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2025-03-21T14:48:31.866851
| 2020-08-26T01:16:25 |
370134
|
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|
Stack Exchange
|
making a random uniform hypergraph linear
Let $\mathcal{H}_{n,p,h}=(V,E)$ be a random $h$-uniform hypergraph on $[n]$, sampled according to the usual binomial distribution. We known that with high probability, the number of edges in $\mathcal{H}_{n,p,h}$ is
$$m = (1+o(1))\binom{n}{h}p$$
Let $\ell$ be given. I would like to delete some edges in order to
have a linear hypergraph (any two edges share at most one vertex)
remove all cycles of length at most $\ell$
I expect that we should be able to do so by deleting with high probabilities $o(m)$ edges, however simple first moment method are failing me... I try to count the number of Berge-cycle of length of length at most $\ell$, but simply looking at potential cycles for each pair of vertices I over-count way too much.
Is there any known upper bound for the number of cycles ? I found some literature on the probability threshold for the appearance of cycles, but not much on counting the cycles.
Edit: I could restrict to very small $p$. For some constant $c>2$,
$$ p = c \cdot n^{1-h+1/\ell}$$
A linear hypergraph cannot have more than $\binom n2/\binom h2$ edges (count pairs of vertices), which is far smaller than $m$ if $p$ is not very small. So usually you can't make it linear by removing $o(m)$ edges.
Hi Brendan, I could restrict to $p\sim n^{1-h+1/\ell}$. So a linear hypergraph would have at most $\frac{n(n-1)}{h(h-1)}$ edges while the random hypergraph has $\binom{n}{h}p \sim n^{1+1/\ell}\cdot h^{-h}$ edges (I think), so $p$ should be small enough.
See Theorem 3 and Lemma 4 here. A combinatorial classic - sparse graphs with
high chromatic number
Note that a 2-cycle is when two hyperedges intersect in at least 2 vertices. So by making the girth greater than $\ell$, this also ensures that the hypergraph is linear.
Thanks @LouisD, I managed to track down the actual result, not trivial, not that difficult.
Note: in order to understand the proof, it was key (at least for me) to see that a cycle of length $t$ in a $k$-uniform hypergraph is set of $t$ edges $(e_1,\ldots,e_t)$ such that (viewing each edge as a $k$-set of vertices)
$$ \left\vert \bigcup_{i=1}^t e_i \right\vert \leq (k-1)t$$
Following @LouisD comment, I followed a trail of references
A combinatorial classic - sparse graphs with high chromatic number by Jaroslav Nesetril, where lemma 4 is the hypergraph version ofthe famous theorem stating that we can find a graph with large girth and large chromatic number. The reference for this lemma is the following,
On a probabilistic graph-theoretic Method, by Nesetril and Rodl, where the lemma page 3 is the same version, without complete proof, referencing the following book,
P. Erdös and J. Spencer, Probabilistic methods in combinatorics, Akadémiai Kiado, Budapest; North-Holland, Amsterdam; Academic Press, New York, 1974. In there (I have no open source link), chapter 11, exercise 4 asks to prove the lemma, giving a final reference,
Erdos,Hajnal, "ON CHROMATIC NUMBER OF GRAPHS AND SET-SYSTEMS" in there, page 96, is the proof of the lemma,
To do so, they introduce $z(H)$ which is for a given $k$-uniform hypergraph $H$ and a given $s$, the number of set of vertices of size exactly $(k-1)t$ for some $t\leq s$, forming a $t$-cycle. They then show that for all but $o\binom{\binom{n}{k}}{m}$ hypergraph on $n$ vertices and $m$ edges,
$$ z(H)\leq \left(\frac{m}{n}\right)^s \log n$$
From there we can conclude that the number of edges in cycles of length of most $s$ is
$$ \binom{(k-1)s}{k}\left(\frac{m}{n}\right)^s \log n$$
Which is the desired property as long as $m< n^{1+1/s}$. However I have one last remark
There is an argument I do not understand in the Erdos-Hajnal article : they consider a subset $V'$ of the $n$ vertices, $V'$ has size $(k-1)t$, and they want to upperbound the number of hypergraph $H$ on $n$ vertices and $m$ edges, with at least $t$ edges in $V'$. They claim (end of page 96) that this is at most
$$ \binom{(k-1)t}{t}\binom{\binom{n}{k}}{m-t}$$
I would have expected rather
$$ \binom{\binom{(k-1)t}{k}}{t}\binom{\binom{n}{k}}{m-t}$$
because we can select $t$ edges among the $k$-uniform edges in $V'$, and then select $m-t$ other edges in any of the $\binom{n}{k}$ edges (we could even substract by $t$ here, but that's okay for an upper bound).
Note that my result also yield $ z(H)\leq \left(\frac{m}{n}\right)^s \log n$, so it's not that important.
|
2025-03-21T14:48:31.867136
| 2020-08-26T03:02:12 |
370137
|
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|
Stack Exchange
|
Fourier relationship between primes and zeros of L functions
In his plenary lecture "L-functions and Automorphic Representations" at the Seoul ICM James Arthur makes the following remark (Proceedings of the ICM 2014, vol. 1, p. 173):
Riemann conjectured that the only zeros of L(s) lie on the vertical line Re(s) = 1/2. This is the famous Riemann hypothesis, regarded by many as the most important unsolved problem in mathematics. Its interest stems from the fact that the zeros {ρ = 1/2 + it} of L(s) on this line are in some sense dual to prime numbers, or more accurately, to logarithms {γ = log pn} of prime powers. We can think of the former as a set of spectral data and the latter as a set of geometric data, which are related to each other by a Fourier transform.
Here $L(s)$ is the completed Riemann zeta function, so $L(s) = L(1-s)$.
(1) What is the "Fourier transform" that relates the primes ("geometric data") to the zeros of $L$ ("spectral data")?
(2) Is there a general Fourier transform relating points of a scheme over ${\mathbb{Z}}$ (resp. motive over $\mathbb{Q}$) to the zeros of its zeta function (resp. motivic $L$-function) of which this is an instance?
This follows simply from contour integration.
See
https://en.wikipedia.org/wiki/Explicit_formulae_for_L-functions
thanks for the link! This answers question (1), in fact for zeta functions of arbitrary number fields and Hecke characters. Do you know of any higher-dimensional generalizations, my question (2)?
This explicit formula can be generalised to hold for any $L$-function of an automorphic representation over $\mathrm{GL}_n(\mathbb{A}_F)$ for a number field $F$.
@PeterHumphries: Thanks! Can you point to a reference?
I'm not sure if it's written down anywhere, but the proof goes through unchanged, since all that is required is a functional equation, an Euler product, and control on the growth of the $L$-function (which follows from Stirling's formula and the Phragmen-Lindelof convexity principle).
Isn't some form of the general case in the Iwaniec-Kowalski book on Analytic Number Theory?
I expect so (though just over $\mathbb{Q}$).
|
2025-03-21T14:48:31.867300
| 2020-08-26T05:17:14 |
370140
|
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|
Stack Exchange
|
On the paper "Patching and the p-adic local Langlands correspondence"
$\DeclareMathOperator\GL{GL}$The question is about the paper: Caraiani, Emerton, Gee, Geraghty, Paskunas, and Shin - Patching and the $p$-adic local Langlands correspondence. Let $F$ be a finite extension of $\mathbb{Q}_p$ and let $E$ be another finite extension of $\mathbb{Q}_p$, which is assumed to be sufficiently large. To a Galois representation $r : \mathrm{Gal}(\overline{F}/F) \to \GL_n(E)$ the authors associate $V(r)$, which is a continuous, unitary admissible $E$-Banach space representation of $\GL_n(F)$. Assume that $r$ is potentially semi-stable, not necessarily generic, and it lies on an automorphic component. By Proposition 4.33 of this paper, the locally algebraic vectors of $V(r)$ are $V(r)^{\text{l.alg}} \simeq \pi \otimes \pi_{\text{alg}}(r)$, where $\pi$ is smooth and admissible.
Question : How to prove that $\pi$ as above has an irreducible generic sub-representation? A representation theoretic argument would be preferred, but any other suggestions are welcome.
Why do you expect that this should be the case? I don't see any such assertion in their paper.
The proof of the proposition 4.33 does not require r to be potentially crystalline and generic.
|
2025-03-21T14:48:31.867426
| 2020-08-26T09:38:07 |
370145
|
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|
Stack Exchange
|
For globally conformally flat surfaces, does radial symmetry of conformal factor imply the surface is a sphere?
We know that for the unit sphere in $\mathbb{R}^3$, the standard metric on such a sphere can be written as $$g=\frac{4}{(1+x_1^2+x_2^2)^2}(dx_1^2+dx_2^2)$$by a stereographic projection map from the sphere to the complex plane.
I'm curious about the following question:
Given a closed Riemannian surface in $\mathbb{R}^3$, if the metric is globally conformal to $\mathbb{R}^2$, that is, $g=e^{2u}(dx_1^2+dx_2^2)$, and if we further assume that $u$ is radially symmetric, then does the surface have to be a sphere? Are there any other possible shapes except spheres? Can anyone give some specific examples?
Any comments or ideas would be really appreciated.
Your question is not very well formulated because the surface is supposed to be closed while the plane is not.
@Sebastian, I'm thinking about given a compact Riemannian surface $(M,g)$ embedded in $\mathbb{R}^3$, is there a way to choose a coordinate chart $\phi: U\subset M \rightarrow \mathbb{R}^2$ such that $g\Big|_U=\phi^*e^{2u}(dx_1^2+dx_2^2)$ for some radially symmetric function $u$ other than the sphere case?
Still, it is not well-defined. Is your chart $\phi$ surjective? If so, the answer to your question is included in the answers to a former question of yours. Or do you mean that such a chart centered at $p$ should exist at every point $p\in M?$ This would then imply that $g$ has constant Gaussian curvature (if your surface is smooth), hence the only compact case is that of a round sphere. Or do you mean there exists at a point $p\in M$ and a local chart to some open set in the euclidean plane such that the metric is radially symmetric ? Then, there are other possibilities than spheres.
@Sebastian, This post was partially motivated by your answer. I assume that the chart $\phi$ is surjective, but I'm not sure how to find some explicit examples other than spheres? For example, what can $M$ be? I know that such $M$ is diffeomorphic to a sphere, but how about the exact shape?
If you only suppose that the chart $\phi$ exists at one point, then there are other shapes possible, e.g. a spheroid (rotational-symmetric ellipsoid).
This looks like a purely topological question to me because (orientable) closed surfaces are classified by their genus; the conformal hypothesis looks like a red herring. Please correct me if I misunderstood the premise of your question.
Let $(\Sigma,g)$ be the closed surface in question. Under the hypotheses imposed, there is a point $p \in \Sigma$ and a coordinate chart $\psi: \Sigma \setminus \{ p \} \to \mathbf{R}^2$ so that $\psi^* g_e = e^{-2u} g$ for some function $u: \mathbf{R}^2 \to \mathbf{R}$. Here I write $g_e = \mathrm{d} x_1^2 + \mathrm{d} x_2^2$ for the euclidean metric. In particular $\Sigma \setminus \{ p \}$ and $\mathbf{R}^2$ are homeomorphic. Therefore the Euler characteristic of $\Sigma \setminus \{ p \}$ is $1$, and $\chi(\Sigma) = 2$. But then $\Sigma$ has genus zero, and thus is a sphere.
The slight hiccup in the argument is that it assumes orientability. It was not clear whether you are willing to impose this or not; certainly this would be the case if the surface is an embedded submanifold of $\mathbf{R}^3$. Without this assumption, $\Sigma$ could also be a Klein bottle. However returning to the homeomorphism above, one sees that $\Sigma \setminus \{ p \}$ is contractible, which is not the case for a Klein bottle.
Edit: It looks like I misunderstood your question, and you are instead asking whether a surface $(\Sigma,g)$ with properties as above is necessarily a round sphere. I suspect there exist simple examples that demonstrate this not to be the case; regardless here is my attempt.
By the above we know that $\Sigma$ is diffeomorphic to $\mathbf{S}^2$. Next, by Nirenberg's solution of the Weyl embedding problem, we know that the metric $g$ can be induced from an isometric embedding into $\mathbf{R}^3$ provided it has positive Gauss curvature.
Therefore it is enough to find $u$ for which the Gauss curvature $K = K_g$ is positive but not constant. In terms of $u$ this is $K = - e^{-2u} \Delta u$. Let $u_0 = \frac{1}{2} \ln \frac{4}{(1 + r^2)^2}$ be the `Euclidean' conformal factor, which we perturb by a compactly supported rotationally invariant function $\varphi$, meaning $u = u_0 + \delta \varphi$ for some small $\delta > 0$. (Doing this only changes the metric near the south pole.)
Then $K = -e^{-2u_0 -2 \delta \varphi} (\Delta u_0 + \delta \Delta \varphi) = e^{-2\delta \varphi}(1 - \delta e^{-2u_0} \Delta \varphi)$. This is positive provided $\delta$ is small enough in terms of $\varphi$. `Basically any' perturbation should do the trick, for example picking $\varphi(r) = r^2 - 1$ one finds $K = -e^{-2\delta(r^2-1)}(1 - 4\delta e^{-2u_0})$, which is not constant. (Technically one would need to change $\varphi$ so as to make the metric smooth, but this can be done without changing the curvature near the south pole.)
Thank you very much for your post. I'm not really thinking about this problem topologically. What I truly want to understand is the question that given a compact Riemannian surface $(M,g)$ embedded in $\mathbb{R}^3$, whehter there is a way to choose a coordinate chart $\phi: U\subset M \rightarrow \mathbb{R}^2$ such that $g\Big|_U=\phi^*e^{2u}(dx_1^2+dx_2^2)$ for some radially symmetric function $u$ other than the sphere case?
By sphere, I mean literally the surface is a sphere. For example, an ellipsoid is not a sphere, even though topologically it is.
Let $\gamma(s)$ be a curve in the plane parametrized by arclength. Assume $\gamma(0) = (0,0)$ and $\gamma'(0) = (1,0)$. Rotate $\gamma$ around the $y$-axis and you get a surface of revolution parametrized by $(s,\theta) \in (0,S)\times \mathbb{S}^1$.
The induced metric is
$$ g = ds^2 + \gamma_1(s)^2 d\theta^2 $$
where $\gamma_1$ is the $x$-component of $\gamma$. By construction we have $\lim_{s\to 0} \gamma_1(s) / s = 1$.
Let $r$ be a function that solves
$$\tag{#} \frac{d}{ds} \ln r(s) = \frac{1}{\gamma_1(s)} $$
One can check that necessarily $r(0) = 0$.
Using the $(r,\theta)$ coordinates instead we find
$$ g = \Big(\frac{\gamma_1(s)}{r}\Big)^2 (dr^2 + r^2 d\theta^2) $$
is conformally flat, with conformal factor that is radially symmetric (independent of $\theta$).
Note that this works for any surface of revolution.
In terms of uniformization: it is of course known by the uniformization theorem that that any topological sphere embedded in $\mathbb{R}^3$ has its induced metric conformal to $\mathbb{S}^2$ and hence is conformally flat. You can regard this as trying to solve an elliptic PDE (for the conformal factor) on the surface. If you assumed that your surface is rotationally symmetric, then this symmetry descends to the PDE you are trying to solve and reduce it to an ordinary differential equation. This is essentially equation (#) above.
Any rotationally symmetric metric will do, so any compact surface of revolution of genus zero embedded in 3-dimensional Euclidean space.
On the other hand, suppose we have any such surface as in the problem. To have an open subset which is conformal to the plane, an oriented connected surface with Riemannian metric must fail to be Brody hyperbolic, and clearly its universal covering space has to be the sphere or the plane. So the surface, if compact, is diffeomorphic to the torus or the sphere. But in the case of the torus, the open set in the universal covering space which is conformal to the plane is already the entire covering space (no Fatou Bieberbach domains in complex dimension one), so in the quotient to the torus, there is no isometric symmetry vector field commuting with the covering map, as the fixed points would have to be invariant under the translations. So no rotational symmetry of the metric. So we are down to the sphere with its standard conformal structure. In the universal covering space, a connected open set conformal to the plane must omit precisely one point. So the symmetry vector field is defined except perhaps at that point. Its flow being by isometries, it will extend smoothly to fix that remaining point. So the question reduces to whether a surface of revolution of genus zero has an isometric embedding which is not rotationally symmetric. For positive curvature surfaces, it follows from Nirenberg's theorem that there is only one isometric embedding up to rotation, so it is rotationally symmetric. But I don't know the general case.
|
2025-03-21T14:48:31.867910
| 2020-08-26T09:39:16 |
370146
|
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|
Stack Exchange
|
Nontrivial interpretation of dependent type theory in the category of chain complexes
A category $\mathrm{Ch}(\mathbf{Ab})$ of chain complex has a model category structure, which makes the category to interpret dependent type theory.
E.g. a term of a type $A$ is interpreted as an arrow from a terminal object to $A$.
However, there is no nontrivial such arrow because terminal and initial objects coincide in $\mathrm{Ch}(\mathbf{Ab})$.
Doesn't this fact make the interpretation trivial?
I think you have confused different usages of the word “model” here.
Could you please make it precise? I think model categories can interpret Martin Löf type theory with identity types. (By "dependent type theory", I meant Martin Löf type theory with identity types.) Is my understanding correct?
That is incorrect, or at least very imprecise. You need a lot more than just a model structure to interpret dependent type theory. For instance you need a locally cartesian closed category, or something very close, to interpret $\Pi$-types.
By "Martin Löf type theory with identity types", I was thinking of the fragment without Π and Σ.
Then by Thm 3.1. of arxiv.org/pdf/0709.0248.pdf, I think finitely completeness and weak factorization system is enough to model Martin Löf type theory with identity types. Or isn't the stability condition satisfied?
I suppose if you omit $\Pi$ and $\Sigma$ and $0$ and $+$ you may have a model, yes. But that’s not a very expressive fragment of type theory.
|
2025-03-21T14:48:31.868095
| 2020-08-26T10:04:58 |
370148
|
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|
Stack Exchange
|
Expectation of Brownian motion increment and exponent of it
While reading a proof of a theorem I stumbled upon the following derivation which I failed to replicate myself. Let $\mu$ be a constant and $B(t)$ be a standard Brownian motion with $t > s$. Show that
$$ E\left( (B(t)−B(s))e^{−\mu (B(t)−B(s))} \right) = - \frac{d}{d\mu}(e^{\mu^2(t-s)/2})$$
Many thanks!
The increments $B(t)-B(s)$ have a Gaussian distribution with mean zero and variance $t-s$, for $t>s$. Hence
$$E\left( (B(t)−B(s))e^{−\mu (B(t)−B(s))} \right) =\int_{-\infty}^\infty xe^{-\mu x}e^{-\frac{x^2}{2(t-s)}}\,dx$$
$$=-\mu(t-s)e^{\mu^2(t-s)/2}=- \frac{d}{d\mu}(e^{\mu^2(t-s)/2}).$$
Let $m:=\mu$ and $X:=B(t)-B(s)$, so that $X\sim N(0,t-s)$ and hence
$Ee^{-mX}=e^{m^2(t-s)/2}$. So, in view of the Leibniz_integral_rule, the expectation in question is
$$EXe^{-mX}=-E\frac d{dm}e^{-mX}=-\frac d{dm}Ee^{-mX}=-\frac d{dm}e^{m^2(t-s)/2},$$
by as desired.
|
2025-03-21T14:48:31.868196
| 2020-08-26T10:08:10 |
370149
|
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"Carl-Fredrik Nyberg Brodda",
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|
Stack Exchange
|
A detail about busy beaverly behavior of distortion functions in graph groups
Say a function $\phi : \mathbb{N} \to \mathbb{N}$ is weakly superrecursive if for any total recursive $\phi : \mathbb{N} \to \mathbb{N}$
we have $\phi(n) > \psi(n)$ for infinitely many $n$. Say it is strongly superrecursive if for any total recursive $\phi : \mathbb{N} \to \mathbb{N}$
we have $\phi(n) > \psi(n)$ for all large enough $n$. Not every weakly superrecursive function is strongly superrecursive.
Let $H \leq G$ be groups with finite generating sets $S \subset T$, respectively. Define the distortion function of $H$ inside $G$ as
$$ \Delta^T_S(\ell) = \max \{|w|_S \;|\; w \in H, |w|_T \leq \ell\}, $$
defined up to the equivalence relation $f \approx g$ if for some $C > 0$, $C^{-1}f(n) \leq g(n) \leq Cf(n)$ for large enough $n$, where $|w|_N$ is the word norm of $w$ under generators $N$.
By a result of Mihajlova, $G = F_2 \times F_2$ has undecidable generalized word problem, and from this it follows that there is a finitely-generated
subgroup $H \leq G$ with weakly superrecursive distortion function.
Does this group, or at least some other graph group (i.e. right-angled Artin group), have strongly superrecursive distortion function?
Might follow from computability principles directly (if so, feel free to ignore groups in the answer), but I don't see this. Might also follow from Mihajlova's construction, but while I know the construction I haven't reconstructed a proof, and I don't know the details of how the word problem of f.p. groups is proved undecidable well enough either. Might also be well-known about graph groups, but I didn't find such a statement.
Mikhajlova, K. A., The occurrence problem for free products of groups, Math. USSR, Sb. 4(1968), 181-190 (1969); Translation from Mat. Sb., n. Ser. 75(117), 199-210 (1968). ZBL0214.27403.
A theorem of Olshanskii-Sapir is that the class of Dehn functions of finitely presented groups coincides with the class of distortion functions for finitely generated subgroups of $F \times F$. So as long as you find a f.p. group with strongly superrecursive Dehn function, the answer is yes.
The answer is yes.
Theorem. There exists a finitely-presented group with strongly superrecursive Dehn function.
Theorem. (Therefore) there exists a finitely-generated subgroup of $F_2 \times F_2$ with strongly superrecursive distortion.
These follow from theorems 1.2. The connection pointed out in the comment of @Carl-FredrikNybergBrodda.
Theorem 1.2 (Olshanskii-Sapir '98). The set of distortion functions of finitely generated subgroups of the direct product of two free groups $F_2 \times F_2$ coincides (up to equivalence) with the set of all Dehn functions of finitely presented groups.
Theorem 1.2 (Sapir-Birget-Rips, '08). Let $D_4$ be the set of all Dehn functions $d(n) \geq n^4$ of finitely presented groups. Let $T_4$ be the set of time functions $t(n) \geq n^4$ of arbitrary Turing machines. Let $T^4$ be the set of superadditive functions which are fourth powers of time functions. Then $T^4 \subset D_4 \subset T_4$.
The time function of a (not necessarily deterministic) Turing machine $M$ is $t : \mathbb{N} \to \mathbb{N}$ where $t(n)$ is the smallest number such that for every acceptable word $w$
with $|w| \leq n$ there exists a computation of length $\leq t(n)$ which accepts $w$. A function $f$ is superadditive if $f(m+n) \geq f(m) + f(n)$.
All we need to do is find a function in $T^4$ which is strongly superrecursive.
Let $M$ be a Turing machine that on input $0^n 1^k 2^h$ performs the following $2^{n+k+h}$ times, in a loop: simulate the $n$th Turing machine $M'$ (i.e. the machine with Gödel number $n$) on all unary inputs $1^0, 1^1, ..., 1^{2k}$, until $M'$ has halted on all of them. (And $M$ does not halt on inputs not of this form.) Let $t$ be the Time function of $M$. Letting $a = \max(a,b)$, we have $t(a + b)^4 \geq t(a + 1)^4 \geq (2t(a))^4 \geq t(a)^4 + t(b)^4$, where $t(a + 1) \geq 2t(a)$ follows because if $t(a)$ is given by input $0^n 1^k 2^h$, then the computation on $0^n 1^k 2^{h+1}$ takes at least twice as long. Therefore, $t(n)^4$ is superadditive.
(The justification of $t(a + 1) \geq 2t(a)$ is not completely precise, since there's a lot of bookkeeping going on and kept implicit. To be more exact without going into Turing machine details, you could replace the $2^{n+k+h}$-length loop on $0^n 1^k 2^h$ by two recursive calls of $M$ on $0^n 1^k 2^{h-1}$, when $h > 0$, or alternatively replace $2^{n+k+h}$ by something that grows much faster.)
For a Turing machine $M'$, write $\alpha(M')$ for some Gödel number of it.
Let $f$ be any total recursive function, computed by some Turing machine $M'$, let $\alpha(M') = \ell$. Total recursive functions are the same if we restrict to unary, so we may suppose that on input $1^k$, $M'$ computes $1^{f(k)}$ and halts (and does whatever on other inputs). We may assume $M'$ takes at least $f(k)$ steps to halt on input $1^k$ (indeed this is automatic since it has to write its output).
Now, on input $0^{\ell} 1^k$ our machine $M$ halts (because $M'$ halts on all unary inputs), and takes (much more than) $\max_{i=0}^{2k} f(i)$ steps to do so.
In particular, as soon as $n \leq 2(n-\ell)$, we have $t(n)^4 \geq t(n) \geq f(n)$.
Therefore $t(n)^4$ is strongly superrecursive.
Since $n^4$ is recursive, we also have $t(n)^4 \geq n^4$ (up to equivalence).
Ol’shanskij, Alexander Yu.; Sapir, Mark V., Length and area functions on groups and quasi-isometric Higman embeddings, Int. J. Algebra Comput. 11, No. 2, 137-170 (2001). ZBL1025.20030.
Sapir, Mark V.; Birget, Jean-Camille; Rips, Eliyahu, Isoperimetric and isodiametric functions of groups, Ann. Math. (2) 156, No. 2, 345-466 (2002). ZBL1026.20021.
Nice one! I suspected the answer would be yes. I wouldn't be surprised if something like this can be found in the literature already (though I don't know where to find it), as statements like "the word problem is really really hard!" are always appealing to publish. Nevertheless, it is very nice to have it written up here.
On the other hand, it bears pointing out that the Dehn function is only, of course, an upper bound on the complexity of the word problem!
I see, so indeed maybe I should've gone one step further and proved directly that the word problem takes strongly superrecursive time to solve in a suitable sense. I imagine that's true since the proofs (of course) involve embedding universal computation, but one would probably need to actually read the construction. If someone connects the dots here I'll accept that as an answer, so we get the best result on top (also I'm hesitant of giving myself the points, especially as it was not very difficult from the literature pointer).
It's a shame this question wasn't asked a few months ago before Sapir left MO!
|
2025-03-21T14:48:31.868795
| 2020-08-26T11:22:43 |
370151
|
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|
Stack Exchange
|
explicit computation of fractional Laplacian of a function
For $x\in\mathbb R$ let
$$
u(x)=\begin{cases}
|x|^{2s-1}-1 &\mbox{if } |x|>1,\\
0 & \mbox{otherwise}.
\end{cases}
$$
Is it possible to calculate explicitly the fractional Laplacian $(-\Delta)^{s} u(x)$ for a fixed $s\in (0, 1/2)$?
Yes, it is, as long as you are OK with special functions. See Corollary 3(ii) in my paper with B. Dyda and A. Kuznetsov titled Fractional Laplace operator and Meijer G-function, DOI:10.1007/s00365-016-9336-4. Here one needs to apply this result twice, with $d = 1$, $\alpha = 2 s$, $l = 0$, $\sigma = 0$ and $\rho$ equal to either $s-\tfrac12$ or $0$.
Edit: To clarify, the result mentioned above implies that
$$ (-\Delta)^s \bigl[|x|^{2\rho} (|x|^2-1)_+^\sigma\bigr] = 2^{2s} \Gamma(1+\sigma) G^{1,2}_{3,3}\biggl(\begin{array}{ccc}\tfrac12-s & 1+\rho+\sigma-s & -s \\ 0 & \rho - s & \tfrac12 \end{array} \; \bigg\vert \; |x|^2\biggr) . $$
If we set $\sigma = 0$, we find that
$$ (-\Delta)^s \bigl[|x|^{2\rho} \mathbb{1}_{|x|>1}\bigr] = 2^{2s} G^{1,2}_{3,3}\biggl(\begin{array}{ccc}\tfrac12-s & 1+\rho-s & -s \\ 0 & \rho - s & \tfrac12 \end{array} \; \bigg\vert \; |x|^2\biggr) . $$
For $\rho = 0$, we get
$$ (-\Delta)^s \bigl[\mathbb{1}_{|x|>1}\bigr] = 2^{2s} G^{1,2}_{3,3}\biggl(\begin{array}{ccc}\tfrac12-s & 1-s & -s \\ 0 & -s & \tfrac12 \end{array} \; \bigg\vert \; |x|^2\biggr) , $$
which simplifies to
$$ \frac{\Gamma(2s) \sin(\pi s)}{\pi} ((|x|^2-1)^{-2s} - (|x|^2 + 1)^{-2s}) $$
when $|x| > 1$ (and to a similar expression when $|x| < 1$).
On the other hand, for $\rho = s-\tfrac12$ we find that
$$ (-\Delta)^s \bigl[|x|^{2s-1} \mathbb{1}_{|x|>1}\bigr] = 2^{2s} G^{1,2}_{3,3}\biggl(\begin{array}{ccc}\tfrac12-s & \tfrac12 & -s \\ 0 & -\tfrac12 & \tfrac12 \end{array} \; \bigg\vert \; |x|^2\biggr) = 2^{2s} G^{1,1}_{2,2}\biggl(\begin{array}{ccc}\tfrac12-s & -s \\ 0 & -\tfrac12 \end{array} \; \bigg\vert \; |x|^2\biggr) , $$
which, for $|x| > 1$, is equal to
$$ \frac{\Gamma(2s) \sin(\pi s)}{\pi |x|^2} ((|x|^2-1)^{-2s} + (|x|^2 + 1)^{-2s}) . $$
The result is found by subtracting the above two expressions:
$$ (-\Delta)^s u(x) = \frac{\Gamma(2s) \sin(\pi s)}{\pi |x|^2} ((|x|^2 + 1)^{1-2s} - (|x|^2-1)^{1-2s}) $$
for $|x| > 1$ (unless I made an error when manipulating these terrible expressions).
Of course, this can be found by direct integration, too. The above method is more involved, but quite general. For example, it carries over to higher dimensions.
@GabS: I guess you can divide by $(2s-1)$ and pass to the limit as $s \to \tfrac12$, no? (The $s$ in the definition of $u$ need not be the same as the exponent $s$ in $(-\Delta)^s$, if this matters.)
@ Mateusz Kwaśnicki Yes, I should have been precise. I would like to calculate $(-\Delta)^{1/2}$. Here $u$ look like the fundamental solution except in a neighbourhood of zero
@ Mateusz Kwaśnicki I believe in both the cases the answer is zero as in one dimension, these are fundamental solution away from zero.
@GabS: No, it is definitely not zero, unless $s = 1$. Why do you expect it to be zero?
@ Mateusz Kwaśnicki The calculation is $(-\Delta)^s u(x)= 2^{2s} \frac{\Gamma(s) \Gamma((1/2)}{\Gamma(0)\Gamma ((1-2s)/2) } |x|^{-1}= 2^{2s} \frac{\Gamma(s) \Gamma((1/2)}{\Gamma (1-2s/2) } |x|^{-1}.$ So I am wrong.
If you are saying about $(-\Delta)^{1/2} [\log |x|]$, then this is indeed zero (observe $\Gamma(0)$ in the denominator in your expression). However, $(-\Delta)^{1/2} [\log |x| \mathbb{1}_{|x| > 1}]$ is not identically equal to zero.
I do agree, but I think there is a problem. The function in your paper $u(x)= |x|^{2\rho}(|x|^{2}-1){+}^{\sigma}$ considering $V(x)=1.$ The function what I want is a bit more complicated $(|x|^{2(s-1/2)}-1){+}$ which need $\rho$ to be zero and $\sigma$ to be $s-1/2$ or may be I am making a mistake.
I expanded my answer in response to your last comment. If everything remains unclear, feel free to ask.
Thank you very much for the information.
@GabS: But what are $f$ and $g$?
Simply write $u$ as a linear combination of $|x|^p \mathbb{1}{|x|<a}$ or $|x|^p \mathbb{1}{|x|>a}$ for different $p$ and $a$, and you are good to go, right?
@ Mateusz Kwaśnicki What is the quickest way to compute the Meijur $G$ function. You are using a Meijur function and hypergeometric series. Is there a good book on this topic.
@GabS: "Quickest" in the sense of numerical methods? No idea. Mathematica seems to evaluate it reasonably well, but relatively slow. WolframAlpha also is able to evaluate single values. Regarding general introduction to Meijer G-function, I do not have anything to add to what we wrote in the article. Prudnikov's book is perhaps a good start.
@ Mateusz Kwaśnicki In your calculation, I think $|x|^2$ should be replaced by $|x|.$
@ Mateusz Kwaśnicki Is the fractional Laplacian of the characteristic function $\chi_{(1, \infty)}$ and $\chi_{(1, 2)}$ is the same? Do the end point contribute of the calculation.
If you look at the expression for $(-\Delta)^s u(x)$, then you will clearly see that these are quite different functions for $u = \chi_{(1,\infty)}$ and $u = \chi_{(1,2)}$.
@ Mateusz Kwaśnicki If I consider $u(x)=\chi_{1,2}(x)$ and calculate $(-\Delta)^s u=constant\times [(x+1)^{-2s}-(x+1)^{-2s} -(x-2)^{-2s}+ (x+2)^{-2s}]$. If $x\in (1, 2)$ the third term is a bit tricky. Am I wrong.
Means that $(x-2)$ is negative when $x\in (1,2)$, then what happens to $(-1)^{1-2s}(2-x)^{-2s}.$
Do you mean $\chi_{1,2}(x) = 1$ if $1 < |x| < 2$ and $\chi_{1,2}(x) = 0$ otherwise? The expression you gave looks OK for $x > 2$, for general $x$ you need absolute values of $x\pm1$ and $x\pm2$ and appropriate sign corrections.
Yes, the characteristic function on the interval (1,2).
|
2025-03-21T14:48:31.869181
| 2020-08-26T11:25:33 |
370152
|
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|
Stack Exchange
|
A property for a group generated by involutions
Let $G$ be a group and $x_1,\ldots,x_n,y_1,\ldots,y_n \in G$ involutions
such that
$G = \langle x_1, \ldots , x_n \rangle = \langle y_1 , \ldots ,
y_n \rangle$
$g:=x_1 \cdots x_n = y_1 \cdots y_n$ is of finite order
Now assume that there exists $1 \leq k < n$ such that
$$ g = x_1 \cdots x_k y_{k+1} \cdots y_n.$$
Is it true that
$$ G = \langle x_1, \ldots , x_k , y_{k+1}, \ldots, y_n \rangle ?$$
Why would you expect this to be true? Aren't you just saying taht $y_{1} \ldots y_{k} = x_{1} \ldots x_{k}$? And do you mean to assume this for a single $g$?
Thank you for the comment. You are right. I forgot to impose some conditions. In the meantime I have written a comment underneath the first answer.
No, this is false even when $G$ is abelian and finite. For instance take
$$G = \langle (1,2), (3,4), (5,6), (7,8) \rangle \le \mathrm{Sym}_8.$$
Define $x_1,x_2,x_3,x_4,y_1,y_2,y_3,y_4$ as indicated by
$$G = \bigl\langle (1,2)(5,6), (3,4)(5,6), (5,6), (7,8) \bigr\rangle. $$
and
$$G = \bigl\langle (1,2), (3,4), (1,2)(5,6), (1,2)(7,8) \bigr\rangle $$
The products of generators are
$$x_1x_2x_3x_4 = y_1y_2y_3y_4 = (1,2)(3,4)(5,6)(7,8),$$
which is $x_1x_2y_3y_4 = (1,2)(5,6)\:(3,4)(5,6)\:(1,2)(5,6)\:(1,2)(7,8)$. But
$$\bigl\langle x_1,x_2,y_3,y_4\bigr\rangle = \bigl\langle (1,2)(5,6), (3,4)(5,6), (1,2)(5,6), (1,2)(7,8) \bigr\rangle$$
has index $2$ in $G$.
Thank you very much for the answer. Despite of the elegant counter-example it would be nice to see if the assertion is correct under some further assumptions: For instance, in the symmetric group, if we only consider transpositions and furthermore the words for $x$ and $y$ are reduced, then the assertion is correct.
No, $G$ need not equal $\langle x_1, \dotsc, x_k, y_{k + 1}, \dotsc, y_n\rangle$.
Let $G$ be any group generated by two involutions $a, b$. Let $k = 2, n = 4$ and let $x_1 = x_2 = y_3 = y_4 = a$ and $y_1 = y_2 = x_3 = x_4 = b$, so $G = \langle x_i \rangle = \langle y_i \rangle$. We have $x_1x_2 = x_3x_4 = y_1y_2 = y_3y_4$, so $g = 1$ has finite order. Now $\langle x_1, x_2, y_3, y_4 \rangle = \langle x_1 \rangle$ has two elements, so pick any $G$ with more than two elements, for example the Klein four-group.
I had this as a comment and it got dropped out while converting to answer. Will fix later.
I thank you for that, kind stranger. ("Thanks" was too short.)
With a mathematician's "sense of humor" I note that you could have also gotten around the minimum length by writing: "Thanks. ('Thanks' was too short.)" :)
Clever! (I may steal that later, but not immediately.)
Thank you very much for the answer.
|
2025-03-21T14:48:31.869377
| 2020-08-26T12:01:53 |
370154
|
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|
Stack Exchange
|
Genericity of contact structures all of whose closed Reeb orbits are nondegenerate
First, a contact form $\alpha$ on $M$ with Reeb vector field $R$ is said to be non-degenerate if, for any point $p$ such that $\phi_T^R(p) = p$, we have $\det{(\textrm{id}_{T_pM} - (d \phi_T^R)_p)} \neq 0$.
It seems to be widely known that the set of nondegenerate contact forms is generic (I guess it means dense here) in the set of all contact forms. However, I'm unable to find a "clean" proof in the existing literature, except in Bourgeois' notes "Introduction to Contact Homology" which can be found here: https://www.imo.universite-paris-saclay.fr/~bourgeois/papers/Berder.pdf, Lemma 2 on Page 3. However, there are quite a few things that are unclear to me about the proof written here. I will try to point them out as specifically as possible:
In the first step, for $\gamma$ a degenerate closed orbit through $p$, and $f_p$ a function such that $df_p =0$ along $\gamma$ and supported in a tubular neighborhood of $\gamma$ why is it okay to take $(1 + \epsilon f_p)\alpha$ instead of $\alpha$ for small enough $\epsilon$?
After that, it is said that the linearized flow (of $R_{(1+\epsilon f_p)\alpha}$, I assume) along $\gamma$ has the additional term $$\frac{\epsilon}{(1+\epsilon f_p)^2} J\textrm{Hess}(f_p)$$for a compatible complex structure $J$ on the contact structure $\xi$. I'm not sure that $\gamma$ remains a closed orbit of $R_{(1+\epsilon f_p)\alpha}$, and I suspect that it doesn't. Also, how does one perform this calculation of the linearized flow, and where does a compatible complex structure come up?
Perhaps most importantly, it is said that, for $\epsilon$ sufficiently small, no closed orbit of period $\leq T$ will be created by this perturbation. Why?
I would be grateful for any additional information regarding this folklore-type result - a different reference, some additional explanations of the proof I linked, or anything else that would shed some light on this.
Other (related) proofs exist, such as Appendix A of Albers-Bramham-Wendl ("On nonseparating contact hypersurfaces in symplectic 4–manifolds"), and Lemma 7.1 of Colin-Honda ("Reeb vector fields and open book decompositions"), and Proposition 6.1 of Hofer-Wysocki-Zehnder ("The Dynamics on Three-Dimensional Strictly Convex Energy Surfaces") though that crucially refers to a result of Robinson.
A very useful document when learning this stuff, assuming you can read French-Math, is Jean Gutt's masters thesis "Dynamique de Reeb et Homologie de Contact". Also you should supplement Bourgeois' paper with his PhD thesis ("A Morse-Bott approach to contact homology") which rephrases the proof in Section 2.2 without mentioning $J$, for example.
1st bullet: This is an example of a very small perturbation! 2nd bullet: Remember that $f|\gamma=1$ and $df|\gamma=0$ (also the linearization of the Reeb flow is not the Reeb field). Perform the calculation carefully by invoking the definitions of what it means to be a Reeb vector field, and as for J you're essentially using a metric which is compatible with $d\alpha$ on the contact structure (normal planes to the Reeb orbit) to discuss Hessians/gradients. 3rd bullet: Argue by contradiction and Arzelà-Ascoli theorem (supposing there exists new orbits as $\epsilon\to0$).
A possibly helpful comment (especially for the second bullet point): Suppose we know $R_{\alpha}$, and we take $g$ a nonzero function. Then $R_{g\alpha} = \frac{1}{g}(R_{\alpha} + V_g)$, where $V_g$ is the symplectic gradient of $\ln(g)$, i.e. the unique vector field in $\xi$ defined by $d\alpha(V_g,\cdot) = d\ln(g)$. For example, in the case $g = 1+\epsilon f$, the condition that $df = 0$ along the Reeb orbit $\gamma$ is precisely what ensures that $\gamma$ remains a Reeb orbit $\gamma$, which is clear from this computation (since $V_g = 0$ along $\gamma$).
@ChrisGerig Thanks so much for your comments! They were very helpful, but I hope you don't mind if I annoy you with another question, concerning the second bullet point: I thought I was sure what "linearized flow" meant - if $\phi_t$ is the flow of the perturbed Reeb field, then I assume that this means $d\phi_t$ - now I'm not so sure anymore. If I'm right, how do I start this calculation? Do I write $\phi_t(x+h) - \phi_t(x) = \cdots$ in a chart, or...?
I looked at the additional references you gave me: the Albers-Bramham-Wendl proof seemed great to me, but something about the Reeb field calculation seems off to me: the $X_{f \lambda}$-component of $\hat{X}$ is $-\hat{f}/f$, not $-\hat{f}$, and if that is taken into account, the vector field called $V_{\hat{f}}$ doesn't satisfy the property labelled (A.3). I could be wrong, of course, but I've checked my calculation enough times to say something. I don't know if it matters in the long run... Also, do you know of a place where I could find Jean Gutt's masters thesis? It's not on his website.
|
2025-03-21T14:48:31.869740
| 2020-08-26T12:24:51 |
370155
|
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|
What does this paper have to do with Hilbert's fifth question?
Apparently, there is a paper
M. Sablik, Final part of the answer to a Hilbert's question. Functional Equations - Results and Advances. Edited by Z. Daróczy and Zs. Páles, Kluwer Academic Publishers 2002, 231-242.
however I struggle to understand what does it have to do with Lie groups and Hilbert's fifth problem. Is it something deeper I fail to see here?
The paper in question does not claim to have solved Hilbert's fifth question/problem, as implied by the original title, so I have edited the title.
Here's a translation of part of the address dealing with the Fifth Problem (source):
For infinite groups the investigation of the corresponding question is, I believe, also of interest. Moreover we are thus led to the wide and interesting field of functional equations which have been heretofore investigated usually only under the assumption of the differentiability of the functions involved. In particular the functional equations treated by Abel$^{12}$ with so much ingenuity, the difference equations, and other equations occurring in the literature of mathematics, do not directly involve anything which necessitates the requirement of the differentiability of the accompanying functions. In the search for certain existence proofs in the calculus of variations I came directly upon the problem: To prove the differentiability of the function under consideration from the existence of a difference equation. In all these cases, then, the problem arises: In how far are the assertions which we can make in the case of differentiable functions true under proper modifications without this assumption?
The paper you link to deals with the functional equation of Abel dealt with in the paper
N.H. Abel, Ueber die functionen welche der Gleichung $\varphi x + \varphi y = \psi(xfy+yfx)$ genugthun, Journal für die reine und angewandte Mathematik, herausgegeben von Crelle, Bd. 2, Berlin 1827. (DigiZeit) (French translation Sur les fonctions qui satisfont à l'équation $\varphi x + \varphi y = \psi(xfy+yfx)$ available in Oeuvres complètes de Niels Henrik Abel pp 389–398, available at the Internet Archive).
The famous Cauchy functional equation $f(x+y) = f(x)+f(y)$ is a very special case, and is a baby case of the usual, finite-dimensional version of Hilbert's Fifth Problem. Or, one can view Abel's functional equation as a variant of Cauchy's with an 'exotic' addition, so in some sense a non-standard group operation on the reals. One can interpret, I gather, the group laws for a (local?) Lie group as a bunch of (coordinate) functions obeying a functional equation. Imagine you had no concept of manifold, topological space etc, but could write down what associativity meant direction in terms of coordinate functions, a priori only continuous. I would tentatively guess this is what Hilbert was driving at, allowing in the remarks above something from calculus of variations not obviously coming from groups.
In a sense, the paper in the OP is not just answering a question raised as a special case of one of Hilbert's problems from 1900, but closing off a line of enquiry starting in back in 1827.
Your link refers to an abstract which reads as follows:
We present new results concerning the following functional equation of Abel
$$
ψ(xf(y)+yf(x))=ϕ(x)+ϕ(y)
$$
D. Hilbert in the second part of his fifth problem asked whether it can be solved without differentiability assumption on the unknown functions ψ, f and ϕ. We gave earlier (cf. [9] and [10]) a positive answer assuming however that 0 is either in the domain or the range of f. Now we solve the equation in the remaining case and thus complete the answer to Hilbert’s question.
which explains, as far as I can see, how the paper in question relates to Hilbert's fifth problem; in particular, there is no claim of having solved Hilbert's fifth problem in its full sense as it is usually understood. Or is there something I fail to see?
Sure, but the common understanding of Hilbert's fifth problem is that it deals with topological groups and showing that under a certain algebraic condition you only get Lie groups. I had no idea until I started looking that the original question was phrased purely in terms of functional equations and extended in this way.
|
2025-03-21T14:48:31.870091
| 2020-08-26T12:34:58 |
370158
|
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Stack Exchange
|
Questions to the proof of Lemma 9.3 in Humphreys "Representations of Semisimple Lie algebras in the BGG Category $\mathcal{O}$"
Let $\mathfrak{g}$ be a semisimple Lie algebra over $\mathbb{C}$ with root system $\Phi$, Weyl group $W$ and Cartan decomposition $\mathfrak{g}=\mathfrak{h}\oplus \bigoplus_{\alpha \in \Phi} \mathfrak{g}_\alpha $. Fix a set of positive roots $\Phi^+ \subset \Phi$ and simple roots $\Delta \subset \Phi$. Then $I \subset \Delta$, defines a root system $\Phi_I \subset \Phi$ with positive roots $\Phi_I^+ \subset \Phi^+$ and a Weyl group $W_I \subset W$. Furthermore be $\mathfrak{n}_I^-=\bigoplus_{\alpha \in -\Phi_I^+}\mathfrak{g}_\alpha$.
In Humphreys "Representations of Semisimple Lie algebras in the BGG Category $\mathcal{O}$" we have the following Lemma 9.3:
Let $M \in \mathcal{O}$ have the set of weights $\Pi(M)$. The following conditions are equivalent:
$M$ is locally $\mathfrak{n}_I^-$-finite.
For all $\alpha \in I$ and $\mu \in \Pi(M)$ we have $\dim M_\mu=\dim M_{s_{\alpha}\mu}$.
For all $w \in W_I$ and $\mu \in \Pi(M)$ we have $\dim M_\mu=\dim M_{w\mu}$.
$\Pi(M)$ is stable under $W_I$.
Now I struggle with the main parts of the proof.
In "$1. \Rightarrow 2.$" he argues the following way. For $\alpha \in I$ fixed, consider the action of the subalgebra generated by $x_\alpha \in \mathfrak{g}_\alpha$ and $y_\alpha \in \mathfrak{g}_{-\alpha}$, which is isomorphic to $\mathfrak{sl} (2,\mathbb{C})$, on $M_\mu$. This gives a finite dimensional submodule $N$ of $M$, which is stable under $\mathfrak{h}$. Then the standard theory for finite dimensional representations of $\mathfrak{sl} (2,\mathbb{C})$ yields $2.$
Questions that arise for me:
Does he consider $N$ as a finite dimensional $\mathfrak{sl} (2,\mathbb{C})$-module?
Why is $N$ $\mathfrak{h}$-stable and why is this necessary?
I know how the simple $\mathfrak{sl} (2,\mathbb{C})$-modules look like but how does this imply 2. ?
In "$4. \Rightarrow 1.$" He constructed a weight $\mu'(=w_I\mu)$, such that only finitely many $\mathbb{Z}^+$-linear combinations of $-\Phi_I^+$ can be added to $\mu'$ to lie in $\Pi(M)$. Then he says that $\mu'$ is a typical weight of $M$, forcing $M$ to be locally $\mathfrak{n}_I^-$-finite.
How does this weight forces $M$ to be locally $\mathfrak{n}_I^-$-finite? Have the impression there is kind of standard argument used.
(1) Yes. (2) $h x_\alpha m_\mu = (\mu(h) - \alpha(h))x_\alpha m_\mu$, or something like that, and similarly for $y_\alpha$. (3) By decomposing into irreducible $\mathfrak{sl}$-submodules that intersect $M_\mu$, one sees that the dimension computation can be done for an irreducible submodule. For each such module, one knows that the dimension of the weight spaces behaves as desired. (4) The $\mathfrak n^-$-submodule of $M$ spanned by $M_{\mu'}$ is spanned by the weight spaces as you say. There are finitely many of those and each is finite dimensional.
(My argument for (3) is a little sketchy, but I think it basically works.)
Thanks for the fast answer! (2) Yeah right. But why is it necessary to have $\mathfrak{h}$-stability? (4) Then the argument is that every weight of $M$ can be expressed as such a $\mu'$? Otherwise I don't see how I get it locally everywhere. My definition of locally $\mathfrak{n}_I^-$-finite is that for every $v \in M$ is $U(\mathfrak{n}_I^-)v$ is finite dimensional.
(2) To conclude that $M_\mu$ is the sum of $N_\mu$ as $N$ ranges over such submodules, and analogously for $M_{s_\alpha\mu}$. (4) Not every weight of $M$, but every weight of $U(\mathfrak n^-)m_{\mu'}$ for $m_{\mu'} \in M_{\mu'}$.
So I still don't get how locally $\mathfrak{n}^-$ finiteness is shown. If I start with $v \in M_\lambda \subset M$ and $\lambda$ is not a weight of such a $U{(\mathfrak{n}^-})M_\mu'$, which as far as I understood you can be the case. How does it follows that $U{(\mathfrak{n}^-})v$ is finite dimensional?
What about this: Let $v \in M_\lambda$. Then $w_I\lambda$ is by assumption a weight of $M$ and only finitely many $Z^+$-linear combinations of $\Phi_I^+$ can be added to get a weight of $M$. Thus, as $\lambda=w_I(w_I\lambda)$, only finitely many $Z^+$-linear combinations of $\Phi_I^-$ can be added to $\lambda$ to get a weight of $M$. Hence $U(\mathfrak{n}^-)M_\lambda $ is finite dimensional, especially $U(\mathfrak{n}^-)v$.
Each $v$ lies in a sum of finitely many weight spaces, and the $U(\mathfrak n^-)$-span of each of those weight spaces is f.d., so $U(\mathfrak n^-)v$ lies in the sum of finitely many f.d. spaces. (I don't know why you worry about $v \in M_\lambda$ with $\lambda$ not as above; $\mu$ is just an arbitrary label here, so, if you want to work with $U(\mathfrak n^-)M_\lambda$ instead of $U(\mathfrak n^-)M_\mu$, then just do the entire reasoning with $\lambda$ in place of $\mu$.)
Yeah that's why, as I did, it is enough to consider only elements from a weight space. I have choosen $\lambda$ because $\mu$ was related to $\mu'$, which was a weight Humphreys constructed. I tried to avoid double meanings etc.
Right, but this is my point: you don’t first choose $\mu$ and then try to show that $U(\mathfrak n^-)v$ is finite for $v \in M_\lambda$; you first choose $v$, decompose into weight spaces as above, pick your favourite weight space, call the weight $\mu$, and then carry out the proof.
One point that bothers me. (3) I thought for your argumentation we need that $M_\mu, M_{s_\alpha\mu} \subset N$. Obviously $M_\mu \subset N$. But why holds$ M_{s_\alpha\mu} \subset N$?
(3) A little sketchy, as I said. Neither one necessarily contained in $N$, but $M_\mu = \sum N_\mu$ and $M_{s_\alpha\mu} = \sum N_{s_\alpha\mu}$, over all possible $N$'s. So we just verify it for the piece of the weight space in each individual $N$, and then sum.
Oh you are right, I'm still not very familiar with Lie algebra representations. Mmmh then I'm wondering what do you mean by "over all possible N's"? The $N$'s you get by varying the $\alpha \in I$?
You say "consider the action of … $\mathfrak{sl}(2, \mathbb C)$ on $M_\mu$. This gives … $N$"; but actually it doesn't pin down a particular $N$. Rather, we can decompose $M$ as a sum of various $N$s (due to complete reducibility of complex $\mathfrak{sl}(2, \mathbb C)$-representations). The $N$s that occur in such a decomposition are what I am referring to as "all possible $N$s".
I took it from Humphreys and understood in the way that $N={m \in M \mid m=X.v \text{ with } X \in \langle x_\alpha, y_\alpha \rangle \cong \mathfrak{sl}(2,\mathbb{C}), v \in M_\mu }$. Hence for me there is a particular $N$. Why it shouldn't? Furthermore what is the reason for Humphreys to fix $\alpha \in I $ and proceed as he did?
Is $M$ complete reducible as a $\mathfrak{sl}(2,\mathbb{C})$ representation if it is not finite-dimensional?
Let $v \in M_\mu$ then $\mu(h_\alpha)v=h_\alpha.v=[x_\alpha,y_\alpha].v \in N$. Hence $v\in N$ and we have $M_\mu \subset N$.
Yes, all representations of $\mathfrak{sl}(2, \mathbb C)$ are completely reducible. (The f.-d. case is less interesting.) Your reasoning for $M_\mu \subseteq N$ seems to use $\mu(h_\alpha) \ne 0$. I guess you could use a variant $N = M_\mu + \mathfrak g_\alpha\cdot M_\mu$ if you already know $M_\mu$ is f.-d. Anyway, I think this is enough back and forth for the comments.
https://math.berkeley.edu/~reb/courses/261/35.pdf Excercise 283 let me believe something different. But okay I will think on my own. Thanks for all your comments.
You are right and I am wrong (I was thinking of the compact-group case). Sorry! I agree that your $N = M_\mu + \mathfrak g_\alpha M_\mu$ is a better construction. Since it is f.-d., it's an $\operatorname{SU}(2)$-rep., so stable under the reflexion $s_\alpha$, so contains $s_\alpha M_\mu = M_{s_\alpha\mu}$.
I'm relieved to hear that. I'm wondering if there isn't a way to see that $M_{s_\alpha \mu}\subset N$ without considering any $SU(2)$-action. Thats the impression I got from Humphreys proofs.
Re, I guess the point is that a suitable $x_\alpha^n M_\mu$ or $y_\alpha^n M_\mu$ (depending on $\langle\alpha, \mu\rangle$) should equal $M_{s_\alpha\mu}$.
Thats was my first idea too, but $\langle \alpha, \mu \rangle $ don't has to lie in $\mathbb{Z}$.
Do you have an example where $\langle\alpha, \mu\rangle$ doesn't lie in $\mathbb Z$? (By the way, I also realise that there is a more serious objection than mine to using $\mathfrak g_\alpha\cdot M_\mu$, namely, that it's not necessarily $\mathfrak g_\alpha$-stable! We both meant $\mathcal U(\mathfrak g_\alpha)\cdot M_\mu$.)
After your doubts I went through some basic textbook and you are right- $\langle \mu, \alpha \rangle \in \mathbb{Z}$. Can be found for example in Humphreys "Lie algebras and Representations Theory" section 7.2.
Thanks to the outstanding help of LSpice I present a version of more detailed proof of the two parts above. Do not hesitate to point out mistakes.
"$(1) \Rightarrow (2)$": Fix $\alpha \in I$ and $\mu \in \Pi(M)$. Observe that for $\mu(h_\alpha)=0$, we have $s_\alpha\mu=\mu-\langle \mu, \alpha^{\vee}\rangle \alpha = \mu - \mu(h_\alpha)\alpha=\mu$ and $(2)$ follows trivially. Hence we can assume $\mu(h_\alpha) \neq 0$.
Then by assumption and as $M\in \mathcal{O}$ the action on $M_\mu$ of the subalgebra $\mathfrak{s}_\alpha \cong \mathfrak{sl}(2,\mathbb{C})$, generated by $x_\alpha \in \mathfrak{g}_\alpha$, $y_\alpha \in \mathfrak{g}_{-\alpha}$, produces a finite dimensional $U(\mathfrak{sl}(2,\mathbb{C}))$-submodule $N \subset M$.
For $v \in M_\mu$, we have that $N \ni h_\alpha.v=\mu(h_\alpha)v$. Hence $M_\mu \subset N$. For $x_\alpha,y_\alpha$ exists $n_\alpha \in \mathbb{N}$ such that $x_\alpha^{n_\alpha+1}.N=y_\alpha^{n_\alpha+1}.N=0.$ Define
\begin{align*}
\exp(x_\alpha)&:=\sum_{k=0}^{n_\alpha} x_\alpha^k/k! \in U(\mathfrak{sl},(2,\mathbb{C}))\\
\exp(y_\alpha)&:=\sum_{k=0}^{n_\alpha} y_\alpha^k/k! \in U(\mathfrak{sl}(2,\mathbb{C})),\\
s&:=\exp(x_\alpha)\exp(-y_\alpha)\exp(x_\alpha) \in U(\mathfrak{sl}(2,\mathbb{C}))
\end{align*}
Then as in the proof of Theorem 21.2 of Humphreys "Lie algebras and Representations Theory", we additionally have that $M_{s_\alpha\mu}=s.M_\mu \subset N$.
As $N$ is finite dimensional, $N=\bigoplus N_i$ with $N_i$ simple $U(\mathfrak{sl}(2,\mathbb{C}))$-module. Observe that
\begin{align*}
h.(x_\alpha.v)&=(x_\alpha h+\alpha(h)x_\alpha).v=(\mu(h)+\alpha(h))x_\alpha.v, \\
h.(y_\alpha.v)&=(x_\alpha h-\alpha(h)x_\alpha).v=(\mu(h)-\alpha(h))y_\alpha.v,
\end{align*}
for $v \in M_\mu, h \in \mathfrak{h}$. Thus $N$ is $\mathfrak{h}$-stable and $M_\mu=\bigoplus (N_i)_\mu$ resp. $M_{s_\alpha\mu}=\bigoplus (N_i)_{s_\alpha\mu}$ follows. But for simple $U(\mathfrak{sl}(2,\mathbb{C}))$-module $N_i$ we know that $\dim((N_i)_{s_\alpha\mu})=\dim((N_i)_\mu)$ and the claim follows.
"$(4) \Rightarrow (1)$": We want to show that $U(\mathfrak{n}_I^-).v$ is finite dimensional for every $v \in M$. As $M$ is $\mathfrak{h}$-semisimple, we can assume that $v \in M_\mu$ for some $\mu \in \Pi(M)$. Then by assumption $w_I\mu \in \Pi(M)$ and only finitely many $\mathbb{Z}^+$-linear combinations of $\Phi_I^+$ can be added to get a weight of $M$.Thus, as $\mu=w_I(w_I\mu)$ and $w_I$ interchange $\Phi_I^+$ and $\Phi_I^-$, only finitely many $\mathbb{Z}^+$-linear combinations of $\Phi_I^-$ can be added to $\mu$ to lie in $\Pi(M)$. Hence $U(\mathfrak{n}_I^-)M_\mu$ is finite dimensional, especially $U(\mathfrak{n}_I^-).v$.
By the way, although it is confusing to the extent of being highly regrettable, I think that the usual notation is to use $\mathfrak u_\alpha$ for the root space and $U_\alpha$ for the root group, and then $\mathfrak g_\alpha$ for the Lie algebra of the group $G_\alpha \mathrel{:=} \langle U_\alpha, U_{-\alpha}\rangle$.
Mmmh where have you seen this? For example in Humphreys "Representations of Semisimple Lie algebras in the BGG Category $\mathcal{O}$" it is as I used it and I consider Humphreys as an expert of this topic. I cannot remember to have ever seen $\mathfrak{u}_\alpha$ for the root space.
There's a holy trinity of reductive-group books: Borel, Humphreys, and Springer. Milne is a recent addition to the canon. I find as you say that all use $\mathfrak g_\alpha$ on the Lie algebra—to my astonishment, because they use $U_\alpha$ and $G_\alpha$ as I say, so $\mathfrak g_\alpha$ is not the Lie alg. of $G_\alpha$!
Alright that's consistent with what I know.
I find to my semi-relief that Conrad, Gabber, and Prasad, at least in the 1st edition, while not adopting my suggestion of using $\mathfrak u_\alpha$ for the root space, at least, as far as I can tell, only ever refer to the space in words without ever giving it a symbolic name.
Okay but I prefer $\mathfrak{g}\alpha$, since the u/U of $\mathfrak{u}\alpha$ comes from unipotent which doesn't fit for me in setting of Lie algebras.
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2025-03-21T14:48:31.871382
| 2020-08-26T12:44:03 |
370159
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|
Stack Exchange
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Continuity of the densities of a stochastic process
Let $X=(X_t)_{t\in I}$ ($I\subset\mathbb{R}$ an interval) be a stochastic process with continuous sample paths and such that $X_t$ admits a continuous Lebesgue density $\chi_t\in C(\mathbb{R}^d)$ for each $t\in I$.
Are you aware of (minimal) conditions on $X$ which guarantee that the function $(t,x)\mapsto \chi_t(x)$ is continuous on $I\times\mathbb{R}^d$?
(Could it be that the sample-continuity of $X$ already suffices?)
The truly minimal condition on $X$ that guarantees that the function $(t,x)\mapsto p_t(x):=\chi_t(x)$ is continuous is tautological: $p_t(x)$ is continuous in $(t,x)$ if and only $p_t(x)$ is continuous in $(t,x)$. As far as the minimality is concerned, I don't think you can do much better than this.
However, one can rather easily see that the sample continuity of $X$ is not enough even for the continuity of $p_t(x)$ in $t$ (for fixed $x$). E.g., let
$$p_t(x):=(1+\sin\tfrac xt)f(x)$$
for real $x$ and real $t\ne0$, with $p_0:=f$, where $f$ is the standard normal pdf. Then $p_t$ is a continuous pdf for each $t$ and, by the Riemann–Lebesgue_lemma,
$$F_t(x):=\int_{-\infty}^x p_t(u)\,du$$
is continuous in real $t$ for each real $x$. Moreover, $F_0$ is continuous and strictly increasing (in fact, $F_t$ is so for each real $t$). Hence, the process $(X_t)$ defined by the formula
$$X_t:=F_t^{-1}(U),$$
where $U$ is a random variable uniformly distributed on the interval $(0,1)$, has continuous paths. Also, $p_t$ is the pdf of $X_t$, for each $t$. However, $p_t(x)$ is not continuous in $t$ for any real $x\ne0$.
Many thanks for your answer, Iosif Pinelis. I accept your counterexample (and thus also the one that mike gave in his answer). I would hope though that, maybe, continuity of $(t,x)\mapsto p_t(x)$ could be achieved by imposing sufficient growth conditions on the differences $\mathbb{E}[|X_t-X_s|^\alpha]$ (akin to Kolmogorov's continuity theorem). Do you have any ideas how sufficient `uniformity' conditions of this sort might look like?
@fsp-b : I don't think Kolmogorov-type conditions will help. If needed, you can make $X_t$ even closer to $X_0$ (for small $t$) by replacing $\sin\frac xt$ by something like $\sin(xe^{1/t})$ or $\sin(x\exp(e^{1/t}))$, with very high frequencies near $t=0$.
I do not think sample path continuity suffices. Here is my alleged counterexample. The densities are 1 + .5*sin(x/(1-t)), 0 < t < 1 . As t -> 1 this converges to the uniform by Riemann-Lebesgue, but, of course, it isn't continuous on [0,1]x[0,1]. To get a stochastic process whose densities these are, let F_t be the cumulant and simple take $X_t(x) = F_t^{-1}(x)$. I think the $F_t$'s are continuous enough so that those paths are continuous.$$$$
Two remarks: 1. your densities have to be weakly continuous in t by the path continuity (a.e convergence implies convergence in distribution,and 2, if it bothers you that I have the discontinuity at an endpoint (t=1), just freeze the process to extend past t=1
Thanks for your comment, mike. I agree that sample-path continuity alone is insufficient.
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2025-03-21T14:48:31.871610
| 2020-08-26T14:48:51 |
370165
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Stack Exchange
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Are the densities of a continuous stochastic process locally positive in time?
Let $X=(X_t)_{t\in I}$ ($I\subset\mathbb{R}$ a (non-degenerate) interval) be a stochastic process with continuous sample paths and such that $X_t$ admits a continuous Lebesgue density $\chi_t\in C(\mathbb{R}^d)$ for each $t\in I$.
Notation: Given $\delta>0$ and $t_0\in I$, we write $I_\delta(t_0):= I\, \cap \,(t_0-\delta, t_0+\delta)$.
Are you aware of (minimal) sufficient conditions on $X$ which guarantee that for each $K\subset\mathbb{R}^d$ compact the following implication holds:
$$\tag{$\star$}\min_{x\in K}\chi_{t_0}(x) > 0 \quad(\text{some }t_0\in I)\ \quad \Longrightarrow \quad \exists\,\delta>0 \, : \, \inf_{t\in I_\delta(t_0)}\,\min_{x\in K}\chi_{t}(x) > 0.$$
(A sufficient condition for $(\star)$ would be that $(t,x)\mapsto\chi_t(x)$ is continuous, but it is not clear how this translates to a natural condition on $X$.)
The answer at https://mathoverflow.net/questions/370159/continuity-of-the-densities-of-a-stochastic-process/370163#370163 and discussion there show that your desired condition may not hold even when $X_t$ is however smooth in $t$, pathwise and on an average. So, I think no essentially non-tautological condition will suffice here.
@IosifPinelis Sorry to bother you with this extremely ill-defined question, but would you "feel" from your experience that imposing a given process $X$ to satisfy all of the above is a very restrictive model assumption if $X$ were to model a "generic data stream" (say ECG data or stock prices)? (The continuity of the density would be satisfied at least for some "reasonably-sized" class of time-homogeneous diffusions with continuously distributed initial value.)
I think diffusions like this should cause no problems. However, as the example at https://mathoverflow.net/questions/370159/continuity-of-the-densities-of-a-stochastic-process/370163#370163 shows, distributions can be however close to one another without any closeness of the corresponding densities. So, without any experience in such modeling, I'd guess it's better to avoid using broad classes of densities in modeling, and instead use integral characteristics such as cdf's and/or moments.
Thank you, @IosifPinelis.
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2025-03-21T14:48:31.871753
| 2020-08-26T15:09:31 |
370167
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"Asvin",
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Stack Exchange
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What is $H^1(\mathbb Z, GL_k(R_n))$ for a ring closely related to the cyclotomic rings of integers?
Let us consider $R_n = \mathbb Z_\ell[\theta_n]/(\theta^{\ell^n}-1)$, an auxiliary prime power $q\equiv 1 \pmod \ell$ with an action of $\mathbb Z = \langle \sigma\rangle$ by $\sigma(\theta_n) = \theta_n^q$. It acts through a quotient $\mathbb Z/\ell^{n-n_0}$ for some $n_0$.
Is there a nice description of the cohomology groups $H^1(\mathbb Z, GL_k(R_n))$? At least for $k = 1$? I am having a hard time finding references.
(The question has been edited after I realized that I wanted to solve a slightly different problem.)
I deleted my comment. Perhaps you should edit the title of the question?
I edited my title but I am not sure it's the best for the question.
Personally, I think "What is $\operatorname H^1(\mathbb Z, \operatorname{GL}k(R_n))$ for a certain ring $R_n$" is easier to read. (Also, $\mathbb Z\ell$ is $\ell$-adic integers, right?)
Thanks, I edited the title to incorporate your suggestion. And yes $\mathbb Z_\ell$ is the ring of $\ell$-adic ring of integers.
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2025-03-21T14:48:31.871855
| 2020-08-26T15:11:07 |
370169
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"url": "https://mathoverflow.net/questions/370169"
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Stack Exchange
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a question on the induced morphism of arc schemes
Let $f : X \rightarrow Y$ be a morphism of affine schemes. Denote by $X_{\infty},Y_{\infty}$ the arc schemes of $X$ and $Y$ respectively and $X_{x,\infty}$ be the closed subscheme of $X_{\infty}$ corresponding to arcs based at a closed point $x \in X$. We have an induced morphism
$f_{\infty,x} : X_{x,\infty} \rightarrow Y_{\infty}$. I am trying to understand the scheme theoretic image of $X_{x,\infty}$ under $f_{\infty}$. To begin with it is clear that if $P \in \mathbb{C}[Y]$ is such that $Pf \in \mathfrak{m}_x^{j}$, then $D^r(P)$ vanishes identically on $f_{\infty}(X_{x,\infty})$ for $0 \leq r < j$.
Let $I_{x,\infty}$ be the ideal generated by the set $\{D^iP \mid P \in \mathbb{C}[Y], Pf \in \mathfrak{m}_x^{i+1} \}$, where $D$ is the natural derivation on $\mathbb{C}[Y]$. From what we have seen above we have
$\overline{f_{\infty}(X_{x,\infty})}$ is a closed subscheme of $V(I_{x,\infty})$. But I am not able to show that they coincide.
So is the scheme theoretic image of $f_{\infty,x}$ given by
$V(I_{\infty,x})$?
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2025-03-21T14:48:31.871956
| 2020-08-26T16:02:52 |
370172
|
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"Alexander Campbell",
"Harry Gindi",
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"Tim Campion",
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"url": "https://mathoverflow.net/questions/370172"
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Stack Exchange
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monoidal (∞,1)-categories from weakly monoidal model categories
In Higher Algebra section 4.1.7, Jacob Lurie proves that the underlying $(\infty,1)$-category of a monoidal model category is a monoidal $(\infty,1)$-category.
Dominic Verity and Yuki Maehara have (independently) defined the lax Gray tensor product for model categories $S$ and $T$ for two different models of $(\infty,2)$-categories and proved that their tensor product is "weakly monoidal" in various senses:
Maehara provides functors $\otimes_n \colon T^{\times n} \to T$ for each $n \geq 2$ that define the left adjoint parts of $n$-variable Quillen adjunctions. But these functors are associative in a weak sense, eg with a zig zag of natural weak equivalences
$$ X \otimes_2 (Y \otimes_2 Z) \to \otimes_3(X,Y,Z) \leftarrow (X \otimes_2 Y) \otimes_2 Z.$$
Verity constructs two binary tensor products $\otimes, \otimes' \colon S \times S \to S$ that are naturally weakly equivalent, one of which is a closed left Quillen bifunctor and the other of which is associative up to isomorphism.
Morally, this weakness should not matter for the underlying $(\infty,1)$-category, but has anyone verified that a monoidal structure on the underlying $(\infty,1)$-category can be produced under these weaker hypotheses?
Alternatively, would someone like to take this on? I'd be happy to help publicize a result along these lines whenever it appears.
(FYI: Andrea Gagna, Yonatan Harpaz, Edoardo Lanari have a third version of the lax Gray tensor product that defines a monoidal model category in the usual sense. But I'm still interested in the question of about weak monoidal model categories, which is why I've set this aside.)
Ignoring the model structure, I would like to call the situation of the first bullet a "lax monoidal category" (or maybe "oplax"?). I've only seen lax monoidal categories (or rather their dual) in one source, namely Higher Algebra Definition <IP_ADDRESS> where they're called "corepresentable ($\infty$-)operads" (of course, it's jazzed up to be $\infty$ everything). I think another example of a (op)lax monoidal category (interacting nicely with a model structure) would be Boardman's handicrafted smash products of spectra.
It's conceivable that Lurie's framework might even be helpful here -- if you're interested mostly in the $\infty$-category presented by your model category, then maybe some of Lurie's general results on localization for $\infty$-operads could be used to say things like "there is a monoidal structure on the presented $\infty$-category".
Localization is a product-preserving functor from relative categories to $\infty$-categories. That means it's compatible with the enrichment of each category in itself, which should give you enough structure to say a functor of 2-categories to the 2-category of relative categories gives after localization one to the $(\infty,2)$-category of $\infty$-categories, which would give another approach to the first part. Unfortunately this kind of naturality for enrichments from module structures hasn't been worked out yet, however, though I believe Hadrian Heine is working on this.
Hiya Emily, nice to see you here. I think that this question was explored in the paper by Heuts, Hinich, and Moerdijk that fixes all of the mistakes from Moerdijk-Weiss. I'm not sure if they verified transport of structure to the associated ∞-category, but it's another data point: https://arxiv.org/abs/1305.3658
In the second situation, isn't the full subcategory of pre-complicial sets a genuine monoidal model category?
@TimCampion Yuki's structure is indeed an example of a lax monoidal category (which is the standard name in the literature). Lurie's "corepresentable operads" correspond to (symmetric) colax monoidal categories. Of course, the opposite of a lax monoidal category is a colax monoidal category, so I agree that this a promising approach.
@AlexanderCampbell Great, that's good to know! Regarding pre-complicial sets, Verity indeed showed that the Gray tensor product is monoidal biclosed on them, and Yuki, Chris Kaupulkin and I observed in Theorem 1.33 here that the usual model structures are monoidal with respect to it. Relatedly, in that same preprint, we construct a monoidal biclosed lax Gray tensor product on a category of marked cubical sets (without having to futz around with the category a la precomplicial sets) and show it's a monoidal model category.
@TimCampion Excellent!
There was further discussion in chat. To sum up: the suggestion from my second comment was misguided because Lurie really only deals in reflective localizations, and so can't say much about the localization from a model category to its associated $\infty$-category. But Rune points out that Hinich has results on how DK localization interacts with cocartesian fibrations which could do the trick if generalized to the locally cocartesian case.
|
2025-03-21T14:48:31.872293
| 2020-08-26T16:09:55 |
370173
|
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"Andrea Ferretti",
"Carlo Beenakker",
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|
Stack Exchange
|
Original proof of Hilbert's syzygy theorem
Does anyone know an English reference for the original proof of Hilbert's syzygy theorem? The three proofs that I know use either:
the theory of projective dimension and change of rings (plus a step to go from projective to free resolutions)
the symmetry of the Tor functors
Groebner bases
None of these tools would have been available to Hilbert, and I guess his original proof was much more direct. But, unfortunately, the original reference is in German. Is there an English proof somewhere?
I am unsure if this was Hilbert's original proof, but Arrondo's Introduction to Projective Varieties contains an elementary proof by induction on the number of variables.
Thank you, I had a quick look at it seems like the kind of argument that would have been available to Hilbert
See Theory of Algebraic Invariants
Thank you, I did not know there was an English translation
this is not a translation of Hilbert's publication on the syzygy theorem in the Mathematische Annalen (which I don't think exists), but a translation of lecture notes by Hilbert's student Sophus Marxsen
Actually, there is an English translation of Hilbert's
"Über die Theorie der algebraischen Formen" (Mathematische Annalen 36, 473--530, 1890), where the theorem is in Part III of that five-part paper. The translation is in pages 143--224 of "Hilbert's Invariant Theory Papers", Volume VIII of R. Hermann's "Lie Groups: History, Frontiers and Applications", Math. Sci. Press, Brookline, MA, 1978. I don't know if the book is available on-line.
|
2025-03-21T14:48:31.872443
| 2020-08-26T16:11:48 |
370174
|
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|
Stack Exchange
|
Representation theorem for quadratic form on Hilbert space
I think my question is more suitable for Mathematics Stack Exchange than to MathOverflow but I've already posted two related questions there and I got even more confused, so maybe I can clarify things here. I'm studying spectral theory by myself as part of my research activity and the following question arose.
Let $H$ be a Hilbert space and $Q: \mathcal{H}\to \mathbb{C}$ a function such that:
(1) There exists $C>0$ such that $|Q(x)| \le C||x||^{2}$ for every $x\in H$
(2) $Q(x+y)+Q(x-y) = 2Q(x) + 2Q(y)$ for every $x,y \in H$ and
(3) $Q(\lambda x) = |\lambda|^{2}Q(x)$ for every $x \in H$ and $\lambda \in \mathbb{C}$.
Question: Is there some bounded linear operator $A \in H$ such that $Q(x) = \langle Ax, x\rangle$?
The answer to this question seems to be affirmative and a sketch of a possible approach is given here (page 7, Lemma 12.2.7). The idea is to define:
\begin{eqnarray}
\Psi(x,y) = \frac{1}{4}[Q(x+y)-Q(x-y)+iQ(x+iy)-iQ(x-iy)] \tag{1}\label{1}
\end{eqnarray}
where $\{e_{\alpha}\}_{\alpha \in I}$ is an orthonormal basis of $H$ and then define $A$ by means of the rule:
\begin{eqnarray}
Ax = \sum_{\alpha \in I}\Psi(x,e_{\alpha})e_{\alpha}\tag{2}\label{2}
\end{eqnarray}
But I'm puzzled with this approach since I was not able to prove that $\sum_{\alpha \in I}\Psi(x,e_{\alpha})e_{\alpha}$ converges in the first place. All I could prove was $|\Psi(x,y)| \le K(||x||^{2}+||y||^{2})$ for some $K > 0$. As you can see in my previous post on math stack, it seems that the convergence problem is a bit tricky indeed.
In summary: I don't know how to prove that (\ref{2}) converges and, thus, I don't quite understand the proof of the result. However, I believe it's possible to find a more direct proof, maybe using Riesz Representation Theorem ideas (although $Q$ here is not linear) or something like that. I'd appreciate any help on either way.
There is indeed a simple proof using the Riesz representation theorem. First note that replacing $x$ by $\lambda^{-1}x$ and $y$ by $\lambda y$ in $\lvert \Psi(x,y)\rvert\leq K(\lVert x\rVert^2+\lVert y\rVert^2)$, you get $\lVert \Psi(x,y)\rvert\leq K(\lambda^{-2}\lVert x\rVert^2+\lambda^2\lVert y\rVert^2)$. With $\lambda=\lVert x\rVert^{1/2}\lVert y\rVert^{-1/2}$ this gives
$$
\lvert \Psi(x,y)\rvert\leq 2K\lVert x\rVert \lVert y\rVert.
$$
Thus for every $x\in H$ there exists $A(x)\in H$ such that $\Psi(x,y)=\langle A(x),y\rangle$ for $y\in H$ by the Riesz representation theorem. Since $\Psi$ is sesquilinear, the map $x\mapsto A(x)$ is linear, and moreover,
$$
\lVert A(x)\rVert=\sup_{\lVert y\rVert=1}\lvert \Psi(x,y)\rvert\leq 2K\lVert x\rVert,
$$
so that $A$ is also bounded.
Thanks for your answer! The only point not clear to me yet is the justification to use Riesz Representation Theorem to write $\Psi(x,y) = \langle A(x),y\rangle$. Don't you need $\Psi(x,y)$ to be linear in $y$ to each $x$? I haven't checked if this is the case, but it certainly doesn't seem to be the case from the properties of $Q$. Or the reason to use Riesz is something different?
The map $\Psi$ is sequilinear (linear in one argument, conjugate linear in the other), although it is probably a little painful to check.
Well, this must be it then. I'll try to prove it by myself. It is certainly a nice approach!
By the way, you also need sesquilinearity of $\Psi$ to see that the operator $A$ defined by your formula (2) is linear (to be precise, you only need linearity in the first argument, but conjugate linearity in the second argument then follows from the symmetry of formula (1)).
I think this is a fine question for mathoverflow. There does indeed seem to be a convergence issue. However, it can be finessed by restricting to the span of some finite subset of the basis. Then we are working on a finite dimensional space and convergence is trivial. Next, use the uniqueness of $A$ to show that when we pass to a larger finite subset the values $\langle Ax,x\rangle$ do not change. We can also use (1) to get a uniform bound on the norms of the partial versions of $A$, so that they do ultimately yield a bounded operator on all of $H$.
Hello Nik! Thanks for you answer again. I didn't comment it before because I didn't have time to elaborate the details on my own. But the idea is: take $\mathcal{E}{k}:={e{\alpha_{1}},...,e_{\alpha_{k}}}$ a finite subset of the basis and consider $\mbox{span}\mathcal{E}{k}$. Then, define $Ax$, as in my post, but only for those elements of $\mbox{span}\mathcal{E}{k}$. Now, take $\mathcal{E}{j} :={e{\alpha_{1}},...,e_{\alpha_{j}}}$ such that $\mathcal{E}{k}\subseteq \mathcal{E}{j}$. (continues)
Define $A$ on $\mathcal{E}{j}$ as before. Then, if $x \in \mbox{span}\mathcal{E}{k}$, say $x=\sum_{n=1}^{k}\alpha_{n}e_{\alpha_{n}}$, then $x \in \mbox{span}\mathcal{E}{k}$, say $x = \sum{n=0}^{j}\alpha_{n}e_{\alpha_{n}}$ where $\alpha_{n} = 0$ if $e_{\alpha_{n}} \not\in \mathcal{E}{k}$. Then, I must show that the two definitions agree, using the representation of $x$ on $\mathcal{E}{j}$. I think, so far, this is correct, right?
Next I shall get a uniform bound on $A$ and extend. This is the part where is not entirely clear to me yet.
I should probably take the bound $\sup_{x\in \mbox{span}\mathcal{E}{k}}|\Psi(x,e{\alpha})|$ to bound $A$ uniformly in each step?
Basically right, but if you let $A_k$ and $A_j$ be the desired "$A$" on ${\rm span}, \mathcal{E}_k$ and ${\rm span}, \mathcal{E}_j$, then the point is not that $A_j$ extends $A_k$ (that $A_jx = A_kx$ on their common domain, but rather that $\langle A_jx, x\rangle = \langle A_kx, x\rangle$ on their common domain.
Hm, right! I must work out the details to see if I really got it and I ask you if something goes wrong, ok?. Thanks again so much!!
Sure, no problem.
Too long for a comment. Let:
$$\tilde{\Psi}(x,y) := Q(x+y)-Q(x-y)$$
Fact 1: $\tilde{\Psi}(x+z,y) = \tilde{\Psi}(x,y)+\tilde{\Psi}(z,y)$, for every $x,y,z \in H$.
Proof: Let us evaluate the difference $\tilde{\Psi}(x+z,y)-\tilde{\Psi}(x,y)-\tilde{\Psi}(z,y)$. We have:
$$\tilde{\Psi}(x+z,y)-\tilde{\Psi}(x,y)-\tilde{\Psi}(z,y) = Q(x+z-y)-Q(x+z-y)-Q(x+y)+Q(x-y)-Q(z+y)+Q(z-y)$$
Now, note that:
\begin{align}
Q(x-y)-Q(z-y) &= \frac{1}{2}[Q(x-y+z-y)+Q(x-y-z+y)] \\ &= \frac{1}{2}[Q(x+z-2y)+Q(x-z)]
\end{align}
and also:
\begin{align}
Q(x+y)+Q(z+y) &= \frac{1}{2}[Q(x+y+z+y)+Q(x+y-z-y)] \\&= \frac{1}{2}[Q(x+z+2y)+Q(x-z)]
\end{align}
Thus, we get:
$$\tilde{\Psi}(x+z,y)-\tilde{\Psi}(x,y)-\tilde{\Psi}(z,y) = Q(x+z+y)-Q(x+z-y)+\frac{1}{2}Q(x+z-2y)-\frac{1}{2}Q(x+z+2y)$$
Moreover:
\begin{align}
Q(x+z+y)-\frac{1}{2}Q(\overbrace{x+z+2y}^{=x+z+y+y}) &= \frac{1}{2}Q(x+z+y-y)-Q(y) \\&= \frac{1}{2}Q(x+z)-Q(y)
\end{align}
and also:
\begin{align}
Q(x+z-y)-\frac{1}{2}Q(\overbrace{x+z-2y}^{=x+z-y-y}) &= \frac{1}{2}Q(x+z-y+y)-Q(y) \\&= \frac{1}{2}Q(x+z)-Q(y)
\end{align}
and this proves the result.
Fact 2: $\tilde{\Psi}(-x,y) = -\tilde{\Psi}(x,y)$
Fact 3: $\tilde{\Psi}(y,x) = Q(y+x)-Q(y-x) = Q(x+y)-Q(x-y) = \tilde{\Psi}(x,y)$
Fact 4: In particular, Fact 1 + Fact 2 lead to $\tilde{\Psi}(kx,y) = k\tilde{\Psi}(x,y)$ for every $x,y \in H$ and $k \in \mathbb{Z}$.
Fact 5: Let $b \in \mathbb{Z}\setminus \{0\}$. Then, $\tilde{\Psi}(x,\frac{1}{b}y) = \frac{1}{b}\tilde{\Psi}(x,y)$.
Proof: Note that:
$$\tilde{\Psi}(x,\frac{1}{b}y) = Q(x+\frac{1}{b}y)-Q(x-\frac{1}{b}y) = \frac{1}{b^{2}}[Q(bx+y)-Q(bx-y)] = \frac{1}{b^{2}}\tilde{\Psi}(bx,y) = \frac{1}{b}\tilde{\Psi}(x,y)$$
where, in the last equality, I used fact 4.
Fact 6: $\tilde{\Psi}(x,y+z) = \tilde{\Psi}(x,y)+\tilde{\Psi}(x,z)$
Proof: By fact 3, $\tilde{\Psi}(x,y+z) = \tilde{\Psi}(y+z,x) = \tilde{\Psi}(y,x)+\tilde{\Psi}(z,x) = \tilde{\Psi}(x,y)+\tilde{\Psi}(x,z)$
Fact 7: Set $\hat{\Psi}(x,y) := iQ(x+iy)-iQ(x-iy) = i\tilde{\Psi}(x,iy)$. Then all the above facts also hold for $\hat{\Psi}(x,y)$.
Fact 8: $|\Psi(x,y)|\le K(||x||^{2}+||y||^{2})$ implies $\Psi$ is continuous in the product topology.
Now, according to MaoWao's answer, the result follows from the Riesz Representation Theorem if $\Psi(x,y)$ is sesquilinear. It is easy to see that $-i\Psi(x,y) = \Psi(x,iy)$. Finally, let $\alpha = a+ib \in \mathbb{C}$. Then, we have:
$$\Psi(x,\alpha y) = \Psi(x,ay+iby) = \Psi(x,ay)-i\Psi(x,by)$$
Thus, find sequences $\{a_{n}\}_{n\in \mathbb{N}}$ and $\{b_{n}\}_{n\in \mathbb{N}}$ of rational numbers such that $a_{n}\to a$ and $b_{n}\to b$ and use the continuity of $\Psi$ to prove that it is anti-linear in the $y$ entry. The same reasoning leads us to the linearity in the $x$ entry. This, together with MaoWao's answer should be enough to prove the result.
|
2025-03-21T14:48:31.872916
| 2020-08-26T16:13:14 |
370175
|
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|
Stack Exchange
|
Question regarding the Wick tensor in white noise analysis
I have a question regarding the definition of Wick tensor in the framework of the white noise analysis.
To put some context to the question we start with the following Gel'fand triple
$$S(\mathbb R)\subset L^2(\mathbb R,dx)\subset S'(\mathbb R),$$
where $S$ and $S'$ are the Schwartz space of rapidly decreasing functions and its dual, the space of temperated distributions.
Let $(S',\mathcal B(S'),\mu)$ be the White noise probability space introduced by Hida.
Kuo introduces in this book the following notation where $:x^n:_{\sigma^2}$ stands for the $n$-th Hermite polynomial with parameter $\sigma^2$.
After that the "Wick tensors" are introduced for elements in $x\in S'$.
My main doubt is: Does this construction has something to do with the Wick product as defined by Janson Svante in "Gaussian Hilbert spaces"?
As far as I know the Wick power $:f^n:$ can be defined for random variables $f$ with finite moments (Janson focuses on the case where the r.v. are Gaussian), but $x\in S'$ is not a random variable, actually $x$ is the "chance parameter"! (We can say that the action of $x$ on some test function is Gaussian though)
The formula he mentions above relating the Hermite polynomial and the Wick power is easily derived for the case of a centered Gaussian random variable, but again $x$ is not a Gaussian random variable!
Do you mind giving me some explanation for this? Thanks in advance.
There is a lot of confusion around the concept of "Wick" product. Much of it is due to the following. As you mention, there is a general formula for the Wick product of a collection of random variables. Given a collection $\{X_i\}_{i \in I}$ and an $I$-valued multiindex $\alpha$, it says that $X^{\diamond \alpha}$ is the unique polynomial of degree $\alpha$ in $X$ (in the sense that each homogeneous term is of degree at most $\alpha_i$ in $X_i$ for all $i \in I$) such that $X^{\diamond 0} = 1$, $\partial_i X^{\diamond \alpha} = \alpha_i X^{\diamond \alpha - e_i}$, and $\mathbb{E} X^{\diamond \alpha} = 0$ for all $\alpha \neq 0$.
One can show that this can always be inverted in the sense that every polynomial $Y = \sum_\alpha Y_\alpha X^\alpha$ with $Y_\alpha \in \mathbb{R}$ can be written as a 'Wick' polynomial $\sum_\alpha Y_\alpha' X^{\diamond\alpha}$ of the same degree and vice-versa. This extends to formal power series.
So far, all this doesn't really look like a product, but it is then natural to define a product $\diamond$ on all Wick polynomials of the $X$'s by postulating that $X^{\diamond \alpha} \diamond X^{\diamond \beta} := X^{\diamond (\alpha+\beta)}$. Again, this product actually makes perfect sense not just between Wick polynomials, but between formal power series, at least provided one has suitable control on their growth. This is why you can define the Wick product between some Hida distributions that aren't actually random variables.
Now to the confusing bit. What if we add a new random variable $Y$ to the mix and would like to define for example $Y \diamond X$ where $X$ is one of the preexisting elements of our collection? On one hand, we could simply enlarge our collection to include $Y$ in which case one would set $Y \diamond X = XY - \mathbb{E} XY$ (assuming they are all centred). On the other hand, it may be that $Y$ is itself a polynomial function of the $X$'s or a limit of such functions. In this case, there's no need to extend our collection and we can simply use the previous definition. The two procedures will in general not give the same answer! Furthermore, in case $Y$ isn't a random variable at all but a formal power series in the $X$'s (e.g. a Hida distribution), only the second procedure would make sense while if $Y$ isn't measurable w.r.t. the $X$'s only the first procedure makes sense...
Hey Martin, I recently came back to this question. Sorry for not giving you any feedback, I've been pretty busy with courses and trying to figure out what my thesis will be about haha. I find your answer quite interesting, but I still feel dubious. In particular regarding the concept of Wick tensor. Namely when I write $\langle :x^{\otimes n}:, f^{\otimes n}\rangle$ I know that this is equivalent to write $I(f)\diamond \cdots \diamond I(f)$, where the stochastic integrals (random variables) are evaluated at the fixed $x$ (chance parameter).
This may be a matter of notation but is there a proper meaning for $:x^{\otimes n}:$? Is it the same as $x\diamond \cdots \diamond x$? But this doesn't make sense since $x$ is not a random variable, it's actually a tempered distribution.
|
2025-03-21T14:48:31.873399
| 2020-08-26T16:16:00 |
370176
|
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|
Stack Exchange
|
Contractibility of the category of cosimplicial resolutions
Let $\gamma : \mathcal{C} \to \mathcal{M}$ be a functor and define a cosimplicial resoultion of $\gamma$ as a functor $\Gamma: \mathcal{C} \to \mathcal{M}^{\Delta}$ such that
$\Gamma C$ is Reedy cofibrant for every $C \in \mathcal{C}$
for every $C$ there is a natural weak equivalence $w(C):\Gamma C \xrightarrow{\sim} c^* \gamma C $
We can define a category $\mathcal{R}=\text{coRes}(\gamma)$ where the morphisms are natural transformations $\eta:\Gamma_1 \to \Gamma_2$ such that for all $C$ the obvious triangles commute i.e. we have $w_2(C) \circ \eta_C = w_1(C)$ for all $C.$
I would like to understand why this category, as is well known, is
contractible.
Since I do not understand anything of the proof I found in the text I consulted, I am trying to prove it by myself in the following way:
A resolution exists because for every $C,$ we can find a cofibrant object $X_C$ in $\mathcal{M}^{\Delta}$ and a weak equivalence $X_C \xrightarrow{\sim} c^*\gamma C$ and this defines a functor $X(C)=X_C$ by functorial factorization.
For every $\Gamma \in \mathcal{R},$ by functorial factoriazion there is a morphism $X \to \Gamma.$
If I call weak equivalence in $\mathcal{R}$ a map $\eta$ such that $\eta_C$ is a weak equivalence in the Reedy model structure in $\mathcal{M}^{\Delta}$ for all $C,$ then given any map of resolutions $\eta:\Gamma_1 \to \Gamma_2,$ by commutativity of the triangle we have that $\eta$ is a weak equivalence under this defintion.
Now, my naive intuition is that the contractibility of $\mathcal{R}$ should follow from the fact that if we formally invert all morphisms in $\mathcal{R}=\text{coRes}(\gamma)$, the resulting localization $\mathcal{R}[\mathcal{R}^{-1}]$ is a simply connected groupoid, hence contractible.
I put on $\mathcal{R}$ the equivalence relation given by identifying all parallel morphisms, which is a congruence. In this way, all morphisms become invertible in the quotient so that I can call $\mathcal{R}/{\sim}=\mathcal{R}[\mathcal{R}^{-1}]$ and I have the quotient functor $q:\mathcal{R}\to \mathcal{R}[\mathcal{R}^{-1}].$
For every $\Gamma,$ the arrow category $\Gamma \downarrow q$ is contractible having initial object, so I conclude by Quillen's theorem A.
Is this proof reasonable?
Edit The last bullet point is wrong because when I pass to the comma category I lose the initial object.
Also, apparently we cannot just pass to the quotient without using some extra propery of $\mathcal{R}$: if it were possible to apply the reasoning I wanted to make, it would imply that any category with an object $X$ such that $\text{Hom}(X,A) \neq \emptyset$ and $\text{Hom}(A,X) \neq \emptyset$ for all $A$ would become contractible. And I just found counterexamples to this fact in this other question.
I still wonder if by using some more property of $\mathcal{R}$, for example the fact that the maps I am inverting were all weak equivalences in some model structure, we can still deduce the contractibility of $\mathcal{R}$ from that of $\mathcal{R}[\mathcal{R}^{-1}]$ along the quotient functor in this case.
Isn't this much easier? In coRes($\gamma$), every object comes equipped with a weak equivalence to the terminal object. So it's contractible. Sure, if you like, you could point out that the homotopy category (of the Reedy model structure, say) is a groupoid, by the two out of three property for weak equivalences, but I think the statement is just obvious. Unless there's a subtlety I'm missing?
Why is the object terminal? I don’t see why the morphism is unique, maybe it’s me
@DavidWhite If by terminal object you mean the identity $c^\gamma \to c^\gamma$, then I think this does not belong to $\text{coRes}\gamma$ since $c^*\gamma C$ is not necessarily cofibrant. If instead you mean the object I call $X,$ then I don't see why the morphism should be unique. But it's possible I made a big deal out of something obvious?
@giuseppe A category $\mathcal{C}$ such that the 1-groupoid $\mathcal{C} [\mathcal{C}^{-1}]$ is simply connected may not be contractible. That would be like saying that any simply connected space is contractible.
@ZhenLin I am saying that the groupoid $\mathcal{C}[\mathcal{C}^{-1}]$ is contractible. Maybe I am using the wrong terminology but by simply connected groupoid I mean that parallel arrows are equal. The proof that the category $\mathcal{C}$ is contractible is in the two last bullets.
@ZhenLin Maybe by your comment you wanted to remark that for general $\mathcal{C}$ we can't always deduce contractibility of $\mathcal{C}$ from that of $\mathcal{C}[\mathcal{C}^{-1}]$ ? But my point was that we could do that whenever there was an appropriate functor between the two as in the last two bullets.
There is nothing in your last two bullet points that uses special properties of $\mathcal{R}$, so it must be wrong.
@ZhenLin I don't see how "There is nothing in your last two bullet points that uses special properties..." is a valid mathematical reason to say that "it must be wrong". Many proofs can work in a more general context. And it used functorial factorization to show any two objects where connected through $X$. That said, I realized that my last bullet is not OK because when you pass to the comma category you don't have an initial object anymore.
It's called proving too much. Your argument, as it stood, would have shown that any connected category whatsoever is contractible, which is obviously nonsense, as you realised. It seems you have also realised your other mistake, which is that you have not shown $\mathcal{R} / {\sim}$ is isomorphic to $\mathcal{R} [\mathcal{R}^{-1}]$ or, equivalently, that $\mathcal{R} [\mathcal{R}^{-1}]$ is contractible.
@ZhenLin Can you give me an "obvious" example of a non contractible category $\mathcal{C}$ with an object $X$ which is both weakly initial and weakly terminal and such that $\mathcal{C}[\mathcal{C}^{-1}]$ is a simply connected groupoid?
I don’t think there are any obvious examples, since the condition of having an object that is simultaneously weakly initial and weakly terminal makes analysis of the homotopy type complicated. But there exists a non-contractible simply connected one-object category.
@ZhenLin I don’t understand what definition of simply connected category you are using. Isn’t a simply connected one object category just a point? If any two parallel arrows are equal there is only the identity map.
For the purposes of this question, a simply connected category is one whose nerve is simply connected. Indeed, a simply connected one-object groupoid must be a point, but this is not so for categories.
@ZhenLin Ok so your claim is that there exist a non contractible category with one object and some endomorphism which is not iso, and the nerve of this category is simply connected. But then when I was saying simply connected all this time I was using a different definition: which is that any two parallel arrows must be equal. But then under this definition any simply connected category would be a groupoid. So maybe my definition is only good for groupoids and equivalent to yours in that case
The category of resolutions is not simply connected in that strong sense, so it isn't the appropriate definition in this context.
@ZhenLin sure but I never claimed that. I always used that definition only for the groupoid $\mathcal{C}[\mathcal{C}^{-1}]$
Since you have functorial factorisations you should exploit that to the hilt.
If $\mathcal{M}$ is a model category with functorial factorisations then the category $\mathbf{c}\mathcal{M}$ of cosimplicial objects in $\mathcal{M}$, with the Reedy model structure, is also a model category with functorial factorisations. There is an obvious fully faithful embedding $\mathcal{M} \to \mathbf{c} \mathcal{M}$, so we may as well just forget about cosimplicial objects and just prove the following claim:
For every model category $\mathcal{M}$ with functorial factorisations and every diagram $F: \mathcal{C} \to \mathcal{M}$, the full subcategory $\mathcal{Q} (F)$ of the over-category $[\mathcal{C}, \mathcal{M}]_{/ F}$ spanned by the componentwise cofibrant replacements of $F$ is contractible.
Indeed, let $Q : \mathcal{M} \to \mathcal{M}$ be a functor and let $p : Q \Rightarrow \textrm{id}_\mathcal{M}$ be a natural transformation such that, for every object $M$ in $\mathcal{M}$, $Q M$ is a cofibrant object in $\mathcal{M}$ and $p_M : Q M \to M$ is a weak equivalence in $\mathcal{M}$. Such $Q$ and $p$ exist because $\mathcal{M}$ has functorial factorisations. Then, for every natural transformation $\alpha : F' \Rightarrow F$ and every object $C$ in $\mathcal{C}$, we have the following commutative square in $\mathcal{M}$:
$$\require{AMScd}
\begin{CD}
Q F' C @>{p_{F' C}}>> F' C \\
@V{Q \alpha_C}VV @VV{\alpha_C}V \\
Q F C @>>{p_{F C}}> F C
\end{CD}$$
This is all natural in $C$, so we actually have a commutative square in $[\mathcal{C}, \mathcal{M}]$, hence a zigzag $(Q F, p F) \leftarrow (Q F', \alpha \bullet p F') \rightarrow (F', \alpha)$ in the overcategory $[\mathcal{C}, \mathcal{M}]_{/ F}$. But $(Q F, p F)$ is a componentwise cofibrant replacement of $F$, and this is natural in $F'$, so we have a zigzag of natural transformations connecting the identity functor on $\mathcal{Q} (F)$ and a constant functor. Therefore $\mathcal{Q} (F)$ is contractible.
If you are geometrically inclined, you may think of the above proof as constructing a deformation retract of $\mathcal{Q} (F)$ to a point. Of course, any space with a deformation retract to a point is contractible. The gist of the argument is widely applicable and can be used in contexts where one does not have a model structure per se – this, I think, is the point of Part II of Homotopy limit functors on model categories and homotopical categories.
Thanks but I am not interested in another proof, there's a proof of this result on any book and they are all similar to this one. I wanted to see if my proof could work and I realized there is a point that does not work (thank you for making me think more about the last bullet). I never asked for a different proof so I'm sorry but I don't see how this is an answer to my question.
I'm surprised you say that you "don't understand anything of the proof", then – this is a much easier proof than what you proposed. It doesn't even rely on Quillen's Theorem A, which in my view is a non-trivial result concerning homotopy colimits.
OK. But how do you show that a zig-zag of natural transformations connecting the identity functor and a constant functor is a sufficient condition for contractibility? It seems that is where all the meat of the proof is.
I see the analogy in the sense that if we define a homotopy between functors as a zig-zag of natural transformations, then clearly by your proof we retract $Q(F)$ to a point by deformation. But why is this the right notion of homotopy which agrees with the definition of contractibility that uses the nerve?
Apply the nerve functor. The natural transformations literally become simplicial homotopies. If you prefer to work with actual topological spaces then apply the geometric realisation functor as well.
Sorry if I come back on this but I was looking at the proof with more care and I realized I don't actually understand. I don't see how you built the zig-zag of natural transformations from the identity to the constant functor. For starters, who would be the constant functor? On which object it is constant?
Constant on $Q F$.
In your diagram, you are just restating the fact that what you call $p$ is a natural transformation from $Q$ to the identity on $\mathcal{M}.$ What I do not understand is how does this say that there is "a zigzag of natural transformations connecting the identity functor on $\mathcal{Q} (F)$ and a constant functor". What constant functor?
And where is the zig-zag?
I thought what you were trying to do was to connect any other cofibrant replacement to $QF$ through $F.$ But $F$ lies outside the category, so it can't be what you were doing.
Or maybe you are taking $F' $ already a cofibrant replacement?
The intermediate is $Q F'$. $F'$ is cofibrant for the purposes of this question but it doesn’t really matter.
As I understand it, what you have done is say that for any resolution $\Gamma,$ you have a wk $\Gamma \to X,$ (what I called $X$ in my question) and you claim that this is natural w.r.t. maps $\Gamma_1 \to \Gamma_2,$ which I don't see how.
Sorry I didnt see your reply. So you are taking a cofibrant replacement of an object which is already cofibrant? $F'$ must be cofibrant otherwise it's not even in the category we want to prove is contractible, this is what I was saying.
Do you agree or I’m still not understanding ?
Yes, you take a functorial cofibrant replacement of a cofibrant replacement.
I guess I still have a problem. What you are doing is take the image of the functor $Q$ and take the slice over $F$ and show is contractible. I’m not sure this is the category of resolutions. Because not all maps of resolutions come from functorial factorization. The fact that lets you say that $p$ is a natural transformation is that all maps in your category come from functorial factorization; but this is not the case in the category of resolutions, where there are more maps.
Probably to complete your proof one needs to show that the category you built is a deformation retract of the category of resolutions.
This is more or less what is done in Hirschhorn theorem 14.5.4
The smallest category for which this argument works is indeed the image of the functor $Q$ in the slice category over $F$. But it also works for any other subcategory of the slice over $F$ that contains the image of $Q$.
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2025-03-21T14:48:31.874283
| 2020-08-26T17:00:05 |
370180
|
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|
Stack Exchange
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The quantization problem: modern quantization procedures and current status
The quantization problem is one of the most well-known current problems of theoretical and mathematical physics. It is also part of Hilbert's sixth problem (on the axiomatization of physics - see here and here) and contains the non-perturbative quantization of Yang-Mills theories, which constitute one of the Millennium problems. By the quantization problem I mean the following vague steps:
defining formally what is a classical theory, whose "space" I denote by $\mathbf{Class}$;
defining formally what is a quantum theory, whose "space" I denote by $\mathbf{Quant}$;
building some kind of globally defined rule $Q:\mathbf{Class}\rightarrow \mathbf{Quant}$ preserving expected physical properties and mapping standard classical theories into their well-know quantum version.
There are some standard quantization procedures (such as geometric quantization), which can be applied to the classical theories whose phase space is a symplectic manifold. On the other hand, even in these cases, due to Groenewold-type examples (here), the quantization procedure, if it exists, then it cannot be functorial, in the sense of mapping all classical observables into corresponding quantum observables.
There are other approaches (such as formal deformation quantization) which can be applied more generally to Poisson manifolds, but building only a perturbative quantum theory.
More recently, there are some approaches which intent to extend geometric quantization to "higher symplectic geometric" in the sense of "higher geometry", allowing to quantize a much more ample class of classical theories. This is the case, e.g., of cohomological (or motivic, or pull-push) quantization - see here and here.
On the other hand, there are also approaches to "higher deformation quantization" which extend it from Poisson algebras (i.e, $P_1$-algebras) to $P_n$-algebras, assigning to them Factorization algebras (here and here) and its variations, such as vertex operator algebras, chiral algebras, blob homology, etc. See here.
My questions are:
what are the limitations of the "higher" quantizations described above?
there are other examples of them?
what is the current status of the quantization problem and what are the most recent reviews about that?
Thank you very much. Any contribution or comment will be most appreciated.
P.S: some discussion should also appear here: https://www.researchgate.net/post/The_Quantization_Problem_Modern_Quantization_Procedures_and_Current_Status
the discussion at Researchgate linked to seems very broad and open-ended; I'm unsure there is a focused answer possible here.
Probably this helps.
I dont think having a globally defined rule $Q:Class\to Quant$ is a reasonnable thing to ask. I rather think one should have a globally defined "classical approximation" rule $A:Quant \to Class$. The quantization problem is then about finding a pre-image of a specific (class of) classical theory(ies) by $A$.
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2025-03-21T14:48:31.874522
| 2020-08-26T17:08:00 |
370182
|
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|
Stack Exchange
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On the elliptic curve $y^2 = x^3 + z^{4k}$
Are there any rational numbers $x, y, z$ with $xyz \neq 0$ such that $y^2 = x^3 + z^{4k}$ for some $k \in \mathbb{Z}_{>1}$ ?
Why not just say $k=1$?
$(x,y,z)=(2,3,1)$
The elliptic curve $y^2 = x^3 + 5^4$ and $y^2 = x^3 + 3^8$ both have rank 1, so have infinitely many rational points.
For $k=1$, the surface $y^2=x^3+z^4$ is a rational surface, so it has lots of rational points. The substitution $x=zu$ and $y=z^2v$ leads to $z=u^3/(v^2-1)$, so for almost all $u,v\in\mathbb Q$, the point
$$
\left( \frac{u^4}{v^2-1},\; \frac{u^6v}{(v^2-1)^2},\; \frac{u^3}{v^2-1} \right)
$$
satisfies $y^2=x^3+z^4$. (The same thing should work in general, take $x=z^eu$ and $y=z^{2k}v$, where $3e=4k\pm1$, with the sign chosen so that $e$ is an integer.)
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2025-03-21T14:48:31.874626
| 2020-08-26T18:07:49 |
370185
|
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|
Stack Exchange
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Forcing, constructibility, and random functions
This question is in some ways an offshoot of my recent question about trying to explain forcing to someone (such as Scott Aaronson, whose questions have prompted my questions) encountering it for the first time. Actually, I have two questions.
In Cohen's book Set Theory and the Continuum Hypothesis, he begins not with an arbitrary countable transitive model of $\mathsf{ZFC}$, but with the minimal model. That is, Cohen assumes that there exists a set model for $\mathsf{ZFC}$ where the $\in$ relation is the standard one, and $M = L(\alpha)$ for the smallest $\alpha$ such that $M$ is a model of $\mathsf{ZFC}$ (here $L(\alpha)$ denotes the constructible sets with rank less than $\alpha$). In this case, the generic extension $M[G]$ can also be described as $L(\alpha,G)$, where $L(\alpha,G)$ is defined in terms of the definable power set operation $\mathscr{D}$:
$$\eqalign{ L(0,G) &:= \lbrace G \rbrace \cup \mathrm{tr\, cl}(G) \cr
L(\gamma+1,G) &:= \mathscr{D}\bigl(L(\gamma,G)\bigr)\cr
L(\gamma,G) &:= \bigcup_{\beta<\gamma} L(\beta,G) \quad \mbox{if $\gamma$ is a limit}\cr}$$
Now in general, for any countable transitive model $M$ of $\mathsf{ZFC}$, it is a theorem that $M[G]$ is the smallest transitive model of $\mathsf{ZFC}$ containing both $M$ and $G$. This brings me to my first question.
For an arbitrary countable transitive model $M$, can $M[G]$ always be described in terms of the definable power set operation?
Suppose now that we are trying to create a model that violates $\mathsf{V}=\mathsf{L}$. We can take our poset $P$ to be the poset of finite partial functions from $\omega$ to $\lbrace 0,1\rbrace$. The standard thing to do now is to take a generic filter $G$ in $P$. Scott wondered whether we could instead take a random function $f$ from $\omega$ to $\{0,1\}$. That is, for each natural number $n$, we flip a fair coin and set $f(n)=0$ or $f(n)=1$ accordingly. Given $f$, we can define $G$ to be the set of all restrictions of $f$ to a finite domain; then $G$ is a filter by construction, but $G$ might not be generic. Now comes the second question.
Will $G$ be $P$-generic over $M$ with positive probability?
At first I thought the answer would be yes, but when I tried to prove it, I realized that I was bumping up against the distinction between measure and category. It seems that the answer to this question might depend on $M$. Perhaps for the minimal model $M$ the answer might be yes, and for some other model the answer might be no?
A randomly generated real will almost surely be $M$-generic for random real forcing (if $M$ is countable), and therefore the probability that it will be generic for Cohen forcing is zero (see Jech's Set Theory, Lemma 15.30).
I don't have my copy of Jech's book handy; I apologize if I'm just duplicating Gabe's comment. A Cohen-generic real $g:\omega\to2$ (over any $M$) has the following property: There are infinitely many $n\in\omega$ such that $g$ is identically $0$ on the interval $[n,n!!!]$. (Proof: The conditions that have such a run of $0$'s are dense in the Cohen forcing poset.) In contrast, random reals satisfy the strong law of large numbers.
What is a description in terms of the definable powerset operation? E.g. we will have $$M[G]=\bigcup_{m\in M}L_{M\cap Ord}(m, G),$$ does that count?
Also, the notation "$L(\alpha, X)$" is rather old-fashioned; more common is "$L_\alpha(X)$." This is helpful when we want to consider looking at the set of things constructible from multiple parameters (e.g. as in my previous comment).
@NoahSchweber : That's more awkward than I was hoping for, but I'm prepared to count that. Why is it true?
@TimothyChow For $x\in M[G]$, fix a name $\nu\in M$ with $\nu[G]=x$. Then from $\nu$ and $G$ we can recover $\nu[G]$ by iterating the definable powerset more-or-less $rank(\nu)$-many times; this means, since $M\cap Ord$ is an upper bound on that rank, that $x\in L_{M\cap Ord}(\nu,G)$. Conversely, for any transitive $N\models\mathsf{ZF}$ and any $y\in N$ we have $L_{N\cap Ord}(y)\subseteq N$, so apply this with $y=(\nu,G)$ and note that $M\cap Ord\le M[G]\cap Ord$ (indeed they're equal). (I can add this as an answer if you'd like.)
Basically, I'm using the fact that "iterated definable powerset" is universal in a particular sense for transfinite recursions: if I have some "nicely definable" recursive construction of length $\eta$ on input $x$, then its result already lives in $L_{\eta+1}(x)$. Here that process is name evaluation.
Timothy, the comments of Gabe and Andreas bring me back to the point I made in a comment to one of the answers in the previous question. If you want to talk about "random reals" instead of "generic reals", use Random reals to construct a model where CH is false.
@NoahSchweber : Yes, please add that as an answer! Asaf, yes, I remember that comment. What I was wondering was whether I could have my cake and eat it too. I hadn't appreciated the dichotomy between Cohen reals and random reals. I see from this MO question that there are still some unanswered questions in this area, but certainly what I was hoping for won't work.
Timothy, the only way to have your cake and eat it too is if you have a four-dimensional digestive system which preserves the 3D structure of the cake, and only takes through its temporal properties. Of course, in that case you are probably a 5+D being, so the cake will be left in the past, and you won't be able to eat it again. I don't know how a 4D digestive tract works, it's your body, not mine... :-P
For reference, here's what Jech (3rd Millennium Edition) Lemma 15.30 says. Say that $g$ dominates $f$ (where $f$ and $g$ are functions from $\omega$ to $\omega$) if $f(n)<g(n)$ for all $n\in\omega$. Then in a random real extension $M[G]$, every $f$ is dominated by some $g\in M$, but in a Cohen real extension $M[G]$, there exists some $f$ that is not dominated by any $g\in M$.
Tim, here are my answers (low on technical but hopefully high on intuition):
The answer to the first question is YES, with one provision. You need to update the
$L(\alpha,G)$ with $L(\alpha,G\cup M)$. Here is the core idea: the minimal model is the constructible universe truncated at $\alpha$, where $\alpha$ is defined by you above. That means that $M$ is made of all constructibe sets from the empty set.
Now, if you throw in G, what do you do? You attempt to build the constructible sets from G (think of the similar notion of relative recursibility. It is , mutatis mutandis, just the same: constructibility is a closure operator on sets).
Onto your question: if you start from $M_0$, a transitive model which is not the minimal one, and you add G, you must add all the constructible sets from G AND M. As it turns out, that set is precisely $M_0[G]$.
Scott's idea is quite brilliant, basically it can be summed up as generic=random. The comments above are related to it, but not entirely: they talk about a special type of forcing, the so called random forcing, whereas Scott's (and yours ) idea is broader:
is all forcing nothing but some kind of randomness in disguise?
I think the answer is yes and no, it needs to be made precise: what does it mean to "toss a coin"?
One needs to relativise this basic construct to M (remember the story of Cohen entering M? Let us do it too).
Inside M, we can define formally law-like sequences of zeros and 1s, and therefore stipulate that a sequence is random if there is no law-like description of it in M. In this sense, to be made precise, I believe Scott's intuition is correct:
the function which corresponds to the ultrafilter is always M-random.
ADDENDUM: as per Andreas comment below, I think I overstated my claim. Genericity is definitely stronger than just being random. However, I still think that the other direction, namely that every generic is M-random, still holds.
The proposed notion of randomness, that "there is no law-like description of it in $M$," is too weak to capture genericity. For example, if $M$ is a countable transitive model of ZFC, then there exists (outside $M$) a well-ordering of $\omega$ whose order-type equals that of the ordinals of $M$. A subset of $\omega$ coding such a well-ordering cannot be definable in $M$ but is also not $M$-generic for any notion of forcing $P\in M$.
OK Andreas I hear you. M-random does not imply generic, too weak. Deal. But what about the other side, namely if something is generic, then it is necessarily M-random? It seems to me that this still holds.
The converse is correct as long as the forcing notion is non-trivial, i.e., produces a model strictly including the ground model. The reason is that, if a subset of $\omega$ is definable in $M$ (i.e., is not "random" in your sense), then it is in $M$ because $M$ satisfies the axiom schema of separation.
Precisely. Any attempt to make the sequence law-like from inside would automatically make G M-constructible, so end of game. Ok Andreas, then let us move it up a notch: you convinced me that generic > random. Assume for a moment that Scott 's intuition is correct, that means that random + X = generic, where X is something else. How about we try to find it?
in other words, you subset A (for Andreas) which codes the ordinals is definitely M-random, but as you pointed out is not generic. Then, what is the difference between these two remaining in the real of randomness? Is there a way to say that G is "minimally" random whereas A is not?
Good news and (double) bad news: If I remember correctly, there's an answer to your question "What makes a set generic over a model?" in the book "Theory of Semisets" by Vopenka and Hajek. First bad news: My copy of this book is in my office, and it'll probably be a while before I return to my office. Second bad news: The only way to read that book is straight from the beginning. There's lots of unusual notation and terminology. (I did read it from the beginning, to review it many years ago --- so long ago that I'd have to start over now.)
Let us continue this discussion in chat.
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2025-03-21T14:48:31.875535
| 2020-08-26T18:59:13 |
370187
|
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|
Stack Exchange
|
Equivalence of topological Hochschild homology and Mac Lane homology via an equivalence $QA\simeq HA \wedge_{\mathbb{S}} H\mathbb{Z}$
Mac Lane homology is a homology theory for (not necessarily commutative) rings. Given a ring $A$, Eilenberg and Mac Lane define its cubical construction $QA$ to be a certain connective chain complex, whose homology is isomorphic to the stable homology of Eilenberg-Mac Lane spaces: $$H_i(QA)\cong H_{i+j}(K(A,j)) \text{ for } j \gg i.$$In fact, $QA$ is a dg-ring, and it comes with a ring map $QA \to A$ inducing an isomorphism on $H_0$. You can find the construction and basic properties of the cubical construction in [1] or [2, Chapter 13]. One then defines the Mac Lane homology of $A$ to be the Hochschild homology $$HML_i(A):= HH_i(QA,A)$$and similarly the Mac Lane cohomology is $HML^i(A):=HH^i(QA,A)$.
In 1992, Pirashvili and Waldhausen [3] proved that $HML_i(A)\cong THH_i(A)$, where the right-hand side is topological Hochschild homology. The proof goes by identifying both of them with a functor homology group. Subsequently in 1995, Fiedorowicz-Pirashvili-Schwaenzl-Vogt-Waldhausen [1] outlined a `brave new algebra' proof of this fact. In modern terms, they noticed that the Pirashvili-Waldhausen result would follow if one has a stable equivalence of $H\mathbb{Z}$-algebra spectra $$HQA \simeq HA \wedge_{\mathbb{S}} H\mathbb{Z}.$$Here, $H$ denotes the Eilenberg-Mac Lane functor taking dgas to $H\mathbb{Z}$-algebra spectra, and $\mathbb{S}$ is the sphere spectrum. (For concreteness, my preferred model of spectra is symmetric spectra in simplicial sets.) Indeed, if one has such a stable equivalence, then it follows by base change results that $HML(A)\simeq THH(A)$, for both homology and cohomology.
My question: Is it known that there is a stable equivalence $HQA \simeq HA \wedge_{\mathbb{S}} H\mathbb{Z}$ of $H\mathbb{Z}$-algebra spectra?
This may be a fact known to experts in $THH$, but I was unable to find anything in the literature. One can almost get there: because the homology of $QA$ is the stable cohomology of Eilenberg-Mac Lane spaces, it follows that one has $$\pi_i(HQA)\cong \pi_i(HA \wedge_{\mathbb{S}} H\mathbb{Z}).$$Moreover, both are $H\mathbb{Z}$-module spectra, hence wedges of Eilenberg-Mac Lane spectra, and so an isomorphism on $\pi_*$ actually lifts to a stable equivalence $HQA \simeq HA \wedge_{\mathbb{S}} H\mathbb{Z}$ of $H\mathbb{Z}$-module spectra. This argument already appears in [1], but it wasn't clear to the authors then, and it's certainly not clear to me now, how to lift this to a stable equivalence of $H\mathbb{Z}$-algebra spectra. I guess because $QA$ is connective, in theory one could write down both sides of the equation and match up the algebra structure somehow, but I am not sure if this is very tractable.
Thanks for your time.
References:
[1] Fiedorowicz, Z.; Pirashvili, T.; Schwänzl, R.; Vogt, R.; Waldhausen, F. Mac Lane homology and topological Hochschild homology. Math. Ann. 303 (1995)
[2] Jean-Louis Loday, Cyclic homology, Springer 1998
[3] Pirashvili, Teimuraz; Waldhausen, Friedhelm. Mac Lane homology and topological Hochschild homology. J. Pure Appl. Algebra 82 (1992)
Great question, and great write-up of a question. I would give more than one +1 if I could.
The answer to your question is yes, these two $H\mathbb{Z}$-algebras are equivalent (and if $A$ is commutative, they are equivalent as commutative $H\mathbb{Z}$-algebras). I am in fact in the process of writing this up with Maxime Ramzi. Hopefully the paper will be finished in a week or 2.
Update : The paper has now appeared on arXiv
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2025-03-21T14:48:31.875803
| 2020-08-26T19:01:52 |
370188
|
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"JamalS",
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"abx",
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|
Stack Exchange
|
Descent and Chow groups
One of the features of the $\mathbb{A}^1$-homotopy theory is the existence of the motivic Eilenberg-MacLane space $K(\mathbb{Z}(n),2n)$ such that for $k$-schemes $X$, we have
$$[X,K(\mathbb{Z}(n),2n)]\cong \text{Ch}^{n}(X)$$
where $[\cdot,\cdot]$ denotes the $\text{Hom}$ in the homotopy categroy of $k$-schemes $H(k)$. Hence I'm wondering for which Grothendieck topologies the Chow groups are sheaves? I'm sure there is plenty of good literature, but I haven't managed to find it. I would be very glad if someone could explain it to me or point me the right way.
$CH^1$ is not a sheaf for the Zariski topology (if it were, we would have $\operatorname{Pic}(\mathbb{P}^1)=0 $), so probably not for any reasonable Grothendieck topology.
@abx sorry but could you spell out exactly why it is not a sheaf for the Zariski topology? Physicist here.
@JamalS The argument abx is refering to is that we have a Zariski covering $\mathbb{A}^1\cup \mathbb{A}^1$ of $\mathbb{P}^1$. If $Pic$ were to satisfy the Zariski sheaf condition, then we would have an injection $$0\rightarrow Pic(\mathbb{P}^1)\rightarrow Pic(\mathbb{A}^1)\times Pic(\mathbb{A}^1).$$ However, $Pic(\mathbb{P}^1)=\mathbb{Z}$ and $Pic(\mathbb{A}^1)=0$, so this can not be.
@curious math guy the representability of the chow groups shouldn't give you descent for the set of homotopy classes of maps into it but rather of the entire space of maps. In this case it boils down to the zariski descendability of the motivic complex of weight 2n or, not any particular cohomology of it. Classically this means you have a spectral sequence computing the sections on a union, involving all the cohomologies together.
|
2025-03-21T14:48:31.875997
| 2020-08-26T20:13:42 |
370190
|
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|
Stack Exchange
|
How did Gauss characterize the metrical relations in the uniform (4 4 4) tiling of the hyperbolic unit disk?
My purpose is to verify an historical hypothesis I have on Gauss's tesselation of the unit disk as described in John Stilwell "Mathematics and its history". Looking at the relevant pages in Gauss's Nachlass (volume 8, p.102-105), I read that the commentor (Robert Fricke) on this fragment of Gauss says that Gauss's drawing (the (4 4 4) tesselation) is intended to be a geometrical illustration for composition of substitutions other then the fundamental generators of the modular group. The following sentences are a citation of Fricke about the substitutions Gauss used:
Gauss has repeatedly dealt with composition of other substitutions of the group defined from these generators. In addition to the information in fragment [i], the following formula should also be mentioned $$\frac{[\alpha, \beta,\dots,v]\theta + [\beta,\gamma,\dots,v]i}{-i[\alpha,\beta,\dots,\mu]\theta+[\beta,\gamma,\dots,\mu]}$$
which can be found in a booklet entitled “Cereri Palladi Junoni sacrum, Febr. 1805”. The continued fraction expansions of the two substitutions are given as examples:$$\frac{128\theta + 37i}{-45i\theta + 13}$$ $$ \frac{121\theta+36i}{-84i\theta+25}$$
Just to explain the words of Fricke, "the group defined from these generators" is the modular group (since the generators mentioned before this sentence are addition of $i$ and inversion with respect to the unit circle), and the $[,,\dots,]$ is the "Gaussian Brackets" notation (which Gauss defined in article 27 of Disquisitions Arithmeticae, and is connected directly to continued fractions). I think $\theta$ is just a notation for the complex variable that is transformed under the Mobius transformation.
Checking the determinants of these substitutions gave $-1$ for the first one and $+1$ for the second one, so this made me suspect that these are isometric Mobius transformations (the only thing that doesn't settle is that $a,b,c,d$ in the Mobius transformation should be real integers, not imaginary integers). One should also notice that the second Mobius transformation can be rewritten as:
$$\frac{121\theta+36i}{-84i\theta+25} = \frac{[3,2,1,3,3]\theta+[2,1,3,3]i}{-[3,2,1,3,2]i\theta+[2,1,3,2]}$$
, in exact accordance with Gauss's formula which uses the Gaussian Brackets notation (here $\alpha=3, \beta=2,\gamma=1, v=3, \mu=2$).
In addition, the diagonal elements of the second Mobius transformation ($121$ and $25$) are both reduced to $1$ modulo $12$, while the off-diagonal elements reduce to $0$ modulo $12$ (if one allows imaginary sizes for $b,c$). In other words, this shows that the second Mobius transform belongs to a congruence subgroup of level 12 in $SL_{2}(Z[i])$.
Although it seems at first that these Mobius transformations are just examples of a general principle of certain continued fractions developement, immediately after that Fricke says:
Both to explain the continued fraction development of the substitutions and to draw conclusions from the theory of functions, Gauss made use of the geometrical representation which has become the basis of the more recent theory of module functions. In the booklet just mentioned, Gauss drew the figure shown here. Since the above-mentioned continued fraction expansions of substitutions are also to be found, Gauss must have used the figure as a means to illustrate these continued fraction expansions. In fact, one has here the beginning of the well-known network of circular arc triangles, which is the basis of the theory of modulus functions. It's evident that Gauss generally understood the "principle of the symmetrical multiplication of curved triangles", which comes into consideration here, and even the character of the "natural limit" of a triangular network to be obtained in this way did not remain hidden... These are circular arc triangles of the angles $\frac{\pi}{4},\frac{\pi}{4},\frac{\pi}{4}$, and the orthogonal circle highlighted in the drawing represents their natural limit. In addition to the drawing, the following information was written by Gauss:
"Center of the first circle: $2^{\frac{1}{4}}$, radius of the first circle: $\sqrt{\sqrt{2}-1}$, center of the second circle: $\frac{1}{2}(\sqrt{\sqrt{2}+1}+\sqrt{\sqrt{2}-1})$, radius of the second circle:$\frac{1}{2}(\sqrt{\sqrt{2}+1}-\sqrt{\sqrt{2}-1})$.
Therefore, it seems that these Mobius substitutions are actually the generators for some tiling of the hyperbolic disk. But this conclusion is a result of a very shallow reading of Fricke's comments and I lack the proffesional knowledge needed to verify my reading.
I also found another reference to Gauss's hyperbolic tiling in p.198 of Schlesinger's essay on Gauss's researches in function theory, where he says:
The extent to which Gauss must also be regarded as a precursor to the theory of automorphic functions is shown not only by his construction of the fundamental domain for the modular function, but also by his interpretation of the linear fractional substitution $$t'=\frac{(\alpha +\beta i)t-(\gamma + \delta i)}{(\gamma - \delta i)t+(\alpha-\beta i)}$$ as a rotation of the spherical surface in itself, which is discussed in two notes printed in volume 8, p.354-356 the works, and that relates to a case of transcendent triangular functions that do not correspond to the modular functions, a polygonal network, found in the p.104 of volume 8 of the works, after a statement on a seperate page.
In the introduction to the book "Papers on Fuchsian Functions" (which is available on internet archive), John Stillwell says several things that seem to me very helpful:
Triangle functions arise in connection with the hypergeometric differential equation of Gauss, as Gauss himself seems to have discovered. This can be guessed from Figure 3, which is an undated fragment found in his Nachlass (see Gauss's werke, volume 8, p. 102-105). The figure is a tessellation of the unit disk by curvilinear triangles with angles $\pi/4$, and it reflects the periodicity of the inverse to a certain quotient of solutions to the hypergeometric equation. The group in this case consists of certain linear fractional transformations which leave the unit disc invariant. Invariance under linear fractional transformations arises quite generally from second order linear differential equations as follows. If $y_1,y_2$ are two linearly independent solutions of $$\frac{d^2y}{dx^2}+p(x)\frac{dy}{dx}+q(x)=0$$ then any other solutions $y'_1,y'_2$ are linear combinations of $y_1,y_2$: $$y'_1 = ay_1+by_2$$ $$y'_2 = cy_1+dy_2$$ hence the quotient of the new solutions is related to the quotient of the old by a linear fractional transformation: $$\frac{y'_1}{y'_2} = \frac{a\eta+b}{c\eta+d}$$ where $\eta = \frac{y_1}{y_2}$. By inverting the quotient of solutions one obtains a function automorphic with respect to a certian group of linear transformations. In the case of the hypergeometric equations, one is led to groups of automorphisms of certain triangle tessellations of the unit disc. The Gauss's figure is one such tessellation. Others were found by Riemann in lectures of 1858-1859 (discovered only in 1897 and published in Riemann [1902]), and the general theory was worked out independently by Schwarz [1872].
I don't now on what sources Stillwell bases his conjecture about the origin of Gauss's tessellation. After I asked people at the department for handwritten manuscripts and rare prints of the Gottingen university to search for the location of Gauss's drawing, they managed to find it after many efforts. Gauss's drawing (together with his related notes on the metrical relations) from p.104 of volume 8 of his works appears in an obscure notebook called Cod. Ms. Gauss Varia 2, as can be seen from this link.
After gaining access to this notebook, I saw only one more drawing in it related to this tessellation - but again, with no supporting context.
Therefore, I believe my attempts to answer that question arrived into a deadlock, and I don't understand how Stillwell came with his "hypergeometric function conjecture" about the origin of this tessellation. However, I still believe his notes on the mertical relation exhibit knowledge of the Poincare disk model (simply because I cannot explain their correctness in other ways).
Historical significance of Gauss's results:
Being the first drawing of it's kind, the tessellation drawed by Gauss and his related results have planted some the seeds of Felix Klein's "Erlangen program" (with the other influences being Galois's theory of equations and Riemann's geometric ideas). Klein read Gauss's fragments very closely and seems to have been influenced by them, so I think it's not an exaggeration to say that Gauss's drawing was one of his sources of inspiration.
Therefore I believe that for a correct historic appreciation of the roots of Erlangen program, it's important to know how Gauss arrived at the metrical relations in his tiling (the results stated by Gauss on the location and radiuses of the centres of the first and secondary circles in his tessellation are already confirmed by my posted answer - although it might not be the original method of Gauss).
Questions
Do Gauss's formulas for the radiuses of curvature of the first and second generation circles imply that he essentialy knew the Poincare disk model of the hyperbolic plane? I ask that because the derivation presented makes heavy use of the conception and machinery of this model. I find it unprobable, at least according to traditional accounts of Gauss's mathematical interests (the disk model is essentially a conformal model of the hyperbolic plane, and Gauss didn't think about conformal mappings until at least 1815).
It's more likely that he arrived at this result in the context of his analytic investigations, and not in the context of his "meditations" on non-Euclidean geometry.
This is an answer that is intended to verify the statements in Gauss's fragment - the four statements about the network of curved triangles. In order to facilitate the whole deduction of Gauss's formulas as presented in this answer, I added the original drawing of Gauss (to help visualize the geometric relations).
Radius and distance of the center of the first circles:
Theorem: Each of the eight curved triangles with one vertex at the origin $(0,0)$ of the Poincare unit disk has two straight sides which are portions of diameters of this disk and one curved side which is a portion of a circle orthogonal to the unit circle. The center of curvature of this orthogonal circle is at distance $2^{\frac{1}{4}}$ from the origin and it's radius of curature is $\sqrt{\sqrt{2}-1}$ (as in Gauss's fragment).
Proof:
The length $a$ of a side of an equilateral hyperbolic triangle with angles $\alpha=\pi/4,\beta=\pi/4,\gamma=\pi/4$ in the "real" hyperbolic plane (not the euclidean distance in the Poincare disk model of it) is according to the laws of hyperbolic trigonometry:
$$\cosh(a) = \frac{\cos\alpha}{1-\cos\alpha} = \frac {1}{\sqrt{2}-1} = \sqrt{2}+1\implies \tanh(a) = \sqrt{2}\sqrt{\sqrt{2}-1}\implies a = \frac{1}{2}\ln(\frac{1+\sqrt{2}\sqrt{\sqrt{2}-1}}{1-\sqrt{2}\sqrt{\sqrt{2}-1}})$$
The relation between the "euclidean" distance $d$ between each of the other two vertexes to the origin and the corresponding real hyperbolic distance $a$ is, according to the formulas of Cayley-Klein metrics:
$$\ln(\frac{1+d}{1-d}) = a \implies \frac{1+d}{1-d} = \sqrt{\frac{1+\sqrt{2}\sqrt{\sqrt{2}-1}}{1-\sqrt{2}\sqrt{\sqrt{2}-1}}}\implies d = \sqrt{\sqrt{2}-1}$$
Therefore the coordinates of the two other vertexes, after aligning the x-axis with one of the straight sides of the curved triangle, are:
$$(\sqrt{\sqrt{2}-1},0),(\frac{\sqrt{\sqrt{2}-1}}{\sqrt{2}},\frac{\sqrt{\sqrt{2}-1})}{\sqrt{2}})$$
Now the equation of any circle orthogonal to the unit circle is of the form:
$$x^2+y^2+ax+by+1 = 0$$
Substituting the x,y coordinates of the two points, one gets two linear equations with variables $a,b$, whose solution is:
$a = -\frac{\sqrt{2}}{\sqrt{\sqrt{2}-1}}, b = a(\sqrt{2}-1)$
Since the canonic form of the equation of the orthogonal circle is:
$$(x+a/2)^2+(y+b/2)^2= \frac{a^2+b^2}{4}-1 = r^2$$
where $r$ is the radius of curvature of the orthogonal circle, one gets the desired theorem by the following calculation:
$$r^2 = \frac{a^2+b^2}{4}-1 = \frac{a^2(1+(\sqrt{2}-1)^2)}{4}-1 = \frac{2\cdot(1+(\sqrt{2}-1)^2)}{(\sqrt{2}-1)\cdot 4}-1 = \frac{8-4\sqrt{2}}{(\sqrt{2}-1)\cdot 4}-1 = \sqrt{2}-1 \implies r = \sqrt{\sqrt{2}-1}$$
and the location of the center of curvature of this circle is at $(-\frac{a}{2},-\frac{b}{2})$, so its distance from the origin is:
$$\sqrt{\frac{a^2+b^2}{4}} = \sqrt{r^2+1} = 2^{\frac{1}{4}}$$
Q.E.D
Radius and distance of the center of the secondary circles:
The principle of generation of the curved triangles network is, as with all tesselations of the plane (whatever it's geometry is euclidean, hyperbolic or spherical), the successive reflection of the triangles with respect to their sides; in this way we fill the whole plane with triangles. In euclidean geometry, one can generate the tiling by simply reflecting the triangles with respect to the sides; however, in hyperbolic geometry one needs to generalize the notion of reflection from reflection with respect to line to reflection with respect to a circle.
The notion required is therefore that of inversion with respect to a circle; the secondary circular arcs (of the "second generation triangles") are inversions of the straight sides of the "first generation triangles" with respect to the first circular arcs. The formula for the radius $r'$ of the inversion of a circle with radius $r$ with respect to a smaller circle with radius $k$ whose center is at distance $d$ from the center of the inverted circle is:
$$r' = \frac{k^2r}{d^2-r^2} = \frac{k^2}{d^2/r-r}$$
In our problem: $k = \sqrt{\sqrt{2}-1}, d^2 = r^2+x^2 -2xr\cos(5\pi/8), x = \sqrt{\sqrt{2}}, -\cos(5\pi/8) = \sin(\pi/8) = \sqrt{\frac{1-\frac{1}{\sqrt{2}}}{2}} $.
In this notation, $x$ is the distance of the center of the first circles from the origin, and $d$ is derived using the euclidean cosine theorem. One needs to calculate $r'$ in the limit where $r$ tends to infinity since the straight sides of the first triangle can be thought of as circles with infinite radius.
Therefore:
$$r' = \lim_{r\to \infty}\frac{(\sqrt{\sqrt{2}-1})^2}{(r^2+x^2 -2xr\cos(5\pi/8))/r - r} = \frac {\sqrt{2}-1}{-2x\cos(5\pi/8)} = \frac{\sqrt{2}-1}{(\sqrt{2-\sqrt{2}})\cdot\sqrt{\sqrt{2}}} = \frac{\sqrt{2}-1}{\sqrt{2(\sqrt{2}-1)}} = \frac{\sqrt{\sqrt{2}-1}}{\sqrt{2}}$$
one can easily verify that the result for $r'$ is equal to the formula given by Gauss (simply by squaring both sides). The result for the distances $x'$ of the centers of the secondary circles from the origin is then easily obtained if we keep in mind that, for any $n$th generation circles: $x^2_n-r^2_n = 1$.
Final remarks:
The derivation presented here involves several conceptual/technical steps:
Calculation of the length of sides of the curved triangles in the real hyperbolic plane - this is done by using the laws of hyperbolic geometry (the hyperbolic cosine theorem). This step alone shows Gauss thought of the network of curved triangles as a tiling of the hyperbolic plane by hyperbolic triangle.
The use of Cayley-Klein metric - the whole procedure of generation of this curved triangles network can be viewed through euclidean eyes; one needs the concept of inversion of circle in a smaller circle to get a recursive definition of the network. However, Gauss deliberately gives sizes of the first triangles that correspond to tessellation of the unit disk (and not, for example, to a disk of radius 2), and i can't see any way to calculate the required size of the first triangles (required in order to make the unit circle the natural border of this network) without using the Cayley–Klein metric.
The conception of a "model" of the hyperbolic plane - the final step makes use of the fact that the straight lines of the Poincare disk model are circular arcs orthogonal to the natural border of the network (the unit circle), and this is exactly the defining property of this model (This fact is used in the derivation of the equation of circle orthogonal to the unit circle).
Goettingen State and University Library, Cod. Ms. Gauß Varia 2, p. 5.
I added here another drawing from the notebook in which he drew his $(4\ 4\ 4)$ tessellation - in this drawing the completion of the sides of triangles to full circles orthogonal to the natural border (the unit circle) is shown more clearly. This drawing, together with the calculations involved, certainly proves he conceived of straight lines as circular arcs perpendicular to the unit circle, with exact correspondence with Poincare disk model.
Remaining questions
I never read any claim that Gauss knew this model by any of the reliable sources on the history of mathematics. In addition, those results of Gauss appear isolated from his body of mathematical research, and that is why I think there is a need to place it in the right context.
The remaining questions are:
How he originally found those results?
How he was led to consider this particular tiling?
I know those questions are very hard to answer without access to his original notebook, but maybe an expert in this field can place Gauss's results in the right context by a kind of "extrapolation" on all known sources.
|
2025-03-21T14:48:31.877475
| 2020-08-26T20:32:05 |
370192
|
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|
Stack Exchange
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Heegard diagrams for three-manifolds
I have a basic question about the Heegaard diagrams involved in providing a framework
for calculation of Floer-Homology of three-manifolds.
Typically such diagrams look like Figure 1 and Figure 2 here or these two (Image1 or Image2) from researchgate network.
And I would thankful if anybody could try to explain how to "read" this diagrams
to a non-topologist.
We see a compact surface, which is probably assumed to be the boundary of certain
three-manifold, and theory of compact surfaces states that this surface
is up to homeomorphism always a connect
sum of $g$ tori for $g \ge 1$. Looking at images in Image2 in last two links we
find two sets of $g$ disjoint curves ("team red" $\alpha_0,..., \alpha_g$ and
"team blue" $\beta_0,..., \beta_g$).
Now how does this information provide instructions to build a three-manifold?
My non-expert guess is that this data tells us: Start with two identical disjoint three-manifolds which have two $g$-tori as surfaces and the data provided by these
Heegaard diagrams is nothing else than instructions how to glue the two
three-manifolds along the surfaces. The instruction says probably that
the curve $\alpha_i$ of one surface has to be glued homeomorphically with
$\beta_i$ for other surface. And seemingly if we know all pairs of curves $\alpha_i$ and
$\beta_i$ are glued together, then the gluing of the two surfaces
is already uniquely determined up to homeomorphism and therefore we know how to glue the two disjoint three-manifold also the boundary.
Is this exactly the correct way to read a Heegaard diagram? Does there exist a more
conventional way? Sorry, if the question is too elementary, I'm not an
algebraic topologist and the motivation of this question is pure curiosity.
Start with $\Sigma \times I$. For each $\alpha$ curve (which you ought to thicken into an annulus, which you can do in a way which is unique up too isotopy), you glue on a copy of $D^2 \times I$ by pasting $S^1 \times I$ along the thickened $\alpha$ curve inside of $\Sigma \times {0}$. Do the same thing for the $\beta$ curves inside of $\Sigma \times {1}$. What you have produced is a compact 3-manifold with boundary $S^2 \sqcup S^2$; glue on 3-balls to make this a closed manifold. Gompf and Stipsicz have a good discussion.
I’ve taken the liberty of editing your title, because I don’t think anyone who studies 3-manifolds calls them “3-folds”.
You may also look at the first three chapters of beautiful paper of P. Ozsváth and Z. Szabó: http://math.mit.edu/~petero/Introduction.pdf
@MikeMiller: You mean it in the sense that you glue along each thickened tubulus of every
$\alpha_i \subset \Sigma \times {0}$ a $2$-handle and them the same game
for $\beta_j \subset \Sigma \times {1}$. What I still not understand is why
the $3$-manifold obtained by this construction has boundary $S^2 \sqcup S^2$?
You are at each stage performing a handle attachment, which changes the boundary by a surgery along the attaching sphere. Check that this process reduces the genus by 1 every time you do a handle attachment. You do it g times. Again, G&S is a good source.
@Mike Miller: Hi, sorry for repeated annoying but I would like to know following
aspect regarding your construction: Recently in Sivek's Lecture 24 (http://wwwf.imperial.ac.uk/~ssivek/courses/273notes/lecture24.pdf) I found
a descripion of explicite construction of $3$-folds by the
Heegaard data that works as follows (page 1):
He wrote: Given a Heegaard decompsition $Y = H_{\alpha} \cup_{\Sigma} H_{\beta}$
, we can desribe the decomposition completely in terms of curves on
the surface $\Sigma$. We construct $H_{\alpha}$ by
attaching $g$ $1$-handles to $\Sigma \times [0,1]$ along curves
$\alpha_1 \times {0}, ..., \alpha_g \times {0}$ and filling
in the resulting $S^2$ on the boundary with a ball,
and similarly we construct $H_{\beta}$
by ataching $g$ $2$-handles to$\Sigma \times [0,1]$ along curves
$\beta_1 \times {1}, ..., \beta_g \times {1}$ and
filling in the remaining $S_2$ with a ball. [...]
This construction looks very similar to yours except that
you attatched to $\beta_1 \times {1}, ..., \beta_g \times {1}$
but as well $\alpha_1 \times {0}, ..., \alpha_g \times {0}$
in both cases $2$-handles. (more precisely to the thickings (=annuli) of
$\alpha_i \times {0}, \beta_j \times {1}$). But the important point is that you worked
on both sides on $\Sigma \times {0}$ and
$\Sigma \times {1}$ with $2$-handles, while Sivek
attatched on $\Sigma \times {0}$ $1$-handles and on
$\Sigma \times {1}$ $2$-handles.
And that's what I not understand. Do we obtain in yours and Sivek's
construction the same object (if yes, why?)
or do Sivek's notes contain an error?
Expecially I don't know how Sivek intends to attatch $1$-handles
along $\alpha_1 \times {0}, ..., \alpha_g \times {0}$.
Sivek is thinking from the ground up: you start with a 0-handle, you attach g 1-handles, and then what you have is diffeomorphic to the handlebody $H_\alpha$ (the $\alpha$ curves being the boundary of the core disc of the 1-handle). Using a diffeomorphism $\partial(H_\alpha) \to S_g$ sending those circles to the $\alpha$ curves, you then attach the 2-handles along the $\beta$ curves. This is the perspective that Morse theory suggests. See maybe Oszvath-Szabo intro lecture on HF.
I think you'd be best off finding a topologist who is willing to work through these constructions in person.
Chapter four of "Knots, Links, Braids and 3-Manifolds" by Prasolov and Sossinsky gives a highly readable (and nicely illustrated) introduction to three-manifolds via Heegaard splittings. Another, more classical, reference is chapter two of "Three-manifolds" by Hempel. Note that Hempel calls handlebodies "cubes with handles".
You are probably familiar with definitions and theorems. But I prefer to write those for completeness. And also excuse for a paint-like drawing. I hope that they will be useful.
A handlebody of genus $g$ is a $3$-manifold constructed from the standart $3$-ball $B^3$ by adding $g$ copies of $1$-handles $B^2 \times B^1$. It is denoted by $H$ and $\partial H \approx \Sigma_g$ where $\Sigma_g$ is a genus $g$ surface, see the following figures.
Let $Y$ be a $3$-manifold. A Heegaard splitting of $Y$ is a decomposition of $Y$ such that
$Y=H_0 \cup H_1$ where $H_0$ and $H_1$ are handlebodies,
$\partial
H_0 = \partial H_1 = \Sigma_g$.
Theorem(Singer, 1933): Any closed oriented 3-manifold $Y$ admits a Heegaard splitting.
The genus $g$-surface $\Sigma_g$ is constructed from $S^2 = \mathbb{R}^2 \cup \{ \infty \}$ by attaching $g$ copies of $1$-handles, where we draw attaching spheres as pairs of matching disks.
So the followings are Heegaard splittings of $S^3$ and $S^1 \times S^2$ respectively:
The following is for a Heegaard diagram of lens space $L(5,2)$:
And the last scheme is for the famous Poincaré homology sphere $\Sigma(2,3,5)$:
|
2025-03-21T14:48:31.878121
| 2020-08-26T20:33:44 |
370193
|
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|
Stack Exchange
|
Higher order arithmetic, hierarchies and proof theoretic ordinals
I asked this question on MSE some days ago but I have not received any answer so I have decided to post it here.
I would like to consider a generalization of the notation $\Pi$ and $\Sigma$ used for the arithmetical hierarchy $(\Pi^0_n$, $\Sigma^0_n)$ and the analytical hierarchy $(\Pi^1_n$, $\Sigma^1_n)$ to set of formulas $\Pi^\alpha_\beta$, $\Sigma^\alpha_\beta$, where $\alpha$ and $\beta$ range over ordinal numbers.
I suppose that the set of formulas of $n$-th order arithmetic is the set $\Pi^{n-1}_\omega \vee \Sigma^{n-1}_\omega$.
The superscript ranging over the ordinals corresponds to the transfinite recursion of the power set operator, similarly to the definition of the Von Neumann universe, but starting with the set of the natural numbers instead of the empty set.
About the subscript (which is the number of quantifiers) I am not sure. I know that it can be defined at least to $\omega_1^{CK}$ as you can see in this table where additionally it is written that $\Pi^0_{\omega_1^{CK}} =\Sigma^0_{\omega_1^{CK}}$ corresponds to a set of formulas $(\Delta^1_1)$ of second order arithmetic. This fact is not clear to me, it would be nice if somebody can elaborate about it in the answer.
Can you use ordinals beyond $\omega_1^{CK}$ as subscript?
If the superscript or the subscript is beyond $\omega_1^{CK}$ will you have non recursive theories with non recursive proof theoretic ordinals?
Given any ordinal $\gamma$ is there an ordinal $\alpha$ such that the proof theoretic ordinal of $\Pi^\alpha_1-CA_0$ is higher than $\gamma$. (What about the subscript, what if you replace $\Pi^\alpha_1$ with $\Pi^1_\alpha$)
The same question but restricted to recursive ordinals: given any ordinal $\gamma<\omega_1^{CK}$ is there an ordinal $\alpha<\omega_1^{CK}$ such that the proof theoretic ordinal of $\Pi^\alpha_1-CA_0$ is higher than $\gamma$.
|
2025-03-21T14:48:31.878276
| 2020-08-27T00:04:35 |
370198
|
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|
Stack Exchange
|
High direct image of dualizing sheaf
I'm reading the paper "High direct image of dualizing sheaf" of professor Kollar. I summarizing my questions as follows:
Let $f:X\rightarrow Y$ be surjective projective morphism between smooth projective variety. L be a very ample line bundle on $Y$. Suppose the torsion part of $R^if_*\omega_X$ is supported at a closed point $y\in Y$. We choose a section $y\in D'\in |L|$. Then the map
$$H^0(L\otimes R^if_*\omega_X)\rightarrow H^0(\mathcal O(D')\otimes L\otimes R^if_*\omega_X)$$
induced by $\mathcal O\rightarrow \mathcal O(D')$ is not injective.
(Actually it's false, hence we have torsion freeness ) I don't know why by choosing such a $D'$ then the injectivity of the above map fails.
Thanks advance, maybe I haven't summarizing the conditions in a very precise way! Experts's opinions will help me a lot.
|
2025-03-21T14:48:31.878836
| 2020-08-27T05:12:38 |
370205
|
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|
Stack Exchange
|
Embedding of wreath product
Consider the wreath product $G=\mathbb{Z}_2\wr O_n(\mathbb{R}),$ where $O_n(\mathbb{R})$ is the set of orthogonal groups over reals. Can we show $G$ embeds in a nice enough group (for example, some subgroup of $\mathbb{GL}_{2n}(\mathbb{R})$ perhaps ?).
A theorem of Burnside shows that a subgroup of finite exponent in $\mathrm{GL}(k,\mathbb{R})$ must be finite. Even the lamplighter group $\mathbb{Z}_2 \wr \mathbb{Z}$ is not a subgroup of $\mathrm{GL}(k,\mathbb{R})$.
AGenevois' comment answers the question in any reasonable interpretation, but I'm still not sure exactly what is being asked. What representation of $O_n(\mathbb{R})$ is used to define the wreath product? Why refer to the 'set' of orthogonal groups?
|
2025-03-21T14:48:31.878927
| 2020-08-27T07:06:38 |
370208
|
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"e. sfe",
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|
Stack Exchange
|
How to retrieve back the input using Bussgang theorem?
If we have a non-linear function $f$, that is applied to input $x$,
we have then the output $y=f(x)$
Using Bussgang decomposition we can linearize this nonlinearity and express $y$ as
$y=Bx+ η$,
Where B could be found as per the below formula below:
$B= \frac{E(yx^{*})}{E(|x|^2)}$
We can use input pilots at the beginning, so assume we we know $x,y$ and we get $B$, from the just above formula.
Then we obtain $\eta= y-Bx$,
and this was for the training phase.
But now for the testing , I already have $B$ that I obtained previously from the training phase, but I don’t have $\eta$ nor $x$, so how can I find back the input $x$?
Something sounds strange. $B$ is obtained as the ratio of two expectations (probably, found empirically on many trials, not just on a single one) to minimize the expected square error, so what "input $x$" are you trying to find?
@fedja yeah B is the ratio of two expectations, i did mention the formula in the question. Concerning the input x it is the variable of linearization, may be i am not understanding Bussgang properly, i would appreciate if you can refer me a good source code imlementation, or resource to understanding Bussgang decompostion better :) thanks in advance
|
2025-03-21T14:48:31.879063
| 2020-08-27T08:14:44 |
370212
|
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"Alapan Das",
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|
Stack Exchange
|
Approximating by incrementing positive integers
Let $\mathbb{N}$ denote the set of positive integers. For $\alpha\in \; ]0,1[\;$, let $$\mu(n,\alpha) = \min\big\{|\alpha-\frac{b}{n}|: b\in\mathbb{N}\cup\{0\}\big\}.$$ (Note that we could have written $\inf\{\ldots\}$ instead of $\min\{\ldots\}$, but it is easy to see that the infimum is always a minimum.)
Is there an $\alpha\in \; ]0,1[$ such that for all $n\in\mathbb{N}$ we have $\mu(n+1,\alpha)<\mu(n,\alpha)$?
There is no such $\alpha$.
If $\alpha\in\mathbb Q$, there is $n$ such that $\mu(n,\alpha)=0$, thus $\mu(n+1,\alpha)<\mu(n,\alpha)$ is impossible.
If $\alpha\notin\mathbb Q$, a classical result in Diophantine approximation says that there are infinitely many $n$ such that
$$\mu(n,\alpha)<\frac1{\sqrt5n^2}.$$
If then
$$\mu(n+1,\alpha)<\mu(n,\alpha),$$
let $a/n$ and $b/(n+1)$ be the respective closest approximations of $\alpha$. We have
$$\left|\frac an-\frac b{n+1}\right|<\frac2{\sqrt5n^2}<\frac1{n(n+1)},$$
while
$$\left|\frac an-\frac b{n+1}\right|=\frac{|a(n+1)-bn|}{n(n+1)}\ge\frac1{n(n+1)}$$
unless $a(n+1)=bn$, i.e., the approximating fractions are $0$ or $1$. Since this happens for infinitely many $n$, this is impossible for $\alpha\in(0,1)$.
No. If $\alpha$ is rational, set $n$ to the denominator of $\alpha$. Otherwise set $n$ to the denominator of the third convergent to $\alpha$. In both cases, we get $\mu(n+1,\alpha)>\mu(n,\alpha)$.
It's easy to proof that $\alpha$ shouldn't be a rational number.
Now, let $\frac{1}{n-1}>\alpha>\frac{1}{n}, n>1$ and $\alpha-\frac{1}{n} < \frac{1}{n-1}-\alpha$.
Then, $\mu(\alpha, k+1)< \mu(\alpha ,k)$ for all $k=1,2..., n-1$.
If $\mu(\alpha, n+1)<\mu(\alpha, n)$,
then either,
$\frac{1}{n-1}>\alpha >\frac{b}{n+1} >\frac{1}{n}$ for some $b \in \mathbb N, b>1$.
or,
$\frac{1}{n-1}>\frac{b}{n+1} >\alpha >\frac{1}{n}
$ for some $b \in \mathbb N, b>1$ (With $\frac{b}{n+1}+\frac{1}{n}>2\alpha$).
Both of these implies
$1+\frac{2}{n-1}>b>1+\frac{1}{n} \Rightarrow n=2$
To satisfy $\mu(\alpha, 4)<\mu(\alpha, 3)<\mu(\alpha, 2)<\mu(\alpha, 1)$ we need, $\frac{3}{4}>\alpha>\frac{17}{24}$.
But, $\frac{4}{5}>\frac{3}{4}$ and $\frac{3}{4}-\frac{17}{24}<\frac{17}{24}-\frac{3}{5}$, hence, $\mu(\alpha, 5)>\mu(\alpha, 4)$.
So, there can't be any such $\alpha \in (0,1)$.
Perhaps some kind of an argument like this can be made to work, but as far as I can see, it is not correct as written: if $17/24<\alpha<3/4$, then $n=2$, but $\mu(1,\alpha)>\mu(2,\alpha)>\mu(3,\alpha)>\mu(4,\alpha)$.
Thank you for correcting me. I missed that.
|
2025-03-21T14:48:31.879264
| 2020-08-27T09:11:10 |
370214
|
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|
Stack Exchange
|
RSK correspondence for sum of two matrices
The celebrated RSK correspondence (see Wikipedia page) assigns to each integer $X$ matrix a pair of Young tableaux $P$ and $Q$. Now suppose that we have three integer matrices $X_1$, $X_2$, and $X_3$, with the corresponding pairs $(P_1,Q_1)$, $(P_2,Q_2)$, and $(P_3,Q_3)$.
Let $X_3 = X_1 + X_2$. What is the relationship between $(P_3,Q_3)$ and $\{(P_1,Q_1),(P_2,Q_2)\}$? Can one derive $(P_3,Q_3)$ in a formal way from the other pairs? For example, it easy to see that the content of $P_3$ is the union of the contents of $P_1$ and $P_2$. Is it true that $P_3$ is some formal sum of the Young tableaux $P_1$ and $P_2$?
One place to start is when $X_1$ is just a matrix with a single non-zero entry.
There is a type of commutative formal sum on SSYT, namely to elementwise add the corresponding entries in the GT-patterns, see https://www.math.upenn.edu/~peal/polynomials/gtpatterns.htm#gtpatternsAsSSYT for this correspondence
I don't think they'll be an easy answer, but it might be helpful to use the "toggle" description of RSK for this problem: http://www-users.math.umn.edu/~shopkins/docs/rsk.pdf
|
2025-03-21T14:48:31.879370
| 2020-08-27T09:18:09 |
370215
|
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"Liviu Nicolaescu",
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|
Stack Exchange
|
Directional gradient on sphere
We consider the following function
$$f:(\mathbb S^n)^N \rightarrow \mathbb R^{n+1} \text{ such that } f(x_1,...,x_N)= \sum_{i=1}^N x_i.$$
This function can be written in Cartesian coordinates as $f(x)=(f_1(x),..,f_{n+1}(x))$ and I would like to know if one can find a simple expression for the derivative
$$\nabla_{x_1} \left(\frac{f_1(x)}{\Vert f(x) \Vert_{\mathbb R^{n+1}}}\right)$$
where $\nabla_{x_1}$ is the gradient on $\mathbb S^n$ with respect to $x_1.$
Can one somehow carry out this differentiation? I am a bit struggeling with computing $\nabla_{x_1} f_1(x)$ here.
If $M$ is a smooth submanifold of a Euclidean space $E$, $U\subset E$ is an open subset of $E$ containing $M$ and $f:U\to\mathbb{R}$ is a $C^1$-function, then for every $p\in M$ we have $\nabla^M f(p)= \mathrm{Proj}_p\nabla^E f(p)$, where $\mathrm{Proj}_p:E \to T_p M$ denotes the orthogonal projection onto the tangent space to $M$ at $p$.
$\newcommand\R{\mathbb R}\newcommand\S{\mathbb S}$Take any $(x_1,x_2,\dots,x_N)\in(\S^n)^N$.
Let $(-1,1)\ni t\mapsto X_1(t)\in\S^n$ be any smooth curve such that $$X_1(0)=x_1.$$ For any $t\in(-1,1)$, let
$$X(t):=(X_1(t),x_2,\dots,x_N)$$ and $$S(t):=f(X(t))=X_1(t)+x_2+\dots+x_N[\in\R^{n+1}],$$
so that $X(0)=(x_1,x_2,\dots,x_N)$ and
$$S(0)=s:=x_1+x_2+\dots+x_N.$$ Let $v:=X'_1(0)$, so that $S'(0)=v$. Let
$$S_1(t):=f_1(X(t))=e_1\cdot S(t),$$
where $\cdot$ denotes the dot product and $e_1$ is the first vector of the standard basis of $\R^{n+1}$. So, $S'_1(0)=e_1\cdot v$. So, for
$$r(t):=\frac{f_1(X(t))}{\|f(X(t))\|}=\frac{S_1(t)}{\|S(t)\|}$$
we have
$$r'(0)=\frac1{\|S(0)\|^2}\Big(\|S(0)\|S'_1(0)-S_1(0)\frac{S(0)}{\|S(0)\|}\cdot S'(0)\Big) \\
=\frac1{\|s\|^2}\Big(\|s\|e_1\cdot v-(e_1\cdot s)\frac{s}{\|s\|}\cdot v\Big).$$
Thus, the gradient in question is $g-(g\cdot x_1)x_1$, where
$$ g:=\frac1{\|s\|^2}\Big(\|s\|e_1-(e_1\cdot s)\frac{s}{\|s\|}\Big).$$
$\big($Note: $g-(g\cdot x_1)x_1$ is the orthogonal projection of the vector $g$ onto the tangent hyperplane to the unit sphere $\S^n$ at point $x_1$.$\big)$
|
2025-03-21T14:48:31.879514
| 2020-08-27T09:29:56 |
370216
|
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"Emre Yolcu",
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|
Stack Exchange
|
Does this prove Collatz is a $\Sigma_1$ problem?
So I got an email from one of my colleagues on the Collatz Conjecture with a link to the article Computer Scientists Attempt to Corner the Collatz Conjecture by Kevin Hartnett in Quanta Magazine.
On digging through the conjecture, it seems it is not even known it is a $\Pi_1$ or $\Sigma_1$ statement. See the Math.StackExchange question Is the Collatz conjecture in $Σ_1/Π_1$?. For more information we can find details in the following thesis:
Matthew Alexander Denend, Challenging variants of the Collatz Conjecture, Masters Thesis, The University of Texas at Austin 2018, doi:10.26153/tsw/1559.
Does it mean they have already shown Collatz is a $\Sigma_1$ statement?
It seems I was wrong - see Emre Yolcu's comment below.
My understanding is that that has not been accomplished (although the Quanta article is pretty vague so I could be misunderstanding the situation).
The Quanta article describes the following process:
Whip up a rewriting system which always terminates iff Collatz is true. This has been successfully done - but note that the termination problem is a priori $\Pi^0_2$, just like Collatz.
Try to find a collection of matrices satisfying some complicated constraints relating to that rewriting system. This is the task which SAT solvers are relevant for. However, they have not yet found an appropriate collection of matrices.
(This is where I was wrong:) Even after finding such a collection, we're not done. All that this will accomplish is reduce Collatz to a particular problem about matrix multiplication (which the Quanta article doesn't state - moreover, it doesn't explain why that problem should be more tractable than the rewriting one or the original Collatz conjecture).
Re: that third bulletpoint, I think that there's a particular part of the article which is potentially confusing:
“You try to find matrices that satisfy these constraints,” said Emre Yolcu, a graduate student at Carnegie Mellon who is working with Heule on the problem. “If you can find them, you prove [they’re] terminating,” and by implication, you prove Collatz.
It would have been clearer to write "If you can find them, you then try to prove [they're] terminating, and if you can you prove Collatz." That is, finding a system of matrices satisfying the given constraints - which is indeed $\Sigma^0_1$ - is only the first step, and the remaining fact we need to prove is presumably still $\Pi^0_2$.
Actually it seems I got that exactly wrong!
That said, pending further elaboration from Emre we may only have a $\Sigma_1$ sentence which implies Collatz - I don't know if the nonexistence of an appropriate matrix family would imply that Collatz fails.
I was misquoted in the article—I actually said "If you can find them, this implies that the rewriting system is terminating." This is due to a method called "matrix interpretations" [1], widely used in proving termination of rewriting systems. Finding the matrices does settle the conjecture as true. We will put a preliminary writeup on arXiv soon. [1]: Endrullis, J., Waldmann, J. & Zantema, H. Matrix Interpretations for Proving Termination of Term Rewriting. J Autom Reasoning 40, 195–220 (2008). https://doi.org/10.1007/s10817-007-9087-9
@EmreYolcu Oh, that's fascinating! I didn't know that at all. Do you know if the existence of such matrices is equivalent to Collatz, or merely sufficient for Collatz?
As far as we know, it is merely sufficient. This method is basically searching for proofs of a certain size that fit a template, and if the search does not succeed after some time we modify the constraints to allow larger proofs and keep searching. There exist rewriting systems that are terminating but do not admit a direct matrix interpretation proof. For a given rewriting system, I imagine it might be possible to prove that if it is terminating then there is a matrix interpretation proof of this fact. We know of no such result for the system at hand.
@EmreYolcu Of course it certainly can't be equivalent in general since termination is $\Pi_2$-complete.
"Finding matrices implies Collatz" implies Collatz is in $\Sigma_1$ and given that this has been known for two to three years why no one in mathematics mainstream cares? This seems pretty fundamental and groundbreaking in itself. What other problems fall in this category?
It does not imply that Collatz is in $\Sigma_1$: as Noah says above, there is only a $\Sigma_1$ sentence implying Collatz, not one that is equivalent to it.
|
2025-03-21T14:48:31.879837
| 2020-08-27T09:57:27 |
370222
|
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|
Stack Exchange
|
P-filter property?
Let $\mathcal{F}$ be a $P$-filter on $\omega$. Denote by $\Omega=\bigsqcup \omega_i$ where $\omega_i=\omega$. Consider the $P$-filter $\mathcal{S}$ on $\Omega$ whose base is as follows
$(\bigsqcup_i F_i, F_i\in \mathcal{F})$.
$\mathcal{F}$ filter is isomorphic to $\mathcal{S}$ filter ?
Is $i$ intended to range over a countably infinite set? Are you asking about the usual notion of isomorphism of filters (bijection between the underlying sets, sending one filter to the other), or about order-isomorphism of the filters considered just as partially ordered sets, or about another notion of isomorphism?
Yes, i have range over a countably infinite set. Isomorphism of filters is bijection between the underlying sets, sending one filter to the other.
I've seen "P-filter" defined to mean a filter $\mathcal F$ such that, for any countably many sets $A_n\in\mathcal F$, there exists $B\in\mathcal F$ with all $B-A_n$ finite. I've also seen it defined with some additional requirements, for example that $\mathcal F$ contains all cofinite sets. What additional requirements do you intend your P-filter $\mathcal F$ to satisfy?
P-filter to mean a filter F such that, for any countably many sets An∈F, there exists B∈F with all B−An finite.
After the clarifications in comments of OP (August 28 and today), here's a counterexample. Let $\mathcal F$ be the filter consisting of only $\omega-\{0\}$ and $\omega$. Then $\mathcal S$ (as defined in the question) contains a set whose complement is infinite, whereas $\mathcal F$ does not. So they cannot be isomorphic.
|
2025-03-21T14:48:31.880083
| 2020-08-27T10:16:49 |
370224
|
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|
Stack Exchange
|
Non linear second order PDE involving max operator (Dynamic Programming)
I'm trying to solve the following Dynamic Programming equation in continuous time ($dt \rightarrow 0$)
$$ v(x,t) = \max\Big\{|x|\,,\,v(x,t)+dt\Big(v_t(x,t)+\frac{1}{2(t+1)}v_{xx}(x,t)\Big) \Big\} - \alpha dt$$
for $x \in [-1,1]$, $t \in [0,T]$ with $v(x,T) = |x|$ and $v(1,t)=v(-1,t)=1 \,\,\forall t$.
Is there a way to properly take the limit where $dt \rightarrow 0$ and arrive to a more conventional PDE? Is this an appropriate way of writing the equation? What would be a more appropriate way, if there is?
The approach I have followed is to take the equation as it is and use an explicit first order Euler method with standard centered second order finite differences for $v_{xx}$, which leads to
$$ v_i^j = \max\Big\{|x^j|\,,\,v_{i+1}^j+\frac{\Delta t}{2{\Delta x}^2(t_{i+1}+1)}(v_{i+1}^{j+1}-2v_{i+1}^{j}+v_{i+1}^{j-1}) \Big\} - \alpha \Delta t$$
where I have defined $t_i \equiv i\Delta t$, $x^j \equiv j\Delta x$, and $v_i^j \equiv v(x^j,t_i)$.
I believe it is producing reasonable solutions for small or moderate values of $\alpha$, but when $\alpha$ gets quite large I need to decrease $\Delta x$, which in turn forces me to decrease $\Delta t$ to maintain stability, and becomes impractical. I'm a bit of a loss on how to implement an implicit method here, given the presence of the max operator.
Edit:
I have noticed that the equation can be written as
$$ \alpha = \max\Big\{\frac{|x|-v(x,t)}{dt}\,,\,v_t(x,t)+\frac{1}{2(t+1)}v_{xx}(x,t) \Big\} $$
but I'm not sure if it helps, I don't really know what to do with the first term inside the $\max$
Edit 2:
Following @MichaelRenardy 's advice, I wonder if it could be written just like this:
$$ v_t(x,t) = \min\Big\{0,-\frac{1}{2(t+1)}v_{xx}(x,t) + \alpha \Big\} $$
since $v_t = 0$ implies $v = |x|$, and using the notation of obstacle problems, it's clear that $v = |x|$ in the contact region.
Something seems to be misstated here: If I set dt=0, I just get $v=\max(|x|,v)$, which does not allow us to solve for $v$.
@MichaelRenardy you have to use the inital (or rather final) condition v(T)=|x|. Also, a property of this equation (I'm not quite sure how to show it formally) is that v(x,t)>=|x| for all t. So you would get v=v. But I don't think you can set dt=0, you have to take the limit
Would this be a more conventional form of your problem: $v-|x|\ge 0$, $v_t-\alpha+\frac{1}{2(t+1)}v_{xx})\le 0$ and $(v-|x|)(v_t-\alpha+\frac{1}{2(t+1)}v_{xx})=0$? Problems like this are called obstacle problems and there is an extensive literature about them.
@MichaelRenardy I was innacurate before, what it's true is that $v(x,t)\ge|x|-\alpha dt$. But could you ellaborate a bit more on how you got to the form you wrote, please? I searched a little about obstacle problems and I guess it makes sense since the equation I posted arises from an optimal stopping problem (it's a Bellman equation).
@MichaelRenardy I'm sorry to insist but I haven't managed to understand how you arrived to your expression. If you would be so kind as to enlighten me I would really appreciate it, please.
As you have noted, your original form implies $v\ge |x|-\alpha dt$, which becomes $v\ge |x|$ in the limit $dt\to 0$. If $v$ is strictly greater than $|x|$, we must have $v_t+\frac{1}{2(t+1)}v_{xx}-\alpha=0$, since $v=v$.
@MichaelRenardy do you agree with my proposal at the end of my question? (my 2nd edit)
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2025-03-21T14:48:31.880308
| 2020-08-27T10:28:05 |
370227
|
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|
Stack Exchange
|
Does formation of the derived $\infty$-category preserve pushouts?
Let $B\leftarrow A\to C$ be a diagram of commutative rings, and let $\mathcal{D}(A)$ be the derived $\infty$-category of $A$-modules (as in Lurie's "Higher Algebra"). Then is there an equivalence
$$\mathcal{D}(B\otimes_A^LC):=\operatorname{Mod}_{B\otimes_A^LC}(\mathrm{Sp})\simeq\mathcal{D}(B)\otimes_{\mathcal{D}(A)}\mathcal{D}(C)$$
(with tensor products taken appropriately, perhaps in $\operatorname{Cat}_\infty(\mathcal{K})$ or $\mathrm{Pr}^L_{st}$)?
I am aware that if we omit the base ring $A,$ this holds true. More precisely, for any symmetric monoidal $\infty$-categories $\mathcal{C}$ and $\mathcal{D}$ and commutative algebra objects $B\in\operatorname{CAlg}(\mathcal{C}),$ $C\in\operatorname{CAlg}(\mathcal{D}),$ the natural map
$$
\operatorname{Mod}_{B}(\mathcal{C})\otimes\operatorname{Mod}_C(\mathcal{D})\to\operatorname{Mod}_{B\otimes C}(\mathcal{C}\otimes\mathcal{D})
$$
is an equivalence (where the tensor products are taken in $\operatorname{Cat}_\infty(\mathcal{K})$). In the special case $\mathcal{C} = \mathcal{D} = \mathrm{Sp},$ we get
$$
\operatorname{Mod}_{B}(\mathrm{Sp})\otimes\operatorname{Mod}_C(\mathrm{Sp})\to\operatorname{Mod}_{B\otimes C}(\mathrm{Sp}\otimes\mathrm{Sp})\simeq\operatorname{Mod}_{B\otimes C}(\mathrm{Sp}).
$$
However, I'm not sure how to bootstrap to the relative case from here.
If this is true, I would appreciate a reference or proof, ideally one which is as hands-on as possible so I can understand these objects better. While I want to know if this holds in the generality above, if there's a more intuitive or explicit way to see this at the level of DG-categories, I would also be interested in that.
The functor $A\mapsto\mathrm{Mod}A(\mathrm{Sp})$ from $E\infty$-rings to presentable symmetric monoidal $\infty$-categories commutes with all colimits, since it is left adjoint to the functor sending $\mathcal{C}$ to $\operatorname{End}_{\mathcal{C}}(1)$. The problem now is to find a reference for this "well-known" fact...
A hands-on explanation: Relative tensor products like $B\otimes_AC$ are computed as the colimit of the simplicial object $B\otimes A^{\otimes \bullet} \otimes C$. The functor $\mathsf{Mod}_{(-)}: \mathsf{Alg}(\mathsf{Sp}) \to \mathsf{Pr}^{L, \mathrm{st}}_{\mathsf{Sp}/}$ preserves all colimits and is symmetric monoidal, so apply $\mathsf{Mod}_{(-)}$ everywhere and we get the formula for the relative tensor product of categories of modules.
A 'reference-heavy' explanation:
The reference that $A \mapsto \mathsf{Mod}_A$ is symmetric monoidal is HA.<IP_ADDRESS>. And, the statement about having a right adjoint (when valued in 'presentable stable $\infty$-categories with a distinguished object') is HA.<IP_ADDRESS>.
So now we just need a general thing about when $F: \mathcal{C} \to \mathcal{D}$ being a symmetric monoidal left adjoint induces a colimit-preserving functor on $\mathsf{CAlg}(-)$, but that'll be true when the tensor product in $\mathcal{C}$ and $\mathcal{D}$ distribute over colimits- then we get (i) $\mathsf{CAlg}(F)$ preserves coproducts since those are tensor products, and (ii) $\mathsf{CAlg}(F)$ preserves sifted colimits since those are computed on underlying objects (HA.<IP_ADDRESS>). Thus $\mathsf{CAlg}(F)$ preserves all colimits, and, in particular, pushouts.
Thanks Dylan, this is very helpful! While this certainly answers my question, I must admit I was hoping for something like being able to explicitly show that every module over $B\otimes_A C$ is equivalent in some homotopy appropriate sense to a "compatible pair" $(M,N)\in\operatorname{Mod}{B}\times\operatorname{Mod}{C}.$ Maybe this is as hands-on as it gets in the $\infty$-category world, though...
@Stahl you can identify objects of $\mathsf{Mod}B \otimes{\mathsf{Mod}_A}\mathsf{Mod}_C$ with $B$-$C$-bimodule objects in the category of $A$-modules. Is that more along the lines of what you were looking for? (This is a special case of HA.<IP_ADDRESS>).
|
2025-03-21T14:48:31.880544
| 2020-08-27T10:36:29 |
370229
|
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|
Stack Exchange
|
Steps of the MMP "in family"
Let $\pi\colon X\to Y$ be a morphism between irreducible varieties with terminal singularities (let us say smooth if you want). I suppose that I have an open subset $U$ of $Y$ over which the fibres are smooth, and I then restrict to $X_U\to U$. Is it possible to do steps of the MMP on $X_U\to U$ if you know that they work on each fibre?
More precisely:
a) let $D\subset X_U$ be an irreducible divisor such that the restriction of each fibre $F$ is a union of disjoint divisors, each of them can be contracted by a divisorial contraction. Can we infer that we can contract $D$ and get a morphism $X_U\to Y_U$ over $U$ doing the job on each fibre?
b) Similarly, if $D\subset X_U$ is an irreducible closed subset such that the restriction to each fibre is the disjoint union of subsets that can be flopped, can I do the flop globally?
I would be happy if the answer was yes, even if some conditions, easy to check in precise cases, have to be fullfilled. Indeed, I would like to apply this to as sequence of examples where I am pretty sure that it works, but would like to have a general argument instead of doing case by case the study of $X$, and restrict simply to fibres, which are easy to handle in my case.
Jeremy -- I don't remember the exact answer (and I am not 100% sure what the precise statement you need is) but of the top of my head, but I think this should be fine as long as we have that $(X,D)$ is say lc and log smooth over the base. Then by Corollary 1.4 of https://arxiv.org/pdf/1412.1186.pdf, sections will extend from fibers to the total space. This should give you what you want (eg. components of $D$ are contracted by the mmp iff this happens when restricted to the fibers).
|
2025-03-21T14:48:31.880682
| 2020-08-27T12:19:37 |
370237
|
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|
Stack Exchange
|
Extension to all dimensions of complex line integral
Let $\Gamma$ be a smooth curve in $\mathbb{C}^d$. Since $\mathbb{C}^d$ can be seen as $\mathbb{R}^{2d}$, one can define the line integral of functions $f:\Gamma\to \mathbb{C}$ using for instance arclength parametrization.
However, when $d=1$, one can use the usual complex line integral of complex analysis instead. Is there a natural extension of this definition when $d \geq 2$ ?
The most natural extension would have f mapping to $C^d$ rather than $C$.
Yes, sure. But i'm dealing with functions defined on curves and the arclength definition of integration does not seem adequate.
The line integral of complex analysis is the integral of a 1-form, so we can immediately generalize to integration of 1-forms along curves, which is defined in any manifold. The line integral of complex analysis is not an arc length integral of a function.
Unfortunatlely I'm not familiar with 1-form. Is the function f, defined on $\Gamma$, a 1-form ?
@Chr: no, a function on a curve is not a 1-form. You can read Spivak, Calculus on Manifolds, for a nice introduction to differential forms. But your question does not appear to be a question of research; perhaps this is the wrong website for your question.
Ok, thank you for the reference. However, I can assure you that question rose during my research. I am trying to construct an explicit solution to a differential equation and I am conviced I need to integrate (in some sense) a function along a curve. However, the arc length integral does not work and in the case $d=1$ the line integral of complex analysis would work, hence my question.
|
2025-03-21T14:48:31.880812
| 2020-08-27T12:22:21 |
370238
|
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|
Stack Exchange
|
If it quacks like an abelian variety over a finite field
Consider smooth projective varieties over a finite field. If a curve "looks like" an elliptic curve (i.e. has genus $1$) then it can be made into an elliptic curve.
Is there something similar in higher dimensions, i.e. if we fix the values of some invariants can we guarantee that the variety admits an algebraic group structure? What is sufficient for threefolds?
Ah yeah, right, my bad. I messed up my Hasse-Weil bound :P
@Wojowu There is actually a much easier proof that every genus 1 curve over a finite field has a point, and it generalizes to the statement that if $F$ Is a finite field and $A/F$ is a variety such that $A/\overline{F}$ is an abelian variety, then $A(F)\ne\emptyset$, so $A/F$ is itself already an abelian variety. (I have a recollection this may be due to Lang, but I could be wrong.)
Have you considered the dimension 2 case? Clearly a necessary condition is that the Kodaira dimension be 0. There are just a few such classes of varieties, so you can check to see whether their discrete invariants suffice to pick out the abelian surfaces.
@JoeSilverman the classification of surfaces in positive characteristic is kind of confusing to me but I tried. In higher dimensions I don't know anything
I don't know about finite fields, but over $\mathbb C$ by https://arxiv.org/abs/math/9903184 and https://arxiv.org/abs/math/0011042, if $h^0(2K_X)=1$ and $h^0(\Omega ^1_X)=\dim X$, then $X$ is birational to an abelian variety. If $k$ is an algebraically closed field of char $p>0$, then related results are contained in https://arxiv.org/abs/1703.06631.
Sommese, Quaternion Manifolds (1975) https://link.springer.com/article/10.1007/BF01357140 constructed a complex 3-fold which is diffeomorphic to $(S^1)^6$ but not a complex torus. So there is grounds for caution. I found a readable description of the construction in the first pages of https://projecteuclid.org/euclid.kjm/1291041217
Okay, I read further in the Catanese et al paper, and I retract my warning. According to Theorem 2.3, if $X$ is a compact complex manifold which (1) has the integer cohomology ring of a torus and (2) is bimeromorphic to a compact Kahler manifold, then $X$ is a complex torus. In particular, if (1) holds and $X$ is algebraic then Chow's lemma tells us it is birational to a projective variety, so this should show it is an abelian variety. That doesn't prove anything in char p, but it means that I no longer consider this a reason for caution.
@DavidESpeyer, a funny business! Thanks for the illuminating points, even if not exactly annihilating the original question.
Perhaps if the integral $\ell$-adic cohomology ring looks like the cohomology of an abelian variety for all $\ell$, including crystalline cohomology if $\ell=p$? This could suffice to show that the Albanese map has degree $1$ (by pulling back cohomology classes in $H^1$ from the Albanese and then cupping them to degree $2n$ to calculate the degree) and that it is finite (because the class of any curve that is contracted by the map would have zero cup product with any $2n-1$ classes in degree $1$) which would imply it is an isomorphism.
One possible answer to this could be Lang's theorem: it says that if $G/\mathbb{F}_q$ is a smooth connected algebraic group, then $H^1(\mathbb{F}_q,G)$ is trivial, or otherwise put every $G$-torsor has an $\mathbb{F}_q$-rational point.
This generalizes your example: if $X/\mathbb{F}_q$ is a smooth projective variety such that $X_{\bar{\mathbb{F}}_q}$ is isomorphic to an abelian variety, then $X$ is a torsor under its Albanese variety $A = Alb(X)$.
If $X$ has dimension $1$ then requiring the genus of $X$ to be $1$ is enough.
If $X$ has dimension $2$ then by the classification of surfaces it is enough to assume that e.g. the canonical bundle of $X$ is trivial and $X$ is not simply connected.
Edit: as pointed out below, the previous sentence is correct only if the characteristic of $\mathbb{F}_q$ is $\neq 2,3$. If we additionally assume that the second $l$-adic Betti number equals $6$, then we get a criterion in all characteristics.
the last sentence is incorrect. One can take $X$ to be a non-classical Enriques surface in characteristic $2$ https://arxiv.org/abs/1608.05198
|
2025-03-21T14:48:31.881132
| 2020-08-27T12:26:01 |
370239
|
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|
Stack Exchange
|
Splendid groups
The following definition has arisen naturally in two papers of mine. The papers are on rather unrelated topics; of course they are within my narrow interests, so there's some symbolic dynamics connection, but really in both cases I just needed to find good "generic elements / finite subsets" of the group, and I can't help but wonder if this has appeared elsewhere, or is even something I know but am just not recognizing.
Definition. A topological group $G$ is splendid if for all compact $C \subset G$, there exists $g \in G$ such that $gCg \cap C = \emptyset$.
I'll mention some sufficient conditions for being splendid and give some examples. First, abelian groups are usually splendid unless obviously not splendid, and this passes to preimages, as follows:
Lemma. Let $G$ be a topological group that has a continuous abelian (surjective) quotient $(H,+)$ such that $\{2h \;|\; h \in H\} \leq H$ is non-compact. Then $G$ is splendid.
For simplicity let's now concentrate on discrete infinite groups, so compact means finite. The previous lemma shows that e.g. free abelian groups, free groups and Thompson's $F$ are splendid, since they have positive rank abelianizations. Every finitely-generated infinite discrete left-orderable amenable group is indicable by a result of Morris, thus splendid by the previous lemma. (You can generalize easily to locally indicable, thus all discrete left-orderable amenable groups.)
Not every group is splendid: the infinite dihedral group $\langle a, b \;|\; a^2 = b^2 = e \rangle$ with the discrete topology is not splendid, namely pick $C = \{e, a\}$. Besides the above lemma, a way to show splendedness is that every non-splendid group that has an infinite abelian subgroup is at least a bit little dihedral (I worked this out just for this post, so take with a grain of salt):
Lemma. Suppose $G$ is a non-splendid discrete infinite group. If $H \leq G$ is abelian, then there is a finite-index subgroup $H' \leq H$ and some $a \in G$ such that $h^a = h^{-1}$ for all $h \in H'$.
Proof: Let $C$ be the finite set proving non-splendidness. Then $hCh = C \neq \emptyset$ for all $h \in H$.
Since $H$ is abelian, $H$ acts on $G$ from the left by $h * g = hgh$.
Since $h*C \cap C \neq \emptyset$, writing $S_a \leq H$ for the stabilizer of $a \in C$ and $T_{a,b}$ for the transporter $\{h \in H \;|\; ha = b\}$, we have
$$ H = \bigcup_{a, b \in C} T_{a, b} = \bigcup_{a, b \in C \\ T_{a,b} \neq \emptyset} h_{a, b} S_a $$
for some choices $h_{a,b} \in H$. By a result of Neumann, some $H' = S_a$ has finite index in $H$, and for $h \in H'$ we have $hah = a \implies h^{-1} = aha^{-1}$. Square.
In particular a group is splendid if there is an element of infinite order such that no power of it is conjugate to its inverse.
For example this lemma applies to Thompson's $T$ (which has no nontrivial abelian quotients so the first lemma does not apply): Pick an element $f$ that fixes a unique interval $I \subset S^1$ and has derivative $>1$ after the right endpoint of $I$. Then $\langle f \rangle$ is of infinite order and every positive power $f^i$ has the same property as $f$, while no negative power $f^{-1}$ does. This property is preserved under conjugacy, so no $f^i$ is conjugate to its inverse.
Question. Does this have a more standard, even if possibly less splendid, existing name? Alternative characterizations that are easier to verify?
You can refine the definition as follows, for discrete groups: say $G$ is $k$-splendid if the splendidness condition holds for all $|C| \leq k$. Every infinite group is $1$-splendid, while the dihedral group is not $2$-splendid. A characterization of $k$-splendidness for small $k$ would also be of interest.
A list of my favorite groups and whether or not they are splendid is also of interest to me, I tried to compile one, but after working out those Thompson's's I got stuck on the Grigorchuk group.
Neumann, B. H., Groups covered by finitely many cosets, Publ. Math., Debrecen 3, 227-242 (1954). ZBL0057.25603.
Morris, Dave Witte, Amenable groups that act on the line., Algebr. Geom. Topol. 6, 2509-2518 (2006). ZBL1185.20042.
Just a heads up: @JeremyRickard famously defined the notion of a "splendid equivalence" between blocks of finite groups. You may wish to double-check that this causes no confusion. (You've probably already done this.)
I have not, I have simply picked a synonym for "nice", as I was already using "nice" in the paper where I needed it, probably for some other notion I didn't understand. It didn't actually occur to me it might be in use, I will check this out.
One option to avoid this kind of things is to avoid using random generic non-descriptive names.
@NajibIdrissi: My experience is the opposite, nothing is more dangerous than a natural name, because you may think you know what it means. That said, I was using this term mainly to clarify I don't know what to call this, and am not planning to call it this in a publication.
"Every infinite group is 1-splendid": what about $(\mathbb{Z}/2\mathbb{Z})^{(\mathbb{N})}$ with the discrete topology? (Anyway, this is a splendid question.)
@LucGuyot: I copied that from my notes without checking, is it true for f.g.?
Yes: assuming by contradiction that for some $x \in G$ we have $gxg = x$ for every $g \in G$ infinite and f.g., then $x \cdot x \cdot x = x$ implies $x^2 = 1$ and subsequently $(gx)x(gx) = x$ implies $g^2 = 1$ for every $g \in G$. As a finitely generated group with exponent $2$ is a finite Abelian elementary $2$-group, we are done.
This is just a comment on the "splendid" terminology. These sorts of incredibly heartwarming adjectives seem to pop up everywhere in symbolic dynamics/geometric group theory. I remember a talk by my MSc advisor C. Bleak entitled "Flawless Vigorous Groups"; here a flawless group has properties which makes it both perfect and lawless!
https://i.redd.it/gnfy50c8tg601.png
"A characterization of -splendidness for small would also be of interest.": ($k = 1$) the groups which aren't 1-splendid are the elementary Abelian 2-groups.
I should've mentioned that this definition first arose in discussion with Thibaut Dumont. (I had more the second application in mind when I wrote the post.)
edit January 18th, 2021
Final nail in the f.g. splendid coffin. Apparently, the dihedral group and its variants are indeed the only non-splendid groups. So maybe I should've called non-splendid groups splendid and vice versa, because this would roll off the tongue better then. I of course originally thought non-splendid groups would usually be some crazy monsters, and that the dihedral group was kind of an outlier.
Let's say an automorphism $\phi : H \to H$ of a group $H$ flips it if $\phi(h) = h^{-1}$ for all $h \in H$. Obviously a flipping automorphism can only exist on an abelian group, since $a^{-1}b^{-1} = (ba)^{-1} = \phi(ba) = \phi(b)\phi(a) = b^{-1}a^{-1}$.
Theorem. A finitely-generated group $G$ is non-splendid if and only if it admits an internal automorphism that flips some non-trivial finite-index normal torsion-free abelian subgroup.
Proof. We know from the previous edit (from yesterday) that any f.g. non-splendid group $G$ is virtually abelian, and a non-trivial finite-index normal torsion-free abelian subgroup implies virtual abelianity, so we may restrict to virtually abelian $G$.
Let $H \leq G$ be any non-trivial finite-index normal torsion-free abelian subgroup (so $H \cong \mathbb{Z}^d$ for some $d \geq 1$), which exists by intersecting the conjugates of any non-trivial finite-index torsion-free abelian subgroup. We show that $H$ is flipped by an inner automorphism of $G$ if and only if $G$ is non-splendid. Because the choice of $H$ is arbitrary, this shows the a priori stronger fact $\exists H: P \implies \mbox{non-splendid} \implies \forall H: P$. Let $F \subset G$ be the finitely many coset representatives for $H$.
Suppose first that conjugation by $t \in G$ flips $H$. Write $h^t = t^{-1} h t$. Pick $C = \{ f^{-1} t, tf \;|\; f \in F \}$. Now, for any $g \in G$ we can write $g = h f$ for $h \in H, f \in F$, and then
$$ g C g \ni hf \cdot f^{-1}t \cdot hf = h t h f = t \cdot h^t h \cdot f = tf \in C. $$
So $G$ is not splendid.
Suppose then that for all $g \in G$, conjugation by $G$ does not flip $H$. Then there exists $h \in H$ which is not flipped by any $g \in G$, and in fact we can find $h$ such that conjugation by any $g \in G$ takes $h$ very far from $h^{-1}$. More precisely, for each $R$ there exists $h \in H$ such that $d(h^g, h) \geq R$ in the right-invariant word metric $d$ of $G$.
There are many ways to see the above, I guess you can geometrize and it's a basic lemma about finite reflection groups. A direct proof is also easy: Observe that it's enough to bound the word metric of $H$ since the distortion is bounded since $H$ is finite index. We only need to consider conjugation by each $f \in F$. Now take for each $f \in F$ an element $h_f$ not flipped by $f$, and consider products of the form $h_{f_1}^{n_1} \circ h_{f_2}^{n_2} \circ \cdots \circ h_{f_{|F|}}^{n_{|F|}}$ where the $n_i$ have different scales. Basically, take $h_{f_1}$ not flipped by $f_1$, so $h_{f_1}^{n_1}$ is conjugated very far from $(h_{f_1}^{n_1})^{-1}$, and take a massive $n_1$. If also $f_2$ does not flip $h_{f_1}$, we continue to $f_3$, if $f_2$ does flip $h_{f_1}$ then it does not flip $h_{f_1}^{n_1} \circ h_{f_2}^{n_2}$ where $n_2$ has a smaller scale, but $f_1$ does not flip this either since $n_1$ is much bigger than $n_2$. And so on. (In the geometric situation I suppose you can replace this by Baire or the fact $\mathbb{R^d}$ is not a union of finitely many $(d-1)$-dimensional subspaces.)
We can now use the elements $h$ given by the two paragraphs above to prove splendidness. Let $C$ be arbitrary, and suppose $C \subset B_R$, the ball of radius $R$ in the right-invariant word metric of $G$ around identity. Let $h$ be such that $d(h^a, h^{-1}) > 2R$ for all $a \in G$. If $hCh \cap C \neq \emptyset$ then $h^a h = a^{-1} h a h = a^{-1} b$ for some $a, b \in C$, so $d(h^a, h^{-1}) \leq 2R$, a contradiction. So $G$ is splendid. Square.
edit January 17th, 2021
Theorem. Suppose $G$ is a f.g. discrete group which is not splendid. Then $G$ is virtually abelian.
Proof. We know from the first version of my answer (see below after the text original) that $G$ is virtually nilpotent. Then it's well-known that $G$ has a finite-index torsion-free nilpotent subgroup $H$. Let $C$ be the finite set contradicting splendidness.
First, let's think about what splendidness says when $H$ is a torsion-free subgroup. For any $g \in H$ and $i \in \mathbb{N}$, we have $g^i a g^i = b$ for some $a, b \in C$. Now, pigeon says $g^i a g^i = b$, $g^{i+k} a g^{i+k} = b$ for some $k > 0$, and we get $g^i a g^i = g^{i+k} a g^{i+k}$, in other words $g^{-k} = a g^k a^{-1}$. Clearly this $k$ is bounded, and by taking the least common multiple of the $k$ that occur, we find $a \in C$ such that for all $g \in H$, $(g^k)^a = g^{-k}$ (true for either convention for the $g^a$ notation, because inversion is an involution).
Note that in particular $H$ is now a torsion-free nilpotent group which admits an automorphism such that for all $g \in H$, $(g^k)^a = g^{-k}$. I claim that if $H$ is not abelian, this is impossible, which will conclude the proof.
To see this, suppose $H$ is not abelian. I claim that in this case there is a three-dimensional Heisenberg group inside: We use the convention $[x,y] = xyx^{-1}y^{-1}$ if it matters. Let $(H_i)_i$ be the lower central series of $H$, $H_n \neq 1, [H_n, H] = 1$. Take $1 \neq c = [a,b] \in H_n$ a non-trivial commutator, where $a \in H_{n-1}, b \in H$. Now, $c$ is superlinearly distorted in $\langle a,b,c \rangle$, so certainly $a$ and $b$ cannot satisfy any identity. If $\langle a,b,c \rangle$ satisfies an identity that is not true in the Heisenberg group, then commuting elements to normal form $a^k b^\ell c^m$ we see that $c$ satisfies an identity on its own, contradicting the torsion-freeness assuption.
Now it suffices to show that the three-dimensional Heisenberg group $H = \langle a,b,c \;|\; [a,b] = c, [a,c], [b,c] \rangle$ does not admit an automorphism $\phi : H \to H$ satisfying $\phi(g^k) = g^{-k}$ for all $g \in H$. This is geometrically obvious, go around a big cycle on the $\langle a,b \rangle$-quotient, and you jump up by the area in the central direction. If you apply the inverse elementwise, you get a cycle of the same area in the same direction (just reordered). So you don't flip the direction.
In algebra, take $[a^{km}, b^{km}] = c^{k^2 m^2}$. Now apply $\phi$ to get
$$ [a^{-km}, b^{-km}] = [\phi(a^{km}), \phi(b^{km})] = \phi([a^{km}, b^{km}]) = \phi(c^{k^2 m^2}) = c^{-k^2 m^2} $$
But
$$ [a^{-km}, b^{-km}] = [a^{km}, b^{km}]^{b^{km} a^{km}} = [a^{km}, b^{km}] = c^{k^2 m^2} $$
since $[a^{km}, b^{km}]$ is central. So $c^{-k^2 m^2} = c^{k^2 m^2}$ which is the desired contradiction. Square.
Presumably you could spell out in more detail what the situations are where you get non-splendidness. I suppose it's just that some element flips the torsion-free part of the finite-index abelian subgroup. Also, the proof essentially uses the observation that non-splendidness implies essentially that some element conjugates things to their inverses, giving that inversion is almost an automorphism. This is strange since it is of course a flip automorphism, which sounds like it should ultimately be the reason that non-splendid groups must be virtualy abelian. Maybe you could cut out the middle man (ICC quotients) by working more on the algebra.
original
Joshua Frisch pointed out to me in private communication that ICC $\implies$ splendid, which seems to give a good partial answer to the question. Let me restrict to f.g. infinite discrete groups $G$.
Let $\mathcal{D}$ be the class of quotients of the fundamental group of the Klein bottle $\langle h, a \;|\; h^a = h^{-1} \rangle$, where $h$ is mapped to an element of infinite order and $a$ to a nontrivial element. For example the quotient where $a^2 = e$ is added gives the dihedral group.
Theorem. If $G$ f.g. infinite non-splendid, then $G$ is virtually nilpotent, and contains a group from $\mathcal{D}$.
This is presumably not a characterization, but at least it reduces the problem to virtually nilpotent groups, which are much better understood than general ones. First, there's the following result of [1] (which is indeed a consequence of Neumann's theorem, but there's a nice algebraic trick that I missed).
Theorem. If $G$ is an f.g. ICC group, then for every finite $Z \subset G$ there exists $g$ such that all solutions to $gxg^{\pm} = y$ with $x, y \in Z$ (if they exist) satisfy $x = y = e$.
This gives ICC $\implies$ splendid. There's a minor complication with involutions, so I give a proof:
Theorem. If $G$ is an f.g. ICC group, then it is splendid.
Proof. Suppose ICC, and fix some symmetric generating set. Let $n$ be arbitrary and take $Z = B_N$ for $N = 2n$. By the previous theorem, for some $g \in G$, among $x, y \in Z$ the equation $gxg^{\pm} = y$ can only have solutions with $x = y = e$. If $g$ is not an involution, there aren't any such solutions either, and we are done.
If $g$ is an involution, then take any $h$ of word norm exactly $n$, and consider $gh$ instead. Now if $gh \cdot k \cdot gh = k'$ for $k, k' \in B_n$, then $g \cdot hk \cdot g = k' h^{-1}$ and here $hk, k'h^{-1} \in Z$, and this cannot happen by the assumption on $g$. So we have $gB_ng \cap B_n = \emptyset$. Square.
Next, there's a theorem of [1,2] (see also [3]).
Theorem. Every f.g. infinite group is either virtually nilpotent, or has a non-trivial ICC quotient.
And also let's prove a lemma.
Lemma. If $G$ has a splendid quotient, it is splendid.
Proof. Let $\phi : G \to H$ be the quotient map, and $C \subset G$ be finite. Then $\phi(C) = C'$ is also finite, and thus $hC'h \cap C' = \emptyset$ for some $h \in H$. Letting $\phi(g) = h$, we have $\phi(gCg) \cap \phi(C) = hC'h \cap C' = \emptyset$, so $gCg \cap C = \emptyset$ as well, and this means $G$ is splendid. Square.
We obtain the following:
Theorem. Every non-splendid group is virtually nilpotent.
Proof. If $G$ is not virtually nilpotent, then it has a non-trivial ICC quotient. That quotient is then splendid, and thus $G$ is splendid. Square.
Now we can prove the theorem I stated at the beginning of the post.
Proof. An infinite f.g. nilpotent group contains $H \cong \mathbb{Z}$, which is abelian. By the lemma in my post, there exists $a \in G$ and $h \in H \setminus \{e\}$ such that $h^a = h^{-1}$ for all $h \in H$. Then $h$ and $a$ give the quotient, taking $n$ the order of $a$. Square.
[1] Erschler, A.; Kaimanovitch, V. A., Arboreal structures on groups and the associated boundaries, Arxiv https://arxiv.org/abs/1903.02095]
[2] Duguid, A. M.; McLain, D. H., (FC)-nilpotent and (FC)-soluble groups, Proc. Camb. Philos. Soc. 52, 391-398 (1956). ZBL0071.02204.
[3] McLain, D. H., Remarks on the upper central series of a group, Proc. Glasg. Math. Assoc. 3, 38-44 (1956). ZBL0072.25702.
[4] Frisch, J.; Ferdowsi, P. V., Non-virtually nilpotent groups have infinite conjugacy class quotients, Arxiv https://arxiv.org/abs/1803.05064]
Amazing! Two minor comments: presumably $B_N$ is the ball centred at $e$ with radius $N$ with respect to word metric defined by the fixed symmetric generating set. It is more difficult to guess what $n$ is in your last sentence. (By the way, you certainly realised that you established the splendor of the Grigorchuk.)
Fixed typo in the last sentence. Thanks for the amazing, I don't know which subset of this Frisch already had in mind (he gave the references and told me ICC $\implies$ splendid), so I don't know how much credit is due to me. Anyway I found the result amazing too. Also yes, Grigorchuk is indeed not virtually nilpotent!
Ok I seem to have the full answer now for the discrete case. Yay. If someone does the general locally compact case, I'll probably move the accept there (I think the technology allows that?)
NB. I first thought the generalization to all discrete groups is trivial and left it in the statement, but I forgot to check it, and I now realize it requires a bit more thought. Apologies in advance if there's a third modification later where I do all discrete groups... (I have no idea for compactly generated because I don't know if the ICC technology exists for them, and I'm not exactly an expert on those anyway.)
Very nice. I only see this now, but this reminds me of the following question/conjecture, that would immediately give the implication splendid $\implies$ virtually abelian : if $(g_n)$ is a random walk on a finitely generated group with respect to a symmetric generating probability measure, then either $G$ is virtually abelian, or $g_n^2$ tends in probability to $\infty$ in $G$, that is $\lim_n P(g_n^2=a)=0$ for every $a \in G$.
(the above conjecture arose in a work of mine with Paul-Henry Leemann, that is somewhat related to your question. We finally managed to solve our problem with a more specific proof, but I find the conjecture quite natural).
My question arose in my work on universal gate sets, and my work on horoballs on locally compact groups.
I like your question, could ICC quotients help?
@VilleSalo : I don't know. The real question to which one easily reduces is to prove that if $a$ has infinite conjugacy class, then $\lim_n P(g_n^2=a)=0$. Perhaps adding as a hypothesis that $G$ itself is ICC might help, but I could not see how.
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2025-03-21T14:48:31.882444
| 2020-08-27T16:51:38 |
370252
|
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|
Stack Exchange
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Semigroup theory for non-symmetric Markov processes / complex-valued potentials
Let $X$ be a continuous-time Markov process on a countable state space $E$, and let $V:E\mapsto\mathbb C$ be some complex function. $X$ can be characterized by its transition rates
$(\lambda_{xy})_{x,y\in E,~x\neq y}$, i.e.,
$$\mathbb P[X_{t+\epsilon}=y|X_{t}=x]=\lambda_{xy}\epsilon+o(\epsilon)$$
as $\epsilon\to0$.
If we define the semigroup
$$T_tf(x)=\mathbb E\left[\exp\left(-\int_0^tV(X_s)~ds\right)f(X_t)\bigg|X_0=x\right]
\tag{1}$$
then back-of-the-envelope calculations suggest that
$$\lim_{t\to0}\frac{\langle1_x,(1-T_t)1_y\rangle}{t}=\begin{cases}
-\lambda_{xy}&\text{if }x\neq y,\\
(\sum_{z\neq x}\lambda_{xz})V(x)&\text{if }x=y.
\end{cases}\tag{2}$$
In the case where $X$ is symmetric and $V$ is real-valued, I can easily find a multitude of references proving that, under general conditions on $X$ and $V$ (e.g., Kato class, etc), the semigroup $(1)$ is strongly continuous on $L^2(E,\mathbb R)$, and that its generator $(2)$ is a densely defined self-adjoint operator (see, e.g., this book, which actually treats much more general state spates).
Question. Are there similar references for the non-self-adjoint case? I.e., general conditions on $X$ (which allow for the process to be non-symmetric) and $V$ (which allow for $V$ to be complex-valued) under which $(1)$ is strongly continuous on $L^2(E,\mathbb C)$, and the generator ($2$) is closed and densely defined?
Given that I'm only interested in the case of a countable state space, I expect that this should not be too hard to do from scratch, but I suspect that this must have been done before.
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2025-03-21T14:48:31.882584
| 2020-08-27T17:48:58 |
370254
|
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|
Stack Exchange
|
Is the super square root of $2$ irrational?
The super square root of $n$ is the solution/solutions to $x^x=n$. Is the super square root of $2$ irrational?
If the super square root is $p/q$ with $p$ and $q$ coprime (note that obviously $p,q\neq0$), then $p^p=2^qq^p$, and taking the 2-adic valuation, $p\nu_2(p)=q+p\nu_2(q)$. Since at least one of $\nu_2(p)$ and $\nu_2(q)$ has to be zero, this is a contradiction.
And then, by Gelfond-Schneider, it's also trascendental
Yes.
If $x$ is rational, say $x=p/q$ with coprime integers $p,q$. We know $x$ cannot be an integer, so $q\neq 1$. Then $(p/q)^{p/q}=2$, or $(p/q)^p=2^q$. But the left-hand side cannot be an integer because its simplest fraction representation is $p^p/q^p$ with denominator $q^p\neq 1$. (Note that $p^p$ and $q^p$ are also coprime.)
Extension:
In fact, $x$ is transcendental. Gelfond–Schneider theorem says that the only cases where $a, b$ and $a^b$ are all algebraic numbers are (1) $b$ is rational, or (2) $a$ is $0$ or $1$.
So if $x^x=2$ and $x$ is algebraic, then it must be case (1), so $x$ is rational, a contradiction to what we have proved.
Which famous theorem?
Found it! Gelfond–Schneider theorem. I am going to edit it.
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2025-03-21T14:48:31.882726
| 2020-08-27T17:52:13 |
370255
|
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|
Stack Exchange
|
Using linear logic in algebraic geometry and commutative algebra
In algebraic geometry and commutative algebra we often deal with many categories that are not topoi nor cartesian closed (or even locally CC), but are nevertheless closed monoidal. These include the category $\mathbf{Mod}_R$ of modules over a commutative ring $R$, the category $\mathbf{PShMod}_X$ of presheaves of modules over a scheme $X$ (perhaps also $\mathbf{ShMod}_X$?), and et cetera.
Of course, there are a lot of topoi occuring in algebraic geometry, but others are not. If a category is not a topos, one can't use "normal" intuitionistic reasoning inside the category (because its internal logic does not entail higher-order intuitionistic logic, or the Calculus of Constructions from a TCS point-of-view). Then, as a computer scientist, the natural idea would be: since the internal logic of a monoidal category is some form of linear logic, can we use linear logic fruitfully to study those categories, perhaps in the style of Blechschmidt?
I found this recent manuscript by Paul-André Melliès, which was apparently submitted to LICS 2020 but not accepted. I can't find anything else in this direction, so is this a problem that has been considered in the past? Does anyone have any pointers in this direction and/or perhaps related work? It seems that this has some connections with Tannaka duality, but this is not a topic I'm familiar with at all...
The short paper by Melliès is the only one I know in this direction. I'd be happy to see more though.
Not specifically related to algebraic geometry, but I suspect that a type theory like the one I proposed here might be more useful than the usual sorts of linear logic for practical applications involving monoidal categories.
If Simon Henry has not found anything beyond the references you have already mentioned, I doubt very much there is anything out there (though one never knows, math is full of works kept in drawers, or published in some remote journal..).
So, the only option seems to be: let us reason together and see how far we can go. Before I start, let me tell you that your question rocks. I have also to declare that it is a bit loose, and that perhaps by making it a tad more focused we could find the magic thread to answering it.
Your line of thought is:
1.Monoidal closed categories come equipped with (a fragment of) linear logic, so it makes sense to leverage such a logic to describe "things" and "constructions" within those categories.
2.Moreover, you suggest that some cats such as R-MOD (the archetypal monoidal closed cat) and sheaves/presheaves of modules over a scheme are ubiquitous in Algebraic Geometry. Thus, assuming that linear logic tells us something about such cats, perhaps it is also useful to express meaningful procedures in Algebraic Geometry.
Breaking it down in these two steps has an advantage: if we can validate 1, we have some chance to also address 2.
Let us start with 1 then.
What you stated about linear logic basically says: linear logic is kind of the vanilla internal logic for monoidal closed cats (CMC in what follows). Now, as one would expect, this does not tell much as far as specific monoidal cats, only about general constructions and entities living in any generic CMC.
Take for instance the ring R of polynomials over a field (which is exactly the archetypal algebraic structure for classical AG). Now, consider the CMC of its modules.
Can I describe it by a list of linear logic axioms?
In other words, is there a linear theory which is valid in this cat or cats equivalent to it? I do not know the answer, but this is definitely the very first step I would take.
Assume one gets somewhere with 1, and now tries to tackle 2.
Here is what I would do: there is an entire field called Computational Algebraic Geometry, with gadgets like Grobner Basis and the like.
Can one axiomatize these constructions in the internal linear theory (see previous point) of polynomial rings?*
If that were possible, it would be quite interesting to computer scientists, for instance in that one could develop a sort of "Linear Prolog" to describe computational AG and computational commutative algebra.
There is something else, much bigger than this, but I shall stop here (but see my last question on quantale sub-object classifiers )
One last reference: almost nobody has tried to take seriously Linear Logic as THE logic of constructive math. There are a few refs worth while though, one being Mike Shulman's article here
To follow up on this answer, I started my PhD attempting to do a fairly similar project to the OP - except in this case I was using differential linear logic to look at differential geometry and classical mechanics. To make a long story short, I wrote about a hundred or so pages of unpublishable notes before coming to the same conclusion as your comment.
Thank you. This is an enlightening answer!
Thanks, dunno about enlightening, but I think that at the very least it helps to drive the conversation. Were you thinking of specific use cases when you formulated your question?
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2025-03-21T14:48:31.883409
| 2020-08-27T18:25:18 |
370258
|
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|
Stack Exchange
|
Signed nD curvature
I wondering if there is a model of signed curvature in n-dimensions for a tangent n-ball, the reason being that I would like to investigation a representation of manifolds by reconstructing any closed shape at least in Euclidean space using only information about its curvature and initial conditions.
Is there a theory I can rely on for this or do I not even need signed curvature?
I do not know what the expression "model of signed curvature" means, or even the expression "signed curvature". Could you give a reference to a publication, blog or wiki using these terms where they are explained in more detail.
I don't understand why signed curvature is confusing to specifically mathematicians (and no one else). A vector relative to a closed surface can point either inward or outward, this has been the basis of many advancements in computer graphics and physics for decades. Similarly, a vector drawn from the tangent point to the radius of a tangent n-ball may point inward or outward.
https://en.wikipedia.org/wiki/Curvature
mentions signed curvature.
The mean curvature is sign dependent but it is not an intrinsic invariant. Globally, reconstruction may not be possible. The flat torus and the Euclidean spaces are examples of spaces with identical trivial curvatures
Alright, well is there a set of minimal supplementary conditions that can be added which would guarantee reconstruction?
What's not clear here is how much differential geometry you've already studied. Could you clarify that? It'll be easier to guide you if we get a sense of that.
I don't see how that is relevant to a yes/no question. If you have a good explanation for that, I can provide additional information, otherwise I consider it off topic.
The expression "signed curvature" remains ambiguous because the reference you gave only defines signed curvature for curves in the plane. You are using the term "signed curvature" in reference to manifolds of apparently arbitrary dimension in Euclidean space, and perhaps to more general Riemannian manifolds, but that is not clear in your question.
What is this tangent $n$-ball? Do you mean that you suppose that you know of the existence of a Riemannian manifold, and you know the sectional curvature of all of its metric balls tangent to any given hyperplane, as a function of radius of ball, and you want to know what Riemannian manifold it is, up to isometry, or up to diffeomorphism, or up to homotopy equivalence?
I don't see how there is ambiguity in a tangent $n$-ball. In a 2D plane you have a tangent circle, in 3D you have a tangent sphere. From there you can see the pattern.
in the wikipedia article you cited above is applied to a curve only and not to a higher dimensional geometric space. Looking at what you've written in both your post and comments, it looks to me like what you want is indeed the second fundamental form. This assigns to each unit tangent vector (i.e., an element of the tangent $n$-sphere), a number that could be called the signed curvature in that direction.
As for reconstructing the surface, there is a theorem that says roughly that if you know the first and second fundamental forms of a surface, the position of one point of the surface in Euclidean space, and the position of the tangent plane at that point, then there is a unique embedding of the surface into Euclidean space with the prescribed first and second fundamental form.
Here is a nice explanation of this theorem by Robert Bryant. https://mathoverflow.net/a/155755/613
In the article they definitely include a vector-valued model, that is what the bolded letters are for.
I think you are probably looking to find an analogue of the Serret--Frenet theory for higher dimensional submanifolds of Euclidean space; this is discussed in detail in What is the analog of the "Fundamental Theorem of Space Curves," for surfaces, and beyond?
The essential difficulty in making this question precise is that the curvature has to be somehow "given". But, if we take curvature to mean the Riemann curvature tensor, for example, or one of its components (or the sectional curvature), then it is not clear how, without knowing the manifold $M$, we can explicitly describe a tensor living on that manifold (or living on a Grassmann bundle over that manifold). To write out the curvature tensor, you need to parameterize the manifold, and then you will at least be assuming that you know the diffeomorphism type of the manifold, which seems to be something you want to assume unknown. One approach to avoid knowing the diffeomorphism type of the manifold: assume quasihomogeneity of the curvature tensor (or the Ricci tensor, etc.), i.e. that any two points of the manifold have the same curvature tensor (or Ricci tensor, etc.) under some linear isometry of tangent spaces. But this hypothesis could be too strong; just consider surfaces. Another approach assumes the diffeomorphism type is known, for example the problem of prescribed scalar or Ricci curvature on a given manifold, for which there are results. But I don't think there are results for prescribing the sectional curvature or the entire Riemann curvature tensor. To make your question clearer, please give some details about what you want to assume is known, and how the known information is presented.
Thank you for the link, that looks interesting. So in this sense, a "Frenet frame" is a broader initial condition needed to reconstruct a manifold from only curvature information?
@OneWhoBlends, "Frenet frame" is not a condition. It's something you can define on a surface or higher dimensional submanifold in Euclidean space.
Right, it's obviously broader than the scope of just an initial condition, but much like an initial condition, it is an additional piece of information that precisely resolves the otherwise ambiguous possibilities of a differential solution. The only question remaining is: does that additional information allow one to reconstruct a surface?
If you have an abstract manifold with Riemannian metric and a symmetric tensor that satisfies the Gauss and Codazzi-Mainardi equations, then that determines an isometric embedding (I.e., a reconstruction), unique up to a rigid motion of the manifold into Euclidean space as a hypersurface, such that the symmetric tensor is the second fundamental form. Is that what you’re looking for?
It sounds like what you call signed curvature on the unit sphere in the tangent space is the second fundamental form. Is that right?
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2025-03-21T14:48:31.883863
| 2020-08-27T18:26:41 |
370259
|
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|
Stack Exchange
|
Counting integral points in a rank-2 lattice
Let $\Lambda \subset \mathbb{Z}^n, n \geq 2$ be a lattice of rank 2. It is well-known (see for example Lemma 4.5 in this book) that
$$\displaystyle \#\{\mathbf{x} \in \Lambda : \mathbf{x} \text{ is primitive, } \lVert \mathbf{x} \rVert \leq R\} \ll_n \frac{R^2}{\det(\Lambda)} + O(1).$$
Recall that $\mathbf{x} \in \mathbb{Z}^n$ is primitive if it is not of the form $r \cdot \mathbf{y}$ for $\mathbf{y} \in \mathbb{Z}^n$ and $r \in \mathbb{N}, r \geq 2$.
My question is a refinement where one bounds the components of $\mathbf{x}$ individually. Suppose that (with respect to the standard basis of $\mathbb{R}^n$) the components of $\mathbf{x} = (x_1, \cdots, x_n)^T$ are bounded by $|x_i| \leq R_i$. Can we expect a bound of the shape
$$\displaystyle O_n \left(\frac{(R_1 \cdots R_n)^{2/n}}{\det(\Lambda)} + O(1) \right)?$$
Interestingly enough, such a result is possible for ternary quadratic forms. I would be interested if the above holds for $n = 3$ for lattices.
I don't think that your first display is correct. Let $\Lambda$ be the $\mathbb{Z}$-span of $(1,0,0,\dotsc)^T$ and $(0,R^2,0,\dotsc)^T$. Then the LHS exceeds $R$, while the RHS is bounded.
@GHfromMO sorry, I forgot to add the condition that $\mathbf{x}$ is primitive. I will fix that
The bound you expect (second display) is too strong. Let $\Lambda$ be the $\mathbb{Z}$-span of $(1,0,0,\dots)^T$ and $(0,1,0,\dotsc)^T$. Let $R_1=R_2=R>1$ and $R_3=1$. Then the LHS is $\gg R^2$, while the RHS is $\ll R^{4/3}$.
|
2025-03-21T14:48:31.884024
| 2020-08-27T18:53:05 |
370263
|
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|
Stack Exchange
|
If the volume-ratio of an inscribed convex set to the circumscribing convex set is rational, can anything of consequence be further deduced?
Say, one has two $n$-dimensional convex sets $A$ and $B$, with $B$ being inscribed in the strictly larger set $A$. ($A$ and $B$ have at least one boundary point in common. $B$ “fits snugly” in $A$ Inscribed.) Now, the ratio of the volume of $B$ to that of $A$ is a rational $\in (0,1)$.
Thus, the ratio is, in some sense, exceptional, since the rationals are countable, and the reals $\in (0,1)$, uncountable.
So, can anything further concerning the relation of $A$ and $B$ be drawn from this very general observation? (If so--though, unlikely, I surmise--how might the choice of geometry, Euclidean or otherwise, affect matters?)
Additionally, I would be interested in constructions in which $A$ is not a
hyperball.
You can produce many examples. Say $A$ is the unit disk in the plane. $D$ is the horizontal diameter. Take a continuous concave function $$f:[-1,1]\to[0,\infty),$$ such that $f(x)\leq \sqrt{1-x^2}$. Now, for $t\in (0,1]$,
we denote by $B_t$ the convex region $$B_t:={ (x,y) ;; x\in [0,1],;y|\leq tf(x)}.$$ Then $B_t$ is inscribed in $A$ and $$\mathrm{area}(B_t)=2t\int_{-1}^1 f(x); dx. $$ By varying $t$ we can get the ratio of the area be any arbitrarily small number rational. If you take $f(x)=\sqrt{1-x^2}$ we can make this ratio be any rational number in $(0,1)$.
Thanks, Liviu Nicolaescu! Guess this doesn't readily extend to higher-dimensional ($n>2$) convex sets.
Yes it does. In the space with coordinates $(y,x)$, $y\in\mathbb{R}$, $x=(x_1,\dotsc, x_n)\in \mathbb{R}^n$, $A$ is the closed unit ball in the $(y, x)$ space, $D$ is the closed ball of radius $1$ in the $x$-space. Choose $f:D\to[0,\infty)$ continuous and concave satisfying $f(x)\leq \sqrt{1-|x|^2} $ and $$ B_t={(y,x);;;x\in D,;; |y|\leq tf(x)}. $$
Thanks again, L. N. Perhaps these comments of yours merit a little expansion into an answer, where a couple of examples of $f$ could be employed and illustrated. Also, is your example somewhat related to that in the comment of Joseph O'Rourke to my rather related query https://mathoverflow.net/questions/368908/construct-pairs-of-n-dimensional-convex-bodies-with-given-ratios-p-of-volu ? Further, perhaps I'll append a remark to my question here, asking if examples in which $A$ is not a hyperball can be given.
|
2025-03-21T14:48:31.884241
| 2020-08-27T19:01:55 |
370264
|
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|
Stack Exchange
|
Torsion freeness of direct image of structure sheaf?
I have the following question.
Let $f:X\rightarrow Y$ be a surjective projective morphism between smooth projective varieties.
I learned that if $\dim Y=1$, then $R^if_*\mathcal O_X$ is torsion free $\forall i\geq0$.
I think smoothness of $X,Y$ is important here, and I wonder what about the condition $dim Y=1$?
Q1 Is it still true that $R^if_*\mathcal O$ is torsion free $\forall i\geq0$ when $\dim Y>1$? If false, what's a natural counter example? (To show how dimension of $Y$ matters).
As your questions stand, the answer is no for obvious reasons: just take for $f$ a closed immersion. You probably want $f$ surjective, $X$ connected, ...?
@abx Thanks.Yes I need $f$ surjective.
Q1: Take a normal surface $Y'$ with non-rational singularity. Let $g:X \to Y'$ be a resolution, and let $f:X \to Y = \mathbf{P}^2$ be the composition of $g$ with a finite map $Y' \to \mathbf{P}^2$. Then $R^1 f_* \mathcal{O}_X \neq 0$ and is supported at finitely many points.
@Anonymous Nice example, thanks!
If you strengthen your smoothness conditions, then it is true. Assume that $f$ is is smooth over $U=Y-B$ the complement of a snc divisor. Then $Rf_\omega X=\sum R^if\omega X[-i]$ and each $R^if\omega X$ and $R^if\mathcal O X$ is locally free (Koll'ar's Higher Direct Images II Thm 2.6). Note that the $R^if*\omega _X$ are always torsion free.
@Hacon. Thanks for the comment!
In addition to @Hacon's comment, in [loc. cit. \S 3] Koll'ar has also pinpointed the torsion free and the torsion part of $R^if_\mathcal{O}X$. The torsion free part is the dual of $R^{n-i}f\omega_X$ and hence is also reflexive.
|
2025-03-21T14:48:31.884416
| 2020-08-27T19:37:38 |
370266
|
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|
Stack Exchange
|
Milnor excision for algebraic stacks
Recall that a commutative square of commutative rings
$$\begin{matrix}
A&\to&B\\
\downarrow &&\downarrow\\
A^\prime&\to&B^\prime\end{matrix}$$
is called a Milnor square if the vertical maps are surjective and the square is both a pullback and pushout of rings.
It has been shown that Milnor squares give rise to diagrams that are still both pullbacks and pushouts in the category of all schemes after applying $\operatorname{Spec}$. Since it is standard that the $\operatorname{Spec}$ functor sends pushouts to pullbacks, perhaps a more interesting way to state this fact is that a functor of points representable by a scheme $S$ satisfies Milnor excision:
$$S(A)\simeq S(A^\prime)\times_{S(B^\prime)}S(B).$$
Question: Is it true that Deligne-Mumford stacks or Artin stacks also satisfy Milnor excision? Is there a reference?
Note: Lurie shows that spectral Deligne-Mumford stacks satisfy a weaker condition in SAG chapter 16 called cohesion, which gives excision as above when all of the maps of rings in the square are surjective. I'm primarily interested in the spectral DM case, but a proof in the non-derived case should be enough to suss out what's going on.
Edit: It turns out that this is quite a hard problem in complete generality, but the case I care about is, specifically, quasicompact quasiseparated Deligne Mumford stacks (but no affine/quasiaffine diagonal!). Not sure if that makes things any easier.
The standard proof, as far as I'm aware, that schemes satisfy Milnor excision builds an explicit pushout of affine schemes and shows that it is a locally ringed space and also an affine scheme, which then by categorical generalities implies the result. These pushouts are not preserved by the Yoneda embedding into the huge fppf site, so it's a bit tricky.
In upcoming (now on arXiv:2205.08623) joint work with Jarod Alper, Jack Hall and Daniel Halpern-Leistner:
Artin algebraization for pairs with applications to the local structure of stacks and Ferrand pushouts
we prove more generally the existence of pushouts of affine morphisms along closed immersions in the category of (quasi-separated) algebraic stacks (Thm. 1.8). This in particular implies that Milnor squares are pushouts in the category of (quasi-separated) algebraic stacks. Let me sketch how this is proved:
Let $X=\operatorname{Spec} B$ and $Y=\operatorname{Spec} A$ and similarly for the primes so we have a cartesian square:
$\require{AMScd}$
\begin{CD}
X' @>f'>> Y'\\
@V g' V V @VV g V\\
X @>>f> Y
\end{CD}
with $g$, $g'$ closed immersions.
By assumption, this is co-cartesian in the category of affine schemes.
To show that this is co-cartesian in the category of algebraic stacks, let $Z$ be an algebraic stack together with maps $u\colon X\to Z$ and $v\colon Y'\to Z$ and a $2$-isomorphism $ug'\cong vf'$. We can replace $Z$ with an open quasi-compact neighborhood of the images of $u$ and $v$ and assume that $Z$ is quasi-compact.
Let $p\colon Z_1\to Z$ be an affine smooth presentation. Consider the pull-backs along $u$, $ug'\cong vf'$ and $v$ and call these $X_1\to X$, $X'_1\to X'$ and $Y'_1\to Y'$. The easiest case is if $Z$ has affine diagonal. Then $p$ is affine and $X_1$, $X'_1$, $Y'_1$ are also affine. Then we can take the pushout of the three affine schemes resulting in $Y_1\to Y$. This gives us a map $Y_1\to Z_1\to Z$. One then observes that $Y_1\to Y$ is smooth (flatness is [Fer, Thm 2.2 (iv)] and finite presentation can be proven similarly and smoothness then follows by considering fibers). Then take $X_2=X_1\times_X X_1$ etc. We obtain two maps $Y_2\rightrightarrows Y_1\to Z_1\to Z$. Since $Y_2$ also is a pushout in the category of affine schemes (they are stable under flat base change by [Fer, Thm 2.2 (iv)]) these two maps coincide (*). By descent, we obtain a map $Y\to Z$.
(*) It remains to show that any two maps $Y\to Z$ fitting in the diagram are isomorphic up to unique 2-isomorphism. For this, one takes two maps and pull-back the diagonal of $Z$. This is then turned into an existence question. Again, if the diagonal is affine, it is immediate.
When the diagonal is not affine, then the $X_1$, $X'_1$ and $Y'_1$ above are merely algebraic spaces. One can take an étale affine presentation of $X_1$ and pull this back to $X'_1$. The crucial step is then to extend this to an étale presentation of $Y'_1$. This is where the Artin algebraization alluded to in the title comes in. It is also needed when you want to construct the pushout $Y$ of a diagram $X\leftarrow X'\rightarrow Y'$ of algebraic stacks (affine / closed immersion).
Edit: In [TT], the case where $\Delta_Z$ is (ind-)quasi-affine is handled. The crucial result is [TT, Thm 5.7/5.8] which in the setup above proves that $Y_1$ exists when $X_1$ is (ind-)quasi-affine. This settles the case when $Z$ is an algebraic space or a Deligne–Mumford stack with separated diagonal. The case where $f$ is finite/integral is easier and treated in [Fer] and [R, Thm. A.4]. Also see MO question Ferrand pushouts for algebraic stacks.
[Fer] Daniel Ferrand, Conducteur, descente et pincement, Bull. Soc. Math. France 131 (2003), no. 4, 553–585.
[R] David Rydh, Compactification of tame Deligne–Mumford stacks, preprint, https://people.kth.se/~dary/tamecompactification20110517.pdf
[TT] Michael Temkin and Ilya Tyomkin, Ferrand pushouts for algebraic spaces, Eur. J. Math. 2 (2016), no. 4, 960–983.
This is great, thanks!
Oh, one minor question: Is the quasiseparated hypothesis known to be necessary, in the sense that there exists a counterexample that doesn't satisfy Milnor excision? If there is a counterexample, it would be interesting to know. Thanks again =).
Milnor excision, as you stated it in the question, holds at least for stacks with quasi-separated diagonal (that is, the double-diagonal is quasi-compact). In particular, it holds for algebraic spaces and algebraic stacks with separated diagonals. I think we can tweak our proof to get it to work in general.
Well, the tweaking is probably not so easy after all...
Yeah, this whole question seems quite tricky. I'm trying to see if I can get it working using locally ringed topoi for higher DM stacks until your paper comes out. Trying to see which pushouts are preserved along the wrong adjoints is very counterintuitive! Anyway, excited to see the paper whenever it comes out!
|
2025-03-21T14:48:31.884996
| 2020-08-27T19:43:03 |
370267
|
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|
Stack Exchange
|
Quick ways to compute transition matrices for classical symmetric function bases
I am trying to implement quick algorithms for computing the transition matrices
involving the monomial, power-sum, elementary, complete homogeneous and Schur polynomials.
There are several relations among these, and all have combinatorial interpretations.
The matrix for expanding the $h_\lambda$ into $s_\mu$, given by inverse Kostka coefficients,
can be computed quickly using a recursion, due to Eğecioğlu and Remmel.
From here, one only needs one of the matrices for
$e_\lambda \to p_\mu$, $h_\lambda \to p_\mu$, $s_\lambda \to p_\mu$ or
$m_\lambda \to p_\mu$, (or the inverse). The remaining transition matrices are obtained via simple matrix operations.
There is a recursion for augmented monomials to power-sums, see M. Merca,
which I implemented, but was surprised that it is quite a bit slower compared to the Eğecioğlu and Remmel recursion above. This is surprising, as Schur polynomials seem much more complicated compared to the other bases.
Are more efficient recursions out there, for the coefficients of the transition matrices
$e_\lambda \to p_\mu$, $h_\lambda \to p_\mu$, $s_\lambda \to p_\mu$ or
$m_\lambda \to p_\mu$?
I have also looked into source code for Symmetrica (do not read C-code commented in German that well), and Stembridges SF package (which uses a method that is slower than M. Merca's approach).
Eğecioğlu, Ömer; Remmel, Jeffrey B., A combinatorial interpretation of the inverse Kostka matrix, Linear Multilinear Algebra 26, No. 1-2, 59-84 (1990). ZBL0735.05013.
Did you look into how it is implemented in SAGE?
@DanPetersen I believe Sage uses Symmetrica under the hood, see https://doc.sagemath.org/html/en/reference/libs/sage/libs/symmetrica/symmetrica.html
And yeah, the Symmetrica source code is not super readable...
@DanPetersen I realized that even though Schur polynomials are more complicated, perhaps it is worth computing the p->s coefficients (characters), as these are so important. I found a preprint by R. Holmes, with a very nice recursion: https://www.math.upenn.edu/~peal/polynomials/murnaghanNakayama.htm#holmesChi
It should be quicker than the naive recursion given by the Murnaghan-Nakayama rule.
|
2025-03-21T14:48:31.885180
| 2020-08-27T19:44:37 |
370268
|
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Stack Exchange
|
Is there a formulation of Huygens' principle using the language of algebraic geometry?
Huygens' principle is a way to describe the propagation of light rays in a manifold $\mathcal{M}$. Suppose we start with an infinitesimal ball of light at position $X$ in a manifold. Let $B(X,t)$ be the set of points to which light rays can arrive at times less than or equal to $t$. Then $\partial B(X,t)=\Sigma(X,t)$ is a closed surface that constitutes the set of points to which light arrive precisely at time $t$. Consider both $\Sigma(X,t_-)$ and $\Sigma(X,t_+)$
Consider also the class of surfaces $\mathcal{A}(t_--t_+)=\Sigma(Y,t_+-t_-)$ for each $Y\in \Sigma(X,t_-)$
Then $\Sigma(X,t_+)$ is the unique closed surface in $\mathcal{M}$ that is tangent to each element in $\mathcal{A}(t_--t_+)$
This is Huygens' theorem.
My question is whether this can be reframed in the language of algebraic geometry. My question is vague because I know very little in algebraic geometry. I guess I am trying to ask if there is a generalization of Huygens' principle to more exotic spaces like supermanifolds, for instance.
Check Volume 1 of the book Singularities of differentiable maps by Arnold, Gusein-Zade and Varchenko, especially section 20.6.
While this is not exactly algebraic geometry, Anders Kock has given a description of Huygen's principle in terms of synthetic differential geometry, which does have some connections to algebraic geometry.
An analogue of the Huygens principle (including interference, so more precisely a Huygens-Fresnel principle) in superspace has been formulated by Gomes in arXiv:gr-qc/0602092. The formulation is used to find the analogue of the WKB approximation to the Wheeler DeWitt equation.
To clarify: "superspace" here is not the kind relevant for supersymmetry but rather the kind appearing in GR/cosmology most famously in "minisuperspace" etc.
From your question, I guess that you start out with a "smooth Riemannian manifold $\mathcal{M}$". Algebraic geometry is, very roughly, the realm of zero loci of polynomial (or perhaps rational) equations, and polynomial (rational) maps between them. A general smooth Riemannian manifold need not belong to that realm, hence algebraic geometry would have very little to say about such an $\mathcal{M}$, or the Huygens' principle on it.
That said, I have some brief observations.
The way you describe Huygens' principle corresponds to the notion of taking an envelope of curves or surfaces (the $\mathcal{A}$ family in your case). This is an operation that makes sense in algebraic geometry because the envelope of a polynomial family of algebraic varieties is also a variety, I believe.
There is a couple of famous articles bringing together algebraic geometry and the Huygens' principle (or something very closely related to it) for constant coefficient hyperbolic equations (at order 2, such equations basically generalize the wave equation on Euclidean space, while at higher orders they are no longer describable by simple Riemannian geometry):
Atiyah, Michael F.; Bott, Raoul; Gårding, Lars, Lacunas for hyperbolic differential operators with constant coefficients. I, Acta Math. 124, 109-189 (1970). ZBL0191.11203.
Atiyah, Michael F.; Bott, Raoul; Gårding, Lars, Lacunas for hyperbolic differential operators with constant coefficients. II, Acta Math. 131, 145-206 (1973). ZBL0266.35045.
As was already implied in the previous point, the notion of light ray (or more generally characteristic ray) only sometimes coincides with that of a geodesic in a Riemannian manifold. The notion itself is more closely tied with the finite speed of propagation for a hyperbolic equation. Such equations can be formulated on supermanifolds and for superfields. Roughly speaking, any supermanifold can be seen as a bundle of odd or super directions over a regular manifold (the body, cf. Serre-Swan theorem). Getting one's hands dirty, and thinking as an analyst, a hyperbolic equation in this setting can be represented as a hierarchical system of regular hyperbolic equations on the body, where the super directions appear in a filtered manner at higher orders of the hierarchy. At the base of the hierarchy, you have a regular hyperbolic PDE on the body without any super information entering into it, and it is only this equation that controls the speed of propagation and hence the Huygens' principle. This approach was pioneered by Choquet-Bruhat in the seminal paper
Choquet-Bruhat, Yvonne, The Cauchy problem in classical supergravity, Lett. Math. Phys. 7, 459-467 (1983). ZBL0529.58039.
|
2025-03-21T14:48:31.885520
| 2020-08-22T18:43:39 |
369871
|
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|
Stack Exchange
|
How to prove that a diophantine equation has only finitely many solutions in integers?
In 1976 Tijdeman proved that the Catalan equation
$$
x^{p}-y^{q}=1
$$
has finitely many solutions in integers $x,y,p,q>1$ in his paper
R. Tijdeman, On the equation of Catalan, Acta Arith. 29 (1976) pp 197–209 (EuDML)
He just found the following upper bound for $p$ and $q$ using Baker theorem in linear form in logarithm
\begin{align}
p& <2c_{9}(\log p)^{c_{10}}\\
q& <c_{1}(\log p)^{c_{2}}
\end{align}
I don't understand how these two inequalities for $p$ and $q$ give us that Catalan's equation has only a finite number of solution since he didn't give an upper bound for $x,y$.
Also in 1993, Overholt showed that Brocard equation
$$
n!+1=m^{2}
$$
has finitely solution if Szpiro's conjecture is true.
He just found that $n<4^{\epsilon}e $.
I don't understand how this upper bound for $n$ make us say that the Brocard equation has finitely many solutions?
I ask if a finding of an upper bound for a
least one variable of an arbitrary Diophantine equation is enough to prove that it has only finitely many solutions in $\mathbb{Z}$? If yes does the upper should do not depend on the other variable of that Diophantine equation?
Edit I need to answer the third question just .
https://en.wikipedia.org/wiki/Faltings%27s_theorem
Please restrict to one question per post (standard MO policy). I answered your first question below.
@Ycor which theorem do you mean
YCor meant that Faltings' theorem (that YCor linked for you) also implies that, for given $p$ and $q$, there are only finitely many solutions in $x$ and $y$. Baker's theorem mentioned in my response has the advantage of providing an effective bound for $x$ and $y$ (in terms of $p$ and $q$), unlike Faltings' theorem.
For the Brocard equation, for each $n$, how many values of $m$ are possible? Answer: 2. Hence if you bound $n$, then you know there are finitely many solutions. If you want an explicit bound, just use $m=\sqrt{n!+1}<\sqrt{4^\epsilon e+1}$ (assuming Szpiro).
@Joe Silverman so in the case of Diophantine equation of two variable it is enough to find an upper bound for one variable to show that the equation has finitely many solutions?
If a Diophantine equation has two variables $x$ and $y$, and you find an upper bound for $y$, then in effect you reduced the original equation to an equation in $x$. Now a one-variable Diophantine equation has finitely many solutions (unless it is degenerate like $x=x$), so the answer to your last question is "yes".
Consider the simple example $x^2-2y^2=1$, Abdo. This equation has infinitely many integer solutions, but if you bound either one of the variables, that automatically bounds the other, and results in there being only finitely many solutions.
Of course one needs to take my previous comment with a grain of salt. For example, consider the two-variable Diophantine equation $xy=y$ in positive integers. The only solutions are $x=1$ and $y$ arbitrary. So $x$ is bounded, yet there are infinitely many solutions.
@GH from MO what about the second part of the 3rd equation does the upper bound should depend only on the variable of the original of it doesn't matter?
Please open a new question, keeping in mind that this site is for research level questions. (For general questions in mathematics see http://math.stackexchange.com/)
By the initial remarks in the paper, one can restrict to $p,q\geq 5$. By Theorem A in the paper (which is a result of Baker's from 1969), $x$ and $y$ can be effectively bounded in terms of $p$ and $q$:
$$\max(|x|,|y|)<\exp\exp(5^{10}p^{10} q^{10q^3}).$$
Hence it suffices to bound $p$ and $q$.
but i mean in general is it enough to find an upper bound for only one varible (3rd question )
@Abdo: The answer is "no". For example, the Fermat equation $x^n+y^n=z^n$ in positive integers implies that $n=1$ or $n=2$, but there are infinitely many solutions. BTW it is virtually impossible to say generalities that hold for all diophantine equations (cf. http://www.scholarpedia.org/article/Matiyasevich_theorem).
so for example if we want to prove that Brocard equation has only finitely solution we need to find upper bound for all the solution
@Abdo Yes, in some sense, but given the form of the Brocard equation, once you give an upper bound for one of the variables, it's trivial to get an upper bound for the other one. Similarly, on an elliptic curve, Siegel's theorem say $y^2=x^3+Ax+B$ has only finite many solutions $(x,y)$ in integers, but effective versions often just write down a bound for $|x|$. However, it's possible sometimes to prove finiteness of solutions without bounding the size of the solutions. Indeed, Sigeel's original proof of his theorem has this form. The trick: Show $E(\mathbb Z)$ is a subset of a finite set ...
... whose number of elements can be bounded, but for which you might not be able to find an upper bound for the size of the largest element. This is typical of proofs in Diophantine approximation, which is often used to study solutions to Diophantine equations.
@GH from MO ,Joe Silverman what about the upper bound .does the upper should depend on the variables of the same equation or it doesn't matter if they depend on other variables which have not related with the original equation?
@Joe Silverman what is the difference between finding an upper bound for each variable and finding an upper for $max(|n|, |m|)$?
@Abdo For an equation such as $n!+1=m^2$, an upper bound for one variable trivially gives an upper bound for the other variable. Similarly for an equation such as $x^3-2y^3=107$. But the original proof that $x^3-2y^3=107$ has finitely many solutions did not actually give an upper bound for $x$ and $y$. ...
... Instead, it proved (roughly) that there were effectively computable constants $a$ and $b$ such *if there were a solution with $\max{|x_0|,|y)_0|}\ge a$, then every solution satisfies $\max{|x|,|y|}\le b\cdot\max{|x_0|,|y)_0|}$. This obviously shows that there are finitely many solutions, without given an explicit upper bound for the largest
|
2025-03-21T14:48:31.885946
| 2020-08-22T19:23:03 |
369874
|
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|
Stack Exchange
|
A functional equation in two complex variables
Let $X$ be a compact metric space, or just $X=\mathbb T$, the unit circle, if it helps. We consider only continuous, complex-valued functions on $X$.
Let $\varepsilon >0$. Is there $\delta > 0$ such that for any given functions $f, g$ on $X$ that are nowhere zero and any function $d$ with $\|d\|_\infty\leqslant \varepsilon$ (the supremum norm) there are two functions $a,b$ such that $fa + gb +ab = d$ with $\|a\|_\infty, \|b\|_\infty \leqslant \delta$?
I have an abstract argument for the circle but I feel there should be an elementary solution, maybe modulo Urysohn's lemma.
Yes, it does, I thought that's what you're doing (fix $f,g$, keep $d$ variable).
Hi Tomek: are you putting any restrictions on $\min_{x\in X} |f(x)|$ and $\min_{x\in X} |g(x)|$ in terms of $\delta$?
@YemonChoi, hi Yemon, nope. If we allow $\delta$ to depend already on $|f^{-1}|_\infty$, then this becomes trivial. However for $X = \mathbb T$ I am happy to assume that $f,g$ have summable Fourier series.
$Hello$, Tomasz! (for some reason the MO prohibits saying "Hi" or "Hello" in the normal text mode). Nice to see you back. Apparently you are still asking the same question whether a function $H$ close to the product $fg$ can be represented as a product $FG$ where $F$ is close to $f$ and $G$ is close to $g$ but now just in the continuous category.
The answer is "Not necessarily" even for the closed unit disk $\mathbb D=\{|z|\le 1\}$. The obvious counterexample would be $f(z)=Mz,g(z)=M\bar z$, $H=M^2|z|^2+\varepsilon$ with huge $M$, but you tried to exclude it by demanding that $f$ and $g$ vanish nowhere. However, it doesn't save the day. Indeed, consider any non-negative continuous functions $\varphi, \psi:\mathbb D\to[0,1]$ such that $\varphi=0$ on $[0,1]$ and $\varphi=1$ outside a small neighborhood of $[0,1]$ while $\psi$ has the same properties with respect to the interval $[-1,0]$. Put $f(z)=Mz\varphi(z)-\frac{\varepsilon}{6M}, g(z)=M\bar z\psi(z)+\frac{\varepsilon}{6M}$. Then the product $fg$ is $\varepsilon$-close to a strictly positive function $H(z)=M^2|z|^2\varphi(z)\psi(z)+\frac\varepsilon 2$. However, if $H=FG$ and $F,G$ are $M/10$ close to $f$ and $g$ respectively, then the argument of $F$ should essentially follow that of $f$ on the left semicircle where $|f|$ is large and then that of $1/g$ on the right semicircle where $|g|$ is large (because $F=H/G$ and we control the arguments of both $H$ and $G$), i.e., it cannot deviate much from that of $z$ anywhere, so the winding number will be $1$ and $F$ will be forced to have a zero inside $\mathbb D$, which is impossible since $H>0$ everywhere.
Of course, if $X$ is the circle, this effect is excluded and the answer becomes "Yes", but you said that you knew it yourself, so I'll stop here for now.
Thank you! This indeed suggests that probably there is no cheap argument for $\mathbb T$...
@TomaszKania The cheap argument for $\mathbb T$ exists, but it is a somewhat boring case consideration. Do you want me to post it or you have a reasonably elegant solution yourself?
@fedja here is the reason.
|
2025-03-21T14:48:31.886209
| 2020-08-22T19:46:46 |
369875
|
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|
Stack Exchange
|
Concentration of $\ell_2$ norm of a vector sampled from a distribution
Let $X=(X_1,\ldots,X_n)$, where $X_i \sim P_{p_i}(0,\frac{1}{\lambda})$ are iid, $P_{p_i}$ is sub gaussian distribution for $i^\text{th}$ element, and 0 and $1/\lambda$ are mean and variance.
I'm looking for a result on the concentration of $\|X\|_2^2$ something of the form $E \|X\|_2^2 \leq c$ with $P(\|X\|_2^2 \geq c+\epsilon)\leq f(\epsilon)$. What happens when all the distribution are normals?
I have asked a similar question on https://math.stackexchange.com/questions/3798100/concentration-of-l2-norm-of-a-vector-sampled-from-a-distribution
I agree with Matt F. that your notation is unnecessarily clumsy. In any case, are you aware that $|X|^2$ concentrates around $n\mbox{var}(X_1)$ ?
@MattF. Thank you for your suggestions, I will change it.
@dohmatob I don't know about that concentration.
Isn't this exactly the same question your asked here https://math.stackexchange.com/questions/3798100/concentration-of-l2-norm-of-a-vector-sampled-from-a-distribution ?
@dohmatob It is the same question, that is why I have linked. I have not received a response there so I posted here.
I think it's agains the policy of either of the sites to re-post duplicates (which kinda makes sense...).
WLOG, let $\lambda = 1$ (rescale your problem appropriately, if necessary). Then, it is well-known consequence of Bernstein's inequality (e.g see theorem 3.1.1 of "High-dimensional Probability" book by R. Vershynin) that
$$
\mathbb P\left(\left|\frac{\|X\|^2}{n}-1\right| \le \epsilon\right) \ge 1 - 2e^{-Cn\min(\epsilon,\epsilon^2)},\;\forall \epsilon \ge 0.
$$
Here, $C$ is a constant which is independent of $n$. In other words, $\|X\|^2$ has good (exponential / Gaussian) concentration around the value $n$.
|
2025-03-21T14:48:31.886355
| 2020-08-22T20:19:35 |
369878
|
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|
Stack Exchange
|
When is inverse geodesic distance positive definite (in a compact manifold)?
We work on a closed smooth Riemannian manifold $(M,g)$ and let $K:M\times M\to \mathbb R\cup\{+\infty\}$ be a kernel, which we assume to be integrable and lower semicontinuous. We say it is positive definite (modifying definitions a bit, so that singular kernels are allowed) if
$$
\int_M\int_M K(x,y) f(x) f(y)d\mathrm{vol}_g(x)\ d\mathrm{vol}_g(y) \ge 0
$$
holds for all $f\in C^\infty(M)$.
I am interested to know cases when it is known if (or if not) the kernel $K(d_g(x,y))$ is positive definite, in which $d_g$ is the geodesic distance.
For example, I don't even know this for the "Riesz kernel formula" analogue $1/(d_g(x,y))^s$ on the torus of dimension $d$, for $0<s<d$ (I suspect it may be false).
Often for kernels on manifolds people don't take the formulas for kernels that work on $\mathbb R^d$, rather they find analogues that are positive definite by construction, so I had trouble finding any known cases in the literature. (However I did not know what to search for.)
|
2025-03-21T14:48:31.886447
| 2020-08-22T20:40:03 |
369880
|
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|
Stack Exchange
|
$1$-factorizability for "complete" finite hypergraphs
Let $H=(V,E)$ be a hypergraph such that $V\neq \varnothing$ and $\varnothing \notin E$. A matching is a subset $M\subseteq E$ such that $m_1\neq m_2 \in M$ implies $m_1\cap m_2 = \varnothing$, and $M$ is said to be perfect if $\bigcup M = V$. We say that $H$ is $1$-factorizable if $E$ is the disjoint union of perfect matchings.
If $H$ is $1$-factorizable, it is easy to see that it implies that for any vertices $v_1,v_2\in V$ we have $\text{deg}(v_1) = \text{deg}(v_2)$ where for $v\in V$ we set $\text{deg}(v) = |\{e\in E: v\in e\}|$.
We regard every positive integer $n$ as an ordinal, that is $n = \{0, \ldots, n-1\}$. If $X$ is a set and $\kappa$ is a cardinal, we let $[X]^\kappa$ be the collection of subsets of $X$ having cardinality $\kappa$.
Question. If $k, n> 2$ are integers, is the hypergraph $(k\cdot n, [k\cdot n]^k)$ $1$-factorizable?
(The answer is yes for $k\leq 2$.)
This is Baranyai's theorem. Other than in Baranyai's original paper you can also find a cool proof in the article "Uniform hypergraphs" by Brouwer and Schrijver which uses max-flow min-cut.
|
2025-03-21T14:48:31.886544
| 2020-08-22T21:13:57 |
369882
|
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|
Stack Exchange
|
Lines on an anticanonical K3 on a Fano 3-fold
Let $Y_d$ be a Fano threefold of Picard rank $1$ and index $2$ (eg cubic 3fold). There is a natural anticanonical map $Y_d\to \mathbb{P}^{d+1}$. Smooth sections of the anticanonical bundle are $K3$ surfaces, so we can ask the following
Question: does such a general $K3$ surface contain lines in $\mathbb{P}^{d+1}$?
If $Y$ is a Fano threefold with Picard rank $1$, then its general anticanonical element $X \in |-K_Y|$ is a $K3$ surface with Picard rank $1$. In particular, $X$ contains no lines.
This is explained (for instance) at p. 797 of
C. F. Doran, A. Harder, A. Y. Novoseltsev, A. Thompson: Calabi–Yau threefolds fibred by high rank lattice polarized K3 surfaces, Mathematische Zeitschrift 294 (2020).
|
2025-03-21T14:48:31.886774
| 2020-08-22T22:09:37 |
369887
|
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|
Stack Exchange
|
Non-isomorphic complex Lie groups with the same exceptional Lie algebra for $\mathfrak{g_2,f_4,e_6,e_7,e_8}$?
An exceptional complex Lie algebra is a simple Lie algebra whose Dynkin diagram is of exceptional (nonclassical) type. There are exactly five such Lie algebras: $\mathfrak{g}_{2}$, ${\mathfrak {f}}_{4}$,${\mathfrak {e}}_{6}$,
${\mathfrak {e}}_{7}$, ${\mathfrak {e}}_{8}$; their respective dimensions are 14, 52, 78, 133, 248.
See https://en.wikipedia.org/wiki/Exceptional_Lie_algebra
Usually, given a complex Lie algebra,
there could be non-isomorphic connected complex Lie groups with the same given Lie algebra.
For example, the ${\rm SO}(N)$ and ${\rm Spin}(N)$ can have the same Lie algebra ${\mathfrak {so}}_{n}$, but they are non-isomorphic Lie groups
because ${\rm SO}(N)={\rm Spin}(N)/(\mathbb{Z}/2\mathbb{Z})$ has smaller center than ${\rm Spin}(N)$.
In particular, ${\rm SO}(5)$ and ${\rm Sp}(2)\simeq {\rm Spin}(5)$ are non-isomorphic Lie groups with isomorphic Lie algebras ${\mathfrak {so}}_{5}\simeq{\mathfrak{sp}}_2$.
Questions:
It is commonly said that the Lie groups with given Lie algebra $\mathfrak{g}_{2}$, ${\mathfrak {f}}_{4}$,${\mathfrak {e}}_{6}$,
${\mathfrak {e}}_{7}$, ${\mathfrak {e}}_{8}$ are $G_2$, $F_4$, $E_6$, $E_7$, $E_8$. However, do we have non-isomorphic Lie groups with the same exceptional Lie algebra $\mathfrak{g}_{2}$, ${\mathfrak {f}}_{4}$,${\mathfrak {e}}_{6}$,
${\mathfrak {e}}_{7}$, ${\mathfrak {e}}_{8}$?
What are the centers $Z(G)$ of these Lie groups $G$? For $G_2$, $F_4$, $E_6$, $E_7$, $E_8$ and possibly others with the same given Lie algebra?
What are the homotopy groups $$\pi_d(G)$$ of these Lie groups $G$ for lower dimensions? say $d=0,1,3,4,5,...$?
We already know that $\pi_2(G)=0$ for any Lie group.
I appreciate your patience, comments and answers
These are pretty basic questions which are probably a better fit for math stack exchange. But certainly there are multiple different Lie groups which have the same exceptional Lie algebra as their Lie algebra: see e.g. https://en.wikipedia.org/wiki/List_of_simple_Lie_groups#List for a starting point.
Of course, as questions (2) and (3) seem to hint at, the general rule for finding Lie groups $G'$ with the same Lie algebra as $G$ is (1) find the universal cover $\tilde G$, (2) find the center $Z(\tilde G)$, (3) enumerate all discrete subgroups $\Gamma \subseteq Z(\tilde G)$, (4) take $G' = \tilde G / \Gamma$ for each $\Gamma$, and (5) look for more direct descriptions. How straightforward is it to hunt down all these steps for all these groups in the literature (I'm no expert)? If nothing else, recording sources for each step here would make this information easier to find on the internet.
@TimCampion At least to me it seems to be quite difficult to find much information on the exceptional Lie groups in the standard textbooks. There is the textbook "Lectures on Exceptional Lie Groups" by Adams which might contain most information but it seems to be out of print in my country at the moment. It would be interesting to see whether there is a modern textbook on Lie groups that contains also detailed descriptions and properties for the exceptional Lie groups and their Lie algebras. I have nearly 10 books on Lie groups but none has full detailed information on all exceptional cases.
Not only is this problem subject to explicit enumeration as @TimCampion says; there's not much enumeration to do: $\mathsf E_8$, $\mathsf F_4$, and $\mathsf G_2$ are all simply connected and adjoint (= centreless, for semisimple groups), so they are the unique (linear) Lie groups with their Lie algebras (and can be realised as the automorphism groups of their Lie algebras, if you've already got those).
For $\mathsf E_6$ and $\mathsf E_7$ the centres of the simply connected groups (equivalently, the fundamental groups of their adjoint quotients) are cyclic of prime order (3 and 2, respectively), so you have only the automorphism groups of the Lie algebras and their simply connected covers.
If "aut. Lie alg." counts as explicit but "s.c. cover" doesn't, then the $\mathsf E_6$ and $\mathsf E_7$ sitting inside $\mathsf E_8$ (as derived groups of Levi subgroups) are both simply connected. I think this is how Frank Adams constructs them in the lovely book @Mare references. If I recall correctly, he gets $\mathsf F_4$ and $\mathsf G_2$ by folding (paper by Stembridge, also lovely).
One more and then I'll stop. I thought we had an MO question about how to lift foldings of root systems to groups, but I can't find it (although …); but there is a nice question https://mathoverflow.net/questions/99736/beautiful-descriptions-of-exceptional-groups . (EDIT: And, oops, I forgot about real forms, of which $\mathsf E_6$ and $\mathsf E_7$ have non-split ones but $\mathsf E_8$, $\mathsf F_4$, and $\mathsf G_2$, because both s.c. and adjoint, do not.)
Thanks very much, I think some of you experts can already answer my question directly in an answer form. ,3
I guess in light of all this discussion I agree that it would be nice for someone to post an answer cataloging all the possible groups.
Thanks, I voted up your comment - please feel free to vote mine post up so your answer will get more votes/attentions too.
Relevant old MO question: https://mathoverflow.net/questions/43546/lie-algebras-to-classify-lie-groups
@LSpice What do you mean that $F_4$ does not have a non-split real form? There are three non-isomorphic real forms of type $F_4$. See e.g. https://mathoverflow.net/a/96477/6818
E8 also has 3 real forms according to the list I linked.
@VítTuček (1) and @SamHopkins (2), thanks for the corrections. I guess I was assuming (with the usual results) that $\mathrm H^1(k, G)$ is trivial for $G$ simply connected as in the $p$-adic case.
@LSpice: You should not assume that!
@MikhailBorovoi, right, sorry, I meant my editorial aside "(with the usual results)" to indicate that I had been obviously foolish. I can plead no better excuse than that I almost never think of real groups, so that I am too seldom bit by my over-analogising.
@MikhailBorovoi: Why did you edit the question to refer to complex Lie groups? It's not at all clear that that's what the OP was asking about. In fact, if someone just said "Lie group," I would assume they meant real Lie group.
@SamHopkins: The OP wrote explicitly "complex Lie algebra". I don't think that she wanted to consider a real Lie group whose real Lie algebra is the restriction of scalars from $\Bbb C$ to $\Bbb R$ of a complex Lie algebra.
@SamHopkins: Anyway, if anybody (say, the OP, or you) asks a question about real forms of semisimple complex Lie algebras (or of semisimple complex algebraic groups), I will answer this question.
@MikhailBorovoi: I would instinctively interpret the question as: which real Lie groups have real Lie algebras whose complexifications are the exceptional complex simple Lie algebras $\mathfrak{g}_2, \mathfrak{f}_4, ...$. So there are two issues: real forms, and nontrivial centers.
@SamHopkins: This was not written! Let us wait for comments of OP....
Yes, I agree the OP should clarify.
@SamHopkins: I mentioned it in my answer and gave references.
@LSpice: Concerning real Galois cohomology see the paper of Adams and Taibi and the paper of Borovoi and Evenor. Both have been published. A new version of this peprint will appear by the end of August.
@MikhailBorovoi's, thanks for the references (which maybe could go in your answer?). Titles: Adams and Taïbi - Galois and Cartan cohomology of real groups; Borovoi and Evenor - Real homogenous spaces, Galois cohomology, and Reeder puzzles; Borovoi and Timashev - Galois cohomology of real semisimple groups.
@LSpice: It is not quite related. One should try to avoid self-promotion.
Crosspost from MathSE: https://math.stackexchange.com/questions/3799810/, https://math.stackexchange.com/questions/3799860/, https://math.stackexchange.com/questions/3799800/
I prefer to use the language of algebraic groups.
All algebraic groups and Lie algebras are defined over $\Bbb C$.
1. Let ${\mathfrak g}$ be a semisimple Lie algebra.
Consider the automorphism group ${\rm Aut\,}{\mathfrak g}$,
its identity component $G^{\rm ad}:=({\rm Aut\,}{\mathfrak g})^0$,
and the group of outer automorphisms ${\rm Out\,} {\mathfrak g}:=({\rm Aut\,} {\mathfrak g})/({\rm Aut\,} {\mathfrak g})^0$.
We say that $G^{\rm ad}$ is the adjoint group (or the group of adjoint type) with Lie algebra ${\mathfrak g}$. Note that $Z(G^{\rm ad})=\{1\}$.
2. Starting with a semisimple Lie algebra ${\mathfrak g}$,
one can construct the simply connected group $G^{\rm sc}$ with Lie algebra ${\mathfrak g}$;
see Steinberg, Lectures on Chevalley groups, AMS, 2016.
Note that $\pi_1(G^{\rm sc})=\{1\}$.
This algebraic group $G^{\rm sc}$ has the following universal property:
for any algebraic group $H$ with Lie algebra ${\mathfrak h}$
and for any homomorphism of Lie algebras $\varphi_{\rm Lie}\colon {\mathfrak g}\to{\mathfrak h}$,
there exists a unique homomorphism of algebraic group $\varphi\colon G^{\rm sc}\to H$ inducing $\varphi_{\rm Lie}$.
3. For any connected algebraic group $G$ with Lie algebra ${\mathfrak g}$, there exists a canonical surjective homomorphism
$$\rho\colon G^{\rm sc}\to G $$
inducing the identity isomorphism on ${\mathfrak g}$; see above.
We have
$$\pi_1(G^{\rm sc})=\{1\},\quad \pi_1(G)={\rm ker}\,\rho.$$
On the other hand, we have a canonical surjective homomorphism
$${\rm Ad}\colon G\to G^{\rm ad}\subseteq {\rm Aut\,} {\mathfrak g}$$
with kernel $Z(G)$.
Write
$$C=Z(G^{\rm sc})=\pi_1(G^{\rm ad}).$$
The homomorphism
$$ {\rm Ad}\colon G\to G^{\rm ad}$$
induces a homomorphism
$$i\colon \pi_1(G)\to\pi_1(G^{\rm ad})=C.$$
Moreover, the homomorphism
$$\rho\colon G^{\rm sc}\to G$$
induces a homomorphism
$$j\colon C=Z(G^{\rm sc})\to Z(G).$$
In this way we obtain a short exact sequence
$$1\to\pi_1(G)\overset{i}{\longrightarrow} C\overset{j}{\longrightarrow} Z(G)\to 1.$$
Conversely, for each subgroup $F\subseteq C$ one can associate a connected semisimple group
$ G_F:=G^{\rm sc}/F$ with Lie algebra ${\mathfrak g}$, with fundamental group $\pi_1(G_F)=F$, and with center $Z(G_F)=C/F$.
In this way we obtain a canonical bijection between the set of subgroups of $C$ up to conjugation by ${\rm Out\,} {\mathfrak g}$
and the set of isomorphism classes of connected semisimple algebraic groups with Lie algebra ${\mathfrak g}$.
It is known that ${\rm Out\,} {\mathfrak g}$ is canonically isomorphic to ${\rm Aut\,} {\rm Dyn}({\mathfrak g})$,
where ${\rm Dyn}({\mathfrak g})$ is the canonical Dynkin diagram of ${\mathfrak g}$.
4. Let us return to our exceptional simple Lie algebras.
The group $C=C({\mathfrak g})$ can be found, for instance, in tables in the book by Bourbaki "Lie Groups and Lie Algebras, Chapters 4-6",
or in the book by Onishchik and Vinberg "Lie Groups and Algebraic Groups", Springer-Verlag, 1990.
For ${\mathfrak g}_2$, ${\mathfrak f}_4$, and ${\mathfrak e}_8$ we have $C({\mathfrak g})=\{1\}$.
Thus there is only one (up to isomorphism) algebraic group $G^{\rm sc}({\mathfrak g})=G^{\rm ad}({\mathfrak g})$ with Lie algebra ${\mathfrak g}$.
For ${\mathfrak g}={\mathfrak e}_6$ we have $C({\mathfrak g})\simeq {\Bbb Z}/3{\Bbb Z}$. This group has no nontrivial subgroups.
Thus there are exactly two connected algebraic groups (up to isomorphism) $E_6^{\rm sc}$ and $E_6^{\rm ad}$ with Lie algebra ${\mathfrak e}_6$.
We have
$$Z(E_6^{\rm sc})=\pi_1(E_6^{\rm ad})\simeq{\Bbb Z}/3{\Bbb Z}.$$
For ${\mathfrak g}={\mathfrak e}_7$ we have $C({\mathfrak g})\simeq {\Bbb Z}/2{\Bbb Z}$. This group has no nontrivial subgroups.
Thus there are exactly two connected algebraic groups (up to isomorphism) $E_7^{\rm sc}$ and $E_7^{\rm ad}$ with Lie algebra ${\mathfrak e}_7$.
We have
$$Z(E_7^{\rm sc})=\pi_1(E_7^{\rm ad})\simeq{\Bbb Z}/2{\Bbb Z}.$$
5. The real forms of a connected algebraic group of an exceptional type correspond bijectively to the real forms of
(or real structures on) its Lie algebra.
My favorite way to classify those is via Kac diagrams.
See Table 7 in the book by Onishchik and Vinberg.
The number of real forms is 2 for ${\mathfrak g}_2$, 3 for ${\mathfrak f}_4$, 3 for ${\mathfrak e}_8$, 4 for ${\mathfrak e}_7$, 5 for ${\mathfrak e}_6$.
These real forms are listed also in Table V in Chapter X of Helgason's book "Differential Geometry, Lie Groups, and Symmetric Spaces"
(Helgason lists all non-compact forms).
Helgason classifies real forms using the original method of Kac with infinite dimensional Lie algebras.
Onishchik and Vinbeg use another method, which gives exactly the same answer (the same Kac diagrams).
I think it may be important to note what is paid for by using the language of algebraic groups (which I also prefer). For example, "there is only one algebraic group with Lie algebra $\mathfrak g$" (where $\mathfrak g$ is $\mathfrak e_8$, $\mathfrak f_4$, or $\mathfrak g_2$) is true only to the extent that we speak of groups over alg. closed field; it's not true over $\mathbb R$ (as you say later). Also, the alg. groups that we are calling connected and simply connected need not be either when regarded as real Lie groups (by taking real points). Finally, we miss non-linear groups this way!
I see now that you edited the question to refer only to complex Lie groups, as the wording obliquely suggested and OP may or may not have intended. In this case I agree with you: we get very lucky—I think of it as luck, but perhaps design is a more appropriate way to describe it—and the confusions I worried about do not arise (which, of course, is how those alg.-gp. terms arose!).
@LSpice: All what you say in your first comment - I do know that. If you ask a relevant question, I will answer it when I have time.
Certainly I didn't mean to suggest that you didn't know that, and I apologise if I gave that impression. I just meant that it might be worth it to make it part of the answer for the benefit of people who might not know it.
In particular, if a semisimple algebraic $\Bbb R$- group $\bf G$ is a connected (over $\Bbb C$), and if it either compact or simply connected (simply connected over $\Bbb C$), then the group of $\Bbb R$-points $\bf G(\Bbb R)$ is connected.
I understood what you meant, but this is an answer on two pages! If somebody asks a separate question, I will answer it.
One thing is wrong with this answer: the exceptional Lie algebra $\mathfrak{e}_6$ has a non-trivial outer automorphism.
@SamHopkins: Oops! Thank you for noticing. I have removed the erroneous assertion.
@MikhailBorovoi: I have one last question about this very nice answer. If I want to compute all the real Lie groups associated with a complex Lie algebra, is it true that I can, so to speak, "multiply" the choices from the two steps of choosing a real form and choosing a center? E.g., for $\mathfrak{e}6$ I would get $5\times 2=10$ groups? In other words, will I have $C(\mathfrak{g}{\mathbb{R}})=C(\mathfrak{g})$ for any real form of my Lie algebra?
Looking at https://en.wikipedia.org/wiki/List_of_simple_Lie_groups#List again, I see that what I suggested in my last comment is not true: the fundamental groups of the centerless real groups will be different from in the complex case.
@SamHopkins: And you wrote that this question would be more suitable for Math StackExchange....
@MikhailBorovoi: Haha, indeed. Well I knew that in principle this was all worked out by Cartan 100 years ago, but there are many moving parts to say the least!
@SamHopkins: Real forms were classified by Cartan in this paper of 1914.
@MikhailBorovoi, in @SamHopkins's defence (not that they need it from me), I'm not sure if the fact that there are lots of interesting things to say about this means that it was an appropriate MO question. Yours is obviously a good answer, and I appreciate your writing it; but it's the sort of thing that can be and should be found in the standard references, as you mention, and I wouldn't want to see people come to MO to ask questions when they can and should first check in the standard references (for whatever field their question is in).
I have just dumped the comments in an answer, and hope someone will make a better answer than this. (EDIT: Someone has! See @MikhailBorovoi's answer.) I'll be happy to delete this one—or you can just edit this one, which is CW to avoid reputation (since I'm just compiling comments).
One thing that hasn't been mentioned in the comments yet is Question 3. Asking about $\pi_0$ of a group based on its Lie algebra is in some sense meaningless; the Lie algebra of a group only sees its identity component, so one may make $\pi_0$ as bad as one allows a discrete Lie group to be (countable?) without changing the Lie algebra. For $\mathsf E_8$, $\mathsf F_4$, and $\mathsf G_2$, $\pi_1$ is trivial. For $\mathsf E_6$ and $\mathsf E_7$, the adjoint form (arising as the automorphism group of the Lie algebra) has fundamental group that is cyclic of prime order (3 and 2, respectively). However, all of these statements are about the complex groups, hence for compact forms, and I know that other real forms may have different fundamental groups, but don't know how they differ. Hopefully someone will fill this in.
As you mention, $\pi_2$ is trivial, and @AndréHenriques gives a reference in the comments pointing to Borel's paper An application of Morse theory to the topology of Lie groups proving that $\pi_3$ is infinite cyclic for simple groups, but I'm not sure if that's for algebraically simple groups, or for abstractly simple groups (i.e., is a finite centre allowed?). Again, hopefully someone will edit this answer, or add their own.
@SamHopkins said:
… certainly there are multiple different Lie groups which have the same exceptional Lie algebra as their Lie algebra: see e.g. http://en.wikipedia.org/wiki/List_of_simple_Lie_groups#List for a starting point.
@TimCampion said:
Of course, as questions (2) and (3) seem to hint at, the general rule for finding Lie groups $G'$ with the same Lie algebra as $G$ is (1) find the universal cover $\tilde G$, (2) find the center $Z(\tilde G)$, (3) enumerate all discrete subgroups $\Gamma \subseteq Z(\tilde G)$, (4) take $G' = \tilde G/\Gamma$ for each $\Gamma$, and (5) look for more direct descriptions. How straightforward is it to hunt down all these steps for all these groups in the literature (I'm no expert)? If nothing else, recording sources for each step here would make this information easier to find on the internet."
@Mare said:
@TimCampion At least to me it seems to be quite difficult to find much information on the exceptional Lie groups in the standard textbooks. There is the textbook "Lectures on Exceptional Lie Groups" by Adams which might contain most information but it seems to be out of print in my country at the moment. It would be interesting to see whether there is a modern textbook on Lie groups that contains also detailed descriptions and properties for the exceptional Lie groups and their Lie algebras. I have nearly 10 books on Lie groups but none has full detailed information on all exceptional cases."
I then blathered on for a while (1 2 3 4), but note that first I forgot to think about real forms, and even when I remembered I got it wrong by over-analogy to the $p$-adic case—see important corrections (1 2) of my wrong statements about non-existence of non-split real forms:
Not only is this problem subject to explicit enumeration as @TimCampion says; there's not much enumeration to do: $\mathsf E_8$, $\mathsf F_4$, and $\mathsf G_2$ are all simply connected and adjoint (= centreless, for semisimple groups), so they are the unique (linear) Lie groups with their Lie algebras (and can be realised as the automorphism groups of their Lie algebras, if you've already got those).
For $\mathsf E_6$ and $\mathsf E_7$ the centres of the simply connected groups (equivalently, the fundamental groups of their adjoint quotients) are cyclic of prime order (3 and 2, respectively), so you have only the automorphism groups of the Lie algebras and their simply connected covers.
If "automorphism group of Lie algeba" counts as explicit but "simply connected cover" doesn't, then the $\mathsf E_6$ and $\mathsf E_7$ sitting inside $\mathsf E_8$ (as derived groups of Levi subgroups) are both simply connected. I think this is how Frank Adams constructs them in the lovely book @Mare references. If I recall correctly, he gets $\mathsf F_4$ and $\mathsf G_2$ by folding (paper by Stembridge, also lovely).
I thought we had an MO question about how to lift foldings of root systems to groups, but I can't find it (although …); but there is a nice question Beautiful descriptions of exceptional groups. (EDIT: And, oops, I forgot about real forms, of which $\mathsf E_6$ and $\mathsf E_7$ have non-split ones but $\mathsf E_8$, $\mathsf F_4$, and $\mathsf G_2$ because both s.c. and adjoint, do not. Further EDIT while compiling: This is wrong, as @VítTuček and @SamHopkins pointed out for $\mathsf F_4$ and $\mathsf E_8$, respectively. In fact I was, impressively, wrong on every count; there's also a non-split form for $\mathsf G_2$. I should have known that this uniqueness statement was wrong even before it was pointed out, because every real (linear) Lie group has at least compact form as well as a split one.)
@SamHopkins said:
Relevant old MO question: Lie algebras to classify Lie groups.
@VítTuček said:
@LSpice What do you mean that $\mathsf F_4$ does not have a non-split real form? There are three non-isomorphic real forms of type $\mathsf F_4$. See e.g. https://mathoverflow.net/a/96477/6818.
@SamHopkins said:
$\mathsf E_8$ also has 3 real forms according to the list I linked.
I appreciate your effort and voted up, BUT this is not an answer! Something more precise and direct counts an answer...
I agree that it is not a very good answer, which is why I hope someone will make a better one; but it seems to me that it literally answers your Question 1 and Question 2. I did not want to editorialise, so included pretty much everything mathematical; but feel free to edit to something more streamlined.
I will post an answer very soon. I am typing it.
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2025-03-21T14:48:31.887914
| 2020-08-22T23:36:58 |
369888
|
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|
Stack Exchange
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Tsuchiya-Ueno-Yamada's proof that sheaves of conformal blocks are locally free
I'm referring to Tsuchiya-Ueno-Yamada's (TUY hereafter) celebrated paper Conformal Field Theory on Universal Family of Stable Curves with Gauge Symmetries. One of the main goals of their paper is to prove that sheaves of covacua (whose dual are the sheaves of conformal blocks) for WZW-models are locally free, i.e., can be viewed as vector bundles. One important consequence is that the dimensions of the spaces of conformal blocks do not rely on the complex structures of the complex curves. Now, my question can be summarized as follows:
Is their proof of local freeness (still) valid in the analytic setting?
(I'm NOT asking about the algebraic setting, which I'm sure is correct.)
Background: Let $\mathfrak g$ be a complex simple Lie algebra, and $\hat{\mathfrak g}_l$ the level $l\in\mathbb N$ affine Lie algebra. Let $\mathfrak F:=(\pi:\mathcal C\rightarrow\mathcal B)$ be a family of $N$-pointed compact Riemann surfaces (let's first ignore nodal curves), associated with $N$ irreducible highest weight integrable representations $W_1,\dots,W_N$ of $\hat{\mathfrak g}_l$. Set $W_\bullet=W_1\otimes\cdots\otimes W_N$. Then the sheaf of covacua $\mathcal V$ is a sheaf of $\mathcal O_{\mathcal B}$-module defined to be the quotient of $W_\bullet\otimes\mathcal O_{\mathcal B}$ by a certain subsheaf. Its dual $\mathcal V^*$ is the sheaf of conformal blocks. (Cf. [TUY] Def. 4.1.2.)
The way TUY proved that $\mathcal V$ (and hence $\mathcal V^*$) is locally free is as follows. In Sec. 5.1-5.3, they defined a (projectively flat) connection on $\mathcal V$. A standard result says that an (analytic or algebraic) coherent sheaf equipped with a connection is locally free. Thus, it reduces to proving the coherence of $\mathcal V$. In [TUY], the coherene is proved in Thm. 4.2.4. However, if you check their proof, you will see (cf. Lemma 4.2.2 and the paragraphs before that) that they actually only proved that $\mathcal V$ is finitely-generated, but not proved that the kernal of any morphism $\mathcal O_{\mathcal B}^n|_U\rightarrow\mathcal V|_U$ (where $n\in\mathbb N$ and $U\subset\mathcal B$ is open) is finitely generated, which is also part of the definition of coherent sheaves. This is certainly OK if we assume $\mathfrak F$ is an algebraic family. In the algebraic setting, $\mathcal V$ is clearly a quasi-coherent algebraic sheaf (since it is the cokernal of a morphism between two (infinite rank) free sheaves). So "finitely generated" implies algebraic coherence. However, in the analytic setting, which seems to be the setting TUY is working in, I want to argue that:
"$\mathcal V$ is a finitely generated $\mathcal O_{\mathcal B}$-module" is not enough to show that "$\mathcal V$ is coherent". Indeed, $\mathcal V$ is not a quasi-coherent analytic sheaf. To my knowledge, in the literature of complex geometry, only "quasi-coherent analytic Frechet sheaves" are defined. (See the book of Eschmeier & Putinar.) But $\mathcal V$ clearly does not admit a Frechet structure since $W_1,\dots,W_N$ do not. Also, I'm wondering if there is a proof that "quasi-coherent analytic Frechet sheaf + finitely generated ==> analytic coherent"? See also a related question about quasi-coherent analytic sheaves.
"Finitely generated + existence of connection" is, I believe, not enough to show that $\mathcal V$ is a locally free $\mathcal O_{\mathcal B}$-module. It will only imply that for each point $b\in\mathcal B$, the stalk $\mathcal V_b$ is a free $\mathcal O_{\mathcal B,b}$-module.
Thus, my opion is that TUY's proof that $\mathcal V$ is locally free is OK in the algebraic setting, but needs adjustment in the analytic setting. Am I right or wrong?
I haven't thought about their paper in a while, but I think Oka's coherence theorem can be used on the dual module ... maybe?
@S.Carnahan Hi Scott, I think for the dual module, the difficulty is to show that it is finitely generated. Its sheaves of relations are finitely generated by Oka's theorem, as you said.
Now I tend to believe that the argument of TUY cannot be directly applied to the analytic setting.
As explained in the question, TUY's main idea of proving that the sheaf of coinvariants $\mathcal V$ is locally free is to show that (1) $\mathcal V$ is coherent (2) $\mathcal V$ admits a connection. The standard definition of $\mathcal V$ being coherent is that: (a) $\mathcal V$ is of finite-type, namely, locally it has an epimorphism from a free module $\mathcal O^n_{\mathcal B}$. (b) its sheaves of relations are of finite type, i.e., the kernel of any morphism $\mathcal O^n_{\mathcal B}\rightarrow\mathcal V$ is of finite type.
TUY's argument only shows that $\mathcal V$ is of finite-type. The reasons that in the algebraic setting, the kernel of any $\varphi:\mathcal O^n_{\mathcal B}\rightarrow\mathcal V$ is of finite type (and hence $\mathcal V$ is coherent) are as follows:
(1) $\mathcal V$ is quasicoherent. So if we assume for simplicity that $\mathcal B$ is affine, then the $\mathcal O_{\mathcal B}$-module $\ker \varphi$ is generated by global sections, i.e. by $(\ker \varphi)(\mathcal B)$. Moreover, the latter is the kernel of $\mathcal O(\mathcal B)^n\rightarrow\mathcal V(\mathcal B)$.
(2) More importantly, in the algebraic setting, $\mathcal O(\mathcal B)$ is Noetherian. So the kernel of $\mathcal O(\mathcal B)^n\rightarrow\mathcal V(\mathcal B)$ is finitely generated. Summing up these two points, one sees that $\ker\varphi$ is of finite type.
In fact, in the paper of Nagatomo-Tsuchiya (which studied genus 0 conformal blocks for C2 cofinite rational VOAs), it was clearly mentioned that the proof of coherence is a combination of finite-type and the Noetherian property, see the proof of Theorem 6.2.1:
The above points (1) and (2) clearly indicate why TUY's argument fails in the analytic setting (just as the proofs of many algebraic coherence properties fail for the analytic coherence due to the lack of Noetherian property, e.g. the coherence of structure sheaves, the coherence of direct image sheaves): When $\mathcal B$ is a complex manifold, $\mathcal O(\mathcal B)$ is in general not Noetherian even if $\mathcal B$ is a Stein manifold, the analytic analog of affine scheme.
One may wonder if (2) still works if one uses the coherence of the structure sheaf $\mathcal O_{\mathcal B}$ instead of the Noetherian property. The answer is no: An $\mathcal O_{\mathcal B}$-submodule of the coherent sheaf $\mathcal O_{\mathcal B}^n$ is not necessarily of finite-type. So one cannot prove that the analytic sheaf $\ker\varphi$ is of finite type.
Unfortunately, TUY's paper talks about complex manifolds instead of varieties or schemes. Even the paper of Nagatomo-Tsuchiya applies (the proof of) the algebraic coherence theorem 6.2.1 to analytic manifolds. (For example, Prop. 8.2.6 of NT says that $\mathcal V_U(M_C)$ is a coherent $\mathcal O_U$-module where $U$, introduced before Prop. 8.2.1, is a product of analytic open disks.) These have confused me for quite a long time...
In the analytic setting I would expect that the TUY proof would work when combined with the local integrability (in the analytic topology) of the projectively flat connection -- i.e., instead of using the action of the Virasoro algebra one can use Segal's (projective) action of the semigroup of annuli, which should provide trivializations of the sheaf of conformal blocks on analytic opens in the neighborhood of every point. I would think that this plus the finite generation of the stalks that you mention is sufficient, but maybe that's naive.
Hi David, Thank you for the reply. I think this method will work if we can show that the action of the annuli "converges absolutely". For example, the action is the exponential of a sum of Virasoro operators including negative modes, and one must show that this exponential converges. If the convergence can be addressed then I believe the local freeness can be proved. But anyway this seems to take more effort than the methods provided in TUY.
I thought the fact that Segal build an an action of the semi group on integrable reps of loop groups already gives the desired exponentiation of the infinitesimal action?
Then the range of the exponential is contained in the Hilbert space, not contained in the original (incomplete) vector space on which the conformal blocks are defined. So to make that approach work, maybe we also need to extend the domain of conformal block to the full Hilbert space, which does not seem so obvious...?
Right, you would need to check that conformal blocks as say Segal would define them (using groups) - which I think is more natural from the physics perspective- are the same as the algebraic definition using Lie algebras. I think of that as a loose affine analog of finite dimensional reps of a Lie algebra being integrable..
By the way I think this distinction helps explain (at least to me) how physicists can get monodromy representations of mapping class groups even outside the world of rational CFT - their spaces of conformal blocks there are strictly larger than the algebraic ones, on which the connections are definitely not integrable
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2025-03-21T14:48:31.888640
| 2020-08-23T00:28:51 |
369890
|
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Stack Exchange
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Proving that finite, connected group schemes in characteristic 0 are trivial
I have a question about a specific proof that all finite group schemes in characteristic 0 are etale. The proof is here, Proposition 8 in lecture notes by Andrew Snowden.
In his notation, let $A = k\oplus I$ be a (finite) local group scheme over a field $k$. For now, we are agnostic about the characteristic of the field. Let $x_1,\dots,x_n$ be elements in $I$ that form a basis for $\Omega_{A/k} = I/I^2$. He defines derivations $D_i: A \to A$ by $A \to A\otimes A \to A \otimes I/I^2 \to A$ where the first map is the comultiplication map, the second map is projection on the second factor and the third map sends $x_i \to 1$ and the other $x_j \to 0$ in the second factor.
He claims in the proof of the proposition that $D_i(x_i) = 1$ but I don't see why this has to be true. Let us do an example: Take $A = k[t]/(t-1)^p = \operatorname{Spec} \mu_p$ in characteristic $p$ and we can take $n=1,x_1 = t-1$. Then the comultiplication map sends $x_1 \to x_1\otimes x_1 + x_1\otimes 1 + 1\otimes x_1$ and so $D_1(x_1) = x_1+1$ if I did the computation right.
Question 1: Where did I go wrong?
Question 2: How do I geometrically interpret this proof? What is the $D_i$ doing and why do they only exist for group schemes? I have seen other geometrically motivated proofs of this fact but I don't understand this one.
Wait, doesn't comultiplication in $\mu_n = \operatorname{Spec} k[x_1]/(x_1^n - 1)$ send $x_1$ to $x_1 \otimes x_1$, including if $p \mid n$? That seems to give $D_1 \colon x_1 \mapsto x_1 \otimes x_1 \mapsto x_1 \otimes x_1 \mapsto x_1 \otimes 1 = x_1$, which still isn't $D_1(x_1) = 1$, as you say even in characteristic $0$.
Sorry, I meant to say $ x_1 = t - 1$, the x_i have to lie in I
on a group scheme $G$, the $D_i$ are the derivatives along the $G$-invariant vector fields on $G$. You can see that from the construction as we can take the invariant derivative of a function $f$ at a point $g$ by taking the derivative of $f(g_1g_2)$ at $(g,e)$ in the $g_2$ direction. $f(g_1g_2)$ is the comultiplication, taking the derivative in the $g_2$ direction is the second step, and then specializing to a give tangent vector is the last step. I think the proof is supposed to be doing something like exponentiating the vector fields.
@WillSawin Thanks, that's along the lines of what I was thinking but I didn't quite get there.
I don't understand that claim either.
It seems to me that if you follow the chain rule you get $$D_i (\varphi(f)) = \sum_{j=1}^n D_i(x_j) \varphi \left( \frac{\partial f}{\partial x_j } \right). $$ You have $D_i(x_j) \equiv \delta_{ij} \mod I$ so the matrix with entries $D_i(x_j)$ is invertible which means if $D_i (\varphi(f))=0$ for all $i$ then $\frac{\partial f}{\partial x_j}=0$ for all $j$ which again means that the kernel of $\varphi$ is invariant under $\frac{\partial}{\partial x_j}$ and you can continue from there.
Great, I think that works!
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2025-03-21T14:48:31.888884
| 2020-08-23T00:34:02 |
369891
|
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|
Stack Exchange
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What are some further examples of proper class models of ZF that are contained in their own "self-relativization"?
Let $\Gamma$ and $\Delta$ be theories in the language of set theory (LST), and let $M = \{x: \phi(x)\}$ be a class, where $\phi(x)$ is some formula of LST. Let us say that $M$ is a (standard) class model of $\Gamma$ in $\Delta$ if and only if $\Gamma \vdash \psi$ implies $\Delta \vdash \psi^M$ for all sentences $\psi$ of LST, where $\psi^M$ denotes the relativization of $\psi$ to the class $M$. I'm curious about what can be said of proper class models $M$ of ZF in ZF such that $\mbox{ZF} \vdash \forall x (x \in M \rightarrow \phi^M(x)$), i.e., $\mbox{ZF} \vdash \forall x (\phi(x) \rightarrow \phi^M(x))$.
A class $M = \{x: \phi(x)\}$ satisfying $\forall x (\phi(x) \rightarrow \phi^M(x))$ could be said to be "contained in its self-relativization." Using Definition I.16.5 from Kunen's set theory book (2013), the statement $\forall x (\phi(x) \rightarrow \phi^M(x))$ (i.e., $(\forall x \, \phi(x))^\phi)$) holds if and only if $\phi$ is absolute for $\{x:\phi(x)\}$, written $\{x: \phi(x)\} \preceq_\phi V$.
Examples of such proper class models of ZF satisfying this self-relativization property are $V$ and $L$ (and both satisfy the stronger condition $\mbox{ZF} \vdash \forall x (\phi(x) \leftrightarrow \phi^M(x)$). Note also that the canonical inner models $L[0^\sharp]$, $L[\mu]$, etc., are transitive class models of ZF that satisfy the self-relativization property, but their consistency strength is well beyond that of ZF.
My questions are as follows. What are some other interesting examples and non-examples? Am I correct in thinking that HOD is a non-example? Are there examples besides $V$ and $L$ whose existence is equiconsistent with ZF? And is there an example that is non-transitive?
One interesting thing about such class models $M$ of ZF in ZF is that ZF does not refute $V = M$ if ZF is consistent, for if $\mbox{ZF} \vdash \neg \forall x (\phi(x))$, then $\mbox{ZF} \vdash \neg \forall x (x\in M \rightarrow \phi^M(x))$, and therefore $\mbox{ZF} \nvdash \forall x (x \in M \rightarrow \phi^M(x))$, assuming ZF is consistent. This is what piqued my interest in the question. Also, one of the comments points out that, if ZF proves that $M$ is non-transitive, then ZF refutes $V = M$. Thus, if there is a non-transitive example, then ZF cannot prove that it is non-transitive. But could there be one that isn't transitive according to a theory stronger than ZF? (Maybe no such theory would be very natural or attractive.)
Does not every model of ZF contains its own L, which is just the ordinary L if the model is transitive and contains all ordinals?
Yes, so $M$ is transitive if and only if it contains $L$, if and only if it contains all the ordinals. I guess I can simplify my question now.
If $ZF$ proves that $M$ is non-transitive, then $ZF$ refutes $V=M$.
Ah, so I'd have to have one of stronger consistency strength than ZF. Maybe it's possible to have a stronger theory $\Gamma$ prove such an $M$ non-transitive, but not ZF, so $\Gamma$ refutes $V = M$, but ZF doesn't?
I edited it to take that into account. Maybe it's still a bad question.
Here's an honest question. What's the fascination with non-transitive models? Once you collapse them, you end up with a transitive model. So it's not like you're missing out some kind of information by not allowing transitive models.
One reason I suppose is that one might not want necessarily to assume the axiom of foundation, and without that you can't collapse the model. In the last of my questions, I was wondering if $L$ is the minimal model satisfying the self-relativization property without having to assume transitivity.
After the last comment: You want a self-relativizing class model not containing $L$ in $ZF$. If $ZF$ proves that $M$ does not contain $L$, then it refutes $V=M$, as in my previous comment. So there are no $ZF$-provable examples. If we drop the self-relativization condition, then we can build something like "the universe of sets transitively containing $\emptyset$ as an element". This is a class model (its collapse is just $V$) not containing $L$, but it is not self-relativizing.
I took that as the content of your previous comment (and I was aware the non-self-relativizing example). I still feel like I can't say, "$L$ is the minimal self-relativizing proper class model of ZF," since ZF doesn't seem to prove that;, rather, it can't disprove it. I haven't ruled out the case where an $M$ is provably non-transitive in some extension of ZF, but not in ZF. It would seem to be like concluding "there are no inaccessible cardinals" just because ZF can't prove that there are any.
Let $AC$ stand for the axiom of choice, let $L$ denote the constructible universe and let $L^*$ the universe of constructible sets transitively containing $\emptyset$ as an element.
Although $L^*$ is a proper subclass of $L$, it collapses to $L$, so it is isomorphic to $L$.
Let $\phi(x)$ be the formula
$(AC\rightarrow x\in L)\wedge(\neg AC\rightarrow x\in L^*)$.
I claim that the corresponding class $M$ is a model of $ZFC$ in $ZF$. In fact, $M$ is either $L$ or $L^*$, so it is a model of $ZFC$ in any case.
(This argument can be made precise: $ZF\vdash AC\rightarrow (\psi^M\leftrightarrow \psi^L)$ and $ZF\vdash \neg AC\rightarrow (\psi^M\leftrightarrow \psi^{L^*})$, for any $\psi$. Therefore $\psi^M$ holds for all $ZFC$ axioms).
Now, I claim that $M$ is self-relativizing. Indeed, $ZF\vdash AC^M$, so $ZF\vdash \phi(x)^M\leftrightarrow (x\in L)^M$. But $L^M$ is $M$ in both cases, because $M$ is either $L$ or $L^*$, and $L^*$ is isomorphic to $L$. Therefore, $ZF\vdash \phi(x)^M\leftrightarrow \phi(x)$.
Finally, $ZF$ cannot prove that $L\subseteq M$, and $ZF+\neg AC$ proves that $M$ is a proper sublass of $L$.
Beautiful! Thank you!
|
2025-03-21T14:48:31.889522
| 2020-08-23T03:30:50 |
369902
|
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|
Stack Exchange
|
On sup over boundaries of Sobolev functions
In Gilbarg-Trudinger's section on the maximum principle for weak solutions, the sup of a boundary of a Sobolev function defined as follows:
Let $\Omega$ be a bounded domain in $\mathbb{R}^n$. For $u, v \in H^1(\Omega)$, $u \leq v$ on $\partial \Omega $ if $(u-v)^+ \in H^1_0 (\Omega)$. Define
$$\sup_{\partial \Omega} u = \inf \; \{ k \in \mathbb{R} \; | \; u \leq k \text{ on } \partial \Omega \}, \quad u \in H^1(\Omega).$$
Here, of course, $H^1(\Omega) = W^{1,2}(\Omega)$ and $H^1_0(\Omega) = W^{1,2}_0(\Omega)$.
My simple question is whether the following inequalities are valid.
Let $u, v \in H^1(\Omega)$.
$\sup_{\partial \Omega} (u+v) \leq \sup_{\partial \Omega} u + \sup_{\partial \Omega} v.$
For $k \in \mathbb R$ with $\sup_{\partial \Omega} u \leq k$, $u \leq k$ on $\partial \Omega$.
If $u \leq v$ on $\partial \Omega$, then $\sup_{\partial \Omega} u \leq \sup_{\partial \Omega} v.$
I tried to prove these inequalities by the definition, but it does not seem straightforward. However, if we assume that the domain is at least Lipschitz, then we can use the trace operator. Considering this post on Math.SE, I can produce a quite straightforward proof for these inequalities since the sup over the boundary is the usual essential sup over the boundary.
I initially guessed that the inequalities can be showed even if we do not require any regularity on the boundary, but now I am not sure. Is there any proof without using the trace operator?
Thanks!
P.S I posted the same question on Math.SE. (link)
Too soon, too soon...
Yes, they are true without any assumption on $\Omega$ and follow from the ideal properties of $H^1$. First one shows that for $u, v \in H^1$, then $|u|, u^+, u\wedge v \in H^1$ and these maps (with $v$ fixed) are continuous from $H^1$ into itself. Then one shows that $0 \le u \le v$ with $v \in H^1_0$ yields $u \in H^1_0$ (if $\phi_n \to v$ then $\phi_n \wedge u \to v\wedge u=u$ and $\phi_n \wedge u$ has compact support).
@GiorgioMetafune Thanks for the comment. Actually, I thought that it boils down to the fact that $0 \leq u \leq v$ with $v \in H^1_0(\Omega)$ implies $u \in H^1_0(\Omega)$ for $u, v \in H^1 (\Omega)$. Is this a special property of $H^1$?
It holds in $W^{1,p}_0$ and the proof is sketched in my previous comment.
I was wondering if this "ordering" $u\le v$ on $\partial\Omega$ is transitive?
|
2025-03-21T14:48:31.889704
| 2020-08-23T04:26:26 |
369903
|
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|
Stack Exchange
|
Isomorphic Jacobians for different choices of basis of $1$-forms
In Otto Forster's Lectures on Riemann Surfaces on page 170 Jacobi Variety is defined in 21.6:
Suppose $X$ is a compact
Riemann surface of genus $g$ and $ \omega_1,..., \omega_g $ is a basis
of $\Omega (X) = H^0(X, \Omega_{X/\mathbb{C}})$ . Then
$$Jac(X):= \mathbb{C}^g/ \operatorname{Per}(\omega_1,..., \omega_g)$$
where $\operatorname{Per}(\omega_1,..., \omega_g)$ consists of all vectors
$$(\int_{\alpha} \omega_1, \int_{\alpha} \omega_2, ...
\int_{\alpha} \omega_g)$$
where $\alpha$ runs through the fundamental group $\pi(X)$ (p 168). Moreover
Theorem 21.4 proves that
$ \operatorname{Per}(\omega_1,..., \omega_g) \subset \mathbb{C}^g$
is a lattice.
The last sentence from the following I not understand:
Here we are considering $Jac(X)$ only as an
abelian group. It also has the structure of a compact complex manifold (a
complex $g$-dimensional torus). Note that the definition depends on the
choice of basis $ \omega_1,..., \omega_g $ but the choice of a
different basis leads to an isomorphic $Jac(X)$.
Therefore my question is why two different choices $ \omega_1,..., \omega_g $ and
$ \omega' _1,..., \omega' _g $ of the basis of $\Omega (X)$ gives
isomorphic Jacobi varieties? Regarded as abelian groups? Or what type of isomorphy Forster here impose? Depending on about which kind of isomorphy Forster talks about there should be different characterizations when $\mathbb{C}^g / \operatorname{Per}(\omega_1,..., \omega_g)$ and $\mathbb{C}^g / \operatorname{Per}(\omega' _1,..., \omega' _g)$
are isomorphic.
and depending on which type of isomorphy we require
(I think Forster here means by "$\cong$" iso only as abelian groups)
$\mathbb{C}^g / \operatorname{Per}(\omega_1,..., \omega_g)$ and
$\mathbb{C}^g / \operatorname{Per}(\omega' _1,..., \omega' _g)$
are isomorphic in category $\mathcal{C}$ iff there exist a
$M \in GL_{2g}(\mathbb{C})$ with
$M \operatorname{Per}(\omega_1,..., \omega_g)=
\operatorname{Per}(\omega' _1,..., \omega' _g)$ and
$M \gamma_i = \gamma' _i$. And the important issue is that depending
of which kind of isomorphy we require the matrix $M$ moreover
lives restrictionally in a subgroup $H(\mathcal{C}) \subset GL_{2g}(\mathbb{C})$.
I assume that Forster is talking about isomorphy as abelian groups.
Then, in which subgroup $H(\mathcal{C}) \subset GL_{2g}(\mathbb{C})$ the trafo
matrix $M$ should like? And why it exist?
Clearly, there exist a
$G \in GL_{2g}(\mathbb{C})$ with $M \omega_j = \omega' _j$. But
of course if $\operatorname{Per}(\omega' _1,..., \omega' _g)
= \mathbb{Z} \gamma' _1 + ... + \mathbb{Z} \gamma' _{2g}$ with
$\gamma' _j= G \gamma_j$ then
$\mathbb{C}^g / \operatorname{Per}(\omega_1,..., \omega_g)$ and
$\mathbb{C}^g / \operatorname{Per}(\omega' _1,..., \omega' _g)$
are not isomorphic as abelian groups.
The standard definition of the Jacobian is $\Omega (X)^/H_1(X,\mathbb{Z})$, where $H_1(X,\mathbb{Z})$ embeds into $\Omega (X)^$ by $\gamma \mapsto \int_{\gamma }$. This is of course independent of any choice, and choosing bases leads to the formulation you mention. For this kind of basic question please consider using MSE.
|
2025-03-21T14:48:31.889920
| 2020-08-23T07:40:23 |
369908
|
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|
Stack Exchange
|
Trace entropies
I'm studying relationships between trace entropy functionals and combinatorics and I'm faced with the following problem. Lets $\mathcal {D}$ be the following differential operator $1 -x\cdot \cfrac{d}{dx}$ i.e. $\mathcal {D} g = g - x\cdot g'$.
For $m\ge 0$ integer, if $\Phi_m(x) := x\cdot \log(x)^m$ then $\mathcal {D} \Phi_m(x) = -m\cdot \Phi_{m-1}(x)$ and, (at least formally) for a function $g$ we can write
$g(x) = \sum\limits_{m\ge 0} g_m \cdot \Phi_m(x)$ where $$g_m = \cfrac{(-1)^m}{m!}\cdot \mathcal {D}^m g(x) |_{x=1}\;.$$
I am trying to figure out what kind of functions $g(x)$ can satisfy the following conditions on $(0;1]$
I) $g(x) \ge 0$, $g(0)=0$. (Positivity)
II) $\mathcal {D}^2 g(x) \le 0$. (some sort of $\mathcal {D}$-concavity)
III) $\left(\mathcal {D}^2 - \mathcal {D}\right) g(x) \le 0$ (this is standard concavity expressed with the operator $\mathcal {D}$
IV) $\exists \ 0< \varepsilon \le 1$ such that $g(x) > -x^2\cdot \log(x) \quad \forall \ x \in (0;\varepsilon)$
V) $\exists \ a \in (0;\frac{1}{2}]$ such that $$g(a)+g(1-a) = -a\cdot \log(a) -(1-a)\cdot \log(1-a).$$
If we impose only conditions I),II) and III) there are a lot of functions satisfying them, but
adding IV) I'm not able to find any function except of the following form $g(x) = k\cdot x \cdot \log(x), \ k$ real constant (here $\varepsilon=1$). Note that $g$ do not satisfy V).
adding IV) and V) I'm not able to find any other function except the Boltzman-Gibbs-Shannon entropy trace : $-x\cdot \log(x)$
I "fear" that the Boltzman-Gibbs-Shannon entropy trace is the unique function satisfying I)-V).
Thanks in advance for any point of view.
When you spell out "$\exists\ 0< \varepsilon\le1$", you will get "there exists the double inequality $0< \varepsilon\le1$". In fact, this double inequality always exists, at least in some sense, and it may or may not be satisfied. Clearly, you wanted to say "$\exists\ \varepsilon\in(0,1]$" instead.
@IosifPinelis you've just created the tag [tag:differential-inequalities]; would you create a tag excerpt as well?
@YCor : I have done that -- my first tag excerpt. Do you have any comments on it?
Looks OK, thanks
For any $c\in(0,\log2]$, the function $g$ defined by the formula $g(x)=cx$ for $x\in[0,1]$ satisfies your conditions I)–V), but it is not the Boltzman–Gibbs–Shannon entropy trace.
There are many more functions $g$ satisfying your conditions I)–V) that are not the Boltzman–Gibbs–Shannon entropy trace. In particular, for any $c_1\in(0,\log2)$, any nonnegative continuous function $H$ on $[0,1]$, and any small enough real $c_2\ge0$ and $c_3\ge0$, the function $g$ defined by the formula
$$g(x)=c_1x-c_2 x\log x-c_3 x\int_0^x du\,H(u)\log\frac xu$$
for $x\in[0,1]$ satisfies your conditions I)–V), but it is not the
Boltzmann–Gibbs–Shannon entropy trace.
@YCor : Thank you for editing the m-dashes. How do you make them? (One n-dash remains, though, in place of an m-dash.)
You can produce them by copy-paste now. On my Mac I'm using shift + ⌥ ("option") + dash to get the medium dash –. Using only shift + dash makes a longer dash —.
Hi Iosif,
Thanks a lot for your detailed and quick answer. I have been very surprised that I didn't saw at least the solution $k\cdot x$. I realized that I did a little mistake in my conditions, the conditon IV) is $g(x) > -x\cdot log(x)$ (there is no square). The $k\cdot x$ is then not ok but the second set of functions work very well.
Thanks again
@Gianfranco : So, if you are satisfied with this answer, can we have it marked respectively? https://mathoverflow.net/help/someone-answers https://mathoverflow.net/help/accepted-answer
|
2025-03-21T14:48:31.890166
| 2020-08-23T08:06:31 |
369909
|
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|
Stack Exchange
|
Sums of reciprocals of primes in an arithmetic progression
Let $y>x\geq 1$, $p_0\geq x$. Consider $$S=\mathop{\sum_{x\leq p\leq y}}_{p\equiv a \mod p_0} \frac{1}{p}.$$By Brun-Titchmarsh and (basically) integration by parts, I seem to get that $$S \leq \frac{1}{x} + \frac{2}{p_0-1} \left(\log \log \frac{y}{p_0} + \frac{1}{\log \frac{y}{p_0}}\right) \leq \frac{1}{x} + \frac{2}{x-1} \left(\log \log \frac{y}{x} + \frac{1}{\log \frac{y}{x}}\right) .$$ moreover, the term $1/\log(y/x)$ can be eliminated for $y/x$ larger than a constant.
Bounds like this one should be standard. What is the strongest or cleanest result in the literature? Perhaps $S\leq \frac{1}{x} + \frac{2}{x} \log \log \frac{y}{x}$ for $x>2$ and $y/x\geq e^{e^{1/6}} = 3.258\dotsc$, say? Or $S\leq \frac{2}{x} \log \log y$ for $y\geq x>e^{e^{1/2}}=5.200\dotsc$?
(The naïve guess would be something like $S\leq \frac{1}{x} (1 + \log \frac{\log y}{\log x})$. A very cursory search in the literature on my part has only yielded Remark 1 in section 2 of Pomerance, On the distribution of amicable numbers, 1976.)
|
2025-03-21T14:48:31.890262
| 2020-08-23T08:32:34 |
369910
|
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|
Stack Exchange
|
Integer surgeries along links yielding lens spaces
Does there exist an integer $N$ such that any lens space $L(p,q)$ can be obtained by integer surgery from $S^3$ along a link $L$ with at most $N$ components?
EDIT:
I have worked out the comment by Ian Agol.
Fintushel and Stern have proved the following:
If $K_1$ is a knot in the lens space $L(p,q)$ such that $n_1$-surgery along $K_1$ yields a homology $3$-sphere, then we can represent $K_1$ in the exterior of an unknot $K_0$ in $S^3$ with surgery coefficient $-p/q$. By comparing the ranks of the first homologies we see that $\pm1=pn_1+q\operatorname{lk}_{01}^2$, where $\operatorname{lk}_{01}$ denotes the linking number of $K_0$ and $K_1$ in $S^3$.
It follows that $L(p,q)$ is homeomorphic to $L(p,\operatorname{lk}_{01}^2)$ and in particular an integer $N$ as in the question has to be larger than $1$.
As suggested by Ian Agol this approach does not give an obstruction for links:
Let $K_1\cup K_2$ be a $2$-component link in $L(p,q)$ such that $(n_1,n_2)$-surgery yields a homology $3$-sphere. As above we can compare the ranks of the first homologies and get
$$\pm1=p(n_1n_2+\operatorname{lk}_{12}^2)+q(n_1\operatorname{lk}_{02}^2+n_2\operatorname{lk}_{01}^2-2\operatorname{lk}_{01}\operatorname{lk}_{02}\operatorname{lk}_{12})$$
which seems to have a solution for any $(p,q)$ coprime.
I don't know of any restrictions. Josh Greene has classified the lens spaces obtained by surgery on a non-trivial knot (which must be integral). Proposition 1 of Fintushel-Stern gives a simpler homological necessary condition: $\pm q$ is a quadratic residue $(mod p)$. But I suspect that one won't get a simple condition like this for surgeries on links. https://link.springer.com/article/10.1007/BF01161380
You might try using results of Berge and Gabai to see what lens spaces you can get. They classify surgeries on knots in the solid torus that give back the solid torus. https://mathscinet.ams.org/mathscinet-getitem?mr=991095https://mathscinet.ams.org/mathscinet-getitem?mr=1093862
A braid has many embeddings into $S^3$ by doing full twists. Maybe one can find examples of lens spaces obtained by integral Dehn surgery on these links? But one would have to compute the framings to see which ones are integral.
There are a number of examples for which $N = 2$ suffices: all lens spaces that are rational surgeries on torus knots (since the surgery coefficient is integer±1/n; these were classified by Moser) and all lens spaces that bound rational homology balls (classified by Lisca; this is because each of them bounds a rational homology ball built with one 1-handle and one 2-handle; this was shown partly by Lisca, and partly by Baker, Buck, and Lecuona).
Actually I think that one can show $N = 2$ cannot work for all lens spaces, but not in a very nice way.
Thanks a lot for all the comments. But it would be great if you could give me a few more hints.
@Ian Agol: As far as I can tell the Berge and Gabai papers are only about knots. I am not sure how this can yield a link with a possibly new integer lens space surgery.
Or do you suggest classifying surgeries on links in the solid torus that give back the solid torus and then checking which of those surgeries are integer and which lens spaces one gets?
@Marco Golla: I understand your comment on the examples where $N=2$ suffices. But I do not see how to conclude that $N=2$ cannot work in general.
@MarcKegel: the second comment was not meant to follow from the first. To clarify, I think that one can play around with some specific (families of) examples and obstruct them from being an integral surgery on a 2-component link. Sorry for the mix-up.
@MarcKegel Yes, I meant the second thing you said. But this probably won’t get much more than what Marco suggests.
Got it. Thanks a lot.
Building on Ian's comment: another way to construct lens spaces for which $N=2$ works is to cable the Hopf link with Berge–Gabai patterns. For the right surgery coefficients (i.e. the ones that give solid tori back), after surgery you're just gluing together two solid tori, so you certainly get a lens space, but you need to check the slopes you obtain.
|
2025-03-21T14:48:31.890547
| 2020-08-23T09:05:46 |
369912
|
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|
Stack Exchange
|
Classification of posets that are quotient posets of the Boolean lattice
Quotient posets of the Boolean lattice $B_n$ have interesting properties and are for example discussed in chapter 5 of Stanley's book on algebraic combinatorics.
$B_n/G$ for a subgroup $G$ of the symmetric group $S_n$ (that acts naturally on $B_n$) is defined as the poset of orbits under the natural order (that is one orbit $a$ is $\geq$ another orbit $b$ if and only if there exists elements $a' \in a$ and $b' \in b$ such that $a' \geq b'$).
The posets $B_n /G$ are graded of rank $n$, rank-symmetric, rank-unimodal, and Sperner. See theorem 5.8. in the book of Stanley. One might ask whether one can add some more conditions to have a characterisation of them by natural properties (not mentioning the Boolean lattice in this characterisation).
Question: Is there a characterisation when a given bounded poset $P$ is isomorphic to $B_n /G$ for a subgroup $G$ of the symmetric group?
Can this at least be done when $G$ is assumed to be cyclic?
@EmilJeřábek I added the definition.
My initial reaction was "because any poset automorphism is a boolean algebra automorphism, won't this poset quotient be equally a boolean algebra quotient?". But of course this is wrong. For instance, quotienting $\mathbb 2 \times \mathbb 2$ by the the obvious involution, we obtain the ordinal $\mathbb 3$, which is not a boolean algebra.
There are numerous necessary conditions, but I doubt whether there is a useful necessary and sufficient condition. For instance, if $k<n/2$ then there must be a set of disjoint saturated chains going from every element of rank $k$ to every element of rank $n-k$. See http://math.mit.edu/~rstan/pubs/pubfiles/60.pdf. Another more trivial necessary condition is that the number of elements of rank $k$ cannot exceed ${n\choose k}$.
@RichardStanley Is the number of (isomorphism classes) posets known that appear as a quotient of $B_n$ in this way? The sequence seems to start for $n=2,..6$ with 2,3,8,11,34.
@Mare: this is not known to me and looks impossibly difficult.
What happens if we extend the set to also include quotients of the form $L_n(q)/G$ where $L_n(q)$ is the lattice of linear subspaces of $\mathbb F_q^n$ and $G$ is some subgroup of $GL_n(\mathbb F_q)$. Is there any chance this can make the classification easier?
A further property of $B_n/G$: if $n\geq 12$ and $B_n/G$ has exactly one element of rank 6, then $B_n/G$ is a chain. This is a deep result requiring the classification of finite simple groups. See https://math.stackexchange.com/questions/388488/subgroup-transitive-on-the-subset-with-same-cardinality.
|
2025-03-21T14:48:31.890874
| 2020-08-23T09:38:17 |
369913
|
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|
Stack Exchange
|
Groups that act transitively on $\mathrm{Gr}(k,\Bbb R^n)$ but not transitively on $\mathrm{Gr}(k+1,\Bbb R^n)$
Is it known for which $n, k\in\Bbb N$ there exists a matrix group $\Gamma\subseteq\mathrm{GL}(\Bbb R^n)$ that
acts transitively on $\mathrm{Gr}(k,n)$, i.e., on the $k$-dimensional subspaces of $\Bbb R^n$, but
acts not transitively on $\mathrm{Gr}(k+1,n)$, i.e., on the $(k+1)$-dimensional subspaces of $\Bbb R^n$?
For example, $\mathrm U(n)$ (acting on $\Bbb C^n\cong\Bbb R^{2n}$) acts transitively on $\mathrm{Gr}(1,\Bbb R^{2n})$ but not on $\mathrm{Gr}(2,\Bbb R^{2n})$.
One could first stick to closed subgroups, which boils down to closed connected subgroups. The answer could be quite short, using the classification of groups acting transitively on the 1-Grassmanian (as can be found in §6 in this paper "2-transitive Lie groups" by L. Kramer). The answer would maybe be: yes only for $k=0$, $n\ge 2$, never for $k\ge 2$, and for $k=1$ precisely for even $n\ge 4$ and $n=7$. But there are some details to check to confirm or fix this:
Namely it seems that for $n\ge 3$ a closed connected $G\subset GL_n(R)$ is transitive on the 2-Grassmannian iff contains a conjugate of $SO(n)$. This follows from classification, except from the following cases I haven't checked: $G_2$ in dim 7, $Spin(7)$ in dim 8, $Spin(9)$ and $Spin(9,1)$ in dimension 16. (If they don't act transitively on the 2-Grassmannian, I think my previous message is correct.)
Mm, there's another thing to check (beyond the few uncertainties above in dim 7,8,16), namely that being transitive on the $k$-Grassmannian for $1<k<n-1$ implies being transitive on the 1-Grassmannian. It already implies being irreducible, but that's not enough.
@YCor: Indeed, $\mathrm{G}_2\subset\mathrm{SO}(7)$ acts transitively on the $2$-Grassmannian $\mathrm{Gr}(2,\mathbb{R}^7)$ but not on the $3$-Grassmannian, while $\mathrm{Spin}(7)\subset\mathrm{SO}(8)$ acts transitively on the $3$-Grassmannian $\mathrm{Gr}(3,\mathbb{R}^7)$ but not on the $4$-Grassmannian. Meanwhile $\mathrm{Spin}(9,1)\subset\mathrm{SO}(16)$ does not act transitively on $\mathrm{Gr}(2,\mathbb{R}^{16})$ or any $\mathrm{Gr}(k,\mathbb{R}^{16})$ except for $k=0,1,15,16$.
This is only a partial answer, and it's based on YCor's comment about the groups that act transitively on spheres. What is missing, as YCor commented, is knowing that if $G\subset\mathrm{GL}(n,\mathbb{R})$ acts transitively on $\mathrm{Gr}(k,\mathbb{R}^n)$ for some $k$ with $1<k<n$, then it acts transitively on $\mathrm{Gr}(1,\mathbb{R}^n)$.
Here is an argument that, at least when $k = 2$ or $3$, this statement is true.
I will assume, as YCor does, that $G$ is closed in $\mathbb{GL}(n,\mathbb{R})$
and hence is a connected Lie group.
First, one might as well use the oriented Grassmanians $\mathrm{Gr}^+(k,\mathbb{R}^n)$ since $G$ acts transitively on $\mathrm{Gr}^+(k,\mathbb{R}^n)$ if and only if it acts transitively on $\mathrm{Gr}(k,\mathbb{R}^n)$. (After all, if $G$ acts transitively on $\mathrm{Gr}(k,\mathbb{R}^n)$, then its orbits in $\mathrm{Gr}^+(k,\mathbb{R}^n)$ are all open, and $\mathrm{Gr}^+(k,\mathbb{R}^n)$ is connected.)
Now let $E_k\to \mathrm{Gr}^+(k,\mathbb{R}^n)$ be the tautological oriented $k$-plane bundle, i.e., $$E_k = \{\ (e,v)\ |\ v\in e\in \mathrm{Gr}^+(k,\mathbb{R}^n)\ \}.$$
When $k=2$ and $n>2$, the oriented $2$-plane bundle $E_2$ has nonzero Euler class in $H^2(\mathrm{Gr}^+(k,\mathbb{R}^n),\mathbb{Z})\simeq \mathbb{Z}$. In particular, it is not the sum of two line bundles (which would be trivial).
Now suppose that $G$ acts transitively on $\mathrm{Gr}^+(2,\mathbb{R}^n)$ and let $H\subset G$ be the subgroup that fixes $e_0\in \mathrm{Gr}^+(2,\mathbb{R}^n)$. Then $H$ acts on the $2$-plane $e_0\subset\mathbb{R}^n$ and hence on $\mathrm{Gr}^+(1,e_0)\simeq S^1$.
If the orbits of $H$ on $\mathrm{Gr}^+(1,e_0)$ are open, then there is only one orbit, so that $H$ acts transitively on $\mathrm{Gr}^+(1,e_0)$. Since $e_0$ was arbitrary, it follows that $G$ acts transitively on $\mathrm{Gr}^+(1,\mathbb{R}^n)$, as desired. Meanwhile, if there were a non-open orbit, then fixing one such $H$-orbit $X_0\subset \mathrm{Gr}^+(1,e_0)\subset \mathrm{Gr}^+(1,\mathbb{R}^n)$ and looking at its $G$-orbit, i.e.,
$$
X = \{ ([g v], g(e_0)\ |\ [v]\in X_0, g\in G\ \}
$$
gives a bundle $X\to\mathrm{Gr}^+(2,\mathbb{R}^n)$ with discrete fibers over the simply-connected base $\mathrm{Gr}^+(2,\mathbb{R}^n)$. Hence it is topologically trivial, so that there exists a section $S\subset X$ that is the projectivization of a line bundle $L\subset E_2$, contradicting the fact that $E_2$ has no rank $1$-subbundles. Thus, this cannot occur, and the proof for $k=2$ is complete.
In the case $k=3$, one has to deal with the cases $n=4$ and $5$ separately. When $n=4$, the group $G = \mathrm{SU}(2)$ acts transitively on $\mathrm{Gr}^+(3,\mathbb{R}^4)\simeq S^3$, but does not act transitively on $\mathrm{Gr}^+(2,\mathbb{R}^4)$. When $n=5$, applying duality and the above argument, one sees that if $G$ acts transitively on $\mathrm{Gr}^+(3,\mathbb{R}^5)$, then it acts transitively on $\mathrm{Gr}^+(4,\mathbb{R}^5)$, but then, by the paper YCor cites, this says $G$ must contain a copy of $\mathrm{SO}(5)$ and hence acts transitively on $\mathrm{Gr}^+(k,\mathbb{R}^5)$ for all $k$.
Thus, we can assume that $n\ge 6$. In this case, we have that $H^p(\mathrm{Gr}^+(3,\mathbb{R}^n),\mathbb{Z})=0$ for $p = 1, 2, 3$, but that, not only is $H^4(\mathrm{Gr}^+(3,\mathbb{R}^n),\mathbb{R})\not=0$, we have that $p_1(E_3)\not=0$. By a characteristic class argument, it follows that the bundle $E_3\to\mathrm{Gr}^+(3,\mathbb{R}^n)$ does not split as the sum of a line bundle and a $2$-plane bundle.
Now, I claim that if $G\subset\mathrm{GL}(n,\mathbb{R})$ acts transitively on $\mathrm{Gr}^+(3,\mathbb{R}^n)$ then it acts transitively on $\mathrm{Gr}^+(2,\mathbb{R}^n)$. To see this, suppose that $G$ acts transitively on $\mathrm{Gr}^+(3,\mathbb{R}^n)$, and let $H\subset G$ be the stabilizer of $e_0\in \mathrm{Gr}^+(3,\mathbb{R}^n)$, and consider the action of $H$ on $e_0\simeq\mathbb{R}^3$.
First, suppose that $H$ has only open orbits in $\mathrm{Gr}^+(1,e_0)\simeq S^2$. Then $H$ acts transitively on $\mathrm{Gr}^+(1,e_0)$ and hence $G$ acts transitively on $\mathrm{Gr}^+(1,\mathbb{R}^n)$.
Second, suppose that $H$ has a 0-dimensional orbit $X_0\subset\mathrm{Gr}^+(1,e_0)$. Then, constructing the bundle $X\to\mathrm{Gr}^+(3,\mathbb{R}^n)$ with discrete fibers in the projectivization of $E_3$ as was done above for $E_2$, we see that $X$ has a section and that this can be used to construct a splitting of $E_3$ as the sum of a line bundle and a $2$-plane bundle, which is known to be impossible.
Finally, suppose that all of the orbits of $H$ on $\mathrm{Gr}^+(1,e_0)$ are either $1$- or $2$-dimensional. Not all of the orbits can be $1$-dimensional because $S^2$ has no foliation, and there cannot be more than a finite number of components of the union of the $1$-dimensional orbits (and there must be at least 1 component, otherwise we would be in the first case already dealt with). Thus, there are only a finite number of components of the union of the $2$-dimensional orbits. Each such component, which is homogeneous under a connected, finite dimensional Lie group, must have Euler characteristic either $0$ or $1$. Thus, there must exist exactly two components that have Euler characteristic 1, i.e., are contractible disks, and these must map to a single component $D_0$ in $\mathrm{Gr}(1,e_0)$. Using the contractibility of this component, we can now construct a rank 1 subbundle $L$ of $E_3\to \mathrm{Gr}^+(3,\mathbb{R}^n)$ with the property that the projectivization of $L_e$ lands in the $e$-fiber of the $G$-orbit of the contractible orbit $D_0$. Again, this is impossible because $E_3$ does not split nontrivially as a sum.
Remark: As far as I know, the only time that $G$ acts transitively on $\mathrm{Gr}(k,\mathbb{R}^n)$ with $4\le k\le n/2$ is when $G$ contains a copy of $\mathrm{SO}(n)$. When $k=3$, there is $\mathrm{Spin}(7)\subset\mathrm{SO}(8)$
and, when $k=2$, we have $\mathrm{G}_2\subset\mathrm{SO}(7)$ and $\mathrm{Spin}(7)\subset\mathrm{SO}(8)$. Neither $\mathrm{Spin}(9)$ nor $\mathrm{Spin}(9,1)$
in their $16$-dimensional representations act transitively on any $\mathrm{Gr}^+(k,\mathbb{R}^{16})$ except when $k=0,1,15,16$.
I believe that one could continue the above line of argument, showing that, if $G$ acts transitively on $\mathrm{Gr}^+(4,\mathbb{R}^n)$ for $n\ge 8$, then it must act transitively on $\mathrm{Gr}^+(3,\mathbb{R}^n)$, and so on. However, I think that this argument will get more complicated as $k\le n/2$ increases, and it's probably not the right approach for general $k$.
|
2025-03-21T14:48:31.891330
| 2020-08-23T09:50:40 |
369914
|
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"url": "https://mathoverflow.net/questions/369914"
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|
Stack Exchange
|
Norm relation with induction
Let $\rho: H\to \text{GL}_n(\mathbb{C})$ be a representation, where $H$ is a subgroup of $\text{GL}_2(\mathbb{C}).$ Let $G$ be another subgroup of $\text{GL}_2(\mathbb{C})$ such that $H$ has finite index in $G$. For any element $A\in \text{GL}_n(\mathbb{C})$ we define its norm $||A||$ to be maximum modulus of its entries. Denote $\tilde{\rho}=\text{Ind}_{H}^{G}\rho$. I can see that $||\rho(\gamma)||\leq ||\tilde{\rho}(\gamma)||$ for any $\gamma \in H$. However, I want a result something like $ ||\tilde{\rho}(\gamma)|| \ll ||\rho(\gamma)||,$ where the absolute constant depends only on $G$ or $H,$ but not on $\gamma.$
Note that, $$||\tilde{\rho}(\gamma)||=\max_{g_i's~\mathrm{are~coset~representatives~of}~G/H}\{||\rho(g_i\gamma g_j^{-1})||]\}.$$ For such fixed $g_i,g_j$ is it possible to say $||\rho(g_i\gamma g_j^{-1})||\ll ||\rho(\gamma))||$ ?
|
2025-03-21T14:48:31.891405
| 2020-08-23T10:22:29 |
369915
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/369915"
}
|
Stack Exchange
|
Weakening s-unitality
Recall that a (non-unital, non-commutative) ring $R$ is left s-unital if for every $r\in R$, we have $r\in Rr$.
Consider the following conditions:
There is a nonzero integer $m$ such that for all $r\in R$, $mr\in Rr$.
For each $r\in R$, there is a nonzero integer $m_r$ such that $m_rr\in Rr$.
Question: Do these properties have established names? Have they been studied?
Some observations:
For example, $2\mathbf Z$ has the former property, while $\bigoplus_n n\mathbf Z$ has the latter, but not the former.
For right Noetherian rings, the two are equivalent.
The first condition is satisfied by all rings of finite characteristic, the second one is satisfied by all $\mathbf Z$-torsion rings.
I suppose you could naturally generalise it to modules: if $R,S$ are rings and $M$ is simultaneously a left $R$-module and a left $S$-module, then the corresponding properties would be:
There is a regular/nonzero $s\in S$ such that for all $v\in M$ we have $sv\in Rv$.
For each $v\in M$ there is a regular/nonzero $s\in S$ such that $sv\in Rv$.
|
2025-03-21T14:48:31.891502
| 2020-08-23T10:40:19 |
369917
|
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"LSpice",
"MSMalekan",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/84700"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/369917"
}
|
Stack Exchange
|
About Tarski number 7
Recall that a group $G$ admits a $(m+n)$-paradoxical decomposition if there exist positive integers $m$ and $n$, a partition $\{P_1,\dotsc,P_m,Q_1,\dotsc,Q_n\}$ of $G$ and elements $x_i, y_j$ of $G$ such that
$$ G=\bigcup_{i=1}^m x_iP_i=\bigcup_{j=1}^ny_jQ_i$$
The minimal possible value of $m + n$ in a paradoxical decomposition of $G$ is the Tarski number of $G$.
Let the Tarski number of a group be 7. Does this group admit (2+5)-paradoxical decomposition, (3+4)-paradoxical decomposition or both?
There are obviously infinitely many questions of this sort one could ask, although many of them are surely naïve. Why are the cases $7 = 2 + 5 = 3 + 4$ particularly interesting?
@LSpice That is because I have some ideas to construct a group which have a $(m+n)$-paradoxical decomposition.
@LSpice Also, the question of the existence of a group with tarski number 7 is open yet.
|
2025-03-21T14:48:31.891588
| 2020-08-23T11:39:19 |
369918
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Mirco A. Mannucci",
"Qfwfq",
"https://mathoverflow.net/users/15293",
"https://mathoverflow.net/users/41291",
"https://mathoverflow.net/users/4721",
"მამუკა ჯიბლაძე"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/369918"
}
|
Stack Exchange
|
Quantum surreal numbers
Toward Quantum Combinatorial Games presents the definition of a "quantum game", allowing a superposition of moves rather than a single classical move. This leaves me wondering: Since surreal numbers are a certain class of games, is there a similar class of "quantum surreal numbers" formed from quantum games, but richer than the surreals? What properties or structure do they have?
surreal numbers are special cases of combinatorial games, so if you can define quantum combinatorial games (I assume as superpositions of classical games) you also know what a quantum surreal is: the superposition of two games which happen to be surreal numbers
@MircoA.Mannucci Might be more invloved than that: "numberness" of a quantum game might be a quantity (such as probability, say) rather than a property.
Mamuka, I see your point: surreal numbers are not simply some special games (essentially the silly ones) but serve as "numbers" to gauge them. Ok, things might be more involved, but why not trying first the simple minded approach I mentioned? Conjecture: my simple quantum surreal work well when a quantum game is not entangled, else you need some more sophisticated weapon...
@Mirco: are we superposing games or moves (or both)?
@Qfwfq in the above I was simply generalizing to simple-minded superposition of games, but why stopping there? The interesting beasts are games where the moves are superpositions of classical moves. PS Conway would have loved that...
|
2025-03-21T14:48:31.891718
| 2020-08-23T12:30:50 |
369919
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Francois Ziegler",
"Gerry Myerson",
"asrxiiviii",
"https://mathoverflow.net/users/157984",
"https://mathoverflow.net/users/158000",
"https://mathoverflow.net/users/19276",
"https://mathoverflow.net/users/41291",
"მამუკა ჯიბლაძე"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/369919"
}
|
Stack Exchange
|
Looking for a paper by Vinberg
I have been reading about the Schinzel-Zassenhauss conjecture, and have been looking for the following reference:
Vinberg E.: On some number-theoretic conjectures of V. Arnold, Japanese J. Math., vol 2 (2007)
Unfortunately, despite extensive search, I haven't found the above paper I would really appreciate if someone could provide the paper or a link to the same. I would also like to ask if there is a website where I could find the articles of the Japanese Journal of Mathematics.
https://doi.org/10.1007/s11537-007-0705-4
Interlibrary loan?
@Gerry Myerson Do I need to get one to get access to the paper? How do I do so? Thanks. (Sorry I am not in a financial position to pay the amount mentioned in that link.)
@asrxiiviii Are you sure it is behind paywall? I have no subscription but can access it simply by clicking: https://link.springer.com/content/pdf/10.1007/s11537-007-0705-4.pdf
მამუკა ჯიბლაძე Hi! Thank you for your answer. Through both the links above, I am getting redirected to the link (https://link.springer.com/article/10.1007%2Fs11537-007-0705-4) which a Springer page containing the title, abstract and reference list of the paper, with two Access Options: 34,95 € and 63,02 € for article PDF and journal subscription respectively. Sorry, I am a little new to that site: should I sign in and/or go through some other formalities before getting access? Thank you.
If you are near a library, asr, you go up to a librarian and ask whether the library can get that paper for you. University libraries do this all the time for their faculty and students, other libraries may do it, too, and in my experience there's no charge for the service. Worth a try!
@Gerry Myerson Thank you so much. Unfortunately, I do not have any library nearby and my institute is shut down for the pandemic. Although most such online library-facilities are available for PhDs, postdocs and faculty of my institute (and I am an undergraduate student), I will try to contact those in charge of administration of my library to request for this. Fingers crossed!
|
2025-03-21T14:48:31.891902
| 2020-08-23T13:45:24 |
369922
|
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|
Stack Exchange
|
K-means clustering benchmarks
What benchmarks do you use for evaluating clustering algorithms, especially for evaluating the performance of K-means vs. another algorithm?
I am especially interested in looking at the correctness of results, meaning that I am looking for clustering problems that have a pretty good chance of K-means failing to find the optimum clustering.
I should also say that I am also looking for problems in a $n$-dimensional isotropic Euclidean space.
I have found some discussions in published literature, but I wanted to hear about as many perspectives as possible. If you had an algorithm and you wanted to compare it to K-means in terms of its ability to avoid local minima, you would need to test it with a reasonable probability that K-means would do just that, get stuck in a local minima. What kind of a test case would you suggest to give that reasonable probability?
This post ends mid-sentence.
I have found some discussions in published literature, but I wanted to hear about as many perspectives as possible.
If you had an algorithm and you wanted to compare it to K-Means in terms of its ability to avoid local minima, you would need to test it with a reasonable probability that K-Means would do just that, get stuck in a local minima. What kind of a test case would you suggest to give that reasonable probability?
This question may get closed - it has 4 out of 5 votes at the time of this comment. The underlying reason, I think, is that it is more of an empirical question than a mathematical one: this forum is about mathematical research, and the goal of mathematical research is to prove interesting theorems rather than hit empirical benchmarks. I would recommend either posting your question at the stats or datascience stackexchange sites instead. There might be a formulation of the question which belongs to pure probability theory - that might be appropriate for this forum.
Here is one source:
P. Fränti and S. Sieranoja.
$K$-means properties on six clustering benchmark datasets.
Applied Intelligence, 48 (12), 4743-4759, December 2018.
DOI.
Web link.
|
2025-03-21T14:48:31.892049
| 2020-08-23T13:50:42 |
369924
|
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"authors": [
"Derek Holt",
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"LSpice",
"YCor",
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|
Stack Exchange
|
Problem while multiplying under a set of relators
I have defined $S_4$ (Symmetric group of order 4), and with the base field $Z_5$, groupring $Z_5S_4$ is constructed. Then I have taken two elements of this group ring and I want to multiply them to get the simplest result.
gap> f := FreeGroup( "a", "b","c" );;
gap> G := f / [ f.1^2, f.2^3,f.3^4, f.1*f.2*f.3 ];
<fp group on the generators [ a, b, c ]>
gap> AsList(G);
[ <identity ...>, a, b, b^-1, c, c^-1, a*b^-1, a*c, b*a, b*c^-1, b^-1*c^-1, c*b^-1, c^2,
c^-1*a, a*b^-1*c^-1, a*c*b^-1, b*a*b^-1, b*a*c, b^-1*c^-1*a, c*b^-1*c^-1, c^-1*a*c,
a*b^-1*c^-1*a, b*a*b^-1*c^-1, b^-1*c^-1*a*c ]
gap> a:=G.1;;
gap> a^2;
a^2
gap> a^2=One(G);
true
Question
Why I am not getting simplified form of the group elements? E.g. a^2 is the identity, but is displayed as a^2.
Please help the reader from the beginning by explaining what you're doing, instead of copying your code and just finish with a short sentence.
Please also be sure to post code as code; this was initially almost unreadable (even, presumably, to someone who knows GAP syntax as I don't). I have edited accordingly. This also looks much more like a technical GAP question than a math question ….
@YCor I am sorry for this. Do I need to explain every syntax ?
No, provide some guidelines about what you're doing
@YCor I have defined $S_4$ (Symmetric group of order $4$), and with the base field $Z_{5}$, groupring $Z_5S_4$ is constructed. Then I have taken two elements of this group ring and I want to multiply them to get the simplest result.
Please edit your post rather than comment?
The group ring is not relevant to this question. The question relates only to the finitely presented group $G$, so there was no need to mention the group ting.
Algorithms for finitely presented groups are hard -- generically problems, such as testing whether a word represents the identity (or finding a shortest word expression) do not have (they cannot exist as they are equivalent to the Halteproblem for Turing machines) general algorithmic solutions.
Therefore GAP will by default not reduce word expressions in finitely presented groups.
If you know that your group G is finite and small, you can force a reduction by issuing the following two commands immediately after creating the group:
Size(G);
SetReducedMultiplication(G);
Then, when the first product of generators is formed, GAP will calculate a confluent rewriting system using a Knuth-Bendix algorithm, and use it to bring elements in a lenlex-minimal representation. In your example you then will get:
gap> List(G);
[ <identity ...>, a, b, b^-1, c, c^-1, c^-1*b, a*c, b*a, b*c^-1, c*b, c*b^-1,
c^2, c^-1*a, a*c*b, b*c^-1*a, b*c^-1*b, b*a*c, c*b*a, c^2*b, a^c, a*c*b*a, b*a*c*b, c^-1*a*c*b ]
gap> a:=G.1;;
gap> a^2;
<identity ...>
This however costs time and memory (and for large groups might not be able to succeed), and is not always necessary, thus it is not done by default.
|
2025-03-21T14:48:31.892246
| 2020-08-23T14:55:44 |
369928
|
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|
Stack Exchange
|
Set of null-sequences is not $\sigma$-compact
I am interested in a reference for the following fact (or a similar result).
PROPOSITION. Let $X$ denote the set of real null sequences; i.e., the set of $(a_n)_{n=0}^{\infty}$ with $a_n\to 0$, with the usual (product) topology. Then $X$ is not $\sigma$-compact.
By definition, $X$ is $F_{\sigma \delta}$ in the space of all real sequences. Since $X$ is a subset of the space of all bounded real sequences, which is $\sigma$-compact, the claim is equivalent to saying that $X$ is not $F_{\sigma}$. The proof of the proposition is not difficult (see below). I am using a similar proof in my recent preprint that shows that the escaping set of a transcendental entire function is never $\sigma$-compact. It would therefore be useful to know of a reference for the above, which feels as though it should be classical.
Proof of the Proposition. If $\underline{a} = (a_n)_{n=0}^{\infty}\in X$ is a null sequence, and $\varepsilon>0$, set
$$ n_{\varepsilon}(\underline{a}) := \min\{n\geq 0\colon |a_n| < \varepsilon\} < \infty.$$
Clearly for all sequences $(\varepsilon_k)_{k=0}^{\infty}$ and $(N_k)_{k=0}^{\infty}$ with $\varepsilon_k\to 0$, there is $\underline{a}\in X$ such that
$$ n_{\varepsilon_k}(\underline{a}) > N_k$$
for all $k\geq 0$. (This is just saying that there are sequences that tend to zero arbitrarily slowly.)
If $A\subset X$ is compact, then
$$ n_{\varepsilon}(A) := \max_{\underline{a}\in A} n_{\varepsilon}(\underline{a}) < \infty.$$
Let $(A_k)_{k=0}^{\infty}$ be a sequence of compact subsets of $X$. Let
$\underline{a}\in X$ be such that $$ n_{1/k}(\underline{a}) > n_{1/k}(A_k)$$
for all $k\geq 0$. Then $\underline{a}\notin A_k$ for all $k$, and hence
$$ X \neq \bigcup_{k=0}^{\infty} A_k,$$
as claimed.
This is in Exercise 23.11 in the textbook of Kechris (and follows from the $\mathbf \Pi^0_3$-completeness of this space).
The topological (infinite-dimensional) structure of this space is described in Corollary 5.5.16 of van Mill's book "The infinite-dimensional topology of function spaces".
The space is not even $\sigma$-complete by Exercise 23.11.
Thank you! I'll have to get hold of a copy of that book.
|
2025-03-21T14:48:31.892402
| 2020-08-23T14:59:32 |
369929
|
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|
Stack Exchange
|
Generalized convexity
Let $X$ be a vector space. The positive-homogeneous function $\|\cdot\|$ is said to be a quasinorm
if $\|x+y\|\le K(\|x\|+\|y\|)$, for some $K\ge1$; it is a norm if $K=1$.
Question: 1. (terminology) if we drop the positive-homogeneity assumption and only assume that $f:X\to\mathbb{R}$ is continuous and satisfies $f(\frac12x+\frac12y)\le \frac{K}{2}[f(x)+f(y)]$, one might expect, by analogy, that this property be called "quasiconvexity" but it is not; the latter term is reserved for $f(\frac12x+\frac12y)\le \max\{f(x),f(y)\}$. Q: So what is the accepted term for this property?
2. What is known about maximizing a quasinorm (or, more generally, a "weakly convex" function $f$ in the sense of Q1) over a compact set -- say, in a finite-dimensional space? Are the maxima necessarily achieved at extreme points?
If you are one of those to whom the typographical difference between $||x||$ and $|x|$ is not conspicuous, look at the difference between $||x|| ||y||$ (coded as ||x|| ||y||) and $|x||y|$ (coded as \|x\|\|y\|). I edited accordingly. $\qquad$
Thanks @MichaelHardy -- looks nicer! Any pointers on either of the questions? :)
If I am not mistaken, the condition $f(\frac12x + \frac12y)\le \frac12 (f(x)+f(y))$ is sometimes called midpoint convexity.
Yes, and it's equivalent to classic convexity for continuous functions.
|
2025-03-21T14:48:31.892557
| 2020-08-23T15:18:51 |
369930
|
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|
Stack Exchange
|
Why is uncomputability of the spectral decomposition not a problem?
Below, we compute with exact real numbers using a realistic / conservative model of computability like Type Two Effectivity.
Assume that there is an algorithm that, given a symmetric real matrix $M$, finds an eigenvector $v$ of $M$ of unit length.
Let
$$M(\epsilon) = \begin{cases}
\left[\begin{matrix}1 & \epsilon\\ \epsilon & 1\end{matrix}\right]
,& \epsilon \geq 0 \\
\left[\begin{matrix}1 - \epsilon & 0\\0 & 1 + \epsilon\end{matrix}\right]
,& \epsilon \leq 0
\end{cases}$$
and assume that it's possible to find an eigenvector $v$ of $M(\epsilon)$.
If $\epsilon > 0$ then $v$ must necessarily be $\pm \frac 1 {\sqrt 2}\left[\begin{matrix}1\\1\end{matrix}\right]$ or $\pm \frac 1 {\sqrt 2}\left[\begin{matrix}-1\\1\end{matrix}\right]$. Observe that in all four cases, the $L^1$ norm of $v$ is $\sqrt 2$.
If $\epsilon < 0$, then $v$ must necessarily be $\pm\left[\begin{matrix}1\\0\end{matrix}\right]$ or $\pm\left[\begin{matrix}0\\1\end{matrix}\right]$. Observe that in all four cases, the $L^1$ norm of $v$ is $1$.
It's easily determinable whether the $L^1$ norm of $v$ is less than $\sqrt 2$ or greater than $1$. Therefore we can decide whether $\epsilon \leq 0$ or $\epsilon \geq 0$, which is impossible!
In a way, this is strange, because many sources say that the Singular Value Decomposition (SVD) and Schur Decomposition (which are generalisations of the Spectral Decomposition) are numerically stable. They're also widely used in numerical applications. But I've just tested the examples above for small $\epsilon$ using SciPy and got incorrect results.
So my question is, how do numerical analysts get around this problem? Or why is this apparently not a problem?
I could venture some guesses: While finding eigenvectors of general matrices may be impossible, it is possible to find their eigenvalues. Also, it's possible to "shift" a problematic matrix by some small $\epsilon$ so that its eigendecomposition is computable.
I feel that the first sentence is a bit vague: "Assume that there is an algorthim that..." but you don't say in which computation model you place the question. This seems to me of paramount importance for the rest of the question (BSS will give you a different answer compared to the case where $\epsilon$ must be written down in a real-world computer memory)(also the 0 test is computable in floating-point arithmetic in real-world computers).
@LoïcTeyssier I assume a realistic model of computation like TTE. BSS is not realistic in my view
@LoïcTeyssier See edit
@MattF. See edit
Your matrix function $M : \epsilon\mapsto M(\epsilon)\in M_{2\times2}(\mathbb{R})$ is itself not computable, or is it ?
@LoïcTeyssier It is. It's a continuous function
Not every continuous is computable, I guess
@LoïcTeyssier $M(\epsilon) = \left[\begin{matrix}\min\left(0, \epsilon\right) + 1 & \max\left(0, \epsilon\right)\\max\left(0, \epsilon\right) & 1 - \min\left(0, \epsilon\right)\end{matrix}\right]
$, which is clearly computable
I stand corrected ;)
@MattF. What do you mean by it doesn't help?
@MattF. I don't want to get into a long discussion about this, but BSS is an unrealistic model of computation with real numbers, and TTE is easily implemented physically. And I'm not the first to say that one is more realistic than the other. Your analogy is flawed. I did not say that one was better, I said that one was more realistic
I know nothing about computability theory, but I don't understand why determining if $\epsilon \geq 0$ gives you some kind of contradiction if, as you assert, $\min(0,\epsilon)$ is computable. What am missing?
@JasonDeVito If $x$ and $y$ are approximately equal, then they are both approximately equal to $\min(x, y)$. Thus one can get a good approximation of $\min(x, y)$ from good approximations of $x$ and $y$, even if the approximations aren't good enough to be certain which is actually smaller.
@lambda: That makes sense! Thanks
Can you make this a bit more self-contained by saying what "TTE" is?
@MichaelHardy I've expanded the acronym. The standard reference is Computable Analysis by Weihrauch. I might include a link to that. [UPDATE] Done.
@wlad : Thank you.
The singular value decomposition, when applied to a real symmetric matrix $A = \sum_i \lambda_i(A) u_i(A) u_i(A)^T$, computes a stable mathematical object (spectral measure $\mu_A = \sum_i \delta_{\lambda_i(A)} u_i(A) u_i(A)^T$, which is a projection-valued measure) using a partially unstable coordinate system (the eigenvalues $\lambda_i(A)$ and eigenvectors $u_i(A)$; the eigenvalues are stable, but the eigenvectors are not). The numerical instability of the latter reflects the coordinate singularities of this coordinate system, but does not contradict the stability of the former. But in numerical computations we have to use the latter rather than the former, because standard computer languages have built-in data representations for numbers and vectors, but usually do not have built-in data representations for projection-valued measures.
An analogy is with floating-point arithmetic. The operation of multiplication of two floating point numbers (expressed in binary $x = \sum_i a_i(x) 2^{-i}$ or decimal $x = \sum_i b_i(x) 10^{-i}$) is a stable (i.e., continuous) operation on the abstract real numbers ${\bf R}$, but when viewed in a binary or decimal representation system becomes "uncomputable". For instance, the square of $1.414213\dots$ could be either $1.99999\dots$ or $2.0000\dots$, depending on exactly what is going on in the $\dots$; hence questions such as "what is the first digit of the square of $1.414213\dots$" are uncomputable. But this is an artefact of the numeral representation system used and is not an indicator of any lack of stability or computability for any actual computational
problem that involves the abstract real numbers (rather than an artificial problem that is sensitive to the choice of numeral representation used). In contrast, floating point division when the denominator is near zero is a true singularity; regardless of what numeral system one uses, this operation is genuinely discontinuous (in a dramatic fashion) on the abstract reals and generates actual instabilities that cannot be explained away as mere coordinate singularity artefacts.
Returning back to matrices, whereas the individual eigenvectors $u_i(A)$ of a real symmetric matrix $A$ are not uniquely defined (there is a choice of sign for $u_i(A)$, even when there are no repeated eigenvalues) or continuously dependent on $A$, the spectral measure $\mu_A := \sum_i \delta_{\lambda_i(A)} u_i(A) u_i(A)^T$ is unambiguous; it is the unique projection-valued measure for which one has the functional calculus
$$ f(A) = \int_{\bf R} f(E)\ d\mu_A(E)$$
for any polynomial $f$ (or indeed for any continuous function $f \colon {\bf R} \to {\bf R}$). The spectral measure $\mu_A$ depends continuously on $A$ in the vague topology; indeed one has the inequality
$$ \| f(A) - f(B) \|_F \leq \|f\|_\text{Lip} \|A-B\|_F$$
for any real symmetric $A,B$ and any Lipschitz $f$, where $\|\|_F$ denotes the Frobenius norm (also known as the Hilbert-Schmidt norm or 2-Schatten norm). This allows for the possibility for stable computation of this measure, and indeed standard algorithms such as tridiagonalisation methods using (for instance) the QR factorisation and Householder reflections do allow one to compute this measure in a numerically stable fashion (e.g., small roundoff errors only lead to small variations in any test $\int_{\bf R} f(E)\ d\mu_A(E)$ of the spectral measure $\mu_A$ against a given test function $f$), although actually demonstrating this stability rigorously for a given numerical SVD algorithm does require a non-trivial amount of effort.
The practical upshot of this is that if one uses a numerically stable SVD algorithm to compute a quantity that can be expressed as a numerically stable function of the spectral measure (e.g., the inverse $A^{-1}$, assuming that the spectrum is bounded away from zero), then the computation will be stable, despite the fact that the representation of this spectral measure in eigenvalue/eigenvector form may contain coordinate instabilities. In examples involving eigenvalue collision such as the one you provided in your post, the eigenvectors can change dramatically (while the eigenvalues remains stable), but when the time comes to apply the SVD to compute a stable quantity such as the inverse $A^{-1}$, these dramatic changes "miraculously" cancel each other out and the algorithm becomes numerically stable again. (This is analogous to how a stable floating point arithmetic computation (avoiding division by very small denominators) applied to an input $x = 1.99999\dots$ and an input $x' = 2.00000\dots$ will lead to outcomes that are very close to each other (as abstract real numbers), even though all the digits in the representations of $x$ and $x'$ are completely different; the changes in digits "cancel each other out" at the end of the day.)
[The situation is a bit more interesting when applying the SVD to a non-symmetric matrix $A = \sum_i \sigma_i(A) u_i(A) v_i(A)^T$. Now one gets two spectral measures, $\mu_{(A^* A)^{1/2}} = \sum_i \delta_{\sigma_i(A)} v_i(A) v_i(A)^T$ and $\mu_{(AA^*)^{1/2}} = \sum_i \delta_{\sigma_i(A)} u_i(A) u_i(A)^T$ which are numerically stable, but these don't capture the full strength of the SVD (for instance, they are not sufficient for computing $A^{-1}$). The non-projection-valued spectral measure $\mu_A = \sum_i \delta_{\sigma_i(A)} u_i(A) v_i(A)^T$ does capture the full SVD in this case, but is only stable using the vague topology on the open half-line $(0,+\infty)$, that is to say $\int_0^\infty f(E)\ d\mu_A(E)$ varies continuously with $A$ as long as $f$ is a test function compactly supported in $(0,+\infty)$, but is unstable if tested by functions that do not vanish at the origin. This is ultimately due to a genuine singularity in the polar decomposition of a non-selfadjoint matrix when the matrix becomes singular, which in one dimension is simply the familiar singularity in the polar decomposition of a complex number near the origin.]
Are you sure about the Lipschitz estimate in the operator norm? On $B(\ell^2)$ already the absolute value map is not Lipschitz continuous on the self-adjoint part, so I would expect the constant in this inequality to blow up as the size of the matrices increases. It is fine for the Hilbert-Schmidt norm, though.
Oops, you are right, thanks!
So the purpose of the spectral decomposition is to extend a continuous function $f:\mathbb R \to \mathbb R$ to matrices, by using the identity $f(U\Sigma U^) = U f(\Sigma) U^$. Have I understood that correctly?
In the case of the SVD, the integral $\int_0^\infty f(E)\ d\mu_A(E)$ where $\mu_A = \sum_i \delta_{\sigma_i(A)} u_i(A) v_i(A)^T$ doesn't seem to correspond to any such identity
I would just like to remark to this very nice answer that there is no a priori reason why we could not directly implement the spectral measure on a computer, nor is there any reason, other than historic accident, that the floating-point format uses numerically unstable representation of the reals. Signed binary (with digits $-1, 0, 1$) is numerically stable.
@AndrejBauer I found this page about signed digit representation where it mentions Booth representation. Is that what you're referring to?
@LoïcTeyssier: yes. Although the most efficient exact real arithmetic implementations tend to use sequences of (nested) intervals to represent reals, not streams of signed digits, see for example iRRAM
Perhaps it is helpful to think about some of the standard things you would want to do with an SVD decomposition from this perspective. For example, you often replace a matrix $\sum \lambda_i u_i v_i^T$ by a low rank approximation $\sum_{\lambda_i > \kappa} \lambda_i u_i v_i^T$ for some cut off $\kappa$. This is integrating the spectral measure against $x \chi_{\mu}(x)$ where $\chi_{\mu}$ is the characteristic function of $(\kappa, \infty)$. (continued)
The characteristic function is discontinuous at $\kappa$ but, if we replaced it by some function which was $0$ on $[0, \kappa-\epsilon]$, $1$ on $[\kappa+\epsilon,\infty)$ and continuous between, then this would fit into the framework of this answer. Similarly, if we are doing dimensional reduction by PCA, then we use the projection operator $\sum_{\lambda>\kappa} u_i u_i^T$ and again, if we fix the discontinuity in the characteristic function, we are in the framework of this answer.
That said, it seems to me that we could construct situations where this would cause a problem. For example, spectral clustering algorithms https://en.wikipedia.org/wiki/Spectral_clustering often compute the largest eigenvector. The largest eigenvector (or the projection onto its span) is a discontinuous function of $A$ near matrices with repeated large eigenvalue. Is this a problem in practice? I don't know.
@ogogmad Yes, in the (essentially) self-adjoint case the spectral measure is precisely the Riesz representation of the functional calculus. For non-selfadjoint $A$ one has $\int_0^\infty f(E)\ d\mu_A(E) = A g((A^A)^{1/2}) = g((AA^)^{1/2}) A$ whenever $f(E) = E g(E)$ for a continuous $g$ (this occurs in particular when $f$ is an odd polynomial). With a left-polar decomposition $A = UP$ or right-polar decomposition $A=QV$ of a non-singular $A$ one has $\int_0^\infty f(E)\ d\mu_A(E) = U f(P) = f(Q) V$ for any continuous $f$.
@DavidESpeyer If the spectrum is sufficiently discrete then instead of selecting a cutoff $\kappa$ in advance, one can first scan the relevant portion of the spectrum for a decent spectral gap (or appeal to the pigeonhole principle for the abstract existence of such a gap), and then place the spectral cutoff in this gap, in which case one achieves a level of numerical stability inversely proportional to the width of the gap.
As a variant of this idea, instead of working with gaps that are completely devoid of spectrum, one can also use the pigeonhole principle (or direct numerical search) to locate intervals which contain only a small (but possibly nonzero) amount of "spectral energy". This "energy pigeonholing" idea for instance powers the spectral proof of the Szemeredi regularity lemma https://terrytao.wordpress.com/2012/12/03/the-spectral-proof-of-the-szemeredi-regularity-lemma/ , and was also widely used by Bourgain in several of his papers on PDE and also density Ramsey theory.
The SVD decomposition falls under the family of phenomena where discontinuity implies non-computability. (Intuitively, this is because, at the point of discontinuity, infinite precision is required.)
In this particular case we speak of the (dis)continuity of a multivalued function which takes a matrix to any of its decompositions, or better, the non-existence of a realizer for the statement
"For every matrix $M$ there exist suitable $U$, $\Sigma$, $V$ yielding an SVD decomposition of $M$." I belive this statement has no conitnuous realizer in function realizability, and hence no computable one either.
Some other examples of this phenomenon are:
The sign function $\mathrm{sgn} : \mathbb{R} \to \{-1,0,1\}$ is discontinuous, therefore non-computable. In fact, every computable map $\mathbb{R} \to \{0,1\}$ is constant.
The rank of a matrix is non-computable.
Gaussian elimination (as taught in school) cannot be performed because testing for zero is non-computable.
The number of distinct zeroes of a polynomial is non-computable.
So why are these, along with your observation, a problem? There are several answers, depending on the context.
In floating-point numerics, all computations are done with a fixed finite precision and numerical errors are simply unavoidable. In this setting the non-computability manifests itself as numerical instability. In your case, we might simply compute the wrong decomposition.
In some situations we can restrict to computation in a subring of $\mathbb{R}$ in which the problem disappears. For example, many of the above problems are non-existent when we restrict to $\mathbb{Q}$ or the algebraic numbers.
In exact real-arithmetic there are no numerical errors, as precision always adapts automatically to achieve the desired result. In this setting non-computability really is non-computability. The algorithm will diverge at points of discontinuity. In your case, it will just run forever trying to determine in which of the two cases it is.
There are models of real-number computation that pretend we can perform exact zero-testing, notably the Blum-Shub-Smale model. They are often used in computational geometry to side-step questions about non-computability. There are various theorems guaranteeing that a small perturbation of the input can get us out of trouble, at the price of possibly computing the wrong result.
I've known about the uncomputability of discontinuous maps for a long time...
Also, you haven't answered my question. Read it again. I said that SVD is considered numerically stable by many sources
So numerical stability is not as closely linked to discontinuity as you claim
You cannot compute an eigenvector given an eigenvalue. That's a mistake
I thought your question was, and I quote: "So my question is, how do numerical analysts get around this problem? Or why is this apparently not a problem?" What about SVD? Are you asking whether it is computable?
I know it's not computable. But it's apparently not a problem
Ok, at this point I do not know what I am supposed to answer, sorry. I mean, I can answer "yeah, it seems so", but that's not what you're expecting.
I'm not sure what I'm asking either, to be honest...
I'd say leave the question. If it starts getting negative votes, then maybe delete it. But it's in general useful to let Google slurp it, as the question is a pretty common one.
I really liked this answer.
Again, you cannot compute an eigenvector given an eigenvalue $\lambda$. Lemme find a counterexample
Actually, the $M(\epsilon)$ in my question provides a counterexample
Computing an eigenvector to a given eigenvalue would be possible if one knew the dimension of the corresponding eigenspace, if I recall correctly.
The SVD is indeed perfectly stable and computable for a fixed matrix - in your case, for a fixed value of $\epsilon$. The fact that the singular values and vectors may be discontinuous functions of $\epsilon$ is well known - at least, as an applied mathematician working in industry, it is something that "everybody working in my problem domain knows" if only from practical experience. (It does have practical consequences, but in fact using a more complete model of the underlying physics removes the mathematical discontinuity at $\epsilon = 0$ anyway.
Thanks @Arno, I removed that false claim about eigenvectors.
@alephzero: stability and comptuabilithy don't make sense for a fixed matrix. There needs to be some sort of a parameter somewhere, because for a fixed matrix there is nothing to compute, as the SVD decomposition is fixed as well. Perhaps I misunderstand your point.
@alephzero: I would be amazed if SVD decomposition is really numerically stable (without any hidden assumptions) becuase I have never seen a procedure that is both non-comptuable and numerically stable. Can you provide a reference to the claim that SVD is stable? Just to avoid a misunderstanding: by "stable" I mean "stable under small perturbations".
@AndrejBauer The confusion here is backwards vs. forwards stability. When people say that the SVD is stable, they mean that it is backwards stable, whereas here you are talking about forwards stability.
@NickAlger: Thanks for the clarification.
The first paragraph is wrong. First of all, the SVD is not a function because it's one-to-many. Therefore, its uncomputability is not reducible to discontinuity in a straightforward way
For example, there is no continuous function which given a vector in $\mathbb R^3$ produces a perpendicular vector. But the problem is still computable
I did not mean to be condescending, I apologize if I gave you that impression. I edited the first paragraph and added a bit more information on how exactly this is about discontinuity to address your other remark. But no, I do not remember every single person on the internet that I have communicated with.
This is primarily an issue of backwards vs. forwards stability. Good SVD algorithms are backwards stable in the sense that the computed singular values and singular vectors are the true singular values and singular vectors of a slightly perturbed problem. You may see this by noting that while $P$ may change drastically as you change $\epsilon$, the product $PDP^T$ changes negligibly.
The SVD is not forwards stable when the singular values have small spectral gap, as your example demonstrates and other answers here discuss in more detail.
For more on backwards and forwards stability, see, e.g., this post and the links therein:
https://math.stackexchange.com/a/78907/3060
SCIPY uses LAPACK; some details on the stability of the algorithm are provided here:
https://www.netlib.org/lapack/lug/node97.html
Now I'm confused: I always thought that backwards stability implies forwards stability. (Indeed, with the definitions I'm familiar with, I seem to remember the proof to be quite simple.) Are there different definitions in play here?
@gmvh The forward error is bounded by the backwards error times the condition number of the problem. But here the condition number (of the mapping from matrices to their singular vectors) can become arbitrarily large
|
2025-03-21T14:48:31.894006
| 2020-08-23T15:26:52 |
369931
|
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|
Stack Exchange
|
Does every set have a rigid self-map?
The question was asked on Mathematics Stackexchange
but has remained unanswered so far.
A self-map is a map $f:X\to X$ from a set $X$ to itself. There is an obvious notion of morphism, and thus of isomorphism and automorphism, of self-maps. [A morphism from $f:X\to X$ to $g:Y\to Y$ is a map $\phi:X\to Y$ such that $g\circ\phi=\phi\circ f$.]
A self-map is rigid if it has no non-trivial automorphism.
The question is in the title:
Does every set have a rigid self-map?
Clearly, the existence of a rigid self-map of a given set $X$ depends only on the cardinality $|X|$ of $X$.
There is an obvious notion of the coproduct $f:X\to X$ of a family $f_i:X_i\to X_i$ of self-maps. [The set $X$ is the disjoint union of the $X_i$ and $f$ coincides with $f_i$ on $X_i$.] Any self-map is the coproduct of its indecomposable components, and a self-map is rigid if and only if its indecomposable components are rigid and pairwise non-isomorphic. (In the sequel I use the expression "component" instead of "indecomposable component". Moreover the identity of the empty set does not count as a component.)
We claim:
(1) If $|X|\le2^{2^{\aleph_0}}$, then $X$ has a rigid self-map.
Define $f:\mathbb N\to\mathbb N$ by $f(i)=\max(i-1,0)$. Then $f$ is rigid. Moreover, for each $n\in\mathbb N$ the map $f$ induces a rigid self-map of the set $\{0,1,\ldots,n\}$. This proves that (1) holds for $|X|\le\aleph_0$.
It remains to prove that $X$ has a rigid self-map when $\aleph_0<|X|\le2^{2^{\aleph_0}}$.
This will follow from Lemmas 1, 2 and 3 below.
Lemma 1. Let $X$ be an infinite set and $\Sigma$ a set of non-isomorphic rigid surjective indecomposable self-maps of $X$. Assume $|\Sigma|>|X|$. Then the coproduct of the elements of $\Sigma$ is a rigid surjective self-map of a set of cardinality $|\Sigma|$.
This is obvious.
Lemma 2. Let $f$ be a rigid surjective self-map of an infinite set $X$, and $Y$ a set satisfying $|X|\le|Y|\le2^{|X|}$. Then $Y$ has a rigid self-map.
Proof. Let $X'$ be a set disjoint from $X$ and $\phi:X'\to X$ a bijection. For each subset $S$ of $X'$ put $X_S=X\sqcup S$ (disjoint union) and define $f_S:X_S\to X_S$ by setting $f_S(x)=f(x)$ for $x\in X$ and $f_S(s)=\phi(s)$ for $s\in S$.
It suffices to show that the coproduct $g:Y\to Y$ of the $f_S:X_S\to X_S$ (where $S$ runs over all the subsets of $X'$) is rigid.
Let $h:Z\to Z$ be a component of $g$. Then $h$ is a component of $f_S$ for some $S$. It is easy to see that there is a unique component $f_0:X_0\to X_0$ of $f$ such that, if we set $S_0:=S\cap\phi^{-1}(X_0)$, then $h$ is equal to
$$
f_{0,S_0}:X_{0,S_0}\to X_{0,S_0},
$$
where $f_{0,S_0}$ is defined as $f_S$ was defined above (replacing the bijection $\phi:X'\to X$ with the bijection $\phi^{-1}(X_0)\to X_0$ induced by $\phi$).
Let
$$
f_{1,T_1}:X_{1,T_1}\to X_{1,T_1}
$$
be another component of $g$, corresponding to a subset $T$ of $X'$, and let
$$
\psi:X_{0,S_0}\to X_{1,T_1}
$$
be an isomorphism from $f_{0,S_0}$ to $f_{1,T_1}$. Since $X_0$ and $X_1$ are the respective images of $f_{0,S_0}$ and $f_{1,T_1}$ by surjectivity of $f$, the isomorphism $\psi$ maps $X_0$ onto $X_1$ and $S_0$ onto $T_1$. By rigidity of $f$ we have $X_0=X_1$ and $\psi(x)=x$ for all $x\in X_0$. Let $s$ be in $S_0$. It suffices to show $\psi(s)=s$. Set $x=\phi(s)\in X_0$. Then $\psi$ maps the fiber of $\phi$ above $x$ to itself, but $s$ is the only point in this fiber. This completes the proof of Lemma 2.
Lemma 3. Let $A$ be the set of all increasing self-maps of $\mathbb N$ such that $a(0)\ge1$. Then there is a family of pairwise non-isomorphic rigid surjective indecomposable self-maps
$$
(f_a:X_a\to X_a)_{a\in A},
$$
where each $X_a$ is an infinite subset of $\mathbb N^2$.
Proof. Define the subset $X_a$ of $\mathbb N^2$ by the condition that $(i,j)\in X_a$ if $i\in a(\mathbb N)$ or if $j=0$, and define $f_a:X_a\to X_a$ by setting
$\bullet\ f_a(i,j)=(i,j-1)$ if $j\ge1$,
$\bullet\ f_a(i,0)=(i-1,0)$ if $i\ge1$,
$\bullet\ f_a(0,0)=(0,0)$.
Let us fix $a\in A$ and sketch the proof that $f_a$ is a rigid self-map of $X_a$.
The point $(0,0)$ is the only fixed point. The points of the form $(i,0)$ with $i\ge1$ are characterized by the fact that they have ancestors which have two parents, and any two distinct such points are at different distances to $(0,0)$. Therefore the points $(i,0)$ are fixed by any automorphism of $f_a$. The point $(a(n),j)$ with $j\ge1$ has no ancestor with two parents, its first descendent with two parents is $(a(n),0)$, which is fixed by the automorphisms of $f_a$, the point $(a(n),j)$ is at distance $j$ from $(a(n),0)$, and these properties characterize $(a(n),j)$. Thus $(a(n),j)$ is fixed by the automorphisms of $f_a$.
This argument shows also that the $f_a$ are pairwise non-isomorphic. The other statements are clear.
[If $y=f(x)$ we say that $x$ is a parent of $y$. If $y=f^n(x)$ for $n\in\mathbb N$ we say that $x$ is an ancestor of $y$ and $y$ a descendent of $x$.]
set-theory
"selfmap/self-map" seems more usual. Actually set + self-map is the same as a $G$-set, for $G$ the 1-generated free monoid (or 1-generated free semigroup).
I sometimes find myself saying "endofunction", but I'm not sure whether that's more or less standard than "self-map".
@TimCampion - Not sure what it really means, but I did this Ngram search https://books.google.com/ngrams/graph?smoothing=3&corpus=26&year_start=1800&year_end=2019&content=self-map%2Cendofunction&direct_url=t1%3B%2Cself%20-%20map%3B%2Cc0%3B.t1%3B%2Cendofunction%3B%2Cc0
Yes. Actually, this was part of my first answer to this question, but this was a digression there (and I also posted there another answer to the same question which addressed it and was accepted). So I'm copying this digression here and will delete the initial answer to that question to avoid a duplicate.
Fact. For every set $X$ there exists $f\in X^X$ whose centralizer in $\mathrm{Sym}(X)$ is reduced to $\{\mathrm{id}_X\}$
It relies on the following second fact: there exists (for $X\neq\emptyset$) a rooted tree structure on $X$ whose automorphism group is trivial. Indeed, granting this, and denoting $v_0$ the root, for a vertex $v$ define $f(v)$ as $v_0$ if $v_0=v$, and as the unique vertex in $[v_0,v]$ at distance 1 to $v$ otherwise. Then $f\in X^X$ and its centralizer in $\mathrm{Sym}(X)$ is the automorphism group of the corresponding rooted tree, which is reduced to $\{\mathrm{id}_X\}$.
To prove the second fact, if $X$ is finite just take a linear tree rooted at an extremal vertex. If $X$ is infinite, by an elementary but very tricky argument (see this answer by user "bof"), there actually exist for every infinite cardinal $\kappa$, $2^{\kappa}$ pairwise non-isomorphic trees of cardinal $\kappa$ each with trivial automorphism group. [Interestingly the induction really requires proving that there are $>\kappa$ such trees, and not only a single one.]
I think it's important to note that the term "tree" means something different in set theory from what it means in graph theory (although the two definitions are equivalent when the tree is finite -- I'm not clear on the relationship in the infinite case). It's graph-theoretic trees which are related to endofunctions. And indeed, the linked construction builds graph-theoretic trees rather than set-theoretic ones.
|
2025-03-21T14:48:31.894516
| 2020-08-23T15:30:43 |
369932
|
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}
|
Stack Exchange
|
Looking for citable reference for a well-known fact about tensor product of finite dimensional algebras over an algebraically closed field
Let $K$ be an algebraically closed field and let $A$ and $B$ be finite dimensional algebras over $K$. Let $e_1,\ldots, e_n$ be orthogonal primitive idempotents of $A$ summing to $1$ and $f_1,\ldots, f_m$ be orthogonal primitive idempotents of $B$ summing to $1$. Then the $e_i\otimes f_j$ form a set of orthogonal primitive idempotents of $A\otimes_K B$ summing to $1$. Moreover, $(A\otimes_K B)(e_i\otimes f_j)\cong (A\otimes_K B)(e_{i'}\otimes f_{j'})$ if and only if $Ae_i\cong Ae_{i'}$ and $Bf_j\cong Bf_{j'}$. Note that this need not hold if $K$ is not algebraically closed.
I know how to prove these things but I want to cite this in a paper I'm writing and couldn't find this spelled out in any of my usual references on finite dimensional algebras. I don't really want to put a proof in my paper as it is a bit far a field. Actually, what I want to use this for is to conclude that the basic algebra of $A\otimes_K B$ is the tensor product of the basic algebras of $A$ and $B$ (again with $K$ algebraically closed).
I would greatly appreciate any reference, particularly, to a book.
Perhaps try Pierce's Associative Algebras?
@PedroTamaroff, I didn't see anything in Pierce
A treatment of this can be found in chapter IV.11. in the book "Frobenius algebras I" by Skowronski and Yamagata with proofs in the case $A=B^{op}$. But the proofs work exactly the same in the general case with the exact same proofs so it can be used as a reference for the general case I think.
Thanks. I'll look at the book. It would be nice if the general case was stated since it is annoying to say they prove it in a special case but the same proof works.
If nobody can find a reference for the general case I will go with this one.
|
2025-03-21T14:48:31.895017
| 2020-08-23T16:43:24 |
369940
|
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|
Stack Exchange
|
Boardline case of $W^{2,p}$ estimates on elliptic equations
Suppose $u$ is a strong solution of
$$\begin{cases}\Delta u =f &\quad \text{in} \quad B_1(0)\\ u=0 &\quad \text{on}\quad \partial B_1\end{cases}$$
The well known $W^{2,p}$ estimates says if $f\in L^p(B_1)$, $1<p<\infty$, then $$||u||_{W^{2,p}(B_{\frac12})}\leq C||f||_{L^p(B_1)}$$
It is obtained through Calderon-Zygmund theory.
Does anyone know what will happen for $p=1$? What is the best we can say about $u$ and $\nabla u$? I saw a paper that says $\nabla u\in L^{2,\infty}$ (the weak $L^2$ space) through the knowledge of singular integral. I did not know how they get this. Can anyone give me more reference? Thank you.
I have no clue about these $L^{2,\infty}$ spaces or best results. But I think a duality argument probably gives optimal $L^p$ estimates for $u$ and $ \nabla u$ (and is also a great example to apply this duality stuff)
also you can apply a duality argument to get results even if you weaken $L^1$ condition on $f$ to $ \int_{B_1} |f(x)| \delta(x) dx \le C$ where $\delta(x)$ is the distance to the boundary function.
You get $\nabla u \in L^{N/(N-1), \infty}$.
Giorgio, where can I find such results? It would be great if you can give me more directions to search.
I know how to show $\nabla u\in L^{2,\infty}$. It just follows from Riesz potential argument. when $p=1$.
@Slm2004 Sorry for the late answer but I did not get any notification of your comment (you forgot @ before my name). The proof I know uses the heat semigroup and it is written in a paper by A. Lunardi and myself in some generality, but the result for the Laplacian is surely older. The title is "On domains of elliptic operators in $L^1$". If you don't find it, write an email to<EMAIL_ADDRESS>
@GiorgioMetafune Thank you for your reference. That is exactly what I wanted.
@Slm2004 Another proof came to mind and this should be the "standard" one. If $\Delta u =f \in L^1$, write $u$ as the newtonian potential of $f$ and differentiate under the integral. Then $\nabla u$ is estimated by the Riesz potential of $I_1(|f|)$ and it is well known that $I_1$ maps $L^1$ to weak $L^{N/(N-1)}$.
@GiorgioMetafune Yes. I also realize that.
You may find useful a paper by Perez in Journal of Functional Analysis in 1995
Thank you for your answer
|
2025-03-21T14:48:31.895328
| 2020-08-23T17:00:37 |
369942
|
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|
Stack Exchange
|
Analogue of Lipschitz continuity of $W^{1,\infty}$ for Hölder continuity and Sobolev-Slobodeckij spaces
A function $u : U \rightarrow \mathbb R$ is an element of the Hölder space $C^{\alpha}(U)$ if
$\sup\limits_{x \in U} |u(x)| < \infty$
$\sup\limits_{x,y \in U} \dfrac{|u(x) - u(y)|}{|x-y|^\alpha} < \infty$
A function $u : U \rightarrow \mathbb R$ is an element of the fractional Sobolev space $W^{\alpha,\infty}(U)$ if
$\mathrm{esssup}_{x \in U} |u(x)| < \infty$
$\mathrm{esssup}_{x,y \in U} \dfrac{|u(x) - u(y)|}{|x-y|^\alpha} < \infty$
Here, $\mathrm{esssup}$ denotes the essentially supremum.
Clearly, $C^\alpha(U)$ is a subspace of $W^{\alpha,\infty}(U)$. Does the converse statement hold, that is, $W^{\alpha,\infty}(U)$ is a subspace of $C^\alpha(U)$?
The claim is true if $\alpha = 1$ when $U$ has sufficient boundary regularity, in which case we just deal with Rademacher's theorem: a function is Lipschitz continuous if and only if it is essentially bounded with essentially bounded derivatives.
The generalization is claimed in the Encyclopedia of Math but no reference is given.
Does there exist a short proof or an explicit reference in the literature (paper, textbook)?
Technically one has to identify functions that agree a.e. before one gets the identity between $C^\alpha(U)$ and $W^{\alpha,\infty}(U)$ (even in the $\alpha=1$ case), but yes, this statement is true. Given $u \in W^{\alpha,\infty}(U)$, standard mollifiers $u * \varphi_\varepsilon$ will be uniformly bounded in $C^\alpha(U)$ and converge uniformly to a limit $\tilde u \in C^\alpha(U)$ that is equal a.e. to $u$.
Another way to get this without any boundary regularity is to note that there exists a set of full measure $A$ on which the suprema are finite. The function $u$ is then continuous on $A$. Since $U$ is open, $A$ is dense in $U$ and $u$ has a unique continuous extension to $U$ that satisfies the required properties.
@JeanVanSchaftingen: thanks, but it is not obvious that you get such a set $A$. Let me explain: the function $f(x,y) = |u(x)-u(y)| \cdot |x-y|^{-\alpha}$ is locally integrable over $U \times U$ and hence there exists a dense subset $B \subseteq U \times U$ over which (a representative of) $f$ is bounded. The existence of $A$ is not immediately clear. Is there some handwaving to find it?
|
2025-03-21T14:48:31.895506
| 2020-08-23T17:30:46 |
369944
|
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|
Stack Exchange
|
Extension of positive functionals
Let $X$ be a function space as $C(K)$ or $L^p$, with its usual norm and order, that is $f \le g$ if and only if $f(x) \le g(x)$ for a.e. $x$. If $M$ is a subspace of $X$ and $L:M \to \bf R$ is a positive functional, that is $Lf \ge 0$ if $0 \le f \in M$ (note that this condition can be void if $M$ does not contain non-trivial positive functions) , there are conditions under which $L$ can be extended as a positive functional defined on $X$. These results go back to Bauer, Namioka, Kantorovich and can be found in the book of H. Schaefer, Topological vector spaces. Counteraxmples are easily constructed by choosing positive functionals discontinuous on $M$; they cannot be extended to positive functionals on $X$, since positivity on the whole space implies continuity. An example is integration against $1/x$ wich is a positive functional on bounded functions in $(0,1)$, with support far from the origin, but cannot extended to $L^p(0,1)$.
The question I have is the following: I do not know examples of positive and continuous functionals on subspaces which cannot be extended as positive functionals on $X$. I recall that I am using the natural order. They should exist, and I hope that somebody knows.
Just take $f\mapsto -f(0)$ on the subspace $M$ of $C[0,1]$ of all functions with zero integral (so that the positivity condition on $M$ is void). The only possible extension would be $f\mapsto -f(0)+C\int_0^1 f$ but that one is not positive for any $C$.
@fedja Thank you very much. The example is very nice and simple...but it did not come to mind!
|
2025-03-21T14:48:31.895632
| 2020-08-23T17:40:48 |
369946
|
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|
Stack Exchange
|
Kernel perfect orientations of complete graphs
How can we create a kernel perfect orientation of a complete graph? A kernel of a graph is a set of vertices in a graph $G$, which absorbs other vertices, that is, has all the vertices in its complement have a directed edge towards it in an orientation of $G$. A kernel -perfect orientation of graph is an orientation of the graph in which each subgraph has a kernel.
A kernel in a complete graph consists of just a vertex as it has edges to every other vertex. But, if the tournament (orientation of the complete graph) is kernel perfect, then we should have every sub-clique to have a kernel. So, how do we orient the edges to obtain a kernel perfect tournament. Any hints? Thanks beforehand.
What is wrong with just enumerating the vertices and orienting all edges from the vertex with a smaller number to that with the larger one?
@fedja well, thanks. Your comment solves the question. You could post as an answer
|
2025-03-21T14:48:31.895730
| 2020-08-23T18:25:19 |
369948
|
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"Zhen Lin",
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|
Stack Exchange
|
Kan extensions and restriction
Suppose $A \to B$ be a fully faithful (codense if it helps, i.e. every object in $B$ is the small limit of a diagram in A) embedding of categories. Let $b$ be an object of $B$, $Hom_B(b,-)$ defines (restrict to) a functor $A \to Set$ which is denoted by $Hom_B(b,-)|_A$ . When does the left or right Kan extension of $Hom_B(b,-)|_A$ along $A \to B$ isomorphic to $Hom_B(b,-)$ ?
If $A$ is dense and $b$ is in $A$, then the left Kan extension of the restriction recovers the original hom functor. This is essentially tautological, by the definition of density. Slightly less trivial example: if $b$ is a retract of an object in $A$ then it has the same property.
@ZhenLin Thx for your comment. What if b is not in A? and what about the right Kan extension? BTW I should probably say codense instead of dense, i.e. every object in $B$ is a small limit of a diagram in $A$.
Covariant hom functors preserve all limits (in the strongest possible sense) so if $A$ is codense in $B$ then the right Kan extension of the restriction of any hom functor will recover the original hom functor.
This seems true to me but it's not clear to me how to prove this rigorously. Should I use for example the limit formula to prove this?
Yes. The same limit diagram is used in both the definition of codensity and in the definition of pointwise right Kan extension. (No surprise, because codensity is really talking about a certain pointwise right Kan extension.)
@ZhenLin right, that makes sense. I'm wondering if the same holds for enriched categories and 2 categories? It seems that the limit formula for pointwise right Kan extension in the 2-categorical setting is less commonly documented.
There is a limit formula for enriched categories too: $\mathcal{C} (c, \operatorname{Ran}_F G (b)) \cong [\mathcal{A}, \mathcal{V}](\mathcal{B} (b, F), \mathcal{C} (c, G))$, hence $\operatorname{Ran}_F G (b)$ is a weighted limit in $\mathcal{B}$ of a diagram of shape $\mathcal{A}$.
|
2025-03-21T14:48:31.895904
| 2020-08-23T21:31:55 |
369957
|
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|
Stack Exchange
|
Chernoff-style concentration inequality for k-tuples
I'm looking for a seemingly natural generalization of a Chernoff bound.
In many scenarios, we have a distribution $D$ with support $\mathsf{Supp}(D)$, and some event $E \subset \mathsf{Supp}(D)$ telling us whether a property of a sample from $D$ holds (i.e. $a \in E$ iff $a\sim D$ has the property we want). Denoting $p =\Pr_{a\sim D}[a \in E]$, we use Chernoff to say something of the kind: if I draw $n$ independent samples from $D$, then with probability at least $1-\exp(-\delta^2pn/2)$, my multiset $A = \{a_1, \cdots, a_n\}$ of samples will be "$\delta$-good", where "$\delta$-good" means that if I fix this multiset $A$ once for all and sample (uniformly over the multiset) $a$ from $A$, then $a \in E$ will hold with probability $(1-\delta)p$. This is the standard Chernoff bound for Bernouilli random variables.
In my scenario, I need a generalization of the above, where the event is over $k$-tuples of samples from $D$ (i.e. $E \subset \mathsf{Supp}(D^k)$). Let $p = \Pr_{(a_1, \cdots, a_k)\sim D^k}[(a_1, \cdots, a_k) \in E]$. Suppose that for $i=1$ to $k$, I draw $n$ independent samples $(a_{i,1}, \cdots, a_{i,n})$ from $D$, which form a multiset $A_i$. I want to be able to make statements of the form: with probability at least 'something', the multisets $(A_1, \cdots, A_k)$ will be "$\delta$-good", where "$\delta$-good" means that if I fix $(A_1, \cdots, A_k)$ once for all and uniformly sample a $k$-tuple $(a_1, \cdots, a_k)$ from $A_1 \times \cdots \times A_k$, then $(a_1, \cdots, a_k)\in E$ will hold with probability at least $(1-\delta)p$.
Of course, the standard Chernoff bound does not apply anymore (it would apply if, instead, I had fixed a single multiset $A$ of $n$ random $k$-tuples sampled from $D^k$). Other concentration bounds I'm familiar with, such as Azuma's inequality or McDiarmid's bounded difference inequality, do not seem to apply either.
Question: is any such bound known in the literature, or does it follow from any standard concentration bound? Any pointer would be welcome. To be clear, I crucially need Chernoff-level strength: Markov or anything of the kind wont do. I've tried to derive a bound of this kind, first from standard concentration bounds with limited dependence (e.g. McDiarmid), and I've searched a bit the literature, both without success. Before trying to establish it from first principles, I figured it would be better to ask first, since it looks like something people should have considered before.
--
EDIT - answering the comments of kodlu
Do you have any other constraints on your function $f$? Lipschitz type? Subgaussian type?
Are you referring to the function $f$ that I initially used to define the event $E$? If so, why would this function being Lipschitz or Subgaussian matter in any way? Note that $f$ has nothing to do with the function we want to be Lipschitz when applying e.g. McDiarmid's inequality. For example, if you consider the case $k=1$ (which is the base case I'm trying to generalize), then whatever $f$ is, the resulting bound is exactly a bound on a sum of independent Bernouilli random variables - that is, the function is just a direct sum, and $f$ is just what defines whether the event happened or not. I understand that my choice of notations might have been confusing, I hope switching to $E$ as suggested by dohmatob makes things better.
What makes you think you'll get concentration in such an arbitrary setting in a product space? Do you have any experimental evidence?
My intuition is that there should be such a bound - now, that is barely more than an intuition. I have some kind of experimental evidence, but only for the very specific context I'm actually working on, though I believe that such a bound should hold in a more general setting (which is why I refrained from describing my precise and confusing setting).
In case it helps, anyway (and simplifying a bit): in the concrete setting I'm working on, a sample from $D$ is a length-$t$ vector of bits (for some parameter $t$) where each entry is sampled independently and is biased towards $0$, and the event over a $k$-tuple of samples $(a_1, \cdots, a_k)$ is defined as follows: the fraction of positions $i \in [1, t]$ such that at least one $a_j$ contains a $1$ at position $i$ belongs to $[1/10, 9/10]$. I'm trying to show that this event happens often enough it I fix $k$ multisets of samples as I described above, and sample one entry of the $k$-tuple from each multiset.
In this setting, yes, I have some weak kind of experimental evidence, coming from the fact that this bound captures the hardness of attacking a cryptographic primitive with a restricted family of attacks (well, at least a part of the analysis requires this bound). Since it's a primitive some people tried to break with these attacks and failed, it appears likely that there is such a bound.
The notation $a \leftarrow D$ really looks odd for what it is.
This is the standard notation for "a is sampled from D" in my field of research (cryptography), but I'll be happy to switch to any notation which will look more natural to people here on MathOverflow :) Can you suggest any?
How about about "$a \sim D$" ?
Done, thanks! I realize that indeed, $\gets$ is more of a computer science notation, for "I assign to $a$ a value that was sampled according to the distribution $D$". Feel free to edit if other notations seem odd/inappropriate. Unrelated: would the downvoters care to explain why the downvote? Is there anything that I can improve with the question, or any reason why it does not seem good as it is?
Also, concerning notation. I'd write "$f(a) = 1$ as "$a \in E$", for appropriately defined $E$ (precisely, $E = f^{-1}({1})$). BTW, i didn't downvote...
For example, with this notation, you can rewrite $p$ as $D(E)$, etc, etc. About the downvotes, my wild guess is that some users might have found the question not very "beautiful". That the chaff hasn't yet been separated from the wheat. But this is just a wild guess...
Switched $f$ to $E$, thanks - but for more concise notations like $D(E)$ (never saw that notation), I guess one has to be a bit more familiar than I am with standard notations in probability (but again, anyone should feel free to edit to improve notations). My background is computer science, where the standard notations are quite different. It's also why I formulated the question is a very "wordy" fashion - I feared that using compact and succinct notations would obfuscate the question, given that the compact notations I'm used to might not be so readable for the general 'Mathoverflow crowd'.
Do you have any other constraints on your function $f$? Lipschitz type? Subgaussian type? What makes you think you'll get concentration in such an arbitrary setting in a product space? Do you have any experimental evidence?
Hi kodlu, thanks for your comment. I tried to address the question it contains directly in the body of the post, since 500 characters would not have sufficed.
Is this summary right? You have $nk$ i.i.d. draws from $D$, arranged in $k$ columns each of height $n$. You have that any fixed row falls in $E$ with probability $p$. You want to show that with high probability, about a $p$ fraction of the possible combinations obtained by taking one element from each column fall in $E$.
yes, that's about right, though I'm looking for the slightly more general setting of having a $(1-\delta)p$ (for an arbitrary $\delta$) fraction of combinations in E, and "high" refers more specifically to a bound comparable to the Chernoff bound for Bernouilli random variables
Thanks for the clarifications
Theorem 2 in [1] gives a bound of $1-\frac{4e^{-\delta^2n/8}}{\delta}$. I think you can incorporate $p$ in the bound since the proof of that theorem uses the standard Chernoff bound.
[1] Yakov Babichenko, Siddharth Barman, Ron Peretz (2017) Empirical Distribution of Equilibrium Play and Its Testing Application. Mathematics of Operations Research 42(1):15-29. http://dx.doi.org/10.1287/moor.2016.0794
Many thanks! I will read that carefully and accept your answer afterwards. By the way, for other readers, here is the direct link to the paper without paywalls: https://arxiv.org/pdf/1310.7654.pdf
The arxiv version is slightly different, but i guess you can get the idea from looking at Lemma 1 in the arxiv version.
Yes, that's what I figured. Otherwise, is there a way to get access to the version which is behind a paywall?
|
2025-03-21T14:48:31.896509
| 2020-08-23T22:11:02 |
369959
|
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"url": "https://mathoverflow.net/questions/369959"
}
|
Stack Exchange
|
When are two complex Tori biholomorphic
Let $g \ge 1$ be a natural number and $\mathbb{C}^g$ complex vector space which
is isomorphic to $\mathbb{R}^{2g}$ is real vector space.
An additive subgroup $\Gamma \subset \mathbb{C}^g$ is called a
lattice if there exist $2g$ vectors $\gamma_1,... \gamma_{2g}$,
which are linearly independent over $\mathbb{R}$ such that
$\Gamma= \mathbb{Z} \gamma_1 + ... + \mathbb{Z} \gamma_{2g}$.
Let $\Gamma, \Gamma' \subset \mathbb{C}^g$ two lattices with
$\Gamma= \mathbb{Z} \gamma_1 + ... + \mathbb{Z} \gamma_{2g}$ and
$\Gamma'= \mathbb{Z} \gamma' _1 + ... + \mathbb{Z} \gamma' _{2g}$.
Here and here is discussed a characterization when two tori
$\mathbb{C} / \Gamma$ and $\mathbb{C} / \Gamma'$
with lattices $\Gamma$ and $\Gamma'$ are biholomorphic.
The main result states that
$\mathbb{C} / \Gamma$ and $\mathbb{C} / \Gamma'$ are biholomorphic
if and only if there exist a scalar $\alpha \in \mathbb{C}^*$
with $\alpha \Gamma = \Gamma'$.
I'm searching for a similar result for $g \ge 2$ which characterizes
for two lattices $\Gamma$ and $\Gamma'$ when the tori
$\mathbb{C}^g / \Gamma $ and $ \mathbb{C}^g / \Gamma'$ are
biholomorphic.
Can the quoted result for $g=1$ be generalized?
What is well known is that for arbitrary two lattices $\Gamma, \Gamma' \subset \mathbb{C}^g$ there exist a matix $M \in GL_{2g}(\mathbb{C})$ with $M \cdot \Gamma = \Gamma'$ and $\gamma' _j = M \gamma_j$.
This implies presumably (??) that arbitrary two tori $\mathbb{C}^g / \Gamma $ and $ \mathbb{C}^g / \Gamma'$ as far as $\Gamma$ and $\Gamma'$ are lattices in $\mathbb{C}^g$ are always isomorphic as (abelian) groups and as topological spaces. As far as that's correct, yes?
Now to decide if they are also biholomorphic is more complicated as stated in linked discussions. Is there any neat characterization for $g \ge 2$ in similar manner as for $g=1$?
In the third from last paragraph, you probably intended $M\in GL_{2g}(\mathbb R)$. I suspect you'll need $M\in GL_g(\mathbb C)$ in order to get the tori biholomorphic.
You'll find the answer in any book on the subject — Debarre's Complex Tori and Abelian Varieties is quite accessible. And again, your questions would be more appropriate at MSE.
Also on MSE: https://math.stackexchange.com/questions/3800744/isomorphic-lattices-complex-tori-and-their-relation-to-jacobians/3803463#3803463
|
2025-03-21T14:48:31.896677
| 2020-08-24T00:16:04 |
369963
|
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|
Stack Exchange
|
An explicit solution to the congruence $x^2\equiv 14(\frac 3p)-(\frac p3)-12\pmod {p}$?
Let $p>3$ be a prime, and let $(\frac{\cdot}p)$ be the Legendre symbol. Then
$$14\left(\frac 3p\right)-\left(\frac p3\right)-12=\begin{cases}1&\text{if}\ p\equiv1\pmod{12},
\\-25&\text{if}\ p\equiv5\pmod{12},\\-27&\text{if}\ p\equiv 7\pmod{12},
\\3&\text{if}\ p\equiv 11\pmod{12};\end{cases}$$
this is a quadratic residue modulo $p$. It is interesting to find an explicit solution to the congruence
$$x^2\equiv 14\left(\frac 3p\right)-\left(\frac p3\right)-12\pmod p.\tag{1}$$
For $n=0,1,2,\ldots$ define
$$w_n:=\sum_{k=0}^n\binom nk\binom{n+k}k\binom{2k}k\binom{2(n-k)}{n-k}(-8)^{n-k}.$$
I have the following conjectures related to the congruence equation $(1)$.
Conjecture 1. (i) For every $n=1,2,3,\ldots$, the number
$$\frac1n\sum_{k=0}^{n-1}(-1)^k(4k+1)48^{n-1-k}w_k$$
is a positive integer.
(ii) For any prime $p>3$ we have
$$\bigg(\frac1p\sum_{k=0}^{p-1}\frac{4k+1}{(-48)^k}w_k\bigg)^2\equiv 14\left(\frac 3p\right)-\left(\frac p3\right)-12\pmod p.\tag{2}$$
Conjecture 2. We have the identity
$$\sum_{k=0}^\infty\frac{4k+1}{(-48)^k}w_k=\frac{\sqrt{72+42\sqrt3}}{\pi}.\tag{3}$$
Conjecture 3. For any prime $p>3$, we have
$$\sum_{k=0}^{p-1}\frac{w_k}{(-48)^k}\equiv\begin{cases}4x^2-2p\pmod{p^2}&\text{if}\ p=x^2+4y^2\ (x,y\in\mathbb Z),\\0\pmod{p^2}&\text{if}\ p\equiv3\pmod 4.\end{cases}\tag{4}$$
QUESTION. How to solve the above three conjectures? In particular, how to prove the congruence $(2)$?
Your comments are welcome!
|
2025-03-21T14:48:31.896771
| 2020-08-24T00:28:48 |
369964
|
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|
Stack Exchange
|
A limit question involving Cramer's decomposition of normal random variables
I've come across the following question. Say we have two families of random variables, $X_N$ and $Y_N$, such that $\mathbb{E} X_N=\mathbb{E} Y_N=0$ and $\mathbb{E}X_N^2=1$. Now assume that:
$$|\mathbb{E} e^{\lambda X_N}\mathbb{E} e^{\lambda Y_N}e^{-a_N^2\lambda^2/2}-1|\to 0$$
uniformly on bounded compact sets of $\mathbb{C}$. My question is whether from this you can conclude that $X_N\Rightarrow \cal{N}(0,1)$?
If $a_N^2$ is bounded this follows from Cramer's decomposition theorem, but I cannot either prove or disprove the statement in the case where $a_N^2$ is unbounded.
Any help would be appreciated.
Edit: I'll sketch the proof when $a_N^2$ is bounded. It suffices to show that any subsequence of $X_N$ has a further subsequence that converges in measure to $\cal{N}(0,1)$. As $a_N^2$ is bounded, we may pass to a subsequence such that $a_N^2\to a^2$ for some $a$. On the other hand this shows that the characteristic functions of $X_N+Y_N$ converges to $e^{-a^2\lambda^2/2}$, and so by Levy's continuity theorem, we have that $X_N+Y_N\Rightarrow \cal{N}(0,a^2)$.
Now by Jensen's inequality we have that for $\lambda \in \mathbb{R}$, $1\le \mathbb{E} e^{\lambda X_N}$. Thus for $\lambda \in \mathbb{C}$, we have that $|\mathbb{E}e^{\lambda X_N}|\le \mathbb{E}e^{\Re(\lambda)X_N}\le e^{a_N^2\Re(\lambda)^2/2}$, so that $\mathbb{E} e^{\lambda X_N}$ is a locally uniformly bounded family of analytic functions, so by Montel's theorem, and passing to a subsequence, we can assume that $\mathbb{E}e^{\lambda X_N}$ converges to some analytic function, which by the Levy continuity theorem, must be the moment-generating function of some random variable, say $X$, such that $X_N\Rightarrow X$ in distribution and in moments. One may do the same for $Y$.
Thus we see that $X+Y=\cal{N}(0,a^2)$. By Cramer's theorem we must have that $X$ is normal, and as by the moment assumptions on $X_N$, as well as the convergence in moments, we see that $X=\cal{N}(0,1)$.
@MattF. I added what you suggested I think.
This need not hold if $a_n$ is unbounded. Take $X_n$ to be Rademacher, i.e. $P(X_n = 1) = P(X_n = -1) = 1/2$. Take $Y_n \sim N(0,n)$ and $a_n = \sqrt{n}$. Then $$\left| \mathbb{E} e^{\lambda X_n} \mathbb{E} e^{\lambda Y_n} e^{-a_n^2 \lambda^2 / 2} - 1\right| =
\left| \cosh(\lambda) - 1\right|$$
which is locally uniformly bounded for $\lambda \in \mathbb{C}$.
Ah, I apologize, I made a mistake while editing the question last night. I meant to write $|\mathbb{E} e^{-\lambda X_N}\mathbb{E} e^{\lambda Y_N} e^{-a_N^2\lambda^2/2}-1|=o(1)$ insted of $O(1)$. I apologize for the confusion. I believe the $O(1)$ statement is wrong even in the finite case with roughly your counterexample.
|
2025-03-21T14:48:31.896944
| 2020-08-24T01:54:49 |
369967
|
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|
Stack Exchange
|
birational geometry of moduli spaces: why work on the coarse space?
In studying the birational geometry of $\overline{\mathcal{M}}_g$, it seems standard to work with the coarse space $\overline{M}_g$ rather than the smooth stack $\overline{\mathcal{M}}_g$. Why is this?
More precisely:
(1) In Harris-Mumford's "On the Kodaira Dimension of the Moduli Space of Curves" (1982), it is proven that $\overline{M}_g$ is of general type for odd $g\ge25$ (of course generalized to all $g\ge24$ later by Harris and Eisenbud-Harris). A lot of work is done in the first section showing that differential forms can be extended from the smooth locus of $\overline{M}_g$ to a desingularization of the whole space, so that Kodaira dimension makes sense in the first place, but can one not work with canonical sheaf on the stack $\overline{\mathcal{M}}_g$, which is already smooth? It seems to me that the rest of the proof ought to go through verbatim on the moduli stack, avoiding all the issues with automorphisms.
(2) In Bruno-Verra's "$\mathcal{M}_{15}$ is rationally connected,'' the strategy of proof of the theorem in question is as follows. Let $\Delta_0\subset\overline{\mathcal{M}}_{15}$ be closure of the locus of irreducible nodal curves, or equivalently the image of the clutching map from $\overline{\mathcal{M}}_{14,2}$. Given general points $x_i\in\mathcal{M}_{15}$, $i=1,2$ it is shown that there exist general points $y_i\in \Delta_0$ and rational curves connecting $x_i$ to $y_i$. It is also shown that $M_{14,2}$ is unirational, hence rationally connected, so $y_1$ and $y_2$ are also connected by a rational curve. Thus $x_1$ and $x_2$ are connected by a chain of rational curves.
Some effort is needed to show that the $y_i$ can be taken to be general, in particular, that they are smooth points of $\overline{M}_{15}$, and they correspond to curves with exactly one node (i.e. they don't lie on a self-intersection of $\Delta_0$). But does the argument not go through if the $y_i$ are arbitrary and we work with the moduli stacks? At least on a smooth, projective variety, the condition that a general pair of points are connected by a rational curve implies that an arbitrary pair of points is too. The self-intersections of $\Delta_0$ don't seem to me to be a problem, because a rational curve between arbitrary lifts of $y_1$ and $y_2$ on the smooth stack $\overline{\mathcal{M}}_{14,2}$ maps to a rational curve between $y_1$ and $y_2$ on $\Delta_0$. Is the issue perhaps that the various notions of rational connectedness (rational chain-connectedness, general vs. arbitrary pairs of points being connected, ...) don't all coincide on smooth proper DM stacks?
It is easy to give examples of smooth proper hyperbolic DM stacks, with plenty of effective pluricanonical divisors, whose moduli spaces are rational (this already occurs in dimension 1).
Thanks, this seems to answer (1); the issue seems to be that a rational map P^n - -> X may not extend over codim 1 (without passing to a ramified cover) when X is a stack, so forms on X might not pull back to forms on P^n.
Conversely if one wants to prove M_g rational (unirational, rationally connected) for g>2 it should suffice to do so for the stack, as it's birational to the coarse space. But maybe the issue here is still that the notions of rational (chain-)connectedness are different for stacks as I speculated above...will wait to see if anyone else has thoughts.
@HansSachs one thing to keep in mind is that the ultimate classical motivation for the line of inquiry of which these papers are a part is Severi's false conjecture that $M_g$ is unirational for all $g$. And of course the moduli stack was not a known object in Severi's lifetime. And Harris-Mumford implicitly (or maybe explicitly? it's been a while) works on the stack when computing divisor classes and things of that nature.
|
2025-03-21T14:48:31.897333
| 2020-08-24T01:55:25 |
369968
|
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"Fedor Petrov",
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|
Stack Exchange
|
Property $(\mathcal{L}(\phi),\phi)\geq 0$ about a operator $\mathcal{L}$
Consider the operator $\mathcal{L} : H^2(\mathbb{T}_L) \subset L^2(\mathbb{T}_L) \longrightarrow L^2(\mathbb{T}_L)$ given by
$$\mathcal{L} = -\omega \partial_x^2+3\varphi^2-1,$$
that is
$$\mathcal{L}(f) = -\omega f''+3\varphi^2f-f,$$
where $\mathbb{T}_L:= \mathbb{R}/L\mathbb{Z}$ and $L,w>0$ are fixeds constants and $\varphi$ is a fixed function. I know that $\mathcal{L}$ is linear, self-adjoint and from the Floquet Theory, the operator $\mathcal{}$ has only negative eigenvalue $\lambda <0$, with eigenfunction $\phi$.
Question. Why for any $\psi \in H^1(\mathbb{T}_L)$ such that $\psi \perp \phi$, that is $(\psi, \phi)_1=0$, we have
$$(\mathcal{L}(\psi),\psi)\geq 0? \tag{1}$$
Here $(\cdot, \cdot)$ denote the inner product in $L^2$ and $(\cdot, \cdot)_1$ the inner product in $H^1$.
Do you mean $\psi$ in (1)? Is $(\cdot,\cdot)_1$ the canonical inner product in $H^1$ or an equivalente one?
@GiorgioMetafune I meant $\phi$. Is the canonical inner product.
Thanks. (1) is true when $(\psi, \phi)=0$, inner product in $L^2$, why should it be true if they are orthogonal in $H^1$?
expand $\psi$ in an eigenbasis of $\mathcal{L}$
Why $($1$)$ is true when $(\psi, \phi)=0$?
@GiorgioMetafune And if the inner product $(\cdot,\cdot)_1$ in $H_1 $is a (convenient) inner product equivalent to the canonical, then $(\psi,\phi)_1=0$ implies $(\psi,\phi)=0$?
If $(\psi, \phi)=0$, then do as @Fedor Petrov said. But orthogonality is sensitive to the inner product you choose.
If in $H^1$ you choose the inner product coming from $(L u,v)$, after integration by parts, this is equivalent to the original inner product in $H^1$ and the orthogonality with respect to $\phi$ is the same in this and $L^2$, since $L\phi=\lambda \phi$.
Then $\psi$ should be orthogonal to both eigenfunctions. Take the eigenbasis of $L$, that is $u=\sum_n c_n \phi_n$ and $L\phi_n=\lambda_n \phi_n$. Then $(Lu,u)=\sum_n \lambda_n |c_n|^2$ and you see that the sign depends on the positivity of $\lambda_n$ and vanishing of $c_n$.
@GiorgioMetafune Understand. Is this property of $ (\mathcal{L} (\phi), \phi) \geq 0 $ strongly linked to the fact that $ \lambda $ is the only negative eigenvalue? Or does this property also apply more generally, that is, if there were more negative eigenvalues?
@GiorgioMetafune thank you very much. Your help was very important.
@GiorgioMetafune In the last case, we would still have $(\mathcal{L}(\psi), \psi) \geq 0$ or $(\mathcal{L}(\psi), \psi) > 0$?
Which is the last case?
@GiorgioMetafune If there more eigenvalues. For instance, two eigenvalues.
If there are two negative eigenvalues and the third Is positive, It will be positive. If the third Is zero It can be zero. Everything follows from the espansione.
|
2025-03-21T14:48:31.897533
| 2020-08-24T05:28:27 |
369969
|
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"Gerry Myerson",
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|
Stack Exchange
|
Related to one of the twin prime conjectures (In between squares)
This question is inspired by an answer that I have received for another question [see here]:
One of the twin prime conjectures states that Between the squares of two consecutive odd numbers $[2n+1]^2$ and $[2(n+1)+1]^2$ there is always at least one pair of twin primes .
In my attempts to compare between sets of twin pairs with other sets, I have made a comparison with a
consecutive list of numbers as follow 1, 1+2=3 , 1+2+3=6 , 1+2+3+4=10 , 1+2+3+4+5=15 up till $18413346$ (Sorry for the limit but, My script keeps crashing after that). Let's call the numbers in this set $B$.
If $B < 188806$: There are 35 consecutive numbers that have no twin primes between them.
If $B > 188805$: So far all of the consecutive numbers have twin primes between them.
Is it a known conjecture (If a conjecture at all), and are there any similar conjectures between twin primes and other consecutive sets (especially ones that grow slower)?
See the results here
In a related effort,I have also made a small heuristic showing that there are probably never more than 3 consecutive twin pairs when $B$ being the first of the pair and $B > 43$ and between the range of $B - 15$ to $B + 17$. Also probably never more than 2 consecutive pairs when $B > 13$ and between $B - 7$ to $B + 9$ See Here? Any knowledge about it will be appreciated too.
I think the standard conjecture would be that for any positive $k$ there's always a pair of twin primes between $m$ and $m+k\sqrt m$, provided $m$ is large enough (as a function of $k$). The triangular numbers would correspond to $k=\sqrt2$.
Yes I think you are right.
|
2025-03-21T14:48:31.897666
| 2020-08-24T06:30:25 |
369971
|
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|
Stack Exchange
|
Gromov-Witten invariants of cocharacter closures in toric varieties
$\require{AMScd}$
Let $X$ be a toric projective variety with dense algebraic torus $\iota:(\mathbb{C}^\times)^n \to X$, and let $u:\mathbb{C}^\times \to X$ be a cocharacter, by which I mean a map admitting a factorization of the form
$$\mathbb{C}^\times \xrightarrow{h} (\mathbb{C}^\times)^n \xrightarrow{\iota} X \qquad \text{where}\qquad h\text{ is a group homomorphism}$$
Definition. The closure $\bar{u}:C \to X$ of the cocharacter $u$ is the unique extension of $u$ to a singular toric curve $C$ that commutes with the $\mathbb{C}^\times$-action on $\mathbb{C}^\times$ and $C$.
This construction seems pretty natural to me. Furthermore, cocharacters are abundant since a cocharacter $u$ is equivalent to an element of $\mathbb{Z}^n$ via the map
$$a = (a_1,\dots,a_n) \mapsto u_a \qquad\text{with}\qquad u_a(z) = (z^{a_1},\dots,z^{a_n})$$
However, I am having trouble finding information about these curves. For example, I am interested in the following question.
Question 1. Are there other characterizations of the curves arising from this construction?
I am also interested in the Gromov-Witten theory of these curves. All that I can ask here is the following vague question.
Question 2. Is there some sense in which the curves $\bar{u}_a$ has a "non-trivial count in Gromov-Witten theory"?
I'm lookin for an answer like: for each $a \in \mathbb{Z}^n$, there exists a $0$-dimensional moduli space of stable curves $\overline{\mathcal{M}}_{g,n}(X,A)$ that naturally includes $\bar{u}_a$ (somehow) and where $GW^{X,A}_{g,n} \neq 0 \in H_0(X)$. This is almost certainly too specific, but anything in this general direction would be great.
|
2025-03-21T14:48:31.897801
| 2020-08-24T07:14:32 |
369974
|
{
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"Master",
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}
|
Stack Exchange
|
Can we add $NF$ to Ackermann's set theory?
Can we simply add stratified comprehension $SF$ to axioms of Ackermann's set theory $Ack$?
Is there a clear argument of inconsistency involved with such addition? supposing that $NF$ and $Ack$ are consistent of course.
It is not even known that $NF$ is consistent, right?
Yes but the argument is relative, if we suppose NF and Ack to be consistent, then is their blend consistent. By the way this question supposedly do have a positive answer according to Holmes, although he didn't write up the full formal proof, but the sketch of the proof is there and it seems convincing.
So you want to, from a model of $Ack$ and a model of $NF$, construct a model of $NF+Ack$?
Yes, something like that!
|
2025-03-21T14:48:31.897883
| 2020-08-24T07:21:09 |
369975
|
{
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"site": "mathoverflow.net",
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"url": "https://mathoverflow.net/questions/369975"
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|
Stack Exchange
|
Reference request: $K$-theoretic wrong-way map for a boundary inclusion
Let $W$ be a compact manifold with boundary. Let $i:\partial W\hookrightarrow W$ be the natural inclusion. We have a long exact sequence in complex $K$-theory:
$$\ldots\to K^*(\partial W)\xrightarrow{d} K^{*+1}(W,\partial W)\xrightarrow{j^*}K^{*+1}(W)\to\ldots,$$
where $j^*$ is induced by the inclusion $j:(W,\emptyset)\hookrightarrow(W,\partial W)$.
Let us assume that $W$ is also spin$^c$. Then we have in addition a wrong-way (Gysin) map
$$i_!:K^*(\partial W)\to K^{*+1}(W)$$
defined via the Thom isomorphism.
It seems from the answers and comments to this question, that $i_!=j^*\circ d$ (in particular equal to zero).
Question: Is there is a reference containing a discussion of this seemingly basic result?
|
2025-03-21T14:48:31.897967
| 2020-08-24T07:57:47 |
369977
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Daniel Loughran",
"Dmitry Ezhov",
"Fedor Petrov",
"Geoff Robinson",
"Gerry Myerson",
"Joe Silverman",
"Maciej Ulas",
"Q_p",
"https://mathoverflow.net/users/11926",
"https://mathoverflow.net/users/137021",
"https://mathoverflow.net/users/14450",
"https://mathoverflow.net/users/158000",
"https://mathoverflow.net/users/164119",
"https://mathoverflow.net/users/4312",
"https://mathoverflow.net/users/480516",
"https://mathoverflow.net/users/5101"
],
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/369977"
}
|
Stack Exchange
|
On the equation $a^{2}b^3 + c^{2}d^3 = e^{2}f^3$
Do there exist positive integers $a, b, c, d, e, f$ such that $a^{2}b^3 + c^{2}d^3 = e^{2}f^3$ where $b, d, f$ are pairwise coprime ?
Addendum: From the comments and Matt. F's answer, there clearly are infinitely many solutions. But what are their parametrizations ?
$1^{3}+2^{3} = 3^{2}$ ($a=b = c = f =1$ and $d = 2, e = 3$).
And even infinitely many: take $c=f=x+1,d=e=x, b=1, a=x(x+1) $, for $x=2$ this is Matt F.'s example.
@FedorPetrov, okay, can the solutions be parametrized ?
$a^{2}b^3 + c^{2}d^3 = e^{2}f^3\implies \Big(\dfrac{ab^3}{c}\Big)^2 - (bf)^3\Big(\dfrac{e}{c}\Big)^2 = -(bd)^3$ $$$$ This have Pell form. Some solutions $(a,b,c,d,e,f)$=(137819, 7, 7, 5, 491218, 3),<PHONE_NUMBER>, 7, 27, 4,<PHONE_NUMBER>, 9),<PHONE_NUMBER>487638370, 7, 70, 3,<PHONE_NUMBER>762229557, 10).
$r^3s^2(r+s)^2+r^2s^3(r+s)^2=r^2s^2(r+s)^3$ provided $\gcd(r,s)=1$.
In case you weren't aware, positive integers of the form $a^2b^3$ are called powerful numbers. Equivalently, they are numbers with the property that if a prime $p$ divides them, then $p^2$ divides them (hence the name). You might try doing a search on "sums of powerful numbers" and see what comes up. For example, the following MO post is related to primes that are differences of powerful numbers: https://mathoverflow.net/questions/269080/is-every-powerful-number-the-sum-of-a-powerful-number-and-a-prime
A closely related problems (sums of three squareful numbers) is studied here:https://arxiv.org/pdf/1106.4472.pdf
A related (and more difficult) equation: $a^3b^4+c^3d^4=e^3f^4$ is studied by A. Nitaj in the paper {\it On a conjecture of Erdos on 3-powerful numbers}, Bulletin of the London Mathematical Society 27 (1995), pp. 317-318.
We can get the parametric solutions using known solution $(a,b,c,d,e,f)$ for fixed $(b,d,f).$
For instance, we get a parametric solution using $(a,b,c,d,e,f)=(6,1,3,2,2,3).$
$1^3(48m^2-6n^2-48mn)^2 + 2^3(24m^2-3n^2+12mn)^2 = 3^3(16m^2+2n^2)^2$
$m,n$ are arbitrary.
m n a c e
1 1 2 11 6
1 2 6 3 2
1 3 150 33 34
1 4 10 1 2
1 5 114 3 22
2 1 30 39 22
2 3 150 141 82
2 5 146 47 38
3 1 282 249 146
3 2 30 69 38
3 5 438 321 194
4 1 190 143 86
4 3 138 501 274
4 5 38 61 34
5 1 318 219 134
5 2 58 59 34
5 3 426 753 418
|
2025-03-21T14:48:31.898154
| 2020-08-24T08:44:36 |
369982
|
{
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"authors": [
"Ernie",
"fedja",
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"site": "mathoverflow.net",
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"url": "https://mathoverflow.net/questions/369982"
}
|
Stack Exchange
|
Convergence of infinite linear programming
Suppose we have the following linear program (LP1),
$$\min_{f \in \mathcal{C}} \int_{\mathbb{R}} f(t) \cos(2 \pi x_0 t) dt \\ \text{subject to } \int_{\mathbb{R}}f(t)dt = 1 \\\forall x\in [0,1]: \int_{\mathbb{R}} \cos(2\pi x t) f(t)dt \geq 0 \text{ ,}$$
Where $\mathcal{C}$ is some "nice" family of functions (which I am not sure what it is. In the comments it was mentioned that $f$ has to be an analytic function with bounded spectrum).
Now define for all $n \in \mathbb{N}$ the following "approximate" linear program (LP2):
$$\min \sum_{i=0}^n a_i \cos(2 \pi x_0i) \\ \text{subject to } \sum_{i=0}^n a_i = 1 \\\forall k \in \{1/n, 2/n, \ldots, 1\} : \sum_{i=0}^n \cos(2\pi i k)a_i \geq 0.$$
Let $f^*$ be the optimal function for (LP1), and let $f_n$ be the optimal function for (LP2) (i.e. $f_n(x) = \sum_{i=0}^n a^*_ig(x)$, for $a_i^*$ optimizing the linear program).
My question is: under what conditions (if any) we can prove that $\lim_{n\rightarrow \infty} f_n$ converges and $f_{\infty} = f^*$? What can we say about the rate of convergence (obviously by the way the question is phrased this is impossible to answer, but are there examples where the answer is known)? Are there known results for the case where the linear program has many constraints?
As posed, LP1 is not even bounded unless $g$ is a linear combination of $c1,c2$, so there is nothing to talk about as long as the convergence is concerned. Also, the summation over integers is a too poor approximation to the integral to ensure anything unless you restrict your function space to analytic functions with bounded spectrum. So the answer to the question as currently posed seems to be "pretty much never". However, if you impose some additional restrictions, it may become meaningful, so I abstain from voting to close :-)
@fedja - thank you. I know that as phrased the question is almost meaningless - but part of my question is what conditions I should pose for convergence of the two programs. I will add your conditions, as well as more clarifications.
|
2025-03-21T14:48:31.898297
| 2020-08-24T08:56:13 |
369983
|
{
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"Wojowu",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/369983"
}
|
Stack Exchange
|
Is every homeomorphism approximately a product of homeomorphisms?
Let $\phi$ be a homeomorphism on $\mathbb{R}^{n+m}$, $\epsilon>0$, and $K\subseteq \mathbb{R}^n$ be a non-empty compact. Does there necessarily exist homeomorphisms $\phi_1,\phi_2$ on $\mathbb{R}^n$ and on $\mathbb{R}^m$, respectively, such that
$$
\sup_{x \in K}\left\|
\phi(x,y) -(\phi_1(x),\phi_2(y))
\right\|<\epsilon?
$$
In other words, is every homeomorphism on a product space approximately the product of homeomorphisms on its components?
Let $n=m=1$, $K$ be a unit square, and $\phi(x,y)=(y,x)$. I believe this is then false for $\epsilon<1/2$.
|
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