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2025-03-21T14:48:31.480804
| 2020-07-12T17:48:44 |
365491
|
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"Ali",
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|
Stack Exchange
|
Proof of second incompleteness theorem for Set theory without Arithmetization of Syntax
Is there a proof of the second incompleteness theorem of Godel for set theory which doesn't use Arithmetization of Syntax (Godel numbering)?
I came across a short proof by Thomas Jech (here), but I think he uses Godel numbering for defining "k" in his proof. Nevertheless Bagaria in his article (here)
in referencing Jech, mentioned there is no need of arithmetizing the syntax.
My question is, does his proof crucially depend on Godel numbering?
If the answer is yes, is there a proof of second incompleteness theorem for Set theory(ZFC) which doesn't need some fixed coding of formulas by proving there is no model of ZFC, in ZFC itself?, or the use of such coding is unavoidable?
Won’t some coding of formulas be inevitable? Even if you want it to be model-theoretic, you have to define what is a theory and what it means for a model to satisfy it.
@MonroeEskew : Do we need some coding like Godel numbering to define the notions of model and satisfaction? Aren't those definable via axioms of ZF?
You don’t have to code with natural numbers, but you have to use something.
@MonroeEskew : I don't mean we shouldn't use any coding and prove the incompleteness theorem from scratch, obviously we need at least definition of the numbers in ZF! In my question i ask for a proof with those basic coding needing for defining some notions like model and satisfaction from ZF axioms.By some coding of formula i mean, some extra coding for helping us in the proof like Godel numbering.
You can do the argument referring directly to finite sequences of symbols, meaning some natural coding in hereditarily finite sets. So you can avoid prime factorization etc.
@MonroeEskew : I see... Are there any books or articles covering these arguments or related concerns?
I think the first chapter of Kunen’s Set Theory has a good sketch of the argument.
For strong enough finite fragment $T$ of $\mathsf{ZFC}$ it would be possible to define a model $M$ of $T$ to be a binary relation $(D^M,\in^M)$ such that we have relativizations of all axioms of $T$ to $(D^M,\in^M)$. So for this kind of theories it would be possible to formulate model-theoretic version of Gödel's 2nd incompleteness theorem without reference to enconding of syntax. However, as far as I know, there are no proofs of G2 that avoid encoding of syntax. But, it is still possible to show this way that $T$ doesn't prove that there is a transitive model of $T$ (using regularity).
See also: https://fuchino.ddo.jp/notes/woodin-incompl-e.pdf and also https://andrescaicedo.files.wordpress.com/2010/11/2ndincompleteness1.pdf
More interesting is whether you can avoid self-referenciality (no appeals to Gödel's diagonal lemma, for instance). See for instance MR4009518. Visser, Albert, From Tarski to Gödel—or how to derive the second incompleteness theorem from the undefinability of truth without self-reference. J. Logic Comput. 29 (2019), no. 5, 595–604.
It makes sense to me to discuss the first incompleteness theorem without coding syntax, but I don't see how it makes sense to discuss the second incompleteness without an internal account of syntax, since the second incompleteness theorem asserts that the theory T does not prove its own consistency. But to even express the statement of the second incompletness theorem, one must already have an internal account of the theory and an internal account of the notion of proof.
I understand the question to be about how to represent syntax internally in a theory. The traditional approach in logic is to use natural numbers and Gödel encodings.
Computer science, and in particular programming language theory, have a very rich theory of syntax. The abstract view of syntactic expressions is that they are finite trees, and that is how syntax is actually represented internally in compilers and other programs that process syntactic expressions. Of course, trees can easily be dealt with in a set theory, so you could use them to present syntax.
(A word of warning: if you speak to a computer scientit and state that syntax is about sequences of symbols they'll think you ignorant, or that you teleported from 1950's. Concrete syntax that humans use is indeed made of sequences of symbols, but these are viewed as convenience for humans, and are prompty parsed into abstract syntax trees. Nobody ever writes code that processes syntax by working directly with sequences of symbols.)
Category theorists have something to say about syntax as well. Joyal's arithmetic universes were designed to give an abstract account of Gödel's incompleteness proofs. Joyal never managed to publish this work, but Emilia Maietti kindly wrote it up in Joyal's arithmetic universes via type theory.
This is not the end of the story however. (Some) computer scientists are obsessed with finding the best way to deal with bound variables and binding operators (such as $\forall$, $\exists$, $\int$, etc.). There are abstract mathematical accounts of syntax with binding, for example higher-order abstract syntax. Perhaps the most interesting to this audience is nominal syntax which uses permutation models of ZF. Murdoch Gabbay's publications page has a wealth or resources, perhaps one can start with Foundations of nominal techniques: logic and semantics of variables in abstract syntax (published as https://doi.org/10.2178/bsl/1305810911 in the Bulletin of Symbolic Logic).
As mentioned in the comments, there are semantic proofs in set theory (thus non constructive, i.e they do not provide a recipe for constructing an undecidable sentence). However, Goedel himself first proved his celebrated result without arithmetization - and constructively. He showed the result to von Neumann, who asked him if he could produce an arithmetic undecidable sentence. Goedel then revised his proof for publication using his famous coding, beta-function, etc. The original proof without arithmetization as far as I know is lost. I don't know of a logician who has constructed this simpler proof, but it is possible. You don't need polynomials to talk about undecidability - that is an roundabout device that Goedel introduced simply in order to make the undecidable sentence an arithmetic (diophantine) one. The sentence can also be in ZF set theory for example, which states that a certain sequence does not belong in a certain set - the sequence could be of ordinal numbers, with the coding of each number represents a symbol in the language. 0 = (, 1 = ), 2 = universal quantifier, etc. etc. And then you don't use polynomials at all. Your sentence is just about sequences of ordinals.
|
2025-03-21T14:48:31.481266
| 2020-07-12T18:05:51 |
365493
|
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|
Stack Exchange
|
Higher derivatives of weighted zeta function and continuation to $s=1$
Is it possible to analytically continue to weighted reciprocal zeta function $\frac{1}{\zeta_{a}(s)}$ to the point $s=1$ by bounding the quantity
\begin{equation}
\left|\frac{d^n}{dx^n}\prod_{k\in S}\left(1-\frac{1}{k^x}\right)\right|\tag{1}
\end{equation}
for real $x>1$ and sets $S$ containinnng real numbers greater or equal to $2$?
Here the reciprocal weighted zeta function is defined by
\begin{equation}
\frac{1}{\zeta_{a}(s)}=\mathbb{E}_{n\in\mathbb{N}}\left[a_n\prod_{p|n}\left(1-\frac{1}{p^{s-1}}\right)\right]
\end{equation}
where $a=\{a_1,a_2,a_3...\}$ is a bounded sequence of real numbers. The motivation behind this definition is that using standard methods one can show that for any complex $s$ with $\Re(s)>1$
\begin{equation}
\frac{1}{\zeta(s)}=\mathbb{E}_{n\in\mathbb{N}}\left[\prod_{p|n}\left(1-\frac{1}{p^{s-1}}\right)\right]\tag{2}
\end{equation}
and so by weighting (2) according to some set $a$ we obtain our definition.
I wish to show that $\frac{1}{\zeta_{a}(s)}$ has an analytic continuation to $s=1$, and to do this we can construct the power series of $\frac{1}{\zeta_{a}(s)}$ about $s=1+\epsilon$ some small $\epsilon>0$ and show that the radius of convergence of this power series is greater than $\epsilon$, thus granting a continuation. To construct this power series we see that
\begin{align*}
\left|\frac{d^n}{ds^n}\frac{1}{\zeta_{a}(s)}\right|&=\left|\frac{d^n}{ds^n}\mathbb{E}_{n\in\mathbb{N}}\left[a_n\prod_{p|n}\left(1-\frac{1}{p^{s-1}}\right)\right]\right|\\
&\leq\mathbb{E}_{n\in\mathbb{N}}\left[\left|a_n\frac{d^n}{ds^n}\prod_{p|n}\left(1-\frac{1}{p^{s-1}}\right)\right|\right]\\
&\leq M\mathbb{E}_{n\in\mathbb{N}}\left[\left|\frac{d^n}{ds^n}\prod_{p|n}\left(1-\frac{1}{p^{s-1}}\right)\right|\right]
\end{align*}
where $M=\sup|a|$. I imagine that (1) has nice bounds, from which we could show that our power series has a suitable radius of convergence thus granting a solution. Is this possible?
EDIT: Motivation for analysis of the weighted zeta function
One can relatively easily show that
\begin{equation}\frac{1}{\zeta_a(s)}=\sum_{q=1}^{\infty}\frac{\mu(q)\mathbb{E}_{n\in\mathbb{N}}[a_{qn}]}{q^s}\tag{3}\end{equation}
If $\frac{1}{\zeta_a(s)}$ has an analytic continuation to $s=1$ and if
\begin{equation}\lim_{N\to\infty}\frac{1}{N}\sum_{q=1}^N \mu(q)\mathbb{E}_{n\in\mathbb{N}}[a_{qn}]\tag{4}\end{equation}
then that is sufficient to show that (3) converges at $s=1$, ensuring the identity
$$\sum_{q=1}^{\infty}\frac{\mu(q)\mathbb{E}_{n\in\mathbb{N}}[a_{qn}]}{q}=0$$
for any bounded sequence $a_n$ for any bounded sequence $a_n$. I have ideas on how to show that (4) holds, and so if one could resolve the question of analytic continuation to $s=1$ I would have this nice identity. For specific sequences $a_n$ one gets various interesting results, like for $a_n$ periodic mod $r$ where we get that
$$\sum_{q=1}^{\infty}\frac{\mu(q)f(\gcd(r,q))}{q}=0$$
for $a_n$ multiplicative we can show that if $f(n)$ is a multiplicative funnction with $\mathbb{E}_{n\in\mathbb{N}}[f(n)]\neq0$ then
$$\sum_{n=1}^{\infty}\frac{\mu(n)f(n)}{n}=0$$
having a nice identity to tie all these results together would be the dream and it would be nice if we could establish a large chunk of the criterion by resolving this problem.
|
2025-03-21T14:48:31.481716
| 2020-07-12T19:31:57 |
365494
|
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"Eduardo Longa",
"Llohann",
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|
Stack Exchange
|
Is Gauss map of a free boundary convex disk a diffeomorphism?
I asked this question on MSE, but obtained no answer. Maybe this is the right place to post it.
Let $D$ be a properly embedded free boundary disk in the closed unit ball $\mathbb{B}^3$ of $\mathbb{R}^3$. This means that $D$ is a smooth disk embedded in this ball, $D \cap \partial \mathbb{B}^3 = \partial D$ and this intersection is orthogonal. By orthogonality here I mean this: if $N$ is a unit normal along $D$ (its Gauss map), then $\langle N(x), x \rangle = 0$ for all $x \in \partial D$.
Assume that $D$ is strictly convex, that is to say, the principal curvatures are positive at each point of $D$ with respect to the fixed unit normal $N: D \to \mathbb{S}^2$. Does it follow that $N$ is a diffeomorphism onto its image? Equivalently, is $N$ injective?
The motivation is the following: if $S$ is a closed and connected surface in $\mathbb{R}^3$ which is also convex, then $N : S \to \mathbb{S}^2$ is a local diffeomorphism, hence a covering map. Since $\mathbb{S}^2$ is simply connected, this implies that $N$ is a global diffeomorphism.
What happens when the surface is a disk?
How about doubling the sphere (topologically), have you though about it? If you consider $\mathbb B^3\cup \mathbb B^3$, you get a surface $S=D\cup D$. By the free boundary hypothesis, N is continuous (maybe it can be made smooth) and the local diffeomorphism hypothesis carries over for every point outside $\partial D$.
The answer is yes. To show this one can use the fact that any topological immersion (locally one-to-one continuous map) of an n-dimensional disk into a sphere of the same dimension is an embedding (globally one-to-one) for $n\geq 2$, provided only that the map is one-to-one on the boundary of the disk. A proof may be found in
Gauss map, topology, and convexity of hypersurfaces with nonvanishing curvature,
Topology, 41 (2002) 107-117.
So it remains to show that $N$ is one-to-one on the boundary $\partial D$ of the disk $D$. To see this one can extend $D$ to a complete $\mathcal{C}^1$ convex surface by attaching to $\partial D$ all the rays which are orthogonal to $S^2$ from outside. These rays belong to a convex cone $C$ with apex at the center $o$ of $S^2$. Since $D$ has positive curvature, it follows from basic differential geometry that $\partial D$ has positive geodesic curvature in $S^2$, and hence is strictly convex, which in turn yields that $C$ is strictly convex. So $N$ will be one-to-one along $\partial D$, since $N$ is just the Gauss map of $C\setminus\{o\}$.
Incidentally, it is not necessary to assume that $D$ is convex or even embedded, but it is enough that it have positive curvature and satisfy the free boundary condition; see the following paper with Changwei Xiong
Nonnegatively curved hypersurfaces with free boundary on a sphere,
Calc. Var. Partial Differential Equations, 58 (2019), Art. 94, 20 pp.
How does one know that $C$ is of class $C^1$?
Because $\partial D$ is $C^1$.
|
2025-03-21T14:48:31.481938
| 2020-07-12T19:49:49 |
365495
|
{
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"authors": [
"Peter Humphries",
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|
Stack Exchange
|
On the ''generalised'' Chebyshev psi function
Let $\chi$ be a Dirichlet character mod $q$ and $\Lambda(n)$ be the von Mangoldt function. Let $c(\chi)=1$ if $\chi$ is the principal character, and zero otherwise. Let $\Theta_\chi$ be the supremum of real part of the zeros of the associated $L$-function $L(s, \chi)$. Define the generalised Chebyshev psi function $\psi(x, \chi):=\sum_{n\leq x} \Lambda(n)\chi(n)$. Are there infinitely many $x \rightarrow \infty$ such that $$\psi(x, \chi) - c(\chi)x = \Omega(x^{\Theta_{\chi}-\varepsilon})$$ for each $\chi$ mod $q$ and every $\epsilon>0$ ?
I think this might actually not be known unless one assumes that $L(s,\chi)$ is nonvanishing when $\Im(s) = 0$ and $\Re(s) > 1/2$, in which case the usual proof via Landau's lemma (Lemma 15.1 of Montgomery-Vaughan) works.
|
2025-03-21T14:48:31.482029
| 2020-07-12T20:46:36 |
365499
|
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"Ben Wieland",
"David Corwin",
"https://mathoverflow.net/users/1355",
"https://mathoverflow.net/users/4639"
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|
Stack Exchange
|
What sort of object represents skyscaper sheaves on the etale site of $\mathbb{Z}_p$?
By SGA 4 IX Proposition 2.7, any constructible sheaf $\mathcal{F}$ on a qcqs scheme $X$ can be represented as an equalizer of two etale maps between representable (by schemes) sheaves. This would imply, in particular, that the original sheaf is representable by an algebraic space that is etale over $X$.
What does the representing algebraic space look like if $X=\operatorname{Spec}{\mathbb{Z}_p}$, and $\mathcal{F}$ is the sheaf represented by $\operatorname{Spec}{\mathbb{F}_q}$? More generally, let's take a non-constant skyscraper sheaf represented by $\operatorname{Spec}{\mathbb{F}_q}$ for $q$ a power of $p$.
This is tautological. We define Spaces by declaring all sheaves to be Spaces. Then a sheaf is its own espace etale (though there is a finiteness condition to check in the claim that it is an Algebraic Space). . . . Look at more examples of algebraic spaces. Seek more examples; maybe try Knutson. Anyhow, example 2 is the answer to your question (mutatis mutandis).
@Ben: Yes, I know, but the question is, what kind of space? What properties does it have? What's an explicit description?
Did you follow my link?
Oops sorry - now I see. Thanks!
So if I want the sheaf represented by $\operatorname{Spec}{\mathbb{F}_p}$, I could take the quotient of $\operatorname{Spec}{\mathbb{Z}_p}$ by the étale equivalence relation $\operatorname{Spec}{\mathbb{Q}_p} \to \operatorname{Spec}{\mathbb{Z}_p}$?
$F_p$ reps terminal sheaf. $i_*$ rep by $Z_p$, also terminal. First study skyscraper sheaves in the usual topology or in the Zariski topology. Push forward the coproduct of two copies of the terminal sheaf, represented by the disjoint union of two copies of $\mathrm{Spec}F_p$ produces something weirder. It is represented by "the line with the doubled origin," a scheme where two copies of $\mathrm{Spec}Z_p$ are glued along $\mathrm{Spec}Q_p$. The full equivalence relation is: $\mathrm{Spec}(Z_p\times Q_p)\Rightarrow\mathrm{Spec}(Z_p\times Z_p)$. $F_q$ yields etale twisted version.
So are you saying that it's still a (non-separated) scheme?
Let $F$ be some constant étale sheaf on $Z=\operatorname{Spec} \mathbb{F}_p$. This is the constant sheaf $\Gamma_Z^\ast(S)$ for some set $S$. Let $X=\operatorname{Spec}(\mathbb{Z}_p)$, and let $E=\Gamma^\ast_X(S)$ be the corresponding constant sheaf on $X$. This sheaf is representable by the scheme $X\times S=\coprod_{s\in S} X$ étale over $X$.
Then let $U=X\setminus Z$. We can present $i_\ast(F)=\operatorname{colim}(U\times_X X\times S^2\rightrightarrows X\times S)$, which is an algebraic space because $U\times_X X\times S^2=U\times S^2\to U \to X$ is an étale scheme over $X$ as well, so the two arrows go between étale schemes over $X$ and therefore must be étale (a priori it is a DM stack, but the colimit is a sheaf rather than a stack).
(Jon Pridham just showed me this example earlier today!).
More generally for a local system, you can descend it to a local system on $\mathbb{Z}_p$ because it is Henselian with closed point given by $\mathbb{F}_p$, so their étale fundamental groups agree. Then use the exact same construction as above.
Anyway, this is a completely general fact that every sheaf on the small étale site of a scheme $X$ is representable by some (potentially horrible) algebraic space that is étale over $X$ (note: I do not require any (quasi)separation or quasicompactness conditions on an algebraic space).
|
2025-03-21T14:48:31.482265
| 2020-07-12T21:04:30 |
365501
|
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"Aaron Mazel-Gee",
"David Corwin",
"Denis Nardin",
"Zhen Lin",
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|
Stack Exchange
|
Group cohomology as homotopy groups
Let $G$ be a group and $A$ a group with a $G$-action. Then in general, $H^0(G;A)=A^G$ is a group, and $H^1(G;A)$ is simply a pointed set. If $A$ is an abelian group, then $H^i(G;A)$ exists and is an abelian group for all $i \ge 0$.
This resembles the fact that if $X$ is a pointed space, then $\pi_1(X)$ is always a group, while $\pi_0(X)$ is just a pointed set. But if $X$ has an abelian group structure (more generally, a double loop space structure), then all $\pi_i(X)$ are abelian groups.
We can extend the analogy further to the non-pointed situation if we consider as given not a group $A$ with an action of $G$, but an extension $0 \to A \to E \to G \to G$. Then there is $H^2(G;A)$, which is simply a truth value, depending on whether the sequence splits or not. This is analogous to the fact that for a non-pointed space $X$, $\pi_{-1}(X)$ is a truth value, depending on whether $X$ is empty or not.
The relation can be made more explicit if we let $X$ be the space (groupoid) of sections of $E \to G$. Then $H^i(G;A)$ is naturally $\pi_{1-i}(X)$, at least for $0 \le i \le 2$.
Is there a generalization to all natural numbers $i$ in the case that $A$ is abelian? This would seem to require producing a non-connective spectrum of sections, but I'm not sure how to do this. My best attempt is to deloop and think about $0 \to BA \to BE \to BG \to 0$. However, how do I deal with an infinite delooping of $BA$ in this sequence when $BG$ is not a loop space (because $G$ might be nonabelian)?
Also, is there a group cohomology analogue of the case where $X$ is a single loop space? I.e., where $H^0(G;A)$ is an abelian group, $H^1(G;A)$ is a group, $H^2(G;A)$ is a pointed set, and $H^3(G;A)$ (if you set things up correctly) is a truth value?
Isn't the observation in your second paragraph actually a theorem? Specifically, for a space $X$ and a sheaf $A$ of abelian groups on $X$, $H^n (X, A) \cong \pi_{m - n} \mathbf{R} \Gamma (X, K(A, m))$, where $K(A, m)$ is the simplicial sheaf version of the usual thing. Group cohomology comes about when you have $X = B G$. The existence of $H^1 (G, A)$ for non-abelian $A$ comes about because we can define $K (A, 1) = B A$ for non-abelian $A$ as well, etc.
Do you mean in my fourth paragraph?
I mean second. I'm not familiar with $H^2$ for group extensions.
Can you give a reference for the definition $H^i(G,A)$ when for a group extension? Or do you just mean normal group cohomology and the extension is just an extension with abelian first term?
In paragraph 5, I think you may be looking for the notion of a parametrized spectrum over $BG$. You can think of such a thing as a functor $BG \to Spectra$, or equivalently as an infinite loop object in $Spaces_{/BG}$. More generally, for a base space $X$, if $X$ has an $E_n$-monoidal structure (e.g. an $n$-fold loopspace) you can contemplate $E_n$-maps $X \to Spectra$. This sort of situation (in the case where the functor is valued in invertible spectra) is studied in Antolin-Camarena--Barthel arXiv:1411.7988.
|
2025-03-21T14:48:31.482489
| 2020-07-12T21:05:56 |
365502
|
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|
Stack Exchange
|
Regularity properties of Minakshisundaram–Pleijel zeta function
Let $(M,g)$ be a closed (compact, no boundary) smooth $n$-dimensional Riemannian manifold. The Laplace–Beltrami operator $\Delta_g$ on $M$ has discrete spectrum $(\lambda_j)_j$ (indexed without multiplicity) with corresponding eigenfunctions $(\psi_j)_j$, normalized in $L^2(\rm{vol}_g)$. The MP zeta function of $\Delta_g$ at a point $p$ in $M$ is defined as
$$\zeta^{\Delta_g}_p(s):=\sum_j \lambda_j^{-s} \psi_j(p)^2, \qquad s\in \mathbb C.$$
Question: Is something known about the continuity/differentiability of $p\mapsto \zeta^{\Delta_g}_p(s)$ as a real-valued function on $M$, say, for real $s>n/2$ ?
As far as I understand, the original paper by Minakshisundaram and Pleijel is only concerned with properties of $\zeta^{\Delta_g}_p(s)$ as a meromorphic function of $s$, holomorphic for $\Re(s)>n/2$. However, it seems to me that the authors do not discuss the properties of $p\mapsto \zeta^{\Delta_g}_p(s)$ for fixed $s$ (equivalently, of the polynomials $A$ in Eqn. (28) in their paper).
Based on Hörmander-type estimates on the $L^\infty$-norm of $\psi_j$ and Weyl's asymptotic, one can show that, for large enough $s$ (e.g. $s\geq n$), the function $p\mapsto \zeta^{\Delta_g}_p(s)$ is continuous (by showing the total convergence of the series). This is however unsatisfactory, for I would hope that $p\mapsto \zeta^{\Delta_g}_p(s)$ is continuous (possibly even smooth) for all $s>n/2$, which is the threshold originating in Weyl's asymptotic for the $\lambda_j$'s. Is there any result in this direction which holds for all $s>n/2$?
|
2025-03-21T14:48:31.482614
| 2020-07-12T21:58:14 |
365504
|
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|
Stack Exchange
|
Baffling proof using function convexity
Let the function $f: \mathbb{R} \rightarrow \mathbb{R}$ be convex, differentiable with derivative $f_x$ and Lipschitz continuous with constant $L$. Then, for $a,b,c,d \in \mathbb{R}$ such that $a \ge b\ge d $ and $ a \ge c\ge d$,
\begin{equation*}
\begin{split}
& f(\max\{ b,c\}) - f(a) + f(\min\{ b,c\}) - f(d)\\
& \le f_x(\min\{ b,c\})(b -d + c - a). \\
\end{split}
\end{equation*} Apparently this can be proven easily using $\max\{ b,c\} - a \le 0 $ and the convexity of $f$, but I am stumped about how exactly that is done … almost seems like a mistake! Would really appreciate any pointers or tips.
For reference, this is from the proof of Lemma 3.2 in the paper: Boetius, Frederik, and Michael Kohlmann. "Connections between optimal stopping and singular stochastic control." Stochastic Processes and their Applications 77.2 (1998): 253-281.
Since the statement is symmetric in $b$ and $c$, it's probably easier just to assume that $a \ge b \ge c \ge d$ and re-write it as $f(b) - f(a) + f(c) - f(d) \le f'(c)(b - d + c - a)$.
Hi LSpice - thanks for answering - the problem is that this is an argument used in a proof that $b \ge c$, so can't really make that assumption here. I'm seriously thinking this entire argument is a mistake
It's just notation. If you can prove the result in the form in which I've stated it, then you can switch the labels $b$ and $c$ in case they're in the other order.
As suggested, assume that $a \ge b \ge c \ge d$ and re-write the desired inequality as
$$
f(b) - f(a) + f(c) - f(d) \le f'(c)(b - d + c - a).
$$
We have $f(a) - f(b) \ge f'(b)(a - b)$, $f(c) - f(d) \le f'(c)(c - d)$, and $f'(b) \ge f'(c)$ by convexity, so the result follows.
Thank you LSpice, you are 100% right here
|
2025-03-21T14:48:31.482864
| 2020-07-12T22:40:21 |
365505
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/365505"
}
|
Stack Exchange
|
Equivariant splitting of loop space of a suspension
It is well known, e.g. by Cohen's "A model for the free loop space of a suspension", that there is a stable splitting of the free loop space $\mathcal{L}
\Sigma X $of the suspension $\Sigma X $ given by
$$
\Sigma^{\infty} \mathcal{L} X \simeq \bigoplus_{n>1} \Sigma^{\infty}(S^1_+ \wedge_{C_n} X^n)
$$
where $C_n$ is the cyclic group of order $n$, and $S^1_+ \wedge_{C_n} X^n$ is the $S^1$-space induced by the inclusion of groups $C_n \to S^1$ and viewing $X^n$ as a $C_n$-space.
I would like to know if this can be promoted to an $S^1$-equivariant splitting, at least in the case when $X$ is an $k$-sphere. There seems to be some equivariance going on, as the map is induced by a map (at the level of spaces)
$$
X^n \to \mathcal{L}X
$$
which is equivariant with respect to the $C_n$-action on the left, and the $C_n$-action on the right obtained by restricting the $S^1$ action. Any input would be greatly appreciated!
The following article, which you probably have seen, might be related https://www.jstor.org/stable/2374740
|
2025-03-21T14:48:31.482969
| 2020-07-13T00:18:21 |
365510
|
{
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"David Roberts",
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"Tim Kohl",
"bof",
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"url": "https://mathoverflow.net/questions/365510"
}
|
Stack Exchange
|
bijection mapping a transversal to a transversal
The following must certainly be a standard result, so what I'm looking for is a reference, or the name of this theorem. I don't have any combinatorics books at my fingertips, but I could see this being included therein.
If finite sets $X$ and $Y$ are each partitioned into $n$ equal size subsets $\{X_i\}$ and $\{Y_i\}$ for $i$ from $1$ to $n$, and $f\colon X\to Y$ is a bijection, there exists $x_i$ in $X_i$ and $\sigma\in S_n$ such that $f(x_i)\in Y_{\sigma(i)}$, for $i$ from $1$ to $n$.
I was able to come up with an argument for this myself without too much difficulty, but I'd rather not include a proof of a (very likely) well known result.
Thanks.
The current statement seems false if $n=|X|=|Y|$ since for $n > 1$ there are bijections with no fixed points. Perhaps there is a typo?
More generally, consider $Y=X$ and with the same partition. Consider then a cyclic permutation $f$ of the equivalence classes. Then $f(x_i)\not\in Y_i$ for each $i$. It seems you are using some unstated assumptions
I think what you really want is not $f(x_i)=y_i$ but a bijection $g:{1,\dots,n}\to{1,\dots,n}$ such that $f(x_i)=y_{g(i)}$. There is no loss of generality in assuming that $X=Y$ and $f$ is the identity map. If that's what you meant to ask about, it is indeed a standard theorem about the existenco of a "common transversal" or "common system of representatives" for two partitions. A classical application is that, if $H$ is a finite subgroup of a group $G$, then there is a system of representatives for the left cosets which is also a system of representatives for the right cosets.
Here is a precise statement of what I think you are asking about: if $n$ is a positive integer, and if a set $S$ (not necessarily finite!) is partitioned into $n$-element sets $X_i$ $(i\in I)$, and the same set $S$ is also partitioned into $n$-element sets $Y_j$ $(j\in J)$, then there is a set $R\subseteq S$ such that $|R\cap X_i|=|R\cap Y_j|=1$ for all $i\in I$, $j\in J$. Due to Philip Hall, I believe.
@bof another way to think of this (at least in the finite case) is by considering the projection map $\mathrm{pr}\colon X\times I \to I$ where $X$ and $I$ are finite (note $|I|=n$, not $|X|=n$, in the question. Let $|X| = k$), and $q\colon X\times I \to I$ is another function with $|q^{-1}(i)| = k$ for all $i$. Is there then, for each $i\in I$, an $x_i\in X$ such that $q(x_i,i) = i$?
Yes, my mistake, I want f(x_i)=y_{sigma(i)} for some \sigma in S_n. The proof I have is that if f(X_i)=Y_j for some i,j then one can remove these two and get a bijection from X-X_i to Y-Y_j and argue by induction. Otherwise, each f(X_i) intersects at least two different Y_j and then basically one can recursively find the transversal elements that get mapped to each other. @bof, I looked at Hall's Marriage Theorem, but I thought it only gave conditions on the existence of a transversal in a possibly overlapping collection, but maybe it does also imply my statement. Thanks.
@TimKohl I suggest you edit the question to include your correction, otherwise it is buried in the comments.
Let $S_i={j:X_i\cap Y_j\ne\emptyset}$. Show that your assumptions on $X_i$ and $Y_j$ (two partitions of the same set into $k$-element sets) imply that the overlapping sets $S_i$ satisfy the conditions of Hall's Marriage Theorem, so you get a permutation $\sigma\in S_n$ such that $X_i\cap Y_{\sigma(i)}\ne\emptyset$, and then choose an element $x_i\in X_i\cap Y_{\sigma(i)}$.
You really should update the question. Does your (proof of a) method of recursively finding a transversal use the fact that the sets are all the same size? If not it might not be correct. A curious fact is that after finding the n pairs and removing them you have n equal size sets where the size in one smaller, so you can find a second family of n pairs (perhaps with a different permutation) and continue in this manner.
@AaronMeyerowitz We assume there are an equal number of sets in each. Assume $f(X_j)\cap Y_j\neq\emptyset$ for at least two $j$, for all $i$. If $I={1\dots n}$ then let $R_0={(i,j)\in I\times I\ |\ f(X_i)\cap Y_j\neq\emptyset}$ and let $R_0^{-1}(b)={(i,b)\in R_0}$ and let $R_0(a)={(a,j)\in R_0}$. Pick $l_1$ such that $|R^{-1}(l_1)|$ is minimal and find $k_1$ such that $(k_1,l_1)\in R_0$ and then let $R_1=R_0-(R_0^{-1}(l_1)\cup R_0(k_1))$. Then find $l_2$ such that $R_1^{-1}(l_2)$ is minimal, and find $k_2$ such that $(k_2,l_2)\in R_1$ etc. and this eventually stops after $n$ steps.
This is a consequence of Hall's marriage theorem. Does anyone have the time to explain how? (The first step is to identify $X$ with $Y$ along $f$, so that $f$ becomes the identity map. Then we are looking for a system of common representatives for the partitions $\left{X_i\right}$ and $\left{Y_i\right}$.)
Actually the statement for bijections between sets partitioned into equal numbers of subsets is a consequence of an earlier result due to K{"o}nig. Hall actually acknowledges this special case.
|
2025-03-21T14:48:31.483311
| 2020-07-13T03:41:56 |
365516
|
{
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"authors": [
"Mateusz Kwaśnicki",
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|
Stack Exchange
|
Does Ahlfors–David regularity of a measure imply its Fourier asymptotic behavior?
Let $\mu$ be a Borel probability measure on $R^d$. If $\mu$ satisfies $\mu(B(x,r))\le Cr^\alpha$ for any $x\in R^d$ and $r>0$, then Strichartz (Fourier asymptotics of fractal measures, J. Funct. Anal. 1990) proved that
$$\limsup\limits_{R\to \infty}\frac{1}{R^{d-\alpha}}\int_{|x|\le R}|\widehat{\mu}(x)|^2dx\le C_2.$$
Here $\widehat{\mu}(x)=\int e^{-2\pi itx}d\mu(t)$ is the Fourier transformation of $\mu.$
My question is: if $\mu$ is Ahlfors–David regular, i.e., $C^{-1}r^\alpha\le\mu(B(x,r))\le Cr^\alpha$ for any $x\in R^d$ and $r>0$, can we have
$$C_1\le \liminf\limits_{R\to \infty}\frac{1}{R^{d-\alpha}}\int_{|x|\le R}|\widehat{\mu}(x)|^2dx\le\limsup\limits_{R\to \infty}\frac{1}{R^{d-\alpha}}\int_{|x|\le R}|\widehat{\mu}(x)|^2dx\le C_2?$$
Here $C, C_1, C_2$ are all constants.
Does someone know related results or can prove it? Thanks!
Looks like necessarily $\alpha = d$, right? Or did you mean "for any $x$ in the support of $\mu$" rather than "for any $x \in R^d$"?
Thanks! @ Mateusz Kwaśnicki I mean for any $x$ in the support of $\mu$. I also think if $\alpha=d,$ then the inequalities hold.
(There are some details missing in this answer. Time permitting, I will try to expand it.)
It suffices to find a non-negative continuous function $\phi$ with $\phi(0) > 0$ and Fourier transform bounded and supported in $B(0, 1)$. In dimension one, $\phi(x) = (1 - \cos x)/(\pi x^2)$, $\hat\phi(z) = (1 - |z|)_+$ is a good example.
Then $\phi_R(x) = R^d \phi(R x)$ has Fourier transform $\hat\phi(R^{-1} z)$ supported in $B(0, R)$, and we have
$$ \int_{|z| \le R} |\hat\mu(z)|^2 dz \ge \frac{1}{\|\hat\phi\|_\infty} \int_{|z| \le R} |\hat\phi_R(z) \hat\mu(z)|^2 dz = \frac{2\pi}{\|\hat\phi\|_\infty} \int_{\mathbb{R}^d} |\mu * \phi_R(x)|^2 dx $$
by Plancherel's theorem. Now, $\phi(x) \ge C_1 \mathbb{1}_{B(0, r)}(x)$ for some $C_1, r > 0$, and hence
$$ \int_{|z| \le R} |\hat\mu(z)|^2 dz \ge \frac{2 \pi C_1 R^{2d}}{\|\hat\phi\|_\infty} \int_{\mathbb{R}^d} |\mu * \mathbb{1}_{B(0, r/R)}(x)|^2 dx . $$
Ahlfors regularity implies that $\mu * \mathbb{1}_{B(0, r/R)}(x) \ge C_2 (r/R)^\alpha$ on a set of measure at least $C_3 (r/R)^{d-\alpha}$ (see Edit below). We conclude that
$$ \int_{|z| \le R} |\hat\mu(z)|^2 dz \ge \frac{C_1 R^{2d}}{\|\hat\phi\|_\infty} \, (C_2 (r/R)^{\alpha})^2 C_3 (r / R)^{d - \alpha} = C_4 R^{d - \alpha} , $$
as desired.
Edit: If $r > 0$, then
$$ \int_{\mathbb{R}^d} \mu * \mathbb{1}_{B(0, r)}(x) dx = C_1 r^d \mu(\mathbb R^d) $$
(by Fubini), and
$$ \mu * \mathbb{1}_{B(0, r)}(x) = \mu(B(x, r)) \leqslant C_2 r^\alpha $$
for every $x$ (by Ahlfors regularity). (Well, we assume this for $x$ in the support of $\mu$ only, but extension to general $x$ is standard.) It follows that the Lebesgue measure of
$$ \{ x : \mu(B(x, r)) > 0 \} $$
is at least $C_3 r^{d - \alpha}$. However, if $\mu(B(x, r)) > 0$, then $\mu(B(x, 2r)) \geqslant C_4 r^\alpha$ (again by Ahlfors regularity). We conclude that
$$ \mu * \mathbb{1}_{B(0, 2 r)}(x) \geqslant C_4 r^\alpha \quad \text{on a set of Lebesgue measure at least } C_3 r^{d - \alpha} . $$
This is of course equivalent to the property used in the answer above.
@ Mateusz Kwaśnicki Many thanks! Would you like to give more comments on the claim that $\mu \ast 1_{B(0,r/R)}\ge C_2 (r/R)^\alpha$ on a set of measure at least $C_3(r/R)^{d-\alpha}$?
@ljjpfx: I added a short comment about that. Let me know if anything is unclear (or plainly wrong).
@ Mateusz Kwaśnicki :Thank you very much! It looks nice!
|
2025-03-21T14:48:31.483521
| 2020-07-13T05:13:45 |
365519
|
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|
Stack Exchange
|
Is it possible to prove the Jordan decomposition starting from Schur's decomposition?
Schur's decomposition says any matrix $A$ is similar to a upper triangular matrix $U$ i.e., there exists unitary $Q$ such that $A = Q^{-1}UQ$. If we split $U$ as $D+N$ where $D$ is the diagonal part and $N$ is the off-diagonal part, then we know $N$ is nilpotent. Any Nilpotent matrix can be brought to Jordan form using a basis $P$ i.e., there exists $P$ such that $N = P^{-1}J_NP$ where $J_N$ is the Jordan form of $N$. Thus we have,
\begin{align*}
QAQ^{-1} & = & U \\
& = & D + N \\
& = & D + P^{-1}J_NP \\
\end{align*}
This implies
$$(PQ)A(PQ)^{-1} = PDP^{-1} + J_N$$
If $P$ commutes with $D$, then we get that in the basis given by $PQ$, the matrix splits into diagonal + nilpotent parts. Is this the same as the Jordan decomposition? If so, why should $P$ commute with $D$?
It usually will not be true that the off-diagonal part of a matrix is nilpotent. Consider $\begin{pmatrix} 0 & 1 \ -1 & 0 \end{pmatrix}$, for example.
@LSpice But an upper triangular matrix with zero diagonal entries is always Nilpotent, which is the case in the question.
One of the points of true Jordan decomposition (of a not necessarily nilpotent) matrix is that the diagonal part and the strictly upper triangular part commute. This will not generally be true of D and N in your question (consider the case when D has distinct diagonal entries and N is non-zero, for exmple).
I'm sorry; I missed the "upper triangular" part. @GeoffRobinson's point is the real reason that this is not the Jordan decomposition. However, it is conjugate to the Jordan decomposition.
I am not sure I get what you mean by "brought to Jordan form", but if you don't consider the structure of $D$ while changing basis for $N$ then it won't work. Example:
$$
U =
\begin{bmatrix}
1 & 1& 0 & 0\\
0 & 1& 0 & 0\\
0 & 0 & 2 & 1\\
0 & 0 & 0 & 2
\end{bmatrix}
$$
has
$$
N =
\begin{bmatrix}
0 & 1& 0 & 0\\
0 & 0& 0 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 0 & 0
\end{bmatrix}
$$
which will presumably be transformed (without further information from $D$) into
$$
J_N =
\begin{bmatrix}
0 & 1& 0 & 0\\
0 & 0& 1 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{bmatrix}
$$
and the transformation that does it does not commute with $D$.
Other example:
$$
U =
\begin{bmatrix}
1 & 1& 0 & 0\\
0 & 1& 1 & 0\\
0 & 0 & 2 & 1\\
0 & 0 & 0 & 2
\end{bmatrix}
$$
has
$$
N =
\begin{bmatrix}
0 & 1& 0 & 0\\
0 & 0& 1 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 0 & 0
\end{bmatrix}
$$
which will presumably be left unchanged, but that doesn't produce a Jordan decomposition.
On the other hand, if you mean something else that uses the structure of $D$ by "brought to Jordan form", then sure, there is a change of basis that turns $U$ into Jordan form, but that's just obvious by following the proof of the Jordan decomposition theorem.
Thanks for the wonderful explanation and the examples. I guess I understand it better now. Particularly, if the Jordan form respects the structure of D, then commutativity follows since each block would correspond to a scaled identity block in D and commutativity is obvious.
Is the following observation true? As one can start with an arbitrarily structured D and add an arbitrary upper triangular matrix with zero diagonal entries, does this mean then, every upper triangular matrix with zero diagonal entries can be brought to every possible Jordon block form under a suitable basis?
every upper triangular matrix with zero diagonal entries can be brought to every possible Jordon block under a suitable basis No, there are some rank constraints. You can't transform $$\begin{bmatrix}0 & 1 & 0\ 0 & 0 & 1\ 0 & 0 & 0\end{bmatrix}$$ into You can't transform $$\begin{bmatrix}0 & 0 & 0\ 0 & 0 & 1\ 0 & 0 & 0\end{bmatrix}$$, for instance.
Another comment: the way I see it, there are two "ingredients" needed in the proof: (1) write the matrix as a direct sum (block diagonal matrix) of upper triangular matrices, each one with a different eigenvalue $\lambda_i$ on the diagonal, and (2) transform those diagonal blocks into Jordan form. I don't think you considered (1) here, but that is a tricky step, because the transformations that you need must involve explicitly $\frac{1}{\lambda_i - \lambda_j}$ to clear the off-diagonal block $(i,j)$ (otherwise they would work also if the $\lambda_i = \lambda_j$, which is not the case).
|
2025-03-21T14:48:31.483796
| 2020-07-13T06:34:44 |
365521
|
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|
Stack Exchange
|
Moving on Riemannian manifolds
Let $a,b,c\in\mathbb{R}^n$ such that $c$ is inside the $n$-disk with $a$ and $b$ as south and north poles. Then as $c$ moves toward $a$ through the line segment joining $a$ and $c$, $c$ is also moving away from $b$ (in terms of $d(b,c)$).
Is there an analogy of the above statement on Riemannian manifolds? Say replace 'distance' by 'Riemannian metric' and replace 'line segment' by 'geodesics'?
Or, in other words, fix any $a$ and $b$ on a Riemannian manifold, what is the set of points $c$ such that when $c$ moves to $a$ through geodesics, $c$ is moving away from $b$? On $\mathbb{R}^n$ we know the set contains all points $c$ such that $\angle acb$ is not acute.
Let your space be a line, $c=0,a=1,b=2$. When $c$ moves to $a$ through the line segment $[0,1]$, it is moving towards $b$.
@AlexandreEremenko In this case $c$ is not inside the disk with $a$ and $b$ as south and north poles, where the disk is $[1,2]$.
Then please explain what are the South and North poles for arbitrary Riemannian metric.
@AlexandreEremenko I am not sure, and that is what I am wondering. Put it in another way, fix any $a$ and $b$ on a Riemannian manifold, what is the set of points $c$ such that when $c$ moves to $a$ through geodesics, $c$ is moving away from $b$?
You are talking about the Dirichlet domain / Voronoi diagram of a pair of points. It perhaps has other names, but that's one.
|
2025-03-21T14:48:31.483919
| 2020-07-13T09:55:28 |
365524
|
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|
Stack Exchange
|
Solve linear system with bordered positive definite matrix
I want to solve the usual $A x = b$ system. In block form:
$$ \begin{bmatrix} B & c \\ c^{T} & 0 \end{bmatrix} \begin{bmatrix} x' \\ x_{n+1} \end{bmatrix} = \begin{bmatrix} b' \\ b_{n+1} \end{bmatrix}$$
where
$B \in \mathbb{R}^{n \times n} $ is a positive definite matrix
$c \in \mathbb{R}^{n}$
$x,b \in \mathbb{R}^{n+1}$, so $x',b' \in \mathbb{R}^{n}$ and $x_{n+1},b_{n+1} \in \mathbb{R}$
Matrix $A$ is neither positive definite nor positive semidefinite.
I am not aware of well-known methods such as Cholesky or $LDL^T$ to solve this. Is there an efficient method to tackle this problem?
@RodrigodeAzevedo my original question was related to the vector with all ones, but then I forgot to specify it. I can keep the question I posted and keep it general like it is.
Have you tried using the Schur complement to "triangularize" the block matrix? Then, one would find $x_{n+1}$ and solve the smaller, $n$-dimensional linear system. Perhaps very inefficient.
@RodrigodeAzevedo Normal Schur complement cannot be applied because $0$ and $c$ are not invertible. Probably a Generalized Schur complement may help? But It requires to find the proper generalized inverse for $c$
@RodrigodeAzevedo OK, I see your point, I have to rearrange the matrix a little to obtain the positive definite matrix in the form like here: https://en.wikipedia.org/wiki/Schur_complement#Application_to_solving_linear_equations. Anyway it is also possible to use the pseudo-inverse (https://en.wikipedia.org/wiki/Generalized_inverse#Types) for the Schur complement, but it looks a bit more complicated. I am going to work evaluate your suggestion soon.
You can find $x_{n+1}$ using the Schur complement — of Cramer's rule, if you prefer. Then you have a linear system in $n$ unknowns. No need to invert anything.
If I understood correctly, with Cramer's rule I would need to compute the determinant of the Matrix $A$ and the determinant of the modified $A$ (let's say $A'$) and then compute the ratio (https://www.purplemath.com/modules/cramers.htm) $det(A') / det(A)$. It seems that the complexity of computing the determinant and the inverse of a matrix is the same (see https://en.wikipedia.org./wiki/Computational_complexity_of_mathematical_operations#Matrix_algebra)
No need to compute the determinant. The determinant of $\rm B$ is both in the numerator and in the denominator. It cancels out. You can get the same result via (block) Gaussian elimination — related to the Schur complement.
@RodrigodeAzevedo I am getting confused on your final goal. From your suggestions, using Schur complement and Cramer's rule $det(A) = det(B) det(0 - c^T B^{-1} c)$ and $det(A') = det(B) det(b' - b'' B^{-1} c)$. So, if you calculate the ratio you can simplify $det(B)$, but still you need to compute the inverse of $B$. Pardon my ignorance in this topic, but can you be a bit more explicit? TBH, I have discovered the Schur complement after your suggestions...
Do not compute the inverse. Just solve the linear system $\rm B \eta = c$ using a sensible method. Note that $\rm B$ is symmetric and positive definite.
I'm not sure about efficiency, but let $x'$ denote the $n$-dimensional truncation of the $(n+1)$-vector $x$. Then $Ax = b$ amounts to a pair of equations $Bx' = b' - x_{n+1}c$ and $c \cdot x' = b_{n+1}$. Thus $c \cdot B^{-1}(b' - x_{n+1}c) = b_{n+1}$, and so provided $c \cdot B^{-1}c \neq 0$ (footnote), we may uniquely solve for $x_{n+1}$ from $c \cdot B^{-1}b' - b_{n+1} = x_{n+1}c \cdot B^{-1}c$. With $x_{n+1}$ in hand, we may uniquely solve for $x'$ from $Bx' = b' - x_{n+1}c$. (footnote) Notice $c \cdot B^{-1} c$ is just $c \cdot \eta$ in Rodrigo's last comment.
By combining the useful comments of Rodrigo and Todd, the methodology to solve this system is shown here below. One caveat is that the method is probably not very efficient, since you need to use the decomposition of $B$ two times to find the final solution.
1. Build Equations with Schur Complement
By using Schur Complement on the original problem, one can obtain the following system:
$$
\left\{\tag{1}
\begin{array}{c}
B x' = b' - x_{n+1} c\\
c^T B^{-1} (b' - x_{n+1} c) = b_{n+1} \\
\end{array}
\right.
$$
2. Compute $\eta^T$
Compute $\eta$ using the equation $B \eta = c$. This can be done with Cholesky or similar methods since B is positive definite.
Transposing the equation $B \eta = c$ leads to $\eta^T B = c^T$ since $B$ is symmetric. Now it is possible to rewrite the equation as
$$\eta^T = c^T B^{-1} \tag{2}$$.
3. Combine
Replace $(2)$ in second equation of $(1)$, expand and simplify to compute $x_{n+1}$:
$$
x_{n+1} = \frac{\eta^T b' - b_{n+1}}{\eta^T c}
$$
Replace the value of $x_{n+1}$ in first equation of $(1)$ to find $x'$ with regular decomposition methods.
|
2025-03-21T14:48:31.484223
| 2020-07-09T23:21:59 |
365261
|
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|
Stack Exchange
|
What is wrong with $A^{(2)}_{2n}$?
When dealing with affine Kac-Moody groups, especially geometrically (e.g. by examining their affine flag varieties or affine Grassmannians) I've been taught that time and time again, issues arise in the case of $A^{(2)}_{2n}$ (or from the perspective of Tits, odd unitary ramified groups).
Some symptoms of the problem:
"All affine Schubert varieties are Gorenstein except for when the special parahoric is a parahoric of the of the ramified odd unitary group $SU_{2n+1}$... In this case, no Schubert variety of positive dimension in the affine flag variety is Gorenstein" -Zhu, "Coherence Conjecture of Pappas-Rapoport" page 7
In Kac's construction of the twisted affine algebras from diagram automorphisms of the underlying semisimple $\mathfrak{g}$, (section 8.3 of Kac "Infinite Dimensional Lie Algebras, 3rd edition), the form of the generators is different (i.e. usually we choose $\theta_0=\frac{1}{r}(\overline{\mu}(\theta^0)+ \dots + \overline{\mu}(\theta^0))$, but to build $A_{2n}^{(2)}$ we use $\theta_0=\theta^0$). Moreover in several other places in the book, I think special care is required with calculations involving $A_{2n}^{(2)}$.
Again according to Zhu in his "Coherence" paper, it seems that there have been miscalculations involving linebundles and their central charges for $Gr_{SU_{\tilde{C}/C}(2n)}$.
In what is probably the closest to the "cause" rather than a symptom, we see in Tits "Reductive Groups over Local Fields", $A_{2n}^{(2)}$ (or in Tits' naming, $C-BC_n$), this is the only local Dynkin diagram where there exist special vertices which are not exchanged by a diagram automorphism.
I would be very happy to learn a unified perspective on this series of groups. I suspect that this last point 4) is the most important point, but since I'm not an expert on buildings or Bruhat-Tits theory, I would appreciate an understanding of why this feature of the Dynkin diagram (or of the building) has such serious ramifications (ha) for the associated geometry of flag varieties, Grassmannians and linebundles.
It is my impression that beyond the split cases or certain quasi-split, the affine buildings even for classical groups don't truly admit an approach as unified as one might like. There seems to be some inescapable combinatorial substratum that is not merely a manifestation of general principles. But perhaps I just gave up too soon myself in trying to understand. :)
|
2025-03-21T14:48:31.484407
| 2020-07-09T23:22:34 |
365262
|
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|
Stack Exchange
|
Does a finite procedure for demonstrating truth-functional unsatisfiability count as a deduction method?
Question: Does a semi-effective procedure for demonstrating that a formula is truth-functionally unsatisfiable count as a ``deduction method''?
Background: According to Warren Goldfarb, in his 1928 ``On Mathematical Logic'', Skolem
argues that if a formula [of first-order predicate logic] has a
truth-functionally inconsistent expansion then it leads to a
contradiction in his deduction-method. (Goldfarb, 1979, p. 363)
As Goldfarb acknowledges, the deductive system in question is not at all explicit. Skolem mentions a few inference rules involving quantifiers but does not give what we would call a formal system. The inference rules are not involved in Skolem's example of the method he would use to show a formula to be contradictory.
The method involves his level-by-level construction (same as in his 1923) of instances of the formula created by dropping the quantifiers and substituting integers for the variables. Each level adds new instances of the formula as the universal variables are taken to range over the integers introduced at the previous level. We then consider, at each level, all the possible truth assignments to atomic components that make the formula come out true ("solutions"). By the same argument used in 1923 to show that the sequence of solutions must converge after a finite number of steps (because there are finitely many solutions at each level), Skolem here argues that when the solutions converge a contradiction is revealed, provided the formula is not consistent. This contradiction is evidenced in the fact that there are no solutions of the next level that extend those of preceding levels - the formulas of this level are what Goldfarb calls a "truth-functionally inconsistent expansion" of the original. This concludes Skolem's method for showing a formula to be contradictory in a finite number of steps. But is it a "deduction"?
References:
GOLDFARB, WARREN D
[1979] Logic in the Twenties: The Nature of the Quantifier. Journal of Symbolic Logic, 44, pp. 351-368.
SKOLEM, THORALF
[1923] Einige Bemerkungen zur axiomatischen Begründung der Mengenlehre.
Matematikerkongressen i Helsingfords den 4–7 Juli 1922, Den
femte skandinaviska matematikerkongressen, Redogörelse . Helsinki:
Akademiska Bokhandeln, 1923, pp. 217–232. English translation in van
Heijenoort (ed.) [1967], pp. 290–231.
Why not? ------
@EmilJeřábek of course the truth table method can be used to establish that an argument is deductively valid, but the process described here of checking each level for truth-functional satisfiability does not fit the definition of a deduction as an argument going from premises to conclusion by validity-preserving inference rules.
The idea here is that checking propositional validity is so simple that it can be done by a single application of a rule.
@EmilJeřábek I.e., for the propositional formulas at each level, the "deduction'' would begin by listing the atomic components, systematically constructing the truth table, and the rule would be "If at least one row of the table assigns "T" to the whole formula, write the formula; otherwise , write its negation."
Also, this would be a semantic procedure whereas Goldfarb has in mind something syntactic: 'Skolem seems to be groping for a more syntactical-but still not entirely formal-parsing of the notion "containing a contradiction", having in mind an informal deduction method.' (Goldfarb, 1979, p. 363)
|
2025-03-21T14:48:31.484770
| 2020-07-09T23:36:55 |
365264
|
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|
Stack Exchange
|
Compactly induced representations and the Whittaker model
Let $G$ be the group $GL_n(F)$, where $F$ is a non-archimedean locally compact field. Let us fix a character $\psi$ of $(F,+)$ which is trivial on $\mathcal{P}_{F}$ and non-trivial on $\mathcal{O}_{F}$ ($\mathcal{O}_{F}$ is the ring of integers of $F$ and $\mathcal{P}_{F}$ is the maximal ideal of $\mathcal{O}_{F}$) and define a character $\theta_{\psi}$ of the group $U$ of upper-triangular matrices with $1$ on the diagonal by
$$\displaystyle\theta_{\psi}((u_{i,j})_{1\leqslant i,j\leqslant n})=\psi(\sum_{i=1}^{n-1}u_{i,i+1}).$$
An irreducible smooth representation $(\pi,V)$ of $G$ is called generic if the intertwining space $\rm{Hom}_G(\pi,{\rm Ind}_{U}^{G}\theta_{\psi})$ is nonzero. J.Bernstein and A.Zelevinski proves that for an irreducible and admissible representation of $G$ the dimension of the intertwining space $\rm{Hom}_G(\pi,{\rm Ind}_{U}^{G}\theta_{\psi})$ is at most one. In particular, an irreducible generic representation $(\pi,V)$ of $G$ is isomorphic to a unique subrepresentation of ${\rm Ind}_{U}^{G}\theta_{\psi}$ called the Whittaker model of $\pi$.
My question is : Does the compactly induced representation $\hbox{$c$-ind}_{N}^{G}1_{N}$, where $N=N_G(T)$ is the normalizer in $G$ of the diagonal torus $T$, embeds in ${\rm Ind}_{U}^{G}\theta_{\psi}$ ?
Is there a reason you are interested in $\hbox{$c$-ind}{N}^{G}1{N}$?
|
2025-03-21T14:48:31.484896
| 2020-07-10T00:28:23 |
365267
|
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|
Stack Exchange
|
Does localization at quasi-isomorphisms imply homotopy invariance?
Usually, the derived category of some abelian category $A$ (I'm happy already with $A$-mod) is defined first taking chain complexes up to homotopy, and then localize at quasi-isomorphisms.
My question is, if one begins with $Chain(A)$=complexes in $A$ (instead of complexes up to homotopy equivalence) , and then localize at quasi-isomorphisms, do we get the same?
Denote
$$Chain(A)\overset{\tilde Q}{\to} [qis]^{-1}Chain(A)$$
the functor with the universal property w.r.t. this localization.
It is obvious that an homotopy equivalence is a quasi-isomorphism, but is it also obvious (or is it true?) that if $f\sim g$ are two homotopic maps then $Q(f)=Q(g)$? Is this category the same as the usual derived category?
Since homotopy equivalences are quasiisomorphisms, the localization functor $\tilde Q$ sends them to isomorphisms, hence it factors through homotopy category by its universal property. So these two localizations are the same. The advantage of localizing homotopy category is that one can use Verdier localization procedure instead of Gabriel-Zisman, which is much easier. For example, it's not obivious that the derived category will be triangulated, if you don't go through this intermediate step.
Universal property is proved with the use of a mapping cylinder: a pair of two homotopic maps from $A$ to $B$ is the same as the single map from the mapping cylinder of an identity map of $A$ to $B$. And mapping cylinder of $id_A$ is homotopy equivalent to $A$, in two ways.
@GrishaPapayanov You should post your first comment as an answer.
I would add that this happens for any model category.
Thanks Grisha and Fernando, I was looking for a direct proof ussing only mapping cones and Grisha's answer unlocked me!
Turning a comment into an answer: a pair of homotopic maps $A \longrightarrow B$ is the same as a single map from a cylinder of the identity map of $A$ into $B$. This cylinder is then homotopy equivalent to $A$ in two ways, and using this one can show that any functor that sends homotopy equivalences into isomorphisms (such as localization functor $\tilde Q$) factors uniquely through the homotopy category. In other words, functor from the category of complexes into the homotopy category is a localization. It follows that quasi-isomorphism localizations of the category of complexes and of the homotopy category are equivalent.
The advantege of going through a two-step process is that the homotopy category is already triangulated, and the Verdier localization procedure for triangulated categories is somewhat simpler than the general Gabriel-Zisman. For example, it's not obivious that the derived category will be triangulated, if you won't go through this intermediate step.
The sources for that are, for example, chapter 10 of Weibel or chapter 3 of Gelfand-Manin.
Thanks! Actually in Gelfand-Manin the definition is localizing directly (I took the notation "Q" from the,) (after that, they do all burocracy with the homotopy category too)
|
2025-03-21T14:48:31.485125
| 2020-07-10T00:29:15 |
365268
|
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|
Stack Exchange
|
Schoen and Yau's proof of the higher dimensional positive mass theorem
In April 2017 Schoen and Yau posted on the arxiv their solution of the time-symmetric positive mass theorem in all dimensions, which has been a significant conjecture since the 70s. As of now, July 2020, it hasn't appeared in a journal. Is there any consensus in the geometric analysis community on the veracity of the proof?
https://arxiv.org/abs/1704.05490
It appears to be a concensus that questions like this one, asking for validity of preprints, are considered off-topic for being too broad. See here: https://meta.mathoverflow.net/a/942/30186
It seems like there is a precedent of this rule being broken for notable mathematicians on major conjectures - even in cases where, for what it's worth, the papers are (at least outwardly) "crackpot", which this is not
It is claimed to be done much earlier by Joachim Lohkamp https://cds.cern.ch/record/981383 (also unpublished by some reason).
My (maybe wrong) impression is that Lohkamp's proof is not widely understood, either positively or negatively
It was just published here https://dx.doi.org/10.4310/SDG.2019.v24.n1.a10
It is yet unsatisfactory to see that a major result (over decades) like this is placed in a typically not rigorously peer reviewed collection of invited addresses called "Surveys in Differential Geometry" edited in chief by one of the authors, namely by Yau, and published in a volume on the occasion of his 70's birthday.
It has been utilized in some papers that are already published and whose authors are not directly related to Schoen and Yau (former PhD students, postdocs...), see e.g. https://arxiv.org/abs/1901.05263
This paper seems to cite Lohkamp for the result.
There are certainly some papers citing SY's paper for this result, e.g. by Coley and Ellis, Sakovich and Sormani, and Borghini and Mazzieri. Anyway, Schoen and Yau's approach seems obviously cogent and credible, and and so maybe the best answer to this question would clarify if there are particular parts of their work that are particularly technically demanding (or even possibly unclear). It is disappointing that their paper, along with Lohkamp's, have been out for four years without any (to my knowledge) detailed expository accounts or commentary (whether positive or negative).
May be the validity of the proof is not established. However, it seems to me that the idea presented to solve the problem is bright; In particular, that of Lohkamp, which seems to be closely linked to that of D. Christodoulou in its proof of the formation of trapped surfaces (cf https://arxiv.org/pdf/0805.3880). From my point of view, the Lohkamp attack technique is full of a huge potential, especially for mathematical general relativity.
|
2025-03-21T14:48:31.485359
| 2020-07-10T01:38:37 |
365269
|
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|
Stack Exchange
|
What is the integral of $r \frac{2^{r-1} \log (2) e^{-\frac{\sqrt{2^r-1}}{b}} \left(2^r-1\right)^{\frac{d}{2}-1}}{b^d \Gamma (d)}$?
I have been trying to solve a research problem for a while now and in doing so, I stumbled upon the following integral:
$$\int_0^{\infty } r \frac{2^{r-1} \log (2) e^{-\frac{\sqrt{2^r-1}}{b}} \left(2^r-1\right)^{\frac{d}{2}-1}}{b^d \Gamma (d)} \, dr.$$
However, I've got no idea how to solve that. Therefore, I'd like to figure out the solution for this integral.
Of course, $\frac{\log(2)}{b^d\Gamma(d)}$ is just clutter. You have $2^r - 1$ in two places and $2^{r - 1}$ in one; is that intentional?
Make the obvious substitution $u = \sqrt{2^r-1}$, and one is left with evaluating (up to multiplicative constants and my arithmetic errors) $$\int_0^{\infty} u^{d-1} \exp(-u/b) \log(1+u^2) ;du.$$ (@LSpice, the presence of $2^{r-1} \log(2)$ in the $du$ term suggests that the expression may well be "correct" as is, for whatever the appropriate definition of "correct" is in this problem.) It may not be possible to find a closed form for this integral in terms of $d$, but at the very least one should be able to find a useful series expansion...
@LSpice,the expression is correct. Thanks!
An solution for this integral can be found at Mathmatica.SE, which is reproduced next.
After applying the change of variable technique with $x=2^r-1$ we get
$$f=\frac{e^{-\frac{\sqrt{r}}{b}} r^{\frac{d}{2}-1} \log _2(r+1)}{2 \left(b^d \Gamma (d)\right)} $$
$$\text{Integrate}[f,\{r,0,\infty \},\text{Assumptions}\to d\in \mathbb{R}\land b\in \mathbb{R}\land d>0\land b>0]$$
then, the solution to this integral is
\begin{align}
\frac{1}{\log(4)b^d\Gamma(d)} \left[\frac{2\pi}{d} \csc\left(\frac{\pi d}{2}\right) \, {}_1F_2\left(\frac{d}{2};\frac{1}{2},\frac{d}{2}+1;-\frac{1}{4 b^2}\right)\\
+
\frac{1}{b^2 (d+1)} \left(2 \left\{-\pi b \sec \left(\frac{\pi d}{2}\right) \, {}_1F_2\left(\frac{d}{2}+\frac{1}{2};\frac{3}{2},\frac{d}{2}+\frac{3}{2};-\frac{1}{4 b^2}
\right) \\ +\,(d+1) b^d \Gamma (d-2)\, {}_2F_3\left(1,1;2,\frac{3}{2}-\frac{d}{2},2-\frac{d}{2};-\frac{1}{4 b^2}\right) \\
+ 2(d^3-2 d^2-d+2) b^{d+2} \Gamma (d-2) (\log (b)+\psi ^{(0)}(d))
\right\}\right)\right]
\end{align}
|
2025-03-21T14:48:31.485514
| 2020-07-10T01:49:55 |
365270
|
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|
Stack Exchange
|
Spines of Teichmuller space of a non-orientable surface
Let $S_{g,n}$ be the orientable surface of genus $g$ and $n$ punctures. Denote $\Gamma_{g,n}$ be the mapping class group of $S_{g,n}$ and $\mathcal T_{g,n}$ the Teichmüller space of $S_{g,n}$.
In http://www.math.lsa.umich.edu/~lji/a-tale-of-two-groups.pdf page 34 item (17) it is mentioned that, provided $n>0$, $\mathcal T_{g,n}$ has a $\Gamma_{g,n}$-equivariant spine of dimension equal to the virtual cohomological dimension of $\Gamma_{g,n}$.
Is there an analogous result for a non-orientable surface with at least one puncture? That is, is it true that the Teichmüller space of a non-orientable surface with at least one puncture has a spine of dimension equal to the virtual cohomological dimension of the corresponding mapping class group?
Yes... I think so? See Theorem 6.9 of Ivanov's paper Complexes of curves and the Teichmüller modular group (paywalled). It looks like his computation of the vcd goes via a (deformation?) retraction to the desired spine.
By the way, I found this via Ivanov's article in Farb's book Problems on the mapping class group and related topics. See Chapter 4, Section 10.
I do not know. I have been reading Ivanov's paper and I am not able to see where does he constructs such a spine.
|
2025-03-21T14:48:31.485628
| 2020-07-10T02:48:07 |
365271
|
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|
Stack Exchange
|
Characterize pullback functors among copresheaf-pra's
Let $P$ and $Q$ be categories, and suppose $P$ has a terminal object $\ast_p$. A parametric right adjoint, or pra, $F$ from $P$ to $Q$ is a functor $F:P\to Q$ such that the functor
$$F\,/\,{\ast_p}: P\,/\,{\ast_p}\to Q\,/\,F(\ast_p)$$
induced by slicing over $*_p$ is a right adjoint. In particular, since right adjoints preserve terminal objects, note that every right adjoint $P\to Q$ is a pra.
Suppose that $C$ and $D$ are categories, and let $P:=C{-}\mathsf{Set}$ and $Q:=D{-}\mathsf{Set}$ be the associated copresheaf categories. A functor $f\colon C\to D$ induces a pullback functor
$$\Delta_f\colon D{-}\mathsf{Set}\to C{-}\mathsf{Set},$$
which has both a left adjoint $\Sigma_f$ and right adjoint
$$
\Pi_f\colon C{-}\mathsf{Set}\longrightarrow D{-}\mathsf{Set}.
$$
In particular, each of $\Delta_f:Q\to P$ and $\Pi_f:P\to Q$ is a parametric right adjoint.
Remark: A functor $C{-}\mathsf{Set}\longrightarrow D{-}\mathsf{Set}$ is a pra iff it is isomorphic to one of the form $\Delta_e\,\overset{\circ}{,}\,\Pi_f\,\overset{\circ}{,}\,\Sigma_g$, where
$$
C\xleftarrow{e}\bullet\xrightarrow{f}\bullet\xrightarrow{g}D
$$
are categories and functors, where $(e,f)$ forms a two-sided discrete fibration, and where $g$ is a discrete opfibration. This fact is due to Mark Weber.
Definition: Let $\mathsf{pra}$ denote the category whose objects are categories $\text{Ob}(\mathsf{pra})=\text{Ob}(\mathsf{Cat})$ and for which a morphism $C\to D$ is a pra between their copresheaf categories
$$
\mathsf{pra}(C,D):=\{F\colon C{-}\mathsf{Set}\longrightarrow D{-}\mathsf{Set}\;\mid\;F\text{ is a pra}\}.
$$
We refer to $\mathsf{pra}$ as the category of categories and copresheaf-pras. It is in fact a bicategory whose 2-morphisms are natural transformations.
Question 1: Inside the bicategory $\mathsf{pra}$ of categories and copresheaf-pras, can one characterize those morphisms $F$ that are of the form $F=\Delta_f$ for some functor $f\colon D\to C$?
As pointed out by Simon Henry in an answer below, it is equivalent to ask:
Question 2: Inside the bicategory $\mathsf{pra}$ of categories and copresheaf-pras, can one characterize those morphisms $F$ that are of the form $F=\Pi_f$ for some functor $f\colon C\to D$?
I can't find in Mark Weber's paper the precise statement of this results you mentioned in remarks, but don't you need to assume that the categories under consideration are Cauchy complete for this to hold ? I also feel like Cauchy completeness will probably be helpful to give nice purely categorical characterization of these functors (as any category is isomorphic to its Cauchy completion in the category pra).
Thanks Simon. I was misremembering the result on page 674. I've edited the text.
A first remark is that question 1 and 2 are equivalent as in the category pra you have $\Delta_f \dashv \Pi_f$. So if you have a characterization of one class you characterize the other as their left/right adjoint functors
Now, the copresheaf functor from pra to Cat can be recovered as the global section functor:
Indeed, the terminal object of pra is the small category $\emptyset$, indeed $\emptyset$-Set is the terminal category and the unique functor $C$-Set $\to 1$ is always a (parametric) right adjoint functor.
Moreover, any functor $\emptyset$-Set $ =1\to C$-Set is parametric right adjoint, so you get that pra$(\emptyset$-Set,$C$-Set$) \simeq C$-Set.
So, at least assuming Cauchy completeness you can characterize the $\Delta_f$ and $\Pi_f$ as the adjunction $h \dashv g$ such that the action of $h$ on global section has a further left adjoint.
I don't think you can give a purely categorical characterization of these functor without assuming Cauchy-completness as every small category is isomorphic to its Cauchy-completion in pra: a purely categorical construction cannot distinguish between a $\Delta_f$ functor and something that is a $\Delta_f$ for an $f$ between the Cauchy completion.
|
2025-03-21T14:48:31.485850
| 2020-07-10T03:24:55 |
365272
|
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|
Stack Exchange
|
How to perturb the Chern-Simons functional
Let $Y$ be a $3$-manifold. Suppose that $P \rightarrow Y$ is a principle $SU(2)$-bundle with a choice of trivialization. Then the space of connections over $P$ is identified with $1$-forms taking value in $\mathfrak{su}(2)$. Let $A \in \Omega_{Y}^{1}(\mathfrak{su}(2))$ be a connection, then the Chern-Simons functional is defined by $CS(A) = \int_{Y} tr(A \wedge dA + \frac23 A \wedge A \wedge A)$.
To define Floer theory, people always perturb the Chern-Simons functional by holonomy perturbations, originating from Donaldson and Floer. The perturbation space is usually "abundant" and the compactness arguments go through without much change.
But from the viewpoint of equivariant Morse homology, there should be a larger class of perturbations of the Chern-Simons functional which result in a construction of (equivariant) Floer homology. So the question is: are there any other effective ways to perturb $CS$?
This tends to be very delicate. Most importantly, you need your perturbations to be continuous under the limits appearing in the Uhlenbeck compactification. This concern does not appear in the finite dimensional story and I don't know any constructions of good perturbations that aren't essentially holonomy perturbations.
|
2025-03-21T14:48:31.485968
| 2020-07-10T07:02:46 |
365278
|
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"url": "https://mathoverflow.net/questions/365278"
}
|
Stack Exchange
|
Surjective monoid homomorphism $\text{End}(B)\to \text{End}(A)$ given surjection $g:B\to A$
For any set $A\neq\varnothing$ let $\text{End}(A)$ denote the endomorphism monoid, consisting of all functions $f:A\to A$, together with composition. If $A, B\neq \varnothing$ are sets and $g:B\to A$ is a surjection, is there a surjective monoid homomorphism $\varphi:\text{End}(B)\to \text{End}(A)$?
We may take $B$ as a disjoint union of $A$ and another set $C$. Then for each endomorphism of $A$ we have an extension to $B$ by requiring the map is the identity when restricted to $C$. So the morphism $\varphi$ even splits. Edit: I took it for granted that the map $\varphi$ is already well-defined. I have to think more about it. I think I should keep this incorrect answer.
Malcev proved that for $X$ infinite, $\mathrm{End}(X)$ has no nontrivial finite quotient, so the answer is no in case $B$ is infinite and $A$ is finite of cardinal $\ge 2$. Probably it's also no for $B$ finite and large enough and $1<|A|<|B|$, by the knowledge of normal congruences of $\mathrm{End}(B)$ (other users here know more about this and might confirm of infirm).
The question is actually unrelated to the original surjection, just the assumption is (in ZFC) that $|B|\ge |A|>0$.
It seems that $|B|=3$ and $|A|=2$ is already an interesting case. Perhaps one could develop a program to test (when $|B|$ is finite).
Thanks @YCor for your comments! Is $\text{End}(\omega)$ a quotient of $\text{End}(\omega_1)$?
Probably not. Every monoid homomorphism $f:End(\omega_1)\to End(\omega)$ is trivial. Indeed $f$ induces $f:Sym(\omega_1)\to Sym(\omega)$. Every nontrivial quotient of $Sym(\omega_1)$ contains an uncountable direct sum of nonabelian groups (consequence of Baer), and this can't be mapped injectively in $Sym(\omega)$ (MacKenzie). Hence $f$ is trivial (= constant $\mathrm{id}_\omega$)) on $Sym(\omega)$. I'm not sure how to conclude; this probably follows from Malcev's classification of quotients of $End(X)$.
This is a cleaner rewrite of my original answer. The answer is no (assuming the surjection is not injective and the smaller set does not have cardinality $1$).
Let $T_A$ be the full transformation monoid on the set $A$. Then the set $C_A$ of constant maps is the unique minimal two-sided ideal of $T_A$. Since $C_A$ has the same cardinality as $A$, we have $T_B\cong T_A$ if and only if $A$ and $B$ have the same cardinality.
I claim if $|A|\geq 2$, then the unique minimal non-trivial congruence on $T_A$ is to identify all the constant maps to a zero element (absorbing element). Assuming $T_A$ acts on the left of $A$, we have that $T_A$ acts faithfully on the left of $C_A$ by essentially the same action. So any homomorphism that is injective on $C_A$ is injective on $T_A$. On the other hand, if a congruence identifies elements of $C_A$ then the restriction of the congruence to $C_A$ is a system of imprimitivity for the symmetric group $S_A\leq T_A$ acting on the left of $C_A$ which is just the same as its natural action on $A$. This action is $2$-transitive and hence primitive. Thus any non-trivial congruence on $T_A$ must collapse $C_A$.
In conclusion, every proper quotient of $T_A$ has an absorbing element and so can only be a $T_X$ if $|X|=1$. Combined with the fact that $T_A\cong T_B$ iff $A$ and $B$ have the same cardinality, we get the answer is no.
Very nice answer, thank you Benjamin!
|
2025-03-21T14:48:31.486224
| 2020-07-10T08:19:57 |
365286
|
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|
Stack Exchange
|
Compact operator without eigenvalues?
Consider the operator $M$ on $\ell^2(\mathbb{Z})$ defined by for $u\in \ell^2(\mathbb Z)$
$$Mu(n)=\frac{1}{\vert n \vert+1}u(n).$$ This is a compact operator!
Then, let $l$ be the left-shift and $r$ the right-shift on $\ell^2(\mathbb Z).$
We consider the compact operator on $\ell^2(\mathbb Z;\mathbb C^2)$ defined by
$$T:=\begin{pmatrix} 0 & l M \\ rM & 0 \end{pmatrix}$$
My question is: Even though $T$ is not normal, since $$T^*T= \begin{pmatrix} MlrM & 0 \\ 0 & Mr lM \end{pmatrix}=M^2$$
whereas $$TT^*= \begin{pmatrix} lM^2 r & 0 \\ 0 & rM^2 l \end{pmatrix}\neq M^2$$ does $T$ have eigenvalues?
What is $u$ in the definition of $M$? Where does $T$ act?
@FedorPetrov sorry, $u \in \ell^2(\mathbb Z)$ and $T$ acts on $\ell^2(\mathbb Z; \mathbb C^2).$
Note that $$T^2 = \begin{pmatrix}lMrM&0\\0&rMlM\end{pmatrix},$$ and hence the eigenvectors of $T^2$ are $$v_j = (e_j, 0) , \qquad w_j = (0, e_j),$$ with corresponding eigenvalues $$\lambda_j = \frac{1}{(1 + |j|) (1 + |j+1|)} \, , \qquad \mu_j = \frac{1}{(1 + |j|) (1 + |j-1|)} \, ,$$ respectively. In particular, the eigenspaces of $T^2$ are four-dimensional: for $j \ge 0$, the eigenspace corresponding to $\lambda_j = \mu_{j+1} = \lambda_{-j-1} = \mu_{-j}$ is spanned by $v_j$, $u_{j+1}$, $v_{-j-1}$ and $u_{-j}$.
If $u$ is an eigenvector of $T$, then it is also an eigenvector of $T^2$, and hence it is a linear combination of $v_j$, $u_{j+1}$, $v_{-j-1}$ and $u_{-j}$ for some $j \ge 0$. By inspection, $$ T(a v_j + b u_{j+1} + c v_{-j-1} + d u_{-j}) = \frac{a u_{j+1} + d v_{-j-1}}{1 + |j|} + \frac{b v_j + c u_{-j}}{1 + |j + 1|} $$ corresponds to a simple block $4\times4$ matrix, and it is now an elementary exercise to find the eigenvectors.
interesting, do you see any argument using abstract arguments that there have to be eigenvalues? It seems in this case, the simple structure of the square bailed us out-which I missed. But perhaps it is clear already from something that there has to be at least one?
@Martinique: Of course there are compact operators with no eigenvalues whatsoever. On the other hand, if any power of $T$ is normal, then the above argument works (the cyclic subspace generated by an eigenvector $v$ of some $T^n$, that is, $\operatorname{lin}{v, Tv, T^2v, T^3v,\ldots}$ is finite-dimensional, and so $T$ has an eigenvector within this subspace).
There's a typo in the penultimate paragraph: 'If $u$ is an eigenvector of $T^2$, then it is also an eigenvector of $T^2$', while tautologically true, is certainly not what was meant there... :) I presume the first one is meant to be $T$ rather than $T^2$, since that implication is true and non-trivial...
@StevenStadnicki: Hardly non-trivial, I'd say, but yes, this is what I meant. Thanks!
Small remark: In case that one is only interested in the existence of eigenvalues (as the OP seems to be, according to their comments), one can replace the second half of the argument by just noting that if $\lambda$ is an eigenvalue of $T^2$, than at least one of the complex roots of $\lambda$ is an eigenvalue of $T$ (which is true for every linear operator $T$).
@JochenGlueck: That's true, thanks. (This is essentially what I was trying to write in one of the comments above).
Sure, let $\mu_n = \frac{1}{|n|+ 1}$, then for each $n$ the vector $\sqrt{\mu_{n-1}}e_n\oplus \pm\sqrt{\mu_n}e_{n-1}$ is an eigenvector with eigenvalue $\pm\sqrt{\mu_n\mu_{n-1}}$.
Thank you also for your answer. Since Mateusz answered a few minutes earlier, I just accepted his answer, I hope this is okay. But let me also ask you, if you see any abstract argument that there have to be eigenvalues (so not by explicitly diagonalization) but by some a priori argument?
Well, under what assumptions? My construction works for your operator with any sequence in $c_0(\mathbb{Z})$ in place of $(\mu_n)$. In general every nonzero element of the spectrum of a compact operator is an eigenvalue, but there do exist compact operators with spectrum ${0}$ and no eigenvalues.
|
2025-03-21T14:48:31.486498
| 2020-07-10T09:15:28 |
365295
|
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|
Stack Exchange
|
Nullity of the linking matrix of a framed link $L$ equals the first betti number of the manifold obtained by surgery on $L$
I have asked this on mathstackexchange as well. I'm not necessarily asking for a proof, just a hint or a point to the right direction (although I'm not saying that a proof isn't welcome). I'm studying Witten invariants and I have come across a statement that most books agree that it's not hard to see. Anyway this is the question and my progress so far.
Let $L$ be a framed link of $m$ components and $(L_i,f_i)$ be the component-framing pair. The linking matrix $A=[a_{i,j}]$ where $a_{i,i}=f_i$ and $a_{i,j}=lk(L_i,L_j)$ when $i\ne j$. I have read that this matrix is a presentation matrix for the first homology group $H_1(M; \Bbb Z)$ and that the nullity of $A$ is equal to the first betti number of $M$, where $M$ is the 3-manifold obtained from $S^3$ by surgery on $L$. Can someone explain this fact to me or point me to a proof or a reference?
EDIT: I am reading Nicolaescu's Reidemeister Torsion of 3-manifolds p.77 and there I made some progress but I have some question's as well. Suppose the $c_i$ are the meridians of the solid tori we glue back on the link complement $E$ to obtain $M$. This means that we send $c_i \to f_iμ_i + λ_i$, where $μ_i, λ_i$ are the meridians and longitudes in the regular neighbourhood of $L$. It is clear to me that $μ_i$ form a basis for $H_1(E)$ and taking $K$ to be the free abelian generated by the $c_i$'s and sending them to $f_iμ_i + λ_i$ we have a morphism $K \to H_1(E)$ with matrix the linking matrix mentioned above. There Nicolaescu says that by taking the natural map $a:H_1(E) \to H_1(M)$ we complete the presentation as this map is onto. Here is when it's unclear to me. By $a$ I guess he means the map that sends $μ_i $ in $E$ to $μ_i $ in $M$. If this is the case then I don't understand why it's onto and while it's clear to me that the $f_iμ_i + λ_i$'s will be in the kernel because now they bound a disk in $M$ I don't see why they should generate it.
Hello Amontillado. It appears that you asked this question on math.SE less than 24 hours ago, and I think you should give more time for people to reply there. You could also include the link in your question. Also, I don't think that this question is appropriate here (but I'm very biased).
That's ok, If there is a problem I will delete the question. It's that recently I have not been getting a lot of answers from SE and I was wondering if someone here could help me.
|
2025-03-21T14:48:31.486827
| 2020-07-10T09:32:33 |
365297
|
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|
Stack Exchange
|
Continuity of weak solutions to wave equation with time-dependent coefficients
Consider the following second-order wave equation
$$
u_{tt} - div( a\cdot \nabla u) = f \quad \text{ in } (0,T)\times \Omega
$$
with boundary conditions
$$
u(0)=g, \ u_t(0)=h, \ u|_{\partial \Omega}=0.
$$
Here, $\Omega\subset \mathbb R^d$ is bounded, $a: (0,T)\times \Omega \to \mathbb R^{d,d}$ is smooth (but dependent on time). Under the assumption $f\in L^2(0,T;L^2(\Omega))$, $g\in H^1_0(\Omega)$, $h\in L^2(\Omega)$, there exists a uniquely determined weak solution $u\in L^\infty(0,T;H^1_0(\Omega))$ with $u_t\in L^\infty(0,T;L^2(\Omega))$ and $u_{tt}\in L^2(0,T;H^{-1}(\Omega))$. This is proven, e.g., in Evans' book.
My question: Can you point me to a reference that includes a proof that we can improve the regularity of the weak solution from $L^\infty$ to $C$ (i.e., continuous in time)? Due to Corona-virus related restrictions I have no access to our university's library...
If $a$ does not depend on time then such a proof can be found in John Hunters PDE-notes online. The idea is to show continuity of the energy, where one step is to mollify the wave equation in time. This does not work if $a$ depends on $t$, as then $\rho_\epsilon \ast(a\cdot \nabla u) \ne a\cdot \nabla (\rho_\epsilon \ast u)$, and I cannot figure out how to estimate the difference of these terms.
Crossposted from https://math.stackexchange.com/questions/3746894/continuity-of-weak-solutions-to-wave-equation-with-time-dependent-coefficients , which did not attract much attention over there.
Majda's book is a standard reference. It discusses nonlinear problems, which requires doing time-dependent coefficients first.
@MichaelRenardy Which one exactly?
``Introduction to PDEs and Waves for the Atmosphere and Ocean''?
Compressible Fluid Flow and Systems of Conservation Laws in Several Space Variables
@MichaelRenardy thanks. At first glance the book treats first-order systems, which does not seem to be applicable to the multi-dimensional wave equation. I still have to think about this.
|
2025-03-21T14:48:31.486988
| 2020-07-10T09:41:54 |
365299
|
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|
Stack Exchange
|
Distance of two points in Grassmannian using Plücker coordinate
Let $G(q,D)$ be the Grassmannian of $q$-dimensional vector spaces in $\mathbb{R}^D$, where $q \le D$ are positive integers. In the paper, a distance between two points of $G(q,D)$ are defined as follows. The authors represented a point in $G(q,D)$ by a matrix $X$ of size $D \times q$ such that $X^T X$ is a $q \times q$ identity matrix. For two points $X_1, X_2 \in G(q,D)$, they define $d(X_1,X_2) = 1/\sqrt{2} || X_1X_1^T - X_2X_2^T ||_F$, where $||\cdot ||_F$ is the matrix Frobenius norm. Could $d(X_1,X_2)$ be defined using Plücker coordinates in $\mathbb{C}[G(q,D)]$? Thank you very much.
Clearly this can be done when $q=1$, since the authors' coordinates parametrise lines in $\mathbb{R}^D$ by vectors of Euclidean norm $1$ (so the only choice is a sign) and the Plücker coordinates are just the usual projective coordinates, which can be normalized in the same way.
|
2025-03-21T14:48:31.487338
| 2020-07-10T10:43:25 |
365300
|
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|
Stack Exchange
|
Is there a notion of symplectic maps between spaces of volume forms on phase spaces?
For a $n$ dimensional smooth manifold $M$, I consider the cotangent bundle $T^*M$ with the canonical symplectic form $\omega$.
A symplectic map $\phi : T^*M \to T^* M$ is a map which leaves the symplectic form invariant, i.e. $\phi^* \omega = \omega$.
Question:
Is there a notion of symplectic maps between the spaces of volume forms over the phase space $T^*M$?
edit:
In more detail: Let me denote the space of all volume forms over $T^*M$ as $\Omega^{2n}(T^*M) := \Gamma( \Lambda^{2n} (T^*M) )$ (the notation is from John M. Lee's book 'Intro to smooth manifolds'.)
Then for a map $\psi$ which transforms volume forms, i.e.
$$\psi : \mathrm{\Omega}^{2n}(T^*M) \to \mathrm{\Omega}^{2n}(T^*M),$$
I am looking for a condition which ensures that $\psi$ is compatible with the symplectic structure of $T^*M$?
Example:
A symplectic map $\phi : T^* M \to T^*M$ implies a map $$\psi : \mathrm{\Omega}^{2n}(T^*M) \to \mathrm{\Omega}^{2n}(T^*M) : \eta \mapsto \phi^* \eta.$$
Such a map should be compatible for example.
But I would expect that there are more maps which are compatible and that not all of them are derived from symplectic maps like in the example.
Background:
For numerical simulation of Hamiltonian equations, it is good to use symplectic integrators, such as symplectic Euler.
However, I am interested in solving Liouville's equations and this raised the question what a corresponding symplectic integrator would be in that case?
I suppose you mean by $\Omega^n(T^*M)$ the $n$-th antisymmetric tensor power of $T^*M$. This is a vector bundle of fiber dimension one over $M$. So the total space has dimension $n+1$. This makes it very unlikely to be symplectic (in ca half of all cases)
I want to look at volume forms in the phase space, so it should be the $n$-th antisymmetric tensor power of $T^*(T^*M)$. (Otherwise, I am also happy with an answer for a general symplectic manifold $X$ in the role of $T^*M$.)
I understand that such a map $\varphi$ is never symplectic in the original sense. My question is if there is a condition which ensures that the map is at least "compatible with the symplectic form on $T^*M$". [I edited the question]
But $\Omega^{2n}(T^*M)$ is the space of volume forms on $T^*M$.
Oh, yes, sorry for the mistake!
Is $\psi$ a vector bundle automorphism of the volume form bundle on $T^*M$? Or is it a map on sections of that bundle?
It should be a map on sections of that bundle.
A couple of questions for clarification: First, is it important that the underlying symplectic manifold is the total space of a cotangent bundle? I can't see why you question requires this. Second, since a symplectomorphism preserves the volume form, its induced action on $\Omega^{2n}$ is the identity. So what does it mean for a self-map on $\Omega^{2n}$ to be compatible with the symplectic structure? Perhaps I've missed something.
Ah, I see in the comment above that you don't particularly care that the symplectic manifold is a cotangent bundle.
@PaulReynolds I'm no a differential geometer, it might be trivial! (The feedback already helped me a lot!)
Yes, the cotangent bundle structure is not important for me. 2. As far as I know, symplectomorphisms preserve the canonical volume form $\omega^{\wedge n}$ but not arbitrary volume forms. (In the application, the volume forms represent densities of particles and they are not related to the canonical volume form at all.)
You're correct though, my assertion that a symplectomorphism preserves all volume forms is not right. Perhaps you are interested in the diffeomorphisms that preserve $\omega^n$, with no further conditions, of which there are many.
@PaulReynolds Yes, that's a good point. I was hoping for some stronger conditions, but it could be that it's just this.
|
2025-03-21T14:48:31.487594
| 2020-07-10T11:01:28 |
365301
|
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|
Stack Exchange
|
While solving the 1988 IMO problem 6, I have questions about new solutions without using Vieta Jumping
I think most of you may know the well-known problem:
"Let $x$ and $y$ be positive integers such that $xy + 1$ divides $x^{2} + y^{2}$. Show that $\frac {x^{2} + y^{2}}{xy + 1}$ is the perfect square."
I know that this problem is meant for using the 'Vieta jumping' method.
But I wanted to solve this problem in another way.
First, using Wolfram Alpha, I wanted to make an equation.
$\frac{x^{2} + y^{2}}{xy + 1} = k$
and solved over $x$ (or $y$, it's same)
The result was $\frac {1}{2}$$\sqrt{4 k - 4 x^2 + k^2 x^2} + \frac {1}{2}$$kx$, $-\frac {1}{2}$$\sqrt{4 k - 4 x^2 + k^2 x^2} + \frac {1}{2}$$kx$
(I am first using Tex...)
I thought that if $y$ is an integer, "$4 k - 4 x^2 + k^2 x^2$ has to be a perfect square" (Even though it is a perfect square, because of $\frac {1}{2}$, it could be a rational number, not an integer. But at least $4 k - 4 x^2 + k^2 x^2$ has to be a perfect square).
I have tried many integers with $k$ and $x$, and concluded that $4 k - 4 x^2 + k^2 x^2$ can be a perfect square only when $k$ is a perfect square, which means the initial hypothesis was right.
(For example, if $k$ is $4$, $12x^2 + 16$ can make a perfect square when $x$ is 2, 8, 30....)
So what I want to ask you is that "how to prove $4 k - 4 x^2 + k^2 x^2$ can be a perfect square only when $k$ is a perfect square".
The next thing I am curious about is, it is said that 11 of the IMO participants solved perfectly.
I don't think all of them used the same 'Vieta jumping' method.
What do you think of more "general ways" to solve this problem for other participants?
Thank you.
(Sorry for the bad English)
1988 IMO 6 has been discussed several times on math.stackexchange, probably also on art of problem solving website. Have you checked to see whether any non-Vieta solutions have been posted on those sites?
@GerryMyerson Thank you for the comment. Of course, I have seen several posts about non-Vieta solutions about this problem but I think I have never seen this kind of approach to this problem... or maybe my skills are not good enough to understand other solutions...
The way the last paragraph of your question is written, it seemed to me that you hadn't seen any non-Vieta solutions, or at any rate that you were asking to see more. But from your comment, it appears you aren't interested in other solutions, you're only interested in whether your approach leads to a solution. Is that right? Then it's not clear to me that reducing the problem to deciding whether $4k-4x^2+k^2x^2$ square implies $k$ square is making any progress toward a solution.
|
2025-03-21T14:48:31.487781
| 2020-07-10T11:28:20 |
365302
|
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|
Stack Exchange
|
Scaling invariance for Hartree equation
Consider generalized Hartree Equation
$ i \partial_t u + \Delta u + \left( \frac{1}{|x|^{\gamma}} \ast |u|^p \right)|u|^{p-2}u, \quad u(x, 0)=u_0(x)$
where $0<\gamma<d, p\geq 2, u(x,t)\in \mathbb C.$
When $p=2$ $-$ classical Hartree equation.
Scaling invariance: if $u(t,x)$ solves GHE, then so does
$ u_{\lambda}(t,x)= \lambda^{\frac{d-\gamma +2}{2(p-1)}} u (\lambda^2 t, \lambda x)$
also.
My question is: What is the scaling invariance for fractional generalized Hartree equation:
$ i \partial_t u - (- \Delta)^{\alpha/2} u + \left( \frac{1}{|x|^{\gamma}} \ast |u|^p \right)|u|^{p-2}u, \quad u(x, 0)=u_0(x)$
Carlo already answered your question: but this is essentially just a plug-and-chug question. The linear terms determine the scaling relation between $t$ and $x$. And after that you just solve an algebraic equation for the scaling factor in front of the function. // Are you asking the question because you don't know how to do this computation (in which case Carlo's answer is unlikely to be too helpful), or because you cannot be bothered to do it yourself?
substitution of $u_\mu(t,x)=\text{constant}\times u(\mu^\alpha t,\mu x)$ in the fractional GHE shows that this is a solution if $\text{constant}=\mu^{\frac{d-\gamma+\alpha}{2(p-1)}}$, so the scale invariance relation is
$$u_\mu(t,x)=\mu^{\frac{d-\gamma+\alpha}{2(p-1)}}u(\mu^\alpha t,\mu x).$$
|
2025-03-21T14:48:31.487901
| 2020-07-10T11:36:28 |
365303
|
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|
Stack Exchange
|
Searching for theorems characterizing when $O_p(G)$ is trivial / non-trivial
Let $G$ be a finite group. Let $p$ be a prime.
Let $O_p(G)$ be the $p$-core of $G$.
Are there any theorems known saying something like
$O_p(G)$ is trivial, if and only if ... and
$O_p(G)$ is non-trivial, if and only if ..., respectively ?
I am especially interested in the case $p=2$.
If not, criteria ensuring that $O_p(G)$ is trivial / non-trivial would also be interesting.
Thank you very much.
There are many such theorems. By the way, I would say that your definition of $p$-core may be non-standard if you are using it to denote the largest normal $p$-subgroup of $G$. I think many people would use $p$-core of $G$ to be $O_{p^{\prime}}(G)$, the largest normal subgroup of $G$ of order co prime to $p$. Certainly when $p = 2$, the largest normal subgroup of $G$ of order prime to $2$ is denoted by $O(G)$ and called the core of $G.$
Anyway, assuming you really do mean the largest normal $p$-subgroup of $G$ when you write $O_{p}(G)$, we have: $O_{p}(G)$ is the intersection of all Sylow $p$-subgroups of $G$. If $G$ has an Abelian Sylow $p$-subgroup $P$ then $O_{p}(G) \neq 1$ if and only if $P \cap P^{g} > 1$ for all $g \in G$. This is a theorem of J. Brodkey, and it does not work in general for non-Abelian Sylow $p$-subgroups. A famous theorem of R. Baer and M Suzuki is that if there is a non-idenity element $x \in G$ such that $\langle x,x^{g} \rangle $ is a $p$-group for all $g \in G$, then $x \in O_{p}(G)$, so that $O_{p}(G) \neq 1.$ When $p =2$, a consequence of this when $p =2$ is that if $t$ is an involution (element of order $2$) which does not invert any non-identity element of odd order, then $t \in O_{2}(G)$.
I could go on, but reading a graduate level group theory text should give you many more examples.
Ok, thank you. Yes, I mean the largest normal $p$-subgroup of $G$. Is there one (or more) particular group theory book which you would recommend (since a few don't treat $O_p(G)$) ?
Aschbacher's' book on Finite Groups would be one.
I'm surprised you didn't mention the $Z_p^*$-theorem, @GeoffRobinson! Suppose that $O_{p'}(G)=1$. If $z$ is an element of order $p$ lying in a Sylow $p$-subgroup $P$ of $G$, and $z$ is not $G$-conjugate to any other elements of $P$, then $z$ lies in $Z(G)$. In particular $z\in O_p(G)$. For $p=2$ this is due to Glauberman, but for $p$ odd it only follows from CFSG.
True. I'm not sure that the odd $Z^*$ theorem is in many of them though. It won't be in Aschbacher, Gorenstein, Suzuki, etc.
@Geoff Robinson and @ David A. Craven: Thank you very much.
Theorem 2 in my preprint "Group Orders That Imply Existence of Nontrivial Normal $p$-Subgroups" shows that if $|G| = p^s m$ and $p \nmid \Gamma(m)$ (defined therein), then either $O_p (G) \ne 1$, or $G$ is not solvable.
@AlexM. Thanks for the edit!
Thank you very much for your answer.
|
2025-03-21T14:48:31.488117
| 2020-07-10T13:52:25 |
365310
|
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|
Stack Exchange
|
Concrete example of non-nuclear operator $E \to F$ and isometry $F \hookrightarrow G$ so that the composition $E \to F \hookrightarrow G$ is nuclear
DISCLAIMER: I posted the same question a week ago on Mathematics Stack Exchange.
We know by an abstract argument that there exist Banach spaces $E$, $F$, $G$ and maps $E \to F \hookrightarrow G$ such that $E \to F$ is non-nuclear, $F \hookrightarrow G$ is an isometry (metric injection), and the composition $E \to F \hookrightarrow G$ is nuclear. (In other words, the operator ideal of nuclear operators is not injective.) The typical line of reasoning can be found in the corresponding post on MSE, or in [DF93, §9.7]. However, these proofs are non-constructive, which leads me to the following question:
Question. Can we write down explicit examples of Banach spaces $E$, $F$, $G$ and maps $E \to F \hookrightarrow G$ such that $E \to F$ is non-nuclear, $F \hookrightarrow G$ is an isometry, and the composition $E \to F \hookrightarrow G$ is nuclear?
In the (non-constructive) example given in my MSE post, all spaces have the approximation property, but for the moment I do not care about this requirement.
References.
[DF93]: A. Defant, K. Floret, Tensor Norms and Operator Ideals (1993), Mathematics Studies 176,
North-Holland.
Using the notation/construction of what you wrote on MSE, I think this should be relatively routine once one has explicit witnesses to the fact that the "inclusion" $\ell^2 \tilde{\otimes} \ell^2 \to \ell^2 \tilde{\otimes} L_1$ does not have closed range; you take a sequence of finite-rank tensors in the LHS whose norms become much smaller on the RHS, and then rescale and stack them together.
By "relatively routine" I don't mean that this is a trivial amount of work, I just mean that once one has the finite building blocks, the steps one takes to produce the desired counterexample don't require any new fancy tricks
@YemonChoi ah yes, that is how these things work. But for this I would have to get a grip on the projective norm. I'll think about it some more at a later time; maybe this approach is not as hard as it seems.
Take any sequence $a_n$ of scalars that is square summable but not summable. That is the "hard" (in the technical sense) part of the argument. The rest is "soft". Let $T$ be the diagonal operator on $\ell_2$ with diagonal $a_n$. So $T$ is $2$-summing (Hilbert-Schmidt) but not nuclear (trace class). Let $S: \ell_2 \to L_1$ be the isomorphism you mention in your MSE post that maps the $n$-th unit vector basis of $\ell_2$ to the $n$-th Rademacher function. $S^*$ is an operator from $L_\infty$ to $\ell_2$, so is $2$-summing. $T^*$ is $2$-summing obviously, so $T^*S^*$ is nuclear, so $ST$ is nuclear because every space in sight has even the metric approximation property.
If you want $S$ to be an isometry from $\ell_2$ into $L_1$, map the unit vector basis of $\ell_2$ injectively onto independent Gaussian random variables that are normalized in $L_1$.
Very nice! You make it look so easy. Seeing as you're not a member of MSE, I'll write up a slightly expanded answer there, aiming at the MSE audience. (Although I guess everyone who understands my question can also understand your answer.)
As a side note, I noticed that $S : \ell_2 \to L_1$ itself is not absolutely $p$-summing for any $p$, since it is not completely continuous. (We have $r_n \rightharpoonup 0$, but $\lVert Sr_n \rVert_{L_1} = 1 \not\to 0$.)
|
2025-03-21T14:48:31.488367
| 2020-07-10T14:11:00 |
365313
|
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|
Stack Exchange
|
Counting points in elliptic curves
Given an elliptic curve over $\mathbb Z_n$
Is it $\#P$ hard to compute $\# E(\mathbb Z_n)$?
Is it $PP$-hard to compute $\# E(\mathbb Z_n)\leq\frac n2$?
Is it $\oplus P$ hard to compute $\# E(\mathbb Z_n)\bmod 2$?
Could you define what you call an elliptic curve over $\Bbb{Z}_n$?
Sorry I was thinking on this https://mathoverflow.net/questions/95408/elliptic-curves-over-rings. If not appropriate I can take it down or just use $\mathbb F_q$ since that does not change the nature of the question.
And even define $\mathbb{Z}_n$ as it's both used for $\mathbb{Z}/n\mathbb{Z}$ and for the $n$-adics (projective limit of $\mathbb{Z}/n^k\mathbb{Z}$).
I was just think $\mathbb Z/n\mathbb Z$. I doubt over infinite sets the problem makes sense.
Thanks to the Chinese remainder theorem and the known structure of the formal group of elliptic curves, these questions all boil down to calculating group orders for elliptic curves over $\mathbb{F}_p$. In particular 2 should be easy and 3 is trivial as there is not problem checking if there are $2$-torsion points.
For $n$ prime we have Schoof's algorithm which is polynomial in the number of bits, so all these problems are in $P$. For other $n$ the hardest step might be factoring, which is not believed to be polynomial time, but also not expected to be hard for any of these hardness classes.
For a reference, see the Wikipedia page on "Counting points on elliptic curves" https://en.wikipedia.org/wiki/Counting_points_on_elliptic_curves
@willsawin Does counting points give factors?
I'm just saying you could first factor and then count points on each prime. This means counting points can't be #P-hard unless factoring is.
I understand. I am asking converse question.
|
2025-03-21T14:48:31.488545
| 2020-07-10T15:33:05 |
365317
|
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|
Stack Exchange
|
Example of a C*-algebra whose $K_1$ is uncountable
We know that if $A$ is a separable $C^{*}$-algebra then $K_1(A)$ is countable.
Can anybody give an example of a C*-algebra for which $K_1(A)$ is uncountable?
There must be tons of ways to do this, but a simple one is to start with an uncountable set $X$, equipped with the discrete topology, and consider $c_0(X)$. There are uncountably many pairwise inequivalent minimal projections in this algebra, so its $K_0$ group is uncountable. Now use $K_0(c_0(X)) \cong K_1(Sc_0(X))$ where $SA$ is the suspension of $A$.
Nik's answer nails it but if you prefer something representable on a separable Hilbert space then you may consider the suspension $SM$ of any ${\rm II}_1$-factor $M$. Indeed, as $M$ is tracial, $K_0(M) \cong \mathbb R$ and the suspension simply reverses the $K$-groups.
Actually you may produce further commutative examples that are representable on separable Hilbert spaces: for example $S\ell_\infty$, since $K_0(\ell_\infty)$ comprises $\mathbb{Z}$-valued continuous functions on $\beta \mathbb N$.
Very nice answer!
Excuse me Sir but in the first paragraph, we $K_1(M) = 0$. Am I wrong? @NikWeaver
@PegLegScott, yes, but $K_1(SM) = K_0(M) = \mathbb R$.
oh, I see now. You mean $K_1(SM)$ as an example.
|
2025-03-21T14:48:31.488663
| 2020-07-10T16:04:01 |
365320
|
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|
Stack Exchange
|
How can construct the equilateral $A''B''C''$ such that area of $A''B''C''$ is biggest
Let $ABC$ be arbitrary triangle in a plane. Let $A'B'C'$ and $A''B''C''$ be two equilateral triangles such that $A \in B'C'$, $B \in C'A'$, $C \in A'B'$ and $A \in B''C''$, $B \in C''A''$, $C \in A''B''$
How can construct the equilateral $A'B'C'$ such that $AA'$, $BB'$, $CC'$ are concurrent;
How can construct the equilateral $A''B''C''$ such that area of $A''B''C''$ is biggest;
See also:
Malfatti circles
Cramer–Castillon problem
|
2025-03-21T14:48:31.488734
| 2020-07-10T16:26:41 |
365321
|
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|
Stack Exchange
|
Trees of prescribed ordinal
My question is very imprecise, as I know very little about descriptive set theory.
In a problem I am thinking about I have a family of well-founded trees (finite sequences on $\cup_n X^n$ closed under taking predecessors, such that every path has a maximal element) on a Polish space (separable, completely metrizable) space. These trees have several additional proprieties/constraints. I am trying to construct for each $\alpha<\omega_1$ a tree that has order at least $\alpha$. The order of the tree is defined inductively by removing maximal elements until nothing is left. The least ordinal for which this happens is the order of the tree.
My questions, as I said very imprecise, is what properties of a tree imply that the order is larger than a prescribed ordinal? Are there any standard techniques of constructing such trees? I know this is a long shot as I have several other constraints on these trees, but if there is some relevant literature where I can start looking, that would be appreciated. I know of the standard reference of Kechris book, but I couldn't find something that I can use (or perhaps I did not know where to look in that book).
Thank you and if you think this question is too vague, please just delete it.
Trees of any order $\alpha < \omega_1$ can easily be constructed by induction: For the successor step assume there is a tree $T$ of order $\alpha$ and define $T'$ by simply putting a new root below $T$. Then $T'$ is well-founded and has order $\alpha+1$. Similar for the limit case $\lambda$: Since $\lambda < \omega_1$ there is a sequence $(\alpha_n){n \in \omega}$ such that $\lambda= \sup{n \in \omega} \alpha_n$ and there are trees $T_n$ of order $\alpha_n$. Define $T'$ simply by putting the $T_n$'s side by side and a new root below them. Again, $T'$ is well-founded and has order $\lambda$.
@JohannesSchürz I apologize, could you please explain in a bit more detail what you mean by "a new root below $T$". These trees are all single rooted at $\emptyset$. Perhaps I understand something different by a root?
The order of a tree, as you call it, is more commonly known as its rank, at least among descriptive set theorists. You can find the definition in section I.2.E (Well-founded trees and ranks) in Kechris's book.
Perhaps it will help you to know that the rank of a tree $T$ is completely determined by the partial order $\prec_T$ defined on $T$ by $s\prec_T t$ iff $s\supsetneq t$. Thus the rank of $T$ is fairly robust to changes on $T$ itself.
As for how you put "a new root below $T$": pick an arbitrary point $x_0$ in $X$, and define $T' = {\emptyset} \cup {\langle x_0 \rangle^\frown s ; | ; s\in T}$. Similarly, to put a new root below several trees at once, you'll need to pick an arbitrary point for each of them.
@MariaM The above "$T$ can be constructed by appending roots via transfinite induction up to $\alpha$" implies that the rank of $T$ is less than an ordinal $\alpha$, not greater. In order to give a better picture of which kind of conditions you're looking for (e.g. Ramsey-theoretic, graph-theoretic, etc), do go have examples of some of the several additional properties/constraints?
|
2025-03-21T14:48:31.489100
| 2020-07-10T16:28:52 |
365322
|
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|
Stack Exchange
|
(Translation request) Hypotheses of the Blom-Fredberg bounds on denumerants?
I don't know Swedish and I'm not finding the article "G. Blom and C. E. Froberg, On money changing" translated into English... so I tried to read the original (Swedish) with the help of Google Translate but, from the results, I fear that some semantics could get lost...
The article prove a denumerant's upper and lower bound:
$\frac{n^{k-1}}{(k-1)!\prod_{i=1}^{k}a_i} \leq d(n;a_1,\dots,a_k) \leq \frac{(n+s_k)^{k-1}}{(k-1)!\prod_{i=1}^{k}a_i}$
where
$s_1=1, s_2=a_2$ and $s_k=a_2+\frac{1}{2}(a_3+\dots+a_k), \forall k\geq 3$.
I would like to know: is it required (hypothesis) that all $a_i$ must be coprime i.e. $(a_1,\dots,a_k)=1$?
I first read about this theorem in the book "J. L. R. Alfonsin, The Diophantine Frobenius Problem" (pag. 74) but what I'm asking is not specified.
UPDATE
I would have written in the comments but it was too long.
Despite @Carl-FredrikNybergBrodda's competent reassurance about the translation (i.e. I'm assuming that there are no explicit statements about co-primality) something in the back of my mind was not in peace... unfortunately very often there are implicit hypotheses in math articles (especially dated ones).
The reasons why this happen... well... are different but it is usually because are considered trivial by the context (or by the author).
So yesterday I started reading the mathematical component of the article trying to reconstruct the proof and from what I seemed to understand... is that the implicit hypothesis is stronger than $(a_1,...,a_k)=1$ i.e. it's $a_1=1$.
This idea is supported by these observations:
This article is about coins and there is always the currency of unitary value (at least $"a.s."$).
All the explicit example start with $a_1=1$
pag. 60 the proof is by induction so there are two "base case" because the general expression for $s_i$ does not cover $s_1$ (by the way in the article the sequence $s$ is called $\Delta$), therefore it is necessary to check $m=1$ and $m=2$ before the inductive step $m\geq 3$.
The case $m=2$ check the truthfulness of:
$\frac{1}{(2-1)!}\frac{n^{2-1}}{a_2}\leq D(2,n)\leq \frac{1}{(2-1)!}\frac{(n+s_2)^{2-1}}{a_2}$
but if it was $a_1\neq 1$ should be:
$\frac{1}{(2-1)!}\frac{n^{2-1}}{a_1 a_2}\leq D(2,n)\leq \frac{1}{(2-1)!}\frac{(n+s_2)^{2-1}}{a_1 a_2}$
Maybe I didn't understand the proof or maybe the case $a_1 = 1$ is not restrictive (but I don't see why).
Do you have a pdf of the original Swedish version? I cannot find it online. I can translate the relevant parts for you.
Yes, I have it! in advance a giant THANK YOU! I am new to the forum how can I send you the article?
If you have a link to it, posting it here is probably easiest. Otherwise, you can send it to me via the email on my profile.
I just sent you an email with the article. Thank you very much for helping.
The article is [Blom, G. and Fröberg, C-E., Om Myntväxling, Nordisk Matematisk Tidskrift, 1962, Vol. 10, No. 1/2 (1962), pp. 55-69] for anyone who wishes to sing along.
After reading through the article, no assumption is made on the coprimality of the $a_i$, in the sense that no added assumptions appear to be stated on the $a_i$ before the statement of the theorem.
As an aside, the Swedish used in this article is absolutely gorgeous, and is well worth the read. My only concern is that the authors assume that whoever is reading the article knows the Swedish currency system well, but not everyone is aware that there are $100$ öre in $1$ krona... to further complicate matters for modern readers, the öre (equivalent to a cent) is no longer in use in Sweden.
Unfortunately it was precisely the examples that made me suspicious because they are all coprime cases due to the first value 1 (common to all the examples). The lack of explicit declaration of coprimality and the observation you make on page 57 I would say that they close the discussion making an even more remarkable a beautiful theorem. Thanks.
"their many examples clearly show that they do not assume this" - do you mean they do assume this by any chance?
@MaxAlekseyev No, I mean it as it is written; "their many examples clearly show that they do not assume [coprimality of the $a_i$]".
I'm confused then. Your bullet points do not support this conclusion (only that they assume no pairwise coprimality), and according to @Ramanumpy, all their examples contain 1 and thus are co-prime.
@MaxAlekseyev You are right, of course. I misread his condition as pairwise co-prime. Nevertheless, the main point is that no additional assumptions are added inside the text to the statement of the theorem.
They use $d_m$ as the GCD of $a_1,\dotsc,a_m$. One can then quite easily factor out this from the problem, and assume $a_1=1$ without losing much.
|
2025-03-21T14:48:31.489484
| 2020-07-10T16:37:46 |
365323
|
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|
Stack Exchange
|
Could computing the next prime in a finite Euler product be made rigorous?
It is well known that:
$$\zeta(s):=\prod_{n=1}^{\infty} \frac{1}{1-p_n^{-s}} \qquad \Re(s) \gt 1$$
with $p_n =$ the $n$-th prime. It also known that:
$$\zeta(2n):= \frac{(-1)^{n+1} B_{2n}(2\pi)^{2n}}{2(2n)!}$$
where $B_{2n}$ is the $2n$-th Bernoulli number.
Now define the function:
$$f(k,N,x):= \zeta(2k) - \left(\prod_{n=1}^{N} \frac{1}{1-p_n^{-2k}}\right)\cdot \left(\frac{1}{1-x^{-2k}}\right) \qquad \Re(s) \gt 1$$
where $k, N \in \mathbb{N}$ and $x$ is the unknown next prime ($p_{N+1}$) to be computed.
I found numerically that solving $x$ in $f(k,N,x)=0$ for some $N$, yields an increasingly accurate approximation of $p_{N+1}$ when $k$ increases. For example take the first 6 primes and try to derive the 7th prime (17):
$f(6, 6, x) = 0 \rightarrow x = 16.64054...$
$f(12, 6, x) = 0 \rightarrow x = 16.95214...$
$f(24, 6, x) = 0 \rightarrow x =16.99830...$
The key question is how high $k$ needs to go to ensure that $x=p_{N+1}$ after rounding. In the following Maple code, I simply used $k=2N$ and it already correctly generates all 'next' primes up to $N=60$:
Digits:=600
for N from 1 to 60 do ithprime(N), ithprime(N+1), round(fsolve(f(2*N, N, x), x = 0 .. 300)) end do
I immediately acknowledge that this is a highly inefficient and impractical algorithm to generate primes. However, is there more to say about the minimally required value of $k$ (maybe as a function of $N$) to ensure that rounding $x$ will correctly yield $p_{N+1}$?
this is cool. its like an analytic approach to finding primes
It's not too difficult to show that $k=p_N$ works, and that we need $k\ge c p_N$ for some positive constant $c$ in case $p_{N+2}=p_{N+1}+2$.
$2k=1+p_N$ works for $N>1$, but $2k\le 0.56 \, p_N$ will fail if
$p_{N+2}=p_{N+1}+2$.
With $q=p_{N+1}$, we have
$$
\frac{1}{1-q^{-2k}} < \frac{1}{1-x^{-2k}} = \frac{1}{1-q^{-2k}} \prod_{p>q} \frac{1}{1-p^{-2k}} .
$$
It follows that
$$
q^{-2k} < x^{-2k} < q^{-2k} + \sum_{j\ge 2} (q+j)^{-2k} < q^{-2k} +\frac{1}{(q+1)^{2k-1}(2k-1)}.
$$
Taking logarithms, and using $\log(1+y)\le y$, we get
$$
-2k \log q < -2k \log x < -2k\log q + \frac{q+1}{\exp\{(1-o(1)) 2k/q\} (2k-1)}.
$$
Dividing by $-2k$ and exponentiating, we have
$$
q > x > q - \frac{q(q+1)}{\exp\{(1-o(1))2k/q\} 2k (2k-1)}.
$$
We want the last expression to be less than $1/2$. Since $q/p_N \to 1 $ as $N\to \infty$, we need $k\ge (1+o(1)) c \, p_N$,
where $c=0.45...$ is the solution to $e^{2c}4 c^2=2$.
So $2k=1+p_N$ works for large $N$. We can check with a computer that it also works for small $N$.
Similar calculations show that when $p_{N+2}=p_{N+1}+2$, it is necessary that $k>0.28 \, p_N$ when $N$ is large.
Edit: Using the inequality $\log(1+y)\le y$ was somewhat wasteful and not necessary. Also, as the OP points out in the comments, we can use the ceiling function instead of rounding, since $x<q$. With those two modifications, we find that $k\ge \frac{1}{3}p_N$ works for $N\ge 1$, but $k\le 0.19 \, p_N$ fails at twin primes $p_{N+2}=p_{N+1}+2$.
Many thanks, Andreas! One more observation that could maybe help sharpening these bounds. Numerical evidence suggests (I have no proof) that when $k$ increases, $x$ always approaches $p_{N+1}$ from 'below'. So, if this is true, could using the ceiling instead of rounding $x$ improve the bound by a factor $2$?
Yes, the last display of my answer shows that $x<q=p_{N+1}$. Using the ceiling instead of rounding will always work if $2k \ge 0.71 , p_N$, but will fail at twin primes if $2k\le 0.38 , p_N$.
|
2025-03-21T14:48:31.489733
| 2020-07-10T16:52:55 |
365324
|
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}
|
Stack Exchange
|
Does degree of jump determine degrees of relatively r.e. sets?
I’m mostly interested in this is the case where $A, \hat{A}$ are r.e. but the general case seems worth asking too.
Suppose I have sets $A >_T \hat{A}$ with $A' \equiv_T \hat{A}'$. Does this imply that if $C$ is r.e. in $A$ then there is a $\hat{C}$ r.e. in $\hat{A}$ such that $A \oplus C \equiv_T \hat{A} \oplus \hat{C}$?
The following is motivation and can be skipped if it's confusing.
I’m basically wondering if I'm building 2-REA sets whether the jump of the base degree ($A$ or $\hat{A}$ here) completely determine the degrees of the 2-REA sets I can build over that degree (assuming I'm only considering bases comparable to each other) or is it possible to sorta thing out the range of 2-REA degrees that can be built by moving to a smaller base degree.
The answer is no. Every properly n-REA set for n < 3 (I believe Peter Cholak and I have shown this fails at 3 but could always fall apart in write-up) can be extended to a properly n+1 REA set by adding a relative r.e. set. Now apply this result to a low r.e. set. You can find that result in a paper by Peter Cholak and Peter Hinman but the result for n=1 was one of the S computability theorists (Shore, Soare, etc) and I don't remember who off the top of my head.
|
2025-03-21T14:48:31.489859
| 2020-07-10T17:03:21 |
365326
|
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}
|
Stack Exchange
|
Congruential equidistribution, prime numbers, and Goldbach conjecture
Let $S$ be an infinite set of positive integers, $N_S(z)$ be the number of elements of $S$ less than or equal to $z$, and let
$$D_S(z, n, p)= \sum_{k\in S,k\leq z}\chi(k\equiv p\bmod{n}).$$
Here $\chi$ is the indicator function, and $z, p, n$ are positive integers, with $p<n$ and $n>1$. If
$$\lim_{z\rightarrow\infty} \frac{D_S(z,n,p)}{N_S(z)} = \frac{1}{n}$$
for all $n>1$, regardless of $p$, then the set $S$ is said to be congruentially equidistributed, or in other words, free of congruential restrictions.The exact same concept, referred to as "uniformly distributed in $Z$", is discussed in chapter 5 in the book Uniform Distribution of Sequences by Kuipers and Niederreiter (1974), see here. It is related to the concept of equidistribution modulo 1 in the following way: the sequence $x_k$ is equidistributed modulo 1 if and only if the sequence $\lfloor n x_k\rfloor$ is congruentially equidistributed modulo $n$ for all integers $n\geq 2$. The brackets represent the floor function.
Examples
Here $p_k$ denotes the $k$-th prime, with $p_1=2$. The set $S_1$ of all $k+p_k$ seems to be congruentially equidistributed. But the set of all primes is not. The set of squares and the set of cubes are not. If $\alpha$ is irrational, then the set consisting of all $\lfloor \alpha p_k \rfloor$ is congruentially equidistributed: this is a known result. It is also true for the set of all $\lfloor \alpha \beta^k \rfloor$ if $\alpha$ is a normal number in base $\beta$ (here $\alpha > 0$, $k=1,2,\cdots$ and $\beta>2$ is an integer), and for the set of all $\lfloor k \log k \rfloor$ where $k$ is an integer $>0$ (this set has same density as the set of primes). The set $S_2$ consisting of all $(p_{k+1}+p_{k+2})/2$ is also congruentially equidistributed, it seems.
Question
If $S$ is congruentially equidistributed and contains enough elements, say
$$N_S(z) \sim \frac{a z^b}{(\log z)^c} \mbox{ as } z\rightarrow\infty$$
where $a, b, c$ are non-negative real numbers with $\frac{1}{2}< b \leq 1$, is it true that $S+S=\{x+y,$ with $x, y \in S\}$ contains all the positive integers except a finite number of them?
This statement would be true if $S$ was a random set having the same distribution of elements. More precisely, in that case, as a result of the Borel-Cantelli lemma, $S+S$ almost surely contains all the positive integers but a finite number of them. See the last paragraph in my answer to my previous MO question here, for a justification.
Connection to Goldbach conjecture
If $a=1, b=1, c=1$, we are dealing with numbers that are distributed just like prime numbers, so this is connected to the Goldbach conjecture (GC). The set $S_1$ (see example above) seems congruentially equidistributed, thus proving that every large enough integer is the sum of two elements of $S_1$, might be much less difficult than proving GC.
The set of primes is NOT congruentially equidistributed, presumably making GC harder to prove. Note that $S_1$ is more sparse than the set of primes. Both $S_1$ and $S_2$ (see example) also have $a=1,b=1, c=1$. So an alternative to GC, easier to prove, could be:
All large enough integer $z$ can be written as $z=x+y$ with $x,y\in S_2$.
Even if you replace primes by super-primes in $S_2$, you would still (I guess) keep the congruential equidistribution, and thus the conjecture would still presumably be easier to prove than GC, even though super-primes are far rarer than primes. Note that for super-primes, $a=1, b = 1, c = 2$.
I also posted a shorter version of this question on MSE, here.
For what it's worth, uniform distribution of sequences of integers is discussed at some length in Kuipers and Niederreiter, Uniform Distribution of Sequences.
Chapter 5, section 1 discuss the same kind of equidistribution and call it "uniformly distributed in $Z$". See page 305 at https://web.maths.unsw.edu.au/~josefdick/preprints/KuipersNied_book.pdf. Theorem 1.2 page 306 provides a criterion, similar to Weyl, to chek if a sequence / set of integers is congruentially equidistributed.
Version 18 of this question.
@Gerry: there are little typos I catch over time like I wrote GB instead of GC, that was the last revision. Not sure if I should leave these typos or not, as they count for a large number of the revisions, yet they don't have impact on the understanding on my post. What is your suggestion?
My suggestion is, get it right the first time.
If $S$ is congruentially equidistributed and contains enough elements .... is it true that $S+S$ contains all the positive integers except a finite number of them?
Let $S=\bigcup_{n=1}^\infty \{2^{2n},2^{2n}+1,\dots, 2^{2n+1}-1\}.$ It is easy to show that $S$ is congruentially equidistributed and $S+S\not\ni 2^{2n}$ for each positive integer $n$.
Looks like more conditions must be imposed on $S$ otherwise the result is not true in general.
|
2025-03-21T14:48:31.490195
| 2020-07-10T17:06:21 |
365327
|
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|
Stack Exchange
|
Maximum pairwise-product of a set in $Z_p$
Let $A$ be a subset of $[1,p-1]$ of size $N$, for a prime $p$.
My question is what is the most efficient algorithm to find: $$\max \{(x \cdot x')~mod~p~|~ x,x'\in A\}$$
In other words, how efficiently can we compute the maximum pairwise-product values for a set $A$ in the field $Z_p$?
Seems that https://cs.stackexchange.com/a/53161 answers your question (details are explained here: https://cs.stackexchange.com/a/104818).
The links consider maximum paiwise sum in integers, I was wondering if something is know for maximum paiwise product in Z_p
FFT works fine in $\mathbb Z_p$.
I'll take a stab at this.
As pointed out at the comments, one can obtain the characteristic vector $\chi_{S+T}$ of the multiset
$$S+T = \{s+t \mid s \in S, t \in T\},$$
by defining $\chi_S$ to be the characteristic vector of a set $S$ in $\mathbb{Z}_p$. Then
$$\chi_{S+T} = \chi_S \ast \chi_T,$$
where $\ast$ is the convolution operator. Now using the Fourier transform $\mathcal{F}$ property gives
$$\mathcal{F}(f\ast g) = \mathcal{F}(f) \times \mathcal{F}(g),$$
where $\times$ represents pointwise multiplication of functions. Thus we have
$$\mathcal{F}(\chi_{S+T}) = \mathcal{F}(\chi_S) \times \mathcal{F}(\chi_T).$$
The Fourier transform is just the complex DFT with length $p.$
Now, let $S_1=A,$ $S_k=S_{k-1} + S_1,$ for $k=2,\ldots,\max(S_1).$ Each one of these operations will be performed in the Fourier domain via
$$\mathcal{F}(\chi_{S_{k-1}+S_1}) = \mathcal{F}(\chi_{S_{k-1}}) \times
\mathcal{F}(\chi_{S_1}).$$
By carrying out the convolution $\max(S_1)$ times you ensure that you can allow for all the elements of $S_1$ being used as multipliers.
However, you do not want to multiply by elements in the complement of $S_1.$
Example: Let $p=7,$ and $S_1=A=\{a_1,a_2,a_3\}=\{2,4,5\},$ then you'd perform the convolution 5 times
but you'd want to avoid "multiplying" by 1,3, or 6. A sum of the form
$a_i+a_j$ is valid (2 is in $A$) and can contribute to the maximum, but the single term sum $a_i$ is not (1 is not in $A$), $a_i+a_j+a_k$ is not (3 is not in $A$), etc.
Note that we can always ignore $0$ even if it is in $A$ since it won't give a maximum value if part of a product.
So what you really want to do now, having computed the $S_k$ is to do the following (recall $S_1=A$):
Let $\textrm{maxprod}=0.$
For $k$ ranging between $1$ and $\max(S_1)$ do:
if $k\in S_1$ (i.e., it is a legitimate multiplier which should be taken into account) then update via
$$
\textrm{maxprod}=\max( \max(S_k), \textrm{maxprod})
$$
$~~~$else don't update.
End do:
Thanks!
I don't fully understand the time complexity of your method. An N^2 log(p) time algorithm is trivial, I am wondering if this can be improved.
Usually the FFT based algorithms are not necessarily the most efficient. Since the subset is arbitrary it may not be possible to solve your problem in less than N^2 log p. Without some structure in the set A, One does need to check every pairwise sum.
My goal was to unpack @Seva’s suggestion. Maybe he can comment if he reads this. There may be a shortcut I am not seeing.
@kodlu: not much to comment. I agree that it seems unlikely that, in general, one can do this in less than $\Omega(N^2\log p)$ operations, but I cannot rigorously prove this. Maybe, something like $O(|A\cdot A|\log p)$ is possible, but I cannot do this either. I am not sure now that the FFT is of much help (the $O(N\log N)$ bound claimed in the linked MathStackExchange reply does not look realistic).
@Seva, however it seems to me that $O(N \log N) $ would work if we wanted to find $\max{ (s+t):s \in S, t \in T}.$ Do you agree?
@kodlu: this is quite possible: I was in fact thinking about computing the whole sumset, not just its largest element.
@Seva,thinking a bit more, wouldn't the computation of the FFT be $O(p \log p)$ ($p$ being the length of ground set where the characteristic function is defined) and N the size of the set is strictly less than $p$? So it may not necessarily be $ O(N \log N)$ I think.
|
2025-03-21T14:48:31.490502
| 2020-07-10T17:06:46 |
365328
|
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|
Stack Exchange
|
A series that is rational?
Let $k=\mathbb F_q(T)$. Can one prove (or disprove) that the series $\sum_{n\ge0}(1-TX^{q^n})Y^{q^n}\in k[[X,Y]]$ belongs to $k(X,Y)$? At first, it looked like it was simple. But in fact, I have no clue to attack this question. I thought about Dwork-Polya-Bertrandias theorem, but I did not find a several variables version of this theorem.
If you set $T=0$ or $X=0$ then you get the series $\sum_{n\geq 0} Y^{q^n}$. This cannot be rational because a rational power series in one variable that is not a polynomial cannot have arbitrarily long sequences of 0 coefficients (since the coefficients satisfy a linear recurrence relation with constant coefficients).
Is your argument still true in positive characteristic?
@joaopa It certainly is.
@joaopa letting $k$ be a field of characteristic $p$, the formal power series $f(Y) = \sum_{n \geq 0} Y^{q^n}$ in $k[[Y]]$, where $q$ is a power of $p$, is not in $k(Y)$ since $f^q - f - Y = 0$ and this relation is impossible for $f$ being in $k(Y)$: that $Z^q - Z - Y$ is monic in $Z$ and vanishes at $Z = f$ would force $f$ to be in $k[Y]$ with positive degree, but then $\deg(f^q - f) > 1 = \deg(Y)$, so $f^q - f - Y$ can't be $0$.
The argument that I gave works in any characteristic.
Incidentally, a formal power series in one variable over a finite field is rational if and only if its coefficients are eventually periodic.
|
2025-03-21T14:48:31.490632
| 2020-07-10T18:04:04 |
365335
|
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|
Stack Exchange
|
Probability involving dependent random variables constructed from i.i.d. Gaussians
This is a problem I need to address for a certain computation in my research.
Let $Y_1,\dots,Y_n$ be a sequence of i.i.d. standard normal variables; and let $I\subset[0,+\infty)$ be an interval. In my application, if it helps one can think of $I=[a_n,b_n]$ where both $a_n,b_n$ are positive sequences that are both $o_n(1)$ that is $a_n,b_n\to 0$ as $n\to+\infty$.
Evaluate the following probability:
$$
\mathbb{P}\left(\bigcap_{1\leq i\leq n}\left\{\sum_{1\leq j\leq n,j\ne i}Y_iY_j\le 0 \right\}\cap \left\{\sum_{1\le j\le n}Y_j \in I\right\}\right).
$$
In particular,
(a) Is there a good way to compute this probability?
(b) How does it behave as a function of the interval $I$, equivalently, as a function of $a_n,b_n$.
If we denote the sum by $S$, the condition $\sum_{1\le j\le n,j\ne i}Y_iY_j\le 0$ is equivalent to having $Y_i^2\ge SY_i$, which, on top of $S\ge 0$ (recall $I\subset[0,\infty)$) implies either $Y_i\le 0$ or $Y_i\ge S$. In particular, as $I$ gets larger, $\mathbb{P}(S\in I)$ gets larger, whereas $\mathbb{P}(Y_i\in [0,S]^c)$ gets smaller, which obviously means the size of $I$ brings a compromise.
I could not see a good way of computing this, and appreciate any help.
Do you really need to evaluate the probability (i.e. give a closed-form expression), or you would be satisfied with lower or upper bounds for that quantity?
I certainly would be satisfied with an upper/lower bound (ideally both) that is not too loose, Mateusz.
(This is not an answer, but too long for a comment.)
Let $Z = \frac{1}{\sqrt{n}} \sum_{i=1}^n Y_i$ and $X_i = Y_i - \tfrac{1}{\sqrt{n}} Z$. Then $Z$ and $(X_1, \ldots, X_n)$ are independent, $Z$ has standard normal distribution, and $(X_1, \ldots, X_n)$ has standard multivariate normal distribution on the hypersurface $x_1 + \ldots + x_n = 0$.
The question asks for the probability that $$Z \in \tfrac{1}{\sqrt{n}} I \qquad \text{and} \qquad X_i \notin [-\tfrac{1}{\sqrt{n}} Z, \tfrac{n - 1}{\sqrt{n}} Z] \quad \text{for every } i = 1, \ldots, n . $$
Denote by $p(z)$ the probability of the latter part, conditionally on $Z = z$. The whole story is therefore reduced to an estimate of the probability that a $n-1$-dimensional standard normal random vector belongs to a certain polytope.
For a fixed $n$, this seems doable. The general case, however, seems out of reach. Crude lower and upper bounds for $p(z)$ are
$$ p(z) \geqslant 1 - \sum_{i = 1}^n \mathbb{P}(X_i \in [-\tfrac{1}{\sqrt{n}} z, \tfrac{n - 1}{\sqrt{n}} z]) = 1 - n \mathbb{P}(N \in [-\tfrac{1}{\sqrt{n-1}} z, \tfrac{n - 1}{\sqrt{n-1}} z]) $$
and
$$ \begin{aligned} p(z) & \leqslant 1 - \sum_{i = 1}^n \mathbb{P}(X_i \in [-\tfrac{1}{\sqrt{n}} z, \tfrac{n - 1}{\sqrt{n}} z]) + 2 \sum_{i = 1}^{n - 1} \sum_{j = i+1}^n \mathbb{P}(X_i \in [-\tfrac{1}{\sqrt{n}} z, \tfrac{n - 1}{\sqrt{n}} z]) \\ & = 1 - n \mathbb{P}(N \in [-\tfrac{1}{\sqrt{n-1}} z, \tfrac{n - 1}{\sqrt{n-1}} z]) + n (n - 1) \mathbb{P}((N, M - \tfrac{1}{n - 1} N) \in [-\tfrac{1}{\sqrt{n-1}} z, \tfrac{n - 1}{\sqrt{n-1}} z]^2), \end{aligned} $$
where $N, M$ are auxiliary independent standard normal random variables. These estimates, of course, follow from the exclusion-inclusion principle, and are reasonably sharp when $z = O(n^{-3/2})$. That would require $b_n = O(n^{-1})$, which is likely not what is meant in the question, so I did not attempt to continue this calculation.
|
2025-03-21T14:48:31.490847
| 2020-07-10T18:12:27 |
365336
|
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"zeraoulia rafik"
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|
Stack Exchange
|
Infinity-homotopies
Koszul duality for operads allows for straightforward generalizations of $A$-infinity algebras and $A$-infinity morphisms for the so called Koszul operads $\mathcal{O}$, among which we find the associative operad. Good accounts of this can be found in standard references. However, I've been unable to find a generalization of $A$-infinity homotopies between $A$-infinity morphisms. Does anyone know of any reference(s) where this is treated in the greatest possible generality? Ideally, for Koszul operads over an arbitrary commutative ground ring, but anything is welcome.
I recomond u this book is for Koszul operads
@zeraouliarafik Thanks. This is one of the 'standard references' I had in mind. Have you found anything about infinity-homotopies therein apart from the A-infinity case? I've search with no success, but I may well have overlooked it.
I don't know if you found an answer since you posted the question, but I will write this just in case: there is a "cute" (easy) definition in case of nonsymmetric operads which generalises the A-infinity story rather trivially (derivation homotopy), while for symmetric operads the definition is more involved. There is a comparison of various possibilities in my paper with Poncin https://link.springer.com/article/10.1007/s10485-015-9407-x , and an explanation that these possibilities actually come from a suitable model category structure in the definitive treatment by Bruno Vallette https://aif.centre-mersenne.org/item/AIF_2020__70_2_683_0/
Wow, thanks! I didn't get to your paper while doing my compulsory search. It looks like it's exactly what I was looking for.
|
2025-03-21T14:48:31.490981
| 2020-07-10T18:37:14 |
365339
|
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|
Stack Exchange
|
Product formula for Laplace de-Rham operator
Let $M$ be a Riemannian manifold with Laplace de-Rham operator $\Delta = (d + \delta)^2$. If $g$ is a smooth $k$-form, and $f$ is a smooth function, is there a simple formula for $\Delta(fg)$ when $k > 0$?
Of course, this is a special case of $\Delta (f \wedge g)$. I would expect the formula to involve $\Delta f, \Delta g, \nabla f, \nabla g$ (like the case $k=0$) and something related to the curvature. I suspect that this is a very easy question to answer for people working with Riemannianian manifolds.
This is straightforward if you use the Weitzenbock formula and the product rule.
Yes, and you can find it and its proof as a special case of Proposition 2.5 in the book of Berline-Getzler-Vergne: let $E$ be a hermitian vector bundle with unitary connection $\nabla,$ then the induced Laplacian $\Delta=\Delta^\nabla$ satisfies
$$\Delta (f s)=(\Delta f)s+f\Delta^\nabla s-2\langle\operatorname{grad} f,\nabla s\rangle.$$
Moreover, the Laplacian induced by the Levi-Civita-connection $\nabla$ on the exterior bundle is the standard Laplacian on the exterior bundle.
|
2025-03-21T14:48:31.491198
| 2020-07-10T18:49:42 |
365340
|
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|
Stack Exchange
|
The diameter of random regular graphs
In 1982, B. Bollobas and Vega in the paper gave the configurational model to generate $r$-regular random graphs. They gave the following theorem (Theorem 1 in the paper).
Theorem: Let $r\geq 3$ and $\epsilon> 0$ be fixed and define $d=d(n)$ as the least integer satisfying $$(r-1)^{d-1}\geq (2+\epsilon)rn \ \text{log} \ n.$$
Then almost every $r$-regular graphs of order $n$ has diameter at most $d$.
I have following two questions
Question 1. Does this theorem implies that the diameter of $r$-regular random graphs is of order $\text{log} (n \text{log} n),$ that is, $d(n)=O(\text{log} (n \text{log} n))$?
Question 2. If Question 1 is true, will all the $r$-regular random graphs (not necessary generated using the configuration model) has the diameter of order $\text{log} (n \text{log} n).$
I think random regular graphs are expanded, so should have diameter O(log(n)). But I don’t know if there is a model for regular random graphs in which they are not expanders.
Yes Friedman proved that all the large enough random regular graphs tend to be Ramanujan graphs, which are the best possible expanders. So maybe these expanders does not satisfy the above theorem.
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2025-03-21T14:48:31.491306
| 2020-07-10T19:02:35 |
365342
|
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|
Stack Exchange
|
The (measurable) Riemann mapping theorem
The Riemann mapping theorem says that a strict, nonempty open subset of the complex plane is conformally equivalent to the unit disk.
The measurable Riemann mapping theorem asserts the existence and uniqueness of a quasiconformal homeomorphism $f$ satisfying the Beltrami equation: $$\frac{\partial f}{\partial \overline{z}} = \mu(z)\frac{\partial f}{\partial z} $$ for given $\mu$ with $ \lVert\mu\rVert_{\infty}<1$.
What (if anything) do these two statements have to do with each other? Wikipedia points out in the link above that the latter isn't a direct generalization of the former, although there does seems to be a proof of Riemann mapping theorem from the measurable RMT.
You misstated Riemann's (original) theorem:
a crucial assumption is that your open subset
is simply connected.
Both theorems can be considered as classification theorems
of Riemann surfaces. The Riemann original theorem says that every simply connected domain in the sphere, whose complement contains at least 2 points
is conformally equivalent to the
unit disk.
"Measurable Riemann theorem" says that a sphere equipped with any Riemannian metric, subject to certain condition (that the Beltrami coefficient $\mu$
has norm <1) is conformally equivalent to the Riemann sphere.
It has simple corollaries that a plane or a disk equipped with a Riemannian metric
satisfying the same condition are conformally equivalent to the plane and disk respectively with the standard metric. (A disk with arbitrary Riemannian metric is a generalization of a simply connected domain in
the plane with the usual metric).
The old, classical name of the "Measurable Riemann theorem" was "Existence and uniqueness theorem for Beltrami equation", or it was called simply by the name of an
author (Korn and Lichtenstein, or Morrey or Boyarski, depending on exact conditions, and the taste of the person who refers). Boyarski's contribution is the very important fact that properly normalized $f$ depends on $\mu$ analytically.
The modern name comes from the paper of Ahlfors and Bers, Riemann's mapping theorem for variable metrics, Ann. Math., 72 2 (1960), 385-404, where they restated the result of Boyarski in the spirit that I outlined above, and emphasized this analytic dependence on $\mu$. Besides $\|\mu\|_\infty<1$,
no condition on $\mu$ is imposed except that it is Lebesgue measurable, and the word "measurable" in the name of the theorem comes from this fact.
|
2025-03-21T14:48:31.491484
| 2020-07-10T19:07:28 |
365344
|
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|
Stack Exchange
|
Visualize (co)sketeton of a simplicial set (geometrical intuition)
I want to understand if there is an intuition approchable with
most possible 'elementary geometrical' knowledge for
$n$-(co)skeleta of simplicial sets?
Formally sketleton & coskeleton functions arise as follows: For $\Delta$ the simplex category write $\Delta_{\leq n}$ for its full subcategory on the objects
$[0],[1],\cdots,[n][0], [1], \cdots, [n]$.
The inclusion $\Delta|_{\leq n} \hookrightarrow \Delta$
induces a truncation functor
$$\mathrm{tr}_n: \mathit{sSet}= [\Delta^{\mathrm{op}},Set] \to [\Delta_{\leq n}^{\mathrm{op}},\mathit{Set}]$$
that takes a simplicial set and restricts it
to its degrees $\leq n$.
This functor has a left adjoint, given by left Kan extension
$\mathrm{sk}_n: [\Delta_{\leq n},\mathit{Set}] \to \mathit{SSet}$ called the $n$-skeleton
and a right adjoint, given by right Kan extension
$\mathrm{cosk}_n : [\Delta_{\leq n},Set] \to SSet$ called the $n$-coskeleton.
Now set $F: \Delta^{\mathrm{op}} \to Set, [n] \mapsto X_n$. The picture one conventionally has in mind thinking intuitionally/geometrically about $X$ is
that one thinks $X_n$ as "the set of $n$-simplices/cells of the
"simplicial complex" $X$ (only as geometrical intuition).
How can I think in this naive manner about $\mathrm{sk}_n(X)$ and
$\mathrm{cosk}_n(X)$?
The $\mathrm{sk}_n(X)$ might be considered as a "subcomplex" of $X$
obtained from $X$ by killing all $m$-simplices with $m > n$.
The way all $\ell$-simplices for $\ell \le n$ are "glued together"
stays the same as for $X$, ie for $\ell$-simplices happens nothing.
If we keep thinking about $X$ as a simplicial complex,
which picture should one have thinking about $\mathrm{cosk}_n(X)$?
How it deviates from original $X$?
For $k \le n$, the $k$-simplices in $\mathrm{cosk}_n(X)$ are the same as in $X$. For larger $k$, there is a unique $k$-simplex for every $n$-skeleton of a $k$-simplex you find in $X$, that is, $(\mathrm{cosk}_n(X))_k \cong \mathrm{Hom}(\mathrm{sk}_n \Delta^k, X)$.
You can also think inductively: again, for $k \le n$ the $k$-simplices in $\mathrm{cosk}_n(X)$ are the same as in $X$; then for each $k > n$ if you already know the simplices of dimension less than $k$ in $\mathrm{cosk}_n(X)$, you get the $k$-simplices by filling in uniquely every empty $k$-simplex you find in $\mathrm{cosk}_n(X)$. That is, for $k>n$, $(\mathrm{cosk}_n(X))_k \cong \mathrm{Hom}(\partial \Delta^k, \mathrm{cosk}_n(X))$.
so maybe in more elementary language can we say $cosk_n(X)$ is
constructed & glued in following way: The $k \le n$-simlices
of $cosk_n(X)$ coinside with $X$. And for
$k >n$ a $k$-simplex is contained in $(cosk_n(X))k$ if all
it's faces are contained in $cosk_n(X){k-1}$ and
additionally are glued in compatible way to auch other?
informally, if $cosk_n(X)$ already "contains" completly the boundary
of this $k$-simplex, then it contains this $k$-simplex itself? Is this the correct intuitive "picture"?
Yes, @MortyPB, that's the right idea.
|
2025-03-21T14:48:31.491701
| 2020-07-10T19:25:00 |
365346
|
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|
Stack Exchange
|
Perfect $\mathbb Z_\ell$-modules
Disclaimer : I asked this question on Maths.StackExchange 20 days ago (and started a bounty) here but got no answer, so I'm asking it here now (with no modification).
$\newcommand{\l}{\ell} \newcommand{\Z}{\mathbb Z}$
I'm trying to understand a remark in Weil's conjectures for function fields I, by Gaitsgory-Lurie.
Specifically it's remark <IP_ADDRESS>., which essentially states the following : an object $M$ in $\mathrm{Mod}_{\mathbb Z_\l}$ is perfect if and only if it's $\l$-complete and $M\otimes_{\Z_\l}\Z/\l$ is perfect in $\mathrm{Mod}_{\Z/\l}$ (this latter condition implies that this is also true for $M\otimes_{\Z_\l}\Z/\l^d$, for any $d\geq 0$)
They state it without proof, and it looks like there is no proof elsewhere in the book (as far as I can see, although I haven't checked the whole book).
What is clear to me is the forward direction : indeed $\Z_\l\otimes_{\Z_\l}\Z/\l$ is perfect over $\Z/\l$ and $\Z_\l$ is $\l$-complete, and these conditions are closed under retracts, so the sub-$\infty$-category of those $M$'s that satisfy the latter conditions is a stable subcategory, contains $\Z_\l$ and is closed under retracts so it contains all perfect $\Z_\l$-modules.
The reverse direction, however, does not seem so clear. I think one ought to use the fact that the $\infty$-category of $\l$-complete modules is equivalent to $\underset{d}{\varprojlim} \mathrm{Mod}_{\Z/\l^d}$ but that doesn't seem to be enough since we want $M$ to be compact in the whole $\mathrm{Mod}_{\Z_\l}$, not just in $\l$-complete modules.
Over a PID, you can check perfectness on homology. So the claim is that the homology of an $\ell$-complete $\mathbb{Z}_\ell$-module is finitely generated if and only if it is so mod $\ell$. The homology groups are $\ell$-complete, and so this follows from the long exact sequence of mod $\ell$-reduction.
Thank you for your answer, but I'm not sure I understand the end. What do you call the long exact sequence of mod $\ell$-reduction ?
Well, for any complex $X$, you have a cofiber sequence $X\xrightarrow{\ell} X\to X/\ell$, and this induces a long exact sequence in homology.
This shows that the mod $\ell$-reduction of homology, $H_(X)/\ell$, maps injectively into $H_(X/\ell)$, so if the latter is finitely generated, so is the former.
Right that makes sense - but it's not clear to me how to go from there to finite generation of $H_*(X)$. It seems like this reduces it to the same question about discrete $\mathbb Z_\ell$-modules, but I don't know how to prove that either
That's a completely elementary exercise, you just check that a map between discrete $\ell$-complete $\mathbb{Z}_\ell$-modules is surjective if it is surjective mod $\ell$.
That's not entirely clear to me : if I'm not mistaken, even if the modules are discrete, $\ell$-complete refers to derived $\ell$-completion, not "discrete" $\ell$-completion, right ? (of course if they are "discrete" $\ell$-complete, then I agree it's an elementary statement, but what I must be missing is the derived statement)
(let me make my doubts about the derived statement more apparent : let $M\to N$ be a map of discrete derived $\ell$-complete modules which is surjective mod $\ell$. Let $K$ be its fiber (a priori nondiscrete) - we would like to show that $K$ is connective, in which case $M\to N$ is indeed surjective. By tensoring with $\mathbb Z/\ell$, and then doing the usual extension game, we get that $K\otimes \mathbb Z/\ell^n$ is connective - $K$ is also (derived) $\ell$-complete, so it's a limit of a tower of connective modules, but as far as I can tell, this does not suffice
we would need $\varprojlim^1_n \mathrm{Tor}^{\mathbb Z_\ell}1(\pi{-1}(K), \mathbb Z/\ell^n)$ to vanish or something like that, but this is not clear to me. Maybe it follows because $\pi_{-1}(K)$ is also derived $\ell$-complete, and maybe this implies a Mittag-Leffler condition ? Anyways, I'm still missing part of the argument)
Ah, I see, that's a subtlety I missed, sorry. But I think for a derived $K$-complete complex you can check connectivity mod $\ell$. To see this, assume $K/\ell$ is connective. Then $\pi_k K$ for $k<0$ are divisible. But they are also derived $\ell$-complete, and so they are zero.
Ah, I guess one can directly make this explicit: Your $\mathrm{Tor}1$ is just the $\ell^n$-torsion subgroup of $\pi{-1}(K)$, and the latter is divisible, so the maps in the diagram for your $\lim^1$ are all surjective.
By the way, using that connectivity is detected mod $\ell$ one can also directly do the original question: For $X$ $\ell$-complete with perfect mod $\ell$-reduction, inductively build a finite complex of copies of $\mathbb{Z}_\ell$ with map to $X$, which induces an equivalence mod $\ell$. This is formally similar to how one builds minimal CW approximations etc.
Oh that's neat ! I hadn't realized that divisibility was enough, I thought I needed unique divisibility. Now I think I have everything, thanks ! (and sorry for the prolonged discussion in the comments)
(I don't know what the etiquette is about answers where some more details are added in the comments - maybe you [ or I, if you'd rather I did it ] can edit it to add a word or two ? )
|
2025-03-21T14:48:31.492045
| 2020-07-10T19:53:54 |
365351
|
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|
Stack Exchange
|
Finding the dimension of the intersection of two real algebraic varieties
Suppose we have two polynomials $p, q \in \mathbb{R}^3$ and we are interested in their simultaneous zeros. Parameter counting tells us that the zero set most probably is going to be a one dimensional curve. But how can I make this statement rigourous? Is there any theorem which gives you the Hausdorff dimension of such a zero set? Indeed, we will have to assume that $p$ and $q$ have no common factors.
What exactly is a polynomial in $\mathbb{R}^3$? Do you mean in $\mathbb{R} [x_1, x_2, x_3]$?
because you work over the reals, parameter counting will tell you nothing.
For a postive integer $n$ denote by $P_n$ the vector space of polynomials of degree $\leq n$ in three variables. Sard's theorem shows that if $n$ is sufficiently large, there exists a measure zero subset $Z\subset P_n\times P_n$ such that for $(p,q)\in P_n\times P_n\setminus Z$ the set ${p=q=0}$ has dimension $1$ if nonempty. In fact $Z$ itself is semialgebraic.
As in my comment denote by $P_n$ the subspace of $\newcommand{\bR}{\mathbb{R}}$ $\bR[x_1,x_2,x_3]$ consisting of polynomials of degree $\leq n$. Consider the map
$$ F: \bR^3\times P_n\times P_n\to\bR^2,\;\;(x,p,q)\mapsto (p(x),q(x)).$$
If $n$ is sufficiently large, then for any $x\in\bR^3$ the map
$$ P_n\times P_n\ni (p,q)\mapsto (p(x),q(x))\in\bR^2 $$
is a submersion. We deduce that the set
$$\Lambda=\{(x,p,q)\in\bR^3\times P_n\times P_n;\;\;p(x)=q(x)=0\} $$
is a submanifold of codimension $2$. We have a natural map
$$\pi:\Lambda\times P_n\times P_n\to P_n\times P_n,\;\;(x,p,q)\mapsto(p,q). $$
Sard's theorem shows that most $(p,q)\in P_n\times P_n$ is a regular value of $\pi$. The fibers of $\pi$ will generically have dimension $1$. Note that the fiber of $\pi$ over $(p,q)$ is the set $\{p=q=0\}$.
|
2025-03-21T14:48:31.492193
| 2020-07-10T19:58:21 |
365352
|
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|
Stack Exchange
|
Summing over normalized characters of the permutation group
Let $\chi_\lambda(\mu)$ be the usual characters of the irreducible representations of the permutation group $S_n$. The normalized character is the quotient $\chi_\lambda(\mu)/f^\lambda$, where $f^\lambda=\chi_\lambda(1)$ is the dimension of the representation.
Can I hope for a nice formula expressing their sum
$$ \sum_{\lambda\vdash n}\frac{\chi_\lambda(\mu)}{f^\lambda},$$
in terms of the parts of $\mu$?
Just a comment. If there were such a nice formula, then multiplying by the nice formula for $f^\lambda$ obtains a nice formula for $\chi_\lambda(\mu)$. In other words, there would be a nice, presumably meaning non-recursive, version of the Murnaghan--Nakayama rule.
@DavidCraven I was hoping for a nice formula only for the whole sum, not for individual terms
Ah sorry, I missed that.
The quantity you are asking about is in fact a well-known expression: When multiplied by $n!$, it is the number of ordered pairs $\sigma, \tau \in S_{n}$ such that
$[\sigma, \tau] = \mu$, where $[\sigma, \tau] = \sigma^{-1}\tau^{-1}\sigma \tau$ is the commutator of $\sigma$ and $\tau$. However, I do not know how to relate this to the disjoint cycle structure of $\mu$, except to say that this quantity is clearly zero if $\mu$ is an odd permutation.
This is good enough, and suggests that no simple formula exists
This is just an observation. I normalize your problem by $n!$, to get rid of denominators.
Let $A_n(\mu) := n! \sum_{\lambda \vdash n} \chi^{\lambda}(\mu)/f^\lambda$.
Define $B_n(x) := n! \sum_{\lambda \vdash n} \frac{p_\lambda(x)}{f^\lambda}$.
Then $A_n(\mu) = \langle B_n(x), s_\mu \rangle$.
That is, $A_n(\mu)$ is the coefficient of $s_\mu$ when expanded in the Schur basis.
The Schur expansion of $B_n(x)$ for $n=1,2,\dotsc$
are
\begin{array}{l}
s_{1} \\
4 s_{2} \\
15 s_{3}+6 s_{21}+9 s_{111} \\
76 s_{4}+64 s_{22}+44 s_{31}+76 s_{211}+12 s_{1111} \\
368 s_{5}+628 s_{32}+416 s_{41}+580 s_{221}+792 s_{311}+344 s_{2111}+200 s_{11111}
\end{array}
Perhaps there is some pattern...
There is some information related to this problem in Exercise 7.68 of Enumerative Combinatorics, vol. 2. For instance, let $n$ be odd. Then the number of ways of writing a fixed $n$-cycle in the form $uvu^{-1}v^{-1}$ is $2n\cdot n!/(n+1)$.
|
2025-03-21T14:48:31.492370
| 2020-07-10T20:41:05 |
365357
|
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}
|
Stack Exchange
|
Conformal mapping
Is there a simple construction of a confomral mapping of the half-plane onto a "circular trianagle", i.e. a domain whose sides are the arcs of three circles.
Yes, it is (how simple, is a matter of opinion). You may always assume that $0,1,\infty$ on the boundary of upper half-plane
are preimages of the vertices. Now suppose that the inner angles of your triangle
are $\pi\alpha_j,$ and let us assume that $\sum\alpha_j$ is not an odd integer (Euclidean triangles must be considered separately and this case is in fact simpler).
Then your mapping function is a solution of the Schwarz differential equation
$$\frac{f'''}{f'}-\frac{3}{2}\left(\frac{f''}{f'}\right)^2
=\frac{1-\alpha_1^2}{2z^2}+\frac{1-\alpha_2^2}{2(z-1)^2}+\frac{\alpha_1^2+\alpha_2^2-\alpha_3^2-1}{2z(z-1)}.$$
This is written in Hurwitz-Courant (with a misprint which I corrected), and in
Caratheodory, vol. II.
This is not as scary as it may look, because in fact if $F(z)$ is the right hand-side, then $f=y_1/y_2$ where $y_1$ and $y_2$ are linearly independent solutions of the linear differential equation
$$y''+(F/2)w=0,$$
which in our case is a hypergeometric equation. Its solutions (hypergeometric functions with real parameters) are special functions, and "everything is known" about them. I mean explicit power series, location of zeros, asymptotics, integral representations, explicit analytic continuation, tables, and so on.
Function $f$ contains 3 arbitrary constants which can be determined from positions
of vertices of your triangle.
For the case of Euclidean triangle (bounded by straight lines, and with $\sum\alpha_j=1$) the function is more explicit: it is the Schwarz--Christoffel integral
$$f(z)=C\int_a^z\zeta^{\alpha_1-1}(\zeta-1)^{\alpha_2-1}d\zeta$$
where $a$ and $C$ are constants, and can be determined from positions of two vertices of your triangle.
Refs. A. Hurwitz and R. Courant, Vorlesungen uber allgemeine Funktionentheorie,
(available in German and Russian)
C. Caratheodory, Theory of functions of a complex variable, vol. II (available in German and English). Most of the vol. II is devoted to this subject.
F. Klein, Vorlesungen über die Hypergeometrische Funktion, Berlin 1933 (the whole book is devoted to the subject).
W. Koppenfels and F. Stahlman, The Practice of conformal Mappings (available in German and Russian).
The next case in complexity, circular quadrilaterals, is much more complicated and still remains a research subject.
Edit: A good modern source in English is S. Donaldson, Riemann surfaces, Oxford, 2011, Thm. 29.
The Christoffel-Schwarz formula for polygons bounded by arcs of circles is considered in the PhD thesis
Computation of Conformal Maps by Modified Schwarz-Christoffel Transformations (1990) by Louis H. Howell. This thesis is accessible in the Internet.
|
2025-03-21T14:48:31.492566
| 2020-07-10T23:21:08 |
365361
|
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"authors": [
"https://mathoverflow.net/users/160378",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/365361"
}
|
Stack Exchange
|
A royal road to Coulomb branches of 3D $\mathcal{N}=4$ gauge theories
So, I've been very interested recently with the developements of the (now mathematically precise) theory of Coulomb branches - in particular because of its recent applications on representation theory and symplectic geometry.
They have been considered by mathematical physicists for a time, but without a rigorous definition. The first attempt to give such a definition was made by H. Nakajima, Introduction to a provisional mathematical definition of Coulomb branches of 3-dimensional N=4 gauge theories.
This project was completed by A. Braverma, N. Finkelberg, H. Nakajima in
Towards a mathematical definition of Coulomb
branches of 3-dimensional N=4 gauge theories, II, and their companion papers
Coulomb branches of 3d N=4 quiver gauge theories and slices in the affine Grassmannian. With two appendices by Braverman, Finkelberg, Joel Kamnitzer, Ryosuke Kodera, Nakajima, Ben Webster and Alex Weekes.
Ring objects in the equivariant derived Satake category arising from Coulomb branches. Appendix by Gus Lonergan.
Now, I've become rather convinced that to proceed with my research I need to have a rather firm grasp on the tree papers above by Braverman, Finkelberg and Nakajima.
Now, I unfortunately have no good physics intuition, and I've found that some parts of the above papers are very technical and are motived by a lot of different constructions in geometric representation theory in the range of the past 20 years. The final form of this theory is a remarkable achievement of novel ideas and technical mastery, and I'm feeling a little bit lost on what should be the relevant things to focus on (important previous work and mathematical techiniques and machinery).
Hence, the question: what is the royal road to Coloumb branches?
PS: the two papers I'm trying to fully appreciate are:
J. Kamnitzer, P. Tingley, B. Webster, A. Weekes and O. Yacobi, On category O
for affine Grassmannian slices and categorified tensor products.
and
A. Weekes, Generators of Coulomb branches of quiver gauge theories, arXiv:1903.07734.
As for prerequisite for three papers, I recommend
Chriss-Ginzburg, Representation Theory and Complex Geometry, and Victor Ginzburg, Geometric Methods in Representation Theory of Hecke Algebras and Quantum Groups.
One also needs to know basics on affine Grassmannians and geometric Satake. There are lots of good survey articles on them, as well as the original paper, Mirkovic-Vilonen, Geometric Langlands duality and representations of algebraic groups over commutative rings.
If three papers are too technical, there are a few survey articles:
Hiraku Nakajima, Introduction to a provisional mathematical definition of Coulomb branches of 3-dimensional N=4 gauge theories
Michael Finkelberg, Double affine Grassmannians and Coulomb branches of 3d N=4 quiver gauge theories
Alexander Braverman, Michael Finkelberg, Coulomb branches of 3-dimensional gauge theories and related structures
For a physical intuition, besides my first paper, I recommend to look at
Stefano Cremonesi, Amihay Hanany, Alberto Zaffaroni, Monopole operators and Hilbert series of Coulomb branches of 3d N = 4 gauge theories
This paper is accessible by mathematicians.
Thank you very much!
I've written a survey paper on this topic:
Symplectic resolutions, symplectic duality, and Coulomb branches.
I begin with symplectic resolutions in general, then discuss symplectic duality, and then the Coulomb branch construction of Braverman-Finkelberg-Nakajima. Throughout, I discuss quiver varieties and (generalized) affine Grassmannian slices. I hope this is helpful!
|
2025-03-21T14:48:31.492793
| 2020-07-11T01:35:07 |
365363
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/365363"
}
|
Stack Exchange
|
Box counting dimension and Besov spaces on $\mathbb R^2$
I found a lemma in this paper of Constantin and Wu, stated with no proof:
Lemma
3.2. Let $b=\chi_{D}$ be the characteristic function of a bounded domain $D\subset\mathbb R^2$ whose boundary has box-counting (fractal) dimension not larger than $d<2:$
$
d_{F}(\partial D) \leq d.
$
Then
$b\in B^{\frac{2-d}p }_{p,\infty}$
$
\text {for } 1 \leq p<\infty.
$
Is it easy to prove? For sufficiently smooth curves (so $d_F(\partial D) = 1$) I have proven (by estimating the Gagliardo seminorm) that $b\in H^s$ for $s<1/2$, and I suspect using Theorem 2.36 of Bahouri, Chemin, and Danchin's book "Fourier analysis and nonlinear partial differential equations":
Theorem 2.36 . Let $s$ be in $( 0,1)$ and $(p, r) \in[1, \infty]^{2}$. A constant $C$ exists such that, for any $u$ in $\mathcal{S}_{h}^{\prime}$
$$
C^{-1}\|u\|_{\dot{B}_{p, r}^{s}} \leq\left\|\frac{\left\|\tau_{-y} u-u\right\|_{L^{p}}}{|y|^{s}}\right\|_{L^{r}\left(\mathbb{R}^{d} ; \frac{d y}{|y|^{d}}\right)} \leq C\|u\|_{\dot{B}_{p, r}^{s}}.
$$
I can get a similar result for Besov spaces. But how can I prove the result in the general case?
The modulus of continuity in direction $v$ is $\omega_p(t,v) := \lVert 1_D - 1_D(\cdot-tv)\rVert_p$. Since $\lvert 1_D(x) - 1_D(x-tv)\rvert \le 1_{N(\partial D,t)}$, where $N(\partial D,t)$ is a $t$-neighborhood of $\partial D$, then by the assumption on the dimension of the boundary
$$
\omega_p(t,v) \le C t^\frac{2-d}{p},
$$
so $1_D\in B^{\frac{2-d}{p}}_{p,\infty}$.
|
2025-03-21T14:48:31.493177
| 2020-07-11T01:55:35 |
365364
|
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"LayZ",
"Narutaka OZAWA",
"Suvrit",
"https://mathoverflow.net/users/75323",
"https://mathoverflow.net/users/7591",
"https://mathoverflow.net/users/8430"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/365364"
}
|
Stack Exchange
|
Proving a majorization inequality for the singular value of the product of two matrices without using tensor product
For any two matrices $\mathbf{A},\mathbf{B} \in \mathbb{C}^{n \times n}$, we know that the following majorization inequality holds
$$
\tag{1}
\label{grz}
\sigma^{\downarrow}(\mathbf{A}\mathbf{B}) \prec_w \sigma^{\downarrow}(\mathbf{A})\sigma^{\downarrow}(\mathbf{B}),
$$
where $\sigma^{\downarrow}(\cdot)$ denotes the vector of singular values, ordered in the decreasing order. This is equivalent to the following system of inequalities
$$
\tag{2}
\label{sysineq}
\sum_{i=1}^k\sigma_i^{\downarrow}(\mathbf{A}\mathbf{B}) \leq \sum_{i=1}^k \sigma_i^{\downarrow}(\mathbf{A})\sigma_i^{\downarrow}(\mathbf{B}),
$$
for $k=1,\dots,n$.
Proof:
In all the textbooks or papers that I have seen, the proof of this majorization inequality is as follows. By the sub-multiplicativity of the spectral norm, one has
$$
\sigma_1^{\downarrow}(\mathbf{A}\mathbf{B}) \leq \sigma_1^{\downarrow}(\mathbf{A})\sigma_1^{\downarrow}(\mathbf{B}).
$$
By employing this inequality to the anti-symmetric tensor powers (i.e. the compound matrices) $\wedge^k(\mathbf{A})$ and $\wedge^k(\mathbf{B})$, we have
$$
\sigma_1^{\downarrow}\big((\wedge^k \mathbf{A})(\wedge^k \mathbf{B})\big) \leq \sigma_1^{\downarrow}\big(\wedge^k \mathbf{A}\big)\sigma_1^{\downarrow}\big(\wedge^k \mathbf{B}\big),
$$
for $k=1,\dots,n$. Then using the facts that $\wedge^k(\mathbf{A}\mathbf{B}) = (\wedge^k \mathbf{A})(\wedge^k \mathbf{B})$ and $\sigma_1^{\downarrow}\big(\wedge^k \mathbf{A}\big) = \prod_{i=1}^k \sigma_i^{\downarrow}(\mathbf{A})$, it follows that
$$
\tag{3}
\label{lwm}
\prod_{i=1}^k\sigma_i^{\downarrow}(\mathbf{A}\mathbf{B}) \leq \prod_{i=1}^k \sigma_i^{\downarrow}(\mathbf{A})\sigma_i^{\downarrow}(\mathbf{B}),
$$
for $k=1,\dots,n$. Finally, inequality \eqref{grz} follows using the fact that log-weak majorization inequality \eqref{lwm} implies weak majorization inequality \eqref{grz} [Bhatia, Matrix analysis, Example II.3.5 (vi)].
Question:
Can we prove the majorization inequality \eqref{grz} without resorting to the tensor products and employing no facts about them?
Thanks in advance!
My attempt:
By the maximal characteristic of the singular values, we know that
\begin{equation}
\sigma_i(\mathbf{A}) = \max_{\substack{\|\bf{x}_i\|=\|\bf{y}_i\|=1 \\ \bf{x}_i \bot \text{span}\{\bf{x}_1,\dots, \bf{x}_{i-1}\} \\ \bf{y}_i \bot \text{span}\{\bf{y}_1,\dots, \bf{y}_{i-1}\}}}\big|\langle \mathbf{A}\bf{x}_i,\bf{y}_i \rangle\big|,
\end{equation}
for $i=1,\dots,n$. Using this formula, we can demonstrate that the inequalities \eqref{sysineq} are equivalent to the following system of inequalities:
\begin{equation}
\max_{\substack{\|\bf{x}_i\|=\|\bf{y}_i\|=1, \;i \in [k] \\ \bf{x}_1 \bot \dots \bot \bf{x}_k \\ \bf{y}_1 \bot \dots \bot \bf{y}_k}} \sum_{i=1}^k \big|\langle \mathbf{A}\mathbf{B} \bf{x}_i,\bf{y}_i \rangle\big| \leq \max_{\substack{\|\bf{x}_i\|=\|\hat{\bf{x}}_i\|=1, \;i \in [k] \\ \bf{x}_1 \bot \dots \bot \bf{x}_k \\ \hat{\bf{x}}_1 \bot \dots \bot \hat{\bf{x}}_k}} \max_{\substack{\|\bf{y}_i\|=\|\hat{\bf{y}}_i\|=1, \;i \in [k] \\ \bf{y}_1 \bot \dots \bot \bf{y}_k \\ \hat{\bf{y}}_1 \bot \dots \bot \hat{\bf{y}}_k}} \sum_{i=1}^k\big| \langle \mathbf{B}\bf{x}_i,\hat{\bf{x}}_i \rangle \langle \mathbf{A}\bf{y}_i,\hat{\bf{y}}_i \rangle\big|,
\end{equation}
for $k=1,\dots,n$. All I can show is that for each $i=1,\dots,k$, we have
\begin{equation}
\begin{split}
\big|\langle \mathbf{A}\mathbf{B} \bf{x}_i,\bf{y}_i \rangle\big| &= \big|\langle \mathbf{B} \bf{x}_i, \mathbf{A}^\mathsf{H}\bf{y}_i \rangle\big| \\
& \leq \|\mathbf{B}\bf{x}_i\| \|\mathbf{A}^\mathsf{H}\bf{y}_i\| \\
& = \max_{\|\hat{\bf{x}}_i\|=1} \big|\langle \mathbf{B}\bf{x}_i,\hat{\bf{x}}_i \rangle\big| \max_{\|\hat{\bf{y}}_i\|=1} \big|\langle \mathbf{A}^\mathsf{H}\bf{y}_i,\hat{\bf{y}}_i \rangle\big|,
\end{split}
\end{equation}
where $\mathbf{A}^\mathsf{H}$ is the conjugate transpose of $\mathbf{A}$. The inequality and the last equality follow by the Cauchy-Schwarz inequality. Therefore
\begin{equation}
\max_{\substack{\|\bf{x}_i\|=\|\bf{y}_i\|=1 \\ \bf{x}_1 \bot \dots \bot \bf{x}_k \\ \bf{y}_1 \bot \dots \bot \bf{y}_k}} \sum_{i=1}^k \big|\langle \mathbf{A}\mathbf{B} \bf{x}_i,\bf{y}_i \rangle\big| \leq \max_{\substack{\|\bf{x}_i\|=\|\hat{\bf{x}}_i\|=1 \\ \bf{x}_1 \bot \dots \bot \bf{x}_k}} \max_{\substack{\|\bf{y}_i\|=\|\hat{\mathbf{y}}_i\|=1 \\ \mathbf{y}_1 \bot \dots \bot \mathbf{y}_k}} \sum_{i=1}^k\big| \langle B\mathbf{x}_i,\hat{\mathbf{x}}_i \rangle \langle A\hat{\mathbf{y}}_i,\bf{y}_i \rangle\big|.
\end{equation}
However, these inequalities are weaker than what we want.
Bhatia, Rajendra, Matrix analysis, Graduate Texts in Mathematics. 169. New York, NY: Springer. xi, 347 p. (1996).
Taking the question literally, the answer is "Yes"; am feeling a bit lazy myself right now, but I hope to type up some of that answer in a few days if nobody else does it in the meanwhile.
Since Suvrit seems busy, I'd say it follows from
$$\sum_{i=1}^k \sigma^\downarrow_i(AB)
= \sup_{U}|\mathop{\mathrm{Tr}}(UAB)|
\le \sup_{U,V}|\mathop{\mathrm{Tr}}(UAVB)|
=\sum_{i=1}^k \sigma^\downarrow_i(A)\sigma^\downarrow_i(B),$$
where $U$ and $V$ run over all partial isometries (contractions) of rank (at most) $k$. The only nontrivial is $\le$ part of the rightmost equality.
Thanks. It would be great if you could refer me to a proof of the last equality. To prove the inequality we can assume that $\hat{U}$ is the partial isometry that maximizes the LHS sup. Then, we need to show
$$
|\mathrm{Tr}(B\hat{U}A)| \le \sup_{V}|\mathrm{Tr}(B\hat{U}AV)|.
$$
Now, let $B\hat{U}A$ have the singular value decomposition $R\Sigma Q^$, where $\Sigma$ has at most $k$ non-zero elements, then there exists a rank (at most) $k$ partial isometry $V=Q\Lambda Q^$ s.t. $|\mathop{\mathrm{Tr}}(B\hat{U}A)|=|\mathop{\mathrm{Tr}}(B\hat{U}AV)|$. The inequality follows by taking the sup.
We prove that
$$\sum_{i=1}^k \sigma^\downarrow_i(AB)
= \sup_{U}|\mathrm{Tr}(UAB)|
\le \sup_{U,V}|\mathrm{Tr}(UAV^*B)|
=\sum_{i=1}^k \sigma^\downarrow_i(A)\sigma^\downarrow_i(B),$$
where $U$ and $V$ run over all partial isometries (or contractions) of rank (at most) $k$.
The only nontrivial is $\le$ part of the rightmost equality.
For the proof of this, we may assume that $A$ and $B$ are positive.
Then by the Cauchy--Schwarz inequality, $|\mathrm{Tr}(UAV^*B)|$ attains
the supremum $\mathrm{Tr}(UAU^*B)$ at some rank $k$ partial isometry $U$ (and $V=U$).
Let's denote by $\tilde{A}$ (resp.\ $\tilde{B}$) the truncated operator
$UAU^*$ (resp.\ $B$) on $\mathop{\mathrm{ran}} U$.
Then $\tilde{A}$ and $\tilde{B}$ are positive operators of rank at most $k$
satisfying
$\sigma^\downarrow(\tilde{A})\prec_w\sigma^\downarrow(A)$,
$\sigma^\downarrow(\tilde{B})\prec_w\sigma^\downarrow(B)$, and
$$\mathrm{Tr}(UAU^*B)=\mathrm{Tr}(\tilde{A}\tilde{B}).$$
For the computation of $\mathrm{Tr}(\tilde{A}\tilde{B})$,
we may assume that $\mathop{\mathrm{ran}} U={\mathbb C}^k$ and $\tilde{A}$ is
the diagonal matrix with entries $\sigma^\downarrow(\tilde{A})$.
Let's denote by $\beta$ the diagonal entries of the positive matrix $\tilde{B}$.
Then it satisfies $\beta^\downarrow\prec\sigma^\downarrow(\tilde{B})$.
Hence in conclusion
$$\sup_{U,V}|\mathrm{Tr}(UAV^*B)|
= \mathrm{Tr}(\tilde{A}\tilde{B})
= \sum_{i=1}^k\sigma^\downarrow_i(\tilde{A})\beta_i
\le \sum_{i=1}^k\sigma^\downarrow_i(A)\sigma^\downarrow_i(B).$$
Here, we have used (twice) the following fact.
For any positive eventually-zero sequences $\alpha,\beta,\gamma$ with
$\beta^\downarrow\prec_w\gamma^\downarrow$, one has $\sum_i\alpha^\downarrow_i\beta_i \le \sum_i\alpha^\downarrow_i\gamma^\downarrow_i$, because
$$\sum_i\alpha^\downarrow_i\beta_i
= \sum_i\bigl((\alpha^\downarrow_i-\alpha^\downarrow_{i+1})\sum_{j=1}^i\beta_j\bigr).$$
Thanks for the answer. I have two questions. First, I understand that we are able to assume WLOG $A$ and $B$ are Hermitian, but why can we assume that they are positive at the same time. Second, could you elaborate more on how Cauchy-Schwarz inequality implies that $|\mathrm{Tr}(UAV^*B)|$ attains the supremum when $V$ and $U$ are equal?
@LayZ: (1) Consider the polar decomposition; (2) $(U,V)\mapsto\mathrm{Tr}(UAV^*B)$ is a semi-inner product.
|
2025-03-21T14:48:31.493700
| 2020-07-11T02:25:16 |
365368
|
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"Nate Eldredge",
"Noam D. Elkies",
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|
Stack Exchange
|
For an infinite n-dimensional discrete grid. Each cell has probability $p$ of being on. No infinitely large connected piece?
A piece is called connected if it's made up of on cells, and let C be the set of on cells in the piece, and $\forall c \in C\ \exists \ d \in C$ such that $c$ and $d$ are directly connected, that is their coordinates are the same, except for one value which differs by one.
The fact that there cannot be a connected piece of infinite size seems apparent to me. But it was debated. We agreed that the 1D case is clearly true i.e. you can't have a piece of infinite size. And the "curse of dimensionality" kinda rings a bell here since the higher the dimension, the more sparse your grid and hence even less likely to have a connected piece.
So I am seeking proof or a theorem from somewhere that can settle this once and for all.
This is one kind of problem studied in "percolation theory",
and looks very close to what the Wikipedia article calls "site percolation".
It seems plausible (though not obvious) that the two problems are equivalent;
if so then for each $n \geq 2$ there exists $p_n < 1$ such that
there is an infinite component with probability $1$ once $p > p_n$.
You are right; I will correct.
A good reference is Grimmett's book Percolation. I think Theorems 1.10 and 1.33 are relevant here.
The "plausible though not obvious" issue I had in mind is whether
a connection between arbitrarily far horizontal lines is equivalent
with probability $1$ to the existence of an infinite component. [This is separate from the mistake that @NateEldredge observed in my initial comment, which used to say "bond percolation" instead of "site percolation".]
If I understand the question correctly, you're asking about site percolation on $\mathbb{Z}^d$, $d \ge 2$, and the question is whether there exists an infinite cluster.
The answer is that for $p$ close enough to 1, there will exist an infinite cluster, almost surely (i.e. with probability 1). Indeed, there is a critical probability $p_c(d)$ such that if $p < p_c$ then almost surely there is no infinite cluster, and if $p > p_c$ then almost surely there is at least one (indeed it can be shown that almost surely there is exactly one). This much follows from the Kolmogorov 0-1 law, since existence of an infinite cluster is a tail event. It can be shown for $d \ge 2$ that $0 < p_c < 1$. See for instance Grimmett's book Percolation. Most of the book is about bond percolation, where you switch edges instead of cells, but many results can be converted from one to the other. For this particular result, see Theorem 1.10 for bond percolation, and Theorem 1.33 for the corollary for site percolation.
Note that higher dimension makes it easier to have an infinite cluster, for any given value of $p$; for example, $\mathbb{Z}^3$ contains infinitely many copies of $\mathbb{Z}^2$, any one of which could have an infinite cluster.
There is lots of other research on percolation; again, Grimmett's book is a great place to start.
|
2025-03-21T14:48:31.493923
| 2020-07-11T04:00:23 |
365372
|
{
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"authors": [
"Peter Gerdes",
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|
Stack Exchange
|
Cupping and capping for 0’ relative to a recursively enumerable set
Is there an r.e. set $A$ such that 0’ is cuppable relative to $A$? What about cappable?
This is equivalent to asking if there is an r.e. $A$ such that 0’ is one half of a pair of $A$ r.e. non-$A$ computable sets whose meet is $A$ and similarly if there is an $A$ such that 0’ can be (non-trivially) joined to $A'$ via an $A$ r.e. set.
I’m pretty sure I’ve seen results on this and I’d hazard a guess they might even be in Odifreddi but it’s really hard to search for since 0’ turns up lots of false positives as do cupping and capping plus you have to decide to search for capping, cappable or minimal pair. If this has been asked here before I apologize but same problem.
You could look at the Jockusch and Shore papers on pseudo jump operators. They showed that for every $e$ there is an r.e. $A$ such that $A+W^A_e$ is Turing equivalent to $0’$. So $0’$ can have the behaviors that you mentioned relative to r.e. sets.
I’m sure I’m missing something (and maybe I used wrong terminology) but how does that get me a B r.e. in A (but not equivalent to A') such that B+A + 0' is equivalent to A'? Or so that B+A and 0' form a minimal paid over A?
Ohh, is your suggestion I choose e to be the index of a construction such that H_e(X) and H_j(X) always yield a minimal paid over X or non-trivially join to X' and now I get 0' to be one half of that pair. Hm, ok yah those are both uniform constructions. THANKS!!
Just in case anyone else is a mere mortal and takes a moment to understand Slaman's answer here is a more spelled out version of his argument. I'll just do the capping side as it's the same argument for cupping
The construction of an r.e. minimal pair (for the other side an incompatible pair of r.e. degrees whose join is $0'$) is uniform. Therefore, there are hops $H_e$ and $H_j$ such that $H_e(X) \wedge H_j(X) = X$ (non-trivially). Now, by the theorem in the psudeojump paper we can invert a hop, i.e., we can find r.e. A such that $H_e(A)= 0'$ providing the desired $A$.
|
2025-03-21T14:48:31.494087
| 2020-07-11T05:57:57 |
365374
|
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|
Stack Exchange
|
On $(2,3)$-generation of finite simple classical groups
A group $G$ is called $(a,b)$-generated if $G=\langle x,y\rangle$ for some $x,y\in G$ with $|x|=a$ and $|y|=b$.
I know some of the histories on this problem. For example, in this early paper in 1996 Liebeck and Shalev proved that
Theorem. All but finitely many finite simple classical groups other than $\operatorname{PSp}_4(2^f)$ or $\operatorname{PSp}_4(3^f)$ are $(2,3)$-generated.
In this paper in 2017, King proved that
Theorem. Every finite simple group is $(2,r)$-generated for some prime $r\ge 3$.
Is there any other result on $(2,3)$-generation or $(2,r)$-generation of finite simple classical groups? For example, the $(2,3)$-generation for low-dimensional classical groups? Or is there any lower bound (w.r.t the dimension and the order of field) of $(2,3)$-generation?
Liebeck and Shalev proved a later result that all sufficiently large-rank classical groups are $(r,s)$-generated for any two primes, at least one of which is odd. I have proved, but never got round to publishing, that all low-rank exceptional groups (up to $F_4(q)$) are $(2,p)$-generated for every odd prime $p$ dividing their order.
An interesting example is provided by ${\rm PSU}(3,3)$. This is the only non-Abelian finite simple group which can not be generated by three involutions. Consequently, ${\rm PSU}(3,3)$ is not $(2,3)$-generated, for it were generated by $a$ and $b$ with $a^{2} = b^{3} = 1$, then it would still be generated by ${a,a^{b},a^{b^{2}}}.$
This is not a definitive answer (I doubt there is one), but too long for a comment.
Indeed, it is known that among the finite simple groups there are, apart from $\operatorname{PSp}_4(2^k)$, $\operatorname{PSp}_4(3^k)$ and ${}^2\mathsf{B}_2(2^{2k+1})$, only finitely many which are not (2,3)-generated. The case of classical groups was done by Liebeck and Shalev, while Lübeck and Malle dealt with the exceptional groups. The complete list of non-(2,3)-generated finite (quasi-)simple groups is not known at the moment, and it takes some effort to show for a particular group of Lie type that it is not (2,3)-generated, see, for example, More classical groups which are not (2, 3)-generated by M. Vsemirnov. It is shown there that $\operatorname{PSU}_5(4)$ is not (2,3)-generated, as well as $\operatorname{Sp}_6(q)$ for any odd $q=p^k$.
Another possible direction is the study of (2,3,7)-generation, that is, whether there is a generating pair of an involution $x$ and an element $y$ of order 3, such that $xy$ is of order 7, see Hurwitz Groups and Hurwitz Generation by Tamburini and Vsemirnov.
You need to be a little careful when you start mentioning quasi-simple groups. For example, ${\rm SL}(2,q)$ ($q$ odd) is never $(2,3)$-generated, as its unique involution is central.
|
2025-03-21T14:48:31.494275
| 2020-07-11T06:07:13 |
365375
|
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|
Stack Exchange
|
Lower bound for $\vert \det A \vert $ for the adjacency matrix of regular graphs
Assume $G$ is a simple $k$-regular graph of order $n$ with adjacency matrix $A$ which is non-singular.
Does anyone know some lower bounds for $\vert \det (A) \vert$ with respect to $n$, $k$ or both?
Thanks in advance.
I'm not sure what sort of lower bounds you're looking for here. Except for very extreme values of $k$ and $n$ it's easy to construct a graph where two vertices have identical neighborhoods. $A$ would then have equal rows, so zero determinant.
You are right. Sorry I forgot to say $A$ is non-singular. Now, the question is modified.
Experimentally, it seems that $k$ is a lower bound. Possibly the matrix tree theorem could help.
$k$ is a lower bound, since $k$ is an eigenvalue and the product of other eigenvalues is integer (they are roots of the monic polynomial with integer coefficients). I doubt that it may be improved without further assumptions.
|
2025-03-21T14:48:31.494371
| 2020-07-11T07:16:52 |
365378
|
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|
Stack Exchange
|
Abelian ideals of $\mathfrak{n}_+$
Let $g$ be a finite dimensional simple Lie algebra. Then we have the root space decomposition
$$
g=\mathfrak{h}\oplus \mathfrak{n}_+ \oplus \mathfrak{n}_-.
$$
My question: how to classify the abelian ideals of $\mathfrak{n}_+$ that are closed under $\mathfrak h$-action?
This can be translated into a combinatorial problem as follows. Being $\mathfrak{h}$-submodules, these are graded submodules of $\mathfrak{n}+$, hence given by a subset $F$ of the set of positive roots $\Phi+$ (in particular, there are finitely many). The condition of being abelian means that $x+y\notin\Phi_+$ for all $x,y\in F$. The condition of being an ideal of $\mathfrak{n}+$ means that if $x\in\Phi+$ and $y\in F$ and $x+y\in\Phi_+$ then $x+y\in F$.
(I assumed a split semisimple Lie algebra in char 0, e.g., complex simple)
|
2025-03-21T14:48:31.494455
| 2020-07-11T07:39:27 |
365379
|
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|
Stack Exchange
|
Another way for defining $K_1$ group for a C*-algebra
Thank you for answering my question.
I have another question about the $K_1$ group. As you may know, some books define the $K_1$ group like below:
Also, it defines the $K_0$ group for an arbitrary C*-algebra like below:
Where $K_{00}(A)$ is the Grothendieck group for the abelian semigroup $V(A)$.
My question is how can I define $K_1(A)$ like the way of $K_0(A)$ more precisely how can I define $K_1(A)$ like below:
It seems when we define $K_0$ like above, we add some 0s to the matrices and when we calculate the $Ker$, we nicely remove those 0s.
anyone can help?
This is how $K_1$ is already defined. Since $K_1(\mathbb C) = 0$ you have $\ker (K_1(A^+) \to K_1(\mathbb C)) = K_1(A^+) = K_1(A)$.
|
2025-03-21T14:48:31.494536
| 2020-07-11T07:41:37 |
365380
|
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|
Stack Exchange
|
Relation of two general series
Assume that a positive series $\sum_{k} a_k$ converges (i.e. $\sum_{k=1}^{\infty} a_k < \infty$),
is it always true that $\sum_{k = 1}^n \sqrt{a_k} = o(\sqrt{n})$ ?
(By Cauchy-Schwarz, it is easy to get $O(\sqrt{n})$.)
Of course: for fixed $\varepsilon>0$ choose $M$ such that $\sum_{n>M} a_n<\varepsilon$, then for $n>M$ we have
$$
\sum_{k=1}^n \sqrt{a_k}=\sum_{k=1}^M \sqrt{a_k}+\sum_{k=M+1}^n \sqrt{a_k} \leqslant \sum_{k=1}^M \sqrt{a_k}+\sqrt{(n-M)\varepsilon}
$$
by Cauchy–Schwarz.
Dividing by $\sqrt{n}$ and taking limsup we get
$$
\limsup \frac1{\sqrt{n}} \sum_{k=1}^n \sqrt{a_k} \leqslant \sqrt{\varepsilon}.
$$
Since $\varepsilon>0$ was arbitrary,it follows that
$$
\lim \frac1{\sqrt{n}} \sum_{k=1}^n \sqrt{a_k}=0.
$$
|
2025-03-21T14:48:31.494611
| 2020-07-11T09:02:21 |
365386
|
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|
Stack Exchange
|
Infinite uniform dimension $\Rightarrow$ infinitely many idempotents in a localization of a quotient
Let $R$ be a commutative ring with $1$ such that its uniform dimension is infinity, equivalently, $$\sup\{k \mid R \text{ contains a direct sum of $k$ nonzero ideals }\}=\infty.$$ How can we construct an ideal $I$ of $R$ and an element $r\in R\setminus I$ such that the ring $(R/I)_{r+I}$ has infinitely many idempotents, where $(R/I)_{r+I}$ is the localization of the ring $R/I$ at the subset $\{r^n+I\mid n\in \mathbb{N}\}$ of $R/I$?
In general, it is not possible to produce such an ideal $I$ and element $r \in R \setminus I$ from your assumption.
An apparent obstacle is the class of rings of infinite uniform dimension having exactly one prime ideal. If a ring $R$ has a unique prime ideal, then the same is true of every nonzero quotient and localization. In particular, $R$ doesn't have any idempotents, nor does any ring resulting from $R$ by quotients and localizations.
On the other hand, it's easy to produce rings with a unique prime ideal which also have infinite uniform dimension. You can just adjoin an infinite number of mutually orthogonal nilpotents to a field, e.g. $R = k[x_1, x_2, \ldots]/ (x_{i}x_j)_{0 \leq i \leq j}$ contains $\bigoplus_i x_iR$.
|
2025-03-21T14:48:31.494797
| 2020-07-11T09:10:16 |
365387
|
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|
Stack Exchange
|
Time-saving (technology) tricks for writing papers
I have over the years learned some tricks which saves a lot of time,
and I wish I had known them earlier. Some tricks are LaTeX-specific, but other tricks are more general. Let me start with a few examples:
Use LaTeX macros and definitions for easy reuse. This is particularly useful for when making many similar-looking figures. Another example is to make a macro that includes the $q$ when typing q-binomial coefficients. This ensures consistency.
In documents with many Tikz figures, compilation time can become quite brutal. However, spreading out all figures in many documents is also inconvenient. Solution: Use one standalone file, where each figure appears as a separate .pdf page. Then include the .pdf pages as figures in the main document. All figures are in one .tex-file, making it easy to reuse macros. I find this trick extremely useful, as it does not lead to duplicate code spread over several files.
Use bibtex and .bib files. I prefer to use doi2bib to convert doi's to a .bib entry (some light editing might be needed).
For collaboration, use git. Also, Dropbox or similar for backups. Keeping track of versions has saved me several times.
Learn Regular expressions, for search-and-replace in .tex files. This is useful for converting hard-coded syntax into macros.
Get electronic (local) copies of standard references, and make sure to name them in a sane manner. Then it is easy to quickly search for the correct book. These are available when the wifi is down, or while traveling.
Do file reorganization and cleanup regularly. Get final versions of your published papers, and store in a folder, as you'll need them for job applications. Hunting down (your own!) published papers in pay-walled journals can be surprisingly tedious!
Take the time to move code snippets from project-specific notebooks, and turn into software packages for easy reuse. Also, it is sometimes worth to spend time optimizing code - waiting for code to run does not seem like a big deal, but I have noticed that small improvements in my work-flow can have big impact. I am much more likely to try out a conjecture if it is easy to run the code.
Somewhat similar question from 2019: Your professional ${\rm\LaTeX}$ experiences that saves your time in typesetting. (Although that one was specifically about LaTeX.)
doi2bib!!! Amazing!! Thank you!
Regarding tikz compile times, I find \tikzexternalize very useful. Make sure to give figures names so they don’t get recompiled every time a new figure gets inserted earlier in the file.
Rather a general advice than a trick: Follow basic programming guidelines when typesetting in $\LaTeX$ (e.g., proper use of indents, consistent names for variables/macros, and so on). It's one of the basic lessons in programming, but surprisingly many mathematicians ignore it when they typeset in $\LaTeX$.
https://github.com/dspinellis/latex-advice
There seems to be some folklore I'm missing about proper use of bibtex, as I've never found it time-saving. I have used it for maybe 5 papers so far, and in total have probably lost more time than I saved from it. All sorts of pitfalls, tricky errors that only become visible after compilation, and the general headache of having to deal with an extra file and an extra step in the compilation routine.
I don't quite understand your tikz tip; what do you mean by "standalone file"? Are you suggesting to precompile the tikz figures and then import them as pdf?
@darijgrinberg Yes, exactly. Put the tikz code in a separate .tex file, and compile. But instead of having a file for each figure, one can instead put all figs in one file, where figure appear as separate pages (of different sizes). One can then send an argument to includegraphics, to say which page/figure to import.
@PerAlexandersson: Aaah, that's a great trick! (Though I assume it won't work if you want to reference labels from the paper in the pictures -- that is, if this ever works.)
@darijgrinberg, I guess this is an option for during the rapid edit–compile–preview cycle, and then one can inline the code (as surely journals require anyway) for the final product.
@JochenGlueck: Because they didn't bother to learn programming.
@user21820: Hmm, I've always been under the impression that many mathematicians - in particular, but not solely, the younger ones - have a solid knowledge of programming. (But then again, my personal impressions seem to have a tendency to turn out to be wrong when tested against empirical evidence.)
@JochenGlueck: Younger ones maybe. Those that are above 50, not many, in my experience. But I haven't been able to find a survey on how many mathematicians know programming...
@JochenGlueck There is also a difference between programming and programming. I know some older mathematicians which can make any language looks like FORTRAN shudders.
Have collaborators that write your papers. Or a professional typesetter.
Quiver by Varkor, provides a graphical interface to generate commutative diagrams. I find it extremely useful. Check out his blog: https://varkor.github.io/blog/2020/11/25/announcing-quiver.html
another similar tool is https://www.mathcha.io/editor
The most helpful trick for me when it comes to writing stuff in LaTeX is using vim. Since you can configure macros, abbreviations, plugins, and whatnot, it makes writing very very fast:
(There are also other features not shown in the above GIF, such as being able to write vertically/in columns, using multiple cursors to replace text, etc.)
Edit: Here's a small guide on how to get a setup like the one above; sorry for not including this here before.
(If anything is unclear or missing, feel free to either edit this answer or let me know and I'll add/correct it.)
Vim
Vim is quite infamous for being difficult to learn. While I think this reputation is partly unwarranted, in any case the time you invest learning how to use it might very well turn out to be one of the best time investments of your life, as Fosco said in the comments (it certainly was so for me). A nice guide on how to use vim is this interactive site.
LaTeX Plugins for Vim
Two very good ones are vimtex and LaTeX-suite. If I recall correctly, vimtex is more actively developed and has more features.
Abbreviations
These are instructions for vim to replace a string by another word as you type it. For instance, when I type "lrs " in the above GIF, vim replaces it for "locally ringed space ". Some other examples are:
iow -> in other words
fab -> $f\colon A\longrightarrow B$
letring -> Let $R$ be a ring.
ox -> $\mathrsfso{O}_X$
cala -> $\mathcal{A}$
You can define an abbreviation by adding the following line to your .vimrc, the file which keeps your configuration for vim:
autocmd FileType tex iabbrev wrt with respect to
(abbreviations are the heart of what makes things so fast in the GIF above)
Other Plugins
Some other good plugins for speeding things up are the following:
YouCompleteMe;
vim-multiple-cursors
vim-renamer
You may add them to vim by first installing vim-plug and then adding the following lines to your .vimrc:
" Plugins
call plug#begin('~/.vim/plugged')
Plug 'https://github.com/qpkorr/vim-renamer'
Plug 'terryma/vim-multiple-cursors'
function! BuildYCM(info)
" info is a dictionary with 3 fields
" - name: name of the plugin
" - status: 'installed', 'updated', or 'unchanged'
" - force: set on PlugInstall! or PlugUpdate!
if a:info.status == 'installed' || a:info.force
!python3 ./install.py
endif
endfunction
Plug 'Valloric/YouCompleteMe', { 'do': function('BuildYCM') }
call plug#end()
While it is a complete mess, you can find my vim configuration file (.vimrc) here.
(I'm happy to help if you want to set-up vim+LaTeX, but are having trouble to do so :)
Perhaps it's important to clarify that to get the kind of workflow displayed in the gif requires a lot of practice as well as configuration of vim. I've been using emacs for some time now and my workflow is nowhere as efficient as the one shown here. Really cool stuff, though. I'm also really curious of how does your autocomplete function works, @Emily. I'd love to use something like this in emacs.
yeah I'm pretty sure vim does not come configured for that kind of workflow out of the box, this answer could be more helpful if it did an overview of how to get this setup
Perhaps it's important to clarify that to get the kind of workflow displayed in the gif requires a lot of practice as well as configuration of vim. I agree, but on the other hand those are the best invested 45 minutes of your life.
@user347489 Hi! :) You're of course right, I should have explained how the above configuration works; I'll edit the answer to add that. Regarding the autocompleter, I'm using YouCompleteMe. I really like it, as it gives suggestions based on the content of the files you open. As a result, it autocomplete words like "extranatural", "Noetherian", "\LongLaTeXMacroThatICanOnlyUseBecauseOfTheAutocompleterAndThatWouldGetMeShotByWhoeverISendTheSourceTeXTo"$^\dagger$, etc.
($\dagger$ Less joke-y, I do use macros like \newcommand{\CategoryOfAlgebrasOverAnOperad}[2]{\Alg_{#1}(#2)}. This ends up being very useful if you write 20 pages of LaTeX code and then decides to change notation.)
About Emacs: It seems until December of last year there was an Emacs implementation of YouCompleteMe, but now it isn't maintained anymore :/
@lurscher I'll add that! :)
@user347489 Oh I realised something as I was editing my answer: I actually didn't use the autocompleter at all on the GIF above! These are all vim abbreviations; e.g. fepix --> for each $p\in X$.
Note that other editors have abbreviation expansion / autocompletion, not only Vim.
@Emily thanks for your reply! I looked a bit into it and it seems like abbrev-mode in emacs offers this kind of functionality. I'm looking forward to try it myself!
What's the advantage of using vim for fab -> $f \colon A \longrightarrow B$ over using LaTeX for \newcommand\fab{$f \colon A \longrightarrow B$}? Not asked as a challenge; I use and love vim, and am genuinely curious why I'd prefer to do the trickery in my editor. (I'd also argue for \to in place of \longrightarrow, with \let\to=\longrightarrow if you really feel strongly about it, but that's just me. :-) )
@user347489 Happy to hear you found it helpful! :)
@LSpice I think there aren't many big advantages, but here are some minor ones, I think: 1) Readability: using too many macros might make it difficult for other people to understand your source (see also this comment by Nate Eldredge); 2) It keeps your preamble tidier, and you don't have to define the same macro again and again in every .tex file you [1/3]
create (but then you can also create a preamble.tex and include it in all of your .tex files, so things are fine either way). 3) If you want to type e.g. f\colon A\longrightarrow C you can do that by typing fab<BS><BS>C$ rather than creating a macro \fac, or even something more elaborate like \function{f}{A}{B}. 4) Assuming you decide to set up abbreviations like lrs --> locally ringed space in your .vimrc, why not put them all there, rather than also in your preamble?
[2/3]
Overall, aside from the cautionary tale by Nate Eldredge, these are all minor points, so maybe this is just personal preference. Also, the \longrightarrows are really just a quirk I have; I find the \to arrows too small :P; (By the way, in the end these are actually as fast to type as \to because of a math-mode macro I made which changes jr to \longrightarrow and similarly for jc and \colon (so typing $fjc Ajr B$ gives f\colon A\longrightarrow B))
[3/3]
Another advantage of abbreviations vs macros: have you ever worked with a co-author with a different, equally invasive set of macros?
This is $f\colon A\longrightarrow B$ulous!
The best advice for vi users is to simply stop using vi. Your a butterfly to redirect cosmic rays to write data on your hard drive, or better yet emacs has a mode for that.
When writing notes, or collecting references or writing down small lemmas,
it might be a good idea to share them on your personal webpage.
The fact that someone else might read your stuff, will improve your effort in writing clearly. Moreover, putting things on a web page makes it available from any computer - this can be useful when not bringing your personal laptop.
To add to this: The notes you personally expect to be the least useful to others -- the ones you upload just for the sake of completeness -- may turn out to get the most interesting feedback. (Happened to me.)
If your university has a subscription, Mathscinet provides much better bib files than any other source (including doi2bib and journal sites), handling the details with painstaking precision: diacritics, escaping capitals in titles, consistent journal names...
I can recall only one case in my career when I needed "light editing" with it (and they fixed it when I sent them an e-mail).
The bib files produced by mathscinet aren't that good. For example they always duplicate the DOI file, they completely ignore entry types such as @inproceedings, the number and volume fields are not always used correctly, and so on.
@NajibIdrissi What do you mean with "they always duplicate the DOI file"? Do you have an example of number/volume fields used wrongly?
Sorry, I meant the DOI field (typo): it appears in the DOI field, but also in the URL field. Volumes are generally misused for books in a series: for e.g. the 123rd entry in the "Mathematical Surveys and Monographs" (random example), you should find number = {123} in the entry, but they put volume = {123}. The volume field would actually indicate that you would be citing the 123rd volume of one single enormous book, not the 123rd entry in the series. This can lead to weird formatting.
@NajibIdrissi I see nothing wrong in having the DOI field twice, since it doubles as an URL. In my view, the problem is with bib-styles trying to cram them both inside the reference line. But for a database, I'm happy if it contains as many fields as possible.
So you are OK with a entry displayed as "Doe, J. "The Theory of Stuff". J. Math 123.45. DOI: 10.1234/5678. URL: https://dx.doi.org/10.1234/5678"? The redundancy doesn't bother you? And if it's a matter of databases, you know that having redundant information in a database is a source of errors, there should be a single source of truth.
No, I am not happy with an entry displayed like that. But in my view it is the job of the bibstyle to display only one of them, not of the database. I agree with you on the point of the "single source of truth", though.
@NajibIdrissi I think there are a few journals and stuff that seem to go out of their way to defy bibtex entries and easy, consistent formatting. I have, on occasion, been forced to change a bibtex entry by moving a volume entry to a number entry or some such thing, simply to get the correct formatting on one citation without wrecking others. It was easier to do that than to figure out what weird set of conditional macros to write and/or options to enable to get things to work out.
Note that you don't actually need a MathSciNet subscription to use their BibTeX entries! They are publicly available via the MR lookup tool (which I usually find by typing 'MR lookup' into my favourite search engine). (I still wonder if there's a Mrs Lookup, but I never dared ask...)
Always do the easy bits first, and check them once at the end.
It is tempting to spend a lot of time crafting ones words or getting something just so, and then going back and revising it. More time efficient is to meta-write, e.g. "*** Find a better way to organize these three definitions" , which is not the literal text but a command to yourself to process that section later. This is a way of breaking up the hard parts of writing into manageable pieces and identify those few parts of the paper that really need the time spent on them. Of course, one goal is to replace every metatext piece with the desired text. And then, do only two or three final passes for catching typos that you missed correcting while writing.
Gerhard "Needs To Follow His Advice" Paseman, 2020.07.11.
For the meta texts in LaTeX one could use tools like https://github.com/sunggnus/sfsimpletodo to make them more visible in your draft
The todonotes package is perfect for this as todo's are highlighted. Moreover, you can easily display a list of open todos, which gives a good overview about the open points.
I find it very hard to spellcheck LaTeX documents, because so many of the words in the document are technical commands and not actually text. I've had some success with a (Mac OS only) application called Excalibur designed to spell-check LaTex documents. I would be interested if others had suggestions for this as well, though.
On Linux you can use aspell, which has a build-in option to ignore TeX commands. Usage could be aspell -t -d en_GB_ize main.tex. The -t is the flag to specify TeX mode, -d en_GB_ize would for example specify British spelling with ize (Oxford spelling). Source is man aspell
@Daniel: aspell somehow doesn't seem to understand that inline equations are equations. Is there any trick to this?
I like TeXtidote a lot. It can check spelling, grammar and style. It is based on LanguageTool.
I use aspell since 5 years. One of the most precious findings ever. Just one thing: I prefer -ise over -ize if I write in British English (and I usually try). Am I wrong?
@Fosco This might be hard to get used to, but there is no central authority for what is "correct English". Like Daniel, I am a British -ize speller, and this is the traditional spelling for OUP publications, and is based on etymology (words with $\zeta$ in the original Greek). British newspapers use -ise, and many British people semi-learnedly claim all -ize spellings are American (wrongly, historically speaking). However, "analyze" is an example of a spelling that is used in American English, for good phonetic reasons, but not in Oxford spelling, because the word "analyse" comes from French.
However, when using -ize, non-native speakers will have to memorize that there is no such word as "advertize", "advize" and "surprize" (though these spellings used to exist), even though these words are pronounced that way.
a lazy trick is to do the spell check in the compiled pdf document, rather than the latex. I do this in skim on a mac. It is a bit annoying (it thinks some formulas are spelling mistakes, thinks newlines are spelling mistakes, ...) but as a quick and dirty solution it is reasonable.
SyncTeX. SyncTeX is something that can be used to jump quickly from the LaTeX source to the corresponding location in the PDF file and vice versa. So for example if you are editing a paragraph in LaTeX and you want to see how it actually look, you press a shortcut or something and your PDF reader takes you to the paragraph, or you are rereading your paper and you want to edit a paragraph, you double click on it or something and your editor takes you directly to the emplacement of the paragraph in the source.
I have found this extremely useful but it requires some amount of configuration: both your editor and your PDF reader need to support it. One basic thing you need to do is pass the --synctex=1 to latex so that it produces the appropriate .synctex.gz file. After that, due to the large variation of possible editors/PDF viewers, I can't give you specifics. If you want to look up online how to do it for your editor/viewer, note that it is sometimes called "forward/inverse search".
You can take a look at how I did it for Emacs (look for the "source-correlate" options).
latexmk. If you are not using an editor/IDE that manages compilation for you, you should definitely use latexmk. This is a script that automatically calls (pdf/xe/lua)latex, bibtex, biber... as many times as needed. It's as simple as launching latexmk -pdf article and it will do its magic. It reads the output of latex and checks if external files have been modified.
The script also has a "watch" mode (-pvc option), where you tell it to compile the article, and recompile it whenever there's a change. Then you can just forget about compilation, modify your article, and check in the PDF reader from time to time. This may not be very convenient if your file takes a long time to compile, though, and there may be some annoying issues if you modify your article when it's in the middle of being compiled.
Many TikZ figures. TikZ has an "externalization" feature. It will automatically copy the figure to an external file and compile it. If the figures doesn't change between runs, it will not recompile the file, saving a lot of time. This is basically an automated version of what is suggested in the OP. You don't even need to change your document: all you need to do is write
\usetikzlibrary{external}
\tikzexternalize
in the preamble of your file, and run latex with the -shell-escape option. For more information, read Chapter 55 of the TikZ manual.
Note: There is currently an issue with tikz-cd, if you use that. You'll have to tweak things a little.
Documentation. LaTeX packages are widely documented. If you have installed e.g. TeX Live in a normal way, you can simply run texdoc <package> in a terminal to get the documentation of the package. No need to search for it on CTAN or anything. So if you'd like to read the TikZ manual I mentioned before, simply run texdoc tikz and go wild!
Large documents. In the same vein (this is probably well known), when you have a very large document, you may want to split it in several files. When you do, use \include rather than \input for the content files. Then, in the preamble, use \includeonly. Only the files specified there will be recompiled and included in the PDF, but \include is smart enough to keep the auxiliary files for the other sub-files so that references and page numbers will still work correctly. More info: https://en.wikibooks.org/wiki/TeX/includeonly
Citing arXiv preprints. This is shameless self-promotion: I wrote a web app to automatically extract .bib information from an arXiv search. I encourage you to read the help before using it. As I explain there, I found issues with all the tools available online, especially when it comes with using BibLaTeX. If you are still using legacy bibliography support, you may run into issues.
Note that I literally started on it three weeks ago so it's not exactly polished (the "DIY" feature is incomplete, too). It's available there: https://a2b.idrissi.eu/
Writing responses to referees & external references. From time to time, you need to write a response to a referee and talk about what you changed in the new version. So for example you need to cite Lemma 3.14 to explain that its hypothesis have to be tweaked or whatever. But since you may be changing other things, the lemma's number may change, and since you cannot use \ref to automatically get the number, you have to track changes down manually. This can get painful quickly.
Fear not! Using a package that provides external references, you can reuse the references from your article in your letter to the editor, and things will automatically work. No need to change your article; just include
\usepackage{nameref,zref-xr}
\zxrsetup{toltxlabel}
\zexternaldocument*{article}[article.pdf]
in your letter, where you replace article by your article's file name. Then you can use \ref{key-lemma} in your letter, compile the article first, then compile the letter, and voilà! The lemma number is correct. If you use hyperref, the reference will even be clickable if <article pdf> is in the same folder as the letter's PDF.
Just to be clear, neither nameref nor zref-xr is necessary for referencing external documents; if you just want that functionality, xr (or probably xr-hyper) will do.
@LSpice Well, neither xr nor xr-hyper are necessary for that functionality either, nameref and zref-xr will do. From what I remember I settled on zref because the links to the external document actually worked.
@LSpice After some tests, I can confirm that xr(-hyper) do not produce hyperlinks to the other document even with some basic configuration (you can try). It's possible that they can be configured to do so, but the lack of documentation for xr-hyper (apart from relatively cryptic comments at the top of the source file) are probably what made me give up when I settled on zref the first time. It's probably good to bear in mind that xr-hyper was written in 1997 and barely updated since, while zref is actively maintained.
Sure, I didn't mean anything against zref. Mainly I meant to mention that (as far as I could see) nameref was just a pleasant addition, not something that directly pertained to your tip as written.
@LSpice No, nameref is necessary if you want load both zref-xr and hyperref, otherwise you get a compilation error. I don't know why. (PS: there was a mistake in the "gist" I linked above, the a.tex file should have called hyperref too, of course. Then there is a link in b.dvi but it doesn't work.)
I have a rather different approach to most posting here. I write my papers in LyX, which uses LaTeX on the backend, but presents a friendlier interface overtop. I feel like I can use less of my brain for writing proper LaTeX, and concentrate more on the mathematics. (Of course, one still needs/wants to remember the basic codes like \alpha, etc, but LyX will automatically handle a fair number of fiddly details.) And that you can immediately see what you are typing (where you are typing it) helps avoid many typos.
There was some discussion about spellchecking. I'll particularly comment that LyX includes a spellchecker, and it straightforwardly underlines unknown words in red, just like any modern word-processor.
A drawback of LyX is that negotiation (more than usual) is needed when writing with coauthors.
On managing bibtex, I like the MathSciNet bibtex, which is generally of a reliable quality. No one has yet noticed, but you don't need a MathSciNet subscription to get at the bibtex: you can use the MRLookup interface, which is available from anywhere. It will only return the top three hits of a search, but the interface is simpler than full MathSciNet, and I prefer to use it even from a university IP when I'm looking for something specific.
I use with BibDesk (on macOS) to manage my master bib file.
Even more heretically, use any math-enabled word processor/markdown editor/writing environment that you like and that exports to LaTeX (which is now pretty much all of them). Assuming, of course, that you don't have collaborators who work with your LaTeX source.
LyX generates fairly reasonable LaTeX, which I like, and is a relatively thin layer over LaTeX, which I also like. I've had very little trouble submitting papers to journals. Have you tried it with a LibreOffice LaTeX export? Collaborators are indeed the rub.
I like LyX and I like Libre, but I personally need an interface with a more modular structure so I can manage my disorganized work style. Right now I'm using Scrivener and exporting to LaTeX via Markdown.
LaTeX:
When editing LaTeX by hand, I've gotten used to using \begin{align*} and \end{align*} (as opposed to \[ or $$ or \begin{equation*}) for all equations, even those that only need 1 line. This allows me to easily add another line without having to change the environment. (I wouldn't be surprised if this makes the compilation a bit slower, though.)
Newcommands for mathematical symbols (like \lcm or \NN) are well-known, but there's a lot more that can be useful. For example, I use \newcommand{\nnn}{\nonumber\\} (for align environments that need only one line labelled) and \newcommand{\underbrack}[2]{\underbrace{#1}_{\substack{#2}}} (for algebraic manipulations with justifications provided).
If you print drafts frequently, put \today (this yields the date of compilation) and \thepage (this yields the page number) on the header (or footer) of each page. This way, even if your papers get jumbled, you will always be able to get them back in order.
You might find it useful to compile your tex files in a temporary folder. That is, instead of running "pdflatex paper.tex", you copy paper.tex, then run pdflatex on that copy (for the necessary number of runs), and finally copy the resulting PDF back into your home folder. (This is best done by script of course.) This way, while pdflatex is running, your existing PDF remains a readable file as opposed to temporarily turning into a construction site (which confuses a bunch of PDF readers).
Literature search:
Search for sources using both Google and Google Scholar. The former searches more widely, the latter specifically among what Google believes to be academic literature. Some preprints are easier to find using the former, but published sources generally are easier to find using the latter (particularly if they are given in the laconic old-fashioned reference format: e.g., Google Scholar easily finds "H. O. FOULKES, Quart. Oxf, (2), 2, 1951, p. 67-73", while standard Google gives the wrong paper).
Be in mind that neither Google nor Google Scholar finds papers on Sci-Hub, so you'll often use the former to find out what papers you need, and the latter to get the actual papers. Sci-Hub works best if you provide it with the DOI, not the URL; if you give it the URL, make sure it's the most standard URL you can find for the paper.
If you're looking for a paper in a conference proceedings volume, Library Genesis might have the volume (search for the authors and the name of the volume, not for the paper itself).
Reverse citation lookup (i.e., given a paper A, find all works that cite A) is easy these days -- Google Scholar does it (click on "Cited by [number]" under an article you find using Google Scholar), and so do Mathscinet (click on "Citations" to the right of the review) and arXiv (see the "Bibliographic data" under the abstract). You can use reverse citation lookup to (1) discover whether the paper you are reading has gotten updates or corrections, (2) find out whether the question you are answering has already been answered (just try to think what works such an answer would have cited, and use reverse citation lookup on them), and (3) see what people have done with your work.
Version control:
git has been mentioned as a great way to keep a sane workflow when collaborating. I found git useful even for single-authored papers: It gives me a way to work in parallel on different sets of changes. For example, assume I have a preprint that I want to add a new section to. The section is complicated and I need a week to write it up. During that week, someone else informs me of a typo in the old sections. Using git, I can easily fix that typo and update my preprint without having to include the unfinished new section. This is also useful when you are making a change that you yourself aren't sure is a step in the right direction (so you want to leave yourself the option of tucking it away and returning to the previous version). And of course, git makes it easy to back your stuff up!
I will focus here on only 1 aspect—library organization:
Use bibtex and .bib files. True, but one needs an efficient way to
produce them—I recommend zotero (or any other similar program, e.g. mendeley) to organize the library.
This has an added benefit of efficient full text search through all
articles that you are using.
Another benefit is a possibility to extract bib-data from PDF files
In order to synchronize PDF files in the zotero library among
different computers, use Dropbox.
Zotero is compatible with overleaf (in the case you are using it)
Furthermore, I have a perl script (bibtexformat) to clean up .bib files and make journal name abbreviations.
As a reference manager, I can strongly recommend JabRef. It uses bib files natively, so you don't have to export manually (or clean the bibtex code afterwards). Moreover, many convenient features are built-in: doi2bib (or generate from arXiv ID, ISBN, ...), MathSciNet search and completion of already existing papers, automatic full text download and rename of linked files, journal abbreviations, import from browser with one click etc..
In addition to regular spell checking, I have found proselint to be extremely helpful for writing texts as a non-native English speaker.
It highlights phrases, words and punctuation according to an extensive list of checks (which can be individually enabled/disabled for each paper or project).
Using one of its example texts:
he is very smart
approximately about 5 atm machines
atm machine
we associate together
it's blatantly obvious that this obviously works.
a very unique idea
a more perfect union
the surrounding circumstances
he filled a much-needed gap
To coin a phrase from the movie,
proselint produces the following output:
<stdin>:1:7: weasel_words.very Substitute 'damn' every time you're inclined to write 'very'; your editor will delete it and the writing will be just as it should be. Found once elsewhere.
<stdin>:2:1: redundancy.garner Redundancy. Use 'approximately' instead of 'approximately about'.
<stdin>:3:1: garner.redundancy.ras RAS syndrome. Use 'ATM' instead of 'atm machine'.
<stdin>:4:4: redundancy.garner Redundancy. Use 'associate' instead of 'associate together'.
<stdin>:5:6: redundancy.garner Redundancy. Use 'obvious' instead of 'blatantly obvious'.
<stdin>:6:3: uncomparables.misc Comparison of an uncomparable: 'very unique ' is not comparable.
<stdin>:8:5: redundancy.garner Redundancy. Use 'circumstances' instead of 'surrounding circumstances'.
<stdin>:9:14: misc.illogic 'much-needed gap' is illogical.
<stdin>:10:1: cliches.write_good 'To coin a phrase' is a cliché.
<stdin>:10:2: misc.illogic.coin You can't coin an existing phrase. Did you mean 'borrow'?
This is a very neat idea and it does seem like a good resource, but proselint does not seem to understand mathematics or LaTeX very well, e.g. it just advised me "'contractable' is the preferred spelling."
That seems to be a problem indeed. In one of proselints issues, combining it with textlint and a LaTeX plugin is recommended, though I haven't used it that way.
In terms of backing up documents, I used to send literally everything I write to a friend of mine who would read it and make comments, and in fact I still send almost everything to them even now.
This actually turned out to be useful as some documents which I wanted to read were lost several computers and several years later, but copies still existed in my sent emails. Since I can access my email account on any computer anywhere, this means I always have access to anything I have written in the past: I just click on sent emails and search for it.
I have to admit I'm slightly surprised that other people don't do this and seem to have trouble getting hold of their own articles.
For the goal of not losing them, consider hosting your papers on Dropbox, version control (such as GitHub, private repos are gratis now) and/or on Overleaf.com (nice in-browser editor with real-time collaboration; can also edit offline with your favourite local editor but may need paid account)
These may be controversial, but in my personal experience...
Do not try to use reference managers for managing references. They will get things wrong often enough that you will spend more time tracking down and fixing errors than you will save. They can be great for managing your personal library, though.
Split your longer latex files into several sub-files and use \include (for instance: each section of a paper goes into a separate file). This will make collaboriative work much easier since individual sections can now be edited independently. As a bonus, it will take much less scrolling to find the line that you want to edit.
Interestingly, (2) disagrees with the advice @qwr posted yesterday: "Avoid splitting short documents (such as a journal or conference paper) into multiple files (e.g. introduction.tex, methods.tex, results.tex, conclusions.tex). Such splitting makes it difficult to work perform a number of tasks with your editor …". (Which, of course, doesn't make it wrong—or maybe you disagree about when a paper becomes long!)
I don't agree with your point 1. It really depends on your reference manager. If you use one which deals with bib files directly (jabref, ebib...) then there is nothing wrong for them to get, because what they take as input is exactly what bibtex et al will take as input.
I suppose that everyone does this: commonly used expressions are typed as macros. For example, \st (such that---remember to include a space afterwords, that is, in plain TeX (I refuse to learn/use LaTeX), \def\st{such that }), \wrt and \Wrt (the latter at the beginning of a sentence), \tfae, and some other commonly used expressions that may be specific to your field, e.g., \Hs for Hilbert space. This saves a surprising amount of time (I think). But overall, I think the best thing is to learn to touch type fluently, especially using the top row of the keyboard.
you will probably need or want to remove these macros when you submit your LaTeX file to a journal, it interferes with their work flow (I try to avoid these from the beginning, because it complicates collaboration with other authors on the same file)
Editors with autocomplete can be used to reduce macros. I don't like short macro names, as there is no standard. Moreover, long macro names can be autocompleted, so the length is not really an issue
@CarlBeenaker I've had no problems with submissions to ArXiv; and I don't use LaTeX, so don't submit LaTeX files. I do submit to journals as either plain TeX or AmS-TeX, and these \def s have not been a problem.
I am not the one who downvoted, but I cannot refrain from noting that this suggestion is in contradiction to common best practices for coding (see for instance here). One might argue that things are slightly different for markup languages than for programming languages, but the issues caused by things like "use very brief macros to reduce typing time" (namely reduced readiblity and maintainability of the code) are at least similar.
as an alternative, with the same time-saving benefit but without the portability drawback, a tool such as TextExpander can be trained to automatically expand macros; this goes beyond local abbreviations, you can also use it to expand \eq into $\text{\begin{equation} [cursor] \end{equation}}$ and similar nonlocal abbreviations.
I am of course aware of the difficulties of macros, but one argument in favour of them is that it allows you to easily change your notation globally. For instance, if I decide that I want to write categories like $\mathbf{Ring}$ or $\mathbf{Set}$ instead of $\underline{\operatorname{Ring}}$ or $\underline{\operatorname{Set}}$, then this is an easy fix. (In particular, I use a meta-macro \Cat and then define \Ring as \Cat{Ring}, which I only have to change once if I want to change my notation convention.) Typing these as \Ring and \Set should not diminish readability.
I guess what I'm saying is that macros can attain semantics separate from the underlying LaTeX operation you're performing: for these types of commands, I want to use a consistent notation. Maybe I also typeset the reals ans rationals like $\mathbf R$ and $\mathbf Q$, but the underlying \mathbf commands then serve different purposes in my document than $\mathbf{Ring}$ and $\mathbf{Set}$. If you insist on avoiding all macros, this makes global replace substantially harder when you need to change your notation for whatever reason.
|
2025-03-21T14:48:31.497736
| 2020-07-11T10:05:46 |
365392
|
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|
Stack Exchange
|
Prime gap transform
Let $n$ be a large enough composite integer, and consider an arithmetic function $f$ that maps $n$ to the sum of prime gaps making a closed interval $J_{f}(n)$ containing $n$ whose extremities are prime. In what follows I'll assume the truth of Goldbach's conjecture and consider the map $f:n\mapsto 2r_{0}(n)$ where $r_{0}(n)=\inf\{r>0,(n-r,n+r)\in\mathbb{P}^{2}\}$ but we can set a general theoretical framework.
Let $G_{f}(n)$ be the multiset of prime gaps $g_{i}(n)$ the sum of which is $f(n)$, $w_{g_{i}}$ their respective "weight", that is the number of times $g_{i}$ appears in $G_{f}(n)$ where we omit $n$ to avoid too heavy notations.
Now consider the "prime gap transform of $f$ evaluated at $n$" defined by $\displaystyle{T_{f}(n):=\prod_{g\in G_{f}(n)}p_{g/2}^{w_{g}}}$ that associates to $n$ and $f$ an integer characterizing the "type" of sequence of gaps making $J_{f}(n)$ sorted in increasing order.
Taking $f:n\mapsto 2r_{0}(n)$, can we take advantage of some fixed point theorem to show that the equality$T_{f}(n)=f(n)$ occurs infinitely often? If yes, could we deduce from that the truth of the twin prime conjecture?
Edit: as $f(n)$ is a sum of prime gaps, it is necessarily even. Then it would suffice to show that $T_{f}(n)$ is itself even infinitely often, that is, for infinitely many $n$, to entail the truth of the twin prime conjecture.
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2025-03-21T14:48:31.497879
| 2020-07-11T11:22:37 |
365394
|
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|
Stack Exchange
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Are category of perfect groups and acyclic space equivalent?
Here is my approach to relate two subcategories of $\textbf{hTop}$ and $\textbf{Gp}$ respectively:
Let $\mathcal{C}$ denote the full subcategory of $\textbf{hToP}$ consisting of acyclic( i.e. reduced homology in all degree is zero ) CW complex and $\mathcal{D}$ denote the category of perfect groups (i.e. commutator is itself). There is functor $\mathcal{F} = \pi_{1} : \mathcal{C} \rightarrow \mathcal{D}$ takes space $X$ to its fundamental group $\pi_{1}(X)$. We have also another functor $\mathcal{G} : \mathcal{D} \rightarrow
\mathcal{C}$ given as if $P$ is any perfect group then define $\mathcal{G}$(P) to be the homotopy fiber of the plus construction map $ BP \rightarrow BP^{+}$. The above two functors seems too natural to me. So mine natural questions are the following:
Are $\mathcal{C}$ and $\mathcal{D}$ equivalent categories?
Is $\mathcal{F}$ related to $\mathcal{G}$ (I mean whether they are in adjunction, equivalence pair etc.)?
I computed the composition $\mathcal{FoG}$: Let $P$ be a perfect group and $F(f)$ denote the homotopy fiber of $ BP \rightarrow BP^{+}$. Then $\mathcal{FoG}(P) = \pi_{1}(F(f))$ and the counit map $\mathcal{FoG}(P) = \pi_{1}(F(f)) \rightarrow P $ is the universal central extension map which in particular says that if we again repeat the same process turns out to be identity. So this is in a way saying that $\mathcal{FoG}$ may not be identity but its square is identity. I don't know what it means. Any help would be appreciated.
I think you can deduce that the answer to 1 is negative by using [Dror, Emmanuel. “Acyclic Spaces.” Topology 11, no. 4 (December 1, 1972): 339–48. https://doi.org/10.1016/0040-9383(72)90030-4.] actually, constructing a wide range of explicit examples of why this fails. IMHO your heuristics would be very similar to thinking that the homotopy category of pointed spaces (connected CW-complexes) were equivalent to the category of groups via the fundamental group.
A suitable localization of the category of groups gives an equivalence between the category of perfect groups and the homotopy category of simply connected spaces. The equivalence is given by the quillen plus construction
For more details, take a look to Kan-Thurston article and Baumslag article.
|
2025-03-21T14:48:31.498081
| 2020-07-11T12:22:07 |
365398
|
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|
Stack Exchange
|
Examples of measures that satisfy FKG, but not the FKG lattice condition
Let a percolation measure be a measure on $\{0,1\}^n$. We have a natural partial order on $\{0,1\}^n$ given by comparing all coordinates. An event $A$ is called increasing if for all $ \omega \in A $ we have for every $\eta \geq \omega$ that $\eta \in A$. Now a percolation measure $\mu$ satisfies the FKG inequality if for all increasing events $A, B$ we have that
$$\mu(A \cap B) \geq \mu(A) \mu(B).$$
A criterion for FKG is the FKG-lattice condition. To state it for $v \in \{0,1\}^n$ and $e, f \in \{1, \dots , n \}$ with $e \neq f$ let the configurations $v_{ef}, v^{ef}, v_{e}^f,v^{e}_f$ be the configurations that coincide with $v$ for all but the two entries $e,f$ and there $v_{ef}(e) = 0 = v_{ef}(f) $ , $v_{e}^f(e) = 0, v_{e}^f(f) = 1$ etc.
If we always have that
$$
\mu(v_{ef}) \mu(v^{ef}) \geq \mu(v_{e}^f) \mu(v^{e}_f)
$$
and $\mu(v) > 0 $ for all $v$ (i.e. strictly positive). Then this implies FKG (see Ex. 11.2 here https://www.ihes.fr/~duminil/publi/2017PIMS.pdf).
Now, what are (interesting) examples of percolation measures that do not satisfy the FKG lattice condition, but that do satisfy FKG?
One example is the Wolff measure, as described in this paper; see the remark following the statement of Theorem 3.3.
|
2025-03-21T14:48:31.498210
| 2020-07-11T12:39:48 |
365399
|
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|
Stack Exchange
|
On $B^1$ and $B^2$ almost-periodic functions
The Besicovitch class of $B^p$ almost-periodic functions is defined as the closure of the set of trigonometric polynomials (of the form $t \mapsto \sum_{n=1}^N a_n e^{i \lambda_n t}$ with $\lambda_1, \dots, \lambda_n \in \mathbb R$) under the semi-norm $$||f||_{B^p} := \left(\limsup_{X \to +\infty} \frac{1}{X} \int_0^X |f(t)|^p \,\mathrm{d}t\right)^{1/p}.$$
Let $F \in B^1$ be such that there exists $(a_n)_{n \in \mathbb N} \in \ell^2(\mathbb N)$ and $(\lambda_n)_{n \in \mathbb N} \in \mathbb R^{\mathbb N}$ be such that $$\lim_{N \to +\infty} ||F-P_N||_{B^1} = 0,$$ where $$P_N := t \mapsto \sum_{n=1}^N a_n e^{i \lambda_n t}.$$ Does $F \in B^2$ follow from the fact that $(a_n)_{n \in \mathbb N} \in \ell^2(\mathbb N)$ ?
Litterature on the subject is a bit hard to find since there are so many different classes of almost-periodic functions, of which $B^1$ is one of the largest. Also $|| \cdot ||_{B^p}$ is only a semi-norm so this might complicate things.
The answer depends on what exactly you mean by the question. The subtle thing about $B^p$ is that it represents not functions by classes, and the classes depend on $p$.
This is a problem since if you replace $F$ by another representative $F'$ in its class, you could have $F \in B^2$ but $F' \notin B^2$.
The core of the issue, as you guessed, comes from the fact that the Besicovitch seminorm is not a norm.
Here is what you can prove:
Claim Under the given conditions, there exists a representative $G$ for the class of $F$ (i.e. $\| F-G \|_{B^1}=0$) such that $G \in B^2$.
Proof Since $(a_n) \in L^2$ you get that $P_n$ is a Cauchy sequence in $(B^2, \| \, \|_{B^2})$. Since this space is complete, there exists an element $G \in B^2$ such that
$$
\lim_N \|P_N -G \|_{B^2}=0
$$
Now, the Cauchy--Schwarz inequality gives
$$
\left(\frac{1}{X} \int_0^X |f(t)| \,\mathrm{d}t\right) \leq \left(\frac{1}{X} \int_0^X |f(t)|^2 \,\mathrm{d}t\right)^{1/2} \left(\frac{1}{X} \int_0^X 1^2\,\mathrm{d}t\right)^{1/2}.
$$
which gives
$$
\| \, \|_{B^1} \leq \| \, \|_{B^2} \,.
$$
Side note here: This implies that if $\| G-G'\|_{B^2}=0$ then $\|G-G'\|_{B^1}=0$ but the converse is not true, which is the core of the issue.
From here we get that
$$
\lim_N \|P_N -G \|_{B^1}=0
$$
Since you are given that
$$
\lim_N \|P_N -F \|_{B^1}=0
$$
you get
$$\| F-G \|_{B_1}=0$$
as claimed
\qed
Example where $F \notin B^2$
It is easy to come up with an example of a function $F$ such that $\|F\|_{B^1}=0$ but $\|F\|_{B^2}=\infty$ (for example $F=\sum_{n \in \mathbb N} b_n 1_{[2^n, 2^n+1]}$ will work for the right $b_n$).
This function trivially satisfies the conditions of your statement with $a_n=0 \forall n$, but $\| F \|_{B^2}=\infty$ implies that $F \notin B^2$.
As in the above claim, $F$ does have some $G \in B^2$ in its class, namely $G=0$.
Thank you for the explanation of the underlying subtleties and for the counterexample ! I'm happy they exist, since I proved a result for $B^1$-a.p. functions with $\ell^2$ coefficients which was already known for $B^2$-a.p. functions, so my result is hopefully new.
|
2025-03-21T14:48:31.498437
| 2020-07-13T10:13:41 |
365527
|
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|
Stack Exchange
|
A problem in additive combinatorics
$\color{red}{\mathrm{Problem:}}$
$n\geq3$ is a given positive integer, and $a_1 ,a_2, a_3, \ldots ,a_n$ are all given integers that aren't multiples of $n$ and $a_1 + \cdots + a_n$ is also not a multiple of $n$.
Prove there are at least $n$ different $(e_1 ,e_2, \ldots ,e_n ) \in \{0,1\}^n $ such that $n$ divides $e_1 a_1 +\cdots +e_n a_n$
$\color{red}{\mathrm{My\, Approach:}}$
We can solve this by induction (not on $n$, as we can see in Thomas Bloom's answer). But I approached in a different way using trigonometric sums. Can we proceed in this way successfully?
$\color{blue}{\text{Reducing modulo $n$ we can assume that $1\leq a_j\leq n-1$}.}$
Throughout this partial approach, $i$ denotes the imaginary unit, i.e. $\color{blue}{i^2=-1}$.
Let $z=e^{\frac{2\pi i}{n}}$. Then $\frac{1}{n}\sum_{k=0}^{n-1}z^{mk} =1$ if $n\mid m$ and equals $0$ if $n\nmid m$.
Therefore, if $N$ denotes the number of combinations $e_1a_1+e_2a_2+\cdots+e_na_n$ with $(e_1,e_2,\ldots, e_n)\in\{0,1\}^n$ such that $n\mid(e_1a_1+e_2a_2+\cdots+e_na_n)$, then $N$ is equal to the following sum,
$$\sum_{(e_1,e_2,\ldots, e_n)\in\{0,1\}^n}\left(\frac{1}{n}\sum_{j=0}^{n-1}z^{j(e_1a_1+e_2a_2+\cdots+e_na_n)}\right)$$
By swapping the order of summation we get,
$$N=\frac{1}{n}\sum_{j=0}^{n-1}\prod_{k=1}^{n}(1+z^{ja_k})$$
Clearly, the problem is equivalent to the following inequality:
$$\left|\sum_{j=0}^{n-1}\prod_{k=1}^{n}(1+z^{ja_k})\right|\geq n^2\tag{1}$$
This is actually IMO shortlist $1991$ problem $13$. No proofs are available except using induction. So if we can prove inequality $(1)$, it will be a completely new proof! In fact, inequality $(1)$ is itself very interesting.
$\color{red}{\mathrm{One\, more\, idea\, (maybe\, not\, useful):}}$
Let $\theta_{jk}=\frac{ja_k\pi}{n}$ and $A=\sum_{k=1}^{n}a_k$, then we get,
$$(1+z^{ja_k})=\left(1+\cos\left(\frac{2ja_k\pi}{n}\right)+i\sin\left(\frac{2ja_k\pi}{n}\right)\right)=2\cos(\theta_{jk})(\cos(\theta_{jk})+i\sin(\theta_{jk}))$$
Therefore,
$$\left|\sum_{j=0}^{n-1}\prod_{k=1}^{n}(1+z^{ja_k})\right|=2^n\left|\sum_{j=0}^{n-1}\prod_{k=1}^{n}\cos(\theta_{jk})e^{i\theta_{jk}}\right|$$
So we get one more equivalent inequality,
$$\left|\sum_{j=0}^{n-1}\prod_{k=1}^{n}\cos(\theta_{jk})e^{i\theta_{jk}}\right|=\left|\sum_{j=0}^{n-1}e^{i\frac{\pi Aj}{n}}\prod_{k=1}^{n}\cos(\theta_{jk})\right|\geq\frac{n^2}{2^n}\tag{2}$$
$\color{red}{\text{Remark:}}$ According to the hypothesis of the question, $n\nmid A$. Therefore $e^{i\frac{\pi A}{n}}\neq\pm1$.
Can we prove this inequality?
Any hint or help will be appreciated. Thank you!
It was posted before on Math Stack Exchange
Also posted to m.se, https://math.stackexchange.com/questions/3750621/a-nice-problem-in-additive-number-theory with no notice to either site. Please don't do that.
I cannot see how do you prove this by induction. For the exponential sum approach, the problem is that you seem not to take into account that $a_1+\dotsb+a_n$ is not divisible by $n$, which is essential (consider $a_1=\dotsb=a_n=1$).
@GerryMyerson sorry.
@Seva it is clearly mentioned in the problem that $\sum_{i=1}^{n}a_i\not\equiv0\pmod{n}$.
This is not what I am saying. The problem is not that the assumption $a_1+\dotsb+a_n\not\equiv 0\pmod n$ is missing from the statement; it is that this assumption does not seem to be used - and it is impossible to prove the assertion without using this assumption.
@Seva Should I add the proof of the main problem (leaving all these exponential sums staffs) using induction?
Well, I would be interested to see it.
In my impression, the result is known, and probably due to Olson.
@Zhi-WeiSun can you give me a source where I can find it?
@Subhrajit Bhattacharya , Can you provide a link to the solution by induction method?
@Seva , Alapan: See the official solution on page 555.
I have nothing to add to the Fourier-type approach suggested in the question, but for those curious, thought it useful to outline the combinatorial solution to the problem that I know (I believe this is the same as the IMO official solution, and claim no originality).
One thing to add is that, although induction is a crucial part of the proof, we do not use induction on the problem statement itself, but rather to prove an auxiliary combinatorial fact, given below.
For each $X\subset \{1,\ldots,n\}$ we have an associated sum $S_X=\sum_{i\in X}a_i$. We want to show that there exist at least $n$ many $X$ such that $S_X\equiv 0\pmod{n}$, assuming that $a_i\not\equiv 0\pmod{n}$ for $1\leq i\leq n$ and $S_{\{1,\ldots,n\}}\not\equiv 0\pmod{n}$.
For any permutation $\pi$ of $\{1,\ldots,n\}$ consider the sequence of $n+1$ distinct sets
$$ I_0,\ldots,I_{n+1}=\emptyset, \{ \pi(1)\}, \{\pi(1),\pi(2)\},\ldots,\{\pi(1),\ldots,\pi(n)\}.$$
By the pigeonhole principle, there must exist some $i<j$ such that $I_i$ and $I_j$ induce the same sum modulo $n$. In particular, there exists some non-empty set of consecutive integers $I=\{i+1,\ldots,j\}$ such that $S_{\pi(I)}\equiv 0\pmod{n}$. Note that by our assumptions we must have $2\leq \lvert I\rvert <n$.
The key fact (which can be established by double induction on $k$ and $n$) is that, for any $n\geq 3$, if we have any collection of $1\leq k\leq n-2$ sets $X_1,\ldots,X_k\subset \{1,\ldots,n\}$, each of size $2\leq \lvert X_i\rvert<n$, then there is a permutation $\pi$ of $\{1,\ldots,n\}$ such that none of $\pi(X_i)$ are a consecutive block of integers.
Given the preceding, it is now straightforward to find $n-1$ many distinct non-empty sets $X\subset \{1,\ldots,n\}$ such that $S_{X}\equiv0\pmod{n}$ (and then the empty set gives a trivial solution, producing the requisite $n$ solutions in total).
It remains to prove the key fact. The case $k=1$ and $n\geq 3$ is obvious. Consider the bipartite graph on $[k]\times [n]$ where $i\sim x$ if either $X_i=[n]\backslash \{x\}$ or $X_i=\{x,y\}$ for some $y\in [n]$. There are clearly at most $2k<2n$ edges, and hence some element of $[n]$ has degree at most 1 in this graph, without loss of generality we can say this element is $n$, and suppose further without loss of generality that if $i\sim n$ then $i=k$.
Consider the collection of $k-1$ sets $Y_i=X_i\backslash \{n\}\subset [n-1]$ for $1\leq i<k$. By construction, these sets satisfy $2\leq \lvert Y_i\rvert<n-1$ and hence, by induction, there is a permutation $\pi$ of $[n-1]$ such that none of $\pi(Y_i)$ are consecutive blocks. If $\pi(X_k\backslash \{n\})$ is not a consecutive block, then we extend $\pi$ to a permutation of $[n]$ in the obvious way (so $\pi(n)=n$). An easy case analysis confirms that, if $\pi(X_k\backslash\{n\})$ is a consecutive block, there is always a way to extend the permutation to one on $[n]$ that 'breaks up' the block, and we are done.
I do not know the official solution, but this is the only solution which I know. It is written, in particular, in IMO Compendium book. By the way, it works for any Abelian group of order $n$. Possibly for groups like $\mathbb{F}_p^k$ with large $k$ the bound may be improved.
Thanks Fedor! I had written this up in some personal notes a while ago, and neglected to write down where the argument came from. Now that you say it, I'm sure it must have been the IMO Compendium book itself.
See the official solution on page 555.
The induction argument suggested by Thomas actually goes back to the last paper of Olson, namely J. E. Olson, A problem of Erdős on abelian groups, Combinatorica 7 (1987), 285–289.
|
2025-03-21T14:48:31.499297
| 2020-07-13T11:04:15 |
365529
|
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|
Stack Exchange
|
A Pell like equation
If one takes in general $(\star)\, \,x^2-dy^2=C$ where $d$, $C$ in $\mathbb{N}$.
Taking $d=w^2p^2+p$ with $w\in \mathbb{Q}\ge 1$ and $p\in \mathbb{Z}$ which is verified (explained later), for the matrix $$A=\begin{pmatrix}2w^2p+1&2w(w^2p^2+p)\\2w&2w^2p+1\end{pmatrix}$$ if $X_0$ is a solution to $(\star)$ then $AX_0$ is another one.
Now $w$ could be taken in a cool way basicaly say $d=a^2b^2+cb$ with $c\in\mathbb{Z}$, $|c|<|a|$ and $c$ coprime with $b$ and $a$, letting
$w=\frac{a}{c}$ and $p=cb$ the matrix $A$ is in $\mathbb{Q}$ but can have a power $A^n$ with integer entries. So to say that i didn't find any reference for this idea which is surprising. This is related and known of course as a Pell equation when $w\in \mathbb{N}$.
A question is if there is a related topic discussion to this approach since Pell equations are known, and as a conjecture to give certain family of $A$ with $A^n$ of integer entries. (It appears there are many). Thanks
Edit, i'll illustrate this in an example just for clarity:
$$x^2-2021y^2=d^2$$ one solution is $(d,0)$, i took $2021$ by chance as it is within what i can get, (i don't know if it should work for $2020$) since $2021=\frac{45^2}{4^2}4^2-4$. An easy argument says if the numerator of $w$ here $45$ is $5 \pmod{8}$ Then $A^3\in \mathbb{M}_2(\mathbb{Z})$ so
$$A=\begin{pmatrix}-1011.5&45472.5\\22.5&-1011.5\end{pmatrix}$$ and $$A^3=\begin{pmatrix}-4.139590049\times 10^9&1.8609747948\times 10^{11}\\9.2081880\times 10^7&-4.139590049\times 10^9\end{pmatrix}.$$
Edit. It seems such $A$ has an all integer power $A^n$ if and only if $c$ is a power of two and mainly $|c|= 1, 2, 4$,
I am not sure if there is a question in this post.
A question is implicitly if there is a related topic discussion to this approach since Pell equations $(\star)$ are known, besides a natural question rises as a conjecture to give certain $A$ with $A^n$ of integer entries for some $n$ etc.
Why not edit the question to explicitly ask what you want to know?
If I understand correctly, your question is the following: suppose that for a given positive integer $d$ the equation
$$\displaystyle x^2 - dy^2 = c \text{ } (\ast)$$
has a solution in integers $x,y$ for some integer $c$. Then does there exist an infinite family of solutions generated by $A^k (x,y)^T$ for some $A \in \text{GL}_2(\mathbb{Z})$ having infinite order?
The answer is yes, and was answered completely by Siegel. Indeed, the equation $(\ast)$ has finitely many solutions modulo the action induced by the unit group of the ring of integers of $\mathbb{Q}(\sqrt{d})$, which always has rank one. See the following paper of Siegel: The average measure of quadratic forms with given determinant and signature.
ok but here i gave the $A$ along. It is related to this. I know there are others given in terms of fundamental solutions theory. Thanks.
@ToniMhax Siegel also gave an explicit $A$ in terms of a fundamental solution of the equation. It's in the paper.
but one should compute a fundamental non trivial solution i guess. Here $A$ is in $\mathbb{M}_2(\mathbb{Q})$ that we elevate to a certain power iteration...
@ToniMhax determining whether the equation $x^2 - dy^2 = c$ has a solution in the first place is extremely difficult. Indeed, even in the case of $c = -1$ things are very challenging. We don't know how to count the number of $d \leq X$ for which $x^2 - dy^2 = -1$ has an integer solution for example, and it was only 10 years ago that an upper and lower bound of the correct order of magnitude was established by Fouvry and Kluners.
Yes of course i'll illustrate the simple algo in an exemple just for trial and record.
Thought it possible to simplify in order to be able to write the solutions of the equation. For this we use the decomposition of the number $c$ on the multipliers.
$$Z^2-dR^2=c=ab$$
To record decisions have to know first the solution of the Pell equation $(Z_1;R_1)$. Although to find all the solutions-it is necessary to substitute all the solutions of this Pell equation.
And solving the following equation Pell $(k_0;n_0)$.
$$k^2-dn^2=1$$
Then the formula is as follows.
$$Z_2=k_0Z_1+dn_0R_1$$
$$R_2=n_0Z_1+k_0R_1$$
The problem in finding the first solution for General Pell equation $(Z_1;R_1)$.
The meaning of the solution is that to factor the number. $c=ab$
Then degradable factoring the difference. $xy=a-b$
If the following expression may be a square.
$$s^2=\frac{1}{d}((\frac{y+x}{2})^2-a)$$
Then the first solution is written simply.
$$Z_1=ds^2+\frac{y^2-x^2}{4}$$
$$R_1=ys$$
Such record these formulas will greatly simplify the calculations. Always better to have a formula. It's not always true that it works... but it works.
You can also create other formulas...
https://artofproblemsolving.com/community/c3046h1049910___4
https://artofproblemsolving.com/community/c3046h1048219___2
I will give one example-which shows in which direction it is necessary to look for solutions.
Although it should be mentioned, and the equation: $$aX^2-qY^2=f$$
If the root of the whole: $\sqrt{\frac{f}{a-q}}$
Using equation Pell: $p^2-aqs^2=1$ solutions can be written:
$$Y=(2aps\pm(p^2+aqs^2))\sqrt{\frac{f}{a-q}}$$
$$X=(2qps\pm(p^2+aqs^2))\sqrt{\frac{f}{a-q}}$$
And for that decision have to find double formula.
$$Y_2=Y+2as(qsY-pX)$$
$$X_2=X+2p(qsY-pX)$$
The interesting thing here is that each solution of the Pell equation determines itself the next formula for solutions of the Pell equation. And there are a lot of different special cases for which it is very easy to find solutions.
|
2025-03-21T14:48:31.499664
| 2020-07-13T11:28:54 |
365530
|
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|
Stack Exchange
|
Sifted sets can't accumulate on a curve
Let $f,g,h$ be elements in $\mathbb{Z}[x,y]$, each geometrically integral and at least two of them are distinct. Without loss of generality, suppose that $f$ is not proportional to $g$ over $\mathbb{C}$, so that the variety $V_{f,g} = \{f = g = 0\}$ has co-dimension 2 in $\mathbb{A}^2$.
Let $T$ be a large real number, and fix a prime $T/\log T < p < T$. Define the set
$$\displaystyle S_{f,g}(T) = \{\mathbf{x} \in \mathbb{Z}^2 \cap [-T,T]^2 : \mathbf{x} \pmod{\ell} \in V_{f,g}(\mathbb{F}_\ell) \text{ for some } T/\log T < \ell < T \},$$
where $\ell$ runs over primes. One can show that
$$\displaystyle |S_{f,g}(T)| \asymp T.$$
My question is, is it possible for the intersection of $S_{f,g}(T)$ and the set
$$\displaystyle C_h(p; T) = \{\mathbf{x} \in \mathbb{Z}^2 \cap [-T,T]^2 : h(\mathbf{x}) \equiv 0 \pmod{p}\}$$
to be large? Indeed, the question is as in the title, can the "sifted set" $S_{f,g}(T)$ have large intersection with a curve modulo a fixed prime $p$?
|
2025-03-21T14:48:31.499769
| 2020-07-13T11:47:52 |
365532
|
{
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"Armando j18eos",
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"abx",
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|
Stack Exchange
|
Weak Lefschetz theorem for Lef line bundles
I'm studying
M. A. A. de Cataldo, L. Migliorini - The Hard Lefschetz Theorem and the topology of semismall maps, Ann. sci. École Norm. Sup., Serie 4 35 (2002) 759-772.
The premises are the following.
Let $f:X\to Y$ be a proper holomorphic (non constant) map of irreducible, complex, projective varieties of dimension $n$. For every $k\in\{-\infty,0,\dotsc,\dim X\}$ defined $Y^k=\{y\in Y\mid\dim f^{-1}(y)=k\}$ with the convection $\dim\emptyset=-\infty$. These spaces $Y^k$ are locally closed analytic subvarieties of $Y$ which (disjoint) union is $Y$ as well.
Definition 1. A proper holomorphic map $f:X\to Y$ is called semismall if $\dim Y^k+2k\leq\dim Y$ for any $k$.
From now on, I assume that all semismall maps are proper and surjective.
Definition 2. A line bundle $L$ over $X$ is Lef (Lefschetz effettivamente funziona) if a positive multiple of $L$ is generated by its global sections and the corresponding morphism (onto the image) is semismall; in other words there exist $d,N\gg0,\,f:X\to\mathbb{P}^d$ semismall (onto the image) such that $L^{\otimes N}\cong~f^{*}\mathcal{O}_{\mathbb{P}^d}(1)$.
The authors stated the following Weak Lefschetz theorem for Lef line bundles
Let $L$ be a Lef line bundle over a smooth, complex, projective variety $X$. Assume that $L$ admits a global section $s$ which reduced locus $Y$ is a smooth divisor, and denoted by $i:Y\hookrightarrow X$ the relevant inclusion. The restriction maps $i^{*}:H^k(X)\to H^k(Y)$ are isomorphisms for $k\in\{0,\dotsc,\dim X-2\}$ and a monomorphism for $i=\dim X-1$.
Proof. The proof can be obtained by a use of Leray spectral sequence coupled with the theorem on the cohomological dimension of constructible sheaves on affine varieties. [...] $\Box$
Ignore whether this proof is a standard application of some ideas\techniques, indeed I have no idea on how to explicit it: can someone give me advice, hint, "roadmap", bibliographical sources?
Thanks in advance.
It is based on certain vanishing property of $U= X\backslash Y$.
First you have a long exact sequence (a derived categorical version is given in the end)
$$H^k(X,Y;\mathbb{Q})\rightarrow H^k(X,\mathbb{Q}) \rightarrow H^k(Y,\mathbb{Q}) \rightarrow H^{k+1}(X,Y;\mathbb{Q}).$$
Note that $H^{k}(X,Y;\mathbb{Q})=H^{k}_c(U,\mathbb{Q}_Y) \simeq H^{n-k}(U, \mathbb{Q}_Y)$. The last isomorphism is the Poincaré duality which only requires that $U$ is smooth. This is true since $X$ is smooth. So if we have $H^k(U, \mathbb{Q}_Y) = 0$ for $k>n$, then we have done.
The vanishing is based on the citation [10] in the article, namely Vanishing and non-vanishing theorems
Astérisque, tome 179-180 (1989), p. 97-112.
Let me pick up the key part. Let $f:X \rightarrow \mathbb{P}^N$ be the corresponding morphism of $M$. Then $Y$ is a pullback of a hyperplane $H \subset \mathbb{P}^N$ by $f$ and hence $f$ restricting on $U$ is a map into a affine variety $\mathbb{P}^N\backslash H$.
Definiton 1.1
Let $g : Y \rightarrow Z$ be a morphism of analytic varieties. We define $r(g) = \mathrm{Max}\{\mathrm{dim} \Gamma - \mathrm{dim} g(\Gamma) - \mathrm{codim} \Gamma \}$, $\Gamma$ closed subvariety of $Y$.
In the semismall case, $r(f) = 0$. Now we have
Lemma 1.2 Assume that there exists a proper surjective morphism $g$ from $U$ to an affine variety $W$.
Then $H^k(U,\mathscr{L})=0$ for $k>n+ r(g)$ and $\mathscr{L}$ a local system.
So the vanishing follows.
A derived categorical version
$$ Y \overset{i}{\hookrightarrow} X \overset{j}{\hookleftarrow} U=X\backslash Y$$
$$\rightarrow j_!j^*\mathbb{Q}_X \rightarrow \mathbb{Q}_X \rightarrow i_*i^* \mathbb{Q}_X \rightarrow j_!j^*\mathbb{Q}_X[1]\rightarrow $$
and apply $R^0\Gamma = H^0c_*$, i.e. taking the hypercohomology, where $c_*$ is the pushforward in derived category to a point.
Since $X$ is projective hence proper, $c_* = c_!$. So
$$R^0\Gamma j_!j^* \mathbb{Q}_X[k] = H^0 c_* j_!j^* \mathbb{Q}_X[k] = H^0 c_!j_!\mathbb{Q}_U[k] = H^0 c_{U,!} \mathbb{Q}_U[k] = H^k_c(U,\mathbb{Q})$$
where $c_{U,!}$ is the direct image with proper support pushforward to a point from $U$. So we have long exact sequence
$$H^k_c(U,\mathbb{Q}) \simeq H^{2n-k}(U,\mathbb{Q}) \rightarrow H^k(X,\mathbb{Q}) \rightarrow H^k(Y,\mathbb{Q}) \rightarrow H^{k+1}_c(U,\mathbb{Q})$$
where the first isomorphism is by the Poincare duality. Note that the Poincare duality only requires that $U$ is smooth. This is true if $X$ is smooth or $Y$ contains all singularity. The result follows by the long exact sequence and the vanishing of $H^k_c(U,\mathbb{Q}) \simeq H^{2n-k}(U,\mathbb{Q})$.
You are assuming that $U$ is affine, so this is just the standard Lefschetz theorem.
@abx This is true since a projective variety excising a hyperplane is affine.
Of course, but this is not what the OP is asking for.
@abx Doesn't the OP want an explicit proof? Sorry for my poor English reading ability.
Yes, but in the case he is considering $Y$ is not a hyperplane section.
@abx So a zero locus of section needs not be a hyperplane? I haven't thought about that.
@abx I think I know what happens now. Sorry for my ignorance and I learn a lot from this question. I will improve my answer right now.
@XTChen Maybe, in the "non derived part", did you have wrong to write the Poincaré duality?
@Armandoj18eos Oh sure. Thank you!
No, I thank you, since I have learnt a little trick. ;) Your answer simplify a lot my understanding of the proof of W.L.T. for L.e.f. line bundles.
|
2025-03-21T14:48:31.500229
| 2020-07-13T13:31:17 |
365536
|
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|
Stack Exchange
|
Proper model category for "categories with finite limits"
I'm looking for a Quillen model category which model the $2$-category of 'small category with finite limits (and functors between them preserving finite limits)':
Left proper,
right proper,
Enriched in $\mathbf{Cat}$, the folk model structure on the category of categories.
To give an example, if I only require left proper or right proper but not both at the same time, it is farily standard:
One can work with "chosen limits": One considers the $1$-category of small categories with chosen (up to equality) terminal object and fiber product. It is a strict $2$-category (so enriched in Cat), and it has a model structure lifted along the forgetful functor to $\mathbf{Cat}$. (i.e. fibration, trivial fibration and equivalences are the map whose image by the forgetfull functor are as such). Every object is fibrant, so this model structure is right proper. But I'm pretty sure that it is not left proper.
One can work with a category of "limit sketches": one considers the $1$-category of small categories equiped with a class of marked cocone closed under isomorphisms. We equip it with its "trivial model structure" (weak equivalences are the equivalence that detect the marked cocone, fibrations are the isofibrations) and we take a left Bousfield localization in which the fibrant objects are the category with all small limits and in which the marked cocone are exactly the colimit cocones.
In this one, every object is cofibrant, so it is left proper. But I can show it is not right proper.
I've made many attempt to construct such a model structure which is both left and right proper, and it always seem to fail.
So I'm wondering if there is a homotopy theoretic obstruction to have all these three properties at the same time...
I'm assuming you know this paper by Steve Lack? It's the best technology I know for this kind of question: https://arxiv.org/abs/math/0607646
Yes, I should have linked it as a reference for example (1). But I can't really find an answer in this model. I tend to think that there is an actual homotopy theoretic obstruction to all three properties together, but I can't find it.
|
2025-03-21T14:48:31.500401
| 2020-07-13T13:56:34 |
365538
|
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|
Stack Exchange
|
Are orbit polytopes of rotation subgroup of Coxeter group combinatorially equivalent?
Suppose that $G\subset O(d)$ is a finite reflection (finite Coxeter) group. For any $v\in \mathbb{R}^d$ which is not fixed by any non-trivial $g\in G$, one can consider the orbit polytope (Coxeter) permutahedra \begin{equation} P(G;v)=Conv (G\cdot v) \end{equation} given by the orbit.
Now consider $G^+\subset SO(d)$, the index-two rotation subgroup of $G$. Again one can consider the orbit polytope \begin{equation} P(G^+;v)=Conv(G^+\cdot v) \end{equation} for $v$ as above (i.e., not fixed by any non-trivial $g$ from the original group). Is it necessarily the case that $P(G^+;v)$ is just obtained by $P(G;v)$ by "alternation"?
If $v_1,v_2\in \mathbb{R}^d$ are not fixed by any $g\in G$, it can be shown that $P(G;v_1)$ and $P(G;v_2)$ are combinatorially equivalent. Must the same be true for $P(G^+;v_1)$ and $P(G^+;v_2)$ as well?
This would definitely seem to be the case for a number of examples (e.g., for $G=A_2\times A_2\times A_2$, for which $P(G;v)$ is a box, $P(G^+;v)$ is a tetrahedra).
It depends on what exactly you mean by "alternation". It is the case that $P(G^+;v)$ has exactly half as many vertices as $P(G;v)$. So in this sense $P(G^+;v)$ is obtained from $P(G;v)$ by "skipping every second vertex". Note that $Gv$ is indeed the set of vertices of the polytope $P(G;v)$, because all points in the orbit lie on a sphere, so that none is a convex combination of any others.
The second question is answered in the positive for the symmetric group in Cruickshank, J., Kelly, S. Rearrangement Inequalities and the Alternahedron.
@GrantB.'s reference: Cruickshank and Kelly - Rearrangement inequalities and the alternahedron (MSN).
Great; thank you
My heuristic argument would be the following: the reason that all the $P(G;v)$ have the same combinatorial type is that there is no generic $v$ (i.e. $v$ not fixed by any non-trivial $g$) so that $G\cdot v$ has non-generic affine dependencies (because such must occur when transitioning from one combinatorial type to another).
This obviously translates to no non-generic affine dependencies between the points in $G^+\cdot v$ (since the notion of "generic $v$" stays the same). This can probably be made precise.
@M.Winter I think it's more subtle than that. The statement is not true for every subgroup of $G$; the answer to the OP's recent question gives a counterexample for a cyclic subgroup (a "non-generic" affine dependency). The paper I mentioned above discusses the problem of characterizing subgroups with combinatorial orbit polytopes, but to my knowledge not much is known.
@GrantB. There is certainly something special about $G^+$ that cannot be generalized to other subgroups of $G$. For example, there are no non-generic dependencies as long as $v$ stays in the same Weyl chamber. We know this because there are no non-generic dependencies for $G$. Now, $P(G^+;v)$ has points in exactly half of the Weyl chambers and so we know that for $v$ in any of these we get the same combinatorial type. But putting $v$ in one of the other chambers gives you the mirrored polytope, still combinatorially equivalent to the previous one. Might this already be the subtlety?
@M.Winter I agree that the problem reduces to showing that the face lattice of $P(G^+;v)$ is independent of $v$ for $v$ in a given Weyl chamber. However this statement is already not true for all subgroups of $G$, as the linked post indicates. The problem is that points in the interior of a Weyl chamber are simply not always generic in the sense of "no extra affine dependencies". There can be and are non-generic affine dependencies in non-degenerate permutahedra, they just aren't among vertices comprising a face of the polytope.
@GrantB. Oh I got it, I was under the impression that the reflection groups are too nice to have non-generic dependencies inside a Weyl chamber. I was aware of the linked question but have not interpreted the answer in that way. Interesting.
@M.Winter Yes, I think we are used to polytopes which use a subset of the defining inequalities of permutahedra rather than a subset of vertices, since these tend to be better behaved.
Let me try to give a rigorous definition to "by alternation" such that the answer to your first question is "yes".
Given a polytope $P$ whose vertex-edge graph is bipartite, one might say that there are two polytope $Q$ is obtained from $P$ "by alternation" if there is a bipartition of the graph such that $Q$ is the convex hull of one block of the bipartition.
Under this definition, yes, your graph $P(G^+,v)$ is obtained from $P(G,v)$ by alternation. The requirement that $v$ is not fixed by any (nontrivial) element of $G$ is equivalent to the requirement that $v$ is not fixed by any reflection, or equivalently, not contained in any reflecting hyperplane. (This is standard...See for example Section 1.12 of Humphreys "Reflection Groups and Coxeter groups".) The reflecting hyperplanes cut the ambient space into simplicial cones, so we're just choosing $v$ in the interior of one of the cones. Then the orbit of $v$ contains exactly one point in each of the cones, and this gives a bijection between the orbit and the elements of the group.
The cones define a fan structure on the ambient space (i.e. any two cones intersect each other in faces), and this fan is the normal fan of $P(G,v)$. If two maximal cones in this fan are adjacent (i.e. share a codimension-1 face), then they are related by a reflection in $G$, so exactly one of the corresponding group elements is in $G^+$. Thus the vertices of $P(G^+,v)$ are one block of a bipartition of the vertices of $P(G,v)$.
That's great; thanks. Does it follow that if $v_1$ and $v_2$ are not fixed by any (non-identity) $g\in G$, then $P(G^+,v_1)$ and $P(G^+,v_2)$ have to combinatorially equivalent?
I'm not sure if it follows from the "by alternation" fact, but I suspect it is true. I also suspect that this is known and that someone has written it down. But I have not thought carefully about it to be sure.
Thanks, I should probably put a "reference request" tag on this. I'm pretty certain its true in $\mathbb{R}^3$.
|
2025-03-21T14:48:31.500811
| 2020-07-13T14:09:13 |
365541
|
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|
Stack Exchange
|
Checking existence of proofs of fixed length
This question is a continuation of a related previous question (check here).
Let $\mathcal{L}$ be a recursive first-order theory with the Hilbert-Ackerman's proof calculus, and such that the computational cost of deciding whether or not a formula $\theta$ is an axiom is in P.
Let $\phi$ be a formula, and let $\ell=(\psi_1, \ldots, \psi_n = \phi)$ be a proof of number of lines $n$; that is, each $\psi_i$ is either an axiom or is obtained by a single application of a inference rule involving only formulas that appear previously in $\ell$ (i.e., $\psi_j, j<i$).
Question: what is the asymptotical computational complexity of deciding whether or not $\phi$ has a proof of number of lines $n$? If we change $\mathcal{L}$ to a suitable (recursive) higher order theory and the proof calculus (e.g., Gentzen's proof calculus), how this complexity function can change?
EDIT as discussed in the comments, the question above is undecidable in general.
Let lenght of $\ell$ be the total number of symbols appearing on it, which I will denot by $|\ell|$
Question modified: what is the asymptotical computational complexity of deciding whether or not $\phi$ has a proof $\ell$ with a fixed lenght $|\ell|=n$? If we change $\mathcal{L}$ to a suitable (recursive) higher order theory and the proof calculus (e.g., Gentzen's proof calculus), how this complexity function can change?
SECOND EDIT
As seen in the coments, the modified question has a complexity function that is essentially always NP-complete.
In general, this is an undecidable problem, even if $n$ is a fixed constant.
(Which is related to the fact that what you call length is usually called the number of lines, whereas the length of a proof is defined as the total number of symbols.)
@EmilJeřábek and what if I change number of lines by the total number of symbols?
Anyway, I thank you very much for pointing this, I am editing the question for appropriate terminology
If $n$ is the number of symbols in the proof, and it is given as input in unary, then the problem is obviously in NP. In fact, it is NP-complete as long as the theory proves $\exists x\exists y,(x\ne y)$.
|
2025-03-21T14:48:31.500983
| 2020-07-13T14:20:39 |
365543
|
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|
Stack Exchange
|
What are the sections of an ideal sheaf on a scheme?
Suppose $X$ is a scheme and $f_1,...,f_n\in \Gamma(X,\mathcal O)$ are global sections.
One often reads about the ideal sheaf $\mathcal I=\mathcal (f_1,...,f_n)\subset \mathcal O$, but I have never seen it defined.
The definition should be that $\mathcal I$ is the sheaf associated to the presheaf $\mathcal I^-$ whose value on an open subset $U\subset X$ is $\mathcal I^-(U)=({f_1}{\vert_U},...,{f_n}{\vert_U})\cdot\mathcal O(U)$. Right?
My question is simply: Is there a non-tautological sufficient condition on $U$ and the $f_i$'s guaranteeing that $\mathcal I(U)=\mathcal I^-(U)$?
The only sufficient condition I can think of is that $n=1$ and $U$ is an arbitrary open subset of an integral scheme $X$.
A sufficient condition is that $U$ is affine. Indeed $\mathcal{I}$ fits in a s.e.s. $0\to \mathcal{K}\to \mathcal{O}^n\to \mathcal{I}\to 0$ and $H^1(U,\mathcal{K}) = 0$ for $U$ affine.
@Christ: Thank you very much for this fine answer: +1. Why not transform this comment in a genuine answer?
@Chris's comment, posted as an answer by request (made CW to avoid reputation):
A sufficient condition is that $U$ is affine. Indeed $\mathcal I$ fits in an s.e.s. $0 \to \mathcal K \to \mathcal O^n \to \mathcal I \to 0$ and $H^1(U, \mathcal K) = 0$ for $U$ affine.
|
2025-03-21T14:48:31.501111
| 2020-07-13T14:37:23 |
365544
|
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|
Stack Exchange
|
Elementary inequality generalizing convexity of a function on a segment
I am looking for a proof of the following statement which is known to be true as far as I heard.
Let $g\colon [a,b]\to \mathbb{R}$ be a smooth function. Assume that
$$b-a< \pi.$$
Assume also $$g(a)\geq 0,g(b)\geq 0,$$
$$g''+g\leq 0 \mbox{ on } [a,b].$$
Then $g\geq 0$ on $[a,b]$.
False. Take $g(x)=\sin x$, $a=\pi$, $b=2\pi$.
@Mizar: Thanks, corrected. The inequality $b-a<\pi$ has to be strict.
Write $g=g^+-g^-$ in $[0,\ell]$, multiply $g''+g \le 0$ by $g^-$ (which vanishes at the the endpoints) and integrate. Then we get with $v=g^-$
$$
\int_0^l v'^2- \int_0^l v^2 \le 0.$$
Since the first eigenvalue of the Dirichlet laplacian in $[0,\ell]$ is $\pi^2/\ell^2 $ we have also
$$
\int_0^l v^2 \le \frac{\ell^2}{\pi^2} \int_0^l v'^2 $$ and then $v=g^-=0$, since $\ell <\pi$.
Here's another argument using Sturm-Liouville theory.
Given $b - \pi < a \le x \le b$,
let
$$
f(x) = \sin (x - b + \pi).
$$
Observe that $f > 0$ and $f'' + f = 0$ on $[a,b)$, $f(a) > 0$, $f(b) = 0$, and $f'(b) = -1$.
Since
\begin{align*}
(fg'-f'g)' &= (g'' + g)f - (f''+f)g \le 0\\
(fg'-f'g)(b) &= f(b)g'(b) - f'(b)g(b) = g(b) \ge 0,
\end{align*}
it follows that, on $[a,b]$,
$$
fg' - f'g \ge 0.
$$
and therefore, on $[a,b)$,
\begin{align*}
\left(\frac{g}{f}\right)' &= \frac{fg' - gf'}{f^2} \ge 0.
\end{align*}
Since
$$
\frac{g(a)}{f(a)} \ge 0,
$$
it follows that, on $[a,b)$,
$$
\frac{g}{f} \ge 0.
$$
What is "Sturm comparison theorem"?
@FedorPetrov, it compares solutions to self-adjoint 2nd order ODEs. I never actually remember the precise statement of the theorem, so I adapt the relatively elementary argument to each particular situation.
And where is it used? The argument looks self-contained.
@FedorPetrov, variants of this argument are commonly used to compare the behavior of geodesics on a Riemannian manifold to those on a manifold with constant curvature. But the argument itself was originally developed in what's known as Sturm-Liouville theory. It appears in many ODE textbooks.
my question was: where is it used in your answer? Probably the answer is that you include the proof but not the formulation.
@FedorPetrov, yes. That's correct.
Suppose the contrary, so that $g(s)<0$ for some $s\in(a,b)$. Replacing now $a$ and $b$ by $\max\{t\in[a,s)\colon g(t)\ge0\}$ and $\min\{t\in(s,b]\colon g(t)\ge0\}$, respectively, we see that without loss of generality (wlog)
\begin{equation}
g(a)=0=g(b).
\end{equation}
Also, by a horizontal shift, wlog
\begin{equation}
a=0,\quad b\in(0,\pi).
\end{equation}
Let
\begin{equation}
h:=-(g''+g)\ge0.
\end{equation}
Then
\begin{equation}
G(t):=G_b(t):=g(t)\sin b=\sin t\,\int_0^b du\,h(u)\sin(b-u)
-\sin b\,\int_0^t du\,h(u)\sin(t-u).
\end{equation}
We have to show that $G\ge0$ on $[0,b]$. Because the nonnegative function $h$ can be however closely approximated in $L^1$ by conical combinations of the indicators of intervals, wlog $h=1_{[c,d]}$ for some $c,d$ such that $0<c<d<b$, in which case
\begin{equation}
G(t)=\sin t\, (\cos (b-d)-\cos (b-c))-\sin b\, (\cos (t-\max (c,\min (d,t)))-\cos (c-t)).
\end{equation}
So, if $0\le t\le c$, then $G(t)=\sin t\, (\cos (b-d)-\cos (b-c))\ge0$.
The case $d\le t\le b$ is similar, by the left-to-right symmetry.
It remains to consider the case when $c\le t\le d$. Then
\begin{equation}
G''(t)=\sin t\, (\cos b\, (\cos c-\cos d)-\sin b\, \sin d)-\sin b \cos c\,\cos t
=A\sin(t+C)=:f(t),
\end{equation}
where $A,C$ depend only on $b,c,d$. The function $f$ will be $\le0$ on the interval $[c,d]$ of length $<\pi$ iff $f(c)\le0$ and $f(d)\le0$. In our case, we have
\begin{equation}
2f(c)=\sin (b-c-d)-\sin (b+c-d)-\sin (b-2 c)-\sin (b)
\end{equation}
and hence $(f(c))'_d=-\sin c\, \sin (b-d)\le0$, so that $f(c)$ decreases in $d$. So, wlog $d=c$, in which case $f(c)=-\sin b\le0$. Thus, $f(c)\le0$ always, that is, for all $d\in[c,b]$.
Similarly, by the left-to-right symmetry, $f(d)\le0$, which completes the proof.
This is sometimes called a maximum principle "on thin domains". My answer below is technically equivalent to @Deane Yang's answer in this specific context, but the scope is more general so I thought I'd still give it a shot (in particular the argument applies to higher dimensions as well, see the comments below).
Note first that if the eliptic operator under consideration had nonnegative zeroth-order coefficient we could immediately conclude from the standard maximum principle that $-g''+g\geq 0$ implies $g\geq 0$, given that $g\geq 0$ on the boundaries. The problem here is of course that your operator
$$
\mathcal L[g]=-g''-g
$$
has negative zeorth-order coefficient, $c(x)\equiv -1$.
Here comes the trick now: assume that you can find some particular function $f(x)\geq C>0$ (up to the boundary) such that $\mathcal L[f]\geq 0$.
(The fact that you can find such a function relies on the thinness of the domain, I will come back to this later on. In your specific example you can take $f(x)=\sin(x-b+\pi)$ as in @Deane Yang's answer.)
Think then of $g$ as $uf$ for some $u$. (Here the strict positivity of $f$ is important so that $u$ is smooth enough, no funny business can arise from this change of variables)
Define then
$$
\tilde{\mathcal{L}}[u]:=\mathcal{L}[uf].
$$
Here you can compute expliticly $\mathcal L$, but the key is that in whole generality this new elliptic operator
$$
\tilde{\mathcal{L}}[u] =\Big(\mbox{1st & 2nd order}\Big) + \Big(\mathcal L[f]\Big) u
$$
always has a zeroth-order coefficient with the right sign, i-e $\tilde c(x)=\mathcal L[f](x)\geq 0$ .
Now if $g$ is a supersolution of $\mathcal L[g]\geq 0$ you have by definition that $u:=\frac{g}{f}$ is a supersolution of $\tilde{\mathcal L}[u]\geq 0$. Applying the usual maximum principle (with $\tilde c(x) \geq 0$) you conclude that $u\geq 0$, hence $g\geq 0$.
Comment 1: here you see that the key point is the existence of a well-chosen $f(x)$, which is not to be taken for granted.
The reason why you can actually find such a function is that your domain is small enough in terms of the lowest eigenvalue of the Dirichlet problem for the homogeneous problem, without zeroth-order terms: notice obviously that this mysterious function $f(x)=\sin(x-b+\pi)$ is indeed the principal eigenvalue of $-\frac{d^2}{dx^2}$ on the domain $(b-\pi,b)$ with zero boundary conditions. The trick is here that your domain $(a,b)\subset (b-\pi,b)$ is $b-a>0$ is small enough.
(This reminds me of a classical exercise in elliptic PDEs where one is asked to apply the Lax-Milgram theorem for operator whose zeroth-order term has the wrong sign, but not too large in modulus compared to the first eigenvalue, see also @Giorgio Metafune's answer)
Comment 2:
The trick works exactly the same in higher dimensions.
For example if you were working in two dimensions on an infinite but thin enough strip, then plugging in a well-chosen $\sin$ function depending only on the thin coordinate automatically gives a suitable $f$. I can't remember where I learnt this trick, and also I'm pretty sure that I also read somewhere a completely general statement that, regardless of the initial coefficients, sufficiently narrowing down the domain always gives a suitable $f$ (this is natural if one thinks of resonant frequencies: in the absence of zeroth-order terms, thinness in any direction of a given object/domain gives a high-pitch natural frequency, hence a large principal eigenvalue that can effectively dominate any zeroth-order coefficient that you might want to add to the free resonance).
The trick also works sometimes for nonlinear operators, but the "change of variables" to go from $g$ to $u$ can be very delicate to find, it's not just products and quotients anymore.
Comment 3: this is called "thin domains" because the argument is usually applied as above (in a systematic way, since thinness is a classical sufficient condition), but in fact everything relies on the existence of a "nice" $f$. So even in the broader context of not necessarily thin domains (balls, or whatever) one might get lucky by guessing the right $f$ (if any, of course!)
I like your discussion. I think basic elliptic PDE theory be presented first in the 1-dimensional case.
What do you mean by "think of $g$ as $uf$"?
I mean "change variables" and write $g=uf$, so the new primary variable is $u$
@Deane Yang: Thanks, me too, and this is how I always try to introduce and motivate the theory when I teach elliptic equations. But I guess this is really a matter of taste.
Another approach in the spirit of viscosity solutions/maximum principle:
Assume wlog $a>0$ and $b<\pi$. Let $I:=[a,b]$ and define
$$f(x):=\sin x,\quad g_\lambda:=g+\lambda f.$$
If $g=g_0<0$ somewhere on $I$, then we can then find $\lambda>0$ such that
$\min_I g_\lambda=0$ (observe that $f\ge c$ on $I$ for some $c>0$ and look at $\lambda\mapsto\min_I g_\lambda$).
But then $g_\lambda''\le g_\lambda''+g_\lambda\le 0$ on $I$, so $g_\lambda$ is concave. Since $g_\lambda>g$ is positive at the boundary of $I$, it must be positive on all of $I$, contradiction.
|
2025-03-21T14:48:31.501710
| 2020-07-13T15:51:08 |
365546
|
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|
Stack Exchange
|
Covering a compact subset by a tubular neighborhood of a smooth curve
Let $K$ be a compact subset of the open and connected set $\Omega\subseteq\mathbb{R}^n$, $B_\varepsilon$ the ball centered at the origin of radius $\varepsilon$. Is it true the intuitive fact that there exists an $\varepsilon>0$ and a smooth curve $\gamma_K:I\to\mathbb{R}^n$ such that $K\subseteq\gamma_K+B_\varepsilon\subseteq\Omega$?
Here's a quick answer, but you should make sure that everything I'm saying is justified.
First, since $K$ is compact, there is a finite set of points $\{p_i\}_{i=1}^N$ such that $K$ is covered by the balls $B_{\epsilon/2}(p_i)$.
Second, we can construct a piecewise-linear curve $\gamma$ by connecting $p_1$ to $p_2$ with a line, $p_2$ to $p_3$, and so on.
Third, you can always find a smooth, embedded curve $\gamma_K$ within a distance of $\epsilon/2$ of $\gamma$.
Therefore, any point in $K$ is within a distance of $\epsilon$ of $\gamma_K$, so
$$K \subseteq \gamma_K + B_\epsilon$$
as desired.
This construction can be modified without much effort to make sure that $\gamma_K + B_\epsilon$ lies inside this set $\Omega$ as well (Hint: since $K$ is compact, the distance between $K$ and the boundary of $\Omega$ is bounded away from zero).
Officially you cannot connect by a line $p_1$ and $p_2$ because this line can lie outside of $\Omega$. The idea is clear, the problem is the (maybe combinatorial) formalization of the existence of a path that visit all the centers of these balls without going out of $\Omega$.
Since $\Omega \subset \mathbb{R}^n$ is (path) connected, you can connect the $p_i$ inside it. That should help you construct $\gamma_K$ to satisfy all requirements.
The third point can be simplified: one does not need smooth approximation. Instead, one can re-parametrize each affine segment of $\gamma$, in such a way that derivatives of all order vanish at any node, which makes the whole curve smooth, with the same image as $\gamma$. In other words, you can make any zig-zag curve smooth, if you stop whenever you need to make an angle.
The problem of the proof is not smoothness. It is to write down rigorously an "algorithm" which builds the curve (piecewise linear, even just continuous, is more than fine) given a linked sequence of balls (because one could be obliged to go through the same balls several times). I was hoping someone already faced such a question and could give a rigorous and simple answer :) The answer given by Rohil cover the simple part that anyone trying to prove the result would write down (I don't know how to express this without appearing rude, but it is not my intention)
I suppose the "piecewise-linear" approach is only rigorous if the set is convex, but as mentioned above it is easy to adapt. If you would like an "algorithm", then you can just connect any pair of two points by curves that pairwise do not intersect (as long as $n \geq 3$). This is some complete graph, and then you just apply your favorite graph theory algorithm for extracting a path that visits all vertices exactly once.
I am confident that someone will post a proper answer at some point. Thanks anyhow.
I'm not going to belabor the point, but I don't really see why there is some deeper issue at work in this problem. You should edit your question to make it more clear what you want out of this, especially if there's a specific application in mind.
The question is precise. Your answer gives an intuitive idea of why should be true (in a very specific case), which is nothing but the intuitive idea that brought me to "conjecture" the question (was not taken from somewhere). I posted the question in the hope of some simple (maybe nonconstructive) proof.
|
2025-03-21T14:48:31.502007
| 2020-07-13T16:02:16 |
365548
|
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|
Stack Exchange
|
On the definition of Gysin homomorphism
According to Lawson-Michelsohn's book (p239), the Gysin homomorphism for a continuous map between (compact) manifolds $f:Y\to X$ is defined by setting
$$
f_!=PD_X^{-1}\circ f_*\circ PD_Y
$$
where $PD$ denote the Poincare duality and $f_*$ is the induced map on homology.
The authors also call $f_!$ `integration along fibers', which makes sense to me if $f$ happens to be a bundle projection map $\pi:E\to X$. But I can hardly understand $f_!$ for an arbitrary continuous map, like the inclusion map $i:X\to E$ of zero section of some vector bundle.
Question 1: Is there a nice way to visualize $f_!$ when $f$ is not a vector bundle map (e.g. an embedding map)?
Question 2: Moreover, is there a more direct definition of $f_!$ without using Poincare duality?
Or, can we make a definition starting from the chain level?
For example, if $f=\pi$ is the projection map for an oriented vector bundle, then we do have a well-defined map $\pi_!: \Omega_{vc}^*(E)\to \Omega^*(X)$, see wiki.
By Poincaré duality, you can think of $k$-dimensional cohomology classes as being represented by (singular) submanifolds of codimension $k$. Then the Gysin map just takes the image or this submanifold.
@MarkGrant Thanks! But I'm still confused. So, why not just work with homology groups? what is the good point of using $f_!$ compared to the usual $f_*$?
Well, there are other good reasons for preferring cohomology over homology (e.g. cup products, cohomology operations) and it's sometimes good to mix these with this extra functoriality for proper maps between manifolds.
Also check out https://mathoverflow.net/q/64744/8103 for many different viewpoints on these Gysin maps.
|
2025-03-21T14:48:31.502268
| 2020-07-13T17:57:16 |
365559
|
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|
Stack Exchange
|
On hereditarily reflexive Banach spaces
It was proved by W.B. Johnson and H.P. Rosenthal [Studia Math. 43 (1972), 77–92]
that every Banach space $X$ with $X^{**}$ separable is hereditarily reflexive:
every infinite dimensional closed subspace of $X$ contains an infinite dimensional
reflexive subspace.
Suppose that $X$ is separable and $X^{**}/X$ reflexive. Is $X$ hereditarily reflexive?
Of course, we would have a positive answer if each infinite dimensional closed subspace of such a space $X$ contains an infinite dimensional subspace $Y$ with $Y^{**}$ separable.
Gowers' dichotomy theorem reduces to the case that $$ is HI. Have you asked Argyros whether there exists a hereditarily non reflexive HI space $X$ with $X^{**}/X$ reflexive?
I have looked at Spiros A. Argyros, Alexander D. Arvanitakis, Andreas G. Tolias.
Saturated extensions, the attractors method and Hereditarily James Tree Spaces. pp. 1-90 in "Methods in Banach space theory".
J.M.F. Castillo and W.B. Johnson eds. Cambridge University Press, 2006.
The question has a negative answer:
Following the idea in Bill Johnson's comment, I looked at the work of Argyros. In this paper (see the reference below), there are several examples of hereditarily indecomposable Banach spaces containing no infinite dimensional reflexive subspaces.
One of these spaces $X$ has the additional property that $X^{**}/X$ is isomorphic to $\ell_2(\Gamma)$ with $\Gamma$ uncountable: see Proposition 5.4 in the paper.
[reference] S.A. Argyros, A.D. Arvanitakis, A.G. Tolias.
Saturated extensions, the attractors method and Hereditarily James Tree Spaces.
pp. 1-90 in "Methods in Banach space theory". J.M.F. Castillo and W.B. Johnson eds.
London Math. Soc. Lecture Note Ser. 337, Cambridge University Press, 2006.
Just one of the many amazing examples constructed by Argyros et al.
|
2025-03-21T14:48:31.502409
| 2020-07-13T18:03:41 |
365561
|
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|
Stack Exchange
|
Prime ideals of formal power series ring that are above the same prime ideal
Let $R$ denote a commutative ring with identity and let $R[[X]]$ denote the
ring of formal power series over $R$ in an indeterminate $X$. If $I$ is an ideal of $R$,
then $I[[X]]$, the set of power series of $R[[X]]$ with coefficients all in $I$. Now if $P$ is a prime ideal of $R$, then $P[[X]]$ is a prime ideal of $R[[X]]$ that is contained in the prime ideal $\langle P, x\rangle$ of $R[[X]]$ such that $\langle P, X \rangle\cap R=P$. Now let $Q$ be a prime ideal of $R[[X]]$ such that $P[[X]]\subseteq Q$ and $Q\cap R=P$. Is there any characterization for such a $Q$? (I guess that $\langle P, X\rangle$ is the only ideal with that property.)
In general, $(P, X)$ is not the only prime containing $P[[X]]$ and contracting to $P$. I don't have anything to say about the problem of characterizing such primes, but in general it seems extremely hard. Let's focus on the case $P = 0$.
As a motivating example we can even use the integers. The ring $\mathbb{Z}[[X]]$ is a UFD. For any prime $p$ and power series $F$, it is clear that $p +xF$ is irreducible in $\mathbb{Z}[[X]]$ and hence prime. Moreover if we take $f \in \mathbb{Z}[X]$ to be such that the content $c(f)$ of $f$, that is the ideal generated by the coefficients of $f$ in $\mathbb{Z}$, is coprime to $p \mathbb{Z}$, then additionally $(p + Xf) \cap \mathbb{Z}= 0$. One way to show this would be to appeal to the Dedekind-Mertens content formula$^1$, which asserts that over any ring $R$, if $f$ is a polynomial of degree $n$, $G,H \in R[[X]]$, with $fG=H$, then $c(f)c(G)^{n+1} = c(G)^{n} c(H)$. Here $c(F)$ denotes the content ideal of the power series $F$. From here, if we had $(p+Xf)G = p G_0 \in \mathbb{Z}$ then the D-M formula would imply $\frac{1}{p} c(G)^k \subseteq c(G)^k$ which would in turn imply $p$ is a unit (absurd). For every prime $p$, we have found infinitely many polynomials which are prime in $\mathbb{Z}[[X]]$ and which lie over $0$ in $\mathbb{Z}$. Moreover in this way we can be sure to find lots of distinct primes in $\mathbb{Z}[[X]]$, which follows for example from this old post of mine on stackexchange.
I'm not sure to what extent this way of producing principal primes over $0$ generalizes to other rings. It does work verbatim for any Archimedean GCD domain $D$ for which $D[[X]]$ has its irreducible elements prime. The tough part is that last bit, which is a very delicate property. However, it is sufficient that $D[[X]]$ be a UFD, which is a well-studied problem. So for example this argument applies just as well to any regular UFD.
$^1$ See theorem 3.6 in the paper Zero divisors in power series rings by R. Gilmer, A. Grams, and T. Parker [Journal für die reine und angewandte Mathematik (1975), EUDML Link]
|
2025-03-21T14:48:31.502628
| 2020-07-13T18:32:57 |
365563
|
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|
Stack Exchange
|
Fixed point stack for a torus action
In this paper, M. Romagny defines for an action of a group scheme $G$ on a stack $X$ the fixed point stacks $X^G$ associated
to the group action on a stack and in Theorem 3.3 he proves that if
the group $G$ is proper and flat of finite representation
$X$ is a Deligne-Mumford stack
then $X^G$ is algebraic. Later in this note, he proves that condition 2. can be relaxed to $X$ being algebraic with the diagonal being locally of finite presentation.
I am mostly interested in actions by complex tori on algebraic stacks locally of finite type. In this case, one doesn't have the properness from condition 1. Is it still true that the stack of fixed points $X^G$ is algebraic?
I am aware that it is common to take fixed points of Deligne-Mumford stacks as in Graber--Pandharipand with respect to the torus action, but the same approach doesn't seem to work in the completely general case.
I hadn't seen this question until Arkadij contacts me directly. The answer is yes: if $G$ is a group scheme of multiplicative type then the fixed point stack is algebraic. This is now here : https://arxiv.org/abs/2101.02450.
|
2025-03-21T14:48:31.502740
| 2020-07-13T19:58:31 |
365564
|
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|
Stack Exchange
|
Counting unions of unlabelled connected graphs
My question can be stated as follows: let $X$ be a hereditary family of unlabelled graphs closed under disjoint unions. Suppose we know, for each $n$, the number $c_n$ of connected graphs in X on $n$ vertices. How do we get from that the total number $x_n$ of graphs in $X$ on $n$ vertices?
I am vaguely aware that there are methods involving generating functions that answer (some version of) this question, but the only source I found for that online is a transcription of a set of lecture notes with no references whatsoever. I am not particularly well versed in the theory of generating functions, so any source which goes from the basics to the relevant derivation would be greatly appreciated.
However, if it is at all possible, I would ideally like to find an explicit formula for the coefficients $x_n$ in terms of the $c_i$, or at least a (tight-ish) lower bound for $x_n$ in terms of the $c_i$.
I can see the answer in some very simple cases: if, say, $c_i = 1$ for all $i$ (e.g., $X$ is the class of unions of cliques), then $x_n$ is just the partition function $p(n)$ from number theory (compare this to the labelled case, where we get the Bell numbers). However, I do not see how to generalise this. It seems to me very likely that this question would have been studied in the language of multisets, but once again, I am not particularly familiar with that area.
This is what exponential generating functions do. If you want an online source see e.g. Section 2.3 of these note: https://arxiv.org/abs/1409.2562.
Exponential generating functions are for labeled graphs. For unlabeled graphs you need ordinary generating functions. If $c_n$ is the number of unlabeled connected graphs in $X$ on $n$ vertices then the generating function for all unlabeled graphs in $X$ is
$$\prod_{n=1}^\infty (1-x^n)^{-c_n}=\exp\biggl(\sum_{i=1}^\infty c(x^i)/i\biggr)$$
where $c(x) = \sum_{n=1}^\infty c_n x^n$. The first formula is easier to see but the second is usually more useful for computations. This is well known, and may have been first found by Pólya.
Apologies, Ira is of course correct, I missed the unlabeled nature of the question. (I'll leave a link to those notes of Ardila because they're very good, anyways.)
One online reference is Flajolet and Sedgewick, http://algo.inria.fr/flajolet/Publications/book.pdf, page 34.
|
2025-03-21T14:48:31.502909
| 2020-07-13T20:33:11 |
365567
|
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|
Stack Exchange
|
Is the determinantal ideal of the span of a linearly independent set of rank-one matrices radical?
Let $k$ be an algebraically closed field, and let $X_1,\dots, X_n \in M_m(k)$ be rank-one matrices that are linearly independent over $k$. For a fixed integer $1 \leq r \leq m$, consider the ideal $I \subseteq k[t_1,\dots, t_n]$ generated by the $r \times r$ minors of $t_1 X_1+\dots+t_n X_n.$ Is $I=\sqrt{I}$?
If we drop the assumption that $X_1,\dots, X_n$ are rank one, then this is false. Indeed, let $n=m=r=2$, $X_1=\begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix}$, and $X_2=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$. Then $I=\langle \det(t_1 X_1+t_2 X_2)\rangle=\langle(t_1+t_2)^2\rangle$ is not radical.
do you have some extra-assumptions on the $X_i$? May they be assumed to be generic among rank $1$ matrices in $M_m(k)$ (instead of just linearly independent)?
@Libli No, I am more interested to know whether it holds without any genericity conditions. It seems like a plausible generalization of the fact that the determinantal ideal is radical. I am also interested in whether $I$ is prime
$I$ is certainly not prime without conditions on $n$ (just take $n=m$ and the $X_i = E_{i,i}$, then $I = (t_1t_2...t_n)$.
The reducedness is a much better question in my opinion. It seems however quite hard without any genericity assumption. Basically, you are asking when specific linear sections of the determinantal varieties are reduced.
Either your linear sections are generic among the ones defined by rank $1$ matrices and something can be said using deformation theory. Or you can choose the $X_i$ without rank conditions and then they are not necessarily reduced (as you noticed in your example). But the in-between assumption you're making does not look easily usable.
|
2025-03-21T14:48:31.503088
| 2020-07-13T21:31:26 |
365569
|
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|
Stack Exchange
|
Access to journals during pandemic
I have just graduated from the University of Chicago and no longer have access to online journal resources, but I cannot afford to pay for them directly. Normally, I would be able to access library resources for a small fee in the campus library. However, due to closures due to Covid-19, this is no longer an option.
Previous responses to a similar question often involve physical access to a library, which is impossible for many for the foreseeable future. They also consider some access to online resources, but I was wondering if there were any Covid-specific online resources which have come about given the recent crisis. Even if they are not specific to the crisis, online tools which are accessible now are increasingly important, so would be useful.
Do you know that the UC alumni access doesn't cover what you need? (I've never made much use of it, and can't remember my password, so I can't check right now.) Have you asked the library director (Brenda Johnson) whether they would be willing to grant access to resources for a fee?
There are several resources I have been looking for which I have lost access to (UC online access for students is vast). One is the Mathematical Proceedings of the Cambridge Philosophical Society, which I was not able to find in their alumni resources. I have not reached out to Brenda Johnson; thanks for that! I have inquired with a librarian about whether I will have faculty resources, considering I will be teaching at their affiliated charter school, but have not heard back yet.
Do you still have VPN from UChicago? That will make it appear as if you are on campus and usually grants journal access. If not, do you still have a departmental unix account (assuming Chicago has one)? It's possible to set up a proxy via a ssh tunnel and this also makes it seem like you are accessing the internet from on campus.
RBega2, I had not thought of that! Technically I am not sure if I should have access to UChicago's VPN software (on the website it doesn't say that I do, but also doesn't say that I don't). However, I just tried it out and it appears that at least for now I still have access! Thank you! Now the question is just whether I will still have it in the near future...
@JoePrevidi, I just dug up my password and tried, and I wasn't able to access the VPN or the proxy, so it looks like a time-limited thing.
There is an illegal Hub for Sci-ence that provides a lot of journal access, started in Kazakhstan. You shouldn't check it out, though.
On my opinion, a majority of recent math papers are available on the arXiv. Moreover, NSF requires that all NSF sponsored papers be deposited on another arxiv which is supposed to be available to all. I have never used this NSF source since I find most of what I need on arXiv.
@AlexandreEremenko the arXiv availability of articles varies widely by field. Some areas don't have as much arXiv coverage as you would think. It's very much a sampling bias effect. Also, it doesn't help for old papers! I'd be interested to know the NSF-mandated website. Is this related to CHORUS?
@David Roberts: The reason why some people DO NOT post their papers on the arXiv totally escapes me. Probably they write their papers not to be read but for some other purpose.
Yes, it is Chorus, https://par.nsf.gov/search/term:%22eremenko%22 I just checked: it says that my papers that I downloaded a week ago will be available on June 21 2021. The papers I uploaded a year ago are available.
@AlexandreEremenko I asked the category theorist Mike Barr why he doesn't post on the arXiv, yet posts all his papers and books on his website (and he is not exactly pro-commercial publisher, either). He said he doesn't want people to get their hands on his .tex files. I don't understand it either.
@David Roberts: I don't understand: can you access .tex files via the arXiv?? How? I only see .pdf files available there.
@AlexandreEremenko it's not a .tex file, but it is LaTeX: https://arxiv.org/e-print/1905.02537 so open it in your favourite editor (more generally, change /abs/ to /format/ in the url of any arXiv paper, and if the source is there, you can get it. I've grabbed good styles and macros this way)
Let us not also forget about the library genesis for books. "Slightly" illegal too, but you know...
Elsevier is giving remote access for university email accounts. At least my university email works. You have to just register using your university email
Something else you can try, if the article was published recently and the author is still active, is to email them and simply ask for a copy of the paper.
In my (rather limited) experiences, authors are usually more than happy to provide copies of their papers if you write to them personally. I've done this on a couple of occasions.
Let me try to summarize this long discussion in the comments. There are many free resources.
arXiv. It is true that not all mathematicians post their papers on the arXiv, for various reasons. But some of those who don't, post them on their personal sites. There are also other depositories, for example in Europe (see 5 below).
NSF depository. NSF's stated policy is that all results of NSF-sponsored research "must be available to the public at most in 1 year since their publication". I checked: the site is somewhat confusing but it works.
Many journals are freely available, and many more make their papers available after some time (usually 4-5 years). When you choose a journal to publish your paper, take this into account! Here is a convenient catalog of online journals. I am sure many other libraries have similar catalogs; I find this one convenient. The journals with free access are marked green, with partial access yellow and red. Access may depend on your location or on the date of publication.
Finally there are "pirate" sites. Some of them may have huge collections, larger than many university libraries. They frequently change their names and location. Some keywords may be "bookfi" or "genesis" for books, and "sci-hub" for journal articles. (Some of them may be illegal in some countries).
A simple search on Google and especially on Google Scholar sometimes finds what you need. It could be a place you do not expect. Some saved/cached copy. Some preprint depository that you do not know. Some personal web site, etc.
The very important resource is MathScinet, which unfortunately has no free version. But its German competitor Zentralblatt Math is partially free. Whatever you search there, it gives you only 3 items for free. But by clever choice of search criteria
you can obtain amazing results. For really old items, there is also
Jahrbuch which is free (it is a subset of Zentralblatt). For new papers, Google Scholar is excellent, especially if you know the author's name and title of the paper. It also sometimes finds you a free copy when available.
EDIT. I asked NSF, and they explained that all NSF-supported papers older than 1 year are
really available, though the site is somewhat confusing. One has to click on
the title of the paper, and then on a little square which says "pdf".
EDIT 2: I collected some links to free resources on my web page.
EDIT 3: Zentralblatt Math is now free for all, full version. Thanks to the German government.
Even if these sites frequently change their names and location, you can usually find a valid URL on their Wikipedia pages. (Your ISP may block that URL, though.)
Not just Jahrbuch, but EuDML is excellent for older articles from European journals (hence lots of Springer). It usually gives a link to a free copy if there is one.
According to a recent press release, Zentralblatt Math will become free by 2021.
@JonCC: this is a great news of course. But let us wait and see: I suspect that they may be talking about some reduced free version (of the kind they currently have).
Some are just too shy to actually give the concrete answer, so here it is:
Sci-Hub: https://sci-hub.st
Pro-tip: Go to the WikipediA page of Sci-Hub to keep up with the new domains of Sci-Hub (since they have to constantly change them owing to the cancel culture brought down by academic publishers).
If you're looking for most scientific papers fetched by Sci-Hub check LibGen: https://libgen.is/scimag/
(The pro-tip mentioned above applies here as well)
Pro-tip2: If for whatever reason you can't access that website then you can simply get around this by installing the Tor Browser and accessing it from there.
I wish you fulfilling and safe mathematical readings.
Sci-Hub is already mentioned in a comment, as well as (indirectly) in Alexandre Eremenko's answer. Directly linking copyright-violating sites on MO is also quite a contentious issue.
In this case, the OP using sci-hub would not represent a copyright violation. His university has already paid for subscriptions to the journals in question. In general, you are allowed to possess an electronic copy of a resource (book, article, journal) when you have already possessed or paid for a different copy. For example, if I have a copy of Dummit-Foote on my bookshelf in my office, I can legally download a pdf of it. I think the same applies here.
Here is another pro tip: very often you internet provider will not let you access any of the said "pirate" websites and prompt instead an error message. To circumvent this it's enough to change your DNS server (use e.g. google's one or whichever fits you best). A quick search on internet will teach you how to do that in a very few clicks, it's really simple and a life-changer (mathematically speaking, I mean, I never have anymore to log into any "nonpirate" VPN from my current institution or whatever to access the papers I need).
@leomonsaingeon Thanks for mentioning that. One comment though is that it's not always possible for example when countries like Russia block individual IPs as well. I think the safest bet here is to use the Tor Browser.
Even when you do have paid access, Sci-Hub is usually faster. Making a publisher's website recognize your institutional subscription can be a ridiculous rigmarole. But with Sci-Hub you just paste in the DoI and up comes the PDF. Perfect.
"His university has already paid for subscriptions to the journals in question." What university? He's a graduate, so he's not affiliated with them anymore.
@Carl-FredrikNybergBrodda: The mentions in comments and other answers are euphemistic and unclear; if illegal sites to be discussed at all, it seems better and more professional to do so frankly and explicitly. It’s good to consider the legal and ethical issues, as discussed in the meta question you linked, and to avoid such sites for literature links in ordinary MO questions/answers. But on a question like this one, where the question itself is “what such sites exist”, it seems pretty denialist to avoid them.
@PeterLeFanuLumsdaine I think that's a good way to put it, and aligns with my personal view on the matter. I think the main point of contention lies in linking the sites, rather than avoiding talking about them as if they don't exist. Personally, I don't mind seeing links or discussion, but would probably like to see a disclaimer added about the ethical/legal issues surrounding it. My comment was mostly in this direction.
@Carl-FredrikNybergBrodda FWIW WikipediA also links to them, why should that be a problem to SE when WikipediA already does it without any issues? (Just linking something doesn't mean one is supporting it, endorsing it, encouraging its use, ...etc)
@user161087 why are you capitalizing the final "a" in Wikipedia?
@PeterLeFanuLumsdaine Absolutely. Plus, let's not be too hasty in calling Sci-Hub "illegal". None of us are lawyers. Just because someone like an Elsevier spokesperson claims it's illegal, doesn't mean it is. Just because some ISPs have been pressured to block access, doesn't mean it's illegal. And just because it's illegal in your country, doesn't mean it's illegal in other people's.
@JoePrevidi Sorry I took their logo too literally, yes "Wikipedia" is the correct version of their name.
@user161087 I do not think it is a problem. Others do, as indicated in the link. Others don't. This has been a contentious issue for a while, and I think it is important to mention that it has been.
@TomLeinster I would guess that violating copyright millions of times is illegal in most jurisdictions, and courts have very definitely decided against Sci-Hub in the US. Some hold the position, though, that providing access to the scientific and research literature to those that cannot otherwise get it, and especially at a time of a global health crisis, is the moral thing to do, despite the law.
Also, note there is a fine line between accessing something the library already has (and so issues around legality of the copy become much less clear), and "accidentally" accessing something it doesn't have. If my library has a book in an inaccessible crate in a long-term storage facility (don't ask), and will have to fly in a copy from interstate so I can check if it is what I need, then I feel no compunctions about checking it out on LibGen. It saves money, time and CO2. If I look up something on Sci-Hub my uni does not have access to, then I feel I need to do so knowing that.
@DavidWhite Unless the copyright holder has explicitly granted you the right to other versions, including an electronic version, then you most definitely do not have a right to an electronic copy just from owning a print copy of something. Publishers may give a deal where you get both a print and electronic version, but it's certainly not the default, and there's nothing inherent that grants you the right to a copy in another format. For the specific situation you describe, you may have a right to an electronic copy, but it's not something you, or anyone, should assume you/they have.
@Makyen: I strongly disagree. If you want to change my mind, I'll need more than a MathOverflow comment. Perhaps you have some sources you could cite? I can dig around for the source where I learned the fact I stated in my comment.
@DavidWhite When you purchase something, you obtain ownership of it and have ownership rights in that specific copy. Making that purchase doesn't inherently grant you rights to something else, unless that was part of the purchase contract. You're asserting you automatically get a right to something else. That's the positive position. It's far easier to prove a positive, than the negative, so please find something that at least indicates you're correct for the general case, not just your one purchase. If you want a more detailed answer from from a legal POV, you could ask on [law.se].
@Makyen: I don't come on MathOverflow to argue with strangers about the law. I come here to do mathematics. I am confident that, when I own a copy of a math paper (e.g. from library download), I can freely download as many other copies of that paper as I like, from any source. When my co-author asks a question, and it's answered in a paper, I freely email him a copy of that paper, knowing that he could just as well get it at his library. Slavish fear of publishing companies should not stand in the way of research getting done. I'm done with this interaction now. Have a nice day!
Sci-hub also has a nice telegram bot @scihubot, you just paste doi and it gives a paper. No need to look for currently not closed mirror etc.
The two answers posted summarize most of the comments, but miss two important ones.
Go to the University of Chicago library website. Click "My Account" in the upper left. Login with your University of Chicago credentials. Contact your librarian if you struggle with this. Once you are logged in, you can access many articles electronically (and also MathSciNet, but that's not what you asked about).
Use a VPN. This will make it appear to the University of Chicago servers that you are on campus, and you can then access library resources as usual. Instructions on how to do this at UChicago are here.
All that said, I also think you are not violating copyright if you find pdfs of articles from other sources in this situation, even so-called "pirate" sources, because you already have the rights to the articles in question through your affiliation to UChicago.
None of these options will work: I have just graduated from the University of Chicago
I still have access to my PhD university library, with the same username and password. For grad students, I thought it was standard practice to keep your old login information?
@DavidWhite Alumni access for bachelor's graduates at my university, at least, is nothing more than what the public is allowed; although to be fair the sample size (1) is small. I did not downvote your answer, and do still agree with the idea of asking until someone says no.
I'm not from a related field, but you might also find a lot of stuff quite comfortably via the Firefox add-on 'unpaywall'.
This is just to confirm that, as announced earlier here, https://zbmath.org/ became open in 2021.
|
2025-03-21T14:48:31.504638
| 2020-07-13T22:03:47 |
365570
|
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|
Stack Exchange
|
Using Young tableaux with weights for writing down character formula
I have been learning about Young tableaux and I was thinking about the simplest possible use for Young tableaux with weights.
There's a formula for the character of an irreducible representation of $S_n$ corresponding to a Young tableau of shape $\lambda=(\lambda_1,\dotsc,\lambda_l)$, evaluated on the conjugacy class given by a $\mu=(\mu_1,\dotsc,\mu_m)$ is given by a coefficient of a multinomial expansion involving $\mu$ and $\lambda$.
The formula is $\chi_\lambda(\mu)=$ coefficient of some monomial in $\prod_{1 \leq i<j \leq l} (x_i-x_j) \prod_{i=1}^m (x_1^{\mu_i}+\dotsb+x_l^{\mu_i})$.
This suggests that one might be able to rewrite this formula using weighted Young diagrams where some of the squares are weighted according to $\mu$. (Maybe one of the squares is weighted $\mu_i$ times or something).
Do any of you know how to do this/if there's a place where this is done?
This character is essentially by definition a Schur polynomial, which is well-known to be a generating function for semistandard Young tableaux. I'm confused as to what the question is...
I made a few edits. Just as a note: 'Young' is a name, so it's usually capitalised. 'Tableau' is singular and 'tableaux' is plural. Use $\prod$ $\prod$ rather than $\Pi$ $\Pi$ for products (and similarly $\sum$ $\sum$ rather than $\Sigma$ $\Sigma$ for sums). (Compare $\displaystyle\prod_{i = 1}^m i$ $$\prod_{i = 1}^m i$$ versus $\displaystyle\Pi_{i = 1}^m i$ $\Pi_{i = 1}^m i$.)
In your formula for $\chi_\lambda(\mu)$ , you need to take the coefficient of $x_1^{\lambda_1+l-1}x_2^{\lambda_2+l-2}\cdots x_l^{\lambda_l}$. The combinatorial interpretation of $\chi_\lambda(\mu)$ that follows from this is just the Murnaghan-Nakayama rule. This is explained, for instance, in Section 7.17 of my book Enumerative Combinatorics, vol. 2.
I think the question should be made more precise. The formula is incorrectly stated and it is $\lambda$ not $\mu$ that refers to the partition (not tableau). Have voted to close.
The murnaghan narayana rule is essentially what you ask for. A sum over certain tableaux...
@user61318 Ah, damn autocorrect! In any case, here is the MN-rule: https://www.math.upenn.edu/~peal/polynomials/schur.htm#schurMurnaghanNakaygama
Thank you Richard, Per and Sam for the helpful comments.
And thanks for the patience. This is not my primary field. I have been working mostly in homotopy theory for the past 5-6 years or so.
|
2025-03-21T14:48:31.504846
| 2020-07-13T22:27:49 |
365572
|
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|
Stack Exchange
|
Fast computation of linear equation with row and column removed
Given $A, x, b$ and a linear system $Ax=b$, where $A\in\mathbb{R}^{n\times n}$ is full rank. Denote $A_{\backslash i}$ as a $(n-1)\times(n-1)$ matrix where $A$ removes its $i^{th}$ column and row, similarly $b_{\backslash i}$. What's the most efficient way to find a $y$ for each $i\in[n]$ s.t $A_{\backslash i}y=b_{\backslash i}$? (So in total these are $n$ linear systems, for different $i$ I want a different solution $y$).
I know one way is to use Sherman-Morrison formula, but is there any way that doesn't involve calculating $A^{-1}$? That is, suppose I have $x$ calculated, can $y$ be derived just using $A, x, b$?
When you write, "Given a linear system $Ax=b$," do you mean, given all three of $A,x,b$? When you write "find $A_{\backslash i}y=b_{\backslash i}$", do you mean "find $y$"?
Yes to both questions, I have modified the question.
Why don't you want to use $A^{-1}$ (or an equivalent factorization)? It seems the right tool for the job here.
Such $y$ may not exist.
@FedericoPoloni Because my $A$ is huge and inverse takes too long
@MaxAlekseyev Maybe I wasn't being super clear in the OP. I want to find different $y$ for different $i$. So in total, there are $n$ solutions $y$. They should always exist because $A_{\backslash i}$ is full rank.
So is your matrix sparse or dense? How big is "huge"? How would you solve linear systems with $A$ or $A_{\setminus i}$ normally; with a direct or an iterative method?
How would you use Sherman-Morrison formula?
@user3799934: Consider matrix $A=\begin{bmatrix} 0 & 1\ 1 & 0\end{bmatrix}$ and $x=b=\begin{bmatrix} 1\ 1\end{bmatrix}$. What is $y$, say, for $i=1$?
@MaxAlekseyev I see. For my particular case it's safe to assume that $A_{\backslash i}$ has full rank for any $i$.
@FedericoPoloni it's roughly 80k*80k and dense. The issue is mostly with memory, I assume existing linear system solvers use iterative methods? Taking inverse seems to load the whole matrix into memory.
@VítTuček Taking a row and column off is a rank-2 update so I can write $A_{\backslash i}=A+UV$ for some $U, V$ (it's slightly different because here i'm setting rows/columns to 0 rather than deleting them). Then apply the woodbury matrix identity (or maybe just apply Sherman morrison twice). This concerns me because it seems to require calculating $(I+UA^{-1}V)^{-1}$ $n$ times because $U, V$ vary.
@user3799934 Ouch; that seems a difficult problem by every metric. I don't think you can avoid essentially running a dense $O(n^3)$ inversion algorithm to solve that problem. You wish to solve various $(n-1)\times(n-1)$ linear systems anyway, so there's no getting around that.
@user3799934 Keep in mind that $I+UA^{-1}V$ is $2\times 2$, so that computation is cheap; the SMW update procedure costs $O(n^2)$ per linear solve. I'd guess that you can do the whole procedure (producing $n$ different vectors $y_{\setminus i}$) in something like $6n^3$ flops with SMW. The obvious lower bound is $\frac23 n^3$ flops for solving a single $(n-1)\times(n-1)$ linear system (barring Strassen-like tricks), so there is at most a small multiplicative factor to gain here, at least thinking in terms of flops.
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2025-03-21T14:48:31.505093
| 2020-07-13T22:29:02 |
365573
|
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|
Stack Exchange
|
For an integral domain $R$ when does $a^2 \equiv b^2 \bmod 4R$ imply $a \equiv b \bmod 2R$?
Suppose we have $a^2 \equiv b^2 \bmod 4R$ where $R$ is an integral domain. Under what conditions on $R$ can we conclude that $a \equiv b \bmod 2R$?
This would hold if $2 \in R$ is a prime or the product of distinct comaximal primes. This can fail when $R$ is not integrally closed: For $R = \mathbb{Z}[\sqrt{-3}]$ we have $(4+\sqrt{-3})^2 \equiv 1^2 \bmod 4R$ but $4 + \sqrt{-3} \not\equiv 1 \bmod 2R$.
Does this hold in a GCD domain? Does it hold, more generally, in an integrally closed domain?
Denote $a-b=x$, then $a^2-b^2=x(x+2b)=4z$ for $z\in R$. Assuming that $R$ is integrally closed, we see that $(x/2)^2+b(x/2)-z=0$, so $x/2$ is an algebraic integer, thus $x/2\in R$.
Does this work for cubes and higher powers, or is this one of those examples where squares are special?
Which claim for cubes do you mean? "If 9 divides $a^3-b^3$, then 3 divides $a-b$"?
Yes. For an integrally closed integral domain $R$ do we have $a^3 \equiv b^3 \bmod 9R \implies a \equiv b \bmod 3R$? Do we have more generally for any $n > 2$ there exists an exponent $e$ such that $a^n \equiv b^n \bmod n^e R \implies a \equiv b \bmod nR$.
What about $a=1,b=w=e^{2\pi i/3}$ in $\mathbb{Z}[w]$?
Yes, and this obviously generalizes: For $n > 2$ let $R = \mathbb{Z}[\theta]$ where $\theta = e^{2\pi i/n}$. Now for any $e \geq 1$ we have $1^n \equiv \theta^n \bmod n^e R$ but $1 \not\equiv \theta \bmod nR$. So this is special to squares.
|
2025-03-21T14:48:31.505219
| 2020-07-13T22:43:07 |
365574
|
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|
Stack Exchange
|
Galois representations and pro-étale Site
On a scheme, we can define the pro-étale site. This is an improvement over the étale site in that we can define the $\ell$-adic cohomology as the sheaf cohomology of the constant sheaf $\underline{\mathbb{Z}_\ell }$ instead of as a limit. We can also define $\mathbb{Z}_\ell(1):=\lim \mu_{\ell^n}\in \text{Ab}(X_{\text{proét}})$ (see Remark 5.2.4 in Bhatt and Scholze, The pro-étale topology for schemes, arXiv:1309.1198). Assume that $X$ is defined over some number field $k$ and is smooth and projective. My question is if
$$H^n_{\text{proét}}(X_{\bar{k}},\mathbb{Z}_\ell(i))\cong H^i_{\text{ét}}(X_{\bar{k}},\mathbb{Z}_\ell(i))$$
as Galois modules.
I think you mean $H^i_{\operatorname{(pro)\acute et}}(X_{\bar k}, \mathbf Z_\ell(i))$ ― you need to base change to the algebraic closure to get a Galois action. Also, what are your assumptions on $X$? Finite type; smooth; etcetera?
Oh, yes, you are absolutely right! I'll change the post. I'm happy to assume $X$ to be smooth projective, and of course other conditions if necessary.
It depends on what you mean by the right side. If you mean the classically defined $\ell$-adic cohomology groups (as an inverse limit), then this is true essentially by definition (see section 5.6 of Bhatt-Scholze for a more general statement). If instead you mean the 'etale cohomology of the constant sheaf attached to the abelian group $\mathbf{Z}\ell(i)$, then this is quite false (in degree $1$, this amounts to observing that there are typically many more maps $\pi_1^{et}(X) \to \mathbf{Z}\ell$ if we give the target the $\ell$-adic topology than if we use the discrete topology).
Thanks! In section 5.6 of Bhatt-Scholze, the just have a "fixed scheme". Does that for instance mean that be can also consider a finite field?
Yup, the derived inverse limit construction $H^i(X, R\lim \mathbf{Z}/\ell^n)$ coming out of Jannsen's theory (or the Bhatt-Scholze paper) agrees with the classical inverse limit definition $\lim_n H^i(X, \mathbf{Z}/\ell^n)$ over a finite field: by the Milnor sequence, the obstruction to agreement comes from $\lim^1 H^{i-1}(X, \mathbf{Z}/\ell^n)$, which vanishes over finite fields.
|
2025-03-21T14:48:31.505392
| 2020-07-13T23:45:14 |
365576
|
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|
Stack Exchange
|
Does a rank 1 CAT(0) space with a proper cocompact group action contain a zero width axis?
A geodesic in a proper CAT(0) space is said to be rank 1 if it does not bound a flat half-plane and zero-width if it does not bound a flat strip of any width.
Let $X$ be a geodesically complete CAT(0) space that contains a rank 1 geodesic. Assume it admits a properly discontinuous cocompact action by a group $G$.
Under these conditions we know:
$X$ contains a rank 1 geodesic which is an axis of an isometry in $G$ ([Link, Lemma 4.2]
https://arxiv.org/pdf/1706.00402.pdf).
$X$ contains zero-width geodesic ([Ricks, Theorem 2] https://arxiv.org/abs/1410.3921).
My question: Must $X$ necessarily contain a zero-width geodesic which is the axis of some isometry in $G$?
Maybe this is built in to your definition, but do you want to assume that $X$ is irreducible? Otherwise I think you can just take a product with an interval to get a counterexample (and to Ricks' theorem).
I believe this would not be geodesically complete--not all geodesic segments would be extendible to infinity
Good point, thanks, I assumed I was missing something.
The question has a positive answer at least for CAT(0) cube complexes. Caprace and Sageev constructed isometries skewering pairs of strongly separated hyperplanes, and the axis of such an isometry must have zero width.
|
2025-03-21T14:48:31.505519
| 2020-07-14T00:00:13 |
365578
|
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|
Stack Exchange
|
Arthur's Simple Trace Formula
In Deligne–Kazhdan–Vigneras's "Représentations des groupes réductifs sur un corps local," they use the Simple Trace Formula to prove cases of the local Jacquet–Langlands correspondence over nonarchimedean fields.
Let's recall some setup for the Simple Trace Formula. Let $F$ be a global field, let $G$ be a connected reductive group over $F$, write $Z$ for the center of $G$, and let $\omega$ be a unitary character of $Z(F)\backslash Z(\mathbb{A})$. Write $L^2(G,\omega)$ for the space of $L^2$-functions on $G(F)\backslash G(\mathbb{A})$ where $Z(\mathbb{A})$ acts via $\omega$, and write $L^2_0(G,\omega)$ for the subspace of cusp forms.
In what follows, $v$ will range over places of $F$. Let $f=\prod_vf_v$ be a function on $G(\mathbb{A})$ on which $Z(\mathbb{A})$ acts via $\omega^{-1}$ such that every $f_v$ is a smooth function on $G(F_v)$ with compact support mod $Z(F_v)$. Such $f$ naturally yield linear operators $\rho_0(f)$ on $L^2_0(G,\omega)$. Under extra assumptions on $f$, the Simple Trace Formula says that
$$\operatorname{tr}\rho_0(f) = \sum_\gamma\operatorname{vol}(Z(\mathbb{A})G_\gamma(F)\backslash G_\gamma(\mathbb{A}))\int_{G_\gamma(\mathbb{A})\backslash G(\mathbb{A})}dg\,f(g^{-1}\gamma g),$$
where $\gamma$ ranges over elliptic regular conjugacy classes in $Z(F)\backslash G(F)$, and $G_\gamma$ denotes the centralizer of $\gamma$ in $G$.
Usually, the extra assumptions on $f$ are that $f_v$ is supercuspidal at one place $v$ and $f_{v'}$ is supported on the elliptic regular elements of $G(F_{v'})$ at one place $v'$. However, in part e. of the introduction of "Représentations des groupes réductifs sur un corps local," Deligne–Kazhdan–Vigneras say that Arthur announced the Simple Trace Formula also holds when $f_v$ and $f_w$ are supercuspidal at two places $v$ and $w$.
Question: Does the Simple Trace Formula indeed hold in this case? If so, where could I find a proof? Thank you in advance!
Yes, I have not known Deligne, Kazhdan and Vigneras to lie. A sketch of the proof, at least with the key details for GL(2), is given in Lecture V of
Steve Gelbart, Lectures on the Arthur--Selberg Trace Formula
Added remarks: In that Lecture, Gelbart addresses both two kinds of simple traces formulas. The one you are asking about is essentially Prop 2.1. While Gelbart states the hypotheses in terms of vanishing of hyperbolic orbital integrals at two places, in the proof he explains how this is related to vanishing of unipotent orbital integrals, which is the condition to be a supercusp form at 2 places. As I remember, while technically he doesn't state that being a supercusp form at 2 places suffices, you should be able to work this out from what he does, at least in the case of GL(2). You'll probably need to look at papers of Arthur, Deligne, Kazhdan, Vigneras, etc for details general G. For me personally, reading Rogawski was also useful for understanding these things.
Thanks for the reference! However, it seems the Simple Trace Formula is proved there (Theorem 3.1 of Lecture V) under the assumption that f is supercuspidal at one place and supported on elliptic regular elements at one place. Instead, I'd like the assumption to be that f is supercuspidal at two places.
@CharlesDenis I added more remarks to address your comment. I'm sure I knew more precise references 10-15 years ago, but unfortunately I have no long-term memory for details.
|
2025-03-21T14:48:31.505744
| 2020-07-14T00:31:51 |
365580
|
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|
Stack Exchange
|
Galois embedding question for dihedral groups
Let $D_n$ be the dihedral group of order $2n$. Then all the quotients of $D_n$ are dihedral as well, and of the form $D_k$ with $k \mid n$. So for a field $K/\mathbb{Q}$ with $\operatorname{Gal}(K/\mathbb{Q}) \cong D_n$, there exists, for any $k \mid n$, a subfield $F \subseteq K$ with $\operatorname{Gal}(F/\mathbb{Q}) \cong D_k$.
My question is about the reverse question. Given a number field $F/\mathbb{Q}$ with $\operatorname{Gal}(F/\mathbb{Q}) \cong D_k$, is there a field $K \supset F$ such that $\operatorname{Gal}(K/\mathbb{Q}) \cong D_n$ for any $n$ a multiple of $k$?
I've been told that this is called the "Galois embedding problem" and is not true for many types of groups. I was wondering if anyone could point me in the right direction for what is known about this in the dihedral case.
Thanks,
MC
The answer is "no", in general, since there may be local obstructions. Suppose, for example, that $k$ and $n$ are odd prime powers, and let $L/\mathbb{Q}$ be the unique intermediate quadratic in $F$. A necessary condition for the existence of $K$ is that the cyclic degree $k$ extension $F/L$ embeds into a cyclic degree $n$ extension $K/L$. Every prime $\mathfrak{p}$ of $L$ that is totally ramified in $F$ would have to be totally ramified in $K$, so if the residue characteristic of $\mathfrak{p}$ is coprime to $n$, then you need $n$ to divide the order of the multiplicative group of the residue field of $\mathfrak{p}$. This is a genuine restriction. For concreteness, take $k=3$, $n=9$. Then there are infinitely many $D_3$ extensions of $\mathbb{Q}$ in which $7$ is split in the quadratic and ramified in the cubic, but none of them embed inside a $D_{9}$ extension, because $7$ cannot be totally tamely ramified of degree $9$.
You can upgrade such local conditions to also ensure that the bottom cyclic group of order $2$ normalises the top group and acts on it by $-1$, so that the whole extension is dihedral. If all such local conditions are satisfied, then you can prove with class field theory that the sought-for embedding always exists. In particular if $k$ and $n/k$ are coprime, then the embedding will always exist. See [1, Section 3.1] for some examples of how such constructions work.
[1] A. Bartel, Large Selmer groups over number fields, Math. Proc. Cambridge Philos. Soc. 148 no. 1 (2010), pp. 73–86.
Thanks, that makes a lot of sense. Thank you for the reference as well, that's precisely what I was looking for.
@MC: You are welcome!
|
2025-03-21T14:48:31.505950
| 2020-07-14T00:43:28 |
365581
|
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|
Stack Exchange
|
Algebraic Space: Two equivalent constructions
According to Wikipedia
there are two common ways to define algebraic spaces:
they can be defined as either quotients of schemes by étale
equivalence relations,
or as sheaves on a big étale site that are locally isomorphic
to schemes.
Namely:
I) a la Knutson:
An algebraic space $X$ comprises a scheme $U$ and a closed subscheme
$R \subset U \times U$ satisfying the following two conditions:
$R$ is an equivalence relation as a subset $U \times U$;
the two projections $P_i: R \to U$ onto each factor are étale.
Knutson adds an extra condition that the diagonal map is quasi-compact.
II) as a sheaf:
An algebraic space $\mathfrak {X}$ can be defined as a sheaf of sets
$$\mathfrak {X}:(\operatorname{Sch}/S)^{\text{op}}_{\text{ét}} \to \operatorname{Sets}$$
such that
There is a surjective étale morphism $h_X \to \mathfrak {X}$;
the diagonal morphism $\Delta _{{\mathfrak {X}}/S}:
\mathfrak {X} \to \mathfrak {X} \times \mathfrak {X}$
is representable and quasicompact (thanks to David's careful remark).
(Rmk: in II)1. we identified a scheme $X$ with its image $h_X$ wrt the Yoneda
embedding $X \to \operatorname{Hom}(X,{-})$.)
Two questions:
About construction I). Wikipedia moreover says that if $R$
is the trivial equivalence over each connected conponent of $U$
(i.e. for all $x,y \in U$ lying in same component then
$xRy$ iff $x=y$) then the so defined algebraic space is a scheme
in the usual sense. Why?
Where I can find a proof/ reason that the constructions
I) and II) are indeed equivalent?
Regarding 1. what do you think is the algebraic space associated to a scheme? The word 'is' here is a little misleading. It's really "is the image of a scheme under the inclusion of schemes into algebraic spaces". Regarding the equivalence, I don't have a reference, but if you want the Knutson definition, to correspond exactly, you might need to add $\Delta_{\mathfrak{X}/S}$ quasicompact to II). I assume this is all in the Stacks project somewhere. It looks like most of what you want is in https://stacks.math.columbia.edu/tag/0264
@David Roberts: Hi. One nitpick on your first point: yes indeed if $U$ is a scheme and we want to consider it as algebraic space we endow it with trivial relation $R$. About the 2. condition in I I'm not sure. Why are in this scheme case the projections $p: R \to U$ etale? Sure, set theoretically we have in this case $\vert R \vert = \vert U \vert$ since $R$ is as a set the diagonal. But how we conclude etaleness of $p$ as a morphism schemes?
You take $R = U$ as a scheme, and $R \to U\times U$ the diagonal, so that $R\to U\times U \xrightarrow{pr_i} U$ is the identity map for $i=1,2$. (Also, the connected component stuff is irrelevant.)
I think "$R$ is an equivalence relation as a subset $U\times U$" is misleading. It should be an equivalence relation internal to the category of schemes. If this is equivalent to saying it is so on the underlying set of the locally ringed space(s), then fine. Otherwise, it should be changed.
About the "equivalence relation" you are of course right. So essentially
we consider the affine scheme $R$ as categoretical equivalence relation,
ie for all $T \in (Sch)$ the set $Hom(T,R) \subset Hom(T, U \times U)=
Hom(T,U) \times Hom(T,U)$ is the equivalence relation in usual sense. That's what you meant by "internal relation", right?
About the choice $R:=U$. I'm not sure. It seems that the
image $\Delta(U)$ of $\Delta: U \to U \times U$) aka the "diagonal" (schematic sense)
which is always locally closed subscheme of $U \times U$ looks like a more "natural" choice for $R$. Or do you see a reason why this is a bad choice in order to show the claim that a scheme is always an algebraic space.
Do you maybe know an argument how to show that the projections $\Delta(U)
\subset U \times U \to U$ is etale?
I would check the Stacks Project and go with what is there.
|
2025-03-21T14:48:31.506219
| 2020-07-14T03:24:08 |
365584
|
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|
Stack Exchange
|
Is there a difference between using nats and bits to express entropy?
It seems to me like questions involving decimal vs binary representations of some number are not particularly interesting: for instance $\pi$ or $\sqrt{2}$ are conjectured to be normal in every base, and as far as I know this is open for any particular base.
On the other hand, in calculating entropy there is again a choice of basis. Further, this gives us certain 'distinguished' real numbers: e.g. the entropy of the Gauss-Kuzmin distribution is $3.432527514776...$ bits, while it is $2.379246769061...$ nats.
Are the properties of these digit strings of the same number 'similar' in any way, or is one 'nicer' in some sense?
this is a bit vague. does one of those two numbers you gave look nicer to you in some sense? is that why you asked?
I also don't understand the question.
I understand the question as follows: some properties of numbers (such as normality) are basis-independent, others are basis-dependent (such as entropy). In the latter case, has the mathematical community come up with arguments in favor of some preferred basis in which to work?
|
2025-03-21T14:48:31.506335
| 2020-07-14T07:33:14 |
365588
|
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|
Stack Exchange
|
GAGA for vector bundles over Riemann surfaces
Serre’s GAGA theorem gives an equivalence of categories between algebraic and analytic coherent sheaves over a complex projective variety. The proof relies on the finiteness of the cohomologies of coherent analytic sheaves over compact complex manifolds, which is a non-trivial analytic result.
I was wondering that maybe GAGA theorem could be proven quite more easily in the case of vector bundles over a Riemann surface $X$ by using the existence of meromorphic functions as the underlying analytic result and “Grothendieck” theorem about the classification of vector bundles on the Riemann sphere.
The idea is to consider a meromorphic function $f:X \rightarrow \mathbb{P}^1$ and, for any vector bundle $E\rightarrow X$, consider the pushforward $f_*E$, which is a vector bundle over $\mathbb{P}^1$, which by the Grothendieck theorem is algebraic.
My question is if this idea can be used to show that, since $f_*E$ is algebraic, then $E$ is also algebraic.
Well, the existence of non-zero meromorphic functions is itself a non-trivial analytic result.
@FrancescoPolizzi Of course, but imho is quite more simple than the other one I mention.
$f_*E$ is not locally free in general
Yes it is, since $f$ is finite and flat and so $R^if_*E=0$ for all $i >0$.
You're right, thank you
I think this can work. There is a natural transformation $E \to f^{\ast} f_{\ast} E$, and I think that in the curve case I can show that it is injective and $E$ even splits off as a summand. Your argument easily shows that $f^{\ast} f_{\ast} E$ is algebraic, so we'd want to know that a summand of an algebraic bundle is algebraic.
|
2025-03-21T14:48:31.506491
| 2020-07-14T09:11:28 |
365592
|
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|
Stack Exchange
|
Proof in the realm of $\infty$ categories
I have recently started learning the language of $\infty$-categories. My approach is more to their use, rather than for their own sake. For this reason, as I feel I reached a good understanding of the basic notions (the definition, functors, adjoints, stable categories, etc.), I have now turned my attention to some papers which work in the framework of stable $\infty$-categories to solve the functoriality issues we encounter when we work with triangulated categories. However, I am having problems understanding what can be proved and what can't using techniques similar to the triangulated world. For example, I am reading Noncommutative homological projective duality by Alexander Perry (arXiv:1804.00132), and in the first part some results about $S$-linear categories (where $S$ is a scheme satisfying some assumptions) are proved. The formal definition is that the categories of $S$-linear categories is the categories of module objects $\text{Mod}_{\mathfrak{Perf}(S)}(\text{Cat}_{\text{st}})$, where $\text{Cat}_{\text{st}}$ is the $\infty$-category of stable, idempotent complete $\infty$-categories with exact functors. The author then wants to prove that if we have an $S$-linear functor $\phi : \mathcal{C} \rightarrow \mathcal{D}$ between $S$-linear categories which has a left adjoint when considered as a plain functor of $\infty$-categories, then $\phi^{L}$ is automatically an $S$-linear functor. The proof which is given in the paper uses the internal hom for $S$-linear categories, and the adjunction properties
$$
\mathcal{H}om_{S}(\phi^L(C \otimes F), D) \simeq \mathcal{H}om_{S}(C \otimes F, \phi(D)) \simeq \mathcal{H}om_S(C, \phi(D) \otimes F^{\vee}) \simeq \dots \mathcal{H}om_S(\phi^{L}(C) \otimes F,D)
$$
where $C \in \mathcal{C}, D \in \mathcal{D}, F \in \mathfrak{Perf}(S)$, and $\otimes$ denotes the module structure. Then the proof is concluded using the Yoneda lemma.
This is the proof I would write if I were working in the triangulated setting, and I don't understand why one is allowed to use it in the $\infty$-categorical framework as well. Adjunction in this case as far more tricky (even if they imply the obvious isomorphism on mapping spaces), and the structure of $S$-linear functor implies $\phi^L(C \otimes F) \simeq \phi^L(C) \otimes F$ up to contractible choices, but it is not equivalent to this isomorphism, is it? It would be very helpful is some could help me understanding to which point one can transport proofs from the triangulated world to the stable $\infty$-categorical framework. Thanks.
That proof doesn't give the full details, but it doesn't do so in the 1-categorical world either: being S-linear is additional structure on a functor, not just a property. However, for the left adjoint of an S-linear functor you have canonical oplax structure maps $\phi^L(C \otimes F) \rightarrow \phi^L(C) \otimes F$ (adjoint to the map $C \otimes F \rightarrow \phi\phi^L(C) \otimes F \simeq \phi(\phi^L(C) \otimes F)$ defined by the unit of the adjunction and the linearity of $\phi$). So what you're really showing is that this particular map is an equivalence.
The same is true in the $\infty$-categorical setting, it's just a bit trickier to set up the oplax structure maps in a coherent way. In this particular case (and many others) this is best dealt with using cocartesian fibrations (for the details see Corollary <IP_ADDRESS> in Higher Algebra).
Ok, I see, so we're taking that map and we are showing that it is an equivalence, thus getting the linearity. I know the following might be a too broad question to answer, or it might be nonsense, but how many of these details are generally swept under the rug? Because most of the papers I came across prove these statements as they were in the triangulated world, which probably means there is something behind which ensure things can't be so bad.
This illustrates a common feature of working with $\infty$-categories: if you're checking a property you can often reason just as in the 1-categorical setting, but to obtain structures you usually have to figure out a canonical reason why they exist (though this can often give cleaner proofs for 1-categories too than you get by writing them out explicitly).
I see, thank you for your clear answers!
@Federico My impression is that many details are swept under the rug in the triangulated world as well, it's just that some people feel more confident in doing so in that setup (although it's not really clear to me why...)
@Dennis You're probably right, it might be because triangulated categories have been around for longer, and people are more accostumed to their subtleties. I'm finding the framework of $\infty$-categories hard to get into because there is a huge amount of things one needs to know (even finding your way through the many references to Lurie's books sometimes isn't easy).
@Denis Unfortunately that is true. However one difference that is relevant to me (an outsider) is that claims in the triangulated setting are often strictly weaker, since all issues of homotopy coherence have been erased. These coherence issues are precisely what $\infty$ categories are designed to handle, so if a statement in an $\infty$ category paper implicitly uses a nontrivial coherence result from a late chapter of HA, then I really think it should be referenced. Otherwise, it's unclear where the work of upgrading the result went.
@Federico I don't think its right to say stable $\infty$ categories "solve the functoriality issues we encounter when we work with triangulated categories." They are just two different frameworks, and one is not inherently better than the other. I could just as well say "triangulated categories solve the problem of $\infty$-categories being difficult to learn about" or "Spanish solves the problem of English having all sorts of exceptions to every rule." I hope you see my point. Value judgments should not be used in such situations.
@Federico I would say the thing that is generally swept under the rug is why two maps are the same. i.e. you claim that a map is an isomorphism because its a map between isomorphic objects. This problem clearly appears in the proof you presented, and it is the same to infinity categories as to 1-categories. It is just culturally excepted to ignore it in 1-categorical settings for basically no good reason. Once you note it, the difference between the theories become much smaller in practice I believe.
@S. carmeli Obviously, it is a bad practice in both settings to assume that two different maps $f,g: X \to Y$ are the same. Mistakes of this sort have invalidated papers before! But this cultural practice is more abusive when done in the $\infty$ setting, because it hides significantly more. Maps are not simply equal or unequal, but the datum of equality needs to be specified, and shown to be consistent with all the other structure in play (which in turn requires introducing even more data).
As illustrated in this discussion, the details are often clear to experts. But as $\infty$-categorical methods are (for good reason) adopted by a greater number of non-experts, I (as a novice) worry when these issues are brushed under the rug.
@DavidWhite I see your point. I wrote that phrase because of the way I think of $\infty$-categories and pretriangulated dg-categories, i.e. as frameworks which give us a functorial notion of cone. However, you are right, it is definitely just a choice of framework.
@PhilTosteson In abstract you're right, but in my experience in practice the adjustments one need to make the proof correct are not that different in the 1-categorical and in the ∞-categorical cases. Typically when you cannot write an ∞-categorical proof and you go back to the 1-categorical case to figure out what goes wrong, you'll find that your 1-categorical proof was also incomplete in the same way... And I am totally convinced that there are all sorts of (mostly unimportant) mistakes in the literature while using 1-categories.
|
2025-03-21T14:48:31.507117
| 2020-07-14T09:24:41 |
365593
|
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Stack Exchange
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Coupling times of subordinate Brownian motions
This is a question about coupling times of subordinate Brownian motions.
We fix $y \in \mathbb{R}^d$ with $y \neq x$ and define a map $R_{x,y} \colon \mathbb{R}^d \to \mathbb{R}^d$ by
\begin{align*}
R_{x,y}(z)=z-2 (z-(x+y)/2,x-y)\frac{x-y}{|x-y|^2},\quad z \in \mathbb{R}^d.
\end{align*}
We note that $R_{x,y}$ is the reflection with respect to the hyper plane $H_{x,y}$ such that the vector $x-y$ is normal with respect to $H_{x,y}$ and such that $(x+y)/2 \in H_{x,y}$. We write $B^x=(\{B_t^{x}\}_{t \ge 0}$ for the $d$-dimensional Brownian motion starting at $x \in \mathbb{R}^d$. We define $W^y=(\{W_t^y\}_{t \ge 0})$ by
\begin{align}
W_t^y=
\begin{cases}
R_{x,y}(B_t^x),&\quad t<T_{x,y}:=\inf\{s>0 \mid B_s^x \in H_{x,y}\}, \\
B_t^x,&\quad t \ge T_{x,y}.
\end{cases}
\end{align}
The couple $(B^x,W^y)$ is called the mirror coupling of Brownian motions.
Let $\{S_t\}_{t \ge 0}$ be a subordinator, which is an increasing pure-jump Lévy process starting at zero independent of $(B^x,W^y)$. If we set $X_t^x=B_{S_t}^x$ and $Y_t^y=W_{S_t}^y$, $t \ge 0$. Then, $(X^x,Y^y)$ becomes a coupling of subordinate Brownian motions. Then, we denote by $U_{x,y}$ the coupling time of $(X^x,Y^y)$. By the result of this paper BSW, Theorem 2.1, we obtain that
\begin{align*}
U_{x,y}=\inf\{ t \ge 0 \mid S_t \ge T_{x,y}\}.
\end{align*}
We denote by $P_{x,y}$ the law of $(X^x,Y^y)$ and $f$ the corresponding Bernstein function.
By using the identity $U_{x,y}=\inf\{ t \ge 0 \mid S_t \ge T_{x,y}\}$, we obtain that
\begin{align}
(1)\quad P_{x,y}(U_{x,y} \ge t)\le \frac{|x-y|}{2\sqrt{2} \pi} \int_{0}^{\infty}\frac{e^{-tf(r)}}{\sqrt{r}}\,dr.
\end{align}
See the proof of BSW, Theorem 2.1 for details. In particular, if $X^x$ is a symmetric $\alpha$-stable process,
\begin{align*}
(2)\quad P_{x,y}(U_{x,y} \ge t) \le C|x-y|/t^{1/\alpha},\quad t>0
\end{align*}
Here, $C$ is a explicit constant.
My question
I think the equation $U_{x,y}=\inf\{ t \ge 0 \mid S_t \ge T_{x,y}\}$ (or (1)) is very useful, but I don't think it shows some geometric information such as where and how $X^x$ and $Y^y$ couple.
For some reasons, I study an asymptotic behavior of the probability $I_{x,y}:=P_{x,y}(U_{x,y} \ge \tau_{B(x,|x-y|^{\varepsilon})}^X)$ as $x \to y$ when $X^x$ is a symmetric $\alpha$-stable process. Here, $\varepsilon<1$ is a small number and $\tau_{B(x,r)}^X=\inf\{t>0 \mid |X_t^{x}-x|>r\}$, $r>0$.
We can easily deduce from the equation (1) that $I_{x,y} \lesssim |x-y|^{(\alpha-\epsilon \alpha)/(1+\alpha)}$ as $x \to y$. Just using (1), however, we do not know whether the index $(\alpha-\epsilon \alpha)/(1+\alpha)$ is optimal. Because $I_{x,y}$ should be a potential theoretical quantity, I also think that it should be possible to use another suitable method for a more precise estimate of $I_{x,y}$. Is there such a method?
ADD: By using (2), we have for any $t>0$,
\begin{align*}
I_{x,y} \le P_{x,y}(U_{x,y} >t)+P_{x,y}(\tau_{B(x,r)}^X \le t)\le C_1|x-y|/t^{1/\alpha}+C_2 tr^{-\alpha}.
\end{align*}
Here, $C_1, C_2$ are positive constant.
If we set $t=|x-y|^{\eta}$, $r=|x-y|^{\varepsilon}$, we arrive at $I_{x,y} \le (C_1\vee C_2)(|x-y|^{1-\eta/\alpha}+|x-y|^{\eta-\varepsilon \alpha})$. Thus, if we take $\eta>0$ such that $1-\eta/\alpha=\eta-\varepsilon \alpha$, we have $I_{x,y} \le (C_1\vee C_2)|x-y|^{(\alpha-\epsilon \alpha)/(\alpha+1)}$
There's something fishy going on here: by scale-invariance, if $|z| = 1$, then $I_{x,y} = P_{0,z}(U_{0,z} \ge \tau_{B(0, |x-y|^{\epsilon-1})}) \to 1$ as $|x - y| \to 0$. Am I missing something?
@MateuszKwaśnicki Thank you for your reply. I'm sorry. I assume that $\varepsilon<1$. In this case, $\tau_{B(0,|x-y|^{\varepsilon-1})}$ should go to infinity as $x \to y$. So, $I_{x,y} $ should go to $0$ as $x \to y$ because $(X^x,Y^y)$ is known to be successful.
Ah, sorry, I was seeing the inequality in the opposite direction...
@MateuszKwaśnicki I see. But using the scale-invariance makes $I_{x,y}$ easier to see. Thank you.
@MateuszKwaśnicki Sorry, I missed to calculate the index. So, I modified.
Here is how I would approach the problem.
The coupling time $U_{x,y}$ is not greater than the first exit time from $H_{x,y}^+$, the half-space bounded by $H_{x,y}$ and containing $x$, by the process $X_t^x$. Thus,
$$ I_{x,y} \le \mathbb{P}^x(\tau_{H_{x,y}^+} \ge \tau_{B(x, r)}) , $$
where $r = |x-y|^\epsilon$. By scale-invariance, we find that
$$ I_{x,y} \le \mathbb{P}^z(\tau_{H^+} \ge \tau_{B(z, 2)}) , $$
where $H^+ = \{x : x_1 > 0\}$, $z = (z_1, 0, \ldots, 0)$ and $z_1 = \tfrac{2}{r} |x - y| = 2 |x - y|^{1 - \epsilon}$. It follows that as long as $z_1 < 1$,
$$ I_{x,y} \le \mathbb{P}^z(\tau_{H^+} \ge \tau_{B(0, 1)}) . $$
The right-hand side decays as $z_1^{\alpha/2}$ when $z \to 0$ (it is a positive $\alpha$-harmonic function of $z$ in $B(0, 1) \cap H^+$). Therefore,
$$ I_{x,y} \le C z_1^{\alpha/2} = C' |x - y|^{(1 - \epsilon) \alpha/2} .$$
Remarks:
This bound should be optimal, I believe.
The above remark seems to be in conflict with your claim with exponent $(1 - \epsilon) \alpha / (1 + \alpha)$, so I may have made an error in the above calculation.
Edit: OK, now I think both bounds are sub-optimal, but the minimum of the two could be sharp. Here is a possible approach.
First of all, the problem is essentially one-dimensional: the probability of leaving a ball before coupling time should be comparable with the probability of leaving a strip.
In dimension one, consider the process $X_t$ (started at some $x > 0$) killed at the coupling time (with another process started at $y = -x$). This is a decent "stable-like" process in $(0, \infty)$, with intensity of jumps of the form $$c (|y - x|^{-1-\alpha} - |y + x|^{-1 - \alpha}),$$ and additionally killed with intensity $$c' x^{-\alpha}.$$ Locally this process behaves as the $\alpha$-stable one, but globally the intensity of jumps decays as $|y - x|^{-2 - \alpha}$.
I bet this process has been studied before, and estimates for the probability of hitting $(r, \infty)$ prior to death are known. And even if not, tools are readily available: this is a positive self-similar Markov process, and one can use the Lamperti–Kiu transformation together with fluctuation theory for Lévy processes to study these kind of problems.
I am grateful for your very kind reply. I'll take a closer look at your reply from now on, but as you say, my bound is different from yours (that's why I added the proof of my bound above). Maybe my bound is wrong...
Your wrote "$I_{x,y} \le P^z(\tau_{H^+} \ge \tau_{B(0,1)})$ provided that $z_1<1/2$". However, this should be "$I_{x,y} \le P^z(\tau_{H^+} \ge \tau_{B(0,1/2)})$ provided that $z_1<1/2$", right (of course, this is a minor issue)?. I think your argument is correct and boundary Harnack principle is suitable to derive the bound. However, my bound seems optimal when $\alpha<1$... On the other hand, your bound seems optimal when $\alpha \ge 1$.
@sharpe: Right, I got the inequality wrong the second time today. :-) I added some more thoughts on the problem, but unfortunately I do not have time to further think about it now.
Thank you very much for your very thoughtful comments. Unfortunately, I don't have the ability to carry out your ideas right now, but I would love to study.
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2025-03-21T14:48:31.507562
| 2020-07-14T12:56:18 |
365603
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/365603"
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Stack Exchange
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Infinite composition of continuous functions
Let $f_n:\mathbb{R}\rightarrow \mathbb{R}$ be a sequence of functions and define $F_n:= f_n\circ \dots\circ f_1$. Then $F_n$ is continuous. However, the pointwise limit need not be (consider Mateusz's example of
$$
f_n = \frac{\sqrt{2} x}{\sqrt{1 + 4 x^2}} \qquad F_n \to \operatorname{sign}(x)
$$
the sign function).
In general, what can be said about the limits of an iterated function system? Are such functions studied and if so what are they called?
All I figure at the moment is that they must necessary be of Baire 1 type.
The iterates of $f(x) = \frac{\sqrt{2},x}{\sqrt{1+4x^2}}$ seem to converge to the sign function.
Didn't the original version ask for continuity of the limit? And, in any case, it does hint what the structure of all possible limits is.
Oh no, I don't expect/want the limit to be continuous. Either way, I updated the question (also using your example) and I think what I'm looking for are limits of iterated function systems (which te earlier formulation was a special case of).
(1/2) For the iterates of the same function (as in the initial version of this question), there are entire books on dynamics in one dimension, and it is pretty easy to see that all potential limits are piecewise constant nonincreasing functions with the property $f(I) \subseteq I$ for any maximal interval $I$ on which $f$ is constant (and we allow $I$ to be a degenerate interval ${x}$).
(2/2) On the other hand, for an arbitrary sequence of continuous functions $f_n$, the situation is much more complicated. My guess would be: all possible limits are of the form $f(x) = g(h(x))$ with $h$ continuous and $g$ monotone, but that is just an uneducated guess.
In the example given I think that $F_n$ tends to $\mathrm{sign}/\sqrt{2}$.
There is an introduction to infinite compositions of complex functions here: https://en.wikipedia.org/wiki/Infinite_compositions_of_analytic_functions
Thanks, but I have read this article (its a good first stop)
Since you’re searching for names: there is a rich literature on discrete non-autonomous dynamical systems generated by an infinite family of continuous maps. Here "non-autonomous" means of course that the map you apply depends on the order of iteration, which is exactly your subject.
For instance:
Cánovas, J. S. (2006). On ω-limit sets of non-autonomous discrete systems. Journal of Difference Equations and Applications, 12(1), 95-100.
ISO 690
Shao, H., Chen, G., & Shi, Y. (2020). Some criteria of chaos in non-autonomous discrete dynamical systems. Journal of Difference Equations and Applications, 26(3), 295-308.
Sharma, P., & Raghav, M. (2015). Dynamics of nonautonomous discrete dynamical systems. arXiv preprint arXiv:1512.08868.
You can find a lot more, I suggested those just to make it clear which kind of name/keywords can be useful for you. You'll find lots of results on interval maps, real maps and on more general spaces. It all depends on which specific questions you want to address.
Good news for you: you don't have to consider Baire class 1 functions in their full glory, since given a sequence $\{F_n\}$ of continuous functions which converges pointwise, the functions $f_n$ solving the countable set of functional equations $$F_{n}(x)=f_n(F_{n-1}(x)), \ n\ge 2,$$ even when they exist, aren't necessarily continuous. So you're exploring a (topologically small, I guess) portion of Baire class 1 maps.
|
2025-03-21T14:48:31.507822
| 2020-07-14T13:03:34 |
365605
|
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"Aurelio",
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"k.j."
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"sort": "votes",
"url": "https://mathoverflow.net/questions/365605"
}
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Stack Exchange
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Locally a Deligne-Mumford stack has a finite etale covering
This is a part of the proof of corollary 6.1.1 of Laumon, Moret-Bailly's "Champs Algeriques".
Let $S$ be a scheme and $\mathscr{X}$ a non empty quasi-separated Deligne-Mumford stack over $S$.
Then there exists a non empty open substack $\mathscr{U}$ of $\mathscr{X}$, an affine scheme $X$, and a finite etale morphism of constant degree $X \to \mathscr{U}$.
The authors say that this is clear.
But I have no idea to show it.
For example, if $\mathscr{X}$ has a proper covering, then this is trivially true.
(Since the degree is locally constant function.)
But what about the general case?
@WillSawin Sorry, it's my mistake. This is a part of the proof of 6.1.1. The authors do not use the theorem to state it. I'll edit.
I suppose you refer to condition (i), as stated in Theorem 6.1; notice that a similar fact is used in the proof of Corollary 5.7.2.
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