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https://mathoverflow.net/questions/405577
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4
|
I have been looking through Alan Pears' "Dimension theory of general spaces" recently. In this book Pears references a 1960 paper by Aleksandrov and Ponomarev called "Some classes of $n$-dimensional spaces" that contains a sufficient condition for the large inductive dimension and the covering dimension to coincide for compact Hausdorff spaces. I can find the paper, but only in Russian.
[Link to paper](http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=smj&paperid=4836&option_lang=eng)
Does anyone know what Aleksandrov and Ponomarev's condition is? Moreover, aside from being pseudometrizable, what are some conditions one can impose on a compact Hausdorff space $X$ so that $\operatorname{Ind}(X)=\dim(X)$?
|
https://mathoverflow.net/users/115694
|
Sufficient conditions for the covering dimension and large inductive dimension of compact Hausdorff spaces to coincide
|
You can read a review of the paper [in Zentralblatt](https://www.zbmath.org/?q=an%3A0108.35605), it contains a short description in German.
The review on [MathSciNet](https://mathscinet.ams.org/mathscinet-getitem?mr=124031) is a bit more extensive (but requires a subscription). There is indeed the condition of having an $n+1$-to-one map from a zero-dimensional compact space onto the space itself. There is also a condition on `gratings' (not defined, but my guess is, based on other papers: a grating is a finite closed cover where the interiors of the closed sets are pairwise disjoint).
In [What is a non-metrizable analog of metrizable compacta? (Part I)](https://doi.org/10.1016/j.topol.2011.04.018) Pasynkov defines a class of compacta where the dimensions coincide.
|
1
|
https://mathoverflow.net/users/5903
|
405732
| 166,327 |
https://mathoverflow.net/questions/405673
|
2
|
Is there a name for a pseudo-Riemannian manifold that admits no nonzero null vectors? More precisely: For a pseudo-Riemannian manifold $(R,g)$, a **null vector** is a non-zero vector field $X:M \to TM$ such that
$$
g(X,X)(m) = 0, \forall m \in M.
$$
As this [question][1] shows - vector like this exist. But can there exist manifolds where they do not exist and do such manifolds have a name?
|
https://mathoverflow.net/users/378228
|
A name for a pseudo-Riemannian manifold that admits no nonzero null vectors
|
Note first that every pseudo-Riemmanian manifold admits a null vector field which is not identically $0$ (just construct one locally and multiply it by a bump function). So by "non-zero vector field" I assume you mean "nowhere vanishing".
Let $(M,g)$ be a pseudo-Riemannian manifold of signature $(p,q)$. The tangent bundle $TM$ always admits an orthogonal splitting as $E \overset{\perp}{\oplus} F$, where $E$ and $F$ are respectively positive and negative definite (hence of respective rank $p$ and $q$). Moreover this splitting is unique up to homotopy (because, pointwise, the set of such splittings is the symmetric space of the orthogonal group $O(p,q)$, which is contractible).
**Proposition:** $M$ admits a nowhere vanishing null vector field if and only if $E$ and $F$ both admit nowhere vanishing sections.
*Proof:* Decompose a nowhere vanishing null vector field $X$ as $X\_E + X\_F$. Then $g(X\_E,X\_E) = -g(X\_F,X\_F)$. If this is $0$ at some point then $X\_E$ and $X\_F$ vanish at that point (since $g$ is positive definite on $E$ and negative definite on $F$) contradicting the non-vanishing of $X$. Hence $X\_E$ and $X\_F$ are non-vanishing sections of $E$ and $F$.
Conversely, if $X\_E$ and $X\_F$ are non-vanishing sections of $E$ and $F$ respectively, then up multiplying $X\_F$ them by a positive function, we can assume that $g(X\_E,X\_E) = -g(X\_F,X\_F)$. Hence $X\_E+X\_F$ is a nowhere vanishing null vector field. CQFD
There are thus topological obstructions to the existence of such a vector field (mainly the non-vanishing of the Euler class of $E$ or $F$). For instance, Let $(A,g\_A)$ and $(B,g\_B)$ be Riemannian manifolds, with $A$ of non-zero Euler characteristic, and consider $(M,g) = (A\times B, g\_A \oplus -g\_B)$. Then $M$ does not admit a nowhere vanishing null vector field. Indeed, we have the splitting $TM = TA\oplus TB$, and the projection of a null vector field to TA must vanish somewhere since the Euler class of TA is non-zero.
|
5
|
https://mathoverflow.net/users/173096
|
405740
| 166,330 |
https://mathoverflow.net/questions/403486
|
3
|
After going through an application of the von Neumann entropy(from quantum information theory) to certain problems in computational neuroscience [2], it occurred to me that this entropy might have applications to classical mechanics as it appears to be generally useful for dimensionality reduction.
Motivation:
-----------
Let's suppose we have an ergodic dynamical system with $N$ observables
$x\_i(t) \in \mathbb{R}$ which are sampled using a sequence of $n$ measurements
so we have a dataset $X \in \mathbb{R}^{N \times n}$. Given $X$, we may compute the statistics $X\_i = x\_i - \langle x\_i \rangle$ and
$\sigma\_i^2 = \langle X\_i^2 \rangle$ which allows us to define the Pearson correlation matrix with entries:
\begin{equation}
R\_{i,j} = \frac{X\_i \cdot X\_j}{\sigma\_i \cdot \sigma\_j} \tag{1}
\end{equation}
Given $R \in \mathbb{R}^{N \times N}$, we may define the density matrix $\rho = \frac{R}{N}$ which is positive semi-definite, Hermitian and has unit-trace. Thus, we may calculate the entropy of $R$ using the von Neumann entropy:
\begin{equation}
S(\rho) = -\text{tr}(\rho \cdot \ln \rho) = - \sum\_{i=1}^N \lambda\_i \cdot \ln \lambda\_i \tag{2}
\end{equation}
where $\lambda\_i$ are the eigenvalues of $\rho$.
If $N$ is large, in order to compress the dataset $X$ so that we keep the dimensions that contain $95\%$ of the statistical information it is sufficient to find the discrete subset $S \subset [1,N]$ that maximises:
\begin{equation}
- \sum\_{i \in S} \lambda\_i \cdot \ln \lambda\_i \tag{3}
\end{equation}
subject to the constraint $\frac{- \sum\_{i\in S} \lambda\_i \cdot \ln \lambda\_i}{- \sum\_{i=1}^N \lambda\_i \cdot \ln \lambda\_i} \leq \frac{95}{100}$ which may be done using sorting algorithms such as Quick Sort. Moreover, the cardinality $\lvert S \rvert$ provides us with an approximate upper-bound on the intrinsic dimension of an ergodic dynamical system.
Question:
---------
Has the von Neumann entropy ever been used in classical mechanics?
Note:
-----
It is worth noting that the idea of viewing the Pearson correlation matrix as a density matrix is quite natural and not new [4].
References:
-----------
1. von Neumann, John (1932). Mathematische Grundlagen der Quantenmechanik (Mathematical foundations of quantum mechanics) Princeton University Press., . ISBN 978-0-691-02893-4.
2. H. Felippe et al. The von Neumann entropy for the Pearson correlation matrix:
A test of the entropic brain hypothesis. [Arxiv](https://arxiv.org/abs/2106.05379). 2021.
3. E.T. Jaynes. Information theory and statistical mechanics. The Physical Review. 1957.
4. linello (<https://physics.stackexchange.com/users/10941/linello>), Can a correlation matrix be regarded as a quantum density matrix?, URL (version: 2016-09-28): <https://physics.stackexchange.com/q/282904>
|
https://mathoverflow.net/users/56328
|
Has the von Neumann entropy ever been used in classical mechanics?
|
You are really asking just about the applications of the notion of entropy to an analysis of correlation matrices, so that this question is not necessarily related with classical mechanics. The entropy here is the plain Boltzmann - Planck - Shannon entropy rather than von Neumann's one, as the mere appearance of a positive definite symmetric matrix does not necessarily warrant referring to the whole quantum mechanics machinery. The keyword is "principal component analysis" (PCA). A quick search on "entropy + principal component analysis" returns quite a few articles (overwhelmingly paramathematical though). For example, [Component retention in principal component analysis with
application to cDNA microarray data](https://biologydirect.biomedcentral.com/track/pdf/10.1186/1745-6150-2-2.pdf) where the entropy is used to provide an estimate of the number of interpretable components in a principal component analysis, or [Entropy principal component analysis and its application to nonlinear chemical process fault diagnosis](https://onlinelibrary.wiley.com/doi/epdf/10.1002/apj.1813) where "entropy principal component analysis" appears right in the title.
|
2
|
https://mathoverflow.net/users/8588
|
405748
| 166,332 |
https://mathoverflow.net/questions/405752
|
0
|
When I did some research, I have not found the analytical expression about the eigenvalues and eigenvectors of Laplacian matrix in a chain graph while I only found those in a cycle graph. The Laplcian matrix is as follows.
$$L=\begin{bmatrix}1 & -1 & 0 & 0 & 0\\-1 & 2 & -1 & 0 & 0\\0 & -1 & 2 & -1 & 0\\ 0 & 0 & -1 & 2 & -1\\0 & 0 & 0 & -1 & 1\end{bmatrix}$$
It seems to have some relationship with the Fourier series, but I still found no paper in which the writer gave this analytical expression about the eigenvalues and eigenvectors of Laplacian matrix in a chain graph. Could you give me some advice on it?
|
https://mathoverflow.net/users/394256
|
The eigenvectors and eigenvalues of Laplacian matrix in a chain graph
|
You can find it in [this paper](https://ieeexplore.ieee.org/abstract/document/5717507?casa_token=wzm-Ny2TJyYAAAAA:yoCoqQc-JPp7yPxO9AhtR43atzGbV0L61c4o1teX4A3QxPGlJEv0_traXxbEmajwKjBfgOstKZs).
In case that link doesn't work, search at Google Scholar for "On the observability of path and cycle graphs".
|
2
|
https://mathoverflow.net/users/9025
|
405759
| 166,335 |
https://mathoverflow.net/questions/405731
|
3
|
Let $X$ be a Polish space, i.e. a separable complete metric space. Any Borel probability measure on $X$ must be locally finite, outer regular and tight. Let $\mathcal{P}(X)$ be the set of all Borel probability measures on $X$.
A **rational preference relation** on $\mathcal{P}(X)$ is a binary relation $\precsim$ on $\mathcal{P}(X)$ that satisfies the following axioms:
1. **Completeness**: for any $ \mu, \nu \in \mathcal{P}(X) $, either $ \mu \precsim \nu $ or $ \nu \precsim \mu $
2. **Transitivity**: for any $ \lambda, \mu, \nu \in \mathcal{P}(X) $, if $ \lambda \precsim \mu $ and $ \mu \precsim \nu $, then $ \lambda \precsim \nu $
3. **Continuity**: for any $ \lambda, \mu, \nu \in \mathcal{P}(X) $, the sets $ \{ p \in [0, 1] : p \mu + (1-p) \nu \precsim \lambda \} $ and $ \{ p \in [0, 1] : \lambda \precsim p \mu + (1-p) \nu \} $ are closed in $[0, 1]$
4. **Independence**: for any $ \lambda, \mu, \nu \in \mathcal{P}(X) $ and $ 0 < p \leq 1 $, we have $ \mu \precsim \nu $ if and only if $ p \mu + (1-p) \lambda \precsim p \nu + (1-p) \lambda $
A **von Neumann-Morgenstern utility** is a Borel random variable $ U: X \to \mathbb{R} $ such that for any $ \mu, \nu \in \mathcal{P}(X) $
$$ \mu \precsim \nu \iff \mathbb{E}\_\mu[U] \leq \mathbb{E}\_\nu[U] $$
The [von Neumann-Morgenstern utility theorem](https://en.wikipedia.org/wiki/Von_Neumann%E2%80%93Morgenstern_utility_theorem) asserts that a von Neumann-Morgenstern utility always exists on a finite set:
>
> Let $X$ be a finite set, equipped with the discrete $\sigma$-algebra
> $2^X$. For any rational preference relation $\precsim$ on
> $\mathcal{P}(X)$, there exists a corresponding von Neumann-Morgenstern
> utility. Note that in this case, $\mathcal{P}(X)$ is just the standard
> $|X|-1$ dimensional simplex in $[0, 1]^{|X|}$.
>
>
>
**Question:**
Let $X$ be a Polish space and $\precsim$ a rational preference relation on $\mathcal{P}(X)$. What sufficient conditions can guarantee the existence of a corresponding von Neumann-Morgenstern utility? What about a **continuous** one?
|
https://mathoverflow.net/users/49284
|
Conditions for the existence of von Neumann-Morgenstern utility on a Polish space
|
There exists a continuous, bounded utility function if and only if the relation is continuous in the stronger sense of being closed in $P(X) \times P(X)$, using the weak${}^\*$ topology on each factor. See Section 3.3 of [this paper](https://arxiv.org/abs/2104.11205), for example. (The point of that paper is to find Lipschitz utility functions, but we give an overview of the general situation.)
|
4
|
https://mathoverflow.net/users/23141
|
405764
| 166,337 |
https://mathoverflow.net/questions/405766
|
5
|
Let $p: \mathcal{X} \rightarrow \text{Spec } \mathcal{O}\_K$ be a normal proper Artin stack with finite diagonal. A $K$-rational point is by definition a section $x: \text{Spec}(K) \rightarrow \mathcal{X}$ of $p$ over the generic point of $\text{Spec } \mathcal{O}\_K$, and an integral point is a section $\bar{x}: \text{Spec } \mathcal{O}\_K \rightarrow \mathcal{X}$ of $p$.
Now if $\mathcal{X}$ were to be a proper scheme, then every rational point extends uniquely to an integral point. Indeed, the valuative criterion gives morphisms from every division ring that we can patch up to a morphism from $\text{Spec } \mathcal{O}\_K$.
I am currently reading the work "Heights on stacks and a generalized Batyrev–Manin–Malle
conjecture" and there it is claimed that the above statement is no longer true for stacks, but it is still true after a (possibly) ramified extension of $\text{Spec } \mathcal{O}\_K$. This leads to my two questions:
1. Why is this not true for stacks? Although a specific counterexample would be great, I am also happy with some intuition why this ought to fail for stacks.
2. Why is it still true after a (possibly) ramified extension of $\text{Spec } \mathcal{O}\_K$? In this case I am mostly interested in a precise argument.
|
https://mathoverflow.net/users/96891
|
Extending rational to integral points
|
$\DeclareMathOperator{Spec}{Spec}$ For 1. take $\mathcal{X} = B \mu\_2$ to be the classifying stack of $\mu\_2$ over $\mathbb{Z}$. For any $\mathbb{Z}$-scheme $S$ we have $\mathcal{X}(S) = \mathrm{H}^1(S, \mu\_2)$. Thus we have $\mathcal{X}(K) = K^\times/K^{\times 2}$, but this doesn't equal $\mathcal{X}(\mathcal{O}\_K)$. For example, if $\mathcal{O}\_K$ is a PID then $\mathcal{X}(\mathcal{O}\_K) = \mathcal{O}\_K^\times/\mathcal{O}\_K^{\times2}$.
For 2., this is the valuative criterion for properness of a morphism of stacks (see <https://stacks.math.columbia.edu/tag/0CLY>, <https://stacks.math.columbia.edu/tag/0CQL>, and <https://stacks.math.columbia.edu/tag/0CL9>). Namely, for stacks one only stipulates the existence of an extension of a map from valuation ring after a possible base change of the ring, unlike for the case of schemes where existence of an extension is stipulated for the actual ring.
You can check this property yourself in the above example of $B \mu\_2$.
|
7
|
https://mathoverflow.net/users/5101
|
405767
| 166,338 |
https://mathoverflow.net/questions/392060
|
8
|
My question is about the correct order to read the papers by Arthur on trace formula. Arthur's papers are perfectly well-written, but maybe a little too hard for me to go through easily.
I would like to start with unrefined trace formula, but there are bunches of papers already. There is a paper on trace formula for rank-1 groups (1974), and there are two papers on trace formula for general reductive groups (1978 & 1980). Also there is a paper in Corvallis Proceedings (1979), and a notes on trace formula (2005) which has become the standard introductory notes on this subject. I am seeking for any advice on how to read these references.
I wish to at least go through some details to feel how the formula is proved. Now I have read the notes by Whitehouse on trace formula for $\operatorname{GL}(2)$ (which covers a great many details about computation), and read a small part of Arthur's paper on rank-1 groups (1974). My question is that whether it is better to take 1974 paper first to gain more intuition, or instead start with the more general setting of reductive groups (like 1978 & 1980 papers), where maybe things are actually more clarified.
Any recommendation of other references is also sincerely welcomed. I am not aware of any other introductory references on trace formula on general reductive groups.
Thank you very much for reading this lengthy question! I am totally new to this community, so you are welcomed to point out any mistake.
|
https://mathoverflow.net/users/159973
|
How to read the paper of Arthur on trace formula on general reductive groups
|
One of the best places to learn about trace formula, other than David Whitehouse's wonderful notes, are the notes by Erez Lapid [Introductory notes on the trace formula](https://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.494.5118).
Arthur's trace formula relies on Langlands's work on Eisenstein series and Spectral decomposition from 1962–64. This stuff is extremely difficult for general reductive groups.
It would be near insanity to directly go to Arthur's papers or notes without mastering the spectral decomposition of automorphic forms on GL(2) and the Selberg trace formula. It is not wise to learn trace formula without learning about spectral decomposition.
For the general case, Borel's notes [Automorphic forms on reductive groups](http://homepages.math.uic.edu/%7Ertakloo/papers/borel/borel3.pdf) are a reasonable beginning.
My advisor Paul Garrett's book [Modern analysis of automorphic forms by example](https://www-users.cse.umn.edu/%7Egarrett/m/v/current_version.pdf) is, of course, my favorite place. They treat several examples, but do not do any higher rank case completely.
|
7
|
https://mathoverflow.net/users/303419
|
405772
| 166,340 |
https://mathoverflow.net/questions/405491
|
3
|
In [Salamon's notes on Floer homology](https://people.math.ethz.ch/%7Esalamon/PREPRINTS/floer.pdf), it's claimed that under some non-degenerancy assumptions the operator $$D:= \partial\_s+J\_0\partial\_t+S(s,t): W^{1,p}(\mathbb{R}\times S^1,\mathbb{R}^{2n})\rightarrow L^p(\mathbb{R}\times S^1,\mathbb{R}^{2n})$$ is Fredholm for $1<p<\infty$, and moreover its Fredholm index is given by index $$D=\mu\_{CZ}(\Psi^{+})-\mu\_{CZ}(\Psi^{-}).$$
Now I was trying to see why this index formula holds up. In the lectures notes the author claims that this formula is true for $p=2$, where we are dealing with Hilbert spaces, by using the spectral flow of the family of operators $J\_0\partial\_t+S(s,t)$. However nothing else is said for the case where $p\neq 2$. It's clear that in the end this Fredholm index won't depend on the choice of $p$, but a priori we would need to prove that this is true so that we get the desired result just by proving the case $p=2$. Is this what is happening in this situation ? We can show that $D$ has the same Fredholm index for any $p$ without using the Conley-Zehnder index ?
Any insight is appreciated, thanks in advance.
|
https://mathoverflow.net/users/155363
|
Computing the Fredholm index in Floer theory
|
The idea is to show that the kernel & cokernel consist of smooth sections, and thus are independent of $p$. Since the Fredholm index is the difference between the dimensions of these, it doesn't depend on $p$.
This is proved, admittedly in the case without punctures, in an appendix of McDuff & Salamon's J-holomorphic curve book.
Another good reference, this time with punctures, is from Chris Wendl's lecture notes/book draft. The direct link to the PDF file is here:
[https://www.mathematik.hu-berlin.de/~wendl/Sommer2020/SFT/lecturenotes.pdf](https://www.mathematik.hu-berlin.de/%7Ewendl/Sommer2020/SFT/lecturenotes.pdf)
|
4
|
https://mathoverflow.net/users/477
|
405782
| 166,346 |
https://mathoverflow.net/questions/405754
|
2
|
It is a well-known fact that the range of a positive measure $\mu$ on a measure space $(X,\cal M)$ is the interval $[0,\mu(X)]$ provided $\mu$ is atomless (i.e., there is no measurable set $A\in \cal M$ such that $\mu(A)>0$ and for any measurable set $B\subset A$ we have $\mu(B)=0$ or $\mu(B)=\mu(A)$).
It seems that the 'standard' proof of this fact involves some transfinite induction argument (cf. e.g. Halmos 1947, Wikipedia or else), but I am looking for a proof not using transfinite induction as suggested by some exercises in Bourbaki or Dieudonné Analysis II.
This would entail proving the following:
Let $r\in ]0,\mu(X)[$, and let $\cal C=\{A\in \cal M:\mu(A)\le r\}$. Suppose that $\mu$ has a 'gap' at $r$, meaning that we have $\alpha<r$, where $\alpha=\sup\{\mu(A):A\in \cal C\}$. Then $\mu$ must have an atom.
A simple path leading to that result would be to prove that in the presence of a gap at $r$, $\cal C$ is closed under finite union; in other words, if $A,B$ are such that $\mu(A)\le \alpha$ and $\mu(B)\le \alpha$, then $A$ and $B$ have no other choice but to overlap in such a way that $\mu(A\cup B)\le \alpha$. Equivalently if $A$ and $B$ are supposed also disjoint, we must have $\mu(B)\le \alpha-\mu(A)$.
The intuitive argument is that if neither $A$ nor $B$ contains an atom, then $A\cup B$ cannot contain it. I would suspect the proof to be well-known, but my brain seems unable to produce it.
|
https://mathoverflow.net/users/65954
|
A question about the range of a positive measure
|
(1) Existence of sets of small measure.
Let $(\Omega,\Sigma,\mu)$ an atomless finite measure space with $\mu(\Omega)>0$.
It is easy to show that any measurable set $A$ with $\mu(A)>0$ contains a measurable subset $B\subset A$ with $0<\mu(B)\le \mu(A)/2$, and therefore a measurable set $M$ of measure $0<\mu(M)\le \varepsilon$.
(2) The space and any measurable set is a numerable union of set of small measure.
Now given $\varepsilon>0$ consider the set $\mathcal M$ of measurable sets $M=\bigcup\_n A\_n$
union of a numerable set of measurable sets of measure $\mu(A\_n)\le \varepsilon$
Let $\lambda$ the supremum of the measures $\mu(M)$ of set in $\mathcal M$.
There is a sequence $(M\_n)$ of sets in $\mathcal M$ with $\lim\_n\mu(M\_n)=\lambda$. Then $M\_0:=\bigcup M\_n\in \mathcal M$ (by definition) and $\lambda(M\_0)=\lambda$. But it follows that $\lambda=\mu(\Omega)$, because in other case $\Omega\smallsetminus M\_0$ will contain a set of measure $\le \varepsilon$, which contradict the maximality of $M\_0$.
(3) Given a measurable set $M$ and two numbers $0<a<b<\mu(M)$, there exists a measurable set $A\subset M$ with $a<\mu(A)<b$.
Let $\varepsilon=(b-a)/2$, by (2) the set $M$ can be put as a union of measurable sets of measure less than $\varepsilon$, $M=\bigcup\_n M\_n$ with $\mu(M\_n)\le \varepsilon$. Then $$\lim\_N\mu\Bigl(\bigcup\_{n=1}^N M\_n\Bigr)=\mu(M).$$ Therefore there is some $N$ such that $a<\mu\Bigl(\bigcup\_{n=1}^N M\_n\Bigr)<b$, since each $M\_n$ have measure $\le (b-a)/2$.
(4) Construct the set with measure $\mu(A)=a$ for a given $0<a<\mu(\Omega)$.
We will construct a sequence of measurable sets $(A\_n)$ with $A\_n\le A\_{n+1}$ and such that
$$\frac{a+\mu(A\_n)}{2}<\mu(A\_{n+1})< a.$$
It is clear that in this case $A=\bigcup\_n A\_n$ have measure $a$.
Start with $A\_0=\emptyset$. We must construct $A\_1$ with $a/2<\mu(A\_1)<a$. Sincer $\Omega$ have measure $>a$, we have constructed such a set in (3).
Assume we have constructed $A\_n$, then $\mu(A\_n)<a$, applying (3) we construct $B\_n\subset \Omega\smallsetminus A\_n$ whose measure is greater than half the difference with our objective: $a-\mu(A\_n)>\mu(B\_n)>\frac{a-\mu(A\_n)}{2}$.
Hence $A\_{n+1}=A\_n\cup B\_n$ satisfies our requirement. First $A\_n$ and $B\_n$ are disjoint $\mu(A\_{n+1})=\mu(A\_n)+\mu(B\_n)<a$, and $\mu(A\_n)+\mu(B\_n)>\frac{a+\mu(A\_n)}{2}$.
|
3
|
https://mathoverflow.net/users/7402
|
405783
| 166,347 |
https://mathoverflow.net/questions/405781
|
4
|
Recall that a [door space](https://en.wikipedia.org/wiki/Door_space) is a topological space where every set is either open or closed (or both). A topological space is *finite* if it has finitely many points. I'm interested in learning about finite door spaces.
**Question 0:** What is an example of a finite topological space which is $T\_0$ but not a door space?
**Question 1:** Let $n$ be a natural number. How many door topologies are there on a set with $n$ elements?
I'm interested in both the "labeled" and "unlabeled" versions of question 1.
Recall that via the specialization order, finite topological spaces are equivalent to finite preorders, i.e. quasi-(partially)ordered sets.
**Question 2:** Which finite posets are the specialization order of a door space?
|
https://mathoverflow.net/users/2362
|
How do finite door spaces work?
|
A door space $X$ is $T\_0$, for if $x,y\in X$ are not separated by the
$T\_0$ axiom then the set $\{x\}$ is neither open nor closed. A finite $T\_0$
space is equivalent to a finite poset $P$ (*Enumerative
Combinatorics*, vol. 1, second ed., Exercise 3.3). An open set
corresponds to an order ideal of $P$. Thus we want to count posets on
an $n$-element set such that every subset of $P$ is either an order
ideal or the complement of an order ideal. Such posets can have at
most one connected component $Q$ which is not a single point. Then $Q$
consists of an element $x$ at the bottom and some positive number of
elements $y$ covering $x$ (i.e., $x<y$ with nothing in between), or
the dual of this connected component. There is one choice if $P$ is a
disjoint union of points. There are $n(n-1)$ choices if $x$ is covered
by one element, but then the dual has the same form and should not be
counted again. If $x$ is covered by more than one element, then there
are $n(2^{n-1}-n)$ choices. This must be multiplied by 2 since the
dual is different. Hence the total number of door spaces on an
$n$-element set is
$$ 1+n(n-1)+2n(2^{n-1}-n) = 1-n-n^2+n2^n. $$
By similar reasoning, the number of unlabelled $n$-element door
topologies is $1+1+2(n-2)=2n-2$.
|
19
|
https://mathoverflow.net/users/2807
|
405785
| 166,348 |
https://mathoverflow.net/questions/402822
|
5
|
With the aid of the simple identity
\begin{equation\*}
\sum\_{k=0}^{n}\binom{n+k}{k}\frac{1}{2^{k}}=2^n
\end{equation\*}
in Item (1.79) on page 35 of the monograph
R. Sprugnoli, *Riordan Array Proofs of Identities in Gould’s Book*, University of Florence, Italy, 2006. (Has this monograph been formally published somewhere?)
I proved the combinatorial identity
$$
\sum\_{k=1}^{n}\binom{2n-k-1}{n-1}k2^k=n\binom{2n}{n}, \quad n\in\mathbb{N}.
$$
My question is: how to prove the more general combinatorial identity
$$
\sum\_{k=\ell}^{n}\binom{2n-k-1}{n-1}k2^k=2^\ell n\binom{2n-\ell}{n}
$$
for $n\ge\ell\ge0$?
|
https://mathoverflow.net/users/147732
|
How to prove the combinatorial identity $\sum_{k=\ell}^{n}\binom{2n-k-1}{n-1}k2^k=2^\ell n\binom{2n-\ell}{n}$ for $n\ge\ell\ge0$?
|
For $k\in\mathbb{N}$, let $s\_k$ and $S\_k$ be two sequences independent of $n$ such that $n\ge k\in\mathbb{N}$. The inversion theorem, Theorem 4.4 on page 528 in the freely downloading paper [1] below, reads that
\begin{equation}\label{Qi-Zou-Guo-Inversion-thm}\tag{1}
s\_n=\sum\_{k=1}^{n}\binom{k}{n-k}S\_k
\quad\text{if and only if}\quad
(-1)^nnS\_n=\sum\_{k=1}^{n}\binom{2n-k-1}{n-1}(-1)^kks\_k.
\end{equation}
Applying the inversion theorem in \eqref{Qi-Zou-Guo-Inversion-thm} and considering the identity
\begin{equation\*}
\sum\_{k=1}^{n}\binom{2n-k-1}{n-1}2^kk
=2n\binom{2n-1}{n},
\end{equation\*}
which can be deduced from letting $\ell=1$ in the resulted identity of the above answers, we conclude
\begin{equation\*}
\sum\_{k=1}^{n}(-1)^k\binom{k}{n-k}\binom{2k-1}{k}=(-1)^n2^{n-1}, \quad n\in\mathbb{N}.
\end{equation\*}
See Remark 3.4 on page 11 in the paper [2] below.
References
1. Feng Qi, Qing Zou, and Bai-Ni Guo, *The inverse of a triangular matrix and several identities of the Catalan numbers*, Applicable Analysis and Discrete Mathematics **13** (2019), no. 2, 518--541; available online at <https://doi.org/10.2298/AADM190118018Q>.
2. Feng Qi and Mark Daniel Ward, *Closed-form formulas and properties of coefficients in Maclaurin's series expansion of Wilf's function*, arXiv (2021), available online at <https://arxiv.org/abs/2110.08576v1>.
|
0
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https://mathoverflow.net/users/147732
|
405812
| 166,356 |
https://mathoverflow.net/questions/405806
|
5
|
Let $X$ be a compact Hausdorff space and $C(X)$ its algebra of continuous complex valued functions. The Gelfand-Naimark theorem tells us that we have a duality between commutative $C^\*$-algebras and compact Hausdorff spaces: Send $X$ to its $C^\*$-algebra of continuous functions.
The Swan-Serre theorem tells that there is a duality between finitely generated projective $C(X)$-modules and complex continuous vector bundles $V$ over $X$: Send a vector bundle $V$ to its space of sections $\Gamma$.
If we go further and put an Hermitian metric $g$ on $V$ then the space of sections $\Gamma$ is a Hilbert $C(X)$-module. In the other direction things are less clear to me. If we have a Hilbert module structure on $\Gamma(V)$, does it necessarily come from an Hermitian metric and if so does this give a duality between Hilbert $C(X)$-modules and Hermitian vector bundles over $C(X)$?
|
https://mathoverflow.net/users/153228
|
Hermitian vector bundles and Hilbert $C^*$-modules
|
In addition to Nik Weaver's references, let me just sketch the proof which is in fact not very difficult:
1. A construction of Kaplansky (Rings of operators, Thm 26) shows that if $\mathcal{A}$ is a $\*$-algebra with the property that for every $A \in M\_n(\mathcal{A})$ the matrix $1 + A^\*A$ is invertible, then every idempotent in $M\_n(\mathcal{A})$ is equivalent to a projection. A $C^\*$-algebra and in particular $C(M)$ clearly has this feature by spectral calculus for every $n \ge 1$.
Suppose now that $\mathcal{A}$ is such a $^\*$-algebra satisfying this condition for all $n$ (†).
2. On every finitely generated projective module $\mathcal{E}$ over $\mathcal{A}$, written without restriction as $\mathcal{E} = P \mathcal{A}^n$ with $P = P^2 = P^\* \in M\_n(\mathcal{A})$, the restriction $\langle . , .\rangle$ of the canonical Hermitian algebra-value inner product of the free module $\mathcal{A}^n$ is still positive (in the strongest possible sense) and non-degenerate (in the strongest possible sense that the musical homomorphism is in fact an anti-isomorphism to the dual module). This gives immediately the existence of positive algebra-valued inner products on finitely generated projective modules over such algebras.
3. Now suppose the algebra satisfies an additional feature: suppose $H \in M\_n(\mathcal{A})$ is an invertible positive element. Then $H$ has a positive square root $H = \sqrt{H}^2$ with the property that $\sqrt{H}$ commutes with all matrices which commute with $H$ (‡). Again, a $C^\*$-algebra clearly satisfies this for all $n$ by spectral calculus.
Now suppose that on the fgpm $\mathcal{E} = P \mathcal{A}^n$ you have another positive algebra-valued inner product, denoted by $h$. On the complement $P^\bot = (1 - P)\mathcal{A}^n$ we use the restriction of the canonical algebra-valued inner product to obtain a new inner product, still denoted by $h$, on the direct sum. This is still positive and has all required (very strong) non-degeneracy properties.
4. By matrix calculus with free modules one obtains then a matrix $H \in M\_n(\mathcal{A})$ with
\begin{equation}
h(x, y) = \langle x, Hy\rangle
\end{equation}
for all $x, y \in \mathcal{A}^n$. Since by assumption $H = \sqrt{H}^2$ the square root $U = \sqrt{H}$ provides an isometry between the two inner products on the free module. Since by assumption the square root commutes with everything commuting with $H$ and since by construction $H$ commutes with $P$ (check this!) $U$ restrict to an isometry between $h$ and $\langle ., .\rangle$ on the original $\mathcal{E}$.
In conclusion: for $^\*$-algebras (like e.g. $C^\*$-algebras) satisfying the above two properties (†) and (‡), every finitely generated projective module carries a unique-up-to-isometry positive algebra-valued inner product.
As a side remark: many other types of $^\*$-algebras satisfy these properties as well like e.g. the smooth functions on a manifold etc. So the same proof also implies (via Serre-Swan in the smooth context) that on smooth vector bundles you always have a unique-up-to-isometry positive fiber metric.
Now, to answer you question: the uniqueness gives you the desired result that every algebra-valued inner product comes from a positive fiber metric. Indeed, a positive fiber metric meets the above properties and the isometry $U$ maps fiber metrics to fiber metrics as it is algebra-linear.
|
4
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https://mathoverflow.net/users/12482
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405817
| 166,359 |
https://mathoverflow.net/questions/405746
|
4
|
If $R$ is a commutative ring with identity with a 'nice' action of a finite group $G$, the subring $R^G\subset R$ gives a Galois extension of rings. In this case, S.U. Chase, D.K. Harrison, A. Rosenberg have shown that there exsits the following exact sequence (also knonwn as Chase-Harrison-Rosenberg exact sequence):
$0\rightarrow H^1 (G, R^\*)\rightarrow Pic(R^G)\rightarrow Pic(R)^G \rightarrow H^2(G, R^\*) \rightarrow Br(R/R^G)\rightarrow H^1(G,Pic(R))$ $(\*)$
where $Pic(R)$ denotes the Picard group of $R$, and $Br(R/R^G) := ker(Br(R^G)\rightarrow Br(G))$ comes from the map on Brower groups.
My question is the following: I have come across an exactly similar exact sequence in a paper, but it comes in the context of a finite Galois morphism $\pi: X\rightarrow Y$ of two smooth projective varieties ( see (3.5) of [This paper](https://arxiv.org/pdf/0904.4640.pdf)). There isn't much information in the paper on how it is obtained. I know that a finite Galois morphism of two varieties can be defined affine locally, in which case it essentially becomes Galois extension of rings, but it is not clear to me how to 'patch' the different local sequences $(\*)$ to get a similar exact sequence for schemes as well.
The Picard and Brauer groups are not sheaves in the etale topology, so it is not clear to me if the concept of patching even makes sense here.
Can anyone give me a reference, or let me know why a 'scheme'-version of $(\*)$ should also be true? Thanks in advance.
|
https://mathoverflow.net/users/90911
|
Regarding a 'global' version of Chase-Harrison-Rosenberg exact sequence for rings
|
I'm converting my comment to an answer. Let $\pi:X\to Y$ be a Galois étale cover, with Galois group $G$. One has a Hochschild-Serre spectral sequence
$$E\_2 = H^p(G, H^q(X\_{et},\mathbb{G}\_m))\Rightarrow H^{p+q}(Y\_{et}, \mathbb{G}\_m)$$
(The reference is given in "A User"'s comment.)
The associated exact sequence of low degree terms reads
$$0\to H^1(G, H^0(X, \mathbb{G}\_m))\to Pic(Y)\to Pic(X)^G \to H^2(G,H^0(X, \mathbb{G}\_m))\to Br(Y)$$
where I'm using $Br$, a bit inaccurately, for the 2nd étale cohomology with coefficients in $\mathbb{G}\_m$. In fact, the image should lie in $Gr^2Br(Y)$ wrt the filtration on the abuttment. And this seen to be $\ker[\ker Br(Y)\to Br(X)]\to H^1(G,Pic(X))$.
In the affine case, this is surely the same as the sequence of Chase et. al. you gave, but it would need to be checked.
|
3
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https://mathoverflow.net/users/4144
|
405818
| 166,360 |
https://mathoverflow.net/questions/401720
|
5
|
$\newcommand\P{\mathbb P} \newcommand\C{\mathcal C}$I am a bit confused by a proof I am reading on the fact that a projective algebraic space-curve (i.e. an algebraic curve in $\P^3(k)$, where $k$ is an algebraically closed field) is the intersection of 3 hypersurfaces. I am trying to constructively create the hypersurfaces given the vanishing ideal of the curve.
The proof I am reading is from an old [source](https://doi.org/10.1007/BF01236924) (written in 1960) by Martin Kneser. Since then, many generalizations of this fact exists (e.g. Storch, Forster, Eisenbud–Evans etc.). However, I would like to understand this proof by Kneser because it is quite constructive and supposedly the simplest of all.
The text is written in German. I just don't quite understand how the proof starts.
This is how Kneser starts the proof (I will do my best by adding my interpretation and translation to the original German text):
Let $\C$ be the curve and suppose $I=\langle f\_1,\dotsc,f\_n \rangle$ ($f\_i$ homogenous and nontrivial) is the vanishing ideal of the curve. And suppose that our coordinates are $(t:x:y:z)$ in $\P^3$. Without loss of generality, one may assume that the point $P:=(1:0:0:0)$ is in $\C$. Then Kneser continues by letting $f\in k[x,y,z]$ be a generator of the vanishing (principal) ideal of the "cone" of $\C$ with $P$ as its vertex.
**First question:**
Is a generator of the vanishing ideal of the cone the elimination ideal $I\cap k[x,y,z]$?
Kneser then continues by assuming that $g \in k[t,x,y,z]$ is (my interpretation) a homogenous polynomial among the generators $f\_1,\dotsc,f\_n$ with minimum degree (say $d$) in $t$. One may write
$$g = \sum\_{i=0}^d g\_i t^i$$
for (homogenous) polynomials $g\_i \in k[x,y,z]$ such that $g\_d\not\equiv 0$.
Then comes the most confusing part of the whole construction. Kneser states (without a proof) something like Lemma:
For any $p\in I$ with degree $m$ over $t$ one has the polynomial division (by $g$)
$$g\_d^m p = qg + r$$
where the remainder $r$ is divisible by $f$.
**Second question:** Why must $r$ be divisible by $f$?
From the above "Lemma" Kneser concludes that
$Z(f,g) = \C\cup V$ where $V$ is a subset of the vanishing of $(f,g\_d)$. He denotes the vanishing set $D:=Z(f,g\_d)$ and states that $D$ is the union of finitely many lines passing $P$.
I will not continue the proof here (unless someone needs to see all of it in my broken translation). I am already confused on why $V\subset Z(f,g\_d)$ and that $Z(f,g\_d)$ is the union of lines.
Can anyone who understands this better than me clarify my confusion? The title of the paper (which is only 2 pages) in the original German is:
[*Über die Darstellung algebraischer Raumkurven als Durchnitte von Flächen*](https://doi.org/10.1007/BF01236924),
Arch. Math. 11, 157–158 (1960).
|
https://mathoverflow.net/users/3949
|
Question on a constructive proof that space projective curves are the intersections of three hypersurfaces
|
$\newcommand\C{\mathcal C} \newcommand\spn[1]{\langle #1\rangle}$Using cohomology is interesting when dealing with higher dimensions and more general results but I don't see why it would result into a *shorter* proof. In fact the generalization of this result of Kneser (Eisenbud and Evans) avoids any sheaf cohomology as well and the result is just as short as this specific case.
Here is Kneser's algorithm with some explanation:
The initial assumption is that, without loss of generality, $P=(1:0:0:0)$ is on the curve. We use the same notation $(t:x:y:z)$ as Kneser for the coordinates. Let $\mathbb k$ be your (algebraically closed) ground field. One also assumes that $\C$ is not a line (otherwise the algorithm won't work, but a line is trivially a set-theoretic complete intersection).
1. There exists a $g \in I(\C)$ with the property that:
a.) $d > 0$ where $d=\deg\_t(g)$
b.) $g\_d$ is not in $I(\C)$ (where $g\_d\in \mathbb k[x,y,z]$ is the coefficient of $t^d$).
c.) $d$ is minimum with the property a.) and b.)
2. As you noted $I(\C)\cap \mathbb k[x,y,z]$ is principal, and is generated by say $f$. The vanishing of $f$ is therefore a cone with "vertex" at $P$ and "base" $\C$. Note that $Z(f,g\_d)$ is a finite union of lines because:
If $Q\in Z(f,g\_d)\backslash\{P\}$ then the line $PQ$ is on the surface $Z(g\_d)$ and on the cone $Z(f)$ which means that this line meets $\C$. We also know that $g\_d\notin I(\C)$ so the curve $\C$ cuts the surface $Z(g\_d)$ finitely many times.
3. Kneser then states that for any $p\in I(\C)$ we can find a $m\ge 1$ such that
$g\_d^m p \in \spn{f,g}$.
As pointed out in the [comments](https://mathoverflow.net/questions/401720/question-on-a-constructive-proof-of-space-projective-curve-are-the-intersection#comment1026831_401720) by Kapil, you can use degree (of $t$) argument to prove this.
4. Suppose now $g(Q)=0$ and $f(Q)=0$, then either $Q\in \C$ or $Q\notin \C$ and in this case there is a $p\in I(\C)$ such that $p(Q)\ne 0$. By 3. $Q\in Z(g\_d)$. So,
$$Z(f,g) \subset \C \cup Z(f,g\_d).$$
In words, $Z(f,g)$ is contained in the union of the curve $\C$ and finitely many lines whose common intersection is $P$. This already provides an immediate and nice proof that all irreducible projective space (algebraic) curves are intersections of an analytic surface and an algebraic surface, just perturb the cone (or the surface $Z(g)$) so that it intersect these lines only at $\C$.
5. We want to now find a third algebraic surface $Z(h')$ (the two previous surfaces are $Z(f)$ and $Z(g)$) that avoids these finite lines $Z(f,g\_d)$ except at $\C$. This can be done in several ways, Kneser himself provides an algorithm with a proof. I believe his algorithm is inefficient but didactically the easiest one to follow. I will not explain this part, but you can usually get away using a surface defined by products and sums of linear and quadratic forms avoiding the lines and having some higher multiplicity at $P$.
6. By Hilbert's Nullstellensatz there is therefore an $N\in \mathbb N$ and $h\in I(\C)$ such that
$$h'^N - h \in \spn{g\_d}.$$
You can obtain $N$ by just iterating through the powers of $h'$ and reducing via Groebner basis (ideal-membership test) of $\spn{g\_d}+I(\C)$ and you can obtain $h$ as a by-product. Therefore we choose $h$ and get $\sqrt{\spn{f,g,h}} = I(\C)$.
|
3
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https://mathoverflow.net/users/1245
|
405819
| 166,361 |
https://mathoverflow.net/questions/405834
|
2
|
Suppose $F\subset V\subsetneq \mathbb {R}$ where $F$ is closed and $V$ is open. I want to show that $\exists$ an open set $U\subset \mathbb {R}$ satisfying $F\subset U\subset \overline {U}\subset V$. My proof is below, but I think it's problematic since I never use the condition that $F$ is closed.
For each $x\in F$, pick $r\_{x}>0$ such that $(x-r\_{x},x+r\_{x})\subset V$. Then define $U=\bigcup\{(x-\frac {r\_{x}}{2},x+\frac {r\_{x}}{2})\}\subset V$. Then in fact, $F\subset U$. This is because countable union of open sets is open and $U$ is an open cover of $F$. It's trivial that $U\subset \overline{U}$. Now, what remains is to show that $\overline{U}\subset V$. First of all, $U\subset V$ since $U$ is just a collection of open intervals, and each one of them is in $V$. In fact, under this construction, $U\subsetneq V$. Therefore, for $y\in \overline{U}\setminus U$, $\exists u\in U$ such that $y\in B(u,r\_{u})$. Take $r\_{y}=r\_{u}-d(y,u)$. Then $B(y,r\_{y})$ is an open ball inside of $V$. Thus, $y\in V$ and $\overline{U}\subset V$.
|
https://mathoverflow.net/users/197849
|
Between an open set and its closed subset
|
$\newcommand{\R}{\mathbb{R}}\newcommand{\Z}{\mathbb{Z}}\newcommand{\de}{\delta}\newcommand{\ol}{\overline}$The property of $\R$ that you want to prove is that the topological space $\R$ is [normal](https://en.wikipedia.org/wiki/Normal_space). [All metric spaces are perfectly normal](https://en.wikipedia.org/wiki/Normal_space#Examples_of_normal_spaces) and hence normal. This proves your desired result.
That all metric spaces are normal immediately follows e.g. from Theorems 5.1.3 and 5.1.5 of the book by [Engelking](https://rads.stackoverflow.com/amzn/click/com/3885380064): Theorem 5.1.3 (due to Stone; see also the short proof by [M. E. Rudin](https://www.jstor.org/stable/2035708)) states that every metrizable space is [paracompact](https://en.wikipedia.org/wiki/Paracompact_space), and Theorem 5.1.5 (due to Dieudonné) states that every paracompact space is normal.
---
Here is an elementary, self-contained proof:
For each $n\in\Z$, the set
\begin{equation\*}
F\_n:=F\cap[n,n+1] \tag{1}
\end{equation\*}
is closed and bounded, and hence a compact subset of the open set $V$, so that there is some $\de\_n\in(0,1)$ such that
\begin{equation\*}
\ol{U\_n}\subseteq V, \tag{2}
\end{equation\*}
where
\begin{equation\*}
U\_n:=(F\_n)\_{\de\_n}, \tag{3}
\end{equation\*}
the open $\de\_n$-neighborhood of $F\_n$.
Let finally
\begin{equation\*}
U:=\bigcup\_{n\in Z}U\_n, \tag{4}
\end{equation\*}
so that $U$ is an open set containing $F$.
It remains to show that $\ol U\subseteq V$. Take any $x\in\ol U$, so that $x=\lim\_{k\to\infty}x\_k$ for some sequence $(x\_k)$ in $U$. Without loss of generality, for some natural $N$ and all natural $k$ we have $x\_k\in U\cap[N-1,N+1]$.
The conditions $F\_n\subseteq[n,n+1]$, (3), and $\de\_n\in(0,1)$ imply
\begin{equation\*}
\ol{U\_n}\subseteq[n-1,n+2].
\end{equation\*}
Hence, by (4), for all natural $k$ we have
\begin{equation\*}
x\_k\in\bigcup\_{n\colon\,[n-1,n+2]\cap[N-1,N+1]\ne\emptyset}U\_n\
\subseteq\bigcup\_{n=N-3}^{N+2}U\_n.
\end{equation\*}
So,
\begin{equation\*}
x\in\bigcup\_{n=N-3}^{N+2}\ol{U\_n}\subseteq V,
\end{equation\*}
by (2). Thus, indeed $\ol U\subseteq V$. $\quad\Box$
(The part $V\ne\R$ of your condition $V\subsetneq\R$ is not needed here; looking at this condition, it seems that you are using $\subset$ for $\subseteq$.)
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2
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https://mathoverflow.net/users/36721
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405840
| 166,366 |
https://mathoverflow.net/questions/405133
|
11
|
It is known that the Haefliger trefoil $S^3\hookrightarrow S^6$ is PL trivial but non-trivial smoothly. I wonder, where exactly does the problem come? Consider its tubular neighborhood $T\cong S^3\times D^3$. By the disc bundle theorem its closed complement $C$ is diffeomorphic to $S^2\times D^4$. Therefore, $S^6$ is presented as a union of $S^3\times D^3$ and $S^2\times D^4$ along some non-trivial diffeomorphism $\phi$ of $S^3\times S^2$. But it follows from the smoothing theory (Burghelea-Lashof) that for any closed 5-fold $M$, $\pi\_0 Diff(M)=\pi\_0 PL(M)$, because $\pi\_i PL\_5/O\_5 = \pi\_i PL/O=0$, $i\leq 6$. In our case, as example, since $S^3\times S^2$ is parallelizable, one has a fiber sequence
$$
Diff(S^3\times S^2) \to PL (S^3\times S^2) \to Map(S^3\times S^2,PL\_5/O\_5).
$$
Therefore, the non-trivial self-diffeomorphism of $S^2\times S^3$ is also non-trivial as PL self-homeomorphism. But why we then get a non-trivial knot in the smooth category but not in PL? I thought maybe the problem is that $\phi$ is PL concordant to the identity but is not so smoothly. But these sets of concordance classes are also the same $\pi\_0\widetilde{Diff}(S^3\times S^2)=
\pi\_0\widetilde{PL}(S^3\times S^2)$ thanks to the fiber sequence
$$
\widetilde{Diff}(S^3\times S^2) \to \widetilde{PL} (S^3\times S^2) \to Map(S^3\times S^2,PL/O).
$$
So, can the smooth non-triviality of the Haefliger trefoil be detected in this decomposition? I am puzzled because it also almost looks like a proof that the Haefliger trefoil is smoothly trivial.
|
https://mathoverflow.net/users/9800
|
Haefliger trefoil $S^3\hookrightarrow S^6$
|
The problem comes from the fact that $S^3\subset S^6$ admits more $PL$-framings than smooth ones: $\pi\_3 SO\_3=\mathbb{Z}$, while $\pi\_3 PL\_{6,3}=\mathbb{Z}^2$. (The group $PL\_{6,3}$ is the group of PL self homeomorphisms of $\mathbb{R}^6$ preserving $\mathbb{R}^3\subset\mathbb{R}^6$ pointwise.) Thus, when we PL isotop the Haefliger trefoil identically to the trivial knot, and we lift this isotopy to its tubular neighborhood $T=S^3\times D^3$, the obtained $PL$-framing of the trivial knot is not smoothable i.e., it does not come from an $SO\_3$-framing. In both smooth and PL categories the space of embeddings $T\hookrightarrow S^6$ that send $S^3\times \*$ identically to itself is homotopy equivalent to the space of such immersions, while the latter space by Smale-Hirsch-Haefliger-Poenaru is equivalent to $Map(S^3,SO\_3)$ and $Map(S^3,PL\_{6,3})$, respectively.
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5
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https://mathoverflow.net/users/9800
|
405843
| 166,369 |
https://mathoverflow.net/questions/405829
|
2
|
This question is related to a [question](https://mathoverflow.net/questions/405567/is-this-internalization-of-a-bijection-between-a-set-and-its-powerset-possible?) lately posted to $\cal MO$. Here, we add two partial unary functions $``j,f"$ to the language of $\sf ZF$.
The question is about if we can add the following on top of axioms of $\sf ZF$ [$j,f$ not used in Replacement nor Separation]?
$\exists \alpha: \text{ limit} (\alpha) \land j: V\_\alpha \to V\_{\alpha +1} \land f: V\_\alpha \to V\_\alpha \land j,f \text{ are bijections } \\ \forall S: j[S]; j^{-1} [S] \text { both exist } \\ \forall S \in V\_\alpha: j(S)=f[S] $
Where: $g[S]=\{g(x) \mid x \in S\}$
The first two conditions have already been [proved](https://mathoverflow.net/questions/405567/is-this-internalization-of-a-bijection-between-a-set-and-its-powerset-possible?#comment1039997_405567) consistent, it's the addition of the last condtion that is unsolved?
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https://mathoverflow.net/users/95347
|
Can we internalize a bijection between a set and its powerset in this way?
|
The above theory is inconsistent. Since j is surjective, there is an x∈ with jx=. Then f[x]=. Therefore x=, since f is a bijection. But is not an element of .
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5
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https://mathoverflow.net/users/133981
|
405851
| 166,374 |
https://mathoverflow.net/questions/405830
|
4
|
Are there any cases of finite subgroups of $O\_n(\mathbb{Q})$ ~~not contained in~~ *not isomorphic to any subgroup of* $O\_n(\mathbb{Z})$?
|
https://mathoverflow.net/users/172799
|
Finite subgroups of $O_n(\mathbb{Z})$ versus $O_n(\mathbb{Q})$
|
Let $n=m^2$ be a square. Then the vector $(1,\dots,1)\in\mathbf{Q}^n$ has norm $m$, as does $(0,\dots 0,m)$. Hence by Witt's theorem there exists an element $u$ of $\mathrm{O}\_n(\mathbf{Q})$ mapping $(0,\dots,0,m)$ to $(1,\dots,1)$. Hence $u$ maps orthogonals to orthogonals: it maps the hyperplane $\mathbf{Q}^{n-1}\times\{0\}$ of equation $x\_n=0$ onto the hyperplane of equation $\sum x\_i=0$. Let $S\_n$ be the subgroup of $\mathrm{O}\_n(\mathbf{Q})$ permutating coordinates. Then $S\_n$ fixes $(1,\dots,1)$ and acts faithfully on its orthogonal. Therefore $u^{-1}S\_nu$ acts faithfully on the orthogonal $\mathbf{Q}^{n-1}$.
Hence, for every square $n$, the group $\mathrm{O}\_{n-1}(\mathbf{Q})$ contains a copy of the symmetric group $S\_n$. But for every $n\ge 5$ there is no nontrivial homomorphism $S\_n$ to the Weyl group $S\_{n-1}\ltimes C\_2^{n-1}$ of type $B\_{n-1}$, which is isomorphic to $\mathrm{O}\_{n-1}(\mathbf{Z})$.
This proves that for every square $n\ge 9$, there is a finite subgroup of $\mathrm{O}\_{n-1}(\mathbf{Q})$, isomorphic to $S\_n$, not embedding into $\mathrm{O}\_{n-1}(\mathbf{Z})$.
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4
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https://mathoverflow.net/users/14094
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405854
| 166,376 |
https://mathoverflow.net/questions/404378
|
0
|
Let $Q=(I,\Omega)$ be the $D\_4$ affine quiver. We choose as dimension vector $(2,1,1,1,1)$ (where $2$ is on the central vertex). As this dimension vector is indivisible, we can choose a generic $\theta \in \mathbb{Z}^I$. We have the usual moment map $$\mu: R(\overline{Q},v) \to \mathbb{C}^I $$ from representations of the doubled quiver to $\mathbb{C}^I$.
We have different quiver varieties associated to these data (I do not know whether the notation is standard): $$\mathfrak{M}\_{0,\theta}(v)=\mu^{-1}(0)//\_{\theta}G(v)$$ and $$\mathfrak{M}\_{\theta,\theta}(v)=\mu^{-1}(\theta)/G(v) $$
A computation shows that these varieties are actually of dimension $2$: is there an explicit geometric presentation of such objects? If we choose $\theta$ in an appropriate way, we can think of $\mathfrak{M}\_{0,\theta}(v),\mathfrak{M}\_{\theta,\theta}(v)$ as certain moduli spaces of parabolic Higgs bundles/parabolic connections over trivial vector bundle over $\mathbb{P}^1$.
I know that in this case we have a concrete description of the full moduli space (the so-called Hausel toy model). Also the character variety side is well understood as affine Del Pezzo surfaces. What about quiver side?
|
https://mathoverflow.net/users/146464
|
Quiver varieties associated to D_4
|
It is [Kronhimer's result](https://projecteuclid.org/journals/journal-of-differential-geometry/volume-29/issue-3/The-construction-of-ALE-spaces-as-hyper-K%C3%A4hler-quotients/10.4310/jdg/1214443066.full) that $\mathfrak M\_{\zeta\_{\mathbb R},\zeta\_{\mathbb C}}(\mathbf v)$ is $\mathbb C^2/\Gamma$ ($\zeta\_{\mathbb R}=\zeta\_{\mathbb C}=0$), its deformation ($\zeta\_{\mathbb R}=0$), and the minimal resolution of the deformation (in general). Here $\Gamma$ is the binary dihedral group of type D4.
|
2
|
https://mathoverflow.net/users/3837
|
405865
| 166,380 |
https://mathoverflow.net/questions/405862
|
3
|
Let $A$ be a finite semisimple algebra over $\mathbb{k}$, a **perfect** field. Is true that the second Hochschild cohomology group vanishes, i.e. $$HH^2(A) = 0?$$
In order to make this question a little bit more complete, it would also be interesting to discuss whether $$HH^{\*>0}(A) = 0?$$
|
https://mathoverflow.net/users/105094
|
Hochschild cohomology of finite semisimple algebras
|
By corollary 18 of <https://projecteuclid.org/journals/nagoya-mathematical-journal/volume-12/issue-none/On-the-dimension-of-modules-and-algebras-VIII-Dimension-of/nmj/1118799929.full>
we have (using the perfect condition which ensures the seperability condition needed for corollary 18 to hold) that the enveloping algebra $A^e=A \otimes\_k A^{op}$ is again semisimple (which is equivalent to having global dimension 0).
Thus $Ext\_{A^e}^i(M,N)=0$ for all $A^e$-modules $M,N$ and $i>0$.
Especially: All Hochschild cohomology spaces $Ext\_{A^e}^i(A,A)$ vanish in positive degree.
|
4
|
https://mathoverflow.net/users/61949
|
405870
| 166,381 |
https://mathoverflow.net/questions/405793
|
5
|
**Question:**
Consider the set of continuous, differentiable a.e. functions from $\mathbb R \to \mathbb R$. Can we characterise the subset of these that satisfy $f’(x) = f(x)$ for almost every $x \in \mathbb R$?
*Remarks:*
*1) The problem can be thought of as a weakening of the defining ODE for the exponential function - $f’(x) = f(x)$ everywhere, which is solvable only by piecewise combinations of $Ce^x$ and the zero function.*
*2) If $f$ is a solution, then so is $f + g$ for any $g$ continuous and differentiable a.e. with $g’ = 0$ a.e.*
*3) The full measure subset on which $f’ = f$ can in general be smaller than the full measure set on which $f’$ is defined.*
|
https://mathoverflow.net/users/173490
|
Which continuous, differentiable a.e. functions have $f’(x) = f(x)$ a.e.?
|
Let $g(x) = e^{-x} f(x)$, so that $f(x) = e^x g(x)$. For a given $x$, $f'(x)$ exists if and only if $g'(x)$ exists, and $g'(x) = e^{-x} (f'(x) - f(x))$. In particular, $f'(x) = f(x)$ if and only if $g'(x) = 0$.
It follows that $f$ is necessarily of the form $f(x) = e^x g(x)$, where $g$ satisfies $g'(x) = 0$ almost everywhere.
|
11
|
https://mathoverflow.net/users/108637
|
405880
| 166,385 |
https://mathoverflow.net/questions/403249
|
3
|
Let $Q$ be a finite quiver. As far as I know there's a great amount of work concerning the so-called quiver varieties one can associate to it. Loosely speaking, these are obtained by taking GIT quotients of semistable representations of the (deformed) preprojective algebra one get from $Q$.
These varieties turn out to have a wide range of applications to representation theory and algebraic geometry.
Is there any work going on pursuing the "stacky" approach? Like taking already the stack of semistable representations or maybe even the full stack of representations of the preprojective algebra?
|
https://mathoverflow.net/users/146464
|
Moduli stack of quiver representations
|
There is a differential graded version of the preprojective algebra, called *Ginzburg dg-algebra*, that is a special case of the Calabi-Yau completion introduced by [Keller](https://arxiv.org/abs/0908.3499).
The derived moduli stack of object (as defined by [Toen-Vaquié](https://arxiv.org/abs/math/0503269)) of a Calabi-Yau completion becomes a ($0$-shifted) symplectic derived stack. This is more generally true for the moduli of object of a finite type $2$-Calabi-Yau dg-category, as shown by [Brav-Dyckerhoff](https://arxiv.org/abs/1812.11913), and also [Toen](https://arxiv.org/abs/1401.1044) in a little less general situation.
These ideas have been used in a recent [paper](https://arxiv.org/pdf/2006.01069.pdf) of mine with Bozec and Scherotzke to introduce and study (apparently new) lagrangian subvarieties in the Hilbert scheme of points in the plane.
There are also [two](https://arxiv.org/abs/1612.06352) [papers](https://arxiv.org/abs/1802.05398) of Yeung on related topics, that are very interesting.
|
2
|
https://mathoverflow.net/users/7031
|
405888
| 166,388 |
https://mathoverflow.net/questions/405869
|
1
|
I am studying von Neumann algebras. In the wiki article [abelian von Neumann algebras](https://en.wikipedia.org/wiki/Abelian_von_Neumann_algebra), it mentions that every abelian von Neumann algebras acting on a separable Hilbert space is \*-isomorphic to $L^{\infty}[0,1]$, $l^{\infty}(\mathbb{N})$ or their direct sum. I am wondering why $[0,1]$ is special in this case, why not other measure space such as $S^{1}$ or $[0,1]^{2}$? Also, if we consider $L^{\infty}([0,1]^{2})$ acting on $L^{2}([0,1]^{2})$, isn't that a von Neumann algebra acting on a separable Hilbert space? Does that mean that $L^{\infty}[0,1]$ and $L^{\infty}([0,1]^{2})$ are \*-isomorphic?
|
https://mathoverflow.net/users/172458
|
Questions about Maharam's classification theorem
|
The spaces $[0,1]$, $[0,1]^2$, and $S^1$ are all isomorphic as measurable spaces, including their sets of measure 0, as required by the [Gelfand-type duality for measurable spaces](https://mathoverflow.net/questions/23408/reference-for-the-gelfand-duality-theorem-for-commutative-von-neumann-algebras/360088#360088).
For instance, the isomorphism $[0,1]→S^1$ is given by identifying $S^1=[0,1]/(0∼1)$. Since the points $0$ and $1$ have measure 0,
and maps that differ on a set of measure 0 are identified,
the above map is indeed an isomorphism.
Likewise, the isomorphism $[0,1]^2→[0,1]$ is given by identifying
$[0,1]$ with $\{0,1\}^N$ (e.g., using binary expansions), where $N$ is a countable set,
and then taking the isomorphism $[0,1]^2≅\{0,1\}^{N⊔N}≅\{0,1\}^N≅[0,1]$,
where $N⊔N$ is isomorphic to $N$ since both are countable sets.
Indeed, the very point of Maharam's theorem is that there are very
few measurable spaces up to an isomorphism:
any measurable space is isomorphic to the disjoint union
of measurable spaces of the form $\{0,1\}^S$, where $S$ is a set
of arbitrary cardinality, and this gives a complete classification
of measurable spaces up to isomorphism: simply count the (infinite) number of summands for each possible cardinality of $S$, and also count the number of isolated points.
|
5
|
https://mathoverflow.net/users/402
|
405892
| 166,389 |
https://mathoverflow.net/questions/405895
|
2
|
This question is related to: <https://math.stackexchange.com/q/4270522/168758>
---
Let $H\_n(x) \in \mathbb R[x]$ be the probabilist's $n$th Hermite polynomial. This an $n$th degree polynomial given by the following equivalent formulae (which ever helps)
$$
\begin{split}
H\_n(x) &= n!\sum\_{k=0}^{\lfloor n/2\rfloor}\frac{(-1)^k}{2^kk!(n-2k)!}x^{n-2k}\\
H\_n(x) &= \frac{1}{\sqrt{2\pi}}\int\_{-\infty}^\infty (x+iy)^n e^{-y^2/2}dy\\
H\_n(x) &= e^{-D^2/2}x^n,
\end{split}
$$
where $D^2$ is the second-derivative-w.r.t-$x$ differential operator $\dfrac{d^2}{dx^2}$, and $e^{-D^2/2}$ should be seen as a power series in $D^2$.
Let $d$ be a large positive integer, $a$ and $b$ be fixed vectors on the unit $(d-1)$-dimensional sphere $S\_{d-1}$, and $X$ be uniformly distributed on $S\_{d-1}$. For fixed nonnegative integers $n$ and $m$, define
$$
s\_{n,m} = \mathbb E[H\_n(X^\top a)H\_m(X^\top b)].
$$
Due to rotational-invarfiance of $X$, it is clear that $s\_{n,m}$ is a polynomial in $t:=a^\top b$. Let $c\_{n,m,k}$ be the coefficient of $t^k$ in $s\_{n,m}$.
>
> **Question.**
> For $k \ge 1$, what is a good Big-O upper-bound for $c\_{n,m,k}$ in the limit $d \to \infty$.
>
>
>
|
https://mathoverflow.net/users/78539
|
Integral of product of Hermite polynomials w.r.t marginal distribution of first two-coordinate of random vector on unit-sphere
|
To find the dependence of $s\_{nm}$ on $t=a\cdot b$, we take $a=(t,\sqrt{1-t^2},0,0,\ldots 0)$, $b=(1,0,0,0,\ldots 0)$, so that
$$s\_{nm} = \mathbb E[H\_n(X^\top a)H\_m(X^\top b)]=\mathbb E[H\_n(X\_1 t+X\_2\sqrt{1-t^2})H\_m(X\_1)].$$
The marginal distribution $P(X\_1,X\_2)$ of two elements from a vector that is uniformly distributed on the $d$-dimensional unit sphere is given by (see, for example, this [calculation](https://stats.stackexchange.com/a/520811))
$$P(X\_1,X\_2)=\frac{d-2}{2\pi}(1-X\_1^2-X\_2^2)^{d/2-2},\;\;X\_1^2+X\_2^2<1,\;\;d\geq 3.$$
Hence we have for $s\_{nm}$ the integral expression
$$s\_{nm}=\frac{d-2}{2\pi}\int\_{0}^{1}rdr\int\_0^{2\pi}d\phi\, (1-r^2)^{d/2-2}H\_n\left(rt\cos\phi+r\sqrt{1-t^2}\sin\phi\right)H\_m(r\cos\phi).$$
For large $d$ the Hermite polynomials can be expanded around $r=0$, which gives
$$s\_{nm}\approx \frac{\pi}{d} 2^{\frac{1}{2} (m+n-2)} \left(\frac{4 t}{\Gamma \left(-\frac{m}{2}\right) \Gamma \left(-\frac{n}{2}\right)}-\frac{-2 d+m+n}{\Gamma \left(\frac{1}{2}-\frac{m}{2}\right) \Gamma \left(\frac{1}{2}-\frac{n}{2}\right)}\right),\;\;d\gg 1.$$
|
2
|
https://mathoverflow.net/users/11260
|
405901
| 166,390 |
https://mathoverflow.net/questions/405751
|
5
|
Lets take a closed four manifold $M:=\Sigma\_1\times \Sigma\_2,$ where $\Sigma\_i$s are compact Riemann surfaces. Now if $V$ and $W$ are Spin$^\mathbb{C}$ bundles on $\Sigma\_1$ and $\Sigma\_2$ respectively, then one can define Spin$^\mathbb{C}$ bundle on $M$ with positive Spin$^\mathbb{C}$ bundle defined as
\begin{equation\*}
V\_+\otimes W\_+\oplus V\_-\otimes W\_-
\end{equation\*}
and negative Spin$^\mathbb{C}$ bundle defined as
\begin{equation\*}
V\_+\otimes W\_-\oplus V\_-\otimes W\_+
\end{equation\*}
In general the quadratic term in Seiberg Witten equation acts like this:
For positive spinors $\phi,\psi,q(\phi)(\psi):=\langle\psi,\phi\rangle\phi-\frac{|\phi|^2}{2}\psi$.
In the above situation for $\phi\in V\_+\otimes W\_+,$ I can understand $q(\phi)$ acting on spinors in $V\_+\otimes W\_+$, but as the spaces are perpendicular, I am confused about the action of $q(\phi)$ on spinors in $V\_-\otimes W\_-$. Is it the trivial action ($0$)? Can we say something about what kind of self dual two form it represents?
|
https://mathoverflow.net/users/131004
|
Understanding the quadratic part in Seiberg Witten equation
|
So your $\phi$ in this special case really means $(\phi,0)$ in the direct sum, while $\psi$ means $(0,\psi)$. Then $q(\phi)$ is the 2x2 matrix with vanishing off-diagonal entries and nontrivial diagonal entries ($\frac12|\phi|^2,-\frac12|\phi|^2)$, which is traceless. Particular values are $q(\phi)\phi=\frac12|\phi|^2\phi$ and $q(\phi)\psi=-\frac12|\phi|^2\psi$. (But if we looked at the full endomorphism $\phi\otimes\phi^\*$ then it would vanish on $\psi$.) Now just compute the inverse (or adjoint) of the Clifford multiplication map to get your induced 2-forms.
|
0
|
https://mathoverflow.net/users/12310
|
405903
| 166,391 |
https://mathoverflow.net/questions/405896
|
0
|
I really want to prove the statement in the title but I'm struggling with it. Here my current state:
Proof via contradiction. Let $\mathcal{H}$ be a RKHS with two reproducing kernels $k$ and $\hat{k}$ and let $x \in \mathcal{H}$. Then:
\begin{align}
\|{k\_x - \hat{k}\_x}\|^2 &= \langle k\_x - \hat{k}\_x, k\_x - \hat{k}\_x \rangle \\
&= \langle k\_x - \hat{k}\_x , k\_x \rangle - \langle k\_x - \hat{k}\_x , \hat{k}\_x \rangle \\
&= \color{orange}{\langle k\_x, k\_x \rangle + \langle \hat{k}\_x, \hat{k}\_x \rangle} -
\color{blue}{\langle \hat{k}\_x, k\_x \rangle - \langle k\_x, \hat{k}\_x \rangle}\\
&= ~... \\
&= \color{orange}{k(x,x) - \hat{k}(x,x)} - \color{blue}{k(x,x) + \hat{k}(x,x)} \\
&= 0.
\end{align}
And this would be a contradiction since $\|x-y\| = 0 \Longleftrightarrow x = y$.
So the orange terms look fine but I don't know how to get the blue terms from the third to the fifth line. Please help.
Cheers. :-)
|
https://mathoverflow.net/users/407441
|
Proof: If a reproducing kernel exists for a Hilbert space, then it is unique
|
$\newcommand\tk{\tilde k}\newcommand\ip[2]{\langle #1,#2\rangle}$Let $k$ be a reproducing kernel of a [reproducing kernel Hilbert space (RKHS)](https://en.wikipedia.org/wiki/Reproducing_kernel_Hilbert_space#Definition) $H:=\mathcal H$ of real-valued functions on a set $X$. Then
$$\ip f{k\_x}=f(x)\tag{1}$$ and
$$k(x,y)=\ip{k\_x}{k\_y}=k\_x(y)\tag{2}$$
for all $f\in H$ and all $x$ and $y$ in $X$, where $\ip\cdot\cdot$ is the inner product on $H$.
Let now $\tk$ be another reproducing kernel of $H$. Then, by (1) and (2), for all $x\in X$
$$\begin{aligned}
\|k\_x-\tk\_x\|^2& =\ip{k\_x}{k\_x}+\ip{\tk\_x}{\tk\_x}-\ip{k\_x}{\tk\_x}-\ip{\tk\_x}{k\_x} \\
& =k(x,x)+\tk(x,x)-k\_x(x)-\tk\_x(x) \\
& =k(x,x)+\tk(x,x)-k(x,x)-\tk(x,x)=0,
\end{aligned}$$
whence $k\_x=\tk\_x$ for all $x$, that is, $k=\tk$.
(You mismatched colors.)
|
3
|
https://mathoverflow.net/users/36721
|
405905
| 166,392 |
https://mathoverflow.net/questions/405907
|
0
|
$$\ \int\_0^{\pi} \bigl(\sin(x)\bigr)^{2n-2k+1} e^{a\cos(x)} dx , \qquad a,n,k\in\mathbb Z.$$
I tried to solve this integral by parts, but I didn't get any result. I look forward to your experience.
|
https://mathoverflow.net/users/380650
|
Integration of the product of sin and exponential with power
|
$$\int\_0^{\pi} (\sin x)^{2n-2k+1} e^{a\cos x} dx =\int\_{-1}^1 (1-\xi^2)^{n-k}e^{a\xi}\,d\xi$$
$$\qquad\qquad=\sqrt{\pi }\, 2^{-k+n+\frac{1}{2}} a^{\frac{1}{2} (2 k-2 n-1)} \Gamma (-k+n+1) I\_{\frac{1}{2} (-2 k+2 n+1)}(a),\;\;n-k+1>0,$$
with $I\_\alpha(x)$ the modified Bessel function of the first kind.
|
2
|
https://mathoverflow.net/users/11260
|
405909
| 166,395 |
https://mathoverflow.net/questions/405791
|
3
|
In Constructive Set Theory (CZF) the Power Set axiom is replaced with the Subset Collection axiom which I will state here:
$$ \exists c \forall u [\forall x \in a \exists y \in b (\psi(x,y,u)) \longrightarrow \exists d \in c (\forall x \in a \exists y \in d (\psi(x,y,u)) \land \forall y \in d \exists x \in a (\psi(x,y,u)))] $$
Typical treatments of CZF ([Aczel and Rathjen - CST Book draft](https://www1.maths.leeds.ac.uk/%7Erathjen/book.pdf)) will introduce the notion of fullness to aid in the comprehension of this notably dense axiom. Which is perfectly reasonable. But still I want to get a better feel for what this axiom is saying and what can be done with it in place of the Power Set axiom (for my own research). Is this misguided?
Before discussing about my difficulty in understanding this axiom I will recall the power set axiom.
$$ \forall a \exists y \forall x [x \in y \longleftrightarrow x \subseteq a] $$
where $a \subseteq b$ is shorthand for $\forall z(z \in a \to z \in b)$. This axiom is easy enough to interpret. For any set $a$ there exists a set $y$ (which we call the powerset of $a$) such that for any $x$, $x$ is in the powerset of $a$ if and only if $x$ is a subset of $a$.
The Subset Collection axiom on the other hand asserts the existence of a set $c$ which I assume to be some sort of collection of subsets, before referencing any set for which we wish to take subsets from. Am I thinking about this axiom the wrong way? How would one interpret the variables here?
|
https://mathoverflow.net/users/312621
|
Subset Collection axiom
|
Subset collection is typically used as a stronger version of exponentiation, rather than a weaker version of power set, so that is why fullness is usually the best way of understanding it.
If you want to think of it as a special case of power set, then what it is saying is that given sets $a$ and $b$ it asserts the existence of a collection of subsets of $b$ that contains something like the image of every multivalued function from $a$ to $b$. For example, if we had the powerset axiom, we could take $c$ to be the powerset of $b$.
We can see it as a stronger version of exponentiation as follows. Note that it is possible to define the exponentiation axiom as saying that there is a set $c$ that contains the image of every function from $a$ to $b$. It is clear this follows from exponentiation and replacement and for the converse, note that a function from $a$ to $b$ viewed as a subset of $a \times b$, is the image of a function from $a$ to $a \times b$. However, a priori we cannot quantify over the set of all functions from $a$ to $b$ (because we don't know it exists yet) so instead we use functional relations, i.e. formulas $\psi$ such that $\forall x \in a\,\exists ! y \in b\,\psi(x, y)$. For subset collection, we drop the uniqueness requirement and only require $\forall x \in a\, \exists y \in b\,\psi(x, y)$, and instead of $d$ being the image, we only require that $d$ contains at least one witness $y \in b$ for each $x \in a$.
As Hanul Jeon mentioned, a famous use of subset collection is the construction of the Dedekind reals, since each Dedekind cut $(L, R)$ corresponds to a multivalued relation $R$ from $\{(p, q) \in \mathbb{Q} \times \mathbb{Q} \;|\; p < q\}$ to $2$, where $R((p, q), 0)$ indicates $p \in L$ and $R((p, q), 1)$ indicates $q \in R$. I think there's another example where it's used in Gambino, [*Heyting-valued interpretations for Constructive Set Theory*](https://www.sciencedirect.com/science/article/pii/S0168007205000709), where it's used to show subset collection holds in a Heyting valued model, but if I recall correctly it would still be necessary to assume in the metatheory even if you only want exponentiation in the model. In general it does come up sometimes when you want a collection of sets that each contain enough witnesses to satisfy a certain property, but you have no way to choose a particular witness.
|
3
|
https://mathoverflow.net/users/30790
|
405912
| 166,396 |
https://mathoverflow.net/questions/405921
|
4
|
Start with [this triangle (OEIS A118981)](https://oeis.org/A118981). This triangle is simple to generate with the following recurrence relation (though $T(0,0)$ ends up different from the OEIS version):
$$
T(0,0) = 2;T(1,0) = 1;T(1,1) = 1;
$$
$$T(n,k) = T(n-1, k) + T(n-1, k-1) + T(n-2, k-2), n>1 $$
The sequence in question is generated by taking the largest element from each row, then dividing it by the row number $n$. The resulting sequence is as follows (listed up to $a\_{20}$):
$1, 1.5, 2, 3, 5, 10, 19, 37, 77.\overline 7, 158, 331, 715, 1505, 3287, 7224.4, 15529, 34839, 77453.8 \overline3, 168950, 386050, ...$
My question is: **Why are the majority of these numbers [convoluted convolved Fibonacci numbers?](https://arxiv.org/pdf/math/0311205.pdf)** Even the ones which are not integers become convoluted convolved Fibonacci numbers when rounded down to the nearest integer. Some examples:
$$floor(a\_{15}) = 7224 = G\_{11}^{(5)}$$
$$a\_{16} = 15529 = G\_{12}^{(5)}$$
$$a\_{17} = 34839 = G\_{12}^{(6)}$$
$$a\_{19} = 168950 = G\_{13}^{(7)}$$
$$a\_{20} = 386050 = G\_{14}^{(7)}$$
$$floor(a\_{21}) = 865264 = G\_{15}^{(7)}$$
|
https://mathoverflow.net/users/174962
|
Why do convoluted convolved Fibonacci numbers pop up from this triangle?
|
We have $$T(n,k) = [x^ny^{n-k}]\ \frac{2 - (1+y)x}{1-(1+y)x-x^2}=[y^{n-k}]\ L\_n(1+y),$$
where $L\_n$ is the $n$-th [Lucas polynomial](https://en.wikipedia.org/wiki/Fibonacci_polynomials).
For $k<n$, we have an explicit formula:
\begin{split}
T(n,k) &= \sum\_{i=0}^{\lfloor n/2\rfloor} \frac{n}{n-i}\binom{n-i}i \binom{n-2i}{n-k}\\
&= \frac{n}{n-k}\sum\_{i=0}^k \binom{n-i-1}{k-i} \binom{k-i}i\\
&=\frac{n}{n-k} F\_{k+1}^{(n-k)}.
\end{split}
So, relation to convoluted Fibonacci numbers should come at no surprise.
In the given examples for $G\_{j+1}^{(r)}$ with $3\nmid n$, we have $\gcd(r,j)=1$, implying that $G\_{j+1}^{(r)} = \frac1r F\_{j+1}^{(r)}$.
Numerical evidence suggests that for a fixed $n$, $T(n,k)$ is unimodal and achieves its maximum at $k=n-\left\lfloor \frac{n}3\right\rfloor$, suggesting that
$$\frac1n\max\_k T(n,k) = \frac1{\lfloor n/3\rfloor} F\_{n-\lfloor n/3\rfloor + 1}^{(\lfloor n/3\rfloor)} \stackrel{3\nmid n}{=} G\_{n-\lfloor n/3\rfloor + 1}^{(\lfloor n/3\rfloor)}.$$
The case $3\mid n$ can be addressed similarly.
|
4
|
https://mathoverflow.net/users/7076
|
405928
| 166,401 |
https://mathoverflow.net/questions/405929
|
3
|
Given an arbitrary set $X$, let $\beta X$ be the set of all ultrafilters over $X$. Consider endowing $\beta X$ with a topology consisting of the following open sets:
$$
\{\mathcal{U} \in \beta X : A \in \mathcal{U}\}
$$
where $A$ ranges over subsets of $X$. Now let $\kappa < |X|$ be an infinite cardinal, and assume there exists a $\kappa$-complete ultrafilter over $X$. Let
$$
\lambda X := \{\mathcal{U} \in \beta X : \mathcal{U} \text{ is $\kappa$-complete}\}
$$
My question is (working in $\mathsf{ZFC}$, if that matters):
>
> Is $\lambda X$ a closed subset of $\beta X$ under the topology above?
>
>
>
If this question is false for general $\kappa$, I would also like to know what are the assumptions I should impose for this assertion to be true. I welcome any large cardinal axioms, if necessary.
|
https://mathoverflow.net/users/146831
|
Is the set of $\kappa$-complete ultrafilters closed in $\beta X$?
|
If $\kappa=\aleph\_0$ then yes: every ultrafilter is $\aleph\_0$-complete.
If $\kappa>\aleph\_0$ then no, if $\lambda X$ is nonempty. Split $X$ into countably many sets $\{X\_n:n\in\mathbb{N}\}$, of the same cardinality as $X$ itself.
Take a $\kappa$-complete $u\_n$ on $X\_n$ for each $n$.
For every free ultrafilter $p$ on $\mathbb{N}$ the $p$-limit $u\_p$ of the sequence $\langle u\_n\rangle\_n$ is in the closure of $\lambda X$, but not in $\lambda X$, because $\bigcup\_{k\ge n}X\_k$ is in $u\_p$ for all $n$, yet $\bigcap\_n\bigcup\_{k\ge n}X\_k=\emptyset$.
|
9
|
https://mathoverflow.net/users/5903
|
405932
| 166,403 |
https://mathoverflow.net/questions/405943
|
5
|
I am trying to get a hold on a copy of a 1988 paper by M. Naimi. It appears in MR as [MR0950949](https://mathscinet.ams.org/mathscinet-getitem?mr=0950949) and in zbMATH as [Zbl 0669.10066](https://www.zbmath.org/?q=an%3A0669.10066). Its title is "Les entiers sans facteurs carré $\le x$ dont leurs facteurs premiers $\le y$". It was published in Publications Mathématiques d'Orsay, vol. 88, as part of a book it seems.
On the [Publ. Math. Orsay website](http://sites.mathdoc.fr/PMO/) it seems there is no information on the paper or volume.
Any help on getting a copy would be appreciated. I have tried contacting the author directly but that didn't help.
|
https://mathoverflow.net/users/31469
|
Copy of paper by M. Naimi on squarefree friable integers
|
Simply trying to put [Naimi "Les entiers sans facteurs carré"](https://www.google.com/search?q=Naimi+%22Les%20entiers%20sans%20facteurs%20carr%C3%A9%22) into Google leads to a [PDF file](https://www.imo.universite-paris-saclay.fr/%7Ebiblio/numerisation/docs/39_SEMINAIRE/pdf/39_SEMINAIRE.pdf) which contains the corresponding volume of *Publications Mathématiques d'Orsay* - including this article.
It seems that some other volumes are available on the same site - I have tried a few similar links: [38](https://www.imo.universite-paris-saclay.fr/%7Ebiblio/numerisation/docs/38_SEMINAIRE/pdf/38_SEMINAIRE.pdf), [37](https://www.imo.universite-paris-saclay.fr/%7Ebiblio/numerisation/docs/37_SEMINAIRE/pdf/37_SEMINAIRE.pdf), [26](https://www.imo.universite-paris-saclay.fr/%7Ebiblio/numerisation/docs/26_SEMINAIRE/pdf/26_SEMINAIRE.pdf), [15](https://www.imo.universite-paris-saclay.fr/%7Ebiblio/numerisation/docs/15_SEMINAIRE/pdf/15_SEMINAIRE.pdf), [04](https://www.imo.universite-paris-saclay.fr/%7Ebiblio/numerisation/docs/04_SEMINAIRE/pdf/04_SEMINAIRE.pdf). However, I failed to find a page which contains some kind of archive linking to those pdf-s. (Still, Google indexed them, so the crawler must have discovered them somewhere.)
|
6
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https://mathoverflow.net/users/8250
|
405956
| 166,409 |
https://mathoverflow.net/questions/405968
|
2
|
Let ${\cal P}(\omega)/({\rm fin})$ be the quotient of the Boolean algebra ${\cal P}(\omega)$ where two sets are considered to be equivalent if they differ by a finite number of elements.
It turns out that ${\cal P}(\omega)/({\rm fin})$ is an atomless Boolean algebra. Is the Dedekind-MacNeille completion isomorphic to ${\cal P}(\omega)$?
|
https://mathoverflow.net/users/8628
|
Dedekind-MacNeille completion of ${\cal P}(\omega)/({\rm fin})$
|
No. $\overline{P(\omega)/\text{fin}}$ is not isomorphic to $P(X)$ for any set $X$ because $P(\omega)/\text{fin}$ is atomless.
We shall write $\overline{B}$ for the Dedekind-MacNeille completion of a Boolean algebra $B$, and we shall write $B^{+}$ for $B\setminus\{0\}$. A minimal element in $B^{+}$ is said to be an atom of the Boolean algebra $B$. A Boolean algebra $B$ is said to be atomic if for each $b\in B^{+}$, there is an $a\in B^{+}$ with $a\leq b$.
Proposition: If $B$ is a complete Boolean algebra, then the following are equivalent.
1. $B$ is atomic.
2. $B\simeq P(X)$ for some set $X$.
3. $B$ satisfies the complete distributivity identity
$$\bigwedge\_{i\in I}\bigvee\_{j\in A\_{i}}a\_{i,j}=\bigvee\_{f}\bigwedge\_{i\in I}a\_{i,f(i)}$$ where $f$ ranges over the collection of all choice functions $f:I\rightarrow\bigcup\_{i\in I}A\_{i}$ with $f(i)\in A\_{i}$ for $i\in I$.
To prove $1\rightarrow 2$, let $p$ be the set of all atoms of $B$. Then $p$ is a partition of the Boolean algebra $B$. The mapping $\phi:P(p)\rightarrow B$ defined by letting $\phi(R)=\bigvee R$ is your required Boolean algebra isomorphism.
To prove $3\rightarrow 1$, we observe that from complete distributivity, we can show that the lattice of all partitions of the Boolean algebra $B$ is complete, and the least element $p$ in the lattice of all partitions of $B$ is actually the set of all atoms in the Boolean algebra $B$. One can show that for each $b\in B^{+}$, there must be an $a\in p$ with $a\leq b$.
Fact: A Boolean algebra $B$ is atomic if and only if the completion $\overline{B}$ is atomic, and if $e:B\rightarrow\overline{B}$ is the canonical embedding, then the mapping $e$ restricts to a bijection between the set of all atoms of $B$ and the set of all atoms of $\overline{B}$. Therefore, if $B$ is a Boolean algebra, then $\overline{B}\simeq P(X)$ if and only if $B$ is atomic, and $B$ has $|X|$ many atoms.
|
6
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https://mathoverflow.net/users/22277
|
405971
| 166,411 |
https://mathoverflow.net/questions/405967
|
15
|
Does there exist a concrete example of a finitely presented group that contains an isomorphic copy of $\operatorname{GL}\_n(\mathbb{Z})$ for every $n\in\mathbb{N}$? I think the Higman embedding theorem implies such a group must exist, but probably not in an especially constructive way, so I'm curious if there's a concrete example. I'm also especially interested in whether there exists a type $F\_\infty$ example. (I'm also generally curious if there are any interesting implications of such a group existing.)
|
https://mathoverflow.net/users/164670
|
Finitely presented group containing every $\mathrm{GL}_n(\mathbb{Z})$
|
The idea suggested by Anthony Genevois works.
I'm using that for any set $S$, the finitary linear group $\mathbf{Z}$ has the presentation with generators $e\_{st}$ for distinct $s,t\in S$, and relators $[e\_{pq},e\_{qr}]=e\_{pr}$ for any distinct $p,q,r\in S$, and $[e\_{pq},e\_{st}]=1$ for any distinct $p,q,s,t\in S$. If $\mathrm{St}\_S(\mathbf{Z})$ is the group defined by this presentation (this is called Steinberg group), there is a canonical homomorphism into the finitary group $\mathrm{GL}\_S(\mathbf{Z})$, mapping $e\_{st}$ to the matrix with $1$ on the diagonal and at position $(s,t)$, and $0$ elsewhere. (Almost) by definition, the kernel is $K\_2(\mathbf{Z})$ and is central. A classical theorem, which can be found in Milnor's book on K-theory, is that the latter is of order $2$; let $w$ be the nontrivial element of the kernel, viewed as word on the $e\_{st}$.
Now assume that $S$ is a $G$-set, $G$ finitely presented, and that $G$ has finitely many orbits on $S^4$, and that stabilizers for the $G$-action on $S^2$ minus diagonal are finitely generated. (There are many such examples: action of Thompson's groups on dyadics, natural action of Houghton groups, etc).
So we have the semidirect product $\tilde{H}=\mathrm{St}\_S(\mathbf{Z})\rtimes G$ (and its quotient $H=\mathrm{GL}\_S(\mathbf{Z})\rtimes G$, killing the central subgroup of order $2$ $\langle w\rangle$).
Then under the given assumptions, $H$ is finitely presented. Indeed, for simplicity suppose that $G$ has a single orbit on $S^2$ minus diagonal; fix a point and let $L$ be the stabilizer. Start from a finite presentation of $G\ast\mathbf{Z}$, with $\mathbf{Z}=\langle u\rangle$. This free product can be viewed as a "non-commutative wreath product" $G\ltimes \langle u\rangle^{\*G}$, the action permuting free factors. Then, modding out by the relators $[s,u]$ whenever $s$ ranges over a finite generating subset of $L$, we obtain the "non-commutative permutational product" $G\ltimes \langle u\rangle^{\*G/L}$ (also equal to $G\*\mathbf{Z}/[L,\mathbf{Z}]$). The relators $[e\_{pq},e\_{qr}]=e\_{pr}$ can be written as $[gug^{-1},huh^{-1}]=kuk^{-1}$ for various values of $g,h,k$ in $G$. Since $G$ has finitely many orbits on $G^3$, only finitely many of these are enough and the other ones follow by conjugating. The same remark applies to relators $[e\_{pq},e\_{st}]$, using that there are finitely many orbits on $S^4$.
This shows that $\tilde{H}$ and hence $H$ is finitely presented. And it contains copies of $\mathrm{SL}\_n(\mathbf{Z})$ for all $n$ (and hence $\mathrm{GL}\_n(\mathbf{Z})$, as the latter embeds into $\mathrm{SL}\_{n+1}(\mathbf{Z})$.
Notes:
1. It shouldn't be hard to extend this to the ring $\mathbf{Z}[1/m]$. However, passing to rationals would seem much less easy.
2. All this is, at the level of handling presentation, is indeed quite analogous to the finite presentability of wreath products.
|
12
|
https://mathoverflow.net/users/14094
|
405980
| 166,414 |
https://mathoverflow.net/questions/405866
|
5
|
Original question:
For compact metric spaces, plenty of subtly different definitions converge to the same concept. Overtness can be viewed as a property dual to compactness. So is there a similar story for overt metric spaces?
Edit: Since overtness is trivially true assuming the Law of the Excluded Middle, clearly the question is primarily interesting when we do not assume the LEM.
Edit 2: It looks like it is extremely difficult for a metric space to not be overt even in constructive settings. So editing the question to ask if there is ANY model where metric spaces are not overt.
Edit 3: For these reasons I changed the question again, from "Is there any model of mathematics where there exists a metric space that is not overt?". PT
|
https://mathoverflow.net/users/174368
|
What does overtness mean for metric spaces?
|
David Roberts has rubbed the magic lamp and the genie appears!
Even though the notion of overtness does depend on the strength of the ambient logic,
I believe the question here is with the notion of metric space, rather than the choice of a model of mathematics.
The natural answer is that any metric space has enough points and is therefore necessarily overt, in any sensible logical setting.
Indeed, I am inclined to think that any *whole* space in topology is in practice overt and the interesting question is what **overt *sub*spaces** look like.
Andrej has already pointed out that the set $A\subset{\mathbb N}$ of programs that don't terminate (with the two-valued metric) is not overt.
We can do better than this. [Steve Vickers](https://www.cs.bham.ac.uk/%7Esjv/papersfull.php) has an alternative to the Cauchy completion in locale theory and formal topology. Like any metric topology, it has a basis of *balls* $B(x,r)$, where we may take the radii $r$ to be dyadic rationals and the centres $x$ to be (for example) points with dyadic rational coordinates.
This construction still gives an overt space, because the set (overt discrete space) of centres is dense.
(Since I mention Steve, in general he is interested in the *hyper*spaces of *all* overt or compact subspaces, which are called the lower and upper *powerdomains*. My interest, in constrast, is with *individual* overt subspaces.)
To the general mathematician, the definition of overtness using an *operator* $\lozenge$ that takes unions of open subspaces to the existential quantifier is not very familiar. However, it has a very natural equivalent form when we're working in a metric space constructed in the above way.
Define $(d(x)< r) \equiv \lozenge B(x,r)$. It is easy to show that this satisfies
$$ d(x)< r'< r \Longrightarrow d(x)< r $$
$$ d(x)< r \Longrightarrow \exists r'.d(x)< r'< r $$
$$ d(x,y)< r \;\land\; d(y)< s \Longrightarrow d(x)< r+s $$
$$ d(x)< r \;\land\; \epsilon\gt 0 \Longrightarrow
\exists y.d(x,y)< r \;\land\; d(y)< \epsilon $$
for any $\epsilon>0$
What this means is that $d:X\to\overline{\mathbb R}$ is an *upper semicontinuous* function,
or alternatively one that is valued in the *upper real numbers*.
This is the essence of the equivalence between overt and **located** subspaces (the latter are used in Bishop-style constructive analysis), which was stated by [Bas Spitters](https://arxiv.org/pdf/math/0703561.pdf). Unfortunately, he only considered the case of *closed* overt/located subspaces, which are characterised by $d$ being valued in the ordinary (Euclidean, Dedekind, ...) real numbers.
The more general case is covered in my draft paper [*Overt Subspaces of ${\mathbb R}^n$*](http://www.paultaylor.eu/ASD/overtrn).
The third condition above is the triangle law. Under suitable conditions, the **Newton--Raphson algorithm** yields a function $\Delta(x)\equiv |f(x)/\dot f(x)|$ that satisfies all the other conditions and a $d$ obeying all of them can easily be derived from it.
My intuition is that **an overt subspace is the solution-space of an algorithm**. To justify this we need more examples from numerical analysis like Newton--Raphson, but that is very much not my subject.
On the other hand, Newton--Raphson actually yields more information than the $d$ function.
There are two possible responses to this:
* Maybe we should replace overtness with something more *quantitative*; or
* Maybe an algorithm could be *derived* from the *formula* for $\lozenge$ or $d$ together with the *proof* that it has the appropriate properties.
The second is not completely unreasonable:
An overt subspace is a generalisation of a point defined by a Dedekind cut or a completely prime filter. Andrej Bauer pioneered some ideas for [*Efficient computation with Dedekind reals*](https://mapcommunity.github.io/ictp/presentation_files/Bauer_P.pdf) and had a prototype calculator called Marshall.
Given how widely used the notions of overt, located or recursively enumerable subspaces now are in the different constructive cults, really we ought to have a better story than "overtness is dual to compactness but classically invisible". There ought to be a way of explaining the idea to "ordinary" (classical) mathematicians, in particular numerical analysts.
I have been trying to do this for more than a decade, but I think I'm the wrong person to do it, and probably we can't do it from the constructive side: somehow we have to kidnap a numerical analyst and inculcate them with this idea.
I still have this *draft* paper (above). Probably I should just stop fussing and publish it. Comments towards that are welcome.
|
6
|
https://mathoverflow.net/users/2733
|
405981
| 166,415 |
https://mathoverflow.net/questions/304399
|
15
|
Let $\mathrm{CombModCat}$ be the category of combinatorial model categories with left Quillen functors between them. By Dugger's theorem and the appendix of Lurie's "Higher Topos Theory" it ought to be true that its localization at the class of Quillen equivalences is equivalent to the homotopy category of presentable $\infty$-categories with left adjoint $\infty$-functors between them:
$$
\mathrm{CombModCat}\big[\text{QuillenEquivs}^{-1}\big]
\;\simeq\;
\mathrm{Ho}\big( \mathrm{Presentable}\infty\mathrm{Cat} \big)
$$
Has this been made explicit anywhere, in citable form?
Something close is made explicit in
* Olivier Renaudin, "Theories homotopiques de Quillen combinatoires et derivateurs de Grothendieck" ([arXiv:math/0603339](https://arxiv.org/abs/math/0603339))
(thanks to Mike Shulman for the pointer!), where it is shown that the 2-categorical localization of the 2-category version of $\mathrm{CombModCat}$ is equivalent to the 2-category of presentable derivators with left adjoints between them.
[edit:] By corollary 2.3.8, this implies that the 1-categorical localization of $\mathrm{CombModCat}$ is equivalent to the 1-categorical homotopy category of presentable derivators with left adjoints between them.
The latter clearly ought to be equivalent to the homotopy category of $\mathrm{Presentable}\infty\mathrm{Cat}$, but is that made explicit anywhere?
|
https://mathoverflow.net/users/381
|
Localizing $\mathrm{CombModCat}$ at the Quillen equivalences
|
The answer is affirmative and is provided by the paper
[Combinatorial model categories are equivalent to presentable quasicategories](https://arxiv.org/abs/2110.04679).
Among other things, it proves that the relative categories of combinatorial model categories, left Quillen functors, and left Quillen equivalences;
presentable quasicategories, cocontinuous functors, and equivalences;
and other models for homotopy locally presentable categories and homotopy cocontinuous functors are all weakly equivalent to each other as relative categories.
This also implies the equivalence of underlying quasicategories.
Additionally, combinatorial model categories can be assumed to be left proper and/or simplicial.
There is also an analogous statement for derivators,
which must be formulated as an equivalence of (2,1)-categories
due to the truncated homotopical nature of derivators.
|
6
|
https://mathoverflow.net/users/402
|
405993
| 166,420 |
https://mathoverflow.net/questions/405969
|
1
|
Let $p$ and $q$ be integers.
Let $f(n)$ be [A007814](https://oeis.org/A007814), the exponent of the highest power of $2$ dividing $n$, a.k.a. the binary carry sequence, the ruler sequence, or the $2$-adic valuation of $n$.
Then we have an integer sequence given by
\begin{align}
a(0)=a(1)&=1\\
a(2n)& = pa(n)+qa(2n-2^{f(n)})\\
a(2n+1) &= a(n-2^{f(n)})
\end{align}
which can also be expressed with a product
$$a(n) = [t(n) = 0] + [t(n) > 0]\prod\limits\_{k=0}^{t(n)-1} (q^{k+1} + p\sum\limits\_{j=0}^{k} q^{j})^{g(n, k)}$$
where
$$t(n)=\begin{cases}
[n=2],&\text{if $n<4$;}\\
t(2^{m-1} + k),&\text{if $0 \leqslant k < 2^{m-1}, m > 1$ where $n = 2^m + k$;}\\
t(k) + A010060(k - 2^{m-1}),&\text{if $2^{m-1} \leqslant k < 2^m, m > 1$ where $n = 2^m + k$.}
\end{cases}$$
$$g(n,0)=\begin{cases}
[n=2]+2\cdot [n=4]+[n=6]+[n=7],&\text{if $n<8$;}\\
g(2^{m-1} + k,0) - A010060(k) + 1,&\text{if $0 \leqslant k < 2^{m-1}, m > 2$ where $n = 2^m + k$;}\\
g(2^{m-2} + k,0) + 1,&\text{if $2^{m-1} \leqslant k < 3\cdot 2^{m-2}, m > 2$ where $n = 2^m + k$;}\\
g(2^{m-3} + k,0) + A010060(k - 3\cdot 2^{m-2}) ,&\text{if $3\cdot 2^{m-2} \leqslant k < 7\cdot 2^{m-3}, m > 2$ where $n = 2^m + k$;}\\
1,&\text{if $7\cdot 2^{m-3} \leqslant k < 2^m, m > 2$.}
\end{cases}$$
$$g(n, k) = g(h(n, k), 0), n \geqslant 0, k > 0,$$
$$h(n, k) = h(h(n, 1), k - 1), n \geqslant 0, k > 1,$$
$$h(n , 1) = s(s(n)), n \geqslant 0,$$
$$s(n) = A053645(n), n > 0, s(0) = 0.$$
Here are the links to the sequences: [A010060](https://oeis.org/A010060), [A053645](https://oeis.org/A053645).
I conjecture that $a(\frac{2^{kn}-1}{2^k-1})=(a(2^n-1))^{2k-1}$ for $n \geqslant 0$, $k>0$. This question generalizes the following: [Subsequence of the cubes](https://mathoverflow.net/questions/404108/subsequence-of-the-cubes).
Is there a way to prove it using expression with a product?
|
https://mathoverflow.net/users/231922
|
Subsequences of odd powers
|
Quite similarly to [my answer](https://mathoverflow.net/q/405210) to the previous question, we have that for $n=2^tk$ with odd $k$,
$$
a(n)=\sum\_{i=0}^t \binom{t}{i}p^{t-i}q^i a(2^i(k-1)+1).
$$
It further follows that for $n=2^{t\_1}(1+2^{t\_2+1}(1+\dots(1+2^{t\_s+1}))\dots)$ with $t\_j\geq 0$, we have
\begin{split}
a(n) &= \sum\_{i\_1=0}^{t\_1} \binom{t\_1}{i\_1} p^{t\_1-i\_1}q^{i\_1}
\sum\_{i\_2=0}^{t\_2+t\_3+1+i\_1} \binom{t\_2+t\_3+1+i\_1}{i\_2}p^{t\_2+t\_3+1+i\_1-i\_2}q^{i\_2}
\sum\_{i\_3=0}^{t\_4+t\_5+1+i\_2} \\
&\qquad\dots \sum\_{i\_\ell=0}^{t\_{2\ell-2}+t\_{2\ell-1}+1+i\_{\ell-1}} \binom{t\_{2\ell-2}+t\_{2\ell-1}+1+i\_{\ell-1}}{i\_\ell}p^{t\_{2\ell-2}+t\_{2\ell-1}+1-i\_\ell}q^{i\_\ell} \\
&=\prod\_{j=0}^{\ell-1} \bigg(q^{\ell-j}+p\frac{q^{\ell-j}-1}{q-1}\bigg)^{t\_{2j}+t\_{2j+1}+1},
\end{split}
where we conveniently define $\ell:=\left\lfloor\frac{s+1}2\right\rfloor$ and $t\_0:=-1$.
---
Now, for $n=\frac{2^{kn}-1}{2^k-1}$ we have $s=n$, $\ell=\left\lfloor\frac{n+1}2\right\rfloor$, $t\_1=0$ and $t\_j=k-1$ for all $j\in\{2,3,\dots,s\}$, implying that
$$a(\tfrac{2^{kn}-1}{2^k-1}) = \prod\_{j=1}^{\lfloor (n-1)/2\rfloor} \bigg(q^{\lfloor (n+1)/2\rfloor-j}+p\frac{q^{\lfloor (n+1)/2\rfloor-j}-1}{q-1}\bigg)^{2k-1}.$$
In particular, setting $k=1$, we get
$$a(2^{n}-1) = \prod\_{j=1}^{\lfloor (n-1)/2\rfloor} \bigg(q^{\lfloor (n+1)/2\rfloor-j}+p\frac{q^{\lfloor (n+1)/2\rfloor-j}-1}{q-1}\bigg),$$
and thus
$$a(\tfrac{2^{kn}-1}{2^k-1}) = a(2^{n}-1)^{2k-1}.$$
|
3
|
https://mathoverflow.net/users/7076
|
405996
| 166,421 |
https://mathoverflow.net/questions/405992
|
2
|
Let $J(x)$ be Riemann's prime counting function given by $\frac{1}{2}w(x) + \sum\_{n < x} w(n)$, where $w(p^k) = \frac{1}{k}$ when $p$ is a prime number and $k$ is a positive integer, and $w$ vanishes everywhere else.
Let the logarithmic integral $\newcommand{\li}{\mathrm{li}} \li(x)$ be some antiderivative of $\frac{1}{\log(x)}$.
I have often seen it asserted that for $x > 1$ or perhaps for $x > 2$, we have $J(x) = \li(x) - \sum\_{\rho} \li(x^{\rho})$ or perhaps $J(x) = \li(x) - \sum\_{\rho} \li(x^{\rho}) - \log(2)$, or some such variant. Here, the sum is over all zeros (trivial and nontrivial) of the Riemann zeta function, in some suitable order. The contribution over just the trivial zeros is some antiderivative of $\frac{1}{t (1 - t^2) \log(t)}$. I understand how to morally (non-rigorously) derive these formulas up to additive constants, but the additive constant specifics elude me.
I have written this without pinning down $\li$ beyond up to an additive constant. Clearly, adding any nonzero constant to $\li$ will destroy the convergence of any such series with infinitely many terms, so if there is some particular fixed choice of $\li$ to be used for all the terms which makes this converge, there is a unique such choice.
Is it indeed the case, as I often see suggested, that using the particular offset logarithmic integral with $\li(2) = 0$ makes these series converge? If so, I am very curious, what is so special about $2$ here?
If not, is there some other fixed choice of logarithmic integral I should be using for all these terms, or should I be using varying choices of $\li$, or in general, how should I interpret these often seen formulas?
|
https://mathoverflow.net/users/3902
|
Constant in logarithmic integral in prime counting
|
See Edwards' book *Riemann's Zeta Function*. He introduces $J(x)$ on p. 22 (in fact Edwards created the notation $J(x)$, which Riemann had written as $f(x)$). On p. 26 he defines ${\rm Li}(x)$ to be $\int\_0^x dt/\log t$, with the integral defined across the point 1 using a Cauchy principal value. On p. 33, equation (3) is the formula
$$
J(x) = {\rm Li}(x) - \sum\_{{\rm Im}(\rho)> 0} ({\rm Li}(x^\rho) + {\rm Li}(x^{1-\rho})) + \int\_x^\infty \frac{dt}{t(t^2-1)\log t} - \log 2
$$
for $x > 1$. The terms in the sum over nontrivial zeros $\rho$ with positive imaginary part pair together $\rho$ and $1-\rho$, where the latter are nontrivial zeros with negative imaginary part. These two terms never agree since $\rho = 1-\rho$ only when $\rho = 1/2$ and $\zeta(1/2) \not= 0$.
The sum counts each nontrivial zero with its multiplicity, and note $\rho$ and $1-\rho$ have the same multiplicity as zeros of the zeta-function. The $-\log 2$ at the end is $\log(1/2) = \log(\xi(0))$ where
$\xi(s) = (s-1)\pi^{-s/2}\Gamma(s/2 + 1)\zeta(s)$, so $\xi(0) = -\zeta(0) = 1/2$.
|
5
|
https://mathoverflow.net/users/3272
|
406001
| 166,422 |
https://mathoverflow.net/questions/299365
|
16
|
* Let $Q$ be the homotopical category of combinatorial model categories and left Quillen functors, with left Quillen equivalences for weak equivalences.
* Let $\mathbf Q$ be the corresponding $\infty$-category (in whatever foundations you prefer).
* Let $Pres^L$ be the $\infty$-category of presentable $\infty$-categories and left adjoint functors.
* The usual functor from relative categories to $\infty$-categories (modeled however you prefer) descends to a functor $N: \mathbf Q \to Pres^L$.
**Variation A:** I'm happy to work with simplicial combinatorial model categories rather than ordinary ones.
**Variation B:** It would also be interesting to know the answer to the following questions for combinatorial model $\infty$-categories and left $\infty$-Quillen functors. This is probably easier because one has more flexibility in this setting.
**Question 1:** Is $N: \mathbf Q \to Pres^L$ an equivalence of $\infty$-categories?
This functor is known to be essentially surjective -- for simplicial model categories this is in HTT, I think.
**Question 2:** Can the homotopical category $Q$ be refined to a model category?
There are size issues here; I'm happy with any way of handling them. I might suspect that something like Dugger's universal homotopy theories provide cofibrant resolutions. More specifically, I'd like to know:
**Question 3:** Are there model categories $C, D$ such that every left adjoint functor $NC \to ND$ is modeled by a left Quillen functor?
|
https://mathoverflow.net/users/2362
|
Do combinatorial model categories and Quillen adjunctions model presentable $\infty$-categories?
|
As pointed out in the answer to [Localizing $\mathrm{CombModCat}$ at the Quillen equivalences](https://mathoverflow.net/questions/304399/localizing-mathrmcombmodcat-at-the-quillen-equivalences/405993#405993),
the answer to Question 1
is affirmative and is provided by the paper
[Combinatorial model categories are equivalent to presentable quasicategories](https://arxiv.org/abs/2110.04679).
|
5
|
https://mathoverflow.net/users/402
|
406002
| 166,423 |
https://mathoverflow.net/questions/342840
|
16
|
We know by Karol Szumiło's thesis (<https://arxiv.org/pdf/1411.0303.pdf>) that there is an equivalence between the two fibration categories of cofibration categories on one side and cocomplete $\infty$-categories on the other. By a dual argument, we obtain an equivalence between fibration categories and complete $\infty$-categories (I'm guessing, in the form of cofibration categories).
1. Does either of these constructions restrict to the case of combinatorial model categories vs. presentable $\infty$-categories? In other words, is it an (almost) immediate consequence of the main theorem therein that there is also an equivalence between some model-like structures of combinatorial model categories and presentable $\infty$-categories?
2. Similar question, but considering all model categories (or maybe just categories with both a fibration and a cofibration structure) on one side and complete/cocomplete $\infty$-categories on the other?
|
https://mathoverflow.net/users/134438
|
Correspondence between classes of model categories and classes of $\infty$-categories
|
As pointed out in the answer to [Localizing $\mathrm{CombModCat}$ at the Quillen equivalences](https://mathoverflow.net/questions/304399/localizing-mathrmcombmodcat-at-the-quillen-equivalences/405993#405993),
the answer to Question 1 whether
>
> there is also an equivalence between some model-like structures of combinatorial model categories and presentable ∞-categories?
>
>
>
is affirmative and is provided by the paper
[Combinatorial model categories are equivalent to presentable quasicategories](https://arxiv.org/abs/2110.04679).
The model-like structure used in the paper is that of the relative category.
Concerning Question 2:
>
> Similar question, but considering all model categories (or maybe just categories with both a fibration and a cofibration structure) on one side and complete/cocomplete ∞-categories on the other?
>
>
>
there are pretty severe size issues with defining “all model categories” or “all complete/cocomplete ∞-categories”.
If, however, we use universes to resolve these size issues,
then the answer is positive and can be deduced from Theorem 8.2 in the cited paper, in a manner similar to Proposition 8.3.
|
3
|
https://mathoverflow.net/users/402
|
406004
| 166,424 |
https://mathoverflow.net/questions/406014
|
2
|
Let $H$ be a separable Hilbert space of infinite dimension and let $(e\_n)\_{n \in \mathbb{N}}$ be an orthonormal basis of $H$. For a series $(\alpha\_n)\_{n \in \mathbb{N}} \subset \mathbb{R^+}$ we are interested in whether or not the unit ball $$B\_1 := \{h \in H \mid 1 \geq \|h\|\_H\}$$ is a subset of the closed symmetric convex hull $\overline{\operatorname{sco}}(\alpha\_n \, e\_n \mid n \in \mathbb{N})$ of the series $(\alpha\_n \, e\_n)\_{n \in \mathbb{N}} \subset H$. It is clear, that such a series exists, since we can iteratively construct one, such that for every $n \in \mathbb{N}$ the $n$-dimensional sphere $B\_{1+\frac{1}{n}} \cap \langle e\_1,...,e\_n \rangle$ is in the convex hull $\overline{\operatorname{sco}}(\alpha\_1 \, e\_1,...,\alpha\_{n} \, e\_n)$.
My question is if there are any known bounds on the rate at which such a series $(\alpha\_n)\_{n \in \mathbb{N}}$ must tend to infinity, i.e. if one can construct the series in a way to have it grow relatively slowly.
I would also be interested in any useful literature and maybe a lower bound on the rate, other than $\mathcal{O}(1)$.
Any help is be much appreciated.
|
https://mathoverflow.net/users/409412
|
Geometry in Hilbert spaces / spheres in high dimensions
|
Set $c\_n:=\frac{1}{\alpha\_n}$. Clearly, for any $N>0$ one must have $\DeclareMathOperator{\spa}{span}$ $\DeclareMathOperator{\conv}{conv}$ $\newcommand{\bsB}{\boldsymbol{B}}$
$$
\bsB\_1\cap\spa\{e\_1,\dotsc, e\_N\} \subset\conv\{ \pm\alpha\_1 e\_1,\dotsc,\pm\alpha\_N e\_n\}.
$$
This happens if and only if
$$
f\_N(\alpha):=\min\{ x\_1^2+\cdots +x\_N^2;\;\;c\_1x\_1+\cdots +c\_Nx\_N=1\}\geq 1.
$$
(The hyperplane $c\_1x\_1+\cdots +c\_Nx\_N=1$ cuts the $k$-th axis at the point $\frac{1}{c\_k}=\alpha\_k$.) Using Lagrange multipliers we get
$$
f\_N(\alpha)=\frac{1}{c\_1^2+\cdots +c\_N^2}\geq 1.
$$
Thus we need
$$\sum\_n\frac{1}{\alpha\_n^2}=\sum\_n c\_n^2\leq 1. $$
|
3
|
https://mathoverflow.net/users/20302
|
406029
| 166,430 |
https://mathoverflow.net/questions/405978
|
4
|
Let $X$ be a $\mathbb{Q}$-Gorenstein (isolated) singularity of dimension at least $2$ and $f:Y \to X$ be a resolution of singularities. In this case the canonical sheaf $K\_X$ is not necessarily invertible, it is only reflexive.
>
> **Question.** Is the pull-back $f^\*K\_X$ invertible? If not, can we say that $f^\*K\_X/\mathrm{tors}$ is invertible?
>
>
>
|
https://mathoverflow.net/users/45397
|
Is the pull-back of canonical sheaf invertible (modulo torsion)?
|
I think that $f^\*K\_X$ is not invertible in general. For instance, take as $X$ a quotient surface singularity of type $\frac{1}{4}(1, \, 1)$. Then straightforward computations give $$K\_Y=f^\*K\_X - \frac{1}{2}E,$$
where $E$ is the exceptional divisor. We infer that $$f^\*K\_X = K\_Y + \frac{1}{2}E$$ is not an invertible sheaf on $Y$. In fact, the exceptional divisor $E$ is not $2$-divisible in $\operatorname{Pic}(Y)$. The way I see this is that $X$ is the singularity given by a cone over a rational normal curve of degree $4$ in $\mathbb{P}^4$, and $Y$ is the blow-up at the vertex. Then $E$ is a section of a (rational) fibration on $Y$, hence it cannot be divisible.
|
2
|
https://mathoverflow.net/users/7460
|
406031
| 166,432 |
https://mathoverflow.net/questions/406025
|
1
|
(I asked this [on MSE](https://math.stackexchange.com/questions/4268290/relation-between-two-notions-intermediate-between-pointwise-convergence-and-u) a week ago, but did not get any answers there, so I'm trying here.)
Let $X$ be a topological space. I will define four ways in which a sequence $(f\_n)$ of *continuous* functions $X \to \mathbb{R}$ might converge to a *continuous* function $f\colon X \to \mathbb{R}$, four notions intermediate between pointwise convergence (labeled (1) below) and uniform convergence (labeled (4) below). (Note: I am listing all four below in order to justify why I care about (2) and (3), but only (2) and (3) are involved in the actual question.) Specifically:
1. $f\_n$ tends to $f$ pointwise, namely: for all $x\in X$ we have $f\_n(x) \to f(x)$ (with no assumption of uniformity of any kind), viꝫ., for all $\varepsilon>0$ and all $x\in X$ there is $n\_0$ such that when $n\geq n\_0$ we have $|f\_n(x)-f(x)|<\varepsilon$.
2. We now demand that (for $\varepsilon>0$ fixed), $n\_0$ take the same value on each element of an open cover $(U\_i)\_{i\in I}$ of $X$. More precisely: for all $\varepsilon>0$ there is a covering $(U\_i)\_{i\in I}$ of $X$ by open sets such that for all $i\in I$ there is $n\_0$ such that when $x \in U\_i$ and $n\geq n\_0$ we have $|f\_n(x)-f(x)|<\varepsilon$.
3. We now demand that the same open cover $(U\_i)\_{i\in I}$ work for every $\varepsilon>0$; this is the same as demanding that $f\_n\to f$ uniformly on each $U\_i$, viꝫ. there is a covering $(U\_i)\_{i\in I}$ of $X$ by open sets such that for all $\varepsilon>0$ and all $i\in I$ there is $n\_0$ such that when $x \in U\_i$ and $n\geq n\_0$ we have $|f\_n(x)-f(x)|<\varepsilon$. (I suppose we could say that $f\_n \to f$ “locally uniformly”, but I'm not sure whether this is standard terminology.)
4. We now demand that the covering consist of just $X$, i.e., that $f\_n \to f$ uniformly, viꝫ. there for all $\varepsilon>0$ there is $n\_0$ such that when $x \in X$ and $n\geq n\_0$ we have $|f\_n(x)-f(x)|<\varepsilon$.
To summarize:
1. $\forall \varepsilon>0. \forall x\in X. \exists n\_0\in \mathbb{N}. \forall n≥n\_0. |f\_n(x)-f(x)|<\varepsilon$
2. $\forall \varepsilon>0. \exists (U\_i) \text{ open covering}. \forall i\in I. \exists n\_0\in \mathbb{N}. \forall x\in U\_i. \forall n≥n\_0. |f\_n(x)-f(x)|<\varepsilon$
3. $\exists (U\_i) \text{ open covering}. \forall \varepsilon>0. \forall i\in I. \exists n\_0\in \mathbb{N}. \forall x\in U\_i. \forall n≥n\_0. |f\_n(x)-f(x)|<\varepsilon$
4. $\forall \varepsilon>0. \exists n\_0\in \mathbb{N}. \forall x\in X. \forall n≥n\_0. |f\_n(x)-f(x)|<\varepsilon$
It's clear that $(4)\Rightarrow(3)\Rightarrow(2)\Rightarrow(1)$.
* A counterexample showing that (3) does not imply (4) is given by $X = \mathbb{R}$ and $f\_n(x) = \exp(-(x-n)^2)$, which converges in sense (3) but not uniformly (4) toward $f = 0$.
* A counterexample showing that (1) does not imply (2) is given by $X = [0,1]$ and $f\_n(x) = \max(0, \min(nx, 2-nx))$ ([graphs here](https://twitter.com/gro_tsen/status/1444595905374658566)) which converge pointwise (1) toward $f = 0$ but there is no neighborhood $V$ of $0$ such that $\exists n\_0\in \mathbb{N}. \forall x\in V. \forall n\geq n\_0. |f\_n(x)|<\frac{1}{2}$ so (2) does not hold.
**Question:** Does (2) have a standard name? Are (2) and (3) perhaps in fact equivalent? If yes, what is a proof? If no, what is a counterexample?
|
https://mathoverflow.net/users/17064
|
Relation between two notions intermediate between “pointwise convergence” and “uniform convergence”
|
I think the following is a counter-example for the equivalence of (2) and (3):
Consider the [Baire space](https://en.wikipedia.org/wiki/Baire_space_(set_theory)) $\mathbf N^\mathbf N$ and let $f\_n$ be the following function: If your sequence $x$ starts with $k$ zeroes followed by the entry $n$ then $f\_n(x) = 1/(k+1)$, otherwise $f\_n(x) = 0$.
The $f\_n$ are continuous: away from the sequence that is constantly 0, they are even locally constant. At $0^\infty$ you can easily check continuity using cylinder sets.
The $f\_n$ satisfy (2): If $\epsilon > 1/(k+1)$, we can take as our covering the $k$-cylinder sets: for $v \in \mathbf N^k$, let $U\_v$ be the set of sequences that start with $v$. If $v = 0^k$, then $\sup f\_n(U\_v) = 1/(k+1) < \epsilon$. For $v \not=0^k$, we can wait until n is larger than the first nonzero entry of $v$, and get $f\_n(U\_v) = 0$ for all $n \geq n\_0$ for some $n\_0$. So the $U\_v$ form the required covering.
The $f\_n$ do not satisfy (3): Some set of the open covering must contain $0^\infty$, and so it must contain a cylinder set $U\_v$ with $v = 0^k$ for some $k$. But $\sup f\_n(U\_v) = 1/(k+1)$ for all $n$, so you don't have local uniform convergence.
In my construction I heavily used that $\mathbf N^\mathbf N$ is not locally compact, I am curious whether (2) and (3) are equivalent for locally compact spaces.
|
2
|
https://mathoverflow.net/users/409730
|
406042
| 166,436 |
https://mathoverflow.net/questions/406045
|
1
|
**Question.** What restrictions are known on closed manifolds $(M^3,g)$ that contain a sequence of embedded minimal surfaces $(\Sigma\_j \mid j \in \mathbf{N})$ with
\begin{equation}
\mathrm{area} \, \Sigma\_j \to \infty \, \, \text{ but } \, \, \mathrm{index} \, \Sigma\_j \leq C
\end{equation}
for some constant $C$ and all $j$?
There are no topological restrictions: Colding and Minicozzi [1] show that for any manifold $M$ there is an open set of metrics and a sequence of tori $(T\_j)$ with
\begin{equation}
\mathrm{index} \, T\_j = 0 \, \, \text{ but } \, \, \mathrm{area} \, T\_j \to \infty.
\end{equation}
They also remark that such a sequence could not exist if the metric has positive Ricci curvature; however I am a bit hazy about the link between index and genus. Is there a bound for the genus of a minimal surface $\Sigma \subset M^3$ in terms of $\mathrm{index} \, \Sigma$ in this case? If indeed $\mathrm{Ric}\_g > 0 $ is forbidden, I'd be interested in any information available beyond this.
[1] Tobias Colding and William Minicozzi. Examples of embedded minimal tori without area bounds. *International Mathematics Research Notices.* No. 20 (1999) pp. 1097-1100.
|
https://mathoverflow.net/users/103792
|
Minimal surfaces with increasing area but bounded Morse index
|
Positive scalar curvature implies that if $\textrm{index}(\Sigma\_j)\leq I$ then $\Sigma\_j$ have bounded area and genus. This is proven here <https://arxiv.org/pdf/1509.06724.pdf> (Theorem 1.3). That paper also contains some other examples related to the Colding--Minicozzi looping example.
A natural generalization is whether or not one can generalize this to a statement like
$$
\textrm{area}(\Sigma) + \textrm{genus}(\Sigma) \leq C \, \textrm{index}(\Sigma)
$$
where $C$ is some constant independent of $\Sigma$ (universal or $M$ dependent). Some form of this inequality was conjectured by Schoen and Marques--Neves. There has been a lot of work on this problem, but it is still open. See e.g. <https://arxiv.org/pdf/1911.09166.pdf>, <https://mathscinet.ams.org/mathscinet-getitem?mr=2779062>, <https://mathscinet.ams.org/mathscinet-getitem?mr=3770846> .
|
2
|
https://mathoverflow.net/users/1540
|
406049
| 166,438 |
https://mathoverflow.net/questions/406018
|
0
|
Let $y\in L^{2}(0,1)$ and let $\widetilde{y}$ be its extension on $(0,\infty
).$ Assume that there exist $p\_{0},p\_{1}\in
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
,$ $p\_{0}\neq p\_{1}$ such that
\begin{eqnarray\*}
L(\widetilde{y})(p\_{0}) &=&0\Leftrightarrow y=0\text{ on }(0,\frac{1}{2}), \\
L(\widetilde{y})(p\_{1}) &=&0\Leftrightarrow y=0\text{ on }(\frac{1}{2},1).
\end{eqnarray\*}
What can I say about the problem $L(y)(p)=0\Rightarrow y=0?.$ What are the
additional assumptions that guarantee $y=0?.$ Is there any references about
this problem?. Thank you in advance?
**Edited:**
The problem comes from the following problem: We want to prove that
the unique solution to the following difference equation is the null one:
$$
au(x)+b\mathbf{1}\_{(0,\frac{1}{2})}(x)u(x+\frac{1}{2})+c\mathbf{1}\_{(\frac{1%
}{2},1)}(x)u(x-\frac{1}{2})=0,\text{ }x\in (0,1).
$$
Extending $u$ by zero outside $(0,1)$ and taking the Laplace transform yields
$$
a\int\_{0}^{1}e^{-px}u(x)dx+be^{\frac{p}{2}}\int\_{\frac{1}{2}%
}^{1}e^{-px}u(x)dx+ce^{-\frac{p}{2}}\int\_{0}^{\frac{1}{2}}e^{-px}u(x)dx=0,%
\text{ }p\in
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
,
$$
that is
$$
\left( a+be^{\frac{p}{2}}\right) \int\_{\frac{1}{2}}^{1}e^{-px}u(x)dx+\left(
a+ce^{-\frac{p}{2}}\right) \int\_{0}^{\frac{1}{2}}e^{-px}u(x)dx=0,\text{ }%
p\in
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
.
$$
If we let for instance $p=\gamma +4n\pi i,n\in
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ with $a+ce^{-\frac{\gamma }{2}}=0$ we get
$$
\left( a+be^{\frac{\gamma }{2}}\right) e^{-\gamma }\int\_{\frac{1}{2}%
}^{1}e^{-4n\pi ix}u(x)dx=0,
$$
which yields that $u=0$ on $(\frac{1}{2},1)$ if $a+be^{\frac{\gamma }{2}%
}\neq 0$ which is equivalent to $a^2-bc \neq 0$. With the same manner, by choosing this time $p=\delta +4n\pi
i,n\in
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ with $a+be^{\frac{\delta }{2}}=0$ we get
$$
\left( a+ce^{-\frac{\delta }{2}}\right) e^{-\delta }\int\_{0}^{\frac{1}{2}%
}e^{-4n\pi ix}u(x)dx=0,
$$
so if $a+ce^{-\frac{\delta }{2}}\neq 0$ we get $u=0$ on $(0,\frac{1}{2})$, which is equivalent to $a^2-bc \neq 0$.
I don't know if this kind of reasoning is correct since the Laplace
transform of $u$ is zero on subintervals with different choices of $p.$
|
https://mathoverflow.net/users/106804
|
Laplace transform injectivity for different values of $p$
|
>
> **Claim:** Assuming that $L$ is the Laplace transform, there is no complex $p\_0$ such that for all $y\in L^2(0,1)$ we have the implication $L(\tilde y)(p\_0)=0\implies y=0$ on $(0,1/2)$.
>
>
>
So, your conditions can never be fulfilled, and therefore they imply any statement, be it true or false.
*Proof of the Claim:* Take any complex $p\_0=a+ib$, where $a$ and $b$ are real. Let $x\_1$ and $x\_2$ be the functions in $L^2(0,1/2)$ defined by
$$x\_1(t):=e^{-at}\cos bt,\quad x\_2(t):=e^{-at}\sin bt$$
for $t\in[0,1/2]$, so that $e^{-p\_0t}=x\_1(t)-i x\_2(t)$. Then there is a **nonzero** function $z\in L^2(0,1/2)$ (orthogonal to both $x\_1$ and $x\_2$) such that
$$\int\_0^{1/2}x\_1(t)z(t)\,dt=0=\int\_0^{1/2}x\_2(t)z(t)\,dt.\tag{1}$$
Then, letting $y\in L^2(0,1)$ be defined by the conditions $y:=z$ on $[0,1/2]$ and $y:=0$ on $(1/2,1]$, we have
$$L(\tilde y)(p\_0)=\int\_0^1 e^{-p\_0t}y(t)\,dt \\
=\int\_0^{1/2}x\_1(t)z(t)\,dt-i\int\_0^{1/2}x\_2(t)z(t)\,dt=0.$$
So, $L(\tilde y)(p\_0)=0$ but $y$ is not $0$ on $(0,1/2)$.
Thus, the Claim is proved.
|
1
|
https://mathoverflow.net/users/36721
|
406054
| 166,441 |
https://mathoverflow.net/questions/406052
|
6
|
If $X$ is a smooth, geometrically connected, projective curve of genus $g$
over $\mathbb{F}\_q$, then the [zeta function](https://en.wikipedia.org/wiki/Weil_conjectures) of $X$ is of the form $P(s)/(1 - s)(1 - qs)$, where $P(s)$ is a polynomial of degree $2g$. Denote the $P(s)$ of $X$ by $P\_X$.
By the Riemann hypothesis for curves, the zeroes of $P\_X$ all have complex absolute value $1/\sqrt q$.
**Question:** Is it possible to find a family of curves $\{X\_n\}\_{n=1}^{\infty}$ with $g(X\_n) \rightarrow \infty$ ($g(X\_n)$ is the genus of $X\_n$) such that the zeros of $P\_{X\_n}$ tends to being evenly distributed, i.e. if $Y\_n$ denotes the set of phase angle differences (normalized into $[0,2\pi)$) of consecutive zeros of $P\_{X\_n}$, then $\max{Y\_n}-\min{Y\_n} = o(1/g(X\_n))$?
|
https://mathoverflow.net/users/125498
|
Could the Weil zeroes of curves be evenly distributed?
|
If $q$ is a prime for which $2$ is a primitive root, then I claim that the Frobenius eigenvalues of the curve $y^2-y = x^q$ over $\mathbb{F}\_2$ have spacing exactly $\tfrac{2 \pi}{q-1}$. This, if [Artin's conjecture](https://en.wikipedia.org/wiki/Artin%27s_conjecture_on_primitive_roots) holds, such examples exist over $\mathbb{F}\_2$.
Let $C$ be the affine curve $y^2-y = x^q$ over $\mathbb{F}\_2$. We note that $C$ is smooth, by taking the partial derivative with respect to $y$; that $C$ has one point at $\infty$ and that $C$ has genus $\tfrac{q-1}{2}$. So the number of points on $C$ over $\mathbb{F}\_{2^k}$ is
$$2^k - \sum\_{i=1}^{q-1} \lambda\_i^k$$
where $\lambda\_i$ are the eignvalues of Frobenius. To show that the $\lambda\_i$ are spaced as claimed, we must show that $\#C(\mathbb{F}\_{2^k}) = 2^k$ for $1 \leq k < q-1$.
The hypothesis that $2$ is a primitive root modulo $q$ means that $x \mapsto x^q$ is bijective on $\mathbb{F}\_{2^k}$ for $k$ not divisible by $q-1$. So, for each $y \in \mathbb{F}\_{2^k}$, there is exactly one solution to $y^2-y = x^q$ in $\mathbb{F}\_{2^k}$. $\square$
The same argument works if $p$ is an odd prime which is a primitive root for infinitely many $q$ and we consider the hyperelliptic curves $y^2 = x^q-1$ over $\mathbb{F}\_p$.
|
9
|
https://mathoverflow.net/users/297
|
406059
| 166,443 |
https://mathoverflow.net/questions/406055
|
7
|
In John Cremona's book, he defines the *modular symbol* of an elliptic curve in the following way.
Let $E/\mathbf{Q}$ be an elliptic curve and let $f\_E$ be the modular form associated to $E$. The *modular symbol* associated to $E$ is the map $\mathbf{Q} \to \mathbf{Q}$ given by sending any $r \in \mathbf{Q}$ to the rational number
$$ [r] := \dfrac{2 \pi i}{\Omega\_E} \left( \int\_r^{i \infty} f\_E(z) \, dz + \int\_{-r}^{i \infty} f\_E(z) \, dz \right). $$
These modular symbols contain interesting information about $L$-values of $E$. I have two questions about this (in increasing levels of importance):
1. Where can I find a proof that $[r]$ is indeed a rational number? If anyone has a reference, that would be great.
2. How do we compute the value of $[r]$ *exactly* (i.e: not a numerical approximation)? Cremona's book says that for any $r \in \mathbf{Q}$, the value of $[r]$ can be computed exactly. (And in any case, Sage can compute the values exactly.) But I can't find a theorem anywhere that gives a formula for the exact value of $[r]$. If anyone knows of an algorithm to get an exact value of $[r]$, or better yet, a formula that gives the exact value of $[r]$ in terms of invariants of the elliptic curve $E$, I'd greatly appreciate it.
Thanks for the help!
|
https://mathoverflow.net/users/394740
|
How do you compute modular symbols?
|
There are two ways.
You can calculate a sufficiently good approximation by integrating numerically the modular form. Since the value of the modular symbol is known to be a rational number and we know a bound for the denominator (by the Manin-Drinfeld Theorem), we know to what precision we need to evaluate the integral. The problem is that the integral converges very slowly if you integrate naively all the way down to the real axis. Instead one should use Atkin-Lehner involutions to move the cusps. This results in a very fast method to compute a single $[r]$ for a single curve as long as the conductor $N$ is square-free and not in the billions. This is implemented as [*implementation="num"* in SageMath](https://doc.sagemath.org/html/en/reference/arithmetic_curves/sage/schemes/elliptic_curves/mod_sym_num.html).
The second method is to determine the $\mathbb{Q}$-vector space of modular symbols attached to the isogeny class of the elliptic curve $E$ among the total space of modular symbols for the group $\Gamma\_0(N)$. This is done by taking the quotient by the relations imposed by the Hecke operators. These computations are all done with rational numbers, they are inherently exact. There is an issue about the scaling to make sure that the period map corresponds to the particular curve in the isogeny class; but this can be done with a single comparison with some $L$-value. Once the space is determined, each evaluation $[r]$ for any $r$ is very fast. This is implemented in pari-gp, magma, SageMath and in eclib. The latter is implemented by John Cremona and based on the explanations in [the book you are refering to](https://johncremona.github.io/book/fulltext/index.html) In particular chapter II. Belabas and Perrin-Riou have improve this a bit for pari-gp. The bottle neck of the computation here is the first step involving linear algebra of rather large matrices. For conductors below a million it should work.
If one is interested in computing many values for varying $r$, like for instance when computing the $p$-adic $L$-function of $E$, one should use overconvergent modular symbols instead.
|
15
|
https://mathoverflow.net/users/5015
|
406062
| 166,445 |
https://mathoverflow.net/questions/406060
|
6
|
This is a [crosspost](https://math.stackexchange.com/questions/4272807/connections-between-0-toposes-and-1-toposes-grothendieck-and-elementary) from math.stackexchange.
A Grothendieck $0$-topos is the same as a frame (see [here](https://ncatlab.org/nlab/show/frame#As0Topos)). Another relationship between Grothendieck toposes and frames is the following: whenever $\mathcal O$ is a frame, then $\mathrm{Sh}(\mathcal O)$, the category of sheaves on $\mathcal O$, is a Grothendieck topos.
A elementary $0$-topos is the same as a *Heyting algebra* (see [here](https://ncatlab.org/nlab/show/Heyting+algebra#to_toposes)). Another relationship between elementary toposes and Heyting algebras is the following: if $\mathcal E$ is an elementary topos and $X\in \mathcal E$ is an object, then $\mathrm{Sub}\_\mathcal E(X)$ is a Heyting algebra.
These facts suggest to me these question:
1. If $H$ is a Heyting algebra, does $H$ induce an elementary topos (in the same way a frame $\mathcal O$ induces the Grothendieck topos $\mathrm{Sh}(\mathcal O)$)?
2. If $\mathcal E$ is a Grothendieck topos and $X\in\mathcal E$ an object, is $\mathrm{Sub}\_\mathcal E(X)$ a frame?
3. Do these connections hold in general for (Grothendieck or elementary) $n$-toposes? That is, whenever $\mathcal E$ is a (Grothendieck or elementary) $n$-topos and $X\in\mathcal E$, then $\mathrm{Sub}\_\mathcal E(X)$ is a (Grothendieck or elementary) $n-1$-topos, and whenever $\mathcal E$ is a (Grothendieck or elementary) $n$-topos, then there's an induced $n+1$-topos $\mathrm{Sh}(\mathcal E)$?
The second question was already answered by Mark Kamsma: yes.
I want to add the following question: what's the intuitive reason that a 0-topos being a frame / Heyting algebra implies that each frame induces a topos and each collection of subobjects is a Heyting algebra? Is this just a coincidence?
|
https://mathoverflow.net/users/409911
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Connections between $0$-toposes and $1$-toposes (Grothendieck and elementary)
|
1. Yes and no. As Jonas Frey pointed out on MSE, you can take the category of finite sheaves on any Heyting algebra $H$ to get an elementary topos. However, unlike in the case of frames and Grothendieck toposes, the Heyting algebra $H$ will not in general coincide with the Heyting algebra of subterminal objects in this elementary topos. Roughly speaking, the topos of finite sheaves doesn't "know" anything about the Heyting structure of $H$, only its underlying distributive lattice (whereas for a frame, the Heyting structure is recoverable from the infinite distributive law and the adjoint functor theorem). It might still be an open question whether any Heyting algebra can occur as the Heyting algebra of subterminal objects in an elementary topos; at least, I believe it was open for a while.
2. Yes, as you say.
3. In the Grothendieck case, yes, sort of. On one hand, if $E$ is a Grothendieck $n$-topos and $X\in E$, then the category of $(n-2)$-truncated objects in $E/X$ is an $(n-1)$-topos; but it only really makes sense to call these "subobjects" when $n=1$. On the other hand, if $E$ is a Grothendieck $n$-topos, you can identify it with the category of $n$-sheaves on some site $C$, and then the category of $(n+1)$-sheaves on $C$ will be an $(n+1)$-topos whose underlying category of $(n-1)$-truncated objects in $E$.
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7
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https://mathoverflow.net/users/49
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406063
| 166,446 |
https://mathoverflow.net/questions/405510
|
1
|
Consider a two-state Markov process in continuous time, with states labelled $A$ and $B$. The transition rates for going from state $A$ to $B$, and state $B$ to $A$ are $\alpha$ and $\beta$ respectively. We are interested in the quantity $p\_t(A,x,t\_A|A)$, which denotes the probability that, at time $t$, the Markov process is in state $A$, has made a total of $x$ jumps, and has spent time equal to $t\_A$ in state $A$, given that it started from state $A$.
I was thinking of two approaches (and both are likely to be equivalent). But I seem to be running into hurdles in both of them, and would be grateful for some help.
My first approach was to write a set of coupled recurrence relations in the following way:
\begin{align}
p\_t(A,x,t\_A|A) &= \int\_0^t e^{-\alpha t'}\cdot \alpha \cdot p\_{t-t'}(A,x-1,t\_A-t'|B)\cdot dt'\\
p\_t(A,x,t\_A|B) &= \int\_0^t e^{-\beta t'}\cdot \beta \cdot p\_{t-t'}(A,x-1,t\_A|A)\cdot dt'
\end{align}
The basic idea behind the above equations is that the first jump takes place at time $t'$, and thus, the rest of the $x-1$ jumps have to be made in the remaining $t-t'$ amount of time. Similarly, depending on what the initial state was, the time needed to be spent in state $A$ in the remaining $t-t'$ time is stated in the argument (e.g. if the first $t'$ amount of time was spent in state $A$, then in the remaining amount of time, only $t\_A-t'$ amount of time needs to be spent there). There is a minor issue here that the above equation does not necessarily restrict the possibility of $t'>t\_A$, which is unphysical. But that can be taken care of by changing the upper limit of the integral in the first equation to $t\_A$.
Naively, one would expect that taking a Laplace transform of the two equations with respect to $t$ should simplify matters a lot. For the second equation, it indeed does, as the equation is a simple convolution in $t$. However, the first is not **(please correct me if I am wrong)** and thus, naive Laplace transform with respect to $t$ might not be helpful. Is there any other approach which makes the problem easier?
My second approach was to derive the differential equation satisfied by the quantity $p\_t(A,x,t\_A|A)$, by working in discrete time-steps of $\Delta t$, and finally taking the limit $\Delta t \to 0$:
\begin{align}
p\_{t+\Delta t}(A,x,t\_A|A) = p\_t(A,x,t\_A-\Delta t|A)\cdot (1-\alpha \Delta t) + p\_t(B,x-1,t\_A-\Delta t|A)\cdot \beta \Delta t
\end{align}
which upon rearranging gives us
\begin{align}
\frac{p\_{t+\Delta t}(A,x,t\_A|A) - p\_{t}(A,x,t\_A- \Delta t|A)}{\Delta t} = -\alpha\cdot p\_t(A,x,t\_A-\Delta t|A) + \beta \cdot p\_t(B,x-1,t\_A - \Delta t|A).
\end{align}
In the limit of $\Delta t \to 0$, we get:
$$\frac{\partial{p\_{t}(A,x,t\_A|A)}}{\partial{t}}+\frac{\partial{p\_{t}(A,x,t\_A|A)}}{\partial{t\_A}} = -\alpha\cdot p\_t(A,x,t\_A|A) + \beta \cdot p\_t(B,x-1,t\_A|A). $$
Similarly, one could write equations for $p\_t(B,x,t\_A|A)$. Here (again naively), it seems like one could take a Laplace transform with respect to $t$ and $t\_A$, and define a generating function for $x$ to simplify the problem. However, one thing that scares me is that $t\_A$ cannot take any value. It must always take values less than $t$. Thus, independently taking Laplace transforms with respect to $t$ and $t\_A$ seems shady.
Is there a way to overcome one (or both) of the hurdles?
|
https://mathoverflow.net/users/168672
|
Occupation times for two-state Markov processes
|
Without loss of generality, let the final time to be $t=1$ (if it is not, we can make it so by rescaling time as $\alpha'=t\alpha$ and $\beta'=t\alpha$).
Then, consider a single trajectory of the process that starts and ends on state $A$ and makes $x$ jumps ($x$ has to be even). We can represent any such trajectory in terms of the jump times, $$\vec{t}=(t\_{0}=0\le t\_{1}\le\dots\le t\_{x}\le1) \tag{1}$$
The conditional probability of the trajectory is given by the product of the jump probabilities, multiplied by the probability of not leaving state $A$ after the last jump:
$$p(\vec{t}\vert A)=e^{-\alpha(1-t\_{x})}\prod\_{i=0}^{x-1}p\_{i}(t\_{i+1}\vert t\_{i}).\tag{2}$$
We will assume below that $x\ge 2$ (otherwise $t\_A=1$ and $p(A,x,t\_A|A)=e^{-\alpha}$).
Now, if $i$ is even, then the jump at $t\_{i+1}$ is from state $A$ to state $B$ and has probability density $p\_{i}(t\_{i+1}\vert t\_{i})=\alpha e^{-(t\_{i+1}-t\_{i})\alpha}$. Otherwise the jump is from state $B$ to state $A$ and has probability density $p\_{i}(t\_{i+1}\vert t\_{i})=\beta e^{-(t\_{i+1}-t\_{i})\beta}$. Plugging into $(2)$ and simplifying gives
$$p(\vec{t}\vert A)=e^{-[t\_{A}(\vec{t})\alpha+(1-t\_{A}(\vec{t}))\beta]}\alpha^{x/2}\beta^{x/2}\tag{3},$$
where $t\_A(\vec{t})$ indicates the amount of time that trajectory $\vec{t}$ spends in state $A$.
To get your answer, we integrate over the set of all ordered sequences $\vec{t}$ that obey $(1)$ and spend $t\_A$ time in state $A$, which we indicate as $\Omega$. This gives:
\begin{align}P(A,x,t\_{A}\vert A)=\int\_{\Omega}p(\vec{t}\vert A)d\vec{t}=e^{-[t\_{A}\alpha+(1-t\_{A})\beta]}\alpha^{x/2}\beta^{x/2}\int\_{\Omega} d\vec{t}.\tag{4}\end{align}
It remains to calculate the volume of $\Omega$. Let us indicate this set as
$$\Omega=\{t\in\mathbb{R}\_{+}^{x}:0\le t\_{1}\le\dots t\_{x}\le1,\sum\_{i=1}^{x/2}t\_{2i}-t\_{2i-1}=1-t\_{A}\}$$
Note that $\Omega$ has the same volume as the following subset of the unit $x$-simplex:
$$\Omega'=\{g\in\mathbb{R}\_{+}^{x+1}:\sum\_{i=1}^{x+1} g\_{i}=1,\sum\_{i=1}^{x/2}g\_{2i}=1-t\_{A}\},$$
which follows from the simple transformation $g\_{i}=t\_{i}-t\_{i-1}$ for $i=1..x+1$ (where $t\_0=0$ and $t\_{x+1}=1$). Then, by rearranging the odd and even coordinates, it is easy to see that $\Omega'$ is itself a cross product of two scaled unit simplices:
$$\Omega'=\{g\in\mathbb{R}\_{+}^{x/2+1}:\sum\_{i=1}^{x/2+1}g\_{i}=t\_{A}\}\times\{g\in\mathbb{R}\_{+}^{x/2}:\sum\_{i=1}^{x/2} g\_{i}=1-t\_{A}\}$$
The volume of $\Omega'$ is the product of the volumes of these simplices,
$$\mathrm{Vol}(\Omega') =\mathrm{Vol}(\Omega) = \frac{t\_{A}^{x/2}}{(x/2)!}\frac{(1-t\_{A})^{x/2-1}}{(x/2-1)!}$$
Combined with $(4)$ this gives
\begin{align}P(A,x,t\_{A}\vert A)=e^{-[t\_{A}\alpha+(1-t\_{A})\beta]}\alpha^{x/2}\beta^{x/2}\frac{t\_{A}^{x/2}}{(x/2)!}\frac{(1-t\_{A})^{x/2-1}}{(x/2-1)!}.\end{align}
|
1
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https://mathoverflow.net/users/76565
|
406082
| 166,448 |
https://mathoverflow.net/questions/405356
|
0
|
Let $p$ and $q$ be integers.
Let $f(n)$ be [A007814](https://oeis.org/A007814), the exponent of the highest power of $2$ dividing $n$, a.k.a. the binary carry sequence, the ruler sequence, or the $2$-adic valuation of $n$.
Then we have an integer sequence given by
\begin{align}
a(0)=a(1)&=1\\
a(2n)& = pa(n)+qa(2n-2^{f(n)})\\
a(2n+1) &= a(n-2^{f(n)})
\end{align}
Let
$$s(n) = \sum\limits\_{k=2^{n-1}}^{2^{n}-1}a(k), s(0)=1$$
then I conjecture that
$$\sum\limits\_{k=0}^{\infty}s(k)x^k=\frac{1}{1-x}+\sum\limits\_{k=2}^{\infty}\left(\prod\limits\_{j=2}^{k}(q^{j-1}+p\sum\limits\_{r=0}^{j-2}q^r)\right)\frac{x^{2(k-1)}\left(1-(-1+q^{k-1}+p\sum\limits\_{s=0}^{k-2}q^s)x\right)}{(1-x)\prod\limits\_{i=2}^{k}\left(1-(q^{i-1}+p\sum\limits\_{t=0}^{i-2}q^t)x\right)^2}$$
There also exist finite variant:
$$\sum\limits\_{k=0}^{n}s(k)x^k=\frac{1}{1-x}+\sum\limits\_{k=2}^{\left\lfloor\frac{n}{2}\right\rfloor+1}\left(\prod\limits\_{j=2}^{k}(q^{j-1}+p\sum\limits\_{r=0}^{j-2}q^r)\right)\frac{x^{2(k-1)}\left(1-(-1+q^{k-1}+p\sum\limits\_{s=0}^{k-2}q^s)x\right)}{(1-x)\prod\limits\_{i=2}^{k}\left(1-(q^{i-1}+p\sum\limits\_{t=0}^{i-2}q^t)x\right)^2}$$
Is there a way to prove it?
|
https://mathoverflow.net/users/231922
|
Generating function for partial sums of the sequence
|
As proved in [this answer](https://mathoverflow.net/q/405996), if we represent $n$ as $n=2^{t\_1}(1+2^{t\_2+1}(1+\dots(1+2^{t\_m+1}))\dots)$ with $t\_j\geq 0$, then
\begin{split}
a(n) = P(\ell)^{t\_1}\prod\_{j=1}^{\ell-1} P(\ell-j)^{t\_{2j}+t\_{2j+1}+1},
\end{split}
where $\ell:=\left\lfloor\frac{m+1}2\right\rfloor$ and
$$P(k) := q^k+p\frac{q^k-1}{q-1}.$$
Notice that this does not depend on $t\_m$ when $m$ is even.
Grouping the summands in $s(n)$ by the number of unit bits (very much like in this [other answer](https://mathoverflow.net/q/405210)), for $n\geq 1$ we have
\begin{split}
s(n) &= \sum\_{m=1}^n\ \sum\_{t\_1+t\_2+\dots+t\_{m}=n-m}\ P(\lfloor(m+1)/2\rfloor)^{t\_1} \prod\_{j=1}^{\lfloor(m-1)/2\rfloor} P(\lfloor(m+1)/2\rfloor-j)^{t\_{2j}+t\_{2j+1}+1}\\
&= \sum\_{m=1}^n\ [x^{n-m}]\ \frac1{1-P(\lfloor(m+1)/2\rfloor)x}\cdot\prod\_{j=1}^{\lfloor(m-1)/2\rfloor} \frac{P(j)}{(1-P(j)x)^2}\cdot\frac{1+\frac{(-1)^m-1}2x}{1-x}\\
&= [x^n]\ \sum\_{m=1}^\infty\ \frac{x^m}{1-P(\lfloor(m+1)/2\rfloor)x}\cdot\prod\_{j=1}^{\lfloor(m-1)/2\rfloor} \frac{P(j)}{(1-P(j)x)^2}\cdot\frac{1+\frac{(-1)^m-1}2x}{1-x}.
\end{split}
That is, the generating function for $s(n)$ is
$$\sum\_{n\geq 0} s(n)x^n = 1+\frac1{1-x}\sum\_{m=1}^\infty\ \frac{x^m(1+\frac{(-1)^m-1}2x)}{1-P(\lfloor(m+1)/2\rfloor)x}\cdot\prod\_{j=1}^{\lfloor(m-1)/2\rfloor} \frac{P(j)}{(1-P(j)x)^2}.$$
Combining terms for $m=2k-1$ and $m=2k$, we can rewrite it as
$$\sum\_{n\geq 0} s(n)x^n = 1+\frac{1}{1-x}\sum\_{k=1}^\infty\ \frac{x^{2k-1}}{1-P(k)x}\cdot\prod\_{j=1}^{k-1} \frac{P(j)}{(1-P(j)x)^2},$$
which is quite close to the conjectured formula.
|
0
|
https://mathoverflow.net/users/7076
|
406085
| 166,449 |
https://mathoverflow.net/questions/406021
|
4
|
>
> Show that a perverse sheaf on $\mathbb{A}^1(\mathbb{C})$ (the complex plane with the analytic topology) is a bounded complex $A$ of sheaves of $\mathbb{Q}$-vector spaces with constructible cohomology sheaves $\mathcal{H}^n(A)$ such that the following three conditions hold true:
>
>
> 1. $\mathcal{H}^n(A)=0$ unless $n\in \{-1,0\}$,
> 2. $\mathcal{H}^{-1}(A)$ has no nonzero global sections with finite support,
> 3. $\mathcal{H}^0(A)$ is a finite sum of skyscraper sheaves.
>
>
>
Although it seems to be a well-known fact, I have some trouble proving it.
**My attempt so far.** By definition of a perverse sheaf, the following two conditions hold true: for all integers $q$, we have:
$(i)$ $\dim \operatorname{supp} \mathcal{H}^{-q}(A)\leq q$,
$(ii)$ $\dim \operatorname{supp} \mathcal{H}^{-q}(\mathcal{D}A)\leq q$,
where $\mathcal{D}A$ is the Verdier dual of $A$ (we agree that the dimension of the empty set is $-\infty$). From $(i)$, $q=0$, we deduce point 3. For $q<0$, we deduce half of point 1, that is $\mathcal{H}^{n}(A)=0$ for $n>0$. Because $\mathbb{A}^1\_{\mathbb{C}}$ is smooth of dimension $1$, we have
$$\mathcal{D}A=R\mathcal{Hom}(A,\mathbb{Q}\_{\mathbb{A}^1})[2]$$
where $\mathbb{Q}\_{\mathbb{A}^1}$ is the constant sheaf equal to $\mathbb{Q}$ on $\mathbb{A}^1$ placed in degree $0$. Hence
$$\mathcal{H}^{-q}(\mathcal{D}A)=\mathcal{H}^{2-q}(R\mathcal{Hom}(A,\mathbb{Q}\_{\mathbb{A}^1})).$$
However, $\mathbb{Q}\_{\mathbb{A}^1}$ is not an injective complex and nothing ensures me that $\mathcal{H}^{2-q}$ commutes with $R\mathcal{Hom}(-,\mathbb{Q}\_{\mathbb{A}^1})$. How should I pursue my computation?
Many thanks!
|
https://mathoverflow.net/users/66686
|
Perverse sheaves on the complex affine line
|
It sounds like you are stuck computing the stalks of $\mathcal D A$. To do this, you can use that the homology of the stalk of $\mathcal D A$ at $p$ is the dual of $$H^\*(X, X-p ; A) = H^\*(U,U-p;A)$$ for any neighborhood $U$ of $p$. This follows from $i^\*\mathcal D A = \mathcal D i^! A$ and the exact triangle $i\_\* i^! A \to A \to j\_\* j^\* A,$ where $i$ is the inclusion of $p$ and $j$ is the inclusion of the complement.
|
3
|
https://mathoverflow.net/users/382874
|
406087
| 166,450 |
https://mathoverflow.net/questions/406086
|
9
|
I am considering a classical mechanics problem with a fairly complicated system where I think it might be possible to simplify the calculations using the formalism of screw theory and screw algebras, but I cannot find many resources where such a type of calculation is explained in a clear and general way (for example, how to derive equations of motion).
There is some explanation in a few robotics textbooks, but the exposition is based around complicated robot arms, so the detail is somewhat overwhelming.
Screw theory is usually mentioned in passing in the well-known classical mechanics textbooks along with Chasles' theorem but is dismissed as excessively elaborate and relegated to a footnote.
Edit: I checked some of the references mentioned in the comments and would recommend *Geometric Robotics* by Selig as being a clear read.
|
https://mathoverflow.net/users/119114
|
Resources on screw theory in classical mechanics
|
This is my impression. Maybe I have this wrong.
In the following, $\mathbb D^3$ is the free 3D linear space (or *free module*) over the dual numbers. We may use the terms *infinitesimal part* and *standard part* to refer to the two parts of a dual number. An element of this space is referred to as a *screw*.
A screw is a representation of a linear velocity and an angular velocity, together. Given a screw $\mathbf v \in \mathbb D^3$, the expression $\exp(t \mathbf v)$ represents the affine transformation which given a rigid body $b$, subjects it to a rigid body motion at *constant* linear+angular velocity $\mathbf v$ for $t$ units of time. More formally, if $\mathbb D^3$ is the algebra of screws, then there is a mapping $\exp : \mathbb D^3 \to SE(3)$, where $SE(3)$ is the special Euclidean group in 3 dimensions.
Above, the "algebra" $\mathbb D^3$ can either be understood as:
* The Lie algebra $\mathfrak{se}\_3$. In which case, $\exp$ is simply the usual exponential map sending a Lie algebra to a corresponding Lie group.
* The six-dimensional linear subspace of the dual quaternions of the form $ai + bj + ck + a'\epsilon i + b'\epsilon j + c' \epsilon j$. The mapping $\exp$ is then surjective over the *unit* dual quaternions. The unit dual quaternions are isomorphic to $2SE(3)$, which is the double cover of $SE(3)$. The dual quaternions provide a useful formalism for representing rigid body motions, especially in computing applications, in spite of the unintuitive fact that they double-cover the desired group.
* The logarithms of those $4 \times 4$ affine matrices which represent elements of $SE(3)$.
Now I *think* that if $\mathbf v$ is a velocity screw, and $M$ is a $3 \times 3$ dual number matrix with standard part a multiple of the identity matrix and infinitesimal part symmetric, then $\mathbf p = M \mathbf v$ is a momentum screw. This represents a linear and angular momentum together. A force screw can be defined as well (force and torque together). Then Newton's second and third law are the same as the usual ones, verbatim.
In order to go from the screw velocities $\mathbf v(t)$ at any time $t$, and the initial position+orientation of a rigid body $x(0)$ (as an element of $SE(3)$ and not a screw), to the position+orientation $x(t)$ of the body at time $t$, one would need to perform a kind of *product integration* (as opposed to the usual summation integral) using the idea that $x(t + dt) = \exp(\mathbf v(t)\, dt) x(t)$. If $x(t)$ is understood to be either a $4 \times 4$ affine matrix or a dual quaternion, then this becomes the fairly concrete differential equation $\frac{d}{dt}x(t) = \mathbf v(t) x(t)$. Can somebody confirm this?
Note: If $p(0)$ is a point on the rigid body at time $t=0$, then $p(t) = x(t) \cdot p(0)$ is the same point at time $t$, where we have used the notation for a group action to show that $x(t)$ (which is a member of either the group $SE(3)$ or $2SE(3)$) acts on $p(0)$ as a rigid body motion producing $p(t)$.
|
7
|
https://mathoverflow.net/users/75761
|
406088
| 166,451 |
https://mathoverflow.net/questions/405216
|
0
|
Given a productive set, there is a collection of c.e. sets union of which is the productive set, as we know that every c.e. set is with a c.e. function with a index.
My question: is the set of the indices of c.e.sets that cover a productive set also productive one?
|
https://mathoverflow.net/users/14024
|
Is set of the indices of c.e.sets that cover a productive set also productive one?
|
No, it isn't. This is a consequence of the fact that we have a choice of infinitely many equivalent indices for each c.e. set. Let $E$ be a set of indices such that
$$(e, j \in E \land e \neq j) \implies W\_e \neq W\_j $$
$$ P = \bigcup\_{e \in E} W\_e$$
For any $e$ let $e' \neq e$ with $W\_e = W\_{e'}$ (note that if $e \in E$ then $e'$ isn't by assumption). Now we build $\hat{E}$ by meeting requirements $R\_i$ ensuring that $\phi\_i$ isn't a production function for $\hat{E}$.
Let $e\_i$ be an enumeration of elements of $E$ *in order*. Suppose at the start of stage $s$ we've have $\hat{E}\_s = \lbrace \hat{e}\_0, \ldots, \hat{e}\_{n} \rbrace$ Let $j\_s$ be such that $W\_{j\_s} = \hat{E}\_s$ (must exist as it's finite). Now compute $\phi\_s(j\_s)$. If it doesn't converge or $\phi\_s(j\_s) \in \hat{E}\_s$, or $\phi\_s(j\_s) = e\_k, k \leq n$ then we've met $R\_i$ as it's either not in $\hat{E}$ or is in $W\_{j\_s}$. So set $\hat{e}\_{n+1} = e\_{n + 1}$.
If $\phi\_s(j\_s)$ isn't of the form $e\_k$ or $e'\_k$ for some $k$ it similarly won't be in $\hat{E}$ so do the same.
If $\phi\_s(j\_s) = e\_k$ or $\phi\_s(j\_s) = e'\_k$ for $k > n$ then simply set $\hat{e}\_k$ to be equal to whichever one disagrees with $\phi\_s(j\_s)$ so $\phi\_s(j\_s) \neq \hat{e}\_k$ and set $\hat{e}\_l, n < l < k$ to be $e\_l$ giving us $E\_{s+1}$.
--
However, note that if $E$ is the set of all $e$ with $W\_e$ contained in $P$ then I'm pretty sure it will be productive since we can just consider the set of all indexes for the empty c.e. set and notice that we can m-reduce $\bar{0'}$ to that set.
|
1
|
https://mathoverflow.net/users/23648
|
406093
| 166,453 |
https://mathoverflow.net/questions/406073
|
1
|
The number $p\_1(n)$ of [overpartitions of $n$](https://oeis.org/A015128) is generated by
$$\sum\_{n\geq0}p\_1(n)\,q^n=\prod\_{k=1}^{\infty}\frac{1+q^k}{1-q^k}.$$
Let $t\in\mathbb{N}$. Now, extend this to construct a family of sequences $p\_t(n)$ as generated by the product
$$\sum\_{n\geq0}p\_t(n)\,q^n=\prod\_{k=1}^{\infty}\left(\frac{1+q^k}{1-q^k}\right)^t.$$
I would like to ask:
>
> **QUESTION 1.** For each $t$, is the sequence $p\_t(n)$ log-concave in $n$, i.e. $p\_t(n)^2\geq p\_t(n+1)\,p\_t(n-1)$ for all $n>1$?
>
>
>
This question is now answered by Gjergji Zaimi as shown below.
**ADDED.** More curiosity:
>
> **QUESTION 2.** For each $n$, is the sequence $p\_t(n)$ log-concave in $t$, i.e. $p\_t(n)^2\geq p\_{t+1}(n)\,p\_{t-1}(n)$ for all $t>1$?
>
>
>
|
https://mathoverflow.net/users/66131
|
Log-concavity of sequence related to overpartitions
|
This follows from the fact that $p\_1(n)$ is a log-concave sequence, together with the fact that the convolution of log concave sequences is also a log concave sequence.
The first fact is proven in B. Engel "Log-concavity of the overpartition function", Ramanujan J 43, 229–241 (2017).
The second is proven in S. Hoggar "Chromatic polynomials and logarithmic concavity", J. Combin. Theory Ser. B 16 (1974) 248–254.
|
4
|
https://mathoverflow.net/users/2384
|
406094
| 166,454 |
https://mathoverflow.net/questions/406095
|
5
|
Let $\pi:\mathcal X\to B$ be a deformation of a compact complex manifold $X=\pi^{-1}(0)$, then for any $t\in B$, the first Chern class $c\_1(X)=c\_1(X\_t)$?
I know the Chern class of a manifold depends on the complex structure, for example, the same diffeomorphism type of $\mathbb C^1$ with a different complex structure may have a different Chern class, but also the same differential manifold with different complex sturctures may have the same Chern class, for example, the deformations of a Calabi-Yau manifold have the same Chern class $c\_1=0$ while the complex structures are different. So my question is: is the Chern class a deformation invariant?
|
https://mathoverflow.net/users/99826
|
Deformation invariance of Chern classes
|
This is actually true for *all* Chern classes, but you must first say how you identify $H^\*(X,\mathbb{Z})$ and $H^\*(X\_t,\mathbb{Z})$. There is no problem for small deformations, that is if $B$ is a ball (say). Then the restriction maps $H^\*(\mathscr{X},\mathbb{Z})\rightarrow H^\*(X,\mathbb{Z})$ and $H^\*(\mathscr{X},\mathbb{Z})\rightarrow H^\*(X\_t,\mathbb{Z})$ are isomorphisms, so you can consider everything in $H^\*(\mathscr{X},\mathbb{Z})$. But now from the exact sequence $0\rightarrow \pi ^\*T\_B\rightarrow T\_{\mathscr{X}}\rightarrow T\_{\mathscr{X}/B}\rightarrow 0$ and the triviality of $T\_B$ you get that $c\_i(X)$ and $c\_i(X\_t)$ are the restrictions of the same class, namely $c\_i(T\_{\mathscr{X}})$.
|
8
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https://mathoverflow.net/users/40297
|
406099
| 166,455 |
https://mathoverflow.net/questions/405994
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1
|
It is clear that the Verma modules are indecomposable modules for the Virasoro algebra with the $L\_0$-weights bounded. I am wondering if the Verma modules exhaust all such indecomposable modules. Does there exist other types of indecomposable modules whose weights are bounded?
Any information and comments will be appreciated.
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https://mathoverflow.net/users/164854
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Indecomposable modules for Virasoro algebra whose weights are bounded
|
Yes there are other indecomposable modules where $L\_0$ is bounded from below. Logarithmic theories provide many examples.
The simplest example I can think of: take a Verma module generated by a primary state of conformal dimension $\Delta$, and formally take the derivative of this state with respect to $\Delta$. The derived state generates a logarithmic module (i.e. $L\_0$ is not diagonalizable) where $L\_0$ is still bounded from below.
This example works for any $\Delta$. For special values of $\Delta$ such that Verma modules have singular vectors, more complicated constructions are also possible, including non-logarithmic examples. For example, consider a primary state $v\_0$ of dimension zero, and a state $v\_1$ such that
$$
L\_{n\geq 2} v\_1 = 0 \quad , \quad L\_1 v\_1 = v\_0 \quad , \quad L\_0 v\_1 = v\_1
$$
Then the representation generated by $v\_1$ is not a Verma module, as it is strictly larger than the Verma module generated by $v\_0$. Since $v\_1$ is an $L\_0$ eigenvector, $L\_0$ is diagonalizable in that representation. And our representation is indecomposable thanks to our choice of dimension for $v\_0$, which prevents us from decomposing it using a subrepresentation generated by a linear combination of $v\_1$ and $L\_{-1}v\_0$.
|
1
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https://mathoverflow.net/users/12873
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406109
| 166,458 |
https://mathoverflow.net/questions/406112
|
6
|
I have encountered an algebraic group $G$ over $\mathbb C$ such that there is a Zariski open orbit for the adjoint action of $G$ on the nilpotent radical $\mathfrak n$ of its Lie algebra, but there is an ideal $\mathfrak n'\subset\mathfrak n$ such that the adjoint action of $G$ on $\mathfrak n'$ does not have open orbit.
In my example $G$ has dimension $35$, $\mathfrak n$ has dimension $28$, $\mathfrak n'$ has dimension $25$ and the largest dimension of a $G$-orbit on $\mathfrak n'$ has dimension $24$. Details are sort of unpleasant but I am ready to provide them if needed, just ask about it in a comment.
I would like to understand this phenomenon better, so my question is what is the "smallest" example of it, i. e. of a linear algebraic group $G\subset\operatorname{GL}(V)$ over $\mathbb C$ possessing an open orbit and a subrepresentation $V'\subset V$ without any open $G$-orbits.
Note that such representation cannot be completely reducible, so $G$ must be nonreductive.
|
https://mathoverflow.net/users/41291
|
Non-hereditarily locally transitive linear algebraic groups
|
(1) Let $G$ be the pointwise stabilizer of the first coordinate line $L\subseteq V$ in $\mathrm{GL}\_2$ (so $G$ is 2-dimensional, connected, non-reductive). Then $G$ has an open orbit on the plane $V$ (the complement of $L$) but acts trivially on $L$, hence with no open orbit.
This answer the main question, although this is not the action of an algebraic group on its nilpotent radical.
---
Edit: (2) here's now an example, of dimension 6, in which the action is the $G$-action on its nilpotent radical (which coincides with the unipotent radical, as is necessarily the case in any $G$ having an open orbit on its nilpotent radical).
Let $G$ be the semidirect product of $\mathrm{SL}\_2$ with the 3-dimensional Heisenberg group $U$. Then $G$ fixes pointwise the line $[U,U]$. However it has an open orbit on $U$, namely $U\smallsetminus [U,U]$. Indeed, the $G$-action on $U$ is conjugate to the $G$-action on its Lie algebra. On the Lie algebra, $\mathrm{SL}\_2$ acts on $\mathfrak{u}=\mathfrak{a}\_2\oplus\mathfrak{a}\_1$ ($2+1$ decomposition) as $g\cdot (x,z)=(gx,z)$. While $U/[U,U]$ acts on $\mathfrak{u}$ as $y\cdot (x,z)=(x,z+\langle x,y\rangle)$, where $\langle,\rangle$ is a symplectic form on the plane $\mathfrak{a}\_2$. Since for each $x\neq 0$, the map $x\mapsto \langle x,y\rangle$ is surjective, we deduce that for every $x\neq 0$, $(x,z)$ is in the same orbit as $(x,0)$. Since $\mathrm{SL}\_2$ acts transitively on $\mathfrak{a}\_2\smallsetminus\{0\}$, it follows that the $G$-action on $U\smallsetminus [U,U]$ is transitive.
---
(3) Moreover, (2) gives the unique smallest-dimensional example. Indeed, as already mentioned, the existence of an open orbit in the (connected) nilpotent radical forces the nilpotent radical to be equal to the unipotent radical. So, we write $G=U\rtimes L$, with $L$ reductive. By minimality, we can suppose that $L$ acts faithfully on $U$. Then $U$ is non-abelian, since otherwise by semisimplicity, existence of open orbits passes to subspaces. Moreover, $G$ has an open orbit on $U/[U,U]$, and this action factors through $G/U=L$. Hence $\dim(L)\ge\dim(U/[U,U])$.
So, $U$ being non-abelian, has dimension $\ge 3$. If $U$ has dimension $3$, $L$ embeds into a Levi factor in the automorphism group of $U$, that is $\mathrm{GL}\_2$. To be an example, the action should be trivial on the center, so $L\subset\mathrm{SL}\_2$. The inequality above says $\dim(L)\ge 2$. Since $\mathrm{SL}\_2$ has no 2-dimensional reductive subgroup, we get $L=\mathrm{SL}\_2$, which is precisely the example above.
If we are looking at at most 6-dimensional examples, given that $\dim(U/[U,U])\ge 2$ in all cases, the only remaining possibility a priori is $\dim(U,[U,U])=\dim(L)=2$, $U$ 4-dimensional (so at this stage we have already shown $6$ is the minimal dimension).
In dimension $6$, the only possibility for such $U$ is the 4-dimensional filiform unipotent group (which occurs, for instance, as semidirect product (1-dimensional) $\ltimes$ (3-dimensional) abelian unipotent groups, the action being unipotent generated by a Jordan matrix. But for this group, a Levi factor in the automorphism group is precisely a 2-dimensional torus $T$. But in this case the proper $T$-invariant ideals in $\mathfrak{u}$ are: the terms of the lower central series (of dimension 0, 1 and 2, with an open $T$-orbit), and two hyperplanes, which also have an open $G$-orbit. So this is not an example.
|
9
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https://mathoverflow.net/users/14094
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406116
| 166,459 |
https://mathoverflow.net/questions/406122
|
3
|
What is an example of a ring $R$ and a subring $S \subseteq R$ such that $R$ is flat as a right module but not flat as a left module.
The following question is my motivation:
[Faithful flatness for rings](https://mathoverflow.net/questions/403926/faithful-flatness-for-rings)
But please note that I am looking for a more specific counterexample.
|
https://mathoverflow.net/users/351872
|
RIng that is flat over a subring as a right module but not as a left module
|
Take the associative algebra over a field $k$, with generators $x$ and $y$ subject to the relation $xy=0$. This admits a basis consisting of monomials of the form $y^a x^b$. It thus contains a subring $k[x]$, and is flat (even free) over this subring as a right module, on the basis $y^a$, $a\geq 0$. As a left module, it isn't flat, since it isn't even torsion free (for example, $y$ is annihilated by $(x)$).
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10
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https://mathoverflow.net/users/39747
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406123
| 166,460 |
https://mathoverflow.net/questions/406115
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5
|
$\DeclareMathOperator\Lip{Lip}$Let $X$ be a real Banach space. The dual $X^\*$ is a closed subspace of $\Lip\_0(X)$. ($\Lip\_0(X)$ denotes the space of real-valued Lipschitz functions $f:X\to\mathbb{R}$ satisfying $f(0)=0$ with Lipschitz constant as norm).
Is there anything known about the complementability of $X^\*$ inside $\Lip\_0(X)$? If $X$ is finite dimensional, then $X^\*$ is trivially complemented in $\Lip\_0(X)$. Are there any examples of infinite dimensional spaces $X$ whose duals are or are not complemented in $\Lip\_0(X)$?
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https://mathoverflow.net/users/12248
|
Dual Banach space $X^*$ complemented in $\mathrm{Lip}_0(X)$?
|
Corollary 5.4 in my book (*Lipschitz Algebras*, 2nd edition): Let $V$ be a Banach space. Then there is a norm 1 linear projection from ${\rm Lip}\_0(V)$ onto $V^\*$. If $V$ is separable then there is a weak\* continuous norm 1 linear projection.
This follows from Theorem 2 of "On nonlinear projections in Banach spaces" by Lindenstrauss (*Michigan Math J* **11** (1964), 263-287).
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10
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https://mathoverflow.net/users/23141
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406124
| 166,461 |
https://mathoverflow.net/questions/406125
|
5
|
Let $(X,d)$ be a metric space and let $H^\alpha$ denote the $\alpha$-dimensional Hausdorff measure on $X$, where $\alpha$ is the Hausdorff dimension of $X$. Is there any simple condition on $X$ that allow me to conclude that $H^\alpha$ is locally finite??
|
https://mathoverflow.net/users/321882
|
Hausdorff measure
|
Not true for all $X$.
Taking the gauge function $\phi(t) = t^{1/2}/\log|t|$, construct a Cantor set $X$ in $\mathbb R$ using Hausdorff's original method so that $H^\phi(X) = 1$. Then the Hausdorff dimension of $X$ is $1/2$, but the "natural measure" on it is $H^\phi$, not $H^{1/2}$. Now every open set $U \subseteq X$ has $0 < H^\phi(U)<\infty$ and thus $H^{1/2}(U) = \infty$. The whole space $X$ is compact, yet $H^{1/2}(X) = \infty$.
Reference
*Hausdorff, F.*, [**Dimension und äußeres Maß.**](http://dx.doi.org/10.1007/BF01457179), Math. Ann. 79, 157-179 (1918). [ZBL46.0292.01](https://zbmath.org/?q=an:46.0292.01).
English translation in
*Edgar, Gerald A. (ed.)*, Classics on fractals, Studies in Nonlinearity. Boulder, CO: Westview Press (ISBN 0-8133-4153-1/pbk). xii, 366 p. (2004). [ZBL1062.28007](https://zbmath.org/?q=an:1062.28007).
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7
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https://mathoverflow.net/users/454
|
406130
| 166,462 |
https://mathoverflow.net/questions/406127
|
9
|
Throughout this post $G$ denotes $GL\_{n}(\mathbb{F})$ where $\mathbb{F}$ denotes the finite field of $q$ elements.
I'm currently reading the aforementioned book to understand how the irreducible representations of $G$ are constructed. To get started, we need to understand the conjugacy class of $G$.
In the book he explains how given a vector space $V$, for each $g \in G$ we can equip $V$ with an $\mathbb{F}[t]$ module structure by setting $t \cdot v = g(v)$ and extending it linearly. When we are regarding $V$ as an $\mathbb{F}[t]$ module for a specific $g$, we denote it as $V^{g}$.
From here I understand that $V^{g\_{1}} \cong V^{g\_{2}}$ as $\mathbb{F}[t]$ modules if and only if $g\_{1}$ and $g\_{2}$ belong to the same conjugacy class. Therefore if we only care about $V^{g}$ upto isomorphism, we can denote it as $V^{\mathcal{C}}$ where $\mathcal{C}$ is the conjugacy class of $g$.
Next he goes on to prove that for any given conjugacy class $\mathcal{C}$, we have $V^{\mathcal{C}} \cong \displaystyle \bigoplus\_{i = 1}^{s -1} \mathbb{F}[t]/(f\_{i})^{e\_{i}}$ where $f\_{i}$ are irreducible polynomials of $\mathbb{F}[t]$. The proof of this is basically using the structure theorem for finitely generated modules over a PID.
Here is where my confusion begins. He says that for each $f\_{i}$ in the decomposition of $V^{\mathcal{C}}$, we can assign a partition and if I'm understanding correctly, it looks like we can construct a partition for $n$. Let $d\_{i}$ denote the degree of $f\_{i}$. Based on the setup so far, my questions are as follows.
1. What exactly is this partition based on the above decomposition? Is the partition on $d\_{i}$ or is the partition on $n$?
2. Does the partition have anything to do with
\begin{align\*}
\displaystyle \sum\_{i = 1}^{s} d\_{i}e\_{i} = n
\end{align\*}
Any help is greatly appreciated. Thanks! :)
|
https://mathoverflow.net/users/156604
|
Chapter 4 Section 2 of Macdonald's Symmetric Functions and Hall Polynomials
|
The point is that the same polynomial $f\_i$ may occur many times. So for each polynomial $f$, we have a partition whose parts are the exponents $e\_i$ in all the different cases where $f\_i=f$. So this is not a partition of $n$ or of $d\_i$.
Assume for a moment that $f(u)=u-a$ for some $a$; then the resulting partition is the lengths of the Jordan blocks with $a$ on the diagonal. The other cases are similar, but Jordan decomposition is more complicated over a field that isn't algebraically closed.
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9
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https://mathoverflow.net/users/66
|
406133
| 166,465 |
https://mathoverflow.net/questions/406135
|
0
|
We want to prove that
the unique solution to the following difference equation is the null one:
$$
au(x)+b\mathbf{1}\_{(0,\frac{1}{2})}(x)u(x+\frac{1}{2})+c\mathbf{1}\_{(\frac{1%
}{2},1)}(x)u(x-\frac{1}{2})=0,\text{ }x\in (0,1).
$$
Extending $u$ by zero outside $(0,1)$ and taking the Laplace transform yields
$$
a\int\_{0}^{1}e^{-px}u(x)dx+be^{\frac{p}{2}}\int\_{\frac{1}{2}%
}^{1}e^{-px}u(x)dx+ce^{-\frac{p}{2}}\int\_{0}^{\frac{1}{2}}e^{-px}u(x)dx=0,%
\text{ }p\in
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
,
$$
that is
$$
\left( a+be^{\frac{p}{2}}\right) \int\_{\frac{1}{2}}^{1}e^{-px}u(x)dx+\left(
a+ce^{-\frac{p}{2}}\right) \int\_{0}^{\frac{1}{2}}e^{-px}u(x)dx=0,\text{ }%
p\in
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
.
$$
If we let for instance $p=\gamma +4n\pi i,n\in
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ with $a+ce^{-\frac{\gamma }{2}}=0$ we get
$$
\left( a+be^{\frac{\gamma }{2}}\right) e^{-\gamma }\int\_{\frac{1}{2}%
}^{1}e^{-4n\pi ix}u(x)dx=0,
$$
which yields that $u=0$ on $(\frac{1}{2},1)$ if $a+be^{\frac{\gamma }{2}%
}\neq 0$ which is equivalent to $a^2-bc \neq 0$. With the same manner, by choosing this time $p=\delta +4n\pi
i,n\in
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ with $a+be^{\frac{\delta }{2}}=0$ we get
$$
\left( a+ce^{-\frac{\delta }{2}}\right) e^{-\delta }\int\_{0}^{\frac{1}{2}%
}e^{-4n\pi ix}u(x)dx=0,
$$
so if $a+ce^{-\frac{\delta }{2}}\neq 0$ we get $u=0$ on $(0,\frac{1}{2})$, which is equivalent to $a^2-bc \neq 0$.
I don't know if this kind of reasoning is correct since the Laplace
transform of $u$ is zero on subintervals with different choices of $p.$
|
https://mathoverflow.net/users/106804
|
Unique zero solution to a difference equation via Laplace transform
|
Your condition can be rewritten as the system of three equations:
$$au(x)+bu(x+1/2)=0\ \forall x\in(0,1/2), \tag{1}$$
$$au(x)+cu(x-1/2)=0\ \forall x\in(1/2,1), \tag{2} $$
$$au(1/2)=0. \tag{3} $$
In turn, (2) can be rewritten as
$$cu(x)+au(x+1/2)=0\ \forall x\in(0,1/2). \tag{2a} $$
So, if the determinant $a^2-bc$ of the system (1)--(2a) of linear equations for $u(x),u(x+1/2)$ is nonzero, then $u(x)=u(x+1/2)=0$ $\forall x\in(0,1/2)$, that is, $u(x)=0$ $\forall x\in(0,1)\setminus\{1/2\}$. Together with (3), this yields $u(x)=0$ $\forall x\in(0,1)$ if $a\ne0$.
The cases when $a^2-bc=0$ or $a=0$ are considered similarly.
In particular, if $a^2-bc=0$ but $a\ne0$, then $a\ne0$ and $b\ne0$, then the system (1)--(3) reduces to the conditions
$$u(x)=-(b/a)u(x+1/2)\ \forall x\in(0,1/2)\tag{1a}$$ and $u(1/2)=0$. So, here one may assign any values to $u$ on $(1/2,1)$, and then use (1a) to determine the values of $u$ on $(1/2,1)$.
---
The OP requested in a comment that the solution be given in terms of the Laplace transform, say $L$. This can be done as follows.
Of course, to use the Laplace transform, we have to assume that $u$ is integrable on $(0,1)$. Let
$$U(x):=\begin{cases}u(x)&\text{ if }0<x<1/2,\\
0&\text{ if }x>1/2,\end{cases}$$
$$V(x):=\begin{cases}u(x+1/2)&\text{ if }0<x<1/2,\\
0&\text{ if }x>1/2.\end{cases}$$
Then (1) and (2) imply
$$aL(U)+bL(V)=0,$$
$$cL(U)+aL(V)=0.$$
So, if $a^2-bc\ne0$, then $L(U)=L(V)=0$ and hence $U=V=0$ almost everywhere (a.e.), so that $u=0$ a.e., so that $u=0$ if $u$ is continuous.
As shown above, there is no uniqueness if $a^2-bc=0$.
|
1
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https://mathoverflow.net/users/36721
|
406139
| 166,466 |
https://mathoverflow.net/questions/406129
|
2
|
If $X\neq \emptyset$ is a set, we say ${\cal C}\subseteq {\cal P}(X)\setminus \{\emptyset\}$ is a *cover* if $\bigcup{\cal C} = X$, and ${\cal C}$ is said to be *minimal* if for all $D\in {\cal C}$ , the collection ${\cal C}\setminus\{D\}$ is no longer a cover. (Equivalently: for every member $D$ of a minimal cover, there is $x\in X$ such that $x$ is only covered by $D$.)
We say that a topological space $(X,\tau)$ is $\min$-*compact* if every open cover has an open refinement that is a minimal cover of $X$.
It is clear that any compact space is $\min$-compact. Are there implications between $\min$-compactness and metacompactness?
|
https://mathoverflow.net/users/8628
|
$\min$-compactness vs metacompactness
|
Using Zorn's lemma, one can show that every point-finite cover of a set has a minimal subcover. Therefore meta-compactness implies min-compactness. I claim that there are min-compact spaces that are not meta-compact (my example is not $T\_{1}$).
Suppose that $X$ is a topological space.
Then we say that a subset $A\subseteq X$ is a discrete spanner if
1. for each $a\in A$, there is an open set $U\subseteq X$ with $U\cap A=\{a\}$ (in other words, the subspace topology on $A$ is discrete), and
2. whenever $(U\_{a})\_{a\in A}$ is a system of open sets with $U\_{a}\cap A=\{a\}$ for each $a\in A$, we have $\bigcup\_{a\in A}U\_{a}=X$.
If $A$ is a discrete spanner, then condition 2 can be strengthened to say that every collection of open subsets of $X$ that covers $A$ must also cover $X$.
Proposition: If $X$ has a discrete spanner, then $X$ is min-compact.
Proof: Suppose that $\mathcal{U}$ is a cover of $X$. Then for each $a\in A$, suppose that $a\in U\_{a}\in\mathcal{U}$. Furthermore, suppose that $V\_{a}$ is a set that is open in $X$ where $V\_{a}\cap A=\{a\}$. Then $(U\_{a}\cap V\_{a})\_{a\in A}$ is a cover of $X$, but $(U\_{a}\cap V\_{a})\_{a\in A}$ is minimal since if $a,b\in A,b\in U\_{a}\cap V\_{a}$, then $a=b$. Q.E.D.
Proposition: Every $T\_{1}$-space with a discrete spanner is discrete.
The above proposition will follow from more general results that we shall prove below.
If $X$ is a topological space, then we give $X$ a pre-ordering $\leq$ known as the specialization ordering where $x\leq y$ if and only if $\overline{\{x\}}\subseteq\overline{\{y\}}$. Recall that the specialization ordering $\leq$ is a partial ordering if and only if $X$ is $T\_{0}$, and the specialization ordering $\leq$ is simply equality if and only if $X$ is $T\_{1}$.
Proposition: Let $X$ be a topological space. If $A\subseteq X$ is a discrete spanner for $X$, then $A$ is the set of all minimal elements with respect to the specialization ordering and for each $x\in X$, there is an $a\in A$ with $a\leq x$.
Proof: Suppose that $A$ is a discrete spanner. I first claim that for each $x\in X$, there is an $a\in A$ with $a\leq x$. Suppose to the contrary, $x\in X$ but there is no $a\in A$ with $a\leq x$. Then $(\overline{\{x\}})^{c}$ is an open set with $A\subseteq(\overline{\{x\}})^{c}$ but where $(\overline{\{x\}})^{c}\neq X$ which is a contradiction. Now, each element in $A$ is minimal. Otherwise, there would be $a,b\in A$ with $a<b$, and this would contradict the minimality of $A$. Furthermore, $A$ must contain all minimal elements since if $x\in X$ is minimal, then $x\geq a$ for some $a\in A$, so $x=a$ by minimality. Q.E.D.
Proposition: Suppose that $X$ is a topological space. Then a subset $A\subseteq X$ is a discrete spanner for $X$ if and only if
1. $A$ is discrete,
2. $A$ is the collection of all minimal elements in $X$ with respect to the specialization ordering, and
3. for each $x\in X$, there is some $a\in A$ with $a\leq x$.
Proof: We have already proven $\rightarrow$. For $\leftarrow$, it suffices to show that whenever $U\_{a}\cap A=\{a\}$ for each $a\in A$, we have $\bigcup\_{a\in A}U\_{a}=X$. However, if $x\in X$, then there is some $a\in A$ with $a\leq x$. Therefore, we have $x\in U\_{a}\subseteq\bigcup\_{a\in A}U\_{a}$. We can conclude that $\bigcup\_{a\in A}U\_{a}=X$. Q.E.D.
Suppose that $X$ has a discrete spanner $A$. Then $X$ must be min-compact, but $X$ is not necessarily metacompact. In fact, $X$ is metacompact if and only if
there is a system of open sets $(U\_{a})\_{a\in A}$ with $\{a\}=U\_{a}\cap A$ for each $a\in A$ but where each $x\in X$ is contained in only finitely many sets of the form $U\_{a}$.
For example, if $X$ is a topological space with a discrete spanner $A$, and there exists some $x\in X$ where there are infinitely many $a\in A$ with $a\leq x$, then $X$ may not be metacompact, but $X$ is min-compact.
|
2
|
https://mathoverflow.net/users/22277
|
406140
| 166,467 |
https://mathoverflow.net/questions/406090
|
2
|
Suppose I have a (incomplete) theory $T$ (e.g. PA) which I skolemize to get a theory $T\_S$ in the expanded language. I now build $T'$ by adding to $T\_S$ any sentence $(\forall x)\phi(x)$ where I can prove (say using ZFC as my background set theory) that for all terms $t$ in my skolemized language $ T\_S \vdash\phi(t)$
In some sense $T'$ is kinda what you get if you assume that all objects are constructed via skolem functions.
Is $T'$ in general a conservative extension of $T$? When $T$ is PA?
Obviously, $T'$ has to be consistent as it's true of the model one gets via the standard Skolem construction over $T$ but I have no idea if it's conservative. Since this is a pretty natural idea I presume this has been studied before and I'd love any pointer to where I can read more about the properties that $T'$ or further iterates ($T''$ etc..) would have.
|
https://mathoverflow.net/users/23648
|
Is adding all sentences true of terms in skolemized theory conservative?
|
So I can mark this answered (when I can tmw) I'm posting [Emil Jeřábek's](https://mathoverflow.net/users/12705/emil-je%c5%99%c3%a1bek) comment as an answer but they deserve all the credit.
Suppose that $(\forall x)\phi(x)$ is in $T' - T\_S$. That means we've proved that $T\_S \vdash \phi(t)$ for each term $t$ in our skolemized language. Now define
$$\theta(y) = \left(\lnot\phi(y) \lor (\forall x)\phi(x)\right)$$
$(\exists y)\theta(y)$ is a tautology so it's in $T$ and when we skolemize we introduce a constant $c$ and place $\theta(c)$ into $T\_S$. By above (and assuming background theory doesn't prove false claims) $T\_S \vdash \phi(c)$. Hence, as $\theta(c) \land \phi(c) \implies (\forall x)\phi(x)$ we have that $T\_S \vdash (\forall x)\phi(x)$. Hence, $T'=T\_S$
As $T\_S$ was itself conservative over $T$ this answers the question.
|
2
|
https://mathoverflow.net/users/23648
|
406145
| 166,468 |
https://mathoverflow.net/questions/406146
|
1
|
Whenever we have a continuous map of topological spaces $f: X \to Y$ and a sheaf $\mathcal{F}$ on $Y$ (of abelian groups for example), we get an induced pullback map
$$ H^n(Y, \mathcal{F}) \to H^n(X, f^\* \mathcal{F}), $$
which is described for example in Iversen, *Cohomology of Sheaves*, II.5.1. I would like to have an explicit description of this map in one particular case.
Suppose $X = Y = S^1$ and further suppose that the sheaf is a local system $\mathcal{L}$. Then its pullback $f^\* \mathcal{L}$ is also a local system. We have a nice description of local systems in terms of representations of the fundamental group, and the pullback is easily describable in this language: if
$$ \pi\_1(Y) \to \mathrm{Aut}(\mathcal{L}\_t) $$
is the representation defining $\mathcal{L}$, then $f^\* \mathcal{L}$ is described by the composition
$$ \pi\_1(X) \xrightarrow{f\_\*} \pi\_1(Y) \to \mathrm{Aut}(\mathcal{L}\_t). $$
The cohomology groups of a local system on $S^1$ are also nicely described in terms of representations: if we denote by $T \in \mathrm{Aut}(\mathcal{L}\_t)$ the image of a generator of $\pi\_1 (S^1)$, then
$$ H^0(S^1, \mathcal{L}) \cong \mathrm{Ker}(T - id) $$
$$ H^1(S^1, \mathcal{L}) \cong \mathrm{Coker}(T - id) $$
$$ H^n(S^1, \mathcal{L}) = 0), \quad \text{for any } n>1 $$
This is proved using Cech cohomology.
Now the homotopy type of our map $f: S^1 \to S^1$ is determined by its degree $d$, and the automorphism of the stalk that defines $f^\* \mathcal{L}$ is just $T^d$. Since $\mathrm{Ker}(T - id) \subset \mathrm{Ker}(T^d - id)$ it seems tempting to say that the pullback map on $H^0$ is the inclusion. However I don't know how to prove this, because I haven't found any result about some possible functoriality of Cech cohomology with respect to the covering. Furthermore, I don't know what the map on $H^1$ could look like. The first candidate I can think of would be some projection between the cokernels, but $\mathrm{Im}(T - id) \subset \mathrm{Im}(T^d - id)$ seems unlikely because of the first isomorphism theorem (if the kernels get larger then the images should get smaller).
|
https://mathoverflow.net/users/173545
|
What is the pullback morphism on sheaf cohomology of local systems in terms of representations of the fundamental group?
|
The circle $S^1$ is a $K(\mathbb{Z},1)$ space, so the cohomology of a local system on it can be identified with group cohomology. Let $R=\mathbb{Z}[T,T^{-1}]$ be the group ring of $\mathbb{Z}$. Given a local system $\mathcal{L}$,
$$H^i(S^1,\mathcal{L})= Ext\_R^i(\mathbb{Z}, \mathcal{L})$$
This can be computed as using the formulas you wrote, because
$$0\to R\xrightarrow{T-1}R\to \mathbb{Z}\to 0$$
is free resolution.
The degree $d$ map on the circle corresponds to $f:R\to R$ by $T\mapsto T^d$. This determines a map on group cohomology. This does not seem compatible with the above resolution, bu you can compute it in other ways. Using derivations mod inner derivations for $H^1$ is probably better for what you want.
|
3
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https://mathoverflow.net/users/4144
|
406151
| 166,470 |
https://mathoverflow.net/questions/406149
|
6
|
Consider a sufficiently nice topological space $X$ as well as topological groups $G$ and $H$. Consider the functor $F$ that associates to $X$ the set of all isomorphism classes of all principal $H$-bundles $\pi\_H\colon P\_H\twoheadrightarrow P\_G$ on all principal $G$-bundles $\pi\_G\colon P\_G\twoheadrightarrow X$ on $X$. I think that this would be representable by the [Brown representability theorem](https://en.wikipedia.org/wiki/Brown%27s_representability_theorem), i.e. there is a topological space $C\_{G,H}$ such that $[X,C\_{G,H}]$ is in bijection with the isomorphism classes of such bundles-over-bundles ([modulo pointed-vs-unpointed spaces](https://mathoverflow.net/questions/285533/classifying-spaces-and-browns-representability-theorem)). But is there an explicit description/characterisation of it?
I am specifically interested in the case where $G=\operatorname U(1)^n$ is a torus (and $H$ is compact Lie); the motivation is that this would describe the so-called [Kaluza–Klein reductions](https://en.wikipedia.org/wiki/Kaluza%E2%80%93Klein_theory) of gauge fields on nontrivially fibred torus compactifications.
As a special case, the subset of $F(X)$ that corresponds to $H$-bundles on the trivial $n$-torus bundle $\operatorname U(1)^n\times X$ is classified by the $n$-fold iterated free loop space $\mathcal L^n\mathrm BG$; and in the case $n=1$ this object is [well studied](https://ncatlab.org/nlab/show/free+loop+space+of+classifying+space). But I don't know how to generalise to arbitary principal torus bundles.
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https://mathoverflow.net/users/411120
|
Classifying space of bundles over bundles
|
If I understand the question right, I think the classifying space can be described like this. Let $Map(G,BH)$ be the space of all continuous maps from $G$ to $BH$. Make the group $G$ act continuously on that function space (using the usual free transitive action of $G$ on the space $G$), and form the associated bundle over $BG$:
$$
EG\times\_G Map(G,BH).
$$
|
8
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https://mathoverflow.net/users/6666
|
406155
| 166,471 |
https://mathoverflow.net/questions/406118
|
7
|
Let $\mathcal{C}$ be a category and $S$ a collection of morphisms in $\mathcal{C}$. The localisation $\mathcal{C}[S^{-1}]$ is defined via a functor $F: \mathcal{C} \longrightarrow \mathcal{C}[S^{-1}]$ and the following universal property. For all $\varphi\in S$ the morphism $F(\varphi)$ is an isomorphism. Now let $G: \mathcal{C} \longrightarrow \mathcal{D}$ be another functor such that $G(\varphi)$ is an isomorphism for all $\varphi \in S$. Then $G$ factors via $HF$ for another functor $H: \mathcal{C}[S^{-1}] \longrightarrow \mathcal{D}$. One can explicitly construct $\mathcal{C}[S^{-1}]$ via the calculus of fractions under some conditions on $S$.
Now my question is the following: In the text above replace "isomorphism(s)" by "morphism(s) with right inverse". Are there similar conditions on $S$ such that we can do the same construction via something like calculus of fractions?
|
https://mathoverflow.net/users/145920
|
Localisation of categories, but instead of "isomorphisms" I want "morphisms with right inverse". Construction via calculus of fractions possible?
|
I think there is a relatively good reason why such a thing shouldn't exists.
In general when you freely add right inverse or inverse, the general arrows of the resulting category will be zig-zag in the original category where the arrow in the wrong directions are all in S, and are there to represent composition with the freely added inverse.
The assumption for a calculus of fraction essentially provide a way to "move" these wrong direction arrow in S to one side of the zig-zag to make it just a span or cospan.
The problem in your situation is that you are not putting any relation between the various freely added right inverse that you add, contrary to the case of actual inverse where inverse being automatically unique any relation in the original category will give rise to relations between the inverse in the localized category.
Typically, you'll assume that starting from a cospan $A \to B \leftarrow C $ where the second arrow is in S, then you can complete it to a commutative square with one of the arrow also in S, which allow to replace the cospan by a span (or the other way around depending on if you do a left or right calculus of fraction)
But when you'll try to do the same sort of rewrite with one sidded inverse to put all the arrow of S in a zig-zag on one side, you have no way to ensure that said rewrite is going to be equal to the zig-zag you started from (in the example above that going through the span or the cospan using the right inverse yields the same result).
Of course if you start adding a lot of relation between the inverses then some thing might become possible, but that will give you a very different universal property...
|
6
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https://mathoverflow.net/users/22131
|
406157
| 166,473 |
https://mathoverflow.net/questions/406158
|
-1
|
Let $X$ be a Banach space. Let $\mathcal{F}$ be the family of all the bounded subsets of $X$. If $I$ is the identity map on $X$, we shall denote by $\operatorname{span}\{I\}$ the vector space generated by $I$.
In [Measures of Weak Compactness and Fixed Point Theory](https://arxiv.org/abs/math/0310422), Barroso and O'Regan defined a $\phi$-space. Inspired by their definition we introduce the concept:
>
> Let $\psi\in \mathcal{L}(X)$ be such that $\psi\notin \operatorname{span}\{I\}$. We will say that $X$ is a $\psi$-space if the following condition is satisfied:
>
>
> **if $C \in \mathcal{F}$ and $\psi(C)$ is relatively weakly compact, then $C$ is relatively compact.**
>
>
>
---
I want to study this class of functions and describe it, I'm looking for more characterization/examples of the $\psi$-space.
I will appreciate your help.
|
https://mathoverflow.net/users/102228
|
Definition of a $\psi$-Banach space
|
The definition makes no sense due to the mixing up of "relatively" (weakly) compact and (not relatively) compact.
I guess that what you mean is:
$\psi$ is strongly-weakly proper on closed balls (that is, the intersection of preimages of weakly compact sets with closed balls are compact). For strongly-weakly continuous maps and strong/weak measures of noncompactness $s,w$ on $X$, the latter can be rephrased as
$$w(\psi(C))=0\implies s(C)=0\quad\text{for every bounded set $C$.}$$
A sufficient condition for this is therefore the existence of a constant $c>0$ with
$$w(\psi(C))\ge c\,s(C)\quad\text{for every bounded set $C$.}$$
For the strong topology (and strong measures of noncompactness), both properties are well studied.
BTW, both are properties of the map $\psi$ and not of the Banach space $X$, so I find the notion $\psi$-space very strange.
To be honest, I doubt that there are any useful examples of maps satisfying any of these two properties for the strong-weak topology. It is not even clear whether such a map $\psi$ does exist at all (even if one considers nonlinear $\psi$).
For instance, in a reflexive infinite-dimensional Banach space the image of every bounded open set has to be unbounded which is a weird condition if one requires strong-weak continuity. Maybe in some exotic non-reflexive spaces, it is easier to find an example $\psi$, but to find any example at all is certainly non-trivial (if at all possible).
|
2
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https://mathoverflow.net/users/165275
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406162
| 166,474 |
https://mathoverflow.net/questions/357785
|
5
|
Is there a simple construction of a Sullivan minimal model $\Lambda U \rightarrow V$ in the case that $H^1(V)\neq 0$? Do you have a reference? I envisage a degree-wise construction as in the case of $H^1(V)=0$ explaining how to deal with the new phenomenon. The existence is proven very generally in *Felix, Halperin, Thomas, Rational Homotopy Theory* but I find the proof very complicated and not transparent enough in this simple case (they prove that relative Sullivan models exist and that any relative Sullivan algebra is isomorphic to a product of a minimal relative Sullivan algebra and an acyclic algebra)
|
https://mathoverflow.net/users/43645
|
Sullivan minimal model in the case of $H^1(V)\neq 0$
|
Gelfand and Manin explain it very nicely in their book "Methods of Homological Algebra", last chapter.
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3
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https://mathoverflow.net/users/6635
|
406165
| 166,476 |
https://mathoverflow.net/questions/406164
|
4
|
If the product of two functions is smooth, then how quickly must one function decay when the other is non-smooth? Suppose we have two functions $f,h$ on $\mathbb{R}$ such that:
* $h$ is Lipschitz continuous
* $f$ is smooth (i.e. $C^\infty$), and
* $fh$ is a smooth function.
What can we conclude about $f$ from this requirement? Presumably $f$ must go to zero 'sufficiently fast' at any point where $h$ is non-smooth.
For example, we can compute that a.e. $(fh)' = f'h + fh'$ and since $f'h$ is continuous, $h' = ((fh)' - f'h)/f$ is continuous in any neighborhood where $f$ is non-zero. Therefore $f$ is zero anywhere $h$ is not $C^1$. Must $f^{(k)}$ be zero anywhere $h$ is not locally smooth? Must $f^{(k)}h$ be smooth? Must $f^{(k)}h^{(\ell)}$ be well-defined (i.e. extend to be 0 at the set where $h$ is not $C^\ell$) and smooth?
In my application, $h$ is also smooth in an open, dense set, which may give some context though I doubt it helps with these.
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https://mathoverflow.net/users/112378
|
If $fh$ is smooth and $h$ Lipschitz, what can be said about $f$?
|
$\newcommand\R{\mathbb R}\newcommand\N{\mathbb N}$Here are answers to your three questions (the latter two of them partial).
**Answer 1:** Yes, for any real $a$ and any $k\in\{0,1,\dots\}$,
\begin{equation\*}
\text{if $h$ does not have all the derivatives at $a$, then $f^{(k)}(a)=0$.}\tag{1}
\end{equation\*}
Indeed, say that $h$ is bad at $a$ (or, equivalently, that $a$ is a bad point of $h$0 if $h$ does not have all the derivatives at $a$.
Take any $a\in\R$. Suppose that $h$ is bad at $a$. Take then the smallest $k\in\{0,1,\dots\}$ such that $f^{(k)}(a)\ne0$, if such a $k$ exists. Let
\begin{equation\*}
g:=fh
\end{equation\*}
and
\begin{equation\*}
F(x):=\frac{f(x)}{(x-a)^k}, \quad G(x):=\frac{g(x)}{(x-a)^k}
\end{equation\*}
for real $x\ne a$, with $F(a):=f^{(k)}(a)/k!$ and $G(a):=g^{(k)}(a)/k!$. By a Taylor formula, if $k\ge1$, then for all real $x$
\begin{equation\*}
F(x)=\frac1{(k-1)!}\int\_0^1(1-s)^{k-1}f^{(k)}(a+(x-a)s)\,ds
\end{equation\*}
and hence for all nonnegative integers $l$
\begin{equation\*}
F^{(l)}(x)=\frac1{(k-1)!}\int\_0^1(1-s)^{k-1}s^lf^{(k+l)}(a+(x-a)s)\,ds,
\end{equation\*}
with the similar formulas for $G(x)$ and $G^{(l)}(x)$.
So, if $k\ge1$, then $F$ and $G$ are smooth, $F(a)\ne0$, and hence $F\ne0$ and $h=G/F$ on a neighborhood $V$ of $a$ (the equality $h(a)=G(a)/F(a)$ follows by continuity); the same conclusions obviously hold for $k=0$.
So, $h$ is smooth on $V$, which contradicts the assumption that $h$ is bad at $a$. So, as claimed, there is no $k\in\{0,1,\dots\}$ such that $f^{(k)}(a)\ne0$.
We have actually proved more than (1):
\begin{equation\*}
\text{if $a$ is in the closure of the set of all bad points of $h$, } \\
\text{then $f^{(k)}(a)=0$ for all $k\in\{0,1,\dots\}$.}
\end{equation\*}
---
**Answer 2 (partial):**
Now it follows that $g\_k:=f^{(k)}h$ must be differentiable. Indeed, take any real $a$. If $h$ has all the derivatives at $a$, then so does $g\_k$. If $h$ does not have all the derivatives at $a$, then, by (1), $f'(x)=o(|x-a|)$ as $x\to a$, so that $g\_k(x)=o(|x-a|)$ as $x\to a$, so that $g\_k'(a)=0$.
---
**Answer 3 (partial):**
Finally, let $g\_{k,l}:=f^{(k)}h^{(l)}$ wherever $h^{(l)}$ exists, with $g\_{k,l}:=0$ elsewhere. Take any real $a$ such that $h$ does not have all the derivatives at $a$. Note that $h'$ is bounded on the set where $h'$ exists, since $h$ is Lipschitz.
By (1), $f^{(k)}(x)=o(|x-a|)$ as $x\to a$, so that $g\_{k,1}(x)=o(|x-a|)$ as $x\to a$, and hence $g\_{k,1}'(a)=0$. So, $g\_{k,1}$ is differentiable.
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4
|
https://mathoverflow.net/users/36721
|
406168
| 166,478 |
https://mathoverflow.net/questions/406175
|
6
|
Let $ G $ be a simple linear algebraic group. Let $ G\_\mathbb{R} $ be the real points of $ G $. Let $ G\_\mathbb{Z} $ be the integer points of $ G $. Is $ G\_\mathbb{Z} $ a maximal closed subgroup? In other words, are the only closed subgroups $ H $ of $ G\_\mathbb{R} $ such that
$$
G\_\mathbb{Z} \subset H \subset G\_\mathbb{R}
$$
just $ G\_\mathbb{R} $ and $ G\_{\mathbb{Z}} $?
The answer is yes for for $\operatorname{SO}\_3$. $ \operatorname{SO}\_3(\mathbb{Z}) $ is the 24 element octohedral group (isomorphic to the symmetric group $ S\_4 $) and this group is a maximal finite subgroup of rotations and indeed it turns out that adding any other rotation and taking the closure will generate all of $ \operatorname{SO}\_3(\mathbb{R})$.
The group of integer points is also maximal in $ \operatorname{SL}\_n(\mathbb R) $. See the [answer](https://math.stackexchange.com/a/3390666) by YCor given to [Is $\operatorname{SL}\_2(\mathbb Z)$ a maximal discrete subgroup in $\operatorname{SL}\_2(\mathbb R)$?](https://math.stackexchange.com/questions/3388661/is-sl-2-mathbb-z-a-maximal-discrete-subgroup-in-sl-2-mathbb-r).
Although this only states that $ \operatorname{SL}\_n(\mathbb{Z}) $ is maximal among discrete subgroups I'm guessing that it is in fact maximal among closed subgroups as well in this case.
|
https://mathoverflow.net/users/387190
|
Is the group of integer points of a simple real linear algebraic group a maximal closed subgroup?
|
Your guess is correct: if $H$ is a closed subgroup containing $\operatorname{SL}\_n({\mathbb Z})$,then it is a Lie subgroup. If $\mathfrak h$ is its Lie algebra, then $\mathfrak h$ is stable under the adjoint action of $SL\_n(\mathbb Z)$ and hence under all of $\operatorname{SL}\_n(\mathbb R)$ (since $\operatorname{SL}\_n(\mathbb Z)$ is Zariski dense in $\operatorname{SL}\_n(\mathbb R)$). If $\mathfrak h$ is non-zero, that means, by the simplicity of $\operatorname{SL}\_n(\mathbb R)$, that $\mathfrak h$ is the Lie algebra of $\operatorname{SL}\_n(\mathbb R)$ and hence $H$ is all of $\operatorname{SL}\_n(\mathbb R)$.
If $\mathfrak h =0$, then $H$ is discrete and you have accepted the result in this case.
In the generality that you have asked, $G(\mathbb Z)$ need not be maximal in $G(\mathbb R)$; if $G$ is a simply connected semi-simple algebraic group over $\mathbb Q$ (and is $\mathbb Q$-simple), with $G(\mathbb R)$ non-compact, and $K\_p$ is a maximal compact open subgroup of $G(\mathbb Q\_p)$ for each prime $p$, then the intersection of $G(\mathbb Q)$ with all the $K\_p$ may be "called" $G(\mathbb Z)$ and the same argument goes through to say that $G(\mathbb Z)$ is maximal among closed subgroups of $G(\mathbb R)$.
|
7
|
https://mathoverflow.net/users/23291
|
406177
| 166,482 |
https://mathoverflow.net/questions/406176
|
9
|
Let $F\_n$ be a free group of rank $n \geq 2$. The group $F\_n$ acts on its commutator subgroup $[F\_n,\, F\_n]$ by conjugation. Let $G = [F\_n,\, F\_n] \rtimes F\_n$. It's not hard to see that $G$ is finitely generated.
>
> **Question**: Is $G$ finitely presentable? Presumably not, but I can't seem to prove it.
>
>
>
|
https://mathoverflow.net/users/411409
|
Finite presentability of semi-direct product of free group and its commutator subgroup
|
This group is not finitely presentable. Indeed, write $F$ for the given free group and $F'$ for its derived subgroup. The map
$$F\ltimes F'\to F\times F,\quad (f,g)\mapsto (f,fg)$$
is an injective group homomorphism, and its image is the fibre product of two copies of $F$ over the abelianization map $F\to F/F'$. (Alternatively, start from this fibre product, and observe that it is the semidirect product $H\ltimes K$, where $H\simeq F$ is the diagonal and $K=F'\times\{1\}$.)
It is a result of Baumslag and Roseblade that a subgroup of a direct product of two free groups is "almost never" finitely presented, i.e. not finitely presentable unless one of the projection is injective on the subgroup, or if it is virtually the direct product of intersection with the factors. Since this is not such an exception, the above group is not finitely presentable.
Reference: G. Baumslag and J. E. Roseblade, Subgroups of direct products of free groups, J. London Math. Soc. (2) 30 (1984), 44-52. [link](https://watermark.silverchair.com/s2-30-1-44.pdf?token=AQECAHi208BE49Ooan9kkhW_Ercy7Dm3ZL_9Cf3qfKAc485ysgAAAs4wggLKBgkqhkiG9w0BBwagggK7MIICtwIBADCCArAGCSqGSIb3DQEHATAeBglghkgBZQMEAS4wEQQMwl7Az8Gm-1Aoh0m6AgEQgIICgXL9-Y-fyUaMyul-VSo28p2ygbYHC3OaDDC0tZUL7PTcTHZc0-M_IIjukqs47Fxkhln44ybAh97Y7-QpifPTc0qNqoq5OVpFYyp3Klok_kGXnHS98io2mtVb-b6uXrGrQmnG5uEZVoZrNNQ8m9P5gDbXV2oCuBul6GE6StDbM9nPlOch-r8Bd4GF9_iUWH6UN1ILHT-NvnjhTdhdVblGu7hnULKGq1Jf_U30vCXGAYpNM_xdLN-QznjXjRpMNlgjPlY8-kCtX1RDVSPjqQiCUb8Q8AMDIxRh2-qWIKnZdYD-1djeYYmPz52KQ01eKmdaivOjZDauzeQWMVBGAwPy7b_GA1gOiuOTlHDC3avPA-BTXjwA00s7xsZIDm4AEPbEyD1Sh6zRaYacGDLjlUqQhFm14CunJkO_vYxepGI3o945EQ6InYOmzwMejo9ZplDhPknjCHVAANkVSQCBaBp0PxkwuKI2rq8PRFjEpndzk1VhRQdB91vh9UDiiUUJ1eRkmUO7RfAmpDKOtx37rJMuHb8LgcB5yt2QMizOjhvBhPadamlvlGB5PpqPg-MGQANOK4tzGl5vZTxBPx8YY45dLlw79kyp0vqXm3YwJmRW14oR77rN-h3mluTzzDuVGxl-1TyN23TlH_19trKximYz5dYu1PXG1DLlkAg-vQDEdySLci_gCd7cZ88ZKLh0-8ENJSg9ynlas-WTUO4wkjIDTRmd4Vz9_7KXw9ixUXLFjKFZxkBMDSNpgBiMhXYjob3AMus-QVNwY471ijwFIe2HwVkRQPUia5ZBh51-cwhi_W3FI98Z2DRhcg0XtybqRVfpZpK-xA8K744jVi5Ekl3dATJX) [DOI behind paywall](https://doi.org/10.1112/jlms/s2-30.1.44)
PS: the same argument works equally if $N$ is an arbitrary nontrivial normal subgroup of infinite index in $F$: then $F\ltimes N$ is not finitely presentable.
|
9
|
https://mathoverflow.net/users/14094
|
406187
| 166,484 |
https://mathoverflow.net/questions/406171
|
21
|
I've heard tell the following anecdote involving Pierre Gabriel and [Jacques Tit](https://euromathsoc.org/news/jacques-tits-1930---2021-46) at least twice in a lapse of four years or so:
>
> When P. Gabriel presented the theorem in a conference [sometime around 1970], he said something like this: "OK, this algebra is of finite representation type; thus, the algebra has a linear quiver and you get $n+1$ indecomposable modules (in the case $A\_{n}$). In the case $D\_{n}$, you have [an exact expression depending on $n$] indecomposable modules. There are the exceptional cases $E\_{6}, E\_{7}$, and $E\_{8}$". Before he mentioned the number of indecomposable modules in the case of $E\_{6}$, somebody in his audience said aloud the correct number. Again, before he mentioned the number of indecomposable modules in the case of $E\_{7}$, the same individual in the audience gave the correct figure. This situation prompted P. Gabriel to ask: "Who is answering there?" As it turned out, that person was J. Tits. When P. Gabriel asked him if he already knew the theorem, J. Tits replied thus: "No, I don't know it. I was only giving the number of roots of the Lie algebra associated to the corresponding Dynkin diagram."... What was the relationship between the indecomposable modules over algebras and the roots of the Lie algebras? The answer was not very clear back then...
>
>
>
Even though the previous rendering of the story may be a bit inexact (I heard it this way (approximately) in a talk by [Professor J. A. de la Peña](https://es.wikipedia.org/wiki/Jos%C3%A9_Antonio_de_la_Pe%C3%B1a) in 2017), does anybody know whether an interchange of this sort between P. Gabriel and J. Tits actually took place? If so, would you be so kind as to share with us the most precise version of it whereof you are aware, or failing that, a pointer to the literature wherein one can find a detailed account of this story?
Let me thank you in advance for your attention and knowledgeable replies.
|
https://mathoverflow.net/users/1593
|
Reference request: a tale of two mathematicians
|
Here is a recent talk by Ringel (in German):
[Algebra und Kombinatorik](https://www.math.uni-bielefeld.de/%7Eringel/lectures/unger-text/hagen.htm).
The related part is:
>
> Besonderes Aufsehen hat ein Ergebnis erregt, das meist Satz von Gabriel genannt wird: *Ein zusammenhängender Köcher ist genau dann darstellungsendlich, wenn er vom Typ A\_n, D\_n, E\_6, E\_7, E\_8 ist.* Und es gibt den Zusatz: *In diesem Fall sind die Dimensionsvektoren der unzerlegbaren Darstellungen gerade die postitiven Wurzeln des zugehörigen Dynkin-Diagramms.* Gabriel selbst nannte den Satz **Satz von Yoshii**: Yoshii, ein Schüler von Nakayama hatte ein entsprechendes, aber fehlerhaftes Ergebnis publiziert (er behauptete, dass ein weiterer Fall, nämlich Fall E\_7~, darstellungsendlich sei). Parallel zu Gabriel, oder sogar etwas früher, haben auch Bäckström (Stockholm) und Kleiner (Kiev) die darstellungsendlichen Köcher bestimmt. Gabriels Fassung erregte vor allem deswegen großes Aufsehen, weil er den Zusammenhang zu Wurzelsystemen, also zur Lie-Theorie herausstellte - aber dieser Aspekt der Theorie stammte gar nicht von ihm selbst, sondern von Tits, der damals auch in Bonn lehrte. Das Auftreten der Dynkin-Diagramme hat viele fasziniert. Die Konstruktion der entsprechenden Hall-Algebren lieferte später einen direkten Zusammenhang zwischen der Darstellungstheorie von Köchern und den Kac-Moody Lie-Algebren: eine "Kategorifizierung" der Wurzelsysteme.
>
>
>
My translation skills are bad but roughly it is said that the connection to the root systems is not due to Gabriel but due to Tits and both were teaching at that time in Bonn, where this might have happened. Gabriel himself called this result the "theorem of Yoshii" (Yoshii was a student of Nakayama).
|
18
|
https://mathoverflow.net/users/61949
|
406188
| 166,485 |
https://mathoverflow.net/questions/406043
|
4
|
Let $\Gamma$ be a connected graph, let $\lambda \ge 1$ and $c \ge 0$ be some constants. Recall that a combinatorial path $p$ in $\Gamma$ is said to be *$(\lambda,c)$-quasigeodesic* if for every combinatorial subpath $q$ of $p$ one has $$\ell(q) \le \lambda d(q\_-,q\_+)+c,$$
where $\ell(q)$ is the length of $q$, $q\_-$ and $ q\_+$ are the endpoints of $q$, and $d(\cdot,\cdot)$ is the standard metric on $\Gamma$.
**Question 1:** suppose that a path $s$ in $\Gamma$ has been obtained from a $(\lambda,c)$-quasigeodesic path $p$ by replacing some (combinatorial) subpaths of $p$ with geodesics. Is it true that this new path $s$ is again $(\lambda,c)$-quasigeodesic?
Intuitively, the answer should be "Yes", because replacing subpaths with geodesics ("shortcutting") should only improve the quasigeodesicity constants. However, I do not see how to prove this.
If the answer to Question 1 is negative, the natural next question is the following:
**Question 2:** suppose that a path $s$ in $\Gamma$ has been obtained from a $(\lambda,c)$-quasigeodesic path $p$ by replacing some subpaths of $p$ with geodesics. Is it true that this new path $s$ is $(\lambda',c')$-quasigeodesic, where the constants $\lambda' \ge 1$, $c' \ge 0$ depend only on $\Gamma$, $\lambda$ and $c$?
It's not hard to show that the answer to Question 2 is positive when the graph $\Gamma$ is $\delta$-hyperbolic, but I do not have a proof or a counter-example for more general graphs.
|
https://mathoverflow.net/users/7644
|
Shortcutting quasigeodesics
|
As suggested, I am turning my comments into an answer. The answer to both questions is negative for any $\lambda > 1$, and positive for $\lambda = 1$.
For $\lambda = 1+\epsilon$ note that in $\mathbb Z^2$, the concatenation of the (unique) geodesics from $(0,0)$ to $(0,k)$, from $(0,k)$ to $(n,k)$ and from $(n,k)$ to $(n,0)$ is $(1+\epsilon,0)$-quasigeodesic given $\frac{2k}{n} < \epsilon$.
If we replace the segment between $(0,0)$ and $(n-1,k)$ by the geodesic from $(0,0)$ via $(n-1,0)$ to $(n-1,k)$ then the points $(0,n-1)$ and $(0,n)$ have distance $1$ in $\mathbb Z^2$, but distance $2k+1$ on the path. So this path can't be $(\lambda',c')$-quasigeodesic unless $\lambda'+c' \geq 2k+1$.
For $\lambda = 1$, note that a path $p$ is $(1,c)$-quasigeodesic if and only if $\ell(p) \leq d(p\_-,p\_+)+c$; in other words, it suffices to check the condition for the whole path rather than for all sub-paths. This property is clearly preserved when replacing any subpath by a shorter or equally long one.
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4
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https://mathoverflow.net/users/97426
|
406192
| 166,487 |
https://mathoverflow.net/questions/406197
|
0
|
Let us look at the subspace of smooth complex functions of $L^2(\mathbb{R}^n,\mathbb{C})$, call $H^s$ the Sobolev spaces. By the diamagnetic inequality $\lvert \nabla \lvert\psi\rvert\rvert (x) \le \lvert\nabla \psi\rvert(x)$ (a proof is [here](https://physics.stackexchange.com/questions/100948/is-there-a-physical-intuition-for-diamagnetic-inequality)), we have
\begin{align\*}
\lVert\,\lvert\psi\rvert\,\rVert\_{H^s} \le c\_s \lVert \psi\rVert\_{H^s}
\end{align\*}
for $s=1$ and $c\_1 = 1$, where $c\_1$ does not depend on $\psi$. It is also true for $s=0$ with $c\_0 = 1$. Do we have such a result for $s=-1$, with $c\_{-1} < + \infty$?
|
https://mathoverflow.net/users/nan
|
'Diamagnetic' inequality for negative Sobolev spaces
|
This is not true in the case of $H^{-1}(\Omega) = H\_0^1(\Omega)^\*$ (real spaces), where $\Omega \subset \mathbb R^d$ is bounded:
* In <https://math.stackexchange.com/questions/336834/decomposition-of-functionals-on-sobolev-spaces>, we see that $|\psi|$ cannot be defined for all $\psi \in H^{-1}(\Omega)$.
* Even if $|\psi|$ can be defined (as a measure), it might not belong to $H^{-1}(\Omega)$, see <https://math.stackexchange.com/questions/1402697/decomposition-of-measures-acting-on-sobolev-spaces>
* Even if $|\psi| \in H^{-1}(\Omega)$, we do not get a bound. On $\Omega = (0,1)$, you can consider $\psi\_n(x) = \sin(n x)$ as an element in $H^{-1}(\Omega)$. Then, $\|\psi\_n\|\_{H^{-1}} \to 0$, but $\||\psi\_n|\|\_{H^{-1}} \not\to 0$.
I would expect that the same arguments can be adopted to your situation.
|
0
|
https://mathoverflow.net/users/32507
|
406201
| 166,490 |
https://mathoverflow.net/questions/406190
|
4
|
I am interested in reading about existence and regularity theorems for elliptic equations on manifolds with negative (constant) curvature outside a compact subset. I am aware of some results in this line for the Dirichlet problem at infinity for harmonic functions (see [M. T. Anderson's work on Laplace-Beltrami operator](https://projecteuclid.org/journals/journal-of-differential-geometry/volume-18/issue-4/The-Dirichlet-problem-at-infinity-for-manifolds-of-negative-curvature/10.4310/jdg/1214438178.full)) but I need similar results for more general elliptic equations. I am also aware of some similar results on compact manifolds (for example, the ones in Taylor's 'Partial Differential Equations') but nothing for noncompact ones. Concretely, I am trying to solve the inhomogeneous Helmholtz equation:
$\Delta\_g u+\lambda u=f$, possibly with restrictions over $\lambda$ or the regularity of $f$, with homogeneous Dirichlet conditions at the ideal boundary $\mathbb{S}^{d-1}(\infty)$.
I tend to believe that properties of the ends of this kind of manifold play an important role in solving elliptic PDEs and would really appreciate any reference on this topic that could help me understand PDE in this specific kind of noncompact manifolds. My final aim is to prove the existence of a Green's function for the Dirichlet equation on an Asymptotically Hyperbolic Manifold, which I am trying by changing a little [Li and Tam's construction for the Laplacian](https://www.jstor.org/stable/2374588?seq=1#metadata_info_tab_contents). Therefore, I will also appreciate any literature about Green functions for general elliptic equations on manifolds with negative curvature (outside a compact set).
|
https://mathoverflow.net/users/411616
|
Elliptic equations in asymptotically hyperbolic manifolds
|
The definite reference for this is the monograph by John Lee "Fredholm operators and Einstein metrics on conformally compact manifolds", Mem. Am. Math. Soc. Series Profile 864, 83 p. (2006), that you can also find on the arXiv: <https://arxiv.org/abs/math/0105046>
This is limited to "natural" differential operators but can be easily extended, see e.g. my work with A. Sakovich (where we also introduce a new class of function spaces that is quite convenient in the context of PDEs on AH manifolds): "A large class of non-constant mean curvature solutions of the Einstein constraint equations on an asymptotically hyperbolic manifold", Commun. Math. Phys. 310, No. 3, 705-763 (2012).
|
2
|
https://mathoverflow.net/users/24271
|
406203
| 166,491 |
https://mathoverflow.net/questions/406161
|
6
|
I am very curious to study [arXiv:1310.7930](https://arxiv.org/abs/1310.7930) (henceforth:DCCT) but am not sure if I have the pre-requisites. I am familiar with basic algebraic topology (singular cohomology, classifying spaces, characteristic classes), differential geometry (bundles, connections, Chern--Weil theory) and some elementary homotopy theory (model categories, spectra, generalised cohomology theories). I am comfortable with the Physics aspects (classical and quantum field theories).
However it appears that in order to understand DCCT, I should know the basics of
* Homotopy Type Theory
* Toposes
* Higher Toposes
* Infinity categories
I have no clue about these subjects. A detailed study of each of these subjects appears to be a humongous task. So I wish to study these subjects with a view to learn as much is necessary for DCCT (and possibly slightly more).
My questions are:
**1. What are the subjects which I should familiarise myself with, in order to be able to read DCCT?**
(I am not sure if my listing above is accurate, I am just guessing by a glance at the TOC.)
and
**2. Could you please suggest a roadmap to learning the required basics for a person with my background?**
As far as possible, I prefer to try to learn stuff linearly and minimise back-and-forth. So a roadmap of pre-requisites would be of great help.
It would be very kind and helpful if you could give pointed references to specific chapters instead of whole books. Thanks a lot!
|
https://mathoverflow.net/users/40386
|
Learning roadmap to 'Differential cohomology in a cohesive $\infty$ topos'
|
I haven't read all of DCCT so take this with a grain of salt, but after having spent a lot of time with it, this is how I would recommend getting started on the abstract stuff.
First one must learn classical Grothendieck Topos Theory. Chapter 1.2 of DCCT gives a pretty good motivation and some nice examples of sheaves on $\mathsf{Cart}$, but I would recommend David Carchedi's [course on Topos theory](https://math.gmu.edu/%7Edcarched/topos.html), which is the quickest course that I could find that covered most of the relevant material.
After learning Grothendieck topos theory one must then learn how to combine homotopy theory with topos theory, this was originally achieved using simplicial sheaves or presheaves, and then was abstracted to the definition of a model topos. For this there is
* Jardine's book [Local Homotopy Theory](https://www.springer.com/gp/book/9781493922994) (only the first few chapters are really necessary),
* Dan Dugger's papers: [Hypercovers and Simplicial Presheaves](https://arxiv.org/abs/math/0205027), [Weak Equivalences of Simplicial Presheaves](https://arxiv.org/abs/math/0205025), [Universal Homotopy Theories](https://arxiv.org/abs/math/0007070), and his unfinished expository paper that Dmitri Pavlov linked to.
* Toen & Vessozi's [paper](https://arxiv.org/abs/math/0207028),
* Rezk's [notes](https://faculty.math.illinois.edu/%7Erezk/homotopy-topos-sketch.pdf) on model topoi, and this [master's thesis](https://algant.eu/documents/theses/vergura.pdf) on model topoi. These presentations are often used to provide examples of objects in higher topos theory.
* I'd also recommend these wonderful [notes](https://dmitripavlov.org/notes/2020s-6322-handwritten.pdf) from Dmitri Pavlov's class. I think I would actually start here.
Now one must learn Infinity Topos Theory. This is harder to recommend resources for as there are much fewer. There are many places to find recommendations for resources on learning infinity category theory, but honestly you don't need to delve into too much of the details to understand much for DCCT, you can really take much of infinity category theory as a black box that just works like usual category theory but with equivalences instead of isomorphisms and homotopy type mapping spaces instead of hom sets. I'd recommend Rezk's [notes](https://faculty.math.illinois.edu/%7Erezk/quasicats.pdf) on quasicategories, get comfortable with the basics, and then watch Rezk's [lectures](https://www.youtube.com/watch?v=f3bdYrxTa1E&ab_channel=InstitutdesHautes%C3%89tudesScientifiques%28IH%C3%89S%29) on Youtube, as well as [Joyal's](https://www.youtube.com/watch?v=Ro8KoFFdtS4&t=2517s).
I'm not saying you need to look at all of these resources, but what you need to know is contained in them. I would start off by first looking at some of Schreiber's previous papers that he mixed into DCCT, like [Cech Cocycles for Differential Characterstic Classes](https://arxiv.org/abs/1011.4735) and the Principal Infinity Bundles papers [1](https://arxiv.org/abs/1207.0248) and [2](https://arxiv.org/abs/1207.0249).
**Addendum**:
Morally, the use of higher topos theory in DCCT is as a generalization of the nonabelian cohomology of Grothendieck and Giraud. In DCCT, these generalized cohomology classes are given by higher principal bundles. The use of higher category theory makes the formulation of this theory very elegant, but ultimately it is grounded in the theory of stacks and gerbes. Knowing this theory is not necessary for understanding DCCT or HTT, but it is a great way to build motivation and see what a lot of this theory is actually being used for. Here are some references in the differential geometry setting:
* I'd start with Noohi's really short [notes](http://www.maths.qmul.ac.uk/%7Enoohi/papers/quick.pdf),
* Ginot's [notes](https://webusers.imj-prg.fr/%7Egregory.ginot/papers/DiffStacksIGG2013.pdf),
* Behrand and Xu - [Differentiable Stacks and Gerbes](https://arxiv.org/abs/math/0605694),
* and probably the most relevant to DCCT's formalism is Carchedi's [thesis](https://math.gmu.edu/%7Edcarched/Thesis_David_Carchedi.pdf).
|
5
|
https://mathoverflow.net/users/124010
|
406206
| 166,492 |
https://mathoverflow.net/questions/406204
|
1
|
**Motivation.** At a recent gathering of co-workers, somebody said that he thinks that of all the people present that he knows, about half are happy with the current COVID19 measures, and the other half are not. Which made me put this into a graph question: Given a finite, simple, undirected graph $G = (V,E)$, we say that G is "perfectly balanceable" if there is a set $C\subseteq V$ such that for every $v\in V$, exactly half of its neighbors are in $C$. Clearly, if a vertex has an odd number of neighbors, the graph cannot be perfectly balanceable. So, below I try to introduce a "discrepancy measure" that indicates how far a graph is from being perfectly balanceable.
**Formal version.** Let $G=(V,E)$ be a finite, simple, undirected graph. For $v\in V$, let $N(v)=\{w\in V: \{v,w\}\in E\}$. For $v\in V$ and $C\subseteq V$ let the *discrepancy* of $v$ with respect to $C$ be $$\text{d}\_C(v) = \big|\#\big(N(v) \cap C\big) - \#\big(N(v) \cap(V\setminus C)\big)\big|\;,$$
where $\#(\cdot)$ denotes the size of a (finite) set. The *maximal discrepancy* of any vertex with respect to $C$ is denoted by $$\text{md}\_G(C) = \max\{\text{d}\_C(v): v\in V\}.$$
Clearly, we have $\text{md}\_G(C) = \text{md}\_G(V\setminus C)$ for all $C\subseteq V$. The *overall discrepancy* of $G$ finally is defined by $$\text{discr}(G) = \min\{\text{md}\_G(C):C\subseteq V\}.$$
For example, it is not hard to see chat $\text{discr}(C\_4) = 0$ for the circle on $4$ vertices: If the circle is $0 - 1 - 2 - 3 - 0$, take $C = \{0,1\}$ for getting $\text{md}\_{C\_4}(C) = 0$ and therefore $\text{discr}(C\_4) = 0$. On the other hand, for any graph $G$ with a vertex $v$ such that $N(v)$ has an odd number of elements, we have $\text{discr}(G) > 0$.
**Question.** Given a positive integer $n>0$, is there a graph $G=(V,E)$ such that $\text{discr}(G) = n$?
|
https://mathoverflow.net/users/8628
|
Discrepancy number of a graph
|
Yes, there is such a graph. Let $A = \{1,\dots,2n\}$, let $B = \{b \subseteq A \colon |b|=n\}$, and let $G$ be a graph with vertex set $A \uplus B$ and an edge between $a\in A$ and $b\in B$ if $a \in b$.
For any set $C$ there are either $n$ vertices in $A \cap C$ or $n$ vertices in $A\setminus C$, and thus there is a vertex $b \in B$ with $d\_C(b) = n$.
|
3
|
https://mathoverflow.net/users/97426
|
406209
| 166,494 |
https://mathoverflow.net/questions/406200
|
1
|
Hilbert introduced a construct $\epsilon x. P(x)$ for a predicate $P$ such that
$$\exists x. P(x) \implies P(\epsilon y.P(y))$$
Obviously, this is equivalent to the axiom of global choice. With this operator, we can form complex propositions conveniently:
$$Q(\epsilon x. P(x)) \text{ equiderivable to } \forall x. P(x)\implies Q(x)$$
assuming $\exists x. P(x)$. I'm wondering if the dual can be achieved. I.e. Can there be an operator $\delta x. P(x)$ such that
$$Q(\delta x. P(x)) \text{ equiderivable to } \exists x. P(x)\land Q(x)$$
I suspect that the difficulty encountered in defining this is because we implicitly assume universal closure, but not *existential* closure. Is there any material related to this?
|
https://mathoverflow.net/users/136535
|
A dual to Hilbert's $\epsilon$ operator
|
Nothing like this can happen.
Take $P(x)\equiv\top$, and let $S$ be any formula such that both $\exists x S(x)$ and $\exists x\neg\ S(x)$ are derivable. Then both $$(\*)\_1:\quad\exists x(P(x)\wedge S(x))$$ and $$(\*)\_2:\quad \exists x(P(x)\wedge\neg S(x))$$ are themselves derivable. However, we can't possibly have a single term (or term-like piece of syntax) $\delta$ such that $Q(\delta)$ is equiderivable with $\exists x(P(x)\wedge Q(x))$ for **either** choice of $Q\in\{S,\neg S\}$: each sentence of the latter type is outright derivable, while since $\delta$ is independent of $Q$ we can't have both $S(\delta)$ and $\neg S(\delta)$ be derivable.
For a concrete example, what should your $\delta$ do in the setting of the natural numbers and first-order Peano arithmetic when we take $P(x)\equiv\top$ and $Q(x)\equiv$ "$x$ is even"?
|
4
|
https://mathoverflow.net/users/8133
|
406210
| 166,495 |
https://mathoverflow.net/questions/406225
|
6
|
I'm just studying Lie algebras. If $A$ is a $k$-algebra (not necessarily Lie or associative, just a bilinear law), it is straightforward to check that any derivation algebra of $A$ is a Lie algebra. I suppose that the converse is not true, but I can't find a counterexample.
1. Is there any example of a Lie algebra which is not isomorphic to the derivation algebra of any $k$-algebra (again, not necessarily associative or unital)?
2. Is there any example of a Lie algebra which is not isomorphic to the derivation algebra of any Lie algebra?
For the the second question the user YCor gives a positive answer. However, I am more interested in the first (and actually my original) question. Thank you!
|
https://mathoverflow.net/users/173535
|
Is there any example of a Lie algebra which is not a derivation algebra?
|
$\newcommand{\r}{\mathfrak{r}}\newcommand{\h}{\mathfrak{h}}\newcommand{\g}{\mathfrak{g}}\newcommand{\a}{\mathfrak{a}}$The 2-dimensional abelian Lie algebra $\a\_2$ is not isomorphic to the derivation Lie algebra of any Lie algebra.
Suppose otherwise. Let $\g$ be such a Lie algebra. We discuss according to whether the inner derivation algebra (that is $\g$ modulo its center) has dimension $d$ equal to 2, 1, or 0.
* if $d=2$, $\g$ modulo its center has dimension 2. So the derived subalgebra has dimension 1. If $[\g,\g]$ is contained in the center we deduce that $\g$ is 2-step-nilpotent, and isomorphic to the direct product of the Heisenberg Lie algebra $\h\_3$ with some abelian Lie algebra. In this case, the derivation algebra is much larger (for $\h\_3$ it is 6-dimensional). Otherwise $\g$ is the direct product of the two-dimensional non-abelian Lie algebra $\r$ with an abelian Lie algebra. But then its derivation algebra contains $\r$, hence is not abelian.
* $d=1$ is impossible, because the center in a Lie algebra can't be of codimension 1 (for the same reason the quotient of a group by its center can't be cyclic)
* $d=0$ means that $\g$ is abelian. But then its derivation algebra is either infinite, or has dimension some square.
|
8
|
https://mathoverflow.net/users/14094
|
406227
| 166,499 |
https://mathoverflow.net/questions/406213
|
4
|
**Context and mathematical maturity:** I have knowledge of the usual engineering math courses, meaning differential+integral+vector calculus, linear algebra, probability and statistics, etc. and some pure math courses like analysis. I am a pure mathematics enthusiast, so I have been working through some texts in pure maths like Abbott's analysis and Pinter's Abstract algebra.
**Request:** I would like a recommendation for a geometry class. I know there are many different areas of Geometry, and I have no idea of where to start. Since I don't know much about geometry I would like a relatively introductory book, but much preferably if it is not at a high school level. Hopefully the book would also have some exercises so I can also practice what I read.
I have seen some people recommending Euclid's elements, but to be honest I don't find it to be super easy to read (I guess it is mostly about the language), and I have also seen visual differential geometry by Needham, which seems interesting, but I am not sure whether that is a **good place to start** learning about Geometry.
**Goal:** My dream would be to read Hilbert's Geometry and Imagination and not be completely clueless of what is happening on the book.
Thanks all for your recommendations!
|
https://mathoverflow.net/users/324468
|
Geometry book recommendation
|
Your requirements: "much preferably if it is not at a high school level" and
"My dream would be to read Hilbert's Geometry and Imagination" are contradictive. The book of Hilbert and Cohn-Vossen, Geometry and imagination IS on the high school level.
Other highly recommended geometry books on the same level are
Marcel Berger, Geometry, (in 2 volumes), and Robin Hartshorne, Geometry: Euclid and beyond, and "Foundations of Projective geometry" by the same author. All these books require no prerequisites.
|
2
|
https://mathoverflow.net/users/25510
|
406229
| 166,500 |
https://mathoverflow.net/questions/403966
|
3
|
In a comment made by Gjergji Zaimi to [this](https://mathoverflow.net/q/61615/167834) older question, it is conjectured that $\frac 73$ is the threshold separating countability and uncountability of the sets of infinite binary words having a given [critical exponent](https://en.wikipedia.org/wiki/Critical_exponent_of_a_word). In a later comment, James Currie cites the (remarkable!) paper:
"J. Karhumaki and J. Shallit. Polynomial versus exponential growth in repetition-free binary words. J. Combin. Theory. Ser. A 105 (2004), 335–347."
as the one proving the conjecture. Is it so, though?
In general, it is not true that, if every element of $S\subset\{0,1\}^\omega$ has at most polynomially many finite subwords of length $n$, then $|S|\le \aleph\_0$. In fact, it is sufficient to allow $n+1$ subwords of length $n$ to get an uncountable set, as exemplified by the well-known [Sturmian words](https://en.wikipedia.org/wiki/Sturmian_word). So the conjecture ie either false or there must be a more specific argument involving critical exponents to establish the conjecture. What is it?
|
https://mathoverflow.net/users/167834
|
Words with critical exponent $< \frac 73$
|
[From user Ben's comment]
In the paper "The ubiquitous Prouhet-Thue-Morse sequence" ([DOI link behind paywall](https://doi.org/10.1007/978-1-4471-0551-0_1), [preprint version](https://cs.uwaterloo.ca/%7Eshallit/Papers/ubiq15.pdf)), Allouche and Shallit claim that there are uncountably many overlap-free binary words, which implies that the threshold is 2. I believe this follows from Fife's theorem.
|
1
|
https://mathoverflow.net/users/14094
|
406232
| 166,502 |
https://mathoverflow.net/questions/406224
|
0
|
Let $s, \delta \in (0,1)$. Consider the outer measure on $\mathbb{R}$, $\mu^s\_{\delta}$, defined by
\begin{align\*}
\mu^s\_{\delta}(E):=\inf \left\{\sum\_{j}\lvert I\_{j}\rvert^s: E \subset \bigcup\_{j} I\_{j}: I\_{j} \text { closed intervals, } \lvert I\_j\rvert\leq\delta\right\}.
\end{align\*}
For an interval $I \subset \mathbb{R}$, $\lvert I\rvert$ denotes the length of $I$. I want to prove that if $E$ is an interval and $\delta< \lvert E\rvert$, then
\begin{align\*}
\mu^s\_{\delta}(E) \geq \delta^{s-1}\left|E\right|-\delta^s.
\end{align\*}
I think that by definition, $\exists \epsilon>0$ such that $\mu^s\_{\delta}(E)+\epsilon\geq \sum\_{j}\left|I\_{j}\right|^s\geq \left(\sum\_{j}\left|I\_{j}\right|\right)^s\geq \left|E\right|^s>\delta^s.$ I wonder how to get $\delta^{s-1}$. I feel like I need to show that the delta cover is the smallest one among all the best covers, thankswhich is achieved through infimum. Such a delta cover is a little bit larger than $\delta^{s-1}\left|E\right|-\delta^{s}$.
|
https://mathoverflow.net/users/197849
|
An outer measure defined on $\mathbb {R}$
|
$\newcommand\de\delta\newcommand\ol\overline$Your goal cannot be attained in general. Indeed, suppose that $\de\in(0,1/2)$. Take any interval $E$ of length $\de\_1:=|E|\in(\de,2\de)$. Then for the closure $\ol E$ of $E$ and some closed intervals $I\_1$ and $I\_2$ of lengths $|I\_1|=\de$ and $|I\_2|=\de\_1-\de\le\de$ we have $E=I\_1\cup I\_2$, so that
$$\mu\_\de^s(E)\le|I\_1|^s+|I\_2|^s
=\de^s+(\de\_1-\de)^s\underset{\de\_1\downarrow\de}\longrightarrow
\de^s<\de^{s-1}-\de^s,$$
so that for some interval $E$ with $\de<|E|$ we have
$$\mu\_\de^s(E)\not\ge\de^{s-1}-\de^s.$$
(It is actually easy to see that for any interval $E$ of length $t$ we have $\mu\_\de^s(E)=k\de^s+(t-k\de)^s$, where $k:=\lfloor t/\de\rfloor$.)
|
3
|
https://mathoverflow.net/users/36721
|
406236
| 166,504 |
https://mathoverflow.net/questions/406248
|
0
|
Let $H$ and $K$ be infinite dimensional (separable) Hilbert spaces and $X=B(H,K)$ denote the space of bounded linear operators. For $T\_1, T\_2$ in $X$, we define $D\_{T\_1,T\_2}:X \to X$ as $D\_{T\_1,T\_2}(T)=TT\_1^\*T\_2$. Finally define $V^0=\operatorname{span}\{D\_{T\_1,T\_2}, T\_1, T\_2 \in X\}$. One can check that $V^0$ is a pre-$C^\*$-algebra with involution $D\_{T\_1,T\_2}^\*=D\_{T\_2,T\_1}$. Let $V$ denotes the closure of $V^0$ inside $B(X)$.
>
> Is $V$ isomorphic to some well known $C^\*$-algebra?
>
>
>
P.S: This question was first posted on Math Stackexchange [here](https://math.stackexchange.com/questions/4274449/trying-to-recognise-a-c-algebra). Also, this question is particular case of a more general construction given at [Trying to understand construction of $C^\*$-algebra corresponding to a ternary $C^\*$-ring from a paper](https://mathoverflow.net/questions/402950/trying-to-understand-construction-of-c-algebra-corresponding-to-a-ternary-c).
|
https://mathoverflow.net/users/129638
|
Trying to recognise a $C^*$-algebra
|
Assuming $H$ is separable? If $K$ is finite dimensional and $H$ is infinite dimensional then this gives you the compact operators on $H$. Otherwise you get $B(H)$.
If $H$ can have uncountable dimension then there are more possibilities because there are more closed ideals of $B(H)$.
Some details: $B(H)$ acts on $X = B(H,K)$ by multiplication from the right, and this isometrically embeds $B(H)$ in $B(X)$. If ${\rm dim}(H) \leq {\rm dim}(K)$ then the operators of the form $T\_1^\*T\_2$ with $T\_1, T\_2 \in X$ comprise all of $B(H)$. If $H$ is infinite dimensional and $K$ is finite dimensional then all operators of the form $T\_1^\*T\_2$ are compact, and as long as ${\rm dim}(K) \geq 1$ they include all rank 1 operators, so their closed span equals the compact operators.
|
4
|
https://mathoverflow.net/users/23141
|
406253
| 166,513 |
https://mathoverflow.net/questions/406265
|
2
|
Call a function$$\mathbb{N}\times \mathbb{N}\to \{0, 1\}, \quad (n, m)\to f(n, m)$$computable in polynomial time in $\log n+\log m$ a PTIME family.
Given a PTIME family $f$ call a computable function $g:\mathbb{N}\to \mathbb{N}$ such that $f(n, g(n))=1$ for all $n\in \mathbb{N}$ a solution of $f$.
Is there a PTIME family $f$ that has at least one solution and such that for each solution $g$ and each polynomial $P$ there exists $n\in \mathbb{N}$ such that the running time of $g$ on $n$ is more than $P(\log n+\log g(n))$?
The reason for taking $\log n+\log g(n)$ is to ensure that neither input nor output dominates the running time of the algorithm.
$\log 0=0$ by convention.
|
https://mathoverflow.net/users/nan
|
Family of PTIME sets where it is hard to name elements
|
This is an open problem.
If $\mathrm{TFNP\ne FP}$, then a TFNP problem outside FP directly gives what you call a PTIME family (with solutions polynomially bounded in terms of the input) whose solutions cannot be computed in polynomial time.
If P = NP, then no such problem exists: the assumption implies that given $n$ and $m$, we can compute in polynomial (in $\log n$ and $\log m$) time a solution $y$ such that $f(n,y)=1$ and $y\le m$ if it exists (see e.g. Arora&Barak, Theorem 2.18). Let’s call this algorithm $h(n,m)$. Then successively call $h(n,1)$, $h(n,2)$, $h(n,4)$, $h(n,8)$, ... until you find a solution; this will take time polynomial in $\log n$ and $\log y$ where $y$ is the least solution, because the last call of $h(n,m)$ will be made with $m\le 2y$.
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4
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https://mathoverflow.net/users/12705
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406266
| 166,516 |
https://mathoverflow.net/questions/405670
|
1
|
I've been trying to write a test function for Fibonacci pseudo-primes with large $n$. Fibonacci pseudoprimes are composite numbers such that $V\_n(P,Q) \equiv P \mod n$ for $P>0$ and $Q =\pm 1$, with $V\_n$ Lucas sequence.
As such I need to compute Lucas sequence for large $n$. They are defined by:
$U\_n(P,Q)=\frac{a^n-b^n}{a-b}$ et $V\_n(P,Q)=a^n+b^n$ with $a,b=(P\pm \sqrt{P^2-4Q})/2$
Direct computation is not possible because I need to work with integers and $\sqrt{P^2-4Q}$ may not be one.
There are also binomial formulations of the form $U\_n(P,Q)=2^{1-n}\sum\_{k=0}^{\lfloor (n-1)/2 \rfloor}\binom{n}{2k+1}P^{n-2k-1}(P^2-4Q)^k$. But computing $U\_n$ that way seems to be $O(n^2)$ which is not acceptable for me.
Are there any ways to compute $U\_n$ and $V\_n$ in $O(1)$ or $O(\log n)$ which involve only integers and not floating point ?
Thanks in advance
|
https://mathoverflow.net/users/402326
|
Computing Lucas sequence for large n
|
Following Emil remark, I used other relations to obtain a $O(\log n)$ algorithm:
$\begin{cases}U\_{2n}=U\_nV\_n\\
V\_{2n}=V\_n^2-2Q^n\\
U\_{2n+1} = U\_{n+1}V\_n - Q^n\\
V\_{2n+1} = V\_{n+1}V\_n -PQ^n\end{cases}$
This concluded in the following python code:
```
def lucas_sequence(p, q, n):
if n == 0:
return (0, 2)
if n == 1:
return (1, p)
if n == 2:
return (p, p**2 - 2* q)
ns2 = n // 2
un, vn = lucas_sequence(p, q, ns2)
if n % 2 == 0:
return (un*vn, vn**2 - 2 * (q**ns2))
else:
unp1, vnp1 = lucas_sequence(p, q, ns2 + 1)
return (unp1*vn - q**ns2, vnp1*vn - p * (q**ns2))
```
|
0
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https://mathoverflow.net/users/402326
|
406272
| 166,517 |
https://mathoverflow.net/questions/406270
|
4
|
Context: I'm trying to deal with presentations in the framework of Gonthier et al. formalization of the group theory in the proof assistant Coq. It was used to machine check the Feit-Thompson odd order theorem. In this formalisation, all groups are assumed to be **finite** and it would be a lot of work to remove the assumption. Actually, roughly speaking, all **sets** are assumed to be finite.
Summary question: Is there a finite group $G$ with a presentation where one of the relation can be deduced by the kownledge that $G$ is finite.
I'd like to define a presentation as follows:
Let $I$ be a finite set and $R$ be a finite collection of word in $I$. For any finite group $G$, and any map $g:I\to G$, I say that $(g, R)$ is a presentation of $G$ iff
1. $\{g(i) | i \in I\}$ generate $G$;
2. The relations $R$ holds in $G$, that is for any $r\in R$, then $\prod\_{i : r} g(i) = 1$;
3. For any **finite** group $H$ and any map $h : I \to H$, if $\prod\_{i : r} h(i) = 1$ holds for any $r\in R$, then there is a group morphism $\phi : G \to H$ such that $\phi(g(i))=h(i)$ for all $i\in I$.
Allow me to stress once again that the universal properties of condition 3. only holds for finite groups.
My question is: does this coincide whith the usual definition of presentation ?
In particular, I'd like to prove that any word in the $\{g(i)\}$ is $1$ if it can be written as a concatenation of relators or their conjugates. But in order to do that I usually need to consider the quotient of the free group by the normal subgroup generated by the relators. However to prove that this quotient is finite, I need the exact propoerty I'm trying to prove.
Here is a rephrasing of my question:
Suppose that a finite group $H$ verify some relations $R$ ($R$ finite) on some generators $h\_i$ ($I$ finite). Suppose that a word $w$ in the $h\_i$ is equal to $1$ in $H$ but can't be written as a concatenation of the conjugate of the relators in $R$. Does there always exists a **finite** group $H'$ such that $H$ is a homomorphic image of $H'$ but $w$ is not $1$ in $H'$ ?
**[edit]** added ($R$ and $I$ finite in the above rephasing).
|
https://mathoverflow.net/users/37190
|
Group presentation in the category of finite group
|
A well-known example, due to G. Higman, of an infinite finitely presented group with no nontrivial finite quotients is $$G = \langle a,b,c,d \mid a^{-1}ba=b^2,b^{-1}cb=c^2,c^{-1}dc=d^2,d^{-1}ad=a^2 \rangle.$$
Proving it has no nontrivial finite quotients is elementary and a nice exercise. The idea is to prove that $f(a) < f(b) < f(c) < f(d) < f(a)$, where $f(g)$ is the smallest prime factor of the order of $g$ in a candidate for a nontrivial finite quotient.
Proving it infinite can be done using the fact that the group is an amalgamated product of two HNN extensions: see [here](https://math.stackexchange.com/questions/2394006) for example.
But according to your definition of a presentation of a finite group, this presentation would define the trivial group, because there are no other finite homomorphic images of $G$.
So the answer to your question is no, your definition of a presentation of a finite group is not equivalent to the standard notion of a group presentation.
|
8
|
https://mathoverflow.net/users/35840
|
406273
| 166,518 |
https://mathoverflow.net/questions/405784
|
14
|
Let $\mathcal{E}/\mathbb{Q}(t)$ be given by $$y^2=x^3+A(T)x+B(T)$$ for some $A(T),B(T)\in\mathbb{Q}[T]$ and assume $\mathcal{E}$ is non-isotrivial (the $j$-invariant $\frac{6912 A(T)^3}{4A(T)^3 + 27B(T)^2}$ is not constant). Does there necessarily exist $t\in\mathbb{Q}$ such that the Mordell-Weil group of $\mathcal{E}\_t/\mathbb{Q}:y^2=x^3+A(t)x+B(t)$ has positive rank?
An unconditional proof or explicit counterexample would be wonderful, but if that's not possible, I would be okay with a conditional proof assuming standard conjectures (e.g. BSD), or a proof for as wide a class of curves as possible, or a discussion of some properties a hypothetical counterexample would have.
---
Here are my thoughts so far:
* We expect that for "most" families, at least $50\%$ of all specializations (ordering $t$ by height) are positive rank. One result in this direction is by Helfgott, who [shows](https://arxiv.org/pdf/math/0408141.pdf) (assuming some standard conjectures) that if $\mathcal{E}$ has a finite place of multiplicative reduction, then half of the specializations $\mathcal{E}\_t$ ordered by height have root number $-1$, and therefore have positive rank assuming the parity conjecture.
* In contrast to Helfgott's work, there exist non-isotrivial families of curves with constant root number: for example, $W(\mathcal{E}\_t)\equiv -1$ [due to Rizzo](https://www.cambridge.org/core/journals/compositio-mathematica/article/average-root-numbers-for-a-nonconstant-family-of-elliptic-curves/600186E244EFF6C2F35DF02073358459) and $W(\mathcal{E}\_t)\equiv 1$ [due to Bettin, David, and Delaunay](https://www.archives-ouvertes.fr/hal-01478267/document) (EDIT: These results only hold for all $t\in\mathbb{Z}$, not $t\in\mathbb{Q}$, so are not directly relevant; see comments for discussion). It turns out that in both of these families, $100\%$ of the specializations have positive rank (again assuming the parity conjecture), but it certainly may be possible to have a family with generic rank $0$ and constant root number $1$. Even in this case, though, I would still expect some rank $\geq 2$ specializations.
* Joe Silverman [conjectures](https://mathoverflow.net/a/63970/404359) that in fact any non-isotrivial family should have *infinitely many* positive rank specializations, but notes that it's not clear how one would prove this. My question is weaker (I'm only asking for a single specialization), and perhaps naively I would hope this makes it more tractable.
* For any particular family, it is often possible to explicitly construct a rank $1$ subfamily (as Joe mentions in the answer I cited above, and Siksek [demonstrates](https://mathoverflow.net/a/63856/404359)). There likely isn't any way to turn this into a general construction guaranteed to work for all families (if there were, it would prove Joe's conjecture).
|
https://mathoverflow.net/users/404359
|
Does every non-isotrivial 1-parameter family of elliptic curves have a positive rank specialization?
|
I suspect that this is unknown in general. I would guess that any method which produces at least one elliptic curve of positive rank should also produce infinitely many of positive rank, which as you already mention is an open problem.
Anyway, here is some discussion around this problem and a particular open case which would need to be resolved.
I first recall some facts that you already know to fix notation.
A generalisation of the Mordell-Weil theorem says that $\mathcal{E}(\mathbb{Q}(t))$ is a finitely generated abelian group. We denote by $r$ its rank, which is called the *generic rank* of the family. A theorem of Silverman says that for all but finitely many $t \in \mathbb{Q}$, the rank of $\mathcal{E}\_t$ is at least equal to the generic rank.
Thus if $r \geq 1$ then the answer to your question *yes*. So the crucial case to study is when $r= 0$, and you are looking for infinitely many specialisations for which the rank is larger.
There is a very nice way to interpret this problem in terms of the associated surface. Namely, associated to $\mathcal{E}$ there is a unique smooth projective relatively minimal elliptic surface $\pi: X \to \mathbb{P}^1$. There key result is now the following:
**Theorem**
There are infinitely many fibres of $\pi$ with a rational point if and only if $X(\mathbb{Q})$ is Zariski dense.
The proof is not too difficult. It relies in a crucial way on Mazur's torsion theorem. You can find a more general statement in [1, Lemma 3.2].
So you just have to show that the rational points on your surface are Zariski dense! There are various conjectures about Zariski density in the literature due to e.g. Lang, Vojta, Campana, and others. These say in the first instance for $X(\mathbb{Q})$ to be Zariski dense we must have that $X$ is not of general type (which does indeed hold in our case).
In any case, it suffices to find some elliptic surface for which it's unknown whether $X(\mathbb{Q})$ is Zariski dense. These already exist amongst geometrically rational surfaces (these satisfy $\deg A(T) \leq 4$ and $\deg B(T) \leq 6$). Such a class is provided by Del Pezzo surfaces of degree $1$. These have a special rational point given the base-locus of the anticanonical linear system, and the blow-up of this point is an elliptic surface. It is a big open problem to prove that these always have a Zariski dense set of rational points. I suspect even in this case it isn't known whether there is a fibre of positive rank.
[1] Julian Lawrence Demeio - Elliptic Fibrations and Hilbert Property
|
9
|
https://mathoverflow.net/users/5101
|
406276
| 166,519 |
https://mathoverflow.net/questions/406181
|
1
|
Given a set $U=\{1,\ldots,c\}$, we have $t$-dimensional vector $v\_i=(x\_1,x\_2,\ldots,x\_t)$, where $x\_j$ is a subset of $U$. We can check that there are $2^{ct}$ above vectors.
Then we construct the following matrix:
$A=(a\_{ij})\_{2^{ct},2^{ct}}$, where $a\_{ij}=1$ if $\bigvee\_{1\leq l\leq t}v\_{i}(l)\wedge v\_{j}(l)=\emptyset$; $0$ otherwise.
I want to know an upper bound for the rank of the matrix $A$. I hope this upper bound is as small as possible.
|
https://mathoverflow.net/users/178444
|
Upper bound of rank of a matrix
|
If I understood your question correctly, it can be reformulated as follows:
The vectors $v\_i$ are in bijective correspondence with subsets of $X=\{1,\cdots, c\} \times \{1,\cdots, t\}$ by sending a vector $v\_i=(x\_1,x\_2,\cdots, x\_t)$ to the subset $S=\{(k,l)\in X:k\in x\_l\}$.
You can then see the matrix $A$ as a $2^X \times 2^X$ matrix and your condition become $A\_{S,T}=1\_{S \cap T =\emptyset}$ for all $S,T\in 2^X$.
You can factor $A=CB$ where $B\_{S,T}=1\_{S=X\setminus T}$ and $C\_{S,T}=1\_{S\subseteq T}$. Both matrices are invertible since $B$ is a permutation matrix and $C$ is upper-triangular (with respect to any linear extension of the inclusion ordering on $2^X$) with $1$'s on the diagonal. Hence your matrix $A$ always has full rank.
|
2
|
https://mathoverflow.net/users/160416
|
406283
| 166,523 |
https://mathoverflow.net/questions/406274
|
5
|
**Motivation:**
This is a toy model of how a closed system will always evolve towards the distribution of maximal entropy, where no further transfer of heat/energy is possible.
**Problem set up:**
Fix a positive integer $N$, and denote by $[N]$ the set $\{1, \dots, N\}$.
Let $\mathcal L := [N] \times [N]$ be a 2D lattice.
We model a flow of heat as follows - Initially at time $0$, all heat is concentrated at $(1, 1)$. At each time step $t$, for $t \in \mathbb Z\_+$, an element of $\mathcal L$ is picked uniformly at random. At that stage, it and all its immediate neighbours to its east, west, north and south average their heat.
Formally, we have a sequence of iid uniformly distributed $\mathcal L$-valued random variables $\varepsilon\_n$, for $n \in \mathbb Z$, and a sequence $X\_n$ of functions $\Omega \times \mathcal L \to [0, 1]$, defined as follows:
$X\_0 = \mathbf 1\_{(1, 1)}$, almost surely.
Assume now $X\_0, \dots, X\_n$ have already been defined.
Write $X\_n = \sum\_{z \in \mathcal L} \lambda\_z \mathbf 1\_{z}$ for (random) $\lambda\_z \in [0, 1]$ with $\sum\_{x \in \mathcal L} \lambda\_z = 1$, and $\mathcal N(\varepsilon\_n)$ for the set consisting of $\varepsilon\_n$ and its two to four, depending on the location of $\varepsilon\_n$, neighbours.
Then define $X\_{n+1} := \left[\underset{x \in \mathcal N(\varepsilon\_n)}{\sum} \underset{y \in \mathcal N(\varepsilon\_n)}{\sum} \frac{\lambda\_y}{|\mathcal N(\varepsilon\_n)|} \mathbf 1\_x \right]+ \underset{z \in \mathcal L \setminus \mathcal N(\varepsilon\_n)}{\sum} \lambda\_z \mathbf 1\_z$.
where $|S|$ denotes the cardinality of a finite set $S$.
>
> **Question:** Let $Y$ be the “uniform distribution of heat” given by $Y := \underset{z \in \mathcal L}{\sum} \frac{1}{|\mathcal L|} \mathbf 1\_z$. Is it true that almost surely, $X\_n \to Y$ uniformly?
>
>
>
Thus the system evolves almost surely toward a distribution where no further transfer of heat is possible.
|
https://mathoverflow.net/users/173490
|
A toy model of heat death
|
The Ehrenfest model (in discrete time, for simplicity) is just a Markov chain with the finite state space $\{0,1,\dots, N\}$ and the transition probabilities
$$
p(k,k-1)=k/N, \quad p(k,k+1)=1-k/N
$$
described by a single transition (averaging) operator $P$. Its stationary distribution $m$ is the binomial one with the parameters $\frac12,N$, and $\frac12 \theta (P^n+P^{n+1}) \to m$ for any initial distribution $\theta$. [Since the operator $P$ has period 2, one has to take the average of $\theta P^n$ and $\theta P^{n+1}$.]
In your situation there is a *family* of averaging operators $P\_x$ indexed by the points from the state space $X$ which have a unique common invariant measure $m$ (the uniform distribution on $X$). You take a sequence $\boldsymbol x=(x\_1,x\_2,\dots)$ of iid $X$-valued uniformly distributed random variables, and ask whether, given an initial distribution $\theta$ on $X$, the sequence
$$
\theta P\_{x\_1} P\_{x\_2} \dots P\_{x\_n}
$$
converges to $m$ almost surely. Note that since we are talking about measures on a *finite* state space, all reasonable kinds of convergence are equivalent (in particular, the $\ell^1$ convergence in the total variation norm $\|\cdot\|$ and the $\ell^\infty$ "uniform" convergence).
Let
$$
f\_n(\boldsymbol x) = \| \theta P\_{x\_1} P\_{x\_2} \dots P\_{x\_n} - m \| \;.
$$
The sequence $f\_n$ is non-increasing, and therefore convergent. By Kolmogorov's 0-1 law its limit $f\_\infty$ is almost surely constant. Let $k$ be the minimal number such that for any $x\in X$ there is a sequence $x\_1,x\_2,\dots, x\_k\in X$ with
$$
\text{supp}\,\delta\_x P\_{x\_1} P\_{x\_2} \dots P\_{x\_k} = X \;.
$$
Then there is $\varepsilon > 0$ such that
$$
\mathbf E [ f\_{n+k} | f\_n ] \le (1-\varepsilon) f\_n \qquad \forall\,n\ge 0 \;,
$$
whence $f\_\infty=0$.
EDIT. The fact that the sequence $f\_n$ is non-increasing is a consequence of the following inequality:
$$
\frac1d \sum\_{i=1}^d |\theta\_i - C| \ge \left| \frac1d \sum\_i \theta\_i - C \right| \;.
$$
Here $d$ is the cardinality of the averaging set (i.e., between 3 and 5 in your example), and $C=1/N^2$ is the common value of the weights of the uniform distribution. After removing $C$ and the division by $d$ the above inequality amounts to the well-known
$$
\sum\_i |\theta\_i| \ge \left| \sum\_i\theta\_i \right| \;.
$$
The expectation bound is just a constructive version of this inequality: if $f\_n(\boldsymbol x)=F>0$, then there are two points $z\_1,z\_2\in X$ such that
$$
\theta P\_{x\_1} P\_{x\_2} \dots P\_{x\_n}(z\_i) - m(z\_i) \qquad i=1,2\;,
$$
have absolute values comparable with $F$ and opposite signs. Therefore by the definition of $k$ there is at least one choice of $x\_{n+1},\dots, x\_{n+k}$ with
$$
\begin{aligned}
&\|\theta P\_{x\_1} P\_{x\_2} \dots P\_{x\_n+k} - m \| \\ \\ &< (1-\epsilon) \cdot \|\theta P\_{x\_1} P\_{x\_2} \dots P\_{x\_n} - m \| \;,
\end{aligned}
$$
where $\epsilon$ is an appropriate constant (which only depends on $N$).
|
2
|
https://mathoverflow.net/users/8588
|
406286
| 166,525 |
https://mathoverflow.net/questions/406153
|
3
|
The Dedekind–MacNeille Completion is the generalized way of completing an arbitrary lattice $L$. We will call $C$ the Dedekind–MacNeille completion of $L$ (I will not go into the details of the Completion but see the comment below.) The completion includes an embedding $i: L \to C$ which preserves existing arbitrary meets and joins. The completion is universal in the sense that for any other complete lattice $C'$ and map $f: L \to C'$ that preserves arbitrary meets and joins there exists a unique map $g: C \to C'$ which preserves arbitrary meets and joins such that $f = g \circ i$. (I'm still learning category theory so please let me know if any of the above is incorrect.)
I have found an alternative completion specifically for Heyting algebras, which instead of considering the powerset of a Heyting algebra $H$ only considers the subsets of $H$ which are called complete ideals. Which are essentially ideals that additionally contain the join of any subset of the ideal, if that join exists. This completion must be equivalent to the Dedekind–MacNeille Completion by the universal property, right?
This is of interest to me because I want to peform a completion in a setting that does not have access to the powerset axiom (particularly CZF). The standard approach is to develop lattice theory for classes and give up the fact that the completion of an arbitrary lattice is provably a set. Before I pursue this I want to tie up this loose end in my mind. Because initially I was under the impression that I may be able to collect the complete ideals into a set without use of the powerset axiom, but that seems to not be the case.
|
https://mathoverflow.net/users/312621
|
Constructive lattice completion
|
Joseph Van Name’s answer is correct for the general case. However, the OP indicated in comments that they are interested in the case where $L$ is a Heyting algebra, and then it turns out that the Dedekind–MacNeille completion *can* be constructed by taking the set of complete ideals of $L$ ordered by inclusion. (I’m frankly quite surprised, since for Heyting algebras, typically ideals are broken, and what works are filters. But here it seems to be the opposite.)
Using the notation in Joseph Van Name’s answer, it suffices to show that if $I\subseteq L$ is a complete ideal, and $a\in{\downarrow\uparrow}I$, then $a\in I$. Now, if $c\in L$ is any element such that $a\land b\le c$ for all $b\in I$, then $a\to c\in{\uparrow}I$, hence $a\le a\to c$, i.e., $a\le c$. Thus, we have established
$$a=\bigvee\{a\land b:b\in I\}=\bigvee\{b\in I:b\le a\}.$$
Since $I$ is a complete ideal, it follows that $a\in I$.
It is also easy to check that the Dedekind–MacNeille completion of $L$ is a Heyting algebra, with relative pseudocomplement operation (extending that of $L$) defined for complete ideals $I$, $J$ by
$$I\to J=\{a\in L:I\cap{\downarrow}a\subseteq J\}.$$
In particular, $I\to J$ is itself a complete ideal: if $a=\bigvee\_ia\_i$ with $a\_i\in I\to J$, and $b\in I\cap{\downarrow}a$, then $b\land a\_i\in J$ for each $i$, and $b=\bigvee\_i(b\land a\_i)$, hence $b\in J$.
|
2
|
https://mathoverflow.net/users/12705
|
406287
| 166,526 |
https://mathoverflow.net/questions/405926
|
5
|
Cross-posted from MSE
<https://math.stackexchange.com/questions/4272017/finite-maximal-closed-subgroups-of-lie-groups>
$\newcommand{\G}{\mathcal{G}} \newcommand{\K}{\mathcal{K}} \DeclareMathOperator\SU{SU}\DeclareMathOperator\PSU{PSU}\DeclareMathOperator\SO{SO}$Let $\G$ be a Lie group.
I am interested in finite maximal closed subgroups of $ G $.
I'm guessing that $ \G $ has a finite maximal closed subgroup if and only if $ \G $ is simple and compact. Does anyone have other examples of finite maximal closed subgroups?
|
https://mathoverflow.net/users/387190
|
Finite maximal closed subgroups of Lie groups
|
$\newcommand{\G}{\mathcal{G}}
\newcommand{\K}{\mathcal{K}} $Question: When does $ \G $ admit a finite maximal closed subgroup?
Answer : Must be one of the following two cases
1. $ \G $ is compact and simple
2. $ \G $ is not compact in which case $ \G $ cannot be connected and moreover the component group $ \G/\G^\circ $ does not preserve any nontrivial proper closed subgroup (see [comment](https://mathoverflow.net/questions/405926/finite-maximal-closed-subgroups-of-lie-groups#comment1040850_405926) from YCor about $ C\_5 \ltimes \mathbb{R}^2 $).
From now on I will confine myself to the case that $ \G $ is connected.
In other words I will consider the statement "A connected Lie group $ \G $ has a finite maximal closed subgroup $ G $ if and only if $ \G $ is compact and simple."
The first implication is true.
**Claim 1:
If a connected Lie group $ \G $ has a finite maximal closed subgroup $ G $ then $ \G $ must be compact and simple.**
Proof:
Let $ \G $ be a connected Lie group and $ G $ a finite maximal closed subgroup. Since $ G $ is finite then $ G $ is a compact subgroup of $ \G $ so must be contained in a maximal compact subgroup, call it $ \K $. But $ G $ is a maximal closed subgroup thus we must have that $ \K=\G $ (note that $ \K $ cannot equal $ G $ since $ \K $ is connected (the maximal compact of a connected group is always connected)). So $ \G $ must be compact. If $ \G $ is not simple then there exists some morphism
$$
\pi: \G \to \G\_i
$$
with positive dimensional kernel (here $ \G\_i $ is basically one of the semisimple factors of $ \G $). Then
$$
\pi^{-1}(\pi(G))
$$
is a closed positive dimensional subgroup containing $ G $, contradicting the fact that $ G $ is a finite maximal closed subgroup. Thus if a connected Lie group $ \G $ has a finite maximal closed subgroup then we can conclude that $ \G $ is simple.
However the reverse implication does not hold: $ SU\_{15} $ is an example of a compact connected simple Lie group with no finite maximal closed subgroups.
To see why this is the case it is important to note that
**Claim 2: For a compact connected simple Lie group $ \G $, $ G $ is a finite maximal closed subgroup of $ \G $ if and only if $ G $ is Ad-irreducible and $ G $ is a maximal finite subgroup of $ \G $.**
this follows from Corollary 3.5 of [Sawicki and Karnas - Universality of single qudit gates](https://arxiv.org/abs/1609.05780).
Since a finite subgroup of $ SU\_n $ is Ad-irreducible if and only if it is a unitary 2-design we have
**Claim 3: $ G $ is a finite maximal closed subgroup of $ SU\_n $ if and only if $ G $ is a maximal unitary 2-group in $ SU\_n $.**
By inspecting Theorem 3 of [Bannai, Navarro, Rizo, and Pham Huu Tiep - Unitary $t$-groups](https://arxiv.org/abs/1810.02507) one immediately determines that $ SU\_{15} $ has no finite maximal closed subgroups.
Some of the main examples of finite maximal closed subgroups of $ SU\_n $ include the normalizer in $ SU\_{p^n} $ of an extra-special group $ p^{2n+1} $. Here $ p $ is an odd prime. There is also a similar construction $ p=2 $. These are known as (complex) Clifford groups. Then there are infinite families of examples relating to the Weil module for $ \operatorname{PSp}\_{2n}(3) $ and another family related to $ U\_n(2) $. Plus many exceptional cases.
A similar normalizer construction to the above gives finite maximal closed subgroups of all the $ \operatorname{SO}(2^n) $ as normalizers of an extra-special group $ 2^{2n+1} $. This is known as the real Clifford group. For details about real and complex Clifford groups see [Nebe, Rains, and Sloane - Self-Dual Codes and Invariant Theory](https://doi.org/10.1007/3-540-30731-1).
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1
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https://mathoverflow.net/users/387190
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406300
| 166,533 |
https://mathoverflow.net/questions/406297
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4
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Recall that a prime $p$ is a Wieferich prime if $p^2|2^{p-1}-1$. The only known Wieferich primes are $p=1093$ and $p=3511$. A prime $p$ is a generalized Wieferich prime to base $q$ if $p^2|q^{p-1}-1$.
It is strongly suspected that there are infinitely many Wieferich primes to base $q$ for any $q>1$. In the other direction it is strongly suspected that for any $q>1$ there are infinitely many non-Wieferich primes to base $q$, but this is also open even for $q=2$.
The situation is slightly different if one replaces $p^2$ with $p^3$. Call a prime $p$ a *super-Wieferich* prime to base $q$ if $p^3|q^{p-1}-1$. In particular, it is a standard conjecture (using similar heuristic arguments) that for any $q>1$ there are only finitely many primes $p$ which are super-Wieferich primes to base $q$. (For interesting recent work related to that question see Ram Murty and Francois Seguin's recent paper "Prime divisors of sparse values of cyclotomic polynomials and Wieferich primes" in the Journal of Number Theory.)
This question is about a highly restricted version of Wieferich primes. In particular: define a prime $p$ to be a hyper-Wieferich prime to base $q$ if there is some $m$ such that $$p^m = \frac{q^{p-1}-1}{q-1}$$ and $m>2$. The idea here is that $p$ is not just super-Wieferich to base $q$, but that every potential factor of $q^{p-1}-1$ which isn't one of the obvious factors coming from $q-1$ is in fact a copy of $p$.
**Question:** Can we show that there are only finitely many hyper-Wieferich primes or at least substantially restrict what they can look like?
We at least have the following:
Claim: The ABC conjecture implies that there are only finitely many primes $p$ which are hyper-Wieferich to base $q$.
Proof: Assume that $p$ is hyper-Wieferich to base $q$. So we have $$p^m = \frac{q^{p-1}-1}{q-1}.$$ One easily has that $m \geq p-2$.
We set $A=p^m (q-1)$, $B=1$ and $C= q^{p-1}$. Then $\mathrm{rad}(ABC) = pq(q-1)$. But $q^{p-1}$ is much larger than $pq$.
The above shouldn't be too surprising since it is essentially a very simplified version of an argument of Silverman that the ABC conjecture implies there are infinitely many non-Wieferich primes.
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https://mathoverflow.net/users/127690
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Ruling out an extremely specific class of Wieferich-like primes
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This is impossible from standard results about cyclotomic polynomials and their factors. Note that $\frac{q^{p-1}-1}{p-1}$ is just
$$
\prod\_{d|(p-1), d>1}\Phi\_d(q),
$$
where $\Phi\_d(x)$ is the $d$th cyclotomic polynomial. Taking $p>3$, there are two such divisors $d$; namely $2$ and $p-1$. By an old result attributed to Bang from 1886, these must have "new" prime divisors.
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8
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https://mathoverflow.net/users/3199
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406307
| 166,537 |
https://mathoverflow.net/questions/401499
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1
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Let $\mathfrak{G}$ be the class of all finite directed and undirected graphs. Let $A,B\subseteq \mathfrak{G} $, $A$ and $B$ are closed under graph isomorphisms, and $A \cap B = \varnothing$. Consider the following two player game. On the graph $G \in \mathfrak{G} $ with the starting vertex $u$, play as follows: the first player presents $x\_1 \subseteq V(G)$ such that $\forall v\in x\_1\; uv\in E(G)$. Then the players in turn present the sets of vertices, so that for any different $x\_i$ and $x\_j$ satisfy $x\_i \cap x\_j = \varnothing$, and $\forall v\in x\_k \; \exists w\in x\_{k-1} : wv\in E(G)$.
Conditions for victory and defeat (checked on each turn in the order of the following list):
1. If on turn $n$ there exists $k$ such that the induced subgraph $\bigcup\_\limits{i=k}^n x\_i$ contains a subgraph $H$ isomorphic to a graph from $A$, then the player who made the move $n$ wins.
2. If on turn $n$ there exists $k$ such that the induced subgraph $\bigcup\_\limits{i=k}^n x\_i$ contains a subgraph $H$ isomorphic to a graph from $B$, then the player who made the move $n$ loses.
3. If after some move the player cannot make a move, then he loses.
For example, if condition 2 is satisfied on the last move, then the player who made this move loses, because condition 2 is checked earlier than condition 3. Or if after the player's move a graph appears containing subgraphs from both $A$ and $B$, then he wins, because condition 1 is checked before condition 2. Let call this game $A-B$ game.
Consider the following language:
$$A-B-NG:=\{(G,u): \text{there is a winning strategy for the first player in $A-B$ game} \}$$
The complexity of this language depends on the classes we are considering. For example, if $pt$ is one-vertex graph, then $\{pt\}-B-NG \in \mathrm{DTIME(1)}$, because condition 1 is satisfied for all non-empty graphs. But $GG\leq\_p \varnothing - \varnothing - NG$, therefore $ \varnothing - \varnothing - NG \in \mathrm{PSPACE-complete} $ (because $\varnothing - \varnothing - NG \leq\_p GG $ ). Is it possible to choose $A$,$B$ so that $A-B-NG$ will have an even worse nesting, that is, it will lie in a class strictly above $ \mathrm{PSPACE} $, or containing $ \mathrm{PSPACE} $ ?
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https://mathoverflow.net/users/175589
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Complexity of games with graph classes
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No, it is not possible to go above PSPACE, because all positional games with an exponentially large game tree are in PSPACE; you can just check all the options. And the game when $A=B=\emptyset$ is indeed PSPACE-hard, I've found a gadget to reduce the original Generalized Geography to it.
Update: Some of my students found a simpler gadget: Just replace each directed edge by a path of length 3. If anytime someone picks more than one vertices, then they practically yield their option to pick the next vertex to their opponent.
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1
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https://mathoverflow.net/users/955
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406317
| 166,540 |
https://mathoverflow.net/questions/406301
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4
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$\def\f#1{\text{#1}}$Does $\f{ACA}$ prove that any two internally complete ordered fields are isomorphic?
Here internal completeness is expressed roughly as "every sequence of reals with an upper bound has a least upper bound", where the sequences and reals are coded appropriately.
This is an attempt to express the categoricity of the reals in the context of subsystems of 2nd-order arithmetic, where $\f{ACA}\_0$ is already equivalent to the internal completeness of the (coded) reals over $\f{RCA}\_0$ ([SoSOA](http://www.personal.psu.edu/t20/sosoa)).
To be precise, let $T$ be the 2-sorted FOL theory $\f{ACA}\_0$ with sorts $N,S$ where $N$ is for the naturals and $S$ is for the subsets of $N$, plus the following:
* Predicate-symbol $R$ on $S$ (intended to represent the subsort of reals).
* Constant-symbols $0\_R,1\_R$ of sort $S$.
* Function-symbols $+\_R,·\_R$ from $S^2$ to $S$.
* Predicate-symbol $<\_R$ on $S^2$.
* Axioms stating that $(R,0\_R,1\_R,+\_R,·\_R,<\_R)$ is an ordered field.
* The **internal completeness axiom**, namely $∀f{∈}N{→}R\ ( \ ∃m{∈}R\ ( \ f ≤ m \ ) ⇒ ∃m{∈}R\ ( \ f ≤ m ∧ ∀x{∈}R\ ( \ f ≤ x ⇒ m ≤ x \ ) \ ) \ )$ where "$f ≤ t$" is short-hand for "$∀k{∈}N\ ( \ f(k) ≤\_R t \ )$". (Here each member of $N{→}R$ is of course coded as a member of $S$.)
$T$ is what I mean by "theory of internally complete ordered field". Now $T$ is actually conservative over $\f{ACA}\_0$, by the equivalence of $\f{ACA}\_0$ and 'completeness of reals' over $\f{RCA}\_0$. And so we can work within this theory $T$ to do applied real analysis, and we know that it is no stronger than $\f{ACA}\_0$.
The question is, does $\f{ACA}$ (no subscript zero) prove that every two $ω$-models of $T$ are isomorphic? From the perspective of a set theory, $\f{ACA}$ cannot reason about uncountable sets, but can reason about countable $ω$-models of a 2-sorted FOL theory, which are precisely what I am interested in here. In particular, "for every countable set" translates to "∀X{∈}S" in the language of $\f{ACA}$.
Since $\f{ACA}$ proves the existence of an $ω$-model of $\f{ACA}\_0$, we know that $\f{ACA}$ also proves the existence of an $ω$-model of $T$. The question is whether it knows the uniqueness of such models up to isomorphism.
~ ~ ~
The standard proof seems to go through: Take any models $K,M$. Construct the isomorphism $f$ from rationals $Q(K)$ of $K$ to rationals $Q(M)$ of $M$ via arithmetical comprehension. Then construct $g$ from elements of $K$ to elements of $M$ where $g(x) = \sup\_M( \{ f(w) : w∈Q(K) ∧ w <\_K x \} )$. Here we can use any usual enumeration of $Q(K)$ and apply internal completeness for $M$. Now we simply have to prove that $g$ is an embedding, since self-embedding on $K$ that fixes $Q(K)$ also fixes everything else. Firstly, $g$ is injective since $Q(K)$ is dense in $K$, by internal completeness of $K$. Secondly, $g$ is a homomorphism, which is a bunch of cases but should be similar.
But I am unable to find anything on the reverse mathematical strength of categoricity of the 'reals'. Can anyone confirm what I said here or give any reference? I could slowly check it myself but it would be nice if it was already a well-known result.
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https://mathoverflow.net/users/50073
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Does ACA prove categoricity of the reals?
|
The current version of the question is flawed: the proposed categoricity principle is in fact classically **false**. Roughly speaking, any "interesting" theory of the appropriate type is going to have lots of non-isomorphic countable $\omega$-models. *(For massive overkill let $M,N$ be countable elementary submodels of $V\_{\omega\_{17}}$ with $M\in N$ and $\vert M\vert^N=\aleph\_0$; then the structures we get from $\mathbb{R}\cap M$ and $\mathbb{R}\cap N$ will be non-isomorphic countable $\omega$-models of whatever theory you whip up.)*
I think the right way to tackle a question like this is to get away from the language of "countably coded structures," or **structures as internal objects**, entirely. While this is usually how we implement anything model-theory-flavored in reverse mathematics, it's not at all suitable here due to a "pincer attack" of the observation in the above paragraph on the one hand and the fact that $\mathsf{RCA\_0}$ proves the nonexistence of countable complete ordered fields on the other hand (the latter killing approaches which avoid the issue posed by the former).
To find the right way to pose the question I think it's necessary to take a step back and think semantically. Given $M\models\mathsf{ACA\_0}$, the object we're interested in initially is the ordered field of reals in the sense of $M$, or a bit more precisely the natural *interpretation of* this field into $M$. Call this "$\mathbb{R}^M$." Now $\mathbb{R}^M$ is **not a "structure in $M$"** in the usual sense, nor is it countable from the perspective of $M$ in any good sense. It's more analogous to a proper class than to an internal object. Now intuitively we want to compare $\mathbb{R}^M$ to its "alternatives" - and these alternatives should be objects of exactly the same type. This leads us away from countably coded structures altogether, and suggests the following precisiation of your question:
>
> $(\*)\quad$ Suppose $M\models\mathsf{ACA\_0}$, so that $\mathbb{R}^M$ is "internally complete." Let $\Phi$ be any tuple of formulas such that $\Phi^M$ is also an $M$-"internally complete" ordered field. Must there be an $M$-definable isomorphism between $\mathbb{R}^M$ and $\Phi^M$?
>
>
>
* Note that in fact this phrasing lets us shed the language of reverse mathematics altogether: as a **way-too-ambitious project** we could try to understand when a structure $\mathfrak{A}$ has exactly one internally complete interpretation of an ordered field up to definable isomorphism. In my opinion part of what makes $\mathsf{RCA\_0}$ and its extensions fun to think about is exactly that it carves out a subclass of structures on which this sort of project is *not* too ambitious.
I believe that $(\*)$ is in fact what you are trying to ask here. The answer to $(\*)$ at first glance appears to be that $\mathsf{ACA\_0}$ is already enough, but I'll check in more detail if you clarify that this is indeed a faithful precisiation of the question you're asking.
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6
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https://mathoverflow.net/users/8133
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406325
| 166,544 |
https://mathoverflow.net/questions/406275
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3
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In my research in Quantum Field Theory, I have encountered two questions that involve partial Bell polynomials:
1. Let $u$ and $x\_i$ be indeterminates. I have checked that the following conjectured identity
$$
\sum \_{m=0}^\infty u^m \frac{(m+1)!}{(2m+2)!} B\_{2m+2,m+2}(x\_1,\dots,x\_{m+1})=
\frac{1}{2} \left( \sum\_{n=0}^\infty u^n \frac{n!}{(2n+1)!} B\_{2n+1,n+1}(x\_1,\dots, x\_{n+1}) \right)^2
$$
holds for the first orders in $u$. Is this identity a known result? A proof or any reference would be very welcome.
2. I would like to find the radius of convergence of a similar infinite sum, in the particular case when $x\_j = j^{j-2}$, or at least prove that it is different from zero. Namely, I would like to discuss the convergence of
$$
f(u) = \sum\_{m=1}^\infty u^m \frac{(m-1)!}{(2m)!} B\_{2m,m+1}(1^{1-2},2^{2-2},3^{3-2},\dots,m^{m-2})
$$
near $u=0$, but I don't know how to estimate the asymptotic growth of $B\_{2m,m+1}(1^{1-2},2^{2-2},3^{3-2},\dots,m^{m-2})$ for large $m$. Again, any help or reference would be welcome.
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https://mathoverflow.net/users/98550
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Two questions about Bell polynomials
|
With respect to the first question, I'm not sure about whether the given identity is known, but here is a proof. It also expresses the involved series in terms of reversion of the series $\frac{t}{g(t)}$ as defined below.
It is convenient to express the given identity in terms of [ordinary Bell polynomials](https://en.wikipedia.org/wiki/Bell_polynomials#Ordinary_Bell_polynomials):
$$\hat{B}\_{n,k}(\frac{x\_1}{1!},\frac{x\_2}{2!},\ldots,\frac{x\_{n-k+1}}{(n-k+1)!}) = \frac{k!}{n!}B\_{n,k}(x\_1,x\_2,\ldots,x\_{n-k+1})$$
and omitting the indeterminates $x\_j$ and multiplying by $u^2$ as
$$\sum\_{m=0}^{\infty} \frac{u^{m+2}}{m+2} \hat{B}\_{2m+2,m+2} = \frac{1}{2}\left(\sum\_{n=0}^{\infty} \frac{u^{n+1}}{n+1} \hat{B}\_{2n+1,n+1} \right)^2.$$
From the [generating function for $\hat{B}\_{n,k}$](https://en.wikipedia.org/wiki/Bell_polynomials#Generating_function) it follows that
$$\hat{B}\_{2m+2,m+2}
= [t^{2m+2}]\ \left(\sum\_{j\geq 1} \frac{x\_j}{j!}t^j\right)^{m+2} = [t^{m}]\ g(t)^{m+2}$$
and
$$\hat{B}\_{2n+1,n+1}
= [t^{2n+1}]\ \left(\sum\_{j\geq 1} \frac{x\_j}{j!}t^j\right)^{n+1} = [t^n]\ g(t)^{n+1},$$
where $g(t):=\sum\_{j\geq 1} x\_j\frac{t^{j-1}}{j!}$.
Using [Lagrange–Bürmann formula](https://en.wikipedia.org/wiki/Lagrange_inversion_theorem#Lagrange%E2%80%93B%C3%BCrmann_formula), we get
$$\sum\_{m\geq 0} \hat{B}\_{2m+2,m+2} t^{m+1} = \frac{w(t)w'(t)}t,$$
$$\sum\_{n\geq 0} \hat{B}\_{2n+1,n+1} t^n = w'(t),$$
where function $w(t)$ satisfies the functional equation: $w(t)=tg(w(t))$.
Correspondingly,
$$\sum\_{m\geq 0} \hat{B}\_{2m+2,m+2} \frac{u^{m+2}}{m+2} = \int\_0^u w(t) w'(t)\ {\rm d}t = \frac12 w(u)^2,$$
$$\sum\_{n\geq 0} \hat{B}\_{2n+1,n+1} \frac{u^{n+1}}{n+1} = \int\_0^u w'(t) {\rm d}t = w(u),$$
from where the required identity follows instantly.
---
As for the second question, we first notice that $f'(u) = \frac12 w(u)^2$. Also, when $x\_j=j^{j-2}$, we have $$g(t)=\sum\_{j\geq 1} j^{j-2}\frac{t^{j-1}}{j!},$$
which can be expressed in terms of Lambert W function as
$$g(t) = \frac{1}{2t}\left(1-(1+W\_0(-t))^2\right).$$
Then convergence of $w(t)$ can be imposed from viewing it as a [series reversion](https://en.wikipedia.org/wiki/Lagrange_inversion_theorem) of $\frac{t}{g(t)}$.
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7
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https://mathoverflow.net/users/7076
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406326
| 166,545 |
https://mathoverflow.net/questions/406221
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3
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I have asked [this question](https://math.stackexchange.com/questions/4276271/chebyshev-polynomials-and-ballot-number) a short time ago on mathstackexchange, but it has already fallen into the abyss of answered and uncommented questions. So I take the risk to ask it on mathoverflow.
Playing with the various expressions of the Chebyshev polynomials of the first kind, I find for all $j \leq \lfloor \frac{n}{2} \rfloor$ the formula:
$$ \sum\_{k=j}^{\lfloor \frac{n}{2} \rfloor} \binom{k}{j} \binom{n}{2k} = 2^{n-1-2j} \frac{n}{n-j} \binom{n-j}{j} $$
I would like to know if there is a (preferably simple) combinatorial proof of this formula. This looks like a Vandermonde type formula for some variant of the ballot numbers, but I don't know enough in this area to find a useful interpreatation. Many thanks for your help!
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https://mathoverflow.net/users/37214
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Chebyshev polynomials and ballot numbers
|
Here is a combinatorial proof.
It is more convenient to prove the equivalent formula (obtained by setting $n=m+2j$)
$$\sum\_k \binom{k}{j}\binom{m+2j}{2k}=2^{m-1}\frac{m+2j}{m+j}\binom{m+j}{j}.\tag{$\*$}$$
To motivate the proof we first prove the closely related but slightly easier
formula
$$\sum\_k \binom{k}{j}\binom{m+2j+1}{2k+1} = 2^m\binom{m+j}{j}.\tag{1}$$
Let us first give a generating function proof of $(1)$. We start by multiplying the summand by $x^j y^m$ and summing on $j$ and $m$. We find that
$$
\sum\_{j,m}\binom{k}{j}\binom{m+2j+1}{2k+1} x^j y^m = \frac{(x+y^2)^k}{(1-y)^{2k+2}}
$$
and that
$$\sum\_{k=0}^\infty \frac{(x+y^2)^k}{(1-y)^{2k+2}}=\frac{1}{1-x-2y} = \sum\_{j,m}2^m\binom{m+j}{j}x^j y^m.\tag{2}$$
It is not hard to find objects counted by $2^m\binom{m+j}{j}$. We need to interpret $(x+y^2)^k/(1-y)^{2k+2}$ as counting some of these objects, and this is what the following proof does.
Let $S$ be the set of words in the the letters $x, y\_1, y\_2$ containing $j$ occurrences of $x$ and a total of $m$ occurrences of $y\_1$ and $y\_2$. There are $\binom{m+j}{m}$ words with $j$ occurrences of $x$ and $m$ occurrences of $y$. Each $y$ can be replaced with $y\_1$ or $y\_2$ so there are $2^m\binom{m+j}{m}$ words in $S$.
Now we count the words in $S$ in a different way. Let us define a *stopper* of a word in $S$ to be an occurrence of either $x$ or $y\_2y\_1$, and let us call the subwords before, between, and after the stoppers *segments*, so a word with $k$ stoppers has $k+1$ segments. Thus each segment is of the form $y\_1^a y\_2^b$, where $a$ and $b$ are nonnegative integers. For example, the word $y\_1 y\_2 x y\_2 y\_1 y\_2 y\_2x$ has three stoppers, $x$, $y\_2y\_1$, and $x$, and four segments, $y\_1y\_2$, $\varnothing$, $y\_2y\_2$, and $\varnothing$. (Here $\varnothing$ denotes the empty word.)
Let us count words in $S$ with $k$ stoppers (and thus $k+1$ segments). We must have $k\ge j$, since every $x$ is a stopper. Then there are $k-j$ stoppers of the form $y\_2y\_1$ so there are $m-2(k-j)$ occurrences of $y\_1$ and $y\_2$ among the $k+1$ segments. The segments are of the form $y\_1^{a\_0}y\_2^{b\_0},y\_1^{a\_1}y\_2^{b\_1},\dots,y\_1^{a\_k}y\_2^{b\_k}$, where $a\_0+b\_0+\cdots +a\_k+b\_k=m-2(k-j)$. The number of solutions in nonnegative integers of this equation is $\binom{m+2j+1}{2k+1}$
(since for fixed $r$ and $s$ the number of nonnegative integer solutions of $c\_1+c\_2+\cdots +c\_r = s$ is $\binom{r+s-1}{s}=\binom{r+s-1}{s-1}$).
Once the $a\_i$ and $b\_i$ are determined, the $k$ slots for the stoppers are fixed, and
we may decide which are $x$ and which are $y\_2y\_1$ in $\binom{k}{j}$ ways. Thus the number of elements of $S$ is
$$
\sum\_k \binom{k}{j}\binom{m+2j+1}{2k+1},
$$
which must also be equal to $2^m\binom{m}{j}$.
To prove the original identity $(\*)$, we replace $2k+2$ with $2k+1$ in the denominator of the left side of $(2)$. This corresponds to counting only those words in $S$ for which $a\_0=0$, i.e., the words in $S$ that do not start with $y\_1$. The number of solutions of $a\_0+b\_0+\cdots +a\_k+b\_k=m-2(k-j)$ with $a\_0=0$ is $\binom{m+2j}{2k}$, so the number of these words is the left side of $(\*)$. But (assuming $j$ and $m$ are not both 0) the number of words in $S$ that start with $y\_1$ is $2^{m-1}\binom{m+j-1}{j}$, so the number of words in $S$ that do not start with $y\_1$ is
$$2^j\binom{m+j}{j} -2^{m-1}\binom{m+j-1}{j} = 2^{m-1}\frac{m+2j}{m+j}\binom{m+j}{j}.$$
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7
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https://mathoverflow.net/users/10744
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406329
| 166,546 |
https://mathoverflow.net/questions/406291
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1
|
My question is about part of the Borel Transgression Theorem on spectral sequences, translated from [1, Theorem 13.1] (see also [2] for a translated version of the whole paper):
>
> Let $B^\bullet := \bigoplus\_{p \in \mathbb{N}\_0} B^p$ and $P^\bullet := \bigoplus\_{q=0}^m P^q$ be finite-dimensional, graded vector spaces, with $P^q$ only nonzero in odd degree.
>
>
> Let further $\{E\_r^{p,q}, d\_r\}\_{r \geq 0}$ be a cohomological spectral sequence whose second page has the shape
> \begin{align\*}
> E\_2^{p,q} = B^p \otimes \Lambda^q P,
> \end{align\*}
> and which converges towards a graded vector space $H^\bullet$ with $H^k = 0$ for all $ k \leq n$, where $n$ some number with $n \geq 2 m +1 $.
>
>
> a) Then $\Lambda^\bullet P$ is an exterior algebra of some subspace $P' \subset P$ and $P'$ admits a basis of transgressive, homogeneous elements $x\_1,\dots,x\_l$. Further, $P'$ contains all transgressive elements of $\Lambda^\bullet P$.
>
>
> b) If $y\_i \in B^\bullet$ fulfils $d\_r( 1 \otimes x\_i) = y\_i \otimes 1$ for $i = 1,\dots,l$, then $B^\bullet = \mathbb{K}[y\_1,\dots,y\_l]$ in degree $\leq n$.
>
>
>
**Question:** Can one still say anything when $H^k = 0$ in a smaller range of $k$, like $k \leq m$ only?
---
The reason I am asking is because Gelfand and Fuks use this theorem to analyze a Hochschild-Serre spectral sequence for the Lie algebra of formal vector fields on $\mathbb{R}^n$, denoted $W\_n$, see [3, Proposition 5.1], but it does not fulfill the requirements at all: In this paper, $H^k = H^k(W\_n)$ is only known to be zero for $1 \leq k \leq n$, but $P$ has one generator in every dimension $1,3,\dots,2 n - 1$, so $m = 2n - 1$.
There are some arguments in this paper that by relating the spectral sequences for different dimensions $n$, one still gets transgressive generators, but since *none* of the spectral sequences fulfill the requirements, I am not sure how Borel's theorem can be applied at all. Are there clever cutoff tricks that can salvage this?
[1] *Borel, Armand*, [**Sur la cohomologie des espaces fibrés principaux et des espaces homogènes de groupes de Lie compacts**](http://dx.doi.org/10.2307/1969728), Ann. Math. (2) 57, 115-207 (1953). [ZBL0052.40001](https://zbmath.org/?q=an:0052.40001).
[2] *Novikov, S. P. (ed.); Taimanov, I. A. (ed.)*, Topological library. Part 3: Spectral sequences in topology. Transl. by V. P. Golubyatnikov, Series on Knot and Everything 50. Hackensack, NJ: World Scientific (ISBN 978-981-4401-30-2/hbk). ix, 576 p. (2012). [ZBL1264.55002](https://zbmath.org/?q=an:1264.55002).
[3] *Gel’fand, I. M.; Fuks, D. B.*, [**Cohomology of the Lie algebra of formal vector fields**](http://dx.doi.org/10.1070/IM1970v004n02ABEH000908), Math. USSR, Izv. 4(1970), 327-342 (1971); translation from Izv. Akad. Nauk SSSR, Ser. Mat. 34, 322-337 (1970). [ZBL0216.20302](https://zbmath.org/?q=an:0216.20302).
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https://mathoverflow.net/users/126256
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Borel's transgression theorem for spectral sequences
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Mimura and Toda's statement in *The Topology of Lie Groups*, Theorem VII.2.9 (p. 378), requires less severe degree constraints. They start with a Serre fibration $F \to E \to B$ with $B$ simply-connected, and demand only that
* there be an oddly-graded space $P$ of "generators" of degrees no more than some fixed $N$ such that the induced additive map $\Lambda P \to H^\*(F;K)$ is bijective in degrees $\leq N$ and injective in degree $N+1$ and
* $\tilde{H}{}^\*(E;K)$ be zero in degrees $\leq N + 2$.
They conclude then that transgressive elements $x\_i$ of $H^\*(F;K)$ (generating it up to degree $N$ and such that the associated map $\Lambda[x\_i] \to H^\*(F;K)$ is injective in degree $N+1$) can be chosen such that lifts $y\_i \in H^\*(B;K)$ of their transgressions are generators in the sense that the associated ring map $K[y\_i] \to H^\*(B;K)$ is bijective in degrees $\leq N+1$ and injective in degree $N+2$.
I do not believe their proof uses anything about the Serre spectral sequence beyond the fact it is a spectral sequence of differential graded algebras whose $E\_2$ page is of the form $H^\*(B;K) \otimes H^\*(F;K)$ (when $K$ is a field), so it should give a purely algebraic result about spectral sequences with $E\_2$ pages of this form.
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3
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https://mathoverflow.net/users/5792
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406335
| 166,549 |
https://mathoverflow.net/questions/406083
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3
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I have two questions---the second question (which is what I'm really interested in) is a generalization of the first, but I think the first may be more likely to get an answer. I'll be happy with an answer to either.
First question
--------------
Let $M$ be a real analytic manifold, and let $P$ be a linear partial differential operator on $M$ with real analytic coefficients. Let $f$ be a hyperfunction on $M$. Does the PDE $$Pu=f$$ have a hyperfunction solution $u$ on a neighborhood of every point of $M$?
Second question
---------------
Is the sheaf of hyperfunctions on $M$ injective as a module over the sheaf $\mathcal{D}\_M$ of linear partial differential operators on $M$ with real analytic coefficients?
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https://mathoverflow.net/users/36720
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do hyperfunction solutions always exist?
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The answer to both questions is Yes. It has been known (Grainger, Kohn, Stein, Proc. Nat. Academy USA, Vol. 72, No. 9, 3287-3289 (1975)) that when $f$ is Baire category 1, then the Lewy operator is locally solvable. This means that if the 'size'of the test space is reduced making its dual larger, then solvability occurs. In fact, this led Hormander to formulate his famous hypo-ellipticity condition in terms of the Lie brackets spanning the whole tangent space.
This has a nice interpretation in the hyperfunction setup, noting that the Dirac measure is a distributional limit of hyperfunctions.
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5
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https://mathoverflow.net/users/167228
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406345
| 166,553 |
https://mathoverflow.net/questions/406314
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2
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I am looking at the following convolution integral (with $\theta > 0$):
$$ f(t) = \int\_{-\infty}^t \exp(-(t-s)\theta) g(s) ds = (\textbf{1}\_{\{.\geq 0 \}} \exp(-.\theta) \* g)(t)$$
I want to find conditions on $g$ s.t.
$$ f \in C^{k, \alpha},$$ by which I denote the Hoelder space. That means, f should be $k$-times differentiable and the $k$-th derivative should satisfy (for some $C \in \mathbb{R}, 0 < \alpha \leq 1$:
$$\lvert f^{(k)}(x) - f^{(k)}(y) \rvert \leq C \lvert x-y \rvert^\alpha$$
I found that, if $g$ is bounded (which would be okay to assume in my case), then $f$ is uniformly continuous and bounded also. But what about the Hölder space?
Any ideas, hints, references welcome. Thanks!
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https://mathoverflow.net/users/336584
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Assumptions under which convolution is in Hölder class
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*Edit: In the first part, $C^{k,\alpha}$ is understood locally. For a global result, see the final part.*
The equation reads $$f(t) = \int\_{-\infty}^t e^{-(t - s)\theta} g(s) ds,$$ is equivalent to $$e^{t \theta} f(t) = \int\_{-\infty}^t e^{s \theta} g(s) ds,$$ so if $g$ is $C^{k-1,\alpha}$, then $s \mapsto e^{s \theta} g(s)$ is $C^{k-1,\alpha}$, its integral $t \mapsto e^{t \theta} f(t)$ is thus $C^{k,\alpha}$, and finally $f$ is $C^{k,\alpha}$.
Note that this is optimal: if $f$ is $C^{k,\alpha}$, then, by the same argument, $g$ is $C^{k-1,\alpha}$.
---
*Edit: a global result is given below.*
The first part shows that if $g \in C^{k-1,\alpha}$ on $[0, 1]$, then $f \in C^{k,\alpha}$ on $[0, 1]$, with $$\|f'\|\_{C^{k-1,\alpha}([0,1])} \leqslant C(k,\alpha) \|g\|\_{C^{k-1,\alpha}([0,1])}.$$ But the result is clearly translation-invariant, and so
$$\|f'\|\_{C^{k-1,\alpha}([a,a+1])} \leqslant C(k,\alpha) \|g\|\_{C^{k-1,\alpha}([0,1])} \leqslant C(k,\alpha) \|g\|\_{C^{k-1,\alpha}(\mathbb R)}.$$
It follows that
$$\|f'\|\_{C^{k-1,\alpha}(\mathbb R)} \leqslant C(k,\alpha) \|g\|\_{C^{k-1,\alpha}(\mathbb R)}.$$
It remains to observe that $\|f\|\_\infty \leqslant C \|g\|\_\infty$, and the desired global estimate follows.
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2
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https://mathoverflow.net/users/108637
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406349
| 166,554 |
https://mathoverflow.net/questions/406315
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21
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The typical application of Sylow's Theorem is to count subgroups. This makes it difficult to search the web for other applications, since most hits are in the context of qualifying exams.
>
> What are other uses of Sylow's Theorem? In particular, are there famous/common instances where the goal is to find two non-conjugate $p$-subgroups, implying that a higher power of $p$ divides $|G|$?
>
>
>
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https://mathoverflow.net/users/13923
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Atypical use of Sylow?
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There are many examples of using Sylow $p$-subgroups to understand the structure of general finite groups. KConrad's answer indicates some of these, and others are mentioned in comments. A finite group $G$ with Sylow $p$-subgroup $P$ is said to have a normal $p$-complement if there is a normal subgroup $K$ (necessarily of order prime to $p$) with $G = PK$ and $P \cap K = 1.$ There are many theorems relating so-called $p$-local analysis and the existence of normal $p$-complements in finite groups. One of the earliest was Burnside's normal $p$-complement theorem, which states that if a finite group $G$ has an Abelian Sylow $p$-subgroup $S$ with $N\_{G}(S) = C\_{G}(S)$, then $G$ has a normal $p$-complement. Another powerful theorem due to G. Frobenius is that if a finite group $G$ has a Sylow $p$-subgroup $P$ such that $N\_{G}(Q)/C\_{G}(Q)$ is a $p$-group for each subgroup $Q$ of $P$, then $G$ has a normal $p$-complement. Other so-called transfer theorems emerged in finite group theory in the early to mid 20th century: these are theorems which used the structure of the normalizers of non-trivial $p$-subgroups of $G$ to demonstrate the existence of non-trivial Abelian homomorphic images of $G$ in many circumstances.
Such theorems were taken to new heights in the late 1950s and the 1960s, in particular by work of J.G. Thompson, G. Glauberman and J.L. Alperin. For example, the work of Glauberman and Thompson demonstrated (for odd $p$) the existence of a non-trivial characteristic $p$-subgroup $C(P)$ of the Sylow $p$-subgroup $P$ of $G$ such that $G$ has a normal $p$-complement if and only if $N\_{G}(C(P))$ has a normal $p$-complement. The use of local analysis by Thompson in his $N$-group paper formed a template/guide for the later completion of the classification of finite simple groups ( some refinements were necessary).
Another use of Sylow $p$-subgroups there was in the development of signalizer functor theory. This is an over-simplification, but the general idea here is, given a finite group $G$, to build a non-trivial subgroup $L$ of order prime to $p$ which is normalized by a Sylow $p$-subgroup $P$ of $G$, and then by $N\_{G}(Q)$ for many non-trivial $p$-subgroups $Q$ of $P$, and finally by $G$ itself (so that $G$ is not simple if $P \neq 1$). This line of development again has origins in work of Thompson, later refined by others, such as Gorenstein, Goldschmidt and Glauberman.
The work of R. Brauer relates the $p$-local structure of finite groups to their representation theory in characteristic $p$, and draws many new conclusions about complex characters of finite groups. Here, defect groups play an important roles. These are $p$-subgroups whose order depends on representation-theoretic properties of $G$, and these behave like Sylow $p$-subgroups in the context of Brauer's block theory. Properties of normalizers of non-trivial subgroups of the defect group determine many representation-theoretic invariants of $G$ ( and are conjectured to determine many more).
So, in the context of finite group theory, Sylow's theorem is an indispensable tool whose use goes far beyond counting theorems.
Later edit: Regarding the last question, the idea of "pushing-up" is very important in the classification of finite simple groups. The idea here is that we have a putative finite simple group $G$, and a maximal subgroup $M$ of $G$ with Sylow $p$-subgroup $P$ and with $C\_{M}(O\_{p}(M)) \subseteq O\_{p}(M)$. The question is to determine whether $P$ is also a Sylow $p$-subgroup of $G$. Again, what follows is an over-simplification of the necessary analysis, but the overall goal is to get many $p$-local subgroups "into one place", that is to say, into a single maximal subgroup which contains a given Sylow $p$-subgroup $P$ and many normalizers $N\_{G}(R)$ for $R$ non-trivial subgroups of $P$.
The answer is yes if there is a non-identity characteristic subgroup of $P$ which is normal in $M$. For if $1 \neq C(P)$ is a characteristic subgroup of $P$ which is normal in $M$, take a Sylow $p$-subgroup $Q$ of $G$ which contains $P$. If $Q \neq P$, then $N\_{Q}(P) > P$ and $C(P) {\rm char} P \lhd N\_{Q}(P)$, so $N\_{Q}(P) \leq N\_{G}(C(P))$. But $N\_{G}(C(P)) \geq M$ since $C(P) \lhd M.$ Since $M$ is maximal and $G$ is simple, $N\_{G}(C(P))$ is a proper subgroup of $G$ containing $M$, so that $N\_{G}(C(P)) = M$ as $M$ is maximal. But then $P < N\_{Q}(P) \leq N\_{G}(C(P)) = M$, contrary to the fact that $P$ is a Sylow $p$-subgroup of $M$. Hence $Q$ must be a Sylow $p$-subgroup of $G$.
If no such characteristic subgroup $C(P)$ exists, then more delicate analysis is necessary. This is a crucial dichotomy, which emerged in the 1960s, and was further pursued by many group theorists, including Aschbacher, Baumann, Glauberman, Niles and Thompson.
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23
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https://mathoverflow.net/users/14450
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406350
| 166,555 |
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