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2025-03-21T14:48:30.463570
| 2020-05-01T12:03:29 |
359064
|
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|
Stack Exchange
|
Maximizing the length of a sequence under constraints
Fix $\{w_n\}_n$ a sequence of positive real numbers, fix positive integers $N,K$, and fix $\eta>1$. I'm looking for a sequence of integers $\{k_n\}_n$ optimizing the following problem:
$$
\begin{aligned}
\min \sum_{i=1}^n \frac{w_i}{k_i} \\
\mbox{s.t.}\\
\sum_{i=1}^n k_ik & = K\\
i^{\eta} w_i & \leq k_{i} \qquad \mbox{ for } i=1,\dots,n\\
N & \leq n
\end{aligned}
$$
Are such problems studied? How to go about it? Since usually $n$ is bounded...
Where are $N$ and $\eta$ supposed to be in the optimization problem? Currently they're all $n$'s, which doesn't make sense. Also, what is $\delta$? Is $k$ fixed ahead of time too?
A distinct but related (now deleted) question was asked a few hours ago by user "probably a human", also using a username of the same style. Are you the same person behind these accounts?
Never heard of this person. What did their question ask? (I clicked on the link you posted but I cant see the post)
I think this question is more suited to: https://math.stackexchange.com/
I'll keep it here for a bit and if nothing I'll move it there. Thanks for the comment.
Presuming $k_i$ are constrained to be positive (which I'm assuming to be the case), this can be solved as a Mixed-Integer Second Order Cone Program (MISOCP). Specifically, the reciprocal in the objective function can be handled by use of a rotated Second Order Cone constraint for each term in the objective function.
I will assume that the "k" in the first constraint is either extraneous (i.e., a typo), or is a constant - in either case, that constraint would be linear. Presumably the "constraint" $N \le n$ isn't really a constraint, but a statement about the input values $N$ and $n$; but $N$ doesn't even appear. So I'll handle the essence of what I think the problem is, and assume the rest can be handled by fixing typos in the problem statement.
If a convex optimization modeling tool, such as CVX, CVXPY, or CVXR is used, handling of the reciprocal can be accomplished with a high level function (such as CVX's inv_pos), resulting in an under the hood transformation to SOCP formulation.
Here is CVX code (assume w is a column vector whose ith element is $w_i$), and let kk take the role of k in the 1st constraint, because I am using k as a (column) vector whose ith element is $k_i$.
cvx_begin
variable k(n) integer
minimize(w'*inv_pos(k)) % inv_pos(k) is vector of 1/k_i
subject to
kk*sum(k) == K
(1:n)'.^eta.*w <= k % this is a vector of n constraints
cvx_end
This requires an MISOCP solver, such as Gurobi or Mosek, or if using CVXPY, also CPLEX. Depending on the the input data, it may take a while for the solver to solve it (it is NP-hard, but only because of the integer constraint).
Here is how the reformulation to SOCP constraint works. The term $w_i/k_i$ is handled by introducing a new variable $t_i$, changing the term $w_i/k_i$ to $w_it_i$, and introducing the rotated second order cone constraint (to enforce $t_i = w_i/k_i$, accomp0lished via epigraph formulation due to minimization driving the inequality to be satisfied with equality at the optimum): $$\|1\|_2 \le \sqrt{k_it_i}, t_i \ge 0, k_i \ge 0$$
|
2025-03-21T14:48:30.463819
| 2020-05-01T12:16:26 |
359065
|
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"Dieter Kadelka",
"as1",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/359065"
}
|
Stack Exchange
|
Convergence of the expectation of a random variable when conditioned on its sum with another, independent but not identically distributed
Suppose that for all $n \in \mathbf{N}$, $X_n$ and $Y_n$ are independent random variables with
$$X_n \sim \mathtt{Binomial}(n,1-q),$$
and
$$Y_n \sim \mathtt{Poisson}(n(q+\epsilon_n)),$$
where $q \in (0,1)$, and $(\epsilon_n)$ is a deterministic sequence such that $\epsilon_n \to 0$ as $n \to \infty$.
Aim:
I am looking for a way to solve the following "signal extraction/estimation" problem, namely:
For a sequence $s_n \geq 0$ with $n s_n \in \mathbf{N}$ and $s_n \to 1$ as $n\to\infty$, show that as $n \to \infty$,
$$\frac{\mathbf{E} [ X_n \mid X_n + Y_n = n s_n ]}{n} = 1 - q + O(|s_n - 1|) + O(\epsilon_n).$$
Heuristic:
Here is why I believe it to be true. We know that $n^{-1} X_n$ and $n^{-1} Y_n$ are both approximately Gaussian and furthermore, if $Z_1, Z_2$ are independent Gaussians with means $\mu_1$ and $\mu_2$, and variances $\sigma_1^2$ and $\sigma_2^2$ respectively, then $Z_1 \mid Z_1 + Z_2 = s$ is also Gaussian and
$$\mathbf{E}[Z_1 \mid Z_1 + Z_2 = s] = \mu_1 + \frac{\sigma_1^2}{\sigma_1^2 + \sigma_2^2} ( s - \mu_1 - \mu_2),$$
ie, one apportions the difference between the expectation of the sum and the observed statistic according to the ratio of the variances.
If one naively assumes that this property can be carried over to the limits of $n^{-1} X_n$ and $n^{-1} Y_n$, then we can believe that
\begin{align}
\frac{\mathbf{E} [ X_n \mid X_n + Y_n = n s_n ]}{n} &= \mathbf{E} [ n^{-1} X_n \mid n^{-1}X_n + n^{-1}Y_n = s_n ] \\
&\approx 1 - q + \frac{q(1-q)}{q(1-q)+q+\epsilon_n} (s_n - (1-q) - (q + \epsilon_n) \\
&= 1 - q + O(|s_n - 1|) + O(\epsilon_n).
\end{align}
Attempt(s):
Exploit local limit theorem: my main attempt has been a brute force approach, trying to prove this directly by approximating the probability mass functions of $X_n$ and $Y_n$ by Gaussian densities using the local limit theorem, ie we can write
\begin{equation}
\frac{\mathbf{E}[ X_n \mid X_n + Y_n = n s_n ]}{n} = \frac{1}{n} \sum_{k=0}^n k \frac{\mathbf{P}[X_n = k] \mathbf{P}[Y_n = ns_n - k]}{\mathbf{P}[X_n + Y_n = ns_n]}.
\end{equation}
Each of the probabilities within the sum can be approximated by a Gaussian density with an error term which is $O(n^{-1/2})$ uniformly in $k$. Carrying this through is extremely messy however, and one will need to be extremely careful when being precise about approximating the Riemann sums which will appear with their corresponding integrals.
Try to find relevant tricks/results under theme of "signal extraction/estimation": essentially, the problem here is one of estimating/extracting a signal from an observation with additive independent (and approximately Gaussian) noise. It seems to me that this would be a problem well studied but searches of permutations of my above question yield standard undergraduate results for sums of i.i.d. random variables.
Specific questions:
Is it possible that there's a clever way to use the approximate Gaussian behaviour of $X_n$ and $Y_n$ to prove this result without the brute force approach of the local limit theorem?
Are there some key-words that may lead me to similar results in the signal extraction/estimation literature?
Do you have shown this for the special case $s_n \equiv 1$? $\epsilon_n$ is deterministic?
I haven't shown this for the case $s_n = 1$ -- do you expect this to be simpler? Yes, $\epsilon_n$ is deterministic.
@as1: Please explain your notation: What is $O(|s_n-1|)$ and $O(\epsilon_n)$? First you should replace $O(\epsilon_n)$ by $O(|\epsilon_n|)$. But this don't suffice: With the usual notation your assumption would imply that the conditional expectation is exactly $1-q$ for sufficiently large $n$ if $s_n \equiv 1$ and $\epsilon \equiv 0$. This seems highly improbable. A numerical calculation shows that for instance for $n = 100000$ and $q = 0.7$ the right side is $0.300000527537945$.
|
2025-03-21T14:48:30.464063
| 2020-05-01T13:21:48 |
359067
|
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"YCor",
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}
|
Stack Exchange
|
Affine projection of polynomials for a given set of points
(Not sure this question fits here, I will remove it in case it doesn't)
Let $F_{\mathrm{ML}}[x_1, \ldots,x_n]$ denote the set of multilinear polynomials over a finite field $F=\mathbb{F}_q$ (i.e. a sum of monomials, such that in every monomial the degree of each variable is less than or equal to $1$) and let $S \subset q^n\,(=F^n)$ be a set of vectors that lies in dimension $m$ (there are $m < n$ independent vectors in $S$).
Is it possible to find a projection $A : q^n \rightarrow q^m$ ($A$ depends on $S$) such that for all $f \in F_{\mathrm{ML}}[x_1, \ldots,x_n]$, there exists $g \in F_{\mathrm{ML}}[x_1, \ldots,x_m]$ such that for all $x \in S$, $f(x) = g(Ax)$? What are the conditions that $S$ has to satisfiy in order for such $A$ to exist? What if we remove the constraint about $S$ being in a subspace?
It is not difficult for some types of sets, for example, if $S = \{x \in q^n : x_{i} = 1: m < i \leq n \}$, then for each $f \in F_{\mathrm{ML}}[x_1, \ldots,x_n]$ it is possible to project $f$ to $F_{\mathrm{ML}}[x_1, \ldots,x_m]$ and the requirement holds.
OK, I'm erasing my comments (also this one, in a few minutes)— no need to reply
|
2025-03-21T14:48:30.464175
| 2020-05-01T13:50:49 |
359069
|
{
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"authors": [
"Dabed",
"Denis Nardin",
"GSM",
"Liam Keenan",
"Timothy Chow",
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"https://mathoverflow.net/users/43054",
"lush"
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|
Stack Exchange
|
Open conjectures and expected applications of homotopy theory to arithmetics
I hope this question is not too broad to be asked here; if it is, please feel free to close the question.
I'm currently near the end of my masters studies and subsequently search for a particular field of maths I want to focus on in the near future.
Recently, the idea of mingling homotopy theory with arithmetic geometry has been very inspiring to me; be it through the anabelian conjectures of Grothendieck, their natural generalization to étale topoi or étale homotopy theory...
What I feel I'm still missing though is a more general 'overview' of what maths is part of 'arithmetic homotopy theory/homotopical arithmetic geometry', what expectations people have in those fields et cetera.
What (other) parts of mathematics are part of the circle of ideas trying to inject homotopy theoretic machinery into arithmetic questions?
What kind of applications might one hope to achieve by these methods? I.e., what are some interesting (open) conjectures people are or have been working on within the area?
Given that $\infty$-categories (in the sense of lurie) are kind of made to build homotopy theory into the very foundation mathematics, it seems very plausible to me that they should come up in research of homotopical arithmetic geometry; hence I was wondering:
Are $\infty$-categorical methods entering the field recently and what would be good examples of research in this area using $\infty$-categories.
Interesting examples I'm aware of are the Exodromy paper by Barwick, Glasman and Haine and Schmidt & Stix's paper on Anabelian geometry with etale homotopy types.
I'd be really interested in any thoughts on these questions: (other) related papers, descriptions of conjectures, points to the literature, ...
I'm far from an expert about them, but the applications of THH to p-adic Hodge theory come to mind.
Does the algebraic K-theory of $\mathbb Z$ count?
I was always curious about this kind of questions. Given $L$ a finite extension of rational field $\mathbb{Q}$ and a scheme $X$ over $\mathbb{Q}$, let $X(L)$ be the set of its rational points. I was always wondering if there are homotopical methods to answer the question of (non)emptiness of $X(L)$.
thanks Denis and Timothy!
@GSM : https://arxiv.org/abs/1110.0164 comes to my mind; may be relevant
Could you tell me broadly speaking what roadmap/material you had to follow in your masters to learn this subject? Sorry for asking something unrelated and thank you
@DanielD. I essentially just studied everything related to Algebraic Geometry and Algebraic Number Theory my university offered and at one point just started reading into this direction by myself to be honest
This falls under the umbrella of what @DenisNardin mentions, but here some examples: Bhatt-Morrow-Scholze (https://arxiv.org/abs/1802.03261), Clausen-Mathew-Morrow (https://arxiv.org/abs/1803.10897), and Antieau-Mathew-Morrow-Nikolaus (https://arxiv.org/abs/2003.12541) are worth pointing out. There's currently lots of activity involving trace methods (K-theory, THH, TC) and arithmetic geometry being carried out by the authors I've listed above and many others whom I've neglected to mention.
|
2025-03-21T14:48:30.464425
| 2020-05-01T13:59:36 |
359070
|
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"authors": [
"Matthew Daws",
"Nik Weaver",
"Rupert",
"Uri Bader",
"YCor",
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"https://mathoverflow.net/users/15482",
"https://mathoverflow.net/users/23141",
"https://mathoverflow.net/users/406",
"https://mathoverflow.net/users/89334"
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"url": "https://mathoverflow.net/questions/359070"
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|
Stack Exchange
|
Seeking to understand meaning of "von Neumann spectrum" in a paper of Bader–Furman–Shaker
In attempting to understand the paper "Superrigidity, Weyl groups, and actions on the circle" of Uri Bader, Alex Furman and Ali Shaker (linked at Furman's page)
I find that towards the end of the proof of Lemma 2.2 they have a situation where $G$ is a locally compact second countable group, $X$ is a Lebesgue space on which $G$ has a measure-class-preserving action, and $Y$ is a probability space on which $G$ has a measure-preserving action, and they speak of an isomorphim (a Banach space isomorphism, presumably) between $(L^{\infty}(Y))^{G}$ and $(L^{\infty}(X \times Y))^{G}$, and then say that this induces a Lebesgue space isomorphism between the corresponding von Neumann spectra, and I'm just wondering if anyone can clarify what "von Neumann spectrum" means in this instance. Presumably not just a subset of $\mathbb{C}$ associated with some Banach space operator. It seems as though it's a little difficult to resolve this question just by Googling.
I think I could answer this, but probably @UriBader.is the best person to answer.
@NikWeaver I guess you should answer the question then (I'd be uncomfortable to answer a question explicitly about my own work)
@YCor really? If someone had a question about one of your papers you would be uncomfortable answering?
@NikWeaver I mean answering here on MO; I'm happy with any kind of question on my papers or remarks (including errata as Uri recently did). Maybe "uncomfortable" is exaggerated, but MO is not a place for private communications, so it's good sign that another person is prone to answer.
@YCor I've answered a question about one of my papers on MO, here and it wasn't uncomfortable for me. It seems to me that Uri can give the definitive answer if he wants to ... if not, maybe he can let me know this privately?
To put it differently, I would dislike a question that would be phrased as addressed to a specific user. In the question you link at, there is no way to guess reading the question that one of the authors is a MO user, and this also applies to this very question (even if the name is written down). Still, would find it unfortunate that anybody feels obliged to leave the answer to the author. And certainly it would make me uncomfortable that anybody refrains to answer any question because of my presence here.
Uri has in fact indicated he is happy to have a Skype chat with me in the next few days, and it may well be that he can clarify things, so in that case I guess I can remove the question.
I think it would be a little sad to remove the question, if an answer could otherwise be given. I spent all of 3 minutes thinking about this yesterday, but couldn't see an answer. I'd be curious to see what an answer was! One of the useful things about MO is that if you have a question, often other people have the same question! I like just browsing questions, and answers, and learning things along the way.
Thanks, @NikWeaver and YCor, for calling my attention and for your care. I intend to leave an answer later on and clarify some things related to the question. Unfortunately I am unable to do it right now. In the meantime, anyone (Nik?) who wishes to answer the actual question asked, namely what does the term "von-Neumann spectrum" stands for, you are most welcome to do so. In fact, I do not have any reference handy currently. The definition is probably in Takasaki I, but most likely without giving this specific terminology.
Let me explain what we mean by the term "von Neumann spectrum".
Before doing so, let me recall the better known Gelfand duality:
the functor $X\mapsto C(X)$, from
the category of compact Hausdorff topological spaces to the category of unital commutative C*-algebras, establishes an equivalence of categories. The functor in the other direction is called the Gelfand spectrum. In particular, the compact space associated with a unital commutative C*-algebra $A$ under this functor is called the Gelfand spectrum of $A$.
Consider now the category of Lebesgue spaces.
Here a Lebesgue space is a standard Borel space endowed with a measure class
and a morphism of such is a class of a.e defined measure class preserving Borel maps, where two such are considered equivalent if they agree a.e.
Recall also that a von Neumann algebra is a unital C*-algebra which is a dual space as Banach space. Let us say that a von Neumann algebra is separable if its predual (which is uniquely defined) is separable wrt the Banach space topology (note that an infinite dimensional von Neumann algebra is never separable as a Banach space, so this terminology should cause no confusion).
There is an obvious functor from Lebesgue spaces to separable commutative von Nuemann algebras, $X\mapsto L^\infty(X)$. By the term von Neumann duality (which is not standard, but it should be) I refer to the fact that this functor establishes an equivalence of categories. The functor in the other direction is called the von Neumann spectrum. In particular, the Lebesgue space associated with a separable commutative von Nuemann algebra $A$ under this functor is called the von Neumann spectrum of $A$.
Again, let me stress that the term "von Neumann spectrum" is not completely standard, but it should be. I currently do not have a handy reference for the above discussion, but any book dealing with the foundations of the theory of von Neumann algebras should cover it. In particular, the above should be discussed when one decomposes a general von Neumann algebra as a direct integral of factors over its center. Note that the center is a commutative von Neumann algebra. The measured space carrying the direct integral is its spectrum.
Before concluding this explanation, I want to briefly expand on something that was not explicitly asked, but is very much related. It is a crucial fact that the variety of objects in the category of Lebesgue spaces is quite dull: up to isomorphisms there is a unique atomless Lebesgue space. But this doesn't mean the category itself is dull, only that its reachness in its morphisms. By analogy, think of the category of separable infinite dimensional Hilbert spaces, where again you have a unique class of objects but a reach variety of morphisms. So really, the miracle is in the group of automorphisms $\text{Aut}(X)$ and things get interesting when you consider representations of groups into this target. Studying these is (one aspect of) Ergodic Theory. It turns out that when one studies such representations of locally compact second countable groups, the von Neumann duality extends equivariantly. This fact is known by the slogan "Mackey's point realization theorem". But I guess I went far enough and end this discussion now.
Finally, let me make some comments about the paper under consideration and the relevant framework.
First thing, you should know that this paper was not published, for some unfortunate circumstances. In particular, it was never refereed and it might have some rough edges. You should handle it with some care. Overall it is solid, but there might be some glitches in the presentation.
Second, the specific framework under which we considered lemma 2.2 was changed in time. We found that it is convenient to replace the "ergodic with unitary coefficient" assumption with the "metrically ergodic" assumption,
defined in section 2 here.
The latter is formally stronger, but we find it easier to handle (and generalize) in our later work.
In particular, an analogue of lemma 2.2 is given in lemma 3.7 here.
Uri may give a more detailed answer later, but a short answer is that "von Neumann spectrum" is also called "dynamical spectrum". (And the latter is easier to Google.)
|
2025-03-21T14:48:30.464912
| 2020-05-01T14:26:55 |
359074
|
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"Oskar",
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|
Stack Exchange
|
Literature on linear categories
I am trying to understand Deligne's 'Categories Tensorielles', and therefore I need some knowledge on linear categories. Looking at Wikipedia and nLab, I found some definitions and explanations, but I could not find any references to a book that covers this topic.
Can anyone recommend a good book in which linear categories are discussed?
Linear categories are not an independent subject of study, they usually appear in combination with other types of categories, for example abelian and/or monoidal. For all this stuff see:
P.Etingof, S.Gelaki, D.Nikshych, V.Ostrik, Tensor Categories (2015)
Note, that the concept of a linear category is a special case of that of enriched category. It is rather general concept, but it may be useful to look at the nlab article to understand the general picture.
That's exactly the book I am currently using. But they seem to assume the reader to be familiar with the concept of linear categories in the first chapter already.
@S.Farr I don't think so. See definition 1.2.2.
Thank you, somehow I missed that.
@S.Farr Btw, the difference between (pre)additive categories and linear categories is the same as the difference between rings and algebras. So you may be also interested in literature on additive categories. For this I recommend Borceux, Handbook of categorical algebra, vol 2. It is a good reference for abelian categories too.
Thank you, I will check it out!
|
2025-03-21T14:48:30.465156
| 2020-05-01T14:30:18 |
359075
|
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"Chris Gerig",
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|
Stack Exchange
|
A differential geometric proof of the Riemann--Roch theorem for lines
I am looking for a differential geometric version of the proof of the Riemann--Roch theorem
for Riemann surfaces, that is, $1$-dimensional compact complex manifolds. The proofs one usually finds are given in algebraic geometric terms, and can be seen as special cases of the sheaf theoretic approach of the general Hirzebruch--Riemann--Roch theorem. What I would like is a
specialization of the proof of the Atiyah--Singer theorem to the $1$-dimensional case.
You should ask these questions under my answer in that post, as this current thread is a duplicate.
The details of the heat kernel proof for Atiyah-Singer index theorem become much simpler for Riemann surfaces (using a compatible Riemannian metric): a not too difficult computation shows that the first non-trivial term of the asymptotic expansion of the heat kernel along the diagonal is given by a certain combination of the scalar curvature and the curvature of the line bundle. In dimension two, this is the only relevant term for computing the index. Details can be found in Roe's book on Elliptic operators, and most likely also in Gilkey's book Invariance theory.... . You should also have a look at the answers to the question What do heat kernels have to do with the Riemann-Roch theorem and the Gauss-Bonnet theorem?
Where exactly is this explained in Roe's book?
In the second edition of the book: Proposition 7.19, Example 11.16 and Question 11.21. Roe cites Kotake, An analytic proof of the classical Riemann-Roch Theorem.
|
2025-03-21T14:48:30.465299
| 2020-05-01T14:56:12 |
359078
|
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"Dmitry Vaintrob",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/359078"
}
|
Stack Exchange
|
The kernel of the isogeny $[n]$ of an abelian variety
Let $A$ be an abelian variety of dimension $g$ over a field $k$, and let $n$ be a positive integer such that $\mathop{\mathrm{char}}(k) \nmid n$. Is it true that the kernel of $[n]: A \to A$, considered as a finite flat group scheme over $k$ (i.e., the fiber product of the following diagram
$$
\require{AMScd}
\begin{CD}
\mathop{\mathrm{ker}}[n] @>{}>> A \\
@V{}VV @V{[n]}VV \\
k @>>{\varepsilon}> A
\end{CD}
$$
), is isomorphic to the constant sheaf defined by $(\mathbb Z/n\mathbb Z)^{2g}$?
No! Look up Tate modules.
If $k$ is a finite field or more generally finitely generated over its prime field, this cannot possibly be true for all $n$, because $A(k)$ is finite (resp. finitely generated), hence only has finitely many torsion elements. (The statement $\ker[n] \cong (\mathbf Z/n)^{2g}$ is equivalent to $An = A[n]$.)
It is however true if $k$ is algebraically closed — maybe this is what the OP had in mind?
|
2025-03-21T14:48:30.465402
| 2020-05-01T15:18:13 |
359081
|
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"H A Helfgott",
"Rajesh D",
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"url": "https://mathoverflow.net/questions/359081"
}
|
Stack Exchange
|
Minimizing total variation
Let $f:\mathbb{R}\to \mathbb{R}$ be a function of bounded variation. Define $\overline{f}(x)$ by
$$\overline{f}(x) = \limsup_{\mu(I)\to 0} \frac{1}{\mu(I)} \int_I f(x) dx,$$ where $I$ ranges over intervals containing $x$ and $\mu$ is the Lebesgue measure. It is known that the total variation $|\overline{f}|_{TV}$ of $\overline{f}$ is minimal among those of all functions equal to $f$ almost everywhere; see, e.g., Lemma 3.3 in https://arxiv.org/pdf/math/0601044.pdf.
Now, that should be a very classical result. Does anybody have an older or standard reference?
$I$ is an interval, as I said clearly. $\mu$ is the standard Lebesgue measure.
It doesn't matter whether you let $I$ range over open or closed intervals.
Oh as $I$ dimnishes, its still always be a neighbourhood of $x$?
Yes: "$I$ ranges over intervals containing $x$".
|
2025-03-21T14:48:30.465498
| 2020-05-01T15:19:06 |
359082
|
{
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"authors": [
"Dieter Kadelka",
"LSpice",
"Sergei Akbarov",
"erz",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/359082"
}
|
Stack Exchange
|
Integrable functions as elements of closed absolutely convex hulls of precompact sets of indicator functions
I am not a specialist in measure theory, so excuse me if this is simple.
Let $\mu$ be a finite measure on a set $X$ (for example, the Lebesgue measure on $[0,1]$). Integrable functions on $X$ can be defined as limits (in different senses, in particular, in some definitions as uniform limits, but this does not play an important role in what I want to ask) of summable step functions
$$
f=\lim_{n\to\infty}\sum_{i=1}^\infty\lambda_i^n\cdot\chi_{A_i^n},
$$
where $\lambda_i^n\in{\mathbb C}$, $A_i^n\subseteq X$.
Is it possible to control the set $\{\lambda_i^n\cdot\chi_{A_i^n};\ i,n\in{\mathbb N}\}$ in this equation so that it would not be "too big"?
For example,
Suppose $f\in L_1(\mu)$ and $|f(x)|\le 1$, $x\in X$. Do there exist a sequence of measurable sets $A_i\subseteq X$ such that
1) the set of indicators $\{\chi_{A_i};\ i\in{\mathbb N}\}$ is totally bounded (or, what is the same, precompact) in $L_1(\mu)$, and
2) $f$ belongs to the closed absolutely convex hull of the set $\{\chi_{A_i};\ i\in{\mathbb N}\}$ in $L_1(\mu)$?
Without going into detail: I think the problem is assumption 2). Is it allowed that $f$ belongs to the "closed" absolute convex hull? What about the usual elementary three-step definition of the integral: First define $\int f d\mu$ for simple functions (as your sums, but with finite sum), then for non-negative functions and then for arbitrary quasiintegrable functions?
Dieter, yes, the closed absolutely convex hull, of course! Excuse me, I'll correct this.
You ask first that the set ${\lambda_i^n\cdot\chi_{A_i^n} ; i, n \in \mathbb N}$ be not too big, but your proposed 'bigness' conditions seem to ignore the coefficients $\lambda$. Is that intentional?
@LSpice I had in mind that if the condition $|f(x)|\le 1$ is not fulfilled, then $\chi_{A_i}$ must be multiplied by some $C>0$.
Attempt number 2. Consider the case $f\ge 0$.
For $\alpha\in[0,1]$ let $A_{\alpha}=\{x\in X, f(x)\ge \alpha\}$, which is measurable. For $n\in \mathbb{N}$ define $f_n=\frac{1}{n}\sum_{k=1}^{n}\chi_{A_{\frac{k}{n}}}$. It is easy to see that $f_n\le f\le f_n+ \frac{1}{n}$, from where convex combinations of $\chi_{A_{\frac{k}{n}}}$ converge to $f$. We only have to show that $\mathcal{A}=\{\chi_{A_\alpha}\}$ is totally bounded.
Let $\alpha, \beta\in [0,1]$. If $\alpha\ge\beta$ then $A_{\alpha}\subset A_{\beta}$, from where $\|\chi_{A_\alpha}-\chi_{A_\beta}\|_1=\mu(A_{\beta})-\mu(A_{\alpha})$. Similarly, if $\alpha\le\beta$, then $\|\chi_{A_\alpha}-\chi_{A_\beta}\|_1=\mu(A_{\alpha})-\mu(A_{\beta})$, and so in both cases $\|\chi_{A_\alpha}-\chi_{A_\beta}\|_1=|\mu(A_{\alpha})-\mu(A_{\beta})|$.
Hence, $\chi_{A_\alpha}\to\mu(A_{\alpha})$ isometrically maps $\mathcal{A}$ into a totally bounded space $[0,\mu(X)]$, and so $\mathcal{A}$ is itself totally bounded.
In the case $f$ is complex-valued, decompose $f=g-h+i(p-q)$, where $0\le g,h,p,q\le 1$, and so each of them belongs to the closed convex hulls of a totally bounded collection of indicators.
However, you cannot get general $f$ into a single convex hull. Consider $X=\{-1,1\}$ and $f$ - identity map. There are just three non-zero indicators available: $\chi_{X}, \chi_{\{1\}}, \chi_{\{-1\}}$. Assume that $f$ is in the (closed) absolute convex hull of these three elements. If $|\alpha|+|\beta|+|\gamma|\le 1$, and $f=\alpha\chi_{X}+\beta\chi_{\{1\}}+\gamma\chi_{\{-1\}}$, then $1=f(1)=\alpha+\beta$, and $-1=f(-1)=\alpha+\gamma$, and so $2=f(1)-f(-1)=\beta-\gamma\le |\beta|+|\gamma|\le 1$.
Why $\sum_{i=1}^{m_n}|\lambda_i^n|\le 1$?
erz, I did not understand your reasoning about the mapping $\chi_{A_\alpha}\mapsto\mu(A_\alpha)$, but independently on this, the family ${\chi_{A_\alpha}}$ is totally bounded in $L_1(\mu)$ because in each sequence ${\chi_{A_{\alpha_i}}}$ we can choose a subsequence ${\chi_{A_{\alpha_{i_k}}}}$ that has limit in $L_1(\mu)$.
Are you saying that this is not true when $f$ has different signs at different points?
@SergeiAkbarov I've edited the answer to address both of your comments
erz, isn't it necessary to prove that the image of the map $\chi_{A_\alpha}\mapsto \mu(A_\alpha)$ is closed, for applying what you say about the isometry?
@SergeiAkbarov total boundedness is a metric space property; moreover it is inherited by any subset (see Theorem 4.3.2 in Engelking)
Ah! The distance is not changed. OK, thank you!
|
2025-03-21T14:48:30.465784
| 2020-05-01T15:53:48 |
359085
|
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}
|
Stack Exchange
|
About the current of finite mass
In Demailly's e-book Complex analytic and differential geometry,
chap3-(1.14) Proposition is stated as follows:
Every positive current $T=i^{(n-p)^{2}} \sum T_{I, J} d z_{I} \wedge d \bar{z}_{J}$ in $\mathscr{D}_{p, p}^{\prime+}(X)$ is real and of order 0, i.e. its coefficients $T_{I . J}$ are complex measures and satisfy $\overline{T_{I . J}}=T_{J, I}$ for all multi-indices $|I|=|J|=n-p .$ Moreover $T_{I, I} \geqslant 0,$ and the absolute values $\left|T_{I, J}\right|$
of the measures $T_{I, J}$ satisfy the inequality
$$
\lambda_{I} \lambda_{J}\left|T_{I, J}\right| \leqslant 2^{p} \sum_{M} \lambda_{M}^{2} T_{M, M}, \quad I \cap J \subset M \subset I \cup J
$$
where $\lambda_{k} \geqslant 0$ are arbitrary coefficients and $\lambda_{I}=\prod_{k \in I} \lambda_{k}$
My first question is :what is the definition of the absolute value of distribution $T_{I,J}$. Distribution is a global quantity , what is its pointwise value.
My 2-nd question is: how can we view the mass measure $\|T\|:=\sum\left|T_{I, J}\right|$ of a positive current as? What confuses me is the same reason as the first question(Distribution is a global quantity , what is its pointwise value.)
https://mathworld.wolfram.com/TotalVariation.html
|
2025-03-21T14:48:30.465887
| 2020-05-01T16:07:33 |
359087
|
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"Dmitry Vaintrob",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/359087"
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|
Stack Exchange
|
What is known about the functional square root of the Riemann zeta function?
Let us consider the Riemann zeta function $\zeta(s)$, where $s$ can take on values on the domain $\mathbb{R}_{>1}$:
$$\zeta(s) := \sum_{n=1}^{\infty} \frac{1}{n^{s}} .$$
I wonder what is known about the functional square root(s) of the Riemann zeta function defined on the aforementioned domain (*). In other words, I'm curious about the properties of the function(s) $f$ such that $$f(f(s)) = \zeta(s). \qquad \qquad (1)$$
Questions
Has a closed-form solution been found for $f$ in equation $(1)$ ?
If not (which I expect), have partial results been found for such a function? Properties like existence, (non)uniqueness, continuity, or results about the functional square root of the partial sums? $$f(f(s)) = \sum_{n=1}^{k} \frac{1}{n^{s}} \qquad \qquad k \in \mathbb{Z}_{>0}$$
If so, I would be grateful if you have some pointers to relevant articles or other sources.
(Cross-post from MSE.)
(*) Edit as per Gerald Edgar's answer, this condition should be changed. We must define $f$ on $(0, \infty) \cup X$ for some subset $X \subset \mathbb{R} \setminus (0,\infty) $. Then $f$ must map $(1,\infty)$ bijectively onto $X$, and $X$ iself onto $(0,\infty)$. Under these conditions, there is still a possibility that $f$ is both continuous and real-valued. I am interested in the properties of such an $f$.
Are you sure this is the question you want to ask? The zeta function on the domain you consider takes values outside this domain, so it cannot have a functional square root. In fact, for z with real part greater than 1, $\zeta(z)$ may have negative real part. See e.g. https://arxiv.org/pdf/1112.4910.pdf.
Well, it turns out that one needs to deal with fixpoints. I no longer have all the references on a website, but take a look at https://mathoverflow.net/questions/45608/does-the-formal-power-series-solution-to-ffx-sin-x-converge and my self-answers. I can see that there is a fixpoint in the reals between 1 and, say, 10.
@DmitryVaintrob I should also include that I'm only considering real $s>1$, so the values $\zeta(s)$ take are positive and real too
For real $s > 1$, $\zeta(s)$ maps $(1,\infty)$ bijectively onto itself. It is continuous and decreasing. A functional square root can therefore not be real-valued and continuous.
@GeraldEdgar Could you please elaborate on that? I don't see why such a root can't be real-valued and continuous at once
Note that $\zeta$ maps $(1,\infty)$ bijectively onto itself, $\zeta$ is continuous, $\zeta$ is decreasing on $(1,\infty)$.
Suppose $\zeta = f \circ f$ where $f$ also maps $(1,\infty)$ onto itself. A continuous injective function $f$ from an interval to an interval is either increasing everywhere or decreasing everywhere. (This is from the intermediate value theorem.) But in either case, $f\circ f$ is increasing, so $f \circ f \ne \zeta$.
Now of course we can define $f$ on some larger set, say $(1,\infty) \cup X$, where $X$ has the power of the continuum. Let $f$ map $(1,\infty)$ bijectively onto $X$ and $X$ bijectively onto $(1,\infty)$. Easy.
Ahh, I see. Thank you! I'll modify the question according to the answer you've given, because I'm still curious about the existence of such an $f$ defined on a larger set.
|
2025-03-21T14:48:30.466116
| 2020-05-01T16:18:52 |
359090
|
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"Christian Remling",
"Fedor Petrov",
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|
Stack Exchange
|
A question about comparison of positive self-adjoint operators
I have the following question but have no idea on its proof (one direction is trivial):
Let $A$ and $B$ be (bounded) positive self-adjoint operators on a complex Hilbert space $H$. Prove that
$$\limsup_{n \to \infty} \|A^n x\|^{1/n} \le
\limsup_{n \to \infty} \|B^n x\|^{1/n}$$
holds for every $x \in H$ if and only if $A^n \le B^n$ for each positive integer $n$.
Any suggestion?
Edit: I suspect the result maybe wrong, for example, if the two limits are equal, then it implies that $A=B$, too strong to be true; anyway, I don't know if the limit (in)equality is so strong. Maybe at most we can say $A^n \le B^n$ for large enough integer $n$.
And for the hard part, it suffices to show $A \le B$ by replacing $A$ with $A^n$ etc. and a similar limit inequality holds. A friend of mine using some trick arguments shows this holds when $H=\mathbb{C}^3$, a good evidence.
What is the source of this question?
You can find it here: https://artofproblemsolving.com/community/c7h1976172p13713536
The $\limsup$'s are actually limits, and they are equal to the maximum of the smallest closed support of the spectral measure of $x$ if $|x|=1$.
@Christian Remling Could you be more precise? I don't know how to prove your claim, though I already expected it's actually limits (of course not knowing as the maximum of the smallest closed support of the spectral measure). I forget a lot thing about spectral theory, it would be very helpful if you know and give an answer of my question above.
@Lao-tzu: This follows from the spectral theorem, by writing $|A^n x|^2 = \int t^{2n}, d\rho(t)$, with $d\rho(t) = d| E(t)x|^2$ being the spectral measure of $x$. (In other words, we are computing the $n$ norm of the function $t$ in $L^2(\rho)$, and $|t|p \to |t|{\infty}$ as $p\to\infty$.)
@Lao-tzu: No, I don't think it does, I thought about it and couldn't make any progress. The "spectral measure" is the $\rho$ from above, and what I'm saying is that $|t|_{L^{\infty}(\rho)}$ is the same as $\max S$, with $S$ defined as the smallest closed set with $\rho(\mathbb R\setminus S)=0$ (this exists and is unique).
@Christian Remling And does the original question follow easily from your observation then? Actually I don't understand what you mean by "the maximum of the smallest closed support of the spectral measure"...
@Christian Remling OK, so it seems the limit is equal to the norm of A, as the spectral radius r(A) = ||A||.
@Lao-tzu: No, the limit depends on $x$ (since $\rho$ does). You can say it's the norm of $A$ restricted to the reducing subspace generated by $x$, but that's just restating what I wrote above.
@Christian Remling I made a stupid mistake...
The condition $A^n \leq B^n$ for all $n$ defines the spectral order on the positive part of $B(H)$, usually written $A \preceq B$. It makes the positive part of any von Neumann algebra a complete lattice. It's equivalent to saying that $P_{[0,t]}(B) \leq P_{[0,t]}(A)$ for all $t > 0$, where
$P_S(A)$ is the spectral projection of $A$ for $S$.
Suppose $\limsup \|A^nx\|^{1/n} \leq \limsup \|B^nx\|^{1/n}$ for all $x$. The set of $x$ for which the left limsup is $\leq t$ is precisely the range of $P_{[0,t]}(A)$; this is easy to see if you take $A$ to be a multiplication operator. Thus the inequality implies $P_{[0,t]}(B) \leq P_{[0,t]}(A)$ for all $t$, i.e., $A \preceq B$.
(You can see that $P_{[0,t]}(B) \subseteq P_{[0,t]}(A)$ for all $t > 0$ implies $A \leq B$ by noting that $\langle f(A)x,x\rangle \leq \langle f(B)x,x\rangle$ for any simple function $f$ of the form $\sum a_i\chi_{[t_i, t_i + 1)}$. Taking a limit as $f$ approaches the function $t \mapsto t$ yields $\langle Ax,x\rangle \leq \langle Bx,x\rangle$. Also, $P_{[0,t]}(B) \subseteq P_{[0,t]}(A)$ for all $t$ implies the same for $B^n$ and $A^n$, so we actually get $A^n \leq B^n$ for all $n$.)
Nik, thanks for your answer! I've never heard about "spectral order". I trust your argument makes good sense but I cannot make sense of all of your argument. Any reference I can find the related material? Maybe you can add that in your answer.
Just to record for others' convenience: Lemma 4 and Theorem 3 of this article https://www.ams.org/journals/proc/1971-028-02/S0002-9939-1971-0276788-1/S0002-9939-1971-0276788-1.pdf answer the question completely, but uses other results which I haven't check.
|
2025-03-21T14:48:30.466410
| 2020-05-01T18:20:55 |
359096
|
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|
Stack Exchange
|
Regarding singular points of a fiber bundle
I had asked this question at math.stackexchange but couldn't find any answer; so I'm posting it here.
Let $X$, $Y$ be two projective varieties over $\mathbb{C}$, where $Y$ is smooth, and let $f:X\rightarrow Y$ be an etale locally trivial fiber bundle, with fiber a variety $F$. Is there a relation between the singular locus of $F$ and the singular locus of $X$? i.e. does the relation $F_{sing}= X_{sing}\,\cap\,F$ hold true?
Thanks in advance.
So, your $F$ is a particular fiber of $f$? In this case, yes. Your equality can be checked locally for the étale topology, so you can assume $X=Y\times F$, in which case the result is clear.
@abx : yes, $F$ is a particular fiber. Can you please give some ideas on how to see it, even in the trivial bundle case? I don't understand how to find the singularities of a product .
$Y$ is smooth, so the singular locus of $Y\times F$ is $Y\times \operatorname{Sing(F)} $.
|
2025-03-21T14:48:30.466506
| 2020-05-01T18:34:39 |
359099
|
{
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"Gerry Myerson",
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|
Stack Exchange
|
The equivalent proposition of Legendre's conjecture
Legendre's conjecture, proposed by Adrien-Marie Legendre, state that there is a prime number between $n^2$ and $(n+1)^2$ for every positive integer $n$.
My conjecture: Let $n$ be a positive integer, $p_1$, $p_2$ and $p_3$ be odd prime numbers, and $n\ge2p_1$, ($n\ge6$)
There is at least one prime number $p_2$, so that $n^2 < p_2 \le n^2 + n - p_1$, and there is at least one prime number $p_3$, so that $n^2 + n - p_1 \le p_3 < (n + 1)^2$
Firstly, at the start you seem to fix $n,p_1,p_2,p_3$ but later you assert some $p_2,p_2$ exist, you may want to clarify that. Secondly, what even is your question?
@Wojowu I interpreted their question as "For any $n \geq 6$, there are primes $p_1$, $p_2$ and $p_3$ such that $n^2 < p_2 \leq n^2 +n -p_1$ and there is another prime $p_3$ such that $n^2 + n - p_1 \leq p_3 \leq (n+1)^2$.
@Josh or maybe the question is, for any $n\ge6$, and any odd prime $p_1$ with $n\ge2p_1$, there exist primes $p_2$ and $p_3$ such that etc., etc.
It's not clear why the title, "equivalent proposition of Legendre's conjecture." The conjecture proposed here is clearly implies Legendre's, but I don't see any implication going the other way.
Your question does not have a question mark anywhere.
Your conjecture for sufficiently large $n$ is implied by Cramer's conjecture. In general though, conjectures like this unless they are coming from some specific application aren't that interesting. It is very easy to make many similar conjectures.
I agree with the general sentiment of this question, however Legendre's conjecture is interesting (more so than say the more demanding assertion that for all sufficiently large $n$ the interval $[n, n + n^{1/1000})$ contains a prime, say) because it is the order of magnitude where we can prove that such short intervals contain primes, assuming the Riemann hypothesis. The fact that what you can get from RH is so far from what's predicted by Cramer's conjecture is by itself astounding; somehow primes in short intervals is harder than RH.
|
2025-03-21T14:48:30.466666
| 2020-05-01T20:26:12 |
359106
|
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|
Stack Exchange
|
General upper bound of extinction probability
We consider here a Galton–Watson process with an offspring distribution $X$, where $\mathbb{E}X = \mu$ and $\operatorname{Var} X = \sigma^{2} < \infty$ and $q = \mathbb{P}(\text{extinction})$, i.e., the extinction probability.
I want to prove that the general upper bound of the extinction probability is given by:
$$q \leq 1 - \frac{\mu - 1}{\sigma^{2} + \mu^{2} - \mu}.$$
So if this helps, I can rewrite the above form as follows:
$$q \leq 1 - \frac{G_X'(1) - G_X(1)}{G_X''(1)},$$
where $G_X(s)$ is the probability generating function of the random variable $X$.
But here I am stuck.
I know the question is not well-asked, but hope that anyone can give me some hints.
I think you should change your username to a more neutral one.
Cross-posted at https://math.stackexchange.com/questions/3649820/general-upper-bound-of-extinction-probability . You should disclose your cross-posts to prevent duplication of effort.
@YCor Sure I'll change as soon as I can. Since I already changed my username around a month ago, I am currently not able to change, but 4 days later :)
We have
$$G(s):=G_X(s)=Es^X,$$
with the convention $0^0:=1$, where $X$ is a random variable with values in $\{0,1,\dots\}$ and $EX^2<\infty$. So, $G$ is a nonnegative nondecreasing convex function from $[0,1]$ to $[0,1]$ with nondecreasing $G''$. Also, $G(0)=P(X=0)$ and $G'(0)=P(X=1)$. So, excepting the case when $P(X=0)=0$ and $P(X=1)=1$ (and hence $G''=0$, which makes your inequality devoid of meaning), the extinction probability is the smallest root of the equation
$$G(q)=q. \tag{1}$$
Also, $G(1)=1$.
So, by the convexity of $G$ and (1), we have $G(s)\le s$ for $s\in[q,1]$ and $G(s)\ge s$ for $s\in[0,q]$. So,
$$G'(q)\le1.$$
So,
$$G'(1)-G(1)=G'(1)-1\le G'(1)-G'(q)=\int_q^1 G''(s)\,ds
\le \int_q^1 G''(1)\,ds=(1-q)G''(1),$$
which implies that indeed
$$q\le1-\frac{G'(1)-G(1)}{G''(1)},$$
as desired.
To illustrate this, here are the graphs $\{(s,s)\colon0\le s\le1\}$ (blue) and $\{(s,G(s))\colon0\le s\le1\}$ (gold) for the case when $X$ takes values $0,1,2$ with probabilities $\frac2{10},\frac1{10},\frac7{10}$, respectively, so that here $q=\frac27$.
How does $G''(s) \leq G''(1)$ inequality trivially hold? Is it a property of convexity?
@AsianbutnoChinese : $G''$ is nondecreasing, because $G'''\ge0$ (and actually $G^{(k)}\ge0$ for all $k=0,1,\dots$) on the interval $[0,1)$.
Ahh right it was not clever to ask. Thank you!
|
2025-03-21T14:48:30.466844
| 2020-05-01T20:34:00 |
359107
|
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|
Stack Exchange
|
What is known about computing all binary error correcting codes of given parameters?
Define a binary $(n, M, 2e + 1)$ code to be a code $C$ having $M$ code words in $\mathbb{F}_2^n$ whose minimum distance is $2e + 1$.
Are there any sources about using algorithms to find all given codes of certain parameters? I suspect that there are some ways to reduce the search space or some other clever techniques for finding all codes that are more subtle than a brute force search of all possible $M$ subsets of $\mathbb{F}^n_2$ and then computing the Hamming distance for every such subset. Certainly I would still expect that such a problem cannot be solved in polynomial time. I could not however, find any good sources on this material and was wondering if anyone knew of any. Partial answers or computational results and methods related to special classes of binary codes are welcome as well.
People are generally only interested in "good" codes. Which means maximum minimum distance $d$ for given $n,M$ or minimum length for given $M,d$ etc. Let $A(n,d)$ be the maximum $M$ for which an $(n,M,d)$ code exists.
Even for such codes the computational complexity of the problem is overwhelming. There are too many bad codes (if $d$ is small relative to $(n,M)$ as determined by bounds such as Hamming, Plotkin, Gilbert-Varshamov the search space is way too big).
There is the idea of code equivalence under the automorphism group of the code, which is used to reduce search complexity. At a very basic level if you find a code with certain parameters, and apply a permutation to the coordinates, this gives another code with the same paramaters. If the code has algebraic structure, you can do more.
The site http://www.codetables.de/ maintained by Markus Grassl has some tables of good codes, for example, and links to other tables. If you restrict to, for example, self-dual codes or codes with only two nonzero weights, etc. you can do more.
The tables by Litsyn, Rains and Sloane at http://www.eng.tau.ac.il/~litsyn/tableand/index.html of lower bounds on $A(n,d)$ may not have been updated in a long time. But a snippet from that site gives an idea about how hard the problem is.
Memo to algorithms specialists:
This file contains a large number of clique-finding problems. Construct the graph whose vertices represent binary strings of length $n.$ Join two vertices by an edge if and only if the Hamming distance bewteen the strings is at least $d.$ Then what we are interested in is the quantity $A(n,d),$ the size of the largest clique in this graph.
This file contains a large number of lower bounds on this clique size. If you can improve any of these entries or establish the optimality of any entries that we don't already know are optimal (these are indicated by a period after the number) please let us know (send us the clique too!).
Edit: I haven't used it for this purpose but the Magma package http://magma.maths.usyd.edu.au/calc/can do computations related to automorphism groups of codes. See here for details. There is a publicly accessible calculator at the first link I gave, but the memory and computation time allowed is limited. I am less familiar with Pari/GP but they may have a similar functionality.
|
2025-03-21T14:48:30.467198
| 2020-05-01T21:07:20 |
359111
|
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|
Stack Exchange
|
Finitely presented non-residually amenable groups without free subgroups
Does there exist a finitely presented group that does not contain a nonabelian free group and is not residually amenable?
I want it not to be residually finite.
What I really want is a finitely presented group that is not residually amenable and contains no free subgroup. Of course if the group has property T and is not residually finite then it is also not residually amenable.
Then the group of piecewise homographies constructed by Monod and Lodha-Moore should work. Indeed I think each such group $G$ (which is finitely presented and non-amenable) has all its proper quotients metabelian (and $G''$ is simple), so it's not residually amenable. It's locally indicable, has no infinite Property T subgroup.
And how do you know it doesn’t contain a free group?
Let us continue this discussion in chat.
Yes it has no free subgroup (as the whole group of continuous piecewise homographies of $P^1$ fixing $\infty$) — sorry I thought I wrote it
Let $G$ be the group of piecewise homographic self-homeomorphisms of the real projective line $\mathbb{P}=\mathbb{P}^1(\mathbf{R})$, and $G_\infty$ the stabilizer of $\infty$. So $G_\infty$ can be viewed as the group of self-homeomorphisms of $\mathbf{R}$, piecewise homographic (i.e. piecewise of the form $x\mapsto\frac{ax+b}{cx+d}$) with finitely many breakpoints.
The group $G$, or rather its $C^1$-subgroup was studied a long time ago as automorphism group of a Moulton plane (a fake projective plane); it was also considered by Greenberg.
It is not hard to show that $G_\infty$ has no nonabelian free subgroup (it's an immediate adaptation of the analogous result in the piecewise affine case, e.g., in Thompson's group $F$; see [M, Theorem 14] for a short sketch including all ideas).
Monod discovered that $G_\infty$ is non-amenable. This is quite involved but intuitively, the idea is as follows. Let $A$ be a countable dense unital subring of $\mathbf{R}$, and $G(A)$ the elements of $G$ with breakpoints in $A$ and acting on pieces as $\mathrm{PGL}_2(A)$. Then $G(A)$ induces on $\mathbb{P}$ the same equivalence relation as its subgroup $\mathrm{PGL}_2(A)$. Carrière and Ghys proved that this equivalence relation is non-amenable (in a suitable sense, pertaining to topological dynamics of groups). Then $G_\infty(A)$ also induces the same equivalence, the only difference being $\{\infty\}$ being apart; this tiny difference doesn't affect non-amenability of the action. Now a non-amenable equivalence relation can't be induced by an amenable group. So $G_\infty(A)$ is non-amenable. In this way Monod produced finitely generated subgroups of $G_\infty(A)$ that are not amenable (while without non-abelian free subgroup).
Next Lodha and Moore produced such subgroup that are in addition finitely presented, using explicit generators. I claim their group $\Gamma$ has its third derived subgroup $\Gamma'''$ containing every nontrivial normal subgroup.
Let $\Gamma_{0}$ be the subgroup of compactly supported elements in $\Gamma$ (the support being viewed in $\mathbf{R}$): this is the kernel of the homomorphism consisting in taking germs at $\pm\infty$ (which is valued in a metabelian group). Hence $\Gamma''\subset\Gamma_{0}\subset$ and in particular $\Gamma_0$ is non-trivial.
Their group $\Gamma$ contains a fixed-point-free self-homeomorphism $\rho$, so for every compact subset $K$ there exists $n$ such that $\rho^n(K)\cap K$ is empty. This implies, by a simple commutator trick (see for instance Lemma 3.3.4 in Burillo's book) that every normal subgroup of $\Gamma$ contains $[\Gamma_0,\Gamma_0]$. In particular, $\Gamma'''=[\Gamma_0,\Gamma_0]$.
Thus the intersection of nontrivial normal subgroups of $\Gamma$ is nontrivial. So if $\Gamma$ is not P (P any property passing to subgroups), then it's not residually P. So $\Gamma$ is not residually amenable.
Probably a closer look shows that $[\Gamma_0,\Gamma_0]=\Gamma'''$ is simple, but it would need some more details: the above is enough (it implies that $[\Gamma_0,\Gamma_0]$ has no nontrivial proper conjugacy $\Gamma$-invariant subgroup and in particular is characteristically simple). Maybe $\Gamma''$ itself is simple but this might be significantly more technical.
[CG] Y Carrière, E. Ghys. Relations d'équivalence moyennables sur les groupes de Lie. Comptes Rendus Acad. Sci. t.300 Sér.I no.19, 1985, 677–680. (French) link at Ghys' webpage
[LM] Y. Lodha, J.T. Moore. A nonamenable finitely presented group of piecewise projective homeomorphisms.
Groups Geom. Dyn. 10 (2016), no. 1, 177–200. arXiv link
|
2025-03-21T14:48:30.467755
| 2020-05-01T21:40:03 |
359113
|
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|
Stack Exchange
|
Class of lattices that excludes $M_3$?
It is well known that a lattice is distributive iff it excludes as a sublattice $N_5$ (the pentagon) and $M_3$ (three unordered elements with a top and bottom). Further, a lattice that only excludes $N_5$ is modular, and consequently a lattice excluding $N_5$ but including $M_3$ is modular but not distributive.
What, if anything, is known about the class of lattices that excludes $M_3$? Does it relate to classes of structures of any particular interest? (Note this is not the same as asking about "nonmodular varieties", since those are varieties which include $N_5$).
I found this question while looking for any studied/natural classes of lattices which are not distributive, and which also do not contain $M_3$ (i.e. not necessarily the class all lattices which avoid $M_3$). For anyone else looking for this type of thing: two classes of lattices with this property are join semi-distributive lattices, and meet-distributive lattices. See Paul H. Edelman, Meet-Distributive Lattices and the Anti-Exchange Closure, or for more information one could start at https://arxiv.org/pdf/1810.01528.pdf
This is a (slightly edited) copy of the answer I posted on math.SE to the question What do we call a lattice that does not have a sublattice the shape of the diamond $M_3$?:
Let $\mathbf K$ be the class of all lattices not containing $M_3$ (the $5$-element nondistributive modular lattice) as a sublattice; in other words, lattices in which every modular sublattice is distributive. It is easy to see that $\mathbf K$ can be characterized as the class of all lattices satisfying the following sentence $\varphi$:$$\forall u\forall v\forall a\forall b\forall c[(ab=ac=bc=u)\wedge(a+b=a+c=b+c=v)\rightarrow(u=v)]$$or equivalently$$\forall a\forall b\forall c[(ab=ac=bc)\wedge(a+b=a+c=b+c)\rightarrow(ab=a+b)]$$ or
$$\forall a\forall b\forall c[(ab=ac=bc)\wedge(a+b=a+c=b+c)\rightarrow(a=b)].$$
Inasmuch as $\varphi$ is a universal Horn sentence, it follows that $\mathbf K$ is closed under taking sublattices, direct products and reduced products; it is a quasivariety.
On the other hand, $\mathbf K$ can not be characterized by identities, and is not closed under taking homomorphic images. Bjarni Jónsson [Sublattices of a free lattice, Canad. J. Math. 13 (1961), 256-264, Lemma 2.6(i)] observed that, as a corollary of P. M. Whitman's work, elements $u,a,b,c$ of a free lattice satisfy the condition: if $u=ab=ac$, then $u=a(b+c)$. It follows that $M_3$ is not a sublattice of a free lattice, i.e, the class $\mathbf K$ contains all free lattices. Thus every lattice is a homomorphic image of a member of $\mathbf K$, and every identity which holds in all $M_3$-free lattices is a consequence of the lattice axioms. In view of this, we are not likely to find a much simpler algebraic characterization of $M_3$-free lattices than the one in the previous paragraph.
This is very nice, thanks! I have another argument that follows from covers of varieties that shows this is not a variety, which is why I wrote "class" and not "variety". I was wondering if it had been considered in any literature nonetheless. I'll leave this open in case anyone has an answer to that, but I do appreciate this answer as nice and clean.
Not a variety, but it is a quasivariety.
As bof says, the class $\mathcal K$ of lattices omitting $\mathbf M_3$ as a sublattice is a quasivariety that is not a variety. As bof also says, this implies that $\mathcal K$ is closed under the formation of sublattices, products, and reduced products, but the class is not closed under the formation of homomorphic images.
But $\mathcal K$ is closed under some types of homomorphic images.
A. (Projection onto a direct factor)
If $L\times L'\in \mathcal K$, then $L, L' \in \mathcal K$. Every quasivariety of lattices has this property, but there exist quasivarieties of algebraic structures that do not have this property.
B. (Quotients of finite members) If $L\in \mathcal K$ is finite, then any homomorphic image of $L$ is also in $\mathcal K$. This property typically fails for quasivarieties, even quasivarieties of lattices. It holds for $\mathcal K$ because $\mathbf M_3$ is projective in the class of finite lattices.
I guess A holds because $L$ and $L'$ are sublattices of $L\times L'$ because lattices have trivial sublattices. B seems harder. Is there a literature about $\mathcal K$? Does it have a name?
|
2025-03-21T14:48:30.468047
| 2020-05-01T21:43:46 |
359114
|
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|
Stack Exchange
|
What is the motive of $\operatorname{Bun}_G(X)$?
$\DeclareMathOperator\Bun{Bun}$Let $X$ be a scheme over algebraically closed field $k$, $G$ a reductive group and $\Bun_G(X)$ the stack of $G$ bundles on $X$. Write $[\Bun_G(X)]\in K_\text{st}$ for its class in the Grothendieck ring of stacks (see Ekdahl - The Grothendieck group of algebraic stacks).
What is $[\Bun_G(X)]$ for $X$ a curve? I've been told that this is a known result, but I might be wrong.
Secondly, is $[\Bun_G(X)]$ known when $X$ is a surface? If not in general then how about in “easy” cases, like $X=\mathbf{P}^2$ and $G=\operatorname{GL}_n$?
For the purposes of this question, “known” means expressible in terms of simpler classes like $[X^{(n)}]$.
Edit: It is true for curves (and I've heard for surfaces too but don't have a reference) that the cohomology of $\text{Bun}_{\text{GL}_n}(X)$ is freely generated by the chern classes of the universal bundle $\mathscr{U}$ on $X\times \text{Bun}_{\text{GL}_n}(X)$ (i.e. the components of $c_i(\mathscr{U})$ in $H^*(\text{Bun}_{\text{GL}_n}(X))$). In these cases, the Poincare polynomial is
$$P(\text{Bun}_{\text{GL}_n}(X),t)\ =\ \prod_{i=0}^n \prod_{k=0}^{2i} (1+(-1)^{k+1}t^{2i-k})^{(-1)^{k+1}\dim H^k(X)}$$
which probably gives partial information as to what the motive should be (for instance, replace $t^2$ by $\mathbf{L}$ in the denominator).
TeX-style quotes `` '' don't work in MathJax, unfortunately. I have edited to Unicode “”.
The paper "On the motivic class of the stack of bundles" by Behrend and Dhillon may be a start.
@WillSawin's reference: Behrend and Dhillon - On the motivic class of the stack of bundles (MSN).
Also this might be pedantic, but you need to complete the Grothendieck ring in some way before you make this computation, as $Bun_X(G)$ has infinitely many strata. (This is done in the reference above)
@WillSawin Thanks, I think that paper shows the answer was conjectured but not known in 2005. Hopefully it is known now and there is at least a conjecture for the surface case.
if someone had, it's very likely that they would cite that paper. I looked in the list of papers citing that one and didn't find it, but it's possible it's there. I actually don't think there is a conjecture even for $\mathbb P^2$. The structure of these spaces is just too wild - why would you expect a nice answer?
|
2025-03-21T14:48:30.468236
| 2020-05-01T22:07:48 |
359116
|
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|
Stack Exchange
|
Proof that Littlewood-Paley vertical square function is NOT bounded on L^infinity
The classical heat semigroup on $\mathbb{R}$ is given by
$$
W_t f(x)=\frac{1}{t}\int_{\mathbb{R}}e^{-\pi (\frac{x-y}{t})^2}f(y)dy, \qquad t>0.
$$
Then the Littlewood-Paley vertical square function is defined by setting
$$
g(f)(x)=\|\partial_t W_t f(x)\|_{L^2((0,\infty),\, tdt)}, \qquad x\in\mathbb{R}.
$$
It is known that $g$ is bounded on $L^p(\mathbb{R})$ for $p\in (1,\infty)$ and of weak type (1,1), but it is $\textbf{not}$ bounded on $L^\infty(\mathbb{R})$.
My question is: what is the simplest argument showing that $g$ is not bounded on $L^\infty(\mathbb{R})$.
I would appreciate any hints!
Take $f(x) = \int_{-\infty}^{x} e^{-t^{2}}dt \in L^{\infty}$. Now its square function is $+\infty$.
Paata, thanks so much! :]
|
2025-03-21T14:48:30.468327
| 2020-05-01T23:41:23 |
359121
|
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|
Stack Exchange
|
Why are faithful actions called faithful and who first called them faithful?
Sorry for this question. I asked this on MSE and HSM but no one answered and I decided to post it here that is full of experts.
I want to know why are faithful actions called faithful and who first called them faithful?
Definition: An action $G$ on $X$ is faithful when ${g_1 \neq g_2 \Rightarrow g_1 x \neq g_2 x}$ for some ${x \in X}$ (different elements of $G$ act differently at some point).
In French "fidèle" (= faithful) can be used in "fidèle à la réalité", that is to say, which I find translated into "true-to-life" or "true to the reality". I don't know if fidèle was used before or after faithful in the mathematical context, but to me the use of "fidèle" for injective is very natural (here it's injectivity of the map $G\to X^X$— note that faithful is widely used in category theory, for functors, in a similar meaning).
@YCor In English it can in certain circumstances as well, e.g. in "faithful reproduction".
@YCor “note that faithful is widely used in category theory, for functors, in a similar meaning”
The meaning is not just similar: Faithful group actions are literally a special case of faithful functors.
A group G can be regarded as being just a groupoid with a single object.
A group action of G on a set X is the same thing as a functor from G to Set sending the single object to X.
And the group action is faithful if and only if that functor is faithful in the sense of category theory.
@YCor is there any doubt that in category theory, the term has been adopted from group action theory?
@Kostya_I I'm not sure why you're asking me (I didn't suggest it was or wasn't adopted from group theory). I'd tend to believe it was indeed.
In the old days "representation" in group theory did not always mean linear representation. So it really meant what later came to be called an action of a group on any sort of object. If we use the word "representation" rather than "action", the point of the word "faithful" becomes clear. The sort of representation that is called faithful gives an accurate picture of the group.
Because it doesn't "kill" any non-trivial element? ;)
The German word is treu, and I would look to papers by Hermann Weyl for its introduction. E.g. Quantenmechanik und Gruppentheorie (1927, p. 16):
Da das Gruppenschema aus der Darstellung abstrahiert wurde, ist die Darstellung getreu, d.h. verschiedenen Elementen entsprechen verschiedene Abbildungen $U$, oder, was dasselbe besagt, $U(s)$ ist $= \mathbf1$ nur für $s = \mathsf1$.
or The theory of groups and quantum mechanics (1931, p. 114 — note the scare quotes):
The realization is said to be faithful
when to distinct elements of the group correspond distinct
transformations:
$$
T(a)\ne T(b)\text{ when } a\ne b.
$$
In accordance with the fundamental equation (2.1) the necessary
and sufficient condition for “faithfulness” is that $T(a)$ shall be
the identity only if $a$ is the unit element.
The same definition occurs in Wigner (1931, p. 79), van der Waerden (1931, p. 178; 1932, p. 32), and in French, Bauer (1933, p. 75).
How fitting that this answer was given by someone with a French first name and a German surname!
@Francois thanks a lot. Could you please translate the first German para? (It would be helpful for further viewers and makes your nice answer perfect.)
Sorry, Are you sure it first appeared in Weyl paper?
@C.F.G It translates almost the same as the second: “Since the group structure was abstracted from the representation, the representation is faithful, i.e. to different elements correspond different mappings , or, which means the same thing, () = 1 only for = .” Re: “first” I can’t be positive, but Weyl sounds like he’s introducing the term; earlier papers and books are few (Weyl’s own, Noether, Schur, Speiser, Burnside, Frobenius,...), and I couldn’t find “faithful” in any of them. Maybe you will!
|
2025-03-21T14:48:30.468667
| 2020-05-01T23:50:01 |
359122
|
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Stack Exchange
|
What is special to dimension 8?
Dimension $8$ seems special, as the partial list below might indicate.
Is there any overarching reason that dim-$8$ is "more special" than, say, dim-$9$?
Surely it isn't it, in the end, simply because $8=2^3$, but $9=3^2$? Or that $\phi(8)=4$ but $\phi(9)=6$?
The sphere packing problem in dimension
$8$: arXiv.
Annals Journal.
Quanta article.
Octonians (Wikipedia).
John Baez: The Octonions.
E8. See also Garrett Lisi's $E_8$ Theory (Wikipedia).
Wolchover, N. "The Peculiar Math That Could Underlie the Laws of Nature." Quanta magazine (2018).
De Giorgi's conjecture:
Abstract: "A counterexample for $N\ge9$ has long been believed to exist. ...we prove a counterexample [...] for $N\ge9$."
Del Pino, Manuel, Michal Kowalczyk, and Juncheng Wei. "Annals of Mathematics (2011): 1485-1569.
DP,MKM,JW. "On De Giorgi’s conjecture and beyond." PNAS 109, no. 18 (2012): 6845-6850.
Both the Snake-in-a-Box and the Coil-in-a-box problems have been solved
for $d \le 8$: arXiv abs.
For $d>8$, only lower bounds are known.
Bott Periodicity:
"period-$8$ phenomena"
(as per @Meow's comment).
The Simons minimal cone,
a $7$-dimensional cone in $\mathbb{R}^8$
(as per @DeanYang's comment).
I mean, the construction of $E_8$ uses $8= 2^3$ in a very clear way...
Triality (exceptional automorphisms of $\mathfrak{so}(8)$ over an algebraically closed field), leading to complications in Galois cohomology etc.
Bott periodicity should be in that list.
It's important to notice that, in an important way, $E_8$ is not 8-dimensional, but 248-dimensional. (But, in another sense, it is 8-dimensional; it all depends whether you're counting the dimension of the space in which the root system lives, or the dimension of the associated Lie algebra. Of course there are other ways of counting dimension that give still other answers.)
No big deal, but I think it's Wolchover
@kodlu: Thanks!
There's also the Simons minimal cone, which is a singular hypersurface in $\mathbb{R}^8$.
There is also an exotic 8-sphere and exotic spheres in most higher dimensions, so perhaps Milnor's exotic 7-spheres don't belong in this list?
Isn’t this just the strong law of small numbers? And, please, no Lisi.
Answers in What makes four dimensions special? are pretty good hope this question could get some awesome answers too
$7\not=8$......
In 8 dimensions there exist SIC-POVM that are not a group orbit of the Heisenberg group, namely the Hoggar lines.
the special role of octonions follows from the fact that $8$ solves $2^{n-2}=n^2$, as mentioned on MSE
Here a recent 8-dimensional example of a finite dimensional algebra answering a question on reflexive simple module: https://arxiv.org/pdf/2004.12208.pdf . It might be true that there are no such algebras of dimension smaller than 8, but that is not known it seems.
I think the two most magical numbers in mathematics are $8$ and $24$, and, sadly, I don't think there's likely to be an “explanation” for that.
BTW this question asks (over a a field of characteristic zero) whether isomorphism between two Lie algebras $\mathfrak{so}(q)$, and $\mathfrak{so}(q')$ are isomorphic, implies that $q'$ is equivalent to a nonzero scalar multiple of $q$. Here $q$ and $q'$ are nondegenerate quadratic forms in finite dimension $\ge 3$ (in dimension 2 it's false over most fields). The answer so far solves this positively, with the possible exception of the case of 8-dimensional $q,q'$.
@Mare's reference, abs-ified: Ringel - Simple reflexive modules over finite-dimensional algebras.
@YCor also asked an 8-dimensional follow-up to the linked question.
Even unimodular lattices exist only for signature a multiple of 8. https://en.wikipedia.org/wiki/Unimodular_lattice#Properties
One of the holonomy groups of Riemannian manifolds is Spin(7) which exists in 8 dimensions. https://en.wikipedia.org/wiki/Spin(7)-manifold
@ChuaKS: What is "$E_n$"?
There is the nice fact that del Pezzo surfaces have degree at most 9 = 1+8, where the degree is the self-intersection of the canonical class. If a del Pezzo surface has degree d (with d not equal to 8), then it's the blowup of 9-d points on P^2. You can get the exceptional root systems by studying (-1)-lines on del Pezzos, and in particular, you get the E_8-root system by studying del Pezzos of degree 1. I don't know how to fit this into the picture you're drawing, but maybe it fits somehow!
Quaternionic hyperbolic lattices exist first in dimension 8, as lattices in $Sp(2,1)$. Quaternionic hyperbolic lattices are distinguished in that they satisfy superrigidity and have property (T) even though they are rank 1 (proved by Corlette and Gromov-Schoen).
Bernstein's problem holds up to dimension 8 and is false in dimension 9 and higher (this is related to Simon's minimal stable cone in 8 dimensions).
https://en.wikipedia.org/wiki/Bernstein%27s_problem (see also De Giorgi's conjecture which holds up to dimension 8 and false in dims. > 8).
It's worth noting that the first four items on the list and Bott periodicity are somewhat related. The E8 lattice solves the sphere packing problem and can be realized as a system of integral octonions, and Bott periodicity can also be related to octonions, as Baez explains here.
The smallest finite dimensional $\mathrm{C}^*$-Hopf algebra aka algebra of functions on a finite quantum group that is neither the commutative nor a group algebra is the 8 dimensional algebra of functions on the Kac-Paljutkin quantum group.
(I should include all these wonderful dim-$8$ examples in my list, but I don't want to bump the question to the front page.)
$E_n$ is the set of $n+1$ tuples of integers $x_1,\dots, x_{n+1}$ such that $x_1 + \dots + x_n =3 x_{n+1}$, which carries the quadratic form $x_1^2 + \dots + x_n^2 - x_{n+1}^2$. This quadratic form has determinant $9-n$, so is nondegenerate if $n\neq 9$, and is positive definite if $n<9$. It is also even. Thus if $n=8$ it produces an even unimodular lattice $E_8$, if $n=7$ it produces $E_7$, if $n=6$ it produces $E_6$, if $n=5$ it produces $D_5$, if $n=4$ it produces $A_4$, and if $n=3$ it produces $A_1 \times A_2$.
The root lattices $A_n, D_n, E_n$ have determinants respectively $n+1, 4, 9-n$, so if $9-8=1$ makes $8$ a special dimension, this suggests that $0$ is also a special dimension.
Here is a fun one: there is an isomorphism between $\mathrm{O}(\mathbf{F}_2^6, q)$ and the symmetric group $\Sigma_8$, where $q$ is the quadratic form sending $v = (v_1 \ v_2)$ to $v_1 \cdot v_2$.
@ChuaKS Do you have a reference for the irreducibility of $\phi_{E_n}(x)$ at these particular values of $n$?
8 is even, so why are you even comparing it to 9?
Not specific to dimension 8, but the first irreducible Mutiple Zeta Value, ζ(3,5), appears in weight 8 (see F.Brown Gergen's lectures (part I slide 65) http://www.ihes.fr/%7Ebrown/GergenLectureI.pdf
Some special properties of dimension 8, in addition to the ones you identify:
Bernstein's problem holds up to dimension $n=8$. The only function of $\mathbb{R}^{n-1}$ whose graph in $\mathbb{R}^n$ is minimal is a linear function. This fails in dimension $n=9$, with failure due to the existence of the Simons cone in dimension 8, so it's related to your last bullet point.
There are 4 infinite families of Euclidean reflection groups, with exceptional ones only up to dimension 8. This is related to the existence of the exceptional simplex reflection groups and exceptional Lie algebras.
There are 4 infinite families of holonomy groups of Riemannian manifolds, with two exceptional cases of $G_2$ and $Spin(7)$, the latter being in dimension 8.
As pointed out by @YCor, triality holds for $Spin(8)$. $Spin(8)$ has three 8-dimensional irreducible representations which are permuted by the $S_3$ action associated with the symmetries of the $D_4$ Dynkin diagram.
Cohn and Kumar found various tight simplices including a maximal 15 point tight simplex in $\mathbb{HP}^2$ which is 8 dimensional. A simplex in this case refers to a collection of equidistant points.
There are several other examples in the comments of phenomena where 8 dimensions is the first dimension in which the phenomenon appears (or is known to appear), but I've listed examples that seem to be special to dimension 8 (and most seem to be connected to the phenomena that you've already identified).
The following is too long to leave as a comment, so I post it as an answer.
At least for the cases of the Bernstein problem for minimal graphs, Simons' theorem on stable minimal cones, etc., there is some evidence that dimension 8 is more likely a coincidence rather than having deep intrinsic reasons. In more general settings, that 8 is as magical as 7,6,5, or 4.
If we measure the area of hypersurfaces in Euclidean space with respect to norms other than the quadratic form ones, then the above magic dimension 8 breaks down, even with respect to norms that have high degrees of symmetry. (Of course, there is a lot of ambiguity here as in general there are many different ways to measure the area in say a Finsler norm. To avoid such ambiguity, here we assign different areas to different hyperplanes directly. The key word usually used in geometric measure theory is elliptic integrands/anisotropic area.)
Some references are as follows.
MR0467476 Reviewed Schoen, R.; Simon, L.; Almgren, F. J., Jr. Regularity and singularity estimates on hypersurfaces minimizing parametric elliptic variational integrals. I, II. Acta Math. 139 (1977), no. 3-4, 217–265. (Reviewer: William K. Allard) 49F22
MR1092180 Morgan, Frank The cone over the Clifford torus in R4 is Φ-minimizing. Math. Ann. 289 (1991), no. 2, 341–354. (Reviewer: Harold Parks) 49Q20 (53C42 53C65 58E15)
The anisotropic Bernstein problem by Connor Mooney, Yang Yang, https://arxiv.org/abs/2209.15551
Roughly speaking, the Almgren-Schoen-Simon paper shows that for the area-minimizing hypersurfaces in well-behaved normed spaces, the singular set has vanishing Hausdorff codimension 2 measure. (Note that for quadratic norm, the singular set is of codimension at least 7). Thus, it is imaginable that for general normed areas on Euclidean space, the Simons cone/failure of Bernstein problems should exist in lower dimensions. It is indeed the case by Frank Morgan, who shows that the cone over the Clifford torus is area-minimizing with respect to an $SO(2)\times SO(2)$-invariant normed area in $R^4$. This is sharp in view of the regularity result by Almgren-Schoen-Simon. Then the work of Mooney and Yang gives counterexamples to Bernstein-type theorems, similar in spirit to the connection mentioned by Ian Agol for the quadratic norm.
I think for such normed areas which are $C^\infty$ close to the quadratic area, such phenomena cannot happen, thus retaining the magical 8. However, it is conceivable that one should be able to find fairly symmetric norms so that the corresponding magical number is any of 8, 7, 6, 5, 4. Thus, the number 8, in this case, might not be as magical as one thinks, if one sees it in a more general setting, say Finsler geometry.
Remark 1: Here is an anecdote I heard from William Allard. Before Simons' paper on stable cones, Wendell Fleming once said that heuristically with dimension increasing, the portion of area inside a unit ball is decreasing. Thus, it's conceivable that with the dimension increasing, strange things can happen.
Remark 2: Notably, by Gary Lawlor's classical results on vanishing calibrations (https://bookstore.ams.org/memo-91-446/) a minimal cone of arbitrary codimension is area-minimizing as long as the second fundamental form and the cotangent of the focal radius of its link are relatively small compared to its dimension. Thus, when dimensions increase, area-minimizing cones should become more and more abundant in some sense, indeed verifying Fleming's heuristic. Lawlor's criterion accurately detects all area-minimizing hypercones known up to date, and can even prove the instability in cases of lower dimensional Simons/Lawson cones. Almost all non-special-holonomic area-minimizing cones known up to today, i.e., excluding holomorphic, special Lagrangians, associatives, etc., either can be proven to be minimizing using Lawlor's criterion or are precisely discovered this way.
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2025-03-21T14:48:30.469629
| 2020-05-02T00:15:33 |
359123
|
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|
Stack Exchange
|
Properties of Dirichlet series
I have a question about convergence and properties of Dirichlet series. it seems a bit interesting and different about the convergences of Dirichlet Series to me.
With $c\in [0,1]$,
$$f(n) = \pm 1,\qquad |\sum_{n\le x} f(n)-x^c|\le 2$$
Let A(x)= $\sum_{n\le x} f(n)$ be a Summatory Function.And F(S)=$\sum_n f(n)n^{-s}$ be a Dirichlet series.
So Why and How;
$\sum_n f(n) n^{-s}$ converges for $\Re(s) > c$ and $\sum_n |f(n)|n^{-s}$ converges for $\Re(s) > 1$ ?
Here is a referrence https://math.stackexchange.com/questions/1962712/polynomial-growth. i tried to solve it by using this theorem in this link.I also have looked a lot of books and article , and have searched to find it.
In general , i look at Analytic Number theory and Multiplicative Number Theory, but I couldn't see the form and way of solution of this question. May you help me ? you can also suggest me referrences or hints.Thanks.
You define $A$ and $F$, but you don't do anything with them. Since $|f(n)|=1$ for all $n$, your second series is just $\sum_nn^{-s}$, the Riemann zeta function. Are you really asking "why and how" the series for the zeta function converges for real part of $s$ greater than one?
i am curious about that; You only see that part of all Question? i can already see that the abscissa of absolute convergence is 1 because of Riemann Zeta. i dont have a problem there. Please Look at integrity of question. there is $x^c$ there and says that convergence is $\sigma_c=c$ of F(s) and i especially think that part and said why? we know that always $\sigma_c ≤ \sigma_a$ .so the $\sigma_c≤ 1$. what else have i to think about it?
ok. i see ! we know $0≤ \sigma_a-\sigma_c ≤1$. if we put $\sigma_a=1$ in the inequality, from here $\sigma_c ∈ [0,1]=c$.
Integrating/summing by parts allows you to prove that if $\sum A(n)/n^{s+1}$ converges then $\sum f(n)/n^s$ converges, which gives you the requested convergence. In general, the region where a Dirichlet series is conditionally convergent but not absolutely convergent can be up to width 1, but no larger.
Titchmarsh's Theory of Functions has a thorough chapter on Dirichlet series with many results like these (it also comes up in his Theory of the Riemann Zeta Function)
Very good comment. Thanks a lot:) I will investigate and analyze the book now.
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2025-03-21T14:48:30.469819
| 2020-05-02T00:18:56 |
359125
|
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|
Stack Exchange
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Construction of Kahler Einstein Metric of Poincare Type
I am reading Kobayashi's Kahler-Einstein metric on an open algebraic manifold. In this paper he constructs a Kahler-Einstein of Poincare type on an open manifold X' = X\D, where X is projective and D is a smooth divisor such that $K_{X}\otimes[D]$ is ample. He uses continuity method to do this.
In the first part of the proof, he needs to show the following result: $\tilde{\Delta}-Id$ is an isomorphism between $C^{k,\beta}(X')$ and $C^{k-2,\beta}(X')$, here $\tilde{\Delta}$ is the Dolbeault Laplacian with respect to some Kahler-Einstein metric of which the precise form is not too important. We are mainly interested in why $\tilde{\Delta}-Id$ has a bounded converse. He doesn't write too much here, so I have to guess a bit what he really wants to do here. Here is my guess about his thought.
We give a compact exhaustion $\Omega_{1}\subset \Omega_{2}\subset \cdots \Omega_{n}\subset \cdots$. For any $v\in C^{k-2,\beta}(X')$. we solve the following equations:
$$\tilde{\Delta}h_{i} - h_{i} = v ~~~in ~~\Omega_{i}$$
$$h_{i} = 0 ~~on ~~\partial\Omega_{i}$$
By maximal principle we get $||h_{i}||_{C^{0}(\Omega_{i})}\leq ||v||_{C^{0}(\Omega_{i})}\leq ||v||_{C^{0}(X')}$. For $j>k>i$, after restriction on $\Omega_{i+1}$, we have:
$$\tilde{\Delta}(h_{j}-h_{k}) - (h_{j}-h_{k}) = 0 ~~on ~~\Omega_{i+1}$$
Then by Schauder estimate:
$$||h_{j}-h_{k}||_{C^{k,\beta}(\Omega_{i})}\leq C||h_{j}-h_{k}||_{C^{0}(\Omega_{i+1})}\leq 2C||v||_{C^{0}(X')}$$
So after restriction on $\Omega_{i}$ $\{h_{j}\}$ is a bounded sequence in $C^{k,\beta}(\Omega_{i})$, so we can find a subsequence of $\{h_{j}\}$ which converges in some $C^{k,\beta'}(\Omega_{i})$, here $\beta'<\beta$. By using a diagonal argument we can find a subsequence of $\{h_{j}\}$ which converges to some element in $C^{k,\alpha}(X')$ where $\alpha<\beta$.
But there are some questions.
First I think we can only get an element in $C^{k,\alpha}(X')$ where $\alpha<\beta$ instead of $C^{k,\beta}(X')$.
Second I am not sure if we can get a good Schauder estimate here. Schauder estimate takes the following form:
$$||h||_{C^{k,\beta}(V')}\leq C(||h||_{C^{0}(V)} + ||Lh||_{C^{k-2,\beta}(V)})$$ where $V'\subset\subset V\subset\mathbb{R}^{N}$.
Notice that $C^{k,\alpha}(X')$ is defined to be $\{f|sup||f||_{C^{k,\beta}(V)}<\infty, V\in \mathcal{V}\}$ where $\mathcal{V}$ is a class of coordicate charts which are quite compatible with the Riemannian structure. In order to get a good Schauder estimate of the form:
$$||h||_{C^{k,\beta}(\Omega_{i})}\leq C(||h||_{C^{0}(\Omega_{i+1})} + ||(\tilde{\Delta}-Id)h||_{C^{k-2,\beta}(\Omega_{i+1})})$$
I think we need to go back to the coordinates, that is to say:
$$||h||_{C^{k,\beta}(\Omega_{i}\cap V)}\leq C(||h||_{C^{0}(\Omega_{i+1}\cap V)} + ||(\tilde{\Delta}-Id)h||_{C^{k-2,\beta}(\Omega_{i+1}\cap V)})$$
for $\forall V\in \mathcal{V}$ such that $V\cap \Omega_{i}\neq \emptyset$. But it is totally possible that $V\subset \Omega_{i}$. Then the last inequality is not likely to be true.
The reasoning above is based on my guess of Kobayashi's idea. It is possible that my guess is wrong. I would be really happy if someone could point out the real thought of Kobayashi.
I am not familiar with Holder space and Schauder estimates on open manifolds. Are there any references which talk about Schauder estimates for $V\subset\subset M$ where $V$ is a compact submanifold with boundary in some Riemannian manifold M?
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2025-03-21T14:48:30.470017
| 2020-05-02T03:04:40 |
359130
|
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|
Stack Exchange
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First moments of uniform distribution on a curve from (0,0) to (1,1) in two-space
The curve $\Gamma$ in $\mathbb{R}^2$ is defined by a continuous and monotonically increasing function $f(x)\in\text{C}[0,1]$, where $f(0)=0$, $f(1)=1$.
Let $(X,Y)$ is jointly and uniformly distributed on the curve $\Gamma$. Here we consider the expectation values of $X$ and $Y$, i.e., $E(X)=\int_{\Gamma} \frac{x}{M} ds$ and $E(Y)=\int_{\Gamma}\frac{y}{M}ds$, where $M$ is the length of the curve $\Gamma$.
I want to know the necessary and sufficient conditions of $E(X)$ and $E(Y)$ when such a curve exists.
Dual problems are whether such the curve exists with given values of $E(X)$ and $E(Y)$ and how to construct this curve.
For example, based on my own obeservation, the curve does not exist when both $E(X)$ and $E(Y)$ is stricly less than 0.5.
So, you want to prescribe the barycenter of the curve, right? How is this related to calculus of variations or convex optimization? Nothing is being optimized whatsoever... Why is it that such a cruve cannot exist if $E(X)$ and $E(Y)$ are less than $\frac 12$?
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2025-03-21T14:48:30.470120
| 2020-05-02T03:08:53 |
359131
|
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|
Stack Exchange
|
Write large $n$ as $n_1+\ldots+n_k\ (n_1<\ldots<n_k)$ with $\varphi(n_1),\ldots,\varphi(n_k)\in\{x^k:\ x\in\mathbb Z\}$
Let $\varphi$ denote Euler's totient function.
QUESTION. Is it true that for each positive integer $k$ large integers $n$ can be written as $n_1+\ldots+n_k$ with $n_1,\ldots,n_k$ distinct positive integers such that $\varphi(n_1),\ldots,\varphi(n_k)$ are $k$th powers?
In 2014, I conjectured that the answer is positive (cf. http://oeis.org/A237123).
For example,
$$101=1+15+85\ \ \ \text{with}\ \varphi(1)=1^3,\ \varphi(15)=2^3,\ \varphi(85)=4^3.$$
For any positive integer $k$, let $s(k)$ be the least positive integer for which any integer $n\ge s(k)$ can be written as $n_1+\ldots+n_k$ with $n_1,\ldots,n_k$ distinct positive integers such that $\varphi(n_1),\ldots,\varphi(n_k)$ are $k$th powers. My computation suggests that
$$s(2)=70640,\ \ s(3)=935,\ \ s(4)=3273.$$
Any ideas towards the solution of my above question? Your comments are welcome!
This question essentially asks whether the set of positive integers $n$ with $\varphi(n)$ a $k$th power is an asymptotic additive base of order $k$.
@zz7948 Any progress is welcome!
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2025-03-21T14:48:30.470213
| 2020-05-02T03:22:46 |
359133
|
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"Iosif Pinelis",
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|
Stack Exchange
|
Non-asymptotic results for M-estimators?
Does anyone know if there are any standard non-asymptotic results for M-estimators? I'm looking for finite-sample guarantees. Figured maybe someone here might know.
Uniform Berry--Esseen-type bounds on the rate of convergence to normality for $M$-estimators were obtained by Michel and Phanzagl -- see e.g. Refs. [14, 16, 17] in this paper. In the latter paper, a non-uniform Berry--Esseen-type bound for maximum likelihood estimators was given as well, with sufficiently detailed instructions (on page 1176) on how to the extend this result to $M$-estimators.
That is a very nice and interesting paper. However, are such bounds on convergence rates enough to call these "finite-sample guarantee"? What I mean is, convergence rates are still given in big-O, and so might hide very large constants. This is in contrast to cases where you can construct "exact" p-values (Fisher's exact test or the Wilcoxon rank sum test under a location-shift assumption). I guess I'm asking what the distinction is between "finite sample guarantees" and "exact" tests?
@RHahn : The mentioned results are apparently the state of the art. Yet indeed, the constants in the bounds, if made explicit, will likely be way too large to be useful for small samples. The problem is that it is very hard to obtain even the existing bounds. To put this into the right perspective, let us recall that the best constant in the uniform Berry--Esseen (BE) bound is still unknown even in the simplest case of sums of iid r.v.'s.
Previous comment continued: The best known constant in the nonuniform BE bound, even in this simplest case, is apparently very far from the best possible constant -- see e.g. shorturl.at/jmnwI, https://arxiv.org/abs/1301.2828, https://arxiv.org/abs/1302.0516. One can also see (https://projecteuclid.org/euclid.ejs/1460463653) how much effort went into getting somewhat moderate (and yet likely far from optimal) constants in the uniform BE-type bound for the Pearson correlation coefficient -- which appears much easier to deal with than the $M$-estimators.
Thanks for the extra information!
@IosifPinelis do the results in your paper also hold for vector-valued instead of scalar parameters?
@Glassjawed : No, those results don't hold for vector-valued parameters. Substantial modifications of the method have to be made to achieve that.
Which parts of the method would I have to modify? I figured out how to modify it for M-estimators (result is that I have a ratio of two expectations instead of the Fisher information in my bound).
@Glassjawed : I think modifications, quite possibly nontrivial, will be needed throughout.
@IosifPinelis I see. Yeah when I modified it for a specific M-estimator, I also don't think the results hold for exponential families and some of the distributions listed since the new estimator I have wasn't necessarily concave (although the log-likelihood is). Trying to figure out how to show the convergence holds for some common model families though. Should I just message you privately?
@Glassjawed : Right now I am occupied by working on other problems, but I think the case of a vector-valued parameter is something worth thinking about, and maybe I will consider it in the future.
|
2025-03-21T14:48:30.470441
| 2020-05-02T04:24:10 |
359135
|
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|
Stack Exchange
|
Compatibility between the source and the boundary condition for an Helmholtz-type equation
Let $\Omega$ an open, convex, bounded domain in $\mathbb{R}^3$, and let us fix also $z\in\mathbb{C}\setminus\mathbb{R}$. Given $\phi\in H^{3/2}(\partial\Omega)$, I would like to show the existence of a function $f\in L^2(\mathbb{R}^3)$ such that:
(i) $f(x)=0$ for $x\in\Omega$
(ii) $\|f\|_{L^2(\mathbb{R}^3)}\leq C\|\phi\|_{H^{3/2}(\partial\Omega)}$, for some absolute constant $C$ (independent on $\phi$)
(iii) the equation
$$\begin{cases}
(-\Delta-z)u=f\\
u\,_{|\partial\Omega}=\phi
\end{cases}
$$
admits a solution $u\in H^2(\mathbb{R}^3)$.
EDIT: It is sufficient to prove the following fact, which I think is known but I'm unable to find a reference. Given $\phi\in H^{3/2}(\partial\Omega)$, the Helmholtz equation
$$\begin{cases}
(-\Delta-z)u=0&x\in\Omega\\
u(x)=\phi(x)&x\in\partial\Omega
\end{cases}
$$
admits a solution $u\in H^2(\Omega)$, such that $\|u\|_{H^2(\Omega)}\lesssim\|\phi\|_{H^{3/2}(\partial\Omega)}$.
Thank you for any suggestion.
Concerning your edit, since $\phi\in H^\frac{3}{2}(\partial\Omega)$ then you can find an extension $g\in H^2(\Omega)$; this is the trace Theorem and your domain is not pathological, so I think it is not hard to find a reference, e.g. a book of Triebel. Then you have to solve $(-\Delta - z)v = (\Delta + z)g\in L^2$, where $u=g+v$; here $v$ is unknown and $g$ is given. Existence and regularity is known, e.g. Evans' book on PDEs.
|
2025-03-21T14:48:30.470553
| 2020-05-02T05:43:18 |
359137
|
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|
Stack Exchange
|
Recreational mathematical papers
Sometimes it is nice to get a less technical paper on mathematics to read and learn something different for a change. These papers often make us discover some new curiosity, to think about the process of learning mathematics or even about mathematics itself and this could be a really nice way to quickly 'scape' our daily problems to amuse ourselves with what I will call here recreational mathematics. As an example, I could mention this paper discussing our unability of seeing beyond three dimensions, which is something really curious. What are other interesting recreational mathematical papers to read during our break time?
Note: Maybe the term 'recreational' fits better when applied to puzzles or something like this but I didn't know what else to call it.
the Wikipedia entry on this topic has a great many pointers, including a journal devoted entirely to recreational math papers; for MO this question seems a bit unfocused, don't you think so?
Sounds like a "here's my opinion, what else" post, which somewhat constraints the replying to follow your way of thought. Actually for me the way I like to learn about "recreational" mathematics is to think about them rather than reading (this morning I thought a while about this question, which I can consider as doing "recreational math during my break time"— also assuming that one really wants to draw a line between one's time for research and one's math for break.
There is a series of books entitled The Mathematics of Various Entertaining Subjects, Volume N, where so far $N$ has taken on the values $1,2,3$.
I’m voting to close because this is a very, very wide “big-list” question: there are hundreds of papers out there that fit this bill, and nothing apart from personal preference to make any of them more appropriate to this question than any others.
Two of my favorite recreational papers are:
J-F Mestre, R Schoof, L Washington and D Zagier,
Quotients homophones des groupes libres. Homophonic quotients of free groups
Experiment. Math. 2 (1993), no. 3, 153–155.
Norman Wildberger, Real fish, real numbers, real jobs, The Mathematical Intelligencer 21(2), (June 1999), 4-7. Here is a youtube recording of the author reading this paper.
In fact the Intelligencer and the American math monthly have many papers of this type. Also a mainstream mathematical journal St Peterburg Mathematical Journal has a permanent department "Easy reading for professionals".
|
2025-03-21T14:48:30.470797
| 2020-05-02T05:58:47 |
359138
|
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|
Stack Exchange
|
Number triangle
This question arose just out of curiosity. Note the triangle of 0-1's below, whose construction is as follows. Choose any number, say 53 as done here. The first line of the triangle is the binary representation (without any extra zeros padded). The next row of triangle is obtained by XOR-ing (adding modul0 2) two consecutive digits of above row, and so on till completion.
$$
1~~1~~0~~1~~0~~1\\
0~~1~~1~~1~~1\\
1~~0~~0~~0\\
1~~0~~0\\
1~~0\\
1
$$
Now read off the numbers at the boundary of the triangle, clockwise. In the above example, that would be $1~1~0~1~0~1$, $1~1~0~0~0~1$ and $1~1~1~1~0~1$. That would be the triplet $(53, 49, 61)$. Note that if one had started with binary representation of 49 or 61 as the first row, the resulting triplet will be the same set of three numbers. In some sense, this is well-defined triplet.
Now, consider the number 13. We get the triangle:
$$
1~~1~~0~~1\\
0~~1~~1\\
1~~0\\
1
$$
Reading off numbers clockwise gives $(13,13,13)$, which is somewhat interesting. The same happens for numbers like 11, 39, 57 and so on. I wrote a code on python to get this sequence of special numbers till 500000. The graph below (which looks piecewise linear) depicts the number of integers less than a given integer which have this special property.
Here are a few more observations/questions:
Pasted the first ten numbers of this sequence (11, 13, 39, 57, 83, 91, 101, 109, 151, 233, 543, 599) into OEIS, and it did not result in anything. Is there any literature on this?
The sequence can be seen as a set of solutions to a linear system of equations in $F_2$, as XOR operation is equivalent to addition in $F_2$. Could that be used to prove that this sequence is (in)finite?
The same can be done for a representation in any base. How would that compare to the binary one?
How does one explain the graph?
Is the sequence of any serious research interest, in general?
Writing at the beginning that xoring= adding mod 2 would save time (I googled to find the definition).
Have you counted, among those numbers, the palindromic ones? (i.e., those for which the resulting triangle has a symmetry group of order 6)
@YCor, thanks for the comments. Have edited XOR-ing accordingly. Have not checked for palindromes yet. Will do that too and post result in the next comment.
@YCor, there are no palindromes (in binary representation) till 500000.
Note that reversing the binary expansion induces an involution on the your set of numbers, whose fixed points are the palindromes, exchanging 11—13, 39—57, 83—101, 91—109, etc. Maybe you can test palindromes until much further than 500000, since a palindrome is governed by twice less digits.
@YCor, thanks for that nice observation. Will try that.
OK, thinking more: say you're considering triangles with $n$ vertices on each large edge. You're first counting those 0-1 (mod 2) valued triangles with the property that the sum is zero on each triangle. The number of solutions of this is then a power of 2. This remains true if you add any symmetry condition, and also if you impose an affine condition, such as the condition that extreme vertices are marked with $1$.
For every $n$ the number of solutions with $n$ binary digits is thus a power of 2, so it would make more sense to compute, instead of the number of solutions $\le m$, the number $2^{u_n}$ of solutions in $[2^n,2^{n+1}[$, which should be computable in terms of ranks of some matrices. Maybe these matrices have a homological interpretation, I'm not sure.
@YCor, I am not sure if I completely understand your comment relating to constructing 0-1 triangles with n-vertices, apart from basic similarity to the original question. Guess what you're suggesting is to view the graph in log-scale?
To start with, compute for $n<19$ the number of solutions $k$ such that $2^n\le k<2^m$ (i.e., with exactly $n+1$ binary digits). it should be a power of 2, do you confirm?
BTW I checked there's no palindrome, by arguing around the center of the triangle. Still, if you allow numbers starting with zero, you have a few: the 0 triangle, and only one other for each $n=3m+1$: the triangle (drawn for $n=7$) $$\begin{matrix}0&&1&&1&&0&&1&&1&&0\&1&&0&&1&&1&&0&&1&\&&1&&1&&0&&1&&1&&\&&&0&&1&&1&&0&&&\&&&&1&&0&&1&&&&\&&&&&1&&1&&&&&\&&&&&&0&&&&&&\end{matrix}$$.
@YCor, can you elaborate your logic about non-existence of palindromes?
And, also the logic for the number of solution being power of 2. I am certainly missing some logic here.
It's not logic, it's interpreting the problem as solving a system of linear equations over $F_2$. Hence basic linear algebra.
My reasoning was not correct in excluding palindromes (by mistake I assumed that reversed little triangles also sum to zero). Reasoning from the center remains anyway probably the right way to get classification. Probably the resulting dimensions are related to the homology mod-2 of the space consisting of this triangulation, with only tiling triangles with vertex on the bottom being filled. This would justify that the number $2^{u_n}$ of solutions of your problem in $[2^n,2^{n+1}[$ should have $u_n$ quadratic in $n$ (with discussion on $n$ mod 3), experiments can lead to the right conjecture.
@YCor, thanks for the clarification. I'll think about the reasoning surrounding palindromes. One complete classification of solutions between two powers of 2 is the set of solutions to a set of linear equations, which I've mentioned in the question. Number of solutions also being a power of two, is simple. My bad. :)
My expectation $u_n$ quadratic isn't correct, it's simpler. It seems $u_{n+3}=u_n+3$ for the counting without symmetry. Precisely here $u_n$ is the number of ways to assign to each vertex of a triangulation as in your picture, with $n$ vertices on the top (hence $n(n+1)/2$ vertices), a boolean number so that every upside down small triangle sums to 0. If $v_n$ counts those with rotational symmetry, we get $v_{n+3}=v_n+1$; precisely in this case, the palindromic ones are those with 0 on the corners, and you count the non-palindromic ones. Counting palindromic should involve discussion mod 2.
PS what I asserted in the previous comment implies in particular that the sum (mod 2) of any two solutions is a palindrome. That is, the solutions (with rotational symmetry and 1 on corners) form an affine space whose linear part is the set of palindrome solutions (necessarily with 0 on extremities). For instance, $[1, 0, 1, 0, 0, 1, 1]+[1, 0, 1, 1, 0, 1, 1]=[0,0,0,1,0,0,0]$.
When you work mod 2, addition is the same as subtraction. So the second row gives the first differences of the first row, the third row gives the second differences, and so on. Then you can recover the top row from the left side as a sum of binomial coefficients. E.g., for top row 1101, the left side is 1011, so $1+{n\choose2}+{n\choose3}$ gives $1,1,0,1$, respectively, for $n=0,1,2,3$, respectively (working mod 2). So you can write down what a "special" number is, in terms of binomial coefficients.
A cool question. You should ask how many triangles there are of each size having the properties you care about. This number will be a power of two since these triangles are a vector spaced over $F_2$.
The nonexistence of palindromic solutions is trivial. Let $a$ be the entry immediately to the right of the $1$ in the upper left corner, let $b$ be the entry immediately below that $1$ (so, the first entry in the 2nd row). For a palindrome, $a=b$. But by construction of the triangle, $1+a=b$. Contradiction.
@GerryMyerson, thanks for that nice proof, and the earlier observation.
It seems that solutions do not exist only for length of the form $2^k+1$.
What I checked the other day is that the number of solutions (rotation invariant and allowing 0 on corners) is $2^{u_n}$ with $u_n=n+v_n$, and $v_n$ 3-periodic with $v_0=v_1=1$ and $v_2=-2$. It follows that the number of solutions with $1$ on the corner is either $0$, or $2^{u_n-1}$, say $t_n2^{u_n-1}$ with $t_n\in{0,1}$. I expected $t_n$ to depend on $n$ modulo $2$, but actually it seems to rather depend on $n-1$ being a $2$-power. So the counting problem is completely settled. It's easy to describe an algorithm to enumerate the solutions (allowing 0 on corners) passing from $n-3$ to $n$.
@YCor, your comments look very interesting, but are tough to read as written. Would you be willing to promote them to an answer, so that they can be read full size and all at once?
@LSpice OK I finally did it (time-consuming, this is why I was hesitating at writing down... what I said about palindromes (namely that all rotational-invariant solutions with 0 on corners )are palindromes seems suspicious by the way.
@MaxAlekseyev, thanks for that interesting observation.
@GerryMyerson, guess your logic regarding non-existence of palindromes holds good for any base-representation.
Michael De Vlieger, an OEIS contributor, made a visually stunning image that contains hundreds of these triangles (with top row encoded by $1, 11, 13, 39, 57, 83, \dots$).
@PeterKagey, wow! Didn't expect that. And, thanks to Michael De Vlieger.
$\def\Ker{\operatorname{Ker}}$Let me try to rewrite my answer --- at least, for clarity reasons. Thanks to Pat Devlin and YCor for stimulation.
First, several definitions. A $k$-triangle is a triangle constructed in the described way. We work over $\mathbb F_2$; the $k$-dimensional coordinate space is denoted by $V_k$ (so, any side of a $k$-triangle is a vector in this space). There is a linear operator $\varphi_k\colon V_k\to V_k$ mapping the top vector of a $k$-triangle into its right vector. We have $\varphi_k^3=1$, so $V_k=U_k\oplus W_k$, where $U_k=\Ker(\varphi+1)$ and $W_k=\Ker(\varphi^2+\varphi+1)$ (recall that $1=-1$). Denote $u_k=\dim U_k$ and $w_k=\dim W_k$ (one can see that $w_k$ is always even). For every $v\in V_k$, its $U_k$- and $W_k$-part are $x_u=(\varphi_k^2+\varphi_k+1)x\in U_k$ and $x_w=(\varphi_k^2+\varphi)x\in W_k$ (recall that $x=x_u+x_w$).
We say that vectors in $U_k$ are stable. Vectors starting with $1$ are proper. So, we are interested in the number of proper stable vectors, and it is either $2^{u_k-1}$ or $0$, depeding on whether $U_k$ contains a proper vector or not.
Observation 1. Each vector in $U_k$ has equal first and last coordinates. So, for $k\geq 2$ the space $W_k$ contains a vector with distinct first and last coordinates. The triangles of such vectors have all three sides in $W_k$ and have two ones at the corners. Hence, there are vectors in $W_k$ starting with $1$ and endind with $0$, starting with $0$ and ending with $1$, and also starting and ending with $1$.
Introduce also the following mixing operator. If $a,b\in V_k$, then $\mu_{2k}(a,b)=(a_1,b_1,a_2,b_2,\dots,a_k,b_k)\in V_{2k}$. If $a\in V_{k+1},b\in V_k$, then $\mu_{2k+1}(a,b)=(a_1,b_1,\dots,a_k,b_k,a_{k+1})\in V_{2k+1}$. For $c\in V_n$, denote by $o(c)\in V_{\lceil n/2\rceil}$ and $e(c)\in V_{\lfloor n/2\rfloor}$ the unique vectors such that $\mu_n(o(),e(c))=c$.
The indices will be omitted when they are clear.
Finally, for a vector $c\in V_{k+1}$ we denote by $c\rangle, \langle c\in V_k$ the vector $c$ without the last coordinate and without the first coordinate, respectively. By $\overleftarrow c$ we denote the left cyclic shift of $c$, i.e., $(c_2,c_3,\dots,c_1)$.
Observation 2. The sum of the three digits in the vertices of a $3$-triangle is zero. By induction, the same holds for any $(2^k+1)$-triangle.
Corollary 1. There are no proper stabe vectors in $U_{2^k+1}$.
Corollary 2. Ay $k$-triangle falls into four sparse triangles (of sizes $\lceil k/2\rceil$, $\lfloor k/2\rfloor$, $\lfloor k/2\rfloor$, and $\lceil k/2\rceil-1$).
Corollary 3. (a) Let $a,b\in V_k$, and let $\varphi_{2k}(\mu(a,b))=\mu(c,d)$. Then $c=\varphi_k(b)$ and $d=\varphi_k(a)+c=\varphi_k(a+b)$. The first claim is clear; the second follows from the observation that a sparse $k$-triangle with $d$ on the right has $a+b$ on the top.
(b) Let $a\in V_{k+1}$, $b\in V_k$, and let $\varphi_{2k+1}(\mu(a,b))=\mu(c,d)$. Then $c=\varphi_{k+1}(a)$ and $d=\varphi_k(b+\langle a)=\varphi_k(b)+c\rangle$.
Hugh. Now come to the problem itself.
Assume first that $n=2k$ is even. Then $c=\mu(a,b)\in V_{2k}$ lies in $U_{2k}$ iff $a=\varphi_k(b)$ and $b=\varphi_k(a+b)$, so $b=\varphi_k^2(b)+\varphi_k(b)$, i.e., $b\in W_k$ (and hence $a\in W_k$ as well). Therefore, we have a one-to-one correspondence $W_k\to U_{2k}$ given by $a\mapsto \mu(a,\varphi^2(a))$, and proper vectors correspond to proper ones.
Therefore,
$$
u_{2k}=w_k=k-u_k,
$$
and the number of proper stable $2k$-vectors is $2^{w_k-1}$ by Observation 1.
Now comes a bit harder case when $n=2k+1$ is odd. A vector $c=\mu(a,b)\in V_{2k+1}$ is stable iff $\varphi(a)=a$ and $\varphi(b)=b+a\rangle$, which rewrites as $a\rangle=(\varphi+1)(b)=\varphi^2(b_w)$. Similarly, we get $\langle a(=\overleftarrow{a\rangle})=\varphi(b_w)$.
So, basicaly we need to search for (proper) $a\in U_{k+1}$ such that $a\rangle\in W_k$; each such will lead to $2^{w_k}$ (proper) stable vectors of the form $\mu(a,b)$, where $b_u=\varphi(a\rangle)$. In other words, we need to find the dimendsion of $W_k\cap (U_{k+1}\rangle)$ and check whether it contains proper vectors.
But in fact $U_{k+1}\rangle \subseteq W_k$. Indeed, take the $(k+1)$-triangle with first row $a$ and remove that first row. We will get the $k$-triangle with top vector $\langle a+a\rangle$, right vector $\langle a$ and left vetor $a\rangle$ which sum up to $0$, as desired.
Hence we get
$$
u_{2k+1}=u_{k+1}+u_k.
$$
Now the small values yield the dimensions claimed by YCor, namely $u_k=a,a+1,a$ when, respectively, $k=3a,3a+1,3a+2$. Mpreover, each $U_{2k}$ contains a proper vector, and $U_{2k+1}$ does contains a one iff $U_{k+1}$ does. This yields that proper stable vectors exist for all $n$ except for those of the form $2^t+1$, and their number is exactly $2^{u_n-1}$ in all thoese cases. We are done.
There is an error in your reasoning for the even case. For instance, $n=4$ and $n=6$ and $n=8$ all have the same number of solutions (all have 2 solutions if we require the top left entry to be 1). $n=10$ has 8 solutions (if we require top left entry to be 1).
@PatDevlin: Thanks for pointing that out! I've made some corrections (and introduced one new idea); seems that it is corrected now. Does a new text agree with the numerical data?
Added some Addendum which makes the things more clear.
It seems that you haven't still reached the modulo 3 phenomenon. Namely, define $f(3n)=f(3n+2)=2^{n-1}$, $f(3n+1)=2^{n}$. Then the number of solutions, if nonzero, is equal to $f(n)$. (While $2f(n)$ is, for every $n\ge 0$, the number of solutions allowing $0$ on corners.)
I@YCor: Yes, in these terms I now know that $2f(k)\cdot 2f(2k)=2^k$ which seems to agree with this formula. An advantage is that I know that for any even $n$ there are solutions with leading ones... Need to think more about an odd case.
This seems better, but your assertion (in both cases) that $b = a + c$ doesn’t look like it’s true to me. I agree that in the case that $n$ is even, then the vector $b$ is completely determined by the vector $a$. And in the odd case, $a$ must be a solution to the case $(n+1)/2$. But this looks like it has good ideas that can be worked into a correct solution.
@Pat Perhaps, it is too sketchy; I will try to fill the details or to rewrite all this.
Ah! I’m seeing it now! If I finish sorting out the case of n odd, then I could either (i) edit your answer to include this (and perhaps to be a bit clearer in parts) or (ii) post a follow-up answer. Whichever you’d prefer
@Pat: Wait till I finish editing it in a clearer manner, please. Perhaps I'll also sort it out -- still unsure.
Sounds good. Also. I now see that the dimension of U_k seems to be floor(k/3) when k is not 2 mod 3, and otherwise it seems to be floor(k/3) + 1. In particular, it seems like dim(U_k) = dim(U_{k+3}) - 1
Actually my argument doesn't involve parity of $n$, and just passes from each solution (allowing 0 on corners) for $n-3$ to two solutions for $n$. It just doesn't say, for given $n$, whether all solutions have 0 on corners, or if half of them have 1 on corners).
Done! Could I ask some of you to cjeck?
Oh, great! Nice work on the odd case. That’s a satisfying solution. :-)
@YCor, after this all has been done --- isn't your solution beter in a sense that it could provide more uniformity? The OP's graphs look really piecewise-linear --- is it true that your construction confirms that? Does it preserve some order of the proper vectors, or it rather smashes them all around?
My solution easily describes the dimension (when nonzero) $u_n$ of the solutions of size $n$; nevertheless it does not say how the $2^{u_n}$ numbers thus defined by binary expansion distribute in $[2^{n-1},2^n]$. The picture might suggest it tends to a uniform distribution (this ought to be confirmed, or not, by more precise computations). I haven't thought about it so far.
@IlyaBogdanov, thanks for some brilliant observations. Taking some cues from your thoughts, I guess one can construct infinitely many special numbers from one special number. The idea is: if $(a_1,a_2, \cdots. a_n)$ is a special number, then $(a_1,a_2, \cdots. a_n, a_2, \cdots. a_n )$ must also be special. This can be seen easily by pasting 4 triangles of $(a_1,a_2, \cdots. a_n)$, appropriately. I did check this on Python starting with (1,1,0,1). Let me know if there's an error in logic.
It does not seem so; one of your four triangles is turned upside-down, and such triangles ate governed by different rules. Check out doubling 111001
On the other hand,if a doubling works on a specific word, then you can add an arbitrary number of such runs, and it will work as well.
@IlyaBogdanov, I agree with your logic. I think I just gor lucky with 13, 109 and ignored that intricacy. Thanks for that!
OK, let $V_n$ be the vertex set of the following tiling by triangle with $n$ vertices at each large edge. Let me call "standing triangle" a little tiling triangle with a single vertex on the bottom.
For an abelian group $A$, let $A^{V_n}$ be the group of all maps $V_n\to A$. Let $P_n^1(A)$ be its subgroup consisting of those maps summing to zero on each standing triangle. Let $P_n^3(A)$ be the subgroup of $P_n^1(A)$ consisting of those maps that are invariant under $2\pi/3$ rotation.
The perimeter of $V_n$ being the union of its boundary edges, its complement can be identified to $V_{n-3}$. For instance, this picture depicts $\emptyset=V_0\subset V_3\subset V_6$.
$$\begin{matrix} \circ && \circ && \circ && \circ && \circ && \circ\\
&\circ && \circ && \circ && \circ && \circ &\\
&&\circ && \circ && \circ && \circ &&\\
&&& \circ && \circ && \circ &&&\\
&&&& \circ && \circ &&&&\\
&&&&& \circ &&&&&
\end{matrix}$$
We proceed to describe $P^i_n(A)$ for each $n$ and $i=1,3$, by passing from $n$ to $n+3$. Hence we start with $n=0,1,2$.
$n=0$: $V_0$ is empty, so clearly $P_0^1(A)=P_0^3(A)=\{0\}$.
$n=1$: $V_1$ is a singleton, so clearly $P_0^1(A)=P_0^3(A)=A$.
$n=2$: $V_2$ is a single standing triangle. Let $A_3=\{a\in A:3a=0\}$. Then $P_0^1(A)$ can be identified to triples of $A$ with sum zero, while $P_0^3(A)=A_3$.
To pass from $n-3$ to $n\ge 3$, we describe the restriction map $P_n^i(A)\to P_{n-3}^i(A)$. Start from $f$ defined on $V_{n-3}$. Then propose some values $a$, $b$, $c$ on the "right" of each extremal vertex:
$$\begin{matrix}
&\circ && a && \circ && \circ && \circ &\\
&&\circ && u && v && b &&\\
&&& \circ && w && \circ &&&\\
&&&& c && \circ &&&&\\
&&&&& \circ &&&&&
\end{matrix}$$
Then using the sum zero condition, this imposes by propagation values to the right on each edge:
$$\begin{matrix}
&x\quad && a && -a-u && a+u-v && y &\\
&& c-u+w && u && v && b\quad &&\\
&&& -c-w && w && -b-v &&&\\
&&&& c && b+v-w &&&&\\
&&&&& z &&&&&
\end{matrix}$$
$$\text{with}\qquad (x,y,z)=(-a-b-u+v,\;-b-c-v+w,\;-a-c+u-w)$$
(sorry the for the poor alignment of matrix mode)
No further condition is to be fulfilled. This constructs a splitting of the restriction map $P_n^1(A)\to P_{n-3}^1(A)$, which in particular is surjective, and has its kernel isomorphic to $A^3$ (corresponding to $(a,b,c)$). In addition, this is compatible with the rotational symmetry, in which case we have to impose $a=b=c$. Thus, $P_n^1(A)$ is isomorphic to $P_{n-3}(A)\oplus A^3$ and $P_n^3(A)$ is isomorphic to $P_{n-3}(A)\oplus A$ (as abelian groups).
This proves that, for every $n\ge 0$
$P^1_{3n}(A)$ is isomorphic to $A^{3n}$ and $P^3_{3n}(A)$ is isomorphic to $A^n$;
$P^1_{3n+1}(A)$ is isomorphic to $A^{3n+1}$ and $P^3_{3n+1}(A)$ is isomorphic to $A^{n+1}$
$P^1_{3n+2}(A)$ is isomorphic to $A^{3n+2}$ and $P^3_{3n+2}(A)$ is isomorphic to $A_3\oplus A^n$ (hence to $A^n$ if $A$ has no element of order 3).
The method also described an efficient algorithm to generate bases of these, and hence the boundary numbers which are the original objects of study (when $A=\mathbf{F}_2$).
Let us now assume that $A=K$ is a field. The problem asks about the subset $P^3_n(K)_1$, defined with the condition that the extremal vertices are labeled by $1$. This is either empty or an affine hyperplane. If nonempty, this fully describes its dimension as affine subspace (namely $3n\mapsto n-1$, $3n+1\mapsto n$, $3n+2\mapsto n-1$ [or $n$ iff $K$ has characteristic 3].
Note that all this does not say when precisely $P^3_n(K)_1$ is non-empty. When $K$ has characteristic 2 (which boils down to $\mathbf{F}_2$) it was conjectured in the comments that it's empty iff $n=0$ or $n-1$ is a power of $2$.
The description of palindromic solutions $P^6_n(A)$ (i.e., invariant under rotation and reflection) is not obvious and seems also sensitive to the binary expansion of $n$ in some way. However, in characteristic 2 it is clearly equal to $P^6_n(K)_0$, i.e. each extremal vertex is labeled zero, since it forms a standing triangle with two vertices labeled with the same number. That is, $P^6_n(K)_1$ is empty.
To be more explicit: the above produces, for $n\ge 3$, a way to pass from
$P^3_{n-3}(A)$ to $P^3_n(A)$ explicit in terms of the boundary word. Let me specify to $A=\mathbf{F}_2$ which is the OP's framework: if we have the boundary word $(u_1,\dots,u_{n-3})$, then it has two "successors" of size $n$:
$$\begin{matrix}(v_{n-3},&0,& u_1&,&u_1+u_2
&,\dots,&
&v_{n-3}&,&v_{n-3})\\
(v_{n-3},&1,&1+u_1&,&1+u_1+u_2
&,\dots,&
&1+v_{n-3}&,&v_{n-3})
\end{matrix}$$
In particular, a solution with $1$ on the corners exists in size $n$ if and only if a solution with nonzero sum (that is, $v_{n-3}\neq 0$) —but possibly 0 on the corners— exists in size $n-3$.
|
2025-03-21T14:48:30.472243
| 2020-05-02T08:50:30 |
359145
|
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|
Stack Exchange
|
Covering number for the unit ball in a reproducing kernel Hilbert space
I am looking for a reference for an upper bound on the covering number for the unit ball $\{ f \in \mathcal{H}: ||f||_{\mathcal{H}} || \leq 1\} $, where $\mathcal{H}$ is a reproducing kernel Hilbert space, the Sobolev space $\mathcal{W}^{p,2}([0,1])$ in my application. I have found some references such as
Covering numbers of Gaussian reproducing kernel Hilbert spaces, T. Kuhn, Journal of Complexity,
but since this is solely for Gaussian kernels, it is not exactly what I need. Therefore I would appreciate all pointers. Thank you in advance.
|
2025-03-21T14:48:30.472340
| 2020-05-02T09:36:49 |
359146
|
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"authors": [
"Andrej Bauer",
"Monroe Eskew",
"Todd Trimble",
"Yemon Choi",
"https://mathoverflow.net/users/11145",
"https://mathoverflow.net/users/1176",
"https://mathoverflow.net/users/2926",
"https://mathoverflow.net/users/75761",
"https://mathoverflow.net/users/763",
"wlad"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/359146"
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|
Stack Exchange
|
Monotonic and bounded sequences throughout mathematics
When I refer to the Monotone Convergence Theorem below, I refer to the very simple claim that if a non-decreasing sequence has an upper bound then it converges. I don't refer to the claim from Measure Theory.
One application I know of for this theorem is to recurrence relations. One shows that a recurrence relation like $a_{n+1} = \sqrt{2 + a_n}, a_0 = 0$ is non-decreasing and bounded, from which one concludes that it has a limit as $n$ approaches $\infty$.
Are there any other applications of the theorem?
I ask because I'm curious about the possibility of systematically "constructivising" applications of this theorem. Obviously in general this isn't possible. But surely there are lots of situations in which it may be possible.
I tried constructivising Herschfeld's Convergence Theorem, I believe successfully. The resulting argument is a great deal less straightforward than Herschfeld's original one.
[edit]
There was a slight misunderstanding in the comments. The Cauchy completeness of the real numbers does not imply the Monotone Convergence Theorem, unless one assumes the Law of Excluded Middle.
It’s probably one of the most commonly used lemmas in analysis.
@MonroeEskew Can you give me at least one application? Thank you
Be aware that in a lot of cases you don't need the theorem. For instance, the Direct Comparison Test for series can be proved directly
How about the characterization of compactness on the real line?
@MonroeEskew You claim that MCT is needed to prove the Heine-Borel Theorem? I don't see where
Oh, you're referring to Bolzano Weierstrass
Yep, that’s what i mean
@MonroeEskew Given that there have only been 2 answers so far, only one of which provides a plausible application (infinite series), which turns out to not need the MCT, is a sign that it's not as widespread as you claim.
@ogogmad Some experience in analysis would suggest otherwise. This is one of those propositions that's so deep in the background that people use it implicitly. I even used it in a recent work and I'm not really an analyst. Bolzano-Weierstrass is fundamental. Another omnipresently applied result would be Fatou's Lemma in measure theory, which uses the notion of "liminf". The well-definedness of this notion is a manifestation of MCT.
@MonroeEskew Bolzano-Weierstrass can often be replaced by the Fan Principle, which is constructively admissable. I'll think about what you said about Fatou and $\liminf$
@MonroeEskew This is the Fan Principle (usually misleadingly called the "Fan Theorem"): https://ncatlab.org/nlab/show/fan+theorem
@MonroeEskew Thanks for mentioning Fatou's lemma. I'm going to think deeply about this
@MonroeEskew Fatou's Lemma is an inequality, not an existential claim. So I'm not sure if the MCT (in my sense, not in the measure theory sense) is actually needed for it
Well it says that a certain integral is bounded by a certain number defined by liminf. The existence of the number is an implicit claim, no? In any case, it's conceivable that the underlying "ontology" allows things like inequalities to be proven. Maybe a silly example-- Goodstein's Theorem.
Also, if you unravel the defined terms to the "machine language," you will probably find that many results in analysis that may look like simple equations involving integrals actually have a thicket of hidden quantifiers underneath.
@MonroeEskew I don't think there are any insurmountable difficulties in developing integration theory constructively. I think the Dominated Convergence Theorem can be proved in the same generality as the classical theory, perhaps assuming the Fan Principle. Fatou's Lemma has to assume that the $\liminf$ exists as a hypothesis, I think. The measure-theoretic Monotone Convergence Theorem has to assume that the limit of $f_n$ exists; without this assumption it is equivalent to the MCT for sequences.
@MonroeEskew More basic results like that every continuous function has a Riemann integrable can be proved using the Fan Principle. No problems there.
In classical mathematics, how do you know that every subset of R which is bounded above has a supremum? If you don't assume this as an axiom, one approach is to take the Monotone Convergence Theorem for sequences in R as an axiom, and use it to prove that suprema exist. One can then get Bolzano-Weierstrass in R quite directly from this MCT.
As an analyst I view the MCT for sequences as an alternative axiomatization of R among totally ordered fields (since it implies the axiom of Archimedes in the form "1/n tends to zero", and I expect with excluded middle the AoA implies the MCT). It is also a useful primer for the monotone convergence theorem(s) in measure theory, and hence I suspect that treatments of real analysis which emphasise the MCT for sequences have the MCT for functions in mind. Also MCT-ish arguments are quite natural in e.g setting up spectral theory in Cstar and von Neumann algebras
I am going to speak in intuitionistic mathematics here, as that's what's relevant for this question.
It's worthwhile recalling a bit of background.
There are several notions of completeness of an ordered field:
Cauchy completeness: every Cauchy sequence converges.
Dedekind completeness: every Dedekind cut determines an element of the field.
MacNeille completeness: an inhabited bounded set has a supremum.
The field of rationals may be completeted with respect to any one of these to yield three kinds of reals, the Cauchy reals $\mathbb{R}_C$, the Dedekind reals $\mathbb{R}_D$ and the MacNeille reals $\mathbb{R}_M$. These are related as $\mathbb{R}_C \subset \mathbb{R}_D \subset \mathbb{R}_M$.
The principle "a bounded non-decreasing sequence has a supremum" holds in $\mathbb{R}_M$, but it cannot be shown to hold in $\mathbb{R}_D$ (and even less so in $\mathbb{R}_C$).
The principle does not fail completely for $\mathbb{R}_D$. If $a : \mathbb{N} \to \mathbb{R}_D$ is a non-decreasing sequences, we can define the lower Dedekind cut $L = \{q \in \mathbb{Q} \mid \exists n . q < a_n\}$, but not in general the upper cut. Thus, the supremum $\sup_n a_n$ exists as a lower Dedekind real, which is good enough in some situations.
Many uses of the principle are inessential, especially when with some extra effort we can show that the non-decreasing sequence is Cauchy (I imagine this is what you did to prove Herschfeld's Convergence Theorem). For an essential use, we need to look for applications in which the non-decreasing sequence cannot be shown to be Cauchy. This often happens when the sequence depends on some extra parameters. Let me give one such simple example.
Consider the sequence of functions $f_n : [0,1] \to \mathbb{R}$ where $f_n(x) = x^n$. Using the principle "every bounded non-increasing sequence has an infimum", we can show that $(f_n)_n$ converges point-wise on $[0,1]$. Indeed, given any $x \in [0,1]$, it is easy to see that $x \geq x^2 \geq x^3 \geq \cdots$, therefore $\lim_n f_n(x)$ exists. Of course, the limit map $f(x) = \lim_n x^n$ satisfies $f(1) = 1$ and $f(x) = 0$ for $x \in [0,1)$. Without the principle, we cannot show that $(f_n)_n$ converges pointwise because its limit $f$ would be a discontinuous function.
Incidentally, the above example shows that there are discontinuous maps on MacNeille reals.
Is the sequence $(f_n)_n$ that you defined actually useful in any way?
Nice answer otherwise
Yes, it demonstrates an essential use of the principle :-). But seriously, once you see this example, you should be able to generate more complex ones by using fancier functions. For example, you can use the principle to show that all sorts of discontinuous functions can be written as Fourier series.
I was thinking along the lines of a specific theorem or class of problems that can be solved using the MCT. For instance, Herschfeld's theorem, or finding fixed points of some functions.
What's the point of generating more complex examples if they're all useless?
Given that the only plausible application for the MCT was given by Timothy Chow, and that application doesn't even need the MCT anyway, I don't think it's a particularly useful principle.
I tihnk you're jumping to conclusions. Just because nobody here gave a second and a third application of MCT, that does not mean there isn't one. For instance, wouldn't MCT be useful in measure theory, where we approximate measurable maps from below by sequences of increasing simple maps (you'd have to take piecewise linear, or polynomials)? Another one that comes to mind is a proof that a linear operator $F$ on a Banach space has a norm if it is bounded (in the sense that $|F x| \leq M |x|$ for some $M > 0$).
I'll think about your examples.
The Arzela-Ascoli theorem might be another one. There we use the fact that a bounded sequence has a convergent subsequence (as @MonroeEskew pointed out already, that seems to be equivalent to MCT, at least with Dependent choice) in an essential way. The Arzela-Ascoli theorem is used all over the place in analysis (although likely in an inessential way in many concrete cases).
Interesting. Maybe there's a classically equivalent formulation of Arzela-Ascoli that has more constructive content. Namely, what happens if the sequence of functions does not have a uniformly convergent subsequence? We must then conclude that the sequence is finite
That sounds too strong, LPO strong at first sight.
I'm not sure if this is an answer to your question but it seems like it might be. In enumerative combinatorics one often has a sequence of nonnegative integers $(a_n)$ and wants to estimate its growth rate. A standard way to proceed is to form the generating function $\sum_n a_n x^n$ or $\sum_n a_n x^n\!/n!$ and then show that it converges to an analytic function. Then one can apply methods from complex analysis. The proof of convergence focuses on showing that the sequence $(a_n)$ doesn't grow too fast; one basically takes for granted that this means that the series converges (at least pointwise), because of what you're calling the Monotone Convergence Theorem.
There are lots of examples in the book Analytic Combinatorics by Flajolet and Sedgwick.
But I'm not fluent enough with constructive reasoning to tell if this is a trivial or eliminable use of MCT.
In this case, I think you can get away with using the Direct Comparison Test, which can be proved constructively
The various convergence tests for infinite series are all constructive. So it seems like it's eliminable
Still giving this +1
|
2025-03-21T14:48:30.473352
| 2020-05-02T09:40:26 |
359147
|
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|
Stack Exchange
|
Is there a classification of higher-degree generalisations of confocal conic sections?
The 1-parameter families of ellipses and hyperbolas with a given pair of points in the plane as their foci yield “orthogonal double-foliations” of the plane. That is, once the foci are specified, any point in the plane lies on both a unique ellipse and a unique hyperbola, which are orthogonal at the point of intersection. For a given choice of foci, the family is described by a single polynomial equation in three variables:
$$P(x,y,a)=0$$
where $a$ is the semi-major axis of the conic. When $a$ is greater than half the distance between the foci, the curve is an ellipse, and when it is less, the curve is a hyperbola.
Along with the limiting case of families of parabolas with a given focus and axis of symmetry, there do not seem to be any other 1-parameter families of degree-2 curves with all of the same properties:
They are defined by a single polynomial equation in the coordinates and a single parameter.
They foliate the plane twice, with the two foliations arising from different domains for the parameter.
The two kinds of curve are orthogonal at the point of intersection.
But there must surely be families of curves like this of higher degree.
I’d be grateful for any of the following:
Specific examples of higher degree families.
Any classification of higher degree families.
Relevant terminology that might help with a literature search.
|
2025-03-21T14:48:30.473508
| 2020-05-02T13:26:56 |
359155
|
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"authors": [
"Carlo Beenakker",
"Changkun",
"LSpice",
"Qfwfq",
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|
Stack Exchange
|
Can Gimbal lock occur with extrinsic Euler angles?
The gimbal lock seems to be considered with local coordinates (if I read most of the materials correctly). However, according to the matrix calculation, the final rotation matrix (RxRyRz) is (from Wiki):
The whole calculation process doesn't show any assumption to the reference frame, which alpha and gamma as Euler angles can also be considered as world (extrinsic) reference frame.
My question is: is Gimbal lock generally happens no matter which coordinate frame is chosen?
As Wikipedia says, your final matrix seems to show that gimbal lock does occur here ($\alpha$ and $\gamma$ are 'locked' together). Also, the physical phenomenon of gimbal lock is a physical thing, that doesn't know, and so can't change according to, in which coördinates it is measured. So I think I don't understand the question.
@LSpice Did you mean you could not understand my question or you don't know the answer to the question? The final matrix indeed shows an arbitrary point (x, y, z) will land in (z, ..., ...), which indeed is a gimbal lock phenomenon. But the overall calculation didn't assume in which way to represent \alpha, \beta \gamma. This seems to suggest that the reference frame can be an arbitrary one. Am I understand correctly?
yes, the reference frame is arbitrary.
@LSpice: I would say G.L. is (in accordance with the answer by BS. below) an intrinsic geometric phenomenon, rather than a physical one, in the sense that it involves a change of manifold by which you parametrize angles but doesn't affect the trajectory of motion that you may define using angles described by one manifold or the other. But that's probably just "semantics".
@Qfwfq, as you say, it's just semantics. There is certainly a genuine mathematical phenomenon here, but it's reflecting a physical phenomenon (as @BS. points out in the sentence beginning "Mechanically …"), and so has to show up no matter what mathematics is used to model the physics. At least, I think so ….
@CarloBeenakker I don't understand why you vote the question off-topic. Can you recommend somewhere else that I can ask? So far I think this is a mathematical question other than a physical question.
as I understand it, Gimbal Lock involves the case of an intrinsic mathematical singularity (1 axis with a 90 degree rotation) applied iteratively, so I don't believe this question should be closed as it's about the mathematical cause of Gimbal Lock... extrinsic application is non-iterative, so in theory it should avoid Gimbal Lock, but I'm no mathematician.
Gimbal lock as I understand it is the fact that when you parameterize the group of rotations $SO(3)$ by three angles (Euler's, or any other of your choice), the corresponding map from the 3-dimensional torus $T^3=(R/2\pi Z)^3$ to $SO(3)$ has to have a non-surjective differential at some point of the torus, i.e. for some angles.
Mechanically, this means that you cannot move easily to all closely neighbouring rotations : you have to change angles by at least $c \sqrt{\epsilon}$ in total to effect some rotations which are only $\epsilon$ away.
The reason is topological : A non-gimbal-locking parameterization $T^3\to SO(3)$ would be a local diffeomorphism, hence a covering map by compactness.
This is impossible because the universal covering space of $SO(3)$ is the sphere $S^3$ of quaternions (the origin of the so-called belt trick), whereas that of $T^3$ is $R^3$.
|
2025-03-21T14:48:30.473892
| 2020-05-02T13:47:27 |
359157
|
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A general question on Sato-Tate
What is the most general version of Sato-Tate ? Like, I know when $f$ is an eigenform (lying in the space of new-forms), without CM, of weight $k$ and level $N,$ then $\frac{a(p)}{p^{(k-1)/2}}'s$ are equidistributed in $[0,1]$ with respect to a mesaure. My question is, when $f$ is arbitary modular form (without CM), or at least if it is an arbitary newform, can we say $\frac{a(p)}{p^{(k-1)/2}}'s$ are equidistributed in (possibly) some larger interval ?
No statement about modular forms is going to be the most general form of Sato-Tate, because we now know Sato-Tate for modular forms on other groups, and conjecture it in much greater generality.
For the statement you're looking for, there will also be a gap between theorem and conjecture. Anyways, the method to deduce your desired statement from known theorems and believed conjectures will be the same, so let's talk about that.
You will get a conjectural equidistribution statement whenever $f$ is a cusp form, regardless of newness or CM. (For Eisenstein series the coefficients are much bigger and obviously don't equidistribute.)
To see this, write $$f(q) = \sum_{i=1}^m w_i f_i(q^{d_i})$$ where $f_i$ is a cuspidal eigenform new of level $N/d_i e_i$ for some natural numbers $d_i$ and $e_i$. This is the decomposition you get from the eigenbasis of newforms and from writing oldforms in terms of newforms of lower level. Then for all $p$ not dividing $N$, $$a_p(f) = \sum_{i=1}^m w_i' a_p(f_i) $$ where we set $w_i'=w_i$ if $d_i=1$ and $0$ otherwise.
The usual way to state the relevant case of the generalized Sato-Tate conjecture is as a statement about the joint distribution of the $a_p(f_i)$s. You could then derive the distribution of a linear combination from this, but it would be more complicated.
For each $f_i$ appearing, we know $a_p(f_i)$ is equidistributed according to the Sato-Tate measure if $f_i$ is not CM and according to a different measure, proven by Hecke, if $f_i$ is CM. We conjecture that the distribution of the $f_i$s are independent from each other, except in the cases when $f_i$ is a quadratic twist of $f_j$, or the CM cases when $f_i$ may be a quartic or sextic twist of $f_j$, where we predict the simplest joint distribution that follows that. If the level is squarefree and the nebentypus is trivial, you can't get twists, and so the predicted distribution is simply independent in this case. Then your coefficient will look like the weighted sum of independent random variables. Depending on the weights, it may approximate a Gaussian, but it will certainly be bounded as the summands are bounded.
In the cases we can prove this, the method of proof would be the same as the classical case: express the moments of this probability distribution in terms of the asymptotics as $s \to 1$ of $L$-functions, then prove these $L$-functions have meromorphic continuation and calculate their poles. I think the potential automorphy results used to prove Sato-Tate, together with Rankin-Selberg L-functions and cyclic base change, suffice to prove the equidistribution when at most two of the $f_i$ are non-CM and the rest are CM, but I didn't check.
Hi Sawin, in the last paragraph you mean analytic continuation of associated symmetric L-functions right ? Could you provide the reference where this is done ? I think Taylor (with other people) have more than one papers, but I need the exact reference because I need to cite somewhere.
@dragoboy What exactly do you want a cite for?
I want to cite the the paper where Sato-Tate for modular forms is proved classical (i.e by proving the corresp. symmetric L-functions have analytic continuation to Re(s)\geq 1). This is the classical, right ?
@dragoboy Theorem B.3 of A Family of Calabi–Yau Varieties and
Potential Automorphy II in Publ. RIMS Kyoto Univ. 47 (2011), 29–98
DOI 10.2977/PRIMS/31 by Tom Barnet-Lamb, David Geraghty, Michael Harris and Richard Taylor
Hi, Sawin: "We conjecture that the distribution of the f_is are independent from each other, except in the cases when f_i is a quadratic twist of f_j, or the CM cases when f_i may be a quartic or sextic twist of f_j, where we predict the simplest joint distribution that follows that. ", is this your conjecture ?
@dragoboy This is part of what's called the generalized Sato-Tate conjecture. I'm not sure who exactly's responsible for it. It follows from an appropriate part of the Langlands conjectures, say.
|
2025-03-21T14:48:30.474228
| 2020-05-02T14:01:39 |
359159
|
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Stack Exchange
|
Is a 8-dimensional quadratic form recognized by its Lie algebra, modulo equivalence and scalar multiplication?
Question. Let $K$ be a field of characteristic zero (large characteristic should be fine too). Let $q,q'$ be two non-degenerate quadratic forms on $K^n$ with $n=8$. Suppose that the Lie algebras $\mathfrak{so}(q,K)$ and $\mathfrak{so}(q',K)$ are isomorphic (these are simple of type $D_4$, 28-dimensional). Does it follow that $q$ is equivalent to some nonzero scalar multiple of $q'$?
A restatement of the question is whether $\mathrm{SO}(q)$ and $\mathrm{SO}(q')$ being isogeneous over $K$ implies the same conclusion.
This is asking the converse of an obvious fact (since $\mathfrak{so}(q,K)$ and $\mathfrak{so}(tq,K)$ are equal for every nonzero scalar $t$. By an elementary argument (see this MO answer), the converse holds for $n\ge 3$ with the possible exception $n=8$ (while it fails for $n=2$ as soon as $K$ has a non-square). The difficulty comes from the existence of triality, namely automorphisms $\mathfrak{so}(q,K)$ not induced by $\mathrm{O}(q,K)$.
The argument can be used to give a positive answer if "the" absolute Galois group of $K$ does not admit as quotient a group of order 3 or 6. This applies to the reals, in which case we can also argue using the signature of the Killing form. An ad-hoc argument can also probably be done for $p$-adic fields.
(In the comments to the linked answer, some hints were given towards a positive answer for $n=8$. I don't know if they're enough to conclude but obviously if so they should be promoted to a full answer.)
Yes: Proposition C.3.14 in Brian Conrad's article Reductive group schemes is that $SO(q)$ determines $q$ up to similarity for all $q$ of dimension $> 2$. (This was pointed out by @user74230 in a comment somewhere.)
This could be viewed as a special case of a more general phenomenon where there is a simple algebraic group $G$ acting on a vector space $V$ and a $G$-invariant homogeneous polynomial $f$ on $V$ so that the twisted forms of $G$ are in bijection with twisted forms of $f$ up to similarity, see Bermudez and Ruozzi Classifying forms of simple groups via their invariant polynomials, where the fact about quadratic forms is stated as Proposition 7.2.
Great, thank you. In addition from the Bermudez-Ruozzi statement, the algebraic group version is true for all $n$ (including $n=2$) and for all fields (including characteristic 2).
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2025-03-21T14:48:30.474420
| 2020-05-02T14:09:35 |
359160
|
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|
Stack Exchange
|
Kan fibrant replacement for a sphere
To compute the simplicial homotopy group of a space $X$, we find a Kan fibrant replacement $X \to Y$ and calculate for that for $Y$, which can be implemented in a computer program.
Computing homotopy groups for spheres are fundamentally hard, and I believe the problem lies in the difficulty of finding their Kan fibrant replacement.
Could you please show why it's hard, for the easiest possible nontrivial case?
Computing homotopy groups for spheres are fundamentally hard, and I believe the problem lies in the difficulty of finding their Kan fibrant replacement.
Computing the fibrant replacement for simplicial sets is quite
easy: it is given by the Kan fibrant replacement functor Ex^∞. Explicitly, n-simplices in the fibrant replacement of a simplicial set X are maps Sd^k Δ^n → X, for some k≥0.
Here Sd^k denotes the k-fold barycentric subdivision of a simplicial set.
We allow to increase k by further subdividing, this does not change the simplex.
This description is very simple and can be easily programmed into a computer.
The problem is, however, is that the number of simplices grows exponentially
with k, and we also do not have an efficient way to get an a priori upper
bound for k. So some problems are bound to be computationally undecidable,
such as the problem of computing whether π_1 of a simplicial set is trivial or not.
Unless I'm confused, there is no finite $k$ that works for all maps into the space.
@PhilTosteson: No part of the answer claims that k is fixed at any point. As I write in my answer, "We allow to increase k by further subdividing, this does not change the simplex."
An n-simplex of Ex^∞ X is a simplicial map Sd^k Δ^n → X for some k≥0. Composing this map with the last vertex map Sd^l Δ^n → Sd^k Δ^n for some l≥k produces a map Sd^l Δ^n → X, which defines exactly the same n-simplex of Ex^∞ X.
Wow, thanks for your answer. Some further questions: 1) You answer indicates that $Ex^\infty$ helps, but it gives a mapping space and thus an infinity set at each level. How can a computer handle that? 2) Does the $Ex^\infty$ help compute relative homotopy groups as well? 3) From your answer, it seems that we can actually compute the homotopy groups. The problems are that (a) it taxes computer a lot. (b) a computer can only give a presentation of the group, but can't tell us which easy group it is isomorphic to in general. Is that correct?
@Student: Concerning 1): computers can perform computations with integers, yet the set of integers is infinite. The story for simplicial sets is analogous. Concerning 2): yes, just postulate that the boundary of Sd^k Δ^n maps to the given simplicial subset instead of the basepoint. 3) Yes. Presentations are easy to produce, but extracting invariants from these presentations (e.g., the cardinality of a group) is hard or even undecidable.
Perhaps I am missing something, but I don't think it is straightforward use $Ex^{\infty} X$ to obtain a finite presentation of $\pi_n X$ for $n > 1$. You need an effective bound on how far you need to subdivide to get generators and relations. Just because you can make computations locally, it does not mean that you can compute globally-- and this is what is required to actually compute $\pi_n X$. Note, if you had a presentation, it would be trivial to determine the isomorphism class of $\pi_n X$, because $\pi_n X$ is abelian.
@PhilTosteson: It depends on what you mean by a "presentation". Different types of presentations will have very different computational properties. That the lack of an effective bound for k can lead to undecidability is already pointed out in my answer: "we also do not have an efficient way to get an a priori upper bound for k. So some problems are bound to be computationally undecidable". As a degenerate case, the simplicial set itself can be considered a presentation for its homotopy groups, and this presentation is trivially computable.
Any fibrant replacement for $S^n$, $n \geq 1$ is going to have infinitely many non-degenerate simplices. This is simply because there are infinitely many elements of $\pi_nS^n$. So, even though to a mathematician it seems that we can "compute" a fibrant replacement, it is not actually easy to program it in such a way that we can determine the homotopy groups.
Can the fibrant replacement be finite at each level?
No, that's what the argument in the answer shows. It has to have infinitely many non degenerate $n$-simplices, at the very least
Maxime is right. However, there must be workarounds-- to represent any given map, one only has to subdivide $Y$ finitely many times. I don't know the literature, but I found this on the arxiv https://arxiv.org/abs/1706.00380.
But note that if $X$ is a simply connected simplicial set each of whose homotopy groups is finite (e.g., a Moore space), then it does have a fibrant replacement with finitely many simplices in each dimension. Unfortunately, the number of simplices still grows exponentially by dimension, so this is not very helpful.
@Charles That's a good point. Also you can construct this replacement (up to some level) by constructing a Postnikov tower for $X$-- which is how Brown originally showed that homotopy groups are computable.
|
2025-03-21T14:48:30.474801
| 2020-05-02T14:56:39 |
359165
|
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|
Stack Exchange
|
Compute the proximal of a mapping
Let $ f: \mathbb{R}\longrightarrow \mathbb{R}$: compute
proximal of following mapping
$$ f(x)= \sqrt {1-x^2} $$
for $ x \geq 0 $
I know that the proximal is given by
$$ \operatorname{prox}_{\!f} (x)= \operatorname{argmin}_{u\in \mathbb{R}} \big\{f(u) +(1/2)\Vert u-x\Vert^2\big\}$$
Too long to comment.
I assume that $x$ and $u$ are within the range $[-1,1]$ for $f$ to be well-defined.
Suppose $x=\sin(\theta)$, $0\leq \theta \leq \pi/2$. Let $u=\sin(\phi)$. In that case, the optimization problem is:
$$
\min_{\phi}~~\cos(\phi) + \frac{1}{2}(\sin(\phi)-\sin(\theta))^2.
$$
Differentiating the cost function yields:
$$
-\sin(\phi) + \sin(\phi)\cos(\phi) - \sin(\theta)\cos(\phi).
$$
To find the points at which the gradient vanishes, note that:
$$
(\sin(\phi) - \sin(\phi)\cos(\phi))^2 = \sin^2(\theta)\cos^2(\phi).
$$
Denoting $t=\cos(\phi)$, one gets the quartic polynomial:
$$
t^2\sin^2(\theta) = (1-t^2)(1-t)^2.
$$
This can be solved in closed form using Ferrari's method (https://en.wikipedia.org/wiki/Quartic_function), or any other numerical root finding methods. Choose the roots which lead to zero gradient (not all roots are useful), and among them the one that gives the lowest cost value.
Hope it helps.
thanks very much. is it $ u=sin{\phi}$?
@Farab, sorry for the typo. Have edited that.
|
2025-03-21T14:48:30.474921
| 2020-05-02T15:50:47 |
359168
|
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|
Stack Exchange
|
Physical measures that are not SRB
It is quite easy to construct a dynamical system which has a physical measure with a positive Lyapunov exponent and zero entropy, just a figure $\infty$ system. By Pesin's entropy formula such a measure can not be a Sinai-Ruelle-Bowen measure. Now my question is, if there exists a system with a physical measure with positive entropy that is not a SRB-measure. I think the answer is Yes, but I did not manage the construction of such a system.
(Perhaps I have to add that I am interested in ergodic and hyperbolic physical measures that are not SRB)
Yes. The simplest construction is to let $f$ be the figure-eight system so that $\delta_p$ is a physical non-SRB measure (where $p$ is the saddle point) and let $g$ be an Anosov diffeomorphism with SRB measure $\mu$ (a hyperbolic toral automorphism with Lebesgue measure will do the job); then consider the product system $f\times g$. The product measure $\delta_p\times \mu$ is physical, not SRB, and has positive entropy.
Of course that example feels like cheating and is almost certainly not the sort of thing you had in mind. A more informative example comes from
Hofbauer, Franz; Keller, Gerhard, Quadratic maps without asymptotic measure, Commun. Math. Phys. 127, No. 2, 319-337 (1990). ZBL0702.58034.
Theorem 2 in that paper says that if one considers a full, continuous family of S-unimodal interval maps (for example, the family of quadratic maps $\{f_a\colon [0,1]\to [0,1] : a\in [0,4]\}$ given by $f_a(x) =ax(1-x)$), then for every $0 < h < \log(\frac{1+\sqrt{5}}2)$ there are uncountably many parameter values with an ergodic measure $\nu$ that is singular to Lebesgue, has entropy $h$, and is the limit of the empirical measures for Lebesgue-a.e. $x$ (in other words, it is physical).
I have not read the paper carefully enough to have any insight into how those parameters are chosen, beyond knowing that it has something to do with kneading sequences.
It is then reasonable to ask about diffeomorphisms (as opposed to non-invertible interval maps) and conjecture that similar examples exist in the Hénon family of maps, but I do not know of any results in this direction and did not find any in a quick search.
Many thanks for this detailed answer!
|
2025-03-21T14:48:30.475109
| 2020-05-02T15:57:55 |
359169
|
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|
Stack Exchange
|
Models for computing cohomology of Lie groupoids
Given a Lie groupoid $\mathcal{G}=[\mathcal{G}_1\rightrightarrows \mathcal{G}_0]$, let $\mathcal{G}_\bullet$ be the associated simplicial manifold.
Let $\Omega^\bullet(\mathcal{G}_\bullet)$ be the associated double complex. The $k$-th cohomology of this double complex is defined to be the $k$-th deRham cohomology of the Lie groupoid $[\mathcal{G}_1\rightrightarrows \mathcal{G}_0]$. This I am calling the simplicial model for the cohomology. Please correct me if there is already a name for this model.
Questions:
Where did the notion of deRham cohomology group(s) of a Lie groupoid appear for the first time?
To compute equivariant cohomology of a topological space/manifold $M$ with an action of a topological/Lie group, there are at least three models Weil model/Borel model/Cartan model. Are there other models that compute cohomology of Lie groupoids?
Are there other cohomology theories in the category of Lie groupoids?
I am following https://www.math.ubc.ca/~behrend/CohSta-1.pdf to read about cohomology of Lie groupoids/stacks..
This is 2 years late, but since nobody has answered I'll give it a go.
$1.$ I don't know the answer to this.
$3.$ Yes, you can take cohomolgy of Lie groupoids with respect to sheaves, just like you could for manifolds. You can take cohomology of $\mathcal{O}\,,\mathcal{O}^*\,,\mathbb{Z}$ for example. You can also take cohomology with respect to representations. Analogues of these cohomologies exist for Lie algebroids as well, and there is a map, called the van Est map, which relates them.
$2.$ The models you named can only be used to compute cohomology with respect to certain sheaves (like the constant sheaf $\mathbb{R}\,,$ or de Rham cohomology). In general the cohomology of Lie groupoids is defined by forming the nerve and taking its cohomology, as a simplicial space. This can be difficult to compute, but one way of doing it is by using van Est theorems.
|
2025-03-21T14:48:30.475258
| 2020-05-02T16:03:54 |
359170
|
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|
Stack Exchange
|
Construction of weight function to satisfy condition on given functional
Consider the following function :
$$F(z) = \omega(z){\sin^2\left(\frac{c\Gamma(z)}{z}\right)}$$
Here, $\omega(z)$ is a weight we are going to consider
The following two conditions should meet for $\omega(z)$ :
$$\lim_{ y→∞}|F(x ± iy)|e^{−2πy }= 0$$ uniformly with respect to $x$
$$\int_0^\infty |F(x + iy) − F(x − iy)|e^{−2πy} dy<+\infty$$ for every $x≥1$ and tends to zero as $x\to\infty$.
Question : Explicit construction of $\omega(z)$ .
We can say this question is Focused version of the following question (as asked in the reason for closing down the question) : https://math.stackexchange.com/q/3570663/702232
Related but different:
On properties on a certain functional
All types of suggestive comments and advices are welcome.
These minor edits intended only to bump the question to the top of the list start being annoying.
@MateuszKwaśnicki sorry for that but how could I make it reach to suitable people if it has no activity?
I am afraid you have to accept the fact that no one is able or willing to help. You can safely assume that everyone potentially interested in answering it has already read your question more than once.
@MateuszKwaśnicki Ok, thank you sir
Any updates or progress?
|
2025-03-21T14:48:30.475378
| 2020-05-02T16:18:54 |
359171
|
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|
Stack Exchange
|
Picard numbers of isogenous K3 surfaces over a non-closed field
Let $S_1, S_2$ be K3 surfaces defined over a field $k$ and $\phi\!: S_1 \dashrightarrow S_2$ a dominant rational $k$-map (so-called isogeny). It is known that $\rho(S_1) = \rho(S_2)$ for the complex field $k = \mathbb{C}$, i.e., the Picard numbers (ranks of the Picard $k$-groups) are equal.
Is this true if $k$ is a non-closed field, for example $k = \mathbb{Q}$ or $\mathbb{F}_{q}$, and $S_1, S_2$ are non-supersingular?
|
2025-03-21T14:48:30.475439
| 2020-05-02T16:54:27 |
359174
|
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|
Stack Exchange
|
Approximation to continuous functions over an closed interval
Let $$f\in C[a,b]$$ A triangular system is a series of numbers
\begin{matrix}
x_{11}\\
x_{21}&x_{22}\\
x_{31}&x_{32}&x_{33}\\
\cdots
\end{matrix}
that $$a<x_{n1}<x_{n2}<\cdots<x_{nn}<b$$
Consider the Lagrange interpolation polynomial $p_{n}$ by the points on n-th "layer". I have following questions:
1) Is there a continuous function $f$ such that for any triangular system, we have $$\lim\limits_{n\to\infty}||p_{n}-f||_{\infty}=0$$
Present results: polynomials are definitly what we want. But I want to generalize this results to analytic functions (which I guess is ture). Is there a proof or a counterexample?
2)Similarly, is there a function that is bad enough that any triangular system dose not give a converging approximation?
This one is solved that the Chebyshev equioscillation theorem implies that the polynomial of best approximation is actually an interpolation. And the Weierstrass' theorem implies the convergence.
3)Is there a triangular system that gives approximation to any continuous function?
The present results show that the Chebyshev nodes gives a perfect approximation to $BV[a,b]\bigcap C[a,b]$. What about other conditions? (A diverging example for Chebyshev nodes is equally wanted.)
4) Is there a triangular system and function $f$ such that $$\forall x\in[a,b]\quad\lim\limits_{n\to\infty}(p_{n}(x)-f(x))\neq 0$$
And welcome any other related results or questions.
For 1) see MSE. For 2) you gave the answer. For 3) and 4) see MO
|
2025-03-21T14:48:30.475550
| 2020-05-02T17:27:57 |
359177
|
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|
Stack Exchange
|
A complete metric space with some convex-type property
Let $(X,d)$ be a complete metric space with this property:
for each $x∈X$, $r>0$ and $y∈X$ with $d(x,y)<r$, there exists $z∈X$ such that $d(x,y)+d(y,z)=d(x,z)=r$.
I want to know if the family of complete metric spaces with this property are known or some work have been done on it.
Remark: I studied about metrically convex space which had been introduced by K.Menger(1928) and I am not sure if there is a relation between this family with my mentioned metric spaces.
|
2025-03-21T14:48:30.475613
| 2020-05-02T17:28:52 |
359178
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/359178"
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|
Stack Exchange
|
Hilbert scheme of points concentrated in a given point
It is well known that if $X$ is a smooth surface, then the Hilbert scheme of points $X^{[n]}$ is also smooth. What about the subscheme $S_p$ of $X^{[n]}$ consisting of all schemes of finite length $Z$ such that $Supp \ Z=\{p\}$? (Here $p\in X$ is a fixed point.) Is this scheme still smooth?
That too is smooth. Fine analysis fo these are done in particular by Briancon and Granger.
Hmmm... I guess you have to define what you mean. If you want $S_p$ to represent some functor then my impression is that this is not very well behaved, e.g. probably not a variety. I think the more natural thing is to take $S_p$ to be the fiber of the Hilbert-Chow morphism $X^{[n]}\rightarrow X^{(n)}$ over the point $[np]$. This is known to be reduced irreducible and Cohen Macaulay, but unfortunately (or not depending on what you like) is singular once $n>2$. The first singular example is $S_p\subset X^{[3]}$ which is isomorphic to the cone over the twisted cubic (singular at the cone point!).
I think about 50% of people call $S_p$ the punctual Hilbert scheme and 50% of people call $X^{[n]}$ the punctual Hilbert scheme. But you might find some relevant results in the literature if you search for "punctual Hilbert scheme".
Thanks Mohan! Thanks also @YosemiteStan I'm interested in both interpretations. Do you have a reference for the fiber interpretation?
|
2025-03-21T14:48:30.475746
| 2020-05-02T19:27:14 |
359185
|
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|
Stack Exchange
|
“Total rainbow” trees
Let $G= (V,E)$ be a simple finite graph which is (not necessarily properly) edge-colored. A rainbow spanning tree refers to a spanning tree T of G such that no color appears more than once amongst the edges of T.
Alternatively, define a total rainbow tree to be a (not necessarily spanning!) tree in G so that each color appears exactly once.
While it’s easy to find results in the literature about rainbow spanning trees (where sometimes the term heterochromatic is used instead of rainbow), e.g. necessary and sufficient conditions for existence & efficient algorithms for finding them, I haven’t been able to find any information relating to the latter concept. In particular, I made up the terminology so it’s possible it’s been studied under a different name.
This motivates the following question:
Question: Are there any known necessary and/or sufficient conditions on G and the associated coloring for the existence of a total rainbow tree?
Of course there are trivial cases, like when one uses only one color (then any edge is a total rainbow tree). I’m interested in a family of graphs arising in surface topology for which the number of colors is large, but still much smaller than the number of vertices (I am happy to say more about the problem for anyone who is interested). The problem is solvable if there exists a rainbow spanning tree, but this is impossible when the number of colors is much smaller than the number of vertices. But I realized I can solve the problem in another way if there exists a total rainbow tree which is what led to the above question.
I would also be happy to know something about when there exists a heterochromatic tree so that some definite fraction of the full set of possible colors appear. Thanks for reading!
Edit: it turns out I can simplify the potential solution to the problem of interest, and now I only need the existence of a connected total rainbow subgraph (so, we drop the assumption of being a tree). Thus the new and improved question is:
What are necessary and sufficient conditions on a (not necessarily properly) edge-colored graph G for the existence of a connected total rainbow subgraph?
If you do not care about connectedness, then a necessary and sufficient condition is given by the matroid intersection theorem. Let $G=(V,E)$ be a (not necessarily properly) edge-colored graph. Define a total forest to be the set of edges of a forest which contains every colour exactly once.
Let $M_1$ be the matroid with ground set $E$, where a set is independent if and only if it contains at most one edge of each colour. Let $M_2$ be the matroid with ground set $E$ where a set is independent if and only if it does not contain a cycle. Let $r_1$ and $r_2$ be the rank functions of $M_1$ and $M_2$, and let $t$ be the total number of colours. Observe that a total rainbow forest exists if and only if $M_1$ and $M_2$ have a common independent set of size $t$. By the matroid intersection theorem this is true if and only if for all $A \subseteq E$,
$r_1(A) + r_2(E \setminus A) \geq t$.
Moreover, it is clear that independence testing for $M_1$ and $M_2$ can both be done in polynomial-time. Therefore, by running the matroid intersection algorithm, we can find a total forest (if it exists) in polynomial-time, or a set $A \subseteq E$ such that
$r_1(A) + r_2(E \setminus A) < t$.
Note that such a set $A$ certifies that no total forest exists. To see this, observe that if $T$ is the set of edges of a total forest, then
$t=|T|=|T \cap A| +|T \cap (E \setminus A)| \leq r_1(A)+r_2(E \setminus A) < t$,
which is a contradiction.
Thank you for this! It will take me a little while to absorb it, but for now I’ll just say that I do care about it being a tree and not a forest. However, it would still be interesting to me to know when there is a forest with an a priori bound on the total number of components (I’d like the number of components to be bounded in terms of, let’s say, square root of the number of colors).
In general, the total number of components can be much more than the square root of the number of colours.. To see this, colour the $n$-cycle with $t$ colours where each colour class is a path with $n/t$ edges. If $t \leq n/2$, then the largest connected rainbow subgraph has at most $2$ edges. Therefore, every total forest contains at least $t/2$ components.
Hi Tony, thanks again! And yes, I understand that in general one can’t hope for a bound on the number of components in terms of the number of colors. What I meant was that while what I’d really like is connectedness (i.e., one component), I would be content with a situation where such a bound on the number of components held. While you’re right that this won’t happen in general, I’m asking for some sufficient conditions on the graph and the coloring that would lead to such a bound
For example, I could imagine something like the following being true:
Given N, C> 0, there exists K so that if the min (or perhaps average) chromatic degree of an edge colored graph with at most N vertices and C colors is at least K, there exists a connected total rainbow subgraph.
|
2025-03-21T14:48:30.476178
| 2020-05-02T19:59:06 |
359188
|
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|
Stack Exchange
|
The existence of a fiber sequence involving $\mathrm{Spin}(9)$ and $\mathrm{SU}(2)$
$\newcommand{\SU}{\mathrm{SU}} \newcommand{\Spin}{\mathrm{Spin}}$There is a fiber sequence $G_2\to \Spin(9) \to \Spin(9)/G_2$, and $G_2$ contains $\SU(3)$ as a subgroup. Is there a space (possibly a topological group) $Y$ which sits in a fiber sequence $\SU(2)\to Y\to \Spin(9)/G_2$, such that there is a canonical map $Y\to \Spin(9)$ which fits into a map of fiber sequences
$$\require{AMScd}
\begin{CD}
\SU(2) @>>> Y @>>> \Spin(9)/G_2\\
@VVV @VVV @| \\
G_2 @>>> \Spin(9) @>>> \Spin(9)/G_2?\\
\end{CD}$$
(I would be happy if such a fiber sequence only existed $2$-locally.)
It suffices to show that the classifying map $\Spin(9)/G_2\to BG_2$ lifts through the canonical map $B\SU(2) = \mathbf{H}P^\infty\to BG_2$. The fiber of this map is $G_2/\SU(2)$, but, unfortunately, this isn't a topological group, as far as I can tell. It may be helpful to note that $G_2/\SU(2)$ is the Stiefel manifold $V_2(\mathbf{R}^7)$.
Here is a candidate for $Y$. The space $\Spin(9)/G_2$ is like
$\SU(2;\mathbf{O})$, and sits in a fiber sequence $S^7\to \Spin(9)/G_2\to
S^{15}$. I believe there is a homotopy pullback square
$$\require{AMScd}
\begin{CD}
\Spin(9)/G_2 @>>> B\SU(2)\\
@VVV @VVV \\
S^{15} @>>> B\Spin(5),
\end{CD}$$
where the map $S^{15}\to B\Spin(5)$ detects the generator(?) of $\pi_{14}
\Spin(5) \cong \mathbf{Z}/1680$. Then, the fiber $Y'$ of this map $S^{15}\to
B\Spin(5)$ is a candidate for $Y$. I don't know how to construct a map $Y'\to \Spin(9)$.
You are quite right that $\mathrm{G}_2/\mathrm{SU}(2)$ is not a topological group. Since, as you point out, there is a fibration $S^5\to \mathrm{G}_2/\mathrm{SU}(2)\to S^6$, all the homotopy groups $\pi_i(\mathrm{G}_2/\mathrm{SU}(2))$ for $i =0,1,2,3$ are trivial, so it can't be a compact Lie group. Also, because $\pi_3\bigl(\mathrm{Spin}(9)/\mathrm{G}_2\bigr)$ is finite (probably trivial, but I haven't checked to be sure), it can't be a compact Lie group. For similar reasons, your $Y$, if it exists, cannot be a topological group either.
Yep, thanks! (I also just added a construction of a candidate for Y.)
|
2025-03-21T14:48:30.476332
| 2020-05-02T20:01:32 |
359189
|
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|
Stack Exchange
|
Question on a subcategory being extension-closed
In the article "Homological theory of noetherian rings" by Idun Reiten from 1996, it was stated that it seems to be not known whether the subcategory $\operatorname{Tr}(\Omega^i(\mathrm{mod}\text{-}A))$ (which includes by defintion all projective modules) is extension-closed for $i>1$, where $\operatorname{Tr}$ denotes the Auslander-Bridger transpose and $A$ an Artin algebra.
Question: Is there some more background on this question? Are there partial solutions for special classes of algebras for this question or is it even answered somewhere in the literature?
Using experiments with QPA, it seems a counterexample has been found. It is quite complicated and surprisingly no simple counterexample seems to work. Maybe someone is interested to check it (I hope there is no mistake). There might be a simplificiation or even a simpler counterexample.
Let $A=KQ/I$ be the finite dimensional quiver algebra over a field $K$ where
\begin{tikzcd}
2 \arrow[r, "z", shift left=0.75ex] & 1 \arrow[loop, distance=3em, out=35, in=-35, "x"] \arrow[l, "y", shift left=0.75ex]
\end{tikzcd}
(how to use tikz here?)
$Q$ is the quiver with 2 vertices 1 and 2 and there is a loop x from 1 to 1, an arrow y from 1 to 2 and an arrow z from 2 to 1.
The relations are $I=\langle xy, yz, zx, x^3 \rangle$. The algebra is representation-finite and special biserial with 13 indecomposable modules.
Let $S_i$ denote the simple $A$-modules.
$A$ has vector space dimension 7 over the field $K$. Let $e_i$ denote the primitive idempotents of the quiver algebra $A$ corresponding to the vertices $i$ for $i=1,2$.
The indecomposable projective $A$-module $P_1=e_1 A= \langle e_1, x ,y , x^2 \rangle$ has dimension vector $[3,1]$ and the indecomposable projective $A$-modules $P_2=e_2 A= \langle e_2, z, zy \rangle $ has dimension vector $[1,2]$. We remark that $P_2$ is also injective.
For finite dimensional algebras we have $D Tr=\tau$, the Auslander-Reiten translate and since $D$ is a duality, the subcategory $Tr (\Omega^2(\mod-A))$ will be extension closed if and only if $\tau ( \Omega^2(\mod-A))$ is extension closed. We will thus look at the subcategory $\tau ( \Omega^2(\mod-A))$ in the following.
We use the notation $\tau_i:=\tau(\Omega^{i-1})$ in the following for the higher Auslander-Reiten translates.
Let $I=(x+z)A$ denote the right ideal generated by $x+z$, which has vector basis basis $\langle x+z, x^2, zy\rangle$.
$M_1:=A/I$ is an indecomposable $A$-module with dimension vector $[2,2]$.
Then it is easy to see that $\tau_3(M_1) \cong M_1$ and thus $M_1 \in \tau ( \Omega^2(\mod-A))$.
Now let $M_2:=P_2/S_2= e_2 A/zy A$, which is an indecomposable $A$-module with dimension vector $[1,1]$.
Again it is easy to see that $\tau_3(M_2) \cong M_2$ and thus $M_2 \in \tau ( \Omega^2(\mod-A))$.
Now $\dim(Ext_A^1(M_2,M_1))=1$ and there is (up to isomorphism) a unique short exact sequence that is not split:
$$0 \rightarrow M_1 \rightarrow W \rightarrow M_2 \rightarrow 0.$$
If we show that $W$ is not in $\tau ( \Omega^2(\mod-A))$ then we have shown that the subcategory $\tau ( \Omega^2(\mod-A))$ is not extension closed.
Now $W \cong P_2 \oplus U$, where $U=A/((x+y+z)A)$.
Here the ideal $(x+y+z)A$ has vector space basis $\langle x+y+z,y,x^2,zy \rangle$ and thus $U$ has dimension vector $[2,1]$.
Now let $f_1: P_2 \rightarrow M_2$ be the projective cover of $M_2$ and let
$f_2: U \rightarrow M_2$ be the canonical non-zero map $A/(xA+yA+zA) \rightarrow e_2 A/zyA=M_2$, which is given by the canonical surjections $g: A \rightarrow e_2 A \rightarrow e_2 A/zyA$ and noting that here $(xA+yA+zA)$ is in the kernel of $g$ which induces the map $f_2$.
Let $f:=f_1 \oplus f_2$, then it is easy to see that the kernel of $f$ is isomorphic to $M_1$ and since $W$ is not isomorphic to $M_1 \oplus M_2$ the short exact sequence is not split.
Now we show that the direct summand $U$ of $W$ is not in $\tau ( \Omega^2(\mod-A))$ or equivalently that $K:=\tau^{-1}(U)$ is not in $\Omega^2(\mod-A)$.
Now a module $T$ is a direct summand of a module of the form $X \oplus P$ for $X$ an $n$-th syzygy module and a projective module $P$ if and only if $T$ is a direct summand of a module of the form $P' \oplus \Omega^n(\Omega^{-n}(T))$.
Now for $T=K$, the module $\Omega^2(\Omega^{-2}(T))$ is isomorphic to $S_2 \oplus V$, where $V$ is an indecomposable $A$-module with dimension vector $[2,0]$.
Thus $K$ is not a direct summand of $P' \oplus \Omega^2(\Omega^{-2}(K))$ for any projective module $P'$ and thus $K$ is not a 2-th syzygy module.
Thus the subcategory $\tau ( \Omega^2(\mod-A))$ is not extension-closed.
For what it's worth, I've checked this by hand, and I agree with your conclusion.
@JeremyRickard Thank you very much!
|
2025-03-21T14:48:30.476615
| 2020-05-02T20:51:10 |
359194
|
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|
Stack Exchange
|
Reference request: Donsker's theorem for non-identical, independent random variables
The central limit theorem can be generalized to independent but not iid random variables, provided they satisfy the Lyapunov condition (which looks something like a variance bound), see https://en.wikipedia.org/wiki/Central_limit_theorem#Lyapunov_CLT. The Lindeberg central limit theorem provides another such condition. Meanwhile, Donsker's theorem states that a random walk with iid mean 0 variance 1 increments converges to Brownian motion. I am looking for a version of Donsker's theorem for independent but not iid random variables, subject to something like the Lyapunov or Lindeberg condition, but I've been unable to find it written down so far. Any reference to something like this would be appreciated!
Such results were obtained by A. A. Borovkov and his students. See e.g.
Borovkov, A. A.
Estimates in the invariance principle. (Russian)
Dokl. Akad. Nauk SSSR 206 (1972), 1037–1039.
Borovkov, A. A.
The rate of convergence in the invariance principle. (Russian. English summary)
Teor. Verojatnost. i Primenen. 18 (1973), 217–234.
Borovkov, A. A.
Rate of convergence and large deviations in invariance principle. Proceedings of the International Congress of Mathematicians (Helsinki, 1978), pp. 725–731, Acad. Sci. Fennica, Helsinki, 1980.
Borovkov, A. A.; Sahanenko, A. I.
On the rate of convergence in invariance principle. Probability theory and mathematical statistics (Tbilisi, 1982), 59–66,
Lecture Notes in Math., 1021, Springer, Berlin, 1983.
https://link.springer.com/chapter/10.1007/BFb0072903
Borovkov, K. A.
The rate of convergence in the invariance principle for a Hilbert space. (Russian)
Teor. Veroyatnost. i Primenen. 29 (1984), no. 3, 532–535.
Thanks! I'll also add for anyone else looking for a reference that Thm. 3 of B. M. Brown, "Martingale Central Limit Theorems" https://projecteuclid.org/download/pdf_1/euclid.aoms/1177693494, which I found after asking this question, works as well.
|
2025-03-21T14:48:30.476772
| 2020-05-02T20:55:57 |
359195
|
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|
Stack Exchange
|
Isomorphic free groups have bijective generating sets
Let $F(X)$ be the free group on a set $X$. Classically, we can prove the statement:
$F(X) \cong F(Y)$ if and only if $|X|=|Y|$.
The proofs (that I have seen) consist of turning the group isomorphism into an isomorphism of free abelian groups, then an isomorphism of vector spaces by modding out by some $p$. Where we can deduce that isomorphic vector spaces have bijective basis sets.
The last part of this proof uses AC. I came across Kleppmann - Generating sets of free groups and the axiom of choice, which proves that ZF + boolean prime ideal theorem can prove this, therefore it is weaker than AC.
I am interested in knowing how true this statement is constructively. I.e. are uses of excluded middle and choice (or rather statements weaker than choice) necessary to prove it?
The statement is way too vague and has several interpretations. I guess you mean the statement "if there is an isomorphism from the free group on $X$ and the free group on $Y$, then there is a bijection from $X$ to $Y$".
@YCor What else could it mean?
@MikeShulman when I first read it, I thought of it as "there exists a free generating subset of the first one and...etc." I converged to the wording I'm writing just by elimination. Actually "free" is hopelessly ambiguous since mathematicians use the same word for "free over some given subset" and "free over some subset"; of course often this doesn't matter, but precisely here it matters.
@YCor Hopefully things are more clear now.
@PaulPlummer's reference: Läuchli - Auswahlaxiom in der Algebra (MSN). I think the Kleppman reference is actually to Kleppmann - Generating sets of free groups and the axiom of choice (MSN) (two 'n's).
@LSpice This is Kleppmann's dissertation which I think Paul was referring to.
Your last parenthesis ("more rather statements weaker than choice") doesn't make sense to me. Is it what was intended?
@LSpice Maybe that wasn't the best way to phrase it. BPIT is strictly weaker than AC so the proof I am looking for would use ZF + something weaker than choice.
It's not a big deal, but I just can't parse it, even understanding what you mean. Is 'more' meant to be 'or'?
@LSpice I've changed it to "or" as that sounds better.
Is the case of vector spaces already known? For instance, let $F^2_X$ be the free 2-elementary abelian group on $X$. Is ZF(+DC?) + "there's an isomorphism $F^2_{\mathbf{R}}\to F^2_{\mathbf{R}/\mathbf{Q}}$" + "there is no bijection $\mathbf{R}\to\mathbf{R}/\mathbf{Q}$" consistent (assuming ZF consistent)? (I allow middle choice here, at least to understand my own question)
@YCor: I feel like I asked this before, what's a 2-elementary abelian group? (Too elementary?)
@AsafKaragila a $p$-elementary abelian group, for prime $p$, is a synonym of vector space over the field on $p$ elements (I don't think a terminology question can be too elementary!). Purely group-wise, it means abelian group in which $px=0$ for all $p$.
@YCor: So free 2-elementary abelian = the finite subsets of $X$ (with symmetric difference as addition)? (Also, I had meant that maybe the groups are too elementary... :-P)
@AsafKaragila Yes, this is one realization of 'the' free 2-elementary abelian group over $X$.
@YCor: Well, in that case, you can easily see that the cardinality of $F^2_{\Bbb R}$ is $2^{\aleph_0}$, while that of $F^2_{\Bbb{R/Q}}$ is at least $|\Bbb{R/Q}|$ (and probably equal, but it's too early in the morning for this. So if there is no bijection between $\Bbb R$ and $\Bbb{R/Q}$, the groups will not be isomorphic.
@AsafKaragila thanks, indeed. Does there always exist (say in ZF+DC) a bijection $X\to F_X^2$ for every infinite $X$? (oh,probably not)
@YCor: Of course not. Since that would imply every infinite set has the same cardinality as its finite subsets, which implies $X\times X$ is equipotent with $X$, which implies choice.
OK thanks, so I should restate my question replacing $\mathbf{R}$ and $\mathbf{R}/\mathbf{Q}$ with $X$ and $X^2$ for $X$ such that there is no injection $X^2\to X$. Namely: is it consistent (for some such $X$, and say with ZF+DC) that there is an group isomorphism $F^2_X\to F^2_{X^2}$ (or $F_X\to F_{X^2}$)?
@YCor: Good question. I feel like the answer should be yes. But I have nothing off hand to give you as an example.
|
2025-03-21T14:48:30.477088
| 2020-05-02T21:17:22 |
359196
|
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|
Stack Exchange
|
One part of a bipartite graph has max degree 3. Partition the other part to 3 ~equal subsets s.t. just a fraction of first part see all 3 subsets
Let $d \gg 1$. Let $G:=(U, V, E)$ be some bipartite graph such that deg$(u) \le d$ for all $u\in U$ and deg$(v) \le 3$ for all $v \in V$.
Now, is it possible to color vertices in $U$ with 3 colors such that firstly, size of each color class is roughly $|U|/3$ and secondly, at most a fraction $\beta$ of vertices in $V$ have neighbours from all the 3 color classes?
Specially interesting is the case $d \sim 20000$,
and $\beta \ll 1 /3$.
Motivation The motivation is from computational complexity. I am trying to simplify the proof or improve Thm. 5.1 of [1]. For d=3 (this is not our $d$), in the first paragraph you make instances of 3-SAT with $n$ variables, where each variable appears in at most 20000 clauses. So there are at most $20000/3$ clauses with 3 variables.
Now, let $U$ be the set of variables, and $V$ be the set of clauses, and there be an edge between $u\in U$ and $v \in V$ iff $u \in v$.
If I have the desired coloring, then the second paragraph and its overhead can (by a trick) be avoided.
In the third paragraph each color class corresponds to a block of variables. Therefore, the size of each color class should be roughly $n/3$.
On the other hand, for each clause that has neighbours from all the color classes, (applying another trick) one variable should appear in two blocks, increasing the size of blocks by at most $\beta n$. Therefore we need $\beta \ll 1 /3$.
[1] M. Patrascu and R. Williams. On the Possibility of Faster SAT Algorithms. In SODA, pages 1065–1075, 2010.
Taking the question as in the title (so, without the roughly and the almost), this is clearly impossible for $K_{3,3}$.
true, I need it for an algorithm, so I am interested in a sympathetic behaviour and roughly in almost would be enough.
Let $G=(U,V,E)$ be a bipartite graph where $U=[n], V=\binom{[n]}{3}$, and there is an edge between $u \in U$ and $v \in V$ if and only if $u \in v$. Then $\deg(u)=\binom{n-1}{2}$ for all $u \in V$ and $\deg(v)=3$ for all $v \in V$. However, every colouring of $U$ with $3$ colours, such that each colour class has size roughly $\frac{n}{3}$ has about $\frac{n^3}{27}$ vertices in $V$ with neighbours from all three colour classes. This is about a $\frac{2}{9}$ fraction of all the vertices in $V$.
I do not think that a $\frac{2}{9}$ fraction can be considered almost none for any reasonable definition of almost none, so this is probably impossible.
Give me 1 hour or 2 ...
I just edited the question. In fact $2/9$ can be pretty acceptable. But $d$ can begi
In that case, taking a random colouring by flipping a fair 3-sided coin for each vertex seems to work. The expected number of vertices in $V$ that see all three colours is $\frac{2}{9} |V|$.
It's not straightforward for me to get the $2/9$ bound for the general case with random coloring, still thinking ...
For a fixed vertex of degree 3, there are 27 ways its neighbours can be colored, and 6 of them in which its neighbours are rainbow. Since the colouring is random, the probability the neighborhood is rainbow is 6/27=2/9. Then use linearity of expectation.
Here is an integer programming formulation of your problem. Not sure if it can provide an analytical insight, but hope it helps someway.
Consider a matrix $X\in \{0,1\}^{|U|\times 3}$, with exactly one 1 in each row. Such a matrix would indicate the membership of each $u\in U$ to one of the 3 groups that you seek. The column sums of $X$ is the number of elements in the three groups. Since you want all the groups of size around $N/3$, a good measure of how off you are would be the maximum of the columns sums. Thus, you would want to minimize $\max_{1\leq j \leq 3} \left(\sum_{i=1}^N X(i,j)\right)$.
Secondly, for a $v\in V$, let $I_v=(i ~ |~ (u_i,v)\in E)$. Also let $X_{[I, :]}$ be the sub-matrix obtained by choosing rows in the index set $I$. Note that the $\sum_{j=1}^{3}\max_{1\leq i\leq |I_v|}{X_{[I, :]}(i,j)}$ denotes the number of groups that $v$ is a neighbour to. Ideally, you would want this to minimize this too.
With the above points, the IP takes the final form given by:
$$
\min_{X}~~\max_{1\leq j \leq 3} \left(\sum_{i=1}^N X(i,j)\right) + \lambda\left(\sum_{v\in V} \sum_{j=1}^{3}\max_{1\leq i\leq |I_v|}{X_{[I, :]}(i,j)}\right) \\\\
\mbox{subject to}
\hspace{1cm} X\in \{\{0,1\}\}^{|U|\times 3}.
$$
Here $\lambda>0$ is a trade-off factor, which the user needs to set. You could put the second term in the cost function as a constraint as well.
For smaller instances, you can possibly try with a free IP solver. For very large instances, try solving the relaxed LP of this problem (by changing $\{0,1\}$ to $[0,1]$) and rounding off to get the groups. I'll also admit that the approximation ratio of this method is something that needs further investigation.
|
2025-03-21T14:48:30.477435
| 2020-05-02T21:55:29 |
359197
|
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|
Stack Exchange
|
Spectrum of Laplacian-like operator
Let $\kappa_1, \kappa_2>0$ be fixed.
Consider the unbounded operator $A: D(A) \rightarrow L^2(-1,1)\times\mathbb{R}$ defined by
$$
A\begin{bmatrix} y \\ h \end{bmatrix} = \begin{bmatrix} \partial_x^2 y - \kappa_1y\\ -\partial_xy(1) \end{bmatrix}
$$
with domain $D(A) \subset L^2(-1,1) \times \mathbb{R}$ defined by
$$
D(A) = \Big\{ \begin{bmatrix} y \\ h \end{bmatrix} \in H^2(-1,1)\times \mathbb{R} \colon y(-1) = 0,\, y(1) = -\kappa_2 h \Big\}.
$$
When considered as an unbounded operator on $L^2(-1,1)\times\mathbb{R}$ endowed with the canonical inner product, $A$ does not appear to be symmetric, hence it cannot be self-adjoint.
Edit: Thus up to my knowledge, checking that the resolvent is compact will imply that the spectrum of $A$ consists only of eigenvalues.
My question is the following:
Can it be shown that the spectrum of $A$ consists only of real eigenvalues?
To this end, I imagine one needs to show that any point $\lambda \in \mathbb{C} \setminus \mathbb{R}$ is in the resolvent of $A$, but I don't know how to show this.
Have you tried to find the eigenvalues of $A$ by an explicit computation? Given the structure of the operator, this might not be impossible.
@JochenGlueck Yes, it's a lengthy computation but I managed to obtain an almost explicit form (basically they are like the Dirichlet Laplacian's + $O(k)$). But how can I guarantee that there is no essential spectrum?
The spectrum of operators with compact resolvent consists of eigenvalues (with finite multiplicity) only. That has nothing to do with self-adjointness.
Ok the theorem I know states this for self-adjoint operators. Do you mind giving me to a reference for non self-adjoint operators?
You can for instance find it in Corollary IV.1.19 of [Engel and Nagel: One-Parameter Semigroups for Linear Evolution Equations (2000)] (but it is most likely also stated in much older references). Please beware, however, that "compact resolvent" includes by definition the property that the resolvent set of $A$ is non-empty; hence, it does not suffices to just observe that the domain of $A$ embeds compactly into the surrounding Banach space - one also needs to check that at least one complex number is in the resolvent set.
The resolvent is non-empty in this case because the elliptic problem associated to $A$ is uniquely solvable.
If u omit the BC at 1 there is a standard change of variables to make problems with advection term where advection is gradient if scalar into self adjoint problem...
@bgsk: So, can your question be considered settled, or is there any point that is still open?
@JochenGlueck Yes it's fine, I proved what I wanted.
|
2025-03-21T14:48:30.477639
| 2020-05-02T22:54:01 |
359201
|
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}
|
Stack Exchange
|
Compatibility with multiplication of a cyclic order on a ring
I am copying my question from here: https://math.stackexchange.com/q/3233462/427611.
Is it correct that $\mathbb Z/3\mathbb Z$ and $\mathbb Z/4\mathbb Z$ are the only rings with three or more elements with a non-linear cyclic order compatible with both the addition and the multiplication?
A cyclic order is compatible with addition if $[a, b, c] \implies [a + x, b + x, c + x]$ and $[x + a, x + b, x + c]$ for any elements $a$, $b$, $c$, $x$ of the ring.
A cyclic order is compatible with multiplication if $[a, b, c] \implies [ax, bx, cx]$ and $[xa, xb, xc]$ for any elements $a$, $b$, $c$, and any positive element $x$ of the ring.
A element $x$ of a ring with a cyclic order is positive if $[0, x, -x]$.
A cyclic order on a ring is non-linear if any cut of it is not compatible with the addition or multiplication.
A cut of a cycle order on a ring is a linear order $<$ such that $a < b < c \implies [a, b, c]$ for any elements $a$, $b$, $c$ of the ring.
A cut of a cyclic order is compatible with addition if $a < b \implies a + x < b + x$ and $x + a < x + b$ for any elements $a$, $b$, $x$ of the ring.
A cut of a cyclic order is compatible with multiplication if $a < b \implies ax < bx$ and $xa < xb$ for any elements $a$, $b$, and any positive element $x$ of the ring.
An element $x$ of a cut of a cyclic order is positive if $0 < x$.
Below is a sketch of the proof provided on the referenced page:
A non-linearly cyclically ordered group has four quadrants;
If a non-linearly cyclically ordered group has more than one positive element, then all the four quadrants are not empty;
The multiplication of elements from different quadrants is incompatible with the cyclic order.
One can find the details of my research on the item in here:
Cycle notation for cyclic orders: https://math.stackexchange.com/q/3236651/427611
Positive and negative elements of a cyclically ordered group: https://math.stackexchange.com/q/2213048/427611
Apex of a cyclically ordered group: https://math.stackexchange.com/q/2204247/427611
Quadrants of a cyclically ordered group: https://math.stackexchange.com/q/3230720/427611
Natural cut of a cyclically ordered group: https://math.stackexchange.com/q/3207182/427611
Compatibility with multiplication of a cyclic order on a ring: https://math.stackexchange.com/q/3233462/427611
A property of an Archimedean cyclically ordered group: https://math.stackexchange.com/q/2205470/427611
The rule of three steps for a cyclically ordered group: https://math.stackexchange.com/q/2208205/427611
I think it would be useful to provide less but better focussed and brief information. I added (from your MathSE post) the main definition.
@YCor I added a sketch of the proof. Would it be sufficient?
At least it clarifies the intent of your question (which is a proof verification).
@YCor Or a shorter proof if possible.
Actually I don't understand the definition: what is meant by "positive" in a cyclic order?
@YCor You can find the definition in here: https://math.stackexchange.com/q/2213048/427611
|
2025-03-21T14:48:30.477819
| 2020-05-03T00:06:53 |
359204
|
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"Jasper Liang",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/359204"
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|
Stack Exchange
|
Complex plane minus Cantor set admits non-constant bounded harmonic function
Let $K\subset [0,1]$ denote the usual 1/3 Cantor set. I know that $\mathbb{C}\backslash K$ has no non-constant bounded analytic function, since the singularity $K$ can be removed. However, a statement I am reading says that $\mathbb{C}\backslash K$ admits a non-constant bounded harmonic function. Why is this true? Any help would be appreciated.
Where is the statement that you are reading?
It is actually an exercise asking one to prove the statement above.
Where is the exercise?
Sorry it is on a set of notes from a class I am taking. I am not sure where the exercise is originally from.
Since the Cantor set $K$ has Hausdorff dimension $\log2/\log 3<1$, it is a removable set for bounded analytic functions, and so, as you say, there is no bounded analytic function outside of $K$. But it does not mean that there are no bounded harmonic functions outside of $K$. Actually, any point of $K$ is regular for the Dirichlet problem. So, choosing any non-constant continuous function $u$ on $K$, there exists a unique harmonic function $f$ in $\overline{\mathbb C}\setminus K$ such that
$$
\forall\zeta\in K,~\lim_{z\to\zeta}f(z)=u(\zeta).
$$
Then $f$ is a non-constant bounded harmonic function in $\overline{\mathbb C}\setminus K$. Note that $K$ is of positive capacity so there is no reason that $f$ can be extended to $\mathbb C$ (and actually it cannot be extended).
A proof that any point of $K$ is regular for the Dirichlet problem, that uses Wiener criterion, is given p.100 of
J.B. Garnett, D.E. Marshall, Harmonic measure. New Mathematical Monographs 2. Cambridge University Press, Cambridge, 2005.
Thank you so much for your answer! Would you mind elaborating a bit on why $K$ is regular for the Dirichlet problem?
As far as I know, it is not completely trivial. The reference I know is Theorem III.63 of Tsuji's book, Potential theory in Modern function theory. You can also look to Theorem 1 of this paper.
I have checked out the paper but not the book yet. Thank you for your help and your references!
|
2025-03-21T14:48:30.478095
| 2020-05-03T02:00:07 |
359206
|
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"DSM",
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"url": "https://mathoverflow.net/questions/359206"
}
|
Stack Exchange
|
$\arg\max$ in the dual norm of the nuclear norm
Given a matrix $X \in \mathbb{R}^{m \times n},$ then the spectral norm is defined by
$$\left \| X\right\| := \max\limits_{i \in \{1, \dots, \min\{m,n\}\} }\sigma_i (X)$$
whereas the nuclear norm is defined by
$$\left \| X \right \|_* := \sum\limits_{i=1}^ {\min\{m,n\}} \sigma_i (X)$$
It is a well-known fact that the dual norm of the spectral norm is the nuclear norm. This implies that
$$\|M\| = \sup_{\|X\|_*\leq 1} \langle M, X \rangle$$
where the inner product is defined by $\langle A, B \rangle := \mathop{\textrm{Tr}}(A^TB)$. Given a matrix $M \in \mathbb{R}^{n \times n},$ how to find a matrix $X^*$ such that the following holds?
$$ X^* \in \arg\sup_{\|X\|_*\leq 1} \langle M, X \rangle$$
Solving the problem with CVXPY, as it is a convex program, should help. Are you looking for a more theoretical treatment of the problem?
@DSM both actually. I am comfortable with python but I haven't used that library before. Is this is easy to solve using CVXPY?
import cvxpy as cvx
import numpy as np
N = 5
X = cvx.Variable((N,N))
M = np.random.randn(N,N)
prob = cvx.Problem(cvx.Maximize(cvx.trace((M.T)@X)), [cvx.normNuc(X)<=1])
prob.solve()
print(prob.status)
print(prob.value)
print(X.value)
The answer is $X^* = uv^*$ where $u$ and $v$ are the left and right singular vectors of $M$ associated with the largest singular value. If the largest singular value has multiplicity larger than $1$, the argsup is a convex set whose extreme points are the matrices of the form described above.
Thanks! You are absolutely right! I will add a proof here in case someone is looking for it in the future:
Suppose $X^=uv^$ where $u$ and $v$ are the left and right singular vectors of $M$ associated with the largest singular value. To show that $X^$ attains the optimum, it suffices to show that $\langle M, X^\rangle = |M| = \sigma_{max}$. But this follows from the cyclicity of the trace:
$trace(M^uv^) = trace(V^\Sigma U u v^) = trace(v^V^\Sigma U u) = \sigma_{max}$
|
2025-03-21T14:48:30.478252
| 2020-05-03T03:05:00 |
359209
|
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"authors": [
"Joseph O'Rourke",
"Sungjin Kim",
"Wolfgang",
"Zhi-Wei Sun",
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}
|
Stack Exchange
|
A tree with prime vertices
Let us construct a simple (undirected) graph $T$ as follows:
$\quad$ Let the set of all primes be the vertex set of $T$. For each prime $p$, take the least prime $q>p$ such that $2(p+1)-q$ is prime (such a prime q should exist in view of Goldbach's conjecture), and then set an edge connecting $p$ and $q$.
Clearly the graph $T$ contains no circle. If it is connected then it is a tree.
QUESTION. Is the above graph $T$ a tree?
In Feb. 2013, I constructed the graph $T$ and conjectured that $T$ is indeed a tree. For example, the path connecting $2$ and $191$ is
\begin{align*}2&\to 3\to 5\to 7\to 11\to 13\to 17\to 19\to 23\to 29\to 31\to 41,
\\ &\to43\to 47\to 53\to 61\to 71\to 73\to 89\to 97\to 107\to 109
\\&\to 113\to 127\to 149\to 151\to 167\to 173\to 181\to 191.
\end{align*}
Any ideas towards the solution of the question? Your comments are welcome!
@zz7948 Can you please clarify on spanning tree part? How would repeatedly removing edges and leaving one edge with smallest weight gives still a connected graph?
I think the problem is this: The process of leaving one edge is not necessarily the same as Prim's algorithm.
Not an answer, just a drawing of the tree including the
OP's $2 \rightarrow 191$ path:
So if we call the trajectory starting with 2 the main branch, the question is whether the trajectory of any prime not on the main branch will eventually hit it.
@Wolfgang: Yes. For example, just going a bit further, $223 \rightarrow 251$ is a separate tree, which eventually gets hooked into the main branch.
Yes, eventually, as expected. I am also wondering: associate with each prime p the number b(p) of branchings the trajectory encounters before arriving in the main branch. So for the small primes above, it is either 0 or 1, but I guess that there should be bigger primes for which there are 2, 3, ... branching points. Or maybe not? Could you easily find some with your implementation?
And likewise, as there are numbers like 89 with 3 branches arriving, are there also others with 4 or more?
@ Wolfgang: You are asking great questions! I hope @ZhiWeiSun responds. I will think about implementation... [software will not let me @ two users]
I thank J. O'Rourke for drawing the prime tree.
This does not give a complete answer. This provides a strategy for conditional approach.
Given that the nature of the problem is asking for $q$ and $2(p+1)-q$ being simultaneously primes, I think that the question can be approached conditionally (similar to zz7948).
A slight modification is shifting the focus to finding $p<q$ such that $p$ and $2p+2-q$ are simultaneously prime, for given prime $q$. As suggested in zz7948's comment, we extend the edges to include all pairs $(p,q)$ with both $p$ and $2p+2-q$ are primes.
Let us assume the following version of prime $k$-tuple conjecture.
Conjecture 1
There is an absolute constant $C>0$ and a prime $q_0$ such that the number $N(q)$ of primes $p<q$ such that $p$ and $2p+2-q$ are simultaneously prime satisfies
$$
N(q)\geq C\frac q{\log^2 q}\geq 1, \ \ \mathrm{ if } \ q\geq q_0.
$$
If Conjecture 1 is true, then any prime $q>q_0$ is connected to some smaller prime.
If we can show (computation) that all primes $q\leq q_0$ are connected, then all primes will be connected.
Once we obtain that all primes are connected, we now start to remove the edges to include only the smallest $q$ with $q>p$ and $2p+2-q$ are both primes.
Then, the issue is, whether or not Prim's algorithm of finding spanning tree of a connected graph gives the desired graph.
Not completely sure, but I think the statement about equations in primes is stronger statement than being able to connect all primes.
As my approach is finding smaller prime $p$ using the solution to the equation (Conjecture 1), the Nonexistence of such solution $p$ does not rule out the possibility of being connected to primes larger than $q$. For example, see 37, 67, 101, etc in Joseph's tree.
@ChuaKS Please see my modified answer. The extended rule for edges is allowing to connect any $p,q$ satisfying the configuration. However, It still needs long way to go.
|
2025-03-21T14:48:30.478587
| 2020-05-03T03:15:46 |
359211
|
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"Lev Bahn",
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|
Stack Exchange
|
Existence of two stochastic processes
I am wondering if I can show that
For given $x,y\in \mathbb{R}$ there are two stochastic processes $S_t$ and $B_t$ such that $S_t$ and $B_t$ are two one dimensional Brownian motions starting at $x$ and $y$ respectively and $P\{S_t = B_t\}=1$ for $t>>0$.
I thank in advance for any help!
This is called coalescent brownian motion. Set $T=\inf{t:S_t=B_t}$ and set $W_t = 1_{t\leq T}S+1_{t>T}B_t$.
@RaphaelB4 Thank you! I think I got the idea but, with the notation given by you, I cannot see how it is possible to show $P{S_t=B_t}=1$. Could I have more hint?
@RaphaelB4 If I understand correctly, the $W_t$ in your explanation is the Brownian motion starting from $x$ and coincide with $B_t^y$ for $t >T$. Now, I see why they are equal for large enough $t$. But I am still thinking why then $W_t$ is Brownian motion.
@RaphaelB4 Actually, come to think of it, $T$ does not need to be finite which means it cannot solve my problem.
$T<\infty$ almost surely but it is not bounded. If you want the equallity for a fixed $t$ it is not possible as the gaussian law $\mathcal{N}(x,t)$ and $\mathcal{N}(y,t)$ are different.
Thanks to @RaphaelB4, I could figure out the proof even for general $d-$dimensional case. Here is my proof.
Let $B_t^x$ and $B_t^y$ be two independent $1-$dimensional Brownian motions with initial points $x$ and $y$ in $\mathbb{R}$ respectively. Now, let
\begin{align*}
\tau=\inf\{t\geq 0 : B_t^x=B_t^y\}
\end{align*}
then note that $\tau<\infty$ a.s. because $1D$ Brownian motion is recurrent. Now, consider $W_t^y=B_t^y\chi_{[0,\tau ]}+B_t^x\chi_{(\tau,\infty)}$ then $W_t^y$ is a $1-$dimensional Brownian motion with initial point $y$ and with $P\{B_t^x=W_t^y
\text{ for all $t>\tau$} \}=1.$
Now, we want to construct $d-$dimensional Brownian motion for arbitrary $d\in \mathbb{Z}^+$. Let $x,y\in \mathbb{R}^d$ be given. Then, by the result above, for each $i\in \{1,\dots, d\}$, if $x_i$ and $y_i$ are $i$th entries of vectors $x$ and $y$, there exist two $1-$dimensional Brownian motions $B_t^{x_i}$ and $B_t^{y_i}$ with initial points $x_i$ and $y_i$ respectively such that $P\{B_t^{x_i}=B_t^{y_i} \text{ for all }t\geq\tau_i \}=1$ where $\tau_i=\inf\{t\geq 0 :B_t^{x_i}=B_t^{y_i}\}$. Now, let $B_t^x$ and $B_t^y$ be $\mathbb{R}^d$ valued function such that
\begin{align*}
B_t^x=\begin{bmatrix}
B_t^{x_1}\\
\vdots\\
B_t^{x_d}
\end{bmatrix}\hspace{4cm}
B_t^y=\begin{bmatrix}
B_t^{y_1}\\
\vdots\\
B_t^{y_d}
\end{bmatrix}
\end{align*}
Now, let $T=\max_{ i} \tau_i$. Noting that $P\{\tau_i<\infty\}=1$ we know that
\begin{align*}
P\{T<\infty \}=P\left( \bigcap_i \{\tau_i<\infty\} \right)=1.
\end{align*}
(https://stats.stackexchange.com/a/100576)
Then $B_t^x$ and $B_t^y$ are two $d-$dimensional Brownian motions such that $P\{B_t^x=B_t^y \text{ for all }t>T\}=1$.
@PierrePC Thanks for pointing that out. You are absolutely right. Maybe I need to make the last $T$ specific.
@PierrePC The reason why they are equal in one dimensional case is because brownian motion is recurrent in one dimensional case. In the case of multidimension, the equality get changed to less than equal to.
@PierrePC Ah.... Yeah... Come to think of it.... you are right.
In $d > 1$ case, if you need a minimal coupling time $T$, then look up "mirror coupling" of Brownian motions.
|
2025-03-21T14:48:30.479151
| 2020-05-03T04:03:11 |
359213
|
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|
Stack Exchange
|
Effective action, partition function and the renormalization group
Mayer expansions and the Hamilton–Jacobi equation by D. Brydges and T. Kennedy begins mentioning that many problems in statistical mechanics and QFT center on the analysis of integrals of the form:
\begin{equation}
\int d\mu_{C}(\varphi)e^{-V(\varphi+\varphi')} \tag{1} \label{1}
\end{equation}
where $d\mu_{C}$ is a Gaussian measure with mean zero and covariance $C$. I'm primarily interested in statistical mechanics so I'd like to understand these integrals in this context. I don't know much about QFT but (correct me if I'm wrong) a common object of study which satisfies (\ref{1}) is the so-called effective action $\mathcal{G}_{\lambda, C}$. But this effective action is also useful in statistical mechanics. To be more specific, we define it as:
\begin{equation}
\mathcal{G}_{\lambda,C}(\psi) \mathrel{:=} \log \int d\mu_{C}(\varphi)e^{-\lambda V(\psi+\varphi)} \tag{2}\label{2}
\end{equation}
where $V$ is an interaction function (I'm assuming here that (\ref{2}) is well-defined). Note that (\ref{2}) has the exact same form of (\ref{1}). Now, let us consider:
\begin{equation}
Z_{\lambda}(\psi) \mathrel{:=} \int d\mu_{C}(\varphi) e^{-\lambda V(\varphi)+\langle C^{-1}\psi, \varphi\rangle} \tag{3}\label{3}
\end{equation}
where $\langle \cdot, \cdot \rangle$ is a given inner product. This is called the partition function of the system with source term $C^{-1}\psi$. The partition function (and also its logarithm) is a powerful tool to calculate correlation functions. With a little algebra, we can prove that:
\begin{equation}
\mathcal{G}_{\lambda,C}(\psi) = -\frac{1}{2}\langle \psi, C^{-1}\psi\rangle + \log Z_{\lambda}(\psi) \tag{4}\label{4}
\end{equation}
and thus we can also use the effective action to study correlation functions. It is often the case that the covariance $C$ can be written as a sum of covariances $C= C_{1}+\dotsb + C_{n}$. Taking $n=2$ for simplicity, plugging it into $\mathcal{G}_{\lambda V,C}$ and using the property that a sum of Gaussian random variables is again a Gaussian random we get:
\begin{equation}
\mathcal{G}_{\lambda V,C_{1}+C_{2}}(\psi) = \mathcal{G}_{\mathcal{G}_{\lambda V, C_{1}}, C_{2}}(\psi) \tag{5}\label{5}
\end{equation}
which is the semi-group property of the effective action. Recursively, this also hold for $n>2$. Thus, it is sufficient to study the map:
\begin{equation}
-\log \int d\mu_{C_{i}}(\varphi)e^{-\lambda V_{i}(\varphi+\psi)} \tag{6}\label{6}
\end{equation}
where $V_{i}$ is properly chosen.
The point that is not clear to me is the following. In statistical mechanics, one is primarily interested in evaluating the partition function, even though other quantities (such as the ones defined above) are also relevant. I'd like to better understand the relation between the above objects and the actual partition function of the system (not the one with source terms) and what is the difference between approaching a problem using the effective action or the partition function itself, is there is any.
These correspondences might be helpful: we have the partition function $Z(\beta)$, the effective action $W(\beta)$, the classical action $I_\beta[\phi]$ for physical variables $\phi(x,\tau)$ in Euclidean time $\tau$ with period $\beta$ equal to inverse temperature. These are related by
$$ e^{-W(\beta)}=Z(\beta)=\int D\phi\, e^{-I_\beta[\phi]}$$
$$\qquad\qquad = \int D\phi \exp\left[-\int_0^\beta d\tau\,\int d^3 x\,\bigl(\tfrac{1}{2}(\nabla\phi)^2+V(\phi)\bigr)\right].$$
The thermodynamic free energy is $F(\beta)=\beta^{-1}W(\beta)$, the entropy
\begin{equation}
S(\beta)=\beta^2 \frac{dF(\beta)}{d\beta}=\left(\beta\frac{d}{d\beta}-1\right)W(\beta).
\end{equation}
My answer to the question in the OP, "what is the difference between approaching a problem using the effective action or the partition function itself" would be --- there is no difference, one is the logarithm of the other.
Thanks for the answer. So, if I understood it correctely, the partition function is my effective action (\ref{2}) at $\psi=0$?
(2) is (minus) the effective action, to obtain the partition function you would exponentiate (2) --- indeed at $\psi=0$.
|
2025-03-21T14:48:30.479419
| 2020-05-03T05:27:27 |
359217
|
{
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"LSpice",
"Lagrida Yassine",
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}
|
Stack Exchange
|
Compute a cardinality using Chinese remainder theorem
I posted the question here but I got no response.
I am looking for computing this cardinality:
$$N(q)=\#\Bigg\{n \in \mathbb{N} \ | \ \gcd\bigg(n^2+1, \prod_{\substack{p \leqslant q \\ p\text{ prime}}}p\bigg)=1 , \ \ n^2+1 \leqslant \!\!\!\prod_{\substack{p \leqslant q \\ p\text{ prime}}}\!\!p \Bigg\},$$
by using Chinese remainder theorem.
First, for $p$ odd prime and $m\in\mathbb{Z}/p\mathbb{Z}$, the number of solutions of the equation $m^2 + 1 = 0 \pmod p$ is :
$$
\begin{cases}
0 & \text{ if } p = 3 \pmod 4 \\
2 & \text{ if } p = 1 \pmod 4
\end{cases}.$$
Using the Chinese remainder theorem and the fundamental counting principle, I get this result:
$$N(q) = \bigg(\prod_{\substack{p \leqslant q \\ p \equiv 3[4] \\ p\text{ prime}}}p \bigg)\prod_{\substack{p \leqslant q \\ p \equiv 1[4] \\ p\text{ prime}}}(p-2) \label{1}\tag{1}$$
Formula \eqref{1} seems not correct as when I check $N(q)$ numerically I do not get the same results as by counting.
The true values are :
$$N(7)=5, \ N(11)=15, \ N(13)=45 , \ N(17)=161, \ N(19)=698, \cdots$$
Question: Why my formula \eqref{1} is not correct !? And what is the correct formula ?
Many thanks for any help.
Numerically it's very likely that:
$$N(q) \approx \dfrac{1}{\sqrt{\displaystyle \prod_{\substack{p \leqslant q \\ p\text{ prime}}}p }} \, \bigg(\prod_{\substack{p \leqslant q \\ p \equiv 3[4] \\ p\text{ prime}}}p \bigg)\prod_{\substack{p \leqslant q \\ p \equiv 1[4] \\ p\text{ prime}}}(p-2)$$
You can see yourself looking already at $q = 3$, say, that your solution overcounts: it allows all residue classes modulo $p$, but your condition $n^2 + 1 \le \prod p$ allows only some. For example, the only allowable $n$ when $q = 3$ are $n = 0$ and $n = 2$; since $n = 4$ is too big, we already get $N(3) = 2$, not $N(3) = \prod_{\substack{p \le 3 \ p \equiv 3 \pmod4}} p = 3$ as predicted.
First of all, you forgot $p=2$, secondly you Chinese remainders method works prexisely only for $n\leq \prod p$ but you need thoose $n$ with $n^2+1\leq \prod p$
@LSpice, we have $N(3)=2$, since the only numbers coprime to $6$ and less than $6$ are $1,5$ and both are written as $n^2+1$. As i say formula $(1)$ not correct.
@PavelKozlov, thank you, i check and you you have right.
It sounds like you disagree, but actually I think we (and @PavelKozlov) agree.
@PavelKozlov, I think the error is only in not taking into account the requirement that $n^2 + 1 \le \prod p$, not in leaving out $p = 2$; indeed, because $n^2 + 1$ has to be coprime to $\prod p$, hence in particular to $2$, we're restricted to $n \equiv 0 \pmod2$.
The correct formula is : $$\displaystyle#\Bigg{n \in \mathbb{N} \ | \ \gcd\bigg(n^2+1, \prod_{\substack{p \leqslant q \ p\text{ prime}}}p\bigg)=1 , \ \ n \leqslant !!!\prod_{\substack{p \leqslant q \ p\text{ prime}}}!!p \Bigg} = \displaystyle\bigg(\prod_{\substack{p \leqslant q \ p \equiv 3[4] \ p\text{ prime}}}p \bigg)\prod_{\substack{p \leqslant q \ p \equiv 1[4] \ p\text{ prime}}}(p-2)$$
Is there a factor $e^{-\gamma}$ to pay attention to?
@SylvainJULIEN yes, see my article : here
|
2025-03-21T14:48:30.479616
| 2020-05-03T07:20:35 |
359219
|
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"Coherent Sheaf",
"Igor Khavkine",
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}
|
Stack Exchange
|
On Whitney's paper on real algebraic varieties
I had previously asked this question on math.stackexchange and did not receive an answer and so I decided to reword it and pose it here.
This question is based on Whitney's paper titled "Elementary Structure of Real Algebraic Varieties", published in the Annals of Mathematics, Vol. 66, No. 3, November 1957.
At the end of section 7 of the above-mentioned paper, Whitney says that $g$ is analytic on $M^*$ seen as an open connected subset of the $(x_{\rho+1}',...,x_n')$-space. Since $g$ is a polynomial in $x_1,...,x_n$, it would make sense to say that it is analytic in $M^*$ under the usual chart, but why should it continue to be analytic in the open subset of the $(x_{\rho+1}',...,x_n')$-space?
I am also finding it extremely confusing as to which field Whitney is working with (because he mentions that he will be proving "Theorem 1" which uses the above-mentioned result first in the complex case, but then never discusses the base field after), the real or the complex (a clarification on that would greatly help)? So, I shall consider both cases.
If the field is $\mathbb{C}$:
In this case, it is easy to solve. If $\phi$ is the defined diffeomorphism from $U$ as a subset of $\mathbb{C}^n$ to the new chart, then $\phi$ and $\phi^{-1}$ are both smooth and so $g\circ \phi^{-1}$ is smooth from $M^*$ as an open subset of the $(x_{\rho+1},...,x_n)$-space and hence analytic.
If the field is $\mathbb{R}$:
In this case, I know that the function $g\circ \phi^{-1}$ as defined above is smooth, but I see no reason for it to be analytic.
The only conclusion I could draw is the following (assume $\mathbb{k}$ is either $\mathbb{C}$ or $\mathbb{R}$):
If $h_1,...,h_n\in \mathbb{k}[x_1,..,x_n]$ are polynomials such that for some $p\in \mathbb{k}^n$, there exists an open set $U$ containing $p$ such that $h(U)$ is open in $\mathbb{k}^n$ and diffeomorphic to $U$, where $h=(h_1,...,h_n)$, then, if $h^{-1}=(h_1',...,h_n')$ on $h(U)$, then each $h_i'$ is analytic.
I am not sure if this result is correct. If it is, then it helps give a unified solution for both, the real and the complex case.
It would have helped to know that all the undefined objects in the question are actually introduced in your original math.SE question. The chart $\phi$, as defined, is analytic. The inverse function theorem guarantees that $\phi^{-1}$ is also real analytic, not just smooth. This holds both in the real and complex cases. I don't think you need anything else.
@IgorKhavkine thank you very much. That helps a lot!
|
2025-03-21T14:48:30.479808
| 2020-05-03T09:39:58 |
359226
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/359226"
}
|
Stack Exchange
|
Is there a way to calculate the Picard $\mathbb{F}_q$-number of an (rational or K3) elliptic surface?
Consider a finite field $\mathbb{F}_{q}$ and an elliptic surface
$$
\mathcal{E}\!: y^2 + a_1(t)xy + a_3(t)y = x^3 + a_2(t)x^2 + a_4(t)x + a_6,
$$
where $a_i(t) \in \mathbb{F}_{q}[t]$. Is there a way to calculate the Picard $\mathbb{F}_q$-number of $\mathcal{E}$ (i.e., rank of the Picard $\mathbb{F}_q$-group $\mathrm{Pic}(\mathcal{E})$) if I don't know the Mordell-Weil rank of $\mathcal{E}$?
I am mainly interested in the cases when $\mathcal{E}$ is a rational or K3 surface and $\mathrm{char}(\mathbb{F}_q) > 3$. By abuse of notation I also denote by $\mathcal{E}$ the corresponding Kodaira-Neron model.
|
2025-03-21T14:48:30.479886
| 2020-05-03T10:31:55 |
359231
|
{
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"authors": [
"David Roberts",
"Philippe Gaucher",
"Piotr Pstrągowski",
"Ryan Budney",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/359231"
}
|
Stack Exchange
|
What is known about homotopy groups of spheres?
I'm looking for a list/table/survey of what is known (and what is not known) about homotopy groups of spheres, for example: which are known, which are known stably, which are known primally, non-$0$ etc.
I'm aware of what is on Wikipedia, but I doubt it's exhaustive...
In terms of the stable homotopy groups of spheres at prime $2$, I think the recent work of Isaksen-Wang-Xu (https://arxiv.org/abs/2001.04511) is the current state of the art.
Typically these types of questions are closed on Math Overflow, as the discussion is more appropriate for a Wikipedia entry/chat than this forum. I certainly sympathize with threads like these as I imagine they would have more utility than a Wikipedia page. Anyhow, this isn't the place for that kind of discussion.
The efforts put forward to a thread like this are more appropriate for the Wikipedia page. https://en.wikipedia.org/wiki/Homotopy_groups_of_spheres
"What is known about homotopy groups of spheres?" That it is a difficult and deep problem :-).
@Ryan I view it as a completely reasonable question, since there is a lot "known to experts" in homotopy theory that is not easy to find. And this is the whole point of MO. Pointing someone to a Wikipedia page not intended for such detailed information is not the most helpful thing.
@DavidRoberts: These kinds of questions regularly get closed when they concern other topics. It would be consistent to close the question, but perhaps consistency is not the goal.
Well, when the person explicitly says they are aware of the content at Wikipedia, and judging by their other questions, aren't exactly a random noob, then pointing them twice to Wikipedia in comments seems a bit odd to me. And I don't view the question as asking for a discussion, either. I have also, in the past, wondered about this question and would like a better reference than WP, namely a decent survey paper, if there is one.
|
2025-03-21T14:48:30.480049
| 2020-05-03T11:33:30 |
359235
|
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"abx",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/359235"
}
|
Stack Exchange
|
Relative Chow's theorem
Suppose you have a smooth quasiprojective complex algebraic variety that is not compact. Suppose you have a complex analytic fiber bundle over the algebraic variety where each fiber is a smooth projective variety. Is the bundle itself algebraic? If not, what are some extra conditions we can impose to deduce that the bundle is algebraic?
I guess this is pretty standard but I couldn't find anything useful from a quick search.
The Hopf surface is an analytic fiber bundle over $\mathbb{P}^1$, with fiber an elliptic curve. Just take out a point in $\mathbb{P}^1$ if you don't want a compact base.
|
2025-03-21T14:48:30.480123
| 2020-05-03T13:17:39 |
359240
|
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"Iosif Pinelis",
"https://mathoverflow.net/users/121692",
"https://mathoverflow.net/users/36721",
"πr8"
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|
Stack Exchange
|
Convergence of probability measures which (asymptotically) concentrate along a submanifold
Let $V : (-1, 1)^d \to \mathbf{R}_+$ be a smooth function, and for $\beta > 0$, define
\begin{align}
P_\beta ( dx ) &= \exp \left( - \beta V ( x ) \right) / z (\beta) \, dx\\
z (\beta) &= \int \exp \left( - \beta V ( y ) \right) \, dy
\end{align}
with the assumption that $V$ is sufficiently well-behaved at the boundary that $z (\beta) < \infty$ for all $\beta > 0$.
Now, if $V$ has a unique global minimiser at (wlog) $V \left(\mathbf{0}\right) = 0$, one can usually argue that as $\beta \to \infty$, $P_\beta$ converges in law to a delta measure at $\mathbf{0}$, maybe with some extra conditions in place.
My situation is that $V$ takes its minimum value (again, wlog taken to be $0$) along a codimension-1 submanifold, i.e. along
$$\mathcal{F} = \{ x \in (-1, 1)^d : V(x) = 0 \}.$$
Now, I would like to reason that, as $\beta \to \infty$, $P_\beta$ converges in law to some measure which concentrates along $\mathcal{F}$. I would guess that the answer has something to do with the Hausdorff measure on $\mathcal{F}$, but i) my intuition for such matters is not very strong, and ii) even if it were, I am not sure where I would look for the relevant mathematical tools to prove it.
As such, my questions are:
What is a reasonable conjecture for the limiting behaviour of $P_\beta$ as $\beta \to \infty$, and
How can I prove it?
For 1., I'd like any conjecture to come equipped with some justification, or a relevant example to which I can compare things; for 2., if a full proof would take too long to outline, a relevant reference would be appreciated.
Here is an outline of the proof. Let $n:=d$. Suppose that for some real $\delta>0$ the $\delta$-neighborhood of the set $F:=\mathcal F$ can be covered by pairwise disjoint sets $U_1,\dots,U_k$ such for each $j=1,\dots,k$ the boundary of the set $U_j$ is of zero Lebesgue measure and the closure $\bar U_j$ of $U_j$ can be parameterized by a smooth enough map
$$[-1,1]^{n-1}\times[-1,1]\ni(s,t)\mapsto x_j(s,t)\in\bar U_j$$
so that the map
$$[-1,1]^{n-1}\ni s\mapsto x_j(s,0)\in F\cap\bar U_j$$
is onto, that is, a smooth enough parameterization of the "$j$th piece" $F\cap\bar U_j$ of the set $F$.
Let $$J_j(s,t):=\partial x(s,t)/\partial(s,t)$$
be the corresponding Jacobian determinant.
Next, let
$$W_j(s,t):=V(x_j(s,t)).$$
Then $W_j(s,t)\ge W_j(s,0)$ and $W_j(s,t)\sim W''_{j;tt}(s,0)t^2/2$ as $t\to0$, assuming $W''_{j;tt}(s,0)\ne0$ and hence $W''_{j;tt}(s,0)>0$ for all $s$, where $W''_{j;tt}$ is the second partial derivative of $W_j$ in $t$. So, by standard reasoning, for any smooth enough real-valued function $f$ on $(-1,1)^n$ and
$$g_j(s,t):=g_{f;j}(s,t):=f(x_j(s,t)),$$
we have
$$\int_{U_j}dx\,f(x)\exp\{-b^2 V(x)\} \\
=\int_{[-1,1]^{n-1}}ds\,\int_{[-1,1]}dt\,|J_j(s,t)|g_j(s,t)
\exp\Big\{-\frac{b^2 W''_{j;tt}(s,0)t^2}{2+o(1)}\Big\} \\
\sim\frac{\sqrt{2\pi}}b\,
\int_{[-1,1]^{n-1}}ds\,\frac{|J_j(s,0)|g_j(s,0)}{\sqrt{W''_{j;tt}(s,0)}}
$$
as $b\to\infty$.
It follows that the probability measure $P_{b^2}$ converges to the probability measure $P_\infty$ given by the condition
$$\int f\,dP_\infty=\sum_{j=1}^k\int_{[-1,1]^{n-1}}\frac{ds\,|J_j(s,0)|f(x_j(s,0))}{\sqrt{W''_{j;tt}(s,0)}}\Big/\sum_{j=1}^k\int_{[-1,1]^{n-1}}\frac{ds\,|J_j(s,0)|}{\sqrt{W''_{j;tt}(s,0)}}$$
for all nice enough $f$.
thanks, this is super useful. is there a clean way of rewriting $P_\infty$ as a simple density (in terms of e.g. $V, \nabla V$) with respect to Hausdorff measure on $\mathcal{F}$? (ignoring normalising constants if need be)
@πr8 : The density is a function of the point on $F$. How can you express this function without using a parameterization of $F$? All a parameterization does is solving the equation $V(x)=0$ for $x$. So, a parameterization is determined by $V$, albeit not uniquely. E.g. if $n=2$ and $V(u,v)=u^2+v^2-1$, then one possible parametrization of $F$ is given by equations $v=\sqrt{1-u^2}$ and $v=-\sqrt{1-u^2}$.
Thanks for the response - is it true that you need a parameterisation of $\mathcal{F}$? Perhaps I was unclear: I had in mind something more like being able to write e.g. $P_\infty (dx) = \pi ( x ) \mathcal{H} (dx)$, where $\pi$ is some nonnegative function, and $\mathcal{H} (dx)$ is the Hausdorff measure on $\mathcal{F}$. The guess is that, in this setting, $\pi$ would be expressible in terms of $( V, \nabla V)$, a bit like one sees in the co-area formula. I am not an expert in this area, though; my entry point is this work (https://arxiv.org/abs/1206.6913).
@πr8 : I think it may be possible to express the density in a coordinate-free form. However, you'll need a second derivative of $V$, instead of or rather in addition to, the first ones. On the other hand, to compute $\pi(x)$ and $\mathcal H(dx)$, you'll need coordinates anyway.
|
2025-03-21T14:48:30.480550
| 2020-05-03T13:21:22 |
359241
|
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|
Stack Exchange
|
Graph structure on $S_\omega$ induced by fixed points on compositions
Let $S_\omega$ denote the collection of bijections $f:\omega\to\omega$. We say that $f \in S_\omega$ has a fixed point if there is $x\in \omega$ with $f(x) = x$.
It is a short exercise to show that if $f,g\in S_\omega$ then $g\circ f$ has a fixed point if and only if $f\circ g$ has a fixed point.
Let $$E = \big\{\{f,g\}: f\neq g \in S_\omega \text{ and } f\circ g \text{ has a fixed point}\big\}.$$
Question. If $G$ is a finite graph, is $G$ isomorphic to an induced subgraph of $(S_\omega, E)$?
Just to clarify, here "graph" can be viewed as "set endowed with an anti-reflexive symmetric relation".
To make it a finite combinatorics question, one can also ask a slightly stronger conclusion, namely whether $G$ is isomorphic to an induced subgraph of $(S_n,E)$ for large enough $n$ (depending on $G$).
Yes, and in addition we can choose the elements to be fixed-point-free order 2 elements in a finite symmetric group. For order 2 elements, the incidence relation means $f(x)\neq g(x)$ for every $x$.
Let $V$ be your graph (sorry $G$ sounds too group-wise to me): I view it as a graph structure on $\{1,\dots,n\}$.
Start with the discrete graph: consider the $n$ basis elements $g_1,\dots,g_n$ in the group $C_2^{n}$ since the action is free, they satisfy the condition.
For the general case, we start with the same, but letting $C_2^n$ acting freely on the disjoint union of $n(n-1)/2$ copies $X_{i,j}$ of $C_2^n$ (indexed by pairs $(i,j)$ for $1\le i<j\le n$. Let $g_1,\dots,g_n$ be the corresponding initial elements in the symmetric group over $\frac{n(n-1)}{2}2^n$ elements.
Then, for each edge, say between $i$ and $j$ for $i<j$, choose $x$ in the $(i,j)$-copy $X_{i,j}$. Then the $\langle g_i,g_j\rangle$-orbit of $x$ is of cardinal 4: then define $h_i$ as $g_i$ modified on this orbit to coincide with $g_j$ (and $h_i=g_i$ outside $X_{i,j}$ too). Hence, for $k\notin\{i,j\}$, it still holds that $h_k(y)\neq h_i(y)$ for all $y\in X_{i,j}$.
The resulting elements $h_1,\dots,h_n$ satisfy the requirements.
Note 1: said otherwise: I solved the problem for the graph consisting of a single edge, and then it works by taking a disjoint union over all edges.
Note 2: it produced an action on $2^n\frac{n(n-1)}{2}$ elements (that is a map into a group of size $(2^n\frac{n(n-1)}{2})!$) but this huge size is certainly far from optimal. If for the graph with $n$ vertices and a single edge one can do $u_n$ then one can get $u_n\frac{n(n-1)}{2}$, and $u_n$ can be taken much smaller. Already this maps into a group of size $\ge (n^2/2)!$ and is probably not very optimal either.
Edit: here's an alternative construction.
Given the incidence graph on $\{1,\dots,n\}$, consider the Coxeter group: $$W=\langle x_1\dots x_i\mid x_i^2=1, \text{for }i<j\;(x_ix_j)^2=1 \text{ if }i\,—\,j,\;(x_ix_j)^3=1\text{ otherwise}\rangle.$$
For classical theory, $x_i$ has order 2 in $W$ and $(x_ix_j)$ has order $2$ or $3$ according to the prescribed order. Let $W$ first act on itself on the left, and then let $W$ act on the set of (unordered) pairs in $W$. Then $x_ix_j$ acting on $W$ with only 2-cycles or with only 3-cycles: in the first case it fixes many pairs, while in the second case it fixes no pair. Hence the action satisfies the requirements.
Moreover using that $W$ is residually finite, one can pass to a quotient $W/N$ in which the image of the generators are distinct, and then pass the action on pairs, which has the same properties. This ought to be quantitative (i.e., one such quotient should exist with some reasonably not too large index) but I haven't checked.
|
2025-03-21T14:48:30.480811
| 2020-05-03T14:25:35 |
359244
|
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|
Stack Exchange
|
Which curves and surfaces are realizable by linkages? references?
Ok, so I try to formulate rigorously the question in the title, for which I am asking for references. My definitions may be flawed, so feel free to adjust/correct them! I care about dimensions 2 and 3 below, but feel free to mention general d as well.
Let $d\ge 1$ be an integer, $G=(V,E)$ be a graph, $w:E\to [0, \infty)$ be a weight function. A realization in $\mathbb R^d$ with graph $G$ and weight $w$ is a map $P:V\to\mathbb R^d$ with the further property that $|P(v)-P(v')|=w(v,v')$ whenever $\{v,v'\}\in E$. I will identify such $P$ with its image, I hope it's not a problem.
Edit (May 11, 2020):
As pointed out by Misha, this below definition is not correct. The action of isometries makes the set of all realizations of a linkage always cover all $\mathbb R^d$. He indicates a paper in which a more inclusive definition is formulated in $d=2$.
(Previous "wrong" definition: I say that a set $A\subset \mathbb R^d$ is realizable by linkages if there exists $G,w$ as above and an cover of $A$ by open sets of $\mathbb R^d$ such that for every $U\subset\mathbb R^d$ in the cover there exist $G,w$ such that that the union of all (images of) realizations of $G,w$ in $\mathbb R^d$ intersected with $U$ coincides with $A\cap U$.)
To "fix the problem", following the we will allow a subset of vertices of $G$ to be kept fixed in $\mathbb R^d$. In dimension $2$ this apparently generalizes the definition in the above paper, but I think that the result of the paper still allows to reply positively to the $d=2$ case of the question, with little extra work.
Revised definition: We say that $A\subset \mathbb R^d$ is realizable by linkages if there exists a cover of $A$ by open sets of $\mathbb R^d$ such that for every $U\subset\mathbb R^d$ in the cover there exist $G=(V,E)$ and $w$ as above, a subset $F\subset V$, and a map $\phi: F\to\mathbb R^d$, such that the union of all (images of) those realizations of $G,w$ which restricted to $F$ equal $\phi$, intersected with $U$, coincides with $A\cap U$.
Question: Say $d=2$ or $d=3$. Is it true that all algebraic sets $A\subset\mathbb R^d$ are realizable by linkages? What are references for this?
(Note: as of May 11 2020, it appears to me that the case $d=2$ is nicely treated in the answers given, while the case $d=3$ is not yet treated, possibly due to the previously bad definition.)
I found some mention of this, without references on Branko Grünbaum's "Lectures on lost mathematics", dated around 1975, and he says there that $d=2$ case is known, but does not give references, and $d=3$ case is a question by Hilbert which is open (but again no references there).
Your definition of a realizable set is wrong: With this definition, the only realizable subsets are the open subset of ${\mathbb R}^d$. See here for precise definitions.
Thank you Misha, I will edit the question!
Thurston sketched a proof that any real algebraic set is a component of the configuration space of a planar linkage. Kapovich and Milson gave a full proof. Check out this paper by Henry King which gives some history.
https://arxiv.org/pdf/math/9807023.pdf
Erik Demaine and I also included a proof for $d=2$ in Geometric Folding Algorithms:
Linkages, Origami, Polyhedra, Chapter 3.
There we asked if there is a planar (non-crossing) linkage that
"signs your name" (traces any semi-algebraic region), a question
posed by Don Shimamoto in 2004.
This was recently settled positively by Zachary Abel in his Ph.D. thesis:
any polynomial curve $f(x,y) = 0$ can be traced by a non-crossing linkage.
Abel, Zachary Ryan. "On folding and unfolding with linkages and origami." PhD diss., Massachusetts Institute of Technology, 2016.
MIT link.
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2025-03-21T14:48:30.481082
| 2020-05-03T14:55:27 |
359247
|
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|
Stack Exchange
|
Bounding $L^p$ norms in terms of lower-order $L^q$ norms
Suppose $f,g\in L^q(\Omega)$ ($\Omega\subset \mathbb{R}^n$) for all $1\le q\le p$. Here, $L^p(\Omega)$ is defined with respect to some measure $\mu$ that is absolutely continuous wrt Lebesgue measure. Are there bounds on $\int|f-g|^p$ or $\int(f^p-g^p)$ in terms of $\int|f-g|^q$ for $q<p$?
Update. As pointed out in an answer below, without further assumptions this is false. At a high level, what I am trying to understand is the following: I wish to know the rate of convergence of $f_n\to f$ in $L^p$, but all I know is the rate of convergence in $L^q$ for some $q<p$. Can anything be said? It seems necessary to assume, at least, that $f_n,f\in L^r$ for some $r\ge p$.
For example, assuming sufficient regularity and additionally $L^2$ convergence of the gradients, Ladyzhenskaya's inequality is precisely such a bound for the case $q=2$ and $p=4$.
The interpolation inequalities for a $p$-norm only work if you know the $q$-norm and the $r$-norm with $q\leq p \leq r$. I would guess that taking $h = f-g$ (it does not matter what $f$ and $g$ are) given by $h(x) = \big( \frac{d}{dx} (\frac{1}{\ln^{[k]} x} ) \big)^{1/p}$ will get you a counterexample (here $\ln^{[k]}$ is an $k$-times iterated logartihm). The integrand of the $p$-norm is logarithmic [or an interation of such], while the integrand of q norms will be dominated by the power of $x^{-q/p}$. But given the amount of downvote, I suspect there is a textbook example.
PS: I think it's very unpolite that the downvoters and closers did not leave a comment...
For your updated question: if you know that $f_n, f$ are bounded in $L^r$ for $r > p$, then you can directly interpolate to get convergence in $L^p$. (You don't need to assume $f_n \to f$ in $L^r$.) Iosif's example shows that this is sharp: with just $r = p$ this is not enough.
@WillieWong It seems I don't know enough about interpolation, then! What kind of interpolation inequalities give such explicit bounds? (If there is a standard reference, please feel free to share it.)
If $q < p < r$ there exists $\theta \in (0,1)$ such that $1/p = (1-\theta)/q + \theta/r$. Then $$ \int |f|^p = \int |f|^{(1-\theta)p} |f|^{\theta p} \leq \left( \int |f|^q \right)^{(1-\theta)p/q} \left( \int|f|^r\right)^{\theta p / r} $$ by Holder. So if $f_n \to f$ in $L^q$ and $f_n, f$ are uniformly bounded (say by $M$) in $L^r$, you have that $$ \int |f_n - f|^p \leq \left( \int |f_n - f|^q \right)^{(1-\theta)p/q} \left( 2M \right)^{\theta p} $$ using triangle inequality.
Yes, it seems I had forgotten to try the simplest of interpolation bounds. If @IosifPinelis is willing, I would suggest adding this to his answer, and I will accept this as the correct answer.
Such a bound is impossible in general. E.g., suppose that $\Omega=[0,1]$, $g=0$, and $f=a^{-1/p}1_{[0,a]}$, where $p>0$ and $a\downarrow0$. Then for any $q\in(0,p)$ we have $\int\lvert f-g\rvert^q=a^{1-q/p}\to0$, whereas
$\int\lvert f-g\rvert^p=\int(f^p-g^p)=1\not\to0$.
This is a great counterexample. Do you have any idea if this extends to the case where either (a) $\int f^p-\int g^p \to 0$ or (b) $\int |f-g|^p \to 0$?
@JohnA : I am not sure how to understand your comment. Can you specify what kind of bound you want to prove or disprove?
My comment was vague so my apologies; please see my edits to the "example" in my post. I'm curious if there is any relationship between the rate of convergence in $L^q$ vs $L^p$. Again, I suspect, the answer is no, but I am not sure.
@JohnA : I am still not sure about what specifically you mean by "arbitrarily bad". Can you just state it in formal terms? Also, I see no point in saying $c\alpha_n\to0$, because you can always rescale $f_n$ and $f$ (by replacing them, say, by $f_n/(c\alpha_n)^{1/q}$ and $f/(c\alpha_n)^{1/q}$) to get $c\alpha_n$ replaced by $1$. Alternatively, you can similarly rescale $f_n$ and $f$ to get $\int|f-g|^p=\int|(f^p-g^p)=1$, as was done in my example.
It's probably better think of my question as asking "under what reasonable assumptions is such a bound possible". I edited my question to point out an example in Ladyzhenskaya's inequality, although this assumes a bit more than I wanted. (I am not sure about your point on $\alpha_n$, this is simply the rate of convergence.)
@JohnA: Ladyzhenskaya assumes a lot more. In effect it is assuming that your sequence of function converges in $L^p$ and $L^r$, and so necessarily converges in $L^q$ when $q\in (p,r)$. Additionally, in this setting if you have rates in $L^p$, then by interpolation you always get some rate in $L^q$, even without knowing anything about the rate of convergence in $L^r$.
@JohnA : Your question was fully answered. Your new question, concerning Ladyzhenskaya's inequalities (which also involve norms of the gradient), is a wholly different thing. If you now want to ask a general question as to when we have (say) $|f|_p=O(|f|_q)$ with $q\in(0,p)$, then this question is way too broad to be asked on MathOverflow.
Previous comment continued: In probability, cases of the inequality $|f|_p=O(|f|_q)$ with $q\in(0,p)$ are referred to as the hypercontractivity phenomenon -- see e.g. https://projecteuclid.org/euclid.aop/1176990550 and https://www.goodreads.com/book/show/13708363-introduction-to-random-chaos
Previous comment continued: Yet another, rather different kind of studies where we have $|f|_p=O(|f|_q)$ with $q\in(0,p)$ is represented by so-called reverse Hölder's inequalities; see e.g. inequality (3.6) in https://www.ams.org/journals/tran/1993-340-01/S0002-9947-1993-1124164-0/S0002-9947-1993-1124164-0.pdf
The fault is mine. My original question did not mention the requirement that $\int|f-g|^p\to0$, and while your counterexample is nice, it misses the main intent of my question. As I suggest above, we can merge @WillieWong's comment with your answer and this would be satisfactory.
|
2025-03-21T14:48:30.481467
| 2020-05-03T15:16:06 |
359249
|
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|
Stack Exchange
|
What is a module over a Boolean ring?
Recall that a (unital) Boolean ring is a (unital) commutative ring $A$ where every element is idempotent; it follows that $A$ is of characteristic 2. There is an equivalence of categories between Boolean rings and Boolean algebras; the Boolean algebra corresponding to a Boolean ring $A$ (which I'll continue to call "$A$") has the same elements as $A$, and multiplication corresponds to "AND", while addition corresponds to "XOR".
Recall also that Stone duality gives an equivalence between the opposite of the category of Boolean algebras and totally disconnected compact Hausdorff spaces. Under this equivalence, a Boolean algebra $A$ is sent to the space $Spec A$ of ultrafilters on $A$, and $A$ is recovered as the algebra of clopen subspaces of $Spec A$.
Question: Let $A$ be a Boolean ring. Let $M$ be an $A$-module. How can the data of $M$ be described in terms of the Boolean algebra $A$, or better yet in terms of the topological space $Spec A$?
One thing to say is that $M$ is naturally an $\mathbb F_2$-vector space, and the $A$-module structure on $M$ corresponds to a representation of $A$ as a sublattice of the lattice of $\mathbb F_2$-subspaces of $M$. This is nice as far as it goes, but I'd really like a description which doesn't mention vector spaces at all, just like the usual definitions of Boolean algebras or totally disconnected compact Hausdorff spaces don't mention rings at all. For instance, it would be nice if this could be described as some kind of representation of the Boolean algebra $A$ on the powerset lattice of a set or something like that.
One possible direction: If $M$ is an $A$-module, then there is a natural preorder on $M$ where $m \leq m'$ iff there is $a \in A$ such that $m = am'$. The set $Spec M$ of ultrafilters on this preorder carries a natural topology with subbasis given by the sets $\hat m = \{p \in Spec M \mid m \in p\}$, for $m \in M$. There is a natural continuous map $Spec M \to Spec A$ given by $p \mapsto \{a \in A \mid \exists m \in p (am = m)\}$. This yields a faithful functor $Spec : Mod_A^{op} \to Top_{/Spec A}$. I wonder if there some additional structure / properties on $Spec M$ which can turn this functor in an equivalence?
Is "$M$ is a sheaf of $\mathbb{Z}/2\mathbb{Z}$-vector space on Spec $A$" an acceptable answer ? or do you really want something that is purely geometric and also apply some sort of Stone duality to $M$ ?
In the second, case you could clarify the sort of description you would be happy with in the special case $A= \mathbb{Z}/2 \mathbb{Z}$.
@SimonHenry I think a sheafy description would be great. Is it as simple as $M$ being an arbitrary sheaf of $\mathbb F_2$-vector spaces?
It seems that Boolean algebra is used for "unital Boolean ring". So these categories are not equivalent (for instance in the category of Boolean rings, ${0}$ is both initial and terminal, while in the category of Boolean algebras, ${0}$ is terminal while $\mathbf{Z}/2\mathbf{Z}$ is initial).
@YCor I see. For me, rings are by default unital. I'll edit to clarify.
Then "recall" is not really the right word: the convention I said seems largely used by Boolean algebra people. What you're defining as definition of Boolean ring is what I'd view as definition of Boolean algebra (I used to say "unital Boolean algebra" a few times here and was correct. I agree that for a ring-theoretic point of view it sounds weird to mean algebra = unital ring.
It seems most categorically natural for a representation of a Boolean algebra on a set to be a Boolean algebra homomorphism into the set’s Boolean algebra of endo-relations. Maybe that’s an option here.
@YCor In my experience, a "Boolean ring" always has unity. Boolean algebras, of course, are not rings, since neither $\vee$ nor $\wedge$ are group operations. But it is a standard theorem that the category of Boolean algebras is equivalent to the category of Boolean rings (and making this theorem true is a pretty good motivation for including unity in the definition of Boolean ring, even if you don't think all rings have unity). I'm not aware of a standard name for a rng (ring without unity) in which every element is idempotent, but "Boolean rng" seems reasonable.
@AlexKruckman ah, but this "theorem" is so trivial that I view it as an identification. Even at a categorical level it makes no difference to consider $(A,1,+,\cdot)$ and $(A,\wedge,\vee,\neg)$ since homomorphisms are the same (laws and axioms of each one follow formally from those in the other setting— I don't know the standard universal-algebraic formulation). So I understand that some people use Boolean algebra/ring to distinguish between unital/non-necessarily-unital setting, while others use it to distinguish between the equivalent point of views (complemented lattice)/(unital ring).
@YCor Although the theorem is trivial nowadays, it wasn't obvious to Stone: https://www.pnas.org/content/21/2/103
@AlexKruckman The non-unital meaning of "Boolean ring" (as either a kind of non-unital ring or a kind of lattice with a bottom element but no top) is not uncommonly used in measure theory. For instance, in Halmos's classic text.
@RobertFurber I don't mean the Stone representation theorem. I mean the result essentially stating that (complemented lattice + conditions) = (associative unital ring in which every element is idempotent).
@YCor That paper isn't the Stone representation theorem, which was another paper in the same year. It is Stone's proof of exactly what you say (see the second sentence of the article). Maybe I should have linked to the full 3 page version
@RobertFurber thanks for the reference! I guess Stone was aware that his proof was trivial, but observing that it holds (and that it was a useful new point of view) wasn't.
Theorem: Given $A$ a boolean ring/boolean algebra then there is an equivalence of categories between the category of $A$-modules and the category of sheaves of $\mathbb{F}_2$-vector spaces on Spec $A$. The equivalence sends every sheaf $\mathcal{M}$ of $\mathbb{F}_2$-vector space to its space of section, $\Gamma(\mathcal{M})$ which is a module over $\Gamma(\mathbb{F}_2) = A$.
Proof: Spec $A$ has a basis of clopen given by the elements of $A$. Using Grothendieck comparison lemma, this allows to give a more algebraic description of sheaves on Spec $A$ as:
For each $a \in A$ a set $F(a)$, with (functorial) restriction maps) $F(a) \to F(b)$ when $b \leqslant a$ such that the natural maps:
$$ F(a \cup b) \to F(a) \times F(b) $$
is an isomorphism when $a \cap b = 0$.
This allows to exhibit the inverse construction: Given a $A$-module $M$, we define
$$ \mathcal{M}(a) = aM $$
with the restriction map $aM \to bM$ being given by multiplication by $b$. One easily check that if $a \cap b = 0$ then $aM \times bM \simeq (a \cup b) M$ hence $\mathcal{M}$ is a sheaf of $\mathbb{F}_2$-vector space such that $\Gamma(\mathcal{M}) = \mathcal{M}(1) = M$.
Conversely, starting from any sheaf of $\mathbb{F}_2$-vector spaces $\mathcal{M}$ we have $\mathcal{M}(1) = \Gamma(\mathcal{M}) = \mathcal{M}(a) \times \mathcal{M}(\neg a)$, and the action of $a \in A$ on $\Gamma(\mathcal{M})$ is the identity on $\mathcal{M}(a)$ and zero on $\mathcal{M}(\neg a)$. It follows that $a\Gamma(\mathcal{M}) = \mathcal{M}(a)$, and from there we easily check that the two constructions are inverse of each others.
Note that the exact same argument proves more generally that:
Theorem: If $X$ is a stone space, and $\mathcal{A}$ is a sheaf of rings on $X$, then there is an equivalence of categories between sheaves of $\mathcal{A}$-modules and $\Gamma(\mathcal{A})$-modules.
In particular sheaves of abelian groups on $X$ corresponds to module over the ring of locally constant integer valued functions on $X$.
Even more generally (but the proof is more involved) the same conclusion holds if $X$ is an arbitrary locally compact space, $\mathcal{A}$ is a "c-soft" sheaf of rings and $\Gamma$ is replaced by the "compactly supported section" functor. I prove this as proposition 5.1 of this paper, which is about generalizing this sort of theorem when $X$ is not a space but a topos satisfying apropriate local finiteness assumption, but I'm convince this had been observed before, I just do not know a reference for it.
Thanks! Just a note: we must use that $\mathcal M$ is a sheaf of $\mathbb F_2$-vector spaces rather than a general sheaf of abelian groups is in verifying that $\Gamma(\mathcal M)$ is an $A$-module. And indeed, the place it is needed is in verifying that $(a \mathrm{XOR} b)m = am + bm$ for all $a,b \in A, m \in \Gamma(\mathcal M)$. I will probably "accept" this answer, but I'll hold off for a bit in case there are other interesting descriptions which might come up.
@TimCampion : Absolutely, though we do obtain a result for sheaves of abelian group, see the edit.
More generally if K is any commutative ring and X is a stone space then modules for the ring of locally constant functions X to K correspond to sheaves of K-modules on X. These all are easy cases of Pierce's sheaf representation theory.
|
2025-03-21T14:48:30.482066
| 2020-05-03T16:13:51 |
359251
|
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|
Stack Exchange
|
Solution of the Mason-Weaver equation in cylindrical coordinates
I'd like to know if there is some litterature about the cylindrical Mason-Weaver equation.
The basic Mason-Weaver equation is treated in Wikipedia :
$$\partial_t f(t,x)=\partial_x f(t,x)+\partial_{xx}^2 f(t,x)=\nabla\mathbf{v} $$
and the boundary conditions at $x=x_b$ : $$f(t,x)+\partial_{x} f(t,x)=0 $$
It is treated with the separation of variables : $f(x,t)=F(t)G(x)$
But I'm wondering how to proceed with the cylindrical symmetry case :
$$\partial_t f(t,z,r)=\partial_z f(t,z,r)+\partial_{zz}^2 f(t,z,r) +\frac{1}{r}\partial_r(r\partial_r f(t,z,r)) =\nabla\mathbf{v}$$
and with boundary conditions and the boundary $s(x,y)=0$ :
$$\cos(\theta)(f(t,z,r)+\partial_{z} f(t,z,r) )-\sin(\theta)\partial_r f(t,z,r)=0$$
where the vector $\mathbf{n}=(\sin(\theta),-\cos(\theta))$ is perpendicular to the boundary so that $\mathbf{v}.\mathbf{n}=0$
One can also attack the problem with the separation of variables : $f(t,z,r)=F(t)G(r)H(z)$ and solve each equation.
We then obtain :
$$f(t,z,r)=\sum C_{\beta,k,\lambda} e^{-\beta t} e^{-\frac{1}{2}z}(a\cos(\omega_k z)+b\sin(\omega_k z))\operatorname{J}_0(r/\lambda)$$
where $\omega_k^2=1+4(\kappa-\beta)$ and $\lambda=1/\sqrt{\kappa}$ and $\operatorname{J}_0$ the cylindrical Bessel function (Bessel Function of the First Kind) - whose derivative at $r=0$ is zero.
But I'm then stuck how to find the good constants in order to satisfy the boundary conditions.
Would you have an advice to tackle this problem or recommend articles if this case has already been treated in the litterature please ?
EDIT :
Here is what I thought for a geometry of a well where the boundary is parametrized by : $z=R-\sqrt{R^2-r^2}$ for $0<x<R$.
First at $r=R$, the boundary condition is : $G'(r)=0$, and at $r=0$, $H(z)+H'(z)=0$.
Hence, $R/\lambda$ is a zero of $\operatorname{J}_0'(x)\sim\operatorname{J}_1(x)$.
Moreover, the boundary condition on $r=0$ yields the same result as in $1d$, meaning $\omega_k=\frac{k\pi}{z_{max}-z_{min}}$ (see there)
Now I'm working on how to deal with the other boundary conditions.
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2025-03-21T14:48:30.482340
| 2020-05-03T16:14:40 |
359252
|
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|
Stack Exchange
|
Need help in understanding meaning of a notation and theorem used in research paper due to a reference being in German Language
I thought of utilizing this lockdown period to study research papers in number theory by myself.
I began reading the research paper By T Estermann ->" On Goldbach Problem : Proof that Almost all Even Positive Integers are sum of two primes". I have read and understood all the proof except one equation and that too due to reason that it uses a equation as a reference from German Book " E Landau, Vorlesungen über Zahlentheorie I ( Liepzig, 1927) . I don't understand German.
I thought of posting this question here in hope that someone must have studied this research paper and also in case a researcher understanding German language wishes to help.
Image of the equation I don't understand ( equation 49 on Page 7 ) ! Equation 49]1
The problem-> 1.I am unable to understand what ( 2|m) means in (49) ?(Notation (2|m) is not used in research paper before) 2.what is the statement of theorem is used to derive (49)? . . Everything except this theorem is Understood in paper.
Images from Original German book->
What i did to resolve the issue -> I tried to find English Edition but couldn't. 2.Then I tried to translate whole pdf book which didn't materalized due to book being a bit long. 3.When I split the pdf I wanted to translate theb it Shows Error 403.
4. I changed the screenshot images( from archive.org) to pdf but again error in translation.
It is my humble request if anybody can tell me meaning of notation (2|m) ?
And it will be very much more helpful if someone tell what is the statement of theorem / result used to derive equation 49.
I shall be really thankful for any help offered.
$m \mid n$ means that $m$ divides $n$. In particular, $2 \mid m$ means that $m$ is even. "m gerade" in the original German means "m is even".
Type it in google translator... the words only, not the math. I have done this with some success.
You should better use "deepl.com", which in the opinion of everybody I know of is lightyears ahead of google translate for translations between English, German, and French.
Estermann refers to formula (239) which is just a formula, no words (except the assumption that $m$ is even). Of course you also need to understand how he proves it.
@KlausNiederkrüger thank you
very much for telling about deepl. I am using it now. It is good
Look at the first page of this paper -->
Daniel A. Goldston, Julian Ziegler Hunts, Timothy Ngotiaoco, The Tail of the Singular Series for the Prime Pair and Goldbach Problems, Funct. Approx. Comment. Math. 56, Number 1 (2017) pp 117–141, doi:10.7169/facm/1602,
arXiv:1409.2151.
@user155294 currently I am reading the German edition of this book by translating line by line using deepl. com. I have a question does on page 226 ( please see image 1) D (m, n) > $ c_{4, 3} $ means D( m, n) is greater than some constant?
@Tim Green Yes. Moreover $S(m)\approx D(m,n)$ for $n>c_{4,2}$ and $m$ is even, with some big value for $m$ and $c_{4,2}$ and $0<m\leq n$.
I have two comments for you, putting them as an answer because I don't have enough credits for a comment:
First, the notation $2\mid m$ should be very common, and it is definitely very common in German books. As an example, I own a copy of Algebra from Siegfried Bosch, first edition from 1992, and it uses this notation as well.
Second, I think it is great that you take the extra effort to read the original text and that you are translating it line by line. So the following post on MO could be very interesting for you: Do you read the masters?
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2025-03-21T14:48:30.482631
| 2020-05-03T16:26:57 |
359254
|
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|
Stack Exchange
|
What is the name of the real form corresponding to the quaternionic symmetric space?
Let $G$ be a compact simple Lie group. Choose a system of positive roots, and let $\mathrm{SU}(2) \subset G$ correspond to the highest root, and $\mathbb{Z}/2 \subset \mathrm{SU}(2)$ the centre. The centralizer of this $\mathbb{Z}/2$ inside of $G$ is a subgroup $H \subset G$ of shape $\mathrm{SU}(2) \circ K = (\mathrm{SU}(2) \times K) / (\mathbb{Z}/2)$. The Dynkin diagram for $H$ can be found by drawing the affine dynkin diagram for $G$, and deleting the node(s) adjacent to the affine root. The now-isolated affine root is the copy of $\mathrm{SU}(2)$, and the rest of the Dynkin diagram for $K$. In the type-A case, $H$ is reductive but not simple, picking up a $\mathrm{U}(1)$ factor; this is because in that case the affine root had two neighbours, not just one. The list of $H$'s is available in the Wikipedia article Quaternion-Kähler symmetric space, because the quotient spaces $G/H$ are precisely the quaternionic symmetric spaces.
Standard arguments then say that $G$ has a real form with maximal compact $H$. It is not the compact form (except for $G = H = \mathrm{SU}(2)$), and it is usually not the split real form. Rather, it is a third canonical real form for any group. For the classical series, it is $\mathrm{SU}(2,n-2)$, $\mathrm{SO}(4,n-4)$, and $\mathrm{Sp}(1,n-1)$. If I am reading Wikipedia correctly, then, together with $\mathrm{SO}(3,n-3)$, these are the real forms that admit quaternionic discrete series representations.
Does this canonical real form have a standard name in the literature?
This is a fairly trivial answer, but the term “quaternionic real form” is found by Google in a number of texts and seems to correspond to what you're describing. (Thanks for pointing out that this form is canonical, I had never realized this!)
In the Crelle paper by Gross and Wallach “On quaternionic discrete series representations, and their continuations” (J. Reine Andgew. Math. 481 (1996) 73–123, available here), the form you mention is described in §3 and called “the quaternionic real form”.
There are other references to “quaternionic real form” found by Google, and all seem to refer to the same thing. I have seen it regularly at least to denote the form of $E_8$ with Cartan index $-24$. So I think we can say that “quaternionic real form” is a reasonably standard term.
(There is still at least some possibility of confusion as this question uses the term, albeit with quotes, to denote a different real form of the $D_n$ series.)
Thanks. It feels like a good name. But I didn't know if it had another one.
Actually, I don't know which form the linked question is referring to. I should be able to divine it: it is a form of Spin(4n) in which one of the half-spin representations is real and the other is quaternionic.
@TheoJohnson-Freyd Regarding the form mentioned in the linked question (which, again, is not the one you were asking about), the notation $SO^*(n)$ is fairly standard, and it's the only real form of $SO(n)$ (for $n$ even) that's not a $SO(k,n-k)$. Its maximal compact subgroup is $U_n$ and it's called DIII in Cartan's notation.
Thanks! I do know that real form, I just didn't know its name, and I don't immediately see why it would be called "quaternionic".
|
2025-03-21T14:48:30.482924
| 2020-05-03T16:54:37 |
359256
|
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|
Stack Exchange
|
Characterisation of (integrable) connections on (trivial) principal bundle
Let $M$ be a manifold. Let $G$ be a Lie group and $\mathfrak{g}$ be its Lie algebra.
Let $P(M,G)$ be a principal bundle. Recall that, a connection on $P(M,G)$ is a distribution $\mathcal{H}\subseteq TP$ satisfying certain conditions. Equivalently, a $\mathfrak{g}$-valued differential forms on $P$ satisfying certain conditions.
Question : Is there a characterization of connections on $P(M,G)$; in the sense, a one-one correspondence between the set of connections on $P(M,G)$ and some "well-described set"?
Question : Can we say something if we restrict to the case of integrable connections?
Questions : Can we say something when the bundle is trivial bundle? A characterization for integrable connections on trivial principal bundle?
Edit : May be it is already clear but, I want to specify that I am seeing two connections $\omega, \omega’$ to be same if there exists an isomorphism $\varphi :P\rightarrow P’$ Of vector bundles such that $\varphi^*\omega’=\omega$, that is the pullback of $\omega’$ is equal to $\omega$.
Consider the trivial bundle $pr_1:M\times G\rightarrow M$. See that, for each $(m,g)\in M\times G$, the kernel of $(pr_1)_{*,(m,g)}=T_gG$. The assignment $(m,g)\mapsto T_mM$ then defines a connectionon on the bundle $M\times G\rightarrow M$. This is an integrable connection...
Connections are affine over $\Omega^1(M;\mathfrak g)$, and in particular form an affine space. Integrable connections are the same thing as flat conns, which are the zeroes of the curvature map $F_A$. Such a map gives rise to a holonomy map $\pi_1 M \to G$. The space of all flat connections on $P$ carries the action of the automorphism group of $P$, and the space of homomorphisms to $G$, the action of conjugation. After quotienting this descends to a bijection from the space of flat conns to the subset of $\text{Hom}(\pi_1 M, G)$ so that $\tilde M \times_{\pi_1 M} G \cong P$ as G-bundles.
If your question is how to tell what the topological type of the bundle $\tilde M \times_{\pi_1 M} G$ is given the homomorphism $\rho: \pi_1 M \to G$, you will probably first want to know how to determine the isomorphism classes of principal G-bundles over M, and then to determine it by hand for your representation $\rho$... in particular it's not hard if $G = U(1)$!
What is $\tilde{M}$ here? Is it the universal cover of $M$? @MikeMiller Can you suggest some reference for “After quotienting this descends to a bijection from the space of flat conns to the subset of $Hom(\pi_1 M,G)$”
@MikeMiller any reference regarding your second comment would be useful.
Yes, universal cover. The stuff in the first comment is in most references on connections; I am surprised to not see it in Kobayashi-Nomizu vol 1, but see eg Taubes, "Differential geometry: [...]", Theorem 13.2 (here he does not separate the set of flat connections by the topological type of the bundle, but rather just lists off all flat connections on all principal bundles). I am not sure what it would mean to give a reference for my second comment, unless you mean the comment about $U(1)$, in which case try playing with UCT and the LES in homology coming from SES of the coefficient groups.
@MikeMiller I have checked Kobayashi before asking you.. It was not there.. So, I had to bother you... I see the theorem.... I will read that... Frankly, I do not completely understand the second comment.. I will come back to second comment after reading that theorem 13.2.. Thanks for the reference...
Atleast for trivial Principal $G$ bundles a partial answer to your question is mentioned in the Theorem 1 in https://arxiv.org/pdf/1003.4485.pdf. Also for non trivial Principal $G$ bundles a partial answer can be found in the section "Idea" in https://ncatlab.org/nlab/show/connection+on+a+bundle where you may find a one-one correspondence(not sure about 1 direction) between connections on Principal $G$ bundles over $M$ and an appropriate subset of functors from Path groupoid of $M$ to the Atiyah Lie Groupoid of the principal $G$ bundle.
@AdittyaChaudhuri +1 for Theorem $1$ that gives a one-one correspondence between $\mathfrak{g}$-valued $1$-forms on $M$ and the set of connections on the trivial principal bundle $M\times G\rightarrow M$... I do not understand the other half of the theorem that says there is a one one correspondence between the above mentioned set and the set of "smooth" functors $\mathcal{P}_1(M)\rightarrow G$.. I never understood what is smooth structure on $\mathcal{P}_1(M)$.. Do you want to shre your thoughts on smoothness of these functors?
@PraphullaKoushik Sure! Please checkout this link https://ncatlab.org/nlab/show/smooth+structure+of+the+path+groupoid
@AdittyaChaudhuri I mean to share your understanding of smooth structure, not the link :P I have seen that before... Did not understand much...
@PraphullaKoushik In Generalised smooth space (like diffeologocal space, Chen space..etc) quotient always exist which fails in smooth manifolds. (see https://arxiv.org/pdf/0807.1704.pdf)Though Path groupoid in general do not have standard manifold structure but can be considered as generalised space (According to https://www.jstor.org/stable/pdf/1970846.pdf?seq=1 set of piecewise differentiable paths has a Chen space structure.Now underlying set of Path groupoid is the quotient of set of path(here thin homotopy) ... hence quotient is also Chen space which is an example of gen smooth space.)
:) See if you can make it as an answer by adding as many details as you can..
@PraphullaKoushik Path groupoid can also be treated as a diffeological space (see the definition of path groupoid here https://ncatlab.org/nlab/show/path+groupoid)
@PraphullaKoushik To make it an answer my explanation should be complete. For example I did not talk about "integrable connections" at all(I do not have much idea about those right now.). So it's better if an expert(which I am definitely not) answer your question with full details so that we both can learn from it :).
@PraphullaKoushik Can you please define what do you mean by an integrable connection on a principal $G$ bundle?
A distribution $\mathcal{H}\subseteq TP$ on a manifold $P$ is said to be an integrable distriibution if, for any two vector fields $X,Y:P\rightarrow TP$ that lies in $\mathcal{H}$, in the sense $X(p),y(p)\in \mathcal{H}_pP$ for each $p\in P$, then, the Lie bracket $[X,Y]$ lies in $\mathcal{H}$, in the above sense.. A connection on a principal bundle is given in terms of distribution.. If it is integrable, we call it integrable connection
In this site, at least for my questions, partial answers are welcome... You can add what ever you know.. Not a problem at all..
@PraphullaKoushik Thanks for the notion of integrable connection. Ok I will try to give the answer :).
@PraphullaKoushik on the trivial principal bundle, a $\mathfrak g$-valued form $\omega$ corresponds to an integrable connection if and only if $\omega$ satisfies the Maurer-Cartan equation $d\omega + \frac{1}{2} [\omega,\omega] = 0$.
If I understand your question this paper answered things for me a while back: https://arxiv.org/pdf/0705.0452
This is not an answer. This is in response to Mike Miller's comment.
Let $M$ be a manifold, $\tilde{M}$ to be its associated universal cover (a simply connected covering space over $M$). I do not understand why they are not assuming $M$ to be connected. I am assuming $M$ is a connected manifold.
The following result is from the book Differential Geometry - Bundles, Connections, Metrics and Curvature by Clifford Henry Taubes.
Theorem $13.2$ (classification theorem for flat connections) says that, there is a bijection between the sets $\mathcal{F}_{M,G}$ and the set $\text{Hom}(\pi_1(M),G)/G$ where,
$\mathcal{F}_{M,G}$ denote the set of equivalence classes of pairs $(P,A)$, where $P\rightarrow M$ is a principal $G$ bundle, and $A$ is a flat connection on $P(M,G)$. Two pairs $(P,A)$ and $(P',A')$ are equivalent, if there is an isomorphism of principal $G$-bundles $(\varphi,1_M):(P,M)\rightarrow (P',M)$ such that $\varphi^*A'=A$ (pullback of connection $A'$ is the connection $A$).
$\text{Hom}(\pi_1(M),G)/G$ denote the set of equivalence classes of group homomorphisms $\pi_1(M)\rightarrow G$. Two morphisms $f_1:\pi_1(M)\rightarrow G$ and $f_2:\pi_1(M)\rightarrow G$ are equivalent if there exists $g\in G$ such that $f_1=gf_2g^{-1}:\pi_1(M)\rightarrow G$.
The bijection $\mathcal{F}_{M,G}\rightarrow \text{Hom}(\pi_1(M),G)/G$ is given as follows:
given a principal bundle $P(M,G)$ with flat connection $A$, we get a group homomorphism $\pi(M)\rightarrow G$. Its equivalence class gives an element in $\text{Hom}(\pi_1(M),G)/G$.
Let $\rho:\pi_1(M)\rightarrow G$ be a representative of an element in $\text{Hom}(\pi_1(M),G)/G$. Consider the trivial principal $G$-bundle $\tilde{M}\times G\rightarrow \tilde{M}$. The map $\rho:\pi_1(M)\rightarrow G$ given an action of $\pi_1(M)$ on $G$, which in turn gives an action of $\pi_1(M)$ on $\tilde{M}\times G$. Thus, trivial principal bundle $\tilde{M}\times G\rightarrow \tilde{M}$ induce
$(\tilde{M}\times G)/\pi_1(M)\rightarrow M$. Thus, we get a principal $G$-bundle over $M$, which we denote by $\tilde{M}\times_{\rho}G\rightarrow M$. It turns out that, there exists a $\mathfrak{g}$-valued $1$-form on $\tilde{M}\times_{\rho}G\rightarrow M$ whose pullback to $\tilde{M}\times G\rightarrow \tilde{M}$ is the canonical connection on the trivial bundle. It turns out that this $\mathfrak{g}$-valued $1$-form on $\tilde{M}\times_{\rho}G\rightarrow M$ is a flat connection on the principal bundle $\tilde{M}\times_{\rho}G\rightarrow M$. Thus, we get a principal bundle $(P_\rho,A_{\rho})$. Take its equivalence class to get an element in $\mathcal{F}_{M,G}$.
It is not clear how does this answer the question:
Given a principal bundle $P\rightarrow M$, how does one know for what
$\rho:\pi_1(M)\rightarrow G$, do we get that that there is an
isomorphism of principal bundle $P\cong \tilde{M}\times_{\rho}G$?
I am also interested in only "different" connections in the sense if two connections on $(P,M)$ are related by an isomorphism, in the sense of pullbacks, then I am calling these to be same.
If anyone wants to see more details, who do not have the book, I can give more details...
It sounds like the missing piece for you is this image. As I said in my comment: $\rho$ is in the image of $\mathcal F_P$ iff the $G$-bundle $(\widetilde M \times_\rho G))$ is isomorphic to $P$ as a G-bundle, where the subscript $\rho$ indicates we identify $(x, g) \sim (\gamma x, \rho(\gamma) g)$ for all $\gamma, x, g$, thinking of $\gamma \in \pi_1 M$ as a deck transformation. This is as explicit as you will ever get in full generality. If you have a specific bundle over a specific manifold (or your group G is not complicated) you may be able to get better answers by ad hoc techniques.
@MikeMiller I have clarified that part.. Do you have a procedure to write this down explicitly if I have a principal bundle over a manifold when the group G is reasonably good.. Any reference is welcome.. I enjoyed reading chapter 13 of Taubes book..
|
2025-03-21T14:48:30.483592
| 2020-05-03T17:04:45 |
359257
|
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|
Stack Exchange
|
Do Borel isomorphisms of Polish spaces preserve the Baire property?
Let $X$ and $Y$ be uncountable Polish spaces and let $h\colon X\to Y$ be a Borel isomorphism. Suppose that $A\subset X$ has the Baire property. Must $h[A]$ have the Baire property in $Y$ too?
Just to be sure, a Borel isomorphism means a bijection that is Borel as well as its inverse? [Oh, after reading Noam's answer I'm sure now.]
No, it need not.
Take for example $X=Y=\mathbb{R}$, let $A=[0,1]$, let $B$ be some nowhere dense perfect subset of $[2,3]$, and consider a Borel isomorphism $h:\mathbb{R}\cong \mathbb{R}$ swapping $A$ and $B$ and fixing everything else.
Then for each $U\subseteq A$ the preimage $h^{-1}(U)$ is meager in $\mathbb{R}$, hence has the property of Baire. Taking some $U\subseteq A$ which doesn't have the property of Baire we get that $h$ does not preserve the Baire-ness of $h^{-1}(U)$.
|
2025-03-21T14:48:30.483689
| 2020-05-03T17:22:04 |
359259
|
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|
Stack Exchange
|
Two questions on history of symplectic geometry in the 80's
I have a question about the history of two important results from the eighties in symplectic geometry. In both cases it seems that important results were developed (almost) simultaneously by different people, which I find quiet remarkable and unusual. I would be interested in knowing if the corresponding researchers were aware of each other, working on the same projects, and if so at what stage of their project they did become aware of the other one. (Highly hypothetical: do you think that these days people accept more easily to publish a joint paper in such a situation?)
There is the convexity result by
Atiyah, Michael F., Convexity and commuting Hamiltonians, Bull. Lond. Math. Soc. 14, 1-15 (1982). ZBL0482.58013. and by Guillemin, V.; Sternberg, S., Convexity properties of the moment mapping, Invent. Math. 67, 491-513 (1982). ZBL0503.58017. Both articles have been published in the same year... of course, it takes some time to get a paper published and there was also nothing like the arxiv to maybe have a better idea of the developments.
I'm also surprised because the localization result Duistermaat, J. J.; Heckman, G. J., On the variation in the cohomology of the symplectic form of the reduced phase space, Invent. Math. 69, 259-268 (1982). ZBL0503.58015. was published the same year, followed by Berline, Nicole; Vergne, Michele, Zeros d’un champ de vecteurs et classes characteristiques équivariantes, Duke Math. J. 50, 539-549 (1983). ZBL0515.58007., which seems to describe a similar thing.
I admit that I do not have a very precise question, but if anybody could give some insight on how so many interesting things developed in such a short time by different authors, I would be extremely interested.
We often have migrations suggested to HSM that are resisted because the askers want to get MO's opinion, but this sort of highly opinion-based question seems like it really belongs there much more than here.
The work by Atiyah and Guillemin-Sternberg were indeed done independently. I believe Kostant may have also proved the same theorem independently. I have often told the story about how Atiyah, Guillemin, and Sternberg discovered they had proved the same theorem. Atiyah was giving a seminar talk about the theorem at Harvard. As the seminar progressed, the front row became more and more agitated. Finally, Raoul Bott spoke up, saying something like "Michael! You should know that these two gentlemen proved the same theorem two weeks ago!" and pointed to Sternberg sitting next to him and Guillemin.
@DeaneYang Thank you very much for this story. This is very interesting. I would have expected that somebody had come up with the conjecture, and that different people would have worked in parallel to solve the claim, but not that they also came up independently with the statement. I can imagine what a shocking moment this must be to discover somebody else is just giving a talk about "your" result.
@LSpice I did not know about HSM. Maybe you are right, but I think migrating my question to HSM is just the same as deleting it.
Obligatory mention of the Wikipedia article on multiple discovery. These examples could perhaps be added to Wikipedia's list.
|
2025-03-21T14:48:30.483937
| 2020-05-03T17:25:00 |
359260
|
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|
Stack Exchange
|
Natural $\Pi_1$ sentence independent of PA
Order invariant graphs and finite incompleteness by Harvey Friedman gives an example of a combinatorial/non-metamathematical $\Pi_1$ sentence that is independent of ZFC. Is there a simpler example of a combinatorial/non-metamathematical $\Pi_1$ sentence that is independent of PA?
I would say that the word problem for finitely presented groups is not metamathematical, although it asks for an algorithm. Its history goes back to before anyone was thinking about undecidability much. Its specific instances are $\Pi_1$ and some of them are undecidable. Whether concocting a specific undecidable one is natural is up to you, I guess. Same with Hilbert's tenth problem.
My answer was bad, so I will delete it. But I wanted to save the two enlighting comments to my answer: Goodstein's theorem is a statement of the form $\forall\exists$ (i.e. $\Pi_2$) so unfortunately it's not $\Pi_1$. – user76284 The standard combinatorial statements from the 80s can all be stated as claims that some recursive function is total. Any such statement is $\Pi_2$, and all the famous examples of this kind are properly $\Pi_2$. – Andrés E. Caicedo
Considering that Harvey's goal is to achieve simple and natural examples of undecidability, perhaps you have found the right source already.
As a step in this direction (giving a natural-looking scheme rather than a single sentence), look at Anton Freund's recent A mathematical commitment without computational strength.
@user76284, tbh if you write down a not-too-large system of integer diophantine equations and ask if it has a solution, that seems to wash away the metamathematicality enough that you could possibly convince an unsuspecting 19th century mathematician that it came up in some other context. But of course for suitable coefficients, the insolubility can be unprovable in PA (MRDP theorem).
You may look at Shelah's paper
``On logical sentences in PA''.
For a modern exposition of Shelah's work and an alternative example see ``Independence in Arithmetic: The Method of (L, n)-Models''
|
2025-03-21T14:48:30.484135
| 2020-05-03T17:36:29 |
359262
|
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|
Stack Exchange
|
Gradient condition implies Hörmander condition
We have tempered distribution $K$ in $\mathbb{R}^n$ which coincides with a locally integrable function in $\mathbb{R}^n\setminus \{0\}$. We call the condition
$$\int_{|x|>2|y|}|K(x-y)-K(x)|dx\leq B \hspace{1cm}\forall y\in \mathbb{R}^n$$
for some constant $B$, the Hörmander condition.
I want to show that if
$$|\nabla K(x)|\leq \frac{C}{|x|^{n+1}} \hspace{1cm}\forall x\neq 0$$ then the Hörmander condition holds.
For reference, it is from page 92 of the book 'Fourier Analysis' written by Javier.
What I have tried is that, by mean value theorem, for fixed $y$ and for each $x$, there is $t\in [0,1]$ with such that $z(x)=tx+(1-t)y$ and
$$\int_{|x|>2|y|}|K(x-y)-K(x)|dx =\int_{|x|>2|y|} |\nabla K(z(x))||y|dx\leq \int_{|x|>2|y|}\frac{C}{|z(x)|^{n+1}}|y| dx$$
We are able to say $|z(x)|^{n+1}\geq |tx+(1-t)y|^{n+1}\geq (t|x|-(1-t)|y|)^{n+1}$.
But my problem here is that $t$ is dependent on $x$ so I currently cannot see how I can get further.
Thanks in advance for any helps!
You can write $K(x-y)-K(x) = -\int_0^1\nabla K(x-ty)\cdot y,dt$ and then use Fubini.
@user90189 Thank you!
I figured out proving it by using mean value theorem (thanks to @WillieWong). Observe that
\begin{align*}
\int_{|x|>2|y|}|K(x-y)-K(x)|dx &\leq \int_{|x|>2|y|}|\nabla K(tx+(1-t)(x-y))||y| dx\\
&\leq \int_{|x|>2|y|} \frac{C}{|x-(1-t)y|^{n+1}}|y|dx\\
&\leq \int_{|x|>2|y|} \frac{C}{[|x|-(1-t)|y|]^{n+1}}|y|dx&\because |x|>|y|\\
&\leq \int_{|x|>2|y|} \frac{C}{[|x|-|y|]^{n+1}}|y| dx&\because t\in (0,1)\\
&\leq \int_{|x|>2|y|} \frac{C}{[|x|-\frac{|x|}{2}]^{n+1}}|y| dx &\because \frac{|x|}{2}>|y|\\
&\leq \int_{2|y|}^{\infty} \frac{2^{n+1}C}{r^{n+1}}\sigma(n) r^{n-1}|y|dr\\
&=\frac{2^{n+1}C}{2|y|}\sigma(n)|y|\\
&={2^nC\sigma(n)}
\end{align*}
where $\sigma(n)$ is the measure of unit sphere in $\mathbb{R}^{n}$.
First line is missing a $y$: it should be $y \cdot \nabla K$ using the fundamental theorem of calculus. Third line: I absolutely don't see how you just factored out the vector $y$. Instead, why not $$ |x - ty| \geq |x| - t |y| \geq |x| - t |x| / 2 \geq |x|/2 $$ so the third line should read $$ \leq \int_0^1 \int_{|x| > 2|y|} \frac{C|y|}{|x|^{n+1}} dx ~dt $$ and the result trivially follows.
@WillieWong If we think of fundamental theorem of calculus, the vector $y$ in the comment should be just a directional unit vector.
I think the third line has a problem. Let me try what you suggested.
@WillieWong Thank you! I made it more clear even though the bound is still depend on $y$ which I did not expect.
No, $y$ is not a "unit" vector. All the $y$s that appear in https://mathoverflow.net/questions/359262/gradient-condition-implies-h%c3%b6rmander-condition/359354?noredirect=1#comment904433_359262 mean the same thing, and is an arbitrary vector in $\mathbb{R}^n$. If you integrate $\int f'(ta) dt $ you get $\frac{1}{a} f(ta)$, not $f(ta)$.
@WillieWong You are right. I thought wrongly. Thank you!
|
2025-03-21T14:48:30.484457
| 2020-05-03T19:49:38 |
359273
|
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"authors": [
"Ady Fall",
"Liviu Nicolaescu",
"Ryan Budney",
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|
Stack Exchange
|
Poincare duality-differential geometry
Let $ M $ be a smooth and compact manifold with boundary $\partial M = X \times F $ on which the structure of a smooth locally trivial bundle $$ \pi: \partial M \longrightarrow X $$
where the $ X $ and the fiber $ F $ are smooth compact manifolds without boundary. Consider the equivalence relation on the set M
\begin{equation}
z \sim z^{\prime} \Longleftrightarrow z = z^{\prime} \quad \text {or} \quad (z, z^ {\prime} \in \partial M \quad \text{and} \quad \pi(z) = \pi (z^{\prime})).
\end{equation}
We define the topological space $ N = M / \sim $ as the quotient space of the manifold M with respect to the equivalence relation above.
Informally speaking, $ N $ is obtained from $ M $ (by contracting the fibers of the bundle $ \pi $ to points). The set $ N $ is a disjoint union $ N = X \sqcup M^{\circ} $ of the manifold $ X $ and the interior $ M^{\circ} $ of $ M $. The natural projection of $$ p: M \longrightarrow N $$ coincides with the identity map on $ M ^ {\circ} $ and the projection $ \pi $ on $ \partial M $.
So the manifold $N$ can be not smooth sometimes. The pair $(M,\pi)$ is called a manifold with fibered boundared.
How to define the map $I : H^{n-k}_{dR}(M,\pi)\longrightarrow H_{k}(N)$ when $F$ is not a singleton?
How do you obtain $N$?
I reformuled my question. Compte on you help.
Dear Ryan, I hope it's clear how I define the problem.
Is the locally-trivial bundle $M$ or its boundary? And you have not said what the map $I$ is supposed to be. Is there a situation where your map has a name?
the boundary is the locally trivial bundle. I can want to define any in case where $F$ is not a singleton. Assume that is $F$ is a singleton, then projection pi is trivial and in that case $M=N$. I suppose now the case where $F$ is not a singleton. Then $N$ may be not smooth. I wanna define a map (any map) which holds when is not smooth.
In the first line of your question you say the boundary is a product. This could be confusing.
You are right. I hope the question is clear now.
Hello, Dear Ryan! Any idea about the problem?
Note that $N$ is homeomorphic to the union of $M$ with $\DeclareMathOperator{\Cyl}{Cyl}$ the mapping cylinder $\Cyl(\pi)$ of the bundle projection $\newcommand{\pa}{\partial}$ $\pi:\pa M\to X$. Denote by $M^\circ$ the interior of $M$.
Observe that the Poincare Duality for $M^\circ$ (or equivalently for $(M,\pa M)$) implies
$$
H^{n-k}_{dR}(M)\cong H^k_{dR}(M^\circ) \cong H^k_{cpt}(M^\circ).
$$
The extension by $0$ defines a morphism
$$
H^k_{cpt}(M^\circ)\to H^k_{cpt}(N)\cong H^k(N)\cong \mathrm{Hom}\big(H_k(N),\mathbb{R}\big).
$$
Comment. Above I assumed that $H^\bullet_{dR}(M,\pi)=H^\bullet_{dR}(M)$. Now observe that
$$
H^{n-k}_{dR}(M)\cong H_k(M,\pa M).
$$
From the excision property of homology we deduce that the inclusion
$$
(M,\pa M)\hookrightarrow (N,\Cyl \pi)
$$
induces an isomorphism
$$
H_k(M,\pa M)\cong H_k(N,\Cyl \pi).
$$
If $H_k(X)=H_{k-1}(X)=0$, then $H_k(N)\cong H_k(N,\Cyl \pi)$.
Thanks Liviu. But how can I define exiplicitly the map $I$ when N is not smooth? In the case where $N$ is not smooth, how can I define a map $$I : H^{n-k}{dR}(M, \pi)\longrightarrow H{k}(N)$$
Since this comment is a bit longer I will add it to my answer.
Thanks Liviu. thanks very much. So, could you define the map $I$ please. hint $\omega \in H^{n-k}{dR}(M)$. How to sent $\omega$ to $ H{k}(N)$. what is equal $I(\omega)$?
This map is not always well defined. The natural map is $$H^{n-k}_{dR}(M)\to H^k(N )\cong\text{Hom};\big(; H_k(N),\mathbb{R};\big)$$
|
2025-03-21T14:48:30.484979
| 2020-05-03T20:48:00 |
359276
|
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|
Stack Exchange
|
Why the Komlós theorem is not valid for any sequence of measurable functions?
I read an article, and they use a certain theorem, called Komlós theorem, which says:
Theorem 1 (Komlós theorem)
Let $(E,\mathcal {A}, \mu ) $ be a finite measure space and $ (f_n)_{n\geq 1} \subset \mathcal {L}_{\mathbb {R}}^1$ is a sequence with : $$\sup_n \int_{E}{|f_n| d\mu} < \infty .$$
Then there exist $ h _{\infty} \in \mathcal {L}_{\mathbb {R}}^1 $ and a sub-sequence $ (g_k)_k $ of $(f_n)_n $ such that for every sub-sequence $ (h_m)_m $ of $(g_k)_k$ : $$ \frac{1}{i}\sum_{j=1}^{i}{h_j}\to h _{\infty} \text{ a.s. }$$
The original Komlós theorem concerns $\mathcal{L}^1_\mathbb{R}$-bounded sequence of functions. The following theorem gives a similar result for nonnegative valued measurable functions.
Theorem 2
Let $ (f_n)_{n\geq 1}$ be a sequence of nonnegative valued measurable functions.
Then there exist a sub-sequence $ (g_k)_k $ of $(f_n)_n $ and a measurable function $h _{\infty}$ such that for every sub-sequence $ (h_m)_m $ of $(g_k)_k$ : $$ \frac{1}{i}\sum_{j=1}^{i}{h_j}\to h _{\infty} \text{ a.s. }$$
My problem:
Why the theorem 2 is not valid for any sequence of measurable functions? I am looking for a counterexample and would appreciate any ideas.
I made a small edit to the last sentence in order to slightly soften its tone.
$\newcommand\om{\omega}$ $\newcommand\Om{\Omega}$
We need to construct an example of a finite measure space $(E,\mathcal A,\mu)$ and a sequence $(f_n)$ of real-valued measurable functions on $E$ such that for any subsequence $(g_k)$ of the sequence $(f_n)$ and any measurable function $g_\infty$ with values in $[-\infty,\infty]$ we have $\mu(\{x\in E\colon \frac1K\,\sum_{k=1}^K g_k(x)\not\to g_\infty(x)\})>0$.
Let $(R_n)$ be the sequence of independent Rademacher random variables (r.v.'s) defined on some probability space $(\Om,\mathcal F,P)$, so that $P(R_n=\pm1)=1/2$ for all $n$; such a probability space exists. Let $(E,\mathcal A,\mu):=(\Om,\mathcal F,P)$. Let
$$f_n:=X_n:=n!R_n$$
for all natural $n$. Let $(g_k):=(Y_k)$ be any subsequence of the sequence $(X_n)$, so that
$$Y_k=X_{n_k}$$
for some strictly increasing sequence $(n_k)$ of natural numbers and all natural $k$. Let $Y_\infty$ be any r.v. on the probability space $(\Om,\mathcal F,P)$ with values in $[-\infty,\infty]$. It suffices to show that
$$P\Big(\frac1K\,\sum_{k=1}^K Y_k\to Y_\infty\Big)\overset{\text{(?)}}=0. \tag{1}$$
Note that for some r.v. $U_{1,K}$ with values in $[-1,1]$ we have
$$\sum_{k=1}^K Y_k=\sum_{k=1}^K (n_k)!R_{n_k}
=(n_K)!R_{n_K}+U_{1,K}\sum_{j=1}^{n_K-1}j!.$$
Next, for natural $n$,
$$\sum_{j=1}^{n-1}j!\le(n-2)(n-2)!+(n-1)!=o(n!).$$
So,
$$\frac1K\,\sum_{k=1}^K Y_k\sim\frac{(n_K)!}K\,R_{n_K}.$$
Therefore and because $n_K\ge K$ and $|R_n|=1$, for each $\om\in\Om$, $\frac1K\,\sum_{k=1}^K Y_k(\om)$ may only converge to $\infty$ or $-\infty$; that is,
$$\text{
on the event $\Big\{\frac1K\,\sum_{k=1}^K Y_k\to Y_\infty\Big\}$ we must have $Y_\infty\in\{\infty,-\infty\}$.} \tag{2}$$
Moreover,
$$P\Big(\frac1K\,\sum_{k=1}^K Y_k\to\infty\Big)=P\Big(\bigcup_{K=1}^\infty A_K\Big),$$
where $A_K:=\{R_{n_K}=1,R_{n_{K+1}}=1,\dots\}$. Obviously, $P(A_K)=0$ for each natural $K$. Hence,
$$P\Big(\frac1K\,\sum_{k=1}^K Y_k\to\infty\Big)=0.$$
Similarly,
$$P\Big(\frac1K\,\sum_{k=1}^K Y_k\to-\infty\Big)=0.$$
Now, in view of (2), we see that (1) follows, as desired.
Adeed:
Having now looked at Komlós's paper, I see that Theorem 2 there presents a stronger counterexample, as follows: For any sequence $(a_n)$ of positive real numbers such that $a_n\to\infty$ there exists a sequence $(\eta_n)$ of iid r.v.'s with $E|\eta_1|=1$ such that for the sequence $(\xi_n)$ with $\xi_n:=a_n\eta_n$ and for any of its subsequences the strong law of large numbers is not valid. Thus, the factor $n!$ in my example can be replaced by arbitrarily slowly growing $a_n$, with the Rademacher $R_n$'s replaced by iid r.v.'s $\eta_n$ with a (barely) finite expectation.
Fixed a typo: Of course, $\sum_{j=1}^{n-1}j!$ is $o(n!)$, not $o(n)$.
@losif Theorem 2 is true for non-negative functions (see the proof of Lemma 5.1 on page 243 in the article "Komlós theorem for unbounded random sets G. KRUPA")
@StephanSturm : You are right, thank you for your comment.
As a simple example let $E$ be a one point set and $\mu$ the one point measure. Let then $f_n \equiv 2^n, ~ n \in \mathbb{N}$. Then $\frac{1}{i} \sum_{j=1}^i h_j$ is dominated by the largest element in $\{h_1,\ldots,h_i\}$, in particular necessarily $h_\infty = \infty$. Thus the second assertion is not true with finite $h_\infty$ and for the first $h_\infty \not\in L^1$.
you did not understand my question, I want a counterexample for which theorem 2 is not valid for a sequence of measurable functions, because the theorem 2 is valid for all sequences of measurable functions
Hmm? What is the meaning of your comment? It seems to me that you allow $h_\infty$ to be infinite. I really don't understand your problem.
@DieterKadelka : I too had trouble understanding the question. At this point, I am pretty sure the intended question was this: Is there an example of a sequence of measurable functions $f_n$ for which the main conclusion of the Komlós theorem, about the a.s. convergence of the averages, does not hold? By the Komlós theorem itself, that conclusion holds assuming the $L^1$ boundedness or the nonnegativity of the $f_n$'s. So, in such a counterexample, the $f_n$'s must take values of both signs and must be $L^1$-unbounded.
Previous comment continued: Also, in such a counterexample, the a.s. limit $h_\infty$ must be allowed to take infinite values -- because otherwise the "nonnegative" version of the Komlós theorem would obviously be false in general.
@DieterKadelka I want a counterexample for Theorem 2
|
2025-03-21T14:48:30.485336
| 2020-05-03T21:07:12 |
359281
|
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|
Stack Exchange
|
"The" axiom of induction up to recursive ordinal $\alpha$ in $\mbox{PRA}$
As far as I understand, Kriesel proved that there exists a recursive relation $R$ of order type $\omega$ such that $\mbox{PRA}+TI(R)$ proves $\mbox{Con}(PA)$, and Beklemishev proved that for any recursive ordinal $\alpha$, there is a recursive relation $R$ of order type $\alpha$ such that $\mbox{PRA}+TI(R)$ does not prove $\mbox{Con}(PA)$. This is somehow unsatisfactory that the provability strength of transfinite induction up to recursive ordinal $\alpha$ depends on the specific recursive encoding. But, intuitively, we can formulate "the" axiom $TI(\alpha)$ for any specific recursive ordinal $\alpha$, if we choose somehow natural and not very pathologic recursive relation $R_\alpha$ which encodes order type $\alpha$. For example, we expect that if $R_\omega$ is not very pathologic encoding of ordinal $\omega$ then $\mbox{PRA}+TI(R_\omega)$ does not prove $\mbox{Con}(PA)$. Can we somehow formalize what it means to be "the natural recursive encoding" of the specific recursive ordinal $\alpha$? Can we construct the good recursive encoding $R_\alpha$ for each recursive ordinal $\alpha$?
The key word you might be looking for is Kleene ordinal notations. For certain small ordinals $\alpha$, e.g., at least up to the Bachmann-Howard ordinal and in fact well beyond that, there are indeed explicitly defined “standard” notations (well-defined up to primitive recursive equivalence) given by systems of ordinal collapsing functions and hierarchies. (contd…)
(contd…) As far as I can tell, the largest computable ordinal to have an explicitly defined system of notations published in the literature is described in Jan-Carl Stegert's 2010 thesis, “Ordinal Proof Theory of Kripke-Platek Set Theory Augmented by Strong Reflection Principles” and his paper with Wolfram Pohlers, “Provably Recursive Functions of Reflection”. But it is not known how to define such a standard notation for any given recursive ordinal (and believed that it cannot be done?): I think this is essentially the “subrecursive stumblingblock”.
@Gro-Tsen Thank you! Is there any hope that we can define the property "R is a (proof-theoretically) good recursive encoding of recursive ordinal $\alpha$", and prove (in metatheory) that each recursive ordinal have such a good representation, without direct construction? If would be nice if we could define the theory $\mbox{PRA} + \bigcup_{\alpha < \omega_1^{CK}} TI(\alpha)$ and ask some questions about it. Is it too naive to hope to define such (noncomputable) theory in a natural canonical way (from a proof-theoretical strength perspective, at least)?
|
2025-03-21T14:48:30.485521
| 2020-05-03T22:00:18 |
359286
|
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|
Stack Exchange
|
Homotopy equivalent fibers and Fibrations
If a morphism of topological spaces $X\rightarrow Y$ is a fibration, and the target space is connected, then the fibers of the points $y\in Y$ are homotopy equivalent, i.e. for all $y_1,y_2\in Y$ we have $X_{y_1}\cong X_{y_2}$. My question is whether or not this is a sufficent condition for this map to be a fibration or what the proper counter-example is to keep in mind.
An important weaker notion is that of a quasi-fibration: a map $f: X \to Y$ for which the canonical map $f^{-1}(x) \to \text{hofib}(f;x)$ is a homotopy equivalence. In particular, all fibers are homotopy equivalent. The usual example of a quasifibration which is not a fibration is the map from the letter T to the unit interval I which collapses the top bar to a point.
The identity map $i:X_{dis}\rightarrow X_{ind}$ from the discrete topology on a set $X$ to the indiscrete topology on the same set. Assume the topologies don't coincide. The fibres are all points. $i$ is not a Serre fibration since it doesn't have the HLP with respect to $I^0=\ast$. Hence it is not a Hurewicz fibration (more directly: one such would necessarily be a quotient). $i$ is not a quasifibration since the homotopy fibre is $X_{dis}$ and is not contractible. It is not a Dold fibration since it is not fibre homotopy equivalent to the identity on $X_{ind}$ (use numerable contractibility)
The figure below gives a simple but extreme counterexample, which I think has all the lifting properties one might want except for actually being a true fibration. The map is the identity everywhere except for the linear segment with negative slope, which is mapped to the horizontal segment in the codomain. This map is a continuous bijection so all of the fibers are homeomorphic. It is even a Serre fibration. However, it does not have the homotopy lifting property with respect to the convergent sequence space $Z=\{1,1/2,1/3,\dots ,0\}$, which appears as the middle cross section of both spaces.
|
2025-03-21T14:48:30.485671
| 2020-05-03T22:40:04 |
359288
|
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|
Stack Exchange
|
Prokhorov theorem on non Polish spaces
It is well known that if $X$ is a Polish space and $\mathcal{F} \subset \mathcal{M}_+(X)$ (the set of finite positive Radon measures on $X$) is uniformly tight and bounded in mass, it is relatively compact w.r.t. to the weak topology, i.e. the coarsest topology on $\mathcal{M}_+(X)$ w.r.t. the maps $\mu \mapsto \int_X \varphi \text{ d} \mu$ are continuous for every $\varphi \in C_b(X)$, the continuous bounded real functions on $X$.
I am intersted in a similar statement but in the general case of a Hausdorff topological space $X$.
A possible way to introduce a topology on $\mathcal{M}_+(X)$, when $X$ is a Hausdorff topological space, is to consider the coarsest topology on $\mathcal{M}_+(X)$ w.r.t. the maps $\mu \mapsto \int_X \varphi \text{ d} \mu$ are lower semi continuous for every $\varphi \in LSC_b(X)$, the lower semi continuous and bounded real functions on $X$.
In the book of Schwarz "Radon measures on arbitrary topological spcaes" it is proven that, in this topology, uniform tightness and boundedness in mass together again imply relative compactness.
I am wondering, how much can I enrich the topology on $\mathcal{M}_+(X)$ and still have that Prokhorov theorem holds?
For example, if $X$ is a Hausdorff topological space and I endow $\mathcal{M}_+(X)$ with the coarsest topology w.r.t the maps $\mu \mapsto \int_X\varphi \text{ d}\mu$ are continuous for every $\varphi \in LSC_b(X) \cup USC_b(X)$, what happens?
Here $LSC_b(X)$ (resp. $USC_b(X)$) is the set of the lower (resp. upper) semi continuous bounded real functions on $X$.
I have problems with your text. What do you mean by "continuous on varying of $\phi$? The last sentence is dubious. Why should $\mu \o \int \phi d\mu$ be continuous for semicontinuous $\phi$? Do you know the notes on p. 454 and 455 in Bogachev, Measure Theory 2, Springer 2007?
I have edited the text, I hope it is clear now. I am just considering different initial topologies on $\mathcal{M}_+(X)$. I've just read those notes: I think I will surely find something useful in the papers suggested there!
Much clearer now! But I still have a problem with your last sentence. Being both u.s.c. and l.s.c. implies continuity of a function.
I mean for functions that are upper or lower semi continuous. Edited again, now it should be clear!
Your topology at the end is finer than the weak topology, and may be much much finer. For instance, take $X = [0,1]$; since the functions $1_{{x}}$ are USC, I think the set of Dirac masses is discrete in this topology, and certainly not relatively compact even though uniformly tight.
The category of Hausdorff topological spaces is ust too big for Prokhorov's Theorem to hold. You are working with the wrong level of generality. You need to be not too far from a space with some countable structure built in. Polish spaces have that since they have a countable dense subset. There also spaces which are not too far, Suslin and Lusin, Standard spaces. A notable example where Prokhorov holds for a space which is not Polish (not even metrizable) is the space of distributions $S'$ or $D'$.
@Bremen Your question combines two different generalizations of the Prohorov's theorem on Polish spaces: In one direction you pass to a larger class of spaces, in another direction to a finer topology on measures. Do you really want to generalize in both directions at the same time? The question would be clearer if you asked about one or the other.
This seems to be a pure general topology problem. Enriching the topology will necessarily destroy the Prohorov property whenever the enrichment matters. If you have two nested Hausdorff topologies and a set is relatively compact in the finer topology, then the closure is the same under both topologies and the trace topologies coincide on the closure.
To see this, let $\tau$ and $\tau'$ be Hausdorff topologies on $X$ such that $\tau\subseteq\tau'$ and let $R\subseteq X$ be relatively compact in the topology $\tau'$. Trivially, $R$ is also relatively compact in the coarser topology $\tau$ and the $\tau'$-closure of $R$ is a subset of the $\tau$-closure of $R$. Let $x$ be a point in the $\tau$-closure of $R$ and $\langle x_\alpha\rangle$ be a net in $R$ converging to $x$ under $\tau$. Since $R$ is relatively compact under $\tau'$, a subnet will $\tau'$ converge to a point $x'$. But this subnet will also converge to $x'$ under $\tau$. Since $\tau$ is Hausdorff , $x=x'$ and the closure of $R$ is the same under both topologies. Let $C$ be this compact closure. The identity is a continuous function from $(C,\tau')$ to $(C,\tau)$ and will therefore map compact subsets of $C$ to compact subsets of $C$. But these are exactly the closed subsets of $C$ under both topologies, so both topologies must coincide on $C$.
|
2025-03-21T14:48:30.485994
| 2020-05-03T23:30:08 |
359292
|
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|
Stack Exchange
|
Closed analytic subvariety of $\mathbb C^n$ not defined by global holomorphic functions
Here is the motivation of this question: $\mathbb C^n$ is already "local" in algebraic category. In other words, algebraic subvarieties of $\mathbb C^n$ are affine, so they are common zero locus of finitely many polynomials defined on $\mathbb C^n$.
However, $\mathbb C^n$ is not "local" enough for analytic varieties, since by definition (Griffiths & Harris, page 12), an analytic subvariety $X$ of $\mathbb C^n$ is that for each $x\in X$, there is an open neighborhood $U$ of $x$ in $\mathbb C^n$, such that $X\cap U$ is common zero loci of holomorphic functions $f_1,...,f_k$ defined on $U$.
Of course, globally defined holomorphic functions will do the job, for example, the graph of the entire function $z\mapsto e^z$ produces the simplest analytic subvariety of $\mathbb C^2$ that is not algebraic. However, is there an example of a closed analytic subvariety of $\mathbb C^n$ not defined by global holomorphic functions on $\mathbb C^n$?
I think that you need to be more careful with a definition of analytic subvariety, otherwise a zero set of a function holomorphic in some domain is an example.
@OlegEroshkin Thanks, I think the example I'm looking for is unbounded, so I added closedness assumption.
A closed subset $X$ of $C^n$ satisfying your definition can be defined by global analytic functions. This follows from the solution of the "Second Cousin problem" in $C^n$, and is explained in any text on functions of several complex variables.
Thank you. Vaninghing of $H^1(\mathbb C^n,\mathcal{O})$ implies gluing divisors on $\mathbb C^n$ has no obstruction. This solve the problem when $X$ is codimension one, but I don't see how the why general case follows from it.
@AGlearner You need Cartan's B theorem for arbitrary codimension.
@OlegEroshkin Sorry, it should be $H^1(X,\mathcal{O}_X^)$ controls the divisor (instead of $H^1(X,\mathcal{O}_X)$ as I wrote). However, $\mathcal{O}_X^$ is not a coherent sheaf anymore since it is not even a sheaf of $\mathcal{O}_X$-module, so I guess we cannot apply Cartan B here. But I should thank for your suggestion since it brought me to the fact that "$X$ being closed analytic subvariety of $\mathbb C^n$ implies $X$ is a stein space".
Dear @OlegEroshkin, if I apply Cartan's theorem A to the ideal sheaf $I_X$ on $\mathbb C^n$, there are finitely many global sections $s_1,...,s_k\in H^0(\mathbb C^n, I_X)$ which induce the finite generation $\oplus_{i=1}^k\mathcal{O}_{\mathbb C^n}\twoheadrightarrow I_X$. Therefore, as globally holomorphic functions, ${s_1=0,...,s_k=0}$ defines $X$. Do you think this is correct? Also, could you explain a little bit about your suggestion on applying Cartan B? Thanks!
That was my thinking. But, unless I am mistaken, you will get finite generation only over the arbitrary compact. I think, that over $\mathbb{C}^n$, the ideal sheaf may not be finitely generated. The idea: take a union of varieties, which required at least $k$ functions to define. By translating varieties, we may enforce that the union is locally finite. Then the union is an analytic set. To avoid such issues, you may consider an irreducible analytic set.
@OlegEroshkin: I see, thanks for pointing that out. So in general a closed analytic subset $X$ of $\mathbb C^n$ is supposedly common zero loci of countably many globally holomorphic functions (by choosing a countable open covers). But if we require additionally that $X$ is irreducible, do you think the ideal sheaf $I_X$ can be (globally) finitely generated? Or do you have a counterexample in mind?
I would be very surprised if there is a not finitely generated irreducible analytic set. However, it's not obvious to me how to prove that for an irreducible analytic set, the ideal sheaf is finitely generated.
|
2025-03-21T14:48:30.486243
| 2020-05-04T00:58:57 |
359297
|
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|
Stack Exchange
|
Optimal partition search
Given an integer $n$, and 2 real sequences $\{a_1, \dots, a_n\}$ and $\{b_1, \dots, b_n\}$, with $a_i$, $b_i$ > 0, for all $i$. For a fixed $m < n$ let $\{P_1, \dots, P_m\}$ be a partition of the set $\{1, \dots, n\}$ as in $P_1 \cup \dots \cup P_m$ = $\{1, \dots, n\}$, with the $P_i$'s pairwise disjoint. I wish to find a partition of size $m$ that solves
$$\max_{P=\{P_1, ..., P_m\}}\sum_{j=1}^{m}\frac{(\sum_{i \in P_j}a_i)^2}{\sum_{i \in P_j}b_i}$$
I am really looking for an algorithm which solves the problem in polynomial time, a brute-force solution is not feasible, as it would involve the a Bell number of order $(n, k)$, with $n$ over 1e6 for realistic cases.
I would be happy to prove that the partition is monotonic in increasing values of $a/b$, in the sense that a partition expressed in the indices of the two sequences $a, b$, sorted by increasing values of $a/b$ will contain monotonic, increasing sets of integers in $1, ..., n$. I believe this is the case - can someone provide a proof?
If so, the brute-force search could be improved to an order $n \choose m-1$ algorithm, still long, but a significant savings.
The script below solves the problem by brute-force. For example, a sample run with $n = 12$, $m = 3$, gives an optimal partition of (expressed in indices of the sorted sequence $a/b$):
[[0, 1, 2, 3, 4, 5, 6, 7], [8, 9], [10, 11]]
which is monotonic, as claimed.
import numpy as np
import multiprocessing
import concurrent.futures
from functools import partial
from itertools import chain, islice
# n
NUM_POINTS = 12
# m
PARTITION_SIZE = 4
rng = np.random.RandomState(55)
def knuth_partition(ns, m):
def visit(n, a):
ps = [[] for i in range(m)]
for j in range(n):
ps[a[j + 1]].append(ns[j])
return ps
def f(mu, nu, sigma, n, a):
if mu == 2:
yield visit(n, a)
else:
for v in f(mu - 1, nu - 1, (mu + sigma) % 2, n, a):
yield v
if nu == mu + 1:
a[mu] = mu - 1
yield visit(n, a)
while a[nu] > 0:
a[nu] = a[nu] - 1
yield visit(n, a)
elif nu > mu + 1:
if (mu + sigma) % 2 == 1:
a[nu - 1] = mu - 1
else:
a[mu] = mu - 1
if (a[nu] + sigma) % 2 == 1:
for v in b(mu, nu - 1, 0, n, a):
yield v
else:
for v in f(mu, nu - 1, 0, n, a):
yield v
while a[nu] > 0:
a[nu] = a[nu] - 1
if (a[nu] + sigma) % 2 == 1:
for v in b(mu, nu - 1, 0, n, a):
yield v
else:
for v in f(mu, nu - 1, 0, n, a):
yield v
def b(mu, nu, sigma, n, a):
if nu == mu + 1:
while a[nu] < mu - 1:
yield visit(n, a)
a[nu] = a[nu] + 1
yield visit(n, a)
a[mu] = 0
elif nu > mu + 1:
if (a[nu] + sigma) % 2 == 1:
for v in f(mu, nu - 1, 0, n, a):
yield v
else:
for v in b(mu, nu - 1, 0, n, a):
yield v
while a[nu] < mu - 1:
a[nu] = a[nu] + 1
if (a[nu] + sigma) % 2 == 1:
for v in f(mu, nu - 1, 0, n, a):
yield v
else:
for v in b(mu, nu - 1, 0, n, a):
yield v
if (mu + sigma) % 2 == 1:
a[nu - 1] = 0
else:
a[mu] = 0
if mu == 2:
yield visit(n, a)
else:
for v in b(mu - 1, nu - 1, (mu + sigma) % 2, n, a):
yield v
n = len(ns)
a = [0] * (n + 1)
for j in range(1, m + 1):
a[n - m + j] = j - 1
return f(m, n, 0, n, a)
def Bell_n_k(n, k):
''' Number of partitions of {1,...,n} into
k subsets, a restricted Bell number
'''
if (n == 0 or k == 0 or k > n):
return 0
if (k == 1 or k == n):
return 1
return (k * Bell_n_k(n - 1, k) +
Bell_n_k(n - 1, k - 1))
def slice_partitions(partitions):
# Have to consume it; can't split work on generator
partitions = list(partitions)
num_partitions = len(partitions)
bin_ends = list(range(0,num_partitions,int(num_partitions/NUM_WORKERS)))
bin_ends = bin_ends + [num_partitions] if num_partitions/NUM_WORKERS else bin_ends
islice_on = list(zip(bin_ends[:-1], bin_ends[1:]))
rng.shuffle(partitions)
slices = [list(islice(partitions, *ind)) for ind in islice_on]
return slices
def reduce(return_values, fn):
return fn(return_values, key=lambda x: x[0])
class SimpleTask(object):
def __init__(self, a, b):
self.a = a
self.b = b
def __call__(self):
time.sleep(1)
return '{self.a} * {self.b} = {product}'.format(self=self, product=self.a * self.b)
def __str__(self):
return '{self.a} * {self.b}'.format(self=self)
class Task(object):
def __init__(self, a, b, partition):
self.partition = partition
self.task = partial(Task._task, a, b)
def __call__(self):
return self.task(self.partition)
@staticmethod
def _task(a, b, partitions, report_each=1000):
max_sum = float('-inf')
arg_max = -1
for ind,part in enumerate(partitions):
val = 0
part_val = [0] * len(part)
part_vertex = [0] * len(part)
for part_ind, p in enumerate(part):
part_sum = sum(a[p])**2/sum(b[p])
part_vertex[part_ind] = part_sum
part_val[part_ind] = part_sum
val += part_sum
if val > max_sum:
max_sum = val
arg_max = part
max_part_vertex = part_vertex
# if not ind%report_each:
# print('Percent complete: {:.{prec}f}'.
# format(100*len(slices)*ind/num_partitions, prec=2))
return (max_sum, arg_max, max_part_vertex)
class Worker(multiprocessing.Process):
def __init__(self, task_queue, result_queue):
multiprocessing.Process.__init__(self)
self.task_queue = task_queue
self.result_queue = result_queue
def run(self):
proc_name = self.name
while True:
task = self.task_queue.get()
if task is None:
# print('Exiting: {}'.format(proc_name))
self.task_queue.task_done()
break
result = task()
self.task_queue.task_done()
self.result_queue.put(result)
NUM_WORKERS = multiprocessing.cpu_count() - 1
INT_LIST= range(0, NUM_POINTS)
num_partitions = Bell_n_k(NUM_POINTS, PARTITION_SIZE)
partitions = knuth_partition(INT_LIST, PARTITION_SIZE)
slices = slice_partitions(partitions)
while True:
a0 = rng.uniform(low=-0.0, high=100.0, size=NUM_POINTS)
b0 = rng.uniform(low=-0.0, high=100.0, size=NUM_POINTS)
# sort by increasing a/b, to check claim
ind = np.argsort(a0/b0)
(a,b) = (seq[ind] for seq in (a0,b0))
tasks = multiprocessing.JoinableQueue()
results = multiprocessing.Queue()
workers = [Worker(tasks, results) for i in range(NUM_WORKERS)]
num_slices = len(slices) # should be the same as NUM_WORKERS
for worker in workers:
worker.start()
for i,slice in enumerate(slices):
tasks.put(Task(a, b, slice))
for i in range(NUM_WORKERS):
tasks.put(None)
tasks.join()
allResults = list()
slices_left = num_slices
while not results.empty():
result = results.get()
allResults.append(result)
# print('result: {!r}'.format(result))
slices_left -= 1
r_max = reduce(allResults, max)
c = a/b
part = r_max[1]
endpoints = [(a[-1], b[0]) for a,b in zip(part[:-1], part[1:])]
d = [(c[r]-c[l]) for l,r in endpoints]
r = [(c[r]-c[l])/c[l] for l,r in endpoints]
all_diffs = np.concatenate([[np.nan], np.diff(c)])
all_rets = np.concatenate([np.diff(c), [np.nan]]) / c
max_diffs = sorted(all_diffs)[-(PARTITION_SIZE-1):]
max_rets = sorted(all_rets)[-(PARTITION_SIZE-1):]
print('TRIAL: {} : max: {:4.6f} pttion: {!r}'.format(i, *r_max[:-1]))
# print('TRIAL: {} : max: {:4.6f} {!r} {!r}'.format(i, *r_max[:-1],
# [float(x)
# for x in ['{0:0.2f}'.format(i)
# for i in r_max[2]]], prec=2))
try:
assert all(np.diff(list(chain.from_iterable(r_max[1]))) == 1)
except AssertionError as e:
import pdb
pdb.set_trace()
If you can prove the monotonicity, you can find an optimal partition by solving a shortest path problem in a directed acyclic layered graph with $O(mn)$ nodes and $O(mn^2)$ arcs.
That could be solved in$O((m + n) \log(mn))$ time. I don't see it though, I don't see even a 2nd order algorithm in $mn$.
Given the monotonicity property, here is a shortest-path formulation. The nodes are $(i,k)$, where $i\in\{1,\dots,n\}$ and $k\in\{1,\dots,m\}$, plus a dummy sink node $(n+1,m+1)$. The directed arcs are from $(i,k)$ to $(j,k+1)$, where $i<j$, with the interpretation that items $i,\dots,j-1$ appear in part $P_k$. The arc cost, which depends only on $i$ and $j$, is: $$\frac{-\left(\sum_{r=i}^{j-1} a_r\right)^2}{\sum_{r=i}^{j-1} b_r},$$
negated because your original problem is maximization. The source node is $(1,1)$, and you want to find a shortest path from the source to the sink. Note that the network is acyclic, so you don't even need Dijkstra's algorithm. Bellman's (dynamic programming) equations can be solved in one backwards pass starting from the sink node.
Yes, I agree. We have fewer edges, naive count gives $n(n-1)/2$ but there is also padding - we shouldn't connect $(1, 1)$ to $(2, n-1)$ if $m >= 3$ for example.
Right, and some nodes aren't reachable from $(1,1)$, like $(1,k)$ for $k>1$. But still $O(n)$ nodes and $O(n^2)$ arcs for each of the $m$ layers. I guess you meant $(n-1,2)$ instead of $(2,n-1)$. Each arc increments the layer by exactly 1.
|
2025-03-21T14:48:30.486877
| 2020-05-04T01:02:27 |
359298
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/359298"
}
|
Stack Exchange
|
Distance pairs in labeled directed graph
Suppose we have a simple directed graph with $n$ nodes and $m$ edges, and we label each edge from $1$ to $m$ (with distinct labels). Define the weighted "length" of a directed path to be the maximum of all edge labels on that path (or $0$ for a trivial path), and define the "distance" $d(a, b)$ from node $a$ to $b$ to be the minimum weight of all paths from $a$ to $b$ (or $\infty$ if $b$ is not reachable from $a$).
Define a triple $a, b, c$ of nodes to be defective if $d(a, c) < d(a, b) < \infty$ and $d(b, c) < d(a, b)$.
I have two questions:
A) What is the maximum number of defective triples possible?
B) What is the average number of defective triples if edge labels are assigned at random?
Unfortunately, I haven't been able to make much progress on either one, so I was hoping other people might have insight into the problem.
For A), here is a construction that gives $2\binom{n}{3}$ defective triples, which is almost best possible. Let $D$ be a digraph with vertex set $[n]$ and arcs $(i,i+1)$ and $(i+1, i)$ for all $i \in [n-1]$. Let the label of the arc $(i,i+1)$ be $n-1+i$, and the label of the arc $(i+1, i)$ be $i$.
For all $i<j<k$, I claim that $(i,k,j)$ is a defective triple. To see this, note that
$d(i,j)=n+j-2 < n+k-2=d(i,k)$ and $d(k,j)=k-1 < n+k-2=d(i,k)$.
Similarly, $(j,k,i)$ is also a defective triple.
Thus, this example contains $2\binom{n}{3}$ defective triples.
In general, note that if $(a,b,c)$ is a defective triple, then $(a,c,b)$ cannot be a defective triple. Thus, every digraph on $n$ vertices contains at most $3 \binom{n}{3}$ defective triples, so our bound is tight up to a factor of $\frac{3}{2}$.
|
2025-03-21T14:48:30.487002
| 2020-05-04T01:10:55 |
359299
|
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/359299"
}
|
Stack Exchange
|
On a structural decomposition of polynomials based on integral roots
Given an irreducible polynomial of structure $$f(x,y)=\sum_{\substack{i,j\in\{0,1,2\}\\i+j\leq3}}a_{i, j}x^iy^j\in\mathbb Z[x,y]$$ with $a_{2,1}a_{1,2}a_{1,1}a_{1,0}a_{0,1}a_{0,0}\neq0$ if $f(m,n)=0$ with $m, n\in\mathbb Z$ then is it possible to decompose $f(x,y)$ as $$f_1(x,y)+g_1(x)h_2(y)+h_1(y)g_2(x)$$
where $f_1(x,y)$ is degree $1$ polynomial and $g_i(x)$ and $h_i(y)$ are degree $i\in\{1,2\}$ polynomials in $\mathbb Z[x]$ and $\mathbb Z[y]$ respectively where$g_1(m)=g_2(m)=0$ and $h_1(n)=h_2(n)=0$ still holds?
How difficult is to get such decompositions?
|
2025-03-21T14:48:30.487078
| 2020-05-04T01:45:39 |
359300
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Dabed",
"Gerry Myerson",
"Hollis Williams",
"Schroedinger'sDog",
"Tom Copeland",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/359300"
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|
Stack Exchange
|
Series and sequences in physical systems & closed form expressions
I gave a colloquium a while ago about physics inspiring recent developments in mathematics and as is almost borderline cliche in such talks, I mentioned the Fibonacci sequence with closed form expression,
$$
F_n = \frac{\varphi^n - (-\varphi)^{-n}}{2 \varphi -1}, \ \varphi = \frac{1+\sqrt 5}{2}
$$
and how it shows up in many natural settings because the Fibonacci sequence is the closest integral approximation to the logarithmic spiral series.
Here's my (possibly naive) question: Are there any other series/sequences with closed form expressions which manifest plainly and obviously in nature?
I suppose is a particular case of the Riemann zeta but I like the Madelung constant(s) it appears naturally when discussing crystal configuration between cations and anions and can be expressed without physical considerations for the Na Cl case as
$M= -\sum_{j,k,\ell=-\infty,j,k,\ell=
\neq 0}^\infty{{(-1)^{j+k+\ell}} \over { (j^2 + k^2 + \ell^2)^{1/2}}}$
The exact value will depend on the crystal configuration but in general in needs to be analytical continued.
Not series/sequences related but you might also interested to read about the appearance of various curve counts from enumerative geometry in physics where they are interpreted as counts of supersymmetric states in string theories, very interesting https://arxiv.org/pdf/1801.09818.pdf.
@HollisWilliams: Thanks, I'm familiar with this work. However, in relation to the theme of the question: These DT invariants count BPS states which are in most cases deformation invariant representations of a superalgebra and exist only in theories with extended supersymmetry i.e. more than one superalgebra generator. Such theories are even 'less physical' than commonly studied supersymmetric standard models. I think if I relaxed the question to include "plausible but unconfirmed physical systems", there would be a host of different series which would enter this list due to supersymmetry. ;)
Yes, fair point.
The Casimir effect is a manifestation of
$$1+2^3+3^3+\cdots=-\frac{1}{120}.$$
The vacuum energy $E$ in the space between two metal plates, separated by a distance $a$ equals
$$E = \frac{ \hbar c \pi^2 }{6a^3}\sum_{n=1}^\infty n^3.$$
This divergent sum is regularized by analytic continuation of the Riemann zeta function, to give
$$1+2^s+3^s+\cdots=\zeta(-s)=-\frac{B_{s+1}}{s+1},$$
with $B_s$ a Bernoulli number, hence
$\sum_{n=1}^\infty n^3=-\frac{1}{120}.$
The resulting attractive force $-dE/da$ between the metal plates, the Casimir effect, has been demonstrated in experiments, providing one justification for the zeta-function regularization of the divergent sum.
In the physics context, what is going on is that the unobservable vacuum energy $E$ is infinite, while the observable force $-dE/da$ is finite. One way to obtain this finite answer is to add a third metal plate, at a distance $L$ from the two plates at separation $a$. The third plate introduces a cutoff in the infinite sum, and the limit $\lim_{L\rightarrow\infty} dE/da$ gives the same finite answer as the zeta-function regularization.
The golden ratio is achieved by any sequences of numbers generated by an arbitrary initial seed of two nonzero natural numbers by recursively adding consecutive numbers together, so it is a limiting property of a quite general class of sequences.
The Feigenbaum constants in chaos theory have a similar nature.
Not so simply, the following sequences have important applications in mathematical physics.
The Riemann zeta function evaluated at natural numbers has several physical interpretations. See the responses to the Geometric, physical, probabilistic interpretations of the Riemann Zeta function for zeta(n > 1).
Scattering amplitudes in certain quantum field theories are related to the combinatorics of the associahedra. See references for OEIS A133437.
(The generating functions for Fuss-Catalan numbers and their compositional inverses are particular cases.)
The soliton wave solution to the KdV equation for shallow water waves, which also crops up in string theory, is related to the Eulerian numbers. See the MO-Q "Why is there a connection between enumerative geometry and nonlinear waves?"
The coefficients of the Lagurre and Hermite sequences of polynomials have numerous combinatorial interpretations and are related to probabilty wave functions in quantum mechanics connected to orbitals and the harmonic oscillator.
The combinatorics of phylogenetic trees, used in modelling bifurcations in evolution of biological structures, are related to the Ward numbers. (There are other sequences in the OEIS related to genomics and structures of chemicals.)
Compositional inverse pairs of functions or formal series are related to flow equations characterized by autonomous ODEs, and the relations among series reps of these pairs can be couched in terms of the Euler polynomials (with integer coefficients) of the associahedra, the refined Ward numbers A134685, the refined Eulerian numbers A145271, or the noncrossing partitions A134264.
The question asks for "closed-form expressions". I don't think the Feigenbaum constants have one. I don't know the other examples well enough to tell.
@Gerry, valid point. The integer constants for the inversions have explicit formulas in terms of sums of integer products as well as a simple algorithm iterating a Lie derivative involving only integers--can be translated into summing over different varieties of trees or dissections of polygons or as the antipode of a comb. Hopf alg. Now how is the square root calculated and when does the algorithm stop for an irrational number? I find the set of integers, each calculable exactly through a finite number of additions or multiplications, with diverse comb. interpretations more satisfying.
Classical, Fermi-Dirac, and Bose-Einstein statistics involve integrals that can be shown to be equivalent to polylogarithms. Polylogs have a particular simple series form:
$$ \text{Li}_s(x)=\sum_{n=1}^\infty \frac{x^n}{n^s} $$
For $x=\pm1$ and positive integer $s$ these equate to frequently-used constants. However, in many physics problems, $s$ is half-integer, and $x$ is large and negative. For $x$ large and negative, you can't use the series anymore, but an asymptotic formula or analytic continuation will work for calculation.
The question asks for closed form expressions. Does $\zeta(3)$ count as a closed form?
@GerryMyerson $\zeta(4)$ appears in Stefan-Boltzmann law, and most would agree a rational times a power of $\pi$ is a closed form. $\zeta(3)$ appears in the expected number of black body photons. I wouldn't consider the odd zetas as closed-form, although $\zeta(3)$ might be an exception, since it is provably irrational.
|
2025-03-21T14:48:30.487514
| 2020-05-04T02:21:41 |
359302
|
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|
Stack Exchange
|
A possibly easy question about latent geometry in Collatz sequences
I have a question about some (seemingly unimportant) behavior I noticed in Collatz sequences, which I haven't been able to find a general answer to upon rough scan of the literature (please be aware that this is not my field, so there may be some ignorance here).
A length-$m$ Collatz tuple is an $m$-tuple of the form $(c_0, c_1, \dotsc, c_{m - 1})$, where $c_0 \in \mathbb N$ and $c_{i + 1}$ is the Collatz iterate of $c_i$ for all $i$.
It appears that all length $m$ Collatz tuples lie on one of $m$ distinct lines in $m$-dimensional space, where the angle between each pair of such lines is $\pi/4$. This fact is clear for $m=2$, where all length-2 Collatz tuples lie on the lines $y = 3x+1$ and $y = x/2$ which have an intersection angle of $\pi/4$.
That this necessarily holds for the general $m$ case however is not immediately obvious to me (where we take Collatz orbits of length greater than or equal to $m$). I have been able to come up with a variety of arguments for why this may be the case, but none have been entirely convincing or neat. Can someone elucidate why this appears to be so, and if this property fails for some $m$, why? Can you provide sketch of proof (unless I am missing something basic)?
I think that your definition of a Collatz $m$-tuple can be much simplified, and I made an edit to that effect. Please feel free to revert if you don't like it (in which case I apologise). What are the lines for $m = 3$?
@LSpice you are right, thank you for the edit. I will also edit the question for the m=3 lines
Did you mean each pair of neighbouring/successive lines? I.e. $c_0,c_5$ don't necessarily have $\pi/4$ between them?
@samerivertwice yes
I see the following arithmetical background of your conjecture.
Put $\Lambda=\{1/2,3\}^m$. For each $\lambda=(\lambda_0,\dots,\lambda_{m-1})\in\Lambda$ we define a vector $r'_\lambda=(r_{\lambda,0},\dots, r_{\lambda,m-1})$ as follows. Put $I(\lambda)=\{0\le i\le m-1:\lambda_i=3\}$. If $I(\lambda)$ is empty then put $r'(\lambda)=0$, otherwise put $i=\min I(\lambda)$, $r'_{\lambda,k}=0$ for $k<i$, $r'_{\lambda,i}=1$, and for each $i+1\le k\le m-1$ let $r'_{\lambda,k}$ equals to $r'_{\lambda,k}/2$, if $\lambda_k=1/2$, and equals to $3r'_{\lambda,k}+1$, if $\lambda_k=3$.
It is easy to see that for each Collatz $m$-tuple $c=(c_0,\dots,c_{m-1})$ there exists $\lambda\in\Lambda$ such that $c=\lambda c_0+ r’_\lambda$. That is $c$ belongs to a line $r’_\lambda+\lambda t$, $t\in\Bbb R$. It can be well-known for which $\lambda\in\Lambda$ there exists a Collatz $m$-tuple $c=(c_0,\dots,c_{m-1})$ such that $c=\lambda c_0+ r’_\lambda$. The set of all such $\lambda$ provides a family $L$ of lines, containing all Collatz $m$-tuples. Moreover, if a line $\ell’$ in $\Bbb R^m$ contains at least $2^m+1$ Collatz $m$-tuples then $\ell’$ has two common points with some line $\ell\in L$, so $\ell’=\ell$.
|
2025-03-21T14:48:30.487721
| 2020-05-01T06:11:49 |
359303
|
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|
Stack Exchange
|
Is $\Delta \phi$ monotone operator on $H^1(\mathbb{R}^d)$ for monotone $\phi$
Let $H^1(\mathbb{R}^d)$ be the usual Sobolev space and let $\phi: \mathbb{R} \to \mathbb{R}$ be a non decreasing Lipschitz function with $\phi(0)=0$. Is the operator $\Delta \phi $ on $H^1(\mathbb{R}^d)$ monotone? i.e. Do we have $\langle \Delta \phi(u) -\Delta \phi(v) , u -v \rangle \ge 0$( or $\le$) for $u,v \in H^1(\mathbb{R}^d)$?
This is easily verified for $\phi(x)=x$, but I'm not able to conclude anything for general $\phi$.
$$\langle \Delta \phi(u) -\Delta \phi(v) , u -v \rangle = -\langle \nabla \phi(u)-\nabla \phi(v), \nabla u -\nabla v \rangle = -\langle \phi'(u)\nabla u-\phi'(v)\nabla v, \nabla u -\nabla v \rangle$$
If $\phi'$ is constant then we are done, otherwise I'm not sure where the above equality leads us.
Have you tried integrating by parts? At least we could all see where the problem lies.
I have added few details.
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2025-03-21T14:48:30.487806
| 2020-05-04T03:57:42 |
359304
|
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|
Stack Exchange
|
Maximum value of $\int (aF^2(x)g(x)+G^2(x)f(x))dx$ over all $f,g$ densities satisfying $\int F(x)g(x)dx=1/2$
I want to maximise $$I(f,g):=\int_{-\infty}^\infty (aF^2(x)g(x)+G^2(x)f(x))dx$$ where $a>0$ is a given constant, over all possible probability densities $f,g$ satisfying $$\int_{-\infty}^\infty F(x)g(x)dx=\int_{-\infty}^\infty G(x)f(x)dx=\dfrac{1}{2}$$ where $G(x)=\int_{-\infty}^x g(t)dt$ and $F(x)=\int_{-\infty}^x f(t)dt$.
Recall $f,g$ are densities means $f\geq0,g\geq0,\int_{-\infty}^\infty f(x)dx=\int_{-\infty}^\infty g(x)dx=1$. Note that $F$ is cdf of $f$ and $G$ is cdf of $g$ so in other words $G'(x)=g(x)$ and $F'(x)=f(x)$ for all $x$.
Since $F,G$ are cdf's, if we just have $\int F(x)g(x)=\dfrac{1}{2}$ then it implies $\int G(x)f(x)dx=\dfrac{1}{2}$ by change of variables formula so just one constraint is enough. Also, if it helps, $\int G(x)g(x)dx=\dfrac{1}{2}, \int G^2(x)g(x)dx=\dfrac{1}{3}$, and same for $f,F$.
I worked out some cases and have a strong feeling that the maximum value is $\max(\dfrac{a}{4}+\dfrac{1}{2},\dfrac{a}{2}+\dfrac{1}{4})$. In particular, the maximum is $\dfrac{a}{4}+\dfrac{1}{2}$ if $a<1$ else the maximum is $\dfrac{a}{2}+\dfrac{1}{4}$.
This is achieved when for example, $f$ (or $g$) is supported entirely between two humps of $g$ (or $f$) i.e. for example a situation where $f$ (or $g$) is supported on $[a_1,a_2]$, and $g$ (or $f$) is supported equally on $[b_1,b_2]$ and $[c_1,c_2]$ where $b_2<a_1<a_2<c_1$.
Many of my cases seem to tell me that this is the "extremal" situation. In fact, observe that $\int F^2(x)g(x)dx\leq \int F(x)g(x)dx=\dfrac{1}{2}$ according to the constraint, and equality is achieved precisely when one density lies between two humps of the other, and in this case the value of the other integral is forced to be $\dfrac{1}{4}$. So the value of $I(f,g)$ is $\dfrac{a}{2}+\dfrac{1}{4}$. Analogously $\int G^2(x)f(x)dx \leq \int G(x)f(x)dx =\dfrac{1}{2}$ and equality is achieved iff the other integral equals $\dfrac{1}{4}$ leading to the value of $I(f,g)=\dfrac{a}{4}+\dfrac{1}{2}$. So it seems to me that my guess for the maximum value of $I(f,g)$ is perhaps correct, but I am not able to prove it exactly.
I am also interested in knowing how one would approach such problems in general.
|
2025-03-21T14:48:30.487972
| 2020-05-04T05:30:44 |
359306
|
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|
Stack Exchange
|
Is the tensor product of chain complexes a Day convolution?
Recently, Jade Master asked whether the tensor product of chain complexes could be viewed as a special case of Day convolution. Noting that chain complexes may be viewed as $\mathsf{Ab}$-functors from a certain $\mathsf{Ab}$-category $\mathsf{C}$, Yuri Sulyma suggested¹ that maybe we could obtain the tensor product of two chain complexes as a Day convolution by endowing $\mathsf{C}$ with the monoidal structure given by $[n]\otimes_\mathsf{C}[m]\overset{\mathrm{def}}{=}[n+m-1]$.
Questions: Is this affirmation true? More precisely:
Given two chain complexes $X_\bullet$ and $Y_\bullet$ on an abelian category $\mathcal{A}$, is their Day convolution as $\mathsf{Ab}$-functors from $(\mathsf{C},\otimes_\mathsf{C})$ the usual tensor product of chain complexes $\otimes_{\mathsf{Ch}(\mathcal{A})}$?
If not, is there some other monoidal structure on $\mathsf{C}$ for which Day convolution gives $\otimes_{\mathsf{Ch}(\mathcal{A})}$?
If this fails too, is there perhaps another way to view $\otimes_{\mathsf{Ch}(\mathcal{A})}$ as a special case of some general construction in enriched category theory?
¹Note that his account is protected and hence his reply is not public.
Let DF(A) denote the filtered derived category, Fun(Z, D(A)). Then, DF(A) admits a t-structure, known as the Beilinson t-structure, whose heart is equivalent to the abelian category of chain complexes in A. The category DF(A) admits a monoidal structure, by Day convolution, and this tensor product is compatible with the t-structure (i.e., the connective objects are closed under the tensor product). The identification of the heart of DF(A) with the category of chain complexes is a symmetric monoidal equivalence.
You might be interested in reading Im, Geun Bin, and G. Max Kelly. "A universal property of the convolution monoidal structure." Journal of Pure and Applied Algebra 43.1 (1986): 75-88. In particular, see page 10, §6.
The answer to the question posed in the title of your post is yes, the tensor product of chain complexes is a Day convolution product. The important thing to note is that, to define a Day convolution monoidal structure on the $\mathcal{V}$-enriched functor category $[\mathcal{C},\mathcal{V}]$ (where $\mathcal{V}$ is a complete and cocomplete symmetric monoidal closed category, e.g. $\mathbf{Ab}$), we needn't demand $\mathcal{C}$ to be a monoidal $\mathcal{V}$-category: it suffices for $\mathcal{C}$ to be a promonoidal $\mathcal{V}$-category. This is the generality at which Day convolution was originally defined in Day's thesis, which may be found here (see also his earlier paper in the Reports of the Midwest Category Seminar IV, where the word "premonoidal" was used).
A promonoidal structure on a small $\mathcal{V}$-category $\mathcal{C}$ consists of tensor product and unit "profunctors", i.e. $\mathcal{V}$-functors $P \colon \mathcal{C}^\mathrm{op}\times\mathcal{C}^\mathrm{op} \times \mathcal{C} \to \mathcal{V}$ and $J \colon \mathcal{C} \to \mathcal{V}$, together with associativity and unit constraints subject to the usual two
"pentagon" and "triangle" axioms. Given a promonoidal structure on $\mathcal{C}$, we may construct the Day convolution monoidal structure on $[\mathcal{C},\mathcal{V}]$, whose tensor product is given at a pair of $\mathcal{V}$-functors $F,G \in [\mathcal{C},\mathcal{V}]$ by the coend
$$F\ast G = \int^{A,B \in \mathcal{C}} P(A,B;-) \otimes FA \otimes GB$$ in $\mathcal{V}$,
and whose unit object is the $\mathcal{V}$-functor $J \in [\mathcal{C},\mathcal{V}]$, and so on. This monoidal structure on $[\mathcal{C},\mathcal{V}]$ is biclosed (i.e. the tensor product $\mathcal{V}$-functor has a right $\mathcal{V}$-adjoint -- equivalently, preserves (weighted) colimits -- in each variable). In fact, every biclosed monoidal structure on $[\mathcal{C},\mathcal{V}]$ arises in this way from some promonoidal structure on $\mathcal{C}$. (For instance, one recovers the $\mathcal{V}$-functor $P$ from the tensor product $\ast$ by $P(A,B;C) = (\mathcal{C}(A,-) \ast \mathcal{C}(B,-))C$.)
So, since the $\mathbf{Ab}$-category $\mathbf{Ch}$ of chain complexes is (equivalent to) an $\mathbf{Ab}$-enriched functor category $[\mathcal{C},\mathbf{Ab}]$ (for the $\mathbf{Ab}$-category $\mathcal{C}$ described in the question to which you linked), and since the standard monoidal structure on $\mathbf{Ch}$ is $\mathbf{Ab}$-enriched and biclosed, this monoidal structure must be the Day convolution monoidal structure for some promonoidal structure on $\mathcal{C}$. And it isn't too hard to describe that promonoidal structure. For instance, (presuming I haven't bungled the calculation) the functor $P$ is defined on objects by $$P(i,j;k) = \begin{cases} \mathbb{Z} & \mathrm{if\,\,} i+j=k, \\
\mathbb{Z} \oplus \mathbb{Z} & \mathrm{if\,\,} i+j=k+1, \\
\mathbb{Z} & \mathrm{if\,\,} i+j=k+2, \\
0 & \mathrm{else}.
\end{cases}$$
In the terminology of my answer, this profunctor is the bimodule. I didn't realize this was the original generality of Day convolution! This generality makes more sense because it is Morita invariant.
A chain complex in any $Ab$-enriched category is an $Ab$-enriched functor $\mathcal A$ $\mathsf C \to \mathcal A$. If $\mathcal A$ is additive, then this is equivalently an additive functor $\bar{\mathsf C} \to \mathcal A$, where $\bar{\mathsf C}$ is the additive envelope of $\mathsf C$ (i.e. its completion under direct sums). I haven't given this much thought, but I think that $\bar{\mathsf C}$ is closed under tensor product in $Ch(Ab)$. So by passing to a slightly larger base category, we get $\otimes$ as the Day convolution of a monoidal rather than a promonoidal structure.
This is great! Thanks, Alexander! (And everyone else!)
If you use the category $C$ to represent chain complexes and you mean day convolution using a functor $C \otimes C\to C$ it is not possible. This boils down to whether you can obtain the totalization functor from bi-complexes to chain complexes, as a left adjoint to restriction for some functor $m: C \otimes C \to C$.
You cannot do this because the left adjoint $m_!$ will always take representable projectives to representable projectives. I.e. we will have that $$ m_! (C \otimes C((i,j), -))(r) = C(m(i,j),r).$$
But the totalization of a representable projective of $C \otimes C$ is a direct sum of two different principal projectives of $C$, so no choice of $m$ will work.
What is going wrong is that the totalization functor is given by a unique $(C \otimes C, C)$ bi-module, and this bi-module cannot come from a homomorphism $C \otimes C \to C$, because in some sense it is "multi-valued." To fix this, one could change $C$ to a morita equivalent category, $C'$ for which the bi-module is in fact given by a homomorphism. To construct such a $C'$, we need to choose a collection of generating projectives of ${\rm Ab}^{C}$ which is closed under tensor product. I don't see a particularly nice choice. But skd's comment is that if we use derived Morita equivalence instead of ordinary Morita equivalence, there is a very nice choice of (non-projective) generators, where $C'$ becomes the category $\mathbb N, \leq$.
Note: I'm not sure the construction I give is correct. There is an unresolved problem in proving it is a comonoid. See point 4 in the comonoid section. I don't have the time to resolve this right now.
I want to expand on Alexander Campbell's answer by giving the concrete promonoid as well as the necessary calculations. However my promonoid does not agree with the one he outlined. It's pretty late by now, so I hope I didn't miss anything.
Conventions
Just to get everything straight, let $\mathcal V$ be a closed abelian category with tensor product $\otimes$ and unit $I$. $\mathcal C$ is the free $\mathcal V$-category with objects $\mathbb Z$, morphisms $\partial_n : n \to n - 1$ and relations $\partial_{n-1}\partial_n = 0$. A chain complex is a functor $C \to \mathcal V$.
A $\mathcal V$-profunctor $D \nrightarrow E$ is a $\mathcal V$-functor $E^{op} \otimes D \to \mathcal V$ (the swapped order of $D$ and $E$ is nLab convention). We pretend $\mathcal V\text -\mathbf{Prof}$ is a strict monoidal 2-category with monoid $\otimes$ and unit $\mathcal I$.
We identify objects with their identity arrows.
Definition
We give all higher comultiplications of the comonoid.
For $n \in \{0,1,2,\ldots\}$ we define $(-|-)_n : \mathcal C \nrightarrow \mathcal C^{\otimes n}$, on objects as
$$
(k_1, \ldots, k_n| x)_n := \begin{cases}
I^{\oplus n} & \text{if } k_1 + \cdots + k_n = x + 1 \\
I & \text{if } k_1 + \cdots + k_n = x \\
0 & \text{otherwise.}
\end{cases}$$
For morphisms in each variable there is only one $\partial$ which has both domain and codomain non zero, in which case the domain is always $I$ and the codomain $I^{\oplus n}$. For the $j$th-contravariant variables the image of $\partial$ is then taken to be the $j$th-inclusion into the direct sum:
$$
(k_1,\ldots,\partial_{k_j + 1}, \ldots,k_n|k_1 + \cdots + k_n)_n := \iota_j.
$$
In the covariant variable the image of $\partial$ is the diagonal $I \to I^{\oplus n}$ but with signs:
$$
(k_1,\ldots, k_n| \partial_{k_1 + \cdots + k_n})_n := \iota_1 + (-1)^{k_1}\iota_2 + (-1)^{k_1 + k_2}\iota_3 + \cdots.
$$
We claim that $(-|-)_n$ is a comonoid in $\mathcal V\text-\mathbf{Prof}$ and the day convolution it defines is the tensor product of chain complexes.
Tensor product of chain complexes
Consider chain complexes $C, D : \mathcal C \to \mathcal V$. We have
$$
C\otimes D := x \mapsto \int^{m,n} (m,n|x)\otimes Cm \otimes Cn.
$$
First we show that
$$
(C \otimes D)(x) = \bigoplus_{m + n = x} C(m)\otimes C(n)
$$
by showing the cones over the bases of these two colimits are the same.
Fix $x$ and consider $m,n$ with $m+n = x$. The following diagram is exemplary for the diagram defining the coend $C\otimes D$.
and here is the same diagram again:
Consider any cone over this. The component from the center term is already completely determined by the inclusions of its two summands on the left whose components are again determined by those of the top and bottom terms. Thus all components cones are given by those from terms of the form $(m,n|m+n)\otimes C(m) \otimes D(n)$, while nothing constrains these components. Thus the cones are in bijection with those over the base of $\bigoplus_{m+n=x} C(m) \otimes D(n)$ as claimed.
The image of $C\otimes D(\partial)$ is given at $C(m)\otimes D(n)$ by
$$
(m,n|\partial_{m+n})\otimes C(m) \otimes D(n) =
\begin{cases}
C(m) \otimes C(n) &\to (C(m)\otimes D(n))^{\oplus 2}\\
\iota_1 + (-1)^m\iota_2.
\end{cases}
$$
That is $\partial$ takes a term $C(m) \otimes D(n)$ at the top or bottom of the heart shaped diagram into one of the two term sums at the center of the heart shaped diagram. The two inclusions can then be pulled back and pushed along to the top and bottom term to yield the desired
$$
((C\otimes D)(\partial))_{m,n} = C(\partial_m)\otimes D(n) + (-1)^m C(m)\otimes D(\partial_n).
$$
Comonoid in $\mathcal V\text-\mathbf{Prof}$
First note that $(-|-)_1$ is in fact $\mathcal C(-,-)$.
Let $j, k, n \in \{0,1,2,\ldots\}$ and sadly $k-j \leq 1$. We have
$$
(m_1,\ldots,m_n|x) \cong \int^y (m_1,\ldots,m_{j-1},y,m_{k+1},\ldots,m_n|x)\otimes(m_j,\ldots,m_k|y).
$$
Let $m_1,\ldots,m_n,x \in \mathbb Z$. We distinguish the following cases:
$m_1 + \cdots + m_n < x$: If we choose $y$ bigger than $m_j + \cdots + m_k$ the right term of the tensor product vanishes, otherwise the left term does.
$m_1 + \cdots + m_n = x$: The only non-zero term is $I\otimes I$.
$m_1 + \cdots + m_n = x + 1$: The coend is the colimit of the following diagram:
Which is the desired $I^{\oplus n}$
$m_1 + \cdots + m_n = x + 2$: Here the coend is the colimit of the following diagram:
If $k-j \leq 1$ this vanishes, but for $k-j >1$ I don't think it does. However since one could express any higher multiplication as one of arity 2 associated such that always $k-j < 1$ the term should also vanish for higher multiplications. So somewhere I must miss something, sadly I don't have the time to resolve this right now.
I think it might be possible to get rid of these terms by modifying the construction by taking $(m_1,\ldots,m_n,x)' := \mathcal C(m_1+\cdots+m_n,x)\otimes(m_1,\ldots,m_n,x)$. I think if we removed the condition that $\partial\partial = 0$ and also the signs we would get a tensor product of derivations where $\partial(x\otimes y) = \partial x \otimes y + x \otimes \partial y$, where these ghost terms would serve a purpose, as terms of degree $n$ could be constructed from terms of degrees higher than $n+1$. Tensoring with $\mathcal C(-,-)$ would then also ensure that $(-|-)_1 = \mathcal C(-,-)$ even if we do add relations to $\mathcal C$.
$m_1 + \cdots + m_n > x + 2$: As in 1. at least one side of the tensor product vanishes.
It remains to check naturality, which is straight forward keeping in mind the presentation of $(m_1,\ldots,m_n|m_1+\cdots+m_n-1)$ as the colimit of 3. above.
Hi Julia, welcome to MathOverflow! I hadn't thought about what the higher arity $P$'s looked like, and this is super nice!
@Emily Thank you. I'm super happy someone finds this useful :)
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2025-03-21T14:48:30.488832
| 2020-05-04T05:40:53 |
359308
|
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|
Stack Exchange
|
Orthonormal frame bundles on a manifold
Let $(\mathcal{M},g)$ be a torsion free compact Riemannian manifold of dimension $n$. Hence from the metric we know there is an associated horizontal sub-bundle $H_u F \mathcal{M}$ of the orthonormal frame bundle $F \mathcal{M}$ at a frame $u:\mathbb{R}^n\to T_p\mathcal{M}$. Furthermore there are theorems that state that for each curve $\gamma:[0,1]\to\mathcal{M}$ and initial frame $u_0$ there is a unique horizontal lift $\tilde{\gamma}:[0,1]\to F \mathcal{M}$.
Question 1
What is the expression for the integrand of the energy functional of a curve $\gamma:[0,1]\to\mathcal{M}$, $$ g_{\gamma}(\dot{\gamma},\dot{\gamma}) $$ in terms of its unique horizontal lift $\tilde{\gamma}:[0,1]\to F \mathcal{M}$? Does the answer depend on $u_0$? Since there is also then a uniquely defined curve in $w:[0,1]\to\mathbb{R}^n$ (which I guess is chart independent), what is the value of $g_{\gamma}(\dot{\gamma},\dot{\gamma})$ via $w$? Could it be just simply $$ g_{\gamma}(\dot{\gamma},\dot{\gamma}) = \langle\dot{w},\dot{w}\rangle_{\mathbb{R}^n} $$?
What is an expression for $\gamma$ in terms of $w$? (I know that $ \dot{\gamma} = \tilde{\gamma}\dot{w}$, but how to solve this for $\gamma$ since there is the lift on the RHS?).
Question 2
It seems like in construction stochastic processes on $\mathcal{M}$ the orthonormal frame bundle is heavily used because one can solve stochastic equations in $\mathbb{R}^n$, move them to $F \mathcal{M}$ in a straight forward manner, and then project them down to $\mathcal{M}$. This is, if I understand correctly, the essence of the Eells–Elworthy–Malliavin construction. My question is then, why not define Brownian motion in $\mathcal{M}$ via charts, first down at the Euclidean space and then pulling back to the manifold using the chart? I guess there would be a way to glue the curve back together.
Is the frame bundle really just a chart-free way to talk about the manifold's Euclidean structure in a way that is compatible with the metric?
On the orthonormal frame bundle we have soldering forms $\omega_i$ and connection forms $\omega_{ij}$. A lift is horizontal just when $\omega_{ij}=0$ on it. So the velocity can be described by its $\omega_i$ components: $v_i(t)=i_{\tilde\gamma'(t)}\omega_i$. The energy is $\sum_i v_i^2$. You don't define $w$, but I guess it is the development, which is coordinate independent but depends on a choice of initial point and frame on Euclidean space. Its energy has the same expression, also equal to $\left<\dot w,\dot w\right>$. To solve for $\gamma$ in terms of $w$, first find the components $v_i=i_{\tilde w'}\omega_i$ on the frame bundle of Euclidean space. Then solve $i_{\tilde\gamma'(t)}\omega_i=v_i(t)$ and $i_{\tilde\gamma'(t)}\omega_{ij}=0$. Then project to find $\gamma(t)$. The frame bundle avoids charts, and allows simpler algebraic expressions for curvature.
Here the map $i_v \xi$ is application of a tangent vector $v$ into a $1$-form $\xi$.
Note that since the frame bundle of Euclidean space is a product of Euclidean space with the rotation group, every lift of a curve $w(t)$ in Euclidean space is a curve $\tilde w(t)=(w(t),E)$ where $E$ is any constant rotation matrix. If the matrix $E$ has columns $E_1,\dots,E_n$, then $v_i(t)=\left<E_i,\dot w(t)\right>$.
Question 2: Brownian motion does not stay inside a single chart for any positive time; it has some chance of escaping. You can see this already in Euclidean space, as the scale invariance of Brownian motion.
Thanks for your answer. I hesitate to mark it as accepted because it doesn't really address Question 2, but if you think I should accept it anyway I will. Can you say more about the map $i$ which you refer to? What is a good reference for this?
|
2025-03-21T14:48:30.489088
| 2020-05-04T05:51:17 |
359309
|
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"url": "https://mathoverflow.net/questions/359309"
}
|
Stack Exchange
|
Counting the image of a map of varieties using the trace formula
Suppose $f: X\to Y$ is a finite map of varieties over a finite field $\mathbb F_q$. Is there an etale constructible $\mathbb Q_\ell$ sheaf $\mathscr F$ on $Y$ which counts the number of rational points of the form $f(x)$ for $x$ itself rational (as an application of the trace formula)?
If $f$ is closed, we can just use the pushforward. On the other hand,even if $X,Y$ are both spectra of fields, say of $\mathbb F_{q^n},\mathbb F_q$,then I am not sure what we want.
Edit 1: It seems to me that it might be easier to count $\deg(f)$ times the number of points and that's okay too. For instance, the number of squares in $\mathbb P^1$ is $(q-1)/2 + 2$ but since the eigenvalues are algebraic integers, we can't get a factor of $q/2$ by cohomology calculations. But if we multiply by $2$, it's possible.
Edit 2: In the case of $()^{\ell^n}: \mathbb G_m \to \mathbb G_m$, let us suppose we have a sheaf of the form we want and let the Frobenius have eigenvalues $q\alpha_i$ on degree zero compactly supported cohomology.
We can work out the number $\ell^n$ th powers in $\mathbb F_{q^r}$ and it depends on what power of $\ell$ divides $q^r-1$. If we further assume that $q-1$ is exactly divisible by $\ell$, we get equations of the form $\sum_i \alpha_i^r = \max(\ell^{n-k},\ell)$ for $\ell^{k-1}||r$.
We can solve this using Fourier inversion and it seems the unique solutions give us $\ell^{n-2k}$ copies of $\alpha_i = \zeta_{\ell^k}$ as $\zeta_{\ell^k}$ ranges over all primitive $\ell^k$ roots of unity and $1\leq k \leq n-1$ and $\ell^{n-1}$ copies of $1$.
Unfortunately $\ell^{n-2k}$ is not integral which seems to show that no $\mathbb Q_\ell$ sheaf that is defined by localizing a $\mathbb Z_\ell$ sheaf can work. Is there some other formalism of $\mathbb Q_\ell$ sheaves that can give us something useful?
Alternatively, perhaps we need to assume that the base field is large enough (perhaps large enough to make the galois group of the generic fiber constant?)
Aren't you just trying to count rational points on the graph ${(x,f(x)) \in X \times Y}$?
@user45878 no, I only want to count the projection of that to Y.
|
2025-03-21T14:48:30.489258
| 2020-05-04T11:35:54 |
359321
|
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|
Stack Exchange
|
Calculate the Jacobian of a particular diffeomorphism of parallelizable manifold onto itself
Let $M$ be $d$-dimensional parallelizable manifold. Let $e_k(x)$ for $k=1, \dots , d$ be the smooth vector fields forming orthonormal basis in tangent bundle of , these vector fields exist because of parallelizability of the manifold.
Consider mapping
$$ \phi (x)= \gamma_{x, \sqrt{d}e_k}(t)$$
where $\gamma_{x,v}(t)$ denotes the point which moved for time $t$ through the geodesics starting at time 0 at the point $x \in M$ with initial velocity $v \in T_xM$.
Please help me to calculate the Jacobian of $\phi$.Thank you very much!
|
2025-03-21T14:48:30.489337
| 2020-05-04T13:03:40 |
359328
|
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|
Stack Exchange
|
Has anyone seen this generalization of the snake lemma? Is it useful?
I originally posted this question on MSE (link), but was suggested to post here instead.
While learning about spectral sequences a friend of mine found a proof of the snake lemma using spectral sequences. We noticed that the proof works equally well for larger bicomplexes. Particularly if you have an exact (anti)-commutative diagram
you get an
exact sequence.
We also have a little write-up of the proof here. We looked around, but couldn't find any reference to this anywhere, and no one else we talked to had really thought about it before. While toying with this we realized that the hypothesis is quite strong. That is, it is pretty difficult to find any interesting exact bicomplexes of the right size. I'm starting to suspect that there might not really be any interesting examples of this, and that that is the reason we haven't found anything about it anywhere.
So we're wondering, has anyone seen this before? Can anyone think of any non-trivial examples or applications of this?
(I would call it the long snake lemma, rather. For some reason "anaconda" sounds... weird.)
Spectral sequences are obtained by applying the snake lemma ad nauseam, essentially. You are doing that (finitely many times) in your spectral sequence argument.
Let me illustrate. Consider the case you have a diagram that has three exact rows and four exact columns, and start with the top right corner space $C_{1,4}$, and pick $x$ there.
If $d_v(x)=0$, pull this back to some $x'\in C_{1,3}$ and apply $d_v(x')\in C_{2,3}$. Then $d_h$ of this is zero, so there is some $x''$ in $C_{2,2}$ such that $d_h(x'') = d_v(x')$. Apply $d_v$ again to fall into $C_{3,2}$. Then take the class of this, and this is your map.
So, in general, you start with $x_{1,n} \in C_{1,n}$ in the kernel of $d_v$,
pull it back to $x_{1,n-1}$, apply $d_v$, pull it back, apply $d_v$, keep going, until you reach $C_{n+1,1}$. You can do this because your rows are exact, and this (I think) is all that you need, you don't need your columns to be exact (as in the snake lemma).
You can do this in general for any $n$, with no spectral sequences, its just a longer version of the snake lemma. The spectral sequence formalism is packaging neatly, but this
is what is going on.
Thank you for your response. I do understand what is going on in the theorem. My question is whether it is at all significant or useful. Are you saying that it is not because it is just a trivial consequence of the snake lemma?
My interpretation of Pedro's answer is something like
"""
part of the point of using spectral sequences is to make applications of the snake lemma invisible, and when working with spectral sequences it is already common to "apply finitely many snake lemmas" without comment.
"""
@JalexStark Agreed. That's a good way to put it. :)
|
2025-03-21T14:48:30.489556
| 2020-05-04T13:14:38 |
359331
|
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"Giulio",
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"R. van Dobben de Bruyn",
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"url": "https://mathoverflow.net/questions/359331"
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|
Stack Exchange
|
Is the preimage of a nonreduced subscheme via a proper map nonreduced?
Let $f\colon X \to Y$ a surjective proper map between smooth varieties over an algebraically closed field $k$ of characteristic zero. Let $Z\subset Y$ be a closed non-reduced subscheme. Is the preimage $f^{-1}(Z)$ nonreduced?
(The situation I am interested in is the resolution of an ideal sheaf, but I do not know if this hypothesis helps.)
Notes:
1- this is a refining of this question, asked by Shende and answered by Sawin. In the answer, $Z$ was supported on the singular locus of $Y$. In my question, I am adding smoothness and properness hypotheses.
2- The following I think is an important example. Let $f\colon X\to Y$ be the blow-up of a smooth point $p$ on a surface. Let $Z$ be isomorphic to $Spec(k[x]/x^2)$, supported at $p$, and $x$ goes to a tangent direction $v$. Then $f^{-1}(Z)$ is the exceptional divisor (with reduced scheme structure) union an embedded point at $v$, so it is indeed nonreduced. However, if you remove $v$, then the map is still surjective but $f^{-1}(Z)$ is reduced. So properness is important.
I think that a morphism between smooth varieties is a local complete intersection morphism. So $f^{-1}$ can be constructed as the usual pull-back of cycles, and a pullback of a non-reduced cycle is non-reduced.
@RizaHawkeye: oh right, I had in mind the flat case. Does the specialization in the non-flat case preserve non-reducedness of cycles?
@FrancescoPolizzi moreover, the cycle associated with a subscheme does not see embedded points.
I do not know the answer even if you replace reduced with generically reduced, so I am also happy with an argument which does not see embedded point. I also do not know what happens at the cycle pull-back in the example I have described.
Intersection-theoretically, the pullback is the class of two points on the exceptional divisor (which two points? any two points) so the cycle-theoretic pullback is not so helpful.
@Giulio: You probably mean $f^{-1}(Z)$ is reduced at the end.
If $Z$ is not generically reduced, then its pullback is not reduced. (The argument can show that the pullback is not generically reduced, but not the way I wrote it.)
Lemma: To show that the pullback of $Z$ is not reduced, it suffices to check that there is a smooth curve $C$ mapping to $Y$ such that the pullback of $\mathcal I_Z$ and $\sqrt{\mathcal I_Z}$ to $C$ differ, where $\mathcal I_Z$ is the ideal sheaf of $Z$ and $\sqrt{\mathcal I_Z}$ is its radical.
Proof: By the valuative criterion, you can map some ramified cover of $C$ to $X$, lifting the map to $Y$, and then pulling back to this ramified cover shows that $f^* \sqrt{\mathcal I_Z}\neq f^* \mathcal I_Z$. Now because $f^* \mathcal I_Z$ contains a power of $f^* \sqrt{\mathcal I_Z}$, it follows that $f^* \mathcal I_Z$ is not radical and thus $f^* Z$ is not reduced.
Now let's check that if $Z$ is generically non-reduced, there exists such a $C$. To do this, work locally near the generic point of $Z$, so that the radical just becomes the maximal ideal at this point. Since $Z$ is not reduced, the map from the ideal of $Z$ to the Zariski cotangent space at the generic point (i.e. generators of this maximal ideal) is not surjective, so there exists a nonzero vector in the Zariski tangent space which is perpendicular to the image of the ideal of $Z$.
Pick a smooth curve $C$ whose tangent vector is that nonzero vector. The pullback of the maximal ideal to $C$ will have multiplicity $1$ while the pullback of $I_Z$ will have multiplicity $>1$, so they will be distinct.
However, this approach (showing equality between $\mathcal I_Z$ and $\sqrt{\mathcal I_Z}$) will not work in general. Let $Z$ be the vanishing locus of $$x_1 x_3, x_2 x_4, (x_1 x_2 - x_2 x_3), (x_2 x_3 - x_3 x_4), (x_3 x_4 - x_1 x_4) $$
as well as $x_i x_j x_k$ for all triples $i,j,k$, not all equal. The point is that the induced reduced subscheme of $Z$ is the vanishing locus of $$x_1x_2,x_1x_3,x_1x_4, x_2x_3,x_2x_4, x_3,x_4$$ and $Z$ differs from this by an embedded point.
I claim the pullback of $Z$ and its induced reduced subscheme to the blowup of $\mathbb A^4$ at $0$ are equal (but neither is reduced since they each contain a double neighborhood of the exception divisor).
In a typical affine chart $x_1 =a_1 z, x_2 =a_2 z, x_3 =a_3 z, x_4 = z$, the pullback of the induced reduced is the vanishing locus of $z^2 (a_1,a_2,a_3)$ and the pullback of $Z$ is the vanishing locus of $$z^2 (a_1a_3, a_2, (a_1a_2-a_2a_3), (a_2a_3 - a_3), (a_3 - a_1)) $$
which equals $z^2(a_1,a_2,a_3)$ since we can cancel $a_2a_3$ with $a_2$ and get $a_3$ then cancel $a_3$ and get $a_1$.
By the symmetry rotating the four variables, all the affine charts look like this, so the pullbacks are equal as ideal sheaves.
|
2025-03-21T14:48:30.489886
| 2020-05-04T13:55:21 |
359334
|
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|
Stack Exchange
|
References on integration on non-compact manifolds
I am looking for references on integration on non-compact Riemannian manifolds, specially on the change of variables theorem.
In particular I have non-compact manifold $M$ and I have an integral (in most interesting cases for me improper integral)
$$\int_M f(x) \mu(dx)$$
Where $\mu$ is standard volume measure on $M$ generated by the metric tensor $g$ on $M$. And I need to make change of variables $y=\phi (x)$, where $\phi$ is a diffeomorphism $\phi: M \to M$. Function $f$ may be unbounded, manifold $M$ may be unbounded (say $M= \mathbb{S}^1 \times \mathbb{R}^2$).
I have found reference book only for compact manifolds and bounded functions.
Please help with reference applicable to my case. Thank you very much!
Could you be more specific? The standard change of variables formula is a local one and therefore valid on any manifold without boundary.
Nearly all results for integration on compact manifolds can be transferred to noncompact via partitions of unity.
thank you both!
@NateEldredge which aspects could not be transferred?! a reference would highly be appreciated.
|
2025-03-21T14:48:30.490324
| 2020-05-04T14:47:27 |
359337
|
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|
Stack Exchange
|
Surfaces with non-constant negative curvature
Are there any nice models of surfaces with non-constant negative curvature, analogous to the Poincare disk for constant negative curvature. I have found lots of general results and theory but no nice clean models.
Until you specify what you mean by model, and how exactly you give the curvature function, this question is impossible to answer.
To clarify, something like a subset of the plane with a riemanian metric where the geodesics and horocycles are easy to describe.
This previous question has a nice discussion of related issues and ideas:
https://mathoverflow.net/questions/37651/riemannian-surfaces-with-an-explicit-distance-function/38183
The question remains imprecise. Should it be interpreted as: "what is a classification of simply connected complete negatively curved surfaces"? or maybe those with a cocompact [cofinite volume] isometry group, if one is interested in compact [finite volume] surfaces. I understand that the answer is expected to be given in terms of model, i.e., say, explicit Riemannian structures, say, on explicit simply connected open subsets of the plane.
If you just want examples for which it's not hard to figure out how the geodesics behave, here's a class of examples with negative and non-constant curvauture in the plane where the geodesics are relatively easy to understand:
Let $a$ and $b$ be smooth functions on $\mathbb{R}$ such that $a(x)+b(y)>0$ for all $x,y\in\mathbb{R}$ and consider the metric
$$
g = \bigl(a(x) + b(y)\bigr)(\mathrm{d}x^2 + \mathrm{d}y^2)
$$
on $\mathbb{R}^2$. The curvature of this metric is
$$
K = \frac{a'(x)^2+b'(y)^2-\bigl(a''(x)+b''(y)\bigr)\bigl(a(x)+b(y)\bigr)}
{2\,\bigl(a(x)+b(y)\bigr)^3}
$$
It's easy to choose $a$ and $b$ so that $K<0$. For example, $a(x) = x^2+1$ and $b(y) = y^2+1$ gives a complete metric on $\mathbb{R}^2$ that has non-constant negative curvature $K = -4/(x^2{+}y^2{+}2)^{3}<0$.
Note that, taking $a$ (respectively, $b$) to be a constant gives a metric $g$ that has a Killing vector field, namely $\partial/\partial x$ (respectively, $\partial/\partial y$), but, for generic choices of $a$ and $b$, the metric $g$ will have no Killing vector field.
As for geodesics, the good thing about these metrics (called Liouville metrics in the literature) is that their geodesic flows are integrable: Any unit speed geodesic $(x(t),y(t))$ satisfies
$$
\bigl(a(x)+b(y)\bigr)\bigl(\dot x^2+\dot y^2\bigr) = 1
\quad\text{and}\quad
\bigl(a(x)+b(y)\bigr)\bigl(b(y)\,\dot x^2- a(x)\,\dot y^2\bigr) = c
$$
for some constant $c$. (Note that, when either $a$ or $b$ is constant, this second 'first integral' of the geodesic equations specializes to the well-known 'Clairaut integral' for surfaces of revolution.)
In particular,
$$
\bigl(b(y)-c)\bigr)\,\dot x^2 - \bigl(a(x)+c)\bigr)\,\dot y^2 = 0,
$$
and, assuming that you are in a region when $a(x){+}c$ and $b(y){-}c$ are both positive,
$$
\frac{\mathrm{d}x}{\sqrt{a(x)+c}} \pm \frac{\mathrm{d}y}{\sqrt{b(y)-c}}=0,
$$
which gives two foliations of this region by geodesics, which can be found by quadrature.
In any case, you will have good qualitative control over these geodesics and can draw some nice pictures.
Added remark (12 May 2020): As an example of what one can do with this more explicit information, you might be interested in this answer of mine to an old question about Riemannian surfaces for which one can compute an explicit distance function.
Minimal surfaces without umbilics. They are nice and have negative Gaussian curvature. The catenoid is a particular example.
Hey Sebastian, you're certainly right that generally speaking the Enneper-Weierstrass parameterization is a nice way to generate examples of negatively curved surfaces. But, they do tend to be immersed and therefore extrinsic in nature, so they are quite different than the kind of model that the Poincare disk is. Still, this is a nice class of examples.
Hi Andy, I see your point. On the other hand we do not have an extrinsic model for the hyperbolic ball in 3-space as well, and the explicit form for the metric of a minimal surface is given on the domain of the conformal parametrisation. Of course, one can ask which complete embedded examples without umbilics do exist. But even with the further constraint of finite total curvature we have at least one (class of) example(s).
You don't even need to settle for a model. You can make such a surface. Take any positive strictly convex smooth function ($f(x) > 0$ and $f''(x) > 0$) and rotate it around the x axis.
The single-sheet hyperboloid, $x^2 + 1 = y^2 + z^2$, has some nice geodesics, but also some messier ones: https://math.stackexchange.com/questions/1601158/how-can-we-find-geodesics-on-a-one-sheet-hyperboloid
You may be able to calculate the geodesics with methods given here:
Geodesics on a hyperbolic paraboloid (which is yet another negative curvature surface)
|
2025-03-21T14:48:30.490682
| 2020-05-04T16:03:15 |
359345
|
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|
Stack Exchange
|
Geometrical proof of Noether Theorem
I am reading a very nice Physics book "The standard model in a nutshell" by D.Goldberg and just read there a mention to Noether Theorem. Of course I knew this outstanding theorem very well from previous physics courses : "A symmetry in the system (Lagrangian) implies a conserved physical quantity". The proof is quite short it follows from the Euler Lagrange equation and a few lines of calculus. But now that I think about it, I am not that happy with it. I fell like such a important result should have a more abstract and geometrical understanding.
Do you know another proof of Noether Theorem written in a differential geometry setting and that can be generalized in a more abstract case?
You'll find a quick proof of Noether's theorem, in greater generality than you're likely to need even, in this recent answer.
I suppose the proof in section 2.3.1 of this article https://arxiv.org/pdf/1601.03616.pdf are those few line of calculus. But maybe not, you can glance it
Khavkine's answer is really excellent.
You may also be interested in the discussion here: https://physics.stackexchange.com/questions/19847/whats-the-interpretation-of-feynmans-picture-proof-of-noethers-theorem
|
2025-03-21T14:48:30.490939
| 2020-05-04T16:44:50 |
359349
|
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|
Stack Exchange
|
The eigenvalue of Schrodinger opeartor
Let $\Omega$ be a smooth bounded domain in $\mathbb{R}^n$, consider $L=\Delta+V$ where $\Delta=div\nabla$. According to section 8.12 in Gilbarg and Trudinger's book, if $V\in L^\infty(\Omega)$, then the Dirichlet spectrum of $L$ exists, and the first eigenvalue is simple.
(1) What is the weakest condition on $V$ such that the spectrum still exists?
(2) What is the condition on $V$ such that the first eigenvalue is positive?
I think this should be well studied, but I haven't been able to find any good reference. If you are familiar with this, please let me know. Thanks for your help.
What do you mean by weakest? (for question 1) // For your second question, as you wrote it, your operator $L$ is formally a negative operator when $V \equiv 0$. I just want to double check that you really meant to ask about the first eigenvalue being positive, or if rather you want the opposite direction? In any case, certainly the answer to the second question depends lots on $\Omega$ and details of $V$, if only just by thinking about the family $\lambda V$ of potentials and noting for all sufficiently small $\lambda$ the spectrum should be a small perturbation of that of $\Delta$.
Welcome to MathOverflow! What precisely do you mean by "the spectrum exists"? Every linear operator has a spectrum by definition.
@JochenGlueck: I deleted my previous comment because my interpretation was not correct. Turns out that particular section of G+T is talking about Rayleigh quotients, and bounded coefficients are convenient. // And also, never noticed this before, but there is a typo in the 2001 edition of G+T: the operator and quadratic form on the bottom of page 212 disagree by a sign (which further muddies my second question to the OP).
One possibility (after taking a quick look at G+T) of a reasonable question: what are the conditions on $V$ that guarantees the spectrum is discrete. Perhaps this is what the OP meant by question 1.
@WillieWong Yes, that's what I meant. Can we have weaker conditions like $V\in L^p(\Omega)$ such that the operator has a discrete increasing spectrum? However, I am more interested about the second one.
I would recommend the book by B. Davies: spectral theory and differential operators.
@Student: The question which unbounded types of potentials are admissible depends heavily on the sign of the potential. If you are interested in $V$ that are bounded above (but not necessarily below), things are easier than if you are interested in $V$ that might be unbounded above.
|
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