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https://mathoverflow.net/questions/401522
|
2
|
I asked this question on Math.StackExchange some time ago and got no responses.
Let $G=(V,E)$ be a finite graph with adjacency matrix $A$. Let us consider the associated subshift of finite type
$$
\Sigma\_A=\{(v\_i)\_{i\in\mathbb{Z}} : A\_{v\_iv\_{i+1}}>0, i\in\mathbb{Z}\}\subset V^{\mathbb{Z}}.
$$
For a stochastic matrix $P$ agreed with $A$ (that is $p\_{ij}>0$ iff $a\_{ij}>0$), there is a shift-invariant probability measure $\mu$ on $\Sigma\_A$. Is is well-known that the dynamical system $(\Sigma\_A,\sigma,\mu)$ is strongly mixing, i.e., for all measurable $B,C\subset\Sigma\_A$,
$$
\mu(\sigma^nB\cap C)\rightarrow\mu(B)\mu(C), \ n\rightarrow\infty,
$$
if and only if the graph $G$ is strongly connected and aperiodic.
I am wondering if it is possible to estimate the speed of convergence. The standard proof given in textbooks uses iterations of $P$ to prove mixing for cylindrical sets. Hence one can estimate the speed of convergence for cylindrical sets in terms of the eigenvalues of $P$. Does it hold for any measurable sets? Is it always exponential, where the exponent comes from the eigenvalues of $P$?
|
https://mathoverflow.net/users/143604
|
Exponential mixing for subshifts
|
I don't think that it's ever possible to mandate any rate of mixing if you are looking at all measurable sets. Here's a silly example using the full $2$-shift $\{0,1\}^{\mathbb{Z}}$ and the i.i.d. measure $\mu$ with $\mu([0]) = \mu([1]) = 1/2$ (in some sense the "most mixing" SFT and measure).
Define $B$ to be $[0]$, the set of sequences with $x(1) = 0$, i.e. with $0$ in the first coordinate. Clearly $\mu(B) = 0.5$.
Define $C$ to be the set of sequences $x$ satisfying all of the following conditions: $x(1)$ is not $0$, $x(2)$ and $x(3)$ are not both $0$, $x(4), x(5), x(6)$ are not all $0$, $x(7), x(8), x(9), x(10)$ are not all $0$, etc. Then $\mu(C) = \prod\_{n=1}^{\infty} (1 - 2^{-n}) > 0$.
But for any $k$, $\mu(\sigma^k B \ | \ C)$ is slightly less than $\mu(B) = 0.5$, in fact it's $0.5 - \frac{1}{2^{n+1} - 2}$, where $n$ is the length of the "condition" that $x(k)$ is part of. (For instance, when $k = 8$, $n = 4$, since $x(8)$ is part of the condition "$x(7), x(8), x(9), x(10)$ are not all $0$."
This already has subexponential mixing rate, but you can make the rate as slow as desired by just spacing your conditions out. For instance, you could wait until $x(n!!)$ to start the $n$th condition, and then $\mu(\sigma^{n!!} B \ | \ C)$ is still roughly $2^{-n}$ away from $\mu(B) = 0.5$.
|
3
|
https://mathoverflow.net/users/116357
|
401532
| 164,817 |
https://mathoverflow.net/questions/401339
|
7
|
Suppose that $\ell\_\phi$ is a reflexive Orlicz sequence space such that its dual space $\ell\_\phi^\*$ is isomorphic to $\ell\_\phi$.
Is $\ell\_\phi$ isomorphic to $\ell\_2$?
|
https://mathoverflow.net/users/39421
|
Self-dual Orlicz sequence spaces
|
For a given $1<p$ and $\frac{1}{p}+\frac{1}{q}=1$ you can construct a *universal* Orlicz sequence space $\ell\_M$ so that every Orlicz function $N$ *in between* $p$ and $q$ is equivalent to a function in $E\_M$ which corresponds to the Orlicz subspaces spanned by constant block bases of $\ell\_M$ thus complemented. Such a space is unique up to isomorphism (depending only to $p$), and $E\_{M^\*}$ has the same property so $\ell\_{M^\*}$ is isomorphic to $\ell\_{M}$ by the uniqueness. This construction is given in [LT, Theorem 4.b.12]. See also the remark after.
This answer was first given by Bill Johnson in comments, I just added the reference.
*Lindenstrauss, Joram; Tzafriri, Lior*, Classical Banach spaces. 1: Sequence spaces. 2. Function spaces., Classics in Mathematics. Berlin: Springer-Verlag. xx, 432 p. (1996). [ZBL0852.46015](https://zbmath.org/?q=an:0852.46015).
|
6
|
https://mathoverflow.net/users/3675
|
401534
| 164,818 |
https://mathoverflow.net/questions/401513
|
0
|
Let $L>0$ and $\Omega \subset \mathbb{R}^n$ a bounded Lipschitz domain. Define
$$
B\_{\frac12,L}:=\{f \in L^2((0,1) \times \Omega) : \|f(t,\cdot)-f(s,\cdot)\|\_{L^2(\Omega)} \leq L|t-s|^{\frac12},~ \forall s,t \in [0,1]\}.
$$
I would like to show that, for every fixed $s,t \in [0,1]$, the functional $f \mapsto \int\_\Omega |f(t,x)-f(s,x)|$ is continuous with respect to $L^2((0,1) \times \Omega)$-norm on the set $B\_{\frac12,L}$, . It seems to me that this amounts to the question of continuity of point evaluation on a set of Hölder continuous functions with respect to $L^2$-norm, but I was not able to show it.
Does anyone have any direction: reference, counterexample, proof (hopefully)?
|
https://mathoverflow.net/users/121671
|
Continuity of point evaluation on space of Hölder functions with $L^p$ norm
|
I believe that I found a way to solve the problem. Let us show that the point evaluation $f \mapsto f\_t$ is a bounded operator from $L^2((0,1) \times \Omega)$ to $L^2(\Omega)$ and the rest follows analogously. Namely, from the given condition, for every $s \in [0,1]$ we have $\|f\_t\|\_{L^2(\Omega)}^2 \leq 2(L^2|t-s| + \|f\_s\|\_{L^2(\Omega)}^2) $ and thus
$
\sup\_{\|f\|\_{L^2((0,1)\times \Omega)}=1} \|f\_t\|\_{L^2(\Omega)} \leq \sqrt{2L^2+2}
$, for every $t \in [0,1]$, from where the boundedness follows.
|
0
|
https://mathoverflow.net/users/121671
|
401536
| 164,819 |
https://mathoverflow.net/questions/401500
|
15
|
[This book](https://hott.github.io/book/nightly/hott-online-1287-g1ac9408.pdf) has a section with proofs of the fact $\pi\_1(S^1)=\mathbb Z$ using the univalence axiom. They are a bit too technical for me at the moment to read, but I want to understand the following (vague but conceptual) question:
*What is the role of the univalence axiom in these proofs?*
Usually, univalence is motivated with a slogan like "isomorphic structures are equal". However, it seems the application of univalence in the proofs of $\pi\_1(S^1)=\mathbb Z$ must be of a completely different kind than "treating isomorphic algebraic structures as the same". So I really would like to get a hint at how univalence can be relevant to such concrete questions, as a motivation for reading the proofs in more detail.
|
https://mathoverflow.net/users/336697
|
Role of univalence in homotopy group calculations
|
Let's start with an easier question: How do we know that loop is not equal to refl in $\pi\_1(S^1)$?
By the universal property of $S^1$ this is exactly saying that there exists some type $T$ and some $t:T$ and some loop $p:t=t$ which is not trivial. This means I want some type where it's easy to write down paths, but where it's also easy to tell if paths aren't equal to each other. Univalence gives a type that's great for this, namely the universe! The paths in the universe are just functions and you can show two functions are not equal by just showing they're not equal on some input.
So just look $\mathbb{Z}: U$ and $\mathrm{ua}(\mathrm{succ}): \mathbb{Z} = \mathbb{Z}$. This is clearly not refl because it sends $0$ to $1$ which is not $0$. Now we're done!
I think of checking $\pi\_1(X) = G$ as having four parts, constructing generating loops, constructing relations, checking the generators generate, and checking that there's no additional relations. The first two of these parts can be done without univalence (and they're much nicer to do in Globular than in HoTT implementations). What I explained above is why univalence is important in checking that there's no additional relations. What I find a lot more subtle is how univalence also allows you to check that the generators generate. Why the generators generate is the real magic of the encode/decode method, and I'm not sure I can explain it the way I want to in just a MO answer.
|
15
|
https://mathoverflow.net/users/22
|
401538
| 164,820 |
https://mathoverflow.net/questions/401545
|
4
|
Let $(X,\mathcal F,μ,T)$ be a dynamical system (i.e. μ is a probability measure and Τ is μ-preserving) and $\mathcal S\subset\mathcal F$ be a family of sets such that for any $A \in \mathcal F$ and $ε>0$ there exists a $B \in \mathcal S$ with $μ(A\triangle B)\lt ε$.
Assume that $\lim\_{N\to \infty}\frac{1}{N}\sum\_{n=0}^N μ(Α\cap T^{-n}B)>0$ , for all $A,B\in S$, with positive measure. Can we conclude that the system is Ergodic?
The motivation behind this question is to understand the concept of ergodicity better, by finding "minimal" conditions for it to occur.
I know, for example, that if the above limit is not only positive but equal to $μ(Α)μ(Β)$ then the system is ergodic and vice versa!
So I am thinking that it is natural to either construct a counterexample (which I tried to do, but failed) or to prove that if the limit is positive it necessarily equals $μ(Α)μ(Β)$.
Another fact; Since Von Neumann's mean ergodic theorem implies that
$$\lim\_{N\to \infty}\frac{1}{N}\sum\_{n=0}^N μ(Α\cap T^{-n}B)=\int\_A\mathbf{E}[1\_B|\mathcal{I}]dμ$$
where $\mathbf{E}[1\_B|\mathcal{I}]$ is the conditional expectation of $1\_B$ with respect to the σ-algebra $\mathcal{I}=\{D\in \mathcal F: T^{-1}D=D\}$, then for the implication to be true it would suffice to show that if for all $A,B\in \mathcal S$ with positive measure we have that $\int\_A\mathbf{E}[1\_B|\mathcal{I}]dμ>0$, then we specifically have the equality $\int\_A\mathbf{E}[1\_B|\mathcal{I}]dμ=μ(Α)μ(Β)$.
|
https://mathoverflow.net/users/336624
|
Does the following condition imply ergodicity?
|
The answer is no. Take $X$ to be a disjoint union $X\_1\sqcup X\_2$ of invariant subsystems of positive measure with $T\restriction\_{X\_1}$ and $T\restriction\_{X\_2}$ ergodic. Define $\mathcal{S}$ as
$\{B\in\mathcal{F}\mid \mu(B\cap X\_1),\mu(B\cap X\_2)>0\}$.
Moreover, assume that both $X\_1$ and $X\_2$ have subsets of arbitrarily small positive measures. Then $\mathcal{S}$ clearly satisfies the condition outlined in the question. Now suppose $A,B\in\mathcal{S}$. They can be partitioned as $A\_1\sqcup A\_2$ and $B\_1\sqcup B\_2$ where $A\_i:=A\cap X\_i$ and $B\_i:=B\cap X\_i$ are with positive measure. Now we have
$$
\mu(A\cap T^{-n}B)=\mu\left((A\_1\cap T^{-n}B\_1)\sqcup (A\_2\cap T^{-n}B\_2)\right)
=\mu(A\_1\cap T^{-n}B\_1)+\mu(A\_2\cap T^{-n}B\_2).
$$
The ergodic average then splits into the sum of two ergodic averages for $T\restriction\_{X\_1}$ and $T\restriction\_{X\_2}$. These systems (equipped with rescaled probability measures $\frac{1}{\mu(X\_1)}.\mu\restriction\_{X\_1}$ and $\frac{1}{\mu(X\_2)}.\mu\restriction\_{X\_2}$) are ergodic. Therefore
$$
\lim\_{N\to\infty}\frac{1}{N}\sum\_{n=0}^{N-1}\mu(A\cap T^{-n}B)=
\frac{\mu(A\_1)\mu(B\_1)}{\mu(X\_1)}+\frac{\mu(A\_2)\mu(B\_2)}{\mu(X\_2)}>0.
$$
|
4
|
https://mathoverflow.net/users/128556
|
401549
| 164,827 |
https://mathoverflow.net/questions/401525
|
0
|
The quotient space $SU(k)/SO(k)$ is also a homogeneous space constructed out of the Lie groups (special unitary $SU(k)$ and special orthogonal $SO(k)$).
Because the $SO(k)$ may not be a normal subgroup of $SU(k)$, so $SU(k)/SO(k)$ may not be a quotient group, or may not be any Lie group. However $SU(k)/SO(k)$ may be a manifold?
>
> Question:
>
>
> 1. If $SU(k)/SO(k)$ is a manifold for every $k$, how does this manifold behave?
> 2. Are the different ways to specify the quotient so we may obtain different results? (see $k=2$ below)
>
>
>
Note that
$k=1$, $SU(1)/SO(1)= $ a point.
$k=2$, $SU(2)/SO(2)=SU(2)/U(1)= S^3/S^1= S^2$. However, there are different ways to have $S^1$ fibered over $S^2$, to get $S^3/\mathbf{Z}\_k$. So I am interested in knowing $S^3/S^1$ can be something else other than $S^2$?
$k=3$, $SU(3)/SO(3)$ = Wu manifold as a 5 real dimensional manifold. But how exactly this is a manifold? How is this $SU(3)/SO(3)$ related to a Dold manifold as a $\mathbf{CP}^2$ fibered over $U(1)$? (correct me if I said the fibration the other way around.)
$k=4,\dots$, do we have a general understanding for this manifold?
|
https://mathoverflow.net/users/336737
|
$SU(k)/SO(k)$ as a manifold, for each positive integer $k$
|
$SU(k)/SO(k)$ is the set of real structures on $\mathbb C^k$, i.e. $k$-dimensional real subspaces of $\mathbb C^k$ which generate it as a complex vector space, satisfying two conditions:
(1) The Hermitian form $((z\_1,\dots, z\_n) , (w\_1,\dots, w\_n)) \mapsto z\_1\overline{w}\_1+ z\_2 \overline{w}\_2 + \dots + z\_n \overline{w}\_n$ takes real values on each pair of vectors in the subspace.
(2) Given a basis for the $k$-dimensional real subspace, the determinant of the corresponding $k \times k$ complex matrix is real.
To prove this is $SU(k)$, it suffices to show that $SU(k)$ acts transitively on it, with stabilizer $SO(k)$. The action of $SU(k)$ is by its action on $\mathbb C^k$.
Restricting the Hermitian form to any such real subspace gives a symmetric form. We can choose a basis which is orthonormal respect to this form, and we can further choose the basis to have positive determinant (by negating the last vector if necessary). That will give an orthonormal basis for $\mathbb C$. So the determinant of the corresponding matrix is a complex number of unit norm, and, because it is a positive real, must be $1$. So there is a matrix in $SU(k)$ transferring the standard basis to this basis, thus transferring the standard real subspace to this real subspace.
Finally, for the standard real subspace, an element of the stabilizer is a $k\times k$ real matrix that also lies in $SU(k)$, i.e. an element of $SO(k)$.
[Wikipedia claims](https://en.wikipedia.org/wiki/Symmetric_space#Classification_of_Riemannian_symmetric_spaces) that we should look at real structures only satisfying condition (2), but this seems to me to give the wrong dimension $2k^2 - k^2 -1 = k^2-1$. Possibly (1) is suppsoed to be assumed in the definition of a real structure.
|
3
|
https://mathoverflow.net/users/18060
|
401553
| 164,829 |
https://mathoverflow.net/questions/401544
|
5
|
Is it possible to classify Hopf algebras $H$, over a field $k$, which admit a unique (up to isomorphism) irreducible comodule, namely the trivial $1$-dim comodule
$$
k \to k \otimes H, ~~ k \mapsto k \otimes 1\_H.
$$
|
https://mathoverflow.net/users/326091
|
Classifying Hopf algebras that admit a single irreducible comodule
|
The HAs you are describing are again the connected (=irreducible) ones. I am using the terminology here as in my answer to your previous question: [Name for a Hopf algebra whose only grouplike element is the identity?](https://mathoverflow.net/q/400726/85967).
Their classification, is in general an open project:
* In the cocommutative case, over a field of $chark=0$, they are all universal enveloping algebras of lie algebras. If $k=\mathbb{C}$, this is by an old result of Milnor and Moore. You can find the statement for any field of char zero at Montgomery's book.
* In the finite dimensional case they appear only over fields of positive characteristic. I think this is an old result of Masuoka (but i do not have the exact reference right now). See [arXiv:1309.0286 [math.RA]](https://arxiv.org/abs/1309.0286), [arXiv:1310.7073 [math.RA]](https://arxiv.org/abs/1310.7073) (and the references therein) for recent results on their classification.
|
7
|
https://mathoverflow.net/users/85967
|
401557
| 164,830 |
https://mathoverflow.net/questions/401470
|
8
|
Let $F$ be a smooth rank one foliation on a manifold $M$. Suppose that all leaves of $F$ are compact (that is, circles). Then its leaf space (edit: when additional assumptions are taken) is an orbifold. When the foliation is (transversally) Riemannian, this is proven in the book "Riemannian Foliations" by P. Molino. I suppose that this result is classical without the Riemannian assumption, but I could not find its proof in the literature. Any help with references is highly appreciated.
|
https://mathoverflow.net/users/3377
|
Smooth rank one foliations with closed leaves
|
The result is not true without additional assumptions. See [A counterexample to the periodic orbit conjecture](http://www.numdam.org/item/PMIHES_1976__46__5_0/) by Sullivan.
---
**Added later.** The paper by Sullivan linked above exhibits a foliation by circles on a compact manifold of dimension $5$ with non-Hausdorff leaf-space. Later, Epstein and Vogt constructed a foliation by circles on a compact manifold of dimension $4$ with the same property, see [A Counterexample to the Periodic Orbit Conjecture in Codimension 3](https://doi.org/10.2307/1971187).
The problem makes sense for foliations of arbitrary dimensions. No need to be restricted to foliation by curves. There are also examples of holomorphic foliations having all its leaves compact but with non-Hausdorff leaf-space, on **non-compact** complex manifolds, see [On the stability of holomorphic foliations](https://link.springer.com/chapter/10.1007%2FBFb0097265) by Holmann.
As far as I know, there are no examples in the literature of holomorphic foliations with all its leaves compact and with non-Hausdorff leaf-space on a compact complex manifold.
|
5
|
https://mathoverflow.net/users/605
|
401562
| 164,833 |
https://mathoverflow.net/questions/401512
|
0
|
Given a field $k$ with characteristic $p$, let $G$ be a transitive permutation group on $4p$ points. Let $P$ be a Sylow $p$-subgroup of $G$ and $Q\leq P$ is a $p$-subgroup of $P$ of index $p$. Now denote $H:=N\_G(Q)/Q$. Could anyone provide me with an counterexample suth that the dimension of the projective cover $P\_k$ of the trivial $kH$-module $k$ does not divide $p(p-1)$?
|
https://mathoverflow.net/users/134942
|
Dimension of projective cover of trivial $kG$-module
|
Here is a Magma calculation that shows that the group ${\rm PSL}(2,11)$ is a counterexample to your question
```
> G := PSL(2,11);
> I := AbsolutelyIrreducibleModules(G,GF(3));
> I;
[
GModule of dimension 1 over GF(3),
GModule of dimension 5 over GF(3),
GModule of dimension 5 over GF(3),
GModule of dimension 10 over GF(3),
GModule of dimension 12 over GF(3^2),
GModule of dimension 12 over GF(3^2)
]
//So GF(9) is a splitting field
> P := ProjectiveCover(TrivialModule(G,GF(9)));
> Dimension(P);
12
> CompositionFactors(P);
[
GModule of dimension 1 over GF(3^2),
GModule of dimension 10 over GF(3^2),
GModule of dimension 1 over GF(3^2)
]
```
|
4
|
https://mathoverflow.net/users/35840
|
401572
| 164,835 |
https://mathoverflow.net/questions/401555
|
5
|
Semirings, also called rigs, are rings without negatives: their underlying additive monoids are not groups (in other words, while rings are monoids in $(\mathsf{Ab},\otimes\_{\mathbb{Z}},\mathbb{Z})$, semirings are monoids in $(\mathsf{CMon},\otimes\_{\mathbb{N}},\mathbb{N})$). Examples include all rings, but also objects like
1. The **natural numbers semiring** $(\mathbb{N},+,\cdot,0,1)$, the monoidal unit of $(\mathsf{Semirings},\otimes\_{\mathbb{N}},\mathbb{N})$;
2. The **Boolean semiring** $\mathbb{B}=\{0,1\}$, with semiring structure characterised by $1+1=1$;
3. The **tropical semiring** $\mathbb{T}=(\mathbb{R}\cup\{\infty\},\min,+)$, and its sibling $\mathbb{A}=(\mathbb{R}\cup\{-\infty\},\max,+)$, sometimes called the **Arctic semiring**;
4. The semiring $(\mathrm{Idl}(R),+,\cdot,(0),R)$ of ideals of a ring $R$.
---
When we pass to the $\infty$-world, we replace the symmetric monoidal category $(\mathsf{Sets},\times,\mathrm{pt})$ of sets and maps between them by the symmetric monoidal $\infty$-category $(\mathcal{S},\times,\mathrm{pt})$ of spaces, and also commutative monoids and groups by $\mathbb{E}\_{\infty}$-monoids and $\mathbb{E}\_\infty$-groups.
As a result, the immediate analogues of $\mathsf{CMon}$ and $\mathsf{Ab}$ become the $\infty$-categories $\mathsf{Mon}\_{\mathbb{E}\_{\infty}}(\mathcal{S})$ and $\mathsf{Grp}\_{\mathbb{E}\_{\infty}}(\mathcal{S})$ of
$\mathbb{E}\_\infty$-spaces and grouplike $\mathbb{E}\_{\infty}$-spaces. The latter of these is equivalent to the $\infty$-category of connective spectra $\mathsf{Sp}\_{\geq0}$, which embeds into the $\infty$-category of all spectra $\mathsf{Sp}$, the “true” analogue of $\mathsf{Ab}$ in homotopy theory.
The $\infty$-categories $\mathcal{S}\_\*$, $\mathsf{Mon}\_{\mathbb{E}\_{\infty}}(\mathcal{S})$, $\mathsf{Grp}\_{\mathbb{E}\_{\infty}}(\mathcal{S})$, and $\mathsf{Sp}$ all admit symmetric monoidal structures, determined uniquely by the requirement that the free functors from $\mathcal{S}$ to them can be equipped with a symmetric monoidal structure (see Theorem 5.1 of Gepner–Groth–Nikolaus's *Universality of multiplicative infinite loop space machines*, [arXiv:1305.4550](https://arxiv.org/abs/1305.4550)).
We can thus consider $\mathbb{E}\_{k}$-monoids in these categories:
* For $(\mathsf{Sp},\otimes\_{\mathbb{S}},\mathbb{S})$, we get **$\mathbb{E}\_{k}$-ring spectra**;
* For $(\mathsf{Grp}\_{\mathbb{E}\_{\infty}}(\mathcal{S}),\otimes\_{QS^0},QS^0)$, we get **$\mathbb{E}\_{k}$-ring spaces**, which are equivalent to connective $\mathbb{E}\_{k}$-ring spectra;
* Finally, for $(\mathsf{Mon}\_{\mathbb{E}\_{\infty}}(\mathcal{S}),\otimes\_{\mathbb{F}},\mathbb{F})$, we obtain **$\mathbb{E}\_{k}$-semiring spaces**.
The last of these provides a partial (i.e. connective) analogue of semirings in homotopy theory.
---
However, examples of $\mathbb{E}\_{k}$-semiring spaces (that are not $\mathbb{E}\_{k}$-ring spectra) are harder to come by.
A motivating example of them is given by the $\mathbb{E}\_{\infty}$-space $\mathbb{F}\overset{\mathrm{def}}{=}\coprod\_{n=0}^{\infty}\mathbf{B}\Sigma\_{n}$―the classifying space of the groupoid of finite sets and permutations―which can be given the structure of an $\mathbb{E}\_{\infty}$-semiring space, becoming the monoidal unit of the symmetric monoidal $\infty$-category $\mathsf{Semirings}\_{\mathbb{E}\_{\infty}}(\mathcal{S})$ of $\mathbb{E}\_{\infty}$-semiring spaces. Additionally, $\mathbb{F}$ is the spectral analogue of the semiring $\mathbb{N}$ of natural numbers, and, by the multiplicative Barratt–Priddy–Quillen–Segal theorem, its $\mathbb{E}\_{\infty}$-ring space completion (i.e. $QS^0\otimes\_{\mathbb{F}}\mathbb{F}$) is $QS^0$, corresponding to the sphere spectrum $\mathbb{S}$ under the equivalence $\mathsf{Ring}\_{\mathbb{E}\_{\infty}}(\mathcal{S})\cong\mathsf{RingSp}\_{\geq0}$.
---
**Questions:**
1. What are some other examples of $\mathbb{A}\_{k}$-semiring or $\mathbb{E}\_{k}$-semiring spaces?
2. What are some examples of homotopy associative/commutative semiring spaces, i.e. monoids or commutative monoids in $(\mathsf{Ho}(\mathcal{S}),\otimes\_{\mathbb{F}},\mathbb{F})$?
3. Finally, do we have semiring space analogues of $\mathbb{B}$, $\mathbb{T}$, $\mathbb{A}$, and $\mathrm{Idl}(R)$?
|
https://mathoverflow.net/users/130058
|
Examples of $\mathbb{E}_{k}$-semiring spaces
|
(as a sidenote on terminology, as Jonathan points out in the comments, "spectrum" is not really a good name for your objects precisely because you expect them not to be spectra)
If you look at theorem 8.8 in the GGN paper you cited, you'll see that categories with a nicely behaved tensor product yield examples by taking their groupoid core.
Here are some explicit classical examples:
* For a commutative ring spectrum, $\mathrm{Proj}\_R^\simeq$ is a commutative semiring space, whose ring completion is $K(R)$ when $R$ is connective. More generally, if $R$ is $E\_k$, this semiring space "is" $E\_{k-1}$.
* There is an equivariant version of $\mathbb F$, namely $\mathrm{Fin}\_G^\simeq$ for a (pro)finite group $G$; as well as an equivariant version of the previous example, where for a commutative ring $G$-spectrum $R$, you call "projective" any $R$-module which is a summand of a direct sum of $R\otimes G/H\_+$'s (no shifts allowed)
Here's an example which doesn't directly fit into the context of [Thm 8.8, GGN] (although it can be made to, via e.g. the condensed/pyknotic approach):
* $\mathrm{Vect}\_\mathbb R^\simeq$, $\mathrm{Vect}\_\mathbb C^\simeq$, the groupoids of finite dimensional vector spaces over $\mathbb{R,C}$ respectively, but where you take into account the topology on the hom-sets. Note that this is not literally $(\mathrm{Vect}\_\mathbb K)^\simeq$ for some $\infty$-category of vector spaces and all linear maps obtained by applying the nerve to the topological category - indeed, if you do that then all mapping spaces are contractible and so you get trivial categories; so you have to restrict to isomorphisms *before* passing to $\infty$-categories. Their ring completions are $\mathrm{ko, ku}$ respectively.
* Again, the above example has an equivariant version.
(the equivariant versions can be made into semiring $G$-spaces, but let me not get into that here)
There are probably tons of other examples.
|
3
|
https://mathoverflow.net/users/102343
|
401588
| 164,840 |
https://mathoverflow.net/questions/401552
|
0
|
Let $N$ be the number of degree $d$ monomials in $n$ variables. We can then view each non-zero point in $\mathbb{A}^N\_k$ as a degree $d$ homogeneous form, $k$ an algebraically closed field. Let $X$ be the union of $\mathbf{0}$ and the set of points where the corresponding form is not smooth. I have heard that smoothness is a generic condition. This implies that $X$ is a Zariski closed set, I have not been able to find a proof of this fact. Any reference or explanation how to prove this along with what $\dim X$ is would be highly appreciated.
|
https://mathoverflow.net/users/84272
|
Dimension of the set of singular hypersurfaces
|
Given a smooth projective variety $X\subset\mathbb{P}^{N-1}$, let $I(X)\subset X\times(\mathbb{P}^{N-1})^\*$ denote the locus of pairs $(x,H)$ where $x\in H$ and $T\_x(X)\subset T\_x(H)$; here we are thinking of $(\mathbb{P}^{N-1})^{\*}$ as the collection of hyperplanes in $\mathbb{P}^{N-1}$.
One can show (using the Euler sequence which is the "global" version of the Jacobian criterion mentioned by Mohan) that $I(X)\to X$ is the projective bundle $\mathbb{P}\_X(K\_{X/\mathbb{P}^{N-1}})$, where $K\_{X/\mathbb{P}^{N-1}}$ is the co-normal bundle of $X$ in $\mathbb{P}^{N-1}$. This can be used to show that $I(X)$ is smooth of dimension $N-2$ in $(\mathbb{P}^{n})^{\*}$.
Since $X\times(\mathbb{P}^{N-1})^{\*}\to(\mathbb{P}^{N-1})^{\*}$ is proper, the image $D(X)$ of $I(X)$ is a closed subvariety of dimension $\leq (N-2)$.
This is the closed subvariety the question is looking for when $X\subset\mathbb{P}^{N-1}$ is the embedding of $X=\mathbb{P}^{n-1}$ given by the monomials mentioned in the question. (In other words, $X$ is the Veronese embedding of $\mathbb{P}^{n-1}$ in $\mathbb{P}^{N-1}$.)
There *are* examples where $D(X)$ is not a of dimension $N-2$. However, "in general" one finds that $D(X)$ is a hypersurface called the dual hypersurface of $X$. It appears to be rather difficult to write the equation of $D(X)$ in general.
(*Ref*: ["The topology of complex projective varieties after S. Lefschetz" by Klaus Lamotke](https://www.sciencedirect.com/science/article/pii/0040938381900136).)
|
2
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https://mathoverflow.net/users/124862
|
401589
| 164,841 |
https://mathoverflow.net/questions/395780
|
5
|
I have some problem calculating the value of some specific (but quite common) induced maps I stumbled on while reading some papers on group (co)homology and I would like to know if there are general tricks or routines to avoid calculations over the complexes or, if not, which calculations are customary. But let's first fix the notation.
>
> Let $H$ and $H'$ be groups, let $M$ be an $H$-module and $M'$ be an $H'$-module. If $\phi:H\rightarrow H'$ is a homomorphism and $\psi:M\rightarrow M'$ is an $H$-module homomorphism with $M'$ regarded as an $H$-module via $\phi$, then it induces a map $(\phi,\psi)\_\*:H\_i(H,M)\rightarrow H\_i(H',M')$ and this is done taking two projective resolutions $\textbf{P}$ and $\textbf{P'}$ of $\mathbb{Z}$ over $\mathbb{Z}H$ and $\mathbb{Z}H'$, respectively, making the latter a resolution over $\mathbb{Z}H'$, noticing that it is still acyclic and using this and the projectivity of $\textbf{P}$ to build the map we are looking for. If $M=M'$ and $\psi=Id\_{M'}$, it is customary to write $(\phi,\psi)\_\*$ as $\phi\_\*$.
>
>
>
Now for the specific cases.
1. Let $G=H$, $G'=H\times H$, $M=K=M'$ for a field $K$ on which $G$ acts trivially. Calculate $\phi\_\*$ where $\phi$ is either $\alpha:g\rightarrow (g,1)$ or $\beta:g\rightarrow (1,g)$
or $\gamma:g\rightarrow (g,g)$. In this case a Künneth formula states that
$$H\_n(G',M)=\bigoplus\_{p=0}^n H\_p(G,M)\otimes\_M H\_{n-p}(G,M).$$ How can one get that the $(H\_0(G,M)\otimes\_M H\_n(G,M)\oplus H\_n(G,M)\otimes\_M H\_0(G,M))$-components of $\alpha\_\*(h)$, $\beta\_\*(h)$ and $\gamma\_\*(h)$ for an $h$ in $H\_n(G,M)$ are, resp., $h\otimes 1$, $1\otimes h$ and $h\otimes 1+1\otimes h$?
2. More generally, let $G$, $G'$, $\alpha$, $\beta$ and $\gamma$ be the same as in point 1., $M$ and $N$ be $G$-modules, $M'=M\otimes\_\mathbb{Z} N$ be a $G'$-module by the action $(g,h)\cdot(m\otimes\_\mathbb{Z} n)=gm\otimes\_\mathbb{Z} hn$ and let $\psi:M\rightarrow M'$ be a suitable $G$-module homomorphism (Doubt: For $\alpha$, $\psi$ could just send $m\in M$ to $m\otimes\_\mathbb{Z}\overline{n}$ for a fixed $\overline{n}\in N$ and it seems the same holds for $\gamma$ if $G$ acts trivially on $N$; but for $\beta$?). What is the best way to find the $(H\_0(G,M)\otimes\_M H\_n(G,M)\oplus H\_n(G,M)\otimes\_M H\_0(G,M))$-components of $(\alpha,\psi)\_\*(h)$, $(\beta,\psi)\_\*(h)$ and $(\gamma,\psi)\_\*(h)$ for an $h$ in $H\_n(G,M)$, provided that a formula like the following $$H\_n(G',M\otimes\_\mathbb{Z} N)=\bigoplus\_{p=0}^n H\_p(G,M)\otimes\_\mathbb{Z} H\_{n-p}(G,M')$$ still holds? (This should be possible by choosing $M$ and $H\_i(G,M')$ to be $\mathbb{Z}$-free)
|
https://mathoverflow.net/users/127914
|
On induced maps in group homology and Künneth formula
|
1. The Kunneth formula is natural for products of homomorphisms. Let $\iota:\{1\}\to H$ be the inclusion of the identity element, and consider $\mathrm{Id}\times \iota:H\times\{1\}\to H\times H$ which can be identified with your map $\alpha$. Applying the Kunneth formula, we see that an element $h\in H\_n(H\times\{1\};M)\cong H\_n(H;M)\otimes H\_0(\{1\};M)$ maps to $h\times 1$ under $\alpha\_\*$. Similarly for $\beta\_\*$.
For the diagonal homomorphism $\gamma:H\to H\times H$, we can use functoriality of Kunneth for the projections $p\_1:H\times H\to H\times\{1\}$ and $p\_2:H\times H\to \{1\}\times H$, and observe that $(p\_1)\_\*$ maps an element in $H\_n(H\times H;M)$ to its $H\_n(H;M)\otimes H\_0(H;M)$ component, and $(p\_2)\_\*$ maps an element in $H\_n(H\times H;M)$ to its $H\_0(H;M)\otimes H\_n(H;M)$ component. Then $p\_1\circ \gamma$ and $p\_2\circ \gamma$ are both idetified with $\mathrm{Id}\_H$.
2. You might be able to apply a similar reasoning with twisted coefficients. A reference for the Kunneth formula in this case is
*Greenblatt, Robert*, [**Homology with local coefficients and characteristic classes**](http://dx.doi.org/10.4310/HHA.2006.v8.n2.a5), Homology Homotopy Appl. 8, No. 2, 91-103 (2006). [ZBL1107.55003](https://zbmath.org/?q=an:1107.55003).
|
2
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https://mathoverflow.net/users/8103
|
401600
| 164,847 |
https://mathoverflow.net/questions/401580
|
4
|
**This is a cross-post from [math.stackexchange](https://math.stackexchange.com/questions/4220444/proof-of-derived-tensor-hom-adjunction), since I didn't get any answers there.**
As far as I know, for $R,S,V,W$ rings and $M$ an $(R,W)$-bimodule, $N$ an $(R,S)$-bimodule and $L$ an $(S,V)$-bimodule, we have an isomorphism in $D(V$-lmod$)$
$$
\text{RHom}\_R(N \otimes\_S^L L, M) \cong \text{RHom}\_S(L,\text{RHom}\_R(N,M)).
$$
The only proof I can imagine would take flat / injective resolutions $P^\bullet$ and $Q^\bullet$ of $N$ and $M$ to yield
$$
\text{Hom}^\bullet\_R(P^\bullet \otimes\_S L, Q^\bullet) \cong \text{Hom}^\bullet\_S(L,\text{Hom}^\bullet\_R(P^\bullet,Q^\bullet)).
$$
However, it is not clear to me why there exists a resolution of $N$ consisting of $(R,S)$-bimodules that are flat right-$S$-modules.
For instance let $S=\mathbb Z$, $R = \mathbb Z/p$, in which case all flat $S$-modules are torsion free, so none of them can have an $R$-module structure.
Does the 'derived adjunction' still hold? If yes, how do you prove it? I would prefer a constructive proof which allows me to understand the map.
|
https://mathoverflow.net/users/81957
|
Proof of derived tensor-hom adjunction
|
>
> Does the 'derived adjunction' still hold? If yes, how do you prove it? I would prefer a constructive proof which allows me to understand the map.
>
>
>
Yes. The easiest way to get the derived adjunction is to specialize the ordinary adjunction to appropriate resolutions of the given bimodules.
For example, using projective resolutions, we can
* projectively resolve $N$ as an $R$-$S$-bimodule, i.e., as a left $R⊗S^{\rm op}$-module;
* projectively resolve $L$ as a left $S$-module.
This follows immediately from the fact
that the underlying tensor-hom adjunction in two variables
is a Quillen adjunction in two variables,
using the indicated projective model structures.
Alternatively, using injective resolutions, we can
* injectively resolve $M$ as a left $R$-module;
* projectively resolve $N$ as a right $S$-module or $L$ as a left $S$-module.
|
4
|
https://mathoverflow.net/users/402
|
401605
| 164,850 |
https://mathoverflow.net/questions/401357
|
2
|
Suppose $R$ is a Noetherian ring and $I$ a nontrivial ideal of $R$, and $A\_0\to B\_0$ a finite faithfully flat lci map of smooth $R\_0 := R/I$-algebras.
We fix a smooth $R$-algebra $A$ lifting $A\_0$ and assume $A$ and $A\_0$ integral.
Assume $R$ is $I$-adically complete.
>
> Does there exist a smooth $R$-algebra $B$ with a **finite** $R$-map $A\to B$ lifting $A\_0\to B\_0$?
>
>
>
If a finite $R$-map $A\to B$ exists, then it is finite lci and faithfully flat, because it is an integral extension and so it is a finite surjective lci on spectra.
I'm trying to use [this](https://stacks.math.columbia.edu/tag/07M8) and variations on the theme. I've been able to show that there is a quasi-finite syntomic map $A\to B$ with $B$ not necessarily smooth.
|
https://mathoverflow.net/users/nan
|
Lifting of flat lci maps
|
No, I don't think so.
Lemma: Given $R$, $I$, $R\_0 = R/I$, $A$, $A\_0 = A/I$. Assume $2$ is invertible in $R\_0$. If the answer to the question is "yes" then the image of the map $Pic(A) \to Pic(A\_0)$ contains all $2$-torsion of $Pic(A\_0)$.
Proof: Namely, suppose $L\_0$ is an invertible $A\_0$-module of order $2$ in the Picard group of $A\_0$. Choosing an isomorphism $\psi : L\_0 \otimes\_{A\_0} L\_0 \to A\_0$ we can set $B\_0 = A\_0 \oplus L\_0$ with multiplication defined by $\psi$. Then $A\_0 \to B\_0$ is finite etale because $2$ is invertible in $A\_0$ by the branched covering trick, whence $B\_0$ is smooth. If we can find $B$ as desired, then we see that $\det\_A(B) \in Pic(A)$ lifts $L\_0$. QED
So now we just have to find $R$ and $A$ where the $2$-torsion in $Pic(A\_0)$ does not lift. Read on if you want to see what I came up with. To understand it, you might have to recall some facts about the deformation to the normal cone. See example at the end for an explicit case.
Say $R$ is a complete discrete valuation ring with uniformizer $t$ and residue field $k$. Let $n \geq 2$ and let $C \subset \mathbf{A}^n\_k$ be an irreducible closed smooth subscheme such that $Pic(C)$ has lots of $2$-torsion. Denote
$$
b : T \longrightarrow \mathbf{A}^n\_R
$$
the blowing up of $C$ viewed as a closed subscheme by identifying $\mathbf{A}^n\_k$ with a closed subscheme of $\mathbf{A}^n\_R$. Then we have
$$
V(t) = E + X
$$
where $E$ is the exceptional fibre of $b$ and $X$ is the strict transform of $\mathbf{A}^n\_k$, i.e., $X$ is the blowing up of $C$ in $\mathbf{A}^n\_k$. Since $E$ is an anti-ample divisor, we see that $X$ is an ample divisor on $T$. Hence we see that
$$
U = T \setminus X = \text{Spec}(A)
$$
is an affine scheme (complement of an ample divisor on a scheme projective over an affine scheme is affine). It is smooth over $R$ as the non-smooth locus of $b$ is $E \cap X$. The special fibre $U\_0 = \text{Spec}(A\_0)$ is an $\mathbf{A}^c$-bundle over $C$ where $c$ is the codimension of $C$ in $\mathbf{A}^n\_k$. Thus we see that $Pic(A\_0) = Pic(U\_0) = Pic(C)$ has nonzero $2$-torsion elements.
Finally, the generic fibre of $\text{Spec}(A)$ is $\mathbf{A}^n\_K$ where $K$ is the fraction field of $R$. Since $t$ is a prime element of $A$ this implies that $\text{Pic}(A)$ is trivial.
Example: If $C$ is a hypersurface $C = V(\overline{f})$, then I think you just get $A = R[x\_1, \ldots, x\_n, y]/(ty - f)$ where $f$ is a lift of $\overline{f}$. You can verify all the properties of this directly, I guess.
|
1
|
https://mathoverflow.net/users/152991
|
401618
| 164,854 |
https://mathoverflow.net/questions/401563
|
4
|
In [page 197, Equivalents of the Riemann Hypothesis Vol 1](https://doi.org/10.1017/9781108178228), the following statement caught my eye
>
> There is an editorial comment in [102] that includes an observation by
> the GCHQ Problem Solving Group. They contest that one could replace the
> number 2 in inequality (7.94) by any constant $K > 1$ at the cost of having to
> check more cases by hand when $K$ is close to one. An example is cited: when
> $K = 1.2$ one needs to verify inequality (7.94) for $1 \leq n \leq 10^6$.
>
>
>
referring the inequality 7.94
>
> $\sigma(n) \leq H\_n +2\exp(H\_n) \log(H\_n)$, $n \geq 1$,
>
>
>
and the reference [102](https://www.jstor.org/stable/4145148) being
>
> [102] J. C. Lagarias and W. Janous, A generous bound for divisor sums: problem 10949, Amer. Math. Monthly 111 (2004), 264–265.
>
>
>
According to [An Elementary Problem Equivalent to the Riemann Hypothesis](https://arxiv.org/abs/math/0008177) by J. C. Lagarias, $K = 1$ is equivalent to RH. I understand the claim is not that RH can be proved by checking more cases by hand, but I want to understand what are the methods by which $K$ can be improved to any $1 + \epsilon$ by checking more cases? What are the barriers that prevent it from proving that $K \leq 1$ by checking for more cases?
Edit: Edited my question with more details after looking at the answer by @Charles
|
https://mathoverflow.net/users/502093
|
Question on coefficient of $\exp(H_n).\log(H_n)$ in Lagarias equivalence of RH
|
**1.** By an inequality due to Robin (see (2.2) in Lagarias's paper),
$$\sigma(n)\leq e^\gamma n\log\log n+\frac{n}{\log\log n},\qquad n\geq 3.$$
By Lemma 3.1 in Lagarias's paper, we also know that
$$\exp(H\_n)\log(H\_n)\geq e^\gamma n\log\log n,\qquad n\geq 3.$$
Combining these two estimates, we infer that
$$\sigma(n)\leq\left(1+\frac{1}{(\log\log n)^2}\right)\exp(H\_n)\log(H\_n),\qquad n\geq 3.$$
The fraction on the right hand side tends to zero (effectively), confirming the observation by the GCHQ Problem Solving Group. Explicitly, given any $\varepsilon>0$, we have
$$\sigma(n)\leq(1+\varepsilon)\exp(H\_n)\log(H\_n),\qquad n\geq\exp\exp(\varepsilon^{-1/2}).$$
**2.** The barrier to prove Lagarias's inequality is the same as the barrier to prove the Riemann Hypothesis, since these two statements are equivalent. It is a difficult problem, and perhaps it is even undecidable from the ZFC axioms (of course most of us believe it is decidable).
|
7
|
https://mathoverflow.net/users/11919
|
401620
| 164,855 |
https://mathoverflow.net/questions/401575
|
2
|
Let $A$ and $B$ be $C^{\ast}-$ algebras and $A \otimes B$ denotes minimal(spatial) tensor product.
>
> Is there any classification of primitive ideals of $A \otimes B$? (I'm mainly interested in the case when either $A$ or $B$ is commutative.)
>
>
>
|
https://mathoverflow.net/users/129638
|
Primitive ideals of minimal tensor product
|
I expect that this answer is satisfactory, although it isn't a complete answer. This is really the best result one can hope for.
For a $C^\ast$-algebra $A$ let $Prime(A)$ be the prime ideal space (defined exactly as the primitive ideal space, but with prime (two-sided closed) ideals instead). It is well-known that $Prime(A) = Prim(A)$ when $A$ is separable or when $A$ is Type I.
The following result is due to Proposition 2.16 and 2.17 in [Blanchard, Etienne; Kirchberg, Eberhard Non-simple purely infinite C∗-algebras: the Hausdorff case. J. Funct. Anal. 207 (2004), no. 2, 461–513.] and builds on Kirchberg's deep work on exact $C^\ast$-algebras:
**Theorem:** If $A$ and $B$ are $C^\ast$-algebra and at least one of them is exact, then $Prime(A\otimes\_{\textrm{min}} B)$ is homeomorphic to $Prime(A) \times Prime(B)$.
There is a canonical map $Prime(A) \times Prime(B) \to Prime(A\otimes\_{\mathrm{min}} B)$ given by $(I,J) \mapsto I\otimes\_{\mathrm{min}} B + A \otimes\_{\mathrm{min}} J$, and this is the homeomorphism in the theorem above.
If $I$ and $J$ are primitive ideals in $A$ and $B$ respectively, then $I\otimes\_{\mathrm{min}} B + A\otimes\_{\mathrm{min}} J$ is also primitive. Hence we get the following corollary for primitive ideal spaces instead of prime ideal spaces.
**Corollary:** Let $A$ and $B$ be $C^\ast$-algebras for which all prime ideals are primitive (e.g. separable/Type I/simple), and such that at least one of $A$ and $B$ is exact. Then $Prim(A\otimes\_{\textrm{min}} B)$ is homeomorphic to $Prim(A) \times Prim(B)$.
The results fail in general (even when $A$ is simple, non-exact, and $B=\mathcal B(\ell^2(\mathbb N))$), so you can't really hope for anything better.
As commutative $C^\ast$-algebras are exact, the above theorem is applicable if one of the $C^\ast$-algebras is commutative.
Note that one gets $Prim(A\otimes\_{\mathrm{min}} B) \cong Prim(A) \times Prim(B)$ (with primitive ideal spaces), if, for instance, $A$ is commutative (hence Type I) and $B$ is separable.
|
5
|
https://mathoverflow.net/users/126109
|
401649
| 164,863 |
https://mathoverflow.net/questions/401652
|
1
|
Consider an $h \times w$ binary matrix (a matrix with all entries $a\_{ij}$, $1 \le i \le h$, $1 \le j \le w$, equal to $0$ or $1$) with $w$ even and $h \ge 3$. We know that each row has $\frac{w+2}{2}$ entries equal to $1$ and $\frac{w-2}{2}$ entries equal to $0$. All rows are different. For each $j$, $1 \le j \le w$ there exists at least one $k$, $1 \le k \le h$ such that $a\_{kj} = 0$. One matrix is equivalent to another when equal up to a permutation of columns and/or rows.
I would like to find a function $c(h,w)$ to count how many possible equivalence classes of matrices are there, if possible for generic values of $h$ and $w$, or at least for some values of $h$. If an exact count is not possible, an upper bound can do.
For example, I can get $c(3,6)=1$ and any matrix satisfying the requirements is equivalent to:
$$
\begin{pmatrix}
1 & 1 & 1 & 1 & 0 & 0 \\
0 & 0 & 1 & 1 & 1 & 1 \\
1 & 1 & 0 & 0 & 1 & 1 \\
\end{pmatrix}
$$
However, I have difficulties in going further.
Any hint?
Thank you.
|
https://mathoverflow.net/users/136218
|
Counting the equivalence classes of some binary matrices
|
Equivalently, you are counting unlabelled bicolored graphs that have $h$ red vertices ("rows") and $w$ blue vertices ("columns"); such that red vertices have degree exactly $w/2+1$, and blue vertices have degree at most $h-1$ (each column has at least one zero). And finally, each red vertex has different neighborhood (all rows are different). Your matrices are the biadjacency matrices of such graphs. "Unlabelled" says you can freely permute rows and columns.
A computational solution for generating (and then counting) such bicolored graphs is **genbg** from [Nauty](https://pallini.di.uniroma1.it/). The degree conditions are easy to specify. We have to make the red vertices to be the *second* color class, then "genbg -z" enforces them to have different neighborhoods. (It does not have a corresponding option for the first color class.) The conditions listed above lead to the following shell script:
```
#/bin/bash
h=$1
w=$2
let bluemaxdeg=$h-1
let reddeg=$w/2+1
./genbg -z $w $h -d0:${reddeg} -D${bluemaxdeg}:${reddeg}
```
For example, if that script is named "somebinary",
```
$ ./somebinary 3 6
>A ./genbg n=6+3 e=12:12 d=0:4 D=2:4 z
H??FeW{
>Z 1 graphs generated in 0.00 sec
```
Extracting the biadjacency matrix in SageMath (from red vertices to blue vertices, so the matrix is oriented as in the problem):
```
sage: G=Graph("H??FeW{")
sage: A=G.adjacency_matrix()
sage: A[:6, 6:]
[1 1 1 1 0 0]
[1 1 0 0 1 1]
[0 0 1 1 1 1]
```
Here are the counts for small parameter values. That's a couple of hours of computation.
$$
\begin{array}{r|rrrr}
& w=4 & 6 & 8 & 10\\
\hline
h=3 & 0 & 1 & 1 & 3 \\
4 & 1 & 3 & 14 & 55 \\
5 & 0 & 9 & 115 & 1265 \\
6 & 0 & 15 & 904 & 33425 \\
7 & 0 & 20 & 6052 & 885810 \\
8 & 0 & 22 & 36311 & 21936149 \\
9 & 0 & 20 & 191568 & 492515184 \\
10 & 0 & 14 & 896697 \\
11 & 0 & 9 & 3738372 \\
12 & 0 & 5 & 13989546 \\
13 & 0 & 2 & 47256369 \\
14 & 0 & 1 & 144910788 \\
15 & 0 & 1 \\
16 & 0 & 0 \\
\end{array}
$$
You'll notice that with $w=6$ the rows have two zeros, so there can be at most $\binom{6}{2}=15$ different rows and the counts are zero for $h>15$.
The columns don't seem to match anything in OEIS. (Okay, the $w=4$ column is [A185014](https://oeis.org/A185014).)
|
2
|
https://mathoverflow.net/users/171662
|
401657
| 164,865 |
https://mathoverflow.net/questions/401659
|
2
|
Let $L$ be an ample line bundle in $\mathbb{P}^n$, with at least $n$ global sections. Choose two sets of $n$ linearly independent global sections of $L$, say $S\_1:=\{D\_1,...,D\_n\}$ and $S\_2:=\{E\_1,....,E\_n\}$, where $D\_i$ and $E\_i$ are effective divisors corresponding to global sections on $L$. Does there exist a Cremona / birational transformation $f: \mathbb{P}^n \dashrightarrow \mathbb{P}^n$ sending $D\_i$ to $E\_i$ (birationally) for each $i$?
|
https://mathoverflow.net/users/58203
|
Cremona transformations and divisors
|
In general, I think that the answer is negative. In fact, any birational transformation must preserve the geometric genus of divisors, but this is not fixed in the linear system $|L|$.
For instance, take $n=2$ and $L=\mathcal{O}(3)$. Let $D\_1, \, D\_2$ be smooth plane cubics and let $E\_1$ be a nodal cubic. There is no birational map between $D\_i$ $(i=1,\, 2)$ and $E\_1$, because $D\_i$ has geometric genus $1$, whereas $E\_1$ is rational.
|
4
|
https://mathoverflow.net/users/7460
|
401663
| 164,867 |
https://mathoverflow.net/questions/401662
|
10
|
Let $M$ and $N$ be topological spaces.
Let $\operatorname{Sh}(M)$ denote the presentable $\infty$-category of space-valued sheaves on $M$.
It seems to me that the equivalence
$$\operatorname{Sh}(M) \otimes \operatorname{Sh}(N) = \operatorname{Sh}(M \times N)$$
where $\otimes$ is the standard symmetric monoidal structure of $\operatorname{Pr^L}$, the $\infty$-category of presentable categories with left adjoints, is well-known.
However, I cannot find a proof in the literature.
My question is if there is a standard reference for this statement? Or if there is a "simple proof" of it by assuming some well-known results?
By the way, I believe that I was once told that it's in Lurie but I've failed to find it in either Higher Topos Theory or Higher Algebra.
|
https://mathoverflow.net/users/41259
|
Taking the category of sheaves is symmetric monoidal
|
1. Provided at least one of $M$ and $N$ is locally compact, the $\infty$-topos $\mathrm{Sh}(M \times N)$ is the product of $\mathrm{Sh}(M)$ and $\mathrm{Sh}(N)$ in $\mathrm{RTop}$. This is HTT 7.3.1.11.
2. Products in $\mathrm{RTop}$ can be computed as tensor products in $\mathrm{Pr^L}$. This is HA Example 4.8.1.19.
HA doesn't include a complete proof of the latter, but the case where one of the factors is of the form $\mathrm{Sh}(M)$ is essentially HTT 7.3.3.9, up to the matter of identifying $\mathrm{Sh}(M; \mathrm{Sh}(N))$ with the tensor product $\mathrm{Sh}(M) \otimes \mathrm{Sh}(N)$, for which HA 4.8.1.17 should be useful.
|
12
|
https://mathoverflow.net/users/126667
|
401667
| 164,868 |
https://mathoverflow.net/questions/401626
|
4
|
Let $X$ be a projective variety and $A$ and $B$ are two vector bundles on $X$. Let $C\_{\bullet}$ denote the complex of sheaves
$$
0\rightarrow A\rightarrow B\rightarrow 0
$$
Then we have a cup product in hypercohomology
$$
\mathbb H^i(C\_{\bullet})\otimes \mathbb H^j(C\_{\bullet})\rightarrow \mathbb H^{i+j}(C\_{\bullet}\otimes C\_{\bullet})
$$
Is it possible to describe the cup product in terms of the co-cycles? A modern reference will also be helpful.
|
https://mathoverflow.net/users/nan
|
Cup product of hypercohomologies
|
Although I doubt this counts as a "modern" reference, chapter II section 6 of Godement's *Topologie algébrique...* gives the most detailed account that I know for cup products in sheaf cohomology. This includes explicit formulas in terms of Cech cocycles.He doesn't treat products in hypercohomology, but the formulas are easy to modify:
Given a bounded complex of sheaves $C^\bullet$, choose a good open cover $\mathcal{U}$ of $X$. In your case "good" means affine. One has
$$\mathbb{H}^i(C^\bullet) \cong H^i(\check{C}(\mathcal{U},C^\bullet))$$
where the thing on the right is the cohomology of the total complex of the double complex formed from the Cech complex of $C^\bullet$. You should be able to modify the formulas in Godement to define products in $\check{C}(\mathcal{U},C^\bullet)$. The formulas are a bit messy, otherwise I would write them here.
|
3
|
https://mathoverflow.net/users/4144
|
401668
| 164,869 |
https://mathoverflow.net/questions/401661
|
0
|
I am recently reading the pde book of Evans's. I am reading the chapter 9.5.2 which is a topic about the radial symmetry for the solution of a elliptic equation. In the book there is a method of moving planes. That is when a function is symmetric for all directions, we can have that it is radial symmetric. I do not know why it is true, can you give me some information about it ? Here if we assume that $ u:\mathbb{R}^n\rightarrow \mathbb{R} $ is a function with $ n $ variables, the symmetry for all directions means that for all $ v\in\mathbb{R}^n $, here is a plane $ P\_v\subset\mathbb{R}^n $ such that $ P\_v\perp v $ and for all $ x,y\in\mathbb{R}^n $ that are symmetric about the plane $ P\_v $, we have $ u(x)=u(y) $.
|
https://mathoverflow.net/users/241460
|
Why the symmetry for all directions implies the radial symmetry?
|
Notations:
1. given $P$ a hyperplane, let $r\_P$ denote the reflection operation $\mathbb{R}^d\to\mathbb{R}^d$ about the plane $P$.
2. given $R$ a co-dimension 2 affine subspace of $\mathbb{R}^d$, denote by $\rho\_{R,\theta}$ the rotation by angle $\theta$ that fixes the "axis" $R$.
Basic linear algebra/affine geometry gives you that if $P, Q$ are hyperplanes with non-trivial intersection, then $r\_P\circ r\_Q = \rho\_{P\cap Q, \theta}$ for some $\theta$. (Two reflections generate a rotation.) The angle $\theta$ depends only on the angle between the two hyperplanes.
Now let $P, Q$ be generic non-parallel hyperplanes. Then the corresponding $\theta$ is an irrational multiple of $\pi$. And so if $u$ is a **continuous** function satisfying $u = u\circ r\_P = u\circ r\_Q$, then we must have $u = u \circ \rho\_{P\cap Q, \phi}$ for **any** angle $\phi$. (Because by alternating between $r\_P$ and $r\_Q$, and composing as many times as you want, you find that for all $n\in \mathbb{N}$ that $u = u\circ \rho\_{P\cap Q, n\theta}$. And $\{n\theta \mod 2\pi\}$ is dense on the circle.)
Returning to your setting, start by choosing $d$ generic directions (in particular, you don't want them to be mutually orthogonal, as that is non-generic), and set $P^k$ be the corresponding planes of symmetry. Let $x\_0$ be the common intersection. The above argument shows that about each $P^i\cap P^j$ (forming a set of $d(d-1)/2$ codimension-2 "axes") through $x\_0$, your function $u$ is rotationally symmetric. These rotations however generate the full rotation group about $x\_0$, and hence you are done.
|
3
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https://mathoverflow.net/users/3948
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401672
| 164,870 |
https://mathoverflow.net/questions/401669
|
21
|
Suppose a proof came out (and was verified by credible peer review) of the following statement:
>
> There is a $T\_0$ such that for all $t>T\_0$, all zeros $\zeta(\beta+it)=0$ have $\beta=1/2.$
>
>
>
where $T\_0$ is totally ineffective. What interesting consequences would this partial result have?
---
Of course you could ask this sort of question for all kinds of weakenings/strengthenings/relatives of RH:
* Zero-density estimates (which already has its own questions [here](https://mathoverflow.net/q/137876/6043) and [here](https://mathoverflow.net/q/161442/6043))
* Density Hypothesis
* Lindelöf Hypothesis
* Generalized Riemann hypotheses for various L-functions
* Grand Lindelöf Hypothesis
But so far all the uses I have seen of $\zeta$ zeros has been in the strip $0<T<T\_0$ and I wondered if that was convenience (where we've checked) or more than that.
|
https://mathoverflow.net/users/6043
|
What are the consequences of an ineffective proof of the Riemann Hypothesis?
|
As a strengthening of what @KConrad commented, it would imply that the density of nontrivial zeros on the critical line is 100% in each horizontal strip of height 1, which is not useless: this is equivalent to the Lindelöf Hypothesis, which states that $\zeta \left( \frac{1}{2} + i t \right) = \mathcal{O}\_{\varepsilon} \left( 1 + \lvert t \rvert^{\varepsilon} \right)$.
One example of a consequence of the Lindelöf Hypothesis (which is exactly much easier to prove directly from your non-effective Riemann Hypothesis using the explicit formula) is that the prime gaps satisfy $p\_{n + 1} - p\_n \leq \sqrt{p\_n} \log \left( p\_n \right)^2$, improving the best current unconditional result by Baker-Harman-Pintz of $p\_{n}^{0.525}$.
The exponent of the logarithm might be a bit less, but I decided to err on the side of caution. By the way, this is still very far from the conjectured upper bound $p\_{n}^{\varepsilon}$ (and in fact it is relatively widely believed that the gap is at most $C \log \left( p\_n \right)^2$ for some absolute constant $C$).
However, the Lindelöf Hypothesis appears in estimating arithmetic sums, as many counting problems can be transformed into a zeta integral. Let me illustrate by a simple example. We will prove (conditionally) the following:
$$\sum\_{n = 1}^{N} d (n) = n \log n + (2 \gamma - 1) n + \mathcal{O}\_{\varepsilon} \left( n^{1/2 + \varepsilon} \right)$$
where $d(n)$ is the number of divisors of $n$ and $\gamma$ is the Euler-Mascheroni constant. This is usually proved via the Dirichlet hyperbola method (and with good reason), and we get a slightly weaker error term than usual, but this is just to illustrate the technique. Recall the classical inverse Mellin transform
$$\intop\_{c - i \infty}^{c + i \infty} x^s \frac{\mathrm{d} s}{s} = 1\_{x > 1} + \frac{1}{2} 1\_{x = 1}$$
for any $c > 0, \ x \in \mathbb{R}$. Taking $c > 1$, and interchanging summation and integration we get (up to an error of $\frac{d (n)}{2}$, which is negligible)
$$\sum\_{n = 1}^{N} d(n) = \intop\_{c - i \infty}^{c + i \infty} \sum\_{n = 1}^{\infty} d(n) \left( \frac{N}{n} \right)^{s} \frac{\mathrm{d} s}{s} = \intop\_{c - i \infty}^{c + i \infty} \zeta \left( s \right)^2 \frac{N^s \mathrm{d} s}{s}$$
Now, shift the contour to $\mathrm{Re(s) = \frac{1}{2}}$. The residue picked up at the pole $s = 1$ is exactly $N \log N + (2 \gamma - 1) N$, so all we have left to proveis to show that the expression
$$N^{\frac{1}{2}} \intop\_{-\infty}^{\infty} \zeta \left( \frac{1}{2} + i t \right)^2 N^{i t} \frac{\mathrm{d} t}{\frac{1}{2} + i t}$$
is $\mathcal{O}\_{\varepsilon} \left( N^{1/2 + \varepsilon} \right)$, or equivalently that the integral is $\mathcal{O}\_{\varepsilon} \left( N^{\varepsilon} \right)$.
Here is the point where I tell you that I actually lied beforehand: it turns out that using the full inverse Mellin transform is, although very elegant, not necessarily the best choice to get a good analytic bound. What is usually done is approximate it by integrating not from $c - i \infty$ to $c + i \infty$, but from $c - i N$ to $c + i N$, where $c$ is say something like $1 + \frac{1}{\log N}$. I don't remember the details off the top of my head (they appear for example in Montgomery's book, and in a few expositions of proofs of the Prime Number Theorem), so just trust me here when I say that it is sufficient to bound the integral
$$\intop\_{- N}^{N} \zeta \left( \frac{1}{2} + i t \right)^{2} N^{i t} \frac{\mathrm{d} t}{\frac{1}{2} + i t}$$
But now (and here we finally use the Lindelöf Hypothesis!) we can bound pointwise this integral, and get that it is $\mathcal{O}\_{\varepsilon} \left( N^{\varepsilon} \right)$ as required.
---
This example, although somewhat stupid, shows the power of the Lindelöf Hypothesis. Indeed, see Tao's answer [The relationship between the Dirichlet Hyperbola Method, the prime counting function, and Mertens function](https://mathoverflow.net/questions/97694/the-relationship-between-the-dirichlet-hyperbola-method%2C-the-prime-counting-function%2C-and-mertens-function), where he points out the fundamental difference between arithmetic functions with zeta in their denominator (whose behaviour is controlled very much by the zeroes of zeta) and arithmetic functions with zeta in the numerator. Despite that, we still managed to use information about the zeroes of zeta to get a nontrivial estimate.
|
19
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https://mathoverflow.net/users/88679
|
401678
| 164,873 |
https://mathoverflow.net/questions/259136
|
6
|
I know that there exist Ito formulae to understand
$
f(X),
$
where $f: H\rightarrow \mathbb{R}$ is sufficiently nice, $H$ is a Hilbert space and $X$ is an $H$-valued semi-martingale.
However I'm wondering if there is a generalization or analogue for $f: H \mapsto \tilde{H}$, sufficiently nice operators between Hilbert spaces? I haven't been able to find anything, any references in the right direction are appreciated!
|
https://mathoverflow.net/users/36886
|
Reference Request: Vector-Valued Ito Formula
|
[Curtain and Falb (1970)](https://core.ac.uk/download/pdf/82209342.pdf) treated this case.
|
1
|
https://mathoverflow.net/users/60775
|
401680
| 164,874 |
https://mathoverflow.net/questions/401633
|
12
|
Background: I spent sometime reading about algebraic K-theory and started reading research papers on the subject with relative facility at least I do understand constructions, statements of the Theorems and (some) proofs.
As a student, I feel more comfortable with standard algebraic topology and topological manifolds. I start to read some classical papers on L-theory (essentially Ranicki papers ). The problem is that I can't make a connection with algebraic K-theory (Quillen, Waldhausen, Thomason,...) One example is the multiple variation of The L-Theory (Symmetric, Quadratic, Hermitian,...).
Here is my question: Could someone indicate how can a student be initiated to the subject of L-theory. I need to understand some basic ideas and more importantly I wish to know how to learn L-Theory assuming that I can understand K-Theory.
|
https://mathoverflow.net/users/141114
|
Roadmap for L-Theory
|
I apologize for the self promotion -- I hope the content of this answer can be useful anyway...
---
My favourite introduction to L-theory is Lurie's notes on [Algebraic L-theory and surgery](https://www.math.ias.edu/%7Elurie/287x.html) (warning: aggressively modern).
Working from the ideas in this notes my coauthors and I have built a series of papers ([I](http://arxiv.org/abs/2009.07223), [II](http://arxiv.org/abs/2009.07224), [III](http://arxiv.org/abs/2009.07225)) developing L-theory and hermitian K-theory in a fashion that closely mirrors the development of algebraic K-theory. One of the main point of our work is to find a home for all (well, almost all) variants of L-theory in a single unified framework, that of Poincaré structures.
These papers are probably not very good for a student though -- while we do all the details, this results in many pages of technical results and it's easy to lose the forest for the trees. If you are interested in this circle of ideas some of my coauthors have given a series of minicourses on the topic (Fabian Hebestreit's [I](https://media.ed.ac.uk/media/The+algebraic+theory+of+cobordism+Talk+1+-+Fabian+Hebestreit/1_q8vyjhvr), [II](https://media.ed.ac.uk/media/The+algebraic+theory+of+cobordism+Talk+2+-+Fabian+Hebestreit/1_4jrkdv56), Markus Land [I](https://youtu.be/DfNaci8lDXk), [II](https://youtu.be/FG4h57OToas), [III](https://youtu.be/Ak7Gvpqxk0E), [IV](https://youtu.be/FeenTO41UsU); Yonatan Harpaz [I](https://youtu.be/DGG9ftI9noI), [II](https://youtu.be/GH-2hgbWMz8), [III](https://youtu.be/eTrNY5n2Awk)) which are probably more accessible than the papers.
|
11
|
https://mathoverflow.net/users/43054
|
401689
| 164,876 |
https://mathoverflow.net/questions/401688
|
6
|
I was talking with a non-mathematician the other week at a workshop about the fact that many mathematicians, like myself, are indexed in the [math genealogy database](https://genealogy.math.ndsu.nodak.edu/). We talked a little about how many people tend to have family trees linking back to a few influential/well-known mathematicians (Newton/Gauss/Euler/etc...). I looked online later and this casual observation seems to have been examined closer (in this [paper](https://arxiv.org/pdf/1603.06371.pdf)) and about 65 percent of the +200k nodes of the network fits within 24 families.
I have some hypotheses for the cause of this concentration and expect that sociological factors play in heavily, but I still wondered if a random graph model with similar properties to the genealogy network would explain this effect at least partially.
For simplicity maybe it would be best to assume a forest (collection of trees) structure on the random model $G$ (ignoring the case where someone has more than one advisor). Some potentially useful properties are below:
* It seems reasonable that the maximum out-degree of $G$ increases as one descends down the generations of the tree (corresponding to time), since more Ph.D.'s tend to be awarded now than before.
* Also, there are more Ph.D. students now that do not go onto direct Ph.D. students themselves (since the number of math Ph.D.'s is about as high as ever, but there are only a select number of Ph.D. awarding institutions at which one can supervise Ph. D.'s). I think these first two conditions can be emulated by adjusting the distribution for $G$ at each generation.
* There also should be a much lower probability of someone
getting a Ph.D. supervised by someone who is not a mathematician (a
disconnected node being generated ) vs the standard case of a
descendent being generated in the graph $G$.
* There should probably be a small number of nodes to start with, but this seems less important.
Altogether, there are two questions that I have.
**Question 1**: *Is there a standard random graph model that emulates these properties of the math genealogy graph?*
**Question 2**: *If so, does that model have concentration of the network in a small number of families, if the network is allowed to generate for a sufficient length of time?*
I don't have enough intuition about random graphs to answer the second question with some very simple random graph models, so would be interested if anyone can point out results for a different setting than the one outlined here.
|
https://mathoverflow.net/users/118731
|
Graphs resembling the math genealogy graph must have concentration in a small number of families?
|
Precisely this question was the starting point of the Galton - Watson theory of branching processes. To quote the opening paragraph of their 1875 paper *On the Probability of the Extinction of Families*:
>
> The decay of the families of men who occupied conspicuous positions in past times has been a subject of frequent remark, and has given rise to various conjectures. It is not only the families of men of genius or those of the aristocracy who tend to perish, but it is those of all with whom history deals, in any way, even of such men as the burgesses of towns, ...
>
>
>
|
13
|
https://mathoverflow.net/users/8588
|
401694
| 164,879 |
https://mathoverflow.net/questions/401686
|
6
|
Say that a logic $\mathcal{L}$ is **directed** iff whenever $\mathfrak{A}\equiv\_\mathcal{L}\mathfrak{B}$ there is some $\mathfrak{C}$ with $\mathcal{L}$-elementary substructures $\mathfrak{A}'\preccurlyeq\_\mathcal{L}\mathfrak{C}$, $\mathfrak{B}'\preccurlyeq\_\mathcal{L}\mathfrak{C}$ with $\mathfrak{A}\cong\mathfrak{A}',\mathfrak{B}\cong\mathfrak{B}'$. It's a standard exercise to show that $\mathsf{FOL}$ is directed - or more generally, that every *compact* logic is directed. On the other hand, it's easy to whip up artificial logics demonstrating that this joint embeddability isn't *equivalent* to compactness.
I'm curious about the situation with second-order logic $\mathsf{SOL}$. It's consistent with $\mathsf{ZF}$ that there are $\mathsf{SOL}$-equivalent structures which do not $\mathsf{SOL}$-elementarily embed into the same structure (see below), but I don't see how to get this result outright in $\mathsf{ZFC}$ (much less $\mathsf{ZF}$). However, I recall seeing an easy argument (due to Mostowski?) that in fact this is a $\mathsf{ZF}$-theorem.
>
> **Question**: Does $\mathsf{ZF}$ prove that $\mathsf{SOL}$ is not directed?
>
>
>
Here's a proof that the non-directedness of $\mathsf{SOL}$ is *consistent* with $\mathsf{ZF}$. Suppose there is a family $\mathbb{A}$ of amorphous sets of pairwise incomparable cardinality such that there is no injection from $\mathbb{A}$ into $2^{\aleph\_0}$. Thinking of each element of $\mathbb{A}$ as a structure in the empty language, we must have $X,Y\in\mathbb{A}$ with $X\equiv\_\mathsf{SOL}Y$ but $X\not\cong Y$. But any set into which both $X$ and $Y$ inject must be non-amorphous, hence cannot satisfy $Th\_\mathsf{SOL}(X)=Th\_\mathsf{SOL}(Y)$ since amorphousness is second-order-expressible.
Of course this doesn't help at all without a background assumption of lots of amorphous sets, so it's not really relevant to the question I'm asking here, but it's still neat. Note that a positive answer will have to crucially involve uncountable structures, since [it's consistent with $\mathsf{ZFC}$ that $\equiv\_{\mathsf{SOL}}$ implies $\cong$ for countable structures](https://mathoverflow.net/a/161694/8133).
|
https://mathoverflow.net/users/8133
|
Failure of "directedness" for second-order logic?
|
The answer to the question is yes.
Let $\alpha\_0<\alpha\_1$ be the least ordinals (in the reverse lex order, say) such that $V\_{\alpha\_0}$ and $V\_{\alpha\_1}$ have the same second order theory $T$. Assume towards a contradiction that $V\_{\alpha\_0}$ and $V\_{\alpha\_1}$ are elementarily embeddable into a common structure $M$, which we may assume is transitive. Note that the embeddings fix $T$. Since $V\_{\alpha\_1}$ satisfies that there is an ordinal $\beta$ such that $V\_\beta$ satisfies $T$, so does $M$, and hence so does $V\_{\alpha\_0}$. But this means that some $\beta < \alpha\_0$ has the same theory as $V\_{\alpha\_0}$, contrary to the minimality of the pair $\alpha\_0 < \alpha\_1$.
(Edit: I now see this is very close to what Trevor was doing.)
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8
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https://mathoverflow.net/users/102684
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401696
| 164,880 |
https://mathoverflow.net/questions/401675
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2
|
Let $R$ be a commutative ring with identity and $A$ and $B$ be two proper ideals of $R$ such that $A+B=R$ and for each $r^2=r\in R$ we have either $r\not\in A$ or $r-1\not\in B$. How can we prove the existance of two maximal ideals $m\_1$ and $m\_2$ of $R$ such that $A\subseteq m\_1$, $B\subseteq m\_2$ and $\{r\in m\_1\mid r^2=r\}=\{r\in m\_2\mid r^2=r\}$?
|
https://mathoverflow.net/users/338309
|
The existence of two maximal ideals with the same set of idempotents
|
Sketch: First, if $e$ is an idempotent in $A$, show that $B$ can be replaced with $B+Re$, and the hypotheses still holds. Use this to reduce to the case that $A$ and $B$ contain the same idempotents.
Second, if $e$ is an idempotent of $R$ with $e,1-e\notin A$ (and hence also not in $B$) show that we can replace $A$ and $B$ with the new pair $A+Re$ and $B+Re$, still satisfying the same conditions. Use this to reduce to the case that $A$ and $B$ contain the same idempotents, and either an idempotent or its complement belongs to $A$.
Now, extend to any maximal ideals containing $A$ and $B$, and note that a maximal ideal cannot contain an idempotent and its complement at the same time.
|
2
|
https://mathoverflow.net/users/3199
|
401704
| 164,881 |
https://mathoverflow.net/questions/346647
|
1
|
Is there an algorithmic way to map the natural numbers to unique k-ary trees?
I am familiar with the work of Tychonievich who created a mapping from integers to binary trees. <https://www.cs.virginia.edu/~lat7h/blog/posts/434.html>
Is there something similar for k-ary trees?
|
https://mathoverflow.net/users/148964
|
mapping integers to k-ary trees
|
If you search for "ranking $k$-ary trees" or "ranking $t$-ary trees" you will find several published papers on this. For example:
[This](https://epubs.siam.org/doi/abs/10.1137/0207039?casa_token=L7KsyEFGnGMAAAAA%3AtRJdKQBkcAnPHnV1GQaFhmsMHA1znPEnW942T513tvnhIhfLFposVAp5vi_XK2JmlbSVWBGlDKeS&)
[This](https://epubs.siam.org/doi/abs/10.1137/0207034?casa_token=BRMD50DToJsAAAAA:3HA_MEs4gzr_HGo0uCkqtdIZGD4OzkskBQfPqL7TOg8RQ33coSr4yj1ye8owuRGoacD6r5kXZ938)
|
3
|
https://mathoverflow.net/users/9025
|
401710
| 164,882 |
https://mathoverflow.net/questions/401705
|
2
|
Let $X$ and $Y$ be smooth algebraic varieties over $\mathbb{C}$. Let $f$ be a continuous map from complex points of $X$ to $Y$. Are there Zariski opens $U$ and $V$ inside $X\times \mathbb{A}^1$ and $Y\times \mathbb{A}^1$ respectively such that $U$ contains $X\times\{0\}$ and $X\times \{1\}$ (similarly $V$ contains $Y\times \{0\}$ and $Y\times \{1\}$) with the property that a continuous map $g$ from $U$ to $V$ can be assigned to $f$ in a way that $g|\_{X\times \{0\}}= f$ and $g|\_{X\times \{1\}}$ is a regular morphism?
|
https://mathoverflow.net/users/127776
|
Are continuous maps generically homotopic to a regular map?
|
I think that the answer is negative.
For instance, let $X=Y$ be a smooth curve of genus $g\geq 3$ with $\operatorname{Aut}(X)=\{\mathrm{id} \}$, and take as $f \colon X \to X$ an isotopically non-trivial diffeomorphism.
|
6
|
https://mathoverflow.net/users/7460
|
401711
| 164,883 |
https://mathoverflow.net/questions/401656
|
3
|
Let $P\subseteq \mathbb{N}$ be the set of primes, and for any integer $n>1$ let $L(n) = \max\{p \in P: p \mid n\}$ be the largest prime divisor of $n$. Moreover, for $n \in \mathbb{N}$ with $n>1$ we let $M(n)$ to be the **median** of the set $$\{L(m)/m : m\in \mathbb{N} \land 1 < m \leq n\}.$$
Does $\lim\_{n\to\infty}M(n)$ exist? If yes, is its value known?
|
https://mathoverflow.net/users/8628
|
Behavior of biggest prime divisor of $n$ as $n$ grows large
|
The number of prime divisors of $n$ grows typically as $\log \log n$. Suppose $n$ has $k$ prime factors. Now $n/L(n)$ has only $k-1$ prime factors, so
$$
k-1 \approx \log \log \frac{n}{L(n)} = \log \log n + \log\left( 1-\frac{\log L(n)}{\log n}\right) = k + \log\left( 1-\frac{\log L(n)}{\log n}\right)
$$
and hence
$$
\log\left( 1-\frac{\log L(n)}{\log n}\right) \approx -1
$$
i.e., typically we will have
$$
\log L(n) \approx \left( 1-\frac{1}{e}\right)\log n
$$
or indeed
$$
L(n) \approx n^{1-\frac{1}{e}} = n^{0.632...}.
$$
Note that this means that typically the largest prime divisor of $n$ will be bigger than $\sqrt{n}$, which is a nice isolated statement as well.
As mathworker21 noted, it suffices to show that the average of $L(m)$ for $1 < m \leq n$ tends to $0$ as $n \to \infty$. Let $A(n)$ be this average. Thus in the typical case
$$
A(n) = \frac{1}{n} \sum\_{m=1}^n \frac{L(m)}{m} \approx \frac{1}{n} \sum\_{m=1}^n \frac{1}{m^{\frac{1}{e}}} = \frac{1}{n} H\_{n,\frac{1}{e}}
$$
where $H\_{n,\frac{1}{e}}$ is the $n$th generalized harmonic number of order $\frac{1}{e}$. As $n\to \infty$, an asymptotic expansion of this right-hand side yields
$$
A(n) \approx \frac{1}{n} H\_{n,\frac{1}{e}} = \frac{e}{n^{1/e}(e-1)} + \frac{\zeta(\frac{1}{e})}{n} + O\left( \frac{1}{n^2}\right) \to 0
$$
as $n \to \infty$. Hence also $\lim\_{n \to \infty} M(n) = 0$.
|
3
|
https://mathoverflow.net/users/120914
|
401713
| 164,884 |
https://mathoverflow.net/questions/401683
|
13
|
Are there any known examples of Einstein manifolds $(M, g)$ such that $$\sup\_{x \in M} \|\text{Rm}(x) \| = \infty$$
I'm looking for these examples because they might provide a counter-example to a problem of mine, but I can't think of any. Obviously such a manifold can't be compact (and therefore it can't be complete with positive scalar curvature) but I haven't been able to think of any concrete example.
|
https://mathoverflow.net/users/119418
|
Are there examples of Einstein manifolds with unbounded curvature?
|
If you don't care about completeness, here's a fairly simple way to construct such examples: Start with a compact Einstein manifold $(M^n,g)$ with Einstein constant $1$ (i.e., $\mathrm{Ric}(g) = (n{-}1)\,g$) that is not conformally flat. Now take the sine-cone, i.e., $\bigl(M\times(0,\pi),h\bigr)$ where $h = \mathrm{d}r^2 + (\sin r)^2\,g$. Then one easily computes that this is also an Einstein manifold with Einstein constant $1$, i.e., $\mathrm{Ric}(h) = n\,h$, but the Weyl curvature of $h$ (which is nonzero since $g$ is not conformally flat) blows up as $r$ approaches either $0$ or $\pi$.
(Also, if one just takes the ordinary cone, $\bigl(M\times(0,\infty),h\bigr)$ where $h = \mathrm{d}r^2 + r^2\,g$, then $h$ will be Ricci-flat, but the Weyl curvature of $h$ will blow up as $r\to 0$.)
Finally, as Anton Petrunin pointed out in the comment below, if $(M,g)$ is complete and Ricci-flat (i.e., Einstein with Einstein constant 0), but not conformally flat (equivalently, not flat), the Riemannian manifold $\bigl(M\times\mathbb{R}, h = \mathrm{d}r^2 + \mathrm{e}^{2r}\,g\bigr)$ will be a complete Einstein manifold with Einstein constant $-1$ whose Weyl curvature has unbounded norm as $r\to-\infty$.
|
12
|
https://mathoverflow.net/users/13972
|
401715
| 164,886 |
https://mathoverflow.net/questions/401712
|
0
|
For a convergent sequence $(a\_n)\_n \rightarrow a$ consider the exponential series
\begin{equation\*}
\exp\_{(a\_n)\_n}(-x) := \sum\_{n=0}^{\infty} \frac{(-x)^n a\_n}{n!}.
\end{equation\*}
Can there be anything said about the resulting expression, more than convergence? I presume that, since e.g. changing only $a\_0$ already changes the result, this is not expressable in terms of the limit $a$.
Now, if $a\_n$ is of the form $u^n$ for some constant $y$, this ofcourse results in $e^{-xy}$. Can there be something similar be said if $a\_n$ more generally is a "homogeneous polynomial" of degree $n$ in some variables? By that I mean something like
\begin{equation\*}
a\_n=b\_{0,n} u^n + b\_{1,n} u^{n-1} v + \dots + b\_{n,n} v^n.
\end{equation\*}
This is trying to generalize
\begin{equation\*}
a\_n= {n\choose 0} u^n + {n\choose 1} u^{n-1} v + \dots + {n\choose n} v^n = (u+v)^n.
\end{equation\*}
|
https://mathoverflow.net/users/338748
|
Exponential Series with a sequence
|
The Appell Sheffer polynomial formalism can be used to deal with these types of polynomials for $a\_0 = b\_{0,0} = 1$.
In umbral notation for which, e.g., $(c.)^n = c\_n$,
$$ e^{c.t} \; e^{xt} = e^{(c.+ x)t} = e^{t \; p.(x)},$$
with the Appell Sheffer polynomials
$$p\_n(x) = (c.+x)^n = \sum\_{k=0}^n \; \binom{n}{k} \; c\_{n-k} \; x^{k} \; .$$
Let $x = v$ and $c\_n = d\_n u^n$ to get polynomials of your type. Convergence is guaranteed if $e^{c.t}$ converges, but even if it doesn't converge, the series can be formally multiplicatively inverted degree by degree.
The coefficients for Appell sequences (e.g.f.s) can be regarded as formal classical moments (many are actual moments of distributions, e.g., the Bernoulli numbers) and then related to formal classical cumulants, and, if the sequence is re-normalized and viewed as coefficients of an o.g.f., the renormalized polynomials can be regraded as moments correlated with the free cumulants of free probability theory.
Appell, Turan, Jensen, and Polya, among others, investigated the zeros of Appell sequences. See, e.g., "[Hyperbolicity of Appell Polynomials of Functions in the δ-Laguerre-Pólya Class](https://arxiv.org/pdf/2002.09906.pdf)" by Jonas Iskander and Vanshika.
One of the more recent refs on the topic is "[General Bivariate Appell Polynomials via Matrix Calculus and Related Interpolation Hints](https://www.mdpi.com/2227-7390/9/9/964/pdf)" by
Francesco Aldo Costabile, Maria Italia Gualtieri and Anna Napoli. Another, dealing with a general Sheffer sequences modified to be bivariate, is "[Operatorial methods and two variable Laguerre polynomials](https://www.researchgate.net/publication/265400377_Operatorial_methods_and_two_variable_Laguerre_polynomials)" by Dattoli and Torre. The Bernoulli and Laguerre polynomials are very useful sequences in pure mathematics and mathematical physics. All Sheffer polynomials $S\_n(x)$ have an e.g.f. of the form
$$A(t) \; e^{x \; B(t)} = e^{t \; S.(x)}$$
with $A(t)$ and $B(t)$ analytic about the origin, $A(0) = 1$, $B(0) = 0$, and $B'(t) \neq 0$ and can be modified to be of your type.
|
0
|
https://mathoverflow.net/users/12178
|
401718
| 164,887 |
https://mathoverflow.net/questions/401703
|
6
|
Let $G$ be a finitely generated Fuchsian group, and let $\mathcal{F}$ denote the Dirichlet fundamental domain of $G$ with respect to $0$ in the Poincaré disc model.
Assume throughout that $\mathcal{F}$ is non-compact.
I am interested in properties of $G$, and how these properties are connected with Poincaré's theorem. Here are my questions:
1. Is $G$ the free product of elementary hyperbolic, parabolic and elliptic subgroups? It is true for the modular group PSL(2,Z). Moreover, it holds under the stronger assumptions in 2).
2. Assume that $G$ has no elliptic elements. Then $G$ is free (because $G$ is fundamental group of a non-compact surface) and thus, $\mathcal{F}$ has vertices only at the boundary of hyperbolic space. Can we derive this property of the vertices from Poincaré's theorem?
3. Assume that $G$ is of the second kind (that is, the limit set of $G$ is not equal to the boundary of hyperbolic space). Then in particular $\mathcal{F}$ is non-compact. Is it true that the sides of $\mathcal{F}$ are pairwise disjoint, except (possibly) the sides paired by elliptic elements?
UPDATE:
Sam Nead answered 3) in the negative. Is the answer in 3) also negative if we allow an arbitrary reference point for the Dirichlet fundamental domain? Or does there always exist a suitable reference point for a Dirichlet fundamental domain such that the sides are pairwise disjoint, except (possibly) the sides paired by elliptic elements?
|
https://mathoverflow.net/users/338619
|
Non-compact Dirichlet fundamental domains and free Fuchsian groups
|
For (1) the answer is "yes". Since the surface has finitely generated fundamental group, there is a finite collection of disjoint embedded bi-infinite geodesics that cut the surface into a collection of ideal (or hyperideal) polygons, each with at most one cone point in its interior. (There is a special case when the cone point has angle $\pi$, which I leave as an exercise.) The statement now follows from the Seifert-van Kampen theorem.
For (2) I read your question as "can a Dirichlet domain for a surface (as above) have material vertices?" and the answer is "yes, it may". To see this, consider the surface made by doubling an ideal triangle across its boundary. If we take the origin to be the centre of one of the triangles, then the Dirichlet domain has three ideal vertices and three material vertices.
For (3) the answer is "not in general". To see this, consider the surface of (2) where we "open" the cusps, replacing them by (identical) funnels. Again the surface has a three-fold symmetry about the centres of the (now hyperideal) triangles. The Dirichlet domain has three material vertices, six ideal vertices, six material edges, and three ideal edges.
The property you want - disjoint material edges glued in pairs - feels very similar to "Schottky, with all circles perpendicular to a single circle". Even this only gets you *some* Dirichlet domain with the desired form, not all. I am not sure if this property has a name, but I would look at Marden's book, or Maskit's, as a first reference.
For (3) "updated" the answer is "not in general". That is, there is a hyperbolic surface $S$ with infinite volume, with finitely generated (and free) fundamental group, so that all Dirichlet domains have material vertices.
Consider the surface $T$ given by (3) (first version) above. Let $x$ in $T$ be one of the points with three-fold symmetry. Let $D \subset T$ be a very small round disk about $x$. Let $S$ be the surface obtained by doubling $T - D$ across $\partial D$. Uniformise $S$. Let $\gamma$ be the image of $\partial D$ in $S$. Since $\gamma$ is fixed by a reflection in $R$, it is also a hyperbolic geodesic. Since the radius of $D$ was very small, the geodesic $\gamma$ is very short.
I claim that $S$ has the desired property - that is, all Dirichlet domains have material vertices. The proof is harder than that for (3).
|
2
|
https://mathoverflow.net/users/1650
|
401725
| 164,889 |
https://mathoverflow.net/questions/401726
|
3
|
Let $M\_1(\mathbf{x})$ and $M\_2(\mathbf{x})$ be $m$ by $m$ matrices with each entry a homogeneous form in $\mathbb{C}[x\_1, \ldots, x\_n]$.
I would like to show that
$$
\{ \mathbf{x} \in \mathbb{A}^n\_{\mathbb{C}}:\dim (\ker M\_1(\mathbf{x}) \cap \ker M\_2(\mathbf{x})) \geq C \},$$
for any $C > 0$,
is 1) an affine variety (zero set of some polynomials), 2) it is defined by homogenous forms.
Hence, an affine cone over some projective variety.
This is easy to see if $M\_1 = M\_2$, but I was not sure how to proceed for this more general case. Any comments appreciated!
|
https://mathoverflow.net/users/84272
|
How can I show $\{\mathbf{x}: \dim (\ker M_1(\mathbf{x}) \cap \ker M_2(\mathbf{x})) \geq C \}$ is an affine variety?
|
Evidently, $\mathrm{Ker} M\_1\cap \mathrm{Ker} M\_2=\mathrm{Ker}(M\_1,M\_2)$,
where $(M\_1,M\_2)=:M$ is the $2m\times m$ matrix obtained by putting $M\_1,M\_2$
together ($k$-th column of $M$ consists the of the $k$-th column of $M\_1$ followed by the $k$-th column of $M\_2$).
Then $\mathrm{dim}\,\mathrm{Ker} M=2m-r$, where $r$ is the rank of $M$.
So $\mathrm{dim}\,\mathrm{Ker} M\geq C$ is equivalent to $r\leq 2m-C$, and this means
that determinants of all submartices of size $(2m-C+1)\times(2m-C+1)$ are zero.
These determinants are polynomials defining your affine vriety.
|
6
|
https://mathoverflow.net/users/25510
|
401728
| 164,890 |
https://mathoverflow.net/questions/401721
|
0
|
Let $A= C([0,1])$ and $J= \{f \in A: f(0) = 0\}$. Consider the Hilbert $C^\*$-module
$E:= A \oplus J$ (with the obvious right $A$-action and inner product). I want to prove that
$$q: E \to E: (f,g) \mapsto (f-g, 0)$$
is not adjointable. This is claimed in Lance's book on Hilbert $C^\*$-modules, p22.
Here is what I tried. Assume to the contrary that $q$ is adjointable. Then there is $q^\*: E \to E$ such that
$$ (\overline{f-g})s= \langle q(f,g) , (s,t)\rangle = \langle (f,g), q^\*(s,t)\rangle.$$
In particular, $q^\*(s,t)$ does not depend on $t$ so we have $q^\*(s,t) = q^\*(s,0)$. Then I'm stuck.
|
https://mathoverflow.net/users/216007
|
Why is $q(f,g) = (f-g,0)$ not adjointable?
|
This is just a calculation. Continuning your argument, $q^\*(s,t) = q^\*(s,0) = (s\_1,-s\_2)$ for some $s\_1\in A, s\_2\in J$ (I add the minus sign for convenience later). Then
$$ \overline{f} s - \overline{g}s = \langle (f,g), (s\_1,-s\_2) \rangle
= \overline{f} s\_1 - \overline{g}s\_2, $$
for all $f\in A, g\in J$.
Set $f=1,g=0$ to see that $s = s\_1$; set $f=0$ to see that
$$ \overline{g} s = \overline{g} s\_2, $$
for all $g\in J$. Letting $g$ run through an approximate identity for $J$ (so a net $(g\_i)$ with $g\_i(x)\rightarrow 1$ for each $x>0$) we conclude that $s(x) = s\_2(x)$ for all $x>0$. If for example $s=1\in A\setminus J$ this shows that $s\_2(x)=1$ for all $x>0$, contradicting that $s\_2\in J$.
|
4
|
https://mathoverflow.net/users/406
|
401730
| 164,892 |
https://mathoverflow.net/questions/401732
|
1
|
Suppose the Lie group $G$ contains the Lie group $J$ as a subgroup, so
$$
G \supset J.
$$
If $G$ has a nontrivial first homotopy group $\pi\_1(G) \neq 0$.
If $G$ has a universal cover $\widetilde{G}$, so $\pi\_1(\widetilde{G}) = 0$.
>
> Question: What are the necessary and sufficient conditions to derive that
> $$
> \widetilde{G} \supset J \quad (?)
> $$
> is also true via a lifting?
> $$
> \begin{array}{ccc}
> & & \widetilde{G}\\
> &\nearrow & \downarrow\\
> J & \longrightarrow & G
> \end{array}.
> $$
>
>
>
My thought:
1. Sufficient condition may be $\pi\_1(J)=0$, due to the property $\pi\_1(\widetilde{G})=0$.
2. Necessary condition may be that
$\pi\_1(J) \to \pi\_1({G})$ is a trivial group homomorphism, which maps all $\pi\_1(J)$ to 0,
because this map
can be decomposed by $\pi\_1(J) \to \pi\_1(\widetilde{G})=0$ and $\pi\_1(\widetilde{G})=0 \to \pi\_1({G})$.
3. The above conditions are for universal covering $\widetilde{G}$, what would be the conditions for the (non-universal) covering space $\widetilde{G}$?
|
https://mathoverflow.net/users/336737
|
Necessary and sufficient conditions for the Lie group embedding $G \supset J$ can be lifted to $G$'s covering space
|
Lifting to covers is completely understood; the necessary and sufficient requirement is that the image of the fundamental group lies in the subgroup associated to the cover. So, for example, you are correct that to lift to the universal cover the map on fundamental groups must be trivial. For a reference, see chapter 1 of Hatcher proposition 1.33.
|
3
|
https://mathoverflow.net/users/134512
|
401738
| 164,894 |
https://mathoverflow.net/questions/401736
|
7
|
Let $\mathcal E$ be a topos and $\varphi$ a statement formulating a property of toposes. There are two ways of checking whether $\mathcal E$ satisfies $\varphi$:
1. Consider the first-order language $L$ of a category. Each topos can be considered as an $L$-structure. So, using standard model-theoretic notions, one can consider the satisfaction relation, $\mathcal E\models \varphi$.
2. Using the Kripke-Joyal semantics, one can look at whether $\varphi$ is true in the internal language of $\mathcal E$.
Roughly, one difference between 1. and 2. is that in 1. only the notions "object", "arrow", and "composition" occuring in $\varphi$ are interpreted in $\mathcal E$, while the logic is interpreted on the meta level. However, in 2. also the logic (i.e., the quantifiers $\forall$ and $\exists$ and $\land, \lor, \neg, \dots$) are interpreted in $\mathcal E$.
Strictly speaking, this is *not* an appropriate comparison because in 1. $\varphi$ is a first-order $L$-sentence and in 2. $\varphi$ is a sentence in higher-order logic. However, in spirit 1. and 2. feel similar, because the give a way of looking whether a statement is true in a topos.
*Is there some way of comparing 1. and 2., and can one say anything interesting about this comparison?*
In particular, I wonder whether the following works: given a sentence $\varphi$ in higher-order logic, can one assign to it a sentence $\varphi'$ in first-order logic such that $\varphi$ is true in the internal language of $\mathcal E$ if and only if $\mathcal E\models \varphi'$ in the model-theoretic sense? (And what about the other way round?)
|
https://mathoverflow.net/users/338895
|
Comparing Kripke-Joyal semantics of toposes to model-theoretic satisfaction
|
As you observe yourself, the question does not quite make sense as $\phi$ in 1. is a formula in the first order language of a category and in 2. $\phi$ is a formula in higher order logic (something like the Mitchel-Benabou language).
The only framework I can think of where this question makes perfect sense is if you interpret "internal logic" in the sense of Mike Shulman "Stack semantics" as in his paper <https://arxiv.org/abs/1004.3802>.
Here you interpret "the first-order language of categories" as in def. 3.1 of the paper linked above, and the stack semantics is an extention of the Kripke-Joyal semantics that is also formulated in that language (where the higher order part come from the fact that you can quantify on objects).
In this case, as pointed out by Andrej Bauer, the Kripke-Joyal semantics (or rather its extention called the stack semantics) turns a formula $\phi$ in this language to a formula $\phi'$, so that "$\phi$ hold internally in $\mathcal{E}$" if and only if $\mathcal{E}$ satisfies $\phi'$).
Of course $\phi'$ is in general different from $\phi$, though I believe $\phi'' = \phi'$.
In any case, not every formula arise as a $\phi'$ (i.e. corresponds to an "internal property"). For example, the validity of such formulas are always local properties: if $\mathcal{E}$ satisfies $\phi'$ then every slice $\mathcal{E}/X$ also satisfies $\phi'$; and conversely if $\mathcal{E}/X$ satisfies $\phi'$ for a family of objects covering the terminal, then $\mathcal{E}$ also satisfies $\phi'$.
|
7
|
https://mathoverflow.net/users/22131
|
401742
| 164,896 |
https://mathoverflow.net/questions/401761
|
0
|
Let $X$ be a separable space and let $x^{\*\*}\in X^{\*\*}$. If $x^{\*\*}(x^{\*}\_{n})\rightarrow 0$ for each weak\*-null sequence $(x^{\*}\_{n})\_{n}$ in $X^{\*}$, is $x^{\*\*}$ in $X$ ?
Thank you!
|
https://mathoverflow.net/users/41619
|
weak*-null sequences in the dual space of a separable space
|
Yes, because $(B\_{X^\*},w^\*)$ is compact metrizable. So $x^{\*\*}$ is $w^\*$-continuous at any $x^\*\in B\_{X^\*}$.
|
2
|
https://mathoverflow.net/users/76412
|
401762
| 164,897 |
https://mathoverflow.net/questions/401755
|
3
|
Functional version of the counting hierarchy is $FCH$. It is an open problem whether there a sequence of $poly(log(n))$ number of $+,\times$ operations utilizing the assistance of $O(1)$ number of constants and arbitrary number of integer variables to compute $n!$.
>
> In terms of output size $n!$ is not even in polynomial in $log(n)$ space. So in terms of Boolean complexity where is the computation of $n!$ in? It is clearly not in $FCH$ but in $FEXPSPACE$.
>
>
>
>
> If we are given a prime $p$ is the computation of $n!\bmod p$ in $FCH$?
>
>
>
>
> Will the answers change much if there is a sequence of $poly(log(n))$ number of $+,\times$ operations utilizing the assistance of $O(1)$ number of constants and arbitrary number of integer variables to compute $n!$?
>
>
>
A sequence of arithmetic operations is referred as a straightline program.
|
https://mathoverflow.net/users/10035
|
Is factorial computation known to be in a class smaller than $FEXP$?
|
Yes, $n!\bmod p$ is computable in FCH. More generally, if $f$ is a polynomial-time computable function, then given $n$ and $m$ in binary, we can compute
$$\prod\_{i<n}f(i)\bmod m\tag1$$
in FCH. This follows from the fact that if we are given in *unary* $n$, $m$, and a sequence of numbers $a\_0,\dots,a\_{n-1}$, then we can compute
$$\prod\_{i<n}a\_i\bmod m$$
in uniform $\mathrm{TC}^0$, which was proved by
>
> William Hesse, Eric Allender, and David A. Mix Barrington: *Uniform constant-depth threshold circuits for division and iterated multiplication*, Journal of Computer and System Sciences 65 (2002), no. 4, pp. 695–716, doi [10.1016/S0022-0000(02)00025-9](https://doi.org/10.1016/S0022-0000(02)00025-9).
>
>
>
In fact, we can avoid almost all the intricate machinery of the [HAB02] paper due to a combination of two factors:
* When the algorithms are exponentially scaled to FCH, we only need *quasipolynomial* $\mathrm{TC}^0$. Thus, for example, we can use the trivial polynomial-time algorithm for modular exponentiation by repeated squaring instead of the [HAB02] algorithm (which actually puts it in the linear-time hierarchy).
* Since the result is computed modulo $m$, we don’t need the full force of Chinese Remainder Reconstruction (which is the most complicated and most costly step in [HAB02]).
So, here is an explicit algorithm. First, if $m=p$ is *prime*, we have
$$\prod\_{i<n}f(i)\equiv g^{\sum\_{i<n}d(f(i))}\pmod p,$$
where $g$ is a generator of $\mathbb F\_p^\times$, and $d$ is the inverse of $g^x\bmod p$ (i.e., discrete logarithm). (Let’s consider that $d(0)=-\infty$.) We can compute $g$ and $d(f(i))$ in FPH, thus $\sum\_id(f(i))$ in $\mathrm{\#P^{PH}}$, thus the final result in
$$\mathrm{FP^{\#P^{PH}}=FP^{\#P}=FP^{PP}}.$$
(I’m using here the fact that $\mathrm{\#P^{PH}\subseteq P^{\#P}}$.)
A similar argument works when $m$ is a prime power.
For general $m$, we can compute the prime factorization $m=\prod\_{j<k}p\_j^{e\_j}$ in FPH, thus (for each $j<k\le\log n$) $r\_j=\prod\_{i<n}f(i)\bmod p\_j^{e\_j}$ (or rather, the pair $(p\_j^{e\_j},r\_j)$) in $\mathrm{FP^{\#P^{PH}}=FP^{PP}}$ as above, and we can reconstruct the result modulo $m$ in polynomial time. Thus again, the overall complexity is that we can compute (1) in
$$\mathrm{FP^{PP}}.$$
---
The non-modular function $n!$ as such is an exponential-output-size function whose bit-graph is computable in CH (again, by [HAB02]). There is no common name for this class of functions, as far as I am aware. It is included in FPSPACE, as long as you make sure not to artificially restrict this class to functions with polynomial output size (for space classes, it is a standard definition that space usage only counts work tapes, not the read-only input tape or the write-only output tape).
|
3
|
https://mathoverflow.net/users/12705
|
401767
| 164,899 |
https://mathoverflow.net/questions/238060
|
7
|
It is known that there are characterizations of weak compactness in most of classical non-reflexive spaces (e.g. $L\_{1}$-spaces and $C(K)$-spaces). I wonder whether there are characterizations of weak compactness in James space $J$ or its dual $J^{\*}$. Can we establish a criterion for it if there is no? Thank you!
|
https://mathoverflow.net/users/41619
|
Weak compactness in the James space and its dual
|
James space is a commutative Banach algebra with pointwise operations. $J^{\ast\ast} = J\oplus\mathbb{C}$ is just $J$ with a unit attached [https://doi.org/10.4153/CJM-1980-083-7].
Second, since $J$ contains no copy of $\ell^1$, every bounded sequence $(x\_n)$ in $J$ has a weakly Cauchy subsequence, say $(u\_n)$, that converges weak\* in $J^{\*\*}$ to an element $x+\lambda 1\in J^{\*\*}$. If $\lambda\neq 0$, then let $e\_n = \lambda^{-1}(u\_n-x)$ so that $(e\_n)$ is a bounded approximate identity for $J$. If a subset $C\subseteq J$ is not weakly compact, then it contains a weakly Cauchy sequence $(u\_n)$ with $\lambda\neq 0$.
Consequently, a subset $C\subseteq J$ is weakly compact if and only if $y+\lambda C$ contains no (sequential) bounded approximate identity for all $y\in J$ and all $\lambda\in\mathbb{C}\backslash\{0\}$.
**About the weak$^\ast$ convergence in $J^{\ast\ast}$:** The spectrum $\sigma(J)=\{\delta\_n:n\in\mathbb{N}\}$ consists of coordinate functionals, defined by $\delta\_n(x\_m)\_{m\in\mathbb{N}} = x\_n$ [Proposition 2.7 in https://doi.org/10.4153/CJM-1980-083-7]. Furthermore, $J^{\ast}$ is the closed linear span of $\sigma(J)$ since $J$ is semisimple, $J$ is an ideal (in fact, maximal) in $J^{\ast\ast}$, and $J$ has a bounded approximate identity. Consequently, weak\* convergence in $J^{\ast\ast}$ is just the pointwise convergence.
Thus, $C\subseteq J$ is weakly compact if and only if $C$ is closed under the pointwise convergence of sequences, i.e., whenever $(x\_n)\subseteq C$ is a sequence such that $x\_n\to x$ pointwise, then $x\in C$.
|
0
|
https://mathoverflow.net/users/164350
|
401770
| 164,901 |
https://mathoverflow.net/questions/401775
|
12
|
Let $G$ be a real Lie group.
What conditions must $G$ satisfy so that the following is true:
>
> For any finite group $\Gamma$ there exist finitely many conjugacy classes of subgroups of $G$ that are isomorphic to $\Gamma$.
>
>
>
I believe that for $G=GL(n,\mathbb{R})$ this is true because:
subgroups of $GL(n,\mathbb{R})$ are conjugate if and only if the restriction of the standard representation of $GL(n,\mathbb{R})$ to the subgroups are isomorphic representations.
Finite groups have finitely many irreducible representations, and that proves the claim.
I also believe that for $G=SO(n)$ this is true because of this answer: <https://mathoverflow.net/a/17074/164084>.
It would be enough for my application to know that every real, compact, connected Lie group that has a faithful representation has the property stated above ("For any finite group $\Gamma$...").
This is a crosspost from Stackexchange Mathematics, see here: <https://math.stackexchange.com/questions/4219091/which-lie-groups-have-finitely-many-conjugacy-classes-of-subgroups-of-fixed-isom>.
|
https://mathoverflow.net/users/164084
|
Which Lie groups have finitely many conjugacy classes of subgroups of fixed isomorphism type?
|
A natural condition is that $G$ has finitely many connected components.
One can easily reduce this case to the connected group case, and then to the compact group case, as all [maximal compact subgroups](https://en.wikipedia.org/wiki/Maximal_compact_subgroup) in a connected Lie group are conjugated.
Then the representation variety $\text{Hom}(\Gamma,G)$ is compact and [local rigidity](https://en.wikipedia.org/wiki/Local_rigidity), that is vanishing of $H^1(\Gamma,\mathfrak{g})$, guarantees the finiteness of the number of $G$-orbits. Here $\mathfrak{g}$ denotes the Lie algebra of $G$. The fact that $H^1$ vanishes could be deduced from the fact that every isometric action of $\Gamma$ on $\mathfrak{g}$ has a fixed point, by averaging an orbit.
|
13
|
https://mathoverflow.net/users/89334
|
401777
| 164,904 |
https://mathoverflow.net/questions/401774
|
0
|
**Motivation.** My eldest son starts school tomorrow. His class is split in two groups of $10$ students each. From time to time, the groups are rearranged. I wondered how many rearrangements are needed until every student has been in the same group with every other student at least once. This led to a more general question.
**Question.** $f: \omega \to \{0,1\}$ is said to be *fair* if the preimages $f^{-1}(\{0\})$ and $f^{-1}(\{1\})$ are both infinite. Is there a positive integer $n$ and fair functions $f\_i: \omega \to \{0,1\}$ for $i \in \{0,\ldots,n-1\}$ with the following property?
>
> For all $a,b\in \omega$ there is $i \in \{0,\ldots,n-1\}$ such that $f\_i(a) = f\_i(b)$.
>
>
>
If yes, how small can $n$ be?
|
https://mathoverflow.net/users/8628
|
Putting $\omega$ in two boxes
|
Let $n=3$ and partition $\omega$ into three infinite sets, $X\_0, X\_1, X\_2$. The functions $f\_0,f\_1,f\_2$ defined by $$f\_i(x) = \begin{cases} 1 & \mbox{if $x$ belongs to } X\_i \\ 0 & \mbox{otherwise} \end{cases}$$ are fair functions, and for all $a, b \in \omega$ there is $i \in \{0,1,2\}$ such that $f\_i(a) = f\_i(b) = 0.$
On the other hand, this is impossible when $n=2$. If $f\_0,f\_1$ are any two fair functions from $\omega$ to $\{0,1\}$, assume without loss of generality that $f\_0(0) = f\_1(0) = 0$ and let $a,b \in \omega$ be such that $f\_0(a)=1$ and $f\_1(b)=1$. There must be $i \in \{0,1\}$ such that $f\_i(a)=f\_i(b)$. However, if $i=0$ then we have $f\_0(b) = f\_1(b) = 1$ so there is no $j \in \{0,1\}$ with $f\_j(0) = f\_j(b)$, while if $i=1$ then we have $f\_0(a) = f\_1(a) = 1$ so there is no $j \in \{0,1\}$ with $f\_j(0) = f\_j(a)$.
|
6
|
https://mathoverflow.net/users/8049
|
401781
| 164,905 |
https://mathoverflow.net/questions/401571
|
5
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Let $X$ be a compact complex manifold in Fujiki class $\mathcal C$, that is bimeromorphic to a compact Kähler manifold, let $T$ be a Kähler current of $X$, then we have the De Rham class $[T]\in H^{1,1}(X,\mathbb R)$, pick a smooth form $\tau$ in the same class as $[T]$, then does the wedge map $\tau^q\wedge :H^0(X,\Omega^{n-q})\to H^q(X,\Omega^n)$ induces a surjective map?
This theorem is already known for the Kähler case, for a compact Kähler manifold $X$, replace here $\tau$ by a Kähler form $\omega$, then the wedge map $\omega^q\wedge:H^0(X,\Omega^{n-q})\to H^q(X,\Omega^n)$ induces a surjective map. What if we generalize the Kähler case to Fujiki class $\mathcal C$？ do we have a similar conclusion as stated above? Does anyone knows any reference about this problem?
Added:from [Ang14](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.744.4158&rep=rep1&type=pdf), page7 theorem 0.10, we know
>
> For a compact manifold $X$ endowed with a symplectic structure $\omega$, $X$ satisfies the hard Lefschetz conditon and $X$ satisfies the $dd^{\wedge}$-lemma (namely, every $d$-exact $d^{\wedge}$-closed form being $dd^{\wedge}$-exact) are equivalent.
>
>
>
see the same page for the defintion of $d^{\wedge}$. So this provides a special case of our question, that is if we further assume $\tau$ being a symplectic form and the $dd^{\wedge}$-lemma is satisfied, the map $\tau^q\wedge :H^0(X,\Omega^{n-q})\to H^q(X,\Omega^n)$ is a surjective map.
But our original problem remained open.
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https://mathoverflow.net/users/99826
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Hard Lefschetz theorem for non-Kähler manifolds
|
I think that the answer to your question is no.
In order to construct a counter-example, the idea is the following: let $T$ be a Kahler current, and $E$ a $d$-closed positive current on a compact complex $3$-fold $X$ such that $T^3>0$ and $E^3<0$. Then, for $t>0$, the current $S\_t:=T+tE$ is a Kahler current. Under the above conditions, you can find $t>0$ such that $S\_t^3=0$. Indeed, $S\_t^3=T^3+3tT^2E+3t^2TE^2+t^3E^3$ and for $t=0$, $S\_0^3=T^3>0$ while for $t\to \infty$, $S\_t^3<0$, so there is a $t>0$ so that $S\_t^3=0$.
Now, for this $t$, denote by $\tau\_t$ a smooth representative of the class of $S\_t$. Then $\tau\_t^3=0$. Now for instance, for $q=n=3$, the map $\tau\_t^3\wedge:H^0(X,\Omega^0)\to H^3(X,\Omega^3)$ is the zero map, while $H^0(X,\Omega^0)$ and $H^3(X, \Omega ^3)$ are $1$-dimensional.
In order to fulfill these conditions, you can take $Y$ to be Hironaka's example, it is a modification of $CP^3$, so there is a map $p:Y\to CP^3$, and denote by $\omega$ the FS metric on $CP^3$
Then the $X$ mentioned above is the blow up of $Y$ at a point $x$, denote by $\pi:X\to Y$ the blow-up. If $E$ is the exceptional divisor, then $[E]$ is a positive current, and $E^3=-1<0$, and $T$ is $\pi^\*p^\*\omega+\varepsilon S$, where $S$ is some Kahler current on $X$ ans $\varepsilon$ is such that $T^3>0$. Then, as mentiond above, you can find $t>0$ such that $T+t[E]$ has zero self-intersection.
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1
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https://mathoverflow.net/users/48958
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401799
| 164,909 |
https://mathoverflow.net/questions/401794
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6
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I have been reading about Puiseux series in the context of the Newton–Puiseux algorithm for resolution of singularities of algebraic curves in $\mathbb{C}^2$. Given a curve $f(x,y)=0$ with $f$ a convergent power series and such that the curve has a singularity at the origin, the algorithm produces convergent Puiseux series $\varphi\_1, \dotsc, \varphi\_m$ verifying $f(x,y) = (y-\varphi\_1(x))\dotsm(y-\varphi\_m(x))$.
All of the books I have consulted start by constructing the Puiseux series as a formal power series and then invoking the Implicit Function Theorem to show that there exist analytic solutions to the problem, and therefore the constructed series must be convergent (in a small enough neighbourhood of the origin). Then they continue by studying the properties of the singularity using $\varphi\_1, \dotsc, \varphi\_m$ as parametrizations of the curve.
Here is where I have a problem. Say for instance that one of the resulting Puiseux series is $\varphi\_1(x) = x^{3/2} + x^{7/4}$. This is not a well-defined complex function, is it? For $x=1$ we could have $x^{3/2}$ equal to $1$ or $-1$, and $x^{7/4}$ equal to $\pm1$ or $\pm i$, so there are eight different determinations of this particular Puiseux series. In general, if the Puiseux series has infinitely many terms, there would be infinitely many determinations.
As far as I understand, only one of these determinations of $\varphi\_1$ is the actual function you want to consider, which is the function that the Implicit Function Theorem gives you — but because of the nature of the proof, you can't really know which one it is! It is not clear to me how do these books account for this problem, but their treatment of the series is as if they were well-defined functions (see for instance [Brieskorn–Knörrer](https://doi.org/10.1007/978-3-0348-5097-1)).
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https://mathoverflow.net/users/206706
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How to treat Puiseux series as functions?
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One standard way to bring actual functions in the picture is the following formulation of the existence + convergence results of Puiseux roots: "Given an irreducible power series $f \in \mathbb{C}[[x, y]]$ which is a [Weirstrass polynomial](https://en.wikipedia.org/wiki/Weierstrass_preparation_theorem) in $y$ of degree $d$ (i.e. $f = y^d + \sum\_{i=1}^d f\_i(x)y^{d-i}$ with $f\_i(x) \in x\mathbb{C}[[x]]$), there is $\phi(t) \in \mathbb{C}[[t]]$ such that
$$f(t^d, y) = \prod\_{i = 1}^d (y - \phi(\zeta^it))$$
where $\zeta$ is a $d$-th primitive root of $1$. In addition, and here is where you have actual functions, if $f$ is convergent, then so is $\phi$."
Added later (prompted by OP's comment about choices for $\phi$): in this formulation the uniqueness of Puiseux roots is the statement that if $\psi(t)$ is any other power series such that $y - \psi(t)$ divides $f(t^d, y)$, then $\psi(t) = \phi(\zeta^i t)$ for some $i$. In other words, there are precisely $d$, *not* infinitely many, Puiseux roots of $f$, namely $\phi(\zeta^it)$, $i = 1, \ldots, d$.
This is covered, e.g. in chapter one of [Singularities of Plane Curves](https://doi.org/10.1017/CBO9780511569326) by Casas-Alvero.
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10
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https://mathoverflow.net/users/1508
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401800
| 164,910 |
https://mathoverflow.net/questions/401789
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7
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A well-known conjecture of Berge and Fulkerson says that every bridgeless cubic graph has a collection of six perfect matchings that together cover every edge exactly twice. Is this still open for bridgeless cubic planar graphs?
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https://mathoverflow.net/users/148974
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Berge-Fulkerson conjecture --- the planar case
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The Berge-Fulkerson conjecture holds for planar graphs. Here is a proof.
Let $G$ be a bridgeless cubic planar graph. The dual graph $G^\*$ is a triangulation. By the Four Colour Theorem, $G^\*$ has a 4-colouring $c$. We will use $\mathbb{Z}\_2 \times \mathbb{Z}\_2$ as the set of colours for $c$. Now for each edge $e \in E(G)$, colour $e$ with colour $c'(e):=c(f\_1)+c(f\_2)$ where $f\_1$ and $f\_2$ are the two faces of $G$ incident to $e$. Since $c$ is a proper colouring of $G^\*$, $c'(e)$ is a non-zero element of $\mathbb{Z}\_2 \times \mathbb{Z}\_2$ for all $e \in E(G)$. Moreover, if $v \in V(G)$ and $e\_1, e\_2$, and $e\_3$ are the edges of $G$ incident to $v$, then the dual edges $e\_1^\*, e\_2^\*$, and $e\_3^\*$ are a triangular face $\Delta$ in $G^\*$. Since the vertices of $\Delta$ receive different colours in $c$, $c'(e\_1), c'(e\_2)$, and $c'(e\_3)$ are all distinct. That is, $c'$ is a proper 3-edge colouring of $G$. In other words, $E(G)$ can be partitioned into three perfect matchings. So you can just use each of these perfect matchings twice.
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12
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https://mathoverflow.net/users/2233
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401810
| 164,913 |
https://mathoverflow.net/questions/401606
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2
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Let $G$ be an algebraic group acting on $X$ (a finite type scheme on $k$).
A $G$-invariant $k$-morphism $f : X \rightarrow S$ is a map such that the following commute:
$\require{AMScd}$
\begin{CD}
G \times\_k X @>\rho>> X\\
@V \pi\_2 V V @VV f V\\
X @>>f> S
\end{CD}
Where $\rho$ is the action map and $\pi\_2$ is the projection on the second coordinate.
Every $k$-point $g \in G(k)$ induces a map $\phi\_g : X \rightarrow X$ given by the composition:
$\require{AMScd}$
\begin{CD}
X = \mathop{Spec} k \times\_k X @>{(g, Id)}>> G \times\_k X @>{\rho}>> X
\end{CD}
We now define a second kind of invariance: $f : X \rightarrow S$ is invariant if $f= f \circ \phi\_g$ for all $g \in G(k)$.
Obviously, the first definition implies the second one. **Is it true the inverse implication?**
In my case $k$ is algebraically closed, but I don't know if I need more assumptions. If yes, do you know some counterexamples?
It is true if I suppose $X$ and $G$ reduced, but I hope I can avoid this.
Furtermore: Let's consider the map induced on $k$-points: $ \rho(k) : G(k) \times X(k) \rightarrow X(k)$. Let $k$ be algebraically closed and let $X$, $G$ be reduced. It's well known that $\rho(k)$ determines $\rho$. It is true if I do not assume $G$ to be reduced?
**My real question:**
My question comes from coarse moduli space, in particular from proposition 3.35 of [these notes](https://userpage.fu-berlin.de/hoskins/M15_Lecture_notes.pdf). It isn't clear to me why $\eta\_S(\mathcal{F})$ is $G$-invariant (with reference to pdf notations). The proposition doesn't request that $X$ and/or $G$ are reduced, but it seems that the proof assumes it.
An algebraic group is a group scheme of finite type over $k$.
Every scheme is intended to be of finite type over $k$.
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https://mathoverflow.net/users/175944
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$G$-invariant morphism and coarse moduli spaces
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This is not true without the assumption that $G$ is reduced. Here is a counterexample.
Fix a prime number $p$ and any field $k$ of characteristic $p$. We define the nonreduced algebraic subgroup $\alpha\_p \subset \mathbb{G}\_a$ by $\alpha\_p = Spec(k[t]/(t^p))$ (here $\mathbb{G}\_a$ denotes the additive group over $k$). Then $\alpha\_p$ contains a single $k$-point-- the identity $0$ of the group.
We set $X = \mathbb{G}\_a$. The group $\alpha\_p$ acts on $X$ by addition (include $\alpha\_p$ into $\mathbb{G}\_a$ and then use the group structure).
The identity morphism $id: X \to X$ is not $\alpha\_p$-invariant (the scheme-theoretic image of $\alpha\_p \times 0 \subset \alpha\_p \times X$ is just $0\subset X$ under the projection and $\alpha\_p \subset X$ under the action). However, it is clearly invariant under the unique $k$-point of $\alpha\_p$ (the identity of the group).
If $G$ is reduced and $S$ is separated, then I think that you should be fine (under your assumption that $k$ is algebraically closed).
Edited:
For the last part, you can argue as follows. Consider the morphism
$$ \bigsqcup\_{p \in G(k)} p \to G$$
By the Nullstellensatz, this has schematic image $G$. Using an argument similar to the end of the proof of Prop 3.2.4 (ii) in [http://virtualmath1.stanford.edu/~conrad/249BW16Page/handouts/alggroups.pdf](http://virtualmath1.stanford.edu/%7Econrad/249BW16Page/handouts/alggroups.pdf), you can show by arguing affine locally that the product morphism
$$\bigsqcup\_{p \in G(k)} p \times X \to G \times X$$
has schematic image $G \times X$. (Notice that the original morphism is not quasicompact! so it is not automatic that the schematic image commutes with flat base-change, but in this case it does. This is true when your base is a field because any ring over $k$ is a free $k$-module).
Now, if $S$ is separated, the locus of $G \times X$ where the projection agrees with the action is a closed subscheme $Z \subset G \times X$. By assumption, we have a factorization
$$\bigsqcup\_{p \in G(k)} p \times X \to Z \to G \times X$$
Therefore, by the scheme theoretic density we must have $Z = G \times X$.
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3
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https://mathoverflow.net/users/339730
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401811
| 164,914 |
https://mathoverflow.net/questions/401004
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7
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Let $n \geq 1$ be an integer and consider the symmetric function
$$D\_n = \sum\_{d|n} p\_d^{n/d},$$
where $p\_{d}$ are the power-sum symmetric functions.
It can be checked up to $n=35$ that the symmetric function $D\_n$ is Schur-positive.
The multiplicity of the Schur function $s\_n$ in $D\_n$ is given by the number of divisors of $n$ (sequence [A000005](https://oeis.org/A000005) in the OEIS).
One can also easily compute the multiplicity of the Schur function $s\_{1^n}$ in $D\_n$:
$$\sum\_{d|n} (-1)^{n+d},$$
which is sequence [A112329](https://oeis.org/A112329) in the OEIS.
>
> Q1: Is the Schur-positivity of $D\_n$ for all $n$ known ?
>
>
>
>
> Q2: Has this been considered somewhere in the literature ?
>
>
>
I have alas not much more context for this.
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https://mathoverflow.net/users/10881
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About the sum of rectangular power sums
|
Talking to Sheila Sundram, as suggested in the comments, was a good idea. After some conversation, the following proof became apparent. I don't know of any proof already in the literature.
Let $g$ be an $n$-cycle in $S\_n$. Since the inverse Frobenius characteristic of the given symmetric function is supported only on classes that intersect $\langle g \rangle$ nontrivially, one might hope that it is induced from an actual character of $\langle g \rangle$, and this turns out to be the case.
For each divisor $d$ of $n$, write $\rho\_d$ for the linear character of $\langle g \rangle$ sending $g$ to $e^{2 \pi i/d}$, and consider the induced character $\theta\_d:=\rho\_d\uparrow\_{\langle g \rangle}^{S\_n}$.
For positive integers $a$,$b$, we write $c\_a(b)$ for the sum of the $b^{th}$ powers of all of the primitive complex $a^{th}$ roots of $1$ (a Ramanujan sum).
If $w \in S\_n$ does not have cycle type $m^{n/m}$ for some divisor $m$ of $n$, then $\theta\_d(w)=0$. If $w$ has shape $m^{n/m}$, a direct computation shows that $$\theta\_d(w)=\frac{1}{n}|C\_{S\_n}(w)|c\_{n/d}(m).$$
It follows that the Frobenius characteristic of $\theta\_d$ is
$$ \frac{1}{n}\sum\_{m|n}c\_{n/d}(m)p\_m^{n/m}.$$
So, it suffices to show that there are nonnegative integers $\alpha\_d$ ($d|n$) such that
$$
\sum\_{d} \alpha\_dc\_{n/d}(m)=n
$$
for all divisors $m$ of $n$.
We write $S(n)$ for the matrix with rows and columns indexed by divisors of $n$ with entry $c\_{n/d}(m)$ in the position indexed by $n$ and $d$, and $\alpha$ for the column vector with rows indexed as are the rows of $S(n)$ and entry $\alpha\_d$ in the position indexed by $d$.
It suffices now to show that $S(n)$ is invertible and that the row sums of $nS(n)^{-1}$ are nonnegative integers, as these row sums are the entries of $\alpha$.
We can appeal to results found in the paper ``Ramanujan sums as supercharacters", by C. F. Fowler, S. R. Garcia and G. Karaali. According to Theorem 3.5 of that paper, if one orders the row and column indices appropriately, one gets $S(n)^2=nI$. So, the row sums of $nS(n)^{-1}$ are the row sums of $S$. A special case of Theorem 4.5 of the same paper implies that the row sums of $S$ are non-negative.
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3
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https://mathoverflow.net/users/36466
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401815
| 164,916 |
https://mathoverflow.net/questions/401822
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5
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Assume $f(x)$ is a smooth function on $\mathbb{R}$ and $f$ does not vanish on any interval. In other words, $f$ can have zero points but we cannot find any interval $(a, b)$ such that $f(x)=0$ for all $x \in (a, b)$. Denote by $\mathcal{Z} = \{x \in \mathbb{R}, f(x) = 0\}$ the zero point set of $f$. Assume $\mathcal{Z}$ is non-empty.
**Question:** Is it possible that every point in $\mathcal{Z}$ is an accumulation point of $\mathcal{Z}$?
Here $x$ is an accumulation point of $\mathcal{Z}$ means there exists a sequence $\{x\_n\}$ such that $x\_n \in \mathcal{Z}$, $x\_n \neq x$ and $\lim\_{n\to \infty}x\_n = x$ under the usual Euclidean topology.
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https://mathoverflow.net/users/114951
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Zero points of a smooth function on $\mathbb{R}$
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The answer is yes.
We can make a smooth function whose zero set is the Cantor set. Simply place a smooth function on each middle third segment as you build the Cantor set, so that on each of these segments, the function arches from 0 up to a height that vanishes quickly as the segments become small, and then back down to 0, but smooth, and with all derivatives vanishing on the endpoints.
The zero set of this function will be the Cantor set itself, which is a closed nowhere dense set in which every point is an accumulation point. The function is smooth on each middle-third segment, and it is smooth at the points of the Cantor set, since the heights of the pieces on the segments approaching it vanish quickly enough.
In fact, this method can show that every closed set in the reals is the zero set of a smooth function. One places a little smooth bump into each open interval of the complement, and by making the heights sufficiently low as the intervals become small, you can get smoothness at the limits of these bumps.
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11
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https://mathoverflow.net/users/1946
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401823
| 164,918 |
https://mathoverflow.net/questions/400948
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9
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In the section 3.2 of *Sheaves in Topology* by A. Dimca, the author explains that if $f:X\to Y$ is a continuous map (between locally compact, $\sigma$-compact topological spaces with finite homological dimension) such that $f\_!$ has finite cohomological dimension, then the following holds:
>
> **(Verdier duality, local form)** There is an additive functor of triangulated categories $f^!:\mathsf{D}^+(Y)\to \mathsf{D}^+(X)$ such that there is a functorial isomorphism
> $$\mathsf{R}\underline{\operatorname{Hom}}^\bullet(\mathsf{R}f\_! \mathscr{F}^\bullet,\mathscr{G}^\bullet)\cong \mathsf{R}f\_\*\mathsf{R}\underline{\operatorname{Hom}}^\bullet(\mathscr{F}^\bullet,f^!\mathscr{G}^\bullet)$$
> in $\mathsf{D}^+(Y)$ for any $\mathscr{F}^\bullet\in\mathsf{D}^b(X)$ and $\mathscr{G}^\bullet\in\mathsf{D}^+(Y)$.
>
>
>
Do we really need all those hypotheses? Perhaps we can use Brown's representability theorem to prove it under more general conditions for the unbounded derived category? (There's a post here on MO about using this theorem but there it is under *less* general conditions.)
**Edit:** Let me be clear about my "proposed" proof. Let $f:X\to Y$ be a morphism of (locally compact) ringed spaces and $\mathsf{D}(X),\mathsf{D}(Y)$ be their derived categories of modules. The tag 0F5Y on the Stacks Project implies that any triangulated functor $\mathsf{D}(X)\to\mathsf{D}(Y)$ which preserves direct sums has a right adjoint.
If we could prove that $\mathsf{R}f\_!$ preserves infinite direct sums, then we would conclude the existence of a functor $f^!:\mathsf{D}(Y)\to\mathsf{D}(X)$ such that
$$\hom\_{\mathsf{D}(Y)}(\mathsf{R}f\_!\mathscr{F}^\bullet,\mathscr{G}^\bullet)\cong \hom\_{\mathsf{D}(X)}(\mathscr{F}^\bullet,f^!\mathscr{G}^\bullet)$$
naturally in $\mathscr{F}^\bullet$ and $\mathscr{G}^\bullet$. This yields the local form as usual (for example, prop. 3.1.10 in Sheaves in Manifolds).
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https://mathoverflow.net/users/131975
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Verdier duality under more general conditions
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The status of the following answer is a bit speculative, unfortunately. To be precise, I believe that all of what I say below is true; I also believe that there is no proof in the literature of some of the things I state below, and I am not enough of an expert to supply those proofs. Nevertheless I'm putting it out there in the hope that it can be helpful.
First of all, when dealing with unbounded complexes of sheaves one runs into the issue of *hypercompleteness*, and I have to say something about this. Let's first do the purely psychological change that instead of complexes of sheaves, we think of sheaves of complexes. Chain complexes are most naturally thought of as an $\infty$-category, once we localize at quasi-isomorphisms, so one is then led to thinking about sheaves valued in an $\infty$-category. Now a sheaf on a space $X$ in a complete 1-category $C$ is is a functor $F \colon \mathrm{Op}(X)^{op} \to C$ such that if $\{U\_i \to U\}$ is an open cover of a subset $U$, then $F(U)$ is the equalizer (limit) of the two arrows $\prod\_i F(U\_i) \to \prod\_{i,j} F(U\_i \cap U\_j)$. If $C$ is an $\infty$-category then the limit must be a homotopy limit, taking higher coherences into account, and then the proper definition of a $C$-valued sheaf turns out to be that the natural map from $F(U)$ to the (homotopy) limit of the cosimplicial diagram which in level $n$ is given by $\prod\_{i\_1,\ldots,i\_n} F(U\_{i\_1} \cap \ldots \cap U\_{i\_n})$, is an isomorphism. This specializes to the usual $1$-categorical sheaf axiom when $C$ is a 1-category.
When we take $C$ to be the $\infty$-category of bounded below cochain complexes modulo quasi-isomorphism, then the $\infty$-category of $C$-valued sheaves on $X$ is an $\infty$-categorical enhancement of the derived category $D^+(X)$. But if we consider instead *unbounded* complexes, the analogous statement is *false*. Namely, the $\infty$-categorical sheaf axiom from the previous paragraph describes what's known as descent for *Cech covers*. To recover the classical unbounded derived category one must instead impose descent for *hypercovers*. In a sense this has been known since the 60's, and traditionally this has been interpreted as meaning that Cech descent produces the "wrong" answer, and hypercovers "correct" this deficiency. After Lurie, a more modern perspective is that Cech descent is for many purposes more natural. In any case, the upshot is that for a space $X$ there are two typically inequivalent notions one can consider: $\infty$-sheaves on $X$ valued in unbounded complexes of sheaves, and hypersheaves on $X$ valued in unbounded complexes. The latter category can be recovered from the former via the process of hypercompletion, and the latter produces an $\infty$-categorical enhancement of the classical unbounded derived category.
Now you mention the condition $(\ast)$ in Spaltenstein's paper, which looks like it is just an annoying technicality, and whether it can be removed using more modern homotopical machinery. I believe instead that the result is just plain false without some condition like $(\ast)$. More specifically I think that if a space $X$ satisfies condition $(\ast)$ then this forces Cech descent and hyperdescent to coincide for abelian sheaves, and that this is fundamentally the reason that $(\ast)$ appears in Spaltenstein's paper. Namely, Spaltenstein's Theorem B concerns the classical unbounded derived category, and I believe that all of these results fail in general when one works with hypersheaves. But a very general version of Spaltenstein's Theorem B should hold if one works with $\infty$-sheaves throughout, with no condition like $(\ast)$ or finiteness.
Here's one reason to believe this. Part of Spaltenstein's Theorem B is proper base change. In Higher Topos Theory, Lurie proves a very general nonabelian version of proper base change, which implies the classical one. Crucially, Lurie's version of proper base change is a theorem for $\infty$-sheaves, not hypersheaves (and he gives an example where proper base change fails for hypersheaves). In particular it implies a form of proper base change for $\infty$-sheaves of unbounded complexes on a space $X$, and *not* for hypersheaves. Now I should add that in Higher Topos Theory Lurie only considers proper morphisms, so he works in the setting where $f\_! = f\_\ast$. So he does not introduce the functors $f\_!$ or $f^!$ to state his result.
Another indication that $(\ast)$ is actually about hypercompleteness is that Spaltenstein remarks that locally finite dimensional spaces satisfy $(\ast)$. But locally finite dimensional things should also be hypercomplete. More precisely, Lurie proves this statement in HTT, if "dimension" is interpreted as "homotopy dimension", a notion that he introduces. For paracompact topological spaces, "homotopy dimension" coincides with "covering dimension". Spaltenstein doesn't elaborate on what notion of dimension he's thinking of but I assume cohomological dimension.
In any case, it is true that higher category theory can be used to give constructions of $f\_!$ and $f^!$, under milder hypotheses than what Spaltenstein uses. Namely, $f\_!$ and $f^!$ should exist for any continuous map between locally compact Hausdorff spaces, for $\infty$-sheaves valued in any complete and cocomplete stable $\infty$-category. Unlike Lurie's proper base theorem which is fully nonabelian (ie works for sheaves of spaces), this part of the story uses stability in a crucial way. In Higher Topos Theory, Lurie proves that on a locally compact Hausdorff space $X$, $\infty$-sheaves on $X$ can be described equivalently in terms of functors taking values on open subsets of $X$, or taking values on compact subsets of $X$. If the target category is moreover stable then this can be used to construct an equivalence of $\infty$-categories between *sheaves* and *cosheaves* on $X$. There is a natural pushforward operation on cosheaves, much like the pushforward of sheaves. Translating the pushforward functor via the sheaf-cosheaf equivalence to an operation on sheaves, one recovers precisely the functor $f\_!$. By the adjoint functor theorem one directly gets $f^!$, too. This is sketched in [https://www.math.ias.edu/~lurie/282ynotes/LectureXXI-Verdier.pdf](https://www.math.ias.edu/%7Elurie/282ynotes/LectureXXI-Verdier.pdf)
In your second question it sort of sounds like you are asking about whether versions of the functors $f\_!$ and $f^!$ can be useful in coherent cohomology (and you are not just asking about Grothendieck duality). Something like this is true in the setting of condensed mathematics. One version of six-functors formalism is in the final chapter of Scholze's "Condensed mathematics", but even closer to what you're looking for is I think the material from Dustin Clausen's final lecture in the Copenhagen Masterclass (you can find it on Youtube), which I will not attempt to summarize.
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8
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https://mathoverflow.net/users/1310
|
401863
| 164,933 |
https://mathoverflow.net/questions/401851
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4
|
This is a follow-up question to an [older question](https://mathoverflow.net/questions/401774/putting-omega-in-two-boxes).
Let $\alpha \in \big(\omega\cup\{\omega\}\big) \setminus \{0,1\}$ be an ordinal. We say that a function $f: \omega \to \alpha$ is *fair* if $$|f^{-1}(\{j\})| = \aleph\_0$$ for all $j\in\alpha$.
We call a set ${\frak F}$ of fair functions $f:\omega\to\alpha$ *equalising for $\alpha$* if for all $a,b\in\omega$ there is $f\in{\frak F}$ with $f(a) = f(b)$. Let $B\_\alpha$ be the minimum cardinality that an set equalising for $\alpha$ can have. [This post](https://mathoverflow.net/a/401781/8628) establishes $B\_2 \leq 3$.
**Question.** Can an explicit value for $B\_\alpha$ be given for $\alpha \in \big(\omega\cup\{\omega\}\big) \setminus \{0,1\}$? I am particularly interested in $B\_\omega$.
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https://mathoverflow.net/users/8628
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Putting $\omega$ into $\alpha$ boxes where $\alpha \in \big(\omega\cup\{\omega\}\big)\setminus\{0,1\}$
|
$B\_\alpha = 3$ for every $\alpha \in (\omega \cup \{\omega\}) \setminus \{0,1\}$. Here is a construction that shows $B\_\alpha \le 3$.
1. Let $g : \mathbb{N} \to \alpha$ be any function such that $g^{-1}(\{j\})$ is infinite for all $j \in \alpha$.
2. Let $h$ be any bijection from $\omega$ to $\{0,1,2\} \times \mathbb{N}$.
3. For $i \in \{0,1,2\}$ define $t\_i : \{0,1,2\} \times \mathbb{N} \to \alpha$ as follows: $$t\_i(r,s) = \begin{cases} g(s) & \text{if } i=r \\ 0 & \text{otherwise} \end{cases}$$
and define $f\_i : \omega \to \alpha$ by $f\_i = t\_i \circ h$.
The functions $f\_i$ are fair. In fact, $f\_i^{-1}(\{j\}) = h^{-1} (t\_i^{-1} (\{j\}))$, which is an infinite set because $h$ is a bijection and $t\_i^{-1}(\{j\})$ contains $\{i\} \times g^{-1}(\{j\})$. The set $\{f\_0,f\_1,f\_2\}$ is equalising for $\alpha$ because for all $a,b \in \omega$, if $h(a) = (r\_a,s\_a)$ and $h(b) = (r\_b,s\_b)$, then for any $i \not\in \{r\_a,r\_b\}$ we have $f\_i(a)=f\_i(b)=0$.
The reasoning used to show $B\_2 > 2$ in the answer to [this question](https://mathoverflow.net/a/401781/8049) extends to show $B\_\alpha > 2$ for all $\alpha > 1$. Hence $B\_\alpha = 3$ for all $\alpha \in (\omega \cup \{\omega\}) \setminus \{0,1\}$.
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7
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https://mathoverflow.net/users/8049
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401874
| 164,938 |
https://mathoverflow.net/questions/351800
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8
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Consider the closure $K \subset \overline{\mathcal{M}}\_{0,n}(\mathbb{P}^1, n)$ in the stable maps space of the locus $K\_0$ of maps $(f : C \to \mathbb{P}^1, p\_1, \ldots, p\_n)$ where $C \cong \mathbb{P}^1$ is smooth and the set of marked points is exactly the preimage of $\infty \in \mathbb{P}^1$; $\{p\_1, \ldots, p\_n\} = f^{-1}(\infty)$. I want to understand the boundary strata $K \setminus K\_0$.
There is an obvious necessary condition for a pointed stable map $(f : C \to \mathbb{P}^1, p\_1, \ldots, p\_n)$ to be in $K$, which is that $f(p\_i) = \infty$ for each $i$. In particular, $K$ is contained in the intersection of the evaluation loci $ev\_i^{-1}(\infty)$ for $i = 1, \ldots, n$ where $ev\_i : \overline{\mathcal{M}}\_{0,n}(\mathbb{P}^1, n) \to \mathbb{P}^1$ are the usual evaluation maps. Let us call this **Condition 0.**
However, this condition is not sufficient. For example there is a stratum paramerizing $f : C\_0 \cup\_p C\_1 \to \mathbb{P}^1$ where the restriction $f\_0 : C\_0 \to \mathbb{P}^1$ is a degree $n$ map from a smooth rational curve with $f\_0(p) = \infty$, $f$ constant on $C\_1$, and all the marked points lying on $C\_1$. This stratum is contained $ev\_i^{-1}(\infty)$ for each $i$ but is larger dimensional than $K$.
The reason is that being in the closure of $K\_0$ imposes some extra conditions on the map. For degree reasons, a map $f : C \to \mathbb{P}^1$ in $K\_0$ must be unramified at all the marked points $p\_i$ but the only way marked points can collide is if the map becomes ramified at $\infty$. This gives us a condition on certain nodes lying over $\infty$ which I think can be phrased as follows:
**Condition 1:** Let $C\_0 \subset C$ is a connected union of components lying in $f^{-1}(\infty)$. Suppose $C\_0$ is attached to components $C\_1, \ldots, C\_k$ at points $x\_1, \ldots, x\_k$ where $f|\_{C\_i}$ is non-constant. Then the number of marked points lying on $C\_0$ is the sum of the ramification of $f|\_{C\_i}$ at $x\_i$.
**Condition 2:** For each component $C\_0 \subset C$ with $f|\_{C\_0}$ non-constant, each point of $f|\_{C\_0}^{-1}(\infty)$ is either a marked point or a node of $C$.
These two conditions are related but its not immediately clear to me exactly how. Note that a general point of the stratum described above not contained in $K$ satisfies neither condition $1$ nor $2$.
**Question 1:** Are conditions $0$, $1$ and $2$ sufficient for a stable map to be contained in the closure of $K$? If not, is there some other description of the boundary strata of $K$?
I think a quick dimension count shows that the dimension of the strata satisfying conditions $0$, $1$ and $2$ have dimension at most the dimension of $K$, and in fact strictly smaller than the dimension of $K$ if you exclude $K\_0$ itself. This gives some evidence that Question 1 has a positive answer. For example, if one knew that the expected dimension of the locus satisfying these conditions is the dimension of $K$, then these dimension bounds would be enough. However, I don't know how to show this.
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https://mathoverflow.net/users/12402
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The closure of a locus in $\overline{\mathcal{M}}_{0,n}(\mathbb{P}^1, n)$
|
It turns out that the answer is yes and these conditions are sufficient.
This question was answered in greater generality by Gathmann ([Absolute and relative Gromov-Witten invariants of very ample hypersurfaces.](https://arxiv.org/abs/math/9908054) *Duke
Math. J.*, 2002, Proposition 1.14) building off of earlier work of Vakil ([The enumerative geometry of rational and elliptic curves in projective space.](http://math.stanford.edu/%7Evakil/files/re4.pdf) *J. Reine Angew. Math.*, 2000, Theorems 4.13 and 6.1).
We wrote a direct proof of the special case considered in the question (namely target $(\mathbb{P}^1, \infty)$) in Section 2 of [this preprint](https://arxiv.org/pdf/2108.05324.pdf).
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1
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https://mathoverflow.net/users/12402
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401888
| 164,942 |
https://mathoverflow.net/questions/401899
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10
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GCH for alephs means the statement that, for any aleph $\kappa$, there are no cardinals $\mathfrak{r}$ such that $\kappa<\mathfrak{r}<2^\kappa$.
Does GCH for alephs imply the axiom of choice?
Remark. Lindenbaum and Tarski assert in ``Communication sur les recherches de la théorie des ensembles'' without proof that GCH for alephs is equivalent to Cantor's aleph hypothesis that $2^{\aleph\_\alpha}=\aleph\_{\alpha+1}$ for all ordinals $\alpha$ (see page 188, No. 96). But as we know (although Lindenbaum and Tarski possibly do not kown) Cantor's aleph hypothesis implies AC. This means that Lindenbaum and Tarski in fact also assert that GCH for alephs implies the axiom of choice. But I do not see how to prove it.
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https://mathoverflow.net/users/101817
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Does GCH for alephs imply the axiom of choice?
|
The answer is positive, yes.
Note that $2^\kappa\leq 2^{\kappa^+}$, and therefore $\kappa^+\leq\kappa^++2^\kappa\leq 2^{\kappa^+}$. So either $2^\kappa=2^{\kappa^+}$ or $2^\kappa=\kappa^+$.
In the first case $\kappa^+$, $\kappa<\kappa^+<2^\kappa$ is impossible. So the latter case holds.
Therefore the power set of an ordinal can be well-ordered, and the Axiom of Choice follows in $\sf ZF$.
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9
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https://mathoverflow.net/users/7206
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401902
| 164,948 |
https://mathoverflow.net/questions/401909
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2
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Working with the level set method introduced by *Osher & Sethian* in shape optimization I came across a simple question that I did not succeed to prove. It mainly asserts that the perimeter of the zero level set is continuous with respect to the level set function. The continuity of the area is true (see here [Lebesgue measure of a neighbourhood of a curve](https://mathoverflow.net/questions/385144/lebesgue-measure-of-a-neighbourhood-of-a-curve)). It is well known that the perimeter is a lower semicontinuous functional, so one inequality is checked.
Here is my question:
Let $\Omega\subset\mathbb{R}^2$ be a bounded set with smooth boundary, $\phi\_n:\overline\Omega\to\mathbb{R},\ n\geq 1$ be a sequence of functions such that:
$\bullet\ \phi\_n\to \phi\ \text{in}\ C^1(\overline{\Omega})$ i.e. $\lim\limits\_{n\to\infty} \sup\_{x\in\overline{\Omega}}|\phi\_n(x)-\phi(x)|+\sup\_{x\in\overline{\Omega}} |\nabla\phi\_n(x)-\nabla\phi(x)|=0$
$\bullet$ $\nabla\phi\_n\neq 0$ on $\phi\_n=0$ and $\nabla\phi\neq 0$ on $\phi=0$
$\bullet$ $\{\phi=0\}\Subset\Omega$
Then how can we prove that:
$$\lim\limits\_{n\to\infty} \int\_{\phi\_n=0} 1 \ d\sigma=\int\_{\phi=0} 1\ d\sigma$$
?
For the continuity of the perimeter on level sets with respect to the level you can see other posts here: [Continuity of Hausdorff measure on level sets](https://mathoverflow.net/questions/398167/continuity-of-hausdorff-measure-on-level-sets), [Continuity of surface integrals on level sets](https://mathoverflow.net/questions/401707/continuity-of-surface-integrals-on-level-sets)
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https://mathoverflow.net/users/61629
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Continuity of the perimeter of level sets w.r.t. level function
|
As it is written it's not true, by trivial reasons: take $\Omega\subset \mathbb{R}^2$ the unit open disk, $\phi(x):=1-|x|^2$ and $\phi\_n(x):=\phi(x)+\frac1n$ for $x\in\overline\Omega$. So $\{x\in\overline\Omega:\phi(x)=0\}$ is the unit circle and $\{x\in\overline\Omega:\phi\_n(x)=0\}$ is empty. The measure of *interior* zeros is neither continuous: if $\psi\_n(x):=\phi(x)-\frac1n$ then $\{x\in \Omega:\phi(x)=0\}$ is empty and $\{x\in \Omega:\psi\_n(x)=0\}$ is a circle of radius $1-o(1).$
You need to add the condition $\phi\neq0$ on $\partial\Omega$, then it is true, just parametrizing the zero set of $\phi\_n$ by $N\ge0$ closed curves $\{\gamma\_{j,n}:\mathbb{S}^1\to\Omega\}\_{1\le j\le N}$ converging in $C^1$ respectively to $N$ curves $\gamma\_{j,n}:\mathbb{S}^1\to\Omega$, that parametrize the zero set of $\phi$. Here $N\ge0$ is the number of connected components of the zero set of $\phi$ and $\phi\_n$, which is finite and definitively constant wrto $n$. Then the length of each component $\int\_{\mathbb{S}^1}|\dot\gamma\_{j,n}(s)|ds$ passes to the limit.
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6
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https://mathoverflow.net/users/6101
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401916
| 164,953 |
https://mathoverflow.net/questions/401760
|
14
|
The quantum harmonic oscillator relies on two classical objects, the so-called creation and annihilation operator
$$a ^\* = x- \partial\_x \text{ and }a = x+\partial\_x.$$
Fix two numbers $\alpha,\beta \in \mathbb R.$
Can we explicitly compute the spectrum of
$$H = \begin{pmatrix} aa^\* + \alpha^2 & \alpha a+ \beta a^\* \\ \alpha a^\* + \beta a & aa^\* +\beta^2 \end{pmatrix}?$$
This looks like an innocent problem, but somehow I find it hard to determine what exactly the spectrum of this operator is.
The operator is acting on a suitable domain of $L^2(\mathbb R) \oplus L^2(\mathbb R).$
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https://mathoverflow.net/users/150549
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Spectrum of matrix involving quantum harmonic oscillator
|
The Hamiltonian
$$H=\begin{pmatrix}
\alpha^2+a^\ast a&\alpha a+\beta a^\ast\\
\alpha a^\ast+\beta a&\beta^2+a^\ast a
\end{pmatrix}
$$
is known in the physics literature as the *anisotropic Rabi Hamiltonian*. (In the most general case there is an additional term $\Delta\sigma\_z$.) I give some pointers to the literature in this [Physics SE posting](https://physics.stackexchange.com/q/660360/38462). The eigenvalues can be computed from a recursive scheme, but closed-form expressions for the spectrum only exist for either $\alpha=\beta$ or $\alpha\beta=0$.
$\bullet$ Consider first the case $\alpha=\beta$. A unitary transformation $H'=UHU^\ast$ with $U=e^{i\pi\sigma\_y/4}$ brings the Hamiltonian to the diagonal form
$$H'=\begin{pmatrix}
b\_+^\ast b\_+&0\\
0&b\_-^\ast b\_-
\end{pmatrix},\;\; b\_\pm=a\pm\alpha.
$$
The eigenvalues are the integers $N=0,1,2,\ldots$, each twofold degenerate. The corresponding eigenstates $|N,\pm\rangle$ are obtained from the eigenstates $|N\rangle$ of the harmonic oscillator by acting on these with the displacement operator,
$$| N,\pm\rangle=e^{\pm\alpha(a-a^\ast)}|N\rangle.$$
$\bullet$ At the other extreme, we can take one of the two parameters $\alpha, \beta$ much smaller than the other. Let me set $\beta=0$, $\alpha\neq 0$. (The spectrum is the same for $\alpha=0$, $\beta\neq 0$.) The Hamiltonian
$$H=\begin{pmatrix}
\alpha^2+a^\ast a&\alpha a\\
\alpha a^\ast&a^\ast a
\end{pmatrix}=(\tfrac{1}{2}\alpha^2+a^\ast a)I+\alpha(\sigma\_+a+\sigma\_-a^\ast)+\tfrac{1}{2}\alpha^2\sigma\_z$$
is known in physics as the [Jaynes-Cummings Hamiltonian.](https://en.wikipedia.org/wiki/Jaynes%E2%80%93Cummings_model)
The eigenvalues can be computed exactly, because the Hamiltonian decomposes into an infinite direct product $H\_n$ of $2\times 2$-matrix Hamiltonians in the basis $|g,n+1\rangle$ (two-level system in the lower state, $n+1$ quanta excited in the oscillator) and $|e,n\rangle$ (two-level system in upper state, $n$ quanta excited):
$$H\_n\begin{pmatrix}
|g,n+1\rangle\\
|e,n\rangle
\end{pmatrix}=
\begin{pmatrix}n+1&\alpha\sqrt{n+1}\\
\alpha\sqrt{n+1}&n+\alpha^2
\end{pmatrix}
\begin{pmatrix}
|g,n+1\rangle\\
|e,n\rangle
\end{pmatrix}.$$
The eigenvalues then follow directly,
$$\Omega\_{n,\pm}=\tfrac{1}{2}(\alpha^2+1)+n\pm\tfrac{1}{2}\sqrt{(\alpha^2+1)^2+4\alpha^2 n}.$$
This sequence of eigenvalue pairs $\Omega\_{n,+},\Omega\_{n,-}$ exists for $n\in\{0,1,2,\ldots\}$. In addition, there is an unpaired ground state $|g,0\rangle$ without any excitations, which is annihilated by $H$ so it has eigenvalue $\Omega\_{0}=0$ independent of $\alpha$.
$\bullet$ The general case of arbitrary $\alpha,\beta$ does not have a simple closed form expression for the eigenvalues. For $\alpha,\beta\ll 1$ a perturbative solution is given in Appendix C of [arXiv:1008.1317](https://arxiv.org/abs/1008.1317).
The ground state does have a simple exact result: *the lowest eigenvalue equals zero for any $\alpha,\beta\in\mathbb{R}$.* The zero-mode – the eigenstate which is annihilated by $H$ – is a coherent state $|\xi\rangle=e^{\xi(a^\ast-a)}|0\rangle$ for the harmonic oscillator (such that $a|\xi\rangle=\xi|\xi\rangle$). Substitution of $H{p\choose q}\otimes|\xi\rangle=0$ gives the two zero-modes
$$\Psi\_\pm={{\sqrt\beta}\choose{\mp\sqrt\alpha}}\otimes|\!\pm\!\!\sqrt{\alpha\beta}\rangle.$$
(Thanks to Michal Pacholski for helping me with this.)
---
Notation: $I$ is the $2\times 2$ unit matrix, $\sigma\_x$, $\sigma\_y$, $\sigma\_z$ are the three Pauli matrices, and $\sigma\_\pm=\tfrac{1}{2}(\sigma\_x\pm i\sigma\_y)$. Also please note that the eigenvalues in the OP are shifted by one unit relative to those given here, because I reordered $aa^\ast\mapsto a^\ast a$.
---
Since there was an issue with the appearance of spurious "parabolas" in the comparison with the numerics by yarchik (in a now deleted answer), I show here a numerical check of the spectrum as a function of $\alpha$ for $\beta=0$, using [a slightly modified version](https://ilorentz.org/beenakker/MO/yarchik.nb) of yarchik's code.
Solid curves are the numerical result, dashed and dotted curve are $\Omega\_{n,\pm}$. Note that the ground state is at zero energy for any $\alpha$ when $\beta=0$. The plot below, for $\beta=2\alpha$ shows this also holds for nonzero $\beta$.
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14
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https://mathoverflow.net/users/11260
|
401927
| 164,955 |
https://mathoverflow.net/questions/401921
|
3
|
I was reading today the book of Stephen Wiggins called "Global Bifurcations and Chaos" (the 1988 edition).
On pages 12-13 he writes the following:
>
> Consider the following ordinary differential equation $$\dot{\theta}\_1=\omega\_1 \ \ \ \dot{\theta}\_2=\omega\_2 \ \ \ \ \theta\_i\in (0,2\pi] \ \ \forall i\in\{ 1,2\}$$ where $\omega\_1,\omega\_2$ positive constants. Since $\theta\_1,\theta\_2$ are two angular variables the phase space of the above ode is $\mathbb{S}^1\times\mathbb{S}^1=\mathbb{T}^2$. If we draw the torus as the surface of donut in $\mathbb{R}^3$ the orbits of this ode spiral around the surface and close (i.e, they are periodic) when $\omega\_1/\omega\_2$ is a rational; alternatively they densely fill the surface when $\omega\_1/\omega\_2$ is irrational, see Arnold [1973] for a detailed proof of these statements.
>
>
>
The 1973 book is the ODE book; I cannot find a suitable copy, but I have the 1992 copy — does someone know where in this book the proof of these statements appears?
on which page?
Thanks!
|
https://mathoverflow.net/users/13904
|
Searching for the proof of a certain claim in Arnold's ODE book from 1992
|
See at p. 163 of this PDF link:
[https://eclass.uoa.gr/modules/document/file.php/PHYS289/Βιβλία/Arnold%2C%20V.I.%20-%20Ordinary%20differential%20equations\_Red.pdf](https://eclass.uoa.gr/modules/document/file.php/PHYS289/%CE%92%CE%B9%CE%B2%CE%BB%CE%AF%CE%B1/Arnold%2C%20V.I.%20-%20Ordinary%20differential%20equations_Red.pdf)
Maybe you got confused because between the statement and the proof of the proposition you are interested in, there is a technical Lemma with its "proof". Moreover, this "proof" is willingly flawed (plain Arnold's style!), and just after the flaw is explained (but not corrected; that is left as a problem for the reader). Then the proof of the proposition you want starts (p.164).
|
7
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https://mathoverflow.net/users/167834
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401928
| 164,956 |
https://mathoverflow.net/questions/401269
|
15
|
I'm looking for an open problem in analysis or number theory with just one "genuine" or "second order" quantifier.
E.g.
* "Every continuous function $\mathbb{R} \rightarrow \mathbb{R}$ has the property $\theta$", where $\theta$ is expressible using only quantifiers over rationals.
* "Every set $S$ of natural numbers has the property $\theta$", where $\theta$ is expressible using only quantifiers over rationals.
No cheat examples like "For every real number, Goldbach's conjecture holds"! That's an arithmetical problem.
In technical terms, I'm looking for a $\Pi^1\_1$ sentence that we don't know how to reduce to an arithmetical sentence.
I'd also like it to be easy to state and obviously $\Pi^1\_1$, so that it can be included in a logic paper without requiring much explanation.
|
https://mathoverflow.net/users/170446
|
Open problem in analysis with just one quantifier?
|
The
[Littlewood conjecture](https://en.wikipedia.org/wiki/Littlewood_conjecture) is an example that meets all my requirements.
It is easy to state and obviously $\Pi^1\_1$. Furthermore [this comment by Christian Reiher](https://gowers.wordpress.com/2009/11/17/problems-related-to-littlewoods-conjecture-2/#comment-4395) gives me confidence that it has no known reduction to an arithmetical sentence. (Hopefully it even lacks a known reduction to a $\Sigma^1\_1$ sentence.)
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4
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https://mathoverflow.net/users/170446
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401930
| 164,957 |
https://mathoverflow.net/questions/401924
|
17
|
$\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\Aut{Aut}$Consider the class of finite groups $G$ having a zero entry in each column of its character table (except the first one), i.e. for all $g \neq e$ there is an irreducible character $\chi$ such that $\chi(g) = 0$.
I have been led to consider such character table, here are the examples found (see the GAP codes in Appendix):
* at order less than $384$, this class reduces to $A\_5$, $S\_5$, $\PSL(2,7)$, $\Aut(\PSL(2,7))$ and $A\_6$,
* every simple group of order less than $3000000$ is in this class except $A\_7$, $M\_{22}$,
* every perfect group of order less than $3600$ in this class is simple,
* if $5 \le n \le 19$, then $A\_n$ is not in this class iff $n \in \{ 7, 11, 13, 15, 16, 18, 19\}$. Idem for $S\_n$.
**Question**: What are the finite groups in this class? Which simple groups are not in?
The group $\PSL(2,q)$ is in this class iff it is simple, iff $q \ge 4$ (the generic character table is known).
Observe that all the examples found above are almost simple, but:
**Proposition**: This class is stable by direct product.
*proof*: Immediate by Theorem 4.21 in Isaacs' [Character Theory of Finite Groups](https://mathscinet.ams.org/mathscinet-getitem?mr=2270898) stating that the irreducible characters of a direct product are exactly the product of the irreducible characters of the components. $\square$
Note that $A\_5 \times A\_5$ is a non almost-simple finite group in this class, maybe the smallest one.
---
**Appendix**
*Small groups*:
```
gap> for o in [2..383] do n:=NrSmallGroups(o);; for d in [1..n] do G:=SmallGroup(o,d);; L:=Irr(CharacterTable(G));; l:=Length(L);; a:=0;; for j in [2..l] do LL:= List([1..l], i-> L[i][j]);; if not 0 in LL then a:=1;; break; fi; od; if a=0 then Print([G,[o,d]]); fi; od; od;
[ AlternatingGroup( [ 1 .. 5 ] ), [ 60, 5 ] ][ SymmetricGroup( [ 1 .. 5 ] ), [ 120, 34 ] ][ Group( [ (3,4)(5,6), (1,2,3)(4,5,7) ] ), [ 168, 42 ] ][ Group( [ (1,4,6,8,5,2,7,3), (1,3,8,6,5,4,7) ] ), [ 336, 208 ] ][ AlternatingGroup( [ 1 .. 6 ] ), [ 360, 118 ] ]
```
*Simple Groups*:
```
gap> it:=SimpleGroupsIterator(10,3000000);; for G in it do L:=Irr(CharacterTable(G));; l:=Length(L);; a:=0;; for j in [2..l] do LL:= List([1..l], i-> L[i][j]);; if not 0 in LL then a:=1;; break; fi; od; if a=1 then Print([G]); fi; od;
[ A7 ][ M22 ]
```
*Perfect Groups*:
```
gap> for o in [60..3599] do n:=NumberPerfectGroups(o);; for d in [1..n] do G:=PerfectGroup(o,d);; L:=Irr(CharacterTable(G));; l:=Length(L);; a:=0;; for j in [2..l] do LL:= List([1..l], i-> L[i][j]);; if not 0 in LL then a:=1;; break; fi; od; if a=0 then Print([G,[o,d]]); fi; od; od;
[ A5, [ 60, 1 ] ][ L3(2), [ 168, 1 ] ][ A6, [ 360, 1 ] ][ L2(8), [ 504, 1 ] ][ L2(11), [ 660, 1 ] ][ L2(13), [ 1092, 1 ] ][ L2(17), [ 2448, 1 ] ][ L2(19), [ 3420, 1 ] ]
```
*Alternating groups*:
```
gap> for n in [5..19] do s:=Concatenation("A",String(n));; L:=Irr(CharacterTable(s));; l:=Length(L);; a:=0;; for j in [2..l] do LL:= List([1..l], i-> L[i][j]);; if not 0 in LL then a:=1;; break; fi; od; if a=1 then Print([s]); fi; od;
[ "A7" ][ "A11" ][ "A13" ][ "A15" ][ "A16" ][ "A18" ][ "A19" ]
```
|
https://mathoverflow.net/users/34538
|
The finite groups with a zero entry in each column of its character table (except the first one)
|
Partial answer: the finite group $G$ is clearly in this class if it has a $p$-block of defect zero for every prime $p$ which divides $|G|$. This is a sufficient condition which may not be necessary. No non-trivial finite solvable group satisfies the sufficient condition. Most (but not all) finite simple groups satisfy the sufficient condition (for example, $M\_{22}$ and $M\_{24}$ do not, and many alternating groups do not).
Later edit: In the other direction, if $G$ is a finite group such that $F^{\ast}(G)$ is a $p$-group for some prime $p$, then whenever $S$ is a Sylow $p$-subgroup of $G$, and $z$ is any non-identity element of $Z(S)$, then no irreducible character of $G$ vanishes at $z$, so $G$ does not have the desired property. This is because $G$ has only one $p$-block, the principal block, and no irreducible character in the principal $p$-block vanishes at a central element of a Sylow $p$-subgroup of $G$ by the fundamental congruence of central characters due to R. Brauer.
Third edit: A well-known paper of Granville and Ono,
"[Defect zero $p$-blocks for finite simple groups](https://doi.org/10.1090/S0002-9947-96-01481-X)", Trans AMS, 348,1, January 1996, gives a complete determination of finite non-Abelian simple groups $G$ which do not have a $p$-block of defect zero for some prime $p$. All other non-Abelian finite simple groups $G$ have the property of the question, that is, for each non-identity element $x \in G$, there is an irreducible complex character $\chi$ of $G$ with $\chi(x) = 0$.
|
15
|
https://mathoverflow.net/users/14450
|
401932
| 164,959 |
https://mathoverflow.net/questions/401925
|
1
|
This question is related to [this one](https://mathoverflow.net/questions/401910/effect-of-snowflaking-on-metric-capacity). Let $(X,d)$ be a metric space, let $\epsilon\in [0,1)$ and consider the snowflake $(X,d^{1-\epsilon})$. Suppose that $(X,d)$ has a finite doubling constant, where the doubling constant $\lambda\_{(X,d)}$ of $(X,d)$ is defined by:
>
> There exists $\lambda \in \mathbb{N}$ such that every ball of radius 2r can be covered
> with at most $\lambda$ balls of radius r. The least such constant is $\lambda\_{(X,d)}$, the
> doubling constant of X.
>
>
>
In [Theorem 10.18 of Heinonen's Book](https://link.springer.com/content/pdf/10.1007%2F978-1-4613-0131-8_10.pdf), it is stated that quasi symmetries of doubling metric spaces must be doubling, so by [the discussion circa (10.2)](https://link.springer.com/content/pdf/10.1007%2F978-1-4613-0131-8_10.pdf) $(X,d^{1-\epsilon})$ must also be a doubling metric space. My question is, how do the doubling constants of $(X,d)$ and of $(X,d^{1-\epsilon})$ (defined as above) relate to one another explicitly?
For instance, is it true that:
$$
\lambda\_{(X,d^{1-\epsilon})}^{\frac1{1-\epsilon}}
\leq C\lambda\_{(X,d)} ,
\qquad \forall \epsilon \in (0,1)
$$
for some absolute constant $C>0$?
|
https://mathoverflow.net/users/36886
|
Effect of snowflaking on doubling constants
|
Suppose $(X,d)$ has doubling constant $\lambda$. This means that for every $r$, every $d$-ball of radius $2r$ can be covered by $\lambda$ many $d$-balls of radius $r$. With respect to the metric $d^\alpha$, this means that every $d^\alpha$-ball of radius $(2r)^\alpha$ can be covered by $\lambda$ many $d^\alpha$-balls of radius $r^\alpha$. Letting $s = r^\alpha$, that means that every $d^\alpha$-ball of radius $2^\alpha s$ can be covered by $\lambda$ many balls of radius $s$.
Now putting $t = 2^{\alpha - 1}s$, we get that every $d^\alpha$-ball of radius $2t$ can be covered by $\lambda$ many $d^\alpha$-balls of radius $2^{1 - \alpha}t$. And each of those smaller balls can be covered by $\lambda$ many $d^\alpha$ balls of radius $2^{1 - 2\alpha}t$, and so on. After $k$ steps, with $k \geq 1/\alpha$, we get down to $d^\alpha$-balls of radius $2^{1 - k\alpha}t \leq t$. Conclusion: the original radius $2t$ ball can be covered by $\lambda^k$ radius $t$ balls, i.e., the doubling constant for the snowflaked space is at most $\lambda^{\lceil 1/\alpha\rceil}$.
|
4
|
https://mathoverflow.net/users/23141
|
401937
| 164,962 |
https://mathoverflow.net/questions/401890
|
0
|
In the paper [Dissimilarity in Graph-Based Semi-Supervised Classification](http://pages.cs.wisc.edu/%7Eswright/papers/dissim_final.pdf), there are few things I could not understand.
Given that $x\_1, x\_2,..., x\_n$ are the vector representation of $n$ items, $f : X \rightarrow \mathbb{R}$ is the discriminant function, and $w\_{ij}$ are all non-negative,
why is $ \frac{1}{2} \sum\_{i, j = 1}^n w\_{ij} (f(x\_i)- f(x\_j))^2$ convex w.r.t to $f$?
Also, why does changing any $w\_{ij}$ to negative value make it non-convex?
I read on [wikipedia](https://en.wikipedia.org/wiki/Convex_function) that
>
> "A twice-differentiable function of a single variable is convex if and
> only if its second derivative is nonnegative on its entire domain"
>
>
>
But I don't understand derivative should be take w.r.t to what?
|
https://mathoverflow.net/users/168789
|
Why is this function convex?
|
Call the sum $S(f)$. Let $0<b<1$, and let $f$ and $g$ be two functions mapping $X$ to $\Bbb R$. Then
$$
S(bf+(1-b)g)=b^2S(f)+(1-b)^2S(g)+b(1-b)\sum\_{i,j}w\_{i,j} [f(x\_i)-f(x\_j)]\cdot[g(x\_i-g(x\_j)].
$$
By Cauchy-Schwarz,
$$
\sum\_{i,j}w\_{i,j} [f(x\_i)-f(x\_j)]\cdot[g(x\_i-g(x\_j)]\le 2\sqrt{S(f)S(g)}.
$$
(The non-negativity of $w\_{i,j}$ is used here: $w\_{i,j} = \sqrt{w\_{i,j}}\cdot \sqrt{w\_{i,j}}$.)
Therefore
$$
\eqalign{
S(bf+(1-b)g)
&\le b^2S(f)+(1-b)^2S(g)+2b(1-b)\sqrt{S(f)S(g)}\cr
&= \left[b\sqrt{S(f)}+(1-b)\sqrt{S(g)}\right]^2\cr
&\le bS(f)+(1-b)S(g),
}
$$
the final inequality because the square function is convex. This shows that $f\mapsto S(f)$ is convex. (Intuitively, $f\mapsto [f(x\_i)-f(x\_j)]^2$ is convex because the square is convex; and then $S$ is convex becasue it's a positive-linear combination of such functions of $f$.)
|
1
|
https://mathoverflow.net/users/42851
|
401939
| 164,963 |
https://mathoverflow.net/questions/401934
|
15
|
Let $G$ be a locally compact Hausdorff (LCH) topological group with left Haar measure $\mu$. Given a compact unit neighborhood $U$, consider the function
$$
\Phi: \quad G \to (0,\infty), \quad x \mapsto \mu(U x U).
$$
My question is:
>
> Can one give a natural characterization of the groups for which this function is bounded?
>
>
>
One conjecture (see below): This happens exactly for IN groups, which are groups for which there exists a compact unit neighborhood $U$ satisfying $x U x^{-1} = U$ for all $x \in G$.
Thoughts/observations:
1. The question is independent of the choice of $U$. Indeed, if $U,V$ are both compact unit neighborhoods, then $U \subset \bigcup\_{i=1}^n x\_i V$ and $U \subset \bigcup\_{j=1}^m V y\_j$ for suitable $x\_i,y\_j \in G$, and this easily allows to bound $\mu(U x U)$ in terms of $\mu(V x V)$.
2. If $G$ is not unimodular, then $\Phi$ is *not* bounded, since $\Phi(x) \geq \mu(U x) = \Delta(x) \cdot \mu(U)$, so that $\Phi$ is bounded from below (up to a constant) by the modular function, which is unbounded for non-unimodular groups.
3. If $G$ is an IN group, then $\Phi$ is bounded. Indeed, by the first observation from above we can choose $U$ to satisfy $x U x^{-1} = U$ for all $x$, and then $\Phi(x) = \mu(U x U) = \mu(x U U) = \mu(U U)$ for all $x \in G$.
What I have not been able to show is that if $\Phi$ is bounded, then $G$ needs to be IN. Of course, it could be that this simply does not hold.
|
https://mathoverflow.net/users/59219
|
For what LCH groups is the Haar measure $\mu(U x U)$ bounded?
|
Your conjecture is correct.
Suppose that we have a compact unit neighbourhood $U$ such that $\mu(UxU) \ll 1$ for all $x$. As you have already noted, we can take $\mu$ to be unimodular, and the choice of neighbourhood is not relevant, so we may assume without loss of generality that $U$ is symmetric: $U^{-1} = U$. (This makes $U$ what is called an [approximate group](https://en.wikipedia.org/wiki/Approximate_group) in arithmetic combinatorics, and the intuition that $U$ should behave like a subgroup of $G$ is underlying the arguments below.) We allow implied constants in asymptotic notation to depend on $U$, thus for instance $\mu(U) \asymp 1$.
Note that the conjugate $x U x^{-1}$ of $U$ is commensurate with $U$ in the sense that $\mu( U \cdot x U x^{-1} ) = \mu( U x U ) \ll 1$. (In the language of arithmetic combinatorics, $U$ stays close to its conjugates $xUx^{-1}$ in [Ruzsa distance](https://en.wikipedia.org/wiki/Ruzsa_triangle_inequality).) In the spirit of the isomorphism theorems (or the [Ruzsa covering lemma](https://en.wikipedia.org/wiki/Freiman%27s_theorem#Ruzsa_covering_lemma) in arithmetic combinatorics), one now expects $U$ and $xUx^{-1}$ to have large intersection, and this can be accomplished (at the cost of enlarging $U$ to $U^2$) by the following convolution argument (cf. the double counting argument that shows that $|H \cdot K| = |H| |K| / |H \cap K|$ for finite subgroups $H,K$ of $G$). Observe that the convolution $1\_U \* 1\_{xUx^{-1}}$ has an $L^1(G,\mu)$ norm of $\mu(U) \mu(x U x^{-1}) \asymp 1$ and is supported on $U x U x^{-1}$, which has measure $O(1)$. Thus there must exist a point $y \in G$ where $1\_U \* 1\_{xUx^{-1}}(y) \gg 1$, thus
$$ \mu( yU \cap x U x^{-1} ) \gg 1$$
which implies
$$ \mu( (yU \cap x U x^{-1})^{-1} \cdot (yU \cap x U x^{-1}) ) \gg 1$$
and hence
$$ \mu( U^2 \cap x U^2 x^{-1} ) \gg 1.$$
To put it another way, the inner products of the functions $1\_{x U^2 x^{-1}}$ with $1\_{U^2}$ are uniformly bounded from below.
We can now use an "ergodic" argument to extract an invariant object (in the spirit of the [Alaoglu--Birkhoff ergodic theorem](https://mathscinet.ams.org/mathscinet-getitem?mr=2026)).
Let $S$ be the closed convex hull in $L^2(G,\mu)$ of the functions $1\_{x U^2 x^{-1}}$. By the [Hilbert projection theorem](https://en.wikipedia.org/wiki/Hilbert_projection_theorem), this set has a unique element $f$ of minimal norm. All elements of $S$ have inner product with $1\_{U^2}$ uniformly bounded from below, so $f$ does also; in particular, $f$ is non-trivial. Since $S$ is conjugation-invariant, symmetric, bounded and consists of non-negative functions, $f$ must be non-negative, symmetric, and conjugation-invariant.
At this point one could already extract a conjugation-invariant set of positive finite measure by taking level sets of $f$, but this is not quite regular enough for the conjecture, so we take a convolution to achieve an additional smoothing (in the spirit of the [Steinhaus theorem](https://en.wikipedia.org/wiki/Steinhaus_theorem)). The convolution $f\*f$ is then in $C\_0(G)$ (this follows from a standard limiting argument, approximating $f$ in $L^2(G,\mu)$ by $C\_c(G)$ functions and using Young's inequality), conjugation-invariant, and strictly positive at the origin; taking level sets, we obtain a non-trivial conjugation-invariant compact unit neighbourhood, as required.
|
14
|
https://mathoverflow.net/users/766
|
401943
| 164,964 |
https://mathoverflow.net/questions/401933
|
5
|
Is there an abelian variety $A/\mathbb R$ of dimension $n$ such that $End\_{\mathbb R}(A)\otimes \mathbb Q$ contains a field $K$ of degree $[K:\mathbb Q]=2n$? ($End\_{\mathbb R}(A)$ is the ring of $\mathbb R$-endomorphisms of $A$)
|
https://mathoverflow.net/users/197736
|
Abelian variety with CM defined over real numbers
|
No.
Assume for contradiction that such an $A$ exists. First look at the singular cohomology $H^1(A\_{\mathbb C}, \mathbb Q)$, which admits an action of $K$ and so is a $K$-vector space. It has dimension $2n$ over $\mathbb Q$ and so is a 1-dimensional $K$-vector space.
Tensoring with $\mathbb C$, we see that $H^1(A\_{\mathbb C}, \mathbb C)$, as a vector space with an action of $K$, is a sum of $2n$ eigenspaces of $K$ associated to the $2n$ different embeddings $K \to \mathbb C$.
Now by Hodge theory, $H^1(A\_{\mathbb C}, \mathbb C) = H^1(A\_{\mathbb C}, \mathcal O\_A) + H^0(A\_{\mathbb C}, \Omega^1\_A)$ with the two summands complex conjugates of each other. So for each eigenvector appearing associated to an embedding appears in $H^1(A\_{\mathbb C}, \mathcal O\_A)$, the eigenvector associated to the complex conjugate embedding appears in $H^0(A\_{\mathbb C}, \Omega^1\_A)$, and thus, because the eigenspace is 1-dimensional so there is only one eigenvector up to scaling, no eigenvector associated to the complex conjugate embedding appears in $H^1(A\_{\mathbb C}, \mathcal O\_A)$.
So $H^1(A\_{\mathbb C}, \mathcal O\_A)$ is not isomorphic to its complex conjugate as a complex vector space with an action of $K$.
But $H^1(A\_{\mathbb C}, \mathcal O\_A) = H^1(A\_{\mathbb R}, \mathcal O\_A) \otimes\_{\mathbb R} \mathbb C$ and thus is isomorphic to its complex conjugate. (Here we use that the endomorphisms in $K$ are defined over $\mathbb R$ and thus act on $ H^1(A\_{\mathbb R}, \mathcal O\_A) $.)
This is a contradiction, so no such $A$ exists.
|
10
|
https://mathoverflow.net/users/18060
|
401945
| 164,965 |
https://mathoverflow.net/questions/401904
|
8
|
$\DeclareMathOperator\SO{SO}$At the very begining of [Akbulut and Kalafat - Algebraic topology of $G\_2$ manifolds](https://arxiv.org/abs/1308.2263v2), the authors stated that there is a "canonical fibration" for $G\_2$ of the form
$$G\_2\to \SO(7)\to \mathbb{R}P^7,$$
where the map $G\_2\to \SO(7)$ is obtained by regarding $G\_2$ as the automorphisms of the imaginary octonions, which, as a real vector space, is of dimension $7$.
So, how does one construct a homotopy equivalence $\SO(7)/G\_2\to\mathbb{R}P^7$?
|
https://mathoverflow.net/users/100553
|
The "canonical fibration" for the Lie group $G_2$
|
In *Spinors and Calibrations* by F. Reese Harvey, you can find proof (p. 283) of $S^7 \simeq Spin(7)/G\_2.$ It takes the same approach as Bryant's notes mentioned in the comments but it is much more detailed in this case.
The main idea is to write down the spinor representation of $Spin(7)$ using octonionic multiplication. It takes some work to see that the stabilizer of a point is then isomorphic to the automorphism group of the octonions, which is the compact Lie group $G\_2.$
|
5
|
https://mathoverflow.net/users/6818
|
401947
| 164,966 |
https://mathoverflow.net/questions/401948
|
2
|
Let $(I,\leq)$ be a directed set, that is $\leq$ is reflexive and transitive and for every $a,b\in I$ we find $c\in I$ such that $a,b\leq c$. Now consider the set $M$ consisting of all maps $\sigma:I\longrightarrow I$ such that $a \leq b$ implies $\sigma a≤ \sigma b$. We define a reflexive and transitive order on $M$ as follows: For $\sigma,\tau:I⟶I$ in M we set $\sigma \leq \tau$ if for all $a\in I$ we obtain $\sigma a \leq \tau a$. Now the question is: Is $M$ again a directed set?
|
https://mathoverflow.net/users/145920
|
Is the set of "endomorphisms" of a directed set again a directed set?
|
I claim that the answer is no. This answer is a generalization of the answers to [this previous question](https://mathoverflow.net/q/190464/22277).
If $X,Y$ are posets, then let $\text{Hom}(X,Y)$ denote the set of all mappings $f:X\rightarrow Y$ where if $x\_{1},x\_{2}\in X$ and $x\_{1}\leq x\_{2}$, then $f(x\_{1})\leq f(x\_{2})$.
Theorem: The following are equivalent:
1. $\text{Hom}(X\times X,X\times X)$ is directed.
2. $\text{Hom}(X\times X,X)$ is directed.
3. $\text{Hom}(P,X)$ is directed for each poset $P$.
4. There is a function $L:X\times X\rightarrow X$ such that $x\leq L(x,y),y\leq L(x,y)$ for each $x,y\in X$ and where if $x\_{1}\leq x\_{2},y\_{1}\leq y\_{2}$, then $L(x\_{1},y\_{1})\leq L(x\_{2},y\_{2})$.
Proof:
$3\rightarrow 2$ is trivial.
$4\rightarrow 3$ Suppose that $f,g\in\text{Hom}(P,X)$. Then let $h:P\rightarrow X$ be the function defined by $h(p)=L(f(p),g(p))$. Then $h\in\text{Hom}(P,X)$ and $f\leq h,g\leq h$.
$1\leftrightarrow 2$ Observe that $\text{Hom}(X\times X,X\times X)\simeq
\text{Hom}(X\times X,X)\times\text{Hom}(X\times X,X)$. To get the equivalence observe that two posets $P,Q$ are both directed if and only if the product $P\times Q$ is direted.
$2\rightarrow 4$ Let $\pi\_{0},\pi\_{1}:X\times X\rightarrow X$ be the projections where $\pi\_{0}(x,y)=x,\pi\_{1}(x,y)=y$. Then there is some $L:X\times X\rightarrow X$ where $\pi\_{0}\leq L,\pi\_{1}\leq L$. Q.E.D.
In [the previous question](https://mathoverflow.net/q/190464/22277), we have an example of posets $P,Q$ where $Q$ is directed by where $\text{Hom}(P,Q)$ is not directed. In this case, $\text{Hom}(Q\times Q,Q\times Q)$ is not directed.
As Joel David Hamkins answered in the previous question, if $X$ is a countable directed set, then $\text{Hom}(P,X)$ is always directed. More generally, if $X$ has a linearly ordered cofinal subset or if $X$ is a join semi-lattice, then $\text{Hom}(P,X)$ is directed.
Let me now generalize and slightly modify Emil Jeřábek's counterexample to the previous question and obtain a directed poset $Y$ where $\text{Hom}(P,Y)$ is not directed for some poset $P$.
Let $\lambda,\kappa$ be regular cardinals with $\lambda<\kappa$. If $\Delta$ is a set, then let $P\_{<\lambda}(\Delta)$ be the collection of all subsets of $\Delta$ of cardinality less than $\lambda$, and order $P\_{<\lambda}(\Delta)$ by inclusion. Let $G$ be a two element poset with underlying set $\{r,s\}$ where $r\not\leq s,s\not\leq r$. Let $X=\kappa\times G$ be the cartesian product. Now let $Y=X\cup P\_{<\lambda}(\kappa)$ be the partial ordering where if $(\alpha,t)\in X,A\in P\_{<\lambda}(\kappa)$, then $(\alpha,t)\leq A$ if and only if $\alpha\leq\alpha'$ for some $\alpha'\in A$. Then the poset $Y$ is directed.
Then let $f,g:\kappa\rightarrow Y$ be the mappings where $f(\alpha)=(\alpha,r),g(\alpha)=(\alpha,s)$. Then there cannot be a $h\in\text{Hom}(\kappa,Y)$ with $f\leq h,g\leq h$. Suppose there were such an $h$. Then we would have $h(\alpha)\in P\_{<\lambda}(\kappa)$ for each $\alpha\in\kappa$. Furthermore, since $(\alpha,r)=f(\alpha)\leq h(\alpha)$, we know that $\alpha\leq\alpha'$ for some $\alpha'\in h(\alpha)$. Therefore, the sequence $h$ cannot be eventually constant. Thus, $h$ is a not-eventually constant monotone
sequence in $P\_{<\lambda}(\kappa)$ of cofinality $\kappa$ which is impossible.
|
3
|
https://mathoverflow.net/users/22277
|
401953
| 164,968 |
https://mathoverflow.net/questions/401960
|
7
|
I would like to know if anyone has an electronic copy of the following paper:
>
> **"Holmgren, E.: Über Systeme von linearen partiellen Differentialgleichungen. Översigt Vetensk. Akad. Handlingar 58, 91–105 (1901)"**
>
>
>
In my search, the best result I found was the (possible) statement of the main result of this article which can be found in the following article: [https://people.kth.se/~haakanh/publications/Hed-MZ2.pdf](https://people.kth.se/%7Ehaakanh/publications/Hed-MZ2.pdf). More precisely,
**Theorem (Holmgren)** Suppose $I$ is a real-analytic nontrivial arc of $\partial \Omega$. Then if $u$ is smooth on a planar neighbohood $\mathcal{O}$ of $I$ and $\Delta^N u=0$ holds on $\mathcal{O} \cap \Omega$ with $\partial\_{n}^{j-1}|\_I=0$ for $j=1, \dots, 2N$, then $u(z)=0$ on $\mathcal{O} \cap \Omega$, provided that the open set $\mathcal{O} \cap \Omega$ is connected.
Any information is welcome, for example, if this article is published in a book.
|
https://mathoverflow.net/users/115618
|
Looking for an electronic copy of Holmgren's old paper
|
The full text of the article can be found scanned [here](https://www.biodiversitylibrary.org/page/32299061#page/103/mode/1up).
|
12
|
https://mathoverflow.net/users/120914
|
401963
| 164,971 |
https://mathoverflow.net/questions/401957
|
0
|
In B. Chow and D. Knopf's book "The Ricci Flow: An Introduction", the authors claim that for any dimension $n$ and any Riemannian manifold $M^n$, there is a constant $C\_n$ depending only on $n$ such that $R \leq C\_n \|\text{Rm}\|$. It's not clear whether this inequality should actually be $R(t) \leq C\_n \|\text{Rm}(t)\|$, where $g(t)$ is a solution to the Ricci flow on $M$. It appears that the definition of norm they're using is the pointwise one, i.e $\|\text{Rm}(x)\| = \sqrt{R\_{ijk\ell}(x)R^{ijk\ell}(x)}$.
Why is this true? From the definitions it doesn't look at all obvious why there should exist such a constant.
|
https://mathoverflow.net/users/119418
|
Estimating scalar curvature by norm of Riemannian curvature tensor under the Ricci flow
|
**Remark**
As @OthisChodosh and @WillieWong have pointed out, the existence of a constant $C\_n$ that depends only on the dimension can be proved using only elementary linear algebra. I might as well provide the details. Although I like my first answer, it was overkill.
**Simpler answer**
First, recall that if $V$ is a finite dimensional inner product space, then given any linear function $\ell: V \rightarrow \mathbb{R}$, there exists a constant $C\_\ell > 0$ such that, for any $v \in V$,
$$
|\ell(v)| \le C\_\ell |v|.
$$
Given the tangent space $T$ at a point in a Riemannian manifold, let
$$
V = S^2(\bigwedge^2V^\*).
$$
The inner product on $T$ induces an inner product on $V$, namely the one used here. The definition of scalar curvature defines a linear function $S: V\rightarrow \mathbb{R}$, and therefore there is a constant $C$ such that
$$
|S(v)| \le C|v|.
$$
Now note that the definition of $S$ is canonical, defined using only the dimension and inner product on $T$. The constant has to be independent of the inner product, because, if you write everything out above with respect to an orthonormal basis, the formulas all remain the same, no matter what inner product is, and therefore the constant $C$ does, too. Therefore, the constant $C$ depends on the dimension only.
**Calculation of sharp constant**
You can compute the best possible value of $C\_n$.
First, given two curvature-like tensors $A$ and $B$, let
$$
A\cdot B = g^{is}g^{jt}g^{ku}g^{lv}A\_{ijkl}B\_{stuv}.
$$
Then $|A|^2 = A\cdot A$.
The idea is to decompose the curvature tensor into two terms,
$$
R\_{ijkl} = T\_{ijkl} + U\_{ijkl},
$$
where
$$
T\_{ijkl} = \frac{S}{n(n-1)}(g\_{ik}g\_{jl}-g\_{il}g\_{jk})
$$
and $S$ is the scalar curvature. A straightforward calculation shows that
$$
T\cdot R = T\cdot T = \frac{2S^2}{n(n-1)},
$$
which implies that
$$
T\cdot U = T\cdot(R-T) = 0.
$$
This, in turn, implies that
$$
|R|^2 = |T|^2 + |U|^2 \ge |T|^2 = \frac{2S^2}{n(n-1)}
$$
Therefore,
$$
|S| \le \sqrt{\frac{n(n-1)}{2}}|R|,
$$
with equality holding if and only if
$$
R\_{ijkl} = \frac{S}{n(n-1)}(g\_{ik}g\_{jl}-g\_{il}g\_{jk}).
$$
If this holds everywhere on $M$, then the metric has constant sectional curvature.
|
6
|
https://mathoverflow.net/users/613
|
401973
| 164,975 |
https://mathoverflow.net/questions/401971
|
6
|
On the nlab [page](https://ncatlab.org/nlab/show/Chevalley-Eilenberg+algebra#DefForLieAlg) for Chevalley–Eilenberg algebras, it defines $\operatorname{CE}(\mathfrak g)$ for $\mathfrak g$ finite dimensional, and then says "This has a more or less evident generalization to infinite-dimensional Lie algebras", and provides no more details.
Please could someone outline this generalisation, and if the theory follows through (e.g. is there still a contravariant equivalence of categories?).
|
https://mathoverflow.net/users/163483
|
CE(g) for g infinite dimensional
|
A definition that always works and does agree with that one in the finite-dimensional case is the following: put
$$
C^k(\mathfrak{g})=({\Lambda}^k\mathfrak{g})^\*=\operatorname{Hom}({\Lambda}^k\mathfrak{g}, \mathbb{F}).
$$
(Here $\mathbb{F}$ is the ground field, of course.)
The differential is given by the formula
$$
(d\phi)(g\_1\wedge\dotsb\wedge g\_{k+1})=\sum\_{1\le i<j\le k+1}(-1)^{i+j-1}\phi([g\_i,g\_j]\wedge g\_1\wedge\dotsb \wedge\widehat{g\_i}\wedge\dotsb\wedge \widehat{g\_j}\dotsb\wedge g\_{k+1}),
$$
where the hat means missing the corresponding factor. If you define the product of cochains by the usual shuffle product formula, you will see that the differential is a derivation, and so all nice properties still hold. The main thing that breaks is precisely the identification $({\Lambda}^k\mathfrak{g})^\*\cong {\Lambda}^k(\mathfrak{g}^\*)$ used in the nlab page.
(See, for example, the classical reference *[Cohomology of Infinite-Dimensional Lie Algebras](https://doi.org/10.1007/978-1-4684-8765-7)* by D.B. Fuks.)
The statement about equivalence of categories does not extend, however, precisely because the cochain algebra is not generated by degree one elements. There are many different situations where things can be patched: many infinite-dimensional algebras have additional gradings with finite-dimensional components (and you can work in the corresponding monoidal category), sometimes all you need is continuous cochains on algebras that have some topology etc. Again, the book of Fuks details many of those situations.
|
6
|
https://mathoverflow.net/users/1306
|
401979
| 164,977 |
https://mathoverflow.net/questions/401962
|
3
|
I was reading this paper ([arXiv link](https://arxiv.org/abs/0912.3506))
>
> On the Large Time Behavior of Solutions of the Dirichlet problem for Subquadratic Viscous Hamilton-Jacobi Equations
> Guy Barles (LMPT), Alessio Porretta, Thierry Wilfried Tabet Tchamba (LMPT)
>
>
>
The authors claimed that in general if $f\in W^{1,\infty}(B(0,2)),1<m\leq 2$ and $u\in W^{2,p}\_{loc}(B(0,2))$ is a solution to
$$ |Du|^m - f(x) - \Delta u = 0 \qquad{in}\;B(0,2)$$
then
>
> If $|Du|\_{L^\infty(B(0,1/2))}\leq K$ then $|D^2u|\_{L^\infty(B(0,1/2))}\leq K'$ by standard elliptic theory.
>
>
>
My questions are:
* What is the exact argument did they use to arrive at this conclusion?
* Would the argument hold true if I have
$$ \delta u+ |Du|^m - f(x) - \Delta u = 0 \qquad{in}\;B(0,2)$$
for $\delta > 0$ instead?
|
https://mathoverflow.net/users/124759
|
A regularity estimate for second-derivative
|
The main tools are the following elliptic regularity results: assume that $u \in W^{2,p}\_{loc}$ and let $f=\Delta u$.
a) If $f \in L^q\_{loc}$ with $q>p$, then $u \in W^{2,q}\_{loc}$;
b) If $f \in W^{1,p}\_{loc}$, then $u \in W^{3,p}\_{loc}$.
With this in mind, first note that $|Du|^m \in W^{1,p}\_{loc}$, because $m>1$ and $Du$ is bounded. This gives $u \in W^{3,p}\_{loc}$ and the result follows if $p>N$, by Sobolev embedding. If, instead, $p \leq N$, use first Sobolev embedding to get $u \in W^{2,q}\_{loc}$ with any $q<\infty$ if $p=N$ and $1/q=1/p-1/N$ when $p<N$ and then rerun the argument before with $q$ instead of $p$. In a finite number of steps you get the result.
|
1
|
https://mathoverflow.net/users/150653
|
401982
| 164,978 |
https://mathoverflow.net/questions/401530
|
3
|
Jim Lawrence has a very important paper on the topic of valuations on polyhedra called "*Rational-function-valued valuations on polyhedra*", published in the DIMACS volume *Discrete and computational geometry* of the AMS.
**Does anyone have a scanned copy of this article that can be shared?**
Google books has a preview of almost all the article, except for section 5 an the references, which are the ones that I mostly need. I wrote to Jim Lawrence asking him if he has a copy, but he told me that he surely has a copy in his office, but for obvious reasons (a global pandemic, in case someone reads this in a 100 more years) he is working from home (like the most of us).
PS: I obviously looked already at those sites which start with *lib* and *sci*, without success.
|
https://mathoverflow.net/users/109085
|
Request for an article by Jim Lawrence
|
I have a scan of the chapter and can email it, just let me know the address.
The reference list (missing from the Google books preview) is here:
|
4
|
https://mathoverflow.net/users/11260
|
401990
| 164,981 |
https://mathoverflow.net/questions/401954
|
2
|
The following is an excerpt from Marco Gualtieri's [thesis](https://arxiv.org/abs/math/0401221)
>
> A central theme of this thesis is that classical geometrical
> structures which appear, at first glance, to be completely different
> in nature, may actually be special cases of a more general unifying
> structure. Of course, there is wide scope for such generalization, as
> we may consider structures defined by sections of any number of
> natural bundles present on manifolds. What must direct us in deciding
> which tensor structures to study is the presence of natural
> integrability conditions.
>
>
>
>
> Good examples of such conditions include the
> closure of a symplectic form, the Einstein or special holonomy
> constraint on a Riemannian metric, the vanishing of the Nijenhuis
> tensor of a complex structure, and the Jacobi identity for a Poisson
> bivector, among many others.
>
>
>
I think by closure of symplectic form, it means $d\omega=0$.
The only notion of integrability I know is integrability of a subbundle of $TM$; that is, for every point of $M$, there is an integrable manifold corresponding to the distribution.
Frobenius theorem says $L\subseteq TM$ is integrable if and only if $[X,Y]\in L$ for every $X,Y\in L$.
But, I do not fully understand what is "integrable" in a differential form being closed, or a bivector to satisfy Jacobi identity or for Nijenhuis tensor to be zero.
I would understand if we call it "a smoothly varying condition" or something similar, but, why would some one want to refer them as integrability conditions?
|
https://mathoverflow.net/users/118688
|
Does "integrability condition" have a specific meaning or is it used in a casual way?
|
It turns out there is a well defined notion of integrability of a G-structure on a manifold.
Thanks to the user Thomas Rot who has given the reference [Linear $G$-structures by examples](https://www.few.vu.nl/%7Epasquott/course16.pdf)
Definition $2.1$ is that of $G$-structure on a manifold M.
It defines the notion of integrability of a $G$-structure in Definition $2.4$.
1. A ($p$-dimensional) distribution $\mathcal{F}$ on a ($n$-dimensional) manifold $M$ can be seen as a $GL(p,n-p)$ structure on the manifold $M$. We can talk about integrability of this $G$-structure. It says (Theorem $2.25$) $\mathcal{F}$ is involutive if and only if $\mathcal{F}$ is an integrable G-structure. This goes by the name Frobenius theorem.
2. An almost complex structure J on a manifold $M$ (of dimension $2k$) can be seen as a $GL\_k(\mathbb{C})$-structure on the manifold $M$. We can talk about integrability of this $G$-structure. It says that (Theorem $2.31$) Nijenhuis tensor of $J$ vanishes if and only if $J$ is an integrable $G$-structure. This goes by the name Newlander-Nirenberg theorem.
3. A non degenerate $2$-form $\omega$ on a manifold $M$ (of dimension $2k$) can be seen as a $Sp\_k(\mathbb{R})$-structure on the manifold $M$. We can talk about integrability of this $G$-structure. It says (Theorem $2.41$) $\omega$ is closed if and only if $\omega$ is an integrable $G$-structure. This goes by the name of Darboux theorem.
|
2
|
https://mathoverflow.net/users/118688
|
401994
| 164,982 |
https://mathoverflow.net/questions/401952
|
3
|
Let $M$ be a smooth manifold and $E \to M$ a vector bundle.
I'm reading a text which says:
>
> Recall that there is a one-to-one correspondence between:
>
>
> 1. linear vector fields on $E$, and
> 2. linear operators $D : \Gamma(E) \to \Gamma(E)$ such that there exists a vector field $X$ on $M$ such that
> $$D(fs) = f D(s) + X(f) s$$
> for all $f \in C^\infty(M)$ and $s \in \Gamma(E)$.
>
>
>
Unfortunately, I can't recall ever seeing this. I'm not even sure what is a linear vector field (Google hasn't return anything useful).
What is this correspondence? If this is a standard fact, where is it explained (textbook/lecture notes/paper)?
|
https://mathoverflow.net/users/341298
|
Linear vector fields $\leftrightarrow$ certain differential operators
|
Using google (linear vector field on vecot bundle), I found the following reasonable definition for a linear vector field $\hat X$ on a vector bundle $E\to M:$ it is a vector field $\hat X\in\mathcal X(E)$ which is a vector bundle morphism $\hat X\colon E\to TE$ along a map $X\colon M\to TM$ given by a vector field on $M.$
The one-to-one correspondence is given as follows:
$1\Rightarrow2:$ if you have a linear vector field $\hat X$ on $E$, and a section $s\colon \Gamma(M,E)$ consider the difference $$\hat D\_ps:=T\_p s(X)-\hat X\_{s(p)}\in T\_{s(p)}E,$$ where $T$ denotes the differential of the smooth map $s\colon M\to E$. It can be checked using the definition that $\hat D\_ps$ is in the vertical tangent bundle $\mathcal V$, i.e., in the kernel of the differential of the projection $\pi\colon E\to M.$ On the other handd, there is a canonical isomorphism $$\pi^\*E=\mathcal V\to E.$$ Then, for any section $s\colon M\to E$ we obtain $$s^\*\pi^\*\mathcal V=E.$$ Using this observation, we can identify the map $$p\in M\mapsto \hat D\_ps\in \mathcal V\_{s(p)}$$ with a section $Ds\in\Gamma(M,E).$$
$2\Rightarrow1:$ given a first order differential operator $D$ the prescribed properties, one can define the vector field $\hat X$ by reversing the above process.
Unfortunately, I do not know a detailed reference for the above, but it is closely related to Ehresmann-type treatmeant of linear connections (by replacing directional derivatives with the differential respectively $D$ with a linear connection $\nabla$ on $E$).
|
4
|
https://mathoverflow.net/users/4572
|
401999
| 164,983 |
https://mathoverflow.net/questions/402006
|
2
|
We will work over $\mathbb C$. Let us consider a $n$-dimensional vector space $V$, then we define the $k$-th Grassmannian as
$$
\mathbb G(k,V):=\{W \subset V : \dim W=k\}.
$$
Then consider a non-degenerate quadratic form on $V$, we write $q: V \times V \to \mathbb C$ if it is symmetric, $\omega: V \times V \to \mathbb C$ if it is skew-symmetric (in this case $n$ is even).
Then one can define, respectively, the orthogonal $k$-th Grassmannian
$$
\mathbb O\mathbb G(k, V)=\{q\text{-isotropic }W \subset V : \dim W=k\}
$$
and the isotropic $k$-th Grassmannian
$$
\mathbb I\mathbb G(k, V)=\{\omega\text{-isotropic }W \subset V : \dim W=k\}.
$$
In the "classic" case, we know that the tangent bundle is given by
$$
T\_{\mathbb G(k,V)}=\mathcal S^\vee \otimes \mathcal Q
$$
where $\mathcal S$ and $\mathcal Q$ denote the universal subbundle and the universal quotient bundle, following Eisenbud and Harris. When $\dim V=2m$, then a similar expression is known for $\mathbb O \mathbb G(m,V)$ and $\mathbb I \mathbb G(m,V)$:
$$
T\_{\mathbb O \mathbb G(m,V)}=\wedge^2 \mathcal U^\vee, \quad T\_{\mathbb I \mathbb G(m,V)}=S^2 \mathcal U^\vee
$$
where $\mathcal U$ is given by the restriction of the tautological bundle on $\mathbb G(n,V)$.
Finally the question: are there similar expressions of the tangent bundles for the other orthogonal and isotropic Grassmannians? Also a reference would be enough.
|
https://mathoverflow.net/users/147236
|
Tangent bundle for orthogonal and isotropic Grassmannians
|
The tangent bundle to the orthogonal Grassmannian fits into an exact sequence
$$
0 \to T\_{\mathrm{OG}(k,V)} \to \mathcal{S}^\vee \otimes \mathcal{Q} \to S^2\mathcal{S}^\vee \to 0.
$$
Taking into account an exact sequence
$$
0 \to \mathcal{S}^\perp/\mathcal{S} \to \mathcal{Q} \to \mathcal{S}^\vee \to 0
$$
one can obtain an exact sequence
$$
0 \to \mathcal{S} \otimes (\mathcal{S}^\perp/\mathcal{S}) \to T\_{\mathrm{OG}(k,V)} \to \wedge^2\mathcal{S}^\vee \to 0.
$$
A similar description exists for the symplectic isotropic Grassmannian.
|
4
|
https://mathoverflow.net/users/4428
|
402007
| 164,986 |
https://mathoverflow.net/questions/402010
|
16
|
In this question [Sizes of bases of vector spaces without the axiom of choice](https://mathoverflow.net/questions/93242/sizes-of-bases-of-vector-spaces-without-the-axiom-of-choice?newreg=1a6b5b1d756a41139ae970cb60c32e6a) it is said that "It is consistent [with ZF] that there are vector spaces that have two bases with completely different cardinalities". Now I've had a hard time finding a reference for this fact, so my first question would be whether someone knows a proper sources detailing the construction of a vector space with bases of different cardinalities (in some suitable model of $ZF+\neg AC$).
Secondly, is there a general procedure for generating such spaces? (E.g. by starting with a model $M$ of $ZF+\neg AC$ and a vector space $V$ which has no basis in $M$ and then adding several bases of $V$ to $M$ via iterated forcing.)
And lastly, are there only "artifical/exotic" examples of such vector spaces or are there also models where some "standard" vector spaces such as $\mathbb{F}\_2^\omega$ over $\mathbb{F}\_2$ or $\mathbb{R}$ over $\mathbb{Q}$ have bases of different cardinalities. (Pretty sure that $\mathbb{R}$ over $\mathbb{Q}$ is still an open problem but maybe there are some other "nice" examples.)
|
https://mathoverflow.net/users/341908
|
Examples of vector spaces with bases of different cardinalities
|
This is not a very thoroughly studied problem. So to start from the end, there is no standard procedure for this sort of construction. We know of one, it can maybe be adapted slightly to get a mildly more general result, but it's not something like "let's add a new vector space without a basis" or "let's add an amorphous set" that has a very well-understood and common constructions.
The original construction is due to Läuchli and you can find it in German in his paper
>
> *Läuchli, H.*, [**Auswahlaxiom in der Algebra**](http://dx.doi.org/10.1007/BF02566957), Comment. Math. Helv. 37, 1-18 (1962). [ZBL0108.01002](https://zbmath.org/?q=an:0108.01002).
>
>
>
It also appears as an exercise in Jech "The Axiom of Choice" as Problem 10.5 with an elaborate explanation of the proof.
What you suggest is, in principle, a valid approach. Start with no bases, add one, then add another one with a different cardinality. Unfortunately, we don't really understand the mechanism of adding subsets to models of $\sf ZF$ as well, so it's not clear as to how to do that without:
1. Making the space well-orderable to begin with; or
2. collapsing the two bases to have the same cardinality after all; or
3. giving the space a well-orderable basis; or
4. one of many other unforeseen problems that can crop up when you add sets to your universe.
Frankly, I do not understand the construction, mainly because Jech's explanation of it is a bit wishywashy to my taste (it is marked with an asterisk, which at least says that it's not very obvious).
|
15
|
https://mathoverflow.net/users/7206
|
402012
| 164,989 |
https://mathoverflow.net/questions/402002
|
5
|
Denote $A$ $-$ set of positive numbers with only prime factors of the form $4k+1$ and
$B$ $-$ set of positive numbers that can be represented as sum of two squares. $A$ is a subset of $B$ and [there is](https://math.stackexchange.com/questions/2376879/does-the-interval-x-x-10-x1-4-contain-a-sum-of-two-squares) upper bound $2 \sqrt{2} n^{\frac{1}{4}}$ for maximal gap between consecutive elements of $B$.
**Question:** Is any upper bound for maximal gap between consecutive elements of $A$ known?
|
https://mathoverflow.net/users/37289
|
Upper bound for maximal gap between consecutive numbers consisting only $4k+1$ primes
|
One can prove a bound comparable to the $O(n^{1/4})$ bound for $B$. To derive it, notice that any odd number of the form $a^2+b^2$ with $(a,b)=1$ lies in $A$. Now, for a given large number $N$, choose the largest even $a$ such that $a^2<N$. Then, of course $N-a^2=O(\sqrt{N})$. We now want to choose $b$ coprime to $a$ such that $N-a^2-b^2$ is small. Let $j(a)$ be the Jacobsthal function of $a$, i.e. the smallest $j$ such that any set of $j$ consecutive integers contains a number coprime to $a$. Consider the numbers $[\sqrt{N-a^2}], [\sqrt{N-a^2}]-1,\ldots, [\sqrt{N-a^2}]-j+1$. One of them is coprime to $a$, denote it by $b$. Then we get
$$
N-a^2-b^2\ll j(a)\sqrt{N-a^2}+j(a)^2.
$$
Next, by the result of [Iwaniec (1978)](https://doi.org/10.1515/dema-1978-0121), $j(a)\ll \ln^2 a$, so we proved that the size of gaps in $A\cap [1,n]$ is at most $O(n^{1/4}\ln^2 n)$.
|
6
|
https://mathoverflow.net/users/101078
|
402013
| 164,990 |
https://mathoverflow.net/questions/396809
|
0
|
Problem
=======
I was wondering if there are any theoretical results that tackle the following problem:
>
> Construct the following matrices $\mathbf{\mathcal{S}\_{1}},\mathbf{\mathcal{S}\_{2}},\ldots,\mathbf{\mathcal{S}\_{p}}$ all of size $m\times n$ with $m\geqslant n$. The goal is to compute:
> $$
> \mathbf{\mathcal{S}}=\mathcal{S}\_{1}+\mathcal{S}\_{2}+\ldots+\mathcal{S}\_{p}
> $$
> not explicitly
>
>
>
My goal is to add up these matrices not explicitly but rather by performing QR or SVD on each individual matrix and attempting to take advantage of the properties of the (assuming QR decomposition) orthonormal matrices $\mathbf{Q\_{\mathcal{S}\_{1}}},\mathbf{Q\_{\mathcal{S}\_{2}}},\ldots,\mathbf{Q\_{\mathcal{S}\_{p}}}$ and the upper triangular matrices $\mathbf{R\_{\mathcal{S}\_{1}}},\mathbf{R\_{\mathcal{S}\_{2}}},\ldots,\mathbf{R\_{\mathcal{S}\_{p}}}$ somehow (not by literally adding each pair together). Therefore, are there any theoretical result for matrix-matrix addition using matrix decomposition for the sake of computational feasibility and can the same be said using SVD?
Motivation
==========
Many matrix decomposition allow computations to be simpler by exploiting their special properties. The most important aspect of handling matrices numerically is to reduce numerical errors obtained due to finite precision on computers for instance addition operation of matrices are sometimes subject to the issue of absorption. Therefore, it might/might not be useful to look at the properties of matrix decompositions to tackle this issue which is why I am asking this question and I would be grateful for any comment/answer :)
|
https://mathoverflow.net/users/313939
|
Using QR or SVD to sum up finite number of matrices
|
If your matrices happen to be low-rank, you might want to consider a simultaneous low-rank approximation of your matrices (as in [Inoue and Urahama - Equivalence of Non-Iterative Algorithms for Simultaneous Low Rank Approximations of Matrices](https://doi.org/10.1109/CVPR.2006.112)), i.e. you compute
$$
\min\_{A,M\_i,B} \sum\_{i=1}^p \lVert S\_i - A M\_i B \rVert\_F^2,
$$
where $M\_i$ are potentially smaller than $S\_i$. You could then compute $$ S \approx A \bigl( \sum\_{i=1}^p M\_i \bigr) B.$$
I am however not aware of any result on the error of $S$. I would suspect that the error is equal to the sum of the individual low-rank approximation errors.
|
0
|
https://mathoverflow.net/users/135810
|
402014
| 164,991 |
https://mathoverflow.net/questions/401985
|
11
|
Suppose $X$ is a CW complex and $M$ is a closed manifold and suppose further that there exists a homotopy equivalence $X \simeq M$. Does there exists an embedding of $M$ into $X$ (i.e. an injective (potentially cellular) map)?
If this setting is to broad, I'm specifically interested in the case, where $M$ is a surface and $X$ is also $2$-dimensional (maybe even restrict it to aspherical surfaces).
Edit: mme provided a counterexample in dimension 3 (homotopy equivalent but not homeomorphic lens spaces), which can probably be generalized to higher dimensions. So only the two-dimensional case remains.
|
https://mathoverflow.net/users/89741
|
Can we embed a closed manifold into a homotopy equivalent CW complex?
|
Pick a torus, and add two discs along a meridian and a longitude. You get a 2-complex homotopic to a sphere that does not contain a sphere. This generalises easily to any genus by picking a genus-$g$ surface.
More generally, a finite 2-complex contains finitely many surfaces, and there are some moves (like the Matveev - Piergallini move) that preserve the (simple) homotopy type of the 2-complex, but can modify the surfaces it contains.
|
15
|
https://mathoverflow.net/users/6205
|
402027
| 164,997 |
https://mathoverflow.net/questions/402034
|
8
|
Let $n > 0$ be a positive integer (large) and $p > 2$ a fixed prime number. What is the probability that $$\sum\_{ 1 \leq i < j \leq n} a\_ia\_j = 0 \mod p$$ where $a\_1, a\_2, \dots a\_n$ are chosen uniformly from the set $S = \{-1, 1\}$. Does this sum equidistribute mod $p$ as $n$ goes to infinity? What would be the speed of equidistribution in terms of $n$? Is there any literature in this type of random sums? I would be surprised if not but I am unable to find anything related or similar to this.
One can also ask what is the probability of this sum being actually zero, but I also have no idea how to deal with it and thought that modulo a prime would be simpler.
|
https://mathoverflow.net/users/7894
|
Question about estimating random symmetric sums modulo p
|
The condition
$$\sum\_{ 1 \leq i < j \leq n} a\_ia\_j \equiv 0 \pmod p$$
is equivalent to
$$\left(\sum\_{ 1 \leq i\leq n} a\_i\right)^2 \equiv n \pmod p.$$
So a necessary condition is that $n$ is a quadratic residue modulo $p$ (including the zero residue). If $n$ is divisible by $p$, then the above condition says that the sum of the $a\_i$'s is divisible by $p$. Otherwise, the condition says that the sum of the $a\_i$'s is congruent to one of the two square-roots of $n$ modulo $p$. Now it is easy to see that the sum of the $a\_i$'s is equidistributed modulo $p$ (think about what happens when an $a\_i=1$ is switched to $a\_i=-1$), hence in the first case the probability is $1/p+o(1)$, in the second case it is $2/p+o(1)$, as $n$ tends to infinity.
In fact the probabilities can be calculated explicitly as a linear combination of $n$-th powers of $p$ complex numbers (which only depend on $p$), since the sum of the $a\_i$'s modulo $p$ is determined by $\#\{i:a\_i=1\}$ modulo $p$, and vice versa. Compare with [this post](https://math.stackexchange.com/questions/2005021/sum-of-binomial-coefficients-with-index-divisible-by-4), where the role of $p$ is played by $4$. It follows, in particular, that the $o(1)$ terms above decay exponentially fast. For a more complete reference, see Theorems 8.7.2 & 8.7.3 in Wagner: A first course in enumerative combinatorics (AMS, 2020).
|
14
|
https://mathoverflow.net/users/11919
|
402038
| 165,002 |
https://mathoverflow.net/questions/402043
|
3
|
$\renewcommand{\S}{\mathcal{S}}\newcommand{\l}{\langle}\newcommand{\r}{\rangle}\newcommand{\op}{\mathsf{op}}\newcommand{\fin}{\mathrm{fin}}$Recently I've noticed that the definitions of special $\Gamma$-spaces and spectra are quite close in spirit:
* **$\Gamma$-spaces** are pointed functors $X\colon(\Gamma^\op,\l0\r)\to(\mathcal{S},\*)$ from Segal's category to the category $\mathcal{S}$ of spaces. Moreover, we call $X$ (see Definition 7.1 [here](https://arxiv.org/abs/1701.06459))
1. **special** if, for each $\l n\r,\l m\r\in\mathrm{Obj}(\Gamma)$, the map
$$X\_{\l n\r\vee\l m\r}\to X\_{\l n\r}\times X\_{\l m\r}$$
induced by the inert surjections $\l n\r\vee\l m\r\to\l n\r$ and $\l n\r\vee\l m\r\to\l m\r$ is a weak equivalence.
2. **very special** if it is special and equivalently
+ $\pi\_0(X\_{\l1\r})$ is a group.
+ The map
$$X\_{\l 2\r}\to X\_{\l1\r}\times X\_{\l1\r}$$
induced by the total map $\l2\r\to\l1\r$ and one of the two inert surjections $\l2\r\to\l1\r$ is a weak equivalence.
* **Spectra** are reduced excisive functors $E\colon\mathcal{S}^\fin\_\*\to\S$ of $\infty$-categories, where
1. $E$ is **excisive** if it sends pushouts to pullbacks.
2. $E$ is **reduced** if $E(\*)\simeq \*$;
Since very special $\Gamma$-spaces are equivalent to connective spectra, this made me wonder: can we view nonconnectivity as arising from enlarging Segal's category $\Gamma^{\mathsf{op}}\overset{\mathrm{def}}{=}\mathsf{Sk}(\mathsf{FinSets}\_\*)$ of finite pointed sets into the $\infty$-category $\S^\fin\_\*$ of finite pointed spaces?
---
In particular, two natural intermediate steps to consider between $\Gamma^{\mathsf{op}}$ and $\S^\fin\_\*$ are, for each $n\in\mathbb{N}$, the $\infty$-categories $\S^{\fin}\_{\leq n,\*}$ and $\S^{\fin}\_{>n,\*}$ of $n$-truncated and $n$-connected spaces, which come with natural inclusions of $\infty$-categories $\iota\_{1},\iota\_{2}\colon\S^{\fin}\_{\leq n,\*},\S^{\fin}\_{>n,\*}\hookrightarrow\S^\fin\_\*$.
**Questions:**
1. Can we describe functors $E\colon\S^{\fin}\_{\leq n,\*}\hookrightarrow\S$ or $E\colon\S^{\fin}\_{>n,\*}\hookrightarrow\S$ in terms of spectra? Are the former maybe somehow related to $(-n-1)$-connected spectra?
2. Since $\mathcal{S}$ is co/complete, we can consider left and right Kan extensions along $\iota\_{1}$ and $\iota\_{2}$, giving functors of the form
$$\mathrm{Lan}\_{\iota\_{1}}\colon\mathsf{Exc}\_\*(\mathcal{S}^\fin\_{\leq n,\*})\to\mathsf{Sp}.$$
What are the essential images of $\mathrm{Lan}\_{\iota\_{1}}$, $\mathrm{Lan}\_{\iota\_{2}}$, $\mathrm{Ran}\_{\iota\_{1}}$, and $\mathrm{Ran}\_{\iota\_{2}}$?
---
Lastly, let me mention that this was asked also by Jonathan on the [Homotopy Theory Discord](https://discord.com/channels/801815065165955143/806351863595270184/877221475294269501). There, Rune Haugseng mentioned [this paper](https://arxiv.org/abs/1703.09764) of Harpaz, where one finds a related construction. Harpaz uses spans of $n$-finite spaces (meaning $n$-truncated spaces which additionally have finite homotopy groups), formulating a notion of $n$-commutative monoid for finite $n$ (which has since been extended to the $n=\infty$ case by [Carmeli–Schlank–Yanovski](https://arxiv.org/abs/2007.13089)). These are related to $n$-semiadditivity, and extend the commutativity ladder $\mathbb{E}\_{1}$, $\mathbb{E}\_{2}$, $\ldots$, $\mathbb{E}\_{\infty}$ of monoid objects in an $\infty$-category in such a way that
1. A $(-2)$-commutative monoid is precisely an $\mathbb{E}\_{-1}$-monoid (where $\mathbb{E}\_{-1}\overset{\mathrm{def}}{=}\mathsf{Triv}^\otimes$), meaning just an object;
2. A $(-1)$-commutative monoid is precisely an $\mathbb{E}\_{0}$-monoid, meaning a pointed object;
3. A $0$-commutative monoid is precisely an $\mathbb{E}\_{\infty}$-monoid.
One could imagine also more generally expanding the definition of $\infty$-operad, replacing $\mathrm{N}\_{\bullet}(\mathsf{Fin}\_\*)$ by $\S^\fin\_\*$ and variants. For spans in $m$-finite spaces, this should give a notion of $m$-operad, as pointed out by Shachar Carmeli on the homotopy Discord.
---
**Edit:** This question has been mostly answer by Marc Hoyois and Dmitri Pavlov below (thanks!). The only remaining part is what are reduced excisive functors $\mathcal{S}^{\mathrm{fin}}\_{\*,\leq n}\to\mathcal{S}$, which I've split into a separate question [here](https://mathoverflow.net/questions/402190).
|
https://mathoverflow.net/users/130058
|
Restricting spectra to finite $n$-truncated/$n$-connected pointed spaces
|
>
> $\renewcommand{\S}{\mathcal{S}}\newcommand{\l}{\langle}\newcommand{\r}{\rangle}\newcommand{\op}{\mathsf{op}}\newcommand{\fin}{\mathrm{fin}}$can we view nonconnectivity as arising from enlarging Segal's category $\Gamma^{\mathsf{op}}\overset{\mathrm{def}}{=}\mathsf{Sk}(\mathsf{FinSets}\_\*)$ of finite pointed sets into the $\infty$-category $\S^\fin\_\*$ of finite pointed spaces?
>
>
>
Yes, the easiest way to see this is to observe that
the category $Γ^\op$ of finite pointed sets embeds in the category $\S^\fin\_\*$ by sending a finite pointed set to the corresponding discrete pointed simplicial set.
The restriction functor along this inclusion induces a right Quillen functor from simplicial W-spectra to Γ-spaces.
In order to see this, it suffices to show that local objects are preserved,
i.e., reduced excisive functors restrict to very special Γ-spaces.
Consider some reduced excisive functor $E$ together
with its restriction to finite pointed sets,
which we want to show is a very special Γ-space.
A Γ-space is special if it sends finite coproducts to finite homotopy products.
Finite coproducts are generated by empty coproducts and binary coproducts.
The coproduct of the empty family is the initial pointed set,
which is sent to a weakly contractible space precisely if the original functor $E$ is reduced.
Excisive functors send binary coproducts to homotopy products by definition.
A Γ-space $E$ is very special if $π\_0(E\_1)$ is a group.
This follows immediately from considering the homotopy pushout diagram for $⟨0⟩←⟨1⟩→⟨0⟩$ (with some apex $A$), which is sent to a homotopy pullback diagram with legs $E\_0←E\_1→E\_0$.
Since $E\_0$ is contractible, this homotopy pullback diagram
exhibits $E\_1$ as the loop space of $E(A)$, so $π\_0(E\_1)$ is a group.
|
1
|
https://mathoverflow.net/users/402
|
402050
| 165,004 |
https://mathoverflow.net/questions/401967
|
12
|
This question is about logical complexity of sentences in third order arithmetic. See [Wikipedia](https://en.wikipedia.org/wiki/Arithmetical_hierarchy) for the basic concepts.
Recall that the Continuum Hypothesis is a $\Sigma^2\_1$ sentence. Furthermore (loosely speaking) it can't be reduced to a $\Pi^2\_1$ sentence, as stated in [Emil Jeřábek's answer to *Can we find CH in the analytical hierarchy?*](https://mathoverflow.net/a/218649/170446).
Is there an example of a $\Sigma^2\_2$ sentence with no known reduction to a $\Pi^2\_2$ sentence? (Equivalently, a $\Pi^2\_2$ sentence with no known reduction to a $\Sigma^2\_2$ sentence.) I mean that there should be no known reduction even under large cardinal assumptions.
I'd prefer an example that's either famous or easy to state. But to begin, any example will do.
*Update:* Sentences such as "$\mathfrak{c} \leqslant \aleph\_2$" and "$\mathfrak{c}$ is a successor cardinal" are $\Delta^2\_2$, meaning that they're simultaneously $\Sigma^2\_2$ and $\Pi^2\_2$. The reason is that each such sentence (and also its negation) can be expressed in the form "$\mathbb{R}$ has a well-ordering $W$ such that $\phi(W)$" where $\phi$ is $\Sigma^2\_2$.
|
https://mathoverflow.net/users/170446
|
Example of a $\Pi^2_2$ sentence?
|
The Suslin hypothesis is $\Pi^2\_2,$ and $T = ZFC + GCH + LC$ (LC an arbitrary large cardinal axiom) does not prove it to be equivalent to any $\Sigma^2\_2$ sentence. Suppose toward contradiction $T$ proves SH to be equivalent to $\exists A \subset \mathbb{R} \varphi(A),$ where $\varphi$ is $\Pi^2\_1.$ Assume $V \models T.$
We'll use several results from Chapters VIII and X of Devlin and Johnsbraten's *The Souslin Problem.* There are generic extensions $V[G] \models T+\diamondsuit^\*$ and $V[G][H] \models T+SH$ which do not add reals to or collapse cardinals of $V.$ In $V[G][H],$ there is $A \subset \mathbb{R}$ such that $\varphi(A)$ holds and $A' \subset \omega\_1$ which codes a bijection between $\mathbb{R}$ and $\omega\_1$ as well as $A.$ By downwards absoluteness of $\varphi,$ $A$ witnesses that $V[G][A'] \models SH.$ But we also have $V[G][A'] \models \diamondsuit^\*$ by Lemma 4 (pg. 79), which is a contradiction since $\diamondsuit$ negates SH.
|
9
|
https://mathoverflow.net/users/109573
|
402053
| 165,006 |
https://mathoverflow.net/questions/402056
|
3
|
Assume $\Omega$ is a smoothly bounded open domain in $ \mathbb R^n$ and $B$ is an open ball of equal volume. Let $\Phi:\partial\Omega\to \partial B$ be a diffeomorphism. Is it true that there exists a volume preserving diffeomorphism $\tilde{\Phi}:\overline{\Omega}\to\overline{B}$ such that $\tilde{\Phi}\vert\_{\partial\Omega}=\Phi$?
|
https://mathoverflow.net/users/79956
|
Extension of volume preserving diffeomorphism
|
Let $\Omega$ be a submanifold of $\mathbb R^n$ with boundary $\partial\Omega$ diffeomorphic to $\partial B$. As far as I know, it is not obvious that $\overline\Omega$ should be diffeomorphic to $\overline B$, and my guess would be that this is actually false in all generality. From what I understand, though, it is true if you assume $\Omega$ is contractible and $n\geq5$, see [this question](https://mathoverflow.net/q/324229/).
So there might be a topological obstruction. Once you get past this (for instance if you already assume $\overline\Omega$ is diffeomorphic to $\overline B$), then it reduces to the same question with $\Omega=B$. Now there might be another topological obstruction, which is that the diffeomorphism of the boundary may not extend to a diffeomorphism of the whole ball, see [this answer](https://mathoverflow.net/a/80857/).
Again, if you already know that $\overline\Omega$ is diffeomorphic to $\overline B$, then these obstructions cannot happen, more or less by definition. Then your question becomes: let $\omega\_1,\omega\_2$ be two smooth volume forms on $B$ with the same global volume, is there a diffeomorphism sending one to the other? This is a Moser-type theorem, and the version for manifolds with boundary is proved in [Formes-volumes sur les variétés à bord](https://zbmath.org/?q=an%3A0281.58001) (volume forms on manifolds with boundary) by A. Banyaga (1974).
|
5
|
https://mathoverflow.net/users/129074
|
402058
| 165,008 |
https://mathoverflow.net/questions/332091
|
4
|
Let $(A,A^+)$ be a sheafy Tate-Huber pair, and let $X=\operatorname{Spa}(A,A^+)$. It is well-known that $H^i(X,\mathcal{O}\_X)=0$ for $i>0$. I assume it is generally not true that $H^i(X,\mathcal{O}\_X^+)=0$ for $i>0$, but I don't think that I have seen an explicit counterexample. Is there a simple example of a nonzero cohomology class in some $H^i(X,\mathcal{O}\_X^+)$ for $i>0$?
|
https://mathoverflow.net/users/93798
|
Failure of Tate acyclicity for integral structure sheaves
|
Below is an example of a formal scheme $\mathfrak{X}$ such that its rigid analytic generic fiber $X$ is affinoid, while its special fiber contains a complete elliptic curve $E$. Because $H^1(E,\mathcal{O}\_E) \ne 0$, $H^1(X,\mathcal{O}\_X^+) \ne 0$ as well.
Let $p$ be an odd prime, and let $X$ be the rigid analytic elliptic curve over $\mathbb{Q}\_p$ defined by $$y^2 = x^3-x, \quad |x| \le |p^{-2}|\,.$$
It has a formal model over $\mathfrak{X}$ over $\mathbb{Z}\_p$ that is covered by two affines
$$\mathfrak{U}=\operatorname{Spf}\mathbb{Z}\_p\left<x,y\right>/(y^2-x^3+x)$$
$$\mathfrak{V}=\operatorname{Spf}\mathbb{Z}\_p\left<u,v,w\right>/(v-u^3+uv^2,uw-p)$$
where $u=x/y$, $v=1/y$, $w=py/x$. These correspond to rational subsets $U$ and $V$, defined by the inequalities $|x| \le 1$ and $|y| \ge 1$, respectively.
One can check that the above rings are normal using Serre's criterion. They are also flat over $\mathbb{Z}\_p$. By Theorem 7.4.1 in de Jong's paper [*Crystalline Diedonné module theory via formal and rigid geometry*](http://www.numdam.org/item/PMIHES_1995__82__5_0/), the above rings can be identified with $\mathcal{O}\_X^+(U)$ and $\mathcal{O}\_X^+(V)$.
Now consider $y/x = xu - v \in \mathcal{O}\_X^+(U \cap V)$. I claim that $y/x \notin \mathcal{O}\_X^+(U) + \mathcal{O}\_X^+(V)$. Indeed, consider the elliptic curve defined by $p=0$ on $\mathfrak{U}$ and $w=0$ on $\mathfrak{V}$. On this curve, $y/x$ has simple poles at the points $x=y=0$ and $u=0$, while elements of $\mathcal{O}\_X^+(U) + \mathcal{O}\_X^+(V)$ have either no pole or a pole of order $\ge 2$ at these points. Hence $y/x$ determines a nontrivial element of $H^1(X,\mathcal{O}\_X^+)$.
Since $H^1(X,\mathcal{O}\_X)=0$, the above cocycle must be $p$-power torsion. In fact, it is annihilated by $p$, since $py/x=w$.
|
3
|
https://mathoverflow.net/users/93798
|
402070
| 165,012 |
https://mathoverflow.net/questions/402071
|
5
|
**Motivation.** (Please skip if you are not in the mood for "chitchat".) Last night I listening to a classical radio station, and for the umpteenth time, they played Mendelssohn's [Psalm 42](https://en.wikipedia.org/wiki/Psalm_42_(Mendelssohn)), a composition that I like very much. Luckily, a week ago, when they played it, it was followed by a different piece (Rodeo by Copland) than yesterday (Bach d-minor piano concerto). I wondered how long they can proceed so that piece $X$ is never immediately followed by piece $Y$ two separate times. Which led to the following little problem.
**Formalization.** We regard any positive integer $n$ as the set of its predecessors, so $n = \{0,\ldots,n-1\}$. For positive integers $m, n\in \mathbb{N}$ we say that a map $f : m\to n$ is a *radio-playing* function if whenever $a,b \in m-1$ with $a\ne b$ and $f(a) = f(b)$, then $f(a+1) \neq f(b+1)$.
Using the pigeonhole principle, it is easy to see that if $m > n^2$ there cannot be a radio-playing function $f : m\to n$. So, given $n\in \mathbb{N}\setminus \{0\}$, let $A\_n$ be the largest integer such that there is a radio-playing function $f: A\_n \to n$.
What is the value of $A\_n$ in terms of $n$?
|
https://mathoverflow.net/users/8628
|
Radio-playing sequence
|
I think you're just asking for a de Bruijn sequence of order $2$ on $n$ symbols, in which case the answer is $n^2$ because the non-simple digraph on $n$ vertices where $u \to v$ for every $u, v$ (including $u=v$) and the edges $u \to v$ and $v \to u$ are considered distinct unless $u=v$ is Eulerian.
|
10
|
https://mathoverflow.net/users/46140
|
402072
| 165,013 |
https://mathoverflow.net/questions/402076
|
12
|
I would like to know (as part of an attempt to streamline some calculations in the cohomology of a Morava stabiliser group) whether $1170\sqrt{-3}\sqrt{5}\sqrt{-7}-19110$ is a square in $\mathbb{Q}(\sqrt{-3},\sqrt{5},\sqrt{-7})$. What is an efficient method for this kind of question?
|
https://mathoverflow.net/users/10366
|
Squares in a triquadratic field
|
The field $L=\mathbb Q(\sqrt{-3},\sqrt{5},\sqrt{-7})$ has a lot of intermediate fields which we can exploit. Pick one of its index two subextensions, say $K=\mathbb Q(\sqrt{-3},\sqrt{5})$. The element $\alpha=1170\sqrt{-3}\sqrt{5}\sqrt{-7}-19110$ of $L$ has as its only conjugate over $K$ the element $-1170\sqrt{-3}\sqrt{5}\sqrt{-7}-19110$, and so the norm of $\alpha$ with respect to the extension $L/K$ is $$N\_{L/K}(\alpha)=-1170^2\cdot(-3)\cdot 5\cdot(-7)+19110^2=221457600 = 2^6\cdot 3^2\cdot 5^2\cdot 7\cdot13^3.$$
This norm is not a square in $K$ - for otherwise $K$ would ramify at $7$ and $13$. Therefore $\alpha$ is not a square in $L$.
This method should generalize quite nicely if you consider elements of the form $a+b\sqrt{n}$ inside a multiquadratic extension $L/\mathbb Q$, as picking a subfield $K$ of index $2$ not containing $n$, the norm will be given by $a^2-bn^2$, and if the factorization of this integer contains some primes in odd exponents which do not ramify in $K$, you know it will not be a square in $K$. It probably won't always give you the right answer but it should work in many cases of interest.
|
15
|
https://mathoverflow.net/users/30186
|
402079
| 165,015 |
https://mathoverflow.net/questions/402075
|
3
|
Say you have a 2D broken line you move along, but only some directions are allowed (I give you the angles relative to the usual cartesian plane):
1. (Up-Left): $]\pi, \dfrac{\pi}{2}[$
2. (Down-Left): $]-\pi, \dfrac{-\pi}{2}[$
3. (Down-Right): $ ]\dfrac{-\pi}{2}, 0[$
An additional rule is that you cannot go *Up-left if you just went Down-right*, and vice-versa.
The goal is to prove that if you want to make a cycle out of this line, you always end up with a line that cross it-self.
**Context**
I'm working on unit-square graphs with a clique number of 2. The directions correspond to which corner of some squares lies inside the other. I'm trying to define some order onto the squares.
This is the last point of a proof that would allow the use of some representation of a graph, ending up with very easy to prove and useful small results.
|
https://mathoverflow.net/users/342793
|
Broken line that can go in specific directions: can it end up on its starting point?
|
There has to be a self-intersection; the path cannot be a simple polygon.
Suppose the path were indeed simple. A vertex with maximal $y$ coordinate must connect edges that go up and then down. According to the rules, the only way this can happen is first up-left, then down-left. [For a simple polygon](https://en.wikipedia.org/wiki/Curve_orientation#Orientation_of_a_simple_polygon), this implies the path is oriented counterclockwise. On the other hand a vertex with minimal $y$-coordinate must connect edges that go down-left, then up-left, implying the path is oriented clockwise: a contradiction.
|
3
|
https://mathoverflow.net/users/1227
|
402089
| 165,018 |
https://mathoverflow.net/questions/402099
|
5
|
In the Wikipedia article (<https://en.wikipedia.org/wiki/Quantifier_elimination#cite_note-4>) it is said that every abelian group has quantifier elimination property and a long old paper of W. Szmielew (1955) is given as a reference. But as far as I know, this is true for special classes of abelian groups (like divisible, ordered, ... groups). I am not an expert of model theory but I need to apply quantifier elimination for reduced abelian groups (abelian groups the only divisible subgroup in which is the trivial one), so I need to know if really we have quantifier elimination for any abelain group and if not, what is the most complex quantifier combination of formulas in such groups.
|
https://mathoverflow.net/users/44949
|
Quantifier elimination for abelian groups
|
Abelian groups are the same thing as $\mathbb Z$-modules. In general, for any ring $R$, the theory of left $R$-modules has quantifier elimination down to Boolean combinations of primitive positive formulas and certain sentences (expressing so-called Baur–Monk invariants). This is the Baur–Monk quantifier elimination theorem; see e.g. §A.1 in Hodges, *Model Theory*.
In particular, since the theory of any particular abelian group or $R$-module is complete, it has quantifier elimination down to Boolean combinations of p.p. formulas.
Note that a p.p. formula expresses solvability of a linear system; in the case where $R$ is a PID (including abelian groups with $R=\mathbb Z$), the Smith normal form shows that you can reduce to the case of p.p. formulas with one quantifier. That is, every formula $\phi(\vec x)$ is equivalent to a Boolean combination of invariant sentences and the divisibility formulas $a\mid\sum\_ia\_ix\_i$ for some $a,a\_i\in R$. (For $a=0$, this amounts to the original atomic formulas $\sum\_ia\_ix\_i=0$.)
|
10
|
https://mathoverflow.net/users/12705
|
402100
| 165,020 |
https://mathoverflow.net/questions/402060
|
3
|
Just for fun I was trying to find a formula that calculates the value of the sum of the Riemann zeta non trivial roots raised to a power $n$, $Z(n)$.
$$Z(n) = \sum\_{\rho} ' \frac{1}{\rho ^n}$$
I managed to find one monster of an equation after a while and it seems to work fine for $n=1$ but when I try $n=3$ the sum I have is almost exactly the same as the one given on Wolfram but with a negative sign on the $3\gamma \gamma\_1 $ term.
According to the relation I derived there should be alternating signs in the gamma terms but Wolfram disagrees.
If it is needed I can type down my work, but basically I worked out that for $n>1$
$$
Z(n) = 1 - \frac{2^n - 1}{2^n} \zeta (n) + \sum\_{k=1}^{n} \frac{(-1)^{n-1-k} (k-1)!}{(n-1)!} B\_{n,k} ((-1)^{n-k+1}(n-k+1)\gamma \_{n-k}.
$$
For $n=1$ the equation it's slightly different but I have confirmed that case to be true.
Plugging in $n=3$ for this equation gives
$$
Z(3)=1+\frac{3}{2} \gamma \_2 -3\gamma \gamma\_1 +\gamma^3 - \frac{7}{8} \zeta (3)
$$
Wolfram gives the value
$$
Z(3)=1+\frac{3}{2} \gamma \_2 +3\gamma \gamma\_1 +\gamma^3 - \frac{7}{8} \zeta (3)
$$
instead. Can anybody prove the Wolfram version so at least I can try to find where I went wrong?
The $B\_{n,k}$ are the Bell polynomials that I used in Faa di Bruno's formula to calculate $Z(n)$ and I shorthanded the notation slightly because it was too long.
Link to the page:
<https://mathworld.wolfram.com/RiemannZetaFunctionZeros.html>
|
https://mathoverflow.net/users/342532
|
Proof of the sum of the reciprocal non trivial zeros cubed
|
This comes directly from the Hadamard product given in the Wolfram page you
refer to by taking logarithmic derivatives and identifying powers of $s$.
However, following Harold Stark, the classical formula given in Wolfram should
be replaced by the much simpler formula
$s(s-1)\Lambda(s)=\prod\_{\rho}(1-s/\rho)$, where the product is to be
understood as the limit as $T\to\infty$ of $\prod\_{|\Im(\rho)|<T}$, and
as usual $\Lambda(s)=\pi^{s/2}\Gamma(s/2)\zeta(s)$. Now take logarithmic
derivatives and the formula follows.
One can also easily obtain a recursion (equivalent to your use of Bell
polynomials): define by induction
$$\delta\_{k+1}=(k+1)\dfrac{\gamma\_k}{k!}+\sum\_{j=0}^{k-1}\dfrac{\gamma\_j\delta\_{k-j}}{j!}\quad\text{($\delta\_1=\gamma\_0=\gamma)$}\;.$$
Then $Z(k)=1-(1-1/2^k)\zeta(k)+\delta\_k\;.$
|
5
|
https://mathoverflow.net/users/81776
|
402104
| 165,021 |
https://mathoverflow.net/questions/363297
|
19
|
In his PhD thesis, [Categorical Structure of Continuation Passing Style](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.9.3717&rep=rep1&type=pdf), Thielecke studies the ⊗¬-categories, which are premonoidal categories with structure (namely, a functor ¬ which is adjoint to itself), in order to model the semantics of continuation-passing style languages.
I've been studying his work for some time now, but I'm not really familiar to literature about category theory. So my question is, and it might sound silly: how should I pronounce "⊗¬-category"? In fact, I could rephrase this: if I were to write about those categories in a paper, how should I refer to them? Should one write it in full, or simply just refer to them as ⊗¬-categories all the time and leave the pronunciation as an open problem?
|
https://mathoverflow.net/users/142280
|
How to pronounce "⊗¬-category"?
|
I shared an office with Hayo Thielecke in the late 1990s.
The pronunciation he used was "Tensor-NOT-category".
|
17
|
https://mathoverflow.net/users/170446
|
402107
| 165,023 |
https://mathoverflow.net/questions/402051
|
5
|
Fix a finite set of integers $S$ and a prime number $p$. Let $(a\_1, a\_2, \dotsc, a\_n)$, $(b\_1, b\_2, b\_3, \dotsc, b\_n)$ be two sequences of integers where the numbers $a\_i$ and $b\_i$ are chosen uniformly from the set $S$ (In the case I care about S = {-2,-1,1,2 }). I would like to understand what is the probability that $$\sum\_{i\_1 < i\_2 < \dotsb < i\_k } a\_{i\_1}a\_{i\_2}\dots a\_{i\_k} = 0 \mod p$$ an the probability that $$\sum\_{i\_1 \leq j\_1 < i\_2 \leq j\_2 < \dotsb < i\_k \leq j\_k} a\_{i\_1}b\_{j\_1}\dots a\_{j\_k}b\_{j\_k} = 0 \mod p.$$
I am interested in the asymptotic probabilities as $n$ increases and $k$ is fixed (or $k$ grows much slower than $n$) and expect some type of equidistribution as $n$. One particular case is [Question about estimating random symmetric sums modulo p](https://mathoverflow.net/questions/402034/question-about-estimating-random-symmetric-sums-modulo-p), where one particular sum is reduced to the linear case. Can this trick be for the general case?
Any suggestion, trick or vaguely related information related to this is highly appreciated.
**P.S.** Can the second type of sum (the one involving $a\_i$, $b\_i$) be somehow related to a sum of the first type?, the ordering makes things much more tricky and I am wondering whether there is a procedure to reduce this sum to the first type. (Also the second sum even though not so natural is the one that I really need to understand).
Thanks!
|
https://mathoverflow.net/users/7894
|
Distribution of some sums modulo p
|
Using the [Newton identities](https://en.wikipedia.org/wiki/Newton%27s_identities), one can (in the high characteristic regime $p>k$) express the elementary symmetric polynomial $\sum\_{i\_1 < \dots < i\_k} a\_{i\_1} \dots a\_{i\_k}$ in terms of the moments $\sum\_{i=1}^k a\_i^j$ for $j=1,\dots,k$. The question then boils down to the equidistribution of $\sum\_{i=1}^k (a\_i^j)\_{j=1}^k$ in ${\mathbf F}\_p^k$. There is however an obstruction to equidistribution because the points $(a^j)\_{j=1}^k$ for $a=-2,-1,1,2$ do not span the entirety of ${\mathbf F}\_p^k$ once $k \geq 4$, due to the existence of non-trivial polynomials $P$ of degree $4$ or higher that vanish at $-2,-1,1,2$. For instance because $P(x) = (x+2)(x+1)(x-1)(x-2) = x^4 - 5 x^2 + 4$ vanishes at these points, the vector $(m\_j)\_{j=1}^k := \sum\_{i=1}^k (a\_i^j)\_{j=1}^k$ is surely constrained to the hyperplane $m\_4 - 5 m\_2 + 4k = 0$. This is going to create some distortions to the probability that $\sum\_{i\_1 < \dots < i\_k} a\_{i\_1} \dots a\_{i\_k}$ vanishes mod $p$ that can be explicitly calculated for each $p,k$, but the formula is going to be messy (these errors will be of lower order in the limit $p \to \infty$ holding $k$ fixed though, by the Lang-Weil estimates).
(To put it another way, one can use the Newton identities to express the elementary symmetric polynomial as some polynoimal combination of the frequencies $d\_{-2}, d\_{-1}, d\_1, d\_2$ that count how often the random sequence $a\_1,\dots,a\_k$ attains each of its permitted values $-2, -1, 1, 2$. To count the probability that this polynomial vanishes mod $p$, one has to count the points in some variety over ${\bf F}\_p$ which in general is a task for the Lang-Weil estimate.)
Your second sum seems to be expressible in terms of a symmetric polynomial in a suitable matrix algebra. If one had $i\_k < j\_k$ in place of the constraint $i\_k \leq j\_k$ then one just needs to take the order $2k$ elementary symmetric polynomial of the matrices $\begin{pmatrix} 0 & b\_i \\ a\_i & 0 \end{pmatrix}$ and extract the top left coefficient. With $i\_k \leq j\_k$ the situation is more complicated but I expect there is still some sort of matrix representation. However being non-abelian I doubt there is a reduction to the abelian elementary symmetric polynomial considered earlier, and given how complicated that formula already was I'm afraid it is not going to be fun to try to control this sum (except possibly in the asymptotic regime where $p$ is somewhat large and $k$ goes to infinity; actually the nonabelian case might be substantially more "mixing" than the abelian one and one could conceivably get better asymptotics by using some of the theory of expansion of Cayley graphs, but this looks like a lot of work...).
|
7
|
https://mathoverflow.net/users/766
|
402109
| 165,024 |
https://mathoverflow.net/questions/319745
|
8
|
Recall that the permanent of an $n\times n$ matrix $A=[a\_{i,j}]\_{1\le i,j\le n}$ is defined by
$$\operatorname{per}A=\sum\_{\sigma\in S\_n}\prod\_{i=1}^n a\_{i,\sigma(i)}.$$
In 2004, R. Chapman [Acta Arith. 115(2004), 231-244] determined the value of
$$\det\left[\left(\frac{i+j-1}p\right)\right]\_{1\le i,j\le(p+1)/2}=\det\left[\left(\frac{i+j}p\right)\right]\_{0\le i,j\le(p-1)/2},$$
where $(\frac{\cdot}p)$ is the Legendre symbol. Motivated by this, here I pose the following conjecture involving permanents.
**Conjecture.** For each $n=0,1,2,\ldots$, we have
$$\operatorname{per}\left[\left(\frac{i+j}{2n+1}\right)\right]\_{0\le i,j\le n}>0,\tag{$\*$}$$
where $(\frac{\cdot}{2n+1})$ is the Jacobi symbol.
Let $a\_n$ denote the permanent in $(\*)$. Via Mathematica I find that
\begin{gather}a\_0=a\_1=1,\ a\_2=a\_3=2,\ a\_4=20,\ a\_5=16,\ a\_6=48,\ a\_7=55,
\\a\_8=128,\ a\_9=320,\ a\_{10}=1206,\ a\_{11}=768,\ a\_{12}=406446336,
\\a\_{13}=43545600,\ a\_{14}=141312,\ a\_{15}=2267136,\ a\_{16}=389112,
\\a\_{17}=1624232,\ a\_{18}=138739712,\ a\_{19}=122605392,\ a\_{20}=2262695936,
\\a\_{21}=20313407488,\ a\_{22}=17060393728,\ a\_{23}=189261676544,
\\a\_{24}=374345132371011500507136,\ a\_{25}=669835780976.
\end{gather}
I don't know how to solve the conjecture. Any ideas towards the solution?
|
https://mathoverflow.net/users/124654
|
Is the permanent of the matrix $[(\frac{i+j}{2n+1})]_{0\le i,j\le n}$ always positive?
|
This is the sequence A322898 in OEIS.
I used a program in PARI and calculated the values of a(26) to a(34).
```
a(26) = -7000008163328
a(27) = 22712032822272
a(28) = 2244036651776
a(29) = 4363027965018112
a(30) = 30229121955004416
a(31) = -46693326700068864
a(32) = -23328907207088128
a(33) = 3173005987716005888
a(34) = 136427303851761536
```
The conjecture is **disproved**.
|
11
|
https://mathoverflow.net/users/43683
|
402122
| 165,030 |
https://mathoverflow.net/questions/402102
|
2
|
On p.112-113 of volume 10-1 of Gauss's werke, which contain an unpublished fragment dated to 1805, Gauss states some results on cubic and biquadratic "Gaussian periods" in a trigonometric form. While the result on the cubic period was already published by Gauss in a footnote to article 358 of his Disquisitions Arithmeticae (1801), the result on biquadratic period has apparently not been published by him. Here is a Google translation of this fragment:
>
> **Cubic**: Let $4p = a^2+27b^2$, so that: $\frac{a}{\sqrt{4p}} = cos\varphi, \frac{b\sqrt{27}}{\sqrt{4p}} = sin\varphi$. Then the period of order $\frac{1}{3}(p-1)$ is: $$-\frac{1}{3}+\frac{2}{3}cos\frac{1}{3}\varphi \sqrt{p}$$
> **Biquadratic**: Let $p = a^2+4b^2$, so that: $\frac{a}{\sqrt{p}} = cos\varphi, \frac{2b}{\sqrt{p}} = sin\varphi$. Then the period of order $\frac{1}{4}(p-1)$ is: $$-\frac{1}{4}+\frac{1}{4}\sqrt{p}+\frac{1}{2}\sqrt{p}\cdot cos\frac{1}{2}\varphi = \prod$$, and: $${\prod}^0 + i\prod'-\prod''- i\prod''' = \sqrt{p}\cdot (\frac{a+b\sqrt{-4}}{a-b\sqrt{-4}})^{\frac{1}{4}}$$
>
>
>
As far as i understand, a "Gaussian period" of $\frac{1}{4}(p-1)$ terms is a complex exponential sum $\sum e^\frac{2\pi in}{p}$ when $n$ goes through all the quartic residues modulo $p$. Although
Gauss doesn't define ${\prod}^0, \prod',\prod'',\prod'''$, i believe that ${\prod}^0$ represents a complex exponential sum over values of $n$ which belong to the subgroup of quartic residues modulo $p$ (subgroup of the larger group of invertible residues modulo $p$), and $\prod',\prod'',\prod'''$ are complex exponential sums over its cosets, arranged according to some order. One can see that Gauss weighted ${\prod}^0, \prod',\prod'',\prod'''$ by the four units in the Gaussian integers $+1,+i,-1,-i$, and then summed them up.
Indirectly connected with this result is Gauss's footnote to article 358 of D.A, which includes the result from his 1805 fragment as well as several other corollaries:
>
> Corollary: Let $\epsilon$ be the root of the equation $x^3-1=0$ and we will have $(p+\epsilon p'+{\epsilon}^2p'')^3 = n(M+N\sqrt{-27})/2$. Let $M/\sqrt{4n} = cos\phi, N\sqrt{27}/\sqrt{4n} = sin\phi$ and as a result $$p = -\frac{1}{3}+\frac{2}{3}cos\frac{1}{3}\phi \sqrt{n}$$
>
>
>
Note that here Gauss denotes the prime number by $4n = M^2+27N^2$ (the prime number is $n$), and the Gaussian period by $p$. The result $$p+\epsilon p'+{\epsilon}^2p'' = (n(M+N\sqrt{-27})/2)^{\frac{1}{3}}$$
for the cubic case is analogous to the summation of ${\prod}^0 + i\prod'-\prod''- i\prod'''$ (for the biquadratic case), since $i$ is a root of the equation $x^4-1=0$.
**Questions**
I'm aware that Gauss sums and Gaussian periods are intimately connected, and according to the survey article "THE DETERMINATION OF GAUSS SUMS" by Bruce C. Berndt and Ronald J. Evans, in the D.A and in his two articles on biquadratic residues, Gauss determined cubic and quartic Gauss sums. However, i can't see how the two identities for the sums $p+\epsilon p'+{\epsilon}^2p''$ and ${\prod}^0 + i\prod'-\prod''- i\prod'''$ are connected to the cubic and quartic Gauss sums (which don't involve weighting by $\epsilon$ or $i$), nor did i find any article that references this particular result of Gauss. Therefore, i will sum up the main content of my question:
* Are those identities an easy corollary to the determination of cubic and quartic Gauss sums? or maybe there are additional inherent difficulties to imply them from the values of Gauss sums? In any case, i'd like to get some guidelines on how to move from the values of Gauss sums to those particular identities.
* If the answer is too long, i'd also be glad if anyone can give a link to a source that discusses those identities. For me linking to such a source can be an alternative answer, since i didn't find any article that refers to those identities.
Any useful comment will be blessed!
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https://mathoverflow.net/users/118562
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Explanation of two interrelated identities of Gauss about cubic and biquadratic periods
|
Let $p \equiv 1 \bmod 3$ be a prime number, let $g$ be a be a primitive root
modulo $p$, and $\zeta$ a primitive $p$-th root of unity. The three
cubic periods are
\begin{align\*}
\eta\_0 & = \zeta + \zeta^{g^3} + \zeta^{g^6} + \ldots + \zeta^{g^{p-4}}, \\
\eta\_1 & = \zeta^g + \zeta^{g^4} + \zeta^{g^7} + \ldots + \zeta^{g^{p-3}}, \\
\eta\_2 & = \zeta^{g^2} + \zeta^{g^5} + \zeta^{g^8} + \ldots + \zeta^{g^{p-2}}.
\end{align\*}
We have $$ \eta\_0 + \eta\_1 + \eta\_2 = -1, $$
and
$$ \tau = \eta\_0 + \rho \eta\_1 + \rho^2 \eta\_2 \quad \text{and} \quad
\overline{\tau} = \eta\_0 + \rho^2 \eta\_1 + \rho \eta\_2 $$
are the cubic Gauss sum
$$ \tau = \sum\_{a=1}^{p-1} \chi\_3(a) \zeta^a $$
and its complex conjugate $\overline{\tau}$; here $\chi\_3$ is a suitably chosen nontrivial cubic character on
$({\mathbb Z} /p{\mathbb Z})^\times$. It is known that
$\tau^3 = \pi^2 \overline{\pi}$, where $\pi$ is a prime factor of $p$ in ${\mathbb Z}[\rho]$ ($\rho$ is a primitive cube root of unity; Gauss used $\epsilon$).
We find
\begin{align\*}
\tau + \overline{\tau} & = 2\eta\_0 - \eta\_1 - \eta\_2, \\
\rho \tau + \rho^2 \overline{\tau} & = - \eta\_0 - \eta\_1 + 2\eta\_2,\\
\rho^2 \tau + \rho \overline{\tau} & = - \eta\_0 + 2\eta\_1 - \eta\_2.
\end{align\*}
Adding $\eta\_0 + \eta\_1 + \eta\_2 = -1$ to these equations we get
$$ \eta\_0 = \frac{-1 + \tau + \overline{\tau}}3, \quad
\eta\_1 = \frac{-1 + \rho^2 \tau + \rho \overline{\tau}}3, \quad
\eta\_2 = \frac{-1 + \rho \tau + \rho^2 \overline{\tau}}3. $$
Since $\tau^2 = \pi^2 \overline{\pi} = p\pi$ and $\tau \overline{\tau} = p$ we have
$$ (\tau + \overline{\tau})^3 = p\pi +p \overline{\pi} + 3p(\tau + \overline{\tau}). $$
If we write $\pi = \frac{L + 3M\sqrt{-3}}2$, then $\tau + \overline{\tau} = L$,
hence $\tau + \overline{\tau}$ is a root of the cubic polynomial
$$ f\_p(X) = X^3 - 3pX - pL. $$
Now if $\xi$ is a root of a polynomial $f(x)$,
then $\frac{\xi-1}3$ is a root of $f(3x+1)$. Thus $\eta\_0$ is a root
of the polynomial
$$ P\_3(x) = \frac1{27} f\_p(3x+1)
= x^3 + x^2 - \frac{p-1}3 x - \frac{(L+3)p-1}{27}. $$
Since $4p = L^2 + 27M^2$ we know that $|\frac{L}{2\sqrt{p}}| < 1$ and
$|\frac{3M\sqrt{3}}{2\sqrt{p}}| < 1$; thus there exists a unique angle $\phi$
with
$$ \cos \phi = \frac{L}{2\sqrt{p}} \quad \text{and} \quad
\sin \phi = \frac{3M\sqrt{3}}{2\sqrt{p}}. $$
With $T = \tau + \overline{\tau}$ we have $|T| \le 2 \sqrt{p}$, hence there is
an angle $\alpha$ such that $\cos \alpha = \frac{T}{2\sqrt{p}}$.
Now $T^3 = 3pT + pL$ implies
$$ 4\Big(\frac{T}{2\sqrt{p}}\Big)^3
- 3 \cdot \frac{T}{2\sqrt{p}} = \frac{L}{2\sqrt{p}}, $$
which can be written as
$$ 4 (\cos \alpha)^3 - 3 \cos \alpha = \frac{L}{2\sqrt{p}}. $$
Since $4 \cos(x)^3 - 3 \cos (x) = \cos(3x)$, this implies
$\alpha = 3\phi$ (for some suitable choice of $\alpha$).
The biquadratic case is similar; $\Pi^0$, $\Pi'$ etc. are the four quartic periods, the linear combination on the left side of your equation is the usual quartic Gauss sum, and the equation is the prime factorization of the Gauss sum (up to a fourth root of unity).
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2
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https://mathoverflow.net/users/3503
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402129
| 165,033 |
https://mathoverflow.net/questions/401651
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11
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Let $\pi:Y\rightarrow X$ be a (smooth, finite dimensional) fibred manifold. Since no other fibrations will be considered on $Y$, I will identify $(Y,\pi,X)$ with $Y$. The finite order jet bundles are denoted $J^rY$ ($r\in\mathbb N$, $J^0Y=Y$), and the infinite jet bundle $J^\infty Y$ is - as a topological space - the projective limit of the system $(J^rY,\pi^r\_s)$ where $\pi^r\_s:J^rY\rightarrow J^sY$, $r\ge s$ is the standard projection.
---
In the literature, there exists several ways through which $J^\infty Y$ acquires a smooth manifold-like structure. What is considered a smooth function on $J^\infty Y$ depends heavily on the choice of this smooth structure. I am confused by the wide variety of these definitions and I often have trouble establishing equivalence. First I will list some examples I have encountered.
1. In [Saunders], $J^\infty Y$ is defined as a Fréchet manifold built on the vector space $\mathbb R^\infty$ which arises as the projective limit of $\mathbb R^n$s with respect to the system $p^n\_{n-1}:\mathbb R^n\rightarrow\mathbb R^{n-1}$, $p^n\_{n-1}(x^1,\cdots,x^{n-1},x^n)=(x^1,\cdots,x^{n-1})$. Smooth functions are determined by what smooth functions are on $\mathbb R^\infty$, and their "finite-orderness" is characterized by the fact that such smooth functions have only a finite number of nonzero partial derivatives at each point. I don't know whether this structure admits partitions of unity (Saunders doesn't go into it) but I find it likely that it does.
2. In [Takens], no systematic smooth structure is given on $J^\infty Y$, he takes the smooth functions to be those which are *locally* the pullbacks of smooth functions from a finite order jet bundle, i.e. $f:J^\infty Y\rightarrow\mathbb R$ is smooth if for each $q\in J^\infty Y$ there is an $r$) and an $U^r\in J^rY$ open set with $q\in (\pi^\infty\_r)^{-1}(U^r)$ and a smooth function $f\_r\in C^\infty(U^r)$ such that $f|\_{(\pi^\infty\_r)^{-1}(U^r)}=f\_r\circ\pi^\infty\_r$. Takens then proves that there is a partition of unity subordinate to every open cover of $J^\infty Y$. I also *think* that this definition allows one to define smooth functions over every $O\subseteq J^\infty Y$ open set as the open sets of $J^\infty Y$ are generated by open sets of $J^rY$ for all $r$.
3. In [Anderson], no systematic smooth structure is given either. Instead, he defines smooth functions to be the (globally) finite order ones, i.e. $f:J^\infty Y\rightarrow\mathbb R$ is smooth if and only if there is an $r$ and a smooth function $f\_r\in C^\infty(J^rY)$ such that $f=f\_r\circ\pi^\infty\_r$. Anderson states that if the notion of smooth function was extended to the locally finite order ones then there would be smooth functions that aren't continuous. This definition seems to be the most convenient for calculus of variations, but I have some concerns. For example it seems non-obvious me how to define the space $C^\infty(O)$ where $O$ is an arbitrary open set of $J^\infty Y$. I guess a Takens-like approach can be taken but then it seems the association $O\mapsto C^\infty(O)$ defines only a presheaf as the glueing of finite order functions only need to be locally finite order. Moreover it seems that the association $W\mapsto C^\infty(W^\infty)$ where $W\subseteq Y$ is open and $W^\infty=(\pi^\infty\_0)^{-1}(W)$ also defines only a presheaf over $Y$ for the same reason.
4. In [Güneysu] pro-finite dimensional manifolds are defined and the infinite jet bundle is given as a pro-finite dimensional manifold.The definition of a smooth function seems to be same as in [Takens], with the added difference that a smooth function is demanded to be also continuous (Takens does not postulate continuity and based on the aforementioned remark in [Anderson] it seems that this added assumption is not superfluous).
---
And now come the problems:
* There are a couple of remarks on nlab which seems to contradict many of these. For example in <https://ncatlab.org/nlab/show/jet+bundle#concrete> they say
>
> but one has to decide in which category of infinite-dimensional manifolds to take this limit: 1) one may form the limit formally, i.e. in pro-manifolds. This is what is implicit for instance in Anderson, p.3-5; 2) one may form the limit in Fréchet manifolds, this is farily explicit in (Saunders 89, chapter 7). See at Fréchet manifold – Projective limits of finite-dimensional manifolds. Beware that this is not equivalent to the pro-manifold structure (see the remark here). It makes sense to speak of locally pro-manifolds.
>
>
>
and in the [link](https://ncatlab.org/nlab/show/Fr%C3%A9chet+manifold#DifferenceBetweenProManifoldAndFrecherManifoldStructure) in this remark states
>
> Beware, that infinite jet bundles are also naturally thought of as pro-manifolds. This differs from the Frechet manifold structure of example 4.5: A morphism of pro-manifolds is equivalently a function that is “globally of finite order”, in that [...] But by prop. 4.4 a morphisms of Fréchet manifolds is only restricted to have finite order of partial derivatives at every point. This is a weaker condition. In fact it seems to be also weaker than the condition of being “locally of finite order” considered in Takens 79.
> Hence it makes sense to speak of locally pro-manifolds.
>
>
>
So it seems that nlab's definition of a pro-manifold (I prefer saying pro-finite dimensional manifold but I meant what is supposed to be the same thing) is different from that of [Güneysu], on the other hand the pro-manifold page on nlab has no references, so I don't know what is precisely the framework they are working in.
* The second problem I have is the paper [GMS]. I will try to briefly summarize its contents. The authors seem to be comparing the cohomology of the variational bicomplex in the Takens-like approach (differential forms are locally finite order) and the Anderson-like approach (differential forms are globally finite order). They make the claim that so far only the locally finite order variational bicomplex (LFOVB) had its cohomology calculated and most approaches that work in this case do not work for the globally finite order variational bicomplex (GFOVB). They then proceed to prove that the GFOVB has the same cohomology as the LFOVB by relating the cohomology of the former to the latter. $$ \ $$ Their argument seems to be around the fact that most cohomology calculations (eg. in [Takens] but also in the paper by Anderson and Duchamp which predates [Anderson] and also in [Krupka] for finite order jet bundles) are taken by considering a fine sheaf of differential forms that are defined on a jet bundle but the sheaf is over $Y$, i.e. a sequence of the form $$ 0\longrightarrow\mathbb R\_Y\longrightarrow \mathcal O^0\longrightarrow\cdots\longrightarrow\mathcal O^r\longrightarrow\cdots, $$ where each $\mathcal O^r$ is a sheaf over $Y$ whose sections are differential forms over $J^\infty Y$ or $J^sY$ (for some $s$) and using sheaf cohomology together with the abstract de Rham theorem. However the globally finite order forms form a separated presheaf whose sheafification is the sheaf of locally finite order forms (I think?) and they don't have partitions of unity which throws a wrench in this argument for the GFOVB. $$ \ $$ However [GMS] postdates [Anderson] by quite a few years and Anderson *does* calculate the cohomology of the GFOVB in [Anderson]. He uses techniques other than the aforementioned sheafy approach though. [GMS] never refers to [Anderson] though (they do refer to Anderson/Duchamp). $$ \ $$ In light of the above I question how correct the global calculations are in [Anderson]. Anderson only uses paritions of unity on $Y$ but upon thinking about it I get the feeling his various glueing arguments are faulty as they might result in locally finite order forms rather than globally finite order ones. At any rate it seems to me that [Anderson] and [GMS] contradict one another thus cannot be both correct (unless of course I am missing something).
---
**Questions:**
I'd like to make some sense into this mess so lets see.
1. How many inequivalent definitions of $J^\infty Y$ as a manifold-like structure exists? Based on the nlab remarks at least the Saunders/Takens/Anderson approaches seem to be inequivalent to one another.
2. Are there multiple inequivalent definitions of pro-finite dimensional manifolds in use? The nlab comments seem to imply that whatever definition they are working with contains the Anderson approach to $J^\infty$ as a subcase but the Güneysu approach seems to be different and closer (but not necessarily equivalent) to the Takens approach.
3. Although this is subjective, what structure on $J^\infty Y$ is the best suited for calculus of variations/the variational bicomplex? Basically, I just want to properly formalize the variational bicomplex for myself in a way that it includes only finite-order differential forms (I don't care about Lagrangians with unbounded order) but I got really confused by all these stuff. It also seems to me the Anderson approach might not actually be the best even if it only contains finite order forms if convenient arguments cannot be used. I am basically looking for the most hassle-free and lazy approach possible.
4. Assuming Anderson wasn't being sloppy with the definitions in [Anderson], what is the generalization of finite dimensional manifolds into which Anderson's definition of $J^\infty Y$ fits into? (eg. what nlab calls pro-manifolds. But since this terminology seems to be ambigous I cannot just rely on the name alone) I'd like to read more on that structure, even outside jet bundles.
5. Regarding the [Anderson]-[GMS] conflict, are the global cohomology calculations in [Anderson] correct? What these to publications say seem to be in conflict and I cannot decide which one is correct. Are they even in conflict actually?
**References:**
* [Saunders]: D. Saunders - [The geometry of jet bundles](https://books.google.hu/books/about/The_Geometry_of_Jet_Bundles.html?id=nYqP6XGSQ28C&source=kp_book_description&redir_esc=y)
* [Takens]: F. Takens - [A global version of the inverse problem to the calculus of variations](https://projecteuclid.org/journals/journal-of-differential-geometry/volume-14/issue-4/A-global-version-of-the-inverse-problem-of-the-calculus/10.4310/jdg/1214435235.full)
* [Anderson]: I. M. Anderson - [The Variational Bicomplex](https://ncatlab.org/nlab/files/AndersonVariationalBicomplex.pdf)
* [Güneysu]: B. Güneysu, M. J. Pflaum - [The Profinite Dimensional Manifold Structure
of Formal Solution Spaces of Formally Integrable PDEs](https://www.emis.de/journals/SIGMA/2017/003/sigma17-003.pdf)
* [Krupka]: D. Krupka - [Introduction to Global Variational Geometry](https://books.google.hu/books/about/Introduction_to_Global_Variational_Geome.html?id=A1IpBgAAQBAJ&source=kp_book_description&redir_esc=y)
* [GMS]: Giachetta, Mangiarotti, Sardanashvily - [Cohomology of the variational bicomplex on the infinite order jet space](https://arxiv.org/abs/math/0006074)
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https://mathoverflow.net/users/85500
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Different smooth structures on the infinite jet bundle (for the purposes of calculus of variations)
|
The following remarks are based on having previously gone through the literature that you've mentioned also for the purposes of figuring out these differences. It has been a while since then, but the state of our understanding that we reached with Urs Schreiber at the time was recorded in the following work, to be more precise in Sec 2.2 and possibly other parts referenced there. The remarks that you find on the nLab are based on that work.
>
> [KhS]: Igor Khavkine, Urs Schreiber *Synthetic geometry of differential equations: I. Jets and comonad structure* [[arXiv:1701.06238]](https://arxiv.org/abs/1701.06238)
>
>
>
If you happen to find any inconsistencies, they might have been fixed in a more current version (though I don't think there were many), which hasn't been uploaded to the arXiv. If you think you've found some, just let me know and I can share the updated version with you.
I haven't rechecked all the references that you cite, so these answers are mostly going by memory and what is written in our paper.
1. I think your zoo of manifold structures on $J^\infty Y$ is overly complicated. There are really only two versions, distinguished by the smooth functions they admit: those of locally finite order ([Saunders], [Takens], [KhS], [Güneysu] + private communication from Pflaum) and those of globally finite order ([Anderson], [DF]). The relation of the Fréchet manifold structure with the locally finite order version is captured by Prop 2.29 in [KhS] (which reproduces the proof by Michor from an earlier reference).
>
> [DF]: P. Deligne and D. S. Freed, [“Classical field theory,”](https://web.archive.org/web/20210529040005/http://publications.ias.edu/node/427) in Quantum fields and strings: a course for
> mathematicians, P. Deligne, D. Kazhdan, P. Etingof, J. W. Morgan, D. S. Freed, D. R. Morrison,
> L. C. Jeffrey, and E. Witten, eds., vol. 1, pp. 137–225. AMS, Providence, RI, 1999.
>
>
>
2. The notion of a pro-finite dimensional manifold depends on the ambient category where you take the projective limit. In the standard category of finite dimensional manifolds this limit does not exist. If you formally throw in projective limits, you get the globally finite order version. If you take the limit in the category of Fréchet manifolds (or any larger category where Fréchet manifolds are faithfully included) then you get the locally finite order version.
3. You asked for a subjective opinion. My preference is the locally finite order version.
4. [Anderson] fits into the globally finite order camp, which is the same as treating pro-finite dimensional manifolds as [pro-objects](https://ncatlab.org/nlab/show/pro-object) in a standard categorical way. I think that the literature on general pro-objects (you can use the nLab link as a starting point) is much larger than on pro-finite dimensional manifolds specifically. Your best bet here might be to try to understand the general theory.
5. I can't personally vouch for all arguments in [Anderson], but his book is so well-known that I would expect any known errors to have already been mentioned in the literature. If you find such a claim, it should be evaluated on its own merits. But the cohomology calculations in Anderson should be correct, since they agree with what is reported in [GMS] and [DF] (see the Appendix to Ch 2 specifically). Even if you think Anderson's arguments are not completely precise, I think that the other two references should be detailed enough, so that you should be able to verify them with enough study.
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3
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https://mathoverflow.net/users/2622
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402134
| 165,035 |
https://mathoverflow.net/questions/402120
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3
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Any knot or link can be written in braid notation (with implied closure of strands). Some natural questions:
1. Assume I don't allow inverses - only overcrossings are used as generators. Anything known about the braid index then (might as well be $\infty$, i.e., this representation doesn't always exist)?
2. Assume I additionally allow Temperley-Lieb generators (what's that called, Turaev algebra?). Same question.
3. For all three variants - surely the braid representation has some overhead with respect to the minimum crossing representation of a knot. How large can it get?
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https://mathoverflow.net/users/11504
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"Effectivity" of braid notation for knots
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(1) It is a theorem of Stallings [1978, "Constructions of fibred knots and links"] that closures of positive braids are always fibered links. Thus "most" knots are not realised as the closure of a positive braid.
(2) If you allow positive (say) crossings only, and TL generators, then you can use the latter to rotate the former by 90 degrees. Then the given diagram can be converted with at most a linear growth in complexity.
(3) Interesting question. I'll guess that the blow up is at worst polynomial.
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https://mathoverflow.net/users/1650
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402144
| 165,038 |
https://mathoverflow.net/questions/402131
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1
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Let $f\_n\colon \mathbb{C} \to \mathbb{C}$ be a sequence of entire functions, such that $f\_n$ converges to the zero function on an open dense subset $U$ of $\mathbb{C}$ pointwise (or equivalently normally). Then, does $f\_n$ genuinely converge to the zero function on $\mathbb{C}$?
It seems to me that it does converge, but I am not sure. If the assumption is a convergence on a dense subset of $\mathbb{C}$,then $f\_n$ does not necessarily converge (<https://math.stackexchange.com/questions/3651442/pointwise-convergence-of-holomorphic-functions-on-a-dense-set/3651462#3651462>).
I would appreciate any comments! Thanks.
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https://mathoverflow.net/users/343456
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Convergence of a sequence of entire functions on an open dense subset
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It is not true. Consider the compact sets
$$K\_n=\{ z:|z|\leq n, |\arg z|\geq 1/n\}\cup\{0\}.$$
By Runge's approximation theorem, there exist polynomials $f\_n$,
such that $|f\_n(z)|<1/n,\; z\in K\_n,$ and $f\_n(1)=n.$
This sequence of polynomials evidently converges uniformly on compact subsets of the dense open set $C\backslash[0,+\infty)$ but does not converge at the point $1$.
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2
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https://mathoverflow.net/users/25510
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402145
| 165,039 |
https://mathoverflow.net/questions/401484
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6
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Let $A$ be an $R$-algebra.
In the book "Noncommutative Geometry and Cayley-smooth Orders" by Le Bruyn one can find the notion of "Serre-smooth" in the introduction.
But no formal definition seems to be given in this (very long) introduction.
It is just mentioned that $A$ is Serre-smooth if $A$ has finite global dimension together with some extra features such as Auslander regularity or the Cohen-Macaulay property.
Then it is refered to the article <https://www.cambridge.org/core/journals/glasgow-mathematical-journal/article/some-properties-of-noncommutative-regular-graded-rings/244AE0A8E9C8AB6782515B504F1AC5C0> but there I can not find the word "Serre" (searching with Strg+F).
>
> Question: Is there a reference for the complete definition of being Serre-smooth?
>
>
>
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https://mathoverflow.net/users/61949
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What is a Serre-smooth algebra?
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No, there is no such reference. The introduction to that book is based on a couple of lectures I gave in Luminy and there I had to distinguish between several notions of 'smoothness', formal smoothness a la Kontsevich-Rosenberg, Cayley-smoothness, and Artin-Schelter- or Auslander-Gorenstein-regularity as used by people working in NAG.
For the later category I then used the term 'Serre-smoothness' as it is the noncommutative equivalent of smoothness for commutative affine algebras (finite global dimension) to the noncommutative world.
If one only considers (maximal) orders in central simple algebras which are finite modules over their centers then one can do with less than the full repertoire of these homological conditions, I think.
Please read 'Serre-smoothness of A' as 'A is an Auslander-Gorenstein regular ring'. My apologies for the confusion this ad-hoc terminology may have caused.
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https://mathoverflow.net/users/2275
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402149
| 165,040 |
https://mathoverflow.net/questions/356458
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0
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Claim:
------
let $a,b,c>0$ and $p\geq 1$ then we have :
$$\left(\frac{a^3}{13a^2+5b^2}\right)^p+\left(\frac{b^3}{13b^2+5c^2}\right)^p+\left(\frac{c^3}{13c^2+5a^2}\right)^p\geq 3\left(\frac{a+b+c}{54}\right)^p$$
---
The case $p=1$ have been proved by user RiverLi and I offer a partial proof too in this case (see [here](https://math.stackexchange.com/questions/1777075/inequality-sum-limits-cyc-fraca313a25b2-geq-fracabc18?noredirect=1&lq=1)).
---
The motivation:
---------------
In my partial proof (and a large part is due to user Alex Ravsky [here](https://math.stackexchange.com/questions/3563335/prove-or-disprove-this-statement?noredirect=1&lq=1)) I use the well-know inequality called [Karamata's inequality](https://en.wikipedia.org/wiki/Karamata%27s_inequality) and [Buffalo's way](https://brilliant.org/discussions/thread/the-buffalo-way/).
The fact is : the use of this inequality allow us to a generalisation because I apply the Karamata's inequality with the exponential function so it works for $e^x$ then it works also for $e^{xp}$.
Now and have a look to Alex's proof the only values for wich the inequality is valid is $p\geq 1$ .It's explained by another inequality called Jensen's inequality.
At the origin the inequality is due to Vasil Cirtoaje and strenghened by user Michael Rozenberg who found the coefficient $13$ and $5$.
---
Some details of the proof :
---------------------------
>
>
> >
> > Let $a,b,c>0$ such that $\frac{a^3}{13a^2+5b^2}\geq \frac{b^3}{13b^2+5c^2}\geq \frac{c^3}{13c^2+5a^2}$ and $a\geq b\geq c$ and $13a^2+5b^2\geq 13b^2+5c^2\geq 13c^2+5a^2$ then $\exists n>1$ such that :$$\frac{\frac{a^3}{13a^2+5b^2}}{n}+ \frac{(n-1)(a+b+c)}{54n}\geq \frac{a+b+c}{54}$$
> > $$\Big(\frac{\frac{a^3}{13a^2+5b^2}}{n}+ \frac{(n-1)(a+b+c)}{54n}\Big)\Big(\frac{\frac{b^3}{13b^2+5c^2}}{n}+ \frac{(n-1)(a+b+c)}{54n}\Big)\geq \frac{(a+b+c)^2}{54^2}$$
> > $$\Big(\frac{\frac{a^3}{13a^2+5b^2}}{n}+ \frac{(n-1)(a+b+c)}{54n}\Big)\Big(\frac{\frac{b^3}{13b^2+5c^2}}{n}+ \frac{(n-1)(a+b+c)}{54n}\Big)\Big(\frac{\frac{c^3}{13c^2+5a^2}}{n}+ \frac{(n-1)(a+b+c)}{54n}\Big)\geq \frac{(a+b+c)^3}{54^3}$$
> >
> >
> >
>
>
>
Remains to apply Karamata's inequality to get the desired result .By Karamata's inequality I mean this special case :
>
>
> >
> > If $a\_1\geq a\_2\geq a\_3\geq\cdots\geq a\_n$ and $b\_1\geq b\_2\geq b\_3\geq\cdots\geq b\_n$ are two sequences of positive real numbers then we have $\frac{a\_1}{n}+\frac{(n-1)b\_1}{n}\geq \frac{a\_2}{n}+\frac{(n-1)b\_2}{n} \geq\cdots\geq \frac{a\_n}{n}+\frac{(n-1)b\_n}{n}$and $b\_1\geq b\_2\geq b\_3\geq\cdots\geq b\_n$ satisfying the following conditions(call the conditions $C$):$$\frac{a\_1}{n}+\frac{(n-1)b\_1}{n}\geq b\_1,(\frac{a\_1}{n}+\frac{(n-1)b\_1}{n})(\frac{a\_2}{n}+\frac{(n-1)b\_2}{n})\geq b\_1b\_2,\cdots,(\frac{a\_1}{n}+\frac{(n-1)b\_1}{n})(\frac{a\_2}{n}+\frac{(n-1)b\_2}{n})\cdots (\frac{a\_n}{n}+\frac{(n-1)b\_n}{n})\geq b\_1b\_2\cdots b\_n,$$ Then we have :
> > $$a\_1+a\_2+a\_3+\cdots+a\_n\geq b\_1+b\_2+b\_3+\cdots+b\_n$$
> >
> >
> >
>
>
>
To get the power $p$ we use Jensen's inequality because we have :
Let $u,v>0$ and $p$ a real number such that $p\geq 1$ and $n$ a natural number large enought:
$$u^p\frac{1}{n}+v^p\frac{n-1}{n}\geq \left(u\frac{1}{n}+v\frac{n-1}{n}\right)^p$$
---
To go further :
---------------
We have also a stronger statement :
let $a,b,c>0$ and $p\geq 1$ then we have :
$$\left(\frac{a^3}{13a^2+5b^2}\right)^p+\left(\frac{b^3}{13b^2+5c^2}\right)^p+\left(\frac{c^3}{13c^2+5a^2}\right)^p\geq \frac{a^p+b^p+c^p}{18^p}\geq 3\left(\frac{a+b+c}{54}\right)^p$$
To show it we come back at the end proof where I use Jensen's inequality .The inequality in $u,v$ can be strenghened using strong convexity and a modulus $m$ .I have not try but it seems promising .see [here](https://journalofinequalitiesandapplications.springeropen.com/articles/10.1186/s13660-018-1897-2#:%7E:text=The%20Jensen%20inequality%20for%2) Theorem 2.
Using the idea above we got :Let $a,b,c>0 $ and $m=\min(\frac{a^{3}}{13a^{2}+5b^{2}},\frac{b^{3}}{13b^{2}+5c^{2}},\frac{c^{3}}{13c^{2}+5a^{2}},\frac{a+b+c}{54})$ and $p\geq 2$ a real number we have :
$$\left(\frac{a^{3}}{13a^{2}+5b^{2}}\right)^{p}+\left(\frac{b^{3}}{13b^{2}+5c^{2}}\right)^{p}+\left(\frac{c^{3}}{13c^{2}+5a^{2}}\right)^{p}-\left(3\left(\frac{\left(a+b+c\right)}{54}\right)^{p}+p\left(p-1\right)\left(m\right)^{\left(p-2\right)}\cdot0.5\cdot\left(\sum\_{cyc}\left(\frac{\left(a+b+c\right)}{54}-\left(\frac{a^{3}}{13a^{2}+5b^{2}}\right)\right)^{2}\right)\right)\geq 0$$
Ps:The sign at the end is not a superior strict but superior or equal .
---
Question: How to show the claim properly?
|
https://mathoverflow.net/users/147649
|
Olympiad inequality as a generalizing result due at the origin to Vasile Cirtaoje
|
The case $p>1$ easily follows from the case $p=1$. Indeed, let $M\_p(x,y,z) := \left(\frac{x^p+y^p+z^p}3\right)^{1/p}$. Then for $p>1$ we have
$$M\_p\big( \frac{a^3}{13a^2+5b^2}, \frac{b^3}{13b^2+5c^2}, \frac{c^3}{13c^2+5a^2}\big) \stackrel{(1)}{\geq} M\_1\big( \frac{a^3}{13a^2+5b^2}, \frac{b^3}{13b^2+5c^2}, \frac{c^3}{13c^2+5a^2}\big) \stackrel{(2)}{\geq} \frac{a+b+c}{54},$$
where (1) the [power mean inequality](https://en.wikipedia.org/wiki/Generalized_mean#Generalized_mean_inequality) and (2) is the case $p=1$.
|
1
|
https://mathoverflow.net/users/7076
|
402151
| 165,041 |
https://mathoverflow.net/questions/401635
|
3
|
For any group $G$, the *universal example* for proper $G$-actions, $\underline{E}G$, is a proper $G$-space such that for any other proper $G$-space $X$, there exists a map (unique up to $G$-equivariant homotopy)
$$X\to\underline{E}G.$$
The quotient $\underline{B}G$ may be called the *classifying space* for proper $G$-actions.
**Question 1:** Does a homomorphism of groups $f\colon G\to H$ induce a $G$-equivariant continuous map $f\_\*\colon\underline{E}G\to\underline{E}H$ and hence also a continuous map $\underline{B}G\to\underline{B}H$?
**Question 2:** Under what conditions do we have that $\underline{B}(G\_1\times G\_2)$ is homotopy equivalent/homeomorphic to $\underline{B}G\_1\times\underline{B}G\_2$?
|
https://mathoverflow.net/users/78729
|
Reference request: functoriality of $\underline{E}$ and $\underline{B}$
|
Here is model that is obviously functorial: take for $\underline{E}G$ the simplicial complex with vertex set the finite subsets of $G$ and simplices the finite chains of sets ordered by inclusion. A homomorphism $f:G\rightarrow H$ induces a function from the finite subsets of $G$ to the finite subsets of $H$. You can also think of this model as the barycentric subdivision of the `simplex' with vertex set $G$.
I can't think of anything to add to Fernando Muro's answer to question 2 in the comments.
|
2
|
https://mathoverflow.net/users/124004
|
402157
| 165,044 |
https://mathoverflow.net/questions/402161
|
2
|
This is related to [Symplectic group over $\mathbb{Z}/p\mathbb{Z}$ is generated by its root subgroups](https://mathoverflow.net/questions/401996/symplectic-group-over-mathbbz-p-mathbbz-is-generated-by-its-root-subgroup). There I was told that in general, the symplectic group $\text{Sp}\_{2n}(R)$ is not generated by its root subgroups and a maximal torus.
My question now is: for which commutative rings $R$ is $\text{Sp}\_{2n}(R)$ generated by its root subgroups and a maximal torus?
|
https://mathoverflow.net/users/nan
|
When is the symplectic group over a commutative ring generated by its root subgroups and a maximal torus?
|
Since $\operatorname{Sp}\_{2\ell}$ is simply-connected, there is no need for the maximal torus. So the question is about the triviality of $\operatorname{K\_1}(\mathsf{C}\_\ell, R) = \operatorname{Sp}(2\ell,R)/\operatorname{Ep}(2\ell,R)$, where $\operatorname{Ep}(2\ell,R)$ is the elementrary symplectic group, that is, the subgroup of $\operatorname{Sp}(2\ell,R)$ generated by its root subgroups.
A lot of cases where $\operatorname{K\_1}(\mathsf{C}\_\ell,R)$ is trivial are not specific fot the symplectic case and hold for all Chevalley groups. Below is a(n incomplete) list. I only consider commutative rings.
* $R$ is a ring of stable rank 1 (this includes local and semi-local rings, boolean rings, the ring of algebraic integers, the ring of entire functions, the disc algebra). The proof is a combination of surjective stabiliy for $\operatorname{K}\_1$ (a [paper](https://doi.org/10.4099/math1924.4.77) by M. Stein) and an easy argument for $\operatorname{SL}(2,R)$ (every its element is a product of 4 elementary matrices, this was apparently first noted by H. Bass);
* $R$ is a Dedekind ring of arithmetic type in a number field which is not totally imaginary (this is Theorem 3.6 from [Bass—Milnor—Serre](http://www.numdam.org/articles/PMIHES_1967__33__59_0/));
* $R$ is Euclidean (the same as the stable rank 1 case, but with the Euclidean algorithm for $\operatorname{SL}(2,R)$ and no bound on the number of elementary factors);
* More generally, if $R$ is a ring such that $\operatorname{K}\_1({}\cdot{},R)=1$, it is sometimes possible to show that $\operatorname{K}\_1({}\cdot{},R[x\_1,\ldots,x\_n])=1$, see my answer [here](https://mathoverflow.net/questions/156563/suslins-stability-theorem-for-chevalley-groups/);
* There are many papers dealing with the rings of geometric or analytical origin, for example, in [this paper](https://doi.org/10.1090/proc/14891) by B. Ivarsson, F. Kutzschebauch and E. Løw the authors consider commutative Banach algebras and some rings of continuous functions (they prove that null-homotopic matrices are elementary, but the paper also contains some references for other rings).
|
2
|
https://mathoverflow.net/users/5018
|
402166
| 165,050 |
https://mathoverflow.net/questions/401602
|
8
|
Let $f: \mathbb R \to \mathbb R$ be a $C^1$ function.
We say a point $c \in \mathbb R$ is a *mean value point* of $f$ if there exists an open interval $(a,b)$ containing $c$ such that $f’(c) = \frac{f(b) - f(a)}{b-a}$.
>
> **Question:** Is it true that (Lebesgue) almost every point in $\mathbb R$ is a mean value point of $f$?
>
>
>
|
https://mathoverflow.net/users/173490
|
Converse of mean value theorem almost everywhere?
|
Let $U$ be an open and dense subset of $\mathbb{R}$ with finite measure. Let $g: \mathbb{R} \to \mathbb{R}^{\ge 0}$ be a continuous function with $\{g = 0\} = U^c$. Then define $f: \mathbb{R} \to \mathbb{R}$ by $f(x) = g(0)+\int\_0^x g(t)dt$. Then $f \in C^1$, and $f' \equiv g$ means $f$ is strictly increasing (since any interval $(x,y)$ contains an interval lying in $U$ on which $g$ is strictly positive). So we have a strictly increasing $C^1$ function with zero derivative everywhere except for a finite measure set.
|
7
|
https://mathoverflow.net/users/129185
|
402171
| 165,052 |
https://mathoverflow.net/questions/402108
|
4
|
Based on my previous answer and your help
[Is there a procedure for extracting first integer $q\_0$ from $\sum\limits\_{k=0}^{\infty}\frac{1}{q\_k^z}$, all $0<q\_0<q\_1<...$ integers, $z$ complex?](https://mathoverflow.net/q/300543/113386)
This formula derived in the question
$$ \ln(p\_n)=-\lim\limits\_{\operatorname{Re}(s) \to +\infty} \frac{\ln\left (\zeta(s)\prod\limits\_{k=1}^{n-1}(1-p\_k^{-s})-1 \right )}{s} $$
can actually turn into
$$ p\_n=\lim\limits\_{m \to +\infty} \left ( \frac{|B\_{2m}|(2\pi)^{2m}}{2(2m)!}\prod\limits\_{k=1}^{n-1}(1-p\_k^{-2m})-1 \right )^{-\frac1{2m}} $$
using
$$\zeta(2m) = \frac{(-1)^{m+1}B\_{2m}(2\pi)^{2m}}{2(2m)!}$$
Do you see any problem in using real limit instead of complex one? (I don't.)
This recurrent formula is not known to me and if correct, it looks like something worth some attention.
The formula is more to the point in logarithm form of course:
$$ \ln(p\_n)=-\lim\limits\_{m \to +\infty} \frac1{2m}\ln \left ( \frac{|B\_{2m}|(2\pi)^{2m}}{2(2m)!}\prod\limits\_{k=1}^{n-1}(1-p\_k^{-2m})-1 \right ) $$
with a form that more clearly displays the precision it needs
$$ \ln(p\_n)=-2\pi-\lim\limits\_{m \to +\infty} \frac1{2m} \ln \left ( \frac{|B\_{2m}|}{2(2m)!}\prod\limits\_{k=1}^{n-1}(1-p\_k^{-2m})-\frac{1}{(2\pi)^{2m}} \right ) $$
which gives if we conveniently replace the internal expressions with $a\_{2m}$ and $b\_{2m}$:
$$ \ln(p\_n)=-2\pi-\lim\limits\_{m \to +\infty} \frac1{2m} \ln \left ( a\_{2m}\prod\limits\_{k=1}^{n-1}(1-p\_k^{-2m})-b\_{2m} \right ) $$
|
https://mathoverflow.net/users/nan
|
Recursive formula for n-th prime derived from a previous question
|
Based on
1. [Keller - A recursion equation for prime
numbers](http://arxiv.org/abs/0711.3940)
2. [Kawalec - The recurrence
formulas for primes and non-trivial zeros of the Riemann zeta
function](http://arxiv.org/abs/2009.02640)
it is obvious that they meant to use or even used real values. The novelty here is that the expression has a form that you can split in two parts, one depending on $m \gg 1$ and another on $n$ where it all becomes very illustrative as some a sort of an internal sieving machine.
I would still opt of logarithm because root looks quite clumsy to deal with:
$$ \ln(p\_n)=-2\pi-\lim\limits\_{m \to +\infty} \frac1{2m} \ln \left ( a\_{2m}\prod\limits\_{k=1}^{n-1}(1-p\_k^{-2m})-b\_{2m} \right ) $$
Additionally, the other summable version one of the authors is mentioning you get directly:
$$\ln(p\_n)=-\lim\limits\_{m \to +\infty} \frac1{2m}\ln \left ( \prod\limits\_{k=1}^{n-1}(1-p\_k^{-2m})-\frac{2(2m)!}{|B\_{2m}|(2\pi)^{2m}} \right )$$
which is cute.
|
0
|
https://mathoverflow.net/users/nan
|
402180
| 165,053 |
https://mathoverflow.net/questions/402177
|
1
|
Let $A$ be a $C^\*$-algebra and $E$ be a (right) Hilbert $C^\*$-module over $A$. Assume $F$ is a closed submodule of $E$ such that $F^\perp := \{x \in E: \langle x, F\rangle=0\}$ is orthogonally complemented, i.e. we have $F^\perp \oplus F^{\perp \perp} = E.$ Can we conclude that $F$ is orthogonally complemented, and if this is the case do we have $F= F^{\perp \perp}$?
I know that in general we don't need to have $F= F^{\perp \perp}$. However, when $F$ is orthogonally complemented, i.e. $F\oplus F^\perp = E$ this is obvious. For Hilbert spaces (i.e. $A = \mathbb{C}$), there are no counterexamples.
|
https://mathoverflow.net/users/216007
|
Complemented submodules of a Hilbert C*-module
|
If $F^\perp =\{0\}$ then $F^\perp \oplus F^{\perp\perp} = E$ is automatic, so all you need is an example of this where $F \neq E$. For instance, $C[0,1]$ as a Hilbert module over itself with $F =$ the functions which vanish at 0.
|
2
|
https://mathoverflow.net/users/23141
|
402183
| 165,055 |
https://mathoverflow.net/questions/400879
|
7
|
Given a partition $\lambda$ and its Young diagram $\pmb{Y}\_{\lambda}$, we say $\lambda$ is a $(t,s)$-core partition provided that neither $t$ nor $s$ is a hook length in $\pmb{Y}\_{\lambda}$. We now recall a conjecture (now a theorem) of D. Armstrong which states "the total number $(s,t)$-core partitions is $\frac1{s+t}\binom{s+t}s$".
Focusing on the special case of $(s,s+1)$-core partitions, [Stanley and Zanello](http://math.mit.edu/%7Erstan/papers/core.pdf) proved the enumeration $C\_s=\frac1{2s+1}\binom{2s+1}s$. Their argument utilizes a result of J. Anderson: $(s,s+1)$-cores correspond bijectively to the order ideals of
the poset $P(s,s+1)$ (that is, positive integers that are not contained in the numerical
semigroup generated by $s$ and $s+1$) by associating the lower ideal $\{a\_1,\dots,a\_j\}$, where $a\_1>\cdots>a\_j$, to the $(s,s+1)$-core partition $(a\_1-(j-1), a\_2-(j-2),\dots,a\_{j-1}-1,a\_j)$.
A few years ago, I considered core-partitions [**with distinct parts**](http://dauns01.math.tulane.edu/%7Etamdeberhan/conjectures.pdf) (see page 13) to the effect that the number of such $(s,s+1)$-core partitions equals the Fibonacci number $F\_{s+1}$. This conjecture is now a theorem due to different people ([see this paper](https://www.combinatorics.org/ojs/index.php/eljc/article/view/v25i1p57) and references therein).
>
> **QUESTION.** Viewing $(s,s+1)$-cores of distinct partitions as embedded in the ordered ideals of the poset $P(s,s+1)$ (the above Catalan case), do they have any "interesting" structure?
>
>
>
|
https://mathoverflow.net/users/66131
|
Fibonacci embedded in Catalan?
|
The order ideal corresponding to a core partition with distinct parts cannot contain an element $x \geq s+2$, otherwise it must also contain $x-s, x-s-1$, resulting in two equal parts. For the same reason, it also can't contain two consecutive elements in $\{1,2,\dots, s-1\}$. Therefore the poset in questions is simply the poset of subsets of $\{1,2,\dots, s-1\}$ that do not contain consecutive elements, ordered by inclusion.
The underlying Hasse diagram of this poset is called the [Fibonacci cube](https://en.wikipedia.org/wiki/Fibonacci_cube).
|
6
|
https://mathoverflow.net/users/2384
|
402188
| 165,057 |
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