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2025-03-21T14:48:30.306842
| 2020-04-15T13:54:27 |
357546
|
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"Mingzhou Liu",
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|
Stack Exchange
|
SDE conditional expectation
Let's suppose I have a bidimensional SDE of the form:
\begin{equation} \label{eq:system}
\begin{cases}
dX_t=b(t,X_t,Y_t)dt+\sigma(t,X_t,Y_t)dW_t^1 \\
X_0=x_0 \\
dY_t= B(t,X_t,Y_t)dt+C(t,X_t,Y_t)dW_t^2 \\
Y_0=y_0
\end{cases}
\end{equation}
where $W^1,W^2$ are Brownian motions.
Now take a time $s>0$, since $X_s,Y_s$ are random variables it should make sense to compute $\mathbb{E}[X_{s}|Y_s ]$.
My question are:
can I apply the tower property in this way?
$$\mathbb{E}[X_{s}|Y_s ]=\mathbb{E}[\mathbb{E}[X_{s}|(Y_\tau, \tau \leq s) ]|Y_s ]$$
What does it mean to compute $\mathbb{E}[X_{s}|Y_s ]$ when the process $X_s$ depends (by the sde) on the all history of $Y_\tau, \tau \leq s$?
Meaning I'm conditioning to the last value of $Y_s$ but what about the preceeding values?
You can use the tower property in this way, but unless you have a way of computing the inner conditional expectation (which involves a correlated-noise filtering problem and unless the coefficients of the SDEs are from a narrow class of functions, this is analytically intractable), this won't be very useful. You could also first compute $p(x,y,t)$, the joint density of $X_t$ and $Y_t$, which under some conditions satisfies a Fokker-Planck equation. Then extract the conditional from that. But to give more practical advice, one would need to know what you plan to accomplish.
@S.Surace Hi, there. I am interested in estimating the inner conditional expectation $E[X_s|(Y_{\tau}, \tau\leq s)]$. But in my case, we don't have the correlated-noise problem, that is, the drift of $Y$ is only decided by its own histroy $dY_t = B(Y_t) dt + d W_t^2$. Do you know any method for this problem? Thanks a lot.
Yeah, minor notational caveats aside (square in the dW term looks off) that‘s the classical filtering problem.
@S.Surace Hi, thanks a lot for the help. After reading some materials on Stochastic Filtering, I notice there are some gaps between my problem and the filtering problem. I am not sure whether they can be solved trivially. I have posted my detailed setting here. It would be greatly appreciated if you could take a look and give some advice ^_^.
|
2025-03-21T14:48:30.307008
| 2020-04-15T14:39:39 |
357551
|
{
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"Kafka91",
"https://mathoverflow.net/users/147200",
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"mme"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/357551"
}
|
Stack Exchange
|
Smooth involutions on homotopy 11-spheres or diffeomorphism classification of homotopy projective 11-space
Does anyone know if smooth fixed point free involutions on homotopy 11-spheres have been studied? Or equivalently, is something known about the diffeomorphism classification of homotopy $\mathbb{R}P^{11}$?
And more generally, homotopy (4k-1)-spheres and homotopy $\mathbb{R}P^{4k-1}$?
I suggest looking at López de Medrano, "Involution on Manifolds", and seeing if you can piece together a full answer for $\Bbb{RP}^{11}$ from the results in Chapter VI. The author outlines what the answer will look like in the general case.
@MikeMiller I know this reference, but the only thing I could find for dimension 11 is the existence of smooth involutions on every homotopy 11-sphere (V.6.2, I suppose you were referencing the comment on p.72).
The results of IV.3, V.2, and V.7 combine to show that you only need to determine the set of possible normal invariants (and then the set of homotopy smoothings will be this times $\Bbb Z$ times $\Bbb Z/992$; this uses that all ). The normal invariants are the inverse image of zero in the map $[\Bbb{RP}^{11}, G/O] \to \Bbb Z/2 = L_{11}(\Bbb Z/2)$. What remains is to calculate this set of homotopy classes, and the surgery obstruction. You're right that this isn't contained in de Medrano's book; I don't know before a literature search whether this is known. You see already how complicated it is.
Apologies if I am telling you things you already know.
@MikeMiller I did in fact already know this, so I guess I wanted to see if somebody either knew where this classification is completed (i.e. the normal invariants are computed) or if there was another approach. But thanks for the summary anyway!
|
2025-03-21T14:48:30.307141
| 2020-04-15T14:58:12 |
357552
|
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|
Stack Exchange
|
Argument against Vojta's more general abc conjecture
Confusion is possible, we got argument against Vojta's more general
abc conjecture.
In A more general abc conjecture, p. 7 Paul Vojta conjectures:
If $x_0,\ldots x_{n-1}$ are nonzero coprime integers satisfing $x_0 + \cdots x_{n-1}=0$
$$ \max\{|x_0|,\ldots |x_{n-1}|\} \le C \prod_{p\mid x_0 \cdots x_{n-1}}p^{1+\epsilon}\qquad (1) $$
for all $x_0 , \ldots, x_{n-1}$ as above outside a proper Zariski-closed subset.
Let $rad$ the radical of polynomial be the product of irreducible factors
and the radical of integer is the product of the primes factors.
For natural $D > 2$ and variables $x,y$, define
$A_D(x,y) : x_1=(x+y)^D,x_2=-x^D,x_3=-y^D,x_4= -(x_1+x_2+x_3)=xyF_{D-2}(x,y)$
with $\deg F_{D-2}=D-2$.
We have $\deg x_1=D$ and $\deg rad(x_1 x_2 x_3 x_4) \le D+1$
and in addition $xy$ divides $x_1 x_2 x_3 x_4$.
For small primes $p,q$ and large exponents $n,m$, set $X=p^n,Y=q^m, X>Y$,
then we have the freedom to remove $xy$ from the radical.
Then in $A_D(X,Y)$ we have $x_1 \sim X^D$ and
$rad(x_1 x_2 x_3 x_4)= O(pq X^{D-1})$.
For fixed $D$ we have infinitely many counterexamples which violate the
hypothesis, so they must be in the exceptional set.
For fixed $D$, there is proper algebraic dependency between the $x_i$,
but the algebraic dependencies are different for different $D$.
We believe this shows the exceptional set must be infinite.
Q1 What is wrong with this counterexample?
We asked Vojta and he kindly replied that this is not
counterexample because the exceptional set is allowed
to depend on epsilon and taking $\epsilon > 1/(D-1)$
kills the infinite family of quadruples.
|
2025-03-21T14:48:30.307272
| 2020-04-15T15:11:16 |
357554
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Abdelmalek Abdesselam",
"Igor Khavkine",
"JustWannaKnow",
"Michael Engelhardt",
"https://mathoverflow.net/users/134299",
"https://mathoverflow.net/users/150264",
"https://mathoverflow.net/users/2622",
"https://mathoverflow.net/users/7410"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/357554"
}
|
Stack Exchange
|
Spins in classical statistical mechanics
I'm reading Kupiainen's notes on the renormalization group and also caught my attention. Actually, this is something that often causes my some confusion. On page 43, in the section about Ginzburg-Landau model, Kupiainen introduces a spin configuration as a function $\phi: \mathbb{Z}^{d} \to \mathbb{R}$. Thus, we can view it as a sequence $\phi = (\phi_{x})_{x\in \mathbb{Z}^{d}} \in \mathbb{R}^{\mathbb{Z}^{d}}$. If $\Lambda \subset \mathbb{Z}^{d}$ is finite, we can consider the restriction $\phi_{\Lambda} = (\phi_{x})_{x \in \Lambda} \in \mathbb{R}^{\Lambda}$, where $\phi_{x}:= \phi(x)$. Because $\Lambda$ is finite, we can consider $\phi_{\Lambda}$ as an usual vector on $\mathbb{R}^{n}$ where $n$ is the cardinality of $\Lambda$. For instance, we can define the Gibbs measure on $\mathbb{R}^{\Lambda}$ as given by:
\begin{eqnarray}
d\mu_{\Lambda}(\phi) = \frac{1}{Z_{\Lambda}}e^{-H_{\Lambda}(\phi)}\prod_{j=1}^{n}d\phi_{x} \tag{1}\label{1}
\end{eqnarray}
where $Z_{\Lambda}$ is a normalizing factor, $H_{\Lambda}: \mathbb{R}^{\mathbb{Z}^{d}}\to \mathbb{R}$ is an Hamiltonian with some given boundary conditions and $\prod_{j=1}^{n}d\phi_{x}$ is just the product Borel measure on $\mathbb{R}^{\Lambda}$. On the other hand, on his new set of notes, on page 31 (also about Ginzburg-Landau model) Kupiainen states that "in classical statistical mechanics one considers $\phi(x)$ as a random variable with probability distribution given by (\ref{1})".Now, if I understood it correctly, this means that each $\phi_{x}$ is now a random variable on some underlying probability space. But then, the picture changes a lot, since now instead of simple vectors on $\mathbb{R}^{\Lambda}$, $\phi_{\Lambda}$ is a vector of functions. What does even mean to write $\prod_{j=1}^{n}d\phi_{x}$? Also, if this were the Ising model, we expect $\phi_{\Lambda}$ to be just a vector with entries $\pm 1$ as in the first picture. So, am I missing something here? Why sometimes $\phi_{\Lambda}$ are viewed as vectors and sometimes as vectors of functions? Also, what does $\prod_{j=1}^{n}d\phi_{x}$ mean if each $\phi_{x}$ is a random variable?
Not sure, but could the confusion simply be whether one writes $d\mu(\phi )$ or $\mu (\phi ) d\phi $, which are just different ways of saying the same thing? Why are you saying $\phi_{\Lambda } $ is a vector of functions? What functions?
@MichaelEngelhardt as far as I understand, saying that $\phi(x) = \phi_{x}$ is a random variable means that $\phi_{\Lambda} = (\phi_{x}){x\in \Lambda}$ is a family of random variables indexed by $x \in \Lambda$, and $\phi{x}: \Omega \to \mathbb{R}$ is a measurable function on some underlying probability space. So each entry of $\phi_{\Lambda}$ is a measurable function.
It seems that Kupiainen is guilty of using terminology that (perhaps implicitly) can lead to confusing, but not that uncommon, notation like $\langle \psi \rangle = \int \psi e^{-\psi^2}, d\psi$. On the left $\psi$ must be interpreted as a random variable, while on the right $\psi$ is a coordinate on the probability space $\Omega = \mathbb{R}$, with probability measure $d\mu(\psi) = e^{-\psi^2}, d\psi$. It would be better to write $\langle \Psi \rangle = \int \psi e^{-\psi^2}, d\psi$ and formally distinguish $\Psi$ and $\psi$, but you can just do that in your head.
@IgorKhavkine in summary, $\bf{\Psi}$ would be a random vector on $\mathbb{R}^{\Lambda}$ whose distribution is given by (\ref{1})? He just used the same notation for the random vector and vectors on $\mathbb{R}^{\Lambda}$.
@IamWill Just going by your description, I would say Yes. It is not that uncommon a notational/terminological shortcut.
@IgorKhavkine it really makes sense to me! Actually, this is something that always gets me confused and has ocurred before. I think you are right! Thanks!
It's the same issue than say in $\mathbb{R}^n$. One may talk of $x_i\in\mathbb{R}$ as the "coordinate" of a specific vector $(x_1,\ldots,x_n)\in\mathbb{R}^n$. But one can also talk about the "coordinate function" $x_i$, which is a map from $\mathbb{R}^n\rightarrow\mathbb{R}$. The random variable is $\phi_x$ seen as a coordinate function.
|
2025-03-21T14:48:30.307694
| 2020-04-15T15:18:00 |
357557
|
{
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"authors": [
"Andreas Blass",
"Tomasz Kania",
"https://mathoverflow.net/users/15129",
"https://mathoverflow.net/users/6794"
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|
Stack Exchange
|
What is the smallest cardinality of a maximal ultrafamily of infinite subsets of $\omega$?
A family $\mathcal U$ of infinite subsets of $\omega$ is called an ultrafamily if for any sets $U,V\in\mathcal U$ one of the sets $U\setminus V$, $U\cap V$ or $V\setminus U$ is finite.
By the Kuratowski-Zorn Lemma, each ultrafamily $\mathcal U\subseteq [\omega]^\omega$ can be enlarged to a maximal ultrafamily.
Let $\mathfrak{uf}$ be the smallest cardinality of a maximal ultrafamily. It can be shown that
$$\max\{\mathfrak s,\mathfrak a\}\le\mathfrak {uf}\le\mathfrak c,$$
where $\mathfrak a$ is the smallest cardinality of a maximal infinite almost disjoint family of infinite sets in $\omega$ and $\mathfrak s$ is the smallest cardinality of a family $\mathcal S\subseteq[\omega]^\omega$ such that for every $X\in[\omega]^\omega$ there exists $S\in\mathcal S$ such that the sets $X\cap S$ and $X\setminus S$ are infinite.
Problem 1. Find non-trivial upper (and lower) bounds for the cardinal $\mathfrak{uf}$.
Problem 2. Is $\mathfrak{uf}$ equal to any known cardinal characteristic of the continuum?
Problem 3. Is $\mathfrak{uf}<\mathfrak c$ consistent?
Added in Edit. The diagrams of known small uncountable cardinals in the surveys of Blass and Vaughan do not show up any cardinal characteristic between $\max\{\mathfrak s,\mathfrak a\}$ and $\mathfrak c$. So the answer to Problem 2 is rather ``no'' unless $\mathfrak{uf}=\mathfrak c$ (which would be a bit surprising).
A piece of heuristics: this statement has a very similar flavour to Booth's lemma (that is equivalent to Martin's axiom), so very likely $uf = c$ under MA and probably it is consistent that there is a strict inequality.
@TomekKania That $\mathfrak{uf}=\mathfrak{c}$ under MA follows from the inequalities in the question plus the fact that MA implies $\mathfrak{s}=\mathfrak{c}$.
To my surprise, I found that this my ``new'' cardinal $\mathfrak{uf}$ is equal to $\mathfrak c$.
Theorem. $\mathfrak{uf}=\mathfrak{c}$.
Proof. Fix any maximal ultrafamily $\mathcal U\subseteq[\omega]^{\omega}$. For two sets $A,B$ we write $A\subset^* B$ if $A\setminus B$ is finite but $B\setminus A$ is infinite.
A subfamily $\mathcal L$ will be called
$\bullet$ linearly ordered if for any distinct sets $A,B\in\mathcal L$ we have either $
A\subset^* B$ or $B\subset^* A$;
$\bullet$ densely linearly ordered if for any distinct sets $A,B\in\mathcal L$ with $A\subset^* B$ there exists a set $C\in\mathcal L$ such that $A\subset^* C\subset^* B$.
Claim 1. If $\mathcal U$ contains a set $U\in\mathcal U$ such that the family ${\downarrow}U=\{V\in\mathcal U:V\subset^* U\}$ is linearly ordered, then ${\downarrow}U$ is densely linearly ordered.
Proof. Assuming that ${\downarrow}U$ is not densely linearly ordered, we can find two sets $A,B\in{\downarrow}U$ such that $A\subset^* B$ and the set $\{C\in\mathcal U:A\subset^* C\subset^* B\}$ is empty. Taking into account that $B\setminus A\subset^* U$ and the family ${\downarrow}U$ is linearly ordered, we conclude that $B\setminus A\notin\mathcal U$. By the maximality of $\mathcal U$, there exists a set $W\in\mathcal U$ such that the sets $(B\setminus A)\cap W$, $(B\setminus A)\setminus W$, $W\setminus(B\setminus A)$ are infinite. Then also the sets $B\cap W$, $B\setminus W$ are infinite. Taking into account that $\mathcal U$ is an ultrafamily, we conclude that $W\subseteq^* B\subset^* U$ and hence $W\in{\downarrow}U$. Now the choice of the sets $A,B$ guarantees that $W\subseteq^*A$ and then $(B\setminus A)\cap W$ is finite, which is a desired contradiction. $\quad\square$
Claim 2. If $\mathcal U$ contains a set $U\in\mathcal U$ such that the family ${\downarrow}U=\{V\in\mathcal U:V\subset^* U\}$ is linearly ordered, then $|\mathcal U|=|{\downarrow}U|=\mathfrak c$.
Proof. By Claim 1, the family ${\downarrow}U$ is densely linearly ordered. Consider the countable set $\mathbb Q_2=\{\frac{k}{2^n}:n\in\omega,\;0\le k\le 2^n\}$ of binary rational numbers in the unit interval $[0,1]$. Using the density of the linear order on $\mathcal L$, we can inductively construct a subfamily $\{L_q\}_{q\in\mathbb Q_2}\subseteq\mathcal L$ such that for any rational numbers $r<q$ in $\mathbb Q_2$ we have $L_r\subset^* L_q$.
To see that $|{\downarrow}U|=\mathfrak c$, it remains to prove the following
Claim 3. For every $r\in[0,1]\setminus \mathbb Q_2$ there exists a set $L_r\in\mathcal U$ such that $L_p\subset^* L_r\subset^* L_q$ for every rational numbers $p,q\in\mathbb Q_2$ with $p<r<q$.
Proof. To derive a contradiction, assume that the maximal ultrafamily $\mathcal U$ does not contain such set $L_r$.
Since the poset $\mathcal P(\omega)/\mathrm{Fin}$ contains no $(\omega,\omega)$-gaps, there exists a set $\tilde L_r\subseteq \omega$ such that $L_p\subset^* \tilde L_r\subset^* L_q$ for any $p,q\in\mathbb Q_2$ with $p<r<q$. By our assumption, $\tilde L_r\notin\mathcal U$.
By the maximality of $\mathcal U$ we can find a set $L_r\in\mathcal U$ such that the sets $\tilde L_r\cap L_r$, $\tilde L_r\setminus L_r$ and $L_r\setminus \tilde L_r$ are infinite. The infiniteness of the intersections $\tilde L_r\cap L_r$ and $\tilde L_r\setminus L_r$ implies that for any $q\in\mathbb Q_2$ with $r<q$ the intersections $L_q\cap L_r$ and $L_q\setminus L_r$ are infinite. Taking that $\mathcal U$ is an ultrafamily, we conclude that $L_r\subseteq^* L_q\subset^* U$ and hence $L_r\in{\downarrow}U$. For every $p\in\mathbb Q_2$ with $p<r$ the infiniteness of the set $L_r\setminus \tilde L_r$ and the almost inclusion $L_p\subset^* \tilde L_r$ implies the infiniteness of the set $L_r\setminus L_p$. Since the family ${\downarrow}U$ is linearly ordered, we conclude that $L_p\subseteq^* L_r$. Therefore, we proved that $L_p\subset^* L_r\subseteq^* L_q$ for any rational numbers $p,q\in\mathbb Q_2$ with $p<r<q$. But the existence of the set $L_r$ contradicts our assumption. $\quad\square$
So, Claims 3 and 2 have been proved. $\quad\square$
Claim 2 reduced the proof of the theorem to the case when for every $U\in\mathcal U$ the family ${\downarrow}U$ is not linearly ordered and hence contains two sets $U_0,U_1$ such that $U_0\cap U_1$ is finite. In this case we can inductively construct a family of sets $\{U_s\}_{s\in 2^{<\omega}}\subseteq\mathcal U$ indexed by the elements of the binary tree $2^{<\omega}=\bigcup_{n\in\omega}2^n$ such that $$\mbox{$U_{s\hat{\;}0}\cup U_{s\hat{\;}1}\subseteq^* U_s$ and $U_{s\hat{\;}0}\cap U_{s\hat{\;}1}=^*\emptyset$ for any binary sequence $s\in 2^{<\omega}$}.$$
Claim 4. For every $s\in 2^\omega$ there exists a set $U_s\in\mathcal U$ such that $U_s\subseteq^* U_{s{\restriction}n}$ for every $n\in\omega$.
Proof. To derive a contradicion, assume that for some $s\in 2^\omega$ the sequence $(U_{s{\restriction}n})_{n\in\omega}$ has no pseudointersection in $\mathcal U$. Choose any infinite pseudointersection $\tilde U_s$ of the sequence $(U_{s{\restriction}n})_{n\in\omega}$. By our assumption, $\tilde U_s\notin\mathcal U$ and by the maximality of the ultrafamily $\mathcal U$, there exists a set $U_s\in\mathcal U$ such that the sets $U_s\cap\tilde U_s$, $U_s\setminus\tilde U_s$ and $\tilde U_s\setminus U_s$ are infinite. Then for every $n\in\omega$ the sets $U_{s{\restriction}n}\cap\tilde U_s$ and $U_{s{\restriction}n}\setminus U_s$ are infinite. Taking into account that $\mathcal U$ is an ultrafamily, we conclude that $U_s\subseteq^* U_{s{\restriction}n}$ for every $n\in\omega$, which contradicts our assumption. $\quad\square$
It is easy to see that the family $(U_s)_{s\in 2^\omega}$ given by Claim 4 is almost disjoint and hence $|\mathcal U|\ge|\{U_s\}_{s\in 2^\omega}|=|2^\omega|=\mathfrak c$. $\quad\square$.
|
2025-03-21T14:48:30.308126
| 2020-04-15T15:50:52 |
357560
|
{
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"Iosif Pinelis",
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|
Stack Exchange
|
Local behavior of the Vandermonde convolution
An interesting combinatorial identity is the Vandermonde convolution identity:
$$ \sum_k {n\choose k}{m\choose s-k} = {n+m \choose s},$$
which can be proved by considering the coefficients in $(x+1)^{n+m} = (x+1)^n (x+1)^m$. I am interested in the behavior of the summands: which summands among them contribute the most?.
More specifically, I'd like to look at the behavior of the function
$$ F(k) = \frac{{n\choose k}{m\choose s-k}}{{n+m \choose s}}. $$
In a specific probabilistic problem, intuition suggests that $F(k)$ peaks near $k$, with $\frac{n}{n+m} = \frac{k}{s}$. Indeed, in the limit case $s = n+m$, $k$ had better be $n$ for the value to survive. However, I'm not satisfied with the limit case.
Given that $F$ does peak around $k = k^* := \frac{ns}{n+m}$, I want to know how much it peaks. Thus the following question:
Find the smallest possible non-negative integer $\epsilon$ such that
$$ \sum_{k \in [k^*-\epsilon, k^*+\epsilon]} F(k) > 90\%.$$
The answer $\epsilon$ depends on $n, m, s$, and "$90\%$".
$\newcommand\ep{\epsilon}$
Of course, there is no simple explicit expression for the smallest $\ep$. However, one can give upper bounds on or approximations of this $\ep$.
Indeed, $F$ is the probability mass function of a random variable, say $X$, with the hypergeometric distribution with parameters $n,m,s$. So, by a Hoeffding inequality, for $\ep\in[0,k^*]$
$$S(\ep):=\sum_{k\in[k^*-\ep,k^*+\ep]}F(k)=P(|X-k^*|\le\ep)
\ge1-2e^{-2\ep^2/s}.$$
Solving now the equation $1-2e^{-2\ep^2/s}=0.90$ for $\ep>0$, we get a desired upper bound on the smallest $\ep$ such that $S(\ep)\ge0.90$.
If $s$ is much smaller than $\min(n,m)$ and if $\frac n{n+m}$ is not close to $0$ or $1$, then one can use a normal approximation to hypergeometric distribution.
Thank you! I have no background in statistics, and thought it's quite hard to analyze the entity I gave. Turned out it's just the hypergeometric distribution (textbook-level)..
To my surprise.. this estimate does not depend on $n+m$.. a direct corollary is that no matter how large the black box it, to understand it (to some extent), you only need to sample a little bit (say 50) of them!!!
An edit has been submitted suggesting that your $\epsilon^2/s$'s are supposed to be $\epsilon^2 s$. Is that correct?
Edit was submitted by me. Comparing that in the wikipedia page, the $s$ and $\epsilon$ here correspond to $n$ and $t$.
I wonder why the Hoeffding inequality can be applied onto hypergeometric distributions? On the nose, it seems that the inequality applies to sums of bounded random variables. Coul you please relate?
@Student : (i) "this estimate does not depend on $n+m$": (ii) "your "$\epsilon^2/s$'s are supposed to be "$\epsilon^2s$'s"; (iii) "why the Hoeffding inequality can be applied onto hypergeometric distributions?" First, concerning (iii): this inequality by Hoeffding holds because, by Theorem 4 in his paper at https://en.wikipedia.org/wiki/Hypergeometric_distribution#cite_note-3 , the sampling without replacement is convex-dominated by the sampling with replacement. I have now replaced the link to Hoeffding by a hopefully better one.
Previous comment continued: Now concerning (i): when sampling with replacement, you have $s$ independent Bernoulli trials with success probability $p:=n/(n+m)$, so that $k^*=np$. The simple Hoeffding bound $e^{-2\epsilon^2/s}$, which allows one to easily solve the corresponding equation for $\epsilon$, indeed does not depend on $p$. A bit better but more complicated bound, in terms of the Kullback-Leibler divergence, found at the second link in my answer, does also depend on $p$; however, it is not so easy to deal with.
Previous comment continued: Now concerning (ii): The original expression in the answer was correct. A quick check of that is as follows: take $\epsilon\asymp\sqrt s\asymp$(standard deviation of $X$). Then the bound should be $\asymp1$, which it was and now again is. I have reversed your edits.
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2025-03-21T14:48:30.308413
| 2020-04-15T16:08:15 |
357561
|
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|
Stack Exchange
|
Kontsevich integral on tangles and Fubini
I am reading about the Kontsevich integral, following this text : https://pdfs.semanticscholar.org/635b/c6370e8aba381724eaaa36abefba7f7a5bec.pdf
At some point (page 10 to be exact) the author claims that if you have two composable tangles $T_1$ and $T_2$ then we have $$
Z(T_1 \bullet T_2) = Z(T_1) Z(T_2)$$
where for a tangle $T$, $Z(T)$ denotes its Kontsevich integral.
This may be a somewhat stupid question but for some reason I can not see the reason why.
This is not entirely obvious especially since in this survey they're not completely precise in defining the Kontsevich integral for tangles. You can find a proof here: https://www-fourier.ujf-grenoble.fr/~lescop/preprints/intkon.pdf
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2025-03-21T14:48:30.308494
| 2020-04-15T16:34:43 |
357565
|
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|
Stack Exchange
|
Riemann mapping theorem with boundaries and corners
I was reading this paper by Hollands and Yazadjiev, where on page 760, they claim that since $\hat{M}$ is an orientable simply connected analytic $2$-manifold with boundaries and corners, we may map it analytically to the upper complex half plane $\mathbb{H}^2=\{\zeta\in\mathbb{C}\mid\operatorname{Im}\zeta>0\}$ by the Riemann mapping theorem. My guess is that they are really talking about mapping the interior of $\hat{M}$ to $\mathbb{H}^2$, and that the Riemann mapping theorem refers to the usual uniformization theorem, which in this case, if I understood their argument correctly, states that $\operatorname{int}\hat{M}$ is conformally equivalent to either $\mathbb{S}^2$, $\mathbb{C}$ or $\mathbb{H}^2$. We can certainly rule out the case $\mathbb{S}^2$ since $\operatorname{int}\hat{M}$ is non-compact, but I'm not sure what rules out the case $\mathbb{C}$. Does this have something to do with the fact that $\hat{M}$ has non-empty boundary? Also, why can we conclude that $\operatorname{int}\hat{M}$ is simply connected from the fact that $\hat{M}$ is?
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2025-03-21T14:48:30.308599
| 2020-04-15T16:58:01 |
357568
|
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|
Stack Exchange
|
On tilting and cotilting modules
Let A be an Artin algebra and assume all modules are basic, then a classical result says that tilting modules $T$ are in bijection with complete cotorsion pairs $(T^{\perp}, \check{ add(T)})$ (with some additional assumptions on the cotorsion pairs) , where $T^{\perp}= \{ X | Ext_A^i(T,X)=0 $ for all $i >0 \}$ and $\check{ add(T)}$ denotes the full subcategory of modules $Y$ with a finite $add(T)$ coresolution.
Dually we have the same for cotilting modules $U$ with the dual properties and cotilting modules are in bijection with complete cotorsion pairs $( ^{\perp} U, \hat{add(U)})$ (with some additional assumptions on the cotorsion pairs).
Assume we have a tilting module $T$ and a cotilting module $U$ such that the following two conditions are satisfied:
$T^{\perp} \cap \hat{add(U)} = \hat{add(U)}$
$^{\perp}U \cap \check{add(T)} = \check{add(T)}$.
Is it then true that $A$ is Gorenstein if and only if $U=T$? (or maybe this is true up to adding another condition?)
Note that being Gorenstein is equivalent to all tilting modules being cotilting and also that every module of finite projective dimension has finite injective dimension and vice versa.
So the direction $U=T$ implies Gorenstein is easy.
For the other direction we can use that $A$ being Gorenstein gives us that $T$ is some cotilting module and 1. also gives us that $Ext^i(T,U)=0$ for all $i>0$. Now it would be enough to show that also $Ext^i(U,T)=0$ for all $i>0$ to conclude that $T=U$. But I do not see a good approach to prove that.
We have $^\perp(T^\perp) = \check{\operatorname{add}}\, T$, and $(\check{\operatorname{add}}\, T)^\perp = T^\perp$, and $(^\perp U)^\perp = \hat{\operatorname{add}}\, U$, and $^\perp(\hat{\operatorname{add}}\, U) = {^\perp U}$. Condition 1. and 2. are equivalent to
$\hat{\operatorname{add}}\, U \subseteq T^\perp$
$\check{\operatorname{add}}\, T \subseteq {^\perp U}$.
Taking $^\perp$ on the correct sides and using the above identities, it follows that 1. and 2. are equivalent. So it is enough to have one of the conditions, say 2. Take your favorite Gorenstein algebra which is not selfinjective. Then choose $U = D(A)$ and $T = A$, which implies that $\check{\operatorname{add}}\, T = \mathcal{P}(A)$ and $^\perp U = \operatorname{mod}\, A$, that is, 2. is satisfied. But $T \neq U$. So additional conditions have to be added in order for this to be true.
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2025-03-21T14:48:30.308751
| 2020-04-15T16:58:27 |
357569
|
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|
Stack Exchange
|
Gradient descent in $U(n)^r$
I have a function $f:U(n)^r\rightarrow \mathbb{R}$ which I would like to minimize. Here, $U(n)$ is the set of unitary matrices, and $r$ should be considered to be much bigger than $n$. For instance, $n=3$ and $r=100$. I would like to apply a gradient-descent method. Lets assume that the function satisfies all requirements for the graident-descent method to work.
Question: I have found several papers describing how to do optimization using gradient descent in the space $U(n)$ (eg. this and this). Is there a general way to transform an algorithm for gradient descent on $U(n)$ into an algorithm for gradient descent in $U(n)^r$? Or even better, is there a reference describing gradient descent methods in the space $U(n)^r$?
Of course I could change the original function to be $g:U(n\cdot r) \rightarrow \mathbb{R}$, and additional constraints forcing a solution to be block diagonal, where blocks are nxn matrices. But this would increase substantially the size of the problem, given that $r$ is much bigger than $n$.
Check iout https://www.manopt.org/ . As appropriate, ask for help on https://groups.google.com/forum/#!forum/manopttoolbox as to whether/how to do it in MANOPT.
|
2025-03-21T14:48:30.308848
| 2020-04-15T17:04:34 |
357570
|
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|
Stack Exchange
|
Fractional moments of multivariate normal distributions
Is there an analytic formula for fractional moments of multivariate normal distribution?
$E(\prod_{i=1}^k x_i^{\nu_i})={?}$ where $X=(x_1,\ldots,x_k) \sim N_k(\mu, \Sigma)$, $\nu_i\in \mathcal{R}$ and $\nu_i>0$. I know there is one for univariate normal distributions, but can't find one for the multivariate case.
The formula for the univariate case I know of is
$E[|x-\mu|^{\nu}]=\sigma^{\nu}\frac{2^{\nu/2}\Gamma(\frac{\nu+1}{2})}{\sqrt{\pi}}$. I understand this is about the central absulate moment, but I am more interested in $E(\prod_{i=1}^k x_i^{\nu_i})$ or $E(\prod_{i=1}^k |x_i|^{\nu_i})$.
Thanks! I have edited my question.
Thanks. What about the central absolute case $E(\prod_{i=1}^k |x_i-\mu_i|^{\nu_i})$ then? And what if $\nu_i$ are integers?
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2025-03-21T14:48:30.308930
| 2020-04-15T17:10:20 |
357572
|
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|
Stack Exchange
|
Reduced complete Tate ring which is not uniform?
Recall that a topological ring $A$ is Tate if there is an open subring $A_0$ such that the induced topology on $A_0$ is t-adic for some $t \in A_0$ that becomes a unit in $A.$ One can, given a Tate ring define a notion of powerbounded elements, denote the ring of powerbounded elements by $A^\circ.$ One says that the Tate ring $A$ is uniform if $A^\circ$ is bounded.
Is there an easy example of a Tate ring $A$ which is complete and reduced but not uniform?
While I do not have an idea as to how easy or difficult the following is compared to other examples, here is an example:
Let $A_0$ be the subring of $\mathbb{Z}_p[[T]]$ consisting of all the power series $\sum_i a_iT^i$ satisfying the condition
$$v_p(a_i)\geq \sqrt{i}$$
(here $v_p$ denotes the $p$-adic valuation written additively). Note that this is a ring - to check that $A_0$ is closed under multiplication, one uses the fact that the function $\sqrt{x}$ is concave, hence subadditive. It is easy to see that $A_0$ is $p$-complete (as an additive group, it is $\prod_ip^{\lceil{\sqrt{i}}\rceil}\mathbb{Z}_p$). The Tate ring in question is then $A=A_0[\frac{1}{p}]$, with the topology given by $(A_0, p)$. Note that $A$ contains $\mathbb{Q}_p[T]$ as a subring.
To demonstrate some power-bounded elements, note that $p\mathbb{Z}_p[T] \subseteq A^{\circ}$. This follows from the fact that for a polynomial $f=\sum_{i=0}^n p a_iT^i$ with $a_i \in \mathbb{Z}_p$, we have $f^N \in A_0$ for all large enough $N$. To see that, note that $$f^N=p^N\left(\sum_{i=0}^na_iT^i\right)^N,$$
and so for $f^N$ to be in $A_0,$ it is enough to have $N \geq \sqrt{nN},$ which is achieved by taking any $N \geq n.$
However, the set $p\mathbb{Z}_p[T]$ is not itself bounded, since it contains e.g. the unbounded set
$$pT, pT^2, pT^3, \dots$$
Thus, $A$ is not uniform.
This is great! Does this topological ring have a geometric interpretation?
@CharlesDenis I am not sure. The defining condition seems too rigid, for example, to get some convergence of power series interpretation, I would expect e.g. a condition in terms of Newton polygons, which this is not. I must admit that the way I come up with this was purely algebra: I just wanted a $p$-complete ring with a transparently demonstrable big normalization, and taking some convergence-rate condition seemed like a natural place to look for something like that.
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2025-03-21T14:48:30.309118
| 2020-04-15T17:17:42 |
357573
|
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|
Stack Exchange
|
Control the convex combination of two classes on the boundary of the kahler cone
Let $(X,w)$ be a compact kahler manifold, and $[\eta]$ be a class is on the boundary of the kahler cone. The claim is that one can find another class $[\beta]$ also on the boundary of the kahler cone such that $(1-t)[\beta] + t[\eta] > 0$ when $t \in (0, 1)$. How can one prove this?
Some thoughts:
according to this question, one reasonable way to prove the claim is to find some $k < 0$ such that $[w] + k[\eta]$ not kahler. Then there is some chance for this choice to be on the boundary of the cone. Now consider $(1-t)([w] + k[\eta]) + t[\eta] = (1-t)[w] + (k + (1-k)t)[\eta]$. If $(k + (1-k)t)$ is positive then the resulting sum will be kahler. However, how can one gaurantee that 1).$[w] + k[\eta]$ is on the boundary of the cone 2). $(k + (1-k)t) > 0$?
This is false if $[\eta]=0$, which is definitely on the boundary of the Kahler cone
Assume $[\eta]$ is on the boundary of the Kähler cone and that $[\eta] \neq 0.$
Define $k := -\sup\{t > 0 : [\omega]-t[\eta] \hskip4pt {\rm Kähler}\}$. We claim that $k > -\infty$.
If $k = -\infty,$ then $\frac{1}{t}[\omega]-[\eta]$ is Kähler for all $t > 0,$ so that $-[\eta]$ is nef. The claim will be proved once we show this implies $[\eta]=0.$
Approach 1, suggested by @YangMills in the comments below:
By weak compactness of currents, nef classes are pseudoeffective, i.e. they contain closed positive currents. So there are distributions $,$ such that $\eta+^{c} \geq 0$ and $−{\eta}+^{c}v \geq 0$. So $^{}(+) \geq 0$ i.e. $+$ is plurisubharmonic, hence constant. Thus $\eta$ and $-{\eta}$ differ by $^{}$ of a distribution, which must be smooth by regularity of $^{}$. So $[\eta]$ and $-[\eta]$ are cohomologous, hence $[\eta]=0.$
Approach 2, admittedly overkill but of a more algebro-geometric flavor:
Since both $[\eta]$ and $[-\eta]$ are nef, it follows from the characterization of nefness given as part (iii) of Theorem 4.3 in the Demailly-Paun paper
https://annals.math.princeton.edu/wp-content/uploads/annals-v159-n3-p05.pdf
that for every Kähler class ${\omega}'$ on $X$ we have $$\int_{X}\eta~ \wedge ({\omega}')^{\dim(X)-1} = 0$$
By nondegeneracy we must then have $[\eta]=0.$
We then have that $k > -\infty$ as claimed.
It is clear that $[\omega]+k[\eta]$ lies on the boundary of the Kähler cone. In order to show that $$(1−)([\omega]+[])+[] = (1−)[\omega]+(+(1−))[]$$ is Kähler for all $t \in (0,1)$ it is enough to verify that
$$\frac{k+(1-k)t}{1-t} > k$$ for all $t \in (0,1),$ which is straightforward.
This doesn't work when $[\eta]=0$, in which case you get $k=-\infty$
Thanks--I will edit accordingly.
Your argument is still incomplete, since you did not show that $k>-\infty$. Here is the proof: if we had $k=-\infty$ then by definition $-[\eta]+\frac{1}{t}[\omega]$ is Kahler for all $t>0$, and so $-[\eta]$ is a limit of Kahler classes, i.e. $-[\eta]$ is nef. But you can easily see that if both $[\eta]$ and $-[\eta]$ are nef, then necessarily $[\eta]=0$.
I have incorporated this into the answer. Thanks again!
You don't need the Demailly-Paun (rather difficult) theorem! By weak compactness of currents, nef classes are pseudoeffective, i.e. they contain closed positive currents. So there are distributions $u,v$ such that $\eta+dd^c u\geq 0$ and $-\eta+dd^c v\geq 0$. So $dd^c(u+v)\geq 0$ i.e. $u+v$ is plurisubharmonic, hence constant. Thus $\eta$ and $-\eta$ differ by $dd^c$ of a distribution, which must be smooth by regularity of $dd^c$. So $\eta$ and $-\eta$ are cohomologous, hence $[\eta]=0$.
|
2025-03-21T14:48:30.309359
| 2020-04-12T21:56:41 |
357283
|
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|
Stack Exchange
|
Paths through convergent sequences in $\Delta$-generated spaces
So-called $\Delta$-generated spaces are topological spaces in which paths "determine" the topology of the space. In particular, $X$ is $\Delta$-generated if a set $U\subseteq X$ is open (resp. closed) if and only if $\alpha^{-1}(U)$ is open (resp. closed) in $[0,1]$ for all paths, i.e. continuous functions, $\alpha:[0,1]\to X$. The $\Delta$-generated spaces form a coreflective convenient category of topological spaces that relate to some familiar properties. For instance, it is true in general that:
first countable and LPC $\Rightarrow$ $\Delta$-generated $\Rightarrow$ sequential and LPC
where LPC abbreviates "locally path connected."
Definition: Let's say that that a topological space $X$ is path-sequential if for every convergent sequence $\{x_n\}\to x$ in $X$, there is a path $\alpha:[0,1]\to X$ such that $\alpha(0)=x$ and $\alpha(1/n)=x_n$ for all $n\in\mathbb{N}$.
With some basic arguments, it becomes clear that we have:
first countable and LPC $\Rightarrow$ sequential and path-sequential $\Rightarrow$ $\Delta$-generated $\Rightarrow$ sequential and LPC
The first and third implications are definitely not reversible.
Question: Must every $\Delta$-generated space be path-sequential?
I am really just interested in the case where $X$ is Hausdorff or is at least a US-space, i.e. a space where convergent sequences have unique limits.
Note: it's easy to see that if $X$ is $\Delta$-generated and a US-space, then for every convergent sequence $\{x_n\}\to x$ there exists a path $\alpha:[0,1]\to X$ such that $\alpha(0)=x$ and $\alpha(1/k)=x_{n_k}$ for some subsequence $\{x_{n_k}\}$.
Your $\Delta$-generated spaces are called path-inductive in this paper: https://arxiv.org/pdf/1903.01937.pdf
The answer to the question is negative. To construct a counterexample, choose a maximal almost disjoint infinite family $\mathcal A$ of infinite subsets of $\omega$.
Endow $\mathcal A$ with the discrete topology and consider the product $[0,1]\times \mathcal A$. For every subset $A\subseteq \omega$, let $$2^{-A}=\{0\}\cup\{2^{-n}:n\in A\}.$$
Let $X$ be the topological sum $2^{-\omega}\cup([0,1]\times\mathcal A)$, and $\sim$ be the smallest equivalence relation on the space $X$ such that $0\sim (0,A)$ and $2^{-n}\sim(2^{-n},A)$ for every $A\in\mathcal A$ and $n\in A$. It can be shown that the quotient space $Y=X/_\sim$ is a required counterexample: $Y$ is $\Delta$-generated but not path-sequential (the latter follows from the fact that $S$ is not contained in a path-connected compact subspace of $Y$).
To be sure that everything works, let us write down the proof of the following
Fact. The space $Y$ is $\Delta$-generated.
Proof. The space $Y$ can be identified with the union $$2^{-\omega}\cup\bigcup_{A\in\mathcal A}([0,1]\setminus 2^{-A})\times\{A\},$$ endowed with a suitable topology. Let $q:X\to Y$ be the quotient map.
Take any non-closed set $C\subset Y$. If there exists some $y\in(\bar C \setminus C)\setminus 2^{-\omega}$, then there exists a unique set $A\in\mathcal A$ such that $y\in ([0,1]\setminus 2^{-A})\times\{A\}$. In this case for the map $\gamma_A:[0,1]\to Y$, $\gamma_A(t)\mapsto q(t,A)$, has the desired property: $\gamma_A^{-1}(C)$ is not closed in $[0,1]$.
So, we assume that $\bar C\setminus C\subseteq 2^{-\omega}$. First assume that $2^{-n}\in\bar C\setminus C$ for some $n\in\omega$. Choose two real numbers $a,b$ such that $2^{-n-1}<a<2^{-n}<b<2^{-n+1}$.
Let $\mathcal A_n=\{A\in\mathcal A:n\in A\}$. For every $A\in\mathcal A_n$, let $C_A=C\cap (([a,b]\setminus 2^{-A})\times\{A\})$. If for some $A\in\mathcal A_n$ the set $C_A$ contains $2^{-n}\times\{A\}$ in its closure, then the map $\gamma_A:[a,b]\to Y$, $\gamma_A:t\mapsto q(t,A)$, has the required property: the set $\gamma_A^{-1}(C)$ is not closed in $[a,b]$.
So, assume that for every $A\in\mathcal A_n$ the set $C_A$ does not contain $2^{-n}$ in its closure. By the definition of the quotient topology on $X$, the set $\bigcup_{A\in\mathcal A_n}q((a,b)\setminus \overline C_A)\times\{A\}$ is an open neighborhood of $2^{-n}$ in $Y$, which is disjoint with $C$. But this contradicts $2^{-n}\in\overline{C}$. This contradiction shows that $\bar C\setminus C=\{0\}$.
If $C\cap 2^{-\omega}$ is infinite, then by the maximality of $\mathcal A$, there exists a set $A\in\mathcal A$ such that $C\cap A$ is infinite. In this case for the map $\gamma_A:[0,1]\to Y$, $\gamma_A:t\mapsto q(t,A)$, the preimage $\gamma^{-1}_A(C)\supset C\cap A$ contains zero in its closure and hence is not closed in $[0,1]$.
If the intersection $C\cap 2^{-\omega}$ is finite, then we can find a real number $b\in (0,1]\setminus 2^{-\omega}$ such that the intersection $C\cap [0,b]$ is empty and $\bar C\cap [0,b]=\{0\}$. For every $A\in\mathcal A$ consider the set $C_A=C\cap ([0,b]\setminus 2^{-A})\times\{A\}$. If for some $A\in\mathcal A$ the set $C_A$ contains zero in its closure in $[0,b]$, then for the map $\gamma_A:[0,1]\to Y$, $\gamma_A:t\mapsto q(t,A)$, the preimage $\gamma_A^{-1}(C)=C_A$ contains zero in its closure and hence is not closed in $[0,1]$.
So, we assume that for every $A\in\overline A$ the closure $\overline{C_A}$ does not contain zero. Since $\overline{C_A}\subset \overline C$ and $\overline C\cap [0,b]=\{0\}$, the set $$[0,b)\cup\bigcup_{A\in\mathcal A}(([0,b)\setminus 2^{-A})\setminus \overline C_A)\times\{A\}$$ is an open neighborhood of zero, which is disjoint with the set $C$. But this contradicts $0\in\bar C$. $\quad\square$
This is an interesting construction. I am trying to sort out what properties of a maximal almost disjoint infinite family of subsets of $\omega$ would really be used here.
Also, I'm having a little trouble seeing why $Y$ is even locally path-connected at the image of $0$ (in the given quotient topology).
@JeremyBrazas It nice that you finally commented on this construction. The maximality of the almost disjoint family is necessary for killing all possible convergent subsequences.
Yes, a space is $\Delta$-generated iff it is a quotient of a disjoint union of copies of $[0,1]$. Since quotients of LPC spaces are LPC, every $\Delta$-generated space must be LPC.
I see. Still, it is an intriguing space!
@JeremyBrazas Ok, I will try to write down the proof that $Y$ is $\Delta$-generated (if it is indeed $\Delta$-generated).
@JeremyBrazas I have written the proof of $\Delta$-generatedness of $Y$. It seems that I was wrong claiming that the space $Y$ is not locally path-connected. In fact, it is locally path connected (at least for maximal almost disjoint families $\mathcal A$ whose cardinality has uncountable cofinality, for example, $|\mathcal A|=\mathfrak c$, the space $Y$ is LPC).
Thank you. I will look this over. As an initial question about the update, are you intending to include the "wedge" point $0$ non-uniquely in this representation of $Y$? What if $y=0$?
@JeremyBrazas Observe that my definition of $2^{-A}$ includes zero!
Thank you! The last two paragraphs were a bit confusing to read since it seems like you may be representing $[0,b]\cap 2^{-\omega}$ by just writing $[0,b]$ only in some places. However, the argument makes sense.
|
2025-03-21T14:48:30.309878
| 2020-04-12T22:50:57 |
357286
|
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|
Stack Exchange
|
Consequences of invariant-subspace problem to Li–Yorke chaos
The invariant-subspace problem is probably an open problem for reflexive spaces which asks:
Does every bounded linear operator on an infinite-dimensional separable Hilbert space have a non-trivial closed invariant subspace?
This problem is closely related to properties of the orbits, which are
called $T$-cyclic subspaces.
On the other hand, Li–Yorke chaos is based on infinite-dimensional separable complex Banach spaces and bounded operators. My question here is: If the answer to the above question is yes, what are the consequences for Li–Yorke chaos?
I am far from an expert on invariant subspaces, but I intuitively assume that you want to translate the operators from the invariant-subspace problem translate to the time-evolution operators from dynamical systems. However for chaos, the latter need to be non-linear, thus violating the conditions of the invariant-subspace problem. So, why would you expect there to be consequences in the first place?
Also note that Yorke himself has said that he has no idea what Li–Yorke chaos is.
Li Yorke chaos means there is an uncountable scrambled set. Linear operators can be Li-York chaotic (see the work of A. Peris and his collaborators, for example) and for the case of a linear operator $T:X \to X$, Li-York chaos is equivalent to there being an irregular vector: i.e., a vector $x\in X$ such that $\liminf_n \| T^n x \|=0$ and $\limsup_n \| T^n x \|=\infty$ (see Bermudez, Bonilla, Martinez-Gimenez and Peris, "Li-Yorke and distributionally chaotic operators", J. Math. Anal. Appl. 373 (2011) 83-93).
All that being said, I am aware of no relation of this to the ISP (but maybe there is one?). In general, I know of no really interesting result that depends on the outcome of the ISP. A lot of work has been put into answering this question: it is maddening that we do not know the answer for such a simple query! On the other hand, a lot of interesting techniques have been developed in the search for an answer. I recommend Radjavi and Rosenthal's classic book "Invariant Subspaces" (the second, Dover, edition, has an updated version of the development of the field).
|
2025-03-21T14:48:30.310060
| 2020-04-12T23:01:57 |
357287
|
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|
Stack Exchange
|
Could the 4-color theorem be proven by contracting snarks?
Suppose someone came up with an algorithm that could take any snark and perform edge contraction to result in the Peterson graph. If an inspection of the algorithm reveals that it works as claimed, would the algorithm be sufficient to prove the 4CT?
https://en.wikipedia.org/wiki/Snark_(graph_theory)#Snark_theorem
Yes, but I was also wondering if the "proof by algorithm" approach (as I've worded my question) is a valid approach. I suppose it's valid as long as the algorithm can be proven correct.
@prideout there are plenty of algorithmic proofs where the existence of some desired combinatorial object is demonstrated by giving an algorithm to find it. Some people like these proofs more than non-constructive proofs of existence.
Yes, the 4-colour theorem is true if and only if every snark is non-planar (this is due to Tait).
Showing that a snark has a Petersen minor would be enough to show that it is non-planar.
|
2025-03-21T14:48:30.310171
| 2020-04-12T23:35:33 |
357290
|
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|
Stack Exchange
|
An inequality involving square roots and sums
I've been trying to prove (maybe even disprove) the following inequality:
$$
\sum_{n=1}^{N} \frac{a_n}{\sqrt{\sum_{i=1}^{n}a_i}} \leq C \sqrt{\sum_{n=1}^{N}a_n}
$$
Where $ a_1,...,a_N\geq 0 $ are some non-negative numbers, and $C$ is an absolute constant.
Help will be much appreciated.
$C$ is absolute means that it is independent of $N$, right? Also, although it probably doesn't much matter, you say symbolically that the $a$'s are non-negative, but then write that they are positive.
@LSpice You're right, my bad. As Iosif Pinelis expalined though, it doesn't really matter
For every $n\in\{1,\dotsc,N\}$, we have
$$2\sqrt{\sum_{i\leq n} a_i}-2\sqrt{\sum_{i\leq n-1} a_i}=\frac{2a_n}{\sqrt{\sum_{i\leq n} a_i}+\sqrt{\sum_{i\leq n-1} a_i}}>\frac{a_n}{\sqrt{\sum_{i\leq n} a_i}}.$$
Summing these up, we obtain the inequality with $C=2$. It is also straightforward to see that for $C<2$ the inequality fails, hence $C=2$ is the optimal constant.
Rewrite your inequality as
$$lhs:=\sum_{n=1}^N \frac{s_n-s_{n-1}}{\sqrt{s_n}}\le C\sqrt{s_N},$$
where $s_n:=\sum_{i=1}^n a_i$. Note that
$$\sum_{n=1}^N \frac{s_n-s_{n-1}}{\sqrt{s_n}}$$
is a lower Riemann sum for the integral
$$\int_0^{s_N}\frac{ds}{\sqrt s}=2\sqrt{s_N}.$$
So,
$$lhs\le2\sqrt{s_N},$$
as desired.
In the above proof, it was tacitly assumed that $a_i>0$ for all $i$. This can be obviously extended to the case when we only know that $a_i\ge0$ for all $i$ -- assuming that, by continuity,
$\frac{a_n}{\sqrt{\sum_{i=1}^{n}a_i}}:=0$ whenever $a_n=0$.
|
2025-03-21T14:48:30.310292
| 2020-04-13T00:12:38 |
357294
|
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|
Stack Exchange
|
Injectivity of the cohomology map associated to the pullback of line bundles
Let $f:X\to Y$ be a flat, surjective, smooth morphism between smooth algebraic varieties (over $\mathbb C$). We assume that $f$ has relative dimension $n$ and we assume also that $\dim Y\ge 2$ (just to avoid the case of a curve that might be easier).
Let $L$ be a line bundle on $Y$, then we have a homomorphism in sheaf cohomology:
$$H^p(Y,L)\to H^p(X, f^\ast L) \quad\text{for } p=0,1,\ldots,\dim Y $$
Can we say anything about the injectivity of this map? Do we need some additional condition on $f$?
The unit of adjunction map $L\rightarrow f_\ast f^\ast L$ is an isomorphism if the fibres of $f$ are connected (I assume this in what follows). Then the Leray spectral sequence $E^{pq}_2=H^p(R^q f_\ast f^\ast L)\Rightarrow E^{p+q}=H^{p+q}(f^\ast L)$ has edge maps $e^p:E^{p0}\rightarrow E^p$ which are exactly the maps you are interested in. In particular, these maps are isomorphisms if $f$ is finite. In general $e^0$ is an isomorphism, $e^1$ is injective, but $e^2$ need not be injective (exact sequence of low degree terms). If $f$ is of the form $X\times Y\rightarrow Y$ then the maps $e^p$ are certainly injective (Künneth).
More generally, it happen quite often that $\mathcal{O}Y\to Rf* \mathcal{O}X$ splits in the derived category, and then so does $L\to Rf* f^* L$, so the pull-back maps $H^p(Y, L)\to H^p(X, f^* L)$ are injective.
@Piotr Achinger, what do you mean by "quite often"?
Hello @Samuel, would you please provide a reference for the first statement?
@svelaz 28.1.I. EXERCISE in Vakil's notes.
@Samuel I see. Thank you very much. I was hoping you were referring to some version of it without the proper assumption.
|
2025-03-21T14:48:30.310429
| 2020-04-13T00:38:26 |
357298
|
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|
Stack Exchange
|
Cwikel–Lieb–Rosenbljum inequality including zero resonances
The Cwikel–Lieb–Rosenbljum inequality asserts that, for any potential $V:\mathbb{R}^n\to\mathbb{R}$, we have
$$(\mbox{number of eigenvalue} \leq 0\mbox{ , counted with multiplicity, of }-\Delta+V\,)\leq\\L_n\int_{\mathbb{R}^n}V_{-}(x)^{\frac{n}{2}}dx,$$
for an optimal constant $L_n$ depending only on $n$. To the best of my knowledge, the exact value of $L_n$ is unknown, but there are very sharp bounds.
Now, it seems to me that a similar inequality holds true if in the l.h.s. we count also an eventual zero energy resonance (which roughly speaking is a solution $\psi$ to $(-\Delta+V)\psi=0$, which does not belong to $L^2$ but only to some $L^p$ or to some weighted $L^2$-space, the precise definition depends on the dimension):
$$(\mbox{number of eigenvalue} \leq 0\mbox{ and zero resonances , counted with multipl., of }-\Delta+V\,)\leq \widetilde{L}_n\int_{\mathbb{R}^n}V_{-}(x)^{\frac{n}{2}}dx,$$
for some optimal constant $\widetilde{L}_n$ depending only on $n$.
Is such inequality actually true? If yes, what bounds can be proved for the constant $\widetilde{L}_n$?
|
2025-03-21T14:48:30.310525
| 2020-04-13T01:01:06 |
357300
|
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|
Stack Exchange
|
In which books we can find structure information for finite simple groups and their Schur covering groups?
In which books we can find representations or character tables, Sylow 2-subgroups and conjugacy classes for finite simple groups and their Schur covering groups and properties for Schur multiplier of finite simple groups?
The following websites may be useful to my questions:
https://math.stackexchange.com/questions/785603/what-do-sylow-2-subgroups-of-finite-simple-groups-look-like
Beyl-Tappe [4] say on p. 119:
5.9 REMARK. The Schur multiplicators of all finite simple groups have been found, often by exhibiting a universal perfect cover (representation group). For the results see the summary by GRIESS [2] and follow the references given there.
[1] Griess, Robert L. jun., Schur multipliers of the known finite simple groups, Bull. Am. Math. Soc. 78, 68-71 (1972). ZBL0263.20008.
[2] Griess, Robert L. jun., Schur multipliers of the known finite simple groups. II, Finite groups, Santa Cruz Conf. 1979, Proc. Symp. Pure Math. 37, 279-282 (1980). ZBL0448.20014.
[3] Griess, Robert L. jun., Schur multipliers of the known finite simple groups. III, Proc. Rutgers group theory year, 1983/84, 69-80 (1984). ZBL0648.20017.
[4] Beyl, F. Rudolf; Tappe, Jürgen, Group extensions, representations, and the Schur multiplicator, Lecture Notes in Mathematics. 958. Berlin-Heidelberg-New York: Springer-Verlag. IV, 278 p. DM 33.50 {$} 13.50 (1982). ZBL0544.20001.
Thank you very much!
If you really search for a book on most of those topics (at least character tables and conjugacy classes) for simple groups with some proofs, the best (and most compact) ones might be those of Gerhard Michler : Theory of Finite Simple Groups Part I and II.
Thank you very much!
|
2025-03-21T14:48:30.310648
| 2020-04-13T02:18:08 |
357307
|
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|
Stack Exchange
|
Lie algebra cohomology: $H^i(R,V)=H^i(R,V^R)$ with $R$ reductive and $V$ an $R$-module
Let $R$ be a reductive, finite-dimensional Lie algebra over a field of characteristic 0, and let $V$ be a semisimple $R$-module (also finite dimensional). I have seen a reference to the fact that $H^i(R,V) \cong H^i(R,V^R)$, which seems like a stronger version of Theorem 10 of Hochschild and Serre's paper. The only reference I can find is to an out-of-print book in Romanian, "Introducer in Coomologia Algebrelor Lie," by Verona. Is there an English reference for this fact? Thanks!
(This is meant to be a comment, but I don't have enough reputations) If I'm not mistaken, this follows almost directly from the paper you cited. By semisimplicity of $V$, it suffices to prove for irreducible $V$. If $V$ is not the trivial module then $V^R =0$ and the statement follows from Theorem 10 of the paper. If $V$ is trivial then the statement is obvious.
Wow, I think you're right. I got so bogged down thinking about the literature I forgot to think about the problem.
|
2025-03-21T14:48:30.310743
| 2020-04-13T03:17:03 |
357309
|
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|
Stack Exchange
|
A weakly holomorphic modular form is a harmonic maass form
It is known that for $\Gamma_0(N)$, a weakly holomorhpic modular form is a harmonic maass form. Here is the definitions.
A weakly modular form $f$ for $\Gamma_0(N)$ is a meromorphic function on the half-plane such that
$$
f(\gamma z)=j(\gamma,z)^{-k}f(z)
$$
and the poles are supported in the cusps.
A harmonic Maass form $f$ for $\Gamma_0(N)$ is a real analytic on the half-plane such that
$$
f(\gamma z)=j(\gamma,z)^{-k}f(z)
$$
and
$$
\Delta_k(f)=0,
$$
where $\Delta_k=-y^2\left(\dfrac{\partial^2}{\partial x^2}+\dfrac{\partial^2}{\partial x^2}\right) + iky\left(\dfrac{\partial}{\partial x}+i\dfrac{\partial}{\partial y}\right)$,
and
there exists a polynomial $P(z) \in \mathbb C [q^{-1}]$ such that
$$
f(z)-P(z)=O(e^{-\epsilon y})
$$
as $y \to \infty$ for some $\epsilon>0$. Analogous conditions are required at all cusps.
Since a weakly holomorphic modular form is holomorphic on the half-plane, it vanishes if we takes the hyperbolic Laplacian $\Delta_k$. But how we know that a weakly holomorphic modular form $f$ has a polynomial $P$ with the third condition of the definition of harmonic Maass form. Namely, is there a polynomial $P_x$ for the cusp $x$ of $\Gamma_0(N)$ such that $f|_k \sigma^{-1} (q)-P_x(q) = O(e^{-\epsilon y})$, where $\sigma \in \mathrm{SL}_2(\mathbb R)$ that sends $x$ to $\infty$? Why?
|
2025-03-21T14:48:30.310841
| 2020-04-13T03:55:02 |
357311
|
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|
Stack Exchange
|
Ancient Heat equation and Liouville's theorem
I encounter a difficulty when proving the bounded solution of ancient heat equation implying constant function. Suppose $u(t,x)$ is the solution of ancient heat equation:
\begin{equation}
u_{t} = \Delta u \quad \mathrm{in} ~\ (- \infty, 0] \times \mathbb{R}^{n}
\end{equation}
In order to prove it is constant function, we can mimic the proof of Liouville's theorem in complex analysis by showing that $\nabla_{x} u(t,x)$ goes to zero. In usual heat equation defined in $(0, T] \times \mathbb{R}^{n}$, we have the following gradient estimation:
\begin{equation}
\sup_{x \in \mathbb{R}^{n}}| \nabla_{x} u(T,x)| \leq \frac{C}{\sqrt{T}} \sup_{x \in \mathbb{R}^{n}} |u_{0}|
\end{equation}
where C is a constant relates to its dimension $n$ and $u_{0}$ is the initial value of $u(t,x)$. Originally, I want to change the variable $t \rightarrow -t$ such that I can apply the above gradient estimation to show that the solution of ancient heat equation $u(t,x)$ is constant function. However, there is a problem that $\sqrt{T}$ would be complex under this change of variable. Therefore, how can I combined these two concepts to show bounded solution for ancient heat equation is constant function, similar to Liouville's theorem in Complex analysis?
Yes, this inequality is false if $T<0$ ... when you look at the proof, you will see that if $t<0$, it doesn't work since the disspative term is now no more negative.
To see that $u$ is constant you have to apply the estimates for the derivatives in the following form $$|D_t^n D_x^k u(t_0,x_0)| \le \frac{C_{n,k} |u|_{\infty, Q(t_0,x_0; R)}}{R^{k+2n}},$$ where $Q(t_0,x_0;R)=(t_0-R^2,t_0)\times B(x_0,R)$ adn let $R \to \infty$ for every fixed $(t_0,x_0)$ (you need only $n,k$ to be 0 or 1).
Your approach works fine, just start with negative moments of time:
\begin{equation}
\sup_{x \in \mathbb{R}^{n}}| \nabla_{x} u(t,x)| \leq \frac{C}{\sqrt{T}} \sup_{x \in \mathbb{R}^{n}} |u(t-T,x)| ,\quad t\le 0,\ T>0.
\end{equation}
|
2025-03-21T14:48:30.310992
| 2020-04-13T06:02:37 |
357316
|
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|
Stack Exchange
|
Question about the exit time of a time-homogeneous Itô diffusion
Consider a one-dimensional Itô diffusion:
$$\mathrm{d} X_{t}=b\left(X_{t}\right) \mathrm{d} t+\sigma\left(X_{t}\right) \mathrm{d} B_{t}$$
where $X_0 = 0$ and $B_t$ is the standard Brownian Motion. The exit time $\tau$ is defined as:
$$\tau = \inf (t >0 : X_t \notin (a, b) )$$
Now I want to calculate the probability $P(X_{\tau} = a)$ and the expectation $E \tau$.
Previously, when I solved the questions about exit time, I used Optional Sampling Theorem. In this question, however, since I don't know the explicit expression of $X_t$, I cannot find a reasonable way to handle it. So if anybody can help me.? Thanks so much!
$f(x) = \mathbb{E}^x\tau$ satisfies $f(a) = f(b) = 0$ and $L f(x) = -1$, where $L = \sigma^2 \partial_{xx} + b \partial_x$ is the generator. The probability that $X_\tau = a$, $f(x) = \mathbb{P}^x(X_\tau = a)$, satisfies a similar relation: $f(a) = 1$, $f(b) = 0$, $L f(x) = 0$. This is pretty standard, you will find it in any textbook that discusses one-dimensional diffusions.
@Mateusz Kwaśnicki Thanks so much! Your approach is very concise. I think $L$ should be $\sigma^2/2 \partial_{xx} + b \partial_{x}$. But when I try your method, I still cannot figure out why $L \mathrm{E}^x \tau = - 1$ and $L \mathrm{P}^x (X_{\tau} = a) = 0$? Could you do a favor and give me more details?
Best.
(1) Yes, I forgot $\tfrac{1}{2}$, sorry. (2) What textbook are you following? Once you identify $L$ with Dynkin's characteristic operator, that is $$L u(x) = \lim_{\substack{x_1 \to x^- \ x_2 \to x^+}} \frac{\mathbb{E}^x u(X(\tau_{(x_1, x_2)})) - u(x)}{\mathbb{E}^x \tau_{(x_1, x_2)}} ,$$ both properties follow directly from the strong Markov property. This is how Dynkin derives these kind of equations in his two-volume Markov processes. Modern textbooks usually take a different approach, though.
@Mateusz Kwaśnicki Appreciate your help!!! In fact, I am just taking a stochastic calculus class for undergraduate students, so it may take me a considerable amount of time to get it sussed. Thanks again!
You can solve this by reducing it to a problem of Brownian motion: Define the scale function
$\varsigma(x) = \int_{X_0}^x e^{-2\int_{X_0}^y \frac{b(z)}{\sigma^2(z)} dz} dy$
the process
$M_t = \varsigma(X_t)$
is a local martingale. Therefore, by Dambis-Dubins-Schwarz it is a time changed Brownian motion
$M_t = W_{\langle M, M\rangle_t}$.
Thus, it suffices to study a time-changed Brownian motion, starting at $\varsigma(X_0)$, in the interval $(\varsigma(a), \varsigma(b))$.
Note that to calculate the probability of hitting each boundary, you do not have to calculate $\langle M, M\rangle$ explicitly. As long as $\langle M, M\rangle< \infty$ a.s., the boundary that one path finally hits does not depend on the speed you let run the clock.
This seem to be no longer true for the expectation. Though here you could proceed similarly as outlined by Mateusz in its comments. If $g$ is a solution to
$$ Lg = 1, \quad g(X_0) = 0$$
you can use Dynkin's formula to conclude
$$\mathbb{E}[\tau] = \mathbb{E}\int_0^\tau Lg(s) ds = \mathbb{E}[g(X_\tau)] = g(a) \mathbb{P}[X_\tau=a] + g(b) \mathbb{P}[X_\tau=b]$$
To make the argument rigorous, you will have of course check technical conditions. In particular existence of a solution to an SDE is not enough, you have to make sure that it exits the interval in finite time a.s. (e.g., as extreme example consider $b=\sigma=0$ when the diffusion will never exit the interval.
Thank you so much!!! Since I just studied Dambis-Dubins-Schwarz Theorem in the last class, this is the solution I really want.
But I have another question: how to calculate $<M, M>_t$? (It seems to be somewhat difficult as it includes an integral.)
I tried to address your question by expanding my answer.
Thanks very much!
So sorry for accepting your answer so late. I am a new user, I just know I can accept one's answer.
@香结丁 No issue. Thank you!
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2025-03-21T14:48:30.311329
| 2020-04-13T06:06:14 |
357317
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Stack Exchange
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Results that are widely accepted but no proof has appeared
The background of this question is the talk given by Kevin Buzzard.
I could not find the slides of that talk. The slides of another talk given by Kevin Buzzard along the same theme are available here.
One of the points in the talk is that, people accept some results but whose proofs are not publicly available. (He says this leads to wrong conclusions, but, I am not interested in wrong conclusions as of now. All I am interested is are results which are accepted as true but without a detailed proof, or with only a partial proof.)
What are results that are widely accepted to be true with no detailed proof, or only a partial proof?
I am looking for situations where $A$ has asserted in print that he/she has a proof of $X$, but hasn't published a proof of $X$, and then $B$ publishes a proof of $Y$, where the proof depends on the validity of $X$. For example as on pages 20,21,22 of the slides mentioned above.
Edit: Please give reference for the following:
Where the result is announced?
Where the result is used?
Edit (made after Per Alexandersson's answer): I am not looking for "readily available but not formally published". As mentioned by Timothy Chow, "there are many more examples if "readily available but not formally published" counts.".
This question is not intended to be a debate on whether some result is true or not :) I am only looking for results whose proofs are not published..
I might object that a result without a proof cannot be known to be true as a matter of principle. What other method of verification do we have, other than a proof? Divine revelation? (A previous version of the question asked for results that were "believed" to be true, rather than "known" to be true, then I could generate a vast list from all theorems that assume the Riemann hypothesis.)
@CarloBeenakker Is the question in its present form looks ok for you? “I might object that a result without a proof cannot be known to be true as a matter of principle. What other method of verification do we have, other than a proof? Divine revelation?” Please have a look at the slides in the question for the relevance of this question...
Everybody knows that Goldbach's Conjecture is true, that the Twin Prime Conjecture is true, that there are infinitely many primes of the form $n^2+1$, and so on, and so forth, even though there are no (credible) proofs. Is that the kind of result you are asking for, Praphulla? Or do you mean results which do have a proof, it's just that no one has bothered to write a proof out? Probably no one has written out a proof that $9876543\times9638527=95195326372161$, but I trust that it's true because my calculator says so. Is that the kind of result you mean?
I am looking for statements which people use to create new mathematics, assuming that statement is true, but there is no proof written.. For example as in page 20,21,22.. @GerryMyerson Does it makes it any clearer?
Maybe reword it as "accepted to be true" but "no publicly available proof"?
Personally I do not understand why people do mathematics without understanding fully what they are doing.
OK, so it looks like you are asking for situations where A has asserted in print that he/she has a proof of X, but hasn't published a proof of X, and then B publishes a proof of Y, where the proof depends on the validity of X. Is that it?
@GerryMyerson thank you. Edited :)
I think the intent of the question would've been clearer if it asked (explicitly, especially in the title) for results which were announced to be true, but for which but whose proof has never appeared (yet). Anyway, clarifications considered, interesting question and +1 from me.
I was unclear on one thing myself. Do you want to include partial proofs as an example, or exclude? It now reads like you want to exclude them.
I wanted to include partial proofs as example but that is fine.. :) The question looks more clearer now.. thank you arsmath
Happy to do it. I am very interested in the question. I edited it again to clarify the partial proof point.
What's the relevance of the ag.algebraic-geometry, ct.category theory and dg.differential-geometry tags? Presumably examples can come from any area of mathematics. The big-list tag would certainly be appropriate, and the question should be community wiki.
@arsmath thank you.
@HJRW Yes, examples can come from anywhere.. I am.in particular interested in examples coming from these tags.. Yes, big list tag is suitable...
I feel that questions about things "widely accepted to be true with no detailed proof" will end up being too much opinion-based. I am not sure that it is appropriate for MO.
@VladimirDotsenko If there is a result A published by some one which is used by others but there is no published proof of the result A, that will be an example for the question.. I do not see where opinion is coming here.. :) Please let me know if I am missing something.. I do not want anybody to spend time on something that is opinion based.. That is the whole point of this question.. :) Opinions are not appreciated.. evidence is appreciated..
Several famous results of Hugh Woodin fit the bill.
@MonroeEskew please feel free to make it as an answer.. I know nothing about them :)
@CarloBeenakker “I might object that a result without a proof cannot be known to be true as a matter of principle. What other method of verification do we have, other than a proof?” PA is consistent. It is known by intuition :)
@MonroeEskew “Personally I do not understand why people do mathematics without understanding fully what they are doing.” Von Neumann would like to have a word with you.
@Monroe Which ones do you have in mind? There is the fact that Turing determinacy implies Suslin-coSuslin determinacy (in the presence of DC?), which gives L(R)-determinacy.
@AndrésE.Caicedo Has the proof of that really not appeared anywhere (by Woodin or others)? I've seen at least "Turing determinacy implies L(R)-determinacy" quoted as well-established fact.
@Noah It may be in the process of being written up (by someone writing a book on AD+). Woodin has lectured on it at a seminar at Harvard, so there may be notes. There is no published account; I don't know the details.
I think all the (unsolved) Clay Mathematics Millennial Problems fit this question? The answers are widely believed to be known, and tons of papers rely on one of them, but most have not been proved.
@Blue I think the distinction is between someone writing "if the Birch Swinnerton-Dyer conjecture is true, then....", and someone writing "since the Birch Swinnerton-Dyer conjecture is true ...." and citing a paper claiming, without proof, that it's true.
I think it is widely accepted that Helfgott has proved the ternary Goldbach conjecture, although the proof has not yet been published. I don't know whether any published paper has relied on Helfgott's work ("relied", in the sense of stating a theorem and citing Helfgott's work as an essential part of the proof).
Can only be a comment, but modern lattice coding techniques are based on intuitions whose proofs are at best obscure. For example what are upper and lower asymptotic bounds on the typical log-amount of points a (sequence of scale-normalized) construction-A lattice(s) puts in a ball with normalized volume $V$? Should be $\sim V$ but you never use this intuition directly. Instead you only talk about the amount of points in a shape of similar volume that covers a proportionally large part of the ball. Fine for many applications but inconvenient.
@enthdegree I have absolutely no knowledge of information theory so, I can not give any useful response... It would be nice if you can spend some time on your idea and turn it into an answer for this question..
I'm going to interpret this as a request for examples of results that were announced a while ago but whose proofs have not yet appeared. In other words, people don't doubt that the result is correct and that the author(s) can prove it, and there is an expectation that the current lack of a proof will not be a permanent state of affairs (i.e., a paper with the proof will be written and made public eventually).
One example of this is Rota's Conjecture on excluded minor characterizations of matroids representable over a given finite field. This was announced in 2014 by Geelen, Gerards, and Whittle, but apart from the sketch in that Notices article, no further details have yet appeared.
EDIT: An example of a paper that cites this unpublished work, and relies on it in an essential way, is The matroid secretary problem for minor-closed classes and random matroids by Tony Huynh and Peter Nelson. After stating Theorem 2, Huynh and Nelson write:
To be forthright with the reader, we stress that Theorem 2 relies on a structural hypothesis communicated to us by Geelen, Gerards, and Whittle, which has not yet appeared in print. This hypothesis is stated as Hypothesis 1. The proof of Hypothesis 1 will stretch to hundreds of pages, and will be a consequence of their decade-plus ‘matroid
minors project’. This is a body of work generalising Robertson and Seymour’s graph minors structure theorem to matroids representable over a fixed finite field, leading to a solution of Rota’s Conjecture.
Another example is On the existence of asymptotically good linear codes in minor-closed classes by Peter Nelson and Stefan H. M. van Zwam, IEEE Trans. Info. Theory 61 (2015), 1153–1158. The results of Nelson and van Zwamm have in turn been used in an essential way to prove Theorem 1.4 of Girth conditions and Rota's basis conjecture by Benjamin Friedman and Sean McGuinness.
Thanks for your answer... Do we know by any chance what are the reasons for no news after 2014?
@PraphullaKoushik : I recently asked Geelen about this and he said that there's no specific reason, just that the proof is long and complicated and that the authors have had difficulty finding solid chunks of time to devote to writing it up.
Ok.. I am also looking for some papers which cite this announcement and use that the result is true...If I see anything I will leave as a comment here, you can add it if you think it fits in the answer..
@PraphullaKoushik : I added an example.
Thank you, this is now a complete answer for the question :)
@PraphullaKoushik Something I only just learned is that Grace and Van Zwam found a counterexample to a statement that Geelen, Gerards, and Whittle had claimed was a theorem. This counterexample still leaves most of their other claims intact, but it does make me wish that GGW would write up their proofs ASAP.
Well, in some sense the Classification of Finite Simple Groups is in this state. It most certainly satisfies your second requirement: lots of papers have been published which rely on CFSG. However, a complete proof is (at least in some sense) still work in progress by Lyons, Solomon, Ashbacher, Smith and others.
This is the first example discussed in Buzzard's talk (though he doesn't go over where CFSG is applied)
Ah, didn't recall that.
Isn't there a now complete proof after the publication of Aschbacher and Smith? The point of the book series is to have the proof in one place, and because it can be proven more efficiently once you already know the complete list of simple groups.
Well, Ron Solomon says they’re still finding gaps. I don’t think anyone expects dramatic changes, but it’s hard to argue that the proof is as solid and final as the typical published paper.
@MarkSapir would that contradict what Kevin Buzzard said in the slides?
An example of a (minor) gap that was filled after the Aschbacher-Smith books were published is a 2008 paper by Harada and Solomon. As the introduction to the paper explains, proofs of the theorems in question had been announced more than once before, but either had errors or were unpublished.
I think one example is given in this MO question of mine: a quartic in $\mathbb{P}^3$ with at worst Du Val singularities is a K3 surface (and similar statements for two types of complete intersections in higher-dimensional projective spaces).
Using the excellent answer and comments I was able to piece together a proof, but I could not locate one in the literature, whereas of course the result was "well-known to experts" (to such an extent that I even felt embarrassed for asking about the proof in the first place).
This kind of known but whose proof is not published and you hesitate to ask on MO are what I find useful to Grad students :) :) Please add an outline of the proof of this result when you write your next paper. :) +1
The proof of the theorem of MacPherson that functors out of the exit path category are equivalent to constructible sheaves was not written down, just claimed. Others have since given much more general theorems, but whose reduction to MacPherson's result is not immediate.
Is the theorem ofMacPherson used anywhere or only the general versions?
It's cited a lot, the massively general versions arrived decades after the original. As in, it's a key technical tool in intersection homology IIRC.
It is interesting how can some one cite a result more than once but there is no publication by the author where the result is mentioned... It must have been a very strong result :) :)
Well, there's a number of famous examples. The Grothendieck–Riemann–Roch theorem, due to Grothendieck, appeared in a paper by Borel and Serre after being communicated in a letter. Grothendieck's proof was published about 14 years later, in SGA6. https://en.wikipedia.org/wiki/Grothendieck%E2%80%93Riemann%E2%80%93Roch_theorem#History
Ok. So, Grothendieck mentioned a theorem (with out proof) in a letter to Serre, what is now considered as a GRR theorem. Then, Serre and Borel ran a seminar to understand this, and published (Is it their own proof?) as “Le théorème de Riemann–Roch”.. Then, after some years Grothendieck published his first proof in the “book” SGA 6... Is it correct?
Yes. It happened the other way around, too. For example: Grothendieck asked Serre a question, the latter proved it, then Grothendieck published the theorem and proof. This is not that unusual. It's the complete lack of proof of MacPherson's theorem by anyone that is weird, here.
"It's the complete lack of proof of MacPherson's theorem by anyone that is weird, here" Agreed... :)
Besides the 2-categorical version of Treumann's thesis, isn't this worked out in detail in the 1-categorical case in section 6 of this paper of Curry and Patel?
@R.vanDobbendeBruyn apart from being phrased in the dual setting, which is really a non-issue, I recall thinking that it's not clear that the definition of exit/entrance path category was quite the same. See the second dot-point on page 2.
I just realized that the OP links to a YouTube video and to some slides, but the two don't match—they're two different talks by Buzzard.
For completeness, let me therefore mention some results by James Arthur, which are mentioned in the linked slides but not the linked YouTube video.
On page 13 of Abelian Surfaces over totally real fields are Potentially Modular by George Boxer, Frank Calegari, Toby Gee, and Vincent Pilloni, there is the following remark.
It should be noted that we use Arthur’s multiplicity formula for the discrete spectrum of GSp4, as announced in [Art04]. A proof of this (relying on Arthur’s work for symplectic and orthogonal groups in [Art13]) was given in [GT18], but this proof is only as unconditional as the results of [Art13] and [MW16a, MW16b]. In particular, it depends on cases of the twisted weighted fundamental lemma that were announced in [CL10], but whose proofs have not yet appeared, as well as on the references [A24], [A25], [A26] and [A27] in [Art13], which at the time of writing have not appeared publicly.
Arthur's (unavailable) references [A24] through [A27] are:
[A24] Endoscopy and singular invariant distributions, in preparation.
[A25] Duality, Endoscopy, and Hecke operators, in preparation.
[A26] A nontempered intertwining relation for $GL(N)$, in preparation.
[A27] Transfer factors and Whittaker models, in preparation.
Yes, and, indeed, sadly, this kind of top-heaviness is not good for the subject.
I have edited the question. I could not find the slides of the talk in that YouTube link.. The slides are for the talk given by Kevin Buzzard along the same theme :)
In 1999, Robertson, Sanders, Seymour, and Thomas announced a proof of Tutte's "snark conjecture" (that every snark has a Petersen graph minor), but as far as I am aware the full proof still has not appeared: see this MO question. I don't know if this result has ever been applied anywhere, though. The proof was announced in "Recent Excluded Minor Theorems for Graphs" by Thomas (available as a preprint online here; with citation information at MR1725004): see Theorem 10.2 of that paper specifically. More information about the status of these results seems available on Thomas's webpage.
I just learned that Thomas died very recently (and tragically at age 57 following a long battle with ALS); see: https://mattbaker.blog/2020/04/19/colorings-and-embeddings-of-graphs/.
A long time ago, M. Ajtai, J. Komlós, M. Simonovits, and E. Szemerédi announced a proof (for large $k$) of the Erdős–Sós conjecture that every graph with average degree
more than $k-1$ contains all trees with $k$ edges as subgraphs, but the proof has not yet appeared as of this writing (2022).
What do I mean by "a long time ago"? Reinhard Diestel, in the notes to Chapter 7 of Graph Theory (5th edition), gives a date of 2009. But Václav Rozhoň, in A local approach to the Erdős-Sós conjecture, says that the result was announced in the early 1990's.
EDIT: I found another reference, Local and mean Ramsey numbers for trees, by B. Bollobás, A. Kostochka, and R. H. Schelp (J. Combin. Theory Ser. B 79 (2000), 100–103), which says, "It was
announced recently that M. Ajtai, J. Komlós, and E. Szemerédi confirmed
the Erdős–Sós conjecture for sufficiently large trees."
Interesting to see from the answers that structural graph/matroid theory has many examples (perhaps because of how many technical cases must be addressed in the proofs?)
The comments by Monroe Eskew and Andrés E. Caicedo concerning unpublished results of Hugh Woodin deserve to be made into an answer IMO. As a concrete example, Caicedo wrote:
There is the fact that Turing determinacy implies Suslin-coSuslin determinacy (in the presence of DC?), which gives L(R)-determinacy.
There are various other results by Woodin that may or may not fit the bill; in many (though maybe not all) cases, proofs have been provided by other authors. For more details, see Woodin's unpublished proof of the global failure of GCH and Unpublished works of Woodin on SCH and Radin forcing.
The Schur positivity of LLT polynomials by I. Grojnowski and M. Haiman is widely accepted in the community of algebraic combinatorics, but their preprint has not been published.
It is still a major open problem to give a combinatorial formula for the coefficients in the Schur expansion, which is manifestly positive.
Your answer highlights the distinction between "publicly available" and "published." My interpretation was that the proof needed to be unavailable, not just unpublished, but reading the question carefully, I see that it is ambivalent. Probably there are many more examples if "readily available but not formally published" counts.
It is not clear how to respond to this answer... As mentioned by Timothy Chow, I am looking for "readily available but not formally published"... Thanks for your answer :)
@PraphullaKoushik : You are? Or is that a typo? In most of the other answers, the proof is not readily available.
@TimothyChow Sorry. It is a typo.. I am not looking for "readily available but not formally published".
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2025-03-21T14:48:30.313015
| 2020-04-13T06:28:18 |
357319
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Stack Exchange
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Is there a precise relationship between the goals of moduli theory and the minimal model program?
I want to get into some of the big classification problems in algebraic geometry, but have a very broad question. Ultimately we would like to classify all varieties over some field up to isomorphism, and this is done via moduli theory. I am studying the moduli of elliptic curves at the moment. From the moduli space we construct we are able to obtain information about the relationship between different varieties based on the geometry of the moduli space. For example varieties in the same irreducible component can somehow be deformed to eachother.
On the other hand it seems like the completion of this program is hopelessly difficult for general varieties. Instead it is common to attempt to classify such objects only up to birational equivalence. This is the goal of the minimal model program.
But it seems that these two aren't really running in parallel in the sense that the latter doesn't appear to be a direct stepping stone for the former. I guess what I was hoping for is that classififying up to birational equivalence would somehow be a big step towards classification up to isomorphism. But when we classify curves up to isomorphism we appeal to the genus. But for birational equivalence we just normalize and apply Chow's lemma.
I know this is a broad question, but is there some geometry of the moduli space that would tell us when two varieties are birational equivalent, such as being in the same connected component or some such simple test? Or even more broadly, would the completion of the minimal model program give us any insight into what the moduli space for some family of varieties would look like?
I left this as a soft question since I probably don't even know enough about the subject to ask it in a precise manner. But hopefully some experts understand what I am trying to get at.
In general, the minimal model program does not tell much about isomorphism classes. Just to give an elementary examples, all Hirzebruch surfaces $\mathbb{F}_n$ ($n \neq 1)$ are minimal models of a rational surface. They are all birational, but pairwise non-isomorphic.
When $X$ is not covered by rational curves and the dimension is $\leq 2$, there is a unique minimal model in each birational class, so birational equivalence and isomorphism coincide.
When the dimension is at least $3$ the minimal model is not unique (and you must also allow some mild singularities), but two distinct minimal models are isomorphic outside subsets of codimension at least 2, and more precisely they are related by a sequence of flops.
This is a far reaching question so I kept my answer short. There are also approaches to construct nice moduli spaces of Fano varieties (and of vars of intermediate Kodaira dimension) insprired by the work of Donaldson and others. A very important step towards this is Birkar's work on the boundedness of Fanos.
Koll\'ar's paper "Moduli of varieties of general type" https://arxiv.org/abs/1008.0621 provides an excellent introduction. We now know that moduli spaces of canonical models of varieties of general type exist and once you fix the volume $K_X^{dim X}$, they can be compactified by adding stable models, obtaining a projective moduli space. Note that the volume is the higher dimensional analog of the genus as the volume of a curve is $2g-2$.
BTW, if $X$ is not a canonical model, then the volume is computed by $\lim \frac{h^0(mK_X)}{m^d/d!}$ where $d=dim X$. Since any variety of general type is birational to its canonical model (and hence shares many properties), this is considered a satisfactory answer. For example, canonical models determine the fundamental group and the cohomology groups $h^i(\mathcal O _X)$.
Note that easy examples show that moduli spaces for varieties that are not canonical models tend to be non-separated (see (4.4) of above ref).
There are likely other issues such as the boundedness.
In dimension 2, if we fix the volume and the Picard number $\rho$ of a surface $X$ then this surface is obtained from its canonical model by blowing up at most $\rho$ times. Thus we expect that these surfaces belong to a bounded family. However, in dimension $\geq 3$ if you fix the volume, the Picard number, the topological type, it is still unclear if you can bound the corresponding families (some results exist in dimension 3).
I will end by remarking that many of the most sophisticated results in the MMP play an important role in the construction of these moduli spaces.
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2025-03-21T14:48:30.313371
| 2020-04-13T07:19:52 |
357324
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Stack Exchange
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Adjoining data about singularities to "correct" the category of pure motives?
There are a few well known constructions of potential categories of pure motives for smooth projective varieties over a field. My understanding is that modulo the standard conjectures these should be equivalent and are understood to be the correct definitions in this context.
As I understand it, the contentious part of the story begins when we consider singular varieties. In this case the known constructions of pure motives are poorly behaved and we need to think of a theory of so-called mixed motives.
An analagous situation presents itself in Gorenstein homological algebra, in which the bounded derived category of coherent sheaves on a scheme $X$ does not capture the cohomological data contained in syzygies. This problem is solved by a recollement adjoining the necessary data in acyclic complexes to the bounded derived category. For specifics see the Murfet's thesis.
Is there any hope of this approach working to define an abelian category of mixed motives?
I suppose the idea would be to define some auxiliary abelian category of which is the "motives of singularities" of $X$ and then to show that there is some abelian category which is a recollement of this new abelian category and the abelian category of pure motives.
The category of pure motives is (or should be) abelian and semisimple, while the mixed motives would/should have nontrivial Ext groups. Could you add the Ext's in independently of knowing what the pure motives are? Perhaps I've misunderstood what you have in mind.
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2025-03-21T14:48:30.313530
| 2020-04-13T07:33:25 |
357327
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Stack Exchange
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$\ell$-adic schematic homotopy type
In the groundbreaking paper Champs Affines (DOI), Toen constructs a generalisation of rational homotopy types which he calls schematic homotopy types. This is part of a larger
programme of a theory developed in Toen's Homotopy types of algebraic varieties (different to motivic homotopy theory), which, to my knowledge, has not been carried out in full.
One particular direction suggested in sections 3.5.2 and 3.5.3 of Champs Affines is the development of a theory of crystalline and $\ell$-adic homotopy types.
The first has been developed by Ollson in $F$-isocrystals and homotopy types (DOI).
Has any work been published in the $\ell$-adic direction?
More generally has any more of Toen's programme been worked out in the literature?
Étale homotopy type of scheme is rather classical object, see original Artin–Mazur paper or/and Quillen works on Adams conjecture with brilliant lectures of Sullivan "geometric topology" on the same topic, where localization and completion of homotopy type were first described.
This begs the question of what is the relation between etale and schematic homotopy types. The original question is about the latter.
I was sure that they coincide, and that may be the case. There are papers by Gereon Quick on section conjecture, where the same statement as in Toen's one made.
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2025-03-21T14:48:30.313669
| 2020-04-13T08:42:02 |
357333
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"url": "https://mathoverflow.net/questions/357333"
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|
Stack Exchange
|
Idea behind Carleson's theorem modern proof "intitial reductions"
I'm having troubles to understand the philosophy behind the modern proof of Carleson's theorem. For convenience, let me state precisely what I am asking for.
For any $f \in L^2(\mathbb{R})$, let $\mathcal{Cf}:=\sup_{N \in \mathbb{Z}} \left\vert P_{-}(e^{iN\cdot}f)\right\vert$ be the maximal Carleson operator; where $P_{-}$ is the projection on negative Fourier spectrum $\{\xi <0\}$. The Carleson theorem essentially states that:
Theorem (Carleson). $|\{\mathcal{C}f > \lambda\}|_{L^2} \lesssim \lambda^{-2}\|f\|^2_{L^2}$.
In the modern proof of this theorem [I'm reading this], one instead studies the operators defined by
$Q_{\xi}f:= \sum_{s \in T} \mathbf{1}_{\omega_s^+}(\xi) \langle f, \varphi _s\rangle \varphi _s$, where the notation is as follows:
$T$ denotes the set of all tiles $I_s \times \omega_s$ such that $I_s, \omega_s$ are dyadic intervals, such that the area of the tile $I_s \times \omega_s$ is one.
$\omega_s^+$ stands for the upper half of the interval, and the $\varphi_s$ are functions such that $\hat \phi_s$ have Fourier support inside $\omega_s^{-}$ (the lower half of the interval)
It is not difficult to pass from $Q_{\xi}$ to the Carleson operator, one can take averages and get that the operator:
$$Q:=\lim_{Y \to \infty} \frac{1}{Y^2}\int_{[1,2] \times [0,Y]^2} Dil^2_{2^{-\lambda}}Tr_{-y}Mod_{-\xi}Q_{\xi} Mod_{\xi}Tr_yDil_{2^{\lambda}}^2d\lambda dy d\xi\,,$$
commutes with translations and dilations and its Kernel is made of functions with Fourier support lying on $\{\xi >0\}$, thus this operator is $P_{-}$.
My question is then:
How is it somewhat 'natural' to come up with the operators $Q_{\xi}$? How one can guess that such an operator has a similar behavior to that of $\mathcal C$?
I think there is some 'discretization' idea behind but I do not see how is it natural in any sense. Put in another way my question is: starting from $\mathcal C$ and $P_{-}$ how does one introduces the operators $Q_{\xi}$?
Does anybody has some good insights?
Firstly, I highly recommend studying the proof of the analogous result in the case of Walsh series before studying the Fourier case. In the Walsh case, all of the technical details involving cut-off functions, Schwartz tails, etc. disappear and one can navigate the (truly beautiful) argument without these technicalities obscuring the way. [I will offer the counter-intuitive claim that the pedagogical path to Carleson through first studying the Walsh case will be shorter than the path directly to Carleson despite the deceptively short modern write-ups, like Lacey and Thiele's MRL paper.]
I have my own confusions here, but let me share my thoughts.
As you mention, there is a discretization here. If you want to decompose the operator $P_-$, you use the standard decomposition $\sum_k\hat{\varphi}_k = 1_{(-\infty,0]}$, where $\hat{\varphi}_k(\xi) := \hat{\varphi}(\xi/2^k)$ is supported at frequencies $\vert\xi\vert\sim 2^k$. You can think of $\varphi_k$ as attached to the tile $I_s\times\omega_s = [-2^{-k-1},2^{-k-1}]\times[-2^k,0]$ ---this tile doesn't belong to the mesh $\mathcal{D}$, but let's ignore these "unfortunate technicalities", as Fefferman put it.
Since $P_-^2f = P_-f$, then we get the decomposition
$$
P_- f = \sum_{k,k'} \varphi_k*\varphi_{k'}*f.
$$
We may assume that $\varphi_k*\varphi_{k'} = 0$ unless $k=k'$; you can take Fourier transform to see that this is morally true. For each term in the series we get
$$
\begin{align}
\varphi_k*\varphi_k*f(x) &= \int f(z)\varphi(y-z)\varphi(x-y)\,dydz \\
&= \int \varphi_k(y)\int f(z)\varphi_k(x-z-y)\,dzdy \\
&= \int \textrm{Tr}_y\varphi_k(x)\langle f,\textrm{Tr}_y\varphi_k\rangle\,dy \\
&= \sum_{\vert I\vert = 2^{-k}}\frac{1}{2^{-k}}\int_{-2^{-k-1}}^{2^{-k-1}}2^{-\frac{k}{2}}\textrm{Tr}_{y+c(I)}\varphi_k(x)\langle f,2^{-\frac{k}{2}}\textrm{Tr}_{y+c(I)}\varphi_k\rangle\,dy
\end{align}
$$
In the third identity we used $\overline{\tilde{\varphi}} = \varphi$, where $\tilde{\varphi}(x) = \varphi(-x)$, because $\hat{\varphi}$ is real. In the last term, let's define $\textrm{Tr}_{c(I)}\textrm{Dil}_{2^{-k}}^2\varphi = \varphi_s$, where $s$ denotes the tile $(c(I)+[-2^{-k-1},2^{-k-1}])\times [-2^k,0]$. We rewrite then the last integral as the average
$$
\varphi_k*\varphi_k*f(x) = \frac{1}{2Y}\int_{-Y}^{Y}\sum_{\vert I\vert= 2^{-k}}\textrm{Tr}_y\varphi_s(x)\langle f,\textrm{Tr}_y\varphi_s\rangle\,dy,
$$
where $Y = 2^{-k-1}$; however, you can modify the argument above to see that actually you can take the limit $Y\to\infty$. Summing up we have
$$
P_-f(x) = \lim_{Y\to\infty}\frac{1}{2Y}\int_{-Y}^Y\sum_s \langle f,\textrm{Tr}_y\varphi_s\rangle\textrm{Tr}_y\varphi_s(x)\,dy,
$$
where the tiles $s$ are those here constructed.
We left open the assumption $\varphi_k*\varphi_{k'}=0$ unless $k=k'$. In the paper they took $\hat{\varphi}$ supported in an interval of length $\frac{1}{4}$, but it is then impossible, I think, to get $\sum_k\hat{\varphi}_k = 1_{(-\infty,0]}$. I suspect the the average in dilation helps to solve the problem here; in fact, I would try to find a function $\varphi$ such that $\sum_k \int_2^4\hat{\varphi}_k^2(t\xi)\frac{dt}{t} = 1_{(-\infty,0]}$, and such that the supports of $\hat{\varphi}_k$ and $\hat{\varphi}_{k'}$ are disjoint, but not sure. The square $\hat{\varphi}_k^2$ is to use the same trick $P^2_-$.
In any case, we are reduced to the operator
$$
Tf := \sum_s\langle f,\varphi_s\rangle\varphi_s.
$$
Now if we try to use it in the Carleson operator, we have to deal with
$$
\vert T(e^{iN\cdot}f)\vert = \vert\sum_s\langle f,\textrm{Mod}_{-N}\varphi_s\rangle\textrm{Mod}_{-N}\varphi_s\vert.
$$
Now there is another technicality, for the frequency support of $\textrm{Mod}_{-N}\varphi_s$ doesn't belong to the mesh $(j2^k,(j+1)2^k)$. The way out from this nuisance is to average over translations of the mesh by using $\textrm{Mod}_\xi$, much like we did above for the intervals $I$.
Excuse me if I do not complete all the computations, but I think the idea it is more or less clear. The point is that the operator $Q_\xi$ allows to get rid from many technicalities, but mainly with problems with the relative position of the mesh $\mathcal{D}$.
that is perfectly what I was looking for, you provided a lot of intuition behind this discretization! Thank you.
is it possible to clarify two points? First, how is it "natural" to use the $P_^2$ trick? Instinctively I would only use $P_{-}f= \sum_k \varphi _k * f$ and not get the same as you. This is a very clever trick.
Secondly: where does the $\int_2^4 \hat\varphi_k ^2 (t\xi)dt/dt$ trick comes from? By reverse engineering I would say that the $dt/t$ is invariant under dilations, which is really helpful, but how does it comes out naturally? I mean, it is effective for average on dilations, but at the point of your thought process you just said "average on dilations would help"
and I do not really understand why (again, in the end, it works, but I am really focusing on how one comes up with that). I remarked that in the recent blog post of T. Tao (last set of notes) he uses the very sames tricks as you, so this must not be a coincidence. Thank you!
I do like Tao's post, very clarifying in many respects. How natural $P_{-}^2$ is, I don't know, I just read it in other instances, and it burned into my head; a less sophisticated though perhaps more "natural" trick here is $\phi_1\phi_2 = \phi_2$, where $\phi_1=1$ in the support of $\phi_2$. The measure $dt/t$ is the Haar measure of the multiplicative goup in $\mathbb{R}_+$, so as the Lebesgue measure to $\tau+\xi$, so $dt/t$ to $\lambda\xi$. Thiele and Christ proof appeared only "recently", so if the steps were all natural, it'd be done many years ago.
|
2025-03-21T14:48:30.314302
| 2020-04-13T08:59:44 |
357335
|
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|
Stack Exchange
|
Statistical divergence
Does anyone know about a statistical divergence of this type?
\begin{equation}
\text{D}(P||Q) = \frac{1}{2} \left[\text{KL}(M||P) + \text{KL}(M||Q)\right]
\end{equation}
where $M = \frac{1}{2} [P+Q]$.
This divergence is very similar to the Jensen-Shannon Divergence $\text{D}(P||Q) = \text{KL}(P||M) + \text{KL}(Q||M)$ but where the distributions in the argument of the statistical divergence appear in the second argument of the KL divergences.
I am interested in knowing if such divergence exists in the literature and to properties of such divergence. Thanks!
That is J-divergence, commonly just called 'symmetrised KL divergence.'
Kullback, Solomon, and Richard A. Leibler. "On Information and Sufficiency." The Annals of Mathematical Statistics 22.1 (1951): 79-86.
Jeffreys, Harold. "An Invariant Form for the Prior Probability in Estimation Problems." Proceedings of the Royal Society of London. Series A. Mathematical and Physical Sciences 186.1007 (1946): 453-461.
Taneja, Inder Jeet. "Information and Divergence Measures." On Generalized Information Measures and their Applications. Advances in Electronics and Electron Physics. Vol. 76. Academic Press, 1989. 327-413. (Also available here: http://www.mtm.ufsc.br/~taneja/book/node21.html)
I remember hearing this is the metric K & L originally called 'divergence', its components being called 'discriminations.' First citation seems to confirm this.
This might help as a place to start. Sgarro has generalized the Kullback-Leibler divergence $D_{\rm KL}(Q||P)$ to multiple distributions by introducing the average divergence
$$D’(P_1,\ldots P_k)=\frac{1}{k(k-1)}\sum_{i,j=1}^k D_{\rm KL}(P_i||P_j),$$
so what you have is related to his measure of divergence by
$$
D(P,Q)=3D’(P,Q,M)-\frac12D_{\rm KL}(P||Q)
-\frac12D_{\rm KL}(Q||P)-\frac12D_{\rm KL}(P||M)
-\frac12D_{\rm KL}(Q||M).
$$
where $M=\frac12(P+Q).$
A. Sgarro, “Informational divergence and the dissimilarity of probability distributions”, Calcolo(18): 293–302 (1981).
|
2025-03-21T14:48:30.314458
| 2020-04-13T10:22:50 |
357338
|
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"url": "https://mathoverflow.net/questions/357338"
}
|
Stack Exchange
|
FLT and integral points on elliptic curves
For integers $x,y,z,t,n$ define $S_n : xy(x+y)=t^n$.
For $ n > 2$, Fermat's Last Theorem implies there are no integral
solution on $S_n$ with $x,y$ coprime and $xy(x+y) \ne 0$ since $x,y,x+y$ are
n-th powers.
For fixed $t,n$, $S_n$ is elliptic curve and it might have infinitely
many rational solutions.
Q1 Do we have elementary constraints on the coprime solutions
of $S_n$?
Partial results.
Solving symbolically, for $y=y_0^n$ we have $4t^n y_0^n + y_0^{4n}=y_1^2$
over the integers.
Homogenize $S_n$ to get $S'_n : xy(x+y)=t^n z^3$
and fix $t,n$.
$S'_n$ has the long Weierstrass model $v^2-t^n v = u^3$
and the projective Weierstrass model $E_n : w v( v-t^n w)= u^3$.
The map from $S'_n$ to $E_n$ is $(z,x,(x+y)/t^n)$
and the inverse map is $(v,-v+t^n w,-u)$
The short Weierstrass model is $v^2=u^3+16 t^{2n}$.
Q2 For $n > 6,t >1$, are there coprime integer solutions $u,v$
to the short Weierstrass model?
Related question: https://mathoverflow.net/questions/160350/on-x3-y2-1728-text-unit-in-number-fields?rq=1
|
2025-03-21T14:48:30.314560
| 2020-04-13T11:11:54 |
357342
|
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|
Stack Exchange
|
Lower bounds of kappa class functions
I saw in the paper "Smooth Satabilization Implies Coprime Factorization" of
Eduardo Sontag the following argument: Given a smooth map $a:\mathbb{R}
^{n}\rightarrow\mathbb{R}^{+}$, let $\rho$ be any smooth function of class
$\mathcal{K}_{\infty}$ such that
$\lim_{\xi\rightarrow\infty}a\left( \xi\right) \rho\left( \left\Vert
\xi\right\Vert \right) =+\infty$.
It is claimed that $\rho$ exists. Let $\alpha_{3}^{\ast}$ be any smooth
function of class $\mathcal{K}_{\infty}$ so that
$
\alpha_{3}^{\ast}\left( s\right) \leq\inf\left\{ a\left( \xi\right)
\rho\left( \left\Vert \xi\right\Vert \right) :\left\Vert \xi\right\Vert
=s\right\}
$
for all $s\geq0$.
My main concern is on the existence of such $\alpha_{3}^{\ast
}\in\mathcal{K}_{\infty}$ because a positive-definite function is not always
lower-bounded by a function in the class $\mathcal{K}_{\infty}$. This is used
in that paper to make the following boundedness:
$-a\left( \xi\right) \rho\left( \left\Vert \xi\right\Vert \right)
\leq-\alpha_{3}^{\ast}\left( \left\Vert \xi\right\Vert \right)$
|
2025-03-21T14:48:30.314647
| 2020-04-13T11:13:36 |
357343
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/357343"
}
|
Stack Exchange
|
What is the uncertainty on the (Pearson) correlation coefficient?
Do you know what is the uncertainty on the Pearson correlation coefficient as a function of the uncertainty on the measurement in the data set.
I know of an expression giving the uncertainty related to the limited size of the data set, but I'm looking for the uncertainty related to the measurement of the data itself, which I think must dominate.
Many thanks!
$\newcommand\tsi{\tilde\sigma}
\newcommand\tY{\tilde Y}
\newcommand\tZ{\tilde Z}$
Let $(Y_1,Z_1),\dots,(Y_n,Z_n)$ be iid copies of a pair $(Y,Z)$ of real-valued random variables (r.v.'s) with finite fourth moments and correlation $\rho\in(-1,1)$. Let $R=R_n$ be the Pearson correlation coefficient for the "random sample" $(Y_1,Z_1),\dots,(Y_n,Z_n)$. Then, by the multivariate delta method, the r.v.
$$\frac{R_n-\rho}{\tsi/\sqrt n}$$
converges (as $n\to\infty$) in distribution to a standard normal r.v., where $\tsi:=\sqrt{\tsi^2}$,
$$\tsi^2:=Var\big(\tY\tZ-\tfrac\rho2(\tY^2+\tZ^2)\big),$$
$$\tY:=\frac{Y-EY}{\sqrt{Var\,Y}},\quad \tZ:=\frac{Z-EZ}{\sqrt{Var\,Z}}.$$
Here it is assumed that $Var\,Y$, $Var\,Z$, and $\tsi$ are nonzero.
So, the asymptotic standard deviation $\tsi/\sqrt n$ of $R_n$ can serve as a natural measure of uncertainty of the values of $R_n$ due to the variation in the values of the random points $(Y_1,Z_1),\dots,(Y_n,Z_n)$ and their failure to lie on a straight line.
Details on all this can be found on page 1016 of this paper; see, in particular, Theorem 3.4 (and the paragraph preceding it) and Remark 3.5 on that page.
Theorems 3.4 and Corollary 3.8 in the linked paper also provide (optimal) $O(1/\sqrt n)$ bounds on the rate of the mentioned convergence of the distribution of $R_n$ to normality. See also Remark 3.2 in the linked paper concerning general results on asymptotic expansions of nonlinear functions of the multivariate sample mean due to Bhattacharya and Ghosh.
|
2025-03-21T14:48:30.314812
| 2020-04-13T11:34:36 |
357345
|
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"sort": "votes",
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|
Stack Exchange
|
Is $Tor_A(k,k)$ a bicommutative Hopf algebra?
Let $A$ be a commutative (or graded commutative) algebra over a field $k.$ In some sources, such as Mcleary's book on spectral sequences, Corollary 7.12, pg. 248, it is claimed that $\text{Tor}_A(k,k)$ is a bicommutative Hopf algebra. Is this a typo? Obviously, $\text{Tor}$ has a commutative algebra structure, but is it true that the coaddition is cocommutative?
What is the comultiplication? To get the multiplication we apply $\operatorname{Tor}_A(-,k)$ to $k \otimes k \rightarrow k$ and then precompose with the map $\operatorname{Tor}_A (k,k) \otimes \operatorname{Tor}_A (k,k) \rightarrow \operatorname{Tor}_A(k \otimes k,k)$. But to get the comultiplication it seems like we can't play the same game with $k \rightarrow k \otimes k$ since the canonical map $\operatorname{Tor}_A (k,k) \otimes \operatorname{Tor}_A (k,k) \rightarrow \operatorname{Tor}_A(k \otimes k,k)$ is in the wrong direction.
Proposition 7.10 gives the comultiplication: $[a_1,\dots, a_n] \rightarrow \Sigma [a_1, \dots, a_j] \otimes [a_{j+1},\dots,a_n]$.
This is not true. Consider the algebra $A=T(V)/V^{\otimes 2}$, it is a commutative algebra whose augmentation ideal has zero multiplication. We have $\mathrm{Tor}_A(k,k)\cong T(V[1])$ with the shuffle product and deconcatenation coproduct, so the coproduct is very much noncommutative (it is the coproduct of the cofree conilpotent coalgebra co-generated by $V[1]$).
To me, this coproduct looks like the dual of something like the cup product which makes me think it should be cocommutative. Is this completely wrongheaded?
@ConnorMalin I can't guess your thinking process but one possibility is that you are confusing the concatenation product with the shuffle product?
|
2025-03-21T14:48:30.314948
| 2020-04-13T12:46:22 |
357352
|
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"Iosif Pinelis",
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|
Stack Exchange
|
restriction of a formula with matrix inverse multiplied by a vector
I'm trying to reproduce a proof from this paper but I'm stuck in one point (Lemma 6). The general subject is bayesian model for multi-armed bandit problem solved with Thompson sampling.
I think I don't know any properties / tricks from linear algebra that can be useful here. Maybe you can help?
First there are a couple of coefficients:
$\sigma_1 > 0, \sigma_2 > 0, \sigma_3>0$
$ N $ number of possible vectors of features (arms)
$ d $ dimentionality of vectors of features
$x_i \in R^d, i \in \{1, 2, ... N\}$ - vectors of features
$ T $ number of time steps
$ t \in {1, 2, ..T} $ a selected moment in time
$ n_{i,t} $ - how many times we observed vector $x_i$ up to time t - we know that $ \sum_{i=1}^{N} n_{i,t} = t$ it's a random number (but depend on the algorithm)
$$
A_t = \frac{1}{\sigma_3^2} + \sum_{i=1}^{N} \frac{n_{i,t}}{\sigma_1^2 + \sigma_2^2 * n_{i,t}} * x_i * x_i^T
$$
We need to restrict somehow for all t:
$$
\sqrt{x_i^T * A_t^{-1} x_i}
$$
or restrict
$$
\sum_{t=1}^{T} \sqrt{x_i^T * A_t^{-1} x_i}
$$
in terms of $\sigma_1 > 0, \sigma_2 > 0, N, d, T$. Apparently $\sigma_3$ is not present in the results.
Do you think you can give me some tips what can we used here or how to proceed?
It's possible that we need some more reasonable assumptions that are not stated in the paper directly. (for example about $x_i$).
It's possible that the whole proof should be approached differently. Any comments are welcome
The paper says: "The detailed proof is in supplementary materials." Did you find the proof there?
It's not possible to obtain these materials anywhere - I searched the internet, wrote to authors and someone from AAI - no one has replied to me.
|
2025-03-21T14:48:30.315088
| 2020-04-13T13:55:34 |
357358
|
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|
Stack Exchange
|
Irreducibility of product bicomodules
Let $H$ be a Hopf algebra, and $V$ and $W$ a left, and a right,
$H$-comodule respectively. The tensor product
$$
V \otimes W
$$
has an obvious $H$-$H$-bicomodule structure.
If $V$ and $W$ are irreducible as left and right comodules, then
is $V \otimes W$ irreducible as a $H$-$H$-bicomodule?
No, off course, not. You need a slightly stronger condition. One of your comodules need to be absolutely irreducible.
To prove it, apply the fundamental theorem of coalgebra. The comodules give you simple subcoalgebras $C_M$ and $C_N$ in the coradical of $H$. You are asking for the algebra $(C^{\ast}_M)^{op}\otimes C^{\ast}_N$ to be simple.
For an elementary example let ${\mathbb R}$ be the ground field, $H$ the free cocommutative Hop algebra generated by the trigonometric coalgebra ${\mathbb C}^{\ast}$. Let $M=N={\mathbb C}$ be the $\mathbb C$-module, hence, ${\mathbb C}^\ast$-comodule, hence $H$-comodule. The $H$-bicomodule structure on $M\otimes N$ is essentially its module structure over ${\mathbb C}\otimes_{\mathbb R}{\mathbb C}\cong {\mathbb C}\oplus{\mathbb C}$. This is not irreducible because $dim(M\otimes N)=4$ but its simple modules are 2-dimensional.
Sorry but I don't see why what I am asking for is equivalent to $(C^_M)^{op} \otimes C^_N$ being simple. (Also, I guess $C^_M$ denotes the linear dual of $C^_M$ endowed with convolution multiplication?)
You are asking for $M\otimes N$ to be a simple $(C^{\ast}_M)^{op}\otimes C^{\ast}_N$-module. And you know that $M\otimes N$ is a faithful finite-dimensional $(C^{\ast}_M)^{op}\otimes C^{\ast}_N$-module. Now simplicity of the algebra $(C^{\ast}_M)^{op}\otimes C^{\ast}_N$ implies simplicity of the module $M\otimes N$. In the opposite direction, hit it with Jacobson Density Theorem.
Yes or no, since you have a typo: $C^{\ast}_M$ is the linear dual of coalgebra $C_M$.
Yes, it should read "$C^*_M$ denotes the linear dual of $C_M$".
|
2025-03-21T14:48:30.315233
| 2020-04-13T14:10:26 |
357359
|
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|
Stack Exchange
|
Equivalence of σ-convex hull and closed convex hull
Let $X$ be a locally convex topological space, and let $K \subset X$ be a compact set. Recalling that the standard convex hull is defined as
$$\text{co}(K) = \Big\{ \sum_{i=1}^n a_i x_i : a_i \geq 0,\, \sum_{i=1}^n a_i = 1,\, x_i \in K \Big\},$$
define the $\sigma$-convex hull as
$$\sigma\text{-}\mathrm{co}(K) = \Big\{ \sum_{i=1}^\infty a_i x_i : a_i \geq 0,\, \sum_{i=1}^\infty a_i = 1,\, x_i \in K \Big\},$$
where the summation is to be understood as convergence of the sequence in the topology of $X$.
I would like to understand conditions under which $\sigma\text{-}\mathrm{co}(K)$ is exactly the closure of $\mathrm{co}(K)$. In particular, does this property hold for any separable normed space $X$, or are further constraints on $X$ (and $K$?) required?
The motivation for this question is Choquet's theorem, which allows one to write
$$\overline{\mathrm{co}}(K) = \Big\{ \int x d\mu(x) : \mu \in M(K) \Big\}$$
with $M(K)$ standing for probability measures on $K$ for any compact subset $K$ in a normed space. I would like to understand the "countable" version of this theorem as presented above, but I could not find any references nor do I have an idea about how one could prove it.
Related: this question
Consider $\ J:=(0;1)\subseteq\Bbb R.\ $ Then the sigma closure is $\ J;\ $ it is not the closure, i.e. $\ [0;1].$
@WlodAA: It seems, though, that the OP considers only compacts sets $K$.
@JochenGlueck, thank you.
there are exercises 1.66 and 1.67 in Fabian, Habala Hajek, Montesinos, Zizler - Banach space theory, dedicated to these notions, although they do not adress your specific question
Wlod AA gave a good counterexample for the case when $K$ is not required to be compact, here I give a counterexample $K$ compact, first in a locally convex space, and then for a(n infinite-dimensional) separable normed space, and (after an edit) for all infinite-dimensional Banach spaces.
There is a standard counterexample if $X$ is only required to be locally convex, which is to take $X = C([0,1])^*$ with the weak-* topology, and to take $K$ to be the set of unital ring homomorphisms $C([0,1]) \rightarrow \mathbb{R}$. Making free use of the Riesz representation theorem to consider elements of $C([0,1])^*$ as measures on $[0,1]$, the elements of $K$ are the Dirac $\delta$-measures. Now, for each element $\mu$ of $\sigma\mbox{-}\mathrm{co}(K)$, there exists a countable set $S \subseteq [0,1]$ such that $\mu([0,1]\setminus S) = 0$. However, $\overline{\mathrm{co}}(K)$ consists of $P([0,1])$, the set of all positive unital linear functionals on $C([0,1])$, i.e. all probability measures on $[0,1]$, and so Lebesgue measure is an element of $\overline{\mathrm{co}}(K) \setminus \sigma\mbox{-}\mathrm{co}(K)$.
To get this to happen in a normed space, we will use $\ell^2$, and embed $P([0,1])$ affinely and continuously into it. First, observe that we can affinely embed $P([0,1])$ into $[0,1]^{\mathbb{N}}$, getting each coordinate by evaluating at $x^n$ (including $n = 0$). This is injective because polynomials are norm dense in $C([0,1])$, and continuous by the definition of the weak-* topology. We can then embed $[0,1]^{\mathbb{N}}$ into $\ell^2$ by the mapping:
$$
f(a)_n = \frac{1}{n+1}a_n
$$
this is affine and continuous from the product topology on $[0,1]^\mathbb{N}$ to the norm topology on $\ell^2$ (in fact, it defines a continuous linear map from the bounded weak-* topology on $\ell^\infty$ to the norm topology on $\ell^2$). We use $e$ for the composition of these two embeddings, and it is affine and continuous on $P([0,1])$.
A continuous injective map from a compact Hausdorff space to a Hausdorff space is a homeomorphism onto its image, and as we also preserved convex combinations by making the embedding affine, we have that $\overline{\mathrm{co}}(e(K)) = e(\overline{\mathrm{co}}(K)) = e(P([0,1]))$, while, taking $\lambda$ to be the element of $P([0,1])$ defined by Lebesgue measure, $e(\lambda) \in e(P([0,1]))$, but $e(\lambda) \not\in e(\sigma\mbox{-}\mathrm{co}(K)) = \sigma\mbox{-}\mathrm{co}(e(K))$.
Added in edit:
As Bill Johnson points out, there is an injective bounded map from $\ell^2$ into any infinite-dimensional Banach space $E$. By the same argument used to transfer the example to $\ell^2$, this allows us to transfer the example to $E$.
In the other direction, the convex hull of a compact subset $K$ of a finite-dimensional space is compact (using Carathéodory's theorem we can express the convex hull of $K$ as the continuous image of the compact set $K^{d+1} \times P(d+1)$, where $d$ is the dimension. Therefore the $\sigma$-convex hull and closed convex hull of $K$ coincide.
All together, this means:
If $E$ is a Banach space, the statement "for all compact sets $K \subseteq E$, the closed convex hull equals the $\sigma$-convex hull" is equivalent to "$E$ is finite-dimensional".
There are, however, complete locally convex spaces in which every bounded set, and therefore every compact set, is contained in a finite-dimensional subspace, and for which, therefore, the $\sigma$-convex and closed convex hulls of compact sets coincide. One example is the space $\phi$ of finitely supported functions $\mathbb{N} \rightarrow \mathbb{R}$, topologized as an $\mathbb{N}$-fold locally convex coproduct of $\mathbb{R}$ with itself, or equivalently as the strong dual space of $\mathbb{R}^{\mathbb{N}}$.
Thank you, I'll need to go through this carefully. Do you have any ideas about the least set of assumptions about the space $X$ to make the property hold true for any compact $K$?
@GregoryD. A sufficient condition is that $X$ is finite-dimensional. Then the convex hull of a compact set is compact, so is equal to both the $\sigma$-convex and closed convex hulls (it is easy to prove this using Carathéodory's theorem. It follows that the property does hold in spaces in which every compact set is finite-dimensional, such as an infinite locally convex coproduct of $\mathbb{R}$ with itself.
@GregoryD. It may well be that this property does not hold for any infinite-dimensional Banach space, but my knowledge of this kind of functional analysis has run out, so I hope one of the experts can help on that subject.
@RobertFurber: Beautiful counterexamples! Interestingly, there are also infinite-dimensional compact sets $K$ for which the $\sigma$-convex hull coincides with the closure of the convex hull - for instance $K = {e_n/n: , n \in \mathbb{N}} \subseteq \ell^2$, where $e_n$ denotes the $n$-th canonical unit vector. It really seems interesting how to characterize such sets $K$; but I wasn't able to come up with any ideas, yet.
Once you have an example in $\ell_2$ you can transfer it to any infinite dimensional Banach space $X$ by taking an injective continuous linear mapping from $\ell_2$ into $X$. It is an exercise for students that such a map exists.
@BillJohnson Thank you Bill. That is one of those statements that becomes easy to do once you know it's true.
|
2025-03-21T14:48:30.315693
| 2020-04-13T15:36:22 |
357368
|
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|
Stack Exchange
|
The Angel and Devil problem with a random angel
In the classic version of Conway's Angel and the Devil problem, an angel starts off at the origin of a 2-D lattice and is able to move up to distance $r$ to another lattice point. The devil is able to eat a lattice point, preventing the angel from ever moving to that point. The angel and devil take turns, and the devil wins if the angel is at some point no longer able to move. The question then is for what values of $r$ does the angel win and for what values does the devil win? This problem was essentially solved completely by distinct proofs by Kloster and Máthé that the 2-angel can escape. (It is easy to see that a 1-angel can be trapped.) One can generalize this problem to higher dimensions; note that if for a given choice of $r$ the angel can escape in $d$ dimensions, then the angel will escape for any higher $d$.
What I'm interested in is the situation where the angel moves randomly (uniformly distributed among all possible legal moves) but the devil has a pre-determined strategy (allowed to depend on $r$ and $d$ but not allowed to depend on any choices the angel has made). For what $r$ and $d$ can the devil beat an angel with probability 1?
It isn't too hard to see that if $d=2$ the devil can win with probability 1. Here's the basic strategy the devil uses: Pick a very fast growing sequence of positive integers, $a_1$, $a_2$, $a_3 \cdots$. The devil works in stages: At each stage, the devil eats a square of side length $a_n$ centered about the origin, and with thick walls of thickness $r$. Each such square requires about $4ra_n+r^2 \sim 4ra_n$ moves by the devil. But by the standard result that a random walk is with probability 1 never much more than the square root of the number of steps away from the origin, in the time the devil has taken to eat the $a_n$ square, the angel with probability 1 will have only moved about $\sqrt{4r}\sqrt{a_n}$ steps from the origin. So, the devil just creates larger and larger squares of this sort, and eventually the angel will be trapped. (This by itself will put the angel in a finite region, but trapping in a finite region is essentially the same as being unable to move since the devil can go back and fill in these squares ever so slowly, say eating a single lattice point near the origin before moving on to start making each new large square.
This construction fails for 3 dimensions. To make a cube of that size takes about $6a_n^2$ steps, so the angel has a high probability of being near the boundary.
Question 1: can this strategy or a similar one be modified to work for $d=3$? My guess is yes for $d=3$, but I don't have a proof. I also don't have any intuition for higher dimension.
One standard observation which simplifies the analysis of the original problem is that one may without loss of generality assume that the angel never returns to the same lattice point. If it did, it would have used a suboptimal strategy, since it is back where it was earlier but with the devil having eating out a few lattice point. So, we can define another variant of the problem with an angel which chooses randomly, but only out of lattice points it has not yet reached.
Question 2: Given this non-repeating angel, is there a strategy for the devil to win with probability 1?
I suspect that the answer for $d=2$ is that the same basic strategy should still work; my suspicion here is that with probability 1, the angel's distance at $k$ steps should be bounded by $k^{(\frac{1}{2}+\epsilon)}$ in which case the same proof would go through. But I'm much less certain about what happens here if $d=3$.
Why did you restrict the devil's strategy to not know the angel's position? I can't see a winning strategy for the devil even if it knows where the angel is.
@OscarCunningham , That's a valid question. Presumably if the devil is allowed in this context to react, then it will have an easier time. I'd strongly suspect that if the devil is allowed to know the angel's position than for any r and d, the devil will win with probability 1. But I also don't have a proof for that situation either.
I think you're right. The tesseravore could use the strategy of always eating one of the squares in the angel's movement range diametrically opposed to the origin. My intuition is that this would be enough bias to cause the angel's random walk to always return to the origin. (My intuition isn't strong here though, high dimensional spaces are weird.)
What if the non-repeating angel reaches a dead-end: a point from where there is no move to a non-visited site? (Self-avoiding walks are weird, too...)
@MateuszKwaśnicki I would consider that functionally a loss for the angel in the non-repeating case. But maybe that's not the best response because that means that the non-repeating case isn't strictly better for the angel. Good point though that I hadn't thought about!
@JoshuaZ: This way the angel would always lose on their own! Oh wait, the devil can block moves that would lead the angel to a dead end. Plot twist: can the devil save the (non-repeating) angel with probability one?
I think it should be possible to check that the devil can win in dimension $3$ by building a box, then building a much bigger box, then a much much bigger box... Since the angel's radius is typically on the order of the square root of the time, they have a positive probability of being trapped by each box, which adds up to probability $1$ over time. But I don't see any quick way to do the random-walk-outside-a-box central limit theorem that would prove this. If this is done, only $d=4$ is left.
What if the devil just moves randomly? Wikipedia (the page "Random walk") just informed me that "Paul Erdős and Samuel James Taylor also showed in 1960 that for dimensions less or equal than 4, two independent random walks starting from any two given points have infinitely many intersections almost surely, but for dimensions higher than 5, they almost surely intersect only finitely often." Unless I'm missing something, doesn't this mean that the devil can win in dimensions 3 and 4 by moving randomly?
@WillBrian That would work if one was using the variant of the game in Will Sawin's answer below where the angel concedes if they randomly choose a square that the devil has already eaten (which incidentally given his comment gives a full characterization of that variant), but in this version the angel picks randomly from the non-eaten squares, so it isn't clear what happens. But the fact that a random method apparently does do so well in a very similar game seems like good evidence that the devil will win on d=3 or d=4.
@JoshuaZ: Oops -- I was indeed thinking of the other version of your problem, described by the other Will. You're right that it's not clear what a random devil can do in that version. Although my guess is that the devil never succeeds in trapping the angel in a finite region, and the angel escapes with positive probability.
In dimension $5$ and above, a random angel escapes a blind devil with positive probability, as long as $r$ is sufficiently large.
To see this, let's replace the angel with one that chooses a point to move with from all points within a distance $r$, and if it chooses a point the devil has eaten, it instantly concedes the game. The winning probability of this angel is clearly less than the winning probability of the original angel, since the original angel is just this angel with an extra probability of surviving at certain times and possibly living forever.
Since this angel is just following an ordinary random walk, after $n$ moves it has an $O ( n^{-d/2})$ probability of being on any particular lattice point, by the central limit theorem. Thus it has an $O( n \cdot n^{-d/2})$ probability of touching a lattice point the devil has eaten. Summing over $n$, we get $O(1)$ as long as $d>4$. Because the constant goes to $0$ with $r$, the probability of never touching such a lattice point is positive with $r$ sufficiently large.
So this leaves just dimensions 3 and 4 open.
@JoshuaZ Indeed - in the comments, I sketch a possible approach to $d=3$. I suspect $d=4$ is the trickiest.
I've added a bounty. I'll award it to anyone who either solves d=4 or anyone who can get the details of your d=3 to work.
Even $r=1$ should work for positive probability of escape: let the maximum pointwise probability of the (weakened) angel's location at a specific coordinate (given the devil's moves so far, which only decrease it) be $p_n<1$. Then the statement that $\sum_{i=1}^\infty p_i<\infty$, shown in this answer, is equivalent to $\prod_{i=1}^\infty (1-p_i)>0$, i.e. the devil has nonzero odds of all their moves failing. (Since we're conditioning on each failure, the probabilities at each step are independent.)
This long comment is to show that if the devil can see him, he can trap him. E.g. in 2-d, it takes (5r)^2 - (4r)^2 steps to build a box he can't get out of at a distance of 4 moves from his current local, that would be covering up all the moves between distance 4r and 5r into which he must step to escape.
Wait unitl he has take 5r steps in the same direction into fresh space. The devil can force him to the border of empty space, and then as the 5 moves into it are legal moves, he makes them with some probablity. The point of waiting for him to move into fresh space is to make sure that these attempts to escape are all the same. His time to hit the inner edge of the devils box is not bounded, so with some probability he hasn't hit the inner edge before you are done. If he does, give up and start a new one. You get infinitely many tries.
Excellent. That at least solves @Oscar Cunningham's quesiton above.
Doesn't the non-repeating angel trap itself with probability 1 even without the devil's help? I mean there is always going to be some finite set of moves the angel can make that leave it nowhere to go. Since it has infinitely many tries, it should make this sequence of moves eventually.
I pointed this out in a comment, and asked whether in this case the devil can save the angel. :-)
|
2025-03-21T14:48:30.316515
| 2020-04-13T15:38:31 |
357369
|
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|
Stack Exchange
|
Approximating a projection by a sum of elementary tensors with a certain property
Let $A$ and $B$ be two C$^{*}$-algebras and suppose we have a non-zero projection $p\in A\otimes B$. (We can assume $A$ is nuclear, so that there is only one possible tensor product.)
Does there exist a choice of elements $a_{1},\ldots,a_{n}\in A$ and $b_{1},\ldots,b_{n}\in B$ such that:
$\left\|\left(\sum_{i=1}^{n}a_{i}\otimes
b_{i}\right)-p\right\|<\frac{1}{2}$;
If $\pi$ is a non-zero irreducible representation of $A$ and $(\pi\otimes\operatorname{id})(p)=0$, then $\pi\left(\sum_{i=1}^{n}|a_{i}|\right)=0$?
All I can deduce is that for a given choice of $a_{i}$'s and $b_{i}$'s, and and irreducible representation $\pi$ of $A$, we have
$$
\left\|\sum_{i=1}^{n}\pi(a_{i})\otimes b_{i}\right\|=\left\|(\pi\otimes \operatorname{id})\left(\left(\sum_{i=1}^{n}a_{i}\otimes b_{i}\right)-p\right)\right\|\leq \left\|\left(\sum_{i=1}^{n}a_{i}\otimes b_{i}\right)-p\right\|<\frac{1}{2},
$$
provided that $(\pi\otimes\operatorname{id})(p)=0$.
Are you sure you've stated the problem correctly? What you have written is clearly false, because you can always throw $x\otimes y - x\otimes y$ into the sum. So you would have to have $\pi(|x|) = 0$ for all $x$.
I am not picky about what the $a_{i}$'s and $b_{i}$'s are so long that the approximation for $p$ holds. I guess I am wondering if there is a way to choose them so that the quoted result is always true.
Given your "I guess I am wondering" could you please edit your question to give a more precise statement of what hypotheses you are assuming and what conclusion you want?
I have updated the question.
Could you change to a more specific title?
It's now a sensible question. I think it may be hard. Is any tensor product of projectionless C*-algebras also projectionless? I'm not sure this is known.
Does it help if $B=\mathcal{O}_{2}\otimes\mathcal{K}$?
Yes, this is possible, assuming that $A$ (or $B$) is nuclear. The same argument below (using that exact $C^\ast$-algebras are locally reflexive) also works if $A$ or $B$ is exact and the tensor product is the spatial (aka minimal) tensor product.
The role of nuclearity is that any two-sided closed ideal $J \subseteq A\otimes B$ is the closed linear span of all tensor products $I_A \otimes I_B$ of two-sided closed contained in $J$, see Corollary 9.4.6 in Brown and Ozawa's book. Now, let $J_A \subseteq A$ be the intersection of all two-sided, closed ideals $J \subseteq A$ such that $p \in J \otimes B$ and let $I:= \bigcap J \otimes B$ where the intersection is indexed by such $J$. Clearly $J_A \otimes B \subseteq I$. For the converse (which is not as trivial as it a priori looks), take $I_A \otimes I_B \subseteq I$ where $I_A$ and $I_B$ are two-sided closed ideals in $A$ and $B$ respectively. Then $I_A \subseteq J_A$ and $I_B \subseteq B$, so $I_A \otimes I_B \subseteq J_A \otimes B$. By nuclearity of $A$, $I$ is the closed linear span of all such $I_A\otimes I_B$, so $I = J_A \otimes B$. In particular, $p\in J_A \otimes B$.
Now, let $a_1,\dots, a_n \in J_A$ and $b_1,\dots, b_n \in B$ such that
\begin{equation}
\| \sum_{i=1}^n a_i \otimes b_i - p \| < 1/2.
\end{equation}
Let $\pi \colon A \to \mathcal B(\mathcal H)$ be an irreducible representation such that $(\pi \otimes \mathrm{id}_B)(p) = 0$. The image of $(\pi \otimes \mathrm{id}_B)$ is canonically $\pi(A) \otimes B$, so by nuclearity of $A$ (or $B$) and exactness of maximal tensor products, it follows that
\begin{equation}
0 \to (\ker \pi)\otimes B \to A \otimes B \to \pi(A) \otimes B \to 0
\end{equation}
is exact, so the kernel of $(\pi \otimes \mathrm{id}_B)$ is $(\ker \pi )\otimes B$. Hence $p \in (\ker \pi)\otimes B$ so $a_1,\dots, a_n \in J_A \subseteq \ker \pi$ by construction of $J_A$.
If $A$ or $B$ is exact, we get $\ker (\pi \otimes \mathrm{id}_B)$ either by using that $A$ is locally reflexive or $B$ is exact to get that
\begin{equation}
0 \to (\ker \pi) \otimes_{\min{}} B \to A \otimes_{\min{}} B \to \pi(A) \otimes_{\min{}} B \to 0
\end{equation}
is exact.
Excellent answer!
|
2025-03-21T14:48:30.316782
| 2020-04-13T15:47:26 |
357370
|
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|
Stack Exchange
|
What is a reference for this sort of test set system that avoids all sets of size $\le k$?
My question is: is there a standard name for a structure like the following?
For positive integers $n$, $k < n$ define a "$k$-set-free test for $n$" as a set $C$ of subsets of the integers $\{0, \dots, n-1\}$ such that: for every $i \in \{0, \dots, n-1\}$ and every $S \subset \{0, \dots, n-1\} \setminus \{i\}$ with $|S| \le k$ there exists $T \in C$ such that $i \in T$ and $S \cap T = \emptyset$.
Is this a standard combinatorial structure, or a simple transform of a standard combinatorial structure? Does anybody have a convenient reference to this sort of problem?
Follow up notes and questions.
Of interest is: bounding $|C|$.
Our application is: we have an unknown set $B \subset \{0, \cdots, n-1\}$ with $|B| \le k$ of "bad" integers and a test procedure that says a set $T$ is spoiled iff $T \cap B \neq \emptyset$. Find a small set of test-sets to identify $B$ using the test procedure.
Trivially the set $C = \{ \{ i \} | i \in \{0, \dots, n-1\} \}$ is an $n-1$-set-free test for $n$ of size $n$. However, it is easy to show for small $k$ there are $C$ with $|C|$ small (for example $|C| = 2 k \lceil 1 + (k+1) \log_2(n)\rceil $ will do, link).
The problem is a standard error-correcting coding problem if test was working over a field (i.e. counting intersection sizes instead of checking non-emptiness).
Is there an $O(k \log(n))$ sized solution? Or can one establish a lower bound that is larger than that?
This is an instance of the set cover problem. I used integer linear programming to obtain the minimum $|C|$ for $1\le k < n \le 12$. Do any of these results surprise you?
\begin{matrix}
n\backslash k &1 &2 &3 &4 &5 &6 &7 &8 &9 &10 &11 \\
\hline
2 &2 \\
3 &3 &3 \\
4 &4 &4 &4 \\
5 &4 &5 &5 &5 \\
6 &4 &6 &6 &6 &6 \\
7 &5 &7 &7 &7 &7 &7 \\
8 &5 &8 &8 &8 &8 &8 &8 \\
9 &5 &9 &9 &9 &9 &9 &9 &9 \\
10 &5 &9 &10 &10 &10 &10 &10 &10 &10 \\
11 &6 &9 &11 &11 &11 &11 &11 &11 &11 &11 \\
12 &6 &9 &12 &12 &12 &12 &12 &12 &12 &12 &12 \\
\end{matrix}
Also, this type of problem is called group testing.
The references are great and exactly what I needed. The table results are not surprising, as the method isn't going to work until n is much larger than k. Thanks!
|
2025-03-21T14:48:30.316966
| 2020-04-13T17:14:40 |
357376
|
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Stack Exchange
|
Another kind of primality related to tessellations by polygons
You can define a number $p$ to be prime by "no tessellation of $p$ identical squares forms a convex figure". This suggests what I'll call a t-prime $p$, defined by "no tessellation of $p$ identical equilateral triangles forms a convex figure".
Are there infinitely many t-primes?
I have thought about a "scaffold" in order to prove this analogously to the corresponding assertion for (regular) primes.
For some body who don’t know what tessellation is, it means arrangement.. I didn’t knew, I did google just now :P
For squares, you have to exclude the trivial case of a 1xp rectangle formed by these squares. Similarly, you need to put some similar restriction on the convex figures formed by triangles.
These convex regions are either hexagons, pentagons, parallelograms or trapezoids or equilateral triangles. Have you tried expressing the area of such a figure as a multiple of the area of the base triangle for each case? Likely you'll find restrictions on the associated areas and maybe those are enough to find the answer (whatever it is).
Triangles <-> triangular numbers, (nontrivial) parallelograms <-> multiples of 4, trapezoids <-> differences of two triangular numbers.
@OlivierBégassat I think you made an error with parallelograms - 2 equilateral triangles stuck on each other form a parallelogram.
I think representability by quadratic forms applies here: Take a large triangle and subtract three small triangles to get a representation of many numbers less than the number of unit triangles in the large triangle. I suspect all sufficiently large numbers "look like" irregular hexagons this way. This would mean a no answer to your question. Gerhard "Even Embedded In Two Ways" Paseman, 2020.04.13.
@user44191 I followed Wojowu's comment which I interpreted as "exclude lines of triangles"
@OlivierBégassat I can still get 18, or twice any composite number, with that restriction on parallelograms.
How about convex tesselations by $1\times2$ rectangles? (excluding trivial cases)
@GerryMyerson. 3 is composite as 2 vertical rectangles with a horizontal on top show; then adding more horizontals on top show all larger numbers to be composite too - unless you have some way of defining trivial arrangements that excludes those. But it would be really nice if there were some non trivial way of rescuing this question... I thought about Penrose tilings, but that didn't give anything interesting as far I could see...
I believe the only t-primes are $1,2,3$ and $5$. OP hasn't clarified what kind of "trivial" configurations they wish to exclude, but hopefully the ones below are nontrivial enough.
Here is a visual proof of the fact any number $n\geq 7$ is not t-prime. The idea is to first give constructions of convex figures consisting of $6k+1$ triangles (first row below) and then appending moe triangles to cover other residue classes mod $6$ (second row). Excuse my crude drawings.
|
2025-03-21T14:48:30.317211
| 2020-04-13T17:51:46 |
357378
|
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|
Stack Exchange
|
Cohn's eight diophantine equations
Today I was reading J.H.E. Cohn's Eight diophantine equations (1966). On p. 158
he comes across the equation $y^2 = a^3 + 3a$ for odd values of $a$ and writes that this is equivalent to $x^3 + (x+1)^3 = (\frac y2)^2$ for $x = \frac{a-1}2$. He then claims that it is known that this equation has the only integral solutions $x = 0$ and $x = 1$ and refers to Dickson's History, vol. II, p. 580. Dickson refers to Moret-Blanc (1882), who reduces the claim to the diophantine equation $3u^4 - v^4 = 2$ with the obvious solution $u = 1$. Then he states that Lucas has shown that this is the only solution in integers; Lucas does give infinitely many solutions, one of which is $z = 1$, but he does not solve the equation completely.
sage can compute the integral points on $y^3 = x^3 + 3x$ in a couple of seconds
(there are four, namely $x =0$, $x = 1$, $x = 3$ and $x = 12$), but now I'm wondering a) whether this was known when Cohn wrote his paper in 1966, and b) whether there is a sufficiently nice solution of $x^3 + (x+1)^3 = z^2$ that can be done by hand. The tricks in my bag don't seem to work.
P.S. I now found that in an addendum to his article, Cohn mentions the gap and writes that a full proof of the claim was given by Ljunggren in his thesis, available to readers with Norwegian IPs here.
in your third line, I think the exponent of $y/2$ must be 2
Thank you, Will.
https://artofproblemsolving.com/community/c3046h1048509_
@individ: The problem of writing the difference of two consecutive cubes as a square leads to an equation of genus 0 and therefore can be parametrized. The problem above leads to an elliptic curve.
|
2025-03-21T14:48:30.317379
| 2020-04-13T17:56:13 |
357379
|
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"Lucia",
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|
Stack Exchange
|
How to explain this prime gap bias around last digits?
My question is related to this article by Oliver and Soundararajan (article about a bias in the distribution of the last digits of consecutive prime numbers).
After trying some python experimental researches on the prime gaps, I notice a bias in the gaps around each last digits {1,3,7,9}.
Could this bias be intrinsically linked to the bias found in the distribution of the last digits of consecutive prime numbers?
Experimental results
Here some experimental plot results up to the 3,500,000th prime number.
Two set-ups are calculated:
1. Cumulative and average gaps
2. Cumulative and occurence of particular gap size
1. Cumulative and average gaps
1.1 Cumulative/average prime gaps after last digits:
+ Cumulative/average prime gaps before last digits
1.2 Cumulative/average prime gaps before + after last digits
2. Cumulative and occurence of a particular gap size
2.1 Cumulative/occurence of gaps size of 10 after last digits
- Animated cumulative/occurence per incremented gap sizes after last digits
3. Misc
3.1 Additional research on a waveform script of the sieve of Ératosthène,
colored by last digits:
+ plot 2
The script is on colab.research and is ready to fork
Why the downvotes? Perfectly reasonable question.
This bias probably exists and seems to be on the scale of $1/\log x$, which is different from the bias of size $\log \log x/\log x$ in Lemke Oliver and Soundararajan. You could analyze this in the same way, but I have not carried out the calculations.
Yes, it can be analyzed in the same way. Since the effects you're measuring are similar to each other, I'm only going to address the first one, cumulative prime gaps after primes with a fixed last digit. In Lemke Oliver and Soundararajan's paper, they count pairs of consecutive primes $p$ and $p_{\text{next}}$ up to $X$ satisfying $p \equiv a$ and $p_{\text{next}} \equiv b \pmod{10}$, or a sum of the form
$$\sum_{\substack{h > 0 \\ h \equiv b-a}} \#\{ p \le x : p \equiv a \pmod{10}, p_{\text{next}} - p = h\}.$$
This setting can be described by two variations to the above sum. First, by summing the prime gaps, each term is being weighted by $h$. Second, since it's all prime gaps after primes with a fixed last digit, the expression is also summed over possible choices of $b$.
Armed with these modifications their analysis should follow through very similarly. For example, the proof of Prop. 2.1 in their paper can be carried out, but instead of considering
$$F_{q,\chi}(s) = \sum_{h \ge 1} \frac{\chi(h)}{h^s} \mathfrak S_q(\{0,h\}),$$
weighing each term by a factor of $h$ means that instead we're considering $F_{q,\chi}(s-1)$. The poles of this function have shifted, so the corresponding sum for $S_0(q,0;H)$ in the statement of Prop. 2.1 would have a leading term of linear size in $H$, rather than log size.
As Lucia commented, the bias seems to be on a smaller scale than the bias from the paper. This reduction in bias should be a result of summing over all values of $b$, which has some averaging effect. As a test of this, we can consider the Main Conjecture on page 2 and sum the $\log \log x /\log x$ term over all values of $b$ with fixed $a$; the result is that it disappears entirely, leaving the $1/\log x$ term.
Another crude experiment can be done by taking the data on page 2 of their paper, assuming that the gap is larger than zero but otherwise as small as possible, and computing what the corresponding sum would be in this case. For $a = 1$, this would be the sum $$10 \cdot \pi(x_0;10,(1,1)) + 2 \cdot \pi(x_0;10,(1,3)) + 6 \cdot \pi(x_0;10,(1,7)) + 8 \cdot \pi(x_0;10,(1,9)).$$ To compare to your data, one can take this a step further by treating the total of these sums as a ``total prime gap'' and taking the percentage of the total for each. The result of doing this with the values from page 2 is:
\begin{array}{c|cc}
a & \text{Sum} & \text{Percent of total} \\
\hline
1 & 149655728 & 25.96 \\
3 & 165705632 & 28.75 \\
7 & 125284384 & 21.73 \\
9 & 135805698 & 23.56 \\
\end{array}
A huge limitation here is that these sums are heavily influenced by the chance that numbers coming immediately after $a$ have of being prime (in particular, they are in increasing order beginning at 5 mod 10). Lemke Oliver and Soundararajan explicitly address this effect, showing that it doesn't account for the full bias. And, in fact, this doesn't quite match the data that you have. Nevertheless, some discrepancy is visible even in this crude model.
First of all, I would like to thank you Vivian for your answer which allows me to understand more deeply the distinction to be made among the "full" bias.
Your crude experiment with "virtual" deviations is very interesting.
I now assume that my data displayed as a bar takes into account the full bias because of the averaging effect of summing over all values.
Finally, the observed pattern of your sum calculated from the data on page 2 is not very far from my results in terms of occurrences between the last digits.
And I would be really confused to learn that the discrepancy visible on my 2.1 data (occurence of gaps size of 10 after last digits) is a unique result of $1/\log x$ term.
Does this experimental research on the prime gaps around the last digits make sense to you? Does it need to be further investigated?
For counting gaps of size exactly 10, I'd still expect a $\log \log x/\log x$ term. Here you wouldn't see an averaging effect, since you care about the values (mod 10) of the prime before and the prime after the gap. So, it does make sense that these gaps would be bigger. An analogous "crude experiment" would be to assume that $\pi(x_0; 10,(1,1))$, for example, on page 2, counts only pairs of primes with gaps equal to 10, instead of all that are 0 mod 10. Then you could take their values of $\pi(x_0;10,(a,a))$ for $a = 1,3,7,9$ and compare to your data.
My 2.1 plot is finally exactly what is computed page 2, which is an evidence.
With data from page 2, we have:
$\begin{array}{c|cc}
a & \text{Sum} & \text{Percent of total} \
\hline
1 & 4623042 & 25.50 \
3 & 4442562 & 24.50 \
7 & 4439355 & 24.48 \
9 & 4622916 & 25.50 \
\end{array}$
While my data gives :
$\begin{array}{c|cc}
a & \text{Sum} & \text{Percent of total} \
\hline
1 & 11726 & 28.33 \
3 & 8969 & 21.67 \
7 & 9023 & 21.8 \
9 & 11666 & 28.19 \
\end{array}$
Biases clearly follow the same pattern but diminish on larger $x$ due to $\log \log x/\log x$ term i guess
|
2025-03-21T14:48:30.318173
| 2020-04-13T18:13:41 |
357382
|
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|
Stack Exchange
|
Finding a square integrable dominating function for function class
problem statement
For any $x \in R^d$, consider the function $$\phi(x,a) = \min_{1\leq j \leq k} \|x-a_j\|^2,$$
where $a = (a_1, \dots, a_k) \in R^{kd}$ and $\| \cdot \|$ is the $L_p$ norm for any $p \geq 1$.
Let $X$ be a continuous random vector in $R^d$ with distribution $P$ such that $\int \|x\|^2 dP(x) < \infty$ and assume measurability is taken care of for all functions.
Consider for some fixed $b$ and for some $R>0$, the class of functions:
$$ G_R := \{ g(x) = \phi(x,a) - \phi(x,b) : \| a - b\| \leq R \} $$
I'd like to find a function $F(x)$ and constant $R>0$ such that:
1) $F(x) \geq |g(x)|, \forall g(x) \in G_R$.
2) $\int {F(x)}^2 dP(x) = O(R^2),$ and $\forall \epsilon >0, \int {F(x)}^2 1\{F(x) > \epsilon\} dP(x) = o(R^2)$ as $R \rightarrow 0$.
my attempt
I have a solution for the case of $p=2$ (Euclidean norm) which I have trouble generalizing for other $L_p$ norms.
For $a = (a_1, \dots, a_k)$, let $A_j$ denote the set of points in $R^d$ that are closest to $a_j$ among $\{a_1, \dots, a_k \}$ according to $\| \cdot \|$).
Notice that if $x \in interior({A_j})$ and $h$ sufficiently small,
\begin{equation}
\label{jlinearexpansioneqn}
\phi(x;{a+h}) = \|x-a_j-h_j\|^2 = \|x-a_j\|^2 - 2h_j'(x-a_j) + \|h_j\|^2.
\end{equation}
Hence, since we restrict analysis to continuous densities, boundaries between $A_j$ and $A_i$ have zero measure. So we have the following taylor expansion:
\begin{equation}
\label{linearexpansioneqn}
\phi(x;{a+h}) = \phi(x;{a}) + {h}'\delta(x,{a}) + \|{{h}}\|r(x,{a},{h}),
\end{equation}
where $\delta(x,{a})$ is the gradient of $\phi(x;{a})$, the $k$ vector of $L^2(P)$ functions with element $j$ equal to $-2(x-a_j) 1\{x\in A_j\}$.
Furthermore,
$$ r(x,{a},{h}) = \sum_{j=1}^k \frac{\|{h_j}\|^2 1\{ x \in A_j\} } {{\|{h}\|}} \rightarrow 0 \text{ as } {h} \rightarrow 0 \quad \text{ a.s. in } x .$$
In addition,
\begin{align}
\label{remainderdominationeqn}
\begin{split}
|r(x,{a},{h})| &\leq {\|{h}\|}^{-1} ( |{h}'\delta(x,{a})| + \max_j | \|{x-a_j-h_j}\|^2 - \|{x-a_j}\|^2 | ) \\
&\leq \|{\delta(x,{a})}\| +{\|{h}\|}^{-1} \sum_j | \|{x-a_j-h_j}\|^2 - \|{x-a_j}\|^2 | ) \\
&\leq C(1+\|{x}\|)
\end{split}
\end{align}
for some $C>0$ and ${h}$ sufficiently small.
Let $R>0$ be small enough such that the above domination is true for $a=b,h=a-b$. Then by the expansion above and Cauchy-Schwarz:
\begin{align}
\begin{split}
|g(x) | & = \left| \phi(x;{a}) - \phi(x;b) \right| \\
&= \left| ({a} - b)'\delta(x,b) + { \|{a} - b}\| r(x,b,{a} - b) \right| \\
&\leq {\|{a} - b}\| \|{ \delta(x,b)}\| + C { \|{a} - b}\|(1+\|{x}\|) \\
&= B(x) \|{{a} - b}\|,
\end{split}
\end{align}
where $B(x) = \|{ \delta(x,b)}\| + C(1+\|{x}\|) \in L^2 (P)$.
The dominating function $F(x) = RB(x)$ satisfies 1) and 2).
Thanks for your help!
The introduction of the $p$-norm looks a little bit strange for me, since the rest seems to be for the $2$-norm case. Can you explain where this type of "generalization" comes from? What is the context?
My attempt is for the p=2 case because the gradient and the hessian of $\phi(x, a)$ is simpler. The gradient when $p \neq 2$ is already pretty ugly. As for the hessian, it is the identity matrix for $p=2$ but it changes along the boundary for $p \neq 2$ so it does not seem like the method of taylor expansion would help in $p \neq 2$ cases. I'm interested in this problem because in statistics, we check that a criterion function (in this case $\phi$) satisfies certain properties to derive asymptotic behavior for the minimizer/maximizer of $P\phi(\cdot,a)$.
|
2025-03-21T14:48:30.318541
| 2020-04-13T18:14:06 |
357383
|
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|
Stack Exchange
|
Fourier transform on Minkowski space
Physicists Some people like to define the "Fourier transform" on Minkowski space as $\hat f(\xi) = \int e^{i \eta(x,\xi)} f(x) dx$, where $\eta(x,\xi)$ is the Minkowski form. I'm used to thinking of the Fourier transform as a canonical isomorphism $L^2(K) \to L^2(\hat K)$ where $K$ is a locally compact abelian group and $\hat K$ is its Pontryagin dual. But this "Minkowski-Fourier transform" doesn't seem to arise in this way.
Questions:
Is there an abstract framework in which to understand this "Minkowski-Fourier transform"? For instance, is there a general theory of "Fourier transforms" on spaces equipped with a nondegenerate symmetric bilinear form? Is there a relationship between this "Minkowski-Fourier transform" and the representations of a suitable "Heisenberg group"?
Which properties of the usual Fourier transform on Euclidean space are shared by the "Minkowski-Fourier transform"? For instance, what is a precise statement of the Fourier inversion formula in this context?
Is there a good reference for the mathematical properties of the "Minkowski-Fourier transform"?
Perhaps it's worth adding that physicists seem to be a bit blase about using this "Minkowski-Fourier transform", and treat it as though it were an ordinary Fourier transform.
Up to a harmless sign change isn't this an instance of the Fourier transform of locally compact abelian groups $K$ with $K=\mathbb R^n$? Physically this is just the translations on Minkowski space (including translations in time).
Also I just realised that we do not actually insert the Minkowski metric inside the Fourier transform like that.
@AlexArvanitakis This appears, for instance, on p. xxi of Peskin and Schroeder, where it's just called a "Fourier transform" (note that according to p. xix, they denote the Minkowski form as a dot product). I'm unconvinced that this sign change is harmless -- for instance, it completely changes the symmetries of the situation from the action of the compact group $O(n)$ to the noncompact Lorentz group.
Ah I see what is happening. Implicit in physicist textbooks is the use of the isomorphism between Minkowski space and its dual in every formula that does not contain indices. So for example by default I'd parse the $k\cdot x$ in Peskin and Schroeder as $k_\mu x^\mu$ (as I mentioned above) but you could also write it as $k^\mu x^\nu \eta_{\mu\nu}$ in Einstein summation notation. This simultaneously defines what you'd call a "Euclidean" Fourier transform (as a function of $k_\mu=\eta_{\mu\nu}k^\nu$) or your "Minkowskian" Fourier transform (as a function of $k^\mu=\eta^{\mu\nu}k_\nu$).
Thanks, that's clarifying! This is strange -- I should have noticed that if you write $x^\mu$ for position and $p_\mu$ for momentum, then you don't need any kind of pairing to write down $p_\mu x^\mu$! It's only if (for some reason) you decide to write things in terms of $p^\mu$ rather than $p_\mu$ that a pairing is needed... I have seen the Fourier transform written out explicitly with the signs from the Minkowski form, but it was in a more math-flavored source, so it seems they were just (somewhat perversely!) working with the dual of momentum space!
As AlexArvanitakis explained, it is just the canonical Fourier transform up to a non-canonical identification $K \cong \hat{K}$ of a vector space with its dual using the Lorentzian inner product.
This sort of thing comes up even for Fourier series, because we do not work directly with the Pontryagin dual, which is a space of functions, but we choose an isomorphism with a well-known space. For Fourier series we do not use $\hat{\mathbb{Z}}$ directly, but $\mathbb{R}/\mathbb{Z}$ or $\mathbb{R}/2\pi\mathbb{Z}$, depending on what isomorphism we choose.
Well you seem to have worked it out but I wrote most of this before your comment happened: I claim there isn't any material difference between your "Minkowski space Fourier transform" and the usual Fourier transform on ${\mathbb R}^n$: in fact write $$ \hat f(\xi)\equiv \int e^{i\eta(x,\xi)} f(x) dx $$ for any non-degenerate bilinear form $\eta$. Then there exists another such form $\eta^{-1}$ so $\eta(x,\eta^{-1}\zeta)= \langle x,\zeta\rangle$ where $\zeta \in ({\mathbb R}^n)^\star$. Clearly $$ \hat f(\eta^{-1} \zeta)=({\mathcal F}f)(\zeta)\,,$$ where $\mathcal F$ is the usual --- ``Euclidean'' --- Fourier transform.
In physics texts this $\zeta$ variable is the down-index momentum ($k_\mu$ in e.g. Peskin and Schroeder) while $\xi$ is the up-index momentum ($k^\mu$ in e.g. Peskin and Schroeder). Derivatives play perfectly nice with up/down index notation, which allows one to be blase about whether the Fourier transform involves the up- or down-index $k$.
Mathematically speaking you're taking the Pontryagin dual of Minkowski space seen as a group of translations, which is exactly the same as that of the corresponding Euclidean space. More abstractly the conserved charge in the sense of Noether's theorem corresponding to a translation is the down-index momentum rather than the up-index one.
I should mention that it is essential, and also implicit, in physics contexts that any indexful expression has a "fundamental" definition, from which all others arise via index gymnastics. This isn't very important for quantum field theory --- because all metrics are flat --- but it's crucial in general relativity. For momenta in particular one defaults to the index-down $k_\mu$ (as I did above). This can be important if e.g. you have a space of the form ${\mathbb R}^n\times T^m$, in which case you need to be very precise about what lattice the "momenta along the toroidal directions" lie in.
|
2025-03-21T14:48:30.318936
| 2020-04-13T18:31:17 |
357384
|
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|
Stack Exchange
|
algorithm for convex $C^2$ interpolation
Let $x_0<x_1<\ldots<x_n$ and $f_0,f_1,\ldots,f_n$ be real numbers and
$$s_i=(f_i-f_{i-1})/(x_i-x_{i-1}),~~~c_i=(s_{i+1}-s_i)/(x_{i+1}-x_{i-1}).$$
If $f$ is a convex function defined on $[x_0,x_n]$ with $f(x_i)=f_i$ for $i=0,\ldots,n$ then all $c_i$ are nonnegative. Conversely, this condition guarantees that a convex function $f$ with this property exists, namely the piecewise linear interpolant. Necessary and sufficient condition for realizing a twice continuously differentiable function $f$ are given in
R. Delbourgo,
Shape preserving interpolation to convex data by rational functions with quadratic numerator and linear denominator,
IMA J. Numer. Anal. 9 (1989), 123-136.
with an algorithm for constructing such a $C^2$ function that involves solving a nonlinear system of equations. Is there a simpler algorithm that does not require solving nonlinear equations?
The premise is insufficient to guarantee the existence of a $C^2$ function. For example, $c_2>c_1=c_3=0$ would only permit piece-wise linear interpolant from $x_0$ to $x_4$. $c_i$ are all positive would suffice though.
@Hans: Yes, this is already in Delbourgo's paper, as noted in my statement.
If $c_i$'s are all positive, there are infinitely many such convex $C^2$ functions. As I have pointed out in my comment above, nonnegativity of $c_i$'s is insufficient to guarantee the existence of a $C^2$ function. One very simple construction via Bezier curve is as follows.
Draw a straight line through each point $(x_i,f_i)$ such that all other points lie above the line. In each interval, construct a quartic Bezier curve as follows. Set the control points of the quadratic Bezier curve resulting from the previously drawn straight lines. Then make the midpoint of each line segment a control point (doubling the control number minus one). Draw the quartic Bezier curve from these control points.
The reason for the construction is that the tangent vector (first derivative) with respect to the parameter ($t$ in the Wikipedia article) of a Bezier at an end point is the attached line segment while the second derivative of the curve is the difference of the two closest line segments. We are making the tangent vectors on both sides of the end (data) point coincide and the difference vectors vanish thus equal. You get a $C^2$ curve with a continuous first derivative and a continuous second derivative vanishing at each data point.
The above algorithm, proving the existence of the convex $C^2$ interpolation function, has a vanishing second derivative at each data point. That makes the first derivative run parallel to the $x$ axis each time the curve reaches a data point, making the first derivative wiggly. It does not have to Having proved the existence of an $C^2$ interpolation, we can make the first derivative of the convex $C^2$ interpolating curve smoother by constructing a higher order Bezier curve by connecting neighbouring data points with many small line segments of almost equal lengths each turning almost a constant angle. This will eliminate the horizontal running points from the first derivative and makes it appear smoother.
Could you please explain why this Bezier construction gives a $C^2$ function?
@ArnoldNeumaier: I have added more details as well as the motivation of the construction. Let me know if it is clear.
OK. As I suspected, quadratic Bezier curves as in the previous version do not provide the required smoothness, but quartics do. This solves my problem from a theoretical point of view. Unfortunately, as in the case of BernM's answer, the derivative is very wiggly, hence it is not suitable for visual applications of the derivative.
@ArnoldNeumaier: This is just one of the easier examples to give a solution and a constructive proof. The wiggle in the derivative comes from the second derivative being zero. This is not necessary. You can construct higher order Bezier curves to smooth out the derivative while preserving $C^2$.
I agree that this is a nice costructive proof of sufficiency.
@ArnoldNeumaier: I am saying more than that. Could you please read the rest of my last comment? You can get as smooth a first derivative as you want with higher order Bezier curves.
Higher order smoothness, yes, but the derivative remains wiggly, with a huge number of inflection points when there are lots of data points.
@ArnoldNeumaier: Wiggliness is not a necessary consequence of a higher Bezier curve. As I mentioned in my last comment, I have addressed that issue in my second last comment. To make it clearer, I added a paragraph at the end of my answer describing in more detail the construction algorithm for the interpolation eliminating the inflection points and the subsequent wiggles.
I am not convinced that this eliminates the wiggles. It apparently only produces smaller, but more wiggles.
Have you coded up the algorithm for higher order Bezier curve and seen the results?
No; did you? Your recipe is not yet an algorithm. I need to be convinced first that the second derivative stays close to what one would expect from local interpolation. If this is not the case, wiggles are very likely.
This reference
Mulansky, Bernd; Schmidt, Jochen W. Constructive methods in convex C2
interpolation using quartic splines. Numer. Algorithms 12 (1996), no. 1-2, 111–124
may be helpful, but there certainly are more recent ones.
This paper indeed gives a finite recipe without the need for solving an auxiliary nonlinear problem or optimization problem. Hence it solves my problem as posed. However, the explicit choice given is not visually pleasing (strong oscillations of the curvature) unless the freedom in the placement of the intermediate nodes is used to minimize a shape-quality measure via a quadratic program. Thus from a practical point of view it still uses a nontrivial optimization step.
Of course, you are right.
Here is some theoretical background:
There always exists a (strictly) convex interpolating POLYNOMIAL to strictly convex data (all $c_i$ positive).
Let S be a finite dimensional subspace of $C^1$. Then there exist strictly convex data which do not admit convexity preserving interpolation from S.
Reference: Mulansky, Neamtu: Interpolation and approximation from convex sets. J. Approx. Theory 92 (1998), no. 1, 82–100
Edit: After some thoughts, I have concluded that the natural cubic spline can not guarantee to be convex for any convex data. The following algorithm does produce a unique natural cubic spline going through all the data points but it is not guaranteed to be convex. The regression/smoothing algorithm guarantees $C^2$ and convexity but not going through each data points.
The Bezier curve construction I supply in the other answer however does provide a simple solution.
You can use cubic spline to not only interpolate these discrete data points but also regress them so that the resulting cubic spline is convex. For the regression problem, minimize
\begin{equation}\label{eq:splineLoss}
L[g] = (1-\lambda)\sum_j w_j(g(t_j)-y_j)^2+\lambda\int_a^b g''(t)^2dt
\end{equation}
for
$$g\in C^2[a,b],\quad g''(t)\ge 0, \quad \lambda\in[0,1].$$
It becomes a quadratic programming problem under the constraint that $g''(t_j)\ge0,\,\forall j$.
These are described and proved in detail in
P.J. Green, Bernard. W. Silverman, Nonparametric Regression and Generalized Linear Models: A roughness penalty approach (Chapman & Hall/CRC Monographs on Statistics & Applied Probability Book 58).
The procedure is even more specifically spelled out in
Berwin A. Turlach, Shape constrained smoothing using smoothing splines
Both of these accounts present the regression (smoothing) algorithm which includes the interpolation which is what OP is asking as a special case.
but regression destroys the interpolation property
@ArnoldNeumaier: I am saying this is a more powerful method than just interpolation. It is "not only" good for interpolation which is easier "but also" good for regression.
The problem is that there is no convex cubic spline interpolant to arbitrary strictly convex data, unless one introduces variable intermediate nodes. This turns the proposed optimization into a nonlinear problem.
@ArnoldNeumaier: That is not true. There is a unique natural cubic spline interpolation for any discrete data. If the data is convex, the interpolating function is convex.The optimization problem is indeed nonlinear but it is, as I have said, a quadratic programming problem.
Where can I find a proof of this property of the natural cubic spline?
@ArnoldNeumaier: I have edited my answer to include the references present much more and stronger results than what you are asking for by treating it as a special case.
I don't have access to the book. Will read the article by Turlach and then decide.
I looked at the paper you cited. Your problem is a QP in infinite dimensions, hence not directly tractable by standard QP software.The author reports that the exact solution of your problem may have more knots tnan the original data (hence need not agree with the natural interpolation spline), and that finding these nodes needs the solution of a nonlinear system. Then he proposes a seqential QP algorithm converging to the solution. This is not the kind of answer I was looking for.
@ArnoldNeumaier: Apparently you have completely misread the paper. In fact, for interpolation, it is only a linear algebra problem, i.e. Equation $(2.3)$ of the paper with $R$ being positive definite due to its strictly diagonal dominance and $g$ set to data. There is nothing more to be done for your purpose, no QP whatsoever. All the rest of the paper is for regression. Even for regression, it is not an infinite dimensional problem. The knots are exactly at the data points. The sequential QP is for regressing for general positive function which is not what you are seeking. But all of
@ArnoldNeumaier: Once you get $\gamma$, use Appendix A to obtain the coefficients to Equation $(2.1)$ to compute the cubic spline.
Too long to comment:
Here's another take, restricting to higher order splines. Suppose the polynomials comprised in the spline are given by $\{p_{0,1}(x),\cdots,p_{n-1,n}(x)\}$. Let $p_{i,i+1}(x) = a^{(0)}_{i,i+1} + a^{(1)}_{i,i+1}x + \cdots + a^{(n)}_{i,i+1}x^{n-1}$, $\forall ~i$, $n$ being odd. The variables here would, of course, be the coefficients of these polynomials (degree, we'll talk about it later). The following points are in order now:
(1) Interpolation constraints $\{x_i,f_i\}$ imply linear equality constraints on the coefficients. In particular those will be:
$$
a^{(0)}_{i,i+1} + a^{(1)}_{i,i+1}x_i + \cdots + a^{(n)}_{i,i+1}x_i^{n-1} = f_i, ~\forall ~i\\
a^{(0)}_{i,i+1} + a^{(1)}_{i,i+1}x_{i+1} + \cdots + a^{(n)}_{i,i+1}x_{i+1}^{n-1} = f_{i+1},~\forall ~i.
$$
$C^2$ requirement will imply another bunch of linear constraints. In particular those will be the following set of equations:
$$
a^{(1)}_{i-1,i} + 2a^{(2)}_{i-1,i}x_i + \cdots + (n-1)a^{(n)}_{i-1,i}x_i^{n-2} =
a^{(1)}_{i,i+1} + 2a^{(2)}_{i,i+1}x_i\cdots + (n-1)a^{(n)}_{i,i+1}x_{i}^{n-3},~\forall ~i,
$$
and
$$
2a^{(2)}_{i-1,i} + 6a^{(3)}_{i-1,i}x_i+ \cdots + (n-1)(n-2)a^{(n)}_{i-1,i}x_i^{n-3} =
2a^{(2)}_{i,i+1} + 6a^{(3)}_{i,i+1}+ \cdots + (n-1)(n-2)a^{(n)}_{i,i+1}x_{i}^{n-2},~\forall ~i.
$$
(2) Requirement of convexity over $[x_i,x_{i+1}]$, implies that the polynomial given by $p_{i,i+1}''(x)$ is positive over $[x_i,x_{i+1}]$. A uni-variate polynomial over $[x_i,x_{i+1}]$ is non-negative if and only if (see Victorial Powers and Bruce Reznick, Polynomials that Are Positive on an Interval for a nice explanation):
$$
p_{i,i+1}''(x) = g_{i,i+1}(x) + (x_{i+1}-x)(x-x_i)h_{i,i+1}(x),
$$
where $g_{i,i+1}$ and $h_{i,i+1}$ are SOS polynomials of degree at most $n$ and $n-2$, respectively. Now, $g_{i,i+1}(x)=z^\top G_{i,i+1}z$, where $G\succeq 0$ and $z=[1 ~x \cdots x^{(n-1)/2}]^\top$. Similarly, $h_{i,i+1}(x)=y^\top H_{i,i+1}y$, where $H_{i,i+1}\succeq 0$ and $y=[1 ~x \cdots x^{(n-1)/2-1}]^\top$. The matrices $G_{i,i+1}$ and $H_{i,i+1}$ are additional variables. Comparing coefficients on either side of the equation given by:
$$
2a^{(2)}_{i,i+1} + 6a^{(3)}_{i,i+1}+ \cdots + (n-1)(n-2)a^{(n)}_{i,i+1}x_{i}^{n-2} = g(x) + (x_{i+1}-x)(x-x_i)h(x),
$$
yields linear affine equations in coefficients of $p_{i,i+1}(x)$, $G$ and $H$. For brevity, let these equations be given as $\mathcal{L}_{i,i+1}\left(a^{(0)}_{i,i+1},\cdots,a^{(n)}_{i,i+1},G_{i,i+1},H_{i,i+1}\right)=0$. Note that you get a set of linear equations $\forall ~i$.
(3) An SDP solver (CVXPY, or alike) can now be used to find a feasible solution, i.e.:
$$
\min ~~1~~\mbox{subject to}\\
a^{(0)}_{i,i+1} + a^{(1)}_{i,i+1}x_i + \cdots + a^{(n)}_{i,i+1}x_i^{n-1} = f_i, ~\forall ~i\\
a^{(0)}_{i,i+1} + a^{(1)}_{i,i+1}x_{i+1} + \cdots + a^{(n)}_{i,i+1}x_{i+1}^{n-1} = f_{i+1},~\forall ~i\\
a^{(1)}_{i-1,i} + \cdots + (n-1)a^{(n)}_{i-1,i}x_i^{n-2} =
a^{(1)}_{i,i+1} + \cdots + (n-1)a^{(n)}_{i,i+1}x_{i}^{n-3},~\forall ~i\\
2a^{(2)}_{i-1,i} + \cdots + (n-1)(n-2)a^{(n)}_{i-1,i}x_i^{n-3} =
2a^{(2)}_{i,i+1} + \cdots + (n-1)(n-2)a^{(n)}_{i,i+1}x_{i}^{n-2},~\forall ~i\\
\mathcal{L}_{i,i+1}\left(a^{(0)}_{i,i+1},\cdots,a^{(n)}_{i,i+1},G_{i,i+1},H_{i,i+1}\right)=0\\
G_{i,i+1},H_{i,i+1} \succeq 0, \forall i.
$$
Lack of a feasible point, however, does not mean there does not exist a $C^2$ function, of a spline with degree higher than the one chosen or outside the realm of splines.
(4) The choice of the degree of the polynomial is something I don't have a clear answer for. I would assume that a degree greater than 8 should work -- for every polynomial piece, there are 2 interpolation constraints, 4 for $C^2$, and 2 LMI constraint.
(5) A disadvantage of this method is that its not practical to work with several thousand data points.
Hope it helps.
You have not provided an (operable) algorithm. Neither does your scheme guarantee a $C^2$ convex function interpolation.
@Hans, thanks for the comment. Will type out all the necessary details in the next edit. However, I do not see why it does not guarantee $C^2$ and convexity. To clarify (1)$C^2$ is guaranteed as long as the poly parts and two derivatives are continuous at interpolation points. This translates the linear equalities as I've mentioned in the answer. And (2)a uni-variate function is convex if and only if its second derivative is non-negative at every point over the desired range. This translates to convex LMI, which comes from a SOS poly-rep of uni-variate positive polys over a compact interval.
I am saying you need to guarantee the continuity of $f, f', f''$ at the nodes $x_i$'s. Not all sums of squares polynomials ensure that. You have not provided any algorithm much less a proof to ensure the existence of such sums of squares polynomials as the second derivatives in the intervals and first derivatives at the nodes. Also, it would be great if you could use the full name of a nomenclature before using its acronym.
Have added a lot of more details. Hope it helps. For the acronyms part, had to do because of the character limit. Sorry for the inconvenience.
Thank you for providing the details, particularly the crucial paper. I like the formulation. However, it still lacks a proof of the existence of a feasible $G$ and $H$ for this SDP problem. My answer using the Bezier curve or Bernstein polynomial interestingly provides just such an existence proof.
The proof of existence of a feasible $G$ and $H$ is intricately related to the choice of the degree of the spline, and I still do not have a clear answer for that. Glad that your method has an existence proof. Haven't looked into its details yet, but you seem to have used cubic splines and proved existence. This should serve my purpose as well, just that our formulations are different.
On second thought, my Bezier curve construction (Bernstein polynomials in each coordinate) proves the existence of a $C^2$ convex interpolating function, but not the existence of a polynomial of one coordinate against another. So it can not serve as an existence proof for your assertion using polynomials. You do not have a proof. I doubt this particular algorithm works since the maximal degree of the polynomial is prescribed while the data can force a transition from arbitrarily high to low curvatures in a given interval. That may be hard to achieve with a given polynomial degree.
By the way, you should put @[name] in your comment directed at [name]. I can comment without the [name] because the comments are below the answer where the default [name] is that of the author of the answer.
@Hans, thanks for the insight. I get your point on the polynomial relationship between coordinates of the Bezier curve, and that it cannot prove as a proof of existence. I have added to the point (3) of my answer to highlight the fact of non-existence of a solution (and hence of a proof) for a given degree.
You are welcome. The way the last statement of step (3) is put forth is interesting. But that is alright. However, the ansatz in step (4) seems to contradict the uncertainty of the polynomial.
|
2025-03-21T14:48:30.319970
| 2020-04-13T18:34:38 |
357385
|
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|
Stack Exchange
|
A linear algebra question regarding the eigenvalues of the product of a diagonal matrix and a projection matrix
I need to prove a statement in my research. The statement seems to be fundamental linear algebra, and numerical studies in MATLAB supported this statement, but I wasn't able to prove it after a few days of effort... Any insight would be greatly appreciated!
The problem is as follows:
Let $A$ be a $p$-by-$p$ diagonal matrix with distinct positive diagonal elements $a_1, \dots, a_p$. Let $Z := A^{1/2} 1_p$, where $1_p$ is a $p$-vector of ones. Let $$P := I - Z \left( Z^T Z \right)^{-1}Z^T$$ where $I$ is a $p$-by-$p$ identity matrix. Let $\xi_1,\dots,\xi_{p-1}$ be the $p-1$ nonzero eigenvalues of $AP$. Show that $1+t\xi_i$, $i=1,\dots,p-1$ are $p-1$ nonzero eigenvalues of $P+tAP$, where $t \in \mathbb R$ and $t \neq 0$.
What does $1+t\xi_i$ mean?
1 + t$\xi$ is "the sum of one and the product of t and $\xi_i$"
$\xi$ is a vector and $1$ is a number- that's why I am confused by your notation.
@DarthVader Each $\xi_i$ is a number (one of the $p - 1$ eigenvalues of $AP$).
I see- I mistakenly read $\xi$ as eigenvector :). Sorry,
I was able to prove the statement and generalize it to the following lemma:
Lemma: Let A be a full-rank $p\times p$ symmetric matrix and P be a $p\times p$ projection matrix with rank $r<p$. Let $\xi_1, \dots, \xi_{r}$ be the $r$ nonzero eigenvalues of $AP$. Then for any $t\in\mathbb R$, $1 + t\xi_i$, $i=1, \dots, r$ are the $r$ nonzero eigenvalues of $(I_p + tA)P$.
Proof:
If $t=0$, the result follows immediately from the fact that the eigenvalues of a projection matrix can be either 1 or 0.
If $t\neq 0$, we note that both $A$ and $P$ are symmetric matrices and thus $(AP)^T = PA$. Since transposing a matrix doesn't affect its eigen values, we study the eigenvalues of $PA$ and $P+tPA$.
Let $X_i$ be the eigen-vector of $PA$ corresponding to $\xi_i$, $i=1,\dots, r$. Then $PAX_i=\xi_iX_i$. We see that
$$PX_i = \frac{1}{\xi_i}PPAX_i = \frac{1}{\xi_i}PAX_i=X_i.$$
Therefore
$$(P+tPA)X_i=PX_i+tPAX_i = X_i + t\xi_iX_i=(1 + t\xi_i)X_i$$
which shows that $(1 + t\xi_i)$ for $i=1, \dots,r$ are $r$ eigenvalues of $P+tPA$ with eigenvectors $X_1, \dots, X_{r}$.
If there is a gap of less than 1 hour between posting the problem and solving a more general version, it suggests that it would have been better to reflect on the problem a bit longer (e.g. a couple of days) prior to posting it. Anyway I'm glad you have resolved your problem.
|
2025-03-21T14:48:30.320159
| 2020-04-13T18:38:32 |
357386
|
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|
Stack Exchange
|
How canonical are integral lifts of Hasse invariants and other mod-$p$ modular forms?
Let $p$ be an odd prime. Recall that the mod $p$ Hasse invariant $A$ of an elliptic curve is an $\mathrm{SL}(2,\mathbb Z)$-modular form of weight $p-1$ defined over $\mathbb{F}_p$. Writing $\overline{\mathcal{M}_{ell}}$ for the compactified moduli stack of elliptic curves, I think this means that $A$ is a section of the bundle $\omega^{p-1} / p$. It is "essentially the only" mod-$p$ modular form whose Fourier expansion is $A(q) = 1 \in \mathbb{F}_p[\![q]\!]$.
When $p \geq 5$, $A$ is liftable to an integral modular form. For example, the (normalized) weight-4 Eisenstein series is $E_4 = 1 + 240\sum_{n\geq 1}\frac{n^3q^n}{1-q^n} \equiv 1 \pmod 5$ lifts the mod-5 Hasse invariant, and $E_6 = 1 + 504\sum(\dots) \equiv 1 \pmod 7$ lifts the mod-7 Hasse invariant. These lifts are "canonical" just because there aren't very many integral modular forms of those degrees.
What about the higher primes? Is there a "canonical" lift of the mod-$p$ Hasse invariant? For instance, in $12$, the space of integral modular forms is $2$-dimensional, spanned by $c_4^3$ and $\Delta = q + O(q)$, and so naively "the" integral lift of the mod-$13$ Hasse invariant suffers, at least, an ambiguity of multiples of $13\Delta$.
I would also like to hear about the case of the prime $p=3$. Then the mod-$3$ Hasse invariant $A$ is not liftable (at level $1$ — it is liftable as as $\Gamma_1(N)$-modular form for $N\geq 2$ coprime to $3$). But maybe there are some mod-$3$ modular forms which are liftable, like $AE_4$ or $AE_6$, and maybe they are liftable "canonically"?
No one would accuse me of being a number theorist. Please let me know about any errors of theory or notation.
For any $p \ge 5$ you have the level 1 Eisenstein series $E_{p-1}$, and that's very commonly regarded as the "standard" lifting of the Hasse invariant.
You should only expect to have "canonical" lifts of $A$ to (possibly non-reduced) rings $R$ with $pR = 0$. Note that $E_4$ is a lift of $A^2 \pmod 3$ when $p = 3$ and $A^4 \pmod 4$ when $p = 2$. Of course @DavidLoeffler is completely correct that $E_{p-1}$ is taken as a standard choice but it might be more useful for you to specify what properties you would like your lift to satisfy or what you would want to do with it.
|
2025-03-21T14:48:30.320345
| 2020-04-13T19:18:28 |
357392
|
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Planar duality generalized to embedded simplicial complexes
Let $K$ be a finite $d$-dimensional simplicial complex embedded in $\mathbb{R}^{d+1}$. The setting of this question is simplicial homology with coefficients over $\mathbb{Z}_2$. By Alexander duality $K$ partitions $\mathbb{R}^{d+1}$ into $\beta_d + 1$ connected components, where $\beta_d$ is the $d$th Betti number of $K$. We define $G$ to be the dual graph of $K$. $G$ is the graph whose vertices are the connected components of $\mathbb{R}^{d+1}$ and whose edges are the $d$-dimensional simplices of $K$. Two vertices are adjacent if and only if their corresponding connected components share a common $d$-simplex in the intersection of their boundaries.
It is easy to see that there is a one-to-one correspondence between $d$-dimensional cycles in $K$ and minimal edge cuts in $G$. This is analogous to cycle/cut duality in planar graphs.
What does a cycle in $G$ correspond to in $K$? My intuition is that it should be something analogous to a cut in a graph. In terms of homology a graph cut is a set of edges whose removal increases the rank of the zeroth homology group of the graph (because this homology group counts the number of connected components.)
Is the dual of a cycle in $G$ a set of $d$-dimensional simplices in $K$ whose removal increases the rank of the $(d-1)$th homology group? This seems likely to me, but I do not know how to prove it.
I fixed an error in my answer below. Is anything missing / are you looking for something different?
This answer is exactly what I wanted. Thanks a lot, the answer will be very useful to me.
You're welcome, it was a very nice question. In retrospect I'm surprised that I haven't seen it before.
Like you guessed, the dual of a cycle in $G$ is a set of $d$-simplices in $K$ whose removal increases the rank of the $(d-1)$-th homology group.
In the following I use that $H_i(X;\mathbb{Z}_2) \cong H^i(X;\mathbb{Z}_2)$. It might have been better to keep track of the distinction...
Consider the closures of the connected components of $K$'s complement in $\mathbb{R}^{d+1}$, $\{X_1,\ldots,X_n\}$, where $n=\beta_d(K)+1$.
These are $(d+1)$-dimensional simplicial complexes,and their boundaries are $d$-dimensional.
Edit: It is of course not true that the connected components of the complement are simplicial complexes, although if the embedding is a PL embedding in $S^{d+1}$ instead then they can be triangulated. I mistakenly answered this more combinatorial variant of the question.
Nevertheless the method of proof works if we think of the components of $\mathbb{R}^{d+1}\setminus K$ as being topological subspaces of $\mathbb{R}^{d+1}$, with the following addition: note that no more than two connected components can contain a single $d$-dimensional simplex of $K$ in their common boundary. A reference is Daverman and Venema, "Embeddings in Manifolds," corollary 7.1.2 and the preceding proposition (the section is "Codimension-one separation properties," accessible from Google Books).
Let us consider the boundaries of each pair of distinct $X_i,X_j$ as though they were disjoint: we want to think of them as different simplicial complexes.
Let $C$ be a cycle in $G$. We want to know what happens when the $d$-simplices $``C\cap K"$ is removed from $K$. Since $K$ is the Alexander dual of $\bigsqcup_{i=1}^n X_i,$ we can determine this by looking at what happens to the $\{X_i\}$ instead. Removing a $d$-simplex from $K$ is equivalent to gluing some pair along it, say $X_1,X_2$, so we can use the Mayer-Vietoris sequence for reduced homology:
$$ \ldots\rightarrow H_k(X_1 \cap X_2) \rightarrow H_k(X_1)\oplus H_k(X_2) \rightarrow H_k(X_1 \cup X_2) \rightarrow H_{k-1}(X_1\cap X_2) \rightarrow \ldots $$
Here, the intersection $X_1 \cap X_2$ is the intersection after the gluing. Hence it is a $(d-1)$-dimensional simplex on the boundary, and its reduced homology is zero in all dimensions. Thus the first homology of $X_1 \cup X_2$ is just the direct sum $H_1(X_1)\oplus H_1(X_2)$.
Now we can remove, one by one, the edges (or $d$-simplices) in a simple cycle of $G$. Let us remove the last pair of edges at once; this means we are gluing the connected component $X_t$ dual to the last vertex in the cycle by both of its edges simultaneously. The other piece glued to $X_t$ is the union of components corresponding to the other vertices of the cycle, let's call it $Y$. The reduced Mayer-Vietoris sequence in low degrees gives us
$$ H_1(X_t \cap Y) \rightarrow H_1(X_t)\oplus H_1(Y) \rightarrow H_1(X_t \cup Y) \rightarrow \tilde{H}_0(X_t \cap Y) \overset{j}{\rightarrow} \tilde{H}_0(X_t) \oplus \tilde{H}_0(Y).$$
Here the intersection $X_t \cap Y$ is a disjoint union of two simplices, which has $H_1 = 0$ but $\tilde{H}_0 = \mathbb{Z}_2$. However, the map $j$ in the sequence above is $0$. So the rank of $H_1(X_t \cup Y)$ is exactly
$$\mathrm{rk}(H_1(X_t)\oplus H_1(Y)) + \mathrm{rk}(\tilde{H}_0(X_t \cap Y)) = \mathrm{rk}(H_1(X_t))+\mathrm{rk}(H_1(Y))+1. $$
By induction on the length of the cycle, and using that the first homology of a space with several connected components is the direct sum of their first homologies, we see that the result of cutting out the dual of a cycle of $G$ from $K$ increases the first homology of $\mathbb{R}^{d+1}\setminus K$ by $1$. Alexander duality (this time applied to the complement) gives the result on $H_{d-1}(K)$.
The procedure used in the above can also be used to show that any inclusion-minimal set of edges in $G$ such that the removal of the corresponding simplices in $K$ increases the rank of $H_{d-1}(K)$ is a cycle:
Observe that removing the simplices corresponding to an acyclic set of edges does not change $H_{d-1}(K)$, and that a cycle does. Thus if a set of edges increases the rank of $K$'s $(d-1)$-th homology it contains a cycle; this cycle is inclusion-minimal among such sets.
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2025-03-21T14:48:30.320843
| 2020-04-13T20:21:27 |
357397
|
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"sort": "votes",
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}
|
Stack Exchange
|
ITTMs with higher types
What is the complexity of Infinite Time Turing Machines (ITTMs) augmented with an initially empty set of real numbers, with the ability to add, remove, and test presence of a real number in the set?
More generally, what is the complexity of ITTMs with $n$ higher types?
Definitions and background
An ITTM with $n$ higher types has a finite internal state and a finite number of tapes of types $1$ through $n+1$. A type 1 tape can be viewed as a subset of $ω$, type 2 - $P(ω)$, type 3 - $P(P(ω))$, and so on. A bit $S$ of a type $i>1$ tape $T$ (indicating whether $S∈T$; 0 indicates absence) can be accessed or modified by first storing $S$ in any type $i-1$ tape. A type 1 tape is the usual infinite tape (and a natural number is a type 0 input). All tapes are initially empty except for the type 1 input tape.
Just like an ITTM, the machine runs for an unlimited transfinite ordinal number of steps, with $\liminf$ behavior at limit stages ($\limsup$ gives essentially the same expressiveness). Thus, the contents of a tape $T$ at a limit time $α$ are $T_α = \liminf_{β→α} T_β = ∪_{λ<α} ∩_{λ<β<α} T_β$. Also, at limit stages, the internal state also uses $\liminf$ and the head locations for type 1 tapes are reset (though any reasonable variation gives the same expressiveness).
If there are no type >2 tapes, then the number of type 2 tapes is immaterial (if nonzero), but beyond that, the status is unclear, and it is possible (but not likely) that the above computational model is too restrictive.
Despite the use of infinite sets of higher types, the complexity is $Δ^1_2$, but apparently very high up in the $Δ^1_2$ hierarchy.
Upper bound:
Analogously to ITTMs (for which the below bounds are optimal (using $ω=ω_0$)), we have an upper bound on the complexity of ITTMs with $n$ higher types and input $r$. Let ordinals $β,γ$ be minimal with $L_β(r) ≺_{Σ_2} L_γ(r)$ and $ω_n^{L_β(r)} = ω_n^{L_γ(r)} < β < γ$. We have:
• If the machine halts, then its computational history and output (and the singleton set containing these) are $Δ_1^{L_β(r)}(r)$.
• If sets are never erased from a given tape, its eventual contents are (as a predicate) $Σ_1^{L_β(r)}(r)$.
• If a tape is eventually constant, its eventual contents are $Δ_2^{L_β(r)}(r)$ (for type $n+1$ tape, this is equivalent to being in $L_β(r)$).
• The contents of a tape at time $ω_1$ are $Σ_2^{L_β(r)}(r)$ (since we are using $\liminf$).
• At every point, the state is in $L_γ(r)$.
But are these bounds optimal?
Extensions with transfinite types: We can add fixed transfinite types using infinitely many tapes to access a bit of a tape of a transfinite type. Or we can even use a special tape to allow adding new tapes and types dynamically (with cofinal changes of a type resetting it to 0). However, if we could grow types without limits, then even for the simple model of ordinal register machines, recognizability would equal $Σ^1_2$ (or $Σ^1_2(r)$).
|
2025-03-21T14:48:30.321082
| 2020-04-13T23:24:57 |
357401
|
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"R. van Dobben de Bruyn",
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"url": "https://mathoverflow.net/questions/357401"
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|
Stack Exchange
|
On a condition on ideals viwed as a Zariski open condition on co-tangent space
Let $(R, \mathfrak m,k)$ be a Noetherian local ring such that the residue field $k$ is infinite. Let $n=\mu(\mathfrak m)$. Then $n=\dim_k(\mathfrak m/\mathfrak m^2)$ . By fixing $x_1,...,x_n \in \mathfrak m$ such that $\bar x_1,...,\bar x_n\in \mathfrak m/\mathfrak m^2$ gives a $k$-vector space basis, we can identify $\mathfrak m/\mathfrak m^2$ with $\mathbb A^n(k)$ by sending $\bar x_i \to e_i$. So we can transfer the classical Zariski topology of $\mathbb A^n(k)$ to $\mathfrak m/\mathfrak m^2$.
I have two similar kind of questions :
(1) If $I$ is $\mathfrak m$-primary i.e. $\sqrt I=\mathfrak m$ and the set $\{\bar x\in \mathfrak m/\mathfrak m^2: x\in \mathfrak m \setminus \mathfrak m^2 \space \text{and} \space (\mathfrak mI:x)=I \}$ is non-empty, then how to show that the set is Zariski Open ?
(2) If $I=\overline I$ is $\mathfrak m$-primary i.e. $\sqrt I=\mathfrak m$ and the set $\{\bar x\in \mathfrak m/\mathfrak m^2: x\in \mathfrak m \setminus \mathfrak m^2 \space \text{and} \space (\overline{\mathfrak mI}:x)=I \}$ is non-empty, then how to show that the set is Zariski Open ?
Here for an ideal $I$, by $\overline I$ we denote the integral closure of $I$ https://en.m.wikipedia.org/wiki/Integral_closure_of_an_ideal
You always have $(\mathfrak mI \colon x) \supseteq I$, right? So the question is when is there nothing else. You would like to have access to $R/J$ for any $\mathfrak m$-primary ideal $J$ as an affine space over $k$ (or a successive fibration of affine spaces), and use some sort of structure constants to show that multiplication $m \colon R/I \times \mathfrak m/\mathfrak m I \to R/\mathfrak m I$ is a morphism of varieties. This is certainly ok if $R$ is a $k$-algebra, but probably also works in other settings. Then study the map $m$ geometrically.
@R. van Dobben de Bruyn: I'm having some issues seeing as to how the multiplication $R/I \times \mathfrak m/\mathfrak mI\to R/\mathfrak m I$ being a morphism of schemes would be helpful (and what scheme structure exactly are you talking about on $ \mathfrak m/\mathfrak mI$ ?) ... a priori, I don't see any connection between the usual Zariski topology of Spec $(R/I)$ and the Zariski topology I've defined on $\mathfrak m/\mathfrak m^2$ ...
Sorry, I didn't mean to suggest that this is a full answer, but it is certainly a strategy in case $R$ is a $k$-algebra. Then $R/I$ is a finite dimensional $k$-algebra, so you can view it as a ring object in affine $k$-schemes (in particular, addition and multiplication are given by polynomials; to make this explicit use the structure constants of $R/I$). For example this immediately shows that for a single $f\not\in I$, the set of $x$ with $f\in(\mathfrak mI:x)$ is open, because $x \mapsto fx \in R/\mathfrak mI$ is continuous.
Case 1 was treated carefully in Appendix A of J. Watanabe's paper "$m$-full ideals". Case 2 can be treated the same way, as sketched below.
Let $J$ be either $mI$ or $\overline{mI}$ and $A=R/J$. Since $J:x \supset I$, we have that $l(A/xA)= l(0:_Ax) \geq l(I/J)$. Thus we need to show that the set of $x$ such that the length $A/xA$ is minimal possible is Zariski open. Then the idea is to consider the "universal generic element". Let $S=A[Y_1,\dots, Y_n]$ and $A'=S_{mS}$. Consider the element $X= \sum Y_ix_i$ in $A'$. Prove that $l(A/xA)\geq l(A'/XA')$ with equality happens if and only if the ideal $(Y_1-a_1,\dots Y_n-a_n)$ does not contain certain radical ideal in $k[Y_1,\dots, Y_n]$ where $x = (a_1,\dots, a_n) \in m/m^2=\mathbb A^n_k$
|
2025-03-21T14:48:30.321301
| 2020-04-13T23:33:53 |
357402
|
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|
Stack Exchange
|
Bounds on the second derivative of a natural cubic spline in terms of the data
Suppose we have real numbers $x_1 < \cdots < x_n$ and $v_1, \ldots, v_n$. Let $f$ be the natural cubic spline such that $f(x_i) = v_i$. Is there a simple explicit bound on $\|f''\|_\infty$ in terms of $n, x_i, v_i$?
One can certainly be derived from the definition of the spline, but I thought I'd ask in case there was a nice reference somewhere, and before I bash through the algebra and clutter up my paper!
The first thing to notice is that the second derivative of a cubic spline is itself
a degree-1 spline i.e., a piecewise linear function.
Therefore, the maximal $|f''|$ value is attained at one of the $x_i$ values (the knots).
We denote by $k_i$ for $i=1,.. ,n-1$ the second derivatives at the knots $x_0,..,x_n$, where
$k_0$ and $k_n$ are zero by the spline natural condition.
The second derivatives $k_i$ are then the solution of the following tri-diagonal system of linear equations:
$$
(x_{i-1}-x_i) k_{i-1} + 2(x_{i-1} - x_{i+1}) k_i + (x_i - x_{i+1}) k_{i+1} =
6 (\frac{v_{i-1} - v_i}{x_{i-1} - x_i} - \frac{v_i - v_{i+1}}{x_i - x_{i+1}})
$$
Or in matrix notation we have:
$$
A k = b
$$
Where the matrix $A$ is:
$$
\left[ {\begin{array}{ccc}
2(x_{0} - x_{2}) & (x_1 - x_{2}) & 0 & ... & 0 \\
(x_{1}-x_2) & 2(x_{1} - x_{3}) & (x_2 - x_{3}) & 0 & ... \\
& & ... & & \\
& (x_{i-1}-x_i) & 2(x_{i-1} - x_{i+1}) & (x_i - x_{i+1}) k_{i+1} & \\
& & ... & & \\
... & 0 &(x_{n-3}-x_{n-2}) & 2(x_{n-3} - x_{n-1}) & (x_{n-2} - x_{n-1}) \\
0 &... & 0 &(x_{n-2}-x_{n-1}) & 2(x_{n-2} - x_{n}) \\
\end{array} } \right]
$$
the vector $b$ is:
$$
\left[ {\begin{array}{c}
6 (\frac{v_{0} - v_1}{x_{0} - x_1} - \frac{v_1 - v_{2}}{x_1 - x_{2}}) \\
.. \\
6 (\frac{v_{i-1} - v_i}{x_{i-1} - x_i} - \frac{v_i - v_{i+1}}{x_i - x_{i+1}}) \\
.. \\
6 (\frac{v_{n-2} - v_{n-1}}{x_{n-2} - x_{n-1}} - \frac{v_{n-1} - v_{n}}{x_{n-1} - x_{n}}) \\
\end{array} } \right]
$$
and the unknown vector $k$:
$$
\left[ {\begin{array}{c}
k_1 \\
.. \\
k_i \\
.. \\
k_{n-1}
\end{array} } \right]
$$
These equations (or variants of them) can be derived in several ways (for example, see here or here).
However, I believe the most straightforward way to attain them in our context
is in constructions of cubic splines that begin by integrating the second derivative constraints.
Examples of such a construction can be found, for example, here and here (which is where the above formulation was taken from).
Now, the explicit formula for $|f''|_\inf$ becomes:
$$
|| A^{-1} b ||_\inf = \max_i (A^{-1} b)
$$
The above is the exact solution (not a bound).
However, if you wish to simplify the expression further, we can bound the second derivative using bounds on the
norm of the matrix $A^{-1}$.
Specifically, $|f''|_\inf \leq |A^{-1}|_\inf |b|_\inf$.
Furthermore, $A$ in our problem is a diagonally dominant matrix.
Therefore, we can use the Varah bound to bound its norm.
The Varah bound states:
$$
||A^{-1}||_\inf < \frac{1}{\min_i (|a_{i,i}| - \sum_{j \neq i} |a_{i,j}|)}
$$
Note that (except for the first and last row) for the tri-diagonal entries of our $A$ matrix, we have $|a_{i,i}| / 2 = (|a_{i, i-1}| + |a_{i, i+1}|)$,
so the Varah bound just becomes:
$$
||A^{-1}||_\inf < \frac{1}{\min_i (|a_{i,i}|/2, |a_{1,1}|-|a_{1, 2}|, |a_{n-1,n-1}|-|a_{n-1, n-2}|)} = \frac{1}{\min_i (|x_{i-1} - x_{i+1}|, 2|x_{0} - x_{2}| - |x_{1} - x_{2}|, 2|x_{n-2} - x_{n}| - |x_{n-2} - x_{n-1}|)}
$$
And we get:
$$
|f''|_\inf \leq |A^{-1}|_\inf |b|_\inf \leq \frac{\max_i b_i}{\min_i (|a_{i,i}|/2, |a_{1,1}|-|a_{1, 2}|, |a_{n-1,n-1}|-|a_{n-1, n-2}|)}
$$
So, an explicit simple bound on the second derivative of a natural cubic spline is:
$$
\frac{\max_i |(6 (\frac{v_{i-1} - v_i}{x_{i-1} - x_i} - \frac{v_i - v_{i+1}}{x_i - x_{i+1}}))|} {\min_i (|x_{i-1} - x_{i+1}|, 2|x_{0} - x_{2}| - |x_{1} - x_{2}|, 2|x_{n-2} - x_{n}| - |x_{n-2} - x_{n-1}|)}
$$
This is not necessarily a tight bound as can be seen in the example figures below. The second figure below is a plot of the (piecewise linear) second derivative function of the natural spline from the first figure.
The exact maximal second derivative in this example is 122.3, whereas the bound is 192.
The $x$-values taken for this example were $(0, 1, 2, 3, 4, 5, 6)$, and the $v$-values were $(0, 2, 0, 8, 0, 32, 0)$.
|
2025-03-21T14:48:30.321567
| 2020-04-13T23:47:22 |
357404
|
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"Ben Wieland",
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|
Stack Exchange
|
Looking for an invariant similar to algebraic K-theory
I'm wondering if there is an invariant, similar to algebraic K-theory, topological hochshild homologic, topological cyclic homology etc... that has the following properties:
a) It attach to each small stable $\infty$-category $A$ a spectrum $E(A)$.
b) Its functorial on exact functor.
c) It satisfies an additivity property similar to that of $K$-theory, $THH$, i.e. at the minimum it sends split exact sequence of small stable $\infty$-category to (split) cofiber sequences. (so essentially it is an "additive invariant" in the sense of Blumberg, Gepner, Tabuada)
d) For each object $a \in A$, there is an element $\chi(a) \in E(A)$ that is natural and behave additively on cofiber sequence, (so more formally, there is a natural transformation $K(A) \to E(A)$
e) If $X$ is a space (an $\infty$-groupoid), and $Sp^X$ denotes the full subcategory of compact object in the infinity category $Sh(X,Sp)$ of (locally constant) sheaves of spectrum on $X$, then:
$$E(Sp^X) \simeq \Sigma^\infty_+ X$$ is the suspension spectrum of $X$. Where the identification is functorial in $X$. I would also be happy if $E(Sp^X)$ gives the homology of $X$ instead, but the former would be more general.
Both $K$-theory, $THH$ and $TC$ have all these properties excepte maybe $(e)$. $THH$ fall close with $THH(Sp^X) \simeq \Sigma^\infty_+ \left( X^{S^1} \right)$.
I don't know well enough $TC$ and its variant to know directly if one them has property $(e)$. If it the case I would be very happy with a reference that proves it. Otherwise I'm hoping that some kind of "simplified K-theory" will do the trick, maybe a quotient of $K$-theory, but I don't know these topics well enough to figure it out.
Motivation: Let $I$ and $J$ be two finite direct (1-)categories. I denote by $\widehat{I}$ and $\widehat{J}$ the $\infty$-categories of presheaves of $\infty$-groupoids on them.
Given $F: \widehat{J} \to \widehat{J}$ a left adjoint functor preserving finitely presented objects, we can define a linear map:
$$ |F| : \bigoplus_{i \in Ob(I)} \mathbb{Z} \to \bigoplus_{j \in Ob(J)} \mathbb{Z}$$
that "computes" the levelwise Euler characteristic of $F(X)$ from the levelwise Euler characteristic of $X$, when $X$ is a finitely presentable objects in $\widehat{I}$, in the sense that:
$$ \chi(X(i)) \in \bigoplus_{i \in Ob(I)} \mathbb{Z} \mapsto \chi(F(X)(j)) \in \bigoplus_{j \in Ob(j)}$$
This construction is quite useful when studying for example nice monads on such presheaf categories: it attach easy to compute and quite subtle numerical invariant to certain nice finitary functors.
I would like to upgrade this, by replacing abelian group by connective spectra and allowing $I$ and $J$ to be more general $\infty$-categories, but I'm struggling in proving that what I want to construct has the appropriate functorialy properties.
I realized at some point that functor above could be interpreted as:
$$ \pi_0 THH( Sp^{I^{op}}) = \bigoplus_{i \in I} \mathbb{Z} $$
where $Sp^{I^{op}}$ denotes the category of compact presheaves of spectrum on $I$. Indeed using additivity of $THH$, we can show by induction on height that:
$$ THH( Sp^{I^{op}}) = \Sigma_\infty^+ \left( Ob(I) \right)$$
To some extent taking $THH$ actually is "good enough" for most of the applications I have in mind. But it does not behave quite as I would like, and this makes everything more complicated. Typically, when $I$ is a generalized direct $\infty$-category (that is it can have some automorphisms) satisfying some finiteness condition I won't go into, I would like the invariant to be:
$$ \Sigma^\infty_+ ( core(I)) $$
where $core$ denotes the maximal sub $\infty$-groupoid. While, induction on height gives:
$$THH(Sp^{I^{op}}) = \Sigma^\infty_+ ( core(I)^{S^1}) $$
This of course is exactly because (e) fails for THH. But on the other hand, when I consider the example I'm interested in, it seems consistant that one could get rid of this $S^1$ : The way it appears in THH is because THH is the target of the universal trace map, and an endomorphism in $Sp^{I^{op}}$ can be over automorphisms of $I$ and hence remember a "loop" in $I$, but I' only interested in "characteristic of objects" (see point (d)) not trace of maps, so this "circle" is not relevant for me. So I'm hoping there is a different invariant from THH that would get rid of it.
I started studying the alternative (K-theory, TC etc...) but couldn't really decide if one of them was solving my problem... K-theory do not have this circle showing up, but lots of additional things, that are not relevant to me, (basicaly the K-theory of the point) comes up. And for TC (and its negative or periodic variant), I haven't been be able to understand how it behaves on spaces. I'm hoping an expert on the topic can point me to the right direction...
I'm afraid I don't understand the definition of $|F|$. Could you elaborate? The natural analogue of $\bigoplus_{i\in\mathrm{ob}I}\mathbb{Z}$ would seem to me $\mathcal{C}\mapsto \Sigma^\infty_+(\mathcal{C})$, but that is well known not to be an additive invariant (and indeed its additive approximation is K-theory), and moreover does not have the result you wish on categories of sheaves.
@DenisNardin : I've added lots of details to the "motivation" part". Note that I'm not expecting to obtain $\Sigma^\infty_+ core(C)$ for all all $\infty$-categories $C$, only well behaved one (typically generalized direct categories).
Also the concrete definition of $|F|$ I know of is complicated and involves some some combinatorics. But at the end it can be characterized by the fact that it computes Euler characteristic as explained above.
Let me delete those comments and start again. Do you really need such an abstract and general construction? Do you have an example of something you would detect? I assume by sheaves you mean locally constant sheaves. (If not, factor through them.) Take the fiber / skyscraper adjunction $C\rightleftharpoons Sh(X;C)$. This is parameterized by the space of points, so you get $X\wedge K(C)\to K(Sh(X;C))$ and $K(Sh(X;C))\to Hom(X,K(C))$. I guess the first is the assembly map. I like Dwyer-Weiss-Williams, but I'm not sure how relevant it is.
@BenWieland : I do not need the full generality explained in the first part of the question. But I need a construction that applies to the category of presheaf of spectra on a generlaized direct category. what I'm really after is to extend the functor $I \mapsto \Sigma^\infty_+ Core(I)$ (for $I$ generalized direct category satisfying some finiteness condition), so that it becomes functorial with respect to profunctors (satisfying finiteness conditions).
... I sould add: In the same way that taking THH makes functorial on profunctors the construction $I \mapsto \Sigma^\infty_+ (Core(I)^{S^1})$ (also for $I$ a generalized direct category, satisfying some finiteness conditions, and only for profunctors satisfying some finiteness conditions)
|
2025-03-21T14:48:30.322012
| 2020-04-14T00:44:38 |
357408
|
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"authors": [
"Brendan McKay",
"Thomas Lesgourgues",
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|
Stack Exchange
|
graph built from orthogonal Latin Squares
I've asked the following question on MathExchange site, with a bounty, with no answer or comments. Maybe I would have additional comments here. The problem came to be while reading some articles on finite geometry. I began wondering if anybody had previously studied this.
Reminder on Latin Square: Given a set $S$ of $n$ elements (we will use $[n]$ in the following for simplicity), a Latin square $L$ is a function $L : [n]\times [n] \to S$, i.e., an $n\times n$ array with elements in $S$, such that each element of $S$ appears exactly once in each row and each column. For example,
Let $L_1$ and $L_2$ be two Latin squares over the ground sets $S_1$, $S_2$ respectively. They are called orthogonal if for every $(x_1, x_2) \in S_1 \times S_2$ there exists a unique $(i,j)\in [n] \times [n]$ such that $L_1(i,j) = x_1$ and $L_2(i,j) = x_2$. For example, the following are two orthogonal Latin squares of order 3.
It is known that there at most $n-1$ mutually orthogonal Latin squares of order $n$, and that the bound is achieved if and only there exist an affine plane of order $n$.
The graph definition: I'm building a graph $G_n$ with vertex set the Latin squares of order $n$ and two vertices are adjacent iff the Latin squares are orthogonal.
I want to understand some properties of this graph. For simplicity I consider the squares up to permutation of $[n]$, hence w.l.o.g. all my squares have for first line $\{1,2,\ldots,n\}$. Indeed if I call $H_n$ the graph not up to permutations, then $H_n$ is the $n!$ graph blowup of $G_n$, or using the Tensor product $$ H_n = G_n \times K_{n!}$$
As I'm mainly interested in the chromatic number of my graph, and we know that $\chi(H_n)\leq \min\{\chi(G_n) ; n!\}$, I will study only $G_n$.
For instance $G_2=K_1$, $G_3=K_2$.
I know that :
It's trivial that $G_n$ is not complete.
If there exist an affine plane of order $n$ then $G_n$ contains $K_{n-1}$ as a subgraph, and $\chi(G_n)\geq n-1$.
$G_4$ is made of 2 disjoint $K_3$ and 18 isolated vertices, for a total of 24 Latin squares.
$G_5$ is made of 36 disjoint $K_4$ and 1200 isolated vertices, for a total of 1344 Latin squares.
The case $n=6$ would be the first interesting case, as there are no affine plance of order 6, hence we will find no $K_5$ in $G_6$. It is known since 1901 (from Tarry hand checking all Latin squares of order 6) that no two Latin squares of order 6 are mutually orthogonal. So $G_6$ is made of only isolated vertices (with 1,128,960 vertices).
It is also know that the case $n=2$ and $n=6$ are the only one with only isolated vertices. (see design theory by Beth, Jingnickel and Lenz).
From the article "Monogamous Latin Square by Danziger, Wanless and Webb, available on Wanless website here. The authors show that for all $n > 6$, if $n$ is not of the form $2p$ for a prime $p \geq 11$, then there exists a Latin square of order $n$ that possesses an orthogonal mate but is not in any triple of Mutually Orthogonal Latin Squares. Therefore our graph $G_n$ will have some isolated $K_2$
I wonder the following :
What is the maximum degree of $G_n$ ? We know that we have at most $n-1$ mutually orthogonal latin squares, but to how many squares can one square be orthogonal (still up to permutation)?
Do we have any other info on the chromatic number, not coming from the property $\chi(G_n)\leq \Delta+1$.
Can $G_n$ contains an induce $k$-cycle with $k>3$ (i.e. chordless cycle)?
The stronger statement would be the following conjecture
Conjecture : for any $n$, $G_n$ is the disjoint union of complete subgraphs (of different sizes).
Or said otherwise, the orthogonal relation is transitive (when restricted to our Latin squares with first row fixed at $\{1,2,\ldots,n\}$.
I would welcome any intuition, direction for some articles, or any known additional facts.
On order 10 there are plenty of squares with more than one orthogonal mate, but there are no triples known. So your conjecture is false.
@BrendanMcKay ok thanks for the info. I'm looking at Prof. Wanless website and references in order to get a better understanding of my others questions. Thanks
I just realised that the coloring problem is quite trivial, as you can partition the Latin squares in $n-1$ sets, defined by the value of $L(2,1)$ (or any specific coordinates not on the first line) : this forms a partition and two Latin squares in the same set cannot be orthogonal, hence $\chi(G_n) \leq n-1$.
Reposted answer from MathSE; seems like there's a little more attention here. Brendan McKay's comment settles the conjecture, and you addressed the coloring question. Here I have some comments on maximum degree. There's still the cycle question...
There's more from the heavily-studied 10×10 case relevant to your questions. The maximum degree in the graph is likely unbounded. Here's a relevant excerpt from pp327-328 of Latin Squares and Their Applications by Keedwell and Dénes (2nd ed., North Holland, 2015).
"[Parker in 1962 and 1963] discovered that 10×10 latin squares with orthogonal mates are not, in fact, particularly scarce and he also showed that there exist squares with a large number of alternative orthogonal mates. His most striking result concerns the square displayed in Figure 13.2.1 which has 5504 transversals and an estimated one million alternative orthogonal mates (that is, sets of 10 disjoint transversals). However, Parker was able to show by a partly theoretical argument that no two of these alternative orthogonal mates are themselves orthogonal and so, much to his own disappointment, he was not able to obtain a triad of mutually orthogonal 10×10 latin squares. The existence or non-existence of such triads remains an open question."
In fact, that particular square has 12,265,168 orthogonal mates (Maenhaut and Wanless, J. Combin. Des. 12 (2004) 12-34).
|
2025-03-21T14:48:30.322519
| 2020-04-14T01:25:47 |
357412
|
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|
Stack Exchange
|
Function satisfying $f(x)^{f^{-1}(x)}=x^2$ with $ f^{-1}$ is a compositional inverse of $f$ and $f:\mathbb{R+}\to \mathbb{R+}$?
Let $f$ be a function such that :$f:\mathbb{R+}\to \mathbb{R+}$ and $f^{-1}$ is a compositional inverse of $f$ , I have tried to find solution of the following functional $f(x)^{f^{-1}(x)}=x^2$, I took $f(x)=x$ but it doesn't work it coincide only for $x=1$ and for $f(x)=\exp(x)$ I come up to $x^x=x^2$ then coincide only for $x=1$ and $x=2$ , so $f(x)^{f^{-1}(x)} > x$ for $x >2$ which means no trivial solution exists probably a formel power series exist arround $x=1$ or $x=2$,Then my question here is : How I can solve $f(x)^{f^{-1}(x)}=x^2$ with $ f^{-1}$ is a compositional inverse of $f$ ?
Edit I suspect such trigonometric function would be works for the titled function,like $\tan x$ as shown here but this need a restriction of our Domain of definitions for which $\tan x$ to be defined
Edit I have edited the question according to the answer given by @Robert I without changing the meaning I missed that I search about a real valued function not complex as stated in the below answer
When $f(x) < 0$ (which must happen for uncountably many $x$ if $f^{-1}$ is defined on $\mathbb R$), the fact that $f(x)^{f^{-1}(x)} > 0$ requires $f^{-1}(x)$ to be a rational number (presuming we define $a^b = \exp(b \log(a))$ for some branch of the logarithm; note all branches of logarithm of a negative number have imaginary part an integer multiple of $\pi i$, and $\exp(b k \pi i)$ is a positive real only when $b k$ is an even integer). Since $f$ can't map a countable set onto an uncountable one, we conclude there is no such function.
Or did you want to just define $f$ and $f^{-1}$ on $(0,\infty)$?
@MarkSapir, we are learn from you ,and i don't know if it is not allowed in MO to ask for one day with one question, I have many related research this is the reason let me ask every day , In anyway thanks for your attention
|
2025-03-21T14:48:30.322667
| 2020-04-14T06:28:56 |
357423
|
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|
Stack Exchange
|
"Global" topologies between compact convergence and uniform convergence
Let $X$ and $Y$ be locally compact (but not compact), second countable, Hausdorff spaces with $Y$ metric. It is easy to see that the topology of compact convergence is weaker than the topology of uniform convergence on $C(X,Y)$ defined by the (possibly infinite) metric
$$
d(f,g) = \sup_{x \in X} | g(x)-f(x)|.
$$
The former has a very "local nature" and is very weak while the latter is a bit unruly which makes me look elsewhere. What (studied) topologies lie in between these two? Ideally, the topology should not be "of a local nature".
Edit: Ideally this topology on $C(X,Y)$ should be stronger than the topology of compact-convergence but also have a nice criterion to verify density (like the Stone-Weirestrass theorem) when $Y=\mathbb{R}^d$ and $X$ is "nice".
Given any set $\mathfrak{S}$ of subsets of $X$, Bourbaki (General Topology X, §1) defines the topology (in fact, the uniform structure) of $\mathfrak{S}$-convergence. You can take $\mathfrak{S}={X} $ (giving uniform convergence), $\mathfrak{S}=$ compact, precompact, bounded... your choice.
Ah, these are the set-open topologies no?
Probably (I don't know what the set-open topologies are).
For a discussion of difference between set open topologies with convergence on sets topologies see Mccoy, Ntantu - topological properties of spaces of continuous functions
Just a couple of suggestions. For a question like this, it can be useful to look at special cases for pointers. Here I would suggest firstly the case where $X$ is the integers (then your space is $Y^N$ with the box, resp., the product topology—an interesting further specialisation is where $Y$ is also the integers, i.e., your space is the irrationals). Secondly, where the image spaces is the reals, in which case you get the standard Fréchet space $C(X)$.
It is hard to make a useful comment without knowing what you have in mind, but one method for extrapolating between your two cases is to use so-called weighted seminorms or pseudometrics.
|
2025-03-21T14:48:30.322837
| 2020-04-14T06:30:07 |
357424
|
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"piapple"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/357424"
}
|
Stack Exchange
|
What is the conjugate prior of Multivariate Log Normal distribution?
For the univariate log-normal distribution, when mean is known, the conjugate prior is gamma distribution. But how about the multivariate log-normal distribution ? What is the conjugate prior ? Multivariate Gamma distribution ? What is the posterior distribution ?
I cannot find any information related to this. Hope someone could help me on this. Thanks.
Where do these questions arise? And what is your version of a multivariate Gamma distribution?
Thanks for your comment. The question comes from the fact that log-normal distribution and gamma distribution are conjugate. And now I am working on the log multivariate normal distribution, but I do not know the the conjugate prior of that. It seems something related to gamma or wishart distribution. But I cannot find anything related.
|
2025-03-21T14:48:30.322922
| 2020-04-14T06:47:45 |
357425
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/357425"
}
|
Stack Exchange
|
A infinity structure on Yoneda Ext group
I am currently trying to control an $A_\infty$-algebra of the form $\mathrm{Ext}_X(F\oplus F'[2n-2],F\oplus F'[2n-2])$ where $X$ is a nice enough scheme and $F,F'$ are sheaves that are NOT locally free. Now as this $A_\infty$-structure is quite elusive I was hoping I could control it on the Yoneda-$\mathrm{Ext}$ level, in particular as I feel more comfortable with homological algebra. So I wanted to ask if anyone knows about such a construction and maybe even point me to a paper on it, I sadly could not find one yet.
|
2025-03-21T14:48:30.322992
| 2020-04-14T07:06:05 |
357427
|
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|
Stack Exchange
|
Surfaces of general type with $q=1$
Let $X$ be a smooth projective connected surface of general type over $\mathbb{C}$ with $q(X) = 1$, where $q(X) = \mathrm{h}^1(X,\mathcal{O}_X)$.
Let $E$ be the Albanese variety of $X$, and let $X\to E$ be the Albanese map (having fixed a point). Let $0$ be a closed point of $E$.
Let $F$ be the scheme-theoretic fibre over $0$. Does $F$ have an irreducible reduced component? That is, does $F$ have an irreducible component of multiplicity one?
Of course, for a general $0$, the fiber $F$ is smooth. I am wondering whether the Albanese map has a multiple fibre or not.
It may certainly happen. Take an Enriques surface $S$; it admits an elliptic fibration $S\rightarrow \mathbb{P}^1$ with 2 double fibers, say above $0$ and $1$. Now take a double cover $\pi :E\rightarrow \mathbb{P}^1$ branched along 4 points $\neq 0,1$, and let $T$ be the pull back of $S$ by $\pi $; its Albanese map is the projection $p:T\rightarrow E$ with 4 double fibers. Then let $\rho :X\rightarrow T$ be a double covering branched along a smooth, ample curve in $T$, transversal to the double fibers. Its Albanese map $p\circ \rho $ still has 4 double fibers.
@abx That's a nice example, thank you! But I am wondering whether this always happens. (I think I phrased the question in a confusing manner. My apologies.) Basically, given a surface of general type $X$ with $q=1$, my question is whether we can prove that $X\to Alb(X)$ has a multiple fibre.
No, of course it does not happen always
How can I find an example of a surface with $q=1$ such that $X\to \mathrm{Alb}(X)$ has no multiple fibres? If I start with one of your examples, I could consider ramified coverings of $X$, but I fear this might increase $q$...
For instance, Cartwright-Steger surface (which has $p_g=q=1$, $K^2=9$) has the property you want. See the main theorem here: https://arxiv.org/pdf/1412.4137.pdf
CS surface is rigid. Morally, one expects that, one the surface is not rigid, the "general" element in moduli has reduced Albanese fibres.
You can construct examples where the Albanese fibres have multiple components but also contains components of multiplicity one by considering a variation of the isogenous constructions, where $G$ is allowed to have isolated fixed points on the product. Look at my paper https://arxiv.org/pdf/math/0703066.pdf and at the other related papers of mine on the arXiv.
Beautiful...Many thanks!
You can find plenty of examples with multiple Albanese fibres by considering surfaces isogenous to a product, namely of the form $S=(C \times F)/G$, where $G$ is a finite group acting faithfully on the smooth curves $C$, $F$ and whose diagonal action on the product is free.
For an explicit situation, you can look at Corollary 2.5 of my paper
On surfaces of general type with $p_g=q=1$ isogenous to a product of curves, Communications in Algebra 36 (2008), no. 6, 2023-2053, arXiv:math/0601063.
|
2025-03-21T14:48:30.323197
| 2020-04-14T07:58:39 |
357430
|
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|
Stack Exchange
|
Is the optimum of this problem convex in the constraint parameter?
Let $f:\mathbb R^+ \to \mathbb R$ be a smooth function, satisfying $f(1)=0$, and suppose that
$|f|$ grows with the distance from $1$: $|f(x)|$ is strictly increasing when $x \ge 1$, and strictly decreasing when $x \le 1$.
Suppose also that $\lim_{x \to \infty} |f(x)| = \infty$.
For any $s \in (0,1)$, define
$$
F(s)=\min_{xy=s,x,y>0} f^2(x)+ f^2(y)
$$
(The minimum exists since $|f|$ diverges at infinity.)
Question: Does there exist a convex function $g(s)$ such that $F=g^p$ for some $p \ge 1$? I do not require $g$ to be positive.
Here are two examples where this happens:
Linear penalization: $f(x)=x-1$. In that case
$$F(s) =
\begin{cases}
2(\sqrt{s}-1)^2, & \text{ if }\, s \ge \frac{1}{4} \\
1-2s, & \text{ if }\, s \le \frac{1}{4}
\end{cases}
$$
is convex, since $F'(s)$ is non-decreasing.
Logarithmic penalization: $f(x)=\log x$. In that case
$$ F(s)=2f^2(\sqrt s)=\frac{1}{2}(\log s)^2$$ is not convex.
However, we have $F(s)=g^2(s)$ where $g(s)=-\frac{1}{\sqrt 2}\log s$ which is convex.
Is there a general phenomena lying behind these two examples?
Yes, thank you very much. This was a typo (fixed now).
The answer is no. E.g., let $f(x):=|x-1|^{3/2}$. Then
$$
F(s)=\begin{cases}
F_1(s) &\text{ if } 0<s\le1/9, \\
F_2(s) &\text{ if } 1/9\le s<1,
\end{cases}
$$
where
$$F_1(s):=1 - 3 s - 2s^{3/2},$$
$$F_2(s):=2 + 6 s - 2(3 + s)s^{1/2}.$$
One may note here that $F_1(1/9)=16/27=F_2(1/9)$ and $F'_1(1/9)=-4=F'_2(1/9)$.
From the definition of $F$, it is clear that $F>0$ on $(0,1)$. So, letting $a:=1/p\in(0,1]$, we see that the desired goal was that
$h:=F^a$
be either convex or (if $p$ is even) concave.
(If $h$ is concave and $p$ is even, we can take $g:=-h$. Then $g$ will be convex and we will also have $g^p=h^p=F$.)
So, letting $h_j:=F_j^a$ for $j=1,2$, we see that we must have one of the following cases:
(i) $h_1$ is convex on $(0,1/9]$ and $h_2$ is convex on $[1/9,1)$;
(ii) $h_1$ is concave on $(0,1/9]$ and $h_2$ is concave on $[1/9,1)$.
However, $h_1''(1/9)$ equals $3a-4$ in sign and hence is $<0$ for $a\in(0,1]$, whereas $h_2''(1/9)$ equals $2+3a$ in sign and hence is $>0$ for $a\in(0,1]$. So, neither one of the cases (i) or (ii) can take place.
Here is the graph of $F''$:
On a second thought, I think that the answer could be remedied as follows: Looking at $\text{sgn}(h_1^{''})=\text{sgn}\big((a-1)F_1^{'}+F_1F_1^{''}\big)$ when $s \to 0$ we must have $h_1^{''}(s)<0$ for sufficiently small $s$. Indeed, since $F_1$ and $F_1^{'}$ tend to finite values at zero, but $F_1''$ tend to $-\infty$, the sum $(a-1)F_1^{'}+F_1F_1^{''}$ must be negative (here we also use the fact that $F_1$ is positive-and in fact tends to $1$ at zero). Do you agree with my analysis?
@AsafShachar : My calculation was correct. The mistake is actually in your calculation of the expression for $h''$, which must have $F'^2$ in place of $F'$.
Thank you. Indeed, this was a silly mistake of mine. This is a great answer. I have one more question if you will: Can you say why did you delete your previous example with $f(x)=(x-1)^2$ instead of $f(x)=|x-1|^{\frac{3}{2}}$? In particular, I wonder whether the fact that $f(x)=|x-1|^{\frac{3}{2}}$ is non-smooth was essential in this non-convexity phenomenon, or is it possible to produce counter-examples with smooth $f$. Also, I wonder if you used some program for calculating the minimum $F(s)$ or just calculus...Thanks.
@AsafShachar : I replaced $f(x):=|x-1|^2$ by $f(x):=|x-1|^{3/2}$ to simplify the resulting expressions. Concerning the smoothness condition on $f$: I'd understand that it can be added hoping that it can help avoid technical complications in a proof of a positive answer. However, I don't see a good point in insisting that this condition be satisfied in a counterexample.
Previous comment continued: Yet, if you want to insist on the smoothness of $f$, there is a choice: either (i) go back to $f(x):=|x-1|^2$ (found in previous edits) and then deal with the more complicated expressions or (ii) approximate $f(x):=|x-1|^{3/2}$ near $x=0$ (say uniformly) by a smooth function so that the resulting strict inequalities hold.
Thank you. I agree with you that the insistence on smoothness in a counterexample is not very natural here. I wonder whether there are natural conditions which imply that $F(s)$ would be convex.
By the way, did you use Mathematica or some other program for deriving these expressions for the minimum? (I managed to do it analytically by hand, but this was non-trivial).
At this point, I don't know of natural conditions which imply that $F$ is convex. Yes, I did use Mathematica -- glad to hear you managed to check this by hand.
|
2025-03-21T14:48:30.323755
| 2020-04-14T08:01:26 |
357431
|
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|
Stack Exchange
|
Cohomology of constructible sheaves via exit paths
Let $X$ be a stratified space, with stratification $S$ (we will ignore technicalities).
The category of exit paths $Ex(X,S)$ is a directed refinement of the path groupoid of $X$ accounting for the stratification. More precicely the objects of $Ex(X,S)$ are points of $X$, and morphisms between points are stratified homotopy classes of paths in $X$ which do monotone with respect to strata. By stratified homotopy classes I mean that the homotopies themselves are monotone in strata in a suitable way.
It is observed by MacPherson (unpublished) that the category of $S$ constructible sheaves of sets on $X$ is equivalent to the category from functors from $Ex(X,S)$ to the category of sets. I assume that this extends to constructible sheaves in vector spaces, but I don't know a reference.
This fact, together with the equivalence between local systems and $\pi_{1}(X)$-representations, tells us that a constructible sheaf of vector spaces is the same as a local system for each stratum together with morphisms of local systems encoding how the strata fit together in $X$.
Does this interpretation have any use in the study of cohomology with coefficients in constructible sheaves? More precisely, given a constructible sheaf $F$ on $X$, can you recover the cohomology groups with coefficients in $F$ from the cohomology groups with coefficients in the corresponding local systems together with the induced linear maps between them?
As a side question, is there any hope of a similar interpretation of stale constructible sheaves in algebraic geometry with exit paths replaced by torsors of some kind?
For your side question, I guess the canonical answer is this paper
Have you looked in Lurie's draft book: Higher Algebra in Appendix A. https://www.math.ias.edu/~lurie/papers/HA.pdf and the various works that cite that source?
For $X \to P$ a stratified space that satisfies the exodromy equivalence for constructible $\infty$-sheaves, the terminal map $X \to $ induces the terminal map $\operatorname{Sing}^P(X) \to $ on exit-path $\infty$-categories, which again induces, by left and right Kan extension, an adjoint triple between the categories of constructible sheaves on $X$ and on $$. The latter is however just the coefficient category. I would claim that the left Kan extension $C_$ agrees with homology with coefficients in the constructible sheaf and the right Kan extension $C^*$ with cohomology.
See https://www.math.ias.edu/~lurie/287xnotes/Lecture26.pdf for the non-stratified analogue (use the Dold-Kan correspondence); e.g. the right adjoint is similar to a (automatically derived) global sections functor. Also, note how this construction works for every stratified map, not only the terminal one. Applying Verdier duality further yields versions of Borel-Moore homology and cptly. supp. cohomology.
Finally, your idea about gluing from local systems sounds very reasonable on the level of cochains; I'll try to figure out a precise statement.
|
2025-03-21T14:48:30.323981
| 2020-04-15T18:14:50 |
357575
|
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|
Stack Exchange
|
Relationship between canonical commutation relations and projective representations?
$\DeclareMathOperator\CCR{CCR}\DeclareMathOperator\Im{Im}\DeclareMathOperator\PU{PU}$Let $V$ be a real vector space equipped with an antisymmetric bilinear form $\omega$. Recall that there is a $C^\ast$ algebra
$$\CCR(V,\omega)$$
(for "canonical commutation relations") freely generated by elements
$$\{W(f) \mid f \in V\}$$
subject to the relations
$W(-f) = W(f)^\ast$ and
$W(f)W(g) = e^{i\omega(f,g)} W(g)W(f)$
Let $\pi: \CCR(V,\omega) \to B(H)$ be a representation of $\CCR(V,\omega)$ on a Hilbert space $H$. Then from (1) and (2) we see that $\pi$ induces a projective unitary representation of $V$ on $H$, i.e., a group homomorphism $V \to \PU(H)$, where $V$ is considered as a group under addition.
Question: Let $\phi: V \to \PU(H)$ be a projective representation of the additive group of a real vector space $V$ (perhaps satisfying some continuity conditions?). Then does there exist an antisymmetric form $\omega$ on $V$ such that $\phi$ arises from a representation of $\CCR(V,\omega)$ on $H$ (perhaps with corresponding continuity conditions)?
Background:
My understanding is that in physics, one is interested in the case where $V$ is the underlying real vector space of a Hilbert space which serves as the state space for a single particle of some sort, with $\omega(f,g) = \Im \langle f,g \rangle$; in this case there is an interesting representation of $\CCR(V,\omega)$ on the bosonic Fock space associated to $V$ from which free field operators and creation
/ annihilation operators are constructed.
$\def\Z{\mathbb Z}\def\R{\mathbb R}$Am I right in re-formulating the question as: $\phi$ canonically gives $\overline\omega_\phi : V \otimes_{\Z} V \to \R/2\pi i\Z$ (essentially the cocycle realising the obstruction to liftability), and you're wondering under what conditions this may be lifted to $\omega : V \otimes_{\Z} V \to \R$ that factors through $V \otimes_{\Z} V \to V \otimes_{\R} V$? (EDIT: I guess the re-formulation doesn't matter, in light of @FrancoisZiegler's comment.)
Yes, to an extent. See Baggett-Kleppner (1973) whose Introduction starts: “The classification of the irreducible multiplier representations of a real vector group is an easy consequence of the Stone-von Neumann theorem. With no loss of generality one may assume that the multiplier is the exponential of an alternating bilinear form.” (And they go on to generalize to “quite arbitrary locally compact abelian groups.” That includes discrete [no continuity condition] but not $\infty$-dimensional topological vector groups...)
@FrancoisZiegler, I appreciate your re-posting the edited comment so that I look prophetic.
I am very predictable :-)
@FrancoisZiegler Thanks! For the finite-dimensional case, I'm having trouble seeing how this follows from the Stone-von Neumann theorem. Is it as obvious as they indicate? Regarding infinite dimensions, perhaps I need to be a bit more precise in my question, but I am interested in this case. The Stone-von Neumann theorem isn't available here.
If you have access to the paper, their argument for $\mathbf R^n$ is on p. 314 with reference to Kleppner (1965).
|
2025-03-21T14:48:30.324204
| 2020-04-15T18:23:14 |
357576
|
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|
Stack Exchange
|
How to prove polynomial inequality encoded from 1-factors in $K_{2n}$
Let $G=(V,E)$ be a complete graph $K_{2n}$ and it has $m$ 1-factors $f_{i,(i=1,\dots,m)}$, where $m=\frac{(2n)!}{n!2^n}$.
Some definition:
$F=\{f_{1},f_{2},...,f_{2n-1}\}$ is one 1-factorization in $K_{2n}$.
$S(F)$ is the number of 1-factors in one $F$; and $S(F)=2n-1$.
$N(F)$ is the number of distinct 1-factorizations in $K_{2n}$;
If $N(F)=k$, we use $F_{1}$,$F_{2}$,...,$F_{k}$ to indicate distinct $F$ .
For $K_{2n}$, when $n=1,2,3,4$, we know $N(F)=1,1,6,6240$; see A000438.
For each $f_{i}$ in $K_{2n}$, we label the edges of $f_{i}$ with numbers 1,2,...,(2n-1). For each $f_{i}$ in $F$, none of them are labeled with the same number (which means disjoint 1-factors).
Now we associate with vertex $v$ in $K_{2n}$ with a word (not number) $r$ such that $r$ is the label of the edge incident with $v$ in $f_{i}$ of $K_{2n}$. It's straightforward to check that the resulting code {$r,v\in V$} has desired parameter from $1,2,...,(2n-1)$. We label the coded word in one $f_{i}$ as $W(f_{i})=r_{1}r_{2}...r_{2n}$ (corresponding to the sequence arrangement as $v_{1}v_{2}...v_{2n}$).
see the following two examples for description:
$f_{x}=\{(v_{1}v_{4}),(v_{2}v_{6}),(v_{3}v_{5})\}=\{(11),(22),(33)\}$ : $W(f_{x})=r_{1}r_{2}...r_{6}=123132$
$f_{x}=\{(v_{1}v_{6}),(v_{2}v_{5}),(v_{3}v_{4})\}=\{(22),(44),(55)\}$ : $W(f_{x})=r_{1}r_{2}...r_{6}=245542$
We separate $m$ 1-factors $f_{i}$ into two groups:
group1: {$f_{i|i=1,2,...,2n-1}$} which forms $F$ (disjoint 1-factors in $K_{2n}$)
group2: {$f_{i|i=2n,2n+1,...,m}$} which forms $F'$ (remaining 1-factors in $K_{2n}$)
They form two polynomials: $Y(F)=\sum_{i=1}^{(2n-1)} W(f_{i})$ and $Y(F')=\sum_{i=2n}^m W(f_{i})$. Thus there will be $N(F)$ cases for $Y(F')$, which we can use $Y(F')_{1}$, $Y(F')_{2}$, ..., $Y(F')_{N(F)}$to indicate.
Question:
For 1-factorization $F$, one can permute the numbers from ($1,2,...,2n-1$) to label each $f_{i}$ in $F$, thus there will be $(2n-1)!$ label settings. In addition, there are in total $N(F)$ 1-factorizations in $K_{2n}$. Therefore we know there will be $t=N(F)*(2n-1)!$ cases for $Y(F')$.
$\exists i,j\in [1,t]$ and $i\not\equiv j$, $Y(F')_{i}==Y(F')_{j}$?
|
2025-03-21T14:48:30.324349
| 2020-04-15T18:28:49 |
357577
|
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"Oleg",
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|
Stack Exchange
|
Backward uniqueness for a heat equation with a drift
Consider heat equation with a drift (=reaction-diffusion equation)
$$
\frac{\partial u}{\partial t}=\frac{\partial^2 u}{\partial x^2}+f(t,u(t,x)), \quad t\ge0,\, x\in [0,1]
$$
with periodic or Dirichlet boundary conditions. Here $f$ is globally bounded and Lipschitz in the second argument. Is it true that if $u$,$v$ are two solutions of this PDE with $u(1)=v(1)$, then $u(0)=v(0)$? How can we prove this?
The case $f=0$ is classical; it can be found, e.g., in Evans book. There, the proof goes by showing that the second derivative of log energy is nonnegative. However, it is not clear to me at all how this technique can be adapted here, since $f$ might have no time derivatives.
An equivalent formulation of the above problem is the following: suppose we are given a bounded function $w\colon [0,1]\times \mathbb{R}_+\to\mathbb{R}$ with Dirichlet or periodic boundary conditions. Suppose that $w(0)\equiv 0$ and
$$
\Bigl|\frac{\partial w}{\partial t}+\frac{\partial^2 w}{\partial x^2}\Bigr|\le C|w|
$$
everywhere. Is it true that $w(t)\equiv0$ for all $t\ge0$?
UPD: I was told that one can use an appropriate Carleman bound. Do you know whether there are any Carleman estimates suitable for this problem?
Have you answerd this question?
No, not yet. I found a way how to solve this problem, but it's not so nice and elegant. Do you know how to solve it?
I have an idea but not checked it yet. What way you meant?
There are so-called Carleman inequalities but they are usually formulated for unbounded domains. One can try to obtain a Carleman inequality for a bounded domain, which would then yield the desired result. But the details are not completely clear to me, so if there is an alternative easier method, I would be very happy to hear about it.
The result follows by an extension of the method of logarithmic convexity which is well-known for the heat backward problem.
Let $H$ be a Hilbert space. Consider the following inequality
\begin{equation}
\|\partial_t u + Au\| \leq \alpha\|u\|, \qquad \text{ on } (0,T), \qquad (1)
\end{equation}
with $\alpha=\mathrm{const}>0$.
We assume that $$A=A_+ + A_-, \qquad (2)$$
where $A_+$ is a linear symmetric operator on $H$ with domain $D(A)$ and $A_-$ is skew-symmetric such that:
\begin{align}
\|A_- u\|^2 &\leq c(\|A_+ u\| \|u\|+ \|u\|^2), \qquad \qquad \qquad \qquad (3)\\
\partial_t \langle A_+ u, u\rangle &\leq 2\langle A_+ u, \partial_t u\rangle + c(\|A_+ u\|\|u\|+\|u\|^2) \,\qquad (4).
\end{align}
Theorem. Let $u(t)\in D(A)$, $u\in C^1([0,T];H)$ be a solution of the differential inequality $(1)$, where $A$ satisfies the conditions $(2)$-$(4)$. Then
\begin{equation}
\|u(t)\| \leq C_1 \|u_0\|^{1- \lambda(t)} \|u(T)\|^{\lambda(t)}, \qquad 0\leq t\leq T, \qquad (*)
\end{equation}
for a constant $C_1=C_1(\alpha,T)>0$ and $\lambda(t)=\dfrac{1-e^{-Ct}}{1-e^{-CT}}$ with $C$ depending on $\alpha$.
In your case, $A=A_+=\Delta$,$\; A_-=0$ and it is easy to check that conditions $(1)$-$(4)$ hold. From $(*)$ it is clear that if $w(0)=0$ then $w(t) \equiv 0$ for all $t\ge 0$, which answers your question (second formulation).
This is Theorem 3.1.3 pp 51 in
V. Isakov, Inverse Problems for Partial Differential Equations. Springer, (2017).
It also applies to more general elliptic operators, (see Example 3.1.6 in the same reference).
thank you very much for your solution! let me have a look at it and let me have a look at the book
|
2025-03-21T14:48:30.324572
| 2020-04-15T19:18:01 |
357583
|
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|
Stack Exchange
|
On the level of measure theory, what does it mean for a drift to be deterministic?
Given a drift $F\in W^{1,2}([0,T])$ adapted to the filtration of a Brownian motion $B(t)$ on Wiener space $(C[0,T],\mathcal B(\|\cdot \|_\infty)$ with Wiener measure $\mu_0$, there is another measure $\mu\ll \mu_0$ on Wiener space so that $B(t)$ under $\mu$ is $F(t)+\tilde B(t)$ by Cameron Martin Girsanov.
What does it mean for $F$ to be deterministic on the level of $\mu$? More precisely, is there a convex function $D$ defined on the space of all Girsanov measures so that $D(\mu)=0$ if and only if $\mu$ corresponds to a deterministic drift?
I know of one such function, but it is not convex (I don't think).
Let $$D(\mu)=D_{KL}(\mu||\mu_0)-\frac12\int_0^T\left(\partial_sE_\mu[\omega(s)]\right)^2ds$$
Then this is simply
$$\int_0^T\operatorname{var}_\mu(F'(s))ds$$
Is there a way of making this convex?
I see a suggest edit with a comment: "I'm actually OP" https://mathoverflow.net/review/suggested-edits/130711 You could try to get the two accounts merged - so that you have better access to your question: See, for example, the comments and the answer here: Missing all activity and questions I asked on this site. Or here: https://mathoverflow.net/help/merging-accounts.
|
2025-03-21T14:48:30.324801
| 2020-04-15T19:42:00 |
357586
|
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}
|
Stack Exchange
|
Classifying toposes of theories of rings that aren't local rings
The standard uses of toposes in algebraic geometry come from sites that look roughly like the syntactic sites of theories of local rings that they classify. This isn't particularly surprising, since this corresponds to allowing localizations as coverings in the site. Are there any interesting uses of classifying toposes of theories of rings that are not necessarily local?
Are you asking for example of interesting Grothendieck topology on the category of affine scheme which do not contains the Zariski topology ?
@SimonHenry That, plus some description of the resulting topos as the classifying topos of a theory of rings.
Well the theory of all rings is classified by the category of functors from finitely-presented rings to Sets. I do not know any other example of such topos used in algebraic geometry, but maybe a geometer can comment on this.
There are examples where the rings are not just local, but this is probably not what you want.
@DavidRoberts That's okay. What kinds of examples are you thinking of? This is more of a question of curiosity than necessity.
Check out the work of Ingo Blechschmidt, eg section 21 of https://rawgit.com/iblech/internal-methods/master/notes.pdf, where he looks at what the étale topos classifies, and what the 'surjective' topos classifies. There are others that he's looked at, but I can't find them right now. See also https://mathoverflow.net/questions/330344/classifying-space-of-valuation-ringed-spaces-over-a-topos/330363#330363
Aha, here it is: https://tcsc.lakecomoschool.org/files/2018/07/Blechschmidt.pdf see the last two slides, and see also work of Hutzler (eg https://dmv2019.math.kit.edu/wp-content/uploads/2019/08/Hutzler_IniToposClassTopos.pdf)
Sorry for storm of replies, but here is Hutzler's thesis, Internal language and classified theories of
toposes in algebraic geometry
Hutzler’s description of the infinitesimal topology fits the bill. Please post it as an answer so I can accept it.
|
2025-03-21T14:48:30.324995
| 2020-04-15T20:05:48 |
357588
|
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"Aaron Bergman",
"Abdelmalek Abdesselam",
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|
Stack Exchange
|
QFT and its notations
I know hardly anything about quantum field theory (QFT) but I'm giving a try to understand some ideas of it. As far as I understand, in QFT one is interested in studying measures such as:
\begin{eqnarray}
d\mu(\varphi) \propto e^{-S(\varphi)}d\varphi \tag{1}\label{1}
\end{eqnarray}
where $S$ is a given action and $d\varphi$ an 'a priori' measure. In particular, one of the most interesting actions is given by:
\begin{eqnarray}
S(\varphi) = \int_{\mathbb{R}^{d}}\bigg{[}\frac{1}{2}\langle \varphi, (-\Delta+m^{2})\varphi \rangle + g\varphi(x)^{4} dx\bigg{]}. \tag{2}\label{2}
\end{eqnarray}
In view of (\ref{2}), expressions such as:
\begin{eqnarray}
d\mu(\varphi) \propto e^{-S_{0}(\varphi)}e^{-\int_{\mathbb{R}^{d}}g\varphi(x)^{4}}d\varphi \equiv e^{-g\int_{\mathbb{R}^{d}}g\varphi(x)^{4}dx}d\mu_{G}(\varphi) \tag{3}\label{3}
\end{eqnarray}
are very common in the theory, where the term $e^{-g\int_{\mathbb{R}^{d}}\varphi(x)^{4}dx}$ is usually viewed as a perturbation of a Gaussian measure. This being said, I'd like to clarify something. As far as I understand, the measure $d\mu_{G}$ in (\ref{3}) is a Gaussian measure on $\mathcal{S}'(\mathbb{R}^{d})$ induced by a positive-definite function $e^{-\frac{1}{2}B(f,Gf)}$ where $B$ is a quadratic form induced by a bilinear and continuous map $B$ defined on $\mathcal{S}(\mathbb{R}^{d})\times \mathcal{S}(\mathbb{R}^{d})$ in terms of the Green's function $G$ of the massive Laplacian $-\Delta + m^{2}$. Thus, the representation $-g\int_{\mathbb{R}^{d}}\varphi(x)^{4}dx$ is only formal (what does it mean?), since $\varphi$ is supposed to be a function on $\mathcal{S}'(\mathbb{R}^{d})$. It turns out that expressions such as (\ref{3}) end up being very confusing to non-experienced students like me, once we're often misguided to wrong conclusions or get wrong understandings of the subject. I think, however, that there is a deeper reason for such notations than just being a simpler way of writting things down. I think this has to be with the fact that one is interested in seeing $\varphi(x)$ as 'tempered distributional'-valued random variables, but I'm not sure. So I'd like to clarify these concepts. Why do quantum field theorists use pointwise notation to express distributions and what can one benefit from it?
Your formula for φ does not depend linearly on f. Distributions are by definition linear functionals on a space of test functions, so your formula cannot define a distribution.
@DmitriPavlov you are right! I didn't even noticed it. Thanks. Well, this reinforces the statements in my post, right?
QFT per se doesn't reference random variables. However, one observes that, formally, QFT path integrals, continued to the Euclidean, look like partition functions in Statistical Mechanics, which does reference random variables. This mapping both allows one to use intuition from Statistical Mechanics to think about QFT, as well as provides paths towards computational methods. To do physics, it's enough to be quite informal about this mapping, but by all means flesh it out for yourself in whatever way seems most natural to you.
I’m having trouble understanding the question, but the separation of the $\phi^4$ is not natural from the physics and really is just the statement that you’re doing perturbation theory. The whole thing is really a measure on ... something. Maybe looking at lattice regularizations will help your intuition?
@AaronBergman my point here is basically the pointwise dependence of tempered distributions. It is a very common language (I've encountered it in many notes, papers etc) in QFT and I don't know the real benefit of it, since it is clear formal. Also, you're correct, I'm seeing it as a perturbative theory. About the lattice regularization, I don't know how rigorous those continuum limits can be (as I said, I'm not an expert) but the formulation on the continuum causes me trouble, as I pointed out, when tempered distributions are represented as 'real valued functions'.
If you’re trying to think of the path integral rigorously, you’re going to have a lot of trouble. The case above, for example, likely does not exist as a theory. Unless you want to pursue it as a field of research, you’ll probably have better luck of thinking of it as an integral over some inchoate ‘space’ of ‘functions’ and assuming that the lattice has a continuum limit when the theory makes sense, even if it’s been too hard to prove in most cases as yet.
The obvious answer to your question is that the common conventions and notations in field theory are historical. In the naive approach to path integrals, one only considers continuous (even smooth) field configurations, for which $S(\varphi)$ is well-defined. AFAIK, distributional configurations are forced on you for technical reasons, like the ability to properly define Gaussian integrals, but I don't know much about that. Anyway, that's why people invented normal ordering, to extend local field polynomials to some distributional configurations.
Recognizing that most configurations in the path integral are "rough", or are distributions, if you wish, doesn't solve the problem of defining field theory. Pretty much everything you calculate will still be infinite. You have to implement the renormalization program, i.e., regularize the theory and scale the couplings such as to produce a well-defined limit. However, once you accept that the theory needs to be regulated anyway, the notion of thinking of the field configurations as distributions becomes much less compelling.
I'm happy my question gathered such a nice set of comments. I'm learning a lot already. Let me say some words above to adress you all at once.
As far as I understand (and I might be wrong) if you take $g=0$ in the action (\ref{2}), you can derive a rigorous (Gaussian) measure $\mu_{G}$ on $\mathcal{S}'(\mathbb{R}^{d}$ whose covariance is $-\Delta+m^{2}$, so this is a Gaussian measure associated to the action $S$. Of course, this has a problem which is that moments such $\int_{\mathcal{S}'(\mathbb{R}^{d})}\varphi(f)^{n}d\mu_{G}(\varphi) = \infty$, so we need to regularize the theory. (continues)
(cont) the addition of a term such as $\int_{\mathbb{R}^{d}}\varphi(x)^{4}dx$ which shall be treated as a perturbation of the former Gaussian measure is, in this case, quite strange to me since you worked hard to give precise meaning to the Gaussian measure etc and then you add some formal object to the theory. I mean, to my understanding, this has two possible explanations: (a) this is motivated by the physics behind it and there's not much we can do or (b) this is just a matter of notation
apart from that, I've seen some books where even thins like $\int_{\mathcal{S}'(\mathbb{R}^{d})}\varphi(f)^{n}d\mu_{G}(\varphi)$ were written as $\int \varphi(x)^{n}d\mu_{G}(\varphi)$, so the pointwise notation was used again but just as a matter of notation for something that has meaning. This led me to believe that the $\varphi(x)^{4}$ was also a notation convention or something, as I said, to idk see field variables as random variables or something. But from the content of the comments it seems that this is a formal convention which we cannot avoid, maybe because of the physics of it (?).
The $\phi^{4} $ is motivated by the physics. The real world is not described by a quadratic action. The infinities you note associated with the Gaussian case are quite innocuous compared to what you have to contend with once you introduce interactions such as the $\phi^{4} $.
I think the issue is that you are taking the method of making the Gaussian integral rigorous and trying to extend that to other QFTs. This hasn’t worked, and, as Michael said above, it’s far from obvious that it’s the right direction to go to. The action is a classical quantity, with honest functions that can be point wise multiplied. The path integral is a not rigorous thing based on that classical action that has a plausible path to definition via the lattice using exactly that classical action.
Now, if all you want to do is get the formal perturbation expansion, you could be asking about how to rigorously calculate the n-point functions you bet by expanding the exponential. I’m sure that’s been done on the math side, but from the physics side, as Igor Khavkine said, the buzzwords are normal ordering and Wick’s theorem.
The statement $\int_{\mathscr{S}'(\mathbb{R}^d)}\varphi(f)^n d\mu_G(\varphi)=\infty$ is false.
The quick answer is that (3) written by physicists is not to be taken too seriously by mathematicians. However, it is a statement of a goal or research problem which is to find a rigorous definition/construction of what (3) is trying to say.
One has an injective continuous linear map $\iota:\mathscr{S}(\mathbb{R}^d)\rightarrow \mathscr{S}'(\mathbb{R}^d)$ given by
$$
\iota(\varphi)=\left(f\longmapsto \int_{\mathbb{R}^d}\varphi(x)f(x)\ d^dx\right)
$$
where $f$ is a generic element of $\mathscr{S}(\mathbb{R}^d)$. Via this map it is natural to identify $\mathscr{S}(\mathbb{R}^d)$ with a subset of $\mathscr{S}'(\mathbb{R}^d)$. Moreover, this subset is dense (in fact sequentially dense) in $\mathscr{S}'(\mathbb{R}^d)$. The action $S(\varphi)$ in (2) is perfectly well defined for $\varphi$ in the subset $\mathscr{S}(\mathbb{R}^d)$ but not for $\varphi\in \mathscr{S}'(\mathbb{R}^d)\backslash\mathscr{S}(\mathbb{R}^d)$.
Unfortunately, the Gaussian measure $d\mu_G$ is not supported on $\mathscr{S}(\mathbb{R}^d)$ but on the much bigger space $\mathscr{S}'(\mathbb{R}^d)$.
One can be a bit more precise and work in a weighted Sobolev or Besov space of exponent $\alpha$ but the latter would be negative except for $d=1$.
So one has to introduce a regularization in order to remove the regularization. This is explained in my answer to
A roadmap to Hairer's theory for taming infinities
and to
https://physics.stackexchange.com/questions/372306/what-is-the-wilsonian-definition-of-renormalizability
Thanks for the answer! Just to clarify, when you say (\ref{2}) is perfectly well defined for $\varphi $ in the subset of $\mathcal{S}(\mathbb{R}^{d})$ you mean $\mathcal{S}'(\mathbb{R}^{d})$ via the injection map you defined?
your comment does not make sense as written. S without prime is a subset of S' with prime. The action is well defined for phi in the subset S without prime, but not in general for phi's in the complement of S without prime in S' with prime.
ahh! Sorry I misunderstood it. Now I got it!
To make an interacting (i.e. not purely quadratic) QFT at all meaningful, you have to impose a regulator. The most transparent regulator in many ways, and the only known regulator that allows to address the full non-perturbative content of the QFT, is a lattice regulator. Replacing $\mathbb{R}^d$ by $\mathbb{Z}^d$ (or a finite subset such as $\Lambda=\mathbb{Z}^d\cap[0;L]^d$ with suitable boundary conditions) makes the path integral measure $\mathrm{d}\varphi=\prod_{x\in\Lambda}\mathrm{d}\varphi_x$ unambiguously defined and removes any problems with pointwise operations on the fields appearing inside $S(\varphi)$. Whether a continuum limit then exists is of course a non-trivial (and generally unsolved) problem.
To answer your explicit question: Quantum field theorists (if such a generalization is permitted), at least when writing expressions such as the action of $\varphi^4$ theory, tend to think not so much in terms of distributions as in terms of fields in the regularized theory, for which pointwise products are unproblematic.
It should also be noted that a QFT is the quantization of a classical field theory, and that is governed by an action that involves pointwise products; I'd consider that sufficient motivation to keep the pointwise notation (which moreover makes the Poincaré invariance of the continuum action apparent, something which a more distribution-centric approach would likely obscure).
Perhaps it's worth adding that using exactly the classical action is simply abuse of notation. Even on the lattice, it's better to use an appropriately renormalized interaction (in this case, the normally-ordered product $:\phi^4:$).
Leaving out the renormalization is also an abuse of notation. I think physicists do it because it works in the d=1 case (quantum mechanics), which was the first case where the path integral was really understood. Not really recommended, as doing this confused their thinking about QFT for decades.
In the path integral formalism, the fields aren't operators, so there is nothing to normal order at that level. Renormalization is accounted for by writing the lattice action in terms of bare parameters (typically indicated by a subscript 0), which of course aren't the physical renormalized quantities. QFT on the lattice is really pretty rigorous at finite lattice spacing (the devil is in the continuum limit).
@gmvh: As you know there are two points of view: 1) operators, 2) path integrals. In 1) normal ordering means putting creation ops on the left and annihilation ops on the right. In 2), it is not like there is no normal ordering. There is one but it is expressed differently, i.e., changing monomials like $\phi^4$ to Hermite polynomials $:\phi^4:$. That's what user1504 is referring to.
To be fair to gmvh: calling the renormalized classical lattice interaction a 'normally ordered product' is itself an abuse of notation. My apologies. :)
@gmvh: Also, another comment, physicists use evaluation at points instead of the point of view of Schwartz distributions. The reason is not the presence of a regulator which makes sample fields $\varphi(x)$ smooth. It is because correlations $\langle \varphi(x_1)\cdots\varphi(x_n)\rangle$ have singular support on the diagonal. Namely, one can evaluate them pointwise at noncoincident points. Sometimes, however, one needs to consider contact terms.
@AbdelmalekAbdesselam: Indeed, one needs to consider contact terms. Otherwise, the Dyson-Schwinger equations would be rather different, not to mention that the tree-level two-point function wouldn't be the Green function of the free field equation anymore.
|
2025-03-21T14:48:30.325895
| 2020-04-15T20:07:38 |
357589
|
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Stack Exchange
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If a vertex belongs to many maximal cliques, does it belong to a large star?
I am curious about the following question:
Suppose I have a connected graph $G = (V,E)$, and let $v \in V$. Let $K_v$ denote the set of maximal cliques in $G$ that contain $v$. Now, suppose $v$ is the center of some maximal induced star $K_{1,n}$, for some fixed $n \in \mathbb{N}$. I would like to show that there exists a constant $C$ such that $|K_v| \le Cn$, but I am unsure how to proceed with this claim (and in fact am starting to suspect this claim may be false).
I would like for this claim to be true, but I've been able to construct progressively worse examples e.g. I was able to construct an example of a graph that contains a vertex $v$ that belongs to 12 maximal cliques (all of which are isomorphic to $K_4$), but the largest star centered at $v$ is a $K_{1,2}$. The example is given in the image below, where $v$ is the central vertex.
As a result, I am starting to suspect that if pairs of maximal cliques are allowed to have large intersection, then the claim will not be true (for example in the image above, some maximal cliques intersect on all but a pair of vertices). If it does not hold in general, then does it start to hold when I limit the size of the intersection between two maximal cliques (say no two maximal cliques intersect on more than $m$ vertices)?
If you take a the complete bipartite graph $K_{n,n}$ and add a vertex $v$ adjacent to every vertex of the bipartite graph, then the resulting graph has a star of size $n$ centered at $v$, but is a member of $n^2$ maximal cliques.
Perhaps $\ 2\cdot n,$?
To ask a question when you already have an answer is against the spirit of this Q&A group. Thus, I stilled "plused-1" you here but not the question. Please, next time avoid this kind of combination.
Can be much worse. Take a complete graph on mn+1 vertices for your favourite m,n, let v be one vertex and remove the edges of a collection of m disjoint n-cliques avoiding v. The largest induced star centred at v has n leaves, but v is in n^m maximal cliques.
@WlodAA I agree with the sentiment of what you say; however, when I asked this question, I did not yet have an answer. When I cam upon an answer, I thought I would share the results of this with others who may be interested in the answer
@user36212 Yes, you are right. I actually came across an even more extreme example when investigating this question: Suppose $|V(G)| = n$ is divisible by 3 (for simplicity). Then consider the Turan graph $T(n, n/3)$. Then this is a complete multipartite graph with $n$ stable sets of size 3. Attaching a vertex $v$ adjacent to all vertices in $T(n, n/3)$ then makes $v$ have a largest star of size 3, but who belongs to $3^{n/3}$ maximal cliques. This is in fact as bad as it possibly can get (see the paper Moon and Moser, "On Cliques in Graphs").
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2025-03-21T14:48:30.326127
| 2020-04-15T21:43:14 |
357594
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Stack Exchange
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Attacking a network at minimum cost
A target system is modelled as a giant, undirected, simple graph $G$ (simple meaning no hyperedge) that can be scrutinized in adequate detail to budget and plan the attack: its topology changes slowly w/r to the time it takes to complete the attack. Although the graph has size $|G| \gg 1$, it is sparse: all its nodes have degree $O(1)$.
An attack against the graph consists in carefully choosing a subset of its nodes to disable them; that is, to cut all their incident edges. The attack is successful if the remaining subgraph has all its connected components of size $o(|G|)$. The cost of the attack is the total number of disabled nodes. I seek to establish that the graph will sustain any attack as long as the attackers limit their costs to $o(|G|)$.
Since it would be pointless to disable a node of degree $1$ or $2$, assume WLOG that suitable preprocessing has reduced the graph $G$ to its $3$-kernel: the largest minor containing neither loops, nor redundant (that is, multiple) edges, nor any subgraph that is a rooted tree or a chain. In particular $G$ has only nodes of degree $\ge 3$. To make the problem interesting, assume planarity analysis will not help any further.
Now, my question: is it true that, against a given connected kernel "in general position", successful attacks cost at least $1/4 - o(1)$ of its nodes, a tight estimate when the graph is $3$-regular? Else, is there a non-trivial lower bound on the cost?
By "in general position", I mean the attacker can only collect $O(1)$ statistics about the graph, such as # of nodes of degree $j$ for each $j$ (of which only $O(1)$ are non-zero), or # of edges connecting a node of degree $j$ to a node of degree $j^\prime$; then, they must postulate their assigned target is just any random instance from amongst a parametric family of graphs, one that happens to match the statistics at hand.
A non-trivial lower bound probably follows from looking at the vertex expansion of a random 3-regular graph. This lower bound would also hold even if the attacker knows the entire graph.
@smapers How probably would a lower bound obtained that way turn out $\ge (1/4 - o(1))|G|$ ?
Good question - where did you get this number? Interestingly, exercise 1 here suggests that $(1/4+o(1))|G|$ indeed would be the best bound achievable for 3-regular graphs using an expansion argument.
@smapers Pick a node at random from giant, random, 3-regular graph $G$. A. c., which I suspect means with probability $1 - O(1/|G|)$ in the present case, neither the chosen node nor its 3 neighbors are on a "small cycle", one of length $O(1)$; so, disable the chosen node and recompute the kernel, it is now smaller by 4 nodes and still 3-regular.
You can find a lower bound by considering $d$-regular Ramanujan graphs, which have a spectral expansion $\lambda \leq 2\sqrt{d-1}$, and therefore an edge expansion $h(G) \geq 1/2(d-2\sqrt{d-1})$ (see e.g. here).
Now if the graph is disconnected into components of size $o(|V|)$, then there must be a set $S$ of size $|V|/2 - o(|V|)$ that was disconnected from its complement. If $G$ has edge expansion $h$ then at least $h(|V|/2-o(|V|))$ edges must have been removed, and hence at least $\frac{h}{d}(|V|/2-o(|V|))$ nodes. If $G$ is a Ramanujan graph, then this gives a lower bound of
$$\left(\frac{1}{2}-\frac{\sqrt{d-1}}{d}\right)\left(\frac{|V|}{2}-o(|V|)\right) = \left(\frac{1}{4}-O\left(\frac{1}{\sqrt{d}}\right)\right)|V|.$$
Looks promising, if this is a guaranteed lower bound of $(1/4 - \sqrt 2 / 6)|G|$ for any 3-kernel. Regarding your set $S$ and its complement, is there a way to estimate the sizes of their kernels? I'd love to submit them to the same treatment as you did for $G$.
You can get a bit better: below Theorem 5 here it is mentioned that whp a random 3-regular graph has edge expansion at least $0.18$, which gives you a better lower bound of $\approx 0.03|G|$. Also, it is mentioned that any 3-regular graph has edge expansion at most $h \leq 1$, so that this method cannot prove you anything better than $|G|/6$.
The promising look comes from the fact that any acceptable separator (any set of nodes that, when disabled, let only connected components of size $o(|G|)$ survive) must leave a kernel of size $o(|G|)$ in the surviving subgraph. It is a consequence of the fact that WLOG the surviving kernel, if any, is connected; if it were giant, then by your result, no addition of size $o(|G|)$ to the separator could break it down. Conversely, a remaining subgraph w/o kernel is planar, and can be broken down into small components at a cost o(its size).
Hence my question above, about the size of the surviving kernels w/r to the size of S. Not downplaying the significance of your answer, by the way: in contrast to everything I have attempted so far, your bound is valid even if the attacker is allowed to exploit every peculiarity of the graph they can think of.
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2025-03-21T14:48:30.326482
| 2020-04-15T21:49:33 |
357595
|
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"Wojowu",
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Stack Exchange
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Are numbers which are the product of n primes more common than numbers which are the product of n-1 primes?
In a recent video (https://www.facebook.com/188916357807416/videos/519169035700435/) Stephen Wolfram wonders whether, for every integer n>2, eventually the number of integers which are precisely the product of n primes (not necessarily different) is greater than the number of integers which are precisely the product of n-1 primes. For n = 2, as he shows, this seems to be the case beyond somewhere around 10,000.
I am certain this has been studied and settled (Paul Erdös perhaps?), but don´t know where or by whom.
This is true for all $n\geq 1$. Asymptotics are known and are due to Landau, more details here
The main terms of these asymptotics suggests that the point at which $k+1$-almost primes overtake $k$-almost primes is around $e^{e^k}$.
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2025-03-21T14:48:30.326575
| 2020-04-15T22:08:38 |
357597
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Stack Exchange
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Superdifferentiable and subdifferentiable at $x$ implies differentiable at $x$
Let $(M,g)$ be a compact connected smooth Riemannian manifold without boundary and let $\phi: M \rightarrow \mathbb{R}$ be a function on $M$. We say that $\phi$ is superdifferiantiable at $x$ with super-gradient $p$, if we have that
$$\phi(\exp_{x}(v)) \leq \phi(x)+ g(p,v)_{x}+o(|v|)_{x}$$
for all small $v \in TM_{x}$
and subdifferentiable at $x$ with sub-gradient $q$ if the reverse inequality holds:
$$\phi(\exp_{x}(v)) \geq \phi(x)+ g(q,v)_{x}+o(|v|)_{x}$$
for all small $v \in TM_{x}$
The following fact has been used in various papers I've seen:
If $\phi$ is superdifferentiable at $x$ with supergradient $p$ and $\phi$ is sub differentiable at $x$ with subgradient $q$ then $p=q$ and $\phi$ is differentiable at $x$ with $p=q$.
We can maybe just prove this locally in the Euclidean case and use normal coordinates to generalize. But I'm having trouble doing this in even Euclidean case. Sorry if this is too standard to be asked on here.
Pick any arbitrary super- sub-gradients $p,q$, and an arbitrary direction $v\in T_x \mathbb R^d\cong \mathbb R^d$.
By definition of sub- super-differentiability we get the double inequality
$$
\phi(x)+\langle q,v\rangle+o_1(|v|) \leq \phi(x+v)\leq \phi(x)+\langle p,v\rangle+o_2(|v|)
$$
for two negligible functions $o_1,o_2$.
Hence
$$
\langle p-q,v\rangle\geq o_1(|v|)-o_2(|v|)=o(|v|)
\qquad
\forall\,v\in\mathbb R^d.
$$
Clearly this is only possible if $q=p$, thus necessarily the sub and super-differentials contain a unique common element. Let me denote this element as $p$ from now on.
The same chain of inequalities, with now the new extra information that $q=p$ (the unique "sub=super-gradient"), allows to write in a slightly different fashion
$$
o_1(|v|)\leq \phi(x+v)-\phi(x)-\langle p,v\rangle \leq o_2(|v|),
$$
hence
$$
\phi(x+v)=\phi(x)+\langle p,v\rangle+o(|v|)
$$
for all $v\in\mathbb R^d$.
This is the exact definition the differentiability of $\phi$ at $x$, with gradient $D\phi(x)=p$.
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2025-03-21T14:48:30.326827
| 2020-04-16T01:10:36 |
357605
|
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"url": "https://mathoverflow.net/questions/357605"
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Stack Exchange
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How to find the optimal lines?
Does anyone know anyway or any algorithm that can exactly and/or
numerically find lines $\left\{ l_{i}\right\} _{i=1}^{n_{k}+2}$ that maximizes $$\min_{1\le i<j\le n_{k}+2}\text{angle}\left(l_{i},l_{j}\right)$$ in $\mathbb{R}^{n_{k}}$ for some sequence of integers $1<n_{1}<n_{2}\cdots<n_{k}<\cdots?$ Thanks a lot.
I'm pretty sure this can be turned into some version of a sphere packing problem.
Equivalently, one can talk about points (in place of lines) of real projective n-spaces. (Of course, projective n-space corresponds to $\ \Bbb R^{n+1}$). The metric of the projective n-space comes from the euclidean unit $\ S^n.$
@user44191 Just wondering, suppose that the question were regarding sphere packing instead, would you be able to find the coordinates of the optimal points for infinitely many dimensions? Thanks.
Bukh and Cox provide exact constructions for the infinite sequence $n_k\equiv 1\bmod 3$. Sloane maintains numerical solutions for $n_k\leq 16$. Computer-assisted proofs of optimality have been obtained for $n_k\leq 6$; see this paper and references therein.
Edit: The Bukh–Cox construction is contained in the proofs of their Lemma 16 and Theorem 3. In what follows, I isolate the construction for convenience. (Throughout, I use their notation whenever possible, so your $n_k$ is replaced by $d$, for example.) Put $k=2$ and $N=\binom{k+1}{2}=3$, and let $A$ denote any $k\times N$ equiangular tight frame. See this survey for general information on these objects, but in this case, we may take
$$ A = \left[\begin{array}{rrr}1&-\frac{1}{2}&-\frac{1}{2}\\0&\frac{\sqrt{3}}{2}&-\frac{\sqrt{3}}{2}\end{array}\right].$$
If we denote the columns of $A$ by $x_1,x_2,x_3$, then notice that each $x_i$ has unit norm and $|\langle x_i,x_j\rangle|=\frac{1}{2}=:\beta$ for every $1\leq i<j\leq 3$. Take
$$ C:=\frac{1}{\beta}(A^\top A-I)+I. $$
Then the diagonal entries of $C$ are $1$ and the off-diagonal entries of $C$ are $\pm1$. One may further show that $\lambda:=\lambda_\mathrm{max}(C)=\frac{1}{\beta}(\frac{N}{k}-1)+1$. Since $d\equiv -k\bmod N$ by assumption, we have $b:=\frac{d+k}{N}\in\mathbb{N}$. Put $\epsilon=\frac{1}{b\lambda-1}$ and consider the matrix $G:=(1+\epsilon)I-\epsilon(C\otimes J)$, where $I$ is the identity matrix of size $Nb=d+k$ and $J$ is the all-ones matrix of size $b$. One may show that $G$ is positive semidefinite with rank at most $d$, and so $G=XX^\top$ for some $X\in\mathbb{R}^{d\times (d+k)}$. The columns of $X$ are unit vectors that achieve equality in the Bukh–Cox bound (i.e., Theorem 2(a)). It follows that the lines they span maximize the minimum interior angle.
Thank you for your answer. I'll look into the papers that you refer to. Thanks.
Just wondering, if Bukh and Cox have provided exact, probably implementable, constructions for the infinite sequence, then why does Sloane maintain numerical solutions only for finitely many dimensions, with respect to the question? Thanks a lot.
Sloane posted the numerical solutions in the 90s, and I'm not sure how regularly he updates his table (if at all). Meanwhile, the Bukh--Cox result is quite recent.
Okay, thank you for the information. But it seems in Bukh and Cox's paper that d and k need to satisfy a special modulo equation and that the bound(s) they established rely on another maximin function. So where can one find the optimal lines for the question? Thanks a lot.
@Tanger - Their construction is spread across a couple of proofs in their paper, so I updated my answer with a description.
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2025-03-21T14:48:30.327085
| 2020-04-16T02:41:43 |
357609
|
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"Nick L",
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|
Stack Exchange
|
Mori extremal contraction with small Betti number
Is there example of a smooth, projective, complex $3$-fold $X$, having $b_{2}(X)=2$ a Mori extremal contraction $\phi: X \rightarrow X'$ which contracts a smooth quadric surface $Q \subset X$?
It doesn't matter which of the two possible types it is, i.e. if $\phi(Q)$ is an ODP or $\{xy-z^2-t^3=0\} \subset \mathbb{C}^{4}$.
I am probably missing something, but why don't you take for $X'$ a hypersurface in $\mathbb{P}^3$, smooth except for one ordinary double point, and for $\phi$ the blowing up of that point?
As abx mentioned, the simplest example is the blowup of a cubic 3-fold with an ODP. Alternatively, the same variety can be obtained as the blowup of $\mathbb{P}^3$ along a smooth complete intersection of a smooth quadric and a cubic. The strict transform of the quadric then can be contracted to an ODP.
Thanks for the answer, is it clear that the class of a line in the Quadric is extremal in the Kleiman-Mori Cone?
Yes. I guess its enough to note that the class $3H - E$ (where $H$ is the hyperplane class of $\mathbb{P}^3$ and $E$ is the exceptional divisor of the blowup $X \to \mathbb{P}^3$) is base point free (because the center of the blowup is an intersection of cubics) and has zero intersection with these lines.
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2025-03-21T14:48:30.327224
| 2020-04-16T05:13:18 |
357612
|
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"url": "https://mathoverflow.net/questions/357612"
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|
Stack Exchange
|
Is the intrinsic volume always positive for maximum dimension?
The intrinsic volume functions on $\mathbb{R}^d$ known from the Steiner formula and Hadwiger's Theorem can be extended to the domain of definable sets of an o-minimal structure $\text{Def}(\mathbb{R}^d)$ as:
$$\mu_k(A):=\int_{G_{d,d-k}}\int_{\mathbb{R}^k(L)} \chi(A\cap (L+x)) \text{ } dx \text{ }d\gamma(L) $$
where $x$ spans over the $k$-dimensional subspace isomorphic to $\mathbb{R}^k$ which is perpendicular to $L\in G_{d,d-k}$, and where $G$ denotes the Grassmanian. Here, $\chi$ denotes the o-minimal Euler characteristic (see citation below, page 8 for details).
It can be shown that these are the intrinsic volumes on convex sets by Hadwiger's Theorem, since they are homogeneous, translation-invariant, continuous valuations on convex bodies.
On convex sets, each of these valuations has the nice property of being strictly positive. However, that is not necessarily the case when the definition above is used to extend to definable sets.
However, it is clear that $\mu_d:\text{Def}(\mathbb{R}^d)\to\mathbb{R}$ is the Lebesgue measure. Furthermore, it can be shown that since intrinsic volumes are independent of the dimension of the ambient space, $\mu_{d-k}$ is the $(d-k)-$dimensional Lebesgue measure of any subset of an affine $(d-k)$-plane in $\mathbb{R}^d$. Furthermore, $\mu_i$ for $i>d-k$ must be 0. This can also be easily extended to unions of such sets.
For such nicely-behaved sets, we have the following property: Define $\mu_{max}(A)$ to be $\mu_i(A)$ where $i$ is the largest index for which $\mu_i(A)\neq 0$. Then $\mu_{max}(A)>0$.
My question: Do all definable sets $A\in\text{Def}(\mathbb{R}^d)$ have this property?
For every example I have constructed, this holds true. I would further conjecture that $\mu_{max}(A)$ is the only positive, finite-valued Hausdorff measure of $A$. I have attempted to show that $\chi(A\cap (L+x))$ takes nonnegative values almost everywhere on the maximum index, but don't know quite how to proceed in the general case, as my background in o-minimal structures is (o-)minimal.
Edit: Assuming my below answer contains a correct proof, all of the conjectures I made above do indeed hold.
Wright, Matthew, "Hadwiger Integration of Definable Functions" (2011). Publicly Accessible Penn Dissertations. 391.
https://repository.upenn.edu/edissertations/391
$\DeclareMathOperator\dim{dim}$To answer my own question, I have put below a proof of both my conjectures (positivity and agreement with Hausdorff measure).
Let $A$ be a definable subset of $\mathbb{R}^{m+n}$ where $A$ is of (o-minimal) dimension $m$. We show $\mu_m(A)>0$.
Define the set $A_x=\{y\in \mathbb{R}^n \mid (x,y)\in A\}$ and (for any integer $d$) define $X_d\mathrel{:=}\{x\in \mathbb{R}^m \mid \dim(A_x) = d\}$. These definitions and theorems concerning them can be found in section 3.3 of Michel Coste's An introduction to o-minimal geometry.
According to Coste's Theorem 3.18, $\dim(A\cap (X_d\times \mathbb{R}^n))=\dim(X_d)+d$ for any positive integer $d$. Therefore, $\dim(A)\geq \dim(X_d)+d$ and so for $d>0$, we have the strict inequality $m=\dim(A)>\dim(X_d)$.
Furthermore, as shown on page 14 of Wright - Hadwiger integration of definable functions we have that $$ \mu_m(A)=\int_{G_{m+n,m}}\int_L \chi(\pi^{-1}(\mathbf{x})) \ d\mathbf{x}\ d\gamma (L)$$
where $\pi^{-1}(\mathbf{x})$ denotes the fiber of the projection mapping $\pi:A\to L\cong\mathbb{R}^m$. Where there is risk for confusion, we will explicitly write the domain and codomain as a subscript like $\pi_{X\to Y}:X\to Y$.
Focusing on the inner integral, we can partition $L$ as $X_0\cup \cdots \cup X_{n+m}$ (neglecting $X_{-\infty}$, which is the part of $L$ on which $A_x$ is empty). However, by our prior inequality, $\dim(X_d)<m$ for $d\neq 0$. But the integral is over $L\cong \mathbb{R}^m$, so for $d\neq 0$, $$\int_{X_d} \chi(\pi^{-1}(\mathbf{x})) \ d\mathbf{x}=0.$$
So our equation becomes $$\mu_m(A)=\int_{G_{m+n,m}}\int_{X_0\subset L} \chi(\pi^{-1}(\mathbf{x})) \ d\mathbf{x}\ d\gamma (L),$$ but the zero-dimensional Euler characteristic is just cardinality, so must be nonnegative.
Furthermore, for almost every $L\in G_{m+n,m}$ the projection $\pi(A)$ will be $m$-dimensional and therefore have positive $m$-Lebesgue measure (denoted $\lambda$). Thus, for almost every $L$, we have $\int_{L} \chi(\pi^{-1}(\mathbf{x})) \ d\mathbf{x}\geq \lambda(\pi(A))>0$. Thus, $$\mu_m(A)=\int_{G_{m+n,m}}F(L)\ d\gamma (L)$$ for some almost-everywhere positive $F$, which shows that $\mu_m(A)>0$.
Lastly, we just need to show that $\mu_d(A)=0$ for $d>m$. This is clear enough from the integral formula for $\mu_d$, since $A$ is only $m$-dimensional.
Edit: The proof can be extended to show that $\mu_m(A)$ is equal to the Hausdorff $m$-measure of $A$. Since we have shown that the integrand is the cardinality on all but a negligible set, we have $$ \mu_m(A)=\int_{G_{m+n,m}}\int_L \#(\pi^{-1}(\mathbf{x})) \ d\mathbf{x}\ d\gamma (L).$$
But since $\pi$ is an orthogonal projection, we have $\pi^{-1}(\mathbf{x})=A\cap [L^\perp+\mathbf{x}]$ where $L^\perp$ is the $n$-plane through the origin which is perpendicular to $L$. Change of variables is simple since $\gamma_{m+n,m}(\{K\})=\gamma_{m+n,n}(\{K^\perp\})$ gives a jacobian determinant of 1 for $K\mapsto K^\perp$. Performing this reparameterization with $M=L^\perp$ then shows that $$\mu_m(A)=\int_{G_{m+n,n}} \int_{M^\perp}\# (A\cap [M+\mathbf{x}])\ d\mathbf{x} \ d\gamma(M).$$
But this is an integral over all affine planes of codimension $m$, so by the Cauchy–Crofton Formula (see Fornasiero and Vasquez Rifo - Hausdorff measure on o-minimal structures), $\mu_m(A)$ is the Hausdorff $m$-measure of $A$.
|
2025-03-21T14:48:30.327585
| 2020-04-16T06:05:58 |
357614
|
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|
Stack Exchange
|
What is the smallest cardinality of a base of an ultrafilter on $\omega$ related to an almost disjoint family of cardinality $\mathfrak c$?
Let $(A_\alpha)_{\alpha\in\mathfrak c}$ be an almost disjoint family of infinite subsets of $\omega$. The almost disjointness of the family means that $A_\alpha\cap A_\beta$ is finite for any ordinals $\alpha,\beta\in\mathfrak c$. This almost disjoint family generates a filter $$\mathcal F=\{F\subseteq \omega:|\{\alpha\in\mathfrak c:A_\alpha\not\subseteq^* F\}|<\mathfrak c\}.$$
Let $\mathcal U$ be any ultrafilter on $\omega$ containing the filter $\mathcal F$.
Question. Can $\mathcal U$ have a base of cardinality strictly less than $\mathfrak c$?
Lyubomyr Zdomskyy informed me that the answer to my question is affirmative and follows from a recent (still unpublished) result of Osvaldo Guzman and Damjan Kalajdzievski who proved the consistency of $\mathfrak u<\mathfrak a=\mathfrak c$.
Take any ultrafilter $\mathcal U$ on $\omega$ that has a base $\mathcal B\subseteq\mathcal U$ of cardinality $|\mathcal B|=\mathfrak u$. Next, choose a maximal almost disjoint family of sets $\mathcal A\subseteq [\omega]^\omega\setminus\mathcal U$. Consider the filter $$\mathcal F=\{F\subseteq \omega:|\{A\in\mathcal A:A\not\subseteq^* F\}|<\mathfrak c\}.$$ We claim that $\mathcal F\subseteq\mathcal U$ under $\mathfrak a=\mathfrak c$. Assuming that $\mathcal F\not\subseteq\mathcal U$, we can find a set $F\in\mathcal F\setminus\mathcal U$. For the set $F$ the family $\mathcal A'=\{A\in\mathcal A:A\not\subseteq^* F\}$ has cardinality $|\mathcal A'|<\mathfrak c=\mathfrak a$. Then the family $\{A\setminus F\}_{A\in\mathcal A'}$ is not maximal almost disjoint in $\omega\setminus F\in\mathcal U$. So, we can choose an infinite set $B\subseteq\omega\setminus F$ such that $B\notin \mathcal A$ and the family $\{B\}\cup\{A\setminus F\}_{A\in\mathcal A'}$ is almost disjoint. Replacing $B$ by a smaller infinite subset we can additionally assume that $B\notin\mathcal U$. Then $\{B\}\cup\mathcal A$ is an almost disjoint family in $[\omega]^\omega\setminus\mathcal U$, which is strictly larger than $\mathcal A$. But this contradicts the maximality of $\mathcal A$. This contradiction shows that $\mathcal F\subseteq\mathcal U$.
|
2025-03-21T14:48:30.327739
| 2020-04-16T07:45:44 |
357619
|
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|
Stack Exchange
|
Are there maximum principles related to the third boundary condition?
I am working on this problem
\begin{equation}
\begin{cases}
u''(s)+\frac{2}{s} u'(s)=R^2 f(u) \quad \text{ for } \eta<s<1, \\
u'(\eta)=0, \ u'(1)+\beta R (u(1)-\bar{\sigma})=0,
\end{cases}
\end{equation}
where $R, \beta, \bar{\sigma}>0$, $0 \leq \eta <1$ are constants, $f \in C^1[0,+\infty)$, $f'$ is positive and bounded on $[0,+\infty)$, and $f$ attains $0$ only at the point $\sigma_0 \geq 0$.
It has been proved that this problem has a unique solution $u(s)=U(s,\eta,R)$ with $\sigma_0<u<\bar{\sigma}$ for all $\eta \leq s \leq 1$. What I want to show next is that $U(s,\eta,R)$ is strictly decreasing in $R$ for $\eta \leq s<1$. To do this, denote $v(s)=\frac{\partial U(s,\eta,R)}{\partial R}$, then $v$ solves the following problem
\begin{equation}
\begin{cases}
v''(s)+\frac{2}{s} v'(s)=2Rf(u)+R^2 f'(u)v \quad \text{ for } \eta<s<1, \\
v'(\eta)=0, \ v'(1)+\beta R v(1)=-\beta (u(1)-\bar{\sigma}).
\end{cases}
\end{equation}
Also, it has been proved that $u'(1)>0$, thereby $-\beta (u(1)-\bar{\sigma})=\frac{u'(1)}{R}>0$ follows. With all those prepared, I want to use the maximum principle to prove that $v(s)<0$ for $\eta \leq s<1$, only to see that neither can I find existing results in textbooks or other papers nor conceive a proof by myself.
Could anyone please tell me where to refer to for them, or provide some ideas on how to tackle this kind of problem? Thank you very much.
I believe that there is a typo: Either the first equation should read $\frac{2}{s}u'(s)$ in the first-order term, or the equation for $v(s)$ is wrong and should include a term $-\frac{2u'(s)}{u^2(s)}v(s)$ arising from the linearization of $\frac{2}{u(s)}u'(s)$. I strongly suspect the first option...
@leomonsaingeon Sorry there is a typo in the first equation, should have been $\frac{2}{s} u'(s)$. Thank you very much.
|
2025-03-21T14:48:30.327889
| 2020-04-16T07:50:42 |
357620
|
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"Dennis",
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|
Stack Exchange
|
Minimal good cover of the torus
Recall that an open cover $\mathfrak{U} = \{ U_\alpha \}$ of a manifold $M$ is called a good cover if all possible finite intersections $U_{\alpha_1} \cap ... \cap U_{\alpha_n}$ are contractible.
Question: What is the minimum number of open sets required for a good cover of the 2-dimensional torus?
The picture below provides a good cover of the torus (i.e. opposite sides of the parallelogram identified as usual) using 7 open sets (i.e. take sufficiently small open neighbourhoods of the hexagons). Can one do any better than 7? If not, how does one prove that 7 is optimal?
Would be nice to see a 3d rendering of that ...
You can't do any better than $7$. This follows from
Karoubi, Max; Weibel, Charles A., On the covering type of a space, Enseign. Math. (2) 62, No. 3-4, 457-474 (2016). ZBL1378.55002.
in particular Theorem 5.3 in the arXiv version.
The strict covering type of a space $X$ is the minimal cardinality of a good cover, denoted $\operatorname{sct}(X)$. This is not a homotopy invariant, so Karoubi and Weibel introduce the covering type, defined by
$$
\operatorname{ct}(X)=\min\{\operatorname{sct}(X')\mid X'\simeq X\}
$$
Obviously $\operatorname{ct}(X)\leq \operatorname{sct}(X)$. In Theorem 5.3 they use cohomological arguments to show that $\operatorname{ct}(T^2)=7$.
Great, thanks for the reference!
Noob question: is this at all related to the fact that all maps on a torus are $7$-colorable but some are not $6$-colorable?
@GregMartin: Good question. I don't know of a relationship between covering type and chromatic numbers of embedded graphs, but equally I don't know enough to say that such a thing doesn't exist.
Doesn't the linked paper state the relation of chromatic numbers and covering types on the first page? Or is there some subtle technical detail I missed?
|
2025-03-21T14:48:30.328046
| 2020-04-16T08:13:58 |
357622
|
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"url": "https://mathoverflow.net/questions/357622"
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|
Stack Exchange
|
Error term in Mertens' third theorem
Mertens' third theorem states that:
$$\prod_{\substack{
p \leq x \\
\text{p prime}
}} \left( 1 - \dfrac{1}{p} \right) \sim \dfrac{e^{-\gamma}}{\log(x)}$$
Question: what is the best functions (unconditionally and conditionally) satisfying:
$$\prod_{\substack{
p \leq x \\
\text{p prime}
}} \left( 1 - \dfrac{1}{p} \right) = \dfrac{e^{-\gamma}}{\log(x)} + \mathcal{O}(f(x))$$
The question is posted here : https://math.stackexchange.com/questions/3626427/error-term-in-mertens-third-theorem
But I can't know if the answer there is correct or not without references or an exact answer.
Cross-post from Math.SE. You've already received an answer there so you should inducate here what's wrong with it and why you are asking again.
@Wojowu, i am not sure if the answer in MSE is correct, he didn't give references, and he use terms like "i think", "i am not sure ".
You should mention all of that in the body of the question.
There's been a lot of work on unconditional results of this sort.
Rosser and Schoenfeld showed in a 1962 paper that one can take
$$\dfrac{e^{-\gamma}}{\log x} \left(1- \frac{1}{2\log^2 x} \right) < \prod_{\substack{
p \leq x \\
\text{p prime}
}} \left( 1 - \dfrac{1}{p} \right) < \dfrac{e^{-\gamma}}{\log x} \left(1+ \frac{1}{2\log^2 x} \right).$$ The upper bound is valid for $x>1$ and the lower bound for $x> 285.$
This was subequently improved by Dusart in a 2016 paper that if one has $ x \geq 2278382$ then one has $$\dfrac{e^{-\gamma}}{\log x} \left(1- \frac{1}{5\log^3 x} \right) < \prod_{\substack{
p \leq x \\
\text{p prime}
}} \left( 1 - \dfrac{1}{p} \right) < \dfrac{e^{-\gamma}}{\log x} \left(1+ \frac{1}{5\log^3 x} \right).$$
Tighter bounds which as far as I'm aware are best known today with reasonably tight explicit constants are in a paper of Axler http://math.colgate.edu/~integers/s52/s52.pdf although the bounds as written are a bit uglier in form. Note that there's a typo in inequality 6.3 in that paper where the parenthetical should be $$\left(1+\frac{1}{20 \log^3 x} +\frac{3}{16\log^4 x} + \frac{1.02}{(x-1)\log x}\right). $$ Axler's error term is essentially of the same order as that of Dusart but with an improved constant. Axler's paper also contains full references for Dusart as well as Rosser and Schoenfeld. All of these papers also have similar bounds on other functions related to counting primes such as Chebyshev's functions.
Many thanks for your answer
Unconditionally, the best known error term (f(x) in the original question) is in
E.A. Vasil’kovskaja, Mertens’ formula for an arithmetic progression, Taškent. Gos. Univ. Nauˇcn. Trudy, VoprosyMat. 548 (1977) 14–17, 139–140
See also the introduction of my paper with Zaccagnini on j.number theory, 127 (2007), 37–46,
https://doi.org/10.1016/j.jnt.2006.12.015
Many thanks for your answer
|
2025-03-21T14:48:30.328237
| 2020-04-16T08:30:19 |
357623
|
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"url": "https://mathoverflow.net/questions/357623"
}
|
Stack Exchange
|
Formal definition of CBIP algorithm
I am working on online coloring algorithms and I felt on the implementation using the CBIP (Common boundary Intersection projection) method.
However, there is not much documentation online about it. The best I could find is this lecture note. It assumes it but does not talk about the algorithm itself.
So my question is "What is the formal or informal definition of the CBIP algorithm in the k-coloring optimization problem"
Thanks
The Common Boundary Intersection Projection algorithm can be found in:
https://www.sciencedirect.com/science/article/abs/pii/S0926580516301984
|
2025-03-21T14:48:30.328305
| 2020-04-16T09:36:28 |
357626
|
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"Averroes",
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|
Stack Exchange
|
The existence of a copy of a random variable with conditional expectation constraint
Let there be two random variables and with a certain joint copula. Is it always true that there is another random variable independent from such as the vectors $(X,Y)$ and $(X,Z)$ have the same law?
What do you mean by "joint copula"? "joint distribution"?
I mean that the r.v X and Y are not independent (their joint distribution is not trivial)
No, suppose $X=Y$ a.s. and that they are non-degenerate. If we want $(X, Z)$ to have the same joint distribution as $(X, Y)$, we must also have $X=Z$ a.s. and hence $Y=Z$ a.s. Then $Y$ and $Z$ can obviously not be independent.
|
2025-03-21T14:48:30.328381
| 2020-04-16T10:07:27 |
357629
|
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"Alex M.",
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|
Stack Exchange
|
Weak convergence in $L^1(X,\mu)$ space
I found an interesting property in some lecture notes on weak convergence: lemma 3.2 (2) on page 10: https://www.uio.no/studier/emner/matnat/math/MAT4380/v06/Weakconvergence.pdf . The idea is as follows:
Let $(X,\mu)$ be a $\sigma$ finite measure space and $(f_n)_{n\in\mathbb{N}^*}, (g_n)_{n\in\mathbb{N}^*}, f,g :X\to \overline{\mathbb{R}}$ are $\mu$ measurable functions. Prove that if :
$\bullet$ There is a function $f:X\to\overline{\mathbb{R}}$ such that $\lim\limits_{n\to\infty} f_n(x)= f(x),$ for $\mu$ - almost all $x\in X$. (pointwise convergence);
$\bullet$ There is some positive constant $M>0$ such that: $||f_n||_{\infty}:=\mathrm{ess sup}\ |f_n|<M, \forall n\in\mathbb{N}^*$;
$\bullet$ $g_n\rightharpoonup g$ in $L^1(X,\mu)$,
then $f_n g_n\rightharpoonup fg$ in $L^1(X,\mu)$ i.e. $\lim\limits_{n\to\infty} \displaystyle\int_{X} f_n g_n h\ d\mu=\int_{X} fgh\ d\mu, \forall h\in L^{\infty}(X,\mu)$
I tried to prove this by definition but I cannot make good estimates on $\int_{X} |g_n||f_n-f|\ d\mu$. In the lecture notes sais that this is a simple property and the proof is omitted...Any help is very welcomed.
Crossposted on MSE: https://math.stackexchange.com/q/3627892/164025
The missing argument is a combination of Egorov's theorem and the Dunford-Pettis theorem (for the precise versions of both that we are going to use, see [Brezis, Haim, Functional analysis, Sobolev spaces and partial differential equations, Springer (2011), theorems 4.29 and 4.30 page 115 ]). Roughly speaking, the former tells us that $f_n\to f$ uniformly, up to removing arbitrarily small sets from $X$. And the latter guarantees that $|g_n|$ gives small mass to such small sets, uniformly in $n$ and only depending on the measure of the small set.
Disclaimer: as correctly pointed out by Nik Weaver in his comment, Egorov's theorem crucially requires a finite measure $\mu(X)<+\infty$. Step 1 below gives the key argument in this finite situation. Then Johannes Schürz's comment settles in step 2 the general case, based on the uniform integrability.
Step 1: assume first that $\mu(X)<+\infty$, and pick any $\epsilon>0$.
By Egorov's theorem there exists a subset $X_\epsilon\subset X$ with small complement $\mu(X\setminus X_\epsilon)\leq \epsilon$ such that $f_n\to f$ uniformly in $X_{\epsilon}$, in particular also $f_nh\to fh$. Thus by standard $L^\infty-L^1$ strong-weak convergence the term
$$
\int_{X_\epsilon} f_ng_nhd\mu\to \int _{X_\epsilon}fg hd\mu.
$$
For the remaining term (the integral on $X\setminus X_\epsilon$), the Dunford-Pettis theorem guarantees that $\{g_n\}_n$ is uniformly integrable. Given $\delta>0$, this means that $\int_{X\setminus X_\epsilon} |g_n|\leq \delta$ uniformly in $n$, as soon as $\mu(X\setminus X_\epsilon)\leq \epsilon$ is sufficiently small. Since $|f_n|_\infty\leq M$ and $h\in L^\infty$ this immediately gives
$$
\left|
\int_{X\setminus X_\epsilon} f_ng_nhd\mu
\right|\leq M\|h\|_\infty \delta
$$
Putting everything together and playing a bit with $\epsilon,\delta,n\geq n_0$, and quantifiers gives the desired result (also noticinng that $\int _{X_\epsilon}fg hd\mu\to \int _{X}fg hd\mu$ if $\epsilon\to 0$).
Step 2: assume now that $X$ has infinite measure. Since the sequence $g_n$ is $L^1$-weakly converging, it is uniformly integrable (by the Dunford-Pettis theorem) and therefore for any small $\eta>0$ there exists $X_\eta\subset X$ with $\mu(X_\eta)<+\infty$ and
$$
\int_{X\setminus X_\eta} |g_n|d\mu \leq \eta
\qquad
\forall\, n\geq 0
$$
As a consequence
$$
\left|
\int_{X\setminus X_\eta} f_n g_n hd\mu
\right|
\leq \|f_n\|_{L^\infty(X)} \|h\|_{L^\infty(X)} \|g_n\|_{L^1(X\setminus X_\eta)}\leq M\|h\|_{L^\infty(X)}\eta
$$
can be made arbitrarily small, uniformly in $n$.
The result follows next by applying step 1 on the finite measure set $X_\eta$.
PS: I was not aware of this specific statement, and it may turn out to be quite handy at some point so thank you Maxim Diana!
Doesn't Egorov only hold for finite measure spaces?
Fait point, thanks for pointing this out! At least the argument works for $\mu(X)<+\infty$. Perhaps considering comapctly supported test-functions $h$, and then explointing the $\sigma$-finiteness as usual?
Maybe a change-of-density argument can make one pretend that the measure is finite?
Use that uniformly integrable also gives you that $\forall \varepsilon >0 ,, \exists A \subset X \colon \mu(A) < +\infty \land \forall n \in \mathbb{N} ,, \Vert g_n \cdot \mathbb{1}A\Vert{L^1(X,\mu)} < \varepsilon$. Now you can use Egorov for $L^1(A, \mu)$.
Of course, I mean $\Vert g_n \cdot \mathbb{1}{X\setminus A} \Vert{L^1(X,\mu)}$.
Thank you Johannes, this works indeed!
|
2025-03-21T14:48:30.328818
| 2020-04-16T10:25:18 |
357630
|
{
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|
Stack Exchange
|
$GL_1(\mathcal{E}'(\mathbb{R}))$ open in $\mathcal{E}'(\mathbb{R})$?
Let $\mathcal{E}'(\mathbb{R})$ be algebra of all compactly supported distributions on $\mathbb{R}$, equipped with the strong dual topology $\beta(\mathcal{E}',\mathcal{E})$, and with the usual operations of addition and convolution.
Is the set ${{\textrm{GL}}}_1(\mathcal{E}'(\mathbb{R}))$ of invertible elements open in $\mathcal{E}'(\mathbb{R})$?
(As an example of an element in ${{\textrm{GL}}}_1(\mathcal{E}'(\mathbb{R}))$ which does not have support $\{0\}$, we have that $\delta_n$ belongs to ${{\textrm{GL}}}_1(\mathcal{E}'(\mathbb{R}))$
because $\delta_n \ast \delta_{-n}=\delta_0$.)
Which invertible distributions do you know? Doesn't the Theorem of supports (the convex hull of the support of the convolution is the sum of the convex hulls of the supports of the factors) imply that an invertible distribution hat support ${0}$?
Sorry, not ${0}$ but a singleton.
Perhaps Jochen Wengenroth's comments already give the answer, but here's a direct argument.
By Paley-Wiener for distributions, the Fourier transform $\tilde{\Delta}(k)$ of a distribution of compact support $\Delta(x) \in \mathcal{E}'(\mathbb{R})$ is entire, of exponential type, with at most asymptotic polynomial growth on the real axis. Since convolutions become products under the Fourier transform, $\Delta(x) \in \mathrm{GL}_1(\mathcal{E}'(\mathbb{R}))$ only if $\tilde{\Delta}(k)$ is pointwise invertible and $\tilde{\Delta}(k)^{-1}$ has the same analytic properties. But by the Hadamard factorization theorem this can only be if $\tilde{\Delta}(k) = e^{i k c_1 + c_0}$, with $c_1 \in \mathbb{R}$, $c_0\in \mathbb{C}$. So indeed, $\mathrm{GL}_1(\mathcal{E}'(\mathbb{R}))$ consists only of scaled translates of $\delta_0$.
I'm no big expert on the topology of $\mathcal{E}'(\mathbb{R})$, but I suspect this set is too small to be open.
|
2025-03-21T14:48:30.328945
| 2020-04-16T10:58:26 |
357635
|
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|
Stack Exchange
|
Solution existence for two-dimensional parabolic PDEs
I am looking for a solution $(f,g) \in C^{1,2}([0;T]\times\mathbb R;\mathbb R^2)$ to the following PDE system
$$
f_t(t,x) + a_1(t,x) f_x(t,x) + a_2(t,x) f_{xx}(t,x) - b_1 f(t,x)^2 + c(t) g(t,x) - d(t,x) = 0 \\
g_t(t,x) + a_1(t,x) g_x(t,x) + a_2(t,x) g_{xx}(t,x) + b_2 f(t,x) g(t,x) + e(t,x) g(t,x) + b_3 f(t,x) = 0
$$
where $a_1,a_2,d,e\colon [0;T]\times\mathbb R\to \mathbb R$, $c \colon [0;T]\to\mathbb R$ and $b_1,b_2,b_3 \in \mathbb R$.
Is there a theorem in the literature which I could use to prove existence (and if possible, uniqueness) of such a solution under appropriate assumptions? Thank you!
|
2025-03-21T14:48:30.329013
| 2020-04-16T11:28:34 |
357639
|
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"Dmitri Pavlov",
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"skd"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/357639"
}
|
Stack Exchange
|
Is there a n-category structure on algebras for $e_n$-like operads?
I'm fishing in troubled waters here and therefore the question is vague and meant to be as general as possible. In particular "$e_n$-like operad" can be an algebraic or topological $e_n$ operad, as for example the topological little n-cubes, or the algebraic operad that encodes homotopy Poisson-n algebras over various base fields.
So the question is then, if there is any known structure of an n-category, or $(\infty,n)$-category on a (subset of) algebras for such an operad, that is naturally induced from that operad.
Are you familiar with the work of Haugseng? https://arxiv.org/abs/1412.8459v2
Not yet, but the abstract sounds interesting.
A perhaps more elementary thing to say is that an (oo, n)-category with one object, one 1-morphism, ..., one (n-1)-morphism is an E_n-algebra in spaces.
|
2025-03-21T14:48:30.329358
| 2020-04-16T13:17:42 |
357645
|
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|
Stack Exchange
|
Collection of pairwise non-isomorphic infinite self-complementary graphs
For any set $X$ let $[X]^2 = \big\{\{a,b\}: a\neq b\in X\big\}$. We say that a graph $G$ is self-complementary if $G\cong \bar{G}$ where $\bar{G} = (V, [V]^2\setminus E)$.
Given an infinite cardinal $\kappa$, is there a collection ${\cal C}$ of pairwise non-isomorphic self-complementary graphs on the vertex set $\kappa$ such that ${\cal C} = 2^\kappa$?
Do you have a family of $2^\kappa$ many pairwise no-isomorphic graphs on every cardinal $\kappa$?
Yes, see this thread.
Thank you! More precisely, one should look at this answer of @Wojowu: https://mathoverflow.net/a/281040/61536
As you know, there are $2^\kappa $ nonisomorphic graphs of cardinality $\kappa$ for every infinite cardinal $\kappa$. (In fact there are $2^\kappa$ nonisomorphic trees of cardinality $\kappa$, see this answer.) I will show how to turn them into nonisomorphic self-complementary graphs of the same cardinality.
Given a graph $G$ of cardinality $\kappa$, let $G_1,G_2$ be two copies of $G$, and let $H_1,H_2$ be two copies of the complement of $G$, and add edges joining all vertices of $G_1$ to all vertices of $H_1$, all vertices of $H_1$ to all vertices of $H_2$, and all vertices of $H_2$ to all vertices of $G_2$. (In other words, we take the self-complementary graph $P_4$ and replace the end vertices with copies of $G$, the internal vertices with copies of $\overline G$.) In this way we get a self-complementary graph $S$ of cardinality $\kappa$.
To recover $G$ from $S$, choose a vertex $y_0$ of eccentricity $3$ and let $X$ be the set of all vertices $x$ such that $d(x,y_0)=3$; then the subgraph of $S$ induced by $X$ is a copy of $G$.
|
2025-03-21T14:48:30.329483
| 2020-04-16T13:29:25 |
357647
|
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|
Stack Exchange
|
Degree reduction in decompositions of multivariate polynomials
Is the following statement true?
Let $m,n,d$ be natural numbers. Then there exists a natural number $D=D(d,m,n)$ with the following property: If a polynomial $P(x_1,\dots,x_n)$ of total degree $d$ could be represented as $P=g(h_1,\dots,h_m)$ where $g=g(y_1,\dots,y_m)$ and $h_i=h_i(x_1,\dots,x_n)$ are polynomials, then there exists another representation $P=\tilde{g}(\tilde{h}_1,\dots,\tilde{h}_m)$ in which the total degree of any of the constituent parts of the RHS is at most $D$.
It is possible to come up with trivial examples such as $g\equiv 0$ or $h_1,\dots,h_m\equiv 0$, or $h_1= h_2$ and $g(y_1,y_2)=y_1-y_2$ where the degree of $g(h_1,\dots,h_m)$ is small but $\max\{\deg g,\deg h_1,\dots,\deg h_m\}$ could be arbitrarily large. For this reason I am asking for an alternative decomposition formed by polynomials of bounded degrees.
$D$ exists : $D=d$.
@user111 Can you elaborate?
|
2025-03-21T14:48:30.329567
| 2020-04-16T13:42:43 |
357648
|
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|
Stack Exchange
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Bousfield-Kan and Generalized Eilenberg-Moore spectral sequences
Building on the work of Anderson and Rector, Bousfield's paper "On the homology spectral sequence of a cosimplicial space" constructs a spectral sequence which takes in a cosimplicial space (here space = simplicial set) $X$ and (under certain conditions) converges to the homology of $Tot(X)$, known as the generalized Eilenberg-Moore spectral sequence (GEMSS). Sticking with with integer coefficients, its $E^2$ page is given by $E^2_{s, t} \cong \pi^sH_tX$, i.e. the cohomology of the cosimplicial abelian group $H_tX$.
On the other hand, there is an adjunction between simplicial sets and simplicial abelian groups, with left adjoint the free abelian group functor $\tilde{\mathbb{Z}}$ and right adjoint the forgetful functor $U$. Note that for a simplicial set $Y$, we have $\pi_tU\tilde{\mathbb{Z}}Y \cong H_t(Y)$. Now consider the cosimplicial space $U\tilde{\mathbb{Z}}X$. The Bousfield-Kan spectral sequence (BKSS) associated to $U\mathbb{Z}X$ has $E^2$ page $E^2_{s, t} \cong \pi^s\pi_tU\tilde{\mathbb{Z}}X \cong \pi^sH_t(X)$.
In other words, the $E^2$ page of the GEMSS of $X$ is isomorphic, objectwise, to the $E^2$ page of the BKSS of $U\tilde{\mathbb{Z}}X$. My qusetion is this: Are these in fact the same spectral sequence? Is there a map of spectral sequences which induces this isomorphism on $E^2$ pages?
Some possible evidence that this might be true: Dwyer's ``Strong convergence of the Eilenberg-Moore spectral sequence'' constructs the EMSS in the following way: Given a fibration $F \to E \to B$, one considers the cobar construction $C(E, B, \ast$), which he calls F. He then constructs the EMSS as the BKSS (or homotopy spectral sequence) associated to $U\tilde{\mathbb{Z}}C(E, B, \ast) = U\tilde{\mathbb{Z}}$F. I suspect also this question is related to Why does strong convergence of the EMSS imply that Tot commutes with suspension spectrum? but I was not able to draw a firm conclusion from that earlier post.
|
2025-03-21T14:48:30.329711
| 2020-04-16T14:36:57 |
357650
|
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|
Stack Exchange
|
Quasi-isometric embedding of graphs in non-compact riemannian surfaces
Given a complete riemannian surface $(S,m)$, where $S$ is homeomorphic to $\mathbb{R}^2$, I would like to find a weighted graph $G$ (which means a graph with real non-negative weights on the edges), embedded in $S$, and such that the (weighted) shortest path metric in $G$ is quasi-isometric to $m$ (this means that there are constants $\lambda,\epsilon,C$ such that for any vertices $x,y$ in $G$, $\tfrac1\lambda d_G(x,y)-\epsilon \le d_m(f(x),f(y)) \le \lambda d_G(x,y)+\epsilon$, where $f$ denotes the embedding of $G$ in $S$, and every point of $S$ is at distance at most $C$ from some vertex of $G$).
Note that in particular $G$ will be planar. I need that in addition, $G$ is countable and locally finite (meaning that any bounded region of $S$ contains only a finite number of vertices of $G$).
I have seen such statements proved for compact surfaces (in this case $G$ can be taken to be finite), or non-compact surfaces with some uniform bounds on the curvature or strong convexity radius (in both cases the vertex set is just an $\epsilon$-net on the surface, for a sufficiently small $\epsilon>0$), but I suspect the result holds in much greater generality.
I am not an expert in riemannian geometry and I lack some background in the area. Have you seen such a statement proved in a clean way in an article or a textbook? Thank you in advance!
Ok, thank you for the reference. What suprises me is that I have only seen mentions of this result (with arbitrary $\lambda$) or explicit proofs in fairly specific cases (compact surfaces for instance). If it holds in greater generality it should certainly appear somewhere in the literature.
It's amusing that "weight" is used in a sense that would intuitively be a length...
I might get things wrong in your comment, but how do you make sure that each point on the surface is close from the square lattice? For any fixed size of square, your surface might be arbitrarily wild inside some square face (with points that are arbitrarily far from the boundary of their face), right? In the quasi-isometry the image has to be a net, i.e. each point of the surface has to be close from a vertex (or edge, that's equivalent) of $G$.
Also, a path following the edges of the grid could be arbitrarily bad compared to a path crossing the faces, no?
@MarkSapir the surface is smooth, but close to be euclidean possibly only on small neighborhoods (and the radius of these neighborhoods can tend to 0 as we go to infinity, since the surface is not compact), so you cannot fix the size of your square grid globally, your graph has to be defined more locally on the surface.
@MattF. Here is a good reference for the Euclidean case: https://arxiv.org/pdf/1210.2435.pdf
Take a look at this paper by Saucan and Katchalsky; maybe it contains the result you need or you can use the proof...
That's extremely helpful, thank you !
This follows from the usual "economic covering" method: By Zorn (but alternatively, you can easily do it with your bare hands without using the choice axiom), $S$ admits a maximal family $(x_i)$ of 1-separated points (meaning that the distance between $x_i$ and $x_j$ is at least 1 for $i\neq j$). Then, the balls $B_i$ of center $x_i$ and radius $2$ cover $S$; but consider rather the covering by the larger balls $B'_i$ of center $x_i$ and radius $5/2$. Clearly, every geodesic segment in $S$ of length $\le 1$ lies in a ball $B'_i$.
Let $G_0$ be the $1$-skeleton of this covering $(B'_i)$: its vertices are the $x_i$'s; its edges are the pairs $(x_i,x_j)$ s.t. $B'_i$ intersects $B'_j$. Put the weight $1$ on each edge. By choosing a shortest geodesic between $x_i$ and $x_j$ for each edge, you get a map $f:G_0\to S$ which is clearly a quasi-isometry (indeed, given any shortest geodesic $\gamma$ on $S$ of length $\le n$, cut it into $n$ segments $[y_k,y_{k+1}]$ of length $\le 1$; one has a vertex $v_{i_k}$
of $G_0$ at distance $\le 2$ from each point $y_k$; by the triangle inequality, $B'_{i_k}$ and
$B'_{i_{k+1}}$ intersect; hence $\gamma$ lies at Hausdorff distance $2$ of a simplicial path in $G_0$ of length $\le n$). Of course, $f$ is not an embedding in general; however it is locally finite (any compact subset of $S$ meets only finitely many edges); in particular, once you have added the intersections of the edges as new vertices, you get an embedded, locally finite, quasi-isometric graph $G$.
(Of course, the nature of the problem changes if one adds the extra requirement that $\lambda=1$; which I do not ).
I'm happy with any constant $\lambda$, so I'm accepting this solution. Thank you very much! I completely missed the idea of replacing edge crossings by vertices, that's much simpler than what I had in mind.
Can you choose G so that its face-boundaries have uniformly bounded lengths? (Equivalently, so that every face of G is bounded by a triangle).
Perhaps you can fiddle around a little to get rid of intersections too close to the x_i, so that your edges have lengths bounded below. Thus, for a different λ, you could let all edges of G have length 1.
|
2025-03-21T14:48:30.330074
| 2020-04-16T14:39:14 |
357651
|
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|
Stack Exchange
|
Are etale morphisms "strongly formally etale"?
EDIT: Thanks to Harry Gindi and Anonymous below for insightful comments, I've refined the definitions here. Recall that a formally etale morphism of schemes $Y \to X$ is a morphism which has the unique right lifting property with respect to all nilpotent thickenings $Z \to W$. One salient feature of nilpotent thickenings is that they are universal homeomorphisms.
Definition: Say that a morphism of schemes $Y \to X$ is strongly formally etale if it has the unique right lifting property with respect to all universal homeomorphisms $Z \to W$. That is, for every commutative square as below, there exists a unique diagonal filler $W \to Y$, as indicated, making the two triangles commute.
$$\require{AMScd} \begin{CD} Z @>>> Y \\ @VVV \nearrow @VVV\\ W @>>> X \end{CD}$$
By definition, then, if $Y \to X$ is strongly formally etale, then $Y \to X$ is formally etale. The converse presumably does not hold. However, etale morphisms have an additional finiteness condition (etale = formally etale + locally of finite presentation) which makes me hope for an affirmative answer to the first question below:
Questions:
Let $Y \to X$ be an etale morphism. Then is $Y \to X$ strongly formally etale?
Does there exist standard terminology for "strongly formally etale"?
Is there a characterization of the class of morphisms which have the unique left lifting property with respect to all etale morphisms? How about (strongly) formally etale?
@downvoter I'd welcome any criticism you may have.
I'm not the downvoter, but the notion of 'weakly nilpotent' that you have is too weak. You need the condition to be pullback stable. As soon as you do that, you get the notion of universal homeomorphism, and I think this becomes related to a theorem of Grothendieck and also to some open problems in anabelian geometry.
@HarryGindi Thanks! What does the theorem of Grothendieck say?
A universal homeomorphism induces an equivalence of étale topoi. Barwick has some notes about this whole question: https://www.maths.ed.ac.uk/~cbarwick/papers/trig1.pdf
Being "weakly nilpotent" is a really weak condition: any field extension gives such a map. Corresponding, being "strongly formally etale" is a ridiculously strong condition. For example, given a field $k$, a $k$-algebra domain $A$ is strongly formally etale over $k$ only when $A=k$.
@Anonymous Thanks, point well taken.So the answer to my main question as stated is no. I wonder if anything changes if one substitutes "universal homeomorphism" as Harry suggests...
Then etale maps are strongly formally etale (by the topological invariance of the etale site) and conversely under finite presentation assumption (by the infinitesimal lifting criterion). Not sure about the last question.
@Anonymous Oh wow -- I was feeling pessimistic about this! I'm having difficulty convincing myself of either implication. I'd be grateful if you could elaborate (perhaps this would be easier in the form of an answer rather than a comment).
@TimCampion This is proved as theorem 4.8 in Barwick's notes that I linked. It's also somewhere in SGA or EGA.
@HarryGindi Ah, thanks! I see that 4.8 is the equivalence of etale sites that you mentioned -- 4.4.1 is an affirmative answer to my Question 1 (in the current form). I think I initially missed your link -- maybe it was added as an edit to your comment?
@TimCampion Yeah, I added it in an edit.
This follows from the well-known "topological invariance of the étale site"; see e.g. Tag 04DZ.
Indeed, write $S' \to S$ for $Z \to W$, and consider the pullbacks of the étale map $Y \to X$ to étale maps $T \to S$ and $T' \to S$. We need to show that a section to $T' \to S'$ uniquely arises by pullback from a section of $T \to S$. This is a special case of Tag 0BTY.
Presumably the same should hold for weakly étale affine morphisms, since these are very close to ind-étale; see Tag 097Y. The only "topological invariance of the pro-étale site" I have seen is Lemma 5.4.2 is Bhatt–Scholze, which is only on the level of sheaves (rather than an actual equivalence of sites).
Is it necessary for the weakly étale morphisms to be affine? It seems to me that when a weakly étale morphism $Y\rightarrow X$ gets tested against a universal homeomorphism $Z \rightarrow W$, by shrinking neighborhoods around chosen points repeatedly, one can make all the schemes involved affine (while maintaining that the restriction $Z'\rightarrow W'$ is a universal homeomorphism, using that univ. homeo's are affine). After obtaining the lifts in the affine situation, uniqueness of the lifts should allow them to glue to a global one, no?
@PavelČoupek even in the étale case I'm not sure how to do this. When you choose an affine in $X$, you want its preimage in $Y$ to still be affine, or at least to contain an affine containing the image of $Z$. I see no a priori reason why this should be true. (Like in my answer, you may assume $W = X$ if you want, but I don't see how that helps.)
I thought that after choosing affine $X'$ in $X$, you just choose some affine $Y'$ of $Y$ in the preimage of $X'$. Then $Y' \rightarrow X'$ should be still w. étale since w. étale is local on the source and target. After that, some yoga on the LHS: take similarly $W'$ affine in $W$ in preimage of $X'$, then intersect the preimages of $Y', W'$ in $Z$ and choose affine open $Z'$ inside this intersection. Image of $Z'$ in $W$ is open in $W'$, so after possibly takings even smaller affine $W''$ inside, taking preimage $Z''$ of $W''$; then $Z'' \rightarrow W''$ is univ. homeo. Is this incorrect?
@PavelČoupek: the thing that's unclear to me is how is lifting of this local version going to imply the global version. The hardest part is probably showing that $W$ can be covered by such $W''$. But you might be right that this just works; I just didn't see an obvious argument.
I think that is OK because $Z \rightarrow W$ is bijective. Given $w \in W$, let $z$ be the preimage of $w$ in $Z$, and denote by $y, x$ their images in $Y, X,$ resp. (in particular, $y \mapsto x$). Then replacing "affine open" by "affine neighbourhood of [the respective fixed point]" in the above, one produces $W''$ containing $w$. Of course, you might not cover $Y$ this way, but you cover at least an open subscheme containing the full (set-theoretic) image of $Z \rightarrow Y$, which agrees with the image of any possible lift $W \rightarrow Y$ (again due to $Z \rightarrow W$ bijective).
EDIT: I edited this answer rather substantially since I misinterpreted the question a bit at first. Sorry about that.
Formally étale is indeed very weak with respect to what you want:
For example, let me consider the case of a $k$-algebra $A=k[X^{p^{-\infty}}]$ and the map $A \rightarrow k$ given by killing all the variables. This is formally étale but obviously does not have the unique lifting property with respect e.g. to the thickening $A/(x) \rightarrow k,$ which is again given by killing all the variables. A version of this example (with details showing why this is formally étale) is given in this answer.
One candidate for what strongly formally étale might be is what was called ln-formally étale last year in M. Morrow's project for the Arizona winter school: Notes with project description, it is project C.1, towards the end of the notes. The "ln" refers to the fact that the lifting property considered there is with respect of killing locally nilpotent ideals. (You can also read there that "ln-formally étale" is an ad hoc term. So does not really answer the question about standard terminology.)
Another candidate is the notion of weakly étale morphism, from B. Bhatt's and P. Scholze's The pro-étale topology for schemes (arXiv). A morphism $X \rightarrow Y$ is weakly étale if both itself and its diagonal $\Delta_{X/Y}$ are flat.
It seems that your "strongly formally étale" sits somewhere in between the two (or agrees with one of them):
Clearly strongly formally étale morphism has to be ln-formally étale, since locally nilpotent thickenings are universal homeomorphisms. On the other hand, the argument used in the above linked AWS notes showing "weakly étale $\Rightarrow$ ln-formally étale" should verbatim show that a weakly étale morphism is strongly formally étale: In terms of algebras, if $A \rightarrow B$ is weakly étale, then there is a faithfully flat map $B \rightarrow C$ such that the composite $A \rightarrow C$ is a filtered colimit of étale $A$-algebras (Theorem 1.3 of the Bhatt-Scholze paper linked above). But such algebras have the desired lifting property since étale algebras do, and hence, by fpqc descent of morphisms, so does $A \rightarrow B$.
Unfortunately, it turns out that weakly étale is strictly stronger that ln-formally étale, so there is still some room there. An example demonstrating this is, more or less, the above mentioned example of formally étale algebra, on steroids. A writeup of the example is here.
ADDED: Let me show how to prove, assuming the unique lifting property for weakly étale maps between affine spaces discussed above, that all weakly étale maps have the unique lifting property:
As in the question, assume the commutative square
$$\require{AMScd}(*)\;\;\;\; \begin{CD} Z @>>> Y \\ @VVV \nearrow @VVV\\ W @>>> X \end{CD}$$
where $W \rightarrow Z$ is a universal homeomorphism and $Y \rightarrow X$ is weakly étale. We want to show that the indicated diagonal map, making the whole thing commutative, uniquely exists. (Clearly it uniquely exists as a map of topological spaces since $Z \rightarrow W$ is a homeomorphism - call this map of topological spaces $f$.)
Consider the collection of all commutative squares
$$\require{AMScd}(*')\;\;\;\; \begin{CD} Z' @>>> Y' \\ @VVV @VVV\\ W' @>>> X' \end{CD}$$
such that
1) all the schemes $W'. X', Y', Z'$ are affine open subschemes of $W. X, Y, Z,$ resp., and the maps in the square are obtained by restriction from $(*)$,
2) $Z'$ is the preimage of $W'$, or equivalently, $Z' \rightarrow W'$ is still a universal homeomorphism.
Since the property of $Y \rightarrow X$ being weakly étale is local on the source and the target, it follows that the map $Y' \rightarrow X'$ is weakly étale automatically; thus, assuming (1) and (2), a diagonal map $f': W' \rightarrow Y'$ fitting commutatively into the diagram $(*')$ uniquely exist (and, forgetting the scheme structure, it is clear that this lift as a topological map is obtained by restriction of $f$). So one just needs to glue these maps to a one map $W \rightarrow Z$. To show that this works, one needs the following two claims:
Claim 1: Each point $w \in W$ is contained in $W'$ for some square $(*')$ as described above.
Claim 2: Given any $w \in W$ and squares $(*'),(*'')$ as above such that $w \in W' \cap W'', f(w) \in Y' \cap Y'',$ there exists another square $(*''')$ as above with $w \in W''' \subseteq W' \cap W'', f(w) \in Y''' \subseteq Y' \cap Y'' $.
Assuming both claims, the lifts $f': W' \rightarrow Y' \hookrightarrow Y$ necessarily glue to the desired map $f: W \rightarrow Y$ (as a map of schemes): Claim 1 makes sure that the map is everywhere defined while Claim 2 implies that the lifts agree on overlaps, because of uniqueness of the lifts (this can always be checked on a basis, in our case basis formed by the affine opens $W'''$ as in the claim). Commutativity of the upper triangle then can be checked again locally, so noting that Claim 1 also implies that the full space $Z$ gets covered (thanks to $Z \rightarrow W$ and all $Z' \rightarrow W'$ being homeomorphisms), the upper of the two triangles commutes, and similarly for the lower triangle. Uniquness follows from uniqueness locally.
So what remains is to check the two claims.
For the first one, consider $w \in W$. Since $Z \rightarrow W$ is a homeomorphism, there is a unique $z \in Z$ mapping to $w$. Denote by $y$ the image of $z$ in $Y$ and by $x$ the image of $w$ in $X$ (so by commutativity of $(*),$ $y \mapsto x$). First choose $x \in X' \subseteq X$ an affine open neighborhood, then $y \in Y' \subseteq Y$ an affine open neighborhood contained in the preimage of $X'$. Next, let $W_1$ denote the preimage of $X'$ in $W$, and let $Z_1$ be the intersection of the preimages of $Y'$ and $W_1$ in $Y$. Since the homeomorphism $Z \rightarrow W$ is open, the image $W_2$ of $Z_1$ in W is an open neighborhood of $w$ contained in $W_1$. Finally, take $x \in W'\subseteq W_2$ an affine open neighborhood, and $Z'$ the preimage of $W'$. Universal homeomorphisms are affine, so $Z'$ is also affine, and hence $W', X', Y', Z'$ assemble to the desired square $(*')$.
The proof of Claim 2 is similar: fixing the same notation for the four points $x, y, z, w$ as above, one sees that $f(w)=y.$ So one can perform the same yoga as for the proof of claim 1, except with the extra condition that $Y'''$ (what was denoted by $Y'$ in the above paragraph) should also be contained in the prescribed intersection $Y' \cap Y''$, and similarly when choosing $W'''$, one has to choose it sufficiently small (inside $W' \cap W''$).
|
2025-03-21T14:48:30.330921
| 2020-04-16T14:49:15 |
357653
|
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|
Stack Exchange
|
Geometric Langlands: From D-mod to Fukaya
This post is rather wordy and speculative, but I promise there is a concrete question embedded within. For experts, I'll open with a question:
Question: Given a compact Riemann surface $X$, why does one prefer the category of D-modules on the space/stack $Bun_{X}(G)$ instead of the (Fukaya) category of Langrangians on the space/stack of $G$-local systems $\mathcal{L}_{X}(G).$
Much of what follows is explaining what I mean by this question.
Let $G$ be a connected reductive complex Lie group and $G^{\vee}$ the Langlands dual group. Let $X$ be a compact Riemann surface (smooth projective curve over $\mathbb{C}$). The central players in the geometric Langlands conjecture are the stacks of $G$-bundles on $X,$ denoted $Bun_{X}(G),$ and the stack of $G^{\vee}$-local systems on $X,$ which I'll denote by $\mathcal{L}_{X}(G^{\vee}).$
The best hope (using the words of Drinfeld) is that there is a (derived) equivalence between the category of D-modules on $Bun_{X}(G)$ and the category of quasi-coherent sheaves on $\mathcal{L}_{X}(G^{\vee}).$ This best hope is by now entirely dashed, and, to name one main player, Gaitsgory has expended a tremendous amount of effort to both indicate its failure, and conjecture a solution, along the way writing some very technical and interesting papers to give evidence.
Meanwhile, Witten, Kapustin, and others have made efforts to indicate how to restore some symmetry to this nebulous blob of conjectures, by arguing that they have natural interpretations as dimensional reductions of certain four dimensional super-symmetric gauge theories.
In the course of reading about various aspects of this story, I was struck by a point that Witten has made many times in talks and in print that I want to ask about here. I'm going to speak rather prosaically from here on forward, and welcome an explanation about why my simplifications don't make any sense.
Starting from Hitchin's work in the late 1980's, it became clear that the space of $G$-local systems $\mathcal{L}_{X}(G)$ is basically the cotangent bundle of $Bun_{X}(G).$ If you take stacky language seriously enough, this can be made reasonably precise. Taking this as true, there are many concrete, though not immediately applicable, theorems which indicate that the category of D-modules on $Bun_{X}(G)$ is equivalent to the Fukaya category of Langrangians in its co-tangent bundle: the latter of which is $\mathcal{L}_{X}(G).$
Using this quasi-logic, the best hope geometric Langlands conjecture posits an equivalence of (derived) categories between the Fukaya category of $\mathcal{L}_{X}(G)$ and the category of quasi-coherent sheaves on $\mathcal{L}_{X}(G^{\vee}).$
In other words, geometric Langlands becomes the statement that $\mathcal{L}_{X}(G)$ and $\mathcal{L}_{X}(G^{\vee})$ are mirror partners in the sense of homological mirror symmetry.
Question: Has this interpretation been taken seriously somewhere in the mathematical literature, and if not, is there a good reason it hasn't been?
As one final comment. There is (at least) one major advantage to putting the conjecture into this language. Both the Fukaya category of $\mathcal{L}_{X}(G)$ and the category of quasi-coherent sheaves on $\mathcal{L}_{X}(G^{\vee})$ can be defined without recourse to the complex/algebraic structure on $X.$ This is because both of these spaces/stacks are naturally complex symplectic, and this structure is independent of the complex/algebraic structure on $X.$
With this in mind, it makes more sense to refer to $\mathcal{L}_{\Sigma}(G)$ and $\mathcal{L}_{\Sigma}(G^{\vee})$ where $\Sigma$ is a connected, oriented, smooth surface. An advantage of this is that now topological symmetries (diffeomorphisms) of $\Sigma$ act naturally on these spaces/stacks and the subsequent categories. The study of these symmetries is what people in my field call the study of the mapping class group of $\Sigma,$ and there are many deep open questions about the mapping class group, that might find a natural home in the aforementioned discussion.
Are you familiar with Stoyanovsky's twisted D-module formulation of geometric Langlands? It has a symmetry between $G$ and $G^\vee$ similar to the one you propose, but with more or less reciprocal twistings.
Thanks for the pointer: I'll look into it, but would appreciate a reference if you have one in mind.
The basic reference is Stoyanovsky's paper https://arxiv.org/abs/math/0610974 but I think there are more refined versions now.
This is perhaps the first place I've seen where the misspelling "Langrangians" might actually be appropriate.
To answer [a paraphrase of] your second question first: yes the Kapustin-Witten perspective on geometric Langlands has I think been taken very seriously by a segment of the math community. I find it very misleading though to say (as is often done) that "geometric Langlands is mirror symmetry for the Hitchin space" -- mirror symmetry is a statement about 2d TFTs, while geometric Langlands is one about 4d TFTs which implies a vast amount more structure -- specifically the most important structure for the Langlands story, the action of Hecke operators, is part of the 4d story but not of the mirror symmetry statement.
In any case as Will mentions the Betti Geometric Langlands conjecture formulated in https://arxiv.org/abs/1606.08523 is a direct response to your question - in particular to have a version of geometric Langlands which as you ask should depend only topologically on the Riemann surface (so eg have a mapping class group symmetry). However it is not formulated in Fukaya category language directly. I'm fairly ignorant of Fukaya categories but my impression is that the vast technical difficulties in the subject prevent them currently being rigorously defined on the kind of spaces we are talking about here -- namely both singular and stacky. So the Fukaya-theoretic conjecture you discuss is still more of a guiding principle than a precise question.
Also since the Hitchin space is noncompact you have to decide what KIND of Fukaya category you'd mean (assume say we are dealing with a smooth manifold), i.e. what conditions to put at infinity as Will says - infinitesimal, wrapped, or partially wrapped ("with stops"). The Betti conjecture, which Nadler and I felt captured the spirit of Kapustin-Witten, is morally taking the Fukaya category with stops in the direction of the Hitchin base - i.e. your prototypical Lagrangians allowed are Hitchin fibers, not sections. [By the way for one of probably quite a few papers I can't remember this instant which does treat aspects of GL in a Fukaya perspective there's Nadler's paper on Springer theory https://arxiv.org/abs/0806.4566]
So what to do instead of Fukaya categories? by the microlocal perspective of Nadler, Zaslow, Kontsevich,.... we expect to replace Fukaya categories with categories of microlocal sheaves, eg for cotangent bundles, with constructible sheaves on the base (and you can impose singular support conditions for the "stops" or growth conditions on the Lagrangians). This actually gets you very close to the original characteristic p origin of the geometric Langlands correspondence, which dealt with l-adic sheaves -- via the Grothendieck function-sheaf dictionary those are natural "categorified" substitutes for functions, eg automorphic functions. There are no D-modules in this story.
The beautiful D-module version developed by Beilinson-Drinfeld and Arinkin-Gaitsgory in particular -- the de Rham geometric Langlands correspondence -- has a quite different flavor in many respects, and I would claim is one step further from both the arithmetic (l-adic) origins and from the mirror symmetry story. It is motivated by two (closely related) stories -- the Beilinson-Bernstein realization of representations of Lie algebras as D-modules, which gives it a very close relation to representation theory of affine Kac-Moody algebras; and conformal field theory (eg theory of vertex algebras). This allowed Beilinson-Drinfeld to leverage a crucial result of Feigin-Frenkel to prove a "big chunk" (half-dimensional slice) of the de Rham conjecture, and Gaitsgory and collaborators to develop an amazing program to understand and solve the conjecture in general.
[As a side note I think it's overly dramatic to say the "best hope" is "dashed" -- rather the distance from the original dream to the Arinkin-Gatisgory formulation is technical and not very large and not terribly unexpected - though led to some beautiful math - and is purely about understanding how to match growth conditions on one side with singularity conditions on the other, just as in studying the Fourier transform in different function spaces.]
The de Rham story also has deep relations to physics. The physics I would say is somewhat insensitive to the de Rham vs Betti distinction, which is about different algebraic structures underlying an analytic equivalence, but many of the mathematical questions require you to pick your setting more precisely (except the "core" ones that live in the intersection of the two conjectures). The de Rham story comes up naturally in relation to CFT, to things like gauge theories of Class S and the AGT conjecture, and a whole world that is part of.
OK this is now way too long.
It's going to take me some time to digest this answer David, and also the answers of Will. Thank you for your efforts, I truly appreciate it guys.
Thanks for giving an answer with, unsurprisingly, a lot more detail and background knowledge than mine! If you don’t mind, I want to interrogate a bit your comment about the Betti story being closer to the arithmetic setting. I agree of course that the advantages of $\ell$-adic sheaves because they look closer to the sheaves we use in the sheaf-functions dictionary (and elsewhere in Langlands) are massive, but doesn’t Betti also put heavy restrictions on which sheaves are allowed on the automorphic side?
So in some way Betti Geometric Langlands can see only those parts of the automorphic side that are deformation-invariant in a suitable sense. A concrete example could be sheaves arising from periods as in your work with Venkatesh and Sakellaridis. If we take $i^{G/H}: \operatorname{Bun}H \to \operatorname{Bun}G$ or other maps associated to other period varieties, the complex $i^{G/H}! \mathbb Q\ell$ does not get to participate in Betti geometric Langlands because it does not have the right singular support.
In some ways this is telling us that if we $\operatorname{Rhom} ( i^X_! \mathbb Q_\ell, i^Y_! \mathbb Q_\ell)$ for spherical varieties $X$ and $Y$, might depend on the curve we're working with and not just the combinatorial data behind $X$ and $Y$, right? I think we can indeed construct examples where this depends, at least if we are allowing our spherical variety to degenerate at some special points. In fact I guess it's probably not possible to give an $\ell$-adic story where these objects participate in a full-featured equivalance of categories...
@WillSawin Thanks for the comments! Maybe the easiest thing to say (though I'm not sure it's the most relevant one) is that you can think of nilpotent support not as a restriction but as an operation -- i.e. you can always project any sheaves to the nilpotent singular support category, so "all sheaves get to participate".
Another thing to say is the Betti category is the "right one" in the abelian case (i.e. local systems on Pic, with no finiteness conditions), and one of the key tests it passes (like Arinkin-Gaitsgory) is it's preserved by parabolic induction/Eisenstein series, so that's another reality check. And it contains the "core" categories, things made out of Hecke eigensheaves under finite colimits.
Now for doing actual "harmonic analysis", where you need large sheaves, it will have deficiencies as you point out, but I believe those match up well with the known deficiencies of the spectral side -- namely the lack of existence of an "honest" moduli stack of Galois representations--- the deformation spaces are fine (and match the finite colimit thing I was saying) but we don't [currently?] have a good theory of families on that side, and I think that matches not having "big" sheaves in the Betti automorphic category that don't have nilpotent support.
[MO is telling me to wrap this up] Let me just add that Gaitsgory, Rozenblyum, Kazhdan and Varshavsky (maybe with Arinkin and Raskin?) are now using the l-adic Betti category and I think have a "first principles" justification for it along the above lines (ie something that optimally matches what we can define on the spectral side). I would love to understand though the implications of your comment about dependence on the curve and how that is mirrored on the spectral side..
BTW David, I didn't mean to undersell the depth of anyone involved by using the phrase "dashed." It was an unfortunate choice of words.
@AndySanders no worries I didn't take it that way, I just wanted to emphasize that the distance from the original idea to the current beliefs is one of fine-tuning and details rather than a complete about-face.
One answer to your initial question is that the $D$-modules are supposed to actually do something - they're supposed to analogize to automorphic forms under the sheaf-functions dictionary. Therefore various things we know or believe about automorphic forms translate straightforwardly into expectations about these $D$-modules. We could translate that into the Fukaya category but we would have to pass through the Lagrangians-sheaves dictionary at every step.
And the Lagrangian-sheaves dictionary is not in such great shape - a significant problem is that the concrete, but not immediately applicable theorems you mention are not just not immediately applicable - no one knows how to make them apply. There is an approach to constructing geometric Langlands by passing through these Lagrangians - see this paper of Donagi and Pantev, which takes 213 pages to make the construction work in one special case.
Recall here that, the mirror symmetry here is supposed to come from the fact that the moduli spaces of Higgs bundles on $G$ and on $\hat{G}$ are dual abelian fibrations over the Hitchin base, so a skyscraper sheaf on one side is sent to a (Lagrangian) Hitchin fiber on the other side, with the additional data of a line bundle. So the mirror symmetry step is totally straightforward in this case - we know exactly what the object should correspond to. The entire difficulty is handling the transition from the Fukaya category to $D$-modules.
For your claim at the end about the mapping class group, you have to be careful, because the claims of geometric Langlands are really about complex-algebraic sheaves and not complex-analytic sheaves. In this setting Higgs bundles, local systems, and representations of the fundamental group are not all equivalent. Only representations of $\pi_1$ depend only on topology and admit this mapping class group action. The best hope statement of geometric Langlands uses local systems (i.e. vector bundles with flat connections), not representations of $\pi_1$.
More recently, Ben-Zvi and Nadler have defined Betti geometric Langlands, which does use representations of $\pi_1$. However on the other side they have to work with a different category of sheaves - derived constructible sheaves with singular support on the nilpotent cone. This category is not known to be topologically invariant, but is conjectured to be. I don't know what its Fukaya analogue might look like. I have heard that this correspondence is supposed to be a better reflection of Kapustin and Witten's physical construction, because that physical construction is topologically invariant.
Thanks for your answer Will. I'm very aware of the difference between local systems, Higgs bundles, and representations with respect to complex and algebraic issues, sorry if I elided that point in my question. You did omit one intermediate object, which is smooth local systems, where the mapping class group does act, but only analytically, not algebraically. A really obscene way to state my question, is to say that any mention of Higgs bundles is a red herring, but I'll also admit I'm just asking a different question at that point...
@AndySanders Don't you need Higgs bundles to say "this can be made reasonably precise"? Also doesn't the equivalence require doing strange things to the Lagrangians near infinity?
I'm not sure anyone has written it down, but if a co-tangent vector to a flat $C^{\infty}$ unitary connection isn't a flat complex connection, I'm not sure what it could be. But, I totally agree, the current state of the art makes my question seem very naive. In that respect, you've probably answered the main question of: why haven't mathematicians looked into this.
@AndySanders But the $D$-modules are defined using the complex (algebraic) structure of the moduli space of vector bundles, which $C^{\infty}$ unitary connections don't see.
@AndySanders I'm sure if David Ben-Zvi sees this question, he will have some additional perspective.
I agree Will, thank you very much for your comments. Now we just need to conjure David :).
Got the message, on it :-)
I think the Lagrangians-sheaves dictionary is in a bit better shape than that! See my answer below.
@VivekShende I'm going to have to claim the defense that the paper you reference was posted after my answer!
@VivekShende Still, as you point out in the paper, it doesn't yet reach the full desired equivalence. In particular, from the point of view of classical Langlands, the most interesting eigensheaves are probably the ones corresponding to sheaves on the zero fiber of the Hitchin system, as these are the ones that can carry an arithmetic structure.
@WillSawin I agree, though I thought the sheaves you mention have to do with other groups, and so perhaps you get them by induction.
@VivekShende Don't all variations of Hodge structures correspond to such sheaves? I don't think there's any reason that a variation of Hodge structures must come from a smaller group
@WillSawin most likely you’re right — my knowledge about this stuff is pretty limited. Perhaps we can discuss it offline (or rather online in another forum.)
More a comment than an answer:
I think the lagrangian-to-sheaf dictionary is more “immediately applicable” than is commonly supposed. In particular, it’s possible to immediately apply the dictionary to construct some (plausibly Hecke eigen) sheaves from smooth Hitchin fibers. See the following note:
https://arxiv.org/abs/2108.13571
Thank you very much for this reference!
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2025-03-21T14:48:30.332318
| 2020-04-16T15:14:40 |
357655
|
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|
Stack Exchange
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A sum of two binomial random variables
Let $p\in(0,1)$, $n$ a positive even integer, $k,l\in\{0,\dots,n\}$, and $X_k\sim \text{Binomial}(k,p)$, $Y_{n-k}\sim \text{Binomial}(n-k,1-p)$ independent random variables. I would like to prove that
$$
\Pr(X_k+Y_{n-k}=l)\leq\Pr(X_{n/2}+Y_{n/2}=n/2).
$$
This question can be stated analytically. Setting $c=(1-p)/p$, define:
$$
f_{n,c}(k,l)=c^{l+k}\sum_{i=\max(0,k+l-n)}^{\min(k,l)}\binom{k}{i}\binom{n-k}{l-i}c^{-2i}.
$$
Prove that $f_{n,c}$ attains its maximum at $k=l=n/2$, for any even $n$ and $c>0$.
An idea (I don'the know how to pursue all the way): express $f_{n, c}$ through beta functions and prove that it is concave. Then, by symmetry, the maximum is attained in the middle.
I don't know why people are voting to close. I don't think this question is more trivial than many similar ones, which get detailed answers here.
The result seems to be true, but I don't see any working monotonicity patterns. A mystery! I think it could help if you let us know how this conjecture arose.
It looks like monotonicity fails due to rounding. For a fixed $k$ the maximizing $l$ is the expectation up to rounding. Maybe one could obtain monotonicity by extending the definition to non-integer $l$.
Here is a (surprising) proof using Cauchy-Schwarz and "rearrangement".
The following lemma will be the key.
Lemma
: Let $X,Y$ be independent integer-valued rvs, then \begin{align*}
(a)\; &\mbox{ for any } z: \;\mathbb{P}(X+Y=z)^2\leq \big(\sum_x\mathbb{P}(X=x)^2\big)\,\big(\sum_y \mathbb{P}(Y=y)^2\big)\\
(b)\; &\sum_z\mathbb{P}(X-Y=z)^2=\sum_z\mathbb{P}(X+Y=z)^2\end{align*}
Proof: (a) apply Cauchy-Schwarz to
$\mathbb{P}(X+Y=z)=\sum_x \mathbb{P}(X=x)\mathbb{P}(Y=z-x)$
(b) let $(X^\prime,Y^\prime)$ be distributed as $(X,Y)$, and independent of $(X,Y)$. Then
\begin{align*}
\sum_z\mathbb{P}(X-Y=z)^2=\mathbb{P}(X-Y=X^\prime-Y^\prime)=\mathbb{P}(X-X^\prime=Y-Y^\prime)\\
\sum_z\mathbb{P}(X+Y=z)^2=\mathbb{P}(X+Y=X^\prime+Y^\prime)=\mathbb{P}(X-X^\prime=Y^\prime-Y)
\end{align*}
Since $Y-Y^\prime$ and $Y^\prime-Y$ are identically distributed, and independent of $(X,X^\prime)$ the right hand sides above are equal. End Proof
Now to your question above. Let $n=2m$. We have to show that
\begin{align*} \mathbb{P}(X_k+Y_{2m-k}=\ell)\leq \mathbb{P}(X_m+Y_m=m)\end{align*}For the right hand side above we have (using 1. below)
\begin{align*} \mathbb{P}(X_m+Y_m=m)=\mathbb{P}(X_m=X_m^\prime)=\sum_l \mathbb{P}(X_m=l)^2\end{align*}
To transform the left hand side, observe that well known properties of the binomial distribution give:
\begin{align} 1.\; &Y_k \mbox{ is distributed as}\; k-X^\prime_k, \mbox{where $X_k^\prime$ is distributed as $X_k$,}\\&\mbox{ and independent of $X_k$}\\
2.\; &\mbox{ $X_{m+j}$ resp. $Y_{m+j}$ are distributed as $X_m+X_j$ resp. $Y_m+Y_j$, where}\\
&\mbox{ the summands are independent}
\end{align}
Using 1. gives that $\sum_l \mathbb{P}(X_m=l)^2=\sum_l \mathbb{P}(Y_m=l)^2$ and further that
$$ \mathbb{P}(X_k+Y_{2m-k}=\ell)=\mathbb{P}(X_{2m-k}+Y_k=2m-\ell)\;,$$ so we may w.l.o.g. assume that $k\leq m$.
Using 1. and 2. we have
$$\mathbb{P}(X_k+Y_{2m-k}=\ell)=\mathbb{P}(X_k-X_{m-k}+Y_m=\ell+k-m)$$
where $X_k,X_{m-k}$ and $Y_m$ on the right hand side are independent.
Using part (a) of the lemma (with $X=X_{k}-X_{m-k}$ and $Y=Y_m$)
gives $$\mathbb{P}(X_k+Y_{2m-k}=z)^2 \leq \big(\sum_x \mathbb{P}(X_k-X_{m-k}=x)^2\big)\big(\sum_y\mathbb{P}(Y_m=y)^2\big)$$
Finally using part (b) of the lemma on the first factor gives
$$\sum_{x}\mathbb{P}(X_k-X_{m-k}=x)^2=\sum_x\mathbb{P}(X_k+X_{m-k}=x)^2=\sum_x\mathbb{P}(X_m=x)^2$$
so that ultimately
$$\mathbb{P}(X_k+Y_{n-k}=z)^2\leq \big(\sum_x \mathbb{P}(X_m=x)^2\big)\big(\sum_y\mathbb{P}(Y_m=y)^2\big)=\big(\sum_x \mathbb{P}(X_m=x)^2\big)^2\;,$$
as desired.
Indeed a very surprising proof. There is a small typo in the third row of proving (b): The middle term is $\mathbb{P}(X+Y = X' + Y')$.
@DieterKadelka: Thanks, corrected!
Thanks, @esg! Very slick!
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2025-03-21T14:48:30.332589
| 2020-04-16T15:39:46 |
357659
|
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|
Stack Exchange
|
Bounds for two arithmetic functions, when one assumes that $n$ are odd perfect numbers
For an integer $n>1$ in this post we denote the Dedekind psi function as $\psi(n)=n\prod_{\substack{p\mid n\\p\text{ prime}}}\left(1+\frac{1}{p}\right)$ and the product of distinct primes dividing our integer $n$ as $\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p\text{ prime}}}p$, we've $\psi(1)=\operatorname{rad}(1)=1$. Both are important number theoretic functions in several subjects of mathematics. As reference I add the Wikipedia article Dedekind psi function (or [1]) and the corresponding Wikipedia Radical of an integer.
An integer $n\geq 1$ is said a perfect number if $\sigma(n)=\sum_{1\leq d\mid n}d=2n$. For example $n=6$ is the first perfect number and the only square-free (even) perfect number. It is unknown if odd perfect numbers do exist. Wikipedia has an article for Perfect number.
Using the properties of the aritmetic functions $\psi(n)$ and $\operatorname{rad}(n)$ one can to check easily the veracity of the following claim for even perfect numbers invoking the known as Euclid-Euler theorem.
Claim. A) If $n$ is an even perfect number then the identity
$$\frac{\psi(n)}{n}=\frac{3}{2}+\frac{3(1+\sqrt{1+8n})}{8n}\tag{1}$$
holds.
B) If $n$ is an even perfect number then the identity
$$\sum_{\substack{1\leq d\mid n\\d<\operatorname{rad}(n)}}\frac{1}{d}=2+\frac{3-\sqrt{1+8n}}{4n}\tag{2}$$
holds.
Previous formulas provides the best approximations of the arithmetic functions in the corresponding LHS of $(1)$ and $(2)$ for even perfect numbers because are identities (in terms of $n$ from the corresponding RHS).
Question. 1) What estimation/bounds can be done in terms of $n$ about $\frac{\psi(n)}{n}$ on assumption that $n$ is an odd perfect number? 2) And, what estimation/bounds can be done in terms pf $n$ about $\sum_{\substack{1\leq d\mid n\\d<\operatorname{rad}(n)}}\frac{1}{d}$ on assumption that $n$ is an odd perfect number? Many thanks.
I don't know if the problem to find bounds for the quantity $\prod_{\substack{p\mid n\\p\text{ prime}}}\left(1+\frac{1}{p}\right)$, when one assumes that $n$ is an odd perfect number, is in the literature, in this case feel free to refer the literature answering the Question 1) as a reference request and I try to find and read those statements from the literature.
As motivation for those expressions I refer that the expression $(1)$ arises just from the quotient of the specialization of $\psi(n)$ for even perfect numbers by $n$, and on the other hand the sum in the LHS of $(2)$ evokes in some manner a variant of the sum $\sum_{1\leq d\mid n}\frac{1}{d}$ for perfect numbers or other more specific sums that are in the literature of odd perfect numbers.
References:
[1] Tom M. Apostol, Introduction to Analytic Number Theory, Undergraduate Texts in Mathematics Springer (1976).
I hope this is interesting for you. Please feel free to add your feedback in comments about this post or other of my posts of this MathOverflow.
Even perfect numbers have a very simple form. If there are any odd perfect numbers they would be much more complicated so the even ones are not very suggestive for the form of potential odd ones. If $f(n)$ is the number of prime divisors then $f(n)=2$ for all even perfect numbers while $f(n)>26$ for any odd perfect number.
(1/2) Many thanks @AaronMeyerowitz . I think that it is justified to estimate $\frac{\psi(n)}{n}$ in the question for odd perfect numbers, and also estimate the quantity $\sum_{\substack{1\leq d\mid n\d<\operatorname{rad}(n)}}\frac{1}{d}$
that evokes a substitute of a sum of the type $\sum_{\substack{1\leq d\mid n\d<\sqrt{n}}}\frac{1}{d}$ (for even perfect numbers greater than $6$ these sums are the same) or for sums of the form $\sum_{\substack{p\mid n\p\text{ prime}}}\frac{1}{p}$. I add examples of articles where these sums were studied in the theory of odd perfect numbers.
(2/2) Judy A. Holdener, A Theorem of Touchard on the Form of Odd Perfect Numbers, The American Mathematical Monthly, Vol. 109, No. 7 (2002), pp. 661-663. And for example D. Suryanarayana, On odd perfect numbers. II, Proc. Amer. Math. Soc., 14 (1963), pp. 896-904. And many thanks again for your attention professor.
If it is better @AaronMeyerowitz , with the purpose to draw attention of the Question, I can to edit the post to add the Claim (and the correspondig related paragraphs) as an appendix at the end of the post. I'm agree that it seems discursive about even perfect numbers, and here the question was about odd perfect numbers. I add for your attention the identificator of other of my posts in this MathOverflow 338182 with title On $\sum_{\substack{1\leq d\mid n\d<f(n)}}d$ and odd perfect numbers, for $f(n)$ the greatest prime factor or $\operatorname{rad}(n)$, respectively
What is your reference for your claim that $\omega(n) > 26$, where $\omega(n)$ is the number of distinct prime factors of the (odd) perfect number $n$, @AaronMeyerowitz? AFAIK, the state-of-the-art result on this remains to be Nielsen's $\omega(n) \geq 10$.
My mistake: I guess Wikipedia says at least 101 prime factors and at least 10 distinct. I don’t know if that is correct. Here is where I got 26 (which was a misreading on my part). If none of 3,5,7 is a divisor THEN 27 at least.
Thank you for getting back to us, Professor @AaronMeyerowitz =)
As far as I'm aware, we don't have any substantially non-trivial bounds on the behavior of $\psi(n)$ when $n$ is an odd perfect number.
We can at least prove the following but none of these are difficult. Recall Euler's theorem that an odd perfect number must be of the form $N=q^em^2$ where $q$ is prime, $(q,m)=1$, and $q \equiv e \equiv 1$ (mod 4). Note that there's very little content to Euler's theorem here: it actually applies to any odd number $n$ where $\sigma(n) \equiv 2$ (mod 4). In what follows, we'll write $N$ as an odd perfect number, and reserve $q$ and $e$ and $m$ as the numbers associated to $N$ as defined above. Note that $q$ is sometimes referred to as the "Euler prime" of an odd perfect number and sometimes as the "special prime." In fact, before Euler's result, Descartes proved the weaker result that an odd perfect number must have exactly one prime factor raised to an odd power, so a better name might be the Cartesian prime, but this term seems to be unpopular.
It is not hard to see that $\psi(N)$ and $N$ must share many prime factors. One easily has that for example $(q+1)m|\psi(N)$, and it isn't hard to see that $\frac{q+1}{2}|N$ (although note one cannot conclude that $\frac{q+1}{2}m|N$ since $q+1$ may have a squared prime factor).
In general, obviously $\psi(n) \leq \sigma(n)$ with equality iff $n$ is prime or $n$ is one. We'll write $t(n) = \frac{\sigma(n)}{\psi(n)}$. Note that if $p$ is an odd prime prime and $k >1$ then we have $$t(p^k) = \frac{p^k + p^{k-1} + \cdots + p + 1}{p^k + p^{k-1}} = 1 + \frac{p^{k-2} + \cdots 1}{p^k + p^{k-1}} \leq 1+ \frac{1}{p^2}.$$
Since $\prod_{p} 1+\frac{1}{p^2}$ converges, one never has for any $n$ (perfect or not) $\psi(n)$ much larger than $\sigma(n)$. One can show for an odd perfect number that they must be somewhat closer to each other than one would get just from this sort of argument.
For example, using the fact that an odd perfect number cannot be divisible by all of $3$, $5$ and $7$, one never has all three of those primes in the actually relevant product.
But none of this content is really non-trivial. This is about what one would expect an undergrad who has taken a basic number theory course to prove if they became interested in the subject.
Many thanks for your excellent answer, I'm going to read it now, and study your reasonings and ideas in next days. I have no enough reputation to upvote your answer, but as soon I can I'm going to do it.
Just one minor note, @JoshuaZ. $q + 1 \nmid N$ since LHS is even while RHS is odd. Perhaps you meant $$\frac{q+1}{2} \mid N$$ instead?
@JoseArnaldoBebita-Dris Yes, absolutely right. Fixed. Thanks.
|
2025-03-21T14:48:30.333246
| 2020-04-16T15:42:41 |
357660
|
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|
Stack Exchange
|
idea and intuition behind triangulated category
I have some trouble in understanding the significance of some axiom of triangulated category.
If someone could explain to me each axiom with some intuition, and explain to me the intuition behind the translation functor
I would be
very grateful !!
Thank you in advance
Hello.. Welcome to MO.. First, let’s start with the axiom you are having difficulty in understanding.. what is the axiom?
Thanks you for your answer,TR2-TR4 the most problematic is TR2 for the two other I have some intuition @PraphullaKoushik
What do you mean by problematic?
no intuition about why this axiom ?
I do realize this is probably not helping, but I am strongly convinced that the only way to get reliable intuition for how triangulated categories work in practice is to get at least an intuitive understanding of the concept of stable ∞-category. Trying to formally prove the axioms of a triangulated category from the axioms of a stable ∞-category is a fun little exercise that helps a lot in understanding what's going on...
@Denis I disagree, I think you just need to understand one example of what triangulated categories are trying to axiomatize. The easiest is the category of chain complexes over a ring, considered up to quasi-isomorphism. Anonyme: in this example, the translation functor is just the operation of shifting a chain complex one step to the left.
I think this question is perfectly legitimate for MO, it is not opinion based and very useful for someone starting to learn homological algebra an homotopy theory. I do not think it belongs to MSE.
I think the question is legitimate. Neverthless, @Anonyme could provide a little more information on their background (especially for what purpose they want to learn triangulated categories). And I guess some of the axioms make sense for them; they should list the axioms that they find difficult to understand.
It is risky to give a motivation for any concept in math and worse that of triangulated category that it is in a sense is a transitional concept form usual mathematics to mathematics up to homotopy.
In any case, I will give a few naive observations with the starting point that the prototypes of triangulated categories are basically the stable homotopy category and the derived category of modules over a ring. It is a good idea to have some notion of at least one of these examples to grasp the ideas behind the axioms.
Distinguished triangles pretend to formalize (co)fibration sequences that in a stable world both concepts agree. From the complexes camp, in the usual category of complexes a short exact sequence yields a long exact sequence in homology, therefore one needs a substitute that works after inverting quasi-isomorphims and very few exact sequences remain.
A distinguished triangle (I much prefer this name over the exact triangle terminology). It has to be a diagram like
$$
X \to Y \to Z \overset{+1}\to X
$$
Where the label "$+1$" refers to a shift of degrees. Why this shift? Consider the connecting homomorphism in a long exact sequence. In the case of complexes a triangle should arise for an exact sequence of complexes $$0 \to A \to B \to C \to 0.$$ From these one gets maps in homology $H^i(C) \to H^{i+1}(A)$ or otherwise put $H^0(C[i]) \to H^0(A[i+1])$. Notice that instead of $H$ one may use another homological functor. In topology, this is related to a canonical map from the cone $C$ of the map $A \to B$, to the suspension of $A$, i.e. a map $C \to \Sigma A$. The fibration sequence becomes a triangle
$$
A \to B \to C \to \Sigma A
.$$
This forces the category to be graded: it has to have an automorphism, called "shift" or "suspension", denoted most frequently as $X[1]$ or $\Sigma X$, with the triangles being depicted now as something like
$$
X \to Y \to Z \to X[1]
$$
The fact that the shift is an automorphism automatically gives a grading into the morphisms so you define $\mathrm{Hom}^i(A,B)$ as $\mathrm{Hom}(A,B[i])$.
Now let's go for the axioms.
TR1. The triangle $X \to X \to 0 \to X[1]$ is distinguished: because the (co)fiber of the identity has to be trivial. Think on the third point as a sort of cokernel, that might as well be interpreted as a kernel when you turn the triangle by TR2.
A (co)fiber or (co)kernel exists: any map can be completed to a distinguished triangle.
And finally everything isomorphic to a distinguished triangle is a distinguished triangle. I would call this axioms the sanity axioms, they allow everything to make sense.
TR2. The triangles can be turned forward or backward. I would call this the stability axiom. You can't tell a fiber from a cofiber or a kernel from a cokernel in this stable world. If you have a long exact sequence you can't tell where it starts, this axiom reflects that idea.
TR3. A partial map between triangles may be completed to a map of triangles, otherwise said, if you give two maps between the corresponding objects of two triangles (making the corresponding square commutative), one obtains a third one making everything else commutative. Caveat: this map is not unique, and in a sense this makes part of what makes homotopy difficult. To complete, I'd call this the weak functoriality of cones axiom.
TR4. This has a classical name: the octahedral axiom. It was introduced by Verdier whose motivation was to be able to take fractions. He applied it to go form the homotopy category to the derived category avoiding arbitrary long zig-zags, just simple fractions will work. Octahedra are in a sense higher triangles (one step higher in the ladder) and sometimes people look for even higher such diagrams. It is however noteworthy how far one can go with this four axioms plus the existence of arbitrary coproducts, like for instance Brown representability, Verdier quotients and Bousfield localizations.
A view of the octahedral axiom that is enlightening for me is as a version of Noether isomorphisms as it is depicted in the following diagram:
Very roughly: $Z/Y \cong (Z/X)/(Y/X)$.
I should add that this is not the end of then story. In some cases one needs higher diagrams ($n$-triangles) and even worse, one would need some fix for the non-uniqueness of cones. The remedies for this are basically stable $\infty$-categories or alternatively stable derivators. But this takes us far afield into the world of advanced homotopy theory.
"Where the label "$+1$" refers to a shift of degrees. This forces the category to be graded: it has to have an automorphism, called "shift" or "suspension", denoted most frequently as $X[1]$ or $\Sigma X$"
can you explain what do you mean by shift of degree
@Anonyme I've edited the post with my previous comments, now deleted.
@PeterMay Of course, TR3 is logically redundant. However it may help as a step to understand the axioms. In my presentation I tried to emphasize the conceptual aspects. There are several ways to approach the octahedral axiom, and in my experience several people prefers different presentations to grasp it better.
TR3 is redundant: it is implied by the other axioms. TR4 is confused by the initial octahedral shape: it starts with a square in which one side is the identity. If one instead starts with a triangle given by a pair of composable arrows and their composition and writes it as a braid, one readily sees that it is just describing the behavior of triangles under composition. The description becomes obvious in the topological framework that was Verdier's starting point (he credited work of Puppe). For the redundancy claim, the braid description, and the behavior of distinguished triangles under products, see https://www.sciencedirect.com/science/article/pii/S0001870801919954 or https://www.math.uchicago.edu/~may/PAPERS/AddJan01.pdf.
I was very surprised by the claim that the topological framework was Verdier's starting point: my impression has always been that Verdier and Grothendieck's primary motivation was to abstractly understand sheaf cohomology in general, and Grothendieck's six functors in particular. Looking now in the introduction of "Des catégories dérivées des catégories abéliennes" I do find the following passage:
"Les catégories $\mathsf K(A)$ et $\mathsf D(A)$ ne sont pas nécessairement abéliennes; mais elles sont munies d'une structure supplémentaire, consistant en la donnée d'une famille de diagrammes définis à partir de la construction du cône : les triangles distingués. Les triangles distingués jouent, pour ces catégories, le rôle des suites exactes des catégories abéliennes. Les catégories additives munies de cette structure supplémentaire, mise en évidence et étudiée par Puppe [2], sont appelées catégories triangulées et sont étudiées de notre point de vue au chapitre II."
Other than this citation of Puppe there is no mention of stable homotopy in the text. I would say it is unclear to what extent Verdier was aware of stable homotopy theory at the time!
|
2025-03-21T14:48:30.333884
| 2020-04-16T16:13:20 |
357663
|
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|
Stack Exchange
|
Maximal quotients of the enveloping algebra of a simple Lie algebra
Let $\mathfrak{g}$ be a simple Lie algebra with Cartan subalgebra $\mathfrak{h}$, $U(\mathfrak{g})$ the universal enveloping algebra of $\mathfrak g$, $Z(\mathfrak g)$ the center of $U(\mathfrak{g})$, $\lambda$ a complex weight and $\chi$ the central character of the Verma module with highest weight $\lambda$. Let $A_\lambda\mathrel{:=}U(\mathfrak{g})/(z-\chi(z), \, z\in Z(\mathfrak{g}))$.
Why is it true that, as $\mathfrak{g}$-modules,
$$A_\lambda=\bigoplus_\mu {V_\mu\otimes V_\mu^*[0]}, $$
where $\mu$ runs over the dominant integral weights, $V_\mu$ is the irreducible finite-dimensional representation of $\mathfrak{g}$ with highest weight $\mu$ and $V_\mu^*[0]$ is the $0$-weight subspace of $V_\mu^*$?
For reference see Etingof and Stryker - Short star products for filtered quantizations, I (Prop. 4.4).
@JoseBrox, I believe that it is best practice when proposing substantive changes, even obviously correct ones such as yours, to leave a comment with the clarification and let the original asker make the change themselves.
@LSpice In this very case the change is very focussed and helps the reader (and is precisely justified in the edit history), and the OP technically has the possibly to roll back to the previous version, so I think this change was good practice. In addition the OP has left MO since June, so would most likely have ignored the comments.
This is proved in Chapter 8 of J. Dixmier, Enveloping algebras (MSN).
See also 3.1. in J. C. Jantzen, Einhüllende Algebren halbeinfacher Lie-Algebren (MSN).
|
2025-03-21T14:48:30.334032
| 2020-04-16T16:18:25 |
357664
|
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|
Stack Exchange
|
Inequality on the Kullback-Leibler divergence
Let us define the arithmetic, geometric, and harmonic means of $x,y \in \mathbb{R}$ weighted by $\alpha =(\alpha_x,\alpha_y) \in [0,1]$, respectively as
\begin{equation}
a_\alpha(x,y) = \frac{\alpha_x x+\alpha_y y}{\alpha_x+\alpha_y}
\end{equation}
\begin{equation}
g_\alpha(x,y) = x^\frac{\alpha_x}{\alpha_x+\alpha_y}y^\frac{\alpha_y}{\alpha_x+\alpha_y}
\end{equation}
\begin{equation}
h_\alpha(x,y) = \frac{xy(\alpha_x+\alpha_y)}{\alpha_xy + \alpha_yx}
\end{equation}
Let us define the following information quantities (see https://arxiv.org/pdf/1912.00610.pdf for more details on these quantities)
\begin{equation}
A_\alpha(x,y) = \alpha_x \text{KL}\left(a_\alpha(x,y),x\right) + \alpha_y \text{KL}\left(a_\alpha(x,y),y\right)
\end{equation}
\begin{equation}
G_\alpha(x,y) = \alpha_x \text{KL}\left(g_\alpha(x,y),x\right) + \alpha_y \text{KL}\left(g_\alpha(x,y),y\right)
\end{equation}
\begin{equation}
H_\alpha(x,y) = \alpha_x \text{KL}\left(h_\alpha(x,y),x\right) + \alpha_y \text{KL}\left(h_\alpha(x,y),y\right)
\end{equation}
where $\text{KL}(\cdot)$ is the Kullback-Leibler divergence. The arithmetic-geometric-harmonic mean inequality is a very well known result that states
\begin{equation}
a_\alpha(x,y) \geq g_\alpha(x,y) \geq h_\alpha(x,y)
\end{equation}
$\forall x,y \in \mathbb{R}$.
Given this inequality, and given that $x\geq y$, can we conclude that the the following is true?
\begin{equation}
A_\alpha(x,y) \geq G_\alpha(x,y) \geq H_\alpha(x,y)
\end{equation}
In the case of simple distributions such as Bernoulli, Poisson or exponential distributions, for which the KL admits a close form, I think it should follow directly from the monotonicity of the function $\log(\cdot)$. But is this true for any distribution? How to prove it?
|
2025-03-21T14:48:30.334137
| 2020-04-16T16:19:33 |
357665
|
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|
Stack Exchange
|
Open embedding of non-separable infinite dimensional manifolds
It is well-known (see here) that separable infinite-dimensional topological Hilbert manifolds can be embedded as open sets of the modeling separable Hilbert space. Using that separable Fréchet (in particular Banach) spaces are homeomorphic to separable Hilbert spaces we conclude that topological Fréchet manifolds can also be embedded. The same holds in the smooth setting.
My questions are:
there are some partial generalization to the non-separable case?
If not, what are the main obstructions?
Any help is welcome.
|
2025-03-21T14:48:30.334196
| 2020-04-16T16:31:15 |
357671
|
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|
Stack Exchange
|
Is there a coproduct on the Weyl algebra which gives the coproduct on $\mathcal{U}_q(\mathfrak{gl}_2)$?
In the paper Modular Double of Quantum Group, Fadeev gives a presentation of $\mathcal{U}_q(\mathfrak{gl}_2)$ in terms of a Weyl algebra $\mathcal{C}_q$ with generators $w_i, i \in \mathbb{Z}/4$ and relations $w_n w_{n+1} = q^2 w_{n+1} w_n$, $w_n$ commuting otherwise.
At the end of the paper, he brings up the problem of giving a coproduct on $\mathcal{C}_q$ which agrees with the coproduct on $\mathcal{U}_q(\mathfrak{gl}_2)$. Has anyone constructed this coproduct? Is it known to not exist?
|
2025-03-21T14:48:30.334269
| 2020-04-16T16:35:19 |
357672
|
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|
Stack Exchange
|
When existence of loxodromic, WPD elements implies an action is acylindrical
Definitions
Say that $(X,d)$ is a $\delta$-hyperbolic space and that $G$ is a finitely generated group acting on $X$ by isometries.
Recall that an action of $G$ on $X$ is called acylindrical if the following holds:
For every $\epsilon>0$, there exists $R,N>0$ such that for every $x,y\in X$ satisfying $d(x,y) \geq R$, there are at most $N$ elements $g\in G$ such that $d(x,gx) \leq \epsilon$ and $d(y,gy) \leq \epsilon$.
Also recall that an element $h\in G$ is called a WPD element if the following holds:
For every $\epsilon >0$ and every $x\in X$, there is an $M\in \mathbb{N}$ such that $|\{ g\in G:d(x,gx)<\epsilon, d(h^Mx,gh^Mx)<\epsilon\}|<\infty$.
Via Osin's paper Acylindrically hyperbolic groups, we know that the following are equivalent definitions of being acylindrically hyperbolic:
$G$ admits a non-elementary acylindrical action on a hyperbolic space $X$.
$G$ is not virtually cyclic and admits an action on a hyperbolic space $Y$ such that at least one element of $G$ is loxodromic and satisfies the WPD condition.
Questions
It is clear to me that if $G$ acts acylindrically and non-elementarily on $X$, that 2 follows, with $X=Y$. My first question is the following:
Question 1: Given 2, the hyperbolic space $X$ in 1 need not be the same hyperbolic space $Y$ from 2. Is it known how one can construct $X$ given $Y$?
My second question is related.
Question 2: If we know that 2 holds, is it known when we can take $X=Y$? That is, if we know that 2 holds, what is required to be able to say that the action of $G$ on $Y$ is acylindrical?
Motivation
In the setting of the mapping class group, we can take both $X$ from 1 and $Y$ from 2 to be the curve complex, (with some mild assumptions on the complexity of the surface). The fact that the action is acylindrical is due to Bowditch in the paper Tight geodesics in the curve complex, and the fact that pseudo-Anosovs act as WPD elements on the curve complex is due to Bestvina-Fujiwara in Bounded cohomology of subgroups of mapping class groups.
I am interested in various subcomplexes of the curve complex and subgroups of the mapping class group acting on these subcomplexes and would like to know if these actions are also acylindrical.
Question 2 is vague, but here is an example which, I think, shows that it is not reasonnable to expect positive results in this direction without (strong) additional assumptions.
Consider the infinite free product $G:= \underset{n \geq 2}{\ast} \left( \mathbb{Z} \oplus \mathbb{Z}_n \right)$. You can think of $G$ as the fundamental group of the following graph of groups:
Let $T$ denote the associated Bass-Serre tree on which $G$ acts. Because vertex-stabilisers are finite, every infinite-order element of $G$ is WPD. However, the action of $G$ on $T$ is not acylindrical (because, for every $n \geq 2$, the axis of a generator of the $\mathbb{Z}$ in $\mathbb{Z} \oplus \mathbb{Z}_n$ is fixed by the $\mathbb{Z}_n$). Even worse, $G$ does not admit an acylindrical action on a hyperbolic space for which all its infinite-order elements are WPD (otherwise, we would have a bound on the index of $\langle g \rangle$ in the centraliser $C(g)$ independ on the infinite-order element $g$), so every hyperbolic space on which $G$ would act acylindrically must be quite different from $T$.
The key point is that $G$ admits a universal action (i.e., an action on a hyperbolic space for which every generalised loxodromic element is loxodromic) but it does not admit a universal acylindrical action (i.e., an acylindrical action on a hyperbolic space for which every generalised loxodromic element is loxodromic). There exist finitely generated such example, such as Dunwoody's inaccessible group.
Nice way to answer question 2 ! BTW, the proof that Dunwoody's group does not admit a universal acylindrical action is in C. Abbott's paper http://www.carolynrabbott.com/uploads/3/0/8/1/30819067/abbott_universal_actions.pdf
To answer question 1, there is a description of the construction of $X$ in Section 5 of Osin's paper, which I recalled in this question. Basically, $X$ is obtained coning-off both a maximal virtually cyclic group $H$ containing your WPD element and a group "transverse" to $H$ in some sense.
Taking the example of the free group $G=F(a,b)$ of two generators and considering $a$ to be WPD, then Osin's contructions yields the Cayley graph $Cay(G, \langle a\rangle, \langle b\rangle)$. You see here that $X$ is not quite $Y$ (the group $G$ itself).
I'm not sure how to understand question 2, but if you're asking if given a WPD element, you can deduce the action is acylindrical, this is not true. If you're asking if there are general methods to prove that in particular cases, I don't know any (except using definitions of acylindrical actions of course).
For your motivation, you can indeed take the same space $X=Y$ to be the curve complex, but this is because the action is acylindrical (and 1 implies 2 as you say). To prove that the action of a subgroup on a subcomplex you're interested in is acylindrical, I would suggest without more details to try to reproduce Bowditch's proof.
|
2025-03-21T14:48:30.334622
| 2020-04-16T16:43:01 |
357674
|
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|
Stack Exchange
|
Dense subgroup of a compact connected Lie group generated by two words
Let $G$ be a compact connected Lie group and $w_1$, $w_2$ be two positive words in alphabet $\{a, b\}$ which are not the powers of some another word $w$. Positive means that $a^{-1}$ and $b^{-1}$ can not be used.
For example the pair $w_1=ab$ and $w_2=ba$ is allowed. And the pair $w_1=ababab$ and $w_2=abab$ is not allowed. The pair $w_1=aba^{-1}b^{-1}$, $w_2=a$ is also not allowed, as $w_1$ is not positive.
Question: is it true that for a typical ( with respect to the Haar measure on $G$) pair $(a, b)\in G^2$ the subgroup generated by $w_1(a,b)$ and $w_2(a,b)$ is dense in $G$?
For which pairs do you know this to hold (apart from pairs generating $F_2$)? or for which $G$ (other than the easy abelian cases)?
@YCor, I know only these two easy cases. But I guess for pairs like $w_1=a^2b$, $w_2=b^5$ it should be eassily reducible to these easy cases, even though formally these words do not generate $F_2$. What is of interest is some $\textit{non-trivial pairs}$ whatever "nontrivial" means, like, for example $w_1=ab$, $w_2=ba$.
$SO(3)$ should be easy anyway for $(ab,ba)$, because the list of proper closed subgroups is short. Namely, if $\langle ab,ba\rangle$ is not dense, then $(ab)^{60}$ and $(ba)^{60}$ commute (indeed either $ab,ba$ preserve a common axis hence their squares commute, or $ab,ba$ generate a tetrahedral/cubic/icosahedral group, whose order divides $60$). Just exhibiting a pair $(a,b)$ for which this fails shows that $(ab,ba)$ is generically topologically generating (actually, for all $a,b$ in the Zariski-open subset ${a,b:[(ab)^{60},(ba)^{60}]\neq 1}$).
@YCor, sure, by 'non-trivial pair' I mean a pair for which the statement ( if true) does not immediately follow by some obvious general reasons, like for $G$ abelian or for a pair generating $F_2$.
Well you didn't specify this reason and that $SO(3)$ can be dealt with (and hence its powers too), which you summarize as "a pair generating $F_2$" but is not exactly this. The special fact of $SO(3)$ is that there is a common law satisfied by all proper closed subgroups but not by $SO(3)$ itself. So, in order to converge, I understand that what you call "trivial case", say for $(ab,ba)$, is the case when $G$ is locally isomorphic to $\mathbf{R}^k\times\mathrm{SO}(3)^\ell$ for some $(k,\ell)$.
@YCor The point is that I do not have any specific pairs in my hands. The only control which I have over the pair is the condition specified in the original question. But of course for a precisely given pair ( which unfortunately I do not have) one can try ( as you did right now) to give a specific ( for this pair) proof.
My argument is not specific to this pair: it works for an arbitrary non-commuting pair $w_1,w_2$ in $F_2$. But only for $\mathrm{SO}(3)$ (and hence all $G$ with only $\mathfrak{so}(3)$ simple factors— let's focus on $G$ simple since it's enough). For given $(w_1,w_2,G)$, is the question equivalent to the question of barely asking the existence of $(a,b)\in G^2$ such that $\langle w_1(a,b),w_2(a,b)\rangle$ is dense?
@YCor Interesting, thank you. Even though still the case of general $G$ is of interest. The question is little more than just existence. The question is whether that such pairs $(a, b)$ form a $\textit{full measure set}$ in $G^2$.
I guess that in a compact connected semisimple Lie group $G$, there are finitely many maximal closed proper subgroups up to conjugation, and that the set $\Xi_G\subset G^2$ of pairs $(a,b)$ belonging to one of those is Zariski-closed. If so, one has the alternative, for given $w_1,w_2$: either the induced map $G^2\to G^2$, mapping $(a,b)$ to $(w_1(a,b),w_2(a,b))$ maps into $\Xi_G$, or the inverse image of $\Xi_G$ is a proper Zariski-closed subset. If so (this is maybe known), the question whether the set of pairs has full measure is the same as asking whether it's non-empty.
@YCor: https://projecteuclid.org/euclid.nmj/1118764740
@MoisheKohan Why do you point out this reference? I'm aware that $G$ has a topologically generating pair, but this just says that $\Xi_G\neq G$. I see that Lemma 3 says that $\Xi_G$ is closed in the ordinary topology, but in the compact case this is immediate if we have finiteness of the number of maximal subgroups mod conjugation. Actually $\Xi_G$ can have non-empty interior, e.g., when $G=\mathrm{SL}_2(\mathbf{R})$, and the link doesn't say anything specific to $G$ compact.
@Ycor I see what you mean. Independent remark: As a matter of fact for more delicate applications ( this is related to pseudorandom walks on G, generated by some particular type of dynamical systems) it might be needed somewhat more delicate information on these pairs than just full measure, but so far at least full measure would be enough..
@Ycor actually in the particular situation which I did (for surfaces of genus one and two) the words which appear do generate $F_2$ but I have no idea how to prove it in more general situation ( any genus). So the condition in the question is something I believe I can achieve.
See the result of Gerstenhaber-Rothaus, which says that if the abelianization of the word map has full rank, then the map $G\times G\to G\times G$ has non-zero degree. This is a necessary condition, as one can see if $G$ is abelian or has an abelian quotient (e.g. $U(n)$).
So this won't apply to $\{ab,ba\}$.
Once the map is non-zero degree, the pushforward of the Haar measure on $G\times G$ should be absolutely continuous with respect to Haar measure. This is because the map is also algebraic, and hence the preimage of points are smaller dimension, so the preimage of a set of measure $0$ will be measure $0$.
A theorem of Weyl implies that a compact subgroup of an algebraic group over $\mathbb{R}$ is an algebraic subgroup. Now we follow the argument in Barnea-Larsen, section 3.
Barnea, Y.; Larsen, M., Random generation in semisimple algebraic groups over local fields., J. Algebra 271, No. 1, 1-10 (2004). ZBL1049.20028.
Let's assume that $G$ is semisimple; I think that the general case can be reduced to this case. Since $G$ is compact, we may complexify to get a semisimple algebraic group $G^{\mathbb{C}}$ over $\mathbb{C}$. By Lemma 3.2, there is a countable set $\{X_0,X_1,\ldots\}$ of proper closed subvarieties such that if $\gamma\in G^{\mathbb{C}}- \cup_i X_i(\mathbb{C})$, then the Zariski closure of $\gamma$ is a maximal torus. Passing to $G=G^{\mathbb{R}}$, the real subgroup, we see that the same is true for $G$. Hence with probability $1$, any element $\gamma\in G$ will have closure a maximal torus.
Proposition 3.3 states that there is a proper closed subvariety $X \subset G^{\mathbb{C}}\times G^{\mathbb{C}}$ so that for any proper algebraic subgroup $H$ containing a maximal torus, $H\times H \subset X$.
Now choose a random pair of elements $(\gamma_1,\gamma_2)\in G\times G$ with respect to a measure absolutely continuous with respect to Haar measure. Then with probability $1$, $\overline{\langle\gamma_i\rangle}$ is a maximal torus, since $\cup_i{X_i(\mathbb{R})}$ has measure $0$. Then if $\langle \gamma_1,\gamma_2\rangle$ is not dense in $G$, then $\overline{\langle \gamma_1,\gamma_2\rangle}=H < G$, where $H$ is closed and contains a maximal rank torus. So $(\gamma_1,\gamma_2)\in X$, again occurring with probability $0$.
I think this gives an outline of a proof under these assumptions.
Thank you so much for your informative answer. The problem though is that I have very little control over the possible words. They appear from some dynamics on higher genus surfaces ( and I do not fix the genus) and the conditions formulated in the original question are the only ones which I can currently prove, this is why I formulated it this way. However your answer made me think if I can establish full rank of abelinization. I did not think about it and that might be possible after extra-work.. however not guaranteed..
@DmitriScheglov: Part of the argument goes through in your case, I think: with probability 1, $\gamma_i$ will be Zariski dense in a maximal torus by the Gerstenhaber-Rothaus argument (because $\gamma_i$ has non-zero exponent sum, so we can pair it with another element giving something of non-zero degree to $G\times G$, then project to the first coordinate). So then you need the map to avoid the set $X$ again. This seems quite plausible to me under your assumption.
let me think about it..
You might be able to compute the derivative of the map at the identity (which should be a map between Lie algebras) to show that the image of the map doesn't lie in the subgroups H, and hence not lying in the subvariety X.
I will try to see.. May be it makes sense to see first how it works at least for {ab, ba}..
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2025-03-21T14:48:30.335567
| 2020-04-16T16:43:29 |
357675
|
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"Carlo Beenakker",
"Nemo",
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|
Stack Exchange
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A close formula for a Fourier transform
I would like to calculate "explicitly" the following integral, which is a Fourier transform: let $\alpha>0$ be a parameter, for $x\in \mathbb R$, we define
$$
I(\alpha, x)=\int_\mathbb R \cos(xt) e^{-\alpha \cosh t} dt.
$$
It is easy to see that $I(\alpha, \cdot)$ is a rapidly decreasing (even) smooth function, but I would like to have an explicit formula.
https://dlmf.nist.gov/10.32.E9
@user82588 --- neat find --- Mathematica is unable to evaluate it in closed form.
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