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2025-03-21T14:48:30.167704
| 2020-03-28T16:42:27 |
355976
|
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|
Stack Exchange
|
Mean and logarithmic values for arithmetic function
Define the mean value for function $f$ as $\lim \limits_{x \to \infty} \frac{1}{x} \sum \limits_{n \leq x} f(n)$ if the limit exists denoted as $M_f$
Define the logarithmic value for function $f$ as $\lim \limits_{x \to \infty} \frac{1}{\ln x} \sum \limits_{n \leq x} \frac{f(n)}{n}$ if the limit exists denoted as $L_f$
Its easy to prove that if $M_f$ exist then so does $L_f$ and they are equal, but the other way around is not true in general.
Let $F(s) = \sum \limits_{n=1}^{\infty} f(n) n^{-s}$ for all $s>1$, and we are given that in addition $F(s)$ satisfy $F(s) = \frac{A}{s-1} +o(\frac{1}{s-1})$ for $s \to 1^{+}$ and $A$ is a constant.
In addition assume that $L_f = A$, and use this to prove that $M_f= A$ ?
Note : This question has been uploaded to MSE but without any answer.
Is this a homework problem? (I don't know how to do it, but that very precise condition with no context about where it comes from leads me to ask.)
Link to the post on [math.se]: Relaxation to Mean and Logarithmic values for arithmetic function.
@LSpice not a homework, i im intersted in NT and this question shows up in one of the notes in the subject i am reading.
BTW if you created two different MO accounts by accident, you can find some instructions related to merging here.
The linked question is essentially answered in comments.
|
2025-03-21T14:48:30.167832
| 2020-03-28T16:43:42 |
355977
|
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"Jochen Glueck",
"Jochen Wengenroth",
"LSpice",
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|
Stack Exchange
|
Separability of an algebra is equivalent to separability of its spectrum
Let $A$ be a commutative C*-algebra.
I would like to show that $A$ is separable (i.e. has a countable dense subset) if and only if the spectrum of $A$ (denoted by $\Omega(A)$) is separable.
Notes and reminders:
A C*-algebra is a Banach algebra over $\mathbb{C}$ with involution $*$ s.t. $\|x^*x\| = \|x\|^2$ for every $x \in A$.
Let me also remind that the spectrum $\Omega(A)$ is the set of the characters of $A$, that is $\Omega(A)\mathrel{:=} \{\chi:A\to \mathbb{C} : \text{$\chi$ is algebra homomorphism $\neq 0$}\}$. Also note that the topology on $\Omega(A)$ is the weak* (or w$^*$) topology. So the closure of any set which indicates separability is also taken under weak* topology.
I tend to rather believe that it's equivalent to its spectrum being both metrizable and separable? (i.e., in the unital case, metrizable)? So a separable non-metrizable Hausdorff compact space, e.g., $\beta\omega$, could be a counterexample.
Don't enter and leave math mode for sets. Namely, not $\Omega(A) :=$ {$\chi : A \to \mathbb C$ : $\chi$ is algebra homomorphism $\neq 0$} $\Omega(A) :=$ {$\chi : A \to \mathbb C$ : $\chi$ is algebra homomorphism $\neq 0$} but $\Omega(A) \mathrel{:=} {\chi : A \to \mathbb C \mathrel: \text{$\chi$ is algebra homomorphism $\ne0$}}$ $\Omega(A) := \{\chi : A \to \mathbb C : \text{$\chi$ is algebra homomorphism $\neq 0$}\}$. I have edited accordingly. Also, what do you mean by "the closure of any separable basis on $\Omega(A)$"? Basis is coll'n of open sets, right?
@YCor - As far as i know the spectrum of any algebra as described is metrizable, so I don't understand why what you suggest would give a counter example (also it would be helpful if you could explain what is $\beta \omega$).
If you take the C*-algebra of bounded functions $\omega\to\mathbf{C}$ (i.e., of bounded sequences), the spectrum is the Stone-Cech compactification $\beta\omega$, which is not metrizable.
@LSpice - thanks for editing, and what i meant was that if B is our dense countable set which indicates separability then the closure of it is taken in the w* topology. (maybe calling it basis wast smart).
@YCor - So there is something i don't understand (and correct me if im wrong) but doesn't then both the algebra and the spectrum separable - so where is the contradiction?
@nedgrekerzberg: Concerning you're last two comments: What do you mean by the weak${}^$-topology on a $C^$-algebra?
$\ell^\infty $ is not separable but its spectrum $\beta\mathbb N $ is separable.
@JochenWengenroth this is what I wrote (with different notation: $\omega$ is the ordinal notation for non-negative integers)
@YCor Of course it is your example. I had the impression that the OP did not understand you.
@YCor: Thank you, you're right of course. Stupid typo. I'll delete my comment since the comments by you and Jochen Wengenroth, as well as Nate Eldredge's answer contain the same information without mistakes ;-).
In all this maze of comments, the very simple, elementary and well-known true necessary and sufficient condition for the separability of the algebra is missing. Here it is (for the unital case): $A$ is separable if and only if its spectrum is metrisable. The non unital case is easily deduced by using the standard ploys: adding a unit and taking the one point compactification.
Thank you all for your helpful comments!
It's not true.
For simplicity, suppose $A$ is unital, so that its spectrum is compact Hausdorff.
If $X$ is a compact Hausdorff space and $C(X)$ is separable, then you can show that $X$ is second countable. (Let $\{f_k\}$ be a countable dense subset of $C(X)$, and $\{U_n\}$ a countable basis of open sets in $\mathbb{C}$; then the countable collection of sets $\{f_k^{-1}(U_n)\}$ is a basis for $X$, using Urysohn's lemma.) In particular, such an $X$ would be metrizable (by the Urysohn metrization theorem).
So the forward direction of your claim is true, since a second countable space is always separable, but this also suggests how to find a counterexample to the reverse direction: take a compact Hausdorff $X$ which is separable but not metrizable. One standard example is $X = [0,1]^{[0,1]}$ with its product topology; considering it as the space of all functions from $[0,1]$ to itself, the set of polynomials with rational coefficients is dense. Another, as YCor and Jochen Wengenroth discussed, is $X = \beta \mathbb{N}$, the Stone-Cech compactification of $\mathbb{N}$.
The converse of the above statement is also true, as user131781 points out: if $X$ is a compact metric space, then $C(X)$ is separable.
So the correct version of your claim (including the non-unital case) would be that $A$ is separable iff $\Omega(A)$ is second countable iff $\Omega(A)$ is a separable metrizable space.
Thank you very much!
|
2025-03-21T14:48:30.168148
| 2020-03-28T16:45:04 |
355978
|
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|
Stack Exchange
|
The Zariski topology of a variety is determined by its principal Cartier divisors
Let $X$ be a smooth separated integral variety over an algebraically closed field.
Question:
Is it true that a basis for the Zariski topology is given by the family $$\mathcal{U}=\{X\setminus \mathrm{Supp}(\mathrm{div}(f))\mid f\in K(X)\}$$
where $\mathrm{div}(f)$ is the principal divisor defined by $f$?
Some ideas that I have had
If $X=\mathrm{Spec}(A)$ is an affine integral variety this is true because for $f\in A$ we have that $X\setminus\mathrm{div}(f)$ is equal to the principal open set $D(f)$ and these generate the topology.
The question is equivalent to the following:
For each prime divisor $D$ on $X$ and a point $x$ outside $D$ there is a rational function $f$ vanishing on $D$ and invertible on $x$.
Indeed, if $\mathcal{U}$ is a basis then we can find a function $f$ such that $x\notin \mathrm{div}(f)$ (i.e $f$ is invertible at $x$) and $D$ is a component of $\mathrm{div}(f)$. Then either $f$ or $f^{-1}$ works.
Conversely, consider an open set $U$ and $x\in U$. The complement $F=X\setminus U$ is a closed set with some irreducible components $F_i$. Take a function $h_i\in \mathcal{O}_{X,x}\setminus \mathcal{O}_{X,F_i}$. Then divisor of poles $\mathrm{div}_{\infty}(h_i)$ contains $F_i$ but not $x$. Let $D=\bigcup \mathrm{div}_{\infty}(h_i)$. By the hypothesis, for each component $D_j$ of $D$ there is a function $f_j$ vanishing on $D_j$ but invertible on $x$. Then, $f=\prod_j f_j^{n_j}$ vanish on $D$ and is invertible on $x$ for some appropriate choice of $n_j\in \mathbb{Z}$.
If $X$ is a quasiprojective variety this is also true. We can see this using the idea in the previous point. Given a prime divisor $D$ and a point $x$ not in $D$ take an ample divisor $H$ different from $D$ and not containing $x$. Then for some $n$ big $nH-D$ is very ample (because the ample cone is open) then it is base-point free and so it has a section $f$ not vanishing at $x$ but vanishing at $D$.
I think this question is related to the concept of divisorial variety (cf. the remark after definition 3.1 in Divisorial Varieties, Borelli, 1963). In some sense, a variety satisfying this is like a principally divisorial variety.
If this is not true for smoothness, it would be still nice if we have the result for something less restrictive than quasi-projective. Something that contains smooth proper toric varieties for example.
Let $U$ be a an affine open set of $X$ containing $x$. If $U$ contains $D$, we choose $f$ as in your first bullet point vanishing on $D$ but not on $x$. If $U$ does not contain $D$, there must exist some $f$ on $U$ with a pole at $D$. If $f$ vanishes at $x$, add $1$ to $f$.
Why must there exist some $f$ on $U$ with a pole at $D$? Otherwise, we get a map $\operatorname{Spec} R \to U$ where $R$ is the local ring of $X$ at $D$, which consists of all rational functions on $X$ without a pole at $D$. This means we have two different maps $\operatorname{Spec} R \to X$, both compatible with $\operatorname{Spec} K \to X$, contradicting separatedness.
|
2025-03-21T14:48:30.168370
| 2020-03-28T17:13:44 |
355981
|
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"Greg Martin",
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|
Stack Exchange
|
Is there a connection between the average 'compositeness' of a rational number and $\phi$ (golden ratio)?
Let $n\in N$, where $n = p_{1}^{k_{1}}p_{2}^{k_{2}}...p_{m}^{k_{m}}$ for $p_{i}$ prime.
Define the 'density' of $n$ as:
$d(n) = \frac{(p_{1}+1)^{k_{1}}(p_{2}+1)^{k_{2}}...(p_{m}+1)^{k_{m}}}{n}$
Notice that $d(n)$ gives us a measure of the 'compositeness' of a number - relative to other numbers of a similar size. Notice also that $n_{1} \neq n_{2} \implies d(n_{1}) \neq d(n_{2})$
Now extend the definition to the rational numbers so that for $r \in Q$, where $r=a/b$, and $a=p_{1}^{k_{1}}p_{2}^{k_{2}}...p_{m}^{k_{m}}$, $b=q_{1}^{l_{1}}q_{2}^{l_{2}}...q_{n}^{l_{n}}$
Define the density of $r$ as:
$d(r) = \frac{(p_{1}+1)^{k_{1}}(p_{2}+1)^{k_{2}}...(p_{m}+1)^{k_{m}}}{a}.\frac{b}{(q_{1}+1)^{l_{1}}(q_{2}+1)^{l_{2}}...(q_{n}+1)^{l_{n}}}$
Now order the rational numbers in an $n$ x $n$ grid (the same grid used to prove the countability of the rationals). Denote the average density of the first $n$ rationals in the grid by $D_{n}$
Does $\lim_{n\to\infty} D_{n}$ exist, and if so what is it?
I have computed this for $n > 500,000$ and it turns out to be 1.61806, which is remarkably close to $\phi$. Is there a relationship between the density definition given above and the golden ratio?
Probably not. I can tell you what the limiting value is when taking averages over $m\times m$ grids themselves, rather than diagonal-counting-sequences; but I suspect the averages are the same.
The average of $d(r)$ over the $m\times m$ grid is simply
$$
\frac1{m^2} \sum_{a=1}^m \sum_{b=1}^m d\big( \tfrac ab\big) = \bigg( \frac1m \sum_{a=1}^m d(a) \bigg) \bigg( \frac1m \sum_{b=1}^m \frac1{d(b)} \bigg).
$$
The first sum is a sum over a totally multiplicative function $d$ with the property that $d(p) = 1+\frac1p$. General results about sums of multiplicative functions that are close to $1$ on primes tell us that
$$
\frac1m \sum_{a=1}^m d(a) \sim \prod_p \bigg( 1 + \frac{d(p)}p + \frac{d(p^2)}{p^2} + \cdots \bigg) \bigg( 1-\frac1p \bigg) = \prod_p \bigg( 1 - \frac{d(p)}p \bigg)^{-1} \bigg( 1-\frac1p \bigg)
$$
as $m\to\infty$ (where the products are over all primes $p$); in this case, we obtain
$$
\frac1m \sum_{a=1}^m d(a) \sim \prod_p \bigg( 1 - \frac{1+1/p}p \bigg)^{-1} \bigg( 1-\frac1p \bigg) = \prod_p \bigg( 1 + \frac1{p^2-p-1} \bigg) \approx 2.67411.
$$
Since $1/d$ is also totally multiplicative, the same procedure gives
\begin{align*}
\frac1m \sum_{b=1}^m \frac1{d(b)} &\sim \prod_p \bigg( 1 - \frac{1/d(p)}p \bigg)^{-1} \bigg( 1-\frac1p \bigg) \\
&= \prod_p \bigg( 1 - \frac{1/(1+1/p)}p \bigg)^{-1} \bigg( 1-\frac1p \bigg) = \prod_p \bigg( 1 - \frac1{p^2} \bigg) = \frac6{\pi^2}.
\end{align*}
Therefore the average in equation is
$$
= \frac6{\pi^2} \prod_p \bigg( 1 + \frac1{p^2-p-1} \bigg) \approx \frac6{\pi^2} \cdot 2.67411 \approx 1.62567.
$$
Thanks Greg. Could you point me to some reference material on "General results about sums of multiplicative functions that are close to 1 on primes", and also how you get the value 2.67411 for the first product. I'm not a professional mathematician.
The "Wirsing–Odoni method" suffices for this calculation; see for example Proposition 4 of my paper with Finch and Sebah.
|
2025-03-21T14:48:30.168585
| 2020-03-28T17:21:07 |
355983
|
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"Mike Shulman",
"Tim Campion",
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|
Stack Exchange
|
Examples of strict monoidal categories and monoidal categories with nontrivial associators
What are some "natural" motivating examples of the following:
i) A strict monoidal category,
ii) A monoidal with non-trivial associatots?
For i) the only examples I know are categories which have been strictified, are there any examples occuring "in nature" which are strict, or is strictness in some sense an "unnatural" or artificial requirement?
For ii) I should clarify what I mean by "non-trivial" - basically the examples I consider trivial are tensor products
of vector spaces, bimodules, representations, and so on, where the associator is just the elementary rewritting of brackets.
Regarding (i), every monoidal category is monoidally equivalent to a strict monoidal category. An example of a "naturally occurring" strict monoidal category is the category $\Delta_+$ of finite ordinals, under ordinal sum. This is the free strict monoidal category on a monoid object. I find (ii) more interesting -- in particular, I believe there are interesting examples of categories equipped with a tensor bifunctor which admit more than one possible associator, and the resulting monoidal structures are not equivalent. I hope somebody can provide such examples.
It's hard to tell what you mean by "non-trivial" if you don't know any examples. One might go so far as to claim that the associator always consists of rewriting of brackets, essentially by definition. How about a 2-group constructed from a group cohomology class; would you regard its associator as always "trivial"?
In this paper Mark Hovey studies the problem of classifying symmetric monoidal closed structure on the category $Mod_R$ of $R$-modules for a ring $R$. For example, when $R$ is a field, there is a unique such, and when $R = \mathbb F_2[C_2]$ there are exactly 7 up to equivalence. He doesn't cleanly separate out the problem of understanding non-symmetric monoidal structures, but I think his analysis should probably be enlightening.
i) A monoid
ii) Representations of a quasi-Hopf algebra
|
2025-03-21T14:48:30.168747
| 2020-03-28T17:27:04 |
355985
|
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"Gerry Myerson",
"Julian",
"Per Alexandersson",
"Sam Hopkins",
"https://mathoverflow.net/users/1056",
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"https://mathoverflow.net/users/3684",
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|
Stack Exchange
|
Number of non-crossing sets of intervals
Consider the integers $[1,n]=\{1,\dots,n\}$ and call subsets of the type $[a,b]=\{a,\dots,b\}$ with $1\le a < b\le n$ intervals. We say that two intervals $[a,b],[c,d]$ are crossing if either $a<c<b<d$ or $c<a<d<b$. Otherwise we say that the intervals are non-crossing. So $[1,3],[2,4]$ are crossing, but $[1,3],[2,3]$ or $[1,3],[3,4]$ are non-crossing.
Let $A_n$ denote the collection of pairwise non-crossing sets of intervals. So the first few are
\begin{align}
A_2&=\Bigl\{\emptyset,\{[1,2]\}\Bigr\}\\
A_3&=\Bigl\{ \emptyset,\{[1,2]\},\{[2,3]\},\{[1,2],[2,3]\},\\
&\qquad\{[1,3]\},\{[1,2],[1,3]\},\{[2,3],[1,3]\},\{[1,2],[1,3],[2,3]\} \Bigr\}\\
A_4 &= \Bigl\{ \emptyset,\{[1,2]\},\{[1,3],[3,4]\},\{[1,2],[2,4],[1,4]\},\dots \Bigr\}.
\end{align}
It seems that there is an easy recursion for the number of such sets of non-crossing intervals:
$$|A_n| = 2 \Bigl(2|A_{n-1}||A_2|-|A_2|^2|A_{n-2}|\Bigr)=8(|A_{n-1}| - |A_{n-2}|) \tag{*},$$
so that
$$(|A_2|,|A_3|,|A_4|,\dots)=(2,8, 48, 320, 2176, 14848, 101376,\dots)$$
which seems to be https://oeis.org/A228568.
Questions:
Is there some standard name for sets sets of non-crossing intervals in the literature? Is the number of such sets known?
Is the recursion above correct?
My reasoning for (*) is as follows: Any non-crossing collection of sets on $[1,n]$ can be obtained from non-crossing collections on $[1,n-1]$ and $[n-1,n]$ or $[1,2]$ and $[2,n]$ by optionally adding the interval $[1,n]$. However, this is over-counting the intervals which can be obtained from combining non-crossing collections on $[1,2],[2,n-1],[n-1,n]$.
Edit: My reasoning for (*) was flawed and the conclusion was incorrect. The correct number of sets seems to be
$$(|A_2|,|A_3|,\dots)=(2, 8, 48, 352, 2880, 25216, 231168, \dots)$$
which is https://oeis.org/A054726.
This is reminiscent of non-crossing partitions, but not quite.
Have you done computations to verify $|A_8|=101376$?
@GerryMyerson I got the number just from ($\ast$). But today I noticed that ($\ast$) is actually flawed, and in particular the numbers of sets I got was wrong. I updated the post with an updated guess for the number of sets
The "number of graphs on n nodes on a circle without crossing edges" interpretation is clearly right, FWIW.
@SamHopkins It seems very related but how does one cut the circle to the interval bijectively?
Map the collection of intervals ${ [a_1,b_1], [a_2,b_2],\ldots, [a_k,b_k]}$ to the graph with edge set ${ {a_1,b_1}, {a_2,b_2},\ldots, {a_k,b_k}}$.
@SamHopkins thanks; I got confused with labelled vs unlabelled nodes on the circle.
So if you trace through the links on that OEIS entry (https://oeis.org/A054726) you see that your number in question is $2^n$ times a little Schroeder number (see https://oeis.org/A001003 and https://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Hipparchus_number). That's as good a formula as you can expect- the little Schroeder numbers are well-studied.
@SamHopkins Yes, these sequences indeed seem quite well-studied. Thanks!
Maybe one of you could write this up and post it as an answer.
@GerryMyerson: Done.
As discussed in the comments, the number in question is clearly the same as the number of graphs on $n$ vertices drawn on a circle without crossing edges, which is in the OEIS at https://oeis.org/A054726. As discussed in that OEIS entry, the number is $2^n$ times a "little Schroeder number": recall that one interpretation of the little Schroeder number is as the number of (not necessarily maximal) dissections of a convex $n$-gon; in other words, these count collections of noncrossing diagonals; the only difference between these and graphs without crossing edges is that the graphs may include edges of the form $\{i,i+1\}$, and any such subset is also allowed, hence the factor of $2^n$. Since the little Schroeder numbers are very well-studied, this is the best formula you could hope for.
Thanks! Small addition: The little-Schroeder numbers count the number of non-crossing graphs on the circle without next-neighbour edges. Since there are $n$ such next-neighbour edges it follows that the total number of graphs is $2^n$ times the little-Schroeder numbers. For more details on the generating functions the paper https://mathscinet.ams.org/mathscinet-getitem?mr=345872 is also a good reference.
@Julian: ah, yes, I see- if we think of the little Schroeder numbers as counting (not necessarily maximal) dissections of an $n$-gon then this interpretation is indeed clear.
|
2025-03-21T14:48:30.169034
| 2020-03-28T17:43:27 |
355987
|
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|
Stack Exchange
|
Is $\mathbb{Q}$-factoriality preserved under contraction?
Let ($X$,$\Delta$) be projective klt pair and $f \colon X \rightarrow Z$ be contraction of ($K_X + \Delta$) - negative extremal ray $R$.
If $X$ is $\mathbb{Q}$ -factorial and $\mathrm{dim}Z < \mathrm{dim} X $, then Z is $\mathbb{Q}$- factorial?
I have read proof of above claim in Kollar and Mori "Birational geometry of algebraic varieties" but I have some trouble.The proof is following:
Let $D$ be effective weil divisor on $Z$ and $Z_0 \subset Z$ be smooth locus of $Z$.
$D'$ be closure of $f^{-1}(D \cap Z_0 )$. By $\mathbb{Q}$-factoriality of X, $mD'$ is cartier for some $m \in \mathbb{Z}$.
Since $D'$ does not meet general fibre of $f$ we have $D'.R = 0$
Hence $mD'$ is pullback of some cartier divisor on $Z$ and $D$ is $\mathbb{Q}$ cartier.
My trouble is the last statement.
Why $D$ is $\mathbb{Q}$ cartier since $mD'$ is pullback of some cartier divisor on Z?
Let $D''$ be a Cartier divisor on $Z$ such that $f^*\mathscr O_Z(D'')\simeq \mathscr O_X(mD')$. Then by the projection formula $\mathscr O_Z(D'')\simeq f_*\mathscr O_X(mD')$ (since $f_*\mathscr O_X\simeq \mathscr O_Z$).
On the other hand, from the construction we see that $\mathscr O_Z(D'')|_{Z_0}\simeq f_*\mathscr O_X(mD')|_{Z_0}\simeq \mathscr O_Z(mD)|_{Z_0}$. Finally, because $Z$ is normal, the complement of $Z_0$ has at least codimension $2$ in $Z$. Both $\mathscr O_Z(D'')$ and $\mathscr O_Z(mD)$ are reflexive sheaves of rank $1$, and they are isomorphic on the complement of a codimension $2$ closed set, so they are isomorphic everywhere. Hence $mD\sim D''$ which is Cartier.
|
2025-03-21T14:48:30.169277
| 2020-03-28T19:00:25 |
355990
|
{
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"Carl-Fredrik Nyberg Brodda",
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|
Stack Exchange
|
Locally finite groups with finite exponent
I am interested in infinite groups that are locally finite (finitely-generated subgroups are finite) with finite exponent (there exists integer $m$ such that $g^{m}=1$ for every element $g$).
I can think of only one method for constructing examples: an infinite direct product $G = \prod_{i\in I} H_{i}$, where $I$ is an infinite index set, and each $H_{i}$ is a finite group of size at most $n$ (for some $n$).
Question 1: Are there other methods for constructing infinite locally finite groups of finite exponent?
Question 2: Are there infinite locally finite groups of finite exponent that are not direct products?
Question 3: Are there infinite locally finite groups of finite exponent that are simple?
I believe that Tarski monsters are not locally finite, and that in fact any Tarski monster can be generated by two elements.
Ah, of course not, you're right.
You have subgroups of these direct products. "Not direct products": do you mean directly indecomposable?
Q3: No There's no such locally finite group (the argument appeared recently on the site, I'll look for it)
@YCor Yes, thanks, I would be interested in an example that is directly indecomposable.
Q3: by Hartley/ Meierfrankenfeld, every simple locally finite group has "a Kegel cover", i.e., for every finite subset $F$ there exists subgroups $G,N$ with $N$ normal in $G$ and $F\subset G$, such that the quotient $G/N$ is finite simple and $F$ is mapped injectively into $G/N$. By classification, simple groups of exponent $\le k$ have bounded cardinal, say $\le n_k$. Taking $F$ of cardinal $>n_k$ yields a contradiction.
A directly indecomposable such group is quite immediate: take $C_2$ cyclic of order $2$, $V$ to be an infinite abelian group of exponent $3$, and $G$ the semidirect product $C_2\ltimes_{\pm}V$. This is not very mysterious and just an instance of my previous comment: "you also have subgroups of these direct products": here this is a subgroup of some power of $\mathrm{Sym}(3)=D_6$.
Jeremy Rickard has found the recent post, about the non-existence of infinite simple locally finite groups of finite exponent: https://mathoverflow.net/a/352400/14094. This makes Q3 a duplicate (and Q2 is too immediate to deserve an answer). You might ask whether every locally finite group of finite exponent is isomorphic to a subgroup of a direct product of finite groups with bounded exponent.
@YCor Suppose I define $G$ to be the union $B_{0}(1,n)\cup B_{0}(2,n)\cup B_{0}(3,n)\cup\cdots$. Then $G$ is countably infinite, locally finite, and it has bounded exponent. Could it be the case that $G$ is a subgroup of a direct product of finite groups, or is this a counterexample?
(By $B(m,n)$ I mean the free Burnside group with $m$ generators and exponent $n$. Also $B_{0}(m,n)$ is $B(m,n)/N$ where $N$ is the intersection of all subgroups with finite index in $B(m,n)$. By the solution to the restricted Burnside problem, $B_{0}(m,n)$ is finite.)
@DillonM This group, let's call it $B_0(\omega,n)$ is isomorphic to a subgroup of a direct (unrestricted) product of finite groups (that is, is residually finite). Indeed if $w\neq 1$ belongs to it, say it belongs to $B_0(k,n)$ for some $k$. Since $B_0(k,n)$ is a retract of $B_0(\omega,n)$, $w$ survives in the finite quotient $B_0(k,n)$.
|
2025-03-21T14:48:30.169498
| 2020-03-28T19:51:12 |
355993
|
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"CuriousKid7",
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"url": "https://mathoverflow.net/questions/355993"
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|
Stack Exchange
|
Showing that a certain simplicial set has levelwise small cardinality
My question concerns Section 2 of the article "The Simplicial Model of Univalent Foundations (after Voevodsky)" (https://arxiv.org/pdf/1211.2851.pdf).
Let $\alpha$ be a strongly inaccessible cardinal.
Let $f : X \to Y$ be a map of simplicial sets.
We say that $f$ is well-ordered if it is equipped with a well-ordering of $Y_x := f_n^{-1}(x)$ for each simplex $x\in X_n$.
We say that $f$ is $\alpha$-small if $\left\lvert{Y_x}\right\rvert < \alpha$ for every simplex $x$.
Let $f: X \to Y$ and $g : Z \to Y$ be well-ordered simplicial maps. A morphism $f \to g$ is a fiber-preserving simplicial map $h : X \to Z$
such that $h_n : f_n^{-1}(y) \to g_n^{-1}(y)$ is order-preserving for every natural number $n$ and every $y\in Y_n$.
Define the functor $\mathcal{U}_{\alpha} : \mathbf{sSet}^{\text{op}} \to \mathbf{Set}$ so that $\mathcal{U}_{\alpha}(X)$ consists of all isomorphism classes of $\alpha$-small well-ordered Kan fibrations $Y\to X$. Also, let
$$
\mathrm{U}_{\alpha} = \mathcal{U}_{\alpha}\circ \mathcal{Y}^{\text{op}} : \varDelta^{\text{op}} \to \mathbf{Set}
$$ where $\mathcal{Y} :\varDelta \to \mathbf{sSet}$ denote the Yoneda embedding.
The authors say that if $\beta <\alpha$ is also inaccessible, then the unique map $\mathrm{U}_{\beta} \to 1$ is $\alpha$-small (p. 23, near the bottom). This amounts to saying that the set
$$
\left(\mathrm{U}_{\beta}\right)_n = \mathcal{U}_{\beta}(\Delta[n])
$$
has cardinality $<\alpha$ for each $n$.
I am unable, however, to see a set-theoretic justification for this. I'd be grateful if someone could provide one!
$\mathcal{U}_\beta(\Delta[n])$ is the set of isomorphism classes of $\beta$-small well-ordered fibrations over $\Delta[n]$. Such a thing is uniquely determined by an isomorphism class of $\beta$-small well-orderings over each element of $\Delta[n]$, together with the face and degeneracy maps between them. There are $\beta$ isomorphism classes of $\beta$-small well-orderings, and there are countably many elements of $\Delta[n]$. Since $\beta \cdot \aleph_0 = \beta$, there are $\beta$ choices of these fibers. Now for every face and degeneracy map, the domain and codomain are $\beta$-small, hence (since $\beta$ is inaccessible) there are $<\beta$ possible maps. Since there are countably many face and degeneracy maps, there are no more than $\beta \cdot \aleph_0 = \beta$ choices of the face and degeneracy maps. Thus there are overall $\beta$ choices, and $\beta<\alpha$.
To clarify, do you mean the face and degeneracy maps between the well-ordered fibers?
Yes, that's what I mean.
|
2025-03-21T14:48:30.169674
| 2020-03-28T21:45:50 |
355996
|
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|
Stack Exchange
|
Gurevich's entropy and topological entropy in a countable Markov shift
Good afternoon, I understand that Gurevich's entropy and topological entropy coincide when the countable Markov shift is topologically mixing (right?)
Does anyone know of an example or a reference where Gurevich's entropy and topological entropy in a Markov accounting change do not coincide.
For a definition of Gurevich's entropy see the second section of https://arxiv.org/pdf/1806.00214.pdf (the original paper is in Russian).
You would be hard pressed to find a reference. An example would be a case where the Gurevich entropy is periodic. Therefore there would be periods when it would increase slower than topological entropy even if they are equivalent in the long time steps.
|
2025-03-21T14:48:30.169741
| 2020-03-28T22:04:20 |
355999
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/355999"
}
|
Stack Exchange
|
Fourier dimension of radial set
In his 1967 article "Sur un theoreme de R. Salem", Gatesoupe proved that if a set $A\subset [0,1]$ has Fourier dimension $\alpha$ then the set $\tilde A:=\{x\in \mathbb{R}^n: |x| \in A\}$ has Fourier dimension at least $n-1+\alpha$. Basically he starts from a measure $\mu$ on $A$ whose Fourier transform decays as $|\xi|^{-\alpha/2}$ and proves (via a rather standard argument with Bessel functions) that the measure $d\nu := r^{(1-n)/2} d\mu \otimes d\sigma^{n-1}$ has a Fourier transform that decays as $|\xi|^{-(n-1+\alpha)/2}$.
Is it known whether the equality between the two dimensions hold? i.e. $\dim_F(\tilde A) = n-1+\dim_F(A)$?
I would expect it to be the case but I could not prove it formally. I only have a partial result for some dimensions $n$, but I see no reason why this should be true e.g. for dimension $5$ and not dimension $3$ or $4$.
|
2025-03-21T14:48:30.169826
| 2020-03-28T22:29:46 |
356000
|
{
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"Dietrich Burde",
"Wojowu",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/356000"
}
|
Stack Exchange
|
A quadratic trinomial that generates only prime numbers of the form $4m+1$
It is known that Euler's polynomials $\,n^2+n+p\,$ ($p\,$ prime) represent a prime for $\,n=0,\,...,\,p-2\,$ if and only if the field $\,Q (\sqrt{1-4p})\,$ has class number $\,h=1$.
The best trinomial of such kind is $\,n^2+n+41$.
But what we can say about quadratic trinomials that generate, for
consecutive values of their integer variable, only prime numbers of
the form $\,4m+1$?
The best I could find out is the following trinomial:
$$p(n)=4\cdot(32\cdot(21-n)-n^2)+1$$
that generates prime numbers of the form $\,4m+1\,$ for $\,n=0,\,...,\,14$.
At the $2$-nd place, I would put the trinomial:
$$p(n)=4\cdot(n\cdot(64+n)-1171)+3$$
that generates (eventually in absolute value) prime numbers of the form $\,4m+1\,$ for $\,n=1,\,...,\,14\,$ and of the form $\,4m+3\,$ for $\,n=15,\,...,\,28$.
Every suggestion is well accepted.
Many thanks.
[ This question have been also posted on MathStackExchange ]
Using an extension of Green-Tao Theorem to polynomial patterns (see here) you can show that there are trinomials which express arbitrarily long strings of primes. I suspect a generalization to primes in a fixed arithmetic progression would be immediate from the method of proof, like for standard Green-Tao, but I can't speak with confidence.
Crossposted at MSE.
|
2025-03-21T14:48:30.169940
| 2020-03-28T22:39:42 |
356001
|
{
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"Samir Canning",
"https://mathoverflow.net/users/108274",
"https://mathoverflow.net/users/124840",
"user267839"
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"url": "https://mathoverflow.net/questions/356001"
}
|
Stack Exchange
|
Morphism between jet spaces smooth
In this article "Introduction to Jet Schemes and Arc Spaces" S. Ishii introduces the spaces of $m$-jets:
Let $X$ be a variety over algebraically closed field $k$. The space $X_m$ of $m$-jets represents the functor
$$F_X^m: (\operatorname{Sch}/k) \to (\operatorname{Set}), Z \mapsto \operatorname{Hom}(Z \times \operatorname{Spec} \ k[t]/(t^{m+1}), X)$$
i.e. for every $Z$ we have $F^m_X(Z) \cong \operatorname{Hom}(Z,X_m)$.
By construction the canonical quotient map $k[t]/(t^{m+1}) \to k[t]/(t^{m})$ induces natural morphism between functors $F_X^m \to F_X^{m-1}$. By Yoneda this induces a unique morphism $\psi_{m, m-1}: X_m \to X_{m-1}$.
Question: Assume that $X$ is non singular (=smooth). Why this implies that $\psi_{m, m-1}$ is a smooth map? (compare Example 1.6 on p 3)
In case $X= \mathbb{A}^n$ one can show that $X_m= \mathbb{A}^{(m+1)n}$ and that $\psi_{m, m-1}: \mathbb{A}^{(m+1)n} \to \mathbb{A}^{mn}$ is the canonical projection. It is smooth.
What about the general case if $X$ smooth. How to prove the smothmess of $\psi_{m, m-1}$? Can the claim be reduced to the case above with affine space?
It will follow from the affine case. You should show that if you have an étale morphism $Y\rightarrow X$, then $Y_m\cong X_m\times_X Y$. Use the smoothness hypothesis to come up with an appropriate étale morphism from a neighborhood of every point in $X$ to affine space.
Yes, that's indeed a local problem, so locally we have an etale map $U \to \mathbb{A}^N$ for open affine $U \subset X$. And $Y_m\cong X_m\times_X Y$ follows from description of functor $F_Y^n(-)$. That is if $T= \operatorname{Spec} \ R$ is an affine test scheme we only need to lift a map $\operatorname{Spec} R[t]/(t^{n+1}) \to X$ to a map $\operatorname{Spec} R[t]/(t^{n+1}) \to Y$ but that's the etaleness! And the claim follows as base change preserve smoothness. I see, thank you.
|
2025-03-21T14:48:30.170072
| 2020-03-28T23:25:10 |
356004
|
{
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"Michael Engelhardt",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/356004"
}
|
Stack Exchange
|
restricted three body problem equations of motion using particle distances and one angle variable
If we solve numerically a three (or $N$) body planar problem, it's easy to calculate the distances of the bodies as function of time.
Conversely if we know the interparticle distances as functions of time and the angle of one distance, we can calculate the positions of all three (or $N$) bodies up to a reflection of the configuration (and of course translation). This is known as the distance geometric problem.
So is it possible to write differential equations for the distances $r_{12}, r_{13}$ and $r_{23}$ and say an angle $a$ in case of three bodies or generally $N$ bodies?
In the case of two bodies (for example in the Kepler problem) that of course is possible and well known.
Moreover, might the equations for the distances be of the form
$$
\frac{\mathrm{d}^2r_{12}}{\mathrm{d}t^2} = f(r_{12}, r'_{12};r_{12},r'_{12},r_{23},r'_{23})
$$
(written for distance between bodies 1 and 2) that is, independent of the angle variable a like is the case with 2-body case?
If you know how the old and new coordinates can be expressed in terms of one another, isn't it just a matter of applying the chain rule? Have you tried that?
The equations of motion cannot be written as
$$ \ddot{r}_{ij} = f_{ij}(r_{12},\ldots, r_{n-1,n}, \dot{r}_{12},\ldots, \dot{r}_{n-1,n}) $$
Consider $n$ equally massive particles arranged equidistantly around a circle. If they are all initially
stationary, the system will collapse to the centre of the circle. But if they are rotating at the correct speed, the system will be in equilibrium with all distances remaining constant. At time $0$ both systems have the same $r_{ij}$ and $\dot{r}_{ij}=0$.
Nice argument. I think there's something to be clarified with the OP, though: The OP asserts that such a representation is possible in the 2-body case, which sounds like the OP regards the dependence on the angular momentum as something one can just ignore - but really, it's there. I therefore suspect that the OP would likewise view an $f_{ij} $ that did depend on angular momenta as acceptable. However, in that case, your two scenaria are indeed distinguishable. The OP hasn't really posed the question very well.
|
2025-03-21T14:48:30.170246
| 2020-03-29T00:39:56 |
356010
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/356010"
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|
Stack Exchange
|
Is the Cassels "$x - \theta$" map algebraic in some sense?
Setup: Let $k$ be a field of characteristic $0$, let $f(x) \in k[x]$ be a monic separable polynomial of degree $n \geq 4$, and let $\theta$ denote the image of $x$ under the map $k[x] \to K_f := k[x]/(f(x))$. Consider the hyperelliptic curve with affine equation $y^2 = f(x)$. The Cassels "$x - \theta$" map takes a point $(x,y) \in C(k)$ satisfying $y \neq 0$ to $[x - \theta] \in (K_f^{\times}/K_f^{\times 2})_{\mathrm{N} \equiv 1}$, where the subscript "$\mathrm{N}\equiv 1$" means "consider only those elements whose norm in $k$ is a square"; the map has a more complicated definition on Weierstrass points of $C$. This map can be thought of as a restriction to $C(k)$ of the composite map
$$J(k) \to J(k)/2J(k) \to H^1(\operatorname{Gal}(\overline{k}/k), J[2]) \simeq (K_f^{\times}/K_f^{\times 2})_{\mathrm{N} \equiv 1}$$
obtained by taking Galois cohomology of the exact sequence $0 \to J[2](\overline{k}) \to J(\overline{k}) \to J(\overline{k}) \to 0$, where $J$ denotes the Jacobian of $C$, and $\overline{k}$ denotes the separable closure of $k$.
(Admittedly vague) Question: Is there a way to realize the Cassels map as an algebraic map of some sort?
What I know: A construction of Bhargava in https://arxiv.org/pdf/1308.0395.pdf (sections 2 and 3) can be summarized as follows. For even $n$, there is a representation $V$ of the algebraic group $\operatorname{SL}_n^{\pm}$ such that the following properties hold:
The orbits of $\operatorname{SL}_n^{\pm}(k)$ on $V(k)$ are in bijection with $(K_f^{\times}/K_f^{\times 2})_{\mathrm{N} \equiv 1}$;
There is a rational map of $k$-varieties $\phi \colon C -\to V$, defined away from the points at infinity, such that the $\operatorname{SL}_n^{\pm}(k)$-orbit of $\phi(x,y)$ is equal to the image of $(x,y)$ under the Cassels "$x - \theta$" map for any $(x,y) \in C(k)$ with $y \neq 0$.
I would like to show that the $\operatorname{SL}_n^{\pm}(k)$-orbit of $\phi(x,y)$ is equal to the image of $(x,y)$ under the Cassels "$x - \theta$" map for any $(x,y) \in C(k)$, not just for non-Weierstrass points (this fact is required to prove Theorem 3 in Bhargava's paper, although no proof is given). To do this, it would be helpful to know if the Cassels map had some algebraic realization, so that I could say that it must agree with $\phi$ because it agrees with $\phi$ on a dense set.
|
2025-03-21T14:48:30.170400
| 2020-03-29T04:02:06 |
356016
|
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|
Stack Exchange
|
Measurable extensions of probability measures
Let $X$ be a set, and let $\mathcal G \subset \mathcal F$ be $\sigma$-fields over $X$. Let $\Delta_\mathcal G$ (resp. $\Delta_\mathcal F$) be the set of probability measures on $\mathcal G$ (resp. $\mathcal F$). Let $\mathcal D_{\mathcal G}$ (resp. $\mathcal D_{\mathcal F}$) be the $\sigma$-field over $\Delta_\mathcal G$ (resp. $\Delta_\mathcal F$) generated by the functions of the form $P \mapsto P(A)$, $A \in \mathcal G$ (resp. $A \in \mathcal F$). If $A$ is a measurable subset of $\Delta_\mathcal G$, let $\mathcal D_{\mathcal G}|_A$ denote the trace $\sigma$-field over $A$.
A measurable extension from $A \in \Delta_{\mathcal G}$ is a measurable function $f:(A, \mathcal D_{\mathcal G}|_A) \to (\Delta_{\mathcal F}, \mathcal D_{\mathcal F})$ such that the restriction of $f(P)$ to $\mathcal G$ is identical to $P$.
I am wondering if measurable extensions have been studied in any detail, and, in particular, what can be said about when they exist. I am also wondering whether there are any interesting applications of this concept.
The motivation for the definition comes from a well-known paper by Blackwell and Ryll-Nardzewski ("Non-existence of everywhere proper conditional distributions,'' 1962). They define a proper conditional distribution on $(X, \mathcal F)$ given $\mathcal G$ to be a $\mathcal G$-measurable function $Q$ from $X$ into $\Delta_{\mathcal F}$ with the property that, for all $x \in X$, the restriction of $Q(x)$ to $\mathcal G$ is point mass at $x$, i.e. $\delta_x$. They then show that when $(X, \mathcal F)$ is a standard space a proper conditional distribution exists if and only if a $\mathcal G$-measurable selection function exists, that is, a $\mathcal G$-measurable function $g: X \to X$ such that $g(x) \in G$ for all $x \in G \in \mathcal G$.
Notice that a proper conditional distribution exists if and only if there is a measurable extension from the set of point masses in $\Delta_{\mathcal G}$. Indeed, if the measurable extension $f$ exists, then (because $d: (X, \mathcal G) \to (\Delta_{\mathcal G}, \mathcal D_{\mathcal G}); x \mapsto \delta_x$ is measurable), $Q(x) = f(d(x))$ defines a proper conditional distribution. Conversely, if a proper conditional distribution $Q$ exists, then $f(\delta_x) = Q(x)$ defines a measurable extension.
So, essentially, I am wondering whether this notion of measurable extension is a potentially interesting generalisation of proper conditional distributions.
|
2025-03-21T14:48:30.170546
| 2020-03-29T05:20:32 |
356022
|
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|
Stack Exchange
|
Does ACC for principal ideal plus Krull dimension equal 0 imply DCC for principal ideals
Assume the ring is commutative and with 1.
We know that ACC + $\dim(R)=0$ imply DCC. However, if we only insist on the condition for principal ideals, can we conclude the same?
We know that having DCC on the principal ideals is the same as being a perfect ring but I'm not sure how to proceed from here.
(Note: ACC/DCC= ascending/descending chain condition.)
No. Consider $K$ a field and $R=K[x_n:n\ge 0]/(x_n^2:n\ge 0)$.
Clearly $R$ is local and its nilradical equals its unique maximal ideal, so it has Krull dimension zero. It has the properly descending sequence of ideals $R\supset (x_0)\supset (x_0x_1)\supset\cdots$.
But it has ACC on principal ideals.
Indeed, consider $(a_k)_{k\ge 0}$ such that $a_{k+1}$ divides $a_k$ for all $k$ (say $a_k=a_{k+1}d_k$); we have to show that $a_k$ divides $a_{k+1}$ for large enough $k$. This is clear if some $a_k$ is invertible, hence assume they're all in the maximal ideal. Also the case all $a_k$ zero being trivial, we can suppose (extracting if necessary) $a_0\neq 0$.
Let $I$ be the (nonempty finite) set of $n$ such that $x_n$ occurs in $a_0$. We consider the linear decomposition $R=R_I\oplus J_I$, where $R_I$ is the subring $K[x_i:i\in I]$ and $J_I$ is the ideal generated by the $x_n$ for $n\notin I$. Then the projection $R\to R_I$ (with respect to this decomposition) is a $K$-algebra homomorphism.
Write $a_k=b_k+c_k$ and $d_k=e_k+f_k$ in the above decomposition. Then $b_k=b_{k+1}e_k$ for all $k$. Since it belongs to the noetherian subring $R_I$, for $k$ large enough, $b_k$ is a nonzero scalar multiple of $b_{k+1}$. Up to rescaling, we suppose henceforth that all $b_k$ are equal for large enough $k$, say to $b$. Since $b$ divides $b_0=a_0\neq 0$, we have $b\neq 0$.
So $be_k=b$ for all large enough $k$, that is, $b(e_k-1)=0$. Since $b\neq 0$, this implies that $e_k-1$ is a zero divisor, and hence belongs to the maximal ideal, which implies that $e_k$ is invertible. So $d_k$ is invertible, and hence the sequence $(a_k)$ is stationary.
Why is $b$ nonzero?
The question is then why $b_0 \neq 0$? E.g. why can't $a_0$ be a linear combination of some binomials?
@auniket you're right, I overly simplified. I've edited. Thanks!
|
2025-03-21T14:48:30.170701
| 2020-03-29T07:39:46 |
356024
|
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|
Stack Exchange
|
Complexity of combined trajectory compression algorithm
Is it possible to calculate the complexity of the combined algorithms (TD-TR-SP, TD-SP-TR) provided here:
A new perspective on trajectory compression techniques
In more details, TD-TR-SP algorithm is a combination of TD_TR and TD_SP algorithm. The complexity of TD_TR is $O(n \log n)$ where $n$ is the size of trajectories. The complexity of TD_SP is $O(n^2)$.
Is it possible to obtain the total complexity when combine these algorithms?
I am not familiar with Big-$O$ complexities at all.
Yes, but it depends on what it means "combining": in principle, if you know how (mathematically speaking) the two algorithms interact, you can (and must) use each complexity estimate in order to derive the new algorithm complexity estimate, no matter if they are expressed by using Landau notation (as it is customary) or not.
At first the TD_TR is executed and then the TD_SP (in case of TD-TR-SP). Then the result1 from TD_TR and result2 from TD_SP is putting together like "if result1> max_dist_error or result2 > max_speed_error" do;
Nice! Could you arrange a description of the conditions for the execution and termination of the algorithms in term of the number $n$ of steps already done and put it in the body of your question? This would help other members to understand the problem without having necessarily to access the paper.
Of course, but can you please help me with the complexity? Or you have to see first the description of the algorithms in order to understand?
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2025-03-21T14:48:30.170835
| 2020-03-29T08:36:51 |
356029
|
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|
Stack Exchange
|
Is the inner/outer measure mapping continuous?
Let $\mathcal F$ be a field of subsets of a set $\Omega$. Equip the space $[0,1]^\mathcal F$ of functions from $\mathcal F$ into $[0,1]$ with the product topology. Then, the set $\Delta$ of finitely additive probability measures on $\mathcal F$ is a convex and compact subset of $[0,1]^\mathcal F$.
If $\mu \in \Delta$, define the inner measure $\mu^i: 2^\Omega \to [0,1]$ for $\mu$ by
$$\mu^i(A) = \sup \big\{\mu(F): F \subset A, F \in \mathcal F \big\}, \ A \subset \Omega.$$
We can view the inner measure as a mapping $\mu \mapsto \mu^i$ from $\Delta$ into $[0,1]^{2^\Omega}$.
Question. Is the inner measure mapping continuous? That is, if $\mu_{\alpha}$ is a net in $\Delta$ that converges to $\mu$ (i.e. $\mu_\alpha(F) \to \mu(F)$ for all $F \in \mathcal F$), then is it true that $\mu^i_\alpha \to \mu^i$ (i.e. $\mu_\alpha^i(A) \to \mu^i(A)$ for every subset $A$ of $\Omega$)?
A similar question arises for outer measure, though I assume the answers are the same.
Let $\Omega = \mathbb{N}$ and let $\mathcal{F}$ be the field consisting of all finite and cofinite subsets of $\mathbb{N}$. Let $\mu_n = \delta_{2n}$ be a point mass at the integer $2n$, and let $\mu$ be the finitely additive measure that assigns measure $0$ to every finite set and $1$ to every cofinite set. Then $\mu_n(F) \to \mu(F)$ for every $F \in \mathcal{F}$.
Let $A \subset \mathbb{N}$ be the set of all even integers. Then we have $\mu_n^i(A) = 1$ for every $n$ but $\mu^i(A) = 0$.
Thanks. On the other hand, if $\mathcal F$ is closed under arbitrary unions, then I think the result holds. I wonder if something like that is necessary, though.
|
2025-03-21T14:48:30.170973
| 2020-03-29T09:50:03 |
356033
|
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|
Stack Exchange
|
Roots of the Tits form of a quiver algebra
Assume $A$ is a finite dimensional quiver algebra such that two indecomposable modules are isomorphic iff their dimension vectors are equal. It is known that $A$ is in this case representation-finite (at least when the field is algebraically closed, is this needed for quiver algebras?)
The tits quadratic form of $A$ is defined as in https://www.encyclopediaofmath.org/index.php/Tits_quadratic_form .
Question: In case every dimension vector is a root of the tits form, is there automatically a bijection between the positive roots and the dimension vectors?
This seems to be true for Nakayama algebras.
The Nakayama algebra with Kupisch series [2,3,2,3] gives an example of a non-acyclic quiver algebra for this.
|
2025-03-21T14:48:30.171165
| 2020-03-29T10:24:08 |
356034
|
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|
Stack Exchange
|
Let $n \ge 5$. Show that, among the $s_n$ space parts, there are at least $(2n − 3)/4$ tetrahedra (HMO 1973)
Engel's first problem
Engel's Second problem Proof (problems are from the Hungarian Mathematics Olympiad, 1973).
I'm having problems interpreting the last part of the proof on the second problem. I understand when Engel says that if the fifth plane intersects the line AB outside of the tetrahedron, as shown in figure 3.3, then there is a contradiction. But can't the fifth plane intersect the edge, for example, AC inside of the tetrahedron? Or in other words, the fifth plane intersects the tetrahedron ABCD into two regions. If that could occur then I don't see how there could be a contradiction, and that at most three planes can have a vertex on only one side. Furthermore, in this first part of the second problem, Engel's alludes to this problem. But then provides a contradiction by saying that this could not occur because then there would be a another point that is closer to the plane $\epsilon$.
H is for Hungarian (as can be verified from Engel's book, p. ix).
@RP_ Thanks. I'm sure I wasn't alone in not knowing that, since I don't have Engel's book and there was no link given besides the two pages.
The phrasing in Engel is a little weird, but the argument is correct. The key sentence is, "It cannot intersect all six edges of the tetrahedron $ABCD$ simultaneously." The fifth plane must intersect all six of the lines that are the edges of $ABCD$, but not all six of these intersections ("vertices") can occur along the line segments forming the tetrahedron. In your case one of these vertices must occur within the tetrahedron by assumption. One must also occur outside of the tetrahedron. Then one of the faces of the tetrahedron has vertices on both sides, whence the contradiction.
In fact, one can show that at most four of the vertices can occur along the line segments, and this limit is realized.
Thank you that helped a lot.
|
2025-03-21T14:48:30.171335
| 2020-03-29T12:26:21 |
356036
|
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|
Stack Exchange
|
Foliated circle bundles whose Euler class is torsion
Let $X$ be a closed manifold. By a foliated circle bundle $E \rightarrow X$ we mean a circle bundle over $X$ with total space $E$ and structure group $Diff^+(S^1)$, and a codimension one foliation of $E$ that is transverse to each circle fiber.
Miyoshi states in Remark 2 of `On foliated circle bundles over closed orientable 3-manifolds', Comm. Math. Helv. , 1997 the following as a fact:
For any torsion class $e \in H^2(X; \mathbb{Z})$ of any dimensional
closed manifold $X$, there exists a $C^\infty$ foliated circle bundle
over $X$ whose Euler class is equal to $e$. [i.e. the foliation is transversely $C^\infty$]
Could anyone give me a sketch of a proof, or point me to a reference please?
Inspired by the above remark, a follow-up question is the following:
Let $X$ and $e$ be as above. Is it true that if $a_1 \in H^2(X;
\mathbb{Z})$ is realized as the Euler class of a foliated circle
bundle over $X$, then $a_2:=a_1 + e$ is also realized?
Update: Igor Belegradek kindly mentioned a reference for the first part:
Miyoshi, A remark on torsion Euler classes of circle bundles, Tokyo Journal of Mathematics, 2001.
He also added a reference in the comments which shows that a principal circle bundle has a flat connection if and only if its Euler class is a torsion.
This answers the second question positively if we add the adjective principal to the circle bundle. See e.g. Theorem 3 of Milnor, One the existence of a connection with curvature zero, Comm. Math. Helv. 1958.
Theorem (Milnor): Let $K$ be a finite complex. The $SO(n)$-bundle over $K$ induced by any homomorphism $\pi_1(K) \rightarrow SO(n)$ has trivial Euler class with rational coefficients.
I don't know the answer to the second question for circle bundles which are not necessarily principal.
There are two points: 1) any class in $H^2(X;\mathbb Z)$ is the Euler class of a (unique) oriented principal circle bundle over $X$, and 2) if a principal bundle has torsion Euler class, then it has a flat connection. Having a flat connection is equivalent to being foliated. For some references see my answer https://mathoverflow.net/questions/104974/nontrivial-examples-of-non-trivial-principal-circle-bundles.
I added a reference in the edit.
Thanks, Igor. The reference you mentioned is great (a paper of Miyoshi from 2001 whose content is the proof of the above remark). Do you have any guess for the answer to the second question?
The second reference in my linked answer is to Oprea-Tanré who prove an if and only if statement. The bundle is flat (i.e. foliated) iff its real Euler class is zero. The latter is equivalent to integral Euler class being torsion (at least over compact manifold base). The sum of torsion classes is torsion, which answer your second question.
@IgorBelegradek I think the second reference is studying principal circle bundles [so the group of the bundle is S^1] whereas my question is about smooth circle bundles [so the group of the bundle is Diff(S^1)]. The if and only if statement is not true for circle bundles which are not necessarily principal circle bundles; I'm sure you are aware since you referred to MW inequality in your answer.
You might want to clarify the question then. It is not enough to say "circle bundle". You need to specify the structure group.
@IgorBelegradek Thanks for pointing it out. I just edited the question.
|
2025-03-21T14:48:30.171565
| 2020-03-29T12:45:05 |
356038
|
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|
Stack Exchange
|
A problem about real quadratic field
I have calculated some real quadratic field 's Hilbert class field with class number $2$,and I found they were satisfied $Gal(H_{K}/Q)\cong Z/2Z\oplus Z/2Z$,here $H_{K}$ is the Hilbert class field of a real quadratic field $K$ whose class number is $2$.Is it true for all quadratic field with class number $2$? How to prove it?(ps:I'm a beginner of algebraic number theory ,and I'm very interested in Hlibert class field .)
If there exist a prime $p$,that $Q(\sqrt p)$ has class number $2$,this could be a counterexample of my problem.
This is a consequence of "genus theory". If $K / \mathbf{Q}$ is an abelian extension, then the "genus field" of $K$ is the maximal extension $L / K$ such that $L/K$ is unramified and $L/\mathbf{Q}$ is abelian. You can read more about genus theory here: https://en.wikipedia.org/wiki/Genus_field.
In your case, the fact that $H_K$ is Galois over $\mathbf{Q}$ and $[H_K : \mathbf{Q}] = 4$ implies that $H_K$ must be abelian over $\mathbf{Q}$ (since there are no nonabelian groups of order 4); so $H_K$ coincides with the genus field $L_K$. The genus field of of a quadratic field is always a composite of quadratic fields, so $H_K$ must have Galois group $C_2 \times C_2$.
(More generally, for any quadratic field (real or imaginary), genus theory completely describes the 2-torsion part of the class group.)
|
2025-03-21T14:48:30.171709
| 2020-03-29T13:48:48 |
356042
|
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|
Stack Exchange
|
Spaces with unique endomorphism monoids
If $(X,\tau)$ is a topological space, let $\text{End}(X)$ denote the collection of all continuous maps $f: X\to X$. With composition, this becomes the endomorphism monoid $(\text{End}(X), \circ)$.
We say that the space $X$ has a unique endomorphism monoid if $\text{End}(X) \cong \text{End}(Y)$ as monoids, for some space $Y$, then the spaces $X$ and $Y$ are homeomorphic.
Question. Is there for every infinite cardinality $\kappa$ a space $(X,\tau)$ with unique endomorphism monoid, and $|X| = \kappa$?
My offhand guess would be 'no', but have you settled this for finite spaces?
The 1-point space is my friend :-) Then I believe the Sierpinsky space (on 2 points) has this property, that is $\mathbb{S} = ({0,1}, {\emptyset, {0}, {0,1}) = (2,3)$, if we regard $2,3$ as ordinals!. From this I would proceed with induction, take $\mathbb{S}_n = (n, n+1)$ for $n\in\omega\setminus{0}$ (so $\mathbb{S}_1$ is the 1-point space and $\mathbb{S}_2$ is the Sierpinsky space, etc.
I don't follow. How do you know that for every finite space $Y$, if $\hom(Y, Y) \cong \hom(\mathbb{S}_n, \mathbb{S}_n)$, then $Y \cong \mathbb{S}_n$?
Right - I want to try to prove it for the standard Sierpinski space $\mathbb{S}_2$ first. Or do you see straight away a space that is not homeomorphic to $\mathbb{S}_2$ but has the same endomorphism monoid? I will think about this and get back here
I think you're safe with $\mathbb{S}_2$, since any finite space with at least three points has at least four endomorphisms (constants and identity). But the situation for general $\mathbb{S}_n$ is less clear to me.
As a first, trivial remark, a map $f: \mathbb{S}_n \to \mathbb{S}_n$ is continuous if and only if it is order-preserving (if we order $n$ with the usual ordering $\in$ that it has as an ordinal). Then I was able to prove that if $Y$ is any topological space such that $\text{End}(Y) \cong \text{End}(\mathbb{S}_n)$, then we have $|Y| = n$. Unfortunately, the comment section here is too short to contain this argument. This has implications on the structure of $\text{End}(Y)$ and $\text{End}(\mathbb{S}_n)$ respectively. I don't know what would be a good place to post this very long comment
I wish I had more time for this. Maybe later on Thursday. (However, a short argument that $\text{End}(X) \cong \text{End}(Y)$ for finite $X, Y$ implies they have the same cardinality: the constant functions $X \to X$ are uniquely characterized as those endomorphisms $\phi$ that are left-absorbing, i.e., $\phi = \phi \circ \psi$ for every $\psi$. So $X, Y$ would have the same number of left-absorbing endos.) There might be a Schroeder-Bernstein type argument that I'm missing that would help here.
I'm also busy on Tuesday and Wednesday with other matters and will get back to this on Thursday. Thanks for your comments!
I expanded my comment to an answer. In particular, the Sierpiński spaces $\mathbb S_n$ do indeed have the property.
Thanks @EmilJeřábek! And you also answer the generalization, whether for infinite cardinals $\kappa$ we have that $S_\kappa = (\kappa, \kappa+1)$ has a unique endomorphism monoid
$\DeclareMathOperator\End{End}$As shown in Todd Trimble’s comment, the set of constant maps $X\to X$ is definable in $\End(X,\tau)$, as it consists of exactly the left-absorbing endomorphisms (i.e., $\phi\in\End(X,\tau)$ such that $\phi\circ\psi=\phi$ for all $\psi\in\End(X,\tau)$). Thus, an isomorphism $F\colon\End(X,\tau)\to\End(Y,\sigma)$ induces a bijection $f\colon X\to Y$ such that $F(c_x)=c_{f(x)}$ for all $x\in X$, where $c_x\colon X\to X$ is the constant-$x$ map. But then $F$ is completely determined by $f$ by
$$F(\phi)(f(x))=f(\phi(x))$$
for all $\phi\in\End(X,\tau)$: indeed, we have
$$c_{f(\phi(x))}=F(c_{\phi(x)})=F(\phi\circ c_x)=F(\phi)\circ F(c_x)=F(\phi)\circ c_{f(x)}=c_{F(\phi)(f(x))}.$$
Since we may as well assume that $X=Y$ and $f$ is the identity, it follows that:
Lemma 1. $(X,\tau)$ has a unique endomorphism monoid iff it is homeomorphic to all spaces of the form $(X,\sigma)$ such that $\End(X,\tau)$ and $\End(X,\sigma)$ are literally equal (i.e., a map $X\to X$ is an endomorphism of $(X,\tau)$ iff it is an endomorphism of $(X,\sigma)$.)
This implies
Proposition 2. If $(X,\le)$ is a total order which is isomorphic to its opposite order, then the Alexandrov space $(X,\tau)$ of upper sets of $(X,\le)$ has a unique endomorphism monoid. In particular, there exist such spaces of arbitrary cardinality.
Indeed, let $\sigma$ be a topology on $X$ such that $\End(X,\sigma)$ consists of the order-preserving maps. We may assume $|X|\ge2$. Then $\sigma$ cannot be indiscrete, hence we can fix $V\in\sigma$ and $a$ and $b$ such that $a\notin V$, $b\in V$. Assume first $a<b$. Then $\tau\subseteq\sigma$: consider an upper set $U\in\tau$. The map
$$\phi_{a,b,U}(x)=\begin{cases}b&x\in U,\\a&x\notin U\end{cases}$$
is order-preserving, hence $\phi_{a,b,U}\in\End(X,\sigma)$, and $\phi_{a,b,U}^{-1}[V]=U$, thus $U\in\sigma$. In fact, $\sigma=\tau$: if we assume for contradiction that $W\in\sigma$ is not an upper set, then the argument above shows that $\sigma$ also includes all lower sets, thus for any upper set $U$, $\phi_{b,a,U}\in\End(X,\sigma)$, but $\phi_{b,a,U}$ is not order-preserving if $U\notin\{\varnothing,X\}$.
Dually, if $a>b$, we obtain that $\sigma$ consists of all lower subsets of $X$, hence it is the Alexandrov topology corresponding to the opposite of $\le$, but this is homeomorphic to $(X,\tau)$ by our assumption on $\le$.
One can generalize the argument to a complete characterization for Alexandrov spaces. (Note that in particular, all finite spaces are Alexandrov.) First, a lemma. If $(X,\tau)$ is a topological space, let $x\le_\tau y$ denote the specialization preorder $x\in\overline{\{y\}}$, and $x\sim_\tau y$ the indistinguishability equivalence $x\le_\tau y\land y\le_\tau x$.
Lemma 3. If $\End(X,\tau)=\End(X,\sigma)$, then ${\sim_\tau}={\sim_\sigma}$ unless one space is discrete and the other indiscrete.
Proof: If, say, $a\sim_\tau b$ but $a\nsim_\sigma b$, then all mappings $X\to\{a,b\}$ are in $\End(X,\tau)$, hence in $\End(X,\sigma)$, hence $(X,\sigma)$ is discrete, hence all mappings $X\to X$ are in $\End(X,\tau)$, hence $(X,\tau)$ is indiscrete lest $a\nsim_\tau b$. QED
Proposition 4. If $(X,\tau)$ is an Alexandrov space, then $\End(X,\tau)\simeq\End(Y,\sigma)$ if and only if
$(Y,\sigma)$ is homeomorphic to $(X,\tau)$, or
$(Y,\sigma)$ is homeomorphic to the opposite of $(X,\tau)$ (i.e., the Alexandrov space corresponding to $\ge_\tau$), or
the two spaces are the discrete and indiscrete topologies on sets of the same cardinality.
Consequently, $(X,\tau)$ has a unique endomorphism monoid iff $(X,\le_\tau)\simeq(X,\ge_\tau)$, and $\tau$ is neither discrete nor indiscrete unless $|X|\le1$.
Proof: The right-to-left implication is clear. For the left-to-right implication, we may assume $X=Y$ and $\End(X,\tau)=\End(X,\sigma)$ as above.
Assume first that $\le_\tau$ is an equivalence (i.e., ${\le_\tau}={\sim_\tau}$). By Lemma 3, we may assume that ${\sim_\tau}={\sim_\sigma}$ and $\tau$ is not indiscrete. (If $\tau$ is indiscrete, then either $\sigma$ is discrete and we are done, or ${\sim}_\sigma={\sim}_\tau$, hence $\sigma$ is indiscrete, i.e., $\sigma=\tau$, and we are also done.) Since $\tau$ is the finest topology with indistinguishability relation $\sim_\tau$, this implies $\sigma\subseteq\tau$; on the other hand, if we fix $a\nsim_\tau b$, and w.l.o.g. $a\lnsim_\sigma b$, then $\phi_{a,b,U}$ is in $\End(X,\sigma)$ for all $U\in\tau$, hence $U\in\sigma$, i.e., $\sigma=\tau$.
If $\le_\tau$ is not an equivalence, let us fix $a\lnsim_\tau b$. This also implies we can fix $V\in\tau$ whose complement is not in $\tau$. Then $\phi_{a,b,V}\in\End(X,\tau)$ and $\phi_{b,a,V}\notin\End(X,\tau)$, hence $a\lnsim_\sigma b$ or $b\lnsim_\sigma a$. W.lo.g., we assume the former (the other choice leads to the opposite order). Then for each $U\in\tau$, $\phi_{a,b,U}\in\End(X,\tau)$ implies $U\in\sigma$, i.e., $\tau\subseteq\sigma$. Since $\tau$ is the finest topology with specialization preorder $\le_\tau$, if $\tau\subsetneq\sigma$, then (in view of ${\sim_\tau}={\sim_\sigma}$) there are $x,y$ such that $x\lnsim_\tau y$ and $x\nleq_\sigma y\nleq_\sigma x$. But as above, this contradicts $\phi_{x,y,V}\notin\End(X,\sigma)$ for suitable $V\in\tau$. Thus, $\tau=\sigma$. QED
The characterization can be easily extended to all non-$R_0$ spaces. Recall that $(X,\tau)$ is $R_0$ if $\le_\tau$ is symmetric (i.e., ${\le_\tau}={\sim_\tau}$).
Proposition 5. If $(X,\tau)$ is a non-Alexandrov non-$R_0$ space, then $(X,\tau)$ has a unique endomorphism monoid.
Proof: Assume that $\End(X,\tau)=\End(X,\sigma)$. Let us fix $a\lnsim_\tau b$. There exists $V\in\tau$ whose complement is not in $\tau$ (e.g., any open set separating $b$ from $a$); then $\phi_{a,b,V}\in\End(X,\tau)$ and $\phi_{b,a,V}\notin\End(X,\tau)$, hence (1) $a\lnsim_\sigma b$ or (2) $b\lnsim_\sigma a$. (In particular, $(X,\sigma)$ is not $R_0$.) If (1) holds, then for every $U\in\tau$, $\phi_{a,b,U}\in\End(X,\tau)$ implies $U=\phi_{a,b,U}^{-1}[b]\in\sigma$, i.e., $\tau\subseteq\sigma$. If (2) holds, then the same argument gives $\{X\smallsetminus U:U\in\tau\}\subseteq\sigma$.
Since $(X,\sigma)$ is not $R_0$ either, a symmetric argument implies that (1') $\sigma\subseteq\tau$, or (2') $\{X\smallsetminus U:U\in\sigma\}\subseteq\tau$. It is impossible that (1) and (2') hold together: this would imply that $\tau$ is closed under complement, whence it is $R_0$. Likewise, (2) and (1') are incompatible. Thus, the only two possibilities are that either (1) and (2) hold, in which case $\tau=\sigma$, or (1') and (2') hold, in which case $\tau$ and $\sigma$ are mutually opposite Alexandrov spaces. QED
Notice that $(X,\tau)$ is $R_0$ iff the Kolmogorov quotient $X/{\sim_\tau}$ is $T_1$. It is easy to see that:
Lemma 6. If $(X,\tau)$ and $(X,\sigma)$ are spaces such that ${\sim_\tau}={\sim_\sigma}$, then $\End(X,\tau)=\End(X,\sigma)$ iff $\End((X,\tau)/{\sim_\tau})=\End((X,\sigma)/{\sim_\sigma})$.
In view of Lemma 3, this gives a reduction of the remaining classification to $T_1$ spaces. Observe that an $R_0$ space $(X,\tau)$ is Alexandrov iff $(X,\tau)/{\sim_\tau}$ is discrete.
Corollary 7. If $(X,\tau)$ is an $R_0$ non-Alexandrov space, then $(X,\tau)$ has a unique mononorphism monoid iff the $T_1$ space $(X,\tau)/{\sim_\tau}$ has a unique monomorphism monoid.
Brilliant, Emil, thanks a lot!
You’re welcome.
Fullbright (+1) !
While the OP question ultimately is specific (as it should), it really offers an entire topic:
TOPIC: What are topological spaces
$\ (X\ T)\ $ which are topologically uniquely characterized by monoid $\ \text{End}(X\ T)\,?$
In other words, given an abstract monoid $\ M,\ $ can
we recover topological space $\ (X\ T)\ $ uniquely (if at all) so that $\ M\ $ and $\ \text{End}(X\ T)\ $ are isomorphic (as abstract algebraic monoids).
In this answer, let me provide some tools.
Let $\ \mathbf M:=(M\ \circ\ J)\ $ be an arbitrary monoid. Let
$$ C\ :=\ \{c\in M:\ \forall_{f\in M}\ c\circ f=c\} $$
If $\ \mathbf M\ $ were isomorphic to $\ \text{End}(X\ T)\ $
then $\ C\ $ and $\ X\ $ would be in a canonical 1-1 correspondence as mentioned by @YCor in a comment to Dominic's
real-question. This is the basic starting tool.
Next, let's discuss the next tool, the idempotents
$\ i\in\mathcal I\subseteq M,\ $ where
$$ \mathcal I\ :=\ \{i\in M:\ i\circ i=i\} $$
For instance, the unit $\ J\in M\ $ and constants
$\ c\in C\ $ are all idempotents.
Definition
$$ \forall_{i\ j\,\in\mathcal I}\quad
(\,i\subseteq j\ \Leftarrow:\Rightarrow\
j\circ i=i\,) $$
It follows that:
$\ \forall_{i\in\mathcal I}\quad i\subseteq i;$
$\ \forall_{i\ j\ k\in\mathcal I}\quad(
(i\subseteq j\ \text{and}\ j\subseteq k)\ \Rightarrow
i\subseteq k) $
$\ \forall_{i\in\mathcal I}\,
\forall_{j\in C}\quad
(\ i\subseteq j\ \Rightarrow\ j=i\ ) $
Topological idempotents $\ i:X\to X\ $ are closely related to Karol Borsuk's retractions; such idempotent
$\ i\ $ retract $\ X\ $ retracts $\ X\ $ onto
$\ i(X)\subseteq X. $
By Bourbaki theorem, $\ (X\ T)\ $ is Hausdorff
$\ \Leftrightarrow\ \Delta_X:=\{(x\ x):x\in X\ $ is closed in $\ X\times X.$ It follows that for Hausdorff spaces the said retract $\ i(X)\ $ is closed in $\ X.\ $ Indeed,
$$ i(X)\ :=\ \{x:\ i(x)=x\}\ =
\ (i\triangle \text{Id}_X)^{-1}(\Delta_X) $$
for diagonal product function
$\ i\triangle \text{Id}_X : X\to X\times X.$
Great!. (This is obviously useful for Hausdorff spaces).
Let $\ \pi:\mathcal I\to 2^C\ $ be defined by
$$ \forall_{i\in\mathcal I}\quad \pi(i)\ :=
\ \{c\in C:\ i\circ c = c\} $$
This is how idempotents of $\ \mathbf M\ $ point to the respective subsets of $\ X;\ $ or to closed subsets in the Hausdorff case -- I mean pointing to $\ \pi(i).$
Theorem
$\ \forall_{i\ j\in\mathcal I}\quad(\,i\subseteq j
\ \Rightarrow \pi(i)\subseteq\pi(j)\, ) $
$\ \forall_{i\ j\in\mathcal I}\quad(\,(i\subseteq j
\ \text{and}\ j\subseteq i)
\ \Rightarrow\ \pi(i)=\pi(j)\, ) $
$\ \forall_{i\in\mathcal I}\quad (\, i\in C
\ \Leftrightarrow\ \pi(i)=\{i\} $
Another tool, the uc-morphisms and nuc-morphims, was mentioned in my answer to Dominic's-real-question. In topological language, if $\ i\ $ is a topological idempotent then $\ I(X)\ $ has or has not fpp when $\ i\ $ is an uc-morphism or nuc-morphism respectively.
These tools may serve as a starting point to a discussion of specific topological spaces or their classes.
I enjoyed this answer very much! Thank you
This is really cool. Are there any paper's on this topic? I'm currently looking into cannonical topologies on monoids and the quetion "What are topological spaces (X T)
which are topologically uniquely characterized by monoid End(X T)?" came up natrually. Google then brought me here :)
NOTATION
$$ S_X\ :=\ \{\emptyset\ X\} $$
$$ D_X\ :=\ 2^X\ =\ \{A:\ A\subseteq X\} $$
Given an arbitrary set $\ X,\ $ topology $\ S_X\ $ is the smallest (the weakest) topology in $\ X;\ $ and the discrete topology $\ D_X\ $ is the largest (the strongest) topology in $\ X.$
--------------------
Here is a logically initial modest positive result:
Theorem Let set $\ X\ $ be finite. Then for every non-discrete topological space $\ (Y\ T)\ $ (i.e. $\ T\ne D_Y),\ $ if monoids $\ \text{End}(X\ S_X)\ $ and
$\ \text{End}(Y\ T)\ $ are isomorphic then topological spaces $\ (X\ S_X)\,$ and $\,(Y\ T)\ $ are homeomorphic, i.e. $\,\ |Y|=|X|\ $ and $\ T=S_Y.$
Proof If two monoids are isomorphic than they have the same number (cardinality) of constants (of the left-absorbing elements). Furthermore, the number of points of an arbitrary topological space is equal to the number the constants of its monoid of continuous self-maps.
Let's assume that monoids $\ \text{End}(X\ S_X)\ $ and
$\ \text{End}(Y\ T)\ $ are isomorphic. Then
$$ |X|\ =\ |Y| $$
Also,
$$ |\text{End}(X\ S_X)|\ =\ |\text{End}(Y\ T)|\ $$
hence
$$ |\text{End}(Y\ T)|\ =\ |\text{End}(X\ S_X)|
\ =\ |X^X| $$
so that
$$ |\text{End}(Y\ T)|\ =\ |Y^Y| $$
This means that $\ T=S_Y\ $ or $\ T=D_Y\ $ hence,
by theorem's assumption, $\ T=S_Y\ $ --
otherwise $\ T\ $ would be not discrete nor the
smallest, i.e. there exists $\ G\ \in\ T\setminus S_Y\ $ and non-isolated $\ p\in Y\ $ (i.e. such that
$\ \{p\}\not\in T)$. Then consider $\ f:Y\to Y\ $ such that $ f(p)\in G\ $ and
$\ f(Y\setminus\{p\})\subseteq Y\setminus G.\ $ Such
$\ f\ $ is not continuous in $\ (Y\ T),\ $ hence
$$ |\text{End}(Y\ T)|\ <\ |Y^Y| $$
-- a contradiction. End of PROOF
----------------
Remark For every set $\ X,\ $ monoids
$$ \text{End}(X\ S_X)\quad \text{and}\quad
\text{End}(X\ D_X)\ $$
are isomorphic while the respective topological spaces $\ (X\ S_X)\ $ and $\ (X\ D_X)\ $ are not homeomorphic whenever $\ |X|>1.$
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2025-03-21T14:48:30.172923
| 2020-03-29T14:24:01 |
356045
|
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|
Stack Exchange
|
Can this function be interpolated with a small power series
Does there exist a power series $\sum_i a_i x^i$ that is $1$ at $0$ and $0$ at integers from $1$ to $n$, and such that $\sum_i |a_i|$ is polynomial in $n$?
I feel the answer might be no but I'm not sure how to prove it.
oh wait, the sum of absolute values of the $a_i$'s in your example grows polynomially in $n$ no? This is what I was asking for
Take an entire function $f$ such that $f(0)=1$ and $f(j) = 0$ for all nonzero integers: an example is $f(z) = \sin(\pi z)/(\pi z) $ for $z \ne 0$, $1$ for $z=0$.
The Maclaurin series of $f$ satisfies $\sum_{i} |a_i| < \infty$.
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2025-03-21T14:48:30.173018
| 2020-03-29T14:31:06 |
356046
|
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"Andreas Blass",
"Asaf Karagila",
"James E Hanson",
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|
Stack Exchange
|
Cofinality of infinitesimals
Suppose $\kappa$ is an infinite cardinal and $U$ is a countably incomplete uniform ultrafilter over $\kappa$. Then $\mathbb R^\kappa/U$ is nonstandard. What is the cofinality of the set of infinitesimals of this field? What can we say when $U$ is $\kappa$-regular?
Background information: Recall that $U$ is $\kappa$-regular when there exists a sequence $\langle X_\alpha : \alpha < \kappa \rangle \subseteq U$ such that for any $\beta < \kappa$, $\{ \alpha : \beta \in X_\alpha \}$ is finite. If $U$ is $\kappa$-regular, then I can show that the cofinality of $\mathbb R^\kappa/U$ (rather than infinitesimals) is $>\kappa$. Furthermore, if $\mathbb R^\kappa/U$ is $\delta$-saturated, then the cofinality of the infinitesimals is $\geq\delta$. $\omega_1$-saturation is automatic for ultrapowers by countably incomplete ultrafilters. If the ultrafilter satisfies a property stronger than regularity called goodness, then the ultrapower is $\kappa^+$-saturated.
I imagine it's the cofinality of $\omega^\kappa/U$, as a linear order.
@AsafKaragila This is the answer if we are looking at the set of things below a fixed element $[f]_U$, since for each $\alpha$ can choose a cofinal $\omega$-sequence in $f(\alpha)$. But there is no supremum to the set of infinitesimals, so basically I am asking about possible “gaps.”
Well, if you look at $1/\varepsilon$, then you're looking at the cofinality of the linear order $\Bbb R^\kappa/U$, so the fact they are infinitesimals is irrelevant here.
The thing is I am looking at increasing sequences of infinitesimals, i.e. converging to the gap. So if we take $1/\varepsilon$, then we are looking a decreasing sequence of infinite numbers.
It should be the cofinality of the reverse order on $(\omega^{\kappa}/U) \setminus \omega$, right? Not that that necessarily makes the question easier.
@JamesHanson Right.
As pointed out in a comment by James Hanson, the cofinality of the infinitesimals is the same as the coinitiality (i.e., cofinality or the reverse order) $\mu$ of the nonstandard part of $\omega^\kappa/U$.
Even for $\kappa=\omega$, this coinitiality $\mu$ is not decided by the axioms of set theory. Furthermore, even within a single model of set theory, $\mu$ can depend on the particular ultrafilter $U$.
Specifically, if one starts with a model of CH and adds $\lambda$ Cohen reals, the resulting model has nonprincipal ultrafilters $U$ on $\omega$ for which $\mu$ is any regular uncountable cardinal $\leq\lambda$. (The same holds for the cofinality of the whole ultrapower $\omega^\omega/U$, and in fact this cofinality and $\mu$ can be chosen independently.) Similarly, if one adds $\lambda$ random reals to a model of CH, every regular uncountable cardinal $\leq\lambda$ occurs as $\mu$ for some $U$. (But now the cofinality of $\omega^\omega/U$ is $\aleph_1$ because random forcing is $\omega^\omega$-bounding.)
These results were proved by Mike Canjar in his thesis; the MathSciNet data for the published version are:
MR0924678 (89g:03073) Reviewed
Canjar, Michael
Countable ultraproducts without CH.
Ann. Pure Appl. Logic 37 (1988), no. 1, 1–79.
Thanks! What about for uncountable $\kappa$? Does it hold in ZFC that $\mu>\kappa$ for regular $U$?
@Monroe: Very clearly, if ZFC is consistent, it has a countable model, so all the cardinals there are countable by definition! There is no such thing as uncountable cardinals.
@MonroeEskew When I first started to think about your question, my feeling was that regularity of $U$ is just what you'd need to get $\mu>\kappa$, Unfortunately, my attempt to prove it had a gap (euphemism for "it was wrong"), and I still don't see how to fix it.
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2025-03-21T14:48:30.173409
| 2020-03-29T14:56:19 |
356047
|
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|
Stack Exchange
|
Upperbound on Shannon capacity of graph and strong product of graph
Given a Graph $G = (V=[n],E)$, if a symmetric matrix $B$ fits $G$, it has non-zero diagonal elements and 0 on off-diagonal entries if $\{i,j\}$ are non-edge in $G$.
Let \begin{equation}
R(G) = \min Rank(B) \text{ } B_{ii}\ne 0, B_{ij}=0 (\{i,j\}\in \overline{G})
\end{equation}
show that $R(G*H)\leq R(G)R(H)$, where $R(G*H)$ is the strong product of $G$ and $H$.
show that $\Theta(G) \leq R(G)$.
I am not sure how to prove these, the second statement follows from the first one naturally, but how the ranks are related to each other?
|
2025-03-21T14:48:30.173500
| 2020-03-29T15:07:54 |
356049
|
{
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"Benjamin Steinberg",
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"R. van Dobben de Bruyn",
"Todd Trimble",
"Wlod AA",
"YCor",
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|
Stack Exchange
|
What spaces $X$ do have $\text{End}(X) \cong \text{End}(\mathbb{R})$?
This is a follow-up on the following question. Let $\text{End}(X)$ denote the endomorphism monoid of a topological space $X$ (that is, the collection of all continuous maps $f:X\to X$ with composition).
What is an example of a topological space $X$ with $X\not\cong \mathbb{R}$ but the monoids $\text{End}(X)$ and $\text{End}(\mathbb{R})$ are isomorphic?
A general remark: in $\mathrm{End}(X)$, constants are precisely the left-absorbing elements (i.e., those $g$ such that $gf=g$ for all $f$). The monoid acts on the left on left-absorbing elements, and this is conjugate to the action on $X$. In particular, the condition that $X$ is homogeneous under $\mathrm{Aut}(X)$ can be read on the abstract monoid $\mathrm{End}(X)$. I expect this is might be a starting point to check that every $X$ with $\mathrm{End}(X)\simeq\mathrm{End}(\mathbf{R})$ is homeomorphic to $\mathbf{R}$.
The constant maps are not enough in general to pick up the topology for a general space. For example the endomorphism monoid of X in the discrete and indiscrete topology are the same.
Indeed it sounds tricky, even assuming beforehand that $X$ is Hausdorff.
No such space exists. We actually get the stronger statement that every isomorphism $\operatorname{End}(X) \stackrel\sim\to \operatorname{End}(\mathbf R)$ is induced by an isomorphism $X \stackrel\sim\to \mathbf R$ (unique by Observation 1 below). In contrast, in Emil Jeřábek's beautiful construction in this parallel post there is an 'outer automorphism' $\operatorname{End}(X) \stackrel\sim\to \operatorname{End}(X)$ that does not come from an automorphism $X \stackrel\sim\to X$ of topological spaces (it comes from an anti-automorphism of ordered sets).
I will use the substantial progress by YCor and Johannes Hahn, summarised as follows:
Observation 1 (YCor). For every topological space $X$, the map $X \to \operatorname{End}(X)$ taking $x$ to the constant function $f_x$ with value $x$ identifies $X$ with the set of left absorbing¹ elements of $\operatorname{End}(X)$.
In particular, an isomorphism of monoids $\operatorname{End}(X) \stackrel\sim\to \operatorname{End}(Y)$ induces a bijection $U(X) \stackrel\sim\to U(Y)$ on the underlying sets.
Observation 2 (Johannes Hahn). If $\operatorname{End}(X) \cong \operatorname{End}(\mathbf R)$, then $X$ is $T_1$. Since the closed subsets of $\mathbf R$ are exactly the sets of the form $f^{-1}(x)$ for $x \in \mathbf R$, we conclude that these are closed in $X$ as well, so the bijection $X \to \mathbf R$ of Observation 1 is continuous.
(The asymmetry is because we used specific knowledge about $\mathbf R$ that we do not have about $X$.)
To conclude, we prove the following lemma.
Lemma. Let $\mathcal T$ be the standard topology on $\mathbf R$, and let $\mathcal T' \supsetneq \mathcal T$ be a strictly finer topology. If all continuous maps $f \colon \mathbf R \to \mathbf R$ for $\mathcal T$ are continuous for $\mathcal T'$, then $\mathcal T'$ is the discrete topology.
Note that Observation 2 and the assumption $\operatorname{End}(X) \cong \operatorname{End}(\mathbf R)$ imply the hypotheses of the lemma, so we conclude that either $X = \mathbf R$ or $X = \mathbf R^{\operatorname{disc}}$. The latter is clearly impossible as it has many more continuous self-maps.
Proof of Lemma. Let $U \subseteq \mathbf R$ be an open set for $\mathcal T'$ which is not open for $\mathcal T$. Then there exists a point $x \in U$ such that for all $n \in \mathbf N$ there exists $x_n \in \mathbf R$ with $|x - x_n| \leq 2^{-n}$ and $x_n \not\in U$. Without loss of generality, infinitely many $x_n$ are greater than $x$, and we can throw out the ones that aren't (shifting all the labels, so that $x_0 > x_1 > \ldots > x$). Up to an automorphism of $\mathbf R$, we can assume $x = 0$ and $x_n = 2^{-n}$ for all $n \in \mathbf N$. Taking the union of $U$ with the usual opens $(-\infty,0)$, $(1,\infty)$, and $(2^{-n},2^{-n+1})$ for all $n \in \mathbf N$ shows that
$$Z = \big\{1,\tfrac{1}{2},\tfrac{1}{4},\ldots\big\}$$
is closed for $\mathcal T'$. Consider the continuous function
\begin{align*}
f \colon \mathbf R &\to \mathbf R\\
x &\mapsto \begin{cases}0, & x \leq 0,\\ x, & x \geq 1, \\ 2^nx, & x \in \big(2^{-2n},2^{-2n+1}\big], \\ 2^{-n}, & x \in \big(2^{-2n-1},2^{-2n}\big].\end{cases}
\end{align*}
Then $f^{-1}(Z)$ is the countable union of closed intervals
$$Z' = \bigcup_{n \in \mathbf N} \big[2^{-2n-1},2^{-2n}\big] = \big[\tfrac{1}{2},1\big] \cup \big[\tfrac{1}{8},\tfrac{1}{4}\big] \cup \ldots.$$
By the assumption of the lemma, both $Z'$ and $2Z'$ are closed in $\mathcal T'$, hence so is the union
$$Z'' = Z' \cup 2Z' \cup [2,\infty) = (0,\infty),$$
and finally so is $Z'' \cup (-Z'') = \mathbf R\setminus 0$. Thus $0$ is open in $\mathcal T'$, hence so is every point, so $\mathcal T'$ is the discrete topology. $\square$
¹Elements $f$ such that $fg = f$ for all $g$. (I would probably have called this right absorbing!)
Thanks for this beautiful answer!
I checked before posting: the condition that $f$ satisfies $\forall g,fg=g$ is always called "left-absorbing", or "left zero" (which sounds intuitive to me: $f$ absorbs on the lefts elements from the right).
@YCor: sorry, I wasn't trying to challenge your use. I also checked and saw that this is what it's called. Anyway, left and right are hard.
No problem. In the same vein, in a group $G$ with subgroup $H$, one calls $gH$ left coset and $Hg$ right coset but this is possibly not obvious either.
I feel that YCor's guys are "left absorbing" or "absorbing right" (in that order).
Over in the other thread I said "left-absorbing", but it means of course absorbing to the left!
By @YCor's comment, $X$ has the same number of elements as $\mathbb{R}$ and the actions of $End(X)$ on $X$ is the same as the action of $End(\mathbb{R})$ on $\mathbb{R}$. Now consider the automorphism group and its action on $X$. $Aut(\mathbb{R})$ consists of strictly increasing and strictly decreasing maps and it is easy to see that the point stabiliser subgroups $Aut(\mathbb{R})_x$ act transitively on $\mathbb{R}\setminus\{x\}$. But this action is imprimitive: $\mathbb{R}\setminus\{x\}$ has two blocks, namely $(x,+\infty)$ and $(-\infty,x)$ and the set stabilisers act transitively on those two sets. Phrased differently: There are exactly three equivalence relations on $\mathbb{R}\setminus\{x\}$ invariant under $Aut(\mathbb{R})_x$, the two trivial ones "everything is equivalent" and "nothing is equivalent" as well as a unique non-trivial one with the two equivalence classes $(-\infty,x)$ and $(x,+\infty)$.
Conclusion: We can recover a linear order on $X$ from $Aut(X)$ and thus from $End(X)$. And in particular we can recover the order topology of this ordering. And all elements of $End(X)$ must be continuous w.r.t. this order topology.
This does not necessarily mean that the order topology coincides with the original topology on $X$, but it is very close.
What can we say about the original topology? We know that there are lots of continuous map, but not too many (since $|End(\mathbb{R})|=|\mathbb{R}|$) so that $X$ is not indiscrete. We can say that the only self-maps with finite image are constant. In particular, there cannot be a Sierpinski space inside $X$, because every open set yields a continuous map to the Sierpinski space. Therefore $X$ is at least a $T_1$-space.
That is enough to conclude that the original topology on $X$ is at least as fine as the order topology: For every order-topology-closed subset $A\subseteq\mathbb{R}$ there is a continuous self-map $f:\mathbb{R}\to\mathbb{R}$ with $A=f^{-1}(0)$. Therefore the corresponding order-topology-closed subset of $X$ is also the preimage of a point under a continuous self-map and thus original-topology-closed. In other words: Our canonical bijection $X\to\mathbb{R}$ is continuous w.r.t. the original topology on $X$.
I have the feeling that the other direction is equally easy, but I just don't see it right now.
If you have a continuous bijection onto $\mathbf{R}$, you have the Hausdorff property. I've been wondering if there are strange strong topologies (such as the one generated by the standard topology and the co-countable subsets), maybe the latter has the same homeomorphism group, but not the same monoid of continuous self-maps).
Thanks for this effort that is a key ingredient towards answering the question!
There is a straightforward approach to proving that isomorphism of monoids $\ \text{End}(X\ T)\ $ and
$\ \text{End}(\Bbb R\,\ T_E)\ $ implies homeomorphism of $\ (X\ T)\ $ and Euclidean space
$\ (\Bbb R\,\ T_E):\ $
let $\ (M\ \circ\ J)\ $ be an arbitrary abstract monoid. One may associate with this monoid a set of
"points" $\ C\ $ of "constant" elements, as @YCor did in his first comment, as the left-absorbing elements $\ c\in C,\ $ where
$$ \forall_{f\in M}\quad c\circ f=c $$
Then one selects purely algebraic (monoidal) notions which have the respective topological meaning in $\ (\Bbb R\,\ T_E).\ $ You need just a couple of such notions.
Then you force on the abstract monoid $\ (M\ \circ\ J)\ $ the purely algebraic (monoidal) axioms that induce the properties of reals. That's all.
For instance one may apply idempotents $\ i\in\mathcal I\subseteq M,\ $ where $\ i\circ i=i. \ $ Then the critical notion to me is what I've defined and call universally closed idempotents or uc-morphism $\ i\in\mathcal I,\ $ which satisfy:
$$ \forall_{f\in M}\,\exists_{p\in M}\quad
i\circ f\circ i\circ p\,\ =\,\ i\circ p $$
Observe that we have a canonical map
$$ \pi: \mathcal I\to 2^C, $$
where
$$ \forall_{i\in\mathcal I}\quad \pi(i)\ :=
\ \{c\in C:\ i\circ c=c\} $$
This has nice properties... etc.
---------------
(After I saw the OP question, I hesitated... and I can remove my post, no sweat.)
Also, idempotents that are not universally closed (or nuc-morphisms for short) have their role as well.
Dominic shoots from the hip on MO but this OP-question and his other one, EndMonDom-question are well aimed! The latter offers great scope and allows for a general presentation of the topic. (I saw this latter but earlier question only now).
|
2025-03-21T14:48:30.174149
| 2020-03-29T16:54:53 |
356057
|
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|
Stack Exchange
|
What kind of "skeleton" does a strict $n$-category admit?
The skeleton of a 1-category is a way to "normalize" it, because
Any category is equivalent to its skeleton, and
Any equivalence between skeletal categories is an isomorphism.
For strict $n$-categories with $n \geq 2$, it's not clear whether such as "normal form" exists. For instance, one might define a "skeletal 2-category" to be a 2-category where any two equivalent objects are equal and any two isomorphic 1-morphisms are equal. But it's not clear that every 2-category is 2-equivalent to a skeletal one in this sense.
Here my notion of "$n$-equivalence" is the weak one (as found in the folk model structure): for $n=1$ it's the usual notion, and for $n\geq 2$, an $n$-equivalence of $n$-categories is defined to be an $n$-functor which is surjective on $n$-equivalence classes of objects and such that the functor is an $(n-1)$-equivalence on hom-$(n-1)$-categories (clearly I could start the base case at a lower value of $n$). Here $n$-equivalences of objects are defined in the usual way.
Question: Let $n \geq 2$. Is there some analog of the "skeleton" construction for 1-categories which applies to strict $n$-categories?
Such a notion should have the analogs of properties (1) and (2) above. I'd already be happy to understand the case $n=2$.
Am I right that the tricky part here is the following: even if we wanted to do something a little simpler and just take the "Hom-skeleton" of a 2-category (i.e., form an equivalent 2-category with skeletal Hom-categories), it would require defining the composition functors in a coherent way. Since taking skeletons does not define a functor, it's not at all clear how to make sure our composition functors play nicely.
@MattFeller I think that's exactly right!
It seems to me that for any 2-category, one can construct a biequivalent skeletal bicategory, but not in general a biequivalent skeletal strict 2-category.
For a proof of the positive statement, see Lemma 2.20 in your PhD supervisor's PhD thesis. For the negative statement, recall the cohomological classification of 2-groups (see e.g. HDA5). A 2-group $G$ is equivalent to a strict & skeletal 2-group iff its "associativity cocycle" (or "Sinh invariant") in $H^3(\pi_0(G),\pi_1(G))$ vanishes.
@AlexanderCampbell Do you know of a concrete example of a 2-group in which this Sinh invariant doesn't vanish? Is this nonvanishing a pathological property that most 2-groups in practice don't satisfy?
@MattFeller No, you should think of it as typical. A general group had no reason to have vanishing third cohomology, and barring that there’s no reason for the invariant to vanish. Every nontrivial third cohomology class determines a distinct 2-group, and in essence, conversely.
|
2025-03-21T14:48:30.174348
| 2020-03-29T17:07:03 |
356059
|
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|
Stack Exchange
|
Projective embeddings of quotients of normal varieties
Let $X$ be a normal complex projective variety of dimension $m$, $G$ be a finite subgroup of $\mathrm{Aut}(X)$, and $Y = X / G$ be the quotient. I am particularly interested in the case where $X$ is a normal rational surface (not just smooth). My question is fundamentally about degrees of projective embeddings of $X$ and $Y$, though I want to state it in slightly different language. I would expect an answer to be in the literature somewhere, since the question seems basic, but my search has only found related but ultimately irrelevant work.
Consider a flat, finite map $\Phi : X \to \mathbb{P}^N$ whose image is not contained in a proper linear subspace. Basically, $\Phi$ is a projective holomorphic immersion with $N$ chosen optimally. Given a $(1,1)$-form $\omega$ on $X$ representing a "good" orbifold Kähler metric (it suffices in my setting to assume it is a positive real multiple of the first Chern class of some ample line bundle), we can define:
$$
\mathrm{Vol}(X, \omega, \Phi) = \int_X \Phi^*(\sigma_N) \wedge \omega^{m-1}
$$
where $\sigma_N$ is the Fubini-Study metric on $\mathbb{P}^N$. We then can define the $N$-dimensional projective volume $\mathrm{Vol}_N(X, \omega)$ to be $\infty$ if no such $\Phi$ exists and
$$
\mathrm{Vol}_N(X, \omega) = \inf_{\Phi : X \to \mathbb{P}^N} \mathrm{Vol}(X, \omega, \Phi)
$$
otherwise. Note that this is finite for sufficiently large $N$. One can then define the projective volume
$$
\mathrm{Vol}(X, \omega) = \liminf_{N \to \infty} \mathrm{Vol}_N(X, \omega) \ge 0.
$$
My interest is in how this quantity behaves under finite quotients.
Assume that $\omega$ is $G$-invariant, descending to $\omega_Y$ on $Y$ (my group $G$ acts by isometries of the orbifold Kähler metric). It is not hard to show that
$$
\mathrm{Vol}(X, \omega) \le |G| \mathrm{Vol}(Y, \omega_Y)
$$
using compositions $X \to Y \to \mathbb{P}^N$. The above assumptions imply that the map $X \to Y$ is étale over Zariski open submanifolds of each, hence it's just the degree formula from differential topology.
Main question: What about a bound in the other direction? More specifically, is it true that
$$
\frac{1}{|G|}\mathrm{Vol}(Y, \omega_Y) \le \mathrm{Vol}(X, \omega)?
$$
If not, is it at least the case that
$$
\frac{1}{p(|G|)}\mathrm{Vol}(Y, \omega_Y) \le \mathrm{Vol}(X, \omega)
$$
for some polynomial $p$, say, depending only on $m = \mathrm{dim(X)}$ or the assumption that $X$ is a normal rational surface?
This is at heart a question about degrees of projective embeddings of $Y$ induced by projective embeddings of $X$. I think one can show $p(|G|) = |G|!$ works, but I would really like a better bound if possible. Some weakenings:
Question 2: If Question 1 has a negative answer in general, is there an assumption on $G$ that can ensure a yes answer? For instance, what if $G$ is solvable? What if $G$ contains an abelian subgroup of index at most $N$ for some fixed $N \ge 0$?
Note that the solvable case reduces by induction to the case where $G$ is cyclic. The last part of Question 2 comes from the fact that if $X$ is a normal rational surface, then $G$ can be realized as a finite subgroup of the Cremona group $\mathrm{Cr}(2)$, and all finite subgroups of the Cremona group are known to have an abelian subgroup of uniformly bounded index.
Any references or general insight would be greatly appreciated.
|
2025-03-21T14:48:30.174581
| 2020-03-29T17:28:35 |
356061
|
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|
Stack Exchange
|
Name for a class of languages closed under union, inverse generalised sequential machine mappings and intersection with regular languages
I asked this question on the TCS stackexchange but have so far received no answer:
Is there a name for classes of languages closed under finite union, inverse generalised sequential machine mappings and intersection with regular languages? These have come up naturally in my research and I don't want to invent a new name for them if they already have one.
An example of a class of languages with these closure properties that is not a full trio (i.e. closed under homomorphism, inverse homomorphism and intersection with regular) is the poly-context-free languages (intersections of finitely many context-free languages).
|
2025-03-21T14:48:30.174657
| 2020-03-29T18:17:11 |
356063
|
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|
Stack Exchange
|
about simple non-abelian 2-generated group
Does there exist a simple non-abelian 2-generated group $G$ and two elements $a, b \in G$, such that $\langle \{a, b\} \rangle = G$, $a^2 =1$ and $\forall c, d \in G$ $\langle \{c^{-1}bc, d^{-1}bd \} \rangle \neq G$?
Duplicate of https://math.stackexchange.com/questions/3600219/does-there-exist-a-two-generated-simple-non-abelian-group-with-specific-properti
No. Note that $\langle b ,a^{-1}ba \rangle$ is normalized by $b$, and by $a$. Hence
$\langle b, a^{-1}ba \rangle$ is normalized by $\langle a,b \rangle = G$. Since $G$ is simple non-Abelian, $G = \langle b, a^{-1}ba \rangle .$
Why $\langle b ,a^{-1}ba \rangle$ is normalized by $b$ ?
Any subgroup $H$ containing an element $b$ of an overgroup $G$ is certainly normalized by $b$ since $b^{-1}Hb = H$ (since $b \in H$).
also we must show why $\langle b, a^{-1}ba \rangle \neq {1} $ ? I think because $ b \neq 1$.
Yes, it is of course since $b \neq 1$, but this is clear since $G \neq \langle a \rangle$ (for $G$ is assumed non-Abelian simple).
or If we show for every $n \in \mathbb{N} , b^n \neq 1$ then our subgroup is not identity?
That would show that $b$ has infinite order, which need not be the case.
|
2025-03-21T14:48:30.174757
| 2020-03-29T19:15:26 |
356067
|
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"Will Sawin",
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"joaopa"
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|
Stack Exchange
|
Transcendance in function fields
Denote by $\Omega$ the completion of an algebraic closure of $\mathbb F_q\left(\!\left(\frac1T\right)\!\right)$ for the valuation $-\deg$. Let $(a_n)_n$ be a sequence of $\overline{\mathbb F_q(T)}\subset\Omega$ such that $a_n\ne0$ for all $n\in\mathbb N$, and such that
$$\lim_{n\to+\infty}\limits \deg(a_n)=+\infty.$$ Consider the entire function on $\Omega$ defined by $$f(z)=\prod_{n\ge0}\left(1-\frac z{a_n}\right).$$ Can one assert that for every $$u\in\overline{\mathbb F_q(T)}\setminus\{a_n\mid n\in\mathbb N\},$$ we have $$f(u)\notin\overline{\mathbb F_q(T)}?$$
Thanks in advance for any answer.
What if $a_n = T^n$ or otherwise lies in $\mathbb F_q((1/T))$, and $z$ lies in $\mathbb F_q((1/T))$, then won't the product obviously lie in $\mathbb F_q((1/T))$?
sorry. I read again my question. There is a problem. I edit it.
This obviously won't work either. You can fix $z$ and a target $c$ in $\overline{\mathbb F_q(T)}$ and choose $a_n$ so that the partial product $\prod_{n\geq 0} (1- \frac{z}{a_n})$ is within $T^n$, say, of $c$.
What do you mean by "within $T^n, say, of $c$" ?
Sorry that was unclear. I mean $ \deg (\prod_{n=0}^m (1- \frac{z}{a_n} ) - c ) \leq m$. It's easy to see that you can always choose $a_m$ satisfying that if $a_0,\dots, a_{m-1}$ do and you will have $\deg(a_n)\to\infty$ if you do.
This is still not quite right unfortunately, I mean the degree is $\leq -m$.
|
2025-03-21T14:48:30.174872
| 2020-03-29T21:19:31 |
356070
|
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"Henri Cohen",
"Somos",
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|
Stack Exchange
|
Computing all eta quotients of given weight and level
I have written a rather naive program for finding all holomorphic eta quotients of
given weight and level (and varying character). When the level has few divisors it is
very fast, but incredibly slow otherwise. For instance, if I am not mistaken there are $1224$ holomorphic eta quotients in $M_{3/2}(\Gamma_1(60))$, i.e., weight $3/2$, level (dividing)
$60$ and character $\chi_D$ with $D=1$, $5$, $12$, or $60$. My program requires almost $40$
minutes. Surely one can do better than that ? I would appreciate an algorithm, if not
readable code is OK, thank you.
Edit: in fact, in this special case the 4 characters can be treated together, so the time divided by 4, but this is special and still very slow, so the question remains.
Can you please share your code? I too am interested in eta quotients.
@Somos Sure, it is in Pari/GP, where do you want me to send it ? (my professional e-mail is public, you can reply there).
|
2025-03-21T14:48:30.174967
| 2020-03-29T21:49:25 |
356072
|
{
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"Ben McKay",
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"url": "https://mathoverflow.net/questions/356072"
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|
Stack Exchange
|
When is a regular surface in $\mathbb{R^4}$ contained in a $3$-dimensional vector subspace of $\mathbb{R^4}$?
Introduction: Let $S$ be a regular connected surface embedded in $\mathbb{R^3}$. We know that $S$ is contained in a plane (a $2$-dimensional vector subspace of $\mathbb{R^3}$) iff every point of $S$ is umbilical and the gaussian curvature is constantly zero.
Question: Let $S$ be a regular connected surface embedded in $\mathbb{R^4}$. Do exist some conditions on the first foundamental form or other "intrinsic" properties of the surface that are equivalent to the statement "$S$ is contained in a $3$-dimensional vector subspace of $\mathbb{R^4}$"?
Observation1: $S$ is contained in a $3$-dimensional vector subspace of $\mathbb{R^4}$ iff every tangent plane of $S$ is orthogonal to a fixed non zero vector $a$.
Observation2:
Let $\phi : U \rightarrow S$ be a local chart. $\phi(U)$ is contained in a $3$-dimensional vector subspace of $\mathbb{R^4}$ iff $\langle\phi, a \rangle=0 $, for a fixed non zero vector $a$.
Additional question: Suppose $S$ as in the first question. Is the following implication true?: " if $S$ is not contained in any $3$-dimensional vector subspace of $\mathbb{R^4}$ then there exists a poin $p \in S$ such that for every neighborhood $W$ of $p$, $S\cap W$ is not contained in any $3$-dimensional vector subspace". I'm asking if the property is local or global.
Clearly the shape operator has to have one dimensional image. I suspect that is necessary and sufficient for a connected surface.
|
2025-03-21T14:48:30.175081
| 2020-03-29T22:31:03 |
356074
|
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"L.F. Cavenaghi",
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|
Stack Exchange
|
Sufficient conditions for constant solutions
Let $u : \mathbb{R}^N \to \mathbb{R}$ be a smooth negative function that satisfies
$$-f(x)u(x) + B(x) = \Delta u(x),~\forall x\in \mathbb{R}^N,$$
where $B(x)$ is a smooth positive function and $f$ is a smooth function with at least one zero. In particular, one can decompose $\mathbb{R}^N = \Omega\cup \Omega'$, where $f\Big|_{\Omega} > 0$. One easily sees that $\Delta u(x) > 0$ for all $x\in \Omega,$ i.e, $u$ is subharmonic on $\Omega$. In my particular, it also holds that $x'\in \partial\Omega$,
$\Delta u(x') > 0.$
Is there any sufficient condition on $f$ in order to conclude that $u$ is constant on a nonempty subset of $\mathbb{R}^N?$ What if we instead of $\mathbb{R}^N$ one considers a smooth closed manifold?
EDIT: Some additional info:
Since,
$$ 0<B(x) = \Delta u(x) + f(x)u(x),$$
by the Bochner's formula,
$$\dfrac{1}{2}\Delta|\nabla u|^2 = \langle \nabla B(x),\nabla u\rangle -\langle \nabla (fu),\nabla u\rangle + |\nabla^2 u|^2 + \mathrm{Ric}(\nabla u).$$
If one assumes that instead of $\mathbb{R}^N$ we are on a closed Riemannian manifold with positive sectional curvature, $u$ has a critical point $x^*$ (in fact, it has a maximum and a minimum). Furthermore, on this point,
$$\dfrac{1}{2}\Delta |\nabla u|^2(x^*) = |\nabla^2u| + \mathrm{Ric}(\nabla u).$$
Hence, there is an open neighborhood $U\ni x^*$ such that
$$\Delta |\nabla u|^2(x) > 0 ~\forall x\in U,$$
in particular, $|\nabla u|^2$ is subharmonic on this neighborhood.
But since
$-fu + B = \Delta u$, if $x^*$ is a maximum point, then by possibly reducing $U$, $$\forall x\in U, \Delta u(x) < 0,$$ and hence, $$-fu(x) + B(x)< 0\forall x\in U,$$ in particular, since $u$ is negative, $$f(x) < 0~\forall x \in U \Rightarrow U\subset \Omega'. $$
Since $\nabla u (x^*) = 0$ if one shows that $|\nabla u|$ is constant on $U$, i.e, $|\nabla u|^2$ does not attaches its maximum on $\partial U$, then $|\nabla u|(x) = 0 ~\forall x\in U,$ in particular, $u$ will be constant on $U$. But note that since $\Delta |\nabla u|^2 > 0~\forall x\in U,$ $|\nabla u|^2$ cannot achieves its maximum on the interior of $U$. On the other hand, $\Delta u <0$ in $U$, hence, $u$ is superharmonic in $U$, therefore, $-u$ is subharmonic on $U$.
I don't know how to proceed.
A semi-trivial remark: if $\Omega$ is bounded and $N \geqslant 3$, then $u = -(-\Delta - f(x))^{-1} B(x)$ always exists (the inverse of $-\Delta - f(x)$ has non-negative kernel), and apparently need not be constant anywhere.
@MateuszKwaśnicki, in fact, on $\Omega$ $u$ is never constant since $B(x) >0$, what about on $\Omega^c?$
I might have misunderstood something: you can choose $u$ and $B$ first, and then define $f$ using your equation, so $u$ can well happen to be constant on some set with non-empty interior. For example, take $u = -1 - \phi$ for your favourite bump function $\phi$, and $B = 1$. Then $f = (B - \Delta u) / u$ is obviously well-defined and smooth, and by choosing $\phi$ appropriately, we can require $f$ to change sign.
@MateuszKwaśnicki, thank your for your answer. I do appreciate it, it seems nice. I will think about it, as I see, I cannot tell, would be a problem if $B$ is not constant?
Regarding non-constant $B$: this does not seem to be a problem, an arbitrary function $B$ would work as well. But this is not an answer; just a hint that any reasonable condition on $f$ (or perhaps jointly on $f$ and $B$) is likely to be complicated.
|
2025-03-21T14:48:30.175413
| 2020-03-29T22:55:51 |
356076
|
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|
Stack Exchange
|
augmentation ideal is always finitely generated?
$G$ is a finitely presented group (but not a finite group), and $\mathbb{Z}G$ is the corresponding group ring.
$I$ is the kernel of the augmentation morphism $\mathbb{Z}G\rightarrow \mathbb{Z}$.
Is $I$ (always) a finitely generated $\mathbb{Z}G$-module (let say right module).
It is finitely generated when G is. If S generates G then the s-1 with s in S generate the augmentation ideal.
This works more generally for monoids.
Conversely if the augmentation ideal of a group ring is finitely generated so is the group. Some finite set of elements of the form g-1 must generate the augmentation ideal and by looking at the Cayley graph with respect to that finite set you can see it is connected and hence these elements generate.
@MarkSapir they generate as a left or right ideal. It is easier to see as a left ideal. Then the left ideal is the image of the boundary map from 1-chains to 0-chains for the right Cayley graph
Or algebraically ab-1=a(b-1)+ a-1 shows how to inductively build g-1 from an expression of g as a product of generators
Said differently the augmentation ideal is finitely generated iff the group is FP1 which is the same as finitely generated.
The augmentation ideal is finitely generated as a left (or right) ideal if and only if the group is finitely generated. It is obvious the augmentation ideal is generated as an abelian group by all elements of the form $g-1$ with $g\in G$. Then from the computation $ab-1=a(b-1)+a-1$ one easily deduces by induction on word length that if $S$ generates $G$, then the elements of the form $s-1$ with $s\in S$ generates the augmentation ideal as a left module.
For the converse, note that if the augmentation ideal is finitely generated as a left ideal, then there must be some finite subset $S$ of $G$ such that the elements $s-1$ with $s\in S$ generate the augmentation ideal (just have $S$ be the elements you need to write your finite generating set in terms of the $g-1$ generating set). Now if one takes the right Cayley graph of $G$ with respect to the set $S$, then the augmentation ideal is precisely the image of the boundary map from $1$-chains to $0$-chains and so the Cayley graph has vanishing reduced homology in dimension $0$ and hence is connected. Thus $S$ generates the group.
@MarkSapir, the ideal is generated as an abelian group by the g-1 so my argument gets the entire ideal
|
2025-03-21T14:48:30.175599
| 2020-03-29T22:56:11 |
356077
|
{
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"Chris Wuthrich",
"Guest",
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"Noam D. Elkies",
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"sort": "votes",
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}
|
Stack Exchange
|
Finding $Q(\sqrt{-2})$-rational points on $X_0(33)$
Let $K = Q(\sqrt{-2})$. How can I compute the $K$-rational points on the modular curve $X_0(33)$?
Recall that $X_0(33)$ is of genus $3$ and has the following affine model,
$$y^2 +(-x^4-x^2-1)y = 2x^6-2x^5+11x^4-10x^3+20x^2-11x+8.$$
My attempt at finding $K$-rational points on $X_0(33)$ is as follows: First I find a rational map $f$ from $X_0(33)$ to a quotient curve $E$ of $X_0(33)$ with $E$ an elliptic curve. Second, I determine the preimages of $E(K)$ under $f$. If $E$ is of rank $0$, $E(K)$ is finite. Then I can use a Grobner basis to determine $f^{-1}(x)$ for every $x \in E(K)$. However in my case $E(K)$ is of rank $1$ and as a result it is computationally infeasible to determine a Grobner basis for every $f^{-1}(x)$ with $x \in E(K)$. I am wondering if there is a work-around this issue?
Any help in finding $K$-rational points on $X_0(33)$ would be appreciated.
EDIT: Removed a question after a clarification by Christian Wuthrich.
Not in the same finite set, the points in the preimage of a rational point $x\in E(K)$ will be defined over a larger field than $K$ for all but finitely many $x$.
Thanks, that's a good point. I edited the question accordingly.
P. Bruin and F. Najman have determined the exceptional quadratic points on $X_0(33)$.
See Table 8 of https://arxiv.org/pdf/1406.0655.pdf
This classifies only exceptional points, or am I missing something?
Yes you are right that they only classify the exceptional points. I will edit the answer accordingly. I will just note that the elliptic curve you found is $X_0(33)^+$ (the quotient of $X_0(33)$ under the Atkin--Lehner involution $\omega_{33}$). This can be compute in Magma via ModularCurveQuotient(33,[33]). This interpretation may help you determine the remaining $\mathbb{Q}(\sqrt{-2})$ points.
It appears that in this case we're lucky: non-exceptional points would have
$x$ rational and $-2y_1^2 = P(x)$ where $y_1 = y - \frac12(x^4+x^2+1)$ and
$P(y) = x^8 + 10x^6 - 8x^5 + 47x^4 - 40x^3 + 82x^2 - 44x + 33$
is the discriminant of the quadratic in $y$; but $P(x)$ happens to have
no real roots, and is thus positive for all $x$, so cannot equal $-2y_1^2$
for any rational $y_1$.
The name of the linked paper: Bruin and Najman - Hyperelliptic modular curves $X_0(n)$ and isogenies of elliptic curves over quadratic fields.
|
2025-03-21T14:48:30.175786
| 2020-03-30T00:09:51 |
356080
|
{
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"authors": [
"Noah Riggenbach",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/356080"
}
|
Stack Exchange
|
Describing the THH of function spectra?
Are there any results describing the $THH$ of spectra of the form $F(X, E)$ where $X$ is a space (say, finite CW) and $E$ is a (nice enough) ring spectrum?
I'm happy to put various (further, or other) restrictions on $X$ and $E$. I'm also happy to consider specific examples for $E$ (e.g. $H\mathbb{Z}$, $KU$, $MU$, $\mathbb{S}$ etc).
I'm hoping for something similar-ish to the result that $HH_*S^*X$, the Hochschild homology of the singular cochains on $X$, is equivalent to $H^*LX$, the cohomology of loop space. That already has a $THH$ analog in the Blumberg-Cohen-Schlichtkrull theorem, which describes the $THH$ of certain Thom spectra in terms of Thom spectra over loop space.
I would also be interested in "cyclic" results, i.e. some kind of topological analogs of the fact that the cyclic homology of $S^*X$ is the $S^1$-equivariant cohomology of $LX$.
If $X$ is finite CW, then it is dualizable in spectra, so that $F(X, E)\cong E\tensor DX$, and then $THH(F(X,E))\cong THH(E)\tensor B^{cyc}(DX)$, where $D(X)\tensor D(X)\to D(X)$ is the dual of the diagonal. I don't know what is known about the $B^{cyc}(DX)$, but it seems approachable at least.
This might also give a connection to the Blumberg-Cohen-Schlichtkrull result, since Atiyah duality gives that $\Sigma_+^\infty M$ and $Th(NM)$ are dual.
|
2025-03-21T14:48:30.175894
| 2020-03-30T00:57:23 |
356082
|
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"url": "https://mathoverflow.net/questions/356082"
}
|
Stack Exchange
|
Integral average near a point of dispersion
Let $\Omega\subset\subset\mathbb R^{n}$ be a bounded domain and let $E\subset \Omega$ be a Lebesgue measurable set. Let $f\in L^{1}(\Omega)$ and let $x\in \Omega$ be a point of dispersion of $E$, that is
$$\lim_{r\to0^+}\frac{\lambda^{n}(E\cap B_{r}(x))}{\lambda^{n}(B_r(x))}=0,$$
where $\lambda ^{n}$ is the Lebesgue measure and $B_{r}(x)$ is the Euclidean ball of center $x$ and radius $r$.
Is it true that
$$\limsup_{r\to0^+}\frac{\int_{E\cap B_{r}(x)}|f|d\lambda^n}{\lambda^{n}(B_r(x))}<\infty?$$
If not, then what reasonable assumptions on $f$ (Higher integrability? Sobolev regularity?) would guarantee this? (The boundedness clearly implies that the limit actually exists and is $0$.)
$\newcommand{\tb}{\tilde B}$
Let $d:=n$.
The dispersion condition
\begin{equation*}
\lim_{r\downarrow0}\frac{|E\cap B_r(x)|}{|B_r(x)|}=0
\end{equation*}
is of no help, where $|\cdot|$ denotes the Lebesgue measure on $\mathbb R^d$.
More specifically, the following is true:
Theorem Suppose that $f$ is a nonnegative function in $L^1(B_1)$ such that
\begin{equation*}
\limsup_{r\downarrow0}\frac{\int_{B_r}f}{|B_r|}=\infty, \tag{0}
\end{equation*}
where $B_r:=B_r(0)$, the open ball of radius $r$ centered at $0$. Then one can construct a measurable set $E\subset B_1$ such that
\begin{equation*}
\lim_{r\downarrow0}\frac{|E\cap B_r|}{|B_r|}=0 \tag{1}
\end{equation*}
but
\begin{equation*}
\limsup_{r\downarrow0}\frac{\int_{E\cap B_r}f}{|B_r|}=\infty. \tag{2}
\end{equation*}
So, without any conditions on $E$ in addition to the dispersion condition (1), the best sufficient condition for
\begin{equation*}
\limsup_{r\downarrow0}\frac{\int_{E\cap B_r}f}{|B_r|}<\infty \tag{not-2}
\end{equation*}
is the trivial sufficient condition
\begin{equation*}
\limsup_{r\downarrow0}\frac{\int_{B_r}f}{|B_r|}<\infty. \tag{not-0}
\end{equation*}
To simplify the presentation of the proof of this theorem a bit, assume that $d=2$. By (0), there is a sequence $(r_n)$ decreasing to $0$ such that
\begin{equation*}
\int_{B_{r_n}}f\ge 2^n|B_{r_n}|
\end{equation*}
for all natural $n$. So, passing successively to subsequences, we can construct an increasing sequence $(n_k)$ of natural numbers and a sequence $(S_k)$ of sets such that
\begin{equation*}
n_k\ge2k,
\end{equation*}
\begin{equation*}
\int_{S_k}f\ge 2^{-k}2^{n_k}|\tb_k|\ge2\int_{\tb_{k+1}}f
\end{equation*}
with
\begin{equation*}
\tb_k:=B_{r_{n_k}},
\end{equation*}
and, for each natural $k$, $S_k$ is a sector of the disk $\tb_k$ with the central angle $2\pi/2^k$ such that $S_k\supset S_{k+1}$.
Let now
\begin{equation*}
E:=\bigcup_k(S_k\cap(\tb_k\setminus\tb_{k+1}))
=\bigcup_k(S_k\setminus\tb_{k+1}).
\end{equation*}
Then for any natural $k$ the condition $r_{n_{k+1}}\le r\le r_{n_k}$ implies
$E\cap B_r\subseteq S_k\cap B_r$, so that $|E\cap B_r|\le|S_k\cap B_r|=2^{-k}|B_r|$, which shows that (1) holds.
On the other hand,
$$E\cap\tb_k\supseteq E\cap(\tb_k\setminus\tb_{k+1})=S_k\cap(\tb_k\setminus\tb_{k+1})=S_k\setminus\tb_{k+1},$$ whence
\begin{multline*}
\int_{E\cap\tb_k}f
\ge\int_{S_k\setminus\tb_{k+1}}f
\ge\int_{S_k}f-\int_{\tb_{k+1}}f \\
\ge\frac12\int_{S_k}f
\ge2^{-k-1}2^{n_k}|\tb_k|\ge2^{k-1}|\tb_k|.
\end{multline*}
So, (2) also holds. $\Box$
|
2025-03-21T14:48:30.176083
| 2020-03-30T02:58:17 |
356086
|
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|
Stack Exchange
|
Any simplification of this inequality if it is true? :For $t\geq 1.22$: $|\zeta(0.5+it)|\leq 0.5 \frac{|\Gamma(0.5+it)|}{|\Gamma(-0.5+it)|}$
When I tried to give bounds for $\zeta(0.5+it)$ using some transformations over Gamma function using the function $f(x)=\exp(-n x)$ over the range $(0,+\infty)$ , For $ Re(s)=\frac12 $ and $t >0$ I come up to the final Bounds for $\zeta(0.5+it) $ which is represented by the following formual :For $t\geq 1.22$: $$|\zeta(0.5+it)|\leq 0.5 \frac{|\Gamma(0.5+it)|}{|\Gamma(-0.5+it)|}\tag{1}$$, For Bounds of $\Gamma(s)$ it is found that is monotonic increasing function for $|t|\geq 5/4$ with the respect to the the real part of $s$ and it were false with $|t|\leq 1$ in this paper entitled On the Horizontal Monotonicity of $|\Gamma(s)|$ by Gopala Krishna Srinivasan and P. Zvengrowski, |$\Gamma(s)$| is given in the introduction of that paper for $s=\sigma+ i t$ by this formula :
$|\Gamma(\sigma+ i t)|=\lambda \frac{\Gamma(1+\sigma)}{\sqrt{\sigma^2+t^2}}\sqrt{\frac{2\pi t}{\exp(\pi t)-\exp(-\pi t)}},\lambda
\in(1,1+\sqrt{1+t^2})\tag{2}$, it seem the Right hand side of that formual related to cos hyperbolic function , Now When I tried to plug this formual in the RHS of my bounds it give me a complicated form such that no simple formula for simplification , My question here How I can simplify RHS OF $1$ if it is true ?
Note: The motivation of this question is to look for some connections of primes distribution to Gaussian distribution.
It should be noted that in (2) the $\lambda$ on the RHS is not a constant, but rather depends on $t$.
@Wojowu , Thanks for your attention, you are right ,$\lambda
\in(1,\sqrt{1+t^2})$
The RHS of (1) is $\sim |t|$ by Stirling's formula. The Weyl bound states that there exists a constant $c>0$ (which one can compute, but I won't) such that if $t\in\mathbb{R}$, then $|\zeta(\frac{1}{2}+it)|\leq c(|t|+1)^{1/6}$. One can do better under assuming the Riemann hypothesis, see Chandee and Soundararajan.
You don't need Stirling's formula. For $z=-0.5+it$ the RHS is $0.5\frac{|\Gamma(z+1)|}{|\Gamma(z)|}=0.5|z|$.
@Wojowu Silly of me to not have spent an additional minute to think of that. Good catch.
|
2025-03-21T14:48:30.176265
| 2020-03-30T04:22:51 |
356088
|
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|
Stack Exchange
|
BV analogue with well-factorable function. (Primes in arithmetic progression)
Is this special case known?
For $\lambda(q)$ -- well-factorable function and $q|P(z)$, $\pi(x;q,a)$ $a=1$.
$\displaystyle \sum_{q\leq x^{1-\epsilon}} \lambda(q) ( \pi (x;q,1)-\frac{\pi(x)}{\varphi (q)} )\leq \frac{Cx}{\log^A(x) }$
You should edit your old question https://mathoverflow.net/questions/356043/primes-in-arithmetic-progression and ask for it to be reopened, rather than starting a new post
I do not know how.
When you go to your old question, do you see any options for editing? Perhaps you may need to register as a user first, a move which I would strongly recommend since it prevents the proliferation of multiple user accounts and duplicate questions
I did not register. I remember something similar to my question in some article(for $a=1$). But I don’t remember where and how much.
Does this answer your question? Primes in arithmetic progression
What is a well-factorable function? What is $P(z)$?
I have found the partial answer. If you remove this $\lambda(q)$ function and not take modulo, then you can take $1-\epsilon$. See: Fouvry Sur le problème des diviseurs de Titchmarsh. 1985 But with this function i do not how to get such result.
Both links are broken.
This type was proved by Bombieri, Friedlander and Iwaniec in the paper "Primes in arithmetic progressions to large moduli.
|
2025-03-21T14:48:30.176408
| 2020-03-30T06:04:33 |
356090
|
{
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"authors": [
"Fedor Petrov",
"GH from MO",
"Gerry Myerson",
"Greg Martin",
"LMZ",
"Noam D. Elkies",
"Steven Stadnicki",
"Sylvain JULIEN",
"https://mathoverflow.net/users/105551",
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|
Stack Exchange
|
Estimate for $\sum_{a=1}^{p-1}\sum_{b=1}^{p-1}\frac{b}{a(ab)_p}$, where $p$ is a large prime
Is this estimate true? Can anyone give a proof of it?
$$
\sum_{a=1}^{p-1}\sum_{b=1}^{p-1}\frac{b}{a(ab)_p}=\frac{1}{2}p\ln^2 p+o(p\ln^2 p)\qquad (p\text{ prime, } p\to\infty)
$$
where $
(ab)_p\equiv ab\;(\operatorname{mod}p)$, $0<(ab)_p<p$.
For what range of $p$ have you calculated it?
p is a large enough prime number@ Gerry Myerson
Would it be anyhow useful to rewrite the LHS as a sum over $n=ab$?
This would give something like $\displaystyle{\sum_{n=1}^{(p-1)^2}\sum_{d\mid n,d\leq\sqrt{n}}\dfrac{d^2}{n(n)_{p}}}$.
Which should be $\leq(1+o(1)\dfrac{1}{p-1}\sum_{n=1}^{(p-1)^2}\tau(n)$.
What you write in note as "we have" is exactly what you ask about. So, I am confused: do you have it or not?
Where does such a sum arise?
Tricky sum. It is known that the residues $(ab)_p$ are equidistributed in ${1,\dotsc,p-1}$ when the pair $(a,b)$ ranges in a discrete rectangle $A\times B$ of size at least $p^{3/2+\epsilon}$. See, for example, http://www.nieuwarchief.nl/serie5/pdf/naw5-2000-01-4-380.pdf However, it seems nontrivial to deduce from here the OP's conjecture. I also rewrote the OP's sum in terms of Kloosterman sums, and the main term exhibited the expected asymptotics, but I failed to show that the error term is negligible. Anyways, these are the natural ideas to explore first (in my opinion).
It is plausible that the equidistribution statement mentioned in my previous remark holds for $|A\times B|>p^{1+\epsilon}$. If this is the case, then the OP's conjecture can be deduced in a straightforward way, by localizing $a$ in short intervals and $b$ in long intervals.
@GH from MO: would the result you conjecture follow from (G)RH?
@SylvainJULIEN: I don't know. I think my conjecture is folklore, but its connection to GRH is unclear. My conjecture has more to do with the distribution of Kloosterman sums.
Not an answer but a simple heuristic argument: if you set $r=(ab)_p$, the OP's
sum is equal to
$$\sum_{1\le r\le p-1}\dfrac{1}{r}\sum_{1\le a\le p-1}\dfrac{1}{a}(ra^{-1})_p$$
This proves immediately that the sum is less than $(p-1)H_{p-1}^2$, asymptotically $p\log(p)^2$, and if we assume (heuristic part) that $(ra^{-1})_p$ has average
$(p-1)/2$ we indeed obtain a guess of $p\log(p)^2/2$. Maybe this last part can be made rigorous.
EDIT: if you consider the much simpler SINGLE sum $S(p)=\sum_{1\le a\le p-1}\dfrac{(a^{-1})_p}{a}$, the same heuristic would give an asymptotic of $p\log(p)/2$. However, numerically $S(p)/(p\log(p))$ does NOT seem to tend to a limit, but oscillates between something like $0.38$ and $0.52$. This should be much easier to analyze, and perhaps indicate that there is also some oscillation in the OP's original question, with no limit.
In the limit of large $p$, could we approximate the OP's double sum by a double integral on some square to support your heuristic argument?
The single sum over $a$ is harder to analyze (in my opinion) than the double sum over $a$ and $r$. In general, extra averaging helps not only the outcome, but also the analysis.
To my eye, the numerical data for the single sum is quite consistent with the theory that $S(p) = \frac12p\log p + X(p) p$, where the $X(p)$ defined by that equation looks like a fixed bump-shaped random distribution.
Not an answer, but an argument that your sum is between $(1/4+o(1))p\log^2(p)$ and $(3/4+o(1))p\log^2(p)$.
Write $S(p)$ for your sum. Separate the sum into pieces according to the integer part of $(ab/p)$:
$$
\begin{align*}
S(p)&=\sum_{a=1}^{p-1}\frac{1}{a}\sum_{k=0}^{a-1} \sum_{kp/a<b<(k+1)p/a} \frac{b}{ab-kp}\\
&\sim\sum_{a=1}^{p-1}\frac{1}{a}\sum_{k=0}^{a-1} \frac{kp}{a}\sum_{kp/a<b<(k+1)p/a} \frac{1}{ab-kp}.\\
\end{align*}
$$
The innermost sum is the sum of reciprocals of integers in an arithmetic progression. The first term in the progression is $(-pk)_a$, and the sum of the reciprocals of the other terms in the progression is $\log(p/a)/a+O(1/a)$, so
$$
\begin{align*}
S(p)&\sim\sum_{a=1}^{p-1}\frac{1}{a}\sum_{k=0}^{a-1} \frac{kp}{a}\left[\frac{1}{(-pk)_a}+\frac{1}{a}\log\left(\frac{p}{a}\right)\right]\\
&=p\sum_{a=1}^{p-1}\frac{1}{a^3}\log(p/a)\sum_{k=0}^{p-1}k+p\sum_{a=1}^{p-1}\frac{1}{a^2}\sum_{k=1}^{a-1}\frac{k}{(-pk)_a}\\
&=\frac{1}{4}p\log^2(p)+p\sum_{a=1}^{p-1}\frac{1}{a^2}\sum_{k=1}^{a-1}\frac{k}{(-pk)_a}.
\end{align*}
$$
This gives the claimed lower bound for $S(p)$. For the upper bound, we observe that for fixed $a$, we have $\{(-pk)_a:1\leq k\leq a-1\}=
\{1,\ldots,a-1\}$. So an upper bound for the second term above is
$$
p\sum_{a=1}^{p-1}\frac{1}{a^2}\sum_{k=1}^{a-1}\frac{k}{a-k}\sim \frac{1}{2}p\log^2(p).
$$
Not really an answer, but here is the plot for the first 200 primes:
Maple code:
with(plots):
f := proc(p)
option remember;
return evalf(add(add(b/a/modp(a*b,p),b=1..p-1),a=1..p-1)/p/ln(p)^2);
end:
listplot([seq(f(ithprime(k)),k=10..200)],style=point);
(Obviously this is completely unintelligent, and much more efficient methods are doubtless possible.)
Just to be clear, this isn't the sum, it's the sum divided by $p(\log p)^2$.
I feel obliged to flesh out my comments (and to modify my wrong answer, thanks to GH from MO).
Write $n=ab$, and let $H_{k}$ the $k$-th harmonic number, $\tau(n)$ the number of positive divisors of $n$. LHS is greater than the related sum $\sum_{a=1}^{p-1}\sum_{b=1}^{a}\dfrac{b}{a(ab)_{p}}$, that we'll denote by $S$.
Provided the limit as $\tau(n)$ tends to $\infty$ of $f(n):=\frac{2}{\tau(n)}\sum_{d\mid n,d\leqslant\sqrt{n}}\dfrac{d^{2}}{n}$ exists and equals a positive constant $M$, we have:
$S=\sum_{n=1}^{(p-1)^2}\sum_{d\mid n,d\leqslant\sqrt{n}}\dfrac{d^2}{n(n)_{p}}\sim\sum_{n=1}^{(p-1)^2}\sum_{d\mid n,d\leqslant\sqrt{n}}\dfrac{M}{(n)_{p}}\sim\sum_{n=1}^{(p-1)^2}\sum_{d\mid n,d\leqslant\sqrt{n}}M\dfrac{H_{p-1}}{p-1}\sim M\sum_{n=1}^{(p-1)^2}\sum_{d\mid n,d\leqslant\sqrt{n}}\dfrac{\log(p-1)}{p-1}$.
The idea is to consider all divisors of $n$ less than its square root, and to replace $\dfrac{1}{n_p}$ by its average value, which is $\dfrac{H_{p-1}}{p-1}\sim\dfrac{\log(p-1)}{p-1}$ (in my comment I erroneously took the reciprocal of the average value and not the average value of the reciprocals, hence the missing $\log(p-1)$).
We thus obtain $S\sim M\dfrac{\log(p-1)}{p-1}\sum_{n=1}^{(p-1)^2}\dfrac{\tau(n)}{2}$. As $D(x):=\sum_{n=1}^{x}\tau(n)$ is provably asymptotic to $x(\log x+K)$ where $K$ is a positive constant (see Dirichlet divisor problem on Wikipedia), we end up with:
$S\sim\frac{M}{2}\frac{\log(p-1)}{p-1}(p-1)^2(\log (p-1)^2+K)\sim Mp\log^{2} p$ which provides a lower bound for the sum of the OP of the desired order of magnitude.
Edit: the following link: http://www.les-mathematiques.net/phorum/read.php?5,1967230,1967504#msg-1967504 shows that if $M$ exists, it equals $\frac{1}{2}$.
Can you get a lower bound this way, too?
Your first display only covers the pairs $(a,b)$ satisfying $b\leq a$ (since your $d^2/n$ equals $b/a$).
So a multiplicative factor equal $2$ is needed?
No, a different argument is needed. Your first display talks about a very different sum than the original post.
It would be the same if the sum over $b$ went from $1$ to $a$, right?
Yes, obviously. Anyways, I don't want to chat about this, but you need to be careful.
|
2025-03-21T14:48:30.176863
| 2020-03-30T07:11:25 |
356092
|
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"sort": "votes",
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|
Stack Exchange
|
Small values of $L(\frac{1}{2}, \chi_{-8d})$
In Section 3 of https://arxiv.org/pdf/0708.3990.pdf Soundararajan shows that there exists infinitely many fundamental discriminants $8d$ for which $L(1/2, \chi_{8d})$ is very small.
His argument runs as follows : Define,
$$
M_1(R; X) := \sum_{X/16 \leq d \leq X/8} \mu(2d)^2 R(8d)^2
$$
and
$$
M_2(R; X) := \sum_{X/16 \leq d \leq X/8} \mu(2d)^2 L(\tfrac 12, \chi_{8d}) R(8d)^2
$$
where $R$ are some real-valued weights. On the beginning of p. 11, Soundararajan shows that a suitable choice of $R$ leads to the bound,
$$
\frac{M_2(R; X)}{M_1(R; X)} \ll \exp \Big ( - \Big ( \frac{1}{\sqrt{5}} + o(1) \Big ) \frac{\sqrt{\log X}}{\log\log X} \Big )
$$
and concludes that this implies
$$
L(\tfrac 12, \chi_{8d}) \ll \exp \Big ( - \Big ( \frac{1}{\sqrt{5}} + o(1) \Big ) \frac{\sqrt{\log X}}{\log\log X} \Big ).
$$
I do not understand this implication : If one assumes that $L(\tfrac 12, \chi_{8d}) \geq 0$ then it is certainly correct, but this is not known (although expected) to be true. In particular if $L(\tfrac 12, \chi_{8d})$ exhibits sign cancellations then I don't see how the previous displayed equation leads to the conclusion about small values of $L(1/2, \chi_{8d})$.
This can be corrected by working with $L(\tfrac 12, \chi_{8d})^2$ instead of $L(\tfrac 12, \chi_{8d})$ but leads to a worse numerical value than $\tfrac{1}{\sqrt{5}}$.
I would like to know if this is an oversight in Soundararajan's paper or my misunderstanding.
|
2025-03-21T14:48:30.176971
| 2020-03-30T07:29:47 |
356093
|
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"François Jurain",
"Peter",
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|
Stack Exchange
|
Expected number of bridges in a random subgraph
I am researching connectivity in random subgraphs and have come across the following problem.
A bridge between two vertices $a$ and $b$ of a graph $G$ is an edge $e$ such that removing $e$ from $G$ disconnects $a$ and $b$. Note that $e$ is not a bridge if $a$ and $b$ are not connected in $G$. Denote by $B_G(a, b)$ the number of bridges between $a$ and $b$.
For any graph $G$ on $n$ vertices (possibly with multiple edges) let $G(1/2)$ denote the random subgraph of $G$ obtained by sampling each edge independently with probability $1/2$.
My question is then: Is there any non-trivial bound on $M=\mathbb E[B_{G(1/2)}(a, b)]$ in terms of $n$, which holds for all graphs $G$?
Trivially, $M\leq n-1$ as there can be at most $n-1$ bridges between $a$ and $b$ in any graph on $n$ vertices. However, I hope that either $M\leq n^{1/2}$ or perhaps even $M\leq n^{1/3}$. I have an example where $M=\Theta(\log n)$, which occurs if $a$ and $b$ are at each end of a path of length $n$ and each edge of the path is replaced by $\Theta(\log n)$ edges.
I have not been able to find any litterature addressing questions of this nature, so references would be much appreciated as well!
If your graph $G$ is some kind of giant random graph, then its random subgraph is also just some kind of giant random graph (of an other persuasion than $G$), and your question reduces to the particulars of the connection of $a$ and $b$ to its giant component, if any.
I am currently at grips with a similar question: a parametric family of subgraphs of the complete graph on $n$ vertices. Unfortunately $n$ is very far from $\gg 1$
That is very true. I am, however, looking for a solution for any given graph. What is your problem?
The problem is to quickly find a small vertex separator (not edge separator) for each subgraph, so that they all break down to small connected components. The obvious way would be to solve the problem just once, on the union of the subgraphs... except it exhibits a "giant" component (80% of the vertices, say), when the individual subgraphs do not.
|
2025-03-21T14:48:30.177125
| 2020-03-30T08:09:27 |
356094
|
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"Alexandre Eremenko",
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|
Stack Exchange
|
Convergence in the Caratheodory sense and Hausdorff sense
Among Jordan domains, I understand that Caratheodory convergence is weaker than Hausdorff convergence.
But if a sequence of Jordan domains all have rectifiable boundary whose arc length are all $L$, and their Caratheodory limit is also a Jordan domain with the same boundary arc length $L$, does this necessarily imply this sequence converge in the Hausdorff metric?
To my knowledge, examples of sequences converging in the Caratheodory sense but not in the Hausdorff sense do not preserve arc length.
-----Edit-----
I realize that Caratheodory convergence is a less commonly known concept, so here I am updating with two equivalent definitions.
A sequence of pointed domains $(\Omega_n,x_n)$ is said to converge to $(\Omega,x)$ in the Caratheodory sense if
$x_n \to x$,
for all compact $K \subseteq \Omega$, we have $K \subseteq \Omega_n$ for every $n$ sufficiently large, and
for all open connected $U$ containing $x$, if $U \subseteq \Omega_n$ for infinitely many $n$, then $U \subseteq \Omega$.
The pointed domains $(\Omega_n,x_n)$ converge to $(\Omega,x)$ in the Caratheodory sense if and only if the harmonic measures $\omega(\Omega_n,x_n)$ converge weakly to the harmonic measure $\omega(\Omega,x)$.
The definition of Hausdorff metric can be found at https://en.m.wikipedia.org/wiki/Hausdorff_distance?wprov=sfla1
Maybe add links where these two are defined.
Can you explain what is "Caratheodory convergence" of Jordan domains?
The answer is yes.
The boundaries $\partial\Omega_n$ are uniformly 1-Lipschitz on $[0;L]$ under the arc-length parametrization, hence, by Arzela—Ascoli, you can extract a uniformly convergent sub-sequence, say $\partial\Omega_{n_k}\to\gamma$, where $\gamma$ also has a 1-Lipschitz parametrisation on $[0;L]$. Also, since $\Omega_{n_k}\to\Omega$ in Carathéodory sense, we must have $\partial\Omega\subset \gamma$. So, you have a rectifiable Jordan curve $\partial \Omega$ of length $L$ and also another curve $\gamma$ of length $\leq L$ that contains $\partial \Omega$. This certainly implies $\gamma=\partial\Omega$, say, by interpreting the length as the Hausdorff measure of dimension 1. So, any sub-sequence of your domains has a further sub-sequence whose boundaries uniformly converge to $\partial\Omega$, which implies the claim.
|
2025-03-21T14:48:30.177398
| 2020-03-30T08:40:57 |
356096
|
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|
Stack Exchange
|
The normal cone to a singular subvariety
Let $Y$ be a smooth variety, and let $X\subset Y$ be a subvariety. If $X$ is smooth, then $C_XY=N_XY$ is a quotient of $TY|_X$. Is the same still true if $X$ is singular? I.e. is there always a surjection
$$TY|_X\to C_XY.$$
A related question: for the normal cone I have the following vague picture in mind: it is a union of closures of normal cones to smooth strata in some "equisingular" stratification of $X$. Is there a precise sense in which this is true?
wouldn't such a surjection require the normal cone to be irreducible?
|
2025-03-21T14:48:30.177472
| 2020-03-30T08:58:25 |
356097
|
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"alvarezpaiva",
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|
Stack Exchange
|
Compatibility of the Hausdorff measure with short exact sequences in normed spaces
Let $(E,\|.\|)$ be a finite dimensional normed space and take $F\subset E$ a
subpace, so that we have the canonical short exact sequence
$0\rightarrow F\rightarrow^\iota E\rightarrow^\pi E/F\rightarrow 0$ of normed spaces ($F$
being endowed by the restriction of $\|.\|$ and $E/F$ by the quotient norm
$\|v+F\| = \inf_{f\in F} \|v-f\|$). Write $n=\dim E$ and $k = \dim F$.
We can define the Hausdorff measure on each of these spaces (say $\mu_F, \mu_E$ and $\mu_{E/F}$), deriving from
each of their norms.
Is this definition compatible with the short exact sequence,
in the sense that for a $k$-vector $\mathfrak{f} = f_1\wedge \cdots\wedge f_k\in\bigwedge^k E$, and $\mathfrak{e} = e_{k+1}\wedge\cdots\wedge e_n\in \bigwedge^{n-k} E$, we have:
$$
\mu_E(\mathfrak{f} \wedge \mathfrak{e}) = \mu_F(\mathfrak{f})\mu_{E/F}(\pi(\mathfrak{e}\wedge\mathfrak{f})).
$$
Remark that this is the case when $\|.\|$ is Euclidean, as a direct consequence of the definition of the norm on the exterior algebra.
Another way of looking at the question is whether the comeasure induced by the Hausdorff measure coincidates with the measure induced by the quotient norm.
No it is not. You can find counterexamples in dimension two already. I think the max norm on the plane and almost any $1$-dimensional subspace will be a counterexample. I'll come back later when I have more time.
And what about with a proper renormalization of the measure, let say by a factor A_k depending only the rank k ?
No that won't work either. Volumes in normed spaces are tricky things. See my paper with Thompson on A Sampler of Riemann Finsler Geometry: http://library.msri.org/books/Book50/contents.html
|
2025-03-21T14:48:30.177600
| 2020-03-30T10:18:22 |
356098
|
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|
Stack Exchange
|
Nimber $2^{2^k} - 1$ is a multiplicative generator of $[2^{2^k}]$?
Let $t = 2^{2^k}$, and consider the field $[t]$ of nimbers below $t$. For $k \leq 6$ one can check that $t - 1$ (in the usual arithmetic sense) is a multiplicative generator of $[t] \backslash \{0\}$. Is this true in general?
Nice observation! I've successfully tested it for $k\leq 12$. On the other hand, this may still be just by a chance since roughly half of the elements are generators. Testing for larger $k$ cannot be accomplished since that involves Fermat numbers with incompletely known factorizations.
|
2025-03-21T14:48:30.177671
| 2020-03-30T10:45:32 |
356099
|
{
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"Chris Wuthrich",
"Joey van Langen",
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|
Stack Exchange
|
isogenies between elliptic curves with multiplicative reduction
Let $ K $ be a $ p $-adic field. Suppose we have an isogeny of elliptic curves $ \phi : E \to E' $ defined over $ K $, where $ E $ and $ E' $ both have multiplicative reduction.
1) Is there anything we can say about the structure of the induced map on the Tate modules $ V_l(E) \to V_l(E') $? Mostly I'm interested in the eigenvalues.
I should note that for question 1) it is actually easy to determine the determinant of this linear map using the Weil pairing. I'm only interested in additional results, such as a way to compute the trace.
Note that by possibly enlarging the field K we may assume the multiplicative reduction for both curves is split, hence $ E $ and $ E' $ are isomorphic to Tate curves $ E_q $ and $ E_{q'} $ with $ q, q' \in K^{*} $ of positive valuation. Note that for any finite extension $ L / K $ we have that $ E_q( L ) \cong L^{*} / q^{\mathbb{Z}} $. For a suitably large choice of $ L $ the $ m $-torsion points of the latter have a basis $ \{ \zeta_m , q^{1/m} \} $, where $ \zeta_m $ is an $ m $-th root of unity.
2) Can we say anything about the structure of the induced map $ \phi : E[m] \to E'[m] $ with respect to the bases given by $ \{ \zeta_m, q^{1/m} \} $ and $ \{ \zeta_m, (q')^{1/m} \} $?
Although I'm interested in general results, I would be happy getting a result in case E' is a Galois conjugate of E, so in case q' is a Galois conjugate of q.
I believe this is the answer in the split case: Let $E$ be the Tate curve with parameter $q$. Let $n>1$. We look for isogenies with cyclic kernel of order $n$. We may suppose that $n$ is prime.
First, there is the isogeny to the Tate curve $E'$ with parameter $q' = q^n$ and the map is induced from $K\to K$ sending $x$ to $x^n$. With respect to the basis of $\ell^{n}$ torsion where the first element is an $\ell^n$-th root of unity $\zeta_{\ell^n}$ and the second element is a choice of an $\ell^n$-th root of $q$ (and correpondingly of $q'$), the matrix for this isogeny on $V_{\ell}(E) \to V_{\ell}(E')$ is diagonal with entries $n$, $1$.
All other cyclic isogenies of degree $n$ leaving $E$ have the kernel generated by an $n$-th root $q'$ of $q$ (and the isogeny is only defined over $K$ if this $q'$ belongs to it). The corresponding map $K/q^{\mathbb{Z}}\to K/(q')^{\mathbb{Z}}$ is induced by the identity map. This time the matrix is also diagonal by with diagonal entries first $1$ then $n$.
You seem to use that every isogeny is uniquely determined by its kernel, but I can multiply any isogeny with a degree 1 endomorphism to get an isogeny with the same kernel, but maybe different eigenvalues. For example replace $ \phi $ by $ - \phi $.
You are absolutely right. This is up to automorphisms that fix the kernel. Luckily our curves only have $\pm 1$ as automorphisms.
|
2025-03-21T14:48:30.177861
| 2020-03-30T10:53:19 |
356100
|
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|
Stack Exchange
|
Quantum ergodicity of Eisenstein series on arithmetic quotients of hyperbolic space
Let $E(z,1/2+it)$ be the Eisenstein series furnishing the continuous spectrum of the Laplace operator $\Delta$ on $X=PSL_2(\mathbb{Z})\setminus H^2$ and $dV(z)=y^{-2} \,dx \,dy$ be the volume element of the upper half plane $H^2$. In analogy with quantum mechanics, Luo-Sarnak defined the measure $\mu_t=|E(z,1/2+it)|^2 \,dV(z)$ and showed that it fulfills
\begin{equation}
\lim_{t\to\infty} \frac{\mu_t(K_1) }{\mu_t(K_2)}=\frac{\operatorname{Vol}(K_1)}{\operatorname{Vol}(K_2)}
\end{equation}
for compact, Jordan-measurable subsets $K_1,K_2$ of $X$. In analogy with the case of compact manifolds they called it quantum ergodicity. Another piece of work by Koyama, Sarnak and Petridis showed that this is also the case for certain arithmetic 3-manifolds (e.g. $X=PSL_2(\mathcal{O}_K)\setminus H^3$, where $\mathcal{O}_K$ is the integer ring of an imaginary quadratic field $K$ of class number one).
Now I am wondering how much is known for general arithmetic quotients of n-dimensional hyperbolic space or if the knowledge about these kind of examples ends with dimension 3. I am definitely grateful for any information on the current state of the matter and further references!
I think it's widely open for arithmetic quotients $n$-dimensional hyperbolic space $\mathbb{H}^n \cong \mathrm{SO}(n,1)/\mathrm{SO}(n)$, since we don't know how to relate the integrals of automorphic forms on these groups to $L$-functions except in very specific cases (the Gan-Gross-Prasad conjecture) that only apply to QUE when $n \in {2,3}$.
Thanks for the insights, Peter!
If $K$ is a real quadratic field of degree $n$ Truelsen (see https://arxiv.org/abs/0706.4239) showed QUE for for Eisenstein series on the arithmetic quotient $\text{PSL}_2(O_K)\backslash (H^2)^n$. I am not aware for a reference dealing with arbitrary number fields $K$ even if it should be possible to do this.
A true higher rank example has been worked out by L. Zhang (see https://arxiv.org/abs/1609.01386). Here the Eisenstein series on $\text{SL}_n(\mathbb{Z})\backslash \text{SL}_n(\mathbb{R})/\text{SO}(n,\mathbb{R})$ associated to a maximal parabolic subgroup are considered.
Thank you for your fast reply and the references! Since I've read Zhang's paper I am curious whether there is a paper fully proving QE for SL3(Z)∖SL3(R)/SO(3,R) as there are explicit formulae available for coefficients of the Eisenstein series and the Hecke theory has already been worked out. Concerning Truelsen's paper I had the feeling that he is more or less dealing with n copies of PSL2(Z)∖H2 rather than an arithmetic quotient of a higher-dimensional hyperbolic space.But my understanding of his paper is limited.
Concerning the higher rank groups $\mathrm{SL}_n(\mathbb{Z})\backslash \mathrm{SL}_n(\mathbb{R}) / \mathrm{SO}(n)$, QUE is only known for certain degenerate Eisenstein series because these are the only class of automorphic forms on these higher groups for which an analogue of the Watson-Ichino formula holds.
|
2025-03-21T14:48:30.178415
| 2020-03-30T11:18:48 |
356102
|
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|
Stack Exchange
|
Coercivity of linear operators
I am trying to understand the following argument: Let $\mathcal{L}:L^2(\mathbb{R})\to L^2(\mathbb{R})$ be an essentially self-adjoint unbounded linear operator with domain $D(\mathcal{L})=H^s(\mathbb{R})$ for some $s>0$. Let us assume that $\mathcal{L}$ has only one negative eigenvalue (which is simple) with associated eigenfunction $\chi$. Moreover, assume that zero is also a simple eigenvalue with associated eigenfunction $\phi_c'$ (the derivate of a fixed function $\phi_c$). Finally, assume that the rest of the spectrum is positive and away from zero.
Now, I can prove the following lemma: Under some extra hypothesis (I don't think they are relevant for my question), if a function $y\in H^{s/2}(\mathbb{R})$ satisfies $$
\langle y,\phi_c\rangle=\langle y,\phi_c'\rangle=0,
$$
then $\langle\mathcal{L}y,y\rangle>0$. Here $\langle\cdot,\cdot\rangle$ denotes the inner product in $L^2$. Now my question is, under these conditions over the spectrum of $\mathcal{L}$, does the previous lemma implies that there exists a constant $C>0$ such that for $y\in H^{s/2}$ satisfying the hypothesis of the lemma we have $\langle \mathcal{L}y,y\rangle\geq C\Vert y\Vert_{H^{s/2}} ^2$?
Edit: To give more context to my question, $\mathcal{L}$ is a self-adjoint differential operator which comes from the linearization of certain PDE around $\phi_c$, so that we have $\mathcal{L}\phi_c'=0$. Moreover, the extra hypothesis on my lemma states that if we define $d(c)=E(\phi_c)+cV(\phi_c)$, then $d''(c)>0$. Here it might be important to notice that $E'(\phi_c)+cV'(\phi_c)=0$ and that $\mathcal{L}=E''(\phi_c)+cV''(\phi_c)$. So I think the most important part is that we are assuming that $d''(c)>0$. Of course, this is extremely important to prove the lemma, but I am not sure if we can use it to prove the inequality I am looking for.
Edit2: The functionals $E$ and $V$ are defined as follows $$
V(u)=\dfrac{1}{2}\int_\mathbb{R} u^2dx \quad \hbox{and}\quad E(u)=\int_\mathbb{R}(\tfrac{1}{2}u\partial_x^2u-\tfrac{1}{2}u^2-\tfrac{1}{3}u^3)dx.
$$
Just to see if I get correctly the question. Does $\chi$ enter the game some way ? The inequality you ask is to bound the $L^2$ bilinear form associated to $\mathcal{L}$, right ?
@an_ordinary_mathematician The function $\chi$ seems to play no role on this point (to prove the property I am asking for), just the fact that it is associated to the unique negative direction of $\mathcal{L}$ . I am wondering if from the fact that under both orthogonality conditions we have $\langle \mathcal{L}y,y\rangle>0$, we can deduce the last inequality of my question.
Your edit doesn't clarify much --- nothing in the expression $E(\phi_c) + cV(\phi_c)$ has been defined ...
@NikWeaver Yes, I am very sorry for that, I am trying to avoid stating a too long question. Let me re-edit it, I hope it helps.
That does help, thank you! What about $\phi_c$?
@NikWeaver Is the function around which you are linearizing. It is positive and belongs to $\phi\in H^\infty(\mathbb{R})$. Besides that, I don't really have more information about it. Oh, sorry, I just realize, it may confused you that I have $\phi$ and $\phi_c$ on my question. Both of them are the same $\phi=\phi_c$.
Yes, the inequality is true. Let $\lambda$ be large enough so $(L+\lambda I)$ is positive definite. We have $$((L+\lambda I)y,y)=((L+\lambda I)^{1/2}y,(L+\lambda I)^{1/2}y)\ge C\|y\|^2_{H^{s/2}},$$
since the domain of $(L+\lambda I)^{1/2}$ is $H^{s/2}$ by the general theory of interpolation spaces. It follows that
$$(Ly,y)\ge C\|y\|^2_{H^{s/2}}-\lambda (y,y).$$
Now we distinguish two cases. Either $\|y\|^2_{H^{s/2}}>2\lambda (y,y)$. In that case it follows that
$$(Ly,y)\ge \frac{C}{2} \|y\|^2_{H^{s/2}}.$$ In the opposite case, let $\mu$ be such that $(Ly,y)\ge \mu (y,y)$. Then it follows that
$$(Ly,y)\ge \mu(y,y)\ge \frac{\mu}{2\lambda}\|y\|^2_{H^{s/2}}.$$
Are you assuming here $y$ is orthogonal to $\chi$, the eigenfunction of the negative eigenvalue? Or is it enough that $y$ is orthogonal to $\phi_c$ as the OP stated?
I changed the argument so I am no longer assuming y orthogonal to $\chi$.
@MichaelRenardy Thank you very much for your answer. Just one (maybe silly) question, why does exist this $\mu$ satisfying $(Ly,y)>\mu(y,y)$?
According to your post, you know $(Ly,y)>0$ for nonzero $y$ satisfying the orthogonality conditions. Now suppose you have a sequence $y_n$ with $(y_n,y_n)=1$ and $(Ly_n,y_n)\to 0$. By taking a subsequence, we may assume $y_n$ converges weakly, let $y$ be the limit. Let $z_n$ be the component of $y_n$ orthogonal to $\chi$ and let $z$ be the weak limit of $z_n$. Since $L$ is positive definite on the subspace orthogonal to $\chi$ and $\phi_c'$, we have $(Lz,z)\le \lim(Lz_n,z_n)$. If the inequality is strict, then $(Ly,y)<0$, a contradiction. If $(Lz_n,z_n)$ converges, we can conclude $(z,z)=1$.
|
2025-03-21T14:48:30.178739
| 2020-03-30T13:07:19 |
356110
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/356110"
}
|
Stack Exchange
|
Positive roots of the Tits unit form and dimension vectors
Let $A$ be a finite dimensional quiver algebra such that two indecomposable modules are isomorphic iff their dimension vectors are the same. Let $T_A$ be the tits unit form of $A$ and $r_A$ the set of positive roots of $T_A$ and $L_A$ the set of dimension vectors of indecomposable modules of $A$:
Question: Do we have $|L_A| \leq |r_A|$ and $|L_A|=|r_A|$ iff $L_A=r_A$?
This is tested for Nakayama algebras and true for all such algebras with at most 7 simple modules using QPA (it would even be interesting to ask this question only for Nakayama algebras).
I would expect this to be somewhere in the literature at least for acyclic algebras, but this question is also for general algebras including those with a non-acyclic quiver.
(It seems the tits unit form has been considered only for acyclic quiver algebras yet, or is there literature also for general such algebras?)
|
2025-03-21T14:48:30.178835
| 2020-03-30T13:40:01 |
356112
|
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"Gerry Myerson",
"Manfred Weis",
"Sylvain JULIEN",
"https://mathoverflow.net/users/13625",
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|
Stack Exchange
|
Sequence of least prime-multiples with smallest Hamming weight
It is known that
Almost all primes have a multiple of small Hamming weight, which makes me wonder what is known about the least multiples of primes that have least Hamming weight.
Questions:
what are the values of the sequence $$\mu(n)\quad :=\quad \min\limits_{m\in\mathbb{N}}\left( \min\limits_{k\in\mathbb{N}}:\ \ \frac{\sum\limits_{i=0}^\infty d_i2^i}{p_n}=m\in\mathbb{N},\ \ \sum\limits_{i=0}^\infty d_i=k,\ \ d_i\in\lbrace0,1\rbrace\right)$$,
where $p_n$ is the $n$-th prime number?
$\mu(1)=1,\ \mu(2)=1,\ \mu(3)=1,\ \mu(4)=1,\ \mu(5)=3,\ \mu(6)=5,\ \dots$
for which $n$ is $H(p_n\mu(n)) \gt 6$ if $H()$ is denotes the Hammig weight of the argument ?
Let me see if I understand the question. If you want to calculate $\mu(6)$, say, you first find $p_6=13$, the sixth prime, then you find $13\times5=65=(1000001)_2$ has Hamming weight $2$, and obviously no multiple of $13$ is a power of $2$, so $\mu(6)=5$ (or is it $\mu(6)=2$?). Are the values of $\mu(n)$ tabulated at the Online Encyclopedia of Integer Sequences?
@GerryMyerson $\mu(6)=5$. I chose that on purpose because knowing the smallest factor $m$ that yields the smallest Hamming weight allows for easy calculation of that weight, but knowing the minimal Haming weight gives no way of easily calculating the minimal $m$ that yields that minimal weight, especially if the minimal Haming weight were proven by contradiction, i.e. the assumption that a higher Hamming weight were minimal led to cotradiction.
OK, then $\mu(8)=27$, since the eighth prime is $19$, and $19\times27=513=2^9+1$ has Hamming weight $2$, and $19m$ has Hamming weight greater than $2$ for all $m<27$. It seems to me there will be a lot of $n$ for which $\mu(n)$ will be considerably greater than $6$.
@GerryMyerson shame on me! I wanted to know the $p_n$ with a Hamming weight larger than 6; I edited the question accordingly. Thank you.
$11\times3=33=2^5+1$, so $\mu(5)=3$, so I still don't follow you. $3^2\cdot331$?
@GerryMyerson today is not my day...
$7m=2^k+1$ is impossible, since $2^k+1\equiv0\bmod7$ has no solution, so $\mu(4)=1$.
If I understand correctly, the Hamming weight is the number of $1$ in the binary expansion, right?
@Sylvain yes, that's the definition of Hamming weights at least in the context of this question.
Do you know any value of $n$ for which $H>6$?
@GerryMyerson checking the linked paper for examples I saw that Theorem 5 states that if the number $\Omega(2^n-1) \lt \frac{\log n}{\log k}$ then one of the prime factors of $2^n-1$ has the property that all of its multiples have Hamming weight of at least $k$; $\Omega()$ denotes the number of primefactors counted with multiples. Unfortunately no clue is given as to which of the factors has that property. The linked paper is real gem by the way.
Certainly the inequality holds if $2^n-1$ is prime, even with $k=n-1$, and there's only one prime factor, so that gives about $40$ examples.
Thank you for the interest in my paper.
Here is a criterion for good candidates:
if the order of 2 modulo $p$ is less than $p^{1/6}$, then
six-fold sums $2^{i_1}+ \cdots + 2^{i_6}$ cannot cover all $p$ residue classes. (Often the 0-class mod $p$ is covered only with high iterations, but it is possible, it is covered earlier.
Actually, a slightly weaker condition might suffice, as there are only Binomial(ord_p (2),6) expressions with distinct exponents.)
As the order of $2$ is at least $\log_2(p+1)$,
which is the case of Mersenne primes, $p$ needs
to be at least $3 \times 10^9$.
Non-Mersenne primes with this criterion might be quite a bit larger.
At the time I wrote that paper, I did some experimental tests,
which shows that smaller primes work,
as the sumsets do not always grow with maximal speed.
Let us look at primes (not being Mersenne primes)
with quite small order of 2. Let
$A=(\{2^i: 0\leq i \leq \rm{ord}_p(2)-1\} ) \subset \mathbb{Z}/p\mathbb{Z}$.
For $p=178481$;
$|A|=23, |2A|=276, |3A|=2047, |4A|=10879, |5A|=42711, |6A|=113275,
|7A|=171810, |8A|=178480$,
which means one class is missing.
It is exactly the 0-class, which is needed so that the sumset is a multiple of $p$.
Only the 9-fold sumset contains the 0-class.
|
2025-03-21T14:48:30.179125
| 2020-03-30T14:22:40 |
356114
|
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|
Stack Exchange
|
About submersion and sections
Let $\pi:X \rightarrow Y$ be a surmersion (surjective submersion) between closed manifolds.
1) Is there any obstruction to the existence of a "multi-valued" section $s$ of $\pi$ such that $\pi \circ s$ is a smooth covering of $Y$ ?
By a multi-valued section I was thinking about gluing local section of $\pi$, where by a local section I mean a map $\sigma : U \rightarrow X$ with $U$ and open subset of $Y$ and satisfying $\pi \circ \sigma = id_U$. The multi-valued section should take the form of an immersion $s : Y' \rightarrow X$ with $p: Y' \rightarrow Y$ is a covering map and such that $\pi \circ s = p$.
2) Is it always possible to restrict us to finite covering $p : Y' \rightarrow Y$ ?
A trivial example is given when the fiber bundle is trivial and the covering is just the trivial covering.
If $Y$ is simply connected there are no covers, then the question reduced to the question if a section always exists right? That is not the case. For example the unit sphere bundle $X=T_1S^2$ over $Y=S^2$ does not have a section.
First, since you are dealing with closed manifolds, a submersion $\pi:X\rightarrow Y$ (assuming that the base is connected) is surjective because it is simultaneously open and closed. Moreover, due to Ehresmann's Theorem, the submersion $\pi:X\rightarrow Y$ is a fiber bundle as it is proper.
Here is an approach for constructing non-examples: Let $Y$ be a compact manifold with $\pi_1(Y)$ finite. Denote its universal cover by $p:\tilde{Y}\rightarrow Y$. Pulling back the fiber bundle $\pi:X\rightarrow Y$ with the finite cover $p:\tilde{Y}\rightarrow Y$ results in a new fiber bundle $\tilde{X}\rightarrow\tilde{Y}$. It is not hard to see that the natural map $\tilde{X}\rightarrow X$ is also a finite cover. Hence a multi-valued section $s:\tilde{Y}\rightarrow X$ lifts to a section of the bundle $\tilde{X}\rightarrow\tilde{Y}$ because $\pi_1(\tilde{Y})$ is trivial. So we need to arrange for the pullback of the bundle $\pi:X\rightarrow Y$ with the universal covering map $p:\tilde{Y}\rightarrow Y$ to do not admit a section.
1) Similar to @ThomasRot's comment, you may consider sphere bundles on $Y$: Let $E\rightarrow Y$ be an oriented vector bundle of rank $r$ whose Euler class
$e(E)\in H^{r}(Y,\Bbb{R})$ is non-zero. Now take $X$ to be the sphere bundle $S(E)\rightarrow Y$. If it admits a multi-valued section, from the discussion above the same must be true for its pullback $S(\tilde{E})\rightarrow \tilde{Y}$ where $\tilde{E}\rightarrow\tilde{Y}$ is the pullback of the bundle $E\rightarrow Y$ with the finite (universal) cover $p:\tilde{Y}\rightarrow Y$. But the former bundle cannot admit a section since the Euler class of the vector bundle $\tilde{E}\rightarrow\tilde{Y}$, given by $p^*(e)$, is non-zero because $p^*:H^{r}(Y,\Bbb{R})\rightarrow H^{r}(\tilde{Y},\Bbb{R})$ is injective.
2) Another approach is to take $\pi:X\rightarrow Y$ to be a principal $G$-bundle. Recall that such bundles are trivial if they admit a section, and their isomorphism classes are in bijection with homotopy classes of maps from the base to the classifying space $BG$. Thus any map $Y\rightarrow BG$ with the property that the composition
$\tilde{Y}\rightarrow Y\rightarrow BG$ is not null-homotopic provides you with a principal $G$-bundle $\pi:X\rightarrow Y$ without any multi-valued section.
Here is a simple counterexample. Let $X=(0,2/3)\cup(4/3,2)$ (open intervals),
$Y=(0,1)$ and $\pi(x)=x$ if $x\in(0,2/3)$ and $\pi(x)=x-1$ if $x\in(4/3,2)$.
Note that $Y$ being simply connected has no nontrivial cover.
If you add the condition that the fibres are connected, then you will have counterexamples with higher-dimensional $Y$.
Maybe you can have a look at a paper of mine:
Submersions, fibrations and bundles, Trans. Amer. Math. Soc. 354 (2002), 3771-3787
(avaible here: http://web.univ-ubs.fr/lmba/meigniez/docu/travaux.html)
|
2025-03-21T14:48:30.179369
| 2020-03-30T15:01:32 |
356116
|
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|
Stack Exchange
|
Wigner distribution
The Wigner distribution of $u\in L^2(\mathbb R)$ is defined as a function $W(u)$ on $\mathbb R^2$ given by
$$
W(u)(x,\xi)=\int_\mathbb R u\left(x+\tfrac z2\right) \overline{u\left(x-\tfrac z2\right)} e^{-2π i z\xi} \, dz.
$$
It is easy to see that $W(u)$ belongs to $L^2(\mathbb R^2)$ with
$
\Vert W(u)\Vert_{L^2(\mathbb R^2)}=\Vert u\Vert_{L^2(\mathbb R)}^2
$
since $W(u)$ is the partial Fourier transform wrt $z$ of $(x,z)\mapsto u(x+\frac z2)\overline{u(x-\frac z2)}$.
However I believe that for some $u\in L^2(\mathbb R)$, the function $W(u)$ does not belong to
$L^1(\mathbb R^2)$. Is there an "explicit" $u$ like this?
I haven't checked all the way through by explicit calculation, but unless I'm missing something, I suspect a box function like $u(t) = 1 $ for $|t|<1$, $u=0$ else, should work, maybe you'd like to check ...
@MichaelEngelhardt You are right, thank you so much. With your $u$, I find $\mathcal W(u)(x,\xi)=H(-x)\frac{\sin(4π(1+x)\xi)}{π \xi}+H(x)\frac{\sin(4π(1-x)\xi)}{π \xi}$ with $H=\mathbf 1_{\mathbb R_+}$ and we find easily that $\int_{\mathbb R^2}\vert\mathcal W\vert dx d\xi=+\infty$. I had in mind a quite complicated abstract argument using the Baire Category Theorem, which I was not able to fully complete.
|
2025-03-21T14:48:30.179482
| 2020-03-30T15:07:53 |
356117
|
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"authors": [
"Denis Nardin",
"Johan",
"Lao-tzu",
"Will Sawin",
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|
Stack Exchange
|
Is any constant Zariski sheaf already a Nisnevich sheaf?
Lat $A$ be a set and $\underline{A}$ the associated constant Zariski sheaf on the category $Sm/S$ of schemes which are smooth over $S$ for a fixed base scheme $S$. Is $\underline{A}$ already a (constant) sheaf for the Nisnevich topology on $Sm/S$?
I ask this because constant Zariski sheaves are easier to describe, which only depend on connected components.
Did you try to prove this by hand?
First, I don't know the answer. A possible way to give an answer is to prove that on sections over a Nisnevich distinguished square, it gives a pullback diagram. Roughly this corresponds to analyzing the connected components, but I don't succeed with it.
Yes. This is fine for every topology in which covers are collections of morphism that are open for the Zariski topology and surjective on points.
To see this, because $\underline{A}$ satisfies the sheaf condition for disjoint unions, it suffices to show for $f: Y \to X$ open and surjective on points, $\underline{A}(X) \to \underline{A}(Y) \substack{ \to \\ \to} \underline{A}( Y\times_X Y)$ is a pullback square.
To do this, it is helpful to note that $\underline{A}(X)$ is the set of disjoint $A$-indexed open covers of $X$.
Given a disjoint $A$-indexed open cover $(F_a)_{a \in A}$ of $Y$ in $\underline{A}(Y)$, look at the image $f(F_a)$ of each set in $X$. This gives an $A$-indexed open cover of $Y$. We must check that if $(F_a)_{a \in A}$ satisfies the gluing condition, the cover of $X$ is disjoint.
In other words we must check that if $x \in X$, $y_1,y_2$ lie in the fiber of $Y$ over $X$, and $y_1 \in F_{a_1}$, $y_2 \in F_{a_2}$, then $a_1=a_2$. This follows from the existence of a point in $Y \times_X Y$ that maps to $y_1$ and $y_2$, which follows from the fact that $\operatorname{Spec} \kappa(y_1) \times_{ \operatorname{Spec} \kappa(x)} \operatorname{Spec} \kappa(y_2)$ is nonempty, where $\kappa(x)$ denotes the residue field at $x$.
You said "it is helpful to note that $\underline{A}(X)$ is the set of disjoint $A$-indexed open covers of $X$", do you mean that $\underline{A}(X)$ is in fact $A$-copies disjoint union of $X$?
@Lao-tzu No, $\underline{A}(X)$ is the set of maps from the underlying set of $X$ to $ A$ that are continuous for the discrete topology of $A$ and the usual topology on $X$. In other words the inverse image of each element of $A$ is an open subset of $X$, and these open sets are disjoint.
In fact, $\underline{A}$ also has the sheaf property with respect to fpqc, h, and V coverings.
@Will Sawin A great answer! I would never take this perspective on constant sheaves before I see your unique proof.
@Johan What do you mean by V coverings?
@Lao-tzu I assume coverings for the v-topology
@ Denis Nardin Thanks for the link!
Please allow me to rewrite Will's fantastic answer below, which I hope to be easier to understand and for others' convenience.
Let $(\mathscr{C}, \tau)$ be a Grothendieck site, where $\mathscr{C}$ is a category of schemes and $\tau$ is a topology on $\mathscr{C}$ finer than the Zariski topology, all of whose covers are collections of morphisms that are open for the Zariski topology and surjective on points. Lat $A$ be a set and $\underline{A}$ the associated constant sheaf on $\mathscr{C}$ with the Zariski topology. Then $\underline{A}$ already a (constant) sheaf on $(\mathscr{C}, \tau)$.
Proof. Let $Y\xrightarrow{f}X$ be a $\tau$-cover on the site $\mathscr{C}$. Note that $\underline{A}(X)$ is the set of locally constant functions with values in $A$, hence can be identified with all disjoint $A$-indexed open covers of $X$ (given by mapping $v\in\underline{A}(X)$ to the family $\{v^{-1}(a)\}_{a\in A}$).
Given any $u\in\underline{A}(Y)$ with $up_1=up_2$, we want to find a (unique) locally constant function $v\in\underline{A}(X)$ with $vf=u$. Of course, we have to define $v$ by letting $v^{-1}(a)=f(u^{-1}(a))$ (note that $f$ is an open map), provided that it is well-defined, that is, $f(u^{-1}(a))\cap f(u^{-1}(b))=\varnothing, \forall a\ne b\in A$.
We show this now: Otherwise, there would exist $x\in X$ and $y_1, y_2\in f^{-1}(x)$ with $u(y_1)=a, u(y_2)=b$. By scheme theory, there exists $z\in Y\times_XY$ with $p_1(z)=y_1, p_2(z)=y_2$, thus $a=u(y_1)=up_1(z)=up_2(z)=u(y_2)=b$, a contraction.
|
2025-03-21T14:48:30.179908
| 2020-03-30T15:35:47 |
356119
|
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|
Stack Exchange
|
Online events during the quarantine
With many places on earth subjected to quarantine and large gathering prohibited, there are announcements of online seminars and talks open to people around the world. The talks can be conducted via ZOOM or other platforms. As far as I know, the information about them is spread by the word of mouth, or more precisely personal emails.
Are there some sources/online pages where information about ongoing online events im mathematics is collected? I am interested in probability in particular, but I assume other people would be interested in events related to their research interests.
A somewhat similar question was asked here 10 years ago but the circumstances are different now, and the available technological tools are more advanced.
Aram Dermenjian put together a nice resource for online seminars in algebraic combinatorics speficially: http://dermenjian.com/seminars/
I think the question should be more clearly either focussed on probability, or not focussed at all.
This (more recent) question is also related: Software and ideas for workshops and conferences with long-distance participants.
This is a topic of great current interest. But a list of online seminars and talks is certain to be out-of-date very soon. Perhaps MathOverflow is not the right place for this list; but I hope that there is a good place for it somewhere.
In case you did not have the informations, there is the One World Probability Seminar https://www.wim.uni-mannheim.de/doering/one-world/
@FrançoisBrunault: And an analogous event for the PDE crowd: https://people.bath.ac.uk/mw2319/owpde/
I think https://jaume.dedios.cat/math-seminars/ is a very good list.
I invite you to look at researchseminars.org.
Anyone can create listings for seminars or conferences, either as an organizer or as a curator. These listings will automatically appear once the individual has been endorsed (similar to arXiv endorsing). The goal is to distribute the task of maintaining up to date listings as broadly as possible.
When browsing talks and seminars, you can filter by topic or institution, tag your favorite talks or seminars, and export to your calendar (new talks in a seminar you have subscribed to will automatically show up in your calendar).
Times are automatically translated to a user's local time zone.
We've added a few seminars that we found online. We are in the process of adding more. If you would like to join our efforts, you're very welcome!
This is the most accessible webpage I’ve come across so far. Great job.
One ongoing effort to collect online seminars in all areas of mathematics (indicated by arXiv category) is by Ao Sun (MIT) and Mingchen Xia (Chalmers), on this website.
Dan Isaksen put together a list of online seminars and events on his webpage.
Not just a list of seminars, it's also a list of lists of seminars!
Some seminars in Russia which are now online, mentioned here:
http://www.mathnet.ru/php/seminars.phtml?option_lang=rus
For some zoom/skype-id is shown, for other one should probably ask the organizers.
Terrence Tao shares an online website for seminar. Have a look Here
or Here. I hope i understand the question.
AMS has started a list, organized by mathematical field, at https://www.ams.org/profession/online-talks.
|
2025-03-21T14:48:30.180182
| 2020-03-30T15:57:28 |
356122
|
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|
Stack Exchange
|
On automorphism group
Let $p$ be a prime number, and let $\mathbb{F}_{p}$ be a finite
field of order $p$. Let $U_{n}$ denote the unitriangular group of $n\times n$ upper
triangular matrices with ones on the diagonal, over $\mathbb{F}_{p}$. Are there groups such as their automorphism group is $U_{n}$?.
Any example or reference will be helpful. Thank you all.
If $p=2$, it seems possible that $Aut(U_n)=U_n$... Although maybe I'm just not seeing some outer auts...
@Nick Gill: I thought that Gaschutz proved that ${\rm Out}(P)$ has order divisible by $p$ for any finite $p$-group $P$.
@GeoffRobinson Is this a $p$-group, though?
@StevenStadnicki It has $p^{n(n-1)/2}$ elements, so yes.
@StevenStadnicki :Yes, $U_{n}$ is always a $p$-group. If $p^{k}$ is the smallest power of $p$ greater than or equal to $n$, we have $u^{p^{k}} = I$ for each $u \in U_{n}.$
@GeoffRobinson, What about $C_p$?
D'oh; those make sense. I'm not sure what i was thinking, mea culpa. :/ Can I blame a lack of coffee this morning?
@Nick Gill: Clearly, that is a counterexample, so I mis-remembered the theorem. Maybe these are the only counterexamples, but I need to look up the theorem.
@NickGill: It seems that groups of order $p$ are the only exceptions to Gaschutz's theorem.
$U_3(p=2)\cong D_8\cong Aut(D_8)$.
|
2025-03-21T14:48:30.180308
| 2020-03-30T16:15:38 |
356123
|
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|
Stack Exchange
|
Conditions for certain inclusion functor to preserve internal homs
Suppose that $\mathcal{C}$ is a locally cartesian closed right proper Quillen model category for which all objects are fibrant. Let $x$ be an object of $\mathcal{C}$. Let $\mathbb{F}$ denote the class of fibrations and let $\mathbb{F}_x$ denote the category of all fibrations with codomain $x$. Consider the inclusion functor
$$
i : \mathbb{F}_x \hookrightarrow \mathcal{C}/x.
$$
Is it true that $i$ preserves all exponentials (i.e., internal homs)?
My question arises from page 54 of the paper https://arxiv.org/pdf/1406.3219.pdf.
Here, the author gives examples of categories $\mathcal{C}$ satisfying the conditions of Theorem 32 (page 53). The relevant condition for us is that $\mathcal{C}$ have a closed class $\mathbb{D}$ of morphisms (Definition 27, page 40). In turn, the relevant condition of this definition is that the inclusion functor $ \mathbb{D}_x \hookrightarrow \mathcal{C}/x$ as above preserve all exponentials for any object $x$.
The author gives the following example:
the category of fibrant objects in any locally cartesian closed
model category that is right proper, and in which the cofibrations are
the monos.
So, if we take $\mathbb{D}$ to be the class of fibrations, we arrive at my original question.
You also need to assume that cofibrations are monos since this implies that (trivial) cofibrations are stable under pullbacks along fibrations. In turn, this implies that the exponent of two fibrations over $x$ is also a fibration. It is easy to see that the exponent also has its universal property in $\mathbb{F}_x$. It follows that $\mathbb{F}_x$ is closed under exponents in $\mathcal{C}/x$.
|
2025-03-21T14:48:30.180443
| 2020-03-30T18:05:07 |
356131
|
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|
Stack Exchange
|
Tannakian criterion for reducedness of Tannakian dual group
Given an affine group scheme G over a field of positive characteristic.
Question: Is there a simple criterion for G to be reduced in terms of the neutral Tannakian category of its finite dimensional algebraic representations?
Yes, and one could argue as follows, at least if the group scheme is assumed to be of finite type (so it is an algebraic group), and the base field is perfect. Recall that if $k$ is a field of characteristic zero, then any algebraic group is smooth, and that if $k$ is a perfect field of characteristic $p$, then an algebraic group is reduced if and only if it is smooth.
I'll make some statements using dg-categories, because that language simplifies some statements. Let $k$ be a perfect field, and let $A$ be a $k$-algebra. Then $A$ is smooth (in the usual sense) if and only if $A$ is a perfect object of the (dg-)category $\mathrm{BMod}_A(\mathrm{Mod}_k)$ of $A$-$A$-bimodules in $k$-vector spaces. I don't really have a reference for this, but one could view this result as a jazzed up statement of Serre's regularity criterion. (Since $k$ is perfect, $A$ is smooth if and only if it is regular.) This MathOverflow post has some discussion of such criteria: Smooth dg algebras (and perfect dg modules and compact dg modules).
This implies that if $A$ is a $k$-algebra, then $A$ is smooth if and only if $\mathrm{Mod}_A$ is a dualizable $k$-linear dg-category such that the unit $\eta:\mathrm{Mod}_k\to \mathrm{Mod}_A \otimes_k \mathrm{Mod}_A^\vee$ preserves compact objects. Indeed, the dual of $\mathrm{Mod}_A$ as a $k$-linear dg-category is just the category of modules over the opposite algebra (which is $A$ itself if $A$ is commutative), so $\mathrm{Mod}_A \otimes_k \mathrm{Mod}_A^\vee$ is the category of $A$-$A$-bimodules in $k$-vector spaces. The functor $\eta$ sends the unit $k\in \mathrm{Mod}_k$ to $A$ regarded as a bimodule over itself, so $A$ is smooth by the above discussion.
This gives the desired criterion over a perfect field $k$: an algebraic group $G$ is smooth if and only if the dg-category $\mathrm{Rep}(G)$ is a dualizable object in $k$-linear dg-categories, such that the unit $\eta:\mathrm{Mod}_k\to \mathrm{Rep}(G) \otimes_k \mathrm{Rep}(G)^\vee$ preserves compact objects.
If this is too abstract of a criterion, one could also just stop at the point where we appealed to Serre's homological criterion for regularity. This leads to the statement that an algebraic group $G$ over a perfect field $k$ is smooth if and only if $\mathrm{Rep}(G)$ is of finite global dimension, i.e., $\mathrm{Hom}_{\mathrm{Rep}(G)}(V, W[n])$ vanishes for all finite-dimensional representations $V,W\in \mathrm{Rep}(G)$ and all but finitely many $n$.
Is the dg-category of representations determined by the category of finite dimensional algebraic representations?
Yes: the dg-category of representations of G, i.e., QCoh(BG), is the Ind-completion of the dg-category of perfect complexes over BG, which is determined by the category of finite-dimensional representations of G.
I feel cheated by this answer (despite upvoting it). Suppose one "knows" $Rep(G)$ without knowing $G$. Think of some kind of geometric Satake. Is there a clear pathway for checking smoothness of $G$ in this criterion?
Fair enough. Edited.
This is quite clever! In the application the group scheme isn't algebraic. The calculation of the Ext's (in finitely generated tensor subcategories) looks difficult. I'm stll hoping for something simpler.
I don't understand the application of Serre's criterion. For example, $\mu_p$ is not smooth in characteristic $p$, but the category of representations is semisimple.
I agree with Angelo: the final statement in the above answer is false.
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2025-03-21T14:48:30.180697
| 2020-03-30T18:16:43 |
356134
|
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|
Stack Exchange
|
Completeness of coefficient functionnals
My questions is about Schauder bases and more specifically about coefficient functionals.
Let $(x_n)$ be a Schauder basis of a Banach space $X$. Thus for all $x$ in $X$, $x = \sum f_n(x) x_n$. The $f_n$ are called coefficient functionals.
They are continuous. If $X$ is reflexive, they form a basis of $X^*$ (with Hahn–Banach theorem).
My question is : if $X$ is not reflexive, but if we suppose $X^*$ separable, is $(f_n)$ a basis of $X^*$?
M. Zippin showed that for a Banach space $X$ with a basis, if every basis of $X$ is boundedly complete or if every basis of $X$ is shrinking, then $X$ is reflexive.
The result of Zippin answers you question in the negative (or, rather, the answer is, "not necessarily".) However to complete the picture let us recall the earlier result of R.C. James (alluded to in your question) asserting the following: For a reflexive Banach space $X$ with a basis $(x_n)$, the basis $(x_n)$ is both shrinking and boundedly complete.
Putting the results of Zippin and James together yields the following:
Let $X$ be a Banach space with a basis. The following are equivalent:
$X$ is reflexive;
every basis of $X$ is shrinking; and
every basis of $X$ is boundedly complete.
No, there are counterexamples. For instance, there exists a space $X$ with a basis, $X^*$ separable, and yet $X^*$ fails approximation property. See Lindenstrauss-Tzafriri's book Theorem 1.e.7 for a discussion of this (it is an existence proof).
The dual space $J^*$ of James' space, with the basis formed by the coefficient functionals of the (usual) shrinking basis of $J$, is another example.
Yes, in fact, that is a better answer:) But you mean boundedly complete basis (i.e., `the summing basis') of $J$
And Zippin showed that for a Banach space $X$ with a basis, if every basis of $X$ is boundedly complete or if every basis of $X$ is shrinking, then $X$ is reflexive: https://link.springer.com/article/10.1007/BF02771607
@ Philip Brooker Nice. This should be the answer to the question.
@BunyaminSari Thanks, Bunyamin, I've just cobbled together an answer :)
|
2025-03-21T14:48:30.180870
| 2020-03-30T18:52:22 |
356136
|
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|
Stack Exchange
|
Characterization of extrinsic distance prevserving embedding (see the definition given!) from low dimensional Euclidean spaces to high dimensions
P.S. I asked the question on MSE more than a week ago, but didn't get any desired answer, so asking here.
Let $m < n \in \mathbb{N}$. Let us equip $\mathbb{R}^m, \mathbb{R}^n $ with their canonical Euclidean (Riemannian) metrics. How can we characterize the extrinsic distance preserving embeddings of $\mathbb{R}^m$ into $\mathbb{R}^n $? To be more precise, I'm looking for sufficiently regular, injective, distance preserving transformations from $\Phi: \mathbb{R}^m \to \mathbb{R}^n$, so that: $d_m(x, y)= d_n(\Phi(x), \Phi(y))$, where $d_m, d_n$ represents the distances in $m, n$ dimensional Euclidean spaces respectively, so $d_m(x,y):=||x-y||_{\mathbb{R}^m}, d_n(x,y):=||x-y||_{\mathbb{R}^n}$. Note that this means: we're looking to preserve the extrinsic distances between the two Euclidean spaces, we don't care about if the first one embeds isometricaly as a submanifold in the second.
I think the answer is:
$$x \mapsto A(Bx, O(Bx))$$ where $A$ is an Euclidean isometry of $\mathbb{R}^n$ (i.e. a rigid motion), $B$ is an Euclidean isometry of $\mathbb{R}^m$ (i.e. a rigid motion), and $O:\mathbb{R}^m \to \mathbb{R}^n$ is "zero padding $n-m$" times, namely: $O(x)= (x, 0,\dots 0)$.
If the above correct/incorrect, how do I go about proving it or characterizing the Euclidean embeddings?
Approaches that immediately come to mind are as follows:
(1) Assume $\phi: \mathbb{R}^m \to \mathbb{R}^n$ is such an isometric embedding. Then we can try to consider the canonical projection $\pi: \mathbb{R}^n \to \mathbb{R}^m$ and try to consider the map: $\phi \circ \pi: \mathbb{R}^n \to \mathbb{R}^n $, but this will not be an isometry (not one to one), which is a problem.
(2) Alternately, we can consider: $\pi \circ \phi: \mathbb{R}^m \to \mathbb{R}^m$. We can't guarantee that $\pi \circ \phi$ is an isometry of $\mathbb{R}^m$, which is again a problem.
(3) As someone suggested in the comments of the MSE question, there exists an isometry, say $A$, of $\mathbb{R}^n$ so that $A \circ \phi (\mathbb{R}^m)= \mathbb{R}^m \times \{0\}^{n-m}$. This intuitively seems correct, as the isometry group of $\mathbb{R}^n$ has dimension $\frac{n(n+1)}{2}$, which means it's "high dimensional enough", i.e. "there're enough elements in it" to render the last $n-m$ components to $0$. This could be made rigorous if $\phi$ was a linear or affine map, but if $\phi$ is nonlinear, I can't see how this argument goes through? In this case we need to show first that $\phi$ is affine, i.e. if $\phi: \mathbb{R}^m \to \mathbb{R}^n$ fixes $0$, then it has to be a linear map from $\mathbb{R}^m$ to $\mathbb{R}^n$.(?)
This is incorrect, even for $m=1$ and $n-3$, as well as $m=2$ and $n-3$. In the first case, any arclength parameterization $c: \mathbb{R}\rightarrow\mathbb{R}^2$ is isometric. In the second case, any map of the form $(x,y) \mapsto (a(x),y,b(x))$, where $x\mapsto c(x)=(a(x),b(x))$ is an arclength parameterization of a curve in $\mathbb{R}^2$, is an isometric embedding of $\mathbb{R}^2$ in $\mathbb{R}^3$.
@DeaneYang Sorry but that's not the kind of isometry i'm looking for, as specified in the question. What you described is an isometry, if we consider the intrinsic distance on the image, not the extrinsic Euclidean distance. For example, if $m=1, n=2,$ then $d(\phi(t_1), \phi(t_2))= |t_1 - t_2|$ if $|\phi'|=1$, fir sure; but the intrisic distance $d(\phi(t_1), \phi(t_2)) \ne ||\phi(t_1) - \phi(t_2)||_{\mathbb{R}^2}$.
Oops. Sorry about that. So here's how the argument for your question goes. First show that, given inner product spaces $V$ and $W$ if a map $f: V \rightarrow W$ satisfies $|f(v_1)-f(v_2)| = |v_1-v_2|$, for all $v_1, v_2 \in V$, then $f(v_1)\cdot f(v_2) = v_1\cdot v_2$, for all $v_1, v_2 \in V$. This implies that $f$ maps an orthonormal basis to an orthonormal basis. You can prove that $f$ is linear and therefore a rotation. Finally, given an extrinsic distance preserving map, compose it with a translation so the origin is fixed, This map satisfies the assumptions of $f$ above.
By the way, no assumption on $n$ and $m$ is needed. That $m \le n$ is a consequence of the proof.
@DeaneYang Thank you again! The first argument is a direct consequence of polarization identity, the second is clear from the drefintion of orthonormal basis (ONB) itself. I'm currently stuck at the third: showing that a map that sends ONB to ONB has to be linear. Hmm, may be I'll show that it's derivative is constant? (I'm thinking!) The rest is clear about the translating it in order to fix the origin.
The trick is to observe that you can recover the coefficients of a vector with respect to an orthonormal basis by using the inner product. That plus the fact the inner product is preserved under $f$ implies the coefficients are preserve by $f$.
@DeaneYang Thank you - this is a useful trick indeed!
The answer is yes. Since $\Phi$ is isometric in your sense, it sends a geodesic to a geodesic, i.e. a straight line to a straight line. Then, it is a classical result of affine geometry (see e.g. Artin's wonderful book "Geometric Algebra") that an injective map $R^m\to R^n$ sending straight lines to straight lines is affine (except maybe for $m=1$, but here you have moreover the distance being preserved).
|
2025-03-21T14:48:30.181222
| 2020-03-30T19:05:26 |
356139
|
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|
Stack Exchange
|
Does a homologically bounded dg A-module admit a "locally finite" semi-free resolution
Let $A$ be a bounded dg-algebra whose underlying algebra is Noetherian and such that $H^*(A)$ is Noetherian. Let $M$ be a cohomologically bounded dg-module over $A$, whose cohomology groups are finitely generated over $H^*(A)$ (one can also assume if it helps that $M$ is finitely generated over $A$).
Question: Does $M$ admit a semi-free resolution such that each each of the filtered pieces is a \emph{finitely generated} free A-module?
Feel free to tweak the hypotheses because I'd be grateful for any result in this vain (with reference or proof).
I think what you are looking for might be in Theorem 11.4.40 of the book https://arxiv.org/abs/1610.09640, but I did not check if these are the exact conditions.
That does look close to what I want but it assumes that A has no cohomology in positive degrees which I definitely don't want.
|
2025-03-21T14:48:30.181311
| 2020-03-30T19:40:28 |
356142
|
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"url": "https://mathoverflow.net/questions/356142"
}
|
Stack Exchange
|
Filtrations of motivic spectral sequences
I had a general question about motivic spectral sequences. In order to derive them we first begin with a filtration of the algebraic $K$-theory spectra. Something like this $\cdots \rightarrow W^2(X)\rightarrow W^1(X)\rightarrow K(X)$ for every smooth variety $X$ over a field. Note that after taking the homotopy groups of this filtration, induces a filtration on the algebraic $K$-groups, which I denote it by $Fil_1$. In order to derive the motivic cohomology we check the fibers of the successive maps which happen to be a simplicial group (I'm focusing on the Grayson filtration). By Dold-Kan we can assign a chain complex to it and sheafify it wrt Nisnevich topology. Now the second page of the motivic ss consists of sheaf cohomology groups of this sheafified complex. This ss converges to the algebraic $K$-group so it induces a filtration on it which I denote by $Fil_2$. Let's denote the Nisnevich sheafification of $W^i$ by $W^i_{Nis}$. If I understand the constructions correctly, $Fil_2$ should be the filtration on the homotopy groups induced by $W^i_{Nis}$s.
My question is that do $Fil_1$ and $Fil_2$ coincide? (For example if $W^i$ s do already satisfy the Nisnevich descent that would be the case). Or maybe they are not same. Do they at least coincide rationally?
|
2025-03-21T14:48:30.181431
| 2020-03-30T19:58:36 |
356144
|
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|
Stack Exchange
|
Computing the maximum modulus
For each $a\in \mathbb C$ define $f_a:\mathbb C\to \mathbb C$ by $f_a(z)=\exp(z)+a$. I am primarily interested in real values $a\in (-\infty,-1)$.
For each $r\in [0,\infty)$ define $M_a(r)=\max\{|f_a(z)|:|z|=r\}$. So $M_a(r)=|z'|$ for some point $z'$ in the image of the circle of radius r. Is there a more explicit (numerical) formula for $M_a(r)$ in terms of $a$ and $r$?
A partial solution: As mentioned by @MargaretFriedland, the desired $M_a(r)$ is the absolute maximum of $g(t):=\sqrt{{\rm{e}}^{2r\cos t}+2a{\rm{e}}^{r\cos t}\cos(r\sin t)+a^2}$. Notice that this is an even function, so it suffices to maximize over $[0,\pi]$. If $t\in\left[\frac{\pi}{2},\pi\right]$, then $\cos t<0$. So ${\rm{e}}^{2r\cos t}\in [0,1]$ and, keeping in mind that $a$ is assumed to belong to $(-\infty,-1)$, we have:
$$\sqrt{{\rm{e}}^{2r\cos t}+2a{\rm{e}}^{r\cos t}\cos(r\sin t)+a^2}\leq\sqrt{1-2a+a^2}=1-a.$$
This value is achieved at $t=\pi$. Consequently:
$M_a(r)=\max\left\{1-a,\max_{\,t\in \left[0,\frac{\pi}{2}\right]}g(t)\right\}$. One can elaborate more by conditioning on $r\geq 0$.
If $r\leq\pi$, as $t$ varies in $\left[0,\frac{\pi}{2}\right]$, $r\sin t$ is increasing and takes its values in the interval $[0,\pi]$ on which cosine is decreasing. We conclude that $g(t)=\sqrt{{\rm{e}}^{2r\cos t}+2a{\rm{e}}^{r\cos t}\cos(r\sin t)+a^2}$ is decreasing on $\left[0,\frac{\pi}{2}\right]$ as every non-constant term under the square root is. Hence in this case $$M_a(r)=\max\{1-a,g(0)=|{\rm{e}}^r+a|\}.$$
If $r>\pi$, the absolute maximum $M_a(r)$ is larger than $1-a$. To see this, notice that there exists $t_0\in\left[0,\frac{\pi}{2}\right]$ with $\sin t_0=\frac{\pi}{r}$. We have:
$$g(t_0)=\sqrt{{\rm{e}}^{2r\cos t_0}-2a{\rm{e}}^{r\cos t_0}+a^2}={\rm{e}}^{r\cos t_0}-a={\rm{e}}^{r\sqrt{1-(\frac{\pi}{r})^2}}-a> 1-a.$$ Thus in this case
$$
M_a(r)=\max_{\,t\in \left[0,\frac{\pi}{2}\right]}g(t)\geq{\rm{e}}^{\sqrt{r^2-\pi^2}}-a.
$$
Added: It is also interesting to notice that always $M_r(a)\in\big[\,|{\rm{e}}^r+a|,{\rm{e}}^r-a\big)$; the function $g(t)$ can never attain
${\rm{e}}^r-a$ because that requires to simultaneously have $\cos t=1$ and $\cos(r \sin t)=-1$. Thus the negativity of $a$ is really what makes this problem interesting. As a result: $\big|M_r(a)-({\rm{e}}^r-a)\big|<2a$.
Fixing $a$, the ratio $\frac{M_r(a)}{{\rm{e}}^r-a}$ tends to $1$ as $r\to\infty$.
An answer before the numerics start: First, note that for $w=u+iv \in \mathbb{C}$ and a fixed $a \in \mathbb{R}$ we have $|w+a|=\sqrt{u^2+v^2+2au+a^2}=\sqrt{|w|^2 + 2au+a^2}$ . Next, consider the image under the complex exponential of the circle $|z|=r>0$. Using polar coordinates, we get $|\exp (r\cos t+i r\sin t)|=e^{r\cos t}$. Thus to get $M_a(r)$ we need to maximize the function $g(t)=\sqrt{e^{2r\cos t}+2ae^{r\cos t}\cos (r\sin t)+a^2}$ for $t \in [0,2\pi]$.
For $r=0$ we get $M_a(0)=|\exp 0 +a|=|1+a|$.
Assuming that a closed formula is not accessible, possibly maxmod, which calculates the maximum modulus of a complex polynomial on the unit disk, would speed a numerical attack. (Polynomial order of a few hundred is not a problem.)
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2025-03-21T14:48:30.181743
| 2020-03-30T22:53:50 |
356148
|
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|
Stack Exchange
|
When the local system of coefficients are simple in the Leray-Serre spectral sequence
Let $F\to E\to B$ a fibration and $\{E_{r}^{\ast,\ast},d_{r}\}$ the Leray-Serre Spectral sequence converging to $H^{\ast}(E;R),$ such that
$$E_{2}^{p,q}=H^{p}(B;\mathcal{H}^{q}(F;R))$$
is the cohomology of $B$ with local coefficients in the cohomology of the fiber $F.$ We know that when the action of $\pi_{1}(B)$ induced by fibration on the cohomology of $F$ is trivial then the system of local coefficients is simple. My question: Can anything be said (assuming that $R=\mathbb{Z}_{2}$ and $\pi_{1}(B)=\mathbb{Z}_{2}$) when the action is trivial except for ONE element of the cohomology? To be more specific, there is only one element $c\in H^{1}(F;R)$ such that $\mu(g,c)=gc\neq c,$ for $g\in \mathbb{Z}_{2}=\langle g\rangle$, where $\mu:\pi_{1}(B)\times H^{\ast}(F)\to H^{\ast}(F)$ denotes the induced action.
Surely there are $2$ elements on which the action is non-trivial ($c$ and $gc$)? Anyway, I would suggest if you know the cohomology of $F$ as a $\pi_1(B)=\mathbb{Z}2$-module, then you have a good chance of computing the $E_2$-term (using the definition on cohomology with local coefficients in $H^q(F)$ as the cohomology of the cochain complex $\operatorname{Hom}{\pi_1(B)}(C_*(\tilde{B}),H^q(F))$, where $\tilde{B}$ is the universal cover of $B$).
|
2025-03-21T14:48:30.181864
| 2020-03-31T02:48:02 |
356152
|
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|
Stack Exchange
|
Computing the kernel of some Artin-Map
let $K = \mathbb{Q}(\sqrt{-5})$ and $L = K(i)$. $\mathcal{O}_K$ is the ring of integers of K.
I would like to show that the kernel of the Artin-Map $\phi_{L/K}: I_K \rightarrow Gal(L/K)$ is $P_K$, where $I_K$ denotes the group of all fractional Ideals from the ring of integers of $K$ and $P_K$ denots the group of all prinicipal ideals.
Because I´m new to this topic, I want to do it without class field theory, but recently I read a lot about the frobenius-element/artin-symbol of a prime ideal.
So far I thought about the following:
I guess I can compute the frobenius-element/artin-symbol of every prime $\mathfrak{P}$ from $\mathbb{Q}(\zeta_{20})$ which lies about a integer prime $p \neq 2,5$. I think this can be useful since $K(i) \subset \mathbb{Q}(\zeta_{20})$.
I also know that $2\mathcal{O}_K = (2, 1 +\sqrt{-5})^{2}$ and $5\mathcal{O}_K = (\sqrt(-5))^2$.
Since $Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ is abelian, the Artin-Symbol of $\mathfrak{P}$ depends only on the underlying prime integer p, for which $(p) = \mathfrak{P} \cap \mathbb{Z}$.
The Artin-symbol of every prime integer $p \equiv 1,3,7,9,11,13,17,19\ mod\ 20$ should be the field-automorphism $\sigma_p$ which maps $\zeta_n$ to $\zeta_n^{p}$.
Further, $<\sigma_3>$ fixes $K$ and $<\sigma_9>$ fixes $L$. Now I know
$p$ splits completely in $K$ resp. $L$ iff $\sigma_p$ fixes $K$ resp. $L$.
Now let $p \equiv 11,13,17,19\ mod\ 20$ a prime integer.
Then $p$ doesn´t split in $K$, which mean (in this case) there is a prime ideal $\mathfrak{p}$ from $\mathcal{O}_K$ lying over $p$ with inertial degree $f(\mathfrak{p}|p) = 2$.
Since $p$ doesn´t split completely in $L$ either, it can only split into one or two prime ideals. But $|<\sigma_{p}|_{L}>| = 2$, where $\sigma_{p}|_{L}$ is the restriciton to $L$. That means that $p$ splits into 2 prime ideals in $L$. That means there a prime ideals $\mathfrak{P}_1,\mathfrak{P}_2$ in $L$ with $p\mathcal{O}_L = \mathfrak{P}_1\mathfrak{P}_2$.
But also $p\mathcal{O}_K = \mathfrak{p}$ with $\mathfrak{p}$ is a prime ideal in $K$.
Then I can see that $f(\mathfrak{P}_i|\mathfrak{p}) = 1$, which means that $\mathfrak{p}$ splits completely in $L$. Is this right so far?
Now let $p \equiv 1,3,7,9\ mod\ 20$ be a prime integer.
$p$ splits completely in $K$, which means there a prime ideal $\mathfrak{p}_1,\mathfrak{p}_2$ in $K$ with $p\mathcal{O}_K = \mathfrak{p}_1\mathfrak{p}_2$. So $f(\mathfrak{p}_i|p) = 1$.
Now for $p \equiv 1,9\ mod\ 20$.
$p$ splits completely in $L$, which means there a prime ideal $\mathfrak{P}_1,\mathfrak{P}_2,\mathfrak{P}_3,\mathfrak{P}_4$ in $L$ with $p\mathcal{O}_L = \mathfrak{P}_1\mathfrak{P}_2\mathfrak{P}_3\mathfrak{P}_4$. So $f(\mathfrak{P}_i|p) = 1$ and hence $f(\mathfrak{P}_i|\mathfrak{p}_j) = 1$.
So every prime ideal $\mathfrak{p}$ in $K$, lying over a prime $p \equiv 1,9\ mod\ 20$ splits completely in $L$, is that right?
When I do the same for $p \equiv 3,7\ mod\ 20$ then I get that every prime ideal $\mathfrak{p}$ in $K$, lying over such a prime integer $p$, doesn´t split in $L$, because $f(\mathfrak{p}|p) = 1$, but also $|<\sigma_{p}|_{L}>| = 2$, which means $p\mathcal{O}_L = \mathfrak{P}_1\mathfrak{P}_2$. Therefore $f(\mathfrak{P}|\mathfrak{p}) = 2$.
According to my calculations only the prim ideals $\mathfrak{p}$ in $K$, which lie over a prime integer $p \equiv 3,7\ mod\ p$ doesn´t split in $L$. can somebody confirm this?
Now, except for those prime ideals lying over 2 and 5, I can tell which prime ideals (dependent on the underlying prime integer) in $K$ splits completely in $L$, which means mapping to the 1 in $Gal(L/K).
Now, how can I check that those prime ideals are exactly the principal prime fractional ideals? And which prime ideals generate $P_K$?
Thanks in advance
Julian
Everything boils down to showing that prime ideals with norm $\equiv 1 \bmod 4$ are principal. This is elementary. You have to show that primes $p$ that split satisfy $p = x^2 + 5y^2$ if $p \equiv 1 \bmod 4$ and $2p = x^2 + 5y^2$ if $p \equiv 3 \bmod 4$.
When you talk about norm, do you mean $\mathcal{N}(\mathfrak{p}) = |\mathcal{O}_K/\mathfrak{p}|$ for a prime ideal $\mathfrak{p}$ in $K$?
Yes. Principal ideals $(x,+y\sqrt{-5})$ have norm $p = x^2 + 5y^2$. The nonprincipal primes lie in the class of the prime ideal above $2$ . . .
Ah, now I see: $p\mathcal{O}_K =(x+y\sqrt{-5})(x-y\sqrt{-5}) = (x^2 +5y^2)$ and this is exactly the case iff $p \equiv 1,9\ mod\ 20$. And for $p \equiv 11,13,17,19\ mod\ 20$ $p\mathcal{O}_K$ remains prime and is principal. Now I only need to compute $\varphi(2\mathcal{O}_K)$ and $\varphi(5\mathcal{O}_K)$.
Two another questions: Is $P_K$ generated by the principal prime ideals and how can I see in which ways $2\mathcal{O}_K,5\mathcal{O}_K$ splits in $L$?
And thank you for your help.
a) No: the product of two nonprincipal prime ideals is principal. b) 2 is inert in one subfields, 5 splits in another.
Then in the next step I want to show that the kernel of $\varphi_{L/K}$ is exactly $P_K$. Because I know excatly which prime ideals maps to $1 \in Gal(L/K$, I can see that every product of principal prime ideals will be mapped to the $1$. Every Product of even many non-principal prime ideals will be mapped to the $1$, too. Because the class number of $K$ is $2$ every product of non-principal ideals must be a principal ideal. Can I conclude from this that every non-principal(resp. principal) Ideal is the product of some principal ideals and uneven(resp. even) man non-principal ideals?
|
2025-03-21T14:48:30.182211
| 2020-03-31T03:06:39 |
356154
|
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|
Stack Exchange
|
A Lefschetz style formula for the $\ell^\infty$ torsion of an Abelian variety over a finite field
Let $A/\mathbb F_q$ be an abelian variety over a finite field. Define $A_\ell = A[\ell^\infty](\mathbb F_q)$, the $\ell^n$ ($n\geq 0)$ torsion points defined over the base field. I can assume $\ell \neq 0$ in $\mathbb F_q$ if necessary.
Is there a Lefschetz style trace formula relating $A_\ell$ to the trace of the Frobenius on some cohomology groups? If we let $a_\ell = |A_\ell|$, then note that:
$$|A(\mathbb F_q)| = \prod_{\ell} (a_\ell)$$
where the right hand side sums over all primes and only finitely many terms are non zero. The left hand side has an interpretation in terms of the action of the Frobenius on the Etale cohomology groups and it would be great if any such relation also lifted to the level of cohomology groups.
Shouldn't the formula involve a product of the $a_\ell$'s? Finite abelian groups are products of the $\ell$-primary parts, and the cardinality of a finite set multiplies under products. Once you make that switch, you can recover $a_\ell$ as from the $\ell$-adic valuation of the trace of Frobenius.
Thanks for the correction! Indeed, I did consider the valuations and it is useful in my context but I would still be interested in a lift of this equality to the level of cohomology.
|
2025-03-21T14:48:30.182348
| 2020-03-31T04:05:34 |
356158
|
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}
|
Stack Exchange
|
Sliding block code on irreducible sofic shift
I was looking at the following exercise from Lind/Marcus book An Introduction to Symbolic Dynamics and Coding that I cannot solve. Can someone give me a hint?
Find an example of a pair of irreducible sofic shifts $X,Y$ and a sliding block code $f:X\rightarrow Y$ such that $f$ is not one-to-one, but the restriction of f to periodic points is one-to-one.
The exercise points to other exercises that I also don't know.
Let $X$ be an irreducible shift of finite type and $f:X\rightarrow Y$ a sliding block code. Prove that $f$ is an embedding if and only if it is one-to-one on the periodic points of $X$.
I know the definitions for sofic, finite type, irreducible and sliding block code; but have no idea how to use them. Why things should be different here for sofic shifts and shifts of finite type?
Glossary: a "sliding block code" is just a continuous equivariant map $f$, i.e., such that $f\circ s_X=s_Y\circ f$ where $s_X$ and $s_Y$ are the shift maps.
This is a fun pair of exercises (the first one you mention is 3.2.9 and the second is 2.3.6a)! For 2.3.6a, recode to a $1$-block code $\phi$ on an irreducible edge shift $X$, suppose that $x, x' \in X$ are such that $x \neq x'$ but $\phi(x) = \phi(x')$. Consider two cases: either $x_j \neq x_j'$ for either all $j > i$ or all $j < i$; or there exist $j < i < k$ with $x_j = x_j'$, $x_k = x_k'$. In either case, try to construct a pair of distinct periodic points whose images agree.
Then, for 3.2.9, here's an idea for a construction: try to construct a sofic shift $X$ by labelling an irreducible graph with alphabet $\{ 0, 1, 2 \}$, such that the path presenting $10^m 1$ can be determined by whether $m$ is even or odd, and such that every cycle (other than self-loops) contains a $2$. Then code from $X$ to $Y$ by replacing $2$ by $1$ and leaving the other symbols the same.
EDIT: I've provided an answer because I didn't notice I was on MO rather than Math.SE, but this question should probably be on Math.SE rather than MO. It's an exercise in an introductory textbook, albeit a textbook with some rather tricky exercises.
"Tricky"? I'd call them "instructive".
|
2025-03-21T14:48:30.182512
| 2020-03-31T04:34:50 |
356160
|
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|
Stack Exchange
|
Fokker-Planck equation with zero diffusion coefficients on the boundaries
I am currently working on a Masters project in which I will be looking at numerical methods for a specific PDE. The PDE that I will be considering is a Fokker-Planck equation on a unit hyper-cube.
More specifically I am studying the following PDE (in non-divergence form)
$$
\frac{\partial u}{\partial t} = \frac{x_{d}(1-x_{d})}{2}\frac{\partial^2u}{\partial x_{d}^2}+\left(1-2x_{d}-M_{d}(x)\right)\frac{\partial u}{\partial x_{d}}-(1-\omega_{d})u,
$$
for all $x \in [0,1]^D$ and $t\in[0,\infty)$. The boundary conditions will be zero flux except at the $(0,...,0)$ and $(1,...,1)$ corners.
where summation over $d$ is assumed on the right side of the equation. We also define
$$
M_d(x)=\sum_{k=1}^{D} M_{dk}(x_k-x_d), \ \ \ \text{and}\ \ \ \omega_d = M_d(x)=\sum_{\substack{k=1\\k\neq d}}^{D} M_{dk}.
$$
The flux vector of this equation is given by
$$
F_d(u,x) = \frac{1}{2}\frac{\partial}{\partial x_d}\left[x_\alpha(1-x_d)u\right]+\sum_{k=1}^{D} M_{dk}(x_k-x_d)u.
$$
I will be using the finite volume method (FVM) to approximate a discrete solution to this PDE. Had this been a constant coefficient problem, or even a positive coefficient problem, I would be fine. However, I am not sure how to deal with the fact that the leading order coefficients go to zero on the boundaries of our domain.
Based on my intuition, I have a hunch that we should be able to show that the solution is a smooth function in the interior of the domain with delta distributions on the boundaries. Sadly, I am not sure how to show this or if this is correct.
Any references or suggestions that you think may be helpful would be appreciated.
Thanks in advance for the help!
This model has been intensively studied in the literature and goes back to Wright and Fischer. In the language of semigroup theory you find an analytic treatment in some papers by P. Clement and S. Cerrai. Just to indicate a place where you can find many references. The solution is smooth in the interior because of local parabolic regularity.
You still have an unresolved issue about boundary conditions, even in dimension 1: Sure, wherever the diffusion coefficient does not vanish you can impose zero-flux, but in dimension 1 there is clearly a problem as the whole boundary is "degenerate" in that respect. In fact the very meaning of weak solutions is unclear and you should chose a definition of weak solution according to your needs/application (HINT: the boundaries will have indeed singular Dirac terms, and one can IMPOSE extra conservation laws in order to get a unique solution in some reasonable sense)
|
2025-03-21T14:48:30.182695
| 2020-03-31T04:48:42 |
356161
|
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|
Stack Exchange
|
On Pitt's inequality (weighted Fourier inequality)
One of Pitt's Theorem (from "Theorems on Fourier Series" by H R Pitt, 1937) states that for an integrable periodic function $F$ over $[-\pi,\pi]$,
$$
\sum_{n=1}^{\infty} |a_n|^q n^{-q\lambda} \leq K(p,q,\lambda) \int_{-\pi}^{\pi}|F(\theta)|^p|\theta|^{p\alpha}d\theta,
$$
where, $a_n$'s are the Fourier series coefficients, K is independent of $F$, $1<p\leq q <\infty$, ${1}/{p} + {1}/{p'}=1$, $0\leq \alpha < 1/p'$ and $\lambda = 1/p + 1/q + \alpha -1 \geq 0$.
Another theorem for Fourier transforms (from "Pitts Inequality with sharp convolution estimates" by W. Beckner, 2007) states that for a function $f$ in $L^2(R)$,
$$
\int_{R} \phi\left\{1/|x|\right\}|f(x)|^2 dx \leq C_{\phi} \int_{R} \phi\left\{|x|\right\}|\hat{f}(x)|^2 dx,
$$
where $\hat{f}$ is the FT of $f$, and $\phi$ is an increasing function.
Is there a way to obtain the first theorem from the second?
|
2025-03-21T14:48:30.182792
| 2020-03-31T05:11:01 |
356162
|
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"Bugs Bunny",
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|
Stack Exchange
|
Order of product of group elements
Let $G$ be a finite non-commutative group of order $N$, and let $x, y \in G$. Let $a$ and $b$ be the orders of $x$ and $y$, respectively. Can we say anything non-trivial about the order of $xy$ in terms of $a$ and $b$? If it helps, you may assume that $a$ and $b$ are coprime.
No, it is more-or-less answered here: https://mathoverflow.net/questions/130697/how-to-find-quotients-of-infinite-triangle-groups-or-von-dyck-groups
Only in the comments, to be fair...
To be more explicit: for all three positive integers $a,b,c$, there exist a finite group and elements $x,y$ such that $x,y,xy$ have order $a,b,c$, with the only trivial restriction: if one of $a,b,c$ is $1$, then the other two are equal.
Let me also mention that the case when $a,b,c$ are all even is trivial: one can then find three such elements in the direct product of three dihedral groups $D_a\times D_b\times D_c$, each with generators of order two $u,v$, namely $x=(uv,u,v)$, $y=(v,uv,u)$, $xy=(u,v,vu)$.
Given $a,b,c\ge 2$, estimating/computing the smallest size $n$ such that there is such a pair in the symmetric group $S_n$, looks like an interesting problem. For $a,b,c$ even, the above gives an upper bound of $a+b+c$.
YCor, your example is interesting to me, but I didn't quite understand how $u$ and $v$ are picked. I understand that you can pick $u$ to be a generator of order 2 (this represents a reflection), but how are there two distinct generators of order 2? In the dihedral group of order $2n$, shouldn't there be one generator of order 2 and one of order $n$ (the rotation part)?
The following theorem (which does not take the order $N$ of the group $G$ into account) shows that all possible combinations of $a$, $b$ and the order of $xy$ are possible. See Theorem 1.64 from Milne's course notes on group theory.
For any integers $a,b,c > 1$, there exists a finite group $G$ with elements $x$ and $y$ such that $x$ has order $a$, $y$ has order $b$, and $xy$ has order $c$.
:what can we say about such a group,....it was asked in exam that "consider such group with above property,then G, (1)cannot be cyclic,(2) G has to non abelion......which one is true,1) or 2)?
I strongly suspect that the answer is "very little", unless you are looking for bounds on the order of $xy$ in terms of $N$ or something like that.
Example. Let's focus on the example $a = 2$ and $b = 3$. It is well-known that
\begin{align*}
\mathbf Z/2 * \mathbf Z/3 &\stackrel\sim\to \operatorname{PSL}_2(\mathbf Z)\\
x &\mapsto \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix},\\
y &\mapsto \begin{pmatrix}0 & 1 \\ -1 & -1\end{pmatrix},
\end{align*}
where $x$ and $y$ are the generators with $x^2 = 1$ and $y^3 = 1$. The product $xy$ maps to
$$\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}\begin{pmatrix}0 & 1 \\ -1 & -1\end{pmatrix} = \begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix},$$
which has infinite order. For any $n \in \mathbf N$, the surjection
$$\operatorname{PSL}_2(\mathbf Z) \twoheadrightarrow \operatorname{PSL}_2(\mathbf Z/n)$$
gives a finite group of order roughly $n^3$ in which $xy$ has exact order $n$.
Example. Another example, still with $a = 2$ and $b = 3$, is the group
$$G = \mathbf Z/n \wr S_3 = (\mathbf Z/n)^3 \rtimes S_3,$$
where multiplication is defined by
$$(a_1,a_2,a_3,\sigma)(b_1,b_2,b_3,\tau) = (a_1+b_{\sigma(1)},a_2+b_{\sigma(2)},a_3+b_{\sigma(3)},\sigma\tau).$$
Then one easily checks that $x = (1,-1,0,(12))$ has order $2$ and $y = (0,0,0,(123))$ has order $3$, and $xy = (1,-1,0,(23))$ has order $2n$. This time we get any even number as the order of $xy$, inside a group of order $6n^3$.
|
2025-03-21T14:48:30.183032
| 2020-03-31T07:50:26 |
356171
|
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"Maxime Lucas",
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|
Stack Exchange
|
Cocomplete and finitely complete category with nice pullbacks that is not locally presentable
I have a result that holds for cocomplete and finitely complete categories such that pullbacks preserve directed colimits, by which I mean $A \times_B (\operatorname{colim}_{i \in I} C_i) = colim_{i \in I} (A \times_B C_i)$, where $I$ is a directed category.
Thanks to a result by Adamek and Rosicky [locally presentable and accessible categories, Proposition 1.59] , I know that this holds in any locally (finitely) presentable category.
Is there any other category that satisfies those conditions?
If you look at the article on quasitoposes, which are locally cartesian closed and therefore satisfy your exactness condition, you'll find a number of examples that are not locally presentable. For example, the category of pseudotopological spaces, the category of bornological sets, the category of equilogical spaces, and (I think) the category of quasitopological spaces in the sense of Spanier. I might think of more (preferably a whole class of more) later.
All responses are helpful, but I'll mark this one as accepted because the examples feel more natural to me. Thank you!
Filtered colimits commute with finite limits in any Grothendieck topos. A Grothendieck topos does not need to be locally finitely presentable; the presentability rank of a topos is tightly related to the structure of its site presentation, as shown in Prop. 5.5 of the preprint
Gabriel-Ulmer duality for topoi and its relation with site presentations, Ivan Di Liberti and Julia Ramos González, arXiv:1902.09391.
Indeed I must confess a conflict of interests, as I am one of the authors of that preprint.
This looks interesting, thank you! Do you have an example of a "weakly k-ary site" for k > aleph_0 ?
Actually Shulman's paper contains lots of examples. Thanks!
Every localization (= full reflective subcategory such that the reflector preserves finite limits) of a locally finitely presentable category satisfies this property. More can be found in Localisation of locally presentable categories (Brian Day and Ross Street, 1989).
|
2025-03-21T14:48:30.183190
| 2020-03-31T10:44:34 |
356181
|
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"url": "https://mathoverflow.net/questions/356181"
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|
Stack Exchange
|
Weakly symmetric rings and derived equivalences
A ring $R$ with Jacobson radical $J$ is called Frobenius in case $R/J \cong soc(R)$ as left and right $R$-modules and weakly symmetric in case we even have $R/J \cong soc(R)$ as $R$-bimodules.
Question 1: Is there a classification of representation-finite Frobenius rings (up to some other classification such as division rings)?
In particular, what are the representation-finite local (or commutative) Frobenius rings? (I guess local implies weakly-symmetric?)
Question 2: Is being weakly symmetric closed under derived equivalences?
(this is true for finite dimensional algebras)
|
2025-03-21T14:48:30.183259
| 2020-03-31T11:02:56 |
356183
|
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"url": "https://mathoverflow.net/questions/356183"
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|
Stack Exchange
|
Maximum of sum of exponential function
Let $x_1,\dots,x_n$ be a set of given vectors in $\mathbb{R}_{+}^d$. Let $c_1,\dots,c_n$ be given positive constants. I am interested in finding the vectors $w_1,\dots,w_n$ in $\mathbb{R}_{+}^d$ that solves the optimization problem
\begin{align}
\max_{w_1,\dots,w_n}\sum_{i=1}^{n}c_i\frac{\exp(w_i^Tx_i)}{\sum_{j=1}^{n}\exp(w_j^Tx_i)}
\end{align}
I am not even sure how to start around this and I currently use a out-of-the-box optimization algorithm. Is this type of problem known in literature? What would be some starting point?
Here is an approach which is quite possibly sub-optimal, but I hope it helps.
Let me use a special form of AM-GM inequality (see ``Proofs from the Book'' for a beautiful proof) to get a lower bound on you cost function. Given positive numbers $\{a_1,\cdots, a_n\}$ and positive numbers $\{p_1,\cdots, p_n\}$ such that $\sum^n_{k=1}p_k = 1$, the following is true:
$$
p_1a_1+\cdots+p_na_n \geq a_1^{p_1}a_2^{p_2}\cdots a_n^{p_n}.
$$
The same applied inequality substituting $a_i = \frac{\exp(w_i^Tx_i)}{\sum_{j=1}^{n}\exp(w_j^Tx_i)}$ and $p_i = \frac{c_1}{\sum^n_{k=1}c_k}$, we get:
$$
\sum_{i=1}^{n}\left(\frac{c_1}{\sum^n_{k=1}c_k}\right)\frac{\exp(w_i^Tx_i)}{\sum_{j=1}^{n}\exp(w_j^Tx_i)} = \sum_{i=1}^{n}\tilde{c_i}\frac{\exp(w_i^Tx_i)}{\sum_{j=1}^{n}\exp(w_j^Tx_i)} \geq \prod_{i=1}^n \left(\frac{\exp(w_i^Tx_i)}{\sum_{j=1}^{n}\exp(w_j^Tx_i)}\right)^{\tilde{c_i}}.
$$
Maximizing the last term in the last inequalities would serve as a relaxation to the original problem. Towards that end, we maximize instead the $\log$ of the last term, since composing wtth a monotonic function does not change optima. Thus, the relaxed problem is given by (post $\log$):
$$
\max_{w_1,\dots,w_n} \left\{\left(\sum^n_{i=1}\tilde{c_i}\left(w_i^Tx_i\right)\right) - \sum^n_{i=1}\tilde{c_i}\log\left(\sum_{j=1}^{n}\exp(w_j^Tx_i)\right)\right\}, ~\mbox{subject to}~ w_i\geq 0~\forall i.
$$
The above optimization problem is concave and can be solved easily using CVXPY, or optimizers alike.
As far as the extent of sub-optimality is concerned, note that $\sum_{i=1}^n c_i$ is a natural upper bound for the cost function. Thus the relaxation gap be at most the difference between $\sum_{i=1}^n c_i$ and the value at the $\arg\max$ of the relaxed problem. Also, if the relaxed problem is unbounded, the original problem is unbounded as well.
|
2025-03-21T14:48:30.183413
| 2020-03-31T11:08:29 |
356184
|
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|
Stack Exchange
|
Tangent space to subspace of orbit in jet spaces
I consider a map germ $f: (\mathbb{R}^n,0) \to \mathbb{R} $ which is $k$-determined for some $k \in \mathbb N$, i.e. for all map germs $g: (\mathbb{R}^n,0) \to \mathbb{R} $ having the same $k$-jet as $f$ , there exists a (local) diffeomorphism $\varphi: (\mathbb{R}^n,0) \to (\mathbb{R}^n,0)$ such that $g = f \circ \varphi$. Now consider the set
\begin{align*} L := \{j^{k+1}(f+h) ~|~ h \in m^{k+1} \} \subseteq J^{k+1}_0(\mathbb R^n, \mathbb R),\end{align*}
where $m^{k+1} = \{h : (\mathbb{R}^n,0) \to \mathbb{R}~|~ \frac{\partial^{\alpha}}{\partial x^{\alpha}}h(0) = 0, \forall \alpha \leq k \} $ and $J^{k+1}_0(\mathbb R^n, \mathbb R)$ is the $k$-jet space with source $0 \in \mathbb R^n$.
In the book "Singularities of Mappings" by David Mond and Juan J. Nuno-Ballesteros on page 48 they claim that for $z = j^{k+1}f(0)$ the tangent space of $L$ is
\begin{align*} T_zL = m^{k+1}/m^{k+2}. \end{align*}
I don't see how this equality is obtained. How is the tangent space of $L$ at $z$ the space of monomials of degree $k+1$? Can anybody help me?
Those are not monomials. The space $m^{k+1}/m^{k+2}$ is the space of functions vanishing up to order $k$, modulo those vanishing up to order $k+1$, so Taylor series of homogeneous polynomials of degree exactly $k+1$. The space $L$ is, as you defined it, the translate of $m^{k+1}/m^{k+2}$ by $j^{k+1}f(0)$ inside $J^{k+1}_0$, i.e. $k+1$-jet of $f$ up to translation by $k+1$-jet of $h$.
The letter $m$ stands for maximal: the functions vanishing at the origin form a maximal ideal.
Thank you for your correction and explanation. So I can identify $L$ with the set ${ j^{k+1}f(0)} + m^{k+1}/m^{k+2}$ which itself can then be identified with $m^{k+1}/m^{k+2}$?
Yes, that seems to be it.
Just one last question: Why did you add the explanation for $m$? Is there a need to use the maximality of $m$ here? I am asking since for $k>1$ $m^k$ is not maximal anymore.
Maximality is not used here. I wanted to make clear that m stands for maximal, not for monomial, because I thought that maybe the use of the letter m led to the idea that these might be monomials.
|
2025-03-21T14:48:30.183703
| 2020-03-31T13:24:33 |
356188
|
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|
Stack Exchange
|
What happens if you strip everything but the “between” relation in metric spaces
Given a metric space $(X,d)$ and three points $x,y,z$ in $X$, say that $y$ is between $x$ and $z$ if $d(x,z) = d(x,y) + d(y,z)$, and write $[x,z]$ for the set of points between $x$ and $z$.
Obviously, we have
$x,z\in[x,z]$;
$[x,z]=[z,x]$;
$y \in [x,z]$ implies $[x,y] \subseteq [x,z]$;
$w,y \in [x,z]$ implies: $w\in [x,y]$ iff $y \in [w,z]$.
My question:
Has the family of objects with such an axiomatic “interval structure” $[\bullet,\bullet]:X \times X \to \mathcal{P}(X)$ satisfying approximately the above conditions been studied somewhere?
There is a bit of that in Busemann's Geometry of Geodesics because of its relation to inverse problems: determine all metrics which have the same geodesics (or same "between" structure: set of triplets $(x,y,z) \in X \times X \times X$ such $y$ is between $x$ and $z$). If you allow more structure (manifolds, foliations, etc.) this is basically the study of path geometries.
Thanks for the comment: I'll have a look there too!
You might find Knuth's Axioms and Hulls of interest. Not only betweenness on an interval but also cyclic order as well. For betweenness the main idea is ordered geometry.
I'll havea look at Knuth's book, thanks!
And indeed, this seems similar in flavour to median spaces/algebra, but, well, they usually use all the "medianness" restrictions. I'll also have a look at Bowditch's book, thanks!
Since Bowditch has been mentioned you may well be aware of this, but a special case of these axioms is given by pretrees. These can be defined by the axioms: (a) $[x,y]=[y,x]$; (b) $y\in[x,z]$ and $z\in[x,y]$ implies $y=z$; and (c) $[x,z]\subseteq[x,y]\cup[y,z]$. The last axiom ensures the structure is treelike, and is of course not shared by most metric spaces. See Bowditch's Memoir of the AMS Treelike structures arising from continua and convergence groups and Adeleke and Neumann's Memoir Relations related to Betweenness.
No, I didn't know about that either but it looks interesting too! thanks for the references!
@shane.orourke: I can't tell at a first glance: Is it possible for a pretree not to be metrizable (maybe allowing for infinite-valued metrics)?
@user148575 Pretrees aren't necessarily metrisable, at least not if the distances are required to be real numbers. Any linearly ordered set admits a natural pretree structure, so sufficiently large losets are counterexamples. If you are prepared to generalise the definition of metric space to allow distances in a pre-assigned linearly ordered abelian group, then every pretree admits a metric where each $[x,y]$ consists of the points between $x$ and $y$ in the sense of the question.
There is a wide body of work on this in connection with the classic De Bruijn–Erdős theorem.
De Bruijn–Erdős Theorem. Every set of $n$ points in the
plane (not all lying on the same line) determine at least $n$ lines.
There is a beautiful conjecture of Chen and Chvátal that the De Bruijn–Erdős theorem actually holds in every metric space, where lines are defined using the notion of betweenness that you describe. That is, given a metric space $M$ and two points $a,b \in M$, the line determined by $a$ and $b$ is the set of points $c$ such that $c$ is between $a$ and $b$, or $a$ is between $c$ and $b$, or $b$ is between $a$ and $c$. Note that this definition reduces to the usual notion of lines if we use the Euclidean distance.
Chen-Chvátal Conjecture. Every set of $n$ points in a metric space (not all of which lie on the same line) determine at least $n$ lines.
If you Google 'Chen-Chvátal Conjecture' you will find many results, including some that focus on the combinatorial aspects of betweenness.
This looks to be what I was looking for, thanks!
Here is a link to Chen-Chvátal's paper: https://www.sciencedirect.com/science/article/pii/S0166218X07001710?via%3Dihub
I'm reminded of W.A. Coppel's book which looks at these kinds of structures from a slightly different vantage point, namely closure systems. I can't actually find the book right now, but here's a quick synopsis of what's going on.
Given a set $X$ together with an arbitrary function $[\bullet,\bullet] : X^2 \rightarrow \mathcal{P}(X),$ let's call $A \subseteq X$ closed if and only if for all $a,b \in A$, we have that $[a,b] \subseteq A$. It's easy to show that an arbitrary intersection of closed sets is closed, or in other words the collection of closed sets forms a Moore family. This gives us a closure operator on $\mathcal{P}(X)$, which makes $(X,\mathcal{P}(X))$ into a closure system. This means that the collection of closed sets forms a complete lattice given by $$\bigwedge_{i \in I} C_i := \bigcap_{i \in I} C_i, \qquad \bigvee_{i \in I} C_i := \mathrm{cl}\left(\bigcup_{i \in I} C_i\right).$$
If we furthermore assume that $[a,a] = \{a\},$ which is true in an arbitrary metric space (though I'm not sure whether this follows from your axioms), then we can prove that singletons are closed sets. Thus we're allowed to consider the expression $\{a\} \vee \{b\}$ in the aforementioned lattice. This set clearly includes $[a,b]$, and I think your axioms probably imply that these sets are equal. Ergo the Moore family alone allows us to recover the original between-relation.
I think that if you take the first three axioms you've written down, and add in the axiom $[a,a] = \{a\},$ the resulting axiom list should provide a complete characterization of closure systems $X$ in which firstly, every singleton is closed, and secondly, for all $A \subseteq X$, we have that $A$ is closed if and only if for all $a,b \in A$, we have $\{a\} \vee \{b\} \subseteq A$. But you should check this explicitly rather than taking my word for it, because right now this is just a hunch.
I'll also remark that you might learn interesting things by looking at partially ordered sets and defining $[a,b] := \{x \in X : a \leq x \leq b\} \cup \{x \in X : b \leq x \leq a\}.$
Ah, this seems actually closer to what I had in mind, thank you!
The axiom $[a,a] = {a}$ is indeed natural and, combined with my third one, implies the first, so should probably be used instead.
@user148575, glad I was able to help. These are exactly the kinds of questions I love by the way; looking at old structures (e.g. metric spaces) via new categories (e.g. between-relation spaces), fresh points of view on familiar ideas, and the axiomatic method at its simplest and most appealing. This is how things get better.
This is more of an extended remark than a full answer.
My message is:
The property $[x,x]=\{x\}$ does not necessarily hold, but you can always make it hold by taking a natural quotient.
Your axioms don't imply $[x,x]=\{x\}$, so you have to add it if you want it.
A pseudometric (e.g. a seminorm) will satisfy your axioms, and $[x,x]$ is the set of points distance zero from $x$, and this set can have any size.
In a pseudometric space you can take the quotient by the relation $x\sim y\iff d(x,y)=0$ and you get a metric space.
It turns out the same works with the interval systems.
We can define a relation $\sim$ on $X$ so that $x\sim y\iff x\in[y,y]$.
This is an equivalence relation:
Reflexivity:
Axiom 1.
Symmetry:
By axioms 1 and 2 $x,y\in[y,x]$.
Thus axiom 4 says $x\in[y,y]\iff y\in[x,x]$.
Transitivity:
Suppose $x\sim y\sim z$.
By axiom 3 $x\sim y$ and $y\sim z$ imply $[y,x]\subset [y,y]$ and $[z,y]\subset [z,z]$.
By axioms 2 and 3 $[y,y]\subset[z,y]$.
Thus $x\in[y,x]\subset[y,y]\subset[z,y]\subset[z,z]$.
Then we can take the quotient space $X/{\sim}$.
To avoid confusion, let's denote by $(x)$ the quotient set of $x\in X$.
(In fact, $(x)=[x,x]$, but I find it cleaner to keep the separate notation.)
Then declaring $(x)\in[(y),(z)]\iff x\in[y,z]$ defines an interval structure on $X/{\sim}$.
To see this, we need to check two things:
$[x,y]\subset[x',y]$ when $x\sim x'$.
(The reverse inclusion and varying both $x$ and $y$ follow easily.)
Proof: Since $x\in[x',x']\subset[x',y]$, we have indeed $[x,y]\subset[x',y]$.
$x\in[z,y]\implies[x,x]\subset[z,y]$.
Proof: Since $x\in[z,y]$ and $x\in[x,y]$, applying axiom 3 twice gives $[x,x]\subset[x,y]\subset[z,y]$.
In the quotient structure we evidently have $[(x),(x)]=\{(x)\}$.
Interesting, thanks for the remark!
|
2025-03-21T14:48:30.184490
| 2020-03-31T13:26:44 |
356189
|
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"Dylan Wilson",
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|
Stack Exchange
|
Where can I read about non-principal obstruction theory?
Most treatments of obstruction theory assume a principal Postnikov tower. I have the vague sense that if one uses cohomology with local coefficients, one does not need to make any assumptions on one's Postnikov tower. I've always assumed that if I ever needed this, I could look it up in a standard reference. That day has come, and I can't seem to find this theory in Hatcher, Whitehead, or Spanier.
Whitehead's Elements of Homotopy Theory comes closest -- Chapter VI.5 does some kind of obstruction theory with local coefficients, but even there, right after Theorem 5.2, he makes some kind of simplicity assumption about his fibration.
Questions:
Does there really exist a form of obstruction theory which works for an arbitrary fibration, with no principality or simplicity assumptions?
If so, where can I read about it?
If not, is there at least a form of obstruction theory which works for any fibration with, say, simply-connected fibers? The assumptions I've come across all seem to require that the fundamental group of the codomain acts trivially on some homotopy groups of the fiber, so that connectivity assumptions on the fiber don't seem to be enough for the theory to work.
Ideally, I'd like a full relative obstruction theory -- i.e. a theory of finding homotopy-coherent lifts for any commutative square.
Have you looked at Olum's "Obstructions to extensions and homotopies" paper? Apparently it does obstruction theory for fibrations with non-simple fibers.
@MarkGrant I see, thanks. I now realize that my question is really a duplicate of this one.
it's overkill, but if you just want some reference there is HTT.<IP_ADDRESS>, and its corollary <IP_ADDRESS>, which you can apply in the case $\mathcal{X}=\mathsf{Spaces}{/X}$ to the n-gerbe obtained by pulling back $Y{\le n}$ along an already constructed map $X \to Y_{\le n-1}$.
@DylanWilson Thanks! I think being able to see the ideas packaged in multiple different ways will be very helpful for me!
The book by H.-J. Baues on obstruction theory might also be helpful.
|
2025-03-21T14:48:30.184671
| 2020-03-31T13:32:31 |
356190
|
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|
Stack Exchange
|
Newman's proof of the prime number theorem
I am teaching a graduate course in Complex Analysis and I am covering Newman's proof of the prime number theorem. I have been using the simplified version in the papers of
Zagier and Korevaar. However, I ran into a problem.
Both papers rely on this theorem:
Theorem
Let $f:[0,\infty)\rightarrow\mathbb{C}$ be bounded and
locally integrable and let
$$
g(z):=\int_{0}^{\infty}f(t)e^{-tz}dt,\quad\operatorname{Re}z>0.
$$
Assume that for every $z\in\mathbb{C}$ with $\operatorname{Re}z=0$ there
exists $r_{z}>0$ such that $g$ can be extended holomorphically to $B(z,r_{z}
)$. Then the generalized Riemann integral
\begin{equation}
\int_{0}^{\infty}f(t)\,dt \label{pn1}
\end{equation}
is well-defined and equals $g(0)$.
This theorem is used to prove that the generalized Riemann integral
$$
\int_{1}^{\infty}\frac{\theta(x)-x}{x^{2}}dx
$$
converges. Here, $$
\theta(x):=\sum_{p\text{ prime}\leq x}\log p,\quad x\in\mathbb{R}.
$$
Everything is fine up to this point. Then the authors use the convergence of this integral to prove that
\begin{equation}
\lim_{x\rightarrow\infty}\frac{\theta(x)}{x}=1. \label{pn limit theta}%
\end{equation}
Their proof is as follows:
Assume by contradiction that
$$
\limsup_{x\rightarrow\infty}\frac{\theta(x)}{x}>1.
$$
There there exists an increasing sequence $x_{n}\rightarrow\infty$ such that
$\theta(x_{n})>(1+\varepsilon)x_{n}$ for all $n\in\mathbb{N}$ and for some
$0<\varepsilon<1$. Since $\theta$ is increasing, if $x>x_{n}$, $\theta
(x)\geq\theta(x_{n})>(1+\varepsilon)x_{n}$, and so
\begin{align*}
\int_{x_{n}}^{(1+\varepsilon)x_{n}}\frac{\theta(x)-x}{x^{2}}dx & \geq
\int_{x_{n}}^{(1+\varepsilon)x_{n}}\frac{(1+\varepsilon)x_{n}-x}{x^{2}}dx\\
& =\int_{1}^{(1+\varepsilon)}\frac{(1+\varepsilon)-s}{s^{2}}ds>0
\end{align*}
where we made the change of variables $x=x_{n}s$ so $dx=x_{n}ds$. Since
$x_{n}\rightarrow\infty$, by selecting a subsequence we can assume that
$x_{n+1}\geq2x_{n}$ for all $n$. Hence, by summing all the disjoint integrals
on the left-hand side we obtain that
$$
\int_{\bigcup(x_{n},(1+\varepsilon)x_{n},)}\frac{\theta(x)-x}{x^{2}}%
dx=\sum_{n=1}^{\infty}\int_{x_{n}}^{(1+\varepsilon)x_{n}}\frac{\theta
(x)-x}{x^{2}}dx\\=\sum_{n=1}^{\infty}\int_{1}^{(1+\varepsilon)}\frac
{(1+\varepsilon)-s}{s^{2}}ds=\infty.
$$
The papers claim that this fact contradicts the fact that the integral converges and proves that
$$
\limsup_{x\rightarrow\infty}\frac{\theta(x)}{x}\leq1.
$$
However, this is not the case since all we know is that
$$
\lim_{T\rightarrow\infty}\int_{1}^{T}\frac{\theta(x)-x}{x^{2}}dx=\ell
\in\mathbb{R}%
$$
but this does not prevent that
$$
\int_{1}^{\infty}\frac{(\theta(x)-x)^{+}}{x^{2}}dx=\int_{1}^{\infty}%
\frac{(\theta(x)-x)^{-}}{x^{2}}dx=\infty.
$$
The typical example is
$$
\int_{1}^{\infty}\frac{\sin x}{x}dx,
$$
which exist as an improper Riemann integral but not as Lebesgue integral.
Am I missing something? If not, is there a correct proof?
It doesn't seem to me that either article follows the line of reasoning as you have presented it. Indeed, we do not take the integral over the union of integrals $(x_n,(1+\varepsilon)x_n)$. We do get that the integral over those intervals is infinite, but you correctly note this does not give a contradiction. Instead, the argument goes as follows.
Let me denote
$$F(T)=\int_{1}^{T}\frac{\theta(x)-x}{x^{2}}dx,\quad C=\int_{1}^{(1+\varepsilon)}\frac{(1+\varepsilon)-s}{s^{2}}ds$$
so that $F(T)$ converges and $C$ is a positive constant. Since $F(T)$ converges, it is in particular Cauchy, so for large enough $x,y$ we have $|F(x)-F(y)|<C$. But for any $x_n$ we have $F((1+\varepsilon)x_n)-F(x_n)\geq C$ by the calculation you present, and for large $x_n$ this is the desired contradiction.
You are right. Thanks!
|
2025-03-21T14:48:30.184909
| 2020-03-31T13:45:06 |
356191
|
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|
Stack Exchange
|
A topological vector space $X$ is separable if its dual space $X^*$ is separable?
Let $(X,\tau)$ be a topological vector space such that the associated dual space $X^*$ is separable. Can we say that $X$ is separable ?
I know that this property is valid for Banach spaces but for topological vector spaces, I have no idea.
Just consider a non-separable $X$ with $X^*={0}$. Maybe you want $X$ to be locally convex.
Since we know this is true for Banach spaces and false for a rather exotic lcs example, I am tempted to ask if there is a positive answer for some of the central classes of well-behaved spaces—the obvious candidates are the (lc) Fréchet spaces and $(DF)$-spaces.
YCor has given a counterexample for topological vector spaces. The statement is still false for locally convex spaces. Consider the space $X$ defined to be a locally convex coproduct of $\newcommand{\R}{\mathbb{R}}\R$, $\R$-many times. So $X$ is the space of functions $\R \rightarrow \R$ that are zero except for on a finite subset of $\R$, equipped with the locally convex coproduct topology.
The dual space $X^* \cong \R^\R$ by the usual pairing of an element of $X$ with an element of $\R^\R$ by summation, and for the topology it doesn't matter whether we take the strong dual topology or the weak-* topology, either way we get the usual product topology on $\R^\R$. This space is separable, because the polynomial functions with rational coefficients $\R \rightarrow \R$ forms a countable dense subset.
However, $X$ itself is not separable. If $X$ were separable, we would have some countable set $D \subseteq X$ such that each $f \in \R^\R$ is determined by its values when paired with each $\phi \in D$. But since each element of $D$ has finite support, $D$ only determines countably many "coordinates" in $\R^\R$, so we could pick two $f_1,f_2 \in \R^\R$ agreeing on those elements of $\R$ but not being equal. This contradicts $D$ being dense.
Instead of an $\R$-fold coproduct we could have taken any $\kappa$-fold coproduct where $\aleph_0 < \kappa \leq 2^{\aleph_0}$, by the Hewitt-Marczewski-Pondiczery theorem.
Very cool example!
For $X^*$ to be ${\mathbb R}^{\mathbb R}$, every linear functional on $X$ must be continuous. That is, the initial topology must be locally convex. Is this obvious?
@მამუკაჯიბლაძე That depends on what you mean by "initial topology". It is true that every linear functional is continuous (and it has to be, to be a coproduct). However, the topological vector space initial topology and locally convex initial topology are not identical for uncountable coproducts. This is why I specified the "locally convex coproduct topology" (which should be considered as one word, rather than as saying that the copduct topology is locally convex). See Schaefer's Topological Vector Spaces section II.6. The coproduct is called the "locally convex direct sum" there.
Thank you for the explanation. It is subtler than I expected, I thought you meant that the topological vector space coproduct is locally convex in this case, which is presumably not true.
@მამუკაჯიბლაძე In terms of counterexamples, in Schaefer it is just not stated that the two are the same, and the fact that they are the same for a $\aleph_0$-fold coproduct is set as an exercise. But in this answer and this comment by Paul Garrett links are provided to counterexamples for the case of an uncountable coproduct.
Let $\Gamma$ be a set and consider the product measure $\mu^{\otimes \Gamma}$, where $\mu$ is the Lebesgue measure on the unit interval. For any $p\in (0,1)$, the space $L_p(\mu^{\otimes \Gamma})$ has trivial dual because $\mu^{\otimes \Gamma}$ is atomless. However, when $\Gamma$ is uncountable, the space $L_p(\mu^{\otimes \Gamma})$ is non-separable.
|
2025-03-21T14:48:30.185156
| 2020-03-31T14:10:00 |
356196
|
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|
Stack Exchange
|
More formulas for joint entropy and for trace form entropies
Linked to some applications of entropy to combinatorics I'm looking for formulas expressing the joint entropy of two r. v. as a function of the conditional entropy . For example
For BWS extensive entropy H : $ H(X,Y) = H(X) + H(Y[X) $
For q-Tsallis non extensive entropy H : $ H(X,Y) = H(X) + H(Y|X) + (1-q)\cdot H(X) \cdot H(Y|X)$
These formula having generalisations for $n$ random variables $X_1, X_2,...,X_n$
There are many different variants of trace form non extensive entropies. To name a few:
Abe, Kaniadakis, Curado, Ubriaco, Shafee, etc....
I'm looking for smilar formulas but for other trace form entropies (different from BWS or Tsallis)
Thanks in advance for any advise or web pointers on dedicated papers
Can you give some references to where those other entropies are defined/used?
Yes, here is a good overview of generalized entropies:
JM Amigo
If you want to see basic on. how to use entropy functionals in combinatorics:
David Galvin
|
2025-03-21T14:48:30.185259
| 2020-03-31T15:04:58 |
356197
|
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|
Stack Exchange
|
Translates of abelian subvarieties
Suppose $A$ is an abelian variety over an algebraically closed field $k$. I'm confused about the notion of translates of abelian subvarieties. I looked over a few related papers, but I couldn't find a precise algebro-geometric definition. People simply used $x+B$, and I assume they are interested in functor of points. In my opinion, given a point $x:\mathrm{Spec}k'\rightarrow A$, a translate should live in the base change $A_{k'}$. Namely there is an abelian subvariety $B$ of $A_{k'}$ such that $x+B$ is our translate.
If $x$ is a closed point of $A$ and $B\subset A$, then $x+B$ would be a closed subvariety of $A$.
If $x$ is not closed, a translate through $x$ is really something lives in a base change of $A$ (so that $x$ becomes a closed point in the base change).
I'm wondering about the following question:
Mordell exceptional locus of a closed subvariety $X\subset A$ is defined as the union of images of positive-dimensional translates inside $X$. Do we need to consider the translates through non closed points? Namely, if we take the union of translates through closed points, will that give us the same thing?
Usually, when one speaks of "translate of an abelian subvariety" it really is in the sense "translate of an abelian subvariety by a closed point". To answer your question: Let $X\subset A$ be a closed subvariety over an abelian variety $A$ over $k$. Assume $k$ is algebraically closed (of characteristic zero?). Let $Sp(X)$ be the union of positive-dimensional translates (by closed points of $A$) of abelian subvarieties contained in $X$. Let $L/k$ be an extension of algebraically closed fields. Then $Sp(X)_L = Sp(X_L)$, where $Sp(X_L)$ is the union of ...
...the positive-dimensional translates of abelian subvarieties by closed points of $A_L$ contained in $X_L$. Thus, the Mordell exceptional locus behaves well with respect to field extensions. You can prove this using spreading out and specialization arguments; see also Proposition 3.7 in https://arxiv.org/abs/1909.12187
@AriyanJavanpeykar: That's really helpful! Thanks a lot! So I assume if we want an argument as Proposition 3.7, we would need to know $Sp(X_L)$ is closed？Do you think there is a way to do it without assuming closedness of the set (so $Sp(X)_L$ just means the inverse image of $Sp(X)$ in $X_L$)?
@AriyanJavanpeykar: Also, if you have a nontrivial map from a group variety to $X$, would it factor through $Sp(X)$?
The fact that $Sp(X)$ is closed in $X$ was proven first by Kawamata; see Thm. 4 in Y. Kawamata, On Bloch’s conjecture, Invent. Math. 57 (1980), 97-100. Ueno subsequently proved that $Sp(X)\neq X$ if and only if $X$ is of general type. Similar statements are true for closed subvarieties of semi-abelian varieties; see Abramovich http://www.numdam.org/item/CM_1994__90_1_37_0/ ...cont'd below...
If you want to avoid using the fact that Sp(X) is closed, you will have to think a bit harder. I will look into it once I have some more time if you really want to know if it is possible to argue without using that Sp(X) is closed....con't below...
Finally, I claim that Sp(X) equals the "groupless-exceptional locus" as defined in Definition 12.4. of https://arxiv.org/abs/2002.11981 . To prove this, you have to combine some of the statements in arxiv.org/abs/1909.12187 with the following fact: If $B$ is an abelian variety and $B\dashrightarrow X$ is a rational map, then $B\dashrightarrow X$ is a morphism (i.e., everywhere defined). (This is because $B$ is smooth and $X$ has no rational curves.) ..con'd below
...This means that images of rational maps of abelian varieties are images of morphisms of abelian varieties. But these images are translates of subabelian varieties (by closed points!) of $A $ because morphisms of abelian varieties are homomorphisms up to translation.
Does this answer all your questions? If so, then I can make the above comments into an answer.
@AriyanJavanpeykar: That's very very helpful! Thanks again! Yes please make it into an answer:)
Usually, when one speaks of "translate of an abelian subvariety" it really is in the sense "translate of an abelian subvariety by a closed point".
To answer your question: Let $X\subset A$ be a closed subvariety of an abelian variety $A$ over $k$. Assume that $k$ is algebraically closed of characteristic zero. Let $Sp(X)$ be the union of positive-dimensional translates (by closed points of $A$) of abelian subvarieties contained in $X$. Kawamata proved that $Sp(X)$ is closed in $X$; see Thm. 4 in Y. Kawamata, On Bloch’s conjecture, Invent. Math. 57 (1980), 97-100.
Side remark. Ueno proved that $Sp(X) \neq X$ if and only if $X$ is of general type. Similar statements are true for closed subvarieties of semi-abelian varieties in positive characteristic; see Abramovich numdam.org/item/CM_1994__90_1_37_0
Let $L/k$ be an extension of algebraically closed fields. Then $Sp(X)_L = Sp(X_L)$, where $Sp(X_L)$ is the union of
the positive-dimensional translates of abelian subvarieties of $A_L$ contained in $X_L$. Thus, the "special" locus of $X$ behaves well with respect to field extensions. Let me explain how to prove this in a more general context.
Let $\Delta^{gr}_X$ be the groupless-exceptional locus. That is, $\Delta^{gr}_X$ is the Zariski closure of the union of the images of non-constant morphisms $U\to X$, where $U$ is a dense open subset of a connected finite type group scheme $G$ over $k$ such that $\mathrm{codim}_G(G\setminus U)\geq 2$.
Then, $\Delta^{gr}_X =Sp(X)$; see Theorem 13.1 in https://arxiv.org/pdf/2002.11981.pdf .
To prove this equality of sets, use the following three facts:
1) every rational map $B\dashrightarrow X$ with $B$ an abelian variety extends to a morphism $B\to X$ (use that $X$ has no rational curves).
2) if $G$ is a connected linear algebraic group, then every rational map $G\dashrightarrow X $ is constant. This is because linear algebraic groups are covered by (non-compact) rational curves.
3) The image of a morphism of abelian varieties $B\to A$ is the translate of a abelian subvariety of $A$.
In the absence of an ambient abelian variety, the general statement you are looking for is the following:
Proposition 3.7 in https://arxiv.org/abs/1909.12187
Let $L/k$ be an extension of algebraically closed fields of characteristic zero. Let X be a proper variety over $k$. Then $(\Delta_X^{gr})_L = \Delta^{gr}_{X_L}$.
I'm thinking about the following example: if $X=E\times C$ where $E$ is an elliptic curve and $C$ is a curve of genus $2$. Then $Sp(X)$ is supposed to be $X$. But the union of translates by closed points seems not including the codim $2$ generic point? I'm probably missing a point here. Could you help me out?
@ggttttll Sp(X) is scheme-theoretically really the closure of the closed subset (of the variety $X(k)$) given by the union of translates of positive-dimensional abelian subvarieties. It is then clear that $Sp(X) = X$ as a scheme if $X = E \times C$.
Thanks for all the help! Take care!
|
2025-03-21T14:48:30.185636
| 2020-03-31T15:18:22 |
356198
|
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|
Stack Exchange
|
When is the cut-locus normal coordinate collared
Let $(M,g)$ be a complete $d$-dimensional Riemannian manifold, $p \in M$ be fixed and let $C_p$ be the cut-locus of $p$.
Other than when $M$ is non-positively curved (in which $C_p=
\emptyset$ by Cartan-Hadamard) when is the cut-locus contained in a collared neighbourhood ie: an open subset $U\subseteq M$ containing $C_p$ satisfying and a homeomorphism $f:U\rightarrow C_p \times [0,1)$ such that
$$
f[C_p]=C_p \times \{0\}.
$$
Collared by normal coordinates, or just topologically collared?
Just topologically
Could you define "topologically collared"? The cut locus $C_p$ is not a submanifold. For example on a surface it should be a topologically embedded tree, and so it cannot have a neighborhood homeomorphic to $[0,1]\times C_p$ because the latter is not a manifold.
And the tree need not be locally finite, as was famously shown by Gluck and Singer in "Scattering of Geodesic Fields, I". Actually, in general on a surface, it is a graph not a tree, see https://arxiv.org/abs/1103.1759.
@IgorBelegradek I added in the definition of collared that I'm looking for.
As I explained in my comment above your definition of collared cannot work. Imagine that the cut locus $C$ in a surface is a tripod (the graph with 3 edges joined in a common vertex). Such examples are in the paper "Every graph is a cut locus" linked above. No neighborhood of a tripod in a $2$-manifold is homeomorphic to $C\times [0,1)$ because the latter is not a manifold.
I’ve probably misread the question, but isn’t the answer no for the standard sphere?
@DeaneYang: right, the OP definition of "collared" fails for the round sphere.
This is impossible because $[0,1)$ is not a homology manifold. So $X\times [0,1)$ can never be a topological manifold for any $X$.
@VitaliKapovitch Why doesn't that contradict Theorem 2 of "Locally flat imbeddings of topological manifolds", Morton Brown, 1962 which states that the boundary of any metrizable manifold with boundary $M$ is collared? Doesn't this use the fact that $\partial M \times [0,1)$ is homeomorphic to an open neiberhood of $\partial M$?
I'm assuming that $M$ has no boundary. Could you give an explicit example of what you mean by a collared neighborhood? For example, if $S \subset M$ is a smooth hypersurface, then any subset of $M$ that is homeomorphic to $S\times [0,1)$ is never open. On the other hand, any subset homeomorphic to $S\times (0,1)$ is.
@AIM_BLB that's certainly a correct theorem but you didn't say anything about boundary in your question so I assumed you meant a manifold without boundary. For manifolds without boundary what you ask for is impossible unless the cut locus is empty. It is of course possible for manifolds with boundary (e.g. a closed disk in $R^n$) provided you define cut locus appropriately.
Permit me to include this nice image from Day & Li
to illustrate @Igor's point that "in general on a surface, it [the cut locus] is a graph not a tree."
The source point $p$ is on the cat's forehead,
the other side in this rear-view.
Dey, Tamal K., and Kuiyu Li. "Cut locus and topology from surface point data." In Proceedings of the 25th Symposium on Computational Geometry, pp. 125-134. 2009.
ACM link.
In the paper they compute an approximation to the cut locus
on this model that pretty much follows the smooth curves
drawn above.
|
2025-03-21T14:48:30.186009
| 2020-03-31T16:19:46 |
356204
|
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|
Stack Exchange
|
Applying the Abramov-Rokhlin skew product entropy formula to a bounded-to-one factor
Let $(X, \mathcal{B}, \mu, S)$ and $(Y, \mathcal{C}, \nu, T)$ be invertible probability-measure-preserving systems, with a measurable factor map $\pi: X \to Y$, i.e. $\pi \circ S = T \circ \pi$. Suppose that there is some $N \in \mathbb{N}$ such that $\nu$-almost every $y \in Y$ has at most $N$ preimages under $\pi$, i.e. $\# \pi^{-1}(\{y\}) \leq N$. If necessary, I'm happy to assume that $X$ and $Y$ are compact metric spaces, $\mu$ and $\nu$ are Borel, and $\pi, S, T$ are continuous.
I'm trying to show that $\pi$ preserves measure-theoretic entropy, i.e. $h(\nu) = h(\mu)$. I'd like to apply the Abramov-Rokhlin entropy formula, which expresses the entropy of a skew product transformation as the sum of the entropy of the base and the entropy of the fibre. To do this, I'd like to realize $X$ as the direct product $Y \times S$ (or a subset of this), with $F = \{ 1, \dots, N \}$, together some probability vector $\mathbf{p} \in \mathbb{R}^N$, such that $\mu = \nu \times \mathbf{p}$. I'd then try to realize $S$ as $S(y,k) = (T(y), \alpha(y)(k))$ where for each $y$, $\alpha(y)$ is an invertible measure-preserving transformation of $(F, \mathbf{p})$.
The problem is that if the $\alpha$'s give a transitive action of $S_N$, this would require that $\mathbf{p}$ be uniform. In particular, it seems that the fiber cardinality would have to be constant, not just bounded. Even forgetting about the measure, this is an issue, because the $\alpha$'s would have to be bijections.
Two ideas I've played with:
The number of preimages $\# \pi^{-1}(\{y\})$ is constant along the orbit of $y$, so I could try partitioning $X$ and $Y$ into at most $N$ disjoint subsystems on which $\pi$ is constant-to-one, then applying the Abramov-Rokhlin formula separately on each one, but I don't know how to show that this decomposition would be measurable.
Bögenschutz and Crauel (1990) have a generalization of the Abramov-Rokhlin formula for a skew product transformation (with the notation above) in which we no longer require a fixed measure on the fiber, but only that $\mu$ be invariant with fixed $Y$-marginal $\nu$. This might solve the problem by letting me add a null set of fixed points to $X$, for instance, to make $\pi$ constant-to-one.
I'm also hoping to apply this in the case where $S$ and $T$ are really actions of a countable amenable group $G$, using Ward and Zhang's generalization of the Abramov-Rokhlin formula. I haven't worked through the proofs to see if the Bögenschutz-Crauel and Ward-Zhang generalizations are compatible with each other.
Jisang Yoo's paper: https://arxiv.org/abs/1612.08648 has a lemma (Lemma 3.1) showing measurability of the decomposition by number of pre-images.
Thanks! It looks like this decomposition may not quite be measurable, but it is universally measurable (in particular, it becomes measurable if you replace $\mathcal{B}$ and $\mathcal{C}$ by their completions with respect to $\mu$ and $\nu$, respectively), which is enough. Universal measurability will actually help elsewhere in this project, too, so I'm glad to have learned about it here.
One should always keep in mind that the natural substrate of ergodic theory is Lebesgue-Rokhlin (aka Lebesgue or standard) measure spaces which enjoy a lot of properties not necessarily present in general measure spaces. One of these properties is an explicit description of homomorphisms of such spaces obtained by Rokhlin in his 1949 paper, which, in particular, implies measurability of the cardinality of the preimages. Therefore, in your situation the number of preimages is constant along the ergodic components, and it is an easy exercise to check that the entropy does not decrease for quotient maps of this kind.
Thank you for the suggestion to look at Rokhlin's paper! The idea I was missing was similar to the one in my reply to Anthony Quas's comment on the question: work with a complete $\sigma$-algebra, rather than with Borel sets only.
|
2025-03-21T14:48:30.186282
| 2020-03-31T16:22:53 |
356205
|
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|
Stack Exchange
|
Commutator estimates for $-(-\Delta)^s$, with $s \in (1,2)$
I'm currently trying to work with the non-local operator given by
$$
(-\Delta)^{\frac{s}{2}}f(x)= c_s\text{P.V} \int_{-\infty}^\infty \frac{-f(x+y)-f(x-y)+2f(x)}{|y|^{1+s}} dy,
$$
where $f :\mathbb R \to \mathbb R$ is a smooth function and $c_s$ is a constant. I am focusing on the case $s \in (1,2)$.
I am particularly interested on some sort the approximations (a remainder estimate for an approximate Leibnitz rule) for $(-\Delta)^s (f^n)(x)$ where $f$ is a $C^{1,s^\prime}_\text{loc}(\mathbb R)$, with $s^\prime >s $ that decays faster than any polynomial and $n$ is an arbitrary positive integer.This article seems to suggest that we could take
$$
(-\Delta)^\frac{s}{2} (f^n)(x) \approx (-\Delta)^\frac{s}{2} (f)(x)f^{n-1}(x) + s(n-1)(-\Delta)^\frac{s-1}{2}(f)(x) f^\prime(x) f^{n-2}(x).
$$
Is it possible to quantify the error? I was hoping for a $L^p$ type of estimate.
I read some of professor's Tao lecture notes in harmonic analysis, in which he points out that one does not expect in general a good Chain rule estimate for $(-\Delta)^{s}F(f)$ if $s \in (1,2)$ for general $F$, however, he says one can expect such given enough regularity for $F$. I know that the commutator estimate seems like a step further, but I was hoping this would be possible for the polynomial case.
Any bibliography is well appreciated, I am just starting to navigate on the subject and analysis is not my first area.
You should look for Kato-Ponce commutator estimates and Kenig-Ponce-Vega commutator estimates. For a view of the subject check Li's paper here</>
That was exactly the kind of result I was looking for, thank you very much!
|
2025-03-21T14:48:30.186428
| 2020-03-31T16:33:41 |
356206
|
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|
Stack Exchange
|
De Rham and Koszul complexes
Consider the algebraic de Rham complex of the $n$-dimensional plane: this is merely
$$\ldots\rightarrow Sym(V^*)\otimes\bigwedge^{k}V^*\rightarrow Sym(V^*)\otimes\bigwedge^{k+1}V^*\rightarrow\ldots
$$
with the standard de Rham differential $fdx_I\rightarrow\sum_{i=1}^n\frac{df}{dx_i}dx_i\wedge dx_I$.
On the other hand, there is a Koszul resolution of the ideal $(x_1,\ldots,x_n)\subset Sym(V^*)$ is defined on a chain complex which is termwise the same as above, but with differential going in the opposite direction: e.g. it sends $dx_{i_1}\wedge\ldots\wedge dx_{i_k}$ to $x_{i_1}dx_{i_2}\wedge\ldots\wedge dx_{i_k}-x_{i_2}dx_{i_1}\wedge dx_{i_3}\wedge\ldots\wedge dx_{i_n}+\ldots+(-1)^{k-1}x_{i_k}dx_{i_1}\wedge\ldots dx_{i_{k-1}}$.
If we call the de Rham differential $d$, and the Koszul differential $e$, then $de+ed$ is scaling by the total degree (exterior and symmetric parts) of the form.
Having some familiarity with different types of cohomology groups that one encounters in algebraic geometry, it seems unusual to me that there would exist two different differentials on the same "complex", and that they would be related in such a way. Thus, I am curious whether this is an instance of some more general phenomenon, for example whether this occurs for certain complexes of sheaves on other spaces. Though I am not an expert in this language, it seems one way to phrase what happens in this example is that there is a single graded algebra with two different DGA structures which are related (by this scaling formula). Are there any/many other examples of this happening?
This structure is called "mixed complex", see, e.g., https://ncatlab.org/nlab/show/mixed+complex
Thank you for the reference. From the paper by Kassels referred to on that page, it seems that a mixed complex is a distinct case where $de+ed=0$.
Sorry, I missed the point, my reference is, indeed, irrelevant.
What you have here is a graded version of an $\mathfrak{sl}_2$-triple. Compare for example the proof of the Lefschetz decomposition and hard Lefschetz theorem in Hodge theory, which is essentially the observation that $L$ and $\Lambda$ form the $e$ and $f$ of an $\mathfrak{sl}_2$-triple, because
$$[L,\Lambda] = (k-n)\operatorname{id}$$
on degree $k$ forms (so let $h$ act by $k-n$ in degree $k$).
I will first give a definition of a 'graded $\mathfrak{sl}_2$-triple', and then give a conceptual explanation for the example you gave.
Definition. Let $k$ be a field (algebraically closed of characteristic $0$, for simplicity), and let $C$ be the complex
$$\ldots \to 0 \to k \stackrel 0\to k \to 0 \to \ldots$$
with $k$ in (cohomological) degrees $0$ and $1$. Let $\mathfrak{gl}(C)$ be the graded Lie algebra defined by the graded algebra $\operatorname{Hom}^*(C,C)$ (all endomorphisms of $C = C^0 \oplus C^1$, with its natural grading), and let $\mathfrak{sl}(C)$ be the graded Lie subalgebra obtained by replacing $\operatorname{Hom}^0(C,C)$ by its trace $0$ elements, recalling that
$$\operatorname{tr}\big(f \colon C \to C\big) = \sum_i (-1)^i\operatorname{tr}\big(f^i \colon C^i \to C^i\big).$$
Explicitly, $\mathfrak{sl}(C)$ has a basis $\{e,f,h\}$ where $h \colon C \to C$ is the identity (which indeed has trace $0$), $e \colon C \to C[1]$ is the map
$$\begin{array}{ccccccccc}\ldots & \to &0 & \to & k & \to & k & \to & \ldots \\ & & \downarrow & & || & & \downarrow & & \\ \ldots & \to & k & \to & k & \to & 0 & \to & \ldots,\! \end{array}$$
and $f \colon C \to C[-1]$ is the map
$$\begin{array}{ccccccccc}\ldots & \to &k & \to & k & \to & 0 & \to & \ldots \\ & & \downarrow & & || & & \downarrow & & \\ \ldots & \to & 0 & \to & k & \to & k & \to & \ldots.\! \end{array}$$
With the graded Lie bracket, we get
\begin{align*}
[e,f] &= ef + fe = h,\\
[h,e] &= 0,\\
[h,f] &= 0.
\end{align*}
Viewing the complex $C$ with zero differentials as a graded object $C^0 \oplus C^1$, the map $e$ can be represented by the map $(a,b) \mapsto (b,0)$ and $f$ by $(a,b) \mapsto (0,a)$.
Definition. A graded $\mathfrak{sl}_2$-triple on a graded vector space $V$ is a triple $(e,f,h)$ of elements in $\operatorname{Hom}^*(V,V)$ of degrees $1$, $-1$, and $0$ respectively satisfying the identities
\begin{align*}
[e,f] &= ef + fe = h,\\
[h,e] &= 0,\\
[h,f] &= 0.
\end{align*}
Example. Let $V$ be a finite dimensional vector space, and let $V \otimes C$ be the two term complex $V \oplus V[-1]$ with zero differential. Then
$$\operatorname{Sym}(V \otimes C) = \operatorname{Sym}(V) \otimes \operatorname{Sym}(V[-1]) = \bigoplus_i \operatorname{Sym}(V) \otimes \left(\bigwedge\nolimits^i V\right)[-i],$$
since the sign in the graded swap $K \otimes L \stackrel\sim\to L \otimes K$ replaces the symmetriser on $K^{\otimes n}$ by the antisymmetriser in odd degree. This is the complex you're studying, except it has differentials all $0$ for now.
But $\operatorname{Hom}^*(C,C)$ acts on $V \otimes C$, hence so does $\mathfrak{sl}_2(C)$. Then the latter also acts¹ on $T^*(V \otimes C)$ and its quotient $\operatorname{Sym}(V \otimes C)$.
Lemma. This graded $\mathfrak{sl}_2$-triple is the one described by the OP, with $e$ the Koszul differential, $f$ the de Rham differential, and $h$ multiplication by the total degree.
Proof. Writing
$$T^n(V \otimes C) = \big(V \oplus V[-1]\big)^{\otimes n}$$
with elements $(v_1,w_1) \otimes \ldots \otimes (v_n,w_n)$, the action of $\mathfrak{sl}_2(C)$ is given by
\begin{align*}
e \colon T^n(V \otimes C) \to& T^n(V \otimes C)[1]\\
(v_1,w_1) \otimes \ldots \otimes (v_n,w_n) \mapsto & \sum_{i=1}^n (v_1,-w_1) \otimes \ldots \otimes (v_{i-1},-w_{i-1}) \\ & \otimes (w_i,0) \otimes (v_{i+1},w_{i+1}) \otimes \ldots \otimes (v_n,w_n),\\\\
f \colon T^n(V \otimes C) \to& T^n(V \otimes C)[-1]\\
(v_1,w_1) \otimes \ldots \otimes (v_n,w_n) \mapsto & \sum_{i=1}^n (v_1,-w_1) \otimes \ldots \otimes (v_{i-1},-w_{i-1}) \\ & \otimes (0,v_i) \otimes (v_{i+1},w_{i+1}) \otimes \ldots \otimes (v_n,w_n),
\end{align*}
and $h$ is multiplication by $n$. In the above notation, the quotient map
$$T^n(V \otimes C) \to \operatorname{Sym}^n(V \otimes C) = \bigoplus_{i+j=n} \operatorname{Sym}^i V \otimes \left(\bigwedge\nolimits^j V\right)[-j]$$
is given by
$$(v_1,w_1) \otimes \ldots \otimes (v_n,w_n) \mapsto \sum_{I \amalg J = \{1,\ldots,n\}} v_I \otimes w_J,$$
where $v_I = \prod_{i \in I} v_i$ and $w_J = w_{j_1} \wedge \ldots \wedge w_{j_s}$ if $\{j_1 < \ldots < j_s\} = J$. Then $f$ descends to the de Rham differential $d$, and $e$ descends to the Koszul differential $e$. Indeed, consider an element
$$(v_1,0) \otimes \ldots \otimes (v_r,0) \otimes (0,w_1) \otimes \ldots \otimes (0,w_s)$$
lifting $v_I \otimes w_J$ where $I = \{1,\ldots,r\}$. Then $f$ maps this to
$$\sum_{i=1}^r (v_1,0) \otimes \ldots \otimes (v_{i-1},0) \otimes (0,v_i) \otimes (v_{i+1},0) \otimes \ldots \otimes (v_r,0) \otimes (0,w_1) \otimes \ldots \otimes (0,w_s),$$
which under the quotient $T^n(V \otimes C) \to \operatorname{Sym}^n(V \otimes C)$ maps to
$$\sum_{i=1}^r \left(\prod_{j \neq i} v_j\right) \otimes \big(v_i \wedge w_J\big),$$
which is the de Rham differential of $v_I \otimes w_J$. On the other hand $e$ takes it to
$$\sum_{j=1}^s (v_1,0) \otimes \ldots \otimes (v_r,0) \otimes (0,-w_1) \otimes \ldots (0,-w_{j-1}) \otimes (w_j,0) \otimes (0,w_{j+1}) \otimes \ldots \otimes (0,w_n),$$
which under the quotient $T^n(V \otimes C) \to \operatorname{Sym}^n(V \otimes C)$ maps to
$$\sum_{j=1}^s (-1)^{j-1} v_Iw_j \otimes w_{J \setminus\{j\}},$$
which is the Koszul differential of $v_I \otimes w_J$. Finally, just like on $T^n(V \otimes C)$, on $\operatorname{Sym}^n(V \otimes C)$ the map $h$ is just multiplication by $n$, the total degree of $v_I \otimes w_J$. $\square$
¹ The map $\operatorname{Hom}^*(C,C) \to \operatorname{Hom}^*(C \otimes C, C \otimes C)$ by $f \mapsto f \otimes f$ is not linear (nor does it preserve the grading). So we do not get a natural action of $\operatorname{Hom}^*(C,C)$ on $\operatorname{Sym}(V \otimes C)$. However for a graded Lie algebra $L$ acting on complexes $C$ and $D$, there is an action on $C \otimes D$ by
$$\rho_{C \otimes D}(x)(c \otimes d) = \Big(\rho_C(x) \otimes 1 + (-1)^{\deg(x)\deg(c)} \otimes \rho_D(x)\Big)(c \otimes d)$$
for $x \in L$, where $\rho_C(x)$ and $\rho_D(x)$ are the actions of $x$ on $C$ and $D$ respectively. Similarly one gets actions on $\operatorname{Sym}(C)$, etcetera.
Allow me to be daft: I don't see how to make sense of the trace of these morphisms from C to its shifts. Also, are we assuming the differentials in C to be trivial from the beginning?
You're right, trace $0$ refers only to the degree $0$ part of morphisms $C \to C$, not the degree $1$ part $C \to C[1]$ or the degree $-1$ part $C \to C[-1]$. And indeed I take the differentials to be $0$; I should have specified.
|
2025-03-21T14:48:30.186900
| 2020-03-31T18:02:58 |
356212
|
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|
Stack Exchange
|
An ODE comparison problem
Recently I met an ODE problem but after thinking for quite a while I still could not find an answer. Here is the question, which looks very simple:
Let $y=y(t)$ be a smooth function defined on $[0,\infty)$. If $y(t)$ satisfies the differential inequality
$$(ty')' \le \frac{4t}{(1+t^2)^2}(1-e^{2y})$$with the initial conditions $y(0)=0$ and $y'(0)=0$, then is it true that $y(t) \le 0$ for all $t \ge 0$?
The motivation is that if the above differential inequality becomes equality with same initial conditions, then $y(t) \equiv 0$. I strongly feel the answer is positive, but so far I don't know how to apply any comparison argument. Any ideas and comments will be fully appreciated.
Update: For the equality case, we can show $y(t) \equiv 0$ using the following argument:
Let $u(t)=\ln\left(\frac{2}{1+t^2}\right)$, then the equality case becomes
$$(ty')'= te^{2u}(1-e^{2y})=-(tu')'-te^{2(u+y)}.$$ Let $w=u+y$, then $w$ solves the differential equation $(tw')'+te^{2w}=0$ with the initial condition $w'(0)=0$. Then $w$ must be of the form
$$w(t)=\ln\left(\frac{2c}{1+c^2t^2}\right)$$where $c$ is any positive constant. Hence $$y=w-u=\ln\left(\frac{c(1+t^2)}{1+c^2t^2}\right).$$Using the condition $y(0)=0$, we have $c=1$ and hence $y(t) \equiv 0$.
what about the equality case and doing a substitution y(t)=ln u(t)
@OOESCoupling, for the equality case, yes, one can find an explicit formula for the general solutions, but the inequality case becomes tricky. This substitution does not seem work.
$\newcommand{\de}{\delta}
\newcommand{\vp}{\varepsilon}$
The conjecture is false. The previous answer, using numerics, was convincing enough for me, but perhaps not for others. So, here is a rigorous answer, based on the same general idea.
Let
\begin{equation*}
L(y)(t):=(ty'(t))'-\frac{4t}{(1+t^2)^2}(1-e^{2y(t)}).
\end{equation*}
Then
\begin{equation*}
L(y_1)(t)=0
\end{equation*}
for real $t>0$,
where
\begin{equation*}
y_1(t):= \ln \frac{c_1^2 e^{c_1 c_2} t^{c_1-1}(t^2+1)}
{c_1^2 e^{2 c_1 c_2}+t^{2 c_1}}
\end{equation*}
with $c_1\ge0$ and any real $c_2$. In what follows, set
\begin{equation*}
c_1:=\sqrt{197/199},\quad c_2:=1/192.
\end{equation*}
Then
\begin{equation*}
y_1(3/10)\approx-0.000101955<0,\quad y_1'(3/10)\approx-0.0200123<0.
\end{equation*}
Let now
\begin{align*}
c&:=-\frac{y_1'(3/10)^2}{4y_1(3/10)}\approx0.98203, \tag{1} \\
\de&:=\frac{2y_1(3/10)}{y_1'(3/10)}\approx0.0101893, \tag{2} \\
\vp&:=3/10-\de\approx0.289811, \tag{3}
\end{align*}
so that
\begin{equation*}
c\de^2=-y_1(3/10),\quad 2c\de=-y_1'(3/10). \tag{4}
\end{equation*}
Letting now
\begin{equation*}
y_2(t):=-c(t-\vp)^2, \tag{5}
\end{equation*}
we see that
\begin{equation*}
y_2(\vp)=y_2'(\vp)=0,\quad y_2(3/10)=y_1(3/10),\quad y_2'(3/10)=y_1'(3/10), \tag{6}
\end{equation*}
and
\begin{equation*}
L(y_2)(t)=c(3/5-2 \de-4 t)+
4 t \,\frac{e^{-2 c (\de+t-3/10)^2}-1}{(t^2+1)^2}
<c(3/5-2 \de-4 t)<c(3/5-4 \vp)<-0.544<0 \tag{7}
\end{equation*}
for $t\ge\vp$.
Letting finally
\begin{equation*}
Y:=y_1\,1_{[3/10,\infty)}+y_2\,1_{[\vp,3/10)},
\end{equation*}
we see that
\begin{equation*}
L(Y)\le0
\end{equation*}
on $[0,\infty)$. However,
\begin{equation*}
Y(6)=y_1(6)\approx0.00360808>0.
\end{equation*}
The function $Y$ is the solution $y$ to the problem
$$L(y)=h$$
on $(0,\infty)$ with the initial conditions $y(0+)=y'(0+)=0$, where $h:=L(Y)\le0$ and $Y''$ is understood as the right (or left) derivative of $Y'$. Also, $Y=0$ on $(0,\vp]$ and $Y(6)>0.003>0$.
Let now $\psi$ be a function in $C^\infty((0,\infty))$ such that $\psi=0$ on $(0,\vp/2]$ and $\psi\le0$ on $(0,\infty)$. Let then $Z$ be the solution $z$ to the problem $$L(z)=\psi$$
on $(0,\infty)$ with the initial conditions $z(0+)=z'(0+)=0$ (so that $z(\vp/2)=z'(\vp/2)=0$). Choose now $\psi$ to be so close to $h$ in $L^1((0,\infty))$ that $|Z(6)-Y(6)|<0.001$. Then $Z(6)>0.002>0$, as desired.
Thank you so much for your time and contribution. This is really inspriational and deserves upvote. I observed that the $Y$ constructed above are two functions glued together, which is $C^2$ but not $C^{\infty}$. I'm curious that does the conjecture hold true for any $C^{\infty}$ functions $y$ satisfying the inequality $L(y)\le 0$?
Just realized that the function $Y$ constructed above is $C^1$, not $C^2$, so the second derivative cannot be defined at the point $x=3/10$.
@student : I have added details concerning a $C^\infty$ version.
I still have two questions: First, why the $L^1$ approximation of $L$ lead to $L^{\infty}$ approximation of solutions. Second, is it possible to give a concrete function $y(t)$ such that the $y(t)$ satisfies the differential inequality while positive somewhere?
(i) If you go over a standard proof of the continuous dependence of the solution to an ODE $y'=f(t,y,\theta)$ on a parameter -- see e.g. https://users.math.msu.edu/users/hhu/848/lec_5.pdf, Theorem 5.2 -- you will see that the condition of the $L^\infty$-boundedness of $f$ can be replaced by the $L^1$-boundedness. Here you'll need to use the absolute continuity of the integral: if $\int|g|,d\mu<\infty$ then $\int_A|g|,d\mu\to0$ whenever $\mu(A)\to0$. (This easily follows by dominated convergence.) In your case, $A$ will be the short intervals of a partition of the interval $[0,6]$, say.
(ii) If you want an explicit $C^\infty$ counterexample to your conjecture, just follow all the many steps of the procedure I have described, doing explicit estimates at each step. The result will be an explicit function, though with a huge and cumbersome description. If you want something better than that, you may want to ask this additional question in a separate post.
I still could not see the continuous dependence of the solution from the theorem, because here $L$ involves second order derivatives.
@student : It a standard trick to rewrite higher-order ODEs as first-order vector ODEs, that is, first-order systems of scalar ODEs. E.g., rewrite $y''=f(t,y,y')$ as a system of two first-order ODEs: $y'=z$ and $z'=f(t,y,z)$ -- or, in the vector form, $X'=F(t,X)$, where $X:=(y,z)$ and $F(t,X):=(z,f(t,y,z))$.
Thanks for your reply, however I have one last quesion, which is my main doubt on this solution. The function $Y$ constructed is not $C^2$ since its second derivative has a jump at $x=10/3$ as seen in the construction, then $Y$ is not a classical solution to the ODE. I checked the proof in the notes you posted, but I think that proof does require the solution is a classical one.
@student : When you rewrite the (vector) ODE in the (more general) integral form (as is of course done in the proof at the link I provided), then you'll see that it is enough for the vector solution to the vector ODE to be just continuous -- which corresponds to the solution of the original scalar second-order ODE being in $C^1$, which is exactly what we have.
I do very appreicate your time, but I'm still very confused... That theorem says the solutions depend continuously on the parameters $(t, x, \lambda)$, but in your context, you use a function $\psi$ to $L^1$ approximate $h$. How can the solution continuously depend on the $L^1$ continuity of functions? Could you possibly elaborate this step in the answer? I would be really appreciated.
@student : Just use the linear (say) interpolation $(1-\theta)h+\theta h_\infty$ between the discontinuous right-hand side $h$ and its $C^\infty$ approximation $h_\infty$.
The conjecture is very likely false. Indeed, let $Y$ be the solution of the ODE
$$(ty'(t))'=\frac{4t}{(1+t^2)^2}(1-e^{2y(t)})-1_{[0.3,0.31]}(t)$$
with the initial conditions $y(0)=0$ and $y'(0)=0$.
Then $Y=0$ on $[0,0.3]$, and the shown below numerical solution of the above ODE strongly suggests that $Y(6)>0$. We solve this ODE (numerically) first on the interval $[0.3,0.31]$ and then on the interval $[0.31,8]$.
A rigorous but tedious proof of the claim that $Y(6)>0$ could be obtained by estimating well enough each of the steps of the procedure for the numerical solution of this ODE.
|
2025-03-21T14:48:30.187386
| 2020-03-31T18:06:05 |
356213
|
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|
Stack Exchange
|
Johnson-Lindenstrauss lemma preserves angles
Edit (April 1, 2020):
I formalized the original question (from the reference below) wrong. Please see @AndreasBlass correction.
Also, all the suggestions proposed so far doesn't seems to be related to 2 hints that I found online (but didn't understand how to use), and I'm curious if anyone could explain more about them:
Hint 1:
For every angle, construct a equilateral triangle that its edges are
being preserved by the projection (add the vertices of those triangles
[conceptually] to the point set being embedded). Argue, that this
implies that the angle is being preserved.
Hint 2:
If two triangles lie in the same plane (a 2-dimensional affine space)
in $\mathbb R^n$, then under a linear mapping their areas have the same
multiplicative error. For every triangle, add an extra point to form a
right-angled isosceles triangle in the same plane.
My original post:
As described in https://sarielhp.org/teach/08/a/lec/22_jl.pdf, Exercise 22.6.5 ("Johnson–Lindenstrauss lemma works for angles").
Let $x,y,z\in \mathbb R^n$. Let $0<\epsilon<1$. Denote $\theta\mathrel{:=}\angle xyz$. Let $k=\Theta(\frac{\log n}{\epsilon^2})$.
I'm struggling to show that $\exists$ linear map $f:\mathbb R^n\to\mathbb R^k$ s.t. with high probability $(1-\epsilon)\theta\le\theta'\le(1+\epsilon)\theta$, where $\theta'=\angle x' y' z'$ (and $x'\mathrel{:=}f(x)$,$y'\mathrel{:=}f(y)$,$z'\mathrel{:=}f(z)$).
Would appreciate your help.
PS: I tried to express $\theta$ using the law of cosine, but got stuck:
Denote $a\mathrel{:=}\|x-y\|$, $b\mathrel{:=}\|x-z\|$, $c\mathrel{:=}\|y-z\|$, and $a'\mathrel{:=}\|x'-y' \|$, $b'\mathrel{:=}\|x'-z'\|$, $c'\mathrel{:=}\|y'-z'\|$. By the law of cosines, $\theta=\arccos(\frac{a^2+b^2-c^2}{2ab})$ and $\theta'=\arccos(\frac{a^2+b'^2-c'^2}{2a'b'})$.
I will point out that the tag (geometry) is deprecated on [mathoverflow.se], see the tag-info. Perhaps you (or some other users) might be able to choose other suitable tag.
Isn't this just a continuity argument? The ratio inside the $arccos$ depends continuously on $a$, $b$, and $c$, and $arccos$ itself is continuous, so the result follows from the definition of continuity.
@PaulSiegel, could you please elaborate? Best regards
@MartinSleziak, thanks.
There is a full proof in the paper cited in those notes, "Dimensionality Reductions That Preserve Volumes and Distance to Affine Spaces, and Their Algorithmic Applications" by Magen, 2002.
Put $y$ at the origin and project $z$ onto $x$; call this point $w$. Now $zwy$ is a right triangle. The projection approximately preserves the lengths of all the edges of this triangle. And one can show that the angle at $w$ is still close to 90 degrees, i.e. the dot product of $(y-w)$ and $(z-w)$ is very small. These seem to be key ingredients the paper is using to show the angle at $x$ is approximately preserved.
@usul, thank you for your reply. As far as I understood from "reading" the paper you referred to (challenging for me), it seems like the paper used a right angle isosceles triangle (not only right angle). Is it necessary? Because from your answer I understand that the requirement for the isosceles property isn't a key ingredient of the proof. Best regards
@keyboardAnt I only skimmed it myself. It seems like they need the isosceles argument too, but I didn't get intuition for why.
I'll discuss separately the question you asked and the exercise in the notes you linked to.
The question you asked is trivial as long as $k\geq3$. Just let $f$ be a projection onto the $3$-dimensional subspace spanned by $x,y,z$. You don't even need $\epsilon$, since $x,y,z$, and the angle are preserved exactly. (In fact, even if $k=2$, you can preserve the angle exactly by projecting orthogonally to the subspace spanned by $x-y$ and $z-y$.)
Of course, this isn't what the exercise (or the Johnson-Lindenstrauss lemma) is about. You wanted to approximately preserve one angle. The analog for angles of Johnson-Lindenstrauss wants to approximately preserve all angles between points from some given $q$-element set. To do this, one needs the target space to have dimension of order $(\log q)/(\epsilon^2)$; note the log of the number $q$ of points, not of the dimension of the space in which they lie.
Once the statement of the problem has been corrected to obtain for angles what Johnson-Lindenstrauss gives for distances, I want to use Paul Siegel's comment, that this follows from continuity of arccos. The only catch is that, at the ends $\pm1$ of its domain, arccos, though continuous, isn't differentiable. In other words, when $x,y,z$ are collinear, one can't trivially estimate the error in $\theta$ in terms of the errors in the distances between $x,y,z$. But one can estimate it with a bit of work. I'm too tired to attempt that work now, but it looks to me as if we might need $\epsilon^4$ rather than $\epsilon^2$ in the estimate for the dimension of the target space. (If we're guaranteed that no three of the $q$ points are collinear, then $\epsilon^2$ should suffice but the implicit constant in "of order $(\log q)/(\epsilon^2)$" will depend on how far from collinear the points are.)
Technicality: Worse than collinearity would be coincidence. If $y$ coincides with $x$ or $z$, then the angle at $y$ is undefined and the question disappears.
EDIT: Comments by usul under the question suggest that the collinearity problem can be attacked using the following rough idea, which might underlie the proof he describes, and which I hope is easier to understand (though harder to believe). Fix a small angle $b$, say $\pi/100$. Now given $n$ points and wanting to project to $O(\epsilon^{-2}\ln n)$ dimensions while approximately preserving the angles they determine, we know from Johnson-Lindenstrauss that we can preserve the distances between them, and we know from the earlier part of this answer that that will approximately preserve the angles if they're in the range $[b,\pi-b]$. For each of the remaining "bad" angles, say $\angle xyz$, adjoin to the configuration a new point $w$ chosen so that all the new angles thereby introduced are in the "good" range $[b,\pi-b]$. (See below if no such $w$ exists.) Now the approximate preservation of lengths and preservation of these "good" angles should imply approximate preservation of $\angle xyz$. This process, applied to all the bad angles in the original configuration adds at most $n^3$ new points, so the increase in the number of points is only polynomial, the increase in $\ln n$ is at most a constant factor, and $O(\epsilon^{-2}\ln n)$ remains correct (with a bigger implicit constant in $O$).
Unfortunately, there might not be an appropriate $w$. Suppose $x$ and $y$ are close together and $z$ is far away. To keep $\angle xwy$ in the good range, $w$ needs to be fairly near $x$ and $y$, but then $\angle xzw$ won't be good. That's part of the reason why this is only a rough idea; another part is the "should imply" in the previous paragraph.
Presumably all this roughness is smoothed out in the actual proof that usul referred to.
@ Andreas, thank you for pointing on my mistake; now I understand that I wrote it wrong.
Regarding yours & @PaulSiegel suggestion to use the continuity of arccos: I'm only familiar with a single definition for a continues function, but now I see there are many other equivalent definitions. I would appreciate if anyone could please explain what exactly the "continuity argument" means, and why the result almost follows it directly. Best regards
(1) I don't see that this proof approach will give the theorem because arccos isn't Lipschitz -- as you say, the constant would depend on how far from collinear the points are. (2) Note the projectionis linear, so if the points are collinear then so are their projections.
@usul I agree with part (2) of your comment. I also agree that, because arccos isn't Lipschitz, you can't directly apply the distance result to get the angle result. That's why I expected (and still expect) to need a higher power of $\epsilon$ in the denominator of the target dimension. I don't see how to get by with only $\epsilon^2$ unless we know something about the points not getting too close to collinear. (Once the angles are bounded away from $0$ and $\pi$, that bound will give a Lipschitz constant for arccos, and everything should be OK.)
@AndreasBlass, have you read the paper mentioned by @ usual in his comment to my original post? "Dimensionality Reductions That Preserve Volumes and Distance to Affine Spaces, and Their Algorithmic Applications" by Magen, 2002. It seems like Magen took a different approach than you described, and got to a target dimension $\Theta(\frac{logn}{\epsilon^2})$. Best regards
@keyboardAnt No, but usul's comments suggest a rough idea, which might underlie the actual proof, and which I'll edit into my answer..
|
2025-03-21T14:48:30.188105
| 2020-03-31T18:24:27 |
356214
|
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|
Stack Exchange
|
Is the co-projection of the projection of a Borel set Lebesgue measurable?
Reading the classical book of Kechris "Classical descriptive set theory", I found the following facts
The projection of a Lebesgue measurable set need not be Lebesgue measurable, but the projection of a Borel set is an analytic set (Souslin set, $\bf \Sigma_{1}^{1}$) which is Lebesgue measurable.
If a set is analytic its complement is co-analytic ($\bf \Pi_{1}^{1}$) and hence also Lebesgue measurable.
The co-projection of a co-analytic set is co-analytic and hence Lebesgue measurable.
It easily follows that the composition of two projections or the composition of two coprojections of a Borel set must be Lebesgue measurable.
The question is what happens if we compose these two operations i.e., take the co-projection of a projection of a Borel set. Is the result Lebesgue measurable?
I can prove that the resulting set is a difference of a Lebesgue measurable and a PCA set ($\bf \Sigma^{1}_{2}$) but it is known that within ZFC it can not be decided whether the latter are or are not Lebesgue measurable.
I still believe that I miss something obvious and such sets are measurable.
To simplify notation assume that we are talkig of subsets of the finite dimensional Euclidean space.
The projection is $$\pi: X\times Y \supset A\ni (x,y)\to x\in \pi(A)\subseteq X$$
The co-projection $\pi'$ related to a projection $$\tilde \pi: X=U\times V\ni (u,v)\to u\in U,$$ is defined as $$\pi'(B)=(\tilde \pi (B^{c}))^{c},$$ where $c$ stnads for the complement in the respective space.
How is "co-projection of the projection of a Borel set" different from "complement of the projection of a coanalytic set" (which is just ${\bf\Pi^1_2}$)?
And a set is Lebesgue measurable iff its complement is. So if there is a $\mathbf{\Sigma}_{2}^1$ set $A$ which is not Lebesgue measurable, then its complement is the "co-projection of a projection" of a Borel set and is not Lebesgue measurable either. So this is equivalent to the undecidable question of whether $\mathbf{\Sigma}_2^1$ sets are Lebesgue measurable. Right?
You seem to be right. I hoped that the image of the projection will somehow ''land'' outside the ''bad locus'' in the domain of the co-projection. This was justified by the observation that the co-projection of the projection of the generators of the $\sigma$-algebra of Borel sets are Lebesgue mesurable (say the co-projection of the projection of open sets is measurable).
|
2025-03-21T14:48:30.188298
| 2020-03-31T19:15:39 |
356217
|
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|
Stack Exchange
|
Structure of the big cone and Seshadri constant on Fano manifolds
I would like to know something about the following two questions.
Given $X$ Fano manifold and $L$ an ample line bundle on $X$, we define
\begin{gather}
\sigma(L,x):=\sup\{t>0\, :\, \mu^{*}L-tE \,\, \mbox{is big}\, \},\\
\end{gather}
for any $x\in X$ where $\mu:Y\to X$ is the blow-up at $x$ and $E$ is the exceptional divisor. Is there any known upper bound on
$$
\sigma(L):=\sup_{x\in X}\sigma(L,x)\, ?
$$
In the same setting of point one and with the same notations, letting
\begin{gather}
\epsilon(L,x):=\sup\{t>0:\, \mu^{*}L-tE\, \,\mbox{is ample}\,\}
\end{gather}
to be the one-point Seshadri constant of $L$ at $x\in X$, is there any known lower bound on $\epsilon(L):=\inf_{x\in X} \epsilon(L,x)$?
An uniform bound depending uniquely on $n:=\dim X$,$(L^n)$ and on $(-K_{X}^{n})$ would be great, but I believe it may be too optimistic. So any comment is very welcome, also regarding particular Fano manifolds.
An important remark. It is possible to prove that $\sigma(L)\epsilon(L)^{n-1}\leq (L^{n})$ so the second question is stronger than the first one.
As particular example, consider $X=S_{r}$ to be a surface given by the blow-up of $P^{2}$ at $r$-general point $\{p_{1},\dots,p_{r}\}$ for $1\leq r\leq 9$ and let $L:=3H-E_{1}-\cdots-E_{r-1}-\delta E_{r}$ for $\delta$ rational such that $dL$ is ample for $d$ natural number divisible enough. What is it known about
\begin{gather}
\sigma(dL),\epsilon(dL)?
\end{gather}
Note that $S_{r}$ is a Del Pezzo surface for $r\leq 8$, while $S_{9}$ is not a Fano manifold but I would like to known more about the quantity described above also for this manifold.
Thanks in advance!!!
|
2025-03-21T14:48:30.188443
| 2020-03-31T20:38:13 |
356220
|
{
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"Andreas Blass",
"David Eppstein",
"Gerry Myerson",
"Noah Schweber",
"Richard Stanley",
"Will Brian",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/356220"
}
|
Stack Exchange
|
Monotone subsets of uncountable plane sets
Let $S$ be an uncountable set of points in the Euclidean plane. Define a subset of $S$ to be upward-monotone if every two points determine a line with non-negative slope, and downward-monotone if every two points determine a line with negative or infinite slope. It follows from the Erdős–Dushnik–Miller theorem that one or more of the following three things happens:
$S$ contains an uncountable upward-monotone subset
$S$ contains an uncountable downward-monotone subset
$S$ contains both a countably-infinite upward-monotone subset and a countably-infinite downward-monotone subset
(This can be viewed as an infinite version of the Erdős–Szekeres theorem on the existence of square-root-sized monotone subsets of finite sets of points in the plane.)
My question concerns sets $S$ that satisfy only the third possibility: $S$ does not have an uncountable monotone subset but does have countable monotone subsets in both directions. Is it known that there exist or that there do not exist sets $S$ for which this is true? If they are known to exist, how can they be constructed? Alternatively, are both the existence and the non-existence of such a set $S$ known to be consistent?
I think the Open Coloring Axiom implies that no such set exists? (If we make infinite slope part of upward-monotone instead, we get that downward-monotone is an open condition - and I don't think that rephrasing of the problem alters the answer.)
Euclidean plane has neither up nor down. Cartesian plane ....
I think the continuum hypothesis implies that such an $S$ exists. The key ingredient is that every upward-monotone set in included in the graph of a (non-strictly) increasing function $\mathbb R\to\mathbb R$ that is continuous except for countably many jumps, while downward-monotone sets are included in inverses of decreasing such functions. Since there are (under CH) only aleph_1 such functions, one can build the required $S$ by an induction of $\aleph_1$ steps. Step $\xi$ puts a point into $S$ that isn't on the graph of any of the first $\xi$ of the relevant functions.
@AndreasBlass: Very nice! Your argument also shows (unconditionally) that there is a subset $S$ of $\mathbb R^2$ with $|S| = \mathfrak{c}$ such that any upward-monotone or downward-monotone subset of $S$ has size $<!\mathfrak{c}$.
@WillBrian: you say "unconditionally," but the argument assumes that some uncountable set can be well-ordered.
@RichardStanley: The factory settings for my brain have the Axiom of Choice option activated. Sometimes I am able to turn this option off manually when the need arises, but usually I think of ZFC as my background theory, and an "unconditional" argument means (to me) one that uses no set-theoretic hypotheses beyond ZFC. But your comment raises an interesting question: Is ZF consistent with "every size-$\mathfrak{c}$ subset of $\mathbb R^2$ has a size-$\mathfrak{c}$ subset that is either upward-monotone or downward-monotone"? (I suspect the answer is yes, but I'm not sure.)
@WillBrian: the situation is similar to the result that assuming ZFC, there exists a poset of size $\mathfrak{c}$ such that every chain and every antichain is countable. (Simply take the intersection of the usual ordering of $\mathbb{R}$ with a well-ordering of $\mathbb{R}$.) Is ZF consistent with the statement that no such poset exists? I don't know whether this is known.
@RichardStanley: Yes, this is more or less just the special case of the poset problem for posets of (Dushnik–Miller!) dimension two, and thinking of it that way is how I got to the problem.
|
2025-03-21T14:48:30.188719
| 2020-03-31T21:48:46 |
356225
|
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"ABC",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/356225"
}
|
Stack Exchange
|
Flat augmentation ideal of a group-ring
If $G$ is a group and $I$ the augmentation ideal $I=Ker(\mathbb{Z}G\rightarrow \mathbb{Z})$ suppose that:
$I$ is a flat (right) $\mathbb{Z}G$-module.
$I$ is a finitely generated (right) $\mathbb{Z}G$-module.
If $J$ is a finitely generated flat (right) $\mathbb{Z}G$-submodule of $\mathbb{Z}G$, Does it follow that there exists an other $L$ finitely generated flat (right) $\mathbb{Z}G$-submodule of $\mathbb{Z}G$ such that $J\oplus L\cong I$ as $\mathbb{Z}G$-module?
You are asking about finitely generated groups of homological dimension 1. There is a lot of literature on this.
Do you mean equality or isomorphism? For instance of $G$ is an infinite cyclic group, then the augmentation ideal is free on one generator. If we view the ring as Laurent polynomials in the variable $x$ then the augmentation ideal is generated by $x-1$. If I consider the ideal generated by $(x-1)^2 $ it is free on one generator but is not literally a direct summand in the augmentation ideal but it is isomorphic as a module.
@BenjaminSteinberg i do mean isomorphism!
Do you know the answer for free groups?
Probably the answer is no. Take G to be a free group of rank 2. The augmentation ideal is then a free module of rank 2. If what you want is true then all flat finitely generated ideals are 2-generated. I suspect there are free submodules of any rank. I would guess the kernel of the projection to the group ring of Z/nxZ/n is free and the rank should be finite and increasing in $n$.
@BenjaminSteinberg I do not know the answer for free groups.
|
2025-03-21T14:48:30.188859
| 2020-03-31T22:47:06 |
356227
|
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|
Stack Exchange
|
Involution in a commutative unital real C* algebra
It follows immediately from Gelfand duality that the involution in a commutative unital real C* algebra is the identity. Is there a direct proof from the axioms of C* algebras?
Is that really true? Consider the set of real $2\times 2$ matrices of the form $\left[\begin{matrix}a&b\cr -b&a\end{matrix}\right]$. That's a commutative unital algebra which is stable under transpose, so a real C*-algebra according to the definition I know, but the involution is the transpose operation, which is not the identity here.
|
2025-03-21T14:48:30.188933
| 2020-03-31T23:10:36 |
356229
|
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"Augusto Santi",
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"https://mathoverflow.net/users/51189",
"zeraoulia rafik"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/356229"
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|
Stack Exchange
|
Some interesting experimental results about the distribution of primes
Let's consider the following metric of the gap between consecutive primes
$$m(k)=\frac {p_k^2-p_{k-1}^2} {24}\;\;\;\;\;(k\ge4)$$
Now, let's define the function
$\delta(k)=m(k)\;\;\;\;$ if $\,m(k)\,$ is prime $\;\;\;\;(1)$
$\delta(k)=0\;\;\;\;\;\;\;\;\;\;$ otherwise $\,\;\;\;\;\;\;\;\;\;\;\;\;\;(2)$
If we plot the graph of $\,m(k)\,$ we obtain (up to $\,k=10^4$):
But, if we plot the graph of $\,\delta(k)$, we obtain something completely different (up to $\,k=10^4$):
In the second graph the three different increasing curves seem to have
the following asymptotic behaviors:
$$\sim \frac {p_k} {2}$$
$$\sim \frac {p_k} {3}$$
$$\sim \frac {p_k} {6}$$
Further, the prime numbers are not uniformly distributed among the three curves. The number of primes decreases passing from the highest
to the lowest curve (I'm trying to estimate the number of primes that fall on each curve).
I do not have adequate knowledge of analytic number theory to
explain the phenomenon, therefore I ask for your help.
Many thanks.
The asymptotic behaviors you got is just derived from the Cramer conjecture, and you have multiplied the equality of Cramer conjecture by even integer which is p_k+p_(k-1) , Could you deduce some thing from the plot of Cramer for your graph plot ?
Your question is not of research level. If $p_k-p_{k-1}=2$, then your expression equals $(p_k+p_{k-1})/12$, which is asymptotically $p_k/6$. Similarly, if $p_k-p_{k-1}=4$, then your expression is asymptotically $p_k/3$. Finally, if $p_k-p_{k-1}=6$, then your expression is asymptotically $p_k/2$. So what you observe follows easily from well-known conjectures on gaps between primes.
I read experimental evidence this way: if $,m(k),$ is prime, then it will be $,\sim p_k/2,$ or $,\sim p_k/3,$ or $,\sim p_k/6$, no other possibility given. The question is to estimate the probability of the three cases.
See my response below. It answers your question in full.
Assume that $k\geq 5$ and $m(k)$ is prime. Observe the factorization
$$\frac{p_k-p_{k-1}}{2}\cdot\frac{p_k+p_{k-1}}{2}=6m(k).$$
On the left hand side, the second fraction is greater than $6$, hence the first fraction is smaller than $m(k)$. So the first factor is relatively prime to $m(k)$, whence it divides $6$. So $p_k-p_{k-1}$ equals $2$ or $4$ or $6$ or $12$.
If $p_k-p_{k-1}=2$, then $m(k)=(p_k+p_{k-1})/12\sim p_k/6$.
If $p_k-p_{k-1}=4$, then $m(k)=(p_k+p_{k-1})/6\sim p_k/3$.
If $p_k-p_{k-1}=6$, then $m(k)=(p_k+p_{k-1})/4\sim p_k/2$.
If $p_k-p_{k-1}=12$, then $m(k)=(p_k+p_{k-1})/2$, which is a contradiction, because there is no prime between $p_k$ and $p_{k-1}$.
This explains the "three increasing curves" $p_k/6$, $p_k/3$, $p_k/2$. The density of these three cases (i.e. the validity of two linear equations in the three primes $p_k$, $p_{k-1}$, $m(k)$) is heuristically known, see e.g. Conjecture 1.2 in Green-Tao: Linear equations in primes.
|
2025-03-21T14:48:30.189163
| 2020-03-31T23:26:38 |
356231
|
{
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"Dmitry Vaintrob",
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|
Stack Exchange
|
Localization of symmetric monoidal categories and geometry
I have a series of vague questions, related to localization of symmetric monoidal categories.
Here is the context. Say we are working over a field of characteristic zero. Then the "one category level higher" version of (DG) commutative ring is a (DG) symmetric monoidal category. It is well-known that for $X$ a scheme (or even, IIUC, a Noetherian stack with affine diagonal) $X$ can be recovered from the DG symmetric monoidal category of its quasicoherent sheaves. (Moreover, the functor $\text{Schemes}\to\text{SymMonCat}$ is fully faithful, including in an $\infty$-categorical context.
Now if $R$ is a commutative ring, we say that $S$ is a localization of $R$ if it can be obtained from $R$ by inverting some set of elements. If $R, S$ are both Noetherian, then there is a very nice alternative way of characterizing localizations:
(*) A map $R\to S$ is a localization if and only if the product map $S\otimes_R S\to S$ (derived tensor product) is an equivalence.
Now there are (at least) three interesting notions of localization for a DG symmetric monoidal category $\mathcal{C}$ (note all make sense also for just monoidal categories, and that when taking universal objects in the category of categories I'm going to be vague about what I'm requiring from categories: I'm willing to assume compactly generated, idempotent complete, etc.).
Localization along a morphism $f:X\to Y.$
Localization along an object $X$ (defined as the universal symmetric monoidal DG category admitting a functor from $\mathcal{C}$ where $X$ is $\otimes$-invertible, maybe satisfying some additional conditions).
"Parallel" localization along a morphism: if $f:X\to Y$ is a morphism, I'm defining this to be the initial category in which $X,Y$ are invertible and there exists a map $f':X^{-1}\to Y^{-1}$ with $f\otimes f' = \text{id}:\mathbb{I}\to \mathbb{I},$ where the equality is understood via an appropriate system of coherences.
(Of course 2. is a special case of 3.)
Here are some questions.
Defining $\otimes$ in terms of colimit in the category of symmetric monoidal categories, is there a context where (*) holds for symmetric monoidal categories (i.e., localizations can be characterized by a tensor-idempotence condition)?
Are there "interesting" localizations of this type of the category $\mathcal{C}$ of DG quasicoherent coherent sheaves on a scheme $X$ which do not come from geometric localizations? If no, are there any examples in more general contexts that have been studied or computed in some sense? (For example, what happens if you $\otimes$-localize the category of vector spaces along a two-dimensional vector space?)
By functoriality and universality, localizations of any of the types 1., 2., 3. can be "combined" (in a commutative way), and (by formal nonsense), the result of applying two localizations $\mathcal{C}\to \mathcal{C}_1$ and $\mathcal{C}\to \mathcal{C}_2$ is the colimit $\mathcal{C}_1\otimes_{\mathcal{C}}\mathcal{C}_2$.
**Definition**. For $\mathcal{C}_1\leftarrow\mathcal{C}\to \mathcal{C}_2$ a pair of localizations of $\mathcal{C}$ as above, set $$\mathcal{C}_{12}: = \mathcal{C}_1\otimes_{\mathcal{C}}\mathcal{C}_2$$ for the "combined" localization. Say that $\mathcal{C}_1$ and $\mathcal{C}_2$ *cover* $\mathcal{C}$ if $\mathcal{C}$ is the limit of the diagram $\mathcal{C}_1\to \mathcal{C}_{12}\leftarrow \mathcal{C}_2.$ My question now is: are there interesting examples of covers of DG symmetric monoidal categories in this sense other than Zariski covers in geometry?
Given a symmetric monoidal functor $\mathcal{C} \rightarrow \mathcal{C}'$, the property that $\mathcal{C}' \otimes_{\mathcal{C}} \mathcal{C}' \rightarrow \mathcal{C}'$ be an isomorphism is equivalent to $\mathcal{C} \rightarrow \mathcal{C}'$ being an epimorphism in the category of symmetric monoidal dg categories. This is a purely formal fact: in any cocartesian symmetric monoidal category $\mathcal{E}$ a map out of the initial object $0 \rightarrow Z$ is an epimorphism if and only if the map $Z \coprod Z \rightarrow Z$ is an isomorphism. Our case follows from this by taking $\mathcal{E}$ to be the category of symmetric monoidal categories over $\mathcal{C}$.
The three notions of localization mentioned in the question are epimorphisms, since a map out of the localization is by definition a map out of $\mathcal{C}$ satisfying a property (namely, that a certain arrow becomes invertible, that a certain object becomes invertible, etc). Hence all three notions satisfy the tensor idempotence condition.
Inverting an arrow $f: X \rightarrow Y$ amounts to passing to the quotient by the ideal generated by the cofiber of $f$. A symmetric monoidal functor $F: \mathcal{C} \rightarrow \mathcal{D}$ maps $\operatorname{cofib}(f)$ to zero if and only if it inverts $1_{\mathcal{C}} \oplus \operatorname{cofib}(f)$. Therefore your first notion of localization is a particular case of the second one.
Your third notion of localization is in fact equivalent to the second one. Given a map $f: X \rightarrow Y$ between invertible objects, any such map $f': X^{-1} \rightarrow Y^{-1}$ is necessarily dual to an inverse to $f$. Hence localizing in your third way along an arbitrary map $f:X \rightarrow Y$ is equivalent to first inverting $X, Y$ and then inverting $f$, which we already observed can be reduced to the second notion.
There are epimorphisms that do not arise as quotients by ideals: Consider for instance the category $\operatorname{Sh}(M)$ of sheaves of (complexes of) vector spaces on a manifold $M$. Let $x$ be a point in $M$ and $U$ its complement. The star pullback functor $\operatorname{Sh}(M) \rightarrow \operatorname{Sh}(\lbrace x \rbrace)$ exhibits $\operatorname{Sh}(\lbrace x \rbrace)$ as the quotient of $\operatorname{Sh}(M)$ by the ideal of sheaves with vanishing stalk at $x$. Similarly, $\operatorname{Sh}(U)$ is the quotient of $\operatorname{Sh}(M)$ by the ideal of sheaves supported at $x$. The ideal generated by the union of these two ideals is the whole $\operatorname{Sh}(M)$, so we see that $\operatorname{Sh}(\lbrace x \rbrace) \otimes_{\operatorname{Sh}(M)} \operatorname{Sh}(U) = 0$. It follows that the functor $\operatorname{Sh}(M) \rightarrow \operatorname{Sh}(\lbrace x \rbrace) \times \operatorname{Sh}(U)$ satisfies the tensor-idempotence condition, but it doesn't arise as the quotient by an ideal since its right adjoint $\operatorname{Sh}(\lbrace x \rbrace) \times \operatorname{Sh}(U) \rightarrow \operatorname{Sh}(M)$ (given by star-pushforward in each coordinate) is not fully faithful.
Under tameness conditions all notions of localization agree: I don't know if every epimorphism arises by inverting an object in general, but under certain tameness conditions one can show that this is the case:
Claim: Let $\mathcal{C}$ be a symmetric monoidal dg category compactly generated by its dualizable objects and $\mathcal{C}'
$ be a compactly generated symmetric monoidal dg category equipped with a symmetric monoidal functor $\mathcal{C} \rightarrow \mathcal{C}'$ that preserves compact objects, and such that the map $\mathcal{C}' \otimes_{\mathcal{C}} \mathcal{C}' \rightarrow \mathcal{C}'$ is an isomorphism. Then the functor $\mathcal{C} \rightarrow \mathcal{C}'$ arises by passing to the quotient by an ideal of $\mathcal{C}$.
Sketch of proof: Let $\mathcal{K}$ be the full subcategory of $\mathcal{C}'$ generated under colimits by the image of the functor $\mathcal{C}\rightarrow \mathcal{C}'$. Our conditions guarantee that the right adjoint to the inclusion $\mathcal{C} \rightarrow \mathcal{K}$ is colimit preserving and monadic. Note moreover that $\mathcal{K}$ is a $\mathcal{C}$-module and the functor $\mathcal{C} \rightarrow \mathcal{K}$ is a map of $\mathcal{C}$-modules. Its right adjoint in principle only commutes with the $\mathcal{C}$-action up to natural transformations, but the fact that $\mathcal{C}$ is compactly generated by its dualizable objects guarantees that the natural transformations are isomorphisms, and so the functor $\mathcal{K}\rightarrow \mathcal{C}$ is also a morphism of $\mathcal{C}$-modules. It follows that $\mathcal{K}$ is the category of algebras for a $\mathcal{C}$-linear monad on $\mathcal{C}$, and so we have an identification $\mathcal{K} = A\operatorname{-mod}$ for some algebra $A$ in $\mathcal{C}$. The fact that $\mathcal{C}' \otimes_{\mathcal{C}} \mathcal{C}' = \mathcal{C}'$ implies that the multiplication map $A \otimes A \rightarrow A$ is an isomorphism. This means that $\mathcal{K}$ is in fact the category of algebras for an idempotent $\mathcal{C}$-linear monad, and so it arises as the quotient of $\mathcal{C}$ by an ideal. This whole thing reduces you to understanding the case when $\mathcal{C} \rightarrow \mathcal{C}'$ is fully faithful. Since the canonical map $\mathcal{C}\otimes_{\mathcal{C}}\mathcal{C}' \rightarrow \mathcal{C}'\otimes_{\mathcal{C}} \mathcal{C}'$ is an isomorphism, we have that $\mathcal{C}'/\mathcal{C} \otimes_{\mathcal{C}} \mathcal{C}'$ vanishes. This contains $\mathcal{C}'/\mathcal{C} \otimes_{\mathcal{C}} \mathcal{C} = \mathcal{C}'/\mathcal{C}$ as a full subcategory, so we see that $\mathcal{C}' = \mathcal{C}$ is the trivial localization.
The algebro-geometric case: The above includes for instance the case of $\mathcal{C} = \operatorname{QCoh}(X)$ for $X$ a separated scheme. Moreover, from the proof we see that the resulting localizations are categories of modules for quasicoherent sheaves of algebras $A$ over $X$ such that the multiplication map $A \otimes A \rightarrow A$ is an isomorphism. In the Noetherian case you can use the result stated in the question to deduce that $A$ is locally a localization of the structure sheaf, so you see that all localizations in this sense are classified by collections of points of $X$ closed under specialization.
If you drop the condition that the functor $\mathcal{C} \rightarrow \mathcal{C}'$ preserves compact objects there are more examples, even in the geometric case. Indeed, any ideal of $\operatorname{QCoh}(X)$ provides an example, and these are classified (in the Noetherian case) by arbitrary collections of (non necessarily closed) points of $X$. This classification goes back to Hopkins, Neeman, and by now there is a whole industry about it - key words being tensor triangular geometry and classification of localizing subcategories.
Beyond algebraic geometry: If you are looking for interesting covers beyond Zariski covers in algebraic geometry one source could be topology. If you have a manifold $M$ and $U$ is an open set of $M$, the category $\operatorname{Sh}(U)$ is the category of comodules for an idempotent coalgebra in $\operatorname{Sh}(M)$ with underlying sheaf $k_U$, and so it can be obtained as the colimit
$$\operatorname{Sh}(M) \xrightarrow{\otimes k_U}\operatorname{Sh}(M) \xrightarrow{\otimes k_U} \operatorname{Sh}(M)\xrightarrow{\otimes k_U} \ldots .$$
This is the same diagram you would use to invert $k_U$, and so $\operatorname{Sh}(U)$ in fact results from inverting $k_U$. From this it follows that $\operatorname{Sh}(U)\otimes_{\operatorname{Sh}(M)} \operatorname{Sh}(V) = \operatorname{Sh}(U \cup V)$ for every pair of opens, and so you see that any open cover of $M$ provides a cover of symmetric monoidal categories in your sense.
You can build many more examples by variations of this theme. You could take $M$ to be the union of two manifolds $U, V$ along a closed submanifold (for instance $M$ could be the union of the $x$ and $y$ axes in $\mathbb{R}^2$) and you still have that $\operatorname{Sh}(M)$ is covered by $\operatorname{Sh}(U)$ and $\operatorname{Sh}(V)$. You could even require your sheaves to be constructible with respect to a stratification and $U, V$ to respect the stratification to get covers of categories of modules over quivers.
Thank you German for this insightful and thorough answer! I thought the third localization may follow from the first two but couldn't see how. Does your argument assume rigidity, or does inverting $X$ automatically imply that $X, X^{-1}$ are dual and thus maps $X\to Y$ have dual maps $Y^{-1}\to X^{-1}$? Your implication (2) => (1) is very nice, and thank you for the reference to Ballmer, Hopkins, Neeman and co.
@DmitryVaintrob No rigidity needed. I believe it is a general fact for any symmetric monoidal category that if an object $X$ is invertible then it is dualizable with dual $X^{-1}$. The idea is that for any object $Y$ one has that $X^{-1}\otimes Y$ is an internal Hom between $X$ and $Y$. Therefore the map $[X, 1] \otimes X \rightarrow [X, X]$ is an isomorphism (where brackets denote internal hom), which is equivalent to $X$ being dualizable. Full details seem to be supplied in https://pages.uoregon.edu/ddugger/invcoh.pdf proposition 4.11. As you say, this provides the dual maps that you need.
|
2025-03-21T14:48:30.190262
| 2020-03-31T23:32:50 |
356233
|
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Stack Exchange
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Classification of involutions on $G_{2}$-homogeneous spaces
Are you aware of a systematic classification of involutions on $G_{2}$-homogeneous spaces?
No; but what do you mean by a $G_2$-homogeneous space? Is your $G_2$ an algebraic group, and if yes, then over what field? If the field is $\mathbb R$, then is your $G_2$ compact or not?
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2025-03-21T14:48:30.190323
| 2020-03-31T23:57:50 |
356234
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Stack Exchange
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Applications of the Atiyah-Patodi-Singer eta-function $\eta(s)$
The eta function of a differential operator was used by Atiyah, Patodi and Singer to derive their famous index theorem, and is given by
$$
\eta(s)=\sum_{\lambda\neq 0}(\mathrm{sign}\lambda)|\lambda|^{-s},
$$
where the sum goes over the nonzero eigenvalues of the operator. The limit $\lim_{s\to 0}\eta(s)$ is the eta invariant that appears in the index theorem itself.
My question is: are there known applications for the eta function evaluated at other values of $s$ ?
Let's assume we're talking about the spinor Dirac operator associated to a Riemannian spin manifold. If you multiply the Riemannian metric by some constant then the eigenvalues of the Dirac operator scale by the same constant, and thus the value of the eta function at $s \neq 0$ changes. So the value at $s = 0$ is the only one we can expect to be an interesting global invariant (though if memory serves it is not actually invariant under all isometries).
@PaulSiegel Thank you for your comment, I am aware of the scaling property. I'm not exclusively interested in global invariants, any potential application would be great to hear about.
The point of this observation is that it limits the sort of applications that one can hope for. $\eta(0)$ has interesting applications because it solves an index problem on manifolds with boundary; $\eta(s)$ can't solve any sort of index problem because it depends too sensitively on the metric. This doesn't rigorously prove that $\eta(s)$ is uninteresting - that's why I left this as a comment instead of an answer - but I don't see much cause for optimism.
Thank you for the clarification! This link states that $\eta(1)$ of a Dirac operator is its propagator: https://ncatlab.org/nlab/show/eta+invariant However, I was not able to find more than this.
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2025-03-21T14:48:30.190482
| 2020-04-01T00:17:19 |
356235
|
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}
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Stack Exchange
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What is the number of iterations needed for the message passing algorithm to converge when applied to an acyclic factor graph?
I understand that the message passing algorithm (Belief Propagation algorithm), when applied to a factor graph consists in an exchange in messages between the factor nodes and the variable nodes, these messages keep being updated until a certain number of iterations where we reach convergence. And here is my question: What is the number of iterations needed for the message passing algorithm to converge when applied to a tanner graph (factor graph)? Is there some conditions for the convergence to happen? if yes, what are these conditions? I am not using a loopy graph, my graph is an acyclic factor graph. Thanks in advance!
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2025-03-21T14:48:30.190559
| 2020-04-01T00:49:47 |
356238
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{
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"Watson",
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Stack Exchange
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Another generalization of the Gauss circle problem
In this question I asked for a generalization of the Gauss circle problem, the type of generalization I am asking is to view the Gauss circle problem as one about counting algebraic integers of bounded norm.
A more geometric generalization is the following: let $\Lambda$ be a unimodular lattice in $\mathbb{R}^2$ (that is, it has determinant $\pm 1$). Consider the counting function
$\displaystyle N_\Lambda(X) = \# \{(u,v) \in \Lambda : u^2 + v^2 \leq X \}.$
Then by geometry of numbers, one has
$$\displaystyle N_\Lambda(X) = \pi X + O_\Lambda\left(X^{1/2} \right).$$
Thus a generalized version of Gauss's circle problem is to ask: how does the error term $O_\Lambda\left(X^{1/2}\right)$ depend on $\Lambda$, and can the exponent be made to be $1/4$?
It can be shown that the error term certainly cannot be made to be independent of $\Lambda$, since if $\Lambda$ has an extremely short vector then there can be arbitrarily many multiples of that vector sitting inside the circle $x^2 + y^2 \leq X$.
Equivalently, we can study integral points in ellipsoids, and p. 95 of this PhD thesis has a nice survey.
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