133 | Are there quantifiers that require multiple "steps" to define? | The quantifier $\exists^\infty$ is not definable over $\mathcal{L}\_0$.
To prove it, we show that a sentence $S$ tautologised by $\exists^\infty$ is also a tautology for the "always false" quantifier $\mathsf{F}$.
Let $\mathfrak{M}$ be a model for $S$. From $\mathfrak{M}$, we build a finite model $\mathfrak{M}'$ which gives the same values to the subformulas of $S$ containing no $\mathsf{Q}$ and no bound variable. The model $\mathfrak{M}'$ is constructed by restricting to a finite subset of relevant elements and changing the values of functions on elements that do not appear in the evaluation of the atomic subformulas of $S$. Since $S$ evaluates to true over $\mathfrak{M}'$ for $\exists^\infty$, it also evaluates to true for $\mathsf{F}$ because the two quantifiers are equal on finite models.
The sentence $S$ is a boolean combination of atomic subformulas and quantified subformulas. For $\mathsf{Q}=\mathsf{F}$ over $\mathfrak{M}$ and $\mathfrak{M}'$, the quantified subformulas evaluate to false and the atomic subformulas do not change their values (by contruction of $\mathfrak{M}'$). This implies that $S$ is true for $\mathsf{F}$ over $\mathfrak{M}$ also. Since $\mathfrak{M}$ is arbitrary, $S$ is tautologised by $\mathsf{F}$.
| 3 | https://mathoverflow.net/users/142409 | 434267 | 175,623 |
https://mathoverflow.net/questions/434260 | 1 | A *Fibonacci-type sequence* is a sequence with two seed-values, $F\_1$ and $F\_2$, and which, for all $n>2$, abides by the recurrence relation $F\_n = F\_{n-1} + F\_{n-2}$. If $F\_1 = F\_2 = s$, then the $n$th number is equal to the number of compositions of $n-1$, consisting only of $1$'s and $2$'s, multiplied by $s$:
$$\begin{align} F\_n = \sum\_{k=0}^{\lfloor (n-1)/2\rfloor} \binom{\lfloor (n-1)/2\rfloor+k}{2k+1} s \end{align} $$
I also have a similar kind of formula for when $F\_1 \ne F\_2$ that uses the exact same technique (counting compositions that is), though I chose not include it here for brevity's sake.
So, my question is the following; **has this result been found before?** I know that people know about the link between compositions of $n-1$ and the $n$th Fibonacci number (it's mentioned on Wikipedia), but have people found this formula? I would think so, but I haven't found it when Googling.
**NB:** If you have $F\_0 = 0$, that amounts to $F\_1 = F\_2$, which means the above formula still stands even though, technically, the seed-values aren't equal.
| https://mathoverflow.net/users/493854 | Explicit formula for Fibonacci numbers; compositions of $n$ | Yes, this identity is well known. According to Singh's [The so-called Fibonacci numbers in ancient and medieval India](https://www.sciencedirect.com/science/article/pii/0315086085900217), the $s=1$ case has been known since at least the the 14th century. Since everything in the sequence with $F\_1 = F\_2 = s$ is a multiple of $s$, the formula for that case follows immediately.
Your Fibonacci-type sequence is known in the literature as generalized Fibonacci sequences or, in Art Benjamin & Jenny Quinn's wonderful book [*Proofs that Really Count*](https://bookstore.ams.org/dol-27/), Gibonacci numbers. Their Identity 4 is the $s=1$ case of your formula, proven combinatorially with square and domino tilings, equivalently compositions restricted to parts 1 and 2 (an interpretation Singh suggests was known some time BCE). With $F\_1 = F\_2 = s$, the parts are restricted to $s$ and $2s$ analogously.
| 1 | https://mathoverflow.net/users/14807 | 434268 | 175,624 |
https://mathoverflow.net/questions/434211 | 13 | Equivariant homotopy theory focuses on spaces together with some group action on them. Jeroen van der Meer and Richard Wong have [a paper](https://arxiv.org/abs/2107.06308) where they use equivariant methods to compute the Picard group of the stable module category of representations for certain finite groups. I was wondering if there are more results of a similar flavor, providing applications of equivariant homotopy theory to representation theory (or I suppose group theory in general).
| https://mathoverflow.net/users/489806 | Applications of equivariant homotopy theory to representation theory | There are decades and decades of algebraic results that use techniques from equivariant homotopy theory. Some examples ...
(1) Quillen's work on ring theoretic aspects of the cohomology of finite groups [The spectrum of an equivariant cohomology ring. I, II. Ann. of Math. (2) 94 (1971), 549–572; ibid. (2) 94 (1971), 573–602]. He uses `descent' arguments that leave the world of finite groups: to learn about $BG$, he studies $EG \times\_G X$ for suitable $G$--spaces $X$.
(2) Quillen's calculation of $H^\*(GL\_{\infty}(\mathbb F\_q); k)$ and his use of this to compute the algebraic $K$--theory of finite fields involves serious excursions into homotopy theory.
(3) Quillen's paper (hmm ... do we see a trend here?) [Homotopy properties of the poset of nontrivial p-subgroups of a group. Adv. in Math. 28 (1978), no. 2, 101–128] amounts to a gorgeous use of elementary applications of categorical/homotopical methods to prove things about groups.
And Quillen is not the only person to use homotopy to prove results about groups. Here is an example:
(4) Peter Symonds, in [On the Castelnuovo-Mumford regularity of the cohomology ring of a group. J. Amer. Math. Soc. 23 (2010), no. 4, 1159–1173], proves a conjecture of Benson's about regularity of the group cohomology of a finite group by first using a descent argument, similar to Quillen, and then using a rather delicate results of Jeanne Duflot about actions of elementary abelian $p$-groups on smooth manifolds.
One could go on with many other examples, including results that are just being proved now.
| 13 | https://mathoverflow.net/users/102519 | 434271 | 175,625 |
https://mathoverflow.net/questions/434274 | 5 | For the past year and a half, I have been working my way through Diamond & Shurman's "A First Course in Modular Forms", and I have just finished it. I Have Some Questions.
1. What is so special about two dimensions? One can think about lattices/tori in N dimensions and their moduli space $$SL\_n(\mathbb{Z})\backslash GL\_n(\mathbb{R})/(SO\_n(\mathbb{R})x\mathbb{R})$$
(I use $GL\_n/\mathbb{R}$ to represent the fact that we are scaling one element of the lattice basis to 1 instead of scaling the volume of the cell). One naturally obtains a metric on the symmetric space and one can look for forms on that space which transform appropriately under the action of $SL\_n(\mathbb{Z})$ or a subgroup thereof. This gives you the whole machinery of the torsion groups, the double coset (Hecke) operators etc.
What exactly do you wind up losing? Does it matter if $n$ is even or odd (specifically can we define a mapping of the torus to a complex algebraic variety if $n$ is even)?
2. It's stated (as several versions of the Modularity Theorem) that given an elliptic curve $E$ there is some modular curve $X\_0(N)$ (alt. its Jacobian $J\_0(N)$) that maps onto $E$, with the minimum value of $N$ given by the conductor. For the Jacobian this is a map from a $2g$-dimensional torus to a 2-dimensional torus. Do these maps induce a unique decomposition of $J\_0(N)$? That is, can we represent $J$ as the direct sum of a bunch of inverse images of elliptic curves plus possibly some other $2d$-dimensional torus?
3. I don't know how to work this example. Pick a random small conductor N. Explicitly find all elliptic curves E with mappings from $X\_0(N)/J\_0(N)$.
| https://mathoverflow.net/users/17496 | Lots of questions about modular forms | (1.) This is a very good question and shows you are thinking in the right directions, but it also is asking for a summary of multiple entire fields of mathematics. Some keywords are "automorphic forms", "locally symmetric spaces", "Shimura varieties".
In brief, you can exactly do that. You have to think about what you mean by "forms" because, [as Noam Elkies says](https://mathoverflow.net/questions/434274/lots-of-questions-about-modular-forms#comment1118539_434274), you lose the complex structure, so you can't consider holomorphic differential forms. The simplest thing to do instead is to consider real-valued functions that are eigenfunctions of the Laplace operator and the other Casimir operators. Many of the analytic results about modular forms have analogues in this setting.
These don't parameterize mappings of the torus to complex algebraic varieties, by which I believe you mean abelian varieties. To do that, you just need to replace the general linear group with the symplectic group in your construction. You can go further, and consider arbitrary algebraic groups. All of these objects have nice classes of functions on them and theories of Hecke operators, and all of them play a role in the Langlands program.
The significance of rank 2 is that, thanks roughly to exceptional isomorphisms, it's a special case of many different nice classes simultaneously (both a general linear group and thus a moduli space of lattices with a good theory of Fourier coefficients and many other useful theorems, and a symplectic group and thus a moduli space of abelian varieties, and a one-dimensional algebraic variety and thus admitting a Jacobian, and...) Any one given nice property of the modular curve and modular forms will have higher analogues, but not all of them simultaneously.
(2.) Roughly, yes. The Jacobian splits up to isogeny into a product of abelian varieties associated to eigenforms, with the dimension of the variety depending on the degree of the field of definition, so eigenforms with integral coefficients give elliptic curves. This is much, much easier to prove than the modularity theorem, and goes back to Eichler and Shimura.
Actually, the way you've stated your question, it may just be answered by the general point that the category of abelian varieties under isogeny is semsimple so they all split uniquely into a product (up to isogeny) of simple factors.
(3.) Using modular symbols, one can find all eigenforms of level $N$, extract the ones with integral coefficients, and then integrate those over loops in the modular curve, calculating their period lattice, and then use that to find equations for the modular curve. See [Cremona's book on algorithms for modular elliptic curves](http://homepages.warwick.ac.uk/%7Emasgaj/book/fulltext/index.html), [Chapter 2](http://homepages.warwick.ac.uk/%7Emasgaj/book/fulltext/index.html).
(4.) I'm not sure exactly what you mean, but I think most questions about other bundles over $X\_0(N)$ will either reduce to questions about modular forms, or else be very difficult.
| 9 | https://mathoverflow.net/users/18060 | 434275 | 175,627 |
https://mathoverflow.net/questions/434291 | 5 | Consider a four-dimensional Lorentzian manifold $(\mathcal{M},g)$ and a $3$-dimensional compact manifold $\Sigma$, such that there exists a spacelike embedding $i:\Sigma\to\mathcal{M}$ so that $h:=i^{\ast}g$ becomes a Riemannian metric on $\Sigma$.
In a paper it was, without references, stated the space
$$\mathrm{Emb}^{\infty}(\Sigma,\mathcal{M}):=\{i:\Sigma\to\mathcal{M}\mid i\text{ smooth spacelike embedding such that }h=i^{\ast}g\text{ is Riemannian} \}$$
is a manifold. Does anyone know any references where this fact is proven? Also, what type of infinite-dimensional manifold is it? I would expect that it is an open subset of $C^{\infty}(\Sigma,\mathcal{M})$ (?) and hence, it should be a Frechet manifold. Also, what about "differentiable" structures on this manifold?
| https://mathoverflow.net/users/259525 | Space of spacelike embeddings as infinite-dimensional manifold | A standard reference on infinite dimensional manifolds is
*Kriegl, Andreas; Michor, Peter W.*, [**The convenient setting of global analysis**](http://www.ams.org/online_bks/surv53/), Mathematical Surveys and Monographs. 53. Providence, RI: American Mathematical Society (AMS). x, 618 p. (1997). [ZBL0889.58001](https://zbmath.org/?q=an:0889.58001). [(on author's website)](https://www.mat.univie.ac.at/%7Emichor/apbookh-ams.pdf)
Some notes on the space of embeddings between two manifolds are in §44.1. Your space of spacelike embeddings $\mathrm{Emb}^\infty(\Sigma,\mathcal{M})$ is itself an open subset of the smooth embeddings between $\Sigma$ and $\mathcal{M}$.
| 6 | https://mathoverflow.net/users/2622 | 434294 | 175,631 |
https://mathoverflow.net/questions/434138 | 1 | Let $G =$ PGL$\_{n}(\textbf{C})$ and $T$ be the image in $G$ of the subgroup of the invertible diagonal matrices of $\operatorname{GL}\_{n}(\textbf{C})$. Let $A$ and $B$ be two elementary abelian $2$-subgroups in $T$ of the same rank.
The Kronecker product $\otimes I\_{2}$ embeds $A$ and $B$ in $H$ = $\operatorname{PGL}\_{2n}(\textbf{C})$. If $A$ and $B$ are not conjugate in $G$, will $A\otimes I\_{2}$ and $B\otimes I\_{2}$ not be conjugate in $H$? Intuitively, they are not conjugate in $H$ but I'm not sure of tools to tackle it. Maybe subgroup conjugacy is an equivalent relation and $\otimes I\_{2}$ reserves it?
Edit:
Some thinking made based on the input in the comment.
If $A$ and $B$ are not conjugate in $G$, and we assume $A\otimes I\_{2}$ and $B\otimes I\_{2}$ are conjugate, attempt to get a contradiction. Since $A\otimes I\_{2}$ and $B\otimes I\_{2}$ are both block diagonal matrices with blocks $I\_{2}$ and $-I\_{2}$, if there always exists a "block permutation matrix" with each block $I\_{2}$, then we're done. But I'm not sure about the existence of such a "block permutation matrix"... Any hints would be appreciated.
| https://mathoverflow.net/users/488802 | Kronecker product preserves the conjugacy relation? | If $A$ and $B$ are elementary abelian $2$-subgroups of $\mathrm{PGL}\_n(\mathbf C)$ of rank $r$ then they lift uniquely to elementary abelian $2$-subgroups of $\mathrm{GL}\_n(\mathbf C)$ of rank $r+1$ (take all lifts $a$ of elements of $A$ such that $a^2 = 1$), so let us assume $A, B \le \mathrm {GL}\_n(\mathbf C)$ to begin with.
Consider a representation $\rho:C\_2^r \to \mathrm{GL}\_n(\mathbf C)$ with character $\chi$. Decompose $\chi$ into irreducible characters: $\chi = w\_1 \chi\_1 + \cdots + w\_m \chi\_m$. Here $\chi\_1, \dots, \chi\_m : C\_2^r \to \{\pm1\}$ and $w\_1, \dots, w\_m$ are positive integers. Two representations $\rho$ and $\rho'$ are equivalent (conjugate) if and only if they have the same character, so equivalence classes of representations $\rho:C\_2^r \to \mathrm{GL}\_n(\mathbf C)$ are in bijection with maps $w: D \to \mathbf N = \{0,1,2,\dots\}$ such that $\sum\_{\chi \in D} w(\chi) = n$, where $D \cong C\_2^r$ is the dual group of $C\_2^r$ (the group of characters $\chi : C\_2^r \to \{\pm1\}$ with pointwise product as the group operation). Here $w(\chi\_i)$ records the multiplicity of an irreducible character $\chi\_i$ in $\chi$. The representation is faithful if and only if the support $\mathrm{supp}(w) = \{\chi : w(\chi) > 0\}$ of $w$ generates $D$.
To get conjugacy classes of embedded copies of $C\_2^r$ we must further quotient $\{w:D \to \mathbf N \mid \langle \mathrm{supp}~w\rangle = D\}$ by the action of $\mathrm{Aut}(C\_2^r) \cong \mathrm{GL}\_r(2)$.
In any case the Kronecker product just sends $w$ to $2w$, so the conjugacy relation is preserved.
| 0 | https://mathoverflow.net/users/20598 | 434296 | 175,632 |
https://mathoverflow.net/questions/434150 | 10 | [Zhang 2022](https://arxiv.org/abs/2211.02515) proves a somewhat suspicious formula:
$$L(1,\chi) \gg (\log D)^{-2022}$$
This raises the obvious-but-frivolous question: did he intentionally weaken the constant to get the current year?
| https://mathoverflow.net/users/22930 | Did Zhang weaken the constant in his Landau-Siegel zero paper to get the current year? | According to himself, yes. The following is a link to some of his comments that he posted on a Chinese forum.
<https://www.zhihu.com/question/564799818/answer/2752632822>
>
> Regarding the question of whether the fixed power of logD , which is taken for many parameters in the paper, is to get the number 2022, in terms of the Landau-Siegel zero itself, that should be a power of logD, and the conjectured should actually be -1. My method can lead to an exponent of negative several hundred which I did not calculate carefully, yet I can guarantee -2022. Since this year is 2022, I chose it casually. Often people do this kind of thing, so this also does not have any special meaning, just like the previous 70 million (in the bounded gaps between primes paper).
>
>
>
| 19 | https://mathoverflow.net/users/494533 | 434308 | 175,636 |
https://mathoverflow.net/questions/434276 | 18 | Consider a function $h$ defined on real numbers, which is not of the form $kx+b$ i.e. a linear function. If $h$ maps rational numbers to rational numbers and it maps irrational numbers to irrational numbers, could $h$ be analytic? If so, how to give an example?
| https://mathoverflow.net/users/494497 | Is there an analytic non-linear function that maps rational numbers to rational numbers and it maps irrational numbers to irrational numbers? | Answering a question of Erdos, Barth and Schneider proved that for every countable dense sets $A$ and $B$
in the complex plane, there exists an entire function such that $f(z)\in B$ if and only if $z\in A$.
K. Barth and W. Schneider, Entire functions mapping arbitrary countable dense sets and their complements to each other,
J. London Math. Soc., 4 (1971/72) 482-488.
Another paper of the same authors concerns the case
when $A$ and $B$ are on the real line. They prove
that for every two such countable dense sets, there is a transcendental entire function that maps $A$ into $B$ monotonically.
MR0269834
Barth, K. F.; Schneider, W. J.
Entire functions mapping countable dense subsets of the reals onto each other monotonically.
J. London Math. Soc. (2) 2 (1970), 620–626.
| 22 | https://mathoverflow.net/users/25510 | 434311 | 175,637 |
https://mathoverflow.net/questions/434289 | 1 | Let $(X, \Sigma, \mu)$ be a $\sigma$-finite complete measure space, $(E, |\cdot|)$ a Banach space, and $p \in (1, \infty)$. Let $L\_p := L\_p(X, \mu, E)$ be the [Bochner space](https://en.wikipedia.org/wiki/Bochner_space) of all $\mu$-integrable functions $f:X \to E$. Here we use [Bochner integrals](https://en.wikipedia.org/wiki/Bochner_integral). If $E = \mathbb R$ then $L\_p$ is uniformly convex for $p \in (1, \infty)$.
>
> Is $L\_p$ uniformly convex if $E$ is uniformly convex for $p \in (1, \infty)$?
>
>
>
Any reference for the proof is greatly appreciated.
---
I saw [this](https://math.stackexchange.com/questions/4513522/let-e-be-a-uniformly-convex-banach-space-and-p-in-1-infty-is-l-px?rq=1) question on MSE which received one answer whose author does not remember the title of the reference paper. So I post it here.
| https://mathoverflow.net/users/477203 | Is $L_p(X, \mu, E)$ uniformly convex for $p \in (1, \infty)$ if $E$ is a uniformly convex Banach space? | A reference for this result would be [Some more uniformly convex spaces](https://www.ams.org/journals/bull/1941-47-06/S0002-9904-1941-07499-9/S0002-9904-1941-07499-9.pdf) by Mahlon M. Day, Bull. Amer. Math. Soc. 47(6): 504-507 (June 1941).
(Alternative link at [Project Euclid](https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society-new-series/volume-47/issue-6/Some-more-uniformly-convex-spaces/bams/1183503696.full))
| 2 | https://mathoverflow.net/users/85906 | 434314 | 175,639 |
https://mathoverflow.net/questions/434305 | 1 | For a multidimensional subshift $X$ over $\mathbb Z^d$, the topological full group $[X]$ is the set of homeomorphisms $f$ of $X$ that can be written as $f : x \mapsto \sigma\_{c(x)}(x)$ with $c : X \to \mathbb Z^d$ a continuous function (namely, a cocyle).
My questions would mainly be about embedability of those groups (for which groups $G$ and which type of subshifts - minimal, of finite type, sofic, effective ... - do we have $G$ as a subgroup of $[X]$), realization (which groups G can be realized as topological full groups of subshifts), computability (how "hard" - in a computability theory sense - it is to determine the topological full group of a given subshift), closure properties ...
Some things are known for one-dimensional subshifts, *e.g.*
* Vile Salo, *Graphs and wreath products in topological full groups of full shifts* shows (among other things) how to embed every finitely generated right angled Artin group in $[\{0, 1\}^{\mathbb Z}]$
* Nicolás Matte Bon, *Topological full groups of minimal subshifts with subgroups of intermediate growth*, shows how to embed the Grigorchuk group(s) in the topological full group of minimal one-dimensional subshifts
On the other hand, the cohomology of multidimensional subshifts has been studied, and some work has been done to understand what their cocycles (with values in arbitrary groups, not restricted not $\mathbb Z^d$) look like:
* Klaus Schmidt, *Tilings, fundamental cocycles and fundamental groups of symbolic Zd actions*
* Einsiedler, *Fundamental cocycles of tiling spaces*
and several other articles.
Is there *a priori* anything "interesting" to say in the multidimensional case (say subshifts/tilings over $\mathbb Z^2$), which would - qualitatively or quantitatively - differ from the 1d-case ? For other algebraic conjugacy invariants, such as automorphism groups, there are indeed interesting questions and approaches to solve them both for subshifts over $\mathbb Z$ and over $\mathbb Z^2$, and although the one-dimensional case has been more extensively studied, it is, as far as I can tell, a consequence of the difficulty of the problem rather than because the 2d-case is "not interesting" or "could be reduced to the 1d-case". Is the situation similar for the understanding of topological full groups of subshifts, or is there some reason that I missed for it to be (apparently) only ever studied for 1d subshifts ?
| https://mathoverflow.net/users/494521 | Topological full groups of subshifts: differences between one-dimensional and multi-dimensional subshifts | I don't know that much literature on the multidimensional case (though I'm not sure I'm the one who would if there is literature, either), but I can collect the comments and try to add a few things. Sorry in advance if there are some mistakes in my claims, I never wrote some of these carefully before (and still didn't).
Let me write $[X]$ for the topological full group of a $\mathbb{Z}^d$-subshift.
>
> Theorem. $\mathbb{Z} \wr \mathbb{Z}$ and $\mathbb{Z}\_2 \wr \mathbb{Z}^2$ do not embed in the topological full group of any one-dimensional subshift, but embed in the topological full group of a two-dimensional subshift of finite type.
>
>
>
Proof. The nonembeddability is due to Le Boudec and Matte Bon [1].
For the two-dimensional subshift of finite type for the first, pick an alphabet that allows you to draw "combs" in the sense of [2] (btw, this construction is more or less exactly why I am interested in combs in groups), and the SFT rule is that the spikes of the comb must continue infinitely (just use some directed arrows). Now by marking the parity of how far you are on a spike, you can simulate the standard action of $\mathbb{Z} \wr \mathbb{Z}$ with orbit growth $n^2$, i.e. one generator goes along the comb, and one goes up even positions of a spike and down the odd ones (wrapping at the base of the comb).
As for the second, pick the sunny-side-up $Y = \{y \in \{0,1\}^{\mathbb{Z}^2} \;|\; \sum y \leq 1\}$. The generators on the quotient side shift by $2$ horizontally and $1$ vertically, and the $\mathbb{Z}\_2$ generator swaps the positions at the egg $1$ and the position immediately to the right of it. Square.
There are a lot more nonembeddability results in [1], but I don't know which ones have a corresponding embeddability result in two dimensions.
>
> Theorem. For one-dimensional sofic shifts, the topological full group has decidable torsion problem, but in two-dimensional full shifts you have finitely-generated subgroups where it's undecidable.
>
>
>
Proof. Directly in [3]. Square.
Finally, we have the result of Elek and Monod, let me try to generalize it a little:
>
> Theorem. If $X$ is a one-dimensional subshift with dense periodic points, then $[X]$ embeds in $[Y]$ for a minimal two-dimensional subshift $Y$.
>
>
>
Proof. Let $X \subset A^{\mathbb{Z}}$, assume $X$ is infinite (the finite case is trivial). Enumerate the repeating patterns of periodic points of $X$ as $w\_1, w\_2, ...$. We construct a two-dimensional subshift over alphabet $\bar A \cup A \cup \{\#\}$ where $\bar A = \{\bar a \;|\; a \in A\}$ is a copy of $A$. The rows of the configurations look like $(w\_i' \#)^{\mathbb{Z}}$ where $w'\_i$ is a "conveyor belt version" of $w\_i$, so if $w\_i = u\_iv\_i$ where $|u\_i|=|v\_i|$ then we interlace $u\_i$ with the reverse of $\bar v\_i$ to get $w\_i'$, and if $|w\_i|$ is odd we take $|v\_i| = |u\_i|-1$ and do the same. The point is, if we see some position of $w\_i'$ at the origin, we can "unwrap it" to $w\_i$, and then move along it as if it were the periodic point $w\_i^{\mathbb{Z}}$. Now no matter how you organize the rows into configurations of $Y$, as long as you use all of the $w\_i$, you get a subshift whose topological full group contains a copy of $[X]$. Now you can make $Y$ minimal using standard tricks, just repeat each type of row a lot, and in all sorts of combinations. Square.
References
[1] arxiv.org/abs/2205.11924
[2] [Which groups contain a comb?](https://mathoverflow.net/questions/325791/which-groups-contain-a-comb)
[3] arxiv.org/pdf/1603.08715.pdf
| 3 | https://mathoverflow.net/users/123634 | 434317 | 175,640 |
https://mathoverflow.net/questions/423541 | 7 | For various types of groups, there exist catalogues of those groups of the
particular type which are "small" in a certain sense. — For example:
* The [GAP Small Groups Library](https://www.gap-system.org/Packages/smallgrp.html) catalogizes groups of small order,
* The [GAP Transitive Groups Library](https://www.gap-system.org/Packages/transgrp.html) and the [GAP Primitive Groups Library](https://www.gap-system.org/Packages/primgrp.html)
catalogize transitive, respectively primitive, subgroups of symmetric groups
of small degree,
* The [GAP Perfect Groups Library](https://www.gap-system.org/Datalib/perfect.html) catalogizes perfect groups
of small order,
and so on.
**Question:** Does there exist such data library for groups with "short"
finite presentations?
*Remarks:*
* There is a [12-years-old question](https://mathoverflow.net/questions/15094/) asking for such database.
At that time, apparently such database was not available.
But maybe this has changed in the meantime(?)
* When setting the *length* of a finite presentation equal to
the sum of the lengths of the relators, for every presentation on two
generators of length at most $10$, it is straightforward to decide
whether the corresponding group $G$ is trivial, nontrivial but finite,
or infinite (if $G$ is infinite, this can be seen from the abelian
invariants of $G$, $G'$ or $G''$, and if $G$ is finite, coset
enumeration finishes even with very small limits). Where things start
to get more interesting is length $11$. One of the few examples
for which deciding finiteness seems more tricky is the group
$$
G \ := \ \langle a, b \ | \ a^3 = ab^{-3}a^{-1}b^{-1}a^{-1}b = 1 \rangle.
$$
What is easy to see is that $G/G'' \cong {\rm C}\_{37} \rtimes {\rm C}\_9$,
and that $G''$ is perfect — but beyond that, things seem to get more
difficult. Almost for sure, people have considered this presentation
(and other similar presentations) before. Looking into an appropriate
data library would tell what is known about that (and other similar)
groups immediately.
| https://mathoverflow.net/users/28104 | Catalogue of groups with short finite presentations | I would very much like to have such a database and would like to contribute to its development. Prompted by this question, we talked about what such a database could look like (e.g. in terms of groups covered, functionality etc.) at a discussion session of a workshop in Manchester with Ian Leary, Marco Linton, Saul Schleimer and Henry Wilton. I encourage anyone interested in this to contact me.
As some inspiration, I'll link this [mathoverflow question on atlas-like websites](https://mathoverflow.net/questions/354327).
| 3 | https://mathoverflow.net/users/24447 | 434318 | 175,641 |
https://mathoverflow.net/questions/434239 | 2 | Let the function $f\colon [a,b] \to\mathbb{C}$ be Lipschitz and let $|f(a)| \geq c,$ $|f(b)| = c$ and $\varepsilon > 0.$
It is easy to see that if $\|f\|\_{\infty}< \frac{\varepsilon}{2} =: \delta (\varepsilon)$ then we can find $g$ with followning properties:
1. $$\|f-g\|\_{\infty}< \varepsilon$$
2. $$g(a)=f(a), \,\, g(b)=f(b)$$
3. $$|g| \geq c$$
Indeed, it is enough to take $g$ with the given values in $a$ and $b$, such that $c \leq |g| < \frac{\varepsilon}{2}.$
Is it true if we replace supremum norm by Lipschitz norm? I.e. can we find $\delta (\varepsilon)$ and $g$ such that the above three conditions hold if we replace $\| \cdot \|\_{\infty}$ by Lipschitz norm everywhere above?
There should be some simple counterexample.
| https://mathoverflow.net/users/490101 | Construction of the Lipschitz function with a given Lipschitz constant, given two values and with small Lipschitz norm | I can achieve $L(f - g) \leq (\frac{1}{2} + \frac{\pi}{4})\epsilon = (1.285\ldots)\epsilon$. Two reductions: (1) we can assume $|f(t)| < c$ for all $t \in (a,b)$ and (2) we can take $\epsilon = 1$.
(1) because $C = \{t: |f(t)| \geq c\}$ is a closed subset of $[a,b]$, so its complement is a countable set of disjoint open intervals $[a\_i, b\_i]$ such
that $|f(a\_i)| = |f(b\_i)| = c$ and $|f(t)| < c$ for all $c \in (a\_i,b\_i)$; then we can set $g = f$ on $C$ and handle each of these intervals separately. (2) just by scaling.
Assuming these reductions, define $g: [a,b] \to \mathbb{C}$ by letting $g(a) = f(a)$, $g(b) = f(b)$, and letting $g(t)$ move along the $|z| = c$ circle from $f(a)$ to $f(b)$ at uniform speed. The greatest possible discrepancy between $|f(b) - f(a)|$ and the length of the arc from $f(a)$ to $f(b)$ occurs when $f(a)$ and $f(b)$ are diametrically opposed and the arc length is $\frac{\pi}{2}$ times longer then the secant line. In that case, over any small interval $[t, t + \delta]$ we have $|f(t + \delta) - f(t)| < \frac{\delta}{2}$ since $L(f) < \frac{1}{2}$ by hypothesis, and $|g(t + \delta) - g(t)| \approx \frac{\pi \delta}{4}$ since small segments of a circle are approximately straight lines. Just from this we get $|(f - g)(t + \delta) - (f - g)(t)| \lessapprox \frac{\delta}{2} + \frac{\pi\delta}{2} = \frac{1 + \pi}{2}\delta$, so that $L(f - g) \leq \frac{1 + \pi}{2}$.
| 2 | https://mathoverflow.net/users/23141 | 434323 | 175,643 |
https://mathoverflow.net/questions/434329 | 11 | Suppose $m$ is a positive integer.
I am looking for finite sets with group actions such that the action is transitive on the set of $m$-element subsets, but NOT transitive on the set of $(m+1)$-element subsets.
An example for $m=2$ is a projective space over a finite field.
| https://mathoverflow.net/users/1441 | Not very transitive actions | According to Theorem 4.11 of Peter Cameron's book `Permutation Groups' it follows from the classification of finite simple groups that the only finite 6-transitive groups are (some of the) symmetric and alternating groups in their natural actions, and the only finite 4-transitive groups are symmetric, alternating and the Mathieu groups $M\_{11}$, $M\_{12}$, $M\_{23}$ and $M\_{24}$. Thus there will not be very many examples of what you are looking for when $m\geq 4$.
| 11 | https://mathoverflow.net/users/124004 | 434333 | 175,646 |
https://mathoverflow.net/questions/427125 | 6 | A function $f:X\to Y$ between topological spaces is called
$\bullet$ *$\sigma$-continuous* if there exists a countable cover $\mathcal C$ of $X$ such that for every $C\in\mathcal C$ the restriction $f{\restriction}\_C$ is continuous;
$\bullet$ a *$\sigma$-homeomorphism* if $f$ is bijective and the maps $f$ and $f^{-1}$ are $\sigma$-continuous.
Two topological spaces $X,Y$ are called *$\sigma$-homeomorphic* if there exists a $\sigma$-homeomorphism $h:X\to Y$.
By Theorem $\Delta^0\_3$ from [this MO-post](https://mathoverflow.net/a/426983/61536), any two countable-dimensional uncountable Polish spaces are $\sigma$-homeomorphic. We recall that a topological space is *countable-dimensional* if it is the countable union of zero-dimensional spaces.
By a famous [result of Pol](https://www.jstor.org/stable/2043785), there exists a totally disconnected Polish space $P$ which is not countable-dimensional. Since the countable-dimensionality is preserved by $\sigma$-homeomorphisms, Pol's space $P$ is not $\sigma$-homeomorphic to the Cantor cube $2^\omega$. By the same reason, the Hilbert cube $[0,1]^\omega$ is not $\sigma$-homeomorphic to the Cantor cube. Since $[0,1]^\omega$ is not a countable union of totally disconnected spaces, the Hilbert cube $[0,1]^\omega$ is not $\sigma$-homeomorphic to Pol's space $P$.
Therefore, we have three spaces: $2^\omega$, $[0,1]^\omega$, $P$ which are not pairwise $\sigma$-homeomorphic.
>
> **Problem.** *Are there infinitely (countably, continuum) many uncountable Polish spaces which are not pairwise $\sigma$-homeomorphic?*
>
>
>
**Remark.** Repeating the argument of the proof of Theorem $\Delta^0\_n$ in [this MO-post](https://mathoverflow.net/a/426983/61536), it is possible to show that two Polish space $X,Y$ are $\sigma$-homeomorphic if and only if there exist countable partitions $\{X\_n\}\_{n\in\omega}$ and $\{Y\_n\}\_{n\in\omega}$ of the spaces $X,Y$ such that $X\_n$ and $Y\_n$ are homeomorphic Polish spaces for every $n\in\omega$. This characterization should imply that the maximal number of pairwise non-$\sigma$-homeomorphic Polish spaces belongs to the set $\omega\cup\{\omega,\aleph\_1,\mathfrak c\}$.
| https://mathoverflow.net/users/61536 | Classification of Polish spaces up to a $\sigma$-homeomorphism | There are indeed continuum many. See:
Kihara, T., & Pauly, A. (2022). Point Degree Spectra of Represented Spaces. Forum of Mathematics, Sigma, 10, E31. doi:10.1017/fms.2022.7
| 3 | https://mathoverflow.net/users/94358 | 434334 | 175,647 |
https://mathoverflow.net/questions/434300 | 4 | $\DeclareMathOperator\PGL{PGL}\DeclareMathOperator\Proj{Proj}\DeclareMathOperator\Pic{Pic}$I have a question about an example for a line bundle not admitting a
$G$-linearization from Mumford's GIT, page 33:
We consider the action of $\PGL(n+1)$ on projecive space
$\mathbb{P}^n= \Proj k[X\_0,\dotsc, X\_n]$.
Observe that $\PGL(n+1)$ is given as the open affine
subscheme of
$$ \mathbb{P}^{n^2+2n} = \Proj k[a\_{00},\dotsc, a\_{0n};
a\_{10}, \dotsc , a\_{nn}]$$
complementary to the determinants hypersurface $\det(a\_{ij})=0$. The
action morphism $ \sigma: \PGL(n+1) \times \mathbb{P}^n \to \mathbb{P}^n $ is determined by
\begin{gather\*}
\sigma^\*(\mathcal{O}\_{\mathbb{P}^n}(1)) \cong
p\_1^\*(\mathcal{O}\_{\mathbb{P}^{n^2+2n}}(1)) \otimes
p\_2^\*(\mathcal{O}\_{\mathbb{P}^{n}}(1)) \\
\sigma^\*(X\_i)= \sum\_{j=0}^n p\_1^\*(a\_{ij}) \otimes p\_2^\*(X\_j)
\end{gather\*}
where $p\_1, p\_2$ are
projections canonical projections.
Mumford claims that $\mathcal{O}\_{\mathbb{P}^n}(1) $ admits no
$\PGL(n+1)$-linearization, because
the restriction of $\mathcal{O}\_{\mathbb{P}^{n^2+2n}}(1)$
to the open subscheme $\PGL(n+1)$ has order $n+1$ in
$\Pic[\PGL(n+1)]$, and is therefore not trivial.
My question is why the fact that $\mathcal{O}\_{\mathbb{P}^{n^2+2n}}(1)$
restricted to the affine open $\PGL(n+1)$ is not trivial, implies that
$\mathcal{O}\_{\mathbb{P}^n}(1) $ admits no
$\PGL(n+1)$-linearization?
[Indeed, $\mathcal{O}\_{\mathbb{P}^{n^2+2n}}(1)$ has order $n+1$ in
$V$ because $\Pic(\mathbb{P}^{n^2+2n}) \to
\Pic(\mathbb{P}^{n^2+2n} \backslash V(\det(a\_{ij}))=
\Pic(\PGL(n+1))$ induces an isomorphism
$\Pic(\mathbb{P}^{n^2+2n} \backslash V(\det(a\_{ij})) \cong
\mathbb{Z}/(\deg(\det(a\_{ij}))\mathbb{Z}$.]
To turn it another way round, why if $\mathcal{O}\_{\mathbb{P}^n}(1) $
would admit a $\PGL(n+1)$-linearization, then
the restriction of $\mathcal{O}\_{\mathbb{P}^{n^2+2n}}(1)$
to $\PGL(n+1)$ must be trivial? I conjecture that this argument can somehow reduced to an easy comparison of orders of group elements in groups $\Pic(X)$, $\Pic^G(X)$ but I do not see how it can be directly related.
Maybe it somehow helps to know that we have always a morphism of groups $ \Pic^G(X) \to \Pic(X)$ which is not neccessarily injective.
| https://mathoverflow.net/users/108274 | Example of a line bundle not admitting a $\operatorname{PGL}(n+1)$-linearization in Mumford's GIT | $PGL(n+1)$-linearization of $\mathcal O\_{\mathbb P^n}(1)$ could be used to produce isomorphism $\gamma : p\_2^\*(\mathcal O\_{\mathbb P^n}(1)) \to \sigma^\*(\mathcal O\_{\mathbb P^n}(1))$, hence, taking to attention isomorphism
$$ \sigma^\*(\mathcal{O}\_{\mathbb{P}^n}(1)) \cong
p\_1^\*(\mathcal{O}\_{\mathbb{P}^{n^2+2n}}(1)) \otimes
p\_2^\*(\mathcal{O}\_{\mathbb{P}^{n}}(1)) $$
which you have mentioned, we have equality of Picard classes $([\mathcal{O}\_{\mathbb{P}^{n^2+2n}}(1)],[\mathcal{O}\_{\mathbb{P}^{n}}(1)])=(0,[\mathcal{O}\_{\mathbb{P}^{n}}(1)])$ in $Pic(PGL(n+1))
\times Pic(\mathbb P^n) = Pic(PGL(n+1) \times \mathbb P^n)$, so, $[\mathcal{O}\_{\mathbb{P}^{n^2+2n}}(1)] = 0 $ in $Pic(PGL(n+1))$
How to construct such $\gamma$? $PGL(n+1)$-linearization of $\mathcal O\_{\mathbb P^n}(1)$ is a choice of morphism $\tilde{\sigma} : PGL(n+1) \times L \to L$ (where $L$ is a total space of the line bundle $\mathcal O\_{\mathbb P^n}(1)$) such that the diagram
$$\require{AMScd}
\begin{CD}
p\_2^\*(\mathcal O\_{\mathbb P^n}(1)) = PGL(n+1) \times L @>\tilde{\sigma}>> L\\
@V{}VV @V{}VV \\
PGL(n+1) \times \mathbb P^n @>{\sigma}>> \mathbb P^n
\end{CD}$$
commutes. Total space of the line bundle $\sigma^\*(\mathcal O\_{\mathbb P^n}(1))$ is by definition a fiber product $(PGL(n+1) \times \mathbb P^n) \times\_{\mathbb P^n} L$, hence, by universal property of fiber products the morphism $\gamma : p\_2^\*(\mathcal O\_{\mathbb P^n}(1)) \to \sigma^\*(\mathcal O\_{\mathbb P^n}(1))$ of $PGL(n+1) \times \mathbb P^n$-varieties exists. $\gamma$ is an isomorphism since for every $g \in PGL(n+1)$ induced morphism $\gamma\_g : O\_{\mathbb P^n}(1) \to g^\* O\_{\mathbb P^n}(1)$ is an isomorphism.
| 2 | https://mathoverflow.net/users/54337 | 434335 | 175,648 |
https://mathoverflow.net/questions/433421 | 4 | *Throughout I'm only interested in the **standard** semantics for second-order logic, and all structures/languages are relational for simplicity.*
If defined naively, second-order logic without equality is equivalent to second-order logic, since $x=y$ is equivalent to $\forall A(x\in A\leftrightarrow y\in A)$. However, there is a natural detrivialization of this. Say that a *quasi-isomorphism* between two structures $\mathfrak{A},\mathfrak{B}$ in the same language $\Sigma$ is a binary relation $I\subseteq\mathfrak{A}\times\mathfrak{B}$ such that:
* $dom(I)=\mathfrak{A}$ and $ran(I)=\mathfrak{B}$.
* $I$ preserves-and-reflects all basic relations: for each $n$-ary $R\in\Sigma$, each $a\_1,...,a\_n\in \mathfrak{A}$, and each $b\_1,...,b\_n\in\mathfrak{B}$ with $a\_iIb\_i$ for all $1\le i\le n$, we have $$(a\_1,...,a\_n)\in R^\mathfrak{A}\quad\iff\quad (b\_1,...,b\_n)\in R^\mathfrak{B}.$$
Write $\mathfrak{A}\cong^q\mathfrak{B}$ if there is a quasi-isomorphism between $\mathfrak{A}$ and $\mathfrak{B}$. Quasi-isomorphisms arise as a natural tool for showing non-expressibility in equality-free $\mathsf{FOL}$ or even $\mathcal{L}\_{\infty,\infty}$. This motivates the following "strongly-without-equality" fragment of second-order logic:
>
> Let $\mathsf{SOL}\_\*$ be the set of second-order formulas such that for every pair of structures $\mathfrak{A},\mathfrak{B}$, every quasi-isomorphism $I:\mathfrak{A}\cong^q\mathfrak{B}$, and every pair of appropriate-arity tuples of elements $a\_1,...,a\_n\in\mathfrak{A}$ and $b\_1,...,b\_n\in\mathfrak{B}$ with $a\_iIb\_i$ for all $1\le i\le n$, we have $$\mathfrak{A}\models\varphi(a\_1,...,a\_n)\quad\iff\quad\mathfrak{B}\models\varphi(b\_1,...,b\_n).$$
>
>
>
In some notes on abstract model theory I'm writing I plan to include a few challenge exercises about $\mathsf{SOL}\_\*$ *(e.g. "give a not-too-terrible syntax for it")*, but I have a strong suspicion that it's already present in the literature. So my question is: what's a good source on this logic?
| https://mathoverflow.net/users/8133 | Source on equality-free second-order logic (nontrivially construed) | It currently appears that this logic is not already treated in the literature.
*(I'm posting this answer to move this question off the unanswered queue, but if someone does find a source on it of course please add it!)*
| 0 | https://mathoverflow.net/users/8133 | 434340 | 175,649 |
https://mathoverflow.net/questions/434347 | 3 | Consider a second order gradient-like system with linear damping
$$\ddot{x}+\dot{x}+\nabla f(x)=0, \quad x(0)=x\_0,\quad\dot{x}(0)=0$$
Suppose $f\in C^2(\mathbb{R}^n)$ and $\inf\_{x\in\mathbb{R}^n}f(x)>-\infty$. The solution $x:[0,\infty)\rightarrow \mathbb{R}^n$ is bounded, i.e., $\lVert x(t)\rVert\leq c$ for all $t\in [0,\infty)$, where $c$ is a constant.
If we change the so called "gravity constant" (see [this paper](https://www.researchgate.net/publication/263878249_THE_HEAVY_BALL_WITH_FRICTION_METHOD_I_THE_CONTINUOUS_DYNAMICAL_SYSTEM_GLOBAL_EXPLORATION_OF_THE_LOCAL_MINIMA_OF_A_REAL-VALUED_FUNCTION_BY_ASYMPTOTIC_ANALYSIS_OF_A_DISSIPATIVE_DYNAMICAL_SYSTEM)) from $1$ to any positive constant $g$, i.e.,
$$\ddot{x}+\dot{x}+g\nabla f(x)=0, \quad x(0)=x\_0,\quad\dot{x}(0)=0$$
Is the solution to this new equation still bounded?
| https://mathoverflow.net/users/490600 | Does gravity constant affect boundedness of solution? | $\newcommand\la\lambda$No. E.g., if $n=1$, $x\_0\ne0$, and $f(u)=-u^2/2$ for all real $u$, then the solution
$$x(t)=\frac{x\_0}{2} \, \Big(\frac{e^{\la\_+ t}-e^{\la\_- t}}{\sqrt{4 g+1}}+e^{\la\_- t}+e^{\la\_+ t}\Big)$$
of the problem
$$\ddot{x}+\dot{x}+g\nabla f(x)=0, \quad x(0)=x\_0,\quad\dot{x}(0)=0 \tag{1}\label{1}$$
will be bounded in $t\ge0$ if $0<g\le2$ and unbounded in $t\ge0$ if $g>2$, where
$$\la\_\pm:=\frac{-1\pm\sqrt{4 g+1}}2.$$
---
**Addendum:** The OP later added the condition that $f$ be bounded below. This changes the problem dramatically.
First here, note that, if $x(t)$ is the solution to \eqref{1}, then for the "energy"
$$E(t):=\frac{|\dot x(t)|^2}2+gf(x(t))$$
and all real $t\ge0$ we have
$$E'(t)
=(\ddot x(t)+g\,\nabla f(x(t))\cdot\dot x(t)
=-\dot x(t)\cdot\dot x(t)
=-|\dot x(t)|^2\le0,$$
so that the "energy" is nonincreasing; of course, here $|\cdot|$ is the Euclidean norm and $\cdot$ is the dot product. It follows that for all real $t\ge0$
$$f(x(t))\le c:=E(0)/g<\infty.$$
If now $f$ is coercive (that is, $f(x)\to\infty$ as $|x|\to\infty$), then the set $\{y\in\mathbb R^n\colon f(y)\le c\}$ is bounded, and hence the solution $x(t)$ is bounded (in real $t\ge0$).
In a comment, the OP said "we can assume $f$ to be polynomial".
If $n=1$, then any bounded from below non-constant polynomial is coercive, and the case of a constant polynomial is quite easy. So, this solves the case $n=1$ (still with a polynomial $f$).
Even if $n>1$, it seems possible to show that "almost all" bounded from below non-constant polynomials are coercive, and for such polynomials, the solution $x(t)$ will be of course bounded.
| 4 | https://mathoverflow.net/users/36721 | 434351 | 175,653 |
https://mathoverflow.net/questions/434350 | 2 | Is there a simple proof that there is no Anosov flow on $S^2$? Where can I find it?
| https://mathoverflow.net/users/479780 | Anosov flow on the 2-sphere | The usual definition of Anosov flow requires three invariant sub-bundles, so I guess you are actually asking about the 3-sphere?
Plante and Thurston have proved in
*Plante, J. F.; Thurston, W. P.*, [**Anosov flows and the fundamental group**](https://doi.org/10.1016/0040-9383(72)90002-X), Topology 11, 147-150 (1972). [ZBL0246.58014](https://zbmath.org/?q=an:0246.58014).
that if a manifold admits a codimension 1 Anosov flow, then its fundamental group has exponential growth.
| 6 | https://mathoverflow.net/users/47274 | 434352 | 175,654 |
https://mathoverflow.net/questions/434343 | 6 | For any commutative [Frobenius algebra](https://en.wikipedia.org/wiki/Frobenius_algebra) $A$ there is an associated *window element* $\omega \in A$. If $\mu: A \otimes A \to A$ denotes the multiplication, $1 \in A$ the unit, $b: A \otimes A \to k$ the non-degenerate pairing, and $c: k \to A \otimes A$ the copairing, then the window element is given by
$$\omega = \mu \circ c(1)$$
The window element is important from a TQFT perspective.
Let's suppose we are working over an algebraically closed characteristic zero field.
If the window element is a unit, then the Frobenius algebra is semisimple. I am interested in non-semisimple examples. For example $A = k[x]/x^{n+1}$ is a commutative Frobenius algebra where the trace picks off the coefficient of $x^n$. In this case the window element is $\omega = (n+1)x^n$. This element is nilpotent, but it squares to zero.
I have tried a number of other examples, but in all cases I have tried the window element squares to zero. For example this will be the case if $A$ is a graded algebra which satisfies Poincare duality.
My question is whether this must always be the case? Are there commutative Frobenius algebras where the window element is not a unit, but also does not square to zero?
| https://mathoverflow.net/users/184 | Commutative Frobenius algebra with non-invertible window element, but not square zero | Assume $A$ is a connected (not necessarily commutative) non-semisimple Frobenius algebra that is finite dimensional over a field of characteristic 0 and given by quiver and relations. (for the commutative case all this reduced to be a local commutative Frobenius algebra that is not a field).
We should have $c(1)= \sum\_{i} y\_i \otimes x\_i$, where $x\_i, y\_i$ for $i=1,...,dim A$
are defined by the condition $b (x\_i \otimes y\_j )= \delta\_{i,j}$ (the existence of such $x\_i, y\_i$ is equivaleng to $A$ being a Frobenius algebra, see lemma 2.11 in the book of Lorenz on representation theory)
Now $\mu c(1)= \sum y\_i x\_i$ (see section 9.1.4 in the book by Lorenz) and
$b (c(1))= \sum b(y\_i \otimes x\_i)=dim A$.
This implies (see Proposition 1.10.18 in the book on representation theory by Zimmermann) that $\mu c(1) /dim A$ is in the socle and thus squares to zero.
Thus the statement is true in the case of commutative local Frobenius algebras when the algebra is not a field extension.
In the non-local case it should be wrong by taking for example $K \times K[x]/(x^2)$ as suggested in the comments by Will Sawin.
By the way for local Frobenius algebras this window element has a purely combinatorial interpretation up to a scalar, namely it is the unique longest (in terms of product of arrows) non-zero element in the quiver algebra.
| 2 | https://mathoverflow.net/users/61949 | 434356 | 175,656 |
https://mathoverflow.net/questions/434339 | 15 | The surreal numbers are sometimes called the "universally embedding" ordered field, in that every ordered field embeds into them. What "universally embedding" means seems to be [somewhat complicated](https://mathoverflow.net/questions/410437/what-does-it-mean-for-the-surreal-numbers-partizan-games-to-be-universally-embe), so I am curious if a simpler method of building them will do.
The idea is, suppose we take "all of" the ordered fields, with homomorphisms as embeddings between them. We can then build the direct limit, which "all of" the ordered fields embed into. Is the result equal to the surreal numbers?
I put "all of" in quotes because there are clearly a few quirks that are needed to avoid set-theoretic paradoxes. So for instance, instead of looking at "all of" the ordered fields, we can look at only those of cardinality less than, suppose, some strongly inaccessible cardinal $\kappa$. Then we can build the direct limit without any problem. Is the result an "initial subfield" of the surreal numbers with birthday up to that inaccessible cardinal. (Or do we get something strictly larger?)
I am sure there are other ways to deal with the set theory issues posed by this question, so I will kind of leave it open to whatever makes for the best answer (as is pretty common when asking questions about surreal numbers in general).
EDIT: a few clarifications
1. Clearly sometimes the choice of embedding from one field into another is non-unique. To phrase the question differently: suppose we have *any* arbitrary directed system in which the objects are all of the ordered fields. Is the direct limit of this directed system always isomorphic to the surreal numbers?
2. Whatever the direct limit is, every ordered field embeds into it. Also, every ordered field embeds into the surreal numbers. Does this tell us anything about the relationship between the two fields?
| https://mathoverflow.net/users/24611 | Can you build the surreal numbers as a simple direct limit of ordered fields? | Here is one way to get a positive answer to the title question.
**Theorem.** There is a definable class $\mathcal{F}$ of ordered fields, containing isomorphic copies of any given field, and a directed order $\unlhd$
on them, with a definable commutative system of embeddings between them $\pi\_{F,K}:F\to K$ for $F\unlhd K$, such that the direct limit of the system is the surreal field $\newcommand\No{\text{No}}\No$.
**Proof.**
The surreal field $\No$ itself is definable. Let $\mathcal{F}$ be the class of set-sized subfields $F\subseteq\No$. Define $F\unlhd K$ if and only if $F\subseteq K$, and let $\pi\_{F,K}:F\to K$ be the inclusion map. This is a definable, directed, commutative system of embeddings, and the direct limit is clearly the surreal field $\No$ itself, since every surreal number is an element of some set-sized subfield. Every ordered field is isomorphic to a subfield of $\No$ by the universality property, and so $\mathcal{F}$ contains copies of any given ordered field. $\Box$
I realize that this answer is not achieving the goal you had wanted, which was an alternative route to universality, since I am using universality to prove that the construction has the property of containing copies of all fields.
| 15 | https://mathoverflow.net/users/1946 | 434358 | 175,658 |
https://mathoverflow.net/questions/434360 | 3 | I am looking for non-trivial examples of flat $U(2)$ connections over the complement of a torus link $\mathcal{S}^3-L$ i.e.
$\mathcal{A}:\mathcal{S}^3-L \longrightarrow \mathfrak{U}(2)$ such that $F\_{\mathcal{A}} = 0$ and $Hol\_{\gamma}(\mathcal{A}) \neq 0$ with $\gamma$ a non-trivial element in $\pi\_1(\mathit{S}^3-L)$
Are there any known examples? Perhaps for $SU(2)$ it is known? I guess these connections would be classical solutions to a Euclidean $U(2)$ Chern-Simons theory on $\mathit{S}^3-L$ but when looking for Chern-Simons and knots all one usually finds is the quantum case...
Thus an explicit example of a classical solution with non-trivial holonomy is what I am looking for as well as the references where one may find how to construct explicit examples of this type.
Thanks
| https://mathoverflow.net/users/494010 | Explicit examples of Classical, Flat $U(2)$-connections on a torus link complement with non-trivial holonomy | For torus *knots*, all of the representations into $SU(2)$ were rather explicitly worked out by Eric Klassen (Representations of knot groups in $SU(2)$. Trans. Amer. Math. Soc. 326 (1991), no. 2, 795–828). The starting point is a rather simple presentation of the fundamental group of the complement of a torus knot; since torus links also have similar presentations it seems likely that the same technique would work for more than one component.
If you just want one example, consider $L$, a $(2,2n)$ torus link. Its double branched cover is a lens space $L(2n,1)$ with $\pi\_1 = Z/2n$. Hence there is a dihedral representation of the group $S^3 -L$, which you can think of as an SO(3) representation. I think this lifts to a representation of $\pi\_1(S^3 -L)$ to $SU(2)$ with image a binary dihedral group.
(I wasn't sure if you really meant multi-component links or if examples with knots would be satisfactory.)
| 4 | https://mathoverflow.net/users/3460 | 434362 | 175,660 |
https://mathoverflow.net/questions/434344 | 3 | Let $v\_1 =\lambda\_1 \zeta\_1$ and $v\_2 = \lambda\_2 \zeta\_2$ with $\zeta\_1 = \frac{4\pi i\omega}{3}$ and $\zeta\_2 = \frac{4\pi i\omega^2}{3}$ where $\omega = e^{2\pi i/3}$ is the third root of unity and $\lambda\_1,\lambda\_2$ some positive integers.
I would like to ask if there is an entire function $f$ such that
$$ f(z+ v\_1) = f(z) e^{2\pi i \lambda\_1 Cz/3 }$$
and
$$ f(z+ v\_2) = f(z) e^{2\pi i \lambda\_2 Cz/3 }$$
where $C \in \mathbb R$ is a constant such that $\frac{\lambda\_1 \lambda\_2 C (\zeta\_2 -\zeta\_1)}{3}=1$. This looks a bit like periodic boundary conditions, but since $z \in \mathbb C$ the modulus of these boundary conditions has of course a growing/decaying direction.
This looks pretty similar to something related to theta functions, but I don't quite get it together, as the underlying lattice looks rather different.
| https://mathoverflow.net/users/457901 | Entire function with almost periodic boundary condition? | The answer seems to be negative. Suppose that an entire function $f$
satisfies $f(z+v\_i)=e^{A\_iz}f(z)$, where $v\_1$ and $v\_2$ generate a lattice. Let $\Pi$ be the fundamental parallelogram of this lattice and integrate $f'/f$ over $\partial \Pi$. You obtain the ``Legendre's relation'':
$$v\_2A\_1-v\_1A\_2=2\pi in,$$
where $n$ is the number of zeros of $f$ in $\Pi$.
Substituting your values, we see that $n=1$.
Now $(f'/f)'$ is doubly periodic with respect to our lattice,
having a single double pole per parallelogram. So we may
assume (by shifting a pole to the origin) that
$(f'/f)'=\wp+c,$ and two integrations integrations give
$$f(z)=e^{P(z)}\sigma(z),$$
where $\sigma$ is the Weierstrass sigma function and $P$ is a polynomial of degree at most $2$. This is the general form of your $f$ (modulo a shift of the origin), if it exists.
Now let us try to find $P$. Sigma satisfies
$$\sigma(z+v\_j)=-e^{\eta\_j(z+v\_j)}\sigma(z),$$
where $\eta\_j=\zeta(\omega\_j)$, and $\zeta$ is the Weierstrass zeta function ($\zeta'=-\wp$),
which gives
$$P(z+v\_j)=P(z)-\eta\_j(z+v\_j)+\pi i, \quad j=1,2.$$
Trying to find such a polynomial with your data, we
just set $P(z)=az^2+bz+c$, and try to find $a,b,c$.
Equation for $a$ is satisfied in view of Legendre's relation, and we have
$$a=(A\_j-\eta\_j)/(2v\_j), \quad j=1,2,$$
but for $b$ we obtain
$$av\_j^2+bv\_j+\eta\_jv\_j=\pi i,\quad i=1,2.$$
These two equations with one variable must be consistent,
which is very unlikely, certainly not for all $\lambda\_1,\lambda\_2$. To check this one has to compute $\eta\_j$ for your lattices.
| 4 | https://mathoverflow.net/users/25510 | 434364 | 175,661 |
https://mathoverflow.net/questions/434320 | 4 | In the setting of symbolic dynamics over $\mathbb{Z}^d$, one can define for the $n$-th pattern complexity of a given a subshift $\Omega\subseteq \mathcal{A}^{\mathbb{Z}^d}$ as
$$ c\_n(\Omega):= \Big\vert \{P\in \mathcal{A}^{Q\_n}: P= \omega\vert\_{Q\_n} \; \text{for some} \; \omega\in \Omega \} \Big\vert $$
with $Q\_n:=\{0,...,n-1\}^d$. I know by Morse-Hedlund theorem that if $d=1$ and $\Omega$ contains an aperiodic configuration $\omega\in \mathcal{A}^{\mathbb{Z}}$, then $c\_n(\Omega)\geq n+1$ for all $n$. I found [this recent paper](https://arxiv.org/pdf/2105.00784.pdf), stating that if $d=2$ and $\Omega$ has an aperiodic configuration, then $c\_n(\Omega)\geq n^2+1$.
I was trying to find what is known in general dimensions $d\geq 3$? I am actually interested whether a weaker version of this sort of claim holds. i.e., if there exists a constant $C\_d>0$, such that $\Omega\subseteq \mathcal{A}^{\mathbb{Z}^d}$ is a subshift containing an aperiodic configuration implies that $c\_n(\Omega)\geq C\_d \cdot n^d$?
I am assuming this might be more commonly referred to in a terminology I am not aware of, but I assume that this should be known. Does anyone know of a work or survey in this direction for general $d$?
| https://mathoverflow.net/users/143153 | Lower bounds for pattern complexity of aperiodic subshifts | The answer is no in a very strong sense: there does not exist such $C\_d$ for $d \geq 3$ even for aperiodic minimal subshifts.
As far as finding lower bounds goes, complexities of subshifts containing aperiodic configurations are more or less the same as complexities of individual aperiodic configurations, namely a subshift containing aperiodic $x$ has at least the complexity of $x$, and conversely an aperiodic $x$ will have orbit closure with exactly the complexity of $x$.
In
<https://eventos.cmm.uchile.cl/sdynamics20208/wp-content/uploads/sites/111/2021/01/cassaigne.pdf>
Julien Cassaigne attributes to Lagarias and Peasants the conjecture that $\limsup\_n c\_n(x)/n^d > \infty$ for any $d$-dimensional aperiodic configuration $x$, i.e. that your $C\_d$ would exist in the lim sup sense, and to Lagarias, Peasants, Sander and Tijdeman the result that there exists an $x$ with $\liminf\_n c\_n(x)/n^d = 0$, i.e. your $C\_d$ does not exist in the lim inf sense. I did not follow up on the references.
He then proves the following theorem:
>
> Theorem. Let $f : \mathbb{N} \to \mathbb{N}$ tend to infinity. Then for any $d \geq 3$ there exists an aperiodic uniformly recurrent $d$-dimensional configuration $x \in A^{\mathbb{Z}^d}$ such that $c\_n(x) = O(n^2 f(n))$.
>
>
>
In particular the orbit closure is a minimal subshift where all configurations are aperiodic and the complexity is as above. He leaves open whether $O(n^2)$ complexity is possible in dimension $d \geq 3$.
| 4 | https://mathoverflow.net/users/123634 | 434371 | 175,664 |
https://mathoverflow.net/questions/434195 | 1 | I am concerned with unweighted directed graphs where each node contains exactly one edge pointing to another node, which could be itself. In other words, each row of the adjacency matrix contains one entry equal to 1, and the rest are 0. Im not sure if these graphs have a name, but they could be called a deterministic Markov chain.
I think the eigenvalues of the adjacency matrix will always come from this set: 1,0 and roots of unity. Is this true, and if so, what ensures this?
| https://mathoverflow.net/users/94774 | Eigenvalues of directed graph with one outward edge for each vertex | Here is an alternative (more combinatorial) proof to the one linked to in my comment.
Suppose that the digraph $D$ has a vertex of in-degree zero, which we may assume is vertex $1$. Then letting $\varphi(D)$ denote the characteristic polynomial of the adjacency matrix $A(D)$, we have
$$
\varphi(D) = |xI - A(D)| = \left|\begin{array}{cc}x&\*\\ 0&xI-A(D\backslash 1)\\\end{array} \right| = x \varphi(D\backslash 1).
$$
Now $A(D\backslash 1)$ is a matrix with exactly one 1 per row and hence we can repeat the process.
In other words, the eigenvalues of $D$ are the eigenvalues of $D \backslash 1$ plus an extra $0$.
If there are no vertices of in-degree 0, then the graph is a disjoint union of directed cycles, and hence the matrix is a permutation matrix, and since $A^k = I$ for some $k$, all of its eigenvalues are roots of unity.
| 1 | https://mathoverflow.net/users/1492 | 434373 | 175,665 |
https://mathoverflow.net/questions/434378 | 0 | You are given a series of vectors $u\_0,\ldots,u\_k$ of non-negative entries, each of dimension $n$, for which the sum of the entries of each vector is $1$ (i.e., for all $i\in \{0,\ldots,k\}$, it holds that $\sum\_j u\_{i,j}=1$).
Is it always true that either there exists a vector $v\in R^{n}$, such that $\sum\_j (u\_{0,j}-u\_{i,j})\cdot v\_j > 0$ for all $i=1,\ldots,k$, or there exist $\alpha\_1,\ldots,\alpha\_k\geq 0$, that sum to $1$ (i.e., $\sum\_i \alpha\_i = 1$), such that $u\_0 = \sum\_{i=1}^k \alpha\_i \cdot u\_i$.
| https://mathoverflow.net/users/480123 | Non-convex combination of vectors | Yes, either point $u\_0$ belongs to the convex hull $H$ of the points $u\_1,\dots,u\_k$ (in which case $\alpha\_i$ exist), or there exists a hyperplane separating $u\_0$ and $H$, when we can take $v$ as a normal vector of this hyperplane "pointing" towards $u\_0$.
| 3 | https://mathoverflow.net/users/7076 | 434381 | 175,668 |
https://mathoverflow.net/questions/433707 | 3 | Consider a binary relation $R$ over a finite set $X$ of size $n$. Assume $R$ is antisymmetric and connected but not necessarily transitive. In essence, we are modeling an "option x beats option y" relation, which is not necessarily transitive. It might be the result of a voting process for instance.
It is sometimes possible to find a function $g : X \rightarrow \mathbb{R}^d$ which assigns a $d$ dimensional real vector to each element of $X$, and a continuous function $f: \mathbb{R}^{d + d} \rightarrow \mathbb{R}$ monotonic in each of its coordinates, such that $xRy \Leftrightarrow f(g(x),g(y)) > 0$.
Intuitively, can we explain the relation in terms of the interplay of $d$ latent qualities for each element?
Taking $d = n$, it's easy to construct $f$, by having $g$ map the elements of $X$ to the canonical basis of $\mathbb{R}^d$ (a one-hot encoding).
This is not generally possible for all $d$. If $d = 1$, this is possible iff $R$ is also transitive.
Let's call the smallest $d$ for which $f$ and $g$ exist the dimension of the relation, and write it $D(R)$.
**Is $D$ bounded?**
My guess is *no*. If you consider sets of non-transitive die, it seems intuitive that you cannot, in general, sum up the relationship between them with $d$ much lower than their number of faces.
| https://mathoverflow.net/users/8737 | Representing a binary relation | Every binary relation $R$ has a representation with $d=2$. Enumerate $X=\{x\_i:i\in[n]\}$, and define $g\colon X\to\mathbb R^2$ by $g(x\_i)=(i,-i)$. Since $\{(i,-i,j,-j):i,j\in[n]\}$ is a set of pairwise incomparable elements of $\mathbb R^4$, the Proposition below implies that there exists a continuous increasing function $f\colon\mathbb R^4\to\mathbb R$ such that
$$f(i,-i,j,-j)=\begin{cases}\phantom-1&x\_i\mathrel Rx\_j,\\-1&\text{otherwise,}\end{cases}$$
whence
$$x\_i\mathrel Rx\_j\iff f(g(x\_i),g(x\_j))>0$$
as required.
>
> **Proposition.** Let $d\in\mathbb N$ and $X\subseteq\mathbb R^d$ be finite. Then any nondecreasing (w.r.t. the coordinatewise partial order on $\mathbb R^d$) function $f\colon X\to\mathbb R$ extends to a continuous nondecreasing function $\tilde f\colon\mathbb R^d\to\mathbb R$. If $f$ is strictly increasing, $\tilde f$ can be taken strictly increasing as well.
>
>
>
(More generally, the argument below can be adapted to $X\subseteq\mathbb R^d$ whose projection to each coordinate is a subset of $\mathbb R$ with no accummulation point.)
I will outline a proof in the strictly increasing case; the nondecreasing case is completely analogous.
First, we can extend $f$ to any countable set by iterating the following observation:
>
> **Lemma.** If $X\subseteq\mathbb R^d$ is finite and $y\in\mathbb R^d$, then any increasing $f\colon X\to\mathbb R$ extends to an increasing function $g\colon X\cup\{y\}\to\mathbb R$.
>
>
>
**Proof:** Assume $y\notin X$. Putting $r=\max\{f(x):x\in X,x<y\}$ and $s=\min\{f(x):x\in X,y<x\}$, it suffices to define $g(y)$ so that $r<g(y)<s$, which is possible as $r<s$. If $r$ or $s$ does not exist due to the relevant set being empty, we ignore the corresponding constraint. QED
>
> **Corollary.** If $X\subseteq Y\subseteq\mathbb R^d$, $X$ is finite, and $Y$ is countable, then any increasing $f\colon X\to\mathbb R$ extends to an increasing function $g\colon Y\to\mathbb R$.
>
>
>
**Proof:** Write $Y=\{y\_n:n\in\omega\}$. Using the Lemma, construct a sequence of $f=f\_0\subseteq f\_1\subseteq f\_2\subseteq\cdots$ of increasing functions $f\_n\colon X\cup\{y\_i:i<n\}\to\mathbb R$, and put $ g=\bigcup\_nf\_n$. QED
At this point, we could extend a given $f\colon X\to\mathbb R$ to an increasing $\bar f\colon Y\to\mathbb R$ for a countable dense set $Y$, and define $\tilde f\colon\mathbb R^d\to\mathbb R$ by $\tilde f(x)=\sup\{\bar f(y):y\le x,y\in Y\}$; this will be an increasing function, but may well be discontinuous. Thus, we need to work a bit harder.
Let $Z=\{z\_n:n\in\mathbb Z\}\subseteq\mathbb R$ be a set such that $n<m\implies z\_n<z\_m$, $\inf Z=-\infty$, $\sup Z=+\infty$, and $Z$ contains all coordinates of all elements of $X$. Using the Corollary, we can extend any increasing $f\colon X\to\mathbb R$ to an increasing $f\_0\colon Z^d\to\mathbb R$. By applying a suitable continuous increasing bijection $\mathbb R\to\mathbb R$, we may assume $Z=\mathbb Z$ without loss of generality.
We now extend the increasing function $f\_0\colon\mathbb Z^d\to\mathbb R$ to a function $f\_1\colon(\frac12\mathbb Z)^d\to\mathbb R$ as follows. Any $x\in(\frac12\mathbb Z)^d$ can be written uniquely as $x=(x\_0+x\_1)/2$, where $x\_0,x\_1\in\mathbb Z^d$, $x\_0\le x\_1$, and $\|x\_1-x\_0\|\_\infty\le1$; we define $f\_1(x)=\frac12(f\_0(x\_0)+f\_0(x\_1))$.
Let $x,y\in(\frac12\mathbb Z)^d$ be neighbouring lattice points, i.e., $x<y$ and $\|y-x\|\_1=1/2$. We have either $x\_0=y\_0$, $x\_1<y\_1$, and $\|y\_1-x\_1\|\_1=1$, or $x\_0<y\_0$, $\|y\_0-x\_0\|\_1=1$, and $x\_1=y\_1$. In the former case, $f\_1(y)-f\_1(x)=\frac12(f\_0(y\_1)-f\_0(x\_1))$, while in the latter case, $f\_1(y)-f\_1(x)=\frac12(f\_0(y\_0)-f\_0(x\_0))$. In particular, $f\_1(y)>f\_1(x)$. Since we can get from any $x$ to any $y>x$ by an increasing sequence of lattice neighbour hops, this implies that $f\_1$ is increasing.
But the argument gives more. Assume $n\in\mathbb N$ and $t>0$ are such that $f\_0$ is $t$-Lipschitz on $[-n,n]^d$ w.r.t. the $\|\cdot\|\_1$ norm:
$$x,y\in(\mathbb Z\cap[-n,n])^d\implies|f\_0(y)-f\_0(x)|\le t\|y-x\|\_1.$$
Then we obtain $|f\_1(y)-f\_1(x)|\le t/2$ when $y,x\in[-n,n]^d$ are neighbours in the $(\frac12\mathbb Z)^d$ lattice, which implies that $f\_1$ is $t$-Lipschitz on $[-n,n]^d$ w.r.t. $\|\cdot\|\_1$ as well.
Thus, if we iterate this process to construct a sequence $f\_0\subseteq f\_1\subseteq f\_2\subseteq\cdots$ of increasing functions $f\_i\colon(2^{-i}\mathbb Z)^d\to\mathbb R$, they will have the property that for any $n\in\mathbb N$ and $t>0$, if $f\_0$ is $t$-Lipschitz on $[-n,n]^d$, then all $f\_i$ are $t$-Lipschitz on $[-n,n]^d$.
Let $\bar f=\bigcup\_if\_i$. Then $\bar f$ is an increasing function $\mathbb D^d\to\mathbb R$, where $\mathbb D=\bigcup\_i2^{-i}\mathbb Z$ is the set of dyadic rationals, and for every $n\in\mathbb N$, there exists $t>0$ such that $\bar f$ is $t$-Lipschitz, hence uniformly continuous, on $[-n,n]^d$. It follows that $\bar f$ has a unique continuous extension $\tilde f\colon\mathbb R^d\to\mathbb R$, and this has to be increasing as well.
EDIT: A more explicit description of the extension of $f\_0\colon\mathbb Z^d\to\mathbb R$ to $\tilde f\colon\mathbb R^d\to\mathbb R$ above can be given as follows. We consider a triangulation of $\mathbb R^d$ whose faces are the simplices $\operatorname{Conv}\{x\_0,\dots,x\_k\}$ where $x\_0,\dots,x\_k\in\mathbb Z^d$ are such that $x\_0<x\_1<\dots<x\_k$ and $\|x\_k-x\_0\|\_\infty\le1$. (One needs to check that this is indeed a triangulation.) We then define $\tilde f$ such that its restriction to each simplex in the triangulation is the unique affine function that agrees with $f\_0$ at the vertices of the simplex. This is a continuous extension of $f\_0$, and one can check easily that it is increasing.
| 2 | https://mathoverflow.net/users/12705 | 434387 | 175,670 |
https://mathoverflow.net/questions/434369 | 1 | We know that the extension operator on paraboloids $\widehat{fd\sigma}(t,x)=\int\_\mathbb{R}^nf(\xi)e^{i(t|\xi|^2+x\cdot\xi)}d\xi$ is a solution to the homogeneous Schrodinger equation with initial data $f$; that on cones (change $|\xi|^2$ to $|\xi|$) a solution to the wave equation with the same initial data; and on (part of) a spheres (change $|\xi|^2$ to $\sqrt{1-|\xi|^2}$) a solution to the Hamiltonian equation.
Is every extension operator on some smooth hypersurface (say, quadratic surfaces like hyperbolic hyperboloids, hyperbolic paraboloids, or more general surfaces that are not graphs of any quadratic or even polynomial functions) a solution to some PDE?
| https://mathoverflow.net/users/494230 | Is the extension (dual restriction) operator on any smooth hypersurface a solution to some PDE? | Your question was basically answered by David Roberts in the comments, but let me write a few more words.
Given a **constant coefficient linear differential operator** of degree $N$
$$ L = \sum\_{|\alpha| \leq N} c\_\alpha \partial^\alpha $$
(here I use multi-index notation for $\alpha$), we can formally take the Fourier transform of the equation $L u = 0$ to get
$$ \widehat{(Lu)}(\xi) = \sum\_{|\alpha| \leq N} c\_\alpha i^{|\alpha|} \xi^\alpha \hat{u}(\xi) = P(\xi) \hat{u}(\xi) = 0 $$
Here $P(\xi)$ is a polynomial function. So within the space of tempered distributions, you have that solving $Lu = 0$ is the same as finding a distribution satisfying $P(\xi)\hat{u}(\xi) = 0$. This requires that the support of $\hat{u}$ be on the 0 set of $P$.
So this gives you the general correspondence between Fourier extensions on algebraic varieties and solutions to PDEs.
---
However, this correspondence is not one-to-one in general.
Take the simplest case of an ODE on $\mathbb{R}$.
Let $L\_1 = \frac{d}{dx}$ and $L\_2 = \frac{d^2}{dx^2}$. Their corresponding polynomials are $i \xi$ and $- \xi^2$. Both have the same zero set.
But in general the solutions to $L\_1 u = 0$ are just the constant distributions (whose Fourier transform are multiples of $\delta\_0$). But solutions to $L\_2 u = 0$ include all linear functions (Fourier transform is a linear combination of $\delta\_0$ and $\delta\_0'$).
This shows that not all constant coefficient linear PDEs have their solutions expressible, in Fourier space, as $\hat{f}$ times the surface measure of some surface.
---
Finally, your question also asked about smooth hypersurfaces which may not be algebraic varieties. Let $\phi$ be a defining function of your hypersurface, then every surface measure $f d\sigma$ has the property that $\phi(\xi) \cdot f d \sigma = 0$, and so is (by definition) the Fourier transform of a distributional solution of the *pseudo-differential* equation $\phi(D)u = 0$.
In general the pseudodifferential operator $\phi(D)$ may fail to be a local operator, and so the corresponding equation may fail to be a PDE.
| 6 | https://mathoverflow.net/users/3948 | 434399 | 175,673 |
https://mathoverflow.net/questions/434328 | 0 | Let $X$ be a metric space, $\mu$ a $\sigma$-finite non-negative Borel measure on $X$, and $(E, |\cdot|)$ a Banach space. Let $\mathcal L\_p := \mathcal L\_p (X, \mu, E)$ and $\|\cdot\|\_{\mathcal L\_p}$ be its semi-norm. Here we use [Bochner integral](https://math.stackexchange.com/questions/4298588/dominated-convergence-theorem-for-banach-space). Let $\mathcal C\_c :=\mathcal C\_c(X, E)$ be the space of all $E$-valued continuous functions on $X$ with compact supports. It is well-known that
>
> [Theorem](https://math.stackexchange.com/questions/4573158/if-x-is-locally-compact-separable-then-mathcal-c-cx-is-dense-in-big) If $X$ is locally compact separable, then $\mathcal C\_c$ is dense in $\big (\mathcal L\_p, \|\cdot\|\_{\mathcal L\_p} \big)$ for all $p \in [1, \infty)$.
>
>
>
I would like to ask if above result can be further strengthened, i.e.,
>
> Let $X$ be locally compact separable. For each $f \in \mathcal L\_p$ with $p \in [1, \infty)$ and $\varepsilon>0$, there is $g \in \mathcal C\_c$ such that
> $$
> \color{red}{|g| \le |f|} \quad \mu\text{-a.e.}
> \quad \quad \text{and} \quad \quad
> \| f-g \|\_{\mathcal L\_p} < \varepsilon.
> $$
>
>
>
This result, if true, generalizes [this lemma](https://math.stackexchange.com/a/4573202/1019043) which itself generalizes [another lemma](https://math.stackexchange.com/q/3967941/1019043).
Thank you so much for your elaboration!
| https://mathoverflow.net/users/477203 | A generalization about the density of $\mathcal C_c(X, E)$ in $\mathcal L_p (X, \mu, E)$ when $E$ is a Banach space | Below is a counter-example taken from [this thread](https://math.stackexchange.com/q/4574262/1019043). It works even when $\mathcal C\_c$ is replaced by $\mathcal C$, the space of all continuous functions from $X$ from $E$.
---
Let $X:=[0, 1]$, $E:=\mathbb R$, and $\mu$ the Lebesgue measure on $[0, 1]$. Let $C$ be the [fat Cantor set](https://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set). Then $C$ is closed and $\mu(C) =\frac{1}{2}$. Let $f :=1\_C$. Then $\| f \|\_{\mathcal L\_1} = \frac{1}{2}$.
Let $g \in \mathcal C$ such that $|g| \le |f|$ $\mu$-a.e. This implies there is a $\mu$-null subset $N$ of $X$ such that $|g| \le |f|$ on $X \setminus N$. Assume $g (a) \neq 0$ for some $a \in X$. By continuity of $g$, there is an open interval $I$ of $X$ such that $g\neq0$ on $I$. Then $f>0$ on $I \setminus N$. So $(I \setminus N) \subset C$ and thus $I \subset (C \cup N)$.
Notice that $I \setminus C$ is non-empty and open, so there is a non-empty open interval $J$ such that $J \subset I$ and $J \cap C = \emptyset$. As such, $J \subset N$ and thus $\lambda (N)>0$. This is a contradiction. As such, $g =0$ everywhere. Hence any $g \in \mathcal C$ can not approximate $f$ from below.
| 0 | https://mathoverflow.net/users/477203 | 434400 | 175,674 |
https://mathoverflow.net/questions/434368 | 9 | Let $n > 13$ be a positive integer. Is there any $n\times n$ circulant $(-1,1)$-matrix $A$ satisfying the following property:
$$AA^T=(n-1)I+J$$
where $I$ is the $n\times n$ identity matrix and $J$ is the $n\times n$ matrix of ones.
I conjecture that the answer is no. But I can't prove it.
| https://mathoverflow.net/users/369335 | One question on circulant $(-1,1)$-matrices | This is a question about a sequence $a(t)\in \{\pm 1\}$ of period $n$ with 2 level periodic autocorrelations, with the nontrivial autocorrelations identically equal to 1. All these problems have a design theoretic aspect as well. For the relationship to the $\{0,1\}$ alphabet see the question [here](https://mathoverflow.net/q/410006) and its answer.
Here we define the periodic autocorrelation as
$$
C\_a(\tau)=\sum\_{t=0}^{n-1} a(t) a(t+\tau)
$$
where the shift by $\tau$ is modulo $n.$
Jungnickel and Pott have a paper on perfect and almost perfect autocorrelation sequences where related questions are discussed
[here](https://doi.org/10.1016/S0166-218X(99)00085-2).
**Edit:** As @MaxAlekseyev points out, Corollary 2.5 in the Jungnickel and Pott paper actually rules out the existence of a circulant matrix as desired by the OP for lengths $13<n\le 20201.$
Maximal length sequences obtained from finite fields give rise to circulant matrices which satisfy
$$
A A^T = (n+1)I-J
$$
and they exist for $n=2^m-1,$ for all $m\geq 1.$ Thus they have
$$
C\_a(\tau)=-1,\quad \forall \tau \neq 0 \pmod n
$$
Legendre sequences (terminology used in coding and cryptography regarding sequences derived from quadratic characters, see comments to this answer) obtained from multiplicative characters exist for odd prime lengths with the same property.
Another conjecture of similar form, which also (curiously) is open for $n>13$ is the existence of a Barker sequence $a(t) \in \{\pm 1\}$ (or Barker code according to engineers) of length $n$ whose *aperiodic* autocorrelation
$$
C\_a(\tau)=\sum\_{t=0}^{n-1-\tau} a(t)a(t+\tau)
$$
satisfies
$$
|C\_a(\tau)| \leq 1, \forall \tau \neq 0.
$$
This has been checked up to $n$ in the thousands. A good general reference for these problems is the chapter by Helleseth and Kumar in the Handbook of Coding Theory, Vol. 2.
| 5 | https://mathoverflow.net/users/17773 | 434406 | 175,676 |
https://mathoverflow.net/questions/424965 | 5 | *(For simplicity, all languages are relational.)*
In analogy with *first-order* languages, say that a **second-order language** is a set of relation symbols of two kinds: *first-order* relation symbols and *second-order* relation symbols. Both types of symbols have a notion of arity; the arity of a first-order relation symbol is a natural number, and the arity of a second-order relation symbol is a finite string of natural numbers.
Given a second-order language $\Sigma=\langle\Sigma\_1,\Sigma\_2\rangle$ (with $\Sigma\_i$ denoting the set of $i$th-order symbols of $\Sigma$), a **$\Sigma$-structure** is defined in the obvious way: as a *triple* $\mathcal{A}=(A; \mathfrak{I}\_1,\mathfrak{I}\_2)$ such that
* $A$ is a (nonempty) set,
* $\mathfrak{I}\_1$ is a map sending each $R\in\Sigma\_1$ with arity $n$ to a subset of $A^n$, and
* $\mathfrak{I}\_2$ is a map sending each $S\in\Sigma\_2$ with arity $\alpha$ to a subset of $\mathcal{P}(A^{\alpha(1)})\times...\times\mathcal{P}(A^{\alpha(\vert\alpha\vert)})$.
There is an obvious semantics for second-order logic over a structure in a second-order language (note that this extends the "full"/"standard" semantics, rather than the Henkin semantics, for second-order logic in the context of first-order languages). I'm curious whether Craig interpolation holds in this broader framework:
>
> Suppose $\varphi,\theta$ are $\mathsf{SOL}$-sentences in second-order languages $\Sigma,\Pi$ respectively such that $\varphi\models\theta$. Must there be a $\mathsf{SOL}[\Sigma\cap\Pi]$-sentence $\delta$ such that $\varphi\models\delta$ and $\delta\models\theta$?
>
>
>
Craig interpolation for $\mathsf{SOL}$ is trivial in the context of *first-order* languages, since we can appropriately quantify out the non-shared symbols. But that doesn't work here. In fact, I suspect that the answer is **negative** via an easy construction, but I don't immediately see it.
*(Incidentally, it's possible that my restriction to relational languages is less benign for these purposes than it appears. I'm happy to generalize in the obvious way to allow function symbols!)*
| https://mathoverflow.net/users/8133 | Does second-order logic satisfy Craig interpolation for second-order languages? | *This is just an expansion of Emil Jerabek's comments above; I've made it CW to avoid reputation gain, and will delete this if he posts an answer of his own.*
Craig interpolation can be rephrased as a "syntactic separation" property: the statement $$\varphi\models\psi$$ can be rephrased as $$\emptyset\models\exists\mathfrak{R}(\varphi[\mathfrak{R}])\rightarrow\exists\mathfrak{S}(\varphi[\mathfrak{S}])$$ for some appropriate objects $\mathfrak{R},\mathfrak{S}$ corresponding to the *un*common parts of the languages of $\varphi,\psi$ respectively. Now if our "base logic" is first-order then this is broadly speaking where the story ends, but using second-order logic as our "base" gives us a lot more power.
Specifically, since we can pin down $\mathcal{N}=(\mathbb{N};+,\times)$ up to isomorphism, we can "localize" the syntactic separation result above into a separation result for sets of natural numbers: if "fully-second-order" logic had the Craig interpolation property then any two disjoint $\Sigma^2\_1$ subsets of $\mathcal{N}$ could be separated by a second-order-definable set of naturals, but this is impossible (by Tarski) since the usual-second-order theory of $\mathcal{N}$ is $\Delta^2\_1$.
---
This raises a natural follow-up question:
* Is there a "reasonably natural" and "not too strong" extension of second-order logic which has the interpolation property for second-order languages?
The point of the "not too strong" requirement is that there is an obvious extension of $\mathsf{SOL}$ which has the interpolation property for second-order languages - **third**-order logic (which then loses interpolation for third-order languages, and so forth)!
| 2 | https://mathoverflow.net/users/8133 | 434414 | 175,678 |
Subsets and Splits