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2025-03-21T14:48:29.627060
| 2020-01-14T20:24:50 |
350441
|
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"authors": [
"Pádua",
"Willie Wong",
"https://mathoverflow.net/users/129014",
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}
|
Stack Exchange
|
Fourier transform for $H^2(\mathbb{R}^N)$, $N\geq 5$
How i can prove that if $u\in H^2(\mathbb{R}^N)$ then $u\in \mathcal{F}(L^{p^*}(\mathbb{R}^N))$, where $1/p+1/{p^*}=1,$ $2\leq p<2N/(N-4)$?
If $u \in H^2(\mathbb{R}^N)$, then its Fourier transform satisfies $\hat{u} \in L^2$ and $(1 + |\xi|^2) \hat{u} \in L^2$. By Holder inequality you have
$$ \|\hat{u}\|_{q} \leq \| (1 + |\xi|^2)^{-1} \|_{r} \|(1 + |\xi|^2) \hat{u} \|_{2} $$
for appropriate $q^{-1} = r^{-1} + 2^{-1}$. For the $L^r$ integration to be bounded you need $2r > N$. Work through the algebra you get what you want.
Sorry, @WillieWong! I dind't understand how $\hat{u}\in L^q$ implies $u\in \mathcal{F}(L^q)$?
@Pádua maybe I misunderstood your notation, but I thought you meant by $u\in \mathcal{F}(L^q)$ that $u$ has a Fourier transform that is in $L^q$. Do you mean something else?
Yes, @WillieWong! I wanted to say that $u=\mathcal{F}(v)$ for some $v \in L^{q}\left(\mathbb{R}^{N}\right)$, $1\leq q\leq 2$.
@Pádua: then is it not what I just stated? You can easily go from the Fourier transform $\hat{u}$ of the function $u$ to the function $v$ (which is just the inverse Fourier transform of $u$).
Sorry @WillierWong, but i don't know how go from the Fourier transform to inverse. There exists some relation between them?
@Pádua: you just compose with reflection. See eg https://en.wikipedia.org/wiki/Fourier_inversion_theorem
I think I understand @WillieWong. Thank you!
|
2025-03-21T14:48:29.627296
| 2020-01-14T21:46:45 |
350445
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Sam Hopkins",
"https://mathoverflow.net/users/25028"
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"site": "mathoverflow.net",
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"url": "https://mathoverflow.net/questions/350445"
}
|
Stack Exchange
|
Determinantal formula for plane partitions of shifted shape
For $\lambda = (\lambda_1,\ldots,\lambda_{\ell})$ a partition, a (weak) plane partition of shape $\lambda$ is a filling of the Young diagram of $\lambda$ with nonnegative integers such that entries weakly decrease in rows and columns. Let me use $\mathrm{PP}^{m}(\lambda)$ to denote the set of plane partitions of shape $\lambda$ with entries $\leq m$. MacMahon (see pg. 367 of "Memoir on the theory of the partitions of numbers. Part VI. Partitions in two-dimesional space, to which is added an adumbration of the theory of the partitions in three-dimensional space") proved that
$$ \# \mathrm{PP}^{m}(\lambda) = \mathrm{det} \left( \binom{\lambda_i + m}{i-j+m} \right)_{i,j=1,\ldots,\ell}$$
Remark 1: Actually, MacMahon proved a $q$-analog; namely, that
$$ \sum_{\pi \in \mathrm{PP}^{m}(\lambda)} q^{\textrm{sum of entries in $\pi$}} = \mathrm{det} \left( q^{f(i,j)} \binom{\lambda_i + m}{i-j+m}_q \right)_{i,j=1,\ldots,\ell}$$
where $\binom{a}{b}_q$ is the usual $q$-binomial and $f(i,j) := \begin{cases} \binom{j-i}{2} &\textrm{if $j>i$}, \\ \binom{i-j+1}{2} &\textrm{if $i \leq j$}. \end{cases}$
Remark 2: Kreweras ("Sur une classe de problèmes de dénombrement liés au treillis des partitions des entiers") extended MacMahon's result to skew shapes; namely,
$$ \# \mathrm{PP}^{m}(\lambda/\mu) = \mathrm{det} \left( \binom{\lambda_i-\mu_j + m}{i-j+m} \right)_{i,j=1,\ldots,\ell}$$
Remark 3: A modern way to prove this result is as follows: first note that by doing $m-$ to every entry, we can count bounded reverse plane partitions of shape $\lambda$ instead of plane partitions; then note that by adding $i$ to all entries in row $i$, we can count semistandard Young tableaux of shape $\lambda$ with a certain flag of bounds on each row. These flagged SSYT can then be enumerated via the Gessel-Viennot method of nonintersecting lattice paths.
Now let $\lambda$ be a strict partition (i.e., one whose parts strictly decrease). The shifted Young diagram corresponding to $\lambda$ is obtained from the Young diagram by indenting each row by one box from the row above it. There are often shifted analogs of all the nice combinatorial results concerning unshifted Young diagrams. My question concerns such an analog.
A shifted plane partition of shape $\lambda$ is a filling of the shifted Young diagram of $\lambda$ with nonnegative integers such that entries weakly decrease in rows and columns. Let me use $\mathrm{SPP}^{m}(\lambda)$ to denote the set of shifted plane partitions of shape $\lambda$ with entries $\leq m$.
Question: Is there a formula expressing $\mathrm{SPP}^{m}(\lambda)$ as a determinant of binomial coefficients analogous to the MacMahon/Kreweras formula?
Remark 4: Stembridge ("Nonintersecting paths, pfaffians, and plane partitions") gave a Pfaffian formula for shifted plane partitions. To me, a Pfaffian is just as good as a determinant. The problem is, the entries of the matrix that Stembridge constructs are not simply binomial coefficients; rather, they are sums of differences of products of binomial coefficients. I see no obvious way to simplify the matrix Stembridge constructs. EDIT: In case anyone might need it, the Pfaffian formula is stated clearly as Theorem 2.2 in https://arxiv.org/abs/2007.05381 (it does not explicitly appear in Stembridge's paper). I do believe now that this is the best general formula for counting bounded shifted plane partitions of arbitrary shape.
My motivation for this question is that it could be useful for answer my other question. (EDIT: That question has been answered by other means.)
Having gone through more literature, my guess is that the answer is no: Stembridge's Pfaffian formula (where the matrix has entries of the form $\sum \binom{a}{b}\binom{c}{d} - \binom{a'}{b'}-\binom{c'}{d'}$) is the best known general formula for counting shifted plane partitions.
Previous comment should say "... of the form $\sum \binom{a}{b}\binom{c}{d} - \binom{a'}{b'}\binom{c'}{d'}$"
|
2025-03-21T14:48:29.627527
| 2020-01-14T22:35:58 |
350447
|
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"LSpice",
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|
Stack Exchange
|
Approximate constant function
Let $f:[0,1]^2 \rightarrow \mathbb C$ be an $H^1$ function with the property that $f(x,x)=0$ and $\Vert f \Vert_{L^2[0,1]}=1.$
Does there exist a constant $c>0$ such that any such function satisfies
$$ \Vert f-1 \Vert_{H^1}>c?$$
I was thinking that the Fourier series could help to prove or disprove something like this, but I did not get far so far.
It would be clearly possible in $L^2$ norm let's say, but I find it tricky in Sobolev norms.
Should $|f|{L^2[0, 1]}$ be $|f|{L^2[0, 1]^2}$?
For all $x$ and $y$ in $[0,1]^2$
$$f(x,y)=
\left\{
\begin{aligned}
\int_x^y f_y(x,z)\,dz&\text{ if }x\le y, \\
-\int_y^x f_y(x,z)\,dz&\text{ if }x\ge y,
\end{aligned}
\right.
$$
where $f_y(x,z):=\frac{\partial f(x,y)}{\partial y}|_{y=z}$, so that
$$|f(x,y)|\le\int_0^1|f_y(x,z)|\,dz\le\sqrt{\int_0^1|f_y(x,z)|^2\,dz}.
$$
Hence,
$$
\begin{split}
1 & = \iint_{[0,1]^2}dx\,dy\,|f(x,y)|^2 \\
& \le \iiint_{[0,1]^3}dx\,dy\,dz\,|f_y(x,z)|^2\\
& = \iint_{[0,1]^2}dx\,dz\,|f_y(x,z)|^2
\le\|f-1\|_{H^1}^2.
\end{split}
$$
So, any $c\in(0,1)$ will do.
|
2025-03-21T14:48:29.627619
| 2020-01-14T22:55:37 |
350448
|
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"authors": [
"Padraig Ó Catháin",
"Seva",
"https://mathoverflow.net/users/27513",
"https://mathoverflow.net/users/9924"
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"url": "https://mathoverflow.net/questions/350448"
}
|
Stack Exchange
|
Sidon sets in finite groups
Suppose $G$ is a group, $S \subset G$. Let’s call $S$ a Sidon subset iff $\forall$ quadruples $(a, b, c, d)$ of distinct elements of $S$ we have $ab \neq cd$ (named after Simon Sidon who studied such subsets in $C_\infty$). Let’s define $Sid(G)$ as the maximal possible cardinality of a Sidon subset in $G$. Do there exist such $0<c<C<+\infty$, such that $c \leq \frac{Sid(G)}{\sqrt{|G|}} \leq C$ for any finite group $G$?
The existence of $C$ can be proved the following way:
If $S$ is a Sidon subset of $G$, then $|G| \geq |S^2| \geq \frac{|S|(|S| - 1)}{2}$, thus $C = \sqrt{2} + 1$ will work.
The only thing I know about $c$ is that Erdos proved such constant exist for finite cyclic groups $G$. But does it exist for all finite $G$?
This question on MSE
The paper of Babai and Sos "Sidon Sets in Groups and Induced Subgraphs of Cayley Graphs" (European J. Comb. (1985) 6, pp.101-114) is highly relevant.
@MarkSapir: seems it does not.
Sets which meet the upper bound with equality are called planar difference sets, and are closely related to finite projective planes. I am not aware of work which addresses your question, but this might give you some additional terms to search for.
Not necessarily, though the character theory works out much more nicely when the group is abelian, so difference sets have been studied more intensively in that case.
|
2025-03-21T14:48:29.627744
| 2020-01-14T22:57:18 |
350449
|
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"authors": [
"Christian Remling",
"De vinci",
"Robert Furber",
"https://mathoverflow.net/users/151115",
"https://mathoverflow.net/users/48839",
"https://mathoverflow.net/users/61785"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/350449"
}
|
Stack Exchange
|
Two independent function when considered as random variable over $([0,1],\mathrm{Lebesgue},\mathcal{B}_{[0,1]})$
Does there exists two non-constant continuous functions $f,g:[0,1]\rightarrow \mathbb{R}$ so that they are in independent (in probability sense) when viewed as random variables over the measure space $([0,1],\mathrm{Lebesgue},\mathcal{B}_{[0,1]})$, where $\mathcal{B}_{[0,1]}$ is the Borel sigma field of $[0,1]$?
I was trying the following thing. Let $U(x)=x$ be be a function on $[0,1]$. This follows uniform distribution. Now I am taking binary exapnsion of $U$. It can be proved that
$$U = \sum_{k\geq 1}B_k/2^k,$$
where $B_k$'s are i.i.d. $\mathrm{Bernoulli}(1/2)$. In particular $$B_k(x)=\mathbb{1}(k^{\text{th}}\;\text{binary digit of $x$ after decimal point is 1}).$$ Here $\mathbb{1}(\cdot)$ is the indicator function. Now I am defining
$$f=\sum_{k\;\text{is odd}}B_k/2^k,\quad g=\sum_{k\;\text{is even}}B_k/2^k.$$
Thus $f$ and $g$ are independent but I can not prove they are continuous, though they are almost surely continuous w.r.t lebesgue measure. Any help will be appreciated. Please provide any other example if available. Thank you.
They are not continuous, for example $f$ jumps at $x=1/2$ (it's also not completely clear if you even defined $f$ at $x=1/2$, but of course that's the smaller problem).
Also, you probably want to add the requirement that $f,g$ are not constant a.e.
@ChristianRemling You are right. Constant functions should not be allowed here.
We can take a Peano curve $\varphi: [0,1]\to [0,1]^2$. More precisely, we want a construction like the one here that spends its fair share of the time in each dyadic square when traced out at unit speed. This will make sure that the image measure of Lebesgue measure on $[0,1]$ under $\varphi$ is (two-dimensional) Lebesgue measure again.
Of course, on the square we have random variables as desired (the coordinates), and we can now pull back to $[0,1]$ and take $f(x)=\varphi_1(x)$, $g(x)=\varphi_2(x)$.
The binary digits $\epsilon_1(x),\epsilon_2(x),\ldots$ of a number $x$ in $[0,1]$ are independent under Lebesgue measure, so the two random variables defined by
$$
f(x) := \sum_{k=1}^\infty {\epsilon_{2k}(x)\over 2^k}
$$
and
$$
g(x):=\sum_{k=1}^\infty{\epsilon_{2k-1}(x)\over 2^k}
$$
are independent, and they are evidently continuous functions of $x$. Both are uniformly distributed.
The function $f$ is not continuous. Consider the sequence $(x_n)$ where $x 1 = 0.01, x_2= 0.011, x_3 = 0.0111, \ldots$, and so on, in binary. This converges to $0.0111.... = 0.1$. But $f(0.1) = 0$, while $f(x_n)$ converges to $\sum{k=1}^\infty \frac{1}{2^k} = 1$.
A successful implementation of your strategy would amount to the same as Christian Remling's answer - defining a space-filling curve $[0,1] \rightarrow [0,1]^2$ whose projections are independent random variables. Though it may not be apparent from versions that are drawn using diagrams, Peano's original paper describes his curve using (ternary) digits.
|
2025-03-21T14:48:29.627934
| 2020-01-15T00:15:40 |
350451
|
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|
Stack Exchange
|
Showing existence of a solution to an underdetermined system of equations with non-negativity constraints
Let $K$ be a positive integer, let $p\in (0,1)$, and let $\{W(k,i),W^B(k,i), \varphi_k(i)\}_{1\leq i\leq k\leq K}$ be variables.
I need to prove that there exists a solution to the following system of equations:
\begin{equation} \label{eq1}
\begin{split}
W(k,i) & =\; & (1-p)\cdot W^B(k,i) &\\
&+ & p\cdot\sum_{j=1}^{i-1}\varphi_{k}(j)\cdot\left(W\left(k-1,i-1\right)+1\right) & \forall 1 \leq i\leq k\leq K\\
&+ & p\cdot\sum_{j=i+1}^{k}\varphi_{k}(j)\cdot\left(W\left(k-1,i\right)+1\right)& \\
W^{B}\left(k,i\right) & = &\; (1-p) \cdot\left(W\left(k,i\right)+1\right) + p\cdot W^{B}\left(k+1,i\right) & \;\; \forall 1 \leq k<K\\
W^{B}\left(k,i\right) & = &\; W\left(k,i\right)+1 & \text{ if } k=K \\
\sum_{i=1}^k \varphi_k(i) &=& 1 & \forall 1 \leq k \leq K \\
W^{B}\left(k,k\right) &=& W^{B}\left(1,1\right) & \forall 1 \leq k\leq K\\
\varphi_k(i) &\geq & 0 & \forall 1 \leq i \leq k \leq K
\end{split}
\end{equation}
I could manually calculated solutions for $K\in {2,3,4}$, for example there is a unique solution for $K=2$ in which $\varphi_2(1) = \frac{1}{3-p} $. As $K$ becomes larger the problem becomes underdetermined, and I could manually solve to find a family of solution for $K=3$, but I do not know how to show that a solution exists for any $K$.
I also know (from Little's law) that for valid solutions $W^B(k,k) = \frac{K+1}{2p}$.
Any help or direction will be greatly appreciated!
Background:
The first three sets of equations equations characterize the expected waiting time of an agent who joins a queue with maximal size $K$. The variables $\{\varphi_k(i)\} _{1\leq i \leq k}$ determine the queue policy. The question is equivalent to showing that there exists a queueing policy that randomizes assignments so that the expected wait for agents who join the queue is the same regardless of the position they join.
Here is the description of the stochastic system. Time is discrete and every period starts with an arrival of an item. With probability $p$ the item is of kind $A$, in which case the item is assigned to an agent on the queue, and the period ends. With probability $1-p$ the item is of kind $B$, in which case agents may join the queue while the system will try to assign the $B$ item. The system draws agents who might take the $B$ item. With probability $1-p$ a drawn agent (of type $\beta$) takes the $B$ item and the period ends; With probability $p$ a drawn agent (of type $\alpha$) joins the queue and the system continues to draw agents. If there are $K$ agents on the queue the next agent takes the $B$ (regardless of whether it is $\alpha$ or $\beta$), and the period ends.
The policy can select which agent on the queue should receive the arriving $A$ item. For a queue of maximal size $K$, we set the probabilities $\left\{ \varphi\left(i,k\right)\right\} _{1\leq i\leq k\leq K}$ where $\varphi_{k}\left(i\right)$ is the probability that the agent in position $i$ will receive an arriving $A$ item when there are $k$ agents on the queue. Agents move to the next position if someone ahead of them is assigned. Some agent must always be assigned, so $\sum_{i\leq k}\varphi_{k}\left(i\right)=1$.
Denote the expected wait until assignment of an agent in position $i$ out of $k$ agents at the beginning of the period by $W(i,k)$. Denote expected wait until assignment of an agent in position $i$ out of $k$ agents when the system is looking to allocate a $B$ item by $W^{B}(i,k)$.
The first three sets of equations correspond to the possible transitions, and determine the expected waiting times. The remaining equations ask that assignment probabilities sum to $1$ and are non-negative, and that all agents have the same expected wait when they join the queue.
Here is a link to the paper from which the question arises. Appendix C contains additional examples.
|
2025-03-21T14:48:29.628198
| 2020-01-15T05:54:10 |
350462
|
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"Robert Furber",
"Santi Spadaro",
"Timothy Chow",
"Todd Eisworth",
"Wlod AA",
"Wojowu",
"YCor",
"Zuhair Al-Johar",
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|
Stack Exchange
|
Necessary use of large cardinals in mathematics
There are some statements, whose consistency (or the consistency of their negation) require the existence of large cardinals (in the sense that if the statement (or its negation) is consistent, then it is consistent that there are some large cardinals). I know many of such examples inside set theory.
Question. What are examples of statements in mathematics (other than set theory), where it is known we need large cardinals to prove their consistency
(or the consistency of their negation)?
For each example, giving what kind of large cardinals are sufficient and what kind of large cardinals are necessary is appreciated. Please also give references.
It might be hard to distinguish "set theory" and "mathematics other than set theory". Also I guess "require the existence of large cardinals" is rather "require the consistency of the existence of large cardinals".
Do you count the Axiom of Infinity as a large cardinal axiom?
@MyNinthAccount It is part of ZFC. By large cardinal , I essentially mean a cardinal $\kappa$ which is at least inaccessible
Accessible categories come to mind.
Hmm, I was hoping for a mathematical question about theorems outside the set theory which are proved by involving a set of large cardinality.
@WlodAA This is that question.
Near-duplicate: https://mathoverflow.net/q/129575/30186 (it has narrower scope but its answers also answer this question). The short answer is: Harvey Friedman's work
What about Solovays model where all sets in $\mathbb{R}$ are Lebesgue measurable. It is not only set theory. See: Jech (1978), Set Theory, p. 537 ff. Must also be in the 3rd edition.
The normal Moore space conjecture in general topology is a classical example of such a statement. Its consistency has been proven from the consistency of a strongly compact cardinal (Nyikos) and implies an inner model with a measurable cardinal (Fleissner). But there are many more examples...maybe this question should be community wiki.
Well Fermat's last theorem was proved originally in a system that has large cardinals, so it was at least heuristically needed, although later on the proof was re-formulated within ZFC.
As a sort of trivial example, the question "does there exist a measurable cardinal" could itself be interpreted as a question from measure theory.
I find it hard to consider these general topology / measure theory facts to be "outside set theory".
A rather anecdotal result outside of set theory: there exists a Borel graph $G$ such that $ZF+DC+$"there is no maximal independent set in $G$" is equiconsistent with $ZFC+$inaccessible.
I don't think this is a duplicated question, it is more general, noting that two of the answers given here do not apply for the other question.
The dual of an abelian group $A$ is defined to be the group $\text{Hom}(A,\mathbb Z)$ of homomorphisms to the infinite cyclic group. As usual with such dualities, there's a canonical homomorphism from $A$ to its double dual
$$
A\to A^{**}:a\mapsto(h\mapsto h(a)).
$$
If this is an isomorphism, $A$ is said to be reflexive.
Question: Are all free abelian groups reflexive?
Answer: Yes if and only if there are no measurable cardinals.
This is amazing! Where is this proved?
@TimothyChow I'm sure it's in the book "Almost Free Modules" by Eklof and Mekler, but I can't check immediately because I'm away from home. The result is, if I remember correctly, due to Los, and it says in more detail that the free abelian group on a set $S$ of generators is reflexive iff there is no measurable cardinal $\leq|S|$.
It's the same situation with "All discrete spaces are realcompact", probably for the same reasons!
Another related fact with a similar proof is that a product of $\kappa$ bornological locally convex spaces is bornological iff $\kappa$ is less than the first measurable cardinal. A locally convex space $E$ is bornological iff for all locally convex spaces $F$ and bounded linear maps $f : E \rightarrow F$, $f$ is continuous. The proof goes by reducing to products of $\mathbb{R}$ (see Schaefer's Topological Vector Spaces, Chapter II, Exercise 19). The statement in II.8 only mentions inaccessible cardinals, I think because Hanf's work was very recent when it was published.
This is the shadow of the following more precise statement: the free abelian group $\mathbf{Z}^{(I)}$ is non-reflexive iff $I$ carries a ${0,1}$-valued probability defined an all subsets and vanishing on singletons. (This means $I\ge\kappa_0$ for $\kappa_0$ the first measurable cardinal, and that $\kappa_0$ exists.) So this is a result where the set $I$ is part of the input.
I think that the theorem first appeared in Fuchs' Abelian Groups of 1958, as Theorem 47.2. Fuchs attributes the theorem to Łoś.
Recall that the character of a point in a topological space is the smallest cardinality of a local base for that point. (So, for example, "first countable" = "every point has character $\aleph_0$".)
Question: Is there a compact Hausdorff space, containing more than one point, in which any two different points have different character?
Steve Watson found an answer to this question [S. Watson, "Using prediction principles to construct ordered continua," Pacific Journal of Mathematics 186, pp. 251-256 (link)] by showing that
Theorem: There is such a space if and only if there is a cardinal $\kappa$ and a set of cardinals $E \subseteq \kappa$ such that $\diamondsuit_\kappa(E)$ holds.
Watson points out in his paper that this condition is implied by $V=L$ plus the existence of a Mahlo cardinal, and it implies the existence of a weakly inaccessible cardinal. Thus
the consistency strength of a positive answer to the above question lies somewhere between an inaccessible cardinal and a Mahlo cardinal.
I do not know whether better bounds for this statement about $\diamondsuit_\kappa(E)$ have been found in the last 34 years. If you know of any improvements to the bounds Watson gives, please feel free to edit this post.
Finally, Watson shows from ZFC alone that there is an infinite $\sigma$-compact Hausdorff space in which any two different points have different character. Thus removing "compact" from the question makes it much easier, and removes the need for any large cardinals.
It was a traditional question of descriptive set theory (a question which can be formulated in the language of second order arithmetic) whether all projective sets are Lebesgue measurable. This remained an open problem for many decades, and for a good reason: it turned out that the statement is independent even of the full ZFC set theory (see Solovay 1970). Only by postulating the existence of some extremely large cardinals (so-called Woodin cardinals) can the hypothesis that all projective sets are Lebesgue measurable be proved (this was achieved as a consequence of their work on so-called projective determinacy by Woodin, Martin and Steel; see Woodin 1988; Martin & Steel 1988, 1989).
I didn't downvote, but I suspect that this may not meet the "outside set theory" criterion.
"Descriptive set theory" is a field of mathematics related to topology and was initiated by the French semi-intuitionists (Lebesgue, Baire, Borel). It studies sets which possess relatively simple definitions (in contradistinction to the ideas of arbitrary sets and various higher power-sets, which the semi-intuitionists rejected as meaningless) called projective or analytic sets.
It is common to count it, in the literature on the foundations of mathematics, the name notwithstanding, as "ordinary mathematics", in contradistinction to set theory.
|
2025-03-21T14:48:29.628735
| 2020-01-15T06:23:42 |
350463
|
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|
Stack Exchange
|
How to generating all flats of the cycle matroid of a graph?
If $M$ is a matroid, I can use M.flats(k) in SageMath to list all the flats of rank $k$. But I hope that there is an algorithm or program to list all flats of the cycle matroid of a graph. And do not use the language of matroids in this process.
In the paper "GRAPH COMPOSITIONS I: BASIC ENUMERATION", a composition of G is defined as a partition of V(G) into vertex sets of connected induced subgraphs of G. All the compositions of a graph G exactly corresponds to the flats of M(G)。 So my problem may be changed to "how to list all the compositions of a graph G".
Since the answer can be exponentially large as a function of the number of vertices, not clear what you mean (there’s an obvious algorithm that checks for each partition if it is a composition of G, but it’s very inefficient).
This works previewed the graph is 2-connected otherwise, reduce the problem to the blocks.
https://stackoverflow.com/questions/20530128/how-to-find-all-partitions-of-a-set
Give the edges of the graph distinct weights. Then the flats are in 1-1 correspondence with the minimal forests, where a forest is defined to be minimal if no other lower-weight forest spans the same components.
You can define a tree structure on the minimal forests, where the parent of a minimal forest is obtained by deleting its heaviest edge. The root of this tree structure is the forest of single-vertex trees. Then the children of each minimal forest F are formed by choosing two components of F, and adding the lightest edge between them, as long as it is heavier than the previous heaviest edge.
Finding the parent of a given minimal forest is obviously easy. And you can list all the children of a given minimal forest in linear time, by labeling each heavy-enough edge by the pair of components it connects, radix sorting the edges by these pairs of labels, and for each set of edges with the same pair of labels, keeping only the lightest one.
One can then find all minimal forests by traversing the tree structure starting from the root. At any given tree, the next tree is found by going to the first child (if there are any children) or following parent links back up the tree structure until finding a parent for which we are not backtracking from the last child, and then going to the next child.
Each link in the tree structure takes time linear in the input graph size to traverse, so the total time is linear per generated minimal forest. You only need enough space to remember two forests (the one you are at now and the one you just came from).
In the same linear time per generated object you can generate the flats themselves, not just the minimal forest. For any given minimal forest, the corresponding flat is the set of edges that connect pairs of vertices in the same tree of the forest. So label all vertices with which tree they are in and loop through all the edges checking which of them have the same label at both endpoints.
(This is an instance of "reverse search"; for the general design of enumeration algorithms with this technique, see Avis & Fukuda, "Reverse search for enumeration", Disc. Appl. Math. 1996. I don't know offhand of references for the specific problem of enumerating flats of graphs.)
Matroids: A Geometric Introduction page 55 gives the theoretic result in purely graph theoretic terms in the last two sentences of this quote:
While it is pretty easy to pick out the flats from the picture of the geometry, it is a bit more difficult to find the flats of a graphic matroid from its graphic representation. To describe the flats of a graphic matroid, we consider a graph G = (V, E) and a subset F of the edges E. Note that the graph (V,F) has various connected components. Then, loosely speaking, F forms a flat in a graphic matroid if adding any edge to F reduces this number of connected components. More precisely, we let Π be a partition of the vertices of G, and then let FΠ be those edges of the graph both of whose endpoints are contained in the same part of the partition. Then FΠ forms a flat of M(G).
It shouldn't be hard for you to implement this for graphs. But doing that misses the point of matroid theory which tells us that we can write the code once in the more general setting and then apply it to graphs as a particular case.
|
2025-03-21T14:48:29.629190
| 2020-01-15T06:30:19 |
350464
|
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"Emil Jeřábek",
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|
Stack Exchange
|
Is every true $\Pi^0_1$ statement entailed from a consistency statement of $PA$?
I want to prove the following.
For every $\Pi^0_1$ statement $\forall x\phi(x)$, where $\phi(x)$ is a $\Delta^0_1$ formula, there is $e\in\mathbb{N}$ such that $\forall x\phi(x)$ implies $W_e=PA$* and $PA+\text{$W_e$ is consistent}\vdash\forall x\phi(x)$.
The argument is inspire by an argument of Turing in his PhD dissertation (lately rephrased in Feferman 1962) to show that the transfinite progressions of adding consistency statements is sensitive on which branch we choose at the limit stage of Kleene's $\mathcal{O}$, and it is as follow.
Fix a computable function $\sigma$ such that the range of $\sigma$ is $PA$. By recursion theorem, we can construct a partial computable function $\varphi_e$ such that
$\varphi_e(n)=\begin{cases}\sigma(n),&\text{if $\forall x<n\phi(n)$},\\
\sigma(n)\wedge\text{$W_e$ is consistent*},&\text{o.w.}\end{cases}$
Since $\forall x\phi(x)$ holds, $W_e=PA$. So the statement "$W_e$ is consistent" is really "$PA$ is consistent". In the next paragraph, we prove in $PA+$"$W_e$ is consistent".
Assume $\forall x\phi(x)$ fails, than $W_e=PA+$"$W_e$ is consistent". By Goedel's second incompleteness theroem, $W_e$ is not consistent. A contradiction.
This finishes the argument.
My question is (1) is this argument valid, or if something is missed or misunderstood? (2) is there other reference on this and related issues?
*$W_e$ is the range of $\Phi_e$ by fixing an enumeration of Turing machines $\{\Phi_e\}_{e\in\mathbb{N}}$. $PA$ stands for Peano Arithmetic. Please forgive my abusing of notation.
I don’t understand the notation. What is $W_e$, $\Sigma_e$, and $PA^*$?
$W_e$ is the range of $\Phi_e$ by fixing an enumeration of Turing machines ${\Phi_e}_e$. $\Sigma_e$ is a typo (fixed), which should be $W_e$. $PA$ stands for Peano Arithmetic. I am sorry for the ambiguities.
I think the "PA$^*$" in your opener should just be PA.
The statement you want to prove is true, and your argument works - but there's a simpler one: drop all reference to the recursion theorem and Godelian incompleteness, and just change the second clause in your definition of $\varphi_e(n)$ to "$\sigma(n)\wedge\exists x(x\not=x)$." That is, introduce an inconsistency directly rather than playing with deeper facts about the theory and computability.
As you say, the only way $W_e$ can be consistent is if that second case doesn't occur, and if $\forall x\phi(x)$ is indeed true then that second case never occurs and $W_e$ "is" PA.
Given the directness of this construction, I suspect that there is no explicit reference.
|
2025-03-21T14:48:29.629404
| 2020-01-15T06:58:15 |
350465
|
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"Amir Mafi",
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|
Stack Exchange
|
Cohen-Macaulay monomial ideal
Let $R=K[x_1,...,x_n]$ be the polynomial ring over a field $K$ and $I[x_1,...,x_n]=(u_1,...,u_t)$ be a Cohen-Macaulay monomial ideal of $R$. If $m<n$, could we say that $I[x_1,...,x_m,0,0,...,0]$ is Cohen-Macaulay in $K[x_1,...,x_m]$?
Not all of us are ring-theory specialists. Would you provide the C-M ideal definition?
@WlodAA An ideal $I$ of $R$ is Cohen-Macaulay if $R/I$ is a Cohen-Macaulay ring.
Oh, well, a reader here has to be either young or a specialist,
No, we can't say that. For a counterexample, take $n = 3$, $m = 2$, and let $I = (x_2^2, x_1x_2, x_1x_3)$.
To see that $I$ is Cohen-Macaulay, note that the Krull dimension of $R/I$ is 1. Cohen-Macaulayness is then equivalent to $I$ having no associated prime $P$ with dimension $R/P$ equal to 0. Since $I$ is homogeneous, its associated primes are homogeneous, and so the only possible example of such an associated prime is $P = (x_1,x_2,x_3)$, which is easily seen not to annihilate any element of $R/I$, hence is not associated with $I$.
On the other hand the restricted ideal $I' = (x_2^2, x_1x_2) \subset K[x_1,x_2]$ is not Cohen-Macaulay, since $x_2$ is annihilated by $(x_1,x_2)$ in $K[x_1,x_2]/I'$, and so $(x_1,x_2)$ is an associated prime of $I'$.
Thank you so much for your nice example.
For another example, take $I=(x_1x_3,x_1x_4,x_2x_4)$. This is Cohen-Macaulay since it is the face ring (aka "Stanley-Reisner ring") of a connected graph. (It's also easy to check Cohen-Macaulayness directly.) But $I=(x_1x_3,x_1x_4)$ is not Cohen-Macaulay since it's the face ring of a nonpure simplicial complex. (The face ring of any nonpure simplicial complex has depth 1.)
|
2025-03-21T14:48:29.629554
| 2020-01-15T09:03:26 |
350466
|
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"CNS709",
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|
Stack Exchange
|
CW complexes obtained by attaching cells not with increasing dimension
CW-complexes are defined by attaching cell with increasing dimension: you start with a set of points, then attach 1-cells, then 2-cells and so on. Why are defined so? My question is: why is it necessary to attach cells ordered by dimension and not to attach a 2-cell, then a 1-cell, then a 3-cell...? The proofs I have seen about CW-complexes are mainly done inductively on the dimension of the complex, but I guess that they could be done also inductively on the number of cells attached (for example I am thinking about the uniqueness theorem for homology theories). Are there theorems that holds only for standard CW-complexes and not for these?
Up to homotopy these two kinds of spaces are the same, so if you only care about homotopy invariant notions there's no difference (arranging the cells by dimension is just a lot more convenient for some arguments)
@DenisNardin Thanks, that's what I was looking for. If you post this as answer, I'll accept it.
The words "convenient for some arguments" apply often to the construction of continuous functions and homotopies on a CW-complex.
The key idea is the cellular approximation theorem. If you attach a cell of lower dimension, you can make the attaching map cellular and then it is easy to see that you could also have attached the cells in the different order. Changing the attaching maps up to homotopy does not change the homotopy type.
There is a name for the kind of space you are describing: a cell complex.
A CW complex is a cell complex which has cell attachments in the increasing order of dimension.
The main advantage of having a CW complex over a mere cell complex is that the filtration by skeleta defines a finite chain complex model for its singular homology: if $X$ is a CW complex with $k$-skeleton $X^{(k)}$ then the homology of the pair $H_\ast(X^{(k)},X^{(k-1)})$ is
concentrated in degree $k$ and is free abelian in that degree. The composition
$$
H_k(X^{(k)},X^{(k-1)}) \overset{\partial}\to H_{k-1}(X^{(k-1)}) \to H_{k-1}(X^{(k-1)},X^{(k-2)})
$$
defines the boundary operator for cellular homology.
Howeover, in the cell complex case, there is no such filtration giving rise to a finite chain complex. The best thing you can do is use the partial ordering of cell attachements to define a filtration indexed by a poset whose relative homology defines a spectral sequence. This is more cumbersome to work with.
There is a thirty year old work by Igusa and Waldhausen which studies families of cell complexes--defining a sort of moduli space called the expansion space. The work was never published. Roughly speaking, they prove that the moduli space of cell complexes which are pointed and contractible give a model for the $h$-cobordism space of a disk.
|
2025-03-21T14:48:29.629788
| 2020-01-15T09:25:31 |
350467
|
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"Dieter Kadelka",
"Xin Wang",
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|
Stack Exchange
|
Non-convergence to a Gaussian
Let $f_n: \mathbb R^2 \rightarrow \mathbb R$ be a family of probability distributions with the property that they vanish on the diagonal $f_n(x,x)=0.$
I would like to know: Can we show that a function like this can never converge to a standard Gaussian $f(x,y) = \frac{1}{2\pi} e^{- \frac{\vert x \vert^2+ \vert y \vert^2}{2}}?$
Of course, one has to measure non-convergence in a norm that "sees" the diagonal. Since the Fourier transform might be useful, I was thinking about showing
$$\Vert \sqrt{f_n}-\sqrt{f} \Vert_{H^1} > \varepsilon$$
for $\varepsilon>0$ independent of $f_n$ where $H^1$ is the Sobolev space. I take square roots in order to give $f$ and $f_n$ unit mass in the $L^2$ sense.
EDIT: I assume it to be true, as $H^1$ decomposes into the direct sum $H^1_0$ and the harmonic functions on the zero set (which is in our case the diagonal). But I am wondering whether there is a very direct way of showing this.
If you want to prove this directly, you need to be more precise what you mean by a norm that "sees" the diagonal. Otherwise the sup-norm has this property. I think that such a norm is not allowed.
@DieterKadelka well, I specified the $H^1$ norm in this post. For the sup norm the question is of course v easy.
@GeraldEdgar sorry, a typo
Your conjecture is true.
Indeed, let $g_n:=\sqrt{f_n}$ and $g:=\sqrt{f}$. Let
$$v:=\|g\|,
$$
where $\|h\|:=\|\,h|_J\,\|_{L^2(J)}$ for $h\in L^2(\mathbb R^2)$, $J:=I^2$, $I:=[-u,u]$, and $u\in(0,1/20)$ is small enough so that
$$v>u/10;
$$
such a number $u$ exists, because $g(0,0)^2=1/(2\pi)>1/400$ and $g$ is continuous. For instance, one may take $u=1/21$, and then
$$v>0.037[>u/10].$$
(The bounds below are numerically very loose, so that the above lower bound on $v$ is easy to significantly improve.)
One of the following two cases must occur.
Case 1: $\|g_n\|\le\|g\|/2$. Then
$$\|\sqrt{f_n}-\sqrt f\,\|_{H^1}=\|g_n-g\|_{H^1}\ge\|g_n-g\|\ge\|g\|/2=v/2.
$$
So, Case 1 is good.
Case 2: $\|g_n\|>v/2$. In this case, use the condition $g_n(x,x)\equiv0$ to note that for all $x$ and $y$ in $I$ we have
$g_n(x,y)=\int_x^y(D_2g_n)(x,z)\,dz$, where $D_2$ is the partial derivative wrt the second argument and $\int_x^y:=-\int_y^x$ if $y<x$, whence
$$g_n(x,y)^2\le\Big(\int_I|(D_2g_n)(x,z)|\,dz\Big)^2
\le\int_I(D_2g_n)(x,z)^2\,dz.
$$
So,
$$\frac{v^2}4<\|g_n\|^2=\int_{I^2}g_n^2
\le\int_{I^3}dx\,dy\,dz\,(D_2g_n)(x,z)^2\le\|D_2g_n\|^2,
$$
so that
$$\|D_2g_n\|>v/2.
$$
On the other hand,
$$\|D_2g\|^2\le\frac1{2\pi}\,\int_{I^2}y^2\,dx\,dy<u^4/4<(u/40)^2<(v/4)^2,
$$
whence $\|D_2g\|<v/4$. So,
$$\|\sqrt{f_n}-\sqrt f\,\|_{H^1}=\|g_n-g\|_{H^1}\ge\|D_2g_n-D_2g\|\ge v/2-v/4=v/4.
$$
So, Case 2 is good as well. $\Box$
|
2025-03-21T14:48:29.629997
| 2020-01-15T09:58:54 |
350468
|
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|
Stack Exchange
|
Several definitions of approximate continuity of real functions
I found the definition of approximate continuity stated as follows:
A function $f:\mathbb R \to \mathbb R$ is approximately continuous at $x_0$ iff there exists a set $A\in \mathcal L$ such that $x_0\in \Phi(A)$ and $$\lim\limits_{x\to x_0,\ x\in A}f(x)=f(x_0)$$
where, $\mathcal L$ is the set of all Lebesgue measurable subsets in $\mathbb R$, and $\Phi(A)$ is the set of all density points of $A\subset \mathbb R$.
Question1: Can we write the above definition in $\epsilon$-$\delta$ form as follows?
A function $f:\mathbb R \to \mathbb R$ is approximately continuous at $x_0$ if and only if for each $\epsilon>0$ there exist $\delta>0$ and $A\in \mathcal L$ with $x_0\in \Phi(A)$ such that $$|f(x)-f(x_0)|<\epsilon\quad \text{whenever}\quad x\in (x_0-\delta, x_0+\delta)\cap A$$
Question2: Also, can we write the definition of "Approximate continuity" as follow?
A function $f:\mathbb R \to \mathbb R$ is approximately continuous at $x_0$ if and only if for each $\epsilon>0$ the set $\{y\in \mathbb R: |f(y)-f(x_0)|<\epsilon\}$ has $x_0$ as a density point.
Related: Functions that are approximately differentiable a.e
I will refer to the three definitions of approximate continuity given as AC0, AC1 and AC2 respectively and show they are all equivalent.
AC1 iff AC2: Given a set $A$ by AC1, $A$ has density 1, $A$ is a subset of the set in AC2, and so the set in definition 2 has density $1$ at $x_0$ Conversely, given such a set in AC2, take $A$ to be that set.
AC0 implies AC1: Take $A$ to be the set given in AC0 for all $\epsilon$.
AC2 implies AC0: Assume that $f$ satisfies AC2. Thus for every positive integer $k$, there exists a set $A_k$ of density $1$ at $x_0$ such that the set of all $x$ in $A_k$ such that $|f(x_0) - f(x)| < \frac{1}{2^k}$ has $x_0$ as a density point.
Now, since the finite intersection of sets with density $1$ at a point again has density $1$ there, given any finite natural $n$, there exists an $r_n > 0$ such that the intersection of the $A_k$ from $1$ to $n$, denoted $C_n$ satisfies $m(C_n \cap B_r (x0))/2r > 1 - 1/2^n$ for all $r \leq r_n$. Let us further choose $r_n$ to be monotonically decreasing to $0$ and $r_{n-1} < r_n/2^n$.
Denote by $D_n$ the set of $x$ such that $r_{n+1} \leq |x - x0| \leq r_n$.
Then the set $A := \bigcup_i C_i \cap D_i$ is the set required in AC0.
Indeed, given $\epsilon$, choose $n$ so large such that $1/2^{n-2} < \epsilon$. Then for all $r < r_n$, the measure of $A$ in $B_r (x)$ is at least $2r(1 - \epsilon)$, so $A$ has density $1$ at $x_0$.
It is easy to see that $A$ satisfies the limit condition, so $A$ is the required set.
Let me know if anything needs clearing up!
Sorry, but I do not see how the final assertion on the measure of $A$ in $B_r(x_0)$ follows for every $r<r_n$, with your construction (it follows immediately only for SOME $r$). Moreover, the intersection of the sets $A_k$, for $k=1,...,n$ is simply $A_n$ (considering simply $A_k$ the set where $|f(x)−f(x_0)|<1/2^k$, or am I wrong?).
|
2025-03-21T14:48:29.630223
| 2020-01-15T10:15:55 |
350469
|
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|
Stack Exchange
|
Existence of a finite group with a given decomposition for a tensor square of one irreducible complex representation
In this post, irrep and dim mean "irreducible complex representation" and "dimension", respectively.
It would be helpful (in a problem of monoidal category) to find a finite group $G$ with (at least) two irreps of dim $5$ (denoted $5_1$ and $5_2$) and (at least) two irreps of dim $7$ (denoted $7_1$ and $7_2$) with $$ 5_1 \otimes 5_1 \simeq 1 \oplus 5_1 \oplus 5_2 \oplus 7_1 \oplus 7_2$$Question: Is there such a finite group $G$?
Remark: such group should be of order a multiple of $35$ and should admit irreps of dims $5$ and $7$, which is helpful to rule out the following cases with GAP on a laptop:
simple, of order less than $10^6$,
perfect, of order less than $15120$,
general, of order less than $2240$.
Let $a_n$ be the smallest order of a group with an irrep of dim $n$ (oeis.org/A220470): $1, 6, 12, 20, 55, 42, 56, 72, 144, 110, 253, 156, 351, 336, 240, 272,\dots$
In particular, $a_5=55$ (given by $C_{11} : C_5$) and $a_7=56$ (given by $C_2^3 : C_7$). Now, I don't even know if there exists a group $G$ with $|G|<55 \times 56=3080$, and with irreps of dims $5$ and $7$.
I would think that if the presence of irreps of dim 5 and 7 is enough to rule out simple groups of order less than $10^6$, then it is enough to rule out ALL simple groups: larger simple groups will not have any irreducible representations that small (checking Landazuri-Seitz would be enough to confirm this). The same is probably true for perfect groups.
@NickGill: the first perfect group with irreps of dims $5$ and $7$ is $A_5 \times \mathrm{PSL}(2,7)$ (of order $10080$) but it does not satisfy the expected decomposition.
@NickGill Confirmation that there is no such simple group by the paper of Hiss-Malle Low-dimensional Representations of Quasi-simple Groups and corrigenda.
Ah, yes, I was in error about perfect groups, but glad to hear that is confirmed for simples.
I think that there is indeed no such finite group $G$, whether simple or otherwise. Note first that the representation $5_{1}$ can be assumed to be faithful ( for if $K$ is its kernel, then the group $G/K$ has the same property), so from now on, we assume it faithful.
Note next that $Z(G) = 1$, since if $5_{1}$ lies over a linear character $\lambda$ of $Z(G)$, then we must have $\lambda^{2} =1 $ since the trivial character occurs in $5_{1} \otimes 5_{1}$. However $\lambda = \lambda^{2}$ since $5_{1}$ also occcurs in $5_{1} \otimes 5_{1}$. Hence $\lambda$ is trivial.
Now $ N = O_{5^{\prime}}(G)$ is Abelian by Clifford's Theorem. If $N$ is non-trivial, then it is also non-central, sincee $Z(G) = 1$, and it follwws from Clifford's Theorem that the representation $5_{1}$ is (up to equivalence) monomial. Then $G$ has an Abelian normal subgroup $A$ such that $G/A$ is isomorphic to a subgroup of $S_{5}$. But in that case, $G$ has an Abelian normal Sylow $7$-subgroup, and $G$ has no irreducible character of degree $7$ (by a theorem of Ito, the degree of a complex irreducible character of $G$ divides $[G:A]$ whenever $A \lhd G$ is Abelian).
Hence it follows that $N = 1$. More generally, this argument shows that the representation $5_{1}$ is primitive, ie not (equivalent to one) induced from any proper subgroup of $G$.
There are several ways to finish from here. One is to invoke Brauer's classification of the finite primitive subgroups of ${\rm GL}(5, \mathbb{C})$ and note that none of these has tthe order of $G/Z(G)$ divisible by $7$.
Another is to note that if $O_{5}(G)$ is non-trivial, then it is irreducibly represented by $5_{1}$, in which case $Z(G)$ has order divisible by $5$, a contradiction.
Now we are reduced to the case $F(G) = 1$ and we continue until we see that $M = F^{\ast}(G)$ is a finite simple subgroup of ${\rm GL}(5,\mathbb{C})$ of orer divisible by $35$. But by a theorem of Feit, if $G$ is a finite simple subgroup (of order divisible by the prime $p$) of ${\rm GL}(p-2,\mathbb{C})$ for some prime $p$, then $p$ is a Fermat prime and $G = {\rm SL}(2,p-1)$ ( we may apply this with $p = 7$, so obtain a contradiction).
My comment about Hiss-Malle is only for the simple groups. Your answer works for every group, right? Does your answer prove a more general statement than what expected? (because you seem to use just partially the assumption) If so, what is it? and to what could it be extended?
Yes, this argument shows that there is no such finite $G$, simple or not. As for a more general statement, I am not sure: arguments about complex linear groups of low dimension tend to be rather ad hoc, and some arguments above are very specific to your hypotheses.
I meant a weaker assumption on the decomposition:
Firstly, perhaps something like: the existence of (at least) one irrep of dim $5$ and of dim $7$ and $$5_1 \otimes 5_1 \ge 1 \oplus 5_1 \oplus 7_1$$ would be enough for your argument, correct?
And secondly, perhaps the couple $(5,7)$ can be generalized to a class of couples $(n,m)$, isn't it?
I think it is probably true that if $p >3$ is a prime such that $q = p+2$ is also prime, then there is no finite group $G$ with a complex irreducible characters $\chi,\mu$ of respective degrees $p$ and $q$ such that $\chi^{2} = 1 + \chi + \mu + \theta$, where $\theta$ is a character ( or $0$). The argument (using Feit's Theorem rather than Brauer's) goes through more or less unchanged, after noting that $(3,5)$ is the only prime pair $(p,q)$ such that $p+1 = q-1$ is power of $2$. But this seems very specialized.
@MarkSapir: a fusion subcategory of $\mathrm{Rep}(G)$ is isomorphic to $\mathrm{Rep}(G/N)$ with $N \unlhd G$ (and reciprocally).
@MarkSapir: $N=G$ corresponds to the trivial subcategory (i.e. given by the trivial irrep).
@MarkSapir : Because if $5_{1}$ has $K$ in its kernel, so does every representation on the right side, because $5_{1} \otimes 5_{1}$ has $K$ in its kernel.
@MarkSapir: the isomorphic classes of irreps form a ring under the tensor product (it has the strucutre of a fusion ring, and it is the Grothendieck ring of the fusion category $\mathrm{Rep}(G)$, but it does not matter). Now let consider $5_1$ as an element of this fusion ring, it generates a fusion subring which, as every fusion subring, is isomorphic to the Grothendieck ring of $\mathrm{Rep}(G/N)$ with $N$ a normal subgroup depending on the subring (here $N=ker(5_1)$).
@MarkSapir: Dietmar and Vaughan know all that, so if you and (one of) them are currently at Vanderbilt, it could be a good opportunity to discuss.
@MarkSapir: As I explained above, it is because the normal subgroup $K$ is the kernel of $5_{1}$, the statement would not be true in general for an arbitrary normal subgroup. Once $K$ is in thhe kernek of $5_{1}$, $K$ is also in the kernel of everything in sight in the original equation, so, as I said, $G/K$ has the same property (regarding $5_{1}$ as a representation of $G/K)$.
|
2025-03-21T14:48:29.630700
| 2020-01-15T11:08:01 |
350473
|
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|
Stack Exchange
|
Two equivalent irreducible representations given by integer matrices
Let $G$ be a finite group, and $\rho_1, \rho_2: G\to GL_n(\mathbb C)$ be two representations. Suppose that $\rho_1$ and $\rho_2$ are equivalent (i.e. conjugate over $\mathbb C$), and suppose that both groups $\rho_1(G)$, $ \rho_2(G)$ belong to $GL(n,\mathbb Z)$. Is it true that these two groups are conjugate in $GL(n,\mathbb Z)$?
If not, is this at least true in the case when $G$ is a symmetric group $S_n$ and the representation $\rho$ is irreducible? The motivation for this question is the following: I know that all complex irreducible representations of $S_n$ can be defined over integers. I wonder whether there is somehow a canonical choice.
Specht modules afford the irreducible complex characters and are defined over the integers. But in general Specht modules are not self-dual (even though their characters are real valued) so even here the choice is not entirely canonical. In general they are many different $\mathbb{Z}$-forms inside an integral Specht module: see for instance https://wildonblog.wordpress.com/2016/10/30/semisimple-modular-reductions/.
The smallest counterexample involving irreducible representations of symmetric groups is the $2$-dimensional irreducible module for $\mathbb{C}S_3$. It can be defined over the integers as the submodule $U = \langle e_2-e_1, e_3-e_1\rangle_\mathbb{Z}$ of the natural integral permutation module $\langle e_1, e_2, e_3 \rangle_\mathbb{Z}$. Then $U \otimes_\mathbb{Z} \mathbb{C}$ is irreducible and affords the ordinary character labelled $(2,1)$. The dual $U^\star = \mathrm{Hom}_{\mathbb{Z}}(U,\mathbb{Z})$ is isomorphic to the quotient of $\langle e_1,e_2,e_3 \rangle$ by the trivial submodule $\langle e_1+e_2+e_3\rangle$. The corresponding homomorphisms $\rho, \rho^\star : S_3 \rightarrow \mathrm{GL}_2(\mathbb{Z})$ are such that $\rho(S_3)$ and $\rho^\star(S_3)$ are conjugate in $\mathrm{GL}_2(\mathbb{C})$ but not in $\mathrm{GL}_2(\mathbb{Z})$.
To prove the final claim: if the representations are $\mathbb{Z}$-equivalent then the modules $U \otimes_\mathbb{Z} \mathbb{F}_3$ and $U^\star \otimes_\mathbb{Z} \mathbb{F}_3$ are isomorphic. The first has a trivial submodule spanned by
$$(e_2-e_1) +(e_3-e_1) = e_1+e_2+e_3;$$
the quotient by this submodule is the sign module. The second is its dual, with the factors in the opposite order. Since both are indecomposable, they are not isomorphic.
Thank you @Jeremy Rickard
No, it is not even true for matrices, e.g., $\left(\begin{array}{rr}0 & 1\\1&0\end{array}\right)$ and $\left(\begin{array}{rr}1 & 0\\0&-1\end{array}\right)$.
Thanks! Do you think however that for irreducible representations of the symmetric group $S_n$ this is true?
No, I see no reason for that. I can see an example below already....
An instructive example (for general $G$, not for symmetrc groups) is provided by the case that $G$ is a dihedral group with eight elements.
Then $G$ has a unique complex irreducible character $\chi$ which may be expressed as ${\rm Ind}_{U}^{G}(\lambda) $ and ${\rm Ind}_{V}^{G}(\mu)$, where $U$ and $V$ are the two Klein $4$-subgroups of $G$, and $\lambda, \mu$ are non-trivial linear characters of $U,V$ respectively.
These representations exhibit $G$ as an absolutely irreducible subgroup of ${\rm GL}(2, \mathbb{Q})$ with all matrix entries in $\mathbb{Z}$ (even in $\{0,1,-1\}$). The two given representations are equivalent over $\mathbb{C}$, but they are not equivalent as integral representations.
One way to explain this is via J.A. Green's theory of vertices and sources : both these integral representations are indecomposable on reduction ( mod $2$). One of the reductions has vertex $U$ and one has vertex $V$. Since $U \lhd G$ and $V \lhd G$ , but $U \neq V$, we see that $U$ and $V$ are not $G$-conjugate.
Since (by Green's theory) the vertex of an indecomposable module is unique up to conjugacy, these two modules are not isomorphic on reduction (mod $2$), so they are certainly not isomorphic as $\mathbb{Z}G$-modules
|
2025-03-21T14:48:29.630979
| 2020-01-15T12:13:41 |
350476
|
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|
Stack Exchange
|
Upper bound for an expression for distributive lattices
Let $L$ be a finite distributive lattice with minimum $0$ and Maximum $1$ and join-irreducible elements $j_1,...,j_l$ and meet irreducible elements $m_1,...,m_l$.
Let $J_L:= \sum\limits_{i=1}^{l}{| [j_i,1]|}$ and $M_L:= \sum\limits_{i=1}^{l}{| [0,m_i]|}$.
Set $X_L$:= $l-|J_L-M_L|$.
Question 1: Is there an easy proof that $X_L >0 $ ? This would prove Frankl's conjecture for distributive lattices (which is already known).
Question 2: What are the distributive lattices with $X_L=l$? Their number starts for $n \geq 3$ with 1,2,1,3,2,7,4. Examples are Boolean lattices.
Question 3: Let $U_n:= \sum\limits_{L \in \mathcal{L}_n}^{}{|J_L-M_L|}$, where $\mathcal{L}_n$ is the set of all distributive lattices on $n$ elements.
$U_n/2$ starts for $n \geq 3$ with 0,0,1,2,6,12,34 (could it be https://oeis.org/A088808 ?). What is $U_n$?
It seems to me that there can't be any good description of lattices with $X_L = \ell$. We have $X_L = \ell$ whenever $L$ is isomorphic to its opposite lattice. If $P$ is any finite poset isomorphic to its opposite poset, then the lattice of lower ideal of $P$ is a lattice isomorphic to its opposite lattice. And there are tons of posets isomorphic to their opposites.
I'm also skeptical of a good answer to Q3, because the cardinality of $\mathcal{L}_n$ is intractable and therefore I would guess that summing more complicated quantities over it is as well. I realize this isn't a strong argument.
The proposed inequality $X_L>0$ is false. Using Birkhoff's representation theorem, identify $L$ with the lattice of lower ideals in a poset $P$. Then $|P| = \ell$.
The join irreducible elements are the principal lower ideals $(p)$ for $p \in P$. Given a join irreducible element $j=(p)$, the interval $[j,1]$ is the set of ideals $I$ containing $p$. Thus,
$$J_L = \sum_{p \in P} \#\{ I : I \ni p \} = \sum_{I \in L} \#(I).$$
Similarly,
$$M_L = \sum_{I \in L} (\ell - \#(I)).$$
So the proposed inequality is
$$\left| \ell \#(L) - 2 \sum_{I \in L} \#(I) \right| < \ell.$$
Let the poset $P$ have $a+b$ elements $x_1$, $x_2$, ..., $x_a$, $y_1$, $y_2$, ..., $y_b$ and the relations that the $x$'s are incomparable, the $y$'s form a chain $y_1 < y_2 < \cdots < y_b$ and $x_i < y_j$ for all $i$, $j$. So the lower ideals are of two types:
(1) Any subset of the $x$'s.
(2) $\{x_1, x_2, \ldots, x_a, y_1, y_2, \ldots, y_k \}$ for $1 \leq k \leq b$.
We have $\ell = a+b$, $|L| = 2^a + b$ and $\sum_{I \in L} \#(I) = (a/2) 2^a + b (a+b/2+1/2)$. So
$$\ell \#(L) - \sum_{I \in L} \#(I) = (a+b)(2^a+b) - (a 2^a + b(2a+b+1))$$
$$=b 2^a - b(a+1).$$
If we choose $a$ and $b$ roughly the same size, then $|b 2^a - b(a+1)|$ will be much larger than $a+b$.
|
2025-03-21T14:48:29.631173
| 2020-01-15T12:43:06 |
350478
|
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|
Stack Exchange
|
Topological mapping class groups of 4-manifolds
It is a classical result of Quinn that for a simply-connected
closed $4$-manifold $X$ the isometries of its intersection form
are in one-to-one correspondence with
$\pi_0 \text{Homeo}(X)$. (Isotopy of 4-manifolds, 1986)
Let $X$ have some simple fundamental group, say $\mathbb{Z}_2$,
and let $h\colon X \to X$ be a diffeomorphism which acts trivially
on $H_2(X;\mathbb{Z})$.
Is $h$ isotopic to the identity? (through $\text{Homeo}(X)$)
Edit: let $\tilde{X}$ denote the universal cover of $X$, and let $s \colon \tilde{X} \to \tilde{X}$ be the covering involution so that $\tilde{X}/s \cong X$. One can take a lift $\tilde{h} \colon \tilde{X} \to \tilde{X}$ of $h$ (there are two, but take any of them.) Another reasonable assumption on $h$ is that we want $\tilde{h}$ to be isotopic to either $\text{id}$ or to $s$.
Edit: The second question has been removed.
I suspect that at you need to look also at the induced action of $h$ on $H_2(X;\Bbb Z_-)$, where this denotes twisted coefficients with fiber $\Bbb Z$ and nonzero $\pi_1$ action. (Whether this action being trivial is enough for (2) I don't know off the top of my head.)
I don't quite understand what you say, yet I’m going to rephrase it.
One can take the universal cover $\tilde{X} \to X$ and consider a lift $\tilde{h} \colon \tilde{X} \to \tilde{X}$. If $h$ is isotopic to $\text{id}$ then at least $\tilde{h}$ is isotopic to either $\text{id}$ or to the covering involution. What you suggest is to check this property for $\tilde{h}$?
That is not precisely what I said, homology with local coefficients is not the same as homology of a covering space. However, suppose $h: X \to X$ is a diffeomorphism which induces an isomorphism on integer homology. Then any two of the following three conditions implies the third. (1) $h$ induces an isomorphism on $\Bbb Z/2$ homology. (2) The cover $\tilde h$ induces an isomorphism on the integer homology of $\tilde X$. (3) $h$ induces an isomorphism on the local coefficient homology $H_*(X;\Bbb Z_-)$.
If $X$ is non-orientable then (3) I believe is automatic by Poincare duality and UCT.
Let $X = (S^2\times S^2)/\mathbb{Z}_2$ where the $\mathbb{Z}_2$ action is generated by $(x, y) \mapsto (-x, -y)$. Note that $H_2(X; \mathbb{Z}) \cong \mathbb{Z}_2$, so every diffeomorphism acts trivially.
Consider the diffeomorphism $f : S^2\times S^2 \to S^2\times S^2$ given by $(x, y) \mapsto (x, -y)$. This descends to a diffeomorphism $g : (S^2\times S^2)/\mathbb{Z}_2 \to (S^2\times S^2)/\mathbb{Z}_2$. Letting $\pi : S^2\times S^2 \to (S^2\times S^2)/\mathbb{Z}_2$ be the universal covering map, we have a commutative diagram
$$\require{AMScd}
\begin{CD}
S^2\times S^2 @>{f}>> S^2\times S^2\\
@V{\pi}VV @VV{\pi}V \\
(S^2\times S^2)/\mathbb{Z}_2 @>{g}>> (S^2\times S^2)/\mathbb{Z}_2
\end{CD}$$
Taking $\pi_2$ of this diagram, we get a commutative diagram of abelian groups
$$\require{AMScd}
\begin{CD}
\mathbb{Z}\oplus\mathbb{Z} @>{f_*}>> \mathbb{Z}\oplus\mathbb{Z}\\
@V{\pi_*}VV @VV{\pi_*}V \\
\mathbb{Z}\oplus\mathbb{Z} @>{g_*}>> \mathbb{Z}\oplus\mathbb{Z}
\end{CD}$$
Note that $\pi_* = \operatorname{id}$ as $\pi$ is a covering map, but $f_*$ is given by $(a, b) \mapsto (a, -b)$. By commutativity, the same is true of $g_*$. In particular, $g_* \neq \operatorname{id}$ and therefore $g$ is not homotopic to the identity map.
Alternatively, note that $(S^2\times S^2)/\mathbb{Z}$ is orientable. As $\pi\circ f = g\circ\pi$, the maps $f$ and $g$ have the same degree, and it is easy to see that $f$ has degree $-1$. Again we see that $g$ is not homotopic to the identity map.
Geometrically, $X = \operatorname{Gr}(2, 4)$, the Grassmannian of unoriented two-planes in $\mathbb{R}^4$, and $S^2\times S^2 = \operatorname{Gr}^+(2, 4)$ the corresponding oriented Grassmannian. The diffeomorphisms $f$, $g$ are the maps given by $P \mapsto P^{\perp}$.
Thank you very much for your answer!
You posted your counter-example while I was editing my question (the very same minute!).
In the edited version, I want my homeomorphism to have a homotopically trivial lift.
@Enumerator: You might not get much attention given that you accepted my answer. I think it would be better to either (1) unaccept my answer, or (2) revert your question back to the original form and then ask a new question in the form it is now, linking to this one.
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2025-03-21T14:48:29.631601
| 2020-01-15T13:50:33 |
350483
|
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|
Stack Exchange
|
Representations of tensor products of algebras
For two associative unital algebras $A$ and $B$, defined over $\mathbb{K} = \mathbb{R}, \mathbb{C}$, is it possible to have an irreducible representation of $A \otimes_{\mathbb{K}}B$ which is not of the form $V \otimes W$, where $V$ is a representation of $A$ and $W$ is a representation of $B$?
Yes, very easily. For example, $(V \otimes W) \oplus (V \otimes W)$ does not in general factorize as $V' \otimes W'$. Did you want some irreducibility condition?
Yes, I want irreducibility. I have now written this.
K=R, A=B=C is a standard example where C/R is any nontrivial field extension.
@PeterMcNamara Why writing answers in the comment section? (I ask this on MO since 10 years ...)
@MartinBrandenburg, when I do something like that, it's because I suspect that the question might change. For example, probably the question should have been not just about irreducible representations (as @MarkWildon suggested) but about absolutely irreducible representations (as indicated in @Mare's answer).
In case your two algebras $A,B$ are finite dimensional and the field is algebraically closed (or more generally the two algebras are split over the field), then all simple modules over $A \otimes_K B$ are indeed of the form $V \otimes_K W$ for a simple $A$-module $V$ and a simple $B$-module $W$.
This is not true when the algebras are not split: Let $K= \mathbb{R}$ and $A=B=\mathbb{C}= \mathbb{R}[x]/(x^2+1)$.
Then $A \otimes_K B= \mathbb{C}[x]/(x^2+1)=\mathbb{C}[x]/(x+i) \times \mathbb{C}[x]/(x-i)=\mathbb{C} \times \mathbb{C}$.
Thus $A \otimes_K B$ has a simple modules of $K$-dimension two, while all non-zero $A \otimes_K B$-modules of the form $V \otimes_K W$ have $K$-dimension at least 4.
What does it mean for an algebra to be split? Isomorphic to a direct power $K^{\oplus n}$?
Also, note that @PeterMcNamara gave a variant of your counterexample in the comments.
@LSpice It means that the algebra modulo its jacobson radical is a products of matrix rings over the field $K$. What you say would be split and basic. Being split is also often called elementary.
What happens in the infinite dimensional case?
|
2025-03-21T14:48:29.631772
| 2020-01-15T13:51:10 |
350484
|
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"მამუკა ჯიბლაძე"
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|
Stack Exchange
|
Sum from combinatorics on nonnegative integer numbers
Let $n_1,n_2,\ldots,n_k\in\{0,1,2,\ldots\}$. Can you calculate the sum
$$
\sum_{n_1,n_2,\ldots,n_k\geqslant0}\mathbb{1}_\left\{n_1+\frac{n_2}{2}+\ldots+\frac{n_k}{k}<1\right\}?
$$
If it's helpful, I can see that
$$
\sum_{n_1,n_2,\ldots,n_k\geqslant0}\mathbb{1}_\left\{n_1+n_2+\ldots+n_k<k\right\}=\binom{2k-1}{k}.
$$
For the number of nonnegative integer solutions to $n_1 + n_2/2 + \ldots + n_k/k \le 1$, see OEIS sequence A212658.
@RobertIsrael and number of solutions for $n_1+n_2/2+...+n_k/k=1$ is A020473, so the question is about difference of these two sequences
As noted in the comments The sequence you want is the third row below.
The first gives the number of solutions where the sum is less than or equal to $1$. The initial entry is for the empty sum when $k=0$ and the fourth entry is for $k=3$ counting the $8$ cases $0,\frac13,\frac12,\frac23,\frac12+\frac13,\frac11,\frac22,\frac33$
The second row is the cases where the sum is exactly $1$ so the $3$ below the $8$ is the things you don't want.
The difference is what you want
$\begin{array}{r}
1 & 2 & 4 & 8 & 17 & 37 & 86 & 199 & 475 & 1138 & 2769 & 6748 & 16613 & 40904 & 101317 & 251401 & 624958 & 1555940 & 3882708 \\
& 1 & 2 & 3 & 5 & 6 & 13 & 14 & 24 & 34 & 60 & 61 & 168 & 169 & 252 & 627 & 1011 & 1012 & 2430\\
&1&2&5&12&31&73 &185&451&1104&2709 &6687& 16445 &\cdots
\end{array}$
Observations:
Allowing the variable $n_1$ is irrelevant for your particular problem since you want a sum less than $1$. Is is a matter of choice if it should be considered in counting the other two sequences. The we can have $n_1=1$ exactly once, when all the rest are $0.$
The sequence you want is not in the OEIS (at this moment)
The numbers in the second row are very small compared to the first row so the last thing I showed is over $98.9\%$ of the corresponding entry in the first row.
The first sequence counts the number of lattice points in the ($k$ dimensional) right pyramid with corners at the origin and $(1,0,0,0,\cdots),(0,2,0,0,\cdots), \cdots,(0,0,\cdots,k).$ That solid has volume $1$ (or $k$ If we forget about $n_1$ and go down a dimension.)
In the OEIS the first sequence is given up to $k=36$ with a comment that the calculations get large. The ratios of consecutive entries do increase, but rather slowly:
$\frac{199}{86}=2.314, \frac{475}{199}=2.387$ The last three entries given $$10002686041923,<PHONE_NUMBER>7818 ,<PHONE_NUMBER>3012 $$ have ratios about $2.528795$ and $2.529789.$
|
2025-03-21T14:48:29.631978
| 2020-01-15T15:31:21 |
350491
|
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"Gerald Edgar",
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|
Stack Exchange
|
Polylogarithm : reference request for proof of integral representation
On page 494 of the book Integrals and series, volume I : elementary functions, Gordon and Breach, 1986, by A. P. Prudnikov, Y. A. Brychkov, and O.I. Marichev, (perhaps translated from Russian), the following integral representation of the polylogarithm $Li_{s+1}$ is given without proof :
\begin{equation}
Li_{s+1}(z) = \frac{z}{\Gamma(s+1)} \int_{1}^{\infty} \frac{\log ^s(t)}{t(t-z)} dt
\end{equation}
for $z,s \in \mathbb{C}$ such that $|arg(1-z)| < \pi$, Re$(s)>-1$. $\Gamma$ is the Gamma function.
Does anyone know where a proof can be found ? Thanks in advance
Wikipedia https://en.wikipedia.org/wiki/Polylogarithm cites in case $s$ is a nonnegative integer: "See equation (4) in section 2 of Borwein, Borwein and Girgensohn's article Explicit evaluation of Euler sums (1994)."
Thanks Mr. Edgar. In fact I was just calculating, it is not very hard to show by differentiating that the function defined as above satisfies the recurrence relation $\frac{\partial Li_{s+1}}{\partial z} = \frac{Li_s(z)}{z}$ which is a defining property of the polylogarithms...
True, it is not hard to prove this (write $\frac{1}{t(t-z)}$ as a power series in $z$, then integrate term-by-term, using $$\int_1^\infty \frac{\log(t)^s}{t^k},dt = \frac{\Gamma(s+1)}{(k-1)^{s+1}}$$). But the question is not about how to prove it, but about where a proof can be found.
Since the question is "where can a proof be found", here is one reference (but there must be many others):
An Integral Representation for the Polylogarithm Function and Some Special Values
thank you very much for the reference'
|
2025-03-21T14:48:29.632111
| 2020-01-15T16:11:36 |
350494
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"Stephen McKean",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350494"
}
|
Stack Exchange
|
Could a motivic spectrum have a "zeta function"?
I'm currently learning about zeta functions, so I apologize in advance if this is riddled with nonsense. Suppose you have a sequence $E=(E_0,E_1,...)$ of motivic spaces along with structure maps $s_i:\mathbb{P}^1\wedge E_i\to E_{i+1}$ that induce weak equivalences $\Sigma_{\mathbb{P}^1}E_i\simeq E_{i+1}$, so that $E$ is a motivic spectrum. Is there a sequence of zeta functions $\zeta_E=(\zeta_{E_0},\zeta_{E_1},...)$ that respects the structure maps?
There are several pieces of this question that I'm not sure about:
If $E_i$ is a projective variety over $\mathbb{F}_q$, then we can assign a local zeta function $\zeta_{E_i}$ to it. Is it possible to do the same for a more general motivic space?
If $X$ and $Y$ are motivic spaces that have zeta functions $\zeta_X$ and $\zeta_Y$, is there a zeta function $\zeta_{X\wedge Y}$ for their smash product? If so, is there any way to express $\zeta_{X\wedge Y}$ in terms of $\zeta_X$ and $\zeta_Y$?
If $X$ and $Y$ are motivic spaces that have zeta functions $\zeta_X$ and $\zeta_Y$, and if $f:X\to Y$ is a sufficiently nice morphism, is there a way to make sense of $f_*\zeta_X$ (for example, would this be $\zeta_{f(X)}$)? If so, is there a way to relate $f_*\zeta_X$ and $\zeta_Y$?
Etale cohomology factors through the stable $\mathbb A^1$ homotopy type. So you should simply consider your local zeta factor in terms of traces of frobenius on etale cohomology. Since smashing with $\mathbb P^1$ corresponds to tensoring with $\mathbb Q_l(q)$ the eigenvalues that give you the local factor will be multiplied by $q$ each time. This means that the local factor will undergo the transformation $f(t) \mapsto f(qt)$. In terms of the global zeta function, this should correspond to translating by $1$.
I suppose this is answered by your first sentence, but does this mean that two weakly equivalent spaces will have the same global zeta function? I was worried that contracting $\mathbb{A}^1$ would change the set of points over which the Euler product is taken.
@StephenMcKean Of course $\mathbb A^1$ and $*$ don't have the same zeta function. Two proper spaces that are weakly equivalent have the same zeta function. The zeta function depends on the compactly supported cohomology, which does not factor through the $\mathbb A^1$ homotopy category. But for proper varieties, compactly supported cohomology agrees with ordinary cohomology, hence factors through the homotopy category.
Great, thanks for your clear explanations!
|
2025-03-21T14:48:29.632291
| 2020-01-15T16:48:40 |
350497
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/350497"
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|
Stack Exchange
|
Problem related to inequality of sum of digits of power sum
Let $D$ be the function define as $D(b,n)$ be the sum of the base-$b$ digits of $n$.
Example: $D(2,7)=3$ means $7=(111)_2\implies D(2,7)=1+1+1=3$
Define $S(a,m)=1^m+2^m+3^m+...+a^m$ where $a,m\in\mathbb{Z}_+$
Can it be shown that
For $a,m>1$,
$(a-1)m>D(a,S(a-1,m))$?
More on observation
$(1)\space (a-1)m=D(a,S(a-1,m)) \iff m=1,a\equiv0\pmod2$
$(2)$ if $m\ge 5,a\ge3$ then $D(a,S(a-1,m))>a-1$
Can someone please give me reference to understand this pattern, thanks.
Source code Pari/GP
for(m=1,50,for(a=2,100,if(sumdigits(sum(i=1,a-1,i^m),a)>=(a-1)*m,print([m,a,sumdigits(sum(i=1,a-1,i^m),a)]))))
Note: For $a,m>1$
● $a^m<S(a,m)<a^{m+1}$
● $1\le D(a,S(a,m))\le(a-1)(m+1)$
● $D(a,S(a,m))=1+D(a,S(a-1,m))$proof
●maybe this post helpful https://math.stackexchange.com/q/3595166/647719
The question was posted in MSE(10/1/20) but no answer hence posting in MO check here
This is surely the case if $m\geq a-1$, when we have
$$S(a-1,m)<\int_1^a x^m\ dx<\frac{a^{m+1}}{m+1}\leq a^m.$$
|
2025-03-21T14:48:29.632385
| 2020-01-15T17:55:30 |
350501
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/350501"
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|
Stack Exchange
|
How does the principal value affects to the limit here?
In Córdoba and Gancedo - Contour dynamics of incompressible 3-D fluids in a porous medium with different densities (page 4) I read that if
$$ v (x_1,x_2,x_3,t)=-\frac{\rho_2-\rho_1}{4\pi} \operatorname{PV}\int_{\mathbb{R}^2} \frac{(y_1,y_2,\nabla f(x-y,t)\cdot y)}{[\lvert y\rvert^2 + (x_3 - f(x-y,t)^2)]^{3/2}}\ dy$$
and for $\varepsilon $ positive and $x=(x_1,x_2)$, we define
$$ v^1(x_1,x_2,f(x,t),t) =\lim_{\varepsilon \rightarrow 0} v\bigl(x_1-\varepsilon \partial_{x_1} f(x,t) , x_2 - \varepsilon \partial_{x_2} f (x,t), f(x,t) + \varepsilon , t\bigr),$$
then they say
\begin{align*}
v^1(x_1,x_2,f(x,t),t) ={} & v(x_1,x_2,f(x,t),t) \\
& {}+ \frac{\rho_2-\rho_1}{2}\frac{\partial_{x_1} f(x,t)(1,0,\partial_{x_1}f(x,t))}{1+(\partial_{x_1}f(x,t))^2 +(\partial_{x_2}f(x,t))^2 } \\
& {}+ \frac{\rho_2-\rho_1}{2}\frac{\partial_{x_2} f(x,t)(0,1,\partial_{x_2}f(x,t))}{1+(\partial_{x_1}f(x,t))^2 +(\partial_{x_2}f(x,t))^2 }.
\end{align*}
I don't understand how they get that expression for $v^1$. In other words, my question is how does PV affects to computing the limit?
I asked this question some time ago, but I got no answers, that is why I decided to try one more time.
Please, any help or idea is welcome.
|
2025-03-21T14:48:29.632733
| 2020-01-15T17:57:01 |
350502
|
{
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"authors": [
"Stephen McKean",
"https://mathoverflow.net/users/146401"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/350502"
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|
Stack Exchange
|
Resultant in many variables
Suppose $P_1,\ldots,P_n$ are homogeneous polynomials in $\mathbb C[x_0,\ldots,x_n]$ of degrees $d_1,\ldots,d_n\ge 1$. These define hypersurfaces $H_1,\ldots,H_n\subset\mathbb P^n$. Is there a nonzero polynomial in the coefficients of $P_1,\ldots,P_n$ which vanishes whenever when the common intersection $H_1\cap\cdots\cap H_n$ is not zero dimensional? If so, what is an explicit method to determine this polynomial? Any references dealing with this would also be appreciated.
There are several helpful references here: https://mathoverflow.net/q/51534/146401
Let $F_i$ be a homogeneous polynomial of degree $d_i$ defining $H_i$. Let $U(x)$ be a linear form. Then consider the multivariate resultant ${\rm Res}(F_1,\ldots,F_n,U)$ which, for fixed $F_i$'s is a homogeneous polynomial of degree $d_1\cdots d_n$ in the coefficients of $U$. You condition is equivalent to the vanishing of this polynomial identically in $U$. For information about this multivariate resultant see, e.g., "Explicit formulas for the multivariate resultant" by D'Andrea and Dickenstein, and references therein.
Computing them explicitly is very hard in general.
|
2025-03-21T14:48:29.632834
| 2020-01-15T18:04:03 |
350503
|
{
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"authors": [
"Jeff Strom",
"Mike Shulman",
"Tim Campion",
"Valery Isaev",
"https://mathoverflow.net/users/2362",
"https://mathoverflow.net/users/3634",
"https://mathoverflow.net/users/49",
"https://mathoverflow.net/users/62782"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/350503"
}
|
Stack Exchange
|
Does Farjoun's "fiberwise localization" have a universal property?
Let $\mathcal S_L$ be any accessible reflective subcategory of the $\infty$-category of spaces. In his book, Farjoun discusses "fiberwise $L$-localization" of a map of spaces $E \to B$, i.e. a factorization $E \to \bar E \to B$ such that the fiber of $\bar E \to B$ is in $\mathcal S_L$ and $E \to \bar E$ is $\mathcal S_L$-local. Chataur and Scherer show (Thm 4.3) that a fiberwise localization always exists.
Farjoun claims (paragraph 1.F.5) that any fiberwise localization enjoys a universal property, but he's not quite precise on what that means -- he says (with slight paraphrase) "the map $E \to \bar E$ as a map over $B$ is universal among all maps of $E$ to spaces over $B$ that have homotopy fiber in $\mathcal S_L$".
I'm not quite clear on what "universal" means here, but surely it should at least imply that $\bar E$ is unique up to equivalence over $B$. But I'm having trouble seeing that. So I ask,
Question: Let $\mathcal S_L$ be any accessible reflective subcategory of the $\infty$-category of spaces. Let $E \to B$ be a map and let $\bar E \to B$, $\bar E' \to B$ be two fiberwise $L$-localizations of $E \to B$. Then are $\bar E \to B$, $\bar E' \to B$ equivalent over $B$ (and under $E$)? More strongly, is the space of fiberwise localizations of $E \to B$ contractible?
Isn't $E \to \bar E \to B$ just the initial object in the $\infty$-category of factorizations of $E \to B$ in which the second map is in $S_L$?
Is this the same notion of fiberwise localization that appears in part I of Hirschhorn's book on model categories?
@ValeryIsaev That's what I'd like to be true, but I'm not sure that actually follows from the definition -- i.e. from the condition that the fiber of $\bar E \to B$ be in $\mathcal S_L$ and $E \to \bar E$ be $\mathcal S_L$-local.
$E \to \bar E \to B$ has the universal property with respect to factorizations satisfying stronger property that the second map is $S_L$-local in the category over $B$ (where I assume that $S_L$ is define as localization with respect to a set of maps). It is not enough to require that its fiber is $S_L$-local. For example, the fiber of the Hopf fibration $f : S^3 \to S^2$ is $f$-local, but the actual $f$-localization of $f$ over $S^2$ is the identity map.
I know that Vandembroucq proved a uniqueness theorem for fiberwise localizations of unpointed functors in MR1923222
|
2025-03-21T14:48:29.633011
| 2020-01-15T18:16:48 |
350505
|
{
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"authors": [
"Uri Bader",
"YCor",
"geodude",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/30366",
"https://mathoverflow.net/users/89334"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/350505"
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|
Stack Exchange
|
Why are coarse maps required to be proper?
In the context of coarse spaces, a map between coarse spaces $f:X\to Y$ is called coarse if it is bornologous (it maps controlled sets to controlled sets), and proper, in the sense that preimages of bounded subsets are bounded.
(A subset $B$ is bounded if $B\times B$ is controlled, or equivalently if $B\times\{x\}$ is for some point $x$.)
Why this last requirement? Why, for example, don't we want the constant map $\Bbb{R}\to\Bbb{Z}$ given by $x\mapsto 0$ to be a coarse map? What breaks down if we allow those?
It's a strange and unpractical convention, indeed. I encourage you not to include properness in the definition.
I agree with @YCor. I am currently writing* a text taking a categorical perspective on the subject in which I omit the proneness property from the definition.
"writing" in my comment above should be taken in the weak sense: this project is kept being postponed due to other administrative duties and more urgent mathematical projects. I am not sure this process is converging for me :(
@UriBader I'd love to read that once it's ready. (In the meantime, is there any reference which does not use this convention?)
@geodude, none that I am aware of.
|
2025-03-21T14:48:29.633128
| 2020-01-15T19:53:46 |
350510
|
{
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"authors": [
"Gerhard Paseman",
"Seva",
"Wojowu",
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"https://mathoverflow.net/users/9924"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350510"
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|
Stack Exchange
|
Quadratic reciprocity for three primes?
The quadratic reciprocity law states that for $p_1\ne p_2$ prime, the product $\left(\frac{p_1}{p_2}\right)\left(\frac{p_2}{p_1}\right)$ takes values $1$ or $-1$ depending on whether $p_1$ and $p_2$ satisfy some set of restrictions mod $4$.
Is there a "quadratic reciprocity law for three primes"? I suspect that the answer is negative.
Is it true that for any integers $M>0$, $\varepsilon\in\{-1,1\}$, and $r_1,r_2,r_3$ coprime with $M$, there exist primes $p_1\equiv r_1\pmod M$, $p_2\equiv r_2\pmod M$, and $p_3\equiv r_3\pmod M$ such that
$$ \left(\frac{p_1}{p_2}\right)\left(\frac{p_2}{p_3}\right)\left(\frac{p_3}{p_1}\right) = \varepsilon ? $$
What about, say, five primes?
I suspect you can take $p_3$ arbitrary, say $1\pmod 4$, and the pick $p_1,p_2$ with $p_1p_2=1\pmod p_3$. I will let someone else work outtge details though
@Wojowu: I thought of this, but working out the details seems messy (and not clear whether this will eventually work).
Working out the details may be messy, but you guys with CAS systems and number theory packages should be able to churn out data and then bin it according to residue by computer in moments. That should point you in the right direction. Gerhard "Assuming There Is A Direction" Paseman, 2020.01.15.
@GerhardPaseman: I actually did some computations, which show that $M=48$ does not work: for any given $r_1,r_2,r_3$ co-prime with $48$, and any $\varepsilon\in{-1,1}$ there exist primes $p_1,p_2,p_3$ satisfying the condition.
Turns out the details are easy so I worked them out myself :) The highlighted statement is true.
Let me assume $4\mid M$. Pick $p_2,p_3$ arbitrary satisfying the congruence modulo $M$ (they exist by Dirichlet). Take any $p_1$ which is congruent to $r_1\pmod M$, congruent to $p_2^{-1}\pmod{p_3}$, and such that
$$\left(\frac{p_1}{p_2}\right)=\varepsilon\cdot(-1)^{\frac{r_1-1}{2}\frac{r_3-1}{2}}$$
(which exists by Dirichlet, CRT, and existence of (non)residues modulo $p_2$.) We have
$$\begin{align*}\left(\frac{p_2}{p_3}\right)\left(\frac{p_3}{p_1}\right)&=\left(\frac{p_2}{p_3}\right)\left(\frac{p_1}{p_3}\right)\cdot(-1)^{\frac{p_1-1}{2}\frac{p_3-1}{2}}\\
&=\left(\frac{p_1p_2}{p_3}\right)\cdot(-1)^{\frac{r_1-1}{2}\frac{r_3-1}{2}}\\
&=\left(\frac{1}{p_3}\right)\cdot(-1)^{\frac{r_1-1}{2}\frac{r_3-1}{2}}\\
&=(-1)^{\frac{r_1-1}{2}\frac{r_3-1}{2}}
\end{align*}$$
(second inequality follows since $p_1\equiv r_1,p_3\equiv r_3\pmod 4$), so multiplying by $\left(\frac{p_1}{p_2}\right)$ leaves $\varepsilon$.
|
2025-03-21T14:48:29.633288
| 2020-01-15T20:16:13 |
350511
|
{
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"authors": [
"Joseph O'Rourke",
"aglearner",
"alesia",
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|
Stack Exchange
|
Finding a not too slim triangulation with prescribed vertices on $\mathbb R^2$
Let us fix a constant $r>1$. Let $d(x,y)$ denote the distance between points $x,y\in \mathbb R^2$. Suppose we have a discreet subset $X\subset \mathbb R^2$ such that
1) For any two points $x,x'\in X$ we have $d(x,x')\ge 1$.
2) For any point $y\in \mathbb R^2$ there is $x\in X$ such that $d(x,y)\le r$.
Question. Is it true that there exist two constants $R>0$ and $\alpha>0$ (depending on $r$), such that for every set satisfying conditions 1 and 2, one can find a triangulation of $\mathbb R^2$ with vertices in $X$, so that each triangle has diameter $\le R$ and all three angles $\ge \alpha$?
Yes, the Delaunay triangulation will have this property. From 2) the circumradius of any triangle will be at most $r$, so diameter will be at least $2r$. Also a too small angle would either imply an edge of length less than $1$ or a circumradius larger than $r$.
Very interesting Alesia! Could you please give me more details? What is the definition of the Delaunay triangulation? (I read wiki before posting the question but have not understood it). Where is it proven that such a triangulation exists?
the Delaunay triangulation is well defined when no four points lie on circle (if that happens you can get away with it for your purpose). It is characterized by the property that circumcircles of all the triangles in it contain no point in their interior
Yes, I think I understood the definition, but I want to know where is it proven that a Delaunay triangulation exists? (Let's even assume that no 4 points lie on the same circle.)
you can look for example at page 37-38 of http://www.lix.polytechnique.fr/~pilaud/enseignement/MPRI/notesCoursMPRI16.pdf where it is shown that Delaunay triangulation are obtained as projections of 3D polytopes
Thank you. I had a look into this text, it is a bit informal. Do you think that there is some reference in pure maths literature, preferably a book?
this is basic computational geometry, should be essentially in any book on the subject
@aglearner: An attractive approach is via edge-flipping, which replaces thin triangles with fatter triangles, and remarkably leads to the Delaunay triangulation just via edge-flipping in any order according to a circumcircle calculation.
The Delaunay triangulation produces a triangulation with the desired the guarantees. 2) ensures that the circumradius of any triangle in the triangulation is at most $r$. A circle of radius $r$ containing the triangle means the diameter of any triangle is at most $2r$, i.e., the diameter of any triangle is at most $R = 2r$.
1) ensures the length of any edge of any triangle is at least $1$. For any Delaunay triangle $T$ with circumradius $r_T$, the law of sines can be applied to any angle $A_T$ and opposite edge $a_T$,
$\sin A_T = \frac{a_T}{2r_T} \leq \frac{1}{2r}.$
So all angles of the triangulation are at least $\alpha = \arcsin\left(\frac{1}{2r}\right)$.
Note that while a unique Delaunay triangulation is only guaranteed to exist if the pointset is in general position (i.e., no four parts are cocircular), these guarantees hold for any set of input points by using the Delaunay construction and picking any arbitrary triangulation for any sets of four of more cocircular points.
|
2025-03-21T14:48:29.633629
| 2020-01-15T20:49:44 |
350514
|
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"Elise",
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|
Stack Exchange
|
moving from sphere spectrum to finite spectrum
I am reading Hatcher's treatment of the Adam's spectral sequence. http://pi.math.cornell.edu/~hatcher/SSAT/SSch2.pdf
On page 20, he states "Thus for each $i$ the groups $\pi_i(Z^k)$ are zero for all sufficiently large $k$. The same is true for the groups $\pi^Y_i (Z^k)$ when $Y$ is a finite spectrum, since a map
$\Sigma^i Y \to Z^k$ can be homotoped to a constant map one cell at a time if all the groups $\pi_j (Z^k)$ vanish for $j$ less than or equal to the largest dimension of the cells of $\Sigma^i Y$.''
Context: $Z^k$ is a CW spectrum of finite type. Also, $\pi^Y_n(Z) = [ \Sigma^n Y, Z ] = colim_k [ \Sigma^{n} Y_k, Z_k]$, where the $Y_k$ and $Z_k$ are spaces.
I don't really understand the details of his statement. So my questions are
1) Can someone explain the details of this more?
2) (I'd prefer) Can we say instead that this holds because we can write $Y$ as a finite limit of sphere spectra, and finite limits and filtered colimits commute? And if so, how does one write $Y$ as a finite limit of sphere spectra?... Does this work exactly the same as if we had spaces?
Let $\{Y,Z\}$ denote the homotopy group of maps between a spectrum $Y$ and a spectrum $Z$. We are assuming that one has a sequence of spectra $Z^1, Z^2, \dots$ such that for any fixed $i$, $\{S^i,Z^k\} = 0$ for $k>>0$, and we want to show that, if $Y$ is a finite spectrum, then $\{Y,Z^k\} = 0$ for $k>>0$.
An easy way to see this is by induction on the number of cells of $Y$, with inductive step as follows: if $Y$ is obtained from $X$ by attaching an $n$--cell, then, by basic homotopy theory, the cofibration sequence $X \rightarrow Y \rightarrow S^n$ induces $\{S^n,Z^k\} \rightarrow \{Y,Z^k\} \rightarrow \{X,Z^k\}$ exact in the middle. By assumption and inductive hypothesis, the first and last of these are 0 for $k>>0$, and thus the so is the middle term. [This is clearer than Hatcher, who is sort of saying the same thing much more awkwardly.]
So basically, exactly the same as if everything were spaces, except that since we are in the stable homotopy category, fibrations and cofibrations are the same so that we can get the exact sequence?
@Elise Yes. But even better: the cofibration is in the first variable, so it is exact in the world of spaces too. No fibrations needed here.
For convenience I will set $Z = Z^k$, where $Z^k$ is as in your notation.
Case 1: $Y$ is the suspension spectrum of a based finite complex $U$ having dimension $n$. Then
$\pi^Y_\ast(Z)$ is given by the homotopy groups of the
function space of based maps $F(U,\Omega^\infty Z)$, where $\Omega^\infty Z$ is the infinite loop space associated with $Z$.
Filter $U$ by its skeleta $U^{(j)} \subset U$. Then we have a sequence of fibrations
$$
F(U,\Omega^\infty Z) = F(U^{(n)},\Omega^\infty Z) \to F(U^{(n-1)},\Omega^\infty Z)\to \cdots \to
F(U^{(0)},\Omega^\infty Z)\, .
$$
The fibers of the map at stage $j$ in the sequence are
identified with
$F(U^{(j)}/U^{(j-1)},\Omega^\infty Z)$ and the latter is a finite product of copies of $F(S^j,\Omega^\infty Z)$.
The homotopy groups of $F(S^j,\Omega^\infty Z)$ are just the homotopy groups of $Z$ shifted by $j$. So, by your assumptions, for any index $i$ there is a $k$ with $Z= Z^k$ such that $\pi_\ell$ of $F(S^j,\Omega^\infty Z)$ vanishes for $\ell\le i$. Consequently, the fiber of the $j$th fibration is $i$-connected. Furthermore, a similar argument shows that $F(U^{(0)},\Omega^\infty Z)$ is $i$-connected.
It follows that $F(U,\Omega^\infty Z)$
is also $i$-connected.
Case 2: $Y$ is an arbitrary finite spectrum. Then there is a non-negative integer $t$ such that $\Sigma^t Y$ is the suspension spectrum of a finite complex $U$. So, $F(Y,Z) = F(U,\Sigma^t Z)$. The argument then works as above with $Z$ replaced by $\Sigma^t Z$.
Remark: What the above in effect shows is this: let $Y$ be a finite dimensional cell spectrum of dimension $s\in \Bbb Z$ and let $Z$ be an $r$-conected spectrum. Then the function spectrum $F(Y,Z)$ is $(r-s)$-connected.
|
2025-03-21T14:48:29.633882
| 2020-01-15T21:46:15 |
350517
|
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|
Stack Exchange
|
Does an exact $(p,q)$-form on a non-compact Kähler manifold admit a primitive with at least $p$ $dz$'s?
Motivation: On a compact Kähler manifold $Y$, $\partial\bar{\partial}$-lemma tells us that an exact $(p,q)$-form $\omega$ has a primitive $\eta$ with $p$ $dz$'s, namely $\eta\in F^p\mathcal{A}^{p+q-1}_Y$. Here $\eta$ is a primitive of $\omega$ means $d\eta=\omega$. This is because $\omega=\bar{\partial}\partial\sigma$ for some $\sigma$ of type $(p-1,q-1)$, so we can just take $\eta=\partial\sigma$.
I'm looking for something similar for a non-compact Kähler manifold. However, we don't have $\partial\bar{\partial}$-lemma anymore.
The case I'm interested in is when $X$ is the total space of a smooth family $\pi:X\to \Delta$ of compact Kähler manifolds, where $\Delta$ is a holomorphic disk. Let $\omega$ be an exact smooth form of type $(p,q)$. Let $X_0=\pi^{-1}(0)$. I'd like to know
Question: Is there a smooth form $\eta$ such that $d\eta=\omega$ and $\eta_{\rvert X_0}\in F^p\mathcal{A}_{X_0}^{p+q-1}$?
I'm not sure if similar questions were considered. I would appreciate it if someone could provide a reference or share some ideas.
|
2025-03-21T14:48:29.634012
| 2020-01-15T23:08:28 |
350520
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"Hempelicious",
"Josh Howie",
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"Ryan Budney",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/350520"
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|
Stack Exchange
|
Generalized Schoenflies - formalizing step in proof?
[Sorry if the level here is wrong, I asked this on math.SE, but even with a bounty, it got no attention.]
I am currently reading Hatcher's 3-Manifolds notes, the part proving Alexander's theorem, which is a specific case of the generalized Schoenflies theorem:
Every smoothly embedded $S^2\subset \mathbb{R}^3$ bounds a smooth 3-ball.
The proof seems to rely on intuition for these low-dimensional arguments, which I find disconcerting, because I have not yet developed that intuition, so I am trying to give actual formal proofs for the statements in Hatcher's proof.
The proof begins with a generic smoothly embedded closed surface $S\subset\mathbb{R}^3$. I have been able to prove that I can isotope $S$ so that projection on the last coordinate $\pi:\mathbb{R}^3\rightarrow\mathbb{R}$ is a Morse function on $S$. Hatcher then argues that if $t$ is a regular value for $\pi$, then $\pi^{-1}(t)\cap S$ is a finite collection of circles.
The proof continues by taking an innermost circle $C\subset \pi^{-1}(t)\cap S$, which by 2-dimensional Schoenflies bounds a disk $D$, and $D\cap S=\partial D=C$. Hatcher then uses surgery to cut away a neighborhood of $C$ in $S$, and cap the cuts with two disks.
This last part is what I want to formalize. It seems we are finding a small-enough tubular neighborhood $C\times(-\epsilon,\epsilon)\subset S$, and then removing that, leaving $S_-=C\times\{-\epsilon\}$ and $S_+=C\times\{\epsilon\}$. Again by 2-dimensional Schoenflies, these bound disks $D_-$ and $D_+$. What I don't get is: why are $S_-$ and $S_+$ still innermost? Or, put differently, why is $D_-\cap S=S-$ (and similarly for $S_+$)?
Intuitively, this seems obvious, and it seems like some sort of "continuity" argument would work, but I cannot figure out how to make this formal. I tried proving that in fact all the disks, "stacked" together for the tubular neighborhood, gave a smooth $D\times [-\epsilon,\epsilon]$, but again I find it hard to make topological arguments when one step of the construction is "apply Schoenflies to get a disk". In particular, I can't prove the projection of this "solid neighborhood" to $D$ is continuous.
Does anyone know how to formalize this? Or, even better, a reference where this type of surgery is discussed? I checked a few places, but only found surgery on a single manifold, not the type discussed here, where we're surgering an embedded submanifold in some ambient manifold.
Hatcher's notes assumes (roughly) that the reader is familiar with the proof of the h-cobordism theorem and the main content of Milnor's book on Morse Theory book. The style and conventions in the types of arguments he's using are built-up when studying those topics.
@RyanBudney: do you think reading through Milnor's books would give me the background to formalize this argument? Is there some key step or convention that I'm missing ?
Maybe Guillemin and Pollack's "Differential Topology" would be the more appropriate place to start. The transversality (stability) arguments there would help you with this particular argument.
For an interval $[a,b]\subset{\mathbb R}$ in which the height function $f:S\to {\mathbb R}$ has no critical values one obtains a product structure on $f^{-1}([a,b])$ by following flow lines of the gradient vector field of $f$. This vector field on $f^{-1}([a,b])$ can be extended to a vector field on ${\mathbb R}^2\times [a,b]$ with positive $z$-coordinate everywhere in ${\mathbb R}^2\times [a,b]$. To do this one can first extend the gradient vector field on the surface to a tubular neighborhood of the surface via a projection of this neighborhood onto the surface, then use a smooth partition of unity subordinate to the cover of ${\mathbb R}^2\times [a,b]$ by the tubular neighborhood and the complement of the surface to combine the vector field on the neighborhood with the vertical vector field $(0,0,1)$ on the complement of the surface. In formulas the combined vector field would have the form $v=\phi_1 v_1+\phi_2 v_2$ where $v_1$ is the vector field on the neighborhood and $v_2$ is the vector field on the complement of the surface, with the partition of unity functions $\phi_1$ supported in the neighborhood and $\phi_2$ supported in the complement of the surface. The flow lines of this extended vector field $v$ then give a new product structure on ${\mathbb R}^2\times [a,b]$ extending the product structure on $f^{-1}([a,b])$. In other words one has a level-preserving diffeomorphism of pairs $({\mathbb R}^2\times [a,b],f^{-1}([a,b]))\approx ({\mathbb R}^2\times [a,b], f^{-1}(a)\times [a,b])$.
This is a special case of the isotopy extension theorem which says that an isotopy of a submanifold can always be extended to an ambient isotopy of the whole manifold. The proof is essentially the same.
I love that on this site you can ask for help with a textbook’s exposition, and get an answer directly from the textbook’s author. If this answer doesn’t merit the bounty, I can’t imagine what would!
Thank you! Extending things to the ambient slice is the piece I was missing.
If $t$ is a regular value, then it is a property of Morse functions that there is some small open neighborhood $U$ of $t$ in $\mathbb{R}$ such that $u$ is also a regular value for all $u\in U$. In particular, we can take $U=(t-\delta,t+\delta)$ for some $\delta>0$. But then $\pi^{-1}(U)\cap S\cong(\pi^{-1}(t)\cap S)\times U$, ie. the surface is a product between any two successive critical levels. Now simply choose $\epsilon<\delta$. Then $S_-=(C\times U)\cap\pi^{-1}(t-\epsilon)$ and $S_+=(C\times U)\cap\pi^{-1}(t+\epsilon)$ will be innermost since $C$ is innermost.
Sorry, I don't follow the final sentence at all. Why would there be a critical value?
I deleted that sentence. Is it clear now?
I think it's intuitively clear, and does help me understand things better, but I'm trying to avoid intuition. In particular, your diffeomorphism with the product does not take into account the ambient slices, so I don't see why we can say anything about a circle being innermost or not. Here's an example of what I'm worried about: if all we know at each slice is there are two circles, there's nothing a priori preventing one moving "inside" another. Intuitively, that can't happen because the enclosed disks would change how they intersect. It's this last part I can't formalize.
That cannot happen because there is also a product $D\times U$.
I think that's the Crux of the issue: how do we know that? How do we know the disks we get from Jordan curve theorem can be aligned as a product via the map you've provided?
You have a collection of circles which move smoothly (hence homotopically, see below) with the height (due to regularity and implicit function theorem). Let us define what means "innermost": It means that if you take a point $x$ from the inner circle then every outer circle will have winding number 1 or -1 around $x$. Or if you take a point $x$ from an outer circle then the inner circle will have winding number 0 around $x$. Now simply use the fact that the winding number is invariant under homotopies, that is, if $H\colon[a,b]\times[0,1]\to\mathbb R^2$ and $x\colon[0,1]\to\mathbb R^2$ are continuous with $H(t,0)=H(t,1)$ and such that $x(t)\notin H(\{t\}\times[0,1])$ for all $t\in[a,b]$ then the winding number $n(H(t,\cdot),x(t))$ is independent of $t\in[a,b]$.
|
2025-03-21T14:48:29.634481
| 2020-01-16T01:54:35 |
350524
|
{
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"authors": [
"Iosif Pinelis",
"LSpice",
"Landauer",
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|
Stack Exchange
|
Properties of convolutions
Consider the function
$$f_{n}(x)=e^{-x^2}x^n.$$
and the function
$$h_p(x):=e^{-\vert x \vert^p}.$$
My goal is to analyze
$$ F_p(y):=\frac{(f_2*h_p)(y)}{(f_0*h_p)(y)}- \left(\frac{(f_1*h_p)(y) }{(f_0*h_p)(y)}\right)^2$$
Question: Can we show that $F_p$ has a global maximum at zero for $p>2$ and a global minimum at zero for $p<2$?
Here are graphs from Mathematica for
$F_1$, with a unique minimum
$F_4$, with a unique maximum
$F_{2+10^{-4}}$, i.e. with an exponent slightly above $2$
$F_{2-10^{-4}}$, i.e. with an exponent slightly below $2$
Further observations:
Iosif Pinelis showed that $F_2(y)$ is constant, see this earlier question of mine question I asked some days ago.
All functions $F_p$ are positive by the Cauchy-Schwarz inequality.
The functions $F_p$ are not log-convex or log-concave in general.
Finally, I computed the first derivative for convolving with $e^{-\vert x \vert^p}$ at $p=2$ by numerically differentiating
If you have any other conjectures you would like to me to verify, I am happy to do so. Just leave a comment!
$x \mapsto e^{-x^2}$ isn't just some random medium-decaying function, though; it's an eigenfunction for the Fourier transform, and that's got to be significant.
Have you tried exponents $p=2\pm\epsilon$ for small (maybe infinitesimally small) real $\epsilon$, instead of exponents $p\in{1,4}$ in $e^{-|x|^p}$? I'd think that a "phase transition" at $p=2$ would not be very surprising, since the resulting function is constant for $p=2$ and thus is on the border between the (log-)convexity and (log-)concavity. Also, what I see as a lack of a "global" "phase transition" is that the resulting function is $>0$ for $p=2$ but seems to be $<0$ for $p\in{1,4}$ -- whereas $1<2<4$.
@IosifPinelis Sorry, $2+10^{-5}$ was all I could do numerically but I gives the desired results. I also checked whether the functions are log-convex/log-concave in general. It seems they are not. So trying to verify some log-convexity/log-concavity here seems to be a misleading approach, at least as far as the numerics is concerned. The negativity was because I plotted logs. I clarified this now and plotted the actual function and one log. The actual functions are all positive by Cauchy-Schwarz (sorry for the confusion about what is plotted I hope it is much clearer now)
@IosifPinelis if you want me to test some other conjectures numerically, please let me know. Your help is greatly appreciated.
@Martinique : I don't understand what you got for $2+10^{-5}$. Did you mean for $p=2+10^{-5}$? Anyhow, what were the "desired results" you got? Did you try something like $p=2\pm10^{-3}$ or $p=2\pm10^{-2}$? (I mean both $+$ and $-$.) Did you try the derivative in $p$ at $p=2$?
@IosifPinelis Yes, it is $p=2+10^{-5}$, I clarified it now. Let me check for the derivative.
@Martinique : Your graph for the derivative in $p$ at $p=2$ (showing a lack of a phase transition for all $y$ at once) seems to contradict the graphs for $p=2\pm10^{-5}$. Have you tried to take the derivative analytically and then evaluate the involved integrals numerically? You can also try to use asymptotics to find the sign at the tails of the derivative, for large $|y|$. Also, can you label the axes (especially the vertical one)?
@IosifPinelis okay, now the derivative is much more stable
@Martinique : In a couple of instances, you labeled the vertical axis by something like $p=...$. I think the vertical axis carries the values of your resulting function, whereas something like $p=...$ is a plot label (rather than an axis label). Also, it is unclear to me how you approximated the derivative at $p=2$: (i) numerically or (ii) analytically and then evaluated the involved integrals numerically.
Yes, $p=$ means I plotted the function $F_p$ with the respective label of $p.$ I agree it looks like a plot label but it is an axis label in mathematica for the $y$ axis. The derivative is still numerically, but I was more careful this time. The problem is if I differentiate it analytically and then do it numerically, it is more difficult for the numerics to compute, as one has many more integrals.
@MattF. it is to verify the numerical findings. But I emphasized the question now more clearly.
Global minimum at $0$ for $p<2$ is trivial in the sense that you can see it without writing a single equation or inequality (i.e., a non-trivial formula with two sides and a sign in between comparing them).
Observation 1 $e^{-|x|^p}$ is some weighted average of $e^{-ax^2}$ with positive $a$.
Observation 2 The thingy you are interested in is just the variance of $x$ with respect to the probability measure $\mu_y$ whose density is proportional to $e^{-x^2}e^{-|x+y|^p}$. By observation 1, this measure is a mixture of the probability measures $\mu_{a,y}$ with densities proportional to $e^{-x^2}e^{-a(x+y)^2}$. The weight of $\mu_{a,y}$ in that mixture is proportional to something independent of $y$ times $e^{-\frac a{a+1}y^2}$, i.e., when we move $y$ away from the origin, the measures $\mu_{a,y}$ with lower $a$ gain more weight in the composition.
Observation 3. The variance of $x$ with respect to $\mu_{a,y}$ is independent of $y$ and decreases in $a$. To be exact, it is just inversely proportional to $1+a$.
Observation 4. The variance in the mixture is at least the mixture of the variances, which is minimized at $0$ by the independence of individual variances of $y$, observation 3, and the last sentence of observation 2. Also at $y=0$ we have equality because all means are at $0$ by symmetry.
The end.
I wish I could come up with an equally simple argument for $p>2$, but, alas, I don't have one at the moment.
Edit OK, I guess I finally figured it out. We'll prove the following statement. Let $\varphi$ be an even convex function such that $\varphi''$ increases on $[0,+\infty)$. Let $p_y(x)$ be the probability density proportional to $e^{yx}e^{-\varphi(x)}$. Then the variance of $x$ with respect to the corresponding probability measure is a non-increasing function of $y$ for $y>0$.
Indeed, let's differentiate in $y$. We have $p_{y+\delta y}(x)$ proportional to $p_y(x)(1+x\delta y)$ (the linearization of $e^{x\delta y}$), but if we leave it at that, the mass will change to $1+\delta y\int p_y(x)xdx=1+c\delta y$ where $c$ is the expectation of $x$ with respect to $p_y$, so we need to compensate by dividing by that factor, which will result in the linearization
$$
p_{y+\delta y}(x)=(1+(x-c)\delta y)p_y(x)
$$
Taking the linearization of the variance of $x-c$ now (which is the same as the variance of $x$ but easier to compute), we see that what we need to show is that
$$
\int (x-c)^3p_y(x)dx\le 0.
$$
We will show even that $\int_{x:|x-c|>a}(x-c)p_y(x)\le 0$ for all $a\ge 0$. The equality obviously holds for $a=0$ (the definition of $c$) and for $a=+\infty$. The derivative in $a$ is just $a(p_y(c-a)-p_y(c+a))$. I claim now that it can change sign only once for $a>0$ and that change is from $-$ to $+$. That is equivalent to saying that $\Phi(a)=\varphi_y(c-a)-\varphi_y(c+a)$, where $\varphi_y(x)=\varphi(x)-yx$, can change sign only once for $a>0$ and that change is from $+$ to $-$.
We clearly have $c>0$ for $y>0$. Therefore, the point $c-a$ is always closer to the origin than $c+a$ for $a>0$ whence, by the assumed property of the second derivative of $\varphi$ (the linear term coming from $yx$ can do nothing with the second derivative), $\Phi''<0$. Thus, if we start from $\Phi(0)=0$ in the positive direction, we change sign once from $+$ to $-$ as promised. If we had started in the negative direction, we would never be able to change sign at all and the integral we are interested in would be monotone all the way, which is ridiculous, because its values at $0$ and $+\infty$ are both $0$, so that case is impossible.
That takes care of $p>2$. If the second derivative of $\varphi$ were decreasing, then all inequalities would be reversed, so we can cover $p<2$ by this method as well and even get unimodality you see on the pictures, not just the global minimum at $0$.
I hope I haven't made any stupid mistake in the computations but, since it is nearly midnight here now, you'd better check them carefully :-)
I still had to open the parenthesis in the quadratic polynomial and complete the square, I have to admit that :-)
@Martinique OK, it looks like it is even simpler than I thought (provided I do not hallucinate after 11PM, of course)
@Martinique "At least at first glance, it does not match the expression I wrote down in the question" Really? Your density is proportional to $e^{-|x|^p-(x-y)^2}=e^{-y^2}e^{-(|x|^p+x^2)}e^{2yx}$ (I prefer the shift in the convolution to be placed on the Gaussian this time). Nobody cares about $e^{-y^2}$ that is killed by the normalization and the rest is just of the kind I considered, isn't it?
@Martinique The variance of $x$ is the same as of $x-y$. The individual terms are, indeed, not translation invariant, but the whole expression is.
@Martinique You are welcome but now I'm perplexed a bit. If you have not known the probabilistic interpretation from the start, where did that monster come from?
@Martinique Yes, I do the same computation that you've made and notice that the term that you squared is quadratic in $\delta y$, so we ignore it in the linearization.
"Assuming you had not computed the variance of $X-c$ but only of $X$" What do you mean? They are the same. If you had followed this way, then the second term would give a non-trivial contribution and you would have to combine the two (with the same final result, of course, but more algebraic transformations on the way). I feel like you want to ask something interesting here but I'm not sure what exactly :-)
@Martinique Yeah, but I have never claimed that $\int x^2(x-c)p_y(x)dx\le 0$. Are you sure it is true?
@Martinique No, you haven't opened the parentheses correctly: the second term produces also $2E_{p_y}[X]E_{p_y}[X(X-c)]\delta y$. Then you have to use that $E_{p_y}X=c$ to get the expression with $(X-c)^3$. Algebra is a hard subject... :-)
|
2025-03-21T14:48:29.635169
| 2020-01-16T02:15:53 |
350525
|
{
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"Mateusz Kwaśnicki",
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"url": "https://mathoverflow.net/questions/350525"
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|
Stack Exchange
|
If a real-valued bivariate function on the unit square is integrable along each line, is it integrable on the square?
Let $f(x,y)$ be a real-valued function on the unit square $[0,1]^2$. Suppose that $f(x,y)$ is Riemann integrable along each straight line. Does this imply that $f$ Riemann integrable on the square? Does the answer change if we only suppose that $f$ is integrable along all vertical lines $x=c$ and integrable along all horizontal lines $y=c$?
No, consider the function
$$g(x,y) = \cases {0,&if $(x,y)=(0,0)$\\
xy^2/(x^2+y^6) &otherwise}$$
taken from Exercise 4.7 of Baby Rudin (3ed) via this Math.SE post. Along every line it is continuous and hence Riemann integrable, but it is unbounded along the curve $x=y^3$, and an unbounded function cannot be Riemann integrable.
The OP deleted a follow-up question asking the same question for hyperplanes in the $n$-cube, but for the record the answer is still no. A simple counterexample suggested by @Nate Eldredge's answer: Let $f$ be the function that takes $(1/k, 1/k^2, 1/k^3, \ldots, 1/k^n)$ to $k$ for each positive integer $k$, and takes everything else to zero. Each hyperplane meets at most $n$ of those points, so the integral of $f$ on any hyperplane vanishes; but again the function is not bounded on the cube, hence is not Riemann integrable.
Thank you for the answer and reference, Nate. I actually had meant to include the condition that $f$ is bounded. As I forgot and you answered the question that I did ask, I accepted.
Do you know though of an example or reference for the case when we add the condition that f is bounded?
If not, would it be ok if I edit this question to ask what I had intended, or is it better to ask in a separate thread?
Also sorry for deleting the follow up. I wasn’t sure if I should leave it since I accepted an answer that focused on the main question.
Thanks, @NoamD.Elkies for the answer to the $n$-cube question. If I may ask, are these types of unbounded functions the only way to construct counterexamples? As I mentioned in my comment above, I had intended to ask about bounded functions. The type of counterexample I was imagining (but couldn't come up with) was a function that had discontinuities on a set $E\subset [0,1]^2 $ that has a positive ("area") measure in $[0,1]^2$ but the intersection of $E$ and any straight lines would have zero ("length") measure.
@YacoubKureh: You can find a countable set $A$ which is dense in $[0,1]^2$, and which intersects any line in at most two points (recursively: choose a sequence of balls which is a basis for the topology in $[0, 1]^2$, and in $n$-th step add a point which is in the $n$-th ball and which does not lie on any line containing any two of previously chosen points). Then the characteristic function of $A$ does the job: it is discontinuous everywhere, but its restriction to any line is zero except at no more than two points.
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2025-03-21T14:48:29.635398
| 2020-01-16T04:16:22 |
350530
|
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|
Stack Exchange
|
LU decomposition for orthogonal or unitary matrices?
Is there any references on LU decomposition for orthogonal or unitary matrices?
It seems to me that the diagonal entries of $U$ has some nice structure regarding to the Euler angles of the original matrix. As one can easily see under a Euler parametrisation:
$$\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}=\begin{bmatrix}1&0\\\tan\theta&1\end{bmatrix}\begin{bmatrix}\cos\theta&\sin\theta\\0&1/\cos\theta\end{bmatrix}.$$
And for the $3\times 3$ case, the diagonal entries for $U$ should be something similar to
$$\cos\theta_1\cos\theta_2, \cos\theta_3/\cos\theta_1, 1/\cos\theta_3\cos\theta_2.$$
Is there any previous work on these?
Here is one study of the analogue of the Cholesky decomposition for orthogonal matrices: Unconstrained representation of orthogonal matrices with application to common principle components (2019).
Recall that the Cholesky decomposition is a LU decomposition of a Hermitian matrix, where $U$ is the conjugate transpose of the lower-triangular matrix $L$. The analogue for an orthogonal matrix $O$ is
$$O=PLR^{-1}$$
where $P$ is a permutation matrix, $L$ is lower triangular, and $R$ is such that $PL=QR$ with $Q$ orthogonal and $R$ upper-triangular.
So, up to a permutation $P$ of the columns of $O$, this orthogonal matrix is fully determined by an unconstrained lower-triangular matrix $L$ --- in this sence the "PLR-decomposition" is the analogue of the Cholesky decomposition.
Thanks. Any relationship with the Euler angles?
none that I know of.
|
2025-03-21T14:48:29.635708
| 2020-01-12T08:31:53 |
350255
|
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|
Stack Exchange
|
Is $δ=δ(x)$ a continuous function
If a real function $f:ℝ→ℝ$ is twice differentiable at a point $x$, then the first derivative must be continuous at $x$, and assuming $f′(x)>0$, then there exist $δ>0$ such that $f′(y)>0 $ for all $y∈(x−δ,x+δ)$, then on this interval $f$ must be increasing. Repeating this process for all $x$, we conclude that $δ$ is a function of $x$
Assuming that this function is analytic (the stronger form). I am asking if $δ=δ(x)$ is a continuous function in $x$.
If no, then what are the conditions on the function $f$ such that the above property holds true.
You have not defined $\delta$ uniquely as a function of $x$. For example, if $f(x)=x$, then one can take $\delta=1$ for $x\in\mathbb{Q}$ and $\delta=2$ for $x\not\in\mathbb{Q}$, which is not continuous (obviously).
@GHfromMO: Did you mean that we must add some thing about the set of $x$.
Note also that on any compact interval $I\subset\mathbb{R}$ you can choose $\delta=\delta(I)$ to be constant, since $f'(x)$ is positive and uniformly continuous on $I$.
No, I meant that for a given $x$, infinitely many positive $\delta$'s work (if $\delta$ is ok than any smaller $\delta$ is also ok), so it is not clear what you mean by $\delta(x)$.
@GHfromMO: It can be considered as one of those reals which I assume depends on $x$
Then trivially the answer to your question is no. For any positive valued function, there is a smaller positive valued non-continuous function. See my first remark.
You may want to see also MR1745893 (2000m:54014) Enayat, Ali $\delta$ as a continuous function of $x$ and $\epsilon$. Amer. Math. Monthly 107 (2000), no. 2, 151–155.
And MR1837868 De Marco, Giuseppe For every $\epsilon$ there continuously exists a $\delta$.Amer. Math. Monthly 108 (2001), no. 5, 443–444.
$\delta(x)$ (defined as the maximum of appropriate $\delta$'s) is even 1-Lip (for any $f$). Indeed, if $|x-y|=a$, then $\delta(y)\geqslant \delta(x)-a$, since $(y-c,y+c)\subset (x-\delta(x),x+\delta(x))$ for $c=\max(\delta(x)-a,0)$. Analogously $\delta(x)\geqslant \delta(y)-a$ and therefore $|\delta(x)-\delta(y)|\leqslant a$.
I guess you meant that $\delta(x)$ is the supremum of all $\delta$'s that work for a given $x$. The problem is that the OP has not defined $\delta(x)$.
@GHfromMO yes, I understood her this way
Smoothness of $f$ is a distraction here. Let us assume that $f$ is continuously differentiable (and we only need this because we want to talk about $f'$, not for any good reason), so that $f'$ exists and is continuous. Then $B : =\{x \mid f'(x) \leq 0\}$ is a closed set, and the largest valid choice for $\delta(x)$ is just $d(x,B)$.
For any set $A$, the function $x \mapsto d(x,A)$ is a $1$-Lipschitz function, by the triangle inequality for metrics. So, essentially, we already get that $\delta$ can be chosen as nice as possible if we only require enough about $f$ to make $\delta$ well-definable.
Just for completeness, if we don't require $\delta(x)$ to be the maximal feasible value, we could make $\delta$ as nasty as we want, e.g. by moving to $\bar\delta$ where $\bar\delta(x) = \delta(x)$ if $x \in Z$ and $\bar\delta(x) = \frac{1}{2}\delta(x)$ if $x \notin Z$ for some horrible $Z$.
Probably better not to use $\delta'$ for a new function when we're discussing differentiability.
@LSpice Bah, non-ambiguous notation is for pansies. Joking. I fixed this.
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2025-03-21T14:48:29.636019
| 2020-01-12T09:39:33 |
350258
|
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|
Stack Exchange
|
Example of an associative unital ring R with stable range 1 and Jac(R)=0 that is not an exchange ring
Rings are supposed to be associative and unital, but not necessarily commutative.
Some definitions:
(Bass) A ring $R$ is said to have stable range $1$ if for all $a,b \in R$, whenever $Ra+Rb=R$, then there exists $x\in R$ with $a+xb$ being a unit.
(Warfield) A ring $R$ is said to be an exchange ring if it has the finite exchange property or equivalently (Nicholson) if for all $a\in R$ there exists an idempotent $e\in R$ with $e\in Ra$ and $1-e \in R(1-a)$.
Examples: Semiperfect rings have stable range $1$ and are exchange.
My question is whether someone has an example of a ring $R$ with stable range $1$ and Jacobson radical zero, that is not exchange?
I know at least two:
The integral closure of $\mathbb Z$ in $\mathbb C$
The ring of holomorphic functions on $\mathbb C$.
Each nontrivial ideal of an exchange ring with Jacobson radical zero must contain a nonzero idempotent, but since these are both domains, this is clearly not the case for them.
I found these using this DaRT query.
Nice! Thanks a lot. Yes, for an exchange ring that is a domain, an element x is either invertible or 1-x is invertible.
|
2025-03-21T14:48:29.636163
| 2020-01-12T11:14:41 |
350264
|
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|
Stack Exchange
|
A combinatorics question: $\lim\limits_{n \to \infty} \frac1{2^{2n}} \sum\limits_{k=1}^n \sum\limits_{i=0}^{k-1} \binom nk \binom ni = \frac12$
Am trying to show that $\lim_{n \rightarrow \infty} \frac{1}{2^{2n}} \sum_{k=1}^n \sum_{i=0}^{k-1} \binom{n}{k} \binom{n}{i} =0.5.$
I think that the above result is true but am not sure how to prove this. Any help on this will be greatly appreciated. Thanks in advance!
The general term is the probability that if one picks a pair of subsets from ${1,2,\ldots,n}$, the second has smaller size than the first, so you just need to show that the probability that the two sets have the same size tends to zero.
We can rewrite the sum as
\begin{align*}
\sum_{k=1}^n \sum_{i=0}^{k-1} \binom{n}{k} \binom{n}{i}
&=\sum_{0\leq i<k\leq n}\binom{n}{k} \binom{n}{i}\\
&=\frac{1}{2}\sum_{\substack{0\leq i,k\leq n\\i\neq k}}\binom{n}{k}\binom{n}{i}\\
&=\frac{1}{2}\left(\sum_{0\leq i\leq n}\binom{n}{i}\right)^2-
\frac{1}{2}\sum_{0\leq i\leq n}\binom{n}{i}^2\\
&=2^{2n-1}-\frac{1}{2}\binom{2n}{n}.
\end{align*}
It is well-known that
$$\binom{2n}{n}\sim\frac{2^{2n}}{\sqrt{\pi n}},$$
hence
$$\sum_{k=1}^n \sum_{i=0}^{k-1} \binom{n}{k} \binom{n}{i}=\left(\frac{1}{2}-o(1)\right)2^{2n}.$$
The proof is complete.
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2025-03-21T14:48:29.636274
| 2020-01-12T12:01:49 |
350268
|
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|
Stack Exchange
|
Can the axiom of choice or its weaker versions be (dis)proved using reflection principles?
In On the Question of Absolute Undecidability, Peter Koellner investigates whether it is possible to prove or disprove $V = L$ using (EDIT: both first and second-order) reflection principles, ie. statements of the form*
$$V \vDash \varphi(A) \to \exists \alpha \ V_\alpha \vDash \varphi^\alpha(A^\alpha)$$
and shows that it cannot be done.
Starting from $ZF$, can these principles be used to prove or disprove the Axiom of choice or some of its weaker variants (eg. Dependent, Countable choice)?
* definition on page 13.
Although this is unrelated to the question on AC, perhaps the Robert's paper "A strong reflection principle" would be relevant. In this paper, there is a formulation of a reflection principle, of similar form, that implies the existence of 1-extendible cardinals (so it's incompatible with V=L).
That seems to be an interesting paper. Bookmarked and saved in the Wayback Machine, thanks.
The question seems to me asking if sufficiently large cardinals defined by indescribability properties will prove the axiom of choice or its weak variants hold below such cardinals. (Disproving is moot since these are consistent with $V=L$.)
The answer is negative, but more complicated. First of all, we can violate any sort of choice "on a small set", well below our large cardinal, and that will preserve the reflection properties. So the question now is if we can make the failure in some sense large. And indeed we can. With Yair Hayut we developed a basic method for lifting elementary embeddings to symmetric extensions and we showed that it's consistent relative to large cardinals that there is a critical point whose successor is singular.
It's not hard to check that this reflects down, and that indeed this critical point satisfies any reflection principle we wanted (it was previously a supercompact cardinal, after all).
We continue research in this direction, and I hope we will have new results to announce soon.
This doesn't make sense to me. If reflection principles were a consequence of ZF, why would anybody bother trying to prove $V = L$ with them? Also, on page 14, Koellner says "This [reflection] principle yields inaccessible cardinals, Mahlo cardinals, weakly compact cardinals and more". Maybe, in my question, I mischaracterized the reflection principles? (But I've taken the definition from the paper.) Or maybe you only considered first-order reflection principles? Is that I also ask about second-order reflection something that I should have included in my question?
I will take a closer look. But just like Koellner says, you need to specify the language and logic in order to determine the extent of your Reflection principles. You specified nothing so I assumed you meant first order (which indeed is too weak to decide something like V=L). Ask, and ye shall receive an answer; ask a precise question, and ye shall receive a precise answer.
I've edited to reflect what I think you're asking.
And I've edited the question. Thank you.
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2025-03-21T14:48:29.636535
| 2020-01-12T12:02:12 |
350269
|
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|
Stack Exchange
|
Some properties of fractional Dirichlet heat kernel
Let $\Omega\subset \mathbb{R}^n (n\geq 2)$ be a bounded open domain with smooth boundary $\partial\Omega$. Consider the fractional heat equation with Dirichlet boundary condition：
\begin{equation}
\left\{
\begin{array}{ll}
\partial_{t}u(x,t)+(-\triangle)^{s}u(x,t)=0 , & \hbox{in $\Omega$, $t>0$;} \\
u(x,t)=0, & \hbox{in~$\mathbb{R}^n\setminus\Omega, t\geq 0$;} \\
u(x,0)=f_{0}(x)\in L^2(\Omega), & \hbox{in $\Omega$, for $t=0$.}
\end{array}
\right.
\end{equation}
Here $s\in (0,1)$, $(-\triangle)^{s}$ is the fractional Laplacian given by
\begin{equation}
(-\triangle)^{s}u(x)=\mathcal{F}^{-1}(|\xi|^{2s}\hat{u}(\xi))(x),
\end{equation}
with $\hat{u}(\xi)=\mathcal{F}u(\xi)=(2\pi)^{-\frac{n}{2}}\int_{\mathbb{R}^n}u(x)e^{-ix\cdot
\xi}dx$ is the Fourier transform of $u$. In paper [1], the author presented a Dirichlet kernel $h_{D}(x,y,t)$ of $(-\triangle)^{s}$ on $\Omega$ and a global heat kernel $h(x,y,t)$ of $(-\triangle)^{s}$ in $\mathbb{R}^n$. Then, the author claimed that (page 221 in [1]) we can deduce from the maximum principle that
$$ 0\leq h_{D}(x,y,t)\leq h(x,y,t)=\int_{\mathbb{R}^n}e^{-t|\xi|^{2s}}e^{i \xi\cdot (x-y)}\frac{d\xi}{(2\pi)^{n}}~~\mbox{for all}~x,y\in \Omega. $$
Then, we have
$$ \sum_{j=1}^{+\infty}e^{-t\lambda_{j}}|\phi_{j}(x)|^{2}\leq \frac{\omega_{n}}{(2\pi)^{n}}\Gamma\left(1+\frac{n}{2s}\right)t^{-\frac{n}{2s}}~~\mbox{for all}~~x\in \Omega,$$
where $\omega_{n}$ is the Lebesgue of unit ball in $\mathbb{R}^n$, $\lambda_{j}$ and $\phi_{j}$ are denoted by the $j$-th Dirichlet eigenvalue and Dirichlet eigenfunction of $(-\triangle)^{s}$ on $\Omega$.
Here is my Question:
I feel very confused about how can we deduce that $h_{D}(x,y,t)\leq h(x,y,t)$ by the maximum principle of fractional Laplacian. Because the regularity results and basic properites of fractional Dirichlet kernel $h_{D}(x,y,t)$ are not clear for me. How can we deduce that $h_{D}(x,y,t)\leq h(x,y,t)$ by the maximum principle? What is the maximum principle for the fractional heat equation? (I found some versions of maximum principles in [4], but it seems not worked.)
Since the fractional Laplacian $(-\triangle)^{s}$ is a non-local operator, it seems we cannot use the classical approach as classical Laplacian to establish the Dirichlet kernel. I found some papers such as [2] [3]，but all of them use the symmetric $\alpha$-stable process to establish the fractional Dirichlet heat kernel for $(-\triangle)^{s}$, and little PDE properties was involved. Can someone give an approach (or a detail reference) to establish the fractional Dirichlet heat kernel in view of classical PDE sense? What is the complete definition of fractional Dirichlet heat kernel?
Can we claim that the series
$$ h_{D}(x,y,t)=\sum_{j=1}^{\infty}e^{-\lambda_{j}t}\phi_{j}(x)\phi_{j}(y)$$
converges uniformly on $\overline{\Omega}\times\overline{\Omega}\times [\varepsilon,+\infty)$ for any $\varepsilon>0$?
For the third question, it seems we can deduce from the fractional Sobolev embedding inequality (cf. [4] and [5]) that $\|\phi_{k}\|_{L^{\infty}(\Omega)}\leq C\cdot \lambda_{k}^{\frac{n}{4s}}$ and from [4] we know the eigenfunctions $\phi_{k}\in C^{\infty}(\Omega)\cap C^{s}(\overline{\Omega})$.
Can someone help me? Thank you very much!
Reference:
[1] Frank, Rupert L., Eigenvalue bounds for the fractional Laplacian: a review, Palatucci, Giampiero (ed.) et al., Recent developments in nonlocal theory. Berlin: De Gruyter Open (ISBN 978-3-11-057155-4/hbk; 978-3-11-057156-1/ebook). 210-235 (2018). ZBL1404.35303.
[2] Bañuelos, Rodrigo; Kulczycki, Tadeusz; Siudeja, Bartłomiej, On the trace of symmetric stable processes on Lipschitz domains, J. Funct. Anal. 257, No. 10, 3329-3352 (2009). ZBL1189.60100.
[3] Chen, Zhen-Qing; Kim, Panki; Song, Renming, Heat kernel estimates for the Dirichlet fractional Laplacian, J. Eur. Math. Soc. (JEMS) 12, No. 5, 1307-1329 (2010). ZBL1203.60114.
[4] Fernández-Real, Xavier; Ros-Oton, Xavier, Boundary regularity for the fractional heat equation, Rev. R. Acad. Cienc. Exactas Fís. Nat., Ser. A Mat., RACSAM 110, No. 1, 49-64 (2016). ZBL1334.35386.
[5] Brasco, L.; Lindgren, Erik; Parini, Enea, The fractional Cheeger problem, Interfaces Free Bound. 16, No. 3, 419-458 (2014). ZBL1301.49115.
The answer to your question 3 is affirmative: this follows from ultracontractivity of the heat operators.
In fact, even more is true: the series $$\frac{1}{\phi_1(x) \phi_1(y)} \sum e^{-t \lambda_j} \phi_j(x) \phi_j(y)$$ converges uniformly to $h_\Omega(t,x,y)/(\phi_1(x) \phi_1(y))$ on $[\epsilon, \infty) \times \Omega \times \Omega$ with no regularity assumptions on $\Omega$, other than it has finite Lebesgue measure. This follows from intrinsic ultracontractivity of the heat semigroup, a result proved independently by Tadeusz Kulczycki (for general domains), and Zhen-Qing Chen and Renming Song (for smooth domains, I believe) in 1997–98.
Ultracontractivity asserts that the heat operator $H(t)$ is bounded from $L^2(\Omega)$ to $L^\infty(\Omega)$. Since the series $$h_\Omega(t, x, \cdot) = \sum e^{-t \lambda_j} \phi_j(x) \phi_j(\cdot)$$ converges in $L^2(\Omega)$ for $t = \epsilon/2$, uniformly with respect to $x$, applying $H(\epsilon/2)$ to both sides proves that the above series converges uniformly with respect to $(x, y) \in \Omega \times \Omega$ for $t = \epsilon$. Another application of $H(t-\epsilon)$ extends this to all $t \in [\epsilon, \infty)$.
Intrinsic ultracontractivity works as above, but for the intrinsic semigroup, with kernel $h_\Omega(t,x,y) / (\phi_1(x) \phi_1(y))$, and with $L^2(\Omega)$ replaced by the weighted space $L^2((\phi_1(x))^2 dx)$.
Unfortunately I cannot help much with regard to questions 1 and 2: probability is my mother tongue.
For the maximum principle mentioned in 1, Proposition 7.1 (Appendix A) of the following reference is enough: Estimates of heat kernels for non-local regular Dirichlet forms by Grigor'yan, Hu and Lau. In the notation of Proposition 7.1, apply the parabolic maximum principle to the function $u(t,x)= (H_D)_t f(x)- H_tf(x)$ where $(H_D)_t, H__t$ denote the heat semigroups for the Dirichlet type and the whole space respectively and $f$ is an arbitrary non-negative function in $L^2$.
|
2025-03-21T14:48:29.637315
| 2020-01-12T13:22:27 |
350276
|
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|
Stack Exchange
|
Visualization of the disintegration theorem
Where can I find a picture that gives a visualization of the disintegration theorem?
If such reference does not exist, what would a nice visualization of this fundamental result look like?
Wikipedia article you refer to gives a simple example with Lebesgue measure on the unit square. What other visualization is needed?
@AlexandreEremenko Sorry, I was unclear: I meant if there exists a picture to illustrate that example drawn in some book.
Picture of the unit square partitioned into horizontal segments? Can't you make this picture yourself?
@AlexandreEremenko Of course I can do it in a very basic way, I was hoping that there is a more "artistic" version in some measure theory book.
There is no need for any "visualization" of the disintegration theorem other than the classical Fubini setup (with the unit square projected on the horizontal unit interval). The point (alas, missing in most measure theory courses) is that not only all purely non-atomic Lebesgue spaces (aka standard probability spaces; all Polish topological spaces endowed with a Borel probability measure are in this class) are isomorphic to the unit interval endowed with the Lebesgue measure, but also that any measure preserving map from one Lebesgue space onto another Lebesgue space with purely non-atomic conditional measures is isomorphic to the Fubini map.
Thank you for your answer. I guess I was a bit unclear: I meant to ask if in some book there exists a picture to illustrate the simple example you mentioned.
BTW. could you please give a reference to th result wchich you have mentioned (about uniqueness of the Fubini map)? I would be grateful
For instance, see Section 2.3 (in particular, Corollary 1) in https://arxiv.org/pdf/1705.06619.pdf.
@RW Actually, figure 1 of that paper may be exactly what I was looking for.
|
2025-03-21T14:48:29.637453
| 2020-01-12T14:11:15 |
350278
|
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|
Stack Exchange
|
Weak convergence of $\mathcal{L}^2$ valued random variables
Consider two continuous functions $f,g: \mathbb{R}^{2} \rightarrow \mathbb{R}$ with $f(x,\cdot), g(x,\cdot) \in \mathcal{L}^2(\mathbb{R},\mathcal{B},\lambda)$ for all $x \in \mathbb{R}$ and a sequence of real random variables $(X_n)_{n \in \mathbb{N}}$ and another real random variable $X$. Then $T_n :=f(X_n,\cdot)$ and $V :=g(X,\cdot)$ can be interpreted as $\mathcal{L}^2$-valued random variables. Denote the distribution of $T_n$ by $\mathbb{P}_n$ and the distribution of $V$ by $\mathbb{P}$ (then $\mathbb{P}_n$ and $\mathbb{P}$ are defined on the Borel-$\sigma$-field on $\mathcal{L}^2$).
Now if we know that $(\mathbb{P}_n)_{n \in \mathbb{N}}$ is tight and that $(f(X_n,t_1), \ldots ,f(X_n,t_k)) \longrightarrow (g(X,t_1), \ldots ,g(X,t_k))$ in distribution (as $\mathbb{R}^k$-valued random variables) for all $t_1,\ldots,t_k \in \mathbb{R},k \in \mathbb{N}$, can we conclude that $\mathbb{P}_n$ is converging weakly to $\mathbb{P}$ (in the Levy-Prokhorov metric)?
I know that similar results hold for the spaces $C[0,1]$, $D[0,1]$ and the Frechet space $C[0,\infty)$, but I just don't know how to deal with this one.
And how do you define convergence in distribution for $L^2$-valued random variables? (Is it just the convergence of $\mathbb P_n$ in the Levy-Prokhorov metric or something weaker or stronger than that?)
Yes, I mean the convergence in the Levy-Prokhorov metric.
Can you give references to those "similar results [that] hold for the spaces $C[0,1]$, $D[0,1]$ and the Frechet space $C(\mathbb{R})$"?
$C[0,1]$: Billingsley, "Convergence of probability measures", Theorem 8.1
$D[0,1]$: Billingsley, "Convergence of probability measures", Theorem 15.1
$C[0,\infty)$ Whitt, "Weak Convergence of Probability Measures on the Function Space $C[ 0, \infty)$", Theorem 3
Sorry, I thought the last one was about $C(\mathbb{R})$, but it probably holds as well. I corrected it.
The difficulty here is that we have to deal with $L^2(\mathbb R)$, and $\mathbb R$ is not compact.
And how would you proceed in the compact case? The prooves of the three Theorems mentioned above are all similar, but adjusting the arguments there doesn't work, because with $L^2$ we have to deal with a space of equivalence classes of functions and so the projections $f \rightarrow f(t)$ are not even well-defined. I forgot to mention: There is also such a result for reproducing kernel hilbert spaces (see Theroem 93 in Thomas-Agnan "Reproducing kernel hilbert spaces in probability and statistics"), but the proof is also the same as for $C[0,1]$.
@esner1994 : What will work in the compact case is your assumption that $g$ is continuous and hence $g(x,\cdot)$ is continuous for each $x$.
|
2025-03-21T14:48:29.637747
| 2020-01-12T14:44:31 |
350280
|
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|
Stack Exchange
|
Complexity of tour-expansion heuristic for the planar Euclidean TSP
This is a followup question to this one: Computational Geometric Aspects of Greedy Tour Expansion.
Assume that the candidate point, whose insertion into current incurs the least tour-length increase, is determined as follows:
identify for each candidate point $p_i$ the nearest tour point $t_j$ and take as the minimal tour-elongation incurred by inserting $p_i$ into the tour $$\min\left(\|p_i-t_{j-1}\|+\|t_j-p_i\|-\|t_j-t_{j-1}\|,\ \|p_i-t_j\|+\|t_{j+1}-p_i\|-\|t_{j+1}-t_j\|\right)$$
insert $p_{i_0}$ that yields the minimal tour-elongation according to the above calculation by replacing the tour-edge for which that minimum is attained.
The nearest tour-point query can be efficiently answered if the Voronoi diagram of the tour-points is available and the containing Voronoi cells of the candidate points have been identified; updating the Voronoi diagram after inserting a point into the tour can also be done efficiently, cf. e.g. Incremental Voronoi diagrams and identifying the candidate points that lie within the newly generated Voronoi cell amounts to scrutinizing the canidate points in the two adjacent Voronoi cells of the tour-vertices adjacent to the replaced tour-edge.
Questions:
Does the sketched "nearest tour-point" method of identifying the optimal candidate for tour expansion actually yield the optimal combination of candidate-point and tour-edge for insertion?
What would the worst and average case complexities of tour-expansion heuristics be, that are based on the nearest tour-point method of identifying optimal tour expansion and utilizing computational geometry concepts as sketched above?
Is the reduction of distance calculations, resp. the reduction of memory for storing calculated distances of interest for very large instances of Euclidean TSPs?
|
2025-03-21T14:48:29.637889
| 2020-01-12T15:12:11 |
350282
|
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|
Stack Exchange
|
Non real eigenvalues for elliptic equations
I am looking for an example of a pure second order uniformly elliptic operator
$L=\sum_{i,j=1}^da_{ij}(x)D_{ij}$ in a bounded domain $\Omega$ (with Dirichlet boundary conditions, for example) having a non-real eigenvalue in $L^2(\Omega)$.
Under the above assumptions, the operator $L$ has a discrete spectrum and the principal eigenvalue is real; however it is non-symmetric. Note that if $a_{ij}=a$ (in particular if $d=1$), then every eigenvalue is real since the measure $d\mu=a^{-1}(x)\, dx$ symmetrizes $L$.
Does anybody know such an example?
Small anecdote: I remember Wolfgang Arendt mentioning the same question a few years ago during coffee. He said that he already had discussed this with several people who know quite a deal about elliptic operators, but none of them knew the answer.
Thank you for the comment. Let me point out that the spectrum is real also for operators like $\Delta+\nabla \phi \cdot \nabla$, since it is symmetric with respect to the the measure $d\mu=e^{\phi}, dx$. In particular the spectrum is always real in 1d and this perhaps explains why it is not easy to produce counteraxamples.
Here is a construction. It elaborates from perturbation analysis of eigenvalues. However it starts from the situation of a non-simple eigenvalue.
So, let me start with the standard self-adjoint $L_0=-\Delta$. I assume that the Dirichlet problem admits an eigenvalue $\lambda_0$ of multiplicity $2$ exactly. I denote $(u,v)$ an orthonormal basis of the eigenspace. A well-known example is $\lambda_0=5$ for the domain $K=(0,\pi)\times(0,\pi)$ and
$$u(x)=\frac2\pi\,\sin x_1\sin2x_2,\quad v=\frac2\pi\,\sin2x_1\sin x_2.$$
Let $M=\sum a_{ij}D_{ij}$ be given with bounded (smooth) functions $a_{ij}$, and form $L_\epsilon=L+\epsilon M$. Perturbation analysis tells us that $L_\epsilon$ admits a stable plane $\Pi_\epsilon$, which depends smoothly upon $|\epsilon|<\!<1$, and $P_0={\rm vec}(u,v)$. In addition, the restriction of $L_\epsilon$ over $\Pi_\epsilon$ is similar to a matrix $C_\epsilon$ such that on the one hand $C_0=\lambda_0I_2$, and on the other hand
$$\left.\frac{d}{d\epsilon}\right|_{\epsilon=0}C_\epsilon=\begin{pmatrix} \langle Mu,u\rangle & \langle Mu,v\rangle \\ \langle Mv,u\rangle & \langle Mv,v\rangle \end{pmatrix}=:X_0.$$
Suppose now that $M$ has been chosen so that
$$(\dagger)\qquad
( \langle Mu,u\rangle - \langle Mv,v\rangle )^2+4 \langle Mu,v\rangle \langle Mv,u\rangle <0.
$$
Then its eigenvalues are complex conjugated. Therefore, for $\epsilon\ne0$ small enough, $C_\epsilon$ will have non-real eigenvalues. Since these are eigenvalues of the Dirichlet problem for $L_\epsilon$, this provides an example.
There remains to find coefficients $a_{ij}$ satisfying ($\dagger$). Let me just choose in my example (on the square $K$) $M=aD_{12}$. The functions $uD_{12}u-vD_{12}v$, $uD_{12}v$ and $vD_{12}u$ are linearly independent. Thus there exists a function $a$ such that
$$\int_Ka(uD_{12}u-vD_{12}v)\,dx=0,\qquad\int_KauD_{12}v\,dx=1,\qquad\int_KavD_{12}u\,dx=-1.$$
This ends the construction.
Edit. It is known that one-dimensional elliptic second-order operators have a real spectrum. Actually the eigenvalues are simple, therefore the construction described above cannot be implemented.
Redit. In the construction above, the eigenvalues depend smoothly upon $\epsilon$. There is no branching phenomenon. This happens because the unperturbed operator $L_0$ is self-adjoint, thus diagonalisable.
Thank you very much.
|
2025-03-21T14:48:29.638137
| 2020-01-12T15:35:39 |
350283
|
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|
Stack Exchange
|
Artificial intelligence simulating mathematicians (what a distopia!)
This is kind of soft and naive question, so feel free to shame on me :)
I start from the fact that, in my opinion, what humans are interested in about mathematics are things that we find deep and yet simple - somehow, that we find beautiful or enlightening. There are for sure out there useful concepts that takes 10 000 pages to be defined and theorems that requires 100 years to be read, but fortunately we will never get them. We are smarter!
So now let's take a logical system like ZFC. This yields a graph with well formed and provable sentences as vertices and simple deductions as edges (I mean deductions that are obtained by applying just one inference rule). Let me set some basic concepts:
--A deduction $a$ is a path on this graph, also denoted as $v \to\to w$ (starting and ending point).
-- We say that $w$ is provable from $v$ if there exist a deduction $v \to \to w$.
-- Also, we say that $w$ is provable from $v$ through $x \to \to y$, if there exist deductions $v \to \to x \to \to y \to \to w$.
-- If $a:=x \to \to y$, and $w$ is provable from $v$ through $a$, set $\ell_a(v,w)$ the length of the shortest path from $v$ to $w$ factoring through $a$, minus the length of the deduction a.
-- As well $\ell_0(v,w)$ is the length of the shortest path from $v$ to $w$.
-- A deduction $a$ is said to be $\gamma$-useful, if for every $w$ provable from $v$ through $a$, $\ell_a(v,w) \le \gamma \ell_0(v,w)$.
Now everybody would agree that an AI which is able to discover 3-or-so-useful deductions would be great. I'd like to have mine if someone invented it. Basically it output theorems that appears in short proofs.
My first question is so purely mathematical:
Can the definition be a bit weakened to say that what I said holds "on average" or most of the times? How this could be formalized? Also, maybe it's good to relativize this concept to "areas". It's very unlikely for us that a big theorem in measure theory is applied many times in category theory. Maybe this is already cut down by limiting to proofs that factors through our theorem.
Does this definition really make sense? For example, set for a deduction $a$
$$\gamma(a) = \sup \{ \frac{\ell_a(v,w)}{\ell_0(v,w)} : w \text{ is provable from } v \text{ through } a \} $$
Is this number really giving some information? I am not even sure it's not constant. It would be great if you could compute it for a class of deductions in a toy model logical system, to see if it captures the idea of usefulness we have.
An idea for point one to get a natural filtration in finite graphs would be to show that the graphs $G_x = \{ y: x \text{ is provable from } y\}$ have a finite number of equivalence classes. A part of this statement is saying that there cannot be arbitrarily long deductions from axioms to a given statement, where implications are strict.
This seems true to me under mild assumptions on the system.
Now let's get to second question. Say also that my definitions above are not the best. Beside this, mathematicians are great in finding useful stuff, or to invent concepts that appears over and over in a lot of proofs (given the specialization of the subject). Intuitively, the bad thing about inventing an AI for proofs is that it does not have a "local" reward system; it's not clear (even for us) if some deductions will eventually get to a solution, and the number of trial-and-errors is exponential in the length as we have an homogeneous degree given by the number of inference rules.
In some chess machines, this is overcome by training machines at the beginning to play with human opponents. Later on, they let different instances of the same AI playing between them. Here come my second question:
Has a statistical learning machine that "learns from mathematicians" (so they become the "local reward system") ever been tried? Like someone that stays there and says "bad idea"/"good idea", and when the machine is a bit trained you let them say bad/good among them in a kind of AI-mathematicians-coomunity.
I think that most of us has experienced exposing an idea to a professor which misteriously says "could work" or "could not", when even himself didn't know precisely why. But then most of the times he was right! XD
Hope to have answers from both AI and logic experts. Both technical and "soft" answers are welcome!
The graph model is inadequate: modus ponens takes two arguments.
Ok, so maybe is better to have an operadic model? i.e. a bunch of operations (= inferences) subjected to some constraints?
However, I don't think it is necessary to think it as two variable thing. Modus ponens is $\phi(A,B,C ) \Rightarrow \psi(A,C)$, and it is a way to pass from a single sentence to another single sentence. Here $\phi(x,y,z) = (x \Rightarrow y) \land (y \Rightarrow z)$ while $\psi(x,z) = x \Rightarrow z$.
The graph is not locally finite. The elimination of a universal quantifier can be applied to any term, so it will have infinitely many successors. If you're going to talk about statistics, you'll need some measures or distributions.
But more importantly, I don't see any evidence whatsoever that your graph has anything to do with how mathematical knowledge is actually structured.
I don't know, I am in representation theory (so neither logic nor AI), but it seems to me that when I learn a theorem is a way to conceptualize a lot of steps in one. When I use theorem I then refer to that idea. Shortly seems like we mathematicians use just some paths=theorems and the resulting length is computed by how many paths we have used. What do you see missing here?
@andrej Bauer: also, I don't see why your argument yields non local finiteness. Unless I am misunderstanding, to eliminate a universal quantifier you need one in your sentence, so it can't be applied to any term. Even if you have an inference rule that you can apply to any sentence and you have $x \to Fx \to F^2x \to ... $ it is not evident to me that exist a y such that $F^nx $ implies y for every $n$. This holds in case by locally finite you mean that the freely generated category is not locally finite.
The theorem $\forall n \in \mathbb{N} . n + 1 = 1 + n$ has infinitely many successors by elimination of $\forall$ (i.e., plug in a value for $n$), namely $0 + 1 = 1 + 0$, $0 + 2 = 2 + 0$, $0 + 3 = 3 + 0$, etc.
Well, I understood that automatic proof is harder than I thought. I found a thesis from JP Bridge in Cambridge "Machine learning and automated theorem proving" that probably address my curiosity in a more general and well established framework. Thanks for the clues!
|
2025-03-21T14:48:29.638850
| 2020-01-12T15:54:46 |
350284
|
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|
Stack Exchange
|
Cliques in overlap graphs for words
Let $\Sigma$ be a finite alphabet, and consider the free monoid $\Sigma^*$. Given $w, w' \in \Sigma^*$ we say that $w$ overlaps $w'$ if there exist non-empty words $u, v, u'$ such that $w = uv$ and $w' = vu'$. Given a finite set of words $S$, define the overlap graph $OG(S)$ to be the simple graph with vertex set $S$ and an edge between two words in $S$ if one overlaps the other. In general this is defined as a weighted directed graph but in this case I do not care about direction or size of the overlaps.
I would like to understand better the cliques in $OG(S)$. In particular, I would like to know:
Given a large enough clique, can we find a large subclique with some nicer properties?
Is there a bound on the chromatic number $\chi(OG(S))$ in terms of the clique number $\omega(OG(S))$ and $| \Sigma |$?
These questions can be asked with some additional hypotheses on the words in $S$. In particular:
What if all words in $S$ are unbordered? A word $w$ is bordered if there exist words $u, v$ such that $w = uvu$ and $u \neq \epsilon$.
What if all words in $S$ are Lyndon words? A word $w$ is Lyndon for a given total order $\preceq$ on $\Sigma$ if it is primitive (i.e., not a positive power of another word) and smaller than all of its proper suffixes for the corresponding lexicographic order. All Lyndon words are unbordered.
All partial answers, as well as links to relevant literature, are welcome.
At least one of the questions admit counterexamples. Namely, let $\Sigma = \{a, b, c, d \}$, and $S = \{ ab^nc d ab^mc : n > m \}$. Then all words in $S$ are Lyndon, $OG(S)$ is triangle-free and has infinite chromatic number. I can give more details if anyone is interested.
|
2025-03-21T14:48:29.639007
| 2020-01-12T17:27:53 |
350289
|
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|
Stack Exchange
|
Mac Lane's proof of coherence for symmetric monoidal categories
This question only concerns the final part of the proof, so I assume that the symmetric monoidal category is a strict monoidal category $\mathsf{C}$ with the braiding $s$.
Let $X_1,...,X_n$ be elements of $\mathsf{C}$ and let $\sigma \in S_n$. Mac Lane proves that any two morphisms $f\colon X_1\otimes ... \otimes X_n \to X_{\sigma(1)}\otimes ... \otimes X_{\sigma(n)}$ which are compositions of braidings possibly tensored with identity morphism on the left or on the right $k$-times are equal. To this end, he realizes every such path a composition $s^{\pm}_{X_{i_1}, X_{i_1 + 1}} \circ ... \circ s^{\pm}_{X_{i_m}, X_{i_m + 1}}$ (I ignore $-\otimes 1_X$ and $1_X\otimes -$ here as far as the notation goes). The key part of the proof in an apparently intereseting connection between $s_{X_i,X_i + 1}$ and $(i,i+1)$.
Mac Lane states and any "closed path" (I suppose he's refering to said morphism with codomain $X_1\otimes ... \otimes X_n$) corresponds to a relations between generators $(i,i+1)$ of the symmetric group $S_n$. On the other hand, he mentions that a symmetric group has the presentation $$\langle \tau_i, i = 1,...n-1 \mid \tau^2_i = 1, (\tau_i\tau_{i+1})^3 = 1, \tau_i\tau_j = \tau_j\tau_i \text{ for }|i - j| > 1 \}$$
where $\tau_i = (i,i+1)$. He then claims that to prove the statement it suffices to show that these relations hold for $s$.
I've been thinking for a while about this and I can't understand what is the precise connection between braidings and permutations. I see that applying a braiding $s_{X_i,X_{i + 1}}$ to $X_1\otimes ... \otimes X_n$ given b$X_{\sigma(1)} \otimes ... \otimes X_{\sigma(n)}$ where $\sigma = (i,i+1)$, but no better than that yet.
Also, connecting relations in $S_n$ with those among $s_{X,Y}$ and deducing the statement of the theorem from that reminds me of the universal property of a presentation:
Let $\langle X \mid R \rangle$ be a presentation and $G$ a group. Let $f\colon X\to G$ be a map such that for every $(u,v) \in R$ we have $f(u) = f(v)$. Then there is a unique group homomorphism $\phi\colon\langle X \mid R \rangle \to G$ such that $\phi(x) = f(x)$ for all $x \in X$.
But I don't understand how this can be applied here as I don't see a resonable binary operation for paths consisting of braidings.
So what Mac Lane really means here, and how it can be made precise?
I think that this is a great question, since it is quite common to just say at this point "each relation between the transpositions follows from the relations in the group presentation of the symmetric group, so it is enough to prove the relations for the symmetry automorphisms", but as you correctly point out, this is not really precise! One can make it precise by saying more about what a deduction of $a=b$ between two words in a group presentation actually is, and how in our case every such deduction can be transported to a deduction of an equality between the associated isomorphisms between the tensor products. But I won't go into that here, since I have actually found a cute trick.
Consider the special case $X_1=\dotsc=X_n$ first, let's call this unique object $X$. Then the task is to define a (well-defined) map
$$S_n \to \mathrm{Aut}(X^{\otimes n}),\, \sigma \mapsto \sigma_*$$
which will actually be constructed as a homomorphism. By the presentation
$$S_n = \langle \tau_1,\dotsc,\tau_n : \tau_i^2=1,\, (\tau_i \tau_{i+1})^3=1, \,\tau_i \tau_j = \tau_j \tau_i \text{ if } |i-j|>1 \rangle$$
it is therefore enough to define elements $s_i = (\tau_i)_* \in \mathrm{Aut}(X^{\otimes n})$ satisfying the mentioned relations. This is what Mac Lane actually proves.
The general case can be deduced from the special case:
Consider $\mathcal{C}_{\sqcup}$, the cocompletion of $\mathcal{C}$ under finite coproducts. Objects are finite (formal) coproducts $\coprod_{i \in T} X_i$ with $X_i \in \mathcal{C}$ and finite sets $T$, the morphisms are
$$\hom\bigl(\coprod_{i \in T} X_i,\coprod_{j \in S} Y_j\bigr) = \prod_{i \in T} \coprod_{j \in S} \hom(X_i,Y_j).$$
There is an evident symmetric monoidal structrue on $\mathcal{C}_{\sqcup}$ extending the one on $\mathcal{C}$, which commutes with finite coproducts in each variable, thus given by
$$\coprod_{i \in T} X_i \otimes \coprod_{j \in S} Y_j := \coprod_{(i,j) \times T \times S} X_i \otimes Y_j.$$
The symmetry $\coprod_{i \in T} X_i \otimes \coprod_{j \in S} Y_j \to \coprod_{j \in S} Y_j \otimes \coprod_{i \in T} X_i$ is induced by the symmetries $X_i \otimes Y_j \to Y_j \otimes X_i$. All coherence axioms immediately follow from the ones in $\mathcal{C}$.
Now consider for $X_1,\dotsc,X_n \in \mathcal{C}$ the object $X := X_1 \sqcup \dotsc \sqcup X_n$ in $\mathcal{C}_{\sqcup}$ with the coproduct inclusions $\iota_1,\dotsc,\iota_n$. We know that every $\sigma \in S_n$ induces a well-defined automorphism $X^{\otimes n} \to X^{\otimes n}$ in $\mathcal{C}_{\sqcup}$. Composing this with $\iota_1 \otimes \dotsc \otimes \iota_n$ yields a well-defined morphism $X_1 \otimes \dotsc \otimes X_n \to X^{\otimes n}$, still in $\mathcal{C}_{\sqcup}$. But if we write $\sigma$ as a product of neighbor transpositions and hence write $X^{\otimes n} \to X^{\otimes n}$ as a product of symmetry automorphisms, we see that the morphism actually factors over $X_{\sigma^{-1}(1)} \otimes \dotsc \otimes X_{\sigma{-1}(n)}$. So we get a morphism $X_1 \otimes \dotsc \otimes X_n \to X_{\sigma^{-1}(1)} \otimes \dotsc \otimes X_{\sigma^{-1}(n)}$ in $\mathcal{C}$. Once again, it is independent from any product decomposition of $\sigma$ since $X^{\otimes n} \to X^{\otimes n}$ is.
This reduction can also be informally phrased as follows: The objects are in their right place anyway, so we might as well just pretend that they are equal.
Thank you for your attention. If you don't mind, a I have a couple of questions. 1) Do "formal coproducts" mean families $(X_i){i \in I}$ for $I$ finite? I assume they will be real coproducts in $C{\sqcup}$? 2) The composition of $(i_j, f\colon X_i\to Y_{j_i}){i \in I} \in \prod{i \in I} \coprod_{j \in J} \mathrm{Hom}(X_i,Y_j)$ and $(k_j, g\colon Y_j\to Z_{k_j}){j \in J} \in \prod{j \in J} \coprod_{k \in K} \mathrm{Hom}(Y_j,Z_k)$ is $(i, k_{j_i})$, right? The last request is the one I'm most uncomfortable writing, but could you, please, provide more details as to why the morphism
$X_1\otimes ... \otimes X_n \to X^{\otimes n}$ factors over $X_{\sigma(1)}\otimes ... \otimes X_{\sigma(n)}$, and why it would imply the uniqueness?
Sorry, in the question about the composition it should read as: $(j_i,f_i\colon X_i\to X_j){i \in I}$ and $(i,g{j_i}\circ f_i)$ instead of $(i_j, f\colon X_i\to Y_{j_i})$ and $(i,k_{j_i})$.
Yes, 2) Yes,
I just give you an example to get a better feeling. Let $\sigma = (1,2,3)$, thus $\sigma = (1,2) \circ (2,3)$ for example. Then $\sigma_* :X^{\otimes 3} \to X^{\otimes 3}$ is given by $s_1 \circ s_2$, where $s_1 = S_{X,X} \otimes X$ and $s_2 = X \otimes S_{X,X}$. If $X = X_1 \sqcup X_2$, then $s_2$ maps $X_1 \otimes X_2 \otimes X_3$ to $X_1 \otimes X_3 \otimes X_2$, and $s_1$ maps $X_1 \otimes X_3 \otimes X_2$ to $X_3 \otimes X_1 \otimes X_2$, which is actually $X_{\sigma^{-1}(1)} \otimes \dotsc \otimes X_{\sigma^{-1}(n)}$. Ok I will correct my answer.
Regarding uniqueness: We can define $X_1 \otimes \cdots \otimes X_n \to X_{\sigma^{-1}(1)} \otimes \cdots \otimes X_{\sigma^{-1}(n)}$ as the unique morphism which makes the following diagram commutative: $$\begin{array}{ccc}
X_1 \otimes \cdots \otimes X_n & \longrightarrow & X_{\sigma^{-1}(1)} \otimes \cdots \otimes X_{\sigma^{-1}(n)} \
\downarrow && \downarrow \
X^{\otimes n} & \xrightarrow{~\sigma_*~} & X^{\otimes n}
\end{array}$$
So this is well-defined.
Sorry, but I don't quite get your example. First, where did $X_3$ come from (a typo?) If $X = X_1\sqcup X_2$, then $s_1$ maps $(X_1\sqcup X_2)\otimes (X_1\sqcup X_2)$ to itself, and I think $(X_1\sqcup X_2)\otimes (X_1\sqcup X_2) = (X_1,X_1)\sqcup(X_1,X_2)\sqcup (X_2,X_1)\sqcup (X_2,X_2)$. What am I missing?
Regarding uniqueness: my lack of experience shows, but I don't see at the moment what the existence of such morphism $X_1\otimes ... \otimes X_n\to X_{\sigma^{-1}(1)}\otimes ... \otimes X_{\sigma^{-1}(n)}$ has to do with the original statement of uniqueness of such morphism.
I meant $X = X_1 \sqcup X_2 \sqcup X_3$, sorry. For the rest, please try to think more about it.
I'm not sure, but it appears that it is not as hard as it seemed. Martin's answer is interesting, but I still wanted to try something elementary before fully committing to powerful machinery.
Let $F$ be the free group on $\{ \tau_i \mid i = 1,...,n - 1 \}$ where $\tau_i = (i,i+1)$ and let $\phi\colon F\to S_n$ be a canonical homomorphism where $S_n = F/N$ with $N$ is the normal closure of the subset of $F$ consisting of elements $\tau_i\tau_i, (\tau_i\tau_{i+1})^3$ and $\tau_i\tau_j\tau^{-1}_i\tau^{-1}_j$ (for $|i - j| > 1$).
Let $M$ be the set of all morphisms with domain $X_1\otimes ... \otimes X_n$. We define a map $f\colon F\to M$ given by $f(e) = 1_{X_1\otimes ... \otimes X_n}$ and $f(\tau^{\pm}_{i_n},...,\tau^{\pm}_{i_1}) = s_{X_{\tau_{i_1}\left(...\left(\tau_{i_{n-1}}(i_n)\right)\right)}, X_{\tau_{i_1}\left(...\left(\tau_{i_{n-1}}(i_n + 1)\right)\right)}}\circ...\circ s_{X_{\tau_{i_1}(i_2)}, X_{\tau_{i_1}(i_2 + 1)}}s_{X_{i_1},X_{i_1 + 1}}$ which is a morphism $X_1\otimes ... \otimes X_n \to X_{\sigma^{-1}(1)}\otimes ... \otimes X_{\sigma^{-1}(n)}$ with $\sigma = \tau_{i_n},...,\tau_{i_1}$. Since coherence states that only all "formal" diagrams commute, every path $X_1\otimes ... \otimes X_n \to X_{\sigma^{-1}(1)}\otimes ... \otimes X_{\sigma^{-1}(n)}$ consisting of symmetries is of this form.
If a path of this form from $X_1\otimes ... \otimes X_n \to X_1\otimes ... \otimes X_n$, then $\sigma = 1_{\{1,...,n\}}$, so $\tau^{\pm}_{i_n},...,\tau^{\pm}_{i_1} = g_1n_1g^{-1}_1...g_mn_mg^{-1}_m \in N$
Now the problem is that $g_1n_1g^{-1}_1...g_mn_mg^{-1}_m$ is not necessarily reduced, so it's possible that it doesn't translate to a product of conjugates of $n_1,...,n_m$ in $M$. We only need to note that, for example, $f(\tau_{i_1}\tau_{i_1}^{-1}) = s_{X_{\tau_{i_1}(i_1)}, X_{\tau_{i_1}(i_1 + 1)}} \circ s_{X_{i_1},X_{i_1 + 1}} = s_{X_{i_1 + 1}, X_{i_1}} \circ s_{X_{i_1}, X_{i_1 + 1}} = 1$ and obersve that it generalizes nicely when such an unreduced instance of juxtaposition sits between two reduced words in $F$ (possibly empty).
Then $(\tau^{\pm}_{i_n},...,\tau^{\pm}_{i_1}) = g_1n_1g^{-1}_1...g_mn_mg^{-1}_m$ where $f(n_i) = 0$ (this is what Mac Lane proves) implies that $s_{X_{\tau_{i_1}\left(...\left(\tau_{i_{n-1}}(i_n)\right)\right)}, X_{\tau_{i_1}\left(...\left(\tau_{i_{n-1}}(i_n + 1)\right)\right)}}\circ...\circ s_{X_{\tau_{i_1}(i_2)}, X_{\tau_{i_1}(i_2 + 1)}}s_{X_{i_1},X_{i_1 + 1}} = 0$.
I hope I didn't miss anything, but wouldn't by surprised if I did since it seems quite easy. Please, tell me if I'm wrong.
The key difficulty seems that $M$ is not a group, so we need to work with relevant properties of would-be group homomorphism $f\colon F\to M$ by hand.
You also need $g_i n_i^{\pm 1} g_i^{-1}$, right?
I don't understand the proof from "so it's possible that it doesn't translate to a product of conjugates ... " onwards.
@MartinBrandenburg Yeah, you need $n^{\pm}$, but I don't think that it causes probelms here. As for your misiunderstanding, it's better to give an example. Suppose that the reduced word $(\tau^{\pm}{i_n},...,\tau^{\pm})$ equals the conjugate product $(\tau^{\pm}{j_m},...,\tau^{\pm}{j_1})(\tau_i)(\tau{i+1})(\tau_i)(\tau_{i + 1})(\tau_i)(\tau_{i+1})((\tau^{\pm}{j_1},...,\tau^{\pm}{j_m})$ where $(\tau^{\pm}{i_n},...,\tau^{\pm})$ is reduced. Since $\tau_i \neq \tau{i + 1}$ in $F$, $(\tau_i)(\tau_{i+1})(\tau_i)(\tau_{i + 1})(\tau_i)(\tau_{i+1}) = $(contd)
@MartinBrandenburg $= (\tau_i,\tau_{i+1},\tau_i,\tau_{i + 1},\tau_i,\tau_{i+1})$ is also reduced, thus the product is $(\tau^{\pm}{j_m},...,\tau^{\pm}{j_1})(\tau_i,\tau_{i + 1},\tau_i,\tau_{i + 1},\tau_i,\tau_{i+1})(\tau^{\pm}{j_1}...\tau^{\pm}{j_m})$. But it is possible that $\tau^{\pm}_{j_1} = \tau^{-1}i$ or $\tau^{\pm}{j_m} = \tau_i$.
@MartinBrandenburg What I showed in the example with $f(\tau_i,\tau^{-1}i)$ shows that this causes no problems since $f(\tau_i,\tau^{-1}i) = f() = 1{X_1\otimes ... \otimes X_n} = s{X_{i_1 + 1}, X_{i_1}} \circ s_{X_{i_1}, X_{i_1 + 1}} = s_{X_{\tau_{i_1}(i_1)}, X_{\tau_{i_1}(i_1 + 1)}} \circ s_{X_{i_1},X_{i_1 + 1}}$ which is the value of $(\tau_i,\tau^{-1}_i)$ under $f$ as if $(\tau_i,\tau^{-1}_i)$ were reduced. I checked on the paper that it trivially generalizes to the case where such "unreduced" instances in $g_1n^{\pm}g^{-1}_1...g_mn^{\pm}g^{-1}_m$ apper between the reduced ones.
@MartinBrandenburg Please, tell me if you have any more questions or suggestions. I really like to know if the proof is correct.
@MartinBrandenburg I'm sorry to bother you, but if you do understand (if not, I can try again to explain), does it seem correct to you?
I will look at it in a few days, please be patient. :) Also, all here can read your answer, and upvotes show that you might be right. :)
@MartinBrandenburg Just one upvote though :(
|
2025-03-21T14:48:29.639828
| 2020-01-12T17:43:47 |
350291
|
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|
Stack Exchange
|
How often a number can be conductor of an elliptic curve
There are several upper bounds for number of elliptic curves (over Q, say) upto-isomorphism with a given conductor N. Probably the best one is given by Helfgott-Venkatesh of order N^{0.22} (or may be some improvement is possible knowing improved bounds for 3-torsion in class groups).
My question is whether there is any lower bound ? I do not how much of this question make sense because, it feels like most of the possible candidate for conductor is not a conductor. Is there a related result ?
Naively like this the lower bound must be zero. For all number divisible by $5^3$ there cannot be an elliptic curve of that conductor. Similar for all $p^3$ with $p\geq 5$ and sufficiently high powers of $2$ and $3$, too.
This is why I said, whether "all possible candidates" are conductor. I know from definition $p^3$ is not a conductor. By possible candidates, I really meant whether all cube free (forget powers of 2,3). I hope my question is clear now ?
Well, $N=2$ is a famous example for which there are no curves. In the current Cremona tables of all curves with $N$ below half a million, there are plenty of $N$ for which there are no curve or just one. Among the 303957 squarefree $N$ there are 149006 without a curve and 52436 with just one curve.
That sounds interesting. Is density of such set known to be zero ? (again, I mean density in the set of cube free numbers). I think I saw some conjecture like that, but unable to find the reference now.
It is conjectured that there are roughly $X^{5/6}$ elliptic curves with conductor less than $X$. This would imply density zero. The corresponding result for heights is due to Bhargava-Shankar. Arul Shankar has thought about the conjecture for conductors quite a lot. If I wanted to know the state of the art about this question, I would probably read this paper (https://arxiv.org/pdf/1904.13063.pdf) and then if I still had questions, I would ask Arul.
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2025-03-21T14:48:29.639982
| 2020-01-12T18:08:57 |
350292
|
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|
Stack Exchange
|
Viterbo restriction map surjective on Weinstein neighbourhood
In a Liouville manifold $M$ having a Liouville subdomain $i: N \hookrightarrow M$, there is the so-called Viterbo restriction map in symplectic cohomology $$SH^*(i): SH^*(M)\rightarrow SH^*(N).$$
In particular, given an exact Lagrangian $i_L: L\hookrightarrow M$, its Weinstein neighborhood $T^*L$ is a Liouville subdomain of $M,$ hence there is a map
$$SH^*(i_L):SH^*(M)\rightarrow SH^*(T^*L).$$ Is there a statement that says that, under certain conditions on $L$, the map $SH^*(i_L)$ is surjective? The examples which I have in mind are when $M$ is Weinstein and $L$ is a component of its Liouville skeleton.
This is already false for the inclusion of a circle in a punctured genus 1 surface. What kind of condition do you have in mind?
Hmm, that is true -- maybe I should be a bit more restrictive, then. I had in mind a setup of holomorphic symplectic manifold $M$ with a Lagrangian skeleton $L$ which is a holomorphic projective variety. Its irreducible components $L_i$ are exact holomorphic Lagrangians, if smooth. In particular, the ambient M could be even hyperkahler, e.g. ADE plumbings of $T^*S^2$. Here the skeleta are ADE trees of Lagrangian $S^2$-spheres, and I think (though I did not prove) that this surjectivity statement is true for those spheres.
++ The fact that makes me think this surjectivity to be true in these ADE plumbings examples is that the rank of $SH^i(M)$ is exactly the sum of the ranks of $SH^i(T^S^2)$, summing by all components of skeleton, for all $i\in \mathbb{Z}$, hence one may think that Viterbo restrictions are sending different subspaces of $SH^(M)$ isomorphically to $SH^*$ of Weinstein neighborhoods of different spheres.
This is almost never true in general, although it's obviously true for boundary connected sums. For example, it is usually the case that Weinstein handle attachment will kill (non-trivial) invertible elements in $\mathit{SH}^0(T^\ast L)$, so the corresponding Viterbo restriction map can never be surjective. The counterexample proposed by Mohammed belongs to such a case. More generally, I think $L$ should never be a torus, because otherwise there are a lot of non-trivial units in $\mathit{SH}^0(T^\ast T^n)$ to be killed by attaching handles.
This would force you to look at Weinstein manifolds $M$ which do not contain exact Lagrangian tori, which is the case, for example, when $M$ admits a dilation in the sense of Seidel-Solomon. More generally, one could consider smooth affine varieties $M$ with log Kodaira dimension $-\infty$, which should be manifolds whose first Gutt-Hutchings capacities are finite. For Milnor fibers, this means that the configuration of vanishing cycles is very "sparse", in other words, it's "close" to a boundary connected sum of $D^\ast S^n$'s.
For the case of 4-dimensional $A_m$ Milnor fibers, explicit calculations of Hochschild cohomologies imply that
$\mathit{SH}^0(M)\cong\mathbb{K}[x]/(x^{m+1}),$
so it really makes sense to expect that the Viterbo restriction map is actually surjective. Similar computations can be done in the $D_m$ and $E_m$ case under the assumption that $\mathrm{char}(\mathbb{K})=0$.
Also a simple observation is that $1\pm x\in\mathit{SH}^0(M)^\times$ doesn't get killed under obvious handle attachments in these examples, so I think probably what you should look at is under which circumstances non-trivial units in $\mathit{SH}^0(M)$ are preserved under handle attachments. Unfortunately the description of the product structure on $\mathit{SH}^\ast(M)$ under Legendrian surgery is quite involved, so it's difficult to extract a simple and explicit condition. I'll update this answer once I have any new ideas.
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2025-03-21T14:48:29.640224
| 2020-01-12T19:40:16 |
350296
|
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|
Stack Exchange
|
A generalization of competitive systems
We consider the following standard partial order relation on $\mathbb{R}^n$:
We say $X=(x_1,x_2,\ldots,x_n)\leq (y_1,y_2,\ldots,y_n)=Y$ iff $\sum_{i=1}^k x_i \leq \sum_{i=1}^k y_i,\quad \forall k: 1\leq k<n,\quad \sum_{i=1}^n x_i=\sum_{i=1}^n y_i$
Is there a vector field $Z$ on $\mathbb{R}^n$ with the monoton property $X\leq Y\implies \phi_t (X) \leq \phi_t(Y),\quad \forall t>0$ where $\phi$ is the flow of $Z$?
What is a classification of all vector fields with this monoton property? Can this property be equivalent to certain algebraic or semi algebraic conditions onthe Jacobian of vector field $Z$?
Note: If the above equivalent relation would be replaced with the relation $X\leq Y \iff x_i\leq y_i$ then the problrm is fully known, the competitive system introduced by M.W. Hirsch.
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2025-03-21T14:48:29.640313
| 2020-01-12T19:44:33 |
350297
|
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|
Stack Exchange
|
Prime analogue of Champernowne's constant
Are any non-trivial properties known about the constant 0.2357111317192329... that is obtained by catenating the digits of sequence of prime numbers in base 10 or in other bases, especially whether it is normal?
In base 10 this is called the Copeland–Erdős constant; it is normal.
https://en.wikipedia.org/wiki/Copeland–Erdős_constant and http://mathworld.wolfram.com/Copeland-ErdosConstant.html and https://projecteuclid.org/euclid.bams/1183509721
@FredRohrer your comment would be an acceptable answer to me.
An immediate consequence of the Dirichlet's Theorem on arithmetic progression is that for each $n \geq 1$ there are infinitely many prime numbers starting with $n$ as the first digits.
@NickS With "first digits" do you mean the leftmost or the rightmost digits? Citing Dirichlet I would assume you mean rightmost, in which case this is only true for $n$ not divisible by $2$ or $5$.
@Wojowu Yes, sorry I was imprecise... What I meant, you add some digits after $n$ and apply the DT, you can easily do this and keep the new number relatively prime to 10.
In base 10 this is called the Copeland–Erdős constant; it is normal. See this article.
Actually, much more general results are known for the normality of numbers which are obtained by concatenating certain function values (at integers, or at primes). See for example Nakai, Y., Shiokawa, I.: Normality of numbers generated by the values of polynomials at primes. Acta Arith. 81 (1997), no. 4, 345–356.
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2025-03-21T14:48:29.640453
| 2020-01-12T20:03:12 |
350298
|
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|
Stack Exchange
|
Symetrical simplex category
The simplex category has for objects totally ordered sets $[n]$ , and for morphisms order-preserving functions between those sets.
We can see the totally ordered set $[n]$ of size $n$ of the simplex category as a very simple form of category (skeletal), for which between 2 elements, there is at most one arrow, which witnesses the fact that X <= Y.
By anti-symetry, if X < Y, witnessed by an element in $Hom[n](X,Y)$ then the set $Hom[n](Y,X)$ is empty since X <> Y. By totality, either one homset has one element.
(Interestingly, viewing ordered sets as a basic form of (skeletal) category, then the simplex category is a basic form of the 2-category $Cat$.)
Now, what if we have, instead of those skeletal categories $[n]$, that between two objects X and Y, there is always a pair of arrow, in opposite direction, inverse of each other, so that, calling XY what was previously witnessing the fact that X < Y, XY = -YX. This is still a category, even simpler to describe : with exactly one morphism in each homset. period.
And my question is : does this "symetric simplex" category has a name ?
https://ncatlab.org/nlab/show/indiscrete+category. This construction is so general and ignores all the interesting aspects of simplices that it is useless. Is it really what you want to ask?
I see what you mean. up to iso of course. Physicists like to "degenerate" things to have a better understanding by gradually adding finer features : to go from this to Simp is a smaller step than to go from nothing to Simp, with all its goodness (nerve and all ...), though it might be just a waste in that case...(?)
You haven't specified what the morphisms in your "symmetric simplex category" are supposed to be, and there are two natural choices:
Functors, or
Natural isomorphism classes of functors.
In the first case, you get a category equivalent to FinSet, and the presheaves on it are called symmetric sets, and AFAIK these also model homotopy theory.
In the second case, all your categories are equivalent, so that your "symmetric simplex category" itself has exactly one morphism in each direction between any two objects, i.e. it is indiscrete and therefore equivalent to the terminal category, which (as Martin says) makes it uninteresting.
I was indeed looking at a really applied case, where some presheaf seemed to make sense, to explicit some homotopy. as a category it is equivalent to FinSet because we map sets to indiscrete categories, but maybe 2-functors are more relevant here (?)
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2025-03-21T14:48:29.640643
| 2020-01-12T21:19:28 |
350303
|
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"Noah Schweber",
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|
Stack Exchange
|
"Robinson arithmetic" for (some) levels of $L$?
I'll write "$\mathcal{L}_\alpha$" for the fragment $\mathcal{L}_{\infty,\omega}\cap L_\alpha$.
Say that a countable admissible $\alpha$ is Robinsonian if there is some sentence $\varphi\in\mathcal{L}_\alpha$ such that $L_\alpha\models\varphi$ and there is no $T\subseteq\mathcal{L}_\alpha$ which is consistent, complete with respect to $\mathcal{L}_\alpha$, and $\Delta_1$ over $L_\alpha$. Intuitively, such a $\varphi$ is the "$L_\alpha$-analogue" of Robinson arithmetic.
By Barwise completeness, if $\alpha$ is a limit of admissibles then the set of satisfiable $\mathcal{L}_\alpha$-sentences is $\Delta_1$ over $L_\alpha$. Hence via a Henkinization argument we have that limits of admissibles are not Robinsonian. On the other hand, $\omega$ is clearly Robinsonian and it's not hard to show that $\omega_1^{CK}$ is Robinsonian as well.
My question is:
What are the Robinsonian ordinals?
I'd love it if the answer were exactly the successor admissibles, but I suspect it isn't; the stumbling point seems to be the non-Gandy ordinals (at a glance I think we do get that every successor admissible of a Gandy ordinal is Robinsonian by generalizing the argument for $\omega_1^{CK}$, but I haven't checked the details).
Note that it's not hard to show that for every $\alpha$ which is either admissible or a limit of admissibles, an analogue of Godel's first incompleteness theorem does hold: there is a $\Sigma_1$-over-$L_\alpha$ theory $T\subseteq\mathcal{L}_\alpha$ such that $L_\alpha\models T$ but $T$ has no $\Delta_1$-over-$L_\alpha$ consistent completion with respect to $\mathcal{L}_\alpha$. Moreover, there is a single $\Sigma_1$ formula which describes such a $T$ in every $L_\alpha$ with $\alpha$ pre-admissible. So it's plausible that there are lots of Robinsonian ordinals.
This question came up in the course of thinking about what the analogue of provability logic for appropriate $\mathcal{L}\alpha$-theories of $L\alpha$ might be; separately, I'd love to be pointed towards any sources about that. Also, the OP can be generalized to countable admissible sets or countable "limit-admissible" sets (= countable transitive sets satisfying Pairing + "Every set is an element of an admissible set"), but that seems infeasibly general; that said, note that for general admissible sets we don't have a good well-ordering - and so the Henkinization argument above breaks down.
EDIT: to my chagrin, the notion of "$n$-admissibility" is not what I thought it was! What I wanted was $\Sigma_n$-admissibility. You can find the definition of $n$-admissibles here; they are vastly smaller than their $\Sigma_n$ counterparts, and indeed for each $n$ the least $n$-admissible is less than the least $\Sigma_2$-admissible. Now $n$-admissibility is a rare notion these days and I've seen "$n$-admissible" used for "$\Sigma_n$-admissible" before, but given the relevance of older papers to this topic it's probably a good idea for me to not butcher this distinction.
Embarrassingly, I think I was overthinking this: I believe that the Robinsonian admissibles are exactly the successor admissibles.
The idea is to lift the following argument for the essential undecidability of $Q$ in the FOL-context to $\mathcal{L}_\alpha$: "If $T\supseteq Q$ is recursive then there is some $\psi$ such that $\psi^N\cap\mathbb{N}=T$ for all $N\models Q$, and if $T\supseteq Q$ is complete and consistent there is some $M\models T$; putting this together we get an $M\models Q$ with $Th(M)$ the standard part of a parameter-freely-definable set in $M$, contradicting (a version of) Tarski's theorem."
So suppose $\alpha$ is the next admissible above some admissible $\beta$ ...
Below, by "definable$_\eta$" I mean "definable by a parameter-free $\mathcal{L}_\eta$-formula," and "$Th_\eta(K)$" is the parameter-free $\mathcal{L}_\eta$-theory of $K$ - thought of as a subset of $L_\eta$. Note that it does make sense to ask whether a structure satisfies an $\mathcal{L}_\eta$-sentence even when that structure is not in $L_\eta$: $\mathcal{L}_\eta$ is just a sublogic of $\mathcal{L}_{\infty,\omega}$. Also, I'll conflate transitive sets with the corresponding $\{\in\}$-structures and conflate $\mathcal{L}_\alpha$-formulas with sets in $L_\alpha$ in some appropriate way.
First, define by recursion a formula $\sigma_s$ assigned to each set $s$ as follows: $$\sigma_s(x): \forall y(y\in x\leftrightarrow\bigvee_{t\in s}\sigma_t(s)).$$ Intuitively, $\sigma_s$ defines $s$ in a parameter-free way.
For $s$ a set, let $\theta_s$ be the sentence $\bigwedge_{t\in s\cup\{s\}}\exists!y(\sigma_t(y))$. The point of all this is that if $M\models$ Extensionality + $\theta_s$, then there is a unique embedding of $tc(\{s\})$ as an initial segment of $M$.
Now consider the $\mathcal{L}_\alpha$-sentence $(*)$ = "KP + Inf + V=L + $\theta_\beta$." I claim that $(*)$ witnesses the Robinsonian-ness of $L_\alpha$.
We observe the following: for every $M\models(*)$ there is a unique end-embedding $l_M: L_\alpha\subseteq_{end}M$, and every element of $im(l_M)$ is definable$_\alpha$ in $M$. The second half of this is trivial given the first half, and the first half combines the initial segment observation from the previous section with the fact that the well-founded part of an admissible set is admissible.
That last bit is what I was missing when I was worrying about non-Gandy-ness. I think it's worth elaborating on:
First, note that it fails for $\Sigma_2$-admissibility, since by the Gandy Basis Theorem there is a model of $KP2$ with wellfounded part having height $\omega_1^{CK}$.
The reason it works for ($\Sigma_1$-)admissibility is the upwards absoluteness of $\Sigma_1$ formulas. Let $M\models KP$ and $N$ be the wellfounded part of $M$. Let $a,\varphi$ be a $\Sigma_1$-Replacement instance in $N$: that is, $\varphi$ is $\Sigma_1$ and for each $b\in a$ there is exactly one $c\in N$ such that $N\models\varphi(b,c)$. Then in $M$ we can apply absoluteness to argue that $a,\hat{\varphi}$ is also a $\Sigma_1$-Replacement instance with the same solution class, where $\hat{\varphi}(x,y)$ is the formula "$\varphi(x,y)$ and no $z$ of rank $<rk(y)$ has $\varphi(x,z)$."
For $M\models (*)$ and $X\subseteq M$, let $st_M(X)=X\cap im(l_M)$.
The next key point is an analogue of Tarski's undefinability theorem:
Suppose $M\models (*)$. Then there is no definable$_\alpha$ $D\subseteq M$ such that $$st_M(D)=Th_\alpha(M).$$
In the interest of length I'll omit the proof; it's just the usual argument, though.
We put all this together as follows. Recapitulating the usual arguments for arithmetic, every $\Delta_1$-over-$L_\alpha$ set $X$ has an invariant definition (a la Kreisel, see also Moschovakis): there is a parameter-free $\Sigma_1$-formula $\varphi\in\mathcal{L}_\alpha$ such that whenever $M\models (*)$ we have $st_M(\varphi^M)=X$.
The "parameter-free" bit may seem like cheating; the point is that we can essentially fold the parameter into the structure of the formula itself via the $\sigma_s$-construction above.
Now if $T$ were a consistent complete $\Sigma_1$-over-$L_\alpha$ extension of $(*)$ in the sense of $\mathcal{L}_\alpha$, by fixing a $\varphi$ as above and an $M\models T$ we would have $T=Th_\alpha(M)=st_M(\varphi^M)$, contradicting the Tarskian result above.
|
2025-03-21T14:48:29.641111
| 2020-01-12T21:54:17 |
350305
|
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|
Stack Exchange
|
Properties of a function based on binomial distribution
$f_{n,p}(k)$ is the probability mass function of a binomial distribution with parameters $n$ and $p$ i.e, for $k \in \{0,1,2, \cdots,n\}$, $f_{n,p}(k) = \binom{n}{k}p^k(1-p)^{n-k}$. Let $F_{n,p}$ be the corresponding cumulative distribution function, i.e. $F_{n, p}(k) = \sum_{i=0}^k f_{n,p}(k)$ for $k \in \{0,1,2,\cdots,n\}.$ Consider the function $h_n(p)$ defined for $0\leq p \leq 1$ as follows:
\begin{equation}
h_n(p) = \sum_{k=1}^n f_{n,p}(k)F_{n,p}(k-1).
\end{equation}
After plotting the function for different values of $n,$ I think that the following two properties hold for $h_n(p):$
$h_n(p) \leq h_{n+1}(p)$ for all $0\leq p \leq 1$ and $n= 1,2,3, ...\, .$
$h_n(p)$ is a concave function for $n= 1,2,3, ...\,.$
Am trying to prove the above two statements analytically. I would really appreciate any help with the proofs.
Let $g_n(p) := \sum_{k=0}^n \binom{n}{k}^2 \cdot p^{2k} \cdot (1-p)^{2(n-k)}$. Note that $h_n(p) = h_n(1-p)$ and $1 = h_n(p) + h_n(1-p) + g_n(p)$. (Write 1 as double sum.) Thus a) is equivalent to $g_n(p) \geq g_{n+1}(p)$ and b) to the convexity of $g_n$. I think I have seen this elsewhere (Marshall/Olkin ?).
Thanks a lot for the hint. Could you please let me know which paper/book of Marshall/Olkin to look at ?
The book is: A.W. Marshall, I. Olkin: Inequalities: Theory of Majorization and its Applications,, Academic Press, 1979. A further comment (surely known to you): Let $X,Y$ be independent random variables $\sim Bin(n,p)$, then $h_n(p) = P(X<Y) = P(X>Y)$ and $P(X=Y) = g_n(p)$,
I glanced through the entire book but could not come across something related to my question. Could you please let me know which section to look at ? I initially thought it should be straightforward to prove convexity of the function $g_n(p)$. But even that is turning out to be more difficult than I had imagined. Any hints/references please. Thanks!
Sorry, I had chapters 11 (Stochastic Majorizations) and 12 (Probabilistic and Statistical Applicatona) in mind, but the content has nothing to do with your problem. At first sight your problem seems to be simple and I still think there must be an elementary solution. I will think about it, now its of interest to me.
|
2025-03-21T14:48:29.641271
| 2020-01-12T22:32:00 |
350307
|
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"Ivan Di Liberti",
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|
Stack Exchange
|
Factorizing "compact" morphisms through "reflexive" objects (with respect to a given monad)
The following question arises from a hope that something true in a particular category might be true for general categorical reasons, coupled with a need to test this by looking for counterexamples.$\newcommand{\cC}{{\mathcal C}}$ I am hoping that if the question is posed in a general setting, removed from the original context, then someone who can look at it with fresh eyes will spot something that I'm not seeing.
Let $\cC$ be a category which has as all small limits and colimits and let $T$ be a monad on $\cC$ where the components of the unit, $\eta_a: a \to Ta$, are monic for each object $a$. (In the example I have in mind, $\eta_a$ is actually a regular monomorphism for each $a$.) Furthermore, although it's not needed for (1) and (2) below, let us assume that $T$ preserves and reflects monomorphisms.
Call a morphism $f:a \to b$ $T$-compact if $T f: Ta \to Tb$ factors through $\eta_b : b\to Tb$ (such a factorization, if it exists, is unique by monicity of $\eta_b$).Then we have a couple of easy observations:
(1) Naturality of $\eta$ shows that pre- or post-composing a $T$-compact morphism with an arbitrary morphism in $\cC$ still results in something which is $T$-compact.
(2) Let $c\in \cC$. Then the identity morphism $1_c$ is $T$-compact if and only if $\eta_c: c\to Tc$ is an isomorphism. (This uses monicity of $\eta_c$, again.) Call such an object $c$ $T$-reflexive.
Putting 1) and 2) together: if a morphism $f:a\to b$ factors through some $T$-reflexive object $c$, then $f$ is $T$-compact.
Question. Suppose $f:a\to b$ is $T$-compact. Does $f$ factor through some $T$-reflexive object $c$?
My instinct is that in general this should be far too optimistic. I do know a particular category where this works (where not every morphism is $T$-compact) but the usual construction of the factorization requires "external technology". So I am asking for counterexamples, to better understand why naive approaches to this construction (in the category where the desired result is true) are unlikely to work.
(In a certain sense the example of $\cC$ where I know things work is "not too big", and so I'd prefer counterexamples which don't rely on cardinality arguments. However I appreciate that this is a very vague comment/request.)
If $C$ has a proper factorization system and $T$ preserves it, then I think there is hope to show that $\text{Im}(f)$ is reflexive. I hope I will have time to check it tomorrow, but I might forget.
@IvanDiLiberti That would be interesting. In the setting where I know this works, mono+epi need not imply iso, so dealing with Im(f) is a bit subtle, and I don't think Im(f) as a subobject of codomain(f) will be reflexive. However, the factorization that I know to exist by "outside reasons", does have something to do with Coim(f) and Im(f), so maybe your idea can be modified/adapted...
@YemonChoi Factorization systems in non-balanced categories need not lead to any problems with images. The most common case is regular epi-mono.
@KevinCarlson perhaps I was being too vague or conciliatory. In the category I have in mind, the image of a $T$-compact morphism will not in general be $T$-reflexive. I am deliberately trying not to name the category and monad in question as I want to avoid preconceptions, but it is not abelian: the canonical map from the coimage to the image is mono and epi but not iso. in general
@YemonChoi Maybe it wasn't clear that I meant "(regular epi)-mono" as opposed to "regular [as in ordinary] epi-mono?" I'm not sure what the relevance of abelianness or of the bimorphism from coimage to image would be otherwise...In any case, yes, this is probably an unhelpfully vague exercise until one gets closer to addressing your question.
@IvanDiLiberti Any further thoughts?
Unfortunately not, I should be sitting with pen and paper for some time to say something relevant. :(
|
2025-03-21T14:48:29.641547
| 2020-01-13T00:53:34 |
350310
|
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"Harry Gindi",
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|
Stack Exchange
|
When is the étale topos of a fibre product the fibre product of étale toposes?
In what follows, all schemes are qcqs. Also, let $\operatorname{\acute{E}t}(X)$ denote the petit étale topos of a scheme $X$.
Let $Y\to X$ be an $X$-scheme. Say that $Y$ is a special $X$-scheme if for any $X$-scheme $Z\to X$, the universal geometric morphism
$$\operatorname{\acute{E}t}(Y\times_X Z)\to \operatorname{\acute{E}t}(Y)\times_{\operatorname{\acute{E}t}(X)} \operatorname{\acute{E}t}(Z)$$
is an equivalence of toposes.
Is there a known characterization of the special $X$-schemes for a scheme $X$?
I know of two classes of $X$-schemes that satisfy this property:
If $Y\to X$ is a closed immersion, then $Y$ is a special $X$-scheme
If $Y\to X$ is pro-étale and qcqs, then $Y$ is a special $X$-scheme (for qcqs étale morphisms, this is obvious, as it follows from slicing of sites, and to get pro-étale morphisms, apply some version of Noetherian approximation)
Are there any other kinds of maps that satisfy this property? What about in the case where $X$ is Noetherian?
Edit:
I have a guess. If anyone can think of an obvious counterexample, it would be much appreciated. If we restrict to the case where $X$ is Noetherian and try to classify only the special $X$-schemes $Y\to X$ of finite type, then it seems like from Will Sawin's idea, the condition should be something like quasi-finite.
At least when $X$ is the spectrum of a field, the quasi-finite (in fact finite) $X$-schemes will be (possibly empty) finite disjoint unions of finite field extensions. So reducing to the connected case, notice that a finite field extension will be the composite of a separable finite field extension and a purely inseparable finite field extension. But finite separable field extensions are étale, and purely inseparable field extensions are universal homeomorphisms.
I'm not sure if this is true, but I wonder if one can play a similar game in the case of a more general base scheme than a field, first using the fact that quasi-finite maps are (étale) locally finite maps, then factoring a finite map as some kind of étale map followed by a 'totally ramified' one that is a universal homeomorphism.
Warning: This is the fibre product of toposes, which does not agree with the fibre product of categories!
One could first try to characterize the special $X$-schemes for $X$ a point. I imagine these will be essentially just unions of points. This would then give a necessary condition as special schemes are clearly preserved by base change so the fibers must be special over points.
|
2025-03-21T14:48:29.641842
| 2020-01-13T00:59:42 |
350311
|
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|
Stack Exchange
|
Status of the Salmon Conjecture
The set-theoretic version of the Salmon Conjecture (that is, finding the equations that cut out the fourth secant variety of the Segre embedding of $\mathbb P^3 \times \mathbb P^3 \times \mathbb P^3$ set-theoretically) has been resolved by S. Friedland and E. Gross in https://arxiv.org/abs/1104.1776
In view of this, I am wondering if resolving the scheme-theoretic version of the Salmon conjecture (that is, finding the defining ideal of the above variety) 1) is much harder/needing new ideas compared to the set-theoretic solution and 2) carries more information for the biological applications that motivated the conjecture than the set-theoretic version.
Also, is the problem of finding the ideals of other secant varieties of Segre embeddings 1) expected to be resolved in the near future and 2) relevant for biological or other applications? What are some approachable/interesting problems here?
The scheme-theoretic version is still open, so it seems that it may need a new idea. Note, that is distinct from finding the defining ideal (scheme-theoretic is the ideal up to a radical). It seems difficult to guess how much harder it will be, or whether the "near future" might hold resolutions for more general versions of the question. I have no idea if ideals or schemes matter for biological applications, but there are connections with geometric complexity theory, see e.g., https://arxiv.org/abs/1305.7387, or the book that Landsberg wrote called "Geometry and Complexity Theory".
Thanks a lot! I am not sure that the question (or its meaning) about equations defining secant varieties of Segre varieties appears in the above Landsberg's writings explicitly. I understand that we want to study the rank or border rank of the matrix multiplication tensor, so we want to understand which secant variety of the Segre variety of $\mathbb P^n \times \mathbb P^n \times \mathbb P^n$ it lies in, is this why we want to study the equations of these secant varieties? Even if we knew the equations, probably it would not be easy to see if the matrix multiplication tensor satisfies them?
Oops, scheme-theoretic means finding the defining ideal up to saturation, not up to radical. Sorry. ... Besides that: Landsberg has definitely written about equations of secant varieties (and so have many others) but maybe not in his GCT writing. And I think you're right, only finding equations isn't necessarily enough - they have to also be ones that you can evaluate on the matrix multiplication tensor, and that's a nontrivial problem.
|
2025-03-21T14:48:29.642039
| 2020-01-13T03:53:42 |
350314
|
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"Gilles Mordant",
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|
Stack Exchange
|
Approximate Multivariate Gaussian Integration by Parts
When $Z$ is a $\mathcal{N}(0,1)$ random variable, $f$ smooth from $\mathbb{R} \to \mathbb{R}$ we have the Gaussian integration by parts formula
$$
\mathbb{E}(Zf(Z)) = \mathbb{E} f'(Z).
$$
One analog of the above in the multivariate case is when $Z$ is $\mathcal{N}(0, \Sigma)$, $f: \mathbb{R}^n \to \mathbb{R}$ in which case
$$
\mathbb{E}[Z_i f(Z)] = \sum_j \Sigma_{ij} \mathbb{E}\left[ \frac{\partial f}{\partial x_j}(Z)\right].
$$
Now suppose $Z$ is an arbitrary mean 0 variance 1 random variable. In the one-dimensional case we now have the "approximate" integration by parts formula
$$
|\mathbb{E}(Zf(Z) - f'(Z))| < C \|f''\|_{\infty} \mathbb{E}(|Z|^3).
$$
Is there any known multivariate generalization of this formula? Crucially, in the proof I know of the multivariate gaussian case we use the fact that if the covariance of a Gaussian random vector is diagonal the components are independent to reduce to the 1-D case, but this doesn't seem possible in the general case.
I guess that Equation 14 in the paper https://arxiv.org/pdf/0902.0333.pdf may be a good starting point for your needs. More generally, the question that you ask is related to the so-called 'Stein method'. I however believe that my comment is useless and that you know this already.
|
2025-03-21T14:48:29.642147
| 2020-01-13T04:16:59 |
350315
|
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|
Stack Exchange
|
Positive matrix and diagonally dominant
There is a well-known criterion to check whether a matrix is positive definite which asks to check that a matrix $A$ is
a) hermitian
b) has only positive diagonal entries and
c) is diagonally dominant.
This is a sufficient condition to ensure that $A$ is hermitian.
A classical counterexample where this criterion fails is the matrix
$$\left(\begin{matrix} 0.1 & 0.2 \\ 0.2 & 10 \end{matrix} \right).$$
This matrix is positive definite but does not satisfy the diagonal dominance.
I think this matrix illustrates well the issue with the diagonal dominance: It does not take into account if you have a rapidly growing diagonal elements that overshadow the failure of the diagonal dominance in each row.
I am therefore looking for a criterion to ensure that a matrix is positive definite that asks for a matrix $A$ to be
a) hermitian
b) has only positive diagonal entries and
c) its diagonal grows sufficiently fast compared to all off-diagonal elements. We might want to assume some sparsity as well on our matrix. So ideally instead of investigating a single row, this improved criterion I would have in mind checks let's say an individual row but instead of comparing the row sum with its single diagonal entry, also takes other diagonal elements into account.
PS: Of course you might say that Sylvester's criterion takes care of this, but this criterion is far too cumbersome for sufficiently sparse matrices, I find. Just like the diagonal dominance is really apparent.
EDIT: I found the so-called Brauer's ovals of Cassini which seem to generalize the notion of Gershgorin circles used to prove the diagonally dominance criterion. This allows me already to take into account pairs of two diagonal entries. Can this be generalized to arbitrarily many entries?
See for background.
Please let me know if you have questions.
This is very much loosely related to your question (in that it doesn't require the Hermitian part ... or the strict diagonal dominance ... actually it is not that related), but the underlying idea in the Hawkins-Simon condition might be somewhat helpful to some specific problems you might consider. See https://en.wikipedia.org/wiki/Hawkins%E2%80%93Simon_condition.
If you have a matrix with a rapidly growing diagonal, you can check alternatively if $DAD$ is diagonally dominant with a positive diagonal, where $D= \operatorname{diag}(1,\rho, \rho^2, \dots, \rho^{n-1})$, for a suitably chosen $\rho$. If that is the case, then $DAD$ is positive definite, and since positive definiteness is preserved by congruence, so $A$ (note that $D^T=D$). In your case, for instance $\rho=1/\sqrt{10}$ works.
that's a nice idea, thank you.
|
2025-03-21T14:48:29.642340
| 2020-01-13T06:31:27 |
350320
|
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|
Stack Exchange
|
With only two characters allowed, is it possible to efficiently reference a 256 character alphabet in a string?
Let us use 0 and 1 for the binary parallel.
You have 256 characters you need to reference, imagining a 256 character alphabet. You can only use a string to reference them that contains 0 and 1. The string can be any length, and your reference can have multiple levels of compression (as long as they would work for any combination of those 256 characters).
Is it possible to efficiently reference this 256 character alphabet this way?
This is not number theory, I'm afraid. I'm re-tagging. I'm not sure this is even on-topic; there is always math.stackexchange.com if this gets closed here.
Thanks for the retag, I was super not certain of which category to put this in.
Your question is imprecise: what do you mean by "efficient"? However I feel there could be a good question with a clean answer behind the current formulation. If you want to ensure frequent characters are assigned a short string (e.g. for compression purposes), an important keyword is entropy. It gives a lower bound on the average length you can achieve. E.g. if the characters all have the same frequency, you cannot do better than encode them all with 8 bits.
Look at Huffman coding, arithmetic coding, or Lempel-Ziv(-Welch) coding depending on what you want to assume about the source.
|
2025-03-21T14:48:29.642486
| 2020-01-13T06:58:38 |
350321
|
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"vidyarthi"
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|
Stack Exchange
|
Generalization of independence complex of graphs
Let $G$ be an undirected graph with no multiple edges or loops. Recall that the independece system $\mathcal{I}(G)$ consists of all those subsets $A$ of the vertex set such that the induced subgraph $G[A]$ is totall disconnected.
The independence system is an abstract simplicial complex and a lot of its topological invariants are related to combinatorial properties (like, chromatic number, domination number etc.) of the underlying graph.
I have following two generalization of the independence system in mind. A simple Google search did not yield any work that explores these complexes.
For a natural number $j\geq 2$ denote by $\mathcal{C}_j(G)$ the collection of subsets of vertices of $G$ such that for $A$ in that collection the induced subgraph $G[A]$ does not contain a $j$-clique.
For a natural number $r\geq 1$ denote by $\mathcal{E}_r(G)$ the collection such that for every $A\in \mathcal{E}_r(G)$ each connectedd component of $G[A]$ has at most $r$ vertices.
It is clear from above definitions that $\mathcal{I}(G) = \mathcal{C}_2(G) = \mathcal{E}_1(G)$. Can anybody point me to a reference where either of these generalizations are studied? I would like to know if any of topological invariants of these complexes relate to graph theoretic information.
You can see clique complex for the first complex you mention
Although I am not sure if these concepts been studied much as simplicial complexes, at least the second has been studied quite extensively in graph theory and theoretical computer science (including by me).
The earliest studies I am aware of are
Sampathkumar. Discrete Math., 1993. https://doi.org/10.1016/0012-365X(93)90493-D
Edwards and Farr. JCTB, 2001. https://doi.org/10.1006/jctb.2000.2018
These come at the problem from independent perspectives, but the concept is natural enough that maybe it gets reintroduced regularly.
For instance, it has some connection to vaccination protocols:
Britton, Janson, Martin-Löf. Adv. Appl. Prob., 2007. https://doi.org/10.1239/aap/1198177233
There is a rather extensive study of efficient partitions of the vertex set of a graph into such (and related) types of sets, a partial (dynamic) survey of which is given by Wood in the Electr. J. Comb.
As an act of pure vanity, my own study of this type of independence concerns the binomial random graph:
Broutin and Kang. J. Comb., 2019. http://dx.doi.org/10.4310/JOC.2018.v9.n3.a1
As for the first concept, it has also been studied a bit I believe, but less so (probably because it is much harder to make progress on).
The notion of $\mathcal{C}_j(G)$ is natural in terms of hypergraphs where is has been studied. If you rephrase in terms of a hypergraphs, then $\mathcal{C}_j(G)$ becomes the independence complex of the hypergraph. These have been studied a lot in (combinatorial) commutative algebra.
An independent set in a hypergraph is any subset of vertices which does not contain any hyperedge. The independence complex of a hypergraph is (just like for graphs) the simplicial complex whose faces are the independent sets. Now for any graph $G$ let $H_j(G)$ be the hypergraph on the same vertex set whose hyperedges are $j$-cliques. Now $\mathcal{C}_j(G)$ is the independence complex of $H_j(G)$. Many references can be found by searching for "independence complex hypergraph" and also changing "hypergraph" to "clutter" is useful in search and you'll find more articles.
The minimal non-faces of the independence complex are the hyperedges. Thus, the square-free monomial ideal with a generator corresponding to each hyperedge is the Stanley–Reisner ideal for the independence complex. As such the topology along with properties like shellability and vertex decomposability are of great interest. Searching things like "Cohen-Macaulay clutters" will turn up more results.
The hypergraphs obtained as $H_j(G)$ will be $j$-uniform (i.e. all hyperedges contain $j$ vertices) and in fact all $j$-uniform hypergraphs are $H_j(G)$ for some $G$. So, the study of the complexes $\mathcal{C}_j(G)$ is the same of the study of independence complexes of $j$-uniform hypergraphs. Also, the map $G \mapsto H_j(G)$ is not one-to-one; so, it makes more sense to consider relations of the topology of the complex and properties of the hypergraph. One such property that arises is chordality and generalizations of it:
Chordal and Sequentially Cohen-Macaulay Clutters by Woodroofe
A class of hypergraphs that generalizes chordal graphs by Emtander
An answer for 1. would definitely include keywords "simplicial complex" and most probably "triangle-free" or some other "-free subgraphs". There's not many results and the only relevant one is this article, which discusses methods for the complex of $\mathcal{F}$-free induced subgraphs in general, where $\mathcal{F}$ is a family of graphs to be excluded as subgraphs (not induced). In your case you would take, respectively, $\mathcal{F}=\{K_j\}$ and $\mathcal{F}=$ $(r+1)$-vertex trees. Motivations are given for other $\mathcal{F}$ only, though, and there's no follow-up work on those devoid complexes.
What has been studied a bit more is simplicial complexes of all graphs on $n$ vertices with a given monotone property, meant as subsets of $\binom{n}{2}$. In other words, all subsets of edges of the $n$-clique, with some property. See the book Simplicial complexes of graphs by Jakob Jonsson, in particular
sections 26.7 Triangle-Free Graphs (p. 351-352) and 18.2 Graphs with No Large Components (pp. 247-258).
To add to my previous answer, a paper of Szabó and Tardos from 2006 ("Extremal problems for transversals in graphs with bounded degree") has a result related to the complex you defined in 2.
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2025-03-21T14:48:29.642873
| 2020-01-13T08:47:44 |
350323
|
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|
Stack Exchange
|
Sliding puzzle related path finding
Let a sliding puzzle of size $m\times n$, with $p$ empty positions be given. Further, let $(x,y)$ and $(x',y')$ be two arbitrary positions inside the puzzle. What conditions have to be fulfilled s.t. it is possible to clear a path between $(x,y)$ and $(x',y')$, if each block of the puzzle is restricted to move not further than one position (distance one)?
This question concerns me for a while now. I listed my findings below, but I am struggling to prove them. I don't know how to find publications on such a topic if they even exists? Consequently my questions are:
Is the formulation of the problem comprehensible? If not, please ask!
Is there some literature on such a topic, or something similar?
How could a proof look like?
I found that the following inequality is a necessary condition for the solution of the problem:
$p \geq |x - x'| + |y-y'| + 1$
It states that the number of empty positions has to be at least as big as the number of positions on the shortest path from $(x,y)$ to $(x',y')$.
For my needs it would suffice to show, that there exists a path between all source/target-pairs, if the number of blocked positions ($mn - p$) is bounded by the minimum of m and n minus one.
$mn - p \leq \min (m,n) - 1$
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2025-03-21T14:48:29.642984
| 2020-01-13T09:17:13 |
350324
|
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Stack Exchange
|
Category of finite models of a $\tau$-sentence
Let $\varphi$ be an $\tau$-sentence, we define the generalized spectrum of $\varphi$ as the class of its finite models, $$\text{GenSpec}(\varphi):=\{\mathcal{A}; \mathcal{A} \models \varphi, \lvert A\rvert < \aleph_0\}$$
and the spectrum of $\varphi$ as the set of cardinalities of finite models
$$\text{Spec}(\varphi):=\{\lvert A \rvert; \mathcal{A}\in\text{GenSpec}(\varphi)\}.$$
It is well known that coding $\text{Spec}(\varphi)$ using binary strings results in precisely all $\text{NE}$ sets, this follows from the Fagin's theorem which states that the set of binary encoded linearly ordered structures of $\text{GenSpec}(\varphi)$ are precisely the $\text{NP}$ sets of binary encoded strctures.
Equip $\text{GenSpec}(\varphi)$ with all $\tau$-structure morphisms and call this category $\text{MOD}_{\text{fin}}(\varphi)$.
The question: Is it known whether for every $\varphi$ an $\tau$-sentence there exists $\psi$ a $\tau$-sentence such that $$\text{Spec}( \varphi)=\text{Spec}(\psi),$$
and $\text{MOD}_{\text{fin}}(\psi)$ is a grupoid? In other words, do morphisms matter while considering spectra of FO sentences?
A correction, to start: "$\mathcal{C}$ is discrete" does not mean "$\text{Mor}(\mathcal{C}) = \emptyset$". Instead, a discrete category has only identity arrows. And even with that correction, your question as written has a somewhat trivial negative answer: The class $\text{GenSpec}(\varphi)$ is always closed under isomorphism, and the category $\text{MOD}_{\text{fin}}(\varphi)$ will always contain all isomorphisms. So as long as $\varphi$ has any nonempty models, $\text{MOD}_{\text{fin}}(\varphi)$ will not be discrete.
I suspect you really want to ask whether you can find $\psi$ such that $\text{MOD}_{\text{fin}}(\psi)$ is a groupoid, i.e. such that every arrow is an isomorphism. So I'll go forward with that interpretation of the question.
If you fix a signature $\tau$, then the answer is no in general. For example, suppose $\tau$ is the empty signature (so a $\tau$-structure is a pure set). For any $\tau$-structure $M$ with $|M|\geq 2$, there is a $\tau$-homomorphism $M\to M$ which is not an isomorphism (e.g. any constant function). So if $\varphi$ is any $\tau$-sentence with $\text{Spec}(\varphi)\not\subseteq \{0,1\}$, and $\psi$ is any $\tau$-sentence with $\text{Spec}(\psi) = \text{Spec}(\varphi)$, then $\text{MOD}_{\text{fin}}(\psi)$ is not a groupoid.
On the other hand, if you allow the signature to change, the answer to your question is positive. Let $\varphi$ be a $\tau$-sentence, and let $\tau'$ be $\tau$ together with some new symbols:
A binary relation $<$.
A binary relation $S$.
Two unary relations $F$ and $L$.
For each relation symbol $R$ in $\tau$, a relation symbol $R_\lnot$ of the same arity.
Let $\psi$ be the conjunction of $\varphi$ with sentences expressing the following:
$<$ is a strict linear order.
$S$ is the successor relation for $<$.
$F$ defines the first element of the order $<$.
$L$ defines the last element of the order $<$.
$R_\lnot$ defines the complement of the relation $R$.
Since any finite model of $\varphi$ can be expanded to a model of $\psi$, we have $\text{Spec}(\varphi) = \text{Spec}(\psi)$. And you can check that any $\tau'$-homomorphism between models of $\psi$ is an isomorphism. So $\text{MOD}_{\text{fin}}(\psi)$ is a groupoid.
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2025-03-21T14:48:29.643203
| 2020-01-13T11:06:02 |
350331
|
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"sort": "votes",
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|
Stack Exchange
|
Which set of functions/measures has range $\mathrm{L}^\infty$ under Fourier transformation
I have a question concerning the Fourier transformation. What I know is that $\mathrm{L}^{\infty}=\{\hat{u}:\ u\in Y\}$ for some space $Y$. Now, I want to specify the space $Y$. The question is, is there any explicit description of the space $Y$, i.e., of which form must $Y$ be such that the image under the Fourier transform of $\mathrm{L}^{\infty}$? I think $Y$ must contain Radon measures but otherwise I have no clue. Any suggestion? Thank you very much for your help.
It is most unlikely that there is such an explicit characterisation. In fact, I would be so bold as to claim that if anybody were to find one, he would achieve instant fame ( well, at least amongst the mathematical community).
You also need to specify which group you are implicitly considering in your question. $L^\infty$ of what?
Moreover, there is a delicate question concerning what you are taking as the domain of your Fourier transform. A priori, the Fourier transform is defined for finite Radon measures on a LCA group; but then the range of the Fourier transform is almost never all of $L^\infty$ of the dual group. If you want to enlarge the domain of definition of the Fourier transform then you need to work out how you define the Fourier transform of something that isn't a measure ...
@YemonChoi : I am looking for $\mathrm{L}^{\infty}(\mathbb{R})$. However, it seems hopeless to get some nice answer to the question.
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2025-03-21T14:48:29.643341
| 2020-01-13T11:24:30 |
350335
|
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|
Stack Exchange
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Approximation of multipliers by multipliers of a smaller set
Let $X$ be a compact metric space, and let $B$ be a convex balanced bounded set in $C(X)$ such that for every $x\in X$ there is $f\in B$ with $f(x)\ne 0$.
Let $M=\{u\in C(X),~ uf\in B,~\forall f\in B\}$ and let $N=\{u\in C(X),~ uf\in \overline{B},~\forall f\in \overline{B}\}$.
Since multiplication $(f,g)\to fg$ is a continuous operation on $C(X)$, it follows that $N$ is closed in $C(X)$ and $M\subset N$. Thus, $\overline{M}\subset N$.
Is it true that $N=\overline{M}$?
Let $X=[0,1]$, let $p_n(x)=x^n$ for $n\in{\mathbb N}$, let $a(x) =e^{-x}$.
Let $V_0 = \operatorname{lin}\{p_n \colon n\in {\mathbb N}\}$ and let $V = {\mathbb C}a + V_0$.
Claim 1: $V_0 = \{ f \in V \colon f(0)=0\}$.
Proof: the LHS is obviously contained in the RHS; for the converse inclusion, note that $a(0)=1$ while $p_n(0)=0$ for all $n\in{\mathbb N}$. $\qquad\Box$
Claim 2. Let $u\in C(X)$. If $up_1\in V$ and $ua\in V$ then $u$ is constant.
Proof. Consider $w=up_1$. By assumption $w\in V$, and $w(0)=0$ by construction, so $w\in V_0$ by Claim 1. Hence $u$ is a polynomial, and we write it as $u = u(0)1+ h$ where $h\in V_0$. We need to show that $h=0$.
Consider $ua$; by assumption, $ua\in V$, and $(ua)(0)=u(0)$. So, using (the proof of) Claim 1 again, we must have $ha \in V_0$. But the set
$$\{p_n \colon n\in{\mathbb N}\cup\{p_n a \colon n\in {\mathbb N}\}$$ is linearly independent: one way to see this is to note that if $g$ is any non-zero polynomial then $ga$ has at most finitely many zeros on the interval $[0,1]$. It follows that $ha=0$ and since $a(x)=e^{-x}$ this forces $h=0$ as required. $\qquad\Box$
Claim 3. $p_1a$ belongs to the closed, balanced, convex hull of $\{3p_n \colon n\in {\mathbb N}\}$.
Proof. Considering the Taylor series for $xe^{-x}$, we have
$$ p_1a = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{(n-1)!} p_n $$
which converges absolutely as a series in $C[0,1]$; note that
$$ \sum_{n=1}^\infty \left\vert \frac{(-1)^{n-1}}{(n-1)!} \right\vert = e < 3. \qquad\qquad\Box$$
The counterexample. Let $B$ be the convex balanced hull of $\{a\}\cup\{3p_n \colon n\in{\mathbb N}\}$. Then $p_1 B \subseteq \overline{B}$ by Claim 3, and since multiplication by $p_1$ is a continuous norm-1 operator on $C[0,1]$, it follows that $p_1\overline{B}\subseteq \overline{B}$. On the other hand, if $u \in C[0,1]$ and $uB\subseteq B$ then in particular we have $up_1\in V$ and $ua\in V$, which by Claim 2 forces $u$ to be constant. Clearly $p_1$ cannot be approximated by constant functions, and so — in the notation/perspective of the original question — the multipliers of $B$ are not dense in the multipliers of $\overline{B}$.
|
2025-03-21T14:48:29.643528
| 2020-01-13T11:56:42 |
350336
|
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|
Stack Exchange
|
Tight estimates for binomial summation
Is there tight estimates for the following logarithmic summation ($\gamma\in(0,1)$)
$$\ln\Bigg(\sum_{t=\frac{n^{}}2-\gamma n^\gamma}^{\frac{n^{}}2+\gamma n^\gamma}\sum_{\ell=\frac{n^{}}2-\gamma n^\gamma}^{\frac{n^{}}2+\gamma n^\gamma}\sum_{k=0}^t\binom{\ell}{k}\binom{n-\ell}{t-k}\Bigg)?$$
Is it roughly bounded above by $$\ln\Bigg(poly(n)\sum_{k=0}^t\binom{\ell}{k}\binom{n-\ell}{t-k}\Bigg)$$ for some minimal choice of $$\sum_{k=0}^t\binom{\ell}{k}\binom{n-\ell}{t-k}?$$ The summand though seems amenable to Cauchy-Schwarz gives very loose bounds.
By Vandermonde's identity, $$\sum_{k=0}^t\binom{\ell}{k}\binom{n-\ell}{t-k}=\binom{n}{t},$$
so the triple sum reduces to
$$
\sum_{t=n/2-\gamma n^\gamma}^{n/2+\gamma n^\gamma}\sum_{\ell=n/2-\gamma n^\gamma}^{n/2+\gamma n^\gamma}\binom{n}{t}
=\sum_{t=n/2-\gamma n^\gamma}^{n/2+\gamma n^\gamma}(2\gamma n^\gamma+1)\binom{n}{t}
=(2\gamma n^\gamma+1)\sum_{t=n/2-\gamma n^\gamma}^{n/2+\gamma n^\gamma}\binom{n}{t}.
$$
Now $\sum_t \binom{n}{t} \le 2^n$, so
$$\ln\left(\sum_{t,\ell,k}\right) =\ln(2\gamma n^\gamma+1)+\ln\left(\sum_{t=n/2-\gamma n^\gamma}^{n/2+\gamma n^\gamma}\binom{n}{t}\right) \le \ln(2\gamma n^\gamma+1)+n\ln 2.$$
If we had $k=t-\gamma t^\gamma$ to $k=t$ would we be able to achieve $n\ln2- o(n)$?
@Robplatt https://mathoverflow.net/questions/350373/tight-sublinear-estimates-for-partial-binomial-summation.
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2025-03-21T14:48:29.643618
| 2020-01-13T12:45:26 |
350337
|
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"Denis Serre",
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|
Stack Exchange
|
Phase transition in matrix
Playing around with Matlab I noticed something very peculiar:
Take the symmetric matrix $A \in \mathbb R^{n \times n}$ defined by
$$A_{ij}= i \delta_{ij} - \frac{\varepsilon}{\sqrt{i}\sqrt{j}}\,.$$
Here $\delta_{ij}$ is the Kronecker delta.
We first note that this matrix is not diagonally dominant if $n$ is large enough.
This is because $\lim_{n \rightarrow \infty} \sum_{i=1}^{n} \vert A_{i,1}\vert=\infty >\vert A_{1,1} \vert.$
It is obvious that we require $\varepsilon<1$ in order for $A$ to be positive definite, since otherwise $A_{1,1}\le 0.$
However, I noticed that for let's say $\varepsilon=0.1$ one can make the dimension as large as one wants and the matrix remains positive definite.
Question: How can one show that $A$ is positive definite independent of the dimension if $\varepsilon$ is sufficiently small but fixed ?
It looks like a consequence of Gershgorin disks theorem. But precise computations have to be done.
Hi Sascha, does your transition occur at $\epsilon = 6/\pi^2$?
@RaphaelB4 it is hard to say numerically to be honest.
@JeanMarieBecker mhmm, not sure. I think Gershgorin's disk theorem is somewhat equivalent to being diagonally dominant for positive definite matrices.
The claim is true with $\epsilon=\frac6{\pi^2}\,$.
To see this, remark that by changing variable $x_i=y_i\sqrt i\,$, this is equivalent to proving that
$$\epsilon\left(\left(\frac1{ij}\right)\right)_{1\le i,j}\le I_\infty.$$
The first (infinite) matrix is $V\otimes V$ with $V=(1,\frac12\,,\ldots,\frac1n\,,\ldots)$. It is symmetric, rank-one, with eigenvalues $0$ (infinitely many times) and ${\rm Tr}(V\otimes V)=\frac{\pi^2}6$ (simple).
what a nice argument this is. Just out of curiosity. Is it also obvious to see that the random signs do not matter? Cause this would somehow break the multiplicative structure.
@Sascha. This is not the same situation, because then the corresponding matrix is not any more rank-one.
I agree, let me just accept this wonderful answer and ask a follow-up question.
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2025-03-21T14:48:29.643895
| 2020-01-13T13:30:43 |
350339
|
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|
Stack Exchange
|
Gluing together dense subset of Projective Limit in $Ban_1$
Let $(X_n,\pi_n^{m})$ be a countable projective system in the category Ban$_1$ of Banach spaces and short linear maps (is (continuous) linear constructions). Then (co)-completeness of this category implies that $\projlim X_n$ is a well-defined Banach space.
Suppose further that, a strict variant of the Mittag-Leffer condition holds true: for all $n \in \mathbb{N}$, there exists $m\geq n$ such that
$$
\overline{\pi_m^n(X_n)} \subseteq X_k \qquad \forall k \geq n,
$$
and that $U\subseteq \projlim X_n$. What reasonable conditions would imply that $U$ is dense in $\projlim X_n$
What I propose (so far...):
Condition:
$$
\bigcap_{k\leq n; \, k,n \in \mathbb{N}} \pi^k_n(X_n)
\subseteq
U
$$
Proof:
Then, by Bourbaki's Mittag-Leffer theorem, the first condition implies that $\cap_{k\leq n; \, k,n \in \mathbb{N}} \pi^k_n(X_n)$ is dense in $\projlim X_n$. The second condition then implies that $U$ is dense in $\cap_{k\leq n; \, k,n \in \mathbb{N}} \pi^k_n(X_n)$. The transitivity of density implies that $U$ is dense in $\projlim X_n$.
Comment: I think the assumption is much too restrictive to be interesting.
I‘n not sure what you mean but consider the case of $\ell^\infty$ and its non-dense subspace $c_0$. The former is the projective limit of the finite dimensional spaces $\ell^\infty_n$ and this would seem to provide a counterexample.
Oh, then how could one (non-trivially) modify the hypothesis so that $U$ is dense?
I think this example shows that you can‘t, at least not in any simple way. The classical abstract Mittag-Leffler theorem (which, to my knowledge, is due to Bourbaki and works in the context of metric spaces) applies to the „normal“ projective limit, i.e., with no restrictions on the Lipschitz constants of the morphisms. You can save the density property but only at the cost of abandoning the category of Banach spces, e.g., by using strict or mixed topologies but that is another story entirely.
That is the whole point of them. Let me give you an example (for the case of continuous functions—this is more transparent). If you have a locally compact space $X$—let‘s say $\sigma$– compact to simplify things, then the family ${C(K)}$ of continuous functions on its compacta forms a projective spectrum and you can take its projective limit in three senses: as lcs‘s, as Banach spaces and in the sense of mixed topologies. This gives you successively the Fréchet space $C(X)$, the Banach space $C^b(X)$ and the strict topology on the latter (R.C. Buck) ....
I'll read up on this now; it seems very interesting/useful and I never heard of such a thing before... In the meantime, since it seems appropriate to consider the modification to this question in a separate poste; I've posted the variant on MIXTOP here (while I read the Buck article and a few others I've found on a google search).
https://mathoverflow.net/questions/350344/gluing-together-mixed-normed-ector-spaces-with-mixed-topologies
The third space has better density and duality properties than the other two ( which is why Buck introduced them). Similar situations arise for spaces of measurable or holomorphic functions, resp. for of opererators, say on Hilbert spaces and they have all been provided with appropriate strict topologies to deal with such problems.
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2025-03-21T14:48:29.644144
| 2020-01-13T13:51:50 |
350341
|
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Stack Exchange
|
Books on exotic structures
The second half of the XX-th century has witnessed an explosion of results on the existence of smooth structures on topological manifolds. Following various sources in Wikipedia, a rough timeline goes like this (I am forgetting many other important results, but these are the ones I am most interested in).
In 1952, Moise proved that topological 3-manifolds admit exactly one smooth structure
In 1956, Milnor proved the existence of exotic 7-spheres, believing at first that they could be counterexamples to the generalized Poincaré conjecture
He eventually would classify the exotic 7-spheres together with Kervaire in 1963
In the meantime, Smale had proved the the generalized Poincaré conjecture in dimension $\geq 5$ in 1961, and the h-cobordism theorem, which would simplify Milnor's construction, in 1962
In 1962, Stallings proved that $\mathbb{R}^n$ has a single smooth structure for $n \neq 4$
The generalized Poincaré conjecture in dimension 4 was then proved by Freedman in 1982. Together with Kirby, he also proved the existence of an exotic $\mathbb{R}^4$
The results of Donaldson on the intersection form of 4-manifolds of 1983, together with work by Freedman, could be used to construct examples of topological 4-manifolds without any smooth structure
As an algebraic geometer, I only know about these important developments from folklore, but I have never taken the time to study them in detail. I am trying to find a reference which would recap the whole story on exotic structures. So far I am aware of
Topology of 4-manifolds, Freedman and Quinn, 1990
The wild world of 4-manifolds, Scorpan, 2005
Differential Algebraic Topology: From Stratifolds to Exotic Spheres, Kreck, 2010
The first two seem to be very specific on dimension 4, while the latter does not talk about dimension 4 at all, and also takes a rather non-standard approach through stratifolds.
Is there any book going through all (or most) of the above results?
Gompf and Stipsicz "4-manifolds and kirby calculus" is one of the better recent-ish 4-manifolds texts. The field is still bubbling-along without any major landmark perspective, so a book hasn't really been fully needed. I imagine in the near future someone will make something like Gompf-Stipsicz combined with the more recent observations coming from Floer theory. But such a book does not exist, yet, as far as I know.
I think part of the issue is that your "exotic structure story" is really 3 different stories, split into the cases that the dimension is less than, equal to, or greater than 4.
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2025-03-21T14:48:29.644678
| 2020-01-13T14:52:25 |
350344
|
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|
Stack Exchange
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Gluing together mixed normed vector spaces with mixed topologies
This is a variant of this question.
Definitions/Facts
$Ball_1(X)$ denotes the unit ball (about $0$) in a normed vector space $X$.
MixTop of triples of pairs $(X,\tau)$ of normed vector spaces $X$ equipped with a LC (locally convex) topology $\tau$ on $X$ making $Ball_1(X)$ complete and bounded together with short linear maps whose restriction to the unit ball is $\tau$-continuous.
MixTop contains the category $Ban_1$ of Banach spaces and continuous short linear maps and similarly it is complete, see this article.
Question
Let $(X_n,\pi_n^{m})$ be a countable projective system in MixTop,thus $\projlim X_n$ is well-defined therein. Suppose further that, a strict variant of the Mittag-Leffer condition holds true:
$$
\pi_m^n(Ball_1(X_n)) \subseteq \overline{\pi_k^n(Ball_1(X_k))} \qquad \forall k \geq n,
$$
and that $U\subseteq \projlim X_n$ such that:
$$
\overline{\pi_n(U)}=X_n,
$$
then is $U$ dense in $\projlim X_n$ itself?
You will find a more comprehensive treatment inthe book „Saks spaces and applications to functional analysis“ by the same author. The appropriate assumptions for a theorem of this type would be a projective limit of Banach spaces so that the images of the unit balls are dense in the balls of the image space.
I'm reading the suggested monograph, however I can't find the said theorem. Is is known?
Not to my knowledge. I guess that it follows from the Bourbaki version, resp. its proof but I have no access at the moment. Best to work on a proof yourself.
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2025-03-21T14:48:29.644814
| 2020-01-13T15:16:59 |
350345
|
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|
Stack Exchange
|
When is $\mathcal{D}(\mathcal{F}):\mathcal{D}(\mathcal{A})\to \mathcal{D}(\mathcal{B})$ fully faithful?
Let $\mathcal{A}$ and $\mathcal{B}$ be two abelian categories and let $\mathcal{F}:\mathcal{A}\to \mathcal{B}$ be an additive functor. Assume that $\mathcal{F}$ is exact and let $\mathcal{D}(\mathcal{F}):\mathcal{D}(\mathcal{A})\to \mathcal{D}(\mathcal{B})$ denotes its (right and left) derived functor.
Under which assumptions on $\mathcal{F}$ is $\mathcal{D}(\mathcal{F})$ fully faithful?
(Surely, $\mathcal{F}$ would be fully faithful, but is it enough?)
Many thanks!
I suspect that it is fully faithful if and only if $\mathcal F$ preserves $Ext^n$-s between objects. The "only if" part is clearly necessary (see below). I am slightly hesitant about the "if" part, especially for the unbounded categories.
On the other hand, you can write many explicit sufficient conditions for that for bounded or bounded on one side categories. It is slightly harder for unbounded categories but I can think of the following sufficient condition: every object of $A$ has an injective envelope in $B$, which is also in $A$, and $B$ has finite injective dimension...
Having $\mathcal F$ fully faithful is not enough. Let $B=mod({\mathbb C}[X])$ and $A=mod({\mathbb C}[X]/(X))$ or $A=mod({\mathbb C}[X]/(X^2))$. In both cases, the hom-s in the derived categories are different.
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2025-03-21T14:48:29.644933
| 2020-01-13T16:04:42 |
350351
|
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|
Stack Exchange
|
presentability rank of categories of coalgebras
The following theorem is relatively classical:
Theorem: Given an accessible endofunctor, (co)pointed endofunctor or (co)monad $T$ on a locally presentable category $C$, then the category of $T$-(co)algebra is also locally presentable.
The proof goes as follows: in each case the category of (co)algebra can be written as a certain weighted bilimits in the category of accessible categories and accessible functors hence it is an accessible category. Moreover it is well know that categories of algebras are complete and categories of co-algebras are co-complete (in both case either limits or colimits are created by the forgetfull functor) so in both case they are locally presentable categories.
Unfortunately the argument above give very little control on the presentability rank of the category of (co)algebras. And this what this question is about: can we give a good bound on the presentability rank of the categories of (co)algebras ?
In the special case of algebra on a monad it is easy to see explicitly that if $C$ is locally $\lambda$-presentable and $T$ is $\lambda$-accessible then the category of $T$ algebras is locally $\lambda$-presentable, by showing that the free algebra on $\kappa$-presentable objects form a dense subcategory of $\kappa$-presentable objects. This done for example in Bird's Phd thesis (and probably in other places as well).
I convinced myself that the following was true:
Conjecture: Given $\kappa$ an uncountable regular cardinal. If in the theorem above $C$ is locally $\kappa$-presentable and $T$ is $\kappa$-accessible then the category of $T$-(co)-algebras is locally $\kappa$-presentable.
Assuming it is correct, I like would to know if it was proved somewhere, or if some other result of this kind is known (or if on the contrary counter-example where known) or not.
I'm stating both the case of algebras and coalgebras, but I am considerably more interested by the case of coalgebras.
Do you have a good example of a failure of this when $\kappa =\omega$?
@SaalHardali : I would be interested to have one ! What I have so far is the following: If you look at the identity endofunctor I on the category of set, the category of I-coalgebra is $\omega$-presentable, but the forgetfull functor do not send finitely presentable object to finitely presentable object (which is always true and essential part of the proof when $\kappa > \omega$). I would be interested to see an example with a co-monad and one where the category of coalgebra fails to be $\omega$-presentable...
That's a good counterexample for the case of coalgebras over just endofunctors. But I must say I have little to no experience with this notion so I can't decide if it counts as positive or negative evidence towards the statement for Comonads... Which I would be delighted to have a proof/counterexample for.
I just found the following exercise (page 127 ex. 2.l.) in Adámek-Rosicky's book: "Prove that a lax limit of locally $\lambda$-presentable categories and limit-preserving $\lambda$-accessible functors is locally $\lambda$-presentable". I believe the formation of the category of coalgebras over a comonad is a special case of a lax limit (a sort of lax equalizer). So at the very least even if this result is false (which now seems much less likely) a counterexample was not known to them when they wrote the book.
@SaalHardali : I don't think this results solve our problem. One problem is in the word "limit-preserving" the comonad is in general not limit preserving. Also I think there might be a problem of lax vs colax to express categories of coalgebras...
Oh you're right! I realized I was quoting the wrong thing. A page before there's exercise 2.j:
"Prove that if $F_1, F_2: \mathcal{K} \to \mathcal{L}$ are $\lambda$-accessible functors between locally $\lambda$-presentable categories and $F_2$ preserves limits, then the inserter category $Ins(F_1, F_2)$ is locally $\lambda$-presentable." but as you say the category of $S$-coalgebras is $Ins(Id,S)$ and that's the wrong direction. The next exercise is the same only instead of $F_2$ preserving limits they require $F_1$ to preserve colimits and no bound on the cardinal of presentability :/
I think I have an example for failure when $\kappa = \omega$. If we take $C = \otimes^{\infty}_{k=0} E{x_k}$ a countable tensor product of simple coalgebras cogenerated by a single primitive element (over $\mathbb{F}_2$). Then in the $\infty$-category of $C$-comodules (in $\mathbb{F}_2$-complexes) there are no non-trivial compact objects. Essentially because for any compact comodule there's an non-nilpotent element in the Ext of the identity functor which acts on it's underlying complex by 0.
I believe this is a really standard example which is probably known to many but I only realized now. Probably you were already aware in which case I apologize for the cluttering this comment section
I wasn't aware of this example, and too be honest do not completely understand it at this point. But I'm glad you found one, I'll to figure it out.
Is there a good reference for the "relatively classical theorem" mentioned in the beginning of the question?
@LeonidPositselski : I don't know one that does all the cases, but I think for each special case I can give you a reference. It follows from the existence of bilimits in the category of accessible category and accessible functor between them (as explained in the paragraph just below the theorem) and this results can be for example found as theorem 5.1.6 in Makkai & Parré "Accessible categories", or a bit in pieces in Rosicky-Adamek locally presentable catgories (they treat lax limits in section 2.H and other case as exercice 2.n and 2.m, which putting everything together give all bilimits.
@SimonHenry Thank you. I came via Jiri Rosicky's link from https://mathoverflow.net/questions/353853/coreflective-subcategories-in-grothendieck-locally-presentable-categories , and so I had in mind the category of coalgebras over a comonad. Now I see that Exercise 2.m in Adamek-Rosicky seems to be a good reference.
The case of algebras for a monad is discussed explicitly in Gregory Bird's thesis (see theorem 6.9). The case of the categories of algebras for an endofunctor or pointed endofunctor can be deduced from the fact that if $F$ is a (pointed) endofunctor on $C$, then $F$-Alg $\rightarrow C$ obviously satisfies the condition of Beck's monadicity theorem, and the induced monad preserves $\lambda$-filtered colimits if $F$ does. All this works for any regular $\lambda$, even $\omega$.
For the case of coalgebras, Jiří Rosický pointed out the key references to me by email:
The following theorem is due to Adámek and Porst in On tree coalgebras and coalgebra presentations as their Theorem 4.2.
We fix $\lambda$ an uncountable regular cardinal.
Theorem: Let $A$ be a $\lambda$-accessible category that admits colimits of $\omega$-chains, and let $F: A \rightarrow A$ be a $\lambda$-accessible endofunctor. Then:
The category of $F$-coalgebra is $\lambda$-accessible.
A $F$-coalgebra is $\lambda$-presentable if and only if its underlying object is $\lambda$-presentable in $A$.
Corollary: If $A$ is a locally $\lambda$-presentable category and $F$ is a $\lambda$-accessible endofunctor on $A$ then the category of $F$-coalgebra is locally $\lambda$-presentable.
The corollary follows immediately: as $A$ is cocomplete it has colimits of $\omega$-chains, and the forgetful functor $F$-coalg $\rightarrow A$ create colimits, so $F$-coalg is $\lambda$-accessible and cocomplete, hence $\lambda$-presentable.
We can immediately deduce that:
Theorem: If $F$ is a $\lambda$-accessible copointed endofunctor or comonad on a locally $\lambda$-presentable $A$, then:
The category of $F$-coalgebras is locally $\lambda$-presentable.
An $F$-coalgebra is $\lambda$-presentable if and only if its underlying object is $\lambda$-presentable.
Indeed, this can be deduced from the corollary above using that (for $\lambda$ an uncountable cardinal) the category of $\lambda$-presentable categories and left adjoint functors between them preserving $\lambda$-presentable objects is closed under $\lambda$-small cat weighted pseudo-limits. The category of $M$-coalgebras for a copointed endofunctor $M$ can be constructed as a full subcategory of the category of $M_0$-coalgebra where $M_0$ is the underlying endofunctor of $M$ as the equifier of $Id,v:U \rightrightarrows U$ where $U:M_0\text{-Coalg} \rightarrow C$ is the forgetful functor, and $v$ is the natural transformation which on each $M_0$-coalgebra $X$ is the composite $X \rightarrow M(X) \rightarrow X$.
When $M$ is a comonad this is a bit more complicated as we would like to take the equifier of the two natural transformation $X \rightrightarrows M_0^2(X)= M_0(M_0(X))$ corresponding to the two side of the usual square, but as $F^S$ is not a left adjoint functor we cannot directly conclude using 2-limits of diagrams of left adjoint functors.
Instead we consider the category:
$$E=\{X \in C, v_1,v_2:X \rightrightarrows M_0^2(X) \}$$
which is the category of coalgebra for the endofunctor:
$$ X \mapsto M_0^2(X) \times M_O(X)^2$$
which is indeed $\lambda$-accessible, so $E$ is locally $\lambda$-presentable and its $\lambda$-presentable objects are these whose underlying object $X$ is $\lambda$-presentable.
One has a natural functor $M_0$-coalg to $E$ which sends each $M_0$-algebra to the pair of maps $X \rightrightarrows M_0^2 $ corresponding to square defining $M$-algebras and another functor from $M_0^2$-Coalg to $E$ that sends each $f:X \rightarrow M^2_0(X)$ to $(X,f,f)$. taking the (pseudo)pullback of these two functors give us exactly the category of $M_0$-coalgebras compatible with the comultiplication of $M$. Both these functors clearly preserve all colimits and $\lambda$-presentable objects, so by the results mentioned above, this category is locally $\lambda$-presentable. Combining this with the case of copointed endofunctors we obtain the result.
I've included this material with a bit more details and other related results in appendix A of this paper.
Regarding relaxing the assumption that $\lambda$ is uncountable, Adámek and Porst show in their paper that the endofunctor:
$$ \mathcal{P}_f(X) = \{ F \subset X | F \text{ is finite} \} $$
as an endofunctor of the category of sets (with the direct image functoriality) is a counter example to the first theorem in the case $\lambda=\omega$. That is the category of $\mathcal{P}_f$ coalgebra is not finitely accessible. For the case of comonads, there seems to be a counter-example in the comments of the question.
This is super cool! Why didn't I know of it?!
Do you know if a similar result has been written down/proved for $\infty$-categories ?
@MaximeRamzi : I'm relatively confident it has not. I couldn't find a 1-categorical reference when I wrote it (which is why I wrote one in the paper mentioned). I don't know an analogue of Adámek and Porst for infinity categories, and the proof is fairly technical and don't generalize in an obvious way. The rest of the argument would also be significantly more complicated in $\infty$-category theory due to the fact that monads (and comonads) are not supper easy to manipulate...
However, if you are not interested in the precise presentability rank then the argument highlighted in the OP should be easier to adapt - I think the theorem about limits of accessible categories appears in HTT.
I'm indeed mostly interested in the question without presentability rank, and that's a good point, I'll try to see if I can work something out :) Thanks !
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2025-03-21T14:48:29.645643
| 2020-01-13T16:41:16 |
350352
|
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Stack Exchange
|
Minimizing along independent directions, nonlinear programming
Good afternoon, I am studying the book Nonlinear Programming: Theory and Algorithms (by Mokhtar S. Bazaraa, Hanif D. Sherali, C. M.) particularly the Theorem $7.3.5$. I'm not sure I understand this theorem, Do any of you know how to apply the theorem or some methods based on that method?
I really appreciate the references you can give me.
Theorem $7.3.5$: Let $f:\mathbb{R}^n\to\mathbb{R}$ be differentiable, and consider the problem to minimize $f(x)$ subject to $x\in\mathbb{R}^n$. Consider an algorithm whose map $\bf{A}$ is defined as follows. The vector $y\in {\bf{A}}(x)$ means that $y$ is obtained by minimizing $f$ sequentially along
the directions $d_1,d_2,\ldots,d_n$ starting from $x$. Here, the search directions $d_1,d_2,\ldots,d_n$ may depend on $x$, and each has norm $1$. Suppose that the following properties are
true:
i.) There exists an $\varepsilon>0$ such that $\det[D(x)]\geq\varepsilon$ for each $x\in\mathbb{R}^n$. Here $D(x)$ is the $n\times n$ matrix whose columns are the search directions generated by the algorithm, and $\det[D(x)]$ denotes the
determinant of $D(x)$.
ii.) The minimum of $f$ along any line in $\mathbb{R}^n$ is unique.
Given a starting point $x_1$, suppose that the algorithm generates the
sequence $\{x_k\}_{k\geq 1}$ as follows. If $\nabla f(x_k)=0$, then the algorithm stops with $x_k$; otherwise, $x_{k+l}\in{\bf{A}}(x_k)$, $k$ is replaced by $k+1$, and the process is repeated. If the sequence $\{x_k\}_{k\geq 1}$ is contained in a compact subset of $\mathbb{R}^n$, then each accumulation
point $x$ of the sequence $\{x_k\}_{k\geq 1}$ must satisfy $\nabla f(x)=0$.
The theorem generates a succession $\{x_{k_j}\}_{j\geq 1}$ such that $$x_{k_j}\to x$$
and $\nabla f(x)=0$. You can see that the previous theorem is a more general version that the Gradient Descent Algorithm, For this theorem, I don't know how to take the direction vectors.
I'm not sure I understand this theorem, Do any of you know how to apply the theorem or some methods based on that method?
I really appreciate the references you can give me.
Thanks
Use \det. Please use quotes to separate the transcription of the book from your own text.
I fixed the former but couldn't guess where the transcription finishes
Thank you, I copied how the theorem appears, I didn't like the way it is written either. It seems to me that it is basically to apply "n" times the fastest descent algorithm.
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2025-03-21T14:48:29.645834
| 2020-01-13T17:22:28 |
350358
|
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|
Stack Exchange
|
Random matrix is positive
This is a follow up question on my previous question here that was on solved in the deterministic setting by Denis Serre, when the perturbation can be separated. Therefore, I decided to split the deterministic case from the random question initially posed in the very same question. So let me state the random case:
We consider a symmetric matrix $A \in \mathbb R^{n \times n}$ defined by
$$A_{ij}= i \delta_{ij} - X_{ij}\varepsilon \,.$$
Here $\delta_{ij}$ is the Kronecker delta and $X_{ij}$ are iid Bernoulli (but of course $X_{ij}=X_{ji}$ to make it hermitian).
We first note that this matrix is not diagonally dominant if the matrix dimension $n$ is large enough, independent of what $\varepsilon$ is.
This is because $\lim_{n \rightarrow \infty} \sum_{i=1}^{n} \vert A_{i,1}\vert=\infty >\vert A_{1,1} \vert.$
Numerical experiments with matrix size $n=50$ or $n=200$ show that the lowest eigenvalue stays far above $0$, if $\varepsilon>0$ is sufficiently small.
Question: How can one show that $A$ is positive definite independent of the dimension if $\varepsilon$ is sufficiently small but fixed ?
To elaborate on the numerical experiments. We find that the lowest eigenvalue of $A$ in some interval $[0,96,1.08]$ if we choose $\varepsilon=0.1$ and $n=50$(upper plot) and $n=200$(lower plot), as I illustrate in the following plot where I sampled the lowest eigenvalue for 100 realizations:
Please let me know if you have any questions!
Let $u\in \mathbb{R}^n$ as $u_i=i^{-1}$. Then $$ \mathbb{E}\big[\langle u Au\rangle \big]=\sum_{i=1}^n i^{-1}-\frac{\epsilon}{2}\sum_{i,j\leq n}\frac{1}{ij}\approx \log n -\frac{\epsilon}{2}(\log n)² $$
Which is stricly negative for $\log n > 2\epsilon^{-1}$.
(Remark : In order to see it numerically one should choose $n$ exponentially large with $\epsilon^{-1}$.)
You can find other counter-examples along the vein of $v\in \mathbb{R}^n$ $v_i=i^a$ for certain $a$, making sure that the sum of the negative terms outweighs that of the positive components of the diagonal elements. Also, a matrix may have all positive eigenvalues yet not be positive definite; see <https://math.stackexchange.com/questions/4336/if-eigenvalues-are-positive-is-the-matrix-positive-definite>.
|
2025-03-21T14:48:29.645991
| 2020-01-13T18:37:23 |
350360
|
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Stack Exchange
|
Vanishing of Chow groups of $\mathcal{M}_{g,n}$
In this question, I am working over $\mathbb{C}$ and with rational Chow groups.
I am interested in the state of the art for vanishing theorems of the form $A^{k}(\mathcal{M}_{g,n})=0$ for some choies of $k,g,n$. (Note that I have restricted to the space of smooth curves.) One specific question I am especially interested in is: for which values of $k$ is $A^k(\mathcal{M}_{1,n})=0$?
Some example results in this direction: due to Mumford ($g=2$), Faber ($g=3,4$), Izadi ($g=5$), and Penev-Vakil ($g=6$), we know that all classes in $A^k(\mathcal{M}_g)$ are tautological, and hence by Looijenga ("On the tautological ring of $\mathcal{M}_g$") we have $A^{k}(\mathcal{M}_g)=0$ for $k\ge g-1$. It is also shown along the way to some of these calculations that $A^{k}(\mathcal{M}_{g,n})=0$ for certain values of $k$ when $g,n$ are small. For arbitrary $n$, Keel has computed the full Chow ring of $\overline{M}_{0,n}$; I think it follows that all Chow groups of $M_{0,n}$ vanish.
Additionally, if there is a good summary in the literature of what is known for singular cohomology too, I would love to know where to find it.
For $\mathcal M_{1,n}$ I would think the singular cohomology is basically known in terms of modular forms, but maybe not if the spectral sequence is tricky.
Can you clarify what you mean by "the spectral sequence"?
I guess I mean iteratively applying the Leray spectral sequence to the fibrations $\mathcal M_{1,n} \to \mathcal M_{1,n-1}$, but there are a few different geometric strategies one could try.
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2025-03-21T14:48:29.646233
| 2020-01-13T19:40:43 |
350364
|
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Stack Exchange
|
Does $\mathbb{Q}$ embed into a finitely generated solvable group?
Does $\mathbb{Q}$ embed into a finitely generated solvable group?
I've checked that $\mathbb{Q}$ is not a subgroup of any finitely generated metabelian group. I don't know how to show this (or whether it is true) for solvable groups of higher step.
Tiny note: no one-relator group contains $\mathbb{Q}$ as a subgroup, a result due to B. B. Newman, but this does not stray far at all from what you already have in the metabelian case.
Yes, it's due to Ph. Hall. It embeds into a f.g. 3-step solvable group.
Let $s:\mathbf{Z}\to\mathbf{Q}^*$ be a map (thought as an bi-infinite word) such that every finite sequence of nonzero rational numbers occurs as subword. Define two automorphisms $u,v$ of $\mathbf{Q}^{(\mathbf{Z})}$ (vector space over $\mathbf{Q}$ with basis $(e_n)_{n\in\mathbf{Z}}$) as follows: $u$ is the shift $e_n\mapsto e_{n+1}$, and $v$ is the diagonal map $e_n\mapsto s(n)e_n$. Since $u^nvu^{-n}$ is also diagonal for every $n$ it commutes with $v$, and hence the pair $\langle u,v\rangle$ generates a quotient of the wreath product $\mathbf{Z}\wr\mathbf{Z}$ (actually a copy of it). Moreover, as $\mathbf{Z}[\langle u,v\rangle$]-module, $\mathbf{Q}^{(\mathbf{Z})}$ is readily seen to be simple. Hence $\langle u,v\rangle\ltimes \mathbf{Q}^{(\mathbf{Z})}$ is generated by 3 elements, contains a copy of $\mathbf{Q}$ and is 3-step solvable. (Reference of roughly the same construction: Ph. Hall, On the finiteness of certain soluble groups, Proc. London Math. Soc. (3) 9 (1959).)
It was also proved by Neumann-Neumann that every countable $k$-step solvable group embeds into a finitely generated $(k+2)$-step solvable group (reference: B. H. Neumann, H. Neumann. Embedding Theorems for Groups, J. London Math. Soc. 34 (1959), 465-479).
Edit: Ph. Hall earlier (Finiteness conditions for soluble groups, Proc. London Math. Soc. (3) 4 (1954)) checked that there exists a 3-step solvable finitely generated group, whose center is free abelian of infinite rank (such a group can be viewed as group of upper triangular $3\times 3$ matrices over $\mathbf{Z}[t^{\pm 1}]$). It is straightforward that every abelian group can be embedded into (the center of) a quotient of the latter group (I guess he was aware of this, and I think I remember that Neumann-Neumann claim to generalize Hall's result but don't have their paper right now).
Thank you! Do you happen to have the name on hand for the Neumann-Neumann paper?
|
2025-03-21T14:48:29.646423
| 2020-01-13T21:31:21 |
350368
|
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"GH from MO",
"H A Helfgott",
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"url": "https://mathoverflow.net/questions/350368"
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|
Stack Exchange
|
Divisor problem: find the fallacy!
The following approach to the divisor problem (that is, the problem of estimating $D(x) = \sum_{n\leq x} d(n)$, where $d(n)$ is the number of divisors of $n$; more precisely, we are meant to bound the error term $\Delta(x)$ in $D(x) = x \log x + x (2 \gamma - 1) + \Delta(x)$) gives a result that is too good, and must hence be mistaken. However, I can't see where the error is. Can you?
We will start from the following result of Wigert's (Thm 7.15 in Titchmarsh's Theory of the Zeta Function): for $z\to 0$ with $|\arg z|\leq \alpha$, $\alpha < \pi/2$,
$$\sum_{n=1}^\infty d(n) e^{-n z} = \frac{\gamma - \log z}{z} + \frac{1}{4} + O(|z|).$$
(In fact, Wigert gave a more precise expression - a series development $\sum_{n=0}^{N-1} b_n z^{2 n + 1} + O(|z|^{2 N})$ instead of $O(|z|)$).
We will proceed as follows. Let $g(x)$ be a $C^2$ (or smooth) function equal to $e^x$ for $0\leq x\leq 1$, equal to $0$ for $x>1+\epsilon$, and going down from $e$ to $0$ as $x$ goes from $1$ to $1+\epsilon$. (What $g(x)$ equals for $x<0$ is immaterial.) Clearly we can bound
$$\sum_{n=1}^x d(n) \leq D_g(x) := \sum_{n=1}^{\infty} d(n) e^{-n/x} g(n/x).$$
We can get a lower bound in exactly the same way by changing the scale by a factor of $(1+\epsilon)$; the gap between the two estimates (to be obtained in what follows) should be small (namely, $O(\epsilon x \log x)$).
Now $\widehat{g}(t)$ is a nice function, with $\widehat{g}(t) = O(\min(1,1/t))$ for every $t$, and $\widehat{g}(t)$ decaying very rapidly for $|t|>10/\epsilon$, say. We should be safely able to invert summation and integration:
$$D_g(x) = \int_{-\infty}^\infty \widehat{g}(t) \sum_{n=1}^\infty d(n) e^{-n/x} e(t n/x) dt.$$
By Wigert's result with $z = (1-2\pi i t)/x$,
$$ \sum_{n=1}^\infty d(n) e^{n/x} e(t n/x) = \frac{x}{1 - 2 \pi i t} (\gamma + \log x - \log(1-2\pi i t)) + \frac{1}{4} + O\left(\frac{1+|t|}{x}\right).$$
Using our bounds on $\widehat{g}(t)$, we obtain, from right to left:
An error term of size $O(1/\epsilon x)$,
A term $$\int_{-\infty}^\infty \frac{\widehat{g}(t)}{4} dt = g(0) = 1,$$
The main terms: the larger one is
$$x (\log x + \gamma) \int_{-\infty}^\infty \frac{\widehat{g}(t)}{1 - 2 \pi i t} dt,$$ which, by Plancherel and the fact that the Fourier transform of $1/(1-2\pi i t)$ is the Heaviside function times $e^{-x}$, equals
$$x (\log x + \gamma) \int_0^\infty g(x) e^{-x} dx =
x ( \log x + \gamma) \cdot (1+O(\epsilon)).$$
The term $$- x \int_{-\infty}^\infty \widehat{g}(t) \frac{\log(1-2\pi i t)}{1 - 2\pi i t} dt$$ is a little harder to estimate (the Fourier transform of $h(t) = \log(1-2\pi i t)/(1-2\pi i t)$ is not trivial to figure out, and diverges at the origin -- logarithmically it would seem) but the following detour seems to work: by Plancherel,
$$\int_{-\infty}^\infty \widehat{g}(t) \frac{\log(1-2\pi i t)}{1 - 2\pi i t} dt = \int_0^\infty g(x) \widehat{h}(x) dx = \int_0^1 e^x \widehat{h}(x) dx + \int_1^{1+\epsilon} (g(x)-1) \widehat{h}(x) dx.$$
By Plancherel again,
$$\int_0^1 e^x \widehat{h}(x) dx = \int_{-\infty}^\infty \frac{e^{1-2\pi i t} - 1}{(1-2\pi i t)^2} \log(1-2\pi i t) dt,$$
and this integral would seem to be $1-\gamma$ (confession: I integrated numerically). Clearly $$\int_1^{1+\epsilon} (g(x)-1) \widehat{h}(x) dx = O(\epsilon).$$
Thus we end up obtaining $$D_g(x) = x (\log x + 2 \gamma - 1) \cdot (1+ O(\epsilon)) + 1 + O\left(\frac{1}{\epsilon x}\right),$$
and thus
$$D(x) = x (\log x + 2 \gamma - 1) \cdot (1+ O(\epsilon)) + 1 + O\left(\frac{1}{\epsilon x}\right).$$
Then we could set $\epsilon=1/x$, and obtain $$D(x) = x (\log x + 2 \gamma - 1) + O(\log x),$$ which is insane, and provably wrong: it is known that the error term cannot be that small.
Where is the error?
Mustn't $\epsilon$ be independent on $x$?
Not really - though one thing that I can see is that the quality of the bound from Wigert may depend on $\alpha$, and we should take $\alpha = \pi/2 - \epsilon/10$. That may be the key - but, beyond that, do you see a reason why $\epsilon$ has to be independent of $x$?
From scanning the proof of the Wigert bound, I think the dependence on $\alpha$ is something like $O(\lvert z\rvert+(\pi/2-\alpha)^{-1}\lvert z\rvert^2)$. A quick look-over suggests that this means your final bound should actually be $O(1/\epsilon^2x)$, so take $\epsilon=x^{-2/3-\epsilon}$ and a final bound of $O(x^{1/3+\epsilon})$, which looks more right, as it matches the Voronoi bound.
If it is really $O(|z|+(\pi/2-\alpha)^{-1} |z|^2)$, how is the final bound not $O(\epsilon x \log x) + O(\epsilon^3/x^2)$, which, for $\epsilon = x^{-3/4}$, would yield the manifestly insane bound $O(x^{1/4} \log x)$?
I think the proof in Titchmarsh yields $O(c|z|)$, where $c=\int_{-\infty}^\infty|\Gamma(-1+it)\zeta^2(-1+it)|e^{|t|(\alpha-\pi/2)},dt$. Standard bounds give that $c\asymp(\pi/2-\alpha)^{-5/2}$.
Well, my previous comment needs a small fix as $\Gamma(-1+it)$ blows up at $t=0$. Restrict the integral to $t>1$, say.
It would be clearer to put an index on the different quantities denoted by $\epsilon$.
I'm getting the same as @GH for $\sigma=-1$. More generally, when the line of integration is shifted to $\sigma = - r$, I get a term $O(\epsilon^{-(3/2+r)} |z|^r)$,(in addition to a polynomial on $z$ whose coefficients are indepdendent of $\epsilon$) and hence ultimately an error term $O(\epsilon^{-3/2-2r}/x^r)$. Adding it to the error term $O(\epsilon x \log x)$ (and to the polynomial on $z$, we see that, in the limit as $r\to \infty$, the exponent of $x$ tends to $1/2$, and thus we obtain $O_\epsilon(x^{1/2+\epsilon})$.
... which of course is fine, but also disappointing. I was hoping that the argument, once fixed, would match the Voronoi bound. Perhaps that's still doable by being a little cleverer. Bedtime for me now.
Is there really nothing in the literature with a better error dependence than Wigert's bound?
Well, there's Voronoi, as in section 4.5 of Iwaniec-Kowalski. Obviously (say I to myself).
I just double-checked. If you use Voronoi instead of Wigert, you get a final result in the order of $x^{1/3}$. That's nice, as it matches... Voronoi :).
Ah, you don't need Voronoi at all. Modularity (facepalm).
|
2025-03-21T14:48:29.646815
| 2020-01-13T21:53:44 |
350370
|
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"Libli",
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|
Stack Exchange
|
Characteristic polynomial of an $8 \times 8$ symmetric matrix with indeterminate entries related to octonionic multiplication
I consider $1,i,j,k,l,m,n,o$ the standard basis of the (complexified if you like) octonions ($\mathbb{O}$ for the octonions).
Let $a = x_1.1 +\ldots + x_8.o$, $b = x_9.1+ \ldots + x_{16}.o$ and $c = x_{17}.1+ \ldots +x_{24}.o$, where $x_1, \ldots, x_{24}$ are indeterminates over the base field (say $\mathbb{C}$).
I denote by $L_a$ the $8 \times 8$ matrix which represents the left multiplication by $a$ in $\mathbb{O} \simeq \mathbb{C}^8$ and $R_a$ the $8 \times 8$ matrix which represents the right multiplication by $a$. Similar notations for $b$ and $c$. I would like to compute the characteristic polynomial of the symmetric matrix:
$$ S = \dfrac{1}{2}(R_a L_b L_c + {}^{t}(R_a L_b L_c)),$$
where ${}^{t} X$ is the transpose of $X$.
I have tried with Macaulay2 and this computation seems to be far beyond the possibilities my machine (which is supposed to be a quite powerful portable workstation).
A simple reforrmulation of the eigenvalue problem in a well-chosen basis (namely let $\mathbb{H}$ be the quaternionic subalgebra generated by $b$ and $c$, split $\mathbb{O}$ as $\mathbb{H} \bigoplus \mathbb{H}.e$, where $e$ is orthogonal to $\mathbb{H}$ and take a basis adapted to this decomposition) shows that:
$$ (T - \mathrm{Re}((bc)\overline{a}))^4 \ \textrm{divides} \ det(S- T.id),$$
where $\mathrm{Re}(z)$ is the real part of $z \in \mathbb{O}$.
I put $f(T) = \dfrac{det(S- T.id)}{(T - \mathrm{Re}((bc)\overline{a})^4}$. A vast number of computations over finite fields and specializing the $x_i$ to random values suggests that $f(T)$ is in fact a square, say $f(T) = g(T)^2$, where $g$ is a quadratic polynomial in $T$.
I would like to get a closed expression of $g(T)$. May it be a clean formula involving $a,b$ and $c$ or a dirty "in coordinates" polynomial. I would really appreciate any suggestion. I would also be interested in a theoretical argument which shows that $f(T)$ is indeed a square.
Thanks a lot!
Let $a,b,c\in\mathbb{O}$ be octonions and consider the linear map $L:\mathbb{O}\to\mathbb{O}$ defined by
$$
L(x) = (b(cx))a = R_aL_bL_c(x).
$$
One desires a formula for the characteristic polynomial of $S$, the symmetric part of $L$, i.e.,
$$
S(x) = \tfrac12\bigl(R_aL_bL_c + {}^t(R_aL_bL_c) \bigr).
$$
(I note that the OP seems to have inadvertently omitted the factor of $\tfrac12$ in the formula for $S$; this becomes apparent when one compares the claimed formula for the characteristic polynomial of $S$ when $a = b = c = 1$.)
This is equivalent to knowing the symmetric functions of the eigenvalues of the quadratic form
$$
Q(x) = L(x)\cdot x = (b(cx))a\cdot x
$$
relative to the quadratic form $Q_0(x) = x\cdot x$.
First, note that, if any of $a$, $b$ or $c$ vanishes, then, of course,
$L$ and $Q$ vanish identically, and all of the eigenvalues of $S$ are equal to zero. Thus, we can assume that none of $a$, $b$, or $c$ vanishes. Then, dividing by
$|abc|\not=0$, we can assume that $|a| = |b| = |c| = 1$.
In this case, since $L$, being a product of elements of $\mathrm{SO}(8)$, belongs to $\mathrm{SO}(8)$, it follows that $L$ is conjugate in $\mathrm{SO}(8)$ to an element of the maximal torus ${\mathrm{SO}(2)}^4$, i.e., a blocked diagonal matrix
where the diagonal elements are the $2$-by-$2$ rotation matrices $R(\theta_i)$ for $1\le i\le 4$. The matrix $S$ (the symmetric part of $L$) is then diagonal with double eigenvalues $\cos(\theta_i)$, and hence the characteristic polynomial of $S$ is
$$
p(t) = (t-\cos\theta_1)^2(t-\cos\theta_2)^2(t-\cos\theta_3)^2(t-\cos\theta_4)^2.
$$
When $p(t)= t^8 + r_1\,t^7 + r_2\,t^6 + \cdots + r_8$, we have $p(t) = q(t)^2$ where
$$
q(t) = t^4 + \tfrac12\,r_1\,t^3 + \tfrac18(4r_2-{r_1}^2)\,t^2 + \cdots
$$
(I leave it to the interested reader to work out the formulae for the $t$ and constant coefficients of $q$ as polynomials in $r_1,r_2,r_3,r_4$).
Now, by a theorem of Dickson, $b$ and $c$ lie in a quaternion subalgebra $\mathbb{A}\subset\mathbb{O}$. Also, we can write $a = \cos\theta\,a_0 + \sin\theta\, u$ where $a_0\in\mathbb{A}$ is a unit vector, $u\in\mathbb{A}^\perp$ is a unit octonion, and $0\le \theta\le \tfrac12\pi$. [This expression for $a$ will be unique relative to $\mathbb{A}$ if $0<\theta<\tfrac12\pi$.]
In what follows, it will be useful to remember that elements of $\mathbb{O}$ satisfy $xy\cdot z = x\cdot z\bar y = y\cdot \bar x z$. Recalling that $\mathrm{Re}(x) = x\cdot\mathbf{1}$ (where $\mathbf{1}\in\mathbb{O}$ is the multiplicative unit), these identities imply that $\mathrm{Re}(xy) = \mathrm{Re}(yx)$ and that $\mathrm{Re}\bigl(a(bc)\bigr) = \mathrm{Re}\bigl((ab)c\bigr)$, so that $\mathrm{Re}(abc)$ is unambiguous, even though $\mathbb{O}$ is not associative. We also have
$$
\mathrm{Re}(abc) = \mathrm{Re}(bca) = \mathrm{Re}(cab)
= \mathrm{Re}(\bar c\,\bar b\,\bar a),
$$
but note that $\mathrm{Re}(abc)\not=\mathrm{Re}(bac)$ in general.
While $\mathbb{O} = \mathbb{A}\oplus \mathbb{A} u$
(note the orthogonal direct sum) is not associative,
we have the product formula of Cayley and Dickson:
$$
(a+b\,u)(c+d\,u) = \bigl( ac - \bar d b\bigr) + (da + b\bar c)\,u
$$
for all $a,b,c,d\in\mathbb{A}$.
Writing $x\in\mathbb{O}$ as $x = x_0 + x_1\,u$ where $x_i\in\mathbb{A}$
and using the Cayley-Dickson formula several times, we have
$$
\begin{aligned}
Q(x)
&= (b(cx))a\cdot x = b(cx)\cdot x\bar a \\
&= b(c(x_0+x_1\,u))\cdot (x_0+x_1\,u)(\cos\theta\,\overline{a_0} -\sin\theta\,u)\\
&= \bigl(bcx_0 + (x_1cb)\,u\bigr)\cdot\bigl((\cos\theta\,x_0\overline{a_0}+\sin\theta\,x_1) + (\cos\theta\,x_1a_0-\sin\theta\,x_0)\,u\bigr)\\
&=\cos\theta\,bcx_0\cdot x_0\overline{a_0}
+ \sin\theta(bcx_0\cdot x_1-x_1cb\cdot x_0) + \cos\theta\,x_1cb\cdot x_1a_0\\
&= \cos\theta\,bcx_0\cdot x_0\overline{a_0}
+ \sin\theta(bcx_0-x_0\bar b\bar c)\cdot x_1
+ \cos\theta\,x_1\cdot x_1a_0\bar b\bar c\\
&= \cos\theta\,bcx_0a_0\cdot x_0
+ \sin\theta(bcx_0-x_0\bar b\bar c)\cdot x_1
+ x_1{\cdot}x_1\,\mathrm{Re}(\cos\theta\,a_0\bar b\bar c)
\end{aligned}
$$
(Note that I have used $x_1\cdot x_1a_0\bar b\bar c = \overline{x_1}x_1\cdot a_0\bar b\bar c = |x_1|^2 \mathrm{Re}(a_0\bar b\bar c)$.
This expression can be further simplified. Since $b$ and $c$ are unit vectors in
$\mathbb{A}$, we can write $bc = w$, which implies that $b = w\bar c$ and hence that $\bar b\bar c = c\bar w\bar c$. Making the substitution $x_0 = y_0\bar c$
and $x_1 = y_1\bar c$ then yields $Q_0(x) = |x_0|^2+|x_1|^2 = |y_0|^2+|y_1|^2$ while setting $v = \bar c a_0 c$ yields
$$
\begin{aligned}
Q(x) &= \cos\theta\,wy_0{\bar c}a_0 \cdot y_0\bar c
+ \sin\theta\,(wy_0-y_0\bar w)\cdot y_1
+ |y_1|^2\,\mathrm{Re}(\cos\theta\,a_0\bar b\bar c)\\
& = \cos\theta\,wy_0v \cdot y_0
+ \sin\theta\,(wy_0-y_0\bar w)\cdot y_1
+ |y_1|^2\,\mathrm{Re}(\cos\theta\,v\bar w)
\end{aligned}
$$
Using this reduced form, it is straightforward to compute that the characteristic polynomial of $S$ is
$$
\bigl(t-\mathrm{Re}(\cos\theta\,v\bar w)\bigr)^4
\bigl(t^2-2\cos\theta\,\mathrm{Re}(v)\mathrm{Re}(w)\,t
+ \cos^2\theta\,\mathrm{Re}(v)^2w\bar w - |\mathrm{Im}(w)|^2\bigr)^2,
$$
which can also be written as
$$
\bigl(t-\mathrm{Re}(\cos\theta\,v\bar w)\bigr)^4
\bigl((t-\cos\theta\,\mathrm{Re}(v)\mathrm{Re}(w))^2
- |\mathrm{Im}(w)|^2(1-\cos^2\theta\,\mathrm{Re}(v)^2)\bigr)^2,
$$
so that its roots are $t = \mathrm{Re}(\cos\theta\,v\bar w)$ with multiplicity $4$
and
$$
t = \cos\theta\,\mathrm{Re}(v)\mathrm{Re}(w)
\pm |\mathrm{Im}(w)|\bigl(1-\cos^2\theta\,\mathrm{Re}(v)^2)^{1/2},
$$
each with multiplicity $2$.
Finally, tracing back through the definitions and normalizations, we see that the characteristic polynomial of $S$ can be written in the form
$$
\bigl(t-\mathrm{Re}(a\bar b\bar c)\bigr)^4
\bigl(\bigl(t-\mathrm{Re}(a)\,\mathrm{Re}(bc)\bigr)^2-\mathrm{Im}(a)^2\,\mathrm{Im}(bc)^2\bigr)^2.
$$
Note on a computation: Since the OP asked, here is a bit of
detail of the computation of the characteristic polynomial.
We can choose a basis of $\mathbb{A}\simeq\mathbb{H}$ as $(\bf{1},\bf{i},\bf{j},\bf{k})$
in such a way that $w = w_0\,{\bf{1}}+w_1\,\bf{i}$ while $v = v_0\,{\bf{1}}+v_1\,{\bf{i}}+v_2\,{\bf{j}}+v_3\,{\bf{k}}$. Now write out $y_0$
in the orthonormal basis $(\bf{1},\bf{i},\bf{j},\bf{k})$ and $y_1$
in the orthonormal basis $(-\bf{i},\bf{1},\bf{j},\bf{k})$.
Set $\lambda_1 = c(v_{{0}}w_{{0}}{-}v_{{1}}w_{{1}})$ and $\lambda_2=c(v_{{0}}w_{{0}}{+}v_{{1}}w_{{1}})$ where $c = \cos\theta$ and $s=\sin\theta$
(to make the matrix below more readable). Then the matrix of
the quadratic form $Q$ in this basis is
$$
\left(
\begin{array}{cccccccc}
\lambda_1&0&-cw_{{1}}v_{{3}}&cw_{{1}}v_{{2}}&sw_{{1}}&0&0&0\\
0&\lambda_1&-cw_{{1}}v_{{2}}&-cw_{{1}}v_{{3}}&0&sw_{{1}}&0&0\\
-cw_{{1}}v_{{3}}&-cw_{{1}}v_{{2}}&\lambda_2&0&0&0&0&0\\
cw_{{1}}v_{{2}}&-cw_{{1}}v_{{3}}&0&\lambda_2&0&0&0&0\\
sw_{{1}}&0&0&0&\lambda_2&0&0&0\\
0&sw_{{1}}&0&0&0&\lambda_2&0&0\\
0&0&0&0&0&0&\lambda_2&0\\
0&0&0&0&0&0&0&\lambda_2
\end{array}
\right)
$$
Now tell MAPLE to compute the characteristic polynomial of this matrix and tell it to factor the result. (I confess that it's a bit surprising that $\lambda_2$ turns out to be a root of multiplicity $4$ instead of just $2$. I don't have a 'theoretical' understanding of this. Looking at the matrix, you can see that by choosing the basis of $\mathbb{A}$ a bit more carefully, you could arrange that $v_2=0$, and that makes it clear that the characteristic polynomial will be a square, but we knew that already.)
Great, I feel somehow comforted that you found out the characteristic polynomial with Maple. Computing directly the matrix of $S$ wia the splitting $\mathbb{O} = \mathbb{H} \oplus \mathbb{H}.e$, I get the following: $\dfrac{1}{2} \begin{pmatrix} R_{a_0} L_b L_c + {}^{t}(R_{a_0} L_b L_c) & P \ {}^{t} P & 2\mathrm{Re}(bc \overline{a}) \end{pmatrix}$, where $P = R_{a_1} {}^{t} L_c {}^{t} L_b - R_{a_1}{}^{t} L_c R_{b}$ and $a = a_0 + a_1.e$.
Up to a change of coordinates, this seems to be exactly the same matrix as for the quadratic form $Q$ you exhibited. I was more or less able to understand that $2 \mathrm{Re}(bc\overline{a})$ is an eigenvalue of even multiplicity (though $4$ was alos a big surprise for me). But I couldn't get my hands on the missing part of the characterisitic polynomial. I thought that perhaps there is theoretical way to do it. Looking at the first version of your answer, I had the impression you knew one.
Anyway, even if you found it with Maple, this is very interesting for me and a nice answer. Thanks!
@Libli: If you do go ahead and choose the basis of $\mathbb{A}$ so that $v_2=0$, then the matrix I wrote down above clearly can be decomposed as the orthogonal direct sum of two copies of the same $3$-by-$3$ symmetric matrix plus a $2$-by-$2$ matrix that is a multiple of the identity. Since one can easily compute the characteristic polynomial of the $3$-by-$3$ matrix by hand, the use of MAPLE was not actually necessary. Thus, one could do the computation by hand, but I felt it was better to state that I used MAPLE, since I didn't see the by-hand argument until I used MAPLE to get the result.
|
2025-03-21T14:48:29.647446
| 2020-01-14T00:49:34 |
350373
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/350373"
}
|
Stack Exchange
|
Tight sublinear estimates for a triple partial binomial summation
Is there tight estimates for the following logarithmic summation ($\gamma,\gamma'\in(0,1)$ and $\mu,\mu'>0$)
$$\log_2\Bigg(\sum_{t=\frac{n^{}}2-n^\gamma\sqrt{\mu\ln n}}^{\frac{n^{}}2+n^\gamma\sqrt{\mu\ln n}}\quad\sum_{\ell=\frac{n^{}}2-n^\gamma\sqrt{\mu\ln n}}^{\frac{n^{}}2+n^\gamma\sqrt{\mu\ln n}}\quad\sum_{k=\frac t2-n^{\gamma'}\sqrt{\mu'\ln n}}^{\frac t2+n^{\gamma'}\sqrt{\mu'\ln n}}\binom{\ell}{k}\binom{n-\ell}{t-k}\Bigg)?$$
I am hoping it might give $n - f(n)$ bound where $f(n)$ is $\omega(\ln n)$ or at least $\Omega(1)$ for diagonal case of $\gamma=\gamma'=\frac12$ at some $\mu,\mu'>0$.
Relevant problem is in Tight estimates for binomial summation (and perhaps an upper bound possible might be $$\log_2\Bigg(\underbrace{n^{2\gamma}\mu(\ln n)}_{\substack{\mbox{coming from}\\\mbox{outer two}\\\mbox{summations}}}\binom{n}{n/4}\Bigg)<n H(\frac14+\epsilon)<0.82n$$ at any $\epsilon>0$ since $t/2=n/4$ might be close to the value that yields the bound (together with loose upper bound from Vandermonde's identity for inner sum)).
what does at least $\gamma=\gamma'=1/2$ mean? $\min{\gamma,\gamma'}\geq 1/2$?
No just for the diagonal case at 1/2.
This conjecture does not hold even in the case $\gamma=\gamma'=1/2$.
Indeed, consider the values of $\ell,t,k$ such that
$$|\ell-n/2|\ll\sqrt n,\ |t-n/2|\ll\sqrt n,\ |k-t/2|\ll\sqrt n,$$
where $A\ll B$ or, equivalently, $B\gg A$ means that $|A|\le CB$ for some universal real constant $C>0$; as usual, $A\asymp B$ means that $A\ll B\ll A$.
By what was shown in this answer,
$$\binom\ell k\asymp\frac{2^\ell}{\sqrt\ell}\,e^{-u^2/2},$$
where
$$u:=\frac{k-\ell/2}{\sqrt{\ell/2}}\ll1, $$
so that
$$\binom\ell k\asymp\frac{2^\ell}{\sqrt n}.$$
Similarly,
$$\binom{n-\ell}{t-k}\asymp\frac{2^{n-\ell}}{\sqrt n},$$
whence
$$\binom\ell k\binom{n-\ell}{t-k}\asymp\frac{2^n}n.$$
Hence (in the case $\gamma=\gamma'=1/2$), your big triple sum is
$\gg \dfrac{2^n}n\,n^{3/2}$
and hence
$$\log_2(\text{the triple sum})-n\gg\ln n.$$
What is the correct asymptotic?
@VS. : I think the correct asymptotics for your $f(n)$ is $\asymp-\ln n$ if $\gamma=\gamma'=1/2$; otherwise, it will much depend on $\gamma$ and $\gamma'$. However, please ask any additional questions in separate posts.
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2025-03-21T14:48:29.647609
| 2020-01-14T03:49:53 |
350376
|
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|
Stack Exchange
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Long non-deformable hyperbolic fillings
The title and question have been edited in light of Ian Agol's comment. The previous question was stated in terms of the wrong notion of length to discuss deformations:
What is the longest slope $\gamma$ in the complement of Dehn surgery space of a cusped hyperbolic 3-manifold $M$? Here Dehn surgery space is the space of fillings such that the hyperbolic structure on the filling $M(\gamma)$ can be realized as a deformation of the original $M$.
To keep things focused, consider only fillings of one cusped manifolds.
This question is related to Ken Baker’s question:
Hyperbolic exceptional fillings of cusped hyperbolic 3-manifolds
However, Ken’s question is interested in the total number of slopes in this complement. This question is focusing on the longest such slope measured using normalized length, where the normalized length of $\gamma$, $\mathcal{L}(\gamma)$ is as defined in:
Hodgson, Craig D.; Kerckhoff, Steven P., Universal bounds for hyperbolic Dehn surgery, Ann. Math. (2) 162, No. 1, 367-421 (2005). ZBL1087.57011.
Namely $\mathcal{L}(\gamma)=length(\gamma)/\sqrt{Area(\partial T)}$. Here, $length(\gamma)$ the translation length of $\gamma$ in a cusp neighborhood and $Area(\partial T)$ is the area of that torus in cusp neighborhood.
Of course, it is possible that this question as stated does not have a realizable answer, because there is no longest slope.
Here is a more carefully stated version:
What is the largest $\mathcal{L}_{max}$ such that there exist a family of slopes $\gamma_i$ in (1-cusped hyperbolic) manifolds $M_i$ such that $$\lim_{i \to \infty} \mathcal{L}(\gamma_i) = \mathcal{L}_{max}$$
and each $M_i(\gamma_i)$ is a hyperbolic manifold such that the hyperbolic structure cannot be realized as a deformation of the hyperbolic structure of $M_i$?
Hodgson and Kerckhoff give an upper bound of $\mathcal{L}_{max}\leq C\approx 7.515$.
To give context how normalized length affects length, the (3,3,3) pretzel knot has slope of length 6 yielding a torus filling. However, the normalized length of this slope is $\mathcal{L}=\frac{6}{\sqrt{A}}\approx 1.91673$, $A=\frac{8\sqrt{3}}{(1+3\sqrt{57})^{1/3}}+\sqrt{3}(1+3\sqrt{57})^{1/3}$.
This example appears in
Adams, Colin; Bennett, Hanna; Davis, Christopher; Jennings, Michael; Kloke, Jennifer; Perry, Nicholas; Schoenfeld, Eric, Totally geodesic Seifert surfaces in hyperbolic knot and link complements. II, J. Differ. Geom. 79, No. 1, 1-23 (2008). ZBL1158.57004.
and as a filling of the manifold defined by Figure 6:
Agol, Ian, Bounds on exceptional Dehn filling, Geom. Topol. 4, 431-449 (2000). ZBL0959.57009.
However, the cusp area appears to grow for (n,n,n)-pretzels, so I chose the (3,3,3)-pretzel to seemingly get the longest slope (in terms of normalized length) in this family.
It's not clear to me that $L\leq 2\pi$ if you require there to be a deformation. You should look at Hodgson-Kerckhoff's universal bounds on Dehn surgery space.
Thanks Ian. I missed that The 2-$\pi$ Theorem just guarantees a negatively curved metric can be extend to across the surgery solid torus, but this extension is not claimed to have constant curvature nor be a globally defined deformation of the original metric.
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2025-03-21T14:48:29.647827
| 2020-01-14T06:44:35 |
350381
|
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|
Stack Exchange
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Properties of Mod $\ell^m$ Galois representation associated to modular form
(Sorry for my poor english..)
Let $F(z)\in S_{2k}(SL_2(\mathbb{Z})$) be a newform and $\ell$ be a prime larger than $3$. Let $K$ be a some number field and $v$ be a prime of $K$ over $\ell$. Let $K_v$ be a $v$-adic completion of $K$ and $O_{K,v}$ be a ring of integers of $K_v$. Let $v=(\pi)$. I already know that Serre and Deligne proved that there exists a Galois representation $\rho_{F,v}$ such that
\begin{equation}
\rho_{F,v} : Gal(\overline{\mathbb{Q}}/\mathbb{Q})\to GL_2(\mathcal{O}_{K,v})
\end{equation}
with for primes $p\neq \ell$,
\begin{equation}
\text{Tr}(\text{Frob}_p)=\lambda_p ,\quad \det(\text{Frob}_p)=p^{2k-1}
\end{equation}
where $\text{Frob}_p$ is a Frobenius element.
Let $\rho_{F,v}^{m}$ be a reduction of $\rho_{F,v}$ modulo $(\pi^{m})$. In other words,
\begin{equation}
\rho_{F,v}^{m} : Gal(\overline{\mathbb{Q}}/\mathbb{Q})\to GL_2(\mathcal{O}_{K,v}/(\pi^m))
\end{equation}
with for primes $p\neq \ell$,
\begin{equation}
\text{Tr}(\text{Frob}_p)\equiv \lambda_p, \quad \det(\text{Frob}_p)\equiv p^{2k-1} \pmod{(\pi^m)}.
\end{equation}
Does there exist $N\in \mathbb{N}$ such that for $p_1\equiv p_2 \pmod{N}$ implies that $\rho_{F,v}^m(\text{Frob}_{p_1})\equiv \rho_{F,v}^m(\text{Frob}_{p_2})$?
Write $L$ for the finite Galois extension of $\mathbb Q$ with Galois group $G_{\mathbb Q}/\operatorname{Ker}\rho_{F,v}^m$. Then $\rho_{F,v}^m(\operatorname{Frob}_p)$ is the identity in $\operatorname{GL}_2(\mathcal O_{K,v}/\pi^m)$ if and only if $\operatorname{Frob}_p$ is the identity in $\operatorname{Gal}(L/\mathbb Q)$ if and only if $p$ splits completely in $L/\mathbb Q$.
If an $N$ as in your question exists, then choosing it large enough so as to eliminate at most finitely many exceptions, we can arrange that there exists $a$ such that $p$ splits completely in $L/\mathbb Q$ if $p\equiv a$ modulo $N$. This implies that the extension $L/\mathbb Q$ is abelian, or equivalently that $\rho_{F,v}^m$ has abelian image. For $m$ large enough, this is never true for $\rho_{F,v}^m$. So the $N$ in your question does not exist.
The statement that $L/\mathbb Q$ must be abelian if $p\equiv a$ modulo $N$ implies that $p$ splits completely in $L/\mathbb Q$ is closely related to classical questions in class field theory, but is not quite stated in the usual form (for instance, it is much easier to prove that $L/\mathbb Q$ must be abelian if $p\equiv a$ modulo $N$ is equivalent to the statement that $p$ splits completely in $L/\mathbb Q$). Nevertheless, a complete proof can be found for instance at the following MO answer.
Why do congruence conditions not suffice to determine which primes split in non-abelian extensions?
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2025-03-21T14:48:29.648113
| 2020-01-14T06:47:53 |
350382
|
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|
Stack Exchange
|
Concise expression for a specific holomorphic map $f:D\times D\longrightarrow \mathbb{B}^{2}$
Let $\mathbb{B}^{2}\subset \mathbb{C}^{2}$ be the unit ball and $D=\{z\in\mathbb{C}:|z|<1\}$. $f=(f_{1},f_{2}):D\times D\longrightarrow \mathbb{B}^{2}$ is a holomorphic map satisfying $\{det df=0\}\subset{\{z_{1}z_{2}=0\}}$. We can assume that $f(0,0)=(0,0)$. So can we choose another appropriate coordinate system centered at $(0,0)$ so that $f$ has a concise expression under this coordinate system? For example, can we find new coordinates $z_{1}, z_{2}$ such that $f_{1}(z_{1},z_{2})=z_{1}^{m}$, $f_{1}(z_{1},z_{2})=z_{1}^{s}z_{2}^{t}$, where $m$, $s$ and $t$ are integers? This problem comes from our consideration on local expressions of high-dimensional complex hyperbolic metrics with singularities along a simple normal crossing divisor.
Unfortunately, it doesn't look like the condition $\det(df)\subset(z_1z_2=0)$ is strong enough. Consider, for example, the map
$$f:(z_1,z_2)\to (z_1, z_2^3+z_1z_2).$$
Clearly, the differential of the map is vanishing along the smooth curve $3z_2^2+z_1=0$ (and so one can change the coordinates in the source to make this curve the axis $z_1=0$). However, the image of this curve $\det(df)=0$ has parameterisation $(-3t^2, -2t^3)$. I.e, the curve $f(\{\det(df)=0\})$ has a cusp at $(0,0)$. So it is rather clear, that there is no change of coordinates that would make the image smooth.
|
2025-03-21T14:48:29.648232
| 2020-01-14T08:02:10 |
350384
|
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|
Stack Exchange
|
Anderson localization for Bernoulli potentials on half-line
Anderson localisation for (discrete) Schrödinger operators with Bernoulli potentials on $l^2(\mathbb{Z})$ was proven in
https://link.springer.com/article/10.1007/BF01210702
I am wondering if there is a similar reference for the corresponding result on $l^2(\mathbb{N})$, assuming it is still true?
All the technics to prove Anderson localisation on $l^2(\mathbb{Z})$ also work for $l^2(\mathbb{N})$. It even easier as you only have to prove the exponential decay in one direction.
I guess everything one should know about 1d Anderson localisation are in the book of Carmona and Lacroix "Spectral Theory of Random Schrodinger Operators"
I agree that at least most of the techniques generalise, but is there a reference for this particular case? I think it is much cleaner to reference a paper than argue a "small" generalisation.
I'm not sure it's that easy. What is straightforward (given the usual machinery) is localization for a typical realization for almost all boundary conditions at the left endpoint (and there's always an exceptional set of bc's with no point spectrum). This is basically the same technical problem that's responsible for Bernoulli models being harder than those with an ac component in the distribution.
|
2025-03-21T14:48:29.648338
| 2020-01-14T08:36:17 |
350385
|
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|
Stack Exchange
|
Surjectivity of colimit maps for topological spaces
From this post and to (co)completness of the category Top of topological spaces and continuous functions we know that for any diagram $B_i$ and an object $A$ in Top, there are natural maps of sets:$\newcommand{\colim}{\operatorname{colim}}$
$\colim C(A,B_i) \to C(A, \colim B_i)$
$\colim C(B_i,A) \to C(\lim B_i, A)$
For any topological spaces, equip $C(X,Y)$ with the compact-open topology; then $C(X,Y)$ is itself a topological space. When is the map of $1$ (resp. $2$) a surjective(or at-least has dense range) continuous map?
Related: This post on the dual question and this post on naturality of the aforementioned maps.
I guess it rarely happens, but for example, if the diagram is filtered and A is compact, 1 can be homeomorphism.
Oh, would you happen to have a reference to this? I'd like to read-up on it.
No, I don't, but the idea would be, if A is compact, and we have a filtered colimit, and, let's say, the colimit of "good" injective maps (whatever that means), for any map from A to the colimit, the image has to be contained in finite stage. And, if A is not a compact, you can take $B_i$'s to be compact approximation, in which case one sees clearly that the map can't be surjective.
|
2025-03-21T14:48:29.648458
| 2020-01-14T08:56:54 |
350387
|
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Stack Exchange
|
1-connected infinity groupoids, groupoids and 1-connected spaces
I am exploring a bit the world of groupoids. What I have in mind is that infinity groupoids correspond to spaces. So my first question is the following:
Consider the model category $\infty-Grpd$ of infinity groupoids, meant as the full subcategory of Kan complexes in $sSet$, with Quillen model structure. On the other side, consider the model category $Top$ with the Quillen model structure.
Is it true that the two model categories are Quillen equivalent under Sing and geometric realization?
In case of 1, the Quillen equivalence restricts to 1-connected groupoids and yields 1-connected spaces. This is kind of tautological because I define homotopy groups of a grpd as the homotopy groups of its geometric realization. The real question is: does the functors (Nerve, homotopy category) induce a Quillen equivalence between classical groupoids (with the model structure induced by the classical one on categories) and 1-connected infinity groupoids ?
It would help maybe if you could provide references, many thanks!
Kan complexes do not have finite (co)limits, so cannot have a model structure. Simplicial sets do have a model structure, whose fibrant objects are Kan complexes. Furthermore, |−| and Sing form a Quillen equivalence. This answers (1).
If 1-connected really means 1-truncated (e.g., 2-coskeletal simplicial sets), then indeed the fundamental groupoid functor and the nerve functor induce a Quillen equivalence between 2-coskeletal simplicial sets and groupoids. This answers (2).
I am curious though as to what source on model categories does not mention the Quillen equivalence between sSet and Top.
Yes,Indeed your first comment answers me. I knew about the classical Quillen equivalence, and it seemed straightforward to me that it restricts to Kan complexes (as sing has image in Kan complexes ). But as you pointed out the problem is that they do not have colimits. 1 connected here means that the homotopy groups of the geometric realization are trivial for n greater than 2. Not sure if the 2-coskeletization of a 1-comnected sset is weakly equivalent to himself, which would conclude.
@AndreaMarino That's what's commonly known as 1-truncated, not 1-connected (that'd be $\pi_0=\pi_1=\ast$)
@AndreaMarino: What Denis said. The 2-coskeletization of a 1-truncated simplicial set is indeed weakly equivalent to the original simplicial set.
Ah well, thank you :)
Regarding part 1, I was probably thinking the following. Even if $\infty-Grpd$ is not a model category, it still make sense to consider its derived category to be the essential image of $\infty-Grpd \to sSet \to D(sSet)$. Geometric realization induces a functor $D(\infty-Grpd) \to D(Top)$ which is fully faithful (as it is a restriction), and it also essentially sujective, because $D(Sing) : D(Top) \to D(sSet)$ factors through $D(\infty-Grpd)$, so that $X \simeq |Sing(X)|$ in the derived category. Can one make sense of $D(\infty-Grpd)$ in a more elegant way?
|
2025-03-21T14:48:29.648668
| 2020-01-14T10:09:21 |
350391
|
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|
Stack Exchange
|
Restriction of non-metrizable topology to dense subset is non-metrizable
Let $(X,\tau)$ be a non-metrizable topological space which is not first-countable and let $\emptyset \neq Y\subset X$ be a proper dense subset. Is it possible for $(Y,\tau_Y)$ (where $\tau_Y$ is the relativisation of $\tau$ to $Y$) be metrizable?
Since second countability wouldn't work I edited your question to require first countability.
You're right, that does indeed make sense.
Did you mean to ask whether it's possible for $Y$ to be metrizable? As you've phrased it, the answer to your question seems obvious: of course it's possible for $Y$ to be non-metrizable, e.g., if we take $Y = X$.
@WillBrian Indeed, I had made an English hickup.
Another class of examples is to let $Y$ be any non-compact metrizable space, and $X = \beta Y$ its Stone-Čech compactification.
@NateEldredge Thanks this is a very nice example.
Yes: the order topology on $\omega_1+1$ (the first uncountable successor ordinal) is an example. It is not first countable, because it's "top" point $\omega_1$ has no countable neighborhood base. But the set of all isolated points of this space is dense in it, and the relative topology on this set is discrete (hence metrizable).
On the other hand, let me point out that if $x \in X$ has no countable neighborhood base, and if $X$ is $T_3$, then $x$ will still fail to have a countable neighborhood base in any dense $Y \subseteq X$ with $x \in Y$. Therefore there are plenty of spaces $X$ for which the answer to your question would be negative. For example, no dense subset of $[0,1]^{\kappa}$ is metrizable (for uncountable $\kappa$), because every point of this space witnesses the fact that the space is not first-countable, and this will continue to be true in any dense subspace.
Do you have a reference to the second point?
I needed a version of this lemma one time in an old paper: see lemma 4.5 in https://wrbrian.files.wordpress.com/2012/01/neight2.pdf. But I'm sure I wasn't the first person to notice this fact -- if you're willing to look hard enough, I wouldn't be surprised if you'd find it in Engelking.
Also, for a slightly smaller example, the order topology on $\omega_1$ is not metrizable.
The one-point compactification of an uncountable discrete space is an even simpler example. More generally, every non-first countable weakly compact subspace of a Banach space is an example.
Good point @SantiSpadaro.
@FrançoisG.Dorais: The OP asks that the original space not be first countable. While $\omega_1$ is not metrizable, it is first countable.
|
2025-03-21T14:48:29.648860
| 2020-01-14T11:14:46 |
350392
|
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Stack Exchange
|
Domain of $(-\Delta)^\alpha$, $\alpha \in (0,1)$
Is there any simple way to characterize explicitly the domain of fractional powers for a given operator? For example, the domain of Dirichlet Laplacian on a bounded nice domain $\Omega \subset \mathbb{R}^n$:
$$(-\Delta)^\alpha, \quad \alpha \in (0,1).$$
I know that there is a characterization using interpolation theory, so one can find $H^{2\alpha}(\Omega)$ or $H^{2\alpha}_0(\Omega)$. But I'm asking if there is a systematic and simple way to do that, so we can use it for more complicated operators, for example matrix valued operators.
Any reference which gives the proof of the above characterization or a related topic will be very helpful.
I do not think there is a simple answer, but Chapter 6: Domains, Uniqueness and the Cauchy Problem in [Martínez, Sanz, The Theory of Fractional Powers of Operators, Elsevier, 2001] might be a good starting point.
Thank you! I will check it.
It depends a lot on the situation. For example, you can look at this paper https://www.sciencedirect.com/science/article/pii/S0022123618301046 where the operator consider is a delta-like perturbation of the Laplacian.
The problem can be, in general, very hard. A famous example is the Kato's conjecture (Wikipedia link) concerning the domain of the square root of certain elliptic operators, which was solved only in 2001, almost a half-century after Kato posed the question.
|
2025-03-21T14:48:29.648982
| 2020-01-14T11:59:13 |
350394
|
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Stack Exchange
|
Alexandrov's rigidity in higher dimensions
If $\Phi_1,\Phi_2$ are convex polyhedra in $\mathbb{R}^3$ such that the sets of outer normals to facets coincide, but $\Phi_1$ is not a translate of $\Phi_2$, then there exist two corresponding facets $F_1,F_2$ (with the same outer normal) such that one of them is a translate of a proper subset of another.
This is A. D. Alexandrov's theorem, which generalizes the theorem of H. Minkowski which assumes that the areas of corresponding facets are always equal.
If I remember well, then in dimensions greater than 3 this is no longer true (while Minkowski theorem holds true in any dimension.) The request is a reference to counterexamples.
Here is a counterexample in dimension four.
Consider positive numbers $x_1,x_2,x_3,x_4\in\Bbb R$ with $x_1<x_2$ and $x_3<x_4$ and construct the two 4-orthotopes (cartesian products of intervals)
\begin{align}
O_1:=[0,x_1]\times [0,x_2]\times[0,x_3]\times[0,x_4]\\
O_2:=[0,x_2]\times [0,x_1]\times[0,x_4]\times[0,x_3]
\end{align}
A pair of parallel facets is defined by a 3-element subset $I=\{i_1,i_2,i_3\}\subset \{1,2,3,4\}$:
\begin{align}
F_1&:=[0,x_{i_1}]\times[0,x_{i_2}]\times[0,x_{i_3}] \subset O_1 \\
F_2&:=[0,x_{\sigma(i_1)}]\times[0,x_{\sigma(i_2)}]\times[0,x_{\sigma(i_3)}] \subset O_2
\end{align}
where $\sigma$ is the permutation $(12)(34)$. (strictly spoken, the inclusions are wrong, but I hope the idea is clear).
For $F_1$ to be (parallel to) a subset of $F_2$ it must hold $(*)\,x_i\le x_{\sigma(i)}$ for all $i\in I$.
But each 3-element subset $I\subset\{1,2,3,4\}$ contains either $\{1,2\}$ or $\{3,4\}$, and so $(*)$ cannot be satisfied.
The easiest example is probably $(x_1,x_2,x_3,x_4)=(1,2,1,2)$.
Great, and I had a look at Alexandrov's book and found the same example there.
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2025-03-21T14:48:29.649118
| 2020-01-14T12:35:24 |
350398
|
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|
Stack Exchange
|
Annihilator of idempotent elements
Let $R $ be a commutative ring with 1 and $s $ a nonnilpotent element of $R $ such that for each idempotent $e $ of $R $ there exists a natural number $n_e $ such that either $s^{n_e}e=0$ or $s^{n_e}(1-e)=0$. Can we deduce that for each idempotent $e$ of $R$ either $s^me=0$ or $s^m(1-e)=0$ for a fix natural number $m $?
No. Consider the subring $R\subseteq\prod_{n\geq 1}\mathbb R[x]/(x^n)$ consisting of sequences $(a_n)$ whose constant terms converge to some limit. Let $s=(x,x,x,\dots)$. Then $s$ is nonnilpotent, since $s^n$ has nonzero coefficient in $\mathbb R[x]/(x^{n+1})$.
Now every idempotent $e$ in $R$ is equal to $0$ or $1$ in each coordinate, and is eventually constant. Replacing $e$ with $1-e$ if necessary, assume $e=(e_n)$ satisfies $e_n=0$ for $n>N$. Then $s^Ne=0$.
However, for $e=(1,\dots,1,0,0,\dots)$ with $n$ ones we have $s^{n-1}e\neq 0,s^{n-1}(1-e)\neq 0$, so we cannot pick one exponent for all idempotents.
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2025-03-21T14:48:29.649201
| 2020-01-16T05:24:52 |
350532
|
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|
Stack Exchange
|
Conditions on a ring which make the kernel of certain matrices over the power series ring trivial
Let $R$ be a (unital) commutative ring and let $\varphi$ be an $n\times n$ matrix over $R[[x]]$. Are there any conditions on $R$ which guarantee that $\ker(\varphi)\cap R[x]^n=0$ implies $\ker(\varphi)=0$? Certainly some conditions are necessary. By the example at the very end of this paper, there are $1\times 1$ matrices for which the implication fails. However, by Theorem 5 of the same paper, this issue is fixed if $R$ is assumed to be Noetherian (in fact, the theorem gives the much stronger implication $\ker(\varphi)\cap R=0\Rightarrow\ker(\varphi)=0$). This does not give an answer for the case that $\varphi$ is not $1\times 1$ though.
I am not sure whether the implication holds even in the case that $R$ is a field. I am particularly interested in the case $R=k[[x_1,\ldots,x_m]]$ for a field, $k$, but there will be no hope on that front if it does not work when $R$ is a field. If the implication does not hold under any nondegenerate conditions, will the answer change if we assume the entries of $\varphi$ are all polynomials?
Edit: As per Mohan's suggestion, in the case that $R$ is Noetherian and the entries of $\varphi$ are polynomials, the implication is true by flatness. Written more explicitly, suppose the entries of $\varphi$ are polynomials and $\ker(\varphi)\cap R[x]^n=0$. Then $\varphi:R[[x]]^n\to R[[x]]^n$ is the unique extension of the $R[x]$-linear map/matrix $\psi:R[x]^n\to R[x]^n$ whose entries are the same as $\varphi$. However, since $R$ (and hence $R[x]$) is Noetherian, $R[[x]]$ is a flat $R[x]$-module. This implies that the induced map $$\psi\otimes id_{R[[x]]}:R[x]^n\otimes_{R[x]} R[[x]]\to R[x]^n\otimes_{R[x]} R[[x]]$$ is injective. But since $R[[x]]^n\cong R[x]^n\otimes_{R[x]} R[[x]]$, we have $\psi\otimes id_{R[[x]]}=\varphi$.
I am still curious to know what happens when the entries of $\varphi$ are not polynomials, but this is sufficient for what I am currently working on.
Of course, if $\varphi$ consists of polynomials, then faithful flatness will do what you want.
@Mohan Of course! I don't know why flatness didn't cross my mind. I'll edit my question to expand on your suggestion. Thank you!
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2025-03-21T14:48:29.649367
| 2020-01-15T03:52:19 |
350536
|
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|
Stack Exchange
|
Seeing what gets Harvey Friedman's "tangible incompleteness" principles into large cardinal territory
I'm trying to wrap my head around some of Harvey Friedman's recent, unpublished work on his tangible incompleteness project, and I'm trying to see the link between his "tangible statements" (propositions about subsets of $\mathbb{Q}[0,n]^k:=\{x\in\mathbb{Q}\;|\;0\leq x\leq n\}^k$ for $n,k\in\mathbb{N}$) and the large cardinals in the subtle/$k$-SRP range.
An example of the sort of statement he considers is the following (from this document):
Definition 1. $S\subseteq\mathbb{Q}[0,n]^k$ is an emulator of $E\subseteq\mathbb{Q}[0,n]^k$ if each element of $S^2$ is order equivalent to some member of $E^2$ (considered as subsets of $\mathbb{Q}[0,n]^{2k}$), where $x,y\in\mathbb{Q}[0,n]^k$ are said to be order equivalent if for any $1\leq i,j\leq k$, $x_i<x_j\leftrightarrow y_i<y_j$.
Definition 2. $A\subseteq\mathbb{Q}[0,n]^k$ is stable if and only if for all $p<1$, $\langle p,1,\ldots,k-1\rangle\in A\leftrightarrow\langle p,2,\ldots,k\rangle\in A$.
Example statement (called "MES" by Friedman). For all $k$, and all finite $E\subseteq\mathbb{Q}[0,k]^k$, there is a stable, $\subseteq$-maximal emulator of $E$.
Apparently MES implies $\mathrm{Con}(\mathsf{SRP})$ (where $\mathsf{SRP}$ is $\mathsf{ZFC}$+"there is a $k$-SRP cardinal" as an axiom scheme in $k$), so somehow there is a lot of strength happening in stable maximal emulators. And the relation to $k$-SRP cardinals, specifically, makes me think that emulators must be hiding a Ramsey condition under the hood, but I haven't been able to spot it yet.
How can I see what is happening with stable maximal emulators of finite subsets of $\mathbb{Q}[0,n]^k$ ($n,k>2$) that their guaranteed existence implies the consistency of the existence of large cardinals? [Edited to reflect JDH's correction in the comments.]
(For a reference on $k$-SRP cardinals, see Friedman, Harvey. Subtle cardinals and linear orderings. (English summary)
Ann. Pure Appl. Logic 107 (2001), no. 1-3, 1–34.)
This might be more appropriate at mathoverflow - this is some seriously technical material (and I think people capable of answering it might be more likely to see it there).
@NoahSchweber - Noted. What's good etiquette on moving it over?
Not sure, honestly - crossposting immediately is generally frowned on, but in this case I would imagine deleting here and asking there should be fine (since it hasn't been up here for long enough for a move to be construed as wasting someone's time). But that's just my opinion; certainly there's no harm asking it here and moving to MO after a bit if no answers show up.
"...their guaranteed existence implies the existence of large cardinals" is not right, because no arithmetic statement can imply the existence of large cardinals. You mean to imply the consistency of the existence of the large cardinals.
@JoelDavidHamkins - I suspected I was misstating that; this is still deeply new territory for me. But that does help me start to see where I'm going wrong...
Your best bet might be to contact Friedman directly by email. One question I have is whether you've worked through any of Friedman's earlier work in this area? Are you just trying to get a feeling for Friedman's general methodology for proving statements of this kind? Or do you have a good grasp of his earlier work and are just having trouble with $k$-SRP cardinals specifically?
@TimothyChow - I'm still working my way back through his previous papers, so that's a work in progress. Right now the earliest I'm passably familiar with is Subtle Cardinals and Linear Orderings, and I'm working on a couple of papers from '96, but I've by no means exhausted his bibliography in between. Perhaps I will email him, though.
@TimothyChow - I suppose I didn't totally answer your question: I'm most interested in the line of thought behind formulating statements of this kind, less with k-SRP cardinals specifically. There's clearly something that leads Friedman to choose stable maximal emulators (or various other similar objects) as good objects for bumping up consistency strength, but I have no clue what the guiding idea is.
@MaliceVidrine I'm currently working through Finite Functions and the Necessary Use of Large Cardinals, which Friedman published around the same time. Broadly speaking the argument there involves using a strong finitary Ramsey's theorem and compactness to build a structure that has the ability to code some basic set theory and has a special cofinal indiscernible sequence of elements. This indiscernible sequence then gives you enough reflection to show that the interpreted model of basic set theory satisfies ZFC. I would guess that the story in this paper is broadly similar.
|
2025-03-21T14:48:29.649684
| 2020-01-16T08:36:04 |
350542
|
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|
Stack Exchange
|
Can $H_{\omega_1}$ and $H_{\omega_2}$ be in bi-interpretation synonymy?
This question concerns the possibility of the bi-interpretation synonymy of the structure
$\langle
H_{\omega_1},\in\rangle$, consisting of the hereditarily countable sets, and the structure $\langle H_{\omega_2},\in\rangle$, consisting of sets of hereditary
size at most $\aleph_1$. These are both models of
Zermelo-Fraenkel set theory $\text{ZFC}^-$, without the power set
axiom. The structure $\langle
H_{\omega_1},\in\rangle$ is of course a definable submodel of $\langle H_{\omega_2},\in\rangle$, which provides one direction of interpretation.
Depending on the set-theoretic background, it is also possible that
there is a converse interpretation in the other direction. Indeed,
in my recent paper with Afredo Roque Freire,
Joel David Hamkins and Alfredo Roque Freire, Bi-interpretation in weak set theories, blog post, arXiv:2001.05262
we prove that it is relatively consistent
with ZFC that the structures $\langle H_{\omega_1},\in\rangle$ and
$\langle H_{\omega_2},\in\rangle$ are bi-interpretable (see theorem
17). This is true, specifically, in the Solovay-Tennenbaum model,
obtained by c.c.c. forcing over $L$ to achieve
$\text{MA}+\neg\text{CH}$. What is needed is (i) $H_{\omega_1}$ has
a definable almost disjoint $\omega_1$-sequence of reals; and (ii)
every subset $A\subseteq\omega_1$ is coded by a real via
almost-disjoint coding with respect to that sequence. The basic
idea is that objects in $H_{\omega_2}$ are coded by a well-founded
relation on $\omega_1$, which is in turn coded by a real, and so in
$H_{\omega_1}$ we can define the class $U$ of reals coding a set in
this manner and an equivalence relation on those reals $x\equiv y$
for when they code the same set, and a relation $\bar\in$ on those
reals, so that $\langle H_{\omega_2},\in\rangle$ is isomorphic to
the quotient structure $\langle U,\bar\in\rangle/\equiv$.
The argument seems to use the equivalence relation in a fundamental
manner, and the question I have about this here is whether one can
omit the need for the equivalence relation. This ultimately amounts
to the following, which is question 18 in the paper:
Question. Is it relatively consistent with ZFC that there is a
binary relation $\bar\in$ that is definable in $H_{\omega_1}$ such
that $$\langle H_{\omega_1},\bar\in\rangle\cong \langle
H_{\omega_2},\in\rangle?$$
This is what it would mean for these structures to form a
bi-interpretation synonymy.
For a positive answer, it would be enough to show the consistency with ZFC of the existence of a definable global well-order in $H_{\omega_1}$, together with the almost-disjoint coding of hypothesis (ii) above. Is that possible?
Apart from $H_{\omega_1}$ and $H_{\omega_2}$ specifically, a
related question we have is whether one can prove any instance of
interpretation in a model of $\text{ZFC}^-$ that requires the
quotient by an equivalence relation.
Question. Is there a structure that is interpretable in a model
of $\text{ZFC}^-$, but only by means of a nontrivial equivalence
relation?
This is question 9 in the paper.
A theorem of Harrington (in his paper "Long projective wellorders") says MA + $\neg \text{CH}$ is consistent with a projective wellorder of the reals, hence a wellorder of $H_{\omega_1}$ definable over $H_{\omega_1}$. Is that what you're looking for in your first question?
That would do it! Kindly post an answer with the reference.
It seems to be this: https://doi.org/10.1016/0003-4843(77)90004-3. And the main theorem B is just what you say.
A theorem of Harrington (Theorem B of his paper "Long projective wellorders") says $\text{MA} + \neg\text{CH}$ is consistent with a projective wellorder of the reals, hence a wellorder of $H_{\omega_1}$
definable over $H_{\omega_1}$. Since $\text{MA}_{\omega_1}$ implies a strong form of almost disjoint coding (i.e., any almost disjoint family $\langle A_\alpha : \alpha < \omega_1\rangle$ almost-disjoint-codes any subset of $\omega_1$ relative to some real), this gives an answer to your first question.
For the second question, assuming $\text{AD}^{L(\mathbb R)}$ and $\delta^1_2 = \omega_2$, there is a prewellorder of $\omega^\omega$ with rank $\omega_2$ definable over $H_{\omega_1}$. On the other hand, there is no wellorder of a subset of $H_{\omega_1}$ in ordertype $\omega_2$ definable over $H_{\omega_1}$, because, I believe, there is no injection from $\omega_2$ into $H_{\omega_1}$ in $L(\mathbb R)$. (I am looking for a reference for this, but I am pretty sure it is true; maybe a ZF/AD expert can help me.) So $(\omega_2,<)$ is interpretable in $(H_{\omega_1},\in)$ via the prewellorder, but not via a trivial equivalence relation.
UPDATE: Here is a proof that $\omega_2$ does not inject into $H_{\omega_1}$ under AD + $V = L(\mathbb R)$. In fact, we use the weaker hypothesis of $\text{AD}^+ + V = L(P(\mathbb R))$. We cite the following theorems:
Theorem 1 (Woodin, $\text{AD}^+ + V= L(P(\mathbb R))$). For any set $X$, either $X$ can be wellordered or there is an injective function from $\mathbb R$ to $X$.
This is Theorem 1.4 of Caicedo-Ketchersid's A trichotomy theorem in natural models of $\text{AD}^+$. I think the $\text{AD}^+ + V= L(P(\mathbb R))$ version is due to Woodin, but it's not clear from the paper.
Theorem 2 (AD) There is no prewellorder of $\mathbb R$ whose proper initial segments are countable.
This follows from the Kuratowski-Ulam Theorem since every set of reals has the Baire property. See Moschovakis's Descriptive Set Theory Exercise 5A.10 for a more general result.
Using these we obtain the following corollary:
Corollary ($\text{AD}^+ + V= L(P(\mathbb R)$). Suppose $X$ is a set and there is no injection from $\omega_1$ to $X$. Then there is no injection from $\omega_1$ to the set $P_{\aleph_1}(X)$ of all countable subsets of $X$.
Proof. Suppose towards a contradiction that $f : \omega_1\to P_{\aleph_1}(X)$ is an injection. It follows that $A = \bigcup_{\alpha < \omega_1} f(\alpha)$ is an uncountable subset of $X$. Hence $A$ is not wellorderable. Moreover, there is a prewellorder of $A$ all of whose initial segments are countable: let $\varphi(x) = \min \{\alpha : x\in f(\alpha)\}$, and set $x \leq y$ if $\varphi(x)\leq \varphi(y)$. Since $A$ is not wellorderable, Theorem 1 yields an injection $g : \mathbb R \to A$. We define a prewellorder of $\mathbb R$ by setting $u \prec w$ if $g(u) < g(w)$. Its initial segments are countable since $g$ is injective, and this contradicts Theorem 2. THis proves the corollary.
The corollary allows us to build a rank hierarchy for $H_{\omega_1}$. Define $R_0 = \emptyset$, $R_{\alpha+1} = P_{\aleph_1}(R_\alpha)$, and for limit ordinals $\gamma$, $R_\gamma = \bigcup_{\alpha < \gamma} R_\alpha$. There are now two key observations:
(1) A trivial induction using the corollary shows that for any ordinal $\alpha < \omega_1$, there is no injection from $\omega_1$ to $R_\alpha$.
(2) $R_{\omega_1} = H_{\omega_1}$. On the one hand, clearly every element of $R_{\omega_1}$ is hereditarily countable so $R_{\omega_1}\subseteq H_{\omega_1}$. On the other hand, $P_{\aleph_1}(R_{\omega_1}) = R_{\omega_1}$, and so by $\in$-induction one can prove $H_{\omega_1}\subseteq R_{\omega_1}$.
Now suppose towards a contradiction that $f : \omega_2\to H_{\omega_1}$ is an injection. Since $\omega_2$ is not the union of $\omega_1$ countable sets, we can find an ordinal $\alpha < \omega_1$, such that $f[A]\subseteq R_\alpha$ for an uncountable set $A\subseteq \omega_2$. But this obviously yields an injection from $\omega_1$ to $H_\alpha$, contradicting the previous paragraph.
This is enough for a positive answer to the first question, because one uses the well order to pick representatives, and then uses Shroeder-Bernstein to make a synonymy. Thanks!
And thank you very much for the second part of your answer. I am definitely interested if you could track down a suitable reference for the claim about whether $\omega_2$ injects into $H_{\omega_1}$.
I figured out a proof (although this was probably known in the 70s).
Ah, yes! Cute. ${}$
:) I've been wondering how to prove the Corollary for a long time. I think it works with $\omega_1$ and $\aleph_1$ replaced by any cardinal $\kappa < \Theta$.
|
2025-03-21T14:48:29.650530
| 2020-01-16T10:31:40 |
350545
|
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|
Stack Exchange
|
Lucas Theorem on a Sector for a special class of polynomials
I had asked a question sometime back Lucas Theorem on a sector. But it seems that for the below subclass of polynomials it may be true. Here is my claim;
Let $P(z)$ be a polynomial of degree $n,$ with real and non-negative coefficients having all its zeros in an open sector region $S=\{re^{i\theta}|\theta\in(0,\phi), r\geq 0\}$ where $\phi\in[0, 2\pi]$ then $p'(z)$ has all its zeros in $S.$
If the sector angle $\phi$ is less than or equal to $\pi,$ (convex region) we are done as it follows directly from Lucas Theorem. We need to prove it for the case $\phi>\pi$ (non-convex region). Any hint or suggestion?
|
2025-03-21T14:48:29.650615
| 2020-01-16T11:08:03 |
350547
|
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|
Stack Exchange
|
Closed cobounded additive submonoid of $\mathbb{R}^n$
Let $M$ be a closed additive submonoid of $\mathbb{R}^n$ with $n\geq1$. Suppose also that there exists $r>0$ such that every ball of radius $r$ intersects $M$. I wonder if we can obtain more information on $M$. For example, if $n=1$, it is easy to see that $M$ has to be a subgroup of $\mathbb{R}$, thanks to the classical characterization of such subgroups. Is it still true when $n\geq2$?
"Closed" in the topological sense or is it a terminology about monoids?
Yes in the topological sense!
The geometric property you mentioned is known as "cobounded" (I added it in the title, you can revert if you like)
Yes, it has to be a subgroup. Fix $v\in M$. We need to prove that $-v\in M$. It is sufficient to find an element of $M$ arbitrarily close to $-v$.
Choose $u_1,\ldots,u_n\in \mathbb{R}^n$ so that $v,u_1,\ldots,u_n$ are the vertices of a regular simplex with center at the origin. Choose large $N$ and consider the points $w_i\in M$ such that $\|Nu_i-w_i\|\leqslant r$. We may choose $N$ so large that the coordinates of $v$ in the basis $\{w_1,\ldots,w_n\}$ are negative: $v=-\sum_{i=1}^n t_i w_i,0<t_i$ (this is because $w_i/N$ is close to $u_i$ for all $i$, and the coordinates of $v$ in the basis $\{u_1,\ldots,u_n\}$ are all equal to -1.)
Now by Kronecker approximation theorem we may find a positive integer $s$ so that $st_1,\ldots,st_n$ are almost integers, denote $st_i=k_i+\varepsilon_i$, where $k_i$ are non-negative integers and $\varepsilon_i$ are small. Then $$M\ni (s-1)v+\sum k_iw_i=-v-\sum \varepsilon_i w_i$$
is close to $v$.
The closed subgroups of $\mathbb{R}^n$ are direct sums of subspaces and lattices. And the non-empty ball condition reads simply as being full-rank.
|
2025-03-21T14:48:29.650750
| 2020-01-16T11:08:35 |
350548
|
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|
Stack Exchange
|
Approximation of multipliers by multipliers of a smaller set 2
This question is a refinement of my previous question.
Let $X$ be a compact metric space, and let $B$ be a bounded Banach Disk in $C(X)$ such that for every $x\in X$ there is $f\in B$ with $f(x)\ne 0$.
Let $M=\{u\in C(X),~ uf\in B,~\forall f\in B\}$ and let $N=\{u\in C(X),~ uf\in \overline{B},~\forall f\in \overline{B}\}$.
Since multiplication $(f,g)\to fg$ is a continuous operation on $C(X)$, it follows that $N$ is closed in $C(X)$ and $M\subset N$. Thus, $\overline{M}\subset N$.
Is it true that $N=\overline{M}$?
A Banach Disk in a vector space is a convex balanced set such that its Minkowski functional is a complete norm on its span.
|
2025-03-21T14:48:29.650826
| 2020-01-16T12:02:03 |
350551
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"url": "https://mathoverflow.net/questions/350551"
}
|
Stack Exchange
|
Extension Operator for $W^{1,\infty}(U,X)$
I am reading through some lectures on Sobolev spaces and the vector-valued (or Banach space valued) version of them. At this moment I am very interested in extension operators for the vector-valued Sobolev space $W^{1,\infty}(I,X)$ where $X$ is a Banach space and I is a compact interval. For $p\neq\infty$ there are many references for extension operator for $W^{1,p}(I,X)$, however for $p=\infty$ I can not find good results. My question is, is there an extension operator $E:W^{1,\infty}(I,X)\to W^{1,\infty}(\mathbb{R},X)$ under the general assumption that $X$ is a Banach space? What I thought was the following: it is known that $W^{1,\infty}(I,X)$ coincides with the space of Lipschitz functions. By the Kirszbraun theorem there exists a Lipschitz extension of such functions but one has to assume that $X$ is a Hilbert space and even then one is not ready because $W^{1,\infty}(\mathbb{R},X)$ is bigger than the space of Lipschitz functions on $\mathbb{R}$, isn't it? Can maybe someone assists with this question?
Say $I:=[a,b]$. We may extend $u$ simply putting it constant for $t\ge b$ and $t\le a$, that is $Eu(t):=u((t\vee a)\wedge b)$. This defines a norm-$1$ linear extensor $E$.
|
2025-03-21T14:48:29.650930
| 2020-01-16T12:06:08 |
350552
|
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"Armin Weiß",
"Geoff Robinson",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350552"
}
|
Stack Exchange
|
Normal subgroup of Fitting length i contained in i-th term of upper Fitting series?
Let $G$ be a finite solvable group of Fitting length $n$ with upper Fitting series
$1= U_0G \leq U_1G \leq \cdots \leq U_nG = G$.
Is it true that every normal subgroup of Fitting length $i$ is contained in $U_iG$? (for $i=1$ it is true because the Fitting subgroup contains all nilpotent normal subgroups)
Yes, it is true. Use induction on the Fitting length of $N$: ,you have done the case of Fitting length $1$. Now note that $N/F(N)$ is isomorphic to a normal subgroup of $G/F(G)$.
I see. Thanks! Do you know a good standard reference for these kind of things?
Almost any book which treats finite solvable groups in any depth would contain such results (though the proofs might be relegated to the exercises in some cases).
You have already done the case, when $i = 1$.
Now suppose it i true for $i$. Let's prove it for $i+1$. Suppose $H \triangleleft G$ has Fitting length $i+1$. That means $\frac{H}{U_iH}$ is nilpotent. Now let's define $\phi_i$ as a natural homomorphism between $G$ and $\frac{G}{U_iG}$. Then $\phi_i(H)$ is a nilpotent normal subgroup of $\frac{G}{U_iG}$ because $U_iH$ lies in $U_iG$ as a normal subgroup (characteristic subgroup of a normal subgroup is always normal) of Fitting length $i$ by induction step. That means, that $\phi_i(H) \triangleleft U_1\frac{G}{U_iG}$, which results in $H \triangleleft \phi_i^{-1}(U_1\frac{G}{U_iG})=U_{i+1}G$.
Thus, $\forall i \in \mathbb{N}$ every normal subgroup of $G$ of fitting length $i$ is contained in $U_iG$.
|
2025-03-21T14:48:29.651051
| 2020-01-16T12:30:04 |
350553
|
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"authors": [
"András Kovács",
"Mario",
"Max New",
"aws",
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"url": "https://mathoverflow.net/questions/350553"
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|
Stack Exchange
|
Is $\prod_{X : \mathcal{U}} (X \to X) \cong 1$ consistent with type theory?
Assume we work in some minimalistic version of Martin-Löf type theory. Does it break consistency to postulate that the function that selects the identity function has an inverse?
$$\prod_{X : \mathcal{U}} (X \to X) \cong 1.$$
From "Parametricity, automorphisms of the universe, and excluded middle", I understand that if it could be proved, excluded middle would not be consistent.
More generally, can we assume this form of Yoneda reduction for $F : \mathcal{U} \to \mathcal{U}$ without breaking consistency?
$$
\left(\prod_{X : \mathcal{U}} (A \to X) \to FX\right) \cong FA,
$$
Or this form of coYoneda?
$$
\left(\sum_{X : \mathcal{U}} (X \to A) \times FX \right) \cong FA,
$$
The same question, but with paths instead of functions, points me to work by Rijke and later by Escardó relating the J-elimination rule and the Yoneda lemma.
I think "admissible" is the wrong word here. You probably mean consistent?
Oh, that's totally true. Thank you.
$\prod_{X : \mathcal{U}} (X \to X) \cong 1$ is consistent. It follows from parametricity and function extensionality. Usual parametric models also support function extensionality. The simplest one which suffices would be the Fam model where every closed type is a set together with a family of sets over it, functions are predicate-preserving functions and the universe is the set of sets together with the family which maps each $A : \mathsf{Set}$ to $A \to \mathsf{Set}$. Here's a reference for a model which also works for our purpose but which is more complicated than necessary.
$(\prod_{X : \mathcal{U}} (A \to X) \to FX) \cong FA$ is provably false in plain MLTT. Let $F\,X := X \to \bot$ and $A := \bot$. Now the statement simplifies to $(\prod_{X : \mathcal{U}} X \to \bot) \cong \top$, which is evidently false. Same for coYoneda. Pick $A := \top$ and $F$ as before, and coYoneda simplifies to
$(\sum_{X : \mathcal{U}} X \to \bot) \cong \bot$. The problem is that $F$ is just an $\mathcal{U}\to \mathcal{U}$ function, and not a functor with respect to functions-as-morphisms in $\mathcal{U}$.
Presheaves over $\cdot \to \cdot$ doesn't quite work. It is true that for every closed term $\vdash M : \Pi_{X : U} X \to X$ in type theory, $\Pi_{X : U} M X = 1_X$ holds in the model, but this is weaker than the internal statement $\Pi_{M : \prod_{X : U} X \to X} M X = 1_X$. E.g. the internal statement implies the law of excluded middle is false, but the double negation of LEM does hold in presheaves over $\cdot \to \cdot$.
@aws I'll believe that, edited. Are functions and universes the crucial difference in the presheaf version?
I don't know enough about parametric models to say how they compare with the Hofmann-Streicher presheaf interpretation of extensional type theory in presheaves. The universe you described for $\mathsf{Fam}$ looks like it works out the same as the Streicher universe in $\mathsf{Set}^\to$. I would have guessed that exponentials were the same as well.
|
2025-03-21T14:48:29.651267
| 2020-01-16T12:34:02 |
350554
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350554"
}
|
Stack Exchange
|
For human proofs of two novel combinatorial identities
For $n=0,1,2,\ldots$, let us define the polynomial
$$S_n(x):=\sum_{k=0}^n\binom{x/2}k\binom{(x-1)/2}k\binom{-(x+1)/2}{n-k}\binom{-(x+2)/2}{n-k}.$$
Such polynomials occur in some series for $1/\pi$ discoveried by me in 2011, see Conjecture 4 of my paper List of conjectural series for powers of $\pi$ and other constants.
In 2011, I found the following two novel identities for the polynomials $S_n(x)$:
$$S_n(x)=\binom{-1/2}n\sum_{k=0}^n(-1)^k\binom{x}k^2\binom{-1-x}{n-k}\tag{1}$$
and
$$\begin{aligned}&\sum_{k=0}^n\binom nk(-1)^kS_k(x)
\\=&\sum_{k=0}^n\binom{x/2}k\binom{-(x+1)/2}k\binom{(x-1)/2}{n-k}\binom{-(x+2)/2}{n-k}.\end{aligned}\tag{2}$$
Note that $(1)$ implies the symmetric identity
$$\sum_{k=0}^n(-1)^k\binom xk^2\binom{-1-x}{n-k}=\sum_{k=0}^n(-1)^k\binom{-1-x}k^2\binom x{n-k},$$
which was proved without computer in my paper Supercongruences involving dual sequences, Finite Fields Appl. 46(2017), 179-216.
By the Zeilberger algorithm, if $u_n$ is the left-hand side or the right-hand side of $(1)$ then we have the recurrence relation:
\begin{align}4(n+2)^3u_{n+2}=&2(2n+3)(2n^2+6n+x^2+x+5)u_{n+1}
\\&-(n+1)(2n+1)(2n+3)u_n.\end{align}
Similarly, if $v_n$ denotes the left-hand side or the right-hand side of $(2)$, then we have the recurrence
\begin{align}4(n+2)^3v_{n+2}=&2(2n+3)(2n^2+6n-x^2-x+5)v_{n+1}
\\&-(n+1)(2n-2x+1)(2n+2x+3)v_n.\end{align}
So both $(1)$ and $(2)$ have proofs via a computer.
Question. How to provide human proofs of the identities $(1)$ and $(2)$?
Your comments are welcome!
|
2025-03-21T14:48:29.651378
| 2020-01-16T13:40:48 |
350556
|
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"Robert Israel",
"afshi7n",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350556"
}
|
Stack Exchange
|
Extreme points of an intersection of convex set with countably many linear spaces
Let $V$ be some `nice' vector space and let $T: V\to \mathbb{R}$ be a linear functional over $V$.
Define
\begin{align}
M= K \cap \bigcup_{i \in \mathbb{N} } \{ v \in V: T(v)=c_i \}
\end{align}
where $K$ is some compact and convex subset of $V$. Moreover, $K$ has at most $n$ extreme points.
That is, $M$ is an intersection of $K$ with countably many hyperplanes.
The question I have is, can we say something about the extreme points of $M$?
The general answer, I suspect, is that it is impossible to say something without extra assumptions on $T$. So, we would have to make some assumptions on $T$.
Some motivation: The following result can be shown when the intersection is finite.
Let $\tilde{M}=K \cap \bigcup_{i=1}^m \{ v \in V: T(v)=c_i \} $.
Then, one can show, with little restriction on $T$, that the extreme points of $ \tilde{M}$ can be represented as a convex combination of at most $m$ extreme pints of $K$.
The last paragraph is wrong. Consider the case $m=1$. $\tilde{M}$ is the intersection of $K$ with one hyperplane. In general this will not contain any extreme point of $K$, so its extreme points will not be convex combinations of $m=1$ extreme points of $K$.
What is true is that every extreme point of $M$ is an
extreme point of the intersection of $K$ with one hyperplane,
and this is a convex combination of two extreme points of $K$.
Namely, suppose $p = \sum_{i=1}^r t_i p_i$, $t_i \in (0,1)$, $\sum_i t_i = 1$, is a convex combination of $r > 2$ extreme points of $K$. Say $T(p) = c$. If any $T(p_j) = c$, then
$p$ is a convex combination of $p_j$ and $(p - t_j p_j)/(1-t_j)$ which are both in $M$, so not an extreme point. Otherwise some
$T(p_i) > c$ and some $< c$. Relabelling, suppose $T(p_1) > c$
and $T(p_2) < c$. Then
$$q = \frac{T(p_1) - c}{T(p_1)-T(p_2)} p_2 + \frac{c - T(p_2)}{T(p_1)- T(p_2)} p_1$$ is a nontrivial convex combination of $p_1$ and $p_2$ which is in $M$, and $p$ is a convex combination of this and some other member of $M$, thus not an extreme point.
where is it used that $r>2$? It seems like the same argument works for $r=2$..
If $r=2$ there's nothing to prove, as $p$ is already a convex combination of two extreme points of $K$.
Thanks! Yes, I understand that. I’m saying your argument seems to be generating a contradiction even for r=2 (as it doesn’t use the fact that r>2). But that would mean the argument is flawed—as you said the statement is trivially correct for r=2
|
2025-03-21T14:48:29.651554
| 2020-01-16T15:56:03 |
350562
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/350562"
}
|
Stack Exchange
|
Gaussian Property of the Renormalization Group
Let $\Lambda \subset \mathbb{Z}^{d}$ be a finite set and $\varphi = (\varphi_{x})_{x\in \Lambda} \in \mathbb{R}^{|\Lambda|}$. Let $F^{\Lambda}=F^{\Lambda}(\varphi)$ be a real-valued global function, because it depends on every entry of $\varphi$. The renormalization group setup goes like this: first, for a fixed $N$, we define $Z_{0}: \overbrace{\mathbb{R}^{|\Lambda|}\times\cdots\times\mathbb{R}^{|\Lambda|}}^{\mbox{$N$ times}} \to \mathbb{R}$ by $Z_{0}(\xi_{1},...,\xi_{N}) := F^{\Lambda}(\xi_{1}+\cdots+\xi_{N})$. Now, suppose $\mu_{C}$ is a Gaussian measure on $\mathbb{R}^{|\Lambda|}$ associated to the (positive-definite) matrix $C$, which has a decomposition $\sum_{j=1}^{N}C_{j}$, where each $C_{j}$ is again positive-definite and let $\mu_{j}$ be the Gaussian measure associated to $C_{j}$. Then, because $\xi_{j} \sim N(C_{j}) \Rightarrow \sum_{j=1}^{N}\xi_{j} \sim N(C)$, we have:
\begin{eqnarray}
\int d\mu_{C}(\varphi)F^{\Lambda}(\varphi) = \int d\mu_{N}(\xi_{N})\int d\mu_{N-1}(\xi_{N-1})\cdots \int d\mu_{1}(\xi_{1})Z_{0}(\xi_{1},...,\xi_{N}) \tag{1}\label{1}
\end{eqnarray}
Now, the left hand side of (\ref{1}) is well-defined provided $F^{\Lambda}$ is measurable and $\mu_{C}$-integrable. In this case, $Z_{0}$ is again measurable because it is the composition of $F^{\Lambda}$ with the continuous function $\mathbb{R}^{\Lambda}\times\cdots\times \mathbb{R}^{\Lambda} \to \mathbb{R}^{\Lambda}$ given by $(\xi_{1},...,\xi_{N}) \mapsto \xi_{1}+\cdots+\xi_{N}$.
My question is: Is the right hand side of (\ref{1}) well-defined? Is $Z_{0}$ automatically $\mu_{j}$-integrable (for each $j$) given that $F^{\Lambda}$ is measurable and $\mu_{C}$-integrable? Do I need more assumptions?
The following (between the horizontal lines) is a statement of what one may call the abstract change of variable theorem. It is taken from some exercise I gave my students a while ago:
Let $(X,\mathcal{M},\mu)$ be a measure space and let $(Y,\mathcal{N})$ be a measurable space.
Let $f:X\rightarrow Y$ be an $(\mathcal{M},\mathcal{N})$-measurable map.
For $B\in\mathcal{N}$ define
$$
f_{\ast}\mu(B)=\mu(f^{-1}(B))\ .
$$
a) Show that $f_{\ast}\mu$ is well defined and gives a measure on $(Y,\mathcal{N})$.
b) Let $\phi:Y\rightarrow\mathbb{R}$ be a nonnegative simple function.
Show that
$$
\int_X \phi\circ f\ {\rm d}\mu= \int_Y \phi\ {\rm d}(f_{\ast}\mu)
$$
c) Using the Monotone Convergence Theorem, prove that the last equality holds without the assumption of $\phi$ being a simple function.
d) Show that a non necessarily nonnegative measurable function $\phi$ from $Y$ to $\mathbb{R}$
is $f_{\ast}\mu$-integrable iff $\phi\circ f$ is $\mu$-integrable. In this case, show that the equality in b) still holds.
Here $f$ is the map $(\xi_1,\ldots,\xi_N)\mapsto \xi_1+\cdots+\xi_N$. The measure $\mu$ is the product measure
$$
\otimes_{j=1}^{N}d\mu_j(\xi_j)
$$
on $X=\mathbb{R}^{|\Lambda|}\times\cdots\times\mathbb{R}^{|\Lambda|}$, $N$ times.
Finally the push-forward measure is $\mu_C$. So you see that there is no need for extra hypotheses on $F^{\Lambda}$. Morover, being able to integrate successively over the $\xi_j$ is a consequence of Fubini's Theorem which, in particular, says that integrability over the product space implies the iterated integral is well defined and gives the same result as the integral over the product space.
|
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