added
stringdate 2025-03-12 15:57:16
2025-03-21 13:32:23
| created
timestamp[us]date 2008-09-06 22:17:14
2024-12-31 23:58:17
| id
stringlengths 1
7
| metadata
dict | source
stringclasses 1
value | text
stringlengths 59
10.4M
|
|---|---|---|---|---|---|
2025-03-21T14:48:31.717901
| 2020-08-08T10:30:18 |
368633
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"HJRW",
"Moishe Kohan",
"https://mathoverflow.net/users/1463",
"https://mathoverflow.net/users/159398",
"https://mathoverflow.net/users/39654",
"rvdaele"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631878",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368633"
}
|
Stack Exchange
|
Gromov Hausdorff distance to tubular neighborhood
Let $M$ be a compact path metric space in $\mathbb{R}^d$, and for $\sigma>0$,
$$
M_\sigma:=\{y\in\mathbb{R}^d:\min_{x\in M}\|x-y\|\leq\sigma\}
$$
the $\sigma$-tube around $X$ in $\mathbb{R}^d$. I consider both $M$ and $M_\sigma$ metric spaces with respect to the shortest path metric (geodesic, not necessarily Euclidean distances) induced by $\mathbb{R}^d$, with possibly distinct intrinsic dimensions. We are furthermore given constants $s,\epsilon>0$, such that for $x,y\in M$, $\|x-y\|<s\implies d_M(x,y)<\epsilon$. Is there any bound we can provide on the Gromov Hausdorff distance $d_{GH}(M, M_\sigma)$ in terms of $s,\epsilon$, and the diameter of $M$, when $\sigma$ is sufficiently small?
The tubular neigborhood can significantly alter the metric, e.g., the tubular neighborhood of a nearly closed circle can suddenly include the circle itself. However, I suspect that such information would be encoded by $s$ and $\epsilon$, and that for $\sigma$ sufficiently small (according to these parameters), the path from $x$ to $y$ in $M_\sigma$ travels `near' the path from the (not necessarily unique) projections of $x$ and $y$ on $M$, and the length of these paths will then be similar.
I could believe that similar problems have been investigated before, but I don't find any helpful references. It would be great if someone could point out some possible directions on this problem.
Your intuition that such problems have been investigated before is certainly correct. You could try looking up the notions of "quasi-isometry" and "coarse embedding".
More specifically, I don't quite understand the question. It's clear that ${}(,\sigma)$ is bounded above by $\sigma$, so for $\sigma$ sufficiently small compared to $s,\epsilon$ and the diameter of $M$, that appears to provide the bound you are looking for. But presumably I misunderstood something...
It is certainly easy to show that there exists a correspondence under which $d_{M_\sigma}$ is not much larger than $d_M$ (more specifically, at most 2$\sigma$). However, showing that $d_M$ is not much larger than $d_{M_\sigma}$ is another story.
Consider a circle $C$ with radius $r$ and a very small angle $\theta$. Now discard any segment in the circle defined by $\theta$. This is now a space homeomorphic to the line, with diameter $r(2\pi - \theta)$. When we let $\sigma = r\sin(\theta / 2)$, i.e., the length of the chord corresponding to the discarded segmented, then $M_\sigma$ includes the entire circle $C$, and its diameter (distance between the two furthest points) is now closer to half the original diameter of $M$. Hence, $d_{GH}(M, M_{\sigma})$ is then at least close $\pi r$.
However, when $\sigma$ is less then two times the length of the chord defined by $r\sin(\theta/2)$, $M_{\sigma}$ would not include the entire circle $C$, and their metrics of $M$ and $M_\sigma$ should now be much closer. The roles of $s$ and $\epsilon$ are here fulfilled by the length of a chord and the length of a segment in the circle, respectively.
I see your point — I think I was getting the restricted metric and the induced path metric mixed up. Anyway, these flavours of problems are often studied in geometric group theory, and the keywords I suggested above should get you started.
Do you mean intrinsic or extrinsic diameter? Equivalently, do you fix the dimension d of the ambient Euclidean space?
Intrinsic diameter and fixed dimension d of the ambient space.
So $M$ could be a smooth curve and $M_{\sigma}$ a bended and rounded cylindrical shape in the same space $\mathbb{R}^d$, but I'm considering both metric spaces for their geodesic distances.
OK, from your comment, diameter is understood intrinsically. Then $diam(M_\sigma)\le diam(M)+2\sigma$. Hence, setting $D=diam(M)$, you trivially obtain $dist_{GH}(M, M_\sigma)\le D+2\sigma$ regardless of $s, \epsilon$ and other assumptions. The genuinely interesting question is about the extrinsic diameter, when dimension $d$ is not fixed. Then $dist_{GH}(M, M_\sigma)$ is cannot be bounded in terms of $diam, s, \epsilon, \sigma$.
I'm actually looking for a less trivial bound on $d_{GH}(M, M_{\sigma})$ that converges to 0 if $\sigma$ converges to 0. I suspect this to be possible. To give some context: I'm mainly interested in this question in a data science setting. More specifically, given a data set $X$ with underlying model $M$ which has geometric properties constrained through $s$ and $\epsilon$, what is an acceptable amount of noise $\sigma$ such that the space $M_{\sigma}$ in which $X$ resides in has similar geometric properties as $M$.
E.g., if $M$ is the unit circle in $\mathbb{R}^2$, then if I'm correct it holds that $d_{GH}(M,M_{\sigma})=\pi\sigma=diam(M)\sigma\rightarrow_{\sigma\rightarrow 0} 0$. I believe that in general, $s$ will determine an initial upper bound on $\sigma$, before we can actually guarantee that $d_{GH}(M,M_{\sigma})$ is small according to $\sigma$.
I see. Then you should modify the question accordingly. However, the answer is negative since $M_\sigma$ need not converge to $M$ in GH-topology. An example is given by the comb space.
I think I have figured this out. More specifically, it should hold that
$$
d_{GH}(M, M_\sigma) \leq \max\left\{2\sigma, \left(\frac{\epsilon}{s-2\sigma}-1\right)(\mathrm{diam}(M)+2\sigma)+\epsilon\right\},
$$
whenever $\sigma < s/2$.
Sketch of the proof:
Define the correspondence $C$ as
$$
(x,y)\in C\leftrightarrow y\in \overline{B}_{\mathbb{R}^d}(x,\sigma)
$$
Clearly it holds that $(x,y),(x',y')\in C$ implies that $d_{M_\sigma}(y,y')\leq d_M(x,x')+2\sigma.$
For the more difficult direction, take $0<\delta\leq s-2\sigma$ and split up the path from $y$ to $y'$ in $M_\sigma$ into $k$ parts of length at most length $s - 2\sigma - \delta$. This can be done with $k\leq \frac{d_{M_\sigma}(y,y')}{s - 2\sigma - \delta}+1$ segments. Each of these segments corresponds to a segment to a segment in $M$ with length at most $\epsilon$. We find that $d_M(x, x')\leq \left(\frac{d_{M_\sigma}(y,y')}{s - 2\sigma - \delta}+1\right)\epsilon$. Now subtract $d_{M_\sigma}(y,y')$ from both sides, and bound $d_{M_\sigma}(y,y')$ in the right hand side by $\mathrm{diam}(M_\sigma)\leq\mathrm{diam}(M)+2\sigma$. Finally, let $\delta\rightarrow 0$.
With the clarification given in the comments, what you are asking is if there is a bound on $d_{GH}(M, M_\sigma)$ which tends to $0$ as $\sigma\to 0$. This question has negative answer. An example is given by $M$ which is the comb space:
The thing is that for each $\sigma=1/n$, $M$ contains a pair of points $p=(0,1), q=(\frac{1}{n},1)$ such that
$$
d_{M_\sigma}(p,q)=1/n,
$$
while $d_M(p,q)= 2+ \frac{1}{n}$. The existence of these pairs of points prevents the GH-convergence $M_\sigma\to M$. (The space satisfies other conditions in your question: $diam(M)=3$, one can take $s=\sqrt{2}, \epsilon=3$.)
One way to define the GH distance is via distortion of bisurjective correspondences:
Definition. Let $A, B$ be compact metric spaces and $R\subset A\times B$ be a bisurjective correspondence meaning that its projection to both $A$ and $B$ is surjective: For every $a\in A$ there is $(a,b)\in R$ and for every $b\in B$ there is $(a,b)\in R$. Define the distortion of $R$ by:
$$
dis(R):= \sup_{(a,b), (a',b')\in R} |d(a, a')- d(b,b')|.
$$
Then $$d_{GH}(A,B)= \frac{1}{2}\inf_R dis(R)$$ where the infimum is taken over all bisurjective correspondences as above. Up to a uniform factor (which is irrelevant for our purposes), $d_{G}$ can be defined using $\epsilon$-surjective maps:
$$
\inf \{dis(f), f: A\to B \ \hbox{is $\epsilon$-surjective}\},
$$
where $dis(f)= \sup \{ |d(f(a), f(a'))- d(a,a')| : a, a'\in A\}$ and $f$ is $\epsilon$-surjective if each $b\in B$ is within distance $\epsilon$ from some $f(a)$.
In other words, if $dis(f)\le \epsilon$ then $f$ is a $(1,\epsilon)$-quasiisometry:
$$
d(a,a')-\epsilon \le d(f(a), f(a'))\le d(a,a')+\epsilon,
$$
Therefore, a sequence of compact metric spaces $M_n$ converges to a metric space $M$ if and only if there is a sequence of $(1,\epsilon_n)$-quasiisometries
$$
f_n: M_n\to M,
$$
which are $\epsilon_n$-surjective and $\lim_{n\to\infty}\epsilon_n=0$.
What you get in your setting is different: The inclusion map $f: M\to M_\sigma$ defines (when $\sigma\le s/3$) a quasi-isometry $M\to M_\sigma$:
$M$ is $\epsilon$-dense in $M_\sigma$ and $f$ satisfies (for all $a, a'\in M$)
$$
\frac{\sigma}{\epsilon}d_M(a, a') - \sigma\le d_{M_\sigma}(f(a), f(a'))\le d_M(a, a').
$$
The multiplicative (Lipschitz) factor $\frac{\sigma}{\epsilon}\ne 1$ in the LHS makes all the difference. This is the difference between the GH distance and quasi-isometries mentioned by in Henry's comment. The attempt to estimate (from above) the GH distance made in your post will also result in a map with such multiplicative factor $\ne 1$ and that just is not good enough. One can define a measure of closeness between compact metric spaces using quasi-isometries instead of the GH-distance. I do not know if it is useful for anything. If you are content with, say, $C^2$-smooth compact submanifolds $M$ instead of general compact subspaces then, indeed, you get GH-convergence $M_\sigma\to M$.
For more on the topic, see this question and
Burago, D.; Burago, Yu.; Ivanov, S., A course in metric geometry, Graduate Studies in Mathematics. 33. Providence, RI: American Mathematical Society (AMS). xiv, 415 p. (2001). ZBL0981.51016.
There are several other questions one can ask along the lines of your post. The more interesting of these is:
Do not fix the dimension of the ambient Euclidean space, but assume that the extrinsic diameter of $M$ is $\le D$. Is there a uniform upper bound on
$$
\liminf_{\sigma\to 0+} d_{GH}(M, M_\sigma)
$$
in terms of $s, \epsilon$ and $D$? This question also has a negative answer but examples are harder; they use the comb space as one of the building blocks.
The reason to use the extrinsic diameter is that if the intrinsic diameter is bounded by $D$ then, trivially,
$$
d_{GH}(M, M_\sigma)\le D+\sigma,
$$
which you find uninteresting. If the extrinsic diameter of $M$ is bounded by $D$ and the ambient dimension $n$ is fixed, one again obtains an upper bound on the intrinsic diameter of $M$ in terms of $D$ and $n$.
Edit. Here is a correct phrasing of your question:
Suppose that $M\subset {\mathbb R}^n$ is a rectifiably-connected subset, such that, when equipped with the intrinsic path-metric $d_M$, $M$ is compact. Does it follow that the family of neighborhoods $M_\sigma$ of $M$ (also equipped with the intrinsic path-metrics) converge to $M$ in the GH topology?
Suppose that $M$ is a compact connected $C^1$-smooth submanifold in ${\mathbb R}^n$. Is $M$ still compact with respect to its intrinsic path-metric?
Suppose that $M$ is a compact connected $C^2$-smooth submanifold in ${\mathbb R}^n$. Can one estimate $d_{GH}(M, M_\sigma)$ in terms of intrinsic and extrinsic differential-geometric invariants of $M$?
Now, this question has positive answer:
Consider the identity embeddings $f_\sigma: M\to M_\sigma$. Then each $f_\sigma$ is $\sigma$-surjective and 1-Lipschitz. Thus (see the interpretation of GH distance above in terms of maps), we just need to prove that
$$
\lim\sup_{\sigma\to 0+} \sup_{p,q\in M} |d_M(p,q)- d_{M_\sigma}(p,q)|=0.
$$
A proof is by contradiction: If this limit is $\delta>0$, then (by compactness!) there are sequences $p_i, q_i\in M$ converging to $p, q\in M$ (with respect to the topology given by its path-metric) such that
$$
\lim_{i\to\infty} (d_{M_{1/i}}(p_i,q_i) - d_M(p_i,q_i))=\delta.
$$
Let $c_i: [0,1]\to M_{1/i}$ be nearly geodesic paths connecting $p_i$ to $q_i$. These paths can be taken uniformly Lipschitz (with respect to the Euclidean metric) since the diameter of $M_{1/i}$ is $\le diam(M)+ 2$. By applying Arzela-Ascoli theorem combined with the Lebesgue dominant convergence theorem, we obtain a limit path $c$ in $M$ connecting $p$ to $q$ whose length is $\le d_M(p, q)-\delta$. A contradiction.
For $C^2$-smooth submanifolds, it is a classical fact proven in pretty much every Riemannian geometry textbook that for a $C^2$-smooth Riemannian metric, the manifold topology agrees with the topology given by the Riemannian distance function. For a $C^1$-smooth submanifold, you can argue instead as follows. It suffices to show that $(M, d_M)$ is sequentially compact. By the compactness of $M$ (with the subspace topology), it suffices to show that if $p_i\to p$ in the subspace topology of $M$, then $d_M(p_i, p)\to 0$. Writing the induced Riemannian metric in local $C^1$-coordinates, it becomes merely continuous but this is enough.
(Actually, one needs even less than continuity.) The proof now becomes just a calculus computation:
$$
\lim_{i\to\infty} \int_{0}^{\epsilon_i} \sqrt{g(c_i'(t), c_i'(t))}dt \le \lim_{i\to\infty} K \epsilon_i =0,
$$
where $g$ is a continuous Riemannian metric on a domain in ${\mathbb R}^k$, $c_i: [0, \epsilon_i]\to {\mathbb R^k}$ are arc-length parameterizations of line segments (emanating from the origin) of length $\epsilon_i$, satisfying $\epsilon_i\to 0$. The constant $K$ is an upper bound on the $g$-norm of unit vectors in ${\mathbb R}^k$ near the origin. (Hence, all what you need is that, in the local coordinate, the metric $g$ is measurable and locally bounded on unit vectors, where unit is understood with respect the Euclidean norm.)
An estimate for $C^2$-smooth compact submanifolds can be given in terms of the 2-nd fundamental form (you need it for submanifolds of arbitrary codimension):
If $\sigma$ is sufficiently small (less than the normal injectivity radius of $M$ in ${\mathbb R^n}$), you have a well-defined nearest-point projection $r_\sigma: M_\sigma\to M$. You need is to estimate the Lipschitz constant $L$ of $r_\sigma$. The estimate is essentially the same as the one for the circle example:
$$
L^{-1} \ge 1- \sigma C,
$$
where, up to some multiplicative constant depending only on the dimension $n$, $C$ is the supremum-norm of the 2nd fundamental form of $M$. (In the circle example, $1/C$ is the radius of the circle.) Thus, for $p, q\in M$, you have
$$
0\le d_M(p, q)- d_{M_\sigma}(p,q)\le CD\sigma.
$$
Thus (up to a uniform multiplicative constant depending only on $n$),
$$
d_{GH}(M, M_\sigma)\le CD\sigma,
$$
if $\sigma$ is less than the normal injectivity radius of $M$.
Thanks for your clear clarification. You are correct that $d_{GH}$ does not converge to 0 under my stated assumptions, for your given example.
As you might have supsected, I'm indeed content with more restrictive spaces, such as $C^2$ smooth compact submanifolds. In matter of fact, for my purpose, though these are not the exact spaces I'm working with, it would be sufficient to assume $M$ is a smooth curve in $\mathbb{R}^d$. Do you know any references where your stated fact that GH convergence is achieved for these spaces is proven? I'll look into some of your current references as well.
|
2025-03-21T14:48:31.719256
| 2020-08-08T11:20:52 |
368634
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Andreas Blass",
"Konstantinos Kanakoglou",
"Sam Hopkins",
"Ted Jh",
"https://mathoverflow.net/users/163464",
"https://mathoverflow.net/users/25028",
"https://mathoverflow.net/users/6794",
"https://mathoverflow.net/users/85967"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631879",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368634"
}
|
Stack Exchange
|
Representation theory in braided monoidal categories
The crux of what I wish to know is what results from representation theory, a subject usually framed within the category $\text{Vect}_\mathbb{k}$, follow in more general braided monoidal categories? I would be very satisfied with references to texts that cover this.
I'll try to be more specific. Let $\mathcal{C}$ be monoidal, abelian, complete under arbitrary countable biproducts, and enriched over $\text{Vect}_\mathbb{C}$. An algebra in $\mathcal{C}$ is an object $A$ with morphisms $m:A\otimes A\rightarrow A, u:1\rightarrow A$ satisfying suitable axioms, and a left $A$-module in $\mathcal{C}$ is an object $V$ with morphism $a_V:A\otimes V\rightarrow V$, again satifying some conditions. If $\mathcal{C}$ has a braiding $\psi$, we can define commutative algebras as those algebras such that $m\psi=m$.
For instance take $\mathcal{C}$ to be $H\text{-Mod}$, the category of (finite-dimensional) modules over quasitriangular Hopf algebra $H$ (the quasitriangular structure on $H$ makes $H\text{-Mod}$ braided monoidal). Recall the following classical (i.e. in $\text{Vect}_\mathbb{C}$) result from representation theory: for commutative algebra $A$, every simple finite-dim'l $A$-module is $1$-dim'l. Is there an analogue of this statement for braided commutative algebras in $H\text{-Mod}$?
Probably you are aware of this textbook?: http://www-math.mit.edu/~etingof/egnobookfinal.pdf
That's a great book, but I don't think it tackles my question, unless I missed something (quite probable)?
Yes, you're probably right, I don't know the specifics, but in general it addresses "studying representation theory in symmetric/braided monoidal categories"
If we consder the special case of $H=\mathbb{CZ}_2$, i.e. the group hopf algebra with its non-trivial quasitriangular structure then we get the super-commutative superalgebras (that is $\mathbb{Z}_2$-graded-commutative assoc algebras). I think we can get an analogous statement but the reps have to have dim $\leq 2$.
I'm confused by "arbitrary countable biproducts". I've seen "biproduct" used to mean "simultaneously a product and coproduct", like what you get with finite products and coproducts in any abelian category. But even in $\text{Vect}_k$, countable products and coproducts are generally different. So what are countable biproducts here?
I will try to provide an answer for a particular case of your last question: Let us consider (following my comment above) the case of $H=\mathbb{CZ}_2$ i.e. the group hopf algebra equipped with its non-trivial quasitriangular structure, that is the $R$-matrix $R=\frac{1}{2}(1\otimes 1+1\otimes g+g\otimes 1-g\otimes g)$.
Its category of representations $\mathbb{CZ}_2\text{-Mod}$, consists of the $\mathbb{Z}_2$-graded vector spaces, commonly refered to as super-vector spaces, with morphisms the even linear maps and an algebra $A$ in the category $\mathbb{CZ}_2\text{-Mod}$ is a $\mathbb{Z}_2$-graded associative algebra (or: associative superalgebra). The braiding of the category is provided by the $R$-matrix.
Consider a f.d. super-module ${}_{A}M$ and the super version of the interwiner, that is an homogeneous linear map $g\in \mathcal{E}nd_\beta(M)$,of degree $\beta(=0,1)$ which super-commutes with the $A$-action on $M$, that is:
$$
g(a\cdot m)=(-1)^{\beta\gamma}a\cdot g(m)
$$
for all $a\in A_\gamma$, $\gamma=0,1$, $m\in M$. Notice that if $\beta=0$ then $g$ is a usual super module morphism (that is an even super module homomorphism) while if $\beta=1$ then we get $\mathbb{C}$-linear, odd maps which are antilinear in the $A$-action. The set of all super-interwiners forms a superalgebra $$\mathcal{E}nd(M)=\mathcal{E}nd_0(M)\oplus\mathcal{E}nd_1(M)$$
In case ${}_{A}M$ is simple (in the super sense, i.e. it contains no proper $\mathbb{Z}_2$-graded submodules), we also have a super-version ($\mathbb{Z}_2$-graded version) of Schur's lemma:
$\mathcal{E}nd_0(M)=\mathbb{C}\cdot Id$ and $\mathcal{E}nd_1(M)=\mathbb{C}\cdot\theta$, where $\theta=0$ or $\theta$ is an odd linear map with $\theta^2=Id$
Considering the usual definitions for the super-center $Z_s(A)$, the super-commutative superalgebra $A$ and the super version of Schur's lemma, we get that:
If $A$ is a superalgebra, ${}_{A}M$ is a f.d. simple super-module and $z_\xi$ an homogeneous element in the supercenter $Z_s(A)$ then there exists some $\lambda=\lambda(z_\xi)\in\mathbb{C}$ such that either $z_\xi\cdot m=\lambda m$ (if $\xi=0$ i.e. $z$ is even) or $z_\xi\cdot m=\lambda\theta(m)$ (if $\xi=1$ i.e. if $z$ is odd), for all $m\in M$.
Finally, using the last lemma, we can get that:
Any f.d., simple super-module ${}_{A}M$ over the super-commutative superalgebra $A$ is at most $2$-dimensional.
P.S.: If you want similar results for more general braided monoidal categories (let say over the category of modules of some more general quasitriangular hopf algebra -even for some other group hopf algebra), i think we will need a braided generalization of Schur's lemma.
I am not aware if something like that exists in the literature (and i would be very interested to know if somebody else knows of such results).
When the underlying category is a quantum group at a root of unity, understanding commutative algebras has been studied extensively by Ocneanu and others under the name “quantum subgroups.” For an explanation in the algebraic language you’ve phrased this question in for the special case of quantum SU(2), see Ostrik-Kirillov. There’s a very nice ADE classification there, which was originally studied in subfactor language by Jones, Ocneanu, etc.
As for your specific question, that basically never happens outside of the case of Vec. Just think about the case where the algebra is trivial, then the category of modules will again be the original category.
|
2025-03-21T14:48:31.719636
| 2020-08-08T14:05:44 |
368638
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631880",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368638"
}
|
Stack Exchange
|
Critical Smoothness on Besov Spaces $B^s_{p}$: how does it evolved with $p$?
We denote by $B_{p}^s(\mathbb{T}) := B_{p,p}^s(\mathbb{T})$ the Besov space over the circle $\mathbb{T}$ with parameters $p=q \in (0, \infty]$ and smoothness $s \in \mathbb{R}$.
For $p>0$ fixed and $f \in \mathcal{S}'(\mathbb{T})$ a generalized function, set
\begin{equation}
s_p(f) = \sup \{s \in \mathbb{R} , \ f \in B_p^s(\mathbb{T})\} \in \mathbb{R} \cup \{+\infty\},
\end{equation}
which can be interpreted as the critical smoothness of $s$ for the $L_p$-scale.
The function $p \mapsto s_p(f)$ is obtained by characterizing in which Besov spaces $f$ is and is not. It is known for some functions. For instance, if $f = \delta$ is the Dirac distribution, we have $s_p(\delta) = \frac{1}{p} - 1$, from which we deduce $s_p ( 1_{[0,1]} ) = \frac{1}{p}$.
The Besov regularity of the Gaussian white noise $W$ is also known (see for instance this paper), from which we deduce that $s_p(W) = - 1/2$. Considering other classes of random processes or fractal-type functions, it is possible to find functions for which
\begin{equation}
s_p(f) = \min ( 1/ p , a) + b
\end{equation}
for any $a , b \in \mathbb{R}$.
Question: Is there some functions $f \in \mathcal{S}'(\mathbb{T})$ for which $s_p(f)$ can take a different form? All the examples I encountered so far suggest that it is not the case.
The answer to my question is actually no: there exists generalized functions such that $s_p(f)$ is not of the proposed form. Stéphane Jaffard discusses the possible forms of the functions $s_p(f)$, denoted by $\eta(p)$, in his paper On the Frisch-Parisi Conjecture.
This question has also been addressed in this paper, which provides some other references. For instance, it includes functions such that $$s_p(f) = \frac{\alpha}{p} + b$$
for any $\alpha \in (0,1)$ and $ b\in \mathbb{R}$.
|
2025-03-21T14:48:31.719783
| 2020-08-08T14:09:51 |
368639
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631881",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368639"
}
|
Stack Exchange
|
How many points of a sequence can we catch with an analytic disc?
Let $X\subset \mathbb{C}^{n}$ be a domain. You can assume that it is nice (e.g. bounded convex balanced ). Let $\{x_n\}$ be a sequence of points that does not have a limit point in $X$.
Let $D$ be the (unit) disc on the plane.
Is there a holomorphic $\varphi:D\to X$ such that $\varphi(D)$ contains an infinite number of $x_n$?
Of course there are further questions: can we catch all the points? Or maybe all but finite number?
The answer to your quesiton is yes.
In particular in Discs in Stein manifolds containing given discrete sets B. Drinovec Drnovšek proves that given a discrete subset $S$ of a connected Stein manifold $M$ (every domain of holomorphy in $\mathbb{C}^m$ is if this kind) there is a proper holomorphic map $f:D \rightarrow M$ such that $S\subset f(D)$. Moreover if dim $M\geq$3 the map f can be chosen to be an embedding.
|
2025-03-21T14:48:31.719881
| 2020-08-08T14:19:14 |
368640
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631882",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368640"
}
|
Stack Exchange
|
Good reference for topological Hochschild homology
I want to start reading topological Hochschild homology(THH) as well as topological cyclic homology (TC).
I have read the Hochschild homology and cyclic homology from the book Cyclic homology by J. Loday. This is very fantastic written book. Can someone suggest me some good reference like this for THH/TC?
As a first introduction I like these notes by Achim Krause and Thomas Nikolaus. They do require some familiarity with spectra and stable homotopy theory though.
Ib Madsen's survey
MR1474979 (98g:19004) Madsen, Ib Algebraic K-theory and traces.
Current developments in mathematics, 1995 (Cambridge, MA), 191–321,
Int. Press, Cambridge, MA, 1994.
was written to be such an introduction, but is 25 years old. Its technical foundations are given in terms of FSPs (functors with smash product). A more recent source is the book
MR3013261 Dundas, Bjørn Ian; Goodwillie, Thomas G; McCarthy, Randy
The local structure of algebraic K-theory. Algebra and Applications,
18. Springer-Verlag London, Ltd., London, 2013. xvi+435 pp. ISBN: 978-1-4471-4392-5; 978-1-4471-4393-2
which works with $\Gamma$-rings (= monoids in $\Gamma$-spaces). Both of these predate the connection to crystalline and syntomic cohomology.
|
2025-03-21T14:48:31.719988
| 2020-08-08T14:28:55 |
368642
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"0xbadf00d",
"Daniele Tampieri",
"https://mathoverflow.net/users/113756",
"https://mathoverflow.net/users/91890"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631883",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368642"
}
|
Stack Exchange
|
Shape derivative of boundary integrals and differentiability of the integrand on a tubular neighborhood
Let $d\in\mathbb N$, $U\subseteq\mathbb R^d$ be open, $$\mathcal A:=\{\Omega\subseteq\mathbb R^d:\Omega\text{ is bounded and open},\overline\Omega\subseteq U\text{ and }\partial\Omega\text{ is of class }C^1\}$$ and $$\mathcal F(\Omega):=\int f\:{\rm d}\sigma_{\partial\Omega}\;\;\;\text{for }\Omega\in\mathcal A$$ for some Borel measurable $f:U\to\mathbb R$ (what do we need to assume to ensure that $f$ is $\sigma_{\partial\Omega}$-integrable for all $\Omega\in\mathcal A$?
Given a family $(T_t)_{t\in[0,\:\tau)}$ of $C^1$-diffeomorphisms from $U$ onto $U$, where $\tau>0$, I want to consider the limit of $$\frac{\mathcal F(\Omega_t)-\mathcal F(\Omega_0)}t\tag1$$ as $t\to0$, where $\Omega\in\mathcal A$ and $\Omega_t:=T_t(\Omega)$ for $t\in[0,\tau)$.
The resulting expression will be the $\sigma_{\partial\Omega}$-integral of an expression involving the gradient of $f$. The question is: What kind of differentiability assumption of $f$ do we need to impose?
In general, $f$ would be called $C^1$-differentiable on $\partial\Omega$ if there is an $\mathbb R^d$-open neighborhood $O$ of $\partial\Omega$ and some $\tilde f\in C^1(O)$ with $$\left.f\right|_{\partial\Omega}=\left.\tilde f\right|_{\partial\Omega}\tag2.$$ I guess the problem with this is that it doesn't yield a well-defined notion of the derivative of $f$ on $\partial\Omega$, since $\nabla\tilde f(x)$ should depend on the choice of $\tilde f$ for $x\in\partial\Omega$.
I've seen frequently the assumption that $f\in C^1(T)$ for some tubular neighborhood $T$ of $\partial\Omega$, but I don't understand why this is a suitable (or even necessary) assumption. Why do we not simply assume that $f\in C^1(O)$ for some $\mathbb R^d$-open neighborhood $O$ of $\partial\Omega$ or even (which might be too restrictive, I guess) $f\in C^1(D)$?
Hi, there’s perhaps a closing curly bracket missing in the above formulas, which prevents them to shows up correctly. Could yo please verify my hypothesis?
@DanieleTampieri Thanks for noting. Didn't saw that.
|
2025-03-21T14:48:31.720151
| 2020-08-08T14:37:49 |
368643
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ann",
"Fernando Muro",
"S. carmeli",
"https://mathoverflow.net/users/115052",
"https://mathoverflow.net/users/12166",
"https://mathoverflow.net/users/136978"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631884",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368643"
}
|
Stack Exchange
|
Module spectrum maps up to stable homotopy
Let $R$ be a commutative ring spectrum, $M$ and $N$ be a $R$-module spectra.
Let us consider $R$-module maps from $M$ to $N$ up to stable homotopy, that is maps $M \to N$ such that the composites $R \wedge M \to M \xrightarrow{f} N$ and $R \wedge M \xrightarrow{1 \wedge f} R \wedge N \to N$ are equal in the stable homotopy category.
Now suppose that $M = R \wedge X$ is a free $R$-module.
Is it true that the set of stable homotopy classes of $R$-module maps up to homotopy in the previous sense, $[M, N]_R = [R \wedge X, N]_R$, is in the natural bijection with stable homotopy classes of all maps $[X, N]$?
I believe that something like that (or even stronger) holds for genuine $R$-module maps, but what about homotopy $R$-module maps if we are interested in the stable homotopy classes of maps only?
What do you mean by "homotopy $R$-module maps$?
Maps which are $R$-module maps only up to stable homotopy, as I said in the body of the question. The square diagram of the R-module condition is commutative up to stable homotopy, i. e. in stable homotopy category. The last question is just a less formal reformulation of the first.
This is true in every symmetric monoidal category. The Mai in one direction is composition with the unit of R and in the other tensoring with R followed by the action map of R on N. It is a routine to check they are inverse.
Thank you very much!
In any symmetric monoidal category $C$, the free $R$-module functor $R\otimes -: C\to Mod_R(C)$ is left adjoint to the forgetful functor $U:Mod_R(C)\to C$. Hence, $Hom_{Mod_R(C)}(R\otimes X,N) \cong Hom_C(X,U(N))$. Since the stable homotopy category is symmetric monoidal, the result you asked for follows.
The question is ambiguous since it doesn't say whether $\wedge$ stands for the monoidal structure in the stable homotopy category or in a model. In the latter case, you'd need $X$ to be cofibrant so that both coincide. Another ambiguity is the use of "stable homotopy classes of maps". If it means morphisms in the stable homotopy category, no extra assumptions are needed, but if it means honest homotopy classes of maps (defined by means of cylinders or paths) you need in addition $N$ to be fibrant.
|
2025-03-21T14:48:31.720320
| 2020-08-08T14:59:12 |
368645
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631885",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368645"
}
|
Stack Exchange
|
On some bounds on two constants concerning the disconnectedness of the spectra of small perturbations of operators
Let $H$ be a separable, infinite dimensional, complex Hilbert space. In the book:
Jiang, C. L.; Wang, Z. Y. (1998). Strongly Irreducible Operators on Hilbert
Space. CRC press
above the statement of Theorem 5.17 the authors say that given $T \in B(H)$ with disconected spectrum, there exists $\delta >0$ such that $\sigma(T+B)$ is disconnected for all $B$ with $\| B \| < \delta$. If an operator has disconnected spectrum, then it is strongly reducible (i.e. there exists a nontrivial projection commuting with it), while the converse is not necessarily true. Moreover, in the introduction of the paper:
Herrero, D. A.; Jiang, C. L. Limits of strongly irreducible operators, and the Riesz decomposition theorem. Michigan Math. J. 37 (1990), no. 2, 283-291
it is stated that if $T$ has disconnected spectrum, all the operators which are "close enough" to $T$ commute with a nontrivial (Riesz) projection, and hence are strongly reducible.
I think that "close enough" here means 'the perturbation $B$ has a small norm'. In what follows, we will consider these two constants to be the largest possible.
As a first question, I wonder if there is any relation between the two constants in the above paper (I think that they may be equal, but I'm not sure about this).
In any case, is there any known way to evaluate these two constants? Or is there any known upper/lower bound?
|
2025-03-21T14:48:31.720457
| 2020-08-08T15:17:10 |
368647
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Caleb Briggs",
"Gerry Myerson",
"John Jiang",
"Wojowu",
"fedja",
"geocalc33",
"https://mathoverflow.net/users/1131",
"https://mathoverflow.net/users/123449",
"https://mathoverflow.net/users/146528",
"https://mathoverflow.net/users/158000",
"https://mathoverflow.net/users/30186",
"https://mathoverflow.net/users/4923"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631886",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368647"
}
|
Stack Exchange
|
What is the analytic continuation of $\varphi(s)=\sum_{n \ge 1} e^{-n^s}?$
My research has lead me to the following function that I'm trying to continue. 3 Months ago I posted this question to MSE, and have placed 3 bounties on the question, but haven't received an answer, so I've decided to ask here.
$\varphi(s)=\sum e^{-n^s}=e^{-1}+e^{-2^s}+e^{-3^s}+\cdot\cdot\cdot $
A natural question might be:
What is the analytic continuation of $\varphi(s)?$
User @reuns noticed that $\sum_n (e^{-n^{-s}}-1)=\sum_{k\ge 1} \frac{(-1)^k}{k!} \zeta(sk).$
And an analytic continuation is indeed possible using the Cahen-Mellin integral to obtain the formula:
$$\varphi(s)=\Gamma\left(1+\frac1s\right)+\sum_{n=0}^\infty\frac{(-1)^n}{n!}\zeta(-ns)$$
which is valid for $0<s<1.$
I noticed that:
$$e^{\frac{1}{\ln(x)}}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{2K_1(2\sqrt{z})}{\sqrt{z}}x^{-z}~dz$$
valid for $0<x<1$ and $\Re(z)>0$ if I'm not mistaken. Here $K_1$ is a modified Bessel function of the second kind.
Letting $x=e^{-n^{-s}}$ we obtain:
$$\varphi(s)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{2K_1(2\sqrt{z})}{\sqrt{z}}\bigg(\sum_{n=1}^\infty e^{zn^{-s}}\bigg)~dz$$
I think the evaluation of this will give a new formula for $\varphi(s).$ Potentially we could use the distributional version of the kernel to evaluate the integral if one exists.
Does anyone see how to accomplish this?
Do you have a reason to believe this function odes have an analytic continuation? If so, what do you mean with "what is it"? I really doubt there is an expression for this analytic continuation in terms of more classical functions.
To lay the groundwork for this question, in what region of the complex plane does the given series already converge to an analytic function?
@GerryMyerson Re s > 0 or am I missing something obvious?
I dunno, I just thought it was an important piece of information that should be on record here.
@JohnJiang You are missing the fact that the exponents $n^s$ not only grow in size but also rotate for complex $s$, so the minus sign becomes quite useless and you get arbitrarily huge terms. I suspect that there is no continuation from the real line anywhere though I cannot offer a proof off hand.
@geocalc33 I discussed the problem with Misha Sodin and you can find the result of this discussion in the set of handwritten notes at https://drive.google.com/file/d/191PhSQzr5Q-MbfuMzmuiogJZrk2bh9Ko/view?usp=sharing .It is supposed to show that the sum of the series is not real analytic at any point $s>1$. I hope there is no mistake, but the argument is a bit involved. Maybe I'll post it as a proper answer later. The possibility still remains that you can go through the boundary of the half-plane $\Re z<1$ somewhere far from the real line but it is another story.
I see nothing wrong with it (though I haven't checked all the statements about fast decay carefully, I'm pretty sure that @metamorphy will supply all details if you request clarification). At least, it is in good agreement with the notes I linked to: the sum of the series is real analytic for $0<s<1$ and extends to the half-plane $\Re z<1$ but it loses the real analyticity property on the line for $s>1$, so no extension from the real line is possible there. My proof promptly breaks down when $s<1$, metamorphy's series promptly diverges for $s>1$, so everything seems to fit together :-)
@fedja Do you think it would be appropriate to ask a follow up question, about a possible further continuation for Re(z)<0?
Sure. If you need to know more, just state clearly what you are now interested in and link to what has been done already to give everybody a clear picture of where the things stand. Follow-ups are pretty common on this site.
This is nowhere near a formal answer, but it might contain some useful starting points for doing computation.
The problematic aspect of the function
$$ \varphi(s) = \Gamma\left(1+\frac{1}{s}\right) + \sum_{n=0}^\infty \frac{(-1)^n}{n!} \zeta(-ns)$$
is that $\zeta(-ns)$ grows faster than the factorial function when $s>1$, so the series cannot converge. However, the series is alternating, so from a certain viewpoint, it should 'morally' cancel out.
From a divergent series regularization point of the view, the simplest way to obtain a finite value from the function $\varphi(x)$ is truncate the series early to approximate its true value. Thus, defining
$$\varphi(s) \approx \Gamma\left(1+\frac{1}{s}\right)+ \sum_{n=0}^{N} \frac{(-1)^n}{n!} \zeta(-ns) $$
gives a good approximation to the true value of the function near $s=1$. The optimal place to truncate the series is generally at the point where the the size of the term is smallest. Here is a graph of this approximation with $N=10$ on the real line, with the infinite series shown in orange and the finite series in black
For some values of $\mathfrak{R}(s)>1$, the cancellation automatically happens on its own if we look at the integral representation rather than the sum of the residues. Thus, the integral
$$ \varphi(s) = \frac{1}{2 \pi i s} \int_{c - i N}^{c + i N} \Gamma\left(\frac{t}{s}\right) \zeta(s)dt $$
provides another way to approximate the values of $\varphi(x)$ outside of its usual realm of convergence. The two methods unsuprisingly agree with each other, but they tend to converge well in different areas.
If we want to obtain the value of $\varphi(s)$ somewhere far from $s=1$, or we want to get an arbitrarily good approximation, we can rewrite zeta using its functional equation to obtain
$$\varphi(s) = \Gamma(1+\frac{1}{s}) - \frac{1}{2} + \sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{n!}\left(\frac{\left(2\pi\right)^{-ns}}{\pi}\sin\left(-\frac{\pi ns}{2}\right)\left(ns\right)!\zeta\left(1+ns\right)\right)$$
The part that causes it to diverge is the factorial, so we can replace it by its integral representation and simplify to obtain
$$\varphi(s) = \Gamma(1+\frac{1}{s}) - \frac{1}{2} - \frac{1}{\pi}\int_{0}^{N}e^{-t_{2}}\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{n!}\left(\frac{t_{2}}{2\pi}\right)^{ns}\sin\left(\frac{\pi ns}{2}\right)\zeta\left(1+ns\right)dt_{2}$$
This integral agrees with the other two methods in areas where they converge. There are likely ways to simplify this last integral, though I'm not sure if such simplifications will actually make the function easier to compute.
Update: Here is a different integral that is only valid for $\mathfrak{R}(s) >1$
$$\varphi(s) = \Gamma\left(1+\frac{1}{s}\right)-\frac{1}{2} - \\ \frac{1}{\pi} \int_0^\infty \frac{e^{-\left(\frac{t}{2\pi}\right)^{s}\cos\left(-\frac{\pi s}{2}\right)}}{e^{t}-1}\left(\sin\left(\ \left(\frac{t}{2\pi}\right)^{s}\sin\left(-\frac{\pi s}{2}\right)\right)\right) dt$$
thanks for the useful starting points. Just wanted to ask your thoughts on the integral $\varphi(s)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{2K_1(2\sqrt{z})}{\sqrt{z}}\bigg(\sum_{n=1}^\infty e^{zn^{-s}}\bigg)~dz$ providing a new formula
and also what do you think about continuing the function for real $s>1$ using quasi analytic continuation, since the function is smooth and in a quasi analytic class in this domain
@geocalc33 Doing some basic numerical calculations, your integral appears to agree with the ones I have presented here, which would suggest there is some sort of consistency in the value to assign the function when $\mathfrak{R}(s)>1$. I'll look more into the integral and update my answer if I discover anything
My integral agrees with $\varphi(s) = \Gamma\left(1+\frac{1}{s}\right) + \sum_{n=0}^\infty \frac{(-1)^n}{n!} \zeta(-ns)$ on $0<s<1$?
|
2025-03-21T14:48:31.721052
| 2020-08-08T16:07:47 |
368651
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Consider Non-Trivial Cases",
"Daniel Loughran",
"R.P.",
"Stanley Yao Xiao",
"Timothy Chow",
"https://mathoverflow.net/users/10898",
"https://mathoverflow.net/users/134689",
"https://mathoverflow.net/users/17907",
"https://mathoverflow.net/users/3106",
"https://mathoverflow.net/users/5101"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631887",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368651"
}
|
Stack Exchange
|
Basic prerequisite (topics) to read current research in Diophantine equation for an independent researcher
I have completed studying Galois theory, Fermat's Last Theorem for Regular prime and some number theoretic complex analysis (prime number theorem), and basic linear forms in logarithm.
What else should I complete reading to be eligible to read contemporary literature in Diophantine Equation (Exponential) ?
My graduation was in engineering, so I am from non-math (I mean serious math!) background, and I am not going to a university in near future, but wish to conduct research as an independent researcher.
There might be a lack of specification in my question, so if possible, adjust your answer according to that, also feel free to edit.
Try looking at papers at a number theory journal, such as Journal of Number Theory, Acta Arithmetica, International Journal of Number Theory, and Algebra and Number Theory and see if you can understand them. If not, then check their references and read those, continue until you reach foundational texts written in a language that you can understand
The modern study is based a lot upon algebraic geometry. So the more of this you know the better.
This book (Rational Points on Varieties by the irrepressible Bjorn Poonen) contains a lot of the prerequisites you would need in case you go the algebro-geometric route: http://www-math.mit.edu/~poonen/papers/Qpoints.pdf
However, the learning curve for the geometric track is both steep and long (a true via longissima). The "pay-off" of your studies (meaning concrete results and papers of your own) would probably be greater if you wouldn't concentrate exclusively on all that high-falutin stuff, and set your sights more on analytic number theory, additive combinatorics, recreational problem solving... There is always a severe danger of involving yourself in massive propaedeutic studies, without the slightest guarantee of the possibility of contributing something of your own. Just my two cents...
@crispr Okay, in that case, plz suggests some specific ways from your point of view.
@RP_ Thanks for your valuable comment, could u know of any blog that helps for this specific subject for building up basics to study the topic or to know about the topic?
Well I was basically saying that, to my mind, there are better ways of spending your time, especially with regard to the algebro-geometric route. But if you do want to press ahead against all odds, you could look at this question here: https://mathoverflow.net/questions/369433/reference-request-diophantine-equations
@RP_ what better ways, please share.. liked your comment, guess it would be helpful.
@Andrew That is not for me to decide. I was just saying: this particular area (let's call it Diophantine geometry) requires massive preparatory studies, and it's far from guaranteed that you'll get your investment back. So depending on your own taste, the conclusion to draw from this could be: (1) disregard what I'm saying completely and persevere; (2) still study Diophantine equations, but from the analytic or combinatoric (or whatever) side; (3) do something else in mathematics; (4) do something else entirely. It's really up to you.
If I understand correctly, the keywords for the type of number theory you're interested in are "Diophantine approximation" and "transcendental number theory." Are you aware of Florian Luca's article on Exponential Diophantine Equations? If not, that might be a good place to start. You might also like M. Ram Murty and Purusottam Rath's book Transcendental Numbers.
Getting to the research frontier is always a tricky business. Michel Waldschmidt's Diophantine Approximation on Linear Algebraic Groups will give you a sense of what some of the major open problems in the subject are, but that's a very demanding book. There undoubtedly exist lower-hanging fruit, but it may be hard to find such fruit without an experienced advisor to guide you.
+1 But I don't know how I feel about referring an independent (that is unpaid) researcher to an article behind a paywall, sold by price-gouging publishing houses. I would try to e-mail the author and request a preprint copy from him, if I were the OP.
@RP_ : Is it behind a paywall? I thought it wasn't, because I am able to view it from a machine that isn't registered to my institution, but maybe I have some cookies in my browser or something.
@TimothyChow I think that must be the case. The page I'm getting says I can buy the chapter for 27,20 EUR (which isn't that bad comparatively speaking, but still).
Is Google Books any better?
|
2025-03-21T14:48:31.721379
| 2020-08-08T16:52:23 |
368654
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"LSpice",
"SHS",
"https://mathoverflow.net/users/163481",
"https://mathoverflow.net/users/2383"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631888",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368654"
}
|
Stack Exchange
|
Constructing representations of a topological group from characteristic polynomials of a generating set
Given a topological group $G$ and a subset $S$ of $G$ that topologically generates it, what are the conditions under which an $n$-dimensional continuous linear representation of $G$ over an algebraically closed topological field $k$ can be constructed once we have specified a collection of degree $n$ polynomials with coefficients in $k$ as the characteristic polynomials of the images of elements of $S$?
In other words, when can a continuous homomorphism $\rho: G \rightarrow GL_n(k)$ be constructed from the data $\{c(\rho (s)) \in k[x]: s \in S\}$ where $c(a)$ is the characteristic polynomial of $a$?
Under which further conditions if any is such a representation uniquely determined? Is there an algorithm for constructing it from the collection of characteristic polynomials?
Finally what if we don't require that $k$ be algebraically closed?
Assuming that $k$ is a field of characteristic 0 would be fine, although I am also interested in the characteristic $p$ case.
What does it mean to "specif[y] a degree $n$ characteristic polynomial … for each element of $S$"? (EDIT: Oh, you mean a polynomial that must be satisfied by the image of that element?) In addition to the algebraic closure, it might matter whether or not $k$ has characteristic $0$, so you may want to specify that.
@LSpice, thanks for asking for clarification! I've edited the question to take your comment into account as well as added the continuity condition on the representation which seems natural and necessary.
|
2025-03-21T14:48:31.721503
| 2020-08-08T18:04:51 |
368656
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"YCor",
"https://mathoverflow.net/users/14094"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631889",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368656"
}
|
Stack Exchange
|
Jacobson-style Galois theory on perfect closure
Promoted from stack.exchange since I received no response:
Jacobson developed a 'Galois' correspondence for purely inseperable extensions of exponent 1 (only consisting of pth roots) $K/k$, where he showed there was a correspondence between its subfields $k\subset{F}\subset{K}$ and finite restricted Lie sub-algebras of $k$-derivations given below:
$$F\mapsto \mathcal{D}_F(K)$$
$$\mathcal{D'}\mapsto F, \text{the field of constants of} \space \mathcal{D'}$$
Davis extends this to exponent 2 extensions, and conjectures that higher derivation Lie-algebras should correspond to higher exponents.
Firstly, I'd like to ask for a list of books or articles that explore these ideas, especially the research done in 1968-1981, or more recently. I found the original papers in the American Mathematical Society very inaccessible, and have mostly been working off Lindsay Childs' survey of this research, but would like to learn more.
The other question I have, is if this theory is applicable, or can be extended, to the perfect closure of a field (in its algebraic closure, or more generally): is there a corresponding notion of an 'absolute lie-algebra of derivations' to that of the absolute galois group? If so please provide me with references to this too.
isn't it rather Galois-style Jacobson theory?
|
2025-03-21T14:48:31.721629
| 2020-08-08T18:32:53 |
368658
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"D.S. Lipham",
"Daniel Robert-Nicoud",
"Gjergji Zaimi",
"Ishan Deo",
"https://mathoverflow.net/users/140681",
"https://mathoverflow.net/users/2384",
"https://mathoverflow.net/users/44134",
"https://mathoverflow.net/users/95718"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631890",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368658"
}
|
Stack Exchange
|
Arc connectedness of product spaces
Is arc connected-ness well-behaved with respect to products?
That is -
$\prod X_\alpha$ is arc connected iff $X_\alpha$ is arc connected $\forall \alpha$
In this question on MathStackexchange, an answer is provided only for the reverse implication, that is -
If $X_\alpha$ is arc connected $\forall \alpha$, then $\prod X_\alpha$ is arc connected
However, I wasn't able to get an answer for the forward implication, nor have I been able to find it in any book or from Googling. So, is the forward implication true, or is there a counterexample for the same?
They are equivalent when $X_\alpha$ is Hausdorff because in this context path-connected implies arc-connected. I will now assume that you are asking about T$_1$ spaces.
In fact, I'm asking for general $X_\alpha$, which need not even be $T_0$.
This is false when the spaces are not Hausdorff. Let $X$ be the line with two origins $\{O_1,O_2\}$ and $Y$ be the usual line. Then $X\times Y$ is arc connected because you can pick an arc that starts at $(O_1,y_1)$ travels outside $\{O_1,O_2\}\times Y$ and then comes back to $(O_2,y_2)$, but $X$ itself is not arc connected.
Sorry it's early in the morning and I might be missing something obvious, but writing $X=\mathbb{R}\sqcup\mathbb{R}/\sim$ why can't I take the path $[0,\frac{1}{2}]\to[0,\frac{1}{2}]$ on the first copy of $\mathbb{R}$ and $[\frac{1}{2},1]\to[\frac{1}{2},0]$ on the second copy to link the two origins?
@DanielRobert-Nicoud that shows that it is path connected, which is different from arc connected. You would need your map from $[0,1]\to X$ to be injective.
Ah I see! Thanks, I had never come across this notion tbh.
Also since the projection of a path is still a path, the implication of the OP is true for path connected..
|
2025-03-21T14:48:31.721790
| 2020-08-08T18:39:30 |
368659
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"LSpice",
"Mikhail Borovoi",
"YCor",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/4149"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631891",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368659"
}
|
Stack Exchange
|
Showing subgroups with equal Lie algebras are equal
Let $k$ be a field. It might as well be algebraically closed, but I do not want to assume that it has characteristic $0$. I will write "group" for "affine group scheme over $k$", not assuming smoothness.
Two groups can have the same Lie algebras without being equal. For example, if $k$ has characteristic $2$, then every maximal torus in $\operatorname{SL}_2$ has the same Lie algebra as the centre $\mu_2$. Even two smooth groups can have the same Lie algebras without being equal: for example, all maximal tori in $\operatorname{SL}_2$ have the same Lie algebra. At least it is true that, if a smooth group $H$ is contained in a connected group $G$, and their Lie algebras are equal, then $H$ equals $G$; and so, if two connected subgroups $H_1$ and $H_2$ of $G$ have equal Lie algebras and smooth intersection, then they are equal.
I'm looking more for a result in line with Borel - Linear algebraic groups, Theorem 13.18(4)(d): given a maximal torus $T$ in a smooth, reductive group $G$, and a root $\alpha$ of $T$ in $G$, there is a unique smooth, connected subgroup of $G$ that is normalised by $T$ and whose Lie algebra is the $\alpha$-weight space of $T$ on $\operatorname{Lie}(G)$. The key ingredients here are reductivity and the torus action.
So I'm interested in any more general results of this sort that allow one to deduce equality of groups from equality of their Lie algebras. If that's too broad, I'll focus a bit: suppose that $G$ is a smooth, reductive group; $H_1$ and $H_2$ are smooth, connected, reductive subgroups; and $T$ is a torus in $H_1 \cap H_2$ that is not necessarily maximal in $G$, but is maximal in both $H_1$ and $H_2$. In this setting, if the Lie algebras of $H_1$ and $H_2$ are equal, then can we conclude that the groups are equal?
EDIT: I forgot to add, in case it helps, that, in my situation, $\operatorname C_G(T)^\circ$ (the connectedness automatic if $G$ itself is connected) is a torus.
I suggest the following question, that seems to be equivalent to yours: Let $U,,U'$ be two 1-dimensional unipotent subgroups in a smooth connected reductive group $G$. Assume that ${\rm Lie,}U={\rm Lie,}U'$. Does it follow that $U=U'$? In char. 0 the answer is Yes...(I have no intuition in positive characteristic.)
@MikhailBorovoi, I agree that question is natural. It is clear that it would suffice, but not clear to me that it is equivalent to mine. (I think that the torus action has got to play some role.) I originally hoped the torus action would be enough: if $T$ acts with "large orbits" on $U$ and $U'$, then it acts so on $U \cap U'$, so it must be smooth, right?—but it isn't. (The group often called $\alpha_p$, the $p$th-order neighbourhood of the identity, has such a torus action.)
However, in line with @MikhailBorovoi's ideas, a reduction in the spirit of Borel's proof may be applied: by considering a fixed root $\alpha$ of $T$ in $\DeclareMathOperator\Lie{Lie}\Lie(H_1) = \Lie(H_2)$ and working inside $\operatorname C_G(\ker(\alpha)^\circ)$, we may assume that $H_1$ and (therefore) $H_2$ have semisimple rank $1$.
The answer to my question is No, see this answer of Will Sawin.
In char 0 two perfect connected groups with isomorphic Lie algebras are quotients of a common one by finite central subgroups.
@MikhailBorovoi, thanks! So I guess one natural approach, though not necessarily the only one, is to see if one can construct a counterexample to my question in the spirit of @WillSawin's elegant counterexample (i.e., intentionally incorporating "Frobenius garbage", in @MikeCrumley's memorable phrasing).
@YCor, thanks! Here the Lie algebras are not just isomorphic but equal. I think that my question is trivial (and the answer is 'yes') in characteristic $0$.
Indeed sorry I didn't read carefully. Possibly it would help writing "subgroups" instead of "groups" in the title.
@YCor, that is a good suggestion. I have done so.
1st try at counterexample using construction pointed out by @MikhailBorovoi failed: consider $G = \operatorname{GL}_3$, $T = \operatorname{diag}(1, t, t^p)$, with positive root group $U_1 = \begin{pmatrix} 1 & u & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix}$ in $H_1$ and $U_2 = \begin{pmatrix} 1 & u & u^p \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix}$ in $H_2$. Problem: $U_2$ isn't obviously root subgroup of an $\operatorname{SL}_2$.
2nd try using construction pointed out by @MikhailBorovoi failed: consider $G = \operatorname{GL}_4$, $H_1 = \operatorname{GL}_2$ embedded in the upper left-hand corner, $H_2 = \begin{pmatrix} A & 0 \ 0 & A^{[p]} \end{pmatrix}$. Problem: equal Lie algebras, but not common maximal torus; $H_1$ can't be fattened to have exactly that torus.
$\DeclareMathOperator\Ad{Ad}\DeclareMathOperator\Cent{C}\DeclareMathOperator\GL{GL}\DeclareMathOperator\Lie{Lie}$The key point is not, as I expected, whether $\Cent_G(T)^\circ$ is a torus, but whether it equals $\Cent_G(\Lie(T))^\circ$. Certainly it is contained in the latter group, so this is the same as asking whether $T$ centralises $\Cent_G(\Lie(T))^\circ$.
If we do not require this, then we may adapt a construction by @WillSawin, pointed out by @MikhailBorovoi, to give a counterexample that is quite close to the one I attempted in the comments. Specifically, we give connected, reductive subgroups $H_1$ and $H_2$ of $G = \GL_4$ that contain a common maximal torus $T$ (for which $\Cent_G(T)^\circ$ is itself a maximal torus in $G$), and satisfy $\Lie(H_1) = \Lie(H_2)$, but $H_1 \ne H_2$. Namely, let $t$ be any non-scalar diagonal matrix in $\GL_2$, and put $H_1 = \left\{\begin{pmatrix} g & 0 \\ 0 & g^{[p]} \end{pmatrix} \mathrel\colon g \in \GL_2\right\}$ and $H_2 = \left\{\begin{pmatrix} g & 0 \\ 0 & t g^{[p]}t^{-1} \end{pmatrix} \mathrel\colon g \in \GL_2\right\}$, where $g^{[p]}$ is the matrix obtained by raising every entry of $g$ to the $p$th power.
Next we prove that, if $H_1$ and $H_2$ are connected, reductive subgroups of a common group $G$ that contain a common maximal torus $T$, and satisfy $\Lie(H_1) = \Lie(H_2)$, and if in addition $T$ centralises $\Cent_G(\Lie(T))^\circ$, then $H_1$ must equal $H_2$. As suggested by @MikhailBorovoi, it suffices to show that, for every root $b$ of $T$ in $\Lie(H_1) = \Lie(H_2)$, the corresponding root subgroups of $b$ in $H_1$ and $H_2$ are equal. Let $\mathfrak u$ be the common $b$-root subspace of $\Lie(H_1) = \Lie(H_2)$. Then we have $T$-equivariant isomorphisms $e_{i\,b} \colon \mathfrak u \to H_i$ such that $\Ad(e_{i\,b}(X))Y$ equals $Y - \mathrm db(Y)X$ for all $X \in \mathfrak u$ and all $Y \in \Lie(T)$. That is, $e_{1\,b}(X)e_{2\,b}(X)^{-1}$ lies in $\Cent_G(\Lie(T))$ for all $X \in \mathfrak u$, and hence, since $\mathfrak u$ is connected, in $\Cent_G(\Lie(T))^\circ$. Since this group is centralised by $T$, we see upon conjugating $e_{1\,b}(X)e_{2\,b}(X)^{-1}$ by $t$ that it equals $e_{1\,b}(b(t)X)e_{2\,b}(b(t)X)^{-1}$, for all $X \in \mathfrak u$ and all $t \in T$. In particular, $e_{1\,b}(X)e_{2\,b}(X)^{-1}$, as a function of $X$, is constant on $\mathfrak u \setminus \{0\}$, and hence, since it is continuous, is constant on $\mathfrak u$; but its value at $X = 0$ is the identity.
Could you please state explicitly, what assertions you prove or disprove in the second and third paragraphs of your answer?
@MikhailBorovoi, thanks for the suggestion. I have tried to make it clearer.
You write: " Then we have isomorphisms $e_{i,b} \colon \mathfrak u \to H_i$ such that ${\rm Ad}(e_{i,b}(X))Y$ equals $Y - \mathrm db(Y)X$ for all $X \in \mathfrak u$ and all $Y \in {\rm Lie}(T)$". What are these isomorphisms $e_{i,b}$ ?
@MikhailBorovoi, they are the automorphisms coming from a choice of Chevalley basis (see Carter - Simple groups of Lie type, p. 64). They are defined at first over $\mathbb Z$, so make sense over any field. A priori they exist only on the adjoint group, but the adjoint-quotient map is an isomorphism on unipotent subgroups, so they may be lifted uniquely to $H_i$.
Could you please add details to the following sentence? "Since this group is centralised by $T$, we have that $e_{1,b}(b(t)X)e_{2,b}(b(t)X)^{-1}$ equals $e_{1,b}(X)e_{2,b}(X)^{-1}$ for all $X \in \mathfrak u$ and all $t \in T$."
To see that, just act by $T$. (Ah, the point is that the isomorphisms are $T$-equivariant. This follows from the explicit construction in Carter (see my reference). I will add that.)
The proof looks correct! What is ${\rm C}_G(T)^\circ$ in your counter-example?
@MikhailBorovoi, $\operatorname C_G(T)^\circ = \operatorname C_G(T)$ in my counterexample is a maximal torus (the standard diagonal torus in $G = \operatorname{GL}_4$), whereas $\operatorname C_G(\operatorname{Lie}(T))^\circ = \operatorname C_G(\operatorname{Lie}(T))$ is $\operatorname C_G(T)^\circ\cdot(1 \times \operatorname{GL}_2)$.
|
2025-03-21T14:48:31.722295
| 2020-08-08T18:41:55 |
368660
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlo Beenakker",
"Math604",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/137457",
"https://mathoverflow.net/users/66623",
"sharpend"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631892",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368660"
}
|
Stack Exchange
|
name of elliptic pde with a power law nonlinearity
Consider an equation like
$$-\Delta u = |u|^p u $$ in $\Omega$ with $u=0$ on $ \partial \Omega$ where $\Omega$ a domain in $ R^N$ and $ u:\Omega \rightarrow R^N$. Here $p$ is arbitrary or maybe $p=2$. Or consider Neumann problems like this with a zero order term $u$ added to the left.
Do these equations have a name? (I am interested in what kind of results are known and I have no clue what to google). thank you very much.
I would search "semilinear elliptic equation." There are many classical results due to Gidas, Nirenberg, Spruck, Wei-Ming Ni, etc.
The case $p=2$ is the nonlinear Schrödinger equation, more generally written as
$$Eu=-\Delta u+\kappa|u|^2 u,$$
with coefficients $E,\kappa\in\mathbb R$. It describes the propagation of light in nonlinear optical fibers and is also a model for a superfluid. In the one-dimensional case the equation can be solved exactly for $\kappa<0$. It supports socalled "soliton" solutions, localized in space.
As written the differential equation applies to harmonic solutions $\propto e^{-iEt}$ in the time domain; alternatively, consider functions of both space and time and replace the left-hand-side of the equation by $i\partial u/\partial t$.
Generalizations with $p$ an even integer are also studied in this context, see for example Schrödinger equation with a power-law nonlinearity.
is $u$ here a scalar or vector?
both cases are considered in the literature, for the vectorial nonlinear Schrödinger equation, see for example https://arxiv.org/abs/1110.2990
do you have any references for the purely elliptic problem on a bounded domain with $u=0$ on $ \partial \Omega$. (ie. $-\Delta u= |u|^p u$ in $\Omega$) where $u$ a vector and where they discuss existence of nonzero solutions and $p$.
|
2025-03-21T14:48:31.722454
| 2020-08-08T18:56:26 |
368662
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alexandre Eremenko",
"https://mathoverflow.net/users/25510"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631893",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368662"
}
|
Stack Exchange
|
Why do we study biholomorphically invariant pseudodistances/metrics
It is said that pseudodistances/metrics which are invariant under biholomorphic maps are used to determine whether domains in $\mathbb{C}^n$ are biholomorphically equivalent or not.
Suppose $\Omega_1$ and $\Omega_2$ are domains in $\mathbb{C}^n$, and $\rho_1$ and $\rho_2$ are pseudodistances on it such that $\rho_2(f(a),f(b))=\rho_2(a,b)$ for every biholomorphic mapping $f:\Omega_1\longrightarrow\Omega_2$.
Then how exactly does one use the above fact to show that the two domains are biholomorphically equivalent or not? Do we take one map at a time and check if it is a isometry under these pseudodistances? But isnt that a tedious process and inefficient?
I also wanted to know as to why do we study domains on which certain pseudodistances are equal? Like the Lempert's Theorem states that the Caratheodory-Reiffen Pseudometric is equal to the Kobayashi-Royden pseudometric on Convex domains.
Invariant metrics and pseudometrics are used to study holomorphic maps between complex manifolds. The model is the use of the hyperbolic metric in dimension 1.
For example, a manifold is called hyperbolic if its Kobayashi pseudometric is a true metric. Then every holomorphic map from a line to a hyperbolic manifold is constant.
This is a generalization of Picard's Little Theorem. Remarkably, for compact complex manifolds there is also the converse statement (Brody's Lemma).
There are also generalizations of Picard's Great Theorem, normality criteria and generalizations of removable singularity theorem.
One major application is to the proof that the group of automorphisms of a compact hyperbolic space is finite.
All this has a lot of further applications, though much less than for the hyperbolic metric in one-dimensional case.
I recommend an excellent survey by Kobayashi himself, as well as his books:
Hyperbolic manifolds and holomorphic mappings, 1970 and
Hyperbolic complex spaces, 1998, and the book by S. Lang,
Introduction to complex hyperbolic spaces.
@Ben McKay: yes, thanks. I corrected.
Just as in Riemannian (and Finsler) geometry, we can often distinguish domains by metric information, such as determining the behaviour of geodesics, or finding some analogue of curvature to compute, so we can often distinguish domains by such data arising from invariant metrics. For example, we can distinguish the ball from a product of lower dimensional balls, since we can find explicitly the Kobayashi metric of the ball to be the complex hyperbolic metric (which is Riemannian) and see that its curvature is not that of a product metric.
Such a procedure is not possible for complicated domains, for which we don't really know the metric, so we can't always do this. But that is a familiar phenomenon in mathematics.
|
2025-03-21T14:48:31.722670
| 2020-08-08T21:40:42 |
368667
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Emil Jeřábek",
"François G. Dorais",
"Mirco A. Mannucci",
"https://mathoverflow.net/users/12705",
"https://mathoverflow.net/users/15293",
"https://mathoverflow.net/users/2000"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631894",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368667"
}
|
Stack Exchange
|
An internalized version of Tennenbaum's Theorem
Tennenbaum's celebrated 1959 theorem (see here for a reference) is certainly one of the key theorems in mathematical logic. Not so much for its proof, but because it helps "isolating" $N$ as something very special: it is the only countable model of Peano arithmetic which is recursive.
But, does it?
Suppose I live inside any countable model $M$ of Peano: I think I am actually living in the standard natural numbers, and addition/multiplication are the standard ones (and so are all the recursive function, etc). So, it would seem that from the point of view of $M$, other PA-models are not recursive. To be a bit more precise, let me state this:
Internalized Tennenbaum Theorem (ITT): Let $M$ be a countable model of Peano. Then, for any other countable model $N$ not isomorphic to $M$, $N$ is not recursive in $M$, ie it is not $\Delta_1$-definable in M.
Question: Can ITT be proved in, say, ZFC? If not, what is the
obstruction?
Post Scriptum.
Thanks to Emil Jerabek for his suggestion: rather than the original misleading name, Derived Tennenbaum, use Internal (or Internalized).
The original name generated some confusion, see the comments of Francois Dorais,
thus I decided to rename the question. The recursivity required is IN the model, not FROM the model.
The main obstruction would be that the statement is false. Maybe you meant to relativize "Peano" as well? If so, please explain how. (See the well-studied notion of PA-degrees.)
@FrançoisG.Dorais pardon me, perhaps my question is entirely dumb, but can you please elaborate on the falsehood of this statement? Obviously, recursivity in M is not computable in the real world, but from the point of view of M I can still express the notion of being recursive, or not? So where does exactly break down?
To be a bit more precise: M and N are both countable. Take ANY injective set map f : N into M . RTT would be the statement that the images of N-addition and N-multiplication via f are not M-recursive
The RTT statement is false because if $M$ is nonstandard then the standard model is not isomorphic to $M$ but the standard model is recursive.
I’m pretty sure Tennenbaum’s theorem is provable in PA, which, translated to model-theoretic terms, means that no proper extension $M'$ satisfying PA of a model $M$ of PA is $\Sigma_1$-definable in $M$. (Note that such internally definable models are automatically end-extensions.) This should also hold for much weaker theories than PA. Perhaps there might be a version that applies to proper end-extensions that are merely externally relatively computable rather than $\Sigma_1$-definable, but I don’t think that any such thing stands a chance to hold when $M'$ is not an extension of $M$.
(In principle, I should have written “proper extension $M'$ of $M$ such that $M$ thinks $M'$ is a model of PA” rather than “proper extension $M'$ satisfying PA”, but since, as I mentioned, this should hold even for much weaker, finitely axiomatizable fragments of PA, the distinction does not matter.)
Rereading the question, it's starting to dawn on me that when you write "$M$-recursive", you perhaps do in fact mean " $\Sigma_1$-definable in $M$" rather than "computable with oracle $M$" (which is how François and me interpreted it). If so, you really, really shouldn't call it "relative" or "relativized" Tennenbaum theorem, but something like "internal" or "internalized" TT. "Relativized" unambiguously refers to computability with an oracle.
@FrançoisG.Dorais the standard model is recursive, but it is not M-recursive (ie Σ1-definable in M). I would suggest you read what Emil just commented.
@Emil: as usual, your comments are pointed and informative. And, as usual, I invite you to expand them (at your leisure, of course) into a complete answer. PS YES, I did not mean computable with oracle M, rather sigma_1 definable in M. Accepted the change from "relative" to "internalized": the Internalized Tennebaum Theorem, or, how every M believes it is the only recursive model of PA. I wish Stan were still alive, I am sure he would have laughed loud....
@EmilJeřábek re-christened the question
Ok, I will try to write it up as an answer, but it may take a while, as I’m out of office at the moment.
To move this off the unanswered queue, let me summarize the situation as correctly explained by the comments above:
The standard proof of Tennenbaum's theorem goes through inside $\mathsf{PA}$: $\mathsf{PA}$ proves that there is no $\Delta_1$ description of a model of $I\Sigma_1$. (As usual, $\mathsf{PA}$ can be replaced with something vastly weaker here; at a glance, already $I\Sigma_1$ should be enough.)
One key point here is that $\mathsf{PA}$ can quantify over $\Delta_1$ descriptions since this only involves reference to a bounded truth predicate. (Something like "There is no definable structure such that [stuff]" would have to be expressed as a scheme, but that's not an issue here.)
On the other hand, $\mathsf{PA}$ cannot express "the structure defined by the formula tuple $\Phi$ satisfies $\mathsf{PA}$" (at least not as cleanly as one might hope - we'd need to talk about explicitly-Skolemized structures, and that's a whole annoying rabbit hole to go down). This is why I've used $I\Sigma_1$ as the "Tennenbaum target:" as a finitely axiomatizable theory, $\mathsf{PA}$ has no trouble talking about its satisfaction (or not) in a defined structure, again by virtue of the satisfactoriness of a bounded truth predicate.
let me process what you wrote (meanwhile you got already my vote, at the very least for trying to sum up the above). . The last comment of Emil was "Ok, I will try to write it up as an answer, but it may take a while,".
|
2025-03-21T14:48:31.723198
| 2020-08-08T21:52:40 |
368669
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Amir Sagiv",
"Ben McKay",
"Incnis Mrsi",
"Kamel",
"LSpice",
"https://mathoverflow.net/users/13268",
"https://mathoverflow.net/users/163488",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/42864",
"https://mathoverflow.net/users/56921"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631895",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368669"
}
|
Stack Exchange
|
Measure in $\mathbb {C} ^p$
If we have a non-constant holomorphic map $ f: \mathbb C ^ p \to X $, where $ X $ is a complex manifold. Let $ \omega $ be a metric on $X$, so $ \omega $ is a positive definite $ (1,1) $-form.
Is $ f ^ * (\omega ^ p) $ a measure over $ \mathbb C ^ p $?
The grammar of what you wrote didn't make sense to me, so I made what seemed like a small change that hopefully preserved meaning. I think that I didn't change the meaning, but please revert if I did. (In that case, though, you might want to make sure to use $\mathbb C$ $\mathbb C$ rather than $\ C$ $\ C$ (a $C$ with a forced space beforehand) as you originally had.)
If $f$ is measurable, why not?
The fact that $f$ is not constant is sufficient or it is also necessary either of rank $=p$
Zero is a measure.
@Ben Mackay, Okay, if $f^*(\omega ^p) =0$, it implies that $f$ is a constant? Or just $f$ is with rank $<p$?
@Kamel: The pullback $f^*\omega^p$ vanishes just when $f$ has rank less than $p$ at every point of $\mathbb{C}^p$ mapping to a point of $X$ where $\omega\ne 0$.This is just determinant of a product is the product of determinants in local coordinates, or in a holomorphic basis of local sections of the cotangent bundle.
@Ben Mackay thank you
A construction impossible for $p > 1$ giving only zero for $\dim_\mathbb{C} X < p$ [see comments], for whichever reasonable meaning of “$\omega^p$” and pullback.
Denote by $z\in\mathbb{C}$ and $w\in\mathbb{C}^{p-1}$ coordinates on $\mathbb{C}^p$.
Let $f := z$ with $X = \mathbb{C}$ (a standard projection).
Consider $(p\!-\!1)$-dimensional (over $\mathbb{C}$) affine transformations $A$ acting on $\mathbb{C}^p$ as
$$(z,w)\quad\mapsto\quad (z,Aw).$$
Does the action preserve $f$? Obviously. May the resulting measure be changed? Impossibly. Which measure can be invariant under all affine transformations of $w$? Only the zero measure.
$\omega$ may be $i(dz\wedge d\bar z)$ or whatever you like – the argument stands for any non-zero form. Ī̲ am puzzled how can this be a research-level math question.
The construction is not impossible; in fact it always well defined and always produces a measure, perhaps the zero measure. Your example produces the zero measure, since $\omega^2=0$ on $\mathbb{C}$. But on the other hand, other examples produce nonzero measures: if $f$ is any holomorphic map of $\mathbb{C}^p$, somewhere of full rank, and $X=\mathbb{C}^p$ and we take $\omega$ any Kaehler form, we get a measure $f^*\omega^p$ which is positive on any open set.
But I agree that the question is too elementary.
Oops, Ī̲ missed that $X$ is not (necessarily) one-dimensional.
|
2025-03-21T14:48:31.723413
| 2020-08-08T23:58:23 |
368672
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Breakfastisready",
"Ian Agol",
"Kimball",
"LSpice",
"https://mathoverflow.net/users/1345",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/6518",
"https://mathoverflow.net/users/94546"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631896",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368672"
}
|
Stack Exchange
|
Does the standard arithmetic subgroup of a closed $\mathbb{Q}$-algebraic groups have non-trivial $\mathbb{Q}$-characters?
I am trying to understand the Borel-Harish Chandra theorem about arithmetic subgroups being lattices.
Suppose $G$ is an algebraic group inside $GL_n(\mathbb{C})$ such that it is definable as a zero set of polynomials with coefficients in $\mathbb{Q}$. Let $G_\mathbb{R} = G \cap G_\mathbb{R}$ and $G_\mathbb{Z} = {G} \cap GL_n(\mathbb{Z})$.
The Borel-Harish Chandra theorem says that $X_\mathbb{Q}(G^0) = \{ e \} \Leftrightarrow G_\mathbb{R} / G_\mathbb{Z}$ has a finite natural measure. Here $X_\mathbb{Q}(G^0)$ is the multiplicative group of $\mathbb{Q}$-morphisms of the algebraic group $G^0$ to $GL_1(\mathbb{C})$.
I am wondering if the "$\Leftarrow$" part can be proven shown easily under the assumption that $G$ is closed. Here is my attempt:
Since $G$ is closed, any $\mathbb{Q}$-morphism $\chi:G^0 \rightarrow GL_1(\mathbb{C})$ can be written as a polynomal, instead of a regular function. However, on the points of $G_\mathbb{Z}^0$, this polynomial will actually take values in $N^{-1} \mathbb{Z}$ for some $N \in \mathbb{N}$ (the lcm of the denominators of all the rational coefficients) but since the image is a multiplicatively closed set of $\mathbb{Q}^*$, $\chi$ can only be $\{ \pm 1\}$ on $G_\mathbb{Z}^0$. But then, $\chi:G^0_\mathbb{R}/G^0_\mathbb{Z} \rightarrow \mathbb{R}^*$ is a continuous map that might induce a finite Haar measure on $\mathbb{R}^*$.
This approach might work, but saying that all $\mathbb{Q}$-characters on $G_\mathbb{Z}^0$ are trivial seems like a big deal and I could not find any such statements in Borel's "Intro. to Arithmetic groups". Do you have any counterexamples that show that the integer points of closed connected $\mathbb{Q}$ algebraic groups may admit a non-trivial $\mathbb{Q}$-character?
While your definition of $G_{\mathbb Z}$ is legal, I have a feeling that your hypotheses are too weak to guarantee that it is a sensible construction. (For example, I think that you can arrange that $G_{\mathbb Z}$ is trivial even for very large groups $G$.)
This is not the question I expected from the title.
@Kimball I am sorry. I have fixed the title.
@LSpice Can you please give me an example of what you are saying?
Not off the top of my head, no.
I think the example to keep in mind is the trivial algebraic group $GL_1(\mathbb{C})$. The integer points are $\pm 1$, so it is not a lattice.
But I don't know the proof, except in some special cases (certain lattices can be proved to be cocompact using the Mahler compactness theorem, which is more elementary than the proof for non-uniform lattices). A reductive Lie group is semisimple iff it has no characters. The construction of arithmetic lattices that I'm familiar with just proves it for semisimple algebraic groups. See Theorem 1.3.9 of Witte-Morris' book. http://deductivepress.ca/
|
2025-03-21T14:48:31.723633
| 2020-08-09T00:17:07 |
368673
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"LSpice",
"S. Carnahan",
"https://mathoverflow.net/users/121",
"https://mathoverflow.net/users/2383"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631897",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368673"
}
|
Stack Exchange
|
Notation on supergeometry — parity
I know that given a manifold $M$ and its corresponding tangent bundle $TM$ we can call $\Pi TM$ the space of forms parametrized (via charts) by $\{x_i\}_{i=1,\dotsc,n}$ and its corresponding cotangent basis $\{dx_i\}_{i=1,\dotsc,n}$. The notation is due to the fact that this space is a $n|n$-supermanifold, since the coordinates of cotangent bundle is anti-commutative.
My question is: Given a Lie algebra $\mathfrak{g}$, what would be $\Pi\mathfrak{g}$? I currently see this notation and don't know what it means.
Where do you see this notation?
Suppose $\mathfrak{g}$ is identified with the tangent space of a Lie group $G$ at identity. In this case, $\Pi \mathfrak{g}$ is the fiber of $\Pi TG$ over the identity element.
Since a Lie algebra is a vector space, $\Pi\mathfrak{g}$ is just the vector space $\mathfrak{g}$ with a $\mathbb{Z}_2$-grading where every element is odd. In particular, if $(t_a)$ is a basis of $\mathfrak{g}$ and $(c^a)$ is the dual basis, then we can take $c^a$ as coordinates on $\Pi\mathfrak{g}$. Then functions on $\Pi\mathfrak{g}$ are polynomials in the $c$ variables where we identify $c^ac^b$ with $-c^bc^a$. A good resource for these matters is D. Fiorenza's paper An introduction to the Batalin-Vilkovisky
formalism (arXiv link).
|
2025-03-21T14:48:31.723757
| 2020-08-09T01:48:28 |
368674
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"anon",
"https://mathoverflow.net/users/161984"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631898",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368674"
}
|
Stack Exchange
|
The size of endomorphism rings and the relation to ordinariness of Abelian surfaces
For Elliptic curves over a finite field, there is a very useful characterization of ordinary elliptic as those with commutative, quadratic endomorphism rings and of supersingular curves as those with Endomorphism ring a non commutative division algebra of rank $4$.
Question: Is there any such characterization in the case of (simple) Abelian surfaces over a finite field? How about in higher dimensions?
There are three possibilities for supersingularity in terms of the p-adic Tate module - it can be $0,1$ or $2$ dimensional (over $\mathbb Z_p)$.
Similarly, there are three possibilities for the endomorphism ring- it can be a commutative (order in a) number ring of dimension $4$ or a non commutative division algebra of order $8$ or $16$ (dimensions over $\mathbb Z$). In the $8$ dimensional case, the center is a quadratic number ring and in the $16$ dimensional case, the center is $\mathbb Z$.
What I know: This mathoverflow question says that ordinary abelian surfaces over finite fields are always commutative. Why is this true and is this also true for higher dimensional abelian varieties?
In the case that the p-adic Tate module is one dimensional, we know that the endomorphism ring has to act faithfully on it and hence it has to be commutative.
In the case of a $16$ dimensional algebra, since the Frobenius is in the center and the center is $\mathbb Z$, the frobenius and it's dual are both just multiplication by some power of $p$ and so the surface is supersingular.
Conversely, I can show that in the supersingular case, the algebra is definitely not commutative.
Whether a simple abelian variety over a finite field is ordinary etc. can be read off from its Weil number, and its endomorphism algebra can also be read off from its Weil number using Tate's formula. Combine the two to answer your questions. See, for example, Tate's 1968 Bourbaki Seminar on the work of Honda.
The general reference for this sort of questions is Waterhouse, Abelian varieties over finite fields. Your question is answered in: Theorem 7.2. If $A$ is ordinary (and simple), then $\mathop{End}(A)$ is commutative and does not change by base field extension.
Furthermore, Theorem 7.4 shows that any order in $\mathop{End}^0(A)$ containing $\pi$ and $\overline{\pi}=q \pi^{-1}$ occurs as an endomorphism ring of an Abelian variety.
|
2025-03-21T14:48:31.724191
| 2020-08-09T02:51:27 |
368675
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gerhard Paseman",
"Jose Arnaldo Bebita",
"https://mathoverflow.net/users/10365",
"https://mathoverflow.net/users/3402"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631899",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368675"
}
|
Stack Exchange
|
If $n$ is a multiperfect number, then necessarily does one of its prime factors $p$ satisfy $p \parallel n$?
My question is as in the title:
If $n$ is a multiperfect number, then necessarily does one of its prime factors $p$ satisfy $p \parallel n$?
I quote from an answer by Varun Vejalla to a closely related MSE question:
I checked all the multiperfect numbers from this site, which include all the multiperfect numbers in the b-file of A007691, as well as additional ones. The largest number that was in the list was $\approx 10^{34850339}$. None of those values were perfect powers.
What I did find for all multiperfect numbers is that there was at least one prime factor with an exponent of exactly $1$ in the prime factorization. If this could be proved for all multiperfect numbers, then the conjecture that there is no number that is both multiperfect and a perfect power could be proven.
While likely to be true, it is not known if multi perfect numbers are never powerful. Indeed, if a perfect number is a perfect power, Euler tells us that the number is odd and is a (4k+1) power for some k. It might be useful to see what is known about primitive factors of sigma(p^e) for increasing exponents e and primes p. Even such a result for perfect numbers is unknown, let alone multi perfects. Gerhard "One Is Too Multi Perfect" Paseman, 2020.08.08.
@GerhardPaseman, it is known that perfect numbers cannot be perfect powers. (See this MSE question for the details.)
What details? There is only an assertion. Gerhard "Not Ready To Suspend Disbelief" Paseman, 2020.08.08.
|
2025-03-21T14:48:31.724332
| 2020-08-09T06:19:07 |
368678
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alexandre Eremenko",
"Claude Leibovici",
"GA316",
"MathCrawler",
"Max Alekseyev",
"https://mathoverflow.net/users/161310",
"https://mathoverflow.net/users/25510",
"https://mathoverflow.net/users/33047",
"https://mathoverflow.net/users/42185",
"https://mathoverflow.net/users/7076"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631900",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368678"
}
|
Stack Exchange
|
Roots of polynomials of particular type
How to find the solutions $x $ of the following equation: $$\frac{n_1}{x + n_1} + \frac{n_2}{x + n_2} + \cdots +\frac{n_k}{x + n_k} = 1$$ where $n_i$s are natural numbers.
For the case $k=2$, I get the solutions $\pm \sqrt{n_1n_2}$. I was trying to use Vieta's formulas to simplify the expression, but I am unable to do make any progress. Kindly share your thoughts. Thank you.
I was trying induction on $k$ but didn't lead anywhere.
I tried to represent it as some indefinite integral. But no success.
Also, is there any theoretical significance of these polynomials? Kindly share some references. Thanks again.
What do you mean by "to find"? Numerically?
@AlexandreEremenko I want to find a closed-form for the roots. Numerical solution also helps me and appreciated. Thank you.
"Closed form", whatever it can mean, is unlikely.
Thank you. For a suitable choice of $n_i$s can we say something?
Not $2$ years but $27$ !!!
@Alexandre "Unlikely" is the word. Setting MAPLE on the case n=3 gives you goosepimples.
This equation is a particular case of the so-called Underwood equation
$$\sum_{i=1}^n \frac{\alpha_i\, z_i}{\alpha_i- \theta}=1-q$$ where the $\alpha_i> 0$ and $z_i >0$ and $n$ can be very large (potentially up to thousands) and $q$ is given.
With my former research group we spent decades to find efficient numerical methods to solve it since, in chemical engineering, it has to be solved zillions of times in a single simulation.
For sure, as soon as $n>4$, there is no nalytical solutions and numerical methods are required. To me, the key point is to avoid its transform into a polynomial.
Our most recent work was published in $2014$ in this paper (you can also find it here) where we proposed rapid and robust solution methods using convex transformations. Beside, and this is a key point, for any root, we proposed simple and efficient starting guesses which,typically, make that very few iterations are required (this is illustrated in the first figure showing that the starting guess is almost the solution).
If you are not too concerned by computing time, there are simple things you could do. Using you equation, for the root between, say, $n_1$ and $n_2$ which correspond to the vertical asymptotes, transform
$$f(x)=\sum_{k=1}^p\frac{n_k}{x + n_k} - 1$$ as
$$g(x)=(x+n_1)(x+n_2)f(x)$$ which is the most basic form of the so-called Leibovici & Neoschil method which has been widely used for this class of problems during the last $27$ years.
$$g(x)=n_1(x+n_2)+n_2(x+n_1)-(x+n_1)(x+n_2)+(x+n_1)(x+n_2)\sum_{k=3}^p\frac{n_k}{x + n_k}$$ which gives
$$g(-n_1)=n_1(n_2-n_1)\qquad \text{and}\qquad g(-n_2)=-n_2(n_2-n_1)$$ Then, a linear interpolation gives, as an estimate,
$$x_0=-\frac {2\,n_1\,n_2}{n_1+n_2}$$
The problem is that, in the range, $g(x)$ go through an extremum and using directly Newton method is dangerous. A very good way to avoid any problem is to use a method which combine Newton steps and bisection steps (when Newton method tends to lead outside the range).
I would recommend for example subroutine rtsafe from "Numerical Recipes" which does it. The code is here.
This works quite well without any convergence problem (but it is much less efficient than what was proposed in our paper). For $99.9$% of the zillions of cases we worked, only one bisection step is used.
Edit
Revisiting all the things we tried, another possibility is, for each interval, to change variable :
$$x=-n_{1}+\frac{n_{1}-n_2} {1+e^{-t}}$$ and use Newton method for $f(t)$ using
$$t_0=\log \left(\frac{n_2}{n_1}\right)$$
For example, using $p=5$ and $n_i=p_{i+3}$ Newton iterates will be
$$\left(
\begin{array}{cc}
n & t_n \\
0 & 0.4519851 \\
1 & 1.4953866 \\
2 & 1.3605234 \\
3 & 1.3532976 \\
4 & 1.3532792
\end{array}
\right)$$
Thank you very much. It is an interesting direction (but I am not very familiar with so I need some time to go through it). I will look at your papers and go through the details. If I have some doubts, I will ask you here. Thanks again for the nice answer.
@GA316. You are very welcome ! You made me (feeling) younger with this problem. Do not hesitate to contact me for any clarification or whatever. I did not recat to the first version of the problem when the $n_k$ are complex since this is a very, very difficult problem. Cheers :-)
These are zeros of $g'(x)$, where
$$g(x) :=\frac{(x+n_1)\cdots(x+n_k)}{x^{k-1}}.$$
Or, switching to $y:=\frac1x$, they correspond to zeroes of $f'(y)$, where
$$f(y) := \frac{(1+n_1y)\cdots(1+n_ky)}y.$$
Thank you. I like the expression $g(x)$ more. But I am unable to see how to proceed further. Any idea, please?
E.g., one can use Rolle's theorem to localize the zeros of the derivative.
We write $n_1 < n_2 < \cdots < n_k$ then between $-n_{i+1}$ and $-n_i$ there is a root of $g^{'}(x)$? Any other method to locate the roots more accurately? Thank you.
Perhaps you should guide us benevolent commenters by providing more background to your question: what is the motivation? why do you think there miight be sensible answers? why could these answers be interesting? etc.
|
2025-03-21T14:48:31.724702
| 2020-08-09T06:38:53 |
368680
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Struggler",
"https://mathoverflow.net/users/51105"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631901",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368680"
}
|
Stack Exchange
|
Question related to complete vertex, interior vertex and boundary vertex of a graph
All the definition and results which I am using here has been taken from the Garry Chartrand book $\textbf{The Introduction of Graph Theory}$ and the paper $\textbf{On the Commuting Graph of Dihedral Group}$. Here I am denoting the set N$(v)$ is the collection of all vertices which are adjacent to $v$ in a simple graph $\Gamma$ and the symbol $u \sim v$ means $u$ and $v$ are adjacent.
A vertex $v$ in a graph $\Gamma$ is a $\textbf{boundary vertex}$ of a vertex $u$ if $d(u, w) \leq d(u, v)$ for $w \in$N$(v)$, while a vertex $v$ is a boundary vertex of a graph $\Gamma$ if $v$ is a boundary vertex of some vertex of $\Gamma$.
A vertex $v$ is said to be a $\textbf{complete vertex}$ if the subgraph induced by the neighbors of $v$ is complete.
A vertex $v$ is said to be an $\textbf{interior vertex}$ of a graph $\Gamma$ if for each $u \ne v$, there exists a vertex $w$ and a path $u-w$ such that $v$ lies in that path at the same distance from both $u$ and $w$. A subgraph induced by the interior vertices of $\Gamma$ is called \emph{interior} of $\Gamma$ and it is denoted by $Int(\Gamma)$.
The following results are on the page of 337 and 339 of the above book.
Let $\Gamma$ be a connected graph and $v \in V(\Gamma)$. Then $v$ is a complete vertex of $\Gamma$ if and only if $v$ is a boundary vertex of $x$ for all $x \in V(\Gamma) \setminus \{v\}$.
Let $\Gamma$ be a connected graph and $v \in V(\Gamma)$. Then $v$ is a boundary vertex of $\Gamma$ if and only if $v$ is not an interior vertex of $\Gamma$.
If $\Gamma$ is a complete graph of size $n$ and $v \in V(\Gamma)$, then by the definition of complete vertex, $v$ is a complete vertex. By using the above two results, $v$ is not an interior vertex.
when I am applying the definition of an interior vertex, for each $u \ne v \in V(\Gamma)$ and chose $w \in V(\Gamma) \setminus \{u, v\}$, we have a path $u \sim v \sim w$. Thus, $v$ is an interior vertex.
I am a little bit confused so please help me where I am doing wrong. I would be thankful for your kind help.
I suspect the issue is with the definition of interior you are using.
This should say that $v$ is an interior vertex of $\Gamma$ if for each $u\neq v$, there exists a vertex $w$ and a $u-w$ geodesic containing $v$.
With this definition, the path $u\sim v\sim w$ in $K_n$ does not cause any problems, because it is not a geodesic (the single edge $u\sim w$ is a shorter path).
Thanks, Brandon for your kind help.
|
2025-03-21T14:48:31.725001
| 2020-08-09T07:49:05 |
368684
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alapan Das",
"Michael Stoll",
"https://mathoverflow.net/users/156029",
"https://mathoverflow.net/users/21146"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631902",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368684"
}
|
Stack Exchange
|
On the Diophantine equation $x^{5} + y^5 = z^p$
Let $x, y, z$ be pairwise coprime positive integers. Does one have $x^5 + y^5 = z^p$ for any prime $p \geq 2$ ?
If Beal's conjecture is true, then $p$ must have to be $2$ so that $\text{gcd}(x,y,z)=1 \Rightarrow x,y,z$ are mutually co-prime. Hence, the problem becomes $x^5+y^5=z^2$
In 1998 B Poonen had proven that the equation $x^5+y^5=z^2$ has no integer solution for $x,y,z$ co-prime.
To the best of my knowledge, this is open for general $p$.
As mentioned by Alapan Das, Bjorn Poonen has solved the case
$p = 2$ and also $p = 3$ [B. Poonen, Some diophantine equations
of the form $x^n + y^n = z^m$, Acta Arith. 86 (1998), 193-205].
The case $p = 5$ is part of FLT. Sander Dahmen and Samir Siksek
[Perfect powers expressible as sums of two fifth or seventh
powers, Acta Arith. 164 (2014), 65-100] solve the cases
$p = 7$ and $p = 19$, and also, assuming GRH, $p = 11, 13$.
In my paper "Chabauty without the Mordell-Weil group"
[In: G. Böckle, W. Decker, G: Malle (Eds.): Algorithmic and
Experimental Methods in Algebra, Geometry, and Number Theory,
Springer Verlag (2018)] I remove the GRH assumption on these
two cases and do also $p = 17$,
and I extend the range of primes for which the
equation can be solved under GRH to $p \le 53$. In all these
cases, no nontrivial solutions exist.
This is, in fact, quite usual in this business: many of these results (for small exponents) involve the computation of some Selmer group, which in turn depends on determining class groups (and unit groups) of certain algebraic number fields. Assuming GRH, class groups can be computed in reasonable time for much larger fields (in terms of degree and discriminant) than unconditionally. In the concrete case, what you need is that the class group of $\mathbb Q(\sqrt[p]{2})$ has odd order (and $p$ is not a Wieferich prime).
|
2025-03-21T14:48:31.725278
| 2020-08-09T08:32:47 |
368685
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"ABIM",
"Pietro Majer",
"Taras Banakh",
"https://mathoverflow.net/users/36886",
"https://mathoverflow.net/users/6101",
"https://mathoverflow.net/users/61536"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631903",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368685"
}
|
Stack Exchange
|
Critical Growth of Dimension for Dense Cover by Linear Subspaces
Let $X$ be a separable Banach space of dimensional $>2$. When does there exist a sequence positive integers $\{N_n\}_{n \in \mathbb{n}}$ such that
For any sequence of distinct finite-dimensional subspaces $\{L_n\}_{n \in \mathbb{N}}$ of $X$ we have:
$$
N_n\leq \dim(L_n) (\forall n \in \mathbb{N})\,\Rightarrow \,\overline{\bigcup_{n \in \mathbb{N}} L_n} =X.
$$
By distict, I simply mean that $L_n\neq L_m$ if $n\neq m$ but we can have containement.
There exists a sequence of subspaces $\{\tilde{L}_n\}$ of $X$ for which $$
Dim(\tilde{L}_n)<N_n \mbox{ and } \overline{\bigcup_{n \in \mathbb{N}} \tilde{L}_n} \neq X.
$$
Let's call such $\{N_n\}$ critical for $X$.
Finite-Dimensional Intuition:
If $X$ is finite-dimensional with dimension at-least $2$, then WLOG assume $X=\mathbb{R}^D$ for some positive integer $D>2$. Then $N_n=\min\{n,D\}$ works. But if $\tilde{L}_n$ are rotations of $\mathbb{R}^{D-1}\subseteq \mathbb{R}^D$ by $\pi_n$ about the origin, then $\tilde{L}_n$ satisfy the second condition. So $N_n=\min\{n,D\}$ is critical in the above sense.
Note: We can always construct a dense linearly independent subset of $X$ by from a topological basis (since X is infinite-dimensional). So we can find $L_n$ with $1$-dimension each. However, here, I'm looking for a "critical" integer $N$ above which any collection of $\{L_n\}$ must satisfy $\overline{\bigcup_{n \in \mathbb{N}} L_n} =X$.
**Edit: ** I removed the requirement of each $L_n$ to be distinct so that I can give my (updated) intuition in the finite-dimensional setting.
Thanks. But your question should be more concrete anyway. Because you can just take a collection of 1-dimensional subspaces with dense union, in which case the second condition holds and the first does not. Also $N$ can be taken to be equal to 1 and then the first part is trivial.
@TarasBanakh Yes I just noticed the bad formulation. I meant to ask if there is a "critical dimension" above which any collection of subspaces of that dimension must have dense union.
I think nothing like you wish cannot hold because of the existence of basic sequences in any infinite-dimensional Banach space. Using a basic sequence you can form many finite-dimensional subspaces with desirable properties.
@TarasBanakh Yes, instead if we ask if there is a sequence of dimensions of these subspaces must grow at then I think its possible. I put down the example in the finite-dimensional case also.
How can the first condition be satisfied, if X is infinite dimensional? We can pick each $L_n$ within a fixed closed hyperplane, with any finite dimension $N_n$ we like.
|
2025-03-21T14:48:31.725457
| 2020-08-09T08:34:24 |
368686
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631904",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368686"
}
|
Stack Exchange
|
Entries of matrix iterates
We consider a matrix
$$A:=\begin{pmatrix} 0 & b & 0 &f \\a & 0 & e & 0 \\ 0 & d & 0 & h \\ c& 0 & g & 0 \end{pmatrix}.$$
This matrix has the interesting property that if you multiply it once with iself, it will have entries only where at the moment it has zero entries, in fact
$$A^2= \left(
\begin{matrix}
a b+c f & 0 & b e+f g & 0 \\
0 & a b+d e & 0 & a f+e h \\
a d+c h & 0 & d e+g h & 0 \\
0 & b c+d g & 0 & c f+g h \\
\end{matrix}
\right)$$
and by multiplying with $A$ again
$$A^3=\left(
\begin{array}{cccc}
0 & b (a b+c f)+d (b e+f g) & 0 & f (a b+c f)+h (b e+f g) \\
a (a b+d e)+c (a f+e h) & 0 & e (a b+d e)+g (a f+e h) & 0 \\
0 & b (a d+c h)+d (d e+g h) & 0 & f (a d+c h)+h (d e+g h) \\
a (b c+d g)+c (c f+g h) & 0 & e (b c+d g)+g (c f+g h) & 0 \\
\end{array}
\right)$$
the entries are back to where we started from.
I am wondering now, if there is a way to write an explicit (so not recursive) formula for the entries $A_{21}$ and $A_{41}$ of $A^{2n+1}$ and $n \in \mathbb N$?
Swap the second and third row and column (which is a conjugation by the orthogonal permutation matrix
$$
\Pi = \begin{bmatrix}
1 & 0 &0 & 0\\
0 & 0 &1 & 0\\
0 & 1 &0 & 0\\
0 & 0 &0 & 1
\end{bmatrix} = \Pi^*.
$$
Then, working with $2\times 2$ blocks,
$$
\Pi A \Pi^* = \begin{bmatrix}0 & B\\ C & 0\end{bmatrix},
$$
with
$$
B = \begin{bmatrix}
b & f\\
d & h
\end{bmatrix}, \, C = \begin{bmatrix}a & e\\ c & g\end{bmatrix}.
$$
So
$$
\Pi A^{2n+1} \Pi^* = (\Pi A \Pi^*)^{2n+1} = ((\Pi A \Pi^*)^2)^n (\Pi A \Pi^*)=
\begin{bmatrix}
BC & 0\\
0 & CB
\end{bmatrix}^n \begin{bmatrix}0 & B\\ C & 0\end{bmatrix}
=\begin{bmatrix}0 & (BC)^n B\\ (CB)^nC & 0\end{bmatrix}
$$
If you expand out the entries of $CB$ and $BC$ in terms of $a,b,\dots,h$ you get closed formulas for all entries.
|
2025-03-21T14:48:31.725586
| 2020-08-09T08:47:38 |
368687
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gjergji Zaimi",
"Mare",
"Martin Rubey",
"Sam Hopkins",
"https://mathoverflow.net/users/2384",
"https://mathoverflow.net/users/25028",
"https://mathoverflow.net/users/3032",
"https://mathoverflow.net/users/61949"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631905",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368687"
}
|
Stack Exchange
|
Distributive lattices with periodic Coxeter matrix
Let $L$ be a finite distributive lattice and $U$ its incidence matrix with entries $u_{i,j}=1$ iff $i \leq j$ and $u_{i,j}=0$ else.
Then $U^{-1}$ is the Moebius matrix of $L$ and $C_L:=- U^{-1} U^{T}$ is the Coxeter matrix of $L$.
Question: For which $L$ is the Coxeter matrix periodic, that is $C_L^n$ is the identity matrix for some $n$?
Call such a lattice (or its incidence algebra) periodic. For example any divisor lattice (products of chains) is periodic.
For distributive lattices on $r \geq 2$ points, the number of periodic distributive lattices starts with 1,1,2,3,5,7,11,8,15 for $r \leq 10$ and oeis finds this : https://oeis.org/A053724 .
Question: Is this still true for $r=11$?
Algebras with periodic coxeter matrix are studied in representation theory of quivers, see for example https://www.sciencedirect.com/science/article/pii/S0024379505001709 and several other articles.
One might also ask what the period is for a given lattice. For example for the Boolean lattice of $n$ points it is 3 if n is odd and 6 if n is even.
Recall that the free distributive lattice is the lattice of order ideals of the Boolean lattice.
Question: For which $n$ is the free distributive lattice on $n$ points periodic and what is the period in case it is? It is periodic with period 4 for $n=1$, period 10 for $n=2$ and with period 6 for $n=3$ and with period 42 for $n=4$.
Some tests suggest that the lattice of order ideals of a divisor lattice is also periodic.
If these two sequences were equal it would be really strange...
@GjergjiZaimi Yes, probably one or two more cases might be enough to rule it out. But sadly I can not generate all distributive lattices with $r=11$ points with SAGE anymore, since SAGE filters those out of all posets which is too big.
Is there a particular significance (e.g. in algebra) to the Coxeter matrix being periodic?
@SamHopkins I added a reference.
How do you test for periodicity?
@MartinRubey I wrote you an email with the code. It can be done with SAGE (coxeter transformation exists for SAGE) or with GAP, which I used for the data here.
Apparently, the sequence begins with $1,1,2,3,5,7,11,8,15,19$, which is not in the OEIS. (I checked periodicity with an upper bound of 1000).
@MartinRubey Thank you very much! Too bad.
|
2025-03-21T14:48:31.725772
| 2020-08-09T09:35:18 |
368691
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alexander Gruber",
"Alexandre Eremenko",
"Andrei Smolensky",
"Harry Wilson",
"Hollis Williams",
"Michael Hardy",
"Terry Tao",
"Timothy Chow",
"https://mathoverflow.net/users/119114",
"https://mathoverflow.net/users/14717",
"https://mathoverflow.net/users/25494",
"https://mathoverflow.net/users/25510",
"https://mathoverflow.net/users/3106",
"https://mathoverflow.net/users/43395",
"https://mathoverflow.net/users/5018",
"https://mathoverflow.net/users/6316",
"https://mathoverflow.net/users/766",
"user14717"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631906",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368691"
}
|
Stack Exchange
|
Recent uses of applied mathematics in pure mathematics
In this answer Yves de Cornulier mentioned a talk about the possible uses of persistent homology in geometric topology and group theory. Persistent homology is a tool from the area of topological data analysis, specifically designed to extract information from empirical data and used for various applications, ranging from changes in brain function under drugs to the study of fluid flows to the merging of maps of distinct scales, along with many others. So this definitely belongs to the realm of applied mathematics, and it being used in pure mathematics is very interesting.
It goes without saying that many applications inspired a lot of research in pure mathematics, both in order to establish the foundations for the tools used in applies mathematics and just for the sake of studying interesting objects that appear in such interactions. I am talking specifically about the applied tools themselves being used in research in pure mathematics.
As an example, interval arithmetics was used in the solution of Smale's 14th problem and in the proof of Kepler conjecture (the latter also used a lot of linear programming).
Going back in time, we find that a lot of methods that were initially developed mainly for some specific application, such as celestial mechanics, the stereometry of wine barrels or heat transfer, became the standard tools in pure mathematics. Now it seems that the flow of methods and techniques is mostly one-way, from pure mathematics to the applied. But not completely one-way, hence the question:
What are the recent uses of the tools from applied mathematics to the problems in pure mathematics?
If one requires a more specific indication what does "recent" mean, let's say last 30 years (but I would be delighted to hear about older examples as well).
Is Theoretical or Mathematical physics counted as "applied mathematics"? Enormous part of "pure mathematics" can be traced back to physics in its origin.
Wine barrels,not vine barrels :-)
@HarryWilson Oups! Fixed.
If theoretical computer science or information theory counts as "applied mathematics" then there are too many examples to list.
For certain values of "recent", the problem of finding the antiderivative of the secant function came from cartography. $\qquad$
The question seems too broad to me. "Applied math" and "pure math" need to be defined more narrowly or else a large fraction of all mathematics could qualify.
I think you need to narrow it down in time more than the last 30 years.
Another recent use of persistent homology with group theory was my dissertation, which I defended in 2019. I didn't publish it because I went into industry, but it should be available on University of Florida library's dissertation database under Inference and Classification of Symmetry in Point Clouds. A quick summary is that I show how persistent homology can identify point clouds that are the result of group actions (similar to G-equivariant point clouds), even if they're very noisy.
I recall lots of numerical quadrature in the polymath project getting the bound for the de Bruijn–Newman constant $\Lambda < 0.22$.
If mathematical developments in physics count as "applied mathematics" there are many examples --- as requested by the OP here is a recent one (< 30 years old) and an older one:
Gauge theory spawned fundamental resuls in
differential geometry, see for example A proof via the Seiberg-Witten
moduli space of Donaldson's theorem on smooth 4-manifolds with
definite intersection forms (1995)
Dirac delta functions were introduced to simplify the normalization of quantum mechanical wave functions, and were formalized by Laurent Schwartz in the theory of distributions.
Applications of physics to pure math seem too numerous to list. Mirror symmetry, supersymmetric proofs of index theorems, applications of statistical mechanics to enumerative combinatorics...where does one stop?
|
2025-03-21T14:48:31.726088
| 2020-08-09T09:52:49 |
368693
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alexandre Eremenko",
"Anna",
"https://mathoverflow.net/users/163507",
"https://mathoverflow.net/users/25510"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631907",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368693"
}
|
Stack Exchange
|
Entire function which modulus grows only along the real axis
My research in theoretical physics led to the necessity of constructing an entire function with modulus decaying in the significant part of the complex plane. I wonder whether this is possible because all my attempts to combine known examples of entire functions show that the result is decaying at most for the half of the complex plane.
Let me put the sharp version of the question. Can the modulus of entire function decay for all directions in the complex plane except the positive real axis? I don't see the obvious contradiction of such a behaviour with Picard's little theorem and Liouville's theorem. However, I failed to construct an example. I would be grateful for references about stronger results constraining the behaviour of entire functions.
For my research I seek an entire function with somewhat weaker condition which decays everywhere except the arguments of complex numbers $z$ satisfying the condition $\Re z > a (\Im z)^2$ with some real positive $a$.
This question was answered here: https://mathoverflow.net/questions/190837/entire-function-bounded-at-every-line If this is not enough, state your question more precisely.
Thank you very much for the reference! Anver Kiro suggested such an example in the explicit form of the Taylor series here: https://mathoverflow.net/q/191663. However, I am confused with the fact that this series is divergent for |z|>1 if the notation log^n (...) is assumed to be (log(...))^n. But an entire function must be represented by the Tailor series which are convergent everywhere. If I misunderstand the notation I would be grateful to learn what does this mean.
You are right: that answer with Taylor series makes no sense. The correct one is the first answer.
|
2025-03-21T14:48:31.726230
| 2020-08-09T12:23:02 |
368696
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Dylan Wilson",
"Harry Gindi",
"curious math guy",
"https://mathoverflow.net/users/1353",
"https://mathoverflow.net/users/152554",
"https://mathoverflow.net/users/6936"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631908",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368696"
}
|
Stack Exchange
|
"Universal" triangulated category
Let $\mathcal{C}$ be some category. One way to map this category into a triangulated category is to take the category of simplicial objects $s\mathcal{C}$ (which is an $\infty$-category), take its stabilization $\text{Stab}(s\mathcal{C})$ and take the homotopy category $\text{Ho}(s\mathcal{C})$ of the simplicial category (which is triangulated since it is the homotopy category of a stable $\infty$-category). Then we get a natural functor
$$\mathcal{C}\rightarrow \text{Ho}(\text{Stab}(s\mathcal{C})).$$
My question is:
Does $\text{Ho}(\text{Stab}(s\mathcal{C}))$ satisfy some universal property?
That is, is it the "universal triangulated category" associated to $\mathcal{C}$ in some sense, i.e. if $\mathcal{T}$ is a triangulated category and $\mathcal{C}\rightarrow \mathcal{T}$ satisfying some properties, does this factor through $\mathcal{C}\rightarrow \text{Ho}(\text{Stab}(s\mathcal{C}))?$
If this isn't a "universal triangulated category," does there exist such a construction?
I think there's some confusions here that are worth sorting through. (1) 'associated simplicial category' would usually mean C back again, but you seem to use it later to mean the category of simplicial objects in C, with its simplicial structure? (2) the htpy category of a stable $\infty$-category has a canonical triangulated structure, not just any old $\infty$-category, (3) the homotopy category of sA, for A abelian, is not $D(A)$, or triangulated, but rather $D_{\ge 0}(A)$... I think there are some other issues, but that's a start
@DylanWilson Thank you very much! I'll change my question accordingly.
more stuff: if you'd like to take an ordinary category $C$ and then view $sC$ as simplicially enriched, I think you need $C$ to be complete or cocomplete (otherwise how do you define the simplicial structure on hom-objects?). Also, another issue you might run into is that the construction 'sC' isn't very universal, as far as I know... for specific examples of C it ends up having a universal property as a sifted cocompletion of a certain subcategory of C, but I don't know a good invariant description in general
@DylanWilson The example I'm thinking of here is where C is something that is already abelian or at least additive, so I think the point is that CMG is implicitly thinking of the construction Ho Stab Ch^+(C) (identifying Ch^+(C) with sC, in which case this does have a universal property I think.
I will give a partial answer. I note that the OP has asked a LOT of questions recently (I count 12 so far in the first 9 days of August), and many of them are good questions on which much research has already been done. I would encourage the OP to slow the rate of question-asking, to spend more time reading the references that have been provided, and to think carefully in future questions to avoid easily avoided problems like those that have been raised in the comments.
Now to the answer. Morally, what the OP is suggesting is exactly the kind of thing we love to do as homotopy theorists, but the devil is in the details. Specifically, in this case, the devil is in "...and $\mathcal{C} \to \mathcal{T}$ satisfying some properties..." The issue is that there could be multiple "obvious" ways to stabilize $\mathcal{C}$, and the functor $F:\mathcal{C} \to \mathcal{T}$ has to know that you mean the one you suggested. For example, suppose $\mathcal{C}$ is the empty category. Then whatever conditions you have in mind will probably be satisfied vacuously, and you're asking for a triangulated category $Ho(Stab(C))$ that is supposed to admit a map from every triangulated category $\mathcal{T}$. That's probably not what you really meant.
That said, homotopy theorists have thought long and hard along the direction you have in mind. I recommend the following papers:
Dugger's Universal homotopy theories: given any small category $C$, create a "universal" model category $UC$, which is essentially the free homotopy theory generated by $C$.
Hovey's Spectra and symmetric spectra in general model categories: given a model category $C$ and a Quillen endofunctor $G$, create the stabilization $Sp(C,G)$ where $G$ becomes a Quillen equivalence, just like if $C = Top$, $G$ is the suspension functor, and $Sp(C,G)$ is spectra.
Lurie's Higher Topos Theory, which expands the work above into the realm of infinity categories, e.g., so that $Sp(C,G)$ is a stable $\infty$-category, when you start with a presentable $\infty$-category $C$. Similarly, Dugger's construction can be made to work to produce a presentable $\infty$-category $UC$.
Hovey's book on model categories: chapter 6 shows how to start from a pointed model category and produce a pre-triangulated category in a universal way. But note that when Hovey says "pre-triangulated" that doesn't mean the same as when other authors say "pre-triangulated." Here the structure that would need to be preserved by $F$ has to do with fiber and cofiber sequences, and you need $C$ to be pointed to define these.
Beligiannis and Reiten's Homological and homotopical aspects of torsion theories: Section 5 shows how, given a left, right, or pretriangulated category, there is a universal stabilization that preserves the old partial triangulated structure.
There are many, many other papers in this vein. For example, building on Beligiannis's paper Relative homological algebra and purity in triangulated categories (from 2000), Balmer and Stevenson just this year showed how to take a triangulated category, take a quotient (making it no longer triangulated), and then stabilize in a nice way.
The point is that you have to know what structure of $\mathcal{C}$ to you want to be preserved by $F: \mathcal{C} \to \mathcal{T}$. For example, if you want to assume that $\mathcal{T}$ can be realized by a stable $\infty$-category, then you can probably get a positive answer by making sure $F$ plays nicely with the eventual $\infty$-category structure on $sC$.
However, it's not true that every triangulated category comes from a stable $\infty$-category (Muro and others have constructed counterexamples), so the universal properties shown by Lurie do not provide an affirmative answer to your question in general. Broadly speaking, the collection of triangulated categories breaks down into two types: those that are "geometric/topological" (e.g., the homotopy category of a stable $\infty$-category) and those that are algebraic. Unless you have some way to connect the geometric-type triangulated categories to the algebraic-type (a problem that many have thought about and none have solved to my knowledge) via whatever conditions you place on $F$, then your question is unlikely to have a positive answer in the generality in which you asked it. Hope this helps!
|
2025-03-21T14:48:31.726652
| 2020-08-09T13:09:43 |
368701
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631909",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368701"
}
|
Stack Exchange
|
What is the most efficient algorithm for calculating $\Phi_q(b) \operatorname{mod} N$?
From Hurwitz's theorem about irreducible factor $F_{n-1}(x)$ of degree $\varphi(n-1)$ of $x^{n-1}-1$ we can deduce the following criterion for the primality of $N=2^m \cdot p_1^{n_1} \cdot p_2^{n_2} \cdot \ldots \cdot p_k^{n_k} +1$ :
Let $N=2^m \cdot p_1^{n_1} \cdot p_2^{n_2} \cdot \ldots \cdot p_k^{n_k} +1$ , $m>0 , n_1>0 ,n_2>0, \ldots ,n_k>0$ with $k$ fixed odd primes $\{p_1,p_2, \ldots ,p_k \}$ , $p_i<p_j$ for $i<j$. Let $q$ be a product of $p_1,p_2, \ldots ,p_k $ and let $\Phi_n(x)$ be the nth cyclotomic polynomial . If there exists an integer $a$ such that $$\Phi_q\left(-a^{(N-1)/2q}\right)\equiv 0 \pmod{N}$$ then $N$ is a prime.
You can run this test here.
My question is: What is the most efficient algorithm for calculating $\Phi_q(b) \operatorname{mod} N$ , where $b$ is an integer?
I know that $\Phi_q(b)$ can be calculated by using repeated polynomial division (see Algorithm 1 in this paper). I also know that there is an algorithm (see Algorithm 2 in the paper mentioned above) for calculating $\Phi_{mp}(x) \operatorname{mod} M$ , where $m$ is an odd squarefree integer , $p$ is a prime not dividing $m$ and $M$ is a prime , but I am not sure if these two algorithms can be used for efficiently calculation of $\Phi_q(b) \operatorname{mod} N$ .
|
2025-03-21T14:48:31.726762
| 2020-08-09T13:40:41 |
368702
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlo Beenakker",
"Dattier",
"Eric Towers",
"Fedor Petrov",
"Gerhard Paseman",
"Zach Hunter",
"https://mathoverflow.net/users/110301",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/130484",
"https://mathoverflow.net/users/3402",
"https://mathoverflow.net/users/4312",
"https://mathoverflow.net/users/75890"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631910",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368702"
}
|
Stack Exchange
|
A sequence of primes?
Does there exist a prime integer $q$ and $P$ a polynomial, such that the sequence $x_{n + 1} = P (x_n)$ and $x_0 = q$ that is to say a sequence of distincts primes integers?
No, P have rational coefficients. Example P(x) =1+x*(x-1)/2
Say, if $P(x)=x^2-x+1$ (that guarantees that all terms of this sequence are mutually coprime), can they all be prime? Most probably this was asked before. It is definitely open whether there are infinitely many Fermat composite numbers. If there are only finitely many, then the sequence given by $q=2^{2^k}+1$, $P(x)=(x-1)^2+1$ produces an infinite sequence of primes if $k$ is large enough. But nobody believes this, I think. Heuristically your question also has negative answer.
When $P$ has integer coefficients, there is infinitely many positive integers $n$ such that $P(n)$ is composite. Here, if P(x)=y$, then $y\mid P(x+ky)$. Not sure how to do this for rational coefficients though.
"That is, there is no known polynomial of degree >1 which takes infinitely many prime values." from https://math.stackexchange.com/questions/2019744/on-polynomials-taking-infinitely-many-prime-values Your polynomial would be one taking infinitely many prime values.
If you allow the sequence to be finite, producing $M$ distinct primes, then such a polynomial exists for any $M$, no matter how large; the polyomial is simply $P_M(x)=x+a_M$. This is the Green-Tao theorem.
An example for $M=23$: $x_0=56211383760397$, $a_{23}=44546738095860$.
The objectif of this question, is to build very big prime number.
More generally, choose the finite sequence (of distinct primes) in advance, and then build the polynomial using Lagrange interpolation. Gerhard "No High-Powered Number Theory Needed" Paseman, 2020.08.09.
I think Gerhard talk about finite sequence.
For $(x,y)$ use $(q_k,q_{k+1})$. Gerhard "See How Easy It Is?" Paseman, 2020.08.09.
@CarloBeenakker https://en.m.wikipedia.org/wiki/Largest_known_prime_number
An answer for the case $P(x)=x^2-x+1=x(x-1)+1$
In this case $x_{n+1}=x_n\times ...\times x_0+1$ for $n>2$
Suppose $x_i$ is prime number, with $x_{i+1}>x_{i}$ so $x_0=2$
But we can see $x_0=2$ is not solution.
? this gives the sequence $2, 3, 7, 43, 1807,\ldots$, but 1807 is not prime...
Yes, it's not a solution.
|
2025-03-21T14:48:31.726946
| 2020-08-09T13:52:53 |
368703
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Piotr Achinger",
"https://mathoverflow.net/users/3847"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631911",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368703"
}
|
Stack Exchange
|
Infinititesimal Automorphisms intuition (algebraic stacks)
Let $F$ be a category cofibered in groupoids over category $C$. Given a morphism $x'\to x$ in $F$ lying over a morphism $A′\to A$ in $C$, there is an induced homomorphism
$\operatorname{Aut} A'(x')\to \operatorname{Aut} A(x)$.
The kernel $\operatorname{Inf}(x'/x):=\operatorname{ker}(\operatorname{Aut} A'(x') \to \operatorname{Aut} A(x))$
is also called the group of infinitesimal automorphisms. That's the defition from stacks project
Question: What is the origin and motivation of the name 'infinitesimal' here? How is it connected to the naive geometrical/analytic intuition of infinitesimality (of course if there exist a connection at all)?
Note that in your reference, the base category $C_\Lambda$ is not just some abstract category, but the category of Artinian local $\Lambda$-algebras with residue field $k$ (Tag 06GC). A surjective $A'\to A$ corresponds to an infinitesimal thickening ${\rm Spec}(A)\to {\rm Spec}(A')$. If $F(A)$ is the category of flat schemes over $A$ (or some other type of geometric object), then ${\rm Inf}(X'/X)$ (where now $X$ is a flat scheme over $A$ etc.) is the group of automorphisms of $X'/A'$ restricting to the identity on $X = X'_A$.
For example, for $A=k$, $A'=k[\varepsilon]/(\varepsilon^2)$, $X/A$, $X' = X\otimes_A A' = X[\varepsilon]$, the group ${\rm Inf}(X'/X)$ is the group of $k$-derivations $\mathcal{O}_X\to \mathcal{O}_X$, and hence the familiar "infinitesimal automorphisms of first order = vector fields".
|
2025-03-21T14:48:31.727083
| 2020-08-09T14:31:51 |
368705
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Kimball",
"LSpice",
"Mirco A. Mannucci",
"https://mathoverflow.net/users/118562",
"https://mathoverflow.net/users/15293",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/35959",
"https://mathoverflow.net/users/6518",
"user2554",
"user64494"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631912",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368705"
}
|
Stack Exchange
|
Question about the correspondence between unitary Möbius transformations and quaternions
One of the main theorems about the classification of Möbius transformations states that pure rotations of the Riemann sphere (without translation and dilatation) correspond to unitary Möbius transformation; that is, Möbius transformations of the form:
$$f(z) = \frac{\alpha z +\beta}{-\bar{\beta}z+\bar{\alpha}} = \frac{(a+bi) z +(c+di)}{-(c-di)z+(a-bi)}$$
where $\lvert\alpha\rvert^2+\lvert\beta\rvert^2 = a^2+b^2+c^2+d^2 = 1$.
I was wondering how to correspond a unit quaternion to a given unitary Mobius transformation, as both can be identified with rotation of the Riemann sphere. To calculate the corresponding quaternion, I outlined the following procedure:
First, find the fixed points $z_1$, $z_2$ of the given unitary Mobius transformation by solving the quadratic equation: $z=f(z)\implies -\bar{\beta}z^2+(\bar{\alpha}-\alpha)z-\beta=0$. This step actually enables to find the axis of rotation of the Riemann sphere, which correspond to the vector part of the desired quaternion $q$. The reason for this is that the fixed points of $f(z)$ are exactly the images of the intersection of the axis with the Riemann sphere (the poles of the rotation), under the stereographic projection.
Secondly, as explained in the first step, one needs to find the pre-images of $z_1$, $z_2$ under the stereograpic projection. This can be done by the equation: $z_1 = \cot(\frac{1}{2}\theta)e^{i\phi}\implies \lvert z_1\rvert=\cot(\frac{1}{2}\theta), arg(z_2) = \phi$, where $\theta$, $\phi$ are the zenith angle and the azimuth of the pre-image. This completes the calculation of the axis direction (which we denote by $\eta_1$), and the vector part of $q$.
Now the problem is to find the angle of rotation $\gamma$, which correspond to the real part of the $q$. To do this, apply $f(z)$ to the simplest complex number $z = 1$ to get $f(1) = \frac{\alpha +\beta}{-\bar{\beta}+\bar{\alpha}}$. Now apply the inverse stereographic projection (which we denote $S^{-1}$) to find the point on the sphere that is projected to $f(1)$. We now have three known vectors: $\eta_1$, $\eta_2 = (1,0,0)$, $\eta_3 = S^{-1}(f(1))$.
As a last step, calculate the length of difference vector: $l = \lvert\eta_3-\eta_2\rvert$. Since $l$ is a chord in a circle whose radius is $r = \sqrt{1-(\eta_1 \cdot \eta_2)^2}$ and central angle $\gamma$, one can solve for $\gamma$ by the equation $l = 2r\sin(\gamma/2)$. The desired quaternion is: $q = \cos(\gamma/2)+(\eta_{1x}i+\eta_{1y}j+\eta_{1z}k)\sin(\gamma/2)$.
Although this procedure might appear practical, the algebra I got is so complicated that I didn't succeed in finding a closed form expression for $q$ in terms of $a$, $b$, $c$, and $d$. Therefore, my questions are:
Are there any known results on the correspondence between unitary Möbius transformation and quaternions?
Is there a conceptually more transparent, and simpler algebraic, way of deriving $q$? If the answer is yes, what is the resulting formula?
Remark:
I've already asked this question on Math StackExchange and didn't get any responses except a comment that the angle of rotation is easily computed by the derivative of the Mobius transformation at one of the fixed points $f'(z_1)$.
See L. Ahlfors, Mobius transformations in several dimensions, 1981, Ch. II. General case, to this end.
@user64494's reference doesn't seem obviously available online, but there's a snippet view on Google Books: Ahlfors - Möbius transformations in several dimensions (MSN).
@LSpice: Thank you for the link. It's kind of you.
try this: https://www.researchgate.net/publication/267012326_Clifford_algebras_Mobius_transformations_Vahlen_matrices_and_B-loops
I've already asked this question on Math StackExchange and didn't get any responses - I think the norm is that you should wait a few days, and not crosspost the same day.
@Kimball - you're definitly right, but i saw that my question recieved low number of views and no likes at all, so i thought that i won't get answers on math stackexchange for this question anymore. Anyway, eventually i was wrong, and the greatest help came from a user from math stackexchange, as i wrote in my answer.
After getting some very useful comments by user10354138 from math stackexchange, i succeeded in finding a solution, which is remarkably simple. According to his comments, the angle of rotation $\gamma$ can be computed by taking the derivative of Mobius transformation $f(z)$ at one of the fixed points of the transformation; that is: $e^{i\gamma} = f'(z_1)$. First, we need to find the fixed point:
the resulting quadratic equation is: $-(c-di)z^2-2biz-(c+di)=0\implies z_{1,2} = \frac{-d + ci}{c^2+d^2} (-b\pm \sqrt{1-a^2})$.
The calculation of the derivative yields:
$ e^{i\gamma}=f'(z_1)= \frac{2a+\sqrt{(2bi)^2-4(c^2+d^2)}}{2a-\sqrt{(2bi)^2-4(c^2+d^2)}} = \frac{2a+2i\sqrt{1-a^2}}{2a-2i\sqrt{1-a^2}} = \frac{a+i\sqrt{1-a^2}}{a-i\sqrt{1-a^2}}$.
According to several trigonometric identities, this means that $cos(\gamma/2) = a$. So the real part of the desired quaternion $q$ is $a$. Now, the norm of $z_1$ is:
$cot(\theta/2)=Norm(z_1) = \frac{-b+\sqrt{1-a^2}}{\sqrt{c^2+d^2}}\implies cos\theta = \frac{cot^2(\theta/2)-1}{cot^2(\theta/2)+1} = -\frac{b}{\sqrt{1-a^2}}$. Since the representation of $q$ is:
$$q = cos(\gamma/2)+(\eta_{1x}i+\eta_{1y}j+\eta_{1z}k)sin(\gamma/2) = a+(\eta_{1x}i+\eta_{1y}j+\eta_{1z}k)\sqrt{1-a^2}$$, one gets that the $k$-component of $q$ is $-b$. In addition, since $\frac{\eta_{1y}}{\eta_{1x}} = -\frac{d}{c}$, and $\eta_{1x}^2+\eta_{1y}^2+\eta_{1z}^2 = 1$, one gets the final result that the desired quaternion is:
$$q = a + ci - dj -bk$$.
L. Ahlfors writes in "Mobius transformations in several dimension" that the quaternion technics works only for $M(H^3)$, where $H$ is the upper half-space and is not particularly useful.
This question was not intended to demonstrate the usefulness of quaternions, but just to identify which rotation of the Riemann sphere (which can be described by unit quaternion) corresponds to a given unitary Mobius transformation. Amazingly, the quaternions coefficients are just $a, c, -d, - b$.
|
2025-03-21T14:48:31.727573
| 2020-08-09T14:35:38 |
368706
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"David Handelman",
"Ira Gessel",
"LSpice",
"https://mathoverflow.net/users/10744",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/35959",
"https://mathoverflow.net/users/42278",
"https://mathoverflow.net/users/51189",
"user64494",
"zeraoulia rafik"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631913",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368706"
}
|
Stack Exchange
|
An expression for the $k$-th derivative of $f(x)=x^n\exp(-x)$
Is there a finite expression for $k$-th derivative of
\begin{align}
f(x)={x^n}{e^{ - x}}
\end{align}
Have you tried differentiating a number of times, then guessing the pattern: e.g., $f^{(n)} = P_n (x)e^{-x}$, and working out the recursive condition on the polynomials $P_n$ (which depend on $N$ as well)?
This is not research level, so it doesn't belong at MO. It could probably do well at MSE, although (a) I'd be surprised if it's not there already, and (b) as @DavidHandelman says, it's something for which you can at least gather experimental data (and show that you have done so before asking others to do it).
Have u tried to use Leibniz's formula? am affraid that is not a question for MO website ,MO website is not for standard questions but for question in high level of research
The command of Maple
diff(x^n*exp(-x), x $ k);
produces $$\sum _{k_1=0}^{k}{k\choose {k_1}}{\it pochhammer} \left( n
-{k_1}+1,{k_1} \right) {x}^{n-{k_1}}{{\rm e}^{-x}}
\left( -1 \right) ^{k-{k_1}}
$$
and the code of Mathematica
D[x^n*Exp[-x], {x, k}]
performs $$e^{-x} k! \binom{n}{k} x^{n-k} \, _1F_1(-k;-k+n+1;x) .$$
Just use Leibniz's formula
$(fg)^{(k)}= \sum_{i=0}^k \binom{k}{i} f^{(i)} g^{(k-i)}$.
@Ira Gessel: Disagreed since CASes simplify it, e.g. for $k=5$ we have $${{\rm e}^{-x}} \left( \left( {n}^{5}-10,{n}^{4}+35,{n}^{3}-50,{n}^
{2}+24,n \right) {x}^{n-5}+ \left( -5,{n}^{4}+30,{n}^{3}-55,{n}^{2
}+30,n \right) {x}^{n-4}+ \left( 10,{n}^{3}-30,{n}^{2}+20,n
\right) {x}^{n-3}+ \left( -10,{n}^{2}+10,n \right) {x}^{n-2}+5,{x}
^{n-1}n-{x}^{n} \right)
.$$
With what are you disagreeing? Your first formula, from Maple, is literally just @IraGessel's suggestion.
@LSpice, I repeat that both CASes may find simplified expressions for concrete values of $k$. Those expressions are too long to do it by hand.
|
2025-03-21T14:48:31.727737
| 2020-08-09T15:14:55 |
368709
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Harry Gindi",
"Matthieu Romagny",
"https://mathoverflow.net/users/108274",
"https://mathoverflow.net/users/1353",
"https://mathoverflow.net/users/17988",
"user267839"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631914",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368709"
}
|
Stack Exchange
|
Stabilizer $G_x$ of a $k$-valued point of an algebraic Stack
An algebraic stack or Artin stack is a stack in
groupoids $\mathcal{X}$ over the étale site such that the diagonal
map of $\mathcal{X}$ is representable and there exists a smooth
surjection from (the stack associated to) a scheme to $\mathcal{X}$.
In Wikipedia's article on stacks I found in the excerpt a statement on local structure of algebraic stacks I do not understand:
[…] Given a quasi-separated algebraic stack $\mathcal{X}$ locally of
finite type over an algebraically closed field $k$ whose stabilizers are
affine, and $x \in \mathcal{X} (k)$ a smooth and closed point
with linearly reductive stabilizer group $G_x$, there exists an
etale cover of the GIT quotient […].
My question is what is here the stabilizer $G_x$ of $x$ at all? Recall we did not assume that $\mathcal{X}$ is a quotient stack, therefore it seems to me not to make any sense to speak about a "stabilizer group" of $x \in \mathcal{X} (k)= \operatorname{Hom}(\operatorname{Spec} k, \mathcal{X})$.
The point is that in order to talk about a
stabilizer group $G_x$ of $x$ it is necessary to require the existence of a group $G$ acting on the set $\mathcal{X} (k)$ of $k$-valued points.
But for general algebraic stacks there is no reason that there is no reason that such group $G$ acting on $\mathcal{X} (k) $ such that $G_x \subset G$, right? Could somebody help me to resolve my confusion?
You still have a groupoid action you're killing as every algstack is a groupoid quotient, and the automorphisms at a point of a groupoid still form a group. The stabilizers are still subgroups of automorphism groups as well.
I think I understand your argument. The (pre) stack $X$ is endowed with a fibred functor $F: X \to C$ to certain category $C$ and every fiber $F^{-1}(c)$ is a groupoid by definition. That means that for every $x \in F^{-1}(c)$ the set $Hom_{F^{-1}(c)}(x,x)$ equals $Aut(x)$. So for every $x \in X$ the group $Aut(x)$ acts on subset ${f \in X(k) \ f(Spec(k)) =x } \subset X(k)$ by composition $Spec(k) \to x \to x$. Is this the action you mean?
Something like that. Loop spaces are always groups, and the stabilizer is the based loop space at a point. If you pull inertia back to a presentation, you literally get stabilizers wrt the smooth groupoid.
@Harry Gindi: What do you mean by a "loop space" in this context? I looked up for a formal definition of a stabilizer $G_x$ of a $1$-morphism $x: Spec(k) \to X$. By definition it is defined as pullback (of stacks) of the morphisms $(x,x): Spec(k) \to X$ and the diagonal map $\Delta: X \to X \times_S X$. Also one can show that if $I_x$ is the inertia stack of $x: Spec(k) \to X$ then $G_x \cong I_x \times Spec(k)$.
What I not completely understand is why and how this $G_x$ can be identified as subgroup of group $Hom_{F^{-1}(c)}(x,x)= Aut_{F^{-1}(c)}(x)$. Do you know literature where this identification is proved or give sketch of the construction?
This was getting a little bit long for a comment, so I'll just write it here:
Let $X\simeq S//R$ be an algebraic stack presented by a smooth surjective map $S\to X$ with $S$ a scheme, then $R=S\times_X S$, and the pair of maps $R\rightrightarrows S$ has the canonical structure of a groupoid in algebraic spaces (with the additional structure coming from the higher simplices of the Cech nerve). Choosing a point $x$ in $X$ classified by some Zariski geometric point in $\operatorname{Spec}(k)\to S$, form the following big fibre square
$$
\begin{matrix}
G_x & \to & S\times_X \operatorname{Spec}(k)&\to& \operatorname{Spec}(k)\\
\downarrow&&\downarrow&&\downarrow\\
\operatorname{Spec}(k)\times_X S&\to& R &\to & S\\
\downarrow &&\downarrow&&\downarrow\\
\operatorname{Spec}(k)&\to&S&\to&X
\end{matrix}
$$
In this case, the maps $G_x\to S\times_X \operatorname{Spec}(k)$ and $S\times_X \operatorname{Spec}(k)$ are injective, being pullbacks of injective maps, which gives an injective map $G_x\to R$, including as the literal stabilizer of the point $x\in S$ by the 'action' of $R$, it is including as the subgroupoid of automorphisms fixing $x\in S$.
One point isn't completely clear to me. In your comment you wrote: "every algstack is a groupoid quotient." Could your give a reference where an algebraic stack is introduced / constructed in that way? Or that this construction is equivalent to that one from Stacks project: https://stacks.math.columbia.edu/tag/026N
As explaned in e.g. https://stacks.math.columbia.edu/tag/04TJ (here only for fppf) we can associate an alg stack to a smooth groupoid in algebraic spaces as groupoid quotient. But I not see why as you said every alg stack is obtained in that way.
@katalaveino By choosing a smooth cover by a scheme or algebraic space, you can regenerate a morita-equivalent groupoid by taking the Cech complex. I don't know where this is in the stacks project, but it's easy to show that this always works.
If you happen to find when you stumble upon a literature source that explains this construction in detail, could you name it? So far I have searched without successful in Olson's Alg Stacks and Laumon's & Moret-Bailly's Champs algebrques.
@katalaveino It's more or less by the definition of an effective epimorphism, that is, a cover.
@katalaveino : see https://stacks.math.columbia.edu/tag/04T3.
Sorry for digging out this topic again but recently
I noticed an aspect in your answer I not understand.
You wrote that the pair $R\rightrightarrows S$ which has structure
of groupoid in alg spaces moreover has additional structure
enconded in higher simplices of the Cech nerve. The last
statement I not understand. Does this Cech nerve contain
a proper additional structure which cannot be derived intrinsically from
$R\rightrightarrows S$ if we forget the Cech nerve?
If yes, which additional structure is encoded here by Cech nerve?
The reason that confuses me a bit is that generally if we start
with groupoid $R\rightrightarrows S$ (without this
additional sructure) we can always canonically
associate to $R\rightrightarrows S$ a Cech nerve.
So seemingly this higher simplices of the Cech nerve
not contain additional information about
$R\rightrightarrows S$, or do I miss an important point?
@katalaveino The third term encodes associativity. The term to look into is "coskeletal". The Cech nerve of an internal groupoid is always 2-coskeletal.
That's interesting, up to now I thought that the main
role of a Cech nerve associated to a stack or it's corresponding
groupoid was to provide neccessary framework for building
cohomology of this stack. About that it encodes 'associavity'
in that context I haven't heard anything up to now.
Could you briefly explain which kind of 'associativity'
the Cech nerve encodes here? Maybe you mean in the sense
that we implicitely consider the algebraic stack $X$ as a
$\infty$-category, and the data $R\rightrightarrows S$ resp it's
associated Cech nerve determines the weakened associativities
of $X$ in sense of the concept of equivalences of morphisms
in higher category theory? Or do you mean by 'encoded associativity'
something else?
@katalaveino yes, I meant the first thing.
Now I'm a bit confused about considering (algebraic) stacks
as $\infty$-categories. Classically, stacks can be introduced
in two equivalent ways:
as a (usual) category $F$ fibered in groupoids over $(Sch)$
or as a sheaf in groupoids, ie a
contravariant $2$-functor.
How do you associate to an (alg) stack a structure of
$\infty$-categery concretely, which's weak associativity
on higher morphisms is exactly
encoded in the Cech nerve of $R\rightrightarrows S$ as you said? Clearly, if we use the first description interpreting
a stack as an usual category, then (as every usual cat)
it can be interpreted as $\infty$-category where all $k$-morphisms
for $k>1$ are strict identities. But I doubt that these are
exactly associativity equivalences encoded by Cech nerve
since for usual categories these are trivial, here
the Cech nerve is 'rich' in structure.
Could you explain or give a source where the (alg)
stacks are introduced as higher categoretical object
such that it contains nontrivial associative equivalences
encoded exactly by this coskeletal structure of Cech nerve?
@katalaveino It's the geometric realization of an internal Kan complex (in the category of (higher) fppf sheaves on Aff). The higher structure in the Cech nerve is not rich, it is coskeletal. You can see this from the fact that the diagonal of an algebraic stack is a relative algebraic space. I suggest looking through Jon Pridham's paper on simplicial schemes and algebraic stacks if you're still confused.
|
2025-03-21T14:48:31.728268
| 2020-08-09T16:29:22 |
368717
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"A413",
"darij grinberg",
"https://mathoverflow.net/users/163521",
"https://mathoverflow.net/users/2530",
"https://mathoverflow.net/users/41291",
"მამუკა ჯიბლაძე"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631915",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368717"
}
|
Stack Exchange
|
Associativity equation for topological rings and logarithms
Let $R$ be a topological ring of characteristic zero. Assume that $R$ is commutative. Let $G :R \times R \to R$ be a continuous function satisfying $G(G(x,y),z)=G(x,G(y,z))$ and $G(0,x)=x$, for all $x,y,z \in R$.
If we assume that $G$ is a convergent power series then $G$ is also a formal group law and the Honda's argument(cf. Hazewinkel, "Formal Groups and Applications", Chapter 1, 5.8, p.37) proves that G admits a logarithm, i.e. $G(x,y)=h^{-1}(h(x)+h(y))$.
Question : Does $G$ admit a logarithm if it is only continuous?
EDIT Assuming that $G$ is continuous, what are the additional assumptions that should be made on $R$ and $G$ to guarantee the existence of logarithms?
But if $G$ is only continuous then the ring structure plays no rôle at all
As @მამუკაჯიბლაძე says: If $R $ has the discrete topology, then continuity means nothing, and we are just requiring that $G $ is a binary operation that makes $R $ a semigroup with left identity $0$. Such operations exist in abundance; not all of them are groups, so not all of them have logarithms.
What about $R=Q_p$?
For $Q_p$ you get arbitrary topological semigroup structure on the Cantor set with a point removed.
|
2025-03-21T14:48:31.728392
| 2020-08-09T16:34:25 |
368718
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"ABIM",
"Carlo Beenakker",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/36886"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631916",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368718"
}
|
Stack Exchange
|
Reference Request: Randomly Generated Contraction
Let $n_1>n_2\geq 1$ be integers. Are there a known algorithms for generating $n_2\times n_1$-dimensional random matrices $A$ such that
$$
\|Ax - Ay\|<\|x-y\| \mbox{ if $x\neq y$}?
$$
On some properties of contracting matrices:
For $n_1=n_2$ and if the norm is the $\|\cdots\|_\infty$ norm, then the contractive property (with $\leq$ instead of $<$) is satisfied if the matrix is a Markov matrix (nonnegative real matrix elements with each row summing to 1) and moreover for every pair of rows there exists a column with nonzero entries in those two rows. With these restrictions a random matrix can be readily generated.
But how to algorithmically generate such a matrix?
well, the condition on nonzero elements is trivially satisfied by choosing all matrix elements $>0$ instead of $\geq 0$; then what remains is to fix the row sum at unity; you could start by generating all elements uniformly from $(0,1)$ and then divide each element by the row sum.
|
2025-03-21T14:48:31.728505
| 2020-08-09T16:52:32 |
368720
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"M. Winter",
"Max Alekseyev",
"Seva",
"Ville Salo",
"fedja",
"https://mathoverflow.net/users/108884",
"https://mathoverflow.net/users/1131",
"https://mathoverflow.net/users/123634",
"https://mathoverflow.net/users/17773",
"https://mathoverflow.net/users/7076",
"https://mathoverflow.net/users/9924",
"kodlu"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631917",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368720"
}
|
Stack Exchange
|
Finding zero-one vectors in the row space of a matrix
Suppose that $M$ is a square matrix with all elements on its main diagonal equal to $1$, and every row containing exactly two off-diagonal elements equal to $-1$; all other elements are equal to $0$. The kernel of $M$ is nonzero and, indeed, contains a vector with all its coordinates nonzero. Does it follow that the row space of $M$ contains a (nonzero) zero-one vector?
In case it matters, the sum of all elements in every column of $M$ is nonpositive, and $M_{ij}M_{ji}=0$ whenever $i\ne j$.
Given a vector space, is it decidable whether it contains a nonzero zero-one vector?
@VilleSalo If the vector space is given as a subspace of $\Bbb R^n$ by some basis vectors, you can just iterate over all $2^n-1$ non-zero 0-1-vectors an check whether one is in there.
Heh, good point. Is it in P?
Is the condition of nonempty kernel essential?
@MaxAlekseyev: If the kernel is trivial, then everything is in the row space of the matrix. I am not sure as to how important is to have the kernel not trapped in a coordinate hyperplane.
So, it is inessential. Btw, it may be the case that the existence of a nonzero 01-vector holds even for the lattice spanned by the rows.
@MaxAlekseyev: 1) Well, we cannot say that the condition "the kernel contains a nowhere-zero vector" is inessential. 2) Absolutely, I am also trying to think in this direction.
Why don't you just ask a very well known open question the way it has been initially posed (in a finite set of numbers each number is a sum of some other two; does it follow that there is a subset with sum $0$?)
@fedja, do you have a name/reference for the problem when posed the way you have stated it?
@kodlu: https://mathoverflow.net/q/16857/9924
@fedja: Because it is not equivalent. My question concerns with a stronger statement, with the goal is to evaluate a certain direction in solving the "very well known open question".
You just mean that the numbers can repeat in your formulation? I heard the question with the same assumption (repeated entries allowed), but OK.
@fedja: No, this is mostly not about repetitions. If we identify our set with a vector $a$, and $M$ is the associated matrix, then the original problem is to prove that $a^\perp$ contains a nonzero vector $\varepsilon\in{0,1}^n$. This is, in general, not equivalent to "the row space of $M$ contains a nonzero vector $\varepsilon\in{0,1}^n$".
But with all quantifiers: "For every $a$ in the kernel of $M$, $a^\perp$ contains a 0-1 vector" it is, isn't it?
@fedja: Ok, but this is already not the original question. Anyway, the source of the discrepancy is that the kernel of $M$ can contain vectors other than $a$.
"but this is already not the original question": I perceive the original question as "for every $a$ in the kernel of every such $M$", but OK, this discussion is not very interesting; we'd better just try to solve it :-)
This is true:
For any square matrix $M$ with all elements on its main diagonal equal to 1, and every row containing exactly two off-diagonal elements equal to −1 (with all other elements are equal to 0), the row space of $M$ contains a nonzero zero-one vector. Moreover, there is a linear combination of the rows with the coefficients $0$ and $-1$ only which yields such a vector.
This is in fact the main lemma of this preprint; see, on the other hand, this MO problem for the explanations.
|
2025-03-21T14:48:31.728748
| 2020-08-09T17:06:55 |
368721
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"https://mathoverflow.net/users/163525",
"visitor"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631918",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368721"
}
|
Stack Exchange
|
Is the logistic map $x_{n+1}=r x_n (1-x_n)$ exactly solvable for any $r$ other than $-2,2,4$?
It is known that for $r=-2,2,4$ the logistic map $x_{n+1}=r x_n (1-x_n)$ has exact solutions of the form
$$
x_n=\frac12 \left\{ 1- f\left(r^n f^{-1}(1-2x_0)\right)\right\} \qquad \qquad{(*)}
$$
for suitable functions $f$. The same source further claims, with reference to a private communication, that $r=-2,2,4$ are the only values of $r$ for which the logistic map has exact solutions of the form $(*)$.
However, one wonders whether
are there other values of $r$ for which the logistic map is exactly solvable?
Of course, for these other values of $r$, if any, the exact solutions would likely be of the form other than (*); alas so far I found in the literature nothing that would answer the above question.
Explicit solutions for arbitrary $r$ exist in various forms:
Logistic map: an
analytical solution (1995) represents the solution as a power of
a transfer matrix.
An explicit solution
for the logistic map (1999) gives a functional integral solution.
A note on
exact solutions of the logistic map (2020, paywall) gives the
solution in terms of a power series.
Thank you. Exploring the citations to the first of your references I have also found a paper giving the exact solution in the form of a determinant for any $r$:
M Bruschi, Determinantal solution of the logistic map, J. Phys. A: Math. Gen. 31 (1998) L153-L155 https://doi.org/10.1088/0305-4470/31/7/003
|
2025-03-21T14:48:31.728879
| 2020-08-09T17:19:31 |
368722
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alexandre Eremenko",
"HeMan",
"Sam Hopkins",
"https://mathoverflow.net/users/25028",
"https://mathoverflow.net/users/25510",
"https://mathoverflow.net/users/62218"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631919",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368722"
}
|
Stack Exchange
|
Gromov's articles suitable for master students
I'm a master student and I have read "Monotonicity of the volume of intersection of balls" by Gromov and it was a great experience. When trying to fill the gaps, I often end up finding some very beautiful ideas. I want to keep reading Gromov because he inspires me and was delighted by the ideas that he show on that article, however, I don't know what articles of him are also suitable for a master student. I know some functional analysis, have taken a graduate course on differential geometry, measure theory, complex analysis, commutative algebra, algebraic topology, algebraic geometry, dynamical systems, fourier analysis on number fields and embeddings of finite spaces.
What Gromov's articles do you recommend to read?
Are you interested in research articles proving a specific theorem, or more sweeping expository accounts? His essay on manifolds (https://www.ihes.fr/~gromov/wp-content/uploads/2018/08/manifolds-Poincare.pdf) fits the latter bill.
I'm interested in research articles proving a specific theorem.
Gromov's articles are hard reading. You can probably read some parts of his book Metric structures of Riemannian and non-Riemannian spaces.
I agree with Alexandre Eremenko in that most can be hard to read. But I think you can get a lot out of trying to understand them, as long as you're willing to black-box certain parts which may be inaccessible. Here's a summary of a few major articles. At the least, they all have significant parts which may be understandable, depending on your specific background.
"Curvature, diameter and Betti numbers" is a foundational contribution to Riemannian manifolds of positive sectional curvature, saying essentially that they cannot be arbitrarily topologically complicated. The tool is a Morse theory for the distance function. If you take the Toponogov theorem as given, it should be accessible with some understanding of homology; there is a review of spectral sequences in an appendix. I think it doesn't require too much Riemannian geometry other than the statement of the Toponogov theorem.
"Groups of polynomial growth and expanding maps" has some technical content, but the core idea (in section 6) is purely to do with metric spaces. It is applied by viewing finitely generated groups as examples of metric spaces. I don't have a good understanding of all of the group theory needed to prove the "main theorem", such as the Tits alternative, but the essential ideas of the proof, and many of the technical parts, are interesting (in and of themselves) and accessible. The key idea is to rescale the metric space structure of a finitely generated group to get a new metric space whose group of isometries is a Lie group; this "discrete" to "continuous" passage is somewhat unexpected and fundamental, and is the main point of interest for me.
"A.D. Aleksandrov spaces with curvatures bounded below" (with Yuri Burago and Grigori Perelman) is also largely to do with metric spaces. The idea is to reconstruct some of the theory of sectional curvature in Riemannian geometry by only using a metric space structure. Many parts of it should be accessible, especially with the book "A course in metric geometry" by Burago-Burago-Ivanov as a companion.
"Filling Riemannian manifolds" is very long, and I've never tried to read the whole thing, as it can be read in different parts, and there are already interesting and accessible ideas in the first few sections, to do with "filling radius" and "filling volume" and its relation to systolic inequalities; the only main prerequisite to begin with is simplicial homology. Interestingly a similar filling radius was studied at the same time in a different context in Schoen & Yau's "The existence of a black hole due to condensation of matter"
"The classification of simply connected manifolds of positive scalar curvature" (with Blaine Lawson) proves that certain topological operations on a manifold preserve the existence of Riemannian metrics with positive scalar curvature. Gromov and Lawson prove it by hand by basically elementary means. I've heard that it may have an error in it, but I don't know where it's supposed to be (I've never read it carefully). Schoen & Yau earlier proved the same result in "On the structure of manifolds with positive scalar curvature" by PDE methods that are simpler for me. Taking that theorem as given, Gromov and Lawson give some topological corollaries in cobordism theory that don't appear in Schoen & Yau - this part is too advanced for me.
"Convex symplectic manifolds" (with Yakov Eliashberg) is quite accessible, I recall it not needing too much past standard differential geometry. I think it could definitely be read with any of the standard introductory symplectic geometry textbooks as a companion. The main theorem says that if there is some weak algebraic equivalence between symplectic manifolds, then provided that they are "convex" in Eliashberg and Gromov's sense, one can construct certain symplectomorphisms which realize the equivalence.
The survey article "Spaces and questions" is also interesting to look through. There are also a number of interesting articles on Gromov's website in the non-pure math sections, such as "Mendelian Dynamics and Sturtevant’s Paradigm" and "Crystals, proteins, stability and isoperimetry". Based on your interest in "inspiring and delightful," and distinct from the standard math genre, I would also recommend some of his "ergo" articles like https://www.ihes.fr/~gromov/wp-content/uploads/2018/08/ergo-cut-copyOct29.pdf.
|
2025-03-21T14:48:31.729387
| 2020-08-09T17:34:42 |
368725
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631920",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368725"
}
|
Stack Exchange
|
Möbius function and polynomials
Let $\mu$ be the Möbius function. It is well known that $\sum_{n|k} \mu(n) = 0$ for $k>1$. What could be said about the polynomials $R_k = \sum_{n|k} \mu(n) x^n$ for $x \in [0,1]$? There does not seam to an easy answer to that question. It could be a good idea to take the average. Let $Q_k=1/k \sum_{n=1}^{k} \mu(n)[k/n]x^n$, where $[x]$ denotes the largest integer smaller then $x$. We have $(k+1)Q_{k+1} - k Q_k = R_{k+1}$. Set $Q_1=R_1=x$. Then by induction $Q_k=1/k \sum_{n=1}^{k} R_n$, is indeed the average. Let $p$ be a prime number then $R_p=x-x^p$ and it is concave and positive on $[0,1]$.
The plot of $Q_k$ on $[0,1]$, for a finite number of values of $k$, suggests that $Q_k$ is positive and could be concave. Moreover for all $c<1$, $Q_k=\sum_{n=1}^{+\infty} (\mu(n)[k/n]x^n)/k$ converges uniformly on $[0,c]$ towards the function $g(x)=\sum_{n=1}^{+\infty} (\mu(n)/n)x^n$. Indeed, $|\mu(n)[k/n](x^n)/k|<c^n$, for all $x \in [0,c]$. As a consequence of Abel summation and the prime number theorem the function $g$ is continuous on $[0,1]$. The function $g$ and its derivative were already studied in the literature, for example Fröberg,Numerical studies of the Möbius power series.
According to the paper above, the function g is concave on $[0,1-\delta]$, and then it starts to oscillate infinitely often, where $\delta = 10^{-18}$. So the $Q_k$'s can not be concave on the whole [0,1], in general.
I was wondering how good of approximation is $Q_k$ of $g$? I believe that $\max_{x \in [0,1]}(|Q_k(x)-Q_{k+1}(x)|)=O(1/k^2)$. However I do know how to prove it.
Here is an "explanation" why $O(1/k^2)$. Since the convergence of $Q_k$ towards $g$ is uniform on every compact of $[0,1)$, "all the points converge at approximately the same rate/speed". But $|Q_k(1)-Q_{k+1}(1)|=1/(k(k+1))$. Then for any point in the small enough neighborhood of 1, the same bound "should hold by continuity" of the polynomial function. By uniform convergence property on every compact, "this bound should propagate on the whole [0,1]". Of course this is not a proof. The statements in quotes in this paragraph are intended to convey a rough idea.
Let $||f||=\max_{x\in[0,1]}|f(x)|$. The function $Q_k$ appears to be "almost concave" on $[0,1]$. We could compare it to something concave via Legendre transform. This transform is defined as follows, for a function $f$ defined on $[0,1]$ let $f^{\ast}(s)=\sup_{x\in[0,1]} ({sx-f(x)})$ (cf. p.36 of Convex analysis and minimization algorithms. I, Grundlehren der Mathematischen Wissenschaften 305). Then by the properties of $()^{\ast}$, described on page 44 of that book, we can say that $C_k:=-(-Q_k)^{\ast\ast}$ is the smallest concave function majorizing $Q_k$ on $[0,1]$. So that we have $C_k \geq Q_k$ and $C_k(0)=Q_k(0)=0$.
Let me suggest the following. By comparing $Q_k$ and $C_k$, I hope that we could prove that $||Q_k-Q_{k+1}||=O(1/k^2)$. I believe that the following statements should hold:
1 $C_k$ converges uniformly on every compact of $[0,1)$ to $-(-g)^{\ast\ast}$ and $||C_k-C_{k+1}||=O(1/k^2)$, for all k large enough.
2 $||Q_k-Q_{k+1}||=O(||C_k-C_{k+1}||)$, for all $k$ large enough.
Question Do we have $||Q_k-Q_{k+1}||=O(1/k^2)$, for $k$ large enough?
|
2025-03-21T14:48:31.729618
| 2020-08-09T17:57:59 |
368729
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Benjamin Steinberg",
"https://mathoverflow.net/users/15934",
"https://mathoverflow.net/users/160156",
"user625452"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631921",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368729"
}
|
Stack Exchange
|
how do I find eigenvalues of Cayley graph for one subset given a different subset
How do I find eigenvalues for the adjacency matrix of Cayley graph $X(S_n,S)$ where $S_n$ is the symmetric group of order $n$ and $S$ is the set of transpositions $(i,i+1)$, if the eigenvalues of the graph $X(S_n,T)$ are given by $|T|\chi(1,2)/\chi(i,j)$ with $T$ being the set of transpositions $(i,j)$, $1<i<j<n$.
Thank you
The eigenvalues are much easier to compute for generating sets closed under conjugation like $T$.
Thank you. But how do I find the eigenvalues of the reduced set S?
I don’t think there is any way to get the eigenvalues for the set S from T but I wouldn't be surprised if they are known. The set S of generators is the set of Coxeter or Coxeter-Moore generators. You can't immediately get the eigenvalues just using character theory because S is not in the center of the group algebra
Many many thanks
|
2025-03-21T14:48:31.730069
| 2020-08-09T18:23:58 |
368730
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"David Ben-Zvi",
"JeCl",
"S. Carnahan",
"Theo Johnson-Freyd",
"https://mathoverflow.net/users/105094",
"https://mathoverflow.net/users/121",
"https://mathoverflow.net/users/582",
"https://mathoverflow.net/users/78"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631922",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368730"
}
|
Stack Exchange
|
Factorization homology of a braided (n-1)-category on an (n-1)-sphere
Let $\mathcal{B}$ be a braided ($n-1$)-category. I will assume that $\mathcal{B}$ is a fully-dualizable object in some $n+1$-category of braided ($n-1$)-categories. Hence, from $\mathcal{B}$, using the cobordism hypothesis, one gets a TQFT $\int_{\Box}\mathcal{B}$.
I have read recently that $$\int_{S^{n-1}_b}\mathcal{B} = \Omega^{n-1}\mathcal{B},$$
where $S^{n-1}_b$ denotes the $(n-1)$-sphere with its blackboard framing, and $\Omega^{n-1}\mathcal{B}$ denotes the $1$-category of $(n-2)$- and $(n-1)$-endomorphisms of the monoidal unit of $\mathcal{B}$.
Why is that true?
Where have you read this?
This is not true. The only case I've thought of, n=2, ie an ordinary braided tensor category, integrated over the circle in the blackboard framing (aka the annulus), is the topic of the paper https://arxiv.org/pdf/1606.04769.pdf and is definitely not the loop of the category, which is trivial eg for B=Rep_q G.
Ah sorry I think what you mean by looping is a little confusing at least to me. Do you mean the endomorphism category of the unit in the [in fact monoidal] 2-category of B-modules? AKA the categorical Drinfeld center? in which case this is correct
Very informally at least this is true since you can write the sphere as two discs glued together, which suitably interpreted gives endomorphisms of the disc, considered as an object in the higher category attached to its boundary, which is what this looping operation is (or must be - I think it's important to emphasize you're not thinking of the braided category as a [n-1]category but as a higher category using the monoidal structure). You also need enough duality to make this picture work.
This is claimed on page 16 of: https://arxiv.org/abs/2003.06663
Oh, I see! I think that's also where my confusion came from! @DavidBen-Zvi your assumptions are spot on. So if I'm understanding you correctly, this seems true, but has not been formally proven yet?
Thank you for checking the details, JeCl. I agree something is not parsing in that section of my paper. If nothing else, my notation is poor.
I think this is an error in my paper. Thank you for finding it. The overall result is correct, but the proof is wrong as written. To correct it, I need to replace "TQFT" with "relative TQFT", and replace $S^{n-1}_b$ with the pair $(D^n, S^{n-1}_b)$, and the rest is correct. In detail:
$\newcommand\cB{\mathcal{B}}\newcommand\cX{\mathcal{X}}$
I want to appeal to a "state-operator correspondence", which for an absolute (aka nonanomalous) $(n+1)$-dimensional TQFT $\cX$ says that the $m$-category of operators of dimension $\leq m$ is $\cX(S_b^{n-m})$. Note that, counting dimensions, this is indeed an $m$-category.
But I wanted to apply this to the relative (aka anomalous) $(n+1)$-dimensional TQFT $\cB$, and its compactification $\cB^2 = \int_{S^1_b}\cB$. This $(n+1)$-dimensional TQFT is relative to an $(n+2)$-dimensional TQFT which depends only on the Morita equivalence class of $\cB$, and my notation did not distinguish these.
So I need to tell you about the state-operator correspondence for relative TQFTs. To say it, let me remind that a relative TQFT is the type of thing that can eat a "cobordism with boundary". This is a manifold with two types of boundaries: you can stitch cobordisms together along one of the types of boundaries, and the other type of boundary is marked by a boundary condition.
I don't have good notation. I will write $\cB_\partial$ for the relative TQFT in question. Suppose $M^m$ is an $m$-dimensional cobordism and I decide that $N^{m-1} \subset \partial M$ is where I will place the boundary condition (and I leave the rest of $\partial M$ as stitchable); then I will write $\cB_\partial(M, N)$, and I won't try to use footnotes. Any absolute $(n+1)$-dimensional TQFT $\cX$ gives a relative $(n+1)$-dimensional TQFT $\cX_{\partial}$ in which $\cX_\partial(M^m,N^{m-1}) = \cX(N^{m-1})$ depends only on $N$. On the other hand, any relative TQFT gives an absolute TQFT in which you only use cobordisms without boundary.
With this all said, the state-operator correspondence for relative $(n+1)$-dimensional TQFTs asserts that the boundary operators of dimension $\leq m$ in $\cX_\partial$ is $\cX_\partial(D^{n+1-m}, S^{n-m}_b)$, where $D^{n+1-m}$ is the disk and $S^{n-m}$ is its full boundary. So this is a cobordism from $\emptyset$ to $\emptyset$.
Ok, now I can compute. I want to show that the multifusion $(n-1)$-category $\cB^e$ is fusion, which is to say that the relative $n$-dimensional TQFT $\cB^e_\partial$ satisfies $\cB^e_\partial(D^n , S^{n-1}) = \mathbb{C}$. As in my paper, use that $\cB^e_\partial$ is a compactification on $S^1$ on $\cB_\partial$, so that what we want to compute is $\cB_\partial(D^n \times S^1, S^{n-1} \times S^1)$. But I can do this by first compactifying on $(D^n, S^{n-1})$ to get an absolute 2D TQFT defined by the 1-category $\cB_\partial(D^n, S^{n-1})$, and we already said that that is precisely the 1-category of boundary lines, i.e. what I called $\Omega^{n-1}\cB$.
Thank you very much for your answer, this has helped me a lot! If I may ask two more questions: I do not really understand what you mean by $\mathcal{B}_{\partial}$? What morphisms does it correspond to? And, to make sure I understand you correctly, the state-operator correspondence is some physical principle you use in order to justify defining $\Omega^m\mathcal{B}$ the way you do?
As I said, I don't have good notation. I think the correct proof is now on arXiv. In any case, what I called $B_\partial$ is supposed to be the relative TQFT corresponding to "$B$ as a $B$-module". This is a relative TQFT valued in $Mor_2((n-1)KarCat)$. Or you could suspend (you have enough dualizability) to think about it as "$\Sigma B$ as a $\Sigma B$ module" which selects a relative TQFT valued in $Mor_1(nKarCat)$.
As for the "state operator correspondence", on the one hand it is a physical principle. On the other hand, I am asserting that if you contemplate (relative) TQFTs and factorization algebras enough, then it will be manifestly true.
|
2025-03-21T14:48:31.730477
| 2020-08-09T18:58:40 |
368732
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"MathCrawler",
"Mohan Swaminathan",
"https://mathoverflow.net/users/110236",
"https://mathoverflow.net/users/161310"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631923",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368732"
}
|
Stack Exchange
|
English reference for Douady/Grauert construction of versal deformations of compact complex spaces
I'm trying to learn about the deformation theory of compact complex spaces. I'm familiar with the case of compact complex manifolds from the paper "On the Locally Complete Families of Complex Analytic Structures" by M. Kuranishi. The original references for the case of compact complex spaces seem to be the papers "Le problème des modules locaux pour les espaces $\mathbb C$-analytiques compacts" by A. Douady and "Der Satz von Kuranishi für kompakte komplexe Räume" by H. Grauert which are in French and German respectively.
Are there any English language references for the construction of versal deformations of compact complex spaces?
Try this one as a starter (in particular Chapter 4).
The title of Grauert's paper should read "Der Satz von Kuranishi für kompakte komplexe Räume".
Thanks for the correction and the reference!
D. Barlet and J. Magnusson, Complex Analytic Cycles II. Unfortunately, the book is not in press at the moment, so you have to wait. While you wait, you can read Complex Analytic Cycles I.
|
2025-03-21T14:48:31.730588
| 2020-08-09T19:09:57 |
368733
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"D.-C. Cisinski",
"Jeremy Rickard",
"Marco Farinati",
"https://mathoverflow.net/users/1017",
"https://mathoverflow.net/users/163478",
"https://mathoverflow.net/users/22989",
"https://mathoverflow.net/users/98863",
"physician"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631924",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368733"
}
|
Stack Exchange
|
Homotopy equivalences and Mapping Cones
Edit: Version 2:
Suppose that $A,B,C$ are chain complexes and $f: A \rightarrow B$ is a chain map. Suppose that there is a homotopy equivalence
$$ \text{Cone}(f: A \rightarrow B) \simeq C.$$
The chain map $u: \text{Cone}(f) \rightarrow C$ provided by the homotopy in particular means that I am provided with a chain map $g: B \rightarrow C$.
Consider now $\text{Cone}(g: B \rightarrow C)$. I would like to say that there is a homotopy equivalence between $A[-1]$ and $\text{Cone}(g)$.
Similarly, the map in the other direction $v: C\rightarrow \text{Cone}(f)$ provides me with a chain map $h: C[1] \rightarrow A$ and I would like to say that $\text{Cone}(h)$ and $B$ are homotopy equivalent.
Are these statements true and if so how can one prove them?
What are $g$ and $h$? As it stands, 2 and 3 are not statements about $A$, $B$, $C$ and $f$.
Yes, I wasnt entirely clear. I mean "the data of f: A --> B + a homotopy equivalence between Cone(f) and C is equivalent to the data of a chain map g:B --> C + a homotopy equivalence between Cone(g) and A[-1]" and so on...
It is true. It is not trivial (this is an opinion), however it is standard. In any triangulated category, two objects and a map determine the third, up to (usually non unique) isomorphism. And the category of Chain complexes with maps up to homotopy is a classical example of triangulated category. A reference for those properties could be Happel's book (triang cat in rep. of finite dim alg).
The relevant fact is the following: (Axiom 2) if $(X\overset{u}{\to}Y\overset{v}{\to}Z\overset{w}{\to}X[-1])$ is a triangle (for chain complexes, I use the convention that $d(C_n)\subseteq C_{n-1}$), then $(Y\overset{v}{\to}Z\overset{w}{\to}X[-1]
\overset{-u}{\longrightarrow}Y[-1]) $ and $(Z[1]\overset{-w}{\longrightarrow}
X\overset{u}{\longrightarrow}
Y\overset{v}{\longrightarrow}Z)$ are triangles too. (Hence, if and only if.)
You should do the exercise that in the category of chain complexes with maps up to homotopy, the class of triangles being the u-ples that are isomorphic (in the homotopy category) to ones of the form $(X\overset{u}{\to}Y\overset{i}{\to}Co(u)\overset{p}{\to}X[-1])$ actually satisfies Axiom 2. This exercise is very instructive because you will find, doing it, that it is not possible in general to have all commutative squares in the category of complexes. When trying to compare the ''rotated'' triangle with a standard one, one is forced to choose maps, and there are two squares to look at: if one of them is commutative, the other is not... but there is an ''obvious'' homotopy and the square commute up to homotopy. In this way, you will learn why the category of chain complexes and chain map is not triangulated, and why the ''up to homotopy'' helps.
After that, you should do the exercise that in a triangulated category, a map $u:X\to Y$ is an isomorphism if and only if $X\overset{u}{\to} Y\to 0\to X[-1]$ is a triangle. Next, you do as exercise that if
$A\overset{f}{\to} B\overset{g}{\to} C\to A[-1]$ and
$A\overset{f}{\to} B\overset{h}{\to} D\to A[-1]$ are two triangles in a triangulated category, then $C\cong D$ in that category. Actually, you can do this in any pre-triangulated category, that is, you don't need to use the octahedral axiom. Once you are happy with this result, you should do the exercise that the category of complexes and maps up to homotopy satifies the rest of the axioms of a triangulated category (octahedral axiom is optional for this question). Also notice that proving the validity of the axioms is not the same as proving directly what you asked at the first time, to check the validity of the axioms is a kind of easy and direct.
I think this should answer your question, or if not, should give you the way that your answer should be re-formulated, and then answered..
I disagree: it would look like a reasonable statement if 1. meant "there exists $f:A\to B$ such that the cone of $f$ is chain homotopic to $C$". But, since $f$ is given a priori, and since there are no relations between $f$ and the data appearing in conditions 2. or 3., there is no way we may prove that these statements are equivalent. Condition 2. or 3. could be true with $f=0$, for instance.
Well, I didn 't check everything.. may be my answer lack of some coherence. In any case, Í think the question should definitely be directioned to the language of the homotopy category as a triangulated one. Plade take my answer a
Please take my answer as a sugestion into that directioin
Sure, the direction you suggest is the right one!
Please check the edit to the original question where I've clarified some of the ambiguities. $g, h$ are specific maps.
I have expanded my answer, I hope it helps. Just as final comment, this is more an ME question than an MO one :)
Thank you very much, it's very clear now. Indeed part of the question was meant to ask "how to think of this property conceptually". Triangulated categories seems to be the answer as you very nicely explain. (I doubt I would've gotten that suggestion on SE :) )
|
2025-03-21T14:48:31.730957
| 2020-08-09T19:38:21 |
368736
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlo Beenakker",
"LSpice",
"Michael Hardy",
"efs",
"https://mathoverflow.net/users/109085",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/6316"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631925",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368736"
}
|
Stack Exchange
|
Reference request: Dictionary of the Leibniz notation
Is there any published, somewhat comprehensive, list of (almost?) all the many ways in which the Leibniz notation ($dx,$ $P(dx),$ $d\mu(x),$ $du\wedge dv,$ etc., etc.) gets used in the various areas of mathematics?
(I posted this question here on m.s.e. and it was closed although nobody verbally expressed any specific objections to it or even hinted at such.)
from what I read on MSE it was closed as not being clear enough --- I'm not sure that asking the same question here will be helpful.
It's also clearly not research level (although lots of also clearly-not-research-level historical questions do well on MO anyway).
@CarloBeenakker : Do you have any guesses as to what someone might find unclear about this?
At least for me, the question is not clear since there is a distinction between, for example: the original notation used by Leibniz; the formalization of this notation (Robinson et al.); the derived notations inspired by the concept of differential in other branches, etc. Perhaps you could be more specific in what aspects are you interested about this.
@EFinat-S : Well, you've managed to surprise me at least. That there are great differences among those something I had to be acutely aware of in order to think of asking this question. And if I were to specify something like either the original usage of Leibniz or the formalization by Robinson, etc., then nothing would be left of the question at all. It's as if you had told someone who's driving to a destination that it's possible to run the car's engine without going anywhere, so why not do that instead?
My intention was not to offend you. Also, I'm not a native speaker of english, so it is sometimes difficult to make me understand. Perhaps its the phrase "gets used" what does not seems clear (at least to me): historically? formally? simply as an "abuse" of notation?
@EFinat-S : "Abuse of notation" is a bit of jargon that mathematicians use too much. I wonder whether there are any "abuses of notation" that are not in fact good notation. They are called abuses only because codified formalisms do not countenance them. The possibility that those formalisms are less than the whole truth should be considered. At any rate what I had in mind has more to do with what Leibniz and Robinson have in common.
"At any rate what I had in mind has more to do with what Leibniz and Robinson have in common". If you have that in mind, well, ask that.
@EFinat-S : As far as I know, I did.
@EFinat-S : Maybe something I said above in comments can be made clearer: It is precisely the fact that there are such great differences as the one example that you cite that is the occasion for this question.
Not quite sure in which direction you are hoping for an answer, but to set a first data point I offer the Pantheon of Derivatives
Part 1 – Directional Derivatives
Part 2 – Manifolds
Part 3 – Vector Bundles
Part 4 – Lie Theory
Part 5 – Theorems
Maybe this is similar to what I had in mind. (Every day I am appalled by the frequent crude typesetting solecisms in MathJax and LaTeX by otherwise literate mathematicians, and these pages are consistent with that. At one point I see $2\mathrm{max}\Big{ \cdots\cdots\Big}$ instead of $2\max\Big{\cdots \cdots \Big}$ and wonder why people don't know about things like this. But the content looks very interesting otherwise.)
|
2025-03-21T14:48:31.731244
| 2020-08-09T20:02:54 |
368738
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Federico Poloni",
"P Emt",
"https://mathoverflow.net/users/163524",
"https://mathoverflow.net/users/17773",
"https://mathoverflow.net/users/1898",
"kodlu"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631926",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368738"
}
|
Stack Exchange
|
Combining quadratic and linear matrix terms into a quadratic term
Given
$$ C = AFA^T + A\bar{F} $$
where $A = [A_1 A_2]$, $F = \begin{bmatrix}
F_1 & F_2 \\
F_2^T & F_3
\end{bmatrix}$, $\bar{F} = 2 \begin{bmatrix}
\bar{F}_1 \\
\bar{F}_2
\end{bmatrix}$ such that $A_i \in \mathbb{R}^{n \times k}, F_i = F_i^T \in \mathbb{R}^{k \times k}, \bar{F}_i \in \mathbb{R}^{k \times n}$ and further $F_i, C = C^T \succeq 0$.
Is it possible to represent $C \succeq 0$ in the following compact-quadratic form:
$$ C = A_M F_M A_M^T $$
for some $A_M$ with $F_M = F_M^T \succeq 0$?
Why are you using F for two different matrices?
Do you have a requirement that you do not want to "mix" $A$ and $F$ terms? Otherwise that is trivially true; every $C \succeq 0$ can be written as $BB^T$ for some $B$.
@FedericoPoloni hi - yes, I need to keep the $A, F$ terms separate. he closest i came across was a DT Lyapunov eqn here en.wikipedia.org/wiki/Lyapunov_equation but of course I have an ¯ sitting in the second term so am not sure ... i just needed a way to combine the second piece into the first one so was thinking if the second-linear term could be re-written as a quadratic in somehow via sq. root etc but couldn't arrive at a firm concl. also, $k \leq n$.
Then I would suggest that you edit your requirements in the question. Your form for $C$ can only produce terms of the form $(A_M){ij}(F_M){jk}(A_M)_{lk}$, so if you want to have a term not in this form you will have to include some nonzero constants inside $A$...
thank you - i am totally open to any additions to the $A, F, \bar{F}$ matrices once I can collapse these into the quadratic expression. Would appreciate any assistance with this.
|
2025-03-21T14:48:31.731386
| 2020-08-09T20:29:41 |
368741
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Andy Putman",
"Calvin McPhail-Snyder",
"Joshua Grochow",
"Qiaochu Yuan",
"Shruthi Sridhar",
"https://mathoverflow.net/users/113402",
"https://mathoverflow.net/users/163534",
"https://mathoverflow.net/users/290",
"https://mathoverflow.net/users/317",
"https://mathoverflow.net/users/38434"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631927",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368741"
}
|
Stack Exchange
|
Is there a knot invariant that is not the sum of finite type invariants
It is an open question whether finite type invariants separate knots, but do we know any knot invariant that is NOT the "sum" of finite type invariants? For instance, many knot polynomials, while not finite type themselves, are a "sum" of finite type invariants of all degrees.
(Disclaimer, I get this isn't a direct sum, so what I'm asking is if the group of knot invariants is direct product of finite type invariants of all degrees in some sense)
How about the knot group?
I don't think this is a well-posed question, largely because of the vagueness of the term "invariants". Qiaochu has already pointed out that many useful invariants don't take numerical values. If you insist on numerical invariants, a random bijection between the set of knots and $\mathbb{Z}$ is a perfectly fine invariant.
@Andy: well, we might insist on an invariant explicitly computable from a knot diagram.
@Andy that's a fair point. Perhaps I can amend my question to be: Are there any well known knot Z valued invariants that can't be broken up as a sum like that.
@QiaochuYuan: That's still not much of a restriction. We've known how to solve the isomorphism problem for knots since Ancient Times (more precisely: we learned how to solve the homeomorphism problem for Haken manifolds like knot complements in the 70's and learned that knots are determined by their complement in the 80's), so all you're ruling out are non-computable enumerations. There really isn't a way to define "invariants" except for ways that make this question "obviously false" or "obviously true".
@AndyPutman: I think it becomes much more interesting if one asks for efficiently computable invariants (say, in polynomial time, or $\mathsf{NP}$, or $\mathsf{PSPACE}$, or others...). But then rather than "obviously false" or "obviously true," it becomes more like "obviously a hard question at the intersection of complexity and topology".
Is there some nice category of knots where it is an interesting question to ask about certain restricted types of functors from that category to abelian categories? (And where it would capture many of the known knot invariants.)
If you specify something like "knot invariants satisfying a skein relation" the answer might even be yes. But that's fairly restrictive, since AFAIK every such invariant is a quantum invariant. You would still have to be careful about what you mean by "skein relation."
Here's an expansion on Qiaochu's objection: homomorphism-counting invariants.
If $G$ is a finite group and $K$ is a knot, the size of the set $N(G,K) := |\hom(\pi_1(S^3 \setminus K), G)|$ is a $\mathbb Z$-valued invariant of $K$.
For example, $N(D_{2n},K)$ is the number of Fox $n$-colorings of $K$ (for $n$ odd.) I read somewhere that one of the best invariants for knot tabulation before the discovery of the quantum invariants was $N(G,K)$ for groups like $SL_n(F_p)$ (or similar, I don't remember at the moment.)
As far as I know, these invariants aren't a sum of finite-type invariants.
This is somewhat surprising from a physical perspective, because $N(G,K)$ is the value of the Dijkgraaf-Witten theory with gauge group $G$ on the knot complement. Dijkgraaf-Witten theory is the discrete version of Chern-Simons theory, which is what underlies quantum knot invariants, hence most finite-type invariants.
|
2025-03-21T14:48:31.731749
| 2020-08-09T21:21:51 |
368744
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631928",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368744"
}
|
Stack Exchange
|
Does the finitely additive integral preserve convergence for non-negative measurable functions?
Let $(X, \mathcal X)$ be a measurable space. Say that a net $(\mu_\alpha)$ of finitely additive probability measures converges to a finitely additive probability measure $\mu$ if and only if $\mu_\alpha(A) \to \mu(A)$ for all $A \in \mathcal X$.
If $f$ is an extended-real-valued simple $\mathcal X$-measurable function of the form $f = \sum_{j=1}^n a_j 1_{A_j}$, then the integral of $f$ with respect to a finitely additive probability measure is defined in the usual way:
$$\int fd\mu = \sum_{j=1}^n a_j \mu(A_j).$$
If $f: X \to [0,\infty]$ is non-negative, then define
$$\int f d\mu = \sup\Big\{ \int gd \mu: g \ \text{simple}, \ 0 \leq g \leq f \Big\}.$$
Question. Is it the case that if $\mu_\alpha \to \mu$, then $\int f d\mu_\alpha \to \int f d\mu$ for all non-negative $\mathcal X$-measurable $f: X \to [0,\infty]$?
If $f$ is bounded (and not necessarily non-negative), then the result holds. The motivation for the question is that I'm wondering to what extent the "usual properties" of the finitely additive integral extend from bounded functions to non-negative ones. For example, in this post it is shown that the finitely additive integral remains linear on non-negative functions.
No, not even for sequences of countably additive measures.
Take $X = \mathbb{N} = \{0,1,2,\dots\}$ with its discrete $\sigma$-algebra, and let $\mu_n$ put mass $1/n$ at the point $n$ and mass $1-1/n$ at $0$. Let $\mu$ put mass $1$ at $0$. Then it is clear that $\mu_n(A) \to \mu(A)$ for every set $A$ (consider the cases $0 \in A$ and $0 \notin A$).
Set $f(n) = n$. Then we have $\int f\,d\mu_n = 1$ for all $n$ but $\int f\,d\mu = 0$.
|
2025-03-21T14:48:31.731892
| 2020-08-09T21:22:31 |
368745
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631929",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368745"
}
|
Stack Exchange
|
Classifying Galois elements using cardinalities of their orbits
Let $g \in G_\mathbb{Q}$, the absolute Galois group of $\mathbb{Q}$. Assume $g$ is not a complex conjugation.
We can look at the orbit of any $q \in \bar{\mathbb{Q}}$ under the action of $g$: $O_g(q) = \{g^n(q) : n \in \mathbb{Z}\}$.
If $O_g(q)$ is a finite set of cardinality $k$, we can say that $q$ is a periodic point of $g$ of period $k$. Otherwise, $p$ is an aperiodic point of $g$.
We can count the number $N_g(k)$ of orbits of each cardinality $k = 1, 2, \dots $. If all these numbers are finite, we can even put them together into some generating function $\nu_g (s)$ attached to $g$ and try to understand its properties.
Can the conjugacy classes of elements of $G_\mathbb{Q}$ be distinguished or classified using the data $\{N_g(k) : k = 1, 2, \dots, \infty\}$, their orbit signatures as it were?
If there are no periodic points at all, or on the other extreme, if there are infinitely many periodic orbits for all possible periods, then the question won't make sense as stated, but could it then be salvaged, say using densities in place of cardinalities or some other normalization device?
First, I claim, there are no "aperiodic" points at all. Indeed, if $q \in \overline{\mathbb{Q}}$, then necessarily $q \in K$ for some finite (wlog normal) extension $K/\mathbb{Q}$. So the open subgroup $G_K \subseteq G_{\mathbb{Q}}$ stabilizes $q$, so the $G_{\mathbb{Q}}$-orbit of $q$ is in fact just a $G_{K/\mathbb{Q}}$-orbit, and hence finite. In particular, some power $g^n$ of $g$ ($n>0$) acts trivially on $q$, and hence $O_g(q)$ is finite. So there are no "aperiodic" points.
Next, I claim, for each $k$, $N_g(k)$ is either $0$ or infinite. Indeed, suppose $N_g(k)>0$. Then there is some $\alpha \in \overline{\mathbb{Q}}$, such that the orbit of $\alpha$ under $g$ has exactly $k$ elements. Then also $\alpha + x$ for any $x \in \mathbb{Z}$ has the same property, i.e., $N_g(x)$ is infinite.
This said, I also don't see a way to rescue the question by using any kind of densities.
|
2025-03-21T14:48:31.732054
| 2020-08-09T22:06:36 |
368750
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Robert Bruner",
"https://mathoverflow.net/users/6872"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631930",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368750"
}
|
Stack Exchange
|
Functoriality of Atiyah-Hirzebruch spectral sequence - Reference Request
I'm interested in a text book reference on the functoriality of the Atiyah–Hirzebruch spectral sequence. The only reference I found are these lecture notes by Kupers (link should lead to the target page 52). Help would be greatly appreciated.
A reference for the claim is Theorem 2.1 in the paper A Generalization of the Atiyah–Hirzebruch
Spectral Sequence by Mehdi Hakim-Hashemi and Donald W. Kahn.
I expect it is in Dyer's book "Cohomology Theories":
MR0268883
Dyer, Eldon .
Cohomology theories.
Mathematics Lecture Note Series
W. A. Benjamin, Inc., New York-Amsterdam 1969
but my copy is in my office, and I don't get there very often right now.
|
2025-03-21T14:48:31.732141
| 2020-08-09T22:41:31 |
368755
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Arno",
"D.S. Lipham",
"Nate Eldredge",
"https://mathoverflow.net/users/15002",
"https://mathoverflow.net/users/4832",
"https://mathoverflow.net/users/95718"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631931",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368755"
}
|
Stack Exchange
|
Mapping $\mathbb P$ onto $\mathbb Q ^\omega$
Let $\mathbb P$ denote the space of irrationals. Is there a continuous bijection (one-to-one and onto) $f:\mathbb P\to \mathbb Q ^\omega$ that maps each closed subset of $\mathbb P$ to a $G_\delta$-subset of $\mathbb Q ^\omega$?
Remark 1. Suppose that $f:\mathbb P\to \mathbb Q ^\omega$ is a continuous bijection mapping closed sets to $G_{\delta}$ sets. Then $f^{-1}:\mathbb Q^\omega\to \mathbb P$ is a Baire class $1$ functions, i.e. $(f^{-1})^{-1}(U)=f(U)$ is an $F_{\sigma}$-subset of $\mathbb Q ^\omega$ for every open set $U\subseteq \mathbb P$. By Theorem 4.1 in the reference below, either there are countably many sets $X_n\subseteq \mathbb Q ^\omega$ such that $\mathbb Q ^\omega=\bigcup \{X_n:n<\omega\}$ and $f^{-1}\restriction X_n$ is continuous, or $f^{-1}$ contains Pawlikowski's function $P:(\omega+1)^\omega\to \omega^\omega$.
Remark 2. While trying to solve this problem, I discovered an example involving complete Erdos space $\mathfrak E_c$. There is a continuous bijection $f:\mathfrak E_c^\omega\to \mathbb Q ^\omega$ which maps closed sets to $G_{\delta}$ sets and such that $f^{-1}$ is not a countable union of continuous functions. So by Theorem 4.1 it must contain $P$. I proved that all similar examples, including the one in @Arno's answer also must contain $P$.
My feeling now is that my question probably has a positive answer, although the zero-dimensionality of $\mathbb P$ makes things interesting.
Solecki, Sławomir, Decomposing Borel sets and functions and the structure of Baire class 1 functions, J. Am. Math. Soc. 11, No. 3, 521-550 (1998). ZBL0899.03034.
Is $\mathbb{P}$ the set of irrationals, or something else?
The "canonical" continuous bijection works. We start by observing that $\mathbb{P}$ is homeomorphic to $\mathbb{N}^\omega$. We pick some bijection $\tau : \mathbb{N} \to \mathbb{Q}$, which is trivially continuous, and has a Baire class 1 inverse. We can then lift $\tau$ to obtain a continuous bijection $\tau^\omega : \mathbb{N}^\omega \to \mathbb{Q}^\omega$ with Baire class 1 inverse $(\tau^\omega)^{-1}$. As $(\tau^\omega)^{-1}$ is Baire class 1, the preimage of a closed set under it is $\Pi^0_2$, hence $\tau$ maps closed sets to $\Pi^0_2$-sets as desired.
D.S. Lipham gave some more details in the comments for checking that the inverse is Baire class 1. We can directly show that $\tau^\omega$ maps open sets to $F_\sigma$-sets. Each basic open subset of $\mathbb{N}^\omega$ maps to a product of $F_\sigma$-subsets of $\mathbb{Q}$ whose factors are eventually all of $\mathbb{Q}^\omega$. Hence, the image is $F_\sigma$ in $\mathbb{Q}^\omega$. Each open subset of $\mathbb{N}^\omega$ is a countabe union of basic open sets, so its image is a countable union of $F_\sigma$-sets.
@D.S.Lipham Any sequence of rationals appears in the range as the image of the sequence of the indices of those rationals.
I see. Looks correct then.
To check that the inverse is Baire class 1, we should show $\tau ^\omega$ maps open sets to $F_\sigma$-sets. Each basic open subset of $\mathbb N^\omega$ maps to a product of $F_{\sigma}$-subsets of $\mathbb Q$ whose factors are eventually all of $\mathbb Q ^\omega$. So that image is $F_{\sigma}$ in $\mathbb Q ^\omega$. Each open subset of $\mathbb N ^\omega$ is a countably union of basic open sets, so its image is a countable union of $F_{\sigma}$-sets.
|
2025-03-21T14:48:31.732353
| 2020-08-09T23:06:28 |
368757
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631932",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368757"
}
|
Stack Exchange
|
Theta Function Associated to Kummer Lattice
This is something which I feel must be out in the literature somewhere, but I have been unable to find anything.
If we let $\text{Km}(A)$ be the Kummer $K3$ surface associated to an abelian surface $A$, then we define the Kummer lattice $K$ to be the minimal primitive sublattice of $H^{2}(\text{Km}(A), \mathbb{Z})$ containing the 16 exceptional curves. The Kummer lattice is negative-definite, even of rank 16 and determinant $2^{6}$. For some details, see page 4 of (https://arxiv.org/pdf/1305.3514.pdf).
The theta function associated to $K$ is of course
$$\theta_{K}(q, \vec{w}) = \sum_{v \in K} q^{v \cdot v} \, \vec{w}^{v}.$$
Has $\theta_{K}(q, \vec{w})$ been studied or described anywhere in the literature? For the life of me, I haven't been able to find a single mention, despite this seeming like a natural question. My guess is that $\theta_{K}(q, \vec{w})$ is some sort of lattice Jacobi form of weight $\frac{16}{2} =8$. It would be great if this were spelled out or computed somewhere.
For what it's worth, I'm actually interested in the generalized theta function with characteristic
$$\theta_{K}^{(a)}(q, \vec{w}) = \sum_{v \in K} q^{(v+\frac{a}{2}) \cdot (v+\frac{a}{2})} \, \vec{w}^{v+ \frac{a}{2}}$$
for some fixed element $a \in K$. If by chance there was also something out there about this more general case that would be great.
|
2025-03-21T14:48:31.732477
| 2020-08-09T23:16:18 |
368758
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Connor Malin",
"Denis Nardin",
"Maxime Ramzi",
"Tim Campion",
"https://mathoverflow.net/users/102343",
"https://mathoverflow.net/users/102390",
"https://mathoverflow.net/users/134512",
"https://mathoverflow.net/users/2362",
"https://mathoverflow.net/users/43054",
"https://mathoverflow.net/users/43326",
"skd",
"user43326"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631933",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368758"
}
|
Stack Exchange
|
One periodic cohomology theories?
Is there a classification of one periodic cohomology theories? In other words, spaces $X$ such that $\Omega X \simeq X$? For example, $X=*$ and $X= K(G,0) \times K(G,1) \times \dots$ are examples. More generally, if $\{Y_n\}$ is an Omega spectrum $X=\dots \times Y_1 \times Y_0 \times \Omega Y_0 \times \dots$ is an example.
I ask because I have a space (related to stable block bundles) which does not seem like it should be contractible but appears to be its own loop space.
Is there any additional structure you're putting on these? I'm skeptical that there's a classification of such things without any additional conditions imposed. For what it's worth, a sort of "universal" example can be obtained as the Thom spectrum of the inclusion Z -> Pic(S^0) = Z x BGL_1(S^0). The connective cover of this spectrum is ΩS^2_+, and so it is an E_1-ring spectrum, but I don't think it has more structure than that. (That's one reason why 2-periodic things are nicer: the Thom spectrum of Z = Ω^2 CP^oo -> Ω^2 BU = BU x Z is an E_2-ring spectrum, whose connective cover is ΩS^3_+.)
@skd I wasn't aware of any nontrivial examples, so I mostly wanted to know if anyone was automatically boring. Interestingly, it did come up in conjunction with $Pic(S^0)$ which I don't know anything about. Do you have any reference for basic facts about $Pic(S^0)$. Particularly its relation to the J-homomorphism.
You can think of the J-homomorphism as the geometric realization of the (symmetric monoidal) functor N(Vect_R) -> Pic(S^0), where Vect_R is the category of finite-dimensional real vector spaces and linear isomorphisms, sending V to its one-point compactification S^V. There is a weak equivalence N(Vect_R) = Z x BO. Note that Pic(S^0) = Z x BGL_1(S^0), and the J-homomorphism sends the Z to Z (because the one-point compactification of R is S^1, the generator of the Picard group of Sp), and BO -> BGL_1(S^0) sends a bundle to the associated spherical fibration. Not sure of a reference, though.
@skd : aren't you supposed to group-complete N(Vect_R) before you get Z x BO ?
These "1-periodic spectra" are exactly the modules over $S[t^\pm1]$, the free associative ring spectrum on one invertible generator in degree 1. Dunno how helpful this observation is
@MaximeRamzi Yes, sorry about that. Note that Denis' associative ring spectrum is exactly the Thom spectrum in my first comment.
This doesn't help you to get a classification, but a simple way to construct a 1-periodic spectra is to take a 2-periodic spectrum $X$ and take wedge with its suspension.
The question reminds me of the fact that real-oriented cohomology theories are much less interesting than complex-oriented ones, owing to the fact that $MO$ is an $H\mathbb F_2$ module. I think maybe a reasonable precisification of the question would be to ask whether every 1-periodic spectrum is of the form described by @user43326 above. I think this is equivalent to asking something like whether $\mathbb S[t_1^{\pm 1}]$ splits as $\mathbb S[t_2^{\pm 1}] \vee\Sigma \mathbb S[t_2^{\pm 1}]$, where $t_i$ is a free generator in degree $i$.
|
2025-03-21T14:48:31.732701
| 2020-08-09T23:44:00 |
368759
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Branimir Ćaćić",
"https://mathoverflow.net/users/6999"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631934",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368759"
}
|
Stack Exchange
|
Is there a Swan-style description of topological K-homology?
A celebrated result of Swan [1] states that, on a compact Hausdorff space $X$, the category of finite rank complex vector bundles is equivalent to the category of finitely generated projective $\mathcal{C}(X)$-modules, where $\mathcal{C}(X)$ is the ring of complex-valued continuous functions on $X$ (analogous statements are true over the reals and the quaternions (loc.cit.), but these will not concern us here). One can look upon this in the following ways:
interesting geometric objects and their morphisms-- viz. complex vector bundles of finite rank and vector bundle maps -- have a description by interesting algebraic objects and their morphisms viz. finitely generated projective modules and their morphisms;
just the other way round: interesting algebraic objects and their morphisms have a description by interesting geometric objects and their morphisms;
as a synthesis of these views one obtains that a topological invariant, the complex $K$-theory $K(X)$ of $X$, has both a geometric description by complex vector bundles and an algebraic description by projective modules.
Pondering a while upon these issues the following questions come to mind:
The category of finite rank vector bundles is additive, but not abelian. This may lead one to consider linear fibrations over $X$ which are locally modelled on the kernels or cokernels of vector bundle maps of finite rank vector bundles; these have fibres which are vector spaces of possibly varying finite dimension.
To which algebraic objects do these correspond?
Are there further algebraic categories of $\mathcal{C}(X)$-modules with an interesting geometric description? In particular, do the finitely generated $\mathcal{C}(X)$-modules have an interesting geometric description?
Does the topological $K$-homology $K_*(X)$ of $X$ have interesting geometrical and algebraic descriptions? The $K$-theory of finiteley generated $\mathcal{C}(X)$-modules appears to fail here, since pushing forward does not seem to work.
[1] Swan, R.G. --
Vector bundles and projective modules. Trans. Amer. Math. Soc. 105, (1962), 264–277.
With regard to your third question, there's is a “geometric” picture of $K$-homology due to Baum and Douglas. If you don't mind a little analysis, though, the pairing between $K$-homology and $K$-theory can be beautifully understood in terms of the Atiyah–Singer index theorem and its various generalisations in index theory and noncommutative geometry.
|
2025-03-21T14:48:31.732889
| 2020-08-10T00:00:11 |
368760
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Lev Bahn",
"https://mathoverflow.net/users/127918"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631935",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368760"
}
|
Stack Exchange
|
Inequality regarding a probability measure
First of all, I am sorry for the ''not clear title' for this question but I cannot find a better way to describe this seemingly very simple and standard inequality,
So.. I am reading a paper 'Two-dimensional Navier-Stokes Equation Driven by a space time white noise' by Daprato and Debussche. And I came across an inequality regarding a probability measure.(It is in the proof of main theorem p.198) It seems very strandard but I cannot see why it is.
$$\mathbb{P}\left[ \sup_{t\in [0,T]} \left| u_N(t,u_0) \right|_{\mathcal{B}^\sigma_{p,\rho}} \geq M \right]\leq \sum_{k=0}^{[T/t^*_M]} \mathbb{P}\left[\sup_{t\in [kt^*_M,(k+1)t^*_M]} \left| u_N(t,u_0) \right|_{\mathcal{B}^\sigma_{p,\rho}}\geq M \right]$$
Here, $\mathcal{B}^\sigma_{p,\rho}$ is a Besov space and $\mathbb{P}$ is a probability measure.
I initially thought it is a typo and the $M$ in the right hand side should be replaced with $M/[T/t^*_M]$ but I realized that there is a possibility that I am missing something. So I wanted to hear from someone else.
I thank in advance for any help with this.
It looks fine to me.
If we let $I_k = [k t^\ast_M, (k+1) t^\ast_M]$ be the relevant subintervals of $[0,T]$, then the supremum of $|u_n|$ over $[0,T]$ must be almost attained along some sequence of points, and by pigeonhole infinitely many of them must be in one of the $I_k$, call it $I_{k_0}$, so that $\sup_{I_{k_0}} |u_n| = \sup_{[0,T]} |u_n|$. So if the sup over $[0,T]$ is at least $M$, then the sup over some $I_k$ must also be at least $M$ (and conversely). In other words, the event $\{\sup_{[0,T]} |u_n| \ge M\}$ equals the union of the events $A_k = \{\sup_{I_k} |u_n| \ge M\}$.
Now we just use the union bound.
Thank you for the answer! I got your point and it seems clear to me.
|
2025-03-21T14:48:31.733025
| 2020-08-10T00:12:57 |
368761
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Boby",
"Iosif Pinelis",
"https://mathoverflow.net/users/36721",
"https://mathoverflow.net/users/69661"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631936",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368761"
}
|
Stack Exchange
|
Bounds on cumulants in terms of moments
I am interested in finding bounds on cumulants in terms of moments.
For example, this paper alludes to the bound
\begin{align}
|\kappa_n|\le n^n E[|X-E[X]|^n]
\end{align}
where $\kappa_n$ is the $n$-th cumulant.
However, due to the language barrier, tracking down the proof of this is difficult.
I would like to see the proof of this bound or something similar to this.
Let $k_n:=\kappa_n$, $a_n:=E(X-EX)^n$, $b_n:=E|X-EX|^n$, so that $|a_n|\le b_n$. We have to show that
$$|k_n|\le n^n b_n$$
for natural $n$. For $n=1,2$ this is obvious. The key is the recursion
$$k_n=a_n-\sum_{m=1}^{n-1}\binom{n-1}{m-1}k_m a_{n-m}$$
at the end of this section , which implies
$$|k_n|\le b_n+\sum_{m=1}^{n-1}\binom{n-1}{m-1}|k_m| b_{n-m}.$$
This allows us to use the induction on $n$, which implies
$$|k_n|\le l_n b_n,$$
where
$$l_n:=1+\sum_{m=1}^{n-1}\binom{n-1}{m-1}m^m;$$
here we used the log-convexity of $b_j$ in $j$ together with the fact $b_0=1$ (or, equivalently, Hölder's inequality), which implies $b_m b_{n-m}\le b_n$.
It remains to show that $l_n\le n^n$ for $n\ge2$. This is trivial for $n=2$. For $n\ge3$, proceed again by induction:
\begin{align*}
l_n&=1+\sum_{m=1}^{n-1}\binom{n-2}{m-1}m^m+\sum_{m=1}^{n-1}\binom{n-2}{m-2}m^m \\
&=1+\sum_{m=1}^{n-2}\binom{n-2}{m-1}m^m+(n-1)^{n-1}+\sum_{m=2}^{n-1}\binom{n-2}{m-2}m^m \\
&=l_{n-1}+(n-1)^{n-1}+\sum_{j=1}^{n-2}\binom{n-2}{j-1}(j+1)^{j+1} \\
&\le l_{n-1}+(n-1)^{n-1}+\sum_{j=1}^{n-2}\binom{n-2}{j-1}j^j c_n \\
&=l_{n-1}+(n-1)^{n-1}+(l_{n-1}-1) c_n \\
&\le (n-1)^{n-1}+(n-1)^{n-1}+(n-1)^{n-1}c_n \\
&=(n-1)^{n-1}\Big(2+\frac{(n-1)^{n-1}}{(n-2)^{n-2}}\Big)=:r_n
\end{align*}
where
$$c_n:=\max_{1\le j\le n-2}\frac{(j+1)^{j+1}}{j^j}=\frac{(n-1)^{n-1}}{(n-2)^{n-2}},$$
because
$$\frac{(j+1)^{j+1}}{j^j}=\Big(1+\frac1j\Big)^j(j+1)$$
is increasing in $j\ge1$.
It remains to check that $r_n\le n^n$ for all real $n>2$, which, by substitution $n=1+\frac1t$, can be rewritten as
\begin{equation}
f_1(t)f_2(t)\le1, \tag{1}
\end{equation}
where
\begin{equation*}
f_1(t):=e \left(\frac{1}{t+1}\right)^{\frac{1}{t}+1},\quad
f_2(t):=\frac{\left(\left(\frac{1}{t}\right)^{\frac{1}{t}}
\left(\frac{1}{t}-1\right)^{\frac{t-1}{t}}+2\right) t}{e};
\end{equation*}
everywhere here, $t\in(0,1]$.
To prove (1), it is enough to prove
\begin{equation}
f_1\le g_1 \tag{2},
\end{equation}
\begin{equation}
f_2\le g_2 \tag{3},
\end{equation}
\begin{equation}
g_1g_2\le1 \tag{4}
\end{equation}
on $(0,1]$, where
\begin{equation*}
g_1(t):=\frac{7 t^2}{24}-\frac{t}{2}+1,\quad
g_2(t):=\left(\frac{2}{e}-\frac{1}{2}\right) t+1.
\end{equation*}
To prove (2), consider
\begin{equation*}
d_1:=\ln f_1-\ln g_1
\end{equation*}
and then $d_{11}(t):=t^2d_1'(t)$ and $d_{11}'$, which latter is a rational function, which is easily seen to be $<0$. So, $d_{11}$ decreases. Also, $d_{11}(0+)=0$. So, $d_{11}<0$ (on $(0,1]$) and hence $d_1$ decreases. Also, $d_1(0+)=0$. So, $d_1<0$, which yields (2).
To prove (3), rewrite it as
\begin{equation*}
d_2(t):=\ln \left(\left(\frac{1}{t}-1\right)^{\frac{t-1}{t}}
\left(\frac{1}{t}\right)^{\frac{1}{t}}\right)-\ln \left(\frac{e (2-t)}{2 t}\right)\le0
\end{equation*}
(for $t\in(0,1]$).
Consider then $d_{21}(t):=t^2d_2'(t)$ and $d_{21}'$, which latter is a rational function, which is easily seen to be $<0$ on $(0,1)$. So, $d_{21}$ decreases. Also, $d_{21}(0+)=0$. So, $d_{21}<0$ (on $(0,1]$) and hence $d_2$ decreases. Also, $d_2(0+)=0$. So, $d_2<0$, which yields (3).
Finally, (4) is elementary, since $g_1g_2$ is a polynomial (of degree $3$). $\Box$
Thanks. Very nice. Did you see this proof before?
@Boby : No, I have not seen a proof of this. I tried to find one, but was unable to. Do you know where the original proof of this can be found, in whatever language?
I found a remark about this in the paper the I linked. They refer to the paper Yu. V. Prokhorov and Yu. A. Rozanov, Probability Theory [in Russian], Nauka, Moscow (1973)
Yu. V. Prokhorov and Yu. A. Rozanov, Probability Theory [in Russian], Nauka, Moscow (1973) is a book. There is no proof of this bound or reference to a proof there.
In the paper, the author say this ``..one can only find implicit inequalities (see, for example, [8]):" where refence [8] is the book.
|
2025-03-21T14:48:31.733254
| 2020-08-10T05:49:48 |
368768
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Asaf Karagila",
"Dominic van der Zypen",
"Haim",
"Noah Schweber",
"bof",
"https://mathoverflow.net/users/37613",
"https://mathoverflow.net/users/43266",
"https://mathoverflow.net/users/7206",
"https://mathoverflow.net/users/8133",
"https://mathoverflow.net/users/8628"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631937",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368768"
}
|
Stack Exchange
|
Does every maximal almost disjoint family have the same chromatic number?
If $H=(V,E)$ is a hypergraph such that $V\neq\varnothing\neq E$ and $|e| > 1$ for all $e\in E$, and $\kappa\neq\varnothing$ is a cardinal, we say that a map $c:V\to\kappa$ is a coloring if the restriction $c\restriction_e: e\to \kappa$ is non-constant for each $e\in E$. We denote by $\chi(H)$ the smallest cardinal $\kappa$ such that there is a coloring $c:V \to \kappa$.
Assume the Axiom of Choice. If ${\cal A, B}$ are infinite maximal almost disjoint families on $\omega$, do we necessarily have $\chi((\omega, {\cal A}))=\chi((\omega,{\cal B}))$?
If I understand the definitions here, the members of the MAD family are the hyperedges, and the vertices are integers. So clearly there is a coloring in $\aleph_0$ colors, so it's really about asking it you can have a MAD family with finite chromatic number, which to me sounds like that it's impossible.
@AsafKaragila I think it does happen - in fact, any MAD family can be turned into one with chromatic number $2$. Suppose $\mathfrak{X}\subset\mathcal{P}(\omega)$ is MAD; I claim that the new family $$\mathfrak{X}[2]:={{2a: a\in A}\cup{2a+1: a\in A}: A\in\mathfrak{X}}$$ is also mad. Suppose $Y\subseteq\omega$ is an infinite set such that $Y\not\in\mathfrak{X}[2]$ but $Y\cap B$ is finite for each $B\in\mathfrak{X}[2]$. One of $Y\cap{Evens}$ and $Y\cap{Odds}$ is infinite; WLOG, it's the former. Then ${x: 2x\in Y}$ is a counterexample to the MADness of $\mathfrak{X}$. Or did I mess up?
@DominicvanderZypen Well even if I'm right, my comment doesn't answer the question.
Thanks @AsafKaragila for your comment, I edited the question accordingly to make proper hypergraphs. - Noah thanks for your comments, will read them after walking the kids to school
I think this paper by Erdos and Shelah may have the answer to your question: https://shelah.logic.at/files/95606/19.pdf
Right @bof thanks for your addendum,will amend
It looks like a completely separable mad family can’t have a finite chromatic number: otherwise, there is a color n such that the set X of integers getting that color is in the coideal generated by the family, and by complete separability, X contains a monochromatic member of the mad family.
@Haim Could you be so kind as to have a look at that Erdos-Shelah paper and tell us whether and how they answer the current Question, perhaps in the form of an Answer?
@bof I'll hopefully find some time later today or tomorrow to have a closer look at the Erdos-Shelah paper. Regardless of their paper, the following seems like an interesting question to me: does the non-existence of a 2-coloring imply complete separability? the definitions look very close to each other, and a positive answer will imply the equivalence of complete separability, the non-existence of a 2-coloring and the non-existence of a finite coloring.
@bof It looks like Theorem 1.1 in the Erdos-Shelah paper gives exactly what you said, a mad family that is $(n+1)$-colorable but not $n$-colorable (which also provides a negative answer to my question). I'm still not completely sure what the definition of $n$-separability is, but the construction in their proof obviously gives what we want.
A negative answer to the question follows by the proof of Theorem 1.1 in the following paper of Erdős and Shelah, where for every $n<\omega$ they construct a mad family that is $(n+1)$-colorable but not $n$-colorable:
Erdős, Paul; Shelah, Saharon, Separability properties of almost-disjoint families of sets, Isr. J. Math. 12, 207-214 (1972). ZBL0246.05002.
It should also be noted that a completely separable mad family can't have a finite chromatic number: given any finite coloring, there is a color $n$ such that the set $x$ of integers getting that color belongs to the coideal generated by the mad family. By complete separability, $x$ contains a monochromatic member of the family. I won't be surprised if there are also $ZFC$ constructions of such families, but I haven't thought about it enough.
Suppose you partition $\omega$ into infinitely many infinite sets $A_n$ and on each $A_n$ you construct a MAD family with chromatic number $n$, then the union will be an AD family with chromatic number $\aleph_0$, and if you extend it to a maximal AD family, the chromatic number won't get any smaller, isn't that right?
Oops! I corrected the answer now. Yes, the Erdos-Shelah construction is done in ZFC. I didn't read the entire paper, but just going through the statements of the theorems (and keeping in mind that we don't know what n-separability is), it doesn't look like they construct a mad family with an infinite chromatic number (the more I think about it, the less trivial it seems. Even a consistency result would be nice). One of the reasons that I write "mad" is to reflect the fact that we use the Hebrew word for "mad/crazy" when we talk about those families. :)
And I suppose "mad" families are related somehow to "happy" families, which I've heard of before though I don't remember the definition.
@bof It looks like your suggested construction works. Indeed, mad families are related to happy families: the coideal corresponding to an infinite almost disjoint family is a happy family. Both terms were introduced by Mathias (though, sadly, he chose to write "MAD" instead of "mad" in his "Happy families" paper).
|
2025-03-21T14:48:31.733729
| 2020-08-10T08:41:15 |
368776
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"AlexArvanitakis",
"David E Speyer",
"G. Gallego",
"Sebastian Goette",
"https://mathoverflow.net/users/143492",
"https://mathoverflow.net/users/22757",
"https://mathoverflow.net/users/297",
"https://mathoverflow.net/users/70808"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631938",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368776"
}
|
Stack Exchange
|
Different definitions for integral de Rham cohomology classes
Suppose that $S$ is a compact orientable surface. In this case, the top de Rham cohomology space $H^2(S)\cong \mathbb{R}$, with the isomorphism given by integration on $2$-forms along $S$.
Now, one can define integral cohomology classes as those cohomology classes $a$ so that $\int_S a \in \mathbb{Z}$. On the other hand, one can also define integral cohomology classes as those classes corresponding to integral Cech cohomology classes $\check{H}^2(S,\underline{\mathbb{Z}})$ in the following way:
If $\omega$ is a closed $2$-form, we can find an open cover $\mathfrak{U}$ of $S$ and $1$-forms $\alpha_U$ on each $U$ so that $\omega|_U = d\alpha_U$. Now, choose functions $f_{UV}$ so that $df_{UV} = \alpha_U - \alpha_V$. The cocycle
$$
f_{UVW} = f_{UV} + f_{VW} - f_{UW}
$$
satisfies that $df_{UVW}=0$, so $f_{UVW} \in \check{H}^2(S,\underline{\mathbb{R}})$.
With this in mind, I claim that $[\omega]$ is integral if and only if one can choose $(\alpha_U)$ and $(f_{UV})$ such that $(f_{UVW}) \in \check{H}^2(S,\underline{\mathbb{Z}})$.
I want to see why these two definitions coincide. More precisely, I would like to see an explicit proof on why, given that $\int_S \omega \in \mathbb{Z}$, I can choose the $f_{UV}$ so that $f_{UVW} \in \mathbb{Z}$. Please, I prefer an explicit proof of this fact, rather than invoking Poincaré Duality/de Rham theorem.
The statement looks wrong: maybe you want $[\omega]$ integral if and only if the data above can be chosen such that $f_{UVW}$ is integral. Otherwise, adding some constant to one of the $f_{UV}$ easily destroys integrality. More important: you seem to ask for a proof of a special case of de Rham's theorem here, but you want to avoid the theorem itself. I would suggest to have a look into the book by Bott and Tu, where de Rham's theorem is explained, if you have a copy at hand.
Exactly, in the last paragraph I explain that what I want to know is why if $\omega$ is integral the $f_{UV}$ can be chosen so that $f_{UVW}$ is an integer. I think this already gives a proof of de Rham’s, since you can recover $\omega$ from the $f_{UVW}$ using a partition of unity. The problem is how to deal with integrality. I already looked at Boot-Tu and I do not think they do that there (though maybe I should look again). For context, this question came to me while studying prequantization in Woodhouse’s book on Geometric Quantization.
This fact is also claimed in Kobayashi’s paper on “Principal fibre bundles with the 1 dimensional toroidal group” (page 35) and in this post on Math StackExchange https://math.stackexchange.com/questions/6099/integral-classes-in-de-rham-cohomology (The comment by David Bar Moshe. My question is essentially the same as the one by Juan Rojo).
$\def\RR{\mathbb{R}}\def\ZZ{\mathbb{Z}}$I've considered assigning this when I've taught sheaf cohomology but it always seemed a little too hard. Let's see if I can do it. I'll be a little more general while I am at it and do the case of a smooth compact oriented $n$-fold. Choose a triangulation $S$ of the $n$-fold; let $F_j$ be the set of $j$-dimensional faces.
For each vertex $u \in F_0$, let $U(u)$ be the star shaped open neighborhood of $u$ as in the OP's answer. The $U(u)$ give an open cover of $X$. For any $u_0$, $u_1$, ..., $u_j$ in $F_0$, the intersection $U(u_0) \cap \cdots \cap U(u_j)$ is empty if $(u_0, \dots, u_j)$ are not the vertices of a face, and this intersection is a contractible open set which I'll call $U(\sigma)$ if $(u_0, \dots, u_j)$ are the vertices of a face $\sigma$ in $F_j$. Thus, the Cech complex of $\underline{\RR}$ is identified with the simplicial cohomology complex
$$\RR^{F_0} \to \RR^{F_1} \to \cdots \RR^{F_{n-1}} \overset{d_{n-1}}{\longrightarrow} \RR^{F_n}.$$
For any $n-1$ dimensional face $\tau$, there are two $n$-faces $\sigma_1$ and $\sigma_2$ containing $\tau$. Letting $e_{\tau}$ be the basis function corresponding to $\tau$, we have $d_{n-1}(e_{\tau}) = e_{\sigma_1} - e_{\sigma_2}$. (I am being sloppy about signs, but the fact that we are on an oriented manifold will make it all work out in the end.) So (using that our manifold is connected) the cokernel of $d_{n-1}$ is clearly $\RR$, and an explicit map from $\RR^{F_n}$ to the cokernel sends a function $f \in \RR^{F_n}$ to $\sum_{\sigma \in F_n} f(\sigma)$.
Let $\Omega^p$ be the sheaf of smooth $p$-forms, and let $Z^p$ be the subsheaf of closed $p$-forms. Note that $Z^0 = \underline{\RR}$, so we have just computed that $H^n(X, Z^0) \cong \RR$. The Poincare lemma gives short exact sequences $Z^p \to \Omega^p \to Z^{p+1}$ for $0 \leq p \leq n$, so we get boundary maps
$$H^0(X, Z^n) \to H^1(X, Z^{n-1}) \to \cdots \to H^n(X, Z^0) \cong \RR.\quad (\ast)$$
In the case of a surface, the OP has given explicit descriptions of these maps in his answer.
By the usual argument with partitions of unity, $H^q(X, \Omega^p)$ vanishes for $q>0$, so all these maps are isomorphisms except the first one. The first map, in turn, is surjective with kernel $d H^0(X, \Omega^{n-1})$. So the image of the first map is $H^n_{DR}(X)$, and all the other $H^q(X, Z^{n-q})$ are isomorphic to $H^n_{DR}(X)$. Our goal, given an $n$-form $\omega$, is to show that the composition of all these maps gives $\int_X \omega$.
Note that a class in $H^q(X, Z^{n-q})$ is given by a Cech representative $( \eta_{\sigma} )_{\sigma \in F_q}$, where $\eta_{\sigma}$ is a closed $(n-q)$-form on $U(\sigma)$.
Choose a regular CW subdivision $S^{\perp}$ of $X$ dual to the triangulation. That means the poset of faces of $S^{\perp}$ is dual to that of $S$ and each $j$-face $\sigma$ in $S$ crosses the dual $n-j$ face $\sigma^{\perp}$ transversely in one point. An explicit way to do this is to take the barycentric subdivision of $S$ and draw the "obvious" dual faces. If we choose an ordering of $F_0$, that gives an orientation to every face $\sigma$ of $F_q$, and then we can use the global orientation of $X$ to orient $\sigma^{\perp}$.
I claim that the composite isomorphism $(\ast)$ from $H^{q}(X, Z^{n-q})$ sends $(\eta_{\sigma})_{\sigma \in F_q}$ to
$$\sum_{\sigma \in F_q} \int_{\sigma^{\perp}} \eta_{\sigma}.$$
Let's see what this means for $q=n$. Each $\eta_{\sigma}$ is a closed $0$-form on $U(\sigma)$. A closed $0$-form is locally constant function and $U(\sigma)$ is connected, so we just have a real number for each $\sigma$ in $F_n$ and we can thus think of $\eta$ as a vector in $\RR^{F_n}$. Each $\sigma^{\perp}$ is just a point in the interior of $\sigma$. So we are just summing up the values of $\eta$ on the $n$-faces, and this is the map $\RR^{F_n} \to \RR$ that we described before.
Let's next see what this means for $q=0$. Each $\eta_{\sigma}$ is an $n$-form on $\sigma$, and the condition that $(\eta_{\sigma})$ is a Cech co-cycle says that $\eta_{\sigma}$ is the restriction of a global $n$-form $\omega$ on $X$. The $n$-faces $\sigma^{\perp}$, for $\sigma \in F_0$, partition $X$. So
$$\sum_{\sigma \in F_q} \int_{\sigma^{\perp}} \eta_{\sigma} = \sum_{\sigma \in F_q} \int_{\sigma^{\perp}} \omega|_{\sigma^{\perp}} = \int_X \omega.$$
Thus, we just need to show that, if $(\eta_{\sigma})$ represents a class in $H^q(X, Z^{n-q})$ and $\delta_q$ is the boundary map $H^q(X, Z^{n-q}) \to H^{q+1}(X, Z^{n-q-1})$, then
$$\sum_{\sigma \in F_q} \int_{\sigma^{\perp}} \eta_{\sigma} = \sum_{\tau \in F_{q+1}} \int_{\tau^{\perp}} \delta_q(\eta)_{\tau}. \quad (\dagger)$$
Whew! Okay, let's remember how the boundary map in sheaf cohomology works. Let $(\eta_{\sigma})$ be a cocycle for $H^q(X, Z^p)$. Since each $\sigma$ is contractible, we can lift each $\eta_{\sigma}$ to a $p-1$ form $\theta_{\sigma}$ with $d(\theta_{\sigma}) = \eta_{\sigma}$. Let $\tau$ be a $q+1$ face of our triangulation. Then
$$\delta_q(\eta)_{\tau} = \sum_{\sigma \subset \tau} \pm \theta_{\sigma},$$ where the sign involves the relative orientation of $\sigma$ and $\tau$.
We want to show $(\dagger)$. Plugging in the above description of the Cech co-boundary, the right hand side is
$$\sum_{\tau \in F_{q+1}} \int_{\tau^{\perp}} \sum_{\sigma \subset \tau} \pm \theta_{\sigma}.$$
Pulling the sum out of the integral and switching order of summation, we have
$$\sum_{\sigma \in F_q} \sum_{\tau \supset \sigma} \int_{\tau^{\perp}} \pm \theta_{\sigma}.\quad (\heartsuit)$$
Now, the subdivisions $S$ and $S^{\perp}$ are dual, so $\tau \supset \sigma$ if and only if $\sigma^{\perp} \subset \tau^{\perp}$ or, in other words, $\tau^{\perp} \subset \partial(\sigma^{\perp})$. All the signs work out perfectly, so that $(\heartsuit)$ is
$$ \sum_{\sigma \in F_q} \int_{\partial(\sigma^{\perp})} \theta_{\sigma}.$$
By Stokes' theorem,
$$\int_{\partial(\sigma^{\perp})} \theta_{\sigma} = \int_{\sigma^{\perp}} d(\theta_{\sigma}) = \int_{\sigma^{\perp}} \eta_\sigma.$$
We have now recovered the left hand side of $(\dagger)$.
The OP only asked for top cohomology, but I think other cohomological degrees are similar. Once again, we have maps
$$H^0(X, Z^k) \to H^1(X, Z^{k-1}) \to \cdots \to H^k(X, Z^0)$$
giving isomorphisms
$$H^k_{DR}(X) \cong H^1(X, Z^{k-1}) \cong \cdots \cong H^k(X, Z^0) \cong H^k(X, \RR). \quad (\diamondsuit)$$
We'd like to know that a class $\omega$ in $H^k_{DR}(X)$ is represented by a class in $H^k(X, \ZZ)$ if and only if $\omega$ pairs to an integer against every integer chain in $H_k(X, \ZZ)$; it is enough to test against chains coming from the triangulation $S$. Let $c = \sum_{\rho \in F_k} c_{\rho} \rho$ be a $k$-chain. We want to how to pair all the spaces in $(\diamondsuit)$ against $c$. Let $\eta$ be a $q$-cocycle for $Z^{k-q}$. I believe the same argument as before shows that $\langle c, \eta \rangle$ is
$$\sum_{\sigma \in F_q} c_{\rho} \sum_{\rho \in F_k} \int_{\sigma^{\perp} \cap \rho} \eta_{\sigma}. $$
In particular, if $q=k$, then $\sigma^{\perp} \cap \rho$ is a single point when $\rho = \sigma$ and otherwise $0$. So, viewing the Cech cohomology $H^k(X,\mathbb{R})$ as the cohomology of
$$\RR^{F_0} \to \RR^{F_1} \to \cdots \RR^{F_{n-1}} \overset{d_{n-1}}{\longrightarrow} \RR^{F_n}$$
and the simplicial cohomology $H_k(X, \ZZ)$ as the homology of
$$\ZZ^{F_0} \leftarrow \ZZ^{F_1} \leftarrow \cdots \leftarrow \ZZ^{F_n},$$
the pairing between $H^k(X,\mathbb{R})$ and $H_k(X, \ZZ)$ is induced by the obvious pairing between $\RR^{F_k}$ and $\ZZ^{F_k}$.
We then want to show that, if a cocycle in $\RR^{F_k}$ pairs integrally against all cycles in $\ZZ^{F_k}$, then that cocyle is cohomologous to one in $\ZZ^{F_k}$. That sounds like some easy linear algebra, although I don't see a one line proof.
Awesome answer! Just one little question. If we call $\psi$ the map sending an element in $\mathbb{R}^{F_n}$ to the sum of all the faces, it is obvious that $\mathbb{im} d_{n-1} \subset \ker \psi$, is the other inclusion also true? I guess the two are isomorphic, but are they equal? I haven’t been able to construct a preimage for $d_{n-1}$ given that the sum is 0. It is not obvious to me.
Ok, I guess page 41 in Munkres answers my question.
Glad you found an answer! Let me point out that you asked to avoid Poincare duality, but it was almost impossible to involve proving it: The same complex computes $H_k$ with respect to $S$ and $H^{n-k}$ with respect to $S^{\perp}$. So the Munkres reference talks about $H_0$, but the exact same computation comes up in your setting where you wanted to compute $H^n$.
I think I have an answer, although there is one step that still bugs me.
First one has to pick a triangulation of $S$. Let us call $V$, $E$ and $F$ the sets of vertices, edges and faces. By choosing now the open covering $\mathfrak{U}$ associated to this triangulation (i.e. the one given by the "stars" of the vertices, see p. 42 at Griffiths-Harris), we get that, while the open sets are in bijection with the vertices, the double intersections correspond to the edges and the triple to the faces.
Thus, in this case the Cech complex looks like this
$$
\mathbb{R}^{|V|} \overset{d_1}{\longrightarrow} \mathbb{R}^{|E|} \overset{d_2}{\longrightarrow} \mathbb{R}^{|F|},
$$
where $d_1( (k_v)_{v \in V} ) = (k_{v} - k_{w})_{vw\in E}$ and $d_2( (k_{vw})_{vw\in E} ) = (k_{vw} + k_{wu} - k_{vu})_{vwu \in F} .$
By carefully writing the matrix of $d_2$ one gets that $\mathrm{coker} d_2 \cong \mathbb{R}$. On the other hand, if one defines $\psi: \mathbb{R}^{|F|} \rightarrow \mathbb{R}$ in such a way that
$$
\psi( (k_{vwu})_{vwu \in F} ) = \sum_{vwu \in F} k_{vwu},
$$
one checks that $\mathrm{im} d_2 \subset \ker \psi$ and $\ker \psi \neq \mathbb{R}^{|F|}$, so $\ker \psi = \mathrm{im} d_2$. What this implies is that $\psi$ gives an isomorphism
$$
\psi: H^2(\mathfrak{U},\underline{\mathbb{R}}) \longrightarrow \mathbb{R}.
$$
The key now is to find an explicit isomorphism
$$
I: H^2(S,\mathbb{R}) \longrightarrow H^2(\mathfrak{U}, \underline{\mathbb{R}})
$$
so that $\psi(I(a)) = \int_S a$.
This would show that the $a$ with $\int_S a \in \mathbb{Z}$ are precisely those that $I(a) \in H^2(\mathfrak{U}, \underline{\mathbb{Z}})$.
I think that now the way to construct explicitly the isomorphism $I$ is like in Woodhouse's "Geometric Quantization" (A6) and also in Weil's paper "Sur les theoremes de de Rham". On one hand, the way one gets a cocycle from a $2$-form is like in the statement of my question. On the other hand, the way to recover $\omega$ from the $f_{UVW}$ is by defining
$$
\omega_f = \sum_{U,V,W} f_{UVW} h_{W} dh_U \wedge dh_V,
$$
where the $h_U$ are a partition on unity on the covering $\mathfrak{U}$. In Woodhouse it is shown that $[\omega_f]=[\omega]$.
However, I still do not understand why (or ever if it is true that) $\int_S \omega_f = \sum_{U,V,W} f_{UVW}$.
(BTW, also check out this question: https://mathoverflow.net/questions/329836/integral-of-top-forms-in-terms-of-Čech-representative ).
I'd think using Stokes' patchwise repeatedly should work, no? There is also the question of normalisation which will be fixed by the same calculation.
@AlexArvanitakis Well, I think that should be the idea, but to be honest I don’t know how to do it exactly...
I think it's a bit fiddly in general (in cases your cover is misbehaving). Try doing a sphere first with the obvious cover (with exactly three (EDIT: two + the overlap) open sets)
Well, in my case is for a triangulation, which is fairly simple. I am trying it for a tetrahedron with no success. The cover you suggest has no triple intersections and thus no 2-cocycles.
indeed actually that's a bad one--noncontractible, sorry (so in fact $\omega$ would not be exact on each open set). You can get a contractible one with, I count, 4 open sets
|
2025-03-21T14:48:31.734676
| 2020-08-10T09:15:21 |
368779
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"ABIM",
"RobPratt",
"https://mathoverflow.net/users/141766",
"https://mathoverflow.net/users/36886"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631939",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368779"
}
|
Stack Exchange
|
Optimal Directed Graph with Maximum Number of Interactions
Definitions:
The number of interactions of a directed graph $(V,E)$, denoted by $\operatorname{inter}(V,E)$ is defined by
$$
\operatorname{inter}(V,E) = \sum_{e,f\in V} I_{\gamma{[e,f]}},
$$
where $I_{\gamma[e,f]}=1$ if there exists a path in $(V,E)$ beginning at $e$ and ending at $f$, and takes value $0$ otherwise.
We say that two elements $e,f\in V$ are connected if there is a path beginning at $e$ and ending at $f$; i.e.: $I_{[e,f]}=1$.
Question:
Let $V_1,V_2$ be non-empty finite sets of the same cardinalty and $N\geq |V|$ be a fixed positive integer. If we're allowed to add nodes to $V_1\cup V_2$ (so suppose $V_1\cup V_2$ is contained in a very large but finite set $V$) how can we construct the graph $(V^{\star},E^{\star})$ with:
Exactly N edges,
Every member of $V_1$ is connected to some member of $V_2$,
There is no path from $V_2$ to $V_1$,
If a graph $(V,E)$ satisfies $1-3$ then
$$
\operatorname{inter}(V,E) \leq \operatorname{inter}(V^{\star},E^{\star})
.
$$
Apparently, $e$ and $f$ are nodes, so $E$ is a node set. Is $V$ an edge set? Normally, the symbols $V$ and $E$ have the reverse meanings.
Oh I didn't know. I fixed this. But yes
$V$ for vertex, and $E$ for edge. Also sometimes $N$ for node, and $A$ for arc.
If you are allowed to modify $V_1$ and $V_2$, it is better to give the new ones different names, like $V_1^$ and $V_2^$. Do you require $|V_1^|=|V_2^|$? What is the relationship between $E$ and $E^*$?
Oh no we cannot modify $V_1$ and $V_2$ but only $V$. The goal is to place N edges between (possibly more vertices) connecting $V_1$ to $V_2$. So A priori there is no relationship between $E$ and $E^{\star}$ other than they both connect $V_1$ and $V_2$.
|
2025-03-21T14:48:31.734845
| 2020-08-10T09:19:08 |
368780
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Dirk",
"Mark L. Stone",
"Shijie Pan",
"fedja",
"https://mathoverflow.net/users/1131",
"https://mathoverflow.net/users/163558",
"https://mathoverflow.net/users/75420",
"https://mathoverflow.net/users/9652"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631940",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368780"
}
|
Stack Exchange
|
What is the convergence rate of the iterative optimization method?
For the following optimization problem:
$$
\mathrm{min}_{A,B} \|I-A^{T}XB\|^2 + \lambda\|B\|^2,
$$
where $A$ and $B$ are the two variables ($\|A\|^2 \le \rho$ where $\rho$ is a constant, e.g. 1), the paper of He uses a method that iteratively computing A and B to solve it.
Here is the method:
Denote the objective function by $\phi$.
STEP 1:
Suppose $A$ is given. By requiring the gradient of
$\phi$ respect to B vanish, we have
$$
\frac{\partial \phi}{\partial B}=0 \rightarrow B=(X^TAA^TX+\lambda I)^{-1}X^TA.
$$
STEP 2:
Suppose $B$ is given. By requiring the gradient of
$\phi$ respect to A vanish, we have
$$
\frac{\partial \phi}{\partial A}=0 \rightarrow A=(XBB^TX)^{-1}XB.
$$
Let $\|A\|^2 \le \rho$ by deviding by a constant.
Thus by iteratively executing STEP 1 and STEP 2 until convergence, we can solve the optimization problem.
The question is what the rate of convergence?
This is an instance of block coordinate descent.
This problem is definitely not convex. But the step 1 and step 2 problems in which one of A or B is fixed are convex.
@Dirk wrote "This is an instance of block coordinate descent". Well, maybe of something called block coordinate descent, except I don't think this algorithm necessarily even descends, i.e., false advertising..
This is all very strange. Consider $1\times 1$ matrices (a.k.a. numbers) with $x=\lambda=1$. You want to minimize $(1-ab)^2+b^2$ but it is the classical example of a polynomial whose infimum is $0$ but is not attained, so what "rate of convergence" are we talking about? In general, multiply $A$ by a huge number and divide $B$ by the same number to get a better pair from any given pair.
@fedja Thanks for your insightful comment. I miss a constrain in the variable $A$. And I have corrected it and fixed the description of the solution.
@Mark L. Stone thanks for your helpful comments. This problem is not convex, I have fixed the description.
What is the basis for your claim that iteratively executing STEP 1 and STEP 2 results in convergence to anything, let alone to a global, or even local, minimum of your problem? Your algorithm does not employ any safeguards, such as line search or trust region, in order to guarantee convergence to a stationary point from an arbitrary starting point. Perhaps that is addressed in the paper, but I don't have access to it. Analysis of rate of convergence is premature until convergence has been established.
The $A$ update as written now is outright ridiculous. It does not diminish the value of the functional in general. Check once more that you haven't misunderstood anything in the paper. The scheme, as you presented it now, does converge but to something that has very little to do with the minimizer.
@fedja Thanks for your helpful comments. Yes, the method mentioned above seems to lack some significant proof to ensure it finally converges to the minimum. You mentioned that it does converge. If we don't care about the position of convergence, we only care about the rate of convergence, can you give me some advice?
@MarkL.Stone Thanks for your helpful comments. I agree with you but cannot find the answer in the original paper. If I only care about the rate of convergence, but not care about where it converges to, can you give me some advice?
What is the rate of convergence defined to be when it doesn't converge? My suggestion is to worry first about whether it converges, and to the solution,, not only in idealized theory, but in practice, with all the vagaries of finite precision floating point computation. Many papers are rather cavalier about this. Very few algorithms in published papers, even in top tier journals, actually work worth a damn in practice. If you want to make this algorithm reliable and robust, use a line search or trust region, which are absent from your description.
I guess you can talk about rate of convergence to the wrong answer. That's like an airline or railroad bragging that we now get you quicker to the wrong place.
@MarkL.Stone Thanks. I guess maybe converge to the "wrong answer" also makes sense, since we can easily determine the solution is whether a local minimum or local maximum.
The answer might be neither kocal min or local max. Could be a saddle point. or maybe even something else?
@Mark L.Stone The original paper hasn't discussed that the answer is a local min or local max, or even a saddle point. I think all three types of points are possible.
|
2025-03-21T14:48:31.735164
| 2020-08-10T10:28:01 |
368787
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gerry Myerson",
"Mare",
"https://mathoverflow.net/users/158000",
"https://mathoverflow.net/users/61949"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631941",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368787"
}
|
Stack Exchange
|
Finite fast tests for periodicity of certain matrices
Let $M=- U^{-1} U^T$ be an $n \times n$-integer matrix, where $U$ is an upper triangular 0-1-matrix where all diagonal entries are equal to one. $M$ is called periodic if $M^r=id$ for some $r \geq 1$.
The question is about whether there is a fast finite test to check whether a matrix is periodic, so that one just has to look for "small" $r$. See Distributive lattices with periodic Coxeter matrix for a motivation.
Question 1: Given $n$, is there a good bound for the minimal number $a(n)$ such that if $M^r \neq id$ for all $r=1,...,a(n)$, then $M$ is not periodic?
Question 2: Given $n$, is there a good bound for the minimal number $b(n)$ such that if $M$ has an entry of absolute value at least $b(n)$, then $M$ is not periodic?
Of course one might ask those questions also for integer matrices $M$ with more general properties. (there it does not work for question 2 as Gerry Myerson showed)
Given $n$, there are only finitely many $n\times n$ 0-1 matrices $U$, hence only finitely many candidate matrices $M$. A fortiori, the minimal numbers $a(n)$ and $b(n)$ exist.
@GerryMyerson Thanks. Right, I first formulated my questions for general integer matrices $M$ but then specified it. I changed the question.
For general integer matrices $M$, there is no $b(n)$. E.g., for $n=2$, if $a(a+1)+1=bc$, and $$M=\pmatrix{a&-b\cr c&-a-1\cr}$$ then $M^3$ is the identity matrix.
EDIT: for general matrices, Theorem 2.7 of James Kuzmanovich and Andrey Pavlichenkov, Finite groups of matrices whose entries are integers, The American Mathematical Monthly Vol. 109, No. 2 (Feb., 2002), pp. 173-186 shows there's a bound on $a(n)$ in terms of $n$. Let $m=p_1^{e_1}p_2^{e_2}\cdots p_t^{e_t}$ with $p_1<p_2<\cdots<p_t$. Then there is an $n\times n$ integer matrix with order $m$ if and only if
$\sum_{i=1}^t(p_i-1)p_i^{e_i-1}-1\le n$ for $p_1^{e_1}=2$, or
$\sum_{i=1}^t(p_i-1)p_i^{e_i-1}\le n$ otherwise.
|
2025-03-21T14:48:31.735334
| 2020-08-10T11:14:11 |
368790
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631942",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368790"
}
|
Stack Exchange
|
Isomorphism preserving transformation graph to graph of logarithmic boolean width and bounded degeneracy
The paper On graph classes with logarithmic boolean-width
claims that some graph problems are fixed parameter tractable with parameter
the boolean width.
In particular, boolean-width of the complement of $k$-degenerate graph
is at most $k\log(n)$.
We got isomorphism preserving transformation
graph $G$ to graph $G_m$ of logarithmic boolean width: set $G'_m$ to be the $m$-subdivision
(i.e. subdivide each edge $m$ times) and set $G_m$ the complement
of $G'_m$. $G'_m$ is $2$-degenerate.
$G_m$ is of logarithmic boolean width since its
order is $O(m n^2)$ and its complement is $2$-degenerate.
For odd $m$, $G_m$ can be partitioned in two cliques (it is co-bipartite).
What is the intuition about so low boolean width for all transformed graphs?
Are there islands of tractable problems on $G_m$ related to problems on $G$?
Are there graph classes for which the transformation gives bounded
boolean width?
|
2025-03-21T14:48:31.735442
| 2020-08-10T11:23:06 |
368792
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"A Z",
"Amir Sagiv",
"Gerry Myerson",
"Mr. Palomar",
"Pierre PC",
"https://mathoverflow.net/users/129074",
"https://mathoverflow.net/users/151698",
"https://mathoverflow.net/users/158000",
"https://mathoverflow.net/users/163415",
"https://mathoverflow.net/users/42864"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631943",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368792"
}
|
Stack Exchange
|
What OEIS sequence is this?
I've come up with an idea of an integer sequence. It can be formulated (perhaps a bit loosely) as follows: For n points N(n) is the number of configurations where each point either lies on some circle or is a center of some circle. Each point lying on a circle can belong to only 1 circle and each center point can be the center of only 1 circle.
Then N(1) = 2 N(2) = 5 N(3) = 10 N(4) = 20
N(5) = 36
(I double checked N(5) but there is no guarantee that it is the right number)
Depending on how N(0) is defined (N(0)=0 or N(0)=1) this sequence very well can be A000712. However it appears that my description is new, so it easily can be another sequence.
all configurations for N(1), N(2), N(3) are shown here
https://oeis.org/search?q=2%2C+5%2C+10%2C+20%2C+36&sort=&language=english&go=Search
Number of configurations of what, exactly? of points? of circles?
configurations of points.
I'm not sure I understand your construction, it's a little too-loosely defined. Can you be more precise?
Frankly I think it is more or less intuitively clear if you look at the examples: https://i.sstatic.net/gdrBi.png Another problem that arises from this construction is to count the number of circles used in all the configurations for each n. This sequence goes like 2,8,22,54... and it is not in the OEIS. I do not in any way insist that it should be added though.
@Amir Sagiv. I reformulated it in this way: For n points N(n) is the number of configurations where each point either lies on some circle or is a center of some circle. Each point lying on a circle can belong to only 1 circle and each center point can be the center of only 1 circle.
It should be possible to understand a question here from what's written here, without needing to go off to stack.imgur
I would add something along the line of "given an infinite collection of unlabelled disjoint circles...". I think what induces confusion is we feel like we can pick circles later, which is another question (for instance, for three points, either they are aligned and they can be put on two circles, or they are not so they can be put on just one circle, so I would have thought N(3) is at most 2).
Your sequence is the same as the linked OEIS sequence. This is the
Number of partitions of $n$ into parts of two kinds.
In your case, the two kinds are circles for which the centre is occupied and circles for which the centre is not occupied. See the "example" section in the OEIS entry where you can match with your worked out example.
|
2025-03-21T14:48:31.735914
| 2020-08-10T11:38:02 |
368793
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"S.D.",
"Tobias Diez",
"https://mathoverflow.net/users/152042",
"https://mathoverflow.net/users/17047"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631944",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368793"
}
|
Stack Exchange
|
Logarithmic Darboux theorem
Let $X$ be a smooth complex analytic manifold and $D$ be a normal crossing divisor. Suppose that there is a complex analytic logarithmic symplectic structure on $X$.
Is there a Darboux like theorem that says that the complex analytic log symplectic form on a neighbourhood of a point is $\sum_{i=1}^r \frac{dx_i}{x_i}\wedge dx_{i+r}+\sum_{j=1}^{2(n-r)-1} dx_{2r+j}\wedge dx_{2r+j+1}$, where $\{x_i\}$'s are some local complex coordinates of $X$ and the divisor is given by the vanishing locus of $x_1\cdots x_r$?
If this is a known result an explanation or a reference of the proof will be very helpful.
For b-symplectic manifolds (i.e. $r=1$), a Darboux theorem is established in https://arxiv.org/abs/1512.05303
@Tobias Diez thanks
|
2025-03-21T14:48:31.736116
| 2020-08-10T12:31:05 |
368796
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631945",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368796"
}
|
Stack Exchange
|
Deriving integral in Gaiotto-Tommasiello theory
I was looking at a paper by Takao Suyama on GT theory, and I couldn't figure out how he derived his formula (3.59):
$$\frac{1}{\pi}\int_a^bdx\frac{1}{z-x}\frac{\sqrt{(z-a)(z-b)}}{\sqrt{|(x-a)(x-b)|}}\frac{\log(e^{-t_1}x)}{2}=\frac{1}{2}\log\left[\frac{e^{-t_1}}{2\sqrt{ab}+a+b}\left(z+\sqrt{ab}-\sqrt{(z-a)(z-b)}\right)^2\right],$$
where $0<a<b$, $[a,b]\subset\mathbb{R}$, $t_1\in\mathbb{R}$, and $z\in\mathbb{C}\setminus[a,b]$. It's a physics paper, but my question is just how would one do the integral?
I tried expanding everything as a power series, and then using the fact that the integral of $x^n\log x$ is known, but then I couldn't figure out how to resum the resulting series, so I'm a bit confused how one would solve this integral.
Okay, so I think I may have found the answer myself. So, really, the absolute value symbol is a trick. You can get rid of it by pulling out an $i$, and then you have
$$\mathcal{I}:=\frac{1}{2\pi i}\int_a^b dx \frac{\log(x e^{-t_1})}{z-x}\frac{\sqrt{(z-a)(z-b)}}{\sqrt{(x-a)(x-b)}}$$
What you have to do is take a dumbbell contour around taking a clockwise "dumbbell" contour $\mathcal{C}$ around the region $[a,b]$ such that we integrate over a region $[a+i0,b+i0]$ from left to right and a region $[a-i0,b-i0]$ from right to left. Since there is a branch cut along $[a,b]$, we have a sign flip as we cross the branch cut, and hence we have
\begin{equation}
2\mathcal{I}= \frac{1}{2\pi i}\int_{\mathcal{C}}dx \frac{\log(x e^{-t_1})}{z-x}\frac{\sqrt{(z-a)(z-b)}}{\sqrt{(x-a)(x-b)}}
\end{equation}
We now deform the contour to infinity, making note of the branch cut from the log and the pole at $x=z$. Then we pick up a residue at $z$
\begin{equation}
2\pi i\text{Res}_z\left(\frac{1}{2\pi i}\frac{\log (xe^{-t_1})}{z-x}\frac{\sqrt{(z-a)(z-b)}}{\sqrt{(x-a)(x-b)}}\right)=\log (z e^{-t_1})
\end{equation}
There is also a contribution from integrating along the log branch cut that can be easily done. Summing all these together and doing a painful amount of simplification yields
\begin{equation}
\mathcal{I}=\frac{1}{2}\log\left[\frac{e^{-t_1}}{2\sqrt{ab}+a+b}\left(z+\sqrt{ab}-\sqrt{(z-a)(z-b)}\right)^2\right]
\end{equation}
as desired.
|
2025-03-21T14:48:31.736277
| 2020-08-10T12:39:16 |
368798
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631946",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368798"
}
|
Stack Exchange
|
What types of semigroups have a Laplacean type operator as infinitesimal generator?
Let $\Omega\subseteq\mathbb{R}^N$ be an open, bounded connected set having Lipschitz uniform boundary. Moreover let $d\in L^{\infty}(\Omega,\mathbb{R}^M),\ d_1(x),d_2(x),\dots, d_M(x)>d>0,\ \forall\ x\in\Omega$ and the linear operator:
$$
A:D(A)\subseteq X\to X,\ A(y)=d\Delta y,\ \forall\ y\in D(A),
$$
where $X\subseteq \{f:\Omega\to\mathbb{R}^M\ |\ f\text{ is a function}\}$ is a Banach (or reflexive Banach, or Hilbert space over $\mathbb{R}$ or $\mathbb{C}$). We can have $$
X=
\begin{cases}
L^2(\Omega;\mathbb{R}^M), & \text{or}\\
H^1(\Omega,\mathbb{R}^M), & \text{or}\\
L^p(\Omega;\mathbb{R}^M), & \text{or}\\
L^{\infty}(\Omega;\mathbb{R}^M), & \text{or}\\
W^{k,p}(\Omega;\mathbb{R}^M) & \text{or} \\\text{etc}\ldots\\
\end{cases}
$$
and
$$
D(A)=\left\{y\in X\ \Big| \ d\Delta y\in X,\ \alpha_iy_i+\beta_i\dfrac{\partial y_i}{\partial\nu}=0,\ \forall\ i\in\overline{1,M}\right\}.$$
Here $\alpha_i\in C(\partial\Omega),\beta_i\in\{0,1\},\ \forall\ i\in\overline{1,M}$.
My question is: what type of semigroup has $A$ as an infinitesimal generator? Is it a strongly continuous semigroup (or $C_0$ semigroup), or analytic semigroup, or sectorial, or compact... Are there any references that deal with this problem with a Robin type boundary condition in a clear way, or at least with a Neumann boundary condition?
Somebody told me that if we have Neumann Boundary conditions ($\alpha_i=0,\beta_i=1,\ \forall\ i\in \overline{1,M}$) then the semigroup $(T(t))_{t\geq 0}$ that generates $A$ is contractive and maps $L^1$ to $L^p$ with norm less than $Ct^{\frac{-N}{p'}}$, $1/p+1/p'=1$. What does that mean exactly, and is there a reference for that topic?
Thanks a lot.
|
2025-03-21T14:48:31.736411
| 2020-08-10T13:01:50 |
368799
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Christian Remling",
"Rajesh D",
"gmvh",
"https://mathoverflow.net/users/14414",
"https://mathoverflow.net/users/45250",
"https://mathoverflow.net/users/48839"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631947",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368799"
}
|
Stack Exchange
|
Convergence of the eigenvector matrix for an analytic perturbation of a singular matrix
Let $A$ be an $n\times n$ matrix of all ones. Consider the analytic perturbation of $A$ as $$\tilde{A} = A + \epsilon H_1 + \epsilon^2 H_2 + \epsilon^3 H_3 + ... $$ All matrices are symmetric. Assume $\tilde{A}$ to be positive definite. Let $E = [e_0,e_1,e_2,...e_{n-1}]$ be the eigenvectors matrix of $A$ and $\tilde{E} = [\tilde{e}_0,\tilde{e}_1,\tilde{e}_2,...\tilde{e}_{n-1}]$ be the eigenvectors matrix of the matrix $\tilde{A}$.
I have seen David Kahan theorem bounds the sine of the angle between the eigenvectors stating $\lim\limits_{\epsilon\to 0}\sin\Theta(e_i,\tilde{e}_i) = 0$ and also another theorem which says $\lim\limits_{\epsilon\to 0}\|E-\tilde{E}\|_{\infty} = 0$ or to this effect in this paper and also this paper. I am sure from these references that these statements hold when eigenvalues of $A$ are distinct.
I have seen some vague references but not sure if the above statements hold when there are eigenvalues of multiplicity greater than 1. (It is the case here, as eigenvalues of $A$ are $n$ with multiplicity $1$ and $0$ with multiplicity $n-1$.
So in this case can we say that $\lim\limits_{\epsilon\to 0} \|\tilde{E}-E'\|_{\infty} = 0$ where $E' = E$ up to a permutation of columns.
PS : I myself am not doing research in perturbation theory, but this situation arise in my work and I referred to these papers to see if the eigenvectors matrix converges. The message in those papers are a bit cryptic and was not sure if these papers apply in this case where there are eigenvalues of multiplicity higher than $1$.
The standard reference for such questions is Kato, Perturbation theory of linear operators. You could try to take a look at this.
Because of the degeneracy in the eigenvalues of $A$, "the eigenvectors matrix of $A$" is far from well-defined. Rather, there is a one-dimensional eigenspace of $A$ for eigenvalue $n$ (spanned by $(1,\ldots,1)^T$), and its orthogonal complement is the eigenspace for eigenvalue $0$. For any $\eta > 0$, there is $\delta > 0$ such that
if $|\epsilon| < \delta$, $\tilde{A}$ has one eigenvalue within distance $\eta$ of $1$, all the others within distance $\eta$ of $0$, and the orthogonal projections for $A$ and $\tilde{A}$ on the span of eigenspaces for eigenvalues within $\eta$ of $0$ (or $1$) are within distance $\eta$ (for whichever matrix norm you prefer).
Thank you for the enlightening answer. I am trying to see what best can be said about $\tilde{E}$. Neverthless $\tilde{E}$ is not well defined, irrespective of how we choose the eigenvector matrix $\tilde{E}$, can we always say the following : Let $\Sigma$ is any constant diagonal matrix, as $\epsilon\to 0$, $\tilde{E}\Sigma\tilde{E}^T$ always converges to some matrix.
I have also encountered the problem of eigenvalue multiplicity with Davis-Kahan sin theorem. I believe I've found an answer with this paper (see Lemma 4, p21).
Welcome to MO! Please note that it is always a good idea to summarize the main point of an answer contained in an off-site resource instead of just providing a link, since links can change or become invalid. Thanks!
|
2025-03-21T14:48:31.736651
| 2020-08-10T13:07:19 |
368800
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Eusebio Gardella",
"LSpice",
"Rohil Prasad",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/29566",
"https://mathoverflow.net/users/43158"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631948",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368800"
}
|
Stack Exchange
|
Local cross-sections for free actions of finite groups
Let $G$ be a finite group, let $X$ be a locally compact Hausdorff space, and let $G$ act freely on $X$. It is well-known that the canonical quotient map $\pi\colon X\to X/G$ onto the orbit space $X/G$ admits local cross-sections. More precisely, for every $z\in X/G$ there are an open set $U$ in $X/G$ containing $z$, and a continuous function $s\colon U\to X$ such that $\pi\circ s$ is the identity on $U$. In particular, there is an open cover of $X/G$ consisting of sets where a local cross-section can be defined.
Question: is there a finite open cover of $X/G$ consisting of sets where a local cross-section can be defined?
(This is the same as asking whether the Schwarz genus of the fiber map $X\to X/G$ is finite.)
The answer is "yes" if $X$ (or at least $X/G$) is finitistic, so in particular whenever $X$ has finite covering dimension, and clearly also whenever $X$ is compact. I wonder if it is true in general.
Here's an idea for making a counterexample. $X$ is a principal $G$-bundle over $X/G$. If there is such a finite cover, then I believe that the $G$-bundle $X$ should extend to a $G$-bundle over the one-point compactification $(X/G)^$ of $X/G$. Therefore, to find a counterexample, it suffices to find a locally compact, Hausdorff space $Y$ and a $G$-bundle $P \to Y$ such that $P$ does not extend over the one-point compactification $Y^$ of $Y$.
Thanks for your comment. I realize that I misstated my question: I actually want to know if an open cover with finite order exists. I edited the question.
What does it mean for a cover to have finite order? It's locally finite, or uniformly locally finite, or something else?
I think a finite cover as desired exists iff a finite-order cover exists. So I went back to the original formulation, this time adding a connection to the Schwarz genus.
Let $X=[-1,1]^\infty\setminus\{0\}$, which is a metrizable, locally compact space. Consider the two-element group $G$, and the free $G$-action on $X$ given by $(x_j)_{j=1}^\infty\mapsto (-x_j)_{j=1}^\infty$. We show that the fibration $X\to X/G$ has infinite Schwarz genus.
Consider the $n$-sphere $S^n$ with the antipodal $G$-action. Then $S^n$ can be embedded equivariantly into $X$ for all $n$. (Use an equivariant map $S^n\to [-1,1]^{n+1}\setminus\{0\}$.) By the Lusternik–Schnirelmann theorem (a strengthening of the Borsuk-Ulam theorem), $S^n$ cannot be covered by $n+1$ closed sets that do not contain antipodal points. It follows that the Schwarz genus of $S^n\to S^n/G$ is at least $n+2$. Since the Schwarz genus of $X\to X/G$ is an upper bound for the Schwarz genus of $S^n\to S^n/G$, it follows that $X\to X/G$ has infinite Schwarz genus.
There is a general cohomological lower bound for the Schwarz genus of a map $p:E\to B$. Namely, if there are cohomology classes $x_1,\ldots , x_k\in H^*(B)$ such that $0=p^*(x_i)\in H^*(E)$ for all $i=1,\ldots , k$ and $x_1\cup\cdots \cup x_k \neq 0$, then the genus of $p$ is greater than $k$. Here the coefficients are completely arbitrary, in particular can be twisted. (This is a generalisation of the cup-length lower bound for Lusternik-Schnirelmann category, since the LS-category of a space $X$ is equal to the genus of any fibration over $X$ with contractible total space.)
So you can get many counter-examples using this cohomological criterion. In fact, whenever $X$ is a contractible CW-complex then it is a model for $EG$, and $X/G$ is a model for $BG$. The cup-length of $BG$ is always infinite for a finite group $G$ (with appropriately chosen, possibly twisted coefficients). This generalises the example in Hannes Thiel's answer.
|
2025-03-21T14:48:31.736901
| 2020-08-10T13:12:20 |
368801
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Cranium Clamp",
"Harry Gindi",
"https://mathoverflow.net/users/1353",
"https://mathoverflow.net/users/152391"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631949",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368801"
}
|
Stack Exchange
|
Applications of the Infinitesimal Lifting Property
This question was posted to MSE but didn't get any answers, so I am posting it here. Original post
Hartshorne in his book gives the 'Infinitesimal Lifting Property' as an exercise in chapter 2, section 8 and mentions this to be very important in the deformation theory of nonsingular varieties. For completeness, I record the statement below:
Let $ k $ be an algebraically closed field and $ A $ a finitely generated $ k $-algebra such that $ \operatorname{Spec} A$ is a nonsingular variety over $ k $. Let $ B $ be a $ k $-algebra and $ B' $ a square-zero extension of $ B $ by $ I $, i e., there is an exact sequence $$ 0 \rightarrow I \rightarrow B' \xrightarrow{\pi} B \rightarrow 0 $$ where $ B' $ is a $ k $-algebra and $ I $ is an ideal such that $ I^2 = 0 $. Let $ f : A \rightarrow B $ be a $ k $-algebra homomorphism. Then there is a lift $ g : A \rightarrow B' $, i.e. a $ k $-algebra homomorphism $ g $ such that $ \pi \circ g = f $.
As someone starting with deformation theory, I would like to know how/why this result is very important as well as some applications of this result. Does this result have applications while studying moduli problems/moduli spaces?
For instance, Hartshorne gives one application: For $ X $ a nonsingular variety over $ k $ and $ \mathcal{F} $ a coherent sheaf on $ X $, the set of Infinitesimal extensions of $ X $ by $ \mathcal{F} $ upto isomorphism is in one to one correspondence with $ H^1(X, \mathcal{F} \otimes \mathcal{T}_X) $ where $ \mathcal{T}_X $ is the tangent sheaf.
There's a really nice derived picture extending the mentioned result of Hartshorne in Adeel Khan's notes https://www.preschema.com/lecture-notes/kdescent/lect5.pdf .
This is known as the formal criterion for "formal smoothness." In this stacks project entry they prove that a morphism of schemes (in your case $\text{Spec }A \to \text{Spec }k$) is smooth if and only if it's formally smooth and locally finite presentation.
Aside from philosophical importance, it's often easier/more intuitive to check this formal criterion than to check the dimension of $\Omega_{\text{Spec }A/\text{Spec }k}$ or use a Jacobian. In the same stacks project tag, they say:
Michael Artin's position on differential criteria of smoothness (e.g., Morphisms, Lemma 01V9) is that they are basically useless (in practice).
Let's suppose $\text{Spec }A$ were instead a moduli space $\overline{M}$, e.g. of curves. Then to check $\overline{M}$ is smooth (if we know finite presentation), we need only consider an infinitesimal extension $S \subseteq S'$ coming from $B' \to B$ as in your question, and try to extend a curve over $S$ to a curve over $S'$.
If you want to build the cotangent complex of $X = \text{Spec } A$ (or equivalently the "normal sheaf") but $A$ is not smooth, the first step is to replace $A$ by a smooth $k$-algebra mapping to it, say $k[A]$. Even if some $A \to B$ doesn't factor through $B'$, $k[A] \to A \to B$ certainly will (by choosing a set-theoretic preimage in $B'$ of the image of $A$ in $B$) and so the "problem" obstructing a factorization of $A \to B$ can be traced to the kernel of $k[A] \to A$. I highly recommend this stacks project article that carries this out as concretely as possible.
You can get specific cohomological obstructions to such a factorization by continuing with a simplicial resolution: $\cdots k[k[A]] \rightrightarrows k[A] \to A$. One can even think of this as "covering $A$ by smooth algebras" in a topological sense using this Jonathan Wise article.
EDIT: I can be a bit more precise about the connection to ordinary differentials. Suppose $f : A \to B$ is fixed and we're trying to find a map $\widetilde{f} : A \to B'$ such that $A \overset{\widetilde{f}}{\to} B' \to B$ is $f$. First, pullback along $f$ to get another squarezero extension
$$0 \to I \to B' \times_B A \to A \to 0.$$
Our old search for $\widetilde{f}$ translates to finding a section of the map $B' \times_B A \to A$. We've reduced to the case $f = id_B : A = B$.
Given one section $s$ of $B' \to B$, this splits the underlying sequence of modules and lets us write $B' =_{modules} B \oplus I$. But what's the ring structure? One computes $(b, i)*(b', i') = (bb', bi' + b'i)$, i.e. $B' = B + I\epsilon$ is the trivial squarezero extension!
If I have two sections $s, t$ of $B' \to B$, they'll give very different isomorphisms $B' \simeq B + I \epsilon$, inducing an automorphism $\varphi$ of $B + I \epsilon$ over $B$. Such automorphisms are precisely derivations! Indeed $\varphi$ must be the identity on $B$ and send $I$ to itself, but $\varphi - id_{B'}$ will be a map from $B \to I$ (using co/kernel univ props on the short exact sequence) which you can check is a derivation.
If sections exist, they all differ by a unique derivation $s-t$. A fancy way to say this is that "sections form a pseudo-torsor under $\text{Der}(B, I)$." If sections really do (locally) exist, you call it a plain torsor.
Thank you for the answer, I will take some time to understand it. I will wait to see if there are more answers before accepting.
|
2025-03-21T14:48:31.737240
| 2020-08-10T13:31:53 |
368803
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"GH from MO",
"Iosif Pinelis",
"bryanjaeho",
"https://mathoverflow.net/users/11919",
"https://mathoverflow.net/users/149093",
"https://mathoverflow.net/users/36721"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631950",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368803"
}
|
Stack Exchange
|
The cotangent sum $\sum_{k=0}^{n-1}(-1)^k\cot\Big(\frac{\pi}{4n}(2k+1)\Big)=n$
On the Wolfram Research Reference page for the cotangent function (https://functions.wolfram.com/ElementaryFunctions/Cot/23/01/), I saw the following partial sum formula
$$\sum_{k=0}^{n-1}(-1)^k\cot\Big(\frac{\pi}{4n}(2k+1)\Big)=n.$$
I was unable to find a reference for it but eventually proved it as described below.
My Question
A reference for the equation above?
Naturally, I wondered what was known about the regular sum of the cotangents above, not the alternating sums. I found no related equation but instead was able to empirically derive the following limit
\begin{align*}
\lim_{n\to\infty}\Big[\sum_{k=0}^{n-1}\frac{1}{2k+1}-\sum_{k=0}^{n-1}\frac{\pi}{4n}\cot\Big(\frac{\pi}{4n}(2k+1)\Big)\Big]&=\lim_{n\to\infty}\Big[\Big(H_{2n}-\frac{1}{2}H_{n}\Big)-\sum_{k=0}^{n-1}\frac{\pi}{4n}\cot\Big(\frac{\pi}{4n}(2k+1)\Big)\Big]\\
&=\ln(\sqrt{\pi/2}).
\end{align*}
Yet, I am stuck on proving this limit and would appreciate any advice on where to begin.
My Proof for the Equation in the Beginning:
Consider the well-known partial fraction expansion for the cotangent function,
$$\pi\cot(\pi x)=\frac{1}{x}+\sum_{k=1}^{\infty}\frac{1}{x-k}+\frac{1}{x+k}$$
excluding integral $x$. Rather unconventionally, let us expand the summation of the RHS and reindex it, getting
\begin{align*}
\frac{1}{x}+\frac{1}{x-1}+\frac{1}{x+1}+\frac{1}{x-2}+\frac{1}{x+2}+\cdots&=\frac{1}{x}-\frac{1}{1-x}+\frac{1}{x+1}-\frac{1}{2-x}+\frac{1}{x+2}-\cdots\\
&=\sum_{k=0}^{\infty}\frac{1}{k+x}-\frac{1}{k+1-x}.
\end{align*}
By letting $x=b/a$ for positive integers $a,b$ such that $a>b\geq1,$ we get
$$\frac{\pi}{a}\cot\Big(\pi\cdot\frac{b}{a}\Big)=\sum_{k=0}^{\infty}\frac{1}{ak+b}-\frac{1}{ak+a-b}.$$
Consider the equation for $a=4n$ and $b$ from $1$ to $2n-1$. Then,
$$\frac{\pi}{4n}\cot\Big(\pi\cdot\frac{1}{4n}\Big)=\sum_{k=0}^{\infty}\frac{1}{(4n)k+1}-\frac{1}{(4n)k+4n-1},$$
$$\frac{\pi}{4n}\cot\Big(\pi\cdot\frac{3}{4n}\Big)=\sum_{k=0}^{\infty}\frac{1}{(4n)k+3}-\frac{1}{(4n)k+4n-3},$$
$$\cdots$$
$$\frac{\pi}{4n}\cot\Big(\pi\cdot\frac{2n-1}{4n}\Big)=\sum_{k=0}^{\infty}\frac{1}{(4n)k+2n-1}-\frac{1}{(4n)k+2n+1}.$$
By summing up the equations in an alternating fashion (adding the first, subtracting the second, and so on,) the summed $LHS$ equals
$$\frac{\pi}{4n}\sum_{k=0}^{n-1}(-1)^k\cot\Big(\frac{\pi}{4n}(2k+1)\Big)$$
while the summed RHS is simply the alternating series of the reciprocals of odd numbers, which is well-known by Leibniz to be $\frac{\pi}{4}.$ Hence, we get the initial formula.
These kind of identities are well-known (folklore). For example, one of the homework problems in my introductory complex analysis course is the following. Let $n=2m+1$ be an odd positive integer. Then $\sum_{k=-m}^m\tan(z+k\pi/n)=n\tan(nz)$. (Specialize this to $z=\pi/(4n)$.)
The expression under the limit sign in question is just a Riemann sum for
$$\int_0^{\pi/2} \frac12\,\Big(\frac1x-\cot x\Big)\,dx=\frac12\,\ln\frac\pi2,$$
which therefore is the value of the limit.
Just to make sure, the definite integral on the left gives $-\ln(\sin(1))$?
I believe the integral should be over $\frac{1}{2}\Big(\frac{1}{x}-\frac{\pi}{2}\cdot \cot(\pi/2\cdot x)\Big)$ ranging from $0$ to $1$. Thank you for the approach!
@bryanjaeho : Oops! The integral is of course from$0$ to $\pi/2$, not to $1$. This typo is now corrected
|
2025-03-21T14:48:31.737480
| 2020-08-10T14:07:38 |
368806
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Dieter Kadelka",
"Gro-Tsen",
"Max Alekseyev",
"catbow",
"https://mathoverflow.net/users/100904",
"https://mathoverflow.net/users/163571",
"https://mathoverflow.net/users/17064",
"https://mathoverflow.net/users/7076"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631951",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368806"
}
|
Stack Exchange
|
Probability of a given string being a substring of another string
I am in interested into the following problem. We are given an alphabet $\Sigma$ of $k$ letters and a fixed string $S_1$ of length $l$ defined over $\Sigma$. Given a probability distribution $D$ over $\Sigma$, we sample another string $S_2$ with length $n$. Each letter is $S_2$ is sampled independently. We consider that $S_1$ is a substring of $S_2$ if there exists two strings $p$ and $q$ such that: $S_2 = p S_1 q$. What is the probability that $S_1$ is a substring of $S_2$? Additional assumptions could be made about D.
For a uniform distribution, it is a combinatoric problem which solution has been studied for example in: https://stackoverflow.com/questions/6790620/probability-of-3-character-string-appearing-in-a-randomly-generated-password
Would you please have any insight about how it generalizes to any distribution $D$ over $\Sigma$?
yes, assuming the probability of each letter in $\Sigma$ is known might be more suited. Thanks for pointing this out
“Substring” has two different possible meanings: “factor” (obtained by removing letters at the start and end only) or “subsequence” (obtained by removing words anywhere within the word). You should clarify which one you mean.
Thanks, it is indeed confusing. I mean factor, ie there exists two strings $p$ and $q$ such that: $S_2 = p S_1 q$. I have clarified in the question.
As noted by Matt F. your question does not make much sense. You should edit it explicitely. Further, without assuming independence of the letters not much can be said. In case of independence I think the accepted answer of the original problem with uniform distribution of the letters can be generalized.
thanks, assuming the independence of the letters is a good point. I have edited the question accordingly and as noted by Matt F.
Penney's game is relevant.
I'm going to give complete solutions for the case when $T$, the target word (which OP calls $S_1$), has $1$ or $2$ letters in it. These should suggest the difficulties involved in finding a solution for $T$ of length $3$ or more, both because of the number of different patterns I can have for $T$, and the proliferation of recursion which is needed.
Without loss of generality, I may assume $\Sigma$ is either equal to the number of unique letters which appear in $T$ (that is, every randomly generated letter could potentially form part of $T$) or $\Sigma$ is the number of unique letters in $T$, plus $1$ (there is a random letter that can only block $T$ from forming, and never aid in forming $T$). We don't need more than one additional letter, as which irrelevant letter is picked doesn't matter to our analysis; we can lump them all together into a single irrelevant letter. So in our analysis, $T$ will be (WLOG) one of $a$, $aa$, or $ab$; and $\Sigma$ will be either $\{a, x \},$ $\{a, b \}$, or $\{a, b, x\}$, where $x$ stands for the irrelevant letter.
We will also have a probability distribution $\Bbb{P}: \Sigma \rightarrow [0, 1]$, and denote $\Bbb{P}(L) = p_L$ when $L = a, b,$ or $x$.
Finally, our $n$-letter word $S_n$ (which OP calls $S_2$) will be randomly chosen as $S_n = L_1 L_2 ... L_n$, where $L_1, L_2, L_3, ...$ are independent, identically $\Bbb{P}$-distributed, $\Sigma$-valued discrete random variables. We will also abuse the notation $S_k = L_1 L_2 ... L_k$ to refer to the first $k$ letters of $S_n$, that is, consider $S_k$ as a substring of $S_n$ when $k < n$.
The case $T = a$: In this case, clearly $\Sigma = \{ a, x \}$ and either we get $a$ at the very beginning of $S_n$, or we get a string of $x$'s terminating in an $a$ as the first few letters of $S_n$. If there are $n$ letters in my word, then this gives me $$p_a + p_x p_a + ... + p_x^{n-1} p_a = p_a \frac{1 - p_x^n}{1 - p_x} = 1 - p_x^n,$$ since $p_a + p_x = 1$. This makes sense; the only way I can avoid having an $a$ in my word $S_n$ is by using an $x$ for every letter.
The case $T = aa$: Once again, $\Sigma = \{a, x \}$. Call $q_n$ the probability that an $n$-letter string $S$ does not contain $T = aa$ (so the probability we want is $1 - q_n$). Then this string either ends in $x$ or in $xa$, so we have the recurrence $$q_n = p_x q_{n-1} + p_a p_x q_{n-2},$$ with initial conditions $q_1 = 1$, $q_2 = 1 - p_a^2$. This is a linear homogeneous recurrence relation, and unfortunately the solutions to the characteristic equation $\lambda^2 - p_x \lambda - p_x p_a = 0$ are rather messy: $$\lambda = \frac{p_x \pm \sqrt{p_x^2 + 4p_x p_a}}{2} = \frac{p_x \pm \sqrt{4p_x - 3p_x^2}}{2},$$ which are, in general, not rational. If we let $\lambda_+$ denote the root with the plus sign and $\lambda_-$ denote the root with the minus sign, then $$q_n = c_+ \lambda_+^n + c_- \lambda_-^n,$$ where $c_+, c_-$ are the solutions to the system of linear equations $c_+ + c_- = 1$, $c_+ \lambda_+ + c_- \lambda_- = 1 - p_a^2$. Then $$\Bbb{P}(S_n = pTq) = 1 - q_n = 1 - (c_+ \lambda_+^n + c_- \lambda_-^n).$$
The case $T = ab$, $\Sigma = \{a, b\}$: As before, let $q_n$ be the probability that $S_n$ does not contain $ab$. The only way $S_n$ ends with $b$ and does not contain $ab$ is if $S_n$ is a string of all $b$'s; otherwise, $S_n$ ends in $a$. So we get $q_n = p_a q_{n-1} + p_b^n$, and expanding out we find $$q_n = p_a^n + p_a^{n-1} p_b + ... + p_a p_b^{n-1} + p_b^n,$$ so $q_n = n p_a^n$ if the letters $a, b$ are equally likely and $$q_n = \frac{p_a^{n+1} - p_b^{n+1}}{p_a - p_b}$$ when they are not. The probability that $S_n$ contains $T$ is then $1 - q_n$, as before.
The case $T = ab$, $\Sigma = \{a, b, x\}$: This is as close as we get to the general case, and we are going to see some really obnoxious recursion for $q_n$, which is as in Cases 2 and 3. If $S_n$ does not contain $T$, then either $S_n$ ends in $a$ or $x$, $S_n$ ends in $xbb...bb$, or $S_n = bbbbbbb...bbbb$ is a string of $n$ $b$'s. So our recurrence is now $$q_n = (p_a + p_x) q_{n-1} + p_b p_x q_{n-2} + p_b^2 p_x q_{n-3} + ... + p_b^{n-2} p_x q_1 + p_b^{n-1} p_x + p_b^n,$$ which can be solved by similar techniques as before, but is rather messy.
From reading this account the complexity of giving a general answer for $T$ of arbitrary length and structure should, I hope, be clear.
Let $A_t$ be the event that $S_1$ is a substring of $S_2$, $S_2=pS_1q$, where the length of $p$ is $t$.
Then the probability of $\cup_t A_t$ can be found by inclusion-exclusion as
$$\sum P(A_t)-\sum P(A_{t_1}\cap A_{t_2}) + \sum_{t_1,t_2,t_3} P(A_{t_1}\cap A_{t_2}\cap A_{t_3})-\dots$$
Terms like $P(A_{t_1}\cap A_{t_2})$ have probabilities that depend on how far apart the $t_i$ are, and on the structure of $S_1$. For instance,
the probability that $01$ is a substring of a 3-letter string $xyz$ is
$$P(x=0,y=1)+P(y=0,z=1)$$
which by your independence assumption is
$2p_0p_1$.
The probability that $01$ is a substring of $xyzw$ is
$$P(x=0,y=1)+P(y=0,z=1)+P(z=0,w=1)-P(x=0,y=1,z=0,w=1)$$
$$=3p_0p_1-p_0^2p_1^2.$$
A further complication arises from the cases where $S_1$ is 111 or 0101 or in general any string that is a power of another string. The probability that 111 is a substring of $xyzw$ is
$$P(x=y=z=1)+P(y=z=w=1)-P(x=y=z=w=1)=2p_1^3-p_1^4.$$
In the most generic setting, the problem may be solved with Knuth-Morris-Pratt algorithm.
Consider a graph $G$ on $l+1$ vertices numbered from $0$ to $l$, such that the vertex $k$ corresponds to the prefix of $S_1$ of size $k$. Each such vertex has $\Sigma$ outgoing arcs from it, each marked by some alphabet letter. Essentially, this graph is a deterministic finite automaton, such that after going with the string $S_2$ through its arcs you end up in a vertex that corresponds to the largest suffix of $S_2$ that is also a prefix of $S_1$.
This automaton could as well be perceived as a Markov chain, that is transition with the letter $c$ has a fixed known probability to happen. Then, the problem is equivalent to finding the probability of not ever getting into the last vertex after having a walk that starts in the vertex $0$ and has a length $n$.
Let $A$ be the stochastic matrix of this chain. Then the $j$-th element of the $i$-th row of $A^n$ equates to the probability of ending up in $j$-th state after making $n$ steps starting in the state $i$. If we redirect all transitions from the state $l$ into itself, we will turn it into "devil state" which forever traps you once you get there.
Now, the question essentially asks to compute the probability of not getting into the devil state after $n$ steps. This probability is equal $1-p$, where $p$ is the $l$-th element of the $0$-th row of $A^n$.
It's not really a formula, but it's a universal and somewhat efficient way to computationally solve these kind of problems for any specific string.
P.S. I know that there is an explicit formula for the expected time of finding $S_1$ to be a substring of $S_2$ for the first time. Unfortunately, I'm not sure whether it can be used for probabilities.
|
2025-03-21T14:48:31.738138
| 2020-08-10T14:42:44 |
368809
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Jon Pridham",
"https://mathoverflow.net/users/103678"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631952",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368809"
}
|
Stack Exchange
|
Can an $\infty$-action on a derived affine scheme by an affine group scheme always be strictified?
Let $X$ be an affine derived scheme, say $X = \operatorname{Spec} A$, for $A$ a simplicial commutative ring. Let $G$ be an affine group scheme (classical), say $G = \operatorname{Spec}B$, and let an $\infty$-action of $G$ on $X$ be given in the form of a simplicial diagram $X \leftleftarrows X \times G \cdots$ in the $\infty$-category of derived schemes. This diagram corresponds to a cosimplicial diagram $D$ of the form $A \rightrightarrows A \otimes B \cdots$ in the $\infty$-category of simplicial commutative rings.
Say that we can strictify $D$ if it is equivalent to a strictly commutative diagram $D'$ in the model category of simplicial commutative rings such that $D'$ corresponds to a levelwise coaction of $B$ on a simplicial commutative ring $A'$. In other words, we can strictify $D$ if we can, up to equivalence, assume that the equivalences and homotopies which exhibit $D$ as a coaction of $B$ on $A$ are in fact isomorphisms and identities.
The question is: can we always strictify $D$? I suspect the answer is: no, not always, but we can for certain $G$. For example, the answer is yes for $G = \mathbb{G}_m$. I hope to find a counterexample to see that it does not work for general $G$.
Since it works for $\pi_0A$, have you tried working up the Postnikov tower of $A$ and comparing obstructions at each stage?
In characteristic $0$, another way to approach it would be to look for obstructions as you work up the lower central series of the pro-unipotent radical of $G$, since strictification should work for pro-reductive groups.
|
2025-03-21T14:48:31.738263
| 2020-08-10T15:26:29 |
368811
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"abx",
"https://mathoverflow.net/users/124705",
"https://mathoverflow.net/users/40297",
"konoa"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631953",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368811"
}
|
Stack Exchange
|
Weights on the linearization
Consider, just as an example, an action of $\mathbb{C}^*$ on $\mathbb{P}^2$ of the form
$$t\cdot p=[p_0:tp_1:t^2p_2]$$
There are $3$ fixed points, namely $e_1,e_2,e_3$. If I consider a $\mathbb{C}^*$-linearizable line bundle -like $L=\mathcal{O}(1)$-, then I have an induced action
$$\phi:\mathbb{C}^*\times L\to L,$$
which is linear along the fibers and equivariant with respect to the previous action. If we consider for example $e_1=[1:0:0]$, I have a linear action $$\phi:\mathbb{C}^*\times L_{e_1}\to L_{e_1}, \text{ i.e. } \mathbb{C}^*\times\mathbb{C}\to\mathbb{C}$$
and I would like to understand what is the weight of the $\mathbb{C}^*$-action here. I'm pretty confident there must be a way to recover the weight of the action from the action on $\mathbb{P}^2$, but I've no idea how to do it and I'm curious (I considered a specific example just to for better understanding).
Any hint, help or reference would be much appreciate, thanks in advance.
If your $\mathbb{P}^2$ is $\mathbb{P}(\mathbb{C}^3)$, you can identify the complement of the zero section in $L^{-1}$ with $\mathbb{C}^3\smallsetminus 0$. One possible way to extend your action is to have $t\in\mathbb{C}^*$ acts by $(x,y,z)\mapsto(x,ty,t^2z)$. Then $t$ acts trivially on $L_{e_0}$, as $t$ on $L_{e_1}$ and as $t^2$ on $L_{e_2}$, so that the weights on $L$ are $0,-1,-2$. Note however that you are free to add a fixed integer to these (the linearization is not unique).
Dear @abx, thanks a lot for the comment! I'm sorry, but I don't follow exactly your passages. You consider $\mathcal{O}(-1)$, and indeed $s_0(\mathbb{P}^2)^c=\mathbb{C}^3\setminus 0$. The you define an action on $\mathbb{C}^3\setminus 0$ with weights $(0,1,2)$, but I don't get the next passage, that is for example $t$ acts precisely as $t^2$ on $L_{e_2}$,I'm sorry
The other points are fine, you swap sign since you're working with the dual bundle, and yes, I 'm aware of the last sentence. If you want to write it as an answer I'll surely accept it, thanks again for the patience!
I have written my comment as an answer with some more details, let me know if this not completely clear.
If your $\mathbb{P}^2$ is $\mathbb{P}(\mathbb{C}^3)$, you can identify the complement of the zero section in $L^{-1}$ with $\mathbb{C}^3\smallsetminus 0$, viewed as a bundle over $\mathbb{P}^2$ via the projection $p:\mathbb{C}^3\smallsetminus 0\rightarrow \mathbb{P}^2$. One possible way to extend your action is to have $t\in \mathbb{C}^*$ acts on $\mathbb{C}^3\smallsetminus 0$ by $(x,y,z)\mapsto (x,ty,t^2z)$. Then $t$ acts trivially on $L_{e_0}\smallsetminus 0=p^{-1}(e_0)= \{(x,0,0)\} $, as $t$ on $L_{e_1}\smallsetminus 0= \{(0,y,0)\}$, and as $t^2$ on $L_{e_2}\smallsetminus 0= \{(0,0,z)\}$, so that the weights on $L$ are $0,-1,-2$. Note however that you are free to add a fixed integer to these weights (the linearization is not unique).
Dear @abx, thanks again for your patience, It's clearer now. What I was wondering is why you need to eliminate the zero section in your construction: intuitively I think you're removing the limit $\lim_{t\to 0}tp$, but maybe there's a deeper justification, Anyway thanks again, hope to receive other answer from you! Have a nice day
Just because the fibration $V\smallsetminus 0\rightarrow \mathbb{P}(V)$, for $V$ a vector space, is not a line bundle but a $\mathbb{C}^*$-bundle, naturally isomorphic to the complement of the zero section in $L^{-1}$. You can also blow up $0$ in $V$, then you get a natural isomorphism with $L^{-1}$.
|
2025-03-21T14:48:31.738502
| 2020-08-10T15:52:40 |
368815
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alice M.",
"Greg Friedman",
"archipelago",
"https://mathoverflow.net/users/151523",
"https://mathoverflow.net/users/32022",
"https://mathoverflow.net/users/6646"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631954",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368815"
}
|
Stack Exchange
|
Novikov-Wall non-additivity theorem with twisted coefficients
Let $Y$ be a compact manifold and let $\pi_1(Y) \to \mathbb{Z}^n= \langle t_1,\ldots,t_n\rangle$ be a homomorphism. Extend it to the group rings $\mathbb{Z}[\pi_1(Y)] \to \mathbb{Z}[ t_1,\ldots,t_n]$ and evaluate it in a certain $\omega \in (S^1)^n \subset \mathbb{C}^n$. We obtain a homomorphism
$$ \psi: \mathbb{Z}[\pi_1(Y)] \to \mathbb{C}$$ which endows $\mathbb{C}$ of a $(\mathbb{C},\mathbb{Z}[\pi_1(Y)])$-bimodule structure. I will indicate this bimodule as $\mathbb{C}^\omega$.
What I call $\omega$-twisted homology of $Y$ will be the homology of the complex $$ \mathbb{C}^\omega \otimes_{\mathbb{Z}[ \pi_1(Y)]} C^{ \text{CW}}_*(\widetilde{Y}) $$
and I denote it with $H_*(Y;\mathbb{C}^\omega)$, where $\widetilde{Y}$ is the universal cover.
The cohomology of the cochain complex $$\text{Hom}_{\text{Mod}-\mathbb{Z}[\pi_1(Y)]}( \text{inv}(C_{*}(\widetilde{Y})), \mathbb{C}^\omega)$$
is the $\omega$-twisted cohomology of $Y$. Here $ \text{inv}(C_{*}(\widetilde{Y}))$ indicates $C_{*}(\widetilde{Y})$ with the same additive structure, but the action of $\mathbb{Z}[\pi_1(Y)]$ is now on the right because we precompose it with $g \mapsto g^{-1}$ for $g \in \pi_1(Y)$.
If $Y$ is of even dimension $2k$ there is an intersection form on $H_k(Y;\mathbb{C}^\omega)$, defined more or less as usual:
$$\phi:H_k(Y;\mathbb{C}^\omega) \to H_k(Y, \partial Y;\mathbb{C}^\omega) \xrightarrow{\text{PD}} H^k(Y;\mathbb{C}^\omega) \xrightarrow{\text{ev}} \text{inv}(\text{Hom}_{\mathbb{C}}(H_k(W;\mathbb{C}^\omega),\mathbb{C})) $$
The Poincaré Duality is an isomorphism in this context as well, and it is defined starting from the following isomorphism:
denote $Y'$ the space $Y$ endowed with the dual cell decomposition w.r.t. $Y$. Then there is a chain complex isomorphism:
$$ C_{n-*}(\widetilde{Y}) \to \text{Hom}_{\text{Mod}-\mathbb{Z}[\pi_1(Y)])}(\text{inv}(C_*(\widetilde{Y'}, \widetilde{\partial Y'})), \mathbb{Z}[\pi_1(Y)])$$
$c' \mapsto [-,c']$ and
$$[c,c']= \sum_{\gamma \in \pi_1(Y)} (c \cdot \gamma c') \gamma $$
where $(c \cdot \gamma c')$ is the integer intersection number of $c$ and $\gamma c'$.
I am trying to adapt the proof of Wall for the non-additivity theorem of signatures (from his article "Non Additivity of the Signature" of 1969) to the case of homology with twisted coefficients in the $(\mathbb{Z}[\pi_1(Y)],\mathbb{C})$-bimodule $\mathbb{C}$.
Most of the proof works exactly the same thanks to the properties of twisted homology, but I am really having difficulties in adapting the final geometric argument to this setting. The setting of the theorem is:
Let $Y$ be an oriented connected compact $4k$-manifold and let $X_0$ be an oriented compact $4k-1$-manifold, properly embedded into $Y$ so that $\partial X_0= X_0 \cap \partial M$. Suppose that $X_0$ splits $Y$ into two manifolds $Y_-$ and $Y_+$. For $\varepsilon= \pm$, denote by $X_\varepsilon$ the closure of $\partial Y_\varepsilon \setminus X_0$, which is a compact $4k-1$-manifold. Let $Z$ denote the compact $4k-2$-manifold $$Z= \partial X_0 = \partial X_+ = \partial X_-. $$
The manifolds $Y_+$ and $Y_-$ inherit an orientation from $Y$. Orient $X_0$, $X_+$ and $X_-$ such that
$$ \partial Y_+ = X_+ \cup (-X_0)$$ and $$ \partial Y_- = X_0 \cup (-X_-)$$
and orient $Z$ such that $$Z= \partial X_- = \partial X_+ = \partial X_0. $$
I would like to prove
Novikov-Wall non additivity theorem:
In the situation above,
$$\text{sign}_\omega(Y)= \text{sign}_\omega(Y_+) + \text{sign}_\omega(Y_-) + \text{Maslov}(L_-,L_0, L_+) $$
where $L_\varepsilon= \ker (H_{2k-1}(Z; \mathbb{C}^\omega) \to H_{2k-1}(X_\varepsilon ; \mathbb{C}^\omega)) $ for $\varepsilon=-, +, 0$.
Here $\text{sign}_\omega$ means the signature of the twisted intersection form.
In the final part of the proof we need to calculate the signature (which should give us the Maslov index term) of the twisted intersection form restricted to a subspace $L$ of $H_{2k}(Y, \partial Y; \mathbb{C}^\omega)$ which is isomorphic to $$\frac{L_0 \cap ( L_-+L_+)}{(L_0 \cap L_+)+ (L_0\cap L_-)} .$$
The idea is to calculate the signature of the intersection form on $L$ by expressing it in terms of $Z$'s own skew-hermitian intersection pairing.
Here is where I get stuck. I understand how to represent an element $b \in L_0 \cap ( L_-+L_+)$ with a $2k$-cycle $ \xi + \eta + \zeta$ in $Y$, where $\xi, \eta, \zeta$ belong respectively to $Z_{2k}(X_+,Z; \mathbb{C}^\omega), Z_{2k}(X_0,Z; \mathbb{C}^\omega), Z_{2k}(X_-,Z; \mathbb{C}^\omega)$ and I know that given $b, b' \in L_0 \cap ( L_-+L_+)$ I want to calculate the twisted intersection form of $\xi + \eta + \zeta $ and $\xi'+\eta'+ \zeta'$, however, I have no idea how to compute it.
I'm not completely sure I understand how you're defining signatures with twisted coefficients. The way I usually see such things involves some sort of self-dual sheaf of coefficients. Can you say more about the definition of your twisted intersection form?
I edited the question, adding some information. I hope it is more clear now!
Thanks. So then isn't the intersection form you're looking for on Z given by essentially the same sort of formula you give on Y, which is pretty explicit?
There is of course an intersection form on Z, but what you want to do is to express the form on the subspace L in terms of the intersection form on Z, but they are not going to be the same.
In a paper of W. Neumann, I noticed a reference to the thesis of W. Meyer, Die Signatur von lokalen Koeffizientensystemen und Faserbündeln. Neumann says that Meyer discusses some details Wall's non-additivity result with local coefficients. I don't have access to Meyer's work (it's in the series Bonner Math. Schriften). You could try Meyer's paper, Die Signatur von Flächenbündeln. Math. Ann. 201 (1973), 239–264 but there's not much on the subject in there.
Perhaps this will be of some use.
Meyer's thesis is available at https://www.maths.ed.ac.uk/~v1ranick/papers/meyer1.pdf
Thanks a lot. I am definitely going to try to read Meyer's papers.
|
2025-03-21T14:48:31.738872
| 2020-08-10T18:00:52 |
368825
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631955",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368825"
}
|
Stack Exchange
|
Albanese morphism induces an isomorphism on global $1$-forms
Let $X$ be a smooth projective variety over a field $k$ of characteristic zero equipped with a point $e\in X(k)$. There is Albanese morphism $a:X\to \mathrm{Alb}\,X$ which is initial among pointed maps to abelian varieties.
Question. Is there a purely algebraic proof of the fact that the induced map $$a^*:H^0(\mathrm{Alb}\, X,\Omega^1)\to H^0(X,\Omega^1)$$ is an isomorphism?
This map is always an injection (in arbitrary characteristic this is a theorem of Igusa, but in characteristic zero this can be proven easily by further restricting 1-forms to сurves in $X$ obtained as hyperplane sections).
The dimension of $H^0(\mathrm{Alb}\, X,\Omega^1)$ is equal to $\dim \mathrm{Alb}X$. Since $\mathrm{Alb}X$ is dual to the Picard variety, $\dim \mathrm{Alb}X=\dim \mathrm{Pic}^0 X=\dim_k H^1(X,\mathcal{O})$ where the last equality comes from the fact that $T_0\mathrm{Pic}^0X\simeq H^1(X,\mathcal{O})$ by deformation theory. Therefore, the Hodge symmetry $\dim_k H^0(X,\Omega^1)=\dim_k H^1(X,\mathcal{O})$ would imply that the map is an isomorphism by comparing dimensions. Technically, there is a "purely algebraic" proof of Hodge symmetry using $p$-adic Hodge theory, but I'm hoping that there is a simpler geometric argument that answers the original question.
|
2025-03-21T14:48:31.738989
| 2020-08-10T18:37:45 |
368828
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Donu Arapura",
"Li Li",
"abx",
"https://mathoverflow.net/users/148748",
"https://mathoverflow.net/users/40297",
"https://mathoverflow.net/users/4144"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631956",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368828"
}
|
Stack Exchange
|
Pushforward of a very ample line bundle on a curve to $\mathbb{P}^1$
Let $p:C\to\mathbb{P}^1$ be a degree $k$ morphism from a smooth projective curve $C$ to the projective line and $L$ a very ample line bundle on $C$. We know that $p_*\mathcal{O}_C(L)$ is a rank $k$ locally free sheaf on $\mathbb{P}^1$ and hence is in the form $\mathcal{O}(e_1)\oplus\cdots\oplus\mathcal{O}(e_k)$ by Birkhoff–Grothendieck theorem. In this case, are all $e_1,...,e_k$ positive?
No, not in general. Take $C=\mathbb{P}^1$, $L=\mathcal{O}(1)$, $p$ to be map $x\mapsto x^2$ in affine coordinates. Then $p_*L$ has rank $2$, but
$$2=h^0(L)=h^0(p_*L)=h^0(\mathcal{O}(e_1))+h^0(\mathcal{O}(e_2))$$
If $e_1$ and $e_2$ were both positive, then term on the right would be at least $4$. So this is impossible.
Added in response to comment. If you are allowed to pick $\deg L\gg 0$ relative to $k$, then I think it's probably true. Here's a result in that direction.
Lemma. If $\deg L\gg 0$ relative to $k$, then all $e_i\ge 0$.
Sketch. We can assume $L=\omega_{C/\mathbb{P}^1}(M)$ with $M$ globally generated. By a standard trick, we can find a cyclic cover $\pi:\tilde C\to C$ such that $L$ is a direct summand of $\pi_*\omega_{\tilde C/\mathbb{P}^1}$. Then $p_*L$ is a summand of $(p\circ \pi)_*\omega_{\tilde C/\mathbb{P}^1}$. The last sheaf is semipositive by a theorem of Fujita.
I suspect with more work, you can make the $e_i$ positive, but I leave that to you.
Thanks for your answering. But I'm still wondered whether this is true if the condition is strenghened such that the degree of $L$ is sufficiently large.
A vector bundle $E$ on $\mathbb{P}^1$ is of the form $\ \bigoplus \mathscr{O}{\mathbb{P}^1}(e_i),$ with the $e_i$ positive if and only if $H^1(E(-2))=0$. If $E=pL$, this is equivalent to $H^1(L(-2p^[0]))=0$. This will hold in particular as soon as $\deg L > 2g(C)-2+2k$.
Good. That does it.
|
2025-03-21T14:48:31.739139
| 2020-08-10T18:47:35 |
368829
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631957",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368829"
}
|
Stack Exchange
|
intuition behind shape optimization using Hadamard's method
I'm trying to understand the intuition behind shape optimization using Hadamard's method. Please consider the following simple example:
Let $\lambda$ denote the Lebesgue measure on $\mathcal B(\mathbb R)$, $d\in\mathbb N$, $U\subseteq\mathbb R^d$ be open, $$\mathcal F(\Omega):=\lambda^{\otimes d}(\Omega)\;\;\;\text{for }\Omega\in\mathcal B(U)$$ and $$\mathcal G(\Omega):=\sigma_{\partial\Omega}(\partial\Omega)\;\;\;\text{for }\Omega\in\mathcal A,$$ where $\sigma_{\partial\Omega}$ denotes the surface measure on $\mathcal B(\partial\Omega)$ for $\Omega\in\mathcal A$ and $$\mathcal A:=\{\Omega\subseteq\mathbb R^d:\Omega\text{ is bounded and open},\overline\Omega\subseteq U\text{ and }\partial\Omega\text{ is of class }C^1\}.$$
Consider the simple problems of minimizing $\mathcal F$ and $\mathcal G$ over their respective domains.
If I got it right, the problem is that there is no canonical notion of a "derivative" of $\mathcal F$ or $\mathcal G$. So, the idea is take a look at what happens to those functionals if their argument is subject to a small perturbation. The question is what we can infer from that, but let me be precise:
Let $\tau>0$ and $T_t$ be a $C^1$-diffeomorphism from $U$ onto $U$ for $t\in[0,\tau)$ with $T_0=\operatorname{id}_U$. Assume $$[0,\tau)\ni T_t(x)\tag1$$ is continuously differentiable for all $x\in U$. Moreover, assume that $$[0,\tau)\times U\ni(t,x)\mapsto{\rm D}T_t(x)\tag2$$ is continuous in the first argument uniformly with respect to the second so that we may assume that $$\det{\rm D}T_t(x)>0\;\;\;\text{for all }(t,x)\in[0,\tau)\times U\tag3$$ by taking $\tau$ sufficiently small. (I guess, intuitively, this ensures that the transformations are orientation-preserving; maybe someone can comment on this.) Let $$v_t(x):=\frac\partial{\partial t}T_t(T_t^{-1}(x))\;\;\;\text{for }(t,x)\in[0,\tau)\times U.$$
Now let $\Omega^{(1)}\in\mathcal B(U)$, $\Omega^{(2)}\in\mathcal A$ and $$\Omega^{(i)}_t:=T_t\left(\Omega^{(i)}\right)\;\;\;\text{for }t\in[0,\tau).$$ We can show that $$\frac{\mathcal F\left(\Omega^{(1)}_t\right)-\mathcal F\left(\Omega^{(1)}_0\right)}t\xrightarrow{t\to0+}\int_{\Omega^{(1)}}\nabla\cdot v_0\:{\rm d}\lambda^{\otimes d}\tag4$$ and if $\Omega^{(1)}\in\mathcal A$, then the right-hand side is equal to $$\int_{\Omega^{(1)}}\nabla\cdot v_0\:{\rm d}\lambda^{\otimes d}=\int\langle v_0,\nu_{\partial\Omega^{(1)}}\rangle\:{\rm d}\sigma_{\partial\Omega^{(1)}}\tag5,$$ where $\nu_{\partial\Omega^{(1)}}$ denotes the unit outer normal field on $\partial\Omega^{(1)}$. Moreover, $$\frac{\mathcal G\left(\Omega^{(2)}_t\right)-\mathcal G\left(\Omega^{(2)}_0\right)}t\xrightarrow{t\to0+}\int\nabla_{\partial\Omega^{(2)}}\cdot v_0\:{\rm d}\sigma_{\partial\Omega^{(2)}}\tag6,$$ where $\nabla_{\partial\Omega^{(2)}}\cdot v_0$ denotes the tangential divergence of $v_0$.
The question is: What can we infer from that? How do $(4)$, $(5)$ and $(6)$ help us to find a (local) minimum of $\mathcal F$ or $\mathcal G$?
Here is what my guess is: I think the implications are less helpful in deriving a theoretical (local) minimum, but they help to come up with a numerical "gradient descent"-like algorithm. It's trivial to observe that if $\mathcal H$ is any functional on a system $\mathcal B$ of sets contained in $2^U$ and $\Omega_t:=T_t(\Omega_0)\in\mathcal B$ for $t\in[0,\tau)$, then $$\frac{\mathcal H(\Omega_t)-\mathcal H(\Omega_0)}t\xrightarrow{t\to0+}a\tag7$$ for some $a<0$ tells us that $$\mathcal H(\Omega_t)<\mathcal H(\Omega_0)\;\;\;\text{for all }t\in[0,\delta)\tag8$$ for some $\delta>0$.
Now, we know that the family $v$ of "velocity" fields and the family $(T_t)_{t\in[0,\:\tau)}$ of transformations are in a one-to-one relationship. As the results infer (and please correct me, if this is a bad conclusion) the derivatives in above only depend on $v_0$ and hence we could show a family of transformations corresponding to a "time-independent" velocity field.
The results above tell us that if we should choose $v_0$ such that the right-hand side of $(4)$, $(5)$ or $(6)$, respectively, is negative.
But is there a "steepest descent" direction? I don't know whether we can show this, but intuitively, it seems like $v_0=-\nu_{\partial\Omega}$ would yield the "steepest descent" for $\mathcal F$. This not only yields a negative right-hand side of $(5)$, but it is even somehow plausible that we shrink the volume of the shape by squeezing it at every point in the opossite direction of the normals. Can this be made rigorous? It would obviously end in the emptyset which is clearly a global minimum of $\mathcal F$ ...
|
2025-03-21T14:48:31.739416
| 2020-08-10T18:59:59 |
368831
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Feng Hao",
"Jason Starr",
"abx",
"https://mathoverflow.net/users/13265",
"https://mathoverflow.net/users/40297",
"https://mathoverflow.net/users/63996"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631958",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368831"
}
|
Stack Exchange
|
Dimension of highest discriminants of a morphism
Let $f: X\to Y$ be a flat morphism between smooth complex affine varieties. Let $Z$ be the closed set of most singular points of $f$ (in the sense: $p$ is a most singular point of $f$ if the tangent map $Tf_p$ at $p$ has the lowest rank among the closed points in $X$). Denote $r:=\text{rank} Tf_p$ for $p\in Z$. Is it true that $\text{dim}f(Z)=r$ (generic smooth theorem tells us it is $\leq r$). If not in general, is there a quick example for $\text{dim}f(Z)<r$?
Newly updated: Is there a counterexample for proper flat morphism $f: X\to Y$ between smooth complex manifolds $X$ and $Y$?
Am I missing something? Take the standard blowup $f:\mathbb{C}^2\rightarrow \mathbb{C^2}$, $f(x,y)=(x,xy)$. Then $Z$ is the line $x=0$, $\operatorname{rk} T_p(f)=1$ for $p\in Z$ (and 2 otherwise), but $f(Z)=(0,0)$.
@abx Sorry I overlooked the trivial examples. I just added the necessary flatness condition.
This answer is cribbed from Johan de Jong's blog (if he would like to post this as an answer, I will delete this comment). Consider the morphism $f:\mathbb{A}^3\to \mathbb{A}^2$ given by $f(x,y,z)=(x,xz+y^2)$. This is flat of relative dimension $1$. The singular locus in $X$ equals $\text{Zero}(x,y)$, and the rank of the derivative map equals $1$ on this locus. Yet the image in $\mathbb{A}^2$ of the critical locus is the origin, which has dimension $0$.
@JasonStarr Thank you for pointing out the interesting example on de Jong's blog! I just voted for your comment as a right example.
|
2025-03-21T14:48:31.739551
| 2020-08-10T19:07:12 |
368832
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"GG1",
"Iosif Pinelis",
"https://mathoverflow.net/users/124312",
"https://mathoverflow.net/users/36721"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631959",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368832"
}
|
Stack Exchange
|
Is Sommerfeld radiation condition invariant under translations?
A smooth function $U:\mathbb{R}^3\setminus B_{r_0}(0)\to\mathbb{C}$ (for some $r_0>0$) satisfies the Sommerfeld Radiation Condition with index $k$, denoted $U\in \texttt{SRC}$, whenever
$$
\lim_{r\to\infty}\;\max_{|x|=r}r\,\Big|\partial_r U(x)-ikU(x)\Big|\,=\,0.
$$
Equivalently, if $U(x)=e^{ik|x|}V(x)$, then
$$
U\in \texttt{SRC} \iff \lim_{r\to\infty}\;\max_{|x|=r}r\,\Big|\partial_r V(x)\Big|\,=\,0.
$$
Here the radial derivative $\partial_r$ is defined as usual by $\partial_r U(x)=\nabla U(x)\cdot x/|x|$.
Question: does $U\in\texttt{SRC}$ imply that $U(x-x_0)\in \texttt{SRC}$, for all $x_0\in\mathbb{R}^3$?
So far, I only know a positive answer when $U$ is radial, that is $U(x)=e^{ik|x|}h(|x|)$.
Any suggestion or reference for the general case? Thanks in advance...
$\newcommand{\x}{\mathbf{x}}
\renewcommand{\a}{\mathbf{a}}
\renewcommand{\b}{\mathbf{b}}
\renewcommand{\d}{\mathbf{d}}
\newcommand{\0}{\mathbf{0}}
\newcommand{\n}{\nabla}
\newcommand{\R}{\mathbb R}
\newcommand{\tth}{\theta}
\newcommand{\thh}{\theta}$
The question can be restated as follows:
Suppose $U$ is a smooth complex-valued function defined outside a neighborhood of $\0\in\R^3$ such that for some real $k>0$
\begin{equation*}
\n U(\x)\cdot\frac\x{|\x|}-ikU(\x)=o\Big(\frac1{|\x|}\Big) \label{1}\tag{1}
\end{equation*}
(as $|\x|\to\infty$). Does it then follow that
\begin{equation*}
\n U(\x)\cdot\frac{\x+\a}{|\x+\a|}-ikU(\x)=o\Big(\frac1{|\x|}\Big) \label{2}\tag{2}
\end{equation*}
for each $\a\in\R^3$?
The answer to this question is no.
Indeed, suppose that
\begin{equation*}
U=e^{i\tth},
\end{equation*}
where $\tth$ is a smooth real-valued function defined outside a neighborhood of $\0\in\R^3$. Conditions \eqref{1} and \eqref{2} can then be rewritten as
\begin{equation*}
\n\thh(\x)\cdot\x-k|\x|=o(1) \label{1a}\tag{1a}
\end{equation*}
and
\begin{equation*}
\n\thh(\x)\cdot\d(\x,\a)\,|\x|=o(1), \label{2a}\tag{2a}
\end{equation*}
where
\begin{equation*}
\d(\x,\a):=\frac{\x+\a}{|\x+\a|}-\frac\x{|\x|}.
\end{equation*}
We need to construct a function $\tth$ satisfying \eqref{1a} but not \eqref{2a}. For $\x=(x,y,z)\ne\0$ and $u:=\sqrt{x^2+y^2}$, let
\begin{equation*}
\tth_0(\x):=\frac{1}{u^2}\,\exp\Big\{-\frac{\ln^2 |z|}{u^2}\Big\} \label{*}\tag{*}
\end{equation*}
if $uz\ne0$, with $\tth_0(\x):=0$ otherwise. The function $\tth_0$ is smooth outside any neighborhood of $\0\in\R^3$. Also,
\begin{equation*}
\n\thh_0(\x)\cdot\x=\frac{2 e^{-t^2} t^2}{u^2}-\frac{2 e^{-t^2} t}{u^3}-\frac{2 e^{-t^2}}{u^2}, \label{**}\tag{**}
\end{equation*}
where $t:=(\ln |z|)/u$. So, if $u\to\infty$, then $\n\thh_0(\x)\cdot\x=o(1)$. If $u$ is bounded but $|\x|^2(=u^2+z^2)\to\infty$, then $|z|\to\infty$ and hence $\ln |z|\to\infty$, so that
\begin{equation*}
\n\thh_0(\x)\cdot\x=\frac{2 e^{-t^2} t^4}{\ln^2 |z|}-\frac{2 e^{-t^2} t^4}{\ln^3 |z|}
-\frac{2 e^{-t^2}t^2}{\ln^2 |z|}=o(1).
\end{equation*}
So, $\n\thh_0(\x)\cdot\x=o(1)$ whenever $|\x|\to\infty$. That is, \eqref{1a} holds with $\tth_0$ and $0$ in place of $\tth$ and $k$, respectively.
However, with $\tth_0$ in place of $\tth$, \eqref{2a} does not hold for $\a=(0,0,1)$ and $\x=(x,0,e^{-x})$ as $x\to\infty$, because then
\begin{equation*}
\n\thh_0(\x)=(0,0,2e^{x-1}/x^3),
\end{equation*}
\begin{equation*}
\d(\x,\a)=\Big(-\frac{1+o(1)}{2 x^2},0,\frac{1+o(1)}{x}\Big),
\end{equation*}
and $|\x|\sim x$.
On the other hand, letting
\begin{equation*}
\tth_1(\x):=k|\x|,
\end{equation*}
we get $\n\tth_1(\x)=k\x/|\x|$, so that \eqref{1a} obviously holds with $\tth_1$ in place of $\tth$. Moreover, it is not hard to see that for any $\a$ \eqref{2a} holds as well with $\tth_1$ in place of $\tth$.
Letting finally
\begin{equation*}
\tth:=\tth_0+\tth_1,
\end{equation*}
we see that $\tth$ satisfies \eqref{1a} but not \eqref{2a}. This concludes the proof.
The initial idea behind the construction of the function $\tth_0$ is to make it a nonzero constant on the surface $z=e^{-u}$ (say), which is fast converging to the plane $z=0$ as $u\to\infty$, which will create a large gradient in a direction almost perpendicular to the plane $z=0$ at points of this surface with large $u$, provided that $\tth_0=0$ on the plane $z=0$. On the other hand, $\d(\x,\a)$ is almost perpendicular to $\x$ if $|\x|$ is large and $\x$ almost perpendicular to a fixed vector $\a$. So, by letting $\a$ be a normal vector to the plane $z=0$, we will hopefully get a violation of condition \eqref{2a}. So, one could try -- cf. \eqref{*} --
\begin{equation*}
\tth_0(\x):=\exp\Big\{-\frac{\ln^2 |z|}{u^2}\Big\}
\end{equation*}
if $uz\ne0$, with $\tth_0(\x):=0$ otherwise.
However, then instead of \eqref{**} we would have
\begin{equation*}
\n\thh_0(\x)\cdot\x=2 e^{-t^2} t^2-\frac{2 e^{-t^2} t}{u},
\end{equation*}
and therefore would be unable to say that $\n\thh_0(\x)\cdot\x=o(1)$ whenever $u\to\infty$.
The final fix is then to introduce a slowly enough decreasing factor $\frac1{u^2}$, as was done in \eqref{*}. Then everything will work as desired.
@DanieleTampieri : Thank you for hyperlinking the formulas.
Thanks so much Iosif. This is certainly a clever construction, and an interesting example.
|
2025-03-21T14:48:31.739954
| 2020-08-10T19:32:34 |
368833
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gerhard Paseman",
"Sylvain JULIEN",
"https://mathoverflow.net/users/13625",
"https://mathoverflow.net/users/3402"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631960",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368833"
}
|
Stack Exchange
|
What is the growth rate of the sum of powers of distinct primes closest to a given a integer?
Let $n$ be a positive integer, and
$$2 = p_1 < p_2 < \dots < p_m \le n$$
be the sequence of all primes less than or equal to $n$.
For each index $j$ let $p_j^{e_j}$ be the largest power of $p_j$ still less than or equal to $n$.
Define
$$S_n = p_1^{e_1} + p_2^{e_2} + \dots + p_{m}^{e_m} $$
to be the sum of these prime powers. What is the growth rate of this series $S_n$?
We can get an upper bound of
$$S_n \le nm \sim \frac{n^2}{\ln n}$$
by the prime number theorem, and a lower bound of
$$S_n \ge \lfloor n/p_1\rfloor + \lfloor n/p_2\rfloor + \dots + \lfloor n/p_m\rfloor \sim n\ln \ln n $$
using the fact that each term $p_j^{e_j}$ is within a factor of $p_j$ of $n$ and asymptotics for the sum of the reciprocals of the primes.
However, there's still some gap between these two bounds. Is the precise asymptotic growth rate of $S_n$ known?
If you just pick the primes bigger than n/2, that sum gets you close to the upper bound. You can improve upon this by taking the sum of all primes bigger than square root of n to get close to optimal. There will still be uncertainty of size sqrt(n)lnln n/in n, unless you do some hard work. Gerhard "Like Some Really Serious Arithmetic" Paseman, 2020.08.10.
Perhaps one can try to figure out the index $j$ that maximizes the quantity $n-p_{j}^{e_{j}}$ to refine the aforementioned bounds.
Also, the quantities $\delta_{j}:=n-p_{j}^{e_{j}}$ must be all different, so you can get this way a new upper bound $nm-m(m-1)/2$.
Would this be related to Sylvester Schur theorem? Gerhard "Is Really Interested In Motivation" Paseman, 2020.08.10.
A better lower bound is $S(n)$, the sum of all primes below $n$, and this lower bound makes a good asymptotic value. One can tweak this by observing that for every term corresponding to a prime less than $\sqrt{n}$ that term is at least $n^{2/3}$, so a tighter lower bound like $S(n) - S(\sqrt{n}) + n^{7/6}/\log n$ is available. Since $S(n)$ is like $O(n^2/\log n)$, one wonders how good an asymptotic is desired.
Gerhard "Wonders What This Is For" Paseman, 2020.08.10.
As observed in another comment thread, we have S(n) + O(n^{3/2}/\log n) as an upper bound, so there is not much room for improvement. Gerhard "Unless The Application Needs It?" Paseman, 2020.08.10.
Only a partial answer for now.
Denote by $M$ the quantity $\max\{n-p_{j}^{e_{j}}\}$. Then $nm-m(m-1)/2\geq S_{n}\geq nm-M-(M-1)-(M-2)-\cdots\geq nm-mM+m(m-1)/2$. So determining $M$ would get us closer to the real order of growth of $S_{n}$.
Edited after Gerhard's contributions: let now $x_n$ be the solution of $x_n=\frac{\log n}{\log x_n}$. Then a lower bound for $S_{n}$ is $\sum_{p\leq x_{n}}p^{\lfloor\frac{\log n}{\log p}\rfloor}+S(n)-S(x_n)$ but this must be difficult to estimate.
Unfortunately M is almost n minus sqrt(n), so this will lead to a poor estimate. Better to just take sum of primes below n. Gerhard "But Many Have Same Parity" Paseman, 2020.08.10.
Kinda curiously one has $S_{100}=1230=\pi(100^2)+1$. So maybe one can expect to have $S_{n}\sim \frac{n^{2}}{2\log n}$.
Indeed, one has that already using S(n) as a lower bound (as in my posted answer), and the poster's sum differs from that by less than n^3/2. Gerhard "The Powers Aren't Very Powerful" Paseman, 2020.08.10.
Also $x_n$ is the "square root for tetration" of $n$, so this might provide a "natural" explanation for the factor $1/2$ as "tetration power" (which I admit is extremely speculative).
For $n=100$, the error between $S_n$ and my lower bound is around $\frac{4}{3}\sqrt{n}\log n$, which is, up the implied constant, the error term for the PNT under RH.
Yes, but log 100 and sqrt 100 are not that far off. If you set x_n to sqrt n you will get better results for larger values of n. Gerhard "Avoid Small Problems Getting Big" Paseman , 2020.08.10.
I saw your other post relating to this one. While your quantity there may be of some theoretical interest, really using sqrt n for x_n gives you a tighter result, and although you have more powers to approximate, you can still do some good estimates of the subsum. Gerhard "Really, Go With The Root" Paseman, 2020.08.10.
|
2025-03-21T14:48:31.740260
| 2020-08-10T21:28:35 |
368840
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alexandre Eremenko",
"Ben McKay",
"Benjamin Steinberg",
"Deane Yang",
"Hollis Williams",
"Ian Agol",
"LSpice",
"Nate Eldredge",
"Pace Nielsen",
"Sergei Akbarov",
"Stanley Yao Xiao",
"Timothy Chow",
"Tom Copeland",
"darij grinberg",
"https://mathoverflow.net/users/10898",
"https://mathoverflow.net/users/119114",
"https://mathoverflow.net/users/12178",
"https://mathoverflow.net/users/13268",
"https://mathoverflow.net/users/1345",
"https://mathoverflow.net/users/15934",
"https://mathoverflow.net/users/18943",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/2530",
"https://mathoverflow.net/users/25510",
"https://mathoverflow.net/users/3106",
"https://mathoverflow.net/users/3199",
"https://mathoverflow.net/users/4832",
"https://mathoverflow.net/users/613"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631961",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368840"
}
|
Stack Exchange
|
Example of a mathematician who had problems with peer review system?
I am wondering if there is some example of a famous or well-known mathematician who often had trouble with peer review, or who often had to publish in obscure journals because referees didn't 'get' what they were saying (there could be any reasons for these troubles).
Now, you might say this is paradoxical (a well-known mathematician is probably well-known because they typically don't have serious problems with referees), but there is a definite example in physics, in the shape of Hannes Alfven. Alfven was renowned especially for his work on plasma physics and was awarded the Nobel Prize for Physics, but even after winning the Nobel, he typically had problems with the peer review process (especially in the US), complaining that referees would automatically reject his papers because they could not understand the formalism he was using, forcing him to publish many of his articles in obscure journals.
Mochizuki springs to mind.
This sort of thing is always in the eye of the beholder: on one side, "the referees are too ignorant to understand the author's brilliant ideas", and on the other, "the author doesn't write clearly and/or their ideas are incoherent".
Although I'm interested from a gossipy point of view in the answers, I think that this is not a research-level math question, and am voting to close. (If the answers need not be limited to mathematicians, then AcademiaSE might be a better fit. If they do need to be limited, then still AcademiaSE might do; or perhaps HSMSE?)
From various conversations with many mathematicians, from very senior/top of their field to graduate students, it seems that the answer to this question is roughly "everyone", except perhaps people at the Fields Medalist level.
On a trivial level, the answer is ''everyone'' and we can all vouch for that I suspect, but I mean, ''to an unusual degree''. For example, the troubles which Alfven had with peer review compared to his peers were extremely unusual, given his stature and that he was a Nobel Prize winner.
The answer is specifically limited to mathematicians.
Fourier's treatise on expanding periodic functions in terms of trigonometric functions apparently took a long time to publish, due to issues that the referees (Laplace and Lagrange) found with it. https://www.aps.org/publications/apsnews/201003/physicshistory.cfm
De Branges had trouble getting anyone to take seriously his proof of the Bieberbach conjecture.
Galois had trouble getting his work carefully read, in particular by Cauchy.
I'm not voting to close, but I'm not very hopeful about this question: Firstly, it's a spectrum (as you noticed yourself, "everyone" could be an answer), and secondly, the question is often "how deservedly", as there is lots of mathematical work (even by first-rate mathematicians) that should not get through peer review because it is unfinished, confusing or plainly wrong. So whatever answers this will get will likely be contested.
I'm sorry, I thought it was a relatively innocent question and my wording was quite specific. I simply gave the example of Alfven (a Nobel Prize winner in Physics) who had some issues with peer review peculiar only to him, and was curious what would be the closest comparison in pure mathematics,.
I suspect that most mathematicians had some negative experience with the peer review system, independently of their standing and level (this includes Fields medalists). My knowledge of this is based on private conversations, and I am not sure that it is proper to cite specific examples here.
Again, I am absolutely not asking people to divulge knowledge of private conversations. I am not sure what people have failed to understand about my question. I asked for an interesting historical example to parallel the example of Alfven of some renowned mathematician who complained openly in writing about the peer review system and had trouble with referees to a very unusual degree. I am sure I could ask this question on Physics Stack Exchange about other physicists and the answers would not be so cagey, it is strange to me that pure mathematicians are more sensitive about peer review.
It seems a lot easier to come up with examples of incorrect papers that were accepted by top journals than correct significant papers that were rejected as being wrong.
Doron Zeilberger in his opinions https://sites.math.rutgers.edu/~zeilberg/OPINIONS.html sometimes makes some complaints but it might be in a different spirit.
I find this question interesting. Wouldn't it be better to move it to another site instead of closing (if this is possible)?
@DeaneYang : There is an MO question devoted to rejections of correct papers.
Do what Rota did--start your own journal.
There does not seem to be a direct mathematical analogue of Alfven. Nobody who has won a Fields Medal or an Abel Prize has made well-publicized complaints about how they have had an unduly difficult time with the peer-review system.
Some partial analogues have been mentioned in the comments. Louis de Branges had trouble getting people to take his proof of the Bieberbach conjecture seriously, but there was a clear reason: de Branges had a well-deserved reputation for making repeated but incorrect claims to have solved famous open problems. Fourier's work met with resistance, but there were some legitimate objections that some crucial arguments were not fully clear; while the Mochizuki saga is still in progress, it seems that the clarity of certain crucial arguments is still in dispute. Galois also ran into difficulties, although the story has been exaggerated, and Galois was not a recognized figure at the time.
Why there is no direct analogue of Alfven is an interesting question. There could be a difference in culture. Mathematicians will often complain privately of getting an unfair rejection, but complaining too loudly or publicly about the lack of recognition of one's own work is generally regarded as whining, and is frowned upon. Mathematicians also take the general attitude that there are three primary reasons for rejection: (1) the work is wrong; (2) the writing is unclear; (3) the result is not regarded by community as sufficiently interesting. The first two are considered to be the fault of the author, and mathematicians tend to take a somewhat fatalistic attitude toward the third. Under these prevailing assumptions, it is hard to be viewed as taking the moral high ground when voicing a complaint about a rejection.
One example of someone making a public complaint was Friedrich Wehrung (search for the word "rejection"). His primary complaint, however, was that the entire subfield of lattice theory was being looked down upon unfairly. Similarly, you can find public discussion about how the peer-review system unfairly treats certain groups of people. But again, if you're looking for an individual analogue of Alfven, I don't think there exists a very good example.
Rota just started his own journal--cunning.
|
2025-03-21T14:48:31.740758
| 2020-08-10T23:25:34 |
368848
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alec Rhea",
"Andreas Blass",
"Andrés E. Caicedo",
"Frank Quinn",
"Monroe Eskew",
"https://mathoverflow.net/users/11145",
"https://mathoverflow.net/users/124645",
"https://mathoverflow.net/users/6085",
"https://mathoverflow.net/users/6794",
"https://mathoverflow.net/users/92164"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631962",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368848"
}
|
Stack Exchange
|
Universal ZFC set theory?
What would consequences be of a model $M$ of Zermelo-Fraenkel-Choice set theory that is universal in the sense that any other model is isomorphic to a transitive subclass of $M$? Any impact on large cardinals or forcing?
Added (2020,8,16): It is known that a model with a measurable cardinal has an elementary extension. The universal model has no extensions, so presumably no measurable cardinal.
What exactly do you mean by one theory being a transitive subclass of another theory? It sounds as if you're conflating theories with classes.
@AndreasBlass I think he means a model M of ZFC such that any other model is isomorphic to a transitive subclass of M.
What about ill-founded models?
He could ask for a universal well-founded model and a universal ill-founded model, no?
@Alec No.${{}}$
@AndrésE.Caicedo I don’t follow? Is ${{}}$ a counterexample to asking for one of each?
@AndrésE.Caicedo Ah, the ${{}}$ was just to get over the character limit — could you please elaborate on why we can’t ask for one of each?
It may also be interesting to ask for a 'minimal' universal model, i.e. a model $N$ of ZFC such that any other model $N'$ contains an isomorphic copy of $N$ as a transitive subclass. (we may also want elementary embeddings so first-order statements are preserved)
Today's edit of the question makes it even less clear. The obvious elementary extension of a model $M$ with a measurable cardinal, namely the ultrapower of $M$ with respect to a countably complete ultrafilter, is isomorphic to a transitive submodel of $M$. The edit does suggest that the question was intended to be about well-founded models; otherwise measurability wouldn't be relevant to the existence of an elementary extension.
@AndreasBlass The question is about well-founded models. It was my understanding that "foundation" is one of the axioms of ZFC (cf Jech 2003 book)
"Foundation" is indeed an axiom of ZFC, but it doesn't imply that all models of ZFC are well-founded. It only ensures that the sets that lack minimal elements (the counterexamples to well-foundedness) are not elements of the model.
|
2025-03-21T14:48:31.740927
| 2020-08-10T23:35:56 |
368849
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631963",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368849"
}
|
Stack Exchange
|
Formally etale algebras over fields of characteristic 0
I was wondering if anyone might have a non-trivial example of a formally etale algebra over a field of characteristic 0 which is not ind-etale (i.e. a union of etale extensions).
For some motivation, over fields of characteristic $p$ there are many such: just take the limit or colimit perfection of any $\mathbf{F}_p$-algebra: e.g. $\mathbf{F}_p[t^{1/p^{\infty}}].$
I have not been able to construct similar examples over $\mathbf{Q}$ say and am wondering if anyone here has come across such an example. Perhaps there are formal reasons why it is not possible?
I think the answer is no if you stick to noetherian rings: Assume $f\colon \mathbb{Q} \to S$ is formally etale. Let $K_1, \dotsc, K_n$ denote the fraction fields of $S$ (= localizations at generic points of the irreducible components; note that $S$ is reduced since $f$ is regular (Stacks 07EL)).
Now since localizations are ind-etale the maps $\mathbb{Q} \to K_i$ are formally etale. Using Stacks 00UO $$\Omega^1_{K_i/\mathbb{Q}} = 0$$ and so $K_i/\mathbb{Q}$ is separable algebraic. In particular $\mathbb{Q} \to K_1 \times \dotsb \times K_n$ is integral. This implies that $S \to K_1 \times \dotsb \times K_n$ is also integral. Hence, $S$ is also zero-dimensional so coincides with $K_1 \times \dotsb \times K_n$.
I don't know about the non-noetherian case but generally some more pathologies can occur then (see for instance, Mukhopadhyay and Smith - Reducedness of formally unramified algebras over fields for an example of a non-reduced formally unramified $\mathbb{Q}$-algebra). Also I realize they also prove what I said above (with weaker assumptions) in Corollary 3.3 of that paper (formal unramifiedness is enough).
2022/9/21 Edit: There is a paper on the arxiv now which answers the question negatively if one also assumes $A$ reduced: Mondal and Mukhopadhyay - Ind-étale vs Formally étale
Theorem 1.3 of the paper says
Let $k$ be a field of characteristic zero and $A$ be a reduced
$k$-algebra – not assumed to be noetherian. If $\Omega_{A/k} = 0$, then $A$ is ind-etale.
Note that if $k \to A$ is formally etale, then $\Omega_{A/k} = 0$ (Stacks 04FE).
|
2025-03-21T14:48:31.741088
| 2020-08-10T23:59:36 |
368850
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"F. C.",
"https://mathoverflow.net/users/10881"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631964",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368850"
}
|
Stack Exchange
|
Bernoulli-like polynomials
Let $\psi_0 (x,t)=\frac{te^{xt}}{1-e^{-t}}$. Then
$$\psi_0(0,t)=\frac{t}{1-e^{-t}};$$
$$\psi_0(x,t)=1+\sum_{n=1}^\infty \frac{t^n}{n!} B_n(x)$$
where $B_n$ is a monic polynomial of degree $n.$
Now let
$$\psi_1(y,t)=\int ( \frac{1}{(1-e^{-t})^2 (y-1)} (y+(1-y)e^{-t})\log (y+(1-y)e^{-t})-\frac{1}{1-e^{-t}})dt$$
normalized so that $\Psi_1(y,0)=1.$ Then
$$\psi_1(0,t)=\frac{t}{1-e^{-t}};$$
$$\psi_1(x,t)=1+\sum_{n=1}^\infty \frac{(-1)^n}{n^2(n+1)}{t^{2n}} P_n(x)$$
where $P_n$ is a monic polynomial of degree $n.$
Also, let
$$\psi_2(y,t)=\int \frac{1}{\sinh(\frac{t}{2})}(1-\frac{\tanh^{-1}((y+1)\tanh(\frac{t}{2}))}{(y+1)\tanh(\frac{t}{2})})dt$$
Then
$$\psi_2(0,t)=\frac{t}{1-e^{-t}};$$
$$\psi_2(x,t)=1-\sum_{n=1}^\infty \frac{t^{2n}}{2^{2n+1}n(2n+1)} Q_{2n}(x)$$
where $Q_{2n}$ is a monic polynomial of degree $2n,$ even with respect to $y+1.$
Questions. Are these polynomials known? Do they appear in any proof of index/Riemann-Roch theorems? Perhaps in relation to gamma functions for associators?
Is the power $t^{2n}$ before $P_n$ correct ? Maybe it is rather $t^{n}$ ?
|
2025-03-21T14:48:31.741174
| 2020-08-11T02:59:05 |
368854
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Qiuhai Zeng",
"https://mathoverflow.net/users/163586"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631965",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368854"
}
|
Stack Exchange
|
Is this $(\Bbb R^{n \times n})^n \to \Bbb R$ function convex?
Let $W := (W_1, W_2,\dots, W_n)$, where $W_i \in \Bbb R^{n \times n}$. Let $x$ be a constant vector. Is the following function convex?
$$f(W) := x^TW_1^TW_2^T \cdots W_n^TW_n \cdots W_2W_1x $$
Function $W \mapsto f(W)$ is not convex when $n \ge 2$.
Take for example when $n=2$ and fix $x=1$.
If function $f(W_1,W_2)$ is convex then it must also be convex when restricted to a linear subspace.
Restricting to $W_1 = \begin{pmatrix} u & 0 \\ 0 & 0 \end{pmatrix}$ and $W_2 = \begin{pmatrix} v & 0 \\ 0 & 0 \end{pmatrix}$ then gives $f(W_1,W_2) = u^2 v^2$, which is not convex in $(u,v)$ (this can be checked by computing its Hessian matrix).
Thank you! It's very helpful!
|
2025-03-21T14:48:31.741510
| 2020-08-11T04:34:34 |
368857
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"AGenevois",
"Carl-Fredrik Nyberg Brodda",
"HJRW",
"YCor",
"https://mathoverflow.net/users/120914",
"https://mathoverflow.net/users/122026",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/1463",
"https://mathoverflow.net/users/15728",
"qkqh"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631966",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368857"
}
|
Stack Exchange
|
Term and theories about "relation-free" elements in a group?
For a group $G$, there are two elements a, b which are "relation-free",
i.e., there is no nonempty, reduced word $W(X,Y)$ such that $W(a,b)=1$ in $G$.
Is there any terminologies or theories for that?
(My group is a free product of finite cyclic groups, and I'm considering one element is not contained in a factor, but is a product of generators of each factors.)
In other words, $\langle a,b \rangle$ is a free subgroup with ${a,b}$ as a basis, no?
In addition to the comment of @AGenevois, as your setting is in the subgroups of a free product of finite cyclic groups you may be interested in the Kurosh subgroup theorem.
I'm calling this a "free pair", and more generally $(x_1,\dots,x_n)$ a free $n$-tuple if the corresponding map $F_n\to G$ is injective.
@AGenevois;@Carl-Fredrik Nyberg Brodda Yes, it is. Maybe, I should ask more, I want to know some about a normally generated subgroup with the two elements. For instance, $\langle X^n, Y\rangle \subsetneq \langle X,Y\rangle$ for any $n>1$, but I don't know whether $\langle\langle X^n, Y\rangle\rangle_G \subsetneq \langle\langle X,Y \rangle\rangle_G$.
Understanding normal closures may be quite difficult. But for free products of finite groups, powerful tools are available. What is your question precisely?
@AGenevois What is the powerful tools for normal closures in a free product of finite groups? Precisely, I want to know whether $\langle\langle X^n, Y \rangle\rangle_G \neq \langle\langle X, Y \rangle\rangle_G$ for $G$ the free product of 4 finite cyclic groups. Here, $X$ can be considered as a product of generators of each factor, and $X$ and $Y$ are "relation-free" (i.e. $\langle X,Y \rangle\cong \mathbb{Z}\ast\mathbb{Z}$).
Usual tools are: normal form, Bass-Serre tree, small cancellation. From small cancellation, a positive answer to your question can be proved if the maximal length of a common subword in (conjugates of) $X$ and $Y$ is sufficiently small compared to the lengths of $X$ and $Y$. In full generality, I don't know. But it should be true that there exists some $N$ such that $\langle \langle X^n,Y^m \rangle \rangle = \langle \langle X^p,Y^q\rangle \rangle$ iff $(n,m)=(p,q)$ for all $n,m,p,q \geq N$.
@AGenevois Thank you. I tried using normal form, but I don't know how for the normal closure. I wonder the case that you gave a positive answer. Isn't it depend on the relators of G (not only the word corresponding to $X$ and $Y$)? Can it be independent with G? Also, on your last statement.
@qkqh: Yes, the argument is essentially independent of the finite groups used to defined $G$. One way to justify this could be to say that small cancellation only deals with loxodromic isometries in the Bass-Serre tree while finite groups are elliptic. So there is no interaction between the two.
As @AGenevois says, small-cancellation techniques can probably be used to show that $\langle\langle X^n,Y\rangle\rangle\neq \langle\langle X,Y\rangle\rangle$ for all sufficiently large $n$. For smaller $n$ (i.e. $n>1$), this is likely to be closer to the theory of one-relator groups with torsion. I'd recommend checking out the papers of Jim Howie, who has done a lot of work on one-relator quotients of free products, IIRC.
@HJRW Sorry, but I could not find theorems to obtain it (for sufficiently large n). Could you tell me what I refer to more specifically?
@qkqh: there’s a vast literature on small-cancellation techniques, and you will need to learn some of it. Perhaps you should start by looking at Denis Osin’s paper on combinatorial Dehn filling.
|
2025-03-21T14:48:31.741758
| 2020-08-11T07:27:13 |
368860
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Christoph Mark",
"Hugh Denoncourt",
"LSpice",
"darij grinberg",
"https://mathoverflow.net/users/22379",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/2530",
"https://mathoverflow.net/users/66288"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631967",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368860"
}
|
Stack Exchange
|
Words that give rise to an enumeration of elements of the symmetric group
Let $\mathbb{S}_m$ be the symmetric group on $m$ letters. Let $n=m-1$. Let $\mathbf{w}=a_1\cdots a_r$ be a word on the alphabet $\{1,\ldots,n\}$. We say that $\mathbf{w}$ gives rise to an enumeration of elements of the symmetric group $\mathbb{S}_m$ if $1,s_{a_r},s_{a_{r-1}}s_{a_r},\ldots,s_{a_2}\cdots s_{a_r},s_{a_1}\cdots s_{a_r}$ are all distinct elements of the symmetric group $\mathbb{S}_m$.
For example, $32323132323132323$ gives rise to an enumeration of elements of the symmetric group $\mathbb{S}_4$, and I think it is (one of) the longest possible such words.
Q. What is the length $r(n)$ of the longest word which gives rise to an enumeration of elements of the symmetric group $\mathbb{S}_m$. Is it always possible to find such a word of length $m!$, or only for $\mathbb{S}_3$?
You are trying to list all $m!$ permutations in $\mathbb{S}_m$ in such a way that each permutation in the list is obtained from the previous one by interchanging two adjacent entries. This is done in Knuth's TAoCP, volume 4A, Section <IP_ADDRESS>, under the subheading "Adjacent interchanges". (This is one of the things classically referred to as "change ringing".)
An old version of Section <IP_ADDRESS> is available at https://www-cs-faculty.stanford.edu/~knuth/fasc2b.ps.gz .
Thanks @darijgrinberg, it seems to be a wellknown, wellstudied issue. Do you want to make your comment an answer? So that you get the points... Do you immediately see the, presumably also known, relation to the Bruhat graph of all paths from $1$ to $v$ where $v$ is a permutation whose reduced expression has precisely $m$ distinct simple reflections? - the relation between that graph of paths where $v$ is such in some $\mathbb{S}_{m'}$ with $m'>m$ and the enumeration in $\mathbb{S}_m$?
Based on your questions list, you might like Conway's "Gray Codes for Reflection Groups" which generalizes this to all finite reflection groups.
Is $s_a$ the transposition $(a, a + 1)$?
@LSpice Yes, it is the simple transposition.
|
2025-03-21T14:48:31.741915
| 2020-08-11T08:30:26 |
368862
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631968",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368862"
}
|
Stack Exchange
|
2D Fourier transform of log function
I am studying the paper found here. Halfway in the paper (Equation 6), the inverse 2D Fourier transform of $1/(k_x^2+k_y^2)$ needs to be determined. Is is stated that this is straightforward, and that the inverse is given by $-\log(r)=-\log(\sqrt{x^2+y^2})$. I am having trouble verifying this; how can this be derived?
If $v(r)=\log r$, then $\Delta v=\delta$ and $\hat {(\Delta v)}=1$, that is $-(\xi^2+\eta^2)\hat {v}(\xi, \eta)=1$. However this yields $-\hat{v}=1/(\xi^2+\eta^2)$ which has no sense since this function is not locally integrable. A precise computation, where a principal value of $1/(\xi^2+\eta^2)$ appears, can be found in Chapter 9 of the book Vladimirov, Equations of mathematical physics.
|
2025-03-21T14:48:31.742000
| 2020-08-11T11:42:22 |
368866
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"JamalS",
"Jeremy Rouse",
"JoshuaZ",
"https://mathoverflow.net/users/127690",
"https://mathoverflow.net/users/48142",
"https://mathoverflow.net/users/50447"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631969",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368866"
}
|
Stack Exchange
|
On $x^2 + y^3 = z!$ with $y \ge 0$
The largest solution that I found to equation $x^2 + y^3 = z!$ with $y \ge 0$ as below.
$$6389296200^2 + 2173500^3 = 21!$$
It is obvious that if $z!/t^{6} = x_{t}^2 + y_{t}^3$ then $z! = (x_{t}t^3)^2 + (y_{t}t^2)^3$. So for a large value of $z$, there can be certain set of $t$ values to check existence of a smaller solution that provides $x^2 + y^3 = z!$ is soluble.
In that direction, I believe that there can be search to other solutions to this equation.
Question. Are there other solutions to $x^2 + y^3 = z!$ for $z > 21$? (It is checked up to $33!$).
I don't have any reference on that equation except https://oeis.org/A273553. So any reference is very welcome if this is well-known equation.
Thanks.
Correct me if I'm wrong but technically you can't call this a Diophantine equation because of the factorial, surely?
Heuristically, this should have finitely many solutions, since the numbers of numbers of the form $x^2 +y^3$ which are at most t, is $O(t^{5/6})$, and the set of factorials grows faster than exponential. I'd be surprised if there are any larger ones than the solution for 21!. On the other hand, I'm a bit surprised that that one exists at all.
From the perspective of elliptic curves, the restriction that $y \geq 0$ is somewhat unnatural (and removing it finds more solutions, for example with $n = 32$). If you remove this restriction, there are in fact $9$ representations of $21!$ in the form $x^{2} + y^{3}$ (including a second with $y > 0$: $5462360064^{2} + 2769984^{3}$).
|
2025-03-21T14:48:31.742252
| 2020-08-11T12:27:09 |
368869
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Fedor Petrov",
"Null_Space",
"https://mathoverflow.net/users/192600",
"https://mathoverflow.net/users/4312"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631970",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368869"
}
|
Stack Exchange
|
Self-complementary graph on 4k + 1 vertices
I am self-studying the graph theory book by Bondy and Murty.
On page 31, the author suggests:
Every self-complementary graph on $4k + 1$ vertices has a vertex of degree $2k$.
Here what I have so far:
Given a self-complementary graph $G = (V, E)$ where $n = 4k + 1, k \geq 1$. We have
$$
m = \frac{4k \cdot (4k + 1)}{4} = k \cdot (4k + 1).
$$
Furthermore, let $d_G$ denote the sum of degrees of all vertices, then
$$
d_G = 2k \cdot (4k + 1) = \frac{n-1}{2} \cdot n
$$
So we need to prove that there must exist a vertex with the degree $(n-1)/2$, given $d_G = (n-1)/{2} \cdot n$.
Why cannot we have the degrees of all vertices different from $(n-1) / 2$, but still achieving the $(n-1) / 2$ average degree?
Because if we have more than $2k$ vertices of degree larger than $2k$, then the complement graph has more than $2k$ vertices of degree smaller than $2k$.
Such an elegant answer
As the graph is self-complementary, by definition there exists some isomorphism $f$ from the graph to its complement. Consider $f$ as a bijection from the vertex-set of the graph to itself (i.e. a permutation on the vertex-set), and consider the cycle structure of this permutation. As $4k + 1$ is odd, there must be at least one odd cycle. Every vertex in this cycle has the same degree and complementary degree, i.e. $2k$.
|
2025-03-21T14:48:31.742377
| 2020-08-11T12:37:51 |
368870
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"JamalS",
"https://mathoverflow.net/users/50447"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631971",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368870"
}
|
Stack Exchange
|
Zhu's $V/C_2(V)$ algebra
As an example, take the Virasoro algebra, i.e. $V$ is spanned by elements of the form $L_{-2}^{k_1} \cdots L_{-n}^{k_{n-1}} \Omega$ where $\Omega$ is the vacuum and $n \geq 2$. As I understand, we define
$$C_2(V) = \{\psi^i_{-h_i-1}v | v \in V\}$$
where $i$ labels all the operators strongly generating the VOA, and $h_i$ are the conformal dimensions. So, for the Virasoro algebra one has,
$$C_2(V) = \{L_{-3}v | v \in V\}.$$
Now it is claimed that $V/C_2(V) \cong \mathbb C[x]$, where the isomorphism maps $L_{-2}\Omega \mapsto x$. However, I do not see how this is the case. We are modding out all elements of the form $L_{-3}v$, i.e. all those that contain a factor of $L_{-3}$. Then say, $L_{-10}\Omega$ cannot be part of an equivalence class with a representative given by a power of $L_{-2}\Omega$, so I do not see how $V/C_2(V)$ is just generated by $L_{-2}\Omega$.
For reference, see page 14 of Arakawa's paper.
You have underdefined the $C_2(V)$ subspace.
It is as you wrote but with $\psi^i$ being any element of $V$. You can intuitively think of this as being the subspace of iterated normally ordered products of vertex operators for which at least one is a derivative.
Thus in your Virasoro example, $L_{-10}\Omega\in C_2(V)$ by setting $\psi^i=\partial^7T$ where $T$ is the stress tensor.
Thank you for your reply, and I would just add thank you for your chiral algebra construction for 4D N=2 SCFTs which I based my MSc thesis on last year :)
|
2025-03-21T14:48:31.742505
| 2020-08-11T12:52:13 |
368873
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"https://mathoverflow.net/users/62562",
"user62562"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631972",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368873"
}
|
Stack Exchange
|
Are K_t-minor free graphs on small vertex sets understood?
In a paper on Hadwiger's conjecture, https://web.math.princeton.edu/~pds/papers/hadwiger/paper.pdf, Seymour explains various results on excluding the complete graph as a minor.
In particular, there is a nice bound on the number of edges, due to Mader, as follows:
If $t\leq 7$ and $n\geq t-2$ then every $K_t$-minor free graph $G$ on $n$ vertices has
$$ |E(G)|\leq (t-2)n-\frac{(t-1)(t-2)}{2}.$$
(The obvious application to Hadwiger, as Seymour describes, is that this bound controls the average degree and hence gives low minimum degree to which you can use induction to get a colouring result.)
Sadly the pattern fails for $t\geq 8$. Jorgensen and then Song and Thomas describe the exceptions explicitly for $t=8$ and $t=9$ respectively but as far as I can see no larger $t$ is understood. To be more precise Jorgensen's result says that if there is no $K_8$-minor and the displayed inequality fails then the exact number of edges is known and $G$ can be built from $K_{2,2,2,2,2}$ in a simple way (by "pure 5-sums").
However, in general the average degree in a $K_t$-minor free graph can be large - Kostochka, Thomason, ... - so the "exceptions" will get really bad for large $t$.
I'm interested in this from a slightly different perspective. Are explicit descriptions known (or potentially tractable) of $K_t$-minor free graphs for general $t$ provided that $n$ is quite small relative to $t$?
So a desired result would be something like:
For any natural number $t$ and any $t-2\leq n \leq 2t$, every $K_t$-minor free graph $G$ on $n$ vertices has
$$ |E(G)|\leq (t-2)n-\frac{(t-1)(t-2)}{2},$$
unless $G$ is ... some exceptions like those in Jorgensen/Song and Thomas ...
If this is somehow easy then replace the vertex bound of $2t$ with something bigger.
There is no known straightforward answer, but pseudorandom graphs must come into the answer. See the paper by Myers and Thomason.
[In response to the comment below] Look at recent papers by Postle--Norine--Song, plus earlier work by Reed--Kawarabayashi, all on Hadwiger's Conjecture. You will see that the difficulty with K_t-minor-free graphs often occurs when the number of vertices is small when compared to t. In particular, for a suitable choice of c a random graph with ct sqrt(log t) vertices has no K_t minor. This says that small K_t minor-free graphs are wild! Another important conjecture (of Seymour and Thomas) is that the above edge bound does hold for sufficiently large highly connected graphs. Here "sufficiently large" is essential, otherwise random graphs provide counterexample.
Thanks. Could you elaborate a bit though? I glanced at the paper by Myers and Thomason but it's extremal - my question is wondering if restricting the number of vertices, in terms of the size of the minor, makes things sufficiently easier for general complete graphs that something nice can be said.
Thanks. Given the info in the edit, I've accepted the answer. For the application I had in mind it's bad news but Norine-Postle-Song's Theorem 2.2 (of Woodall) might help instead.
|
2025-03-21T14:48:31.742720
| 2020-08-11T13:44:37 |
368874
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631973",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368874"
}
|
Stack Exchange
|
Support of the fundamental class of irreducible components of partial Springer fibres corresponding to flag signatures
The (partial) Springer resolution is defined as a map $\mu: T^*\mathcal{F} \to \mathcal{N}$, where $\mathcal{F}$ is the partial flag variety consisting of $n$-step partial flags of $\mathbb{C}^d$, and $\mathcal{N}$ a set of nilpotent operators $\mathcal{N} = \{X \in \text{End}(\mathbb{C}^d) : X^n = 0\}$. A partial Springer fibre is the set $\mathcal{F}^X := \mu^{-1}(X) = \{F_\bullet : X F_i \subset F_{i-1}\}$.
Ginzburg has a construction of certain irreps of $\mathfrak{sl}_n$ on the (top Borel-Moore homology) of these fibres $H(\mathcal{F}^X)$ (the vector space spanned by the fundamental classes of the irreducible components of $\mathcal{F}^X$). In particular, they are irreducible, with highest weight given by the Jordan type of $X$ (a partition consisting of the sizes of the Jordan blocks in its decomposition), see Chriss-Ginzburg Theorem 4.2.3 for example.
In the proof of this theorem, the first part goes (below is an abridged version):
For any partition (or sequence/composition) $\lambda$ the fundamental class of the diagonal $T_\Delta^*(\mathcal{F}_\lambda \times \mathcal{F}_\lambda)$ acts as the identity map on any cycle in $H(\mathcal{F}^X)$ supported only on $\mathcal{F}_\lambda$, annihilating any other cycle.
I am looking hopefully for a reference to why this statement should be true. I can think of an equivalent statement that is (at least) easier for me to understand:
Let $F$ be an irreducible component of the partial Springer fibre $\mathcal{F}^X$, and let $[F]$ be its fundamental class in the top Borel-Moore homology. Then the support of $F$ is contained in $\mathcal{F}_\mu$ if and only if $F \subseteq \mathcal{F}_\mu$.
A point to make here is that the general definition (that I'm comfortable with) of the support of a cycle (as a sum of singular simplices) is the union of the images of each of the simplices.
In the case of the Borel-Moore homology, it seems that the support of a cycle (as a linear combination of fundamental classes - the distinguished basis) is the union of the irreducible components whose fundamental classes appear with nonzero coefficient in the expression of that cycle in the vector space $H(\mathcal{F}^X)$.
I am unable to see how to reconcile these two definitions, but surely they must be equivalent in the BM homology.
In lieu of a reference to the previous statement/s or even the proof in Ginzburg's book, a hint on why they may or may not be true (even just morally), or even better a reference/outline, would be desired in the answer.
Finally, this is a repost of the question I asked on MSE found here.
This is also my first time posting on MO, by advice of my supervisor - please let me know if there are any improvements I can make to the question.
|
2025-03-21T14:48:31.742923
| 2020-08-11T14:06:09 |
368875
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"LSpice",
"MathCrawler",
"Will Sawin",
"https://mathoverflow.net/users/108274",
"https://mathoverflow.net/users/161310",
"https://mathoverflow.net/users/18060",
"https://mathoverflow.net/users/2383",
"user267839"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631974",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368875"
}
|
Stack Exchange
|
Algebraic spaces as quotients of schemes (Definition from wikipedia)
I think that wikipedia article on Algebraic spaces contains a serious content error in the part on the definition of Algebraic spaces as quotients of schemes and I would like to discuss if it is indeed an error or I just missing something. The definition says:
An algebraic space $X$ comprises a scheme $U$ and a closed (!) subscheme
$R \subset U \times U$ satisfying the following two conditions:
$R$ is an equivalence relation as a subset $U \times U$;
the two projections $p_i: R \to U$ onto each factor are étale maps
Some authors, such as Knutson, add an extra condition that an algebraic space has to be quasi-separated, meaning that the diagonal map is quasi-compact.
One can always assume that $R$ and $U$ are affine schemes. Doing so means that the theory of algebraic spaces is not dependent on the full theory of schemes, and can indeed be used as a (more general) replacement of that theory.
I think that in this definition is a subtle problem hidden, since it is important to distinguish strictly if $U$ may be always assumed to be affine or not when we work with the condition that the $R \subset U \times U$ is assumed to be a closed subscheme as I will explain below why. So I'm not sure why it can "one can always assume that $R$ and $U$ are affine". That's doesn't matter only in the case if $U$ is moreover separated as we will see in following example:
It is well know that an arbitrary scheme $S$ is canonically an algebraic space. If we want to verify it using the characterization above we have to recognize it as a quotient of schemes, so we need a pair $(U,R)$ representing $S$ and satisfying axioms $1$ and $2$.
For $U$ we set $U:=S$ and for $R$ we need a closed subscheme of $U \times U$ and as canonical choice for $R$ might be choosen the closure $R:=\overline{\Delta(U)}$ of the image with respect the diagonal map $\Delta: U \to U \times U$. Here I'm not sure if that is a correct choice for $R$ in the case when $S$ is not separated, but it seems to be the only "canonical" choice. And if this choice for $R$ is correct, we have a problem: Why is the restriction of the projection map $p \vert _{\overline{\Delta(U)}}: \overline{\Delta(U)} \to U$ etale? (take the closure bar into account)
Clearly $p \vert _{\Delta(U)}: \Delta(U) \to U$ is etale since $U \cong \Delta(U)$ and $p \circ \Delta= id_U$, but in general if $f: X \to Y$ is a morphism of schemes, $Z \subset X$ an locally closed immersion (but not cl imm) and the restriction $f \vert _Z:Z \to Y$ is etale, then in general $f \vert _{\overline{Z}}: \overline{Z} \to Y$ is not more etale.
Thus there is no reason if we have a scheme $S (=U)$ with $\overline{\Delta(U)} \not \cong \Delta(U)$ for restriction $p \vert _{\overline{\Delta(U)}}: \overline{\Delta(U)} \to U$ to be etale. On the other hand affine schemes are separated and therefore $\overline{\Delta(U)} = \Delta(U)$, but for general scheme $S$ the requirement that $R \subset U \times U$ is closed is stronger than to say one may always assume that $R$ and $U$ are affine schemes. Or do I confuse something here? Is there maybe a better choice for $R$ involved?
In summary the concern of my question can be reduced to two question:
A: Is for arbitrary scheme $S$ the restriction of projection $p \vert _{\overline{\Delta(S)}}: \overline{\Delta(S)} \to U$ etale?
B: If A is true, then B is clear. If A is wrong, why it is sufficient to work with this definition of algebraic spaces with affine schemes in viewpoint the described problemwith closedness of $R$ in $U \times U$?
I don't know the answer to this question, or if it is an error, but I hope very much that MO doesn't become a place to go to report mathematical errors in Wikipedia. (Which is to say, I think that the post would be better if it foregrounded the question and perhaps gave a low-key mention of Wikipedia as the motivation.)
Since Wikipedia contains a lot of useful information and is consulted by many people many times, I think there is some benefit in discussing possible errors here in this forum.
In the modification I tried to point out the key questions arising from my concern on the discussed definition. I understand your concern, but I think that it highly depends on the specific case what kind of error in considered. A simple typo or a wrongly quoted basic definition from an undergrade lecture is one thing. But here the question treats a subtle problem. The question is if a whole theory can be made up from affine pieces or not.
Here the definitions would become completely different in dependence of the requirement that $R$ is assumed to be closed or locally closed in $U \times U$. If the $R$ would be assumed as locally closed then indeed the theory can be reduced to $U$ affine. But if $R$ assumed to be closed, can the theory nevertheless reduced to affine "pieces"?
If we take $R$ to be the closure of the diagonal, I don't understand why the quotient of this equivalence relation is $S$, even before reducing to the affine case.
@Will Sawin: For a non separated scheme $S$ I'm not sure if the choice $R:=\overline{Δ(S)}$ is the right one, it seems just to be somehow "canonical", that is it also not sure why the obtained quotient equals S. In case $S$ affine or more general separated, the reason why this quotient equals $S$ should be clear, right?
For a non separated scheme $S$ there is actually no closed equivalence relation whose quotient is $S$. Such equivalence relations will always produce separated algebraic spaces. (I think it's not so hard to check the valuative criterion of separatedness from the definition). So the Wikipedia definition is only a definition of separated algebraic spaces, and in particular doesn't generalize schemes.
|
2025-03-21T14:48:31.743294
| 2020-08-11T14:31:41 |
368877
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Jochen Wengenroth",
"erz",
"https://mathoverflow.net/users/21051",
"https://mathoverflow.net/users/53155"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631975",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368877"
}
|
Stack Exchange
|
Equicontinuity-like property of a convex compact set
Let $X$ be a Tychonoff topological space and let $x\in X$. Let $B\subset C(X)$ be convex and compact in the topology of pointwise convergence, and such that $f(x)=1$, for every $f\in B$.
Is there an open neighborhood $U$ of $x$ such that $f(y)\ne 0$, for every $f\in B$ and $y\in U$?
Without convexity a counterexample would be $X=\mathbb R$ and $B={f_n:n\in\mathbb N} \cup{1}$ where $f_n$ are $1$ outside $(0,2/n)$, $f_n(1/n)=0$ and affine linear and on the two halfs of $[0,2/n]$ so that they are continuous. Then $f_n\to 1$ pointwise which implies compactness.
@JochenWengenroth yeah, perhaps I should have put this example in the question itself. Luckily (?), in my situation I have additional convexity assumption
Since $C(X)$ is not complete one cannot take the closed convex hull of the example in the comment. But what about this:
Let $g_n=1-f_n$ with $f_n$ as in my comment.
Since the $g_n$ are bounded by one, the linear map $T:\ell^1\to C(X)$, $\lambda\mapsto \sum\limits_{n=1}^\infty \lambda_ng_n$ is well defined. We would like to have $T$ continuous as a map $(\ell^1,\sigma(\ell^1,c_0)) \to (C(X),pw)$ (where $c_0$ is the space of all null sequences so that $\ell^1$ is its dual). This follows from the pointwise convergence to $0$ of all sequences $(g_n(x))_{n\in\mathbb N}$. The unit ball $K$ of $\ell^1$ is weak$^*$ compact à la Alaoglu and hence $T(K)$ is compact and convex in $(C(X),pw)$. Then $B=\{1-g: g\in T(K)\}$ is convex and compact in the pointwise topology, it satisfies $f(0)=1$ for all $f\in B$ (because $g_n(0)=0$ for all $n$), and it contains all $f_n=1-g_n=1-T(e_n)$ with the standard unit vectors $e_n$ of $\ell^1$.
|
2025-03-21T14:48:31.743438
| 2020-08-11T14:46:24 |
368880
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Abdelmalek Abdesselam",
"MathMath",
"Yvan Velenik",
"https://mathoverflow.net/users/152094",
"https://mathoverflow.net/users/5709",
"https://mathoverflow.net/users/7410"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631976",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368880"
}
|
Stack Exchange
|
What is the role of Gibbs states with free boundary conditions in the theory of Gibbs measure?
This is actually a more elaborated version of a previous question of mine, which is now deleted. First, some quick notations:
(1) $\Omega_{0} := \{-1,1\}$ and $\mathcal{F}_{0} := 2^{\Omega_{0}}$ are, respectivelly, the single particle configuration space and its associated $\sigma$-algebra.
(2) If $\Lambda \subset \mathbb{Z}^{d}$ is finite, $\Omega_{\Lambda} := \{-1,1\}^{\Lambda}$ and $\mathcal{F}_{\Lambda} := \bigotimes_{x\in \Lambda}\mathcal{F}_{0}$ is its associated product $\sigma$-algebra.
(3) $\Omega := \{-1,1\}^{\mathbb{Z}^{d}}$, $\mathcal{F} := \bigotimes_{x\in \mathbb{Z}^{d}}\mathcal{F}_{0}$.
In what follows, I'm following Friedli and Velenik's book, chapter 3. For finite-volume systems, we can define Gibbs distributions with free boundary conditions (according to Definition 3.1 of the mentioned reference) by:
\begin{eqnarray}
\mu_{\Lambda,\beta,h}^{\emptyset}(\{\omega\}):= \frac{1}{Z_{\Lambda,\beta, h}^{\emptyset}}e^{-\beta H_{\Lambda,\beta,h}^{\emptyset}(\omega)} \tag{1}\label{1}
\end{eqnarray}
this is a discrete measure on $\Omega_{\Lambda}$. Also, if we fix $\eta \in \Omega$, we can define (according to Definition 3.3) Gibbs states with $\eta$-boundary conditions:
\begin{eqnarray}
\mu_{\Lambda,\beta,h}^{\eta}(\{\omega\}):= \frac{1}{Z_{\Lambda,\beta, h}^{\eta}}e^{-\beta H_{\Lambda,\beta,h}^{\eta}(\omega)} \tag{2}\label{2}
\end{eqnarray}
This, on the other hand, is a discrete measure on $\Omega_{\Lambda}^{\eta}:= \{\omega \in \Omega: \hspace{0.1cm} \mbox{$\omega_{x} = \eta_{x}$ for all $x \in \Lambda^{c}$}\}$ (which is equipped with its discrete $\sigma$-algebra).
The main ideia of the theory is to study thermodynamic limits, both for thermodynamic quantities and Gibbs states. In the case of Gibbs states, thermodynamic limits means weak convergence of finite-volume Gibbs states (at least in the present context, where $\Omega_{0}=\{-1,1\}$. This is, in fact, the main reason for which $\mu_{\Lambda,\beta,h}^{\eta}$ is conveniently defined on $\Omega_{\Lambda}^{\eta}$ instead of $\Omega_{\Lambda}$. Once we'd like to study Gibbs states on $\Omega$ using weak-convergence of finite volume Gibbs states, we need to extend our finite volume Gibbs states to equivalent notions that live in 'the whole space' $\Omega$. It is easy to do that with the measure (\ref{2}), since we can define (with abuse of notation):
\begin{eqnarray}
\mu_{\Lambda, \beta,h}^{\eta}(\{\omega\}) = \begin{cases}
\displaystyle \frac{1}{Z_{\Lambda,\beta, h}^{\eta}}e^{-\beta H_{\Lambda, h}^{\eta}(\Pi_{\Lambda,\eta}\omega)} \quad \mbox{if $\omega_{x} = \eta_{x}$ for all $x\in \Lambda^{c}$} \\
\displaystyle 0 \quad \mbox{otherwise}
\end{cases}
\tag{3}\label{3}
\end{eqnarray}
where $\Pi_{\Lambda,\eta}$ is the canonical projection $\Omega \hookrightarrow \Omega_{\Lambda}^{\eta}$. Note that this simple procedure does not work for extending $\mu_{\Lambda,\beta,h}^{\emptyset}$, since this would require to define $\eta$ as zero outside $\Lambda$, which is not consistent with our definition of $\Omega_{0}$.
Now, in this context, R. Ellis defines (I think this is standard, tho) $\mathcal{G}_{0}(\beta,h)$ as the set of all weak-limits of measures $\mu_{\Lambda_{n},\beta,h}^{\eta_{n}}$, where $(\forall n)$ $\eta_{n}\in \Omega$ and $\Lambda_{n}\to \mathbb{Z}^{d}$ is an increasing sequence of finite subsets of $\mathbb{Z}^{d}$. Furthermore, we define:
\begin{eqnarray}
\mathcal{G}(\beta,h) := \overline{\mbox{conv}\mathcal{G}_{0}(\beta,h)} \tag{4}\label{4}
\end{eqnarray}
where $\mbox{conv}\mathcal{G}_{0}(\beta,h)$ stands for the convex hull of $\mathcal{G}_{0}(\beta,h)$.
Question(s): First, I'd like to know how to extend $\mu_{\Lambda,\beta,h}^{\emptyset}$ to 'the whole space' $\Omega$ as done in (\ref{3}), since we cannot take zero values on $\Omega_{0}$. Also, I'd like to understand the role of these measures $\mu_{\Lambda,\beta,h}^{\emptyset}$ on $\Omega$: there are some results on weak-convergence of these measures in the theory, which seems to indicate they play some important role after all, but they seem not being considered in the definition of $\mathcal{G}_{0}(\beta,h)$, since we're only taking $\eta_{n}\in \Omega$ as boundary conditions of the sequence of Gibbs states $\mu_{\Lambda_{n},\beta,h}^{\eta_{n}}$. So, what am I missing here? Can we go on and study all (or at least almost all) relevant issues of the theory without considering infinite volume measures $\mu_{\Lambda,\beta,h}^{\emptyset}$? I know these measures are relevant to study, say, correlation inequalities and so on, but these do not demand them to be defined in the whole $\Omega$.
The free b.c. measure on $\Omega$ is obtained by 1) defining the product measure in infinite volume, then 2) multiplying this infinite product measure (spins seen as independent Rademacher random variables) by the Radon-Nikodym weight $e^{-H_{\Lambda}}/Z_{\lambda}$. Intuitively this means the eta's outside the finite volume are random (independent $\pm 1$ spins) instead of deterministic as in the description of $\mu^{\eta}$.
One way to construct the thermodynamic limit of the states $\mu_{\Lambda,\beta,h}^\varnothing$ is to observe that, for any local function $f$ and any increasing sequence of sets $\Lambda_n\uparrow\mathbb{Z}^d$, the support of $f$ will be included inside $\Lambda_n$ for all large enough $n$. In particular, for any local function $f$, one can prove that the limit $\lim_{n \to\infty}\mu_{\Lambda_n,\beta,h}^\varnothing(f)$ is well defined and independent of the sequence $(\Lambda_n)$ (this is Exercise 3.16 in our book). Then, one shows that there is a single probability measure on $(\Omega,\mathcal{F})$ that satisfies
$$
\mu(f) = \lim_{n\to\infty} \mu_{\Lambda_n,\beta,h}^\varnothing(f)
$$
for all local functions $f$ (this is Theorem 6.5 in our book). Finally, one proves that the measure $\mu$ indeed belongs to $\mathcal{G}(\beta,h)$ (this is Exercise 6.14 in our book); here $\mathcal{G}(\beta,h)$ is defined as the set of all probability measures satisfying the DLR equations (see the beginning of Chapter 6 in our book), which coincides with the definition you state (by Theorem 6.63 in the book).
Now, regarding the relevance of the free boundary condition. In my opinion, for the Ising model on $\mathbb{Z}^d$ (or other amenable graphs), the interest of this boundary condition is mostly technical (it's one of the few boundary conditions for which one can explicitly prove convergence, without resorting to compactness arguments). Moreover, the finite-volume measures enjoy nice properties that carry on to the limiting state and can occasionally be useful.
On nonamenable graphs, it can play a more important role. For instance, on trees it is known that, under some conditions, the state obtained by taking the thermodynamic limit using free boundary condition is extremal for a range of temperatures below the critical temperature.
Note that this boundary condition can be more interesting in other models. For instance, in the Potts model on $\mathbb{Z}^d$ with $q$ colors, when the phase transition is of first order (that is, when $d=2$ and $q\geq 5$, or when $d\geq 3$ and $q\geq 3$), then, at the phase transition temperature, the $q$ low-temperature pure states coexist with the (unique) high-temperature state. While the former can be selected by taking the thermodynamic limit using the corresponding monochromatic boundary condition, the latter can be selected using free boundary condition.
Amazing answer! Always good to talk to an expert! Just to clarify, the construction of limit states of $\mu_{\Lambda,\beta,h}^{\emptyset}$ is not precisely the same as for $\eta$ boundary conditions, since not all $\mu_{\Lambda,\beta,h}^{\emptyset}$ is defined to be in the same measurable space, right? One uses the convergence of real numbers $\langle f\rangle_{\Lambda_{n},\beta,h}^{\emptyset}$ to a real number $\langle f \rangle_{\beta,h}^{\emptyset}$ for each local $f$ and then one uses the Riesz-Markov-Kakutani representation theorem. Is that right?
Yes, exactly. Of course, nothing prevents you from extending a configuration in $\Omega_\Lambda$ to a configuration in $\Omega$ by fixing the value of all spins outside $\Lambda$ to be $+$, for example. Since the measure $\mu_{\Lambda,\beta,h}^\varnothing$ does not feel at all what's happening outside $\Lambda$, this does not change anything, but it makes $f$ (and the expectations) to be always well defined. But I find it simpler to just consider a sequence of large enough boxes ;) .
great! Perfect! I got it now! Thank you so much!
|
2025-03-21T14:48:31.744013
| 2020-08-11T15:38:30 |
368883
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Dag Oskar Madsen",
"David Roberts",
"Jan Peter Schäfermeyer",
"Martin Sleziak",
"Ovi",
"daemon",
"esg",
"https://mathoverflow.net/users/109828",
"https://mathoverflow.net/users/18756",
"https://mathoverflow.net/users/36586",
"https://mathoverflow.net/users/4177",
"https://mathoverflow.net/users/48831",
"https://mathoverflow.net/users/8250",
"https://mathoverflow.net/users/96921"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631977",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368883"
}
|
Stack Exchange
|
Math history research: a copy of "Zur relativen Wertbemessung der Turnierresultate" , eigenvector centrality by Edmund Landau
I'm searching for a copy of an old paper made by Edmund Landau:
Zur relativen Wertbemessung der Turnierresultate, Deutsches Wochenschach, 11. Jahrgang (1895), 366–369.
However, I can't find it anywhere. I looked at some old books written by Landau and did not succeed in my quest.
The Princeton University has the volume 50-1895 of the magazine "Deutsche Schachzeitung" where this paper should appear: HathiTrust record.
Landau published this early work in a Chess magazine. However, it seems to be that paper has several implications in the theory of Ranking and positive matrices and worked as a motivation for the next works of Landau.
Perhaps you can find help from a librarian. Tracking down old materials is part of their expertise.
I'm thinking about that. Perhaps some big and old library has the book where this article was published. However, I don't know how can I achieve that. I'm just a student in Brazil.
For example, the Princeton University seems to have the specific volume where this paper appears, v50-1895 https://catalog.hathitrust.org/Record/009008387
Right, but I think your university librarians might have a way to help. Libraries are often part of international networks which allow them to share books with each other. One such network is WorldCat https://www.worldcat.org/ You can try going through them yourself, but in any case, remember that librarians have masters or PhD's in Library Science, so they really know a lot.
Neither MathSciNet nor zbMath know of a paper from 1895 by Landau. But then it's a Chess magazine, so I'm not sure why you are asking here... Google books gives these snippets for 1895: https://books.google.com.au/books?redir_esc=y&id=kykCAAAAYAAJ&pg=PR1&sig=ACfU3U3pck7_uWtkOhTm0BWSVdZsxSrn7g&focus=searchwithinvolume&q=landau
And here's the WorldCat page for the periodical: https://www.worldcat.org/title/deutsches-wochenschach-und-berliner-schachzeitung/oclc/50283180 (maybe the title changed at some point, it's not clear). Unfortunately the Internet Archive only has the years 1889 and 1890
If you manage to get a copy, it's out of copyright, so I recommend hosting a scan in a stable location, or, even better, since it's not that long, transcribe it to LaTeX if you at all can, and host it on the arXiv.
I already tried the google book. As can been seen the scan of google book is incomplete. In archve.org there are some volumes of this magazine, but the specific year or 1895 is missing https://archive.org/search.php?query=Deutsches%20Wochenschach.
Ok, the paper was published in a chess magazine. However, "Zur relativen Wert Messung der Turnierresultate" is referenced in several works about ranking problem and graph theory.
You could also try to ask at chess.stackexchange.com ,
Thanks for the sugestion @DagOskarMadsen! It's a realy good idea.
Meanwhile the paper has been digitized by google: https://books.google.de/books/about/Deutsches_Wochenschach.html?id=rDr8AmfYCFkC&redir_esc=y
The paper which actually was Landau's first scientific paper written at the tender age of 18, was published in his Collected Works, vol. 1. In it, he proposes to rank chess players having played a round robin tournament according to an eigenvector of the results matrix.
https://books.google.nl/books?id=rDr8AmfYCFkC&pg=PA366
A much more comprehensive analysis of this method with the help of the (then new) Perron-Frobenius theorem is given in his 1915 paper "Über Preisverteilung bei Spielturnieren". It was published in the "Zeitschrift für Mathematik und Physik", vol. 63, p.192. See link below.
It seems very hard to get a copy from this Collected Works . But I'm happy now with this 1915 paper. Thanks.
It's sad, those papers deserves an english translation.
The introduction of his 1915 paper is a restatement of his earlier paper, i.e. the proposal to rate the players according to an eigenvector of the results matrix, which in chess contains the values 1, 1/2 and 0. In §1 he says that all is well if the results matrix is irreducible. §2 shows the possibility of a nonunique rating if the matrix is reducible. In §3 he shows that in certain cases inconsistencies can occur, namely that the score of a player can decrease if the outcome of a certain match is changed in his or her favor, which ultimately leads him to reject his own method.
The link no longer works, but judging from the Wayback Machine, it was a link to Zeitschrift für Mathematik und Physik. Tome 63.
Yes, on page 192. Thanks for the correction.
This volume (1895) of Deutsches Wochenschach has been digitised by Google books. You will find the digital copy at part 25, page 366.
https://books.google.nl/books?id=rDr8AmfYCFkC
As pointed out by @esg in the comments.
A working link for
Landau, Edmund (1915), "Über Preisverteilung bei Spielturnieren" in
Zeitschrift für Mathematik und Physik, 63: 192–202
can be found here:
https://lillonum.univ-lille.fr/s/lillonum/ark:/72505/bIVFka
If you have an account in springer you can see here:
https://link.springer.com/chapter/10.1007%2F978-1-4615-4819-5_23
It's a great compendium about some works of E. Landau. However, "Zur relativen Wertbemessung der Turnierresultate" was not published in this chapter
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.